Student solutions manual to accompany : Calculus, single and multivariable [Seventh edition.] 9781119138549, 111913854X

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Student solutions manual to accompany : Calculus, single and multivariable [Seventh edition.]
 9781119138549, 111913854X

Table of contents :
Cover
Title Page
Copyright
Preface
Acknowledgements
Contents
1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS
1.1 FUNCTIONS AND CHANGE
1.2 EXPONENTIAL FUNCTIONS
1.3 NEW FUNCTIONS FROM OLD
1.4 LOGARITHMIC FUNCTIONS
1.5 TRIGONOMETRIC FUNCTIONS
1.6 POWERS, POLYNOMIALS, AND RATIONAL FUNCTIONS
1.7 INTRODUCTION TO LIMITS AND CONTINUITY
1.8 EXTENDING THE IDEA OF A LIMIT
1.9 FURTHER LIMIT CALCULATIONS USING ALGEBRA
2 KEY CONCEPT: THE DERIVATIVE
2.1 HOW DO WE MEASURE SPEED?
2.2 THE DERIVATIVE AT A POINT
2.3 THE DERIVATIVE FUNCTION
2.4 INTERPRETATIONS OF THE DERIVATIVE
2.5 THE SECOND DERIVATIVE
2.6 DIFFERENTIABILITY
3 SHORT-CUTS TO DIFFERENTIATION
3.1 POWERS AND POLYNOMIALS
3.2 THE EXPONENTIAL FUNCTION
3.3 THE PRODUCT AND QUOTIENT RULES
3.4 THE CHAIN RULE
3.5 THE TRIGONOMETRIC FUNCTIONS
3.6 THE CHAIN RULE AND INVERSE FUNCTIONS
3.7 IMPLICIT FUNCTIONS
3.8 HYPERBOLIC FUNCTIONS
3.9 LINEAR APPROXIMATION AND THE DERIVATIVE
3.10 THEOREMS ABOUT DIFFERENTIABLE FUNCTIONS
4 USING THE DERIVATIVE
4.1 USING FIRST AND SECOND DERIVATIVES
4.2 OPTIMIZATION
4.3 OPTIMIZATION AND MODELING
4.4 FAMILIES OF FUNCTIONS AND MODELING
4.5 APPLICATIONS TO MARGINALITY
4.6 RATES AND RELATED RATES
4.7 L’HOPITAL’S RULE, GROWTH, AND DOMINANCE
4.8 PARAMETRIC EQUATIONS
5 KEY CONCEPT: THE DEFINITE INTEGRAL
5.1 HOW DO WE MEASURE DISTANCE TRAVELED?
5.2 THE DEFINITE INTEGRAL
5.3 THE FUNDAMENTAL THEOREM AND INTERPRETATIONS
5.4 THEOREMS ABOUT DEFINITE INTEGRALS
6 CONSTRUCTING ANTIDERIVATIVES
6.1 ANTIDERIVATIVES GRAPHICALLY AND NUMERICALLY
6.2 CONSTRUCTING ANTIDERIVATIVES ANALYTICALLY
6.3 DIFFERENTIAL EQUATIONS AND MOTION
6.4 SECOND FUNDAMENTAL THEOREM OF CALCULUS
7 INTEGRATION
7.1 INTEGRATION BY SUBSTITUTION
7.2 INTEGRATION BY PARTS
7.3 TABLES OF INTEGRALS
7.4 ALGEBRAIC IDENTITIES AND TRIGONOMETRIC SUBSTITUTIONS
7.5 NUMERICAL METHODS FOR DEFINITE INTEGRALS
7.6 IMPROPER INTEGRALS
7.7 COMPARISON OF IMPROPER INTEGRALS
8 USING THE DEFINITE INTEGRAL
8.1 AREAS AND VOLUMES
8.2 APPLICATIONS TO GEOMETRY
8.3 AREA AND ARC LENGTH IN POLAR COORDINATES
8.4 DENSITY AND CENTER OF MASS
8.5 APPLICATIONS TO PHYSICS
8.6 APPLICATIONS TO ECONOMICS
8.7 DISTRIBUTION FUNCTIONS
8.8 PROBABILITY, MEAN, AND MEDIAN
9 SEQUENCES AND SERIES
9.1 SEQUENCES
9.2 GEOMETRIC SERIES
9.3 CONVERGENCE OF SERIES
9.4 TESTS FOR CONVERGENCE
9.5 POWER SERIES AND INTERVAL OF CONVERGENCE
10 APPROXIMATING FUNCTIONS USING SERIES
10.1 TAYLOR POLYNOMIALS
10.2 TAYLOR SERIES
10.3 FINDING AND USING TAYLOR SERIES
10.4 THE ERROR IN TAYLOR POLYNOMIAL APPROXIMATIONS
10.5 FOURIER SERIES
11 DIFFERENTIAL EQUATIONS
11.1 WHAT IS A DIFFERENTIAL EQUATION?
11.2 SLOPE FIELDS
11.3 EULER’S METHOD
11.4 SEPARATION OF VARIABLES
11.5 GROWTH AND DECAY
11.6 APPLICATIONS AND MODELING
11.7 THE LOGISTIC MODEL
11.8 SYSTEMS OF DIFFERENTIAL EQUATIONS
11.9 ANALYZING THE PHASE PLANE
11.10 SECOND-ORDER DIFFERENTIAL EQUATIONS: OSCILLATIONS
11.11 LINEAR SECOND-ORDER DIFFERENTIAL EQUATIONS
12 FUNCTIONS OF SEVERAL VARIABLES
12.1 FUNCTIONS OF TWO VARIABLES
12.2 GRAPHS AND SURFACES
12.3 CONTOUR DIAGRAMS
12.4 LINEAR FUNCTIONS
12.5 FUNCTIONS OF THREE VARIABLES
12.6 LIMITS AND CONTINUITY
13 A FUNDAMENTAL TOOL: VECTORS
13.1 DISPLACEMENT VECTORS
13.2 VECTORS IN GENERAL
13.3 THE DOT PRODUCT
13.4 THE CROSS PRODUCT
14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES
14.1 THE PARTIAL DERIVATIVE
14.2 COMPUTING PARTIAL DERIVATIVES ALGEBRAICALLY
14.3 LOCAL LINEARITY AND THE DIFFERENTIAL
14.4 GRADIENTS AND DIRECTIONAL DERIVATIVES IN THE PLANE
14.5 GRADIENTS AND DIRECTIONAL DERIVATIVES IN SPACE
14.6 THE CHAIN RULE
14.7 SECOND-ORDER PARTIAL DERIVATIVES
14.8 DIFFERENTIABILITY
15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA
15.1 CRITICAL POINTS: LOCAL EXTREMA AND SADDLE POINTS
15.2 OPTIMIZATION
15.3 CONSTRAINED OPTIMIZATION: LAGRANGE MULTIPLIERS
16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES
16.1 THE DEFINITE INTEGRAL OF A FUNCTION OF TWO VARIABLES
16.2 ITERATED INTEGRALS
16.3 TRIPLE INTEGRALS
16.4 DOUBLE INTEGRALS IN POLAR COORDINATES
16.5 INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES
16.6 APPLICATIONS OF INTEGRATION TO PROBABILITY
17 PARAMETERIZATION AND VECTOR FIELDS
17.1 PARAMETERIZED CURVES
17.2 MOTION, VELOCITY, AND ACCELERATION
17.3 VECTOR FIELDS
17.4 THE FLOW OF A VECTOR FIELD
18 LINE INTEGRALS
18.1 THE IDEA OF A LINE INTEGRAL
18.2 COMPUTING LINE INTEGRALS OVER PARAMETERIZED CURVES
18.3 GRADIENT FIELDS AND PATH-INDEPENDENT FIELDS
18.4 PATH-DEPENDENT VECTOR FIELDS AND GREEN’S THEOREM
19 FLUX INTEGRALS AND DIVERGENCE
19.1 THE IDEA OF A FLUX INTEGRAL
19.2 FLUX INTEGRALS FOR GRAPHS, CYLINDERS, AND SPHERES
19.3 THE DIVERGENCE OF A VECTOR FIELD
19.4 THE DIVERGENCE THEOREM
20 THE CURL AND STOKES’ THEOREM
20.1 THE CURL OF A VECTOR FIELD
20.2 STOKES’ THEOREM
20.3 THE THREE FUNDAMENTAL THEOREMS
21 PARAMETERS, COORDINATES, AND INTEGRALS
21.1 COORDINATES AND PARAMETERIZED SURFACES
21.2 CHANGE OF COORDINATES IN A MULTIPLE INTEGRAL
21.3 FLUX INTEGRALS OVER PARAMETERIZED SURFACES
READY REFERENCE
ANSWERS TO ODD-NUMBERED PROBLEMS
INDEX
EULA

Citation preview

Lines

Distance and Midpoint Formulas

Slope of line through (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ):

Distance 𝐷 between (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ): √ 𝐷 = (𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2

𝑚=

𝑦2 − 𝑦1 𝑥2 − 𝑥1

Point-slope equation of line through (𝑥1 , 𝑦1 ) with slope 𝑚: 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) Slope-intercept equation of line with slope 𝑚 and 𝑦-intercept 𝑏: 𝑦 = 𝑏 + 𝑚𝑥

Midpoint of (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ): ( ) 𝑥1 + 𝑥2 𝑦1 + 𝑦2 , 2 2

Quadratic Formula If 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0, then √ −𝑏 ± 𝑏2 − 4𝑎𝑐 𝑥= 2𝑎

Factoring Special Polynomials 𝑥2 − 𝑦2 = (𝑥 + 𝑦)(𝑥 − 𝑦) 𝑥3 + 𝑦3 = (𝑥 + 𝑦)(𝑥2 − 𝑥𝑦 + 𝑦2 ) 𝑥3 − 𝑦3 = (𝑥 − 𝑦)(𝑥2 + 𝑥𝑦 + 𝑦2 )

Rules of Exponents 𝑎𝑥 𝑎𝑡 = 𝑎𝑥+𝑡 𝑎𝑥 = 𝑎𝑥−𝑡 𝑎𝑡 (𝑎𝑥 )𝑡 = 𝑎𝑥𝑡

Circles

Definition of Natural Log

Center (ℎ, 𝑘) and radius 𝑟: (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟2

𝑦

𝑦 = ln 𝑥 means 𝑒 = 𝑥 ex: ln 1 = 0 since 𝑒0 = 1

Ellipse 𝑥2 𝑦2 + =1 𝑎2 𝑏2

𝑦 𝑥

𝑦=𝑒 1

𝑦 = ln 𝑥 𝑥

𝑦 𝑏

1 −𝑎

𝑎

Identities

𝑥

−𝑏

ln 𝑒𝑥 = 𝑥 𝑒ln 𝑥 = 𝑥

Rules of Natural Logarithms

Hyperbola 𝑥2 𝑦2 − =1 𝑎2 𝑏2 𝑦

ln(𝐴𝐵) = ln 𝐴 + ln 𝐵 ( ) 𝐴 = ln 𝐴 − ln 𝐵 ln 𝐵 𝑝 ln 𝐴 = 𝑝 ln 𝐴

𝑦 = 𝑏𝑥∕𝑎 𝑎

𝑥 𝑦 = −𝑏𝑥∕𝑎

Geometric Formulas Conversion Between Radians and Degrees: 𝜋 radians = 180◦ Triangle Circle Sector of Circle 𝐴 = 𝜋𝑟2 𝐴 = 12 𝑟2 𝜃 (𝜃 in radians) 𝐴 = 12 𝑏ℎ 𝑠 = 𝑟𝜃 (𝜃 in radians) 1 𝐶 = 2𝜋𝑟 = 𝑎𝑏 sin 𝜃 2

𝑟

𝑎

𝑟

𝑠





𝜃

𝜃



𝑏

𝑟

Sphere 𝑉 = 43 𝜋𝑟3 𝐴 = 4𝜋𝑟2

Cylinder 𝑉 = 𝜋𝑟2 ℎ

Cone 𝑉 = 13 𝜋𝑟2 ℎ

Trigonometric Functions 𝑦 𝑟 𝑥 cos 𝜃 = 𝑟 𝑦 tan 𝜃 = 𝑥 sin 𝜃 tan 𝜃 = cos 𝜃

(𝑥, 𝑦)



sin 𝜃 =

sin(𝐴±𝐵) = sin 𝐴 cos 𝐵±cos 𝐴 sin 𝐵

𝑦

cos(𝐴±𝐵) = cos 𝐴 cos 𝐵∓sin 𝐴 sin 𝐵

𝑟





sin(2𝐴) = 2 sin 𝐴 cos 𝐴 ✛

𝜃 𝑥



cos(2𝐴) = 2 cos2 𝐴−1 = 1−2 sin2 𝐴



cos2 𝜃 + sin2 𝜃 = 1 𝑦

𝑦

𝑦 𝑦 = tan 𝑥

1

𝑦 = sin 𝑥

1

𝑦 = cos 𝑥

𝑥 𝜋

𝑥 𝜋

2𝜋

−1

The Binomial Theorem

2𝜋

𝑥 −𝜋

𝜋

−1

𝑛(𝑛 − 1) 𝑛−2 2 𝑛(𝑛 − 1)(𝑛 − 2) 𝑛−3 3 𝑥 𝑦 + 𝑥 𝑦 + ⋯ + 𝑛𝑥𝑦𝑛−1 + 𝑦𝑛 1⋅2 1⋅2⋅3 𝑛(𝑛 − 1) 𝑛−2 2 𝑛(𝑛 − 1)(𝑛 − 2) 𝑛−3 3 𝑥 𝑦 − 𝑥 𝑦 + ⋯ ± 𝑛𝑥𝑦𝑛−1 ∓ 𝑦𝑛 (𝑥 − 𝑦)𝑛 = 𝑥𝑛 − 𝑛𝑥𝑛−1 𝑦 + 1⋅2 1⋅2⋅3

(𝑥 + 𝑦)𝑛 = 𝑥𝑛 + 𝑛𝑥𝑛−1 𝑦 +

CALCULUS Seventh Edition

We dedicate this book to Andrew M. Gleason. His brilliance and the extraordinary kindness and dignity with which he treated others made an enormous difference to us, and to many, many people. Andy brought out the best in everyone. Deb Hughes Hallett for the Calculus Consortium

CALCULUS Seventh Edition Produced by the Calculus Consortium and initially funded by a National Science Foundation Grant.

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Problems from Calculus: The Analysis of Functions, by Peter D. Taylor (Toronto: Wall & Emerson, Inc., 1992). Reprinted with permission of the publisher. This book was set in Times Roman by the Consortium using TEX, Mathematica, and the package ASTEX, which was written by Alex Kasman. It was printed and bound by 2VBE(SBQIJDT7FSTBJMMFT. The cover was printed by 2VBE(SBQIJDT7FSTBJMMFT. This book is printed on acid-free paper. Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulfill their aspirations. Our company is built on a foundation of principles that include responsibility to the communities we serve and where we live and work. In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social, economic, and ethical challenges we face in our business. Among the issues we are addressing are carbon impact, paper specifications and procurement, ethical conduct within our business and among our vendors, and community and charitable support. For more information, please visit our website: www.wiley.com/go/citizenship. Copyright ©2017, 2013, 2009, 2005, 2001, and 1998 John Wiley & Sons, Inc. All rights reserved.

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PREFACE

Calculus is one of the greatest achievements of the human intellect. Inspired by problems in astronomy, Newton and Leibniz developed the ideas of calculus 300 years ago. Since then, each century has demonstrated the power of calculus to illuminate questions in mathematics, the physical sciences, engineering, and the social and biological sciences. Calculus has been so successful both because its central theme—change—is pivotal to an analysis of the natural world and because of its extraordinary power to reduce complicated problems to simple procedures. Therein lies the danger in teaching calculus: it is possible to teach the subject as nothing but procedures— thereby losing sight of both the mathematics and of its practical value. This edition of Calculus continues our effort to promote courses in which understanding and computation reinforce each other. It reflects the input of users at research universities, four-year colleges, community colleges, and secondary schools, as well as of professionals in partner disciplines such as engineering and the natural and social sciences.

Mathematical Thinking Supported by Theory and Modeling The first stage in the development of mathematical thinking is the acquisition of a clear intuitive picture of the central ideas. In the next stage, the student learns to reason with the intuitive ideas in plain English. After this foundation has been laid, there is a choice of direction. All students benefit from both theory and modeling, but the balance may differ for different groups. Some students, such as mathematics majors, may prefer more theory, while others may prefer more modeling. For instructors wishing to emphasize the connection between calculus and other fields, the text includes: • A variety of problems from the physical sciences and engineering. • Examples from the biological sciences and economics. • Models from the health sciences and of population growth. • Problems on sustainability. • Case studies on medicine by David E. Sloane, MD.

Active Learning: Good Problems As instructors ourselves, we know that interactive classrooms and well-crafted problems promote student learning. Since its inception, the hallmark of our text has been its innovative and engaging problems. These problems probe student understanding in ways often taken for granted. Praised for their creativity and variety, these problems have had influence far beyond the users of our textbook. The Seventh Edition continues this tradition. Under our approach, which we call the “Rule of Four,” ideas are presented graphically, numerically, symbolically, and verbally, thereby encouraging students to deepen their understanding. Graphs and tables in this text are assumed to show all necessary information about the functions they represent, including direction of change, local extrema, and discontinuities. Problems in this text include: • Strengthen Your Understanding problems at the end of every section. These problems ask students to reflect on what they have learned by deciding “What is wrong?” with a statement and to “Give an example” of an idea. • ConcepTests promote active learning in the classroom. These can be used with or without personal response systems (e.g., clickers), and have been shown to dramatically improve student learning. Available in a book or on the web at www.wiley.com/college/hughes-hallett. • Class Worksheets allow instructors to engage students in individual or group class-work. Samples are available in the Instructor’s Manual, and all are on the web at www.wiley.com/college/hughes-hallett. • Data and Models Many examples and problems throughout the text involve data-driven models. For example, Section 11.7 has a series of problems studying the spread of the chikungunya virus that arrived v

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Preface

in the US in 2013. Projects at the end of each chapter of the E-Text (at www.wiley.com/college/hugheshallett) provide opportunities for sustained investigation of real-world situations that can be modeled using calculus. • Drill Exercises build student skill and confidence.

Enhancing Learning Online This Seventh Edition provides opportunities for students to experience the concepts of calculus in ways that would not be possible in a traditional textbook. The E-Text of Calculus, powered by VitalSource, provides interactive demonstrations of concepts, embedded videos that illustrate problem-solving techniques, and built-in assessments that allow students to check their understanding as they read. The E-Text also contains additional content not found in the print edition: • Worked example videos by Donna Krawczyk at the University of Arizona, which provide students the opportunity to see and hear hundreds of the book’s examples being explained and worked out in detail • Embedded Interactive Explorations, applets that present and explore key ideas graphically and dynamically— especially useful for display of three-dimensional graphs • Material that reviews and extends the major ideas of each chapter: Chapter Summary, Review Exercises and Problems, CAS Challenge Problems, and Projects • Challenging problems that involve further exploration and application of the mathematics in many sections • Section on the 𝜖, 𝛿 definition of limit (1.10) • Appendices that include preliminary ideas useful in this course

Problems Available in WileyPLUS Students and instructors can access a wide variety of problems through WileyPLUS with ORION, Wiley’s digital learning environment. ORION Learning provides an adaptive, personalized learning experience that delivers easy-to-use analytics so instructors and students can see exactly where they’re excelling and where they need help. WileyPLUS with ORION features the following resources: • Online version of the text, featuring hyperlinks to referenced content, applets, videos, and supplements. • Homework management tools, which enable the instructor to assign questions easily and grade them automatically, using a rich set of options and controls. • QuickStart pre-designed reading and homework assignments. Use them as-is or customize them to fit the needs of your classroom. • Intelligent Tutoring questions, in which students are prompted for responses as they step through a problem solution and receive targeted feedback based on those responses. • Algebra & Trigonometry Refresher material, delivered through ORION, Wiley’s personalized, adaptive learning environment that assesses students’ readiness and provides students with an opportunity to brush up on material necessary to master Calculus, as well as to determine areas that require further review.

Flexibility and Adaptability: Varied Approaches The Seventh Edition of Calculus is designed to provide flexibility for instructors who have a range of preferences regarding inclusion of topics and applications and the use of computational technology. For those who prefer the lean topic list of earlier editions, we have kept clear the main conceptual paths. For example, • The Key Concept chapters on the derivative and the definite integral (Chapters 2 and 5) can be covered at the outset of the course, right after Chapter 1.

Preface

vii

• Limits and continuity (Sections 1.7, 1.8, and 1.9) can be covered in depth before the introduction of the derivative (Sections 2.1 and 2.2), or after. • Approximating Functions Using Series (Chapter 10) can be covered before, or without, Chapter 9. • In Chapter 4 (Using the Derivative), instructors can select freely from Sections 4.3–4.8. • Chapter 8 (Using the Definite Integral) contains a wide range of applications. Instructors can select one or two to do in detail. To use calculus effectively, students need skill in both symbolic manipulation and the use of technology. The balance between the two may vary, depending on the needs of the students and the wishes of the instructor. The book is adaptable to many different combinations. The book does not require any specific software or technology. It has been used with graphing calculators, graphing software, and computer algebra systems. Any technology with the ability to graph functions and perform numerical integration will suffice. Students are expected to use their own judgment to determine where technology is useful.

Content This content represents our vision of how calculus can be taught. It is flexible enough to accommodate individual course needs and requirements. Topics can easily be added or deleted, or the order changed. Changes to the text in the Seventh Edition are in italics. In all chapters, problems were added and others were updated. In total, there are more than 1300 new problems.

Chapter 1: A Library of Functions This chapter introduces all the elementary functions to be used in the book. Although the functions are probably familiar, the graphical, numerical, verbal, and modeling approach to them may be new. We introduce exponential functions at the earliest possible stage, since they are fundamental to the understanding of realworld processes. The content on limits and continuity in this chapter has been revised and expanded to emphasize the limit as a central idea of calculus. Section 1.7 gives an intuitive introduction to the ideas of limit and continuity. Section 1.8 introduces one-sided limits and limits at infinity and presents properties of limits of combinations of functions, such as sums and products. The new Section 1.9 gives a variety of algebraic techniques for computing limits, together with many new exercises and problems applying those techniques, and introduces the Squeeze Theorem. The new online Section 1.10 contains the 𝜖, 𝛿 definition of limit, previously in Section 1.8.

Chapter 2: Key Concept: The Derivative The purpose of this chapter is to give the student a practical understanding of the definition of the derivative and its interpretation as an instantaneous rate of change. The power rule is introduced; other rules are introduced in Chapter 3.

Chapter 3: Short-Cuts to Differentiation The derivatives of all the functions in Chapter 1 are introduced, as well as the rules for differentiating products; quotients; and composite, inverse, hyperbolic, and implicitly defined functions.

Chapter 4: Using the Derivative The aim of this chapter is to enable the student to use the derivative in solving problems, including optimization, graphing, rates, parametric equations, and indeterminate forms. It is not necessary to cover all the sections in this chapter.

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Chapter 5: Key Concept: The Definite Integral The purpose of this chapter is to give the student a practical understanding of the definite integral as a limit of Riemann sums and to bring out the connection between the derivative and the definite integral in the Fundamental Theorem of Calculus. The difference between total distance traveled during a time interval is contrasted with the change in position.

Chapter 6: Constructing Antiderivatives This chapter focuses on going backward from a derivative to the original function, first graphically and numerically, then analytically. It introduces the Second Fundamental Theorem of Calculus and the concept of a differential equation.

Chapter 7: Integration This chapter includes several techniques of integration, including substitution, parts, partial fractions, and trigonometric substitutions; others are included in the table of integrals. There are discussions of numerical methods and of improper integrals.

Chapter 8: Using the Definite Integral This chapter emphasizes the idea of subdividing a quantity to produce Riemann sums which, in the limit, yield a definite integral. It shows how the integral is used in geometry, physics, economics, and probability; polar coordinates are introduced. It is not necessary to cover all the sections in this chapter. Distance traveled along a parametrically defined curve during a time interval is contrasted with arc length.

Chapter 9: Sequences and Series This chapter focuses on sequences, series of constants, and convergence. It includes the integral, ratio, comparison, limit comparison, and alternating series tests. It also introduces geometric series and general power series, including their intervals of convergence. Rearrangement of the terms of a conditionally convergent series is discussed.

Chapter 10: Approximating Functions This chapter introduces Taylor Series and Fourier Series using the idea of approximating functions by simpler functions. The term Maclaurin series is introduced for a Taylor series centered at 0. Term-by-term differentiation of a Taylor series within its interval of convergence is introduced without proof. This term-by-term differentiation allows us to show that a power series is its own Taylor series.

Chapter 11: Differential Equations This chapter introduces differential equations. The emphasis is on qualitative solutions, modeling, and interpretation.

Chapter 12: Functions of Several Variables This chapter introduces functions of many variables from several points of view, using surface graphs, contour diagrams, and tables. We assume throughout that functions of two or more variables are defined on regions with piecewise smooth boundaries. We conclude with a section on continuity.

Chapter 13: A Fundamental Tool: Vectors This chapter introduces vectors geometrically and algebraically and discusses the dot and cross product. An application of the cross product to angular velocity is given.

Preface

ix

Chapter 14: Differentiating Functions of Several Variables Partial derivatives, directional derivatives, gradients, and local linearity are introduced. The chapter also discusses higher order partial derivatives, quadratic Taylor approximations, and differentiability.

Chapter 15: Optimization The ideas of the previous chapter are applied to optimization problems, both constrained and unconstrained.

Chapter 16: Integrating Functions of Several Variables This chapter discusses double and triple integrals in Cartesian, polar, cylindrical, and spherical coordinates.

Chapter 17: Parameterization and Vector Fields This chapter discusses parameterized curves and motion, vector fields and flowlines. Additional problems are provided on parameterizing curves in 3-space that are not contained in a coordinate plane.

Chapter 18: Line Integrals This chapter introduces line integrals and shows how to calculate them using parameterizations. Conservative fields, gradient fields, the Fundamental Theorem of Calculus for Line Integrals, and Green’s Theorem are discussed.

Chapter 19: Flux Integrals and Divergence This chapter introduces flux integrals and shows how to calculate them over surface graphs, portions of cylinders, and portions of spheres. The divergence is introduced and its relationship to flux integrals discussed in the Divergence Theorem. We calculate the surface area of the graph of a function using flux.

Chapter 20: The Curl and Stokes’ Theorem The purpose of this chapter is to give students a practical understanding of the curl and of Stokes’ Theorem and to lay out the relationship between the theorems of vector calculus.

Chapter 21: Parameters, Coordinates, and Integrals This chapter covers parameterized surfaces, the change of variable formula in a double or triple integral, and flux though a parameterized surface.

Appendices There are online appendices on roots, accuracy, and bounds; complex numbers; Newton’s method; and vectors in the plane. The appendix on vectors can be covered at any time, but may be particularly useful in the conjunction with Section 4.8 on parametric equations.

Supplementary Materials and Additional Resources Supplements for the instructor can be obtained online at the book companion site or by contacting your Wiley representative. The following supplementary materials are available for this edition: • Instructor’s Manual containing teaching tips, calculator programs, overhead transparency masters, sample worksheets, and sample syllabi. • Computerized Test Bank, comprised of nearly 7,000 questions, mostly algorithmically-generated,which allows for multiple versions of a single test or quiz.

x

Preface

• Instructor’s Solution Manual with complete solutions to all problems. • Student Solution Manual with complete solutions to half the odd-numbered problems. • Graphing Calculator Manual, to help students get the most out of their graphing calculators, and to show how they can apply the numerical and graphing functions of their calculators to their study of calculus. • Additional Material, elaborating specially marked points in the text and password-protected electronic versions of the instructor ancillaries, can be found on the web at www.wiley.com/college/hughes-hallett.

ConcepTests ConcepTests, modeled on the pioneering work of Harvard physicist Eric Mazur, are questions designed to promote active learning during class, particularly (but not exclusively) in large lectures. Our evaluation data show students taught with ConcepTests outperformed students taught by traditional lecture methods 73% versus 17% on conceptual questions, and 63% versus 54% on computational problems.

Advanced Placement (AP) Teacher’s Guide The AP Guide, written by a team of experienced AP teachers, provides tips, multiple-choice questions, and free-response questions that align to each chapter of the text. It also features a collection of labs designed to complement the teaching of key AP Calculus concepts. New material has been added to reflect recent changes in the learning objectives for AB and BC Calculus, including extended coverage of limits, continuity, sequences, and series. Also new to this edition are grids that align multiple choice and free-response questions to the College Board’s Enduring Understandings, Learning Objectives, and Essential Knowledge.

Acknowledgements First and foremost, we want to express our appreciation to the National Science Foundation for their faith in our ability to produce a revitalized calculus curriculum and, in particular, to our program officers, Louise Raphael, John Kenelly, John Bradley, and James Lightbourne. We also want to thank the members of our Advisory Board, Benita Albert, Lida Barrett, Simon Bernau, Robert Davis, M. Lavinia DeConge-Watson, John Dossey, Ron Douglas, Eli Fromm, William Haver, Seymour Parter, John Prados, and Stephen Rodi. In addition, a host of other people around the country and abroad deserve our thanks for their contributions to shaping this edition. They include: Huriye Arikan, Pau Atela, Ruth Baruth, Paul Blanchard, Lewis Blake, David Bressoud, Stephen Boyd, Lucille Buonocore, Matthew Michael Campbell, Jo Cannon, Ray Cannon, Phil Cheifetz, Scott Clark, Jailing Dai, Ann Davidian, Tom Dick, Srdjan Divac, Tevian Dray, Steven Dunbar, Penny Dunham, David Durlach, John Eggers, Wade Ellis, Johann Engelbrecht, Brad Ernst, Sunny Fawcett, Paul Feehan, Sol Friedberg, Melanie Fulton, Tom Gearhart, David Glickenstein, Chris Goff, Sheldon P. Gordon, Salim Haïdar, Elizabeth Hentges, Rob Indik, Adrian Iovita, David Jackson, Sue Jensen, Alex Kasman, Matthias Kawski, Christopher Kennedy, Mike Klucznik, Donna Krawczyk, Stephane Lafortune, Andrew Lawrence, Carl Leinert, Daniel Look, Andrew Looms, Bin Lu, Alex Mallozzi, Corinne Manogue, Jay Martin, Eric Mazur, Abby McCallum, Dan McGee, Ansie Meiring, Lang Moore, Jerry Morris, Hideo Nagahashi, Kartikeya Nagendra, Alan Newell, Steve Olson, John Orr, Arnie Ostebee, Andrew Pasquale, Scott Pilzer, Wayne Raskind, Maria Robinson, Laurie Rosatone, Ayse Sahin, Nataliya Sandler, Ken Santor, Anne Scanlan-Rohrer, Ellen Schmierer, Michael Sherman, Pat Shure, David Smith, Ernie Solheid, Misha Stepanov, Steve Strogatz, Carl Swenson, Peter Taylor, Dinesh Thakur, Sally Thomas, Joe Thrash, Alan Tucker, Doug Ulmer, Ignatios Vakalis, Bill Vélez, Joe Vignolini, Stan Wagon, Hannah Winkler, Debra Wood, Deane Yang, Bruce Yoshiwara, Kathy Yoshiwara, and Paul Zorn. Reports from the following reviewers were most helpful for the sixth edition: Barbara Armenta, James Baglama, Jon Clauss, Ann Darke, Marcel Finan, Dana Fine, Michael Huber, Greg Marks, Wes Ostertag, Ben Smith, Mark Turner, Aaron Weinberg, and Jianying Zhang. Reports from the following reviewers were most helpful for the seventh edition: Scott Adamson, Janet Beery, Tim Biehler, Lewis Blake, Mark Booth, Tambi Boyle, David Brown, Jeremy Case, Phil Clark, Patrice Conrath, Pam Crawford, Roman J. Dial, Rebecca Dibbs, Marcel B. Finan, Vauhn

Preface

xi

Foster-Grahler, Jill Guerra, Salim M. Haidar, Ryan A. Hass, Firas Hindeleh, Todd King, Mary Koshar, Dick Lane, Glenn Ledder, Oscar Levin, Tom Linton, Erich McAlister, Osvaldo Mendez, Cindy Moss, Victor Padron, Michael Prophet, Ahmad Rajabzadeh, Catherine A. Roberts, Kari Rothi, Edward J. Soares, Diana Staats, Robert Talbert, James Vicich, Wendy Weber, Mina Yavari, and Xinyun Zhu. Finally, we extend our particular thanks to Jon Christensen for his creativity with our three-dimensional figures. Deborah Hughes-Hallett

David O. Lomen

Douglas Quinney

Andrew M. Gleason

David Lovelock

Karen Rhea

William G. McCallum Eric Connally

Guadalupe I. Lozano Jerry Morris

Ayşe Şahin Adam Spiegler

Daniel E. Flath Selin Kalaycıoğlu

David O. Mumford Brad G. Osgood

Jeff Tecosky-Feldman Thomas W. Tucker

Brigitte Lahme Patti Frazer Lock

Cody L. Patterson

Aaron D. Wootton

To Students: How to Learn from this Book • This book may be different from other math textbooks that you have used, so it may be helpful to know about some of the differences in advance. This book emphasizes at every stage the meaning (in practical, graphical or numerical terms) of the symbols you are using. There is much less emphasis on “plug-andchug” and using formulas, and much more emphasis on the interpretation of these formulas than you may expect. You will often be asked to explain your ideas in words or to explain an answer using graphs. • The book contains the main ideas of calculus in plain English. Your success in using this book will depend on your reading, questioning, and thinking hard about the ideas presented. Although you may not have done this with other books, you should plan on reading the text in detail, not just the worked examples. • There are very few examples in the text that are exactly like the homework problems. This means that you can’t just look at a homework problem and search for a similar–looking “worked out” example. Success with the homework will come by grappling with the ideas of calculus. • Many of the problems that we have included in the book are open-ended. This means that there may be more than one approach and more than one solution, depending on your analysis. Many times, solving a problem relies on common sense ideas that are not stated in the problem but which you will know from everyday life. • Some problems in this book assume that you have access to a graphing calculator or computer. There are many situations where you may not be able to find an exact solution to a problem, but you can use a calculator or computer to get a reasonable approximation. • This book attempts to give equal weight to four methods for describing functions: graphical (a picture), numerical (a table of values) algebraic (a formula), and verbal. Sometimes you may find it easier to translate a problem given in one form into another. The best idea is to be flexible about your approach: if one way of looking at a problem doesn’t work, try another. • Students using this book have found discussing these problems in small groups very helpful. There are a great many problems which are not cut-and-dried; it can help to attack them with the other perspectives your colleagues can provide. If group work is not feasible, see if your instructor can organize a discussion session in which additional problems can be worked on. • You are probably wondering what you’ll get from the book. The answer is, if you put in a solid effort, you will get a real understanding of one of the most important accomplishments of the millennium— calculus—as well as a real sense of the power of mathematics in the age of technology.

xii

Preface

CONTENTS

1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

1

1.1 FUNCTIONS AND CHANGE 2 1.2 EXPONENTIAL FUNCTIONS 13 1.3 NEW FUNCTIONS FROM OLD 23 1.4 LOGARITHMIC FUNCTIONS 32 1.5 TRIGONOMETRIC FUNCTIONS 39 1.6 POWERS, POLYNOMIALS, AND RATIONAL FUNCTIONS 1.7 INTRODUCTION TO LIMITS AND CONTINUITY

49

58

1.8 EXTENDING THE IDEA OF A LIMIT 67 1.9 FURTHER LIMIT CALCULATIONS USING ALGEBRA

75

1.10 OPTIONAL PREVIEW OF THE FORMAL DEFINITION OF A LIMIT REVIEW PROBLEMS ONLINE PROJECTS

ONLINE

ONLINE

2 KEY CONCEPT: THE DERIVATIVE

83

2.1 HOW DO WE MEASURE SPEED? 84 2.2 THE DERIVATIVE AT A POINT 91 2.3 THE DERIVATIVE FUNCTION 99 2.4 INTERPRETATIONS OF THE DERIVATIVE 2.5 THE SECOND DERIVATIVE

108

115

2.6 DIFFERENTIABILITY 123 REVIEW PROBLEMS ONLINE PROJECTS

ONLINE

3 SHORT-CUTS TO DIFFERENTIATION

129

3.1 POWERS AND POLYNOMIALS 130 3.2 THE EXPONENTIAL FUNCTION 140 3.3 THE PRODUCT AND QUOTIENT RULES

144

For online material, see www.wiley.com/college/hughes-hallett.

Preface

3.4 3.5 3.6 3.7 3.8 3.9 3.10

THE CHAIN RULE 151 THE TRIGONOMETRIC FUNCTIONS 158 THE CHAIN RULE AND INVERSE FUNCTIONS 164 IMPLICIT FUNCTIONS 171 HYPERBOLIC FUNCTIONS 174 LINEAR APPROXIMATION AND THE DERIVATIVE 178 THEOREMS ABOUT DIFFERENTIABLE FUNCTIONS 186 REVIEW PROBLEMS ONLINE PROJECTS ONLINE

4 USING THE DERIVATIVE 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8

USING FIRST AND SECOND DERIVATIVES 192 OPTIMIZATION 203 OPTIMIZATION AND MODELING 212 FAMILIES OF FUNCTIONS AND MODELING 224 APPLICATIONS TO MARGINALITY 233 RATES AND RELATED RATES 243 L’HOPITAL’S RULE, GROWTH, AND DOMINANCE PARAMETRIC EQUATIONS 259 REVIEW PROBLEMS ONLINE PROJECTS ONLINE

191

252

5 KEY CONCEPT: THE DEFINITE INTEGRAL 5.1 5.2 5.3 5.4

xiii

HOW DO WE MEASURE DISTANCE TRAVELED? 272 THE DEFINITE INTEGRAL 283 THE FUNDAMENTAL THEOREM AND INTERPRETATIONS THEOREMS ABOUT DEFINITE INTEGRALS 302 REVIEW PROBLEMS ONLINE PROJECTS ONLINE

271

292

6 CONSTRUCTING ANTIDERIVATIVES 6.1 ANTIDERIVATIVES GRAPHICALLY AND NUMERICALLY 316 6.2 CONSTRUCTING ANTIDERIVATIVES ANALYTICALLY 322

315

xiv

Preface

6.3 DIFFERENTIAL EQUATIONS AND MOTION 329 6.4 SECOND FUNDAMENTAL THEOREM OF CALCULUS REVIEW PROBLEMS ONLINE PROJECTS ONLINE

335

7 INTEGRATION 7.1 7.2 7.3 7.4 7.5 7.6 7.7

341

INTEGRATION BY SUBSTITUTION 342 INTEGRATION BY PARTS 353 TABLES OF INTEGRALS 360 ALGEBRAIC IDENTITIES AND TRIGONOMETRIC SUBSTITUTIONS NUMERICAL METHODS FOR DEFINITE INTEGRALS 376 IMPROPER INTEGRALS 385 COMPARISON OF IMPROPER INTEGRALS 394 REVIEW PROBLEMS ONLINE PROJECTS ONLINE

8 USING THE DEFINITE INTEGRAL 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8

AREAS AND VOLUMES 402 APPLICATIONS TO GEOMETRY 410 AREA AND ARC LENGTH IN POLAR COORDINATES DENSITY AND CENTER OF MASS 429 APPLICATIONS TO PHYSICS 439 APPLICATIONS TO ECONOMICS 450 DISTRIBUTION FUNCTIONS 457 PROBABILITY, MEAN, AND MEDIAN 464 REVIEW PROBLEMS ONLINE PROJECTS ONLINE

401

420

9 SEQUENCES AND SERIES 9.1 9.2 9.3 9.4 9.5

SEQUENCES 474 GEOMETRIC SERIES 480 CONVERGENCE OF SERIES 488 TESTS FOR CONVERGENCE 494 POWER SERIES AND INTERVAL OF CONVERGENCE REVIEW PROBLEMS ONLINE PROJECTS ONLINE

366

473

504

Preface

10 APPROXIMATING FUNCTIONS USING SERIES 10.1 TAYLOR POLYNOMIALS 10.2 TAYLOR SERIES 523

513

514

10.3 FINDING AND USING TAYLOR SERIES 530 10.4 THE ERROR IN TAYLOR POLYNOMIAL APPROXIMATIONS 10.5 FOURIER SERIES

xv

539

546

REVIEW PROBLEMS PROJECTS ONLINE

ONLINE

11 DIFFERENTIAL EQUATIONS

561

11.1 WHAT IS A DIFFERENTIAL EQUATION?

562

11.2 SLOPE FIELDS 567 11.3 EULER’S METHOD 575 11.4 SEPARATION OF VARIABLES 11.5 GROWTH AND DECAY 586

580

11.6 APPLICATIONS AND MODELING 11.7 THE LOGISTIC MODEL 606

597

11.8 SYSTEMS OF DIFFERENTIAL EQUATIONS

616

11.9 ANALYZING THE PHASE PLANE 626 11.10 SECOND-ORDER DIFFERENTIAL EQUATIONS: OSCILLATIONS 11.11 LINEAR SECOND-ORDER DIFFERENTIAL EQUATIONS REVIEW PROBLEMS ONLINE PROJECTS

640

ONLINE

12 FUNCTIONS OF SEVERAL VARIABLES 12.1 FUNCTIONS OF TWO VARIABLES 12.2 GRAPHS AND SURFACES 660

651

652

12.3 CONTOUR DIAGRAMS 668 12.4 LINEAR FUNCTIONS 682 12.5 FUNCTIONS OF THREE VARIABLES 12.6 LIMITS AND CONTINUITY 695 REVIEW PROBLEMS PROJECTS

632

ONLINE

ONLINE

689

xvi

Preface

13 A FUNDAMENTAL TOOL: VECTORS 13.1 13.2 13.3 13.4

701

DISPLACEMENT VECTORS 702 VECTORS IN GENERAL 710 THE DOT PRODUCT 718 THE CROSS PRODUCT 728 REVIEW PROBLEMS ONLINE PROJECTS ONLINE

14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8

739

THE PARTIAL DERIVATIVE 740 COMPUTING PARTIAL DERIVATIVES ALGEBRAICALLY 748 LOCAL LINEARITY AND THE DIFFERENTIAL 753 GRADIENTS AND DIRECTIONAL DERIVATIVES IN THE PLANE 762 GRADIENTS AND DIRECTIONAL DERIVATIVES IN SPACE 772 THE CHAIN RULE 780 SECOND-ORDER PARTIAL DERIVATIVES 790 DIFFERENTIABILITY 799 REVIEW PROBLEMS ONLINE PROJECTS ONLINE

15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA

805

15.1 CRITICAL POINTS: LOCAL EXTREMA AND SADDLE POINTS 806 15.2 OPTIMIZATION 815 15.3 CONSTRAINED OPTIMIZATION: LAGRANGE MULTIPLIERS 825 REVIEW PROBLEMS ONLINE PROJECTS ONLINE

16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES 16.1 16.2 16.3 16.4

THE DEFINITE INTEGRAL OF A FUNCTION OF TWO VARIABLES ITERATED INTEGRALS 847 TRIPLE INTEGRALS 857 DOUBLE INTEGRALS IN POLAR COORDINATES 864

839 840

Preface

16.5 INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 16.6 APPLICATIONS OF INTEGRATION TO PROBABILITY 878 REVIEW PROBLEMS ONLINE PROJECTS ONLINE

869

17 PARAMETERIZATION AND VECTOR FIELDS 17.1 17.2 17.3 17.4

PARAMETERIZED CURVES 886 MOTION, VELOCITY, AND ACCELERATION VECTOR FIELDS 905 THE FLOW OF A VECTOR FIELD 913 REVIEW PROBLEMS ONLINE PROJECTS ONLINE

885 896

18 LINE INTEGRALS 18.1 18.2 18.3 18.4

921

THE IDEA OF A LINE INTEGRAL 922 COMPUTING LINE INTEGRALS OVER PARAMETERIZED CURVES 931 GRADIENT FIELDS AND PATH-INDEPENDENT FIELDS 939 PATH-DEPENDENT VECTOR FIELDS AND GREEN’S THEOREM 949 REVIEW PROBLEMS ONLINE PROJECTS ONLINE

19 FLUX INTEGRALS AND DIVERGENCE 19.1 19.2 19.3 19.4

xvii

THE IDEA OF A FLUX INTEGRAL 962 FLUX INTEGRALS FOR GRAPHS, CYLINDERS, AND SPHERES THE DIVERGENCE OF A VECTOR FIELD 982 THE DIVERGENCE THEOREM 991 REVIEW PROBLEMS ONLINE PROJECTS ONLINE

20 THE CURL AND STOKES’ THEOREM 20.1 THE CURL OF A VECTOR FIELD

961 973

999 1000

xviii

Preface

20.2 STOKES’ THEOREM

1008

20.3 THE THREE FUNDAMENTAL THEOREMS REVIEW PROBLEMS ONLINE PROJECTS

1015

ONLINE

21 PARAMETERS, COORDINATES, AND INTEGRALS 21.1 COORDINATES AND PARAMETERIZED SURFACES

1021 1022

21.2 CHANGE OF COORDINATES IN A MULTIPLE INTEGRAL 1033 21.3 FLUX INTEGRALS OVER PARAMETERIZED SURFACES 1038 REVIEW PROBLEMS PROJECTS ONLINE

ONLINE

Online

APPENDICES A ROOTS, ACCURACY, AND BOUNDS B COMPLEX NUMBERS C NEWTON’S METHOD

ONLINE

ONLINE ONLINE

D VECTORS IN THE PLANE ONLINE E DETERMINANTS ONLINE

READY REFERENCE

1043

ANSWERS TO ODD-NUMBERED PROBLEMS

1061

INDEX

1131

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Chapter One

FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

Contents 1.1 Functions and Change. . . . . . . . . . . . . . . . . . . . . . . . 2 The Rule of Four . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Examples of Domain and Range . . . . . . . . . . . . . . . 3 Linear Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Families of Linear Functions . . . . . . . . . . . . . . . . . . 5 Increasing versus Decreasing Functions . . . . . . . . . 6 Proportionality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.2 Exponential Functions. . . . . . . . . . . . . . . . . . . . . . . 13 Concavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Elimination of a Drug from the Body . . . . . . . . . . 15 The General Exponential Function . . . . . . . . . . . . 15 Half-Life and Doubling Time . . . . . . . . . . . . . . . . 16 The Family of Exponential Functions . . . . . . . . . . 16 Exponential Functions with Base e . . . . . . . . . . . . 17 1.3 New Functions from Old . . . . . . . . . . . . . . . . . . . . . 23 Shifts and Stretches . . . . . . . . . . . . . . . . . . . . . . . . 23 Composite Functions . . . . . . . . . . . . . . . . . . . . . . . 24 Odd and Even Functions: Symmetry . . . . . . . . . . . 25 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 26 1.4 Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . 32 Logarithms to Base 10 and to Base e. . . . . . . . . . . 32 Solving Equations Using Logarithms . . . . . . . . . . 33 1.5 Trigonometric Functions. . . . . . . . . . . . . . . . . . . . . 39 Radians. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 The Sine and Cosine Functions . . . . . . . . . . . . . . . 40 The Tangent Function . . . . . . . . . . . . . . . . . . . . . . 43 The Inverse Trigonometric Functions . . . . . . . . . . 44 1.6 Powers, Polynomials, and Rational Functions . . . 49 Power Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . 49 Dominance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . 53 1.7 Introduction to Limits and Continuity . . . . . . . . . 58 The Idea of Continuity . . . . . . . . . . . . . . . . . . . . . . 58 The Idea of a Limit . . . . . . . . . . . . . . . . . . . . . . . . 59 Definition of Limit . . . . . . . . . . . . . . . . . . . . . . . . . 60 Definition of Continuity. . . . . . . . . . . . . . . . . . . . . 60 The Intermediate Value Theorem . . . . . . . . . . . . . . . 60 Finding Limits Exactly Using Continuity and Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 1.8 Extending the Idea of a Limit. . . . . . . . . . . . . . . . . 67 One-Sided Limits . . . . . . . . . . . . . . . . . . . . . . . . . . 67 Limits and Asymptotes . . . . . . . . . . . . . . . . . . . . . 68 1.9 Further Limit Calculations using Algebra . . . . . . 75 Limits of Quotients . . . . . . . . . . . . . . . . . . . . . . . . 75 Calculating Limits at Infinity . . . . . . . . . . . . . . . . . 78 The Squeeze Theorem . . . . . . . . . . . . . . . . . . . . . . 79

2

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

1.1

FUNCTIONS AND CHANGE In mathematics, a function is used to represent the dependence of one quantity upon another. Let’s look at an example. In 2015, Boston, Massachusetts, had the highest annual snowfall, 110.6 inches, since recording started in 1872. Table 1.1 shows one 14-day period in which the city broke another record with a total of 64.4 inches.1

Table 1.1

Daily snowfall in inches for Boston, January 27 to February 9, 2015

Day

1

2

3

4

5

6

7

8

9

10

11

12

13

14

Snowfall

22.1

0.2

0

0.7

1.3

0

16.2

0

0

0.8

0

0.9

7.4

14.8

You may not have thought of something so unpredictable as daily snowfall as being a function, but it is a function of day, because each day gives rise to one snowfall total. There is no formula for the daily snowfall (otherwise we would not need a weather bureau), but nevertheless the daily snowfall in Boston does satisfy the definition of a function: Each day, 𝑡, has a unique snowfall, 𝑆, associated with it. We define a function as follows: A function is a rule that takes certain numbers as inputs and assigns to each a definite output number. The set of all input numbers is called the domain of the function and the set of resulting output numbers is called the range of the function. The input is called the independent variable and the output is called the dependent variable. In the snowfall example, the domain is the set of days {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14} and the range is the set of daily snowfalls {0, 0.2, 0.7, 0.8, 0.9, 1.3, 7.4, 14.8, 16.2, 22.1}. We call the function 𝑓 and write 𝑆 = 𝑓 (𝑡). Notice that a function may have identical outputs for different inputs (Days 8 and 9, for example). Some quantities, such as a day or date, are discrete, meaning they take only certain isolated values (days must be integers). Other quantities, such as time, are continuous as they can be any number. For a continuous variable, domains and ranges are often written using interval notation: The set of numbers 𝑡 such that 𝑎 ≤ 𝑡 ≤ 𝑏 is called a closed interval and written [𝑎, 𝑏]. The set of numbers 𝑡 such that 𝑎 < 𝑡 < 𝑏 is called an open interval and written (𝑎, 𝑏).

The Rule of Four: Tables, Graphs, Formulas, and Words Functions can be represented by tables, graphs, formulas, and descriptions in words. For example, the function giving the daily snowfall in Boston can be represented by the graph in Figure 1.1, as well as by Table 1.1. snowfall (inches)

25 20 15 10 5 0

2

4

6

8

10

12

14

day

Figure 1.1: Boston snowfall, starting January 27, 2015

As another example of a function, consider the snowy tree cricket. Surprisingly enough, all such crickets chirp at essentially the same rate if they are at the same temperature. That means that the chirp rate is a function of temperature. In other words, if we know the temperature, we can determine 1 http://w2.weather.gov/climate/xmacis.php?wfo=box.

Accessed June 2015.

1.1 FUNCTIONS AND CHANGE

3

𝐶 (chirps per minute) 400 300 200

𝐶 = 4𝑇 − 160

100 40

100

140

𝑇 (◦ F )

Figure 1.2: Cricket chirp rate versus temperature

the chirp rate. Even more surprisingly, the chirp rate, 𝐶, in chirps per minute, increases steadily with the temperature, 𝑇 , in degrees Fahrenheit, and can be computed by the formula 𝐶 = 4𝑇 − 160 to a fair level of accuracy. We write 𝐶 = 𝑓 (𝑇 ) to express the fact that we think of 𝐶 as a function of 𝑇 and that we have named this function 𝑓 . The graph of this function is in Figure 1.2. Notice that the graph of 𝐶 = 𝑓 (𝑇 ) in Figure 1.2 is a solid line. This is because 𝐶 = 𝑓 (𝑇 ) is a continuous function. Roughly speaking, a continuous function is one whose graph has no breaks, jumps, or holes. This means that the independent variable must be continuous. (We give a more precise definition of continuity of a function in Section 1.7.)

Examples of Domain and Range If the domain of a function is not specified, we usually take it to be the largest possible set of real numbers. For example, we usually think of the domain of the function 𝑓 (𝑥) = 𝑥2 as all real numbers. However, the domain of the function 𝑔(𝑥) = 1∕𝑥 is all real numbers except zero, since we cannot divide by zero. Sometimes we restrict the domain to be smaller than the largest possible set of real numbers. For example, if the function 𝑓 (𝑥) = 𝑥2 is used to represent the area of a square of side 𝑥, we restrict the domain to nonnegative values of 𝑥. Example 1

The function 𝐶 = 𝑓 (𝑇 ) gives chirp rate as a function of temperature. We restrict this function to temperatures for which the predicted chirp rate is positive, and up to the highest temperature ever recorded at a weather station, 134◦F. What is the domain of this function 𝑓 ?

Solution

If we consider the equation 𝐶 = 4𝑇 − 160 simply as a mathematical relationship between two variables 𝐶 and 𝑇 , any 𝑇 value is possible. However, if we think of it as a relationship between cricket chirps and temperature, then 𝐶 cannot be less than 0. Since 𝐶 = 0 leads to 0 = 4𝑇 − 160, and so 𝑇 = 40◦ F, we see that 𝑇 cannot be less than 40◦ F. (See Figure 1.2.) In addition, we are told that the function is not defined for temperatures above 134◦ . Thus, for the function 𝐶 = 𝑓 (𝑇 ) we have Domain = All 𝑇 values between 40◦F and 134◦F = All 𝑇 values with 40 ≤ 𝑇 ≤ 134 = [40, 134].

Example 2

Find the range of the function 𝑓 , given the domain from Example 1. In other words, find all possible values of the chirp rate, 𝐶, in the equation 𝐶 = 𝑓 (𝑇 ).

Solution

Again, if we consider 𝐶 = 4𝑇 − 160 simply as a mathematical relationship, its range is all real 𝐶 values. However, when thinking of the meaning of 𝐶 = 𝑓 (𝑇 ) for crickets, we see that the function predicts cricket chirps per minute between 0 (at 𝑇 = 40◦F) and 376 (at 𝑇 = 134◦F). Hence, Range = All 𝐶 values f rom 0 to 376 = All 𝐶 values with 0 ≤ 𝐶 ≤ 376 = [0, 376].

4

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

In using the temperature to predict the chirp rate, we thought of the temperature as the independent variable and the chirp rate as the dependent variable. However, we could do this backward, and calculate the temperature from the chirp rate. From this point of view, the temperature is dependent on the chirp rate. Thus, which variable is dependent and which is independent may depend on your viewpoint.

Linear Functions The chirp-rate function, 𝐶 = 𝑓 (𝑇 ), is an example of a linear function. A function is linear if its slope, or rate of change, is the same at every point. The rate of change of a function that is not linear may vary from point to point.

Olympic and World Records During the early years of the Olympics, the height of the men’s winning pole vault increased approximately 8 inches every four years. Table 1.2 shows that the height started at 130 inches in 1900, and increased by the equivalent of 2 inches a year. So the height was a linear function of time from 1900 to 1912. If 𝑦 is the winning height in inches and 𝑡 is the number of years since 1900, we can write 𝑦 = 𝑓 (𝑡) = 130 + 2𝑡. Since 𝑦 = 𝑓 (𝑡) increases with 𝑡, we say that 𝑓 is an increasing function. The coefficient 2 tells us the rate, in inches per year, at which the height increases. Table 1.2

Men’s Olympic pole vault winning height (approximate) Year

1900

1904

1908

1912

Height (inches)

130

138

146

154

This rate of increase is the slope of the line in Figure 1.3. The slope is given by the ratio Rise 146 − 138 8 = = = 2 inches/year. Run 8−4 4 Calculating the slope (rise/run) using any other two points on the line gives the same value. What about the constant 130? This represents the initial height in 1900, when 𝑡 = 0. Geometrically, 130 is the intercept on the vertical axis. Slope =

𝑦 (height in inches)

𝑦 = 130 + 2𝑡

150 140





✻Rise = 8 ❄

Run = 4

130

4

8

12

𝑡 (years since 1900)

Figure 1.3: Olympic pole vault records

You may wonder whether the linear trend continues beyond 1912. Not surprisingly, it does not exactly. The formula 𝑦 = 130 + 2𝑡 predicts that the height in the 2012 Olympics would be 354 inches or 29 feet 6 inches, which is considerably higher than the actual value of 19 feet 7.05 inches. There is clearly a danger in extrapolating too far from the given data. You should also observe that the data in Table 1.2 is discrete, because it is given only at specific points (every four years). However, we have treated the variable 𝑡 as though it were continuous, because the function 𝑦 = 130 + 2𝑡 makes

1.1 FUNCTIONS AND CHANGE

5

sense for all values of 𝑡. The graph in Figure 1.3 is of the continuous function because it is a solid line, rather than four separate points representing the years in which the Olympics were held. As the pole vault heights have increased over the years, the time to run the mile has decreased. If 𝑦 is the world record time to run the mile, in seconds, and 𝑡 is the number of years since 1900, then records show that, approximately, 𝑦 = 𝑔(𝑡) = 260 − 0.39𝑡. The 260 tells us that the world record was 260 seconds in 1900 (at 𝑡 = 0). The slope, −0.39, tells us that the world record decreased by about 0.39 seconds per year. We say that 𝑔 is a decreasing function.

Difference Quotients and Delta Notation We use the symbol Δ (the Greek letter capital delta) to mean “change in,” so Δ𝑥 means change in 𝑥 and Δ𝑦 means change in 𝑦. The slope of a linear function 𝑦 = 𝑓 (𝑥) can be calculated from values of the function at two points, given by 𝑥1 and 𝑥2 , using the formula 𝑚=

𝑓 (𝑥2 ) − 𝑓 (𝑥1 ) Δ𝑦 Rise = = . Run Δ𝑥 𝑥2 − 𝑥1

The quantity (𝑓 (𝑥2 ) − 𝑓 (𝑥1 ))∕(𝑥2 − 𝑥1 ) is called a difference quotient because it is the quotient of two differences. (See Figure 1.4.) Since 𝑚 = Δ𝑦∕Δ𝑥, the units of 𝑚 are 𝑦-units over 𝑥-units. 𝑦

𝑦 = 𝑓 (𝑥)

(𝑥2 , 𝑓 (𝑥2 ))

✻ (𝑥1 , 𝑓 (𝑥1 ))

Rise = 𝑓 (𝑥2 ) − 𝑓 (𝑥1 )







Run = 𝑥2 − 𝑥1

𝑥 𝑥1

𝑥2

Figure 1.4: Difference quotient =

𝑓 (𝑥2 ) − 𝑓 (𝑥1 ) 𝑥2 − 𝑥1

Families of Linear Functions A linear function has the form 𝑦 = 𝑓 (𝑥) = 𝑏 + 𝑚𝑥. Its graph is a line such that • 𝑚 is the slope, or rate of change of 𝑦 with respect to 𝑥. • 𝑏 is the vertical intercept, or value of 𝑦 when 𝑥 is zero. Notice that if the slope, 𝑚, is zero, we have 𝑦 = 𝑏, a horizontal line. To recognize that a table of 𝑥 and 𝑦 values comes from a linear function, 𝑦 = 𝑏 + 𝑚𝑥, look for differences in 𝑦-values that are constant for equally spaced 𝑥-values. Formulas such as 𝑓 (𝑥) = 𝑏 + 𝑚𝑥, in which the constants 𝑚 and 𝑏 can take on various values, give a family of functions. All the functions in a family share certain properties—in this case, all the

6

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

graphs are straight lines. The constants 𝑚 and 𝑏 are called parameters; their meaning is shown in Figures 1.5 and 1.6. Notice that the greater the magnitude of 𝑚, the steeper the line. 𝑦 = −𝑥

𝑦 = −2𝑥

𝑦

𝑦 = 2𝑥

𝑦 = −0.5𝑥

𝑦 𝑦=𝑥

𝑦=𝑥

𝑦 = −1 + 𝑥

𝑦 = 0.5𝑥

𝑥

𝑥 𝑦=2+𝑥 𝑦=1+𝑥

Figure 1.5: The family 𝑦 = 𝑚𝑥 (with 𝑏 = 0)

Figure 1.6: The family 𝑦 = 𝑏 + 𝑥 (with 𝑚 = 1)

Increasing versus Decreasing Functions The terms increasing and decreasing can be applied to other functions, not just linear ones. See Figure 1.7. In general, A function 𝑓 is increasing if the values of 𝑓 (𝑥) increase as 𝑥 increases. A function 𝑓 is decreasing if the values of 𝑓 (𝑥) decrease as 𝑥 increases. The graph of an increasing function climbs as we move from left to right. The graph of a decreasing function falls as we move from left to right. A function 𝑓 (𝑥) is monotonic if it increases for all 𝑥 or decreases for all 𝑥. Decreasing

Increasing

Figure 1.7: Increasing and decreasing functions

Proportionality A common functional relationship occurs when one quantity is proportional to another. For example, the area, 𝐴, of a circle is proportional to the square of the radius, 𝑟, because 𝐴 = 𝑓 (𝑟) = 𝜋𝑟2 . We say 𝑦 is (directly) proportional to 𝑥 if there is a nonzero constant 𝑘 such that 𝑦 = 𝑘𝑥. This 𝑘 is called the constant of proportionality. We also say that one quantity is inversely proportional to another if one is proportional to the reciprocal of the other. For example, the speed, 𝑣, at which you make a 50-mile trip is inversely proportional to the time, 𝑡, taken, because 𝑣 is proportional to 1∕𝑡: ( ) 50 1 = . 𝑣 = 50 𝑡 𝑡

7

1.1 FUNCTIONS AND CHANGE

Exercises and Problems for Section 1.1 EXERCISES 1. The population of a city, 𝑃 , in millions, is a function of 𝑡, the number of years since 2010, so 𝑃 = 𝑓 (𝑡). Explain the meaning of the statement 𝑓 (5) = 7 in terms of the population of this city. 2. The pollutant PCB (polychlorinated biphenyl) can affect the thickness of pelican eggshells. Thinking of the thickness, 𝑇 , of the eggshells, in mm, as a function of the concentration, 𝑃 , of PCBs in ppm (parts per million), we have 𝑇 = 𝑓 (𝑃 ). Explain the meaning of 𝑓 (200) in terms of thickness of pelican eggs and concentration of PCBs. 3. Describe what Figure 1.8 tells you about an assembly line whose productivity is represented as a function of the number of workers on the line.

13. Match the graphs in Figure 1.10 with the following equations. (Note that the 𝑥 and 𝑦 scales may be unequal.) (a) 𝑦 = −2.72𝑥 (c) 𝑦 = 27.9 − 0.1𝑥 (e) 𝑦 = −5.7 − 200𝑥 (I)

𝑦

(II)

(b) 𝑦 = 0.01 + 0.001𝑥 (d) 𝑦 = 0.1𝑥 − 27.9 (f) 𝑦 = 𝑥∕3.14

𝑦

𝑥

(IV)

𝑦

(III)

𝑥

𝑦

(V)

𝑦

𝑥

𝑦

(VI)

productivity

𝑥

𝑥

𝑥

Figure 1.10 number of workers

14. Estimate the slope and the equation of the line in Figure 1.11.

Figure 1.8

𝑦

For Exercises 4–7, find an equation for the line that passes through the given points. 4. (0, 0) and (1, 1)

5. (0, 2) and (2, 3)

6. (−2, 1) and (2, 3)

7. (−1, 0) and (2, 6)

4 2 𝑥 5

For Exercises 8–11, determine the slope and the 𝑦-intercept of the line whose equation is given. 8. 2𝑦 + 5𝑥 − 8 = 0

11. 12𝑥 = 6𝑦 + 4

12. Match the graphs in Figure 1.9 with the following equations. (Note that the 𝑥 and 𝑦 scales may be unequal.) (a) 𝑦 = 𝑥 − 5 (c) 5 = 𝑦 (e) 𝑦 = 𝑥 + 6 (I)

𝑦

(b) −3𝑥 + 4 = 𝑦 (d) 𝑦 = −4𝑥 − 5 (f) 𝑦 = 𝑥∕2 (II)

𝑦

𝑥

(IV)

𝑦

(III)

(V)

𝑦

𝑥

𝑥

Figure 1.9

𝑥

(VI)

16. Find a linear function that generates the values in Table 1.3. Table 1.3 𝑥

5.2

5.3

5.4

5.5

5.6

𝑦

27.8

29.2

30.6

32.0

33.4

For Exercises 17–19, use the facts that parallel lines have equal slopes and that the slopes of perpendicular lines are negative reciprocals of one another.

𝑦

𝑥

Figure 1.11 15. Find an equation for the line with slope 𝑚 through the point (𝑎, 𝑐).

9. 7𝑦 + 12𝑥 − 2 = 0

10. −4𝑦 + 2𝑥 + 8 = 0

10

17. Find an equation for the line through the point (2, 1) which is perpendicular to the line 𝑦 = 5𝑥 − 3. 18. Find equations for the lines through the point (1, 5) that are parallel to and perpendicular to the line with equation 𝑦 + 4𝑥 = 7.

𝑦

𝑥

19. Find equations for the lines through the point (𝑎, 𝑏) that are parallel and perpendicular to the line 𝑦 = 𝑚𝑥 + 𝑐, assuming 𝑚 ≠ 0.

8

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

For Exercises 20–23, give the approximate domain and range of each function. Assume the entire graph is shown. 20.

21.

𝑦 5

24. 𝑦 = 𝑥2 + 2

𝑦 6

25. 𝑦 =

1 𝑥2 + 2

√ 26. If 𝑓 (𝑡) = 𝑡2 − 16, find all values of 𝑡 for which 𝑓 (𝑡) is a real number. Solve 𝑓 (𝑡) = 3.

4

3

Find the domain and range in Exercises 24–25.

𝑦 = 𝑓 (𝑥) 𝑦 = 𝑓 (𝑥)

2 1

𝑥

𝑥 1

3

1

5

𝑦

22.

𝑥 2 𝑦 = 𝑓 (𝑥)

27. The volume of a sphere is proportional to the cube of its radius, 𝑟. 28. The average velocity, 𝑣, for a trip over a fixed distance, 𝑑, is inversely proportional to the time of travel, 𝑡.

5

2

5

𝑦

23.

−2

3

In Exercises 27–31, write a formula representing the function.

𝑦 = 𝑓 (𝑥)

3 1 𝑥

−2 1

3

5

29. The strength, 𝑆, of a beam is proportional to the square of its thickness, ℎ. 30. The energy, 𝐸, expended by a swimming dolphin is proportional to the cube of the speed, 𝑣, of the dolphin. 31. The number of animal species, 𝑁, of a certain body length, 𝑙, is inversely proportional to the square of 𝑙.

PROBLEMS 32. In December 2010, the snowfall in Minneapolis was unusually high,2 leading to the collapse of the roof of the Metrodome. Figure 1.12 gives the snowfall, 𝑆, in Minneapolis for December 6–15, 2010.

(b) Sketch a possible graph of 𝑉 against 𝑎. Is 𝑓 an increasing or decreasing function? Explain. (c) Explain the significance of the horizontal and vertical intercepts in terms of the value of the car.

(a) How do you know that the snowfall data represents a function of date? (b) Estimate the snowfall on December 12. (c) On which day was the snowfall more than 10 inches? (d) During which consecutive two-day interval was the increase in snowfall largest?

34. Which graph in Figure 1.13 best matches each of the following stories?3 Write a story for the remaining graph. (a) I had just left home when I realized I had forgotten my books, so I went back to pick them up. (b) Things went fine until I had a flat tire. (c) I started out calmly but sped up when I realized I was going to be late.

𝑆 (inches) (I)

15

distance from home

distance from home

(II)

10 5

time

𝑡 (date)

(III)

6 7 8 9 10 11 12 13 14 15

distance from home

(IV)

time distance from home

Figure 1.12 time

33. The value of a car, 𝑉 = 𝑓 (𝑎), in thousands of dollars, is a function of the age of the car, 𝑎, in years. (a) Interpret the statement 𝑓 (5) = 6.

time

Figure 1.13 In Problems 35–38 the function 𝑆 = 𝑓 (𝑡) gives the average annual sea level, 𝑆, in meters, in Aberdeen, Scotland,4

2 http://www.crh.noaa.gov/mpx/Climate/DisplayRecords.php 3 Adapted

from Jan Terwel, “Real Math in Cooperative Groups in Secondary Education.” Cooperative Learning in Mathematics, ed. Neal Davidson, p. 234 (Reading: Addison Wesley, 1990). 4 www.gov.uk, accessed January 7, 2015.

1.1 FUNCTIONS AND CHANGE

9

as a function of 𝑡, the number of years before 2012. Write a mathematical expression that represents the given statement.

you pass through Kalamazoo, Michigan. Sketch a graph of your distance from Kalamazoo as a function of time.

35. In 2000 the average annual sea level in Aberdeen was 7.049 meters.

46. US imports of crude oil and petroleum have been increasing.5 There have been many ups and downs, but the general trend is shown by the line in Figure 1.15.

36. The average annual sea level in Aberdeen in 2012. 37. The average annual sea level in Aberdeen was the same in 1949 and 2000. 38. The average annual sea level in Aberdeen decreased by 8 millimeters from 2011 to 2012. Problems 39–42 ask you to plot graphs based on the following story: “As I drove down the highway this morning, at first traffic was fast and uncongested, then it crept nearly bumperto-bumper until we passed an accident, after which traffic flow went back to normal until I exited.”

(a) Find the slope of the line. Include its units of measurement. (b) Write an equation for the line. Define your variables, including their units. (c) Assuming the trend continues, when does the linear model predict imports will reach 18 million barrels per day? Do you think this is a reliable prediction? Give reasons. US oil imports (million barrels per day)

14 13 12 11 10 9 8 7 6 5 4

39. Driving speed against time on the highway 40. Distance driven against time on the highway 41. Distance from my exit vs time on the highway 42. Distance between cars vs distance driven on the highway 43. An object is put outside on a cold day at time 𝑡 = 0. Its temperature, 𝐻 = 𝑓 (𝑡), in ◦ C, is graphed in Figure 1.14. (a) What does the statement 𝑓 (30) = 10 mean in terms of temperature? Include units for 30 and for 10 in your answer. (b) Explain what the vertical intercept, 𝑎, and the horizontal intercept, 𝑏, represent in terms of temperature of the object and time outside. 𝐻 (◦ C) 𝑎

year

1992 1996 2000 2004 2008

Figure 1.15 Problems 47–49 use Figure 1.16 showing how the quantity, 𝑄, of grass (kg/hectare) in different parts of Namibia depended on the average annual rainfall, 𝑟, (mm), in two different years.6 quantity of grass (kg/hectare)

𝑡 (min) 𝑏

Figure 1.14 44. A rock is dropped from a window and falls to the ground below. The height, 𝑠 (in meters), of the rock above ground is a function of the time, 𝑡 (in seconds), since the rock was dropped, so 𝑠 = 𝑓 (𝑡). (a) Sketch a possible graph of 𝑠 as a function of 𝑡. (b) Explain what the statement 𝑓 (7) = 12 tells us about the rock’s fall. (c) The graph drawn as the answer for part (a) should have a horizontal and vertical intercept. Interpret each intercept in terms of the rock’s fall. 45. You drive at a constant speed from Chicago to Detroit, a distance of 275 miles. About 120 miles from Chicago 5 http://www.theoildrum.com/node/2767. 6 David

138–144.

6000 5000 4000 3000 2000 1000

1939 1997

rainfall (mm)

100 200 300 400 500 600

Figure 1.16 47. (a) For 1939, find the slope of the line, including units. (b) Interpret the slope in this context. (c) Find the equation of the line. 48. (a) For 1997, find the slope of the line, including units. (b) Interpret the slope in this context. (c) Find the equation of the line. 49. Which of the two functions in Figure 1.16 has the larger difference quotient Δ𝑄∕Δ𝑟? What does this tell us about grass in Namibia?

Accessed May 2015. Ward and Ben T. Ngairorue, “Are Namibia’s Grasslands Desertifying?”, Journal of Range Management 53, 2000,

10

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

50. Marmots are large squirrels that hibernate in the winter and come out in the spring. Figure 1.17 shows the date (days after Jan 1) that they are first sighted each year in Colorado as a function of the average minimum daily temperature for that year.7 (a) Find the slope of the line, including units. (b) What does the sign of the slope tell you about marmots? (c) Use the slope to determine how much difference 6◦ C warming makes to the date of first appearance of a marmot. (d) Find the equation of the line. day of first marmot sighting

150 140 130 120 110 100 90

mean minimum

◦ 8 10 12 14 16 18 20 22 24 temperature ( C)

Figure 1.17 51. In Colorado spring has arrived when the bluebell first flowers. Figure 1.18 shows the date (days after Jan 1) that the first flower is sighted in one location as a function of the first date (days after Jan 1) of bare (snowfree) ground.8 (a) If the first date of bare ground is 140, how many days later is the first bluebell flower sighted? (b) Find the slope of the line, including units. (c) What does the sign of the slope tell you about bluebells? (d) Find the equation of the line. day of first bluebell flower

52. On March 5, 2015, Capracotta, Italy, received 256 cm (100.787 inches) of snow in 18 hours.9 (a) Assuming the snow fell at a constant rate and there were already 100 cm of snow on the ground, find a formula for 𝑓 (𝑡), in cm, for the depth of snow as a function of 𝑡 hours since the snowfall began on March 5. (b) What are the domain and range of 𝑓 ? 53. In a California town, the monthly charge for waste collection is $8 for 32 gallons of waste and $12.32 for 68 gallons of waste. (a) Find a linear formula for the cost, 𝐶, of waste collection as a function of the number of gallons of waste, 𝑤. (b) What is the slope of the line found in part (a)? Give units and interpret your answer in terms of the cost of waste collection. (c) What is the vertical intercept of the line found in part (a)? Give units and interpret your answer in terms of the cost of waste collection. 54. For tax purposes, you may have to report the value of your assets, such as cars or refrigerators. The value you report drops with time. “Straight-line depreciation” assumes that the value is a linear function of time. If a $950 refrigerator depreciates completely in seven years, find a formula for its value as a function of time. 55. Residents of the town of Maple Grove who are connected to the municipal water supply are billed a fixed amount monthly plus a charge for each cubic foot of water used. A household using 1000 cubic feet was billed $40, while one using 1600 cubic feet was billed $55. (a) What is the charge per cubic foot? (b) Write an equation for the total cost of a resident’s water as a function of cubic feet of water used. (c) How many cubic feet of water used would lead to a bill of $100?

180

56. A controversial 1992 Danish study10 reported that men’s average sperm count decreased from 113 million per milliliter in 1940 to 66 million per milliliter in 1990.

170 160 150 140 day of first 130 110 120 130 140 150 160 170 bare ground

Figure 1.18

(a) Express the average sperm count, 𝑆, as a linear function of the number of years, 𝑡, since 1940. (b) A man’s fertility is affected if his sperm count drops below about 20 million per milliliter. If the linear model found in part (a) is accurate, in what year will the average male sperm count fall below this level?

7 David W. Inouye, Billy Barr, Kenneth B. Armitage, and Brian D. Inouye, “Climate change is affecting altitudinal migrants and hibernating species”, PNAS 97, 2000, 1630–1633. 8 David W. Inouye, Billy Barr, Kenneth B. Armitage, and Brian D. Inouye, “Climate change is affecting altitudinal migrants and hibernating species”, PNAS 97, 2000, 1630–1633. 9 http://iceagenow.info/2015/03/official-italy-captures-world-one-day-snowfall-record/ 10 “Investigating the Next Silent Spring,” US News and World Report, pp. 50–52 (March 11, 1996).

1.1 FUNCTIONS AND CHANGE 𝑦

57. Let 𝑓 (𝑡) be the number of US billionaires in year 𝑡. 80

(a) Express the following statements11 in terms of 𝑓 . (i) In 2001 there were 272 US billionaires. (ii) In 2014 there were 525 US billionaires. (b) Find the average yearly increase in the number of US billionaires from 2001 to 2014. Express this using 𝑓 . (c) Assuming the yearly increase remains constant, find a formula predicting the number of US billionaires in year 𝑡. 58. The cost of planting seed is usually a function of the number of acres sown. The cost of the equipment is a fixed cost because it must be paid regardless of the number of acres planted. The costs of supplies and labor vary with the number of acres planted and are called variable costs. Suppose the fixed costs are $10,000 and the variable costs are $200 per acre. Let 𝐶 be the total cost, measured in thousands of dollars, and let 𝑥 be the number of acres planted. (a) Find a formula for 𝐶 as a function of 𝑥. (b) Graph 𝐶 against 𝑥. (c) Which feature of the graph represents the fixed costs? Which represents the variable costs? 59. An airplane uses a fixed amount of fuel for takeoff, a (different) fixed amount for landing, and a third fixed amount per mile when it is in the air. How does the total quantity of fuel required depend on the length of the trip? Write a formula for the function involved. Explain the meaning of the constants in your formula. 60. For the line 𝑦 = 𝑓 (𝑥) in Figure 1.19, evaluate (a)

𝑓 (423) − 𝑓 (422)

(b) 𝑓 (517) − 𝑓 (513)

𝑦 8000 6000 4000

11

70 60 50 𝑥 3000 4000 5000 6000

Figure 1.20 62. An alternative to petroleum-based diesel fuel, biodiesel, is derived from renewable resources such as food crops, algae, and animal oils. The table shows the recent annual percent growth in US biodiesel exports.12 (a) Find the largest time interval over which the percentage growth in the US exports of biodiesel was an increasing function of time. Interpret what increasing means, practically speaking, in this case. (b) Find the largest time interval over which the actual US exports of biodiesel was an increasing function of time. Interpret what increasing means, practically speaking, in this case. Year

2010

2011

2012

2013

2014

% growth over previous yr

−60.5

−30.5

69.9

53.0

−57.8

63. Hydroelectric power is electric power generated by the force of moving water. Figure 1.21 shows13 the annual percent growth in hydroelectric power consumption by the US industrial sector between 2006 and 2014. (a) Find the largest time interval over which the percentage growth in the US consumption of hydroelectric power was an increasing function of time. Interpret what increasing means, practically speaking, in this case. (b) Find the largest time interval over which the actual US consumption of hydroelectric power was a decreasing function of time. Interpret what decreasing means, practically speaking, in this case. percent growth over previous year

2000 20 𝑥 400

500

600

700

10 year

Figure 1.19 −10

61. For the line 𝑦 = 𝑔(𝑥) in Figure 1.20, evaluate (a)

𝑔(4210) − 𝑔(4209)

(b) 𝑔(3760) − 𝑔(3740)

11 www.statista.com,

2008

2010

2012

2014

−20

Figure 1.21

accessed March 18, 2015. accessed March 29, 2015. 13 Yearly values have been joined with line segments to highlight trends in the data; however, values in between years should not be inferred from the segments. From www.eia.doe.gov, accessed March 29, 2015. 12 www.eia.doe.gov,

12

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

64. Solar panels are arrays of photovoltaic cells that convert solar radiation into electricity. The table shows the annual percent change in the US price per watt of a solar panel.14

67. A company rents cars at $40 a day and 15 cents a mile. Its competitor’s cars are $50 a day and 10 cents a mile.

Year

2005

2006

2007

2008

2009

2010

% growth over previous yr

6.7

9.7

−3.7

3.6

−20.1

−29.7

(a) Find the largest time interval over which the percentage growth in the US price per watt of a solar panel was a decreasing function of time. Interpret what decreasing means, practically speaking, in this case. (b) Find the largest time interval over which the actual price per watt of a solar panel was a decreasing function of time. Interpret what decreasing means, practically speaking, in this case. 65. Table 1.4 shows the average annual sea level, 𝑆, in meters, in Aberdeen, Scotland,15 as a function of time, 𝑡, measured in years before 2008. Table 1.4 𝑡

0

25

50

75

100

125

𝑆

7.094

7.019

6.992

6.965

6.938

6.957

(a) What was the average sea level in Aberdeen in 2008? (b) In what year was the average sea level 7.019 meters? 6.957 meters? (c) Table 1.5 gives the average sea level, 𝑆, in Aberdeen as a function of the year, 𝑥. Complete the missing values. Table 1.5 𝑥

1883

?

1933

1958

1983

2008

𝑆

?

6.938

?

6.992

?

?

(a) For each company, give a formula for the cost of renting a car for a day as a function of the distance traveled. (b) On the same axes, graph both functions. (c) How should you decide which company is cheaper?

68. A $25,000 vehicle depreciates $2000 a year as it ages. Repair costs are $1500 per year. (a) Write formulas for each of the two linear functions at time 𝑡, value, 𝑉 (𝑡), and repair costs to date, 𝐶(𝑡). Graph them. (b) One strategy is to replace a vehicle when the total cost of repairs is equal to the current value. Find this time. (c) Another strategy is to replace the vehicle when the value of the vehicle is some percent of the original value. Find the time when the value is 6%. 69. A bakery owner knows that customers buy a total of 𝑞 cakes when the price, 𝑝, is no more than 𝑝 = 𝑑(𝑞) = 20 − 𝑞∕20 dollars. She is willing to make and supply as many as 𝑞 cakes at a price of 𝑝 = 𝑠(𝑞) = 11 + 𝑞∕40 dollars each. (The graphs of the functions 𝑑(𝑞) and 𝑠(𝑞) are called a demand curve and a supply curve, respectively.) The graphs of 𝑑(𝑞) and 𝑠(𝑞) are in Figure 1.22. (a) Why, in terms of the context, is the slope of 𝑑(𝑞) negative and the slope of 𝑠(𝑞) positive? (b) Is each of the ordered pairs (𝑞, 𝑝) a solution to the inequality 𝑝 ≤ 20 − 𝑞∕20? Interpret your answers in terms of the context. (60, 18)

66. The table gives the required standard weight, 𝑤, in kilograms, of American soldiers, aged between 21 and 27, for height, ℎ, in centimeters.16 (a) How do you know that the data in this table could represent a linear function? (b) Find weight, 𝑤, as a linear function of height, ℎ. What is the slope of the line? What are the units for the slope? (c) Find height, ℎ, as a linear function of weight, 𝑤. What is the slope of the line? What are the units for the slope?

(120, 12)

(c) Graph in the 𝑞𝑝-plane the solution set of the system of inequalities 𝑝 ≤ 20 − 𝑞∕20, 𝑝 ≥ 11 + 𝑞∕40. What does this solution set represent in terms of the context? (d) What is the rightmost point of the solution set you graphed in part (c)? Interpret your answer in terms of the context. 𝑝 25 20 15 10 5

𝑑(𝑞) = 20 − 𝑞∕20

𝑠(𝑞) = 11 + 𝑞∕40 𝑞

ℎ (cm)

172

176

180

184

188

192

196

𝑤 (kg)

79.7

82.4

85.1

87.8

90.5

93.2

95.9

40

80

120 160 200

Figure 1.22

14 We use the official price per peak watt, which uses the maximum number of watts a solar panel can produce under ideal conditions. From www.eia.doe.gov, accessed March 29, 2015. 15 www.decc.gov.uk, accessed June 2011. 16 Adapted from usmilitary.about.com, accessed March 29, 2015.

1.2 EXPONENTIAL FUNCTIONS

70. (a) Consider the functions graphed in Figure 1.23(a). Find the coordinates of 𝐶. (b) Consider the functions in Figure 1.23(b). Find the coordinates of 𝐶 in terms of 𝑏. 𝑦

(a)

falling under gravity. He originally thought that the velocity of such a falling body was proportional to the distance it had fallen. What do the experimental data in Table 1.6 tell you about Galileo’s hypothesis? What alternative hypothesis is suggested by the two sets of data in Table 1.6 and Table 1.7?

𝑦

(b)

𝑦 = 𝑥2

13

𝑦 = 𝑥2

𝐶

𝐶

Table 1.6 (0, 2)

(0, 𝑏) (1, 1)

(1, 1) 𝑥

𝑥

Figure 1.23

Distance (ft)

0

1

2

3

4

Velocity (ft/sec)

0

8

11.3

13.9

16

Table 1.7 Time (sec)

0

1

2

3

4

Velocity (ft/sec)

0

32

64

96

128

71. When Galileo was formulating the laws of motion, he considered the motion of a body starting from rest and

Strengthen Your Understanding In Problems 72–76, explain what is wrong with the statement.

79. For any two points in the plane, there is a linear function whose graph passes through them.

72. For constants 𝑚 and 𝑏, the slope of the linear function 𝑦 = 𝑏 + 𝑚𝑥 is 𝑚 = Δ𝑥∕Δ𝑦. 73. The lines 𝑥 = 3 and 𝑦 = 3 are both linear functions of 𝑥. 74. The line 𝑦 − 3 = 0 has slope 1 in the 𝑥𝑦-plane. 75. Values of 𝑦 on the graph of 𝑦 = 0.5𝑥 − 3 increase more slowly than values of 𝑦 on the graph of 𝑦 = 0.5 − 3𝑥. 76. The equation 𝑦 = 2𝑥 + 1 indicates that 𝑦 is directly proportional to 𝑥 with a constant of proportionality 2.

80. If 𝑦 = 𝑓 (𝑥) is a linear function, then increasing 𝑥 by 1 unit changes the corresponding 𝑦 by 𝑚 units, where 𝑚 is the slope. 81. The linear functions 𝑦 = −𝑥 + 1 and 𝑥 = −𝑦 + 1 have the same graph. 82. The linear functions 𝑦 = 2 − 2𝑥 and 𝑥 = 2 − 2𝑦 have the same graph. 83. If 𝑦 is a linear function of 𝑥, then the ratio 𝑦∕𝑥 is constant for all points on the graph at which 𝑥 ≠ 0.

In Problems 77–78, give an example of: 77. A linear function with a positive slope and a negative 𝑥-intercept. 78. A formula representing the statement “𝑞 is inversely proportional to the cube root of 𝑝 and has a positive constant of proportionality.” Are the statements in Problems 79–84 true or false? Give an explanation for your answer.

1.2

84. If 𝑦 = 𝑓 (𝑥) is a linear function, then increasing 𝑥 by 2 units adds 𝑚 + 2 units to the corresponding 𝑦, where 𝑚 is the slope. 85. Which of the following functions has its domain identical with its range? √ (b) 𝑔(𝑥) = 𝑥 (a) 𝑓 (𝑥) = 𝑥2 (c) ℎ(𝑥) = 𝑥3

(d) 𝑖(𝑥) = |𝑥|

EXPONENTIAL FUNCTIONS

Population Growth The population of Burkina Faso, a sub-Saharan African country,17 from 2007 to 2013 is given in Table 1.8. To see how the population is growing, we look at the increase in population in the third column. If the population had been growing linearly, all the numbers in the third column would be the same. 17 dataworldbank.org,

accessed March 29, 2015.

14

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

Table 1.8

Population of Burkina Faso (estimated), 2007–2013 Year

𝑃 (population in millions)

Population

Change in

(millions)

population (millions)

2007

14.235

2008

14.660

2009

15.095

2010

15.540

2011

15.995

2012

16.460

2013

16.934

𝑃 = 14.235(1.03)𝑡

60

0.425

40

0.435

20

0.445 0.455

𝑡 (years since 2007) −10 10 20 30 40 50 Figure 1.24: Population of Burkina Faso (estimated): Exponential growth

0.465 0.474

Suppose we divide each year’s population by the previous year’s population. For example, Population in 2008 14.660 million = = 1.03 Population in 2007 14.235 million Population in 2009 15.095 million = = 1.03. Population in 2008 14.660 million The fact that both calculations give 1.03 shows the population grew by about 3% between 2007 and 2008 and between 2008 and 2009. Similar calculations for other years show that the population grew by a factor of about 1.03, or 3%, every year. Whenever we have a constant growth factor (here 1.03), we have exponential growth. The population 𝑡 years after 2007 is given by the exponential function 𝑃 = 14.235(1.03)𝑡 . If we assume that the formula holds for 50 years, the population graph has the shape shown in Figure 1.24. Since the population is growing faster and faster as time goes on, the graph is bending upward; we say it is concave up. Even exponential functions which climb slowly at first, such as this one, eventually climb extremely quickly. To recognize that a table of 𝑡 and 𝑃 values comes from an exponential function, look for ratios of 𝑃 values that are constant for equally spaced 𝑡 values.

Concavity We have used the term concave up19 to describe the graph in Figure 1.24. In words: The graph of a function is concave up if it bends upward as we move left to right; it is concave down if it bends downward. (See Figure 1.25 for four possible shapes.) A line is neither concave up nor concave down.

Concave down

Concave up

Figure 1.25: Concavity of a graph 19 In

Chapter 2 we consider concavity in more depth.

1.2 EXPONENTIAL FUNCTIONS

15

Elimination of a Drug from the Body Now we look at a quantity which is decreasing exponentially instead of increasing. When a patient is given medication, the drug enters the bloodstream. As the drug passes through the liver and kidneys, it is metabolized and eliminated at a rate that depends on the particular drug. For the antibiotic ampicillin, approximately 40% of the drug is eliminated every hour. A typical dose of ampicillin is 250 mg. Suppose 𝑄 = 𝑓 (𝑡), where 𝑄 is the quantity of ampicillin, in mg, in the bloodstream at time 𝑡 hours since the drug was given. At 𝑡 = 0, we have 𝑄 = 250. Since every hour the amount remaining is 60% of the previous amount, we have 𝑓 (0) = 250 𝑓 (1) = 250(0.6) 𝑓 (2) = (250(0.6))(0.6) = 250(0.6)2, and after 𝑡 hours, 𝑄 = 𝑓 (𝑡) = 250(0.6)𝑡. This is an exponential decay function. Some values of the function are in Table 1.9; its graph is in Figure 1.26. Notice the way in which the function in Figure 1.26 is decreasing. Each hour a smaller quantity of the drug is removed than in the previous hour. This is because as time passes, there is less of the drug in the body to be removed. Compare this to the exponential growth in Figure 1.24, where each step upward is larger than the previous one. Notice, however, that both graphs are concave up. Drug elimination

Table 1.9

𝑄 (mg) 250

𝑡 (hours)

𝑄 (mg)

200

0

250

150

1

150

100

2

90

50

3

54

4

32.4

5

19.4

1

2

3

4

5

𝑡 (hours)

Figure 1.26: Drug elimination: Exponential decay

The General Exponential Function We say 𝑃 is an exponential function of 𝑡 with base 𝑎 if 𝑃 = 𝑃0 𝑎𝑡 , where 𝑃0 is the initial quantity (when 𝑡 = 0) and 𝑎 is the factor by which 𝑃 changes when 𝑡 increases by 1. If 𝑎 > 1, we have exponential growth; if 0 < 𝑎 < 1, we have exponential decay. Provided 𝑎 > 0, the largest possible domain for the exponential function is all real numbers. The reason we do not want 𝑎 ≤ 0 is that, for example, we cannot define 𝑎1∕2 if 𝑎 < 0. Also, we do not usually have 𝑎 = 1, since 𝑃 = 𝑃0 1𝑡 = 𝑃0 is then a constant function. The value of 𝑎 is closely related to the percent growth (or decay) rate. For example, if 𝑎 = 1.03, then 𝑃 is growing at 3%; if 𝑎 = 0.94, then 𝑃 is decaying at 6%, so the growth rate is 𝑟 = 𝑎 − 1.

16

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

Example 1

Suppose that 𝑄 = 𝑓 (𝑡) is an exponential function of 𝑡. If 𝑓 (20) = 88.2 and 𝑓 (23) = 91.4: (a) Find the base. (b) Find the growth rate. (c) Evaluate 𝑓 (25).

Solution

(a) Let 𝑄 = 𝑄0 𝑎 𝑡 . Substituting 𝑡 = 20, 𝑄 = 88.2 and 𝑡 = 23, 𝑄 = 91.4 gives two equations for 𝑄0 and 𝑎: 88.2 = 𝑄0 𝑎20

and 91.4 = 𝑄0 𝑎23 .

Dividing the two equations enables us to eliminate 𝑄0 : 23 91.4 𝑄0 𝑎 = = 𝑎3 . 88.2 𝑄0 𝑎20

Solving for the base, 𝑎, gives ) 91.4 1∕3 = 1.012. 88.2 (b) Since 𝑎 = 1.012, the growth rate is 1.012 − 1 = 0.012 = 1.2%. (c) We want to evaluate 𝑓 (25) = 𝑄0 𝑎25 = 𝑄0 (1.012)25. First we find 𝑄0 from the equation 𝑎=

(

88.2 = 𝑄0 (1.012)20. Solving gives 𝑄0 = 69.5. Thus, 𝑓 (25) = 69.5(1.012)25 = 93.6.

Half-Life and Doubling Time Radioactive substances, such as uranium, decay exponentially. A certain percentage of the mass disintegrates in a given unit of time; the time it takes for half the mass to decay is called the half-life of the substance. A well-known radioactive substance is carbon-14, which is used to date organic objects. When a piece of wood or bone was part of a living organism, it accumulated small amounts of radioactive carbon-14. Once the organism dies, it no longer picks up carbon-14. Using the half-life of carbon-14 (about 5730 years), we can estimate the age of the object. We use the following definitions: The half-life of an exponentially decaying quantity is the time required for the quantity to be reduced by a factor of one half. The doubling time of an exponentially increasing quantity is the time required for the quantity to double.

The Family of Exponential Functions The formula 𝑃 = 𝑃0 𝑎𝑡 gives a family of exponential functions with positive parameters 𝑃0 (the initial quantity) and 𝑎 (the base, or growth/decay factor). The base tells us whether the function is increasing (𝑎 > 1) or decreasing (0 < 𝑎 < 1). Since 𝑎 is the factor by which 𝑃 changes when 𝑡 is increased by 1, large values of 𝑎 mean fast growth; values of 𝑎 near 0 mean fast decay. (See Figures 1.27 and 1.28.) All members of the family 𝑃 = 𝑃0 𝑎𝑡 are concave up.

1.2 EXPONENTIAL FUNCTIONS

𝑃 40

𝑡

10

5𝑡

3𝑡

17

𝑃

2𝑡

1 0.8

30

(0.95)𝑡

0.6 (1.5)𝑡

20

(0.9)𝑡

0.4 10

0.2 (0.1)𝑡 𝑡 1

2

3

4

5

6

0

7

Figure 1.27: Exponential growth: 𝑃 = 𝑎𝑡 , for 𝑎 > 1

(0.5)𝑡

(0.8)𝑡 𝑡

2

4

6

8

10

12

Figure 1.28: Exponential decay: 𝑃 = 𝑎𝑡 , for 0 < 𝑎 < 1

Example 2

Figure 1.29 is the graph of three exponential functions. What can you say about the values of the six constants 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, 𝑞?

Solution

All the constants are positive. Since 𝑎, 𝑐, 𝑝 represent 𝑦-intercepts, we see that 𝑎 = 𝑐 because these graphs intersect on the 𝑦-axis. In addition, 𝑎 = 𝑐 < 𝑝, since 𝑦 = 𝑝 ⋅ 𝑞 𝑥 crosses the 𝑦-axis above the other two. Since 𝑦 = 𝑎 ⋅ 𝑏𝑥 is decreasing, we have 0 < 𝑏 < 1. The other functions are increasing, so 1 < 𝑑 and 1 < 𝑞. 𝑦 𝑦 = 𝑝 ⋅ 𝑞𝑥

𝑦 = 𝑐 ⋅ 𝑑𝑥 𝑦 = 𝑎 ⋅ 𝑏𝑥 𝑥 Figure 1.29: Three exponential functions

Exponential Functions with Base 𝒆 The most frequently used base for an exponential function is the famous number 𝑒 = 2.71828 … . This base is used so often that you will find an 𝑒𝑥 button on most scientific calculators. At first glance, this is all somewhat mysterious. Why is it convenient to use the base 2.71828 …? The full answer to that question must wait until Chapter 3, where we show that many calculus formulas come out neatly when 𝑒 is used as the base. We often use the following result: Any exponential growth function can be written, for some 𝑎 > 1 and 𝑘 > 0, in the form 𝑃 = 𝑃0 𝑎𝑡

or 𝑃 = 𝑃0 𝑒𝑘𝑡

and any exponential decay function can be written, for some 0 < 𝑎 < 1 and −𝑘 < 0, as 𝑄 = 𝑄0 𝑎 𝑡

or 𝑄 = 𝑄0 𝑒−𝑘𝑡 ,

where 𝑃0 and 𝑄0 are the initial quantities. We say that 𝑃 and 𝑄 are growing or decaying at a continuous19 rate of 𝑘. (For example, 𝑘 = 0.02 corresponds to a continuous rate of 2%.) 19 The

reason that 𝑘 is called the continuous rate is explored in detail in Chapter 11.

18

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

Example 3

Convert the functions 𝑃 = 𝑒0.5𝑡 and 𝑄 = 5𝑒−0.2𝑡 into the form 𝑦 = 𝑦0 𝑎𝑡 . Use the results to explain the shape of the graphs in Figures 1.30 and 1.31. 𝑄

𝑃 5

30

4 3

20 0.5𝑡

𝑃 =𝑒

10

1

1

𝑡

𝑡 1

2

3

4

5

7

6

2

Figure 1.30: An exponential growth function

Solution

𝑄 = 5𝑒−0.2𝑡

2

4

6

8

10

Figure 1.31: An exponential decay function

We have 𝑃 = 𝑒0.5𝑡 = (𝑒0.5 )𝑡 = (1.65)𝑡. Thus, 𝑃 is an exponential growth function with 𝑃0 = 1 and 𝑎 = 1.65. The function is increasing and its graph is concave up, similar to those in Figure 1.27. Also, 𝑄 = 5𝑒−0.2𝑡 = 5(𝑒−0.2 )𝑡 = 5(0.819)𝑡, so 𝑄 is an exponential decay function with 𝑄0 = 5 and 𝑎 = 0.819. The function is decreasing and its graph is concave up, similar to those in Figure 1.28.

Example 4

The quantity, 𝑄, of a drug in a patient’s body at time 𝑡 is represented for positive constants 𝑆 and 𝑘 by the function 𝑄 = 𝑆(1 − 𝑒−𝑘𝑡 ). For 𝑡 ≥ 0, describe how 𝑄 changes with time. What does 𝑆 represent?

Solution

The graph of 𝑄 is shown in Figure 1.32. Initially none of the drug is present, but the quantity increases with time. Since the graph is concave down, the quantity increases at a decreasing rate. This is realistic because as the quantity of the drug in the body increases, so does the rate at which the body excretes the drug. Thus, we expect the quantity to level off. Figure 1.32 shows that 𝑆 is the saturation level. The line 𝑄 = 𝑆 is called a horizontal asymptote. Q (quantity of drug)

Saturation level



𝑆

1

2

3

4

5

𝑡 (time in hours)

Figure 1.32: Buildup of the quantity of a drug in body

Exercises and Problems for Section 1.2 EXERCISES In Exercises 1–4, decide whether the graph is concave up, concave down, or neither. 2. 1.

3.

4.

𝑥 𝑥

𝑥

𝑥

1.2 EXPONENTIAL FUNCTIONS

The functions in Exercises 5–8 represent exponential growth or decay. What is the initial quantity? What is the growth rate? State if the growth rate is continuous. 5. 𝑃 = 5(1.07)𝑡

6. 𝑃 = 7.7(0.92)𝑡

7. 𝑃 = 3.2𝑒0.03𝑡

8. 𝑃 = 15𝑒−0.06𝑡

19

17. For which pairs of consecutive points in Figure 1.33 is the function graphed: (a) (b) (c) (d)

Increasing and concave up? Increasing and concave down? Decreasing and concave up? Decreasing and concave down?

𝐼

Write the functions in Exercises 9–12 in the form 𝑃 = 𝑃0 𝑎𝑡 . Which represent exponential growth and which represent exponential decay?

𝐵 𝐹

9. 𝑃 = 15𝑒0.25𝑡

10. 𝑃 = 2𝑒−0.5𝑡

𝐺

𝐶

𝐻

𝐸

11. 𝑃 = 𝑃0 𝑒0.2𝑡

𝑥

12. 𝑃 = 7𝑒−𝜋𝑡 𝐷 𝐴

In Exercises 13–14, let 𝑓 (𝑡) = 𝑄0 𝑎𝑡 = 𝑄0 (1 + 𝑟)𝑡 . (a) Find the base, 𝑎. (b) Find the percentage growth rate, 𝑟. 13. 𝑓 (5) = 75.94 and 𝑓 (7) = 170.86 14. 𝑓 (0.02) = 25.02 and 𝑓 (0.05) = 25.06

Figure 1.33

18. The table gives the average temperature in Wallingford, Connecticut, for the first 10 days in March. (a) Over which intervals was the average temperature increasing? Decreasing? (b) Find a pair of consecutive intervals over which the average temperature was increasing at a decreasing rate. Find another pair of consecutive intervals over which the average temperature was increasing at an increasing rate.

15. A town has a population of 1000 people at time 𝑡 = 0. In each of the following cases, write a formula for the population, 𝑃 , of the town as a function of year 𝑡. (a) The population increases by 50 people a year. (b) The population increases by 5% a year. 16. An air-freshener starts with 30 grams and evaporates over time. In each of the following cases, write a formula for the quantity, 𝑄 grams, of air-freshener remaining 𝑡 days after the start and sketch a graph of the function. The decrease is: (a)

Day

1

2

3

4

5

6

7

8

9

10

◦F

42◦

42◦

34◦

25◦

22◦

34◦

38◦

40◦

49◦

49◦

(b) 12% a day

2 grams a day

PROBLEMS 19. (a) Which (if any) of the functions in the following table could be linear? Find formulas for those functions. (b) Which (if any) of these functions could be exponential? Find formulas for those functions.

In Problems 20–21, find all the tables that have the given characteristic. (A) (B)

𝑥

𝑓 (𝑥)

𝑔(𝑥)

ℎ(𝑥)

−2

12

16

37

−1

17

24

34

0

20

36

31

1

21

54

28

2

18

81

25

(C) (D)

𝑥

0

40

80

160

𝑦

2.2

2.2

2.2

2.2

𝑥

−8

−4

0

8

𝑦

51

62

73

95

𝑥

−4

−3

4

6

𝑦

18

0

4.5

−2.25

𝑥

3

4

5

6

𝑦

18

9

4.5

2.25

20. 𝑦 could be a linear function of 𝑥. 21. 𝑦 could be an exponential function of 𝑥.

20

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

22. Table 1.10 shows some values of a linear function 𝑓 and an exponential function 𝑔. Find exact values (not decimal approximations) for each of the missing entries.

26. Figure 1.36 shows 𝑄 = 50(1.2)𝑡 , 𝑄 = 50(0.6)𝑡 , 𝑄 = 50(0.8)𝑡 , and 𝑄 = 50(1.4)𝑡 . Match each formula to a graph. 𝑄 (I)

Table 1.10 𝑥

0

1

2

3

4

𝑓 (𝑥)

10

?

20

?

?

𝑔(𝑥)

10

?

20

?

?

(II)

(III) (IV)

𝑡

23. Match the functions ℎ(𝑠), 𝑓 (𝑠), and 𝑔(𝑠), whose values are in Table 1.11, with the formulas 𝑦 = 𝑎(1.1)𝑠 , 𝑦 = 𝑏(1.05)𝑠 , 𝑦 = 𝑐(1.03)𝑠 ,

Figure 1.36 In Problems 27–32, give a possible formula for the function.

assuming 𝑎, 𝑏, and 𝑐 are constants. Note that the function values have been rounded to two decimal places.

𝑦

27.

𝑦

28. (2, 12)

Table 1.11

18

3 𝑠

ℎ(𝑠)

𝑠

𝑓 (𝑠)

𝑠

𝑔(𝑠)

2

1.06

1

2.20

3

3.47

3

1.09

2

2.42

4

3.65

4

1.13

3

2.66

5

3.83

5

1.16

4

2.93

6

4.02

6

1.19

5

3.22

7

4.22

𝑥

29.

𝑦

8 (1, 4)

(1, 6)

𝑥

𝑥 −2

𝑦

31.

(c)

Figure 1.34

4

𝑡

𝑔(𝑡)

ℎ(𝑡)

𝑘(𝑡)

1

23

10

2.2

2

24

20

2.5

3

26

29

2.8

4

29

37

3.1

5

33

44

3.4

6

38

50

3.7

25. Each of the functions in Table 1.13 decreases, but each decreases in a different way. Which of the graphs in Figure 1.35 best fits each function? Table 1.13 (a)

(b)

(c)

Figure 1.35

𝑦

32.

Table 1.12 (b)

𝑦

30. (2, 18)

24. Each of the functions 𝑔, ℎ, 𝑘 in Table 1.12 is increasing, but each increases in a different way. Which of the graphs in Figure 1.34 best fits each function?

(a)

(2, 8)

𝑥

(−1, 8)

(1, 2)

(1, 2) 𝑥

𝑥

33. The table gives the number of North American houses (millions) with analog cable TV.20 (a) Plot the number of houses, 𝐻, in millions, with cable TV versus year, 𝑌 . (b) Could 𝐻 be a linear function of 𝑌 ? Why or why not? (c) Could 𝐻 be an exponential function of 𝑌 ? Why or why not?

𝑥

𝑓 (𝑥)

𝑔(𝑥)

ℎ(𝑥)

Year

2010

2011

2012

2013

2014

2015

1

100

22.0

9.3

Houses

18.3

13

7.8

3.9

1

0.5

2

90

21.4

9.1

3

81

20.8

8.8

4

73

20.2

8.4

5

66

19.6

7.9

6

60

19.0

7.3

20 http://www.statista.com.

Accessed May 2015.

34. When a new product is advertised, more and more people try it. However, the rate at which new people try it slows as time goes on. (a) Graph the total number of people who have tried such a product against time. (b) What do you know about the concavity of the graph?

1.2 EXPONENTIAL FUNCTIONS

35. Sketch reasonable graphs for the following. Pay particular attention to the concavity of the graphs. (a) The total revenue generated by a car rental business, plotted against the amount spent on advertising. (b) The temperature of a cup of hot coffee standing in a room, plotted as a function of time. 36. (a) A population, 𝑃 , grows at a continuous rate of 2% a year and starts at 1 million. Write 𝑃 in the form 𝑃 = 𝑃0 𝑒𝑘𝑡 , with 𝑃0 , 𝑘 constants. (b) Plot the population in part (a) against time. 37. A 2008 study of 300 oil fields producing a total of 84 million barrels per day reported that daily production was decaying at a continuous rate of 9.1% per year.21 Find the estimated production in these fields in 2025 if the decay continues at the same rate. 38. In 2014, the world’s population reached 7.17 billion22 and was increasing at a rate of 1.1% per year. Assume that this growth rate remains constant. (In fact, the growth rate has decreased since 2008.) (a) Write a formula for the world population (in billions) as a function of the number of years since 2014. (b) Estimate the population of the world in the year 2020. (c) Sketch world population as a function of years since 2014. Use the graph to estimate the doubling time of the population of the world. 39. Aircraft require longer takeoff distances, called takeoff rolls, at high-altitude airports because of diminished air density. The table shows how the takeoff roll for a certain light airplane depends on the airport elevation. (Takeoff rolls are also strongly influenced by air temperature; the data shown assume a temperature of 0◦ C.) Determine a formula for this particular aircraft that gives the takeoff roll as an exponential function of airport elevation.

21

41. The decrease in the number of colonies of honey bees, essential to pollinate crops providing one quarter of US food consumption, worries policy makers. US beekeepers say that over the three winter months a 5.9% decline in the number of colonies per month is economically sustainable, but higher rates are not.23 (a) Assuming a constant percent colony loss, which function I–III could describe a winter monthly colony loss trend that is economically sustainable? Assume 𝑦 is the number of US bee colonies, 𝑡 is time in months, and 𝑎 is a positive constant. I. 𝑦 = 𝑎(1.059)𝑡 II. 𝑦 = 𝑎(0.962)𝑡 III. 𝑦 = 𝑎(0.935)𝑡 (b) What is the annual bee colony trend described by each of the functions in part (a)? 42. A certain region has a population of 10,000,000 and an annual growth rate of 2%. Estimate the doubling time by guessing and checking. 43. According to the EPA, sales of electronic devices in the US doubled between 1997 and 2009, when 438 million electronic devices sold.24 (a) Find an exponential function, 𝑆(𝑡), to model sales in millions since 1997. (b) What was the annual percentage growth rate between 1997 and 2009? 44. (a) Estimate graphically the doubling time of the exponentially growing population shown in Figure 1.37. Check that the doubling time is independent of where you start on the graph. (b) Show algebraically that if 𝑃 = 𝑃0 𝑎𝑡 doubles between time 𝑡 and time 𝑡 + 𝑑, then 𝑑 is the same number for any 𝑡. population

80,000 60,000 40,000 20,000

Elevation (ft) Takeoff roll (ft)

Sea level

1000

2000

3000

4000

670

734

805

882

967

1 2 3 4 5 6 7 8 9

time (years)

Figure 1.37 45. A deposit of 𝑃0 into a bank account has a doubling time of 50 years. No other deposits or withdrawals are made.

40. One of the main contaminants of a nuclear accident, such as that at Chernobyl, is strontium-90, which decays exponentially at a continuous rate of approximately 2.47% per year. After the Chernobyl disaster, it was suggested that it would be about 100 years before the region would again be safe for human habitation. What percent of the original strontium-90 would still remain then? 21 International

(a) How much money is in the bank account after 50 years? 100 years? 150 years? (Your answer will involve 𝑃0 .) (b) How many times does the amount of money double in 𝑡 years? Use this to write a formula for 𝑃 , the amount of money in the account after 𝑡 years.

Energy Agency, World Energy Outlook, 2008. accessed June 14, 2015. 23 “A Key to America’s Crops Is Disappearing at a Staggering Rate,” www.businessinsider.com, accessed January 2016. 24 http://www.epa.gov/osw/conserve/materials/ecycling/docs /summarybaselinereport2011.pdf. Accessed DATE 22 www.indexmundi.com,

22

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

46. A 325 mg aspirin has a half-life of 𝐻 hours in a patient’s body. (a) How long does it take for the quantity of aspirin in the patient’s body to be reduced to 162.5 mg? To 81.25 mg? To 40.625 mg? (Note that 162.5 = 325∕2, etc. Your answers will involve 𝐻.) (b) How many times does the quantity of aspirin, 𝐴 mg, in the body halve in 𝑡 hours? Use this to give a formula for 𝐴 after 𝑡 hours. 47. (a) The half-life of radium-226 is 1620 years. If the initial quantity of radium is 𝑄0 , explain why the quantity, 𝑄, of radium left after 𝑡 years, is given by 𝑄 = 𝑄0 (0.99572)𝑡 . (b) What percentage of the original amount of radium is left after 500 years? 48. In the early 1960s, radioactive strontium-90 was released during atmospheric testing of nuclear weapons and got into the bones of people alive at the time. If the half-life of strontium-90 is 29 years, what fraction of the strontium-90 absorbed in 1960 remained in people’s bones in 2010? [Hint: Write the function in the form 𝑄 = 𝑄0 (1∕2)𝑡∕29 .] 49. Food bank usage in Britain has grown dramatically over the past decade. The number of users, in thousands, of the largest food bank in year 𝑡 is estimated to be 𝑁(𝑡) = 1.3𝑒0.81𝑡 , where 𝑡 is the number of years since 2006.25 (a) What does the 1.3 represent in this context? Give units. (b) What is the continuous growth rate of users per year? (c) What is the annual percent growth rate of users per year? (d) Using only your answer for part (c), decide if the doubling time is more or less than 1 year. Problems 50–51 concern biodiesel, a fuel derived from renewable resources such as food crops, algae, and animal oils. The table shows the percent growth over the previous year in US biodiesel consumption.26

(b) Graph the points showing the annual US consumption of biodiesel, in millions of gallons of biodiesel, for the years 2009 to 2014. Label the scales on the horizontal and vertical axes. 51. (a) True or false: The annual US consumption of biodiesel grew exponentially from 2008 to 2010. Justify your answer without doing any calculations. (b) According to this data, during what single year(s), if any, did the US consumption of biodiesel at least triple? 52. Hydroelectric power is electric power generated by the force of moving water. The table shows the annual percent change in hydroelectric power consumption by the US industrial sector.27 Year

2009

2010

2011

2012

2013

2014

% growth over previous yr

6.3

−4.9

22.4

−15.5

−3.0

−3.4

(a) According to the US Department of Energy, the US industrial sector consumed about 2.65 quadrillion28 BTUs of hydroelectric power in 2009. Approximately how much hydroelectric power (in quadrillion BTUs) did the US consume in 2010? In 2011? (b) Graph the points showing the annual US consumption of hydroelectric power, in quadrillion BTUs, for the years 2009 to 2014. Label the scales on the horizontal and vertical axes. (c) According to this data, when did the largest yearly decrease, in quadrillion BTUs, in the US consumption of hydroelectric power occur? What was this decrease? Problems 53–54 concern wind power, which has been used for centuries to propel ships and mill grain. Modern wind power is obtained from windmills that convert wind energy into electricity. Figure 1.38 shows the annual percent growth in US wind power consumption29 between 2009 and 2014. percent growth over previous year

30 20

Year

2008

2009

2010

2011

2012

2013

2014

% growth

−14.1

5.9

−19.3

240.8

1.0

56.9

1.1

10 year

2010

50. (a) According to the US Department of Energy, the US consumed 322 million gallons of biodiesel in 2009. Approximately how much biodiesel (in millions of gallons) did the US consume in 2010? In 2011?

2012

2014

Figure 1.38

25 Estimates for the Trussell Trust: http://www.bbc.com/news/education-30346060 and http://www.trusselltrust.org/stats. Accessed November 2015. 26 www.eia.doe.gov, accessed March 29, 2015. 27 From www.eia.doe.gov, accessed March 31, 2015. 28 1 quadrillion BTU=1015 BTU. 29 Yearly values have been joined with line segments to highlight trends in the data. Actual values in between years should not be inferred from the segments. From www.eia.doe.gov, accessed April 1, 2015.

1.3 NEW FUNCTIONS FROM OLD

53. (a) According to the US Department of Energy, the US consumption of wind power was 721 trillion BTUs in 2009. How much wind power did the US consume in 2010? In 2011? (b) Graph the points showing the annual US consumption of wind power, in trillion BTUs, for the years 2009 to 2014. Label the scales on the horizontal and vertical axes. (c) Based on this data, in what year did the largest yearly increase, in trillion BTUs, in the US consumption of wind power occur? What was this increase? 54. (a) According to Figure 1.38, during what single year(s), if any, did the US consumption of wind power energy increase by at least 25%? Decrease by at least 25%? (b) True or false: The US consumption of wind power energy doubled from 2008 to 2011?

23

55. (a) The exponential functions in Figure 1.39 have 𝑏, 𝑑, 𝑞 positive. Which of the constants 𝑎, 𝑐, and 𝑝 must be positive? (b) Which of the constants 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, and 𝑞 must be between 0 and 1? (c) Which two of the constants 𝑎, 𝑏, 𝑐, 𝑑, 𝑝, and 𝑞 must be equal? (d) What information about the constants 𝑎 and 𝑏 does the point (1, 1) provide? 𝑦 𝑦 = 𝑐 ⋅ 𝑑𝑥 𝑦=𝑝

𝑦 = 𝑎 ⋅ 𝑏𝑥

⋅ 𝑞𝑥 (1, 1)

𝑥

Figure 1.39

Strengthen Your Understanding In Problems 56–59, explain what is wrong with the statement.

64. An exponential function that grows slower than 𝑦 = 𝑒𝑥 for 𝑥 > 0.

56. A quantity that doubles daily has an exponential growth rate of 200% per day. 57. The function 𝑦 = 𝑒−0.25𝑥 is decreasing and its graph is concave down. 58. The function 𝑦 = 2𝑥 is increasing, and its graph is concave up. 59. The points (0, 1), (1, 𝑒), (2, 2𝑒), and (3, 3𝑒) are all on the graph of 𝑦 = 𝑒𝑥

Are the statements in Problems 65–72 true or false? Give an explanation for your answer.

In Problems 60–64, give an example of: 60. A decreasing exponential function with a vertical intercept of 𝜋. 61. A formula representing the statement “𝑞 decreases at a constant percent rate, and 𝑞 = 2.2 when 𝑡 = 0.” 62. A function that is increasing at a constant percent rate and that has the same vertical intercept as 𝑓 (𝑥) = 0.3𝑥 + 2. 63. A function with a horizontal asymptote at 𝑦 = −5 and range 𝑦 > −5.

1.3

65. The function 𝑓 (𝑥) = 𝑒2𝑥 ∕(2𝑒5𝑥 ) is exponential. 66. The function 𝑦 = 2 + 3𝑒−𝑡 has a 𝑦-intercept of 𝑦 = 3. 67. The function 𝑦 = 5 − 3𝑒−4𝑡 has a horizontal asymptote of 𝑦 = 5. 68. If 𝑦 = 𝑓 (𝑥) is an exponential function and if increasing 𝑥 by 1 increases 𝑦 by a factor of 5, then increasing 𝑥 by 2 increases 𝑦 by a factor of 10. 69. If 𝑦 = 𝐴𝑏𝑥 and increasing 𝑥 by 1 increases 𝑦 by a factor of 3, then increasing 𝑥 by 2 increases 𝑦 by a factor of 9. 70. An exponential function can be decreasing. 71. If 𝑎 and 𝑏 are positive constants, 𝑏 ≠ 1, then 𝑦 = 𝑎+𝑎𝑏𝑥 has a horizontal asymptote. 72. The function 𝑦 = 20∕(1 + 2𝑒−𝑘𝑡 ) with 𝑘 > 0, has a horizontal asymptote at 𝑦 = 20.

NEW FUNCTIONS FROM OLD

Shifts and Stretches The graph of a constant multiple of a given function is easy to visualize: each 𝑦-value is stretched or shrunk by that multiple. For example, consider the function 𝑓 (𝑥) and its multiples 𝑦 = 3𝑓 (𝑥) and 𝑦 = −2𝑓 (𝑥). Their graphs are shown in Figure 1.40. The factor 3 in the function 𝑦 = 3𝑓 (𝑥) stretches each 𝑓 (𝑥) value by multiplying it by 3; the factor −2 in the function 𝑦 = −2𝑓 (𝑥) stretches 𝑓 (𝑥) by multiplying by 2 and reflects it across the 𝑥-axis. You can think of the multiples of a given function as a family of functions.

24

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

𝑦 𝑦

3

𝑦 = 𝑥2 + 4

𝑦 𝑦 = 𝑥2

𝑦 = −2𝑓 (𝑥) 𝑦 = 𝑥2 𝑥

𝑦 = (𝑥 − 2)2

𝑦 = 𝑓 (𝑥)

4 𝑥

𝑥 −3

𝑦 = 3𝑓 (𝑥)

Figure 1.40: Multiples of the function 𝑓 (𝑥)

2 Figure 1.41: Graphs of 𝑦 = 𝑥2 with 𝑦 = 𝑥2 + 4 and 𝑦 = (𝑥 − 2)2

It is also easy to create families of functions by shifting graphs. For example, 𝑦 − 4 = 𝑥2 is the same as 𝑦 = 𝑥2 + 4, which is the graph of 𝑦 = 𝑥2 shifted up by 4. Similarly, 𝑦 = (𝑥 − 2)2 is the graph of 𝑦 = 𝑥2 shifted right by 2. (See Figure 1.41.) • Multiplying a function by a constant, 𝑐, stretches the graph vertically (if 𝑐 > 1) or shrinks the graph vertically (if 0 < 𝑐 < 1). A negative sign (if 𝑐 < 0) reflects the graph across the 𝑥-axis, in addition to shrinking or stretching. • Replacing 𝑦 by (𝑦 − 𝑘) moves a graph up by 𝑘 (down if 𝑘 is negative). • Replacing 𝑥 by (𝑥 − ℎ) moves a graph to the right by ℎ (to the left if ℎ is negative).

Composite Functions If oil is spilled from a tanker, the area of the oil slick grows with time. Suppose that the oil slick is always a perfect circle. Then the area, 𝐴, of the oil slick is a function of its radius, 𝑟: 𝐴 = 𝑓 (𝑟) = 𝜋𝑟2 . The radius is also a function of time, because the radius increases as more oil spills. Thus, the area, being a function of the radius, is also a function of time. If, for example, the radius is given by 𝑟 = 𝑔(𝑡) = 1 + 𝑡, then the area is given as a function of time by substitution: 𝐴 = 𝜋𝑟2 = 𝜋(1 + 𝑡)2 . We are thinking of 𝐴 as a composite function or a “function of a function,” which is written 𝐴 = 𝑓 (𝑔(𝑡)) = 𝜋(𝑔(𝑡))2 = 𝜋(1 + 𝑡)2 . ⏟⏟⏟ Composite function; 𝑓 is outside function, 𝑔 is inside function

To calculate 𝐴 using the formula 𝜋(1 + 𝑡)2 , the first step is to find 1 + 𝑡, and the second step is to square and multiply by 𝜋. The first step corresponds to the inside function 𝑔(𝑡) = 1 + 𝑡, and the second step corresponds to the outside function 𝑓 (𝑟) = 𝜋𝑟2 . Example 1

If 𝑓 (𝑥) = 𝑥2 and 𝑔(𝑥) = 𝑥 − 2, find each of the following: (a) 𝑓 (𝑔(3)) (b) 𝑔(𝑓 (3)) (c) 𝑓 (𝑔(𝑥)) (d) 𝑔(𝑓 (𝑥))

Solution

(a) Since 𝑔(3) = 1, we have 𝑓 (𝑔(3)) = 𝑓 (1) = 1. (b) Since 𝑓 (3) = 9, we have 𝑔(𝑓 (3)) = 𝑔(9) = 7. Notice that 𝑓 (𝑔(3)) ≠ 𝑔(𝑓 (3)). (c) 𝑓 (𝑔(𝑥)) = 𝑓 (𝑥 − 2) = (𝑥 − 2)2 . (d) 𝑔(𝑓 (𝑥)) = 𝑔(𝑥2 ) = 𝑥2 − 2. Again, notice that 𝑓 (𝑔(𝑥)) ≠ 𝑔(𝑓 (𝑥)). Notice that the horizontal shift in Figure 1.41 can be thought of as a composition 𝑓 (𝑔(𝑥)) = (𝑥 − 2)2 .

1.3 NEW FUNCTIONS FROM OLD

Example 2

Express each of the following functions as a composition: (a) ℎ(𝑡) = (1 + 𝑡3 )27

Solution

25

(b) 𝑘(𝑦) = 𝑒−𝑦

2

(c) 𝑙(𝑦) = −(𝑒𝑦 )2

In each case think about how you would calculate a value of the function. The first stage of the calculation gives you the inside function, and the second stage gives you the outside function. (a) For (1 + 𝑡3 )27 , the first stage is cubing and adding 1, so an inside function is 𝑔(𝑡) = 1 + 𝑡3 . The second stage is taking the 27th power, so an outside function is 𝑓 (𝑦) = 𝑦27 . Then 𝑓 (𝑔(𝑡)) = 𝑓 (1 + 𝑡3 ) = (1 + 𝑡3 )27 . In fact, there are lots of different answers: 𝑔(𝑡) = 𝑡3 and 𝑓 (𝑦) = (1 + 𝑦)27 is another possibility. 2 (b) To calculate 𝑒−𝑦 we square 𝑦, take its negative, and then take 𝑒 to that power. So if 𝑔(𝑦) = −𝑦2 and 𝑓 (𝑧) = 𝑒𝑧 , then we have 2 𝑓 (𝑔(𝑦)) = 𝑒−𝑦 . (c) To calculate −(𝑒𝑦 )2 , we find 𝑒𝑦 , square it, and take the negative. Using the same definitions of 𝑓 and 𝑔 as in part (b), the composition is 𝑔(𝑓 (𝑦)) = −(𝑒𝑦 )2 . Since parts (b) and (c) give different answers, we see the order in which functions are composed is important.

Odd and Even Functions: Symmetry There is a certain symmetry apparent in the graphs of 𝑓 (𝑥) = 𝑥2 and 𝑔(𝑥) = 𝑥3 in Figure 1.42. For each point (𝑥, 𝑥2 ) on the graph of 𝑓 , the point (−𝑥, 𝑥2 ) is also on the graph; for each point (𝑥, 𝑥3 ) on the graph of 𝑔, the point (−𝑥, −𝑥3) is also on the graph. The graph of 𝑓 (𝑥) = 𝑥2 is symmetric about the 𝑦-axis, whereas the graph of 𝑔(𝑥) = 𝑥3 is symmetric about the origin. The graph of any polynomial involving only even powers of 𝑥 has symmetry about the 𝑦-axis, while polynomials with only odd powers of 𝑥 are symmetric about the origin. Consequently, any functions with these symmetry properties are called even and odd, respectively. Even function

Odd function

𝑓 (𝑥) = 𝑥2 (𝑥, 𝑥2 )

2

(−𝑥, 𝑥 )

𝑔(𝑥) = 𝑥3 (𝑥, 𝑥3 )

−𝑥

𝑥 𝑥

3

𝑥 −𝑥

(−𝑥, −𝑥 )

𝑥

Figure 1.42: Symmetry of even and odd functions

For any function 𝑓 , 𝑓 is an even function if 𝑓 (−𝑥) = 𝑓 (𝑥) for all 𝑥. 𝑓 is an odd function if 𝑓 (−𝑥) = −𝑓 (𝑥) for all 𝑥. 2

For example, 𝑔(𝑥) = 𝑒𝑥 is even and ℎ(𝑥) = 𝑥1∕3 is odd. However, many functions do not have any symmetry and are neither even nor odd.

26

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

Inverse Functions On August 26, 2005, the runner Kenenisa Bekele30 of Ethiopia set a world record for the 10,000meter race. His times, in seconds, at 2000-meter intervals are recorded in Table 1.14, where 𝑡 = 𝑓 (𝑑) is the number of seconds Bekele took to complete the first 𝑑 meters of the race. For example, Bekele ran the first 4000 meters in 629.98 seconds, so 𝑓 (4000) = 629.98. The function 𝑓 was useful to athletes planning to compete with Bekele. Let us now change our point of view and ask for distances rather than times. If we ask how far Bekele ran during the first 629.98 seconds of his race, the answer is clearly 4000 meters. Going backward in this way from numbers of seconds to numbers of meters gives 𝑓 −1 , the inverse function31 of 𝑓 . We write 𝑓 −1 (629.98) = 4000. Thus, 𝑓 −1 (𝑡) is the number of meters that Bekele ran during the first 𝑡 seconds of his race. See Table 1.15, which contains values of 𝑓 −1 . The independent variable for 𝑓 is the dependent variable for 𝑓 −1 , and vice versa. The domains and ranges of 𝑓 and 𝑓 −1 are also interchanged. The domain of 𝑓 is all distances 𝑑 such that 0 ≤ 𝑑 ≤ 10000, which is the range of 𝑓 −1 . The range of 𝑓 is all times 𝑡, such that 0 ≤ 𝑡 ≤ 1577.53, which is the domain of 𝑓 −1 . Table 1.14

𝑡 = 𝑓 (𝑑) (seconds)

𝑑 (meters)

Distance run by Bekele

Table 1.15

Bekele’s running time

𝑑 = 𝑓 −1 (𝑡) (meters)

𝑡 (seconds)

0

0.00

0.00

0

2000

315.63

315.63

2000

4000

629.98

629.98

4000 6000

6000

944.66

944.66

8000

1264.63

1264.63

8000

10000

1577.53

1577.53

10000

Which Functions Have Inverses? If a function has an inverse, we say it is invertible. Let’s look at a function which is not invertible. Consider the flight of the Mercury spacecraft Freedom 7, which carried Alan Shepard, Jr. into space in May 1961. Shepard was the first American to journey into space. After launch, his spacecraft rose to an altitude of 116 miles, and then came down into the sea. The function 𝑓 (𝑡) giving the altitude in miles 𝑡 minutes after lift-off does not have an inverse. To see why it does not, try to decide on a value for 𝑓 −1 (100), which should be the time when the altitude of the spacecraft was 100 miles. However, there are two such times, one when the spacecraft was ascending and one when it was descending. (See Figure 1.43.) The reason the altitude function does not have an inverse is that the altitude has the same value for two different times. The reason the Bekele time function did have an inverse is that each running time, 𝑡, corresponds to a unique distance, 𝑑. 𝑑 (miles)

𝑦 = 𝑓 (𝑥)

116 100

𝑦 𝑓 (𝑡)

Inverse function

𝑓 (𝑥) ✛



✻ Original function

𝑡1

𝑡2

𝑡 (min)

Figure 1.43: Two times, 𝑡1 and 𝑡2 , at which altitude of spacecraft is 100 miles 30 kenenisabekelle.com/, 31 The

❄ 𝑥

𝑓 −1 (𝑦)

Figure 1.44: A function which has an inverse

accessed January 11, 2011. notation 𝑓 −1 represents the inverse function, which is not the same as the reciprocal, 1∕𝑓 .

1.3 NEW FUNCTIONS FROM OLD

27

Figure 1.44 suggests when an inverse exists. The original function, 𝑓 , takes us from an 𝑥-value to a 𝑦-value, as shown in Figure 1.44. Since having an inverse means there is a function going from a 𝑦-value to an 𝑥-value, the crucial question is whether we can get back. In other words, does each 𝑦-value correspond to a unique 𝑥-value? If so, there’s an inverse; if not, there is not. This principle may be stated geometrically, as follows: A function has an inverse if (and only if) its graph intersects any horizontal line at most once. For example, the function 𝑓 (𝑥) = 𝑥2 does not have an inverse because many horizontal lines intersect the parabola twice.

Definition of an Inverse Function If the function 𝑓 is invertible, its inverse is defined as follows: 𝑓 −1 (𝑦) = 𝑥

means 𝑦 = 𝑓 (𝑥).

Formulas for Inverse Functions If a function is defined by a formula, it is sometimes possible to find a formula for the inverse function. In Section 1.1, we looked at the snowy tree cricket, whose chirp rate, 𝐶, in chirps per minute, is approximated at the temperature, 𝑇 , in degrees Fahrenheit, by the formula 𝐶 = 𝑓 (𝑇 ) = 4𝑇 − 160. So far we have used this formula to predict the chirp rate from the temperature. But it is also possible to use this formula backward to calculate the temperature from the chirp rate. Example 3

Find the formula for the function giving temperature in terms of the number of cricket chirps per minute; that is, find the inverse function 𝑓 −1 such that 𝑇 = 𝑓 −1 (𝐶).

Solution

Since 𝐶 is an increasing function, 𝑓 is invertible. We know 𝐶 = 4𝑇 − 160. We solve for 𝑇 , giving 𝑇 =

𝐶 + 40, 4

so 𝑓 −1 (𝐶) =

𝐶 + 40. 4

Graphs of Inverse Functions The function 𝑓 (𝑥) = 𝑥3 is increasing everywhere and so has an inverse. To find the inverse, we solve 𝑦 = 𝑥3 for 𝑥, giving 𝑥 = 𝑦1∕3 . The inverse function is 𝑓 −1 (𝑦) = 𝑦1∕3 or, if we want to call the independent variable 𝑥, 𝑓 −1 (𝑥) = 𝑥1∕3 .

28

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

The graphs of 𝑦 = 𝑥3 and 𝑦 = 𝑥1∕3 are shown in Figure 1.45. Notice that these graphs are the reflections of one another across the line 𝑦 = 𝑥. For example, (8, 2) is on the graph of 𝑦 = 𝑥1∕3 because 2 = 81∕3 , and (2, 8) is on the graph of 𝑦 = 𝑥3 because 8 = 23 . The points (8, 2) and (2, 8) are reflections of one another across the line 𝑦 = 𝑥. In general, we have the following result. If the 𝑥- and 𝑦-axes have the same scales, the graph of 𝑓 −1 is the reflection of the graph of 𝑓 across the line 𝑦 = 𝑥. 𝑦 𝑦 = 𝑥3

𝑦=𝑥

8 6 4 𝑦 = 𝑥1∕3

2 −2

𝑥 2 −2

𝑥1∕3

4

6

8

𝑥3 Figure 1.45: Graphs of inverse functions, 𝑦 = 𝑥3 and 𝑦 = 𝑥1∕3 , are reflections across the line 𝑦 = 𝑥

Exercises and Problems for Section 1.3 EXERCISES For the functions 𝑓 in Exercises 1–3, graph:

6. 𝑘(𝑡) = 𝑚(𝑡 + 1.5)

(a) 𝑓 (𝑥 + 2) (b) 𝑓 (𝑥 − 1) (d) 𝑓 (𝑥 + 1) + 3 (e) 3𝑓 (𝑥)

7. 𝑤(𝑡) = 𝑚(𝑡 − 0.5) − 2.5

1.

𝑓 (𝑥)

4

(c) 𝑓 (𝑥) − 4 (f) −𝑓 (𝑥) + 1

2.

3

3

2

2

1

1 0

1

(a) 𝑦 = 𝑓 (𝑥) + 3 (c) 𝑦 = 𝑓 (𝑥 + 4)

𝑓 (𝑥)

2

𝑥

2

−2 −1

(b) 𝑦 = 2𝑓 (𝑥) (d) 𝑦 = 4 − 𝑓 (𝑥) 𝑦

𝑥 −2 −1

8. Use Figure 1.47 to graph each of the following. Label any intercepts or asymptotes that can be determined.

4

0

1

2 1

3.

4 𝑓 (𝑥)

𝑥

2

−5

−3

−1

2

𝑥 −2

5

−1

2 −2

Figure 1.47

In Exercises 4–7, use Figure 1.46 to graph the functions. For the functions 𝑓 and 𝑔 in Exercises 9–12, find

2

(a) 𝑓 (𝑔(1)) (d) 𝑔(𝑓 (𝑥))

𝑚(𝑡) −3

(b) 𝑔(𝑓 (1)) (e) 𝑓 (𝑡)𝑔(𝑡)

𝑡 5 −1

Figure 1.46 4. 𝑛(𝑡) = 𝑚(𝑡) + 2

5. 𝑝(𝑡) = 𝑚(𝑡 − 1)

9. 𝑓 (𝑥) = 𝑥2 , 𝑔(𝑥) = 𝑥 + 1 √ 10. 𝑓 (𝑥) = 𝑥 + 4, 𝑔(𝑥) = 𝑥2

11. 𝑓 (𝑥) = 𝑒𝑥 , 𝑔(𝑥) = 𝑥2

12. 𝑓 (𝑥) = 1∕𝑥, 𝑔(𝑥) = 3𝑥 + 4

(c) 𝑓 (𝑔(𝑥))

29

1.3 NEW FUNCTIONS FROM OLD

13. If 𝑓 (𝑥) = 𝑥2 + 1, find and simplify: (b) 𝑓 (𝑡2 + 1) (c) (e) (𝑓 (𝑡))2 + 1

(a) 𝑓 (𝑡 + 1) (d) 2𝑓 (𝑡)

For Exercises 29–31, use a graph of the function to decide whether or not it is invertible.

𝑓 (2)

29. 𝑓 (𝑥) = 𝑥2 + 3𝑥 + 2

30. 𝑓 (𝑥) = 𝑥3 − 5𝑥 + 10

2

14. For 𝑔(𝑥) = 𝑥 + 2𝑥 + 3, find and simplify: (a) 𝑔(2 + ℎ) (c) 𝑔(2 + ℎ) − 𝑔(2)

31. 𝑓 (𝑥) = 𝑥3 + 5𝑥 + 10

(b) 𝑔(2)

Simplify the quantities in Exercises 15–18 using 𝑚(𝑧) = 𝑧2 . 15. 𝑚(𝑧 + 1) − 𝑚(𝑧)

16. 𝑚(𝑧 + ℎ) − 𝑚(𝑧)

17. 𝑚(𝑧) − 𝑚(𝑧 − ℎ)

18. 𝑚(𝑧 + ℎ) − 𝑚(𝑧 − ℎ)

Are the functions in Exercises 19–26 even, odd, or neither? 19. 𝑓 (𝑥) = 𝑥6 + 𝑥3 + 1

20. 𝑓 (𝑥) = 𝑥3 + 𝑥2 + 𝑥

21. 𝑓 (𝑥) = 𝑥4 − 𝑥2 + 3

22. 𝑓 (𝑥) = 𝑥3 + 1

32. Let 𝑝 be the price of an item and 𝑞 be the number of items sold at that price, where 𝑞 = 𝑓 (𝑝). What do the following quantities mean in terms of prices and quantities sold? (b) 𝑓 −1 (30)

(a) 𝑓 (25)

33. Let 𝐶 = 𝑓 (𝐴) be the cost, in dollars, of building a store of area 𝐴 square feet. In terms of cost and square feet, what do the following quantities represent? (b) 𝑓 −1 (20,000)

(a) 𝑓 (10,000)

𝑥2 −1

23. 𝑓 (𝑥) = 2𝑥

24. 𝑓 (𝑥) = 𝑒 2

34. Let 𝑓 (𝑥) be the temperature (◦ F) when the column of mercury in a particular thermometer is 𝑥 inches long. What is the meaning of 𝑓 −1 (75) in practical terms?

26. 𝑓 (𝑥) = 𝑒𝑥 − 𝑥

25. 𝑓 (𝑥) = 𝑥(𝑥 − 1)

For Exercises 27–28, decide if the function 𝑦 = 𝑓 (𝑥) is invertible. 𝑦 𝑦 28. 27. 𝑓 𝑓 𝑥 𝑥

35. (a) Write an equation for a graph obtained by vertically stretching the graph of 𝑦 = 𝑥2 by a factor of 2, followed by a vertical upward shift of 1 unit. Sketch it. (b) What is the equation if the order of the transformations (stretching and shifting) in part (a) is interchanged? (c) Are the two graphs the same? Explain the effect of reversing the order of transformations.

PROBLEMS 36. How does the graph of 𝑄 = 𝑆(1 − 𝑒−𝑘𝑡 ) in Example 4 on page 18 relate to the graph of the exponential decay function, 𝑦 = 𝑆𝑒−𝑘𝑡 ?

39. 𝑢(𝑣(10))

40. 𝑢(𝑣(40))

41. 𝑣(𝑢(10))

42. 𝑣(𝑢(40))

For Problems 43–48, use the graphs in Figure 1.49. In Problems 37–38 find possible formulas for the graphs using shifts of 𝑥2 or 𝑥3 . 𝑦 𝑦 37. 38.

3

3 𝑓 (𝑥)

𝑔(𝑥)

𝑥

(−1, 3)

−3 𝑥

3

𝑥 −3

3

𝑥 −3

(2, −1)

−3

Figure 1.49 In Problems 39–42, use Figure 1.48 to estimate the function value or explain why it cannot be done. 100

𝑢(𝑥)

100 𝑣(𝑥)

43. Estimate 𝑓 (𝑔(1)).

44. Estimate 𝑔(𝑓 (2)).

45. Estimate 𝑓 (𝑓 (1)).

46. Graph 𝑓 (𝑔(𝑥)).

47. Graph 𝑔(𝑓 (𝑥)).

48. Graph 𝑓 (𝑓 (𝑥)).

For Problems 49–52, determine functions 𝑓 and 𝑔 such that ℎ(𝑥) = 𝑓 (𝑔(𝑥)). (Note: There is more than one correct answer. Do not choose 𝑓 (𝑥) = 𝑥 or 𝑔(𝑥) = 𝑥.) 𝑥 50

Figure 1.48

𝑥 50

49. ℎ(𝑥) = (𝑥 + 1)3 √ 51. ℎ(𝑥) = 𝑥2 + 4

50. ℎ(𝑥) = 𝑥3 + 1 52. ℎ(𝑥) = 𝑒2𝑥

30

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

53. A tree of height 𝑦 meters has, on average, 𝐵 branches, where 𝐵 = 𝑦−1. Each branch has, on average, 𝑛 leaves, where 𝑛 = 2𝐵 2 − 𝐵. Find the average number of leaves on a tree as a function of height. 54. A spherical balloon is growing with radius 𝑟 = 3𝑡 + 1, in centimeters, for time 𝑡 in seconds. Find the volume of the balloon at 3 seconds. 55. Complete the following table with values for the functions 𝑓 , 𝑔, and ℎ, given that: (a) 𝑓 is an even function. (b) 𝑔 is an odd function. (c) ℎ is the composition ℎ(𝑥) = 𝑔(𝑓 (𝑥)). 𝑓 (𝑥)

𝑔(𝑥)

−3

0

0

−2

2

2

−1

2

2

0

0

0

𝑥

In Problems 62–66, interpret the expression in terms of carbon footprint, a measure of the environmental impact in kilograms of green house gas (GHG) emissions. Assume that a bottle of drinking water that travels 𝑑 km from its production source has a carbon footprint 𝑓 (𝑑) kg of GHGs.32 62. 𝑓 (150)

63. 𝑓 (8700) = 0.25

64. 𝑓 −1 (1.1)

65. 𝑓 (150) + 𝑓 (0)

66.

𝑓 (1500) − 𝑓 (150) 1500 − 150

In Problems 67–70 the functions 𝑟 = 𝑓 (𝑡) and 𝑉 = 𝑔(𝑟) give the radius and the volume of a commercial hot air balloon being inflated for testing. The variable 𝑡 is in minutes, 𝑟 is in feet, and 𝑉 is in cubic feet. The inflation begins at 𝑡 = 0. In each case, give a mathematical expression that represents the given statement.

ℎ(𝑥)

67. The volume of the balloon 𝑡 minutes after inflation began.

1 2

68. The volume of the balloon if its radius were twice as big.

3

56. Write a table of values for 𝑓 −1 , where 𝑓 is as given below. The domain of 𝑓 is the integers from 1 to 7. State the domain of 𝑓 −1 . 𝑥

1

2

3

4

5

6

7

𝑓 (𝑥)

3

−7

19

4

178

2

1

57. (a) Use Figure 1.50 to estimate 𝑓 −1 (2). (b) Sketch a graph of 𝑓 −1 on the same axes. 𝑦

69. The time that has elapsed when the radius of the balloon is 30 feet. 70. The time that has elapsed when the volume of the balloon is 10,000 cubic feet. 71. The cost of producing 𝑞 articles is given by the function 𝐶 = 𝑓 (𝑞) = 100 + 2𝑞. (a) Find a formula for the inverse function. (b) Explain in practical terms what the inverse function tells you.

4

72. Figure 1.51 shows 𝑓 (𝑡), the number (in millions) of motor vehicles registered33 in the world in the year 𝑡.

𝑓 (𝑥) 𝑥 −4

4

−4

Figure 1.50

(a) Is 𝑓 invertible? Explain. (b) What is the meaning of 𝑓 −1 (400) in practical terms? Evaluate 𝑓 −1 (400). (c) Sketch the graph of 𝑓 −1 . (millions)

1000

For Problems 58–61, decide if the function 𝑓 is invertible. 58. 𝑓 (𝑑) is the total number of gallons of fuel an airplane has used by the end of 𝑑 minutes of a particular flight. 59. 𝑓 (𝑡) is the number of customers in Saks Fifth Avenue at 𝑡 minutes past noon on December 18, 2014. 60. 𝑓 (𝑛) is the number of students in your calculus class whose birthday is on the 𝑛th day of the year. 61. 𝑓 (𝑤) is the cost of mailing a letter weighing 𝑤 grams. 32 For

800 600 400 200 (year) ’65 ’70 ’75 ’80 ’85 ’90 ’95 ’00 ’05 ’10

Figure 1.51

the data in Problem 63, see www.triplepundit.com. Accessed March 1, 2015. accessed June 5, 2011. In 2000, about 30% of the registered vehicles were in the US.

33 www.earth-policy.org,

31

1.3 NEW FUNCTIONS FROM OLD

73. Figure 1.52 is a graph of the function 𝑓 (𝑡). Here 𝑓 (𝑡) is the depth in meters below the Atlantic Ocean floor where 𝑡 million-year-old rock can be found.34 (a) Evaluate 𝑓 (15), and say what it means in practical terms. (b) Is 𝑓 invertible? Explain. (c) Evaluate 𝑓 −1 (120), and say what it means in practical terms. (d) Sketch a graph of 𝑓 −1 . Time (millions of years)

0

5

74. Figure 1.53 shows graphs of 4 useful functions: the step, the sign, the ramp, and the absolute value. We have { 0 if 𝑥 < 0 step(𝑥) = . 1 if 𝑥 ≥ 0 Match the remaining 3 graphs with the following formulas in terms of the step function. (a) 𝑥 step(𝑥) (b) step(𝑥) − step(−𝑥) (c) 𝑥 step(𝑥) − 𝑥 step(−𝑥) sign(𝑥)

step(𝑥)

10 15 20 25 30 35 40 1

1

20

𝑥 −100

40

Depth below 60 ocean floor 80 (m)

100

𝑥 −100

100

−1

−1

100

100 ramp(𝑥)

100 120

abs(𝑥)

𝑥

140

−100

Figure 1.52

100

𝑥 −100

100

Figure 1.53

Strengthen Your Understanding In Problems 75–79, explain what is wrong with the statement.

85. The graph of 𝑓 (𝑥) = 100(10𝑥 ) is a horizontal shift of the graph of 𝑔(𝑥) = 10𝑥 .

75. The graph of 𝑓 (𝑥) = −(𝑥 + 1)3 is the graph of 𝑔(𝑥) = −𝑥3 shifted right by 1 unit.

86. If 𝑓 is an increasing function, then 𝑓 −1 is an increasing function.

76. The functions 𝑓 (𝑥) = 3𝑥 and 𝑔(𝑥) = 𝑥3 are inverses of each other.

87. If a function is even, then it does not have an inverse.

77. 𝑓 (𝑥) = 3𝑥+5 and 𝑔(𝑥) = −3𝑥−5 are inverse functions of each other.

88. If a function is odd, then it does not have an inverse. 2

78. The function 𝑓 (𝑥) = 𝑒𝑥 is its own inverse.

89. The function 𝑓 (𝑥) = 𝑒−𝑥 is decreasing for all 𝑥.

79. The inverse of 𝑓 (𝑥) = 𝑥 is 𝑓 −1 (𝑥) = 1∕𝑥.

90. If 𝑔(𝑥) is an even function then 𝑓 (𝑔(𝑥)) is even for every function 𝑓 (𝑥).

In Problems 80–83, give an example of:

91. If 𝑓 (𝑥) is an even function then 𝑓 (𝑔(𝑥)) is even for every function 𝑔(𝑥).

80. An invertible function whose graph contains the point (0, 3). 81. An even function whose graph does not contain the point (0, 0). 82. An increasing function 𝑓 (𝑥) whose values are greater than those of its inverse function 𝑓 −1 (𝑥) for 𝑥 > 0. 83. Two functions 𝑓 (𝑥) and 𝑔(𝑥) such that moving the graph of 𝑓 to the left 2 units gives the graph of 𝑔 and moving the graph of 𝑓 up 3 also gives the graph of 𝑔. Are the statements in Problems 84–93 true or false? Give an explanation for your answer. 84. The composition of the exponential functions 𝑓 (𝑥) = 2𝑥 and 𝑔(𝑥) = 3𝑥 is exponential.

92. There is a function which is both even and odd. 93. The composition of two linear functions is linear. In Problems 94–97, suppose 𝑓 is an increasing function and 𝑔 is a decreasing function. Give an example of functions 𝑓 and 𝑔 for which the statement is true, or say why such an example is impossible. 94. 𝑓 (𝑥) + 𝑔(𝑥) is decreasing for all 𝑥. 95. 𝑓 (𝑥) − 𝑔(𝑥) is decreasing for all 𝑥. 96. 𝑓 (𝑥)𝑔(𝑥) is decreasing for all 𝑥. 97. 𝑓 (𝑔(𝑥)) is increasing for all 𝑥.

34 Data of Dr. Murlene Clark based on core samples drilled by the research ship Glomar Challenger, taken from Initial Reports of the Deep Sea Drilling Project.

32

1.4

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

LOGARITHMIC FUNCTIONS In Section 1.2, we approximated the population of Burkina Faso (in millions) by the function 𝑃 = 𝑓 (𝑡) = 14.235(1.03)𝑡 , where 𝑡 is the number of years since 2007. Now suppose that instead of calculating the population at time 𝑡, we ask when the population will reach 20 million. We want to find the value of 𝑡 for which 20 = 𝑓 (𝑡) = 14.235(1.03)𝑡 . We use logarithms to solve for a variable in an exponent.

Logarithms to Base 10 and to Base 𝒆 We define the logarithm function, log10 𝑥, to be the inverse of the exponential function, 10𝑥 , as follows: The logarithm to base 10 of 𝑥, written log10 x, is the power of 10 we need to get 𝑥. In other words, log10 𝑥 = 𝑐 means 10𝑐 = 𝑥. We often write log 𝑥 in place of log10 𝑥. The other frequently used base is 𝑒. The logarithm to base 𝑒 is called the natural logarithm of 𝑥, written ln 𝑥 and defined to be the inverse function of 𝑒𝑥 , as follows: The natural logarithm of 𝑥, written ln 𝑥, is the power of 𝑒 needed to get 𝑥. In other words, ln 𝑥 = 𝑐

means

𝑒𝑐 = 𝑥.

Values of log 𝑥 are in Table 1.16. Because no power of 10 gives 0, log 0 is undefined. The graph of 𝑦 = log 𝑥 is shown in Figure 1.54. The domain of 𝑦 = log 𝑥 is positive real numbers; the range is all real numbers. In contrast, the inverse function 𝑦 = 10𝑥 has domain all real numbers and range all positive real numbers. The graph of 𝑦 = log 𝑥 has a vertical asymptote at 𝑥 = 0, whereas 𝑦 = 10𝑥 has a horizontal asymptote at 𝑦 = 0. One big difference between 𝑦 = 10𝑥 and 𝑦 = log 𝑥 is that the exponential function grows extremely quickly whereas the log function grows extremely slowly. However, log 𝑥 does go to infinity, albeit slowly, as 𝑥 increases. Since 𝑦 = log 𝑥 and 𝑦 = 10𝑥 are inverse functions, the graphs of the two functions are reflections of one another across the line 𝑦 = 𝑥, provided the scales along the 𝑥and 𝑦-axes are equal. 𝑦 Table 1.16

Values for log 𝑥 and

𝑥

log 𝑥

𝑥

0

undefined

1

0

2 3

10𝑥

10 𝑥

10

8

0

1

6

1

10

4

0.3

2

100

0.5

3

103 4

4

0.6

4

10









10

1

10

1010

𝑦 = 10𝑥 (1, 10)



Exponential: grows quickly Log: grows slowly

(10, 1)

2 (0, 1) -

? 62 (1, 0)

4

6

8

10

𝑦 = log 𝑥 𝑥

Figure 1.54: Graphs of log 𝑥 and 10𝑥

1.4 LOGARITHMIC FUNCTIONS

33

The graph of 𝑦 = ln 𝑥 in Figure 1.55 has roughly the same shape as the graph of 𝑦 = log 𝑥. The 𝑥-intercept is 𝑥 = 1, since ln 1 = 0.The graph of 𝑦 = ln 𝑥 also climbs very slowly as 𝑥 increases. Both graphs, 𝑦 = log 𝑥 and 𝑦 = ln 𝑥, have vertical asymptotes at 𝑥 = 0. 𝑦

𝑦 = ln 𝑥 1 1

10

𝑥

Figure 1.55: Graph of the natural logarithm

The following properties of logarithms may be deduced from the properties of exponents:

Properties of Logarithms Note that log 𝑥 and ln 𝑥 are not defined when 𝑥 is negative or 0. 1. ln (𝐴𝐵) = ln 𝐴 + ln 𝐵 1. log(𝐴𝐵) = log 𝐴 + log 𝐵 ( ) ( ) 𝐴 𝐴 = ln 𝐴 − ln 𝐵 2. ln 2. log = log 𝐴 − log 𝐵 𝐵 𝐵 3. ln (𝐴𝑝 ) = 𝑝 ln 𝐴 3. log (𝐴𝑝 ) = 𝑝 log 𝐴 𝑥 4. ln 𝑒𝑥 = 𝑥 4. log (10 ) = 𝑥 5. 𝑒ln 𝑥 = 𝑥

5. 10log 𝑥 = 𝑥

In addition, log 1 = 0 because 100 = 1, and ln 1 = 0 because 𝑒0 = 1.

Solving Equations Using Logarithms Logs are frequently useful when we have to solve for unknown exponents, as in the next examples. Example 1

Find 𝑡 such that 2𝑡 = 7.

Solution

First, notice that we expect 𝑡 to be between 2 and 3 (because 22 = 4 and 23 = 8). To calculate 𝑡, we take logs to base 10 of both sides. (Natural logs could also be used.) log(2𝑡 ) = log 7. Then use the third property of logs, which says log(2𝑡 ) = 𝑡 log 2, and get: 𝑡 log 2 = log 7. Using a calculator to find the logs gives 𝑡=

log 7 ≈ 2.81. log 2

Example 2

Find when the population of Burkina Faso reaches 20 million by solving 20 = 14.235(1.03)𝑡 .

Solution

Dividing both sides of the equation by 14.235, we get 20 = (1.03)𝑡 . 14.235

34

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

Now take logs of both sides:

) 20 = log(1.03𝑡 ). 14.235 Using the fact that log(𝐴𝑡 ) = 𝑡 log 𝐴, we get ( ) 20 log = 𝑡 log(1.03). 14.235 log

(

Solving this equation using a calculator to find the logs, we get 𝑡=

log(20∕14.235) = 11.5 years, log(1.03)

which is between 𝑡 = 11 and 𝑡 = 12. This value of 𝑡 corresponds to the year 2018.

Example 3

Traffic pollution is harmful to school-age children. The concentration of carbon monoxide, CO, in the air near a busy road is a function of distance from the road. The concentration decays exponentially at a continuous rate of 3.3% per meter.36 At what distance from the road is the concentration of CO half what it is on the road?

Solution

If 𝐶0 is the concentration of CO on the road, then the concentration 𝑥 meters from the road is 𝐶 = 𝐶0 𝑒−0.033𝑥 . We want to find the value of 𝑥 making 𝐶 = 𝐶0 ∕2, that is, 𝐶0 𝑒−0.033𝑥 =

𝐶0 . 2

Dividing by 𝐶0 and then taking natural logs yields

so

( ) ) ( 1 = −0.6931, ln 𝑒−0.033𝑥 = −0.033𝑥 = ln 2 𝑥 = 21 meters.

At 21 meters from the road the concentration of CO in the air is half the concentration on the road. In Example 3 the decay rate was given. However, in many situations where we expect to find exponential growth or decay, the rate is not given. To find it, we must know the quantity at two different times and then solve for the growth or decay rate, as in the next example. Example 4

The population of Mexico was 100.3 million in 2000 and 120.3 million in 2014.37 Assuming it increases exponentially, find a formula for the population of Mexico as a function of time.

Solution

If we measure the population, 𝑃 , in millions and time, 𝑡, in years since 2000, we can say 𝑃 = 𝑃0 𝑒𝑘𝑡 = 100.3𝑒𝑘𝑡 , where 𝑃0 = 100.3 is the initial value of 𝑃 . We find 𝑘 by using the fact that 𝑃 = 120.3 when 𝑡 = 14, 36 Rickwood, P. and Knight, D. (2009). “The health impacts of local traffic pollution on primary school age children.” State of Australian Cities 2009 Conference Proceedings, www.be.unsw.edu.au, accessed March 24, 2016. 37 www.indexmundi.com, accessed January 28, 2016.

1.4 LOGARITHMIC FUNCTIONS

so

35

120.3 = 100.3𝑒𝑘⋅14 .

To find 𝑘, we divide both sides by 100.3, giving 120.3 = 1.1994 = 𝑒14𝑘 . 100.3 Now take natural logs of both sides: ln(1.1994) = ln(𝑒14𝑘 ). Using a calculator and the fact that ln(𝑒14𝑘 ) = 14𝑘, this becomes 0.182 = 14𝑘. So 𝑘 = 0.013, and therefore 𝑃 = 100.3𝑒0.013𝑡 . Since 𝑘 = 0.013 = 1.3%, the population of Mexico was growing at a continuous rate of 1.3% per year. In Example 4 we chose to use 𝑒 for the base of the exponential function representing Mexico’s population, making clear that the continuous growth rate was 1.3%. If we had wanted to emphasize the annual growth rate, we could have expressed the exponential function in the form 𝑃 = 𝑃0 𝑎𝑡 . Example 5

Give a formula for the inverse of the following function (that is, solve for 𝑡 in terms of 𝑃 ): 𝑃 = 𝑓 (𝑡) = 14.235(1.029)𝑡 .

Solution

We want a formula expressing 𝑡 as a function of 𝑃 . Take logs: log 𝑃 = log(14.235(1.029)𝑡 ). Since log(𝐴𝐵) = log 𝐴 + log 𝐵, we have log 𝑃 = log 14.235 + log((1.029)𝑡 ). Now use log(𝐴𝑡 ) = 𝑡 log 𝐴:

log 𝑃 = log 14.235 + 𝑡 log 1.029.

Solve for 𝑡 in two steps, using a calculator at the final stage: 𝑡 log 1.029 = log 𝑃 − log 14.235 𝑡=

log 14.235 log 𝑃 − = 80.545 log 𝑃 − 92.898. log 1.029 log 1.029

Thus, 𝑓 −1 (𝑃 ) = 80.545 log 𝑃 − 92.898. Note that 𝑓 −1 (20) = 80.545(log 20) − 92.898 = 11.89, which agrees with the result of Example 2.

36

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

Exercises and Problems for Section 1.4 EXERCISES In Exercises 1–6, simplify the expression completely. ln(1∕2)

2. 10log(𝐴𝐵)

1. 𝑒

3. 5𝑒ln(𝐴

2)

5. ln (1∕𝑒) + ln(𝐴𝐵)

4. ln(𝑒2𝐴𝐵 ) ( ) 6. 2 ln 𝑒𝐴 + 3 ln 𝐵 𝑒

For Exercises 7–18, solve for 𝑥 using logs. 7. 3𝑥 = 11

11. 7 = 5𝑒0.2𝑥 −0.4𝑥

21. 𝑄 = 𝑄0 𝑎𝑛𝑡

22. 𝑃0 𝑎𝑡 = 𝑄0 𝑏𝑡

23. 𝑎 = 𝑏𝑒𝑡

24. 𝑃 = 𝑃0 𝑒𝑘𝑡

26. 𝑃 = 10(1.7)𝑡

10. 4 ⋅ 3𝑥 = 7 ⋅ 5𝑥

27. 𝑃 = 174(0.9)𝑡

28. 𝑃 = 4(0.55)𝑡

12. 2𝑥 = 𝑒𝑥+1

Find the inverse function in Exercises 29–31.

3𝑥

5𝑥

14. 2𝑒

15. 7𝑥+2 = 𝑒17𝑥

16. 10𝑥+3 = 5𝑒7−𝑥

17. 2𝑥 − 1 = 𝑒

20. 𝑃 = 𝑃0 𝑎𝑡

25. 𝑃 = 15(1.5)𝑡

13. 50 = 600𝑒

ln 𝑥2

19. 𝑎 = 𝑏𝑡

In Exercises 25–28, put the functions in the form 𝑃 = 𝑃0 𝑒𝑘𝑡 .

8. 17𝑥 = 2

9. 20 = 50(1.04)𝑥

For Exercises 19–24, solve for 𝑡. Assume 𝑎 and 𝑏 are positive, 𝑎 and 𝑏 ≠ 1, and 𝑘 is nonzero.

= 4𝑒

2𝑥−3

18. 4𝑒

29. 𝑝(𝑡) = (1.04)𝑡

30. 𝑓 (𝑡) = 50𝑒0.1𝑡

31. 𝑓 (𝑡) = 1 + ln 𝑡

−5=𝑒

PROBLEMS 32. The exponential function 𝑦(𝑥) = 𝐶𝑒𝛼𝑥 satisfies the conditions 𝑦(0) = 2 and 𝑦(1) = 1. Find the constants 𝐶 and 𝛼. What is 𝑦(2)? For Problems 33–34, find 𝑘 such that 𝑝 = 𝑝0 𝑒𝑘𝑡 has the given doubling time. 33. 10

34. 0.4

35. A culture of bacteria originally numbers 500. After 2 hours there are 1500 bacteria in the culture. Assuming exponential growth, how many are there after 6 hours? 36. One hundred kilograms of a radioactive substance decay to 40 kg in 10 years. How much remains after 20 years? 37. The population of the US was 281.4 million in 2000 and 316.1 million in 2013.37 Assuming exponential growth, (a) In what year is the population expected to go over 350 million? (b) What population is predicted for the 2020 census? 38. The population of a region is growing exponentially. There were 40,000,000 people in 2005 (𝑡 = 0) and 48,000,000 in 2015. Find an expression for the population at any time 𝑡, in years. What population would you predict for the year 2020? What is the doubling time? 37 data.worldbank.org,

39. Oil consumption in China grew exponentially38 from 8.938 million barrels per day in 2010 to 10.480 million barrels per day in 2013. Assuming exponential growth continues at the same rate, what will oil consumption be in 2025? 40. The concentration of the car exhaust fume nitrous oxide, NO2 , in the air near a busy road is a function of distance from the road. The concentration decays exponentially at a continuous rate of 2.54% per meter.39 At what distance from the road is the concentration of NO2 half what it is on the road? 41. For children and adults with diseases such as asthma, the number of respiratory deaths per year increases by 0.33% when pollution particles increase by a microgram per cubic meter of air.40 (a) Write a formula for the number of respiratory deaths per year as a function of quantity of pollution in the air. (Let 𝑄0 be the number of deaths per year with no pollution.) (b) What quantity of air pollution results in twice as many respiratory deaths per year as there would be without pollution?

accessed April 1, 2015. on www.eia.gov/cfapps/ipdbproject, accessed May 2015. 39 Rickwood, P. and Knight, D. (2009). “The health impacts of local traffic pollution on primary school age children”, State of Australian Cities 2009 Conference Proceedings, www.be.unsw.edu.au, accessed March 24, 2016. 40 Brook, R. D., Franklin, B., Cascio, W., Hong, Y., Howard, G., Lipsett, M., Luepker, R., Mittleman, M., Samet, J., and Smith, S. C., “Air pollution and cardiovascular disease.” Circulation, 2004;109:2655–2671. 38 Based

37

1.4 LOGARITHMIC FUNCTIONS

42. The number of alternative fuel vehicles41 running on E85, a fuel that is up to 85% plant-derived ethanol, increased exponentially in the US between 2005 and 2010. (a) Use this information to complete the missing table values. (b) How many E85-powered vehicles were there in the US in 2004? (c) By what percent did the number of E85-powered vehicles grow from 2005 to 2009? Year

2005

2006

2007

2008

2009

2010

No. E85 vehicles

246,363

?

?

?

?

618,505

43. A cup of coffee contains 100 mg of caffeine, which leaves the body at a continuous rate of 17% per hour. (a) Write a formula for the amount, 𝐴 mg, of caffeine in the body 𝑡 hours after drinking a cup of coffee. (b) Graph the function from part (a). Use the graph to estimate the half-life of caffeine. (c) Use logarithms to find the half-life of caffeine. 44. Persistent organic pollutants (POPS) are a serious environmental hazard. Figure 1.56 shows their natural decay over time in human fat.42 (a) How long does it take for the concentration to decrease from 100 units to 50 units? (b) How long does it take for the concentration to decrease from 50 units to 25 units? (c) Explain why your answers to parts (a) and (b) suggest that the decay may be exponential. (d) Find an exponential function that models concentration, 𝐶, as a function of 𝑡, the number of years since 1970. concentration units

100 75 50 25 year

1970 1975 1980 1985 1990 1995 2000

Figure 1.56 45. At time 𝑡 hours after taking the cough suppressant hydrocodone bitartrate, the amount, 𝐴, in mg, remaining in the body is given by 𝐴 = 10(0.82)𝑡 . (a) What was the initial amount taken? (b) What percent of the drug leaves the body each hour? (c) How much of the drug is left in the body 6 hours after the dose is administered?

(d) How long is it until only 1 mg of the drug remains in the body? 46. Different isotopes (versions) of the same element can have very different half-lives. With 𝑡 in years, the decay of plutonium-240 is described by the formula 𝑄 = 𝑄0 𝑒−0.00011𝑡 , whereas the decay of plutonium-242 is described by 𝑄 = 𝑄0 𝑒−0.0000018𝑡 . Find the half-lives plutonium-242.

of

plutonium-240

47. The size of an exponentially growing bacteria colony doubles in 5 hours. How long will it take for the number of bacteria to triple? 48. Air pressure, 𝑃 , decreases exponentially with height, ℎ, above sea level. If 𝑃0 is the air pressure at sea level and ℎ is in meters, then 𝑃 = 𝑃0 𝑒−0.00012ℎ . (a) At the top of Denali, height 6194 meters (about 20,320 feet), what is the air pressure, as a percent of the pressure at sea level? (b) The maximum cruising altitude of an ordinary commercial jet is around 12,000 meters (about 39,000 feet). At that height, what is the air pressure, as a percent of the sea level value? 49. With time, 𝑡, in years since the start of 1980, textbook prices have increased at 6.7% per year while inflation has been 3.3% per year.43 Assume both rates are continuous growth rates. (a) Find a formula for 𝐵(𝑡), the price of a textbook in year 𝑡 if it cost $𝐵0 in 1980. (b) Find a formula for 𝑃 (𝑡), the price of an item in year 𝑡 if it cost $𝑃0 in 1980 and its price rose according to inflation. (c) A textbook cost $50 in 1980. When is its price predicted to be double the price that would have resulted from inflation alone? 50. In November 2010, a “tiger summit” was held in St. Petersburg, Russia.44 In 1900, there were 100,000 wild tigers worldwide; in 2010 the number was 3200. (a) Assuming the tiger population has decreased exponentially, find a formula for 𝑓 (𝑡), the number of wild tigers 𝑡 years since 1900. (b) Between 2000 and 2010, the number of wild tigers decreased by 40%. Is this percentage larger or smaller than the decrease in the tiger population predicted by your answer to part (a)?

41 www.eia.gov, 42 K.C. Jones,

and

accessed April 1, 2015. P. de Voogt, “Persistent organic pollutants (POPs): state of the science,” Environmental Pollution 100, 1999,

pp. 209–221. 43 Data from “Textbooks headed for ash heap of history”, http://educationtechnews.com, Vol 5, 2010. 44 “Tigers would be extinct in Russia if unprotected,” Yahoo! News, Nov. 21, 2010.

38

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

51. In 2014, the populations of China and India were approximately 1.355 and 1.255 billion people,45 respectively. However, due to central control the annual population growth rate of China was 0.44% while the population of India was growing by 1.25% each year. If these growth rates remain constant, when will the population of India exceed that of China? 52. The revenue of Apple® went from $54.5 billion in 2013 to $74.6 billion46 in 2015. Find an exponential function to model the revenue as a function of years since 2013. What is the continuous percent growth rate, per year, of revenue? 53. The world population was 6.9 billion at the end of 2010 and is predicted to reach 9 billion by the end of 2050.47 (a) Assuming the population is growing exponentially, what is the continuous growth rate per year? (b) The United Nations celebrated the “Day of 6 Billion” on October 12, 1999, and “Day of 7 Billion” on October 31, 2011. Using the growth rate in part (a), when is the “Day of 8 Billion” predicted to be? 54. In the early 1920s, Germany had tremendously high inflation, called hyperinflation. Photographs of the time show people going to the store with wheelbarrows full of money. If a loaf of bread cost 1∕4 marks in 1919 and 2,400,000 marks in 1922, what was the average yearly inflation rate between 1919 and 1922? 55. In 2010, there were about 246 million vehicles (cars and trucks) and about 308.7 million people in the US.48 The number of vehicles grew 15.5% over the previous decade, while the population has been growing at 9.7% per decade. If the growth rates remain constant, when will there be, on average, one vehicle per person?

mine, requiring the poison be removed, as in the following table, where 𝑡 is in years since 2012. (a) Find an exponential model for 𝑐(𝑡), the concentration, in parts per million, of cyanide in the groundwater. (b) Use the model in part (a) to find the number of years it takes for the cyanide concentration to fall to 10 ppm. (c) The filtering process removing the cyanide is sped up so that the new model is 𝐷(𝑡) = 𝑐(2𝑡). Find 𝐷(𝑡). (d) If the cyanide removal was started three years earlier, but run at the speed of part (a), find a new model, 𝐸(𝑡). 𝑡 (years)

0

1

2

𝑐(𝑡) (ppm)

25.0

21.8

19.01

59. In 2015, Nepal had two massive earthquakes, the first measuring 7.8 on the Richter scale and the second measuring 7.3. The Richter scale compares the strength, 𝑊 , of the seismic waves of an earthquake with the earthquake givstrength, 𝑊0 , of the waves of a standard ( ) ing a Richter rating of 𝑅 = log 𝑊 ∕𝑊0 . By what factor were the seismic waves in Nepal’s first earthquake stronger than the seismic waves in (a) A standard earthquake? (b) The second Nepal earthquake? 60. Find the equation of the line 𝑙 in Figure 1.57. 𝑓 (𝑥) = 10𝑥 𝑙

56. Tiny marine organisms reproduce at different rates. Phytoplankton doubles in population twice a day, but foraminifera doubles every five days. If the two populations are initially the same size and grow exponentially, how long does it take for (a) The phytoplankton population to be double the foraminifera population. (b) The phytoplankton population to be 1000 times the foraminifera population. 57. A picture supposedly painted by Vermeer (1632–1675) contains 99.5% of its carbon-14 (half-life 5730 years). From this information decide whether the picture is a fake. Explain your reasoning.

𝑥 log 2

Figure 1.57 61. Without a calculator or computer, match the functions 𝑒𝑥 , ln 𝑥, 𝑥2 , and 𝑥1∕2 to their graphs in Figure 1.58. 𝐴𝐵

𝐶 𝐷

58. Cyanide is used in solution to isolate gold in a mine.49 This may result in contaminated groundwater near the

𝑥 45 www.indexmundi.com,

Figure 1.58

accessed April 11, 2015. accessed April 2, 2015. 47 “Reviewing the Bidding on the Climate Files”, in About Dot Earth, New York Times, Nov. 19, 2010. 48 http://www.autoblog.com/2010/01/04/report-number-of-cars-in-the-u-s-dropped-by-four-million-in-20/ and http://2010.census.gov/news/releases/operations/cb10-cn93.html. Accessed February 2012. 49 www.www.miningfacts.org/environment/what-is-the-role-of-cyanide-in-mining/r. Accessed June 9, 2015. 46 techcrunch.com,

1.5 TRIGONOMETRIC FUNCTIONS

39

62. Is there a difference between ln(ln(𝑥)) and ln2 (𝑥)? (Note: ln2 (𝑥) is another way of writing (ln 𝑥)2 .)

66. If 𝑓 (𝑥) = 𝑎 ln(𝑥 + 2), what is the effect of increasing 𝑎 on the vertical asymptote?

63. If ℎ(𝑥) = ln(𝑥 + 𝑎), where 𝑎 > 0, what is the effect of increasing 𝑎 on

67. If 𝑔(𝑥) = ln(𝑎𝑥 + 2), where 𝑎 ≠ 0, what is the effect of increasing 𝑎 on the vertical asymptote?

(a)

The 𝑦-intercept?

(b) The 𝑥-intercept?

64. If ℎ(𝑥) = ln(𝑥 + 𝑎), where 𝑎 > 0, what is the effect of increasing 𝑎 on the vertical asymptote? 65. If 𝑔(𝑥) = ln(𝑎𝑥 + 2), where 𝑎 ≠ 0, what is the effect of increasing 𝑎 on (a)

The 𝑦-intercept?

(b) The 𝑥-intercept?

68. Show that the growth rate 𝑘 of the exponential function 𝑓 (𝑡) = 𝑃0 𝑒𝑘𝑡 , with 𝑃0 > 0, can be computed from two values of 𝑓 by a difference quotient of the form: 𝑘=

ln 𝑓 (𝑡2 ) − ln 𝑓 (𝑡1 ) . 𝑡2 − 𝑡1

Strengthen Your Understanding In Problems 69–74, explain what is wrong with the statement.

76. A function 𝑓 (𝑥) such that ln(𝑓 (𝑥)) is only defined for 𝑥 < 0.

69. The function − log |𝑥| is odd.

77. A function with a vertical asymptote at 𝑥 = 3 and defined only for 𝑥 > 3.

70. For all 𝑥 > 0, the value of ln(100𝑥) is 100 times larger than ln 𝑥. 71. ln 𝑥 > log 𝑥. 72. ln(𝐴 + 𝐵) = (ln 𝐴)(ln 𝐵). 73. ln(𝐴 + 𝐵) = ln 𝐴 + ln 𝐵. 1 74. = 𝑒𝑥 ln 𝑥 In Problems 75–77, give an example of: 75. A function that grows slower than 𝑦 = log 𝑥 for 𝑥 > 1.

1.5

Are the statements in Problems 78–81 true or false? Give an explanation for your answer. 78. The graph of 𝑓 (𝑥) = ln 𝑥 is concave down. 79. The graph of 𝑔(𝑥) = log(𝑥 − 1) crosses the 𝑥-axis at 𝑥 = 1. 80. The inverse function of 𝑦 = log 𝑥 is 𝑦 = 1∕ log 𝑥. 81. If 𝑎 and 𝑏 are positive constants, then 𝑦 = ln(𝑎𝑥 + 𝑏) has no vertical asymptote.

TRIGONOMETRIC FUNCTIONS Trigonometry originated as part of the study of triangles. The name tri-gon-o-metry means the measurement of three-cornered figures, and the first definitions of the trigonometric functions were in terms of triangles. However, the trigonometric functions can also be defined using the unit circle, a definition that makes them periodic, or repeating. Many naturally occurring processes are also periodic. The water level in a tidal basin, the blood pressure in a heart, an alternating current, and the position of the air molecules transmitting a musical note all fluctuate regularly. Such phenomena can be represented by trigonometric functions.

Radians There are two commonly used ways to represent the input of the trigonometric functions: radians and degrees. The formulas of calculus, as you will see, are neater in radians than in degrees.

An angle of 1 radian is defined to be the angle at the center of a unit circle which cuts off an arc of length 1, measured counterclockwise. (See Figure 1.59(a).) A unit circle has radius 1.

An angle of 2 radians cuts off an arc of length 2 on a unit circle. A negative angle, such as −1∕2 radians, cuts off an arc of length 1∕2, but measured clockwise. (See Figure 1.59(b).)

40

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

(a)



(b)



Arc length = 2

Arc length = 1

1

1

2 radians

1 radian

− 12 rad

Arc length =



1 2

Figure 1.59: Radians defined using unit circle

It is useful to think of angles as rotations, since then we can make sense of angles larger than 360◦; for example, an angle of 720◦ represents two complete rotations counterclockwise. Since one full rotation of 360◦ cuts off an arc of length 2𝜋, the circumference of the unit circle, it follows that 360◦ = 2𝜋 radians,

so

180◦ = 𝜋 radians.

In other words, 1 radian = 180◦∕𝜋, so one radian is about 60◦ . The word radians is often dropped, so if an angle or rotation is referred to without units, it is understood to be in radians. Radians are useful for computing the length of an arc in any circle. If the circle has radius 𝑟 and the arc cuts off an angle 𝜃, as in Figure 1.60, then we have the following relation: 𝑟 𝜃

Arc length = 𝑠 = 𝑟𝜃.

𝑠

Figure 1.60: Arc length of a sector of a circle

The Sine and Cosine Functions The two basic trigonometric functions—the sine and cosine—are defined using a unit circle. In Figure 1.61, an angle of 𝑡 radians is measured counterclockwise around the circle from the point (1, 0). If 𝑃 has coordinates (𝑥, 𝑦), we define cos 𝑡 = 𝑥

and

sin 𝑡 = 𝑦.

We assume that the angles are always in radians unless specified otherwise. Since the equation of the unit circle is 𝑥2 +𝑦2 = 1, writing cos2 𝑡 for (cos 𝑡)2 , we have the identity cos2 𝑡 + sin2 𝑡 = 1. As 𝑡 increases and 𝑃 moves around the circle, the values of sin 𝑡 and cos 𝑡 oscillate between 1 and −1, and eventually repeat as 𝑃 moves through points where it has been before. If 𝑡 is negative, the angle is measured clockwise around the circle.

Amplitude, Period, and Phase The graphs of sine and cosine are shown in Figure 1.62. Notice that sine is an odd function, and cosine is even. The maximum and minimum values of sine and cosine are +1 and −1, because those are the maximum and minimum values of 𝑦 and 𝑥 on the unit circle. After the point 𝑃 has moved around the complete circle once, the values of cos 𝑡 and sin 𝑡 start to repeat; we say the functions are periodic.

1.5 TRIGONOMETRIC FUNCTIONS

41

For any periodic function of time, the • Amplitude is half the distance between the maximum and minimum values (if it exists). • Period is the smallest time needed for the function to execute one complete cycle. The amplitude of cos 𝑡 and sin 𝑡 is 1, and the period is 2𝜋. Why 2𝜋? Because that’s the value of 𝑡 when the point 𝑃 has gone exactly once around the circle. (Remember that 360◦ = 2𝜋 radians.) (0, 1)

𝑃



𝑥 = cos 𝑡 𝑦 = sin 𝑡 1

𝑦

cos 𝑡



sin 𝑡

Amplitude = 1

𝑡

❄ ✛𝑥✲

❄𝑡

(1, 0) −3𝜋

−2𝜋

−𝜋

𝜋

−1 Figure 1.61: The definitions of sin 𝑡 and cos 𝑡

2𝜋

3𝜋

✛ Period = 2𝜋 ✲

Figure 1.62: Graphs of cos 𝑡 and sin 𝑡

In Figure 1.62, we see that the sine and cosine graphs are exactly the same shape, only shifted horizontally. Since the cosine graph is the sine graph shifted 𝜋∕2 to the left, cos 𝑡 = sin(𝑡 + 𝜋∕2). Equivalently, the sine graph is the cosine graph shifted 𝜋∕2 to the right, so sin 𝑡 = cos(𝑡 − 𝜋∕2). We say that the phase difference or phase shift between sin 𝑡 and cos 𝑡 is 𝜋∕2. Functions whose graphs are the shape of a sine or cosine curve are called sinusoidal functions. To describe arbitrary amplitudes and periods of sinusoidal functions, we use functions of the form 𝑓 (𝑡) = 𝐴 sin(𝐵𝑡) and 𝑔(𝑡) = 𝐴 cos(𝐵𝑡), where |𝐴| is the amplitude and 2𝜋∕|𝐵| is the period. The graph of a sinusoidal function is shifted horizontally by a distance |ℎ| when 𝑡 is replaced by 𝑡 − ℎ or 𝑡 + ℎ. Functions of the form 𝑓 (𝑡) = 𝐴 sin(𝐵𝑡) + 𝐶 and 𝑔(𝑡) = 𝐴 cos(𝐵𝑡) + 𝐶 have graphs which are shifted vertically by 𝐶 and oscillate about this value.

Example 1

Solution

Find and show on a graph the amplitude and period of(the) functions 𝑡 (a) 𝑦 = 5 sin(2𝑡) (b) 𝑦 = −5 sin 2

(c) 𝑦 = 1 + 2 sin 𝑡

(a) From Figure 1.63, you can see that the amplitude of 𝑦 = 5 sin(2𝑡) is 5 because the factor of 5 stretches the oscillations up to 5 and down to −5. The period of 𝑦 = sin(2𝑡) is 𝜋, because when 𝑡 changes from 0 to 𝜋, the quantity 2𝑡 changes from 0 to 2𝜋, so the sine function goes through one complete oscillation. (b) Figure 1.64 shows that the amplitude of 𝑦 = −5 sin (𝑡∕2) is again 5, because the negative sign

42

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

reflects the oscillations in the 𝑡-axis, but does not change how far up or down they go. The period of 𝑦 = −5 sin (𝑡∕2) is 4𝜋 because when 𝑡 changes from 0 to 4𝜋, the quantity 𝑡∕2 changes from 0 to 2𝜋, so the sine function goes through one complete oscillation. (c) The 1 shifts the graph 𝑦 = 2 sin 𝑡 up by 1. Since 𝑦 = 2 sin 𝑡 has an amplitude of 2 and a period of 2𝜋, the graph of 𝑦 = 1 + 2 sin 𝑡 goes up to 3 and down to −1, and has a period of 2𝜋. (See Figure 1.65.) Thus, 𝑦 = 1 + 2 sin 𝑡 also has amplitude 2. 𝑦 5

𝑦 𝑦 = 5 sin 2𝑡

5



𝑦 𝑦 = −5 sin(𝑡∕2)



Amplitude



Amplitude

2𝜋

❄ −𝜋

𝑦 = 1 + 2 sin 𝑡

3

𝜋

𝑡

Amplitude

❄ 𝑡

−𝜋

2𝜋



1

4𝜋

𝑡 𝜋

✛Period✲ Figure 1.63: Amplitude = 5,



−1 ✛

Figure 1.64: Amplitude = 5, period = 4𝜋

period = 𝜋

Example 2



Period

2𝜋

Period



Figure 1.65: Amplitude = 2, period = 2𝜋

Find possible formulas for the following sinusoidal functions. (a)

3

−6𝜋

(b)

𝑔(𝑡)

6𝜋 −3

𝑡 12𝜋

2

(c)

𝑓 (𝑡) 𝑡

−1 −2

1 2 3

4

3

−5𝜋

ℎ(𝑡)

𝜋 7𝜋 13𝜋

𝑡

−3

Solution

(a) This function looks like a sine function with amplitude 3, so 𝑔(𝑡) = 3 sin(𝐵𝑡). Since the function executes one full oscillation between 𝑡 = 0 and 𝑡 = 12𝜋, when 𝑡 changes by 12𝜋, the quantity 𝐵𝑡 changes by 2𝜋. This means 𝐵 ⋅ 12𝜋 = 2𝜋, so 𝐵 = 1∕6. Therefore, 𝑔(𝑡) = 3 sin(𝑡∕6) has the graph shown. (b) This function looks like an upside-down cosine function with amplitude 2, so 𝑓 (𝑡) = −2 cos(𝐵𝑡). The function completes one oscillation between 𝑡 = 0 and 𝑡 = 4. Thus, when 𝑡 changes by 4, the quantity 𝐵𝑡 changes by 2𝜋, so 𝐵 ⋅ 4 = 2𝜋, or 𝐵 = 𝜋∕2. Therefore, 𝑓 (𝑡) = −2 cos(𝜋𝑡∕2) has the graph shown. (c) This function looks like the function 𝑔(𝑡) in part (a), but shifted a distance of 𝜋 to the right. Since 𝑔(𝑡) = 3 sin(𝑡∕6), we replace 𝑡 by (𝑡 − 𝜋) to obtain ℎ(𝑡) = 3 sin[(𝑡 − 𝜋)∕6].

Example 3

On July 1, 2007, high tide in Boston was at midnight. The water level at high tide was 9.9 feet; later, at low tide, it was 0.1 feet. Assuming the next high tide is at exactly 12 noon and that the height of the water is given by a sine or cosine curve, find a formula for the water level in Boston as a function of time.

Solution

Let 𝑦 be the water level in feet, and let 𝑡 be the time measured in hours from midnight. The oscillations have amplitude 4.9 feet (= (9.9 − 0.1)∕2) and period 12, so 12𝐵 = 2𝜋 and 𝐵 = 𝜋∕6. Since the water is highest at midnight, when 𝑡 = 0, the oscillations are best represented by a cosine function. (See Figure 1.66.) We can say ( ) 𝜋 Height above average = 4.9 cos 𝑡 . 6

1.5 TRIGONOMETRIC FUNCTIONS

43

Since the average water level was 5 feet (= (9.9 + 0.1)∕2), we shift the cosine up by adding 5: ( ) 𝜋 𝑡 . 𝑦 = 5 + 4.9 cos 6 𝑦 9.9

𝑦 = 5 + 4.9 cos 5

24

12 12 mid.

6 am

12 noon

6 pm

(

) 𝜋 𝑡 6

𝑡

12 mid.

Figure 1.66: Function approximating the tide in Boston on July 1, 2007

Example 4

Of course, there’s something wrong with the assumption in Example 3 that the next high tide is at noon. If so, the high tide would always be at noon or midnight, instead of progressing slowly through the day, as in fact it does. The interval between successive high tides actually averages about 12 hours 24 minutes. Using this, give a more accurate formula for the height of the water as a function of time.

Solution

The period is 12 hours 24 minutes = 12.4 hours, so 𝐵 = 2𝜋∕12.4, giving ) ( 2𝜋 𝑡 = 5 + 4.9 cos(0.507𝑡). 𝑦 = 5 + 4.9 cos 12.4

Example 5

Use the information from Example 4 to write a formula for the water level in Boston on a day when the high tide is at 2 pm.

Solution

When the high tide is at midnight, 𝑦 = 5 + 4.9 cos(0.507𝑡). Since 2 pm is 14 hours after midnight, we replace 𝑡 by (𝑡 − 14). Therefore, on a day when the high tide is at 2 pm, 𝑦 = 5 + 4.9 cos(0.507(𝑡 − 14)).

The Tangent Function If 𝑡 is any number with cos 𝑡 ≠ 0, we define the tangent function as follows

tan 𝑡 =

sin 𝑡 . cos 𝑡

Figure 1.61 on page 41 shows the geometrical meaning of the tangent function: tan 𝑡 is the slope of the line through the origin (0, 0) and the point 𝑃 = (cos 𝑡, sin 𝑡) on the unit circle.

44

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

The tangent function is undefined wherever cos 𝑡 = 0, namely, at 𝑡 = ±𝜋∕2, ±3𝜋∕2, . . . , and it has a vertical asymptote at each of these points. The function tan 𝑡 is positive where sin 𝑡 and cos 𝑡 have the same sign. The graph of the tangent is shown in Figure 1.67.

tan 𝑡

3 tan 𝑡✲

10

10 tan 𝑡 ✲ 𝑡

−𝜋

𝜋

𝑡

−𝜋

−10

𝜋 −10

✛ Period ✲

✛ Period ✲

Figure 1.67: The tangent function

Figure 1.68: Multiple of tangent

The tangent function has period 𝜋, because it repeats every 𝜋 units. Does it make sense to talk about the amplitude of the tangent function? Not if we’re thinking of the amplitude as a measure of the size of the oscillation, because the tangent becomes infinitely large near each vertical asymptote. We can still multiply the tangent by a constant, but that constant no longer represents an amplitude. (See Figure 1.68.)

The Inverse Trigonometric Functions On occasion, you may need to find a number with a given sine. For example, you might want to find 𝑥 such that sin 𝑥 = 0 or such that sin 𝑥 = 0.3. The first of these equations has solutions 𝑥 = 0, ±𝜋, ±2𝜋, . . . . The second equation also has infinitely many solutions. Using a calculator and a graph, we get 𝑥 ≈ 0.305, 2.84, 0.305 ± 2𝜋, 2.84 ± 2𝜋, … . For each equation, we pick out the solution between −𝜋∕2 and 𝜋∕2 as the preferred solution. For example, the preferred solution to sin 𝑥 = 0 is 𝑥 = 0, and the preferred solution to sin 𝑥 = 0.3 is 𝑥 = 0.305. We define the inverse sine, written “arcsin” or “sin−1 ,” as the function which gives the preferred solution.

For −1 ≤ 𝑦 ≤ 1, means

arcsin 𝑦 = 𝑥 𝜋 𝜋 sin 𝑥 = 𝑦 with − ≤ 𝑥 ≤ . 2 2

Thus the arcsine is the inverse function to the piece of the sine function having domain [−𝜋∕2, 𝜋∕2]. (See Table 1.17 and Figure 1.69.) On a calculator, the arcsine function50 is usually denoted by sin−1 . 50 Note

that sin−1 𝑥 = arcsin 𝑥 is not the same as (sin 𝑥)−1 = 1∕ sin 𝑥.

45

1.5 TRIGONOMETRIC FUNCTIONS

Table 1.17 𝑥

Values of sin 𝑥 and sin−1 𝑥 sin 𝑥

𝜋 2 −1

sin

𝑥

𝑥

− 𝜋2

−1.000

−1.000

− 𝜋2

−1.0

−0.841

−0.841

−1.0

−0.5

−0.479

−0.479

−0.5

0.0

0.000

0.000

0.0

0.5

0.479

0.479

0.5

1.0

0.841

0.841

1.0

𝜋 2

1.000

1.000

𝜋 2

𝑦 = sin−1 𝑥

1

𝑦 = sin 𝑥 𝑥

− 𝜋2 −1

𝜋 2

1 −1 − 𝜋2

Figure 1.69: The arcsine function

The inverse tangent, written “arctan” or “tan−1 ,” is the inverse function for the piece of the tangent function having the domain −𝜋∕2 < 𝑥 < 𝜋∕2. On a calculator, the inverse tangent is usually denoted by tan−1 . The graph of the arctangent is shown in Figure 1.71. For any 𝑦, arctan 𝑦 = 𝑥 tan 𝑥 = 𝑦 with

means



𝜋 𝜋 0, increasing the value of 𝐵 increases the period. 77. The function 𝑦 = sin 𝑥 cos 𝑥 is periodic with period 2𝜋. 78. For positive 𝐴, 𝐵, 𝐶, the maximum value of the function 𝑦 = 𝐴 sin(𝐵𝑥) + 𝐶 is 𝑦 = 𝐴.

84. The function 𝑓 (𝑡) = sin(0.05𝜋𝑡) has period 0.05. 85. If 𝑡 is in seconds, 𝑔(𝑡) = cos(200𝜋𝑡) executes 100 cycles in one second. 86. The function 𝑓 (𝜃) = tan(𝜃 − 𝜋∕2) is not defined at 𝜃 = 𝜋∕2, 3𝜋∕2, 5𝜋∕2 …. 87. sin |𝑥| = sin 𝑥 for −2𝜋 < 𝑥 < 2𝜋

88. sin |𝑥| = | sin 𝑥| for −2𝜋 < 𝑥 < 2𝜋

89. cos |𝑥| = | cos 𝑥| for −2𝜋 < 𝑥 < 2𝜋 90. cos |𝑥| = cos 𝑥 for −2𝜋 < 𝑥 < 2𝜋

In Problems 79–80, give an example of:

91. The function 𝑓 (𝑥) = sin(𝑥2 ) is periodic, with period 2𝜋.

79. A sine function with period 23. 80. A cosine function which oscillates between values of 1200 and 2000. Are the statements in Problems 81–97 true or false? Give an explanation for your answer. 81. The family of functions 𝑦 = 𝑎 sin 𝑥, with 𝑎 a positive constant, all have the same period.

92. The function 𝑔(𝜃) = 𝑒sin 𝜃 is periodic. 93. If 𝑓 (𝑥) is a periodic function with period 𝑘, then 𝑓 (𝑔(𝑥)) is periodic with period 𝑘 for every function 𝑔(𝑥). 94. If 𝑔(𝑥) is a periodic function, then 𝑓 (𝑔(𝑥)) is periodic for every function 𝑓 (𝑥).

82. The family of functions 𝑦 = sin 𝑎𝑥, 𝑎 a positive constant, all have the same period.

95. The function 𝑓 (𝑥) = | sin 𝑥| is even.

83. The function 𝑓 (𝜃) = cos 𝜃 − sin 𝜃 is increasing on 0 ≤ 𝜃 ≤ 𝜋∕2.

97. The function 𝑓 (𝑡) = sin−1 (sin 𝑡) is periodic with period 2𝜋.

1.6

96. sin−1 (sin 𝑡) = 𝑡 for all 𝑡.

POWERS, POLYNOMIALS, AND RATIONAL FUNCTIONS

Power Functions A power function is a function in which the dependent variable is proportional to a power of the independent variable: A power function has the form 𝑓 (𝑥) = 𝑘𝑥𝑝 ,

where 𝑘 and 𝑝 are constant.

50

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

For example, the volume, 𝑉 , of a sphere of radius 𝑟 is given by 4 3 𝜋𝑟 . 3 As another example, the gravitational force, 𝐹 , on a unit mass at a distance 𝑟 from the center of the earth is given by Newton’s Law of Gravitation, which says that, for some positive constant 𝑘, 𝑘 𝐹 = 2 or 𝐹 = 𝑘𝑟−2 . 𝑟 We consider the graphs of the power functions 𝑥𝑛 , with 𝑛 a positive integer. Figures 1.81 and 1.82 show that the graphs fall into two groups: odd and even powers. For 𝑛 greater than 1, the odd powers have a “seat” at the origin and are increasing everywhere else. The even powers are first decreasing and then increasing. For large 𝑥, the higher the power of 𝑥, the faster the function climbs. 𝑉 = 𝑔(𝑟) =

𝑥5 𝑥3

𝑥4

10 5 −2

𝑥2

10 𝑥

5

−1

𝑥 1

2 𝑥

−5

−3 −2 −1

−10

1

2

3

−5

Figure 1.81: Odd powers of 𝑥: “Seat” shaped for 𝑛 > 1

Figure 1.82: Even powers of 𝑥:

⋃ -shaped

Exponentials and Power Functions: Which Dominate? In everyday language, the word exponential is often used to imply very fast growth. But do exponential functions always grow faster than power functions? To determine what happens “in the long run,” we often want to know which functions dominate as 𝑥 gets arbitrarily large. Let’s consider 𝑦 = 2𝑥 and 𝑦 = 𝑥3 . The close-up view in Figure 1.83(a) shows that between 𝑥 = 2 and 𝑥 = 4, the graph of 𝑦 = 2𝑥 lies below the graph of 𝑦 = 𝑥3 . The far-away view in Figure 1.83(b) shows that the exponential function 𝑦 = 2𝑥 eventually overtakes 𝑦 = 𝑥3 . Figure 1.83(c), which gives a very far-away view, shows that, for large 𝑥, the value of 𝑥3 is insignificant compared to 2𝑥 . Indeed, 2𝑥 is growing so much faster than 𝑥3 that the graph of 2𝑥 appears almost vertical in comparison to the more leisurely climb of 𝑥3 . We say that Figure 1.83(a) gives a local view of the functions’ behavior, whereas Figure 1.83(c) gives a global view. In fact, every exponential growth function eventually dominates every power function. Although an exponential function may be below a power function for some values of 𝑥, if we look at large enough 𝑥-values, 𝑎𝑥 (with 𝑎 > 1) will eventually dominate 𝑥𝑛 , no matter what 𝑛 is. 𝑦

(a)

𝑦

(b)

20

3

2

𝑥

𝑥

2

3

𝑥

1,000

10

𝑥 2

3

4

4

10

5,000

Close up (Local)

1

𝑦

(c)

2,000

𝑥

2𝑥

Far away (Global)

𝑥3

𝑥 5

10

𝑥 5

Figure 1.83: Comparison of 𝑦 = 2𝑥 and 𝑦 = 𝑥3 : Notice that 𝑦 = 2𝑥 eventually dominates 𝑦 = 𝑥3

10

15

1.6 POWERS, POLYNOMIALS, AND RATIONAL FUNCTIONS

51

Polynomials Polynomials are the sums of power functions with nonnegative integer exponents: 𝑦 = 𝑝(𝑥) = 𝑎𝑛 𝑥𝑛 + 𝑎𝑛−1 𝑥𝑛−1 + ⋯ + 𝑎1 𝑥 + 𝑎0 . Here 𝑛 is a nonnegative integer called the degree of the polynomial, and 𝑎𝑛 , 𝑎𝑛−1 , … , 𝑎1 , 𝑎0 are constants, with leading coefficient 𝑎𝑛 ≠ 0. An example of a polynomial of degree 𝑛 = 3 is 𝑦 = 𝑝(𝑥) = 2𝑥3 − 𝑥2 − 5𝑥 − 7. In this case 𝑎3 = 2, 𝑎2 = −1, 𝑎1 = −5, and 𝑎0 = −7. The shape of the graph of a polynomial depends on its degree; typical graphs are shown in Figure 1.84. These graphs correspond to a positive coefficient for 𝑥𝑛 ; a negative leading coefficient turns the graph upside down. Notice that the quadratic “turns around” once, the cubic “turns around” twice, and the quartic (fourth degree) “turns around” three times. An 𝑛th -degree polynomial “turns around” at most 𝑛 − 1 times (where 𝑛 is a positive integer), but there may be fewer turns.

Quadratic

Cubic

(𝑛 = 2)

Quartic

(𝑛 = 3)

Quintic

(𝑛 = 4)

(𝑛 = 5)

Figure 1.84: Graphs of typical polynomials of degree 𝑛

Example 1

Find possible formulas for the polynomials whose graphs are in Figure 1.85. 4

(a)

𝑓 (𝑥)

(b)

(c)

𝑔(𝑥)

ℎ(𝑥)

𝑥 𝑥 −2

2

−3

1

2

−12

𝑥 −3

2

Figure 1.85: Graphs of polynomials

Solution

(a) This graph appears to be a parabola, turned upside down, and moved up by 4, so 𝑓 (𝑥) = −𝑥2 + 4. The negative sign turns the parabola upside down and the +4 moves it up by 4. Notice that this formula does give the correct 𝑥-intercepts since 0 = −𝑥2 + 4 has solutions 𝑥 = ±2. These values of 𝑥 are called zeros of 𝑓 . We can also solve this problem by looking at the 𝑥-intercepts first, which tell us that 𝑓 (𝑥) has factors of (𝑥 + 2) and (𝑥 − 2). So 𝑓 (𝑥) = 𝑘(𝑥 + 2)(𝑥 − 2). To find 𝑘, use the fact that the graph has a 𝑦-intercept of 4, so 𝑓 (0) = 4, giving 4 = 𝑘(0 + 2)(0 − 2), or 𝑘 = −1. Therefore, 𝑓 (𝑥) = −(𝑥 + 2)(𝑥 − 2), which multiplies out to −𝑥2 + 4. Note that 𝑓 (𝑥) = 4 − 𝑥4 ∕4 also has the same basic shape, but is flatter near 𝑥 = 0. There are many possible answers to these questions. (b) This looks like a cubic with factors (𝑥 + 3), (𝑥 − 1), and (𝑥 − 2), one for each intercept: 𝑔(𝑥) = 𝑘(𝑥 + 3)(𝑥 − 1)(𝑥 − 2).

52

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

Since the 𝑦-intercept is −12, we have −12 = 𝑘(0 + 3)(0 − 1)(0 − 2). So 𝑘 = −2, and we get the cubic polynomial 𝑔(𝑥) = −2(𝑥 + 3)(𝑥 − 1)(𝑥 − 2). (c) This also looks like a cubic with zeros at 𝑥 = 2 and 𝑥 = −3. Notice that at 𝑥 = 2 the graph of ℎ(𝑥) touches the 𝑥-axis but does not cross it, whereas at 𝑥 = −3 the graph crosses the 𝑥-axis. We say that 𝑥 = 2 is a double zero, but that 𝑥 = −3 is a single zero. To find a formula for ℎ(𝑥), imagine the graph of ℎ(𝑥) to be slightly lower down, so that the graph has one 𝑥-intercept near 𝑥 = −3 and two near 𝑥 = 2, say at 𝑥 = 1.9 and 𝑥 = 2.1. Then a formula would be ℎ(𝑥) ≈ 𝑘(𝑥 + 3)(𝑥 − 1.9)(𝑥 − 2.1). Now move the graph back to its original position. The zeros at 𝑥 = 1.9 and 𝑥 = 2.1 move toward 𝑥 = 2, giving ℎ(𝑥) = 𝑘(𝑥 + 3)(𝑥 − 2)(𝑥 − 2) = 𝑘(𝑥 + 3)(𝑥 − 2)2 . The double zero leads to a repeated factor, (𝑥 − 2)2 . Notice that when 𝑥 > 2, the factor (𝑥 − 2)2 is positive, and when 𝑥 < 2, the factor (𝑥 − 2)2 is still positive. This reflects the fact that ℎ(𝑥) does not change sign near 𝑥 = 2. Compare this with the behavior near the single zero at 𝑥 = −3, where ℎ does change sign. We cannot find 𝑘, as no coordinates are given for points off of the 𝑥-axis. Any positive value of 𝑘 stretches the graph vertically but does not change the zeros, so any positive 𝑘 works. Example 2

Using a calculator or computer, graph 𝑦 = 𝑥4 and 𝑦 = 𝑥4 − 15𝑥2 − 15𝑥 for −4 ≤ 𝑥 ≤ 4 and for −20 ≤ 𝑥 ≤ 20. Set the 𝑦 range to −100 ≤ 𝑦 ≤ 100 for the first domain, and to −100 ≤ 𝑦 ≤ 200,000 for the second. What do you observe?

Solution

From the graphs in Figure 1.86 we see that close up (−4 ≤ 𝑥 ≤ 4) the graphs look different; from far away, however, they are almost indistinguishable. The reason is that the leading terms (those with the highest power of 𝑥) are the same, namely 𝑥4 , and for large values of 𝑥, the leading term dominates the other terms. 𝑦

𝑦

𝑦 = 𝑥4

100

100 𝑦 = 𝑥4 − 15𝑥2 − 15𝑥

Close-up or Local

4

𝑥 −4

4

𝑥

−4

−100

−100 𝑦

𝑦

2 ⋅ 105

2 ⋅ 105

𝑦 = 𝑥4

𝑦 = 𝑥4 − 15𝑥2 − 15𝑥 Far away or Global

𝑥 −20

20

𝑥 −20

20

Figure 1.86: Local and global views of 𝑦 = 𝑥4 and 𝑦 = 𝑥4 − 15𝑥2 − 15𝑥

1.6 POWERS, POLYNOMIALS, AND RATIONAL FUNCTIONS

53

Rational Functions Rational functions are ratios of polynomials, 𝑝 and 𝑞: 𝑓 (𝑥) =

𝑝(𝑥) . 𝑞(𝑥)

1 . +4

Example 3

Look at a graph and explain the behavior of 𝑦 =

Solution

The function is even, so the graph is symmetric about the 𝑦-axis. As 𝑥 gets larger, the denominator gets larger, making the value of the function closer to 0. Thus the graph gets arbitrarily close to the 𝑥-axis as 𝑥 increases without bound. See Figure 1.87.

𝑥2

𝑦

𝑥 Figure 1.87: Graph of 𝑦 =

1 𝑥2 +4

In the previous example, we say that 𝑦 = 0 (i.e. the 𝑥-axis) is a horizontal asymptote. Writing “→” to mean “tends to,” we have 𝑦 → 0 as 𝑥 → ∞ and 𝑦 → 0 as 𝑥 → −∞. If the graph of 𝑦 = 𝑓 (𝑥) approaches a horizontal line 𝑦 = 𝐿 as 𝑥 → ∞ or 𝑥 → −∞, then the line 𝑦 = 𝐿 is called a horizontal asymptote.55 This occurs when 𝑓 (𝑥) → 𝐿

as 𝑥 → ∞

or

𝑓 (𝑥) → 𝐿

as 𝑥 → −∞.

If the graph of 𝑦 = 𝑓 (𝑥) approaches the vertical line 𝑥 = 𝐾 as 𝑥 → 𝐾 from one side or the other, that is, if 𝑦 → ∞ or 𝑦 → −∞ when 𝑥 → 𝐾, then the line 𝑥 = 𝐾 is called a vertical asymptote. The graphs of rational functions may have vertical asymptotes where the denominator is zero. For example, the function in Example 3 has no vertical asymptotes as the denominator is never zero. The function in Example 4 has two vertical asymptotes corresponding to the two zeros in the denominator. Rational functions have horizontal asymptotes if 𝑓 (𝑥) approaches a finite number as 𝑥 → ∞ or 𝑥 → −∞. We call the behavior of a function as 𝑥 → ±∞ its end behavior. Example 4

Look at a graph and explain the behavior of 𝑦 =

Solution

Factoring gives 𝑦=

3𝑥2 − 12 , including end behavior. 𝑥2 − 1

3𝑥2 − 12 3(𝑥 + 2)(𝑥 − 2) = (𝑥 + 1)(𝑥 − 1) 𝑥2 − 1

so 𝑥 = ±1 are vertical asymptotes. If 𝑦 = 0, then 3(𝑥 + 2)(𝑥 − 2) = 0 or 𝑥 = ±2; these are the 55 We

are assuming that 𝑓 (𝑥) gets arbitrarily close to 𝐿 as 𝑥 → ∞.

54

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

𝑥-intercepts. Note that zeros of the denominator give rise to the vertical asymptotes, whereas zeros of the numerator give rise to 𝑥-intercepts. Substituting 𝑥 = 0 gives 𝑦 = 12; this is the 𝑦-intercept. The function is even, so the graph is symmetric about the 𝑦-axis. 𝑦 Table 1.18

𝑦=

Vertical asymptote

Values of

𝑥 = −1

3𝑥2 −12 𝑥2 −1

𝑥=1

Horizontal asymptote

𝑦=

𝑥

Vertical asymptote

20

𝑦=3

3𝑥2 −12

𝑥

𝑥2 −1

−4

±10

2.909091

±100

2.999100

±1000

2.999991

4 −10 Figure 1.88: Graph of the function 𝑦 =

3𝑥2 −12 𝑥2 −1

To see what happens as 𝑥 → ±∞, look at the 𝑦-values in Table 1.18. Clearly 𝑦 is getting closer to 3 as 𝑥 gets large positively or negatively. Alternatively, realize that as 𝑥 → ±∞, only the highest powers of 𝑥 matter. For large 𝑥, the 12 and the 1 are insignificant compared to 𝑥2 , so 𝑦=

3𝑥2 − 12 3𝑥2 ≈ =3 𝑥2 − 1 𝑥2

for large 𝑥.

So 𝑦 → 3 as 𝑥 → ±∞, and therefore the horizontal asymptote is 𝑦 = 3. See Figure 1.88. Since, for 𝑥 > 1, the value of (3𝑥2 − 12)∕(𝑥2 − 1) is less than 3, the graph lies below its asymptote. (Why doesn’t the graph lie below 𝑦 = 3 when −1 < 𝑥 < 1?)

Exercises and Problems for Section 1.6 Online Resource: Additional Problems for Section 1.6 EXERCISES

13. 100𝑥5

For Exercises 1–2, what happens to the value of the function as 𝑥 → ∞ and as 𝑥 → −∞? 1. 𝑦 = 0.25𝑥3 + 3

4

14. 2𝑥

2. 𝑦 = 2 ⋅ 104𝑥

or

25 − 40𝑥2 + 𝑥3 + 3𝑥5

(a) What is the minimum possible degree of the polynomial? (b) Is the leading coefficient of the polynomial positive or negative?

4. 𝑓 (𝑥) = 3𝑥5 5. 𝑓 (𝑥) = 5𝑥4 − 25𝑥3 − 62𝑥2 + 5𝑥 + 300 6. 𝑓 (𝑥) = 1000 − 38𝑥 + 50𝑥2 − 5𝑥3

10. 𝑓 (𝑥) = 𝑒𝑥

10𝑥3 + 25𝑥2 + 50𝑥 + 100

17. Each of the graphs in Figure 1.89 is of a polynomial. The windows are large enough to show end behavior.

3. 𝑓 (𝑥) = −10𝑥4

3𝑥2 + 5𝑥 + 6 𝑥2 − 4 10 + 5𝑥2 − 3𝑥3 8. 𝑓 (𝑥) = 2𝑥3 − 4𝑥 + 12 9. 𝑓 (𝑥) = 3𝑥−4

or

1.05𝑥

15. 20𝑥4 + 100𝑥2 + 5𝑥 √ 16. 𝑥 or ln 𝑥

In Exercises 3–10, determine the end behavior of each function as 𝑥 → +∞ and as 𝑥 → −∞.

7. 𝑓 (𝑥) =

or

(I)

(II)

(IV)

(III)

(V)

In Exercises 11–16, which function dominates as 𝑥 → ∞? 11. 1000𝑥4

or

0.2𝑥5

12. 10𝑒0.1𝑥

or

5000𝑥2

Figure 1.89

1.6 POWERS, POLYNOMIALS, AND RATIONAL FUNCTIONS

Find cubic polynomials for the graphs in Exercises 18–19. 18.

19.

𝑥2 − 2 𝑥2 + 2 II. 𝑦 = 2 2 𝑥 +2 𝑥 −2 III. 𝑦 = (𝑥−1)(1−𝑥)(𝑥+1)2 IV. 𝑦 = 𝑥3 − 𝑥 1 V. 𝑦 = 𝑥 − VI. 𝑦 = (𝑥2 − 2)(𝑥2 + 2) 𝑥

𝑥 −2

1

In Exercises 33–40, which of the following functions have the given property? I.

4

2

𝑥

5 −2

55

2

𝑦=

Find possible formulas for the graphs in Exercises 20–23.

33. A polynomial of degree 3.

20.

34. A polynomial of degree 4.

21.

35. A rational function that is not a polynomial. 𝑥 −3

1

𝑥

4

−3

4

36. Exactly two distinct zeros. 37. Exactly one vertical asymptote.

22.

38. More than two distinct zeros.

23.

39. Exactly two vertical asymptotes. 40. A horizontal asymptote.

𝑥 −2

1

3

5

𝑥 −2

2

5

In Exercises 24–26, choose the functions that are in the given family, assuming 𝑎, 𝑏, and 𝑐 are positive constants. √ 𝑔(𝑥) = 𝑎𝑥23 𝑓 (𝑥) = 𝑥4 + 16 𝑎3 𝑏𝑥 1 𝑝(𝑥) = ℎ(𝑥) = − 𝑥−2 𝑐 5 √ 𝑎𝑏2 𝑟(𝑥) = −𝑥 + 𝑏 − 𝑐𝑥4 𝑞(𝑥) = 𝑐 24. Exponential

25. Quadratic

31. 𝑗(𝑥) = 𝑎𝑥−1 +

Figure 1.90

26. Linear

In Exercises 27–32, choose each of the families the given function is in, assuming 𝑎 is a positive integer and 𝑏 and 𝑐 are positive constants. I. Exponential III. Polynomial 𝑎𝑥 +𝑐 27. 𝑓 (𝑥) = 𝑏( ) 𝑥 𝑎 29. ℎ(𝑥) = 𝑏 𝑐

For Exercises 41–44, assuming the window is large enough to show end behavior, identify the graph as that of a rational function, exponential function or logarithmic function. 41. 42.

Figure 1.91 44.

43.

II. Power IV. Rational 28. 𝑔(𝑥) = 𝑎𝑥2 +

𝑏 𝑥

𝑏 𝑥2

Figure 1.92

Figure 1.93

30. 𝑘(𝑥) = 𝑏𝑥𝑎 ( ) 𝑎 + 𝑏 2𝑥 32. 𝑙(𝑥) = 𝑐

PROBLEMS 𝑦

45. How many distinct roots can a polynomial of degree 5 have? (List all possibilities.) Sketch a possible graph for each case. 46. Find a calculator window in which the graphs of 𝑓 (𝑥) = 𝑥3 + 1000𝑥2 + 1000 and 𝑔(𝑥) = 𝑥3 − 1000𝑥2 − 1000 appear indistinguishable. 47. A cubic polynomial with positive leading coefficient is shown in Figure 1.94 for −10 ≤ 𝑥 ≤ 10 and −10 ≤ 𝑦 ≤ 10. What can be concluded about the total number of zeros of this function? What can you say about the location of each of the zeros? Explain.

10 5 𝑥 −10

−5

5

−5 −10

Figure 1.94

10

56

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

48. (a) If 𝑓 (𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐, what can you say about the values of 𝑎, 𝑏, and 𝑐 if:

where 𝑣0 is the initial velocity and 𝑔 is the acceleration due to gravity.

(i) (1, 1) is on the graph of 𝑓 (𝑥)? (ii) (1, 1) is the vertex of the graph of 𝑓 (𝑥)? (Hint: The axis of symmetry is 𝑥 = −𝑏∕(2𝑎).) (iii) The 𝑦-intercept of the graph is (0, 6)?

(a) At what height is the object initially? (b) How long is the object in the air before it hits the ground? (c) When will the object reach its maximum height? (d) What is that maximum height?

(b) Find a quadratic function satisfying all three conditions. 49. A box of fixed volume 𝑉 has a square base with side length 𝑥. Write a formula for the height, ℎ, of the box in terms of 𝑥 and 𝑉 . Sketch a graph of ℎ versus 𝑥. 50. A closed cylindrical can of fixed volume 𝑉 has radius 𝑟. (a) Find the surface area, 𝑆, as a function of 𝑟. (b) What happens to the value of 𝑆 as 𝑟 → ∞? (c) Sketch a graph of 𝑆 against 𝑟, if 𝑉 = 10 cm3 . 51. The DuBois formula relates a person’s surface area 𝑠, in m2 , to weight 𝑤, in kg, and height ℎ, in cm, by 𝑠 = 0.01𝑤0.25 ℎ0.75 . (a) What is the surface area of a person who weighs 65 kg and is 160 cm tall? (b) What is the weight of a person whose height is 180 cm and who has a surface area of 1.5 m2 ? (c) For people of fixed weight 70 kg, solve for ℎ as a function of 𝑠. Simplify your answer. 52. According to Car and Driver, an Alfa Romeo going at 70 mph requires 150 feet to stop.56 Assuming that the stopping distance is proportional to the square of velocity, find the stopping distances required by an Alfa Romeo going at 35 mph and at 140 mph. 53. Poiseuille’s Law gives the rate of flow, 𝑅, of a gas through a cylindrical pipe in terms of the radius of the pipe, 𝑟, for a fixed drop in pressure between the two ends of the pipe. (a) Find a formula for Poiseuille’s Law, given that the rate of flow is proportional to the fourth power of the radius. (b) If 𝑅 = 400 cm3 /sec in a pipe of radius 3 cm for a certain gas, find a formula for the rate of flow of that gas through a pipe of radius 𝑟 cm. (c) What is the rate of flow of the same gas through a pipe with a 5 cm radius? 54. A pomegranate is thrown from ground level straight up into the air at time 𝑡 = 0 with velocity 64 feet per second. Its height at time 𝑡 seconds is 𝑓 (𝑡) = −16𝑡2 + 64𝑡. Find the time it hits the ground and the time it reaches its highest point. What is the maximum height? 55. The height of an object above the ground at time 𝑡 is given by 𝑔 𝑠 = 𝑣0 𝑡 − 𝑡2 , 2 56 http://www.caranddriver.com/alfa-romeo/4c.

56. The rate, 𝑅, at which a population in a confined space increases is proportional to the product of the current population, 𝑃 , and the difference between the carrying capacity, 𝐿, and the current population. (The carrying capacity is the maximum population the environment can sustain.) (a) Write 𝑅 as a function of 𝑃 . (b) Sketch 𝑅 as a function of 𝑃 . In Problems 57–61, the length of a plant, 𝐿, is a function of its mass, 𝑀. A unit increase in a plant’s mass stretches the plant’s length more when the plant is small, and less when the plant is large.57 Assuming 𝑀 > 0, decide if the function agrees with the description. 57. 𝐿 = 2𝑀 1∕4

58. 𝐿 = 0.2𝑀 3 + 𝑀 4

59. 𝐿 = 2𝑀 −1∕4

60. 𝐿 =

4(𝑀 + 1)2 − 1 (𝑀 + 1)2

2

61. 𝐿 =

10(𝑀 + 1) − 1 (𝑀 + 1)3

In Problems 62–64, find all horizontal and vertical asymptotes for each rational function. 62. 𝑓 (𝑥) =

5𝑥 − 2 2𝑥 + 3

64. 𝑓 (𝑥) =

5𝑥3 + 7𝑥 − 1 𝑥3 − 27

63. 𝑓 (𝑥) =

𝑥2 + 5𝑥 + 4 𝑥2 − 4

65. For each function, fill in the blanks in the statements: as 𝑥 → −∞, 𝑓 (𝑥) → 𝑓 (𝑥) → as 𝑥 → +∞. (a) 𝑓 (𝑥) = 17 + 5𝑥2 − 12𝑥3 − 5𝑥4 3𝑥2 − 5𝑥 + 2 (b) 𝑓 (𝑥) = 2𝑥2 − 8 (c) 𝑓 (𝑥) = 𝑒𝑥 66. A rational function 𝑦 = 𝑓 (𝑥) is graphed in Figure 1.95. If 𝑓 (𝑥) = 𝑔(𝑥)∕ℎ(𝑥) with 𝑔(𝑥) and ℎ(𝑥) both quadratic functions, give possible formulas for 𝑔(𝑥) and ℎ(𝑥). 𝑦

1

𝑦=2 𝑦 = 𝑓 (𝑥) 𝑥

Figure 1.95

Accessed February 2016. K. and Enquist, B., “Invariant scaling relationships for interspecific plant biomass production rates and body size”, PNAS, Feb 27, 2001. 57 Niklas,

1.6 POWERS, POLYNOMIALS, AND RATIONAL FUNCTIONS

67. After running 3 miles at a speed of 𝑥 mph, a man walked the next 6 miles at a speed that was 2 mph slower. Express the total time spent on the trip as a function of 𝑥. What horizontal and vertical asymptotes does the graph of this function have? 68. Which of the functions I–III meet each of the following descriptions? There may be more than one function for each description, or none at all. (a) (b) (c) (d) (e)

57

71. Consider the point 𝑃 at the intersection of the circle 𝑥2 +𝑦2 = 2𝑎2 and the parabola 𝑦 = 𝑥2 ∕𝑎 in Figure 1.97. If 𝑎 is increased, the point 𝑃 traces out a curve. For 𝑎 > 0, find the equation of this curve. 𝑦 𝑦 = 𝑥2 ∕𝑎 𝑃

Horizontal asymptote of 𝑦 = 1. The 𝑥-axis is a horizontal asymptote. Symmetric about the 𝑦-axis. An odd function. Vertical asymptotes at 𝑥 = ±1.

𝑥

𝑥2 + 𝑦2 = 2𝑎2

𝑥2 − 1 𝑥2 + 1 𝑥−1 II. 𝑦 = III. 𝑦 = 𝑥2 + 1 𝑥2 + 1 𝑥2 − 1 69. Values of three functions are given in Table 1.19, rounded to two decimal places. One function is of the form 𝑦 = 𝑎𝑏𝑡 , one is of the form 𝑦 = 𝑐𝑡2 , and one is of the form 𝑦 = 𝑘𝑡3 . Which function is which?

Figure 1.97

I. 𝑦 =

72. When an object of mass 𝑚 moves with a velocity 𝑣 that is small compared to the velocity of light, 𝑐, its energy is given approximately by

Table 1.19 𝑡

𝑓 (𝑡)

𝑡

𝑔(𝑡)

𝑡

ℎ(𝑡)

2.0

4.40

1.0

3.00

0.0

2.04

2.2

5.32

1.2

5.18

1.0

3.06

2.4

6.34

1.4

8.23

2.0

4.59

2.6

7.44

1.6

12.29

3.0

6.89

2.8

8.62

1.8

17.50

4.0

10.33

3.0

9.90

2.0

24.00

5.0

15.49

𝐸≈

If 𝑣 is comparable in size to 𝑐, then the energy must be computed by the exact formula

𝐸 = 𝑚𝑐

70. Use a graphing calculator or a computer to graph 𝑦 = 𝑥4 and 𝑦 = 3𝑥 . Determine approximate domains and ranges that give each of the graphs in Figure 1.96. (a)

𝑦

(b)

𝑥4

𝑦

(c)

𝑦

𝑥4

3𝑥

3𝑥

3𝑥 𝑥

1 2 𝑚𝑣 . 2

𝑥4

𝑥

Figure 1.96

2

(

1

)

−1 . √ 1 − 𝑣2 ∕𝑐 2

(a) Plot a graph of both functions for 𝐸 against 𝑣 for 0 ≤ 𝑣 ≤ 5 ⋅ 108 and 0 ≤ 𝐸 ≤ 5 ⋅ 1017 . Take 𝑚 = 1 kg and 𝑐 = 3 ⋅ 108 m/sec. Explain how you can predict from the exact formula the position of the vertical asymptote. (b) What do the graphs tell you about the approximation? For what values of 𝑣 does the first formula give a good approximation to 𝐸?

𝑥

73. If 𝑦 = 100𝑥−0.2 and 𝑧 = ln 𝑥, explain why 𝑦 is an exponential function of 𝑧.

Strengthen Your Understanding In Problems 74–79, explain what is wrong with the statement. 74. The graph of a polynomial of degree 5 cuts the horizontal axis five times. 75. A fourth-degree polynomial tends to infinity as 𝑥 → ∞.

In Problems 80–85, give an example of: 80. A polynomial of degree 3 whose graph cuts the horizontal axis three times to the right of the origin. 81. A rational function with horizontal asymptote 𝑦 = 3.

76. A rational function has a vertical asymptote.

82. A rational function that is not a polynomial and that has no vertical asymptote.

77. 𝑥5 > 𝑥3 for 𝑥 > 0

83. A function that has a vertical asymptote at 𝑥 = −7𝜋.

78. Every rational function has a horizontal asymptote.

84. A function that has exactly 17 vertical asymptotes.

79. A function cannot cross its horizontal asymptote.

58

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

85. A function that has a vertical asymptote which is crossed by a horizontal asymptote. Are the statements in Problems 86–89 true or false? Give an explanation for your answer.

89. For 𝑥 > 0 the graph of the rational function 𝑓 (𝑥) = 5(𝑥3 − 𝑥) is a line. 𝑥2 + 𝑥 90. List the following functions in order from smallest to largest as 𝑥 → ∞ (that is, as 𝑥 increases without bound).

86. Every polynomial of even degree has a least one real zero. 87. Every polynomial of odd degree has a least one real zero. 88. The composition of two quadratic functions is quadratic.

1.7

(a) 𝑓 (𝑥) = −5𝑥

(b) 𝑔(𝑥) = 10𝑥

(c) ℎ(𝑥) = 0.9𝑥

(d) 𝑘(𝑥) = 𝑥5

(e) 𝑙(𝑥) = 𝜋

𝑥

INTRODUCTION TO LIMITS AND CONTINUITY In this section we switch focus from families of functions to an intuitive introduction to continuity and limits. Limits are central to a full understanding of calculus.

The Idea of Continuity Roughly speaking, a function is continuous on an interval if its graph has no breaks, jumps, or holes in that interval. A function is continuous at a point if nearby values of the independent variable give nearby values of the function. Most real world phenomena are modeled using continuous functions.

The Graphical Viewpoint: Continuity on an Interval A continuous function has a graph which can be drawn without lifting the pencil from the paper. Example: The function 𝑓 (𝑥) = 3𝑥3 −𝑥2 +2𝑥−1 is continuous on any interval. (See Figure 1.98.) Example: The function 𝑓 (𝑥) = 1∕𝑥 is not defined at 𝑥 = 0. It is continuous on any interval not containing 0. (See Figure 1.99.) Example: A company rents cars for $7 per hour or fraction thereof, so it costs $7 for a trip of one hour or less, $14 for a trip between one and two hours, and so on. If 𝑝(𝑥) is the price of trip lasting 𝑥 hours, then its graph (in Figure 1.100) is a series of steps. This function is not continuous on any open interval containing a positive integer because the graph jumps at these points. 𝑓 (𝑥)

5

𝑓 (𝑥) =

1 𝑥

𝑥

𝑥 −2 −1

1

dollars

2

14 7

−5 Figure 1.98: The graph of 𝑓 (𝑥) = 3𝑥3 − 𝑥2 + 2𝑥 − 1

𝑝(𝑥)

21

Figure 1.99: Not continuous on any interval containing 0

1

2

3

𝑥 (hours)

Figure 1.100: Cost of renting a car

The Numerical Viewpoint: Continuity at a Point In practical work, continuity of a function at a point is important because it means that small errors in the independent variable lead to small errors in the value of the function. Conversely, if a function is not continuous at a point, a small error in input measurement can lead to an enormous error in output. Example: Suppose that 𝑓 (𝑥) = 𝑥2 and that we want to compute 𝑓 (𝜋). Knowing 𝑓 is continuous tells us that taking 𝑥 = 3.14 should give a good approximation to 𝑓 (𝜋), and that we can get as accurate an approximation to 𝑓 (𝜋) as we want by using enough decimals of 𝜋.

1.7 INTRODUCTION TO LIMITS AND CONTINUITY

59

Example: If 𝑝(𝑥) is the price of renting a car graphed in Figure 1.100, then 𝑝(0.99) = 𝑝(1) =$7, whereas 𝑝(1.01) =$14, because as soon as we pass one hour, the price jumps to $14. So a small difference in time can lead to a significant difference in the cost. Hence 𝑝 is not continuous at 𝑥 = 1. As we see from its graph, this also means it is not continuous on any open interval including 𝑥 = 1. We express continuity at a point by saying that if 𝑓 (𝑥) is continuous at 𝑥 = 𝑐, the values of 𝑓 (𝑥) approach 𝑓 (𝑐) as 𝑥 approaches 𝑐. Example 1

Investigate the continuity of 𝑓 (𝑥) = 𝑥2 at 𝑥 = 2.

Solution

From Table 1.20, it appears that the values of 𝑓 (𝑥) = 𝑥2 approach 𝑓 (2) = 4 as 𝑥 approaches 2. Thus 𝑓 appears to be continuous at 𝑥 = 2. Values of 𝑥2 near 𝑥 = 2

Table 1.20 𝑥

1.9

1.99

1.999

2.001

2.01

2.1

𝑥2

3.61

3.96

3.996

4.004

4.04

4.41

Which Functions Are Continuous? Most of the functions we have seen are continuous everywhere they are defined. For example, exponential functions, polynomials, and the sine and cosine are continuous everywhere. Rational functions are continuous on any interval in which their denominators are not zero. Functions created by adding, multiplying, or composing continuous functions are also continuous.58

The Idea of a Limit Continuity at a point describes behavior of a function near a point, as well as at the point. To find the value of a function at a point we can just evaluate the function there. To focus on what happens to the values of a function near a point, we introduce the concept of limit. First some notation: We write lim 𝑓 (𝑥) = 𝐿 if the values of 𝑓 (𝑥) approach 𝐿 as 𝑥 approaches 𝑐. 𝑥→𝑐

How should we find the limit 𝐿, or even know whether such a number exists? We look for trends in the values of 𝑓 (𝑥) as 𝑥 gets closer to 𝑐, but 𝑥 ≠ 𝑐. A graph from a calculator or computer often helps. Example 2

Use a graph to estimate lim

𝜃→0

(

sin 𝜃 𝜃

)

. (Use radians.)

1 𝑓 (𝜃) =

−2𝜋

−𝜋

𝜋

sin 𝜃 𝜃

2𝜋

Figure 1.101: Find the limit as 𝜃 → 0 58 For

more details, see Theorem 1.8 in Section 1.8.

𝜃

60

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

Solution

Figure 1.101 shows that as 𝜃 approaches 0 from either side, the value of sin 𝜃∕𝜃 appears to approach 1, suggesting that lim (sin 𝜃∕𝜃) = 1. Zooming in on the graph near 𝜃 = 0 provides further support 𝜃→0

for this conclusion. Notice that sin 𝜃∕𝜃 is undefined at 𝜃 = 0. Figure 1.101 suggests that lim (sin 𝜃∕𝜃) = 1, but to be sure we need to be more precise about 𝜃→0

words like “approach” and “close.”

Definition of Limit By the beginning of the 19th century, calculus had proved its worth, and there was no doubt about the correctness of its answers. However, it was not until the work of the French mathematician Augustin Cauchy (1789–1857) that calculus was put on a rigorous footing. Cauchy gave a formal definition of the limit, similar to the following: A function 𝑓 is defined on an interval around 𝑐, except perhaps at the point 𝑥 = 𝑐. We define the limit of the function 𝑓 (𝑥) as 𝑥 approaches 𝑐, written lim 𝑓 (𝑥), to be a number 𝐿 (if one 𝑥→𝑐 exists) such that 𝑓 (𝑥) is as close to 𝐿 as we want whenever 𝑥 is sufficiently close to 𝑐 (but 𝑥 ≠ 𝑐). If 𝐿 exists, we write lim 𝑓 (𝑥) = 𝐿. 𝑥→𝑐

Note that this definition ensures a limit, if it exists, cannot have more than one value.

Finding Limits As we saw in Example 2, we can often estimate the value of a limit from its graph. We can also estimate a limit by using a table of values. However, no matter how closely we zoom in on a graph at a point or how close to a point we evaluate a function, there are always points closer, so in using these techniques for estimating limits we are never completely sure we have the exact answer. In Example 4 and Section 1.9 we show how a limit can be found exactly using algebra.

Definition of Continuity In Example 2 we saw that the values of sin 𝜃∕𝜃 approach 1 as 𝜃 approaches 0. However, sin 𝜃∕𝜃 is not continuous at 𝜃 = 0 since its graph has a hole there; see Figure 1.101. This illustrates an important difference between limits and continuity: a limit is only concerned with what happens near a point but continuity depends on what happens near a point and at that point. We now give a more precise definition of continuity using limits. The function 𝑓 is continuous at 𝑥 = 𝑐 if 𝑓 is defined at 𝑥 = 𝑐 and if lim 𝑓 (𝑥) = 𝑓 (𝑐).

𝑥→𝑐

In other words, 𝑓 (𝑥) is as close as we want to 𝑓 (𝑐) provided 𝑥 is close enough to 𝑐. The function is continuous on an interval [𝑎, 𝑏] if it is continuous at every point in the interval.59

The Intermediate Value Theorem Continuous functions have many useful properties. For example, to locate the zeros of a continuous function, we can look for intervals where the function changes sign. 59 If

𝑐 is an endpoint of the interval, we can define continuity at 𝑥 = 𝑐 using one-sided limits at 𝑐; see Section 1.8.

1.7 INTRODUCTION TO LIMITS AND CONTINUITY

Example 3

What do the values in Table 1.21 tell you about the zeros of 𝑓 (𝑥) = cos 𝑥 − 2𝑥2 ? Table 1.21

Solution

61

𝑓 (𝑥) = cos 𝑥 − 2𝑥2

1

𝑥

𝑓 (𝑥)

0

1.00

0.2

0.90

0.4

0.60

0.6

0.11

0.8

−0.58

1.0

−1.46

𝑥 0.2 0.4 0.6 0.8 1 −1 Figure 1.102: Zeros occur where the graph of a continuous function crosses the horizontal axis

Since 𝑓 (𝑥) is the difference of two continuous functions, it is continuous. Since 𝑓 is positive at 𝑥 = 0.6 and negative at 𝑥 = 0.8, and its graph has no breaks, the graph must cross the axis between these points. We conclude that 𝑓 (𝑥) = cos 𝑥 − 2𝑥2 has at least one zero in the interval 0.6 < 𝑥 < 0.8. Figure 1.102 suggests that there is only one zero in the interval 0 ≤ 𝑥 ≤ 1, but we cannot be sure of this from a graph or a table of values. In the previous example, we concluded that 𝑓 (𝑥) = cos 𝑥 − 2𝑥2 has a zero between 𝑥 = 0 and 𝑥 = 1 because it changed from positive to negative without skipping values in between—in other words, because it is continuous. If it were not continuous, the graph could jump across the 𝑥-axis, changing sign but not creating a zero. For example, 𝑓 (𝑥) = 1∕𝑥 has opposite signs at 𝑥 = −1 and 𝑥 = 1, but no zeros for −1 ≤ 𝑥 ≤ 1 because of the break at 𝑥 = 0. (See Figure 1.99.) More generally, the intuitive notion of continuity tells us that, as we follow the graph of a continuous function 𝑓 from some point (𝑎, 𝑓 (𝑎)) to another point (𝑏, 𝑓 (𝑏)), then 𝑓 takes on all intermediate values between 𝑓 (𝑎) and 𝑓 (𝑏). (See Figure 1.103.) The formal statement of this property is known as the Intermediate Value Theorem, and is a powerful tool in theoretical proofs in calculus.

Theorem 1.1: Intermediate Value Theorem Suppose 𝑓 is continuous on a closed interval [𝑎, 𝑏]. If 𝑘 is any number between 𝑓 (𝑎) and 𝑓 (𝑏), then there is at least one number 𝑐 in [𝑎, 𝑏] such that 𝑓 (𝑐) = 𝑘. (𝑏, 𝑓 (𝑏)) 𝑘 (𝑎, 𝑓 (𝑎)) 𝑥 𝑐 𝑏 Figure 1.103: The Intermediate Value Theorem guarantees that a continuous 𝑓 takes on the value 𝑘 𝑎

Finding Limits Exactly Using Continuity and Algebra The concept of limit is critical to a formal justification of calculus, so an understanding of limit outside the context of continuity is important. We have already seen how to use a graph or a table of values to estimate a limit. We now see how to find the exact value of a limit.

Limits of Continuous Functions In Example 1, the limit of 𝑓 (𝑥) = 𝑥2 at 𝑥 = 2 is the same as 𝑓 (2). This is because 𝑓 (𝑥) is a continuous function at 𝑥 = 2, and this is precisely the definition of continuity:

62

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

Limits of Continuous Functions If a function 𝑓 (𝑥) is continuous at 𝑥 = 𝑐, the limit is the value of 𝑓 (𝑥) there: lim 𝑓 (𝑥) = 𝑓 (𝑐).

𝑥→𝑐

Thus, to find the limits for a continuous function: Substitute 𝑐.

Limits of Functions Which Are not Continuous If a function is not continuous at a point 𝑥 = 𝑐 it is still possible for the limit to exist, but it can be hard to find. Sometimes such limits can be computed using algebra. 𝑥2 − 9 and then use algebra to find the limit exactly. 𝑥→3 𝑥 − 3

Example 4

Use a graph to estimate lim

Solution

Evaluating (𝑥2 − 9)∕(𝑥 − 3) at 𝑥 = 3 gives us 0∕0, so the function is undefined at 𝑥 = 3, shown as a hole in Figure 1.104. However, we see that as 𝑥 approaches 3 from either side, the value of (𝑥2 − 9)∕(𝑥 − 3) appears to approach 6, suggesting the limit is 6. To find this limit exactly, we first use algebra to rewrite the function. We have 𝑥2 − 9 (𝑥 + 3)(𝑥 − 3) = . 𝑥−3 𝑥−3 Since 𝑥 ≠ 3 in the limit, we can cancel the common factor 𝑥 − 3 to see (𝑥 + 3)(𝑥 − 3) 𝑥2 − 9 = lim = lim (𝑥 + 3). 𝑥→3 𝑥 − 3 𝑥→3 𝑥→3 𝑥−3 lim

Since 𝑥 + 3 is continuous, we have 𝑥2 − 9 = lim (𝑥 + 3) = 3 + 3 = 6. 𝑥→3 𝑥→3 𝑥 − 3 lim

𝑓 (𝑥) =

𝑥2 −9 𝑥−3

6

𝑥 3 Figure 1.104: Find the limit as 𝑥 → 3

If a function is continuous at a point, then its limit must exist at that point. Example 4 illustrates that the existence of a limit is not enough to guarantee continuity. More precisely, as we can see from Figure 1.104, the function 𝑓 (𝑥) = (𝑥2 − 9)∕(𝑥 − 3) is not continuous at 𝑥 = 3 as there is a hole in its graph at that point. However, Example 4 shows that the limit lim 𝑓 (𝑥) does exist and is equal to 6. This is because for a function 𝑓 (𝑥) to be continuous at 𝑥→3

a point 𝑥 = 𝑐, the limit has to exist at 𝑥 = 𝑐, the function has to be defined at 𝑥 = 𝑐 and the limit has to equal the function. In this case, even though the limit does exist, the function does not have a value at 𝑥 = 3, so it cannot be continuous there.

1.7 INTRODUCTION TO LIMITS AND CONTINUITY

63

When Limits Do Not Exist Whenever there is no number 𝐿 such that lim 𝑓 (𝑥) = 𝐿, we say lim 𝑓 (𝑥) does not exist. The fol𝑥→𝑐 𝑥→𝑐 lowing three examples show some of the ways in which a limit may fail to exist. |𝑥 − 2| does not exist. 𝑥−2

Example 5

Use a graph to explain why lim

Solution

As we see in Figure 1.105, the value of 𝑓 (𝑥) =

𝑥→2

|𝑥 − 2| approaches −1 as 𝑥 approaches 2 from the left 𝑥−2 |𝑥 − 2| and the value of 𝑓 (𝑥) approaches 1 as 𝑥 approaches 2 from the right. This means if lim =𝐿 𝑥→2 𝑥 − 2 then 𝐿 would have to be both 1 and −1. Since 𝐿 cannot have two different values, the limit does not exist.

1 𝑥

𝑥

1 − 2𝜋

𝑥

2

1 2𝜋

−1

Figure 1.105: Graph of 𝑓 (𝑥) = |𝑥 − 2|∕(𝑥 − 2)

Figure 1.106: Graph of 𝑔(𝑥) = 1∕𝑥2

Figure 1.107: Graph of ℎ(𝑥) = sin (1∕𝑥)

1 does not exist. 𝑥2

Example 6

Use a graph to explain why lim

Solution

As 𝑥 approaches zero, 𝑔(𝑥) = 1∕𝑥2 becomes arbitrarily large, so it cannot approach any finite number 𝐿. See Figure 1.106. Therefore we say 1∕𝑥2 has no limit as 𝑥 → 0.

𝑥→0

Since 1∕𝑥2 gets arbitrarily large on both sides of 𝑥 = 0, we can write lim 1∕𝑥2 = ∞. The limit 𝑥→0

still does not exist since it does not approach a real number 𝐿, but we can use limit notation to describe its behavior. This behavior may also be described as “diverging to infinity.” ( ) 1 does not exist. 𝑥

Example 7

Explain why lim sin

Solution

The sine function has values between −1 and 1. The graph of ℎ(𝑥) = sin (1∕𝑥) in Figure 1.107 oscillates more and more rapidly as 𝑥 → 0. There are 𝑥-values approaching 0 where sin(1∕𝑥) = −1. There are also 𝑥-values approaching 0 where sin(1∕𝑥) = 1. So if the limit existed, it would have to be both −1 and 1. Thus, the limit does not exist.

𝑥→0

Notice that in all three examples, the function is not continuous at the given point. This is because continuity requires the existence of a limit, so failure of a limit to exist at a point automatically means a function is not continuous there.

64

Chapter 1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS

Exercises and Problems for Section 1.7 EXERCISES 1. (a) Using Figure 1.108, find all values of 𝑥 for which 𝑓 is not continuous. (b) List the largest open intervals on which 𝑓 is continuous.

5.

3 2 1

𝑥 −3−2−1 −2 −3 −4 −5 −6

𝑓 (𝑥) 𝑥 −3 −2 −1

1

2

𝑓 (𝑥)

3

1 2 3 4 5 6

Figure 1.112

Figure 1.108 2. (a) Using Figure 1.109, find all values of 𝑥 for which 𝑓 is not continuous. (b) List the largest open intervals on which 𝑓 is continuous.

6. Assume 𝑓 (𝑥) is continuous on an interval around 𝑥 = 0, except possibly at 𝑥 = 0. What does the table of values suggest as the value of lim 𝑓 (𝑥)? Does the limit defi𝑥→0 nitely have this value?

𝑓 (𝑥)

𝑥

𝑥

−0.1

−0.01

0.01

0.1

𝑓 (𝑥)

1.987

1.999

1.999

1.987

7. Assume 𝑔(𝑡) is continuous on an interval around 𝑡 = 3, except possibly at 𝑡 = 3. What does the table of values suggest as the value of lim 𝑔(𝑡)? Does the limit def𝑡→3 initely have this value?

1 2 3 4 5 6

Figure 1.109 𝑡

3. Use the graph of 𝑓 (𝑥) in Figure 1.110 to give approximate values for the following limits (if they exist). If the limit does not exist, say so. (a)

lim 𝑓 (𝑥)

𝑥→−3

(d) lim 𝑓 (𝑥) 𝑥→0

lim 𝑓 (𝑥)

(b)

𝑥→−2

(e) lim 𝑓 (𝑥) 𝑥→1

lim 𝑓 (𝑥)

(c)

𝑥→−1

lim 𝑓 (𝑥)

(f)

𝑥→3

𝑔(𝑡)

2.9

2.99

3.01

3.1

0.948

0.995

1.005

1.049

In Exercises 8–9, (a) Make a table of values of 𝑓 (𝑥) for 𝑥 = −0.1, −0.01, 0.01, and 0.1. (b) Use the table to estimate lim 𝑓 (𝑥). 𝑥→0

𝑓 (𝑥)

3

sin(5𝑥) 𝑒3𝑥 − 1 9. 𝑓 (𝑥) = 𝑥 𝑥 10. Use a table of values to estimate lim (5 + ln 𝑥).

2

8. 𝑓 (𝑥) =

1

𝑥→1

𝑥 −4 −3 −2 −1 −1

1

2

3 4

In Exercises 11–16, is the function continuous on the interval?

Figure 1.110 In Exercises 4–5, the graph of 𝑦 = 𝑓 (𝑥) is given. (a) Give the 𝑥-values where 𝑓 (𝑥) is not continuous. (b) Does the limit of 𝑓 (𝑥) exist at each 𝑥-value where 𝑓 (𝑥) is not continuous? If so, give the value of the limit. 4. 5 4 3 2 1 𝑥 −3 −2 −1 −1 −2 −3 −4

1 2 3 4

Figure 1.111

𝑓 (𝑥)

11.

1 on [−1, 1] 𝑥−2

12.

1 on [0, 3] 𝑥−2

1 13. √ on [3, 4] 2𝑥 − 5

14. 2𝑥 + 𝑥−1 on [−1, 1]

15.

16.

1 on [0, 𝜋] cos 𝑥

𝑒sin 𝜃 on [− 𝜋4 , 𝜋4 ] cos 𝜃

17. Are the following functions continuous? Explain. { 𝑥 𝑥≤1 (a) 𝑓 (𝑥) = 2 1 0 and 𝑟′ (𝑡) < 0 for 0 ≤ 𝑡 ≤ 12.

4

𝑟(𝑡) 𝑑𝑡 or

∫0

58. Give a reasonable overestimate of

61. I.

𝑟(𝑡) 𝑑𝑡 or 4𝑟(4)?

∫5

12

∫0

𝑟(𝑡) 𝑑𝑡

III. 𝑅(10)+𝑟(10)⋅2

11

𝑟(𝑡) 𝑑𝑡

II.

∫8

𝑟(𝑡) 𝑑𝑡 III. 𝑅(12)−𝑅(9)

Strengthen Your Understanding In Problems 62–63, explain what is wrong with the statement.

In Problems 64–65, give an example of: 64. A function 𝑓 (𝑥) and limits of integration 𝑎 and 𝑏 such 𝑏 that ∫𝑎 𝑓 (𝑥) 𝑑𝑥 = 𝑒4 − 𝑒2 .

62. If 𝑓 (𝑡) represents the rate, in lbs per year, at which a 4 dog gains weight 𝑡 years after it is born, then ∫0 𝑓 (𝑡)𝑑𝑡 represents the weight of the dog when the dog is four years old. √ 63. If 𝑓 (𝑥) = 𝑥 the Fundamental Theorem of Calculus √ √ 9√ states that ∫4 𝑥 𝑑𝑥 = 9 − 4.

5.4

65. The graph of a velocity function of a car that travels 200 miles in 4 hours. 66. True or False? The units for an integral of a function 𝑓 (𝑥) are the same as the units for 𝑓 (𝑥).

THEOREMS ABOUT DEFINITE INTEGRALS

Properties of the Definite Integral 𝑏

For the definite integral ∫𝑎 𝑓 (𝑥) 𝑑𝑥, we have so far only considered the case 𝑎 < 𝑏. We now allow 𝑏

𝑎 ≥ 𝑏. ∑ We still set 𝑥0 = 𝑎, 𝑥𝑛 = 𝑏, and Δ𝑥 = (𝑏 − 𝑎)∕𝑛. As before, we have ∫𝑎 𝑓 (𝑥)𝑑𝑥 = lim𝑛→∞ 𝑛𝑖=1 𝑓 (𝑥𝑖 )Δ𝑥.

Theorem 5.2: Properties of Limits of Integration If 𝑎, 𝑏, and 𝑐 are any numbers and 𝑓 is a continuous function, then 𝑎

1.

𝑏

𝑓 (𝑥) 𝑑𝑥 = −

∫𝑏 𝑐

∫𝑎

𝑓 (𝑥) 𝑑𝑥.

𝑏

𝑓 (𝑥) 𝑑𝑥 +

𝑏

𝑓 (𝑥) 𝑑𝑥. ∫𝑐 ∫𝑎 ∫𝑎 In words: 1. The integral from 𝑏 to 𝑎 is the negative of the integral from 𝑎 to 𝑏. 2.

𝑓 (𝑥) 𝑑𝑥 =

2. The integral from 𝑎 to 𝑐 plus the integral from 𝑐 to 𝑏 is the integral from 𝑎 to 𝑏. By interpreting the integrals as areas, we can justify these results for 𝑓 ≥ 0. In fact, they are true for all functions for which the integrals make sense. 11 “A

catastrophic rupture hits region’s water system,” The Boston Globe, May 2, 2010.

5.4 THEOREMS ABOUT DEFINITE INTEGRALS

𝑓 (𝑥)

𝑓 (𝑥)

𝑥 𝑎

303

𝑥

𝑐

𝑎

𝑐 𝑏 Figure 5.62: Additivity of the definite integral (𝑎 < 𝑏 < 𝑐)

𝑏 Figure 5.61: Additivity of the definite integral (𝑎 < 𝑐 < 𝑏)

Why is ∫𝒃𝒂 𝒇 (𝒙) 𝒅𝒙 = − ∫𝒂𝒃 𝒇 (𝒙) 𝒅𝒙? ∑ By definition, both integrals are approximated by sums of the form 𝑓 (𝑥𝑖 )Δ𝑥. The only difference 𝑎 𝑏 in the sums for ∫𝑏 𝑓 (𝑥) 𝑑𝑥 and ∫𝑎 𝑓 (𝑥) 𝑑𝑥 is that in the first Δ𝑥 = (𝑎 − 𝑏)∕𝑛 = −(𝑏 − 𝑎)∕𝑛 and in the second Δ𝑥 = (𝑏 − 𝑎)∕𝑛. Since everything else about the sums is the same, we must have 𝑎 𝑏 ∫𝑏 𝑓 (𝑥) 𝑑𝑥 = − ∫𝑎 𝑓 (𝑥) 𝑑𝑥.

Why is ∫𝒂𝒄 𝒇 (𝒙) 𝒅𝒙 + ∫𝒄𝒃 𝒇 (𝒙) 𝒅𝒙 = ∫𝒂𝒃 𝒇 (𝒙) 𝒅𝒙? 𝑐

𝑏

𝑏

Suppose 𝑎 < 𝑐 < 𝑏. Figure 5.61 suggests that ∫𝑎 𝑓 (𝑥) 𝑑𝑥 + ∫𝑐 𝑓 (𝑥) 𝑑𝑥 = ∫𝑎 𝑓 (𝑥) 𝑑𝑥 since the area under 𝑓 from 𝑎 to 𝑐 plus the area under 𝑓 from 𝑐 to 𝑏 together make up the whole area under 𝑓 from 𝑎 to 𝑏. This property holds for all numbers 𝑎, 𝑏, and 𝑐, not just those satisfying 𝑎 < 𝑐 < 𝑏. (See Figure 5.62.) For example, the area under 𝑓 from 3 to 6 is equal to the area from 3 to 8 minus the area from 6 to 8, so 6

∫3

Example 1

8

𝑓 (𝑥) 𝑑𝑥 =

∫3

8

𝑓 (𝑥) 𝑑𝑥 −

1.25

∫6

8

𝑓 (𝑥) 𝑑𝑥 =

∫3

6

𝑓 (𝑥) 𝑑𝑥 +

∫8

𝑓 (𝑥) 𝑑𝑥.

1

Given that ∫0 cos(𝑥2 ) 𝑑𝑥 = 0.98 and ∫0 cos(𝑥2 ) 𝑑𝑥 = 0.90, what are the values of the following integrals? (See Figure 5.63.) 1.25

(a)

∫1

1

cos(𝑥2 ) 𝑑𝑥

(b)

∫−1

−1

cos(𝑥2 ) 𝑑𝑥

(c)

∫1.25

cos(𝑥2 ) 𝑑𝑥

1 𝑥 −1 ✻

1 ✻ 1.25

−1.25 −1

Figure 5.63: Graph of 𝑓 (𝑥) = cos(𝑥2 )

Solution

(a) Since, by the additivity property, 1.25

∫0

1

cos(𝑥2 ) 𝑑𝑥 =

∫0

1.25

cos(𝑥2 ) 𝑑𝑥 +

∫1

we get 1.25

0.98 = 0.90 +

∫1

cos(𝑥2 ) 𝑑𝑥,

cos(𝑥2 ) 𝑑𝑥,

304

Chapter 5 KEY CONCEPT: THE DEFINITE INTEGRAL

so

1.25

∫1

cos(𝑥2 ) 𝑑𝑥 = 0.08.

(b) By the additivity property, we have 1

∫−1

cos(𝑥2 ) 𝑑𝑥 =

0

∫−1

cos(𝑥2 ) 𝑑𝑥 +

1

∫0

cos(𝑥2 ) 𝑑𝑥.

By the symmetry of cos(𝑥2 ) about the 𝑦-axis, 0

∫−1 So

cos(𝑥2 ) 𝑑𝑥 =

1

∫−1

1

∫0

cos(𝑥2 ) 𝑑𝑥 = 0.90.

cos(𝑥2 ) 𝑑𝑥 = 0.90 + 0.90 = 1.80.

(c) Using both properties in Theorem 5.2, we have −1

1.25 2

∫1.25

cos(𝑥 ) 𝑑𝑥 = −

2

∫−1

cos(𝑥 ) 𝑑𝑥 = −

(

0

∫−1

1.25 2

cos(𝑥 ) 𝑑𝑥 +

2

∫0

cos(𝑥 ) 𝑑𝑥

)

= −(0.90 + 0.98) = −1.88.

Theorem 5.3: Properties of Sums and Constant Multiples of the Integrand Let 𝑓 and 𝑔 be continuous functions and let 𝑐 be a constant. 𝑏

1.

𝑏

(𝑓 (𝑥) ± 𝑔(𝑥)) 𝑑𝑥 =

∫𝑎 𝑏

∫𝑎

𝑏

𝑓 (𝑥) 𝑑𝑥 ±

∫𝑎

𝑔(𝑥) 𝑑𝑥.

𝑏

𝑓 (𝑥) 𝑑𝑥. ∫𝑎 ∫𝑎 In words: 1. The integral of the sum (or difference) of two functions is the sum (or difference) of their integrals. 2.

𝑐𝑓 (𝑥) 𝑑𝑥 = 𝑐

2. The integral of a constant times a function is that constant times the integral of the function.

Why Do These Properties Hold? Both can be visualized by thinking of the definite integral as the limit of a sum of areas of rectangles. For property 1, suppose that 𝑓 and 𝑔 are positive on the interval [𝑎, 𝑏] so that the area under 𝑓 (𝑥) + 𝑔(𝑥) is approximated by the sum of the areas of rectangles like the one shaded in Figure 5.64. The area of this rectangle is (𝑓 (𝑥𝑖 ) + 𝑔(𝑥𝑖 ))Δ𝑥 = 𝑓 (𝑥𝑖 )Δ𝑥 + 𝑔(𝑥𝑖 )Δ𝑥. Since 𝑓 (𝑥𝑖 )Δ𝑥 is the area of a rectangle under the graph of 𝑓 , and 𝑔(𝑥𝑖 )Δ𝑥 is the area of a rectangle under the graph of 𝑔, the area under 𝑓 (𝑥) + 𝑔(𝑥) is the sum of the areas under 𝑓 (𝑥) and 𝑔(𝑥). For property 2, notice that multiplying a function by 𝑐 stretches or shrinks the graph in the vertical direction by a factor of 𝑐. Thus, it stretches or shrinks the height of each approximating rectangle by 𝑐, and hence multiplies the area by 𝑐.

5.4 THEOREMS ABOUT DEFINITE INTEGRALS



𝑓 (𝑥) + 𝑔(𝑥)

❄ ✻

𝑓 (𝑥)

305

𝑔(𝑥𝑖 )

𝑓 (𝑥𝑖 )

❄ ✲ ✛

𝑎

𝑥

Δ𝑥

𝑏

𝑏

𝑏

𝑏

Figure 5.64: Area = ∫𝑎 [𝑓 (𝑥) + 𝑔(𝑥)] 𝑑𝑥 = ∫𝑎 𝑓 (𝑥) 𝑑𝑥 + ∫𝑎 𝑔(𝑥) 𝑑𝑥

2

Example 2

Evaluate the definite integral

Solution

We can break this integral up as follows:

∫0

(1 + 3𝑥) 𝑑𝑥 exactly.

2

2

(1 + 3𝑥) 𝑑𝑥 =

∫0

∫0

2

1 𝑑𝑥 +

∫0

2

3𝑥 𝑑𝑥 =

∫0

2

1 𝑑𝑥 + 3

𝑥 𝑑𝑥.

∫0

From Figures 5.65 and 5.66 and the area interpretation of the integral, we see that 2

∫0

Area of

1 𝑑𝑥 =

rectangle

2

=2

and

𝑥 𝑑𝑥 =

∫0

Area of triangle

=

1 ⋅ 2 ⋅ 2 = 2. 2

Therefore, 2

∫0

2

(1 + 3𝑥) 𝑑𝑥 =

2

1 𝑑𝑥 + 3

∫0

∫0

𝑥 𝑑𝑥 = 2 + 3 ⋅ 2 = 8. 𝑦

𝑦 2

2 𝑦=1

𝑦=𝑥

1 2

∫0

2

1 𝑑𝑥 = 2 ∫0

𝑥

𝑥 𝑑𝑥 = 2 𝑥 2

2 2

2

Figure 5.65: Area representing ∫0 1 𝑑𝑥

Figure 5.66: Area representing ∫0 𝑥 𝑑𝑥

Area Between Curves Theorem 5.3 enables us to find the area of a region between curves. We have the following result: If the graph of 𝑓 (𝑥) lies above the graph of 𝑔(𝑥) for 𝑎 ≤ 𝑥 ≤ 𝑏, then Area between 𝑓 and 𝑔 for 𝑎 ≤ 𝑥 ≤ 𝑏

𝑏

=

∫𝑎

(𝑓 (𝑥) − 𝑔(𝑥)) 𝑑𝑥.

306

Chapter 5 KEY CONCEPT: THE DEFINITE INTEGRAL

Example 3

Find the area of the shaded region in Figure 5.67. 𝑔(𝑥) = 𝑥2 − 4𝑥 + 5

𝑓 (𝑥) = −𝑥2 + 4𝑥 − 1 𝑥 Figure 5.67: Area between two parabolas

Solution

The curves cross where 𝑥2 − 4𝑥 + 5 = −𝑥2 + 4𝑥 − 1 2𝑥2 − 8𝑥 + 6 = 0 2(𝑥 − 1)(𝑥 − 3) = 0 𝑥 = 1, 3. Since 𝑓 (𝑥) = −𝑥2 + 4𝑥 − 1 is above 𝑔(𝑥) = 𝑥2 − 4𝑥 + 5 for 𝑥 between 1 and 3, we find the shaded area by subtraction: 3

Area =

∫1

3

𝑓 (𝑥) 𝑑𝑥 −

3

𝑔(𝑥) 𝑑𝑥 =

∫1

(𝑓 (𝑥) − 𝑔(𝑥)) 𝑑𝑥

∫1 3

=

((−𝑥2 + 4𝑥 − 1) − (𝑥2 − 4𝑥 + 5)) 𝑑𝑥

∫1 3

=

∫1

(−2𝑥2 + 8𝑥 − 6) 𝑑𝑥 = 2.667.

Using Symmetry to Evaluate Integrals Symmetry can be useful in evaluating definite integrals. An even function is symmetric about the 𝑦-axis. An odd function is symmetric about the origin. Figures 5.68 and 5.69 suggest the following results for continuous functions: 𝑎

If 𝑓 is even, then

∫−𝑎

𝑎

𝑓 (𝑥) 𝑑𝑥 = 2

∫0

𝑎

𝑓 (𝑥) 𝑑𝑥.

If 𝑔 is odd, then

∫−𝑎

𝑔(𝑥) 𝑑𝑥 = 0.

𝑔(𝑥)

𝑓 (𝑥) −𝑎 −𝑎

𝑎

𝑥

Figure 5.68: For an even function, 𝑎 𝑎 ∫−𝑎 𝑓 (𝑥) 𝑑𝑥 = 2 ∫0 𝑓 (𝑥) 𝑑𝑥

𝑥 𝑎

Figure 5.69: For an odd function, 𝑎 ∫−𝑎 𝑔(𝑥) 𝑑𝑥 = 0

307

5.4 THEOREMS ABOUT DEFINITE INTEGRALS

𝜋 𝜋

𝜋

Example 4

Given that ∫0 sin 𝑡 𝑑𝑡 = 2, find (a)

Solution

Graphs of sin 𝑡 and | sin 𝑡| are in Figures 5.70 and 5.71. (a) Since sin 𝑡 is an odd function

∫−𝜋

sin 𝑡 𝑑𝑡

(b)

∫−𝜋

| sin 𝑡| 𝑑𝑡

𝜋

∫−𝜋 (b) Since | sin 𝑡| is an even function

sin 𝑡 𝑑𝑡 = 0.

𝜋

∫−𝜋

𝜋

| sin 𝑡| 𝑑𝑡 = 2

∫0

| sin 𝑡| 𝑑𝑡 = 4. 𝑦

𝑦 1

1 −𝜋

𝑡

𝑡

−𝜋

𝜋

𝜋

−1

−1

Figure 5.70

Figure 5.71

Comparing Integrals Suppose we have constants 𝑚 and 𝑀 such that 𝑚 ≤ 𝑓 (𝑥) ≤ 𝑀 for 𝑎 ≤ 𝑥 ≤ 𝑏. We say 𝑓 is bounded above by 𝑀 and bounded below by 𝑚. Then the graph of 𝑓 lies between the horizontal lines 𝑦 = 𝑚 and 𝑦 = 𝑀. So the definite integral lies between 𝑚(𝑏 − 𝑎) and 𝑀(𝑏 − 𝑎). See Figure 5.72. Suppose 𝑓 (𝑥) ≤ 𝑔(𝑥) for 𝑎 ≤ 𝑥 ≤ 𝑏, as in Figure 5.73. Then the definite integral of 𝑓 is less than or equal to the definite integral of 𝑔. This leads us to the following results: 𝑦 𝑀

𝑔(𝑥)

𝑓 (𝑥)

𝑓 (𝑥)

𝑚

𝑏

Total shaded area

=

Dark shaded area

=

𝑏

𝑥 𝑎

∫𝑎

𝑓 (𝑥) 𝑑𝑥

𝑥

𝑏 Figure 5.72: The area under the graph of 𝑓 lies between the areas of the rectangles

𝑎

𝑏 𝑏

𝑏

Figure 5.73: If 𝑓 (𝑥) ≤ 𝑔(𝑥) then ∫𝑎 𝑓 (𝑥) 𝑑𝑥 ≤ ∫𝑎 𝑔(𝑥) 𝑑𝑥

Theorem 5.4: Comparison of Definite Integrals Let 𝑓 and 𝑔 be continuous functions. 𝑏

1. If 𝑚 ≤ 𝑓 (𝑥) ≤ 𝑀 for 𝑎 ≤ 𝑥 ≤ 𝑏, then

𝑚(𝑏 − 𝑎) ≤

∫𝑎

𝑓 (𝑥) 𝑑𝑥 ≤ 𝑀(𝑏 − 𝑎).

∫𝑎

𝑏

2. If 𝑓 (𝑥) ≤ 𝑔(𝑥) for 𝑎 ≤ 𝑥 ≤ 𝑏, then

𝑔(𝑥) 𝑑𝑥

∫𝑎

𝑏

𝑓 (𝑥) 𝑑𝑥 ≤

∫𝑎

𝑔(𝑥) 𝑑𝑥.

308

Chapter 5 KEY CONCEPT: THE DEFINITE INTEGRAL

√ 𝜋

√ 𝜋.

sin(𝑥2 ) 𝑑𝑥 ≤

Example 5

Explain why

Solution

Since sin (𝑥2 ) ≤ 1 for all 𝑥 (see Figure 5.74), part 2 of Theorem 5.4 gives

∫0

∫0



𝜋 2

sin(𝑥 ) 𝑑𝑥 ≤

∫0



𝜋

1 𝑑𝑥 =

√ 𝜋.

𝑦 1

𝑥 √

𝜋

𝑦 = sin(𝑥2 )

−1

√ 𝜋

Figure 5.74: Graph showing that ∫0

sin(𝑥2 ) 𝑑𝑥
0. ∫ 𝑥 If 𝑥 < 0, then ln 𝑥 is not defined, so it can’t be an antiderivative of 1∕𝑥. In this case, we can try ln(−𝑥): 𝑑 1 1 ln(−𝑥) = (−1) = 𝑑𝑥 −𝑥 𝑥 so 1 𝑑𝑥 = ln(−𝑥) + 𝐶, for 𝑥 < 0. ∫ 𝑥 This means ln 𝑥 is an antiderivative of 1∕𝑥 if 𝑥 > 0, and ln(−𝑥) is an antiderivative of 1∕𝑥 if 𝑥 < 0. Since |𝑥| = 𝑥 when 𝑥 > 0 and |𝑥| = −𝑥 when 𝑥 < 0, we can collapse these two formulas into: An antiderivative of

Therefore

1 is ln |𝑥|. 𝑥

1 𝑑𝑥 = ln |𝑥| + 𝐶. ∫ 𝑥 Since the exponential function is its own derivative, it is also its own antiderivative; thus



𝑒𝑥 𝑑𝑥 = 𝑒𝑥 + 𝐶.

Also, antiderivatives of the sine and cosine are easy to guess. Since 𝑑 sin 𝑥 = cos 𝑥 𝑑𝑥

and

𝑑 cos 𝑥 = − sin 𝑥, 𝑑𝑥

we get



cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝐶

and



sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝐶.

(3𝑥 + 𝑥2 ) 𝑑𝑥.

Example 1

Find

Solution

We know that 𝑥2 ∕2 is an antiderivative of 𝑥 and that 𝑥3 ∕3 is an antiderivative of 𝑥2 , so we expect ( 2) 𝑥3 𝑥 2 + + 𝐶. (3𝑥 + 𝑥 ) 𝑑𝑥 = 3 ∫ 2 3



You should always check your antiderivatives by differentiation—it’s easy to do. Here ( ) 𝑑 3 2 𝑥3 3 3𝑥2 𝑥 + + 𝐶 = ⋅ 2𝑥 + = 3𝑥 + 𝑥2 . 𝑑𝑥 2 3 2 3

6.2 CONSTRUCTING ANTIDERIVATIVES ANALYTICALLY

325

The preceding example illustrates that the sum and constant multiplication rules of differentiation work in reverse:

Theorem 6.1: Properties of Antiderivatives: Sums and Constant Multiples In indefinite integral notation, 1.



(𝑓 (𝑥) ± 𝑔(𝑥)) 𝑑𝑥 =



𝑓 (𝑥) 𝑑𝑥 ±



𝑔(𝑥) 𝑑𝑥

𝑐𝑓 (𝑥) 𝑑𝑥 = 𝑐 𝑓 (𝑥) 𝑑𝑥. ∫ ∫ In words, 1. An antiderivative of the sum (or difference) of two functions is the sum (or difference) of their antiderivatives. 2. An antiderivative of a constant times a function is the constant times an antiderivative of the function. 2.

These properties are analogous to the properties for definite integrals given on page 304 in Section 5.4, even though definite integrals are numbers and antiderivatives are functions. Example 2

Find

Solution

We break the antiderivative into two terms:



(sin 𝑥 + 3 cos 𝑥) 𝑑𝑥.

(sin 𝑥 + 3 cos 𝑥) 𝑑𝑥 =





sin 𝑥 𝑑𝑥 + 3



cos 𝑥 𝑑𝑥 = − cos 𝑥 + 3 sin 𝑥 + 𝐶.

Check by differentiating: 𝑑 (− cos 𝑥 + 3 sin 𝑥 + 𝐶) = sin 𝑥 + 3 cos 𝑥. 𝑑𝑥

Using Antiderivatives to Compute Definite Integrals As we saw in Section 5.3, the Fundamental Theorem of Calculus gives us a way of calculating definite 𝑏 integrals. Denoting 𝐹 (𝑏) − 𝐹 (𝑎) by 𝐹 (𝑥)||𝑎 , the theorem says that if 𝐹 ′ = 𝑓 and 𝑓 is continuous, then 𝑏 |𝑏 𝑓 (𝑥) 𝑑𝑥 = 𝐹 (𝑥)|| = 𝐹 (𝑏) − 𝐹 (𝑎). ∫𝑎 |𝑎 𝑏

To find ∫𝑎 𝑓 (𝑥) 𝑑𝑥, we first find 𝐹 , and then calculate 𝐹 (𝑏) − 𝐹 (𝑎). This method of computing definite integrals gives an exact answer. However, the method only works in situations where we can find the antiderivative 𝐹 (𝑥). This is not always easy; for example, none of the functions we have encountered so far is an antiderivative of sin(𝑥2 ). 2

3𝑥2 𝑑𝑥 using the Fundamental Theorem.

Example 3

Compute

Solution

Since 𝐹 (𝑥) = 𝑥3 is an antiderivative of 𝑓 (𝑥) = 3𝑥2 ,

∫1

2

∫1

|2 3𝑥2 𝑑𝑥 = 𝐹 (𝑥)|| = 𝐹 (2) − 𝐹 (1), |1

326

Chapter 6 CONSTRUCTING ANTIDERIVATIVES

gives 2

∫1

|2 3𝑥2 𝑑𝑥 = 𝑥3 || = 23 − 13 = 7. |1

Notice in this example we used the antiderivative 𝑥3 , but 𝑥3 + 𝐶 works just as well because the constant 𝐶 cancels out: 2 |2 3𝑥2 𝑑𝑥 = (𝑥3 + 𝐶)|| = (23 + 𝐶) − (13 + 𝐶) = 7. ∫1 |1 𝜋∕4

1 𝑑𝜃 exactly. cos2 𝜃

Example 4

Compute

Solution

We use the Fundamental Theorem. Since 𝐹 (𝜃) = tan 𝜃 is an antiderivative of 𝑓 (𝜃) = 1∕ cos2 𝜃, we get 𝜋∕4 ( ) |𝜋∕4 1 𝜋 | − tan(0) = 1. 𝑑𝜃 = tan 𝜃 = tan | 2 ∫0 4 cos 𝜃 |0

∫0

Exercises and Problems for Section 6.2 Online Resource: Additional Problems for Section 6.2 EXERCISES

1. If 𝑝′ (𝑥) = 𝑞(𝑥), write a statement involving an integral sign giving the relationship between 𝑝(𝑥) and 𝑞(𝑥). 2. If 𝑢′ (𝑥) = 𝑣(𝑥), write a statement involving an integral sign giving the relationship between 𝑢(𝑥) and 𝑣(𝑥). 𝑥∕2

3. Which of (I)–(V) are antiderivatives of 𝑓 (𝑥) = 𝑒 𝑥∕2

I. 𝑒 IV. 2𝑒𝑥∕2 + 1

?

II. 2𝑒 2 V. 𝑒𝑥 ∕4

III. 2𝑒

II. −1∕𝑥2 V. ln(𝑥 + 1)

III. ln 𝑥 + ln 3

5. Which of (I)–(V) are antiderivatives of 𝑓 (𝑥) = 2 sin 𝑥 cos 𝑥? I. −2 sin 𝑥 cos 𝑥 III. sin2 𝑥 V. 2 sin2 𝑥 + cos2 𝑥

II. 2 cos2 𝑥 − 2 sin2 𝑥 IV. − cos2 𝑥

8. 𝑓 (𝑥) = 𝑥 √ 10. 𝑔(𝑧) = 𝑧

1 12. 𝑟(𝑡) = 2 𝑡 1 14. 𝑔(𝑧) = 3 𝑧

19. 𝑓 (𝑥) = 5𝑥 −

20. 𝑓 (𝑡) =

𝑡2 + 1 𝑡

21. 𝑝(𝑡) = 𝑡3 −

√ 𝑥

𝑡2 −𝑡 2

9. 𝑔(𝑡) = 𝑡2 + 𝑡 1 11. ℎ(𝑧) = 𝑧

22. 𝑓 (𝑡) = 6𝑡 24. 𝑓 (𝑥) = 𝑥2 − 4𝑥 + 7

23. ℎ(𝑥) = 𝑥3 − 𝑥 √ 25. 𝑔(𝑡) = 𝑡

26. 𝑟(𝑡) = 𝑡3 + 5𝑡 − 1

27. 𝑓 (𝑧) = 𝑧 + 𝑒𝑧

28. 𝑔(𝑥) = sin 𝑥 + cos 𝑥

29. ℎ(𝑥) = 4𝑥3 − 7

5 𝑥3 1 32. 𝑝(𝑡) = √ 𝑡

31. 𝑝(𝑡) = 2 + sin 𝑡 33. ℎ(𝑡) =

7 cos2 𝑡

34. 𝑓 (𝑥) = 3

35. 𝑓 (𝑥) = 2𝑥

36. 𝑓 (𝑥) = −7𝑥

37. 𝑓 (𝑥) = 2 + 4𝑥 + 5𝑥2

38. 𝑓 (𝑥) =

13. ℎ(𝑡) = cos 𝑡 15. 𝑞(𝑦) = 𝑦4 +

In Exercises 22–33, find the general antiderivative.

In Exercises 34–41, find an antiderivative 𝐹 (𝑥) with 𝐹 ′ (𝑥) = 𝑓 (𝑥) and 𝐹 (0) = 0. Is there only one possible solution?

7. 𝑓 (𝑡) = 5𝑡 2

18. 𝑓 (𝑡) = 2𝑡2 + 3𝑡3 + 4𝑡4

30. 𝑔(𝑥) =

In Exercises 6–21, find an antiderivative. 6. 𝑓 (𝑥) = 5

17. 𝑔(𝑡) = sin 𝑡

(1+𝑥)∕2

𝑥∕2

4. Which of (I)–(V) are antiderivatives of 𝑓 (𝑥) = 1∕𝑥? I. ln 𝑥 IV. ln(2𝑥)

16. 𝑓 (𝑧) = 𝑒𝑧

1 𝑦

1

𝑥 4 √ 40. 𝑓 (𝑥) = 𝑥

39. 𝑓 (𝑥) = 𝑥2 41. 𝑓 (𝑥) = sin 𝑥

6.2 CONSTRUCTING ANTIDERIVATIVES ANALYTICALLY 1

In Exercises 42–55, find the indefinite integrals. 42. 44. 46. 48.

50.

52.

54.

∫ ∫ ∫ ∫







(5𝑥 + 7) 𝑑𝑥

43.

(2 + cos 𝑡) 𝑑𝑡

45.

(3𝑒𝑥 + 2 sin 𝑥) 𝑑𝑥

47.

(

√ ) 5𝑥2 + 2 𝑥 𝑑𝑥

8 √ 𝑑𝑥 𝑥

51.

(𝑒𝑥 + 5) 𝑑𝑥 ( √

𝑥3 −

49.

2 𝑥

53. )

𝑑𝑥

55.

(



4𝑡 +

1 𝑡

62.

)

2

64.

7𝑒𝑥 𝑑𝑥



(4𝑒𝑥 − 3 sin 𝑥) 𝑑𝑥



(𝑥 + 3)2 𝑑𝑥



66. (







3 𝑡



2 𝑡2

)

𝑑𝑡

67.

𝑡3 (𝑡2 + 1) 𝑑𝑡

68.

(

69.

𝑥+1 𝑥

)

𝑑𝑥

In Exercises 56–65, evaluate the definite integrals exactly [as in ln(3𝜋)], using the Fundamental Theorem, and numerically [ln(3𝜋) ≈ 2.243]:

71. 72.

3

3

(𝑥2 + 4𝑥 + 3) 𝑑𝑥

∫0

57.

∫1

sin 𝑥 𝑑𝑥

∫0

59.

∫0

2𝑒𝑥 𝑑𝑥

∫0

2

60.

1 𝑑𝑡 𝑡

𝑥3 3

)

+ 2𝑥

1 + 𝑦2

∫1

𝑑𝑦

𝑦 𝜋∕4

65.

𝑑𝑥

∫0

(sin 𝑡 + cos 𝑡) 𝑑𝑡

73.

∫ ∫ ∫ ∫ ∫ ∫

(8𝑥2 − 4𝑥 + 3) 𝑑𝑥 =

) ( √ 5 5 3 𝑥 + 2 𝑑𝑥 = 2𝑥3∕2 + + 𝐶 𝑥 𝑥 √ ( )3∕2 𝑥2 + 1 𝑑𝑥 = 𝑥2 + 1 +𝐶 𝑥−2 𝑑𝑥 = −

1 +𝐶 𝑥

𝑥−1 𝑑𝑥 = −1𝑥0 + 𝐶 = −1 + 𝐶 𝑒2𝑥 𝑑𝑥 = 2𝑒2𝑥 + 𝐶 2

∫ ∫

8 3 𝑥 − 2𝑥2 + 3𝑥 + 𝐶 3

2

𝑒𝑥 𝑑𝑥 = 2𝑥 ⋅ 𝑒𝑥 + 𝐶 3 cos 𝑥 𝑑𝑥 = 3 sin 𝑥 + 𝐶

1

𝜋∕4

58.

∫0

(

63.

In Exercises 66–75, decide if the statement is True or False by differentiating the right-hand side.

70.

56.

∫0

𝑑𝑡

2

sin 𝜃 𝑑𝜃

327

74.



5 sin

( ) ( ) 𝑥 𝑥 𝑑𝑥 = 5 cos +𝐶 5 5

5

3𝑒𝑥 𝑑𝑥

61.

∫2

(𝑥3 − 𝜋𝑥2 ) 𝑑𝑥

75.



𝑥 cos 𝑥 𝑑𝑥 =

𝑥2 sin 𝑥 + 𝐶 2

PROBLEMS 76. Use the Fundamental Theorem to find the area under 𝑓 (𝑥) = 𝑥2 between 𝑥 = 0 and 𝑥 = 3. 77. Find the exact area of the region bounded by the 𝑥-axis and the graph of 𝑦 = 𝑥3 − 𝑥. 78. Calculate the exact area above the graph of 𝑦 = ( )2 1 3 𝑥 and below the graph of 𝑦 = cos 𝑥. The curves 2 𝜋 intersect at 𝑥 = ±𝜋∕3. 79. Find the exact area of the shaded region in Figure 6.25 between 𝑦 = 3𝑥2 − 3 and the 𝑥-axis.

80. (a) Find the exact area between 𝑓 (𝑥) = 𝑥3 −7𝑥2 +10𝑥, the 𝑥-axis, 𝑥 = 0, and 𝑥 = 5. 5 (b) Find ∫0 (𝑥3 − 7𝑥2 + 10𝑥) 𝑑𝑥 exactly and interpret this integral in terms of areas. 81. Find the exact area between the curve 𝑦 = 𝑒𝑥 − 2 and the 𝑥-axis for 0 ≤ 𝑥 ≤ 2. 82. Find the exact area between the curves 𝑦 = 𝑥2 and 𝑦 = 2 − 𝑥2 . 83. Find the exact area between the 𝑥-axis and the graph of 𝑓 (𝑥) = (𝑥 − 1)(𝑥 − 2)(𝑥 − 3).

𝑦 = 3𝑥2 − 3

𝑥 1

Figure 6.25

3

√ 84. The area under 1∕ 𝑥 on the interval 1 ≤ 𝑥 ≤ 𝑏 is equal to 6. Find the value of 𝑏 using the Fundamental Theorem. 85. Use the Fundamental Theorem to find the value of 𝑏 if the area under the graph of 𝑓 (𝑥) = 8𝑥 between 𝑥 = 1 and 𝑥 = 𝑏 is equal to 192. Assume 𝑏 > 1.

328

Chapter 6 CONSTRUCTING ANTIDERIVATIVES

86. Find the exact positive value of 𝑐 which makes the area under the graph of 𝑦 = 𝑐(1 − 𝑥2 ) and above the 𝑥-axis equal to 1.

94. A helicopter rotor slows down at a constant rate from 350 revs/min to 260 revs/min in 1.5 minutes. (a) Find the angular acceleration (i.e. change in rev/min) during this time interval. What are the units of this acceleration? (b) Assuming the angular acceleration remains constant, how long does it take for the rotor to stop? (Measure time from the moment when speed was 350 revs/min.) (c) How many revolutions does the rotor make between the time the angular speed was 350 revs/min and stopping?

87. Sketch the parabola 𝑦 = 𝑥(𝑥 − 𝜋) and the curve 𝑦 = sin 𝑥, showing their points of intersection. Find the exact area between the two graphs. √ 88. Find the exact average value of 𝑓 (𝑥) = 𝑥 on the interval 0 √ ≤ 𝑥 ≤ 9. Illustrate your answer on a graph of 𝑓 (𝑥) = 𝑥. 89. (a) What is the average value of 𝑓 (𝑡) = sin 𝑡 over 0 ≤ 𝑡 ≤ 2𝜋? Why is this a reasonable answer? (b) Find the average of 𝑓 (𝑡) = sin 𝑡 over 0 ≤ 𝑡 ≤ 𝜋.

90. Let ∫ 𝑞(𝑥) 𝑑𝑥 = 𝑄(𝑥)+𝐶 where 𝑄(3) = 12. Given that 8 ∫3 𝑞(𝑥) 𝑑𝑥 = 5, find 𝑄(8). 91. Water is pumped into a cylindrical tank, standing vertically, at a decreasing rate given at time 𝑡 minutes by 3

𝑟(𝑡) = 120 − 6𝑡 ft ∕min

for 0 ≤ 𝑡 ≤ 10.

95. Use the fact that (𝑥𝑥 )′ = 𝑥𝑥 (1+ln 𝑥) to evaluate exactly: 3

∫1

96. Assuming that ∫ 𝑔(𝑥) 𝑑𝑥 = 𝐺(𝑥) + 𝐶, where 𝐺(4) = 9, 𝐺(6) = 4, and 𝐺(9) = 6, evaluate the definite integral: 4

The tank has radius 5 ft and is empty when 𝑡 = 0. Find the depth of water in the tank at 𝑡 = 4.

(a)

92. A car moves along a straight line with velocity, in feet/second, given by

(c)

𝑣(𝑡) = 6 − 2𝑡

for 𝑡 ≥ 0.

(a) Describe the car’s motion in words. (When is it moving forward, backward, and so on?) (b) The car’s position is measured from its starting point. When is it farthest forward? Backward? (c) Find 𝑠, the car’s position measured from its starting point, as a function of time. 93. In drilling an oil well, the total cost, 𝐶, consists of fixed costs (independent of the depth of the well) and marginal costs, which depend on depth; drilling becomes more expensive, per meter, deeper into the earth. Suppose the fixed costs are 1,000,000 riyals (the riyal is the unit of currency of Saudi Arabia), and the marginal costs are 𝐶 ′ (𝑥) = 4000 + 10𝑥 riyals/meter, where 𝑥 is the depth in meters. Find the total cost of drilling a well 𝑥 meters deep.

𝑥𝑥 (1 + ln 𝑥) 𝑑𝑥.

9

(b)

𝑔(𝑥) 𝑑𝑥

∫6

∫6

7𝑔(𝑥) 𝑑𝑥

9

∫4

(𝑔(𝑥) + 3) 𝑑𝑥

For Problems 97–99, let ∫ 𝑔(𝑥) 𝑑𝑥 = 𝐺(𝑥) + 𝐶. Which of (I)–(III), if any, is equal to the given integral? 97.



𝑔(2𝑥) 𝑑𝑥

I. 0.5𝐺(0.5𝑥) + 𝐶

II. 0.5𝐺(2𝑥) + 𝐶

III. 2𝐺(0.5𝑥) + 𝐶 98.



cos (𝐺(𝑥)) 𝑔(𝑥) 𝑑𝑥

I. sin (𝐺(𝑥)) 𝑔(𝑥) + 𝐶

II. sin (𝐺(𝑥)) 𝐺(𝑥) + 𝐶

III. sin (𝐺(𝑥)) + 𝐶 99.



𝑥𝑔(𝑥) 𝑑𝑥

( ) I. 𝐺 𝑥2 + 𝐶

II.

𝑥𝐺(𝑥) + 𝐶

III.

1 2 𝑥 𝐺(𝑥)+𝐶 2

Strengthen Your Understanding In Problems 100–101, explain what is wrong with the statement. 100.



𝑥3 + 𝑥 3𝑥2 + 1 𝑑𝑥 = +𝐶 2𝑥 𝑥2

101. For all 𝑛,



𝑥𝑛 𝑑𝑥 =

𝑥𝑛+1 + 𝐶. 𝑛+1

In Problems 102–103, give an example of: 102. Two different functions 𝐹 (𝑥) and 𝐺(𝑥) that have the

same derivative. 103. A function 𝑓 (𝑥) whose antiderivative 𝐹 (𝑥) has a graph which is a line with negative slope. Are the statements in Problems 104–112 true or false? Give an explanation for your answer. √ 104. An antiderivative of 3 𝑥 + 1 is 2(𝑥 + 1)3∕2 .

105. An antiderivative of 3𝑥2 is 𝑥3 + 𝜋.

106. An antiderivative of 1∕𝑥 is ln |𝑥| + ln 2.

6.3 DIFFERENTIAL EQUATIONS AND MOTION 2

2

107. An antiderivative of 𝑒−𝑥 is −𝑒−𝑥 ∕2𝑥. 108. ∫ 𝑓 (𝑥) 𝑑𝑥 = (1∕𝑥) ∫ 𝑥𝑓 (𝑥) 𝑑𝑥. 109. If 𝐹 (𝑥) is an antiderivative of 𝑓 (𝑥) and 𝐺(𝑥) = 𝐹 (𝑥)+2, then 𝐺(𝑥) is an antiderivative of 𝑓 (𝑥). 110. If 𝐹 (𝑥) and 𝐺(𝑥) are two antiderivatives of 𝑓 (𝑥) for −∞ < 𝑥 < ∞ and 𝐹 (5) > 𝐺(5), then 𝐹 (10) > 𝐺(10).

6.3

329

111. If 𝐹 (𝑥) is an antiderivative of 𝑓 (𝑥) and 𝐺(𝑥) is an antiderivative of 𝑔(𝑥), then 𝐹 (𝑥)⋅𝐺(𝑥) is an antiderivative of 𝑓 (𝑥) ⋅ 𝑔(𝑥). 112. If 𝐹 (𝑥) and 𝐺(𝑥) are both antiderivatives of 𝑓 (𝑥) on an interval, then 𝐹 (𝑥) − 𝐺(𝑥) is a constant function.

DIFFERENTIAL EQUATIONS AND MOTION An equation of the form 𝑑𝑦 = 𝑓 (𝑥) 𝑑𝑥 is an example of a differential equation. Finding the general solution to the differential equation means finding the general antiderivative 𝑦 = 𝐹 (𝑥) + 𝐶 with 𝐹 ′ (𝑥) = 𝑓 (𝑥). Chapter 11 gives more details.

Example 1

Find and graph the general solution of the differential equation 𝑑𝑦 = 2𝑥. 𝑑𝑥

Solution

We are asking for a function whose derivative is 2𝑥. One antiderivative of 2𝑥 is 𝑦 = 𝑥2 . The general solution is therefore 𝑦 = 𝑥2 + 𝐶, where 𝐶 is any constant. Figure 6.26 shows several curves in this family. 𝑦 𝐶=4

20

𝐶=2 𝐶=0

15

𝐶 = −4

10 5

𝑥 −4

−2

2

4

−4 Figure 6.26: Solution curves of 𝑑𝑦∕𝑑𝑥 = 2𝑥

𝒅𝒚 = 𝒇 (𝒙) ? 𝒅𝒙 Picking one antiderivative is equivalent to selecting a value of 𝐶. To do this, we need an extra piece of information, usually that the solution curve passes through a particular point (𝑥0 , 𝑦0 ). The differential equation plus the extra condition 𝑑𝑦 = 𝑓 (𝑥), 𝑦(𝑥0 ) = 𝑦0 𝑑𝑥 is called an initial value problem. (The initial condition 𝑦(𝑥0 ) = 𝑦0 is shorthand for 𝑦 = 𝑦0 when 𝑥 = 𝑥0 .) An initial value problem usually has a unique solution, called the particular solution.

How Can We Pick One Solution to the Differential Equation

330

Chapter 6 CONSTRUCTING ANTIDERIVATIVES

Example 2

Find the solution of the initial value problem 𝑑𝑦 = 2𝑥, 𝑑𝑥

Solution

𝑦(3) = 5.

We have already seen that the general solution to the differential equation is 𝑦 = 𝑥2 + 𝐶. The initial condition allows us to determine the constant 𝐶. Substituting 𝑦(3) = 5 gives 5 = 𝑦(3) = 32 + 𝐶, so 𝐶 is given by 𝐶 = −4. Thus, the (unique) solution is 𝑦 = 𝑥2 − 4 Figure 6.26 shows this particular solution, marked 𝐶 = −4.

Equations of Motion We now use differential equations to analyze the motion of an object falling freely under the influence of gravity. It has been known since Galileo’s time that an object moving under the influence of gravity (ignoring air resistance) has constant acceleration, 𝑔. In the most frequently used units, its value is approximately 𝑔 = 9.8 m∕sec2 , or 𝑔 = 32 f t∕sec2 . Thus, if 𝑣 is the upward velocity and 𝑡 is the time, we have the differential equation 𝑑𝑣 = −𝑔. 𝑑𝑡 The negative sign represents the fact that positive velocity is measured upward, whereas gravity acts downward. Example 3

A stone is dropped from a 100-foot-high building. Find, as functions of time, its position and velocity. When does it hit the ground, and how fast is it going at that time?

Solution

Suppose 𝑡 is measured in seconds from the time when the stone was dropped. If we measure distance, 𝑠, in feet above the ground, then the velocity, 𝑣, is in ft/sec upward, and the acceleration due to gravity is 32 ft/sec2 downward, so we have the differential equation 𝑑𝑣 = −32. 𝑑𝑡 From what we know about antiderivatives, the general solution is 𝑣 = −32𝑡 + 𝐶, where 𝐶 is some constant. Since 𝑣 = 𝐶 when 𝑡 = 0, the constant 𝐶 represents the initial velocity, 𝑣0 . The fact that the stone is dropped rather than thrown tells us that the initial velocity is zero, so the initial condition is 𝑣0 = 0. Substituting gives 0 = −32(0) + 𝐶

so

Thus, 𝑣 = −32𝑡. But now we can write a second differential equation: 𝑣=

𝑑𝑠 = −32𝑡. 𝑑𝑡

𝐶 = 0.

6.3 DIFFERENTIAL EQUATIONS AND MOTION

331

The general solution is 𝑠 = −16𝑡2 + 𝐾, where 𝐾 is another constant. Since the stone starts at the top of the building, we have the initial condition 𝑠 = 100 when 𝑡 = 0. Substituting gives 100 = −16(02) + 𝐾, so 𝐾 = 100, and therefore 𝑠 = −16𝑡2 + 100. Thus, we have found both 𝑣 and 𝑠 as functions of time. The stone hits the ground when 𝑠 = 0, so we must solve 0 = −16𝑡2 + 100 giving 𝑡2 = 100∕16 or 𝑡 = ±10∕4 = ±2.5 sec. Since 𝑡 must be positive, 𝑡 = 2.5 sec. At that time, 𝑣 = −32(2.5) = −80 ft/sec. (The velocity is negative because we are considering moving up as positive and down as negative.) After the stone hits the ground, the differential equation no longer applies.

Example 4

An object is thrown vertically upward with a speed of 10 m/sec from a height of 2 meters above the ground. Find the highest point it reaches and the time when it hits the ground.

Solution

We must find the position as a function of time. In this example, the velocity is in m/sec, so we use 𝑔 = 9.8 m/sec2 . Measuring distance in meters upward from the ground, we have the differential equation 𝑑𝑣 = −9.8. 𝑑𝑡 As before, 𝑣 is a function whose derivative is constant, so 𝑣 = −9.8𝑡 + 𝐶. Since the initial velocity is 10 m/sec upward, we know that 𝑣 = 10 when 𝑡 = 0. Substituting gives 10 = −9.8(0) + 𝐶

so

𝐶 = 10.

Thus, 𝑣 = −9.8𝑡 + 10. To find 𝑠, we use 𝑣=

𝑑𝑠 = −9.8𝑡 + 10 𝑑𝑡

and look for a function that has −9.8𝑡 + 10 as its derivative. The general solution is 𝑠 = −4.9𝑡2 + 10𝑡 + 𝐾, where 𝐾 is any constant. To find 𝐾, we use the fact that the object starts at a height of 2 meters above the ground, so 𝑠 = 2 when 𝑡 = 0. Substituting gives 2 = −4.9(0)2 + 10(0) + 𝐾,

so

and therefore 𝑠 = −4.9𝑡2 + 10𝑡 + 2.

𝐾 = 2,

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Chapter 6 CONSTRUCTING ANTIDERIVATIVES

The object reaches its highest point when the velocity is 0, so at that time 𝑣 = −9.8𝑡 + 10 = 0. This occurs when 𝑡=

10 ≈ 1.02 sec. 9.8

When 𝑡 = 1.02 seconds, 𝑠 = −4.9(1.02)2 + 10(1.02) + 2 ≈ 7.10 meters. So the maximum height reached is 7.10 meters. The object reaches the ground when 𝑠 = 0: 0 = −4.9𝑡2 + 10𝑡 + 2. Solving this using the quadratic formula gives 𝑡 ≈ −0.18 and 𝑡 ≈ 2.22 sec. Since the time at which the object hits the ground must be positive, 𝑡 ≈ 2.22 seconds.

History of the Equations of Motion The problem of a body moving freely under the influence of gravity near the surface of the earth intrigued mathematicians and philosophers from Greek times onward and was finally solved by Galileo and Newton. The question to be answered was: How do the velocity and the position of the body vary with time? We define 𝑠 to be the position, or height, of the body above a fixed point (often ground level), 𝑣 is the velocity of the body measured upward, and 𝑎 is the acceleration. The velocity and position at time 𝑡 = 0 are represented by 𝑣0 and 𝑠0 respectively. We assume that the acceleration of the body is a constant, −𝑔 (the negative sign means that the acceleration is downward), so 𝑑𝑣 = 𝑎 = −𝑔. 𝑑𝑡 Problem 34 asks you to show that the motion satisfies 𝑣 = −𝑔𝑡 + 𝑣0 , 𝑔 𝑠 = − 𝑡 2 + 𝑣0 𝑡 + 𝑠 0 . 2 Our derivation of the formulas for the velocity and the position of the body hides an almost 2000year struggle to understand the mechanics of falling bodies, from Aristotle’s Physics to Galileo’s Dialogues Concerning Two New Sciences. Though it is an oversimplification of his ideas, we can say that Aristotle’s conception of motion was primarily in terms of change of position. This seems entirely reasonable; it is what we commonly observe, and this view dominated discussions of motion for centuries. But it misses a subtlety that was brought to light by Descartes, Galileo, and, with a different emphasis, by Newton. That subtlety is now usually referred to as the principle of inertia. This principle holds that a body traveling undisturbed at constant velocity in a straight line will continue in this motion indefinitely. Stated another way, it says that one cannot distinguish in any absolute sense (that is, by performing an experiment) between being at rest and moving with constant velocity in a straight line. If you are reading this book in a closed room and have no external reference points, there is no experiment that will tell you, one way or the other, whether you are at rest or whether you, the room, and everything in it are moving with constant velocity in a straight line. Therefore, as Newton saw, an understanding of motion should be based on change of velocity rather than change of position. Since acceleration is the rate of change of velocity, it is acceleration that must play a central role in the description of motion.

6.3 DIFFERENTIAL EQUATIONS AND MOTION

333

Newton placed a new emphasis on the importance of forces. Newton’s laws of motion do not say what a force is, they say how it acts. His first law is the principle of inertia, which says what happens in the absence of a force—there is no change in velocity. His second law says that a force acts to produce a change in velocity, that is, an acceleration. It states that 𝐹 = 𝑚𝑎, where 𝑚 is the mass of the object, 𝐹 is the net force, and 𝑎 is the acceleration produced by this force. Galileo demonstrated that a body falling under the influence of gravity does so with constant acceleration. Assuming we can neglect air resistance, this constant acceleration is independent of the mass of the body. This last fact was the outcome of Galileo’s famous observation around 1600 that a heavy ball and a light ball dropped off the Leaning Tower of Pisa hit the ground at the same time. Whether or not he actually performed this experiment, Galileo presented a very clear thought experiment in the Dialogues to prove the same point. (This point was counter to Aristotle’s more common-sense notion that the heavier ball would reach the ground first.) Galileo showed that the mass of the object did not appear as a variable in the equation of motion. Thus, the same constantacceleration equation applies to all bodies falling under the influence of gravity. Nearly a hundred years after Galileo’s experiment, Newton formulated his laws of motion and gravity, which gave a theoretical explanation of Galileo’s experimental observation that the acceleration due to gravity is independent of the mass of the body. According to Newton, acceleration is caused by force, and in the case of falling bodies, that force is the force of gravity.

Exercises and Problems for Section 6.3 Online Resource: Additional Problems for Section 6.3 EXERCISES

1. Show that 𝑦 = 𝑥𝑒−𝑥 + 2 is a solution of the initial value problem 𝑑𝑦 = (1 − 𝑥)𝑒−𝑥 , 𝑑𝑥

𝑦(0) = 2.

2. Show that 𝑦 = sin(2𝑡), for 0 ≤ 𝑡 < 𝜋∕4, is a solution to the initial value problem √ 𝑑𝑦 = 2 1 − 𝑦2 , 𝑑𝑡

𝑦(0) = 0.

In Exercises 3–8, find the general solution to the differential equation. 3.

𝑑𝑦 = 2𝑥 𝑑𝑥

4.

𝑑𝑦 = 𝑡2 𝑑𝑡

5.

𝑑𝑦 = 𝑥3 + 5𝑥4 𝑑𝑥

6.

𝑑𝑦 = 𝑒𝑡 𝑑𝑡

7.

𝑑𝑦 = cos 𝑥 𝑑𝑥

8.

𝑑𝑦 1 = , 𝑑𝑥 𝑥

𝑥>0

In Exercises 9–12, find the solution of the initial value problem. 𝑑𝑦 𝑑𝑥 𝑑𝑦 10. 𝑑𝑥 𝑑𝑦 11. 𝑑𝑥 𝑑𝑦 12. 𝑑𝑥 9.

= 3𝑥2 ,

𝑦(0) = 5

= 𝑥5 + 𝑥6 , = 𝑒𝑥 , = sin 𝑥,

𝑦(1) = 2

𝑦(0) = 7 𝑦(0) = 3

PROBLEMS 13. A rock is thrown downward with velocity 10 ft∕sec from a bridge 100 ft above the water. How fast is the rock going when it hits the water? 14. A water balloon launched from the roof of a building at time 𝑡 = 0 has vertical velocity 𝑣(𝑡) = −32𝑡 + 40 feet/sec at time 𝑡 seconds, with 𝑣 > 0 corresponding to upward motion. (a) If the roof of the building is 30 feet above the ground, find an expression for the height of the water balloon above the ground at time 𝑡. (b) What is the average velocity of the balloon between 𝑡 = 1.5 and 𝑡 = 3 seconds? (c) A 6-foot person is standing on the ground. How

fast is the water balloon falling when it strikes the person on the top of the head? 15. A car starts from rest at time 𝑡 = 0 and accelerates at −0.6𝑡 + 4 meters/sec2 for 0 ≤ 𝑡 ≤ 12. How long does it take for the car to go 100 meters? 16. Ice is forming on a pond at a rate given by √ 𝑑𝑦 = 𝑘 𝑡, 𝑑𝑡 where 𝑦 is the thickness of the ice in inches at time 𝑡 measured in hours since the ice started forming, and 𝑘 is a positive constant. Find 𝑦 as a function of 𝑡. 17. A revenue 𝑅(𝑝) is obtained by a farmer from selling

334

Chapter 6 CONSTRUCTING ANTIDERIVATIVES

grain at price 𝑝 dollars/unit. The marginal revenue is given by 𝑅′ (𝑝) = 25 − 2𝑝. (a) Find 𝑅(𝑝). Assume the revenue is zero when the price is zero. (b) For what prices does the revenue increase as the price increases? For what prices does the revenue decrease as price increases? 18. A firm’s marginal cost function is 𝑀𝐶 = 3𝑞 2 + 6𝑞 + 9. (a) Write a differential equation for the total cost, 𝐶(𝑞). (b) Find the total cost function if the fixed costs are 400. 19. A tomato is thrown upward from a bridge 25 m above the ground at 40 m/sec. (a) Give formulas for the acceleration, velocity, and height of the tomato at time 𝑡. (b) How high does the tomato go, and when does it reach its highest point? (c) How long is it in the air? 20. A car going 80 ft/sec (about 55 mph) brakes to a stop in five seconds. Assume the deceleration is constant. (a) Graph the velocity against time, 𝑡, for 0 ≤ 𝑡 ≤ 5 seconds. (b) Represent, as an area on the graph, the total distance traveled from the time the brakes are applied until the car comes to a stop. (c) Find this area and hence the distance traveled. (d) Now find the total distance traveled using antidifferentiation. 21. An object is shot vertically upward from the ground with an initial velocity of 160 ft/sec. (a) At what rate is the velocity decreasing? Give units. (b) Explain why the graph of velocity of the object against time (with upward positive) is a line. (c) Using the starting velocity and your answer to part (b), find the time at which the object reaches the highest point. (d) Use your answer to part (c) to decide when the object hits the ground. (e) Graph the velocity against time. Mark on the graph when the object reaches its highest point and when it lands. (f) Find the maximum height reached by the object by considering an area on the graph. (g) Now express velocity as a function of time, and find the greatest height by antidifferentiation. 22. A stone thrown upward from the top of a 320-foot cliff at 128 ft/sec eventually falls to the beach below. (a) How long does the stone take to reach its highest point? (b) What is its maximum height? (c) How long before the stone hits the beach? (d) What is the velocity of the stone on impact?

23. A 727 jet needs to attain a speed of 200 mph to take off. If it can accelerate from 0 to 200 mph in 30 seconds, how long must the runway be? (Assume constant acceleration.) 24. A cat, walking along the window ledge of a New York apartment, knocks off a flower pot, which falls to the street 200 feet below. How fast is the flower pot traveling when it hits the street? (Give your answer in ft/sec and in mph, given that 1 ft/sec = 15/22 mph.) 25. An Acura NSX going at 70 mph stops in 157 feet. Find the acceleration, assuming it is constant. 26. (a) Find the general solution of the differential equation 𝑑𝑦∕𝑑𝑥 = 2𝑥 + 1. (b) Sketch a graph of at least three solutions. (c) Find the solution satisfying 𝑦(1) = 5. Graph this solution with the others from part (b). 27. (a) Find and graph the general solution of the differential equation 𝑑𝑦 = sin 𝑥 + 2. 𝑑𝑥 (b) Find the solution of the initial value problem 𝑑𝑦 = sin 𝑥 + 2, 𝑑𝑥

𝑦(3) = 5.

28. On the moon, the acceleration due to gravity is about 1.6 m/sec2 (compared to 𝑔 ≈ 9.8 m/sec2 on earth). If you drop a rock on the moon (with initial velocity 0), find formulas for: (a) Its velocity, 𝑣(𝑡), at time 𝑡. (b) The distance, 𝑠(𝑡), it falls in time 𝑡. 29. (a) Imagine throwing a rock straight up in the air. What is its initial velocity if the rock reaches a maximum height of 100 feet above its starting point? (b) Now imagine being transplanted to the moon and throwing a moon rock vertically upward with the same velocity as in part (a). How high will it go? (On the moon, 𝑔 = 5 ft/sec2 .) 30. An object is dropped from a 400-foot tower. When does it hit the ground and how fast is it going at the time of the impact? 31. The object in Problem 30 falls off the same 400-foot tower. What would the acceleration due to gravity have to be to make it reach the ground in half the time? 32. A ball that is dropped from a window hits the ground in five seconds. How high is the window? (Give your answer in feet.) 33. On the moon the acceleration due to gravity is 5 ft/sec2 . An astronaut jumps into the air with an initial upward velocity of 10 ft/sec. How high does he go? How long is the astronaut off the ground? 34. Assume the acceleration of a moving body is −𝑔 and its initial velocity and position are 𝑣0 and 𝑠0 respectively. Find velocity, 𝑣, and position, 𝑠, as a function of 𝑡.

6.4 SECOND FUNDAMENTAL THEOREM OF CALCULUS

335

𝑣

35. A particle of mass, 𝑚, acted on by a force, 𝐹 , moves in a straight line. Its acceleration, 𝑎, is given by Newton’s Law: 𝐹 = 𝑚𝑎. The work, 𝑊 , done by a constant force when the particle moves through a displacement, 𝑑, is

𝑡 𝑡1

𝑊 = 𝐹 𝑑.

𝑡2

𝑡3

𝑡4

Figure 6.27

The velocity, 𝑣, of the particle as a function of time, 𝑡, is given in Figure 6.27. What is the sign of the work done during each of the time intervals: [0, 𝑡1 ], [𝑡1 , 𝑡2 ], [𝑡2 , 𝑡3 ], [𝑡3 , 𝑡4 ],[𝑡2 , 𝑡4 ]?

Strengthen Your Understanding In Problems 36–39, explain what is wrong with the statement.

43. If 𝐹 (𝑥) is an antiderivative of 𝑓 (𝑥), then 𝑦 = 𝐹 (𝑥) is a solution to the differential equation 𝑑𝑦∕𝑑𝑥 = 𝑓 (𝑥).

36. A rock dropped from a 400-foot cliff takes twice as long to hit the ground as it would if it were dropped from a 200-foot cliff. 37. The function 𝑦 = cos(𝑡2 ) is a solution to the initial value problem 𝑑𝑦 = − sin(𝑡2 ), 𝑦(0) = 1. 𝑑𝑡

44. If 𝑦 = 𝐹 (𝑥) is a solution to the differential equation 𝑑𝑦∕𝑑𝑥 = 𝑓 (𝑥), then 𝐹 (𝑥) is an antiderivative of 𝑓 (𝑥).

38. Two solutions to a differential equation 𝑑𝑦∕𝑑𝑥 = 𝑓 (𝑥) have graphs which cross at the initial value. 39. A differential equation cannot have a constant solution. In Problems 40–42, give an example of: 40. Two different solutions to the differential equation 𝑑𝑦 = 𝑡 + 3. 𝑑𝑡 41. A differential equation that has solution 𝑦 = cos (5𝑥). 42. A differential equation that has a solution that is decreasing for all 𝑡. Are the statements in Problems 43–51 true or false? Give an explanation for your answer.

6.4

45. If an object has constant nonzero acceleration, then the position of the object as a function of time is a quadratic polynomial. 46. In an initial value problem for the differential equation 𝑑𝑦∕𝑑𝑥 = 𝑓 (𝑥), the value of 𝑦 at 𝑥 = 0 is always specified. 47. If 𝑓 (𝑥) is positive for all 𝑥, then there is a solution of the differential equation 𝑑𝑦∕𝑑𝑥 = 𝑓 (𝑥) where 𝑦(𝑥) is positive for all 𝑥. 48. If 𝑓 (𝑥) > 0 for all 𝑥 then every solution of the differential equation 𝑑𝑦∕𝑑𝑥 = 𝑓 (𝑥) is an increasing function. 49. If two solutions of a differential equation 𝑑𝑦∕𝑑𝑥 = 𝑓 (𝑥) have different values at 𝑥 = 3 then they have different values at every 𝑥. 50. If the function 𝑦 = 𝑓 (𝑥) is a solution of the differential equation 𝑑𝑦∕𝑑𝑥 = sin 𝑥∕𝑥, then the function 𝑦 = 𝑓 (𝑥) + 5 is also a solution. 51. There is only one solution 𝑦(𝑡) to the initial value problem 𝑑𝑦∕𝑑𝑡 = 3𝑡2 , 𝑦(1) = 𝜋.

SECOND FUNDAMENTAL THEOREM OF CALCULUS Suppose 𝑓 is an elementary function, that is, a combination of constants, powers of 𝑥, sin 𝑥, cos 𝑥, 𝑒𝑥 , and ln 𝑥. Then we have to be lucky to find an antiderivative 𝐹 which is also an elementary function. But if we can’t find 𝐹 as an elementary function, how can we be sure that 𝐹 exists at all? In this section we use the definite integral to construct antiderivatives.

Construction of Antiderivatives Using the Definite Integral 2

Consider the function 𝑓 (𝑥) = 𝑒−𝑥 . Its antiderivative, 𝐹 , is not an elementary function, but we would still like to find a way of calculating its values. Assuming 𝐹 exists, we know from the Fundamental

336

Chapter 6 CONSTRUCTING ANTIDERIVATIVES

Theorem of Calculus that

𝑏 2

𝐹 (𝑏) − 𝐹 (𝑎) =

𝑒−𝑡 𝑑𝑡.

∫𝑎

Setting 𝑎 = 0 and replacing 𝑏 by 𝑥, we have 𝑥 2

𝐹 (𝑥) − 𝐹 (0) =

𝑒−𝑡 𝑑𝑡.

∫0

Suppose we want the antiderivative that satisfies 𝐹 (0) = 0. Then we get 𝑥

𝐹 (𝑥) =

2

𝑒−𝑡 𝑑𝑡.

∫0

This is a formula for 𝐹 . For any value of 𝑥, there is a unique value for 𝐹 (𝑥), so 𝐹 is a function. For any fixed 𝑥, we can calculate 𝐹 (𝑥) numerically. For example, 2

𝐹 (2) =

∫0

2

𝑒−𝑡 𝑑𝑡 = 0.88208 … .

Notice that our expression for 𝐹 is not an elementary function; we have created a new function using the definite integral. The next theorem says that this method of constructing antiderivatives works in general. This means that if we define 𝐹 by 𝑥

𝐹 (𝑥) =

𝑓 (𝑡) 𝑑𝑡,

∫𝑎

then 𝐹 must be an antiderivative of 𝑓 .

Theorem 6.2: Construction Theorem for Antiderivatives (Second Fundamental Theorem of Calculus) If 𝑓 is a continuous function on an interval, and if 𝑎 is any number in that interval, then the function 𝐹 defined on the interval as follows is an antiderivative of 𝑓 : 𝑥

𝐹 (𝑥) =

∫𝑎

𝑓 (𝑡) 𝑑𝑡.

Proof Our task is to show that 𝐹 , defined by this integral, is an antiderivative of 𝑓 . We want to show that 𝐹 ′ (𝑥) = 𝑓 (𝑥). By the definition of the derivative, 𝐹 ′ (𝑥) = lim

ℎ→0

𝐹 (𝑥 + ℎ) − 𝐹 (𝑥) . ℎ

To gain some geometric insight, let’s suppose 𝑓 is positive and ℎ is positive. Then we can visualize 𝑥

𝐹 (𝑥) =

∫𝑎

𝑥+ℎ

𝑓 (𝑡) 𝑑𝑡

and

𝐹 (𝑥 + ℎ) =

∫𝑎

as areas, which leads to representing 𝑥+ℎ

𝐹 (𝑥 + ℎ) − 𝐹 (𝑥) =

∫𝑥

𝑓 (𝑡) 𝑑𝑡

𝑓 (𝑡) 𝑑𝑡

6.4 SECOND FUNDAMENTAL THEOREM OF CALCULUS

337

as a difference of two areas. From Figure 6.28, we see that 𝐹 (𝑥 + ℎ) − 𝐹 (𝑥) is roughly the area of a rectangle of height 𝑓 (𝑥) and width ℎ (shaded darker in Figure 6.28), so we have 𝐹 (𝑥 + ℎ) − 𝐹 (𝑥) ≈ 𝑓 (𝑥)ℎ, hence

𝐹 (𝑥 + ℎ) − 𝐹 (𝑥) ≈ 𝑓 (𝑥). ℎ More precisely, we can use Theorem 5.4 on comparing integrals on page 307 to conclude that 𝑥+ℎ

𝑚ℎ ≤

∫𝑥

𝑓 (𝑡) 𝑑𝑡 ≤ 𝑀ℎ,

where 𝑚 is the greatest lower bound for 𝑓 on the interval from 𝑥 to 𝑥 + ℎ and 𝑀 is the least upper bound on that interval. (See Figure 6.29.) Hence 𝑚ℎ ≤ 𝐹 (𝑥 + ℎ) − 𝐹 (𝑥) ≤ 𝑀ℎ, so

𝐹 (𝑥 + ℎ) − 𝐹 (𝑥) ≤ 𝑀. ℎ Since 𝑓 is continuous, both 𝑚 and 𝑀 approach 𝑓 (𝑥) as ℎ approaches zero. Thus 𝑚≤

𝑓 (𝑥) ≤ lim

ℎ→0

𝐹 (𝑥 + ℎ) − 𝐹 (𝑥) ≤ 𝑓 (𝑥). ℎ

𝑓

𝑓

Area = 𝐹 (𝑥)

Area = 𝐹 (𝑥)





𝑀 𝑚

✛ Area = 𝐹 (𝑥 + ℎ) − 𝐹 (𝑥) ≈ 𝑓 (𝑥)ℎ

𝑓 (𝑥)

𝑓 (𝑥)

❄ 𝑎







𝑥 𝑥+ℎ

𝑎

Figure 6.28: 𝐹 (𝑥 + ℎ) − 𝐹 (𝑥) is area of roughly rectangular region

𝑥 𝑥+ℎ

Figure 6.29: Upper and lower bounds for 𝐹 (𝑥 + ℎ) − 𝐹 (𝑥)

Thus both inequalities must actually be equalities, so we have the result we want: 𝑓 (𝑥) = lim

ℎ→0

𝐹 (𝑥 + ℎ) − 𝐹 (𝑥) = 𝐹 ′ (𝑥). ℎ

Relationship Between the Construction Theorem and the Fundamental Theorem of Calculus If 𝐹 is constructed as in Theorem 6.2, then we have just shown that 𝐹 ′ = 𝑓 . Suppose 𝐺 is any other antiderivative of 𝑓 , so 𝐺′ = 𝑓 , and therefore 𝐹 ′ = 𝐺′ . Since the derivative of 𝐹 − 𝐺 is zero, the Constant Function Theorem on page 187 tells us that 𝐹 − 𝐺 is a constant, so 𝐹 (𝑥) = 𝐺(𝑥) + 𝐶. Since we know 𝐹 (𝑎) = 0 (by the definition of 𝐹 ), we can write 𝑏

∫𝑎

𝑓 (𝑡) 𝑑𝑡 = 𝐹 (𝑏) = 𝐹 (𝑏) − 𝐹 (𝑎) = (𝐺(𝑏) + 𝐶) − (𝐺(𝑎) + 𝐶) = 𝐺(𝑏) − 𝐺(𝑎). 𝑏

This result, that the definite integral ∫𝑎 𝑓 (𝑡) 𝑑𝑡 can be evaluated using any antiderivative of 𝑓 , is the (First) Fundamental Theorem of Calculus. Thus, we have shown that the First Fundamental Theorem of Calculus can be obtained from the Construction Theorem (the Second Fundamental Theorem of Calculus).

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Chapter 6 CONSTRUCTING ANTIDERIVATIVES

Using the Construction Theorem for Antiderivatives The construction theorem enables us to write down antiderivatives of functions that do not have elementary antiderivatives. For example, an antiderivative of (sin 𝑥)∕𝑥 is 𝑥

𝐹 (𝑥) =

∫0

sin 𝑡 𝑑𝑡. 𝑡

Notice that 𝐹 is a function; we can calculate its values to any degree of accuracy.1 This function already has a name: it is called the sine-integral, and it is denoted Si(𝑥). Example 1

Construct a table of values of Si(𝑥) for 𝑥 = 0, 1, 2, 3.

Solution

Using numerical methods, we calculate the values of Si(𝑥) = ∫0 (sin 𝑡)∕𝑡 𝑑𝑡 given in Table 6.2. Since the integrand is undefined at 𝑡 = 0, we took the lower limit as 0.00001 instead of 0.

𝑥

Table 6.2

A table of values of Si(𝑥)

𝑥

0

1

2

3

Si(𝑥)

0

0.95

1.61

1.85

The reason the sine-integral has a name is that some scientists and engineers use it all the time (for example, in optics). For them, it is just another common function like sine or cosine. Its derivative is given by 𝑑 sin 𝑥 Si(𝑥) = . 𝑑𝑥 𝑥

Example 2

Find the derivative of 𝑥 Si(𝑥).

Solution

Using the product rule, ( ) ( ) 𝑑 𝑑 𝑑 (𝑥 Si(𝑥)) = 𝑥 Si(𝑥) + 𝑥 Si(𝑥) 𝑑𝑥 𝑑𝑥 𝑑𝑥 sin 𝑥 = 1 ⋅ Si(𝑥) + 𝑥 𝑥 = Si(𝑥) + sin 𝑥.

Exercises and Problems for Section 6.4 Online Resource: Additional Problems for Section 6.4 EXERCISES

1. For 𝑥 = 0, 0.5,√ 1.0, 1.5, and 2.0, make a table of values 𝑥 for 𝐼(𝑥) = ∫0 𝑡4 + 1 𝑑𝑡.

2. Assume that 𝐹 ′ (𝑡) = sin 𝑡 cos 𝑡 and 𝐹 (0) = 1. Find 𝐹 (𝑏) for 𝑏 = 0, 0.5, 1, 1.5, 2, 2.5, and 3. 3. (a) Continue the table of values for Si(𝑥) = 𝑥 ∫0 (sin 𝑡∕𝑡) 𝑑𝑡 on page 338 for 𝑥 = 4 and 𝑥 = 5. (b) Why is Si(𝑥) decreasing between 𝑥 = 4 and 𝑥 = 5?

In Exercises 4–6, write an expression for the function, 𝑓 (𝑥), with the given properties. 4. 𝑓 ′ (𝑥) = sin(𝑥2 ) and 𝑓 (0) = 7 5. 𝑓 ′ (𝑥) = (sin 𝑥)∕𝑥 and 𝑓 (1) = 5 6. 𝑓 ′ (𝑥) = Si(𝑥) and 𝑓 (0) = 2

1 You may notice that the integrand, (sin 𝑡)∕𝑡, is undefined at 𝑡 = 0; such improper integrals are treated in more detail in Chapter 7.

6.4 SECOND FUNDAMENTAL THEOREM OF CALCULUS 𝑥

Find the derivatives in Exercises 11–16.

8.

11.

In Exercises 7–10, let 𝐹 (𝑥) = ∫0 𝑓 (𝑡) 𝑑𝑡. Graph 𝐹 (𝑥) as a function of 𝑥.

𝑑 7.

𝑓 (𝑡)

𝑓 (𝑡) 𝑡

𝑡

9.

10.

𝑓 (𝑡)

𝑑𝑥 ∫0 𝑑

13.

15.

𝑡

𝑡

cos(𝑡2 ) 𝑑𝑡

12.

√ 𝑑 sin( 𝑥) 𝑑𝑥 𝑑𝑡 ∫4

(1 + 𝑡)200 𝑑𝑡

14.

𝑑 ln(𝑡2 + 1) 𝑑𝑡 𝑑𝑥 ∫2

arctan(𝑡2 ) 𝑑𝑡

16.

𝑑 [ ] Si(𝑥2 ) 𝑑𝑥

𝑥

𝑑𝑥 ∫1 𝑑

𝑓 (𝑡)

𝑥

𝑥

𝑥

𝑑𝑥 ∫0.5

339

𝑡

PROBLEMS 𝑥

2

6

17. Find intervals where the graph of 𝐹 (𝑥) = ∫0 𝑒−𝑡 𝑑𝑡 is concave up and concave down.

𝑓 (𝑡)

3

−𝑥

18. Use properties of the function 𝑓 (𝑥) = 𝑥𝑒 to determine the number of values 𝑥 that make 𝐹 (𝑥) = 0, given 𝑥 𝐹 (𝑥) = ∫1 𝑓 (𝑡) 𝑑𝑡.

𝑡 5

10

25

Figure 6.32

𝑥

19. Approximate 𝐹 (𝑥) for 𝑥 = 0, 0.5, 1, 1.5, 2, 2.5. 20. Using a graph of 𝐹 ′ (𝑥), decide where 𝐹 (𝑥) is increasing and where 𝐹 (𝑥) is decreasing for 0 ≤ 𝑥 ≤ 2.5. 21. Does 𝐹 (𝑥) have a maximum value for 0 ≤ 𝑥 ≤ 2.5? If so, at what value of 𝑥 does it occur, and approximately what is that maximum value?

𝑥

25. Let 𝑔(𝑥) = ∫0 𝑓 (𝑡) 𝑑𝑡. Using Figure 6.33, find (b) 𝑔 ′ (1)

(a) 𝑔(0)

(c) The interval where 𝑔 is concave up. (d) The value of 𝑥 where 𝑔 takes its maximum on the interval 0 ≤ 𝑥 ≤ 8.

𝑥

22. Use Figure 6.30 to sketch a graph of 𝐹 (𝑥) = ∫0 𝑓 (𝑡) 𝑑𝑡. Label the points 𝑥1 , 𝑥2 , 𝑥3 .

𝑓 (𝑡)

3 2 1

𝑡

𝑓 (𝑥) 𝑥 𝑥2

20

−3

For Problems 19–21, let 𝐹 (𝑥) = ∫0 sin(𝑡2 ) 𝑑𝑡.

𝑥1

15

𝑥3

−1 −2

2

4

6

8

Figure 6.33 Figure 6.30 𝑥

26. Let 𝐹 (𝑥) = ∫0 sin(2𝑡) 𝑑𝑡.



23. The graph of the derivative 𝐹 of some function 𝐹 is given in Figure 6.31. If 𝐹 (20) = 150, estimate the maximum value attained by 𝐹 . 𝐹 ′ (𝑥)

20 10

60 20

−10

(a) Evaluate 𝐹 (𝜋). (b) Draw a sketch to explain geometrically why the answer to part (a) is correct. (c) For what values of 𝑥 is 𝐹 (𝑥) positive? negative? 𝑥

𝑥

40

Figure 6.31

27. Let 𝐹 (𝑥) = ∫2 (1∕ln 𝑡) 𝑑𝑡 for 𝑥 ≥ 2. (a) Find 𝐹 ′ (𝑥). (b) Is 𝐹 increasing or decreasing? What can you say about the concavity of its graph? (c) Sketch a graph of 𝐹 (𝑥).

𝑥

24. (a) Let 𝐹 (𝑥) = ∫5 𝑓 (𝑡) 𝑑𝑡, with 𝑓 in Figure 6.32. Find the 𝑥-value at which the maximum value of 𝐹 occurs. (b) What is the maximum value of 𝐹 ?

In Problems 28–29, find the value of the function with the given properties. 2

28. 𝐹 (1), where 𝐹 ′ (𝑥) = 𝑒−𝑥 and 𝐹 (0) = 2 29. 𝐺(−1), where 𝐺′ (𝑥) = cos(𝑥2 ) and 𝐺(0) = −3

340

Chapter 6 CONSTRUCTING ANTIDERIVATIVES

In Problems 30–33, estimate the value of each expression, 𝑡 𝑡 given 𝑤(𝑡) = ∫0 𝑞(𝑥) 𝑑𝑥 and 𝑣(𝑡) = ∫0 𝑞 ′ (𝑥) 𝑑𝑥. Table 6.3 gives values for 𝑞(𝑥), a function with negative first and second derivatives. Are your answers under- or overestimates?

In Problems 38–41, find the given quantities. The error function, erf(𝑥), is defined by 𝑥

2 2 𝑒−𝑡 𝑑𝑡. erf (𝑥) = √ 𝜋 ∫0

Table 6.3 𝑥

0.0

0.1

0.2

0.3

0.4

0.5

𝑞(𝑥)

5.3

5.2

4.9

4.5

3.9

3.1

30. 𝑤(0.4)

31. 𝑣(0.4)

32. 𝑤′ (0.4)

33. 𝑣′ (0.4)

𝑑 (𝑥 erf (𝑥)) 𝑑𝑥 ( 3 ) 𝑑 2 𝑥 ∫0 𝑒−𝑡 𝑑𝑡 40. 𝑑𝑥

38.

√ 𝑑 (erf ( 𝑥)) 𝑑𝑥 ( 3 ) 𝑑 2 𝑥 ∫𝑥 𝑒−𝑡 𝑑𝑡 41. 𝑑𝑥 39.

42. In year 𝑥, the forested area in hectares of an island is 𝑓 (𝑥) = 820 +

In Problems 34–37, use the chain rule to calculate the derivative. 𝑥2

34.

𝑑 𝑑𝑥 ∫0

36.

√ 𝑑 sin( 𝑥) 𝑑𝑥 𝑑𝑡 ∫2𝑡

ln(1 + 𝑡2 ) 𝑑𝑡

sin 𝑡

35.

𝑑 𝑑𝑡 ∫1

37.

2 𝑑 𝑒𝑡 𝑑𝑡 𝑑𝑥 ∫−𝑥2

cos(𝑥2 ) 𝑑𝑥

𝑥2

4

𝑥

∫0

𝑟(𝑡) 𝑑𝑡,

where 𝑟(𝑡) has only one zero, at 𝑡 = 40. Let 𝑟′ (40) < 0, 40 𝑓 (40) = 1170 and ∫30 𝑟(𝑥) 𝑑𝑥 = 82. (a) What is the maximum forested area? (b) How large is the forested area in year 𝑡 = 30? 𝑥

𝑥

43. Let 𝐹 (𝑥) = ∫0 𝑝(𝑡) 𝑑𝑡 and 𝐺(𝑥) = ∫3 𝑝(𝑡) 𝑑𝑡, where 𝐹 (3) = 5 and 𝐺(5) = 7. Find each of the following. (a) 𝐺(0) (c) 𝐹 (5)

(b) 𝐹 (0) (d) 𝑘 where 𝐺(𝑥) = 𝐹 (𝑥)+𝑘

Strengthen Your Understanding In Problems 44–46, explain what is wrong with the statement. 5

𝑑 𝑡2 𝑑𝑡 = 𝑥2 . 44. 𝑑𝑥 ∫0 45. 𝐹 (𝑥) =

𝑥

∫−2

𝑡2 𝑑𝑡 has a local minimum at 𝑥 = 0.

46. For the function 𝑓 (𝑥) shown in Figure 6.34, 𝐹 (𝑥) = 𝑥 ∫0 𝑓 (𝑡) 𝑑𝑡 has a local minimum at 𝑥 = 2.

𝑓 (𝑥)

In Problems 47–48, give an example of: 47. A function, 𝐹 (𝑥), constructed using the Second Fundamental Theorem of Calculus such that 𝐹 is a nondecreasing function and 𝐹 (0) = 0. 48. A function 𝐺(𝑥), constructed using the Second Fundamental Theorem of Calculus, such that 𝐺 is concave up and 𝐺(7) = 0. Are the statements in Problems 49–54 true or false? Give an explanation for your answer. 49. Every continuous function has an antiderivative. 𝑥

𝑥 −1

1 2

3

50. ∫0 sin(𝑡2 ) 𝑑𝑡 is an antiderivative of sin(𝑥2 ). 𝑥

5

51. If 𝐹 (𝑥) = ∫0 𝑓 (𝑡) 𝑑𝑡, then 𝐹 (5) − 𝐹 (3) = ∫3 𝑓 (𝑡)𝑑𝑡. 𝑥

52. If 𝐹 (𝑥) = ∫0 𝑓 (𝑡) 𝑑𝑡, then 𝐹 (𝑥) must be increasing. Figure 6.34:

𝑥

𝑥

𝑥

𝑥

53. If 𝐹 (𝑥) = ∫0 𝑓 (𝑡) 𝑑𝑡 and 𝐺(𝑥) = ∫2 𝑓 (𝑡) 𝑑𝑡, then 𝐹 (𝑥) = 𝐺(𝑥) + 𝐶. 54. If 𝐹 (𝑥) = ∫0 𝑓 (𝑡) 𝑑𝑡 and 𝐺(𝑥) = ∫0 𝑔(𝑡) 𝑑𝑡, then 𝑥 𝐹 (𝑥) + 𝐺(𝑥) = ∫0 (𝑓 (𝑡) + 𝑔(𝑡)) 𝑑𝑡.

Online Resource: Review problems and Projects

Chapter Seven

INTEGRATION

Contents 7.1 Integration by Substitution. . . . . . . . . . . . . . . . . . The Guess-and-Check Method . . . . . . . . . . . . . . The Method of Substitution . . . . . . . . . . . . . . . . . Definite Integrals by Substitution . . . . . . . . . . . . More Complex Substitutions . . . . . . . . . . . . . . . . 7.2 Integration by Parts. . . . . . . . . . . . . . . . . . . . . . . . The General Formula for Integration by Parts . . . 7.3 Tables of Integrals . . . . . . . . . . . . . . . . . . . . . . . . . Using the Table of Integrals . . . . . . . . . . . . . . . . . Transforming the Integrand . . . . . . . . . . . . . . . . .

7.4

7.5

7.6

7.7

342 342 343 346 347 353 353 360 360 362 Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362 Long Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362 Completing the Square. . . . . . . . . . . . . . . . . . . . . . . . 362 Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 Algebraic Identities and Trigonometric Substitutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366 Method of Partial Fractions . . . . . . . . . . . . . . . . . 366 Trigonometric Substitutions. . . . . . . . . . . . . . . . . 369 Sine Substitutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . 369 Tangent Substitutions. . . . . . . . . . . . . . . . . . . . . . . . . 372 Completing the Square. . . . . . . . . . . . . . . . . . . . . . . . 373 Numerical Methods for Definite Integrals. . . . . . 376 The Midpoint Rule. . . . . . . . . . . . . . . . . . . . . . . . 376 The Trapezoid Rule . . . . . . . . . . . . . . . . . . . . . . . 378 Over- or Underestimate? . . . . . . . . . . . . . . . . . . . 378 The Midpoint Rule. . . . . . . . . . . . . . . . . . . . . . . . 379 Error in Left and Right Rules. . . . . . . . . . . . . . . . 379 Error in Trapezoid and Midpoint Rules . . . . . . . . 380 Simpson’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 381 An Alternate Approach: Approximating by Lines and Parabolas . . . . . . . . . . . . . . . . . . . 381 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . 385 When the Limit of Integration Is Infinite. . . . . . . 385 When the Integrand Becomes Infinite . . . . . . . . . 389 Comparison of Improper Integrals . . . . . . . . . . . 394 Making Comparisons . . . . . . . . . . . . . . . . . . . . . . 394 How Do We Know What to Compare With? . . . . . . . 395

342

Chapter 7 INTEGRATION

7.1

INTEGRATION BY SUBSTITUTION In Chapter 3, we learned rules to differentiate any function obtained by combining constants, powers of 𝑥, sin 𝑥, cos 𝑥, 𝑒𝑥 , and ln 𝑥, using addition, multiplication, division, or composition of functions. Such functions are called elementary. In this chapter, we introduce several methods of antidifferentiation. However, there is a great difference between looking for derivatives and looking for antiderivatives. √ Every elementary function has elementary derivatives, but most elementary functions—such as 𝑥3 + 1, (sin 𝑥)∕𝑥, and 2 𝑒−𝑥 —do not have elementary antiderivatives. All commonly occurring antiderivatives can be found with a computer algebra system (CAS). However, just as it is useful to be able to calculate 3 + 4 without a calculator, we usually calculate some antiderivatives by hand.

The Guess-and-Check Method A good strategy for finding simple antiderivatives is to guess an answer (using knowledge of differentiation rules) and then check the answer by differentiating it. If we get the expected result, then we’re done; otherwise, we revise the guess and check again. The method of guess-and-check is useful in reversing the chain rule. According to the chain rule, Inside ⏞⏞⏞ 𝑑 (𝑓 (𝑔(𝑥))) = 𝑓 ′ ( 𝑔(𝑥) ) ⋅ 𝑔 ′ (𝑥) . 𝑑𝑥 ⏟⏟⏟ ⏟⏟⏟ Derivative of outside↗

↖ Derivative of inside

Thus, any function which is the result of applying the chain rule is the product of two factors: the “derivative of the outside” and the “derivative of the inside.” If a function has this form, its antiderivative is 𝑓 (𝑔(𝑥)). 3𝑥2 cos(𝑥3 ) 𝑑𝑥.

Example 1

Find

Solution

The function 3𝑥2 cos(𝑥3 ) looks like the result of applying the chain rule: there is an “inside” function 𝑥3 and its derivative 3𝑥2 appears as a factor. Since the outside function is a cosine which has a sine as an antiderivative, we guess sin(𝑥3 ) for the antiderivative. Differentiating to check gives



𝑑 (sin(𝑥3 )) = cos(𝑥3 ) ⋅ (3𝑥2 ). 𝑑𝑥 Since this is what we began with, we know that ∫

3𝑥2 cos(𝑥3 ) 𝑑𝑥 = sin(𝑥3 ) + 𝐶.

The basic idea of this method is to try to find an inside function whose derivative appears as a factor. This works even when the derivative is missing a constant factor, as in the next example. √ 𝑥3 𝑥4 + 5 𝑑𝑥.

Example 2

Find

Solution

Here the inside function is 𝑥4 + 5, and its derivative appears as a factor, with the exception of a missing 4. Thus, the integrand we have is more or less of the form



√ 𝑔 ′ (𝑥) 𝑔(𝑥),

7.1 INTEGRATION BY SUBSTITUTION

with 𝑔(𝑥) = 𝑥4 + 5. Since 𝑥3∕2 ∕(3∕2) is an antiderivative of the outside function guess that an antiderivative is (𝑔(𝑥))3∕2 (𝑥4 + 5)3∕2 = . 3∕2 3∕2

343

√ 𝑥, we might

Let’s check and see: 𝑑 𝑑𝑥 so

(

(𝑥4 + 5)3∕2 3∕2

)

=

3 (𝑥4 + 5)1∕2 ⋅ 4𝑥3 = 4𝑥3 (𝑥4 + 5)1∕2 , 2 3∕2

(𝑥4 + 5)3∕2 is too big by a factor of 4. The correct antiderivative is 3∕2 1 (𝑥4 + 5)3∕2 1 = (𝑥4 + 5)3∕2 . 4 3∕2 6

Thus

√ 1 𝑥3 𝑥4 + 5 𝑑𝑥 = (𝑥4 + 5)3∕2 + 𝐶. ∫ 6

As a final check: 𝑑 𝑑𝑥

(

) 1 4 1 3 (𝑥 + 5)3∕2 = ⋅ (𝑥4 + 5)1∕2 ⋅ 4𝑥3 = 𝑥3 (𝑥4 + 5)1∕2 . 6 6 2

As we see in the preceding example, antidifferentiating a function often involves “correcting for” constant factors: if differentiation produces an extra factor of 2, antidifferentiation will require a factor of 21 .

The Method of Substitution When the integrand is complicated, it helps to formalize this guess-and-check method as follows:

To Make a Substitution Let 𝑤 be the “inside function” and 𝑑𝑤 = 𝑤′ (𝑥) 𝑑𝑥 =

𝑑𝑤 𝑑𝑥. 𝑑𝑥

Let’s redo the first example using a substitution. 3𝑥2 cos(𝑥3 ) 𝑑𝑥.

Example 3

Find

Solution

As before, we look for an inside function whose derivative appears—in this case 𝑥3 . We let 𝑤 = 𝑥3 . Then 𝑑𝑤 = 𝑤′ (𝑥) 𝑑𝑥 = 3𝑥2 𝑑𝑥. The original integrand can now be completely rewritten in terms of the new variable 𝑤:





3𝑥2 cos(𝑥3 ) 𝑑𝑥 =



cos (𝑥3 ) ⋅ 3𝑥2 𝑑𝑥 = cos 𝑤 𝑑𝑤 = sin 𝑤 + 𝐶 = sin(𝑥3 ) + 𝐶. ⏟⏟⏟ ⏟⏟⏟ ∫ 𝑤

𝑑𝑤

By changing the variable to 𝑤, we can simplify the integrand. We now have cos 𝑤, which can be antidifferentiated more easily. The final step, after antidifferentiating, is to convert back to the original variable, 𝑥.

Why Does Substitution Work? The substitution method makes it look as if we can treat 𝑑𝑤 and 𝑑𝑥 as separate entities, even canceling them in the equation 𝑑𝑤 = (𝑑𝑤∕𝑑𝑥)𝑑𝑥. Let’s see why this works. Suppose we have an integral

344

Chapter 7 INTEGRATION

of the form ∫ 𝑓 (𝑔(𝑥))𝑔 ′ (𝑥) 𝑑𝑥, where 𝑔(𝑥) is the inside function and 𝑓 (𝑥) is the outside function. 𝑑 If 𝐹 is an antiderivative of 𝑓 , then 𝐹 ′ = 𝑓 , and by the chain rule 𝑑𝑥 (𝐹 (𝑔(𝑥))) = 𝑓 (𝑔(𝑥))𝑔 ′ (𝑥). Therefore, 𝑓 (𝑔(𝑥))𝑔 ′(𝑥) 𝑑𝑥 = 𝐹 (𝑔(𝑥)) + 𝐶. ∫ Now write 𝑤 = 𝑔(𝑥) and 𝑑𝑤∕𝑑𝑥 = 𝑔 ′ (𝑥) on both sides of this equation: 𝑓 (𝑤)



𝑑𝑤 𝑑𝑥 = 𝐹 (𝑤) + 𝐶. 𝑑𝑥

On the other hand, knowing that 𝐹 ′ = 𝑓 tells us that ∫

𝑓 (𝑤) 𝑑𝑤 = 𝐹 (𝑤) + 𝐶.

Thus, the following two integrals are equal: ∫

𝑓 (𝑤)

𝑑𝑤 𝑑𝑥 = 𝑓 (𝑤) 𝑑𝑤. ∫ 𝑑𝑥

Substituting 𝑤 for the inside function and writing 𝑑𝑤 = 𝑤′ (𝑥)𝑑𝑥 leaves the indefinite integral unchanged. Let’s revisit the second example that we did by guess-and-check. √ 𝑥3 𝑥4 + 5 𝑑𝑥.

Example 4

Find

Solution

The inside function is 𝑥4 + 5, with derivative 4𝑥3 . The integrand has a factor of 𝑥3 , and since the only thing missing is a constant factor, we try



𝑤 = 𝑥4 + 5. Then 𝑑𝑤 = 𝑤′ (𝑥) 𝑑𝑥 = 4𝑥3 𝑑𝑥, giving 1 𝑑𝑤 = 𝑥3 𝑑𝑥. 4 Thus, ∫

√ 𝑥3 𝑥4 + 5 𝑑𝑥 =



√ 1 1 1 1 𝑤3∕2 + 𝐶 = (𝑥4 + 5)3∕2 + 𝐶. 𝑤 𝑑𝑤 = 𝑤1∕2 𝑑𝑤 = ⋅ ∫ 4 4 4 3∕2 6

Once again, we get the same result as with guess-and-check.

Warning We saw in the preceding example that we can apply the substitution method when a constant factor is missing from the derivative of the inside function. However, we may not be able to use substitution if anything other than a constant factor is missing. For example, setting 𝑤 = 𝑥4 + 5 to find √ ∫

𝑥2 𝑥4 + 5 𝑑𝑥

does us no good because 𝑥2 𝑑𝑥 is not a constant multiple of 𝑑𝑤 = 4𝑥3 𝑑𝑥. Substitution works if the integrand contains the derivative of the inside function, to within a constant factor.

7.1 INTEGRATION BY SUBSTITUTION

345

Some people prefer the substitution method over guess-and-check since it is more systematic, but both methods achieve the same result. For simple problems, guess-and-check can be faster. 𝑒cos 𝜃 sin 𝜃 𝑑𝜃.

Example 5

Find

Solution

We let 𝑤 = cos 𝜃 since its derivative is − sin 𝜃 and there is a factor of sin 𝜃 in the integrand. This gives 𝑑𝑤 = 𝑤′ (𝜃) 𝑑𝜃 = − sin 𝜃 𝑑𝜃,



so −𝑑𝑤 = sin 𝜃 𝑑𝜃. Thus 𝑒cos 𝜃 sin 𝜃 𝑑𝜃 =





𝑒𝑤 (−𝑑𝑤) = (−1)



𝑒𝑤 𝑑𝑤 = −𝑒𝑤 + 𝐶 = −𝑒cos 𝜃 + 𝐶.

𝑒𝑡 𝑑𝑡. ∫ 1 + 𝑒𝑡

Example 6

Find

Solution

Observing that the derivative of 1 + 𝑒𝑡 is 𝑒𝑡 , we see 𝑤 = 1 + 𝑒𝑡 is a good choice. Then 𝑑𝑤 = 𝑒𝑡 𝑑𝑡, so that 1 1 𝑒𝑡 𝑑𝑤 = ln |𝑤| + 𝐶 𝑑𝑡 = 𝑒𝑡 𝑑𝑡 = ∫ 1 + 𝑒𝑡 ∫ 𝑤 ∫ 1 + 𝑒𝑡 = ln |1 + 𝑒𝑡 | + 𝐶 = ln(1 + 𝑒𝑡 ) + 𝐶.

(Since (1 + 𝑒𝑡 ) is always positive.)

Since the numerator is 𝑒𝑡 𝑑𝑡, we might also have tried 𝑤 = 𝑒𝑡 . This substitution leads to the integral ∫ (1∕(1+𝑤))𝑑𝑤, which is better than the original integral but requires another substitution, 𝑢 = 1+𝑤, to finish. There are often several different ways of doing an integral by substitution. Notice the pattern in the previous example: having a function in the denominator and its derivative in the numerator leads to a natural logarithm. The next example follows the same pattern. Example 7

Find

Solution

Recall that tan 𝜃 = (sin 𝜃)∕(cos 𝜃). If 𝑤 = cos 𝜃, then 𝑑𝑤 = − sin 𝜃 𝑑𝜃, so



tan 𝜃 𝑑𝜃.



tan 𝜃 𝑑𝜃 =

−𝑑𝑤 sin 𝜃 𝑑𝜃 = = − ln |𝑤| + 𝐶 = − ln | cos 𝜃| + 𝐶. ∫ 𝑤 ∫ cos 𝜃

One way to think of integration is in terms of standard forms, whose antiderivatives are known. Substitution is useful for putting a complicated integral in a standard form. Example 8

Give a substitution 𝑤 and constants 𝑘, 𝑛 so that the following integral has the form ∫ 𝑘𝑤𝑛 𝑑𝑤: ∫

𝑒𝑥 cos3 (𝑒𝑥 ) sin(𝑒𝑥 ) 𝑑𝑥.

346

Chapter 7 INTEGRATION

Solution

We notice that one of the factors in the integrand is (cos(𝑒𝑥 ))3 , so if we let 𝑤 = cos(𝑒𝑥 ), this factor is 𝑤3 . Then 𝑑𝑤 = (− sin(𝑒𝑥 ))𝑒𝑥 𝑑𝑥, so ∫

𝑒𝑥 cos3 (𝑒𝑥 ) sin(𝑒𝑥 ) 𝑑𝑥 =

cos3 (𝑒𝑥 )(sin(𝑒𝑥 ))𝑒𝑥 𝑑𝑥 =





𝑤3 (−𝑑𝑤).

Therefore, we choose 𝑤 = cos(𝑒𝑥 ) and then 𝑘 = −1, 𝑛 = 3.

Definite Integrals by Substitution 2

2

𝑥𝑒𝑥 𝑑𝑥.

Example 9

Compute

Solution

To evaluate this definite integral using the Fundamental Theorem of Calculus, we first need to find 2 an antiderivative of 𝑓 (𝑥) = 𝑥𝑒𝑥 . The inside function is 𝑥2 , so we let 𝑤 = 𝑥2 . Then 𝑑𝑤 = 2𝑥 𝑑𝑥, so 1 𝑑𝑤 = 𝑥 𝑑𝑥. Thus, 2

∫0

2



𝑥𝑒𝑥 𝑑𝑥 =



1 1 1 2 𝑒𝑤 𝑑𝑤 = 𝑒𝑤 + 𝐶 = 𝑒𝑥 + 𝐶. 2 2 2

Now we find the definite integral 2

∫0

2

2

𝑥𝑒𝑥 𝑑𝑥 =

1 𝑥2 || 1 1 𝑒 = (𝑒4 − 𝑒0 ) = (𝑒4 − 1). 2 ||0 2 2

There is another way to look at the same problem. After we established that 2



𝑥𝑒𝑥 𝑑𝑥 =

1 𝑤 𝑒 + 𝐶, 2

our next two steps were to replace 𝑤 by 𝑥2 , and then 𝑥 by 2 and 0. We could have directly replaced the original limits of integration, 𝑥 = 0 and 𝑥 = 2, by the corresponding 𝑤 limits. Since 𝑤 = 𝑥2 , the 𝑤 limits are 𝑤 = 02 = 0 (when 𝑥 = 0) and 𝑤 = 22 = 4 (when 𝑥 = 2), so we get 𝑥=2

∫𝑥=0

2

𝑥𝑒𝑥 𝑑𝑥 =

4 𝑤=4 ) 1 1 1 | 1( 4 𝑒 − 𝑒0 = (𝑒4 − 1). 𝑒𝑤 𝑑𝑤 = 𝑒𝑤 || = 2 ∫𝑤=0 2 |0 2 2

As we would expect, both methods give the same answer.

To Use Substitution to Find Definite Integrals Either • Compute the indefinite integral, expressing an antiderivative in terms of the original variable, and then evaluate the result at the original limits, or • Convert the original limits to new limits in terms of the new variable and do not convert the antiderivative back to the original variable.

𝜋∕4

Example 10

Evaluate

∫0

tan3 𝜃 𝑑𝜃. cos2 𝜃

7.1 INTEGRATION BY SUBSTITUTION

Solution

347

To use substitution, we must decide what 𝑤 should be. There are two possible inside functions, tan 𝜃 and cos 𝜃. Now 𝑑 1 𝑑 (tan 𝜃) = (cos 𝜃) = − sin 𝜃, and 2 𝑑𝜃 𝑑𝜃 cos 𝜃 and since the integral contains a factor of 1∕ cos2 𝜃 but not of sin 𝜃, we try 𝑤 = tan 𝜃. Then 𝑑𝑤 = (1∕ cos2 𝜃)𝑑𝜃. When 𝜃 = 0, 𝑤 = tan 0 = 0, and when 𝜃 = 𝜋∕4, 𝑤 = tan(𝜋∕4) = 1, so 1 𝜋∕4 𝜋∕4 1 tan3 𝜃 1 1 | 1 𝑑𝜃 = 𝑑𝜃 = (tan 𝜃)3 ⋅ 𝑤3 𝑑𝑤 = 𝑤4 || = . 2 2 ∫0 ∫ ∫ 4 |0 4 cos 𝜃 cos 𝜃 0 0 3

𝑑𝑥 . ∫1 5 − 𝑥

Example 11

Evaluate

Solution

Let 𝑤 = 5 − 𝑥, so 𝑑𝑤 = −𝑑𝑥. When 𝑥 = 1, 𝑤 = 4, and when 𝑥 = 3, 𝑤 = 2, so 3

∫1

2 ( ) |2 −𝑑𝑤 𝑑𝑥 4 = = − ln |𝑤||| = − (ln 2 − ln 4) = ln = ln 2 = 0.693. ∫ 5−𝑥 𝑤 2 |4 4

Notice that we write the limit 𝑤 = 4 at the bottom of the second integral, even though it is larger than 𝑤 = 2, because 𝑤 = 4 corresponds to the lower limit 𝑥 = 1.

More Complex Substitutions In the examples of substitution presented so far, we guessed an expression for 𝑤 and hoped to find 𝑑𝑤 (or some constant multiple of it) in the integrand. What if we are not so lucky? It turns out that it often works to let 𝑤 be some messy expression contained inside, say, a cosine or under a root, even if we cannot see immediately how such a substitution helps. √

√ 𝑥 𝑑𝑥.

Example 12

Find

Solution

This time, the derivative of the inside function is nowhere to be seen. Nevertheless, we try 𝑤 = 1 + √ Then 𝑤 − 1 = 𝑥, so (𝑤 − 1)2 = 𝑥. Therefore 2(𝑤 − 1) 𝑑𝑤 = 𝑑𝑥. We have



1+





1+

√ 𝑥 𝑑𝑥 =



√ 𝑤 2(𝑤 − 1) 𝑑𝑤 = 2



√ 𝑥.

𝑤1∕2 (𝑤 − 1) 𝑑𝑤

) 2 5∕2 2 3∕2 𝑤 − 𝑤 +𝐶 ∫ 5 3 ) ( √ √ 2 2 = 2 (1 + 𝑥)5∕2 − (1 + 𝑥)3∕2 + 𝐶. 5 3 =2

(𝑤3∕2 − 𝑤1∕2 ) 𝑑𝑤 = 2

(

Notice that the substitution in the preceding example again converts the inside of the messiest function into something simple. In addition, since the derivative of the inside function is not waiting for us, we have to solve for 𝑥 so that we can get 𝑑𝑥 entirely in terms of 𝑤 and 𝑑𝑤. Example 13

Find



√ 3 (𝑥 + 7) 3 − 2𝑥 𝑑𝑥.

348

Chapter 7 INTEGRATION

Solution

The inside function is 3 − 2𝑥, with derivative −2. However, instead of a factor of −2, the integrand contains the factor (𝑥 + 7), but substituting 𝑤 = 3 − 2𝑥 turns out to help anyway. Then 𝑑𝑤 = −2 𝑑𝑥, so (−1∕2) 𝑑𝑤 = 𝑑𝑥. Now we must convert everything to 𝑤, including 𝑥 + 7. If 𝑤 = 3 − 2𝑥, then 2𝑥 = 3 − 𝑤, so 𝑥 = 3∕2 − 𝑤∕2, and therefore we can write 𝑥 + 7 in terms of 𝑤. Thus ∫

)√ ( ) 3 𝑤 1 − +7 3 𝑤 − 𝑑𝑤 ∫ 2 2 2 ) ( 1 17 𝑤 =− − 𝑤1∕3 𝑑𝑤 2∫ 2 2 1 =− (17 − 𝑤)𝑤1∕3 𝑑𝑤 4∫ 1 =− (17𝑤1∕3 − 𝑤4∕3 ) 𝑑𝑤 4∫ ( ) 𝑤4∕3 𝑤7∕3 1 17 − +𝐶 =− 4 4∕3 7∕3 ( ) 1 51 3 (3 − 2𝑥)4∕3 − (3 − 2𝑥)7∕3 + 𝐶. =− 4 4 7

√ 3 (𝑥 + 7) 3 − 2𝑥 𝑑𝑥 =

(

√ 3 Looking back over the solution, the reason this substitution works is that it converts 3 − 2𝑥, the √ messiest part of the integrand, to 3 𝑤, which can be combined with the other factor and then integrated.

Exercises and Problems for Section 7.1 Online Resource: Additional Problems for Section 7.1 EXERCISES

1. Use substitution to express each of the following inte𝑏 grals as a multiple of ∫𝑎 (1∕𝑤) 𝑑𝑤 for some 𝑎 and 𝑏. Then evaluate the integrals. 1

(a)

𝑥 𝑑𝑥 ∫0 1 + 𝑥2

𝜋∕4

(b)

∫0

sin 𝑥 𝑑𝑥 cos 𝑥

2. (a) Find the derivatives of sin(𝑥2 + 1) and sin(𝑥3 + 1). (b) Use your answer to part (a) to find antiderivatives of: (i) 𝑥 cos(𝑥2 + 1)

(ii) 𝑥2 cos(𝑥3 + 1)

15. 17. 19. 21. 23.

(c) Find the general antiderivatives of: (i) 𝑥 sin(𝑥2 + 1)

(ii) 𝑥2 sin(𝑥3 + 1)

Find the integrals in Exercises 3–50. Check your answers by differentiation. 3. 5. 7. 9. 11. 13.

∫ ∫ ∫ ∫ ∫ ∫

𝑒3𝑥 𝑑𝑥

4.

𝑒−𝑥 𝑑𝑥

6.

sin(2𝑥) 𝑑𝑥 sin(3 − 𝑡) 𝑑𝑡 (𝑟 + 1)3 𝑑𝑟 𝑥2 (1 + 2𝑥3 )2 𝑑𝑥

8. 10. 12. 14.

∫ ∫

27.

𝑡𝑒𝑡 𝑑𝑡

29.

25𝑒−0.2𝑡 𝑑𝑡

31.

2



25.

𝑡 cos(𝑡2 ) 𝑑𝑡

33.

∫ ∫

𝑥(𝑥2 + 3)2 𝑑𝑥

16.

𝑦2 (1 + 𝑦)2 𝑑𝑦

18.

3 +1



𝑥2 𝑒𝑥

𝑑𝑥

1 𝑑𝑥 √ ∫ 4−𝑥 ∫ ∫ ∫ ∫

24.

sin 𝜃(cos 𝜃 + 5)7 𝑑𝜃

26.

sin6 𝜃 cos 𝜃 𝑑𝜃

28.

sin6 (5𝜃) cos(5𝜃) 𝑑𝜃

30.

𝑑𝑧

𝑧 (𝑡 + 1)2



𝑑𝑡

𝑡2

32.

34.

2

∫ ∫ ∫

∫ ∫

𝑑𝑥 ∫ 1 + 2𝑥2 √ cos 𝑥 37. 𝑑𝑥 √ ∫ 𝑥

𝑦(𝑦2 + 5)8 𝑑𝑦 𝑡2 (𝑡3 − 3)10 𝑑𝑡

(2𝑡 − 7)73 𝑑𝑡 (𝑥2 + 3)2 𝑑𝑥

∫ 𝑦+5 ∫ ∫ ∫ ∫

36.

tan(𝜃 + 1) 𝑑𝜃 √ cos 3𝑡 sin 3𝑡 𝑑𝑡 sin3 𝛼 cos 𝛼 𝑑𝛼 tan(2𝑥) 𝑑𝑥 𝑒𝑡 + 1

∫ 𝑒𝑡 + 𝑡

𝑑𝑡

𝑦 𝑑𝑦 ∫ 𝑦2 + 4

𝑥𝑒−𝑥 𝑑𝑥 35.

𝑥(𝑥2 − 4)7∕2 𝑑𝑥

𝑑𝑦 22.

𝑒−0.1𝑡+4 𝑑𝑡

(ln 𝑧)2 ∫

20.





𝑑𝑥 √ 1 − 4𝑥2 √

𝑒 𝑦 𝑑𝑦 38. ∫ √𝑦

7.1 INTEGRATION BY SUBSTITUTION

39.

41.

43. 45. 47. 49.

1 + 𝑒𝑥 𝑑𝑥 √ ∫ 𝑥 + 𝑒𝑥 𝑥+1

∫ 𝑥2 + 2𝑥 + 19 𝑒𝑥 − 𝑒−𝑥 ∫ 𝑒𝑥 + 𝑒−𝑥 ∫ ∫ ∫

𝑒𝑥 40.

∫ 2+

4

𝑑𝑥 𝑒𝑥

63.

∫1

𝑡 42.

𝑑𝑥

44.

𝑑𝑥

sinh 3𝑡 𝑑𝑡

46.

cosh(2𝑤 + 1) 𝑑𝑤

48.

cosh2 𝑥 sinh 𝑥 𝑑𝑥

50.

4

∫ 1 + 3𝑡2

𝑑𝑡

𝑥 cos(𝑥2 ) 𝑑𝑥 √ ∫ sin(𝑥2 ) cosh 𝑥 𝑑𝑥

∫ ∫

65.

∫1

∫−1

𝑥 cosh 𝑥2 𝑑𝑥

𝑥 𝑑𝑥 ∫0 (1 + 𝑥2 )2

∫−1 ∫1 2

71.

∫−1

68.

1 𝑑𝑦 ∫−1 1 + 𝑦2 3

1 𝑑𝑥 𝑥

70.

√ 𝑥 + 2 𝑑𝑥

72.

𝑑𝑡 ∫1 (𝑡 + 7)2 2

∫1

51. 𝑝(𝑡) = 𝜋𝑡3 + 4𝑡

52. 𝑓 (𝑥) = sin 3𝑥

53. 𝑓 (𝑥) = 2𝑥 cos(𝑥2 )

54. 𝑟(𝑡) = 12𝑡2 cos(𝑡3 )

Find the integrals in Exercises 73–80.

55. 𝑓 (𝑥) = sin(2 − 5𝑥)

56. 𝑓 (𝑥) = 𝑒sin 𝑥 cos 𝑥

73.

58. 𝑓 (𝑥) =

3 cos2 (2𝑥)

For Exercises 59–66, use the Fundamental Theorem to calculate the definite integrals. 59.

cos(𝑥 + 𝜋) 𝑑𝑥

∫0

60.

2

𝑒− cos 𝜃 sin 𝜃 𝑑𝜃

∫0

cos(𝜋𝑥) 𝑑𝑥

∫0

𝜋∕2

61.

75.

77.

1∕2

𝜋

62.

∫1



1

𝑥 𝑥2 + 1

𝑑𝑡

𝑡+2

1

(𝑥3 + 5𝑥) 𝑑𝑥

3

69.

57. 𝑓 (𝑥) =

1

2

66.

3

67.

For the functions in Exercises 51–58, find the general antiderivative. Check your answers by differentiation.

64.

For Exercises 67–72, evaluate the definite integrals. Whenever possible, use the Fundamental Theorem of Calculus, perhaps after a substitution. Otherwise, use numerical methods.

(sinh 𝑧)𝑒cosh 𝑧 𝑑𝑧



𝑒−2



𝑒 𝑥 √ 𝑑𝑥 𝑥 √ cos 𝑥 𝑑𝑥 √ 𝑥

349

79.





sin 𝑡 𝑑𝑡 𝑡

√ 𝑦 𝑦 + 1 𝑑𝑦

74.

𝑡2 + 𝑡 𝑑𝑡 √ 𝑡+1 √ 𝑥2 𝑥 − 2 𝑑𝑥

𝑑𝑥 √ ∫ 2+2 𝑥 √ 78. (𝑧 + 2) 1 − 𝑧 𝑑𝑧 ∫ 76.

𝑡 𝑑𝑡 √ 𝑡+1



2

2𝑥𝑒𝑥 𝑑𝑥

𝑧(𝑧 + 1)1∕3 𝑑𝑧



80.

3𝑥 − 2 𝑑𝑥 √ 2𝑥 + 1



PROBLEMS 81. Show that the following integral can be calculated by multiplying the numerator and denominator by 𝑒𝑡 and using a substitution: 1 𝑑𝑡. ∫ 1 + 𝑒−𝑡

87. Using the substitution 𝑤 = 𝑒𝑥 , find a function 𝑔(𝑤)

In Problems 88–94, evaluate the integral. Your answer should not contain 𝑓 , which is a differentiable function with the following values:

In Problems 82–85, show the two integrals are equal using a substitution. 𝜋∕3

82.

𝜋

3 sin2 (3𝑥) 𝑑𝑥 =

∫0

∫0

2

83.

4 2

2 ln(𝑠 + 1) 𝑑𝑠 =

∫1

1

𝑒

84.

3

∫1

∫1

(ln 𝑤) 𝑑𝑤 =

3 𝑧

∫0

∫0

sin2 (𝑦) 𝑑𝑦 88.

∫0 3

90.

∫1

𝜋

𝑥 cos(𝜋 − 𝑥) 𝑑𝑥 =

∫0

𝑒

(𝜋 − 𝑡) cos 𝑡 𝑑𝑡

86. Using the substitution 𝑤 = 𝑥2 , find a function 𝑔(𝑤) √ 𝑏

such that ∫√𝑎 𝑑𝑥 =

𝑏 ∫𝑎

𝑥

0

1

𝜋/2

e

3

𝑓 (𝑥)

5

7

8

10

11

𝑓 ′ (𝑥)

2

4

6

9

12

1

ln(𝑡 + 1) 𝑑𝑡 √ 𝑡

𝑧 𝑒 𝑑𝑧

𝜋

85.

𝑒𝑏

𝑏

such that ∫𝑎 𝑒−𝑥 𝑑𝑥 = ∫𝑒𝑎 𝑔(𝑤) 𝑑𝑤 for all 𝑎 < 𝑏.

𝑔(𝑤) 𝑑𝑤 for all 0 < 𝑎 < 𝑏.

92.

∫1

3

𝑓 ′ (𝑥) sin 𝑓 (𝑥) 𝑑𝑥

89.

𝑓 ′ (𝑥) 𝑑𝑥 𝑓 (𝑥)

91.

𝑓 ′ (ln 𝑥) 𝑑𝑥 𝑥

93.

𝜋∕2

94.

∫0

sin 𝑥 ⋅ 𝑓 ′ (cos 𝑥) 𝑑𝑥

𝑓 ′ (𝑥)𝑒𝑓 (𝑥) 𝑑𝑥

∫1 1

𝑒𝑥 𝑓 ′ (𝑒𝑥 ) 𝑑𝑥

∫0 1

∫0

𝑓 ′ (𝑥)(𝑓 (𝑥))2 𝑑𝑥

350

Chapter 7 INTEGRATION

In Problems 95–99, explain why the two antiderivatives are really, despite their apparent dissimilarity, different expressions of the same problem. You do not need to evaluate the integrals. 𝑒𝑥 𝑑𝑥 ∫ 1 + 𝑒2𝑥 ln 𝑥 96. 𝑑𝑥 ∫ 𝑥

95.

97. 98.

99.



cos 𝑥 𝑑𝑥 ∫ 1 + sin2 𝑥

and and

𝑒sin 𝑥 cos 𝑥 𝑑𝑥

and

3





(sin 𝑥) cos 𝑥 𝑑𝑥 √ 𝑥 + 1 𝑑𝑥

and

and

𝑒arcsin 𝑥 𝑑𝑥 ∫ √1 − 𝑥2 3

∫ √



3 2

(𝑥 + 1) 𝑥 𝑑𝑥

√ 1+ 𝑥 𝑑𝑥 √ 𝑥

In Problems 100–103, find an expression for the integral which contains 𝑔 but no integral sign. 100. 102.

∫ ∫

𝑔 ′ (𝑥)(𝑔(𝑥))4 𝑑𝑥

101.

𝑔 ′ (𝑥) sin 𝑔(𝑥) 𝑑𝑥

103.

∫ ∫

𝑔 ′ (𝑥)𝑒𝑔(𝑥) 𝑑𝑥 √ 𝑔 ′ (𝑥) 1 + 𝑔(𝑥) 𝑑𝑥

In Problems 104–106, find a substitution 𝑤 and constants 𝑘, 𝑛 so that the integral has the form ∫ 𝑘𝑤𝑛 𝑑𝑤. 104. 106.





𝑥2 1 − 4𝑥3 𝑑𝑥

105.

1 √ 𝑑𝑥 𝑥 1 𝑑𝑥 (c) √ ∫ 𝑥+1

(a)

cos 𝑡 𝑑𝑡 ∫ sin 𝑡

(a) (c) (e)



𝑥 sin(𝑥2 ) 𝑑𝑥

5

108.

𝑥3 𝑒𝑥 𝑑𝑥

In Problems 119–125, find the exact area. 2

119. Under 𝑓 (𝑥) = 𝑥𝑒𝑥 between 𝑥 = 0 and 𝑥 = 2. 120. Under 𝑓 (𝑥) = 1∕(𝑥 + 1) between 𝑥 = 0 and 𝑥 = 2. 121. Under 𝑓 (𝑥) = sinh(𝑥∕2) between 𝑥 = 0 and 𝑥 = 2. 122. Under 𝑓 (𝜃) = (𝑒𝜃+1 )3 for 0 ≤ 𝜃 ≤ 2. 123. Between 𝑒𝑡 and 𝑒𝑡+1 for 0 ≤ 𝑡 ≤ 2. 124. Between 𝑦 = 𝑒𝑥 , 𝑦 = 3, and the 𝑦-axis. 125. Under one arch of the curve 𝑉 (𝑡) = 𝑉0 sin(𝜔𝑡), where 𝑉0 > 0 and 𝜔 > 0. 126. Find the area inside the asteroid in Figure 7.1 given that 1

∫0

∫0

3𝑥 𝑑𝑥 , √ 5𝑥2 + 7 2𝑥 𝑑𝑥 , 2𝑥 + 3

𝑥2 sin 𝑥 𝑑𝑥



𝑥 𝑑𝑥 ∫ (1 + 𝑥2 )2 sin 𝑥 𝑑𝑥 (f) ∫ 2 + cos 𝑥

3𝜋 . 32

(1 − 𝑥2∕3 )3∕2 𝑑𝑥 = 𝑦 𝑎

0

∫1

1 𝑑𝑥 √ 𝑥+1



(d)

2



In Problems 107–109, find constants 𝑘, 𝑛, 𝑤0 , 𝑤1 so the the 𝑤 integral has the form ∫𝑤 1 𝑘𝑤𝑛 𝑑𝑤. 107.

(b)

𝑥2 𝑑𝑥 ∫ 1 + 𝑥2

2𝑥 𝑑𝑥 ∫ (𝑥2 − 3)2

5

(b)



118. If appropriate, evaluate the following integrals by substitution. If substitution is not appropriate, say so, and do not evaluate.

𝑥 𝑑𝑥



117. Integrate:

𝑥2∕3 + 𝑦2∕3 = 𝑎2∕3

2

𝑤 = 5𝑥 + 7 𝑥 −𝑎

𝑤 = 2𝑥 + 3

𝑎

𝜋∕4

109.

∫𝜋∕12

sin7 (2𝑥) cos(2𝑥) 𝑑𝑥,

In Problems 110–114, find a substitution 𝑤 and a constant 𝑘 so that the integral has the form ∫ 𝑘𝑒𝑤 𝑑𝑤. 110. 112. 114.

2

∫ ∫ ∫

𝑥𝑒−𝑥 𝑑𝑥

111.

√ 𝑒𝑟 𝑑𝑟

113.



𝑒sin 𝜙 cos 𝜙 𝑑𝜙

𝑧2 𝑑𝑧 ∫ 𝑒−𝑧3

𝑒2𝑡 𝑒3𝑡−4 𝑑𝑡

127. Find the exact average value of 𝑓 (𝑥) = 1∕(𝑥 + 1) on the interval 𝑥 = 0 to 𝑥 = 2. Sketch a graph showing the function and the average value. 128. Let 𝑔(𝑥) = 𝑓 (2𝑥). Show that the average value of 𝑓 on the interval [0, 2𝑏] is the same as the average value of 𝑔 on the interval [0, 𝑏]. 2

7

∫3

Figure 7.1

129. Suppose ∫0 𝑔(𝑡) 𝑑𝑡 = 5. Calculate the following:

In Problems 115–116, find a substitution 𝑤 and constants 𝑏 𝑎, 𝑏, 𝐴 so that the integral has the form ∫𝑎 𝐴𝑒𝑤 𝑑𝑤. 115.

−𝑎

𝑤 = sin 2𝑥

1

𝑒2𝑡−3 𝑑𝑡

116.

∫0

𝑒cos(𝜋𝑡) sin(𝜋𝑡) 𝑑𝑡

4

(a)

∫0

2

𝑔(𝑡∕2) 𝑑𝑡

(b)

∫0

𝑔(2 − 𝑡) 𝑑𝑡

7.1 INTEGRATION BY SUBSTITUTION 1

130. Suppose ∫0 𝑓 (𝑡) 𝑑𝑡 = 3. Calculate the following: 0.5

(a)

(b)

𝑓 (1 − 𝑡) 𝑑𝑡

∫0



1.5

(c)

∫1

𝜋

(a) Do the multiplication first, and then antidifferentiate. (b) Use the substitution 𝑤 = 𝑥2 + 1. (c) Explain how the expressions from parts (a) and (b) are different. Are they both correct? 133. (a) Find ∫ sin 𝜃 cos 𝜃 𝑑𝜃. (b) You probably solved part (a) by making the substitution 𝑤 = sin 𝜃 or 𝑤 = cos 𝜃. (If not, go back and do it that way.) Now find ∫ sin 𝜃 cos 𝜃 𝑑𝜃 by making the other substitution. (c) There is yet another way of finding this integral which involves the trigonometric identities 2

cos(2𝜃) = cos2 𝜃 − sin 𝜃. Find ∫ sin 𝜃 cos 𝜃 𝑑𝜃 using one of these identities and then the substitution 𝑤 = 2𝜃. (d) You should now have three different expressions for the indefinite integral ∫ sin 𝜃 cos 𝜃 𝑑𝜃. Are they really different? Are they all correct? Explain. For Problems 134–137, find a substitution 𝑤 and constants 𝑏 𝑎, 𝑏, 𝑘 so that the integral has the form ∫𝑎 𝑘𝑓 (𝑤) 𝑑𝑤. 5

9

136.

∫1 5

137.

∫2

𝜋

( ) 𝑓 𝑥2 𝑥 𝑑𝑥

135.

𝑓 (cos 𝑥) sin 𝑥 𝑑𝑥

∫0

( √ )√ 𝑓 6𝑥 𝑥 𝑥 𝑑𝑥 ( ( )) 𝑓 ln 𝑥2 + 1 𝑥 𝑥2 + 1

𝑑𝑥

In Problems 138–141, given that 𝑓 (𝑥) = 𝐹 ′ (𝑥), use Table 7.1 to evaluate the expression correct to three decimals places. Table 7.1

138.

√ 3

∫1

𝑥

0

0.5

1

1.5

2

2.5

3

𝐹 (𝑥)

2

3

16

21

24

26

31

𝜋∕2

( ) 𝑓 𝑥2 𝑥 𝑑𝑥

139.

𝑒𝐹 (𝑥) 𝑓 (𝑥) 𝑑𝑥

141.

∫𝜋∕6

0.5

140.

∫0

3

1 Based

∫1

143. Find possible formulas for 𝑓 (𝑥) and 𝑔(𝑥) given that ( √ ) 1 𝑓 (𝑔(𝑥)) 𝑔(𝑥) ⋅ √ 𝑑𝑥 = sin 𝑒 𝑥 + 𝐶. ∫ 𝑥 144. Find the solution of the initial value problem 𝑦′ = tan 𝑥 + 1,

𝑦(0) = 1.

1

145. Let 𝐼𝑚,𝑛 = ∫0 𝑥𝑚 (1−𝑥)𝑛 𝑑𝑥 for constant 𝑚, 𝑛. Show that 𝐼𝑚,𝑛 = 𝐼𝑛,𝑚 . 146. Let 𝑓 (𝑡) be the velocity in meters/second of a car at time 𝑡 in seconds. Give an integral for the change in the position of the car (a) Over the time interval 0 ≤ 𝑡 ≤ 60. (b) In terms of time 𝑇 in minutes, over the same time interval.

sin(2𝜃) = 2 sin 𝜃 cos 𝜃

∫−2

𝑔 (𝑓 (𝑥)) 𝑔(𝑥) 𝑑𝑥 = sin (sin 𝑥) + 𝐶.

𝑓 (3 − 2𝑡) 𝑑𝑡

131. (a) Calculate exactly: ∫−𝜋 cos2 𝜃 sin 𝜃 𝑑𝜃. (b) Calculate the exact area under the curve 𝑦 = cos2 𝜃 sin 𝜃 between 𝜃 = 0 and 𝜃 = 𝜋. 132. Find ∫ 4𝑥(𝑥2 + 1) 𝑑𝑥 using two methods:

134.

142. Find possible formulas for 𝑓 (𝑥) and 𝑔(𝑥) given that

1

𝑓 (2𝑡) 𝑑𝑡

∫0

351

𝑓 (sin 𝑥) cos 𝑥 𝑑𝑥 𝑓 (𝑥) 𝑑𝑥 𝐹 (𝑥)

147. With 𝑡 in years since 1790, the US population in millions can be approximated by 𝑃 =

192 . 1 + 48𝑒−0.0317𝑡

Based on this model, what was the average US population between 1900 and 1950? 148. Over the past fifty years the carbon dioxide level in the atmosphere has increased. Carbon dioxide is believed to drive temperature, so predictions of future carbon dioxide levels are important. If 𝐶(𝑡) is carbon dioxide level in parts per million (ppm) and 𝑡 is time in years since 1950, three possible models are:1 I 𝐶 ′ (𝑡) = 1.3 II 𝐶 ′ (𝑡) = 0.5 + 0.03𝑡 III 𝐶 ′ (𝑡) = 0.5𝑒0.02𝑡 (a) Given that the carbon dioxide level was 311 ppm in 1950, find 𝐶(𝑡) for each model. (b) Find the carbon dioxide level in 2020 predicted by each model. 149. Let 𝑓 (𝑡) be the rate of flow, in cubic meters per hour, of a flooding river at time 𝑡 in hours. Give an integral for the total flow of the river (a) Over the 3-day period 0 ≤ 𝑡 ≤ 72. (b) In terms of time 𝑇 in days over the same 3-day period.

on data from www.esrl.noaa.gov/gmd/ccgg. Accessed March 13, 2013.

352

Chapter 7 INTEGRATION

150. With 𝑡 in years since April 20th, 2015, the population, 𝑃 , of the world in billions2 can be modeled by 𝑃 = 7.17𝑒0.01064𝑡 . (a) What does this model predict for the world population on April 20th, 2020? In 2025? (b) Use the Fundamental Theorem to predict the average population of the world between April 2015 and April 2025. 151. Oil is leaking out of a ruptured tanker at the rate of 𝑟(𝑡) = 50𝑒−0.02𝑡 thousand liters per minute. (a) At what rate, in liters per minute, is oil leaking out at 𝑡 = 0? At 𝑡 = 60? (b) How many liters leak out during the first hour? 152. At the start of 2014, the world’s known copper reserves were 690 million tons. With 𝑡 in years since the start of 2014, copper has been mined at a rate given by 17.9𝑒0.025𝑡 million tons per year.3 (a) Assuming that copper continues to be extracted at the same rate, write an expression for the total quantity mined in the first 𝑇 years after the start of 2014. (b) Under these assumptions, when is the world predicted to run out of copper? 153. Throughout much of the 20th century, the yearly consumption of electricity in the US increased exponentially at a continuous rate of 7% per year. Assume this trend continues and that the electrical energy consumed in 1900 was 1.4 million megawatt-hours. (a) Write an expression for yearly electricity consumption as a function of time, 𝑡, in years since 1900. (b) Find the average yearly electrical consumption throughout the 20th century. (c) During what year was electrical consumption closest to the average for the century? (d) Without doing the calculation for part (c), how could you have predicted which half of the century the answer would be in?

154. An electric current, 𝐼(𝑡), flowing out of a capacitor, decays according to 𝐼(𝑡) = 𝐼0 𝑒−𝑡 , where 𝑡 is time. Find the charge, 𝑄(𝑡), remaining in the capacitor at time 𝑡. The initial charge is 𝑄0 and 𝑄(𝑡) is related to 𝐼(𝑡) by 𝑄′ (𝑡) = −𝐼(𝑡).

155. (a) Between 2005 and 2015, ACME Widgets sold widgets at a continuous rate of 𝑅 = 𝑅0 𝑒0.125𝑡 widgets per year, where 𝑡 is time in years since January 1, 2005. Suppose they were selling widgets at a rate of 1000 per year on January 1, 2005. How many widgets did they sell between 2005 and 2015? How many did they sell if the rate on January 1, 2005 was 1,000,000 widgets per year? (b) In the first case (1000 widgets per year on January 1, 2005), how long did it take for half the widgets in the ten-year period to be sold? In the second case (1,000,000 widgets per year on January 1, 2005), when had half the widgets in the ten-year period been sold? (c) In 2015, ACME advertised that half the widgets it had sold in the previous ten years were still in use. Based on your answer to part (b), how long must a widget last in order to justify this claim? 156. The rate at which water is flowing into a tank is 𝑟(𝑡) gallons/minute, with 𝑡 in minutes. (a) Write an expression approximating the amount of water entering the tank during the interval from time 𝑡 to time 𝑡 + Δ𝑡, where Δ𝑡 is small. (b) Write a Riemann sum approximating the total amount of water entering the tank between 𝑡 = 0 and 𝑡 = 5. Write an exact expression for this amount. (c) By how much has the amount of water in the tank changed between 𝑡 = 0 and 𝑡 = 5 if 𝑟(𝑡) = 20𝑒0.02𝑡 ? (d) If 𝑟(𝑡) is as in part (c), and if the tank contains 3000 gallons initially, find a formula for 𝑄(𝑡), the amount of water in the tank at time 𝑡.

Strengthen Your Understanding In Problems 157–159, explain what is wrong with the statement. 157. ∫ (𝑓 (𝑥))2 𝑑𝑥 = (𝑓 (𝑥))3 ∕3 + 𝐶 . 158. ∫ cos(𝑥2 ) 𝑑𝑥 = sin(𝑥2 )∕(2𝑥) + 𝐶. 159.

𝜋∕2 ∫0

cos(3𝑥) 𝑑𝑥 =

𝜋∕2 (1∕3) ∫0

cos 𝑤 𝑑𝑤.

In Problems 160–161, give an example of: 160. A possible 𝑓 (𝜃) so that the following integral can be integrated by substitution: ∫

𝑓 (𝜃)𝑒cos 𝜃 𝑑𝜃.

2 www.indexmundi.com, 3 Data

161. An indefinite integral involving sin(𝑥3 − 3𝑥) that can be evaluated by substitution. In Problems 162–164, decide whether the statements are true or false. Give an explanation for your answer. 162. ∫ 𝑓 ′ (𝑥) cos(𝑓 (𝑥)) 𝑑𝑥 = sin(𝑓 (𝑥)) + 𝐶. 163. ∫ (1∕𝑓 (𝑥)) 𝑑𝑥 = ln |𝑓 (𝑥)| + 𝐶.

164. ∫ 𝑡 sin(5 − 𝑡2 ) 𝑑𝑡 can be evaluated using substitution.

accessed April 20, 2015. from http://minerals.usgs.gov/minerals/pubs/commodity/ Accessed February 8, 2015.

7.2 INTEGRATION BY PARTS

7.2

353

INTEGRATION BY PARTS The method of substitution reverses the chain rule. Now we introduce integration by parts, which is based on the product rule. 𝑥𝑒𝑥 𝑑𝑥.

Example 1

Find

Solution

We are looking for a function whose derivative is 𝑥𝑒𝑥 . The product rule might lead us to guess 𝑥𝑒𝑥 , because we know that the derivative has two terms, one of which is 𝑥𝑒𝑥 :



𝑑 𝑑 𝑑 (𝑥𝑒𝑥 ) = (𝑥)𝑒𝑥 + 𝑥 (𝑒𝑥 ) = 𝑒𝑥 + 𝑥𝑒𝑥 . 𝑑𝑥 𝑑𝑥 𝑑𝑥 Of course, our guess is wrong because of the extra 𝑒𝑥 . But we can adjust our guess by subtracting 𝑒𝑥 ; this leads us to try 𝑥𝑒𝑥 − 𝑒𝑥 . Let’s check it: 𝑑 𝑑 𝑑 𝑥 (𝑥𝑒𝑥 − 𝑒𝑥 ) = (𝑥𝑒𝑥 ) − (𝑒 ) = 𝑒𝑥 + 𝑥𝑒𝑥 − 𝑒𝑥 = 𝑥𝑒𝑥 . 𝑑𝑥 𝑑𝑥 𝑑𝑥 It works, so



𝑥𝑒𝑥 𝑑𝑥 = 𝑥𝑒𝑥 − 𝑒𝑥 + 𝐶.

Example 2

Find

Solution

We guess the antiderivative is 𝜃 sin 𝜃 and use the product rule to check:



𝜃 cos 𝜃 𝑑𝜃.

𝑑(𝜃) 𝑑 𝑑 (𝜃 sin 𝜃) = sin 𝜃 + 𝜃 (sin 𝜃) = sin 𝜃 + 𝜃 cos 𝜃. 𝑑𝜃 𝑑𝜃 𝑑𝜃 To correct for the extra sin 𝜃 term, we must subtract from our original guess something whose deriva𝑑 tive is sin 𝜃. Since 𝑑𝜃 (cos 𝜃) = − sin 𝜃, we try: 𝑑 𝑑 𝑑 (𝜃 sin 𝜃 + cos 𝜃) = (𝜃 sin 𝜃) + (cos 𝜃) = sin 𝜃 + 𝜃 cos 𝜃 − sin 𝜃 = 𝜃 cos 𝜃. 𝑑𝜃 𝑑𝜃 𝑑𝜃 Thus,



𝜃 cos 𝜃 𝑑𝜃 = 𝜃 sin 𝜃 + cos 𝜃 + 𝐶.

The General Formula for Integration by Parts We can formalize the process illustrated in the last two examples in the following way. We begin with the product rule: 𝑑 (𝑢𝑣) = 𝑢′ 𝑣 + 𝑢𝑣′ 𝑑𝑥 where 𝑢 and 𝑣 are functions of 𝑥 with derivatives 𝑢′ and 𝑣′ , respectively. We rewrite this as: 𝑑 (𝑢𝑣) − 𝑢′ 𝑣 𝑢𝑣′ = 𝑑𝑥 and then integrate both sides: ∫ Since an antiderivative of

𝑢𝑣′ 𝑑𝑥 =

𝑑 (𝑢𝑣) 𝑑𝑥 − 𝑢′ 𝑣 𝑑𝑥. ∫ ∫ 𝑑𝑥

𝑑 (𝑢𝑣) is just 𝑢𝑣, we get the following formula: 𝑑𝑥

354

Chapter 7 INTEGRATION

Integration by Parts 𝑢𝑣′ 𝑑𝑥 = 𝑢𝑣 −





𝑢′ 𝑣 𝑑𝑥.

This formula is useful when the integrand can be viewed as a product and when the integral on the right-hand side is simpler than that on the left. In effect, we were using integration by parts in the previous two examples. In Example 1, we let 𝑥𝑒𝑥 = (𝑥) ⋅ (𝑒𝑥 ) = 𝑢𝑣′ , and choose 𝑢 = 𝑥 and 𝑣′ = 𝑒𝑥 . Thus, 𝑢′ = 1 and 𝑣 = 𝑒𝑥 , so (𝑒𝑥 ) 𝑑𝑥 = (𝑥) (𝑒𝑥 ) − (1) (𝑒𝑥 ) 𝑑𝑥 = 𝑥𝑒𝑥 − 𝑒𝑥 + 𝐶. (𝑥) ∫ ⏟⏟⏟ ⏟⏟⏟ ⏟⏟⏟ ⏟⏟⏟ ∫ ⏟⏟⏟ ⏟⏟⏟ 𝑢

𝑣′

𝑢

𝑣

𝑢′

𝑣

So 𝑢𝑣 represents our first guess, and ∫ 𝑢′ 𝑣 𝑑𝑥 represents the correction to our guess. Notice what would have happened if, instead of 𝑣 = 𝑒𝑥 , we took 𝑣 = 𝑒𝑥 + 𝐶1 . Then ∫

𝑥𝑒𝑥 𝑑𝑥 = 𝑥(𝑒𝑥 + 𝐶1 ) −

(𝑒𝑥 + 𝐶1 ) 𝑑𝑥 ∫ = 𝑥𝑒𝑥 + 𝐶1 𝑥 − 𝑒𝑥 − 𝐶1 𝑥 + 𝐶 = 𝑥𝑒𝑥 − 𝑒𝑥 + 𝐶,

as before. Thus, it is not necessary to include an arbitrary constant in the antiderivative for 𝑣; any antiderivative will do. What would have happened if we had picked 𝑢 and 𝑣′ the other way around? If 𝑢 = 𝑒𝑥 and ′ 𝑣 = 𝑥, then 𝑢′ = 𝑒𝑥 and 𝑣 = 𝑥2 ∕2. The formula for integration by parts then gives 𝑥2 𝑥 𝑥2 𝑥 𝑒 − ⋅ 𝑒 𝑑𝑥, ∫ 2 ∫ 2 which is true but not helpful since the integral on the right is worse than the one on the left. To use this method, we must choose 𝑢 and 𝑣′ to make the integral on the right easier to find than the integral on the left. 𝑥𝑒𝑥 𝑑𝑥 =

How to Choose 𝒖 and 𝒗′ • Whatever you let 𝑣′ be, you need to be able to find 𝑣. • It helps if 𝑢′ is simpler than 𝑢 (or at least no more complicated than 𝑢). • It helps if 𝑣 is simpler than 𝑣′ (or at least no more complicated than 𝑣′ ). If we pick 𝑣′ = 𝑥 in Example 1, then 𝑣 = 𝑥2 ∕2, which is certainly “worse” than 𝑣′ . There are some examples which don’t look like good candidates for integration by parts because they don’t appear to involve products, but for which the method works well. Such examples often involve ln 𝑥 or the inverse trigonometric functions. Here is one: 3

Example 3

Find

Solution

This does not look like a product unless we write ln 𝑥 = (1)(ln 𝑥). Then we might say 𝑢 = 1 so 𝑢′ = 0, which certainly makes things simpler. But if 𝑣′ = ln 𝑥, what is 𝑣? If we knew, we would not need integration by parts. Let’s try the other way: if 𝑢 = ln 𝑥, 𝑢′ = 1∕𝑥 and if 𝑣′ = 1, 𝑣 = 𝑥, so

∫2

ln 𝑥 𝑑𝑥.

7.2 INTEGRATION BY PARTS

3

∫2

355

3 ( ) |3 1 (ln 𝑥) (1) 𝑑𝑥 = (ln 𝑥) (𝑥) || − ⋅ (𝑥) 𝑑𝑥 ∫ 𝑥 ⏟⏟⏟ ⏟⏟⏟ ⏟⏟⏟ ⏟⏟⏟ |2 ⏟⏟⏟ 2 ⏟⏟⏟ ′ 𝑢 𝑣 𝑣 𝑢 𝑣

𝑢′

|3

|3 = 𝑥 ln 𝑥|| − 1 𝑑𝑥 = (𝑥 ln 𝑥 − 𝑥)|| |2 ∫2 |2 = 3 ln 3 − 3 − 2 ln 2 + 2 = 3 ln 3 − 2 ln 2 − 1. 3

Notice that when doing a definite integral by parts, we must remember to put the limits of integration (here 2 and 3) on the 𝑢𝑣 term (in this case 𝑥 ln 𝑥) as well as on the integral ∫ 𝑢′ 𝑣 𝑑𝑥. 𝑥6 ln 𝑥 𝑑𝑥.

Example 4

Find

Solution

View 𝑥6 ln 𝑥 as 𝑢𝑣′ where 𝑢 = ln 𝑥 and 𝑣′ = 𝑥6 . Then 𝑣 = 17 𝑥7 and 𝑢′ = 1∕𝑥, so integration by parts gives us:





𝑥6 ln 𝑥 𝑑𝑥 =

(ln 𝑥)𝑥6 𝑑𝑥 = (ln 𝑥)

∫ 1 = 𝑥7 ln 𝑥 − 7 1 = 𝑥7 ln 𝑥 − 7

(

) 1 7 1 1 7 𝑥 − 𝑥 ⋅ 𝑑𝑥 ∫ 7 7 𝑥

1 𝑥6 𝑑𝑥 7∫ 1 7 𝑥 + 𝐶. 49

In Example 4 we did not choose 𝑣′ = ln 𝑥, because it is not immediately clear what 𝑣 would be. In fact, we used integration by parts in Example 3 to find the antiderivative of ln 𝑥. Also, using 𝑢 = ln 𝑥, as we have done, gives 𝑢′ = 1∕𝑥, which can be considered simpler than 𝑢 = ln 𝑥. This shows that 𝑢 does not have to be the first factor in the integrand (here 𝑥6 ). 𝑥2 sin 4𝑥 𝑑𝑥.

Example 5

Find

Solution

If we let 𝑣′ = sin 4𝑥, then 𝑣 = − 14 cos 4𝑥, which is no worse than 𝑣′ . Also letting 𝑢 = 𝑥2 , we get 𝑢′ = 2𝑥, which is simpler than 𝑢 = 𝑥2 . Using integration by parts: ) ( ) ( 1 1 2𝑥 − cos 4𝑥 𝑑𝑥 𝑥2 sin 4𝑥 𝑑𝑥 = 𝑥2 − cos 4𝑥 − ∫ ∫ 4 4 1 2 1 = − 𝑥 cos 4𝑥 + 𝑥 cos 4𝑥 𝑑𝑥. 4 2∫



The trouble is we still have to grapple with ∫ 𝑥 cos 4𝑥 𝑑𝑥. This can be done by using integration by parts again with a new 𝑢 and 𝑣, namely 𝑢 = 𝑥 and 𝑣′ = cos 4𝑥: ) ( 1 1 sin 4𝑥 − 1 ⋅ sin 4𝑥 𝑑𝑥 𝑥 cos 4𝑥 𝑑𝑥 = 𝑥 ∫ ∫ 4 4 ( ) 1 1 1 = 𝑥 sin 4𝑥 − ⋅ − cos 4𝑥 + 𝐶 4 4 4 1 1 cos 4𝑥 + 𝐶. = 𝑥 sin 4𝑥 + 4 16

356

Chapter 7 INTEGRATION

Thus, ∫

1 𝑥2 sin 4𝑥 𝑑𝑥 = − 𝑥2 cos 4𝑥 + 4 1 = − 𝑥2 cos 4𝑥 + 4 1 = − 𝑥2 cos 4𝑥 + 4

1 𝑥 cos 4𝑥 𝑑𝑥 2∫ ( ) 1 1 1 𝑥 sin 4𝑥 + cos 4𝑥 + 𝐶 2 4 16 1 1 𝑥 sin 4𝑥 + cos 4𝑥 + 𝐶. 8 32

Notice that, in this example, each time we used integration by parts, the exponent of 𝑥 went down by 1. In addition, when the arbitrary constant 𝐶 is multiplied by 12 , it is still represented by 𝐶.

cos2 𝜃 𝑑𝜃.

Example 6

Find

Solution

Using integration by parts with 𝑢 = cos 𝜃, 𝑣′ = cos 𝜃 gives 𝑢′ = − sin 𝜃, 𝑣 = sin 𝜃, so we get





cos2 𝜃 𝑑𝜃 = cos 𝜃 sin 𝜃 +



sin2 𝜃 𝑑𝜃.

Substituting sin2 𝜃 = 1 − cos2 𝜃 leads to ∫

cos2 𝜃 𝑑𝜃 = cos 𝜃 sin 𝜃 + = cos 𝜃 sin 𝜃 +

∫ ∫

(1 − cos2 𝜃) 𝑑𝜃 1 𝑑𝜃 −



cos2 𝜃 𝑑𝜃.

Looking at the right side, we see that the original integral has reappeared. If we move it to the left, we get 2



cos2 𝜃 𝑑𝜃 = cos 𝜃 sin 𝜃 +



1 𝑑𝜃 = cos 𝜃 sin 𝜃 + 𝜃 + 𝐶.

Dividing by 2 gives ∫

cos2 𝜃 𝑑𝜃 =

1 1 cos 𝜃 sin 𝜃 + 𝜃 + 𝐶. 2 2

Problem 59 asks you to do this integral by another method. The previous example illustrates a useful technique: Use integration by parts to transform the integral into an expression containing another copy of the same integral, possibly multiplied by a coefficient, then solve for the original integral. 𝑒2𝑥 sin(3𝑥) 𝑑𝑥.

Example 7

Use integration by parts twice to find

Solution

Using integration by parts with 𝑢 = 𝑒2𝑥 and 𝑣′ = sin(3𝑥) gives 𝑢′ = 2𝑒2𝑥 , 𝑣 = − 13 cos(3𝑥), so we get



1 2 𝑒2𝑥 sin(3𝑥) 𝑑𝑥 = − 𝑒2𝑥 cos(3𝑥) + 𝑒2𝑥 cos(3𝑥) 𝑑𝑥. 3 3∫ On the right side we have an integral similar to the original one, with the sine replaced by a cosine. Using integration by parts on that integral in the same way gives ∫



𝑒2𝑥 cos(3𝑥) 𝑑𝑥 =

2 1 2𝑥 𝑒 sin(3𝑥) − 𝑒2𝑥 sin(3𝑥) 𝑑𝑥. 3 3∫

7.2 INTEGRATION BY PARTS

357

Substituting this into the expression we obtained for the original integral gives ( ) 1 2 1 2𝑥 2 𝑒 sin(3𝑥) − 𝑒2𝑥 sin(3𝑥) 𝑑𝑥 = − 𝑒2𝑥 cos(3𝑥) + 𝑒2𝑥 sin(3𝑥) 𝑑𝑥 ∫ 3 3 3 3∫ 2 4 1 𝑒2𝑥 sin(3𝑥) 𝑑𝑥. = − 𝑒2𝑥 cos(3𝑥) + 𝑒2𝑥 sin(3𝑥) − 3 9 9∫ The right side now has a copy of the original integral, multiplied by −4∕9. Moving it to the left, we get ( ) 4 1 2 1+ 𝑒2𝑥 sin(3𝑥) 𝑑𝑥 = − 𝑒2𝑥 cos(3𝑥) + 𝑒2𝑥 sin(3𝑥). ∫ 9 3 9 Dividing through by the coefficient on the left, (1+4∕9) = 13∕9 and adding a constant of integration 𝐶, we get ( ) 1 9 2 − 𝑒2𝑥 cos(3𝑥) + 𝑒2𝑥 sin(3𝑥) + 𝐶 13 3 9 1 2𝑥 𝑒 (2 sin(3𝑥) − 3 cos(3𝑥)) + 𝐶. = 13

𝑒2𝑥 sin(3𝑥) 𝑑𝑥 =



Example 8

Use a computer algebra system to investigate ∫ sin(𝑥2 ) 𝑑𝑥.

Solution

It can be shown that sin(𝑥2 ) has no elementary antiderivative. A computer algebra system gives an antiderivative involving a non-elementary function, the Fresnel Integral, which you may not recognize.

Exercises and Problems for Section 7.2 Online Resource: Additional Problems for Section 7.2 EXERCISES

1. Use integration by parts to express ∫ 𝑥2 𝑒𝑥 𝑑𝑥 in terms of (a)



𝑥3 𝑒𝑥 𝑑𝑥

(b)



𝑥𝑒𝑥 𝑑𝑥

15. 17.

∫ ∫

(ln 𝑡)2 𝑑𝑡

16.

√ 𝑦 𝑦 + 3 𝑑𝑦

18.

(𝜃 +1) sin(𝜃 +1) 𝑑𝜃

20.

2. Write arctan 𝑥 = 1 ⋅ arctan 𝑥 to find ∫ arctan 𝑥 𝑑𝑥. 19. Find the integrals in Exercises 3–38. 3. 5. 7. 9. 11. 13.



𝑡 sin 𝑡 𝑑𝑡 5𝑡

∫ ∫ ∫ ∫ ∫

𝑡𝑒 𝑑𝑡

4. 6.

𝑝𝑒−0.1𝑝 𝑑𝑝

8.

𝑥 ln 𝑥 𝑑𝑥

10.

𝑞 5 ln 5𝑞 𝑑𝑞

12.

sin2 𝜃 𝑑𝜃

14.



𝑡2 sin 𝑡 𝑑𝑡

21.



ln 𝑥 ∫

𝑑𝑥

𝑥2

22.

2 5𝑡

∫ ∫ ∫ ∫ ∫

𝑡 𝑒 𝑑𝑡 23. (𝑧 + 1)𝑒2𝑧 𝑑𝑧 𝑥3 ln 𝑥 𝑑𝑥

25.

𝜃 2 cos 3𝜃 𝑑𝜃

27.

cos2 (3𝛼 + 1) 𝑑𝛼

29.

𝑡+7 𝑑𝑡 √ ∫ 5−𝑡 ∫ ∫ ∫

24.

√ 𝑥 ln 𝑥 𝑑𝑥

26.

𝑟(ln 𝑟)2 𝑑𝑟

28.

arctan 7𝑧 𝑑𝑧

30.

∫ ∫

ln(𝑥2 ) 𝑑𝑥 √ (𝑡 + 2) 2 + 3𝑡 𝑑𝑡

𝑧 𝑑𝑧 ∫ 𝑒𝑧 𝑦 𝑑𝑦 √ ∫ 5−𝑦 ∫

∫ ∫ ∫

𝑥(ln 𝑥)4 𝑑𝑥

√ 𝑦 1 − 𝑦 𝑑𝑦 arcsin 𝑤 𝑑𝑤 𝑥 arctan 𝑥2 𝑑𝑥

358 31. 33. 35.

37.

Chapter 7 INTEGRATION 2

∫ ∫ ∫



𝑥3 𝑒𝑥 𝑑𝑥

32.

𝑥 sinh 𝑥 𝑑𝑥 √

𝑒

𝑥

34. 36.

𝑑𝑥

𝑥5 cos 𝑥3 𝑑𝑥

∫ ∫

(𝑥 − 1) cosh 𝑥 𝑑𝑥 𝜃 5 cos 𝜃 3 𝑑𝜃



(2𝑥 + 1)2 ln(2𝑥 + 1) 𝑑𝑥

1

43.

∫0

1

40.

3

𝑧𝑒−𝑧 𝑑𝑧

42.

∫1

∫0

𝑢 arcsin 𝑢2 𝑑𝑢

𝑥𝑒𝑥 𝑑𝑥



(d)

1 𝑑𝑥 ∫ √3𝑥 + 1

(e) (g)

𝑥2 𝑑𝑥 ∫ 1 + 𝑥3

(b)

2

(c)

𝑥 cos 𝑥 𝑑𝑥

∫3

𝑥 sin 𝑥 𝑑𝑥



5

ln 𝑡 𝑑𝑡 10

∫0

46.

(a)

5

41.

arcsin 𝑧 𝑑𝑧

ln(1 + 𝑡) 𝑑𝑡

∫0

47. For each of the following integrals, indicate whether integration by substitution or integration by parts is more appropriate. Do not evaluate the integrals.

Evaluate the integrals in Exercises 39–46 both exactly [e.g. ln(3𝜋)] and numerically [e.g. ln(3𝜋) ≈ 2.243].

∫1

44.

1

45.

𝑒1∕𝑥 38. 𝑑𝑥 ∫ 𝑥3

39.

∫0

5

arctan 𝑦 𝑑𝑦

(f)

𝑥2 cos(𝑥3 ) 𝑑𝑥

∫ ∫

𝑥2 sin 𝑥 𝑑𝑥

ln 𝑥 𝑑𝑥

∫ 2

𝑡 ln 𝑡 𝑑𝑡

2

48. Find ∫1 ln 𝑥 𝑑𝑥 numerically. Find ∫1 ln 𝑥 𝑑𝑥 using antiderivatives. Check that your answers agree.

PROBLEMS In Problems 49–51, using properties of ln, find a substitution 𝑤 and constant 𝑘 so that the integral has the form ∫ 𝑘 ln 𝑤 𝑑𝑤. ) ( ( ) 1 2 𝑑𝑥 49. ln (5 − 3𝑥) 𝑑𝑥 50. ln √ ∫ ∫ 4 − 5𝑥 ) ( ln (ln 𝑥)3 51. 𝑑𝑥 ∫ 𝑥

63. Use the results from Problems 60 and 61 and integration by parts to find ∫ 𝜃𝑒𝜃 cos 𝜃 𝑑𝜃. 64. If 𝑓 is a twice differentiable function, find 𝑓 (𝑥) 𝑑𝑥 𝑥2 (Your answer should contain 𝑓 , but no integrals.) 65. If 𝑓 is a twice differentiable function, find ∫ 𝑥𝑓 ′′ (𝑥) 𝑑𝑥. (Your answer should contain 𝑓 , but no integrals.) ∫

𝑓 ′′ (𝑥) ln 𝑥 𝑑𝑥 +



In Problems 66–69, derive the given formulas. In Problems 52–57, find the exact area. −𝑡

52. Under 𝑦 = 𝑡𝑒 for 0 ≤ 𝑡 ≤ 2. 53. Under 𝑓 (𝑧) = arctan 𝑧 for 0 ≤ 𝑧 ≤ 2. 54. Under 𝑓 (𝑦) = arcsin 𝑦 for 0 ≤ 𝑦 ≤ 1. 55. Between 𝑦 = ln 𝑥 and 𝑦 = ln(𝑥2 ) for 1 ≤ 𝑥 ≤ 2. 56. Between 𝑓 (𝑡) = ln(𝑡2 − 1) and 𝑔(𝑡) = ln(𝑡 − 1) for 2 ≤ 𝑡 ≤ 3. 57. Under the first arch of 𝑓 (𝑥) = 𝑥 sin 𝑥. 58. In Exercise 13, you evaluated ∫ sin2 𝜃 𝑑𝜃 using integration by parts. (If you did not do it by parts, do so now!) Redo this integral using the identity sin2 𝜃 = (1 − cos 2𝜃)∕2. Explain any differences in the form of the answer obtained by the two methods. 59. Compute ∫ cos2 𝜃 𝑑𝜃 in two different ways and explain any differences in the form of your answers. (The identity cos2 𝜃 = (1 + cos 2𝜃)∕2 may be useful.) 60. Use integration by parts twice to find ∫ 𝑒𝑥 sin 𝑥 𝑑𝑥. 61. Use integration by parts twice to find ∫ 𝑒𝜃 cos 𝜃 𝑑𝜃. 62. Use the results from Problems 60 and 61 and integration by parts to find ∫ 𝑥𝑒𝑥 sin 𝑥 𝑑𝑥.

66.



𝑥𝑛 𝑒𝑥 𝑑𝑥 = 𝑥𝑛 𝑒𝑥 − 𝑛



𝑥𝑛−1 𝑒𝑥 𝑑𝑥

1 𝑛 𝑛 𝑥𝑛−1 sin 𝑎𝑥 𝑑𝑥 𝑥 sin 𝑎𝑥 − 𝑎 𝑎∫ 1 𝑛 𝑥𝑛−1 cos 𝑎𝑥 𝑑𝑥 68. 𝑥𝑛 sin 𝑎𝑥 𝑑𝑥 = − 𝑥𝑛 cos 𝑎𝑥 + ∫ 𝑎 𝑎∫ 1 𝑛−1 cos𝑛−2 𝑥 𝑑𝑥 69. cos𝑛 𝑥 𝑑𝑥 = cos𝑛−1 𝑥 sin 𝑥 + ∫ 𝑛 𝑛 ∫

67.



𝑥𝑛 cos 𝑎𝑥 𝑑𝑥 =

70. Integrating 𝑒𝑎𝑥 sin 𝑏𝑥 by parts twice gives ∫

𝑒𝑎𝑥 sin 𝑏𝑥 𝑑𝑥 = 𝑒𝑎𝑥 (𝐴 sin 𝑏𝑥 + 𝐵 cos 𝑏𝑥) + 𝐶.

(a) Find the constants 𝐴 and 𝐵 in terms of 𝑎 and 𝑏. [Hint: Don’t actually perform the integration.] (b) Evaluate ∫ 𝑒𝑎𝑥 cos 𝑏𝑥 𝑑𝑥 by modifying the method in part (a). [Again, do not perform the integration.] 5

71. Use the table with 𝑓 (𝑥) = 𝐹 ′ (𝑥) to find

𝑥𝑓 ′ (𝑥) 𝑑𝑥.

∫0

0

1

2

3

4

5

𝑓 (𝑥)

2

−5

−6

−1

10

27

𝐹 (𝑥)

10

8

2

−2

2

20

𝑥

7.2 INTEGRATION BY PARTS 3

3

72. Find ∫2 𝑓 (𝑥)𝑔 ′ (𝑥) 𝑑𝑥, given that ∫2 𝑓 ′ (𝑥)𝑔(𝑥) 𝑑𝑥 = 1.3 and the values in the table: 𝑥

𝑓 (𝑥)

𝑓 ′ (𝑥)

𝑔(𝑥)

𝑔 ′ (𝑥)

2

5

−1

0.2

3

3

7

−2

0.1

2

73. Find possible formulas for 𝑓 (𝑥) and 𝑔(𝑥) given that ∫

𝑥3 𝑔 ′ (𝑥) 𝑑𝑥 = 𝑓 (𝑥)𝑔(𝑥) −



𝑥2 cos 𝑥 𝑑𝑥.

359

78. The concentration, 𝐶, in ng/ml, of a drug in the blood as a function of the time, 𝑡, in hours since the drug was administered is given by 𝐶 = 15𝑡𝑒−0.2𝑡 . The area under the concentration curve is a measure of the overall effect of the drug on the body, called the bioavailability. Find the bioavailability of the drug between 𝑡 = 0 and 𝑡 = 3. 79. Given ℎ(𝑥) = 𝑓 (𝑥) ln |𝑥| and 𝑔 ′ (𝑥) = ∫

𝑓 (𝑥) , rewrite 𝑥

𝑓 ′ (𝑥) ln |𝑥| 𝑑𝑥. in terms of ℎ(𝑥) and 𝑔(𝑥).

80. The error function, erf(𝑥), is defined by 74. Let 𝑓 be a function with 𝑓 (0) = 6, 𝑓 (1) = 5, and 1 𝑓 ′ (1) = 2. Evaluate the integral ∫0 𝑥𝑓 ′′ (𝑥) 𝑑𝑥. √ √ 75. Given ℎ(𝑥) = 𝑓 (𝑥) 𝑥 and 𝑔 ′ (𝑥) = 𝑓 (𝑥)∕ 𝑥, rewrite in terms of ℎ(𝑥) and 𝑔(𝑥): ∫

√ 𝑓 ′ (𝑥) 𝑥 𝑑𝑥.

(a) Let 𝑢 = erf(𝑥). Use integration by parts to write ∫

Your answer should not include integrals, 𝑓 (𝑥), ℎ′ (𝑥), or 𝑔 ′ (𝑥). 7

76. Given that 𝑓 (7) = 0 and ∫0 𝑓 (𝑥) 𝑑𝑥 = 5, evaluate 7

∫0

𝑥

2 2 𝑒−𝑡 𝑑𝑡. erf (𝑥) = √ ∫ 𝜋 0

𝑣 𝑢′ 𝑑𝑥. Give 𝑢′ and 𝑣′ .

𝑣 𝑢′ 𝑑𝑥 from part (a) by ∫ making a substitution 𝑤. Give the values of 𝑤 and 𝑑𝑤. (c) Use your answers to parts (a) and (b) to find ∫

(a) Find a formula for 𝐹 (𝑎). (b) Is 𝐹 an increasing or decreasing function? (c) Is 𝐹 concave up or concave down for 0 < 𝑎 < 2?



(b) Evaluate the integral

𝑥𝑓 ′ (𝑥) 𝑑𝑥.

77. Let 𝐹 (𝑎) be the area under the graph of 𝑦 = 𝑥2 𝑒−𝑥 between 𝑥 = 0 and 𝑥 = 𝑎, for 𝑎 > 0.

erf(𝑥) 𝑑𝑥 = 𝑢𝑣 −

erf(𝑥) 𝑑𝑥. Your answer may involve erf(𝑥).

81. The Eulerian logarithmic integral Li(𝑥) is defined4 as 𝑥 1 𝑑𝑡. Letting 𝑢 = Li(𝑥) and 𝑣 = ln 𝑥, use Li(𝑥) = ∫2 ln 𝑡 integration by parts to evaluate



Li(𝑥)𝑥−1 𝑑𝑥. Your

answer will involve Li(𝑥).

Strengthen Your Understanding In Problems 82–84, explain what is wrong with the statement.

86. An integral that requires three applications of integration by parts.

82. To integrate ∫ 𝑡 ln 𝑡 𝑑𝑡 by parts, use 𝑢 = 𝑡, 𝑣′ = ln 𝑡. 83. The integral ∫ arctan 𝑥 𝑑𝑥 cannot be evaluated using integration by parts since the integrand is not a product of two functions. 84. Using integration by parts, we can show that

87. An integral of the form ∫ 𝑓 (𝑥)𝑔(𝑥) 𝑑𝑥 that can be evaluated using integration by parts either with 𝑢 = 𝑓 (𝑥) or with 𝑢 = 𝑔(𝑥).



𝑓 (𝑥) 𝑑𝑥 = 𝑥𝑓 ′ (𝑥) −



In Problems 88–90, decide whether the statements are true or false. Give an explanation for your answer.

𝑥𝑓 ′ (𝑥) 𝑑𝑥. 88. ∫ 𝑡 sin(5 − 𝑡) 𝑑𝑡 can be evaluated by parts.

In Problems 85–87, give an example of:

89. The integral ∫ 𝑡2 𝑒3−𝑡 𝑑𝑡 can be done by parts.

85. An integral using only powers of 𝜃 and sin 𝜃 which can be evaluated using integration by parts twice.

90. When integrating by parts, it does not matter which factor we choose for 𝑢.

4 http://en.wikipedia.org/wiki/Logarithmic_integral_function#Offset_logarithmic_integral,

accessed February 17, 2011.

360

7.3

Chapter 7 INTEGRATION

TABLES OF INTEGRALS Today, many integrals are done using a CAS. Traditionally, the antiderivatives of commonly used functions were compiled in a table, such as the one in the back of this book. Other tables include CRC Standard Mathematical Tables (Boca Raton, Fl: CRC Press). The key to using these tables is being able to recognize the general class of function that you are trying to integrate, so you can know in what section of the table to look. Warning: This section involves long division of polynomials and completing the square. You may want to review these topics!

Using the Table of Integrals Part I of the table inside the back cover gives the antiderivatives of the basic functions 𝑥𝑛 , 𝑎𝑥 , ln 𝑥, sin 𝑥, cos 𝑥, and tan 𝑥. (The antiderivative for ln 𝑥 is found using integration by parts and is a special case of the more general formula III-13.) Most of these are already familiar. Part II of the table contains antiderivatives of functions involving products of 𝑒𝑥 , sin 𝑥, and cos 𝑥. All of these antiderivatives were obtained using integration by parts. Example 1

Find

Solution

Since the integrand is the product of two sines, we should use II-10 in the table,



sin 7𝑧 sin 3𝑧 𝑑𝑧.



sin 7𝑧 sin 3𝑧 𝑑𝑧 = −

1 (7 cos 7𝑧 sin 3𝑧 − 3 cos 3𝑧 sin 7𝑧) + 𝐶. 40

Part III of the table contains antiderivatives for products of a polynomial and 𝑒𝑥 , sin 𝑥, or cos 𝑥. It also has an antiderivative for 𝑥𝑛 ln 𝑥, which can easily be used to find the antiderivatives of the product of a general polynomial and ln 𝑥. Each reduction formula is used repeatedly to reduce the degree of the polynomial until a zero-degree polynomial is obtained. (𝑥5 + 2𝑥3 − 8)𝑒3𝑥 𝑑𝑥.

Example 2

Find

Solution

Since 𝑝(𝑥) = 𝑥5 + 2𝑥3 − 8 is a polynomial multiplied by 𝑒3𝑥 , this is of the form in III-14. Now 𝑝′ (𝑥) = 5𝑥4 + 6𝑥2 and 𝑝′′ (𝑥) = 20𝑥3 + 12𝑥, and so on, giving





(𝑥5 + 2𝑥3 − 8)𝑒3𝑥 𝑑𝑥 = 𝑒3𝑥

1 5 1 1 (𝑥 + 2𝑥3 − 8) − (5𝑥4 + 6𝑥2 ) + (20𝑥3 + 12𝑥) 3 9 27 ) 1 1 1 2 (120𝑥) − ⋅ 120 + 𝐶. − (60𝑥 + 12) + 81 243 729

(

Here we have the successive derivatives of the original polynomial 𝑥5 + 2𝑥3 − 8, occurring with alternating signs and multiplied by successive powers of 1/3. Part IV of the table contains reduction formulas for the antiderivatives of cos𝑛 𝑥 and sin𝑛 𝑥, which can be obtained by integration by parts. When 𝑛 is a positive integer, formulas IV-17 and IV-18 can be used repeatedly to reduce the power 𝑛 until it is 0 or 1. Example 3

Find



sin6 𝜃 𝑑𝜃.

7.3 TABLES OF INTEGRALS

Solution

361

Use IV-17 repeatedly: 1 5 sin6 𝜃 𝑑𝜃 = − sin5 𝜃 cos 𝜃 + sin4 𝜃 𝑑𝜃 6 6∫ 1 3 sin4 𝜃 𝑑𝜃 = − sin3 𝜃 cos 𝜃 + sin2 𝜃 𝑑𝜃 ∫ 4 4∫ 1 1 sin2 𝜃 𝑑𝜃 = − sin 𝜃 cos 𝜃 + 1 𝑑𝜃. ∫ 2 2∫



Calculate ∫ sin2 𝜃 𝑑𝜃 first, and use this to find ∫ sin4 𝜃 𝑑𝜃; then calculate ∫ sin6 𝜃 𝑑𝜃. Putting this all together, we get ∫

15 1 5 15 sin6 𝜃 𝑑𝜃 = − sin5 𝜃 cos 𝜃 − sin3 𝜃 cos 𝜃 − sin 𝜃 cos 𝜃 + 𝜃 + 𝐶. 6 24 48 48

The last item in Part IV of the table is not a formula: it is advice on how to antidifferentiate products of integer powers of sin 𝑥 and cos 𝑥. There are various techniques to choose from, depending on the nature (odd or even, positive or negative) of the exponents. cos3 𝑡 sin4 𝑡 𝑑𝑡.

Example 4

Find

Solution

Here the exponent of cos 𝑡 is odd, so IV-23 recommends making the substitution 𝑤 = sin 𝑡. Then 𝑑𝑤 = cos 𝑡 𝑑𝑡. To make this work, we’ll have to separate off one of the cosines to be part of 𝑑𝑤. Also, the remaining even power of cos 𝑡 can be rewritten in terms of sin 𝑡 by using cos2 𝑡 = 1−sin2 𝑡 = 1 − 𝑤2 , so that





cos3 𝑡 sin4 𝑡 𝑑𝑡 =



cos2 𝑡 sin4 𝑡 cos 𝑡 𝑑𝑡

(1 − 𝑤2 )𝑤4 𝑑𝑤 = (𝑤4 − 𝑤6 ) 𝑑𝑤 ∫ ∫ 1 1 1 1 = 𝑤5 − 𝑤7 + 𝐶 = sin5 𝑡 − sin7 𝑡 + 𝐶. 5 7 5 7 =

cos2 𝑥 sin4 𝑥 𝑑𝑥.

Example 5

Find

Solution

In this example, both exponents are even. The advice given in IV-23 is to convert to all sines or all cosines. We’ll convert to all sines by substituting cos2 𝑥 = 1 − sin2 𝑥, and then we’ll multiply out the integrand:





cos2 𝑥 sin4 𝑥 𝑑𝑥 =



(1 − sin2 𝑥) sin4 𝑥 𝑑𝑥 =



sin4 𝑥 𝑑𝑥 −



sin6 𝑥 𝑑𝑥.

In Example 3 we found ∫ sin4 𝑥 𝑑𝑥 and ∫ sin6 𝑥 𝑑𝑥. Put them together to get ∫

3 1 3 cos2 𝑥 sin4 𝑥 𝑑𝑥 = − sin3 𝑥 cos 𝑥 − sin 𝑥 cos 𝑥 + 𝑥 4 8 8 ) ( 15 5 15 1 sin3 𝑥 cos 𝑥 − sin 𝑥 cos 𝑥 + 𝑥 + 𝐶 − − sin5 𝑥 cos 𝑥 − 6 24 48 48 3 1 5 1 3 3 sin 𝑥 cos 𝑥 − sin 𝑥 cos 𝑥 + 𝑥 + 𝐶. = sin 𝑥 cos 𝑥 − 6 24 48 48

362

Chapter 7 INTEGRATION

The last two parts of the table are concerned with quadratic functions: Part V has expressions with quadratic denominators; Part VI contains square roots of quadratics. The quadratics that appear in these formulas are of the form 𝑥2 ± 𝑎2 or 𝑎2 − 𝑥2 , or in factored form (𝑥 − 𝑎)(𝑥 − 𝑏), where 𝑎 and 𝑏 are different constants. Quadratics can be converted to these forms by factoring or completing the square.

Preparing to Use the Table: Transforming the Integrand To use the integral table, we often need to manipulate or reshape integrands to fit entries in the table. The manipulations that tend to be useful are factoring, long division, completing the square, and substitution.

Factoring 3𝑥 + 7 𝑑𝑥. ∫ 𝑥2 + 6𝑥 + 8

Example 6

Find

Solution

In this case we factor the denominator to get it into a form in the table: 𝑥2 + 6𝑥 + 8 = (𝑥 + 2)(𝑥 + 4). Now in V-27 we let 𝑎 = −2, 𝑏 = −4, 𝑐 = 3, and 𝑑 = 7, to obtain 1 3𝑥 + 7 𝑑𝑥 = (ln |𝑥 + 2| − (−5) ln |𝑥 + 4|) + 𝐶. ∫ 𝑥2 + 6𝑥 + 8 2

Long Division 𝑥2 𝑑𝑥. ∫ 𝑥2 + 4

Example 7

Find

Solution

A good rule of thumb when integrating a rational function whose numerator has a degree greater than or equal to that of the denominator is to start by doing long division. This results in a polynomial plus a simpler rational function as a remainder. Performing long division here, we obtain: 𝑥2 4 =1− . +4 𝑥2 + 4

𝑥2 Then, by V-24 with 𝑎 = 2, we obtain:

𝑥 1 1 𝑥2 𝑑𝑥 = 1 𝑑𝑥 − 4 𝑑𝑥 = 𝑥 − 4 ⋅ arctan + 𝐶. ∫ ∫ 𝑥2 + 4 ∫ 𝑥2 + 4 2 2

Completing the Square to Rewrite the Quadratic in the Form 𝒘𝟐 + 𝒂𝟐 1 𝑑𝑥. ∫ 𝑥2 + 6𝑥 + 14

Example 8

Find

Solution

By completing the square, we can get this integrand into a form in the table: 𝑥2 + 6𝑥 + 14 = (𝑥2 + 6𝑥 + 9) − 9 + 14 = (𝑥 + 3)2 + 5.

7.3 TABLES OF INTEGRALS

363

Let 𝑤 = 𝑥 + 3. Then 𝑑𝑤 = 𝑑𝑥 and so the substitution gives 1 1 1 𝑤 1 𝑥+3 𝑑𝑥 = 𝑑𝑤 = √ arctan √ + 𝐶 = √ arctan √ + 𝐶, ∫ 𝑥2 + 6𝑥 + 14 ∫ 𝑤2 + 5 5 5 5 5

where the antidifferentiation uses V-24 with 𝑎2 = 5.

Substitution Getting an integrand into the right form to use a table of integrals involves substitution and a variety of algebraic techniques. 𝑒𝑡 sin(5𝑡 + 7) 𝑑𝑡.

Example 9

Find

Solution

This looks similar to II-8. To make the correspondence more complete, let’s try the substitution 𝑤 = 5𝑡 + 7. Then 𝑑𝑤 = 5 𝑑𝑡, so 𝑑𝑡 = 15 𝑑𝑤. Also, 𝑡 = (𝑤 − 7)∕5. Then the integral becomes





𝑒𝑡 sin(5𝑡 + 7) 𝑑𝑡 = =



𝑑𝑤 5

𝑒−7∕5 𝑒𝑤∕5 sin 𝑤 𝑑𝑤. 5 ∫

Now we can use II-8 with 𝑎 =



𝑒(𝑤−7)∕5 sin 𝑤

1 5

(Since 𝑒(𝑤−7)∕5 = 𝑒𝑤∕5 𝑒−7∕5 and 𝑒−7∕5 is a constant)

and 𝑏 = 1 to write

𝑒𝑤∕5 sin 𝑤 𝑑𝑤 =

) ( sin 𝑤 1 − cos 𝑤 + 𝐶, 𝑒𝑤∕5 5 (1∕5)2 + 12

so ∫

Example 10

Solution

( )) ( 𝑒−7∕5 25 (5𝑡+7)∕5 sin(5𝑡 + 7) 𝑒 − cos(5𝑡 + 7) +𝐶 5 26 5 ( ) 5𝑒𝑡 sin(5𝑡 + 7) − cos(5𝑡 + 7) + 𝐶. = 26 5

𝑒𝑡 sin(5𝑡 + 7) 𝑑𝑡 =

Find a substitution 𝑤 and constants 𝑘, 𝑛 so that the following integral has the form ∫ 𝑘𝑤𝑛 ln 𝑤 𝑑𝑤, found in III-13: ln(𝑥 + 1) + ln(𝑥 − 1) 𝑥 𝑑𝑥 √ ∫ 𝑥2 − 1 First we use properties of ln to simplify the integral:

ln((𝑥 + 1)(𝑥 − 1)) ln(𝑥2 − 1) ln(𝑥 + 1) + ln(𝑥 − 1) 𝑥 𝑑𝑥 = 𝑥 𝑑𝑥 = 𝑥 𝑑𝑥. √ √ ∫ ∫ √ 2 ∫ 𝑥2 − 1 𝑥2 − 1 𝑥 −1

Let 𝑤 = 𝑥2 − 1, 𝑑𝑤 = 2𝑥 𝑑𝑥, so that 𝑥 𝑑𝑥 = (1∕2) 𝑑𝑤. Then

ln(𝑥2 − 1) 1 −1∕2 1 𝑤 ln 𝑤 𝑑𝑤, 𝑥 𝑑𝑥 = 𝑤−1∕2 ln 𝑤 𝑑𝑤 = √ ∫ ∫ ∫ 2 2 𝑥2 − 1 so 𝑘 = 1∕2, 𝑛 = −1∕2.

364

Chapter 7 INTEGRATION

Exercises and Problems for Section 7.3 Online Resource: Additional Problems for Section 7.3 EXERCISES

For Exercises 1–14, say which formula, if any, to apply from the table of integrals. Give the values of any constants. 1. 3. 5.

∫ ∫ ∫

𝑥1∕2 𝑒3𝑥 𝑑𝑥

2.

sin4 𝑥 𝑑𝑥

4.

𝑒𝑥 ln 𝑥 𝑑𝑥

6.

1 𝑑𝑥 ∫ 𝑥2 − 3𝑥 − 4 √ 9. 𝑥2 − 9 𝑑𝑥 ∫ 7.

11. 13.



sin 𝑒𝑥 𝑑𝑥

8. 10. 12.

4𝑥 − 2 𝑑𝑥 ∫ 𝑥2 + 9

14.

∫ ∫ ∫ ∫ ∫ ∫

cos 3𝑥 sin 4𝑥 𝑑𝑥 𝑒5𝑥 sin 2𝑥 𝑑𝑥 ln 𝑥 𝑑𝑥 cos 𝑥 𝑑𝑥 𝑥 √ 𝑥3 + 8 𝑑𝑥

17. 19. 21. 23. 25. 27.

29. 31. 33.

∫ ∫ ∫ ∫ ∫ ∫

𝑥5 ln 𝑥 𝑑𝑥

16.

𝑥3 sin 5𝑥 𝑑𝑥

18.

(𝑥3 + 5)2 𝑑𝑥.

20.

sin4 𝑥 𝑑𝑥

22.

𝑥2 𝑒3𝑥 𝑑𝑥

24.

𝑥4 𝑒3𝑥 𝑑𝑥 1

∫ 3 + 𝑦2

26.

𝑑𝑦

30.

√ ∫ 25 − 16𝑥2

sin 3𝜃 cos 5𝜃 𝑑𝜃 1



∫ ∫

32. 34.

𝑑𝑥

cos3 𝑥

38.

39.

1 𝑑𝑥 ∫ cos4 7𝑥

40.

41.

1 𝑑𝜃 ∫ sin2 2𝜃

42.

43.

1 𝑑𝑥 ∫ 𝑥2 + 4𝑥 + 3

44.

1 𝑑𝑥 ∫ 𝑥2 + 4𝑥 + 4

45.

𝑑𝑧 ∫ 𝑧(𝑧 − 3)

46.

𝑑𝑦 ∫ 4 − 𝑦2



49. 51.

1 ∫ 1 + (𝑧 + 2)2 ∫ ∫

48.

𝑑𝑧

sin3 𝑥 𝑑𝑥

50.

sinh3 𝑥 cosh2 𝑥 𝑑𝑥

52.

sin3 3𝜃 cos2 3𝜃 𝑑𝜃

54.



cos 2𝑦 cos 7𝑦 𝑑𝑦 𝑥3 sin 𝑥2 𝑑𝑥

1 𝑑𝜃 ∫ sin3 3𝜃 1 ∫

𝑑𝑥.

cos5 𝑥

1 ∫ 𝑦2 + 4𝑦 + 5 ∫ ∫

𝑑𝑦

tan4 𝑥 𝑑𝑥 sinh2 𝑥 cosh3 𝑥 𝑑𝑥 2

𝑧𝑒2𝑧 cos(2𝑧2 ) 𝑑𝑧

53.

(𝑥2 + 3) ln 𝑥 𝑑𝑥.

For Exercises 55–64, evaluate the definite integrals. Whenever possible, use the Fundamental Theorem of Calculus, perhaps after a substitution. Otherwise, use numerical methods.

sin 𝑤 cos4 𝑤 𝑑𝑤 𝑥3 𝑒2𝑥 𝑑𝑥





𝜋∕12

55.

𝜋

sin(3𝛼) 𝑑𝛼

56.

(𝑥 − 2𝑥3 ) ln 𝑥 𝑑𝑥

58.

∫0

∫−𝜋

sin 5𝑥 cos 6𝑥 𝑑𝑥

3

∫ ∫

𝑥2 𝑒𝑥 𝑑𝑥 𝑢5 ln(5𝑢) 𝑑𝑢

∫1

1

1

𝑑𝑥 √ ∫ 9𝑥2 + 25 ∫

2

57.

59.

61.

∫0 𝑥2 + 2𝑥 + 1

𝑡2 + 1 𝑑𝑡 ∫ 𝑡2 − 1

𝑥 𝑑𝑥 √ 1 − 𝑥4

𝜋∕3

𝑑𝑥

∫𝜋∕4

In Problems 65–66, using properties of ln, find a substitution 𝑤 and constants 𝑘, 𝑛 so that the integral has the form 𝑘𝑤𝑛 ln 𝑤 𝑑𝑤.

65. 66.

∫ ∫

∫0

√ 3 − 𝑥2 𝑑𝑥

1

60.

𝑑𝑥

√ 1∕ 2

∫0

sin 3𝜃 sin 5𝜃 𝑑𝜃 63.

1

(𝑥 + 2) 𝑑𝑥 ∫0 (𝑥 + 2)2 + 1 −1

64.

3

sin 𝑥

(2𝑥 + 1)3 ln(2𝑥 + 1) 𝑑𝑥 (2𝑥 + 1)3 ln √

𝑑𝑥 ∫0 𝑥2 + 2𝑥 + 5 1

62.

PROBLEMS





𝑒−3𝜃 cos 𝜃 𝑑𝜃

𝑑𝑥 28. ∫ 9𝑥2 + 16

𝑑𝑥





𝑦2 sin 2𝑦 𝑑𝑦

37.

47.

4𝑥 − 2 𝑑𝑥 ∫ 𝑥2 − 9



36.



𝑥2 sin 2𝑥 𝑑𝑥

For Exercises 15–54, antidifferentiate using the table of integrals. You may need to transform the integrand first. 15.

𝑒5𝑥 sin 3𝑥 𝑑𝑥

35.

1 2𝑥 + 1

𝑑𝑥

∫−3

𝑑𝑥 √ 𝑥2 + 6𝑥 + 10

7.3 TABLES OF INTEGRALS

In Problems 67–69, find constants 𝑎, 𝑏, 𝑐, 𝑚, 𝑛 so that the integral is in one of the following forms from a table of integrals.5 Give the form (i)–(iii) you use. 𝑚𝑥 + 𝑛 𝑑𝑥 (ii) 𝑑𝑥 (i) ∫ 𝑎𝑥2 + 𝑏𝑥 + 𝑐 ∫ 𝑎𝑥2 + 𝑏𝑥 + 𝑐 𝑑𝑥 (iii) ) ,𝑛>0 ( ∫ 𝑎𝑥2 + 𝑏𝑥 + 𝑐 𝑛 67.

69.

𝑑𝑥 𝑥 𝑥2 5− − 4 6



68.

𝑑𝑥 ∫

5 2𝑥 + 7 + 3𝑥

𝑑𝑥 ∫ (𝑥2 − 5𝑥 + 6)3 (𝑥2 − 4𝑥 + 4)2 (𝑥2 − 6𝑥 + 9)2

In Problems 70–71, find constants 𝑎, 𝑏, 𝜆 so that the integral has the form found in some tables of integrals:6 𝑒2𝜆𝑥 𝑑𝑥. ∫ 𝑎𝑒𝜆𝑥 + 𝑏

70.

𝑒6𝑥 𝑑𝑥 ∫ 4 + 𝑒3𝑥+1

71.

𝑒8𝑥 𝑑𝑥 ∫ 4𝑒4𝑥 + 5𝑒6𝑥

72. According to a table of integrals,7 ( 2 ) 𝑥 2 2𝑥 𝑥2 𝑒𝑏𝑥 𝑑𝑥 = 𝑒𝑏𝑥 − 2 + 3 + 𝐶. ∫ 𝑏 𝑏 𝑏

365

(a) Find a substitution 𝑤 and constant 𝑘 so that the in2 tegral ∫ 𝑥5 𝑒𝑏𝑥 𝑑𝑥 can be rewritten in the form



𝑘𝑤2 𝑒𝑏𝑤 𝑑𝑤.

(b) Evaluate the integral in terms of 𝑥. Your answer may involve the constant 𝑏. 73. Show that for all integers 𝑚 and 𝑛, with 𝑚 ≠ ±𝑛, 𝜋 ∫−𝜋 sin 𝑚𝜃 sin 𝑛𝜃 𝑑𝜃 = 0. 74. Show that for all integers 𝑚 and 𝑛, with 𝑚 ≠ ±𝑛, 𝜋 ∫−𝜋 cos 𝑚𝜃 cos 𝑛𝜃 𝑑𝜃 = 0. 75. Water pipelines sometimes spring leaks, and water escapes until the leak is repaired. (a) At 𝑡 days after a leak is detected, water leaks from a pipeline at an estimated rate of 1 thousands of gallons per day. 𝑟(𝑡) = 1 − √ 𝑡2 + 1

By finding a derivative or looking at a graph, explain why this rate could represent a new leak. In the long run, if the leak is not fixed, what happens to 𝑟(𝑡)? (b) What is the shape of the graph of 𝑉 (𝑡), the total volume of water that has escaped by time 𝑡? (c) Find a formula for 𝑉 (𝑡).

Strengthen Your Understanding In Problems 76–80, explain what is wrong with the statement. 𝑑𝑡 . ∫ 7 − 𝑡2 77. If 𝑎 > 0, then ∫ 1∕(𝑥2 + 4𝑥 + 𝑎) 𝑑𝑥 always involves arctan. 76. The table of integrals cannot be used to find

78. By Formula II-8 of the table with 𝑎 = 1, 𝑏 = 1, 𝑒𝑥 sin 𝑥 𝑑𝑥 =



1 𝑥 𝑒 (sin 𝑥 − cos 𝑥) + 𝐶. 2

1 (𝑏 sin(𝑎𝑥) sin(𝑏𝑥) + 𝑎 cos(𝑎𝑥) cos(𝑏𝑥)) + 𝐶. 𝑏2 − 𝑎2 80. The table can be used to evaluate ∫ sin 𝑥∕𝑥 𝑑𝑥 . In Problems 81–82, give an example of: 81. An indefinite integral involving a square root that can be evaluated by first completing a square. 82. An indefinite integral involving sin 𝑥 that can be evaluated with a reduction formula

Therefore ∫

𝑒2𝑥+1 sin(2𝑥 + 1) 𝑑𝑥 = 1 2𝑥+1 𝑒 (sin(2𝑥 + 1) − cos(2𝑥 + 1)) + 𝐶. 2

79. The integral ∫ sin 𝑥 cos 𝑥 𝑑𝑥 with 𝑎 = 1, 𝑏 = 1 is undefined according to Table Formula II-12 since, for 𝑎 ≠ 𝑏, ∫

sin(𝑎𝑥) cos(𝑏𝑥) 𝑑𝑥 =

In Problems 83–86, decide whether the statements are true or false. Give an explanation for your answer. 83. ∫ sin7 𝜃 cos6 𝜃 𝑑𝜃 can be written as a polynomial with cos 𝜃 as the variable. 84. ∫ 1∕(𝑥2 + 4𝑥 + 5) 𝑑𝑥 involves a natural logarithm. 85. ∫ 1∕(𝑥2 + 4𝑥 − 5) 𝑑𝑥 involves an arctangent. 86. ∫ 𝑥−1 ((ln 𝑥)2 + (ln 𝑥)3 ) 𝑑𝑥 is a polynomial with ln 𝑥 as the variable.

5 http://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions,

page accessed February 24, 2010. page accessed May 5, 2010. 7 http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions, page accessed February 17, 2011. 6 http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions,

366

Chapter 7 INTEGRATION

7.4

ALGEBRAIC IDENTITIES AND TRIGONOMETRIC SUBSTITUTIONS Although not all functions have elementary antiderivatives, many do. In this section we introduce two powerful methods of integration which show that large classes of functions have elementary antiderivatives. The first is the method of partial fractions, which depends on an algebraic identity and allows us to integrate rational functions. The second is the method of trigonometric substitutions, which allows us to handle expressions involving the square root of a quadratic polynomial. Some of the formulas in the table of integrals can be derived using the techniques of this section.

Method of Partial Fractions The integral of some rational functions can be obtained by splitting the integrand into partial fractions. For example, to find 1 𝑑𝑥, ∫ (𝑥 − 2)(𝑥 − 5) the integrand is split into partial fractions with denominators (𝑥 − 2) and (𝑥 − 5). We write 1 𝐴 𝐵 = + , (𝑥 − 2)(𝑥 − 5) 𝑥 − 2 𝑥 − 5 where 𝐴 and 𝐵 are constants that need to be found. Multiplying by (𝑥 − 2)(𝑥 − 5) gives the identity 1 = 𝐴(𝑥 − 5) + 𝐵(𝑥 − 2) so 1 = (𝐴 + 𝐵)𝑥 − 5𝐴 − 2𝐵. Since this equation holds for all 𝑥, the constant terms on both sides must be equal.8 Similarly, the coefficients of 𝑥 on both sides must be equal. So −5𝐴 − 2𝐵 = 1 𝐴 + 𝐵 = 0. Solving these equations gives 𝐴 = −1∕3, 𝐵 = 1∕3. Thus, −1∕3 1∕3 1 = + . (𝑥 − 2)(𝑥 − 5) 𝑥 − 2 𝑥 − 5 (Check the answer by writing the right-hand side over the common denominator (𝑥 − 2)(𝑥 − 5).) 1 𝑑𝑥. ∫ (𝑥 − 2)(𝑥 − 5)

Example 1

Use partial fractions to integrate

Solution

We split the integrand into partial fractions, each of which can be integrated: ( ) −1∕3 1∕3 1 1 1 𝑑𝑥 = + 𝑑𝑥 = − ln |𝑥 − 2| + ln |𝑥 − 5| + 𝐶. ∫ ∫ (𝑥 − 2)(𝑥 − 5) 𝑥−2 𝑥−5 3 3 You can check that using formula V-26 in the integral table gives the same result.

This method can be used to derive formulas V-29 and V-30 in the integral table. A similar method works on rational functions whenever the denominator of the integrand factors into distinct linear factors and the numerator has degree less than the denominator. Example 2

Find

𝑥+2 𝑑𝑥. ∫ 𝑥2 + 𝑥

8 We

have not shown that the equation holds for 𝑥 = 2 and 𝑥 = 5, but these values do not affect the argument.

7.4 ALGEBRAIC IDENTITIES AND TRIGONOMETRIC SUBSTITUTIONS

Solution

367

We factor the denominator and split the integrand into partial fractions: 𝑥+2 𝐴 𝐵 𝑥+2 = + . = 2 𝑥(𝑥 + 1) 𝑥 𝑥 +1 𝑥 +𝑥 Multiplying by 𝑥(𝑥 + 1) gives the identity 𝑥 + 2 = 𝐴(𝑥 + 1) + 𝐵𝑥 = (𝐴 + 𝐵)𝑥 + 𝐴. Equating constant terms and coefficients of 𝑥 gives 𝐴 = 2 and 𝐴 + 𝐵 = 1, so 𝐵 = −1. Then we split the integrand into two parts and integrate: 𝑥+2 𝑑𝑥 = ∫ 𝑥2 + 𝑥 ∫

(

1 2 − 𝑥 𝑥+1

)

𝑑𝑥 = 2 ln |𝑥| − ln |𝑥 + 1| + 𝐶.

The next example illustrates what to do if there is a repeated factor in the denominator. 𝐵 𝐴 𝐶 10𝑥 − 2𝑥2 , . 𝑑𝑥 using partial fractions of the form , 2 ∫ (𝑥 − 1)2 (𝑥 + 3) 𝑥 − 1 (𝑥 − 1) 𝑥 + 3

Example 3

Calculate

Solution

We are given that the squared factor, (𝑥 − 1)2 , leads to partial fractions of the form: 𝐵 10𝑥 − 2𝑥2 𝐴 𝐶 + . = + (𝑥 − 1)2 (𝑥 + 3) 𝑥 − 1 (𝑥 − 1)2 𝑥 + 3 Multiplying through by (𝑥 − 1)2 (𝑥 + 3) gives 10𝑥 − 2𝑥2 = 𝐴(𝑥 − 1)(𝑥 + 3) + 𝐵(𝑥 + 3) + 𝐶(𝑥 − 1)2 = (𝐴 + 𝐶)𝑥2 + (2𝐴 + 𝐵 − 2𝐶)𝑥 − 3𝐴 + 3𝐵 + 𝐶. Equating the coefficients of 𝑥2 and 𝑥 and the constant terms, we get the simultaneous equations: 𝐴 + 𝐶 = −2 2𝐴 + 𝐵 − 2𝐶 = 10 −3𝐴 + 3𝐵 + 𝐶 = 0. Solving gives 𝐴 = 1, 𝐵 = 2, 𝐶 = −3. Thus, we obtain three integrals which can be evaluated: ( ) 1 2 3 10𝑥 − 2𝑥2 + 𝑑𝑥 𝑑𝑥 = − ∫ ∫ (𝑥 − 1)2 (𝑥 + 3) 𝑥 − 1 (𝑥 − 1)2 𝑥 + 3 2 − 3 ln |𝑥 + 3| + 𝐾. = ln |𝑥 − 1| − (𝑥 − 1)

For the second integral, we use the fact that ∫ 2∕(𝑥 − 1)2 𝑑𝑥 = 2 ∫ (𝑥 − 1)−2 𝑑𝑥 = −2(𝑥 − 1)−1 + 𝐾. If there is a quadratic in the denominator which cannot be factored, we need an expression of the form 𝐴𝑥 + 𝐵 in the numerator, as the next example shows.

Example 4

Find

𝐴𝑥 + 𝐵 𝐶 2𝑥2 − 𝑥 − 1 . 𝑑𝑥 using partial fractions of the form 2 and ∫ (𝑥2 + 1)(𝑥 − 2) 𝑥−2 𝑥 +1

368

Chapter 7 INTEGRATION

Solution

We are given that the quadratic denominator, (𝑥2 + 1), which cannot be factored further, has a numerator of the form 𝐴𝑥 + 𝐵, so we have 2𝑥2 − 𝑥 − 1 𝐴𝑥 + 𝐵 𝐶 . = + 𝑥−2 (𝑥2 + 1)(𝑥 − 2) 𝑥2 + 1 Multiplying by (𝑥2 + 1)(𝑥 − 2) gives 2𝑥2 − 𝑥 − 1 = (𝐴𝑥 + 𝐵)(𝑥 − 2) + 𝐶(𝑥2 + 1) = (𝐴 + 𝐶)𝑥2 + (𝐵 − 2𝐴)𝑥 + 𝐶 − 2𝐵. Equating the coefficients of 𝑥2 and 𝑥 and the constant terms gives the simultaneous equations 𝐴+𝐶 = 2 𝐵 − 2𝐴 = −1 𝐶 − 2𝐵 = −1. Solving gives 𝐴 = 𝐵 = 𝐶 = 1, so we rewrite the integral as follows: ( ) 𝑥+1 1 2𝑥2 − 𝑥 − 1 𝑑𝑥. 𝑑𝑥 = + ∫ ∫ (𝑥2 + 1)(𝑥 − 2) 𝑥2 + 1 𝑥 − 2 This identity is useful provided we can perform the integration on the right-hand side. The first integral can be done if it is split into two; the second integral is similar to those in the previous examples. We have 2𝑥2 − 𝑥 − 1 𝑥 1 1 𝑑𝑥. 𝑑𝑥 = 𝑑𝑥 + 𝑑𝑥 + ∫ (𝑥2 + 1)(𝑥 − 2) ∫ 𝑥2 + 1 ∫ 𝑥2 + 1 ∫ 𝑥−2 To calculate ∫ (𝑥∕(𝑥2 + 1)) 𝑑𝑥, substitute 𝑤 = 𝑥2 + 1, or guess and check. The final result is 1 2𝑥2 − 𝑥 − 1 𝑑𝑥 = ln |𝑥2 + 1| + arctan 𝑥 + ln |𝑥 − 2| + 𝐾. ∫ (𝑥2 + 1)(𝑥 − 2) 2 The next example shows what to do if the numerator has degree larger than the denominator. 𝑥3 − 7𝑥2 + 10𝑥 + 1 𝑑𝑥 using long division before integrating. ∫ 𝑥2 − 7𝑥 + 10

Example 5

Calculate

Solution

The degree of the numerator is greater than the degree of the denominator, so we divide first: 1 𝑥3 − 7𝑥2 + 10𝑥 + 1 𝑥(𝑥2 − 7𝑥 + 10) + 1 = =𝑥+ . 𝑥2 − 7𝑥 + 10 𝑥2 − 7𝑥 + 10 𝑥2 − 7𝑥 + 10 The remainder, in this case 1∕(𝑥2 − 7𝑥 + 10), is a rational function on which we try to use partial fractions. We have 1 1 = 2 𝑥 − 7𝑥 + 10 (𝑥 − 2)(𝑥 − 5) so in this case we use the result of Example 1 to obtain ( ) 1 𝑥2 1 1 𝑥3 − 7𝑥2 + 10𝑥 + 1 𝑥+ 𝑑𝑥 = − ln |𝑥 − 2| + ln |𝑥 − 5| + 𝐶. 𝑑𝑥 = ∫ ∫ (𝑥 − 2)(𝑥 − 5) 2 3 3 𝑥2 − 7𝑥 + 10

7.4 ALGEBRAIC IDENTITIES AND TRIGONOMETRIC SUBSTITUTIONS

369

Many, though not all, rational functions can be integrated by the strategy suggested by the previous examples.

Strategy for Integrating a Rational Function,

𝑷 (𝒙) 𝑸(𝒙)

• If degree of 𝑃 (𝑥) ≥ degree of 𝑄(𝑥), try long division and the method of partial fractions on the remainder. • If 𝑄(𝑥) is the product of distinct linear factors, use partial fractions of the form 𝐴 . (𝑥 − 𝑐) • If 𝑄(𝑥) contains a repeated linear factor, (𝑥 − 𝑐)𝑛 , use partial fractions of the form 𝐴2 𝐴1 𝐴𝑛 + +⋯+ . (𝑥 − 𝑐) (𝑥 − 𝑐)2 (𝑥 − 𝑐)𝑛 • If 𝑄(𝑥) contains an unfactorable quadratic 𝑞(𝑥), try a partial fraction of the form 𝐴𝑥 + 𝐵 . 𝑞(𝑥) To use this method, we must be able to integrate each partial fraction. We can integrate terms of the form 𝐴∕(𝑥 − 𝑐)𝑛 using the power rule (if 𝑛 > 1) and logarithms (if 𝑛 = 1). Next we see how to integrate terms of the form (𝐴𝑥 + 𝐵)∕𝑞(𝑥), where 𝑞(𝑥) is an unfactorable quadratic.

Trigonometric Substitutions Section 7.1 showed how substitutions could be used to transform complex integrands. Now we see how substitution of sin 𝜃 or tan 𝜃 can be used for integrands involving square roots of quadratics or unfactorable quadratics.

Sine Substitutions Substitutions involving√ sin 𝜃 make use of the Pythagorean identity, cos2 𝜃 + sin2 𝜃 = 1, to simplify an integrand involving 𝑎2 − 𝑥2 . 1 𝑑𝑥 using the substitution 𝑥 = sin 𝜃. ∫ √ 1 − 𝑥2

Example 6

Find

Solution

If 𝑥 = sin 𝜃, then 𝑑𝑥 = cos 𝜃 𝑑𝜃, and substitution converts 1 − 𝑥2 to a perfect square:



1 cos 𝜃 1 𝑑𝑥 = 𝑑𝜃. cos 𝜃 𝑑𝜃 = √ √ √ ∫ ∫ 2 2 1−𝑥 cos2 𝜃 1 − sin 𝜃

√ √ Now either cos2 𝜃 = cos 𝜃 or cos2√𝜃 = − cos 𝜃 depending on the values taken by 𝜃. If we choose −𝜋∕2 ≤ 𝜃 ≤ 𝜋∕2, then cos 𝜃 ≥ 0, so cos2 𝜃 = cos 𝜃. Then cos 𝜃 cos 𝜃 𝑑𝜃 = 1 𝑑𝜃 = 𝜃 + 𝐶 = arcsin 𝑥 + 𝐶. 𝑑𝜃 = √ ∫ ∫ cos 𝜃 cos2 𝜃 The last step uses the fact that 𝜃 = arcsin 𝑥 if 𝑥 = sin 𝜃 and −𝜋∕2 ≤ 𝜃 ≤ 𝜋∕2. ∫

370

Chapter 7 INTEGRATION

From now on, when we substitute sin 𝜃, we assume that the interval −𝜋∕2 ≤ 𝜃 ≤ 𝜋∕2 has been chosen. Notice that the previous example is the case 𝑎 = 1 of formula VI-28. The next example illustrates how to choose the substitution when 𝑎 ≠ 1. 1 𝑑𝑥. √ 4 − 𝑥2

Example 7

Use a trigonometric substitution to find

Solution

This time we choose 𝑥 = 2 sin 𝜃, with −𝜋∕2 ≤ 𝜃 ≤ 𝜋∕2, so that 4 − 𝑥2 becomes a perfect square: √ √ √ √ 4 − 𝑥2 = 4 − 4 sin2 𝜃 = 2 1 − sin2 𝜃 = 2 cos2 𝜃 = 2 cos 𝜃.



Then 𝑑𝑥 = 2 cos 𝜃 𝑑𝜃, so substitution gives ( ) 𝑥 1 1 2 cos 𝜃 𝑑𝜃 = 1 𝑑𝜃 = 𝜃 + 𝐶 = arcsin + 𝐶. 𝑑𝑥 = √ ∫ ∫ ∫ 2 cos 𝜃 2 2 4−𝑥 The general rule for choosing a sine substitution is:

To simplify

√ 𝑎2 − 𝑥2 , for constant 𝑎, try 𝑥 = 𝑎 sin 𝜃, with −𝜋∕2 ≤ 𝜃 ≤ 𝜋∕2.

√ Notice 𝑎2 − 𝑥2 is only defined on the interval [−𝑎, 𝑎]. Assuming that the domain of the integrand is [−𝑎, 𝑎], the substitution 𝑥 = 𝑎 sin 𝜃, with −𝜋∕2 ≤ 𝜃 ≤ 𝜋∕2, is valid for all 𝑥 in the domain, because its range is [−𝑎, 𝑎] and it has an inverse 𝜃 = arcsin(𝑥∕𝑎) on [−𝑎, 𝑎]. Example 8

Find the area of the ellipse 4𝑥2 + 𝑦2 = 9.

Solution

Solving for 𝑦 shows that 𝑦 = by symmetry that

√ 9 − 4𝑥2 gives the upper half of the ellipse. From Figure 7.2, we see Area = 4

∫0

3∕2 √

9 − 4𝑥2 𝑑𝑥.

To decide which trigonometric substitution to use, we write the integrand as √ √ ( )2 √ 9 3 2 2 −𝑥 =2 9 − 4𝑥 = 2 − 𝑥2 . 4 2 This suggests that we should choose 𝑥 = (3∕2) sin 𝜃, so that 𝑑𝑥 = (3∕2) cos 𝜃 𝑑𝜃 and √ ( )2 ( )2 ( )√ √ 3 3 3 9 − 4𝑥2 = 2 1 − sin2 𝜃 = 3 cos 𝜃. − sin2 𝜃 = 2 2 2 2 When 𝑥 = 0, 𝜃 = 0, and when 𝑥 = 3∕2, 𝜃 = 𝜋∕2, so 4

∫0

3∕2 √

𝜋∕2

9 − 4𝑥2 𝑑𝑥 = 4

∫0

3 cos 𝜃

( ) 𝜋∕2 3 cos 𝜃 𝑑𝜃 = 18 cos2 𝜃 𝑑𝜃. ∫0 2

Using Example 6 on page 356 or formula IV-18 we find ∫

cos2 𝜃 𝑑𝜃 =

1 1 cos 𝜃 sin 𝜃 + 𝜃 + 𝐶. 2 2

7.4 ALGEBRAIC IDENTITIES AND TRIGONOMETRIC SUBSTITUTIONS

371

So we have Area = 4

∫0

3∕2 √

9 − 4𝑥2 𝑑𝑥 =

) ( |𝜋∕2 9𝜋 𝜋 18 = . =9 0+ (cos 𝜃 sin 𝜃 + 𝜃)|| 2 2 2 |0

𝑦

𝑦=

√ 9 − 4𝑥2

𝑥 − 32

3 2

√ 𝑦 = − 9 − 4𝑥2 Figure 7.2: The ellipse 4𝑥2 + 𝑦2 = 9

In Example 8, we did not return to the original variable 𝑥 after making the substitution because we had also converted the limits of the definite integral. However, if we are calculating an indefinite integral, we have to return to the original variable. In the next example, we see how a triangle representing the substitution can be useful. √ 9 − 4𝑥2 𝑑𝑥 corresponding to Example 8.

Example 9

Find the indefinite integral

Solution

From Example 8, we know if 𝑥 = (3∕2) sin 𝜃, then 𝑑𝑥 = (3∕2) cos 𝜃 𝑑𝜃, so ( ) √ 9 1 1 3 cos 𝜃 sin 𝜃 + 𝜃 + 𝐶. 9 − 4𝑥2 𝑑𝑥 = 3 cos 𝜃 ⋅ cos 𝜃 𝑑𝜃 = ∫ ∫ 2 2 2 2



To rewrite the antiderivative in terms of the original variable 𝑥, we use the fact that sin 𝜃 = 2𝑥∕3 to write 𝜃 = arcsin(2𝑥∕3). To express cos 𝜃 in terms of 𝑥, we draw the right triangle in Figure 7.3 with opposite side√2𝑥 and hypotenuse 3, so sin 𝜃 = 2𝑥∕3. Then we use the Pythagorean Theorem to see that cos 𝜃 = 9 − 4𝑥2 ∕3, so √ 9 9 9 − 4𝑥2 𝑑𝑥 = cos 𝜃 sin 𝜃 + 𝜃 + 𝐶 ∫ 4 4 √ √ 𝑥 9 − 4𝑥2 9 9 − 4𝑥2 9 9 2𝑥 2𝑥 2𝑥 = ⋅ ⋅ + arcsin +𝐶 = + arcsin + 𝐶. 4 3 3 4 3 2 4 3

3

2𝑥

𝜃 √ 9 − 4𝑥2 Figure 7.3: Triangle with sin 𝜃 = 2𝑥∕3

372

Chapter 7 INTEGRATION

Tangent Substitutions Integrals involving 𝑎2 + 𝑥2 may be simplified by a substitution involving tan 𝜃 and the trigonometric identities tan 𝜃 = sin 𝜃∕ cos 𝜃 and cos2 𝜃 + sin2 𝜃 = 1. 1 𝑑𝑥 using the substitution 𝑥 = 3 tan 𝜃. ∫ 𝑥2 + 9

Example 10

Find

Solution

If 𝑥 = 3 tan 𝜃, then 𝑑𝑥 = (3∕ cos2 𝜃) 𝑑𝜃, so ( ) )( 1 1 3 1 1 𝑑𝑥 = 𝑑𝜃 = 𝑑𝜃 ( 2 ) ∫ ∫ 𝑥2 + 9 sin 𝜃 3∫ cos2 𝜃 9 tan2 𝜃 + 9 2𝜃 + 1 cos cos2 𝜃 ( ) 1 𝑥 1 1 1 1 + 𝐶. 𝑑𝜃 = 1 𝑑𝜃 = 𝜃 + 𝐶 = arctan = 3 ∫ sin2 𝜃 + cos2 𝜃 3∫ 3 3 3

To simplify 𝑎2 + 𝑥2 or

√ 𝑎2 + 𝑥2 , for constant 𝑎, try 𝑥 = 𝑎 tan 𝜃, with −𝜋∕2 < 𝜃 < 𝜋∕2.

√ Note that 𝑎2 + 𝑥2 and 𝑎2 + 𝑥2 are defined on (−∞, ∞). Assuming that the domain of the integrand is (−∞, ∞), the substitution 𝑥 = 𝑎 tan 𝜃, with −𝜋∕2 < 𝜃 < 𝜋∕2, is valid for all 𝑥 in the domain, because its range is (−∞, ∞) and it has an inverse 𝜃 = arctan(𝑥∕𝑎) on (−∞, ∞). Example 11

Use a tangent substitution to show that the following two integrals are equal:

∫0

1√

𝜋∕4

1 + 𝑥2 𝑑𝑥 =

∫0

1 𝑑𝜃. cos3 𝜃

What area do these integrals represent? Solution

We put 𝑥 = tan 𝜃, with −𝜋∕2 < 𝜃 < 𝜋∕2, so that 𝑑𝑥 = (1∕ cos2 𝜃) 𝑑𝜃, and √ √ √ 2 𝜃 cos2 𝜃 + sin2 𝜃 1 sin . = = 1 + 𝑥2 = 1 + 2 2 cos 𝜃 cos 𝜃 cos 𝜃 When 𝑥 = 0, 𝜃 = 0, and when 𝑥 = 1, 𝜃 = 𝜋∕4, so

∫0 1

The integral ∫0

1√

1 + 𝑥2 𝑑𝑥 =

∫0

𝜋∕4 (

1 cos 𝜃

)(

1 cos2 𝜃

)

𝜋∕4

𝑑𝜃 =

∫0

1 𝑑𝜃. cos3 𝜃

√ 1 + 𝑥2 𝑑𝑥 represents the area under the hyperbola 𝑦2 − 𝑥2 = 1 in Figure 7.4. 𝑦

𝑦=

√ 1 + 𝑥2

𝑥 1 √ 𝑦 = − 1 + 𝑥2

Figure 7.4: The hyperbola 𝑦2 − 𝑥2 = 1

7.4 ALGEBRAIC IDENTITIES AND TRIGONOMETRIC SUBSTITUTIONS

373

Completing the Square to Use a Trigonometric Substitution To make a trigonometric substitution, we may first need to complete the square. 3 𝑑𝑥. ∫ √ 2𝑥 − 𝑥2

Example 12

Find

Solution

To use a sine or tangent substitution, the expression under the square root sign should be in the form 𝑎2 + 𝑥2 or 𝑎2 − 𝑥2 . Completing the square, we get 2𝑥 − 𝑥2 = 1 − (𝑥 − 1)2 . This suggests we substitute 𝑥 − 1 = sin 𝜃, or 𝑥 = sin 𝜃 + 1. Then 𝑑𝑥 = cos 𝜃 𝑑𝜃, and ∫

3 3 3 𝑑𝑥 = 𝑑𝑥 = cos 𝜃 𝑑𝜃 √ √ √ ∫ ∫ 2 1 − (𝑥 − 1) 2𝑥 − 𝑥2 1 − sin2 𝜃 3 cos 𝜃 𝑑𝜃 = 3 𝑑𝜃 = 3𝜃 + 𝐶. = ∫ ∫ cos 𝜃

Since 𝑥 − 1 = sin 𝜃, we have 𝜃 = arcsin(𝑥 − 1), so 3 𝑑𝑥 = 3 arcsin(𝑥 − 1) + 𝐶. ∫ √ 2𝑥 − 𝑥2

1 𝑑𝑥. ∫ 𝑥2 + 𝑥 + 1

Example 13

Find

Solution

Completing the square, we get 1 𝑥 +𝑥+1= 𝑥+ 2 2

(

)2

( ) 1 2 3 + = 𝑥+ + 4 2

( √ )2 3 . 2

√ √ This √ suggests we substitute 𝑥 + 1∕2 = ( 3∕2) tan 𝜃, or 𝑥 = −1∕2 + ( 3∕2) tan 𝜃. Then 𝑑𝑥 = ( 3∕2)(1∕ cos2 𝜃) 𝑑𝜃, so ( ) ⎛ ⎞ √ 3 1 1 1 ⎜ ⎟ 𝑑𝑥 = 𝑑𝜃 ∫ ⎜ (𝑥 + 1 )2 + 3 ⎟ ∫ 𝑥2 + 𝑥 + 1 2 cos2 𝜃 ⎝ 2 4⎠ √ ) ⎞( ⎛ 3 2 1 1 1 ⎟ ⎜ = 𝑑𝜃 = √ 𝑑𝜃 2 ∫ 2 ∫ ⎜ 3 tan2 𝜃 + 3 ⎟ cos2 𝜃 (tan 𝜃 + 1) cos2 𝜃 3 ⎝4 4⎠ 1 2 2 2 1 𝑑𝜃 = √ 𝜃 + 𝐶. =√ 𝑑𝜃 = √ 2 ∫ ∫ 2𝜃 sin 𝜃 + cos 3 3 3 √ √ √ Since 𝑥 + 1∕2 = ( 3∕2) tan 𝜃, we have 𝜃 = arctan((2∕ 3)𝑥 + 1∕ 3), so ( ) 2 2 1 1 𝑑𝑥 = √ arctan √ 𝑥 + √ + 𝐶. ∫ 𝑥2 + 𝑥 + 1 3 3 3

374

Chapter 7 INTEGRATION

Alternatively, using a computer algebra system gives9 ( ) 2𝑥 + 1 −1 2 tan √ 3 , √ 3

essentially the same as we obtained by hand. You can check either answer by differentiation.

Exercises and Problems for Section 7.4 Online Resource: Additional Problems for Section 7.4 EXERCISES

Split the functions in Exercises 1–7 into partial fractions. 20 25 − 𝑥2 2𝑦 4. 3 𝑦 − 𝑦2 + 𝑦 − 1

𝑥+1 6𝑥 + 𝑥2 1 3. 4 𝑤 − 𝑤3

2.

1.

5.

8 𝑦3 − 4𝑦

7.

2 𝑠4 − 1

18.

6.

2(1 + 𝑠) 𝑠(𝑠2 + 3𝑠 + 2)

19.

8. Exercise 1

9. Exercise 2

10. Exercise 3

11. Exercise 4

12. Exercise 5

13. Exercise 6

14. Exercise 7



𝑥4 + 3𝑥3 + 2𝑥2 + 1 𝑑𝑥; use division. 𝑥2 + 3𝑥 + 2

In Exercises 20–22, use the substitution to find the integral. 20.

In Exercises 8–14, find the antiderivative of the function in the given exercise.

𝑥4 + 12𝑥3 + 15𝑥2 + 25𝑥 + 11 𝑑𝑥; ∫ 𝑥3 + 12𝑥2 + 11𝑥 𝐴 𝐵 𝐶 use division and + + . 𝑥 𝑥 + 1 𝑥 + 11

21.

22.



1 𝑑𝑥, √ 9 − 4𝑥2

1 𝑑𝑥, ∫ √4𝑥 − 3 − 𝑥2 1 𝑑𝑥, ∫ 𝑥2 + 4𝑥 + 5



In Exercises 15–19, evaluate the integral. 𝐵 𝐶 𝐴 3𝑥2 − 8𝑥 + 1 + + . 𝑑𝑥; use ∫ 𝑥3 − 4𝑥2 + 𝑥 + 6 𝑥−2 𝑥+1 𝑥−3 𝐵 𝑑𝑥 𝐴 𝐶 . 16. ; use + 2 + ∫ 𝑥3 − 𝑥2 𝑥 𝑥 𝑥−1 𝐴 𝐵𝑥 + 𝐶 10𝑥 + 2 𝑑𝑥; use . + 2 17. ∫ 𝑥3 − 5𝑥2 + 𝑥 − 5 𝑥−5 𝑥 +1

3 sin 𝑡 2 𝑥 = sin 𝑡 + 2

𝑥 = tan 𝑡 − 2

23. Which of the following integrals are best done by a trigonometric substitution, and what substitution? (a)

15.

𝑥=

√ 9 − 𝑥2 𝑑𝑥

(b)



√ 𝑥 9 − 𝑥2 𝑑𝑥

24. Give a substitution (not necessarily trigonometric) which could be used to compute the following integrals: (a)



𝑥 𝑑𝑥 √ 2 𝑥 + 10

(b)



1 𝑑𝑥 √ 2 𝑥 + 10

PROBLEMS 25. Find a value of 𝑘 and a substitution 𝑤 such that 12𝑥 − 2 𝑑𝑤 𝑑𝑥 = 𝑘 . ∫ (3𝑥 + 2)(𝑥 − 1) ∫ 𝑤

27. Write the integral

2𝑥 + 9 𝑑𝑥 in the form ∫ (3𝑥 + 5)(4 − 5𝑥)

𝑐𝑥 + 𝑑 𝑑𝑥. Give the values of the constants ∫ (𝑥 − 𝑎)(𝑥 − 𝑏) 𝑎, 𝑏, 𝑐, 𝑑. You need not evaluate the integral.

26. Find values of 𝐴 and 𝐵 such that 𝐴 𝑑𝑥 𝐵 𝑑𝑥 12𝑥 − 2 𝑑𝑥 = + . ∫ 3𝑥 + 2 ∫ 𝑥 − 1 ∫ (3𝑥 + 2)(𝑥 − 1)

9 wolframalpha.com,

accessed January 11, 2011.

28. Write the integral

𝑑𝑥 ∫ √12 − 4𝑥2

in the form

𝑘 𝑑𝑥 . Give the values of the positive constants ∫ √𝑎2 − 𝑥2 𝑎 and 𝑘. You need not evaluate the integral.

7.4 ALGEBRAIC IDENTITIES AND TRIGONOMETRIC SUBSTITUTIONS

3𝑥 + 6 𝑑𝑥 by partial fractions. ∫ 𝑥2 + 3𝑥 (b) Show that your answer to part (a) agrees with the answer you get by using the integral tables.

29. (a) Evaluate

Complete the square and give a substitution (not necessarily trigonometric) which could be used to compute the integrals in Problems 30–37. 30. 32. 34.

36. 37.

1 𝑑𝑥 ∫ 𝑥2 + 2𝑥 + 2

31.

𝑑𝑦 ∫ 𝑦2 + 3𝑦 + 3

33.



∫ ∫

4 𝑑𝑧 √ 2𝑧 − 𝑧2

35.

1 𝑑𝑥 ∫ 𝑥2 + 6𝑥 + 25



𝑧−1 𝑑𝑧 √ 2𝑧 − 𝑧2

(2 − 𝜃) cos(𝜃 2 − 4𝜃)𝑑𝜃

Calculate the integrals in Problems 38–53. 38.

1 𝑑𝑥 ∫ (𝑥 − 5)(𝑥 − 3)

39.

1 𝑑𝑥 ∫ (𝑥 + 2)(𝑥 + 3)

40.

1 𝑑𝑥 ∫ (𝑥 + 7)(𝑥 − 2)

41.

𝑥 𝑑𝑥 ∫ 𝑥2 − 3𝑥 + 2

42.

44. 46.

∫ 𝑧2 + 𝑧 𝑑𝑃 ∫ 3𝑃 − 3𝑃 2 ∫

2𝑦2

𝑦+2 𝑑𝑦 + 3𝑦 + 1

𝑑𝑥 43. ∫ 𝑥2 + 5𝑥 + 4

1 𝑑𝑥 ∫ (16 − 𝑥2 )3∕2 √ 𝑥2 + 4 64. 𝑑𝑥 ∫ 𝑥4 62.

1 𝑑𝑥 √ ∫ 𝑥 1 + 16𝑥2

61.

1 𝑑𝑥 ∫ (25 + 4𝑥2 )3∕2

63.

𝑥2 𝑑𝑥 ∫ (1 + 9𝑥2 )3∕2

65.



𝑥3 𝑑𝑥 √ 4 − 𝑥2

Find the exact area of the regions in Problems 66–71. 66. Bounded by 𝑦 = 3𝑥∕((𝑥 − 1)(𝑥 − 4)), 𝑦 = 0, 𝑥 = 2, 𝑥 = 3. 67. Bounded by 𝑦 = (3𝑥2 + 𝑥)∕((𝑥2 + 1)(𝑥 + 1)), 𝑦 = 0, 𝑥 = 0, 𝑥 = 1. √ 68. Bounded by 𝑦 = 𝑥2 ∕ 1 − 𝑥2 , 𝑦 = 0, 𝑥 = 0, 𝑥 = 1∕2. √ √ 69. Bounded by 𝑦 = 𝑥3 ∕ 4 − 𝑥2 , 𝑦 = 0, 𝑥 = 0, 𝑥 = 2. √ 70. Bounded by 𝑦 = 1∕ 𝑥2 + 9, 𝑦 = 0, 𝑥 = 0, 𝑥 = 3. √ 71. Bounded by√𝑦 = 1∕(𝑥 𝑥2 + 9), 𝑦 = 0, 𝑥 = 3, 𝑥 = 3. Calculate the integrals in Problems 72–74 by partial fractions and then by using the indicated substitution. Show that the results you get are the same.

3𝑥 + 1 𝑑𝑥 ∫ 𝑥2 − 3𝑥 + 2

72.

𝑑𝑥 ; substitution 𝑥 = sin 𝜃. ∫ 1 − 𝑥2

47.

𝑥+1 𝑑𝑥 ∫ 𝑥3 + 𝑥

73.

2𝑥 𝑑𝑥; substitution 𝑤 = 𝑥2 − 1. ∫ 𝑥2 − 1

𝑦2 49. 𝑑𝑦 ∫ 25 + 𝑦2

𝑑𝑧 50. ∫ (4 − 𝑧2 )3∕2

10 51. 𝑑𝑠 ∫ (𝑠 + 2)(𝑠2 + 1)

1 𝑑𝑥 ∫ 𝑥2 + 4𝑥 + 13

1 𝑑𝑥 √ ∫ 𝑥2 4 − 𝑥2

59.

45.

𝑥−2 48. 𝑑𝑥 ∫ 𝑥2 + 𝑥4

52.

60.

1 𝑑𝑥 √ ∫ 𝑥 9 − 4𝑥2

𝑥+1 𝑑𝑥 ∫ 𝑥2 + 2𝑥 + 2

(𝑡 + 2) sin(𝑡2 + 4𝑡 + 7) 𝑑𝑡

𝑑𝑧

58.

53.



(𝑒𝑥

𝑒𝑥 𝑑𝑥 − 1)(𝑒𝑥 + 2)

In Problems 54–65, evaluate the indefinite integral, using a trigonometric substitution and a triangle to express the answer in terms of 𝑥.

74.



3𝑥2 + 1 𝑑𝑥; substitution 𝑤 = 𝑥3 + 𝑥. 𝑥3 + 𝑥

1 1 + 𝐶. 𝑑𝜃 = − ∫ sin2 𝜃 tan 𝜃 𝑑𝑦 . (b) Calculate √ ∫ 𝑦2 5 − 𝑦2

75. (a) Show

Solve Problems 76–78 without using integral tables. 1 𝑑𝑥 for ∫ (𝑥 − 𝑎)(𝑥 − 𝑏) (b) 𝑎 = 𝑏

76. Calculate the integral (a) 𝑎 ≠ 𝑏

𝑥 𝑑𝑥 for ∫ (𝑥 − 𝑎)(𝑥 − 𝑏) (b) 𝑎 = 𝑏

77. Calculate the integral 1 𝑑𝑥 ∫ 𝑥2 √1 + 𝑥2 √ 1 − 4𝑥2 𝑑𝑥 56. ∫ 𝑥2 54.

2

𝑥 𝑑𝑥 ∫ √9 − 𝑥2 √ 25 − 9𝑥2 𝑑𝑥 57. ∫ 𝑥 55.

375

(a) 𝑎 ≠ 𝑏

1 𝑑𝑥 for ∫ 𝑥2 − 𝑎 (b) 𝑎 = 0 (c)

78. Calculate the integral (a) 𝑎 > 0

𝑎 0, the time 𝑡 at which a fraction 𝑝 of the school population has heard the rumor is given by 𝑝

𝑡(𝑝) =

𝑏 𝑑𝑥. ∫𝑎 𝑥(1 − 𝑥)

(a) Evaluate the integral to find an explicit formula for 𝑡(𝑝). Write your answer so it has only one ln term.

Strengthen Your Understanding In Problems 80–81, explain what is wrong with the statement. 80. To integrate 1 𝑑𝑥 ∫ (𝑥 − 1)2 (𝑥 − 2) using a partial fraction decomposition, let

85. An integral that can be made easier to evaluate by using the trigonometric substitution 𝑥 = 32 sin 𝜃. In Problems 86–87, decide whether the statements are true or false. Give an explanation for your answer. 1 𝑑𝑡 can be made easier to eval√ ∫ 9 − 𝑡2 uate by using the substitution 𝑡 = 3 tan 𝜃.

86. The integral

1 𝐴 𝐵 = . + (𝑥 − 1)2 (𝑥 − 2) (𝑥 − 1)2 𝑥 − 2 81. Use the substitution 𝑥 = 2 sin 𝜃 to integrate the following integral: 1 𝑑𝑥. ∫ (𝑥2 + 4)3∕2

87. To calculate

1 𝑑𝑥, we can split the integrand ∫ 𝑥3 + 𝑥2

into ∫

(

𝐴 𝐵 𝐶 + + 𝑥 𝑥2 𝑥 + 1

)

𝑑𝑥

In Problems 82–85, give an example of: 82. A rational function whose antiderivative is not a rational function. 83. An integral whose evaluation requires factoring a cubic. 84. A linear polynomial 𝑃 (𝑥) and a quadratic polynomial 𝑄(𝑥) such that the rational function 𝑃 (𝑥)∕𝑄(𝑥) does not have a partial fraction decomposition of the form 𝑃 (𝑥) 𝐴 𝐵 = + 𝑄(𝑥) 𝑥−𝑟 𝑥−𝑠 for some constants 𝐴, 𝐵, 𝑟, and 𝑠.

7.5

For Problems 88–89, which technique is useful in evaluating the integral? (a) Integration by parts (c) Long division (e) A trig substitution

88.

𝑥2 𝑑𝑥 ∫ √1 − 𝑥2

(b) Partial fractions (d) Completing the square (f) Other substitutions

89.

𝑥2 𝑑𝑥 ∫ 1 − 𝑥2

NUMERICAL METHODS FOR DEFINITE INTEGRALS Many functions do not have elementary antiderivatives. To evaluate the definite integrals of such functions, we cannot use the Fundamental Theorem; we must use numerical methods. We know how to approximate a definite integral numerically using left- and right-hand Riemann sums; in this section, we introduce more accurate methods.

The Midpoint Rule In the left- and right-hand Riemann sums, the heights of the rectangles are found using the left-hand or right-hand endpoints, respectively, of the subintervals. For the midpoint rule, we use the midpoint of each of the subintervals. 2 For example, in approximating ∫1 𝑓 (𝑥) 𝑑𝑥 by a Riemann sum with two subdivisions, we first divide the interval 1 ≤ 𝑥 ≤ 2 into two pieces. The midpoint of the first subinterval is 1.25 and the midpoint of the second is 1.75. The heights of the two rectangles are 𝑓 (1.25) and 𝑓 (1.75), respec-

7.5 NUMERICAL METHODS FOR DEFINITE INTEGRALS

377

tively. (See Figure 7.5.) The Riemann sum is 𝑓 (1.25)0.5 + 𝑓 (1.75)0.5. Figure 7.5 shows that evaluating 𝑓 at the midpoint of each subdivision often gives a better approximation to the area under the curve than evaluating 𝑓 at either end. ✛0.5✲ ✛0.5✲

𝑓 (𝑥)



✻ 𝑓 (1.75)

𝑓 (1.25)



❄𝑥 1 1.25

1.75 2

Figure 7.5: Midpoint rule with two subdivisions

Thus, we have three ways of estimating an integral using a Riemann sum: 1. The left rule uses the left endpoint of each subinterval. 2. The right rule uses the right endpoint of each subinterval. 3. The midpoint rule uses the midpoint of each subinterval. We write LEFT(𝑛), RIGHT(𝑛), and MID(𝑛) to denote the results obtained by using these rules with 𝑛 subdivisions.

2

1 𝑑𝑥, compute LEFT(2), RIGHT(2) and MID(2), and compare your answers with the exact ∫1 𝑥 value of the integral.

Example 1

For

Solution

For 𝑛 = 2 subdivisions of the interval [1, 2], we use Δ𝑥 = 0.5. Then, to four decimal places, 1 1 (0.5) + (0.5) = 0.8333 1 1.5 1 1 (0.5) + (0.5) = 0.5833 RIGHT(2) = 𝑓 (1.5)(0.5) + 𝑓 (2)(0.5) = 1.5 2 1 1 (0.5) + (0.5) = 0.6857. MID(2) = 𝑓 (1.25)(0.5) + 𝑓 (1.75)(0.5) = 1.25 1.75 LEFT(2) = 𝑓 (1)(0.5) + 𝑓 (1.5)(0.5) =

All three Riemann sums in this example are approximating 2

∫1

|2 1 𝑑𝑥 = ln 𝑥|| = ln 2 − ln 1 = ln 2 = 0.6931. 𝑥 |1

With only two subdivisions, the left and right rules give quite poor approximations but the midpoint rule is already fairly close to the exact answer. Figures 7.6(a) and (b) show that the midpoint rule is more accurate than the left and right rules because the error to the left of the midpoint tends to cancel the error to the right of the midpoint.

378

Chapter 7 INTEGRATION

(a)

(b)

𝑓 (𝑥) =

(c)

1 𝑥

𝑓 (𝑥) =

Left rule Right rule

1 𝑥

𝑓 (𝑥) =

1 𝑥

Trapezoid rule

Midpoint rule

❄ ❄

❄ 𝑥 1

❄ 𝑥

2

1

𝑥

2

1 2 1 𝑥

Figure 7.6: Left, right, midpoint, and trapezoid approximations to ∫1

2 𝑑𝑥

The Trapezoid Rule We have just seen how the midpoint rule can have the effect of balancing out the errors of the left and right rules. Another way of balancing these errors is to average the results from the left and right rules. This approximation is called the trapezoid rule: LEFT(𝑛) + RIGHT(𝑛) . 2 The trapezoid rule averages the values of 𝑓 at the left and right endpoints of each subinterval and multiplies by Δ𝑥. This is the same as approximating the area under the graph of 𝑓 in each subinterval by a trapezoid (see Figure 7.7). TRAP(𝑛) =

𝑓 (𝑥)

✛ 𝑓 (𝑥0 )



Δ𝑥



Area =

𝑓 (𝑥0 )+𝑓 (𝑥1 ) Δ𝑥 2

𝑓 (𝑥1 ) 𝑥

𝑥0

𝑥1

Figure 7.7: Area used in the trapezoid rule

2

Example 2

Solution

1 𝑑𝑥, compare the trapezoid rule with two subdivisions with the left, right, and midpoint ∫1 𝑥 rules. For

In the previous example we got LEFT(2) = 0.8333 and RIGHT(2) = 0.5833. The trapezoid rule is the average of these, so TRAP(2) = 0.7083. (See Figure 7.6(c).) The exact value of the integral is 0.6931, so the trapezoid rule is better than the left or right rules. The midpoint rule is still the best, in this example, since MID(2) = 0.6857.

Is the Approximation an Over- or Underestimate? It is useful to know when a rule is producing an overestimate and when it is producing an underestimate. In Chapter 5 we saw that if the integrand is increasing, the left rule underestimates and the right rule overestimates the integral. If the integrand is decreasing, the roles reverse. Now we see how concavity relates to the errors in the trapezoid and midpoint rules.

7.5 NUMERICAL METHODS FOR DEFINITE INTEGRALS

379

The Trapezoid Rule If the graph of the function is concave down on [𝑎, 𝑏], then each trapezoid lies below the graph and the trapezoid rule underestimates. If the graph is concave up on [𝑎, 𝑏], the trapezoid rule overestimates. (See Figure 7.8.)

𝑓 concave up: Trapezoid overestimates

𝑓 concave down: Trapezoid underestimates

Figure 7.8: Error in the trapezoid rule

The Midpoint Rule To understand the relationship between the midpoint rule and concavity, take a rectangle whose top intersects the curve at the midpoint of a subinterval. Draw a tangent to the curve at the midpoint; this gives a trapezoid. See Figure 7.9. (This is not the same trapezoid as in the trapezoid rule.) The midpoint rectangle and the new trapezoid have the same area, because the shaded triangles in Figure 7.9 are congruent. Hence, if the graph of the function is concave down, the midpoint rule overestimates; if the graph is concave up, the midpoint rule underestimates. (See Figure 7.10.) If the graph of 𝑓 is concave down on [𝑎, 𝑏], then 𝑏

TRAP(𝑛) ≤

∫𝑎

𝑓 (𝑥) 𝑑𝑥 ≤ MID(𝑛).

If the graph of 𝑓 is concave up on [𝑎, 𝑏], then 𝑏

MID(𝑛) ≤

Figure 7.9: Midpoint rectangle and trapezoid with same area

∫𝑎

𝑓 (𝑥) 𝑑𝑥 ≤ TRAP(𝑛).

𝑓 concave down: Midpoint overestimates

𝑓 concave up: Midpoint underestimates

Figure 7.10: Error in the midpoint rule

When we compute an approximation, we are always concerned about the error, namely the difference between the exact answer and the approximation. We usually do not know the exact error; if we did, we would also know the exact answer. We take Error = Actual value − Approximate value. The errors for some methods are much smaller than those for others. In general, the midpoint and trapezoid rules are more accurate than the left or right rules. Comparing the errors in the midpoint and trapezoid rules suggests an even better method, called Simpson’s rule.

Error in Left and Right Rules 2

We work with the example ∫1 (1∕𝑥) 𝑑𝑥 because we know the exact value of this integral (ln 2) and we can investigate the behavior of the errors.

380

Chapter 7 INTEGRATION

Let us see what happens to the error in the left and right rules as we increase 𝑛. The results are in Table 7.2. A positive error indicates that the Riemann sum is less than the exact value, ln 2. Notice that the errors for the left and right rules have opposite signs but are approximately equal in magnitude. (See Figure 7.11.) This leads us to want to average the left and right rules; this average is the trapezoid rule. Table 7.2 Errors for the left and right rule 2 approximation to ∫1 𝑥1 𝑑𝑥 = ln 2 ≈ 0.6931471806 Error in left rule

Error in right rule

2

−0.1402

0.1098

10

−0.0256

0.0244

50

−0.0050

0.0050

250

−0.0010

0.0010

𝑛

Right rule underestimate





Left rule overestimate

𝑓 (𝑥) =

1 𝑥

Figure 7.11: Errors in left and right sums

There is another pattern to the errors in Table 7.2. If we compute the ratio of the errors in Table 7.3, we see that the error10 in both the left and right rules decreases by a factor of about 5 as 𝑛 increases by a factor of 5. There is nothing special about the number 5; the same holds for any factor. To get one extra digit of accuracy in any calculation, we must make the error 1∕10 as big, so we must increase 𝑛 by a factor of 10. In fact, for the left or right rules, each extra digit of accuracy requires about 10 times the work. Table 7.3

2 1 𝑥

Ratio of the errors as 𝑛 increases for ∫1

/ Error(2) Error(10) / Error(10) Error(50) / Error(50) Error(250)

𝑑𝑥

Ratio of errors in left rule

Ratio of errors in right rule

5.47

4.51

5.10

4.90

5.02

4.98

Error in Trapezoid and Midpoint Rules Table 7.4 shows that the trapezoid and midpoint rules generally produce better approximations to 2 ∫1 (1∕𝑥) 𝑑𝑥 than the left and right rules. Again there is a pattern to the errors. For each 𝑛, the midpoint rule is noticeably better than the trapezoid rule; the error for the midpoint rule, in absolute value, seems to be about half the error of the trapezoid rule. To see why, compare the shaded areas in Figure 7.12. Also, notice in Table 7.4 that the errors for the two rules have opposite signs; this is due to concavity.

Table 7.4 The errors for the trapezoid and 2 midpoint rules for ∫1 𝑥1 𝑑𝑥 𝑛

Error in trapezoid rule

Error in midpoint rule

2

−0.0152

0.0074

10

−0.00062

0.00031

50

−0.0000250

0.0000125

250

−0.0000010

0.0000005

Midpoint error

✲ ✛

Trapezoid error

𝑓 (𝑥) =

1 𝑥

Figure 7.12: Errors in the midpoint and trapezoid rules

10 The values in Table 7.2 are rounded to 4 decimal places; those in Table 7.3 were computed using more decimal places and then rounded.

7.5 NUMERICAL METHODS FOR DEFINITE INTEGRALS

381

We are interested in how the errors behave as 𝑛 increases. Table 7.5 gives the ratios of the errors for each rule. For each rule, we see that as 𝑛 increases by a factor of 5, the error decreases by a factor of about 25 = 52 . In fact, it can be shown that this squaring relationship holds for any factor, so increasing 𝑛 by a factor of 10 will decrease the error by a factor of about 100 = 102 . Reducing the error by a factor of 100 is equivalent to adding two more decimal places of accuracy to the result. In other words: In the trapezoid or midpoint rules, each extra 2 digits of accuracy requires about 10 times the work. Table 7.5

2 1 𝑥

Ratios of the errors as 𝑛 increases for ∫1

/ Error(2) Error(10) / Error(10) Error(50) / Error(50) Error(250)

𝑑𝑥

Ratio of errors in trapezoid rule

Ratio of errors in midpoint rule

24.33

23.84

24.97

24.95

25.00

25.00

Simpson’s Rule Observing that the trapezoid error has the opposite sign and about twice the magnitude of the midpoint error, we may guess that a weighted average of the two rules, with the midpoint rule weighted twice the trapezoid rule, has a smaller error. This approximation is called Simpson’s rule11 : 2 ⋅ MID(𝑛) + TRAP(𝑛) . 3 Table 7.6 gives the errors for Simpson’s rule. Notice how much smaller the errors are than the previous errors. Of course, it is a little unfair to compare Simpson’s rule at 𝑛 = 50, say, with the previous rules, because Simpson’s rule must compute the value of 𝑓 at both the midpoint and the endpoints of each subinterval and hence involves evaluating the function at twice as many points. We see in Table 7.6 that as 𝑛 increases by a factor of 5, the errors decrease by a factor of about 600, or about 54 . Again this behavior holds for any factor, so increasing 𝑛 by a factor of 10 decreases the error by a factor of about 104 . In other words: In Simpson’s rule, each extra 4 digits of accuracy requires about 10 times the work. SIMP(𝑛) =

Table 7.6 The errors for Simpson’s rule and the ratios of the errors 𝑛

Error

2

−0.0001067877

10

−0.0000001940

50

−0.0000000003

Ratio 550.15 632.27

Alternate Approach to Numerical Integration: Approximating by Lines and Parabolas These rules for numerical integration can be obtained by approximating 𝑓 (𝑥) on subintervals by a function: • The left and right rules use constant functions. • The trapezoid and midpoint rules use linear functions. • Simpson’s rule uses quadratic functions. Problems 70 and 71 (available online) show how a quadratic approximation leads to Simpson’s rule. 11 Some books and computer programs use slightly different terminology for Simpson’s rule; what we call 𝑛 = 50, they call 𝑛 = 100.

382

Chapter 7 INTEGRATION

Exercises and Problems for Section 7.5 Online Resource: Additional Problems for Section 7.5 EXERCISES

In Exercises 1–6, sketch the area given by the following ap𝑏 proximations to ∫𝑎 𝑓 (𝑥)𝑑𝑥. Identify each approximation as an overestimate or an underestimate. (a) LEFT(2) (b) RIGHT(2) (c) TRAP(2) (d) MID(2) 1.

6

7. Calculate the following approximations to ∫0 𝑥2 𝑑𝑥. (a) LEFT(2)

(b) RIGHT(2)

(c) TRAP(2)

(d) MID(2)

2. 𝑓 (𝑥)

𝑓 (𝑥)

6

8. Estimate ∫0 𝑥2 𝑑𝑥 using SIMP(2). 4

𝑥 𝑎

3.

𝑥 𝑎

𝑏

𝑏

4.

𝑓 (𝑥)

9. (a) Find LEFT(2) and RIGHT(2) for ∫0 (𝑥2 + 1) 𝑑𝑥. (b) Illustrate your answers to part (a) graphically. Is each approximation an underestimate or overestimate?

𝑓 (𝑥) 4

𝑥

5.

𝑎

𝑏

𝑎

𝑏

𝑥

6.

𝑎

𝑏

𝑎

𝑏

𝑥

𝑥

10. (a) Find MID(2) and TRAP(2) for ∫0 (𝑥2 + 1) 𝑑𝑥. (b) Illustrate your answers to part (a) graphically. Is each approximation an underestimate or overestimate? 𝜋

11. Calculate the following approximations to ∫0 sin 𝜃 𝑑𝜃.

𝑓 (𝑥) 𝑓 (𝑥)

(a) LEFT(2)

(b) RIGHT(2)

(c) TRAP(2)

(d) MID(2)

PROBLEMS 2

12. Use Table 7.7 to estimate ∫1 𝑔(𝑡) 𝑑𝑡 by MID(5).

0.6

14. Estimate ∫0.2 𝑔(𝑡) 𝑑𝑡 using a left-hand sum with 𝑛 = 4. 0.9

Table 7.7

15. Estimate ∫0

𝑡

1.0

1.1

1.2

1.3

1.4

1.5

𝑔(𝑡)

−2.1

−2.9

−3.4

−3.7

−3.6

−3.2

𝑡

1.6

1.7

1.8

1.9

2.0

2.1

𝑔(𝑡)

−2.5

−1.7

−0.7

0.5

2.1

4.1

2

13. Compute MID(4) for the integral ∫0 𝑓 (𝑥) 𝑑𝑥 using the values in Table 7.8. Table 7.8 0

𝑥

0.25

0.50

0.75

1.00

1.25

𝑓 (𝑥)

2.3

5.8

7.8

9.3

10.3

10.8

𝑥

1.50

1.75

2.00

2.25

2.50

2.75

𝑓 (𝑥)

10.8

10.3

9.3

7.8

5.8

3.3

16. Estimate 3.

0.6 ∫0

𝑔(𝑡) 𝑑𝑡 using a right-hand sum with 𝑛 = 3. 𝑔(𝑡) 𝑑𝑡 using the midpoint rule with 𝑛 =

17. Assuming 𝑔(𝑡) is increasing and does not change con0.9 cavity, which methods underestimate ∫0 𝑔(𝑡) 𝑑𝑡? I. Left-hand sum III. Midpoint rule In Problems 3 ∫2 (1∕𝑥2 ) 𝑑𝑥.

18–19,

18. TRAP(2)

𝑡

0.0

0.1

0.2

0.3

0.4

𝑔(𝑡)

1.87

2.64

3.34

3.98

4.55

𝑡

0.5

0.6

0.7

0.8

0.9

𝑔(𝑡)

5.07

5.54

5.96

6.35

6.69

compute

approximations

to

19. MID(2) 1

In Problems 14–17, use Table 7.9. Table 7.9

II. Right-hand sum IV. Trapezoid rule

20. (a) Estimate ∫0 1∕(1 + 𝑥2 ) 𝑑𝑥 by subdividing the interval into eight parts using: (i) the left Riemann sum (ii) the right Riemann sum (iii) the trapezoidal rule (b) Since the exact value of the integral is 𝜋∕4, you can estimate the value of 𝜋 using part (a). Explain why your first estimate is too large and your second estimate too small.

383

7.5 NUMERICAL METHODS FOR DEFINITE INTEGRALS

21. Using the table, estimate the total distance traveled from time 𝑡 = 0 to time 𝑡 = 6 using LEFT, RIGHT, and TRAP. Time, 𝑡

0

1

2

3

4

5

6

Velocity, 𝑣

3

4

5

4

7

8

11

2

27. The table shows approximations to ∫1 𝑓 (𝑥) 𝑑𝑥 for an increasing 𝑓 . (a) The value of TRAP(5) or MID(5) in the table is incorrect. Which one? Find the correct value. (b) The concavity of the graph of 𝑓 does not change on [1, 2]. Is it concave up or down? (c) Find the value SIMP(5).

22. Using Figure 7.13, order the following approximations 3 to the integral ∫0 𝑓 (𝑥)𝑑𝑥 and its exact value from smallest to largest: LEFT(𝑛), RIGHT(𝑛), MID(𝑛), TRAP(𝑛), Exact value.

LEFT(5)

RIGHT(5)

TRAP(5)

MID(5)

0.3153

0.4539

0.3890

0.3871

SIMP(5)

In Problems 28–31, decide which approximation—left, right, trapezoid, or midpoint—is guaranteed to give an over5 estimate for ∫0 𝑓 (𝑥) 𝑑𝑥, and which is guaranteed to give an underestimate. (There may be more than one.)

𝑓 (𝑥)

𝑥 0

28.

3

29.

Figure 7.13

𝑓 (𝑥)

𝑓 (𝑥) 2

23. (a) What is the exact value of ∫0 (𝑥3 + 3𝑥2 ) 𝑑𝑥? (b) Find SIMP(𝑛) for 𝑛 = 2, 4, 100. What do you notice? 24. The results from the left, right, trapezoid, and midpoint 1 rules used to approximate ∫0 𝑔(𝑡) 𝑑𝑡, with the same number of subdivisions for each rule, are as follows: 0.601, 0.632, 0.633, 0.664. (a) Using Figure 7.14, match each rule with its approximation. (b) Between which two consecutive approximations does the true value of the integral lie?

𝑔(𝑡)

𝑡 1

Figure 7.14 2 ∕2

25. The graph of 𝑦 = 𝑒−𝑥 1.

𝑥

𝑥

5

30.

5

31.

𝑓 (𝑥)

𝑓 (𝑥) 𝑥

𝑥 5

5

4 √ 32. Consider the integral ∫0 3 𝑥𝑑𝑥.

(a) Estimate the value of the integral using MID(2). (b) Use the Fundamental Theorem of Calculus to find the exact value of the definite integral. (c) What is the error for MID(2)? (d) Use your knowledge of how errors change and your answer to part (c) to estimate the error for MID(20). (e) Use your answer to part (d) to estimate the approximation MID(20).

is concave down for −1 < 𝑥
TRAP(80).

63. If LEFT(𝑛) = RIGHT(𝑛) for all 𝑛, then 𝑓 is a constant function. 64. The trapezoid estimate TRAP(𝑛) = (LEFT(𝑛) + 6 RIGHT(𝑛))∕2 is always closer to ∫2 𝑓 (𝑥)𝑑𝑥 than LEFT(𝑛) or RIGHT(𝑛).

In Problems 55–56, decide whether the statements are true or false. Give an explanation for your answer.

6

65. ∫2 𝑓 (𝑥) 𝑑𝑥 lies between LEFT(𝑛) and RIGHT(𝑛).

1

55. The midpoint rule approximation to ∫0 (𝑦2 − 1) 𝑑𝑦 is always smaller than the exact value of the integral. 56. The trapezoid rule approximation is never exact.

66. If LEFT(2) 𝑏 ∫𝑎 𝑓 (𝑥) 𝑑𝑥.

The left and right Riemann sums of a function 𝑓 on the interval [2, 6] are denoted by LEFT(𝑛) and RIGHT(𝑛), respectively, when the interval is divided into 𝑛 equal parts.

67. If 0 < 𝑓 ′ < 𝑔 ′ everywhere, then the error in approxi𝑏 mating ∫𝑎 𝑓 (𝑥) 𝑑𝑥 by LEFT(𝑛) is less than the error in 𝑏 approximating ∫𝑎 𝑔(𝑥) 𝑑𝑥 by LEFT(𝑛).

7.6


0 and if ∞ lim𝑥→∞ 𝑓 (𝑥) = 0, then ∫0 𝑓 (𝑥) 𝑑𝑥 converges.

59. A function 𝑓 (𝑥), continuous at 𝑥 = 2 and 𝑥 = 5, such 5 that the integral ∫2 𝑓 (𝑥) 𝑑𝑥 is improper and divergent.

63. If 𝑓 (𝑥) is continuous and positive for 𝑥 > 0 and if ∞ lim𝑥→∞ 𝑓 (𝑥) = ∞, then ∫0 (1∕𝑓 (𝑥)) 𝑑𝑥 converges.

394

Chapter 7 INTEGRATION ∞



64. If ∫0 𝑓 (𝑥) 𝑑𝑥 and ∫0 𝑔(𝑥) 𝑑𝑥 both converge, then ∞ ∫0 (𝑓 (𝑥) + 𝑔(𝑥)) 𝑑𝑥 converges. ∞

which are false. Give an explanation for your answer. ∞

66. ∫0 𝑎𝑓 (𝑥) 𝑑𝑥 converges.



65. If ∫0 𝑓 (𝑥) 𝑑𝑥 and ∫0 𝑔(𝑥) 𝑑𝑥 both diverge, then ∞ ∫0 (𝑓 (𝑥) + 𝑔(𝑥)) 𝑑𝑥 diverges.



67. ∫0 𝑓 (𝑎𝑥) 𝑑𝑥 converges. ∞

Suppose that 𝑓 is continuous for all real numbers and that ∞ ∫0 𝑓 (𝑥) 𝑑𝑥 converges. Let 𝑎 be any positive number. Decide which of the statements in Problems 66–69 are true and

7.7

68. ∫0 𝑓 (𝑎 + 𝑥) 𝑑𝑥 converges. ∞

69. ∫0 (𝑎 + 𝑓 (𝑥)) 𝑑𝑥 converges.

COMPARISON OF IMPROPER INTEGRALS

Making Comparisons Sometimes it is difficult to find the exact value of an improper integral by antidifferentiation, but it may be possible to determine whether an integral converges or diverges. The key is to compare the given integral to one whose behavior we already know. Let’s look at an example. ∞

1 𝑑𝑥 converges. √ 𝑥3 + 5

Example 1

Determine whether

Solution

First, let’s see what this integrand does as 𝑥 → ∞. For large 𝑥, the 5 becomes insignificant compared with the 𝑥3 , so 1 1 1 ≈ √ = 3∕2 . √ 𝑥 𝑥3 + 5 𝑥3

∫1

Since



∫1

1 √ 𝑑𝑥 = ∫1 𝑥3



1 𝑥3∕2

𝑏

𝑑𝑥 = lim

𝑏→∞ ∫1



1 𝑥3∕2

( ) |𝑏 𝑑𝑥 = lim −2𝑥−1∕2 || = lim 2 − 2𝑏−1∕2 = 2, 𝑏→∞ 𝑏→∞ |1

the integral ∫1 (1∕𝑥3∕2 ) 𝑑𝑥 converges. So we expect our integral to converge as well. In order to confirm this, we observe that for 0 ≤ 𝑥3 ≤ 𝑥3 + 5, we have 1 1 ≤√ . √ 𝑥3 + 5 𝑥3

and so for 𝑏 ≥ 1, 𝑏

∫1

𝑏

1 1 𝑑𝑥 ≤ √ √ 𝑑𝑥. ∫ 1 𝑥3 + 5 𝑥3

𝑦 𝑦=

1 √ 𝑥3

∞ 𝑑𝑥 √ 𝑥3

Total shaded area = ∫1



𝑦=

1 √ 𝑥3 +5

Dark shaded area = ∫1

1 √ 𝑥3 +5

𝑑𝑥

𝑥 1 ∞

Figure 7.23: Graph showing ∫1

1 √ 𝑥3 +5

∞ 𝑑𝑥 √ 𝑥3

𝑑𝑥 ≤ ∫1

7.7 COMPARISON OF IMPROPER INTEGRALS

395

√ 𝑏 (See Figure 7.23.) Since ∫1 (1∕ 𝑥3 + 5) 𝑑𝑥 increases as 𝑏 approaches infinity but is always smaller √ 𝑏 ∞ ∞ than ∫1 (1∕𝑥3∕2 ) 𝑑𝑥 < ∫1 (1∕𝑥3∕2 ) 𝑑𝑥 = 2, we know ∫1 (1∕ 𝑥3 + 5) 𝑑𝑥 must have a finite value less than 2. Thus, ∞ 𝑑𝑥 converges to a value less than 2. ∫1 √𝑥3 + 5 Notice that we first looked at the behavior of the integrand as 𝑥 → ∞. This is useful because the convergence or divergence of the integral is determined by what happens as 𝑥 → ∞. ∞

The Comparison Test for

∫𝒂

𝒇 (𝒙) 𝒅𝒙

Assume 𝑓 (𝑥) is positive. Making a comparison involves two stages: 1. Guess, by looking at the behavior of the integrand for large 𝑥, whether the integral converges or not. (This is the “behaves like” principle.) 2. Confirm the guess by comparison with a positive function 𝑔(𝑥): ∞ ∞ • If 𝑓 (𝑥) ≤ 𝑔(𝑥) and ∫𝑎 𝑔(𝑥) 𝑑𝑥 converges, then ∫𝑎 𝑓 (𝑥) 𝑑𝑥 converges. ∞



• If 𝑔(𝑥) ≤ 𝑓 (𝑥) and ∫𝑎 𝑔(𝑥) 𝑑𝑥 diverges, then ∫𝑎 𝑓 (𝑥) 𝑑𝑥 diverges.



𝑑𝑡 converges or diverges. (ln 𝑡) − 1

Example 2

Decide whether

Solution

Since ln 𝑡 grows without bound as 𝑡 → ∞, the −1 is eventually going to be insignificant in comparison to ln 𝑡. Thus, as far as convergence is concerned,

∫4



∫4

1 𝑑𝑡 (ln 𝑡) − 1



behaves like

∫4

1 𝑑𝑡. ln 𝑡



Does ∫4 (1∕ ln 𝑡) 𝑑𝑡 converge or diverge? Since ln 𝑡 grows very slowly, 1∕ ln 𝑡 goes to zero very slowly, and so the integral probably does not converge. We know that (ln 𝑡) − 1 < ln 𝑡 < 𝑡 for all positive 𝑡. So, provided 𝑡 > 𝑒, we take reciprocals: 1 1 1 > > . (ln 𝑡) − 1 ln 𝑡 𝑡 ∞

Since ∫4 (1∕𝑡) 𝑑𝑡 diverges, we conclude that ∞

∫4

1 𝑑𝑡 diverges. (ln 𝑡) − 1

How Do We Know What to Compare With? In Examples 1 and 2, we investigated the convergence of an integral by comparing it with an easier integral. How did we pick the easier integral? This is a matter of trial and error, guided by any information we get by looking at the original integrand as 𝑥 → ∞. We want the comparison integrand to be easy and, in particular, to have a simple antiderivative.

396

Chapter 7 INTEGRATION

Useful Integrals for Comparison ∞



∫1

1 𝑑𝑥 converges for 𝑝 > 1 and diverges for 𝑝 ≤ 1. 𝑥𝑝

1



1 𝑑𝑥 converges for 𝑝 < 1 and diverges for 𝑝 ≥ 1. ∫0 𝑥𝑝 ∞



∫0

𝑒−𝑎𝑥 𝑑𝑥 converges for 𝑎 > 0.

Of course, we can use any function for comparison, provided we can determine its behavior. ∞

(sin 𝑥) + 3 𝑑𝑥. √ 𝑥

Example 3

Investigate the convergence of

Solution

Since it looks difficult to find an antiderivative of this function, we try comparison. What happens to this integrand as 𝑥 → ∞? Since sin 𝑥 oscillates between −1 and 1,

∫1

2 −1 + 3 (sin 𝑥) + 3 1 + 3 4 ≤ ≤ √ =√ , √ = √ √ 𝑥 𝑥 𝑥 𝑥 𝑥

√ √ the integrand oscillates between 2∕ 𝑥 and 4∕ 𝑥. (See Figure 7.24.) √ √ ∞ ∞ What do ∫1 (2∕ 𝑥) 𝑑𝑥 and ∫1 (4∕ 𝑥) 𝑑𝑥 do? As far as convergence is concerned, they certainly do the integral does it too. It is important to √ same thing, and whatever that is, the original √ notice that 𝑥 grows very slowly. This means that 1∕ 𝑥 gets small slowly, which means that con√ vergence is unlikely. Since 𝑥 = 𝑥1∕2 , the result in the preceding box (with 𝑝 = 12 ) tells us that √ ∞ ∫1 (1∕ 𝑥) 𝑑𝑥 diverges. So the comparison test tells us that the original integral diverges. 𝑦 𝑏 (sin 𝑥)+3 √ 𝑥

Total shaded area = ∫1 Dark shaded area =

𝑦=

4 √ 𝑥

𝑦=



(sin 𝑥)+3 √ 𝑥



𝑏 ∫1 √2𝑥

𝑦=

𝑑𝑥

𝑑𝑥

2 √ 𝑥

✠ 𝑥

1

𝑏 𝑏

Figure 7.24: Graph showing ∫1

2 √ 𝑥

𝑏 (sin 𝑥)+3 √ 𝑥

𝑑𝑥 ≤ ∫1

𝑑𝑥, for 𝑏 ≥ 1

Notice that there are two possible comparisons we could have made in Example 3: (sin 𝑥) + 3 (sin 𝑥) + 3 4 2 or ≤√ . √ ≤ √ √ 𝑥 𝑥 𝑥 𝑥 √ √ ∞ ∞ Since both ∫1 (2∕ 𝑥) 𝑑𝑥 and ∫1 (4∕ 𝑥) 𝑑𝑥 diverge, only the first comparison is useful. Knowing that an integral is smaller than a divergent integral is of no help whatsoever!

397

7.7 COMPARISON OF IMPROPER INTEGRALS

The next example shows what to do if the comparison does not hold throughout the interval of integration. ∞

2 ∕2

𝑒−𝑥

Example 4

Show

Solution

We know that 𝑒−𝑥 ∕2 goes very rapidly to zero as 𝑥 → ∞, so we expect this integral to converge. Hence we look for some larger integrand which has a convergent integral. One possibility ∞ ∞ is ∫1 𝑒−𝑥 𝑑𝑥, because 𝑒−𝑥 has an elementary antiderivative and ∫1 𝑒−𝑥 𝑑𝑥 converges. What is the 2 relationship between 𝑒−𝑥 ∕2 and 𝑒−𝑥 ? We know that for 𝑥 ≥ 2,

∫1

𝑑𝑥 converges. 2

𝑥≤

𝑥2 2



so

𝑥2 ≤ −𝑥, 2

and so, for 𝑥 ≥ 2 2 ∕2

𝑒−𝑥

≤ 𝑒−𝑥 .

Since this inequality holds only for 𝑥 ≥ 2, we split the interval of integration into two pieces: ∞

∫1 2

2 ∕2

𝑒−𝑥

2

𝑑𝑥 =

∫1

2 ∕2

𝑒−𝑥



𝑑𝑥 +



2

∫2

2 ∕2

𝑒−𝑥

𝑑𝑥. ∞

2

Now ∫1 𝑒−𝑥 ∕2 𝑑𝑥 is finite (it is not improper) and ∫2 𝑒−𝑥 ∕2 𝑑𝑥 is finite by comparison with ∫2 𝑒−𝑥 𝑑𝑥. 2 ∞ Therefore, ∫1 𝑒−𝑥 ∕2 𝑑𝑥 is the sum of two finite pieces and therefore must be finite. The previous example illustrates the following general principle:

If 𝑓 is positive and continuous on [𝑎, 𝑏], ∞

∫𝑎



𝑓 (𝑥) 𝑑𝑥 and

𝑓 (𝑥) 𝑑𝑥

∫𝑏

either both converge or both diverge.



In particular, when the comparison test is applied to ∫𝑎 𝑓 (𝑥) 𝑑𝑥, the inequalities for 𝑓 (𝑥) and 𝑔(𝑥) do not need to hold for all 𝑥 ≥ 𝑎 but only for 𝑥 greater than some value, say 𝑏.

Exercises and Problems for Section 7.7 EXERCISES In Exercises 1–9, use the box on page 396 and the behavior of rational and exponential functions as 𝑥 → ∞ to predict whether the integrals converge or diverge.



5.

∫1

𝑥2

𝑥 𝑑𝑥 + 2𝑥 + 4

6.

𝑥4

5𝑥 + 2 𝑑𝑥 + 8𝑥2 + 4

8.

𝑥4

𝑥2 + 4 𝑑𝑥 + 3𝑥2 + 11

∞ ∞

1.

∫1

𝑥 𝑑𝑥 𝑥4 + 1



3.

∫1



2

𝑥3

𝑥2 + 1 𝑑𝑥 + 3𝑥 + 2

2.

∫2

3

𝑥 𝑑𝑥 𝑥4 − 1



4.

∫1

𝑥2

1 𝑑𝑥 + 5𝑥 + 1

7.

∫1



9.

∫1



𝑥2 − 6𝑥 + 1 𝑑𝑥 𝑥2 + 4



1 𝑑𝑡 𝑒5𝑡 + 2

∫1

∫1

398

Chapter 7 INTEGRATION

In Exercises 10–13, what, if anything, does the comparison tell us about the convergence of the integral? ∞

10.

∫1 ∞

11.

∫1 1

12.

∫0 1

13.

∫0

sin2 𝑥 𝑑𝑥, compare with 𝑥2 sin2 𝑥 𝑑𝑥, compare with 𝑥

1

18.

1 𝑥2 1 𝑥

20.

sin2 𝑥 1 𝑑𝑥, compare with 2 𝑥2 𝑥

22.

sin2 𝑥 1 √ 𝑑𝑥, compare with √ 𝑥 𝑥

24.

∫0 ∞

∫1 1

∫0

In Exercises 14–29, decide if the improper integral converges or diverges. ∞

14.



𝑑𝑧



15.

∫50 𝑧3

∫1

𝑑𝑥



16.

∫1

8

𝑑𝑥 17.

𝑥3 + 1

∫1

1+𝑥 6



21.

∫0

23.

25.

2 + cos 𝜙

𝑑𝑢 𝑢 + 𝑢2

∫2 ∞

𝑑𝜃 √ 𝜃3 + 𝜃

2 − sin 𝜙

∫1 ∞

𝑑𝜃 √ 𝜃2 + 1

𝑑𝑡

∫−1 (𝑡 + 1)2

𝑑𝑢

∫0

𝑑𝜃 √ 𝜃3 + 1 𝑑𝑦 1 + 𝑒𝑦



27.

𝑑𝜙

𝜙2 𝜋

28.

19.

𝑑𝑥

∫−∞ 1 + 𝑢2 ∞

26.

5

1 𝑥19∕20

∫0 ∞

29.

𝑑𝜙

𝜙2

𝑑𝑧 𝑒𝑧

∫4

+ 2𝑧

3 + sin 𝛼

𝑑𝛼

𝛼

𝑑𝑡 ∫5 √𝑡 − 5

PROBLEMS 30. The graphs of 𝑦 = 1∕𝑥, 𝑦 = 1∕𝑥2 and the functions 𝑓 (𝑥), 𝑔(𝑥), ℎ(𝑥), and 𝑘(𝑥) are shown in Figure 7.25. (a) Is the area between 𝑦 = 1∕𝑥 and 𝑦 = 1∕𝑥2 on the interval from 𝑥 = 1 to ∞ finite or infinite? Explain. (b) Using the graph, decide whether the integral of each of the functions 𝑓 (𝑥), 𝑔(𝑥), ℎ(𝑥) and 𝑘(𝑥) on the interval from 𝑥 = 1 to ∞ converges, diverges, or whether it is impossible to tell.

𝑓 (𝑥) 𝑔(𝑥)





𝑥

𝑎

Figure 7.26 𝑘(𝑥) ℎ(𝑥)



32. Suppose ∫𝑎 𝑓 (𝑥) 𝑑𝑥 converges. What does Figure 7.27 ∞ suggest about the convergence of ∫𝑎 𝑔(𝑥) 𝑑𝑥?

✛ 1∕𝑥 𝑔(𝑥) ✛ 1∕𝑥2

𝑓 (𝑥)

𝑔(𝑥)

1

𝑓 (𝑥)

Figure 7.25

✠ 𝑥 𝑎 ∞ ∫𝑎

𝑓 (𝑥) 𝑑𝑥 and 31. For 𝑓 (𝑥) in Figure 7.26, both 𝑎 ∫0 𝑓 (𝑥) 𝑑𝑥 converge. Decide if the following integrals converge, diverge, or could do either. Assume that 0 < 𝑔(𝑥) < 𝑓 (𝑥) for 𝑥 > 𝑎 and 0 < 𝑓 (𝑥) < 𝑔(𝑥) for 𝑥 < 𝑎. ∞

(a)

∫𝑎 𝑔(𝑥) 𝑑𝑥

(c)

∞ ∫0

(e)

∞ ∫𝑎 (𝑓 (𝑥)

(g)



𝑓 (𝑥) 𝑑𝑥

𝑎

(b) ∫0 𝑔(𝑥) 𝑑𝑥 (d)

− 𝑔(𝑥)) 𝑑𝑥 (f)

∫𝑎 (𝑓 (𝑥))2 𝑑𝑥

∞ ∫0 𝑔(𝑥) 𝑑𝑥 ∞

∫𝑎

1 𝑑𝑥 𝑓 (𝑥)

Figure 7.27

For what values of 𝑝 do the integrals in Problems 33–34 converge or diverge? ∞

33.

∫2

𝑑𝑥 𝑥(ln 𝑥)𝑝

2

34.

𝑑𝑥 ∫1 𝑥(ln 𝑥)𝑝

7.7 COMPARISON OF IMPROPER INTEGRALS

35. (a) Find an upper bound for ∞

2

𝑒−𝑥 𝑑𝑥.

∫3

(a) Explain why a graph of the tangent line to 𝑒𝑡 at 𝑡 = 0 tells us that for all 𝑡 1 + 𝑡 ≤ 𝑒𝑡 .

2

[Hint: 𝑒−𝑥 ≤ 𝑒−3𝑥 for 𝑥 ≥ 3.] (b) For any positive 𝑛, generalize the result of part (a) to find an upper bound for ∞

∫𝑛

(b) Substituting 𝑡 = 1∕𝑥, show that for all 𝑥 ≠ 0

2

𝑒−𝑥 𝑑𝑥 𝑒1∕𝑥 − 1 >

by noting that 𝑛𝑥 ≤ 𝑥2 for 𝑥 ≥ 𝑛. 36. In Planck’s Radiation Law, we encounter the integral ∞

∫1

399

𝑑𝑥 𝑥5 (𝑒1∕𝑥 − 1)

.

1 . 𝑥

(c) Use the comparison test to show that the original integral converges.

Strengthen Your Understanding In Problems 37–40, explain what is wrong with the statement.

In Problems 43–48, decide whether the statements are true or false. Give an explanation for your answer. ∞



37. ∫1 1∕(𝑥3 + sin 𝑥) 𝑑𝑥 converges by comparison with ∞ ∫1 1∕𝑥3 𝑑𝑥. ∞

√ 2

38. ∫1 1∕(𝑥

43. The integral

∫0

1 𝑑𝑥 converges. 𝑒𝑥 + 𝑥

1

+ 1) 𝑑𝑥 is divergent.

1 𝑑𝑥 diverges. ∫0 𝑥2 − 3 45. If 𝑓 (𝑥) < 𝑔(𝑥) for all 𝑥 and 𝑎 < 𝑏, then 44. The integral

∞ ∫0

39. If 0 ≤ 𝑓 (𝑥) ≤ 𝑔(𝑥) and 𝑔(𝑥) 𝑑𝑥 diverges then by ∞ the comparison test ∫0 𝑓 (𝑥) 𝑑𝑥 diverges. ∞

40. Let 𝑓 (𝑥) > 0. If ∫1 𝑓 (𝑥) 𝑑𝑥 is convergent then so is ∞ ∫1 1∕𝑓 (𝑥) 𝑑𝑥.

𝑏

𝑏

𝑓 (𝑥) 𝑑𝑥
0. (b) If this region is rotated about the 𝑥-axis, find the volume of the solid of revolution in terms of 𝑏.

53. The solid whose base is the given region and whose cross-sections perpendicular to the 𝑥-axis are squares.

For Problems 41–43, sketch the solid obtained by rotating each region around the indicated axis. Using the sketch, show how to approximate the volume of the solid by a Riemann sum, and hence find the volume.

In Problems 55–58, the region is rotated around the 𝑦-axis. Write, then evaluate, an integral or sum of integrals giving the volume.

54. The solid whose base is the given region and whose cross-sections perpendicular to the 𝑥-axis are semicircles.

55. Bounded by 𝑦 = 2𝑥, 𝑦 = 6 − 𝑥, 𝑦 = 0. 41. Bounded by 𝑦 = 𝑥3 , 𝑥 = 1, 𝑦 = −1. Axis: 𝑦 = −1. √ 42. Bounded by 𝑦 = 𝑥, 𝑥 = 1, 𝑦 = 0. Axis: 𝑥 = 1.

43. Bounded by the first arch of 𝑦 = sin 𝑥, 𝑦 = 0. Axis: 𝑥 axis.

56. Bounded by 𝑦 = 2𝑥, 𝑦 = 6 − 𝑥, 𝑥 = 0. √ √ 57. Bounded by 𝑦 = 𝑥, 𝑦 = 2 − 𝑥, 𝑦 = 0. √ √ 58. Bounded by 𝑦 = 𝑥, 𝑦 = 2 − 𝑥, 𝑥 = 0.

59. Consider the hyperbola 𝑥2 − 𝑦2 = 1 in Figure 8.37.

2

Problems 44–49 concern the region bounded by 𝑦 = 𝑥 , 𝑦 = 1, and the 𝑦-axis, for 𝑥 ≥ 0. Find the volume of the solid. 44. The solid obtained by rotating the region around the 𝑦axis.

(a) The shaded region 2 ≤ 𝑥 ≤ 3 is rotated around the 𝑥-axis. What is the volume generated? (b) What is the arc length with 𝑦 ≥ 0 from 𝑥 = 2 to 𝑥 = 3?

418

Chapter 8 USING THE DEFINITE INTEGRAL 𝑦

𝑥 −3 −2

−1 1

2

3

Figure 8.37

Figure 8.40

60. Rotating the ellipse 𝑥2 ∕𝑎2 + 𝑦2 ∕𝑏2 = 1 about the 𝑥-axis generates an ellipsoid. Compute its volume. 61. Rotating the asteroid 𝑥2∕3 + 𝑦2∕3 = 𝑎2∕3 in Figure 8.38 about the 𝑥-axis generates a star-shaped solid. Compute its volume. 𝑦

𝑎

𝑥 −𝑎

𝑎

64. A tree trunk has a circular cross-section at every height; its circumference is given in the following table. Estimate the volume of the tree trunk using the trapezoid rule. Height (feet)

0

20

40

60

80

100

120

Circumference (feet)

26

22

19

14

6

3

1

65. The circumference of a tree at different heights above the ground is given in the table below. Assume that all horizontal cross-sections of the tree are circles. Estimate the volume of the tree. Height (inches)

−𝑎

Figure 8.38 62. (a) A pie dish with straight sides is 9 inches across the top, 7 inches across the bottom, and 3 inches deep. See Figure 8.39. Compute the volume of this dish. (b) Make a rough estimate of the volume in cubic inches of a single cut-up apple, and estimate the number of apples needed to make an apple pie that fills this dish.

Circumference (inches)

0

20

40

60

80

100

120

31

28

21

17

12

8

2

66. Compute the perimeter of the region used for the base of the solids in Problems 50–54. 67. Write an integral that represents the arc length of the portion of the graph of 𝑓 (𝑥) = −𝑥(𝑥−4) that lies above the 𝑥-axis. Do not evaluate the integral. 68. Find a curve 𝑦 = 𝑔(𝑥), such that when the region between the curve and the 𝑥-axis for 0 ≤ 𝑥 ≤ 𝜋 is revolved around the 𝑥-axis, it forms a solid with volume given by 𝜋

∫0

𝜋(4 − 4 cos2 𝑥) 𝑑𝑥.

[Hint: Use the identity sin2 𝑥 = 1 − cos2 𝑥.]

Figure 8.39 63. The design of boats is based on Archimedes’ Principle, which states that the buoyant force on an object in water is equal to the weight of the water displaced. Suppose you want to build a sailboat whose hull is parabolic with cross-section 𝑦 = 𝑎𝑥2 , where 𝑎 is a constant. Your boat will have length 𝐿 and its maximum draft (the maximum vertical depth of any point of the boat beneath the water line) will be 𝐻. See Figure 8.40. Every cubic meter of water weighs 10,000 newtons. What is the maximum possible weight for your boat and cargo?

In Problems 69–70, a hemisphere of radius 𝑎 has its base on the 𝑥𝑦-plane, centered at the origin; the 𝑧-axis is vertically upward. Using the given slices, (a) Write an expression for the volume of a slice. (b) Write an integral giving the volume of the hemisphere. (c) Calculate the integral. 69. Vertical slices perpendicular to the 𝑥-axis. 70. Horizontal slices perpendicular to the 𝑧-axis. In Problems 71–72, find the volume of the solid whose base is the region in the first quadrant bounded by 𝑦 = 4 − 𝑥2 , the 𝑥-axis, and the 𝑦-axis, and whose cross-section in the given direction is an equilateral triangle. Include a sketch of the region and show how to find the area of a triangular crosssection. 71. Perpendicular to the 𝑥-axis. 72. Perpendicular to the 𝑦-axis.

8.2 APPLICATIONS TO GEOMETRY

73. For 𝑘 > 0, the volume of the solid created by rotating the region bounded by 𝑦 = 𝑘𝑥(𝑥 − 2) and the 𝑥-axis between 𝑥 = 0 and 𝑥 = 2 around the 𝑥-axis is 192𝜋∕5. Find 𝑘. 74. A particle moves in the 𝑥𝑦-plane along the elliptical path 𝑥 = 2 cos 𝑡, 𝑦 = sin 𝑡, where 𝑥 and 𝑦 are in cm and 0 ≤ 𝑡 ≤ 𝜋. Find the distance traveled by the particle. 75. A projectile moves along a path 𝑥 = 20𝑡 , 𝑦 = 20𝑡 − 4.9𝑡2 , where 𝑡 is time in seconds, 0 ≤ 𝑡 ≤ 5, and 𝑥 and 𝑦 are in meters. Find the total distance traveled by the projectile. 76. A small plane flies in a sinusoidal pattern at a constant altitude over a river. The path of the plane is 𝑦 = 0.1 sin(30𝑥), where 0 ≤ 𝑥 ≤ 10 for 𝑥 and 𝑦 in km. Find the total distance traveled by the plane during the flight. 77. The arc length of 𝑦 = 𝑓 (𝑥) from 𝑥 = 2 to 𝑥 = 12 is 20. Find the arc length of 𝑔(𝑥) = 4𝑓 (0.25𝑥 + 1) from 𝑥 = 4 to 𝑥 = 44.

82. An airplane takes off at 𝑡 = 0 hours flying due north. It takes 24 minutes for the plane to reach cruising altitude, and during this time its ground speed (or horizontal velocity), in mph, is 𝑑𝑥 1050 = 1250 − 𝑑𝑡 𝑡+1 and its vertical velocity, in mph, is 𝑑𝑦 = −625𝑡2 + 250𝑡. 𝑑𝑡 (a) What is the cruising altitude of the plane? (b) What is the ground distance (or horizontal distance) covered from take off until cruising altitude? (c) Find the total distance traveled by the plane from take off until cruising altitude. 2

83. Rotate the bell-shaped curve 𝑦 = 𝑒−𝑥 ∕2 shown in Figure 8.41 around the 𝑦-axis, forming a hill-shaped solid of revolution. By slicing horizontally, find the volume of this hill. 𝑦 1

78. A particle starts at the origin and moves along the curve 𝑦 = 2𝑥3∕2 ∕3 in the positive 𝑥-direction at a speed of 3 cm/sec, where 𝑥, 𝑦 are in cm. Find the position of the particle at 𝑡 = 6.

2 ∕2

𝑦 = 𝑒−𝑥

79. A particle’s position along a circular path at time 𝑡 with 0 ≤ 𝑡 ≤ 3 is given by 𝑥 = cos(𝜋𝑡) and 𝑦 = sin(𝜋𝑡). (a) Find the distance traveled by the particle over this time interval. (b) How does your answer in part (a) relate to the circumference of the circle? (c) What is the particle’s displacement between 𝑡 = 0 and 𝑡 = 3? 80. A particle moves with velocity 𝑑𝑥∕𝑑𝑡 in the 𝑥-direction and 𝑑𝑦∕𝑑𝑡 in the 𝑦-direction at time 𝑡 in seconds, where 𝑑𝑥 = − sin 𝑡 𝑑𝑡

and

𝑑𝑦 = cos 𝑡. 𝑑𝑡

(a) Find the change in position in the 𝑥 and 𝑦coordinates between 𝑡 = 0 and 𝑡 = 𝜋. (b) If the particle passes through (2, 3) at 𝑡 = 0, find its position at 𝑡 = 𝜋. (c) Find the distance traveled by the particle from time 𝑡 = 0 to 𝑡 = 𝜋. 81. A particle moves with velocity 𝑑𝑥∕𝑑𝑡 in the 𝑥-direction and 𝑑𝑦∕𝑑𝑡 in the 𝑦-direction at time 𝑡 in seconds, where 𝑑𝑥 = 3𝑡2 𝑑𝑡

and

𝑥

Figure 8.41 84. A bowl has the shape of the graph of 𝑦 = 𝑥4 between the points (1, 1) and (−1, 1) rotated about the 𝑦-axis. When the bowl contains water to a depth of ℎ units, it flows out through a hole √ in the bottom at a rate (volume/time) proportional to ℎ, with constant of proportionality 6. (a) Show that the water level falls at a constant rate. (b) Find how long it takes to empty the bowl if it is originally full to the brim.

85. The hull of a boat has widths given by the following table. Reading across a row of the table gives widths at points 0, 10, . . . , 60 feet from the front to the back at a certain level below waterline. Reading down a column of the table gives widths at levels 0, 2, 4, 6, 8 feet below waterline at a certain distance from the front. Use the trapezoidal rule to estimate the volume of the hull below waterline. Front of boat ⟶ Back of boat

𝑑𝑦 = 12𝑡. 𝑑𝑡

(a) Find the change in position in the 𝑥 and 𝑦coordinates between 𝑡 = 0 and 𝑡 = 3. (b) If the particle passes through (−7, 11) at 𝑡 = 0, find its position at 𝑡 = 3. (c) Find the distance traveled by the particle from time 𝑡 = 0 to 𝑡 = 3.

419

0

10

20

30

40

50

60 10

0

2

8

13

16

17

16

Depth

2

1

4

8

10

11

10

8

below

4

0

3

4

6

7

6

4

waterline

6

0

1

2

3

4

3

2

(in feet)

8

0

0

1

1

1

1

1

420

Chapter 8 USING THE DEFINITE INTEGRAL

86. (a) Write an integral which represents the circumference of a circle of radius 𝑟. (b) Evaluate the integral, and show that you get the answer you expect. 87. Find a curve 𝑦 = 𝑓 (𝑥) whose arc length from 𝑥 = 1 to 𝑥 = 4 is given by 4√ √ 1 + 𝑥 𝑑𝑥. ∫1 88. Write a simplified expression that represents the arc length of the concave-down portion of the graph of 2 𝑓 (𝑥) = 𝑒−𝑥 . Do not evaluate your answer.

89. Write an expression for the arc length of the portion of the graph of 𝑓 (𝑥) = 𝑥4 − 8𝑥3 + 18𝑥2 + 3𝑥 + 7 that is concave down. Do not simplify or evaluate the answer. 90. With 𝑥 and 𝑏 in meters, a chain hangs in the shape of the catenary cosh 𝑥 = 12 (𝑒𝑥 + 𝑒−𝑥 ) for −𝑏 ≤ 𝑥 ≤ 𝑏. If the chain is 10 meters long, how far apart are its ends? 91. The path of a robotic camera inspecting a suspension bridge cable is 𝑦 = 𝑎(𝑒𝑥∕𝑎 + 𝑒−𝑥∕𝑎 )∕2 for −5 ≤ 𝑥 ≤ 5 with 𝑎 constant. Find, but do not evaluate, an integral that gives the distance traveled by the camera.

Strengthen Your Understanding In Problems 92–95, explain what is wrong with the statement.

98. Two different curves from (0, 0) to (10, 0) that have the same arc length.

92. The solid obtained by rotating the region bounded by the curves 𝑦 = 2𝑥 and 𝑦 = 3𝑥 between 𝑥 = 0 and 𝑥 = 5 5 around the 𝑥-axis has volume ∫0 𝜋(3𝑥 − 2𝑥)2 𝑑𝑥.

99. A function 𝑓 (𝑥) whose graph passes through the points (0, 0) and (1, 1) and whose √ arc length between 𝑥 = 0 and 𝑥 = 1 is greater than 2.

93. The arc length of√the curve 𝑦 = sin 𝑥 from 𝑥 = 0 to 𝜋∕4 1 + sin2 𝑥 𝑑𝑥. 𝑥 = 𝜋∕4 is ∫0

94. The arc length of the curve 𝑦 = 𝑥5 between 𝑥 = 0 and 𝑥 = 2 is less than 32. 95. The circumference of a circle with parametric equations 𝑥 = cos(2𝜋𝑡), 𝑦 = sin(2𝜋𝑡) is 2

∫0

√ (−2𝜋 sin(2𝜋𝑡))2 + (2𝜋 cos(2𝜋𝑡))2 𝑑𝑡.

In Problems 96–99, give an example of:

Are the statements in Problems 100–103 true or false? If a statement is true, explain how you know. If a statement is false, give a counterexample. 100. Of two solids of revolution, the one with the greater volume is obtained by revolving the region in the plane with the greater area. 101. If 𝑓 is differentiable on the interval [0, 10], then the arc length of the graph of 𝑓 on the interval [0, 1] is less than the arc length of the graph of 𝑓 on the interval [1, 10].

96. A region in the plane which gives the same volume whether rotated about the 𝑥-axis or the 𝑦-axis.

102. If 𝑓 is concave up for all 𝑥 and 𝑓 ′ (0) = 3∕4, then the arc length of the graph of 𝑓 on the interval [0, 4] is at least 5.

97. A region where the solid obtained by rotating the region around the 𝑥-axis has greater volume than the solid obtained by revolving the region around the 𝑦-axis.

103. If 𝑓 is concave down for all 𝑥 and 𝑓 ′ (0) = 3∕4, then the arc length of the graph of 𝑓 on the interval [0, 4] is at most 5.

8.3

AREA AND ARC LENGTH IN POLAR COORDINATES Many curves and regions in the plane are easier to describe in polar coordinates than in Cartesian coordinates. Thus their areas and arc lengths are best found using integrals in polar coordinates. A point, 𝑃 , in the plane is often identified by its Cartesian coordinates (𝑥, 𝑦), where 𝑥 is the horizontal distance to the point from the origin and 𝑦 is the vertical distance.2 Alternatively, we can identify the point, 𝑃 , by specifying its distance, 𝑟, from the origin and the angle, 𝜃, shown in Figure 8.42. The angle 𝜃 is measured counterclockwise from the positive 𝑥-axis to the line joining 𝑃 to the origin. The labels 𝑟 and 𝜃 are called the polar coordinates of point 𝑃 .

2 Cartesian

coordinates can also be called rectangular coordinates.

8.3 AREA AND ARC LENGTH IN POLAR COORDINATES

421

𝑦

𝑦 𝑃

(𝑥, 𝑦)



6

𝑃

4

𝑈

𝑟 𝑄

𝑅

𝑦

𝑥 −5

𝜃



𝑥

❄𝑥 ✲

3 −5

Figure 8.42: Cartesian and polar coordinates for the point 𝑃

𝑉

Figure 8.43: Points on the plane for Example 1

Relation Between Cartesian and Polar Coordinates From the right triangle in Figure 8.42, we see that ∙ 𝑥 = 𝑟 cos 𝜃 and 𝑦 = 𝑟 sin 𝜃 √ 𝑦 ∙ 𝑟 = 𝑥2 + 𝑦2 and tan 𝜃 = , 𝑥 ≠ 0 𝑥 √ √ The angle 𝜃 is determined by the equations cos 𝜃 = 𝑥∕ 𝑥2 + 𝑦2 and sin 𝜃 = 𝑦∕ 𝑥2 + 𝑦2 . Warning: In general 𝜃 ≠ tan−1 (𝑦∕𝑥). It is not possible to determine which quadrant 𝜃 is in from the value of tan 𝜃 alone. Example 1

(a) Give Cartesian coordinates for the points with polar coordinates (𝑟, 𝜃) given by 𝑃 = (7, 𝜋∕3), 𝑄 = (5, 0), 𝑅 = (5, 𝜋). (b) Give polar coordinates for the points with Cartesian coordinates (𝑥, 𝑦) given by 𝑈 = (3, 4) and 𝑉 = (0, −5).

Solution

(a) See Figure 8.43. Point 𝑃 is a distance of 7 from the origin. The angle 𝜃 is 𝜋∕3 radians (60◦). The Cartesian coordinates of 𝑃 are √ 7 3 7 𝜋 𝜋 and 𝑦 = 𝑟 sin 𝜃 = 7 sin = . 𝑥 = 𝑟 cos 𝜃 = 7 cos = 3 2 3 2 Point 𝑄 is located a distance of 5 units along the positive 𝑥-axis with Cartesian coordinates 𝑥 = 𝑟 cos 𝜃 = 5 cos 0 = 5

and

𝑦 = 𝑟 sin 𝜃 = 5 sin 0 = 0.

For point 𝑅, which is on the negative 𝑥-axis, 𝑥 = 𝑟 cos 𝜃 = 5 cos 𝜋 = −5 and 𝑦 = 𝑟 sin 𝜃 = 5 sin 𝜋 = 0. √ (b) For 𝑈 = (3, 4), we have 𝑟 = 32 + 42 = 5 and tan 𝜃 = 4∕3. A possible value for 𝜃 is 𝜃 = arctan 4∕3 = 0.927 radians, or about 53◦ . The polar coordinates of 𝑈 are (5, 0.927). The point 𝑉 falls on the negative 𝑦-axis, so we can choose 𝑟 = 5, 𝜃 = 3𝜋∕2 for its polar coordinates. In this case, we cannot use tan 𝜃 = 𝑦∕𝑥 to find 𝜃, because tan 𝜃 = 𝑦∕𝑥 = −5∕0 is undefined. Because the angle 𝜃 can be allowed to wrap around the origin more than once, there are many possibilities for the polar coordinates of a point. For the point 𝑉 in Example 1, we can also choose 𝜃 = −𝜋∕2 or 𝜃 = 7𝜋∕2, so that (5, −𝜋∕2), (5, 7𝜋∕2), and (5, 3𝜋∕2) are all polar coordinates for 𝑉 . However, we often choose 𝜃 between 0 and 2𝜋.

422

Chapter 8 USING THE DEFINITE INTEGRAL

Example 2

Give three possible sets of polar coordinates for the point 𝑃 in Figure 8.44. 𝜋∕2 3𝜋∕4

𝜋∕4 𝑃

𝜋 2

5𝜋∕4

3𝜋∕2

4

0

7𝜋∕4

Figure 8.44

Solution

Because 𝑟 = 3 and 𝜃 = 𝜋∕4, one set of polar coordinates for 𝑃 is (3, 𝜋∕4). We can also use 𝜃 = 𝜋∕4 + 2𝜋 = 9𝜋∕4 and 𝜃 = 𝜋∕4 − 2𝜋 = −7𝜋∕4, to get (3, 9𝜋∕4) and (3, −7𝜋∕4).

Graphing Equations in Polar Coordinates The equations for certain graphs are much simpler when expressed in polar coordinates than in Cartesian coordinates. On the other hand, some graphs that have simple equations in Cartesian coordinates have complicated equations in polar coordinates. Example 3

(a) Describe in words the graphs of the equation 𝑦 = 1 (in Cartesian coordinates) and the equation 𝑟 = 1 (in polar coordinates). (b) Write the equation 𝑟 = 1 using Cartesian coordinates. Write the equation 𝑦 = 1 using polar coordinates.

Solution

(a) The equation 𝑦 = 1 describes a horizontal line. Since the equation 𝑦 = 1 places no restrictions on the value of 𝑥, it describes every point having a 𝑦-value of 1, no matter what the value of its 𝑥-coordinate. Similarly, the equation 𝑟 = 1 places no restrictions on the value of 𝜃. Thus, it describes every point having an 𝑟-value of 1, that is, having a distance of 1 from the origin. This set of points is the unit circle. See Figure 8.45. 𝑦

𝑦 5𝜋 2

𝑦=1

𝑟=𝜃 𝑥

𝜋 2

𝜋

𝑥 2𝜋

𝑟=1 Figure 8.45: The graph of the equation 𝑟 = 1 is the unit circle because 𝑟 = 1 for every point regardless of the value of 𝜃. The graph of 𝑦 = 1 is a horizontal line since 𝑦 = 1 for any 𝑥

−3𝜋 2

Figure 8.46: A graph of the Archimedean spiral 𝑟 = 𝜃

√ √ (b) Since 𝑟 = 𝑥2 + 𝑦2 , we rewrite the equation 𝑟 = 1 using Cartesian coordinates as 𝑥2 + 𝑦2 = 1, or, squaring both sides, as 𝑥2 + 𝑦2 = 1. We see that the equation for the unit circle is simpler in polar coordinates than it is in Cartesian coordinates. On the other hand, since 𝑦 = 𝑟 sin 𝜃, we can rewrite the equation 𝑦 = 1 in polar coordinates as 𝑟 sin 𝜃 = 1, or, dividing both sides by sin 𝜃, as 𝑟 = 1∕ sin 𝜃. We see that the equation for this horizontal line is simpler in Cartesian coordinates than in polar coordinates.

423

8.3 AREA AND ARC LENGTH IN POLAR COORDINATES

Example 4

Graph the equation 𝑟 = 𝜃. The graph is called an Archimedean spiral after the Greek mathematician Archimedes who described its properties (although not by using polar coordinates).

Solution

To construct this graph, use the values in Table 8.1. To help us visualize the shape of the spiral, we convert the angles in Table 8.1 to degrees and the 𝑟-values to decimals. See Table 8.2. Table 8.1

Table 8.2

Points on the Archimedean spiral 𝑟 = 𝜃, with 𝜃 in radians 𝜃

0

𝑟

0

𝜋 6 𝜋 6

𝜋 3 𝜋 3

𝜋 2 𝜋 2

2𝜋 3 2𝜋 3

5𝜋 6 5𝜋 6

7𝜋 6 7𝜋 6

𝜋 𝜋

4𝜋 3 4𝜋 3

3𝜋 2 3𝜋 2

Points on the Archimedean spiral 𝑟 = 𝜃, with 𝜃 in degrees

𝜃

0

30◦

60◦

90◦

120◦

150◦

180◦

210◦

240◦

270◦

𝑟

0.00

0.52

1.05

1.57

2.09

2.62

3.14

3.67

4.19

4.71

Notice that as the angle 𝜃 increases, points on the curve move farther from the origin. At 0◦ , the point is at the origin. At 30◦, it is 0.52 units away from the origin, at 60◦ it is 1.05 units away, and at 90◦ it is 1.57 units away. As the angle winds around, the point traces out a curve that moves away from the origin, giving a spiral. (See Figure 8.46.) In our definition, 𝑟 is positive. However, graphs of curves in polar coordinates are traditionally drawn using negative values of 𝑟 as well, because this makes the graphs symmetric. If an equation 𝑟 = 𝑓 (𝜃) gives a negative 𝑟-value, it is plotted in the opposite direction to 𝜃. See Examples 5 and 6 and Figures 8.47 and 8.49. Example 5

For 𝑎 > 0 and 𝑛 a positive integer, curves of the form 𝑟 = 𝑎 sin 𝑛𝜃 or 𝑟 = 𝑎 cos 𝑛𝜃 are called roses. Graph the roses (a) 𝑟 = 3 sin 2𝜃 (b) 𝑟 = 4 cos 3𝜃

Solution

(a) Using a calculator or making a table of values, we see that the graph is a rose with four petals, each extending a distance of 3 from base to tip. See Figure 8.47. Negative values of 𝑟 for 𝜋∕2 < 𝜃 < 𝜋 and 3𝜋∕2 < 𝜃 < 2𝜋 give the petals in Quadrants II and IV. For example, 𝜃 = 3𝜋∕4 gives 𝑟 = −3, which is plotted 3 units from the origin in the direction opposite to 𝜃 = 3𝜋∕4, namely in Quadrant IV. (b) The graph is a rose with three petals, each extending 4 from base to tip. See Figure 8.48. 𝑦

𝑦 𝑟 = 3 sin 2𝜃

𝑟 = 4 cos 3𝜃

𝑥

Figure 8.47: Graph of 𝑟 = 3 sin 2𝜃 (petals in Quadrants II and IV have 𝑟 < 0)

𝑥

Figure 8.48: Graph of 𝑟 = 4 cos 3𝜃

424

Chapter 8 USING THE DEFINITE INTEGRAL

Example 6

Curves of the form 𝑟 = 𝑎 + 𝑏 sin 𝜃 or 𝑟 = 𝑎 + 𝑏 cos 𝜃 are called limaçons. Graph 𝑟 = 1 + 2 cos 𝜃 and 𝑟 = 3 + 2 cos 𝜃.

Solution

See Figures 8.49 and 8.50. The equation 𝑟 = 1 + 2 cos 𝜃 leads to negative 𝑟 values for some 𝜃 values between 𝜋∕2 and 3𝜋∕2; these values give the inner loop in Figure 8.49. For example, 𝜃 = 𝜋 gives 𝑟 = −1, which is plotted 1 unit from the origin in the direction opposite to 𝜃 = 𝜋, namely on the positive 𝑥-axis. The equation 𝑟 = 3 + 2 cos 𝜃 does not lead to negative 𝑟-values. 𝑦

𝑦 𝑟 = 3 + 2 cos 𝜃

𝑟 = 1 + 2 cos 𝜃

𝑥

𝑥

Figure 8.49: Graph of 𝑟 = 1 + 2 cos 𝜃 (inner loop has 𝑟 < 0)

Figure 8.50: Graph of 𝑟 = 3 + 2 cos 𝜃

Polar coordinates can be used with inequalities to describe regions that are obtained from circles. Such regions are often much harder to represent in Cartesian coordinates. Example 7

Using inequalities, describe a compact disc with outer diameter 120 mm and inner diameter 15 mm.

Solution

The compact disc lies between two circles of radius 7.5 mm and 60 mm. See Figure 8.51. Thus, if the origin is at the center, the disc is represented by 7.5 ≤ 𝑟 ≤ 60

and

0 ≤ 𝜃 ≤ 2𝜋.

𝑦

𝑦

9′′ 60 mm

■7.5 mm

𝜋∕6

𝑥

Figure 8.51: Compact disc

𝑥

Figure 8.52: Pizza slice

Example 8

An 18-inch pizza is cut into 12 slices. Use inequalities to describe one of the slices.

Solution

The pizza has radius 9 inches; the angle at the center is 2𝜋∕12 = 𝜋∕6. See Figure 8.52. Thus, if the origin is at the center of the original pizza, the slice is represented by 0≤𝑟≤9

and

0≤𝜃≤

𝜋 . 6

425

8.3 AREA AND ARC LENGTH IN POLAR COORDINATES

Area in Polar Coordinates We can use a definite integral to find the area of a region described in polar coordinates. As previously, we slice the region into small pieces, construct a Riemann sum, and take a limit to obtain the definite integral. In this case, the slices are approximately circular sectors. To calculate the area of the sector in Figure 8.53, we think of the area of the sector as a fraction 𝜃∕2𝜋 of the area of the entire circle (for 𝜃 in radians). Then Area of sector =

𝜃 1 ⋅ 𝜋𝑟2 = 𝑟2 𝜃. 2𝜋 2 𝑦 Δ𝜃



𝜃

𝜃 𝑟

Figure 8.53: Area of shaded sector = 21 𝑟2 𝜃 (for 𝜃 in radians)

𝑥

Figure 8.54: Finding the area of the limaçon 𝑟 = 3 + 2 cos 𝜃

Example 9

Use circular sectors to set up a definite integral to calculate the area of the region bounded by the limaçon 𝑟 = 3 + 2 cos 𝜃, for 0 ≤ 𝜃 ≤ 2𝜋. See Figure 8.54.

Solution

The slices are not exactly circular sectors because the radius 𝑟 depends on 𝜃. However, Area of sector ≈

1 1 2 𝑟 Δ𝜃 = (3 + 2 cos 𝜃)2 Δ𝜃. 2 2

Thus, for the whole area, Area of region ≈

∑1 2

(3 + 2 cos 𝜃)2 Δ𝜃.

Taking the limit as 𝑛 → ∞ and Δ𝜃 → 0 gives the integral 2𝜋

Area =

∫0

1 (3 + 2 cos 𝜃)2 𝑑𝜃. 2

To compute this integral, we expand the integrand and use integration by parts or formula IV-18 from the table of integrals: 2𝜋

1 (9 + 12 cos 𝜃 + 4 cos2 𝜃) 𝑑𝜃 2 ∫0 ( )|2𝜋 4 1 9𝜃 + 12 sin 𝜃 + (cos 𝜃 sin 𝜃 + 𝜃) || = 2 2 |0 1 = (18𝜋 + 0 + 4𝜋) = 11𝜋. 2

Area =

426

Chapter 8 USING THE DEFINITE INTEGRAL

The reasoning in Example 9 suggests a general area formula.

For a curve 𝑟 = 𝑓 (𝜃), with 𝑓 (𝜃) ≥ 0, the area in Figure 8.55 is given by 𝛽

Area of region enclosed =

1 𝑓 (𝜃)2 𝑑𝜃. 2 ∫𝛼

𝑦 𝑟 = 𝑓 (𝜃)

𝛽

𝛼

𝑥

Figure 8.55: Area in polar coordinates

Example 10

Find the area of one petal of the four-petal rose 𝑟 = 3 sin 2𝜃 in Figure 8.56. 𝑦 𝑟 = 3 sin 2𝜃

𝑥

Figure 8.56: One petal of the rose 𝑟 = 3 sin 2𝜃 with 0 ≤ 𝜃 ≤ 𝜋∕2

Solution

The petal in the first quadrant is described by 𝑟 = 3 sin 2𝜃 for 0 ≤ 𝜃 ≤ 𝜋∕2, so Area of shaded region =

1 2 ∫0

𝜋∕2

(3 sin 2𝜃)2 𝑑𝜃 =

9 2 ∫0

𝜋∕2

sin2 2𝜃 𝑑𝜃.

Using the substitution 𝑤 = 2𝜃, we rewrite the integral and use integration by parts or formula IV-17 from the table of integrals: 𝜋∕2

𝜋

9 9 sin2 2𝜃 𝑑𝜃 = sin2 𝑤 𝑑𝑤 2 ∫0 4 ∫0 ( ) 1 1 |𝜋 9 𝜋 9𝜋 9 − cos 𝑤 sin 𝑤 + 𝑤 || = ⋅ = . = 4 2 2 |0 4 2 8

Area =

8.3 AREA AND ARC LENGTH IN POLAR COORDINATES

427

Slope in Polar Coordinates For a curve 𝑟 = 𝑓 (𝜃), we can express 𝑥 and 𝑦 in terms of 𝜃 as a parameter, giving 𝑥 = 𝑟 cos 𝜃 = 𝑓 (𝜃) cos 𝜃

and

𝑦 = 𝑟 sin 𝜃 = 𝑓 (𝜃) sin 𝜃.

To find the slope of the curve, we use the formula for the slope of a parametric curve 𝑑𝑦∕𝑑𝜃 𝑑𝑦 = . 𝑑𝑥 𝑑𝑥∕𝑑𝜃

Example 11

Find the slope of the curve 𝑟 = 3 sin 2𝜃 at 𝜃 = 𝜋∕3.

Solution

Expressing 𝑥 and 𝑦 in terms of 𝜃, we have 𝑥 = 3 sin(2𝜃) cos 𝜃

and

𝑦 = 3 sin(2𝜃) sin 𝜃.

The slope is given by 𝑑𝑦 6 cos(2𝜃) sin 𝜃 + 3 sin(2𝜃) cos 𝜃 = . 𝑑𝑥 6 cos(2𝜃) cos 𝜃 − 3 sin(2𝜃) sin 𝜃 At 𝜃 = 𝜋∕3, we have √ √ √ 6(−1∕2)( 3∕2) + 3( 3∕2)(1∕2) 3 𝑑𝑦 || . = = √ √ 𝑑𝑥 ||𝜃=𝜋∕3 6(−1∕2)(1∕2) − 3( 3∕2)( 3∕2) 5

Arc Length in Polar Coordinates We can calculate the arc length of the curve 𝑟 = 𝑓 (𝜃) by expressing 𝑥 and 𝑦 in terms of 𝜃 as a parameter 𝑥 = 𝑓 (𝜃) cos 𝜃 𝑦 = 𝑓 (𝜃) sin 𝜃 and using the formula for the arc length of a parametric curve: 𝛽

Arc length =

∫𝛼

√ (

𝑑𝑥 𝑑𝜃

)2

+

(

𝑑𝑦 𝑑𝜃

)2

𝑑𝜃.

The calculations may be simplified by using the alternate form of the arc length integral in Problem 45. Example 12

Find the arc length of one petal of the rose 𝑟 = 3 sin 2𝜃 for 0 ≤ 𝜃 ≤ 𝜋∕2. See Figure 8.56.

Solution

The curve is given parametrically by 𝑥 = 3 sin(2𝜃) cos 𝜃

and

𝑦 = 3 sin(2𝜃) sin 𝜃.

Thus, calculating 𝑑𝑥∕𝑑𝜃 and 𝑑𝑦∕𝑑𝜃 and evaluating the integral on a calculator, we have: Arc length =

𝜋∕2 √

∫0 = 7.266.

(6 cos(2𝜃) cos 𝜃 − 3 sin(2𝜃) sin 𝜃)2 + (6 cos(2𝜃) sin 𝜃 + 3 sin(2𝜃) cos 𝜃)2 𝑑𝜃

428

Chapter 8 USING THE DEFINITE INTEGRAL

Exercises and Problems for Section 8.3 EXERCISES Convert the polar coordinates in Exercises 1–4 to Cartesian coordinates. Give exact answers. √ 1. (1, 2𝜋∕3) 2. ( 3, −3𝜋∕4) √ 3. (2 3, −𝜋∕6) 4. (2, 5𝜋∕6)

In Exercises 17–19, give inequalities for 𝑟 and 𝜃 which describe the following regions in polar coordinates. 17.

𝑦 𝑦=𝑥 Circular ✕ arcs



(3, 3) (2, 2)

Convert the Cartesian coordinates in Exercises 5–8 to polar coordinates. 5. (1, 1) √ √ 7. ( 6, − 2)

6. (−1, 0) √ 8. (− 3, 1)

9. (a) Make a table of values for the equation 𝑟 = 1 − sin 𝜃. Include 𝜃 = 0, 𝜋∕3, 𝜋∕2, 2𝜋∕3, 𝜋, …. (b) Use the table to graph the equation 𝑟 = 1 − sin 𝜃 in the 𝑥𝑦-plane. This curve is called a cardioid. (c) At what point(s) does the cardioid 𝑟 = 1 − sin 𝜃 intersect a circle of radius 1/2 centered at the origin? (d) Graph the curve 𝑟 = 1 − sin 2𝜃 in the 𝑥𝑦-plane. Compare this graph to the cardioid 𝑟 = 1 − sin 𝜃.

𝑥

18.

𝑦

√ ( 3, 1)



√ ( 3, −1)

10. Graph the equation 𝑟 = 1 − sin(𝑛𝜃), for 𝑛 = 1, 2, 3, 4. What is the relationship between the value of 𝑛 and the shape of the graph? 11. Graph the equation 𝑟 = 1 − sin 𝜃, with 0 ≤ 𝜃 ≤ 𝑛𝜋, for 𝑛 = 2, 3, 4. What is the relationship between the value of 𝑛 and the shape of the graph? 12. Graph the equation 𝑟 = 1−𝑛 sin 𝜃, for 𝑛 = 2, 3, 4. What is the relationship between the value of 𝑛 and the shape of the graph? 13. Graph the equation 𝑟 = 1 − cos 𝜃. Describe its relationship to 𝑟 = 1 − sin 𝜃. 14. Give inequalities that describe the flat surface of a washer that is one inch in diameter and has an inner hole with a diameter of 3/8 inch. 15. Graph the equation 𝑟 = 1 − sin(2𝜃) for 0 ≤ 𝜃 ≤ 2𝜋. There are two loops. For each loop, give a restriction on 𝜃 that shows all of that loop and none of the other loop. 16. A slice of pizza is one eighth of a circle of radius 1 foot. The slice is in the first quadrant, with one edge along the 𝑥-axis, and the center of the pizza at the origin. Give inequalities describing this region using: (a)

Polar coordinates

(b) Rectangular coordinates

Circular arc

𝑥

19.

𝑦

1



Circular arc

𝑥 1

2

Note: Region extends indefinitely in the 𝑦-direction.

20. Find the slope of the curve 𝑟 = 2 at 𝜃 = 𝜋∕4. 21. Find the slope of the curve 𝑟 = 𝑒𝜃 at 𝜃 = 𝜋∕2. 22. Find the slope of the curve 𝑟 = 1 − cos 𝜃 at 𝜃 = 𝜋∕2. 23. Find the arc length of the curve 𝑟 = 𝑒𝜃 from 𝜃 = 𝜋∕2 to 𝜃 = 𝜋. 24. Find the arc length of the curve 𝑟 = 𝜃 2 from 𝜃 = 0 to 𝜃 = 2𝜋.

PROBLEMS 25. Sketch the polar region described by the following integral expression for area: 1 2 ∫0

𝜋∕3

sin2 (3𝜃) 𝑑𝜃.

26. Find the area inside the spiral 𝑟 = 𝜃 for 0 ≤ 𝜃 ≤ 2𝜋.

27. Find the area between the two spirals 𝑟 = 𝜃 and 𝑟 = 2𝜃 for 0 ≤ 𝜃 ≤ 2𝜋.

8.4 DENSITY AND CENTER OF MASS

28. Find the area inside the cardioid 𝑟 = 1 + cos 𝜃 for 0 ≤ 𝜃 ≤ 2𝜋. 29. (a) In polar coordinates, write equations for the line 𝑥 = 1 and the circle of radius 2 centered at the origin. (b) Write an integral in polar coordinates representing the area of the region to the right of 𝑥 = 1 and inside the circle. (c) Evaluate the integral. 30. Show that the area formula for polar coordinates gives the expected answer for the area of the circle 𝑟 = 𝑎 for 0 ≤ 𝜃 ≤ 2𝜋. 31. Show that the arc length formula for polar coordinates gives the expected answer for the circumference of the circle 𝑟 = 𝑎 for 0 ≤ 𝜃 ≤ 2𝜋. 32. Find the area inside the circle 𝑟 = 1 and outside the cardioid 𝑟 = 1 + sin 𝜃. 33. Find the area inside the cardioid 𝑟 = 1 − sin 𝜃 and outside the circle 𝑟 = 1∕2. 34. Find the area lying outside 𝑟 = 2 cos 𝜃 and inside 𝑟 = 1 + cos 𝜃. 35. (a) Graph 𝑟 = 2 cos 𝜃 and 𝑟 = 2 sin 𝜃 on the same axes. (b) Using polar coordinates, find the area of the region shared by both curves. 36. For what value of 𝑎 is the area enclosed by 𝑟 = 𝜃, 𝜃 = 0, and 𝜃 = 𝑎 equal to 1? 37. (a) Sketch the bounded region inside the lemniscate √ 𝑟2 = 4 cos 2𝜃 and outside the circle 𝑟 = 2. (b) Compute the area of the region described in part (a).

429

38. Using Example 11 on page 427, find the equation of the tangent line to the curve 𝑟 = 3 sin 2𝜃 at 𝜃 = 𝜋∕3. 39. Using Example 11 on page 427 and Figure 8.47, find the points where the curve 𝑟 = 3 sin 2𝜃 has horizontal and vertical tangents. 40. For what values of 𝜃 on the polar curve 𝑟 = 𝜃, with 0 ≤ 𝜃 ≤ 2𝜋, are the tangent lines horizontal? Vertical? 41. (a) In Cartesian coordinates, write an equation for the tangent line to 𝑟 = 1∕𝜃 at 𝜃 = 𝜋∕2. (b) The graph of 𝑟 = 1∕𝜃 has a horizontal asymptote as 𝜃 approaches 0. Find the equation of this asymptote. 42. Find the maximum value of the 𝑦-coordinate of points on the limaçon 𝑟 = 1 + 2 cos 𝜃. Find the arc length of the curves in Problems 43–44. 43. 𝑟 = 𝜃, 0 ≤ 𝜃 ≤ 2𝜋 44. 𝑟 = 1∕𝜃, 𝜋 ≤ 𝜃 ≤ 2𝜋 45. For the curve 𝑟 = 𝑓 (𝜃) from 𝜃 = 𝛼 to 𝜃 = 𝛽, show that 𝛽

Arc length =

∫𝛼



(𝑓 ′ (𝜃))2 + (𝑓 (𝜃))2 𝑑𝜃.

46. Find the arc length of the spiral 𝑟 = 𝜃 where 0 ≤ 𝜃 ≤ 𝜋. 47. Find the arc length of part of the cardioid 𝑟 = 1 + cos 𝜃 where 0 ≤ 𝜃 ≤ 𝜋∕2.

Strengthen Your Understanding In Problems 48–51, explain what is wrong with the statement. 48. The point with Cartesian coordinates (𝑥, 𝑦) has polar √ coordinates 𝑟 = 𝑥2 + 𝑦2 , 𝜃 = tan−1 (𝑦∕𝑥). 49. All points of the curve 𝑟 = sin(2𝜃) for 𝜋∕2 < 𝜃 < 𝜋 are in quadrant II. 50. If the slope of the curve 𝑟 = 𝑓 (𝜃) is positive, then 𝑑𝑟∕𝑑𝜃 is positive. 51. Any polar curve that is symmetric about both the 𝑥 and 𝑦 axes must be a circle, centered at the origin.

8.4

In Problems 52–55, give an example of: 52. Two different pairs of polar coordinates (𝑟, 𝜃) that correspond to the same point in the plane. 53. The equation of a circle in polar coordinates. 54. A polar curve 𝑟 = 𝑓 (𝜃) that is symmetric about neither the 𝑥-axis nor the 𝑦-axis. 55. A polar curve 𝑟 = 𝑓 (𝜃) other than a circle that is symmetric about the 𝑥-axis.

DENSITY AND CENTER OF MASS

Density and How to Slice a Region The examples in this section involve the idea of density. For example, • A population density is measured in, say, people per mile (along the edge of a road), or people per unit area (in a city), or bacteria per cubic centimeter (in a test tube). • The density of a substance (e.g. air, wood, or metal) is the mass of a unit volume of the substance and is measured in, say, grams per cubic centimeter. Suppose we want to calculate the total mass or total population, but the density is not constant over a region.

430

Chapter 8 USING THE DEFINITE INTEGRAL

To find total quantity from density: Divide the region into small pieces in such a way that the density is approximately constant on each piece, and add the contributions of the pieces.

Example 1

The Massachusetts Turnpike (“the Pike”) starts in the middle of Boston and heads west. The number of people living next to it varies as it gets farther from the city. Suppose that, 𝑥 miles out of town, the population density adjacent to the Pike is 𝑃 = 𝑓 (𝑥) people/mile. Express the total population living next to the Pike within 5 miles of Boston as a definite integral.

Solution

Divide the Pike into segments of length Δ𝑥. The population density at the center of Boston is 𝑓 (0); let’s use that density for the first segment. This gives an estimate of People living in first segment ≈ 𝑓 (0) people∕ mile ⋅ Δ𝑥 mile = 𝑓 (0)Δ𝑥 people. Population ≈ 𝑓 (𝑥)Δ𝑥 Points west

5

❄ ✛ ✲✛

0



𝑥

Boston

Δ𝑥 Figure 8.57: Population along the Massachusetts Turnpike

Similarly, the population in a typical segment 𝑥 miles from the center of Boston is the population density times the length of the interval, or roughly 𝑓 (𝑥) Δ𝑥. (See Figure 8.57.) The sum of all these estimates gives the estimate ∑ Total population ≈ 𝑓 (𝑥) Δ𝑥. Letting Δ𝑥 → 0 gives

Total population = lim

Δ𝑥→0



5

𝑓 (𝑥) Δ𝑥 =

∫0

𝑓 (𝑥) 𝑑𝑥.

The 5 and 0 in the limits of the integral are the upper and lower limits of the interval over which we are integrating.

Example 2

The air density ℎ meters above the earth’s surface is 𝑓 (ℎ) kg/m3 . Find the mass of a cylindrical column of air 2 meters in diameter and 25 kilometers high, with base on the surface of the earth.

Solution

The column of air is a circular cylinder 2 meters in diameter and 25 kilometers, or 25,000 meters, high. First we must decide how we are going to slice this column. Since the air density varies with altitude but remains constant horizontally, we take horizontal slices of air. That way, the density will be more or less constant over the whole slice, being close to its value at the bottom of the slice. (See Figure 8.58.)

Figure 8.58: Slicing a column of air horizontally

8.4 DENSITY AND CENTER OF MASS

431

A slice is a cylinder of height Δℎ and diameter 2 m, so its radius is 1 m. We find the approximate mass of the slice by multiplying its volume by its density. If the thickness of the slice is Δℎ, then its volume is 𝜋𝑟2 ⋅ Δℎ = 𝜋12 ⋅ Δℎ = 𝜋 Δℎ m3 . The density of the slice is given by 𝑓 (ℎ). Thus, Mass of slice ≈ Volume ⋅ Density = (𝜋Δℎ m3 )(𝑓 (ℎ) kg∕m3 ) = 𝜋 Δℎ ⋅ 𝑓 (ℎ) kg. Adding these slices up yields a Riemann sum: Total mass ≈



𝜋𝑓 (ℎ) Δℎ kg.

As Δℎ → 0, this sum approximates the definite integral: 25,000

Total mass =

∫0

𝜋𝑓 (ℎ) 𝑑ℎ kg.

In order to get a numerical value for the mass of air, we need an explicit formula for the density as a function of height, as in the next example. Example 3

Find the mass of the column of air in Example 2 if the density of air at height ℎ is given by 𝑃 = 𝑓 (ℎ) = 1.28𝑒−0.000124ℎ kg/m3 .

Solution

Using the result of the previous example, we have ) ( |25,000 −1.28𝜋 −0.000124ℎ | 𝑒 𝜋1.28𝑒 𝑑ℎ = Mass = | ∫0 0.000124 |0 ( ) 1.28𝜋 = 𝑒0 − 𝑒−0.000124(25,000) ≈ 31,000 kg. 0.000124 25,000

−0.000124ℎ

It requires some thought to figure out how to slice a region. The key point is that you want the density to be nearly constant within each piece. Example 4

The population density in Ringsburg is a function of the distance from the city center. At 𝑟 miles from the center, the density is 𝑃 = 𝑓 (𝑟) people per square mile. Ringsburg is circular with radius 5 miles. Write a definite integral that expresses the total population of Ringsburg.

Solution

We want to slice Ringsburg up and estimate the population of each slice. If we were to take straightline slices, the population density would vary on each slice, since it depends on the distance from the city center. We want the population density to be pretty close to constant on each slice. We therefore take slices that are thin rings around the center. (See Figure 8.59.) Since the ring is very thin, we can approximate its area by straightening it into a thin rectangle. (See Figure 8.60.) The width of the rectangle is Δ𝑟 miles, and its length is approximately equal to the circumference of the ring, 2𝜋𝑟 miles, so its area is about 2𝜋𝑟 Δ𝑟 mi2 . Since Population on ring ≈ Density ⋅ Area, we get Population on ring ≈ (𝑓 (𝑟) people/mi2 )(2𝜋𝑟Δ𝑟 mi2 ) = 𝑓 (𝑟) ⋅ 2𝜋𝑟 Δ𝑟 people. Adding the contributions from each ring, we get ∑ Total population ≈ 2𝜋𝑟𝑓 (𝑟) Δ𝑟 people.

432

Chapter 8 USING THE DEFINITE INTEGRAL

So

5

Total population =

∫0

2𝜋𝑟𝑓 (𝑟) 𝑑𝑟 people.

5✸ 𝑟✲

✻ Δ𝑟

Width = Δ𝑟

✛ Figure 8.59: Ringsburg

2𝜋𝑟



Figure 8.60: Ring from Ringsburg (straightened out)

Note: You may wonder what happens if we calculate the area of the ring by subtracting the area of the inner circle (𝜋𝑟2 ) from the area of the outer circle (𝜋(𝑟 + Δ𝑟)2 ), giving Area = 𝜋(𝑟 + Δ𝑟)2 − 𝜋𝑟2 . Multiplying out and subtracting, we get Area = 𝜋(𝑟2 + 2𝑟 Δ𝑟 + (Δ𝑟)2 ) − 𝜋𝑟2 = 2𝜋𝑟 Δ𝑟 + 𝜋(Δ𝑟)2 . This expression differs from the one we used before by the 𝜋(Δ𝑟)2 term. However, as Δ𝑟 becomes very small, 𝜋(Δ𝑟)2 becomes much, much smaller. We say its smallness is of second order, since the power of the small factor, Δ𝑟, is 2. In the limit as Δ𝑟 → 0, we can ignore 𝜋(Δ𝑟)2 .

Center of Mass The center of mass of a mechanical system is important for studying its behavior when in motion. For example, some sport utility vehicles and light trucks tend to tip over in accidents, because of their high centers of mass. In this section, we first define the center of mass for a system of point masses on a line. Then we use the definite integral to extend this definition.

Point Masses Two children on a seesaw, one twice the weight of the other, will balance if the lighter child is twice as far from the pivot as the heavier child. Thus, the balance point is 2/3 of the way from the lighter child and 1/3 of the way from the heavier child. This balance point is the center of mass of the mechanical system consisting of the masses of the two children (we ignore the mass of the seesaw itself). See Figure 8.61. To find the balance point, we use the displacement (signed distance) of each child from the pivot to calculate the moment, where Moment of mass about pivot = Mass × Displacement from pivot. A moment represents the tendency of a child to turn the system about the pivot point; the seesaw balances if the total moment is zero. Thus, the center of mass is the point about which the total moment is zero.

433

8.4 DENSITY AND CENTER OF MASS

Heavy child mass 2𝑚

Seesaw



𝑙

✛ (1∕3)𝑙 ✲✛

2𝑚

child ✲ Light mass 𝑚



✛ 𝑥̄ ✲✛



(2∕3)𝑙

𝑚

Balance point

𝑙 − 𝑥̄



Center of mass

Figure 8.61: Children on seesaw

Figure 8.62: Center of mass of point masses

Example 5

Calculate the position of the center of mass of the children in Figure 8.61 using moments.

Solution

Suppose the center of mass in Figure 8.62 is at a distance of 𝑥̄ from the left end. The moment of the left mass about the center of mass is −2𝑚𝑥̄ (it is negative since it is to the left of the center of mass); the moment of the right mass about the center of mass is 𝑚(𝑙 − 𝑥). ̄ The system balances if −2𝑚𝑥̄ + 𝑚(𝑙 − 𝑥) ̄ =0

𝑚𝑙 − 3𝑚𝑥̄ = 0

or

1 𝑙. 3

𝑥̄ =

so

Thus, the center of mass is 𝑙∕3 from the left end. We use the same method to calculate the center of mass, 𝑥, ̄ of the system in Figure 8.63. The sum of the moments of the three masses about 𝑥̄ is 0, so 𝑚1 (𝑥1 − 𝑥) ̄ + 𝑚2 (𝑥2 − 𝑥) ̄ + 𝑚3 (𝑥3 − 𝑥) ̄ = 0. Solving for 𝑥, ̄ we get 𝑚1 𝑥̄ + 𝑚2 𝑥̄ + 𝑚3 𝑥̄ = 𝑚1 𝑥1 + 𝑚2 𝑥2 + 𝑚3 𝑥3 ∑3 𝑚𝑖 𝑥 𝑖 𝑚1 𝑥 1 + 𝑚 2 𝑥 2 + 𝑚 3 𝑥 3 . = ∑𝑖=1 𝑥̄ = 3 𝑚1 + 𝑚 2 + 𝑚 3 𝑚 𝑖=1

𝑖

Generalizing leads to the following formula:

The center of mass of a system of 𝑛 point masses 𝑚1 , 𝑚2 , … , 𝑚𝑛 located at positions 𝑥1 , 𝑥2 , … , 𝑥𝑛 along the 𝑥-axis is given by ∑ 𝑥𝑚 𝑥 = ∑ 𝑖 𝑖. 𝑚𝑖 The numerator is the sum of the moments of the masses about the origin; the denominator is the total mass of the system. Example 6

Show that the definition of 𝑥 gives the same answer as we found in Example 5.

Solution

Suppose the origin is at the left end of the seesaw in Figure 8.61. The total mass of the system is 2𝑚 + 𝑚 = 3𝑚. We compute ∑ 𝑥𝑚 𝑚𝑙 𝑙 1 (2𝑚 ⋅ 0 + 𝑚 ⋅ 𝑙) = = . 𝑥= ∑𝑖 𝑖 = 3𝑚 3𝑚 3 𝑚𝑖

𝑚1 𝑥1

0

𝑚2

𝑚3

𝑥2

𝑥3

Figure 8.63: Discrete masses 𝑚1 , 𝑚2 , 𝑚3

434

Chapter 8 USING THE DEFINITE INTEGRAL

Mass 𝑚𝑖 ≈ 𝛿(𝑥𝑖 )Δ𝑥

Δ𝑥

✛ ✲

✠ 𝑎

𝑥 𝑏

𝑥𝑖

Figure 8.64: Calculating the center of mass of an object of variable density, 𝛿(𝑥)

Continuous Mass Density Instead of discrete masses arranged along the 𝑥-axis, suppose we have an object lying on the 𝑥-axis between 𝑥 = 𝑎 and 𝑥 = 𝑏. At point 𝑥, suppose the object has mass density (mass per unit length) of 𝛿(𝑥). To calculate the center of mass of such an object, divide it into 𝑛 pieces, each of length Δ𝑥. On each piece, the density is nearly constant, so the mass of the piece is given by density times length. See Figure 8.64. Thus, if 𝑥𝑖 is a point in the 𝑖th piece, Mass of the 𝑖th piece, 𝑚𝑖 ≈ 𝛿(𝑥𝑖 )Δ𝑥. ∑ ∑ Then the formula for the center of mass, 𝑥̄ = 𝑥𝑖 𝑚𝑖 ∕ 𝑚𝑖 , applied to the 𝑛 pieces of the object gives ∑ 𝑥 𝛿(𝑥 )Δ𝑥 𝑥̄ = ∑ 𝑖 𝑖 . 𝛿(𝑥𝑖 )Δ𝑥

In the limit as 𝑛 → ∞ we have the following formula:

The center of mass 𝑥 of an object lying along the 𝑥-axis between 𝑥 = 𝑎 and 𝑥 = 𝑏 is 𝑏

𝑥=

∫𝑎 𝑥𝛿(𝑥) 𝑑𝑥 𝑏

,

∫𝑎 𝛿(𝑥) 𝑑𝑥 where 𝛿(𝑥) is the density (mass per unit length) of the object.

As in the discrete case, the denominator is the total mass of the object. Example 7

Find the center of mass of a 2-meter rod lying on the 𝑥-axis with its left end at the origin if: (a) The density is constant and the total mass is 5 kg. (b) The density is 𝛿(𝑥) = 15𝑥2 kg/m.

Solution

(a) Since the density is constant along the rod, we expect the balance point to be in the middle, that is, 𝑥̄ = 1. To check this, we compute 𝑥. ̄ The density is the total mass divided by the length, so 𝛿(𝑥) = 5∕2 kg/m. Then 2

5 2 Moment ∫0 𝑥 ⋅ 2 𝑑𝑥 1 5 𝑥2 || 𝑥̄ = = = ⋅ ⋅ | = 1 meter. Mass 5 5 2 2 |0 (b) Since more of the mass of the rod is closer to its right end (the density is greatest there), we expect the center of mass to be in the right half of the rod, that is, between 𝑥 = 1 and 𝑥 = 2. We have 2 2 Total mass = 15𝑥2 𝑑𝑥 = 5𝑥3 ||0 = 40 kg. ∫0 Thus, 2 2 2 Moment ∫0 𝑥 ⋅ 15𝑥 𝑑𝑥 15 𝑥4 || = = ⋅ | = 1.5 meter. 𝑥̄ = Mass 40 40 4 |0

8.4 DENSITY AND CENTER OF MASS

435

Two- and Three-Dimensional Regions For a system of masses that lies in the plane, the center of mass is a point with coordinates (𝑥, ̄ 𝑦). ̄ In three dimensions, the center of mass is a point with coordinates (𝑥, ̄ 𝑦, ̄ 𝑧). ̄ To compute the center of mass in three dimensions, we use the following formulas in which 𝐴𝑥 (𝑥) is the area of a slice perpendicular to the 𝑥-axis at 𝑥, and 𝐴𝑦 (𝑦) and 𝐴𝑧 (𝑧) are defined similarly. In two dimensions, we use the same formulas for 𝑥̄ and 𝑦, ̄ but we interpret 𝐴𝑥 (𝑥) and 𝐴𝑦 (𝑦) as the lengths of strips perpendicular to the 𝑥- and 𝑦-axes, respectively.

For a region of constant density 𝛿, the center of mass is given by 𝑥̄ =

∫ 𝑥𝛿𝐴𝑥 (𝑥) 𝑑𝑥 Mass

𝑦̄ =

∫ 𝑦𝛿𝐴𝑦 (𝑦) 𝑑𝑦 Mass

𝑧̄ =

∫ 𝑧𝛿𝐴𝑧 (𝑧) 𝑑𝑧 . Mass

The expression 𝛿𝐴𝑥 (𝑥)Δ𝑥 is the moment of a slice perpendicular to the 𝑥-axis. Thus, these formulas are extensions of the formula for the one-dimensional case. In the two- and three-dimensional case, we are assuming that the density 𝛿 is constant. If the density is not constant, finding the center of mass may require a double or triple integral from multivariable calculus. Example 8

Find the coordinates of the center of mass of the isosceles triangle in Figure 8.65. The triangle has constant density and mass 𝑚. 𝑦

𝑦 1 2

1 2 1 (1 2



− 𝑥)



𝑥 1

− 12

1



− 12

Figure 8.65: Find center of mass of this triangle

Solution

𝑥

𝑥

Δ𝑥

Figure 8.66: Sliced triangle

Because the mass of the triangle is symmetrically distributed with respect to the 𝑥-axis, 𝑦̄ = 0. We expect 𝑥̄ to be closer to 𝑥 = 0 than to 𝑥 = 1, since the triangle is wider near the origin. The area of the triangle is 21 ⋅ 1 ⋅ 1 = 12 . Thus, Density = Mass/Area = 2𝑚. If we slice the triangle into strips of width Δ𝑥, then the strip at position 𝑥 has length 𝐴𝑥 (𝑥) = 2 ⋅ 12 (1 − 𝑥) = (1 − 𝑥). See Figure 8.66. So Area of strip = 𝐴𝑥 (𝑥)Δ𝑥 ≈ (1 − 𝑥)Δ𝑥.

Since the density is 2𝑚, the center of mass is given by 𝑥̄ =

∫ 𝑥𝛿𝐴𝑥 (𝑥) 𝑑𝑥 Mass

1

=

∫0 2𝑚𝑥(1 − 𝑥) 𝑑𝑥 𝑚

=2

(

𝑥2 𝑥3 − 2 3

So the center of mass of this triangle is at the point (𝑥, ̄ 𝑦) ̄ = (1∕3, 0).

Example 9

)

|1 1 | = . | |0 3

Find the center of mass of a hemisphere of radius 7 cm and constant density 𝛿.

436

Chapter 8 USING THE DEFINITE INTEGRAL

Solution

Stand the hemisphere with its base horizontal in the 𝑥𝑦-plane, with the center at the origin. Symmetry tells us that its center of mass lies directly above the center of the base, so 𝑥̄ = 𝑦̄ = 0. Since the hemisphere is wider near its base, we expect the center of mass to be nearer to the base than the top. To calculate the center of mass, slice the hemisphere into horizontal disks, as in Figure 8.67. A disk of thickness Δ𝑧 at height 𝑧 above the base has Volume of disk = 𝐴𝑧 (𝑧)Δ𝑧 ≈ 𝜋(72 − 𝑧2 )Δ𝑧 cm3 . So, since the density is 𝛿, 𝑧=

∫ 𝑧𝛿𝐴𝑧 (𝑧) 𝑑𝑧 Mass

7

=

∫0 𝑧𝛿𝜋(72 − 𝑧2 ) 𝑑𝑧

.

Mass

Since the total mass of the hemisphere is ( 23 𝜋73 ) 𝛿, we get 7

𝑧̄ =

𝛿𝜋 ∫0 (72 𝑧 − 𝑧3 ) 𝑑𝑧 Mass

( )|7 𝛿𝜋 72 𝑧2 ∕2 − 𝑧4 ∕4 | |0 = = Mass

74 𝛿𝜋 4 2 𝜋73 𝛿 3

=

21 = 2.625 cm. 8

The center of mass of the hemisphere is 2.625 cm above the center of its base. As expected, it is closer to the base of the hemisphere than its top.

Figure 8.67: Slicing to find the center of mass of a hemisphere

Exercises and Problems for Section 8.4 EXERCISES 1. Find the mass of a rod of length 10 cm with density 𝛿(𝑥) = 𝑒−𝑥 gm/cm at a distance of 𝑥 cm from the left end. 2. A plate occupying the region 0 ≤ 𝑥 ≤ 2, 0 ≤ 𝑦 ≤ 3 has density 𝛿 = 5 gm/cm2 . Set up two integrals giving the mass of the plate, one corresponding to strips in the 𝑥-direction and the other corresponding to strips in the 𝑦-direction. 3. A rod has length 2 meters. At a distance 𝑥 meters from its left end, the density of the rod is given by 𝛿(𝑥) = 2 + 6𝑥 gm∕m. (a) Write a Riemann sum approximating the total mass of the rod. (b) Find the exact mass by converting the sum into an integral.

4. If a rod lies along the 𝑥-axis between 𝑎 and 𝑏, the mo𝑏 ment of the rod is ∫𝑎 𝑥𝛿(𝑥) 𝑑𝑥, where 𝛿(𝑥) is its density in grams/meter at a position 𝑥 meters. Find the moment and center of mass of the rod in Exercise 3. 5. The density of cars (in cars per mile) down a 20-mile stretch of the Pennsylvania Turnpike is approximated by )) ( ( √ 𝛿(𝑥) = 300 2 + sin 4 𝑥 + 0.15 , at a distance 𝑥 miles from the Breezewood toll plaza.

(a) Sketch a graph of this function for 0 ≤ 𝑥 ≤ 20. (b) Write a Riemann sum that approximates the total number of cars on this 20-mile stretch. (c) Find the total number of cars on the 20-mile stretch.

8.4 DENSITY AND CENTER OF MASS

6. (a) Find a Riemann sum which approximates the total mass of a 3 × 5 rectangular sheet, whose density per unit area at a distance 𝑥 from one of the sides of length 5 is 1∕(1 + 𝑥4 ). (b) Calculate the mass. 7. A point mass of 2 grams located 3 centimeters to the left of the origin and a point mass of 5 grams located 4 centimeters to the right of the origin are connected by a thin, light rod. Find the center of mass of the system.

437

8. Find the center of mass of a system containing three point masses of 5 gm, 3 gm, and 1 gm located respectively at 𝑥 = −10, 𝑥 = 1, and 𝑥 = 2. 9. Find the mass of the block 0 ≤ 𝑥 ≤ 10, 0 ≤ 𝑦 ≤ 3, 0 ≤ 𝑧 ≤ 1, whose density 𝛿, is given by 𝛿 = 2 − 𝑧 for 0 ≤ 𝑧 ≤ 1.

PROBLEMS Problems 10–12 refer to a colony of bats which flies out of a cave each night to eat insects. To estimate the colony’s size, a naturalist counts samples of bats at different distances from the cave. Table 8.3 gives 𝑛, her count per hectare, at a distance 𝑟 km from the cave. For instance, she counts 300 bats in one hectare at the cave’s mouth, and 219 bats in one hectare one kilometer from the cave. The bat count 𝑟 km from the cave is the same in all directions. [Note that 1 km2 = 100 hectares, written 100 ha.] Table 8.3 𝑟

0

1

2

3

4

5

𝑛

300

219

160

117

85

62

10. Give an overestimate of the number of bats between 3 and 4 km from the cave. 11. Give an underestimate of the number of bats between 3 and 4 km from the cave. 12. Letting 𝑛 = 𝑓 (𝑟), write an integral in terms of 𝑓 representing the number of bats in the cave. Assume that bats fly no farther away than 5 km from the cave. Do not evaluate the integral. 13. Find the total mass of the triangular region in Figure 8.68, which has density 𝛿(𝑥) = 1 + 𝑥 grams/cm2 . 𝑦 (cm) 1

𝑥 (cm) −1

1

Figure 8.68 14. A rectangular plate is located with vertices at points (0, 0), (2, 0), (2, 3) and (0, 3) in the 𝑥𝑦-plane. The density of the plate at point (𝑥, 𝑦) is 𝛿(𝑦) = 2 + 𝑦2 gm∕cm2 and 𝑥 and 𝑦 are in cm. Find the total mass of the plate. 15. A cardboard figure has the shape shown in Figure 8.69. The region is bounded on the left by the line 𝑥 = 𝑎, on the right by the line 𝑥 = 𝑏, above by 𝑓 (𝑥), and below by 𝑔(𝑥). If the density 𝛿(𝑥) gm/cm2 varies only with 𝑥, find an expression for the total mass of the figure, in terms of 𝑓 (𝑥), 𝑔(𝑥), and 𝛿(𝑥).

𝑓 (𝑥) 𝑔(𝑥) 𝑥 𝑎

𝑏

Figure 8.69 16. Circle City, a typical metropolis, is densely populated near its center, and its population gradually thins out toward the city limits. In fact, its population density is 10,000(3 − 𝑟) people/square mile at distance 𝑟 miles from the center. (a) Assuming that the population density at the city limits is zero, find the radius of the city. (b) What is the total population of the city? 17. The density of oil in a circular oil slick on the surface of the ocean at a distance 𝑟 meters from the center of the slick is given by 𝛿(𝑟) = 50∕(1 + 𝑟) kg/m2 . (a) If the slick extends from 𝑟 = 0 to 𝑟 = 10,000 m, find a Riemann sum approximating the total mass of oil in the slick. (b) Find the exact value of the mass of oil in the slick by turning your sum into an integral and evaluating it. (c) Within what distance 𝑟 is half the oil of the slick contained? 18. The soot produced by a garbage incinerator spreads out in a circular pattern. The depth, 𝐻(𝑟), in millimeters, of the soot deposited each month at a distance 𝑟 kilometers from the incinerator is given by 𝐻(𝑟) = 0.115𝑒−2𝑟 . (a) Write a definite integral giving the total volume of soot deposited within 5 kilometers of the incinerator each month. (b) Evaluate the integral you found in part (a), giving your answer in cubic meters. 19. The concentration of silt in a cylindrical well 2 m across and 20 m deep is 10 g/m3 at the surface (depth 𝑠 = 0) and increases linearly to 50 g/m3 at the bottom (depth 𝑠 = 20). (a) Find a formula for the silt concentration 𝐶(𝑠) as a function of depth 𝑠. What is the silt concentration in g/m3 at a depth of 𝑠 = 10?

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Chapter 8 USING THE DEFINITE INTEGRAL

(b) How much silt would there be in the bottom meter of the well assuming concentration does not change from 50 g/m3 . Is this an over or underestimate? (c) Approximate the silt in a horizontal slice of water of thickness Δ𝑠 at depth 𝑠. (d) Set up and evaluate an integral to calculate the total amount of silt in the well.

𝑦 𝑏 2

𝑥 𝑎 − 2𝑏

Figure 8.70 20. Three point masses of 4 gm each are placed at 𝑥 = −6, 1 and 3. Where should a fourth point mass of 4 gm be placed to make the center of mass at the origin? 21. A rod of length 3 meters with density 𝛿(𝑥) = 1 + 𝑥2 grams/meter is positioned along the positive 𝑥-axis, with its left end at the origin. Find the total mass and the center of mass of the rod. 22. A rod with density 𝛿(𝑥) = 2 + sin 𝑥 lies on the 𝑥-axis between 𝑥 = 0 and 𝑥 = 𝜋. Find the center of mass of the rod. 23. A rod of length 1 meter has density 𝛿(𝑥) = 1 + 𝑘𝑥2 grams/meter, where 𝑘 is a positive constant. The rod is lying on the positive 𝑥-axis with one end at the origin. (a) Find the center of mass as a function of 𝑘. (b) Show that the center of mass of the rod satisfies 0.5 < 𝑥̄ < 0.75. 24. A rod of length 2 meters and density 𝛿(𝑥) = 3−𝑒−𝑥 kilograms per meter is placed on the 𝑥-axis with its ends at 𝑥 = ±1. (a) Will the center of mass of the rod be on the left or right of the origin? Explain. (b) Find the coordinate of the center of mass. 25. One half of a uniform circular disk of radius 1 meter lies in the 𝑥𝑦-plane with its diameter along the 𝑦-axis, its center at the origin, and 𝑥 > 0. The mass of the halfdisk is 3 kg. Find (𝑥, ̄ 𝑦). ̄

29. Find the center of mass of a cone of height 5 cm and base diameter 10 cm with constant density 𝛿 gm/cm3 . 30. A solid is formed by rotating the region bounded by the curve 𝑦 = 𝑒−𝑥 and the 𝑥-axis between 𝑥 = 0 and 𝑥 = 1, around the 𝑥-axis. It was shown in Example 1 on page 411 that the volume of this solid is 𝜋(1−𝑒−2 )∕2. Assuming the solid has constant density 𝛿, find 𝑥̄ and 𝑦. ̄ 31. (a) Find the mass of a pyramid of constant density 𝛿 gm/cm3 with a square base of side 40 cm and height 10 cm. [That is, the vertex is 10 cm above the center of the base.] (b) Find the center of mass of the pyramid. 32. The storage shed in Figure 8.71 is the shape of a halfcylinder of radius 𝑟 and length 𝑙. (a) What is the volume of the shed? (b) The shed is filled with sawdust whose density (mass/unit volume) at any point is proportional to the distance of that point from the floor. The constant of proportionality is 𝑘. Calculate the total mass of sawdust in the shed.

26. A metal plate, with constant density 2 gm∕cm2 , has a shape bounded by the curve 𝑦 = 𝑥2 and the 𝑥-axis, with 0 ≤ 𝑥 ≤ 1 and 𝑥, 𝑦 in cm. (a) Find the total mass of the plate. (b) Sketch the plate, and decide, on the basis of the shape, whether 𝑥̄ is less than or greater than 1∕2. (c) Find 𝑥. ̄ 2 27. A metal plate, with constant density √ 5 gm∕cm , has a shape bounded by the curve 𝑦 = 𝑥 and the 𝑥-axis, with 0 ≤ 𝑥 ≤ 1 and 𝑥, 𝑦 in cm.

Figure 8.71

(a) Find the total mass of the plate. (b) Find 𝑥̄ and 𝑦. ̄ 28. An isosceles triangle with uniform density, altitude 𝑎, and base 𝑏 is placed in the 𝑥𝑦-plane as in Figure 8.70. Show that the center of mass is at 𝑥̄ = 𝑎∕3, 𝑦̄ = 0. Hence show that the center of mass is independent of the triangle’s base.

33. Water leaks out of a tank through a square hole with 1inch sides. At time 𝑡 (in seconds) the velocity of water flowing through the hole is 𝑣 = 𝑔(𝑡) ft/sec. Write a definite integral that represents the total amount of water lost in the first minute.

8.5 APPLICATIONS TO PHYSICS

34. An exponential model for the density of the earth’s atmosphere says that if the temperature of the atmosphere were constant, then the density of the atmosphere as a function of height, ℎ (in meters), above the surface of the earth would be given by 𝛿(ℎ) = 1.28𝑒−0.000124ℎ kg/m3 . (a) Write (but do not evaluate) a sum that approximates the mass of the portion of the atmosphere from ℎ = 0 to ℎ = 100 m (i.e., the first 100 meters above sea level). Assume the radius of the earth is 6400 km. (b) Find the exact answer by turning your sum in part (a) into an integral. Evaluate the integral.

439

35. The following table gives the density 𝐷 (in gm/cm3 ) of the earth at a depth 𝑥 km below the earth’s surface. The radius of the earth is about 6370 km. Find an upper and a lower bound for the earth’s mass such that the upper bound is less than twice the lower bound. Explain your reasoning; in particular, what assumptions have you made about the density?

𝑥

0

1000

2000

2900

3000

4000

5000

6000

6370

𝐷

3.3

4.5

5.1

5.6

10.1

11.4

12.6

13.0

13.0

Strengthen Your Understanding In Problems 36–39, explain what is wrong with the statement. 36. A 10 cm rod can have mass density given by 𝑓 (𝑥) = 𝑥2 − 5𝑥 gm∕cm, at a point 𝑥 cm from one end. 37. The center of mass of a rod with density 𝑥2 gm∕cm for 10 0 ≤ 𝑥 ≤ 10 is given by ∫0 𝑥3 𝑑𝑥. 38. If the center of mass of a rod is in the center of the rod, then the density of the rod is constant. 39. A disk with radius 3 cm and density 𝛿(𝑟) = 3 − 𝑟 gm∕cm2 , where 𝑟 is in centimeters from the center of the disk, has total mass 27𝜋 gm. In Problems 40–42, give an example of: 40. A mass density on a rod such that the rod is most dense at one end but the center of mass is nearer the other end. 41. A rod of length 2 cm, whose density 𝛿(𝑥) makes the center of mass not at the center of the rod. 42. A rod of length 2 cm, whose density 𝛿(𝑥) makes the center of mass at the center of the rod. In Problems 43–49, are the statements true or false? Give an explanation for your answer. 43. To find the total population in a circular city, we always slice it into concentric rings, no matter what the population density function.

8.5

44. A city occupies a region in the 𝑥𝑦-plane, with population density 𝛿(𝑦) = 1 + 𝑦. To set up an integral representing the total population in the city, we should slice the region parallel to the 𝑦-axis. 45. The population density in a circular city of radius 2 depends on the distance 𝑟 from the center by 𝑓 (𝑟) = 10−3𝑟, so that the density is greatest at the center. Then the population of the inner city, 0 ≤ 𝑟 ≤ 1, is greater than the population of the suburbs, 1 ≤ 𝑟 ≤ 2. 46. The location of the center of mass of a system of three masses on the 𝑥-axis does not change if all the three masses are doubled. 47. The center of mass of a region in the plane cannot be outside the region. 48. Particles are shot at a circular target. The density of particles hitting the target decreases with the distance from the center. To set up a definite integral to calculate the total number of particles hitting the target, we should slice the region into concentric rings. 49. A metal rod of density 𝑓 (𝑥) lying along the 𝑥-axis from 𝑥 = 0 to 𝑥 = 4 has its center of mass at 𝑥 = 2. Then the two halves of the rod on either side of 𝑥 = 2 have equal mass.

APPLICATIONS TO PHYSICS Although geometric problems were a driving force for the development of the calculus in the seventeenth century, it was Newton’s spectacularly successful applications of the calculus to physics that most clearly demonstrated the power of this new mathematics.

Work In physics the word “work” has a technical meaning which is different from its everyday meaning. Physicists say that if a constant force, 𝐹 , is applied to some object to move it a distance, 𝑑, then the force has done work on the object. The force must be parallel to the motion (in the same or the opposite direction). We make the following definition:

440

Chapter 8 USING THE DEFINITE INTEGRAL

Work done = Force ⋅ Distance

or

𝑊 = 𝐹 ⋅ 𝑑.

Notice that if we walk across a room holding a book, we do no work on the book, since the force we exert on the book is vertical, but the motion of the book is horizontal. On the other hand, if we lift the book from the floor to a table, we accomplish work. There are several sets of units in common use. To measure work, we will generally use the two sets of units, International (SI) and British, in the following table. Force

Distance

Work

Conversions

International (SI) units

newton (nt)

meter (m)

joule (j)

1 lb = 4.45 nt

British units

pound (lb)

foot (ft)

foot-pound (ft-lb)

1 ft = 0.305 m 1 ft-lb = 1.36 joules

One joule of work is done when a force of 1 newton moves an object through 1 meter, so 1 joule = 1 newton-meter. Example 1

Solution

Calculate the work done on an object when (a) A force of 2 newtons moves it 12 meters. (a) Work done = 2 nt ⋅ 12 m = 24 joules.

(b) A 3-lb force moves it 4 feet. (b) Work done = 3 lb ⋅ 4 ft = 12 ft-lb.

In the previous example, the force was constant and we calculated the work by multiplication. In the next example, the force varies, so we need an integral. We divide up the distance moved and sum to get a definite integral representing the work. Example 2

Wall

Hooke’s Law says that the force, 𝐹 , required to compress the spring in Figure 8.72 by a distance 𝑥, in meters from its equilibrium position, is given by 𝐹 = 𝑘𝑥, for some constant 𝑘. Find the work done in compressing the spring by 0.1 m if 𝑘 = 8 nt/m. Wall

Equilibrium position



✛ 𝑥 ✲

𝑥 ✲ Δ𝑥 Figure 8.73: Work done in compressing spring a small distance Δ𝑥 is 𝑘𝑥Δ𝑥 0.1

Figure 8.72: Compression of spring: Force is 𝑘𝑥

Solution

✛✲✛

Since 𝑘 is in newtons/meter and 𝑥 is in meters, we have 𝐹 = 8𝑥 newtons. Since the force varies with 𝑥, we divide the distance moved into small increments, Δ𝑥, as in Figure 8.73. Then Work done in moving through an increment ≈ 𝐹 Δ𝑥 = 8𝑥Δ𝑥 joules. So, summing over all increments gives the Riemann sum approximation ∑ Total work done ≈ 8𝑥Δ𝑥. Taking the limit as Δ𝑥 → 0 gives 0.1

Total work done =

∫0

|0.1 8𝑥 𝑑𝑥 = 4𝑥2 || = 0.04 joules. |0

8.5 APPLICATIONS TO PHYSICS

441

In general, if force is a function 𝐹 (𝑥) of position 𝑥, then in moving from 𝑥 = 𝑎 to 𝑥 = 𝑏, 𝑏

Work done =

∫𝑎

𝐹 (𝑥) 𝑑𝑥.

The Force Due to Gravity: Mass Versus Weight When an object is lifted, work is done against the force exerted by gravity on the object. By Newton’s Second Law, the downward gravitational force acting on a mass 𝑚 is 𝑚𝑔, where 𝑔 is the acceleration due to gravity. To lift the object, we need to exert a force equal to the gravitational force but in the opposite direction. In International units, 𝑔 = 9.8 m/sec2 , and we usually measure mass, 𝑚, in kilograms. In British units, mass is seldom used. Instead, we usually talk about the weight of an object, which is the force exerted by gravity on the object. Roughly speaking, the mass represents the quantity of matter in an object, whereas the weight represents the force of gravity on it. The mass of an object is the same everywhere, whereas the weight can vary if, for example, the object is moved to outer space where gravitational forces are smaller. When we are given the weight of an object, we do not multiply by 𝑔 to find the gravitational force as it has already been done. In British units, a pound is a unit of weight. In International units, a kilogram is a unit of mass, and the unit of weight is a newton, where 1 newton = 1 kg ⋅ m/sec2 . Example 3

Solution

How much work is done in lifting (a) A 5-pound book 3 feet off the floor?

(b) A 1.5-kilogram book 2 meters off the floor?

(a) The force due to gravity is 5 lb, so 𝑊 = 𝐹 ⋅ 𝑑 = (5 lb)(3 ft) = 15 foot-pounds. (b) The force due to gravity is 𝑚𝑔 = (1.5 kg)(𝑔 m/sec2 ), so 𝑊 = 𝐹 ⋅ 𝑑 = [(1.5 kg)(9.8 m/sec2 )] ⋅ (2 m) = 29.4 joules.

In the previous example, work is found by multiplication. In the next example, different parts of the object move different distances, so an integral is needed. Example 4

A 28-meter uniform chain with a mass density of 2 kilograms per meter is dangling from the roof of a building. How much work is needed to pull the chain up onto the top of the building?

Solution

Since 1 meter of the chain has mass density 2 kg, the gravitational force per meter of chain is (2 kg)(9.8 m/sec2 ) = 19.6 newtons. Let’s divide the chain into small sections of length Δ𝑦, each requiring a force of 19.6 Δ𝑦 newtons to move it against gravity. See Figure 8.74. If Δ𝑦 is small, all of this piece is hauled up approximately the same distance, namely 𝑦, so Work done on the small piece ≈ (19.6 Δ𝑦 newtons)(𝑦 meters) = 19.6𝑦 Δ𝑦 joules. The work done on the entire chain is given by the total of the work done on each piece: ∑ Work done ≈ 19.6𝑦 Δ𝑦 joules. As Δ𝑦 → 0, we obtain a definite integral. Since 𝑦 varies from 0 to 28 meters, the total work is 28

Work done =

∫0

|28 (19.6𝑦) 𝑑𝑦 = 9.8𝑦2 || = 7683.2 joules. |0

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Chapter 8 USING THE DEFINITE INTEGRAL

Top of building

✻ 𝑦 Δ𝑦 ❄





Figure 8.74: Chain for Example 4

Example 5

Calculate the work done in pumping oil from the cone-shaped tank in Figure 8.75 to the rim. The oil has density 800 kg/m3 and its vertical depth is 10 m.

Figure 8.75: Cone-shaped tank containing oil

Solution

Figure 8.76: Slicing the oil horizontally to compute work

We slice the oil horizontally because each part of such a slice moves the same vertical distance. Each slice is approximately a circular disk with radius 𝑤∕2 m, so, with ℎ in meters, ( )2 𝑤 𝜋 Volume of slice ≈ 𝜋 Δℎ = 𝑤2 Δℎ m3 . 2 4 𝜋 Force of gravity on slice = Density ⋅ 𝑔 ⋅ Volume = 800𝑔 𝑤2 Δℎ = 200𝜋𝑔𝑤2 Δℎ nt. 4 Since each part of the slice has to move a vertical distance of (20 − ℎ) m, we have Work done on slice ≈ Force ⋅ Distance = 200𝜋𝑔𝑤2 Δℎ nt ⋅ (20 − ℎ) m = 200𝜋𝑔𝑤2 (20 − ℎ)Δℎ joules. To find 𝑤 in terms of ℎ, we use the similar triangles in Figure 8.76: 𝑤 25 = ℎ 20

so

𝑤=

5 ℎ = 1.25ℎ. 4

Thus, Work done on strip ≈ 200𝜋𝑔(1.25ℎ)2(20 − ℎ)Δℎ = 312.5𝜋𝑔ℎ2(20 − ℎ)Δℎ joules. Summing and taking the limit as Δℎ → 0 gives an integral with upper limit ℎ = 10, the depth of the oil. 10 ∑ 312.5𝜋𝑔ℎ2 (20 − ℎ)Δℎ = 312.5𝜋𝑔ℎ2 (20 − ℎ) 𝑑ℎ joules. Total work = lim ∫0 Δℎ→0

8.5 APPLICATIONS TO PHYSICS

443

Evaluating the integral using 𝑔 = 9.8 m/sec2 gives Total work = 312.5𝜋𝑔

(

20

ℎ 3 ℎ4 − 3 4

)|10 | | = 1,302,083𝜋𝑔 ≈ 4.0 ⋅ 107 joules. | |0

In the following example, information is given in British units about the weight of the pyramid, so we do not need to multiply by 𝑔 to find the gravitational force. Example 6

It is reported that the Great Pyramid of Egypt was built in 20 years. If the stone making up the pyramid has a density of 200 pounds per cubic foot, find the total amount of work done in building the pyramid. The pyramid is 481 feet high and has a square base 756 feet by 756 feet. Estimate how many workers were needed to build the pyramid.

Volume of slice ≈ 𝑠2 Δℎ



❄ 481 ft



✻ ❄

𝑠



756 ft Figure 8.77: Pyramid for Example 6

Solution

We assume that the stones were originally located at the approximate height of the construction site. Imagine the pyramid built in layers as we did in Example 5 on page 405. By similar triangles, the layer at height ℎ has a side length 𝑠 = 756(481 − ℎ)∕481 ft. (See Figure 8.77.) The layer at height ℎ has a volume of 𝑠2 Δℎ ft3 , so its weight is 200𝑠2 Δℎ lb. This layer is lifted through a height of ℎ, so Work to lift layer = (200𝑠2 Δℎ lb)(ℎ ft) = 200𝑠2 ℎ Δℎ ft-lb. Substituting for 𝑠 in terms of ℎ and summing over all layers gives Total work ≈



200𝑠2 ℎΔℎ =



200

(

756 481

)2

(481 − ℎ)2 ℎ Δℎ ft-lb.

Since ℎ varies from 0 to 481, as Δℎ → 0, we obtain 481

Total work =

∫0

200

(

756 481

)2

(481 − ℎ)2 ℎ 𝑑ℎ ≈ 2.2 ⋅ 1012 foot-pounds.

We have calculated the total work done in building the pyramid; now we want to estimate the total number of workers needed. Let’s assume every laborer worked 10 hours a day, 300 days a year, for 20 years. Assume that a typical worker lifted ten 50-pound blocks a distance of 4 feet every hour, thus performing 2000 foot-pounds of work per hour (this is a very rough estimate). Then each laborer performed (10)(300)(20)(2000) = 1.2 ⋅ 108 foot-pounds of work over a twenty-year period. Thus, the number of workers needed was about (2.2 ⋅ 1012 )∕(1.2 ⋅ 108 ), or about 18,000.

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Chapter 8 USING THE DEFINITE INTEGRAL

Force and Pressure We can use the definite integral to compute the force exerted by a liquid on a surface, for example, the force of water on a dam. The idea is to get the force from the pressure. The pressure in a liquid is the force per unit area exerted by the liquid. Two things you need to know about pressure are: • At any point, pressure is exerted equally in all directions—up, down, sideways. • Pressure increases with depth. (That is one of the reasons why deep sea divers have to take much greater precautions than scuba divers.) At a depth of ℎ meters, the pressure, 𝑝, exerted by the liquid, measured in newtons per square meter, is given by computing the total weight of a column of liquid ℎ meters high with a base of 1 square meter. The volume of such a column of liquid is just ℎ cubic meters. If the liquid has density 𝛿 (mass per unit volume), then its weight per unit volume is 𝛿𝑔, where 𝑔 is the acceleration due to gravity. The weight of the column of liquid is 𝛿𝑔ℎ, so Pressure = Mass density ⋅ 𝑔 ⋅ Depth

𝑝 = 𝛿𝑔ℎ.

or

Provided the pressure is constant over a given area, we also have the following relation: Force = Pressure ⋅ Area. The units and data we will generally use are given in the following table: Density of water 3

SI units

1000 kg/m (mass)

British units

3

62.4 lb/ft (weight)

Force newton (nt) pound (lb)

Area meter foot

2

2

Pressure

Conversions 2

pascal (nt/m ) lb/ft

1 lb = 4.45 nt 1ft2 = 0.093 m2

2

1 lb∕ft2 = 47.9 pa

In International units, the mass density of water is 1000 kg/m3 , so the pressure at a depth of ℎ meters is 𝛿𝑔ℎ = 1000 ⋅ 9.8ℎ = 9800ℎ nt/m2 . See Figure 8.78. In British units, the density of the liquid is usually given as a weight per unit volume, rather than a mass per unit volume. In that case, we do not need to multiply by 𝑔 because it has already been done. For example, water weighs 62.4 lb/ft3 , so the pressure at depth ℎ feet is 62.4ℎ lb/ft2 . See Figure 8.79. Surface of water



1 square meter

❘ ✠

Surface of water



ℎm

1 square foot

❘ ✠

ℎ ft Pressure here = 9800ℎ nt/m2





Figure 8.78: Pressure exerted by column of water (International units)

Pressure here = 62.4ℎ lb/ft2





Figure 8.79: Pressure exerted by a column of water (British units)

If the pressure is constant over a surface, we calculate the force on the surface by multiplying the pressure by the area of the surface. If the pressure is not constant, we divide the surface into small pieces in such a way that the pressure is nearly constant on each one to obtain a definite integral for the force on the surface. Since the pressure varies with depth, we divide the surface into horizontal strips, each of which is at an approximately constant depth.

8.5 APPLICATIONS TO PHYSICS

445

Example 7

In 1912, the ocean liner Titanic sank to the bottom of the Atlantic, 12,500 feet (nearly 2.5 miles) below the surface. Find the force on one side of a 100-foot square plate at the depth of the Titanic if the plate is: (a) Lying horizontally (b) Standing vertically.

Solution

(a) When the plate is horizontal, the pressure is the same at every point on the plate, so Pressure = 62.4 lb/ft3 ⋅ 12,500 ft = 780,000 lb/ft2 . To imagine this pressure, convert to pounds per square inch, giving 780,000∕144 ≈ 5400 lb/in2 . For the horizontal plate Force = 780,000 lb/ft2 ⋅ 1002 ft2 = 7.8 ⋅ 109 pounds. (b) When the plate is vertical, only the bottom is at 12,500 feet; the top is at 12,400 feet. Dividing into horizontal strips of width Δℎ, as in Figure 8.80, we have Area of strip = 100Δℎ ft2 . Since the pressure on a strip at a depth of ℎ feet is 62.4ℎ lb/ft2 , Force on strip ≈ 62.4ℎ ⋅ 100Δℎ = 6240ℎΔℎ pounds. Summing over all strips and taking the limit as Δℎ → 0 gives a definite integral. The strips vary in depth from 12,400 to 12,500 feet, so Total force = lim

Δℎ→0



12,500

6240ℎΔℎ =

∫12,400

6240ℎ 𝑑ℎ pounds.

Evaluating the integral gives 12,500

Total force = 6240

ℎ2 || = 3120(12,5002 − 12,4002) = 7.77 ⋅ 109 pounds. 2 ||12,400

Notice that the answer to part (b) is smaller than the answer to part (a). This is because part of the plate is at a smaller depth in part (b) than in part (a). ✛ ✲ 100 ft ✲ Depth = 12,400 ft ✻ ❄

Δℎ

✻ 100 ft Depth = 12,500 ft



❄Bottom

of ocean

Figure 8.80: Square plate at bottom of ocean; ℎ measured from the surface of water

Example 8

Figure 8.81 shows a dam approximately the size of Hoover Dam, which stores water for California, Nevada, and Arizona. Calculate: (a) The water pressure at the base of the dam. (b) The total force of the water on the dam.

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Chapter 8 USING THE DEFINITE INTEGRAL





400 m



✛ ✻

✻ ✛

220 m



𝑤



220 m





200 m



Figure 8.81: Trapezoid-shaped dam

Solution



400 m





200 m

❄ Δℎ ❄ ✻



Figure 8.82: Dividing dam into horizontal strips

(a) Since the density of water is 𝛿 = 1000 kg/m3 , at the base of the dam, Water pressure = 𝛿𝑔ℎ = 1000 ⋅ 9.8 ⋅ 220 = 2.156 ⋅ 106 nt/m2 . (b) To calculate the force on the dam, we divide the dam into horizontal strips because the pressure along each strip is approximately constant. See Figure 8.82. Since each strip is approximately rectangular, Area of strip ≈ 𝑤Δℎ m2 . The pressure at a depth of ℎ meters is 𝛿𝑔ℎ = 9800ℎ nt/m2 . Thus, Force on strip ≈ Pressure ⋅ Area = 9800ℎ𝑤Δℎ nt. To find 𝑤 in terms of ℎ, we use the fact that 𝑤 decreases linearly from 𝑤 = 400 when ℎ = 0 to 𝑤 = 200 when ℎ = 220. Thus 𝑤 is a linear function of ℎ, with slope (200−400)∕220 = −10∕11, so 10 𝑤 = 400 − ℎ. 11 Thus ) ( 10 Force on strip ≈ 9800ℎ 400 − ℎ Δℎ nt. 11 Summing over all strips and taking the limit as Δℎ → 0 gives ( ) ∑ 10 9800ℎ 400 − ℎ Δℎ Total force on dam = lim Δℎ→0 11 ) 220 ( 10 = 9800ℎ 400 − ℎ 𝑑ℎ newtons. ∫0 11 Evaluating the integral gives ( )|220 10 Total force = 9800 200ℎ2 − ℎ3 || = 6.32 ⋅ 1010 newtons. 33 |0

In fact, Hoover Dam is not flat, as the problem assumed, but arched, to better withstand the pressure.

Exercises and Problems for Section 8.5 Online Resource: Additional Problems for Section 8.5 EXERCISES

1. Find the work done on a 40 lb suitcase when it is raised 9 inches. 2. Find the work done on a 20 kg suitcase when it is raised 30 centimeters.

3. A particle 𝑥 feet from the origin has a force of 𝑥2 + 2𝑥 pounds acting on it. What is the work done in moving the object from the origin a distance of 1 foot? In Exercises 4–6, the force, 𝐹 , required to compress a spring

8.5 APPLICATIONS TO PHYSICS

by a distance 𝑥 meters is given by 𝐹 = 3𝑥 newtons. 4. Find the work done in compressing the spring from 𝑥 = 1 to 𝑥 = 2. 5. Find the work done to compress the spring to 𝑥 = 3, starting at the equilibrium position, 𝑥 = 0. 6. (a) Find the work done in compressing the spring from 𝑥 = 0 to 𝑥 = 1 and in compressing the spring from 𝑥 = 4 to 𝑥 = 5. (b) Which of the two answers is larger? Why? 7. A circular steel plate of radius 20 ft lies flat on the bottom of a lake, at a depth of 150 ft. Find the force on the plate due to the water pressure.

447

8. A fish tank is 2 feet long and 1 foot wide, and the depth of the water is 1 foot. What is the force on the bottom of the fish tank? 9. A child fills a bucket with sand so that the bucket and sand together weigh 10 lbs, lifts it 2 feet up and then walks along the beach, holding the bucket at a constant height of 2 ft above the ground. How much work is done on the bucket after the child has walked 100 ft? 10. The gravitational force on a 1 kg object at a distance 𝑟 meters from the center of the earth is 𝐹 = 4 ⋅ 1014 ∕𝑟2 newtons. Find the work done in moving the object from the surface of the earth to a height of 106 meters above the surface. The radius of the earth is 6.4 ⋅ 106 meters.

PROBLEMS 11. How much work is required to lift a 1000-kg satellite from the surface of the earth to an altitude of 2 ⋅ 106 m? The gravitational force is 𝐹 = 𝐺𝑀𝑚∕𝑟2 , where 𝑀 is the mass of the earth, 𝑚 is the mass of the satellite, and 𝑟 is the distance between them. The radius of the earth is 6.4 ⋅ 106 m, its mass is 6 ⋅ 1024 kg, and in these units the gravitational constant, 𝐺, is 6.67 ⋅ 10−11 . 12. A worker on a scaffolding 75 ft above the ground needs to lift a 500 lb bucket of cement from the ground to a point 30 ft above the ground by pulling on a rope weighing 0.5 lb/ft. How much work is required? 13. An anchor weighing 100 lb in water is attached to a chain weighing 3 lb/ft in water. Find the work done to haul the anchor and chain to the surface of the water from a depth of 25 ft. 14. A 1000-lb weight is being lifted to a height 10 feet off the ground. It is lifted using a rope which weighs 4 lb per foot and which is being pulled up by construction workers standing on a roof 30 feet off the ground. Find the work done to lift the weight. 15. A 2000-lb cube of ice must be lifted 100 ft, and it is melting at a rate of 4 lb per minute. If it can be lifted at a rate of one foot every minute, find the work needed to get the block of ice to the desired height. 16. A cylindrical garbage can of depth 3 ft and radius 1 ft fills with rainwater up to a depth of 2 ft. How much work would be done in pumping the water up to the top edge of the can? (Water weighs 62.4 lb/ft3 .) 17. A rectangular swimming pool 50 ft long, 20 ft wide, and 10 ft deep is filled with water to a depth of 9 ft. Use an integral to find the work required to pump all the water out over the top. 18. A water tank is in the form of a right circular cylinder with height 20 ft and radius 6 ft. If the tank is half full of water, find the work required to pump all of it over the top rim. 19. The tank in Problem 18 is full of water. Find the work required to pump all of it to a point 10 ft above the top of the tank.

20. Water in a cylinder of height 10 ft and radius 4 ft is to be pumped out. Find the work required if (a) The tank is full of water and the water is to pumped over the top of the tank. (b) The tank is full of water and the water must be pumped to a height 5 ft above the top of the tank. (c) The depth of water in the tank is 8 ft and the water must be pumped over the top of the tank. 21. A water tank is in the shape of a right circular cone with height 18 ft and radius 12 ft at the top. If it is filled with water to a depth of 15 ft, find the work done in pumping all of the water over the top of the tank. (The density of water is 𝛿 = 62.4 lb/ft3 .) 22. A cone with height 12 ft and radius 4 ft, pointing downward, is filled with water to a depth of 9 ft. Find the work required to pump all the water out over the top. 23. A hemispherical bowl of radius 2 ft contains water to a depth of 1 ft at the center. Let 𝑦 be measured vertically upward from the bottom of the bowl. Water has density 62.4 lb/ft3 . (a) Approximately how much work does it take to move a horizontal slice of water at a distance 𝑦 from the bottom to the rim of the bowl? (b) Write and evaluate an integral giving the work done to move all the water to the rim of the bowl. 24. A bucket of water of mass 20 kg is pulled at constant velocity up to a platform 40 meters above the ground. This takes 10 minutes, during which time 5 kg of water drips out at a steady rate through a hole in the bottom. Find the work needed to raise the bucket to the platform. 25. A gas station stores its gasoline in a tank under the ground. The tank is a cylinder lying horizontally on its side. (In other words, the tank is not standing vertically on one of its flat ends.) If the radius of the cylinder is 4 feet, its length is 12 feet, and its top is 10 feet under the ground, find the total amount of work needed to pump the gasoline out of the tank. (Gasoline weighs 42 lb/ft3 .)

448

Chapter 8 USING THE DEFINITE INTEGRAL

26. (a) The trough in Figure 8.83 is full of water. Find the force of the water on a triangular end. (b) Find the work to pump all the water over the top.

34. Set up and calculate a definite integral giving the total force on the dam shown in Figure 8.84, which is about the size of the Aswan Dam in Egypt.



3600 m

✲ ✻ 100 m



✛ 3000 m ✲ Figure 8.84

Figure 8.83 27. The dam in Hannawa Falls, NY, on the Raquette River is 60 feet across, 25 feet high, and approximately rectangular. Find the water force on the dam. 28. What is the total force on the bottom and each side of a full rectangular water tank that has length 20 ft, width 10 ft, and depth 15 ft? 29. A rectangular dam is 100 ft long and 50 ft high. If the water is 40 ft deep, find the force of the water on the dam. 30. A lobster tank in a restaurant is 4 ft long by 3 ft wide by 2 ft deep. Find the water force on the bottom and on each of the four sides. 31. The Three Gorges Dam started operation in China in 2008. With the largest electrical generating capacity in the world, the dam is about 2000 m long and 180 m high, and has created a lake longer than Lake Superior.3 Assume the dam is rectangular in shape. (a) Estimate the water pressure at the base of the dam. (b) Set up and evaluate a definite integral giving the total force of the water on the dam. 32. On August 12, 2000, the Russian submarine Kursk sank to the bottom of the sea, 350 feet below the surface. Find the following at the depth of the Kursk. (a) The water pressure in pounds per square foot and pounds per square inch. (b) The force on a 5-foot square metal sheet held (i) Horizontally. (ii) Vertically. 33. The ocean liner Titanic lies under 12,500 feet of water at the bottom of the Atlantic Ocean. (a) What is the water pressure at the Titanic? Give your answer in pounds per square foot and pounds per square inch. (b) Set up and calculate an integral giving the total force on a circular porthole (window) of diameter 6 feet standing vertically with its center at the depth of the Titanic. 3 en.wikipedia.org/wiki/Three_Gorges_Dam.

35. A climbing plane has vertical velocity 𝑣(ℎ) meters∕second, where ℎ is its altitude in meters. Write a definite integral for the time it takes the plane to climb 7000 meters from the ground. 36. Climbing a ladder at a constant rate, a painter sprays 1∕4 kg of paint per meter onto a pole from the ground up to a height of 8 meters. Her sprayer starts on the ground with 3 kg of paint and weighs 2 kg when empty. (a) Find a formula for the mass of the sprayer with paint as a function of ℎ, the height of the sprayer above the ground. (b) Approximate the work done by the painter in lifting the sprayer from height ℎ to ℎ + Δℎ. (c) Find total work done lifting the sprayer for one coat of paint. 37. Old houses may contain asbestos, now known to be dangerous; removal requires using a special vacuum. A contractor climbs a ladder and sucks up asbestos at a constant rate from a 10 m tall pipe covered by 0.2 kg/m using a vacuum weighing 14 kg with a 1.2 kg capacity. (a) Let ℎ be the height of the vacuum from the ground. If the vacuum is empty at ℎ = 0, find a formula for the mass of the vacuum and the asbestos inside as a function of ℎ. (b) Approximate the work done by the contractor in lifting the vacuum from height ℎ to ℎ + Δℎ. (c) At what height does the vacuum fill up? (d) Find total work done lifting the vacuum from height ℎ = 0 until the vacuum fills. (e) Assuming again an empty tank at ℎ = 0, find the work done lifting the vacuum when removing the remaining asbestos. 38. A skyrocket firework burns fuel as it climbs. Its mass, 𝑚(ℎ) kg, at height ℎ m is given by 𝑚(ℎ) = 0.96 +

0.065 . 1+ℎ

(a) What is the mass of the rocket at ℎ = 0 just before it is launched?

Accessed April 2012.

8.5 APPLICATIONS TO PHYSICS

39. We define the electric potential at a distance 𝑟 from an electric charge 𝑞 by 𝑞∕𝑟. The electric potential of a charge distribution is obtained by adding up the potential from each point. Electric charge is sprayed (with constant density 𝜎 in units of charge/unit area) on to a circular disk of radius 𝑎. Consider the axis perpendicular to the disk and through its center. Find the electric potential at the point 𝑃 on this axis at a distance 𝑅 from the center. (See Figure 8.85.)



40. A uniform, thin, circular disk of radius 𝑎 and mass 𝑀 lies on a horizontal plane. The point P lies a distance 𝑦 directly above O, the center of the disk. Calculate the gravitational force on a mass 𝑚 at the point P. (See Figure 8.86.) Use the fact that the gravitational force exerted on the mass 𝑚 by a thin horizontal ring of radius 𝑟, mass 𝜇, and center O is toward O and given by

𝐹 =

𝐺𝜇𝑚𝑦 , where 𝐺 is constant. (𝑟2 + 𝑦2 )3∕2

P

𝑦

Radius = 𝑎



O ✛ 𝑟



𝑎

𝑃 𝑅



(b) Show that the mass decreases as the rocket climbs. (c) What work would be required to lift the rocket 10 m if its mass did not decrease? (d) Approximate the work done as the rocket goes from height ℎ to ℎ + Δℎ. (e) The rocket explodes at ℎ = 250 m. Find total work done by the rocket from launch to explosion.

449

Figure 8.86 Figure 8.85

Strengthen Your Understanding In Problems 41–43, explain what is wrong with the statement. 41. A 20 meter rope with a mass of 30 kg dangles over the edge of a cliff. Ignoring friction, the work required to pull the rope to the top of the cliff is ( ) m Work = (30 kg) 9.8 (20 m) . 2 sec

45. Two cylindrical tanks 𝐴 and 𝐵 such that it takes less work to pump the water from tank 𝐴 to a height of 10 meters than from tank 𝐵. Both tanks contain the same volume of water and are less than 10 meters high. In Problems 46–51, are the statements true or false? Give an explanation for your answer. 46. It takes more work to lift a 20 lb weight 10 ft slowly than to lift it the same distance quickly. 47. Work can be negative or positive.

42. A cylindrical tank is 10 meters deep. It takes twice as much work to pump all the oil out through the top of the tank when the tank is full as when the tank is half full. 43. Lifting a 10 kg rock 2 meters off the ground requires 20 joules of work. In Problems 44–45, give an example of: 44. A situation where work can be computed as Force × Distance without doing an integral.

48. The force on a rectangular dam is doubled if its length stays the same and its depth is doubled. 49. To find the force of water on a vertical wall, we always slice the wall horizontally, no matter what the shape of the wall. 50. The force of a liquid on a wall can be negative or positive. 51. If the average value of the force 𝐹 (𝑥) is 7 on the interval 1 ≤ 𝑥 ≤ 4, then the work done by the force in moving from 𝑥 = 1 to 𝑥 = 4 is 21.

450

8.6

Chapter 8 USING THE DEFINITE INTEGRAL

APPLICATIONS TO ECONOMICS

Present and Future Value Many business deals involve payments in the future. If you buy a car or furniture, for example, you may buy it on credit and pay over a period of time. If you are going to accept payment in the future under such a deal, you obviously need to know how much you should be paid. Being paid $100 in the future is clearly worse than being paid $100 today for many reasons. If you are given the money today, you can do something else with it—for example, put it in the bank, invest it somewhere, or spend it. Thus, even without considering inflation, if you are to accept payment in the future, you would expect to be paid more to compensate for this loss of potential earnings. The question we will consider now is, how much more? To simplify matters, we consider only what we would lose by not earning interest; we will not consider the effect of inflation. Let’s look at some specific numbers. Suppose you deposit $100 in an account which earns 3% interest compounded annually, so that in a year’s time you will have $103. Thus, $100 today will be worth $103 a year from now. We say that the $103 is the future value of the $100, and that the $100 is the present value of the $103. Observe that the present value is smaller than the future value. In general, we say the following: • The future value, $𝐵, of a payment, $𝑃 , is the amount to which the $𝑃 would have grown if deposited in an interest-bearing bank account. • The present value, $𝑃 , of a future payment, $𝐵, is the amount which would have to be deposited in a bank account today to produce exactly $𝐵 in the account at the relevant time in the future. With an interest rate of 𝑟 per year, compounded annually, and a time period of 𝑡 years, a deposit of $𝑃 grows to a future balance of $𝐵, where 𝐵 = 𝑃 (1 + 𝑟)𝑡 ,

or equivalently, 𝑃 =

𝐵 . (1 + 𝑟)𝑡

Note that for a 3% interest rate, 𝑟 = 0.03. If instead of annual compounding, we have continuous compounding, we get the following result: 𝐵 = 𝑃 𝑒𝑟𝑡 ,

or equivalently, 𝑃 =

𝐵 = 𝐵𝑒−𝑟𝑡 . 𝑒𝑟𝑡

Example 1

You win the lottery and are offered the choice between $1 million in four yearly installments of $250,000 each, starting now, and a lump-sum payment of $920,000 now. Assuming a 6% interest rate per year, compounded continuously, and ignoring taxes, which should you choose?

Solution

We do the problem in two ways. First, we assume that you pick the option with the largest present value. The first of the four $250,000 payments is made now, so Present value of first payment = $250,000. The second payment is made one year from now, so Present value of second payment = $250,000𝑒−0.06(1).

8.6 APPLICATIONS TO ECONOMICS

451

Calculating the present value of the third and fourth payments similarly, we find: Total present value = $250,000 + $250,000𝑒−0.06(1) + $250,000𝑒−0.06(2) + $250,000𝑒−0.06(3) ≈ $250,000 + $235,441 + $221,730 + $208,818 = $915,989. Since the present value of the four payments is less than $920,000, you are better off taking the $920,000 right now. Alternatively, we can compare the future values of the two pay schemes. The scheme with the highest future value is the best from a purely financial point of view. We calculate the future value of both schemes three years from now, on the date of the last $250,000 payment. At that time, Future value of the lump sum payment = $920,000𝑒0.06(3) ≈ $1,101,440. Now we calculate the future value of the first $250,000 payment: Future value of the first payment = $250,000𝑒0.06(3). Calculating the future value of the other payments similarly, we find: Total future value = $250,000𝑒0.06(3) + $250,000𝑒0.06(2) + $250,000𝑒0.06(1) + $250,000 ≈ $299,304 + $281,874 + $265,459 + $250,000 = $1,096,637. The future value of the $920,000 payment is greater, so you are better off taking the $920,000 right now. Of course, since the present value of the $920,000 payment is greater than the present value of the four separate payments, you would expect the future value of the $920,000 payment to be greater than the future value of the four separate payments. (Note: If you read the fine print, you will find that many lotteries do not make their payments right away, but often spread them out, sometimes far into the future. This is to reduce the present value of the payments made, so that the value of the prizes is much less than it might first appear!)

Income Stream When we consider payments made to or by an individual, we usually think of discrete payments, that is, payments made at specific moments in time. However, we may think of payments made by a company as being continuous. The revenues earned by a huge corporation, for example, come in essentially all the time and can be represented by a continuous income stream, written 𝑃 (𝑡) dollars/year. Notice that 𝑃 (𝑡) is the rate at which deposits are made (its units are dollars per year, for example) and that this rate may vary with time, 𝑡.

Present and Future Values of an Income Stream Just as we can find the present and future values of a single payment, so we can find the present and future values of a stream of payments. We assume that interest is compounded continuously. Suppose that we want to calculate the present value of the income stream described by a rate of 𝑃 (𝑡) dollars per year, and that we are interested in the period from now until 𝑀 years in the future. We divide the stream into many small deposits, each of which is made at approximately one instant. We divide the interval 0 ≤ 𝑡 ≤ 𝑀 into subintervals, each of length Δ𝑡: ✛ 0

𝑡

✲✛ 𝑡

(𝑀 − 𝑡) 𝑡 + Δ𝑡

✲ 𝑀

452

Chapter 8 USING THE DEFINITE INTEGRAL

Assuming Δ𝑡 is small, the rate, 𝑃 (𝑡), at which deposits are being made will not vary much within one subinterval. Thus, between 𝑡 and 𝑡 + Δ𝑡: Amount deposited ≈ Rate of deposits × Time ≈ (𝑃 (𝑡) dollars/year)(Δ𝑡 years) = 𝑃 (𝑡)Δ𝑡 dollars. Measured from the present, the deposit of 𝑃 (𝑡)Δ𝑡 is made 𝑡 years in the future. Thus, with an interest rate 𝑟 per year, we have Present value of money deposited in interval 𝑡 to 𝑡 + Δ𝑡

≈ 𝑃 (𝑡)Δ𝑡𝑒−𝑟𝑡 .

Summing over all subintervals gives Total present value ≈



𝑃 (𝑡)𝑒−𝑟𝑡 Δ𝑡 dollars.

In the limit as Δ𝑡 → 0, we get the following integral: 𝑀

Present value =

𝑃 (𝑡)𝑒−𝑟𝑡 𝑑𝑡 dollars.

∫0

In computing future value, the deposit of 𝑃 (𝑡)Δ𝑡 has a period of (𝑀 − 𝑡) years to earn interest, and therefore Future value of money deposited ≈ (𝑃 (𝑡)Δ𝑡) 𝑒𝑟(𝑀 −𝑡) . in interval 𝑡 to 𝑡 + Δ𝑡 Summing over all subintervals, we get: Total future value ≈



𝑃 (𝑡)Δ𝑡𝑒𝑟(𝑀 −𝑡) dollars.

As the length of the subdivisions tends toward zero, the sum becomes an integral: 𝑀

Future value =

∫0

𝑃 (𝑡)𝑒𝑟(𝑀 −𝑡) 𝑑𝑡 dollars.

In addition, by writing 𝑒𝑟(𝑀 −𝑡) = 𝑒𝑟𝑀 ⋅ 𝑒−𝑟𝑡 and factoring out 𝑒𝑟𝑀 , we see that Future value = 𝑒𝑟𝑀 ⋅ Present value. Example 2

Find the present and future values of a constant income stream of $1000 per year over a period of 20 years, assuming an interest rate of 10% per year, compounded continuously.

Solution

Using 𝑃 (𝑡) = 1000 and 𝑟 = 0.1, we have 20

Present value =

∫0

( −0.1𝑡 ) 20 | 𝑒 | = 10,000(1 − 𝑒−2 ) ≈ 8646.65 dollars. 1000𝑒−0.1𝑡𝑑𝑡 = 1000 − 0.1 ||0

There are two ways to compute the future value. Using the present value of $8646.65, we have Future value = 8646.65𝑒0.1(20) = 63,890.58 dollars. Alternatively, we can use the integral formula:

8.6 APPLICATIONS TO ECONOMICS

20

Future value =

∫0

1000𝑒0.1(20−𝑡)𝑑𝑡 =

20

∫0

453

1000𝑒2 𝑒−0.1𝑡 𝑑𝑡

( −0.1𝑡 ) 20 | 𝑒 | = 10,000𝑒2(1 − 𝑒−2 ) ≈ 63,890.58 dollars. = 1000𝑒 − 0.1 ||0 Notice that the total amount deposited is $1000 per year for 20 years, or $20,000. The additional $43,895.58 of the future value comes from interest earned. 2

Supply and Demand Curves In a free market, the quantity of a certain item produced and sold can be described by the supply and demand curves of the item. The supply curve shows the quantity of the item the producers will supply at different price levels. It is usually assumed that as the price increases, the quantity supplied will increase. The consumers’ behavior is reflected in the demand curve, which shows what quantity of goods are bought at various prices. An increase in price is usually assumed to cause a decrease in the quantity purchased. See Figure 8.87. 𝑝 (price/unit) 𝑝1 Supply

𝑝∗ 𝑝0

Demand

𝑞∗

𝑞1

𝑞 (quantity)

Figure 8.87: Supply and demand curves

It is assumed that the market settles to the equilibrium price and quantity, 𝑝∗ and 𝑞 ∗ , where the graphs cross. At equilibrium, a quantity 𝑞 ∗ of an item is produced and sold for a price of 𝑝∗ each.

Consumer and Producer Surplus Notice that at equilibrium, a number of consumers have bought the item at a lower price than they would have been willing to pay. (For example, there are some consumers who would have been willing to pay prices up to 𝑝1 .) Similarly, there are some suppliers who would have been willing to produce the item at a lower price (down to 𝑝0 , in fact). We define the following terms: • The consumer surplus measures the consumers’ gain from trade. It is the total amount gained by consumers by buying the item at the current price rather than at the price they would have been willing to pay. • The producer surplus measures the suppliers’ gain from trade. It is the total amount gained by producers by selling at the current price, rather than at the price they would have been willing to accept. In the absence of price controls, the current price is assumed to be the equilibrium price. Both consumers and producers are richer for having traded. The consumer and producer surplus measure how much richer they are. Suppose that all consumers buy the good at the maximum price they are willing to pay. Divide the interval from 0 to 𝑞 ∗ into subintervals of length Δ𝑞. Figure 8.88 shows that a quantity Δ𝑞 of items are sold at a price of about 𝑝1 , another Δ𝑞 are sold for a slightly lower price of about 𝑝2 , the next Δ𝑞 for a price of about 𝑝3 , and so on. Thus, ∑ 𝑝𝑖 Δ𝑞. Consumers’ total expenditure ≈ 𝑝1 Δ𝑞 + 𝑝2 Δ𝑞 + 𝑝3 Δ𝑞 + ⋯ =

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Chapter 8 USING THE DEFINITE INTEGRAL

If 𝐷 is the demand function given by 𝑝 = 𝐷(𝑞), and if all consumers who were willing to pay more than 𝑝∗ paid as much as they were willing, then as Δ𝑞 → 0, we would have 𝑞∗

Consumer expenditure =

𝐷(𝑞)𝑑𝑞 =

∫0

Area under demand curve from 0 to 𝑞 ∗ .

Now if all goods are sold at the equilibrium price, the consumers’ actual expenditure is 𝑝∗ 𝑞 ∗ , the area of the rectangle between the axes and the lines 𝑞 = 𝑞 ∗ and 𝑝 = 𝑝∗ . Thus, if 𝑝∗ and 𝑞 ∗ are equilibrium price and quantity, the consumer surplus is calculated as follows:

Consumer surplus =

(

𝑞∗

∫0

𝐷(𝑞)𝑑𝑞

)

− 𝑝∗ 𝑞 ∗ =

Area under demand curve above 𝑝 = 𝑝∗ .

𝑝 (price/unit) 𝑝 (price/unit)

𝑝1 𝑝2 𝑝3

Consumer surplus Producer surplus

Supply: 𝑝 = 𝑆(𝑞)



𝑝∗ 𝑝



Demand: 𝑝 = 𝐷(𝑞)

Supply: 𝑝 = 𝑆(𝑞)

☛ Demand: 𝑝 = 𝐷(𝑞)

Δ𝑞✲ ✛

...

𝑞

𝑞 (quantity)



𝑞∗

Figure 8.88: Calculation of consumer surplus

𝑞 (quantity)

Figure 8.89: Consumer and producer surplus

See Figure 8.89. Similarly, if the supply curve is given by the function 𝑝 = 𝑆(𝑞) and 𝑝∗ and 𝑞 ∗ are equilibrium price and quantity, the producer surplus is calculated as follows:

Producer surplus = 𝑝∗ 𝑞 ∗ −

(

∫0

𝑞∗

𝑆(𝑞)𝑑𝑞

)

=

Area between supply curve and line 𝑝 = 𝑝∗ .

Exercises and Problems for Section 8.6 EXERCISES In Exercises 1–7 give an expression that represents the statement. Do not simplify your expression. 1. The future value of single $𝐶 deposit, after 20 years, at a 2% interest rate per year, compounded annually. 2. The present value of $𝐶 deposited 20 years from now, at a 2% interest rate per year, compounded annually. 3. The present value of a deposit of $𝐶, made 10 years from now, with a 2% interest rate per year, compounded continuously. 4. The present value of an income stream paying 𝐶 dollars/year for a period of 15 years, at a 2% interest rate per year, compounded continuously.

5. The future value at the end of 15 years of an income stream paying 𝐶 dollars/year throughout the 15 years, at a 2% interest rate per year, compounded continuously. 6. The future value, at the end of 𝐶 years, of a series of three $500 deposits, where the first deposit is made now, the second a year from now, and the third two years from now. Assume a 2% interest rate per year, compounded continuously. 7. The continuous interest rate per year for a deposit $𝐶 that will grow to $25,000 in 30 years. 8. Find the future value of an income stream of $2000 per year, deposited into an account paying 2% interest per year, compounded continuously, over a 15-year period.

8.6 APPLICATIONS TO ECONOMICS

9. Find the present and future values of an income stream of $5000 a year, for a period of 5 years, if the continuous interest rate is 4% per year. 10. A person deposits money into a retirement account, which pays 5% interest per year compounded continuously, at a rate of $5000 per year for 15 years. Calculate: (a) The balance in the account at the end of the 15 years. (b) The amount of money actually deposited into the account. (c) The interest earned during the 15 years. 11. (a) Find the present and future values of a constant income stream of $100 per year over a period of 20 years, assuming a 10% annual interest rate per year, compounded continuously. (b) How many years will it take for the balance to reach $5000?

455

Exercises 12–14 concern a single investment of $10,000. Find the continuous interest rate per year, yielding a future value of $20,000 in the given time period. 12. 60 years

13. 20 years

14. 5 years

Exercises 15–17 concern a constant income stream that pays a total of $20,000 over a certain time period with an interest rate of 2% per year, compounded continuously. Find the rate at which money is paid, in dollars/year, and the future value of the stream at the end of the given time period. 15. 5 years

16. 10 years

17. 20 years

PROBLEMS 18. A company expects a factory to make $200 million/year for ten years, beginning in year 𝑡 = 5. With a continuous interest rate of 1% per year, what is the present value of this income stream from 𝑡 = 5 to 𝑡 = 15? 19. With a continuous interest rate of 1.5% per year, a company expects the present value (in $ millions) of the income stream from a new factory to be given by the integral 18

∫3

300𝑒−0.015𝑡 𝑑𝑡.

(a) What income stream (in $ millions/year) does the company expect the factory to make, once it comes online? (b) For how many years does the company expect the factory to generate an income stream? 20. Draw a graph, with time in years on the horizontal axis, of what an income stream might look like for a company that sells sunscreen in the northeast United States. 21. On March 6, 2007, the Associated Press reported that Ed Nabors had won half of a $390 million jackpot, the largest lottery prize in US history at the time. Suppose he was given the choice of receiving his $195 million share paid out continuously over 20 years or one lump sum of $120 million paid immediately. Based on present value: (a) Which option is better if the interest rate is 6% per year, compounded continuously? An interest rate of 3% per year? (b) If Mr. Nabors chose the lump-sum option, what assumption was he making about interest rates? 22. Find a constant income stream (in dollars per year) which after 10 years has a future value of $20,000, assuming a continuous interest rate of 3% per year. 23. (a) A bank account earns 2% interest per year compounded continuously. At what (constant, continu-

ous) rate must a parent deposit money into such an account in order to save $100,000 in 15 years for a child’s college expenses? (b) If the parent decides instead to deposit a lump sum now in order to attain the goal of $100,000 in 15 years, how much must be deposited now? 24. (a) If you deposit money continuously at a constant rate of $4000 per year into a bank account that earns 2% interest per year, how many years will it take for the balance to reach $10,000? (b) How many years would it take if the account had $1000 in it initially? 25. A business associate who owes you $3000 offers to pay you $2800 now, or else pay you three yearly installments of $1000 each, with the first installment paid now. If you use only financial reasons to make your decision, which option should you choose? Justify your answer, assuming a 3% interest rate per year, compounded continuously. In Problems 26–29 find the continuous interest rate per year that yields a future value of $18,000 in 20 years for each $9000 investment. 26. A single $9000 deposit. 27. An initial $6000 deposit plus a second $3000 deposit made three years after the first. 28. An initial $3000 deposit plus a second $6000 deposit made three years after the first. 29. An income stream of $300 per year. 30. A family wants to save for college tuition for their daughter. What continuous yearly interest rate 𝑟% is needed in their savings account if their deposits of $4800 per year are to grow to $100,000 in 15 years? Assume that they make deposits continuously throughout the year.

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Chapter 8 USING THE DEFINITE INTEGRAL

31. Big Tree McGee is negotiating his rookie contract with a professional basketball team. They have agreed to a three-year deal which will pay Big Tree a fixed amount at the end of each of the three years, plus a signing bonus at the beginning of his first year. They are still haggling about the amounts and Big Tree must decide between a big signing bonus and fixed payments per year, or a smaller bonus with payments increasing each year. The two options are summarized in the table. All values are payments in millions of dollars. Signing bonus

Year 1

Year 2

Year 3

Option #1

6.0

2.0

2.0

2.0

Option #2

1.0

2.0

4.0

6.0

(a) Big Tree decides to invest all income in stock funds which he expects to grow at a rate of 10% per year, compounded continuously. He would like to choose the contract option which gives him the greater future value at the end of the three years when the last payment is made. Which option should he choose? (b) Calculate the present value of each contract offer. 32. Sales of Version 6.0 of a computer software package start out high and decrease exponentially. At time 𝑡, in years, the sales are 𝑠(𝑡) = 100𝑒−𝑡 thousands of dollars per year. After two years, Version 7.0 of the software is released and replaces Version 6.0. Assuming that all income from software sales is immediately invested in government bonds which pay interest at a 4% rate per year, compounded continuously, calculate the total value of sales of Version 6.0 over the two-year period. 33. The value of some good wine increases with age. Thus, if you are a wine dealer, you have the problem of deciding whether to sell your wine now, at a price of $𝑃 a bottle, or to sell it later at a higher price. Suppose you know that the amount a wine-drinker is willing to pay for √ a bottle of this wine 𝑡 years from now is $𝑃 (1 + 20 𝑡). Assuming continuous compounding and a prevailing interest rate of 5% per year, when is the best time to sell your wine? 34. An oil company discovered an oil reserve of 100 million barrels. For time 𝑡 > 0, in years, the company’s extraction plan is a linear declining function of time as follows: 𝑞(𝑡) = 𝑎 − 𝑏𝑡, where 𝑞(𝑡) is the rate of extraction of oil in millions of barrels per year at time 𝑡 and 𝑏 = 0.1 and 𝑎 = 10. (a) How long does it take to exhaust the entire reserve? (b) The oil price is a constant $20 per barrel, the extraction cost per barrel is a constant $10, and the market interest rate is 10% per year, compounded continuously. What is the present value of the company’s profit?

35. You are manufacturing a particular item. After 𝑡 years, the rate at which you earn a profit on the item is (2−0.1𝑡) thousand dollars per year. (A negative profit represents a loss.) Interest is 3% per year, compounded continuously, (a) Write a Riemann sum approximating the present value of the total profit earned up to a time 𝑀 years in the future. (b) Write an integral representing the present value in part (a). (You need not evaluate this integral.) (c) For what 𝑀 is the present value of the stream of profits on this item maximized? What is the present value of the total profit earned up to that time? 36. In May 1991, Car and Driver described a Jaguar that sold for $980,000. At that price only 50 have been sold. It is estimated that 350 could have been sold if the price had been $560,000. Assuming that the demand curve is a straight line, and that $560,000 and 350 are the equilibrium price and quantity, find the consumer surplus at the equilibrium price. 37. Using Riemann sums, explain the economic signifi𝑞∗ cance of ∫0 𝑆(𝑞) 𝑑𝑞 to the producers. 38. Using Riemann sums, give an interpretation of pro𝑞∗ ducer surplus, ∫0 (𝑝∗ − 𝑆(𝑞)) 𝑑𝑞, analogous to the interpretation of consumer surplus. 39. In Figure 8.89, page 454, mark the regions representing the following quantities and explain their economic meaning: 𝑞∗

(a) 𝑝∗ 𝑞 ∗

(b)

𝑞∗

(c)

∫0

𝑆(𝑞) 𝑑𝑞

(e) 𝑝∗ 𝑞 ∗ −

(d)

𝑞∗

∫0

𝑆(𝑞) 𝑑𝑞 (f)

∫0 ∫0 ∫0

𝐷(𝑞) 𝑑𝑞 𝑞∗

𝐷(𝑞) 𝑑𝑞 − 𝑝∗ 𝑞 ∗ 𝑞∗

(𝐷(𝑞) − 𝑆(𝑞)) 𝑑𝑞

40. The dairy industry is an example of cartel pricing: the government has set milk prices artificially high. On a supply and demand graph, label 𝑝+ , a price above the equilibrium price. Using the graph, describe the effect of forcing the price up to 𝑝+ on: (a) The consumer surplus. (b) The producer surplus. (c) The total gains from trade (Consumer surplus + Producer surplus). 41. Rent controls on apartments are an example of price controls on a commodity. They keep the price artificially low (below the equilibrium price). Sketch a graph of supply and demand curves, and label on it a price 𝑝− below the equilibrium price. What effect does forcing the price down to 𝑝− have on: (a) The producer surplus? (b) The consumer surplus? (c) The total gains from trade (Consumer surplus + Producer surplus)?

8.7 DISTRIBUTION FUNCTIONS

457

Strengthen Your Understanding In Problems 42–45, explain what is wrong with the statement.

In Problems 46–49, give an example of: 46. Supply and demand curves where producer surplus is smaller than consumer surplus.

42. The future value of an income stream of $2000 per year after 10 years is $15,000, assuming a 3% continuous interest rate per year. 43. The present value of a lump-sum payment 𝑆 dollars one year from now is greater with an interest rate of 4% per year, compounded annually, than with an interest rate of 3% per year compounded annually. 44. Payments are made at a constant rate of 𝑃 dollars per year over a two-year period. The present value of these payments is 2𝑃 𝑒−2𝑟 , where 𝑟 is the continuous interest rate per year. 45. Producer surplus is measured in the same units as the quantity, 𝑞.

8.7

47. A continuous interest rate such that a $10,000 payment in 10 years’ time has a present value of less than $5000. 48. An interest rate, compounded annually, and a present value that correspond to a future value of $5000 one year from now. 49. An interest rate, compounded annually, and a table of values that shows how much money you would have to put down in a single deposit 𝑡 years from now, at 𝑡 = 0, 1, 2, 3, or 4, if you want to have $10,000 ten years from now (ignoring inflation).

DISTRIBUTION FUNCTIONS Understanding the distribution of various quantities through the population is important to decision makers. For example, the income distribution gives useful information about the economic structure of a society. In this section we look at the distribution of ages in the US. To allocate funding for education, health care, and social security, the government needs to know how many people are in each age group. We will see how to represent such information by a density function.

US Age Distribution The data in Table 8.4 shows how the ages of the US population were distributed in 2012.4 To represent this information graphically we use a type of histogram, with a vertical bar above each age group constructed so that the area of each bar represents the fraction of the population in that age group.5 The total area of all the rectangles is 100% = 1. We only consider people who are less than 100 years old.6 For the 0–20 age group, the base of the rectangle is 20, and we want the area to be 0.27, so the height must be 0.27∕20 = 0.0135. We treat ages as though they were continuously distributed. The category 0–20, for example, contains people who are just one day short of their twentieth birthday. (See Figure 8.90.) Distribution of ages in the US in 2012

Table 8.4

fraction of population per year of age

0.015

Age group

Fraction of total population

0–20 20–40

27% = 0.27 27% = 0.27

0.01

40–60

28% = 0.28

60–80 80–100

15% = 0.15 3% = 0.03

0.005

27% = 0.27

27% = 0.27

20

28% = 0.28

40

15% = 0.15 60

3% = 0.03 80

100

Figure 8.90: How ages were distributed in the US in 2012 4 www.census.gov,

accessed April 22, 2015. are other types of histogram which have frequency on the vertical axis. 6 In fact, 0.02% of the population is over 100, but this is too small to be visible on the histogram. 5 There

age (years)

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Chapter 8 USING THE DEFINITE INTEGRAL

Example 1

In 2012, estimate what fraction of the US population was: (a) Between 20 and 60 years old. (b) Less than 10 years old. (c) Between 75 and 80 years old. (d) Between 80 and 85 years old.

Solution

(a) We add the fractions, so 0.27 + 0.28 = 0.55; that is, 55% of the US population was in this age group. (b) To find the fraction less than 10 years old, we could assume, for example, that the population was distributed evenly over the 0–20 group. (This means we are assuming that babies were born at a fairly constant rate over the last 20 years, which is probably reasonable.) If we make this assumption, then we can say that the population less than 10 years old was about half that in the 0–20 group, that is, 0.135 of the total population. Notice that we get the same result by computing the area of the rectangle from 0 to 10. (See Figure 8.91.) (c) To find the population between 75 and 80 years old, since 0.15 of Americans in 2012 were in the 60-80 group, we might apply the same reasoning and say that 41 (0.15) = 0.0375 of the population was in this age group. This result is represented as an area in Figure 8.91. The assumption that the population was evenly distributed is not a good one here; certainly there were more people between the ages of 60 and 65 than between 75 and 80. Thus, the estimate of 0.0375 is certainly too high. (d) Again using the (faulty) assumption that ages in each group were distributed uniformly, we would find that the fraction between 80 and 85 was 14 (0.03) = 0.0075. (See Figure 8.91.) This estimate is also poor—there were certainly more people in the 80–85 group than, say, the 95–100 group, and so the 0.0075 estimate is too low. fraction of population per year of age

0.015

0.01 0.135

0.27

0.0375

0.28



0.005

0.0075

✠ ✠ 10

20

40

60

75 80 85

100

age (years)

Figure 8.91: Ages in the US in 2012—various subgroups (for Example 1)

Smoothing Out the Histogram We could get better estimates if we had smaller age groups (each age group in Figure 8.90 is 20 years, which is quite large). The more detailed data in Table 8.5 leads to the new histogram in Figure 8.92. As we get more detailed information, the upper silhouette of the histogram becomes smoother, but the area of any of the bars still represents the percentage of the population in that age group. Imagine, in the limit, replacing the upper silhouette of the histogram by a smooth curve in such a way that area under the curve above one age group is the same as the area in the corresponding rectangle. The total area under the whole curve is again 100% = 1. (See Figure 8.92.)

8.7 DISTRIBUTION FUNCTIONS

Ages in the US in 2012 (more detailed)

459

Table 8.5

Age group

Fraction of total population

0–10

13% = 0.13

10–20

14% = 0.14

20–30

14% = 0.14

30–40

13% = 0.13

40–50

14% = 0.14

50–60

14% = 0.14

60–70

10% = 0.10

70–80

5% = 0.05

80–90

2% = 0.02

90–100

1% = 0.01

Shaded areas equal, so area under curve ≈ area of rectangle

fraction of population per year of age



0.015 0.01

0.14 0.14

0.13

0.13



0.05

0.14 0.14 70

80

0.1

0.005

0.02 0.01

0.05

✠ 20

40

60



80

100

age (years)

Figure 8.92: Smoothing out the age histogram

The Age Density Function If 𝑡 is age in years, we define 𝑝(𝑡), the age density function, to be a function which “smooths out” the age histogram. This function has the property that Fraction of population between ages 𝑎 and 𝑏

=

Area under graph of 𝑝 between 𝑎 and 𝑏

𝑏

=

∫𝑎

𝑝(𝑡)𝑑𝑡.

If 𝑎 and 𝑏 are the smallest and largest possible ages (say, 𝑎 = 0 and 𝑏 = 100), so that the ages of all of the population are between 𝑎 and 𝑏, then 100

𝑏

∫𝑎

𝑝(𝑡)𝑑𝑡 =

∫0

𝑝(𝑡)𝑑𝑡 = 1.

What does the age density function 𝒑 tell us? Notice that we have not talked about the meaning 𝑏 of 𝑝(𝑡) itself, but only of the integral ∫𝑎 𝑝(𝑡) 𝑑𝑡. Let’s look at this in a bit more detail. Suppose, for example, that 𝑝(10) = 0.015 per year. This is not telling us that 0.015 of the population is precisely 10 years old (where 10 years old means exactly 10, not 10 12 , not 10 14 , not 10.1). However, 𝑝(10) = 0.015 does tell us that for some small interval Δ𝑡 around 10, the fraction of the population with ages in this interval is approximately 𝑝(10) Δ𝑡 = 0.015 Δ𝑡.

The Probability Density Function Suppose we are interested in how a certain characteristic, 𝑥, is distributed through a population. For example, 𝑥 might be height or age if the population is people, or might be wattage for a population of light bulbs. Then we define a general density function with the following properties: The function, 𝑝(𝑥), is a probability density function, or pdf, if Fraction of population for which 𝑥 is between 𝑎 and 𝑏

=

Area under graph of 𝑝 between 𝑎 and 𝑏

𝑏

=

∫𝑎

𝑝(𝑥)𝑑𝑥.



∫−∞

𝑝(𝑥) 𝑑𝑥 = 1

and

𝑝(𝑥) ≥ 0 for all 𝑥.

The density function must be nonnegative because its integral always gives a fraction of the population. Also, the fraction of the population with 𝑥 between −∞ and ∞ is 1 because the entire population has the characteristic 𝑥 between −∞ and ∞. The function 𝑝 that was used to smooth

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Chapter 8 USING THE DEFINITE INTEGRAL

out the age histogram satisfies this definition of a density function. We do not assign a meaning to the value 𝑝(𝑥) directly, but rather interpret 𝑝(𝑥) Δ𝑥 as the fraction of the population with the characteristic in a short interval of length Δ𝑥 around 𝑥. The density function is often approximated by formulas, as in the next example. Example 2

Find formulas to approximate the density function, 𝑝, for the US age distribution. To reflect Figure 8.92, use a continuous function, constant at 0.013 up to age 60 and then dropping linearly.

Solution

We have 𝑝(𝑡) = 0.013 for 0 ≤ 𝑡 < 60. For 𝑡 ≥ 60, we need a linear function sloping downward. Because 𝑝 is continuous, we have 𝑝(60) = 0.013. Because 𝑝 is a density function we have 100 ∫0 𝑝(𝑡) 𝑑𝑡 = 1. Suppose 𝑏 is as in Figure 8.93; then 100

∫0

60

𝑝(𝑡) 𝑑𝑡 =

100

𝑝(𝑡) 𝑑𝑡 +

∫0

∫60

1 𝑝(𝑡) 𝑑𝑡 = 60(0.013) + (0.013)𝑏 = 1, 2

100

where ∫60 𝑝(𝑡) 𝑑𝑡 is given by the area of the triangle. Solving for 𝑏 gives 𝑏 = 33.85. Thus the slope of the line is −0.013∕33.85 = −0.00038, so for 60 ≤ 𝑡 ≤ 60 + 33.85 = 93.85, we have 𝑝(𝑡) − 0.013 = −0.00038(𝑡 − 60), 𝑝(𝑡) = 0.0358 − 0.00038𝑡. According to this way of smoothing the data, there is no one over 93.85 years old, so 𝑝(𝑡) = 0 for 𝑡 > 93.85. fraction of population per year of age

0.013

𝑝(𝑡) = 0.013 here

✠ 𝑝(𝑡) 100

✛ 𝑏 ✲

𝑡 (age in years)

60 Figure 8.93: Age density function

Cumulative Distribution Function for Ages Another way of showing how ages are distributed in the US is by using the cumulative distribution function 𝑃 (𝑡), defined by 𝑃 (𝑡) =

𝑡 Fraction of population = 𝑝(𝑥) 𝑑𝑥. ∫0 of age less than 𝑡

Thus, 𝑃 is the antiderivative of 𝑝 with 𝑃 (0) = 0, and 𝑃 (𝑡) gives the area under the density curve between 0 and 𝑡. Notice that the cumulative distribution function is nonnegative and increasing (or at least nondecreasing), since the number of people younger than age 𝑡 increases as 𝑡 increases. Another way of seeing this is to notice that 𝑃 ′ = 𝑝, and 𝑝 is positive (or nonnegative). Thus the cumulative age distribution is a function which starts with 𝑃 (0) = 0 and increases as 𝑡 increases. 𝑃 (𝑡) = 0 for 𝑡 < 0 because, when 𝑡 < 0, there is no one whose age is less than 𝑡. The limiting value of 𝑃 , as 𝑡 → ∞, is 1 since as 𝑡 becomes very large (100 say), everyone is younger than age 𝑡, so the fraction of people with age less than 𝑡 tends toward 1. (See Figure 8.94.) For 𝑡 less than 60, the graph of 𝑃 is a line, because 𝑝 is constant there. For 𝑡 > 60, the graph of 𝑃 levels off as 𝑝 tends to 0.

461

8.7 DISTRIBUTION FUNCTIONS

fraction of population per year of age

0.013

fraction of population

𝑡

Area= 𝑃 (𝑡) = ∫0 𝑝(𝑥) 𝑑𝑥

1



𝑃 (𝑡)

𝑝(𝑥)

60

𝑡

100

𝑥 (age in years)

60

100

𝑡 (age in years)

Figure 8.94: 𝑃 (𝑡), the cumulative age distribution function, and its relation to 𝑝(𝑥), the age density function

Cumulative Distribution Function A cumulative distribution function, or cdf, 𝑃 (𝑡), of a density function 𝑝, is defined by 𝑡

𝑃 (𝑡) =

∫−∞

𝑝(𝑥) 𝑑𝑥 =

Fraction of population having values of 𝑥 below 𝑡.

Thus, 𝑃 is an antiderivative of 𝑝, that is, 𝑃 ′ = 𝑝. Any cumulative distribution has the following properties: • 𝑃 is increasing (or nondecreasing). • lim 𝑃 (𝑡) = 1 and lim 𝑃 (𝑡) = 0. 𝑡→∞



𝑡→−∞

Fraction of population having values of 𝑥 between 𝑎 and 𝑏

𝑏

=

∫𝑎

𝑝(𝑥) 𝑑𝑥 = 𝑃 (𝑏) − 𝑃 (𝑎).

Exercises and Problems for Section 8.7 EXERCISES 1. Match the graphs of the density functions (a), (b), and (c) with the graphs of the cumulative distribution functions I, II, and III. (a)

(I)

Decide if the function graphed in Exercises 5–10 is a probability density function (pdf) or a cumulative distribution function (cdf). Give reasons. Find the value of 𝑐. Sketch and label the other function. (That is, sketch and label the cdf if the problem shows a pdf, and the pdf if the problem shows a cdf.) 5. 4 6. 𝑐

(II) (b)

𝑥

𝑥 𝑐 (c)

(III)

4

7. 𝑐

8. 2𝑐 𝑐 𝑥

In Exercises 2–4, graph a density function and a cumulative distribution function which could represent the distribution of income through a population with the given characteristics. 2. A large middle class. 3. Small middle and upper classes and many poor people. 4. Small middle class, many poor and many rich people.

𝑥

5

0.5

1

10. 𝑐

9. 3𝑐 𝑐

𝑥

𝑥 2

4

1

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Chapter 8 USING THE DEFINITE INTEGRAL

11. Let 𝑝(𝑥) be the density function for annual family income, where 𝑥 is in thousands of dollars. What is the meaning of the statement 𝑝(70) = 0.05?

12. Find a density function 𝑝(𝑥) such that 𝑝(𝑥) = 0 when 𝑥 ≥ 5 and when 𝑥 < 0, and is decreasing when 0 ≤ 𝑥 ≤ 5.

PROBLEMS 13. Figure 8.95 shows the distribution of kinetic energy of molecules in a gas at temperatures 300 kelvins and 500 kelvins. At higher temperatures, more of the molecules in a gas have higher kinetic energies. Which graph corresponds to which temperature?

fraction of plants

Unfertilized

1

❄ A

B

𝑥 height (meters) 1

𝐴

2

Figure 8.97

16. The cumulative distribution function for heights (in meters) of trees in a forest is 𝐹 (𝑥). 𝐵 energy

Figure 8.95

14. A large number of people take a standardized test, receiving scores described by the density function 𝑝 graphed in Figure 8.96. Does the density function imply that most people receive a score near 50? Explain why or why not.

(a) Explain in terms of trees the meaning of the statement 𝐹 (7) = 0.6. (b) Which is greater, 𝐹 (6) or 𝐹 (7)? Justify your answer in terms of trees. 17. The density function for heights of American men, in inches is 𝑝(𝑥). What is the meaning of the statement 𝑝(68) = 0.2? 18. The fraction of the US population of age less than 𝑡 is 𝑃 (𝑡). Using Table 8.5 on page 459, make a table of values for 𝑃 (𝑡). 19. Figure 8.98 shows a density function and the corresponding cumulative distribution function.7

fraction of students per test score

(a) Which curve represents the density function and which represents the cumulative distribution function? Give a reason for your choice. (b) Put reasonable values on the tick marks on each of the axes. 𝑥 test scores

10 20 30 40

50 60 70

Figure 8.96

15. An experiment is done to determine the effect of two new fertilizers A and B on the growth of a species of peas. The cumulative distribution functions of the heights of the mature peas without treatment and treated with each of A and B are graphed in Figure 8.97. (a) About what height are most of the unfertilized plants? (b) Explain in words the effect of the fertilizers A and B on the mature height of the plants. 7 Adapted

Figure 8.98

20. The density function and cumulative distribution function of heights of grass plants in a meadow are in Figures 8.99 and 8.100, respectively. (a) There are two species of grass in the meadow, a short grass and a tall grass. Explain how the graph of the density function reflects this fact.

from Calculus, by David A. Smith and Lawrence C. Moore (Lexington, D.C. Heath, 1994).

8.7 DISTRIBUTION FUNCTIONS

(b) Explain how the graph of the cumulative distribution function reflects the fact that there are two species of grass in the meadow. (c) About what percentage of the grasses in the meadow belong to the short grass species?

463

fraction of students per GPA

fraction of plants per meter of height

GPA

0

1

2

3

4

Figure 8.101

0.5

1

1.5

2

height (meter)

Figure 8.99

23. Figure 8.1028 shows the distribution of elevation, in miles, across the earth’s surface. Positive elevation denotes land above sea level; negative elevation shows land below sea level (i.e., the ocean floor). (a) Describe in words the elevation of most of the earth’s surface. (b) Approximately what fraction of the earth’s surface is below sea level?

fraction of plants

1 0.75 0.5

fraction of earth’s surface per mile of elevation

0.25 0.5

1

1.5

2

height (meter)

Figure 8.100 21. After measuring the duration of many telephone calls, the telephone company found their data was well approximated by the density function 𝑝(𝑥) = 0.4𝑒−0.4𝑥 , where 𝑥 is the duration of a call, in minutes. (a) What percentage of calls last between 1 and 2 minutes? (b) What percentage of calls last 1 minute or less? (c) What percentage of calls last 3 minutes or more? (d) Find the cumulative distribution function. 22. Students at the University of California were surveyed and asked their grade point average. (The GPA ranges from 0 to 4, where 2 is just passing.) The distribution of GPAs is shown in Figure 8.101.8 (a) Roughly what fraction of students are passing? (b) Roughly what fraction of the students have honor grades (GPAs above 3)? (c) Why do you think there is a peak around 2? (d) Sketch the cumulative distribution function.

elevation (miles)

−4

−2

0

2

4

Figure 8.102 24. Consider a population of individuals with a disease. Suppose that 𝑡 is the number of years since the onset of the disease. The death density function, 𝑓 (𝑡) = 𝑐𝑡𝑒−𝑘𝑡 , approximates the fraction of the sick individuals who die in the time interval [𝑡, 𝑡 + Δ𝑡] as follows: Fraction who die ≈ 𝑓 (𝑡)Δ𝑡 = 𝑐𝑡𝑒−𝑘𝑡 Δ𝑡 where 𝑐 and 𝑘 are positive constants whose values depend on the particular disease. (a) Find the value of 𝑐 in terms of 𝑘. (b) If 40% of the population dies within 5 years, find 𝑐 and 𝑘. (c) Find the cumulative death distribution function, 𝐶(𝑡). Give your answer in terms of 𝑘.

Strengthen Your Understanding In Problems 25–31, explain what is wrong with the statement. 25. If 𝑝(𝑥) is a probability density function with 𝑝(1) = 0.02, then the probability that 𝑥 takes the value 1 is 0.02. 26. If 𝑃 (𝑥) is a cumulative distribution function with 𝑃 (5) = 0.4, then the probability that 𝑥 = 5 is 0.4. 27. The function 𝑝(𝑡) = 𝑡2 is a density function. 8 Adapted

28. The function 𝑝(𝑥) = 𝑥2 𝑒𝑥 is a density function. 29. The function 𝑃 (𝑥) = 𝑥2 𝑒𝑥 is a cumulative distribution function. 2

30. The function 𝑃 (𝑡) = 𝑒−𝑡 is a cumulative distribution function. 31. A probability density function is always increasing.

from Statistics, by Freedman, Pisani, Purves, and Adikhari (New York: Norton, 1991).

464

Chapter 8 USING THE DEFINITE INTEGRAL

In Problems 32–35, give an example of:

In Problems 36–37, are the statements true or false? Give an explanation for your answer.

32. A density function that is greater than zero on 0 ≤ 𝑥 ≤ 20 and zero everywhere else.

2

36. If 𝑝(𝑥) = 𝑥𝑒−𝑥 for all 𝑥, then 𝑝(𝑥) is a probability density function.

33. A cumulative distribution function that is piecewise linear.

2

37. If 𝑝(𝑥) = 𝑥𝑒−𝑥 for all 𝑥 > 0 and 𝑝(𝑥) = 0 for 𝑥 ≤ 0, then 𝑝(𝑥) is a probability density function.

34. A probability density function which is nonzero only between 𝑥 = 2 and 𝑥 = 7. 35. A cumulative distribution function with 𝑃 (3) = 0 and 𝑃 (7) = 1.

8.8

PROBABILITY, MEAN, AND MEDIAN

Probability Suppose we pick a member of the US population at random and ask what is the probability that the person is between, say, the ages of 70 and 80. We saw in Table 8.5 on page 459 that 5% = 0.05 of the population is in this age group. We say that the probability, or chance, that the person is between 70 and 80 is 0.05. Using any age density function 𝑝(𝑡), we can define probabilities as follows: Probability that a person is between ages 𝑎 and 𝑏

=

Fraction of population between ages 𝑎 and 𝑏

𝑏

=

∫𝑎

𝑝(𝑡) 𝑑𝑡.

Since the cumulative distribution function gives the fraction of the population younger than age 𝑡, the cumulative distribution can also be used to calculate the probability that a randomly selected person is in a given age group. Probability that a person is younger than age 𝑡

=

Fraction of population younger than age 𝑡

𝑡

= 𝑃 (𝑡) =

∫0

𝑝(𝑥) 𝑑𝑥.

In the next example, both a density function and a cumulative distribution function are used to describe the same situation. Example 1

Suppose you want to analyze the fishing industry in a small town. Each day, the boats bring back at least 2 tons of fish, and never more than 8 tons. (a) Using the density function describing the daily catch in Figure 8.103, find and graph the corresponding cumulative distribution function and explain its meaning. (b) What is the probability that the catch will be between 5 and 7 tons? fraction of days per ton of caught fish

0.24

𝑝(𝑥)

0.12 0.08 2

5 6 7 8

𝑥 (tons of fish)

Figure 8.103: Density function of daily catch

8.8 PROBABILITY, MEAN, AND MEDIAN

Solution

465

(a) The cumulative distribution function 𝑃 (𝑡) is equal to the fraction of days on which the catch is less than 𝑡 tons of fish. Since the catch is never less than 2 tons, we have 𝑃 (𝑡) = 0 for 𝑡 ≤ 2. Since the catch is always less than 8 tons, we have 𝑃 (𝑡) = 1 for 𝑡 ≥ 8. For 𝑡 in the range 2 < 𝑡 < 8, we must evaluate the integral 𝑡

𝑃 (𝑡) =

𝑡

∫−∞

𝑝(𝑥)𝑑𝑥 =

∫2

𝑝(𝑥)𝑑𝑥.

This integral equals the area under the graph of 𝑝(𝑥) between 𝑥 = 2 and 𝑥 = 𝑡. It can be calculated by noting that 𝑝(𝑥) is given by the formula 𝑝(𝑥) =

{

0.04𝑥 for 2 ≤ 𝑥 ≤ 6 −0.06𝑥 + 0.6 for 6 < 𝑥 ≤ 8

and 𝑝(𝑥) = 0 for 𝑥 < 2 or 𝑥 > 8. Thus, for 2 ≤ 𝑡 ≤ 6, 𝑡

𝑡

𝑃 (𝑡) =

∫2

0.04𝑥 𝑑𝑥 = 0.04

And for 6 ≤ 𝑡 ≤ 8, 6

𝑡

𝑃 (𝑡) =

∫2

𝑥2 || = 0.02𝑡2 − 0.08. 2 ||2

𝑝(𝑥) 𝑑𝑥 =

∫2

𝑡

𝑝(𝑥) 𝑑𝑥 +

∫6

𝑝(𝑥) 𝑑𝑥

) 𝑡 ( | 𝑥2 (−0.06𝑥 + 0.6) 𝑑𝑥 = 0.64 + −0.06 + 0.6𝑥 || ∫6 2 |6 𝑡

= 0.64 +

= −0.03𝑡2 + 0.6𝑡 − 1.88. Thus 𝑃 (𝑡) =

{

for 2 ≤ 𝑡 ≤ 6 0.02𝑡2 − 0.08 −0.03𝑡2 + 0.6𝑡 − 1.88 for 6 < 𝑡 ≤ 8.

In addition 𝑃 (𝑡) = 0 for 𝑡 < 2 and 𝑃 (𝑡) = 1 for 8 < 𝑡. (See Figure 8.104.) fraction of days

fraction of days per ton of caught fish

0.24

1 0.8

𝑝(𝑥)

𝑃 (𝑡)

0.6

0.12

0.4

0.08

0.2 2

5

6

7

8

𝑡 (tons of fish)

Figure 8.104: Cumulative distribution of daily catch

𝑥 (tons of fish) 2 5 6 7 8 Figure 8.105: Shaded area represents the probability that the catch is between 5 and 7 tons

(b) The probability that the catch is between 5 and 7 tons can be found using either the density function, 𝑝, or the cumulative distribution function, 𝑃 . If we use the density function, this probability can be represented by the shaded area in Figure 8.105, which is about 0.43: Probability catch is between 5 and 7 tons

7

=

∫5

𝑝(𝑥) 𝑑𝑥 = 0.43.

466

Chapter 8 USING THE DEFINITE INTEGRAL

The probability can be found from the cumulative distribution as follows: Probability catch is between 5 and 7 tons

= 𝑃 (7) − 𝑃 (5) = 0.85 − 0.42 = 0.43.

The Median and Mean It is often useful to be able to give an “average” value for a distribution. Two measures that are in common use are the median and the mean.

The Median A median of a quantity 𝑥 distributed through a population is a value 𝑇 such that half the population has values of 𝑥 less than (or equal to) 𝑇 , and half the population has values of 𝑥 greater than (or equal to) 𝑇 . Thus, a median 𝑇 satisfies 𝑇

∫−∞

𝑝(𝑥) 𝑑𝑥 = 0.5,

where 𝑝 is the density function. In other words, half the area under the graph of 𝑝 lies to the left of 𝑇 .

Example 2

Find the median age in the US in 2012, using the age density function given by { 0.013 for 0 ≤ 𝑡 ≤ 60 𝑝(𝑡) = 0.0358 − 0.00038𝑡 for 60 < 𝑡 ≤ 93.85.

Solution

We want to find the value of 𝑇 such that 𝑇

∫−∞

𝑇

𝑝(𝑡) 𝑑𝑡 =

∫0

𝑝(𝑡) 𝑑𝑡 = 0.5.

Since 𝑝(𝑡) = 0.013 up to age 60, we have Median = 𝑇 =

0.5 ≈ 38.5 years. 0.013

(See Figure 8.106.) fraction of population per year of age

Median



0.013

𝑝(𝑡) 0.5

38.5

60

𝑡 (age in years)

Figure 8.106: Median of age distribution

8.8 PROBABILITY, MEAN, AND MEDIAN

467



𝑝(𝑡)



Area= 𝑝(𝑡)Δ𝑡

❄ 𝑡

𝑡 (age)

𝑡 + Δ𝑡

Figure 8.107: Shaded area is percentage of population with age between 𝑡 and 𝑡 + Δ𝑡

The Mean Another commonly used average value is the mean. To find the mean of 𝑁 numbers, you add the numbers and divide the sum by 𝑁. For example, the mean of the numbers 1, 2, 7, and 10 is (1 + 2 + 7 + 10)∕4 = 5. The mean age of the entire US population is therefore defined as ∑ Ages of all people in the US Mean age = . Total number of people in the US Calculating the sum of all the ages directly would be an enormous task; we will approximate the sum by an integral. The idea is to “slice up” the age axis and consider the people whose age is between 𝑡 and 𝑡 + Δ𝑡. How many are there? The fraction of the population between 𝑡 and 𝑡 + Δ𝑡 is the area under the graph of 𝑝 between these points, which is well approximated by the area of the rectangle, 𝑝(𝑡)Δ𝑡. (See Figure 8.107.) If the total number of people in the population is 𝑁, then Number of people with age between 𝑡 and 𝑡 + Δ𝑡

≈ 𝑝(𝑡)Δ𝑡𝑁.

The age of all of these people is approximately 𝑡: Sum of ages of people

≈ 𝑡𝑝(𝑡)Δ𝑡𝑁.

between age 𝑡 and 𝑡 + Δ𝑡 Therefore, adding and factoring out an 𝑁 gives us Sum of ages of all people ≈

(∑

) 𝑡𝑝(𝑡)Δ𝑡 𝑁.

In the limit, as we allow Δ𝑡 to shrink to 0, the sum becomes an integral, so ( ) 100

Sum of ages of all people =

∫0

𝑡𝑝(𝑡)𝑑𝑡 𝑁.

Therefore, with 𝑁 equal to the total number of people in the US, and assuming no person is over 100 years old, Mean age =

100 Sum of ages of all people in US = 𝑡𝑝(𝑡)𝑑𝑡. ∫0 𝑁

We can give the same argument for any9 density function 𝑝(𝑥). 9 Provided

all the relevant improper integrals converge.

468

Chapter 8 USING THE DEFINITE INTEGRAL

If a quantity has density function 𝑝(𝑥), ∞

Mean value of the quantity =

∫−∞

𝑥𝑝(𝑥) 𝑑𝑥.

It can be shown that the mean is the point on the horizontal axis where the region under the graph of the density function, if it were made out of cardboard, would balance. Example 3

Find the mean age of the US population, using the density function of Example 2.

Solution

The formula for 𝑝 is ⎧0 ⎪ 0.013 𝑝(𝑡) = ⎨ ⎪ 0.0358 − 0.00038𝑡 ⎩0

for 𝑡 < 0 for 0 ≤ 𝑡 ≤ 60 for 60 < 𝑡 ≤ 93.85 for 𝑡 > 93.85.

Using these formulas, we compute 100

Mean age =

∫0

60

𝑡𝑝(𝑡)𝑑𝑡 =

93.85

𝑡(0.013)𝑑𝑡 +

∫0

𝑡(0.0358 − 0.00038𝑡)𝑑𝑡

∫60

60 93.85 93.85 𝑡2 | 𝑡2 | 𝑡3 | = 0.013 || + 0.0358 || − 0.00038 || ≈ 39.3 years. 2 |0 2 |60 3 |60

The mean is shown is Figure 8.108.

fraction of population per year of age

𝑝(𝑡)

39.3 𝑡 Mean = Balance point

Figure 8.108: Mean of age distribution

Normal Distributions How much rain do you expect to fall in your home town this year? If you live in Anchorage, Alaska, the answer is something close to 15 inches (including the snow). Of course, you don’t expect exactly 15 inches. Some years have more than 15 inches, and some years have less. Most years, however, the amount of rainfall is close to 15 inches; only rarely is it well above or well below 15 inches. What does the density function for the rainfall look like? To answer this question, we look at rainfall data over many years. Records show that the distribution of rainfall is well approximated by a normal distribution. The graph of its density function is a bell-shaped curve which peaks at 15 inches and slopes downward approximately symmetrically on either side. Normal distributions are frequently used to model real phenomena, from grades on an exam to the number of airline passengers on a particular flight. A normal distribution is characterized by

8.8 PROBABILITY, MEAN, AND MEDIAN

469

its mean, 𝜇, and its standard deviation, 𝜎. The mean tells us the location of the central peak. The standard deviation tells us how closely the data is clustered around the mean. A small value of 𝜎 tells us that the data is close to the mean; a large 𝜎√tells us the data is spread out. In the following formula for a normal distribution, the factor of 1∕(𝜎 2𝜋) makes the area under the graph equal to 1. A normal distribution has a density function of the form 2 2 1 𝑝(𝑥) = √ 𝑒−(𝑥−𝜇) ∕(2𝜎 ) , 𝜎 2𝜋

where 𝜇 is the mean of the distribution and 𝜎 is the standard deviation, with 𝜎 > 0. To model the rainfall in Anchorage, we use a normal distribution with 𝜇 = 15 and 𝜎 = 1. (See Figure 8.109.)

(15,

1 √ ) 2𝜋

𝑝(𝑥) =

2 1 √ 𝑒−(𝑥−15) ∕2 2𝜋

𝑥 13

15

17

Figure 8.109: Normal distribution with 𝜇 = 15 and 𝜎 = 1

Example 4

For Anchorage’s rainfall, use the normal distribution with the density function with 𝜇 = 15 and 𝜎 = 1 to compute the fraction of the years with rainfall between (a) 14 and 16 inches, (b) 13 and 17 inches, (c) 12 and 18 inches.

Solution

(a) The fraction of the years with annual rainfall between 14 and 16 inches is ∫14

16

Since there is no elementary antiderivative for 𝑒−(𝑥−15) value is about 0.68. Fraction of years with rainfall between 14 and 16 inches

16

=

∫14

(b) Finding the integral numerically again: Fraction of years with rainfall between 13 and 17 inches

17

=

∫13

(c) Fraction of years with rainfall between 12 and 18 inches

18

=

∫12

2 ∕2

2 1 √ 𝑒−(𝑥−15) ∕2 𝑑𝑥. 2𝜋

, we find the integral numerically. Its

1 −(𝑥−15)2 ∕2 𝑑𝑥 ≈ 0.68. √ 𝑒 2𝜋 1 −(𝑥−15)2 ∕2 𝑑𝑥 ≈ 0.95. √ 𝑒 2𝜋 1 −(𝑥−15)2 ∕2 𝑑𝑥 ≈ 0.997. √ 𝑒 2𝜋

Since 0.95 is so close to 1, we expect that most of the time the rainfall will be between 13 and 17 inches a year. Among the normal distributions, the one having 𝜇 = 0, 𝜎 = 1 is called the standard normal distribution. Values of the corresponding cumulative distribution function are published in tables.

470

Chapter 8 USING THE DEFINITE INTEGRAL

Exercises and Problems for Section 8.8 EXERCISES 1. Show that the area under the fishing density function in Figure 8.103 on page 464 is 1. Why is this to be expected?

(b) Explain how the graphs confirm that 𝜇 is the mean of the distribution and that 𝜎 is a measure of how closely the data is clustered around the mean.

2. Find the mean daily catch for the fishing data in Figure 8.103, page 464.

In Exercises 4–6, a density function 𝑝(𝑥) satisfying (I)–(IV) gives the fraction of years with a given total annual snowfall (in m) for a city.

3. (a) Using a calculator or computer, sketch graphs of the density function of the normal distribution 𝑝(𝑥) =

1 −(𝑥−𝜇)2 ∕(2𝜎 2 ) . √ 𝑒 𝜎 2𝜋

0.5

I.

∫0

III.

2

𝑝(𝑥) 𝑑𝑥 = 0.1

II.

2.72

𝑝(𝑥) 𝑑𝑥 = 0.5

∫0

∫0

IV.

𝑝(𝑥) 𝑑𝑥 = 0.3 ∞

∫0

𝑥𝑝(𝑥) 𝑑𝑥 = 2.65

(i) For fixed 𝜇 (say, 𝜇 = 5) and varying 𝜎 (say, 𝜎 = 1, 2, 3).

4. What is the median annual snowfall (in m)?

(ii) For varying 𝜇 (say, 𝜇 = 4, 5, 6) and fixed 𝜎 (say, 𝜎 = 1).

6. What is the probability of an annual snowfall between 0.5 and 2 m?

5. What is the mean snowfall (in m)?

PROBLEMS 7. A screening test for susceptibility to diabetes reports a numerical score between 0 to 100. A score greater than 50 indicates a potential risk, with some lifestyle training recommended. Results from 200,000 people who were tested show that : • 75% received scores evenly distributed between 0 and 50. • 25% received scores evenly distributed between 50 and 100. The probability density function (pdf) is in Figure 8.110. (a) Find the values of 𝐴 and 𝐵 that make this a probability density function. (b) Find the median test score. (c) Find the mean test score. (d) Give a graph of the cumulative distribution function (cdf) for these test scores.

fraction of years per meter of snow

1

𝑝(𝑥)

0.5 0

snowfall (m)

1

2

3

4

Figure 8.111

8. Which of the following events is most likely? I. A winter has 3 m or more of snow? II. A winter has 2 m or less of snow? III. A winter with between 2 and 3 m of snow? 9. The shaded area is 0.8. What percentage of winters see more than 3 meters of total snowfall?

fraction of population per test score point

𝐴

10. What appears to be the smallest and largest total annual snowfall?

𝐵 𝑥 (test score) 50

11. If 𝑝(2.8) = 1.1, approximately what percentage of winters see snowfall totals between 2.8 m and 3.0 m?

100

Figure 8.110

In Problems 8–11, use Figure 8.111, a graph of 𝑝(𝑥), a density function for the fraction of a region’s winters with a given total snowfall (in m).

12. A quantity 𝑥 has density function 𝑝(𝑥) = 0.5(2 − 𝑥) for 0 ≤ 𝑥 ≤ 2 and 𝑝(𝑥) = 0 otherwise. Find the mean and median of 𝑥. 13. A quantity 𝑥 has cumulative distribution function 𝑃 (𝑥) = 𝑥 − 𝑥2 ∕4 for 0 ≤ 𝑥 ≤ 2 and 𝑃 (𝑥) = 0 for 𝑥 < 0 and 𝑃 (𝑥) = 1 for 𝑥 > 2. Find the mean and median of 𝑥.

471

8.8 PROBABILITY, MEAN, AND MEDIAN

14. The probability of a transistor failing between 𝑡 = 𝑎 𝑏 months and 𝑡 = 𝑏 months is given by 𝑐 ∫𝑎 𝑒−𝑐𝑡 𝑑𝑡, for some constant 𝑐.

19. The distribution of IQ scores can be modeled by a normal distribution with mean 100 and standard deviation 15.

(a) If the probability of failure within the first six months is 10%, what is 𝑐? (b) Given the value of 𝑐 in part (a), what is the probability the transistor fails within the second six months?

(a) Write the formula for the density function of IQ scores. (b) Estimate the fraction of the population with IQ between 115 and 120.

15. Suppose that 𝑥 measures the time (in hours) it takes for a student to complete an exam. All students are done within two hours and the density function for 𝑥 is { 3 𝑝(𝑥) = 𝑥 ∕4 if 0 < 𝑥 < 2 0 otherwise. (a) What proportion of students take between 1.5 and 2.0 hours to finish the exam? (b) What is the mean time for students to complete the exam? (c) Compute the median of this distribution.

20. The speeds of cars on a road are approximately normally distributed with a mean 𝜇 = 58 km/hr and standard deviation 𝜎 = 4 km/hr. (a) What is the probability that a randomly selected car is going between 60 and 65 km/hr? (b) What fraction of all cars are going slower than 52 km/hr? 21. Consider the normal distribution, 𝑝(𝑥). (a) Show that 𝑝(𝑥) is a maximum when 𝑥 = 𝜇. What is that maximum value? (b) Show that 𝑝(𝑥) has points of inflection where 𝑥 = 𝜇 + 𝜎 and 𝑥 = 𝜇 − 𝜎. (c) Describe in your own words what 𝜇 and 𝜎 tell you about the distribution.

16. In 1950 an experiment was done observing the time gaps between successive cars on the Arroyo Seco Freeway.10 The data show that the density function of these time gaps was given approximately by 𝑝(𝑥) = 𝑎𝑒−0.122𝑥 where 𝑥 is the time in seconds and 𝑎 is a constant. (a) (b) (c) (d)

Find 𝑎. Find 𝑃 , the cumulative distribution function. Find the median and mean time gap. Sketch rough graphs of 𝑝 and 𝑃 .

17. Consider a group of people who have received treatment for a disease such as cancer. Let 𝑡 be the survival time, the number of years a person lives after receiving treatment. The density function giving the distribution of 𝑡 is 𝑝(𝑡) = 𝐶𝑒−𝐶𝑡 for some positive constant 𝐶. (a) What is the practical meaning for the cumulative 𝑡 distribution function 𝑃 (𝑡) = ∫0 𝑝(𝑥) 𝑑𝑥? (b) The survival function, 𝑆(𝑡), is the probability that a randomly selected person survives for at least 𝑡 years. Find 𝑆(𝑡). (c) Suppose a patient has a 70% probability of surviving at least two years. Find 𝐶. 18. While taking a walk along the road where you live, you accidentally drop your glove, but you don’t know where. The probability density 𝑝(𝑥) for having dropped the glove 𝑥 kilometers from home (along the road) is 𝑝(𝑥) = 2𝑒−2𝑥

22. For a normal population of mean 0, show that the fraction of the population within one standard deviation of the mean does not depend on the standard deviation. [Hint: Use the substitution 𝑤 = 𝑥∕𝜎.] 23. Which of the following functions makes the most sense as a model for the probability density representing the time (in minutes, starting from 𝑡 = 0) that the next customer walks into a store? { cos 𝑡 0 ≤ 𝑡 ≤ 2𝜋 (a) 𝑝(𝑡) = 𝑒𝑡−2𝜋 𝑡 ≥ 2𝜋 (b) 𝑝(𝑡) = 3𝑒−3𝑡 for 𝑡 ≥ 0 (c) 𝑝(𝑡) = 𝑒−3𝑡 for 𝑡 ≥ 0 (d) 𝑝(𝑡) = 1∕4 for 0 ≤ 𝑡 ≤ 4 24. Let 𝑃 (𝑥) be the cumulative distribution function for the household income distribution in the US in 2009.11 Values of 𝑃 (𝑥) are in the following table:

Income 𝑥 (thousand $) 𝑃 (𝑥) (%)

for 𝑥 ≥ 0.

(a) What is the probability that you dropped it within 1 kilometer of home? (b) At what distance 𝑦 from home is the probability that you dropped it within 𝑦 km of home equal to 0.95? 10 Reported

20

40

60

75

100

29.5

50.1

66.8

76.2

87.1

(a) What percent of the households made between $40,000 and $60,000? More than $100,000? (b) Approximately what was the median income? (c) Is the statement “More than one-third of households made between $40,000 and $75,000" true or false?

by Daniel Furlough and Frank Barnes.

11 http://www.census.gov/hhes/www/income/income.html,

accessed January 7, 2012.

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Chapter 8 USING THE DEFINITE INTEGRAL

25. If we think of an electron as a particle, the function 2

−2𝑟

𝑃 (𝑟) = 1 − (2𝑟 + 2𝑟 + 1)𝑒

is the cumulative distribution function of the distance, 𝑟, of the electron in a hydrogen atom from the center of the atom. The distance is measured in Bohr radii. (1 Bohr radius = 5.29 × 10−11 m. Niels Bohr (1885–1962) was a Danish physicist.) For example, 𝑃 (1) = 1 − 5𝑒−2 ≈ 0.32 means that the electron is within 1 Bohr radius from the center of

the atom 32% of the time. (a) Find a formula for the density function of this distribution. Sketch the density function and the cumulative distribution function. (b) Find the median distance and the mean distance. Near what value of 𝑟 is an electron most likely to be found? (c) The Bohr radius is sometimes called the “radius of the hydrogen atom.” Why?

Strengthen Your Understanding In Problems 26–27, explain what is wrong with the statement. 26. A median 𝑇 of a quantity distributed through a population satisfies 𝑝(𝑇 ) = 0.5 where 𝑝 is the density function.

29. A distribution with a mean of 1∕2 and median 1∕2. In Problems 30–34, a quantity 𝑥 is distributed through a population with probability density function 𝑝(𝑥) and cumulative distribution function 𝑃 (𝑥). Decide if each statement is true or false. Give an explanation for your answer.

27. The following density function has median 1: 30. If 𝑝(10) = 1∕2, then half the population has 𝑥 < 10. ⎧ 0 ⎪ 𝑝(𝑥) = ⎨ 2(1 − 𝑥) ⎪ ⎩ 0

for 𝑥 < 0 for 0 ≤ 𝑥 ≤ 1 for 𝑥 > 1.

31. If 𝑃 (10) = 1∕2, then half the population has 𝑥 < 10. 32. If 𝑝(10) = 1∕2, then the fraction of the population lying between 𝑥 = 9.98 and 𝑥 = 10.04 is about 0.03.

In Problems 28–29, give an example of:

33. If 𝑝(10) = 𝑝(20), then none of the population has 𝑥 values lying between 10 and 20.

28. A distribution with a mean of 1∕2 and standard deviation 1∕2.

34. If 𝑃 (10) = 𝑃 (20), then none of the population has 𝑥 values lying between 10 and 20.

Online Resource: Review problems and Projects

Chapter Nine

SEQUENCES AND SERIES

Contents 9.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Numerical, Algebraic, and Graphical Viewpoint . . . . . . . . . . . . . . . . . . . . . Defining Sequences Recursively . . . . . . . . . . . . . Convergence of Sequences. . . . . . . . . . . . . . . . . . Convergence and Bounded Sequences. . . . . . . . . 9.2 Geometric Series . . . . . . . . . . . . . . . . . . . . . . . . . . Repeated Drug Dosage . . . . . . . . . . . . . . . . . . . .

474

474 475 476 477 480 481 What Happens as n → ∞? . . . . . . . . . . . . . . . . . . . . . 482 The Geometric Series in General. . . . . . . . . . . . . 482 Sum of a Finite Geometric Series . . . . . . . . . . . . 482 Sum of an Infinite Geometric Series . . . . . . . . . . 482 Regular Deposits into a Savings Account . . . . . . 484 What Happens as n → ∞? . . . . . . . . . . . . . . . . . . . 485 9.3 Convergence of Series . . . . . . . . . . . . . . . . . . . . . . 488 Partial Sums and Convergence of Series . . . . . . . 488 Visualizing Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 489 Comparison of Series and Integrals . . . . . . . . . . . 490 9.4 Tests for Convergence . . . . . . . . . . . . . . . . . . . . . . 494 Comparison of Series. . . . . . . . . . . . . . . . . . . . . . 494 Limit Comparison Test. . . . . . . . . . . . . . . . . . . . . . . . 496 Series of Both Positive and Negative Terms . . . . 497 Comparison with a Geometric Series: The Ratio Test . . . . . . . . . . . . . . . . . . . . . . . . . . 498 Alternating Series. . . . . . . . . . . . . . . . . . . . . . . . . 499 Absolute and Conditional Convergence. . . . . . . . 500 9.5 Power Series and Interval of Convergence . . . . . 504 Numerical and Graphical View of Convergence . . . . . . . . . . . . . . . . . . . . . . . . . 505 Intervals of Convergence . . . . . . . . . . . . . . . . . . . 506 What Happens at the Endpoints of the Interval of Convergence? . . . . . . . . . . . . . . . . . . 508 Series with All Odd, or All Even, Terms . . . . . . . 509

474

Chapter 9 SEQUENCES AND SERIES

9.1

SEQUENCES A sequence1 is an infinite list of numbers 𝑠1 , 𝑠2 , 𝑠3 , … , 𝑠𝑛, …. We call 𝑠1 the first term, 𝑠2 the second term; 𝑠𝑛 is the general term. For example, the sequence of squares, 1, 4, 9, … , 𝑛2, …, can be denoted by the general term 𝑠𝑛 = 𝑛2 . Thus, a sequence is a function whose domain is the positive integers, but it is traditional to denote the terms of a sequence using subscripts, 𝑠𝑛 , rather than function notation, 𝑠(𝑛). In addition, we may talk about sequences whose general term has no simple formula, such as the sequence 3, 3.1, 3.14, 3.141, 3.1415, …, in which 𝑠𝑛 gives the first 𝑛 digits of 𝜋.

The Numerical, Algebraic, and Graphical Viewpoint Just as we can view a function algebraically, numerically, graphically, or verbally, we can view sequences in different ways. We may give an algebraic formula for the general term. We may give the numerical values of the first few terms of the sequence, suggesting a pattern for the later terms. Example 1

Give the first six terms of the following sequences: 𝑛(𝑛 + 1) 𝑛 + (−1)𝑛 (a) 𝑠𝑛 = (b) 𝑠𝑛 = 2 𝑛

Solution

(a) Substituting 𝑛 = 1, 2, 3, 4, 5, 6 into the formula for the general term, we get 1⋅2 2⋅3 3⋅4 4⋅5 5⋅6 6⋅7 , , , , , = 1, 3, 6, 10, 15, 21. 2 2 2 2 2 2 (b) Substituting 𝑛 = 1, 2, 3, 4, 5, 6 into the formula for the general term, we get 1−1 2+1 3−1 4+1 5−1 6+1 3 2 5 4 7 , , , , , = 0, , , , , . 1 2 3 4 5 6 2 3 4 5 6

Example 2

Give a general term for the following sequences: (a) 1, 2, 4, 8, 16, 32, …

Solution

(b)

7 7 7 7 1 7 , , , , , ,… 2 5 8 11 2 17

Although the first six terms do not determine the sequence, we can sometimes use them to guess a possible formula for the general term. (a) We have powers of 2, so we guess 𝑠𝑛 = 2𝑛 . When we check by substituting in 𝑛 = 1, 2, 3, 4, 5, 6, we get 2, 4, 8, 16, 32, 64, instead of 1, 2, 4, 8, 16, 32. We fix our guess by subtracting 1 from the exponent, so the general term is 𝑠𝑛 = 2𝑛−1 . Substituting the first six values of 𝑛 shows that the formula checks. (b) In this sequence, the fifth term looks different from the others, whose numerators are all 7. We can fix this by rewriting 1∕2 = 7∕14. The sequence of denominators is then 2, 5, 8, 11, 14, 17. This looks like a linear function with slope 3, so we expect the denominator has formula 3𝑛 + 𝑘 for some 𝑘. When 𝑛 = 1, the denominator is 2, so 2 = 3⋅1+𝑘

giving

𝑘 = −1

and the denominator of 𝑠𝑛 is 3𝑛 − 1. Our general term is then 𝑠𝑛 =

7 . 3𝑛 − 1

To check this, evaluate 𝑠𝑛 for 𝑛 = 1, … , 6. 1 In everyday English, the words “sequence” and “series” are used interchangeably. In mathematics, they have different meanings and cannot be interchanged.

9.1 SEQUENCES

475

There are two ways to visualize a sequence. One is to plot points with 𝑛 on the horizontal axis and 𝑠𝑛 on the vertical axis. The other is to label points on a number line 𝑠1 , 𝑠2 , 𝑠3 , . . . . See Figure 9.1 for the sequence 𝑠𝑛 = 1 + (−1)𝑛 ∕𝑛. 𝑠𝑛

1 0 𝑛 2

4

6

8

10

1 𝑠3 𝑠5

𝑠1

2 𝑠6 𝑠4

𝑠2

𝑠𝑛

Figure 9.1: The sequence 𝑠𝑛 = 1 + (−1)𝑛 ∕𝑛

Defining Sequences Recursively Sequences can also be defined recursively, by giving an equation relating the 𝑛th term to the previous terms and as many of the first few terms as are needed to get started. Example 3

Give the first six terms of the recursively defined sequences. (a) 𝑠𝑛 = 𝑠𝑛−1 + 3 for 𝑛 > 1 and 𝑠1 = 4 (b) 𝑠𝑛 = −3𝑠𝑛−1 for 𝑛 > 1 and 𝑠1 = 2 (c) 𝑠𝑛 = 21 (𝑠𝑛−1 + 𝑠𝑛−2 ) for 𝑛 > 2 and 𝑠1 = 0, 𝑠2 = 1 (d) 𝑠𝑛 = 𝑛𝑠𝑛−1 for 𝑛 > 1 and 𝑠1 = 1

Solution

(a) When 𝑛 = 2, we obtain 𝑠2 = 𝑠1 +3 = 4+3 = 7. When 𝑛 = 3, we obtain 𝑠3 = 𝑠2 +3 = 7+3 = 10. In words, we obtain each term by adding 3 to the previous term. The first six terms are 4, 7, 10, 13, 16, 19. (b) Each term is −3 times the previous term, starting with 𝑠1 = 2. We have 𝑠2 = −3𝑠1 = −3⋅2 = −6 and 𝑠3 = −3𝑠2 = −3(−6) = 18. Continuing, we get 2, −6, 18, −54, 162, −486. (c) Each term is the average of the previous two terms, starting with 𝑠1 = 0 and 𝑠2 = 1. We get 𝑠3 = (𝑠2 + 𝑠1 )∕2 = (1 + 0)∕2 = 1∕2. Then 𝑠4 = (𝑠3 + 𝑠2 )∕2 = ((1∕2) + 1)∕2 = 3∕4. Continuing, we get 1 3 5 11 0, 1, , , , . 2 4 8 16 (d) Here 𝑠2 = 2𝑠1 = 2 ⋅ 1 = 2 so 𝑠3 = 3𝑠2 = 3 ⋅ 2 = 6 and 𝑠4 = 4𝑠3 = 4 ⋅ 6 = 24. Continuing gives 1, 2, 6, 24, 120, 720.

The general term of part (d) of the previous example is given by 𝑠𝑛 = 𝑛(𝑛 − 1)(𝑛 − 2) ⋯ 3 ⋅ 2 ⋅ 1, which is denoted 𝑠𝑛 = 𝑛! and is called 𝑛 factorial. We define 0! = 1. We can also look at the first few terms of a sequence and try to guess a recursive definition by looking for a pattern. Example 4

Give a recursive definition of the following sequences. (a) 1, 3, 7, 15, 31, 63, …

(b) 1, 4, 9, 16, 25, 36, …

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Chapter 9 SEQUENCES AND SERIES

Solution

(a) Each term is twice the previous term plus one; for example, 7 = 2 ⋅ 3 + 1 and 63 = 2 ⋅ 31 + 1. Thus, a recursive definition is 𝑠𝑛 = 2𝑠𝑛−1 + 1 for 𝑛 > 1 and 𝑠1 = 1. There are other ways to define the sequence recursively. We might notice, for example, that the differences of consecutive terms are powers of 2. Thus, we could also use 𝑠𝑛 = 𝑠𝑛−1 + 2𝑛−1

for 𝑛 > 1 and 𝑠1 = 1.

(b) We recognize the terms as the squares of the positive integers, but we are looking for a recursive definition which relates consecutive terms. We see that 𝑠2 = 𝑠1 + 3 𝑠3 = 𝑠2 + 5 𝑠4 = 𝑠3 + 7 𝑠5 = 𝑠4 + 9, so the differences between consecutive terms are consecutive odd integers. The difference between 𝑠𝑛 and 𝑠𝑛−1 is 2𝑛 − 1, so a recursive definition is 𝑠𝑛 = 𝑠𝑛−1 + 2𝑛 − 1, for 𝑛 > 1 and 𝑠1 = 1. Recursively defined sequences, sometimes called recurrence relations, are powerful tools used frequently in computer science, as well as in differential equations. Finding a formula for the general term can be surprisingly difficult.

Convergence of Sequences The limit of a sequence 𝑠𝑛 as 𝑛 → ∞ is defined the same way as the limit of a function 𝑓 (𝑥) as 𝑥 → ∞; see also Problem 80 (available online). The sequence 𝑠1 , 𝑠2 , 𝑠3 , … , 𝑠𝑛 , … has a limit 𝐿, written lim 𝑠𝑛 = 𝐿, if 𝑠𝑛 is as close to 𝐿 as 𝑛→∞ we please whenever 𝑛 is sufficiently large. If a limit, 𝐿, exists, we say the sequence converges to its limit 𝐿. If no limit exists, we say the sequence diverges. To calculate the limit of a sequence, we use what we know about the limits of functions, including the properties in Theorem 1.2 and the following facts: • The sequence 𝑠𝑛 = 𝑥𝑛 converges to 0 if |𝑥| < 1 and diverges if |𝑥| > 1 • The sequence 𝑠𝑛 = 1∕𝑛𝑝 converges to 0 if 𝑝 > 0 Example 5

Do the following sequences converge or diverge? If a sequence converges, find its limit. (a) 𝑠𝑛 = (0.8)𝑛

Solution

(b) 𝑠𝑛 =

1 − 𝑒−𝑛 1 + 𝑒−𝑛

(c) 𝑠𝑛 =

𝑛2 + 1 𝑛

(d) 𝑠𝑛 = 1 + (−1)𝑛

(a) Since 0.8 < 1, the sequence converges by the first fact and the limit is 0. (b) Since 𝑒−1 < 1, we have lim 𝑒−𝑛 = lim (𝑒−1 )𝑛 = 0 by the first fact. Thus, using the properties 𝑛→∞ 𝑛→∞ of limits from Section 1.8, we have 1 − 𝑒−𝑛 1−0 = 1. = lim 𝑛→∞ 1 + 𝑒−𝑛 1+0 (c) Since (𝑛2 + 1)∕𝑛 grows without bound as 𝑛 → ∞, the sequence 𝑠𝑛 diverges. (d) Since (−1)𝑛 alternates in sign, the sequence alternates between 0 and 2. Thus the sequence 𝑠𝑛 diverges, since it does not get close to any fixed value.

9.1 SEQUENCES

477

Convergence and Bounded Sequences A sequence 𝑠𝑛 is bounded if there are numbers 𝐾 and 𝑀 such that 𝐾 ≤ 𝑠𝑛 ≤ 𝑀 for all terms. If lim 𝑠 = 𝐿, then from some point on, the terms are bounded between 𝐿 − 1 and 𝐿 + 1. Thus we 𝑛→∞ 𝑛 have the following fact: A convergent sequence is bounded. On the other hand, a bounded sequence need not be convergent. In Example 5, we saw that 1 + (−1)𝑛 diverges, but it is bounded between 0 and 2. To ensure that a bounded sequence converges we need to rule out this sort of oscillation. The following theorem gives a condition that ensures convergence for a bounded sequence. A sequence 𝑠𝑛 is called monotone if it is either increasing, that is 𝑠𝑛 < 𝑠𝑛+1 for all 𝑛, or decreasing, that is 𝑠𝑛 > 𝑠𝑛+1 for all 𝑛.

Theorem 9.1: Convergence of a Monotone, Bounded Sequence If a sequence 𝑠𝑛 is bounded and monotone, it converges. To understand this theorem graphically, see Figure 9.2. The sequence 𝑠𝑛 is increasing and bounded above by 𝑀, so the values of 𝑠𝑛 must “pile up” at some number less than or equal to 𝑀. This number is the limit.2 𝑠1

𝑠2

𝑠3

𝑠4 𝑠5

𝑀

Figure 9.2: Values of 𝑠𝑛 for 𝑛 = 1, 2, ⋯ , 10

Example 6

The sequence 𝑠𝑛 = (1∕2)𝑛 is decreasing and bounded below by 0, so it converges. We have already seen that it converges to 0.

Example 7

The sequence 𝑠𝑛 = (1 + 1∕𝑛)𝑛 can be shown to be increasing and bounded (see Project 2 (available online)). Theorem 9.1 then guarantees that this sequence has a limit, which turns out to be 𝑒. (In fact, the sequence can be used to define 𝑒.)

Example 8

If 𝑠𝑛 = (1 + 1∕𝑛)𝑛 , find 𝑠100 and 𝑠1000 . How many decimal places agree with 𝑒?

Solution

We have 𝑠100 = (1.01)100 = 2.7048 and 𝑠1000 = (1.001)1000 = 2.7169. Since 𝑒 = 2.7183 …,we see that 𝑠100 agrees with 𝑒 to one decimal place and 𝑠1000 agrees with 𝑒 to two decimal places.

Exercises and Problems for Section 9.1 Online Resource: Additional Problems for Section 9.1 EXERCISES

For Exercises 1–6, find the first five terms of the sequence from the formula for 𝑠𝑛 , 𝑛 ≥ 1. 1. 2𝑛 + 1 3.

2𝑛 2𝑛 + 1

5. (−1)𝑛+1

( )𝑛−1 1 2 2 See

2. 𝑛 + (−1)𝑛 ( )𝑛 1 4. (−1)𝑛 2 )𝑛+1 ( 1 6. 1 − 𝑛+1 the online supplement for a proof.

In Exercises 7–12, find a formula for 𝑠𝑛 , 𝑛 ≥ 1. 7. 4, 8, 16, 32, 64, …

8. 1, 3, 7, 15, 31, …

9. 2, 5, 10, 17, 26, …

10. 1, −3, 5, −7, 9, …

11. 1∕3, 2∕5, 3∕7, 4∕9, 5∕11, … 12. 1∕2, −1∕4, 1∕6, −1∕8, 1∕10, …

478

Chapter 9 SEQUENCES AND SERIES

PROBLEMS 13. Match formulas (a)–(d) with graphs (I)–(IV). (a) 𝑠𝑛 = 1 − 1∕𝑛 (c) 𝑠𝑛 = 1∕𝑛

(b) 𝑠𝑛 = 1 + (−1) ∕𝑛 (d) 𝑠𝑛 = 1 + 1∕𝑛

𝑠𝑛

(I)

Do the sequences in Problems 16–27 converge or diverge? If a sequence converges, find its limit.

𝑛

18. 3 + 𝑒−2𝑛

19. (−0.3)𝑛

20.

10 𝑛 + 10 𝑛

21.

2𝑛 3𝑛

22.

2𝑛 + 1 𝑛

23.

(−1)𝑛 𝑛

24.

1 + ln 𝑛 𝑛

25.

2𝑛 + (−1)𝑛 5 4𝑛 − (−1)𝑛 3

26.

sin 𝑛 𝑛

27. cos(𝜋𝑛)

1 𝑛

𝑛

10

10

𝑠𝑛

(III)

17. (0.2)𝑛

𝑠𝑛

(II)

1

16. 2𝑛

𝑠𝑛

(IV)

1

1 𝑛

𝑛

10

10

In Problems 28–31, find the first six terms of the recursively defined sequence. 28. 𝑠𝑛 = 2𝑠𝑛−1 + 3 for 𝑛 > 1 and 𝑠1 = 1

14. Match formulas (a)–(e) with graphs (I)–(V). (a) (b) (c) (d) (e)

𝑠𝑛 𝑠𝑛 𝑠𝑛 𝑠𝑛 𝑠𝑛

29. 𝑠𝑛 = 𝑠𝑛−1 + 𝑛 for 𝑛 > 1 and 𝑠1 = 1 ( )𝑛−1 30. 𝑠𝑛 = 𝑠𝑛−1 + 12 for 𝑛 > 1 and 𝑠1 = 0

= 2 − 1∕𝑛 = (−1)𝑛 2 + 1∕𝑛 = 2 + (−1)𝑛 ∕𝑛 = 2 + 1∕𝑛 = (−1)𝑛 2 + (−1)𝑛 ∕𝑛

31. 𝑠𝑛 = 𝑠𝑛−1 + 2𝑠𝑛−2 for 𝑛 > 2 and 𝑠1 = 1, 𝑠2 = 5 In Problems 32–33, let 𝑎1 = 8, 𝑏1 = 5, and, for 𝑛 > 1, 𝑎𝑛 = 𝑎𝑛−1 + 3𝑛

(I)

𝑠𝑛 −2

0

2

−2

0

2

−2

0

2

(II)

𝑠𝑛

(III)

𝑠𝑛

(IV)

𝑠𝑛 −2

0

2

−2

0

2

(V)

𝑠𝑛

15. Match formulas (a)–(e) with descriptions (I)–(V) of the behavior of the sequence as 𝑛 → ∞. (a) (b) (c) (d) (e) (I) (II) (III) (IV) (V)

𝑠𝑛 𝑠𝑛 𝑠𝑛 𝑠𝑛 𝑠𝑛

= 𝑛(𝑛 + 1) − 1 = 1∕(𝑛 + 1) = 1 − 𝑛2 = cos(1∕𝑛) = (sin 𝑛)∕𝑛

Diverges to −∞ Diverges to +∞ Converges to 0 through positive numbers Converges to 1 Converges to 0 through positive and negative numbers 3 http://en.wikipedia.org/wiki/Tribonacci_numbers,

𝑏𝑛 = 𝑏𝑛−1 + 𝑎𝑛−1 . 32. Give the values of 𝑎2 , 𝑎3 , 𝑎4 . 33. Give the values of 𝑏2 , 𝑏3 , 𝑏4 , 𝑏5 . 34. Suppose 𝑠1 = 0, 𝑠2 = 0, 𝑠3 = 1, and that 𝑠𝑛 = 𝑠𝑛−1 + 𝑠𝑛−2 + 𝑠𝑛−3 for 𝑛 ≥ 4. The members of the resulting sequence are called tribonacci numbers.3 Find 𝑠4 , 𝑠5 , … , 𝑠10 . In Problems 35–40, find a recursive definition for the sequence. 35. 1, 3, 5, 7, 9, …

36. 2, 4, 6, 8, 10, …

37. 3, 5, 9, 17, 33, …

38. 1, 5, 14, 30, 55, …

39. 1, 3, 6, 10, 15, …

3 5 8 13 40. 1, 2, , , , , … 2 3 5 8

In Problems 41–43, show that the sequence 𝑠𝑛 satisfies the recurrence relation. 41. 𝑠𝑛 = 3𝑛 − 2 𝑠𝑛 = 𝑠𝑛−1 + 3 for 𝑛 > 1 and 𝑠1 = 1 42. 𝑠𝑛 = 𝑛(𝑛 + 1)∕2 𝑠𝑛 = 𝑠𝑛−1 + 𝑛 for 𝑛 > 1 and 𝑠1 = 1 43. 𝑠𝑛 = 2𝑛2 − 𝑛 𝑠𝑛 = 𝑠𝑛−1 + 4𝑛 − 3 for 𝑛 > 1 and 𝑠1 = 1 accessed March 16, 2011.

9.1 SEQUENCES

In Problems 44–47, use the formula for 𝑠𝑛 to give the third term of the sequence, 𝑠3 . 44. 𝑠𝑛 = (−1)𝑛 2𝑛−1 ⋅ 𝑛2

45. 𝑠𝑛 =

𝑛 ∑

3 ⋅ 2𝑘

𝑘=0

1

46. 𝑠𝑛 = 2 + 3𝑠𝑛−1 , where 𝑠0 = 5.

47. 𝑠𝑛 =

1 𝑑𝑥 ∫1∕𝑛 𝑥2

Problems 48–50 concern analog signals in electrical engineering, which are continuous functions 𝑓 (𝑡), where 𝑡 is time. To digitize the signal, we sample 𝑓 (𝑡) every Δ𝑡 to form the sequence 𝑠𝑛 = 𝑓 (𝑛Δ𝑡). For example, if 𝑓 (𝑡) = sin 𝑡 with 𝑡 in seconds, sampling 𝑓 every 1∕10 second produces the sequence sin(1∕10), sin(2∕10), sin(3∕10), …. Give the first 6 terms of a sampling of the signal every Δ𝑡 seconds. 48. 𝑓 (𝑡) = (𝑡 − 1)2 , Δ𝑡 = 0.5 49. 𝑓 (𝑡) = cos 5𝑡, Δ𝑡 = 0.1 sin 𝑡 , Δ𝑡 = 1 50. 𝑓 (𝑡) = 𝑡 In Problems 51–53, we smooth a sequence, 𝑠1 , 𝑠2 , 𝑠3 , …, by replacing each term 𝑠𝑛 by 𝑡𝑛 , the average of 𝑠𝑛 with its neighboring terms 𝑡𝑛 =

(𝑠𝑛−1 + 𝑠𝑛 + 𝑠𝑛+1 ) for 𝑛 > 1. 3

Start with 𝑡1 = (𝑠1 + 𝑠2 )∕2, since 𝑠1 has only one neighbor. Smooth the given sequence once and then smooth the resulting sequence. What do you notice? 51. 18, −18, 18, −18, 18, −18, 18 … 52. 0, 0, 0, 18, 0, 0, 0, 0 … 53. 1, 2, 3, 4, 5, 6, 7, 8 … In Problems 54–57, for the function 𝑓 define a sequence recursively by 𝑥𝑛 = 𝑓 (𝑥𝑛−1 ) for 𝑛 > 1 and 𝑥1 = 𝑎. Depending on 𝑓 and the starting value 𝑎, this sequence may converge to a limit 𝐿. If 𝐿 exists, it has the property that 𝑓 (𝐿) = 𝐿. For the functions and starting values given, use a calculator to see if the sequence converges. [To obtain the terms of the sequence, push the function button repeatedly.] 54. 𝑓 (𝑥) = cos 𝑥, 𝑎 = 0 56. 𝑓 (𝑥) = sin 𝑥, 𝑎 = 1

55. 𝑓 (𝑥) = 𝑒−𝑥 , 𝑎 = 0 √ 57. 𝑓 (𝑥) = 𝑥, 𝑎 = 0.5

58. Let 𝑉𝑛 be the number of new SUVs sold in the US in month 𝑛, where 𝑛 = 1 is January 2016. In terms of SUVs, what do the following represent? (a) 𝑉10 (b) 𝑉𝑛 − 𝑉𝑛−1 ∑12 ∑𝑛 (c) 𝑖=1 𝑉𝑖 and 𝑖=1 𝑉𝑖

479

59. (a) Let 𝑠𝑛 be the number of ancestors a person has 𝑛 generations ago. (Your ancestors are your parents, grandparents, great-grandparents, etc.) What is 𝑠1 ? 𝑠2 ? Find a formula for 𝑠𝑛 . (b) For which 𝑛 is 𝑠𝑛 greater than 6 billion, the current world population? What does this tell you about your ancestors? 60. For 1 ≤ 𝑛 ≤ 10, find a formula for 𝑝𝑛 , the payment in year 𝑛 on a loan of $100,000. Interest is 5% per year, compounded annually, and payments are made at the end of each year for ten years. Each payment is $10,000 plus the interest on the amount of money outstanding. 61. (a) Cans are stacked in a triangle on a shelf. The bottom row contains 𝑘 cans, the row above contains one can fewer, and so on, until the top row, which has one can. How many rows are there? Find 𝑎𝑛 , the number of cans in the 𝑛th row, 1 ≤ 𝑛 ≤ 𝑘 (where the top row is 𝑛 = 1). (b) Let 𝑇𝑛 be the total number of cans in the top 𝑛 rows. Find a recurrence relation for 𝑇𝑛 in terms of 𝑇𝑛−1 . (c) Show that 𝑇𝑛 = 12 𝑛(𝑛 + 1) satisfies the recurrence relation. 62. You are deciding whether to buy a new or a two-yearold car (of the same make) based on which will have cost you less when you resell it at the end of three years. Your cost consists of two parts: the loss in value of the car and the repairs. A new car costs $20,000 and loses 12% of its value each year. Repairs are $400 the first year and increase by 18% each subsequent year. (a) For a new car, find the first three terms of the sequence 𝑑𝑛 giving the depreciation (loss of value) in dollars in year 𝑛. Give a formula for 𝑑𝑛 . (b) Find the first three terms of the sequence 𝑟𝑛 , the repair cost in dollars for a new car in year 𝑛. Give a formula for 𝑟𝑛 . (c) Find the total cost of owning a new car for three years. (d) Find the total cost of owning the two-year-old car for three years. Which should you buy? 63. The Fibonacci sequence, first studied by the thirteenthcentury Italian mathematician Leonardo di Pisa, also known as Fibonacci, is defined recursively by 𝐹𝑛 = 𝐹𝑛−1 + 𝐹𝑛−2 for 𝑛 > 2 and 𝐹1 = 1, 𝐹2 = 1. The Fibonacci sequence occurs in many branches of mathematics and can be found in patterns of plant growth (phyllotaxis). (a) Find the first 12 terms. (b) Show that the sequence of successive ratios 𝐹𝑛+1 ∕𝐹𝑛 appears to converge to a number 𝑟 satisfying the equation 𝑟2 = 𝑟 + 1. (The number 𝑟 was known as the golden ratio to the ancient Greeks.) (c) Let 𝑟 satisfy 𝑟2 = 𝑟 + 1. Show that the sequence 𝑠𝑛 = 𝐴𝑟𝑛 , where 𝐴 is constant, satisfies the Fibonacci equation 𝑠𝑛 = 𝑠𝑛−1 + 𝑠𝑛−2 for 𝑛 > 2.

480

Chapter 9 SEQUENCES AND SERIES

For example, 𝑓 (12∕5) = 2 + (1 − 2∕5) = 13∕5 and 𝑓 (3) = 3 + (1 − 0) = 4.

64. This problem defines the Calkin-Wilf-Newman sequence of positive rational numbers. The sequence is remarkable because every positive rational number appears as one of its terms and none appears more than once. Every real number 𝑥 can be written as an integer 𝐴 plus a number 𝐵 where 0 ≤ 𝐵 < 1. For example, for 𝑥 = 12∕5 = 2 + 2∕5 we have 𝐴 = 2 and 𝐵 = 2∕5. For 𝑥 = 3 = 3 + 0 we have 𝐴 = 3 and 𝐵 = 0. Define the function 𝑓 (𝑥) by

(a) Evaluate 𝑓 (𝑥) for 𝑥 = 25∕8, 13∕9, and 𝜋. (b) Find the first six terms of the recursively defined Calkin-Wilf-Newman sequence: 𝑠𝑛 = 1∕𝑓 (𝑠𝑛−1 ) for 𝑛 > 1 and 𝑠1 = 1.

𝑓 (𝑥) = 𝐴 + (1 − 𝐵).

Strengthen Your Understanding 73. If the sequence 𝑠𝑛 of positive terms is unbounded, then the sequence has an infinite number of terms greater than a million.

In Problems 65–67, explain what is wrong with the statement. 3𝑛 + 10 , which begins with the 7𝑛 + 3 13 16 19 22 , , , , … converges to 0 because the terms 10 17 24 31 terms of the sequence get smaller and smaller.

65. The sequence 𝑠𝑛 =

74. If a sequence 𝑠𝑛 is convergent, then the terms 𝑠𝑛 tend to zero as 𝑛 increases. 75. A monotone sequence cannot have both positive and negative terms.

66. A convergent sequence consists entirely of terms greater than 2, then the limit of the sequence is greater than 2.

76. If a monotone sequence of positive terms does not converge, then it has a term greater than a million.

67. If the convergent sequence has limit 𝐿 and 𝑠𝑛 < 2 for all 𝑛, then 𝐿 < 2.

77. If all terms 𝑠𝑛 of a sequence are less than a million, then the sequence is bounded.

In Problems 68–69, give an example of:

78. If a convergent sequence has 𝑠𝑛 ≤ 5 for all 𝑛, then lim 𝑠𝑛 ≤ 5. 𝑛→∞

68. An increasing sequence that converges to 0.

79. Which of the sequences I–IV is monotone and bounded for 𝑛 ≥ 1? 1 I. 𝑠𝑛 = 10 − 𝑛 10𝑛 + 1 II. 𝑠𝑛 = 𝑛 III. 𝑠𝑛 = cos 𝑛 IV. 𝑠𝑛 = ln 𝑛

69. A monotone sequence that does not converge. Decide if the statements in Problems 70–78 are true or false. Give an explanation for your answer. 70. You can tell if a sequence converges by looking at the first 1000 terms. 71. If the terms 𝑠𝑛 of a convergent sequence are all positive then lim 𝑠𝑛 is positive.

(a) (b) (c) (d)

𝑛→∞

72. If the sequence 𝑠𝑛 of positive terms is unbounded, then the sequence has a term greater than a million.

9.2

I I and II II and IV I, II, and III

GEOMETRIC SERIES Adding the terms of a sequence produces a series. For example, we have the sequence 1, 2, 3, 4, 5, 6, … and the series 1 + 2 + 3 + 4 + 5 + 6 + ⋯. This section introduces infinite series of constants, which are sums of the form 1+

1 2

+

1 3

+

1 4

+⋯

0.4 + 0.04 + 0.004 + 0.0004 + ⋯ . The individual numbers, 1, 21 , 13 , … , or 0.4, 0.04, … , etc., are called terms in the series. To talk about the sum of the series, we must first explain how to add infinitely many numbers. Let us look at the repeated administration of a drug. In this example, the terms in the series represent each dose; the sum of the series represents the drug level in the body in the long run.

9.2 GEOMETRIC SERIES

481

Repeated Drug Dosage A person with an ear infection is told to take antibiotic tablets regularly for several days. Since the drug is being excreted by the body between doses, how can we calculate the quantity of the drug remaining in the body at any particular time? To be specific, let’s suppose the drug is ampicillin (a common antibiotic) taken in 250 mg doses four times a day (that is, every six hours). It is known that at the end of six hours, about 4% of the drug is still in the body. What quantity of the drug is in the body right after the tenth tablet? The fortieth? Let 𝑄𝑛 represent the quantity, in milligrams, of ampicillin in the blood right after the 𝑛th tablet. Then 𝑄1 = 250 250(0.04) 𝑄2 = ⏟⏞⏞⏟⏞⏞⏟

+

= 250 mg = 260 mg

250 ⏟⏟⏟ New tablet

Remnants of first tablet

𝑄3 = 𝑄2 (0.04) + 250 = (250(0.04) + 250) (0.04) + 250 =

250(0.04)2 + 250(0.04) ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

+

Remnants of first and second tablets

= 260.4 mg

250 ⏟⏟⏟ New tablet

) 𝑄4 = 𝑄3 (0.04) + 250 = 250(0.04)2 + 250(0.04) + 250 (0.04) + 250 (

= 250(0.04)3 + 250(0.04)2 + 250(0.04) + 250 ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏟⏟ Remnants of first, second, and third tablets

= 260.416 mg.

New tablet

Looking at the pattern that is emerging, we guess that 𝑄10 = 250(0.04)9 + 250(0.04)8 + 250(0.04)7 + ⋯ + 250(0.04)2 + 250(0.04) + 250. Notice that there are 10 terms in this sum—one for every tablet—but that the highest power of 0.04 is the ninth, because no tablet has been in the body for more than 9 six-hour time periods. Now suppose we actually want to find the numerical value of 𝑄10 . It seems that we have to add 10 terms— fortunately, there’s a better way. Notice the remarkable fact that if you subtract (0.04)𝑄10 from 𝑄10 , all but two terms drop out. First, multiplying by 0.04, we get (0.04)𝑄10 = 250(0.04)10 + 250(0.04)9 + 250(0.04)8 + ⋯ + 250(0.04)3 + 250(0.04)2 + 250(0.04). Subtracting gives 𝑄10 − (0.04)𝑄10 = 250 − 250(0.04)10. Factoring 𝑄10 on the left and solving for 𝑄10 gives ) ( 𝑄10 (1 − 0.04) = 250 1 − (0.04)10 ) ( 250 1 − (0.04)10 . 𝑄10 = 1 − 0.04 This is called the closed-form expression for 𝑄10 . It is easy to evaluate on a calculator, giving 𝑄10 = 260.42 (to two decimal places). Similarly, 𝑄40 is given in closed form by ) ( 250 1 − (0.04)40 𝑄40 = . 1 − 0.04 Evaluating this on a calculator shows 𝑄40 = 260.42, which is the same (to two decimal places) as 𝑄10 . Thus after ten tablets, the value of 𝑄𝑛 appears to have stabilized at just over 260 mg. Looking at the closed forms for 𝑄10 and 𝑄40 , we can see that, in general, 𝑄𝑛 must be given by 𝑄𝑛 =

250 (1 − (0.04)𝑛) . 1 − 0.04

482

Chapter 9 SEQUENCES AND SERIES

What Happens as 𝒏 → ∞? What does this closed form for 𝑄𝑛 predict about the long-run level of ampicillin in the body? As 𝑛 → ∞, the quantity (0.04)𝑛 → 0. In the long run, assuming that 250 mg continue to be taken every six hours, the level right after a tablet is taken is given by 𝑄𝑛 =

250 (1 − (0.04)𝑛) 250(1 − 0) → = 260.42. 1 − 0.04 1 − 0.04

The Geometric Series in General In the previous example we encountered sums of the form 𝑎 + 𝑎𝑥 + 𝑎𝑥2 + ⋯ + 𝑎𝑥8 + 𝑎𝑥9 (with 𝑎 = 250 and 𝑥 = 0.04). Such a sum is called a finite geometric series. A geometric series is one in which each term is a constant multiple of the one before. The first term is 𝑎, and the constant multiplier, or common ratio of successive terms, is 𝑥. A finite geometric series has the form 𝑎 + 𝑎𝑥 + 𝑎𝑥2 + ⋯ + 𝑎𝑥𝑛−2 + 𝑎𝑥𝑛−1 . An infinite geometric series has the form 𝑎 + 𝑎𝑥 + 𝑎𝑥2 + ⋯ + 𝑎𝑥𝑛−2 + 𝑎𝑥𝑛−1 + 𝑎𝑥𝑛 + ⋯ . The “⋯” at the end of the second series tells us that the series is going on forever—in other words, that it is infinite.

Sum of a Finite Geometric Series The same procedure that enabled us to find the closed form for 𝑄10 can be used to find the sum of any finite geometric series. Suppose we write 𝑆𝑛 for the sum of the first 𝑛 terms, which means up to the term containing 𝑥𝑛−1 : 𝑆𝑛 = 𝑎 + 𝑎𝑥 + 𝑎𝑥2 + ⋯ + 𝑎𝑥𝑛−2 + 𝑎𝑥𝑛−1 . Multiply 𝑆𝑛 by 𝑥: 𝑥𝑆𝑛 = 𝑎𝑥 + 𝑎𝑥2 + 𝑎𝑥3 + ⋯ + 𝑎𝑥𝑛−1 + 𝑎𝑥𝑛 . Now subtract 𝑥𝑆𝑛 from 𝑆𝑛 , which cancels out all terms except for two, giving 𝑆𝑛 − 𝑥𝑆𝑛 = 𝑎 − 𝑎𝑥𝑛 (1 − 𝑥)𝑆𝑛 = 𝑎(1 − 𝑥𝑛 ). Provided 𝑥 ≠ 1, we can solve to find a closed form for 𝑆𝑛 as follows: The sum of a finite geometric series is given by 𝑆𝑛 = 𝑎 + 𝑎𝑥 + 𝑎𝑥2 + ⋯ + 𝑎𝑥𝑛−1 =

𝑎(1 − 𝑥𝑛 ) , 1−𝑥

provided 𝑥 ≠ 1.

Note that the value of 𝑛 in the formula for 𝑆𝑛 is the number of terms in the sum 𝑆𝑛 .

Sum of an Infinite Geometric Series In the ampicillin example, we found the sum 𝑄𝑛 and then let 𝑛 → ∞. We do the same here. The sum 𝑄𝑛 , which shows the effect of the first 𝑛 doses, is an example of a partial sum. The first three partial sums of the series 𝑎 + 𝑎𝑥 + 𝑎𝑥2 + ⋯ + 𝑎𝑥𝑛−1 + 𝑎𝑥𝑛 + ⋯ are 𝑆1 = 𝑎 𝑆2 = 𝑎 + 𝑎𝑥 𝑆3 = 𝑎 + 𝑎𝑥 + 𝑎𝑥2 .

9.2 GEOMETRIC SERIES

483

To find the sum of this infinite series, we consider the partial sum, 𝑆𝑛 , of the first 𝑛 terms. The formula for the sum of a finite geometric series gives 𝑆𝑛 = 𝑎 + 𝑎𝑥 + 𝑎𝑥2 + ⋯ + 𝑎𝑥𝑛−1 =

𝑎(1 − 𝑥𝑛 ) . 1−𝑥

What happens to 𝑆𝑛 as 𝑛 → ∞? It depends on the value of 𝑥. If |𝑥| < 1, then 𝑥𝑛 → 0 as 𝑛 → ∞, so 𝑎(1 − 𝑥𝑛 ) 𝑎(1 − 0) 𝑎 = = . 𝑛→∞ 1 − 𝑥 1−𝑥 1−𝑥

lim 𝑆𝑛 = lim

𝑛→∞

Thus, provided |𝑥| < 1, as 𝑛 → ∞ the partial sums 𝑆𝑛 approach a limit of 𝑎∕(1 − 𝑥). When this happens, we define the sum 𝑆 of the infinite geometric series to be that limit and say the series converges to 𝑎∕(1 − 𝑥).

For |𝑥| < 1, the sum of the infinite geometric series is given by 𝑆 = 𝑎 + 𝑎𝑥 + 𝑎𝑥2 + ⋯ + 𝑎𝑥𝑛−1 + 𝑎𝑥𝑛 + ⋯ =

𝑎 . 1−𝑥

If, on the other hand, |𝑥| > 1, then 𝑥𝑛 and the partial sums have no limit as 𝑛 → ∞ (if 𝑎 ≠ 0). In this case, we say the series diverges. If 𝑥 > 1, the terms in the series become larger and larger in magnitude, and the partial sums diverge to +∞ (if 𝑎 > 0) or −∞ (if 𝑎 < 0). When 𝑥 < −1, the terms become larger in magnitude, the partial sums oscillate as 𝑛 → ∞, and the series diverges. What happens when 𝑥 = 1? The series is 𝑎 +𝑎+𝑎+ 𝑎+⋯, and if 𝑎 ≠ 0, the partial sums grow without bound, and the series does not converge. When 𝑥 = −1, the series is 𝑎−𝑎+ 𝑎−𝑎+𝑎 −⋯, and, if 𝑎 ≠ 0, the partial sums oscillate between 𝑎 and 0, and the series does not converge. Example 1

For each of the following infinite geometric series, find several partial sums and the sum (if it exists). (a) 1 +

Solution

1 1 1 + + +⋯ 2 4 8

(b) 1 + 2 + 4 + 8 + ⋯

(c) 6 − 2 +

2 2 2 − + −⋯ 3 9 27

(a) This series may be written 1+

( )2 ( )3 1 1 1 + + +⋯ 2 2 2

which we can identify as a geometric series with 𝑎 = 1 and 𝑥 = 12 , so 𝑆 = check this by finding the partial sums: 𝑆1 = 1 1 1 3 = =2− 2 2 2 1 1 1 7 𝑆3 = 1 + + = = 2 − 2 4 4 4 1 1 1 1 15 =2− . 𝑆4 = 1 + + + = 2 4 8 8 8

𝑆2 = 1 +

1 = 2. Let’s 1 − (1∕2)

484

Chapter 9 SEQUENCES AND SERIES

The sequence of partial sums begins 1 1 1 1, 2 − , 2 − , 2 − , … . 2 4 8 The formula for 𝑆𝑛 gives 𝑆𝑛 =

1 − ( 12 )𝑛 1−

1 2

=2−

( )𝑛−1 1 . 2

Thus, the partial sums are creeping up to the value 𝑆 = 2, so 𝑆𝑛 → 2 as 𝑛 → ∞. (b) The partial sums of this geometric series (with 𝑎 = 1 and 𝑥 = 2) grow without bound, so the series has no sum: 𝑆1 = 1 𝑆2 = 1 + 2 = 3 𝑆3 = 1 + 2 + 4 = 7 𝑆4 = 1 + 2 + 4 + 8 = 15. The sequence of partial sums begins 1, 3, 7, 15, … . The formula for 𝑆𝑛 gives 𝑆𝑛 =

1 − 2𝑛 = 2𝑛 − 1. 1−2

(c) This is an infinite geometric series with 𝑎 = 6 and 𝑥 = − 13 . The partial sums, 𝑆1 = 6.00,

𝑆2 = 4.00,

𝑆3 ≈ 4.67,

𝑆4 ≈ 4.44,

𝑆5 ≈ 4.52,

𝑆6 ≈ 4.49,

appear to be converging to 4.5. This turns out to be correct because the sum is 𝑆=

6 = 4.5. 1 − (−1∕3)

Regular Deposits into a Savings Account People who save money often do so by putting some fixed amount aside regularly. To be specific, suppose $1000 is deposited every year in a savings account earning 5% a year, compounded annually. What is the balance, 𝐵𝑛 , in dollars, in the savings account right after the 𝑛th deposit? As before, let’s start by looking at the first few years: 𝐵1 = 1000 𝐵2 = 𝐵1 (1.05) + 1000 = 1000(1.05) + 1000 ⏟⏟⏟ ⏟⏞⏞⏞⏟⏞⏞⏞⏟ Original deposit

New deposit

2

𝐵3 = 𝐵2 (1.05) + 1000 = 1000(1.05) + 1000(1.05) + 1000 ⏟⏟⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ First two deposits

New deposit

Observing the pattern, we see 𝐵𝑛 = 1000(1.05)𝑛−1 + 1000(1.05)𝑛−2 + ⋯ + 1000(1.05) + 1000.

9.2 GEOMETRIC SERIES

485

So 𝐵𝑛 is a finite geometric series with 𝑎 = 1000 and 𝑥 = 1.05. Thus we have 𝐵𝑛 =

1000 (1 − (1.05)𝑛) . 1 − 1.05

We can rewrite this so that both the numerator and denominator of the fraction are positive: 𝐵𝑛 =

1000 ((1.05)𝑛 − 1) . 1.05 − 1

What Happens as 𝒏 → ∞? Common sense tells you that if you keep depositing $1000 in an account and it keeps earning interest, your balance grows without bound. This is what the formula for 𝐵𝑛 shows also: (1.05)𝑛 → ∞ as 𝑛 → ∞, so 𝐵𝑛 has no limit. (Alternatively, observe that the infinite geometric series of which 𝐵𝑛 is a partial sum has 𝑥 = 1.05, which is greater than 1, so the series does not converge.)

Exercises and Problems for Section 9.2 EXERCISES In Exercises 1–7, is a sequence or a series given? 1. 22 + 42 + 62 + 82 + ⋯ 2. 22 , 42 , 62 , 82 , …

3 1 1 1 1 − + − + +⋯ 2 2 6 18 54 1 2 4 5 + + + +⋯ 21. 2 3 5 6 20.

3. 1, −2, 3, −4, 5, … 4. 1 + 2, 3 + 4, 5 + 6, 7 + 8, …

22.

∞ ∑ (−1)𝑛 𝑛=2

5. 1 − 2 + 3 − 4 + 5 − ⋯

23.

22𝑛

∞ ∑ 1 √ 𝑛 𝑛=1

6. 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + ⋯ 7. −𝑆1 + 𝑆2 − 𝑆3 + 𝑆4 − 𝑆5 + ⋯ In Exercises 8–18, decide which of the following are geometric series. For those which are, give the first term and the ratio between successive terms. For those which are not, explain why not. 8. 5 − 10 + 20 − 40 + 80 − ⋯ 1 1 1 9. 2 + 1 + + + + ⋯ 2 4 8 1 1 1 1 10. 1 + + + + + ⋯ 2 3 4 5 11. 1 + 𝑥 + 2𝑥2 + 3𝑥3 + 4𝑥4 + ⋯ 1 1 1 1 +⋯ 12. 1 − + − + 2 4 8 16 2 3 13. 3 + 3𝑧 + 6𝑧 + 9𝑧 + 12𝑧4 + ⋯ 14. 1 + 2𝑧 + (2𝑧)2 + (2𝑧)3 + ⋯ 15. 1 − 𝑦2 + 𝑦4 − 𝑦6 + ⋯ 16. 1 − 𝑥 + 𝑥2 − 𝑥3 + 𝑥4 − ⋯ 17. 𝑧2 − 𝑧4 + 𝑧8 − 𝑧16 + ⋯ 18. 𝑦2 + 𝑦3 + 𝑦4 + 𝑦5 + ⋯ In Exercises 19–23, state whether or not the series is geometric. If it is geometric and converges, find the sum of the series. 19.

1 4 16 64 + + + +⋯ 3 9 27 81

In Exercises 24–27, say how many terms are in the finite geometric series and find its sum. 24. 2 + 2(0.1) + 2(0.1)2 + ⋯ + 2(0.1)25 25. 2(0.1) + 2(0.1)2 + ⋯ + 2(0.1)10 26. 2(0.1)5 + 2(0.1)6 + ⋯ + 2(0.1)13 1 1 27. 8 + 4 + 2 + 1 + + ⋯ + 10 2 2 In Exercises 28–30, find the sum of the infinite geometric series. 4 4 + +⋯ 3 9 29. −810 + 540 − 360 + 240 − 160 + ⋯ 40 20 80 30. 80 + √ + 40 + √ + 20 + √ + ⋯ 2 2 2 28. 36 + 12 + 4 +

In Exercises 31–34, find a closed-form for the geometric series and determine for which values of 𝑥 it converges. 31.

∞ ∑ (5𝑥)𝑛

32.

𝑛=0 ∞

33.

∑ (1 − 𝑥∕2) 2 𝑛=0

∞ ∑

(1 − 𝑥)𝑛

𝑛=0

𝑛

34.

∞ ∑ (1 − 𝑥∕3)𝑛 3 𝑛=0

486

Chapter 9 SEQUENCES AND SERIES

PROBLEMS In Problems 35–42, find the sum of the series. For what values of the variable does the series converge to this sum? 35. 1 + 𝑧∕2 + 𝑧2 ∕4 + 𝑧3 ∕8 + ⋯ 36. 1 + 3𝑥 + 9𝑥2 + 27𝑥3 + ⋯ 37. 𝑦 − 𝑦2 + 𝑦3 − 𝑦4 + ⋯ 38. 2 − 4𝑧 + 8𝑧2 − 16𝑧3 + ⋯ 39. 3 + 𝑥 + 𝑥2 + 𝑥3 + ⋯ 40. 4 + 𝑦 + 𝑦2 ∕3 + 𝑦3 ∕9 + ⋯ ( ) ( )2 ( )3 41. 8 + 8 𝑥2 − 5 + 8 𝑥2 − 5 + 8 𝑥2 − 5 + ⋯ ( ) ( )2 ( )3 42. 5 + 5 𝑧3 − 8 + 5 𝑧3 − 8 + 5 𝑧3 − 8 + ⋯ In Problems 43–45, for the given value of 𝑥 determine whether the infinite geometric series converges. If so, find its sum: 3 + 3 cos 𝑥 + 3 (cos 𝑥)2 + 3 (cos 𝑥)3 + ⋯ .

43. 𝑥 = 0

44. 𝑥 = 2𝜋∕3

45. 𝑥 = 𝜋

46. Bill invests $200 at the start of each month for 24 months, starting now. If the investment yields 0.5% per month, compounded monthly, what is its value at the end of 24 months? 47. Once a day, eight tons of pollutants are dumped into a bay. Of this, 25% is removed by natural processes each day. What happens to the quantity of pollutants in the bay over time? Give the long-run quantity right after a dump. 48. (a) The total reserves of a non-renewable resource are 400 million tons. Annual consumption, currently 25 million tons per year, is expected to rise by 1% each year. After how many years will the reserves be exhausted? (b) Instead of increasing by 1% per year, suppose consumption was decreasing by a constant percentage per year. If existing reserves are never to be exhausted, what annual percentage reduction in consumption is required? 49. A repeating decimal can always be expressed as a fraction. This problem shows how writing a repeating decimal as a geometric series enables you to find the fraction. (a) Write the repeating decimal 0.232323. . . as a geometric series using the fact that 0.232323 … = 0.23 + 0.0023 + 0.000023 + ⋯. (b) Use the formula for the sum of a geometric series to show that 0.232323 … = 23∕99. 4 Data

50. One way of valuing a company is to calculate the present value of all its future earnings. A farm expects to sell $1000 worth of Christmas trees once a year forever, with the first sale in the immediate future. What is the present value of this Christmas tree business? The interest rate is 1% per year, compounded continuously. 51. Around January 1, 1993, Barbra Streisand signed a contract with Sony Corporation for $2 million a year for 10 years. Suppose the first payment was made on the day of signing and that all other payments were made on the first day of the year. Suppose also that all payments were made into a bank account earning 4% a year, compounded annually. (a) How much money was in the account (i) On the night of December 31, 1999? (ii) On the day the last payment was made? (b) What was the present value of the contract on the day it was signed? 52. Peter wishes to create a retirement fund from which he can draw $20,000 when he retires and the same amount at each anniversary of his retirement for 10 years. He plans to retire 20 years from now. What investment need he make today if he can get a return of 5% per year, compounded annually? 53. In 2013, the quantity of copper mined worldwide was 17.9 million tonnes and had been increasing by 0.025% annually. By the end of that year, the world’s known copper reserves were 690 million tonnes.4 (a) Write and sum a series giving the total quantity of copper mined in the 𝑛 years since 2013. Assume the quantity mined each year continues to increase at 0.025% per year, with the first increase in 2014. (b) In what year are the reserves exhausted? (c) How does the sum constructed in part (a) relate to 𝑛 the integral ∫0 17.9(1.025)𝑡 𝑑𝑡? 54. This problem shows another way of deriving the longrun ampicillin level. (See page 481.) In the long run the ampicillin levels off to 𝑄 mg right after each tablet is taken. Six hours later, right before the next dose, there will be less ampicillin in the body. However, if stability has been reached, the amount of ampicillin that has been excreted is exactly 250 mg because taking one more tablet raises the level back to 𝑄 mg. Use this to solve for 𝑄. 55. On page 481, you saw how to compute the quantity 𝑄𝑛 mg of ampicillin in the body right after the 𝑛th tablet of 250 mg, taken once every six hours. (a) Do a similar calculation for 𝑃𝑛 , the quantity of ampicillin (in mg) in the body right before the 𝑛th tablet is taken.

from http://minerals.usgs.gov/minerals/pubs/commodity, accessed February 8, 2015.

9.2 GEOMETRIC SERIES

(b) Express 𝑃𝑛 in closed form. (c) What is lim𝑛→∞ 𝑃𝑛 ? Is this limit the same as lim𝑛→∞ 𝑄𝑛 ? Explain in practical terms why your answer makes sense. 56. Figure 9.3 shows the quantity of the drug atenolol in the blood as a function of time, with the first dose at time 𝑡 = 0. Atenolol is taken in 50 mg doses once a day to lower blood pressure. (a) If the half-life of atenolol in the blood is 6.3 hours, what percentage of the atenolol present at the start of a 24-hour period is still there at the end? (b) Find expressions for the quantities 𝑄0 , 𝑄1 , 𝑄2 , 𝑄3 , …, and 𝑄𝑛 shown in Figure 9.3. Write the expression for 𝑄𝑛 in closed form. (c) Find expressions for the quantities 𝑃1 , 𝑃2 , 𝑃3 , …, and 𝑃𝑛 shown in Figure 9.3. Write the expression for 𝑃𝑛 in closed form. 𝑞 (quantity, mg) 𝑄1

𝑄2

𝑄3

𝑄4

𝑄0

𝑃1 1

𝑃2 2

𝑃3 3

𝑃4 4

𝑡 (time, days) 5

Figure 9.3 57. Draw a graph like that in Figure 9.3 for 250 mg of ampicillin taken every 6 hours, starting at time 𝑡 = 0. Put on the graph the values of 𝑄1 , 𝑄2 , 𝑄3 , … introduced in the text on page 481 and the values of 𝑃1 , 𝑃2 , 𝑃3 , … calculated in Problem 55. 58. In theory, drugs that decay exponentially always leave a residue in the body. However, in practice, once the drug has been in the body for 5 half-lives, it is regarded as being eliminated.5 If a patient takes a tablet of the same drug every 5 half-lives forever, what is the upper limit to the amount of drug that can be in the body? 59. This problem shows how to estimate the cumulative effect of a tax cut on a country’s economy. Suppose the government proposes a tax cut totaling $100 million. We assume that all the people who have extra money spend 80% of it and save 20%. Thus, of the extra income generated by the tax cut, $100(0.8) million = $80 million is spent and becomes extra income to someone else. These people also spend 80% of their additional income, or $80(0.8) million, and so on. Calculate the total additional spending created by such a tax cut.

5 http://dr.pierce1.net/PDF/half_life.pdf,

487

60. The government proposes a tax cut of $100 million as in Problem 59, but that economists now predict that people will spend 90% of their extra income and save only 10%. How much additional spending would be generated by the tax cut under these assumptions? 61. (a) What is the present value of a $1000 bond which pays $50 a year for 10 years, starting one year from now? Assume the interest rate is 5% per year, compounded annually. (b) Since $50 is 5% of $1000, this bond is called a 5% bond. What does your answer to part (a) tell you about the relationship between the principal and the present value of this bond if the interest rate is 5%? (c) If the interest rate is more than 5% per year, compounded annually, which is larger: the principal or the present value of the bond? Why is the bond then described as trading at a discount? (d) If the interest rate is less than 5% per year, compounded annually, why is the bond described as trading at a premium? 62. A ball is dropped from a height of 10 feet and bounces. Each bounce is 43 of the height of the bounce before. Thus, after the ball hits the floor for the first time, the ball rises to a height of 10( 34 ) = 7.5 feet, and after it hits the floor for the second time, it rises to a height of 7.5( 43 ) = 10( 34 )2 = 5.625 feet. (Assume that there is no air resistance.) (a) Find an expression for the height to which the ball rises after it hits the floor for the 𝑛th time. (b) Find an expression for the total vertical distance the ball has traveled when it hits the floor for the first, second, third, and fourth times. (c) Find an expression for the total vertical distance the ball has traveled when it hits the floor for the 𝑛th time. Express your answer in closed form. 63. You might think that the ball in Problem 62 keeps bouncing forever since it takes infinitely many bounces. This is not true! (a) Show that a ball dropped √ from a height of ℎ feet 1 reaches the ground in 4 ℎ seconds. (Assume 𝑔 = 32 ft∕sec2 .) (b) Show that the ball in Problem 62 stops bouncing after ) √ ( 3 1√ 1√ 1 ≈ 11 seconds. 10+ 10 √ 4 2 4 1 − 3∕4

accessed on May 10, 2003.

488

Chapter 9 SEQUENCES AND SERIES

Strengthen Your Understanding In Problems 64–66, explain what is wrong with the statement. 1 1 4 64. The sequence 4, 1, , , … converges to = 4 16 1 − 1∕4 16 . 3 3 9 65. The sum of the infinite geometric series 1 − + − 2 4 27 1 2 + ⋯ is = . 8 1 + 3∕2 5 66. The following series is convergent: 0.000001 + 0.00001 + 0.0001 + 0.001 + ⋯ .

In Problems 67–72, give an example of: 67. A geometric series that does not converge. 68. A geometric series in which a term appears more than once.

9.3

69. A finite geometric series with four distinct terms whose sum is 10. 70. An infinite geometric series that converges to 10. 71. Two geometric series whose sum is geometric. 72. Two geometric series whose sum is not geometric. 73. Which of the following geometric series converge? (I) (II) (III) (IV) (a) (b) (c) (d) (e)

20 − 10 + 5 − 2.5 + ⋯ 1 − 1.1 + 1.21 − 1.331 + ⋯ 1 + 1.1 + 1.21 + 1.331 + ⋯ 1 + 𝑦2 + 𝑦4 + 𝑦6 + ⋯ , for − 1 < 𝑦 < 1 (I) only (IV) only (I) and (IV) (II) and (IV) None of the other choices is correct.

CONVERGENCE OF SERIES We now consider ∑ general series in which each term 𝑎𝑛 is a number. The series can be written compactly using a sign as follows: ∞ ∑

𝑎𝑛 = 𝑎1 + 𝑎2 + 𝑎3 + ⋯ + 𝑎 𝑛 + ⋯ .

𝑛=1

For any values of 𝑎 and 𝑥, the geometric series is such a series, with general term 𝑎𝑛 = 𝑎𝑥𝑛−1 .

Partial Sums and Convergence of Series As in Section 9.2, we define the partial sum, 𝑆𝑛 , of the first 𝑛 terms of a series as 𝑆𝑛 =

𝑛 ∑

𝑎𝑖 = 𝑎1 + 𝑎2 + ⋯ + 𝑎 𝑛 .

𝑖=1

To investigate the convergence of the series, we consider the sequence of partial sums 𝑆1 , 𝑆2 , 𝑆3 , … , 𝑆𝑛 , … . If 𝑆𝑛 has a limit as 𝑛 → ∞, then we define the sum of the series to be that limit.

If the sequence 𝑆𝑛 of partial sums converges to 𝑆, so lim 𝑆𝑛 = 𝑆, then we say the series 𝑛→∞

∞ ∑

𝑎𝑛 converges and that its sum is 𝑆. We write

𝑛=1

that the series diverges.

∞ ∑

𝑎𝑛 = 𝑆. If lim 𝑆𝑛 does not exist, we say 𝑛→∞

𝑛=1

The following example shows how a series leads to a sequence of partial sums and how we use them to determine convergence.

9.3 CONVERGENCE OF SERIES

Example 1

Investigate the convergence of the series with 𝑎𝑛 = 1∕(𝑛(𝑛 + 1)): ∞ ∑

𝑎𝑛 =

𝑛=1

Solution

489

1 1 1 1 + + + + ⋯. 2 6 12 20

In order to determine whether the series converges, we first find the partial sums: 1 2 1 1 2 𝑆2 = + = 2 6 3 1 3 1 1 = 𝑆3 = + + 2 6 12 4 1 1 4 1 1 + = 𝑆4 = + + 2 6 12 20 5 ⋮

𝑆1 =

It appears that 𝑆𝑛 = 𝑛∕(𝑛 + 1) for each positive integer 𝑛. We check that this pattern continues by assuming that 𝑆𝑛 = 𝑛∕(𝑛 + 1) for a given integer 𝑛, adding 𝑎𝑛+1 , and simplifying 𝑆𝑛+1 = 𝑆𝑛 + 𝑎𝑛+1 =

1 𝑛2 + 2𝑛 + 1 𝑛+1 𝑛 + = = . 𝑛 + 1 (𝑛 + 1)(𝑛 + 2) (𝑛 + 1)(𝑛 + 2) 𝑛 + 2

Thus the sequence of partial sums has formula 𝑆𝑛 = 𝑛∕(𝑛 + 1), which converges to 1, so the series ∑∞ 𝑛=1 𝑎𝑛 converges to 1. That is, we can say that 1 1 1 1 + + + + ⋯ = 1. 2 6 12 20

Visualizing Series We can visualize the terms of the series in Example 1 as the heights of the bars in Figure 9.4. The partial sums of the series are illustrated by stacking the bars on top of each other in Figure 9.5.

1 𝑆3 = 3∕4 𝑆2 = 2∕3 𝑆1 = 1∕2

𝑎 = 1∕2 𝑎 = 1∕6 𝑎 = 1∕12

⋯ 1 2 3 4 5 Figure 9.4: Terms of the series with 𝑎𝑛 = 1∕(𝑛(𝑛 + 1))

𝑆=1

𝑛

⋯ 𝑛 1 2 3 4 5 Figure 9.5: Partial sums of the series with 𝑎𝑛 = 1∕(𝑛(𝑛 + 1))

Here are some properties that are useful in determining whether or not a series converges.

490

Chapter 9 SEQUENCES AND SERIES

Theorem 9.2: Convergence Properties of Series 1. If

∞ ∑

𝑎𝑛 and

𝑛=1 ∞ ∑

• •

∞ ∑

𝑏𝑛 converge and if 𝑘 is a constant, then

𝑛=1

(𝑎𝑛 + 𝑏𝑛 ) converges to

𝑛=1 ∞ ∑

∞ ∑

𝑎𝑛 +

𝑛=1

𝑘𝑎𝑛 converges to 𝑘

𝑛=1

∞ ∑

∞ ∑

𝑏𝑛 .

𝑛=1

𝑎𝑛 .

𝑛=1

2. Changing a finite number of terms in a series does not change whether or not it converges, although it may change the value of its sum if it does converge. ∞ ∑ 𝑎𝑛 diverges. 3. If lim 𝑎𝑛 ≠ 0 or lim 𝑎𝑛 does not exist, then 𝑛→∞

𝑛→∞

𝑛=1

4. If

∞ ∑

𝑎𝑛 diverges, then

𝑛=1

∞ ∑

𝑘𝑎𝑛 diverges if 𝑘 ≠ 0.

𝑛=1

For proofs of these properties, see Problems 39–42. As with improper integrals, the convergence of a series is determined by its behavior for large 𝑛. (See the “behaves like” principle on page 401.) ∑ ∑∞ From Property 2 we see that, if 𝑁 is a positive integer, then ∞ 𝑛=1 𝑎𝑛 and 𝑛=𝑁 𝑎𝑛 either both converge or both diverge. Thus, if all we care about is the convergence of a series, we can omit the ∑ limits and write 𝑎𝑛 . ∑ (1 − 𝑒−𝑛 ) converge?

Example 2

Does the series

Solution

Since the terms in the series 𝑎𝑛 = 1−𝑒−𝑛 tend to 1, not 0, as 𝑛 → ∞, the series diverges by Property 3 of Theorem 9.2.

Comparison of Series and Integrals We investigate the convergence of some series by comparison with an improper integral. The harmonic series is the infinite series ∞ ∑ 1

𝑛 𝑛=1

=1+

1 1 1 1 + + + ⋯+ +⋯. 2 3 4 𝑛

Convergence of this sum would mean that the sequence of partial sums 𝑆1 = 1,

1 𝑆2 = 1 + , 2

𝑆3 = 1 +

1 1 + , 2 3

⋯,

𝑆𝑛 = 1 +

1 1 1 + +⋯+ , 2 3 𝑛



tends to a limit as 𝑛 → ∞. Let’s look at some values: 𝑆1 = 1,

𝑆10 ≈ 2.93,

𝑆100 ≈ 5.19,

𝑆1000 ≈ 7.49,

𝑆10000 ≈ 9.79.

The growth of these partial sums is slow, but they do in fact grow without bound, so the harmonic series diverges. This is justified in the following example and in Problem 60 (available online).

9.3 CONVERGENCE OF SERIES

491

Example 3

Show that the harmonic series 1 + 1∕2 + 1∕3 + 1∕4 + ⋯ diverges.

Solution

The idea is to approximate ∫1 (1∕𝑥) 𝑑𝑥 by a left-hand sum, where the terms 1, 1∕2, 1∕3, … are heights of rectangles of base 1. In Figure 9.6, the sum of the areas of the 3 rectangles is larger than the area under the curve between 𝑥 = 1 and 𝑥 = 4, and the same kind of relationship holds for the first 𝑛 rectangles. Thus, we have



𝑆𝑛 = 1 +

1 1 1 + +⋯+ > 2 3 𝑛 ∫1

𝑛+1

1 𝑑𝑥 = ln(𝑛 + 1). 𝑥

Since ln(𝑛 + 1) gets arbitrarily large as 𝑛 → ∞, so do the partial sums, 𝑆𝑛 . Thus, the partial sums have no limit, so the series diverges. 𝑦

✛ 𝑦 = 1∕𝑥 ✠

Area= 1 Area= 1∕2 Area= 1∕3



Rectangles showing

1+

1 2

+

1 3

4 1 𝑥

> ∫1

𝑑𝑥 = ln 4



𝑥 1

2

3

4 ∞

Figure 9.6: Comparing the harmonic series to ∫1 (1∕𝑥) 𝑑𝑥

Notice that the harmonic series diverges, even though lim 𝑎𝑛 = lim (1∕𝑛) = 0. Although 𝑛→∞ 𝑛→∞ ∑ ∑ Property 3 of Theorem 9.2 guarantees 𝑎𝑛 diverges if lim 𝑎𝑛 ≠ 0, it is possible for 𝑎𝑛 to either 𝑛→∞ converge or diverge if lim 𝑎𝑛 = 0. When we have lim 𝑎𝑛 = 0, we must investigate the series further 𝑛→∞ 𝑛→∞ to determine whether it converges or diverges. Example 4



By comparison with the improper integral ∫1 (1∕𝑥2 ) 𝑑𝑥, show that the following series converges: ∞ ∑ 1 1 1 = 1 + + +⋯. 2 4 9 𝑛 𝑛=1

Solution

Since we want to show that

∞ ∑

1∕𝑛2 converges, we want to show that the partial sums of this series

𝑛=1

tend to a limit. We do this by showing that the sequence of partial sums increases and is bounded above, so Theorem 9.1 applies. Each successive partial sum is obtained from the previous one by adding one more term in the series. Since all the terms are positive, the sequence of partial sums is increasing. ∞ ∑ To show that the partial sums of 1∕𝑛2 are bounded, we consider the right-hand sum repre𝑛=1

sented by the area of the rectangles in Figure 9.7. We start at 𝑥 = 1, since the area under the curve is infinite for 0 ≤ 𝑥 ≤ 1. The shaded rectangles in Figure 9.7 suggest that: 1 1 1 1 + + +⋯+ ≤ 4 9 16 𝑛2 ∫1



1 𝑑𝑥. 𝑥2

The area under the graph is finite, since ∞

∫1

) 𝑏 ( 1 1 1 + 1 = 1. − 𝑑𝑥 = lim 𝑑𝑥 = lim 𝑏→∞ ∫1 𝑥2 𝑏→∞ 𝑏 𝑥2

492

Chapter 9 SEQUENCES AND SERIES

To get 𝑆𝑛 , we add 1 to both sides, giving 𝑆𝑛 = 1 +

1 1 1 1 + + +⋯+ ≤1+ ∫1 4 9 16 𝑛2



1 𝑑𝑥 = 2. 𝑥2

Thus, the increasing sequence of partial sums is bounded above by 2. Hence, by Theorem 9.1 the sequence of partial sums converges, so the series converges. 𝑦 1

Shaded rectangles show

𝑦 = 1∕𝑥2

1 1 1 1 + + + < 4 9 16 25 ∫1



1 𝑑𝑥 𝑥2

Area = 1∕4

1∕4



1∕9



1∕16

Area = 1∕9 Area = 1∕16 ✠ ✠Area = 1∕25

𝑥

1

2

3

4

Figure 9.7: Comparing

⋯ ∑∞

𝑛=1



1∕𝑛2 to ∫1 (1∕𝑥2 ) 𝑑𝑥

Notice that we have shown that the series in the Example 4 converges, but we have not found its sum. The integral gives us a bound on the partial sums, but it does not give us the limit of the partial sums. Euler proved the remarkable fact that the sum is 𝜋 2 ∕6. The method of Examples 3 and 4 can be used to prove the following theorem. See Problem 59 (available online).

Theorem 9.3: The Integral Test Suppose 𝑎𝑛 = 𝑓 (𝑛), where 𝑓 (𝑥) is decreasing and positive. ∞ ∑ 𝑓 (𝑥) 𝑑𝑥 converges, then 𝑎𝑛 converges. • If ∫1 ∞ ∑ • If 𝑓 (𝑥) 𝑑𝑥 diverges, then 𝑎𝑛 diverges. ∫1 Suppose 𝑓 (𝑥) is continuous. Then if 𝑓 (𝑥) is positive and decreasing for all 𝑥 beyond some point, say 𝑐, the integral test can be used. The integral test allows us to analyze a family of series, the 𝑝-series, and see how convergence depends on the parameter 𝑝.

Example 5

For what values of 𝑝 does the series

∞ ∑

1∕𝑛𝑝 converge?

𝑛=1

Solution

If 𝑝 ≤ 0, the terms in the series 𝑎𝑛 = 1∕𝑛𝑝 do not tend to 0 as 𝑛 → ∞. Thus the series diverges for 𝑝 ≤ 0. ∞ ∑ ∞ If 𝑝 > 0, we compare 1∕𝑛𝑝 to the integral ∫1 1∕𝑥𝑝 𝑑𝑥. In Example 3 of Section 7.6 we saw 𝑛=1

that the integral converges if 𝑝 > 1 and diverges if 𝑝 ≤ 1. By the integral test, we conclude that ∑ 1∕𝑛𝑝 converges if 𝑝 > 1 and diverges if 𝑝 ≤ 1.

9.3 CONVERGENCE OF SERIES

493

We can summarize Example 5 as follows:

The 𝑝-series

∞ ∑

1∕𝑛𝑝 converges if 𝑝 > 1 and diverges if 𝑝 ≤ 1.

𝑛=1

Exercises and Problems for Section 9.3 Online Resource: Additional Problems for Section 9.3 EXERCISES

In Exercises 1–3, find the first five terms of the sequence of partial sums. 1.

∞ ∑

2.

𝑛

𝑛=1

∞ ∑ (−1)𝑛 𝑛 𝑛=1

3.

∞ ∑ 𝑛=1

1 𝑛(𝑛 + 1)

In Exercises 4–7, use the integral test to decide whether the series converges or diverges. ∞

4.

∑ 𝑛=1

1 (𝑛 + 2)2



5.

∑ 𝑛=1





∑ 1 6. 𝑒𝑛 𝑛=1

7.

∑ 𝑛=2



8. Use comparison with ∫1 𝑥−3 𝑑𝑥 to show that ∑∞ 1∕𝑛3 converges to a number less than or equal 𝑛=2 to 1∕2. ∞

9. Use comparison with ∫0 1∕(𝑥2 + 1) 𝑑𝑥 to show that ∑∞ 1∕(𝑛2 + 1) converges to a number less than or 𝑛=1 equal to 𝜋∕2. In Exercises 10–12, explain why the integral test cannot be used to decide if the series converges or diverges.

𝑛 𝑛2 + 1 10. 1 𝑛(ln 𝑛)2

∞ ∑

𝑛2

∞ ∑

1 𝑛(1 + ln 𝑛)

30.

1 𝑛2 + 2𝑛 + 2

32.

11.

𝑛=1

∞ ∑ (−1)𝑛 𝑛=1

12.

𝑛

∞ ∑

𝑒−𝑛 sin 𝑛

𝑛=1

PROBLEMS In Problems 13–32, does the series converge or diverge? 13.

∞ ∑ 𝑛=0

3 𝑛+2

14.

𝑛=1

∞ ∑

4 2𝑛 + 1

∞ ∑

2𝑛 1 + 𝑛4

𝑛=0

29.



31.

∑ 𝑛=0

15.

∞ ∑ 𝑛=0

17.

∞ ∑ 𝑛=0

19.

∞ ∑ 𝑛=1 ∞

21.

∑ 𝑛=0 ∞

23.

∑ 𝑛=1

2 √ 2+𝑛 2𝑛 (1 + 𝑛2 )2

16.

𝑛=0

18.

∞ ∑ 𝑛=0

3 (2𝑛 − 1)2

20.

3 𝑛2 + 4

22.

𝑛 𝑛+1

24.

∞ ∑ 𝑛=1

∞ ∑ 𝑛=0

4 (2𝑛 + 1)3

1 diverges. ln(2𝑛 )

34. Show that

∞ ∑

1 converges. (ln(2𝑛 ))2

2 1 + 4𝑛2

∞ ∑ 𝑛+1 2𝑛 +3 𝑛=0

25.

∞ (( )𝑛 ( )𝑛 ) ∑ 2 1 + 2 3 𝑛=1

26.

) ∞ (( )𝑛 ∑ 1 3 + 4 𝑛 𝑛=1

27.

∞ ∑ 𝑛 + 2𝑛 𝑛2𝑛 𝑛=1

28.

∞ ∑ ln 𝑛 𝑛 𝑛=1

𝑛=1

𝑛+1 𝑛2 + 2𝑛 + 2

∞ ∑ 𝑛 ln 𝑛 + 4 𝑛2 𝑛=2

∞ ∑ 𝑛=1

4 + 𝑛2

𝑛=3

33. Show that

2𝑛 √

∞ ∑

∞ ∑

) 𝑛+1 . 𝑛 𝑛=1 (b) Does the series in part (a) converge or diverge?

35. (a) Find the partial sum, 𝑆𝑛 , of

ln

(

36. (a) Show 𝑟ln 𝑛 = 𝑛ln 𝑟 for positive numbers 𝑛 and 𝑟. ∑∞ (b) For what values 𝑟 > 0 does 𝑛=1 𝑟ln 𝑛 converge? 37. Consider the series

∞ ∑ 𝑘=1

1 1 1 = + + ⋯. 𝑘(𝑘 + 1) 1⋅2 2⋅3

1 1 1 = . (a) Show that − 𝑘 𝑘+1 𝑘(𝑘 + 1) (b) Use part (a) to find the partial sums 𝑆3 , 𝑆10 , and 𝑆𝑛 . (c) Use part (b) to show that the sequence of partial sums 𝑆𝑛 , and therefore the series, converges to 1.

494

Chapter 9 SEQUENCES AND SERIES

38. Consider the series ∞ ∑ 𝑘=1

ln

(

𝑘(𝑘 + 2) (𝑘 + 1)2

)

= ln

(

) ( ) 2⋅4 1⋅3 + ln + ⋯. 2⋅2 3⋅3

(a) Show that the partial sum of the first three nonzero terms 𝑆3 = ln (5∕8). ) ( 𝑛+2 . (b) Show that the partial sum 𝑆𝑛 = ln 2(𝑛 + 1) (c) Use part (b) to show that the partial sums 𝑆𝑛 , and therefore the series, converge to ln (1∕2). ∑ ∑ 39. Show that if 𝑎𝑛 and 𝑏𝑛 converge and if 𝑘 is a con∑ ∑ ∑ stant, then (𝑎𝑛 + 𝑏𝑛 ), (𝑎𝑛 − 𝑏𝑛 ), and 𝑘𝑎𝑛 converge.

40. Let 𝑁 be a positive integer. Show that if 𝑎𝑛 = 𝑏𝑛 for ∑ ∑ 𝑛 ≥ 𝑁, then 𝑎𝑛 and 𝑏𝑛 either both converge, or both diverge. ∑ 41. Show that if 𝑎𝑛 converges, then lim 𝑎𝑛 = 0. [Hint: 𝑛→∞

Consider lim𝑛→∞ (𝑆𝑛 − 𝑆𝑛−1 ), where 𝑆𝑛 is the 𝑛th partial sum.] ∑ ∑ 42. Show that if 𝑎𝑛 diverges and 𝑘 ≠ 0, then 𝑘𝑎𝑛 diverges. ∑ 43. The series 𝑎𝑛 converges. Explain, by looking at par∑ tial sums, why the series (𝑎𝑛+1 − 𝑎𝑛 ) also converges. ∑ 44. The series 𝑎𝑛 diverges. Give examples that show the ∑ series (𝑎𝑛+1 − 𝑎𝑛 ) could converge or diverge.

Strengthen Your Understanding In Problems 45–48, explain what is wrong with the statement. ∑ 45. The series (1∕𝑛)2 converges because the terms approach zero as 𝑛 → ∞. ∑∞ ∞ 46. The integral ∫1 (1∕𝑥3 ) 𝑑𝑥 and the series 𝑛=1 1∕𝑛3 both converge to the same value, 1∕2. ∑∞ 47. The series 𝑛=1 𝑛𝑘 converges for 𝑘 > 1 and diverges for 𝑘 ≤ 1. 2 48. Since 𝑒−𝑥 has no elementary antiderivative, the series ∑∞ −𝑛2 𝑒 does not converge. 𝑛=1 In Problems 49–50, give an example of: ∑∞ 49. A series 𝑛=1 𝑎𝑛 with lim𝑛→∞ 𝑎𝑛 = 0, but such that ∑∞ 𝑎 diverges. 𝑛=1 𝑛 ∑∞ 50. A convergent series 𝑛=1 𝑎𝑛 , whose terms are all posi∑∞ √ tive, such that the series 𝑛=1 𝑎𝑛 is not convergent. Decide if the statements in Problems 51–58 are true or false. Give an explanation for your answer.

9.4

TESTS FOR CONVERGENCE

51.

∞ ∑ (1 + (−1)𝑛 ) is a series of nonnegative terms. 𝑛=1

52. If a series converges, then the sequence of partial sums of the series also converges. ∑ ∑ ∑ 53. If |𝑎𝑛 + 𝑏𝑛 | converges, then |𝑎𝑛 | and |𝑏𝑛 | converge. 54. The series

∞ ∑

𝑛

2(−1) converges.

𝑛=1



55. If a series 𝑎𝑛 converges, then the terms, 𝑎𝑛 , tend to zero as 𝑛 increases. 56. If the terms, 𝑎𝑛 , of a series tend to zero as 𝑛 increases, ∑ then the series 𝑎𝑛 converges. ∑ ∑ 57. If 𝑎𝑛 does not converge and 𝑏𝑛 does not converge, ∑ then 𝑎𝑛 𝑏𝑛 does not converge. ∑ ∑ ∑ 58. If 𝑎𝑛 𝑏𝑛 converges, then 𝑎𝑛 and 𝑏𝑛 converge.

Comparison of Series In Section 7.7, we compared two integrals to decide whether an improper integral converged. In Theorem 9.3 we compared an integral and a series. Now we compare two series.

Theorem 9.4: Comparison Test Suppose 0 ≤ 𝑎𝑛 ≤ 𝑏𝑛 for all 𝑛 beyond a certain value. ∑ ∑ • If 𝑏𝑛 converges, then 𝑎𝑛 converges. ∑ ∑ • If 𝑎𝑛 diverges, then 𝑏𝑛 diverges. Since 𝑎𝑛 ≤ 𝑏𝑛 , the plot of the 𝑎𝑛 terms lies ∑ under the plot of the 𝑏𝑛 terms. See ∑ Figure 9.8. The comparison test says that if the total area for 𝑏𝑛 is finite, then the total area for 𝑎𝑛 is finite also. ∑ ∑ If the total area for 𝑎𝑛 is not finite, then neither is the total area for 𝑏𝑛 .

9.4 TESTS FOR CONVERGENCE

495

𝑏1 𝑏3 𝑏2

𝑎3 𝑏4

𝑎1 𝑎2

𝑎4

𝑏5 𝑎5

1 2 3 4 5

𝑎𝑛 𝑏𝑛 ... ❘ ... 𝑛

Figure 9.8: Each 𝑎𝑛 is represented by the area of a dark rectangle, and each 𝑏𝑛 by a dark plus a light rectangle

Example 1

Use the comparison test to determine whether the series

∞ ∑ 𝑛=1

Solution

𝑛3

1 converges. +1

For 𝑛 ≥ 1, we know that 𝑛3 ≤ 𝑛3 + 1, so 0≤

1 1 ≤ . 𝑛3 + 1 𝑛3

∑ Thus, every term in the series ∞ 1∕(𝑛3 + 1) is less than or equal to the corresponding term 𝑛=1 ∑ ∑∞ ∞ 1∕𝑛3 . Since we saw that 𝑛=1 1∕𝑛3 converges as a 𝑝-series with 𝑝 > 1, we know that in ∑∞ 𝑛=1 3 𝑛=1 1∕(𝑛 + 1) converges.

Example 2 Solution

Decide whether the following series converge: (a)

∞ ∑ 𝑛−1 3 𝑛=1 𝑛 + 3

(b)

∞ ∑ 6𝑛2 + 1 𝑛=1

2𝑛3 − 1

.

(a) Since the convergence is determined by the behavior of the terms for large 𝑛, we observe that 𝑛−1 𝑛3 + 3

behaves like

𝑛 1 = 3 𝑛 𝑛2

as

𝑛 → ∞.

∑ ∑ Since 1∕𝑛2 converges, we guess that (𝑛 − 1)∕(𝑛3 + 3) converges. To confirm this, we use the comparison test. Since a fraction increases if its numerator is made larger or its denominator is made smaller, we have 0≤

𝑛−1 𝑛 1 ≤ 3 = 2 3 𝑛 +3 𝑛 𝑛

𝑛 ≥ 1.

for all

∑ ∑ Thus, the series (𝑛 − 1)∕(𝑛3 + 3) converges by comparison with 1∕𝑛2 . (b) First, we observe that 6𝑛2 + 1 2𝑛3 − 1

behaves like

6𝑛2 3 = 𝑛 2𝑛3

as

𝑛 → ∞.

∑ ∑ ∑ Since 1∕𝑛 diverges, so does 3∕𝑛, and we guess that (6𝑛2 + 1)∕(2𝑛3 − 1) diverges. To confirm this, we use the comparison test. Since a fraction decreases if its numerator is made smaller or its denominator is made larger, we have 0≤

6𝑛2 6𝑛2 + 1 , ≤ 3 3 2𝑛 2𝑛 − 1

496

Chapter 9 SEQUENCES AND SERIES

so 0≤ Thus, the series



3 6𝑛2 + 1 ≤ . 𝑛 2𝑛3 − 1

(6𝑛2 + 1)∕(2𝑛3 − 1) diverges by comparison with



3∕𝑛.

Limit Comparison Test The comparison test uses the relationship between the terms of two series, 𝑎𝑛 ≤ 𝑏𝑛 , which can be difficult to establish. However, the convergence or divergence of a series depends only on the longrun behavior of the terms as 𝑛 → ∞; this idea leads to the limit comparison test. Example 3

Predict the convergence or divergence of ∑

Solution

𝑛2 − 5 . +𝑛+2

𝑛3

As 𝑛 → ∞, the highest power terms in the numerator and denominator, 𝑛2 and 𝑛3 , dominate. Thus the term 𝑛2 − 5 𝑎𝑛 = 3 𝑛 +𝑛+2 behaves, as 𝑛 → ∞, like 𝑛2 1 = . 3 𝑛 𝑛 ∑ ∑ Since the harmonic series 1∕𝑛 diverges, we guess that 𝑎𝑛 also diverges. However, the inequality 𝑛2 − 5 1 ≤ 𝑛3 + 𝑛 + 2 𝑛

is in the wrong direction to use with the comparison test to confirm divergence, since we need the given series to be greater than a known divergent series. The following test can be used to confirm a prediction of convergence or divergence, as in Example 3, without inequalities.

Theorem 9.5: Limit Comparison Test Suppose 𝑎𝑛 > 0 and 𝑏𝑛 > 0 for all 𝑛. If lim

𝑛→∞

then the two series



𝑎𝑛 and



𝑎𝑛 =𝑐 𝑏𝑛

where 𝑐 > 0,

𝑏𝑛 either both converge or both diverge.

The limit lim𝑛→∞ 𝑎𝑛 ∕𝑏𝑛 = 𝑐 captures the idea that 𝑎𝑛 “behaves like” 𝑐𝑏𝑛 as 𝑛 → ∞. Example 4

Use the limit comparison test to determine if the following series converge or diverge. ( ) ∑ ∑ 𝑛2 − 5 1 (b) sin (a) 3 𝑛 𝑛 +𝑛+2

9.4 TESTS FOR CONVERGENCE

Solution

(a) We take 𝑎𝑛 = have

497

𝑛2 − 5 𝑛2 1 . Because 𝑎𝑛 behaves like 3 = as 𝑛 → ∞ we take 𝑏𝑛 = 1∕𝑛. We 𝑛 +𝑛+2 𝑛

𝑛3

𝑎𝑛 𝑛2 − 5 1 𝑛3 − 5𝑛 ⋅ = lim = 1. = lim 3 3 𝑛→∞ 𝑛→∞ 𝑏𝑛 1∕𝑛 𝑛 + 𝑛 + 2 𝑛 +𝑛+2 ∑ The limit comparison test applies with 𝑐 = 1. Since 1∕𝑛 diverges, the limit comparison test ∑ 𝑛2 − 5 also diverges. shows that 𝑛3 + 𝑛 + 2 (b) Since sin 𝑥 ≈ 𝑥 for 𝑥 near 0, we know that sin (1∕𝑛) behaves like 1∕𝑛 as 𝑛 → ∞. Since sin(1∕𝑛) is positive for 𝑛 ≥ 1, we can apply the limit comparison test with 𝑎𝑛 = sin (1∕𝑛) and 𝑏𝑛 = 1∕𝑛. We have 𝑎 sin (1∕𝑛) = 1. lim 𝑛 = lim 𝑛→∞ 𝑏𝑛 𝑛→∞ 1∕𝑛 ∑ ∑ Thus 𝑐 = 1 and since 1∕𝑛 diverges, the series sin (1∕𝑛) also diverges. lim

𝑛→∞

Series of Both Positive and Negative Terms ∑ If 𝑎𝑛 has both positive and negative terms, then its plot has rectangles lying both above and below ∑ the horizontal axis. See Figure 9.9. The total area of the rectangles is no longer equal to 𝑎𝑛 . However, it is still true that if the total area of the rectangles above and below the axis is finite, then the series converges. The area of the 𝑛th rectangle is |𝑎𝑛 |, so we have:

Theorem 9.6: Convergence of Absolute Values Implies Convergence If



|𝑎𝑛 | converges, then so does



𝑎𝑛 .

Problem 144 (available online) shows how to prove this result. Example 5

Explain how we know that the following series converges: ∞ ∑ (−1)𝑛−1 𝑛=1

Solution

𝑛2

=1−

1 1 + −⋯. 4 9

Writing 𝑎𝑛 = (−1)𝑛−1 ∕𝑛2 , we have 𝑛−1 | | |𝑎𝑛 | = || (−1) || = 1 . | | | 𝑛2 | 𝑛2 | | ∑ ∑ The 𝑝-series 1∕𝑛2 converges, since 𝑝 > 1, so (−1)𝑛−1 ∕𝑛2 converges.

𝑎1 𝑎3

𝑎𝑛 𝑎5 𝑎4

...

... 𝑛

𝑎2 Figure 9.9: Representing a series with positive and negative terms

498

Chapter 9 SEQUENCES AND SERIES

Comparison with a Geometric Series: The Ratio Test ∑ A geometric series 𝑎𝑛 has the property that the ratio 𝑎𝑛+1 ∕𝑎𝑛 is constant for all 𝑛. For many other series, this ratio, although not constant, tends to a constant as 𝑛 increases. In some ways, such series behave like geometric series. In particular, a geometric series converges if the ratio |𝑎𝑛+1 ∕𝑎𝑛 | < 1. A non-geometric series also converges if the ratio |𝑎𝑛+1 ∕𝑎𝑛 | tends to a limit which is less than 1. This idea leads to the following test.

Theorem 9.7: The Ratio Test For a series



𝑎𝑛 , suppose the sequence of ratios |𝑎𝑛+1 |∕|𝑎𝑛 | has a limit: lim

𝑛→∞

• If 𝐿 < 1, then



𝑎𝑛 converges.

|𝑎𝑛+1 | = 𝐿. |𝑎𝑛 |

∑ • If 𝐿 > 1, or if 𝐿 is infinite,6 then 𝑎𝑛 diverges. ∑ • If 𝐿 = 1, the test does not tell us anything about the convergence of 𝑎𝑛 . |𝑎𝑛+1 | = 𝐿 < 1. Let 𝑥 be a number between 𝐿 𝑛→∞ |𝑎𝑛 | and 1. Then for all sufficiently large 𝑛, say for all 𝑛 ≥ 𝑘, we have

Proof Here are the main steps in the proof. Suppose lim

Then,

|𝑎𝑛+1 | < 𝑥. |𝑎𝑛 |

|𝑎𝑘+1 | < |𝑎𝑘 |𝑥,

|𝑎𝑘+2 | < |𝑎𝑘+1 |𝑥 < |𝑎𝑘 |𝑥2 ,

|𝑎𝑘+3 | < |𝑎𝑘+2 |𝑥 < |𝑎𝑘 |𝑥3 ,

and so on. Thus, writing 𝑎 = |𝑎𝑘 |, we have for 𝑖 = 1, 2, 3, … , |𝑎𝑘+𝑖 | < 𝑎𝑥𝑖 .

∑ Now we can use the comparison test: |𝑎𝑘+𝑖 | converges by comparison with the geometric series ∑ ∑ ∑ 𝑖 𝑎𝑥 . Since |𝑎𝑘+𝑖 | converges, Theorem 9.6 tells us that 𝑎𝑘+𝑖 converges. So, by property 2 of ∑ Theorem 9.2, we see that 𝑎𝑛 converges too. If 𝐿 > 1, then for sufficiently large 𝑛, say 𝑛 ≥ 𝑚, |𝑎𝑛+1 | > |𝑎𝑛 |,

so the sequence |𝑎𝑚 |, |𝑎𝑚+1|, |𝑎𝑚+2 |, …, is increasing. Thus, lim 𝑎𝑛 ≠ 0, so 𝑛→∞



𝑎𝑛 diverges (by The-

orem 9.2, property 3). The argument in the case that |𝑎𝑛+1 |∕|𝑎𝑛 | is unbounded is similar. Example 6

Show that the following series converges: ∞ ∑ 1 1 1 =1+ + +⋯. 𝑛! 2! 3! 𝑛=1

Solution

Since 𝑎𝑛 = 1∕𝑛! and 𝑎𝑛+1 = 1∕ (𝑛 + 1)!, we have

6 That

|𝑎𝑛+1 | 1∕(𝑛 + 1)! 𝑛(𝑛 − 1)(𝑛 − 2) ⋯ 2 ⋅ 1 𝑛! | | = = . = |𝑎 | 1∕𝑛! (𝑛 + 1)𝑛(𝑛 − 1) ⋯ 2 ⋅ 1 + 1)! (𝑛 | 𝑛|

is, the sequence |𝑎𝑛+1 |∕|𝑎𝑛 | grows without bound.

9.4 TESTS FOR CONVERGENCE

499

We cancel 𝑛(𝑛 − 1)(𝑛 − 2) ⋅ ⋯ ⋅ 2 ⋅ 1, giving |𝑎𝑛+1 | 𝑛! 1 = lim = lim = 0. 𝑛→∞ |𝑎𝑛 | 𝑛→∞ (𝑛 + 1)! 𝑛→∞ 𝑛 + 1 lim

Because the limit is 0, which is less than 1, the ratio test tells us that

∞ ∑

1∕𝑛! converges. In Chapter 10,

𝑛=1

we see that the sum is 𝑒 − 1.

Example 7

What does the ratio test tell us about the convergence of the following two series? ∞ ∑ 1 𝑛=1

Solution

and

𝑛

∞ ∑ (−1)𝑛−1 𝑛=1

.

𝑛

Because ||(−1)𝑛 || = 1, in both cases we have lim𝑛→∞ ||𝑎𝑛+1 ∕𝑎𝑛 || = lim𝑛→∞ 𝑛∕(𝑛 + 1) = 1. The first series is the harmonic series, which diverges. However, Example 8 will show that the second series converges. Thus, if the ratio test gives a limit of 1, the ratio test does not tell us anything about the convergence of a series.

Alternating Series A series is called an alternating series if the terms alternate in sign. For example, ∞ ∑ (−1)𝑛−1 𝑛=1

𝑛

=1−

(−1)𝑛−1 1 1 1 + − +⋯+ + ⋯. 2 3 4 𝑛

The convergence of an alternating series can often be determined using the following test:

Theorem 9.8: Alternating Series Test An alternating series of the form ∞ ∑ (−1)𝑛−1𝑎𝑛 = 𝑎1 − 𝑎2 + 𝑎3 − 𝑎4 + ⋯ + (−1)𝑛−1𝑎𝑛 + ⋯ 𝑛=1

converges if 0 < 𝑎𝑛+1 < 𝑎𝑛

for all 𝑛

and

lim 𝑎 𝑛→∞ 𝑛

= 0.

Although we do not prove this result, we can see why it is reasonable. The first partial sum, 𝑆1 = 𝑎1 , is positive. The second, 𝑆2 = 𝑎1 − 𝑎2 , is still positive, since 𝑎2 < 𝑎1 , but 𝑆2 is smaller than 𝑆1 . (See Figure 9.10.) The next sum, 𝑆3 = 𝑎1 − 𝑎2 + 𝑎3 , is greater than 𝑆2 but smaller than 𝑆1 . The partial sums oscillate back and forth, and since the distance between them tends to 0, they eventually converge. ✛𝑎4 ✲ 𝑎3 ✛ 𝑎2 ✲ 𝑎1 𝑥 0

𝑆2

𝑆4

𝑆3

𝑆1

Figure 9.10: Partial sums, 𝑆1 , 𝑆2 , 𝑆3 , 𝑆4 of an alternating series

500

Chapter 9 SEQUENCES AND SERIES

Example 8

Show that the following alternating harmonic series converges: ∞ ∑ (−1)𝑛−1

Solution

.

𝑛

𝑛=1

We have 𝑎𝑛 = 1∕𝑛 and 𝑎𝑛+1 = 1∕(𝑛 + 1). Thus, 𝑎𝑛+1 =

1 1 < = 𝑎𝑛 𝑛+1 𝑛

for all 𝑛,

and

lim 1∕𝑛 = 0.

𝑛→∞

Thus, the hypothesis of Theorem 9.8 is satisfied, so the alternating harmonic series converges. Suppose 𝑆 is the sum of an alternating series, so 𝑆 = lim𝑛→∞ 𝑆𝑛 . Then 𝑆 is trapped between any two consecutive partial sums, say 𝑆3 and 𝑆4 or 𝑆4 and 𝑆5 , so 𝑆 2 < 𝑆4 < ⋯ < 𝑆 < ⋯ < 𝑆 3 < 𝑆1 . Thus, the error in using 𝑆𝑛 to approximate the true sum 𝑆 is less than the distance from 𝑆𝑛 to 𝑆𝑛+1 , which is 𝑎𝑛+1 . Stated symbolically, we have the following result:

Theorem 9.9: Error Bounds for Alternating Series 𝑛 ∑ Let 𝑆𝑛 = (−1)𝑖−1 𝑎𝑖 be the 𝑛th partial sum of an alternating series and let 𝑆 = lim 𝑆𝑛 . 𝑛→∞

𝑖=1

Suppose that 0 < 𝑎𝑛+1 < 𝑎𝑛 for all 𝑛 and lim𝑛→∞ 𝑎𝑛 = 0. Then |𝑆 − 𝑆 | < 𝑎 . 𝑛| 𝑛+1 |

Thus, provided 𝑆𝑛 converges to 𝑆 by the alternating series test, the error in using 𝑆𝑛 to approximate 𝑆 is less than the magnitude of the first term of the series which is omitted in the approximation.

Example 9

∞ ∑ Estimate the error in approximating the sum of the alternating harmonic series (−1)𝑛−1 ∕𝑛 by the 𝑛=1

sum of the first nine terms.

Solution

The ninth partial sum is given by 𝑆9 = 1 −

1 1 1 + − ⋯ + = 0.7456 … . 2 3 9

The first term omitted is −1∕10, with magnitude 0.1. By Theorem 9.9, we know that the true value of the sum differs from 0.7456 … by less than 0.1.

Absolute and Conditional Convergence ∑ We say that the series 𝑎𝑛 is ∑ ∑ • Absolutely convergent if 𝑎𝑛 and |𝑎𝑛 | both converge. ∑ ∑ • Conditionally convergent if 𝑎𝑛 converges but |𝑎𝑛 | diverges.

501

9.4 TESTS FOR CONVERGENCE

Conditionally convergent series rely on cancellation between positive and negative terms for their convergence. ∞ ∑ (−1)𝑛−1 Example: The series is absolutely convergent because the series converges and the 𝑛2 𝑛=1 ∑ 𝑝-series 1∕𝑛2 also converges. ∞ ∑ (−1)𝑛−1 is conditionally convergent because the series converges but Example: The series 𝑛 ∑ 𝑛=1 the harmonic series 1∕𝑛 diverges. Absolutely and conditionally convergent series behave differently when their terms are reordered: • Absolutely convergent series: Rearranging terms does not change the sum. • Conditionally convergent series: Rearranging terms can change the sum to any number.7 See Problem 112 on page 503 and Problems 136 and 137 (available online).

Exercises and Problems for Section 9.4 Online Resource: Additional Problems for Section 9.4 EXERCISES

In Exercises 1–4, use the comparison test to confirm the statement. 1.

∞ ∞ ∑ ∑ 1 1 diverges, so diverges. 𝑛 𝑛 − 3 𝑛=4 𝑛=4

∞ ∞ ∑ ∑ 1 𝑒−𝑛 3. converges, so converges. 2 𝑛 𝑛2 𝑛=1 𝑛=1

4.

𝑛=1

7

∞ ∑ 𝑛 𝑛 2 𝑛=1

16.

∞ ∑ 1 17. 𝑛 𝑛𝑒 𝑛=1

19.

∞ ∑ 𝑛!(𝑛 + 1)! (2𝑛)! 𝑛=1

∞ ∑ (𝑛!)2 18. (2𝑛)! 𝑛=1

20.

15.

∞ ∞ ∑ ∑ 1 1 converges. converges, so 2. 2 2 +2 𝑛 𝑛 𝑛=1 𝑛=1

∞ ( )𝑛 ∑ 3

In Exercises 15–21, use the ratio test to decide whether the series converges or diverges.

∞ ( )𝑛 ∑ 3𝑛 converges, so converges. 7𝑛 + 1 𝑛=1

In Exercises 5–8, use end behavior to compare the series to a 𝑝-series and predict whether the series converges or diverges.

21.

∞ ∑ 𝑛=0

∞ ∑ 𝑛=1

1 (2𝑛)!

∞ ∑ 1 ,𝑟 > 0 𝑟𝑛 𝑛! 𝑛=1

2𝑛 +1

𝑛3

In Exercises 22–32, use the limit comparison test to determine whether the series converges or diverges. ∞ ∞ ∑ ∑ 5𝑛 + 1 1 , by comparing to 2 𝑛 3𝑛 𝑛=1 𝑛=1 ∞ ( ∞ ( )𝑛 ) ∑ ∑ 1+𝑛 𝑛 1 , by comparing to 23. 3𝑛 3 𝑛=1 𝑛=1

22. 5.

∞ ∑ 𝑛=1 ∞

7.

∑ 𝑛=1

𝑛3 + 1 𝑛4 + 2𝑛3 + 2𝑛

6.

1 𝑛4 + 3𝑛3 + 7

8.

∞ ∑ 𝑛=1

∞ ∑ 𝑛=1

𝑛+4 𝑛3 + 5𝑛 − 3 𝑛−4 √

𝑛3 + 𝑛2 + 8

In Exercises 9–14, use the comparison test to determine whether the series converges. 9.

𝑛2 𝑛4 + 1

10.

∞ ∑

1 𝑛4 + 𝑒𝑛

∞ ∑ 1 12. ln 𝑛 𝑛=2

11.

𝑛=1

𝑛=1

∞ ∑ 𝑛 sin2 𝑛 13. 𝑛3 + 1 𝑛=1

1 3𝑛 + 1

∞ ∑ 2𝑛 + 1 14. 𝑛2𝑛 − 1 𝑛=1 7 See

25.

∞ ∑ 𝑛=1

∞ ∑ 𝑛=1

∞ ∑

[Hint: lim𝑛→∞ (1 + 1∕𝑛)𝑛 = 𝑒.] ∞ ( ) ∑ ∑ 1 1 1 − cos 24. , by comparing to 𝑛 𝑛2 𝑛=1 𝑛=1 ∞

27.

∞ ∑ 𝑛3 − 2𝑛2 + 𝑛 + 1 𝑛4 − 2 𝑛=1 ∞

29.

∑ 𝑛=1 ∞

31.

1 𝑛4 − 7

1 √ √ 2 𝑛+ 𝑛+2

∑ 4 sin 𝑛 + 𝑛 𝑛2 𝑛=1

26.

∞ ∑ 𝑛+1 𝑛2 + 2 𝑛=1

28.

∞ ∑ 𝑛=1

30.

∞ ( ∑ 𝑛=1

32.

2𝑛 3𝑛 − 1

∞ ∑ 𝑛=1

1 1 − 2𝑛 − 1 2𝑛

𝑛 cos 𝑛 + 𝑒𝑛

Walter Rudin, Principles of Mathematical Analysis, 3rd ed. (New York: McGraw-Hill, 1976), pp. 76–77.

)

502

Chapter 9 SEQUENCES AND SERIES

(c) Determine whether the following series converges or diverges:

Which of the series in Exercises 33–36 are alternating? ( ) ∞ ∞ ∑ ∑ 1 𝑛 33. 34. (−1) 2 − cos(𝑛𝜋) 𝑛 𝑛=1 𝑛=1 35.

∞ ∑ (−1)𝑛 cos(𝑛𝜋)

36.

𝑛=1

∞ ∑

(−1)𝑛 cos 𝑛

𝑛=1

In Exercises 37–42, use the alternating series test to decide whether the series converges. ∞ ∑ (−1)𝑛−1 37. √ 𝑛 𝑛=1

∞ ∑ (−1)𝑛−1 38. 2𝑛 + 1 𝑛=1

39.

40.

∞ ∑ (−1)𝑛 𝑛2 + 1 𝑛=1 ∞

41.

∑ 𝑛=1

∞ ∑ sin 𝑛 . 𝑛3 𝑛=1

∞ ∑ (−1)𝑛 𝑛! 𝑛=1

In Exercises 44–52, determine whether the series is absolutely convergent, conditionally convergent, or divergent. 44.

46.



(−1)𝑛−1 2 𝑛 + 2𝑛 + 1

42.

∑ (−1)𝑛−1 𝑒𝑛 𝑛=1

∑ sin 𝑛 . 𝑛3 𝑛=1 (b) Use the comparison test to determine whether the following series converges or diverges: ∞ ∑ | sin 𝑛 | | | | 𝑛3 | . | 𝑛=1 |

PROBLEMS

In Problems 53–54, explain why the comparison test cannot be used to decide if the series converges or diverges. 53.

∞ ∑ (−1)𝑛 𝑛=1

54.

𝑛2

∞ ∑

sin 𝑛

𝑛=1

In Problems 55–56, explain why the ratio test cannot be used to decide if the series converges or diverges. 55.

∞ ∑ (−1)𝑛

56.

𝑛=1

∞ ∑

sin 𝑛

48.

∑ (−1)𝑛 𝑛4 + 7 𝑛=1

50.

∞ ∑ cos 𝑛 𝑛2 𝑛=1

51.

∞ ( ) ∑ 1 (−1)𝑛−1 arcsin 𝑛 𝑛=1

52.

∞ ∑ (−1)𝑛−1 arctan(1∕𝑛) 𝑛2 𝑛=1



64.

∞ ∑ 8𝑛 𝑛! 𝑛=1

66.

∞ ∑ (−2)𝑛−1

59.

∑ 1 (−1)𝑛−1 2 − 𝑛 𝑛=1

)

∞ ∑

(−1)𝑛

𝑛=1

49.

𝑛 𝑛+1

∞ ∑ (−1)𝑛−1 𝑛 ln 𝑛 𝑛=2

65.

∞ ∑

67.

∞ ∑ (−1)𝑛−1

𝑛=1

𝑛2

𝑛=1 ∞

𝑛=1

1 4𝑛 + 3 2𝑛 + 1



𝑛

∑ 5+𝑒

69.

70.

∞ ∑ 𝑛2𝑛 3𝑛 𝑛=1

71.

∞ ∑ (0.1)𝑛 𝑛! 𝑛=0

72.

∞ ∑

73.

∞ ∑ 1 + 3𝑛

3𝑛

𝑛=1

𝑛=1

(

47.

In Problems 64–92, determine whether the series converges.

68.

𝑛=1

∞ ∑ (−1)𝑛 2𝑛 𝑛=1

62. 1 − 0.1 + 0.01 − 0.001 + ⋯ + (−1)𝑛 10−𝑛 + ⋯ 1 1 1 1 + − + ⋯ + (−1)𝑛 + ⋯ 63. 1 − 1! 2! 3! 𝑛!

𝑛=1

In Problems 57–60, explain why the alternating series test cannot be used to decide if the series converges or diverges. ∞ ∞ ∑ ∑ 57. (−1)𝑛−1 𝑛 58. (−1)𝑛−1 sin 𝑛

∞ ∑ (−1)𝑛−1 √ 𝑛 𝑛=1

45.



43. (a) Decide whether the following series is alternating: ∞

∞ ∑ (−1)𝑛 2𝑛 𝑛=1

𝑛=1

𝑛2 𝑛2 + 1

∑ 𝑛+2 𝑛2 − 1 𝑛=2

𝑛=1



4𝑛



2 1 2 1 2 1 60. − + − + − +⋯ 1 1 2 2 3 3

74.

∑ (𝑛 − 1)! 𝑛2 𝑛=1

75.

∑ (2𝑛)! (𝑛!)2 𝑛=1

In Problems 61–63, use a computer or calculator to investigate the behavior of the partial sums of the alternating series. Which appear to converge? Confirm convergence using the alternating series test. If a series converges, estimate its sum.

76.

∞ ∑

77.

∞ ∑

𝑒−𝑛

∞ ∑

5𝑛 + 2 2𝑛2 + 3𝑛 + 7

61. 1 − 2 + 3 − 4 + 5 + ⋯ + (−1)𝑛 (𝑛 + 1) + ⋯

∞ ∑ (−1)𝑛−1 80. √ 3𝑛 − 1 𝑛=1

𝑒𝑛

𝑛=1

78.

∞ ∑ 𝑛+1 3 +6 𝑛 𝑛=1

𝑛=0

79.

𝑛=1

∞ ∑ (−1)𝑛−1 2𝑛 81. 𝑛2 𝑛=1

9.4 TESTS FOR CONVERGENCE

82.

∞ ∑

84.

∞ ∑ 2𝑛3 − 1

𝑛=1

𝑛=1

86.

∞ ∑ 𝑛=1

1 √ 𝑛2 (𝑛 + 2)

83.

85.

𝑛3 + 1

𝑛=1

3𝑛2 − 2

∞ ∑ 2𝑛 + 1 87. √ 𝑛=1 3𝑛3 − 2

6 𝑛 + 2𝑛

∞ ∑ 3 88. ln 𝑛2 𝑛=2

∞ ∑ sin 𝑛 89. 𝑛2 𝑛=1

∞ ∑ sin 𝑛2 90. 𝑛2 𝑛=1

92.

∞ ∑ 𝑛(𝑛 + 1) √ 𝑛3 + 2𝑛2 𝑛=1 ∞ ∑ 𝑛+1

94. 𝑎𝑛 =

1 + 𝑒2𝑛



𝑛 96. 𝑎𝑛 =

1 + 𝑛2

1+

𝑛 √

𝑛

4 + 2𝑛 3𝑛

97. 𝑎𝑛 = 2 + (−1)𝑛

In Problems 98–101, the series converges. Is the sum affected by rearranging the terms of the series? ∞

98.

100.



∑ (−1)𝑛 𝑛 𝑛=1

99.

∞ ∑ 1 𝑛 2 𝑛=1

101.

(ln 𝑎)𝑛 , 𝑎 > 0

∞ ∑

(−1)𝑛 arctan

105.

𝑛=1

107.

𝑒𝑛

∞ ∑

𝑛=1

In Problems 93–97, use the expression for 𝑎𝑛 to decide: (a) If the sequence {𝑎𝑛 }∞ converges or diverges. 𝑛=1 ∑∞ (b) If the series 𝑛=1 𝑎𝑛 converges or diverges.

95. 𝑎𝑛 =

104.

∞ ∑ ln 𝑛 𝑛𝑎 𝑛=1

( ) 𝑎 ,𝑎 > 0 𝑛

The series in Problems 107–109 converge by the alternating series test. Use Theorem 9.9 to find how many terms give a partial sum, 𝑆𝑛 , within 0.01 of the sum, 𝑆, of the series.

∞ ( ) ∑ 1 1 tan 𝑛 𝑛2 𝑛=1

93. 𝑎𝑛 =

In Problems 102–106, for what values of 𝑎 does the series converge? ∞ ( )𝑎 ∞ ( )𝑛 ∑ ∑ 2 2 102. 103. ,𝑎 > 0 𝑛 𝑎 𝑛=1 𝑛=1

106.

∞ ∑ cos(𝑛𝜋) 91. 𝑛 𝑛=1

503

∑ (−1)𝑛 𝑛2 𝑛=1 ∞ ∑ (−1)𝑛+1 ln 𝑛 𝑛=2

∞ ∑ (−1)𝑛−1 𝑛 𝑛=1

108.

∞ ( ∞ ) ∑ ∑ (−1)𝑛−1 2 𝑛−1 109. − 3 (2𝑛)! 𝑛=1 𝑛=1

110. Suppose 0 ≤ 𝑏𝑛 ≤ 2𝑛 ≤ 𝑎𝑛 and 0 ≤ 𝑐𝑛 ≤ 2−𝑛 ≤ 𝑑𝑛 for ∑ ∑ ∑ ∑ all 𝑛. Which of the series 𝑎𝑛 , 𝑏𝑛 , 𝑐𝑛 , and 𝑑𝑛 definitely converge and which definitely diverge? ∑ ∑ 111. Given two convergent series 𝑎𝑛 and 𝑏𝑛 , we know ∑ that the term-by-term sum (𝑎𝑛 + 𝑏𝑛 ) converges. What about the series formed by taking the products of the ∑ terms 𝑎𝑛 ⋅ 𝑏𝑛 ? This problem shows that it may or may not converge. ∑ ∑ ∑ ∑ (a) Show that if 𝑎𝑛 = 1∕𝑛2 and 𝑏𝑛 = 1∕𝑛3 , ∑ then 𝑎𝑛 ⋅ 𝑏𝑛 converges. √ ∑ (b) Explain why (−1)𝑛 ∕ 𝑛 converges. √ ∑ (c) Show that if 𝑎𝑛 = 𝑏𝑛 = (−1)𝑛 ∕ 𝑛, then 𝑎𝑛 ⋅ 𝑏𝑛 does not converge. ∑ 112. Let 𝑎𝑛 be a series that converges to a number 𝐿 and whose terms are all positive. We show that any rear∑ rangement 𝑐𝑛 of this series also converges to 𝐿. ∑ (a) Show that 𝑆𝑛 < 𝐿 for any partial sum 𝑆𝑛 of 𝑐𝑛 . (b) Show that for any 𝐾 < 𝐿, there is a partial sum 𝑆𝑛 ∑ of 𝑐𝑛 with 𝑆𝑛 > 𝐾. ∑ (c) Use parts (a) and (b) to show 𝑐𝑛 converges to 𝐿.

Strengthen Your Understanding In Problems 113–115, explain what is wrong with the statement. 113. The series series test.

(−1) ∕𝑛 converges by the alternating

114. The series

∑∞

1∕(𝑛2 + 1) converges by the ratio test.

𝑛=1

115. The series ∑∞ 1∕𝑛2 . 𝑛=1

𝑛=1 ∑∞ 𝑛=1

2

∑∞

𝑛=1

𝑎𝑛 such that 𝑛→∞

|𝑎𝑛+1 | = 3. |𝑎𝑛 |

Decide if the statements in Problems 119–134 are true or false. Give an explanation for your answer.

1∕𝑛3∕2 converges by comparison with

In Problems 116–118, give an example of: 116. A series



lim

∑∞

2𝑛

118. A series

𝑎𝑛 that converges but

∑∞

𝑛=1

|𝑎𝑛 | diverges.

117. An alternating series that does not converge.

119. If the terms 𝑠𝑛 of a sequence alternate in sign, then the sequence converges. ∑ ∑ 120. If 0 ≤ 𝑎𝑛 ≤ 𝑏𝑛 for all 𝑛 and 𝑎𝑛 converges, then 𝑏𝑛 converges. ∑ ∑ 121. If 0 ≤ 𝑎𝑛 ≤ 𝑏𝑛 for all 𝑛 and 𝑎𝑛 diverges, then 𝑏𝑛 diverges.

504

Chapter 9 SEQUENCES AND SERIES

∑ ∑ 122. If 𝑏𝑛 ≤ 𝑎𝑛 ≤ 0 for all 𝑛 and 𝑏𝑛 converges, then 𝑎𝑛 converges. ∑ ∑ 123. If 𝑎𝑛 converges, then |𝑎𝑛 | converges. ∑ 124. If 𝑎𝑛 converges, then lim |𝑎𝑛+1 |∕|𝑎𝑛 | ≠ 1.

130. If

𝑛→∞

𝑛=0

∞ ∑ (−1)𝑛 2𝑛 converges. 𝑛=0

127. If



𝑎𝑛 converges, then

|𝑎𝑛 | converges, then

135. Which test will help you determine if the series converges or diverges?

∑ (−1)𝑛 𝑎𝑛 converges.

128. If an alternating series converges by the alternating series test, then the error in using the first 𝑛 terms of the series to approximate the entire series is less in magnitude than the first term omitted.

∞ ∑ 𝑘=1

1 𝑘3 + 1

(a) Integral test (b) Comparison test (c) Ratio test

129. If an alternating series converges, then the error in using the first 𝑛 terms of the series to approximate the entire series is less in magnitude than the first term omitted.

9.5

∑ (−1)𝑛 |𝑎𝑛 | converges.

131. To find the sum of the alternating harmonic series ∑ (−1)𝑛−1 ∕𝑛 to within 0.01 of the true value, we can sum the first 100 terms. ∑ 132. If 𝑎𝑛 is absolutely convergent, then it is convergent. ∑ 133. If 𝑎𝑛 is conditionally convergent, then it is absolutely convergent. ∑ ∑ 134. If 𝑎𝑛 > 0.5𝑏𝑛 > 0 for all 𝑛 and 𝑏𝑛 diverges, then 𝑎𝑛 diverges.

∞ ∑ (−1)𝑛 cos(2𝜋𝑛) is an alternating series. 125.

126. The series



POWER SERIES AND INTERVAL OF CONVERGENCE ∑ In Section 9.2 we saw that the geometric series 𝑎𝑥𝑛 converges for −1 < 𝑥 < 1 and diverges otherwise. This section studies the convergence of more general series constructed from powers. Chapter 10 shows how such power series are used to approximate functions such as 𝑒𝑥 , sin 𝑥, cos 𝑥, and ln 𝑥. A power series about 𝑥 = 𝑎 is a sum of constants times powers8 of (𝑥 − 𝑎): 2

𝑛

𝐶0 + 𝐶1 (𝑥 − 𝑎) + 𝐶2 (𝑥 − 𝑎) + ⋯ + 𝐶𝑛 (𝑥 − 𝑎) + ⋯ =

∞ ∑

𝐶𝑛 (𝑥 − 𝑎)𝑛 .

𝑛=0

∑ We think of 𝑎 as a constant. For any fixed 𝑥, the power series 𝐶𝑛 (𝑥−𝑎)𝑛 is a series of numbers like those considered in Section 9.3. To investigate the convergence of a power series, we consider the partial sums, which in this case are the polynomials 𝑆𝑛 (𝑥) = 𝐶0 + 𝐶1 (𝑥 − 𝑎) + 𝐶2 (𝑥 − 𝑎)2 + ⋯ + 𝐶𝑛 (𝑥 − 𝑎)𝑛 . As before, we consider the sequence9 𝑆0 (𝑥), 𝑆1 (𝑥), 𝑆2 (𝑥), … , 𝑆𝑛 (𝑥), … . For a fixed value of 𝑥, if this sequence of partial sums converges to a limit 𝑆, that is, if lim 𝑆𝑛 (𝑥) = 𝑆, then we say that the power series converges to 𝑆 for this value of 𝑥. 𝑛→∞

A power series may converge for some values of 𝑥 and not for others. Example 1

Find an expression for the general term of the series and use it to write the series using (𝑥 − 2)4 (𝑥 − 2)6 (𝑥 − 2)8 (𝑥 − 2)10 − + − +⋯. 4 9 16 25 8 For



notation:

𝑛 = 0, when 𝑥 = 𝑎 we define (𝑥 − 𝑎)0 to be 1. we call the first term in the sequence 𝑆0 (𝑥) rather than 𝑆1 (𝑥) so that the last term of 𝑆𝑛 (𝑥) is 𝐶𝑛 (𝑥 − 𝑎)𝑛 .

9 Here

9.5 POWER SERIES AND INTERVAL OF CONVERGENCE

Solution

The series is about 𝑥 = 2 and the odd terms are zero. We use (𝑥 − 2)2𝑛 and begin with 𝑛 = 2. Since the series alternates and is positive for 𝑛 = 2, we multiply by (−1)𝑛 . For 𝑛 = 2, we divide by 4, for 𝑛 = 3 we divide by 9, and in general, we divide by 𝑛2 . One way to write this series is ∞ ∑ (−1)𝑛(𝑥 − 2)2𝑛

𝑛2

𝑛=2

Example 2

Determine whether the power series

∞ ∑ 𝑥𝑛 𝑛=0

2𝑛

.

converges or diverges for

(a) 𝑥 = −1 Solution

505

(b) 𝑥 = 3

(a) Substituting 𝑥 = −1, we have ∞ ∑ 𝑥𝑛 𝑛=0

2𝑛

=

∞ ∑ (−1)𝑛 𝑛=0

2𝑛

=

∞ ( ) ∑ 1 𝑛 − . 2 𝑛=0

This is a geometric series with ratio −1∕2, so the series converges to 1∕(1 − (− 12 )) = 2∕3. (b) Substituting 𝑥 = 3, we have ∞ ∑ 𝑥𝑛 𝑛=0

2𝑛

=

∞ 𝑛 ∑ 3 𝑛=0

2𝑛

=

∞ ( )𝑛 ∑ 3 𝑛=0

2

.

This is a geometric series with ratio greater than 1, so it diverges.

Numerical and Graphical View of Convergence Consider the series (𝑥 − 1) −

(𝑥 − 1)𝑛 (𝑥 − 1)2 (𝑥 − 1)3 (𝑥 − 1)4 + − + ⋯ + (−1)𝑛−1 +⋯. 2 3 4 𝑛

To investigate the convergence of this series, we look at the sequence of partial sums graphed in Figure 9.11. The graph suggests that the partial sums converge for 𝑥 in the interval (0, 2). In Examples 4 and 5, we show that the series converges for 0 < 𝑥 ≤ 2. This is called the interval of convergence of this series. At 𝑥 = 1.4, which is inside the interval, the series appears to converge quite rapidly: 𝑆5 (1.4) = 0.33698 …

𝑆7 (1.4) = 0.33653 …

𝑆6 (1.4) = 0.33630 …

𝑆8 (1.4) = 0.33645 …

Table 9.1 shows the results of using 𝑥 = 1.9 and 𝑥 = 2.3 in the power series. For 𝑥 = 1.9, which is inside the interval of convergence but close to an endpoint, the series converges, though rather slowly. For 𝑥 = 2.3, which is outside the interval of convergence, the series diverges: the larger the value of 𝑛, the more wildly the series oscillates. In fact, the contribution of the twenty-fifth term is about 28; the contribution of the hundredth term is about −2,500,000,000. Figure 9.11 shows the interval of convergence and the partial sums.

506

Chapter 9 SEQUENCES AND SERIES



Interval of convergence



𝑆14 (𝑥) 𝑆 (𝑥) 5

2.3 1

1.9

Partial sums for series in Example 4 with 𝑥 = 1.9 inside interval of convergence and 𝑥 = 2.3 outside

Table 9.1

𝑛

𝑆𝑛 (1.9)

𝑛

𝑥 3

𝑆8 (𝑥)

𝑆𝑛 (2.3)

2

0.495

2

0.455

5

0.69207

5

1.21589

8

0.61802

8

0.28817

11

0.65473

11

1.71710

14

0.63440

14

−0.70701

𝑥 = 2 𝑆11 (𝑥) Figure 9.11: Partial sums for series in Example 4 converge for 0 < 𝑥 ≤ 2

Notice that the interval of convergence, 0 < 𝑥 ≤ 2, is centered on 𝑥 = 1. Since the interval extends one unit on either side, we say the radius of convergence of this series is 1.

Intervals of Convergence Each power series falls into one of three cases, characterized by its radius of convergence, 𝑅. This radius gives an interval of convergence.

• The series converges only for 𝑥 = 𝑎; the radius of convergence is defined to be 𝑅 = 0. • The series converges for all values of 𝑥; the radius of convergence is defined to be 𝑅 = ∞. • There is a positive number 𝑅, called the radius of convergence, such that the series converges for |𝑥 − 𝑎| < 𝑅 and diverges for |𝑥 − 𝑎| > 𝑅. See Figure 9.12. Using the radius of convergence, we make the following definition: • The interval of convergence is the interval between 𝑎 − 𝑅 and 𝑎 + 𝑅, including any endpoint where the series converges.

Series diverges

✛ ✛

𝑅

Interval of convergence

✲ ✲

Series diverges

𝑥 𝑎−𝑅

𝑎

𝑎+𝑅

Figure 9.12: Radius of convergence, 𝑅, determines an interval, centered at 𝑥 = 𝑎, in which the series converges

There are some series whose radius of convergence we already know. For example, the geometric series 1 + 𝑥 + 𝑥2 + ⋯ + 𝑥𝑛 + ⋯ converges for |𝑥| < 1 and diverges for |𝑥| ≥ 1, so its radius of convergence is 1. Similarly, the series 1+

( )2 ( )𝑛 𝑥 𝑥 𝑥 + +⋯+ +⋯ 3 3 3

converges for |𝑥∕3| < 1 and diverges for |𝑥∕3| ≥ 1, so its radius of convergence is 3.

9.5 POWER SERIES AND INTERVAL OF CONVERGENCE

507

The next theorem gives a method of computing the radius of convergence for many series. To ∞ ∑ 𝐶𝑛 (𝑥−𝑎)𝑛 converges, we use the ratio test. Writing find the values of 𝑥 for which the power series 𝑛=0

𝑎𝑛 = 𝐶𝑛 (𝑥 − 𝑎)𝑛 and assuming 𝐶𝑛 ≠ 0 and 𝑥 ≠ 𝑎, we have

|𝑎𝑛+1 | |𝐶𝑛+1 (𝑥 − 𝑎)𝑛+1 | |𝐶𝑛+1 ||𝑥 − 𝑎| |𝐶𝑛+1 | = lim = lim = |𝑥 − 𝑎| lim . 𝑛 𝑛→∞ |𝑎𝑛 | 𝑛→∞ 𝑛→∞ 𝑛→∞ |𝐶𝑛 | |𝐶𝑛 (𝑥 − 𝑎) | |𝐶𝑛 | lim

Case 1. Suppose lim |𝑎𝑛+1 |∕|𝑎𝑛 | is infinite. Then the ratio test shows that the power series converges 𝑛→∞

only for 𝑥 = 𝑎. The radius of convergence is 𝑅 = 0. Case 2. Suppose lim |𝑎𝑛+1 |∕|𝑎𝑛 | = 0. Then the ratio test shows that the power series converges for 𝑛→∞ all 𝑥. The radius of convergence is 𝑅 = ∞. Case 3. Suppose lim |𝑎𝑛+1 |∕|𝑎𝑛 | = 𝐾|𝑥 − 𝑎|, where lim |𝐶𝑛+1 |∕|𝐶𝑛 | = 𝐾. In Case 1, 𝐾 does not 𝑛→∞

𝑛→∞

exist; in Case 2, 𝐾 = 0. Thus, we can assume 𝐾 exists and 𝐾 ≠ 0, and we can define 𝑅 = 1∕𝐾. Then we have |𝑎𝑛+1 | |𝑥 − 𝑎| = 𝐾|𝑥 − 𝑎| = , lim 𝑛→∞ |𝑎𝑛 | 𝑅 so the ratio test tells us that the power series: |𝑥 − 𝑎| • Converges for < 1; that is, for |𝑥 − 𝑎| < 𝑅 𝑅 |𝑥 − 𝑎| • Diverges for > 1; that is, for |𝑥 − 𝑎| > 𝑅. 𝑅 The results are summarized in the following theorem.

Theorem 9.10: Method for Computing Radius of Convergence To calculate the radius of convergence, 𝑅, for the power series

∞ ∑

𝐶𝑛 (𝑥 − 𝑎)𝑛 , use the ratio

𝑛=0

test with 𝑎𝑛 = 𝐶𝑛 (𝑥 − 𝑎)𝑛 . • If lim |𝑎𝑛+1 |∕|𝑎𝑛 | is infinite, then 𝑅 = 0. 𝑛→∞

• If lim |𝑎𝑛+1 |∕|𝑎𝑛 | = 0, then 𝑅 = ∞. 𝑛→∞

• If lim |𝑎𝑛+1 |∕|𝑎𝑛 | = 𝐾|𝑥 − 𝑎|, where 𝐾 is finite and nonzero, then 𝑅 = 1∕𝐾. 𝑛→∞

Note that the ratio test does not tell us anything if lim𝑛→∞ |𝑎𝑛+1 |∕|𝑎𝑛 | fails to exist, which can occur, for example, if some of the 𝐶𝑛 s are zero. A proof that a power series has a radius of convergence and of Theorem 9.10 is given in the online theory supplement. To understand these facts informally, we can think of a power series as being like a geometric series whose coefficients vary from term to term. The radius of convergence depends on the behavior of the coefficients: if there are constants 𝐶 and 𝐾 such that for larger and larger 𝑛, |𝐶𝑛 | ≈ 𝐶𝐾 𝑛 , ∑ ∑ ∑ then it is plausible that 𝐶𝑛 𝑥𝑛 and 𝐶𝐾 𝑛 𝑥𝑛 = 𝐶(𝐾𝑥)𝑛 converge or diverge together. The ∑ 𝑛 geometric series 𝐶(𝐾𝑥) converges for |𝐾𝑥| < 1, that is, for |𝑥| < 1∕𝐾. We can find 𝐾 using the ratio test, because |𝑎𝑛+1 | |𝐶𝑛+1 ||(𝑥 − 𝑎)𝑛+1 | 𝐶𝐾 𝑛+1 |(𝑥 − 𝑎)𝑛+1 | = ≈ = 𝐾|𝑥 − 𝑎|. |𝑎𝑛 | |𝐶𝑛 ||(𝑥 − 𝑎)𝑛 | 𝐶𝐾 𝑛 |(𝑥 − 𝑎)𝑛 |

508

Chapter 9 SEQUENCES AND SERIES

Example 3

Show that the following power series converges for all 𝑥: 1+𝑥+

Solution

𝑥 2 𝑥3 𝑥𝑛 + +⋯+ +⋯. 2! 3! 𝑛!

Because 𝐶𝑛 = 1∕𝑛!, none of the 𝐶𝑛 s are zero and we can use the ratio test: |𝑎𝑛+1 | |𝐶𝑛+1 | 1∕(𝑛 + 1)! 𝑛! 1 = |𝑥| lim = |𝑥| lim = |𝑥| lim = |𝑥| lim = 0. 𝑛→∞ |𝑎𝑛 | 𝑛→∞ |𝐶𝑛 | 𝑛→∞ 𝑛→∞ (𝑛 + 1)! 𝑛→∞ 𝑛 + 1 1∕𝑛! lim

This gives 𝑅 = ∞, so the series converges for all 𝑥. We see in Chapter 10 that it converges to 𝑒𝑥 .

Example 4

Determine the radius of convergence of the series (𝑥 − 1) (𝑥 − 1)2 (𝑥 − 1)3 (𝑥 − 1)4 (𝑥 − 1)𝑛 − + − + ⋯ + (−1)𝑛−1 +⋯. 3 𝑛 ⋅ 3𝑛 2 ⋅ 32 3 ⋅ 33 4 ⋅ 34 What does this tell us about the interval of convergence of this series?

Solution

Because 𝐶𝑛 = (−1)𝑛−1 ∕(𝑛 ⋅ 3𝑛 ) is never zero we can use the ratio test. We have (−1)𝑛

| (𝑛+1)⋅3𝑛+1 | |𝑎𝑛+1 | |𝐶𝑛+1 | |𝑥 − 1| 𝑛 = |𝑥 − 1| lim = |𝑥 − 1| lim = . lim = |𝑥 − 1| lim 𝑛−1 𝑛→∞ |𝑎𝑛 | 𝑛→∞ 3(𝑛 + 1) 𝑛→∞ |𝐶𝑛 | 𝑛→∞ 3 | (−1) | 𝑛⋅3𝑛

Thus, 𝐾 = 1∕3 in Theorem 9.10, so the radius of convergence is 𝑅 = 1∕𝐾 = 3. The power series converges for |𝑥 − 1| < 3 and diverges for |𝑥 − 1| > 3, so the series converges for −2 < 𝑥 < 4. Notice that the radius of convergence does not tell us what happens at the endpoints, 𝑥 = −2 and 𝑥 = 4. The endpoints are investigated in Example 5.

What Happens at the Endpoints of the Interval of Convergence? The ratio test does not tell us whether a power series converges at the endpoints of its interval of convergence, 𝑥 = 𝑎 ± 𝑅. There is no simple theorem that answers this question. Since substituting 𝑥 = 𝑎 ± 𝑅 converts the power series to a series of numbers, the tests in Sections 9.3 and 9.4 are often useful. Example 5

Determine the interval of convergence of the series 𝑛 (𝑥 − 1) (𝑥 − 1)2 (𝑥 − 1)3 (𝑥 − 1)4 𝑛−1 (𝑥 − 1) − + − + ⋯ + (−1) +⋯. 3 𝑛 ⋅ 3𝑛 2 ⋅ 32 3 ⋅ 33 4 ⋅ 34

Solution

In Example 4 we showed that this series has 𝑅 = 3; it converges for −2 < 𝑥 < 4 and diverges for 𝑥 < −2 or 𝑥 > 4. We need to determine whether it converges at the endpoints of the interval of convergence, 𝑥 = −2 and 𝑥 = 4. At 𝑥 = 4, we have the series 1−

(−1)𝑛−1 1 1 1 + − +⋯+ +⋯. 2 3 4 𝑛

This is an alternating series with 𝑎𝑛 = 1∕𝑛, so by the alternating series test (Theorem 9.8), it con-

9.5 POWER SERIES AND INTERVAL OF CONVERGENCE

509

verges. At 𝑥 = −2, we have the series −1 −

1 1 1 1 − − −⋯ − −⋯. 2 3 4 𝑛

This is the negative of the harmonic series, so it diverges. Therefore, the interval of convergence is −2 < 𝑥 ≤ 4. The right endpoint is included and the left endpoint is not.

Series with All Odd, or All Even, Terms The ratio test requires lim |𝑎𝑛+1 |∕|𝑎𝑛 | to exist for 𝑎𝑛 = 𝐶𝑛 (𝑥 −𝑎)𝑛. What happens if the power series 𝑛→∞ has only even or odd powers, so some of the coefficients 𝐶𝑛 are zero? Then we use the fact that an infinite series can be written in several ways and pick one in which the terms are nonzero. Example 6

Find the radius and interval of convergence of the series 1 + 2 2 𝑥2 + 2 4 𝑥4 + 2 6 𝑥6 + ⋯ .

Solution

If we take 𝑎𝑛 = 2𝑛 𝑥𝑛 for 𝑛 even and 𝑎𝑛 = 0 for 𝑛 odd, lim |𝑎𝑛+1 |∕|𝑎𝑛 | does not exist. Therefore, for 𝑛→∞ this series we take 𝑎𝑛 = 22𝑛 𝑥2𝑛 , so that, replacing 𝑛 by 𝑛 + 1, we have 𝑎𝑛+1 = 22(𝑛+1) 𝑥2(𝑛+1) = 22𝑛+2 𝑥2𝑛+2 . Thus,

We have

| | |𝑎𝑛+1 | |22𝑛+2 𝑥2𝑛+2 | | | = |22 𝑥2 | = 4𝑥2 . |=| | | |𝑎 | |22𝑛 𝑥2𝑛 | | | | 𝑛| | | |𝑎𝑛+1 | | = 4𝑥2 . lim | 𝑛→∞ |𝑎 | | 𝑛|

The ratio test guarantees that the power series converges if 4𝑥2 < 1, that is, if |𝑥| < 12 . The radius

of convergence is 21 . The series converges for − 12 < 𝑥


1 2

or 𝑥 < − 21 . At

𝑥 = ± 12 , all the terms in the series are 1, so the series diverges (by Theorem 9.2, Property 3). Thus,

the interval of convergence is − 12 < 𝑥 < 12 .

Example 7

Write the general term 𝑎𝑛 of the following series so that none of the terms are zero: 𝑥−

Solution

𝑥3 𝑥5 𝑥7 𝑥9 + − + −⋯. 3! 5! 7! 9!

This series has only odd powers. We can get odd integers using 2𝑛 − 1 for 𝑛 ≥ 1, since 2 ⋅ 1 − 1 = 1,

2 ⋅ 2 − 1 = 3,

2 ⋅ 3 − 1 = 5, etc.

Also, the signs of the terms alternate, with the first (that is, 𝑛 = 1) term positive, so we include a

510

Chapter 9 SEQUENCES AND SERIES

factor of (−1)𝑛−1 . Thus we get 𝑎𝑛 = (−1)𝑛−1

𝑥2𝑛−1 . (2𝑛 − 1)!

We see in Chapter 10 that the series converges to sin 𝑥. Exercise 24 shows that the radius of convergence of this series is infinite, so that it converges for all values of 𝑥.

Exercises and Problems for Section 9.5 Online Resource: Additional Problems for Section 9.5 EXERCISES

Which of the series in Exercises 1–4 are power series? 1. 𝑥 − 𝑥3 + 𝑥6 − 𝑥10 + 𝑥15 − ⋯ 1 1 1 1 + + + +⋯ 2. 𝑥 𝑥2 𝑥3 𝑥4 3. 1 + 𝑥 + (𝑥 − 1)2 + (𝑥 − 2)3 + (𝑥 − 3)4 + ⋯ 4. 𝑥 + 𝑥 + 2

7. 8. 9. 10.

∞ ∑

𝑛𝑥𝑛

∞ ∑

𝑛3 𝑥𝑛

12.

𝑛=0

1 1⋅3 2 1⋅3⋅5 3 𝑥+ 2 𝑥 + 3 𝑥 +⋯ 2 2 ⋅ 2! 2 ⋅ 3! 𝑝(𝑝 − 1) 2 𝑝(𝑝 − 1)(𝑝 − 2) 3 𝑥 + 𝑥 +⋯ 𝑝𝑥 + 2! 3! (𝑥 − 1)2 (𝑥 − 1)4 (𝑥 − 1)6 1− + − +⋯ 2! 4! 6! (𝑥 − 1)5 (𝑥 − 1)7 (𝑥 − 1)9 + − +⋯ (𝑥 − 1)3 − 2! 4! 6! 𝑥 − 𝑎 (𝑥 − 𝑎)2 (𝑥 − 𝑎)3 (𝑥 − 𝑎)4 + + + +⋯ 1 2 ⋅ 2! 4 ⋅ 3! 8 ⋅ 4! 7 5(𝑥 + 5)9 4(𝑥 + 5) + +⋯ 2(𝑥 + 5)3 + 3(𝑥 + 5)5 + 2! 3!

∞ ∑

(5𝑥)𝑛

∞ ∑

(2𝑛 + 𝑛2 )𝑥𝑛

𝑛=0

14.

𝑛=0

In Exercises 5–10, find an expression for the general term of the series. Give the starting value of the index (𝑛 or 𝑘, for example).

6.

11.

13.

7

5.

In Exercises 11–23, find the radius of convergence.

𝑛=0

15.

∞ ∑ (𝑛 + 1)𝑥𝑛 2𝑛 + 𝑛 𝑛=0

16.

∞ ∑ 2𝑛 (𝑥 − 1)𝑛 𝑛 𝑛=1

17.

∞ ∑ (𝑥 − 3)𝑛 𝑛2𝑛 𝑛=1

18.

∞ ∑

(−1)𝑛

𝑛=0

𝑥2𝑛 (2𝑛)!

𝑥5 𝑥2 𝑥3 𝑥4 + − + −⋯ 4 9 16 25 4𝑥2 8𝑥3 16𝑥4 32𝑥5 1 + 2𝑥 + + + + +⋯ 2! 3! 4! 5! 6!𝑥3 8!𝑥4 10!𝑥5 4!𝑥2 + + + +⋯ 1 + 2𝑥 + 2 2 2 (2!) (3!) (4!) (5!)2 5 7 9 11 3𝑥 + 𝑥2 + 𝑥3 + 𝑥4 + 𝑥5 + ⋯ 2 3 4 5 𝑥3 𝑥5 𝑥7 𝑥− + − +⋯ 3 5 7 Show that the radius of convergence of the power series 𝑥3 𝑥5 𝑥7 𝑥− + − + ⋯ in Example 7 is infinite. 3! 5! 7!

19. 𝑥 − 20. 21. 22. 23. 24.

PROBLEMS 25. (a) Determine the radius of convergence of the series 𝑥−

𝑥𝑛 𝑥2 𝑥3 𝑥4 + − + ⋯ + (−1)𝑛−1 + ⋯ . 2 3 4 𝑛

What does this tell us about the interval of convergence of this series? (b) Investigate convergence at the end points of the interval of convergence of this series. ∞ ∑ (2𝑥)𝑛 converges for |𝑥| < 1∕2. 𝑛 𝑛=1 Investigate whether the series converges for 𝑥 = 1∕2 and 𝑥 = −1∕2.

26. Show that the series

In Problems 27–34, find the interval of convergence. 27.

∞ ∑ 𝑥𝑛 3𝑛 𝑛=0

28.

∞ ∑ (𝑥 − 3)𝑛 𝑛 𝑛=2

29.

∞ ∑ 𝑛2 𝑥2𝑛 22𝑛 𝑛=1

30.

∞ ∑ (−1)𝑛 (𝑥 − 5)𝑛 2𝑛 𝑛2 𝑛=1

31.

∞ ∑ 𝑥2𝑛+1 𝑛! 𝑛=1

32.

∞ ∑

33.

∞ ∑ (5𝑥)𝑛 √ 𝑛 𝑛=1

𝑛!𝑥𝑛

𝑛=0

34.

∞ ∑ (5𝑥)2𝑛 √ 𝑛 𝑛=1

9.5 POWER SERIES AND INTERVAL OF CONVERGENCE

In Problems 35–38, use the formula for the sum of a geometric series to find a power series centered at the origin that converges to the expression. For what values does the series converge? 35.

1 1 + 2𝑧

36.

2 1 + 𝑦2

37.

3 1 − 𝑧∕2

38.

8 4+𝑦

39. For all 𝑧-values for which it converges, the function 𝑓 is defined by the series ) ( ) ) ( ( 𝑧−3 2 𝑧−3 3 𝑧−3 +5⋅ +5⋅ +⋯ . 𝑓 (𝑧) = 5+5 7 7 7 (a) Find 𝑓 (4). (b) Find the interval of convergence of 𝑓 (𝑧). 40. For all 𝑡-values for which it converges, the function 𝑓 is defined by the series ∞ ∑ (𝑡 − 7)𝑛 𝑓 (𝑡) = . 5𝑛 𝑛=0

(a) Find 𝑓 (4). (b) Find the interval of convergence of 𝑓 (𝑡). ∑ 41. The series 𝐶𝑛 𝑥𝑛 converges when 𝑥 = −4 and diverges when 𝑥 = 7. Decide whether each of the following statements is true or false, or whether this cannot be determined. (a) The power series converges when 𝑥 = 10. (b) The power series converges when 𝑥 = 3. (c) The power series diverges when 𝑥 = 1. (d) The power series diverges when 𝑥 = 6. ∑ 42. If 𝐶𝑛 (𝑥 − 3)𝑛 converges at 𝑥 = 7 and diverges at 𝑥 = 10, what can you say about the convergence at 𝑥 = 11? At 𝑥 = 5? At 𝑥 = 0? ∑ 43. The series 𝐶𝑛 𝑥𝑛 converges at 𝑥 = −5 and diverges at 𝑥 = 7. What can you say about its radius of convergence?

511

∑ 44. The series 𝐶𝑛 (𝑥 + 7)𝑛 converges at 𝑥 = 0 and diverges at 𝑥 = −17. What can you say about its radius of convergence? 45. The power series 𝐶0 + 𝐶1 𝑥 + 𝐶2 𝑥2 + 𝐶3 𝑥3 + ⋯ converges to 𝜋 for 𝑥 = 𝑘. Find possible values for 𝑘, 𝐶0 , 𝐶1 , 𝐶2 , 𝐶3 , where 𝐶𝑖 is an integer. [Hint: 𝜋 = 3 + 0.1 + 0.04 + 0.001 + ⋯ .] 46. For constant 𝑝, find the radius of convergence of the binomial power series:10 1 + 𝑝𝑥 +

𝑝(𝑝 − 1)𝑥2 𝑝(𝑝 − 1)(𝑝 − 2)𝑥3 + + ⋯. 2! 3!

47. Show that if 𝐶0 + 𝐶1 𝑥 + 𝐶2 𝑥2 + 𝐶3 𝑥3 + ⋯ converges for |𝑥| < 𝑅 with 𝑅 given by the ratio test, then so does 𝐶1 + 2𝐶2 𝑥 + 3𝐶3 𝑥2 + ⋯. Assume 𝐶𝑖 ≠ 0 for all 𝑖. 48. For all 𝑥-values for which it converges, the function 𝑓 is defined by the series 𝑓 (𝑥) =

∞ ∑ 𝑥𝑛 . 𝑛! 𝑛=0

(a) What is 𝑓 (0)? (b) What is the domain of 𝑓 ? (c) Assuming that 𝑓 ′ can be calculated by differentiating the series term-by-term, find the series for 𝑓 ′ (𝑥). What do you notice? (d) Guess what well-known function 𝑓 is. 49. From Exercise 24 we know the following series converges for all 𝑥. We define 𝑔(𝑥) to be its sum: 𝑔(𝑥) =

∞ ∑ 𝑛=1

(−1)𝑛−1

𝑥2𝑛−1 . (2𝑛 − 1)!

(a) Is 𝑔(𝑥) odd, even, or neither? What is 𝑔(0)? (b) Assuming that derivatives can be computed term by term, show that 𝑔 ′′ (𝑥) = −𝑔(𝑥). (c) Guess what well-known function 𝑔 might be. Check your guess using 𝑔(0) and 𝑔 ′ (0).

Strengthen Your Understanding In Problems 50–52, explain what is wrong with the statement.

In Problems 53–56, give an example of:

50. If lim𝑛→∞ |𝐶𝑛+1 ∕𝐶𝑛 | = 0, then the radius of conver∑ gence for 𝐶𝑛 𝑥𝑛 is 0. ∑ 51. If 𝐶𝑛 (𝑥 − 1)𝑛 converges at 𝑥 = 3, then

54. A power series that converges at 𝑥 = 5 but nowhere else.

52. The series 𝑥 = 3.



| 𝐶 | | | lim | 𝑛 | = 2. | 𝑛→∞ | 𝐶 | 𝑛+1 |

𝐶𝑛 𝑥𝑛 diverges at 𝑥 = 2 and converges at

10 For

an explanation of the name, see Section 10.2.

53. A power series that is divergent at 𝑥 = 0.

55. A power series that converges at 𝑥 = 10 and diverges at 𝑥 = 6. ∑ 56. A series 𝐶𝑛 𝑥𝑛 with radius of convergence 1 and that converges at 𝑥 = 1 and 𝑥 = −1.

512

Chapter 9 SEQUENCES AND SERIES

Decide if the statements in Problems 57–69 are true or false. Give an explanation for your answer. 57.

∞ ∑ (𝑥 − 𝑛)𝑛 is a power series. 𝑛=1

∑ 58. If the power series 𝐶𝑛 𝑥𝑛 converges for 𝑥 = 2, then it converges for 𝑥 = 1. ∑ 59. If the power series 𝐶𝑛 𝑥𝑛 converges for 𝑥 = 1, then the power series converges for 𝑥 = 2. ∑ 60. If the power series 𝐶𝑛 𝑥𝑛 does not converge for 𝑥 = 1, then the power series does not converge for 𝑥 = 2. ∑ ∑ 61. 𝐶𝑛 (𝑥 − 1)𝑛 and 𝐶𝑛 𝑥𝑛 have the same radius of convergence. ∑ ∑ 62. If 𝐶𝑛 𝑥𝑛 and 𝐵𝑛 𝑥𝑛 have the same radius of convergence, then the coefficients, 𝐶𝑛 and 𝐵𝑛 , must be equal. 63. If a power series converges at one endpoint of its interval of convergence, then it converges at the other. 64. A power series always converges at at least one point.

Online Resource: Review problems and Projects

∑ 65. If the power series 𝐶𝑛 𝑥𝑛 converges at 𝑥 = 10, then it converges at 𝑥 = −9. ∑ 66. If the power series 𝐶𝑛 𝑥𝑛 converges at 𝑥 = 10, then it converges at 𝑥 = −10. 67. −5 < 𝑥 ≤ 7 is a possible interval of convergence of a power series. 68. −3 < 𝑥 < 2 could be the interval of convergence of ∑ 𝐶𝑛 𝑥𝑛 .

69. If −11 < 𝑥 < 1 is the interval of convergence of ∑ 𝐶𝑛 (𝑥 − 𝑎)𝑛 , then 𝑎 = −5. ∑ 70. The power series 𝐶𝑛 𝑥𝑛 diverges at 𝑥 = 7 and converges at 𝑥 = −3. At 𝑥 = −9, the series is (a) (b) (c) (d) (e)

Conditionally convergent Absolutely convergent Alternating Divergent Cannot be determined.

Chapter Ten

APPROXIMATING FUNCTIONS USING SERIES

Contents 10.1 Taylor Polynomials . . . . . . . . . . . . . . . . . . . . . . . Linear Approximations . . . . . . . . . . . . . . . . . . . . Quadratic Approximations . . . . . . . . . . . . . . . . . . Higher-Degree Polynomials. . . . . . . . . . . . . . . . . Taylor Polynomials Around x = a . . . . . . . . . . . . 10.2 Taylor Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . Taylor Series for cos x, sin x, ex . . . . . . . . . . . . . . Taylor Series in General. . . . . . . . . . . . . . . . . . . . The Binomial Series Expansion. . . . . . . . . . . . . . Convergence of Taylor Series . . . . . . . . . . . . . . . 10.3 Finding and Using Taylor Series . . . . . . . . . . . . New Series by Substitution . . . . . . . . . . . . . . . . . New Series by Differentiation and Integration . .

514 514 515 516 519 523 524 524 525 526 530 530 531 Convergence of a Series and Differentiation . . . . . . . 531 Multiplying and Substituting Taylor Series . . . . . 532 Applications of Taylor Series. . . . . . . . . . . . . . . . 533 10.4 The Error in Taylor Polynomial Approximations . . . . . . . . . . . . . . . . . . . . . . . . . . 539 Bounds on the Error . . . . . . . . . . . . . . . . . . . . . . . 540 Using the Lagrange Error Bound for Taylor Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . 540

Convergence of the Taylor Series for cos x . . . . . 541 Showing that the Taylor Series for cos x Converges to cos x . . . . . . . . . . . . . . . . . . . . . . . . . . 542

Error Bounds Using the Alternating Series Estimate. . . . . . . . . . . . . . . . . . . . . . . . . . Deriving the Lagrange Error Bound . . . . . . . . . . 10.5 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . Fourier Polynomials. . . . . . . . . . . . . . . . . . . . . . . Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . Harmonics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

543 543 546 547 549 550 Energy and the Energy Theorem . . . . . . . . . . . . . . . . 551 Musical Instruments. . . . . . . . . . . . . . . . . . . . . . . . . . 552 What Do We Do If Our Function Does Not Have Period 2π? . . . . . . . . . . . . . . . . . . . . . 553 Seasonal Variation in the Incidence of Measles. . 554 Informal Justification of the Formulas for the Fourier Coefficients . . . . . . . . . . . . . . . . 555

514

Chapter 10 APPROXIMATING FUNCTIONS USING SERIES

10.1

TAYLOR POLYNOMIALS In this section, we see how to approximate a function by polynomials.

Linear Approximations We already know how to approximate a function using a degree-1 polynomial, namely the tangent line approximation given in Section 3.9: 𝑓 (𝑥) ≈ 𝑓 (𝑎) + 𝑓 ′ (𝑎)(𝑥 − 𝑎). The tangent line and the curve have the same slope at 𝑥 = 𝑎. As Figure 10.1 suggests, the tangent line approximation to the function is generally more accurate for values of 𝑥 close to 𝑎.

True value 𝑓 (𝑥)

Approximate value of 𝑓 (𝑥) Tangent line

☛ ✻𝑓 ′ (𝑎)(𝑥 − 𝑎) ✛ 𝑥 − 𝑎 ✲❄ ✻𝑓 (𝑎) 𝑓 (𝑎) ✻ ❄ ❄ 𝑥 𝑎

𝑥

Figure 10.1: Tangent line approximation of 𝑓 (𝑥) for 𝑥 near 𝑎

We first focus on 𝑎 = 0. The tangent line approximation at 𝑥 = 0 is referred to as the first Taylor approximation at 𝑥 = 0, or first Maclaurin polynomial, and written as follows:

Taylor Polynomial of Degree 1 Approximating 𝒇 (𝒙) for 𝒙 near 0 𝑓 (𝑥) ≈ 𝑃1 (𝑥) = 𝑓 (0) + 𝑓 ′ (0)𝑥

Example 1

Find the Taylor polynomial of degree 1 for 𝑔(𝑥) = cos 𝑥, with 𝑥 in radians, for 𝑥 near 0.

Solution

The tangent line at 𝑥 = 0 is just the horizontal line 𝑦 = 1, as shown in Figure 10.2, so 𝑃1 (𝑥) = 1. We have 𝑔(𝑥) = cos 𝑥 ≈ 1, for 𝑥 near 0. If we take 𝑥 = 0.05, then 𝑔(0.05) = cos(0.05) = 0.998 … , which is quite close to the approximation cos 𝑥 ≈ 1. Similarly, if 𝑥 = −0.1, then 𝑔(−0.1) = cos(−0.1) = 0.995 … is close to the approximation cos 𝑥 ≈ 1. However, if 𝑥 = 0.4, then 𝑔(0.4) = cos(0.4) = 0.921 … ,

10.1 TAYLOR POLYNOMIALS

515

so the approximation cos 𝑥 ≈ 1 is less accurate. For 𝑥 near 0, the graph suggests that the farther 𝑥 is from 0, the worse the approximation, cos 𝑥 ≈ 1, is likely to be. 𝑦

𝑦=1



cos 𝑥 𝑥

−𝜋

𝜋 −1

Figure 10.2: Graph of cos 𝑥 and its tangent line at 𝑥 = 0

The previous example shows that the Taylor polynomial of degree 1 might actually have degree less than 1.

Quadratic Approximations To get a more accurate approximation, we use a quadratic function instead of a linear function. Example 2

Find the quadratic approximation to 𝑔(𝑥) = cos 𝑥 for 𝑥 near 0.

Solution

To ensure that the quadratic, 𝑃2 (𝑥), is a good approximation to 𝑔(𝑥) = cos 𝑥 at 𝑥 = 0, we require that cos 𝑥 and the quadratic have the same value, the same slope, and the same second derivative at 𝑥 = 0. That is, we require 𝑃2 (0) = 𝑔(0), 𝑃2′ (0) = 𝑔 ′ (0), and 𝑃2′′ (0) = 𝑔 ′′ (0). We take the quadratic polynomial 𝑃2 (𝑥) = 𝐶0 + 𝐶1 𝑥 + 𝐶2 𝑥2 and determine 𝐶0 , 𝐶1 , and 𝐶2 . Since 𝑃2 (𝑥) = 𝐶0 + 𝐶1 𝑥 + 𝐶2 𝑥2 𝑃2′ (𝑥) = 𝐶1 + 2𝐶2 𝑥

and

𝑃2′′ (𝑥) = 2𝐶2

𝑔(𝑥) = cos 𝑥 𝑔 ′ (𝑥) = − sin 𝑥 𝑔 ′′ (𝑥) = − cos 𝑥,

we have 𝐶0 = 𝑃2 (0) = 𝑔(0) = cos 0 = 1 𝐶1 = 2𝐶2 =

so

𝑃2′ (0) = 𝑔 ′ (0) = − sin 0 = 0 𝑃2′′ (0) = 𝑔 ′′ (0) = − cos 0 = −1,

𝐶0 = 1 𝐶1 = 0 𝐶2 = − 12 .

Consequently, the quadratic approximation is 𝑥2 1 cos 𝑥 ≈ 𝑃2 (𝑥) = 1 + 0 ⋅ 𝑥 − 𝑥2 = 1 − , 2 2

for 𝑥 near 0.

Figure 10.3 suggests that the quadratic approximation cos 𝑥 ≈ 𝑃2 (𝑥) is better than the linear approximation cos 𝑥 ≈ 𝑃1 (𝑥) for 𝑥 near 0. Let’s compare the accuracy of the two approximations. Recall that 𝑃1 (𝑥) = 1 for all 𝑥. At 𝑥 = 0.4, we have cos(0.4) = 0.921 … and 𝑃2 (0.4) = 0.920, so the quadratic approximation is a significant improvement over the linear approximation. The magnitude of the error is about 0.001 instead of 0.08.

516

Chapter 10 APPROXIMATING FUNCTIONS USING SERIES

𝑃1 (𝑥) = 1

1 − 𝜋2



cos 𝑥

−𝜋

𝜋 2

√ − 2

−1

𝑥

√ 2

𝜋



𝑃2 (𝑥) = 1 −

𝑥2 2

Figure 10.3: Graph of cos 𝑥 and its linear, 𝑃1 (𝑥), and quadratic, 𝑃2 (𝑥), approximations for 𝑥 near 0

Generalizing the computations in Example 2, we define the second Taylor approximation at 𝑥 = 0, or second Maclaurin polynomial, as follows:

Taylor Polynomial of Degree 2 Approximating 𝒇 (𝒙) for 𝒙 near 0 𝑓 (𝑥) ≈ 𝑃2 (𝑥) = 𝑓 (0) + 𝑓 ′ (0)𝑥 +

𝑓 ′′ (0) 2 𝑥 2

Higher-Degree Polynomials Figure 10.3 shows that the quadratic approximation can still bend away from the original function for large 𝑥. We attempt to fix this by approximating 𝑓 (𝑥) near 0 using a polynomial of higher degree: 𝑓 (𝑥) ≈ 𝑃𝑛 (𝑥) = 𝐶0 + 𝐶1 𝑥 + 𝐶2 𝑥2 + ⋯ + 𝐶𝑛−1 𝑥𝑛−1 + 𝐶𝑛 𝑥𝑛 . To find the values of the constants: 𝐶0 , 𝐶1 , 𝐶2 , … , 𝐶𝑛 , we require that the function 𝑓 (𝑥) and each of its first 𝑛 derivatives agree with those of the polynomial 𝑃𝑛 (𝑥) at the point 𝑥 = 0. In general, the higher the degree of a Taylor polynomial, the larger the interval on which the function and the polynomial remain close to each other. To see how to find the constants, let’s take 𝑛 = 3 as an example: 𝑓 (𝑥) ≈ 𝑃3 (𝑥) = 𝐶0 + 𝐶1 𝑥 + 𝐶2 𝑥2 + 𝐶3 𝑥3 . Substituting 𝑥 = 0 gives 𝑓 (0) = 𝑃3 (0) = 𝐶0 . Differentiating 𝑃3 (𝑥) yields

𝑃3′ (𝑥) = 𝐶1 + 2𝐶2 𝑥 + 3𝐶3 𝑥2 ,

so substituting 𝑥 = 0 shows that 𝑓 ′ (0) = 𝑃3′ (0) = 𝐶1 . Differentiating and substituting again, we get 𝑃3′′ (𝑥) = 2 ⋅ 1𝐶2 + 3 ⋅ 2 ⋅ 1𝐶3 𝑥, which gives 𝑓 ′′ (0) = 𝑃3′′ (0) = 2 ⋅ 1𝐶2 , so that 𝐶2 = The third derivative, denoted by 𝑃3′′′ , is

𝑓 ′′ (0) . 2⋅1

𝑃3′′′ (𝑥) = 3 ⋅ 2 ⋅ 1𝐶3 , so 𝑓 ′′′ (0) = 𝑃3′′′ (0) = 3 ⋅ 2 ⋅ 1𝐶3 ,

10.1 TAYLOR POLYNOMIALS

and then 𝐶3 =

517

𝑓 ′′′ (0) . 3⋅2⋅1

You can imagine a similar calculation starting with 𝑃4 (𝑥) and using the fourth derivative 𝑓 (4) , which would give 𝑓 (4) (0) 𝐶4 = , 4⋅3⋅2⋅1 and so on. Using factorial notation,1 we write these expressions as 𝐶3 =

𝑓 ′′′ (0) , 3!

𝐶4 =

𝑓 (4) (0) . 4!

Writing 𝑓 (𝑛) for the 𝑛th derivative of 𝑓 , we have, for any positive integer 𝑛, 𝐶𝑛 =

𝑓 (𝑛) (0) . 𝑛!

So we define the 𝑛th Taylor approximation at 𝑥 = 0, or 𝑛th Maclaurin polynomial, as follows:

Taylor Polynomial of Degree 𝒏 Approximating 𝒇 (𝒙) for 𝒙 near 0 𝑓 (𝑥) ≈ 𝑃𝑛 (𝑥) = 𝑓 (0) + 𝑓 ′ (0)𝑥 +

𝑓 ′′ (0) 2 𝑓 ′′′ (0) 3 𝑓 (4) (0) 4 𝑓 (𝑛) (0) 𝑛 𝑥 + 𝑥 + 𝑥 +⋯+ 𝑥 2! 3! 4! 𝑛!

We call 𝑃𝑛 (𝑥) the Taylor polynomial of degree 𝑛 centered at 𝑥 = 0 or the Taylor polynomial about (or around) 𝑥 = 0.

Example 3

Construct the Taylor polynomial of degree 7 approximating the function 𝑓 (𝑥) = sin 𝑥 for 𝑥 near 0. Compare the value of the Taylor approximation with the true value of 𝑓 at 𝑥 = 𝜋∕3.

Solution

We have 𝑓 (𝑥) =

sin 𝑥

giving

𝑓 ′ (𝑥) = cos 𝑥 𝑓 ′′ (𝑥) = − sin 𝑥

𝑓 (0) =

0

𝑓 ′ (0) = 𝑓 ′′ (0) =

1 0

𝑓 ′′′ (𝑥) = − cos 𝑥 𝑓 (4) (𝑥) = sin 𝑥

𝑓 ′′′ (0) = −1 𝑓 (4) (0) = 0

𝑓 (5) (𝑥) = cos 𝑥 𝑓 (6) (𝑥) = − sin 𝑥

𝑓 (5) (0) = 𝑓 (6) (0) =

𝑓 (7) (𝑥) = − cos 𝑥

𝑓 (7) (0) = −1.

1 0

Using these values, we see that the Taylor polynomial approximation of degree 7 is 0 1 0 1 0 1 ⋅ 𝑥2 − ⋅ 𝑥3 + ⋅ 𝑥4 + ⋅ 𝑥5 + ⋅ 𝑥6 − ⋅ 𝑥7 2! 3! 4! 5! 6! 7! 𝑥3 𝑥5 𝑥7 + − , for 𝑥 near 0. =𝑥− 3! 5! 7!

sin 𝑥 ≈ 𝑃7 (𝑥) = 0 + 1 ⋅ 𝑥 +

1 Recall

that 𝑘! = 𝑘(𝑘 − 1) ⋯ 2 ⋅ 1. In addition, 1! = 1, and 0! = 1.

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Chapter 10 APPROXIMATING FUNCTIONS USING SERIES

Notice that since 𝑓 (8) (0) = 0, the seventh and eighth Taylor approximations to sin 𝑥 are the same. In Figure 10.4 we show the graphs of the sine function and the approximating polynomial of degree 7 for 𝑥 near 0. They are indistinguishable where 𝑥 is close to 0. However, as we look at values of 𝑥 farther away from 0 in either direction, the two graphs move apart. To√ check the accuracy of this approximation numerically, we see how well it approximates sin(𝜋∕3) = 3∕2 = 0.8660254 … . sin 𝑥

𝑃7 (𝑥)

1 𝑥

−𝜋

𝜋 3

−1

𝜋 𝑃7 (𝑥)

sin 𝑥

Figure 10.4: Graph of sin 𝑥 and its seventh-degree Taylor polynomial, 𝑃7 (𝑥), for 𝑥 near 0

When we substitute 𝜋∕3 = 1.0471976 … into the polynomial approximation, we obtain 𝑃7 (𝜋∕3) = 0.8660213 … , which is extremely accurate—to about four parts in a million.

Example 4

Graph the Taylor polynomial of degree 8 approximating 𝑔(𝑥) = cos 𝑥 for 𝑥 near 0.

Solution

We find the coefficients of the Taylor polynomial by the method of the preceding example, giving cos 𝑥 ≈ 𝑃8 (𝑥) = 1 −

𝑥 2 𝑥4 𝑥6 𝑥8 + − + . 2! 4! 6! 8!

Figure 10.5 shows that 𝑃8 (𝑥) is close to the cosine function for a larger interval of 𝑥-values than the quadratic approximation 𝑃2 (𝑥) = 1 − 𝑥2 ∕2 in Example 2 on page 515. 𝑃8 (𝑥)

𝑃8 (𝑥) 1

cos 𝑥

cos 𝑥 𝑥

−𝜋

𝜋 −1

𝑃2 (𝑥)

𝑃2 (𝑥)

Figure 10.5: 𝑃8 (𝑥) approximates cos 𝑥 better than 𝑃2 (𝑥) for 𝑥 near 0

Example 5

Construct the Taylor polynomial of degree 10 about 𝑥 = 0 for the function 𝑓 (𝑥) = 𝑒𝑥 . Check the accuracy of this approximation at 𝑥 = 1.

Solution

We have 𝑓 (0) = 1. Since the derivative of 𝑒𝑥 is equal to 𝑒𝑥 , all the higher-order derivatives are equal to 𝑒𝑥 . Consequently, for any 𝑘 = 1, 2, … , 10, 𝑓 (𝑘) (𝑥) = 𝑒𝑥 and 𝑓 (𝑘) (0) = 𝑒0 = 1. Therefore, the Taylor polynomial approximation of degree 10 is given by 𝑒𝑥 ≈ 𝑃10 (𝑥) = 1 + 𝑥 +

𝑥10 𝑥2 𝑥3 𝑥4 + + +⋯+ , 2! 3! 4! 10!

for 𝑥 near 0.

To check the accuracy of the approximation, we substitute 𝑥 = 1 to get 𝑃10 (1) = 2.718281801 and compare with 𝑒 = 𝑒1 = 2.718281828 …. We see 𝑃10 yields the first seven decimal places for 𝑒.

10.1 TAYLOR POLYNOMIALS

519

Figure 10.6 shows graphs of 𝑓 (𝑥) = 𝑒𝑥 and the Taylor polynomials of degree 𝑛 = 0, 1, 2, 3, 4. Notice that each successive approximation remains close to the exponential curve for a larger interval of 𝑥-values. For any 𝑛, the accuracy decreases for large 𝑥 because 𝑒𝑥 grows faster than any polynomial as 𝑥 → ∞. 𝑒𝑥 𝑃4 𝑃3

𝑃4

𝑃2

20 𝑃2 10

𝑃1

𝑃0

𝑃0

𝑒𝑥

𝑥 −4

𝑃1

−2

2

4

−10

𝑃3

Figure 10.6: For 𝑥 near 0, the value of 𝑒𝑥 is more closely approximated by higher-degree Taylor polynomials

1 for 𝑥 near 0. 1−𝑥

Example 6

Construct the Taylor polynomial of degree 𝑛 approximating 𝑓 (𝑥) =

Solution

Differentiating gives 𝑓 (0) = 1, 𝑓 ′ (0) = 1, 𝑓 ′′ (0) = 2, 𝑓 ′′′ (0) = 3!, 𝑓 (4) (0) = 4!, and so on. This means 1 ≈ 𝑃𝑛 (𝑥) = 1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4 + ⋯ + 𝑥𝑛 , for 𝑥 near 0. 1−𝑥 Let us compare the Taylor polynomial with the formula obtained from the sum of a finite geometric series on page 482: 1 − 𝑥𝑛+1 = 1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4 + ⋯ + 𝑥𝑛 . 1−𝑥 If 𝑥 is close to 0 and 𝑥𝑛+1 is small enough to neglect, the formula for the sum of a finite geometric series gives us the Taylor approximation of degree 𝑛: 1 ≈ 1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4 + ⋯ + 𝑥 𝑛 . 1−𝑥

Taylor Polynomials Around 𝒙 = 𝒂 Suppose we want to approximate 𝑓 (𝑥) = ln 𝑥 by a Taylor polynomial. This function has no Taylor polynomial about 𝑥 = 0 because the function is not defined for 𝑥 ≤ 0. However, it turns out that we can construct a polynomial centered about some other point, 𝑥 = 𝑎. First, let’s look at the equation of the tangent line at 𝑥 = 𝑎: 𝑦 = 𝑓 (𝑎) + 𝑓 ′ (𝑎)(𝑥 − 𝑎). This gives the first Taylor approximation 𝑓 (𝑥) ≈ 𝑓 (𝑎) + 𝑓 ′ (𝑎)(𝑥 − 𝑎)

for 𝑥 near 𝑎.

The 𝑓 ′ (𝑎)(𝑥 − 𝑎) term is a correction that approximates the change in 𝑓 as 𝑥 moves away from 𝑎.

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Chapter 10 APPROXIMATING FUNCTIONS USING SERIES

Similarly, the Taylor polynomial 𝑃𝑛 (𝑥) centered at 𝑥 = 𝑎 is set up as 𝑓 (𝑎) plus correction terms that are zero for 𝑥 = 𝑎. This is achieved by writing the polynomial in powers of (𝑥 − 𝑎) instead of powers of 𝑥: 𝑓 (𝑥) ≈ 𝑃𝑛 (𝑥) = 𝐶0 + 𝐶1 (𝑥 − 𝑎) + 𝐶2 (𝑥 − 𝑎)2 + ⋯ + 𝐶𝑛 (𝑥 − 𝑎)𝑛 . If we require 𝑛 derivatives of the approximating polynomial 𝑃𝑛 (𝑥) and the original function 𝑓 (𝑥) to agree at 𝑥 = 𝑎, we get the following result for the 𝑛th Taylor approximation at 𝑥 = 𝑎:

Taylor Polynomial of Degree 𝒏 Approximating 𝒇 (𝒙) for 𝒙 near 𝒂 𝑓 (𝑥) ≈ 𝑃𝑛 (𝑥) = 𝑓 (𝑎) + 𝑓 ′ (𝑎)(𝑥 − 𝑎) +

𝑓 (𝑛) (𝑎) 𝑓 ′′ (𝑎) (𝑥 − 𝑎)2 + ⋯ + (𝑥 − 𝑎)𝑛 2! 𝑛!

We call 𝑃𝑛 (𝑥) the Taylor polynomial of degree 𝑛 centered at 𝑥 = 𝑎 or the Taylor polynomial about 𝑥 = 𝑎. The formulas for these coefficients are derived in the same way as we did for 𝑎 = 0. (See Problem 42, page 522.)2 Example 7

Construct the Taylor polynomial of degree 4 approximating the function 𝑓 (𝑥) = ln 𝑥 for 𝑥 near 1.

Solution

We have 𝑓 (𝑥) = ln 𝑥 𝑓 ′ (𝑥) = 1∕𝑥

so

𝑓 (1) = ln(1) = 0 𝑓 ′ (1) = 1

𝑓 ′′ (𝑥) = −1∕𝑥2 𝑓 ′′′ (𝑥) = 2∕𝑥3

𝑓 ′′ (1) = 𝑓 ′′′ (1) =

𝑓 (4) (𝑥) = −6∕𝑥4 ,

𝑓 (4) (1) = −6.

−1 2

The Taylor polynomial is therefore (𝑥 − 1)3 (𝑥 − 1)4 (𝑥 − 1)2 +2 −6 2! 3! 4! (𝑥 − 1)2 (𝑥 − 1)3 (𝑥 − 1)4 + − , for 𝑥 near 1. = (𝑥 − 1) − 2 3 4

ln 𝑥 ≈ 𝑃4 (𝑥) = 0 + (𝑥 − 1) −

Graphs of ln 𝑥 and several of its Taylor polynomials are shown in Figure 10.7. Notice that 𝑃4 (𝑥) stays reasonably close to ln 𝑥 for 𝑥 near 1, but bends away as 𝑥 gets farther from 1. Also, note that the Taylor polynomials are defined for 𝑥 ≤ 0, but ln 𝑥 is not. 𝑃3 (𝑥) 𝑃1 (𝑥) ln 𝑥

1

𝑃2 (𝑥) 𝑥 1

𝑃1 (𝑥)

✛ 𝑃4 (𝑥)

𝑃2 (𝑥) 𝑃3 (𝑥)

ln 𝑥

2

3 𝑃4 (𝑥)

Figure 10.7: Taylor polynomials approximate ln 𝑥 closely for 𝑥 near 1, but not necessarily farther away

2A

Taylor polynomial is also called a Maclaurin polynomial if 𝑎 = 0.

521

10.1 TAYLOR POLYNOMIALS

The examples in this section suggest that the following results are true for common functions: • Taylor polynomials centered at 𝑥 = 𝑎 give good approximations to 𝑓 (𝑥) for 𝑥 near 𝑎. Farther away, they may or may not be good. • Typically, a Taylor polynomial of higher degree fits the function closely over a larger interval.

Exercises and Problems for Section 10.1 EXERCISES For Exercises 1–10, find the Taylor polynomials of degree 𝑛 approximating the functions for 𝑥 near 0. (Assume 𝑝 is a constant.) 1. 3.

1 1−𝑥

, 𝑛 = 3, 5, 7

√ 1 + 𝑥, 𝑛 = 2, 3, 4

5. cos 𝑥, 𝑛 = 2, 4, 6

2.

1

,

𝑛 = 4, 6, 8

1+𝑥 √ 3 4. 1 − 𝑥, 𝑛 = 2, 3, 4 6. ln(1 + 𝑥), 𝑛 = 5, 7, 9

For Exercises 11–16, find the Taylor polynomial of degree 𝑛 for 𝑥 near the given point 𝑎. √ 11. 1 − 𝑥, 𝑎 = 0, 𝑛 = 3 12. 𝑒𝑥 , 𝑎 = 1, 𝑛 = 4 1 13. , 𝑎 = 2, 𝑛 = 4 1+𝑥 14. cos 𝑥, 𝑎 = 𝜋∕2, 𝑛 = 4 15. sin 𝑥,

𝑎 = −𝜋∕4, 𝑛 = 3

2

16. ln(𝑥 ), 𝑎 = 1, 7. arctan 𝑥,

𝑛 = 3, 4

8. tan 𝑥,

1 , 𝑛 = 2, 3, 4 9. √ 1+𝑥

𝑛=4

𝑛 = 3, 4

10. (1 + 𝑥)𝑝 , 𝑛 = 2, 3, 4

PROBLEMS 17. The function 𝑓 (𝑥) is approximated near 𝑥 = 0 by the third-degree Taylor polynomial 𝑃3 (𝑥) = 2 − 𝑥 − 𝑥2 ∕3 + 2𝑥3 .

In Problems 23–24, find a simplified formula for 𝑃5 (𝑥), the fifth-degree Taylor polynomial approximating 𝑓 near 𝑥 = 0. 23. Use the values in the table.

Give the value of (b) 𝑓 ′ (0) (d) 𝑓 ′′′ (0)

(a) 𝑓 (0) (c) 𝑓 ′′ (0)

18. Find the second-degree Taylor polynomial for 𝑓 (𝑥) = 4𝑥2 − 7𝑥 + 2 about 𝑥 = 0. What do you notice? 19. Find the third-degree Taylor polynomial for 𝑓 (𝑥) = 𝑥3 + 7𝑥2 − 5𝑥 + 1 about 𝑥 = 0. What do you notice? 20. (a) Based on your observations in Problems 18–19, make a conjecture about Taylor approximations in the case when 𝑓 is itself a polynomial. (b) Show that your conjecture is true. 21. The Taylor polynomial of degree 7 of 𝑓 (𝑥) is given by 𝑃7 (𝑥) = 1 −

𝑥 5𝑥2 𝑥5 + + 8𝑥3 − + 8𝑥7 . 3 7 11

𝑝(𝑥) =

10 ∑ (𝑥 − 1)𝑛 . 𝑛! 𝑛=0

𝑓 ′ (0)

𝑓 ′′ (0)

𝑓 ′′′ (0)

𝑓 (4) (0)

𝑓 (5) (0)

−3

5

−2

0

−1

4

24. Let 𝑓 (0) = −1 and, for 𝑛 > 0, 𝑓 (𝑛) (0) = −(−2)𝑛 . 25. (a) From a graph of 𝑓 (𝑥) = ln(2𝑥+4) decide if the tangent line approximation to 𝑓 at 𝑥 = 0 has a positive or negative slope. (b) Find the tangent line approximation. (c) Find the second and third Taylor approximations to 𝑓 about 𝑥 = 0. For Problems 26–29, suppose 𝑃2 (𝑥) = 𝑎 + 𝑏𝑥 + 𝑐𝑥2 is the second-degree Taylor polynomial for the function 𝑓 about 𝑥 = 0. What can you say about the signs of 𝑎, 𝑏, and 𝑐 if 𝑓 has the graph given below? 26. 27. 𝑓 (𝑥)

𝑓 (𝑥)

𝑥

Find the Taylor polynomial of degree 3 of 𝑓 (𝑥). 22. Find the value of 𝑓 (5) (1) if 𝑓 (𝑥) is approximated near 𝑥 = 1 by the Taylor polynomial

𝑓 (0)

28.

𝑓 (𝑥) 𝑥

𝑥

29. 𝑥 𝑓 (𝑥)

522

Chapter 10 APPROXIMATING FUNCTIONS USING SERIES III

In Problems 30–33, use the third degree Taylor polynomial 𝑃3 (𝑥) = 4 + 2(𝑥 − 1) − 2(𝑥 − 1)2 + (𝑥 − 1)3 ∕2 of 𝑓 (𝑥) about 𝑥 = 1 to find the given value, or explain why you can’t. 31. 𝑓 ′′′ (1)

30. 𝑓 (1)

32. 𝑓 ′′ (0)

33. 𝑓 (4) (1)

II

I

𝐵

𝐴

In Problems 34–36, use a Taylor polynomial with the derivatives given to make the best possible estimate of the value. ′

III

II

I

Figure 10.8

Figure 10.9

′′

34. 𝑓 (1.1), given that 𝑓 (1) = 3, 𝑓 (1) = 2, 𝑓 (1) = −4. 35. 𝑔(1.8), given that 𝑔(2) = −3, 𝑔 ′ (2) = −2, 𝑔 ′′ (2) = 0, 𝑔 ′′′ (2) = 12. 36. ℎ′ (0.1), given that ℎ(0) = 6, ℎ′ (0) = 2, ℎ′′ (0) = −4. 37. Use the Taylor approximation for 𝑥 near 0, 𝑥3 sin 𝑥 ≈ 𝑥 − , 3! sin 𝑥 = 1. 𝑥 38. Use the fourth-degree Taylor approximation for 𝑥 near 0, 𝑥2 𝑥4 + , cos 𝑥 ≈ 1 − 2! 4! 1 − cos 𝑥 1 to explain why lim = . 𝑥→0 2 𝑥2 39. Use a fourth-degree Taylor approximation for 𝑒ℎ , for ℎ near 0, to evaluate the following limits. Would your answer be different if you used a Taylor polynomial of higher degree? to explain why lim 𝑥→0

(a) lim ℎ→0

(b) lim ℎ→0

𝑒ℎ − 1 − ℎ ℎ2 𝑒ℎ − 1 − ℎ −

ℎ2 2

ℎ3

40. If 𝑓 (2) = 𝑔(2) = ℎ(2) = 0, and 𝑓 ′ (2) = ℎ′ (2) = 0, 𝑔 ′ (2) = 22, and 𝑓 ′′ (2) = 3, 𝑔 ′′ (2) = 5, ℎ′′ (2) = 7, calculate the following limits. Explain your reasoning. 𝑓 (𝑥) 𝑓 (𝑥) (a) lim (b) lim 𝑥→2 ℎ(𝑥) 𝑥→2 𝑔(𝑥) 41. One of the two sets of functions, 𝑓1 , 𝑓2 , 𝑓3 , or 𝑔1 , 𝑔2 , 𝑔3 , is graphed in Figure 10.8; the other set is graphed in Figure 10.9. Points 𝐴 and 𝐵 each have 𝑥 = 0. Taylor polynomials of degree 2 approximating these functions near 𝑥 = 0 are as follows: 𝑓1 (𝑥) ≈ 2 + 𝑥 + 2𝑥2 2

𝑔1 (𝑥) ≈ 1 + 𝑥 + 2𝑥2

𝑓2 (𝑥) ≈ 2 + 𝑥 − 𝑥

𝑔2 (𝑥) ≈ 1 + 𝑥 + 𝑥2

𝑓3 (𝑥) ≈ 2 + 𝑥 + 𝑥2

𝑔3 (𝑥) ≈ 1 − 𝑥 + 𝑥2 .

(a) Which group of functions, the 𝑓 s or the 𝑔s, is represented by each figure? (b) What are the coordinates of the points 𝐴 and 𝐵? (c) Match each function with the graphs (I)–(III) in the appropriate figure.

42. Derive the formulas given in the box on page 520 for the coefficients of the Taylor polynomial approximating a function 𝑓 for 𝑥 near 𝑎. 43. (a) Find and multiply the Taylor polynomials of degree 1 near 𝑥 = 0 for the two functions 𝑓 (𝑥) = 1∕(1 − 𝑥) and 𝑔(𝑥) = 1∕(1 − 2𝑥). (b) Find the Taylor polynomial of degree 2 near 𝑥 = 0 for the function ℎ(𝑥) = 𝑓 (𝑥)𝑔(𝑥). (c) Is the product of the Taylor polynomials for 𝑓 (𝑥) and 𝑔(𝑥) equal to the Taylor polynomial for the function ℎ(𝑥)? 44. (a) Find and multiply the Taylor polynomials of degree 1 near 𝑥 = 0 for the two functions 𝑓 (𝑥) and 𝑔(𝑥). (b) Find the Taylor polynomial of degree 2 near 𝑥 = 0 for the function ℎ(𝑥) = 𝑓 (𝑥)𝑔(𝑥). (c) Show that the product of the Taylor polynomials for 𝑓 (𝑥) and 𝑔(𝑥) and the Taylor polynomial for the function ℎ(𝑥) are the same if 𝑓 ′′ (0)𝑔(0) + 𝑓 (0)𝑔 ′′ (0) = 0. 45. (a) Find the Taylor polynomial approximation of de2 gree 4 about 𝑥 = 0 for the function 𝑓 (𝑥) = 𝑒𝑥 . (b) Compare this result to the Taylor polynomial approximation of degree 2 for the function 𝑓 (𝑥) = 𝑒𝑥 about 𝑥 = 0. What do you notice? (c) Use your observation in part (b) to write out the Taylor polynomial approximation of degree 20 for the function in part (a). (d) What is the Taylor polynomial approximation of degree 5 for the function 𝑓 (𝑥) = 𝑒−2𝑥 ? 1

46. The integral ∫0 (sin 𝑡∕𝑡) 𝑑𝑡 is difficult to approximate using, for example, left Riemann sums or the trapezoid rule because the integrand (sin 𝑡)∕𝑡 is not defined at 𝑡 = 0. However, this integral converges; its value is 0.94608 … . Estimate the integral using Taylor polynomials for sin 𝑡 about 𝑡 = 0 of (a) Degree 3

(b) Degree 5

𝑥3 = 0.2. 3! (a) How many solutions does each equation have? (b) Which of the solutions of the two equations are approximately equal? Explain.

47. Consider the equations sin 𝑥 = 0.2 and 𝑥 −

10.2 TAYLOR SERIES

48. When we model the motion of a pendulum, we sometimes replace the differential equation 𝑔 𝑑2𝜃 = − sin 𝜃 𝑙 𝑑𝑡2

by

𝑔 𝑑2𝜃 = − 𝜃, 𝑙 𝑑𝑡2

523

(b) Replace cos 𝑥 by its second-degree Taylor polynomial near 0 and solve the equation. Your answers are approximations to the solutions to the original equation at or near 0.

where 𝜃 is the angle between the pendulum and the vertical. Explain why, and under what circumstances, it is reasonable to make this replacement.

50. Let 𝑃𝑛 (𝑥) be the Taylor polynomial of degree 𝑛 approximating 𝑓 (𝑥) near 𝑥 = 0 and 𝑄𝑛 (𝑥) the Taylor polynomial of degree 𝑛 approximating 𝑓 ′ (𝑥) near 𝑥 = 0. Show that 𝑄𝑛−1 (𝑥) = 𝑃𝑛′ (𝑥).

49. (a) Using a graph, explain why the following equation has a solution at 𝑥 = 0 and another just to the right of 𝑥 = 0: cos 𝑥 = 1 − 0.1𝑥.

51. Let 𝑃𝑛 (𝑥) be the Taylor polynomial of degree 𝑛 approximating 𝑓 (𝑥) near 𝑥 = 0 and 𝑄𝑛 (𝑥) the Taylor polynomial of degree 𝑛 approximating its antiderivative 𝐹 (𝑥) 𝑥 with 𝐹 (0) = 1. Show that 𝑄𝑛+1 (𝑥) = 1 + ∫0 𝑃𝑛 (𝑡) 𝑑𝑡.

Strengthen Your Understanding In Problems 52–54, explain what is wrong with the statement.

58. A function whose Taylor polynomial of degree 1 about 𝑥 = 0 is closer to the values of the function for some values of 𝑥 than its Taylor polynomial of degree 2 about that point.

52. If 𝑓 (𝑥) = ln(2+𝑥), then the second-degree Taylor polynomial approximating 𝑓 (𝑥) near 𝑥 = 0 has a negative constant term. 1 53. Let 𝑓 (𝑥) = . The coefficient of the 𝑥 term of the 1−𝑥 Taylor polynomial of degree 3 approximating 𝑓 (𝑥) near 𝑥 = 0 is −1.

Decide if the statements in Problems 59–66 are true or false. Give an explanation for your answer.

54. If the Taylor polynomial of degree 3 approximating 𝑓 about 𝑥 = 0 is

60. Using sin 𝜃 ≈ 𝜃 − 𝜃 3 ∕3! with 𝜃 = 1◦ , we have sin(1◦ ) ≈ 1 − 13 ∕6 = 5∕6.

𝑃3 (𝑥) =

𝑥 𝑥2 𝑥3 + + , 2 3 4

then the Taylor polynomial of degree 3 approximating 𝑓 about 𝑥 = 1 is 𝑃3 (𝑥) =

(𝑥 − 1) (𝑥 − 1)2 (𝑥 − 1)3 + + . 2 3 4

In Problems 55–58, give an example of: 55. A function 𝑓 (𝑥) for which every Taylor polynomial approximation near 𝑥 = 0 involves only odd powers of 𝑥. 56. A third-degree Taylor polynomial near 𝑥 = 1 approximating a function 𝑓 (𝑥) with 𝑓 ′ (1) = 3. 57. Two different functions having the same Taylor polynomial of degree 1 about 𝑥 = 0.

10.2

59. If 𝑓 (𝑥) and 𝑔(𝑥) have the same Taylor polynomial of degree 2 near 𝑥 = 0, then 𝑓 (𝑥) = 𝑔(𝑥).

61. The Taylor polynomial of degree 2 for 𝑒𝑥 near 𝑥 = 5 is 1 + (𝑥 − 5) + (𝑥 − 5)2 ∕2. 62. If the Taylor polynomial of degree 2 for 𝑓 (𝑥) near 𝑥 = 0 is 𝑃2 (𝑥) = 1+𝑥−𝑥2 , then 𝑓 (𝑥) is concave up near 𝑥 = 0. 63. The quadratic approximation to 𝑓 (𝑥) for 𝑥 near 0 is better than the linear approximation for all values of 𝑥. 64. A Taylor polynomial for 𝑓 near 𝑥 = 𝑎 touches the graph of 𝑓 only at 𝑥 = 𝑎. 65. The linear approximation to 𝑓 (𝑥) near 𝑥 = −1 shows that if 𝑓 (−1) = 𝑔(−1) and 𝑓 ′ (−1) < 𝑔 ′ (−1), then 𝑓 (𝑥) < 𝑔(𝑥) for all 𝑥 sufficiently close to −1 (but not equal to −1). 66. The quadratic approximation to 𝑓 (𝑥) near 𝑥 = −1 shows that if 𝑓 (−1) = 𝑔(−1), 𝑓 ′ (−1) = 𝑔 ′ (−1), and 𝑓 ′′ (−1) < 𝑔 ′′ (−1), then 𝑓 (𝑥) < 𝑔(𝑥) for all 𝑥 sufficiently close to −1 (but not equal to −1).

TAYLOR SERIES In the previous section we saw how to approximate a function near a point by Taylor polynomials. Now we define a Taylor series, which is a power series that can be thought of as a Taylor polynomial that goes on forever.

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Chapter 10 APPROXIMATING FUNCTIONS USING SERIES

Taylor Series for cos 𝒙, sin 𝒙, 𝒆𝒙 We have the following Taylor polynomials centered at 𝑥 = 0 for cos 𝑥: cos 𝑥 ≈ 𝑃0 (𝑥) = 1 𝑥2 2! 𝑥 2 𝑥4 + cos 𝑥 ≈ 𝑃4 (𝑥) = 1 − 2! 4! 𝑥 2 𝑥4 𝑥6 + − cos 𝑥 ≈ 𝑃6 (𝑥) = 1 − 2! 4! 6! 𝑥 2 𝑥4 𝑥6 𝑥8 + − + . cos 𝑥 ≈ 𝑃8 (𝑥) = 1 − 2! 4! 6! 8! Here we have a sequence of polynomials, 𝑃0 (𝑥), 𝑃2 (𝑥), 𝑃4 (𝑥), 𝑃6 (𝑥), 𝑃8 (𝑥), ..., each of which is a better approximation to cos 𝑥 than the last, for 𝑥 near 0. When we go to a higher-degree polynomial (say from 𝑃6 to 𝑃8 ), we add more terms (𝑥8 ∕8!, for example), but the terms of lower degree don’t change. Thus, each polynomial includes the information from all the previous ones. We represent the whole sequence of Taylor polynomials by writing the Taylor series for cos 𝑥: cos 𝑥 ≈ 𝑃2 (𝑥) = 1 −

𝑥2 𝑥4 𝑥6 𝑥8 + − + −⋯. 2! 4! 6! 8! Notice that the partial sums of this series are the Taylor polynomials, 𝑃𝑛 (𝑥). We define the Taylor series for sin 𝑥 and 𝑒𝑥 similarly. It turns out that, for these functions, the Taylor series converges to the function for all 𝑥, so we can write the following: 1−

𝑥2 𝑥3 𝑥4 + + +⋯ 2! 3! 4! 𝑥5 𝑥7 𝑥9 + − + −⋯ 5! 7! 9! 𝑥4 𝑥6 𝑥8 + − + −⋯ 4! 6! 8!

𝑒𝑥 = 1 + 𝑥 + 𝑥3 3! 𝑥2 cos 𝑥 = 1 − 2! sin 𝑥 = 𝑥 −

These series are also called Maclaurin series or Taylor expansions of the functions sin 𝑥, cos 𝑥, and 𝑒𝑥 about 𝑥 = 0. Notice that the series for 𝑒𝑥 contains both odd and even powers, while sin 𝑥 has only odd powers and cos 𝑥 has only even powers.3 The general term of a Taylor series is a formula that gives any term in the series. For example, 𝑥𝑛 ∕𝑛! is the general term in the Taylor expansion for 𝑒𝑥 , since substituting 𝑛 = 0, 1, 2, … gives each of the terms in the series for 𝑒𝑥 . Similarly, (−1)𝑘 𝑥2𝑘 ∕(2𝑘)! is the general term in the expansion for cos 𝑥 because substituting 𝑘 = 0, 1, 2, … gives each of its terms. We call 𝑛 or 𝑘 the index.

Taylor Series in General Any function 𝑓 , all of whose derivatives exist at 0, has a Taylor series. However, the Taylor series for 𝑓 does not necessarily converge to 𝑓 (𝑥) for all values of 𝑥. For the values of 𝑥 for which the series does converge to 𝑓 (𝑥), we have the following formula:

Taylor Series for 𝒇 (𝒙) About 𝒙 = 0 𝑓 (𝑥) = 𝑓 (0) + 𝑓 ′ (0)𝑥 + 3A

constant term is an even power.

𝑓 (𝑛) (0) 𝑛 𝑓 ′′ (0) 2 𝑓 ′′′ (0) 3 𝑥 + 𝑥 +⋯+ 𝑥 +⋯ 2! 3! 𝑛!

10.2 TAYLOR SERIES

525

In addition, just as we have Taylor polynomials centered at points other than 0, we can also have a Taylor series centered at 𝑥 = 𝑎 (provided all the derivatives of 𝑓 exist at 𝑥 = 𝑎). For the values of 𝑥 for which the series converges to 𝑓 (𝑥), we have the following formula:

Taylor Series for 𝒇 (𝒙) About 𝒙 = 𝒂 𝑓 (𝑥) = 𝑓 (𝑎) + 𝑓 ′ (𝑎)(𝑥 − 𝑎) +

𝑓 ′′ (𝑎) 𝑓 ′′′ (𝑎) 𝑓 (𝑛) (𝑎) (𝑥 − 𝑎)2 + (𝑥 − 𝑎)3 + ⋯ + (𝑥 − 𝑎)𝑛 + ⋯ 2! 3! 𝑛!

The Taylor series is a power series whose partial sums are the Taylor polynomials. As we saw in Section 9.5, power series converge on an interval centered at 𝑥 = 𝑎. For a given function 𝑓 and a given 𝑥, even if the Taylor series converges, it might not converge to 𝑓 (𝑥). However, the Taylor series for most common functions, including 𝑒𝑥 , cos 𝑥, and sin 𝑥, do converge to the original function for all 𝑥. See Section 10.4.

The Binomial Series Expansion We now find the Taylor series about 𝑥 = 0 for the function 𝑓 (𝑥) = (1 + 𝑥)𝑝 , with 𝑝 a constant, but not necessarily a positive integer. Taking derivatives: 𝑓 (𝑥) = (1 + 𝑥)𝑝 𝑓 ′ (𝑥) = 𝑝(1 + 𝑥)𝑝−1

so

𝑓 (0) = 1 𝑓 ′ (0) = 𝑝

𝑓 ′′ (𝑥) = 𝑝(𝑝 − 1)(1 + 𝑥)𝑝−2

𝑓 ′′ (0) = 𝑝(𝑝 − 1)

𝑓 ′′′ (𝑥) = 𝑝(𝑝 − 1)(𝑝 − 2)(1 + 𝑥)𝑝−3 ,

𝑓 ′′′ (0) = 𝑝(𝑝 − 1)(𝑝 − 2).

Thus, the third-degree Taylor polynomial for 𝑥 near 0 is 𝑝(𝑝 − 1) 2 𝑝(𝑝 − 1)(𝑝 − 2) 3 𝑥 + 𝑥 . 2! 3! Graphing 𝑃3 (𝑥), 𝑃4 (𝑥), … for various specific values of 𝑝 suggests that the Taylor polynomials converge to 𝑓 (𝑥) for −1 < 𝑥 < 1 for all 𝑝, and in fact this is correct. (See Problems 38–39, page 529.) The Taylor series for 𝑓 (𝑥) = (1 + 𝑥)𝑝 about 𝑥 = 0 is as follows: (1 + 𝑥)𝑝 ≈ 𝑃3 (𝑥) = 1 + 𝑝𝑥 +

The Binomial Series (1 + 𝑥)𝑝 = 1 + 𝑝𝑥 +

𝑝(𝑝 − 1) 2 𝑝(𝑝 − 1)(𝑝 − 2) 3 𝑥 + 𝑥 +⋯ 2! 3!

for −1 < 𝑥 < 1.

In fact, when 𝑝 is a positive integer, the binomial series gives the same result as multiplying (1 + 𝑥)𝑝 out. (Newton discovered that the binomial series can be used for non-integer exponents.) Example 1

Find the Taylor series about 𝑥 = 0 for

Solution

Since

1 . Explain why the result is a geometric series. 1+𝑥

1 = (1 + 𝑥)−1 , use the binomial series with 𝑝 = −1. Then we have 1+𝑥 (−1)(−2) 2 (−1)(−2)(−3) 3 1 = (1 + 𝑥)−1 = 1 + (−1)𝑥 + 𝑥 + 𝑥 +⋯ 1+𝑥 2! 3! for − 1 < 𝑥 < 1. = 1 − 𝑥 + 𝑥2 − 𝑥3 + ⋯

This binomial series is also geometric since each term is −𝑥 times the previous term. The common ratio is −𝑥.

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Chapter 10 APPROXIMATING FUNCTIONS USING SERIES

Example 2

Find the Taylor series about 𝑥 = 0 for

Solution

Since

√ 1 + 𝑥.

√ 1 + 𝑥 = (1 + 𝑥)1∕2 , we use the binomial series with 𝑝 = 1∕2. Then 𝑓 (𝑥) =

√ ( 1 )(− 21 )𝑥2 ( 21 )(− 21 )(− 23 )𝑥3 1 1+𝑥=1+ 𝑥+ 2 + +⋯ 2 2! 3! (− 41 )𝑥2 ( 83 )𝑥3 1 =1+ 𝑥+ + +⋯ 2 2! 3! 𝑥 𝑥2 𝑥3 =1+ − + +⋯ for − 1 < 𝑥 < 1. 2 8 16

Convergence of Taylor Series Let us look again at the Taylor polynomial for ln 𝑥 about 𝑥 = 1 that we derived in Example 7 on page 520. A similar calculation gives the Taylor series (𝑥 − 1)2 (𝑥 − 1)3 (𝑥 − 1)4 (𝑥 − 1)𝑛 + − + ⋯ + (−1)𝑛−1 + ⋯. 2 3 4 𝑛 Using the ratio test, we get convergence on 0 < 𝑥 < 2. Substituting 𝑥 = 0 givs a divergent series (the negative harmonic series); substituting 𝑥 = 2 gives a convergent series by the alternating series test. Thus, this power series has interval of convergence 0 < 𝑥 ≤ 2. However, although we know that the series converges in this interval, we do not yet know that its sum is ln 𝑥. Figure 10.10 shows the polynomials fit the curve well for 0 < 𝑥 < 2, suggesting that the Taylor series does converge to ln 𝑥 for 0 < 𝑥 ≤ 2, as turns out to be the case. However, when 𝑥 > 2, the polynomials move away from the curve and the approximations get worse as the degree of the polynomial increases. For 𝑥 ≤ 0, the function ln 𝑥 is not defined. Thus, the Taylor polynomials are effective only as approximations to ln 𝑥 for values of 𝑥 between 0 and 2; outside that interval, they should not be used. Inside the interval, but near the ends, 0 or 2, the polynomials converge very slowly. This means we might have to take a polynomial of very high degree to get an accurate value for ln 𝑥. ln 𝑥 = (𝑥 − 1) −



Convergence to ln 𝑥

✲ 𝑃7 (𝑥)

𝑃5 (𝑥) ln 𝑥 𝑥

1

𝑃5 (𝑥) ✲ 𝑃6 (𝑥) ✲ 𝑃7 (𝑥) ✲ 𝑃8 (𝑥) ✲

ln 𝑥

2

3

𝑃8 (𝑥)

4

𝑃6 (𝑥)

Figure 10.10: Taylor polynomials 𝑃5 (𝑥), 𝑃6 (𝑥), 𝑃7 (𝑥), 𝑃8 (𝑥), … converge to ln 𝑥 for 0 < 𝑥 ≤ 2 and diverge outside that interval

Proving that the Taylor series converges to ln 𝑥 between 0 and 2, as Figure 10.10 suggests, requires the error term introduced in Section 10.4.

10.2 TAYLOR SERIES

527

Example 3

Find the Taylor series for ln(1 + 𝑥) about 𝑥 = 0, and investigate its convergence to ln(1 + 𝑥).

Solution

Taking derivatives of ln(1 + 𝑥) and substituting 𝑥 = 0 leads, assuming convergence, to the Taylor series 𝑥 2 𝑥3 𝑥4 + − +⋯. ln(1 + 𝑥) = 𝑥 − 2 3 4 Notice that this is the same series that we get by substituting (1 + 𝑥) for 𝑥 in the series for ln 𝑥: ln 𝑥 = (𝑥 − 1) −

(𝑥 − 1)2 (𝑥 − 1)3 (𝑥 − 1)4 + − +⋯ 2 3 4

for 0 < 𝑥 ≤ 2.

Since the series for ln 𝑥 about 𝑥 = 1 converges to ln 𝑥 for 𝑥 between 0 and 2, the Taylor series for ln(1 + 𝑥) about 𝑥 = 0 converges to ln(1 + 𝑥) for 𝑥 between −1 and 1. See Figure 10.11. Notice that the series could not possibly converge to ln(1 + 𝑥) for 𝑥 ≤ −1 since ln(1 + 𝑥) is not defined there. ✛

Convergence to

ln(1 + 𝑥)

𝑃9 (𝑥) 𝑃7 (𝑥)



𝑃5 (𝑥)

ln(1 + 𝑥) 𝑥 −1 𝑃5 (𝑥) 𝑃6 (𝑥) 𝑃7 (𝑥) 𝑃8 (𝑥)

1

✲ ✲ ✲ ✛ 𝑃9 (𝑥) ✲ ✛ ln(1 + 𝑥)

𝑃8 (𝑥)

𝑃6 (𝑥)

Figure 10.11: Convergence of the Taylor series for ln(1 + 𝑥)

The following theorem tells us about the relationship between power series and Taylor series; the justification involves calculations similar to those in Problem 60.

Theorem 10.1: A Power Series Is its Own Taylor Series If a power series about 𝑥 = 𝑎 converges to 𝑓 (𝑥) for |𝑥 − 𝑎| < 𝑅, then the power series is the Taylor series for 𝑓 (𝑥) about 𝑥 = 𝑎.

Exercises and Problems for Section 10.2 EXERCISES For Exercises 1–9, find the first four nonzero terms of the Taylor series for the function about 0. √ 4

1. (1 + 𝑥)3∕2

2.

3. sin(−𝑥)

4. ln(1 − 𝑥)

𝑥+1

5.

1 1−𝑥

√ 7. 3 1 − 𝑦 9. ln(5 + 2𝑥)

6. √

1 1+𝑥

8. tan(𝑡 + 𝜋∕4)

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Chapter 10 APPROXIMATING FUNCTIONS USING SERIES

For Exercises 10–17, find the first four terms of the Taylor series for the function about the point 𝑎. 10. sin 𝑥,

𝑎 = 𝜋∕4

11. cos 𝜃,

𝑎 = 𝜋∕4

12. cos 𝑡,

𝑎 = 𝜋∕6

13. sin 𝜃,

𝑎 = −𝜋∕4

14. tan 𝑥,

𝑎 = 𝜋∕4

15. 1∕𝑥,

𝑎=1

17. 1∕𝑥,

𝑎 = −1

16. 1∕𝑥,

𝑎=2

In Exercises 18–25, find an expression for the general term of the series and give the range of values for the index (𝑛 or 𝑘, for example).

1 = 1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4 + ⋯ 1−𝑥 1 19. = 1 − 𝑥 + 𝑥2 − 𝑥3 + 𝑥4 − ⋯ 1+𝑥 𝑥2 𝑥3 𝑥4 − − −⋯ 20. ln(1 − 𝑥) = −𝑥 − 2 3 4 2 3 4 𝑥 𝑥 𝑥 𝑥5 21. ln(1 + 𝑥) = 𝑥 − + − + −⋯ 2 3 4 5 𝑥3 𝑥5 𝑥7 + − +⋯ 22. sin 𝑥 = 𝑥 − 3! 5! 7! 3 5 𝑥 𝑥 𝑥7 23. arctan 𝑥 = 𝑥 − + − +⋯ 3 5 7 4 6 8 2 𝑥 𝑥 𝑥 24. 𝑒𝑥 = 1 + 𝑥2 + + + +⋯ 2! 3! 4! 6 10 𝑥 𝑥14 𝑥 + − +⋯ 25. 𝑥2 cos 𝑥2 = 𝑥2 − 2! 4! 6!

18.

PROBLEMS 26. Compute the binomial series expansion for (1 + 𝑥)3 . What do you notice? 27. The function 𝑓 is given by its Taylor series 𝑓 (𝑥) =

∞ ∑

(−1)𝑛−1

𝑛=1

32. One of the two sets of functions, 𝑓1 , 𝑓2 , 𝑓3 , or 𝑔1 , 𝑔2 , 𝑔3 is graphed in Figure 10.12; the other set is graphed in Figure 10.13. Taylor series for the functions about a point corresponding to either 𝐴 or 𝐵 are as follows:

2𝑛 𝑥𝑛 . 𝑛!

𝑓1 (𝑥) = 3 + (𝑥 − 1) − (𝑥 − 1)2 + ⋯ 𝑓2 (𝑥) = 3 − (𝑥 − 1) + (𝑥 − 1)2 + ⋯ 𝑓3 (𝑥) = 3 − 2(𝑥 − 1) + (𝑥 − 1)2 + ⋯

(a) Find 𝑓 (0). (b) At 𝑥 = 0, is 𝑓 increasing or decreasing? (c) At 𝑥 = 0, is the graph of 𝑓 concave up or concave down? 28. (a) Find the tangent line approximation to the function 𝑔 whose Taylor series is ∞

𝑔(𝑥) = ln 5 −

∑ 2𝑛 ⋅ (𝑛 − 1)! 𝑥𝑛 . 𝑛 5 𝑛=1

(b) What are the slope and 𝑦-intercept of this line? (c) Is the tangent line above or below the graph of 𝑔?

𝑔1 (𝑥) = 5 − (𝑥 − 4) − (𝑥 − 4)2 + ⋯ 𝑔2 (𝑥) = 5 − (𝑥 − 4) + (𝑥 − 4)2 + ⋯ 𝑔3 (𝑥) = 5 + (𝑥 − 4) + (𝑥 − 4)2 + ⋯ . (a) Which group of functions is represented in each figure? (b) What are the coordinates of the points 𝐴 and 𝐵? (c) Match each function with the graphs (I)–(III) in the appropriate figure. I

29. Using the Taylor series for 𝑓 (𝑥) = 𝑒𝑥 around 0, compute the following limit: lim 𝑥→0

𝑒𝑥 − 1 . 𝑥

I

𝐴 𝐵

𝑥2 −

𝑥6 𝑥10 𝑥14 + − +⋯ 3! 5! 7!

to find 𝑔 ′′ (0), 𝑔 ′′′ (0), and 𝑔 (10) (0). (There is an easy way and a hard way to do this!) 2

31. The Taylor series of 𝑓 (𝑥) = 𝑥2 𝑒𝑥 about 𝑥 = 0 is 𝑥6 𝑥8 𝑥10 + + +⋯. 2! 3! 4! ( )| )| ( 2 𝑑6 𝑑 | 2 𝑥2 | 𝑒 𝑥 𝑥2 𝑒𝑥 | and Find | . | | 𝑑𝑥 𝑑𝑥6 |𝑥=0 |𝑥=0 𝑥2 + 𝑥4 +

II

II

30. Use the fact that the Taylor series of 𝑔(𝑥) = sin(𝑥2 ) is

III

III

Figure 10.12

Figure 10.13

For Problems 33–36, what must be true of of 𝑓 (𝑛) (0) to ensure that the Taylor series of 𝑓 (𝑥) about 𝑥 = 0 has the given property? 33. All coefficients are positive. 34. Only has terms with even exponents. 35. Only has terms with odd exponents. 36. Coefficients for even exponent terms are negative, and coefficients for odd exponent terms are positive.

10.2 TAYLOR SERIES

1 37. By graphing the function 𝑓 (𝑥) = √ and several 1+𝑥 of its Taylor polynomials, estimate the interval of convergence of the series you found in Exercise 6. √ 38. By graphing the function 𝑓 (𝑥) = 1 + 𝑥 and several of its Taylor polynomials, estimate√where the series we found in Example 2 converges to 1 + 𝑥. 1 39. (a) By graphing the function 𝑓 (𝑥) = and sev1−𝑥 eral of its Taylor polynomials, estimate where the series you found in Exercise 5 converges to 1∕(1 − 𝑥) . (b) Compute the radius of convergence analytically. 40. Find the radius of convergence of the Taylor series around 𝑥 = 0 for 𝑒𝑥 .

52. Let 𝑓 (𝑥) = 1−2 cos 𝑥+

By recognizing each series in Problems 43–51 as a Taylor series evaluated at a particular value of 𝑥, find the sum of each of the following convergent series. 2 4 8 2𝑛 + + +⋯+ +⋯ 1! 2! 3! 𝑛! (−1)𝑛 1 1 1 + − +⋯+ +⋯ 44. 1 − 3! 5! 7! (2𝑛 + 1)! ( )𝑛 ( )2 ( )3 1 1 1 1 45. 1 + + + +⋯+ +⋯ 4 4 4 4 (−1)𝑛 ⋅ 102𝑛 100 10000 46. 1 − + +⋯+ +⋯ 2! 4! (2𝑛)! 43. 1 +

1 2 ( 21 )3 ( 21 )4 (−1)𝑛 ⋅ ( 12 )𝑛+1 1 (2) 47. − + − +⋯+ +⋯ 2 2 3 4 (𝑛 + 1)

48. 1 − 0.1 + 0.12 − 0.13 + ⋯ 9 27 81 49. 1 + 3 + + + +⋯ 2! 3! 4! 1 1 1 + − +⋯ 50. 1 − 2! 4! 6! 0.01 0.001 − +⋯ 51. 1 − 0.1 + 2! 3!

(−2)(−3)(−4) (−2)(−3) (cos 𝑥)2 + (cos 𝑥)3 +⋯ . 2! 3!

(a) By letting 𝑧 = cos 𝑥, rewrite 𝑓 (𝑥) as a rational function in 𝑧 by recognizing it as a binomial series in 𝑧. (b) Use part (a) to find the exact value of 𝑓 (𝜋∕3). In Problems 53–54 solve exactly for the variable. 53. 1 + 𝑥 + 𝑥2 + 𝑥3 + ⋯ = 5 1 1 54. 𝑥 − 𝑥2 + 𝑥3 + ⋯ = 0.2 2 3 In Problems 55–57, find the described value from the series

41. Find the radius of convergence of the Taylor series around 𝑥 = 0 for ln(1 − 𝑥). 42. (a) Write the general term of the binomial series for (1 + 𝑥)𝑝 about 𝑥 = 0. (b) Find the radius of convergence of this series.

529

𝐶(𝑥) =

∞ ∑ (−1)𝑛 𝑥4𝑛+1 . (2𝑛)!(4𝑛 + 1) 𝑛=0

55. The integer 𝑎 where 𝑥𝑘 ∕𝑎 is the term of the series corresponding to 𝑛 = 2. 56. The integer 𝑘 where 𝑥𝑘 ∕𝑎 is the term of the series corresponding to 𝑛 = 2. 57. 𝐶 ′ (0) 58. By recognizing the limit as a Taylor series, find the exact value of lim 𝑠𝑛 , given that 𝑠0 = 1 and 𝑠𝑛 = 𝑛→∞

1 𝑠𝑛−1 + ⋅ 2𝑛 for 𝑛 ≥ 0. 𝑛! √ 59. Let 𝑖 = −1. We define 𝑒𝑖𝜃 by substituting 𝑖𝜃 in the Taylor series for 𝑒𝑥 . Use this definition4 to explain Euler’s formula 𝑒𝑖𝜃 = cos 𝜃 + 𝑖 sin 𝜃.

∑ 60. The power series 𝐶𝑛 𝑥𝑛 converges to the function 𝑓 (𝑥) for |𝑥| < 𝑅. Assuming that you can differentiate term by term, show that 𝑓 (𝑛) (0) = 𝑛!𝐶𝑛 for all 𝑛. Thus the coefficients 𝐶𝑛 of the power series are the coefficients 𝑓 (𝑛) (0)∕𝑛! of the Taylor series for 𝑓 (𝑥).

Strengthen Your Understanding In Problems 61–62, explain what is wrong with the statement.

62. The radius of convergence is 2 for the following Taylor series: 1 + (𝑥 − 3) + (𝑥 − 3)2 + (𝑥 − 3)3 + ⋯.

61. Since

In Problems 63–64, give an example of:

1 = 1 + 𝑥 + 𝑥2 + 𝑥3 + ⋯ , 1−𝑥 we conclude that 1 = 1 + 2 + 22 + 23 + ⋯ . 1−2 4 Complex

numbers are discussed in Appendix B.

63. A function with a Taylor series whose third-degree term is zero. 64. A Taylor series that is convergent at 𝑥 = −1.

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Chapter 10 APPROXIMATING FUNCTIONS USING SERIES

Decide if the statements in Problems 65–71 are true or false. Assume that the Taylor series for a function converges to that function. Give an explanation for your answer.

(Assume the pattern of the coefficients continues.) 68. The Taylor series for 𝑓 converges everywhere 𝑓 is defined.

65. The Taylor series for sin 𝑥 about 𝑥 = 𝜋 is (𝑥 − 𝜋) −

(𝑥 − 𝜋)3 (𝑥 − 𝜋)5 + −⋯. 3! 5!

69. The graphs of 𝑒𝑥 and its Taylor polynomial 𝑃10 (𝑥) get further and further apart as 𝑥 → ∞.

66. If 𝑓 is an even function, then the Taylor series for 𝑓 near 𝑥 = 0 has only terms with even exponents. 67. If 𝑓 has the following Taylor series about 𝑥 = 0, then 𝑓 (7) (0) = −8: 3 4 𝑓 (𝑥) = 1 − 2𝑥 + 𝑥2 − 𝑥3 + ⋯ . 2! 3!

10.3

70. If the Taylor series for 𝑓 (𝑥) around 𝑥 = 0 has a finite number of terms and an infinite radius of convergence, then 𝑓 (𝑥) is a polynomial. 71. If 𝑓 (𝑥) is a polynomial, then its Taylor series has an infinite radius of convergence.

FINDING AND USING TAYLOR SERIES Finding a Taylor series for a function means finding the coefficients. Assuming the function has all its derivatives defined, finding the coefficients can always be done, in theory at least, by differentiation. That is how we derived the four most important Taylor series, those for the functions 𝑒𝑥 , sin 𝑥, cos 𝑥, and (1 + 𝑥)𝑝 . For many functions, however, computing Taylor series coefficients by differentiation can be a very laborious business. We now introduce easier ways of finding Taylor series, if the series we want is closely related to a series that we already know.

New Series by Substitution 2

Suppose we want to find the Taylor series for 𝑒−𝑥 about 𝑥 = 0. We could find the coefficients by 2 2 differentiation. Differentiating 𝑒−𝑥 by the chain rule gives −2𝑥𝑒−𝑥 , and differentiating again gives 2 2 −2𝑒−𝑥 + 4𝑥2 𝑒−𝑥 . Each time we differentiate we use the product rule, and the number of terms 2 2 grows. Finding the tenth or twentieth derivative of 𝑒−𝑥 , and thus the series for 𝑒−𝑥 up to the 𝑥10 or 20 𝑥 terms, by this method is tiresome (at least without a computer or calculator that can differentiate). Fortunately, there’s a quicker way. Recall that 𝑒𝑦 = 1 + 𝑦 +

𝑦2 𝑦3 𝑦4 + + +⋯ 2! 3! 4!

for all 𝑦.

Substituting 𝑦 = −𝑥2 tells us that (−𝑥2 )2 (−𝑥2 )3 (−𝑥2 )4 + + +⋯ 2! 3! 4! 𝑥4 𝑥6 𝑥8 − + − ⋯ for all 𝑥. = 1 − 𝑥2 + 2! 3! 4! Using this method, it is easy to find the series up to the 𝑥10 or 𝑥20 terms. It can be shown that this is 2 the Taylor series for 𝑒−𝑥 . 2

𝑒−𝑥 = 1 + (−𝑥2 ) +

1 . 1 + 𝑥2

Example 1

Find the Taylor series about 𝑥 = 0 for 𝑓 (𝑥) =

Solution

The binomial series, or geometric series, tells us that 1 = (1 + 𝑦)−1 = 1 − 𝑦 + 𝑦2 − 𝑦3 + 𝑦4 − ⋯ 1+𝑦

for −1 < 𝑦 < 1.

Substituting 𝑦 = 𝑥2 gives 1 = 1 − 𝑥2 + 𝑥4 − 𝑥6 + 𝑥8 − ⋯ 1 + 𝑥2

for −1 < 𝑥 < 1,

10.3 FINDING AND USING TAYLOR SERIES

which is the Taylor series for

531

1 . 1 + 𝑥2

Notice that substitution can affect the radius of convergence. For example, in Example 1 if we find the Taylor series for 𝑔(𝑥) = 1∕(1 + 4𝑥2 ) by substituting 𝑦 = 4𝑥2 , the interval of convergence changes from −1 < 𝑦 < 1 to −1∕2 < 𝑥 < 1∕2.

New Series by Differentiation and Integration Just as we can get new series by substitution, we can also get new series by differentiation and integration. Term-by-term differentiation of a Taylor series for 𝑓 (𝑥) gives a Taylor series for 𝑓 ′ (𝑥); antidifferentiation works similarly. Example 2

Find the Taylor series about 𝑥 = 0 for

Solution

We know that

1 1 . from the series for 2 1 − 𝑥 (1 − 𝑥)

( ) 1 1 𝑑 = , so we start with the geometric series 𝑑𝑥 1 − 𝑥 (1 − 𝑥)2 1 = 1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4 + ⋯ for − 1 < 𝑥 < 1. 1−𝑥

Differentiation term by term gives the binomial series ( ) 1 1 𝑑 = 1 + 2𝑥 + 3𝑥2 + 4𝑥3 + ⋯ for − 1 < 𝑥 < 1. = 𝑑𝑥 1 − 𝑥 (1 − 𝑥)2

Example 3

Find the Taylor series5 about 𝑥 = 0 for arctan 𝑥 from the series for

Solution

We know that

1 . 1 + 𝑥2

1 𝑑 (arctan 𝑥) = , so we use the series from Example 1: 𝑑𝑥 1 + 𝑥2

1 𝑑 (arctan 𝑥) = = 1 − 𝑥2 + 𝑥4 − 𝑥6 + 𝑥8 − ⋯ 𝑑𝑥 1 + 𝑥2

for −1 < 𝑥 < 1.

Antidifferentiating term by term and assuming convergence gives arctan 𝑥 =

1 𝑥3 𝑥5 𝑥7 𝑥9 + − + −⋯ 𝑑𝑥 = 𝐶 + 𝑥 − ∫ 1 + 𝑥2 3 5 7 9

for − 1 < 𝑥 < 1,

where 𝐶 is the constant of integration. Since arctan 0 = 0, we have 𝐶 = 0, so arctan 𝑥 = 𝑥 −

𝑥3 𝑥5 𝑥7 𝑥9 + − + −⋯ 3 5 7 9

for − 1 < 𝑥 < 1.

Convergence of a Series and Differentiation Example 2 assumed that the series created by term-by-term differentiation converged, and to the function we wanted. The following theorem, which is proved in more advanced courses, justifies the assumption. Problem 59 confirms that the radius of convergence remains the same under term-byterm differentiation. 5 The

series for arctan 𝑥 was discovered by James Gregory (1638–1675).

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Chapter 10 APPROXIMATING FUNCTIONS USING SERIES

Theorem 10.2: Convergence of the Term-by-Term Derivative If a Taylor series for 𝑓 (𝑥) at 𝑥 = 𝑎 converges to 𝑓 (𝑥) for |𝑥 − 𝑎| < 𝑅, then the series found by term-by-term differentiation is the Taylor series for 𝑓 ′ (𝑥) and converges to 𝑓 ′ (𝑥) on the same interval |𝑥 − 𝑎| < 𝑅.

Multiplying and Substituting Taylor Series

We can also form a Taylor series for a product of two functions. In some cases, this is easy; for example, if we want to find the Taylor series about 𝑥 = 0 for the function 𝑓 (𝑥) = 𝑥2 sin 𝑥, we can start with the Taylor series for sin 𝑥, sin 𝑥 = 𝑥 −

𝑥3 𝑥5 𝑥7 + − + ⋯, 3! 5! 7!

and multiply the series by 𝑥2 : ) ( 𝑥3 𝑥5 𝑥7 + − +⋯ 𝑥2 sin 𝑥 = 𝑥2 𝑥 − 3! 5! 7! = 𝑥3 −

𝑥5 𝑥7 𝑥9 + − +⋯. 3! 5! 7!

However, in some cases, finding a Taylor series for a product of two functions requires more work. Example 4

Find the Taylor series about 𝑥 = 0 for 𝑔(𝑥) = sin 𝑥 cos 𝑥.

Solution

The Taylor series about 𝑥 = 0 for sin 𝑥 and cos 𝑥 are 𝑥3 𝑥5 + −⋯ 3! 5! 𝑥2 𝑥4 + −⋯. cos 𝑥 = 1 − 2! 4! sin 𝑥 = 𝑥 −

So we have 𝑔(𝑥) = sin 𝑥 cos 𝑥 =

( )( ) 𝑥3 𝑥 5 𝑥2 𝑥4 𝑥− + −⋯ 1− + −⋯ . 3! 5! 2! 4!

To multiply these two series, we must multiply each term of the series for sin 𝑥 by each term of the series for cos 𝑥. Because each series has infinitely many terms, we organize the process by first determining the constant term of the product, then the linear term, and so on. The constant term of this product is zero because there is no combination of a term from the first series and a term from the second that yields a constant. The linear term of the product is 𝑥; we obtain this term by multiplying the 𝑥 from the first series by the 1 from the second. The degree-2 term of the product is also zero; more generally, we notice that every even-degree term of the product is zero because every combination of a term from the first series and a term from the second yields an odddegree term. To find the degree-3 term, observe that the combinations of terms that yield degree-3 2 3 terms are 𝑥 ⋅ − 𝑥2! = − 21 𝑥3 and − 𝑥3! ⋅ 1 = − 16 𝑥3 , and thus the degree-3 term is − 12 𝑥3 − 61 𝑥3 = − 23 𝑥3 . Continuing in this manner, we find that 2 2 𝑔(𝑥) = 𝑥 − 𝑥3 + 𝑥5 − ⋯ . 3 15 There is another way to find this series. Notice that sin 𝑥 cos 𝑥 = Taylor series about 𝑥 = 0 for the sine function, we get sin(2𝑥) = 2𝑥 −

1 2

sin(2𝑥). Substituting 2𝑥 into the

(2𝑥)3 (2𝑥)5 + −⋯ 3! 5!

10.3 FINDING AND USING TAYLOR SERIES

533

4 4 = 2𝑥 − 𝑥3 + 𝑥5 − ⋯ . 3 15 Therefore, we have 𝑔(𝑥) =

2 2 1 sin(2𝑥) = 𝑥 − 𝑥3 + 𝑥5 − ⋯ . 2 3 15

We can also obtain a Taylor series for a composite function by substituting a Taylor series into another one, as in the next example. Example 5

Find the Taylor series about 𝜃 = 0 for 𝑔(𝜃) = 𝑒sin 𝜃 .

Solution

For all 𝑦 and 𝜃, we know that 𝑒𝑦 = 1 + 𝑦 +

𝑦2 𝑦3 𝑦4 + + +⋯ 2! 3! 4!

and sin 𝜃 = 𝜃 −

𝜃3 𝜃5 + −⋯. 3! 5!

Let’s substitute the series for sin 𝜃 for 𝑦: ) ( )2 ( )3 ( 1 𝜃3 𝜃5 𝜃3 𝜃5 𝜃3 𝜃5 1 𝑒sin 𝜃 = 1 + 𝜃 − + −⋯ + 𝜃− + −⋯ + 𝜃− + −⋯ +⋯. 3! 5! 2! 3! 5! 3! 3! 5! To simplify, we multiply out and collect terms. The only constant term is the 1, and there’s only one 𝜃 term. The only 𝜃 2 term is the first term we get by multiplying out the square, namely 𝜃 2 ∕2!. There are two contributors to the 𝜃 3 term: the −𝜃 3 ∕3! from within the first parentheses and the first term we get from multiplying out the cube, which is 𝜃 3 ∕3!. Thus the series starts ( 3 ) 𝜃 𝜃3 𝜃2 𝑒sin 𝜃 = 1 + 𝜃 + + − + +⋯ 2! 3! 3! =1+𝜃+

𝜃2 + 0 ⋅ 𝜃3 + ⋯ 2!

for all 𝜃.

Applications of Taylor Series Example 6

Use the series for arctan 𝑥 to estimate the numerical value of 𝜋.

Solution

Since arctan 1 = 𝜋∕4, we use the series for arctan 𝑥 from Example 3. We assume—as is the case— that the series does converge to 𝜋∕4 at 𝑥 = 1. Substituting 𝑥 = 1 into the series for arctan 𝑥 gives ) ( 1 1 1 1 𝜋 = 4 arctan 1 = 4 1 − + − + − ⋯ . 3 5 7 9 Table 10.1

Approximating 𝜋 using the series for arctan 𝑥

𝑛

4

5

25

100

500

1000

10,000

𝑆𝑛

2.895

3.340

3.182

3.132

3.140

3.141

3.141

Table 10.1 shows the value of the 𝑛th partial sum, 𝑆𝑛 , obtained by summing the first 𝑛 nonzero terms. The values of 𝑆𝑛 do seem to converge to 𝜋 = 3.141 … . However, this series converges very slowly, meaning that we have to take a large number of terms to get an accurate estimate for 𝜋. So this way of calculating 𝜋 is not particularly practical. (A better one is given in Project 2 (available online).) However, the expression for 𝜋 given by this series is surprising and elegant.

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Chapter 10 APPROXIMATING FUNCTIONS USING SERIES

A basic question we can ask about two functions is which one gives larger values. Taylor series can often be used to answer this question over a small interval. If the constant terms of the series for two functions are the same, compare the linear terms; if the linear terms are the same, compare the quadratic terms, and so on. Example 7

By looking at their Taylor series, decide which of the following functions is largest for 𝑡 near 0. 1 (a) 𝑒𝑡 (b) 1−𝑡

Solution

The Taylor expansion about 𝑡 = 0 for 𝑒𝑡 is 𝑒𝑡 = 1 + 𝑡 +

𝑡3 𝑡2 + +⋯. 2! 3!

Viewing 1∕(1 − 𝑡) as the sum of a geometric series with initial term 1 and common ratio 𝑡, we have 1 = 1 + 𝑡 + 𝑡2 + 𝑡3 + ⋯ for − 1 < 𝑡 < 1. 1−𝑡 Notice that these two series have the same constant term and the same linear term. However, their remaining terms are different. For values of 𝑡 near zero, the quadratic terms dominate all of the subsequent terms,6 so we can use the approximations 𝑒𝑡 ≈ 1 + 𝑡 +

𝑡2 2

1 ≈ 1 + 𝑡 + 𝑡2 . 1−𝑡 Since

1 1 + 𝑡 + 𝑡 2 < 1 + 𝑡 + 𝑡2 , 2 and since the approximations are valid for 𝑡 near 0, we conclude that, for 𝑡 near 0, 𝑒𝑡
0 23. √ 𝑎 𝑎2 + 𝑥2 (b) Explain why the graph of 𝑒𝑥 − 𝑒−𝑥 near 𝑥 = 0 looks like the graph of a cubic polynomial symmetric about the origin. What is the equation for this cubic? 34. Find the first three terms of the Taylor series for 𝑓 (𝑥) = 2 𝑒𝑥 around 0. Use this information to approximate the integral

35. Find the sum of

∞ ∑ 𝑝=1

2

𝑒𝑥 𝑑𝑥.

∫0

𝑝𝑥𝑝−1 for |𝑥| < 1.

36. For values of 𝑦 near 0, put the following functions in increasing order, using their Taylor expansions. (a) ln(1 + 𝑦2 )

(b) sin(𝑦2 )

(c)

1 − cos 𝑦

37. For values of 𝜃 near 0, put the following functions in increasing order, using their Taylor expansions. (a) 1 + sin 𝜃

(b) 𝑒𝜃

(c)

1 √ 1 − 2𝜃

38. A function has the following Taylor series about 𝑥 = 0:

31. Use the series for 𝑒𝑥 to find the Taylor series for sinh 2𝑥 and cosh 2𝑥. 32. (a) Find the first three nonzero terms of the Taylor series for 𝑒𝑥 + 𝑒−𝑥 . (b) Explain why the graph of 𝑒𝑥 + 𝑒−𝑥 looks like a parabola near 𝑥 = 0. What is the equation of this parabola?

(𝑎 +

𝑟 𝑟)2

22.

1

30. The Taylor series for ln(1 − 𝑥) about 𝑥 = 0 converges for −1 < 𝑥 < 1. For each of the following functions, use this interval to find the largest interval on which the function’s Taylor series about 𝑥 = 0 converges.

1

𝑟 in terms of

28. Let 𝑔(𝑧) be the function obtained by substituting 𝑧 = 𝑥 − 1 into 𝑓 (𝑥) = 1∕𝑥. Use a series for 𝑔(𝑧) to get a Taylor series for 𝑓 (𝑥) around 𝑥 = 1. 29. (a) Let 𝑔(𝑧) be the function obtained by substituting 𝑧 = (𝑥 − 1)∕2 into 𝑓 (𝑥) = 1∕(𝑥 + 1). Use a series for ℎ(𝑧) to find a Taylor series for 𝑓 (𝑥) around 𝑥 = 1. (b) Find the Taylor series of 𝑔(𝑥) = (𝑥 − 1)2 ∕(𝑥 + 1) around 𝑥 = 1.

√ ℎ 𝑇 + ℎ in terms of 𝑇

𝑓 (𝑥) =

∞ ∑ 𝑥2𝑛+1 . 2𝑛 + 1 𝑛=0

Find the ninth-degree Taylor polynomial for 𝑓 (2𝑥). 39. Figure 10.16 shows the graphs of the four functions below for values of 𝑥 near 0. Use Taylor series to match graphs and formulas. (a)

1 1 − 𝑥2

(b) (1 + 𝑥)1∕4

10.3 FINDING AND USING TAYLOR SERIES

(c)



1+

𝑥 2

(d)

1 √ 1−𝑥

46. One of Einstein’s most amazing predictions was that light traveling from distant stars would bend around the sun on the way to earth. His calculations involved solving for 𝜙 in the equation

(III)

(I)

sin 𝜙 + 𝑏(1 + cos2 𝜙 + cos 𝜙) = 0,

(IV)

where 𝑏 is a very small positive constant. (a) Explain why the equation could have a solution for 𝜙 which is near 0. (b) Expand the left-hand side of the equation in Taylor series about 𝜙 = 0, disregarding terms of order 𝜙2 and higher. Solve for 𝜙. (Your answer will involve 𝑏.)

(II)

Figure 10.16 40. The sine integral function is defined by the improper in𝑥 sin 𝑡 tegral Si(𝑥) = 𝑑𝑡. Use the Taylor polynomial, ∫0 𝑡 𝑃7 (𝑥), of degree 7 about 𝑥 = 0 for the sine function to estimate Si(2). 41. Write out the first four nonzero terms of the Taylor se𝑥 ( ) sin 𝑡2 𝑑𝑡. ries about 𝑥 = 0 for 𝑓 (𝑥) = ∫0

42. (a) Find the Taylor series for 𝑓 (𝑡) = 𝑡𝑒𝑡 about 𝑡 = 0. (b) Using your answer to part (a), find a Taylor series expansion about 𝑥 = 0 for 𝑥

∫0

𝑡𝑒𝑡 𝑑𝑡.

(c) Using your answer to part (b), show that 1 1 1 1 1 + + + + + ⋯ = 1. 2 3 4(2!) 5(3!) 6(4!) ∞ ∑ 𝑘𝑛−1 −𝑘 43. Find the sum of 𝑒 . (𝑛 − 1)! 𝑛=1

44. Use Taylor series to explain the patterns in the digits in the following expansions: 1 = 1.02040816 … 0.98 ) ( 1 2 = 1.020304050607 … (b) 0.99 (a)

45. Padé approximants are rational functions used to approximate more complicated functions. In this problem, you will derive the Padé approximant to the exponential function. (a) Let 𝑓 (𝑥) = (1 + 𝑎𝑥)∕(1 + 𝑏𝑥), where 𝑎 and 𝑏 are constants. Write down the first three terms of the Taylor series for 𝑓 (𝑥) about 𝑥 = 0. (b) By equating the first three terms of the Taylor series about 𝑥 = 0 for 𝑓 (𝑥) and for 𝑒𝑥 , find 𝑎 and 𝑏 so that 𝑓 (𝑥) approximates 𝑒𝑥 as closely as possible near 𝑥 = 0. 7 http://en.wikipedia.org/wiki/Pendulum.

537

47. A hydrogen atom consists of an electron, of mass 𝑚, orbiting a proton, of mass 𝑀, where 𝑚 is much smaller than 𝑀. The reduced mass, 𝜇, of the hydrogen atom is defined by 𝑚𝑀 . 𝜇= 𝑚+𝑀 (a) Show that 𝜇 ≈ 𝑚. (b) To get a more accurate approximation for 𝜇, express 𝜇 as 𝑚 times a series in 𝑚∕𝑀. (c) The approximation 𝜇 ≈ 𝑚 is obtained by disregarding all but the constant term in the series. The first-order correction is obtained by including the linear term but no higher terms. If 𝑚 ≈ 𝑀∕1836, by what percentage does including the linear term change the estimate 𝜇 ≈ 𝑚? 48. Resonance in electric circuits leads to the expression ) ( 1 2 , 𝜔𝐿 − 𝜔𝐶 where 𝜔 is the variable and 𝐿 and 𝐶 are constants. (a) Find 𝜔0 , the value of 𝜔 making the expression zero. (b) In practice, 𝜔 fluctuates about 𝜔0 , so we are interested in the behavior of this expression for values of 𝜔 near 𝜔0 . Let 𝜔 = 𝜔0 + Δ𝜔 and expand the expression in terms of Δ𝜔 up to the first nonzero term. Give your answer in terms of Δ𝜔 and 𝐿 but not 𝐶. 49. A pendulum consists of a mass, 𝑚, swinging on the end of a string of length 𝑙. With the angle between the string and the vertical represented by 𝜃, the motion satisfies the differential equation 𝑔 𝜃 ′′ + 𝑔 sin 𝜃 = 0. 𝑙 (a) For small swings, we can replace sin 𝜃 by its lowest nonzero Taylor approximation. What does the differential equation become? (b) If the amplitude of the oscillation is 𝜃0 , the solutions to the original differential equation are oscillations with7 √ ( ) 𝑙 1 1 + 𝜃02 + ⋯ . Period = 2𝜋 𝑔 16

Accessed April, 2015.

538

Chapter 10 APPROXIMATING FUNCTIONS USING SERIES

The solutions to the approximate differential equation are oscillations with √ 𝑙 . Period = 2𝜋 𝑔 If 𝜃0 = 20◦ , by what percentage is the more accurate estimate of the period obtained using the solution to the original equation up to the 𝜃02 -term different to the approximate estimate using the solution of the approximate equation? 50. Stefan’s Law says the rate at which a body’s temperature, 𝐻, changes with time is proportional to 𝐴4 − 𝐻 4 , where 𝐴 is the surrounding temperature.8 Use a Taylor series to expand 𝐴4 − 𝐻 4 up to the linear term. 51. The Michelson-Morley experiment, which contributed to the formulation of the theory of relativity, involved the difference between the two times 𝑡1 and 𝑡2 that light took to travel between two points. If 𝑣 is velocity; 𝑙1 , 𝑙2 , and 𝑐 are constants; and 𝑣 < 𝑐, then 𝑡1 and 𝑡2 are given by 2𝑙2 2𝑙 − √ 1 𝑐(1 − 𝑣2 ∕𝑐 2 ) 𝑐 1 − 𝑣2 ∕𝑐 2 2𝑙1 2𝑙 − 𝑡2 = √ 2 . 𝑐(1 − 𝑣2 ∕𝑐 2 ) 2 2 𝑐 1 − 𝑣 ∕𝑐 𝑡1 =

(a) Find an expression for Δ𝑡 = 𝑡1 − 𝑡2 , and give its Taylor expansion in terms of 𝑣2 ∕𝑐 2 up to the second nonzero term. (b) For small 𝑣, to what power of 𝑣 is Δ𝑡 proportional? What is the constant of proportionality? 52. The theory of relativity predicts that when an object moves at speeds close to the speed of light, the object appears heavier. The apparent, or relativistic, mass, 𝑚, of the object when it is moving at speed 𝑣 is given by the formula 𝑚0 𝑚= √ , 1 − 𝑣2 ∕𝑐 2

where 𝑉0 and 𝑟0 are positive constants. (a) Show that if 𝑟 = 𝑟0 , then 𝑉 takes on its minimum value, −𝑉0 . (b) Write 𝑉 as a series in (𝑟 − 𝑟0 ) up through the quadratic term. (c) For 𝑟 near 𝑟0 , show that the difference between 𝑉 and its minimum value is approximately proportional to (𝑟 − 𝑟0 )2 . In other words, show that 𝑉 − (−𝑉0 ) = 𝑉 + 𝑉0 is approximately proportional to (𝑟 − 𝑟0 )2 . (d) The force, 𝐹 , between the molecules is given by 𝐹 = −𝑑𝑉 ∕𝑑𝑟. What is 𝐹 when 𝑟 = 𝑟0 ? For 𝑟 near 𝑟0 , show that 𝐹 is approximately proportional to (𝑟 − 𝑟0 ). 54. Van der Waal’s equation relates the pressure, 𝑃 , and the volume, 𝑉 , of a fixed quantity of a gas at constant temperature 𝑇 : ( ) 𝑛2 𝑎 𝑃 + 2 (𝑉 − 𝑛𝑏) = 𝑛𝑅𝑇 , 𝑉 where 𝑎, 𝑏, 𝑛, 𝑅 are constants. Find the first two nonzero terms of the Taylor series of 𝑃 in terms for 1∕𝑉 . 55. The hyperbolic sine and cosine are differentiable and satisfy the conditions cosh 0 = 1 and sinh 0 = 0, and 𝑑 (cosh 𝑥) = sinh 𝑥 𝑑𝑥

(a) Using only this information, find the Taylor approximation of degree 𝑛 = 8 about 𝑥 = 0 for 𝑓 (𝑥) = cosh 𝑥. (b) Estimate the value of cosh 1. (c) Use the result from part (a) to find a Taylor polynomial approximation of degree 𝑛 = 7 about 𝑥 = 0 for 𝑔(𝑥) = sinh 𝑥. In Problems 56–57, find the value √of the integer 𝑎𝑖 given that the Taylor series of 𝑓 (𝑥) = cos 𝑥 is: 𝑎0 +

where 𝑐 is the speed of light and 𝑚0 is the mass of the object when it is at rest.

(a) Use the formula for 𝑚 to decide what values of 𝑣 are possible. (b) Sketch a rough graph of 𝑚 against 𝑣, labeling intercepts and asymptotes. (c) Write the first three nonzero terms of the Taylor series for 𝑚 in terms of 𝑣. (d) For what values of 𝑣 do you expect the series to converge? 53. The potential energy, 𝑉 , of two gas molecules separated by a distance 𝑟 is given by ( ( )6 ( )12 ) 𝑟 𝑟 𝑉 = −𝑉0 2 0 − 0 , 𝑟 𝑟 8 http://en.wikipedia.org/wiki/StefanBoltzmann_law.

𝑑 (sinh 𝑥) = cosh 𝑥. 𝑑𝑥

1 1 1 𝑥 + 𝑥2 + 𝑥3 + ⋯ . 𝑎1 𝑎2 𝑎3 57. 𝑎3

56. 𝑎2

𝑔(𝑥)

58. Find 𝑓 (42) for 𝑓 (𝑥) =

∫1

(1∕𝑡) 𝑑𝑡 where

𝑥

𝑔(𝑥) = lim

𝑏→−∞ ∫𝑏

ℎ(𝑡) 𝑑𝑡,

and

ℎ(𝑥) =

∞ ∑ 𝑥𝑘 . 𝑘! 𝑘=0

∑ 59. The ratio test applied to the power series 𝐶𝑛 𝑥𝑛 gives 𝑅 as the radius of convergence. Show that the term-byterm derivative of the power series has the same radius of convergence. Accessed May 2015.

10.4 THE ERROR IN TAYLOR POLYNOMIAL APPROXIMATIONS

539

Strengthen Your Understanding In Problems 60–61, explain what is wrong with the statement.

65. The Taylor series for 𝑥3 cos 𝑥 about 𝑥 = 0 has only odd powers.

60. Within its radius of convergence, ( )2 ( )3 𝑥 𝑥 𝑥 1 =1− + − + ⋯. 2+𝑥 2 2 2

66. The Taylor series for 𝑓 (𝑥)𝑔(𝑥) about 𝑥 = 0 is

61. Using the Taylor series for 𝑒𝑥 = 1 + 𝑥 + we find that 𝑒−𝑥 = 1 − 𝑥 −

𝑓 (0)𝑔(0) + 𝑓 ′ (0)𝑔 ′ (0)𝑥 +

𝑥2 𝑥3 + +⋯, 2! 3!

𝑥2 𝑥3 − −⋯. 2! 3!

In Problems 62–63, give an example of: 62. A function with no Taylor series around 0. 63. A function 𝑓 (𝑥) that does not have a Taylor series around 0 even though 𝑓 (0) is defined. Decide if the statements in Problems 64–68 are true or false. Assume that the Taylor series for a function converges to that function. Give an explanation for your answer.

67. If 𝐿1 (𝑥) is the linear approximation to 𝑓1 (𝑥) near 𝑥 = 0 and 𝐿2 (𝑥) is the linear approximation to 𝑓2 (𝑥) near 𝑥 = 0, then 𝐿1 (𝑥) + 𝐿2 (𝑥) is the linear approximation to 𝑓1 (𝑥) + 𝑓2 (𝑥) near 𝑥 = 0. 68. If 𝐿1 (𝑥) is the linear approximation to 𝑓1 (𝑥) near 𝑥 = 0 and 𝐿2 (𝑥) is the linear approximation to 𝑓2 (𝑥) near 𝑥 = 0, then 𝐿1 (𝑥)𝐿2 (𝑥) is the quadratic approximation to 𝑓1 (𝑥)𝑓2 (𝑥) near 𝑥 = 0. 69. Given that the Taylor series for tan 𝑥 = 𝑥 + 𝑥3 ∕3 + 21𝑥5 ∕120 + ⋯, then that of 3 tan(𝑥∕3) is (a) (b) (c) (d)

64. To find the Taylor series for sin 𝑥 + cos 𝑥 about any point, add the Taylor series for sin 𝑥 and cos 𝑥 about that point.

10.4

𝑓 ′′ (0)𝑔 ′′ (0) 2 𝑥 + ⋯. 2!

3𝑥 + 𝑥3 + 21𝑥5 ∕120 + ⋯ 3𝑥 + 𝑥3 + 21𝑥5 ∕40 + ⋯ 𝑥 + 𝑥3 ∕27 + 7𝑥5 ∕3240 + ⋯ 𝑥 + 𝑥3 ∕3 + 21𝑥5 ∕120 + ⋯

THE ERROR IN TAYLOR POLYNOMIAL APPROXIMATIONS In order to use an approximation with confidence, we need to know how big the error could be. The error is the difference between the exact answer and the approximate value. If 𝑃𝑛 (𝑥) is the 𝑛th -degree Taylor polynomial, the error in approximating 𝑓 (𝑥) is 𝐸𝑛 (𝑥) = 𝑓 (𝑥) − 𝑃𝑛 (𝑥). We have seen that the higher the degree of the Taylor polynomial, in general the larger the interval on which the function and the polynomial are close to each other. For example, Figure 10.17 shows the first three Taylor polynomial approximations about 0 for the function 𝑓 (𝑥) = 𝑒𝑥 . If we pick a particular Taylor polynomial, say 𝑃2 , the closer 𝑥 is to 0, the better 𝑃2 (𝑥) approximates 𝑒𝑥 . 𝑓 (𝑥) = 𝑒𝑥 𝑃2 4 𝑃1

2 𝑃0 𝑥 −2

2

Figure 10.17: Graph of 𝑓 (𝑥) = 𝑒𝑥 and first three Taylor polynomials around 𝑥 = 0

540

Chapter 10 APPROXIMATING FUNCTIONS USING SERIES

Example 1

Compare the errors in the approximations 𝑒0.1 ≈ 1 + 0.1 +

Solution

1 (0.1)2 2!

and

𝑒0.05 ≈ 1 + (0.05) +

1 (0.05)2. 2!

We are approximating 𝑒𝑥 by its second-degree Taylor polynomial about 0. At 𝑥 = 0.1 and 𝑥 = 0.05 the errors are: ) ( 1 𝐸2 (0.1) = 𝑒0.1 − 1 + 0.1 + (0.1)2 = 1.105171 … − 1.105000 = 0.000171 2! ) ( 1 𝐸2 (0.05) = 𝑒0.05 − 1 + 0.05 + (0.05)2 = 1.051271 … − 1.051250 = 0.000021. 2! As expected, the error in the approximation is smaller for 𝑥 = 0.05 than for 𝑥 = 0.1. In Example 1 we obtained a better approximation by taking 𝑥 closer 0. Example 2 shows if we are interested in a particular value of 𝑥, we can often improve the approximation by choosing a higher-degree polynomial.

Example 2

Compare the errors in the approximations for 𝑒0.1 using 𝑃2 and 𝑃3 .

Solution

In the previous example we saw 𝐸2 (0.1) = 0.000171. We now compute ) ( 1 1 𝐸3 (0.1) = 𝑒0.1 − 𝑃3 (0.1) = 𝑒0.1 − 1 + 0.1 + (0.1)2 + (0.1)3 = 1.105171 … − 1.1051666 … 2! 3! = 0.0000043 …

Bounds on the Error In practice, we usually want to find a bound on the magnitude of the error—we want to figure out how big it could be—without calculating the error exactly. Lagrange found an expression for an error bound 𝐸𝑛 based on the degree-(𝑛 + 1) term in the Taylor series that is used to make such estimates:

Theorem 10.3: The Lagrange Error Bound for 𝑷𝒏 (𝒙) Suppose 𝑓 and all its derivatives are continuous. If 𝑃𝑛 (𝑥) is the degree-𝑛 Taylor polynomial for 𝑓 (𝑥) about 𝑎, and 𝐸𝑛 (𝑥) = 𝑓 (𝑥) − 𝑃𝑛 (𝑥) is the error function, then |𝐸𝑛 (𝑥)| ≤

𝑀 |𝑥 − 𝑎|𝑛+1 , (𝑛 + 1)!

| | where 𝑀 is an upper bound on |𝑓 (𝑛+1) | on the interval between 𝑎 and 𝑥. | |

To find 𝑀 in practice, we often find the maximum of |𝑓 (𝑛+1) | on the interval and pick any larger value for 𝑀. See page 543 for a justification of Theorem 10.3.

Using the Lagrange Error Bound for Taylor Polynomials Example 3

Give a bound on the error, 𝐸4 , when 𝑒𝑥 is approximated by its fourth-degree Taylor polynomial about 0 for −0.5 ≤ 𝑥 ≤ 0.5.

10.4 THE ERROR IN TAYLOR POLYNOMIAL APPROXIMATIONS

Solution

541

Let 𝑓 (𝑥) = 𝑒𝑥 . We need to find a bound for the fifth derivative, 𝑓 (5) (𝑥) = 𝑒𝑥 . Since 𝑒𝑥 is positive and increasing, its largest value is at the right endpoint of the interval and: √ |𝑓 (5) (𝑥)| ≤ 𝑒0.5 = 𝑒 for −0.5 ≤ 𝑥 ≤ 0.5. Since



𝑒 < 2, we can take 𝑀 = 2 (or any larger value). Then |𝐸4 | = |𝑓 (𝑥) − 𝑃4 (𝑥)| ≤

2 5 |𝑥| . 5!

This means, for example, that on −0.5 ≤ 𝑥 ≤ 0.5, the approximation 𝑒𝑥 ≈ 1 + 𝑥 + has an error of at most

2 (0.5)5 120

𝑥2 𝑥3 𝑥4 + + 2! 3! 4!

< 0.0006.

The Lagrange error bound for Taylor polynomials can be used to see how the accuracy of the approximation depends on the value of 𝑥 and the value of 𝑛. Observe that the error bound for a Taylor polynomial of degree 𝑛 is proportional to |𝑥 − 𝑎|𝑛+1 . That means, for example, with a Taylor polynomial of degree 𝑛 centered at 0, if we decrease 𝑥 by a factor of 2, the error bound decreases by a factor of 2𝑛+1 . We saw this happen in Example 1 where 𝐸2 (0.1) = 0.000171 and 𝐸2 (0.05) = 0.000021. Since 𝐸2 (0.1) 0.000171 = = 8.1, 𝐸2 (0.05) 0.000021 we see that as the value of 𝑥 decreases by a factor of 2, the error decreases by a factor of about 8 = 23 .

Convergence of the Taylor Series for cos 𝒙 We have already seen that the Taylor polynomials centered at 𝑥 = 0 for cos 𝑥 are good approximations for 𝑥 near 0. (See Figure 10.18.) In fact, for any value of 𝑥, if we take a Taylor polynomial centered at 𝑥 = 0 of high enough degree, its graph is nearly indistinguishable from the graph of the cosine function near that point. 𝑃8 (𝑥)

𝑃8 (𝑥) 1

cos 𝑥

cos 𝑥 𝑥

−𝜋

𝜋 −1

𝑃2 (𝑥)

𝑃2 (𝑥)

Figure 10.18: Graph of cos 𝑥 and two Taylor polynomials for 𝑥 near 0

Let’s see what happens numerically. Let 𝑥 = 𝜋∕2. The successive Taylor polynomial approximations to cos(𝜋∕2) = 0 about 𝑥 = 0 are 𝑃2 (𝜋∕2) =

1 − (𝜋∕2)2 ∕2!

= −0.23370 …

𝑃4 (𝜋∕2) = 𝑃6 (𝜋∕2) =

1 − (𝜋∕2)2 ∕2! + (𝜋∕2)4 ∕4! ⋯

= 0.01997 … = −0.00089 …

𝑃8 (𝜋∕2) =



= 0.00002 … .

542

Chapter 10 APPROXIMATING FUNCTIONS USING SERIES

It appears that the approximations converge to the true value, cos(𝜋∕2) = 0, very rapidly. Now take a value of 𝑥 somewhat farther away from 0, say 𝑥 = 𝜋; then cos 𝜋 = −1 and 𝑃2 (𝜋) = 1 − (𝜋)2 ∕2! = −3.93480 … 𝑃4 (𝜋) = 𝑃6 (𝜋) =

⋯ ⋯

= 0.12391 … = −1.21135 …

𝑃8 (𝜋) = 𝑃10 (𝜋) =

⋯ ⋯

= −0.97602 … = −1.00183 …

𝑃12 (𝜋) = 𝑃14 (𝜋) =

⋯ ⋯

= −0.99990 … = −1.000004 … .

We see that the rate of convergence is somewhat slower; it takes a 14th -degree polynomial to approximate cos 𝜋 as accurately as an 8th -degree polynomial approximates cos(𝜋∕2). If 𝑥 were taken still farther away from 0, then we would need still more terms to obtain as accurate an approximation of cos 𝑥. Exercise 18 on page 510 uses the ratio test to show that the Taylor series for cos 𝑥 converges for all values of 𝑥. To show that its sum is indeed cos 𝑥, we use Theorem 10.3. This justifies our writing the equality: 𝑥2 𝑥4 𝑥6 𝑥8 cos 𝑥 = 1 − + − + −⋯ for all 𝑥. 2! 4! 6! 8! Showing that the Taylor Series for cos 𝒙 Converges to cos 𝒙 The Lagrange error bound in Theorem 10.3 allows us to see if the Taylor series for a function converges to that function. In the series for cos 𝑥, the odd powers are missing, so we assume 𝑛 is even and write ) ( 𝑛 𝑥2 𝑛∕2 𝑥 𝐸𝑛 (𝑥) = cos 𝑥 − 𝑃𝑛 (𝑥) = cos 𝑥 − 1 − + ⋯ + (−1) , 2! 𝑛! giving 𝑥𝑛 𝑥2 + ⋯ + (−1)𝑛∕2 + 𝐸𝑛 (𝑥). cos 𝑥 = 1 − 2! 𝑛! Thus, for the Taylor series to converge to cos 𝑥, we must have 𝐸𝑛 (𝑥) → 0 as 𝑛 → ∞. Showing 𝑬𝒏 (𝒙) → 0 as 𝒏 → ∞ Proof Since 𝑓 (𝑥) = cos 𝑥, the (𝑛 + 1)st derivative, 𝑓 (𝑛+1) (𝑥), is ± cos 𝑥 or ± sin 𝑥, no matter what 𝑛 is. So for all 𝑛, we have |𝑓 (𝑛+1) (𝑥)| ≤ 1 on the interval between 0 and 𝑥. By the Lagrange error bound with 𝑀 = 1, we have |𝑥|𝑛+1 for every 𝑛. (𝑛 + 1)! To show that the errors go to zero, we must show that for a fixed 𝑥, |𝐸𝑛 (𝑥)| = | cos 𝑥 − 𝑃𝑛 (𝑥)| ≤

|𝑥|𝑛+1 → 0 as 𝑛 → ∞. (𝑛 + 1)! To see why this is true, consider the ratio of successive terms of this sequence. We have |𝑥|𝑛+2∕(𝑛 + 2)! |𝑥| . = 𝑛+1 𝑛 +2 |𝑥| ∕(𝑛 + 1)!

Therefore, we obtain the (𝑛 + 1)st term of this sequence, |𝑥|𝑛+2 ∕(𝑛 + 2)!, by multiplying the previous term, |𝑥|𝑛+1 ∕(𝑛 + 1)!, by |𝑥|∕(𝑛 + 2). Since |𝑥| is fixed and 𝑛 + 2 is increasing, for sufficiently large 𝑛, this ratio is less than 1∕2 (or any other constant between 0 and 1). Thus eventually each term in the sequence is less than 1∕2 the previous term, so the sequence of errors approaches zero. Therefore, the Taylor series 1 − 𝑥2 ∕2! + 𝑥4 ∕4! − ⋯ does converge to cos 𝑥.

10.4 THE ERROR IN TAYLOR POLYNOMIAL APPROXIMATIONS

543

Problems 27 and 28 ask you to show that the Taylor series for sin 𝑥 and 𝑒𝑥 converge to the original function for all 𝑥. In each case, you again need the following limit: 𝑥𝑛 = 0. 𝑛→∞ 𝑛! lim

Error Bounds Using the Alternating Series Estimate We saw a method of bounding the error in an alternating series in Theorem 9.9 on page 500. This method can be used for a Taylor series if we know the series converges to the function 𝑓 (𝑥) and that the series is alternating and has terms whose magnitudes decrease monotonically to zero. Example 4

As we have shown, the Taylor series for cos 𝑥 converges to the function for all 𝑥. Use an alternating series to bound the errors in approximating cos 1 by 𝑃6 (1) and 𝑃9 (1).

Solution

The Taylor series is cos 𝑥 = 1 −

𝑥2 𝑥4 𝑥6 𝑥8 𝑥10 + − + − + ⋯, 2! 4! 6! 8! 10!

so, substituting 𝑥 = 1, we have 𝑃6 (1)

⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞ 1 1 1 1 1 cos 1 = 1 − + − + − +⋯. 2! 4! 6! 8! 10! ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ 𝑃9 (1)

This series is alternating and the magnitude of the terms is monotonically decreasing to zero. Thus we can use the alternating series error bound, which tells us that the error between cos 1 and 𝑃𝑛 (1) is bounded by the magnitude of the next term in the series. So we have | cos 1 − 𝑃6 (1)| ≤ 1 = 0.0000248, | | 8! | cos 1 − 𝑃9 (1)| ≤ 1 = 0.0000003. | | 10!

Deriving the Lagrange Error Bound Recall that we constructed 𝑃𝑛 (𝑥), the Taylor polynomial of 𝑓 about 0, so that its first 𝑛 derivatives equal the corresponding derivatives of 𝑓 (𝑥). Therefore, 𝐸𝑛 (0) = 0, 𝐸𝑛′ (0) = 0, 𝐸𝑛′′ (0) = 0, ⋯, 𝐸𝑛(𝑛) (0) = 0. Since 𝑃𝑛 (𝑥) is an 𝑛th -degree polynomial, its (𝑛 + 1)st derivative is 0, so 𝐸𝑛(𝑛+1) (𝑥) = | | 𝑓 (𝑛+1) (𝑥). In addition, suppose that |𝑓 (𝑛+1) (𝑥)| is bounded by a positive constant 𝑀, for all positive | | values of 𝑥 near 0, say for 0 ≤ 𝑥 ≤ 𝑑, so that −𝑀 ≤ 𝑓 (𝑛+1) (𝑥) ≤ 𝑀

for 0 ≤ 𝑥 ≤ 𝑑.

This means that −𝑀 ≤ 𝐸𝑛(𝑛+1) (𝑥) ≤ 𝑀 for 0 ≤ 𝑥 ≤ 𝑑. Writing 𝑡 for the variable, we integrate this inequality from 0 to 𝑥, giving 𝑥



∫0

𝑥

𝑀 𝑑𝑡 ≤

∫0

𝑥

𝐸𝑛(𝑛+1) (𝑡) 𝑑𝑡 ≤

∫0

𝑀 𝑑𝑡

for 0 ≤ 𝑥 ≤ 𝑑,

so −𝑀𝑥 ≤ 𝐸𝑛(𝑛) (𝑥) ≤ 𝑀𝑥

for 0 ≤ 𝑥 ≤ 𝑑.

544

Chapter 10 APPROXIMATING FUNCTIONS USING SERIES

We integrate this inequality again from 0 to 𝑥, giving 𝑥



∫0

𝑥

𝑀𝑡 𝑑𝑡 ≤

∫0

𝑥

𝐸𝑛(𝑛) (𝑡) 𝑑𝑡 ≤

∫0

𝑀𝑡 𝑑𝑡

for 0 ≤ 𝑥 ≤ 𝑑,

so

1 1 − 𝑀𝑥2 ≤ 𝐸𝑛(𝑛−1) (𝑥) ≤ 𝑀𝑥2 2 2 By repeated integration, we obtain the following bound: −

for 0 ≤ 𝑥 ≤ 𝑑.

1 1 𝑀𝑥𝑛+1 ≤ 𝐸𝑛 (𝑥) ≤ 𝑀𝑥𝑛+1 (𝑛 + 1)! (𝑛 + 1)!

for 0 ≤ 𝑥 ≤ 𝑑,

which means that 1 𝑛+1 |𝐸𝑛 (𝑥)| = |𝑓 (𝑥) − 𝑃𝑛 (𝑥)| ≤ | | | | (𝑛 + 1)! 𝑀𝑥

for 0 ≤ 𝑥 ≤ 𝑑.

When 𝑥 is to the left of 0, so −𝑑 ≤ 𝑥 ≤ 0, and when the Taylor series is centered at 𝑎 ≠ 0, similar calculations lead to Theorem 10.3.

Exercises and Problems for Section 10.4 EXERCISES In Exercises 1–8, use Theorem 10.3 to find a bound for the error in approximating the quantity with a third-degree Taylor polynomial for the given function 𝑓 (𝑥) about 𝑥 = 0. Compare the bound with the actual error. 1. 𝑒0.1 , 𝑓 (𝑥) = 𝑒𝑥

In Exercises 9–11, the Taylor polynomial 𝑃𝑛 (𝑥) about 0 approximates 𝑓 (𝑥) with error 𝐸𝑛 (𝑥) and the Taylor series converges to 𝑓 (𝑥). Find the smallest constant 𝐾 given by the alternating series error bound such that |𝐸4 (1)| ≤ 𝐾. 9. 𝑓 (𝑥) = 𝑒−𝑥

10. 𝑓 (𝑥) = cos 𝑥

11. 𝑓 (𝑥) = sin 𝑥

2. sin(0.2), 𝑓 (𝑥) = sin 𝑥 3. cos(−0.3), 𝑓 (𝑥) = cos 𝑥 √ √ 4. 0.9, 𝑓 (𝑥) = 1 + 𝑥

In Exercises 12–14, the Taylor polynomial 𝑃𝑛 (𝑥) about 0 approximates 𝑓 (𝑥) with error 𝐸𝑛 (𝑥) and the Taylor series converges to 𝑓 (𝑥) on |𝑥| ≤ 1. Find the smallest constant 𝐾 given by the alternating series error bound such that |𝐸6 (1)| ≤ 𝐾.

5. ln(1.5), 𝑓 (𝑥) = ln(1 + 𝑥) √ 6. 1∕ 3, 𝑓 (𝑥) = (1 + 𝑥)−1∕2

12.

7. tan 1, 𝑓 (𝑥) = tan 𝑥

∞ ∑ 𝑥𝑛 (−1)𝑛 𝑛 2 𝑛=0

13.

∞ ∑ (−𝑥)𝑛 2 +1 𝑛 𝑛=0

14.

∞ ∑ (−1)𝑛+1 𝑥𝑛 (𝑛 + 1)𝑛 𝑛=0

8. 0.51∕3 , 𝑓 (𝑥) = (1 − 𝑥)1∕3

PROBLEMS 15. (a) Using a calculator, make a table of values to four decimal places of sin 𝑥 for 𝑥 = −0.5, −0.4, … , −0.1, 0, 0.1, … , 0.4, 0.5. (b) Add to your table the values of the error 𝐸1 = sin 𝑥 − 𝑥 for these 𝑥-values. (c) Using a calculator or computer, draw a graph of the quantity 𝐸1 = sin 𝑥 − 𝑥 showing that |𝐸1 | < 0.03

for

− 0.5 ≤ 𝑥 ≤ 0.5.

16. Find a bound √ on the magnitude of the error if we approximate 2√using the Taylor approximation of degree three for 1 + 𝑥 about 𝑥 = 0. 17. (a) Let 𝑓 (𝑥) = 𝑒𝑥 . Find a bound on the magnitude of the error when 𝑓 (𝑥) is approximated using 𝑃3 (𝑥), its Taylor approximation of degree 3 around 0 over the interval [−2, 2]. (b) What is the actual maximum error in approximating 𝑓 (𝑥) by 𝑃3 (𝑥) over the interval [−2, 2]?

10.4 THE ERROR IN TAYLOR POLYNOMIAL APPROXIMATIONS

18. Let 𝑓 (𝑥) = cos 𝑥 and let 𝑃𝑛 (𝑥) be the Taylor approximation of degree 𝑛 for 𝑓 (𝑥) around 0. Explain why, for any 𝑥, we can choose an 𝑛 such that |𝑓 (𝑥) − 𝑃𝑛 (𝑥)| < 10−9 .

19. Consider the error in using the approximation sin 𝜃 ≈ 𝜃 on the interval [−1, 1].

545

26. For |𝑥| ≤ 0.1, graph the error

𝐸0 = cos 𝑥 − 𝑃0 (𝑥) = cos 𝑥 − 1.

Explain the shape of the graph, using the Taylor expansion of cos 𝑥. Find a bound for |𝐸0 | for |𝑥| ≤ 0.1.

27. Show that the Taylor series about 0 for 𝑒𝑥 converges to 𝑒𝑥 for every 𝑥. Do this by showing that the error 𝐸𝑛 (𝑥) → 0 as 𝑛 → ∞.

(a) Reasoning informally, say where the approximation is an overestimate and where it is an underestimate. (b) Use Theorem 10.3 to bound the error. Check your answer graphically on a computer or calculator.

28. Show that the Taylor series about 0 for sin 𝑥 converges to sin 𝑥 for every 𝑥.

20. Repeat Problem 19 for the approximation sin 𝜃 ≈ 𝜃 − 𝜃 3 ∕3!.

(a) Using the fact that 𝑑(arctan 𝑥)∕𝑑𝑥 = 1∕(1 + 𝑥2 ) and arctan 1 = 𝜋∕4, approximate the value of 𝜋 using the third-degree Taylor polynomial of 4 arctan 𝑥 about 𝑥 = 0. √ (b) Using the fact that 𝑑(arcsin 𝑥)∕𝑑𝑥 = 1∕ 1 − 𝑥2 and arcsin 1 = 𝜋∕2, approximate the value of 𝜋 using the third-degree Taylor polynomial of 2 arcsin 𝑥 about 𝑥 = 0. (c) Estimate the maximum error of the approximation you found in part (a). (d) Explain the problem in estimating the error in the arcsine approximation.

21. You approximate 𝑓 (𝑡) = 𝑒𝑡 by a Taylor polynomial of degree 0 about 𝑡 = 0 on the interval [0, 0.5]. (a) Reasoning informally, say whether the approximation is an overestimate or an underestimate. (b) Use Theorem 10.3 to bound the error. Check your answer graphically on a computer or calculator. 22. Repeat Problem 21 using the second-degree Taylor approximation to 𝑒𝑡 . 23. (a) Use the graphs of 𝑦 = cos 𝑥 and its Taylor polynomials, 𝑃10 (𝑥) and 𝑃20 (𝑥), in Figure 10.19 to bound: (i) The error in approximating cos 6 by 𝑃10 (6) and by 𝑃20 (6). (ii) The error in approximating cos 𝑥 by 𝑃20 (𝑥) for |𝑥| ≤ 9. (b) If we want to approximate cos 𝑥 by 𝑃10 (𝑥) to an accuracy of within 0.1, what is the largest interval of 𝑥-values on which we can work? Give your answer to the nearest integer. 𝑃20 (𝑥)

𝑃20 (𝑥)

29. To approximate 𝜋 using a Taylor polynomial, we could use the series for the arctangent or the series for the arcsine. In this problem, we compare the two methods.

In Problems 30–31, you will compare the Lagrange and alternating series error bounds at 𝑥 = 1∕2 in approximating the function 𝑓 (𝑥) by 𝑃𝑛 (𝑥), its Taylor polynomial about 𝑥 = 0. The Taylor series converges to 𝑓 (𝑥) at 𝑥 = 1∕2. (a) Find the smallest bound for the error using Lagrange’s method. (b) Find a bound for the same error using the alternating series bound. (c) Compare your answers to parts (a) and (b). What do you notice? 30. 𝑓 (𝑥) = 1∕(1 + 𝑥)

31. 𝑓 (𝑥) = 𝑒−𝑥 ,

3 𝑥 −12

−6

−3

3

6

12

−3 𝑃10 (𝑥)

𝑃10 (𝑥)

Figure 10.19

24. Give a bound for the error for the 𝑛th -degree Taylor polynomial about 𝑥 = 0 approximating cos 𝑥 on the interval [0, 1]. What is the bound for sin 𝑥? 25. What degree Taylor polynomial about 𝑥 = 0 do you need to calculate cos 1 to four decimal places? To six decimal places? Justify your answer using the results of Problem 24.

32. To use the alternating series error bound, the magnitude of the terms must decrease monotonically to 0. Here we use the alternating series error bound on the Taylor series for cos 𝑥, whose convergence to cos 𝑥 is known. (a) Find the ratio of the magnitudes of successive terms in the Taylor series for cos 𝑥. (b) Using your answer to part (a), if 𝑥 = 1.1, show that the magnitudes of the terms decrease monotonically to zero. (c) Bound the error in approximating cos 𝑥 by its degree-4 Taylor polynomial when 𝑥 = 1.1. (d) If 𝑥 = 10, after which term in the Taylor series do the terms of the series decrease monotonically to zero? (e) Assuming the alternating series error bound applies, bound the error 𝐸12 (10) using an alternating series.

546

Chapter 10 APPROXIMATING FUNCTIONS USING SERIES

Strengthen Your Understanding In Problems 33–34, explain what is wrong with the statement. 33. Let 𝑃𝑛 (𝑥) be a Taylor approximation of degree 𝑛 for a function 𝑓 (𝑥) about 𝑎, where 𝑎 is a constant. Then |𝑓 (𝑎) − 𝑃𝑛 (𝑎)| > 0 for any 𝑛.

34. Let 𝑓 (𝑥) be a function whose Taylor series about 𝑥 = 0 converges to 𝑓 (𝑥) for all 𝑥. Then there exists a positive integer 𝑛 such that the 𝑛th -degree Taylor polynomial 𝑃𝑛 (𝑥) for 𝑓 (𝑥) about 𝑥 = 0 satisfies the inequality |𝑓 (𝑥) − 𝑃𝑛 (𝑥)| < 1 for all values of 𝑥.

Decide if the statements in Problems 38–42 are true or false. Assume that the Taylor series for a function converges to that function. Give an explanation for your answer. 38. Let 𝑃𝑛 (𝑥) be the 𝑛th Taylor polynomial for a function 𝑓 near 𝑥 = 𝑎. Although 𝑃𝑛 (𝑥) is a good approximation to 𝑓 near 𝑥 = 𝑎, it is not possible to have 𝑃𝑛 (𝑥) = 𝑓 (𝑥) for all 𝑥. 39. If |𝑓 (𝑛) (𝑥)| < 10 for all 𝑛 > 0 and all 𝑥, then the Taylor series for 𝑓 about 𝑥 = 0 converges to 𝑓 (𝑥) for all 𝑥.

In Problems 35–37, give an example of: 35. A function 𝑓 (𝑥) whose Taylor series converges to 𝑓 (𝑥) for all values of 𝑥. 36. A polynomial 𝑃 (𝑥) such that |1∕𝑥 − 𝑃 (𝑥)| < 0.1 for all 𝑥 in the interval [1, 1.5].

37. A function 𝑓 (𝑥) and an interval [−𝑐, 𝑐] such that the value of 𝑀 in the error of the second-degree Taylor

10.5

polynomial of 𝑓 (𝑥) centered at 0 on the interval could be 4.

40. If 𝑓 (𝑛) (0) ≥ 𝑛! for all 𝑛, then the Taylor series for 𝑓 near 𝑥 = 0 diverges at 𝑥 = 0. 41. If 𝑓 (𝑛) (0) ≥ 𝑛! for all 𝑛, then the Taylor series for 𝑓 near 𝑥 = 0 diverges at 𝑥 = 1. 42. If 𝑓 (𝑛) (0) ≥ 𝑛! for all 𝑛, then the Taylor series for 𝑓 near 𝑥 = 0 diverges at 𝑥 = 1∕2.

FOURIER SERIES We have seen how to approximate a function by a Taylor polynomial of fixed degree. Such a polynomial is usually very close to the true value of the function near one point (the point at which the Taylor polynomial is centered), but not necessarily at all close anywhere else. In other words, Taylor polynomials are good approximations of a function locally, but not necessarily globally. In this section, we take another approach: we approximate the function by trigonometric functions, called Fourier approximations. The resulting approximation may not be as close to the original function at some points as the Taylor polynomial. However, the Fourier approximation is, in general, close over a larger interval. In other words, a Fourier approximation can be a better approximation globally. In addition, Fourier approximations are useful even for functions that are not continuous. Unlike Taylor approximations, Fourier approximations are periodic, so they are particularly useful for approximating periodic functions. Many processes in nature are periodic or repeating, so it makes sense to approximate them by periodic functions. For example, sound waves are made up of periodic oscillations of air molecules. Heartbeats, the movement of the lungs, and the electrical current that powers our homes are all periodic phenomena. Two of the simplest periodic functions are the square wave in Figure 10.20 and the triangular wave in Figure 10.21. Electrical engineers use the square wave as the model for the flow of electricity when a switch is repeatedly flicked on and off.

𝑦

𝑦 𝑓 (𝑥)

1

1

𝑔(𝑥)

𝑥 −1

0

1

2

3

Figure 10.20: Square wave

4

𝑥 −1

0

1

2

3

4

Figure 10.21: Triangular wave

10.5 FOURIER SERIES

547

Fourier Polynomials We can express the square wave and the triangular wave by the formulas ⋮ ⋮ ⎧ ⋮ ⎧⋮ ⎪ −𝑥 ⎪0 −1 ≤ 𝑥 < 0 −1 ≤ 𝑥 < 0 ⎪ 𝑥 ⎪1 0≤𝑥 0, find and graph solutions of 𝑑𝐻 = −𝑘(𝐻 − 20). 𝑑𝑡

Solution

The slope field in Figure 11.39 shows the qualitative behavior of the solutions. To find the equation of the solution curves, we separate variables and integrate: 1 𝑑𝐻 = − 𝑘 𝑑𝑡. ∫ ∫ 𝐻 − 20 This gives Solving for 𝐻 leads to: or

ln |𝐻 − 20| = −𝑘𝑡 + 𝐶.

|𝐻 − 20| = 𝑒−𝑘𝑡+𝐶 = 𝑒−𝑘𝑡 𝑒𝐶 = 𝐴𝑒−𝑘𝑡 𝐻 − 20 = (±𝐴)𝑒−𝑘𝑡 = 𝐵𝑒−𝑘𝑡 𝐻 = 20 + 𝐵𝑒−𝑘𝑡 . 𝑦 𝐵=3

𝐵=2

𝐵=1

𝑥

𝐵 = −1

𝐵 = −3

𝐵 = −2

Figure 11.38: Graphs of 𝑦 = 𝐵𝑒𝑘𝑥 , which are solutions to 𝑑𝑦∕𝑑𝑥 = 𝑘𝑦, for some fixed 𝑘 > 0

582

Chapter 11 DIFFERENTIAL EQUATIONS

Again, 𝐵 = 0 also gives a solution. Graphs for 𝑘 = 1 and 𝐵 = −10, 0, 10, with 𝑡 ≥ 0, are in Figure 11.39. 𝐻 30

𝐻 = 20 + 10𝑒−𝑡

✠ ✛ 𝐻 = 20 ■

20

𝐻 = 20 − 10𝑒−𝑡

10

𝑡 Figure 11.39: Slope field and some solution curves for 𝑑𝐻∕𝑑𝑡 = −𝑘(𝐻 − 20), with 𝑘 = 1

This differential equation can be used to represent the temperature, 𝐻(𝑡), in ◦ C at time 𝑡 of a cup of water standing in a room at 20◦ C. As Figure 11.39 shows, if the initial temperature is 10◦ C, the water warms up; if the initial temperature is 30◦ C, the water cools down. If the initial temperature is 20◦ C, the water remains at 20◦C. Example 3

Find and sketch the solution to 𝑑𝑃 = 2𝑃 − 2𝑃 𝑡 𝑑𝑡

Solution

satisfying 𝑃 = 5 when 𝑡 = 0.

Factoring the right-hand side gives 𝑑𝑃 = 𝑃 (2 − 2𝑡). 𝑑𝑡 Separating variables, we get ∫

𝑑𝑃 = (2 − 2𝑡) 𝑑𝑡, ∫ 𝑃

so

Solving for 𝑃 leads to

ln |𝑃 | = 2𝑡 − 𝑡2 + 𝐶. 2 +𝐶

|𝑃 | = 𝑒2𝑡−𝑡

2

2

= 𝑒𝐶 𝑒2𝑡−𝑡 = 𝐴𝑒2𝑡−𝑡

with 𝐴 = 𝑒𝐶 , so 𝐴 > 0. In addition, 𝐴 = 0 gives a solution. Thus the general solution to the differential equation is 2

𝑃 = 𝐵𝑒2𝑡−𝑡

for any 𝐵.

To find the value of 𝐵, substitute 𝑃 = 5 and 𝑡 = 0 into the general solution, giving 2

5 = 𝐵𝑒2⋅0−0 = 𝐵, so 2

𝑃 = 5𝑒2𝑡−𝑡 .

11.4 SEPARATION OF VARIABLES

583

𝑃 𝑃 = 5𝑒2𝑡−𝑡

2

5 𝑡 0.5

1

1.5

2

Figure 11.40: Bell-shaped solution curve

The graph of this function is in Figure 11.40. Since the solution can be rewritten as 2

2

2

𝑃 = 5𝑒1−1+2𝑡−𝑡 = 5𝑒1 𝑒−1+2𝑡−𝑡 = (5𝑒)𝑒−(𝑡−1) , 2

the graph has the same shape as the graph of 𝑦 = 𝑒−𝑡 , the bell-shaped curve of statistics. Here the maximum, normally at 𝑡 = 0, is shifted one unit to the right to 𝑡 = 1.

Justification for Separation of Variables A differential equation is called separable if it can be written in the form 𝑑𝑦 = 𝑔(𝑥)𝑓 (𝑦). 𝑑𝑥 Provided 𝑓 (𝑦) ≠ 0, we write 𝑓 (𝑦) = 1∕ℎ(𝑦), so the right-hand side can be thought of as a fraction, 𝑔(𝑥) 𝑑𝑦 = . 𝑑𝑥 ℎ(𝑦) If we multiply through by ℎ(𝑦), we get ℎ(𝑦)

𝑑𝑦 = 𝑔(𝑥). 𝑑𝑥

Thinking of 𝑦 as a function of 𝑥, so 𝑦 = 𝑦(𝑥), and 𝑑𝑦∕𝑑𝑥 = 𝑦′ (𝑥), we can rewrite the equation as ℎ(𝑦(𝑥)) ⋅ 𝑦′ (𝑥) = 𝑔(𝑥). Now integrate both sides with respect to 𝑥: ∫

ℎ(𝑦(𝑥)) ⋅ 𝑦′ (𝑥) 𝑑𝑥 =



𝑔(𝑥) 𝑑𝑥.

The form of the integral on the left suggests that we use the substitution 𝑦 = 𝑦(𝑥). Since 𝑑𝑦 = 𝑦′ (𝑥) 𝑑𝑥, we get ∫

ℎ(𝑦) 𝑑𝑦 =



𝑔(𝑥) 𝑑𝑥.

If we can find antiderivatives of ℎ and 𝑔, then this gives the equation of the solution curve. Note that transforming the original differential equation, 𝑑𝑦 𝑔(𝑥) = , 𝑑𝑥 ℎ(𝑦) into ∫

ℎ(𝑦) 𝑑𝑦 =



𝑔(𝑥) 𝑑𝑥

584

Chapter 11 DIFFERENTIAL EQUATIONS

looks as though we have treated 𝑑𝑦∕𝑑𝑥 as a fraction, cross-multiplied, and then integrated. Although that’s not exactly what we have done, you may find this a helpful way of remembering the method. In fact, the 𝑑𝑦∕𝑑𝑥 notation was introduced by Leibniz to allow shortcuts like this (more specifically, to make the chain rule look like cancellation).

Exercises and Problems for Section 11.4 EXERCISES 1. Determine which of the following differential equations are separable. Do not solve the equations. (a) (c) (e) (g) (i) (k)

𝑦′ = 𝑦 𝑦′ = 𝑥𝑦 𝑦′ − 𝑥𝑦 = 0 𝑦′ = ln (𝑥𝑦) 𝑦′ = (sin 𝑥)(cos 𝑥𝑦) 𝑦′ = 2𝑥

(b) (d) (f) (h) (j) (l)

𝑦′ = 𝑥 + 𝑦 𝑦′ = sin(𝑥 + 𝑦) 𝑦′ = 𝑦∕𝑥 𝑦′ = (sin 𝑥)(cos 𝑦) 𝑦′ = 𝑥∕𝑦 𝑦′ = (𝑥 + 𝑦)∕(𝑥 + 2𝑦)

𝑑𝑢 = 𝑢2 , 𝑢(0) = 1 𝑑𝑡 𝑑𝑧 = 𝑧𝑦, 𝑧 = 1 when 𝑦 = 0 16. 𝑑𝑦

15. 2

17. 18. 19.

For Exercises 2–5, determine if the differential equation is separable, and if so, write it in the form ℎ(𝑦) 𝑑𝑦 = 𝑔(𝑥) 𝑑𝑥. 2. 𝑦′ = 𝑥𝑒𝑦 ′

𝑦

4. 𝑦 = 𝑥𝑒 − 1

3. 𝑦′ = 𝑥𝑒𝑦 − 𝑥 ′

7. 8. 9. 10. 11. 12. 13. 14.

21.



5. 𝑦 − 𝑥𝑦 = 𝑦

In Exercises 6–32, use separation of variables to find the solution to the differential equation subject to the initial condition. 6.

20.

𝑑𝑃 = −2𝑃 , 𝑃 (0) = 1 𝑑𝑡 𝑑𝑃 = 0.02𝑃 , 𝑃 (0) = 20 𝑑𝑡 𝑑𝐿 𝐿 = , 𝐿(0) = 100 𝑑𝑝 2 𝑄 𝑑𝑄 = , 𝑄 = 50 when 𝑡 = 0 𝑑𝑡 5 𝑑𝑃 𝑃 = 1, 𝑃 (0) = 1 𝑑𝑡 𝑑𝑚 = 3𝑚, 𝑚 = 5 when 𝑡 = 1 𝑑𝑡 𝑑𝐼 = 0.2𝐼, 𝐼 = 6 where 𝑥 = −1 𝑑𝑥 1 𝑑𝑧 = 5, 𝑧(1) = 5 𝑧 𝑑𝑡 𝑑𝑚 = 𝑚, 𝑚(1) = 2 𝑑𝑠

22. 23. 24. 25. 26. 27. 28. 29. 30. 31.

𝑑𝑦 𝑦 + = 0, 𝑦(0) = 10 𝑑𝑥 3 𝑑𝑦 = 0.5(𝑦 − 200), 𝑦 = 50 when 𝑡 = 0 𝑑𝑡 𝑑𝑃 = 𝑃 + 4, 𝑃 = 100 when 𝑡 = 0 𝑑𝑡 𝑑𝑦 = 2𝑦 − 4, through (2, 5) 𝑑𝑥 𝑑𝑄 = 0.3𝑄 − 120, 𝑄 = 50 when 𝑡 = 0 𝑑𝑡 𝑑𝑚 = 0.1𝑚 + 200, 𝑚(0) = 1000 𝑑𝑡 𝑑𝑅 + 𝑅 = 1, 𝑅(1) = 0.1 𝑑𝑦 𝑑𝐵 + 2𝐵 = 50, 𝐵(1) = 100 𝑑𝑡 𝑦 𝑑𝑦 = , 𝑦(0) = 1 𝑑𝑡 3+𝑡 𝑑𝑧 = 𝑡𝑒𝑧 , through the origin 𝑑𝑡 5𝑦 𝑑𝑦 = , 𝑦 = 3 where 𝑥 = 1 𝑑𝑥 𝑥 𝑑𝑦 = 𝑦2 (1 + 𝑡), 𝑦 = 2 when 𝑡 = 1 𝑑𝑡 𝑑𝑧 = 𝑧 + 𝑧𝑡2 , 𝑧 = 5 when 𝑡 = 0 𝑑𝑡 𝑑𝑤 = 𝜃𝑤2 sin 𝜃 2 , 𝑤(0) = 1 𝑑𝜃 𝑑𝑤 = −𝑤2 tan 𝜓, 𝑤(0) = 2 𝑑𝜓

32. 𝑥(𝑥 + 1)

𝑑𝑢 = 𝑢2 , 𝑢(1) = 1 𝑑𝑥

PROBLEMS 33. (a) Solve the differential equation 𝑑𝑦 4𝑥 = 2. 𝑑𝑥 𝑦 Write the solution 𝑦 as an explicit function of 𝑥. (b) Find the particular solution for each initial con-

dition below and graph the three solutions on the same coordinate plane. 𝑦(0) = 1,

𝑦(0) = 2,

𝑦(0) = 3.

11.4 SEPARATION OF VARIABLES

𝑑𝑃

34. (a) Solve the differential equation

40.

𝑑𝑃 = 0.2𝑃 − 10. 𝑑𝑡

𝑃 (0) = 50,

𝑃 (0) = 60.

35. (a) Find the general solution to the differential equation modeling how a person learns: 𝑑𝑦 = 100 − 𝑦. 𝑑𝑡 (b) Plot the slope field of this differential equation and sketch solutions with 𝑦(0) = 25 and 𝑦(0) = 110. (c) For each of the initial conditions in part (b), find the particular solution and add it to your sketch. (d) Which of these two particular solutions could represent how a person learns? 36. A circular oil spill grows at a rate given by the differential equation 𝑑𝑟∕𝑑𝑡 = 𝑘∕𝑟, where 𝑟 represents the radius of the spill in feet, and time is measured in hours. If the radius of the spill is 400 feet 16 hours after the spill begins, what is the value of 𝑘? Include units in your answer. 37. Figure 11.41 shows the slope field for 𝑑𝑦∕𝑑𝑥 = 𝑦2 . (a) Sketch the solutions that pass through the points (i) (0, 1)

(ii) (0, −1)

𝑑𝑄 =𝑃 −𝑎

41.

= 𝑘(𝑃 − 𝑎)

43.

𝑑𝑃

Write the solution 𝑃 as an explicit function of 𝑡. (b) Find the particular solution for each initial condition below and graph the three solutions on the same coordinate plane. 𝑃 (0) = 40,

𝑑𝑡

(iii) (0, 0)

(b) In words, describe the end behavior of the solution curves in part (a). (c) Find a formula for the general solution. (d) Show that all solution curves except for 𝑦 = 0 have both a horizontal and a vertical asymptote. 𝑦

42.

𝑑𝑡

= 𝑎𝑅 + 𝑏

44.

𝑑𝑃 − 𝑎𝑃 = 𝑏 𝑑𝑡

45.

𝑑𝑦 = 𝑘𝑦2 (1 + 𝑡2 ) 𝑑𝑡

46.

𝑑𝑅 = 𝑎(𝑅2 + 1) 𝑑𝑥

47.

𝑑𝐿 = 𝑘(𝑥 + 𝑎)(𝐿 − 𝑏) 𝑑𝑥

In Problems 48–51, solve the differential equation. Assume 𝑥, 𝑦, 𝑡 > 0. 48. 49. 50. 51. 52.

𝑑𝑦 = 𝑦(2 − 𝑦), 𝑦(0) = 1 𝑑𝑡 𝑥 ln 𝑥 𝑑𝑥 = 𝑑𝑡 𝑡 𝑑𝑥 = (1 + 2 ln 𝑡) tan 𝑥 𝑡 𝑑𝑡 (𝑦) 𝑑𝑦 = −𝑦 ln , 𝑦(0) = 1 𝑑𝑡 2 Figure 11.42 shows the slope field for the equation { 𝑑𝑦 𝑦2 if |𝑦| ≥ 1 = 1 if −1 ≤ 𝑦 ≤ 1. 𝑑𝑥 (a) Sketch the solutions that pass through (0, 0). (b) What can you say about the end behavior of the solution curve in part (a)? (c) For each of the following regions, find a formula for the general solution (i) −1 ≤ 𝑦 ≤ 1 (iii) 𝑦 ≥ 1

(ii) 𝑦 ≤ −1

(d) Show that each solution curve has two vertical asymptotes. (e) How far apart are the two asymptotes of a solution curve?

𝑥

−1

−1

Figure 11.41

Figure 11.42

𝑑𝑄

𝑑𝑅 = 𝑘𝑅

𝑑𝑡

1

In Problems 38–47, solve the differential equation. Assume 𝑎, 𝑏, and 𝑘 are nonzero constants.

𝑑𝑡

= 𝑏−𝑄

𝑑𝑅

𝑥

38.

𝑑𝑡

𝑦

1

39.

𝑑𝑡

𝑄 −

𝑘

=0

585

53. (a) Sketch the slope field for 𝑦′ = 𝑥∕𝑦. (b) Sketch several solution curves. (c) Solve the differential equation analytically. 54. (a) Sketch the slope field for 𝑦′ = −𝑦∕𝑥. (b) Sketch several solution curves. (c) Solve the differential equation analytically.

586

Chapter 11 DIFFERENTIAL EQUATIONS

55. Compare the slope field for 𝑦′ = 𝑥∕𝑦, Problem 53, with that for 𝑦′ = −𝑦∕𝑥, Problem 54. Show that the solution curves of Problem 53 intersect the solution curves of Problem 54 at right angles. 56. Consider the family of differential equations given by 𝑦′ = 𝑘𝑦2 where 𝑘 > 0. (a) Explain why the constant function 𝑦 = 0 is a solution.

(b) Explain why all solutions except 𝑦 = 0 are increasing. (c) Use separation of variables to find the general solution of this family, assuming 𝑦 ≠ 0. Your answer will have both the parameter 𝑘 and an arbitrary constant 𝐶. (d) Does the method of separation of variables give all possible solutions of the family of differential equations? Explain.

Strengthen Your Understanding In Problems 57–59, explain what is wrong with the statement. 57. Separating variables in 𝑑𝑦∕𝑑𝑥 = 𝑥 + 𝑦 gives −𝑦 𝑑𝑦 = 𝑥 𝑑𝑥. 58. The solution to 𝑑𝑃 ∕𝑑𝑡 = 0.2𝑡 is 𝑃 = 𝐵𝑒0.2𝑡 .

Are the statements in Problems 64–69 true or false? Give an explanation for your answer. 64. A differential equation of the form 𝑑𝑦∕𝑑𝑥 = 𝑓 (𝑥) is separable.

59. Separating variables in 𝑑𝑦∕𝑑𝑥 = 𝑒𝑥+𝑦 gives −𝑒𝑦 𝑑𝑦 = 𝑒𝑥 𝑑𝑥.

65. A differential equation of the form 𝑑𝑦∕𝑑𝑥 = 1∕𝑔(𝑦) is separable.

In Problems 60–63, give an example of:

66. For all constants 𝑘, the equation 𝑦′ + 𝑘𝑦 = 0 has exponential functions as solutions.

60. A differential equation that is not separable. 61. An expression for 𝑓 (𝑥) such that the differential equation 𝑑𝑦∕𝑑𝑥 = 𝑓 (𝑥) + 𝑥𝑦 − cos 𝑥 is separable.

67. The differential equation 𝑑𝑦∕𝑑𝑥 = 𝑥 + 𝑦 can be solved by separation of variables.

62. A differential equation all of whose solutions form the family of functions 𝑓 (𝑥) = 𝑥2 + 𝐶.

68. The differential equation 𝑑𝑦∕𝑑𝑥−𝑥𝑦 = 𝑥 can be solved by separation of variables.

63. A differential equation all of whose solutions form the family of hyperbolas 𝑥2 − 𝑦2 = 𝐶.

69. The only solution to the differential equation 𝑑𝑦∕𝑑𝑥 = 3𝑦2∕3 passing through the point (0, 0) is 𝑦 = 𝑥3 .

11.5

GROWTH AND DECAY In this section we look at exponential growth and decay equations. Consider the population of a region. If there is no immigration or emigration, the rate at which the population is changing is often proportional to the population. In other words, the larger the population, the faster it is growing because there are more people to have babies. If the population at time 𝑡 is 𝑃 and its continuous growth rate is 2% per unit time, then we know Rate of growth of population = 2%(Current population), and we can write this as

𝑑𝑃 = 0.02𝑃 . 𝑑𝑡 The 2% growth rate is called the relative growth rate to distinguish it from the absolute growth rate, 𝑑𝑃 ∕𝑑𝑡. Notice they measure different quantities. Since Relative growth rate = 2% =

1 𝑑𝑃 , 𝑃 𝑑𝑡

the relative growth rate is a percent change per unit time, while Absolute growth rate = Rate of change of population = is a change in population per unit time.

𝑑𝑃 𝑑𝑡

11.5 GROWTH AND DECAY

587

We showed in Section 11.4 that since the equation 𝑑𝑃 ∕𝑑𝑡 = 0.02𝑃 is of the form 𝑑𝑃 ∕𝑑𝑡 = 𝑘𝑃 for 𝑘 = 0.02, it has solution 𝑃 = 𝑃0 𝑒0.02𝑡 . Other processes are described by differential equations similar to that for population growth, but with negative values for 𝑘. In summary, we have the following result from the preceding section:

Every solution to the equation 𝑑𝑃 = 𝑘𝑃 𝑑𝑡 can be written in the form 𝑃 = 𝑃0 𝑒𝑘𝑡 , where 𝑃0 is the value of 𝑃 at 𝑡 = 0, and 𝑘 > 0 represents growth, whereas 𝑘 < 0 represents decay.

Recall that the doubling time of an exponentially growing quantity is the time required for it to double. The half-life of an exponentially decaying quantity is the time required for half of it to decay.

Continuously Compounded Interest At a bank, continuous compounding means that interest is accrued at a rate that is a fixed percentage of the balance at that moment. Thus, the larger the balance, the faster interest is earned and the faster the balance grows. Example 1

A bank account earns interest continuously at a rate of 5% of the current balance per year. Assume that the initial deposit is $1000 and that no other deposits or withdrawals are made. (a) Write the differential equation satisfied by the balance in the account. (b) Solve the differential equation and graph the solution.

Solution

(a) We are looking for 𝐵, the balance in the account in dollars, as a function of 𝑡, time in years. Interest is being added to the account continuously at a rate of 5% of the balance at that moment. Since no deposits or withdrawals are made, at any instant, Rate balance increasing = Rate interest earned = 5%(Current balance), which we write as

𝑑𝐵 = 0.05𝐵. 𝑑𝑡 This is the differential equation that describes the process. It does not involve the initial condition $1000 because the initial deposit does not affect the process by which interest is earned. (b) Solving the differential equation by separation of variables gives 𝐵 = 𝐵0 𝑒0.05𝑡 , where 𝐵0 is the value of 𝐵 at 𝑡 = 0, so 𝐵0 = 1000. Thus 𝐵 = 1000𝑒0.05𝑡 and this function is graphed in Figure 11.43.

588

Chapter 11 DIFFERENTIAL EQUATIONS

𝐵 (dollars)

𝐵 = 1000𝑒0.05𝑡 1000 𝑡 (years) Figure 11.43: Bank balance against time

The Difference Between Continuous and Annual Percentage Growth Rates If 𝑃 = 𝑃0 (1 + 𝑟)𝑡 with 𝑡 in years, we say that 𝑟 is the annual growth rate, while if 𝑃 = 𝑃0 𝑒𝑘𝑡 , we say that 𝑘 is the continuous growth rate. The constant 𝑘 in the differential equation 𝑑𝑃 ∕𝑑𝑡 = 𝑘𝑃 is not the annual growth rate, but the continuous growth rate. In Example 1, with a continuous interest rate of 5%, we obtain a balance of 𝐵 = 𝐵0 𝑒0.05𝑡 , where time, 𝑡, is in years. At the end of one year the balance is 𝐵0 𝑒0.05 . In that one year, our balance has changed from 𝐵0 to 𝐵0 𝑒0.05 , that is, by a factor of 𝑒0.05 = 1.0513. Thus the annual growth rate is 5.13%. This is what the bank means when it says “5% compounded continuously for an effective annual yield of 5.13%.” Since 𝑃0 𝑒0.05𝑡 = 𝑃0 (1.0513)𝑡, we have two different ways to represent the same function. Since most growth is measured over discrete time intervals, a continuous growth rate is an idealized concept. A demographer who says a population is growing at the rate of 2% per year usually means that after 𝑡 years the population is 𝑃 = 𝑃0 (1.02)𝑡 . To find the continuous growth rate, 𝑘, we express the population as 𝑃 = 𝑃0 𝑒𝑘𝑡 . At the end of one year 𝑃 = 𝑃0 𝑒𝑘 , so 𝑒𝑘 = 1.02. Thus 𝑘 = ln 1.02 ≈ 0.0198. The continuous growth rate, 𝑘 = 1.98%, is close to the annual growth rate of 2%, but it is not the same. Again, we have two different representations of the same function since 𝑃0 (1.02)𝑡 = 𝑃0 𝑒0.0198𝑡 .

Pollution in the Great Lakes In the 1960s pollution in the Great Lakes became an issue of public concern. We set up a model for how long it would take for the lakes to flush themselves clean, assuming no further pollutants are being dumped in the lakes. Suppose 𝑄 is the total quantity of pollutant in a lake of volume 𝑉 at time 𝑡. Suppose that clean water is flowing into the lake at a constant rate 𝑟 and that water flows out at the same rate. Assume that the pollutant is evenly spread throughout the lake and that the clean water coming into the lake immediately mixes with the rest of the water. We investigate how 𝑄 varies with time. Since pollutants are being taken out of the lake but not added, 𝑄 decreases, and the water leaving the lake becomes less polluted, so the rate at which the pollutants leave decreases. This tells us that 𝑄 is decreasing and concave up. In addition, the pollutants are never completely removed from the lake though the quantity remaining becomes arbitrarily small: in other words, 𝑄 is asymptotic to the 𝑡-axis. (See Figure 11.44.) 𝑄 (quantity of pollutant)

𝑡 (time) Figure 11.44: Pollutant in lake versus time

11.5 GROWTH AND DECAY

589

Setting Up a Differential Equation for the Pollution To model exactly how 𝑄 changes with time, we write an equation for the rate at which 𝑄 changes. We know that ( ) Rate 𝑄 Rate pollutants =− changes leave in outflow where the negative sign represents the fact that 𝑄 is decreasing. At time 𝑡, the concentration of pollutants is 𝑄∕𝑉 , and water containing this concentration is leaving at rate 𝑟. Thus Rate pollutants leave in outflow

Rate of outflow

=

×

Concentration

=

𝑟⋅

𝑄 . 𝑉

So the differential equation is 𝑑𝑄 𝑟 = − 𝑄, 𝑑𝑡 𝑉 and its solution is 𝑄 = 𝑄0 𝑒−𝑟𝑡∕𝑉 . Table 11.4 contains values of 𝑟 and 𝑉 for four of the Great Lakes.3 We use this data to calculate how long it would take for certain fractions of the pollution to be removed. Table 11.4 Volume and outflow in Great Lakes4 𝑉 (thousands of km3 )

𝑟 (km3 /year)

Superior

12.2

Michigan

4.9

158

65.2

Erie

0.46

175

Ontario

1.6

209

Example 2

According to this model, how long will it take for 90% of the pollution to be removed from Lake Erie? For 99% to be removed?

Solution

Substituting 𝑟 and 𝑉 for Lake Erie into the differential equation for 𝑄 gives 𝑑𝑄 𝑟 −175 =− 𝑄= 𝑄 = −0.38𝑄 𝑑𝑡 𝑉 0.46 × 103 where 𝑡 is measured in years. Thus 𝑄 is given by 𝑄 = 𝑄0 𝑒−0.38𝑡 . When 90% of the pollution has been removed, 10% remains, so 𝑄 = 0.1𝑄0 . Substituting gives 0.1𝑄0 = 𝑄0 𝑒−0.38𝑡 . Canceling 𝑄0 and solving for 𝑡, we get 𝑡=

− ln(0.1) ≈ 6 years. 0.38

When 99% of the pollution has been removed, 𝑄 = 0.01𝑄0 , so 𝑡 satisfies 0.01𝑄0 = 𝑄0 𝑒−0.38𝑡 . Solving for 𝑡 gives 𝑡=

− ln(0.01) ≈ 12 years. 0.38

3 Data from William E. Boyce and Richard C. DiPrima, Elementary Differential Equations, 9th Edition (New York: Wiley, 2009), pp. 63–64. 4 www.epa.gov/greatlakes/

590

Chapter 11 DIFFERENTIAL EQUATIONS

Newton’s Law of Heating and Cooling Newton proposed that the temperature of a hot object decreases at a rate proportional to the difference between its temperature and that of its surroundings. Similarly, a cold object heats up at a rate proportional to the temperature difference between the object and its surroundings. For example, a hot cup of coffee standing on the kitchen table cools at a rate proportional to the temperature difference between the coffee and the surrounding air. As the coffee cools, the rate at which it cools decreases because the temperature difference between the coffee and the air decreases. In the long run, the rate of cooling tends to zero, and the temperature of the coffee approaches room temperature. See Figure 11.45. temperature Initial temperature

✶ q

✛ Room temperature time

Figure 11.45: Temperature of two cups of coffee with different initial temperatures

Example 3

When a murder is committed, the body, originally at 37◦ C, cools according to Newton’s Law of Cooling. Suppose that after two hours the temperature is 35◦C and that the temperature of the surrounding air is a constant 20◦ C. (a) Find the temperature, 𝐻, of the body as a function of 𝑡, the time in hours since the murder was committed. (b) Sketch a graph of temperature against time. (c) What happens to the temperature in the long run? Show this on the graph and algebraically. (d) If the body is found at 4 pm at a temperature of 30◦C, when was the murder committed?

Solution

(a) We first find a differential equation for the temperature of the body as a function of time. Newton’s Law of Cooling says that for some constant 𝛼, Rate of change of temperature = 𝛼(Temperature difference). If 𝐻 is the temperature of the body, then Temperature difference = 𝐻 − 20, so

𝑑𝐻 = 𝛼(𝐻 − 20). 𝑑𝑡 What about the sign of 𝛼? If the temperature difference is positive (that is, 𝐻 > 20), then 𝐻 is falling, so the rate of change must be negative. Thus 𝛼 should be negative, so we write: 𝑑𝐻 = −𝑘(𝐻 − 20), 𝑑𝑡

for some 𝑘 > 0.

Separating variables and solving, as in Example 2 on page 581, gives: 𝐻 − 20 = 𝐵𝑒−𝑘𝑡 .

11.5 GROWTH AND DECAY

591

To find 𝐵, substitute the initial condition that 𝐻 = 37 when 𝑡 = 0: 37 − 20 = 𝐵𝑒−𝑘(0) = 𝐵, so 𝐵 = 17. Thus, 𝐻 − 20 = 17𝑒−𝑘𝑡. To find 𝑘, we use the fact that after 2 hours, the temperature is 35◦ C, so 35 − 20 = 17𝑒−𝑘(2). Dividing by 17 and taking natural logs, we get: ( ) 15 = ln(𝑒−2𝑘 ) ln 17 −0.125 = −2𝑘 𝑘 ≈ 0.063. Therefore, the temperature is given by 𝐻 − 20 = 17𝑒−0.063𝑡 or 𝐻 = 20 + 17𝑒−0.063𝑡 . (b) The graph of 𝐻 = 20 + 17𝑒−0.063𝑡 has a vertical intercept of 𝐻 = 37 because the temperature of the body starts at 37◦C. The temperature decays exponentially with 𝐻 = 20 as the horizontal asymptote. (See Figure 11.46.) 𝐻(◦ C) 37

𝐻 = 20 + 17𝑒−0.063𝑡

20

10

20

30

𝑡 (hours)

Figure 11.46: Temperature of dead body

(c) “In the long run” means as 𝑡 → ∞. The graph shows that as 𝑡 → ∞, 𝐻 → 20. Algebraically, since 𝑒−0.063𝑡 → 0 as 𝑡 → ∞, we have 𝐻 = 20 + 17𝑒−0.063𝑡 ⟶ 20 as 𝑡 → ∞. ⏟⏞⏞⏟⏞⏞⏟ goes to 0 as 𝑡 → ∞

(d) We want to know when the temperature reaches 30◦ C. Substitute 𝐻 = 30 and solve for 𝑡: 30 = 20 + 17𝑒−0.063𝑡 10 = 𝑒−0.063𝑡 . 17 Taking natural logs: −0.531 = −0.063𝑡, which gives 𝑡 ≈ 8.4 hours. Thus the murder must have been committed about 8.4 hours before 4 pm. Since 8.4 hours = 8 hours 24 minutes, the murder was committed at about 7:30 am.

592

Chapter 11 DIFFERENTIAL EQUATIONS

Equilibrium Solutions Figure 11.47 shows the temperature of several objects in a 20◦ C room. One is initially hotter than 20◦ C and cools down toward 20◦ C; another is initially cooler and warms up toward 20◦C. All these curves are solutions to the differential equation 𝑑𝐻 = −𝑘(𝐻 − 20) 𝑑𝑡 for some fixed 𝑘 > 0, and all the solutions have the form 𝐻 = 20 + 𝐴𝑒−𝑘𝑡 for some 𝐴. Notice that 𝐻 → 20 as 𝑡 → ∞ because 𝑒−𝑘𝑡 → 0 as 𝑡 → ∞. In other words, in the long run, the temperature of the object always tends toward 20◦ C, the temperature of the room. This means that what happens in the long run is independent of the initial condition. In the special case when 𝐴 = 0, we have the equilibrium solution 𝐻 = 20 for all 𝑡. This means that if the object starts at 20◦ C, it remains at 20◦ C for all time. Notice that such a solution can be found directly from the differential equation by solving 𝑑𝐻∕𝑑𝑡 = 0: 𝑑𝐻 = −𝑘(𝐻 − 20) = 0 𝑑𝑡 giving 𝐻 = 20. Regardless of the initial temperature, 𝐻 always gets closer and closer to 20 as 𝑡 → ∞. As a result, 𝐻 = 20 is called a stable equilibrium5 for 𝐻. 𝐻

𝐵 𝐻 = 20 + 10𝑒−𝑘𝑡

30

✠ ✛

20



10

𝐻 = 20



𝐵 = 10

10



𝐻 = 20 − 10𝑒−𝑘𝑡 𝑡

Figure 11.47: 𝐻 = 20 is stable equilibrium (𝑘 > 0)

𝐵 = 10 + 𝑒𝑘𝑡

𝐵 = 10 − 𝑒𝑘𝑡 𝑡

Figure 11.48: 𝐵 = 10 is unstable equilibrium (𝑘 > 0)

A different situation is displayed in Figure 11.48, which shows solutions to the differential equation

𝑑𝐵 = 𝑘(𝐵 − 10) 𝑑𝑡 for some fixed 𝑘 > 0. Solving 𝑑𝐵∕𝑑𝑡 = 0 gives the equilibrium 𝐵 = 10, which is unstable because if 𝐵 starts near 10, it moves away as 𝑡 → ∞. In general, we have the following definitions. • An equilibrium solution is constant for all values of the independent variable. The graph is a horizontal line. • An equilibrium is stable if a small change in the initial conditions gives a solution that tends toward the equilibrium as the independent variable tends to positive infinity. • An equilibrium is unstable if a small change in the initial conditions gives a solution curve that veers away from the equilibrium as the independent variable tends to positive infinity. 5 In

more advanced work, this behavior is described as asymptotic stability.

11.5 GROWTH AND DECAY

593

Solutions that do not veer away from an equilibrium solution are also called stable. If the differential equation is of the form 𝑦′ = 𝑓 (𝑦), equilibrium solutions can be found by setting 𝑦′ to zero. Example 4

Find the equilibrium solution to the differential equation 𝑑𝑃 ∕𝑑𝑡 = −50 + 4𝑃 and determine whether the equilibrium is stable or unstable.

Solution

In order to find the equilibrium solution, we solve 𝑑𝑃 ∕𝑑𝑡 = −50 + 4𝑃 = 0. Thus, we have one equilibrium solution at 𝑃 = 50∕4 = 12.5. We can determine whether the equilibrium is stable or unstable by analyzing the behavior of solutions with initial conditions near 𝑃 = 12.5. For example, when 𝑃 > 12.5, the derivative 𝑑𝑃 ∕𝑑𝑡 = −50 + 4𝑃 > −50 + 4 ⋅ 12.5 = 0, so P is increasing. Thus, solutions above the equilibrium 𝑃 = 12.5 will increase and move away from the equilibrium. For a solution with initial value 𝑃 < 12.5 we have 𝑑𝑃 ∕𝑑𝑡 = −50 + 4𝑃 < −50 + 4 ⋅ 12.5 = 0. Thus, solutions that have an initial value less than 𝑃 = 12.5 will decrease and veer away from the equilibrium. Since a small change in the initial condition gives solutions that move away from the equilibrium, 𝑃 = 12.5 is an unstable equilibrium. Notice it is possible to find equilibrium solutions and study their stability without finding a formula for the general solution to the differential equation.

Exercises and Problems for Section 11.5 EXERCISES bank balance

1. Match the graphs in Figure 11.49 with the following descriptions.

(II)

(a) The temperature of a glass of ice water left on the kitchen table. (b) The amount of money in an interest-bearing bank account into which $50 is deposited. (c) The speed of a constantly decelerating car. (d) The temperature of a piece of steel heated in a furnace and left outside to cool. (I)

(I)

time

Figure 11.50

(II)

3. The slope field for 𝑦′ = 0.5(1 + 𝑦)(2 − 𝑦) is given in Figure 11.51.

time

time

(III)

(IV) (III)

(IV)

(a) List equilibrium solutions and state whether each is stable or unstable. (b) Draw solution curves on the slope field through each of the three marked points. 𝑦

time

time

Figure 11.49

2. Each curve in Figure 11.50 represents the balance in a bank account into which a single deposit was made at time zero. Assuming continuously compounded interest, find: (a) (b) (c) (d)

The curve representing the largest initial deposit. The curve representing the largest interest rate. Two curves representing the same initial deposit. Two curves representing the same interest rate.

𝑥

Figure 11.51

594

Chapter 11 DIFFERENTIAL EQUATIONS

4. The slope field for a differential equation is given in Figure 11.52. Estimate all equilibrium solutions for this differential equation, and indicate whether each is stable or unstable.

In Exercises 10–13, the number of fish, 𝑃 , in a lake with an initial population of 10,000 satisfies, for constant 𝑘 and 𝑟, 𝑑𝑃 = 𝑘𝑃 − 𝑟. 𝑑𝑡

𝑦

5

10. If 𝑘 = 0.15 and 𝑟 = 1000, is the fish population increasing or decreasing at time 𝑡 = 0?

𝑡 −5

5

Figure 11.52 For Exercises 5–6, sketch solution curves with a variety of initial values for the differential equations. You do not need to find an equation for the solution. 𝑑𝑦

= 𝛼 − 𝑦, 𝑑𝑡 where 𝛼 is a positive constant. 𝑑𝑤 = (𝑤 − 3)(𝑤 − 7) 6. 𝑑𝑡 7. A yam is put in a 200◦ C oven and heats up according to the differential equation 𝑑𝐻 = −𝑘(𝐻 − 200), 𝑑𝑡

12. If 𝑘 = 0.05, what must 𝑟 be in order for the fish population to remain at a constant level? 13. If 𝑘 = 0.10 and 𝑟 = 500, use a tangent line approximation to estimate 𝑃 at 𝑡 = 0.5.

−5

5.

11. If 𝑘 = 0.2 and 𝑟 = 3000, is the fish population increasing or decreasing at time 𝑡 = 0?

for 𝑘 a positive constant.

(a) If the yam is at 20◦ C when it is put in the oven, solve the differential equation. (b) Find 𝑘 using the fact that after 30 minutes the temperature of the yam is 120◦ C. 8. (a) Find the equilibrium solution to the differential equation 𝑑𝑦 = 0.5𝑦 − 250. 𝑑𝑡 (b) Find the general solution to this differential equation. (c) Sketch the graphs of several solutions to this differential equation, using different initial values for 𝑦. (d) Is the equilibrium solution stable or unstable? 9. (a) A cup of coffee is made with boiling water and stands in a room where the temperature is 20◦ C. If 𝐻(𝑡) is the temperature of the coffee at time 𝑡, in minutes, explain what the differential equation 𝑑𝐻 = −𝑘(𝐻 − 20) 𝑑𝑡 says in everyday terms. What is the sign of 𝑘? (b) Solve this differential equation. If the coffee cools to 90◦ C in 2 minutes, how long will it take to cool to 60◦ C degrees?

For Exercises 14–16, the growth of a tumor with size 𝑞 is given by ( ) 𝐿 , 𝑘, 𝐿 > 0. 𝑞 ′ = 𝑘𝑞 ln 𝑞 14. Will the tumor’s size increase or decrease if 𝑞 = 30 and 𝐿 = 50. 15. Will the tumor’s size increase or decrease if 𝑞 = 25 and 𝐿 = 20. 16. Let 𝐿 = 100 and 𝑘 = 0.1. If 𝑞 = 20 at time 𝑡 = 2, use the tangent-line approximation to estimate 𝑞 at 𝑡 = 2.1. 17. In Example 2 on page 589, we saw that it would take about 6 years for 90% of the pollution in Lake Erie to be removed and about 12 years for 99% to be removed. Explain why one time is double the other. 18. Using the model in the text and the data in Table 11.4 on page 589, find how long it would take for 90% of the pollution to be removed from Lake Michigan and from Lake Ontario, assuming no new pollutants are added. Explain how you can tell which lake will take longer to be purified just by looking at the data in the table. 19. Use the model in the text and the data in Table 11.4 on page 589 to determine which of the Great Lakes would require the longest time and which would require the shortest time for 80% of the pollution to be removed, assuming no new pollutants are being added. Find the ratio of these two times. For Exercises 20–22, is the function a solution to the differential equation 𝑑𝑦∕𝑑𝑥 = 𝑔(𝑦) for 𝑔(𝑦) in Figure 11.53? 3 𝑔(𝑦) 𝑦 −2

−1

1 −3

Figure 11.53 20. The constant function 𝑦 = 0. 21. The constant function 𝑦 = 1. 22. The constant function 𝑦 = 1∕2.

2

11.5 GROWTH AND DECAY

23. (a) Find all equilibrium solutions for the differential equation 𝑑𝑦 = 0.5𝑦(𝑦 − 4)(2 + 𝑦). 𝑑𝑥

595

(b) Draw a slope field and use it to determine whether each equilibrium solution is stable or unstable.

PROBLEMS In Problems 24–27, (a) Define the variables. (b) Write a differential equation to describe the relationship. (c) Solve the differential equation. 24. In 2014, the population of India6 was 1.236 billion people and increasing at a rate proportional to its population. If the population is measured in billions of people and time is measured in years, the constant of proportionality is 0.0125. 25. Nicotine leaves the body at a rate proportional to the amount present, with constant of proportionality 0.347 if the amount of nicotine is in mg and time is in hours. The amount of nicotine in the body immediately after smoking a cigarette is 0.4 mg. 26. By 2014, the cumulative world capacity of solar photovoltaic (PV) reached 178,391 megawatts and was growing exponentially at a continuous rate of 30% per year.7 27. In 2007, Grinnell Glacier in Glacier National Park covered 142 acres and was estimated to be shrinking exponentially at a continuous rate of 4.3% per year.8 28. Between 2002 and 2013, textbook prices9 increased at 5.6% per year while inflation was 2.3% per year. Assume both rates are continuous growth rates and let time, 𝑡, be in years since the start of 2002. (a) Write a differential equation satisfied by 𝐵(𝑡), the price of a textbook at time 𝑡. (b) Write a differential equation satisfied by 𝑃 (𝑡), the price at time 𝑡 of an item growing at the inflation rate. (c) Solve both differential equations. (d) What is the doubling time of the price of a textbook? (e) What is the doubling time of the price of an item growing according to the inflation rate? (f) How is the ratio of the doubling times related to the ratio of the growth rates? Justify your answer. 29. A bottle of milk is taken out of a 3◦ C refrigerator and left in a 22◦ C room. (a) Write a differential equation whose solution is the temperature of milk as a function of time. (The equation will include a positive constant 𝑘.) (b) Solve this equation. 6 www.indexmundi.com,

30. A raw chicken is taken from a 3◦ C refrigerator and put in a preheated 190◦ C oven. One hour later, a meat thermometer shows that the internal temperature of the chicken has risen to 40◦ C. If the safe internal temperature for chicken is 74◦ C, what is the total cooking time? 31. With time 𝑡 in years, the population 𝑃 of fish in a lake (in millions) grows at a continuous rate of 𝑟% per year and is subject to continuous harvesting at a constant rate of 𝐻 million fish per year. (a) Write a differential equation satisfied by 𝑃 . (b) What is the equilibrium level of 𝑃 if 𝑟 = 5 and 𝐻 = 15? (c) Without solving the differential equation, explain why the equilibrium you found in part (b) is unstable. (d) Find a formula for the equilibrium for general positive 𝑟 and 𝐻. (e) Are there any positive values of 𝑟 and 𝐻 that make the equilibrium stable? 32. The population of fish in a lake declines at a continuous rate of 10% per year. To curb this decline, wildlife management services continuously restock the lake at a rate of 5 thousand fish per year. Let 𝑃 be the fish population (in thousands) as a function of time 𝑡 in years. (a) Write a differential equation satisfied by 𝑃 . (b) Without solving the differential equation, find the equilibrium fish population. (c) Is the equilibrium stable or unstable? 33. A radioactive substance decays at a rate proportional to the quantity, 𝑄, present at the time, 𝑡. The constant of proportionality is 𝑘. (a) Write a differential equation satisfied by 𝑄. (b) Find the half-life as a function of 𝑘. (c) Is the half-life an increasing or decreasing function of 𝑘? 34. Warfarin is a drug used as an anticoagulant. After administration of the drug is stopped, the quantity remaining in a patient’s body decreases at a rate proportional to the quantity remaining. The half-life of warfarin in the body is 37 hours. (a) Sketch the quantity, 𝑄, of warfarin in a patient’s body as a function of the time, 𝑡, since stopping administration of the drug. Mark the 37 hours on your graph.

accessed April 27, 2015. accessed 1 April, 2016. 8 "Warming climate shrinking Glacier Park’s glaciers", www.usatoday.com, October 15, 2007. 9 www.cnbc.com, accessed April 27, 2015. 7 en.wikipedia.org,

596

Chapter 11 DIFFERENTIAL EQUATIONS

(b) Write a differential equation satisfied by 𝑄. (c) How many days does it take for the drug level in the body to be reduced to 25% of the original level? 35. The rate at which a drug leaves the bloodstream and passes into the urine is proportional to the quantity of the drug in the blood at that time. If an initial dose of 𝑄0 is injected directly into the blood, 20% is left in the blood after 3 hours. (a) Write and solve a differential equation for the quantity, 𝑄, of the drug in the blood after 𝑡 hours. (b) How much of this drug is in a patient’s body after 6 hours if the patient is given 100 mg initially? 36. Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were 1 million barrels of oil in the well; six years later 500,000 barrels remain. (a) At what rate was the amount of oil in the well decreasing when there were 600,000 barrels remaining? (b) When will there be 50,000 barrels remaining? 37. The radioactive isotope carbon-14 is present in small quantities in all life forms, and it is constantly replenished until the organism dies, after which it decays to stable carbon-12 at a rate proportional to the amount of carbon-14 present, with a half-life of 5730 years. Let 𝐶(𝑡) be the amount of carbon-14 present at time 𝑡. (a) Find the value of the constant 𝑘 in the differential equation 𝐶 ′ = −𝑘𝐶. (b) In 1988 three teams of scientists found that the Shroud of Turin, which was reputed to be the burial cloth of Jesus, contained 91% of the amount of carbon-14 contained in freshly made cloth of the same material.10 How old was the Shroud of Turin at the time of this data? 38. The amount of radioactive carbon-14 in a sample is measured using a Geiger counter, which records each disintegration of an atom. Living tissue disintegrates at a rate of about 13.5 atoms per minute per gram of carbon. In 1977 a charcoal fragment found at Stonehenge, England, recorded 8.2 disintegrations per minute per gram of carbon. Assuming that the half-life of carbon14 is 5730 years and that the charcoal was formed during the building of the site, estimate the date when Stonehenge was built. 39. A detective finds a murder victim at 9 am. The temperature of the body is measured at 90.3◦ F. One hour later, the temperature of the body is 89.0◦ F. The temperature of the room has been maintained at a constant 68◦ F. (a) Assuming the temperature, 𝑇 , of the body obeys Newton’s Law of Cooling, write a differential equation for 𝑇 . (b) Solve the differential equation to estimate the time the murder occurred. 10 The

New York Times, October 18, 1988.

40. At 1:00 pm one winter afternoon, there is a power failure at your house in Wisconsin, and your heat does not work without electricity. When the power goes out, it is 68◦ F in your house. At 10:00 pm, it is 57◦ F in the house, and you notice that it is 10◦ F outside. (a) Assuming that the temperature, 𝑇 , in your home obeys Newton’s Law of Cooling, write the differential equation satisfied by 𝑇 . (b) Solve the differential equation to estimate the temperature in the house when you get up at 7:00 am the next morning. Should you worry about your water pipes freezing? (c) What assumption did you make in part (a) about the temperature outside? Given this (probably incorrect) assumption, would you revise your estimate up or down? Why? 41. Before Galileo discovered that the speed of a falling body with no air resistance is proportional to the time since it was dropped, he mistakenly conjectured that the speed was proportional to the distance it had fallen. (a) Assume the mistaken conjecture to be true and write an equation relating the distance fallen, 𝐷(𝑡), at time 𝑡, and its derivative. (b) Using your answer to part (a) and the correct initial conditions, show that 𝐷 would have to be equal to 0 for all 𝑡, and therefore the conjecture must be wrong. 42. (a) An object is placed in a 68◦ F room. Write a differential equation for 𝐻, the temperature of the object at time 𝑡. (b) Find the equilibrium solution to the differential equation. Determine from the differential equation whether the equilibrium is stable or unstable. (c) Give the general solution for the differential equation. (d) The temperature of the object is 40◦ F initially and 48◦ F one hour later. Find the temperature of the object after 3 hours. 43. Hydrocodone bitartrate is used as a cough suppressant. After the drug is fully absorbed, the quantity of drug in the body decreases at a rate proportional to the amount left in the body. The half-life of hydrocodone bitartrate in the body is 3.8 hours, and the usual oral dose is 10 mg. (a) Write a differential equation for the quantity, 𝑄, of hydrocodone bitartrate in the body at time 𝑡, in hours since the drug was fully absorbed. (b) Find the equilibrium solution of the differential equation. Based on the context, do you expect the equilibrium to be stable or unstable? (c) Solve the differential equation given in part (a). (d) Use the half-life to find the constant of proportionality, 𝑘. (e) How much of the 10 mg dose is still in the body after 12 hours?

11.6 APPLICATIONS AND MODELING

44. (a) Let 𝐵 be the balance at time 𝑡 of a bank account that earns interest at a rate of 𝑟%, compounded continuously. What is the differential equation describing the rate at which the balance changes? What is the constant of proportionality, in terms of 𝑟? (b) Find the equilibrium solution to the differential equation. Determine whether the equilibrium is stable or unstable and explain what this means

597

about the bank account. (c) What is the solution to this differential equation? (d) Sketch the graph of 𝐵 as function of 𝑡 for an account that starts with $1000 and earns interest at the following rates: (i)

4%

(ii) 10%

(iii) 15%

Strengthen Your Understanding In Problems 45–47, explain what is wrong with the statement.

In Problems 48–50, give an example of: 48. A differential equation for a quantity that is decaying exponentially at a continuous rate per unit time.

45. The line 𝑦 = 2 is an equilibrium solution to the differential equation 𝑑𝑦∕𝑑𝑥 = 𝑦3 − 4𝑥𝑦. 46. The function 𝑦 = 𝑥2 is an equilibrium solution to the differential equation 𝑑𝑦∕𝑑𝑥 = 𝑦 − 𝑥2 . 47. At time 𝑡 = 0, a roast is taken out of a 40◦ F refrigerator and put in a 350◦ F oven. If 𝐻 represents the temperature of the roast at time 𝑡 minutes after it is put in the oven, we have 𝑑𝐻∕𝑑𝑡 = 𝑘(𝐻 − 40).

11.6

49. A differential equation with an equilibrium solution of 𝑄 = 500. 50. A graph of three possible solutions, with initial 𝑃 values of 20, 25, and 30, respectively, to a differential equation that has an unstable equilibrium solution at 𝑃 = 25.

APPLICATIONS AND MODELING Much of this book involves functions that represent real processes, such as how the temperature of a yam or the population of the US is changing with time. You may wonder where such functions come from. In some cases, we fit functions to experimental data by trial and error. In other cases, we take a more theoretical approach, leading to a differential equation whose solution is the function we want. In this section we give examples of the more theoretical approach.

How a Layer of Ice Forms When ice forms on a lake, the water on the surface freezes first. As heat from the water travels up through the ice and is lost to the air, more ice is formed. The question we will consider is: How thick is the layer of ice as a function of time? Since the thickness of the ice increases with time, the thickness function is increasing. In addition, as the ice gets thicker, it insulates better. Therefore, we expect the layer of ice to form more slowly as time goes on. Hence, the thickness function is increasing at a decreasing rate, so its graph is concave down.

A Differential Equation for the Thickness of the Ice To get more detailed information about the thickness function, we have to make some assumptions. Suppose 𝑦 represents the thickness of the ice as a function of time, 𝑡. Since the thicker the ice, the longer it takes the heat to get through it, we assume that the rate at which ice is formed is inversely proportional to the thickness. In other words, we assume that for some constant 𝑘, Rate thickness

=

is increasing

𝑘 , Thickness

so 𝑑𝑦 𝑘 = where 𝑘 > 0 and 𝑦 > 0. 𝑑𝑡 𝑦 This differential equation enables us to find a formula for 𝑦. Using separation of variables: ∫

𝑦 𝑑𝑦 =



𝑘 𝑑𝑡

𝑦2 = 𝑘𝑡 + 𝐶. 2

598

Chapter 11 DIFFERENTIAL EQUATIONS

𝑦 (thickness) Large 𝑘

𝑦=



2𝑘𝑡

Small 𝑘

𝑡 (time) Figure 11.54: Thickness of ice as a function of time

Extending the solution to 𝑦 = 0 and measuring time from then gives 𝐶 = 0. Since 𝑦 must be nonnegative, we have √ 𝑦 = 2𝑘𝑡 .

Graphs of 𝑦 against 𝑡 are in Figure 11.54. Notice that the larger 𝑦 is, the more slowly 𝑦 increases. In addition, this model suggests that 𝑦 increases indefinitely as time passes. (Of course, the value of 𝑦 cannot increase beyond the depth of the lake.)

The Net Worth of a Company In the preceding section, we saw an example in which money in a bank account was earning interest (Example 1, page 587). Consider a company whose revenues are proportional to its net worth (like interest on a bank account) but that must also make payroll payments. The question is: under what circumstances does the company make money, and under what circumstances does it go bankrupt? Common sense says that if the payroll exceeds the rate at which revenue is earned, the company will eventually be in trouble, whereas if revenue exceeds payroll, the company should do well. We assume that revenue is earned continuously and that payments are made continuously. (For a large company, this is a good approximation.) We also assume that the only factors affecting net worth are revenue and payroll. Example 1

A company’s revenue is earned at a continuous annual rate of 5% of its net worth. At the same time, the company’s payroll obligations amount to $200 million a year, paid out continuously. (a) Write a differential equation that governs the net worth of the company, 𝑊 million dollars. (b) Solve the differential equation, assuming an initial net worth of 𝑊0 million dollars. (c) Sketch the solution for 𝑊0 = 3000, 4000, and 5000.

Solution

First, let’s see what we can learn without writing a differential equation. For example, we can ask if there is any initial net worth 𝑊0 that will exactly keep the net worth constant. If there’s such an equilibrium, the rate at which revenue is earned must exactly balance the payments made, so Rate revenue is earned

=

Rate payments are made.

If net worth is a constant 𝑊0 , revenue is earned at a constant rate of 0.05𝑊0 per year, so we have 0.05𝑊0 = 200 giving 𝑊0 = 4000. Therefore, if the net worth starts at $4000 million, the revenue and payments are equal, and the net worth remains constant. Therefore, $4000 million is an equilibrium solution. Suppose, however, the initial net worth is above $4000 million. Then, the revenue earned is more than the payroll expenses, and the net worth of the company increases, thereby increasing the revenue still further. Thus the net worth increases more and more quickly. On the other hand, if the initial net worth is below $4000 million, the revenue is not enough to meet the payments, and the net worth of the company declines. This decreases the revenue, making the net worth decrease still more

11.6 APPLICATIONS AND MODELING

599

quickly. The net worth will eventually go to zero, and the company goes bankrupt. See Figure 11.55.

𝑊

5000 4000 3000

𝑡 27.7 Figure 11.55: Net worth as a function of time: Solutions to 𝑑𝑊 ∕𝑑𝑡 = 0.05𝑊 − 200

(a) Now we set up a differential equation for the net worth, using the fact that Rate net worth is increasing

=

Rate revenue is earned



Rate payroll payments

.

are made

In millions of dollars per year, revenue is earned at a rate of 0.05𝑊, and payments are made at a rate of 200 per year, so for 𝑡 in years, 𝑑𝑊 = 0.05𝑊 − 200. 𝑑𝑡 The equilibrium solution, 𝑊 = 4000, is obtained by setting 𝑑𝑊 ∕𝑑𝑡 = 0. (b) We solve this equation by separation of variables. It is helpful to factor out 0.05 before separating, so that the 𝑊 moves over to the left-hand side without a coefficient: 𝑑𝑊 = 0.05(𝑊 − 4000). 𝑑𝑡 Separating and integrating gives 𝑑𝑊 = 0.05 𝑑𝑡, ∫ 𝑊 − 4000 ∫ so or This means

ln |𝑊 − 4000| = 0.05𝑡 + 𝐶,

|𝑊 − 4000| = 𝑒0.05𝑡+𝐶 = 𝑒𝐶 𝑒0.05𝑡 .

𝑊 − 4000 = 𝐴𝑒0.05𝑡

where 𝐴 = ±𝑒𝐶 .

To find 𝐴, we use the initial condition that 𝑊 = 𝑊0 when 𝑡 = 0: 𝑊0 − 4000 = 𝐴𝑒0 = 𝐴. Substituting this value for 𝐴 into 𝑊 = 4000 + 𝐴𝑒0.05𝑡 gives 𝑊 = 4000 + (𝑊0 − 4000)𝑒0.05𝑡.

600

Chapter 11 DIFFERENTIAL EQUATIONS

(c) If 𝑊0 = 4000, then 𝑊 = 4000, the equilibrium solution. If 𝑊0 = 5000, then 𝑊 = 4000 + 1000𝑒0.05𝑡. If 𝑊0 = 3000, then 𝑊 = 4000 − 1000𝑒0.05𝑡. If 𝑊 = 0, then 𝑡 ≈ 27.7, so the company goes bankrupt in its twenty-eighth year. These solutions are shown in Figure 11.55. Notice that if the net worth starts with 𝑊0 near, but not equal to, $4000 million, then 𝑊 moves farther away. Thus, 𝑊 = 4000 is an unstable equilibrium.

The Velocity of a Falling Body: Terminal Velocity Think about the downward velocity of a sky-diver jumping out of a plane. When the sky-diver first jumps, his velocity is zero. The pull of gravity then makes his velocity increase. As the sky-diver speeds up, the air resistance also increases. Since the air resistance partly balances the pull of gravity, the force causing him to accelerate decreases. Thus, the velocity is an increasing function of time, but it is increasing at a decreasing rate. The air resistance increases until it balances gravity, when the sky-diver’s velocity levels off. Thus, we expect the the graph of velocity against time to be increasing and concave down with a horizontal asymptote.

A Differential Equation: Air Resistance Proportional to Velocity In order to compute the velocity function, we need to know exactly how air resistance depends on velocity. To decide whether air resistance is, say, proportional to the velocity, or is some other function of velocity, requires either lab experiments or a theoretical idea of how the air resistance is created. We consider a very small object, such as a dust particle settling on a computer component during manufacturing,11 and assume that air resistance is proportional to velocity. Thus, the net force on the object is 𝐹 = 𝑚𝑔 − 𝑘𝑣, where 𝑚𝑔 is the gravitational force, which acts downward, and 𝑘𝑣 is the air resistance, which acts upward, so 𝑘 > 0. (See Figure 11.56.) Then, by Newton’s Second Law of Motion, Force = Mass ⋅ Acceleration, we have

𝑑𝑣 . 𝑑𝑡 This differential equation can be solved by separation of variables. It is easier if we factor out −𝑘∕𝑚 before separating, giving ( 𝑚𝑔 ) 𝑑𝑣 𝑘 =− 𝑣− . 𝑑𝑡 𝑚 𝑘 Separating and integrating gives 𝑚𝑔 − 𝑘𝑣 = 𝑚

𝑘 𝑑𝑣 =− 𝑑𝑡 ∫ 𝑣 − 𝑚𝑔∕𝑘 𝑚∫

Solving for 𝑣, we have

| 𝑚𝑔 || 𝑘 = − 𝑡 + 𝐶. ln ||𝑣 − 𝑘 || 𝑚 |

| | |𝑣 − 𝑚𝑔 | = 𝑒−𝑘𝑡∕𝑚+𝐶 = 𝑒𝐶 𝑒−𝑘𝑡∕𝑚 | 𝑘 || | 𝑚𝑔 = 𝐴𝑒−𝑘𝑡∕𝑚 , 𝑣− 𝑘 where 𝐴 is an arbitrary nonzero constant. We find 𝐴 from the initial condition that the object starts from rest, so 𝑣 = 0 when 𝑡 = 0. Substituting 𝑚𝑔 = 𝐴𝑒0 0− 𝑘 11 Example

suggested by Howard Stone.

11.6 APPLICATIONS AND MODELING

gives 𝐴=−

601

𝑚𝑔 . 𝑘

✻ 𝑣 (velocity) Air resistance, 𝑘𝑣

Terminal velocity

𝑚𝑔 𝑘

𝑣=

𝑚𝑔 (1 𝑘

− 𝑒−𝑘𝑡∕𝑚 )

Force due to gravity, 𝑚𝑔

❄ Figure 11.56: Forces acting on a falling object

𝑡 (time) Figure 11.57: Velocity of falling dust particle assuming that air resistance is 𝑘𝑣

Thus

𝑚𝑔 𝑚𝑔 −𝑘𝑡∕𝑚 𝑚𝑔 − 𝑒 (1 − 𝑒−𝑘𝑡∕𝑚 ). = 𝑘 𝑘 𝑘 The graph of this function is in Figure 11.57. The horizontal asymptote represents the terminal velocity, 𝑚𝑔∕𝑘. Notice that the terminal velocity can also be obtained from the differential equation by setting 𝑑𝑣∕𝑑𝑡 = 0 and solving for 𝑣: 𝑣=

𝑚

𝑑𝑣 = 𝑚𝑔 − 𝑘𝑣 = 0, 𝑑𝑡

so

𝑣=

𝑚𝑔 . 𝑘

Compartmental Analysis: A Reservoir Many processes can be modeled as a container with various solutions flowing in and out—for example, drugs given intravenously or the discharge of pollutants into a lake. We consider a city’s water reservoir, fed partly by clean water from a spring and partly by run-off from the surrounding land. In New England and many other areas with much snow in the winter, the run-off contains salt that has been put on the roads to make them safe for driving. We consider the concentration of salt in the reservoir. If there is no salt in the reservoir initially, the concentration builds up until the rate at which the salt is entering into the reservoir balances the rate at which salt flows out. If, on the other hand, the reservoir starts with a great deal of salt in it, then initially, the rate at which the salt is entering is less than the rate at which it is flowing out, and the quantity of salt in the reservoir decreases. In either case, the salt concentration levels off at an equilibrium value.

A Differential Equation for Salt Concentration A water reservoir holds 100 million gallons of water and supplies a city with 1 million gallons a day. The reservoir is partly refilled by a spring that provides 0.9 million gallons a day, and the rest of the water, 0.1 million gallons a day, comes from run-off from the surrounding land. The spring is clean, but the run-off contains salt with a concentration of 0.0001 pound per gallon. There was no salt in the reservoir initially, and the water is well mixed (that is, the outflow contains the concentration of salt in the tank at that instant). We find the concentration of salt in the reservoir as a function of time. It is important to distinguish between the total quantity, 𝑄, of salt in pounds and the concentration, 𝐶, of salt in pounds/gallon where ( ) Quantity of salt 𝑄 lb Concentration = 𝐶 = = . Volume of water 100 million gal

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Chapter 11 DIFFERENTIAL EQUATIONS

(The volume of the reservoir is 100 million gallons.) We will find 𝑄 first, and then 𝐶. We know that Rate of change of quantity of salt

= Rate salt entering − Rate salt leaving.

Salt is entering through the run-off of 0.1 million gallons per day, with each gallon containing 0.0001 pound of salt. Therefore, Rate salt entering = Concentration ⋅ Volume per day ) ( ) ( million gal lb ⋅ 0.1 = 0.0001 gal day ) ( million lb = 10 lb∕day. = 0.00001 day

Salt is leaving in the million gallons of water used by the city each day . Thus Rate salt leaving = Concentration ⋅ Volume per day ( ) ( ) million gal 𝑄 𝑄 lb ⋅1 = lb∕day. = 100 million gal day 100 Therefore, 𝑄 satisfies the differential equation 𝑄 𝑑𝑄 = 10 − . 𝑑𝑡 100 We factor out −1∕100 = −0.01 and separate variables, giving 𝑑𝑄 = −0.01(𝑄 − 1000) 𝑑𝑡 𝑑𝑄 = − 0.01 𝑑𝑡 ∫ ∫ 𝑄 − 1000 ln |𝑄 − 1000| = −0.01𝑡 + 𝑘 𝑄 − 1000 = 𝐴𝑒−0.01𝑡 .

There is no salt initially, so we substitute 𝑄 = 0 when 𝑡 = 0: 0 − 1000 = 𝐴𝑒0

giving

𝐴 = −1000.

Thus 𝑄 − 1000 = −1000𝑒−0.01𝑡, so 𝑄 = 1000(1 − 𝑒−0.01𝑡 )

pounds.

Therefore 𝑄

1000 (1 − 𝑒−0.01𝑡 ) = 10−5 (1 − 𝑒−0.01𝑡 ) lb/gal. 108 A sketch of concentration against time is in Figure 11.58. Concentration = 𝐶 =

100 million

=

𝐶 (lb/gal) 10−5 ( ) 𝐶 = 10−5 1 − 𝑒−0.01𝑡

𝑡 (days) Figure 11.58: Concentration of salt in reservoir

11.6 APPLICATIONS AND MODELING

603

Exercises and Problems for Section 11.6 Online Resource: Additional Problems for Section 11.6 EXERCISES

1. Match the graphs in Figure 11.59 with the following descriptions.

6. A bank account that earns 10% interest compounded continuously has an initial balance of zero. Money is deposited into the account at a constant rate of $1000 per year.

(a) The population of a new species introduced onto a tropical island. (b) The temperature of a metal ingot placed in a furnace and then removed. (c) The speed of a car traveling at uniform speed and then braking uniformly. (d) The mass of carbon-14 in a historical specimen. (e) The concentration of tree pollen in the air over the course of a year. (I)

(II)

7. At time 𝑡 = 0, a bottle of juice at 90◦ F is stood in a mountain stream whose temperature is 50◦ F. After 5 minutes, its temperature is 80◦ F. Let 𝐻(𝑡) denote the temperature of the juice at time 𝑡, in minutes.

(III)

𝑡

𝑡

(IV)

(a) Write a differential equation that describes the rate of change of the balance 𝐵 = 𝑓 (𝑡). (b) Solve the differential equation to find the balance as a function of time.

𝑡

(V)

𝑡

𝑡

Figure 11.59

(a) Write a differential equation for 𝐻(𝑡) using Newton’s Law of Cooling. (b) Solve the differential equation. (c) When will the temperature of the juice have dropped to 60◦ F? 8. With 𝑡 in years since the start of 2014, copper has been mined worldwide at a rate of 17.9𝑒0.025𝑡 million tons per year. At the start of 2014, the world’s known copper reserves were 690 million tons.12

2. The balance is earning interest at a continuous rate of 5% per year, and payments are being made out of the fund at a continuous rate of $12,000 per year.

(a) Write a differential equation for 𝐶, the total quantity of copper mined in 𝑡 years since the start of 2014. (b) Solve the differential equation. (c) According to this model, when will the known copper reserves be exhausted?

3. The balance is earning interest at a continuous rate of 3.7% per year, and money is being added to the fund at a continuous rate of $5000 per year.

9. The velocity, 𝑣, of a dust particle of mass 𝑚 and acceleration 𝑎 satisfies the equation

In Exercises 2–5, write a differential equation for the balance 𝐵 in an investment fund with time, 𝑡, measured in years.

4. The balance is losing value at a continuous rate of 8% per year, and money is being added to the fund at a continuous rate of $2000 per year. 5. The balance is losing value at a continuous rate of 6.5% per year, and payments are being made out of the fund at a continuous rate of $50,000 per year.

𝑚𝑎 = 𝑚

𝑑𝑣 = 𝑚𝑔 − 𝑘𝑣, 𝑑𝑡

where 𝑔, 𝑘 are constant.

By differentiating this equation, find a differential equation satisfied by 𝑎. (Your answer may contain 𝑚, 𝑔, 𝑘, but not 𝑣.) Solve for 𝑎, given that 𝑎(0) = 𝑔.

PROBLEMS 10. A deposit is made to a bank account paying 2% interest per year compounded continuously. Payments totaling $2000 per year are made continuously from this account.

(c) Write the solution to the differential equation. (d) How much is in the account after 5 years if the initial deposit is (i) $80,000? (ii) $120,000?

(a) Write a differential equation for the balance, 𝐵, in the account after 𝑡 years. (b) Find the equilibrium solution of the differential equation. Is the equilibrium stable or unstable? Explain what happens to an account that begins with slightly more money or slightly less money than the equilibrium value.

11. Dead leaves accumulate on the ground in a forest at a rate of 3 grams per square centimeter per year. At the same time, these leaves decompose at a continuous rate of 75% per year. Write a differential equation for the total quantity of dead leaves (per square centimeter) at time 𝑡. Sketch a solution showing that the quantity of

12 Data

from http://minerals.usgs.gov/minerals/pubs/commodity/ Accessed February 8, 2015.

604

Chapter 11 DIFFERENTIAL EQUATIONS

dead leaves tends toward an equilibrium level. What is that equilibrium level? 12. A stream flowing into a lake brings with it a pollutant at a rate of 8 metric tons per year. The river leaving the lake removes the pollutant at a rate proportional to the quantity in the lake, with constant of proportionality −0.16 if time is measured in years. (a) Is the quantity of pollutant in the lake increasing or decreasing at a moment at which the quantity is 45 metric tons? At which the quantity is 55 metric tons? (b) What is the quantity of pollutant in the lake after a long time? 13. Caffeine is metabolized and excreted at a continuous rate of about 17% per hour. A person with no caffeine in the body starts drinking coffee, containing 130 mg of caffeine per cup, at 7 am. The person drinks coffee continuously all day at the rate of one cup an hour. Write a differential equation for 𝐴, the amount of caffeine in the body 𝑡 hours after 7 am and give the particular solution to this differential equation. How much caffeine is in the person’s body at 5 pm? 14. The rate (per foot) at which light is absorbed as it passes through water is proportional to the intensity, or brightness, at that point. (a) Find the intensity as a function of the distance the light has traveled through the water. (b) If 50% of the light is absorbed in 10 feet, how much is absorbed in 20 feet? 25 feet? 15. In 2012, the world population was 7.052 billion. The birth rate had stabilized to 135 million per year and is projected to remain constant. The death rate is projected to increase from 56 million per year in 2012 to 80 million per year in 2040 and to continue increasing at the same rate.13 (a) Assuming the death rate increases linearly, write a differential equation for 𝑃 (𝑡), the world population in billions 𝑡 years from 2012. (b) Solve the differential equation. (c) Find the population predicted for 2050. 16. A bank account earns 2% annual interest, compounded continuously. Money is deposited in a continuous cash flow at a rate of $1200 per year into the account. (a) Write a differential equation that describes the rate at which the balance 𝐵 = 𝑓 (𝑡) is changing. (b) Solve the differential equation given an initial balance 𝐵0 = 0. (c) Find the balance after 5 years. 17. The rate at which barometric pressure decreases with altitude is proportional to the barometric pressure at that altitude. If the barometric pressure is measured in 13 en.wikipedia.org,

accessed 1 April, 2016.

inches of mercury, and the altitude in feet, then the constant of proportionality is 3.7 ⋅ 10−5 . The barometric pressure at sea level is 29.92 inches of mercury. (a) Calculate the barometric pressure at the top of Mount Whitney, 14,500 feet (the highest mountain in the US outside Alaska), and at the top of Mount Everest, 29,000 feet (the highest mountain in the world). (b) People cannot easily survive at a pressure below 15 inches of mercury. What is the highest altitude to which people can safely go? 18. According to a simple physiological model, an athletic adult male needs 20 Calories per day per pound of body weight to maintain his weight. If he consumes more or fewer Calories than those required to maintain his weight, his weight changes at a rate proportional to the difference between the number of Calories consumed and the number needed to maintain his current weight; the constant of proportionality is 1∕3500 pounds per Calorie. Suppose that a particular person has a constant caloric intake of 𝐼 Calories per day. Let 𝑊 (𝑡) be the person’s weight in pounds at time 𝑡 (measured in days). (a) What differential equation has solution 𝑊 (𝑡)? (b) Find the equilibrium solution of the differential equation. Based on the context, do you expect the equilibrium to be stable or unstable? (c) Solve this differential equation. (d) Graph 𝑊 (𝑡) if the person starts out weighing 160 pounds and consumes 3000 Calories a day. 19. Morphine is often used as a pain-relieving drug. The half-life of morphine in the body is 2 hours. Suppose morphine is administered to a patient intravenously at a rate of 2.5 mg per hour, and the rate at which the morphine is eliminated is proportional to the amount present. (a) Use the half-life to show that, to three decimal places, the constant of proportionality for the rate at which morphine leaves the body (in mg/hour) is 𝑘 = −0.347. (b) Write a differential equation for the quantity, 𝑄, of morphine in the blood after 𝑡 hours. (c) Use the differential equation to find the equilibrium solution. (This is the long-term amount of morphine in the body, once the system has stabilized.) 20. Water leaks out of the bottom of a barrel at a rate proportional to the square root of the depth of the water at that time. If the water level starts at 36 inches and drops to 35 inches in 1 hour, how long will it take for all of the water to leak out of the barrel? 21. When a gas expands without gain or loss of heat, the rate of change of pressure with respect to volume is proportional to pressure divided by volume. Find a law connecting pressure and volume in this case.

11.6 APPLICATIONS AND MODELING

(a) Solve a differential equation for the quantity, 𝑄, in milligrams, of the drug in the body at time 𝑡 hours. Assume there is no drug in the body initially. Your answer will contain 𝑟 and 𝛼. Graph 𝑄 against 𝑡. What is 𝑄∞ , the limiting long-run value of 𝑄? (b) What effect does doubling 𝑟 have on 𝑄∞ ? What effect does doubling 𝑟 have on the time to reach half the limiting value, 21 𝑄∞ ? (c) What effect does doubling 𝛼 have on 𝑄∞ ? On the time to reach 21 𝑄∞ ?

22. A spherical snowball melts at a rate proportional to its surface area. (a) Write a differential equation for its volume, 𝑉 . (b) If the initial volume is 𝑉0 , solve the differential equation and graph the solution. (c) When does the snowball disappear? 23. Water leaks from a vertical cylindrical tank through a small hole in its base at a rate proportional to the square root of the volume of water remaining. If the tank initially contains 200 liters and 20 liters leak out during the first day, when will the tank be half empty? How much water will there be after 4 days? 24. As you know, when a course ends, students start to forget the material they have learned. One model (called the Ebbinghaus model) assumes that the rate at which a student forgets material is proportional to the difference between the material currently remembered and some positive constant, 𝑎. (a) Let 𝑦 = 𝑓 (𝑡) be the fraction of the original material remembered 𝑡 weeks after the course has ended. Set up a differential equation for 𝑦. Your equation will contain two constants; the constant 𝑎 is less than 𝑦 for all 𝑡. (b) Solve the differential equation. (c) Describe the practical meaning (in terms of the amount remembered) of the constants in the solution 𝑦 = 𝑓 (𝑡). 25. An item is initially sold at a price of $𝑝 per unit. Over time, market forces push the price toward the equilibrium price, $𝑝∗ , at which supply balances demand. The Evans Price Adjustment model says that the rate of change in the market price, $𝑝, is proportional to the difference between the market price and the equilibrium price. (a) Write a differential equation for 𝑝 as a function of 𝑡. (b) Solve for 𝑝. (c) Sketch solutions for various different initial prices, both above and below the equilibrium price. (d) What happens to 𝑝 as 𝑡 → ∞? 26. Let 𝐿, a constant, be the number of people who would like to see a newly released movie, and let 𝑁(𝑡) be the number of people who have seen it during the first 𝑡 days since its release. The rate that people first go see the movie, 𝑑𝑁∕ 𝑑𝑡 (in people/day), is proportional to the number of people who would like to see it but haven’t yet. Write and solve a differential equation describing 𝑑𝑁∕ 𝑑𝑡 where 𝑡 is the number of days since the movie’s release. Your solution will involve 𝐿 and a constant of proportionality, 𝑘.

605

28. When people smoke, carbon monoxide is released into the air. In a room of volume 60 m3 , air containing 5% carbon monoxide is introduced at a rate of 0.002 m3 /min. (This means that 5% of the volume of the incoming air is carbon monoxide.) The carbon monoxide mixes immediately with the rest of the air, and the mixture leaves the room at the same rate as it enters. (a) Write a differential equation for 𝑐(𝑡), the concentration of carbon monoxide at time 𝑡, in minutes. (b) Solve the differential equation, assuming there is no carbon monoxide in the room initially. (c) What happens to the value of 𝑐(𝑡) in the long run? 29. (Continuation of Problem 28.) Government agencies warn that exposure to air containing 0.02% carbon monoxide can lead to headaches and dizziness.14 How long does it take for the concentration of carbon monoxide in the room in Problem 28 to reach this level? 30. An aquarium pool has volume 2 ⋅ 106 liters. The pool initially contains pure fresh water. At 𝑡 = 0 minutes, water containing 10 grams/liter of salt is poured into the pool at a rate of 60 liters∕minute. The salt water instantly mixes with the fresh water, and the excess mixture is drained out of the pool at the same rate (60 liters/minute). (a) Write a differential equation for 𝑆(𝑡), the mass of salt in the pool at time 𝑡. (b) Solve the differential equation to find 𝑆(𝑡). (c) What happens to 𝑆(𝑡) as 𝑡 → ∞? 31. In 1692, Johann Bernoulli was teaching the Marquis de l’Hopital calculus in Paris. Solve the following problem, which is similar to the one that they did. What is the equation of the curve which has subtangent (distance 𝐵𝐶 in Figure 11.60) equal to twice its abscissa (distance 𝑂𝐶)?

27. A drug is administered intravenously at a constant rate of 𝑟 mg/hour and is excreted at a rate proportional to the quantity present, with constant of proportionality 𝛼 > 0. 14 www.lni.wa.gov/Safety/Topics/AtoZ/CarbonMonoxide/,

𝑦

Tangent

𝐴

■ Curve



𝑥 𝐵

𝑂

𝐶

Figure 11.60 accessed on June 14, 2007.

606

Chapter 11 DIFFERENTIAL EQUATIONS

32. An object of mass 𝑚 is fired vertically upward from the surface of the earth with initial velocity 𝑣0 . We will calculate the value of 𝑣0 , called the escape velocity, with which the object just escapes the pull of gravity and never returns to earth. Since the object is moving far from the surface of the earth, we must take into account the variation of gravity with altitude. If the acceleration due to gravity at sea level is 𝑔, and 𝑅 is the radius of the earth, the gravitational force, 𝐹 , on the object of mass 𝑚 at an altitude ℎ above the surface of the earth is 𝑚𝑔𝑅2 𝐹 = . (𝑅 + ℎ)2 (a) The velocity of the object (measured upward) is 𝑣 at time 𝑡. Use Newton’s Second Law of Motion to show that 𝑔𝑅2 𝑑𝑣 . =− 𝑑𝑡 (𝑅 + ℎ)2 (b) Rewrite this equation with ℎ instead of 𝑡 as the independent variable using the chain rule 𝑑𝑣 = 𝑑𝑡 𝑑𝑣 𝑑ℎ



𝑑ℎ . 𝑑𝑡

33. A ball is attached to one end of a rod of length 𝑎. A person at the origin holds the other end and moves along the positive 𝑥-axis, dragging the ball along a curve, called a tractrix, in the plane. See Figure 11.61. (a) What is the initial position of the ball? (b) Use the fact that the rod is tangent to the tractrix at every point to find the differential equation that the tractrix satisfies. (c) Show that the parametric equations 𝑥 = 𝑎(𝑡 − tanh 𝑡), 𝑦 = 𝑎∕ cosh 𝑡 satisfy the differential equation and the initial condition.

𝑦 𝑎

(𝑥, 𝑦)

Hence, show that

𝑎

𝑔𝑅2 𝑑𝑣 . =− 𝑣 𝑑ℎ (𝑅 + ℎ)2 (c) Solve the differential equation in part (b). (d) Find the escape velocity, the smallest value of 𝑣0 such that 𝑣 is never zero.

𝑥

Figure 11.61

Strengthen Your Understanding In Problems 34–35, explain what is wrong with the statement. 34. At a time when a bank balance $𝐵, which satisfies 𝑑𝐵∕𝑑𝑡 = 0.08𝐵 − 250, is $5000, the balance is going down. 35. The differential equation 𝑑𝑄∕𝑑𝑡 = −0.15𝑄 + 25 represents the quantity of a drug in the body if the drug is metabolized at a continuous rate of 15% per day and an IV line is delivering the drug at a constant rate of 25 mg per hour. In Problems 36–38, give an example of:

11.7

36. A differential equation for the quantity of a drug in a patient’s body if the patient is receiving the drug at a constant rate through an IV line and is metabolizing the drug at a rate proportional to the quantity present. 37. A differential equation for any quantity which grows in two ways simultaneously: on its own at a rate proportional to the cube root of the amount present and from an external contribution at a constant rate. 38. A differential equation for a quantity that is increasing and grows fastest when the quantity is small and grows more slowly as the quantity gets larger.

THE LOGISTIC MODEL Oil prices have a significant impact on the world’s economies. In the eighteen months preceding July 2008, the price of oil more than doubled from about $60 a barrel to about $140 a barrel.15 The impact of the increase was significant, from the auto industry, to family budgets, to how people commute. Even the threat of a price hike can send stock markets tumbling. Many reasons are suggested for the increase, but one fact is inescapable: there is a finite supply of oil in the world. To fuel its expanding economy, the world consumes more oil each succeeding year. This cannot go on indefinitely. Economists and geologists are interested in estimating the remaining oil reserves and the date at which annual oil production is expected to peak (that is, reach a maximum). 15 http://www.nyse.tv/crude-oil-price-history.htm.

Accessed February 2012.

11.7 THE LOGISTIC MODEL

607

US oil production has already peaked—and the date was predicted in advance. In 1956, geologist M. King Hubbert predicted that annual US oil production would peak some time in the period 1965– 1970. Although many did not take his prediction seriously, US oil production did in fact peak in 1970. The economic impact was blunted by the US’s increasing reliance on foreign oil. In this section we introduce the logistic differential equation and use it, as Hubbert did, to predict the peak of US oil production.16 Problems 38–41 investigate the peak of world oil production.

The Logistic Model The logistic differential equation describes growth subject to a limit. For oil, the limit is the total oil reserves; for a population, the limit is the largest population that the environment can support; for the spread of information or a disease, the limit is the number of people that could be affected. The solution to this differential equation is the family of logistic functions introduced in Section 4.4. Suppose 𝑃 is growing logistically toward a limiting value of 𝐿, and the relative growth rate, (1∕𝑃 )𝑑𝑃 ∕𝑑𝑡, is 𝑘 when 𝑃 = 0. In the exponential model, the relative growth rate remains constant at 𝑘. But in the logistic model, the relative growth rate decreases linearly to 0 as 𝑃 approaches 𝐿; see Figure 11.62. 1 𝑑𝑃 𝑃 𝑑𝑡 𝑘

𝑃 𝐿 Figure 11.62: Logistic model: Relative growth rate is a linear function of 𝑃

So we have

( ) 𝑘 𝑃 1 𝑑𝑃 =𝑘− 𝑃 =𝑘 1− . 𝑃 𝑑𝑡 𝐿 𝐿 The logistic differential equation can also be written ( ) 𝑃 𝑑𝑃 = 𝑘𝑃 1 − . 𝑑𝑡 𝐿 This equation was first proposed as a model for population growth by the Belgian mathematician P. F. Verhulst in the 1830s. In Verhulst’s model, 𝐿 represents the carrying capacity of the environment, which is determined by the supply of food and arable land along with the available technology.

Qualitative Solution to the Logistic Equation Figure 11.63 shows the slope field and characteristic sigmoid, or 𝑆-shaped, solution curve for the logistic model. Notice that for each fixed value of 𝑃 , that is, along each horizontal line, the slopes are the same because 𝑑𝑃 ∕𝑑𝑡 depends only on 𝑃 and not on 𝑡. The slopes are small near 𝑃 = 0 and near 𝑃 = 𝐿; they are steepest around 𝑃 = 𝐿∕2. For 𝑃 > 𝐿, the slopes are negative, so if the population is above the carrying capacity, the population decreases. 𝑑𝑃 𝑑𝑡

𝑃

𝐿

𝑡 Figure 11.63: Slope field for 𝑑𝑃 ∕𝑑𝑡 = 𝑘𝑃 (1 − 𝑃 ∕𝐿) 16 Based

on an undergraduate project by Brad Ernst, Colgate University.

𝐿 2

𝐿

Figure 11.64: 𝑑𝑃 ∕𝑑𝑡 = 𝑘𝑃 (1 − 𝑃 ∕𝐿)

𝑃

608

Chapter 11 DIFFERENTIAL EQUATIONS

We can locate the inflection point where the slope is greatest using Figure 11.64. This graph is a parabola because 𝑑𝑃 ∕𝑑𝑡 is a quadratic function of 𝑃 . The horizontal intercepts are at 𝑃 = 0 and 𝑃 = 𝐿, so the maximum, where the slope is greatest, is at 𝑃 = 𝐿∕2. Figure 11.64 also tells us that for 0 < 𝑃 < 𝐿∕2, the slope 𝑑𝑃 ∕𝑑𝑡 is positive and increasing, so the graph of 𝑃 against 𝑡 is concave up. (See Figure 11.65.) For 𝐿∕2 < 𝑃 < 𝐿, the slope 𝑑𝑃 ∕𝑑𝑡 is positive and decreasing, so the graph of 𝑃 is concave down. For 𝑃 > 𝐿, the slope 𝑑𝑃 ∕𝑑𝑡 is negative, so 𝑃 is decreasing. 𝑃

𝑃 Carrying capacity



𝐿

𝐿

✛ ✛

𝐿 2

𝑃0





Concave down 𝐿 2

Inflection point: Maximum slope Concave up Approximately exponential

𝑃0

𝑡

Figure 11.65: Logistic growth with inflection point

𝑡

Figure 11.66: Solutions to the logistic equation

If 𝑃 = 0 or 𝑃 = 𝐿, there is an equilibrium solution. Figure 11.66 shows that 𝑃 = 0 is an unstable equilibrium because solutions which start near 0 move away from 0. However, 𝑃 = 𝐿 is a stable equilibrium.

The Analytic Solution to the Logistic Equation We have already obtained a lot of information about logistic growth without finding a formula for the solution. However, the equation can be solved analytically by separating variables: ( ) ( ) 𝑑𝑃 𝑃 𝐿−𝑃 = 𝑘𝑃 1 − = 𝑘𝑃 𝑑𝑡 𝐿 𝐿 giving 𝑘 𝑑𝑃 = 𝑑𝑡. ∫ 𝑃 (𝐿 − 𝑃 ) ∫ 𝐿 We can integrate the left side using the integral tables (Formula V 26) or by partial fractions: ( ) 1 𝑘 1 1 + 𝑑𝑃 = 𝑑𝑡. ∫ 𝐿 ∫ 𝐿 𝑃 𝐿−𝑃 Canceling the constant 𝐿, we integrate to get ln |𝑃 | − ln |𝐿 − 𝑃 | = 𝑘𝑡 + 𝐶.

Multiplying through by (−1) and using properties of logarithms, we have

Exponentiating both sides gives

|𝐿 − 𝑃 | | = −𝑘𝑡 − 𝐶. ln || | | 𝑃 |

|𝐿 − 𝑃 | −𝑘𝑡−𝐶 | | = 𝑒−𝐶 𝑒−𝑘𝑡 , | 𝑃 |=𝑒 | |

so

Then, writing 𝐴 = ±𝑒−𝐶 , we have

𝐿−𝑃 = 𝐴𝑒−𝑘𝑡 . 𝑃

𝐿−𝑃 = ±𝑒−𝐶 𝑒−𝑘𝑡 . 𝑃

11.7 THE LOGISTIC MODEL

609

We find 𝐴 by substituting 𝑃 = 𝑃0 when 𝑡 = 0, which, when 𝑃0 ≠ 0, gives 𝐿 − 𝑃0 = 𝐴𝑒0 = 𝐴. 𝑃0 Since (𝐿 − 𝑃 )∕𝑃 = (𝐿∕𝑃 ) − 1, we have 𝐿 = 1 + 𝐴𝑒−𝑘𝑡 , 𝑃 which gives the following result:

The solution to the logistic differential equation: ( ) 𝑑𝑃 𝑃 = 𝑘𝑃 1 − with initial condition 𝑃0 when 𝑡 = 0 𝑑𝑡 𝐿 is, for 𝑃0 ≠ 0, the logistic function 𝑃 =

𝐿 1 + 𝐴𝑒−𝑘𝑡

with 𝐴 =

𝐿 − 𝑃0 . 𝑃0

The parameter 𝐿 represents the limiting value. The parameter 𝑘 represents the relative growth rate when 𝑃 is small relative to 𝐿. The parameter 𝐴 depends on the initial condition 𝑃0 .

Peak Oil: US Production We apply the logistic model to US oil production as Hubbert did in 1956. To make predictions, we need the values of 𝑘 and 𝐿. We calculate these values from the oil production data Hubbert had available to him in the 1950s; see Table 11.5. We define 𝑃 to be the total amount of oil, in billions of barrels, produced in the US since 1859, the year the first oil well was built. See Table 11.5. With time, 𝑡, in years, 𝑑𝑃 ∕𝑑𝑡 approximates the annual oil production in billions of barrels per year. Peak oil production occurs when 𝑑𝑃 ∕𝑑𝑡 is a maximum. Table 11.5

US oil production17 for 1931–1950 (billions of barrels)

Year

𝑑𝑃 ∕𝑑𝑡

𝑃

Year

𝑑𝑃 ∕𝑑𝑡

𝑃

Year

𝑑𝑃 ∕𝑑𝑡

𝑃

1931

0.851

13.8

1938

1.21

21.0

1945

1.71

31.5

1932

0.785

14.6

1939

1.26

22.3

1946

1.73

33.2

1933

0.906

15.5

1940

1.50

23.8

1947

1.86

35.1

1934

0.908

16.4

1941

1.40

25.2

1948

2.02

37.1

1935

0.994

17.4

1942

1.39

26.6

1949

1.84

38.9

1936

1.10

18.5

1943

1.51

28.1

1950

1.97

40.9

1937

1.28

19.8

1944

1.68

29.8

Figure 11.67 shows a scatterplot of the relative growth rate (𝑑𝑃 ∕𝑑𝑡)∕𝑃 versus 𝑃 . If the data follow a logistic differential equation, we see a linear relationship with intercept 𝑘 and slope −𝑘∕𝐿: ( ) 1 𝑑𝑃 𝑃 𝑘 =𝑘 1− = 𝑘 − 𝑃. 𝑃 𝑑𝑡 𝐿 𝐿 17 Data

from http://www.eia.gov/dnav/pet/hist/LeafHandler.ashx?n=PET&s=MCRFPUS1&f=A. Accessed Feb, 2012.

610

Chapter 11 DIFFERENTIAL EQUATIONS

Figure 11.67 shows a line fitted to the data.18 The vertical intercept gives the value 𝑘 = 0.0649. The slope of the line is −𝑘∕𝐿 = −0.00036, so we have 𝐿 = 0.0649∕0.00036 = 180 billion barrels of oil. Thus, the model predicts that the total oil reserves in the US (the total amount in the ground before drilling started in 1859) were 180 billion barrels of oil.19 1 𝑑𝑃 𝑃 𝑑𝑡

(fraction/yr)

0.07 0.06 0.05 0.04

1 𝑑𝑃 = 0.0649 − 0.00036𝑃 𝑃 𝑑𝑡

0.02

10

20

30

40

𝑃 (bn barrels)

Figure 11.67: US oil production 1931–1950: Scatterplot and line for 1∕𝑃 (𝑑𝑃 ∕𝑑𝑡) versus 𝑃

If we let 𝑡 = 0 be 1950, then 𝑃0 = 40.9 billion barrels and 𝐴 = (180 − 40.9)∕40.9 = 3.401. The logistic function representing US oil production is 𝑃 =

180 . 1 + 3.401𝑒−0.0649𝑡

Predicting Peak Oil Production To predict, as Hubbert did, the year when annual US oil production would peak, we use the fact that the maximum value for 𝑑𝑃 ∕𝑑𝑡 occurs when 𝑃 = 𝐿∕2. We derive a formula for the peak year (used again in Problems 38–41 to find peak production in world oil). The crucial observation is that, at the peak, the denominator of the expression for 𝑃 must equal 2. Since 𝑃 =

𝐿 𝐿 = , 2 1 + 𝐴𝑒−𝑘𝑡

we have 𝐴𝑒−𝑘𝑡 = 1.

Using logarithms to solve the equation 𝐴𝑒−𝑘𝑡 = 1, we get 𝑡 = (1∕𝑘) ln 𝐴. Since 𝐴 = (𝐿 − 𝑃0 )∕𝑃0 , we see that the time to peak oil production is an example of the following result:

For a logistic function, the maximum value of 𝑑𝑃 ∕𝑑𝑡 occurs when 𝑃 = 𝐿∕2, and Time to the maximum rate of change =

1 𝐿 − 𝑃0 1 ln 𝐴 = ln . 𝑘 𝑘 𝑃0

Thus, for the US, Time to peak oil production =

180 − 40.9 1 ln ≈ 19 years; 0.0649 40.9

that is, oil production was predicted to peak in the year 1950 + 19 = 1969. That year, 𝑃 = 𝐿∕2 and annual production was expected to be ( ) ( ) 𝑑𝑃 𝑃 180 1 = 𝑘𝑃 1 − = 0.0649 ⋅ 1− ≈ 3 billion barrels. 𝑑𝑡 𝐿 2 2 18 The

line is a least-squares regression line. the same analysis is repeated for other time periods, for example 1900–1950 or 1900–2000, the value for 𝐿 varies between 120 and 220 billion barrels, while the value of 𝑘 varies between 0.060 and 0.075. 19 If

11.7 THE LOGISTIC MODEL

611

The actual peak in US oil production was 3.5 billion barrels in 1970. Repeating the analysis using other time periods gives peak oil years in the range 1965-1970, as Hubbert predicted. Figure 11.68 shows actual annual US production data and the parabola predicting its peak around 1970. Figure 11.69 shows 𝑃 as a logistic function of 𝑡, with the limiting value of 𝑃 = 180 and maximum production at 𝑃 = 90. In fact, the first major oil crisis hit the US in the 1970s, with spiraling gas prices and long lines at service stations. The decline in US oil production since 1970 was partly mitigated by the opening of the Alaskan oil fields, which led to a second but lower peak in 1985. However, the US has increasingly depended on foreign oil. Although Hubbert’s predictions of the peak year proved to be accurate, extrapolation into the future is risky. Figure 11.68 and Figure 11.69 show that since 1970, oil production has slowed, though not as much as predicted. 𝑃 (bn barrels)

𝑑𝑃 ∕𝑑𝑡 (bn barrels/yr)

2010

1970 1975 1980

3 1955

1965 1960

180

1990

1950 1940 1945 𝑑𝑃 ∕𝑑𝑡 = 0.0649𝑃 (1 − 𝑃 ∕180) 1 2

2000 1990 1980

20002010

120

1970 1960

60 1940 𝑃 (bn barrels)

25 50 75 100 125 150 175 200 Figure 11.68: US oil production: 𝑑𝑃 ∕𝑑𝑡 versus 𝑃 , predicted (parabola) and actual

1975

1965

1955 1950

1945 𝑡 (yrs since 1950) −10 30 50 Figure 11.69: US oil production: 𝑃 versus 𝑡, predicted (logistic) and actual

Interestingly, Hubbert used the logistic model only to estimate 𝑘; for 𝐿, he relied on geological studies. It is remarkable that using only annual oil production for 1930-1950, we get an estimate for 𝐿 that is in such close agreement with the geological estimates.

US Population Growth The logistic equation is often used to model population growth. Table 11.6 gives the annual census figures in millions for the US population from 1790 to 2010.20 Since we have the population, 𝑃 (𝑡), of the US at ten-year intervals, we compute the relative growth rate using averages of estimates of the form: 𝑃 (1870) − 𝑃 (1860) 1 𝑑𝑃 1 in 1860 = ⋅ . 𝑃 𝑑𝑡 𝑃 (1860) 10 (See Problem 19 for details.) If we focus on the period 1790–1940, a line fitted to the scatterplot of (1∕𝑃 )𝑑𝑃 ∕𝑑𝑡 versus 𝑃 has intercept 𝑘 = 0.0317 and slope −𝑘∕𝐿 = −0.000165, so 𝐿 ≈ 192. Thus the differential equation modeling the US population during this period is ( ) 𝑑𝑃 𝑃 = 0.0317𝑃 1 − . 𝑑𝑡 192 Using 1790 as 𝑡 = 0, we get 𝐿 − 𝑃0 192 − 3.9 𝐴= ≈ 48. = 𝑃0 3.9 Thus the solution to the logistic differential equation is 192 . 1 + 48𝑒−0.0317𝑡 Table 11.6 shows the actual census data for 1790–1940 with projected values from a logistic model; the largest deviations are 4% in 1840 and 1870 (the Civil War accounts for the second one).21 𝑃 =

20 www.census.gov.Accessed 21 Calculations

February 12, 2012. were done with more precise values of the population data and the constants 𝑘, 𝐿, and 𝐴 than those shown.

612 Table 11.6

Chapter 11 DIFFERENTIAL EQUATIONS

US population, in millions, for 1790–2010, actual data and logistic predictions

Year

Actual

Logistic

Year

Actual

Logistic

Year

Actual

Logistic

Year

Actual

Logistic

1790

3.9

3.9

1850

23.2

23.6

1910

92.2

92.9

1970

203.2

165.7

1800

5.3

5.4

1860

31.4

30.9

1920

106.0

108.0

1980

226.5

172.2

1810

7.2

7.3

1870

38.6

40.1

1930

123.2

122.6

1990

248.7

177.2

1820

9.6

9.8

1880

50.2

51.0

1940

132.2

136.0

2000

281.4

181.0

1830

12.9

13.3

1890

63.0

63.7

1950

151.3

147.7

2010

308.7

183.9

1840

17.1

17.8

1900

76.2

77.9

1960

179.3

157.6

After 1940, the actual figures leave the logistic model in the dust. The model predicts an increase of 9.9 million from 1950 to 1960 versus the actual change of 28 million. By 1970 the actual population of 203.2 million exceeded the predicted limiting population of 𝐿 = 192 million. The unprecedented surge in US population between 1945 and 1965 is referred to as the baby boom.

Exercises and Problems for Section 11.7 Online Resource: Additional Problems for Section 11.7 EXERCISES

1. (a) Show that 𝑃 = 1∕(1 + 𝑒−𝑡 ) satisfies the logistic equation 𝑑𝑃 = 𝑃 (1 − 𝑃 ). 𝑑𝑡 (b) What is the limiting value of 𝑃 as 𝑡 → ∞?

𝑑𝑃 ∕𝑑𝑡

𝑃

2. A quantity 𝑃 satisfies the differential equation ( ) 𝑃 𝑑𝑃 = 𝑘𝑃 1 − . 𝑑𝑡 100 Sketch approximate solutions satisfying each of the following initial conditions: (a)

𝑃0 = 8

(b) 𝑃0 = 70

(c)

𝑃0 = 125

45

Figure 11.70 7. Figure 11.71 shows a slope field of a differential equation for a quantity 𝑄 growing logistically. Sketch a graph of 𝑑𝑄∕𝑑𝑡 against 𝑄. Include a scale on the horizontal axis. 𝑄

3. A quantity 𝑄 satisfies the differential equation 𝑑𝑄 = 𝑘𝑄(1 − 0.0004𝑄). 𝑑𝑡

800

Sketch approximate solutions satisfying each of the following initial conditions: (a)

𝑄0 = 300

(b) 𝑄0 = 1500 (c)

𝑄0 = 3500

4. A quantity 𝑃 satisfies the differential equation ( ) 𝑑𝑃 𝑃 = 𝑘𝑃 1 − , with 𝑘 > 0. 𝑑𝑡 250 Sketch a graph of 𝑑𝑃 ∕𝑑𝑡 as a function of 𝑃 . 5. A quantity 𝐴 satisfies the differential equation 𝑑𝐴 = 𝑘𝐴(1 − 0.0002𝐴), 𝑑𝑡

with 𝑘 > 0.

Sketch a graph of 𝑑𝐴∕𝑑𝑡 as a function of 𝐴. 6. Figure 11.70 shows a graph of 𝑑𝑃 ∕𝑑𝑡 against 𝑃 for a logistic differential equation. Sketch several solutions of 𝑃 against 𝑡, using different initial conditions. Include a scale on your vertical axis.

𝑡

Figure 11.71 8. (a) On the slope field for 𝑑𝑃 ∕𝑑𝑡 = 3𝑃 − 3𝑃 2 in Figure 11.72, sketch three solution curves showing different types of behavior for the population, 𝑃 . (b) Is there a stable value of the population? If so, what is it? (c) Describe the meaning of the shape of the solution curves for the population: Where is 𝑃 increasing? Decreasing? What happens in the long run? Are there any inflection points? Where? What do they mean for the population? (d) Sketch a graph of 𝑑𝑃 ∕𝑑𝑡 against 𝑃 . Where is 𝑑𝑃 ∕𝑑𝑡 positive? Negative? Zero? Maximum? How do your observations about 𝑑𝑃 ∕𝑑𝑡 explain the shapes of your solution curves?

11.7 THE LOGISTIC MODEL 𝑃

12.

1

𝑡 1

2

Figure 11.72 Exercises 9–10 give a graph of 𝑑𝑃 ∕𝑑𝑡 against 𝑃 . (a) What are the equilibrium values of 𝑃 ? (b) If 𝑃 = 500, is 𝑑𝑃 ∕𝑑𝑡 positive or negative? Is 𝑃 increasing or decreasing? 9. 𝑑𝑃 ∕𝑑𝑡

10. 𝑑𝑃 ∕𝑑𝑡

2000

𝑃

613

𝑑𝑃 = 0.1𝑃 − 0.00008𝑃 2 𝑑𝑡

In Exercises 13–16, give the general solution to the logistic differential equation. ( ) 𝑑𝑃 𝑃 13. = 0.05𝑃 1 − 𝑑𝑡 2800 ( ) 𝑃 𝑑𝑃 = 0.012𝑃 1 − 14. 𝑑𝑡 5700 𝑑𝑃 = 0.68𝑃 (1 − 0.00025𝑃 ) 15. 𝑑𝑡 𝑑𝑃 = 0.2𝑃 − 0.0008𝑃 2 16. 𝑑𝑡 In Exercises 17–20, give 𝑘, 𝐿, 𝐴, a formula for 𝑃 as a function of time 𝑡, and the time to the peak value of 𝑑𝑃 ∕𝑑𝑡. 𝑑𝑃 = 10𝑃 − 5𝑃 2 , 𝑃0 = 𝐿∕4 𝑑𝑡 𝑑𝑃 = 0.02𝑃 − 0.0025𝑃 2 , 𝑃0 = 1 18. 𝑑𝑡 ( ) 𝑃 1 𝑑𝑃 = 0.3 1 − , 𝑃0 = 75 19. 𝑃 𝑑𝑡 100 1 𝑑𝑃 20. = 0.012 − 0.002𝑃 , 𝑃0 = 2 10𝑃 𝑑𝑡 17.

400

𝑃

For the logistic differential equations in Exercises 11–12, (a) Give values for 𝑘 and for 𝐿 and interpret the meaning of each in terms of the growth of the quantity 𝑃 . (b) Give the value of 𝑃 when the rate of change is at its peak. ( ) 𝑃 𝑑𝑃 = 0.035𝑃 1 − 11. 𝑑𝑡 6000

In Exercises 21–22, give the solution to the logistic differential equation with initial condition. ( ) 𝑃 𝑑𝑃 = 0.8𝑃 1 − with 𝑃0 = 500 21. 𝑑𝑡 8500 𝑑𝑃 = 0.04𝑃 (1 − 0.0001𝑃 ) with 𝑃0 = 200 22. 𝑑𝑡

PROBLEMS Problems 23–27 are about chikungunya, a disease that arrived in the Americas in 2013 and spread rapidly in 2014. While seldom fatal, the disease causes debilitating joint pain and a high fever. In August 2014, a public challenge was issued to predict the number of cases in each of the affected countries. The winners, Joceline Lega and Heidi Brown, used a logistic model to make their predictions.22 Let 𝑁 be the total number of cases of chikungunya in a country by week 𝑡. From the data given, find approximate values for the country for (a) The number of cases expected in the long run. (b) The maximum number of new cases in a week. 23. In Guadeloupe: use the fitted curve in Figure 11.73.

( ) 𝑁 𝑑𝑁 = 0.31𝑁 1 − . 𝑑𝑡 3771 539,226 25. In the Dominican Republic: 𝑁 = . 1 + 177𝑒−0.35𝑡 1 𝑑𝑁 26. In Haiti: = 0.7 − 0.00001𝑁. 𝑁 𝑑𝑡 27. In Ecuador: use the fitted line in Figure 11.74.

24. In Dominica:

(1∕𝑁)𝑑𝑁∕𝑑𝑡, per week 1.6 1.2 0.8 0.4 0

𝑑𝑁∕𝑑𝑡, cases per week

𝑁 , total cases 10000

8000

20000

30000

Figure 11.74

6000

In Problems 28–29, a population 𝑃 (𝑡) = 𝐿∕(1+10𝑒−𝑘𝑡 ) grows logistically, with 𝑡 in days. Determine how long it takes the population to grow from 10% to 90% of its carrying capacity.

4000 2000 0

𝑁 , total cases 20000 40000 60000 80000

28. 𝑘 = 0.1

29. 𝑘 = 0.2

Figure 11.73 22 http://www.darpa.mil/news-events/2015-05-27. Based on preprint "Data Driven Outbreak Forecasting with a Simple Nonlinear Growth Model" J. Lega, H. Brown, April 2016.

614

Chapter 11 DIFFERENTIAL EQUATIONS

30. A rumor spreads among a group of 400 people. The number of people, 𝑁(𝑡), who have heard the rumor by time 𝑡 in hours since the rumor started is approximated by 400 𝑁(𝑡) = . 1 + 399𝑒−0.4𝑡 (a) Find 𝑁(0) and interpret it. (b) How many people will have heard the rumor after 2 hours? After 10 hours? (c) Graph 𝑁(𝑡). (d) Approximately how long will it take until half the people have heard the rumor? 399 people? (e) When is the rumor spreading fastest? 31. The total number of people infected with a virus often grows like a logistic curve. Suppose that time, 𝑡, is in weeks and that 10 people originally have the virus. In the early stages, the number of people infected is increasing exponentially with 𝑘 = 1.78. In the long run, 5000 people are infected. (a) Find a logistic function to model the number of people infected. (b) Sketch a graph of your answer to part (a). (c) Use your graph to estimate the length of time until the rate at which people are becoming infected starts to decrease. What is the vertical coordinate at this point? 32. The Tojolobal Mayan Indian community in southern Mexico has available a fixed amount of land. The proportion, 𝑃 , of land in use for farming 𝑡 years after 1935 is modeled with the logistic function in Figure 11.75:23 1 . 1 + 2.968𝑒−0.0275𝑡

𝑃 =

(a) What proportion of the land was in use for farming in 1935? (b) What is the long-run prediction of this model? (c) When was half the land in use for farming? (d) When is the proportion of land used for farming increasing most rapidly?

𝑑𝑃 = 0.25𝑃 (1 − 0.0004𝑃 ). 𝑑𝑡 (a) What is the long-term equilibrium population of carp in the lake? (b) A census taken ten years ago found there were 1000 carp in the lake. Estimate the current population. (c) Under a plan to join the lake to a nearby river, the fish will be able to leave the lake. A net loss of 10% of the carp each year is predicted, but the patterns of birth and death are not expected to change. Revise the differential equation to take this into account. Use the revised differential equation to predict the future development of the carp population. 34. Table 11.7 gives values for a logistic function 𝑃 = 𝑓 (𝑡). (a) Estimate the maximum rate of change of 𝑃 and estimate the value of 𝑡 when it occurs. (b) If 𝑃 represents the growth of a population, estimate the carrying capacity of the population. Table 11.7 𝑡

0

10

20

30

40

50

60

70

𝑃

120

125

135

155

195

270

345

385

35. For a population 𝑃 growing logistically, Table 11.8 gives values of the relative growth rate 𝑟=

1 𝑑𝑃 . 𝑃 𝑑𝑡

(a) What is the carrying capacity of this population? (b) What is the maximum value of 𝑑𝑃 ∕𝑑𝑡? Table 11.8 𝑃

1000

2000

3000

4000

5000

6000

𝑟

0.11

0.10

0.09

0.08

0.07

0.06

36. Figure 11.76 shows the spread of the Code-red computer virus during July 2001. Most of the growth took place starting at midnight on July 19; on July 20, the virus attacked the White House, trying (unsuccessfully) to knock its site off-line. The number of computers infected by the virus was a logistic function of time.

𝑃 (proportion of land in use)

1

0.5

40

33. A model for the population, 𝑃 , of carp in a landlocked lake at time 𝑡 is given by the differential equation

80

𝑡 (years 120 160 200 since 1935)

Figure 11.75

(a) Estimate the limiting value of 𝑓 (𝑡) as 𝑡 increased. What does this limiting value represent in terms of Code-red? (b) Estimate the value of 𝑡 at which 𝑓 ′′ (𝑡) = 0. Estimate the value of 𝑛 at this time. (c) What does the answer to part (b) tell us about Code-red? (d) How are the answers to parts (a) and (b) related?

23 Adapted from J. S. Thomas and M. C. Robbins, “The Limits to Growth in a Tojolobal Maya Ejido,” Geoscience and Man 26 (Baton Rouge: Geoscience Publications, 1988), pp. 9–16.

11.7 THE LOGISTIC MODEL 𝑛 (thousands of infected computers)

39. In Problem 38 we used a logistic function to model 𝑃 , total world oil production since 1859, as a function of time, 𝑡, in years since 1993. Use this function to answer the following questions:

40 𝑛 = 𝑓 (𝑡)

30 20 10 8

16

24

32

615

𝑡 (hours since midnight)

Figure 11.76 37. According to an article in The New York Times,24 pigweed has acquired resistance to the weedkiller Roundup. Let 𝑁 be the number of acres, in millions, where Roundup-resistant pigweed is found. Suppose the relative growth rate, (1∕𝑁)𝑑𝑁∕ 𝑑𝑡, was 15% when 𝑁 = 5 and 14.5% when 𝑁 = 10. Assuming the relative growth rate is a linear function of 𝑁, write a differential equation to model 𝑁 as a function of time, and predict how many acres will eventually be afflicted before the spread of Roundup-resistant pigweed halts. In Problems 38–42, we analyze world oil production.25 When annual world oil production peaks and starts to decline, major economic restructuring will be needed. We investigate when this slowdown is projected to occur. 38. We define 𝑃 to be the total oil production worldwide since 1859 in billions of barrels. In 1993, annual world oil production was 22.0 billion barrels and the total production was 𝑃 = 724 billion barrels. In 2013, annual production was 27.8 billion barrels and the total production was 𝑃 = 1235 billion barrels. Let 𝑡 be time in years since 1993. (a) Estimate the rate of production, 𝑑𝑃 ∕𝑑𝑡, for 1993 and 2013. (b) Estimate the relative growth rate, (1∕𝑃 )(𝑑𝑃 ∕𝑑𝑡), for 1993 and 2013. (c) Find an equation for the relative growth rate, (1∕𝑃 )(𝑑𝑃 ∕𝑑𝑡), as a function of 𝑃 , assuming that the function is linear. (d) Assuming that 𝑃 increases logistically and that all oil in the ground will ultimately be extracted, estimate the world oil reserves in 1859 to the nearest billion barrels. (e) Write and solve the logistic differential equation modeling 𝑃 .

(a) When does peak annual world oil production occur? (b) Geologists have estimated world oil reserves to be as high as 3500 billion barrels.26 When does peak world oil production occur with this assumption? (Assume 𝑘 and 𝑃0 are unchanged.) 40. As in Problem 38, let 𝑃 be total world oil production since 1859. In 1998, annual world production was 24.4 billion barrels and total production was 𝑃 = 841 billion barrels. In 2003, annual production was 25.3 billion barrels and total production was 𝑃 = 964 billion barrels. In 2008, annual production was 26.9 billion barrels and the total production was 1100 billion barrels. (a) Graph 𝑑𝑃 ∕𝑑𝑡 versus 𝑃 from Problem 38 and show the data for 1998, 2003 and 2008. How well does the model fit the data? (b) Graph the logistic function modeling worldwide oil production (𝑃 versus 𝑡) from Problem 38 and show the data for 1998, 2003 and 2008. How well does the model fit the data? 41. Use the logistic function obtained in Problem 38 to model the growth of 𝑃 , the total oil produced worldwide in billions of barrels since 1859: (a) Find the projected value of 𝑃 for 2014 to the nearest billion barrels. (b) Use the derivative to estimate the annual world oil production during 2014. (c) How much oil is projected to remain in the ground in 2014? (d) Compare the projected production in part (b) with the actual figure of 28.4 billion barrels. 42. With 𝑃 , the total oil produced worldwide since 1859, in billions of barrels, modeled as a function of time 𝑡 in years since 1993 as in Problem 38: (a) Predict the total quantity of oil produced by 2020. (b) In what year does the model predict that only 300 billion barrels remain?

Strengthen Your Understanding In Problems 43–45, explain what is wrong with the statement. 43. The differential equation 𝑑𝑃 ∕𝑑𝑡 = 0.08𝑃 − 0.0032𝑃 2 has one equilibrium solution, at 𝑃 = 25.

44. The maximum rate of change occurs at 𝑡 = 25 for a quantity 𝑄 growing according to the logistic equation 𝑑𝑄 = 0.13𝑄(1 − 0.02𝑄). 𝑑𝑡

24 http://www.nytimes.com/2010/05/04/business/energy-environment/04weed.html, 25 cta.ornl.gov,

accessed April 29, 2015. 26 www.hoodriver.k12.or.us, accessed April 29, 2015.

accessed May 3, 2010.

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Chapter 11 DIFFERENTIAL EQUATIONS

45. Figure 11.77 shows a quantity growing logistically.

48. A graph of 𝑑𝑄∕𝑑𝑡 against 𝑄 if 𝑄 is growing logistically and has an equilibrium value at 𝑄 = 500.

𝑃

49. A graph of 𝑑𝑃 ∕𝑑𝑡 against 𝑃 if 𝑃 is a logistic function which increases when 0 < 𝑃 < 20 and which decreases when 𝑃 < 0 or 𝑃 > 20.

100 60

𝑡

Are the statements in Problems 50–51 true or false? Give an explanation for your answer.

Figure 11.77

In Problems 46–49, give an example of:

50. There is a solution curve for the logistic differential equation 𝑑𝑃 ∕𝑑𝑡 = 𝑃 (2 − 𝑃 ) that goes through the points (0, 1) and (1, 3).

46. A quantity that increases logistically. 47. A logistic differential equation for a quantity 𝑃 such that the maximum rate of change of 𝑃 occurs when 𝑃 = 75.

51. For any positive values of the constant 𝑘 and any positive values of the initial value 𝑃 (0), the solution to the differential equation 𝑑𝑃 ∕𝑑𝑡 = 𝑘𝑃 (𝐿 − 𝑃 ) has limiting value 𝐿 as 𝑡 → ∞.

11.8

SYSTEMS OF DIFFERENTIAL EQUATIONS In Section 11.7 we modeled the growth of a single population over time. We now consider the growth of two populations that interact, such as a population of sick people infecting the healthy people around them. This involves not just one differential equation, but a system of two.

Diseases and Epidemics Differential equations can be used to predict when an outbreak of a disease will become so severe that it is called an epidemic27 and to decide what level of vaccination is necessary to prevent an epidemic. Let’s consider a specific example.

Flu in a British Boarding School In January 1978, 763 students returned to a boys’ boarding school after their winter vacation. A week later, one boy developed the flu, followed by two others the next day. By the end of the month, nearly half the boys were sick. Most of the school had been affected by the time the epidemic was over in mid-February.28 Being able to predict how many people will get sick, and when, is an important step toward controlling an epidemic. This is one of the responsibilities of Britain’s Communicable Disease Surveillance Centre and the US’s Center for Disease Control and Prevention.

The 𝑺 -𝑰 -𝑹 Model We apply one of the most commonly used models for an epidemic, called the 𝑆-𝐼-𝑅 model, to the boarding school flu example. The population of the school is divided into three groups: 𝑆 = the number of susceptibles, the people who are not yet sick but who could become sick 𝐼 = the number of infecteds, the people who are currently sick 𝑅 = the number of recovered, or removed, the people who have been sick and can no longer infect others or be reinfected. The number of susceptibles decreases with time as people become infected. We assume that the rate at which people become infected is proportional to the number of contacts between susceptible 27 Exactly when a disease should be called an epidemic is not always clear. The medical profession generally classifies a disease an epidemic when the frequency is higher than usually expected—leaving open the question of what is usually expected. See, for example, Epidemiology in Medicine by C. H. Hennekens and J. Buring (Boston: Little, Brown, 1987). 28 Data from the Communicable Disease Surveillance Centre (UK); reported in “Influenza in a Boarding School,” British Medical Journal, March 4, 1978, and by J. D. Murray in Mathematical Biology (New York: Springer Verlag, 1990).

11.8 SYSTEMS OF DIFFERENTIAL EQUATIONS

617

and infected people. We expect the number of contacts between the two groups to be proportional to both 𝑆 and 𝐼. (If 𝑆 doubles, we expect the number of contacts to double; similarly, if 𝐼 doubles, we expect the number of contacts to double.) Thus we assume that the number of contacts is proportional to the product, 𝑆𝐼. In other words, we assume that for some constant 𝑎 > 0, ( ) Rate susceptibles 𝑑𝑆 =− = −𝑎𝑆𝐼. 𝑑𝑡 get sick (The negative sign is used because 𝑆 is decreasing.) The number of infecteds is changing in two ways: newly sick people are added to the infected group, and others are removed. The newly sick people are exactly those people leaving the susceptible group and so accrue at a rate of 𝑎𝑆𝐼 (with a positive sign this time). People leave the infected group either because they recover (or die), or because they are physically removed from the rest of the group and can no longer infect others. We assume that people are removed at a rate proportional to the number of sick, or 𝑏𝐼, where 𝑏 is a positive constant. Thus, 𝑑𝐼 = 𝑑𝑡

Rate susceptibles get sick



Rate infecteds get removed

= 𝑎𝑆𝐼 − 𝑏𝐼.

Assuming that those who have recovered from the disease are no longer susceptible, the recovered group increases at the rate of 𝑏𝐼, so 𝑑𝑅 = 𝑏𝐼. 𝑑𝑡 We are assuming that having the flu confers immunity on a person, that is, that the person cannot get the flu again. (This is true for a given strain of flu, at least in the short run.) In analyzing the flu, we can use the fact that the total population 𝑆 + 𝐼 + 𝑅 is not changing. (The total population, the total number of boys in the school, did not change during the epidemic.) Thus, once we know 𝑆 and 𝐼, we can calculate 𝑅. So we restrict our attention to the two equations 𝑑𝑆 = −𝑎𝑆𝐼 𝑑𝑡 𝑑𝐼 = 𝑎𝑆𝐼 − 𝑏𝐼. 𝑑𝑡 The Constants 𝒂 and 𝒃 The constant 𝑎 measures how infectious the disease is—that is, how quickly it is transmitted from the infecteds to the susceptibles. In the case of the flu, we know from medical accounts that the epidemic started with one sick boy, with two more becoming sick a day later. Thus, when 𝐼 = 1 and 𝑆 = 762, we have 𝑑𝑆∕𝑑𝑡 ≈ −2, enabling us to roughly29 approximate 𝑎: 𝑑𝑆∕𝑑𝑡 2 = = 0.0026. 𝑆𝐼 (762)(1) The constant 𝑏 represents the rate at which infected people are removed from the infected population. In this case of the flu, boys were generally taken to the infirmary within one or two days of becoming sick. About half the infected population was removed each day, so we take 𝑏 ≈ 0.5. Thus, our equations are: 𝑑𝑆 = −0.0026𝑆𝐼 𝑑𝑡 𝑑𝐼 = 0.0026𝑆𝐼 − 0.5𝐼. 𝑑𝑡 𝑎=−

The Phase Plane We can get a good idea of the progress of the disease from graphs. You might expect that we would look for graphs of 𝑆 and 𝐼 against 𝑡, and eventually we will. However, we first look at a graph of 𝐼 against 𝑆. If we plot a point (𝑆, 𝐼) representing the number of susceptibles and the number of infecteds at any moment in time, then, as the numbers of 29 The values of 𝑎 and 𝑏 are close to those obtained by J. D. Murray in Mathematical Biology (New York: Springer Verlag, 1990).

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Chapter 11 DIFFERENTIAL EQUATIONS

susceptibles and infecteds change, the point moves. The 𝑆𝐼-plane on which the point moves is called the phase plane. The path along which the point moves is called the phase trajectory, or orbit, of the point. To find the phase trajectory, we need a differential equation relating 𝑆 and 𝐼 directly. Thinking of 𝐼 as a function of 𝑆, and 𝑆 as a function of 𝑡, we use the chain rule to get 𝑑𝐼 𝑑𝐼 𝑑𝑆 = ⋅ , 𝑑𝑡 𝑑𝑆 𝑑𝑡 giving 𝑑𝐼∕𝑑𝑡 𝑑𝐼 = . 𝑑𝑆 𝑑𝑆∕𝑑𝑡 Substituting for 𝑑𝐼∕𝑑𝑡 and 𝑑𝑆∕𝑑𝑡, we get 0.0026𝑆𝐼 − 0.5𝐼 𝑑𝐼 = . 𝑑𝑆 −0.0026𝑆𝐼 Assuming 𝐼 is not zero, this equation simplifies to approximately 192 𝑑𝐼 = −1 + . 𝑑𝑆 𝑆 The slope field of this differential equation is shown in Figure 11.78. The trajectory with initial condition 𝑆0 = 762, 𝐼0 = 1 is shown in Figure 11.79. Time is represented by the arrow showing the direction that a point moves on the trajectory. The disease starts at the point 𝑆0 = 762, 𝐼0 = 1. At first, more people become infected and fewer are susceptible. In other words, 𝑆 decreases and 𝐼 increases. Later, 𝐼 decreases as 𝑆 continues to decrease. 𝐼 (infecteds) 400

𝐼 (infecteds) 400

200

200

400

𝑆 800 (susceptibles)

Figure 11.78: Slope field for 𝑑𝐼∕𝑑𝑆 = −1 + 192∕𝑆

⑥ 192

400

(762, 1) 𝑆 (susceptibles)

Figure 11.79: Trajectory for 𝑆0 = 762, 𝐼0 = 1

What Does the 𝑺𝑰 -Phase Plane Tell Us? To learn how the disease progresses, look at the shape of the curve in Figure 11.79. The value of 𝐼 increases to about 300 (the maximum number infected and infectious at any one time); then 𝐼 decreases to zero. This peak value of 𝐼 occurs when 𝑆 ≈ 200. We can determine exactly when the peak value occurs by solving 𝑑𝐼 192 = −1 + = 0, 𝑑𝑆 𝑆 which gives 𝑆 = 192. Notice that the peak value for 𝐼 always occurs at the same value of 𝑆, namely 𝑆 = 192. The graph shows that if a trajectory starts with 𝑆0 > 192, then 𝐼 first increases and then decreases to zero. On the other hand, if 𝑆0 < 192, there is no peak as 𝐼 decreases right away. For this example, the value 𝑆0 = 192 is called a threshold value. If 𝑆0 is around or below 192, there is no epidemic. If 𝑆0 is significantly greater than 192, an epidemic occurs.30 30 Here we are using J. D. Murray’s definition of an epidemic as an outbreak in which the number of infecteds increases from the initial value, 𝐼0 . See Mathematical Biology (New York: Springer Verlag, 1990).

11.8 SYSTEMS OF DIFFERENTIAL EQUATIONS

619

The phase diagram makes clear that the maximum value of 𝐼 is about 300. Another question answered by the phase plane diagram is the total number of students who are expected to get sick during the epidemic. (This is not the maximum value reached by 𝐼, which gives the maximum number infected at any one time.) The point at which the trajectory crosses the 𝑆-axis represents the time when the epidemic has passed (since 𝐼 = 0). The 𝑆-intercept shows how many boys never get the flu and thus, how many do get it.

How Many People Should Be Vaccinated? An epidemic can sometimes be avoided by vaccination. How many boys would have had to be vaccinated to prevent the flu epidemic? To answer this, think of vaccination as removing people from the 𝑆 category (without increasing 𝐼), which amounts to moving the initial point on the trajectory to the left, parallel to the 𝑆-axis. To avoid an epidemic, the initial value of 𝑆0 should be at or below the threshold value. Therefore, all but 192 boys would need to be vaccinated.

Graphs of 𝑺 and 𝑰 Against 𝒕 To find out exactly when 𝐼 reaches its maximum, we need numerical methods. A modification of Euler’s method was used to generate the solution curves of 𝑆 and 𝐼 against 𝑡 in Figure 11.80. Notice that the number of susceptibles drops throughout the disease as healthy people get sick. The number of infecteds peaks after about 6 days and then drops. The epidemic has run its course in 20 days.

Analytical Solution for the 𝑺𝑰 -Phase Trajectory The differential equation 192 𝑑𝐼 = −1 + 𝑑𝑆 𝑆 can be integrated, giving 𝐼 = −𝑆 + 192 ln 𝑆 + 𝐶. Using 𝑆0 = 762 and 𝐼0 = 1 gives 1 = −762 + 192 ln 762 + 𝐶, so we get 𝐶 = 763 − 192 ln 762. Substituting this value for 𝐶, we get: 𝐼 = −𝑆 + 192 ln 𝑆 − 192 ln 762 + 763 ( ) 𝑆 𝐼 = −𝑆 + 192 ln + 763. 762 This is the equation of the solution curve in Figure 11.79. number of people

762 Susceptibles

Infecteds

1

5

10

15

20

𝑡 (days)

Figure 11.80: Progress of the flu over time

Two Interacting Populations: Predator-Prey We now consider two populations which interact. They may compete for food, one may prey on the other, or they may enjoy a symbiotic relationship in which each helps the other. We model a predator-prey system using the Lotka-Volterra equations.

620

Chapter 11 DIFFERENTIAL EQUATIONS

Robins and Worms Let’s look at an idealized case31 in which robins are the predators and worms are the prey. There are 𝑟 thousand robins and 𝑤 million worms. If there were no robins, the worms would increase exponentially according to the equation 𝑑𝑤 = 𝑎𝑤 where 𝑎 is a constant and 𝑎 > 0. 𝑑𝑡 If there were no worms, the robins would have no food and their population would decrease according to the equation32 𝑑𝑟 = −𝑏𝑟 where 𝑏 is a constant and 𝑏 > 0. 𝑑𝑡 Now we account for the effect of the two populations on one another. Clearly, the presence of the robins is bad for the worms, so 𝑑𝑤 = 𝑎𝑤 − Effect of robins on worms. 𝑑𝑡 On the other hand, the robins do better with the worms around, so 𝑑𝑟 = −𝑏𝑟 + Effect of worms on robins. 𝑑𝑡 How exactly do the two populations interact? Let’s assume the effect of one population on the other is proportional to the number of “encounters.” (An encounter is when a robin eats a worm.) The number of encounters is likely to be proportional to the product of the populations because the more there are of either population, the more encounters there will be. So we assume 𝑑𝑤 = 𝑎𝑤 − 𝑐𝑤𝑟 𝑑𝑡

and

𝑑𝑟 = −𝑏𝑟 + 𝑘𝑤𝑟, 𝑑𝑡

where 𝑐 and 𝑘 are positive constants. To analyze this system of equations, let’s look at the specific example with 𝑎 = 𝑏 = 𝑐 = 𝑘 = 1: 𝑑𝑤 = 𝑤 − 𝑤𝑟 𝑑𝑡

and

𝑑𝑟 = −𝑟 + 𝑤𝑟. 𝑑𝑡

To visualize the solutions to these equations, we look for trajectories in the phase plane. First we use the chain rule, 𝑑𝑟∕𝑑𝑡 𝑑𝑟 = , 𝑑𝑤 𝑑𝑤∕𝑑𝑡 to obtain

𝑑𝑟 −𝑟 + 𝑤𝑟 = . 𝑑𝑤 𝑤 − 𝑤𝑟

The Slope Field and Equilibrium Points We can get an idea of what solutions of this equation look like from the slope field in Figure 11.81. At the point (1, 1) there is no slope drawn because at this point the rate of change of the worm population with respect to time is zero: 𝑑𝑤 = 1 − (1)(1) = 0. 𝑑𝑡 The rate of change of the robin population with respect to time is also zero: 𝑑𝑟 = −1 + (1)(1) = 0. 𝑑𝑡 31 Based

on ideas from Thomas A. McMahon. might criticize this assumption because it predicts that the number of robins will decay exponentially, rather than die out in finite time. 32 You

11.8 SYSTEMS OF DIFFERENTIAL EQUATIONS

621

𝑟 (predator)

3

2

1

1

2

Figure 11.81: Slope field for

3

𝑤 (prey)

−𝑟 + 𝑤𝑟 𝑑𝑟 = 𝑑𝑤 𝑤 − 𝑤𝑟

Thus 𝑑𝑟∕𝑑𝑤 is undefined. In terms of worms and robins, this means that if at some moment 𝑤 = 1 and 𝑟 = 1 (that is, there are 1 million worms and 1 thousand robins), then 𝑤 and 𝑟 remain constant forever. The point 𝑤 = 1, 𝑟 = 1 is therefore an equilibrium solution. The slope field suggests that there are no other equilibrium points except the origin. At an equilibrium point, both 𝑤 and 𝑟 are constant, so 𝑑𝑤 =0 𝑑𝑡

and

𝑑𝑟 = 0. 𝑑𝑡

Therefore, we look for equilibrium points by solving 𝑑𝑤 𝑑𝑟 = 𝑤 − 𝑤𝑟 = 0 and = −𝑟 + 𝑟𝑤 = 0, 𝑑𝑡 𝑑𝑡 which has 𝑤 = 0, 𝑟 = 0 and 𝑤 = 1, 𝑟 = 1 as the only solutions.

Trajectories in the 𝒘𝒓-Phase Plane Let’s look at the trajectories in the phase plane. Remember that a point on a curve represents a pair of populations (𝑤, 𝑟) existing at the same time 𝑡 (though 𝑡 is not shown on the graph). A short time later, the pair of populations is represented by a nearby point. As time passes, the point traces out a trajectory. The direction is marked on the curve by an arrow. (See Figure 11.82.) How do we figure out which way to move on the trajectory? Approximating the solution numerically shows that the trajectory is traversed counterclockwise. Alternatively, look at the original pair of differential equations. At the point 𝑃0 in Figure 11.83, where 𝑤 > 1 and 𝑟 = 1, 𝑑𝑟 = −𝑟 + 𝑤𝑟 = −1 + 𝑤 > 0. 𝑑𝑡 Therefore, 𝑟 is increasing, so the point is moving counterclockwise around the closed curve. Now let’s think about why the solution curves are closed curves (that is, why they come back and meet themselves). Notice that the slope field is symmetric about the line 𝑤 = 𝑟. We can confirm this by observing that interchanging 𝑤 and 𝑟 does not alter the differential equation for 𝑑𝑟∕𝑑𝑤. This means that if we start at point 𝑃 on the line 𝑤 = 𝑟 and travel once around the point (1, 1), we arrive back at the same point 𝑃 . The reason is that the second half of the path, from 𝑄 to 𝑃 , is the reflection of the first half, from 𝑃 to 𝑄, about the line 𝑤 = 𝑟. (See Figure 11.82.) If we did not end up at 𝑃 again, the second half of our path would have a different shape from the first half.

622

Chapter 11 DIFFERENTIAL EQUATIONS

𝑟

𝑟 3

3

𝑃1 2

2

𝑃

1

1

𝑃2

𝑃0 𝑃3

𝑄 𝑤 1

2

𝑤

3

1

Figure 11.82: Solution curve is closed

2

3

Figure 11.83: Trajectory in the phase plane

The Populations as Functions of Time The shape of the trajectories tells us how the populations vary with time. We start at 𝑡 = 0 at the point 𝑃0 in Figure 11.83. Then we move to 𝑃1 at time 𝑡1 , to 𝑃2 at time 𝑡2 , to 𝑃3 at time 𝑡3 , and so on. At time 𝑡4 we are back at 𝑃0 , and the whole cycle repeats. Since the trajectory is a closed curve, both populations oscillate periodically with the same period. The worms (the prey) are at their maximum a quarter of a cycle before the robins. (See Figure 11.84.) population

Worms

Robins





𝑡 𝑃0 𝑃1 𝑃2

𝑃3 𝑃0 𝑃1 𝑃2

𝑃3 𝑃0 𝑃1 𝑃2

𝑃3 𝑃0

Figure 11.84: Populations of robins (in thousands) and worms (in millions) over time

Exercises and Problems for Section 11.8 Online Resource: Additional Problems for Section 11.8 EXERCISES

For Exercises 1–4, suppose 𝑥 and 𝑦 are the populations of two different species. Describe in words how each population changes with time.

In Exercises 5–8, find all equilibrium points. Give answers as ordered pairs (𝑥, 𝑦). 5.

1.

𝑦

2.

𝑦

7.

𝑥

𝑥

𝑑𝑥 = −3𝑥 + 𝑥𝑦 𝑑𝑡 𝑑𝑦 = 5𝑦 − 𝑥𝑦 𝑑𝑡 𝑑𝑥 = 15𝑥 − 5𝑥𝑦 𝑑𝑡 𝑑𝑦 = 10𝑦 + 2𝑥𝑦 𝑑𝑡

6.

8.

𝑑𝑥 = −2𝑥 + 4𝑥𝑦 𝑑𝑡 𝑑𝑦 = −8𝑦 + 2𝑥𝑦 𝑑𝑡 𝑑𝑥 = 𝑥2 − 𝑥𝑦 𝑑𝑡 𝑑𝑦 = 15𝑦 − 3𝑦2 𝑑𝑡

9. Given the system of differential equations 3.

𝑦

4.

𝑦

𝑑𝑥 = 5𝑥 − 3𝑥𝑦 𝑑𝑡 𝑑𝑦 = −8𝑦 + 𝑥𝑦 𝑑𝑡 determine whether 𝑥 and 𝑦 are increasing or decreasing at the point

𝑥

𝑥

(a) 𝑥 = 3, 𝑦 = 2

(b) 𝑥 = 5, 𝑦 = 1

11.8 SYSTEMS OF DIFFERENTIAL EQUATIONS

10. Given the system of differential equations 𝑑𝑃 = 2𝑃 − 10 𝑑𝑡 𝑑𝑄 = 𝑄 − 0.2𝑃 𝑄 𝑑𝑡

13. Find all equilibrium points of the system. For Exercises 14–18, 𝑣 and 𝑤 are the number of individuals in two interacting populations with 𝑣, 𝑤 > 0 satisfying the system of equations 1 𝑑𝑣 = −0.1 + 0.003𝑤 𝑣 𝑑𝑡 1 𝑑𝑤 = 0.06 − 0.001𝑣. 𝑤 𝑑𝑡

determine whether 𝑃 and 𝑄 are increasing or decreasing at the point (a)

𝑃 = 2, 𝑄 = 3

623

(b) 𝑃 = 6, 𝑄 = 5

For Exercises 11–13, 𝑥 and 𝑦 satisfy the system of differential equations 𝑑𝑥 = 𝑥(3 − 𝑥 + 𝑦), 𝑑𝑡 𝑑𝑦 = 𝑦(6 − 𝑥 − 𝑦). 𝑑𝑡

14. If 𝑣 = 20 and 𝑤 = 50, what is the relative growth rate of 𝑣? 15. If 𝑣 = 20 and 𝑤 = 50 what is 𝑑𝑣∕𝑑𝑡? 16. What must 𝑣 be in order for 𝑤 to remain constant?

11. Starting at 𝑃1 = (3, 2), as 𝑡 increases will (𝑥, 𝑦) approach 𝑃2 = (2.90, 1.96) or 𝑃3 = (3.10, 2.03)? 12. Starting at 𝑄1 = (6, 2), as 𝑡 increases will (𝑥, 𝑦) approach 𝑄2 = (5.90, 1.94) or 𝑄3 = (6.10, 2.07)?

17. Is the presence of the population with 𝑤 members helpful or harmful to the growth of the population with 𝑣 members? 18. Is the presence of the population with 𝑣 members helpful or harmful to the growth of the population with 𝑤 members?

PROBLEMS 19. Figure 11.85 shows the trajectory through the SI phase plane of a 50-day epidemic. (a) Make an approximate table of values for the number of susceptibles and infecteds on the days marked on the trajectory. (b) When is the epidemic at its peak? How many people are infected then? (c) During the course of the epidemic, how many catch the disease and how many are spared?

(b) How many have had the disease by day 20? (c) How many have had the disease by the time the epidemic is over?

susceptibles

4000 3000 2000 1000

infecteds

500

0

22 24

time (days)

10 20 30 40 50 60

20

infecteds

400 300

1000 800 600 400 200 0

28 16

200

32

100

36 40

44 400

12 8

time (days)

10 20 30 40 50 60 susceptibles

Figure 11.86

800 1200 1600 2000

Figure 11.85: Days 8 through 44 of an epidemic

20. Figure 11.86 shows the number of susceptibles and infecteds in a population of 4000 through the course of a 60-day epidemic. (a) How many are infected on day 20? 33 http://humansvszombies.org

21. Humans vs Zombies33 is a game in which one player starts as a zombie and turns human players into zombies by tagging them. Zombies have to “eat" on a regular basis by tagging human players, or they die of starvation and are out of the game. The game is usually played over a period of about five days. If we let 𝐻 represent the

624

Chapter 11 DIFFERENTIAL EQUATIONS

size of the human population and 𝑍 represent the size of the zombie population in the game, then, for constant parameters 𝑎, 𝑏, and 𝑐, we have:

𝑟 (predator) 4

3

𝑑𝐻 = 𝑎𝐻𝑍 𝑑𝑡 𝑑𝑍 = 𝑏𝑍 + 𝑐𝐻𝑍 𝑑𝑡

2

1

(a) Decide whether each of the parameters 𝑎, 𝑏, 𝑐 is positive or negative. (b) What is the relationship, if any, between 𝑎 and 𝑐?

𝑤 (prey) 1

2

Figure 11.87: 22. Four pairs of species are given, with descriptions of how they interact. I. Bees/flowers: each needs the other to survive II. Owls/trees: owls need trees but trees are indifferent III. Elk/buffalo: in competition and would do fine alone IV. Fox/hare: fox eats the hare and needs it to survive Match each system of differential equations with a species pair, and indicate which species is 𝑥 and which is 𝑦. 𝑑𝑥 = −0.2𝑥 + 0.03𝑥𝑦 𝑑𝑡 𝑑𝑦 = 0.4𝑦 − 0.08𝑥𝑦 𝑑𝑡 𝑑𝑥 (b) = 0.18𝑥 𝑑𝑡 𝑑𝑦 = −0.4𝑦 + 0.3𝑥𝑦 𝑑𝑡 𝑑𝑥 (c) = −0.6𝑥 + 0.18𝑥𝑦 𝑑𝑡 𝑑𝑦 = −0.1𝑦 + 0.09𝑥𝑦 𝑑𝑡 (d) Write a possible system of differential equations for the species pair that does not have a match. (a)

23. Show that if 𝑆, 𝐼, and 𝑅 satisfy the differential equations on page 617, the total population, 𝑆 + 𝐼 + 𝑅, is constant. For Problems 24–32, let 𝑤 be the number of worms (in millions) and 𝑟 the number of robins (in thousands) living on an island. Suppose 𝑤 and 𝑟 satisfy the following differential equations, which correspond to the slope field in Figure 11.87. 𝑑𝑤 = 𝑤 − 𝑤𝑟, 𝑑𝑡

𝑑𝑟 = −𝑟 + 𝑤𝑟. 𝑑𝑡

3 𝑑𝑟 𝑑𝑤

=

4 𝑟(𝑤−1) 𝑤(1−𝑟)

24. Explain why these differential equations are a reasonable model for interaction between the two populations. Why have the signs been chosen this way? 25. Solve these differential equations in the two special cases when there are no robins and when there are no worms living on the island. 26. Describe and explain the symmetry you observe in the slope field. What consequences does this symmetry have for the solution curves? 27. Assume 𝑤 = 2 and 𝑟 = 2 when 𝑡 = 0. Do the numbers of robins and worms increase or decrease at first? What happens in the long run? 28. For the case discussed in Problem 27, estimate the maximum and the minimum values of the robin population. How many worms are there at the time when the robin population reaches its maximum? 29. On the same axes, graph 𝑤 and 𝑟 (the worm and the robin populations) against time. Use initial values of 1.5 for 𝑤 and 1 for 𝑟. You may do this without units for 𝑡. 30. People on the island like robins so much that they decide to import 200 robins all the way from England, to increase the initial population from 𝑟 = 2 to 𝑟 = 2.2 when 𝑡 = 0. Does this make sense? Why or why not? 31. Assume that 𝑤 = 3 and 𝑟 = 1 when 𝑡 = 0. Do the numbers of robins and worms increase or decrease initially? What happens in the long run? 32. For the case discussed in Problem 31, estimate the maximum and minimum values of the robin population. Estimate the number of worms when the robin population reaches its minimum. In Problems 33–35, the system of differential equations models the interaction of two populations 𝑥 and 𝑦. (a) What kinds of interaction (symbiosis,34 competition, predator-prey) do the equations describe?

34 Symbiosis takes place when the interaction of two species benefits both. An example is the pollination of plants by insects.

11.8 SYSTEMS OF DIFFERENTIAL EQUATIONS

(b) What happens in the long run? Your answer may depend on the initial population. Draw a slope field. 33.

35.

1 𝑑𝑥 = 𝑦−1 𝑥 𝑑𝑡 1 𝑑𝑦 = 𝑥−1 𝑦 𝑑𝑡

34.

𝑥 𝑦 1 𝑑𝑥 =1− − 𝑥 𝑑𝑡 2 2 1 𝑑𝑦 = 1 − 𝑥 − 𝑦 𝑦 𝑑𝑡

1 𝑑𝑥 = 𝑦 − 1 − 0.05𝑥 𝑥 𝑑𝑡 1 𝑑𝑦 = 1 − 𝑥 − 0.05𝑦 𝑦 𝑑𝑡

1 𝑑𝑝 = 0.01𝑞 − 0.3 𝑝 𝑑𝑡 1 𝑑𝑞 = 0.02𝑝 − 0.2. 𝑞 𝑑𝑡 36. What is the relative rate of change of 𝑝 if 𝑞 = 10? 37. What populations result in an equilibrium? 38. Give a description of how these populations interact. For Problems 39–42, consider a conflict between two armies of 𝑥 and 𝑦 soldiers, respectively. During World War I, F. W. Lanchester assumed that if both armies are fighting a conventional battle within sight of one another, the rate at which soldiers in one army are put out of action (killed or wounded) is proportional to the amount of fire the other army can concentrate on them, which is in turn proportional to the number of soldiers in the opposing army. Thus Lanchester assumed that if there are no reinforcements and 𝑡 represents time since the start of the battle, then 𝑥 and 𝑦 obey the differential equations

𝑎, 𝑏 > 0.

39. Near the end of World War II a fierce battle took place between US and Japanese troops over the island of Iwo Jima, off the coast of Japan. Applying Lanchester’s analysis to this battle, with 𝑥 representing the number of US troops and 𝑦 the number of Japanese troops, it has been estimated35 that 𝑎 = 0.05 and 𝑏 = 0.01. (a) Using these values for 𝑎 and 𝑏 and ignoring reinforcements, write a differential equation involving 𝑑𝑦∕𝑑𝑥 and sketch its slope field.

35 See

(b) Assuming that the initial strength of the US forces was 54,000 and that of the Japanese was 21,500, draw the trajectory which describes the battle. What outcome is predicted? (That is, which side do the differential equations predict will win?) (c) Would knowing that the US in fact had 19,000 reinforcements, while the Japanese had none, alter the outcome predicted? 40. (a) For two armies of strengths 𝑥 and 𝑦 fighting a conventional battle governed by Lanchester’s differential equations, write a differential equation involving 𝑑𝑦∕𝑑𝑥 and the constants of attrition 𝑎 and 𝑏. (b) Solve the differential equation and hence show that the equation of the phase trajectory is

For Problems 36–38, 𝑝 and 𝑞 are the number of individuals in two interacting populations with 𝑝, 𝑞 > 0 satisfying the system of equations

𝑑𝑥 = −𝑎𝑦 𝑑𝑡 𝑑𝑦 = −𝑏𝑥 𝑑𝑡

625

𝑎𝑦2 − 𝑏𝑥2 = 𝐶 for some constant 𝐶. This equation is called Lanchester’s square law. The value of 𝐶 depends on the initial sizes of the two armies. 41. Consider the battle of Iwo Jima, described in Problem 39. Take 𝑎 = 0.05, 𝑏 = 0.01 and assume the initial strength of the US troops to be 54,000 and that of the Japanese troops to be 21,500. (Again, ignore reinforcements.) (a) Using Lanchester’s square law derived in Problem 40, find the equation of the trajectory describing the battle. (b) Assuming that the Japanese fought without surrendering until they had all been killed, as was the case, how many US troops does this model predict would be left when the battle ended? 42. In this problem we adapt Lanchester’s model for a conventional battle to the case in which one or both of the armies is a guerrilla force. We assume that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (a) Give a justification for the assumption that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (b) Write the differential equations which describe a conflict between a guerrilla army of strength 𝑥 and a conventional army of strength 𝑦, assuming all the constants of proportionality are 1. (c) Find a differential equation involving 𝑑𝑦∕𝑑𝑥 and solve it to find equations of phase trajectories. (d) Describe which side wins in terms of the constant of integration. What happens if the constant is zero? (e) Use your solution to part (d) to divide the phase plane into regions according to which side wins.

Martin Braun, Differential Equations and Their Applications, 2nd ed. (New York: Springer Verlag, 1975).

626

Chapter 11 DIFFERENTIAL EQUATIONS

Strengthen Your Understanding In Problems 43–44, explain what is wrong with the statement. 43. If 𝑑𝑥∕𝑑𝑡 = 3𝑥 − 0.4𝑥𝑦 and 𝑑𝑦∕𝑑𝑡 = 4𝑦 − 0.5𝑥𝑦, then an increase in 𝑥 corresponds to a decrease in 𝑦. 44. For a system of differential equations for 𝑥 and 𝑦, at the point (2, 3), we have 𝑑𝑥∕𝑑𝑡 < 0 and 𝑑𝑦∕𝑑𝑡 > 0 and 𝑑𝑦∕𝑑𝑥 > 0.

47. Two diseases 𝐷1 and 𝐷2 such that the parameter 𝑎 in the 𝑆-𝐼-𝑅 model on page 616 is larger for disease 𝐷1 than it is for disease 𝐷2 . Explain your reasoning. Are the statements in Problems 48–49 true or false? Give an explanation for your answer.

In Problems 45–47, give an example of: 45. A system of differential equations for two populations 𝑋 and 𝑌 such that 𝑌 needs 𝑋 to survive and 𝑋 is indifferent to 𝑌 and thrives on its own. Let 𝑥 represent the size of the 𝑋 population and 𝑦 represent the size of the 𝑌 population.

11.9

46. A system of differential equations for the profits of two companies if each would thrive on its own but the two companies compete for business. Let 𝑥 and 𝑦 represent the profits of the two companies.

48. The system of differential equations 𝑑𝑥∕𝑑𝑡 = −𝑥 + 𝑥𝑦2 and 𝑑𝑦∕𝑑𝑡 = 𝑦 − 𝑥2 𝑦 requires initial conditions for both 𝑥(0) and 𝑦(0) to determine a unique solution. 49. Populations modeled by a system of differential equations never die out.

ANALYZING THE PHASE PLANE In the previous section we analyzed a system of differential equations using a slope field. In this section we analyze a system of differential equations using nullclines. We consider two species having similar niches, or ways of living, and that are in competition for food and space. In such cases, one species often becomes extinct. This phenomenon is called the Principle of Competitive Exclusion. We see how differential equations predict this in a particular case.

Competitive Exclusion: Citrus Tree Parasites The citrus farmers of Southern California are interested in controlling the insects that live on their trees. Some of these insects can be controlled by parasites that live on the trees too. Scientists are, therefore, interested in understanding under what circumstances these parasites flourish or die out. One such parasite was introduced accidentally from the Mediterranean; later, other parasites were introduced from China and India; in each case the previous parasite became extinct over part of its habitat. In 1963 a lab experiment was carried out to determine which one of a pair of species became extinct when they were in competition with each other. The data on one pair of species, called A. fisheri and A. melinus, with populations 𝑃1 and 𝑃2 respectively, is given in Table 11.9 and shows that A. melinus (𝑃2 ) became extinct after 8 generations.36 Table 11.9

Population (in thousands) of two species of parasite as a function of time

Generation number

1

2

3

4

5

6

7

8

Population 𝑃1 (thousands)

0.193

1.093

1.834

5.819

13.705

16.965

18.381

16.234

Population 𝑃2 (thousands)

0.083

0.229

0.282

0.378

0.737

0.507

0.13

0

Data from the same experimenters indicates that, when alone, each population grows logistically. In fact, their data suggests that, when alone, the population of 𝑃1 might grow according to the equation ) ( 𝑃 𝑑𝑃1 = 0.05𝑃1 1 − 1 , 𝑑𝑡 20 and when alone, the population of 𝑃2 might grow according to the equation ) ( 𝑃 𝑑𝑃2 = 0.09𝑃2 1 − 2 . 𝑑𝑡 15 Now suppose both parasites are present. Each tends to reduce the growth rate of the other, so each 36 Data adapted from Paul DeBach and Ragnhild Sundby, “Competitive Displacement Between Ecological Homologues,” Hilgardia 34:17 (1963).

11.9 ANALYZING THE PHASE PLANE

627

differential equation is modified by subtracting a term on the right. The experimental data shows that together 𝑃1 and 𝑃2 can be well described by the equations ) ( 𝑃1 𝑑𝑃1 = 0.05𝑃1 1 − − 0.002𝑃1 𝑃2 𝑑𝑡 20 ) ( 𝑃 𝑑𝑃2 = 0.09𝑃2 1 − 2 − 0.15𝑃1 𝑃2 . 𝑑𝑡 15 The fact that 𝑃2 dies out with time is reflected in these equations: the coefficient of 𝑃1 𝑃2 is much larger in the equation for 𝑃2 than in the equation for 𝑃1 . This indicates that the interaction has a much more devastating effect upon the growth of 𝑃2 than on the growth of 𝑃1 .

The Phase Plane and Nullclines We consider the phase plane with the 𝑃1 axis horizontal and the 𝑃2 axis vertical. To find the trajectories in the 𝑃1 𝑃2 phase plane, we could draw a slope field as in the previous section. Instead, we use a method that gives a good qualitative picture of the behavior of the trajectories even without a calculator or computer. We find the nullclines or curves along which 𝑑𝑃1 ∕𝑑𝑡 = 0 or 𝑑𝑃2 ∕𝑑𝑡 = 0. At points where 𝑑𝑃2 ∕𝑑𝑡 = 0, the population 𝑃2 is momentarily constant, so only population 𝑃1 is changing with time. Therefore, at this point the trajectory is horizontal. (See Figure 11.88.) Similarly, at points where 𝑑𝑃1 ∕𝑑𝑡 = 0, the population 𝑃1 is momentarily constant and population 𝑃2 is the only one changing, so the trajectory is vertical there. A point where both 𝑑𝑃1 ∕𝑑𝑡 = 0 and 𝑑𝑃2 ∕𝑑𝑡 = 0 is called an equilibrium point because 𝑃1 and 𝑃2 both remain constant if they reach these values. 𝑃2

𝑑𝑃1 =0 𝑑𝑡

𝑑𝑃2 =0 𝑑𝑡 𝑃1 Figure 11.88: Points on a trajectory where 𝑑𝑃1 ∕𝑑𝑡 = 0 or 𝑑𝑃2 ∕𝑑𝑡 = 0

On the 𝑃1 𝑃2 phase plane: 𝑑𝑃1 = 0, the trajectory is vertical. • If 𝑑𝑡 𝑑𝑃2 • If = 0, the trajectory is horizontal. 𝑑𝑡 𝑑𝑃1 𝑑𝑃2 • If = = 0, there is an equilibrium point. 𝑑𝑡 𝑑𝑡

Using Nullclines to Analyze the Parasite Populations In order to see where 𝑑𝑃1 ∕𝑑𝑡 = 0 or 𝑑𝑃2 ∕𝑑𝑡 = 0, we factor the right side of our differential equations: ) ( 𝑃1 𝑑𝑃1 = 0.05𝑃1 1 − − 0.002𝑃1𝑃2 = 0.001𝑃1(50 − 2.5𝑃1 − 2𝑃2 ) 𝑑𝑡 20 ) ( 𝑃2 𝑑𝑃2 = 0.09𝑃2 1 − − 0.15𝑃1 𝑃2 = 0.001𝑃2 (90 − 150𝑃1 − 6𝑃2 ). 𝑑𝑡 15

628

Chapter 11 DIFFERENTIAL EQUATIONS

𝑃2 25 𝑃1 = 0, 𝑑𝑃1 ∕𝑑𝑡 = 0

Region I



✛ ✠❄

15 Region ✲ II

❄ ❘

Region III

✻ ✒ ✲





✛ 0.6 ✻ 20 𝑃2 = 0, 𝑑𝑃2 ∕𝑑𝑡 = 0

{

50 − 2.5𝑃1 − 2𝑃2 = 0 𝑑𝑃1 ∕𝑑𝑡 = 0

{

90 − 150𝑃1 − 6𝑃2 = 0 𝑑𝑃2 ∕𝑑𝑡 = 0

𝑃1

Figure 11.89: Analyzing three regions in the phase plane using nullclines (axes distorted) with equilibrium points represented by dots

Thus 𝑑𝑃1 ∕𝑑𝑡 = 0 where 𝑃1 = 0 or where 50 − 2.5𝑃1 − 2𝑃2 = 0. Graphing these equations in the phase plane gives two lines, which are nullclines. Since the trajectory is vertical where 𝑑𝑃1 ∕𝑑𝑡 = 0, in Figure 11.89 we draw small vertical line segments on these nullclines to represent the direction of the trajectories as they cross the nullcline. Similarly 𝑑𝑃2 ∕𝑑𝑡 = 0 where 𝑃2 = 0 or where 90 − 150𝑃1 − 6𝑃2 = 0. These equations are graphed in Figure 11.89 with small horizontal line segments on them. The equilibrium points are where both 𝑑𝑃1 ∕𝑑𝑡 = 0 and 𝑑𝑃2 ∕𝑑𝑡 = 0, namely the points 𝑃1 = 0, 𝑃2 = 0 (meaning that both species die out); 𝑃1 = 0, 𝑃2 = 15 (where 𝑃1 is extinct); and 𝑃1 = 20, 𝑃2 = 0 (where 𝑃2 is extinct). Notice that the equilibrium points are located at the intersection of a nullcline with vertical line segments and a nullcline with horizontal line segments.

What Happens in the Regions Between the Nullclines? Nullclines are useful because they divide the plane into regions in which the signs of 𝑑𝑃1 ∕𝑑𝑡 and 𝑑𝑃2 ∕𝑑𝑡 are constant. In each region, the direction of every trajectory remains roughly the same. In Region I, for example, we might try the point 𝑃1 = 20, 𝑃2 = 25. Then 𝑑𝑃1 = 0.001(20)(50 − 2.5(20) − 2(25)) < 0 𝑑𝑡 𝑑𝑃2 = 0.001(25)(90 − 150(20) − 6(25)) < 0. 𝑑𝑡 Now 𝑑𝑃1 ∕𝑑𝑡 < 0, so 𝑃1 is decreasing, which can be represented by an arrow in the direction ←. Also, 𝑑𝑃2 ∕𝑑𝑡 < 0, so 𝑃2 is decreasing, as represented by the arrow ↓. Combining these directions, ✛ we know that the trajectories in this region go approximately in the diagonal direction ✠❄(See Region I in Figure 11.89.) In Region II, try, for example, 𝑃1 = 1, 𝑃2 = 1. Then we have 𝑑𝑃1 = 0.001(1)(50 − 2.5 − 2) > 0 𝑑𝑡 𝑑𝑃2 = 0.001(1)(90 − 150 − 6) < 0. 𝑑𝑡 So here, 𝑃1 is increasing while 𝑃2 is decreasing. (See Region II in Figure 11.89.) In Region III, try 𝑃1 = 0.1, 𝑃2 = 0.1: 𝑑𝑃1 = 0.001(0.1)(50 − 2.5(0.1) − 2(0.1)) > 0 𝑑𝑡 𝑑𝑃2 = 0.001(0.1)(90 − 150(0.1) − 6(0.1)) > 0. 𝑑𝑡

11.9 ANALYZING THE PHASE PLANE

629

So here, both 𝑃1 and 𝑃2 are increasing. (See Region III in Figure 11.89.) Notice that the behavior of the populations in each region makes biological sense. In region I both populations are so large that overpopulation is a problem, so both populations decrease. In Region III both populations are so small that they are effectively not in competition, so both grow. In Region II competition between the species comes into play. The fact that 𝑃1 increases while 𝑃2 decreases in Region II means that 𝑃1 wins.

Solution Trajectories Suppose the system starts with some of each population. This means that the initial point of the trajectory is not on one of the axes, and so it is in Region I, II, or III. Then the point moves on a trajectory like one of those computed numerically and shown in Figure 11.90. Notice that all these trajectories tend toward the point 𝑃1 = 20, 𝑃2 = 0, corresponding to a population of 20,000 for 𝑃1 and extinction for 𝑃2 . Consequently, this model predicts that no matter what the initial populations are, provided 𝑃1 ≠ 0, the population of 𝑃2 is excluded by 𝑃1 , and 𝑃1 tends to a constant value. This makes biological sense: in the absence of 𝑃2 , we would expect 𝑃1 to settle down to the carrying capacity of the niche, which is 20,000. 𝑃2 25

15

0.6

20

𝑃1

Figure 11.90: Trajectories showing exclusion of population 𝑃2 (not to scale)

Exercises and Problems for Section 11.9 EXERCISES 1. Give the coordinates for each equilibrium point.

In Exercises 1–5, use Figure 11.91.

2. At each point, give the signs of 𝑑𝑥∕𝑑𝑡 and 𝑑𝑦∕𝑑𝑡. 𝑦 15

Region II

✛ ✠❄

Region III ✲

❘ ❄

(b) (4, 10)

(c)

(6, 15)

3. Draw a possible trajectory that starts at the point (2, 5). 4. Draw a possible trajectory that starts at the point (2, 10).

10 Region I 5

(a) (4, 7)

Region IV

✛✻ ■

✲ ✒ ✻

𝑥 2

4

6

Figure 11.91: Nullclines in the phase plane of a system of differential equations

5. What is the long-run behavior of a trajectory that starts at any point in the first quadrant?

630

Chapter 11 DIFFERENTIAL EQUATIONS

In Exercises 6–10, use Figure 11.92, which shows four nullclines, two of which are on the axes.

10. What is the long-run behavior of a trajectory that starts at any point in the first quadrant? 11. Figure 11.93 shows a phase plane for a system of differential equations. Draw the nullclines.

𝑦 6

𝑦 4

✛✻ ■ ✻ ✒ ✲

2

5

✛ ✠ ❄

4 3 2 1 𝑥

5

10

15

Figure 11.92: Nullclines in the phase plane of a system of differential equations

6. Give the approximate coordinates for each equilibrium point. 7. At each point, give the signs of 𝑑𝑥∕𝑑𝑡 and 𝑑𝑦∕𝑑𝑡. (a)

(5, 2)

(b) (10, 2)

(c)

𝑥 1

2

3

4

5

6

Figure 11.93

12. (a) Find the equilibrium points for the following system of equations 𝑑𝑥 = 20𝑥 − 10𝑥𝑦 𝑑𝑡

(10, 1)

8. Draw a possible trajectory that starts at the point (2, 5). 9. Draw a possible trajectory that starts at the point (10, 4).

𝑑𝑦 = 25𝑦 − 5𝑥𝑦. 𝑑𝑡 (b) Explain why 𝑥 = 2, 𝑦 = 4 is not an equilibrium point for this system.

PROBLEMS For Problems 13–18, analyze the phase plane of the differential equations for 𝑥, 𝑦 ≥ 0. Show the nullclines and equilibrium points, and sketch the direction of the trajectories in each region. 𝑑𝑥 = 𝑥(2 − 𝑥 − 𝑦) 13. 𝑑𝑡 𝑑𝑦 = 𝑦(1 − 𝑥 − 𝑦) 𝑑𝑡 𝑑𝑥 = 𝑥(2 − 𝑥 − 2𝑦) 𝑑𝑡 𝑑𝑦 = 𝑦(1 − 2𝑥 − 𝑦) 𝑑𝑡 ( 𝑦) 𝑑𝑥 17. =𝑥 1−𝑥− 𝑑𝑡 3) ( 𝑑𝑦 𝑥 =𝑦 1−𝑦− 𝑑𝑡 2 15.

𝑑𝑥 14. = 𝑥(2 − 𝑥 − 3𝑦) 𝑑𝑡 𝑑𝑦 = 𝑦(1 − 2𝑥) 𝑑𝑡 𝑑𝑥 𝑥 = 𝑥(1 − 𝑦 − ) 𝑑𝑡 3 𝑦 𝑑𝑦 = 𝑦(1 − − 𝑥) 𝑑𝑡 2 ( ) 𝑑𝑥 𝑥 18. =𝑥 1− −𝑦 𝑑𝑡 2 ( ) 𝑑𝑦 𝑦 = 𝑦 1− −𝑥 𝑑𝑡 3

16.

19. The equations describing the flu epidemic in a boarding school are 𝑑𝑆 = −0.0026𝑆𝐼 𝑑𝑡 𝑑𝐼 = 0.0026𝑆𝐼 − 0.5𝐼. 𝑑𝑡 (a) Find the nullclines and equilibrium points in the 𝑆𝐼 phase plane.

(b) Find the direction of the trajectories in each region. (c) Sketch some typical trajectories and describe their behavior in words. 20. Use the idea of nullclines dividing the plane into sectors to analyze the equations describing the interactions of robins and worms: 𝑑𝑤 = 𝑤 − 𝑤𝑟 𝑑𝑡 𝑑𝑟 = −𝑟 + 𝑟𝑤. 𝑑𝑡 21. Two companies share the market for a new technology. They have no competition except each other. Let 𝐴(𝑡) be the net worth of one company and 𝐵(𝑡) the net worth of the other at time 𝑡. Suppose that net worth cannot be negative and that 𝐴 and 𝐵 satisfy the differential equations 𝐴′ = 2𝐴 − 𝐴𝐵 𝐵 ′ = 𝐵 − 𝐴𝐵. (a) What do these equations predict about the net worth of each company if the other were not present? What effect do the companies have on each other?

631

11.9 ANALYZING THE PHASE PLANE

(b) Are there any equilibrium points? If so, what are they? (c) Sketch a slope field for these equations (using a computer or calculator), and hence describe the different possible long-run behaviors. 22. In the 1930s L. F. Richardson proposed that an arms race between two countries could be modeled by a system of differential equations. One arms race that can be reasonably well described by differential equations is the US-Soviet Union arms race between 1945 and 1960. If $𝑥 represents the annual Soviet expenditures on armaments (in billions of dollars) and $𝑦 represents the corresponding US expenditures, it has been suggested37 that 𝑥 and 𝑦 obey the following differential equations: 𝑑𝑥 = −0.45𝑥 + 10.5 𝑑𝑡 𝑑𝑦 = 8.2𝑥 − 0.8𝑦 − 142. 𝑑𝑡 (a) Find the nullclines and equilibrium points for these differential equations. Which direction do the trajectories go in each region? (b) Sketch some typical trajectories in the phase plane. (c) What do these differential equations predict will be the long-term outcome of the US-Soviet arms race? (d) Discuss these predictions in the light of the actual expenditures in Table 11.10. Table 11.10 Arms budgets of the United States and the Soviet Union for the years 1945–1960 (billions of dollars) 1945

USSR

USA

14

97

1953

USSR

USA

25.7

71.4

23. In the 1930s, the Soviet ecologist G. F. Gause performed a series of experiments on competition among two yeasts with populations 𝑃1 and 𝑃2 , respectively. By performing population studies at low density in large volumes, he determined what he called the coefficients of geometric increase (and we would call continuous exponential growth rates). These coefficients described the growth of each yeast alone: 1 𝑑𝑃1 = 0.2 𝑃1 𝑑𝑡 1 𝑑𝑃2 = 0.06 𝑃2 𝑑𝑡 where 𝑃1 and 𝑃2 are measured in units that Gause established. He also determined that, in his units, the carrying capacity of 𝑃1 was 13 and the carrying capacity of 𝑃2 was 6. He then observed that one 𝑃2 occupies the niche space of 3 𝑃1 and that one 𝑃1 occupied the niche space of 0.4 𝑃2 . This led him to the following differential equations to describe the interaction of 𝑃1 and 𝑃2 : ) ( 𝑑𝑃1 13 − (𝑃1 + 3𝑃2 ) = 0.2𝑃1 𝑑𝑡 13 ) ( 𝑑𝑃2 6 − (𝑃2 + 0.4𝑃1 ) = 0.06𝑃2 . 𝑑𝑡 6 When both yeasts were growing together, Gause recorded the data in Table 11.11. Table 11.11 Gause’s yeast populations Time (hours)

6

16

24

29

48

53

0.375

3.99

4.69

6.15

7.27

8.30

0.29

0.98

1.47

1.46

1.71

1.84

1946

14

80

1954

23.9

61.6

𝑃1

1947

15

29

1955

25.5

58.3

𝑃2

1948

20

20

1956

23.2

59.4

1949

20

22

1957

23.0

61.4

1950

21

23

1958

22.3

61.4

1951

22.7

49.6

1959

22.3

61.7

1952

26.0

69.6

1960

22.1

59.6

(a) Carry out a phase plane analysis of Gause’s equations. (b) Mark the data points on the phase plane and describe what would have happened had Gause continued the experiment.

37 R. Taagepera, G. M. Schiffler, R. T. Perkins and D. L. Wagner, Soviet-American and Israeli-Arab Arms Races and the Richardson Model (General Systems, XX, 1975).

632

Chapter 11 DIFFERENTIAL EQUATIONS

Strengthen Your Understanding 𝑦

In Problems 24–25, explain what is wrong with the statement.

3

24. A solution trajectory and nullclines for a system of differential equations are shown in Figure 11.94.

2

𝑦

1

𝑥 1 𝑥

2

3

Figure 11.96

192

Figure 11.94

25. The nullclines for a system of differential equations are shown in Figure 11.95. The system has an equilibrium at the point (6, 6).

28. A trajectory for a system of differential equations with nullcline in Figure 11.97 and initial conditions 𝑥 = 1 and 𝑦 = 2. 𝑦 4

𝑦

3 2





1

6

𝑥 1

2

3

4

Figure 11.97

𝑥 6

Figure 11.95 Are the statements in Problems 29–32 true or false? Give an explanation for your answer. In Problems 26–28, give an example of:

29. Nullclines are always straight lines.

26. A graph of the nullclines of a system of differential equations with exactly two equilibrium points in the first quadrant. Label the nullclines to show whether trajectories pass through the nullcline vertically or horizontally.

30. If two nullclines cross, their intersection point is an equilibrium.

27. The nullclines of a system of differential equations with the trajectory shown in Figure 11.96.

32. A system of two differential equations must have at least two nullclines.

11.10

31. A system of two differential equations has at most two nullclines.

SECOND-ORDER DIFFERENTIAL EQUATIONS: OSCILLATIONS

A Second-Order Differential Equation When a body moves freely under gravity, we know that 𝑑2𝑠 = −𝑔, 𝑑𝑡2

11.10 SECOND-ORDER DIFFERENTIAL EQUATIONS: OSCILLATIONS

633

where 𝑠 is the height of the body above ground at time 𝑡 and 𝑔 is the acceleration due to gravity. To solve this equation, we first integrate to get the velocity, 𝑣 = 𝑑𝑠∕𝑑𝑡: 𝑑𝑠 = −𝑔𝑡 + 𝑣0 , 𝑑𝑡 where 𝑣0 is the initial velocity. Then we integrate again, giving 1 𝑠 = − 𝑔𝑡2 + 𝑣0 𝑡 + 𝑠0 , 2 where 𝑠0 is the initial height. The differential equation 𝑑 2 𝑠∕𝑑𝑡2 = −𝑔 is called second order because the equation contains a second derivative but no higher derivatives. The general solution to a second-order differential equation is a family of functions with two parameters, here 𝑣0 and 𝑠0 . Finding values for the two constants corresponds to picking a particular function out of this family.

A Mass on a Spring Not every second-order differential equation can be solved simply by integrating twice. Consider a mass 𝑚 attached to the end of a spring hanging from the ceiling. We assume that the mass of the spring itself is negligible in comparison with the mass 𝑚. (See Figure 11.98.)

✻ Direction of positive 𝑠

𝐹 𝑠=0



𝑠 = 𝑠0 (Initial position)



𝑠=0

(Equilibrium)

(Equilibrium)

Figure 11.98: Spring and mass in equilibrium position and after upward displacement

When the system is left undisturbed, no net force acts on the mass. The force of gravity is balanced by the force the spring exerts on the mass, and the spring is in the equilibrium position. If we pull down on the mass, we feel a force pulling upward. If instead, we push upward on the mass, the opposite happens: a force pushes the mass down.38 What happens if we push the mass upward and then release it? We expect the mass to oscillate up and down around the equilibrium position.

Springs and Hooke’s Law In order to figure out how the mass moves, we need to know the exact relationship between its displacement, 𝑠, from the equilibrium position and the net force, 𝐹 , exerted on the mass. (See Figure 11.98.) We would expect that the further the mass is from the equilibrium position, and the more the spring is stretched or compressed, the larger the force. In fact, provided that the displacement is not large enough to deform the spring permanently, experiments show that the net force, 𝐹 , is approximately proportional to the displacement, 𝑠: 𝐹 = −𝑘𝑠, 38 Pulling down on the mass stretches the spring, increasing the tension, so the combination of gravity and spring force is upward. Pushing up the spring decreases tension in the spring, so the combination of gravity and spring force is downward.

634

Chapter 11 DIFFERENTIAL EQUATIONS

where 𝑘 is the spring constant (𝑘 > 0) and the negative sign means that the net force is in the opposite direction to the displacement. The value of 𝑘 depends on the physical properties of the particular spring. This relationship is known as Hooke’s Law. Suppose we push the mass upward some distance and then release it. After we let go the net force causes the mass to accelerate toward the equilibrium position. By Newton’s Second Law of Motion we have Force = Mass × Acceleration. Since acceleration is 𝑑 2 𝑠∕𝑑𝑡2 and force is 𝐹 = −𝑘𝑠 by Hooke’s law, we have −𝑘𝑠 = 𝑚

𝑑2𝑠 , 𝑑𝑡2

which is equivalent to the following differential equation.

Equation for Oscillations of a Mass on a Spring 𝑑2𝑠 𝑘 =− 𝑠 𝑚 𝑑𝑡2

Thus, the motion of the mass is described by a second-order differential equation. Since we expect the mass to oscillate, we guess that the solution to this equation involves trigonometric functions.

Solving the Differential Equation by Guess-and-Check Guessing is the tried-and-true method for solving differential equations. It may surprise you to learn that there is no systematic method for solving most differential equations analytically, so guesswork is often extremely important. Example 1

By guessing, find the general solution to the equation 𝑑2𝑠 = −𝑠. 𝑑𝑡2

Solution

We want to find functions whose second derivative is the negative of the original function. We are already familiar with two functions that have this property: 𝑠(𝑡) = cos 𝑡 and 𝑠(𝑡) = sin 𝑡. We check that they are solutions by substituting: 𝑑2 𝑑 (cos 𝑡) = (− sin 𝑡) = − cos 𝑡, 𝑑𝑡 𝑑𝑡2 and

𝑑2 𝑑 (sin 𝑡) = (cos 𝑡) = − sin 𝑡. 𝑑𝑡 𝑑𝑡2 The remarkable thing is that starting from these two particular solutions, we can build up all the solutions to our equation. Given two constants 𝐶1 and 𝐶2 , the function 𝑠(𝑡) = 𝐶1 cos 𝑡 + 𝐶2 sin 𝑡 satisfies the differential equation, since 𝑑2 𝑑 (𝐶1 cos 𝑡 + 𝐶2 sin 𝑡) = (−𝐶1 sin 𝑡 + 𝐶2 cos 𝑡) 2 𝑑𝑡 𝑑𝑡

11.10 SECOND-ORDER DIFFERENTIAL EQUATIONS: OSCILLATIONS

635

= −𝐶1 cos 𝑡 − 𝐶2 sin 𝑡 = −(𝐶1 cos 𝑡 + 𝐶2 sin 𝑡). It can be shown (though we will not do so) that 𝑠(𝑡) = 𝐶1 cos 𝑡 + 𝐶2 sin 𝑡 is the most general form of the solution. As expected, it contains two constants, 𝐶1 and 𝐶2 . If the differential equation represents a physical problem, then 𝐶1 and 𝐶2 can often be computed, as shown in the next example. Example 2

If a mass on a spring is displaced by a distance of 𝑠0 and then released at 𝑡 = 0, find the solution to 𝑑2𝑠 = −𝑠 𝑑𝑡2

Solution

The position of the mass is given by the equation 𝑠(𝑡) = 𝐶1 cos 𝑡 + 𝐶2 sin 𝑡. We also know that the initial position is 𝑠0 ; thus, 𝑠(0) = 𝐶1 cos 0 + 𝐶2 sin 0 = 𝐶1 ⋅ 1 + 𝐶2 ⋅ 0 = 𝑠0 , so 𝐶1 = 𝑠0 , the initial displacement. What is 𝐶2 ? To find it, we use the fact that at 𝑡 = 0, when the mass has just been released, its velocity is 0. Velocity is the derivative of the displacement, so | 𝑑𝑠 || = (−𝐶1 sin 𝑡 + 𝐶2 cos 𝑡)|| = −𝐶1 ⋅ 0 + 𝐶2 ⋅ 1 = 0, | 𝑑𝑡 |𝑡=0 |𝑡=0

so 𝐶2 = 0. Therefore the solution is 𝑠 = 𝑠0 cos 𝑡.

Solution to the General Spring Equation Having found the general solution to the equation 𝑑 2 𝑠∕𝑑𝑡2 = −𝑠, let us return to the equation 𝑘 𝑑2𝑠 = − 𝑠. 2 𝑚 𝑑𝑡 We write 𝜔 =

√ 𝑘∕𝑚 (why will be clear in a moment), so the differential equation becomes 𝑑2𝑠 = −𝜔2 𝑠. 𝑑𝑡2

This equation no longer has sin 𝑡 and cos 𝑡 as solutions, since 𝑑2 (sin 𝑡) = − sin 𝑡 ≠ −𝜔2 sin 𝑡. 𝑑𝑡2 However, since we want a factor of 𝜔2 after differentiating twice, the chain rule leads us to guess that sin 𝜔𝑡 may be a solution. Checking this: 𝑑2 𝑑 (𝜔 cos 𝜔𝑡) = −𝜔2 sin 𝜔𝑡, (sin 𝜔𝑡) = 2 𝑑𝑡 𝑑𝑡 we see that sin 𝜔𝑡 is a solution, and you can check that cos 𝜔𝑡 is a solution, too.

636

Chapter 11 DIFFERENTIAL EQUATIONS

The general solution to the equation 𝑑2𝑠 + 𝜔2 𝑠 = 0 𝑑𝑡2 is of the form 𝑠(𝑡) = 𝐶1 cos 𝜔𝑡 + 𝐶2 sin 𝜔𝑡, where 𝐶1 and 𝐶2 are arbitrary constants. (We assume 𝜔 > 0.) The period of this oscillation is 𝑇 =

2𝜋 . 𝜔

Such oscillations are called simple harmonic motion.

Thus the solution to our original equation,

𝑑2𝑠 𝑘 + 𝑠 = 0, is 𝑠 = 𝐶1 cos 𝑑𝑡2 𝑚



𝑘 𝑡+ 𝑚

𝐶2 sin



𝑘 𝑡. 𝑚

Initial Value and Boundary-Value Problems A problem in which the initial position and the initial velocity are used to determine the particular solution is called an initial value problem. (See Example 2.) Alternatively, we may be given the position at two known times. Such a problem is known as a boundary-value problem. Example 3

Find a solution to the differential equation satisfying each set of conditions below 𝑑2𝑠 + 4𝑠 = 0. 𝑑𝑡2 (a) The boundary conditions 𝑠(0) = 0, 𝑠(𝜋∕4) = 20. (b) The initial conditions 𝑠(0) = 1, 𝑠′ (0) = −6.

Solution

Since 𝜔2 = 4, 𝜔 = 2, the general solution to the differential equation is 𝑠(𝑡) = 𝐶1 cos 2𝑡 + 𝐶2 sin 2𝑡. (a) Substituting the boundary condition 𝑠(0) = 0 into the general solution gives 𝑠(0) = 𝐶1 cos(2 ⋅ 0) + 𝐶2 sin(2 ⋅ 0) = 𝐶1 ⋅ 1 + 𝐶2 ⋅ 0 = 𝐶1 = 0. Thus 𝑠(𝑡) must have the form 𝑠(𝑡) = 𝐶2 sin 2𝑡. The second condition yields the value of 𝐶2 : ( ) ) ( 𝜋 𝜋 𝑠 = 𝐶2 sin 2 ⋅ = 𝐶2 = 20. 4 4 Therefore, the solution satisfying the boundary conditions is 𝑠(𝑡) = 20 sin 2𝑡. (b) For the initial value problem, we start from the same general solution: 𝑠(𝑡) = 𝐶1 cos 2𝑡+𝐶2 sin 2𝑡. Substituting 0 for 𝑡 once again, we find 𝑠(0) = 𝐶1 cos(2 ⋅ 0) + 𝐶2 sin(2 ⋅ 0) = 𝐶1 = 1. Differentiating 𝑠(𝑡) = cos 2𝑡 + 𝐶2 sin 2𝑡 gives 𝑠′ (𝑡) = −2 sin 2𝑡 + 2𝐶2 cos 2𝑡,

637

11.10 SECOND-ORDER DIFFERENTIAL EQUATIONS: OSCILLATIONS

and applying the second initial condition gives us 𝐶2 : 𝑠′ (0) = −2 sin(2 ⋅ 0) + 2𝐶2 cos(2 ⋅ 0) = 2𝐶2 = −6, so 𝐶2 = −3 and our solution is 𝑠(𝑡) = cos 2𝑡 − 3 sin 2𝑡.

What do the Graphs of Our Solutions Look Like? Since the general solution of the equation 𝑑 2 𝑠∕𝑑𝑡2 + 𝜔2 𝑠 = 0 is of the form 𝑠(𝑡) = 𝐶1 cos 𝜔𝑡 + 𝐶2 sin 𝜔𝑡 it would be useful to know what the graph of such a sum of sines and cosines looks like. We start with the example 𝑠(𝑡) = cos 𝑡 + sin 𝑡, which is graphed in Figure 11.99. Interestingly, the graph in Figure 11.99 looks like another sine function, and in fact it is one. If we measure the graph carefully, we find that√its period is 2𝜋, the same as sin 𝑡 and cos 𝑡, but the amplitude is approximately 1.414 (in fact, it’s 2), and the graph is shifted along the 𝑡-axis (in fact by 𝜋∕4). If we plot 𝐶1 cos 𝑡 + 𝐶2 sin 𝑡 for any 𝐶1 and 𝐶2 , the resulting graph is always a sine function with period 2𝜋, though the amplitude and shift can vary. For example, the graph of 𝑠 = 6 cos 𝑡−8 sin 𝑡 is in Figure 11.100; as expected, it has period 2𝜋. 𝑠

𝑠(𝑡) = √ cos 𝑡 + sin 𝑡 = 2 sin(𝑡 + 𝜋∕4)

𝑠 = 6 cos 𝑡 − 8 sin 𝑡 = 10 sin(𝑡 + 𝜙)

10

1.5 𝑡 −2𝜋

𝑡 −2𝜋

2𝜋

2𝜋 −10

−1.5 Figure 11.99: Graph√of the sum: 𝑠(𝑡) = cos 𝑡 + sin 𝑡 = 2 sin(𝑡 + 𝜋∕4)

Figure 11.100: Graph of a sum of √sine and cosine. Amplitude 𝐴 = 10 = 62 + 82

These graphs suggest that we can write the sum of a sine and a cosine of the same argument as one single sine function.39 Problem 27 shows that this can always be done using the following relations. If 𝐶1 cos 𝜔𝑡 + 𝐶2 sin 𝜔𝑡 = 𝐴 sin(𝜔𝑡 + 𝜙), the amplitude, 𝐴, is given by √ 𝐴 = 𝐶12 + 𝐶22 . The angle 𝜙 is called the phase shift and satisfies tan 𝜙 =

𝐶1 . 𝐶2

We choose 𝜙 in (−𝜋, 𝜋] such that if 𝐶1 > 0, 𝜙 is positive, and if 𝐶1 < 0, 𝜙 is negative. 39 The

sum of a sine and a cosine can also be written as a single cosine function.

638

Chapter 11 DIFFERENTIAL EQUATIONS

The reason that it is often useful to write the solution to the differential equation as a single sine function 𝐴 sin(𝜔𝑡 + 𝜙), as opposed to the sum of a sine and a cosine, is that the amplitude 𝐴 and the phase shift 𝜙 are easier to recognize in this form.

Warning: Phase Shift Is Not the Same as Horizontal Translation Let’s look at 𝑠 = sin(3𝑡 + 𝜋∕2). If we rewrite this as 𝑠 = sin(3(𝑡 + 𝜋∕6)), we can see from Figure 11.101 that the graph of this function is the graph of 𝑠 = sin 3𝑡 shifted to the left a distance of 𝜋∕6. (Remember that replacing 𝑥 by 𝑥 − 2 shifts a graph by 2 to the right.) But 𝜋∕6 is not the phase shift; the phase shift40 is 𝜋∕2. From the point of view of a scientist, the important question is often not the distance the curve has shifted, but the relation between the distance shifted and the period.

1

sin 3𝑡

sin(3𝑡 + 𝜋2 )

𝑡 2𝜋 3

𝜋 2

𝜋 6

𝜋

✛ Shift to the left by

𝜋 6

Figure 11.101: Phase shift is 𝜋2 ; horizontal translation is

𝜋 6

Exercises and Problems for Section 11.10 EXERCISES 1. Find the amplitude of 𝑦 = 3 sin 2𝑡 + 4 cos 2𝑡. 2. Find the amplitude of 3 sin 2𝑡 + 7 cos 2𝑡. 3. Write 5 sin(2𝑡) + 12 cos(2𝑡) in the form 𝐴 sin(𝜔𝑡 + 𝜓). 4. Write 7 sin 𝜔𝑡 + 24 cos 𝜔𝑡 in the form 𝐴 sin(𝜔𝑡 + 𝜙). 5. (a) Using a calculator or computer, graph 𝑠 = 4 cos 𝑡+ 3 sin 𝑡 for −2𝜋 ≤ 𝑡 ≤ 2𝜋. (b) If 4 cos 𝑡 + 3 sin 𝑡 = 𝐴 sin(𝑡 + 𝜙), use your graph to estimate the values of 𝐴 and 𝜙. (c) Calculate 𝐴 and 𝜙 analytically. 6. Check by differentiation that 𝑦 = 2 cos 𝑡 + 3 sin 𝑡 is a solution to 𝑦′′ + 𝑦 = 0. 7. Check by differentiation that 𝑦(𝑡) = 3 sin 2𝑡 + 2 cos 2𝑡 is a solution of 𝑦′′ + 4𝑦 = 0.

8. Check by differentiation that 𝑦 = 𝐴 cos 𝑡 + 𝐵 sin 𝑡 is a solution to 𝑦′′ + 𝑦 = 0 for any constants 𝐴 and 𝐵. 9. Check by differentiation that 𝑦(𝑡) = 𝐴 sin 2𝑡 + 𝐵 cos 2𝑡 is a solution of 𝑦′′ + 4𝑦 = 0 for all values of 𝐴 and 𝐵. 10. Check by differentiation that 𝑦(𝑡) = 𝐴 sin 𝜔𝑡 + 𝐵 cos 𝜔𝑡 is a solution of 𝑦′′ + 𝜔2 𝑦 = 0 for all values of 𝐴 and 𝐵. 11. What values of 𝛼 and 𝐴 make 𝑦 = 𝐴 cos 𝛼𝑡 a solution to 𝑦′′ + 5𝑦 = 0 such that 𝑦′ (1) = 3? Find a general solution to the differential equations in Exercises 12–13. 𝑑2𝑧 + 𝜋 2𝑧 = 0 𝑑𝑡2 13. 9𝑧′′ + 𝑧 = 0 12.

PROBLEMS 14. A spring has spring constant 𝑘 = 0.8 and mass 𝑚 = 10 grams. (a) Write the second-order differential equation for oscillations of the mass on the spring. (b) Write the general solution to the differential equa40 This

tion. 15. A spring has spring constant 𝑘 = 250 and mass 𝑚 = 100 grams. (a) Write the second-order differential equation for oscillations of the mass on the spring.

definition of phase shift is the one usually used in the sciences.

11.10 SECOND-ORDER DIFFERENTIAL EQUATIONS: OSCILLATIONS

(b) Write the solution to the differential equation if the initial position is 𝑠(0) = 5 and the initial velocity is 𝑠′ (0) = −10. (c) How far down does the mass go?

(I)

(II)

𝑥

𝑥 1

2

𝑡

𝑡

The functions in Problems 16–18 describe the motion of a mass on a spring satisfying the differential equation 𝑦′′ = −9𝑦, where 𝑦 is the displacement of the mass from the equilibrium position at time 𝑡, with upward as positive. In each case, describe in words how the motion starts when 𝑡 = 0. For example, is the mass at the highest point, the lowest point, or in the middle? Is it moving up or down or is it at rest? 16. 𝑦 = 2 cos 3𝑡

−1

−2 (III)

(IV)

𝑥

𝑥

3 1 𝑡

𝑡 −1 −3

17. 𝑦 = −0.5 sin 3𝑡

Figure 11.102

18. 𝑦 = − cos 3𝑡

19. (a) Show that 𝑦 = 𝑐1 sinh 𝑤𝑡 + 𝑐2 cosh 𝑤𝑡 is a solution to 𝑦′′ − 𝑤2 𝑦 = 0. (b) Find a solution of this differential equation such that (i) 𝑦(0) = 0, 𝑦(1) = 6. (ii) 𝑦′ (0) = 0, 𝑦(1) = 6. 20. (a) Find the general solution of the differential equation 𝑦′′ + 9𝑦 = 0. (b) For each of the following initial conditions, find a particular solution. (i) 𝑦(0) = 0, 𝑦′ (0) = 1 (ii) 𝑦(0) = 1, 𝑦′ (0) = 0 (iii) 𝑦(0) = 1, 𝑦(1) = 0 (iv) 𝑦(0) = 0, 𝑦(1) = 1 (c) Sketch a graph of the solutions found in part (b). 21. Consider the motion described by the differential equations: (a) 𝑥′′ + 16𝑥 = 0, (b) 25𝑥′′ + 𝑥 = 0,

639

𝑥(0) = 5, 𝑥′ (0) = 0, 𝑥(0) = −1, 𝑥′ (0) = 2.

In each case, find a formula for 𝑥(𝑡) and calculate the amplitude and period of the motion. 22. Each graph in Figure 11.102 represents a solution to one of the differential equations: (b) 𝑥′′ + 4𝑥 = 0, (a) 𝑥′′ + 𝑥 = 0, (c) 𝑥′′ + 16𝑥 = 0. Assuming the 𝑡-scales on the four graphs are the same, which graph represents a solution to which equation? Find an equation for each graph.

23. The following differential equations represent oscillating springs. (i) 𝑠′′ + 4𝑠 = 0 𝑠(0) = 5,

𝑠′ (0) = 0

(ii) 4𝑠′′ + 𝑠 = 0 𝑠(0) = 10,

𝑠′ (0) = 0

′′

(iii) 𝑠 + 6𝑠 = 0 𝑠(0) = 4, (iv)

6𝑠′′ + 𝑠 = 0 𝑠(0) = 20,

𝑠′ (0) = 0 𝑠′ (0) = 0

Which differential equation represents: (a) The spring oscillating most quickly (with the shortest period)? (b) The spring oscillating with the largest amplitude? (c) The spring oscillating most slowly (with the longest period)? (d) The spring with largest maximum velocity? 24. A pendulum of length 𝑙 makes an angle of 𝑥 (radians) with the vertical (see Figure 11.103). When 𝑥 is small, it can be shown that, approximately: 𝑔 𝑑2𝑥 = − 𝑥, 𝑙 𝑑𝑡2 where 𝑔 is the acceleration due to gravity.

✻ 𝑙 𝑥

❄ Figure 11.103 (a) Solve this equation assuming that 𝑥(0) = 0 and 𝑥′ (0) = 𝑣0 . (b) Solve this equation assuming that the pendulum is let go from the position where 𝑥 = 𝑥0 . (“Let go” means that the velocity of the pendulum is zero when 𝑥 = 𝑥0 . Measure 𝑡 from the moment when the pendulum is let go.)

640

Chapter 11 DIFFERENTIAL EQUATIONS

25. Look at the pendulum motion in Problem 24. What effect does it have on 𝑥 as a function of time if: (a)

𝑥0 is increased?

where 𝐶 is the capacitance, and 𝐿 is the inductance, so 𝐿

(b) 𝑙 is increased?

26. A brick of mass 3 kg hangs from the end of a spring. When the brick is at rest, the spring is stretched by 2 cm. The spring is then stretched an additional 5 cm and released. Assume there is no air resistance.

𝑑2𝑄 𝑄 + = 0. 𝐶 𝑑𝑡2

The unit of charge is the coulomb, the unit of capacitance the farad, the unit of inductance the henry, the unit of current is the ampere, and time is measured in seconds.

(a) Set up a differential equation with initial conditions describing the motion. (b) Solve the differential equation.

Capacitor

𝐶

27. (a) Expand 𝐴 sin(𝜔𝑡+𝜙) using the trigonometric identity sin(𝑥 + 𝑦) = sin 𝑥 cos 𝑦 + cos 𝑥 sin 𝑦. (b) Assume 𝐴 > 0. If 𝐴 sin(𝜔𝑡 + 𝜙) = 𝐶1 cos 𝜔𝑡 + 𝐶2 sin 𝜔𝑡, show that we must have √ 𝐴 = 𝐶12 + 𝐶22 and tan 𝜙 = 𝐶1 ∕𝐶2 . Problems 28–30 concern the electric circuit in Figure 11.104. A charged capacitor connected to an inductor causes a current to flow through the inductor until the capacitor is fully discharged. The current in the inductor, in turn, charges up the capacitor until the capacitor is fully charged again. If 𝑄(𝑡) is the charge on the capacitor at time 𝑡, and 𝐼 is the current, then 𝑑𝑄 𝐼= . 𝑑𝑡 If the circuit resistance is zero, then the charge 𝑄 and the current 𝐼 in the circuit satisfy the differential equation 𝑑𝐼 𝑄 𝐿 + = 0, 𝑑𝑡 𝐶

𝐿 Inductor

Figure 11.104

28. If 𝐿 = 36 henry and 𝐶 = 9 farad, find a formula for 𝑄(𝑡) if (a) 𝑄(0) = 0 𝐼(0) = 2 (b) 𝑄(0) = 6 𝐼(0) = 0 = 𝐼(0) = 4, and the max29. Suppose 𝑄(0) = 0, 𝑄′ (0) √ imum possible charge is 2 2 coulombs. What is the capacitance if the inductance is 10 henry? 30. What happens to the charge and current as 𝑡 goes to infinity? What does it mean that the charge and current are sometimes positive and sometimes negative?

Strengthen Your Understanding In Problems 31–32, explain what is wrong with the statement. 31. A solution to the spring equation 𝑑 2 𝑠∕𝑑𝑡2 = −𝑠 with initial displacement 5 is given in Figure 11.105.

32. The general solution to the second-order differential equation 𝑑 2 𝑦∕𝑑𝑡2 = 𝑘 with constant 𝑘 is 𝑦 = (𝑘∕2)𝑡2 + 𝑡 + 𝐶. In Problems 33–34, give an example of:

𝑠

33. A second-order differential equation which has 𝑦 = 𝑒2𝑥 as a solution.

5 3 𝑡

34. Any second-order boundary value problem.

−5

Figure 11.105

11.11

LINEAR SECOND-ORDER DIFFERENTIAL EQUATIONS

A Spring with Friction: Damped Oscillations The differential equation 𝑑 2 𝑠∕𝑑𝑡2 = −(𝑘∕𝑚)𝑠, which we used to describe the motion of a spring, disregards friction. But there is friction in every real system. For a mass on a spring, the frictional force from air resistance increases with the velocity of the mass. The frictional force is often approx-

11.11 LINEAR SECOND-ORDER DIFFERENTIAL EQUATIONS

641

imately proportional to velocity, and so we introduce a damping term of the form 𝑎(𝑑𝑠∕𝑑𝑡), where 𝑎 is a constant called the damping coefficient and 𝑑𝑠∕𝑑𝑡 is the velocity of the mass. Remember that without damping, the differential equation was obtained from Force = Mass ⋅ Acceleration. With damping, the spring force −𝑘𝑠 is replaced by −𝑘𝑠 − 𝑎(𝑑𝑠∕𝑑𝑡), where 𝑎 is positive and the 𝑎(𝑑𝑠∕𝑑𝑡) term is subtracted because the frictional force is in the direction opposite to the motion. The new differential equation is therefore −

𝑘𝑠 − ⏟⏟⏟ Spring force

𝑑𝑠 𝑎 𝑑𝑡 ⏟⏟⏟ Frictional force

𝑑2𝑠 𝑚 2 𝑑𝑡 ⏟⏟⏟

=

Mass⋅Acceleration

which is equivalent to the following differential equation:

Equation for Damped Oscillations of a Spring 𝑑 2 𝑠 𝑎 𝑑𝑠 𝑘 + 𝑠=0 + 𝑑𝑡2 𝑚 𝑑𝑡 𝑚

We expect the solution to this equation to die away with time, as friction brings the motion to a stop.

The General Solution to a Linear Differential Equation The equation for damped oscillations is an example of a linear second-order differential equation with constant coefficients. This section gives an analytic method of solving the equation, 𝑑2𝑦 𝑑𝑦 + 𝑐𝑦 = 0, +𝑏 𝑑𝑡 𝑑𝑡2 for constant 𝑏 and 𝑐. As we saw for the spring equation, if 𝑓1 (𝑡) and 𝑓2 (𝑡) satisfy the differential equation, then, for any constants 𝐶1 and 𝐶2 , another solution is given by 𝑦(𝑡) = 𝐶1 𝑓1 (𝑡) + 𝐶2 𝑓2 (𝑡) This is called the principle of superposition. It can be shown that the general solution is of this form, provided 𝑓1 (𝑡) is not a multiple of 𝑓2 (𝑡).

Finding Solutions: The Characteristic Equation We now use complex numbers to solve the differential equation 𝑑2𝑦 𝑑𝑦 + 𝑐𝑦 = 0. +𝑏 𝑑𝑡 𝑑𝑡2 The method is a form of guess-and-check. We ask what kind of function might satisfy a differential equation in which the second derivative 𝑑 2 𝑦∕𝑑𝑡2 is a sum of multiples of 𝑑𝑦∕𝑑𝑡 and 𝑦. One possibility is an exponential function, so we try to find a solution of the form: 𝑦 = 𝐶𝑒𝑟𝑡 ,

642

Chapter 11 DIFFERENTIAL EQUATIONS

where 𝑟 may be a complex number.41 To find 𝑟, we substitute into the differential equation. 𝑑𝑦 𝑑2𝑦 + 𝑐𝑦 = 𝑟2 𝐶𝑒𝑟𝑡 + 𝑏 ⋅ 𝑟𝐶𝑒𝑟𝑡 + 𝑐 ⋅ 𝐶𝑒𝑟𝑡 = 𝐶𝑒𝑟𝑡 (𝑟2 + 𝑏𝑟 + 𝑐) = 0. +𝑏 𝑑𝑡 𝑑𝑡2 We can divide by 𝑦 = 𝐶𝑒𝑟𝑡 provided 𝐶 ≠ 0, because the exponential function is never zero. If 𝐶 = 0, then 𝑦 = 0, which is not a very interesting solution (though it is a solution). So we assume 𝐶 ≠ 0. Then 𝑦 = 𝐶𝑒𝑟𝑡 is a solution to the differential equation if 𝑟2 + 𝑏𝑟 + 𝑐 = 0. This quadratic is called the characteristic equation of the differential equation. Its solutions are 1√ 2 1 𝑟=− 𝑏± 𝑏 − 4𝑐. 2 2 There are three different types of solutions to the differential equation, depending on whether the solutions to the characteristic equation are real and distinct, complex, or repeated. The sign of 𝑏2 −4𝑐 determines the type of solutions.

The Case with 𝒃𝟐 − 𝟒𝒄 > 𝟎 There are two real solutions 𝑟1 and 𝑟2 to the characteristic equation, and the following two functions satisfy the differential equation: 𝐶1 𝑒𝑟1 𝑡 and 𝐶2 𝑒𝑟2 𝑡 . The sum of these two solutions is the general solution to the differential equation:

If 𝑏2 − 4𝑐 > 0, the general solution to 𝑑2𝑦 𝑑𝑦 + 𝑐𝑦 = 0 +𝑏 2 𝑑𝑡 𝑑𝑡 is 𝑦(𝑡) = 𝐶1 𝑒𝑟1 𝑡 + 𝐶2 𝑒𝑟2 𝑡 where 𝑟1 and 𝑟2 are the solutions to the characteristic equation. If 𝑟1 < 0 and 𝑟2 < 0, the motion is called overdamped.

A physical system satisfying a differential equation of this type is said to be overdamped because it occurs when there is a lot of friction in the system. For example, a spring moving in a thick fluid such as oil or molasses is overdamped: it will not oscillate. Example 1

A spring is placed in oil, where it satisfies the differential equation 𝑑2𝑠 𝑑𝑠 + 3 + 2𝑠 = 0. 𝑑𝑡 𝑑𝑡2 Solve this equation with the initial conditions 𝑠 = −0.5 and 𝑑𝑠∕𝑑𝑡 = 3 when 𝑡 = 0. 41 See

Appendix B on complex numbers.

11.11 LINEAR SECOND-ORDER DIFFERENTIAL EQUATIONS

Solution

643

The characteristic equation is 𝑟2 + 3𝑟 + 2 = 0, with solutions 𝑟 = −1 and 𝑟 = −2, so the general solution to the differential equation is 𝑠(𝑡) = 𝐶1 𝑒−𝑡 + 𝐶2 𝑒−2𝑡 . We use the initial conditions to find 𝐶1 and 𝐶2 . At 𝑡 = 0, we have 𝑠 = 𝐶1 𝑒−0 + 𝐶2 𝑒−2(0) = 𝐶1 + 𝐶2 = −0.5. Furthermore, since 𝑑𝑠∕𝑑𝑡 = −𝐶1 𝑒−𝑡 − 2𝐶2 𝑒−2𝑡 , we have 𝑑𝑠 || = −𝐶1 𝑒−0 − 2𝐶2 𝑒−2(0) = −𝐶1 − 2𝐶2 = 3. 𝑑𝑡 ||𝑡=0

Solving these equations simultaneously, we find 𝐶1 = 2 and 𝐶2 = −2.5, so that the solution is 𝑠(𝑡) = 2𝑒−𝑡 − 2.5𝑒−2𝑡 . The graph of this function is in Figure 11.106. The mass is so slowed by the oil that it passes through the equilibrium point only once (when 𝑡 ≈ 1∕4) and for all practical purposes, it comes to rest after a short time. The motion has been “damped out” by the oil. 𝑠 𝑠(𝑡) = 2𝑒−𝑡 − 2.5𝑒−2𝑡

0.4 0.2

𝑡 1

2

3

4

5

−0.2 −0.4 Figure 11.106: Solution to overdamped equation 𝑑2𝑠 + 3 𝑑𝑠 + 2𝑠 = 0 𝑑𝑡2 𝑑𝑡

The Case with 𝒃𝟐 − 𝟒𝒄 = 𝟎 In this case, the characteristic equation has only one solution, 𝑟 = −𝑏∕2. By substitution, we can check that both 𝑦 = 𝑒−𝑏𝑡∕2 and 𝑦 = 𝑡𝑒−𝑏𝑡∕2 are solutions.

If 𝑏2 − 4𝑐 = 0, 𝑑2𝑦 𝑑𝑦 + 𝑐𝑦 = 0 +𝑏 2 𝑑𝑡 𝑑𝑡 has general solution 𝑦(𝑡) = (𝐶1 𝑡 + 𝐶2 )𝑒−𝑏𝑡∕2 . If 𝑏 > 0, the system is said to be critically damped.

644

Chapter 11 DIFFERENTIAL EQUATIONS

The Case with 𝒃𝟐 − 𝟒𝒄 < 𝟎 In this case, the characteristic equation has complex roots. Using Euler’s formula,42 these complex roots lead to trigonometric functions which represent oscillations. Example 2

An object of mass 𝑚 = 10 kg is attached to a spring with spring constant 𝑘 = 20 kg/sec2 , and the object experiences a frictional force proportional to the velocity, with constant of proportionality 𝑎 = 20 kg/sec. At time 𝑡 = 0, the object is released from rest 2 meters above the equilibrium position. Write the differential equation that describes the motion.

Solution

As we saw on page 641, the differential equation for damped oscillations of a spring is 𝑑 2 𝑠 𝑎 𝑑𝑠 𝑘 + 𝑠 = 0. + 𝑑𝑡2 𝑚 𝑑𝑡 𝑚 Substituting 𝑚 = 10 kg, 𝑘 = 20 kg/sec2 , and 𝑎 = 20 kg/sec, we obtain the differential equation: 𝑑2𝑠 𝑑𝑠 + 2 + 2𝑠 = 0. 𝑑𝑡 𝑑𝑡2 At 𝑡 = 0, the object is at rest 2 meters above equilibrium, so the initial conditions are 𝑠(0) = 2 and 𝑠′ (0) = 0, where 𝑠 is in meters and 𝑡 in seconds. Notice that this is the same differential equation as in Example 1 except that the coefficient of 𝑑𝑠∕𝑑𝑡 has decreased from 3 to 2, which means that the frictional force has been reduced. This time, the roots of the characteristic equation have imaginary parts which lead to oscillations.

Example 3

Solve the differential equation 𝑑𝑠 𝑑2𝑠 + 2 + 2𝑠 = 0, 𝑑𝑡 𝑑𝑡2 subject to 𝑠(0) = 2, 𝑠′ (0) = 0.

Solution

The characteristic equation is 𝑟2 + 2𝑟 + 2 = 0 giving 𝑟 = −1 ± 𝑖. The solution to the differential equation is 𝑠(𝑡) = 𝐴1 𝑒(−1+𝑖)𝑡 + 𝐴2 𝑒(−1−𝑖)𝑡 , where 𝐴1 and 𝐴2 are arbitrary complex numbers. The initial condition 𝑠(0) = 2 gives 2 = 𝐴1 𝑒(−1+𝑖)⋅0 + 𝐴2 𝑒(−1−𝑖)⋅0 = 𝐴1 + 𝐴2 . Also, 𝑠′ (𝑡) = 𝐴1 (−1 + 𝑖)𝑒(−1+𝑖)𝑡 + 𝐴2 (−1 − 𝑖)𝑒(−1−𝑖)𝑡 , so 𝑠′ (0) = 0 gives 0 = 𝐴1 (−1 + 𝑖) + 𝐴2 (−1 − 𝑖). Solving the simultaneous equations for 𝐴1 and 𝐴2 gives (after some algebra) 𝐴1 = 1 − 𝑖 42 See

Appendix B on complex numbers.

and 𝐴2 = 1 + 𝑖.

11.11 LINEAR SECOND-ORDER DIFFERENTIAL EQUATIONS

645

The solution is therefore 𝑠(𝑡) = (1 − 𝑖)𝑒(−1+𝑖)𝑡 + (1 + 𝑖)𝑒(−1−𝑖)𝑡 = (1 − 𝑖)𝑒−𝑡𝑒𝑖𝑡 + (1 + 𝑖)𝑒−𝑡 𝑒−𝑖𝑡 . Using Euler’s formula, 𝑒𝑖𝑡 = cos 𝑡 + 𝑖 sin 𝑡 and 𝑒−𝑖𝑡 = cos 𝑡 − 𝑖 sin 𝑡, we get 𝑠(𝑡) = (1 − 𝑖)𝑒−𝑡(cos 𝑡 + 𝑖 sin 𝑡) + (1 + 𝑖)𝑒−𝑡 (cos 𝑡 − 𝑖 sin 𝑡). Multiplying out and simplifying, all the complex terms drop out, giving 𝑠(𝑡) = 𝑒−𝑡 cos 𝑡 + 𝑖𝑒−𝑡 sin 𝑡 − 𝑖𝑒−𝑡 cos 𝑡 + 𝑒−𝑡 sin 𝑡 +𝑒−𝑡 cos 𝑡 − 𝑖𝑒−𝑡 sin 𝑡 + 𝑖𝑒−𝑡 cos 𝑡 + 𝑒−𝑡 sin 𝑡 = 2𝑒−𝑡 cos 𝑡 + 2𝑒−𝑡 sin 𝑡. The cos 𝑡 and sin 𝑡 terms tell us that the solution oscillates; the factor of 𝑒−𝑡 tells us that the oscillations are damped. See Figure 11.107. However, the period of the oscillations does not change as the amplitude decreases. This is why a spring-driven clock can keep accurate time even as it is running down. 𝑠 2

𝑠(𝑡) = 2𝑒−𝑡 cos 𝑡 + 2𝑒−𝑡 sin 𝑡

1 𝜋 Figure 11.107: Solution to underdamped equation

2𝜋 𝑑2𝑠 𝑑𝑡2

𝑡 + 2 𝑑𝑠 + 2𝑠 = 0 𝑑𝑡

In Example 3, the coefficients 𝐴1 and 𝐴2 are complex, but the solution, 𝑠(𝑡), is real. (We expect this, since 𝑠(𝑡) represents a real displacement.) In general, provided the coefficients 𝑏 and 𝑐 in the original differential equation and the initial values are real, the solution is always real. The coefficients 𝐴1 and 𝐴2 are always complex conjugates (that is, of the form 𝛼 ± 𝑖𝛽).

If 𝑏2 − 4𝑐 < 0, to solve 𝑑2𝑦 𝑑𝑦 + 𝑐𝑦 = 0, +𝑏 2 𝑑𝑡 𝑑𝑡 • Find the solutions 𝑟 = 𝛼 ± 𝑖𝛽 to the characteristic equation 𝑟2 + 𝑏𝑟 + 𝑐 = 0. • The general solution to the differential equation is, for some real 𝐶1 and 𝐶2 , 𝑦 = 𝐶1 𝑒𝛼𝑡 cos 𝛽𝑡 + 𝐶2 𝑒𝛼𝑡 sin 𝛽𝑡. If 𝛼 < 0, such oscillations are called underdamped. If 𝛼 = 0, the oscillations are undamped.

646

Chapter 11 DIFFERENTIAL EQUATIONS

(a) 𝑦′′ = 9𝑦

(b) 𝑦′′ = −9𝑦.

Example 4

Find the general solution of the equations

Solution

(a) The characteristic equation is 𝑟2 − 9 = 0, so 𝑟 = ±3. Thus the general solution is 𝑦 = 𝐶1 𝑒3𝑡 + 𝐶2 𝑒−3𝑡 . (b) The characteristic equation is 𝑟2 + 9 = 0, so 𝑟 = 0 ± 3𝑖. The general solution is 𝑦 = 𝐶1 𝑒0𝑡 cos 3𝑡 + 𝐶2 𝑒0𝑡 sin 3𝑡 = 𝐶1 cos 3𝑡 + 𝐶2 sin 3𝑡. Notice that we have seen the solution to this equation in Section 11.10; it’s the equation of undamped simple harmonic motion with 𝜔 = 3.

Example 5

Solve the initial value problem 𝑦′′ + 4𝑦′ + 13𝑦 = 0,

Solution

𝑦(0) = 0, 𝑦′ (0) = 30.

We solve the characteristic equation 𝑟2 + 4𝑟 + 13 = 0,

getting 𝑟 = −2 ± 3𝑖.

The general solution to the differential equation is 𝑦 = 𝐶1 𝑒−2𝑡 cos 3𝑡 + 𝐶2 𝑒−2𝑡 sin 3𝑡. Substituting 𝑡 = 0 gives 𝑦(0) = 𝐶1 ⋅ 1 ⋅ 1 + 𝐶2 ⋅ 1 ⋅ 0 = 0,

so

𝐶1 = 0.

Differentiating 𝑦(𝑡) = 𝐶2 𝑒−2𝑡 sin 3𝑡 gives 𝑦′ (𝑡) = 𝐶2 (−2𝑒−2𝑡 sin 3𝑡 + 3𝑒−2𝑡 cos 3𝑡). Substituting 𝑡 = 0 gives 𝑦′ (0) = 𝐶2 (−2 ⋅ 1 ⋅ 0 + 3 ⋅ 1 ⋅ 1) = 3𝐶2 = 30 so The solution is therefore 𝑦(𝑡) = 10𝑒−2𝑡 sin 3𝑡.

Summary of Solutions to 𝒚 ′′ + 𝒃𝒚 ′ + 𝒄𝒚 = 0 If 𝑏2 − 4𝑐 > 0, then 𝑦 = 𝐶1 𝑒𝑟1 𝑡 + 𝐶2 𝑒𝑟2 𝑡 If 𝑏2 − 4𝑐 = 0, then 𝑦 = (𝐶1 𝑡 + 𝐶2 )𝑒−𝑏𝑡∕2 If 𝑏2 − 4𝑐 < 0, then 𝑦 = 𝐶1 𝑒𝛼𝑡 cos 𝛽𝑡 + 𝐶2 𝑒𝛼𝑡 sin 𝛽𝑡

𝐶2 = 10.

647

11.11 LINEAR SECOND-ORDER DIFFERENTIAL EQUATIONS

Exercises and Problems for Section 11.11 EXERCISES For Exercises 1–16, find the general solution to the given differential equation. 1. 𝑦′′ + 4𝑦′ + 3𝑦 = 0 ′′

2. 𝑦′′ + 4𝑦′ + 4𝑦 = 0



′′

3. 𝑦 + 4𝑦 + 5𝑦 = 0

4. 𝑠 − 7𝑠 = 0

′′

′′

5. 𝑠 + 7𝑠 = 0



6. 𝑦 − 3𝑦 + 2𝑦 = 0 2

𝑑 𝑥 7. 4𝑧′′ + 8𝑧′ + 3𝑧 = 0

8.

𝑑𝑡2

𝑑𝑥 +4

𝑑𝑡

+ 8𝑥 = 0

17. 𝑦′′ + 5𝑦′ + 6𝑦 = 0,

𝑦(0) = 1, 𝑦′ (0) = 0.

18. 𝑦′′ + 5𝑦′ + 6𝑦 = 0,

𝑦(0) = 5, 𝑦′ (0) = 1.

19. 𝑦′′ − 3𝑦′ − 4𝑦 = 0,

𝑦(0) = 1, 𝑦′ (0) = 0.

20. 𝑦′′ − 3𝑦′ − 4𝑦 = 0,

𝑦(0) = 0, 𝑦′ (0) = 0.5.

21. 𝑦′′ + 6𝑦′ + 5𝑦 = 0,

𝑦(0) = 1,

𝑦′ (0) = 0

22. 𝑦′′ + 6𝑦′ + 5𝑦 = 0,

𝑦(0) = 5,

𝑦′ (0) = 5

23. 𝑦′′ + 6𝑦′ + 10𝑦 = 0,

𝑦(0) = 0,

𝑦′ (0) = 2

10. 𝑧′′ + 2𝑧 = 0

24. 𝑦′′ + 6𝑦′ + 10𝑦 = 0,

𝑦(0) = 0,

𝑦′ (0) = 0

11. 𝑧 + 2𝑧′ = 0

12. 𝑃 ′′ + 2𝑃 ′ + 𝑃 = 0

For Exercises 25–28, solve the boundary value problem.

13. 9𝑧′′ − 𝑧 = 0

14. 𝑦′′ + 6𝑦′ + 8𝑦 = 0

25. 𝑦′′ + 5𝑦′ + 6𝑦 = 0,

𝑦(0) = 1, 𝑦(1) = 0.

15. 𝑦′′ + 2𝑦′ + 3𝑦 = 0

16. 𝑥′′ + 2𝑥′ + 10𝑥 = 0

26. 𝑦′′ + 5𝑦′ + 6𝑦 = 0,

𝑦(−2) = 0, 𝑦(2) = 3.

2

𝑑 𝑝 9.

𝑑𝑡2

𝑑𝑝 +

𝑑𝑡

+𝑝=0

′′

′′

For Exercises 17–24, solve the initial value problem.



27. 𝑝 + 2𝑝 + 2𝑝 = 0,

𝑝(0) = 0,

𝑝(𝜋∕2) = 20

28. 𝑝′′ + 4𝑝′ + 5𝑝 = 0,

𝑝(0) = 1,

𝑝(𝜋∕2) = 5

PROBLEMS 29. Match the graphs of solutions in Figure 11.108 with the differential equations below. (a) (b) (c) (d)

(I)

𝑦

𝑥′′ + 4𝑥 = 0 𝑥′′ − 4𝑥 = 0 𝑥′′ − 0.2𝑥′ + 1.01𝑥 = 0 𝑥′′ + 0.2𝑥′ + 1.01𝑥 = 0

1 𝑥 4 𝑥

(III)

4

−1 𝑦

(IV)

400 (I)

𝑦

(II)

1

𝑦 1

(II)

𝑥 4 (III)

𝑥 4

(IV)

Figure 11.108

Each of the differential equations (i)–(iv) represents the position of a 1 gram mass oscillating on the end of a damped spring. For Problems 31–35, pick the differential equation representing the system which answers the question. (i) 𝑠′′ + 𝑠′ + 4𝑠 = 0 (ii) 𝑠′′ + 2𝑠′ + 5𝑠 = 0 (iii) 𝑠′′ + 3𝑠′ + 3𝑠 = 0 (iv) 𝑠′′ + 0.5𝑠′ + 2𝑠 = 0 31. Which spring has the largest coefficient of damping?

30. Match the differential equations to the solution graphs (I)–(IV). Use each graph only once. (b) 𝑦′′ + 𝑦′ − 6𝑦 = 0 (a) 𝑦′′ + 5𝑦′ + 6𝑦 = 0 ′′ ′ (d) 𝑦′′ = −9𝑦 (c) 𝑦 + 4𝑦 + 9𝑦 = 0

32. Which spring exerts the smallest restoring force for a given displacement? 33. In which system does the mass experience the frictional force of smallest magnitude for a given velocity? 34. Which oscillation has the longest period?

648

Chapter 11 DIFFERENTIAL EQUATIONS

35. Which spring is the stiffest? [Hint: You need to determine what it means for a spring to be stiff. Think of an industrial strength spring and a slinky.] For each of the differential equations in Problems 36–38, find the values of 𝑐 that make the general solution: (a)

overdamped, (b) underdamped,(c)

36. 𝑠′′ + 4𝑠′ + 𝑐𝑠 = 0

critically damped.

√ 37. 𝑠′′ + 2 2𝑠′ + 𝑐𝑠 = 0

38. 𝑠′′ + 6𝑠′ + 𝑐𝑠 = 0

44. If 𝑦 = 𝑒2𝑡 is a solution to the differential equation 𝑑2𝑦 𝑑𝑦 + 𝑘𝑦 = 0, −5 𝑑𝑡 𝑑𝑡2 find the value of the constant 𝑘 and the general solution to this equation. 45. Assuming 𝑏, 𝑐 > 0, explain how you know that the solutions of an underdamped differential equation must go to 0 as 𝑡 → ∞. 46. Could the graph in Figure 11.109 show the position of a mass oscillating at the end of an overdamped spring? Why or why not? 𝑠 (position)

For each of the differential equations in Problems 39–40, find the values of 𝑏 that make the general solution: (a) overdamped, (c) critically damped. 39. 𝑠′′ + 𝑏𝑠′ + 5𝑠 = 0

(b) underdamped,

40. 𝑠′′ + 𝑏𝑠′ − 16𝑠 = 0

𝑡 (time)

Figure 11.109 47. Find a solution to the following equation which satisfies 𝑧(0) = 3 and does not tend to infinity as 𝑡 → ∞:

41. The motion of a mass on the end of a spring satisfies the differential equation 𝑑𝑠 𝑑2𝑠 + 10𝑠 = 0. +7 𝑑𝑡 𝑑𝑡2 (a) If the mass 𝑚 = 10, what is the spring coefficient 𝑘? What is the damping coefficient 𝑎? (b) Solve the differential equation if the initial conditions are 𝑠(0) = −1 and 𝑠′ (0) = −7. (c) How low does the mass at the end of the spring go? How high does it go? (d) How long does it take until the spring stays within 0.1 unit of equilibrium? 42. The motion of a mass on the end of a spring satisfies the differential equation 𝑑2𝑠 𝑑𝑠 +2 + 3𝑠 = 0. 𝑑𝑡 𝑑𝑡2 (a) Give the general solution to the differential equation. (b) Solve the differential equation if the initial height is +2 and the initial velocity is +5. (c) How low does the mass at the end of the spring go? How high does it go? (d) How long does it take until the spring stays within 0.1 unit of equilibrium? 43. If the spring constant 𝑘 = 500 and the mass 𝑚 = 100, what values of the damping coefficient 𝑎 make the motion (a) Overdamped? (b) Critically damped? (c) Underdamped?

𝑑 2 𝑧 𝑑𝑧 + − 2𝑧 = 0. 𝑑𝑡 𝑑𝑡2 48. Consider an overdamped differential equation with 𝑏, 𝑐 > 0. (a) Show that both roots of the characteristic equation are negative. (b) Show that any solution to the differential equation goes to 0 as 𝑡 → ∞. 49. Consider the system of differential equations 𝑑𝑥 = −𝑦 𝑑𝑡

𝑑𝑦 = −𝑥. 𝑑𝑡

(a) Convert this system to a second order differential equation in 𝑦 by differentiating the second equation with respect to 𝑡 and substituting for 𝑥 from the first equation. (b) Solve the equation you obtained for 𝑦 as a function of 𝑡; hence find 𝑥 as a function of 𝑡. 50. Juliet is in love with Romeo, who happens (in our version of this story) to be a fickle lover. The more Juliet loves him, the more he begins to dislike her. When she hates him, his feelings for her warm up. On the other hand, her love for him grows when he loves her and withers when he hates her. A model for their ill-fated romance is 𝑑𝑗 𝑑𝑟 = 𝐴𝑟, = −𝐵𝑗, 𝑑𝑡 𝑑𝑡 where 𝐴 and 𝐵 are positive constants, 𝑟(𝑡) represents Romeo’s love for Juliet at time 𝑡, and 𝑗(𝑡) represents Juliet’s love for Romeo at time 𝑡. (Negative love is hate.) (a) The constant on the right-hand side of Juliet’s equation (the one including 𝑑𝑗∕𝑑𝑡) has a positive

11.11 LINEAR SECOND-ORDER DIFFERENTIAL EQUATIONS

sign, whereas the constant in Romeo’s equation is negative. Explain why these signs follow from the story. (b) Derive a second-order differential equation for 𝑟(𝑡) and solve it. (Your equation should involve 𝑟 and its derivatives, but not 𝑗 and its derivatives.) (c) Express 𝑟(𝑡) and 𝑗(𝑡) as functions of 𝑡, given 𝑟(0) = 1 and 𝑗(0) = 0. Your answer will contain 𝐴 and 𝐵. (d) As you may have discovered, the outcome of the relationship is a never-ending cycle of love and hate. Find what fraction of the time they both love one another. 51. (a) If 𝑟1 ≠ 𝑟2 and 𝑟1 and 𝑟2 satisfy 𝑟2 + 𝑏𝑟 + 𝑐 = 0 show that 𝑒𝑟1 𝑡 − 𝑒𝑟2 𝑡 𝑦= 𝑟1 − 𝑟2 is a solution to 𝑦′′ + 𝑏𝑦′ + 𝑐𝑦 = 0. (b) If 𝑟1 = 𝑟2 + ℎ, show that the solution in part (a) can be written (𝑒ℎ𝑡 − 1) 𝑦 = 𝑒𝑟2 𝑡 . ℎ (c) Using the Taylor series, show that ℎ𝑡

2

2 3

(𝑒 − 1) ℎ𝑡 ℎ𝑡 =𝑡+ + +⋯. ℎ 2! 3! (d) Use the result of part (c) in the solution from part (𝑒ℎ𝑡 − 1) (b) to show that lim 𝑒𝑟2 𝑡 = 𝑡𝑒𝑟2 𝑡 . ℎ→0 ℎ (e) If there is a double root 𝑟2 = 𝑟1 = −𝑏∕2. By direct substitution, show that 𝑦 = 𝑡𝑒−𝑏𝑡∕2 satisfies 𝑦′′ + 𝑏𝑦′ + 𝑐𝑦 = 0. Recall the discussion of electric circuits on page 640. Just as a spring can have a damping force which affects its motion, so can a circuit. Problems 52–55 involve a damping force caused by the resistor in Figure 11.110. The charge 𝑄 on a capacitor in a circuit with inductance 𝐿, capacitance 𝐶, and resistance 𝑅, in ohms, satisfies the differential equation 𝐿

𝑑2𝑄 𝑑𝑄 1 +𝑅 + 𝑄 = 0. 𝑑𝑡 𝐶 𝑑𝑡2

649

Capacitor

𝐶 𝑅

Resistor

𝐿 Inductor

Figure 11.110

52. If 𝐿 = 1 henry, 𝑅 = 2 ohms, and 𝐶 = 4 farads, find a formula for the charge when (a) 𝑄(0) = 0, 𝑄′ (0) = 2. (b) 𝑄(0) = 2, 𝑄′ (0) = 0. 53. If 𝐿 = 1 henry, 𝑅 = 1 ohm, and 𝐶 = 4 farads, find a formula for the charge when (a) 𝑄(0) = 0, 𝑄′ (0) = 2. (b) 𝑄(0) = 2, 𝑄′ (0) = 0. (c) How did reducing the resistance affect the charge? Compare with your solution to Problem 52. 54. If 𝐿 = 8 henry, 𝑅 = 2 ohm, and 𝐶 = 4 farads, find a formula for the charge when (a) 𝑄(0) = 0, 𝑄′ (0) = 2. (b) 𝑄(0) = 2, 𝑄′ (0) = 0. (c) How did increasing the inductance affect the charge? Compare with your solution to Problem 52. 55. Given any positive values for 𝑅, 𝐿 and 𝐶, what happens to the charge as 𝑡 goes to infinity?

Strengthen Your Understanding In Problems 56–57, explain what is wrong with the statement.

57. Figure 11.112 represents motion of a mass on a spring with initial conditions 𝑠(0) = 5 and 𝑠′ (0) = 2.

56. Figure 11.111 represents motion for which the damping coefficient is zero. 𝑡

𝑠 5 𝑡

𝑠

−5

Figure 11.112 Figure 11.111

650

Chapter 11 DIFFERENTIAL EQUATIONS

In Problems 58–60, give an example of: 58. A linear second-order differential equation representing spring motion that is overdamped. 59. A linear second-order differential equation representing

Online Resource: Review problems and Projects

spring motion that is critically damped. 60. Values of the spring constant 𝑘, the mass 𝑚, and the damping coefficient 𝑎 so that the motion is underdamped and shows damped oscillations.

Chapter Twelve

FUNCTIONS OF SEVERAL VARIABLES

Contents 12.1 Functions of Two Variables . . . . . . . . . . . . . . . . Function Notation . . . . . . . . . . . . . . . . . . . . . . . . Graphical Example: A Weather Map . . . . . . . . . . Numerical Example: Body Mass Index (BMI) . . Algebraic Examples: Formulas . . . . . . . . . . . . . . A Tour of 3-Space . . . . . . . . . . . . . . . . . . . . . . . . Graphing Equations in 3-Space . . . . . . . . . . . . . Distance Between Two Points . . . . . . . . . . . . . . . 12.2 Graphs and Surfaces . . . . . . . . . . . . . . . . . . . . . Visualizing a Function of Two Variables Using a Graph. . . . . . . . . . . . . . . . . . .

652 652 652 653 653 654 655 656 660 660

Plotting the Graph of the Function f (x, y) = x2 + y2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 661

12.3

12.4

12.5

12.6

New Graphs from Old . . . . . . . . . . . . . . . . . . . . . Cross-Sections and the Graph of a Function . . . . Linear Functions . . . . . . . . . . . . . . . . . . . . . . . . . When One Variable Is Missing: Cylinders . . . . . Contour Diagrams . . . . . . . . . . . . . . . . . . . . . . . Topographical Maps . . . . . . . . . . . . . . . . . . . . . . Corn Production . . . . . . . . . . . . . . . . . . . . . . . . . Contour Diagrams and Graphs . . . . . . . . . . . . . . Finding Contours Algebraically . . . . . . . . . . . . . Contour Diagrams and Tables . . . . . . . . . . . . . . . Using Contour Diagrams: The Cobb-Douglas Production Function . . . . . . . . . . . . . . . . . . . . . Formula for a Production Function . . . . . . . . . . . . The Cobb-Douglas Production Model . . . . . . . . . . Linear Functions . . . . . . . . . . . . . . . . . . . . . . . . . What Is a Linear Function of Two Variables? . . . What Makes a Plane Flat? . . . . . . . . . . . . . . . . . . Linear Functions from a Numerical Point of View . . . . . . . . . . . . . . . . . . . . . . . . . . . What Does the Contour Diagram of a Linear Function Look Like? . . . . . . . . . . . . . . . Functions of Three Variables . . . . . . . . . . . . . . . Representing a Function of 3 Variables Using a Family of Level Surfaces . . . . . . . . . . . A Catalog of Surfaces . . . . . . . . . . . . . . . . . . . . . How Surfaces Can Represent Functions of Two Variables and Functions of Three Variables . . . Limits and Continuity . . . . . . . . . . . . . . . . . . . .

661 662 664 664 668 669 670 671 671 673 674 675 675 682 682 682 683 684 689 689 691 692 695

652

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

12.1

FUNCTIONS OF TWO VARIABLES

Function Notation Suppose you want to calculate your monthly payment on a five-year car loan; this depends on both the amount of money you borrow and the interest rate. These quantities can vary separately: the loan amount can change while the interest rate remains the same, or the interest rate can change while the loan amount remains the same. To calculate your monthly payment you need to know both. If the monthly payment is $𝑚, the loan amount is $𝐿, and the interest rate is 𝑟%, then we express the fact that 𝑚 is a function of 𝐿 and 𝑟 by writing: 𝑚 = 𝑓 (𝐿, 𝑟). This is just like the function notation of one-variable calculus. The variable 𝑚 is called the dependent variable, and the variables 𝐿 and 𝑟 are called the independent variables. The letter 𝑓 stands for the function or rule that gives the value of 𝑚 corresponding to given values of 𝐿 and 𝑟. A function of two variables can be represented graphically, numerically by a table of values, or algebraically by a formula. In this section, we give examples of each.

Graphical Example: A Weather Map Figure 12.1 shows a weather map from a newspaper. What information does it convey? It displays the predicted high temperature, 𝑇 , in degrees Fahrenheit (◦ F), throughout the US on that day. The curves on the map, called isotherms, separate the country into zones, according to whether 𝑇 is in the 60s, 70s, 80s, 90s, or 100s. (Iso means same and therm means heat.) Notice that the isotherm separating the 80s and 90s zones connects all the points where the temperature is exactly 90◦ F. Example 1

Estimate the predicted value of 𝑇 in Boise, Idaho; Topeka, Kansas; and Buffalo, New York.

Solution

Boise and Buffalo are in the 70s region, and Topeka is in the 80s region. Thus, the predicted temperature in Boise and Buffalo is between 70 and 80 while the predicted temperature in Topeka is between 80 and 90. In fact, we can say more. Although both Boise and Buffalo are in the 70s, Boise is quite close to the 𝑇 = 70 isotherm, whereas Buffalo is quite close to the 𝑇 = 80 isotherm. So we estimate the temperature to be in the low 70s in Boise and in the high 70s in Buffalo. Topeka is about halfway between the 𝑇 = 80 isotherm and the 𝑇 = 90 isotherm. Thus, we guess the temperature in Topeka to be in the mid-80s. In fact, the actual high temperatures for that day were 71◦ F for Boise, 79◦ F for Buffalo, and 86◦ F for Topeka.

60s

Buffalo

70s

80s

80s 90s 100s

70s

70s

60s Boise

90s

90s Topeka

90s

Figure 12.1: Weather map showing predicted high temperatures, 𝑇 , on a summer day

12.1 FUNCTIONS OF TWO VARIABLES

653

The predicted high temperature, 𝑇 , illustrated by the weather map is a function of (that is, depends on) two variables, often longitude and latitude, or miles east-west and miles north-south of a fixed point, say, Topeka. The weather map in Figure 12.1 is called a contour map or contour diagram of that function. Section 12.2 shows another way of visualizing functions of two variables using surfaces; Section 12.3 looks at contour maps in detail.

Numerical Example: Body Mass Index (BMI) The body mass index (BMI) is a value that attempts to quantify a person’s body fat based on their height ℎ and weight 𝑤. In function notation, we write: BMI = 𝑓 (ℎ, 𝑤). Table 12.1 contains values of this function for ℎ in inches and 𝑤 in pounds. Values of 𝑤 are across the top, values of ℎ are down the left side, and corresponding values of 𝑓 (ℎ, 𝑤) are in the table.1 For example, to find the value of 𝑓 (66, 140), we look in the row corresponding to ℎ = 66 under 𝑤 = 140, where we find the number 22.6. Thus, 𝑓 (66, 140) = 22.6. This means that if an individual is 66 inches tall and weights 140 lbs, their body mass index is 22.6. Table 12.1 Body mass index (BMI) Weight 𝑤 (lbs)

Height ℎ (inches)

120

140

160

180

200

60

23.4

27.3

31.2

35.2

39.1

63

21.3

24.8

28.3

31.9

35.4

66

19.4

22.6

25.8

29.0

32.3

69

17.7

20.7

23.6

26.6

29.5

72

16.3

19.0

21.7

24.4

27.1

75

15.0

17.5

20.0

22.5

25.0

Notice how this table differs from the table of values of a one-variable function, where one row or one column is enough to list the values of the function. Here many rows and columns are needed because the function has a value for every pair of values of the independent variables.

Algebraic Examples: Formulas In the weather map example there is no formula for the underlying function. That is usually the case for functions representing real-life data. On the other hand, for many models in physics, engineering, and economics, there are exact formulas. Example 2

Give a formula for the function 𝑀 = 𝑓 (𝐵, 𝑡) where 𝑀 is the amount of money in a bank account 𝑡 years after an initial investment of 𝐵 dollars, if interest is accrued at a rate of 1.2% per year compounded annually.

Solution

Annual compounding means that 𝑀 increases by a factor of 1.012 every year, so 𝑀 = 𝑓 (𝐵, 𝑡) = 𝐵(1.012)𝑡.

Example 3

A cylinder with closed ends has radius 𝑟 and height ℎ. If its volume is 𝑉 and its surface area is 𝐴, find formulas for the functions 𝑉 = 𝑓 (𝑟, ℎ) and 𝐴 = 𝑔(𝑟, ℎ).

Solution

Since the area of the circular base is 𝜋𝑟2 , we have 𝑉 = 𝑓 (𝑟, ℎ) = Area of base ⋅ Height = 𝜋𝑟2 ℎ.

1 http://www.cdc.gov.

Last accessed January 8, 2016.

654

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

The surface area of the side is the circumference of the bottom, 2𝜋𝑟, times the height ℎ, giving 2𝜋𝑟ℎ. Thus, 𝐴 = 𝑔(𝑟, ℎ) = 2 ⋅ Area of base + Area of side = 2𝜋𝑟2 + 2𝜋𝑟ℎ.

A Tour of 3-Space In Section 12.2 we see how to visualize a function of two variables as a surface in space. Now we see how to locate points in three-dimensional space (3-space). Imagine three coordinate axes meeting at the origin: a vertical axis, and two horizontal axes at right angles to each other. (See Figure 12.2.) Think of the 𝑥𝑦-plane as being horizontal, while the 𝑧-axis extends vertically above and below the plane. The labels 𝑥, 𝑦, and 𝑧 show which part of each axis is positive; the other side is negative. We generally use right-handed axes in which looking down the positive 𝑧-axis gives the usual view of the 𝑥𝑦-plane. We specify a point in 3-space by giving its coordinates (𝑥, 𝑦, 𝑧) with respect to these axes. Think of the coordinates as instructions telling you how to get to the point: start at the origin, go 𝑥 units along the 𝑥-axis, then 𝑦 units in the direction parallel to the 𝑦-axis, and finally 𝑧 units in the direction parallel to the 𝑧-axis. The coordinates can be positive, zero or negative; a zero coordinate means “don’t move in this direction,” and a negative coordinate means “go in the negative direction parallel to this axis.” For example, the origin has coordinates (0, 0, 0), since we get there from the origin by doing nothing at all. Example 4

Describe the position of the points with coordinates (1, 2, 3) and (0, 0, −1).

Solution

We get to the point (1, 2, 3) by starting at the origin, going 1 unit along the 𝑥-axis, 2 units in the direction parallel to the 𝑦-axis, and 3 units up in the direction parallel to the 𝑧-axis. (See Figure 12.3.) To get to (0, 0, −1), we don’t move at all in the 𝑥- and the 𝑦-directions, but move 1 unit in the negative 𝑧-direction. So the point is on the negative 𝑧-axis. (See Figure 12.4.) You can check that the position of the point is independent of the order of the 𝑥, 𝑦, and 𝑧 displacements. 𝑧

𝑧

𝑧 (1, 2, 3)

𝑂

𝑦

𝑦

𝑦 (0, 0, −1)

𝑥

𝑥

Figure 12.2: Coordinate axes in three-dimensional space

𝑥

Figure 12.3: The point (1, 2, 3) in 3-space

Figure 12.4: The point (0, 0, −1) in 3-space

Example 5

You start at the origin, go along the 𝑦-axis a distance of 2 units in the positive direction, and then move vertically upward a distance of 1 unit. What are the coordinates of your final position?

Solution

You started at the point (0, 0, 0). When you went along the 𝑦-axis, your 𝑦-coordinate increased to 2. Moving vertically increased your 𝑧-coordinate to 1; your 𝑥-coordinate did not change because you did not move in the 𝑥-direction. So your final coordinates are (0, 2, 1). (See Figure 12.5.) 𝑧 (0, 2, 1) 𝑦 𝑥

Figure 12.5: The point (0, 2, 1) is reached by moving 2 along the 𝑦-axis and 1 upward

12.1 FUNCTIONS OF TWO VARIABLES

655

It is often helpful to picture a three-dimensional coordinate system in terms of a room. The origin is a corner at floor level where two walls meet the floor. The 𝑧-axis is the vertical intersection of the two walls; the 𝑥- and the 𝑦-axis are the intersections of each wall with the floor. Points with negative coordinates lie behind a wall in the next room or below the floor.

Graphing Equations in 3-Space We can graph an equation involving the variables 𝑥, 𝑦, and 𝑧 in 3-space; such a graph is a picture of all points (𝑥, 𝑦, 𝑧) that satisfy the equation. Example 6

What do the graphs of the equations 𝑧 = 0, 𝑧 = 3, and 𝑧 = −1 look like?

Solution

To graph 𝑧 = 0, we visualize the set of points whose 𝑧-coordinate is zero. If the 𝑧-coordinate is 0, then we must be at the same vertical level as the origin; that is, we are in the horizontal plane containing the origin. So the graph of 𝑧 = 0 is the middle plane in Figure 12.6. The graph of 𝑧 = 3 is a plane parallel to the graph of 𝑧 = 0, but three units above it. The graph of 𝑧 = −1 is a plane parallel to the graph of 𝑧 = 0, but one unit below it.

Figure 12.6: The planes 𝑧 = −1, 𝑧 = 0, and 𝑧 = 3

The plane 𝑧 = 0 contains the 𝑥- and the 𝑦-coordinate axes, and is called the 𝑥𝑦-plane. There are two other coordinate planes. The 𝑦𝑧-plane contains both the 𝑦- and the 𝑧-axis, and the 𝑥𝑧-plane contains the 𝑥- and the 𝑧-axis. (See Figure 12.7.)

Figure 12.7: The three coordinate planes

Example 7

Which of the points 𝐴 = (1, −1, 0), 𝐵 = (0, 3, 4), 𝐶 = (2, 2, 1), and 𝐷 = (0, −4, 0) lies closest to the 𝑥𝑧-plane? Which point lies on the 𝑦-axis?

Solution

The magnitude of the 𝑦-coordinate gives the distance to the 𝑥𝑧-plane. The point 𝐴 lies closest to that plane, because it has the smallest 𝑦-coordinate in magnitude. To get to a point on the 𝑦-axis, we move along the 𝑦-axis, but we don’t move at all in the 𝑥- or the 𝑧-direction. Thus, a point on the 𝑦-axis has both its 𝑥- and 𝑧-coordinates equal to zero. The only point of the four that satisfies this is 𝐷. (See Figure 12.8.)

656

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

In general, if a point has one of its coordinates equal to zero, it lies in one of the coordinate planes. If a point has two of its coordinates equal to zero, it lies on one of the coordinate axes. 𝑧 𝐵

𝐷

𝐶

𝑦 𝐴 𝑥 Figure 12.8: Which point lies closest to the 𝑥𝑧-plane? Which point lies on the 𝑦-axis?

Figure 12.9: The line 𝑥 = 0, 𝑧 = −2

Example 8

You are 2 units below the 𝑥𝑦-plane and in the 𝑦𝑧-plane. What are your coordinates?

Solution

Since you are 2 units below the 𝑥𝑦-plane, your 𝑧-coordinate is −2. Since you are in the 𝑦𝑧-plane, your 𝑥-coordinate is 0; your 𝑦-coordinate can be anything. Thus, you are at the point (0, 𝑦, −2). The set of all such points forms a line parallel to the 𝑦-axis, 2 units below the 𝑥𝑦-plane, and in the 𝑦𝑧-plane. (See Figure 12.9.)

Example 9

You are standing at the point (4, 5, 2), looking at the point (0.5, 0, 3). Are you looking up or down?

Solution

The point you are standing at has 𝑧-coordinate 2, whereas the point you are looking at has 𝑧coordinate 3; hence you are looking up.

Example 10

Imagine that the 𝑦𝑧-plane in Figure 12.7 is a page of this book. Describe the region behind the page algebraically.

Solution

The positive part of the 𝑥-axis pokes out of the page; moving in the positive 𝑥-direction brings you out in front of the page. The region behind the page corresponds to negative values of 𝑥, so it is the set of all points in 3-space satisfying the inequality 𝑥 < 0.

Distance Between Two Points In 2-space, the formula for the distance between two points (𝑥, 𝑦) and (𝑎, 𝑏) is given by √ Distance = (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 . The distance between two points (𝑥, 𝑦, 𝑧) and (𝑎, 𝑏, 𝑐) in 3-space is represented by 𝑃 𝐺 in Figure 12.10. The side 𝑃 𝐸 is parallel to the 𝑥-axis, 𝐸𝐹 is parallel to the 𝑦-axis, and 𝐹 𝐺 is parallel to the 𝑧-axis. Using Pythagoras’ theorem twice gives (𝑃 𝐺)2 = (𝑃 𝐹 )2 + (𝐹 𝐺)2 = (𝑃 𝐸)2 + (𝐸𝐹 )2 + (𝐹 𝐺)2 = (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 + (𝑧 − 𝑐)2 . Thus, a formula for the distance between the points (𝑥, 𝑦, 𝑧) and (𝑎, 𝑏, 𝑐) in 3-space is Distance =

√ (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 + (𝑧 − 𝑐)2 .

12.1 FUNCTIONS OF TWO VARIABLES

657

Figure 12.10: The diagonal 𝑃 𝐺 gives the distance between the points (𝑥, 𝑦, 𝑧) and (𝑎, 𝑏, 𝑐)

Example 11

Find the distance between (1, 2, 1) and (−3, 1, 2).

Solution

Distance =

Example 12

Find an expression for the distance from the origin to the point (𝑥, 𝑦, 𝑧).

Solution

The origin has coordinates (0, 0, 0), so the distance from the origin to (𝑥, 𝑦, 𝑧) is given by √ √ Distance = (𝑥 − 0)2 + (𝑦 − 0)2 + (𝑧 − 0)2 = 𝑥2 + 𝑦2 + 𝑧2 .

Example 13

Find an equation for a sphere of radius 1 with center at the origin.

Solution

The sphere consists of all points (𝑥, 𝑦, 𝑧) whose distance from the origin is 1, that is, which satisfy the equation √ 𝑥2 + 𝑦2 + 𝑧2 = 1.

√ √ (−3 − 1)2 + (1 − 2)2 + (2 − 1)2 = 18 = 4.243.

This is an equation for the sphere. If we square both sides we get the equation in the form 𝑥2 + 𝑦2 + 𝑧2 = 1.

Note that this equation represents the surface of the sphere. The solid ball enclosed by the sphere is represented by the inequality 𝑥2 + 𝑦2 + 𝑧2 ≤ 1.

Exercises and Problems for Section 12.1 Online Resource: Additional Problems for Section 12.1 EXERCISES

1. Which of the points 𝑃 = (1, 2, 1) and 𝑄 = (2, 0, 0) is closest to the origin? 2. Which two of the three points 𝑃1 = (1, 2, 3), 𝑃2 = (3, 2, 1) and 𝑃3 = (1, 1, 0) are closest to each other? 3. Which of the points 𝑃1 = (−3, 2, 15), 𝑃2 = (0, −10, 0), 𝑃3 = (−6, 5, 3) and 𝑃4 = (−4, 2, 7) is closest to 𝑃 = (6, 0, 4)?

4. Which of the points 𝐴 = (1.3, −2.7, 0), 𝐵 = (0.9, 0, 3.2), 𝐶 = (2.5, 0.1, −0.3) is closest to the 𝑦𝑧-plane? Which one lies on the 𝑥𝑧-plane? Which one is farthest from the 𝑥𝑦-plane?

658

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

5. You are at the point (3, 1, 1), standing upright and facing the 𝑦𝑧-plane. You walk 2 units forward, turn left, and walk another 2 units. What is your final position? From the point of view of an observer looking at the coordinate system in Figure 12.2 on page 654, are you in front of or behind the 𝑦𝑧-plane? To the left or to the right of the 𝑥𝑧-plane? Above or below the 𝑥𝑦-plane? 6. On a set of 𝑥, 𝑦 and 𝑧 axes oriented as in Figure 12.5 on page 654, draw a straight line through the origin, lying in the 𝑦𝑧-plane and such that if you move along the line with your 𝑦-coordinate increasing, your 𝑧-coordinate is increasing. 7. What is the midpoint of the line segment joining the points (−1, 3, 9) and (5, 6, −3)? In Exercises 8–11, which of (I)–(IV) lie on the graph of the equation? I. (2, 2, 4) III. (−3, −2, −1)

II. (−1, 1, 0) IV. (−2, −2, 4)

8. 𝑧 = 4 2

9. 𝑥 + 𝑦 + 𝑧 = 0 2

2

10. 𝑥 + 𝑦 + 𝑧 = 14

In Exercises 12–15 sketch graphs of the equations in 3-space. 13. 𝑥 = −3

14. 𝑦 = 1

15. 𝑧 = 2 and 𝑦 = 4

18. Find the equation of the sphere with radius 2 and centered at (1, 0, 0). 19. Find the equation of the vertical plane perpendicular to the 𝑦-axis and through the point (2, 3, 4). Exercises 20–22 refer to the map in Figure 12.1 on page 652. 20. Give the range of daily high temperatures for: (a)

Pennsylvania

(c)

California

(b) North Dakota

21. Sketch a possible graph of the predicted high temperature 𝑇 on a line north-south through Topeka. 22. Sketch possible graphs of the predicted high temperature on a north-south line and an east-west line through Boise. For Exercises 23–25, refer to Table 12.1 on page 653 where 𝑤 is a person’s weight (in lbs), and ℎ their height (in inches).

11. 𝑥 − 𝑦 = 0

12. 𝑧 = 4

17. Find an equation of the sphere with radius 5 centered at the origin.

16. With the 𝑧-axis vertical, a sphere has center (2, 3, 7) and lowest point (2, 3, −1). What is the highest point on the sphere?

23. Compute a table of values of BMI, with ℎ fixed at 60 inches and 𝑤 between 120 and 200 lbs at intervals of 20. 24. Medical evidence suggests that BMI values between 18.5 and 24.9 are healthy values.2 Estimate the range of weights that are considered healthy for a woman who is 6 feet tall. 25. Estimate the BMI of a man who weighs 90 kilograms and is 1.9 meters tall.

PROBLEMS 26. The temperature adjusted for wind chill is a temperature which tells you how cold it feels, as a result of the combination of wind and temperature.3 See Table 12.2. Table 12.2 Temperature adjusted for wind chill (◦ F) as a function of wind speed and temperature Temperature (◦ F)

Wind Speed (mph)

35

30

25

20

15

10

5

0

5

31

25

19

13

7

1

−5

−11

10

27

21

15

9

3

−4

−10

−16

15

25

19

13

6

0

−7

−13

−19

20

24

17

11

4

−2

−9

−15

−22

25

23

16

9

3

−4

−11

−17

−24

(a) If the temperature is 0◦ F and the wind speed is 15 mph, how cold does it feel? 2 http://www.cdc.gov. 3 www.nws.noaa.gov.

Accessed January 10, 2016. Accessed January 10, 2016.

(b) If the temperature is 35◦ F, what wind speed makes it feel like 24◦ F? (c) If the temperature is 25◦ F, what wind speed makes it feel like 12◦ F? (d) If the wind is blowing at 20 mph, what temperature feels like 0◦ F? In Problems 27–28, use Table 12.2 to make tables with the given properties. 27. The temperature adjusted for wind chill as a function of wind speed for temperatures of 20◦ F and 0◦ F. 28. The temperature adjusted for wind chill as a function of temperature for wind speeds of 5 mph and 20 mph. For Problems 29–31, refer to Table 12.3 which contains values of beef consumption 𝐶 (in pounds per week per household) as a function of household income, 𝐼 (in thousands

12.1 FUNCTIONS OF TWO VARIABLES

of dollars per year), and the price of beef, 𝑝 (in dollars per pound). Values of 𝑝 are shown across the top, values of 𝐼 are down the left side, and corresponding values of beef consumption 𝐶 = 𝑓 (𝐼, 𝑝) are given in the table.4 Table 12.3 Quantity of beef bought (pounds/household/week) Price of beef ($/lb)

Household income per year, 𝐼 ($1000)

3.00

3.50

4.00

4.50

20

2.65

2.59

2.51

2.43

40

4.14

4.05

3.94

3.88

60

5.11

5.00

4.97

4.84

80

5.35

5.29

5.19

5.07

100

5.79

5.77

5.60

5.53

29. Give tables for beef consumption as a function of 𝑝, with 𝐼 fixed at 𝐼 = 20 and 𝐼 = 100. Give tables for beef consumption as a function of 𝐼, with 𝑝 fixed at 𝑝 = 3.00 and 𝑝 = 4.00. Comment on what you see in the tables. 30. Make a table of the proportion, 𝑃 , of household income spent on beef per week as a function of price and income. (Note that 𝑃 is the fraction of income spent on beef.) 31. How does beef consumption vary as a function of household income if the price of beef is held constant? For Problems 32–35, a person’s body mass index (BMI) is a function of their weight 𝑊 (in kg) and height 𝐻 (in m) given by 𝐵(𝑊 , 𝐻) = 𝑊 ∕𝐻 2 .

659

𝑠, the number of monthly subscribers it serves, and 𝑚, the total number of special feature movies that its subscribers view. (b) If 𝑅 = 𝑓 (𝑠, 𝑚), find 𝑓 (1000, 5000) and interpret it in terms of revenue. 38. The gravitational force, 𝐹 newtons, exerted on an object by the earth depends on the object’s mass, 𝑚 kilograms, and its distance, 𝑟 meters, from the center of the earth, so 𝐹 = 𝑓 (𝑚, 𝑟). Interpret the following statement in terms of gravitation: 𝑓 (100, 7000000) ≈ 820. 39. A heating element is attached to the center point of a metal rod at time 𝑡 = 0. Let 𝐻 = 𝑓 (𝑑, 𝑡) represent the temperature in ◦ C of a point 𝑑 cm from the center after 𝑡 minutes. (a) Interpret the statement 𝑓 (2, 5) = 24 in terms of temperature. (b) If 𝑑 is held constant, is 𝐻 an increasing or a decreasing function of 𝑡? Why? (c) If 𝑡 is held constant, is 𝐻 an increasing or a decreasing function of 𝑑? Why? 40. The pressure, 𝑃 atmospheres, of 10 moles of nitrogen gas in a steel cylinder depends on the temperature of the gas, 𝑇 Kelvin, and the volume of the cylinder, 𝑉 liters, so 𝑃 = 𝑓 (𝑇 , 𝑉 ). Interpret the following statement in terms of pressure: 𝑓 (300, 5) = 49.2. 41. The monthly payment, 𝑚 dollars, for a 30-year fixed rate mortgage is a function of the total amount borrowed, 𝑃 dollars, and the annual interest rate, 𝑟%. In other words, 𝑚 = 𝑓 (𝑃 , 𝑟).

32. What is the BMI of a 1.72 m tall man weighing 72 kg? 33. A 1.58 m tall woman has a BMI of 23.2. What is her weight? 34. With a BMI less than 18.5, a person is considered underweight. What is the possible range of weights for an underweight person 1.58 m tall? 35. For weight 𝑤 in lbs and height ℎ in inches, a persons BMI is approximated using the formula 𝑓 (𝑤, ℎ) = 703𝑤∕ℎ2 . Check this approximation by converting the formula 𝐵(𝑊 , 𝐻). 36. A car rental company charges $40 a day and 15 cents a mile for its cars. (a) Write a formula for the cost, 𝐶, of renting a car as a function, 𝑓 , of the number of days, 𝑑, and the number of miles driven, 𝑚. (b) If 𝐶 = 𝑓 (𝑑, 𝑚), find 𝑓 (5, 300) and interpret it.

(a) Interpret the following statement in the context of monthly payment: 𝑓 (300,000, 5) = 1610.46. (b) If 𝑃 is held constant, is 𝑚 an increasing or a decreasing function of 𝑟? Why? (c) If 𝑟 is held constant, is 𝑚 an increasing or a decreasing function of 𝑃 ? Why? 42. Consider the acceleration due to gravity, 𝑔, at a distance ℎ from the center of a planet of mass 𝑚. (a) If 𝑚 is held constant, is 𝑔 an increasing or decreasing function of ℎ? Why? (b) If ℎ is held constant, is 𝑔 an increasing or decreasing function of 𝑚? Why? 43. A cube is located such that its top four corners have the coordinates (−1, −2, 2), (−1, 3, 2), (4, −2, 2) and (4, 3, 2). Give the coordinates of the center of the cube.

37. A cable company charges $100 for a monthly subscription to its services and $5 for each special feature movie that a subscriber chooses to watch.

44. Describe the set of points whose distance from the 𝑥axis is 2.

(a) Write a formula for the monthly revenue, 𝑅 in dollars, earned by the cable company as a function of

45. Describe the set of points whose distance from the 𝑥axis equals the distance from the 𝑦𝑧-plane.

4 Adapted from Richard G. Lipsey, An Introduction to Positive Economics, 3rd ed. (London: Weidenfeld and Nicolson, 1971).

660

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

46. Find the point on the 𝑥-axis closest to the point (3, 2, 1). 47. Does the line parallel to the 𝑦-axis through the point (2, 1, 4) intersect the plane 𝑦 = 5? If so, where? 48. Find a formula for the shortest distance between a point (𝑎, 𝑏, 𝑐) and the 𝑦-axis. 49. Find the equations of planes that just touch the sphere (𝑥 − 2)2 + (𝑦 − 3)2 + (𝑧 − 3)2 = 16 and are parallel to (a) The 𝑥𝑦-plane (c) The 𝑥𝑧-plane

(b) The 𝑦𝑧-plane

50. Find an equation of the largest sphere contained in the cube determined by the planes 𝑥 = 2, 𝑥 = 6; 𝑦 = 5, 𝑦 = 9; and 𝑧 = −1, 𝑧 = 3. 51. A cube has edges parallel to the axes. One corner is at 𝐴 = (5, 1, 2) and the corner at the other end of the longest diagonal through 𝐴 is 𝐵 = (12, 7, 4).

(a) What are the coordinates of the other three vertices on the bottom face? (b) What are the coordinates of the other three vertices on the top face?

52. An equilateral triangle is standing vertically with a vertex above the 𝑥𝑦-plane and its two other vertices at (7, 0, 0) and (9, 0, 0). What is its highest point? 53. (a) Find the midpoint of the line segment joining 𝐴 = (1, 5, 7) to 𝐵 = (5, 13, 19). (b) Find the point one quarter of the way along the line segment from 𝐴 to 𝐵. (c) Find the point one quarter of the way along the line segment from 𝐵 to 𝐴.

Strengthen Your Understanding In Problems 54–56, explain what is wrong with the statement.

62. A table for a function 𝑓 (𝑥, 𝑦) cannot have any values of 𝑓 appearing twice.

55. The 𝑥𝑦-plane has equation 𝑥𝑦 = 0.

63. If 𝑓 (𝑥) and 𝑔(𝑦) are both functions of a single variable, then the product 𝑓 (𝑥) ⋅ 𝑔(𝑦) is a function of two variables.

56. The distance from (2, 3, 4) to the 𝑥-axis is 2.

64. The point (1, 2, 3) lies above the plane 𝑧 = 2.

In Problems 57–58, give an example of:

65. The graph of the equation 𝑧 = 2 is a plane parallel to the 𝑥𝑧-plane.

57. A formula for a function 𝑓 (𝑥, 𝑦) that is increasing in 𝑥 and decreasing in 𝑦.

66. The points (1, 0, 1) and (0, −1, 1) are the same distance from the origin.

58. A point in 3-space with all its coordinates negative and farther from the 𝑥𝑧-plane than from the plane 𝑧 = −5.

67. The point (2, −1, 3) lies on the graph of the sphere (𝑥 − 2)2 + (𝑦 + 1)2 + (𝑧 − 3)2 = 25.

54. In 3-space, 𝑦 = 1 is a line parallel to the 𝑥-axis.

Are the statements in Problems 59–72 true or false? Give reasons for your answer. 59. If 𝑓 (𝑥, 𝑦) is a function of two variables defined for all 𝑥 and 𝑦, then 𝑓 (10, 𝑦) is a function of one variable. 60. The volume 𝑉 of a box of height ℎ and square base of side length 𝑠 is a function of ℎ and 𝑠. 61. If 𝐻 = 𝑓 (𝑡, 𝑑) is the function giving the water temperature 𝐻 ◦ C of a lake at time 𝑡 hours after midnight and depth 𝑑 meters, then 𝑡 is a function of 𝑑 and 𝐻.

12.2

68. There is only one point in the 𝑦𝑧-plane that is a distance 3 from the point (3, 0, 0). 69. There is only one point in the 𝑦𝑧-plane that is a distance 5 from the point (3, 0, 0). 70. If the point (0, 𝑏, 0) has distance 4 from the plane 𝑦 = 0, then 𝑏 must be 4. 71. A line parallel to the 𝑧-axis can intersect the graph of 𝑓 (𝑥, 𝑦) at most once. 72. A line parallel to the 𝑦-axis can intersect the graph of 𝑓 (𝑥, 𝑦) at most once.

GRAPHS AND SURFACES The weather map on page 652 is one way of visualizing a function of two variables. In this section we see how to visualize a function of two variables in another way, using a surface in 3-space.

Visualizing a Function of Two Variables Using a Graph For a function of one variable, 𝑦 = 𝑓 (𝑥), the graph of 𝑓 is the set of all points (𝑥, 𝑦) in 2-space such that 𝑦 = 𝑓 (𝑥). In general, these points lie on a curve in the plane. When a computer or calculator graphs 𝑓 , it approximates by plotting points in the 𝑥𝑦-plane and joining consecutive points by line segments. The more points, the better the approximation. Now consider a function of two variables.

12.2 GRAPHS AND SURFACES

661

The graph of a function of two variables, 𝑓 , is the set of all points (𝑥, 𝑦, 𝑧) such that 𝑧 = 𝑓 (𝑥, 𝑦). In general, the graph of a function of two variables is a surface in 3-space.

Plotting the Graph of the Function 𝒇 (𝒙, 𝒚) = 𝒙𝟐 + 𝒚 𝟐 To sketch the graph of 𝑓 we connect points as for a function of one variable. We first make a table of values of 𝑓 , such as in Table 12.4. Table 12.4

Table of values of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 𝑦

𝑥

−3

−2

−1

0

1

2

3

−3

18

13

10

9

10

13

18

−2

13

8

5

4

5

8

13

−1

10

5

2

1

2

5

10

0

9

4

1

0

1

4

9

1

10

5

2

1

2

5

10

2

13

8

5

4

5

8

13

3

18

13

10

9

10

13

18

Now we plot points. For example, we plot (1, 2, 5) because 𝑓 (1, 2) = 5 and we plot (0, 2, 4) because 𝑓 (0, 2) = 4. Then, we connect the points corresponding to the rows and columns in the table. The result is called a wire-frame picture of the graph. Filling in between the wires gives a surface. That is the way a computer drew the graphs in Figures 12.11 and 12.12. As more points are plotted, we get the surface in Figure 12.13, called a paraboloid. You should check to see if the sketches make sense. Notice that the graph goes through the origin since (𝑥, 𝑦, 𝑧) = (0, 0, 0) satisfies 𝑧 = 𝑥2 + 𝑦2 . Observe that if 𝑥 is held fixed and 𝑦 is allowed to vary, the graph dips down and then goes back up, just like the entries in the rows of Table 12.4. Similarly, if 𝑦 is held fixed and 𝑥 is allowed to vary, the graph dips down and then goes back up, just like the columns of Table 12.4. 𝑧

𝑧

𝑥

𝑧

𝑥 𝑦

𝑦 Figure 12.11: Wire frame picture Figure 12.12: Wire frame picture of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 for of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 with more −3 ≤ 𝑥 ≤ 3, −3 ≤ 𝑦 ≤ 3 points plotted

𝑥

𝑦

Figure 12.13: Graph of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 for −3 ≤ 𝑥 ≤ 3, −3 ≤ 𝑦 ≤ 3

New Graphs from Old We can use the graph of a function to visualize the graphs of related functions. Example 1

Let 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 . Describe in words the graphs of the following functions: (a) 𝑔(𝑥, 𝑦) = 𝑥2 + 𝑦2 + 3, (b) ℎ(𝑥, 𝑦) = 5 − 𝑥2 − 𝑦2 , (c) 𝑘(𝑥, 𝑦) = 𝑥2 + (𝑦 − 1)2 .

662

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

Solution

We know from Figure 12.13 that the graph of 𝑓 is a paraboloid, or a bowl, with its vertex at the origin. From this we can work out what the graphs of 𝑔, ℎ, and 𝑘 will look like. (a) The function 𝑔(𝑥, 𝑦) = 𝑥2 + 𝑦2 + 3 = 𝑓 (𝑥, 𝑦) + 3, so the graph of 𝑔 is the graph of 𝑓 , but raised by 3 units. See Figure 12.14. (b) Since −𝑥2 −𝑦2 is the negative of 𝑥2 +𝑦2 , the graph of −𝑥2 −𝑦2 is a paraboloid opening downward. Thus, the graph of ℎ(𝑥, 𝑦) = 5−𝑥2 −𝑦2 = 5−𝑓 (𝑥, 𝑦) looks like a downward-opening paraboloid with vertex at (0, 0, 5), as in Figure 12.15. (c) The graph of 𝑘(𝑥, 𝑦) = 𝑥2 + (𝑦 − 1)2 = 𝑓 (𝑥, 𝑦 − 1) is a paraboloid with vertex at 𝑥 = 0, 𝑦 = 1, since that is where 𝑘(𝑥, 𝑦) = 0, as in Figure 12.16. 𝑧

𝑧

𝑧



(0, 0, 5) 𝑦

𝑥 (0, 0, 3)

q 𝑦

𝑥 Figure 12.14: Graph of 𝑔(𝑥, 𝑦) = 𝑥2 + 𝑦2 + 3

𝑥 Figure 12.15: Graph of ℎ(𝑥, 𝑦) = 5 − 𝑥2 − 𝑦2

2 +𝑦2 )

(0, 1, 0)

𝑦

Figure 12.16: Graph of 𝑘(𝑥, 𝑦) = 𝑥2 + (𝑦 − 1)2

Example 2

Describe the graph of 𝐺(𝑥, 𝑦) = 𝑒−(𝑥

Solution

Since the exponential function is always positive, the graph lies entirely above the 𝑥𝑦-plane. From the graph of 𝑥2 + 𝑦2 we see that 𝑥2 + 𝑦2 is zero at the origin and gets larger as we move farther 2 2 from the origin in any direction. Thus, 𝑒−(𝑥 +𝑦 ) is 1 at the origin, and gets smaller as we move away from the origin in any direction. It can’t go below the 𝑥𝑦-plane; instead it flattens out, getting closer and closer to the plane. We say the surface is asymptotic to the 𝑥𝑦-plane. (See Figure 12.17.) Now consider a point (𝑥, 𝑦) on the circle 𝑥2 + 𝑦2 = 𝑟2 . Since

. What symmetry does it have?

2 +𝑦2 )

𝐺(𝑥, 𝑦) = 𝑒−(𝑥

2

= 𝑒−𝑟 ,

the value of the function 𝐺 is the same at all points on this circle. Thus, we say the graph of 𝐺 has circular symmetry. 𝑧



(0, 0, 1)

𝑥 𝑦 2 +𝑦2 )

Figure 12.17: Graph of 𝐺(𝑥, 𝑦) = 𝑒−(𝑥

Cross-Sections and the Graph of a Function We have seen that a good way to analyze a function of two variables is to let one variable vary while the other is kept fixed.

663

12.2 GRAPHS AND SURFACES

For a function 𝑓 (𝑥, 𝑦), the function we get by holding 𝑥 fixed and letting 𝑦 vary is called a cross-section of 𝑓 with 𝑥 fixed. The graph of the cross-section of 𝑓 (𝑥, 𝑦) with 𝑥 = 𝑐 is the curve, or cross-section, we get by intersecting the graph of 𝑓 with the plane 𝑥 = 𝑐. We define a cross-section of 𝑓 with 𝑦 fixed similarly. For example, the cross-section of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 with 𝑥 = 2 is 𝑓 (2, 𝑦) = 4 + 𝑦2 . The graph of this cross-section is the curve we get by intersecting the graph of 𝑓 with the plane perpendicular to the 𝑥-axis at 𝑥 = 2. (See Figure 12.18.) Surface

Curve

𝑧

𝑧

Surface

✙ 𝑓 (𝑎, 𝑦)

𝑓 (𝑥, 𝑦)

Curve

✙ 𝑓 (𝑥, 𝑏)

𝑓 (𝑥, 𝑦)

𝑦

𝑦

𝑥 Figure 12.18: Cross-section of the surface 𝑧 = 𝑓 (𝑥, 𝑦) by the plane 𝑥 = 2

𝑥

Figure 12.19: The curves 𝑧 = 𝑓 (𝑎, 𝑦) with 𝑎 constant: cross-sections with 𝑥 fixed

Figure 12.20: The curves 𝑧 = 𝑓 (𝑥, 𝑏) with 𝑏 constant: cross-sections with 𝑦 fixed

Figure 12.19 shows graphs of other cross-sections of 𝑓 with 𝑥 fixed; Figure 12.20 shows graphs of cross-sections with 𝑦 fixed. Example 3

Describe the cross-sections of the function 𝑔(𝑥, 𝑦) = 𝑥2 − 𝑦2 with 𝑦 fixed and then with 𝑥 fixed. Use these cross-sections to describe the shape of the graph of 𝑔.

Solution

The cross-sections with 𝑦 fixed at 𝑦 = 𝑏 are given by 𝑧 = 𝑔(𝑥, 𝑏) = 𝑥2 − 𝑏2 . Thus, each cross-section with 𝑦 fixed gives a parabola opening upward, with minimum 𝑧 = −𝑏2 . The cross-sections with 𝑥 fixed are of the form 𝑧 = 𝑔(𝑎, 𝑦) = 𝑎2 − 𝑦2 , which are parabolas opening downward with a maximum of 𝑧 = 𝑎2 . (See Figures 12.21 and 12.22.) The graph of 𝑔 is shown in Figure 12.23. Notice the upward-opening parabolas in the 𝑥-direction and the downward-opening parabolas in the 𝑦-direction. We say that the surface is saddle-shaped. 𝑧

{

✛ ✛ 𝑥

−1

𝑦=0

𝑧

𝑧 4

2

{𝑧 = 𝑥 𝑦 = ±1

𝑧 = 𝑥2 − 1

✛ 1

✛ ✛

{

𝑦 = ±2 𝑧 = 𝑥2 − 4

−4 Figure 12.21: Cross-sections of 𝑔(𝑥, 𝑦) = 𝑥2 − 𝑦2 with 𝑦 fixed



𝑦

{

𝑥 = ±2

{

𝑥 = ±1

𝑧 = 4 − 𝑦2 𝑥 2

{𝑧 = 1−𝑦 𝑥=0

𝑧 = −𝑦2 Figure 12.22: Cross-sections of 𝑔(𝑥, 𝑦) = 𝑥2 − 𝑦2 with 𝑥 fixed

𝑦 Figure 12.23: Graph of 𝑔(𝑥, 𝑦) = 𝑥2 − 𝑦2 showing cross-sections

664

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

Linear Functions Linear functions are central to single-variable calculus; they are equally important in multivariable calculus. You may be able to guess the shape of the graph of a linear function of two variables. (It’s a plane.) Let’s look at an example. Example 4

Describe the graph of 𝑓 (𝑥, 𝑦) = 1 + 𝑥 − 𝑦.

Solution

The plane 𝑥 = 𝑎 is vertical and parallel to the 𝑦𝑧-plane. Thus, the cross-section with 𝑥 = 𝑎 is the line 𝑧 = 1 + 𝑎 − 𝑦 which slopes downward in the 𝑦-direction. Similarly, the plane 𝑦 = 𝑏 is parallel to the 𝑥𝑧-plane. Thus, the cross-section with 𝑦 = 𝑏 is the line 𝑧 = 1 + 𝑥 − 𝑏 which slopes upward in the 𝑥-direction. Since all the cross-sections are lines, you might expect the graph to be a flat plane, sloping down in the 𝑦-direction and up in the 𝑥-direction. This is indeed the case. (See Figure 12.24.)

Figure 12.24: Graph of the plane 𝑧 = 1 + 𝑥 − 𝑦 showing cross-section with 𝑥 = 𝑎

When One Variable Is Missing: Cylinders Suppose we graph an equation like 𝑧 = 𝑥2 which has one variable missing. What does the surface look like? Since 𝑦 is missing from the equation, the cross-sections with 𝑦 fixed are all the same parabola, 𝑧 = 𝑥2 . Letting 𝑦 vary up and down the 𝑦-axis, this parabola sweeps out the troughshaped surface shown in Figure 12.25. The cross-sections with 𝑥 fixed are horizontal lines obtained by cutting the surface by a plane perpendicular to the 𝑥-axis. This surface is called a parabolic cylinder, because it is formed from a parabola in the same way that an ordinary cylinder is formed from a circle; it has a parabolic cross-section instead of a circular one. 𝑧 𝑧

𝑦 𝑥 Figure 12.25: A parabolic cylinder 𝑧 = 𝑥2

Example 5

Graph the equation 𝑥2 + 𝑦2 = 1 in 3-space.

𝑦 𝑥 Figure 12.26: Circular cylinder 𝑥2 + 𝑦2 = 1

665

12.2 GRAPHS AND SURFACES

Although the equation 𝑥2 + 𝑦2 = 1 does not represent a function, the surface representing it can be graphed by the method used for 𝑧 = 𝑥2 . The graph of 𝑥2 + 𝑦2 = 1 in the 𝑥𝑦-plane is a circle. Since 𝑧 does not appear in the equation, the intersection of the surface with any horizontal plane will be the same circle 𝑥2 + 𝑦2 = 1. Thus, the surface is the cylinder shown in Figure 12.26.

Solution

Exercises and Problems for Section 12.2 Online Resource: Additional Problems for Section 12.2 EXERCISES

In Exercises 1–4, which of (I)–(IV) lie on the graph of the function 𝑧 = 𝑓 (𝑥, 𝑦)? √ I. (1, 0, 1) II. ( 8, 1, 3) III. (−3, 7, −3) IV. (1, 1, 1∕2) √ 1. 𝑓 (𝑥, 𝑦) = −3 2. 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 3. 𝑓 (𝑥, 𝑦) = 1∕(𝑥2 + 𝑦2 )

4. 𝑓 (𝑥, 𝑦) = 4 − 𝑦

7. Without a calculator or computer, match the functions with their graphs in Figure 12.29. (a) (c) (e)

1 𝑥2 + 𝑦2 𝑧 = 𝑥 + 2𝑦 + 3 𝑧 = 𝑥3 − sin 𝑦. 𝑧=

2 −𝑦2

(b) 𝑧 = −𝑒−𝑥 (d) 𝑧 = −𝑦2

𝑧

(I)

𝑧

(II)

𝑦

5. Figure 12.27 shows the graph of 𝑧 = 𝑓 (𝑥, 𝑦).

𝑥

(a) Suppose 𝑦 is fixed and positive. Does 𝑧 increase or decrease as 𝑥 increases? Graph 𝑧 against 𝑥. (b) Suppose 𝑥 is fixed and positive. Does 𝑧 increase or decrease as 𝑦 increases? Graph 𝑧 against 𝑦. 𝑥 𝑦 𝑧

(III)

(IV)

𝑦

𝑧

𝑥 𝑦

Figure 12.27 6. Without a calculator or computer, match the functions with their graphs in Figure 12.28. (a) 𝑧 = 2 + 𝑥2 + 𝑦2 (c) 𝑧 = 2(𝑥2 + 𝑦2 ) (e) 𝑧 = 2

𝑥 𝑧

(V)

(b) 𝑧 = 2 − 𝑥2 − 𝑦2 (d) 𝑧 = 2 + 2𝑥 − 𝑦 𝑥 𝑦 (II)

(I)

Figure 12.29

(IV)

(III)

In Exercises 8–15, sketch a graph of the surface and briefly describe it in words. 9. 𝑥2 + 𝑦2 + 𝑧2 = 9

8. 𝑧 = 3 10. 𝑧 = 𝑥2 + 𝑦2 + 4

11. 𝑧 = 5 − 𝑥2 − 𝑦2

12. 𝑧 = 𝑦2

13. 2𝑥 + 4𝑦 + 3𝑧 = 12

2

2

14. 𝑥 + 𝑦 = 4

15. 𝑥2 + 𝑧2 = 4

(V)

Figure 12.28

In Exercises 16–18, find the equation of the surface. √ 16. A cylinder of radius 7 with its (axis along) the 𝑦-axis. √ 17. A sphere of radius 3 centered at 0, 7, 0 . 18. The paraboloid obtained by moving the surface 𝑧 = 𝑥2 + 𝑦2 so that its vertex is at (1, 3, 5), its axis is parallel to the 𝑥-axis, and the surface opens towards negative 𝑥 values.

666

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

PROBLEMS 19. Consider the function 𝑓 given by 𝑓 (𝑥, 𝑦) = 𝑦3 + 𝑥𝑦. Draw graphs of cross-sections with:

𝑦

(a) 𝑓 (4, 𝑡)

𝑦

𝑥 𝑧

(III)

𝑧

(IV)

𝑦

20. Find 𝑓 (3, 2). Give units and interpret in terms of drug concentration. 21. Graph the following single-variable functions and explain their significance in terms of drug concentration.

𝑧

(II)

𝑥

(a) 𝑥 fixed at 𝑥 = −1, 𝑥 = 0, and 𝑥 = 1. (b) 𝑦 fixed at 𝑦 = −1, 𝑦 = 0, and 𝑦 = 1. Problems 20–22 concern the concentration, 𝐶, in mg per liter, of a drug in the blood as a function of 𝑥, the amount, in mg, of the drug given and 𝑡, the time in hours since the injection. For 0 ≤ 𝑥 ≤ 4 and 𝑡 ≥ 0, we have 𝐶 = 𝑓 (𝑥, 𝑡) = 𝑡𝑒−𝑡(5−𝑥) .

𝑧

(I)

𝑥

𝑦

𝑥 𝑧

(V)

𝑧

(VI)

(b) 𝑓 (𝑥, 1)

Problems 23–24 concern the kinetic energy, 𝐸 = 𝑓 (𝑚, 𝑣) = 1 𝑚𝑣2 , in joules, of a moving object as a function of its mass 2 𝑚 ≥ 0, in kg, and its speed 𝑣 ≥ 0, in m/sec.

𝑦

𝑥

22. Graph 𝑓 (𝑎, 𝑡) for 𝑎 = 1, 2, 3, 4 on the same axes. Describe how the graph changes as 𝑎 increases and explain what this means in terms of drug concentration.

𝑥 𝑦 𝑧

(VII)

𝑧

(VIII)

𝑦

23. Find 𝑓 (2, 10). Give units and interpret this quantity in the context of kinetic energy.

𝑥

(a) 𝑓 (6, 𝑣)

(b) 𝑓 (𝑚, 20)

25. Find 𝑓 (2, 12). Give units and interpret this quantity in the context of atmospheric pressure. 26. Graph the following single-variable functions and explain the significance of the shape of the graph in terms of atmospheric pressure. (b) 𝑓 (𝑦, 24)

27. Without a computer or calculator, match the equations (a)–(i) with the graphs (I)–(IX). (√ ) 2 2 (b) 𝑧 = cos 𝑥2 + 𝑦2 (a) 𝑧 = 𝑥𝑦𝑒−(𝑥 +𝑦 ) (c) 𝑧 = sin 𝑦 (e) 𝑧 = cos2 𝑥 cos2 𝑦 (g) 𝑧 = cos(𝑥𝑦) 2 2 (i) 𝑧 = (2𝑥2 + 𝑦2 )𝑒1−𝑥 −𝑦

𝑧

(IX)

In Problems 25–26, the atmospheric pressure, 𝑃 = 𝑓 (𝑦, 𝑡) = (950 + 2𝑡)𝑒−𝑦∕7 , in millibars, on a weather balloon, is a function of its height 𝑦 ≥ 0, in km above sea level after 𝑡 hours with 0 ≤ 𝑡 ≤ 48.

(a) 𝑓 (3, 𝑡)

𝑦

𝑥

24. Graph the following single-variable functions and explain their significance in terms of kinetic energy.

1 𝑥2 + 𝑦2 sin(𝑥2 + 𝑦2 ) (f) 𝑧 = 𝑥2 + 𝑦2 (h) 𝑧 = |𝑥||𝑦|

(d) 𝑧 = −

𝑦

𝑥

28. Decide whether the graph of each of the following equations is the shape of a bowl, a plate, or neither. Consider a plate to be any flat surface and a bowl to be anything that could hold water, assuming the positive 𝑧-axis is up. (a)

𝑧 = 𝑥2 + 𝑦2

(c)

𝑥+𝑦+𝑧 = 1

(e)

𝑧=3

(b) 𝑧 = 1 − 𝑥2 − 𝑦2 √ (d) 𝑧 = − 5 − 𝑥2 − 𝑦2

29. Sketch cross-sections for each function in Problem 28. 30. Without a calculator or computer, match the functions with their cross-sections with 𝑥 fixed in Figure 12.30. (a)

𝑧 = 1∕(1 + 𝑥2 + 𝑦2 ) (b) 𝑧 = 1 + 𝑥 + 𝑦

(c)

𝑧 = 𝑒−𝑥+𝑦

(d) 𝑧 = 𝑒𝑥−𝑦

(e)

𝑧 = sin(𝑥𝑦)

(f) 𝑧 = 𝑥2 .

12.2 GRAPHS AND SURFACES 𝑧

(I)

𝑧

(II)

(I)

667

(II)

1 2 𝑦 −2

2 −2 𝑦 −2 𝑧

(III)

2 𝑧

(IV)

9

6

(III)

(IV)

4 4 2 1

𝑦 −2

2 𝑧

(V)

𝑦

−2

2 𝑧

(VI)

Figure 12.32

6

1

33. For each of the graphs I–IV in Problem 32, draw: 4

(a) Two cross-sections with pizza fixed (b) Two cross-sections with cola fixed.

𝑦 −2

2

2

𝑦 For Problems 34–37, give a formula for a function whose

−1 −2

2

graph is described. Sketch it using a computer or calculator.

2

34. A bowl which opens upward and has its vertex at 5 on the 𝑧-axis.

Figure 12.30 2

31. Without a calculator or computer, for 𝑧 = 𝑥 + 2𝑥𝑦 , determine which of (I)–(II) in Figure 12.31 are crosssections with 𝑥 fixed and which are cross-sections with 𝑦 fixed. 𝑧

(I)

𝑧

(II)

4

38. Sketch cross-sections of 𝑓 (𝑟, ℎ) = 𝜋𝑟2 ℎ, first keeping ℎ fixed, then keeping 𝑟 fixed.

2 −4

36. A parabolic cylinder opening upward from along the line 𝑦 = 𝑥 in the 𝑥𝑦-plane. 37. A cone of circular cross-section opening downward and with its vertex at the origin.

4 −2

35. A plane which has its 𝑥-, 𝑦-, and 𝑧-intercepts all positive.

−2

2 −2

Figure 12.31 32. You like pizza and you like cola. Which of the graphs in Figure 12.32 represents your happiness as a function of how many pizzas and how much cola you have if (a) There is no such thing as too many pizzas and too much cola? (b) There is such a thing as too many pizzas or too much cola? (c) There is such a thing as too much cola but no such thing as too many pizzas?

39. By setting one variable constant, find a plane that intersects the graph of 𝑧 = 4𝑥2 − 𝑦2 + 1 in a: (a) Parabola opening upward (b) Parabola opening downward (c) Pair of intersecting straight lines. 40. Sketch cross-sections of the equation 𝑧 = 𝑦 − 𝑥2 with 𝑥 fixed and with 𝑦 fixed and use them to sketch a graph of 𝑧 = 𝑦 − 𝑥2 . 41. A wave travels along a canal. Let 𝑥 be the distance along the canal, 𝑡 be the time, and 𝑧 be the height of the water above the equilibrium level. The graph of 𝑧 as a function of 𝑥 and 𝑡 is in Figure 12.33. (a) Draw the profile of the wave for 𝑡 = −1, 0, 1, 2. (Put the 𝑥-axis to the right and the 𝑧-axis vertical.)

668

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

(b) Is the wave traveling in the direction of increasing or decreasing 𝑥? (c) Sketch a surface representing a wave traveling in the opposite direction. 𝑧

where 𝑇 is the temperature of the gas, in Kelvin, and 𝑉 is the volume of the cylinder, in liters. Figures 12.34 and 12.35 give cross-sections of the function 𝑓 . (a) Which figure shows cross-sections of 𝑓 with 𝑇 fixed? What does the shape of the cross-sections tell you about the pressure? (b) Which figure shows cross-sections of 𝑓 with 𝑉 fixed? What does the shape of the cross-sections tell you about the pressure? 𝑃

𝑃

𝑡 𝑥

Figure 12.33 42. The pressure of a fixed amount of compressed nitrogen gas in a cylinder is given, in atmospheres, by

Figure 12.34

Figure 12.35

10𝑇 , 𝑃 = 𝑓 (𝑇 , 𝑉 ) = 𝑉

Strengthen Your Understanding In Problems 43–44, explain what is wrong with the statement.

52. The plane 𝑥 + 𝑦 + 𝑧 = 3 intersects the 𝑥-axis when 𝑥 = 3.

43. The graph of the function 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 is a circle.

53. The sphere 𝑥2 +𝑦2 +𝑧2 = 10 intersects the plane 𝑥 = 10.

44. Cross-sections of the function 𝑓 (𝑥, 𝑦) = 𝑥2 with 𝑥 fixed are parabolas.

54. The cross-section of the function 𝑓 (𝑥, 𝑦) = 𝑥 + 𝑦2 with 𝑦 = 1 is a line.

In Problems 45–47, give an example of:

55. The function 𝑔(𝑥, 𝑦) = 1 − 𝑦2 has identical parabolas for all cross-sections with 𝑥 constant.

45. A function whose graph lies above the 𝑥𝑦-plane and intersects the plane 𝑧 = 2 in a single point.

56. The function 𝑔(𝑥, 𝑦) = 1 − 𝑦2 has lines for all crosssections with 𝑦 constant.

46. A function which intersects the 𝑥𝑧-plane in a parabola and the 𝑦𝑧-plane in a line.

57. The graphs of 𝑓 (𝑥, 𝑦) = sin(𝑥𝑦) and 𝑔(𝑥, 𝑦) = sin(𝑥𝑦)+ 2 never intersect.

47. A function which intersects the 𝑥𝑦-plane in a circle.

58. The graphs of 𝑓 (𝑥, 𝑦) = 𝑥2 +𝑦2 and 𝑔(𝑥, 𝑦) = 1−𝑥2 −𝑦2 intersect in a circle.

Are the statements in Problems 48–61 true or false? Give reasons for your answer. 48. The function given by the formula 𝑓 (𝑣, 𝑤) = 𝑒𝑣 ∕𝑤 is an increasing function of 𝑣 when 𝑤 is a nonzero constant. 49. A function 𝑓 (𝑥, 𝑦) can be an increasing function of 𝑥 with 𝑦 held fixed, and be a decreasing function of 𝑦 with 𝑥 held fixed. 50. A function 𝑓 (𝑥, 𝑦) can have the property that 𝑔(𝑥) = 𝑓 (𝑥, 5) is increasing, whereas ℎ(𝑥) = 𝑓 (𝑥, 10) is decreasing. 51. The plane 𝑥 + 2𝑦 − 3𝑧 = 1 passes through the origin.

12.3

59. If all the cross-sections of the graph of 𝑓 (𝑥, 𝑦) with 𝑥 constant are lines, then the graph of 𝑓 is a plane. 60. The only point of intersection of the graphs of 𝑓 (𝑥, 𝑦) and −𝑓 (𝑥, 𝑦) is the origin. 61. The point (0, 0, 10) is the highest point on the graph of the function 𝑓 (𝑥, 𝑦) = 10 − 𝑥2 − 𝑦2 . 62. The object in 3-space described by 𝑥 = 2 is (a) A point (c) A plane

(b) A line (d) Undefined.

CONTOUR DIAGRAMS The surface which represents a function of two variables often gives a good idea of the function’s general behavior—for example, whether it is increasing or decreasing as one of the variables increases. However, it is difficult to read numerical values off a surface and it can be hard to see all of the function’s behavior from a surface. Thus, functions of two variables are often represented by contour diagrams like the weather map on page 652. Contour diagrams have the additional advantage that they can be extended to functions of three variables.

12.3 CONTOUR DIAGRAMS

669

Topographical Maps One of the most common examples of a contour diagram is a topographical map like that shown in Figure 12.36. It gives the elevation in the region and is a good way of getting an overall picture of the terrain: where the mountains are, where the flat areas are. Such topographical maps are frequently colored green at the lower elevations and brown, red, or white at the higher elevations.

Sout h Hami l t on

1300 nt as a Pl e

1500

00 17

1505

1311 l Hi

1600

a Cl

l

rk

0 170

00 13

Figure 12.36: A topographical map showing the region around South Hamilton, NY

300 400 500

300

600 800

500

700 800

The curves on a topographical map that separate lower elevations from higher elevations are called contour lines because they outline the contour or shape of the land.5 Because every point along the same contour has the same elevation, contour lines are also called level curves or level sets. The more closely spaced the contours, the steeper the terrain; the more widely spaced the contours, the flatter the terrain (provided, of course, that the elevation between contours varies by a constant amount). Certain features have distinctive characteristics. A mountain peak is typically surrounded by contour lines like those in Figure 12.37. A pass in a range of mountains may have contours that look like Figure 12.38. A long valley has parallel contour lines indicating the rising elevations on both sides of the valley (see Figure 12.39); a long ridge of mountains has the same type of contour lines, only the elevations decrease on both sides of the ridge. Notice that the elevation numbers on the contour lines are as important as the curves themselves. We usually draw contours for equally spaced values of 𝑧.

500

20 10 0 10 0 20 0 0

10 0

0 20

300

Figure 12.37: Mountain peak

Figure 12.38: Pass between two mountains

Figure 12.39: Long valley

Figure 12.40: Impossible contour lines

Notice that two contours corresponding to different elevations cannot cross each other as shown in Figure 12.40. If they did, the point of intersection of the two curves would have two different elevations, which is impossible (assuming the terrain has no overhangs). 5 In

fact they are usually not straight lines, but curves. They may also be in disconnected pieces.

670

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

Corn Production Contour maps can display information about a function of two variables without reference to a surface. Consider the effect of weather conditions on US corn production. Figure 12.41 gives corn production 𝐶 = 𝑓 (𝑅, 𝑇 ) as a function of the total rainfall, 𝑅, in inches, and average temperature, 𝑇 , in degrees Fahrenheit, during the growing season.6 At the present time, 𝑅 = 15 inches and 𝑇 = 76◦ F. Production is measured as a percentage of the present production; thus, the contour through 𝑅 = 15, 𝑇 = 76, has value 100, that is, 𝐶 = 𝑓 (15, 76) = 100. Example 1

Use Figure 12.41 to estimate 𝑓 (18, 78) and 𝑓 (12, 76) and interpret in terms of corn production.

60

80

80

90

100 110

Present

40

𝑇 (temperature in ◦ F) 76 74

70

50

78

113

72 6

9

12 15 18 𝑅 (rainfall in inches)

21

24

Figure 12.41: Corn production, C, as a function of rainfall and temperature

Solution

The point with 𝑅-coordinate 18 and 𝑇 -coordinate 78 is on the contour 𝐶 = 100, so 𝑓 (18, 78) = 100. This means that if the annual rainfall were 18 inches and the temperature were 78◦ F, the country would produce about the same amount of corn as at present, although it would be wetter and warmer than it is now. The point with 𝑅-coordinate 12 and 𝑇 -coordinate 76 is about halfway between the 𝐶 = 80 and the 𝐶 = 90 contours, so 𝑓 (12, 76) ≈ 85. This means that if the rainfall fell to 12 inches and the temperature stayed at 76◦ , then corn production would drop to about 85% of what it is now.

Example 2

Use Figure 12.41 to describe in words the cross-sections with 𝑇 and 𝑅 constant through the point representing present conditions. Give a common-sense explanation of your answer.

Solution

To see what happens to corn production if the temperature stays fixed at 76◦ F but the rainfall changes, look along the horizontal line 𝑇 = 76. Starting from the present and moving left along the line 𝑇 = 76, the values on the contours decrease. In other words, if there is a drought, corn production decreases. Conversely, as rainfall increases, that is, as we move from the present to the right along the line 𝑇 = 76, corn production increases, reaching a maximum of more than 110% when 𝑅 = 21, and then decreases (too much rainfall floods the fields). If, instead, rainfall remains at the present value and temperature increases, we move up the vertical line 𝑅 = 15. Under these circumstances corn production decreases; a 2◦ F increase causes a 10% drop in production. This makes sense since hotter temperatures lead to greater evaporation and hence drier conditions, even with rainfall constant at 15 inches. Similarly, a decrease in temperature leads to a very slight increase in production, reaching a maximum of around 102% when 𝑇 = 74, followed by a decrease (the corn won’t grow if it is too cold). 6 Adapted from S. Beaty and R. Healy, “The Future of American Agriculture,” Scientific American 248, No. 2, February 1983.

12.3 CONTOUR DIAGRAMS

671

Contour Diagrams and Graphs Contour diagrams and graphs are two different ways of representing a function of two variables. How do we go from one to the other? In the case of the topographical map, the contour diagram was created by joining all the points at the same height on the surface and dropping the curve into the 𝑥𝑦-plane. How do we go the other way? Suppose we wanted to plot the surface representing the corn production function 𝐶 = 𝑓 (𝑅, 𝑇 ) given by the contour diagram in Figure 12.41. Along each contour the function has a constant value; if we take each contour and lift it above the plane to a height equal to this value, we get the surface in Figure 12.42. 100 contour raised 100 units

110





110 contour raised 110 units

0 10

90

10 0

80 70 60 50 40

110

Figure 12.42: Getting the graph of the corn yield function from the contour diagram

Notice that the raised contours are the curves we get by slicing the surface horizontally. In general, we have the following result: Contour lines, or level curves, are obtained from a surface by slicing it with horizontal planes. A contour diagram is a collection of level curves labeled with function values.

Finding Contours Algebraically Algebraic equations for the contours of a function 𝑓 are easy to find if we have a formula for 𝑓 (𝑥, 𝑦). Suppose the surface has equation 𝑧 = 𝑓 (𝑥, 𝑦). A contour is obtained by slicing the surface with a horizontal plane with equation 𝑧 = 𝑐. Thus, the equation for the contour at height 𝑐 is given by: 𝑓 (𝑥, 𝑦) = 𝑐.

Example 3

Find equations for the contours of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 and draw a contour diagram for 𝑓 . Relate the contour diagram to the graph of 𝑓 .

Solution

The contour at height 𝑐 is given by 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 = 𝑐. This is a contour only for 𝑐 ≥ 0, For 𝑐 > 0 it is a circle of radius

√ 𝑐. For 𝑐 = 0, it is a single point (the

672

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

origin). Thus, the contours at an elevation of 𝑐 = 1, 2, 3, 4, … are all circles centered at the origin of √ √ radius 1, 2, 3, 2, …. The contour diagram is shown in Figure 12.43. The bowl–shaped graph of 𝑓 is shown in Figure 12.44. Notice that the graph of 𝑓 gets steeper as we move further away from the origin. This is reflected in the fact that the contours become more closely packed as we move further from the origin; for example, the contours for 𝑐 = 6 and 𝑐 = 8 are closer together than the contours for 𝑐 = 2 and 𝑐 = 4. 𝑦

3 2

8

6

4

2

1

𝑥 2

−1 −2

4

6

8

−3 −3 −2 −1 1 2 3 Figure 12.43: Contour diagram for 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 (even values of 𝑐 only)

Figure 12.44: The graph of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2

Example 4

Draw a contour diagram for 𝑓 (𝑥, 𝑦) =

Solution

The contour at level 𝑐 is given by



𝑥2 + 𝑦2 and relate it to the graph of 𝑓 .

𝑓 (𝑥, 𝑦) =

√ 𝑥2 + 𝑦2 = 𝑐.

√ For 𝑐 > 0 this is a circle, just as in the previous example, but here the radius is 𝑐 instead of 𝑐. For 𝑐 = 0, it is the origin. Thus, if the level 𝑐 increases by 1, the radius of the contour increases by 1. This means the contours are equally spaced concentric circles (see Figure 12.45) which do not become more closely packed further from the origin. Thus, the graph of 𝑓 has the same constant slope as we move away from the origin (see Figure 12.46), making it a cone rather than a bowl. 𝑧 𝑦 3 2 1

𝑥

𝑦

𝑥 Figure 12.45: A contour √ diagram for 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2

Figure 12.46: The √ graph of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2

In both of the previous examples the level curves are concentric circles because the surfaces have circular symmetry. Any function of two variables which depends only (𝑥2 + 𝑦2 ) ) (√ on the quantity 2 2 𝑥2 + 𝑦2 . has such symmetry: for example, 𝐺(𝑥, 𝑦) = 𝑒−(𝑥 +𝑦 ) or 𝐻(𝑥, 𝑦) = sin

673

12.3 CONTOUR DIAGRAMS

Example 5

Draw a contour diagram for 𝑓 (𝑥, 𝑦) = 2𝑥 + 3𝑦 + 1.

Solution

The contour at level 𝑐 has equation 2𝑥 + 3𝑦 + 1 = 𝑐. Rewriting this as 𝑦 = −(2∕3)𝑥 + (𝑐 − 1)∕3, we see that the contours are parallel lines with slope −2∕3. The 𝑦-intercept for the contour at level 𝑐 is (𝑐 − 1)∕3; each time 𝑐 increases by 3, the 𝑦-intercept moves up by 1. The contour diagram is shown in Figure 12.47. 𝑦 3 19

2

16 13

1

10

0 −1

1 −2

−2 −3

𝑥

7 4

−5

1

2

3

4

5

6

Figure 12.47: A contour diagram for 𝑓 (𝑥, 𝑦) = 2𝑥 + 3𝑦 + 1

Contour Diagrams and Tables Sometimes we can get an idea of what the contour diagram of a function looks like from its table. Relate the values of 𝑓 (𝑥, 𝑦) = 𝑥2 − 𝑦2 in Table 12.5 to its contour diagram in Figure 12.48.

Example 6 Table 12.5

𝑦

Table of values of 𝑓 (𝑥, 𝑦) = 𝑥2 − 𝑦2 3 −5

−8

−9

−8

−5

0

−4

2

5

0

−3

−4

−3

0

5

1

8

3

0

−1

0

3

8

0

9

4

1

0

1

4

9

−1

8

3

0

−1

0

3

8

−2

5

0

−3

−4

−3

0

5

−1

−3

0

−5

−8

−9

−8

−5

0

−2

−3

−2

−1

0

1

2

3

−3

−2

4

2

1

4 2

𝑦

0

2

3

𝑥

0

0

𝑥 0

0

−2 −4

−3 −2 −1

1

2

3

Figure 12.48: Contour map of 𝑓 (𝑥, 𝑦) = 𝑥2 − 𝑦2

Solution

One striking feature of the values in Table 12.5 is the zeros along the diagonals. This occurs because 𝑥2 − 𝑦2 = 0 along the lines 𝑦 = 𝑥 and 𝑦 = −𝑥. So the 𝑧 = 0 contour consists of these two lines. In the triangular region of the table that lies to the right of both diagonals, the entries are positive. To the left of both diagonals, the entries are also positive. Thus, in the contour diagram, the positive contours lie in the triangular regions to the right and left of the lines 𝑦 = 𝑥 and 𝑦 = −𝑥. Further, the table shows that the numbers on the left are the same as the numbers on the right; thus, each contour has two pieces, one on the left and one on the right. See Figure 12.48. As we move away from the origin along the 𝑥-axis, we cross contours corresponding to successively larger values. On the saddle-shaped graph of 𝑓 (𝑥, 𝑦) = 𝑥2 − 𝑦2 shown in Figure 12.49, this corresponds to climbing out of the saddle along one of the ridges. Similarly, the negative contours occur in pairs in the top and bottom triangular regions; the values get more and more negative as we go out along the 𝑦-axis. This corresponds to descending from the saddle along the valleys that are submerged below the 𝑥𝑦-

674

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

plane in Figure 12.49. Notice that we could also get the contour diagram by graphing the family of hyperbolas 𝑥2 − 𝑦2 = 0, ±2, ±4, ….

Figure 12.49: Graph of 𝑓 (𝑥, 𝑦) = 𝑥2 − 𝑦2 showing plane 𝑧 = 0

Using Contour Diagrams: The Cobb-Douglas Production Function Suppose you decide to expand your small printing business. Should you start a night shift and hire more workers? Should you buy more expensive but faster computers which will enable the current staff to keep up with the work? Or should you do some combination of the two? Obviously, the way such a decision is made in practice involves many other considerations— such as whether you could get a suitably trained night shift, or whether there are any faster computers available. Nevertheless, you might model the quantity, 𝑃 , of work produced by your business as a function of two variables: your total number, 𝑁, of workers, and the total value, 𝑉 , of your equipment. What might the contour diagram of the production function look like? Example 7

Explain why the contour diagram in Figure 12.50 does not model the behavior expected of the production function, whereas the contour diagram in Figure 12.51 does. 𝑉

𝑉

𝑃 =3 𝑃 =2 𝑃 =1 𝑁 Figure 12.50: Incorrect contours for printing production

Solution

𝑃 =3 𝑃 =2

✠ 𝑃 =1 ✠ ✠ 𝑁 Figure 12.51: Correct contours for printing production

Look at Figure 12.50. Notice that the contour 𝑃 = 1 intersects the 𝑁- and the 𝑉-axes, suggesting that it is possible to produce work with no workers or with no equipment; this is unreasonable. However, no contours in Figure 12.51 intersect either the 𝑁- or the 𝑉-axis. In Figure 12.51, fixing 𝑉 and letting 𝑁 increase corresponds to moving to the right, crossing contours less and less frequently. Production increases more and more slowly because hiring additional workers does little to boost production if the machines are already used to capacity. Similarly, if we fix 𝑁 and let 𝑉 increase, Figure 12.51 shows production increasing, but at a decreasing rate. Buying machines without enough people to use them does not increase production much. Thus Figure 12.51 fits the expected behavior of the production function best.

12.3 CONTOUR DIAGRAMS

675

Formula for a Production Function Production functions are often approximated by formulas of the form 𝑃 = 𝑓 (𝑁, 𝑉 ) = 𝑐𝑁 𝛼 𝑉 𝛽 where 𝑃 is the quantity produced and 𝑐, 𝛼, and 𝛽 are positive constants, 0 < 𝛼 < 1 and 0 < 𝛽 < 1. Example 8

Show that the contours of the function 𝑃 = 𝑐𝑁 𝛼 𝑉 𝛽 have approximately the shape of the contours in Figure 12.51.

Solution

The contours are the curves where 𝑃 is equal to a constant value, say 𝑃0 , that is, where 𝑐𝑁 𝛼 𝑉 𝛽 = 𝑃0 . Solving for 𝑉, we get 𝑉 =

(

𝑃0 𝑐

)1∕𝛽

𝑁 −𝛼∕𝛽 .

Thus, 𝑉 is a power function of 𝑁 with a negative exponent, so its graph has the general shape shown in Figure 12.51.

The Cobb-Douglas Production Model In 1928, Cobb and Douglas used a similar function to model the production of the entire US economy in the first quarter of this century. Using government estimates of 𝑃 , the total yearly production between 1899 and 1922, of 𝐾, the total capital investment over the same period, and of 𝐿, the total labor force, they found that 𝑃 was well approximated by the Cobb-Douglas production function 𝑃 = 1.01𝐿0.75𝐾 0.25 . This function turned out to model the US economy surprisingly well, both for the period on which it was based and for some time afterward.7

Exercises and Problems for Section 12.3 EXERCISES In Exercises 1–4, sketch a possible contour diagram for each surface, marked with reasonable 𝑧-values. (Note: There are many possible answers.) 𝑧 𝑧 2. 1.

𝑦 𝑥

𝑦

In Exercises 5–13, sketch a contour diagram for the function with at least four labeled contours. Describe in words the contours and how they are spaced. 5. 𝑓 (𝑥, 𝑦) = 𝑥 + 𝑦

6. 𝑓 (𝑥, 𝑦) = 3𝑥 + 3𝑦

7. 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2

8. 𝑓 (𝑥, 𝑦) = −𝑥2 − 𝑦2 + 1

9. 𝑓 (𝑥, 𝑦) = 𝑥𝑦 11. 𝑓 (𝑥, 𝑦) = 𝑥2 + 2𝑦2 𝑥

13. 𝑓 (𝑥, 𝑦) = cos 3.

4.

√ 𝑥2 + 𝑦2

14. Let 𝑓 (𝑥, 𝑦) = 3𝑥2 𝑦 + 7𝑥 + 20. Find an equation for the contour that goes through the point (5, 10).

𝑧

𝑥 𝑦

7 Cobb,

10. 𝑓 (𝑥, 𝑦) = 𝑦 − 𝑥2 √ 12. 𝑓 (𝑥, 𝑦) = 𝑥2 + 2𝑦2

15. (a) For 𝑧 = 𝑓 (𝑥, 𝑦) = 𝑥𝑦, sketch and label the level curves 𝑧 = ±1, 𝑧 = ±2. (b) Sketch and label cross-sections of 𝑓 with 𝑥 = ±1, 𝑥 = ±2. (c) The surface 𝑧 = 𝑥𝑦 is cut by a vertical plane containing the line 𝑦 = 𝑥. Sketch the cross-section.

C. and Douglas, P., "A Theory of Production", American Economic Review 18 (1928: Supplement), pp. 139–165.

676

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

I. (1, 0, 2) III. (0, −1, −2) V. (0, 1, 1)

16. Match the surfaces (a)–(e) in Figure 12.52 with the contour diagrams (I)–(V) in Figure 12.53. 𝑧

(a)

𝑧

(b)

II. (1, 1, 1) IV. (−1, 0, −2) VI. (−1, −1, 0)

𝑦

𝑥

𝑦 2 𝑥

2

−2

𝑦

1

1

−1 1

𝑧

(c)

𝑧

(d)

𝑥

0

−2

−1

2

1

2

−1 𝑦

−2

Figure 12.54

𝑦 𝑥

𝑥

18. Match Tables 12.6–12.9 with contour diagrams (I)– (IV) in Figure 12.55.

𝑧 (e)

Table 12.6

𝑥

Table 12.7

𝑦∖𝑥

−1

0

1

𝑦∖𝑥

−1

0

1

−1

2

1

2

−1

0

1

0

0

1

0

1

0

1

2

1

1

2

1

2

1

0

1

0

𝑦

Figure 12.52 𝑦

(I)

Table 12.8

𝑦

(II) −3 −1

𝑥

0

𝑥 1

3

1 3

(I)

𝑦

(III)



𝑦∖𝑥

−1

0

1

𝑦∖𝑥

−1

0

1

−1

2

0

2

−1

2

2

2

0

2

0

2

0

0

0

0

1

2

0

2

1

2

2

2

𝑦

(II)

𝑦

𝑦

(IV)

−4

Table 12.9

1

2



0





𝑥

3

1 6

0

𝑥

0

𝑥

1 2

(III)

𝑦

(V)

1

6

1

𝑦

(IV)

2

𝑥 1

Figure 12.53 17. Figure 12.54 shows the contour diagram of 𝑧 = 𝑓 (𝑥, 𝑦). Which of the points (I)–(VI) lie on the graph of 𝑧 = 𝑓 (𝑥, 𝑦)?

2

3

4

3

32 1

𝑥

𝑥 1

0 1 − 2 −

Figure 12.55

𝑦

0

1 23

𝑥

12.3 CONTOUR DIAGRAMS

677

PROBLEMS 19. Figure 12.56 shows a graph of 𝑓 (𝑥, 𝑦) = (sin 𝑥)(cos 𝑦) for −2𝜋 ≤ 𝑥 ≤ 2𝜋, −2𝜋 ≤ 𝑦 ≤ 2𝜋. Use the surface 𝑧 = 1∕2 to sketch the contour 𝑓 (𝑥, 𝑦) = 1∕2.

22. Figure 12.59 shows a contour plot of job satisfaction as a function of the hourly wage and the safety of the workplace (higher values mean safer). Match the jobs at points 𝑃 , 𝑄, and 𝑅 with the three descriptions. (a) The job is so unsafe that higher pay alone would not increase my satisfaction very much. (b) I could trade a little less safety for a little more pay. It would not matter to me. (c) The job pays so little that improving safety would not make me happier. safety level

20

𝑃

15

Figure 12.56 10

20. Total sales, 𝑄, of a product are a function of its price and the amount spent on advertising. Figure 12.57 shows a contour diagram for total sales. Which axis corresponds to the price of the product and which to the amount spent on advertising? Explain.

𝑅 hourly wage

5

𝑦 6 5 4 3 2 1

𝑄 = 2000

✛ 𝑄 = 4000 𝑄 = 5000 𝑥

grapes

(b)

18

8

21. Each contour diagram (a)–(c) in Figure 12.58 shows satisfaction with quantities of two items 𝑋 and 𝑌 combined. Match (a)–(c) with the items in (I)–(III).

16 14

6

12 10

4

8

𝑌

6

2 10

4 2

50

20

cherries

40

30

1

30 40

20 50

10

𝑋 (c)

20

10

Figure 12.57

𝑌

15

(a) What is the slope of the contours? (b) What does the slope tell you?

1 2 3 4 5 6

(a)

10

Figure 12.59 23. Figure 12.60 shows a contour diagram of Dan’s happiness with snacks of different numbers of cherries and grapes.

✛ 𝑄 = 3000



𝑄

5

𝑋

𝑌

2

3

4

5

Figure 12.60 24. Figure 12.61 shows contours of 𝑓 (𝑥, 𝑦) = 100𝑒𝑥 −50𝑦2 . Find the values of 𝑓 on the contours. They are equally spaced multiples of 10. 𝑦

10

1

20

0.8

30 40

0.6

50

𝑋

Figure 12.58 (I) 𝑋: Income; 𝑌 : Leisure time (II) 𝑋: Income; 𝑌 : Hours worked (III) 𝑋: Hours worked; 𝑌 : Time spent commuting

0.4 0.2 𝑥 0.2 0.4 0.6 0.8

Figure 12.61

1

678

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

25. Figure 12.62 shows contours for a person’s body mass index, BMI = 𝑓 (𝑤, ℎ) = 703𝑤∕ℎ2 , where 𝑤 is weight in pounds and ℎ is height in inches. Find the BMI contour values bounding the underweight and normal regions. ℎ (in) 78

Underweight

75

Normal Overweight

72 69

Obese

66

27. The wind chill tells you how cold it feels as a function of the air temperature and wind speed. Figure 12.63 is a contour diagram of wind chill (◦ F). (a) If the wind speed is 15 mph, what temperature feels like −20◦ F? (b) Estimate the wind chill if the temperature is 0◦ F and the wind speed is 10 mph. (c) Humans are at extreme risk when the wind chill is below −50◦ F. If the temperature is −20◦ F, estimate the wind speed at which extreme risk begins. (d) If the wind speed is 15 mph and the temperature drops by 20◦ F, approximately how much colder do you feel? wind speed, mph

30

63

−80 −60 −40

20

𝑤 (lbs) 120 140 160 180 200 220

10

−20

Figure 12.62

(a) (b) (c) (d) (e) (f)

2

𝑓 (𝑥, 𝑦) = 𝑥 − 𝑦 − 2𝑥 + 4𝑦 − 3 𝑔(𝑥, 𝑦) = 𝑥2 + 𝑦2 − 2𝑥 − 4𝑦 + 15 ℎ(𝑥, 𝑦) = −𝑥2 − 𝑦2 + 2𝑥 + 4𝑦 − 8 2 2 𝑗(𝑥, 𝑦) = −𝑥 √ + 𝑦 + 2𝑥 − 4𝑦 + 3 𝑘(𝑥, 𝑦) = (𝑥 − 1)2 + (𝑦 − 2)2 √ 𝑙(𝑥, 𝑦) = − (𝑥 − 1)2 + (𝑦 − 2)2

(I)

𝑦

(II)

20

air temp, ◦ F

−60 −40 −20

26. Match the functions (a)–(f) with the level curves (I)– (VI): 2

0

0

20

40

Figure 12.63 28. Figure 12.64 shows contour diagrams of 𝑓 (𝑥, 𝑦) and 𝑔(𝑥, 𝑦). Sketch the smooth curve with equation 𝑓 (𝑥, 𝑦) = 𝑔(𝑥, 𝑦). 𝑦 10 10

12

14

16

18

16 14

𝑦

8

12 10

6

8 4

6

2

2 0

4 0.5 1 1.5 2

−4 −5 −6 −7

(III)

𝑥

𝑦

(IV)

Figure 12.64: Black: 𝑓 (𝑥, 𝑦). Blue: 𝑔(𝑥, 𝑦)

𝑦

−2 −1

2 1

0

0

−1 −2

1 2

𝑥 (V)

𝑦

𝑥

29. Figure 12.65 shows the level curves of the temperature 𝐻 in a room near a recently opened window. Label the three level curves with reasonable values of 𝐻 if the house is in the following locations.

𝑥 (VI)

10

𝑥

𝑦

(a) (b) (c) (d)

Minnesota in winter (where winters are harsh). San Francisco in winter (where winters are mild). Houston in summer (where summers are hot). Oregon in summer (where summers are mild).

Room

−0.5 −1 −1.5 −2

𝑥

11 12 13 14

Window

𝑥

Figure 12.65

12.3 CONTOUR DIAGRAMS

679

𝑦

30. You are in a room 30 feet long with a heater at one end. In the morning the room is 65◦ F. You turn on the heater, which quickly warms up to 85◦ F. Let 𝐻(𝑥, 𝑡) be the temperature 𝑥 feet from the heater, 𝑡 minutes after the heater is turned on. Figure 12.66 shows the contour diagram for 𝐻. How warm is it 10 feet from the heater 5 minutes after it was turned on? 10 minutes after it was turned on?

2

−2 −1

1

21

0

0

− −2 1

1

2

𝑥 −2 0 −1

−1

𝑡 (minutes) 60

1 2

−2 −2

85

50

−1

40

0

1

2

Figure 12.68

30

80

20

75

10

70 65

𝑥 (feet) 5

10

15

20

25

30

Figure 12.66 31. Using the contour diagram in Figure 12.66, sketch the graphs of the one-variable functions 𝐻(𝑥, 5) and 𝐻(𝑥, 20). Interpret the two graphs in practical terms, and explain the difference between them. 32. Figure 12.67 shows a contour map of a hill with two paths, 𝐴 and 𝐵. (a) On which path, 𝐴 or 𝐵, will you have to climb more steeply? (b) On which path, 𝐴 or 𝐵, will you probably have a better view of the surrounding countryside? (Assume trees do not block your view.) (c) Alongside which path is there more likely to be a stream? Goal

❃✻

𝑧 = 300 𝑧 = 200

33. (0, 0) and (2, 0)

34. (0, −1) and (1, 0)

35. (−1, 1) and (1, 2)

36. (0, −2) and (0, 2)

37. Figure 12.69 is a contour diagram of the monthly payment on a 5-year car loan as a function of the interest rate and the amount you borrow. The interest rate is 13% and you borrow $6000 for a used car. (a) What is your monthly payment? (b) If interest rates drop to 11%, how much more can you borrow without increasing your monthly payment? (c) Make a table of how much you can borrow without increasing your monthly payment, as a function of the interest rate. loan amount ($)

8,000 7,000 140

6,000

120

5,000

100

4,000

80

60

3,000

𝐴 𝑧 = 100

2,000 1

𝐵

Figure 12.67 In Problems 33–36, for the two given points: (a) Find the distance ℎ from the first to the second point. (b) Use Figure 12.68, the contour diagram of 𝑓 (𝑥, 𝑦), to find Δ𝑓 , the difference between the values of 𝑓 from the first to the second point. (c) Find Δ𝑓 ∕ℎ, the average rate of change of 𝑓 (𝑥, 𝑦) from the first to the second point.

3

5

7

9 11 13 15

interest rate (%)

Figure 12.69 38. Hiking on a level trail going due east, you decide to leave the trail and climb toward the mountain on your left. The farther you go along the trail before turning off, the gentler the climb. Sketch a possible topographical map showing the elevation contours. 39. The total productivity 𝑓 (𝑛, 𝑇 ) of an advertising agency (in ads per day) depends on the number 𝑛 of workers and the temperature 𝑇 of the office in degrees Fahrenheit. More workers create more ads, but the farther the temperature from 75◦ F, the slower they work. Draw a possible contour diagram for the function 𝑓 (𝑛, 𝑇 ).

680

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

(b) After buying 4 units of item 𝐴, how many units of 𝐵 must the shopper buy to obtain the same satisfaction as obtained from buying 8 units of 𝐴 and 2 units of 𝐵? (c) The shopper reduces the purchase of item 𝐴 by 𝑘, a fixed number of units, while increasing the purchase of 𝐵 to maintain satisfaction. In which of the following cases is the increase in 𝐵 largest?

40. Match the functions (a)–(d) with the shapes of their level curves (I)–(IV). Sketch each contour diagram. (a) 𝑓 (𝑥, 𝑦) = 𝑥2 (c) 𝑓 (𝑥, 𝑦) = 𝑦 − 𝑥2

(b) 𝑓 (𝑥, 𝑦) = 𝑥2 + 2𝑦2 (d) 𝑓 (𝑥, 𝑦) = 𝑥2 − 𝑦2

I. Lines III. Hyperbolas

II. Parabolas IV. Ellipses

41. Match the functions (a)–(d) with the shapes of their typical level curves (I)–(IV). 𝑦 1 (b) 𝑓 (𝑥, 𝑦) = 2 (a) 𝑓 (𝑥, 𝑦) = 2 𝑥 +1 𝑥 + 2𝑦2 𝑥 𝑥2 + 1 (d) 𝑓 (𝑥, 𝑦) = 2 (c) 𝑓 (𝑥, 𝑦) = 2 𝑦 +1 𝑥 + 𝑦2 + 1 I. Circles III. Hyperbolas

• Initial purchase of 𝐴 is 6 units • Initial purchase of 𝐴 is 8 units 45. Match each Cobb-Douglas production function (a)–(c) with a graph in Figure 12.71 and a statement (D)–(G).

II. Parabolas IV. Ellipses

(a) (b) (c) (D) (E) (G)

42. Figure 12.70 shows the density of the fox population 𝑃 (in foxes per square kilometer) for southern England.8 Draw two different cross-sections along a north-south line and two different cross-sections along an east-west line of the population density 𝑃 . kilometers north

𝐾

(I)

150

𝐹 (𝐿, 𝐾) = 𝐿 0.25 𝐾 0.25 𝐹 (𝐿, 𝐾) = 𝐿 0.5 𝐾 0.5 𝐹 (𝐿, 𝐾) = 𝐿 0.75 𝐾 0.75 Tripling each input triples output. Quadrupling each input doubles output. Doubling each input almost triples output.

North

2

1.5 2 1.5

120

180

kilometers east

Figure 12.70 43. A manufacturer sells two goods, one at a price of $3000 a unit and the other at a price of $12,000 a unit. A quantity 𝑞1 of the first good and 𝑞2 of the second good are sold at a total cost of $4000 to the manufacturer. (a) Express the manufacturer’s profit, 𝜋, as a function of 𝑞1 and 𝑞2 . (b) Sketch curves of constant profit in the 𝑞1 𝑞2 -plane for 𝜋 = 10,000, 𝜋 = 20,000, and 𝜋 = 30,000 and the break-even curve 𝜋 = 0. 44. A shopper buys 𝑥 units of item 𝐴 and 𝑦 units of item 𝐵, obtaining satisfaction 𝑠(𝑥, 𝑦) from the purchase. (Satisfaction is called utility by economists.) The contours 𝑠(𝑥, 𝑦) = 𝑥𝑦 = 𝑐 are called indifference curves because they show pairs of purchases that give the shopper the same satisfaction. (a) A shopper buys 8 units of 𝐴 and 2 units of 𝐵. What is the equation of the indifference curve showing the other purchases that give the shopper the same satisfaction? Sketch this curve. 8 From

𝐹 =1 𝐿

3

1

2

3

𝐾 𝐹 =4 𝐹 =3 𝐹 =2 𝐹 =1 𝐿

2

1

0.5

60

2

3

1 0.5

2

𝐹 = 1.5

1

𝐹 =1 𝐿 1

(III)

50 1.5

2 𝐹 =2

1

1

100

3

𝐹 =3

2

1.5

0.5

𝐾

(II)

3

1 1

2

3

Figure 12.71

46. A Cobb-Douglas production function has the form 𝑃 = 𝑐𝐿𝛼 𝐾 𝛽

with 𝛼, 𝛽 > 0.

What happens to production if labor and capital are both scaled up? For example, does production double if both labor and capital are doubled? Economists talk about • increasing returns to scale if doubling 𝐿 and 𝐾 more than doubles 𝑃 , • constant returns to scale if doubling 𝐿 and 𝐾 exactly doubles 𝑃 , • decreasing returns to scale if doubling 𝐿 and 𝐾 less than doubles 𝑃 . What conditions on 𝛼 and 𝛽 lead to increasing, constant, or decreasing returns to scale?

“On the spatial spread of rabies among foxes”, Murray, J. D. et al, Proc. R. Soc. Lond. B, 229: 111–150, 1986.

12.3 CONTOUR DIAGRAMS 𝑦

47. (a) Match 𝑓 (𝑥, 𝑦) = 𝑥0.2 𝑦0.8 and 𝑔(𝑥, 𝑦) = 𝑥0.8 𝑦0.2 with the level curves in Figures (I) and (II). All scales on the axes are the same. (b) Figure (III) shows the level curves of ℎ(𝑥, 𝑦) = 𝑥𝛼 𝑦1−𝛼 for 0 < 𝛼 < 1. Find the range of possible values for 𝛼. Again, the scales are the same on both axes.

(I)

𝑦

681

2 1 0 −1 −2

𝑥

𝑦

(II)

Figure 12.72 50. Figure 12.73 shows part of the contour diagram of 𝑓 (𝑥, 𝑦). Complete the diagram for 𝑥 < 0 if 𝑥 (III)

𝑥

𝑦

(a)

𝑓 (−𝑥, 𝑦) = 𝑓 (𝑥, 𝑦)

(b) 𝑓 (−𝑥, 𝑦) = −𝑓 (𝑥, 𝑦)

𝑦 0 1 2 3 4 5

𝑥

𝑥

48. Match the functions (a)–(d) with the contour diagrams in Figures I–IV. (a) (b) (c) (d)

𝑓 (𝑥, 𝑦) = 0.7 ln 𝑥 + 0.3 ln 𝑦 𝑔(𝑥, 𝑦) = 0.3 ln 𝑥 + 0.7 ln 𝑦 ℎ(𝑥, 𝑦) = 0.3𝑥2 + 0.7𝑦2 𝑗(𝑥, 𝑦) = 0.7𝑥2 + 0.3𝑦2 𝑦

(I)

51. The contour at level 0 of 𝑓 (𝑥, 𝑦) = (𝑥+2𝑦)2 −(3𝑥−4𝑦)2 consists of two intersecting lines in the 𝑥𝑦-plane. Find equations for the lines. 52. Let 𝑧 = 𝑓 (𝑥, 𝑦) = 𝑥2 ∕(𝑥2 + 𝑦2 ). (a) Why are there no contours for 𝑧 < 0? (b) Why are there no contours for 𝑧 > 1? (c) Sketch a contour diagram for 𝑓 (𝑥, 𝑦) with at least four labeled contours.

𝑦

(II)

4

4

𝑥

53. Let 𝑓 (𝑥, 𝑦) = 𝑥2 − 𝑦2 = (𝑥 − 𝑦)(𝑥 + 𝑦). Use the factored form to sketch the contour 𝑓 (𝑥, 𝑦) = 0 and to find the regions in the 𝑥𝑦-plane where 𝑓 (𝑥, 𝑦) > 0 and the regions where 𝑓 (𝑥, 𝑦) < 0. Explain how this sketch shows that the graph of 𝑓 (𝑥, 𝑦) is saddle-shaped at the origin.

𝑥

4

4

𝑦

(III)

𝑦

(IV)

4

Figure 12.73

4

54. Use Problem 53 to find a formula for a “monkey saddle” surface 𝑧 = 𝑔(𝑥, 𝑦) which has three regions with 𝑔(𝑥, 𝑦) > 0 and three with 𝑔(𝑥, 𝑦) < 0. 𝑥

𝑥

4

4

49. Figure 12.72 is the contour diagram of 𝑓 (𝑥, 𝑦). Sketch the contour diagram of each of the following functions. (a) 3𝑓 (𝑥, 𝑦)

(b) 𝑓 (𝑥, 𝑦) − 10

(c) 𝑓 (𝑥 − 2, 𝑦 − 2) 9 From

(d) 𝑓 (−𝑥, 𝑦)

55. The power 𝑃 produced by a windmill is proportional to the square of the diameter 𝑑 of the windmill and to the cube of the speed 𝑣 of the wind.9 (a) Write a formula for 𝑃 as a function of 𝑑 and 𝑣. (b) A windmill generates 100 kW of power at a certain wind speed. If a second windmill is built having twice the diameter of the original, what fraction of the original wind speed is needed by the second windmill to produce 100 kW? (c) Sketch a contour diagram for 𝑃 .

www.ecolo.org/documents/documents_in_english/WindmillFormula.htm, accessed October 9, 2011.

682

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

Strengthen Your Understanding 2 +𝑦2 )

In Problems 56–57, explain what is wrong with the statement.

64. If 𝑧 = 𝑒−(𝑥 point (𝑥, 𝑦).

56. A contour diagram for 𝑧 = 𝑓 (𝑥, 𝑦) is a surface in 𝑥𝑦𝑧space. √ 57. The functions 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 and 𝑔(𝑥, 𝑦) = 𝑥2 +𝑦2 have the same contour diagram.

Are the statements in Problems 65–72 true or false? Give reasons for your answer.

In Problems 58–59, give an example of:

66. A weather map can have two isotherms representing the same temperature that do not intersect.

58. A function 𝑓 (𝑥, 𝑦) whose 𝑧 = 10 contour consists of two or more parallel lines. 59. A function whose contours are all parabolas. Decide if the statements in Problems 60–64 must be true, might be true, or could not be true. The function 𝑧 = 𝑓 (𝑥, 𝑦) is defined everywhere. 60. The level curves corresponding to 𝑧 = 1 and 𝑧 = −1 cross at the origin. 2

2

61. The level curve 𝑧 = 1 consists of the circle 𝑥 + 𝑦 = 2 and the circle 𝑥2 + 𝑦2 = 3, but no other points. 62. The level curve 𝑧 = 1 consists of two lines which intersect at the origin. 2 +𝑦2 )

63. If 𝑧 = 𝑒−(𝑥 𝑧.

12.4

, there is a level curve for every value of

, there is a level curve through every

65. Two isotherms representing distinct temperatures on a weather map cannot intersect.

67. The contours of the function 𝑓 (𝑥, 𝑦) = 𝑦2 + (𝑥 − 2)2 are either circles or a single point. 68. If the contours of 𝑔(𝑥, 𝑦) are concentric circles, then the graph of 𝑔 is a cone. 69. If the contours for 𝑓 (𝑥, 𝑦) get closer together in a certain direction, then 𝑓 is increasing in that direction. 70. If all of the contours of 𝑓 (𝑥, 𝑦) are parallel lines, then the graph of 𝑓 is a plane. 71. If the 𝑓 = 10 contour of the function 𝑓 (𝑥, 𝑦) is identical to the 𝑔 = 10 contour of the function 𝑔(𝑥, 𝑦), then 𝑓 (𝑥, 𝑦) = 𝑔(𝑥, 𝑦) for all (𝑥, 𝑦). 72. The 𝑓 = 5 contour of the function 𝑓 (𝑥, 𝑦) is identical to the 𝑔 = 0 contour of the function 𝑔(𝑥, 𝑦) = 𝑓 (𝑥, 𝑦) − 5.

LINEAR FUNCTIONS

What Is a Linear Function of Two Variables? Linear functions played a central role in one-variable calculus because many one-variable functions have graphs that look like a line when we zoom in. In two-variable calculus, a linear function is one whose graph is a plane. In Chapter 14, we see that many two-variable functions have graphs which look like planes when we zoom in.

What Makes a Plane Flat? What makes the graph of the function 𝑧 = 𝑓 (𝑥, 𝑦) a plane? Linear functions of one variable have straight line graphs because they have constant slope. On a plane, the situation is a bit more complicated. If we walk around on a tilted plane, the slope is not always the same: it depends on the direction in which we walk. However, at every point on the plane, the slope is the same as long as we choose the same direction. If we walk parallel to the 𝑥-axis, we always find ourselves walking up or down with the same slope;10 the same is true if we walk parallel to the 𝑦-axis. In other words, the slope ratios Δ𝑧∕Δ𝑥 (with 𝑦 fixed) and Δ𝑧∕Δ𝑦 (with 𝑥 fixed) are each constant. Example 1

A plane cuts the 𝑧-axis at 𝑧 = 5 and has slope 2 in the 𝑥-direction and slope −1 in the 𝑦-direction. What is the equation of the plane?

Solution

Finding the equation of the plane means constructing a formula for the 𝑧-coordinate of the point on the plane directly above the point (𝑥, 𝑦) in the 𝑥𝑦-plane. To get to that point start from the point above the origin, where 𝑧 = 5. Then walk 𝑥 units in the 𝑥-direction. Since the slope in the 𝑥-direction is 2, the height increases by 2𝑥. Then walk 𝑦 units in the 𝑦-direction; since the slope in the 𝑦-direction is −1, the height decreases by 𝑦 units. Since the height has changed by 2𝑥 − 𝑦 units, the 𝑧-coordinate 10 To

be precise, walking in a vertical plane parallel to the 𝑥-axis while rising or falling with the plane you are on.

12.4 LINEAR FUNCTIONS

683

is 5 + 2𝑥 − 𝑦. Thus, the equation for the plane is 𝑧 = 5 + 2𝑥 − 𝑦.

For any linear function, if we know its value at a point (𝑥0 , 𝑦0 ), its slope in the 𝑥-direction, and its slope in the 𝑦-direction, then we can write the equation of the function. This is just like the equation of a line in the one-variable case, except that there are two slopes instead of one. If a plane has slope 𝑚 in the 𝑥-direction, has slope 𝑛 in the 𝑦-direction, and passes through the point (𝑥0 , 𝑦0 , 𝑧0 ), then its equation is 𝑧 = 𝑧0 + 𝑚(𝑥 − 𝑥0 ) + 𝑛(𝑦 − 𝑦0 ). This plane is the graph of the linear function 𝑓 (𝑥, 𝑦) = 𝑧0 + 𝑚(𝑥 − 𝑥0 ) + 𝑛(𝑦 − 𝑦0 ). If we write 𝑐 = 𝑧0 − 𝑚𝑥0 − 𝑛𝑦0 , then we can write 𝑓 (𝑥, 𝑦) in the equivalent form 𝑓 (𝑥, 𝑦) = 𝑐 + 𝑚𝑥 + 𝑛𝑦.

Just as in 2-space a line is determined by two points, so in 3-space a plane is determined by three points, provided they do not lie on a line. Example 2

Find the equation of the plane passing through the points (1, 0, 1), (1, −1, 3), and (3, 0, −1).

Solution

The first two points have the same 𝑥-coordinate, so we use them to find the slope of the plane in the 𝑦-direction. As the 𝑦-coordinate changes from 0 to −1, the 𝑧-coordinate changes from 1 to 3, so the slope in the 𝑦-direction is 𝑛 = Δ𝑧∕Δ𝑦 = (3 − 1)∕(−1 − 0) = −2. The first and third points have the same 𝑦-coordinate, so we use them to find the slope in the 𝑥-direction; it is 𝑚 = Δ𝑧∕Δ𝑥 = (−1 − 1)∕(3 − 1) = −1. Because the plane passes through (1, 0, 1), its equation is 𝑧 = 1 − (𝑥 − 1) − 2(𝑦 − 0)

or 𝑧 = 2 − 𝑥 − 2𝑦.

You should check that this equation is also satisfied by the points (1, −1, 3) and (3, 0, −1). Example 2 was made easier by the fact that two of the points had the same 𝑥-coordinate and two had the same 𝑦-coordinate. An alternative method, which works for any three points, is to substitute the 𝑥, 𝑦, and 𝑧-values of each of the three points into the equation 𝑧 = 𝑐 + 𝑚𝑥 + 𝑛𝑦. The resulting three equations in 𝑐, 𝑚, 𝑛 are then solved simultaneously.

Linear Functions from a Numerical Point of View To avoid flying planes with empty seats, airlines sell some tickets at full price and some at a discount. Table 12.10 shows an airline’s revenue in dollars from tickets sold on a particular route, as a function of the number of full-price tickets sold, 𝑓 , and the number of discount tickets sold, 𝑑. In every column, the revenue jumps by $40,000 for each extra 200 discount tickets. Thus, each column is a linear function of the number of discount tickets sold. In addition, every column has the same slope, 40,000∕200 = 200 dollars/ticket. This is the price of a discount ticket. Similarly, each row is a linear function and all the rows have the same slope, 450, which is the price in dollars of a

684

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

Table 12.10

Revenue from ticket sales (dollars) Full-price tickets (𝑓 )

Discount tickets (𝑑)

100

200

300

400

200

85,000

130,000

175,000

220,000

400

125,000

170,000

215,000

260,000

600

165,000

210,000

255,000

300,000

800

205,000

250,000

295,000

340,000

1000

245,000

290,000

335,000

380,000

full-fare ticket. Thus, 𝑅 is a linear function of 𝑓 and 𝑑, given by: 𝑅 = 450𝑓 + 200𝑑. We have the following general result: A linear function can be recognized from its table by the following features: • Each row and each column is linear. • All the rows have the same slope. • All the columns have the same slope (although the slope of the rows and the slope of the columns are generally different).

Example 3

Solution

The table contains values of a linear function. Fill in the blank and give a formula for the function. 𝑥∖𝑦

1.5

2.0

2

0.5

1.5

3

−0.5

?

In the first column the function decreases by 1 (from 0.5 to −0.5) as 𝑥 goes from 2 to 3. Since the function is linear, it must decrease by the same amount in the second column. So the missing entry must be 1.5 − 1 = 0.5. The slope of the function in the 𝑥-direction is −1. The slope in the 𝑦-direction is 2, since in each row the function increases by 1 when 𝑦 increases by 0.5. From the table we get 𝑓 (2, 1.5) = 0.5. Therefore, the formula is 𝑓 (𝑥, 𝑦) = 0.5 − (𝑥 − 2) + 2(𝑦 − 1.5) = −0.5 − 𝑥 + 2𝑦.

What Does the Contour Diagram of a Linear Function Look Like? The formula for the airline revenue function in Table 12.10 is 𝑅 = 450𝑓 + 200𝑑, where 𝑓 is the number of full fares and 𝑑 is the number of discount fares sold. Notice that the contours of this function in Figure 12.74 are parallel straight lines. What is the practical significance of the slope of these contour lines? Consider the contour 𝑅 = 100,000; that means we are looking at combinations of ticket sales that yield $100,000 in revenue. If we move down and to the right on the contour, the 𝑓 -coordinate increases and the 𝑑-coordinate decreases, so we sell more full fares and fewer discount fares. This is because to receive a fixed revenue of $100,000, we must sell more full fares if we sell fewer discount fares. The exact trade-off depends on the slope of the contour; the diagram shows that each contour has a slope of about −2. This means that for a fixed revenue, we must sell two discount fares to replace one full fare. This can also be seen by comparing prices. Each full fare brings in $450; to earn the same amount in discount fares we need to sell 450∕200 = 2.25 ≈ 2 fares. Since the price ratio is independent of how many of each type of fare we sell, this slope remains constant over the whole contour map; thus, the contours are all parallel straight lines.

685

12.4 LINEAR FUNCTIONS

Notice also that the contours are evenly spaced. Thus, no matter which contour we are on, a fixed increase in one of the variables causes the same increase in the value of the function. In terms of revenue, no matter how many fares we have sold, an extra fare, whether full or discount, brings the same revenue as before. 𝑦

𝑑 3

500

10

2

400

6

1

0 ,00 250

000 50,

0 ,00 100

0 ,00 150

200

0 ,00 200

300

100

8 4 2 0

𝑥

−2

−1

−4 −6

−2

𝑓 100 200 300 400 500 Figure 12.74: Revenue as a function of full and discount fares, 𝑅 = 450𝑓 + 200𝑑

−8 −10

−3 −3 −2 −1 1 2 3 Figure 12.75: Contour map of linear function 𝑓 (𝑥, 𝑦)

Example 4

Find the equation of the linear function whose contour diagram is in Figure 12.75.

Solution

Suppose we start at the origin on the 𝑧 = 0 contour. Moving 2 units in the 𝑦-direction takes us to the 𝑧 = 6 contour, so the slope in the 𝑦-direction is Δ𝑧∕Δ𝑦 = 6∕2 = 3. Similarly, a move of 2 units in the 𝑥-direction from the origin takes us to the 𝑧 = 2 contour, so the slope in the 𝑥-direction is Δ𝑧∕Δ𝑥 = 2∕2 = 1. Since 𝑓 (0, 0) = 0, we have 𝑓 (𝑥, 𝑦) = 𝑥 + 3𝑦.

Exercises and Problems for Section 12.4 Online Resource: Additional Problems for Section 12.4 EXERCISES

Exercises 1–2 each contain a partial table of values for a linear function. Fill in the blanks. 1.

𝑥∖𝑦

0.0

0.0 2.0

2.

1.0

3.0

𝑥∖𝑦

−1.0

1.0

2.0

4.0

5.0

3.0

0.0

1.0

8. Find the linear function whose graph is the plane through the points (4, 0, 0), (0, 3, 0) and (0, 0, 2).

3.0

5.0

9. Find an equation for the plane containing the line in the 𝑥𝑦-plane where 𝑦 = 1, and the line in the 𝑥𝑧-plane where 𝑧 = 2.

In Exercises 3–6, could the tables of values represent a linear function? 3.

4.

𝑦

𝑥

0

1

2

0

0

1

4

1

1

0

1

2

4

1

0

5.

0

1

2

0

10

13

16

1

6

9

12

2

2

5

8

6.

𝑦

𝑥

𝑦

𝑥

0

1

2

0

0

5

10

1

2

7

12

2

4

9

14

𝑦

𝑥

7. Find the equation of the linear function 𝑧 = 𝑐 + 𝑚𝑥 + 𝑛𝑦 whose graph contains the points (0, 0, 0), (0, 2, −1), and (−3, 0, −4).

0

1

2

0

5

7

9

1

6

9

12

2

7

11

15

10. Find the equation of the linear function 𝑧 = 𝑐 + 𝑚𝑥 + 𝑛𝑦 whose graph intersects the 𝑥𝑧-plane in the line 𝑧 = 3𝑥+4 and intersects the 𝑦𝑧-plane in the line 𝑧 = 𝑦+4. 11. Suppose that 𝑧 is a linear function of 𝑥 and 𝑦 with slope 2 in the 𝑥-direction and slope 3 in the 𝑦-direction. (a) A change of 0.5 in 𝑥 and −0.2 in 𝑦 produces what change in 𝑧? (b) If 𝑧 = 2 when 𝑥 = 5 and 𝑦 = 7, what is the value of 𝑧 when 𝑥 = 4.9 and 𝑦 = 7.2? 12. (a) Find a formula for the linear function whose graph is a plane passing through point (4, 3, −2) with slope 5 in the 𝑥-direction and slope −3 in the 𝑦direction. (b) Sketch the contour diagram for this function.

686

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

In Exercises 13–14, could the contour diagram represent a linear function? 13.

14.

𝑦 16

14

𝑦

12

10

8

28

6 4

24

2 20 16

0

12

𝑥

8 4

2

𝑥 0 −4 −8

PROBLEMS 15. An internet video streaming company offers a basic and premium monthly streaming subscription package. Figure 12.76 shows the revenue (in dollars per month) of the company as a function of the number, 𝑐, of basic subscribers and the number, 𝑑, of premium subscribers it has. What is the price of a basic subscription? What is the price of a premium subscription?

(b) Revenue as a function of 𝑞1 and 𝑞2 . (c) Revenue as a function of 𝑝1 and 𝑞1 . Problems 18–20 concern Table 12.11, which gives the number of calories burned per minute for someone rollerblading, as a function of the person’s weight and speed.11

𝑑 400 5000

300

Table 12.11

4000

Calories burned per minute

3000

200

2000

100

1000

𝑐 50

100 150 200

Figure 12.76

Weight

8 mph

9 mph

10 mph

11 mph

120 lbs

4.2

5.8

7.4

8.9

140 lbs

5.1

6.7

8.3

9.9

160 lbs

6.1

7.7

9.2

10.8

180 lbs

7.0

8.6

10.2

11.7

200 lbs

7.9

9.5

11.1

12.6

16. The charge, 𝐶, in dollars, for access to a company’s 4G LTE network is a function of 𝑚, the number of months of use, and 𝑡, the total number of gigabytes used: 𝐶 = 𝑓 (𝑚, 𝑡) = 99 + 30𝑚 + 10𝑡. (a) Is 𝑓 a linear function? (b) Give units for the coefficients of 𝑚 and 𝑡, and interpret them as charges. (c) Interpret the intercept 99 as a charge. (d) Find 𝑓 (3, 8) and interpret your answer.

18. Does the data in Table 12.11 look approximately linear? Give a formula for 𝐵, the number of calories burned per minute in terms of the weight, 𝑤, and the speed, 𝑠. Does the formula make sense for all weights or speeds?

17. A manufacturer makes two products out of two raw materials. Let 𝑞1 , 𝑞2 be the quantities sold of the two products, 𝑝1 , 𝑝2 their prices, and 𝑚1 , 𝑚2 the quantities purchased of the two raw materials. Which of the following functions do you expect to be linear, and why? In each case, assume that all variables except the ones mentioned are held fixed.

19. Who burns more total calories to go 10 miles: A 120lb person going 10 mph or a 180-lb person going 8 mph? Which of these two people burns more calories per pound for the 10-mile trip?

(a) Expenditure on raw materials as a function of 𝑚1 and 𝑚2 . 11 From

20. Use Problem 18 to give a formula for 𝑃 , the number of calories burned per pound, in terms of 𝑤 and 𝑠, for a person weighing 𝑤 lbs roller-blading 10 miles at 𝑠 mph.

the August 28, 1994, issue of Parade Magazine.

687

12.4 LINEAR FUNCTIONS

For Problems 21–22, find a possible equation for a linear function with the given contour diagram. 21.

In Problems 27–34, could the table of values represent a linear function? If so, find a possible formula for the function. If not, give a reason why not.

𝑦

27.

3 10 8

2

1

6

𝑥

4

1 2

𝑥

0

−2 −4 −6

−1

1

2

3

1

5

9

2

2

6

10

3

3

7

11

29.

−8 −10

−2

1

2

3 𝑥

22.

𝑦

-2

0

2

2

2

2

0

5

5

5

2

8

8

8

0 2

1

𝑥

2

0

1

2

1

1

2

3

2

2

3

4

3

2

4

6

1

0

3

6

3

1

4

7

5

4

7

10

1

2

3

1

1

2

3

2

4

5

6

4

7

8

9

0

2

5

0

0

4

10

1

1

5

11

2

4

6

14

32.

0

1

2

0

-5

-7

-9

2

-2

-4

-6

4

1

-1

-3

𝑦

𝑥

6

4

𝑥

1

𝑦

𝑥

𝑦

−2

2

0

30.

-2

31.

−4

3

𝑦

𝑥

𝑦

−3 −3 −2 −1

28.

𝑦

8

−1

33.

−3 −3 −2 −1

1

2

In Problems 23–24, could the contour diagram represent a linear function? If so, find an equation for that function. 𝑦

𝑦

24.

3

2

−3

1 −4

1 1

−3 −2 −1 −1

5

1

0

3

5

0

3

5

8

1

5

7

10

3

9

11

14

𝑥 −3 −2 −1 −1

−2

−2

−3

−3

𝑦

𝑥

𝑦

1

𝑥

2

2

In Problems 35–38, use the contours of the linear function 𝑧 = 𝑓 (𝑥, 𝑦) in Figure 12.77 to create possible contour labels for the linear function 𝑧 = 𝑔(𝑥, 𝑦) satisfying the given condition.

3

2

0

3 𝑥

23.

34.

𝑦

12

10

−2

1

2

3

𝑦

4

50

3 4

𝑓 (𝑥, 𝑦)

40 30

2 1

4

?

3 ?

2

20

1

10

𝑔(𝑥, 𝑦)

? ? ?

𝑥 0.5

For Problems 25–26, find an equation for the linear function with the given values. 25. 𝑥∖𝑦

−1

0

1

2

0

1.5

1

0.5

1

3.5

3

2

5.5

5

3

7.5

7

1

1.5 2

𝑥 0.5

1

1.5 2

Figure 12.77 35. The graph of 𝑔 is parallel but different from the graph of 𝑓 .

26. 𝑥∖𝑦

10

20

30

40

0

100

3

6

9

12

2.5

2

200

2

5

8

11

4.5

4

300

1

4

7

10

6.5

6

400

0

3

6

9

36. The graph of 𝑔 is parallel to the graph of 𝑓 and passes through the point (2, 2, 0). 37. The graph of 𝑔 has the same contour as 𝑓 for the value 𝑧 = 30 but is different from the graph of 𝑓 . 38. The graph of 𝑔 has the same contour as 𝑓 for the value 𝑧 = 40, and a negative slope in the 𝑥-direction.

688

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

In Problems 39–42, graph the linear function by plotting the 𝑥, 𝑦, and 𝑧-intercepts and joining them by a triangle as in Figure 12.78. This shows the part of the plane in the octant where 𝑥 ≥ 0, 𝑦 ≥ 0, 𝑧 ≥ 0. If the intercepts are not all positive, the same method works if the 𝑥, 𝑦, and 𝑧-axes are drawn from a different perspective.

44. For the contour diagrams (I)–(IV) on −2 ≤ 𝑥 ≤ 2, −2 ≤ 𝑦 ≤ 2, pick the corresponding function.

𝑓 (𝑥, 𝑦) = 2𝑥 + 3𝑦 + 10

𝑘(𝑥, 𝑦) = −2𝑥 + 3𝑦 + 12

𝑔(𝑥, 𝑦) = 2𝑥 + 3𝑦 + 60

𝑚(𝑥, 𝑦) = −2𝑥 + 3𝑦 + 60

ℎ(𝑥, 𝑦) = 2𝑥 − 3𝑦 + 12

𝑛(𝑥, 𝑦) = −2𝑥 − 3𝑦 + 14

𝑗(𝑥, 𝑦) = 2𝑥 − 3𝑦 + 60

𝑝(𝑥, 𝑦) = −2𝑥 − 3𝑦 + 60

𝑦

(I)

10 2 1 4 1

Figure 12.78 39. 𝑧 = 2 − 2𝑥 + 𝑦

40. 𝑧 = 2 − 𝑥 − 2𝑦

41. 𝑧 = 4 + 𝑥 − 2𝑦

42. 𝑧 = 6 − 2𝑥 − 3𝑦

43. Figure 12.79 is the contour diagram of a linear function 𝑓 (𝑥, 𝑦) = 𝑚𝑥 + 4𝑦 + 𝑐. What is the value of 𝑚?

8

𝑥

𝑦

(IV)

1 1 6 12 4

𝑦

1 12 0

𝑥

𝑦

(III)

𝑦

(II)

64 2 6 0 6

𝑥

𝑥

3

𝑥 −3

3

−3

Figure 12.79

45. A linear function has the formula 𝑓 (𝑥, 𝑦) = 𝑎+10𝑥−5𝑦, but you don’t know the value of 𝑎. Give a numerical value for the following, if possible. (a) 𝑓 (50, 62) (b) 𝑓 (51, 60) (c) 𝑓 (51, 60) − 𝑓 (50, 62)

Strengthen Your Understanding In Problems 46–47, explain what is wrong with the statement. 46. If the contours of 𝑓 are all parallel lines, then 𝑓 is linear. 47. A function 𝑓 (𝑥, 𝑦) with linear cross-sections for 𝑥 fixed and linear cross-sections for 𝑦 fixed is a linear function. In Problems 48–49, give an example of: 48. A table of values, with three rows and three columns, for a nonlinear function that is linear in each row and in each column.

50. The planes 𝑧 = 3+2𝑥+4𝑦 and 𝑧 = 5+2𝑥+4𝑦 intersect. 51. The function represented in Table 12.12 is linear. Table 12.12 𝑢∖𝑣

1.1

1.2

1.3

1.4

3.2

11.06

12.06

13.06

14.06

3.4

11.75

12.82

13.89

14.96

3.6

12.44

13.58

14.72

15.86

3.8

13.13

14.34

15.55

16.76

4.0

13.82

15.10

16.38

17.66

49. A linear function whose contours are lines with slope 2. 52. Contours of 𝑓 (𝑥, 𝑦) = 3𝑥 + 2𝑦 are lines with slope 3. Are the statements in Problems 50–62 true or false? Give reasons for your answer.

53. If 𝑓 is a non-constant linear function, then the contours of 𝑓 are parallel lines.

12.5 FUNCTIONS OF THREE VARIABLES

54. If 𝑓 (0, 0) = 1, 𝑓 (0, 1) = 4, 𝑓 (0, 3) = 5, then 𝑓 cannot be linear.

59. A linear function 𝑓 (𝑥, 𝑦) = 2𝑥 + 3𝑦 − 5, has exactly one point (𝑎, 𝑏) satisfying 𝑓 (𝑎, 𝑏) = 0.

55. The graph of a linear function is always a plane. 56. The cross-section 𝑥 = 𝑐 of a linear function 𝑓 (𝑥, 𝑦) is always a line.

60. In a table of values of a linear function, the columns have the same slope as the rows. 61. There is exactly one linear function 𝑓 (𝑥, 𝑦) whose 𝑓 = 0 contour is 𝑦 = 2𝑥 + 1.

57. There is no linear function 𝑓 (𝑥, 𝑦) with a graph parallel to the 𝑥𝑦-plane. 58. There is no linear function 𝑓 (𝑥, 𝑦) with a graph parallel to the 𝑥𝑧-plane.

12.5

689

62. If the contours of 𝑓 (𝑥, 𝑦) = 𝑐 + 𝑚𝑥 + 𝑛𝑦 are vertical lines, then 𝑛 = 0.

FUNCTIONS OF THREE VARIABLES In applications of calculus, functions of any number of variables can arise. The density of matter in the universe is a function of three variables, since it takes three numbers to specify a point in space. Models of the US economy often use functions of ten or more variables. We need to be able to apply calculus to functions of arbitrarily many variables. One difficulty with functions of more than two variables is that it is hard to visualize them. The graph of a function of one variable is a curve in 2-space, the graph of a function of two variables is a surface in 3-space, so the graph of a function of three variables would be a solid in 4-space. Since we can’t easily visualize 4-space, we won’t use the graphs of functions of three variables. On the other hand, it is possible to draw contour diagrams for functions of three variables, only now the contours are surfaces in 3-space.

Representing a Function of Three Variables Using a Family of Level Surfaces A function of two variables, 𝑓 (𝑥, 𝑦), can be represented by a family of level curves of the form 𝑓 (𝑥, 𝑦) = 𝑐 for various values of the constant, 𝑐.

A level surface, or level set of a function of three variables, 𝑓 (𝑥, 𝑦, 𝑧), is a surface of the form 𝑓 (𝑥, 𝑦, 𝑧) = 𝑐, where 𝑐 is a constant. The function 𝑓 can be represented by the family of level surfaces obtained by allowing 𝑐 to vary.

The value of the function, 𝑓 , is constant on each level surface. Example 1

The temperature, in ◦ C, at a point (𝑥, 𝑦, 𝑧) is given by 𝑇 = 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 + 𝑧2 . What do the level surfaces of the function 𝑓 look like and what do they mean in terms of temperature?

Solution

The level surface corresponding to 𝑇 = 100 is the set of all points where the temperature is 100◦ C. That is, where 𝑓 (𝑥, 𝑦, 𝑧) = 100, so 𝑥2 + 𝑦2 + 𝑧2 = 100. This is the equation of a sphere of radius 10, with center √ at the origin. Similarly, the level surface corresponding to 𝑇 = 200 is the sphere with radius 200. The other level surfaces are concentric spheres. The temperature is constant on each sphere. We may view the temperature distribution as a set of nested spheres, like concentric layers of an onion, each one labeled with a different temperature, starting from low temperatures in the middle and getting hotter as we go out from the center. (See Figure 12.80.) The level surfaces become more closely spaced as we move farther from the origin because the temperature increases more rapidly the farther we get from the origin.

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Figure 12.80: Level surfaces of 𝑇 = 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 + 𝑧2 , each one having a constant temperature

Example 2

What do the level surfaces of 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 and 𝑔(𝑥, 𝑦, 𝑧) = 𝑧 − 𝑦 look like?

Solution

The level surface of 𝑓 corresponding to the constant 𝑐 is the surface consisting of all points satisfying the equation 𝑥2 + 𝑦2 = 𝑐. Since there is no 𝑧-coordinate in the equation, 𝑧 can take any value. For 𝑐 > 0, this is a circular √ cylinder of radius 𝑐 around the 𝑧-axis. The level surfaces are concentric cylinders; on the narrow ones near the 𝑧-axis, 𝑓 has small values; on the wider ones, 𝑓 has larger values. See Figure 12.81. The level surface of 𝑔 corresponding to the constant 𝑐 is the plane 𝑧 − 𝑦 = 𝑐. Since there is no 𝑥 variable in the equation, these planes are parallel to the 𝑥-axis and cut the 𝑦𝑧-plane in the line 𝑧 − 𝑦 = 𝑐. See Figure 12.82.

Figure 12.81: Level surfaces of 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2

Figure 12.82: Level surfaces of 𝑔(𝑥, 𝑦, 𝑧) = 𝑧 − 𝑦

We say 𝑔(𝑥, 𝑦, 𝑧) = 𝑧 − 𝑦 in Example 2 is a linear function of the three variables 𝑥, 𝑦, 𝑧, whereas 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 and 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 + 𝑧2 are quadratic functions of three variables. Example 3

What do the level surfaces of 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 − 𝑧2 look like?

Solution

In Section 12.3, we saw that the two-variable quadratic function 𝑔(𝑥, 𝑦) = 𝑥2 − 𝑦2 has a saddleshaped graph and three types of contours. The contour equation 𝑥2 − 𝑦2 = 𝑐 gives a hyperbola opening right-left when 𝑐 > 0, a hyperbola opening up-down when 𝑐 < 0, and a pair of intersecting lines when 𝑐 = 0. Similarly, the three-variable quadratic function 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 − 𝑧2 has three types of level surfaces depending on the value of 𝑐 in the equation 𝑥2 + 𝑦2 − 𝑧2 = 𝑐. Suppose that 𝑐 > 0, say 𝑐 = 1. Rewrite the equation as 𝑥2 + 𝑦2 = 𝑧2 + 1 and think of what

12.5 FUNCTIONS OF THREE VARIABLES

691

happens as we cut the surface perpendicular to the 𝑧-axis by holding 𝑧 fixed. The result is a circle, 𝑥2 + 𝑦2 = constant, of radius at least 1 (since the constant 𝑧2 + 1 ≥ 1). The circles get larger as 𝑧 gets larger. If we take the 𝑥 = 0 cross-section instead, we get the hyperbola 𝑦2 − 𝑧2 = 1. The result is shown in Figure 12.86, with 𝑎 = 𝑏 = 𝑐 = 1. Suppose instead that 𝑐 < 0, say 𝑐 = −1. Then the horizontal cross-sections of 𝑥2 + 𝑦2 = 𝑧2 − 1 are again circles except that the radii shrink to 0 at 𝑧 = ±1 and between 𝑧 = −1 and 𝑧 = 1 there are no cross-sections at all. The result is shown in Figure 12.87 with 𝑎 = 𝑏 = 𝑐 = 1. When 𝑐 = 0, we get the equation 𝑥2 + 𝑦2 = 𝑧2 . Again the horizontal cross-sections are circles, this time with the radius shrinking down to exactly√0 when 𝑧 = 0. The resulting surface, shown in Figure 12.88 with 𝑎 =√𝑏 = 𝑐 = 1, is the cone 𝑧 = 𝑥2 + 𝑦2 studied in Section 12.3, together with the lower cone 𝑧 = − 𝑥2 + 𝑦2 .

A Catalog of Surfaces For later reference, here is a small catalog of the surfaces we have encountered.

Figure 12.83: Elliptical 2 2 paraboloid 𝑧 = 𝑥𝑎2 + 𝑦𝑏2

Figure 12.84: Hyperbolic 2 2 paraboloid 𝑧 = − 𝑥𝑎2 + 𝑦𝑏2

Figure 12.85: Ellipsoid 2 2 𝑥2 + 𝑦𝑏2 + 𝑧𝑐 2 = 1 𝑎2

Figure 12.86: Hyperboloid of 2 2 2 one sheet 𝑥𝑎2 + 𝑦𝑏2 − 𝑧𝑐 2 = 1

Figure 12.87: Hyperboloid of two 2 2 2 sheets 𝑥𝑎2 + 𝑦𝑏2 − 𝑧𝑐 2 = −1

Figure 12.88: Cone 2 2 𝑥2 + 𝑦𝑏2 − 𝑧𝑐 2 = 0 𝑎2

Figure 12.89: Plane 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑

Figure 12.90: Cylindrical surface 𝑥2 + 𝑦2 = 𝑎2

Figure 12.91: Parabolic cylinder 𝑦 = 𝑎𝑥2

(These are viewed as equations in three variables 𝑥, 𝑦, and 𝑧)

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How Surfaces Can Represent Functions of Two Variables and Functions of Three Variables You may have noticed that we have used surfaces to represent functions in two different ways. First, we used a single surface to represent a two-variable function 𝑓 (𝑥, 𝑦). Second, we used a family of level surfaces to represent a three-variable function 𝑔(𝑥, 𝑦, 𝑧). These level surfaces have equation 𝑔(𝑥, 𝑦, 𝑧) = 𝑐. What is the relation between these two uses of surfaces? For example, consider the function 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 + 3. Define 𝑔(𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 + 3 − 𝑧 The points on the graph of 𝑓 satisfy 𝑧 = 𝑥2 + 𝑦2 + 3, so they also satisfy 𝑥2 + 𝑦2 + 3 − 𝑧 = 0. Thus the graph of 𝑓 is the same as the level surface 𝑔(𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 + 3 − 𝑧 = 0. In general, we have the following result: A single surface that is the graph of a two-variable function 𝑓 (𝑥, 𝑦) can be thought of as one member of the family of level surfaces representing the three-variable function 𝑔(𝑥, 𝑦, 𝑧) = 𝑓 (𝑥, 𝑦) − 𝑧. The graph of 𝑓 is the level surface 𝑔 = 0. Conversely, a single level surface 𝑔(𝑥, 𝑦, 𝑧) = 𝑐 can be regarded as the graph of a function 𝑓 (𝑥, 𝑦) if it is possible to solve for 𝑧. Sometimes the level surface is pieced together from the graphs of two or more two-variable functions. For example, if 𝑔(𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 + 𝑧2 , then one member of the family of level surfaces is the sphere 𝑥2 + 𝑦2 + 𝑧2 = 1. This equation defines 𝑧 implicitly as a function of 𝑥 and 𝑦. Solving it gives two functions √ √ 𝑧 = 1 − 𝑥2 − 𝑦2 and 𝑧 = − 1 − 𝑥2 − 𝑦2 .

The graph of the first function is the top half of the sphere and the graph of the second function is the bottom half.

Exercises and Problems for Section 12.5 EXERCISES 1. Match the following functions with the level surfaces in Figure 12.92.

2. Match the functions with the level surfaces in Figure 12.93.

(a) 𝑓 (𝑥, 𝑦, 𝑧) = 𝑦2 + 𝑧2 (b) ℎ(𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑧2 .

(I)

𝑧

(II)

(a) 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 + 𝑧2 (b) 𝑔(𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑧2 .

𝑧 (I)

𝑧

𝑧

(II)

𝑦

𝑦

𝑦 𝑥

Figure 12.92

𝑥

𝑥 𝑥

Figure 12.93

𝑦

12.5 FUNCTIONS OF THREE VARIABLES

693

3. Write the level surface 𝑥 + 2𝑦 + 3𝑧 = 5 as the graph of a function 𝑓 (𝑥, 𝑦).

In Exercises 12–15, decide if the given level surface can be expressed as the graph of a function, 𝑓 (𝑥, 𝑦).

4. Find a formula for a function 𝑓 (𝑥, 𝑦, 𝑧) whose level surface 𝑓 = 4 is a sphere of radius 2, centered at the origin. √ 5. Write the level surface 𝑥2 + 𝑦 + 𝑧 = 1 as the graph of a function 𝑓 (𝑥, 𝑦).

12. 𝑧 − 𝑥2 − 3𝑦2 = 0

13. 2𝑥 + 3𝑦 − 5𝑧 − 10 = 0

14. 𝑥2 + 𝑦2 + 𝑧2 − 1 = 0

15. 𝑧2 = 𝑥2 + 3𝑦2

6. Find a formula for a function 𝑓 (𝑥, 𝑦, 𝑧) whose level surfaces are spheres centered at the point (𝑎, 𝑏, 𝑐). 7. Which of the graphs in the catalog of surfaces on page 691 is the graph of a function of 𝑥 and 𝑦? In Exercises 8–11, use the catalog on page 691 to identify the surface. 8. 𝑥2 + 𝑦2 − 𝑧 = 0

9. −𝑥2 − 𝑦2 + 𝑧2 = 1 11. 𝑥2 + 𝑦2 ∕4 + 𝑧2 = 1

10. 𝑥 + 𝑦 = 1

16. Match the functions (a)–(d) with the descriptions of their level surfaces in I–IV. √ (a) 𝑓 (𝑥, 𝑦, 𝑧) = 9 − 𝑥2 − 𝑦2 √ (b) 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 + 𝑧2 1 (c) 𝑓 (𝑥, 𝑦, 𝑧) = 2 𝑥 + 𝑦2 + 𝑧2 (d) 𝑓 (𝑥, 𝑦, 𝑧) = 5 + 𝑦2 + 𝑧2 I. II. III. IV.

Cylinders that get larger as the function value increases Cylinders that get smaller as the function value increases Spheres that get larger as the function value increases Spheres that get smaller as the function value increases

PROBLEMS In Problems 17–19, represent the surface whose equation is given as the graph of a two-variable function, 𝑓 (𝑥, 𝑦), and as the level surface of a three-variable function, 𝑔(𝑥, 𝑦, 𝑧) = 𝑐. There are many possible answers. 17. The plane 4𝑥 − 𝑦 − 2𝑧 = 6 18. The top half of the sphere 𝑥2 + 𝑦2 + 𝑧2 − 10 = 0 19. The bottom half of the ellipsoid 𝑥2 + 𝑦2 + 𝑧2 ∕2 = 1 20. The balance, 𝐵, in dollars, in a bank account depends on the amount deposited, 𝐴 dollars, the annual interest rate, 𝑟%, and the time, 𝑡, in months since the deposit, so 𝐵 = 𝑓 (𝐴, 𝑟, 𝑡). (a) Is 𝑓 an increasing or decreasing function of 𝐴? Of 𝑟? Of 𝑡? (b) Interpret the statement 𝑓 (1250, 1, 25) ≈ 1276. Give units. 21. A person’s basal metabolic rate (BMR) is the minimal number of daily calories needed to keep their body functioning at rest. The BMR (in kcal/day) of a man of mass 𝑚 (in kg), height ℎ (in cm) and age 𝑎 (in years) can be approximated by12

(c) Describe the level surface 𝑃 = 2000 for a woman and explain what the points on this level surface represent. (d) If a 40-year-old man 175 cm tall weighing 77 kg restricts himself to a diet with a daily caloric intake of 1600 kcal, should he expect to lose weight? 22. The monthly payments, 𝑃 dollars, on a mortgage in which 𝐴 dollars were borrowed at an annual interest rate of 𝑟% for 𝑡 years is given by 𝑃 = 𝑓 (𝐴, 𝑟, 𝑡). Is 𝑓 an increasing or decreasing function of 𝐴? Of 𝑟? Of 𝑡? 23. The balance in a bank account, 𝐵 dollars, is given by 𝐵 = 𝑓 (𝑃 , 𝑟, 𝑡) = 𝑃 (1 + 0.01𝑟)𝑡 , where 𝑃 dollars is the principal amount invested, 𝑟% is the annual interest rate, and 𝑡 years is the time since the investment was made. (a) Find a formula for the level surface of 𝑓 containing the point (𝑃 , 𝑟, 𝑡) = (1000, 5, 20), and explain the significance of this surface in terms of balance. (b) Find another point on the level surface in part (a), and explain the significance of this point in terms of balance. 24. The pressure of gas in a storage container, in atmospheres, is given by

𝑃 = 𝑓 (𝑚, ℎ, 𝑎) = 14𝑚 + 5ℎ − 7𝑎 + 66 𝑃 = 𝑓 (𝑛, 𝑇 , 𝑉 ) = and for women by 𝑃 = 𝑔(𝑚, ℎ, 𝑎) = 10𝑚 + 2ℎ − 5𝑎 + 655. (a) What is the BMR of a 28-year-old man 180 cm tall weighing 59 kg? (b) What is the BMR of a 43-year-old woman 162 cm tall weighing 52 kg? 12 www.wikipedia.org,

accessed May 11, 2016.

82𝑛𝑇 , 𝑉

where 𝑛 is the amount of gas, in kilomoles, 𝑇 is the temperature of the gas, in Kelvin, and 𝑉 is the volume of the storage container, in liters. (a) Find a formula for the level surface of 𝑓 containing the point (𝑛, 𝑇 , 𝑉 ) = (1, 270, 20), and explain the significance of this surface in terms of pressure.

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Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

(b) Find another point on the level surface in part (a), and explain the significance of this point in terms of pressure. 25. The mass, in grams, of a rod in the shape of a right circular cylinder, is given by 𝑚 = 𝑓 (𝑟, ℎ, 𝜌) = 𝜋𝑟2 ℎ𝜌, where the rod has a radius of 𝑟 cm, a height of ℎ cm, and a uniform density of 𝜌 gm/cm3 . (a) Find a formula for the level surface of 𝑓 containing the point (𝑟, ℎ, 𝜌) = (2, 10, 3), and explain the significance of this surface in terms of mass. (b) Find another point on the level surface in part (a), and explain the significance of this point in terms of mass. 26. Find a function 𝑓 (𝑥, 𝑦, 𝑧) whose level surface 𝑓 = 1 is the graph of the function 𝑔(𝑥, 𝑦) = 𝑥 + 2𝑦. 27. Find two functions 𝑓 (𝑥, 𝑦) and 𝑔(𝑥, 𝑦) so that the graphs of both together form the ellipsoid 𝑥2 +𝑦2 ∕4+𝑧2 ∕9 = 1.

32. A cone 𝐶, with height 1 and radius 1, has its base in the 𝑥𝑧-plane and its vertex on the positive 𝑦-axis. Find a function 𝑔(𝑥, 𝑦, 𝑧) such that 𝐶 is part of the level √ surface 𝑔(𝑥, 𝑦, 𝑧) = 0. [Hint: The graph of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 is a cone which opens up and has vertex at the origin.] 33. Describe the level surface 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 ∕4 + 𝑧2 = 1 in words. 34. Describe the level surface 𝑔(𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 ∕4 + 𝑧2 = 1 in words. [Hint: Look at cross-sections with constant 𝑥, 𝑦, and 𝑧 values.] 35. Describe in words the level surfaces of the function 𝑔(𝑥, 𝑦, 𝑧) = 𝑥 + 𝑦 + 𝑧. 36. Describe in words the level surfaces of 𝑓 (𝑥, 𝑦, 𝑧) = sin(𝑥 + 𝑦 + 𝑧). 37. Describe the surface 𝑥2 + 𝑦2 = (2 + sin 𝑧)2 . In general, if 𝑓 (𝑧) ≥ 0 for all 𝑧, describe the surface 𝑥2 + 𝑦2 = (𝑓 (𝑧))2 .

28. Find a formula for a function 𝑔(𝑥, 𝑦, 𝑧) whose level surfaces are planes parallel to the plane 𝑧 = 2𝑥 + 3𝑦 − 5.

38. What do the level surfaces of 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 − 𝑦2 + 𝑧2 look like? [Hint: Use cross-sections with 𝑦 constant instead of cross-sections with 𝑧 constant.]

29. Which of the following functions have planes as level surfaces?

39. Describe in words the level surfaces of 𝑔(𝑥, 𝑦, 𝑧) = 2 2 2 𝑒−(𝑥 +𝑦 +𝑧 ) . 40. Describe in words the level surfaces of 𝑓 (𝑥, 𝑦, 𝑧) = 𝑧∕𝑥.

𝑓 (𝑥, 𝑦, 𝑧) = 𝑒𝑥+𝑧

𝑟(𝑥, 𝑦, 𝑧) = 𝑥3

𝑔(𝑥, 𝑦, 𝑧) = 𝑒𝑥 + 𝑧

𝑚(𝑥, 𝑦, 𝑧) = ln (𝑥 + 𝑧)

30. The surface 𝑆 is the graph of 𝑓 (𝑥, 𝑦) =



1 − 𝑥2 − 𝑦2 .

(a) Explain why 𝑆 is the upper hemisphere of radius 1, with equator in the 𝑥𝑦-plane, centered at the origin. (b) Find a level surface 𝑔(𝑥, 𝑦, 𝑧) = 𝑐 representing 𝑆. √ 31. The surface 𝑆 is the graph of 𝑓 (𝑥, 𝑦) = 1 − 𝑦2 .

(a) Explain why 𝑆 is the upper half of a circular cylinder of radius 1, centered along the 𝑥-axis. (b) Find a level surface 𝑔(𝑥, 𝑦, 𝑧) = 𝑐 representing 𝑆.

41. Show that the level surfaces of 𝑔(𝑥, 𝑦, 𝑧) = 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 where 𝑐 ≠ 0 are parallel planes. 42. Sketch and label level surfaces of ℎ(𝑥, 𝑦, 𝑧) = 𝑒𝑧−𝑦 for ℎ = 1, 𝑒, 𝑒2 . 43. Sketch and label level surfaces of 𝑓 (𝑥, 𝑦, 𝑧) = 4 − 𝑥2 − 𝑦2 − 𝑧2 for 𝑓 = 0, 1, 2. 44. Sketch and label level surfaces of 𝑔(𝑥, 𝑦, 𝑧) = 1−𝑥2 −𝑦2 for 𝑔 = 0, −1, −2. 45. What is the relationship between the level surfaces of 𝑔(𝑥, 𝑦, 𝑧) = 𝑓 (𝑥, 𝑦) − 𝑧 and the graph of 𝑧 = 𝑓 (𝑥, 𝑦)? 46. Describe the level surfaces of 𝑔(𝑥, 𝑦, 𝑧) = 𝑦 − 𝑓 (𝑥).

Strengthen Your Understanding In Problems 47–49, explain what is wrong with the statement.

51. A function 𝑓 (𝑥, 𝑦, 𝑧) whose level sets are concentric cylinders centered on the 𝑦-axis.

47. The graph of a function 𝑓 (𝑥, 𝑦, 𝑧) is a surface in 3space.

52. A nonlinear function 𝑓 (𝑥, 𝑦, 𝑧) whose level sets are parallel planes.

48. The level surfaces of 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 − 𝑦2 are all saddleshaped.

53. A function 𝑓 (𝑥, 𝑦, 𝑧) whose level sets are paraboloids.

𝑥2 + 𝑦2 are

Are the statements in Problems 54–64 true or false? Give reasons for your answer.

50. A function 𝑓 (𝑥, 𝑦, 𝑧) whose level surfaces are equally spaced planes perpendicular to the 𝑦𝑧-plane.

54. The graph of the function 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 is the same as the level surface 𝑔(𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 − 𝑧 = 0. √ 55. The graph of 𝑓 (𝑥, 𝑦) = 1 − 𝑥2 − 𝑦2 is the same as the level surface 𝑔(𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 + 𝑧2 = 1.

49. The level surfaces of 𝑓 (𝑥, 𝑦, 𝑧) paraboloids.

=

In Problems 50–53, give an example of:

12.6 LIMITS AND CONTINUITY

56. Any surface which is the graph of a two-variable function 𝑓 (𝑥, 𝑦) can also be represented as the level surface of a three-variable function 𝑔(𝑥, 𝑦, 𝑧). 57. Any surface which is the level surface of a threevariable function 𝑔(𝑥, 𝑦, 𝑧) can also be represented as the graph of a two-variable function 𝑓 (𝑥, 𝑦). 58. The level surfaces of the function 𝑔(𝑥, 𝑦, 𝑧) = 𝑥+2𝑦+𝑧 are parallel planes. 2

2

59. The level surfaces of 𝑔(𝑥, 𝑦, 𝑧) = 𝑥 + 𝑦 + 𝑧 are cylinders with axis along the 𝑦-axis. 60. A level surface of a function 𝑔(𝑥, 𝑦, 𝑧) cannot be a single

12.6

695

point. 61. If 𝑔(𝑥, 𝑦, 𝑧) = 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 + 𝑑, where 𝑎, 𝑏, 𝑐, 𝑑 are nonzero constants, then the level surfaces of 𝑔 are planes. 62. If the level surfaces of 𝑔 are planes, then 𝑔(𝑥, 𝑦, 𝑧) = 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 + 𝑑, where 𝑎, 𝑏, 𝑐, 𝑑 are constants. 63. If the level surfaces 𝑔(𝑥, 𝑦, 𝑧) = 𝑘1 and 𝑔(𝑥, 𝑦, 𝑧) = 𝑘2 are the same surface, then 𝑘1 = 𝑘2 . 64. If 𝑥2 + 𝑦2 + 𝑧2 = 1 is the level surface 𝑔(𝑥, 𝑦, 𝑧) = 1, then 𝑥2 + 𝑦2 + 𝑧2 = 4 is the level surface 𝑔(𝑥, 𝑦, 𝑧) = 4.

LIMITS AND CONTINUITY The sheer face of Half Dome, in Yosemite National Park in California, was caused by glacial activity during the Ice Age. (See Figure 12.94.) As we scale the rock from the west, the height of the terrain rises abruptly by nearly 5000 feet from the valley floor, 2000 feet of it vertical. If we consider the function ℎ giving the height of the terrain above sea level in terms of longitude and latitude, then ℎ has a discontinuity along the path at the base of the cliff of Half Dome. Looking at the contour map of the region in Figure 12.95, we see that in most places a small change in position results in a small change in height, except near the cliff. There, no matter how small a step we take, we get a large change in height. (You can see how crowded the contours get near the cliff; some end abruptly along the discontinuity.) This geological feature illustrates the ideas of continuity and discontinuity. Roughly speaking, a function is said to be continuous at a point if its values at places near the point are close to the value at the point. If this is not the case, the function is said to be discontinuous. The property of continuity is one that, practically speaking, we usually assume of the functions we are studying. Informally, we expect (except under special circumstances) that values of a function do not change drastically when making small changes to the input variables. Whenever we model a one-variable function by an unbroken curve, we are making this assumption. Even when functions come to us as tables of data, we usually make the assumption that the missing function values between data points are close to the measured ones. In this section we study limits and continuity a bit more formally in the context of functions of several variables. For simplicity we study these concepts for functions of two variables, but our discussion can be adapted to functions of three or more variables. One can show that sums, products, and compositions of continuous functions are continuous,

©Clint Spencer/iStockphoto

Figure 12.95: A contour map of Half Dome Figure 12.94: Half Dome in Yosemite National Park

696

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

while the quotient of two continuous functions is continuous everywhere the denominator function is nonzero. Thus, each of the functions cos(𝑥2 𝑦),

ln(𝑥2 + 𝑦2 ),

𝑒𝑥+𝑦 , 𝑥+𝑦

ln(sin(𝑥2 + 𝑦2 ))

is continuous at all points (𝑥, 𝑦) where it is defined. As for functions of one variable, the graph of a continuous function over an unbroken domain is unbroken—that is, the surface has no holes or rips in it. Example 1

From Figures 12.96–12.99, which of the following functions appear to be continuous at (0, 0)? { ⎧ 2 𝑥2 ⎪ 𝑥 𝑦 , (𝑥, 𝑦) ≠ (0, 0), , (𝑥, 𝑦) ≠ (0, 0), (a) 𝑓 (𝑥, 𝑦) = ⎨ 𝑥2 + 𝑦2 (b) 𝑔(𝑥, 𝑦) = 𝑥2 + 𝑦2 ⎪ 0, 0, (𝑥, 𝑦) = (0, 0). (𝑥, 𝑦) = (0, 0). ⎩ 𝑦

1

.45

.45 .35

.35 .25

.15 .05

.05 .15

.25

𝑥 −1

1

−.45

−.35

−.25

−.15

−.05

−.05

−.15

−.25

−.45

−.35

−1 Figure 12.96: Graph of 𝑧 = 𝑥2 𝑦∕(𝑥2 + 𝑦2 )

Figure 12.97: Contour diagram of 𝑧 = 𝑥2 𝑦∕(𝑥2 + 𝑦2 ) 𝑦

.65

.75 .85

.75

.85

.95

−1

.95

.85

5 .5

.05

.35

.25 .15 .05

.65 .25 .15

.3 5

1 .75

.75 .6 5

𝑥

.95 .85

.95

5 .4

.5 5 .4 5

.5 5

.35

.25

.05 .15

5 .6

.05 .15 .25

5 .3

5 .4 5 .5

.4 5

1

−1 Figure 12.98: Graph of 𝑧 = 𝑥2 ∕(𝑥2 + 𝑦2 )

Solution

Figure 12.99: Contour diagram of 𝑧 = 𝑥2 ∕(𝑥2 + 𝑦2 )

(a) The graph and contour diagram of 𝑓 in Figures 12.96 and 12.97 suggest that 𝑓 is close to 0 when (𝑥, 𝑦) is close to (0, 0). That is, the figures suggest that 𝑓 is continuous at the point (0, 0); the graph appears to have no rips or holes there. However, the figures cannot tell us for sure whether 𝑓 is continuous. To be certain we must investigate the limit analytically, as is done in Example 2(a) on page 697.

12.6 LIMITS AND CONTINUITY

697

(b) The graph of 𝑔 and its contours near (0, 0) in Figure 12.98 and 12.99 suggest that 𝑔 behaves differently from 𝑓 : The contours of 𝑔 seem to “crash” at the origin and the graph rises rapidly from 0 to 1 near (0, 0). Small changes in (𝑥, 𝑦) near (0, 0) can yield large changes in 𝑔, so we expect that 𝑔 is not continuous at the point (0, 0). Again, a more precise analysis is given in Example 2(b). The previous example suggests that continuity at a point depends on a function’s behavior near the point. To study behavior near a point more carefully we need the idea of a limit of a function of two variables. Suppose that 𝑓 (𝑥, 𝑦) is a function defined on a set in 2-space, not necessarily containing the point (𝑎, 𝑏), but containing points (𝑥, 𝑦) arbitrarily close to (𝑎, 𝑏); suppose that 𝐿 is a number.

The function 𝑓 has a limit 𝐿 at the point (𝑎, 𝑏), written lim

(𝑥,𝑦)→(𝑎,𝑏)

𝑓 (𝑥, 𝑦) = 𝐿,

if 𝑓 (𝑥, 𝑦) is as close to 𝐿 as we please whenever the distance from the point (𝑥, 𝑦) to the point (𝑎, 𝑏) is sufficiently small, but not zero.

We define continuity for functions of two variables in the same way as for functions of one variable:

A function 𝑓 is continuous at the point (𝑎, 𝑏) if lim

(𝑥,𝑦)→(𝑎,𝑏)

𝑓 (𝑥, 𝑦) = 𝑓 (𝑎, 𝑏).

A function is continuous on a region 𝑅 in the 𝑥𝑦-plane if it is continuous at each point in 𝑅.

Thus, if 𝑓 is continuous at the point (𝑎, 𝑏), then 𝑓 must be defined at (𝑎, 𝑏) and the limit, lim(𝑥,𝑦)→(𝑎,𝑏) 𝑓 (𝑥, 𝑦), must exist and be equal to the value 𝑓 (𝑎, 𝑏). If a function is defined at a point (𝑎, 𝑏) but is not continuous there, then we say that 𝑓 is discontinuous at (𝑎, 𝑏). We now apply the definition of continuity to the functions in Example 1, showing that 𝑓 is continuous at (0, 0) and that 𝑔 is discontinuous at (0, 0). Example 2

Let 𝑓 and 𝑔 be the functions in Example 1. Use the definition of the limit to show that: (a) lim 𝑓 (𝑥, 𝑦) = 0 (b) lim 𝑔(𝑥, 𝑦) does not exist. (𝑥,𝑦)→(0,0)

Solution

(𝑥,𝑦)→(0,0)

To investigate these limits of 𝑓 and 𝑔, we consider values of these functions near, but not at, the origin, where they are given by the formulas 𝑓 (𝑥, 𝑦) =

𝑥2 𝑦 𝑥2 + 𝑦2

𝑔(𝑥, 𝑦) =

𝑥2 . 𝑥2 + 𝑦2

(a) The graph and contour diagram of 𝑓 both suggest that lim(𝑥,𝑦)→(0,0) 𝑓 (𝑥, 𝑦) = 0. To use the definition of the limit, we estimate |𝑓 (𝑥, 𝑦) − 𝐿| with 𝐿 = 0: | 𝑥2 𝑦 | | 𝑥2 | √ | | | | |𝑦| ≤ |𝑦| ≤ 𝑥2 + 𝑦2 . |𝑓 (𝑥, 𝑦) − 𝐿| = | 2 − 0 |=| 2 | 𝑥 + 𝑦2 | | 𝑥 + 𝑦2 || | |

698

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

√ Now 𝑥2 + 𝑦2 is the distance from (𝑥, 𝑦) to (0, 0). Thus, to make |𝑓 (𝑥, 𝑦) − 0| < 0.001, for example, we need only require that (𝑥, 𝑦) be within 0.001 of (0, 0). More generally, for any positive number 𝑢, no matter how small, we are sure that |𝑓 (𝑥, 𝑦) − 0| < 𝑢 whenever (𝑥, 𝑦) is no farther than 𝑢 from (0, 0). This is what we mean by saying that the difference |𝑓 (𝑥, 𝑦) − 0| can be made as small as we wish by choosing the distance to be sufficiently small. Thus, we conclude that 𝑥2 𝑦 lim 𝑓 (𝑥, 𝑦) = lim = 0. (𝑥,𝑦)→(0,0) (𝑥,𝑦)→(0,0) 𝑥2 + 𝑦2 Notice that since this limit equals 𝑓 (0, 0), the function 𝑓 is continuous at (0, 0). (b) Although the formula defining the function 𝑔 looks similar to that of 𝑓 , we saw in Example 1 that 𝑔’s behavior near the origin is quite different. If we consider points (𝑥, 0) lying along the 𝑥-axis near (0, 0), then the values 𝑔(𝑥, 0) are equal to 1, while if we consider points (0, 𝑦) lying along the 𝑦-axis near (0, 0), then the values 𝑔(0, 𝑦) are equal to 0. Thus, within any distance (no matter how small) from the origin, there are points where 𝑔 = 0 and points where 𝑔 = 1. Therefore the limit lim(𝑥,𝑦)→(0,0) 𝑔(𝑥, 𝑦) does not exist, and thus 𝑔 is not continuous at (0, 0). While the notions of limit and continuity look formally the same for one- and two-variable functions, they are somewhat more subtle in the multivariable case. The reason for this is that on the line (1-space), we can approach a point from just two directions (left or right) but in 2-space there are an infinite number of ways to approach a given point.

Exercises and Problems for Section 12.6 EXERCISES In Exercises 1–6, is the function continuous at all points in the given region? 1. 2. 3. 4. 5. 6.

1 on the square −1 ≤ 𝑥 ≤ 1, −1 ≤ 𝑦 ≤ 1 𝑥2 + 𝑦2 1 on the square 1 ≤ 𝑥 ≤ 2, 1 ≤ 𝑦 ≤ 2 𝑥2 + 𝑦2 𝑦 on the disk 𝑥2 + 𝑦2 ≤ 1 𝑥2 + 2 𝑒sin 𝑥 on the rectangle − 𝜋2 ≤ 𝑥 ≤ 𝜋2 , 0 ≤ 𝑦 ≤ 𝜋4 cos 𝑦 tan(𝑥𝑦) on the square −2 ≤ 𝑥 ≤ 2, −2 ≤ 𝑦 ≤ 2 √ 2𝑥 − 𝑦 on the disk 𝑥2 + 𝑦2 ≤ 4

In Exercises 12–15, use the contour diagram for 𝑓 (𝑥, 𝑦) in Figure 12.100 to suggest an estimate for the limit, or explain why it may not exist. 𝑦 3 4

2

0

6

4 6

2

8

8

1

10

−3

𝑥

10

−2

−1

1

2

−1

8

In Exercises 7–11, find the limit as (𝑥, 𝑦) → (0, 0) of 𝑓 (𝑥, 𝑦). Assume that polynomials, exponentials, logarithmic, and trigonometric functions are continuous.

3 8

−2

6 4

6 0

2

2

4

−3

Figure 12.100

7. 𝑓 (𝑥, 𝑦) = 𝑒−𝑥−𝑦 2

2

2

8. 𝑓 (𝑥, 𝑦) = 𝑥 + 𝑦 𝑥 9. 𝑓 (𝑥, 𝑦) = 2 𝑥 +1 𝑥+𝑦 10. 𝑓 (𝑥, 𝑦) = (sin 𝑦) + 2 sin(𝑥2 + 𝑦2 ) 11. 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2

12. 14. sin 𝑡 = 1.] [Hint: lim 𝑡→0 𝑡

lim

(𝑥,𝑦)→(2,1)

lim

𝑓 (𝑥, 𝑦)

(𝑥,𝑦)→(−2,0)

𝑓 (𝑥, 𝑦)

13. 15.

lim

(𝑥,𝑦)→(−1,2)

lim

(𝑥,𝑦)→(0,0)

𝑓 (𝑥, 𝑦)

𝑓 (𝑥, 𝑦)

699

12.6 LIMITS AND CONTINUITY

PROBLEMS In Problems 16–17, show that the function 𝑓 (𝑥, 𝑦) does not have a limit as (𝑥, 𝑦) → (0, 0). [Hint: Use the line 𝑦 = 𝑚𝑥.] 𝑥+𝑦 , 𝑥≠𝑦 𝑥−𝑦 2 2 𝑥 −𝑦 17. 𝑓 (𝑥, 𝑦) = 2 𝑥 + 𝑦2 18. By approaching the origin along the positive 𝑥-axis and the positive 𝑦-axis, show that the following limit does not exist: 2𝑥 − 𝑦2 . lim (𝑥,𝑦)→(0,0) 2𝑥 + 𝑦2 19. Show that 𝑓 (𝑥, 𝑦) has no limit as (𝑥, 𝑦) → (0, 0) if 𝑥𝑦 𝑓 (𝑥, 𝑦) = , 𝑥 ≠ 0 and 𝑦 ≠ 0. |𝑥𝑦| 16. 𝑓 (𝑥, 𝑦) =

27. The function 𝑓 , whose graph and contour diagram are in Figures 12.101 and 12.102, is given by { 𝑥𝑦 , (𝑥, 𝑦) ≠ (0, 0), 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 0, (𝑥, 𝑦) = (0, 0). (a) Show that 𝑓 (0, 𝑦) and 𝑓 (𝑥, 0) are each continuous functions of one variable. (b) Show that rays emanating from the origin are contained in contours of 𝑓 . (c) Is 𝑓 continuous at (0, 0)?

20. Show that the function 𝑓 does not have a limit at (0, 0) by examining the limits of 𝑓 as (𝑥, 𝑦) → (0, 0) along the curve 𝑦 = 𝑘𝑥2 for different values of 𝑘: 𝑥2 + 𝑦 ≠ 0.

for 𝑥 ≠ 0 for 𝑥 = 0. (b) On the 𝑦-axis?

−1

.15 .35

−. 3 5

.35 .15

−.15 1

.35 .15

−. 3 5

−.15

(b) Do your answers to part (a) suggest that 𝑓 is continuous at (0, 0)? Explain your answer.

−.15

5 −. 3

26. (a) Use a computer to draw the graph and the contour diagram of the following function: { 𝑥𝑦(𝑥2 − 𝑦2 ) , (𝑥, 𝑦) ≠ (0, 0), 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 0, (𝑥, 𝑦) = (0, 0).

5 −. 3

25. What value of 𝑐 makes the following function continuous at (0, 0)? { 𝑥2 + 𝑦2 + 1 if (𝑥, 𝑦) ≠ (0, 0) 𝑓 (𝑥, 𝑦) = 𝑐 if (𝑥, 𝑦) = (0, 0)

𝑦 1 −.15

In Problems 22–23, determine whether there is a value for the constant 𝑐 making the function continuous everywhere. If so, find it. If not, explain why not. { 𝑐 + 𝑦, 𝑥 ≤ 3, 22. 𝑓 (𝑥, 𝑦) = 5 − 𝑥, 𝑥 > 3. { 𝑐 + 𝑦, 𝑥 ≤ 3, 23. 𝑓 (𝑥, 𝑦) = 5 − 𝑦, 𝑥 > 3. 24. Is the following function continuous at (0, 0)? { 𝑥2 + 𝑦2 if (𝑥, 𝑦) ≠ (0, 0) 𝑓 (𝑥, 𝑦) = 2 if (𝑥, 𝑦) = (0, 0)

Figure 12.101: Graph of 𝑧 = 𝑥𝑦∕(𝑥2 + 𝑦2 )

.15

⎧ |𝑥| 𝑦 ⎪ 21. Let 𝑓 (𝑥, 𝑦) = ⎨ 𝑥 ⎪0 ⎩ Is 𝑓 (𝑥, 𝑦) continuous (a) On the 𝑥-axis? (c) At (0, 0)?

𝑥2 , +𝑦

𝑥2

.35

𝑓 (𝑥, 𝑦) =

−1

Figure 12.102: Contour diagram of 𝑧 = 𝑥𝑦∕(𝑥2 + 𝑦2 )

𝑥

700

Chapter 12 FUNCTIONS OF SEVERAL VARIABLES

Strengthen Your Understanding In Problems 28–29, explain what is wrong with the statement.

Are the statements in Problems 35–40 true or false? Give reasons for your answer.

28. If a function 𝑓 (𝑥, 𝑦) has a limit as (𝑥, 𝑦) approaches (𝑎, 𝑏), then it is continuous at (𝑎, 𝑏).

35. If the limit of 𝑓 (𝑥, 𝑦) is 1 as (𝑥, 𝑦) approaches (0, 0) along the 𝑥-axis, and the limit of 𝑓 (𝑥, 𝑦) is 1 as (𝑥, 𝑦) approaches (0, 0) along the 𝑦-axis, then

29. If both 𝑓 and 𝑔 are continuous at (𝑎, 𝑏), then so are 𝑓 + 𝑔, 𝑓 𝑔 and 𝑓 ∕𝑔.

lim

(𝑥,𝑦)→(0,0)

In Problems 30–31, give an example of: 30. A function 𝑓 (𝑥, 𝑦) which is continuous everywhere except at (0, 0) and (1, 2). 31. A function 𝑓 (𝑥, 𝑦) that approaches 1 as (𝑥, 𝑦) approaches (0, 0) along the 𝑥-axis and approaches 2 as (𝑥, 𝑦) approaches (0, 0) along the 𝑦-axis.

36. If 𝑓 (1, 0) = 2, then

𝑓 (𝑥, 𝑦) exists.

lim

(𝑥,𝑦)→(1,0)

37. If 𝑓 (𝑥, 𝑦) is continuous and 𝑓 (1, 0) = 2, then lim

(𝑥,𝑦)→(1,0)

In Problems 32–34, construct a function 𝑓 (𝑥, 𝑦) with the given property. 38. If 32. Not continuous along the line 𝑥 = 2; continuous everywhere else.

𝑓 (𝑥, 𝑦) = 2.

lim

(𝑥,𝑦)→(0,0)

𝑓 (𝑥, 𝑦) = 2.

𝑓 (𝑥, 𝑦) = 3, then the limit of 𝑓 (𝑥, 𝑦) is 3 as

(𝑥, 𝑦) approaches (0, 0) along the 𝑥-axis.

33. Not continuous at the point (2, 0); continuous everywhere else.

39. If 𝑓 (𝑥, 𝑦) is continuous at (𝑎, 𝑏), then its limit exists at (𝑎, 𝑏).

34. Not continuous along the curve 𝑥2 +𝑦2 = 1; continuous everywhere else.

40. If

Online Resource: Review problems and Projects

lim

(𝑥,𝑦)→(𝑎,𝑏)

(𝑎, 𝑏).

𝑓 (𝑥, 𝑦) exists then 𝑓 (𝑥, 𝑦) is continuous at

Chapter Thirteen

A FUNDAMENTAL TOOL: VECTORS

Contents 13.1 Displacement Vectors . . . . . . . . . . . . . . . . . . . . . Notation and Terminology . . . . . . . . . . . . . . . . . . Addition and Subtraction of Displacement Vectors . . . . . . . . . . . . . . . . . . . . Scalar Multiplication of Displacement Vectors . . Parallel Vectors . . . . . . . . . . . . . . . . . . . . . . . . . Components of Vectors: I I IDisplacement The Vectors i , j , k . . . . . . . . . . . . . . . . . . . . . . . Unit Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 Vectors in General . . . . . . . . . . . . . . . . . . . . . . . . Velocity Versus Speed . . . . . . . . . . . . . . . . . . . . . Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Properties of Addition and Scalar Multiplication . . . . . . . . . . . . . . . . . . . . . Using Components . . . . . . . . . . . . . . . . . . . . . . . Vectors in n Dimensions . . . . . . . . . . . . . . . . . . . Why Do We Want Vectors in n Dimensions?. . . . . . 13.3 The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . Definition of the Dot Product . . . . . . . . . . . . . . . Why the Two Definitions of the Dot Product Give the Same Result . . . . . . . . . . Properties of the Dot Product . . . . . . . . . . . . . . . Perpendicularity, Magnitude, and Dot Products . . . Using the Dot Product . . . . . . . . . . . . . . . . . . . . . Normal Vectors and the Equation of a Plane . . . . . . The Dot Product in n Dimensions . . . . . . . . . . . . . Resolving a Vector into Components: Projections . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

702 702 702 703 704 704 708 710 711 712 712 713 714 714 715 718 718 719 719 719 720 721 722 723

A Physical Interpretation of the Dot Product: Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 724

13.4 The Cross Product . . . . . . . . . . . . . . . . . . . . . . . The Area of a Parallelogram . . . . . . . . . . . . . . . . Definition of the Cross Product . . . . . . . . . . . . . . Properties of the Cross Product . . . . . . . . . . . . . . The Equation of a Plane Through Three Points . . . . . . . . . . . . . . . . . . . . . . . . . . . Areas and Volumes Using the Cross Product and Determinants . . . . . . . . . . . . . . . . . . . . . . . Volume of a Parallelepiped . . . . . . . . . . . . . . . . . Angular Velocity . . . . . . . . . . . . . . . . . . . . . . . . .

728 728 729 731 731 732 732 733

702

13.1

Chapter 13 A FUNDAMENTAL TOOL: VECTORS

DISPLACEMENT VECTORS Suppose you are a pilot planning a flight from Dallas to Pittsburgh. There are two things you must know: the distance to be traveled (so you have enough fuel to make it) and in what direction to go (so you don’t miss Pittsburgh). Both these quantities together specify the displacement or displacement vector between the two cities. The displacement vector from one point to another is an arrow with its tail at the first point and its tip at the second. The magnitude (or length) of the displacement vector is the distance between the points and is represented by the length of the arrow. The direction of the displacement vector is the direction of the arrow. Figure 13.1 shows a map with the displacement vectors from Dallas to Pittsburgh, from Albuquerque to Oshkosh, and from Los Angeles to Buffalo, SD. These displacement vectors have the same length and the same direction. We say that the displacement vectors between the corresponding cities are the same, even though they do not coincide. In other words, Displacement vectors which point in the same direction and have the same magnitude are considered to be the same, even if they do not coincide.

Buffalo, SD

Oshkosh Pittsburgh

Los Angeles Albuquerque Dallas

Figure 13.1: Displacement vectors between cities

Notation and Terminology The displacement vector is our first example of a vector. Vectors have both magnitude and direction; in comparison, a quantity specified only by a number, but no direction, is called a scalar.1 For instance, the time taken by the flight from Dallas to Pittsburgh is a scalar quantity. Displacement is a vector since it requires both distance and direction to specify it. In this book, vectors are written with an arrow over them, 𝑣⃗ , to distinguish them from scalars. Other books use a bold 𝐯 to denote a vector. We use the notation 𝑃⃖⃖⃖⃖⃖𝑄⃗ to denote the displacement vector from a point 𝑃 to a point 𝑄. The magnitude, or length, of a vector 𝑣⃗ is written ‖⃗ 𝑣 ‖.

Addition and Subtraction of Displacement Vectors

Suppose NASA commands a robot on Mars to move 75 meters in one direction and then 50 meters in another direction. (See Figure 13.2.) Where does the robot end up? Suppose the displacements ⃗ , respectively. Then the sum 𝑣⃗ + 𝑤 ⃗ gives the final position. are represented by the vectors 𝑣⃗ and 𝑤 1 So

named by W. R. Hamilton because they are merely numbers on the scale from −∞ to ∞.

703

13.1 DISPLACEMENT VECTORS

⃗ , of two vectors 𝑣⃗ and 𝑤 ⃗ is the combined displacement resulting from first The sum, 𝑣⃗ + 𝑤 ⃗ . (See Figure 13.3.) The sum 𝑤 ⃗ + 𝑣⃗ gives the same displacement. applying 𝑣⃗ and then 𝑤 Finish

Finish

𝑣⃗

Combined displacement 50 m

⃗ 𝑤

⃗ 𝑣⃗ + 𝑤 ⃗ 𝑤

Start

75 m

𝑣⃗

Start

Figure 13.2: Sum of displacements of robots on Mars

⃗ =𝑤 ⃗ + 𝑣⃗ Figure 13.3: The sum 𝑣⃗ + 𝑤

Suppose two different robots start from the same location. One moves along a displacement ⃗ . What is the displacement vector, 𝑥⃗ , from vector 𝑣⃗ and the second along a displacement vector 𝑤 ⃗ , we define 𝑥⃗ to be the difference the first robot to the second? (See Figure 13.4.) Since 𝑣⃗ + 𝑥⃗ = 𝑤 ⃗ − 𝑣⃗ . In other words, 𝑤 ⃗ − 𝑣⃗ gets you from the first robot to the second. 𝑥⃗ = 𝑤 ⃗ − 𝑣⃗ , is the displacement vector that, when added to 𝑣⃗ , gives 𝑤 ⃗ . That is, The difference, 𝑤 ⃗ = 𝑣⃗ + (𝑤 ⃗ − 𝑣⃗ ). (See Figure 13.4.) 𝑤

Second robot

⃗ − 𝑣⃗ 𝑥⃗ = 𝑤 ⃗ 𝑤 First robot Start

𝑣⃗ ⃗ − 𝑣⃗ Figure 13.4: The difference 𝑤

If the robot ends up where it started, then its total displacement vector is the zero vector, 0⃗ . The zero vector has no direction. The zero vector, 0⃗ , is a displacement vector with zero length.

Scalar Multiplication of Displacement Vectors If 𝑣⃗ represents a displacement vector, the vector 2𝑣⃗ represents a displacement of twice the magnitude in the same direction as 𝑣⃗ . Similarly, −2𝑣⃗ represents a displacement of twice the magnitude in the opposite direction. (See Figure 13.5.)

𝑣⃗

2𝑣⃗ 0.5𝑣⃗

−2𝑣⃗

Figure 13.5: Scalar multiples of the vector 𝑣⃗

704

Chapter 13 A FUNDAMENTAL TOOL: VECTORS

If 𝜆 is a scalar and 𝑣⃗ is a displacement vector, the scalar multiple of 𝑣⃗ by 𝜆, written 𝜆⃗ 𝑣 , is the displacement vector with the following properties: • The displacement vector 𝜆⃗ 𝑣 is parallel to 𝑣⃗ , pointing in the same direction if 𝜆 > 0 and in the opposite direction if 𝜆 < 0. • The magnitude of 𝜆⃗ 𝑣 is |𝜆| times the magnitude of 𝑣⃗ , that is, ‖𝜆⃗ 𝑣 ‖ = |𝜆| ‖⃗ 𝑣‖. Note that |𝜆| represents the absolute value of the scalar 𝜆 while ‖𝜆⃗ 𝑣 ‖ represents the magnitude of the vector 𝜆⃗ 𝑣. Example 1

⃗ − 𝑣⃗ = 𝑤 ⃗ + (−1)𝑣⃗ . Explain why 𝑤

Solution

The vector (−1)𝑣⃗ has the same magnitude as 𝑣⃗ , but points in the opposite direction. Figure 13.6 ⃗ + (−1)𝑣⃗ is the same as the displacement 𝑤 ⃗ − 𝑣⃗ . shows that the combined displacement 𝑤 (−1)𝑣⃗ Finish

⃗ − 𝑣⃗ 𝑤

⃗ 𝑤

⃗ 𝑤 Start

𝑣⃗ Figure 13.6: Explanation for ⃗ − 𝑣⃗ = 𝑤 ⃗ + (−1)𝑣⃗ why 𝑤

Parallel Vectors ⃗ are parallel if one is a scalar multiple of the other, that is, if 𝑤 ⃗ = 𝜆⃗ Two vectors 𝑣⃗ and 𝑤 𝑣 , for some scalar 𝜆.

Components of Displacement Vectors: The Vectors ⃗𝒊, ⃗𝒋, and 𝒌⃖⃗ Suppose that you live in a city with equally spaced streets running east-west and north-south and that you want to tell someone how to get from one place to another. You’d be likely to tell them how many blocks east-west and how many blocks north-south to go. For example, to get from 𝑃 to 𝑄 in Figure 13.7, we go 4 blocks east and 1 block south. If ⃗𝑖 and 𝑗⃗ are as shown in Figure 13.7, then the displacement vector from 𝑃 to 𝑄 is 4⃗𝑖 − 𝑗⃗ . 𝑃

4⃗𝑖 −𝑗⃗ 𝑄

𝑗⃗ ⃗𝑖

Figure 13.7: The displacement vector from 𝑃 to 𝑄 is 4⃗𝑖 − 𝑗⃗

We extend the same idea to 3 dimensions. First we choose a Cartesian system of coordinate axes. The three vectors of length 1 shown in Figure 13.8 are the vector ⃗𝑖 , which points along the ⃗ , along the positive 𝑧-axis. positive 𝑥-axis, the vector 𝑗⃗ , along the positive 𝑦-axis, and the vector 𝑘

13.1 DISPLACEMENT VECTORS

705

𝑧 1 𝑦 ⃗ 𝑘 ⃗𝑖

(3, 2)

2 𝑣⃗

1

𝑗⃗

3⃗𝑖

1

1

𝑦

𝑥

2𝑗⃗

1

⃗ in Figure 13.8: The vectors ⃗𝑖 , 𝑗⃗ and 𝑘 3-space

𝑥 2

3

Figure 13.9: We resolve 𝑣⃗ into components by writing 𝑣⃗ = 3⃗𝑖 + 2𝑗⃗

⃖⃗ Writing Displacement Vectors Using ⃗𝒊, ⃗ 𝒋, 𝒌 Any displacement in 3-space or the plane can be expressed as a combination of displacements in the coordinate directions. For example, Figure 13.9 shows that the displacement vector 𝑣⃗ from the origin to the point (3, 2) can be written as a sum of displacement vectors along the 𝑥- and 𝑦-axes: 𝑣⃗ = 3⃗𝑖 + 2𝑗⃗ . This is called resolving 𝑣⃗ into components. In general: We resolve 𝑣⃗ into components by writing 𝑣⃗ in the form ⃗, 𝑣⃗ = 𝑣1 ⃗𝑖 + 𝑣2 𝑗⃗ + 𝑣3 𝑘 ⃗ the components of 𝑣⃗ . where 𝑣1 , 𝑣2 , 𝑣3 are scalars. We call 𝑣1 ⃗𝑖 , 𝑣2 𝑗⃗ , and 𝑣3 𝑘

An Alternative Notation for Vectors Many people write a vector in three dimensions as a string of three numbers, that is, as 𝑣⃗ = (𝑣1 , 𝑣2 , 𝑣3 ) instead of

⃗. 𝑣⃗ = 𝑣1 ⃗𝑖 + 𝑣2 𝑗⃗ + 𝑣3 𝑘

Since the first notation can be confused with a point and the second cannot, we usually use the second form. Example 2

Resolve the displacement vector, 𝑣⃗ , from the point 𝑃1 = (2, 4, 10) to the point 𝑃2 = (3, 7, 6) into components.

Solution

To get from 𝑃1 to 𝑃2 , we move 1 unit in the positive 𝑥-direction, 3 units in the positive 𝑦-direction, ⃗. and 4 units in the negative 𝑧-direction. Hence 𝑣⃗ = ⃗𝑖 + 3𝑗⃗ − 4𝑘

Example 3

⃗ is parallel to each of the following vectors: Decide whether the vector 𝑣⃗ = 2⃗𝑖 + 3𝑗⃗ + 5𝑘 ⃗, ⃗ = 4⃗𝑖 + 6𝑗⃗ + 10𝑘 𝑤

Solution

⃗, 𝑎⃗ = −⃗𝑖 − 1.5𝑗⃗ − 2.5𝑘

⃗. 𝑏⃗ = 4⃗𝑖 + 6𝑗⃗ + 9𝑘

⃗ = 2𝑣⃗ and 𝑎⃗ = −0.5𝑣⃗ , the vectors 𝑣⃗ , 𝑤 ⃗ , and 𝑎⃗ are parallel. However, 𝑏⃗ is not a multiple Since 𝑤 ⃗ of 𝑣⃗ (since, for example, 4∕2 ≠ 9∕5), so 𝑣⃗ and 𝑏 are not parallel.

Chapter 13 A FUNDAMENTAL TOOL: VECTORS

In general, Figure 13.10 shows us how to express the displacement vector between two points in components:

Components of Displacement Vectors The displacement vector from the point 𝑃1 = (𝑥1 , 𝑦1 , 𝑧1 ) to the point 𝑃2 = (𝑥2 , 𝑦2 , 𝑧2 ) is given in components by ⃖⃖⃖⃖⃖⃖⃖⃖ ⃗ ⃗ ⃗ ⃗ 𝑃 1 𝑃2 = (𝑥2 − 𝑥1 )𝑖 + (𝑦2 − 𝑦1 )𝑗 + (𝑧2 − 𝑧1 )𝑘 .

Position Vectors: Displacement of a Point from the Origin A displacement vector whose tail is at the origin is called a position vector. Thus, any point (𝑥0 , 𝑦0 , 𝑧0 ) ⃗ . (See Figure 13.11.) In in space has associated with it the position vector 𝑟⃗ 0 = 𝑥0 ⃗𝑖 + 𝑦0 𝑗⃗ + 𝑧0 𝑘 general, a position vector gives the displacement of a point from the origin. 𝑧

𝑧 (𝑥0 , 𝑦0 , 𝑧0 )



⃖⃖⃖⃖⃖⃖⃖ ⃗ 𝑃 1 𝑃2

𝑃2 = (𝑥2 , 𝑦2 , 𝑧2 )

𝑧0

𝑟⃗ 0

𝑃1 = (𝑥1 , 𝑦1 , 𝑧1 )

𝑦

𝑦

𝑦0



706

𝑥0

𝑥

𝑥

Figure 13.11: The position vector ⃗ 𝑟⃗ 0 = 𝑥0 ⃗𝑖 + 𝑦0 𝑗⃗ + 𝑧0 𝑘

Figure 13.10: The displacement vector ⃗ ⃖⃖⃖⃖⃖⃖⃖ ⃗ ⃗ ⃗ 𝑃 𝑃 1 2 = (𝑥2 − 𝑥1 )𝑖 + (𝑦2 − 𝑦1 )𝑗 + (𝑧2 − 𝑧1 )𝑘

The Components of the Zero Vector ⃗. The zero displacement vector has magnitude equal to zero and is written 0⃗ . So 0⃗ = 0⃗𝑖 + 0𝑗⃗ + 0𝑘

The Magnitude of a Vector in Components For a vector, 𝑣⃗ = 𝑣1 ⃗𝑖 + 𝑣2 𝑗⃗ , the Pythagorean theorem is used to find its magnitude, ‖⃗ 𝑣 ‖. (See Figure 13.12.) The angle 𝜃 gives the direction of 𝑣⃗ . 𝑦

‖𝑣⃗ ‖ = Length =

√ 𝑣21 + 𝑣22

𝑣⃗

𝜃

𝑣2 𝑣1

𝑥

Figure 13.12: Magnitude, ‖𝑣⃗ ‖, of a 2-dimensional vector, 𝑣⃗

⃗ , we have In three dimensions, for a vector 𝑣⃗ = 𝑣1 ⃗𝑖 + 𝑣2 𝑗⃗ + 𝑣3 𝑘 Magnitude of 𝑣⃗ = ‖⃗ 𝑣 ‖ = Length of the arrow = ⃗ , then ‖⃗ For instance, if 𝑣⃗ = 3⃗𝑖 − 4𝑗⃗ + 5𝑘 𝑣‖ =



𝑣21 + 𝑣22 + 𝑣23 .

√ √ 32 + (−4)2 + 52 = 50.

13.1 DISPLACEMENT VECTORS

707

Addition and Scalar Multiplication of Vectors in Components ⃗ are given in components: Suppose the vectors 𝑣⃗ and 𝑤 ⃗ 𝑣⃗ = 𝑣1 ⃗𝑖 + 𝑣2 𝑗⃗ + 𝑣3 𝑘

⃗. ⃗ = 𝑤1 ⃗𝑖 + 𝑤2 𝑗⃗ + 𝑤3 𝑘 and 𝑤

Then ⃗, ⃗ = (𝑣1 + 𝑤1 )⃗𝑖 + (𝑣2 + 𝑤2 )𝑗⃗ + (𝑣3 + 𝑤3 )𝑘 𝑣⃗ + 𝑤 and ⃗. 𝜆⃗ 𝑣 = 𝜆𝑣1 ⃗𝑖 + 𝜆𝑣2 𝑗⃗ + 𝜆𝑣3 𝑘 ⃗ = 𝑣⃗ + Figures 13.13 and 13.14 illustrate these properties in two dimensions. Finally, 𝑣⃗ − 𝑤 ⃗. ⃗ , so we can write 𝑣⃗ − 𝑤 ⃗ = (𝑣1 − 𝑤1 )⃗𝑖 + (𝑣2 − 𝑤2 )𝑗⃗ + (𝑣3 − 𝑤3 )𝑘 (−1)𝑤 𝑣1 𝑣⃗ 2𝑣⃗ ⃗ 𝑣⃗ + 𝑤

⃗ 𝑤

𝑤2

𝑣1

𝑤1

⃗ 𝑤 𝑣⃗ 𝑣2 𝑣1

⃗ in Figure 13.13: Sum 𝑣⃗ + 𝑤 components

𝑣2

2𝑣2

𝑣⃗

𝑣2

−3𝑣⃗

−3𝑣2

2𝑣1

−3𝑣1

Figure 13.14: Scalar multiples of vectors showing 𝑣⃗ , 2𝑣⃗ , and −3𝑣⃗

How to Resolve a Vector into Components You may wonder how we find the components of a 2-dimensional vector, given its length and direction. Suppose the vector 𝑣⃗ has length 𝑣 and makes an angle of 𝜃 with the 𝑥-axis, measured counterclockwise, as in Figure 13.15. If 𝑣⃗ = 𝑣1 ⃗𝑖 + 𝑣2 𝑗⃗ , Figure 13.15 shows that 𝑣1 = 𝑣 cos 𝜃

and 𝑣2 = 𝑣 sin 𝜃.

Thus, we resolve 𝑣⃗ into components by writing 𝑣⃗ = (𝑣 cos 𝜃)⃗𝑖 + (𝑣 sin 𝜃)𝑗⃗ . Vectors in 3-space are resolved using direction cosines; see Problem 66 (available online). 𝑦

✛ 𝑣

𝑣 sin 𝜃



𝜃

𝑥 𝑣 cos 𝜃

Figure 13.15: Resolving a vector: 𝑣⃗ = (𝑣 cos 𝜃)⃗𝑖 + (𝑣 sin 𝜃)𝑗⃗

Example 4 Solution

Resolve 𝑣⃗ into components if ‖⃗ 𝑣 ‖ = 2 and 𝜃 = 𝜋∕6. We have 𝑣⃗ = 2 cos(𝜋∕6)⃗𝑖 + 2 sin(𝜋∕6)𝑗⃗ = 2

(√

) √ 3∕2 ⃗𝑖 + 2(1∕2)𝑗⃗ = 3⃗𝑖 + 𝑗⃗ .

708

Chapter 13 A FUNDAMENTAL TOOL: VECTORS

Unit Vectors ⃗ are unit vectors in the A unit vector is a vector whose magnitude is 1. The vectors ⃗𝑖 , 𝑗⃗ , and 𝑘 directions of the coordinate axes. It is often helpful to find a unit vector in the same direction as a given vector 𝑣⃗ . Suppose that ‖⃗ 𝑣 ‖ = 10; a unit vector in the same direction as 𝑣⃗ is 𝑣⃗ ∕10. In general, a unit vector in the direction of any nonzero vector 𝑣⃗ is 𝑢⃗ =

𝑣⃗ . ‖⃗ 𝑣‖

Example 5

Find a unit vector, 𝑢⃗ , in the direction of the vector 𝑣⃗ = ⃗𝑖 + 3𝑗⃗ .

Solution

If 𝑣⃗ = ⃗𝑖 + 3𝑗⃗ , then ‖⃗ 𝑣‖ =

√ √ 12 + 32 = 10. Thus, a unit vector in the same direction is given by

𝑣⃗ 1 1 3 𝑢⃗ = √ = √ (⃗𝑖 + 3𝑗⃗ ) = √ ⃗𝑖 + √ 𝑗⃗ ≈ 0.32⃗𝑖 + 0.95𝑗⃗ . 10 10 10 10

Example 6

Find a unit vector at the point (𝑥, 𝑦, 𝑧) that points directly outward away from the origin.

Solution

The vector from the origin to (𝑥, 𝑦, 𝑧) is the position vector ⃗. 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 Thus, if we put its tail at (𝑥, 𝑦, 𝑧) it will point away from the origin. Its magnitude is √ ‖⃗𝑟 ‖ = 𝑥2 + 𝑦2 + 𝑧2 , so a unit vector pointing in the same direction is

⃗ 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 𝑦 𝑟⃗ 𝑥 𝑧 ⃗. ⃗𝑖 + √ =√ =√ 𝑘 𝑗⃗ + √ 2 2 2 2 2 2 2 2 2 2 2 2 ‖⃗𝑟 ‖ 𝑥 +𝑦 +𝑧 𝑥 +𝑦 +𝑧 𝑥 +𝑦 +𝑧 𝑥 +𝑦 +𝑧

Exercises and Problems for Section 13.1 Online Resource: Additional Problems for Section 13.1 EXERCISES

In Exercises 1–6, resolve the vectors into components. 1.

2.

𝑦 3

𝑦 3

𝑑⃗

2

2

𝑏⃗ 𝑎⃗

⃗ 1 𝑤

1 𝑏⃗

𝑐⃗

𝑎⃗

𝑥 𝑥 −2

−1 𝑣⃗

1 −1 −2

2

1

3

2

3

4

𝑒⃗ −1

3. A vector starting at the point 𝑄 = (4, 6) and ending at the point 𝑃 = (1, 2). 4. A vector starting at the point 𝑃 = (1, 2) and ending at the point 𝑄 = (4, 6).

13.1 DISPLACEMENT VECTORS 𝑧

6.



1

𝑒⃗

For Exercises 20–25, perform the indicated operations on the following vectors:

𝑓⃗ 𝑣⃗

𝑑⃗ 𝑏⃗



𝑦

𝑥

2

𝑦 1







𝑐⃗



1

⃗, 𝑎⃗ = 2𝑗⃗ + 𝑘





𝑢⃗

2

3



𝑥

For Exercises 7–14, perform the indicated computation. 7. (4⃗𝑖 + 2𝑗⃗ ) − (3⃗𝑖 − 𝑗⃗ ) 8. (⃗𝑖 + 2𝑗⃗ ) + (−3)(2⃗𝑖 + 𝑗⃗ ) ⃗) 9. −4(⃗𝑖 − 2𝑗⃗ ) − 0.5(⃗𝑖 − 𝑘 ⃗ ⃗) ⃗ ⃗ 10. 2(0.45𝑖 − 0.9𝑗 − 0.01𝑘 ) − 0.5(1.2⃗𝑖 − 0.1𝑘 ⃗ ) − (6⃗𝑖 + 8𝑗⃗ − 𝑘 ⃗) 11. (3⃗𝑖 − 4𝑗⃗ + 2𝑘 ⃗ ) − 2(5⃗𝑖 + 𝑗⃗ − 2𝑘 ⃗) 12. (4⃗𝑖 − 3𝑗⃗ + 7𝑘 ⃗ ) + (0.3⃗𝑖 + 0.3𝑘 ⃗) 13. (0.6⃗𝑖 + 0.2𝑗⃗ − 𝑘 14.

1 ⃗ (2𝑖 2

⃗ ) + 3(⃗𝑖 − − 𝑗⃗ + 3𝑘

1⃗ 𝑗 6

+

1⃗ 𝑘 2

)

In Exercises 15–19, find the length of the vectors. ⃗ 15. 𝑣⃗ = ⃗𝑖 − 𝑗⃗ + 2𝑘

⃗ 19. 𝑣⃗ = 1.2⃗𝑖 − 3.6𝑗⃗ + 4.1𝑘





𝑎⃗

⃗ 18. 𝑣⃗ = 7.2⃗𝑖 − 1.5𝑗⃗ + 2.1𝑘



𝑧

5.

709

⃗ 16. 𝑧⃗ = ⃗𝑖 − 3𝑗⃗ − 𝑘

⃗ 17. 𝑣⃗ = ⃗𝑖 − 𝑗⃗ + 3𝑘

⃗, 𝑏⃗ = −3⃗𝑖 + 5𝑗⃗ + 4𝑘

𝑥⃗ = −2⃗𝑖 + 9𝑗⃗ ,

𝑦⃗ = 4⃗𝑖 − 7𝑗⃗ ,

𝑐⃗ = ⃗𝑖 + 6𝑗⃗ ,

⃗. 𝑧⃗ = ⃗𝑖 − 3𝑗⃗ − 𝑘

20. 4⃗ 𝑧

21. 5𝑎⃗ + 2𝑏⃗

23. 2⃗ 𝑐 + 𝑥⃗

24. 2𝑎⃗ + 7𝑏⃗ − 5⃗ 𝑧 25. ‖𝑦⃗ − 𝑥⃗ ‖

22. 𝑎⃗ + 𝑧⃗

26. (a) Draw the position vector for 𝑣⃗ = 5⃗𝑖 − 7𝑗⃗ . (b) What is ‖𝑣⃗ ‖? (c) Find the angle between 𝑣⃗ and the positive 𝑥-axis.

⃗. 27. Find the unit vector in the direction of 0.06⃗𝑖 − 0.08𝑘 28. Find the unit vector in the opposite direction to ⃗𝑖 − 𝑗⃗ + ⃗. 𝑘 29. √ Find a unit vector in the opposite direction to 2⃗𝑖 − 𝑗⃗ − ⃗. 11𝑘

30. Find a vector with length 2 that points in the same di⃗. rection as ⃗𝑖 − 𝑗⃗ + 2𝑘

PROBLEMS 31. Find the value(s) of 𝑎 making 𝑣⃗ = 5𝑎⃗𝑖 − 3𝑗⃗ parallel to ⃗ = 𝑎2 ⃗𝑖 + 6𝑗⃗ . 𝑤

(b) What is the relation between 𝑢⃗ and 𝑣⃗ ? Why was this to be expected?

32. (a) For 𝑎 = 1, 2, and 3, draw position vectors for ⃗ = 5⃗𝑖 − 𝑎2 𝑗⃗ (i) 𝑣⃗ = 𝑎2 𝑖⃗ + 6𝑗⃗ (ii) 𝑤

𝑦

(b) Explain why there is no value of 𝑎 that makes 𝑣⃗ ⃗ parallel. and 𝑤

3 𝐵

35. Resolve the following vectors into components: (a) The vector in 2-space of length 2 pointing up and to the right at an angle of 𝜋∕4 with the 𝑥-axis. (b) The vector in 3-space of length 1 lying in the 𝑥𝑧plane pointing upward at an angle of 𝜋∕6 with the positive 𝑥-axis. 36. (a) From Figure 13.16, read off the coordinates of the five points, 𝐴, 𝐵, 𝐶, 𝐷, 𝐸, and thus resolve into components the following two vectors: 𝑢⃗ = ⃖⃖⃖⃖⃖⃗ ⃖⃖⃖⃖⃖⃖ ⃗ 𝑣⃗ = (2.5)𝐵𝐴−(−0.8) ⃖⃖⃖⃖⃖⃗ ⃖⃖⃖⃖⃖⃖ (2.5)𝐴𝐵+(−0.8) 𝐶𝐷, 𝐶𝐷⃗

𝐸

2

33. (a) Find a unit vector from the point 𝑃 = (1, 2) and toward the point 𝑄 = (4, 6). (b) Find a vector of length 10 pointing in the same direction. 34. If north is the direction of the positive 𝑦-axis and east is the direction of the positive 𝑥-axis, give the unit vector pointing northwest.

𝐷

4

1 𝐶

𝐴 1

2

3

4

5

6

𝑥

7

Figure 13.16 37. Find the components of a vector 𝑝⃗ that has the same di⃖⃖⃖⃖⃖⃗ in Figure 13.16 and whose length equals rection as 𝐸𝐴 two units. 38. For each of the four statements below, answer the following questions: Does the statement make sense? If yes, is it true for all possible choices of 𝑎⃗ and 𝑏⃗ ? If no, why not? (a) (b) (c) (d)

𝑎⃗ + 𝑏⃗ = 𝑏⃗ + 𝑎⃗ 𝑎⃗ + ‖𝑏⃗ ‖ = ‖𝑎⃗ + 𝑏⃗ ‖ ‖𝑏⃗ + 𝑎⃗ ‖ = ‖𝑎⃗ + 𝑏⃗ ‖ ‖𝑎⃗ + 𝑏⃗ ‖ = ‖𝑎⃗ ‖ + ‖𝑏⃗ ‖.

710

Chapter 13 A FUNDAMENTAL TOOL: VECTORS

39. For each condition, find unit vectors 𝑎⃗ and 𝑏⃗ or explain why no such vectors exist. (a) ‖𝑎⃗ + 𝑏⃗ ‖ = 0 (c) ‖𝑎⃗ + 𝑏⃗ ‖ = 2

(b) ‖𝑎⃗ + 𝑏⃗ ‖ = 1 (d) ‖𝑎⃗ + 𝑏⃗ ‖ = 3

40. Two adjacent sides of a regular hexagon are given as the vectors 𝑢⃗ and 𝑣⃗ in Figure 13.17. Label the remaining sides in terms of 𝑢⃗ and 𝑣⃗ .

⃗ = 43. Find all unit vectors 𝑣⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ parallel to 𝑤 𝑦⃗𝑖 + 𝑥𝑗⃗ . 44. Find all vectors 𝑣⃗ in 2 dimensions having ‖𝑣⃗ ‖ = 5 such that the ⃗𝑖 -component of 𝑣⃗ is 3⃗𝑖 .

45. (a) Find the point on the 𝑥-axis closest to the point (𝑎, 𝑏, 𝑐). (b) Find a unit vector that points from the point you found in part (a) toward (𝑎, 𝑏, 𝑐).

46. Figure 13.18 shows a molecule with four atoms at 𝑂, 𝐴, 𝐵 and 𝐶. Check that every atom in the molecule is 2 units away from every other atom. 𝑣⃗ 𝑢⃗

Figure 13.17

𝑧 √ √ 𝐶(1, 1∕ 3, 2 2∕3)

41. For what values of 𝑡 are the following pairs of vectors parallel? 2 𝑡 3

⃗, ⃗ (a) 2⃗𝑖 + (𝑡 + + 1)𝑗⃗ + 𝑡𝑘 6⃗𝑖 + 8𝑗⃗ + 3𝑘 ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ (b) 𝑡𝑖 + 𝑗 + (𝑡 − 1)𝑘 , 2𝑖 − 4𝑗 + 𝑘 ⃗, ⃗. (c) 2𝑡⃗𝑖 + 𝑡𝑗⃗ + 𝑡𝑘 6⃗𝑖 + 3𝑗⃗ + 3𝑘 2

0 𝐴(2, 0, 0) 𝑥

Figure 13.18

𝑦 √ 𝐵(1, 3, 0)

42. Show that the unit vector 𝑣⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ is not parallel ⃗ = 𝑦⃗𝑖 − 𝑥𝑗⃗ for any choice of 𝑥 and 𝑦. to 𝑤

Strengthen Your Understanding In Problems 47–50, explain what is wrong with the statement. 47. If ‖⃗𝑢 ‖ = 1 and ‖𝑣⃗ ‖ > 0, then ‖⃗𝑢 + 𝑣⃗ ‖ ≥ 1.

48. The vector 𝑐 𝑢⃗ has the same direction as 𝑢⃗ .

49. ‖𝑣⃗ − 𝑢⃗ ‖ is the length of the shorter of the two diagonals of the parallelogram determined by 𝑢⃗ and 𝑣⃗ .

⃗ , if 𝑢⃗ + 𝑤 ⃗ = 𝑢⃗ then it 50. Given three vectors 𝑢⃗ , 𝑣⃗ , and 𝑤 ⃗ ≠ 𝑣⃗ . is possible for 𝑣⃗ + 𝑤 In Problems 51–53, give an example of: ⃗ -component 51. A vector 𝑣⃗ of length 2 with a positive 𝑘 and lying on a plane parallel to the 𝑦𝑧-plane. 52. Two unit vectors 𝑢⃗ and 𝑣⃗ for which 𝑣⃗ − 𝑢⃗ is also a unit vector. ⃗ = 53. Two vectors 𝑢⃗ and 𝑣⃗ that have difference vector 𝑤 2⃗𝑖 + 3𝑗⃗ . Are the statements in Problems 54–63 true or false? Give reasons for your answer.

13.2

54. There is exactly one unit vector parallel to a given nonzero vector 𝑣⃗ . 1 −1 2 ⃗ 55. The vector √ 𝑖⃗ + √ 𝑗⃗ + √ 𝑘 is a unit vector. 3 3 3

56. The length of the vector 2𝑣⃗ is twice the length of the vector 𝑣⃗ . ⃗ are any two vectors, then ‖𝑣⃗ + 𝑤 ⃗‖ = 57. If 𝑣⃗ and 𝑤 ⃗ ‖. ‖𝑣⃗ ‖ + ‖𝑤 ⃗ are any two vectors, then ‖𝑣⃗ − 𝑤 ⃗‖ = 58. If 𝑣⃗ and 𝑤 ⃗ ‖. ‖𝑣⃗ ‖ − ‖𝑤 ⃗ and ⃗𝑖 − 2𝑗⃗ + 𝑘 ⃗ are parallel. 59. The vectors 2⃗𝑖 − 𝑗⃗ + 𝑘

60. The vector 𝑢⃗ + 𝑣⃗ is always larger in magnitude than both 𝑢⃗ and 𝑣⃗ . 61. For any scalar 𝑐 and vector 𝑣⃗ we have ‖𝑐 𝑣⃗ ‖ = 𝑐‖𝑣⃗ ‖.

62. The displacement vector from (1, 1, 1) to (1, 2, 3) is ⃗. −𝑗⃗ − 2𝑘 63. The displacement vector from (𝑎, 𝑏) to (𝑐, 𝑑) is the same as the displacement vector from (𝑐, 𝑑) to (𝑎, 𝑏).

VECTORS IN GENERAL Besides displacement, there are many quantities that have both magnitude and direction and are added and multiplied by scalars in the same way as displacements. Any such quantity is called a vector and is represented by an arrow in the same manner we represent displacements. The length of the arrow is the magnitude of the vector, and the direction of the arrow is the direction of the vector.

13.2 VECTORS IN GENERAL

711

Velocity Versus Speed The speed of a moving body tells us how fast it is moving, say 80 km/hr. The speed is just a number; it is therefore a scalar. The velocity, on the other hand, tells us both how fast the body is moving and the direction of motion; it is a vector. For instance, if a car is heading northeast at 80 km/hr, then its velocity is a vector of length 80 pointing northeast. The velocity vector of a moving object is a vector whose magnitude is the speed of the object and whose direction is the direction of its motion. The velocity vector is the displacement vector if the object moves at constant velocity for one unit of time. Example 1

A car is traveling north at a speed of 100 km/hr, while a plane above is flying horizontally southwest at a speed of 500 km/hr. Draw the velocity vectors of the car and the plane.

Solution

Figure 13.19 shows the velocity vectors. The plane’s velocity vector is five times as long as the car’s, because its speed is five times as great. Velocity vector of car



✻ North

Velocity vector of plane



Figure 13.19: Velocity vector of the car is 100 km/hr north and of the plane is 500 km/hr southwest

The next example illustrates that the velocity vectors for two motions add to give the velocity vector for the combined motion, just as displacements do. Example 2

A riverboat is moving with velocity 𝑣⃗ and a speed of 8 km/hr relative to the water. In addition, the river has a current 𝑐⃗ and a speed of 1 km/hr. (See Figure 13.20.) What is the physical significance of the vector 𝑣⃗ + 𝑐⃗ ? 𝑐⃗ = Velocity of current ‖⃗ 𝑐 ‖ = 1 km/hr

𝑣⃗ + 𝑐⃗ 𝑣⃗ = Velocity relative to water ‖𝑣⃗ ‖ = 8 km/hr

Figure 13.20: Boat’s velocity relative to the river bed is the sum 𝑣⃗ + 𝑐⃗

Solution

The vector 𝑣⃗ shows how the boat is moving relative to the water, while 𝑐⃗ shows how the water is moving relative to the riverbed. During an hour, imagine that the boat first moves 8 km relative to the water, which remains still; this displacement is represented by 𝑣⃗ . Then imagine the water moving 1 km while the boat remains stationary relative to the water; this displacement is represented by 𝑐⃗ . The combined displacement is represented by 𝑣⃗ + 𝑐⃗ . Thus, the vector 𝑣⃗ + 𝑐⃗ is the velocity of the boat relative to the riverbed. Note that the effective speed of the boat is not necessarily 9 km/hr unless the boat is moving in the direction of the current. Although we add the velocity vectors, we do not necessarily add their lengths. Scalar multiplication also makes sense for velocity vectors. For example, if 𝑣⃗ is a velocity vector, then −2𝑣⃗ represents a velocity of twice the magnitude in the opposite direction.

712

Chapter 13 A FUNDAMENTAL TOOL: VECTORS

Example 3

A ball is moving with velocity 𝑣⃗ when it hits a wall at a right angle and bounces straight back, with its speed reduced by 20%. Express its new velocity in terms of the old one.

Solution

The new velocity is −0.8𝑣⃗ , where the negative sign expresses the fact that the new velocity is in the direction opposite to the old. We can represent velocity vectors in components in the same way we did on page 707.

Example 4

Represent the velocity vectors of the car and the plane in Example 1 using components. Take north to be the positive 𝑦-axis, east to be the positive 𝑥-axis, and upward to be the positive 𝑧-axis.

Solution

The car is traveling north at 100 km/hr, so the 𝑦-component of its velocity is 100𝑗⃗ and the 𝑥⃗ . So we have component is 0⃗𝑖 . Since it is traveling horizontally, the 𝑧-component is 0𝑘 ⃗ = 100𝑗⃗ . Velocity of car = 0⃗𝑖 + 100𝑗⃗ + 0𝑘 ⃗ component equal to zero. Since it is traveling southwest, its ⃗𝑖 The plane’s velocity vector also has 𝑘 and 𝑗⃗ components have negative coefficients (north √ and east are positive). Since the plane is traveling at 500 km/hr, in one hour it is displaced 500∕ 2 ≈ 354 km to the west and 354 km to the south. (See Figure 13.21.) Thus,

0 50

✛ 45◦ √ ✛ 500∕ 2

✻ North



Distance traveled by the plane in one hour

✻ ✲ 100 ❄ ✻



Distance traveled by the car in one hour

√ 500∕ 2 ≈ 354





Figure 13.21: Distance traveled by the plane and car in one hour

Velocity of plane = −(500 cos 45◦)⃗𝑖 − (500 sin 45◦ )𝑗⃗ ≈ −354⃗𝑖 − 354𝑗⃗ . ⃗ Of course, if the car were climbing a hill or if the plane were descending for a landing, then the 𝑘 component would not be zero.

Acceleration Another example of a vector quantity is acceleration. Acceleration, like velocity, is specified by both a magnitude and a direction — for example, the acceleration due to gravity is 9.81 m/sec2 vertically downward.

Force Force is another example of a vector quantity. Suppose you push on an open door. The result depends both on how hard you push and in what direction. Thus, to specify a force we must give its magnitude (or strength) and the direction in which it is acting. For example, the gravitational force exerted on an object by the earth is a vector pointing from the object toward the center of the earth; its magnitude is the strength of the gravitational force.

13.2 VECTORS IN GENERAL

Example 5

713

The earth travels around the sun in an ellipse. The gravitational force on the earth and the velocity of the earth are governed by the following laws: Newton’s Law of Gravitation: The gravitational attraction, 𝐹⃗ , of a mass 𝑚1 on a mass 𝑚2 at a distance 𝑟 has magnitude ||𝐹⃗ || = 𝐺𝑚1 𝑚2 ∕𝑟2 , where 𝐺 is a constant, and is directed from 𝑚2 toward 𝑚1 . Kepler’s Second Law: The line joining a planet to the sun sweeps out equal areas in equal times.

(a) Sketch vectors representing the gravitational force of the sun on the earth at two different positions in the earth’s orbit. (b) Sketch the velocity vector of the earth at two points in its orbit. Solution

(a) Figure 13.22 shows the earth orbiting the sun. Note that the gravitational force vector always points toward the sun and is larger when the earth is closer to the sun because of the 𝑟2 term in the denominator. (In fact, the real orbit looks much more like a circle than we have shown here.) (b) The velocity vector points in the direction of motion of the earth. Thus, the velocity vector is tangent to the ellipse. See Figure 13.23. Furthermore, the velocity vector is longer at points of the orbit where the planet is moving quickly, because the magnitude of the velocity vector is the speed. Kepler’s Second Law enables us to determine when the earth is moving quickly and when it is moving slowly. Over a fixed period of time, say one month, the line joining the earth to the sun sweeps out a sector having a certain area. Figure 13.23 shows two sectors swept out in two different one-month time-intervals. Kepler’s law says that the areas of the two sectors are the same. Thus, the earth must move farther in a month when it is close to the sun than when it is far from the sun. Therefore, the earth moves faster when it is closer to the sun and slower when it is farther away. Earth’s orbit Earth’s orbit Earth

Force, 𝐹⃗

Earth Sun

Velocity, 𝑣⃗

Sun

Force, 𝐹⃗ Velocity, 𝑣⃗ Earth

Earth

Figure 13.22: Gravitational force, 𝐹⃗ , exerted by the sun on the earth: Greater magnitude closer to sun

Figure 13.23: The velocity vector, 𝑣⃗ , of the earth: Greater magnitude closer to the sun

Properties of Addition and Scalar Multiplication In general, vectors add, subtract, and are multiplied by scalars in the same way as displacement ⃗ and any scalars 𝛼 and 𝛽, we have the following properties: vectors. Thus, for any vectors 𝑢⃗ , 𝑣⃗ , and 𝑤

Commutativity ⃗ =𝑤 ⃗ + 𝑣⃗ 1. 𝑣⃗ + 𝑤 Distributivity 4. (𝛼 + 𝛽)𝑣⃗ = 𝛼 𝑣⃗ + 𝛽 𝑣⃗ ⃗ ) = 𝛼 𝑣⃗ + 𝛼 𝑤 ⃗ 5. 𝛼(𝑣⃗ + 𝑤

Associativity ⃗ = 𝑢⃗ + (𝑣⃗ + 𝑤 ⃗) 2. (⃗𝑢 + 𝑣⃗ ) + 𝑤 3. 𝛼(𝛽 𝑣⃗ ) = (𝛼𝛽)𝑣⃗ Identity 6. 1𝑣⃗ = 𝑣⃗ 8. 𝑣⃗ + 0⃗ = 𝑣⃗ ⃗ ⃗ + (−1)𝑣⃗ = 𝑤 ⃗ − 𝑣⃗ 7. 0𝑣⃗ = 0 9. 𝑤

Problems 28–35 at the end of this section ask for a justification of these results in terms of displacement vectors.

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Chapter 13 A FUNDAMENTAL TOOL: VECTORS

Using Components Example 6

A plane, heading due east at an airspeed of 600 km/hr, experiences a wind of 50 km/hr blowing toward the northeast. Find the plane’s direction and ground speed.

Solution

We choose a coordinate system with the 𝑥-axis pointing east and the 𝑦-axis pointing north. See Figure 13.24. 𝑦 𝜃



⃗ = Wind velocity 𝑤

⃗ 𝑣⃗ + 𝑤 45◦

𝑥

𝑣⃗ = Velocity relative to air

⃗ Figure 13.24: Plane’s velocity relative to the ground is the sum 𝑣⃗ + 𝑤 The airspeed tells us the speed of the plane relative to still air. Thus, the plane is moving due ⃗. east with velocity 𝑣⃗ = 600⃗𝑖 relative to still air. In addition, the air is moving with a velocity 𝑤 ⃗ in components, we have Writing 𝑤

⃗ = (50 cos 45◦ )⃗𝑖 + (50 sin 45◦ )𝑗⃗ = 35.4⃗𝑖 + 35.4𝑗⃗ . 𝑤 ⃗ represents the displacement of the plane in one hour relative to the ground. The vector 𝑣⃗ + 𝑤 ⃗ is the velocity of the plane relative to the ground. In components, we have Therefore, 𝑣⃗ + 𝑤 ( ) ⃗ = 600⃗𝑖 + 35.4⃗𝑖 + 35.4𝑗⃗ = 635.4⃗𝑖 + 35.4𝑗⃗ . 𝑣⃗ + 𝑤 The direction of the plane’s motion relative to the ground is given by the angle 𝜃 in Figure 13.24, where 35.4 tan 𝜃 = 635.4 so ) ( 35.4 = 3.2◦. 𝜃 = arctan 635.4 The ground speed is the speed of the plane relative to the ground, so √ ⃗ || = 635.42 + 35.42 = 636.4 km/hr. Ground speed = ||𝑣⃗ + 𝑤 Thus, the speed of the plane relative to the ground has been increased slightly by the wind. (This is as we would expect, as the wind has a positive component in the direction in which the plane is traveling.) The angle 𝜃 shows how far the plane is blown off course by the wind.

Vectors in 𝒏 Dimensions Using the alternative notation 𝑣⃗ = (𝑣1 , 𝑣2 , 𝑣3 ) for a vector in 3-space, we can define a vector in 𝑛 dimensions as a string of 𝑛 numbers. Thus, a vector in 𝑛 dimensions can be written as 𝑐⃗ = (𝑐1 , 𝑐2 , … , 𝑐𝑛 ). Addition and scalar multiplication are defined by the formulas ⃗ = (𝑣1 , 𝑣2 , … , 𝑣𝑛 ) + (𝑤1 , 𝑤2 , … , 𝑤𝑛 ) = (𝑣1 + 𝑤1 , 𝑣2 + 𝑤2 , … , 𝑣𝑛 + 𝑤𝑛 ) 𝑣⃗ + 𝑤 and 𝜆⃗ 𝑣 = 𝜆(𝑣1 , 𝑣2 , … , 𝑣𝑛 ) = (𝜆𝑣1 , 𝜆𝑣2 , … , 𝜆𝑣𝑛 ).

13.2 VECTORS IN GENERAL

715

Why Do We Want Vectors in 𝒏 Dimensions? Vectors in two and three dimensions can be used to model displacement, velocities, or forces. But what about vectors in 𝑛 dimensions? There is another interpretation of 3-dimensional vectors (or 3-vectors) that is useful: they can be thought of as listing three different quantities—for example, the displacements parallel to the 𝑥-, 𝑦-, and 𝑧-axes. Similarly, the 𝑛-vector 𝑐⃗ = (𝑐1 , 𝑐2 , … , 𝑐𝑛 ) can be thought of as a way of keeping 𝑛 different quantities organized. For example, a population ⃗ shows the number of children and adults in a population: vector 𝑁 ⃗ = (Number of children, Number of adults), 𝑁 or, if we are interested in a more detailed breakdown of ages, we might give the number in each ten-year age bracket in the population (up to age 110) in the form ⃗ = (𝑁1 , 𝑁2 , 𝑁3 , 𝑁4 , … , 𝑁10 , 𝑁11 ), 𝑁 where 𝑁1 is the population aged 0–9, and 𝑁2 is the population aged 10–19, and so on. A consumption vector 𝑞⃗ = (𝑞1 , 𝑞2 , … , 𝑞𝑛 ) shows the quantities 𝑞1 , 𝑞2 , …, 𝑞𝑛 consumed of each of 𝑛 different goods. A price vector 𝑝⃗ = (𝑝1 , 𝑝2 , … , 𝑝𝑛 ) contains the prices of 𝑛 different items. In 1907, Hermann Minkowski used vectors with four components when he introduced spacetime coordinates, whereby each event is assigned a vector position 𝑣⃗ with four coordinates, three for its position in space and one for time: 𝑣⃗ = (𝑥, 𝑦, 𝑧, 𝑡). Example 7

Suppose the vector 𝐼⃗ represents the number of copies, in thousands, made by each of four copy centers in the month of December and 𝐽⃗ represents the number of copies made at the same four copy centers during the previous eleven months (the “year-to-date”). If 𝐼⃗ = (25, 211, 818, 642), and 𝐽⃗ = (331, 3227, 1377, 2570), compute 𝐼⃗ + 𝐽⃗ . What does this sum represent?

Solution

The sum is 𝐼⃗ + 𝐽⃗ = (25 + 331, 211 + 3227, 818 + 1377, 642 + 2570) = (356, 3438, 2195, 3212). Each term in 𝐼⃗ + 𝐽⃗ represents the sum of the number of copies made in December plus those in the previous eleven months, that is, the total number of copies made during the entire year at that particular copy center.

Example 8

The price vector 𝑝⃗ = (𝑝1 , 𝑝2 , 𝑝3 ) represents the prices in dollars of three goods. Write a vector that gives the prices of the same goods in cents.

Solution

The prices in cents are 100𝑝1 , 100𝑝2, and 100𝑝3 respectively, so the new price vector is (100𝑝1, 100𝑝2, 100𝑝3) = 100𝑝⃗ .

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Chapter 13 A FUNDAMENTAL TOOL: VECTORS

Exercises and Problems for Section 13.2 Online Resource: Additional Problems for Section 13.2 EXERCISES

In Exercises 1–5, say whether the given quantity is a vector or a scalar. 1. The population of the US. 2. The distance from Seattle to St. Louis.

8. A car is traveling at a speed of 50 km/hr. The positive 𝑦-axis is north and the positive 𝑥-axis is east. Resolve the car’s velocity vector (in 2-space) into components if the car is traveling in each of the following directions:

3. The temperature at a point on the earth’s surface.

(a)

East

(b) South

4. The magnetic field at a point on the earth’s surface. 5. The populations of each of the 50 states.

(c)

Southeast

(d) Northwest.

6. Give the components of the velocity vector for wind blowing at 10 km/hr toward the southeast. (Assume north is in the positive 𝑦-direction.) 7. Give the components of the velocity vector of a boat that is moving at 40 km/hr in a direction 20◦ south of west. (Assume north is in the positive 𝑦-direction.)

9. Which is traveling faster, a car whose velocity vector is 21⃗𝑖 +35𝑗⃗ or a car whose velocity vector is 40⃗𝑖 , assuming that the units are the same for both directions? 10. What angle does a force of 𝐹⃗ = 15⃗𝑖 + 18𝑗⃗ make with the 𝑥-axis?

PROBLEMS 11. The velocity of the current in a river is 𝑐⃗ = 0.6⃗𝑖 + 0.8𝑗⃗ km/hr. A boat moves relative to the water with velocity 𝑣⃗ = 8⃗𝑖 km/hr. (a) What is the speed of the boat relative to the riverbed? (b) What angle does the velocity of the boat relative to the riverbed make with the vector 𝑣⃗ ? What does this angle tell us in practical terms? 12. The current in Problem 11 is twice as fast and in the opposite direction. What is the speed of the boat with respect to the riverbed? 13. A boat is heading due east at 25 km/hr (relative to the water). The current is moving toward the southwest at 10 km/hr. (a) Give the vector representing the actual movement of the boat. (b) How fast is the boat going, relative to the ground? (c) By what angle does the current push the boat off of its due east course? 14. A truck is traveling due north at 30 km/hr approaching a crossroad. On a perpendicular road a police car is traveling west toward the intersection at 40 km/hr. Both vehicles will reach the crossroad in exactly one hour. Find the vector currently representing the displacement from the police car to the truck. 15. An airplane heads northeast at an airspeed of 700 km/hr, but there is a wind blowing from the west at 60 km/hr. In what direction does the plane end up flying? What is its speed relative to the ground? 16. Two forces, represented by the vectors 𝐹⃗ 1 = 8⃗𝑖 − 6𝑗⃗ and 𝐹⃗ 2 = 3⃗𝑖 + 2𝑗⃗ , are acting on an object. Give a vector representing the force that must be applied to the object if it is to remain stationary.

17. An airplane is flying at an airspeed of 500 km/hr in a wind blowing at 60 km/hr toward the southeast. In what direction should the plane head to end up going due east? What is the airplane’s speed relative to the ground? 18. The current in a river is pushing a boat in direction 25◦ north of east with a speed of 12 km∕hr. The wind is pushing the same boat in a direction 80◦ south of east with a speed of 7 km∕hr. Find the velocity vector of the boat’s engine (relative to the water) if the boat actually moves due east at a speed of 40 km∕hr relative to the ground. 19. A large ship is being towed by two tugs. The larger tug exerts a force which is 25% greater than the smaller tug and at an angle of 30 degrees north of east. Which direction must the smaller tug pull to ensure that the ship travels due east? 20. An object 𝑃 is pulled by a force 𝐹⃗1 of magnitude 15 lb at an angle of 20 degrees north of east. Give the components of a force 𝐹⃗2 of magnitude 20 lb to ensure that 𝑃 moves due east. 21. An object is to be moved vertically upward by a crane. As the crane cannot get directly above the object, three ropes are attached to guide the object. One rope is pulled parallel to the ground with a force of 100 newtons in a direction 30◦ north of east. The second rope is pulled parallel to the ground with a force of 70 newtons in a direction 80◦ south of east. If the crane is attached to the third rope and can pull with a total force of 3000 newtons, find the force vector for the crane. What is the resulting (total) force on the object? (Assume vector ⃗𝑖 ⃗ points points east, vector 𝑗⃗ points north, and vector 𝑘 vertically up.)

13.2 VECTORS IN GENERAL

717

22. The earth is at the origin, the moon is at the point (384, 0), and a spaceship is at (280, 90), where distance is in thousands of kilometers.

(b) What is its velocity vector 30 seconds after it passes the point (1, 0)? Does your answer change if the object is moving clockwise? Explain.

(a) What is the displacement vector of the moon relative to the earth? Of the spaceship relative to the earth? Of the spaceship relative to the moon? (b) How far is the spaceship from the earth? From the moon? (c) The gravitational force on the spaceship from the earth is 461 newtons and from the moon is 26 newtons. What is the resulting force?

27. An object is attached by a string to a fixed point and rotates 30 times per minute in a horizontal plane. Show that the speed of the object is constant but the velocity is not. What does this imply about the acceleration?

23. A particle moving with speed 𝑣 hits a barrier at an angle of 60◦ and bounces off at an angle of 60◦ in the opposite direction with speed reduced by 20 percent. See Figure 13.25. Find the velocity vector of the object after impact. 𝑦

In Problems 28–35, use the geometric definition of addition and scalar multiplication to explain each of the properties. ⃗ + 𝑣⃗ = 𝑣⃗ + 𝑤 ⃗ 28. 𝑤

29. (𝛼 + 𝛽)𝑣⃗ = 𝛼 𝑣⃗ + 𝛽 𝑣⃗

⃗ ) = 𝛼 𝑣⃗ + 𝛼 𝑤 ⃗ 30. 𝛼(𝑣⃗ + 𝑤

31. 𝛼(𝛽 𝑣⃗ ) = (𝛼𝛽)𝑣⃗

32. 𝑣⃗ + 0⃗ = 𝑣⃗

33. 1𝑣⃗ = 𝑣⃗

⃗ = 𝑣⃗ − 𝑤 ⃗ 34. 𝑣⃗ + (−1)𝑤 ⃗ = 𝑢⃗ + (𝑣⃗ + 𝑤 ⃗) 35. (⃗𝑢 + 𝑣⃗ ) + 𝑤 Before impact After impact

60◦

60◦

𝑥

Figure 13.25 24. There are five students in a class. Their scores on the midterm (out of 100) are given by the vector 𝑣⃗ = (73, 80, 91, 65, 84). Their scores on the final (out of 100) ⃗ = (82, 79, 88, 70, 92). If the final counts are given by 𝑤 twice as much as the midterm, find a vector giving the total scores (as a percentage) of the students. 25. The price vector of beans, rice, and tofu is (1.6, 1.28, 2.60) in dollars per pound. Express it in dollars per ounce. 26. An object is moving counterclockwise at a constant speed around the circle 𝑥2 + 𝑦2 = 1, where 𝑥 and 𝑦 are measured in meters. It completes one revolution every minute. (a) What is its speed?

36. In the game of laser tag, you shoot a harmless laser gun and try to hit a target worn at the waist by other players. Suppose you are standing at the origin of a threedimensional coordinate system and that the 𝑥𝑦-plane is the floor. Suppose that waist-high is 3 feet above floor level and that eye level is 5 feet above the floor. Three of your friends are your opponents. One is standing so that his target is 30 feet along the 𝑥-axis, another lying down so that his target is at the point 𝑥 = 20, 𝑦 = 15, and the third lying in ambush so that his target is at a point 8 feet above the point 𝑥 = 12, 𝑦 = 30. (a) If you aim with your gun at eye level, find the vector from your gun to each of the three targets. (b) If you shoot from waist height, with your gun one foot to the right of the center of your body as you face along the 𝑥-axis, find the vector from your gun to each of the three targets. 37. A car drives northeast downhill on a 5◦ incline at a constant speed of 60 miles per hour. The positive 𝑥-axis points east, the 𝑦-axis north, and the 𝑧-axis up. Resolve the car’s velocity into components.

Strengthen Your Understanding In Problems 38–39, explain what is wrong with the statement.

force vector 𝑅⃗ with a positive ⃗𝑖-component and a negative 𝑗⃗-component.

⃗ 38. Two vectors in 3-space that have equal 𝑘-components and the same magnitude must be the same vector.

41. Nonzero vectors 𝑢⃗ and 𝑣⃗ such that ‖⃗𝑢 + 𝑣⃗ ‖ = ‖⃗𝑢 ‖ + ‖𝑣⃗ ‖.

39. A vector 𝑣⃗ in the plane whose 𝑖⃗-component is 0.5 has ⃗ = 2⃗𝑖 . smaller magnitude than the vector 𝑤

In Problems 42–47, is the quantity a vector? Give a reason for your answer.

In Problems 40–41, give an example of:

42. Velocity

43. Speed

44. Force

40. A nonzero vector 𝐹⃗ on the plane that when combined with the force vector 𝐺⃗ = ⃗𝑖 + 𝑗⃗ results in a combined

45. Area

46. Acceleration

47. Volume

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Chapter 13 A FUNDAMENTAL TOOL: VECTORS

13.3

THE DOT PRODUCT We have seen how to add vectors; can we multiply two vectors together? In the next two sections we will see two different ways of doing so: the scalar product (or dot product), which produces a scalar, and the vector product (or cross product), which produces a vector.

Definition of the Dot Product The dot product links geometry and algebra. We already know how to calculate the length of a vector from its components; the dot product gives us a way of computing the angle between two vectors. ⃗ and 𝑤 ⃗ , shown in Figure 13.26, we ⃗ = 𝑤1 ⃗𝑖 + 𝑤2 𝑗⃗ + 𝑤3 𝑘 For any two vectors 𝑣⃗ = 𝑣1 ⃗𝑖 + 𝑣2 𝑗⃗ + 𝑣3 𝑘 define a scalar as follows: ⃗ , are equivalent: The following two definitions of the dot product, or scalar product, 𝑣⃗ ⋅ 𝑤 • Geometric definition ⃗ = ‖⃗ ⃗ and 0 ≤ 𝜃 ≤ 𝜋. 𝑣⃗ ⋅ 𝑤 𝑣 ‖‖𝑤⃗ ‖ cos 𝜃 where 𝜃 is the angle between 𝑣⃗ and 𝑤

• Algebraic definition ⃗ = 𝑣1 𝑤1 + 𝑣2 𝑤2 + 𝑣3 𝑤3 . 𝑣⃗ ⋅ 𝑤 Notice that the dot product of two vectors is a number, not a vector.

⃗ ? The reason is that both definitions are equally Why don’t we give just one definition of 𝑣⃗ ⋅ 𝑤 important; the geometric definition gives us a picture of what the dot product means and the algebraic definition gives us a way of calculating it. How do we know the two definitions are equivalent—that is, they really do define the same thing? First, we observe that the two definitions give the same result in a particular example. Then we show why they are equivalent in general. 𝑦 2 ⃗ 𝑤 ⃗ 𝑤 𝜃

𝜃

𝑣⃗

Figure 13.26: The vectors 𝑣⃗ and ⃗ 𝑤

𝑣⃗ 𝑥 1

2

Figure 13.27: Calculating the dot product of the vectors 𝑣 = ⃗𝑖 ⃗ = 2⃗𝑖 + 2𝑗⃗ geometrically and algebraically gives the and 𝑤 same result

Example 1

⃗ = 2⃗𝑖 + 2𝑗⃗ . Compute 𝑣⃗ ⋅ 𝑤 ⃗ both geometrically and algebraically. Suppose 𝑣⃗ = ⃗𝑖 and 𝑤

Solution

To use the geometric definition, see Figure 13.27. The angle between the vectors is 𝜋∕4, or 45◦ , and the lengths of the vectors are given by √ ⃗ ‖ = 2 2. ‖⃗ 𝑣 ‖ = 1 and ‖𝑤 Thus,

( ) √ 𝜋 ⃗ = ‖⃗ ⃗ ‖ cos 𝜃 = 1 ⋅ 2 2 cos 𝑣⃗ ⋅ 𝑤 𝑣 ‖‖𝑤 = 2. 4 Using the algebraic definition, we get the same result: ⃗ = 1 ⋅ 2 + 0 ⋅ 2 = 2. 𝑣⃗ ⋅ 𝑤

13.3 THE DOT PRODUCT

719

Why the Two Definitions of the Dot Product Give the Same Result In the previous example, the two definitions give the same value for the dot product. To show that the geometric and algebraic definitions of the dot product always give the same result, we must show ⃗ and 𝑤 ⃗ with an angle 𝜃 between ⃗ = 𝑤1 ⃗𝑖 + 𝑤2 𝑗⃗ + 𝑤3 𝑘 that, for any vectors 𝑣⃗ = 𝑣1 ⃗𝑖 + 𝑣2 𝑗⃗ + 𝑣3 𝑘 them: ‖⃗ 𝑣 ‖‖𝑤⃗ ‖ cos 𝜃 = 𝑣1 𝑤1 + 𝑣2 𝑤2 + 𝑣3 𝑤3 .

One method follows; a method that does not use trigonometry is given in Problem 105 (available online). ⃗ form a Using the Law of Cosines. Suppose that 0 < 𝜃 < 𝜋, so that the vectors 𝑣⃗ and 𝑤 triangle. (See Figure 13.28.) By the Law of Cosines, we have ⃗ ‖2 = ‖⃗ ⃗ ‖2 − 2‖⃗ ⃗ ‖ cos 𝜃. ‖⃗ 𝑣 −𝑤 𝑣 ‖2 + ‖𝑤 𝑣 ‖‖𝑤

This result is also true for 𝜃 = 0 and 𝜃 = 𝜋. We calculate the lengths using components: ‖⃗ 𝑣 ‖2 = 𝑣21 + 𝑣22 + 𝑣23

‖𝑤⃗ ‖2 = 𝑤21 + 𝑤22 + 𝑤23

⃗ ‖2 = (𝑣1 − 𝑤1 )2 + (𝑣2 − 𝑤2 )2 + (𝑣3 − 𝑤3 )2 ‖⃗ 𝑣 −𝑤 = 𝑣21 − 2𝑣1 𝑤1 + 𝑤21 + 𝑣22 − 2𝑣2 𝑤2 + 𝑤22 + 𝑣23 − 2𝑣3 𝑤3 + 𝑤23 . Substituting into the Law of Cosines and canceling, we see that −2𝑣1 𝑤1 − 2𝑣2 𝑤2 − 2𝑣3 𝑤3 = −2‖⃗ 𝑣 ‖‖𝑤⃗ ‖ cos 𝜃.

Therefore we have the result we wanted, namely that:

⃗ ‖ cos 𝜃. 𝑣1 𝑤1 + 𝑣2 𝑤2 + 𝑣3 𝑤3 = ‖⃗ 𝑣 ‖‖𝑤 ⃗ 𝑣⃗ − 𝑤

⃗ 𝑤 𝜃 𝑣⃗

⃗ ‖ cos 𝜃 = 𝑣1 𝑤1 + 𝑣2 𝑤2 + 𝑣3 𝑤3 Figure 13.28: Triangle used in the justification of ‖𝑣⃗ ‖‖𝑤

Properties of the Dot Product The following properties of the dot product can be justified using the algebraic definition; see Problem 101. For a geometric interpretation of Property 3, see Problem 103 (both available online). ⃗ and any scalar 𝜆, Properties of the Dot Product. For any vectors 𝑢⃗ , 𝑣⃗ , and 𝑤 ⃗ =𝑤 ⃗ ⋅ 𝑣⃗ 1. 𝑣⃗ ⋅ 𝑤 ⃗ ) = 𝜆(𝑣⃗ ⋅ 𝑤 ⃗ ) = (𝜆⃗ ⃗ 2. 𝑣⃗ ⋅ (𝜆𝑤 𝑣)⋅𝑤 ⃗ ) ⋅ 𝑢⃗ = 𝑣⃗ ⋅ 𝑢⃗ + 𝑤 ⃗ ⋅ 𝑢⃗ 3. (𝑣⃗ + 𝑤

Perpendicularity, Magnitude, and Dot Products Two vectors are perpendicular if the angle between them is 𝜋∕2 or 90◦. Since cos(𝜋∕2) = 0, if 𝑣⃗ ⃗ are perpendicular, then 𝑣⃗ ⋅ 𝑤 ⃗ = 0. Conversely, provided that 𝑣⃗ ⋅ 𝑤 ⃗ = 0, then cos 𝜃 = 0, so and 𝑤 𝜃 = 𝜋∕2 and the vectors are perpendicular. Thus, we have the following result:

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Chapter 13 A FUNDAMENTAL TOOL: VECTORS

⃗ are perpendicular, or orthogonal, if and only if Two nonzero vectors 𝑣⃗ and 𝑤 ⃗ = 0. 𝑣⃗ ⋅ 𝑤 ⃗ = 0, ⃗𝑖 ⋅ 𝑘 ⃗ = 0. For example: ⃗𝑖 ⋅ 𝑗⃗ = 0, 𝑗⃗ ⋅ 𝑘 If we take the dot product of a vector with itself, then 𝜃 = 0 and cos 𝜃 = 1. For any vector 𝑣⃗ : Magnitude and dot product are related as follows: 𝑣⃗ ⋅ 𝑣⃗ = ‖⃗ 𝑣 ‖2 .

⃗ ⋅𝑘 ⃗ = 1. For example: ⃗𝑖 ⋅ ⃗𝑖 = 1, 𝑗⃗ ⋅ 𝑗⃗ = 1, 𝑘

Using the Dot Product Depending on the situation, one definition of the dot product may be more convenient to use than the other. In Example 2, the geometric definition is the only one that can be used because we are not given components. In Example 3, the algebraic definition is used. Example 2

Suppose the vector 𝑏⃗ is fixed and has length 2; the vector 𝑎⃗ is free to rotate and has length 3. What are the maximum and minimum values of the dot product 𝑎⃗ ⋅ 𝑏⃗ as the vector 𝑎⃗ rotates through all possible positions in the plane? What positions of 𝑎⃗ and 𝑏⃗ lead to these values?

Solution

The geometric definition gives 𝑎⃗ ⋅ 𝑏⃗ = ‖𝑎⃗ ‖‖𝑏⃗ ‖ cos 𝜃 = 3 ⋅ 2 cos 𝜃 = 6 cos 𝜃. Thus, the maximum value of 𝑎⃗ ⋅ 𝑏⃗ is 6, and it occurs when cos 𝜃 = 1 so 𝜃 = 0, that is, when 𝑎⃗ and 𝑏⃗ point in the same direction. The minimum value of 𝑎⃗ ⋅ 𝑏⃗ is −6, and it occurs when cos 𝜃 = −1 so 𝜃 = 𝜋, that is, when 𝑎⃗ and 𝑏⃗ point in opposite directions. (See Figure 13.29.) When 𝑎⃗ is in this 𝑎⃗ position, 𝑎⃗ ⋅ 𝑏⃗ = 6

𝑎⃗

When 𝑎⃗ is in this position, 𝑎⃗ ⋅ 𝑏⃗ = 0

𝑏⃗

When 𝑎⃗ is in this position, 𝑎⃗ ⋅ 𝑏⃗ = −6

𝑎⃗ Figure 13.29: Maximum and minimum values of 𝑎⃗ ⋅ 𝑏⃗ obtained from a fixed vector 𝑏⃗ of length 2 and rotating vector 𝑎⃗ of length 3

Example 3

Solution

Which pairs from the following list of 3-dimensional vectors are perpendicular to one another? √ √ √ ⃗ , 𝑣⃗ = ⃗𝑖 + 3 𝑗⃗ , 𝑤 ⃗. ⃗ = 3 ⃗𝑖 + 𝑗⃗ − 𝑘 𝑢⃗ = ⃗𝑖 + 3 𝑘

The geometric definition tells us that two vectors are perpendicular if and only if their dot product is zero. Since the vectors are given in components, we calculate dot products using the algebraic definition:

13.3 THE DOT PRODUCT

721

√ √ √ ⃗ ) ⋅ (⃗𝑖 + 0𝑗⃗ + 3 𝑘 ⃗ ) = 1 ⋅ 1 + 3 ⋅ 0 + 0 ⋅ 3 = 1, 3 𝑗⃗ + 0𝑘 √ √ √ √ √ ⃗ ) ⋅ ( 3 ⃗𝑖 + 𝑗⃗ − 𝑘 ⃗ ) = 1 ⋅ 3 + 3 ⋅ 1 + 0(−1) = 2 3, ⃗ = (⃗𝑖 + 3 𝑗⃗ + 0𝑘 𝑣⃗ ⋅ 𝑤 √ √ √ √ ⃗ ) ⋅ (⃗𝑖 + 0𝑗⃗ + 3 𝑘 ⃗ ) = 3 ⋅ 1 + 1 ⋅ 0 + (−1) ⋅ 3 = 0. ⃗ ⋅ 𝑢⃗ = ( 3 ⃗𝑖 + 𝑗⃗ − 𝑘 𝑤 𝑣⃗ ⋅ 𝑢⃗ = (⃗𝑖 +



⃗ and 𝑢⃗ . So the only two vectors that are perpendicular are 𝑤

Example 4 Solution

⃗ from Example 3. Compute the angle between the vectors 𝑣⃗ and 𝑤 ⃗ 𝑣⃗ ⋅ 𝑤 ⃗ = ‖⃗ We know that 𝑣⃗ ⋅ 𝑤 𝑣 ‖‖𝑤⃗ ‖ cos 𝜃, so cos 𝜃 = . From Example 3, we know that ⃗ ‖⃗ 𝑣 ‖‖ 𝑤 ‖ √ ⃗ = 2 3. This gives: 𝑣⃗ ⋅ 𝑤 √ √ √ 2 3 3 2 3 cos 𝜃 = =√ =√ √ ( ) ) ( √ √ ⃗ ‖⃗ 𝑣 ‖‖𝑤 ‖ 2 2 5 12 + 3 + 02 3 + 12 + (−1)2 (√ ) 3 so 𝜃 = arccos √ = 39.2315◦. 5

Normal Vectors and the Equation of a Plane In Section 12.4 we wrote the equation of a plane given its 𝑥-slope, 𝑦-slope and 𝑧-intercept. Now we write the equation of a plane using a vector 𝑛⃗ and a point 𝑃0 . The key idea is that all the displacement vectors from 𝑃0 that are perpendicular to 𝑛⃗ form a plane. To picture this, imagine a pencil balanced on a table, with other pencils fanned out on the table in different directions. The upright pencil is 𝑛⃗ , its base is 𝑃0 , the other pencils are perpendicular displacement vectors, and the table is the plane. More formally, a normal vector to a plane is a vector that is perpendicular to the plane, that is, it is ⃗ perpendicular to every displacement vector between any two points in the plane. Let 𝑛⃗ = 𝑎⃗𝑖 +𝑏𝑗⃗ +𝑐 𝑘 be a normal vector to the plane, let 𝑃0 = (𝑥0 , 𝑦0 , 𝑧0 ) be a fixed point in the plane, and let 𝑃 = (𝑥, 𝑦, 𝑧) ⃗ ⃗ ⃗ ⃗ be any other point in the plane. Then 𝑃⃖⃖⃖⃖⃖⃖ 0 𝑃 = (𝑥 − 𝑥0 )𝑖 + (𝑦 − 𝑦0 )𝑗 + (𝑧 − 𝑧0 )𝑘 is a vector whose head ⃖⃖⃖⃖⃖⃖ and tail both lie in the plane. (See Figure 13.30.) Thus, the vectors 𝑛⃗ and 𝑃0 𝑃⃗ are perpendicular, so ⃗ ⃗ 𝑛⃗ ⋅𝑃⃖⃖⃖⃖⃖⃖ ⃗ ⋅𝑃⃖⃖⃖⃖⃖⃖ 0 𝑃 = 0. The algebraic definition of the dot product gives 𝑛 0 𝑃 = 𝑎(𝑥−𝑥0 )+𝑏(𝑦−𝑦0 )+𝑐(𝑧−𝑧0 ), so we obtain the following result:

Figure 13.30: Plane with normal 𝑛⃗ and containing a fixed point (𝑥0 , 𝑦0 , 𝑧0 )

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Chapter 13 A FUNDAMENTAL TOOL: VECTORS

⃗ and containing the point The equation of the plane with normal vector 𝑛⃗ = 𝑎⃗𝑖 + 𝑏𝑗⃗ + 𝑐 𝑘 𝑃0 = (𝑥0 , 𝑦0 , 𝑧0 ) is 𝑎(𝑥 − 𝑥0 ) + 𝑏(𝑦 − 𝑦0 ) + 𝑐(𝑧 − 𝑧0 ) = 0. Letting 𝑑 = 𝑎𝑥0 + 𝑏𝑦0 + 𝑐𝑧0 (a constant), we can write the equation of the plane in the form 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑.

Example 5

⃗ and passing through the point (a) Find the equation of the plane perpendicular to 𝑛⃗ = −⃗𝑖 + 3𝑗⃗ + 2𝑘 (1, 0, 4). (b) Find a vector parallel to the plane.

Solution

(a) The equation of the plane is −(𝑥 − 1) + 3(𝑦 − 0) + 2(𝑧 − 4) = 0, which can be written as −𝑥 + 3𝑦 + 2𝑧 = 7. (b) Any vector 𝑣⃗ that is perpendicular to 𝑛 is also parallel to the plane, so we look for any vector satisfying 𝑣⃗ ⋅ 𝑛⃗ = 0; for example, 𝑣⃗ = 3⃗𝑖 + 𝑗⃗ . There are many other possible vectors.

Example 6

Find a normal vector to the plane with equation (a) 𝑥 − 𝑦 + 2𝑧 = 5

Solution

⃗ in a normal vector are the coefficients of 𝑥, 𝑦, and 𝑧 in the (a) Since the coefficients of ⃗𝑖 , 𝑗⃗ , and 𝑘 ⃗. equation of the plane, a normal vector is 𝑛⃗ = ⃗𝑖 − 𝑗⃗ + 2𝑘 (b) Before we can find a normal vector, we rewrite the equation of the plane in the form

(b) 𝑧 = 0.5𝑥 + 1.2𝑦.

0.5𝑥 + 1.2𝑦 − 𝑧 = 0. ⃗. Thus, a normal vector is 𝑛⃗ = 0.5⃗𝑖 + 1.2𝑗⃗ − 𝑘

The Dot Product in 𝒏 Dimensions The algebraic definition of the dot product can be extended to vectors in higher dimensions. If 𝑢⃗ = (𝑢1 , … , 𝑢𝑛 ) and 𝑣⃗ = (𝑣1 , … , 𝑣𝑛 ) then the dot product of 𝑢⃗ and 𝑣⃗ is the scalar 𝑢⃗ ⋅ 𝑣⃗ = 𝑢1 𝑣1 + ⋯ + 𝑢𝑛 𝑣𝑛 .

Example 7

A video store sells videos, tapes, CDs, and computer games. We define the quantity vector 𝑞⃗ = (𝑞1 , 𝑞2 , 𝑞3 , 𝑞4 ), where 𝑞1 , 𝑞2 , 𝑞3 , 𝑞4 denote the quantities sold of each of the items, and the price vector 𝑝⃗ = (𝑝1 , 𝑝2 , 𝑝3 , 𝑝4 ), where 𝑝1 , 𝑝2 , 𝑝3 , 𝑝4 denote the price per unit of each item. What does the dot product 𝑝⃗ ⋅ 𝑞⃗ represent?

Solution

The dot product is 𝑝⃗ ⋅ 𝑞⃗ = 𝑝1 𝑞1 + 𝑝2 𝑞2 + 𝑝3 𝑞3 + 𝑝4 𝑞4 . The quantity 𝑝1 𝑞1 represents the revenue received by the store for the videos, 𝑝2 𝑞2 represents the revenue for the tapes, and so on. The dot product represents the total revenue received by the store for the sale of these four items.

13.3 THE DOT PRODUCT

723

Resolving a Vector into Components: Projections In Section 13.1, we resolved a vector into components parallel to the axes. Now we see how to resolve a vector, 𝑣⃗ , into components, called 𝑣⃗ parallel and 𝑣⃗ perp , which are parallel and perpendicular, respectively, to a given nonzero vector, 𝑢⃗ . (See Figure 13.31.) (b)

(a)

𝑣⃗

𝑣⃗ perp

𝑢⃗ 𝑣⃗

𝑣⃗ perp

𝑢⃗

𝜃

𝑣⃗ parallel

𝜃

𝑣⃗ parallel Figure 13.31: Resolving 𝑣⃗ into components parallel and perpendicular to 𝑢⃗ (a) 0 < 𝜃 < 𝜋∕2 (b) 𝜋∕2 < 𝜃 < 𝜋

The projection of 𝑣⃗ on 𝑢⃗ , written 𝑣⃗ parallel , measures (in some sense) how much the vector 𝑣⃗ is aligned with the vector 𝑢⃗ . The length of 𝑣⃗ parallel is the length of the shadow cast by 𝑣⃗ on a line in the direction of 𝑢⃗ . To compute 𝑣⃗ parallel , we assume 𝑢⃗ is a unit vector. (If not, create one by dividing by its length.) Then Figure 13.31(a) shows that, if 0 ≤ 𝜃 ≤ 𝜋∕2: ‖⃗ 𝑣 parallel ‖ = ‖⃗ 𝑣 ‖ cos 𝜃 = 𝑣⃗ ⋅ 𝑢⃗

(since ‖⃗𝑢 ‖ = 1).

Now 𝑣⃗ parallel is a scalar multiple of 𝑢⃗ , and since 𝑢⃗ is a unit vector,

𝑣 ‖ cos 𝜃)⃗𝑢 = (⃗ 𝑣 ⋅ 𝑢⃗ )⃗𝑢 . 𝑣⃗ parallel = (‖⃗

A similar argument shows that if 𝜋∕2 < 𝜃 ≤ 𝜋, as in Figure 13.31(b), this formula for 𝑣⃗ parallel still holds. The vector 𝑣⃗ perp is specified by 𝑣⃗ perp = 𝑣⃗ − 𝑣⃗ parallel . Thus, we have the following results:

Projection of 𝒗⃗ on the Line in the Direction of the Unit Vector 𝒖⃗ If 𝑣⃗ parallel and 𝑣⃗ perp are components of 𝑣⃗ that are parallel and perpendicular, respectively, to 𝑢⃗ , then Projection of 𝑣⃗ onto 𝑢⃗ = 𝑣⃗ parallel = (⃗ 𝑣 ⋅ 𝑢⃗ )⃗𝑢 and

Example 8

𝑣⃗ = 𝑣⃗ parallel + 𝑣⃗ perp

provided ‖⃗𝑢 ‖ = 1 𝑣⃗ perp = 𝑣⃗ − 𝑣⃗ parallel .

so

Figure 13.32 shows the force the wind exerts on the sail of a sailboat. Find the component of the force in the direction in which the sailboat is traveling. Wind direction



𝑢⃗

✲ ✲ Boat’s direction of travel ✲ Component of 𝐹⃗ wind

Sail

30◦

in boat’s direction of travel

❥❄

𝐹⃗ wind Figure 13.32: Wind moving a sailboat

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Chapter 13 A FUNDAMENTAL TOOL: VECTORS

Solution

Let 𝑢⃗ be a unit vector in the direction of travel. The force of the wind on the sail makes an angle of 30◦ with 𝑢⃗ . Thus, the component of this force in the direction of 𝑢⃗ is 𝐹⃗ parallel = (𝐹⃗ ⋅ 𝑢⃗ )⃗𝑢 = ‖𝐹⃗ ‖(cos 30◦ )⃗𝑢 = 0.87‖𝐹⃗ ‖⃗𝑢 .

Thus, the boat is being pushed forward with about 87% of the total force due to the wind. (In fact, the interaction of wind and sail is much more complex than this model suggests.)

A Physical Interpretation of the Dot Product: Work In physics, the word “work” has a different meaning from its everyday meaning. In physics, when a force of magnitude 𝐹 acts on an object through a distance 𝑑, we say the work, 𝑊 , done by the force is 𝑊 = 𝐹 𝑑, provided the force and the displacement are in the same direction. For example, if a 1 kg body falls 10 meters under the force of gravity, which is 9.8 newtons, then the work done by gravity is 𝑊 = (9.8 newtons) ⋅ (10 meters) = 98 joules. What if the force and the displacement are not in the same direction? Suppose a force 𝐹⃗ acts on an object as it moves along a displacement vector 𝑑⃗ . Let 𝜃 be the angle between 𝐹⃗ and 𝑑⃗ . First, we assume 0 ≤ 𝜃 ≤ 𝜋∕2. Figure 13.33 shows how we can resolve 𝐹⃗ into components that are parallel and perpendicular to 𝑑⃗ : 𝐹⃗ = 𝐹⃗ parallel + 𝐹⃗ perp . Then the work done by 𝐹⃗ is defined to be 𝑊 = ‖𝐹⃗ parallel ‖ ‖𝑑⃗ ‖.

We see from Figure 13.33 that 𝐹⃗ parallel has magnitude ‖𝐹⃗ ‖ cos 𝜃. So the work is given by the dot product: 𝑊 = (‖𝐹⃗ ‖ cos 𝜃)‖𝑑⃗ ‖ = ‖𝐹⃗ ‖‖𝑑⃗ ‖ cos 𝜃 = 𝐹⃗ ⋅ 𝑑⃗ . 𝐹⃗

𝐹⃗ perp

𝜃

𝑑⃗

𝐹⃗ parallel

Figure 13.33: Resolving the force 𝐹⃗ into two forces, one parallel to 𝑑⃗ , one perpendicular to 𝑑⃗

The formula 𝑊 = 𝐹⃗ ⋅ 𝑑⃗ holds when 𝜋∕2 < 𝜃 ≤ 𝜋 also. In that case, the work done by the force is negative and the object is moving against the force. Thus, we have the following definition: The work, 𝑊 , done by a force 𝐹⃗ acting on an object through a displacement 𝑑⃗ is given by 𝑊 = 𝐹⃗ ⋅ 𝑑⃗ .

Example 9

How much work does the wind do on the sailboat from Example 8 if the boat moves 20 m and the wind’s force is 120 newtons?

13.3 THE DOT PRODUCT

725

From Example 8, we know that the force of the wind 𝐹⃗ makes a 30◦ angle with the boat’s displacement 𝑑⃗ . Since ‖𝐹⃗ ‖ = 120 and ‖𝑑⃗ ‖ = 20, the work done by the wind on the boat is

Solution

𝑊 = 𝐹⃗ ⋅ 𝑑⃗ = ‖𝐹⃗ ‖‖𝑑⃗ ‖ cos 30◦ = 2078.461 joules.

Notice that if the vectors 𝐹⃗ and 𝑑⃗ are parallel and in the same direction, with magnitudes 𝐹 and 𝑑, then cos 𝜃 = cos 0 = 1, so 𝑊 = ‖𝐹⃗ ‖‖𝑑⃗ ‖ = 𝐹 𝑑, which is the original definition. When the vectors are perpendicular, cos 𝜃 = cos(𝜋∕2) = 0, so 𝑊 = 0 and no work is done in the technical definition of the word. For example, if you carry a heavy box across the room at the same horizontal height, no work is done by gravity because the force of gravity is vertical but the motion is horizontal.

Exercises and Problems for Section 13.3 Online Resource: Additional Problems for Section 13.3 EXERCISES

In Exercises 1–4, evaluate the dot product.

In Exercises 21–27, find an equation of a plane that satisfies the given conditions.

⃗ ) ⋅ (⃗𝑖 − 2𝑗⃗ − 3𝑘 ⃗) 1. (3⃗𝑖 + 2𝑗⃗ − 5𝑘 ⃗ ) ⋅ (4⃗𝑖 + 5𝑗⃗ + 6𝑘 ⃗) 2. (⃗𝑖 + 𝑗⃗ + 𝑘

⃗. 21. Through (1, 5, 2) perpendicular to 3⃗𝑖 − 𝑗⃗ + 4𝑘 ⃗. 22. Through (2, −1, 3) perpendicular to 5⃗𝑖 + 4𝑗⃗ − 𝑘

⃗ ) ⋅ (3⃗𝑖 − 2𝑗⃗ − 4𝑘 ⃗) 3. (3⃗𝑖 − 2𝑗⃗ − 4𝑘 ⃗ ) ⋅ 10𝑗⃗ 4. (2𝑖 + 5𝑘

⃗. 23. Through (1, 3, 5) and normal to ⃗𝑖 − 𝑗⃗ + 𝑘 ⃗ and passing through 24. Perpendicular to 5⃗𝑖 + 𝑗⃗ − 2𝑘 (0, 1, −1).

⃗. In Exercises 5–6, evaluate 𝑢⃗ ⋅ 𝑤 ⃗ ‖ = 5; the angle between 𝑢⃗ and 𝑤 ⃗ is 45◦ . 5. ‖⃗𝑢 ‖ = 3, ‖𝑤

⃗ ‖ = 20; the angle between 𝑢⃗ and 𝑤 ⃗ 6. ‖⃗𝑢 ‖ = 10, ‖𝑤 is 120◦ .

For Exercises 7–15, perform the following operations on the given 3-dimensional vectors. ⃗ 𝑎⃗ = 2𝑗⃗ + 𝑘

⃗ 𝑏⃗ = −3⃗𝑖 + 5𝑗⃗ + 4𝑘

𝑦⃗ = 4⃗𝑖 − 7𝑗⃗

𝑐⃗ = ⃗𝑖 + 6𝑗⃗

⃗ 𝑧⃗ = ⃗𝑖 − 3𝑗⃗ − 𝑘

25. Parallel to 2𝑥 + 4𝑦 − 3𝑧 = 1 and through (1, 0, −1). 26. Through (−2, 3, 2) and parallel to 3𝑥 + 𝑦 + 𝑧 = 4. ⃗ and through 27. Perpendicular to 𝑣⃗ = 2⃗𝑖 − 3𝑗⃗ + 5𝑘 (4, 5, −2). In Exercises 28–32, compute the angle between the vectors. ⃗ and ⃗𝑖 − 𝑗⃗ − 𝑘 ⃗. 28. ⃗𝑖 + 𝑗⃗ + 𝑘 ⃗ and 𝑗⃗ − 𝑘 ⃗. 29. ⃗𝑖 + 𝑘 ⃗ and 2 ⃗𝑖 + 3 𝑗⃗ + 𝑘 ⃗. 30. ⃗𝑖 + 𝑗⃗ − 𝑘 ⃗. 31. 𝑖⃗ + 𝑗⃗ and ⃗𝑖 + 2 𝑗⃗ − 𝑘

7. 𝑎⃗ ⋅ 𝑦⃗

8. 𝑐⃗ ⋅ 𝑦⃗

9. 𝑎⃗ ⋅ 𝑏⃗

10. 𝑎⃗ ⋅ 𝑧⃗

⃗. 32. ⃗𝑖 and 2 ⃗𝑖 + 3 𝑗⃗ − 𝑘

11. 𝑐⃗ ⋅ 𝑎⃗ + 𝑎⃗ ⋅ 𝑦⃗

12. 𝑎⃗ ⋅ (⃗ 𝑐 + 𝑦⃗ )

33. Match statements (a)-(c) with diagrams (I)-(III) of vec⃗ in Figure 13.34. tors 𝑢⃗ and 𝑤

13. (𝑎⃗ ⋅ 𝑏⃗ )𝑎⃗

14. (𝑎⃗ ⋅ 𝑦⃗ )(⃗ 𝑐 ⋅ 𝑧⃗ )

(a)

⃗ = 0 (b) 𝑢⃗ ⋅ 𝑤 ⃗ > 0 (c) 𝑢⃗ ⋅ 𝑤

⃗ ‖𝑣⃗ × 𝑤 ⃗‖ 51. ||𝑣⃗ ⋅ 𝑤

⃗ || < ‖𝑣⃗ × 𝑤 ⃗‖ 50. ||𝑣⃗ ⋅ 𝑤

52. Use a parallelepiped to show that 𝑎⃗ ⋅(𝑏⃗ ×𝑐⃗ ) = (𝑎⃗ ×𝑏⃗ )⋅𝑐⃗ for any vectors 𝑎⃗ , 𝑏⃗ , and 𝑐⃗ . 53. Figure 13.47 shows the tetrahedron determined by three vectors 𝑎⃗ , 𝑏⃗ , 𝑐⃗ . The area vector of a face is a vector perpendicular to the face, pointing outward, whose magnitude is the area of the face. Show that the sum of the four outward-pointing area vectors of the faces equals the zero vector.

(𝑥, 𝑦) 𝑣⃗ 𝑟⃗ 𝑥 𝑐⃗ − 𝑎⃗

Figure 13.46 𝑏⃗ − 𝑎⃗

𝑎⃗

44. The points 𝑃1 = (0, 0, 0), 𝑃2 = (2, 4, 2), 𝑃3 = (3, 0, 0), and 𝑃4 = (5, 4, 2) are vertices of a parallelogram. (a) Find the displacement vectors along each of the four sides. Check that these are equal in pairs. (b) Find the area of the parallelogram.

𝑐⃗ 𝑏⃗ − 𝑐⃗ 𝑏⃗

Figure 13.47

In Problems 45–46, find an area vector for the parallelogram with given vertices. 45. 𝑃 = (2, 1, 1), 𝑄 = (3, 3, 0), 𝑅 = (4, 0, 2), 𝑆 = (5, 2, 1) 46. 𝑃 = (−1, −2, 0), 𝑄 = (0, −1, 0), 𝑅 = (−2, −4, 1), 𝑆 = (−1, −3, 1) 47. A parallelogram 𝑃 formed by the vectors 𝑣⃗ = ⃗𝑖 − 2𝑗⃗ + ⃗ and 𝑤 ⃗ has area vector 𝐴⃗ = 𝑣⃗ × 𝑤 ⃗ = ⃗𝑖 + 2𝑗⃗ − 2𝑘 ⃗ = 𝑘 ⃗ . Find area vectors of each of the following 2⃗𝑖 +3𝑗⃗ +4𝑘 parallelograms and explain how they are related to 𝐴⃗ . (a) The parallelogram obtained by projecting 𝑃 onto the 𝑥𝑦-plane. (b) The parallelogram obtained by projecting 𝑃 onto the 𝑥𝑧-plane. (c) The parallelogram obtained by projecting 𝑃 onto the 𝑦𝑧-plane. 4 http://en.wikipedia.org,

accessed May 12, 2016.

In Problems 54–56, find the vector representing the area of a surface. The magnitude of the vector equals the magnitude of the area; the direction is perpendicular to the surface. Since there are two perpendicular directions, we pick one by giving an orientation for the surface. 54. The rectangle with vertices (0, 0, 0), (0, 1, 0), (2, 1, 0), and (2, 0, 0), oriented so that it faces downward. 55. The circle of radius 2 in the 𝑦𝑧-plane, facing in the direction of the positive 𝑥-axis. 56. The triangle 𝐴𝐵𝐶, oriented upward, where 𝐴 = (1, 2, 3), 𝐵 = (3, 1, 2), and 𝐶 = (2, 1, 3).

13.4 THE CROSS PRODUCT

737

57. This problem relates the area of a parallelogram 𝑆 lying in the plane 𝑧 = 𝑚𝑥 + 𝑛𝑦 + 𝑐 to the area of its projection 𝑅 in the 𝑥𝑦-plane. Let 𝑆 be determined by the vectors ⃗ and 𝑣⃗ = 𝑣1 ⃗𝑖 + 𝑣2 𝑗⃗ + 𝑣3 𝑘 ⃗ . See 𝑢⃗ = 𝑢1 ⃗𝑖 + 𝑢2 𝑗⃗ + 𝑢3 𝑘 Figure 13.48. (a) Find the area of 𝑆. (b) Find the area of 𝑅. (c) Find 𝑚 and 𝑛 in terms of the components of 𝑢⃗ and 𝑣⃗ . (d) Show that √ Area of 𝑆 = 1 + 𝑚2 + 𝑛2 ⋅ Area of 𝑅.

Figure 13.48

Strengthen Your Understanding In Problems 58–59, explain what is wrong with the statement.

64. ‖⃗𝑢 × 𝑣⃗ ‖ = ‖⃗𝑢 ‖‖𝑣⃗ ‖.

58. There is only one unit vector perpendicular to two nonparallel vectors in 3-space. 59. 𝑢⃗ × 𝑣⃗ = 0⃗ when 𝑢⃗ and 𝑣⃗ are perpendicular.

⃗ , then 66. If 𝑣⃗ is a nonzero vector and 𝑣⃗ × 𝑢⃗ = 𝑣⃗ × 𝑤 ⃗. 𝑢⃗ = 𝑤

⃗ = ⃗𝑖 ⋅ (𝑗⃗ × 𝑘 ⃗ ). 65. (⃗𝑖 × 𝑗⃗ ) ⋅ 𝑘

⃗ ) is always 0. 67. The value of 𝑣⃗ ⋅ (𝑣⃗ × 𝑤

In Problems 60–61, give an example of:

⃗ is never the same as 𝑣⃗ ⋅ 𝑤 ⃗. 68. The value of 𝑣⃗ × 𝑤

60. A vector 𝑢⃗ whose cross product with 𝑣⃗ = ⃗𝑖 + 𝑗⃗ is ⃗. parallel to 𝑘 61. A vector 𝑣⃗ such that ‖⃗𝑢 × 𝑣⃗ ‖ = 10, where 𝑢⃗ = 3⃗𝑖 +4𝑗⃗ .

69. The area of the triangle with two sides given by ⃗𝑖 + 𝑗⃗ ⃗ is 3∕2. and 𝑗⃗ + 2𝑘

Are the statements in Problems 62–72 true or false? Give reasons for your answer. 62. 𝑢⃗ × 𝑣⃗ is a vector. 63. 𝑢⃗ × 𝑣⃗ has direction parallel to both 𝑢⃗ and 𝑣⃗ .

Online Resource: Review problems and Projects

70. Given a nonzero vector 𝑣⃗ in 3-space, there is a nonzero ⃗ such that 𝑣⃗ × 𝑤 ⃗ = 0⃗ . vector 𝑤 ⃗ =𝑤 ⃗ × 𝑣⃗ . 71. It is never true that 𝑣⃗ × 𝑤 72. Two points are circling an axis at a rate of 5 rad/sec. The point closer to the axis has the greater speed.

Chapter Fourteen

DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

Contents 14.1 The Partial Derivative . . . . . . . . . . . . . . . . . . . . Rate of Change of Temperature in a Rod . . . . . . Rate of Change of Temperature in a Metal Plate . Definition of the Partial Derivative . . . . . . . . . . . Visualizing Partial Derivatives on a Graph . . . . . Estimating Partial from a Contour Diagram . . . . Using Units to Interpret Partial Derivatives . . . . . 14.2 Computing Partial Derivatives Algebraically . . Interpretation of Partial Derivatives . . . . . . . . . . 14.3 Local Linearity and the Differential . . . . . . . . . Zooming In to See Local Linearity . . . . . . . . . . . The Tangent Plane . . . . . . . . . . . . . . . . . . . . . . . . Local Linearization . . . . . . . . . . . . . . . . . . . . . . . The Differential . . . . . . . . . . . . . . . . . . . . . . . . . . 14.4 Gradients and Directional Derivatives in the Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Rate of Change in an Arbitrary Direction. . . Computing Directional Derivatives from Partial Derivatives . . . . . . . . . . . . . . . . . . . The Gradient Vector . . . . . . . . . . . . . . . . . . . . . . . Properties of the Gradient Vector. . . . . . . . . . . . . 14.5 Gradients and Directional Derivatives in Space. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Gradient Vector in Three Variables . . . . . . . . 14.6 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . Composition and Rates of Change. . . . . . . . . . . . The Chain Rule for z = f(x, y), x = g(t), y = h(t) . . How to Formulate a General Chain Rule. . . . . . . An Application to Physical Chemistry. . . . . . . . . 14.7 Second-Order Partial Derivatives . . . . . . . . . . . The Mixed Partial Derivatives Are Equal . . . . . . What Do the Second-Order Partial Derivatives Tell Us? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Taylor Approximations . . . . . . . . . . . . . . . . . . . . Linear and Quadratic Approximations Near (0,0) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Linear and Quadratic Approximations Near (a, b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.8 Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . . Partial Derivatives and Differentiability . . . . . . . . . Continuity and Differentiability . . . . . . . . . . . . . .

740 740 740 741 742 742 743 748 750 753 753 754 755 758 762 762 764 765 766 772 772 780 780 780 782 784 790 791 791 792 792 794 799 799 802

740

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

14.1

THE PARTIAL DERIVATIVE The derivative of a one-variable function measures its rate of change. In this section we see how a two-variable function has two rates of change: one as 𝑥 changes (with 𝑦 held constant) and one as 𝑦 changes (with 𝑥 held constant).

Rate of Change of Temperature in a Metal Rod: a One-Variable Problem Imagine an unevenly heated metal rod lying along the 𝑥-axis, with its left end at the origin and 𝑥 measured in meters. (See Figure 14.1.) Let 𝑢(𝑥) be the temperature (in ◦ C) of the rod at the point 𝑥. Table 14.1 gives values of 𝑢(𝑥). We see that the temperature increases as we move along the rod, reaching its maximum at 𝑥 = 4, after which it starts to decrease. Table 14.1

Temperature 𝑢(𝑥) of the rod

𝑥 (m) 0

1

2

3

4

5

Figure 14.1: Unevenly heated metal rod

𝑥 (m)

0

1

2

3

4

5

𝑢(𝑥) (◦ C)

125

128

135

160

175

160

Example 1

Estimate the derivative 𝑢′ (2) using Table 14.1 and explain what the answer means in terms of temperature.

Solution

The derivative 𝑢′ (2) is defined as a limit of difference quotients: 𝑢′ (2) = lim

ℎ→0

𝑢(2 + ℎ) − 𝑢(2) . ℎ

Choosing ℎ = 1 so that we can use the data in Table 14.1, we get 𝑢′ (2) ≈

𝑢(2 + 1) − 𝑢(2) 160 − 135 = = 25. 1 1

This means that the temperature increases at a rate of approximately 25◦ C per meter as we go from left to right, past 𝑥 = 2.

Rate of Change of Temperature in a Metal Plate Imagine an unevenly heated thin rectangular metal plate lying in the 𝑥𝑦-plane with its lower left corner at the origin and 𝑥 and 𝑦 measured in meters. The temperature (in ◦ C) at the point (𝑥, 𝑦) is 𝑇 (𝑥, 𝑦). See Figure 14.2 and Table 14.2. How does 𝑇 vary near the point (2, 1)? We consider the horizontal line 𝑦 = 1 containing the point (2, 1). The temperature along this line is the cross section, 𝑇 (𝑥, 1), of the function 𝑇 (𝑥, 𝑦) with 𝑦 = 1. Suppose we write 𝑢(𝑥) = 𝑇 (𝑥, 1). 𝑦 (m) 3

𝑥=2

Table 14.2

2 (2, 1)

1 0

1

2

3

𝑦=1

4

5

𝑦 (m)

𝑥 (m)

Temperature (◦ C) of a metal plate 3

85

90

110

135

155

180

2

100

110

120

145

190

170

1

125

128

135

160

175

160

0

120

135

155

160

160

150

0

1

2

3

4

5

𝑥 (m)

Figure 14.2: Unevenly heated metal plate

What is the meaning of the derivative 𝑢′ (2)? It is the rate of change of temperature 𝑇 in the 𝑥-direction at the point (2, 1), keeping 𝑦 fixed. Denote this rate of change by 𝑇𝑥 (2, 1), so that 𝑇𝑥 (2, 1) = 𝑢′ (2) = lim

ℎ→0

𝑇 (2 + ℎ, 1) − 𝑇 (2, 1) 𝑢(2 + ℎ) − 𝑢(2) = lim . ℎ→0 ℎ ℎ

14.1 THE PARTIAL DERIVATIVE

741

We call 𝑇𝑥 (2, 1) the partial derivative of 𝑇 with respect to 𝑥 at the point (2, 1). Taking ℎ = 1, we can read values of 𝑇 from the row with 𝑦 = 1 in Table 14.2, giving 𝑇 (3, 1) − 𝑇 (2, 1) 160 − 135 = = 25◦ C/m. 1 1 The fact that 𝑇𝑥 (2, 1) is positive means that the temperature of the plate is increasing as we move past the point (2, 1) in the direction of increasing 𝑥 (that is, horizontally from left to right in Figure 14.2). 𝑇𝑥 (2, 1) ≈

Example 2

Estimate the rate of change of 𝑇 in the 𝑦-direction at the point (2, 1).

Solution

The temperature along the line 𝑥 = 2 is the cross-section of 𝑇 with 𝑥 = 2, that is, the function 𝑣(𝑦) = 𝑇 (2, 𝑦). If we denote the rate of change of 𝑇 in the 𝑦-direction at (2, 1) by 𝑇𝑦 (2, 1), then 𝑇 (2, 1 + ℎ) − 𝑇 (2, 1) 𝑣(1 + ℎ) − 𝑣(1) = lim . ℎ→0 ℎ→0 ℎ ℎ

𝑇𝑦 (2, 1) = 𝑣′ (1) = lim

We call 𝑇𝑦 (2, 1) the partial derivative of 𝑇 with respect to 𝑦 at the point (2, 1). Taking ℎ = 1 so that we can use the column with 𝑥 = 2 in Table 14.2, we get 𝑇𝑦 (2, 1) ≈

𝑇 (2, 1 + 1) − 𝑇 (2, 1) 120 − 135 = = −15◦C/m. 1 1

The fact that 𝑇𝑦 (2, 1) is negative means that at (2, 1), the temperature decreases as 𝑦 increases..

Definition of the Partial Derivative We study the influence of 𝑥 and 𝑦 separately on the value of the function 𝑓 (𝑥, 𝑦) by holding one fixed and letting the other vary. This leads to the following definitions.

Partial Derivatives of 𝒇 with Respect to 𝒙 and 𝒚 For all points at which the limits exist, we define the partial derivatives at the point (𝐚, 𝐛) by 𝑓𝑥 (𝑎, 𝑏) = 𝑓𝑦 (𝑎, 𝑏) =

Rate of change of 𝑓 with respect to 𝑥 at the point (𝑎, 𝑏) Rate of change of 𝑓 with respect to 𝑦 at the point (𝑎, 𝑏)

𝑓 (𝑎 + ℎ, 𝑏) − 𝑓 (𝑎, 𝑏) , ℎ→0 ℎ 𝑓 (𝑎, 𝑏 + ℎ) − 𝑓 (𝑎, 𝑏) . = lim ℎ→0 ℎ

= lim

If we let 𝑎 and 𝑏 vary, we have the partial derivative functions 𝑓𝑥 (𝑥, 𝑦) and 𝑓𝑦 (𝑥, 𝑦). Just as with ordinary derivatives, there is an alternative notation:

Alternative Notation for Partial Derivatives If 𝑧 = 𝑓 (𝑥, 𝑦), we can write 𝑓𝑥 (𝑥, 𝑦) = 𝑓𝑥 (𝑎, 𝑏) =

𝜕𝑧 𝜕𝑥

𝜕𝑧 || 𝜕𝑥 ||(𝑎,𝑏)

and

𝑓𝑦 (𝑥, 𝑦) =

and

𝑓𝑦 (𝑎, 𝑏) =

𝜕𝑧 , 𝜕𝑦 𝜕𝑧 || . 𝜕𝑦 ||(𝑎,𝑏)

742

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

We use the symbol 𝜕 to distinguish partial derivatives from ordinary derivatives. In cases where the independent variables have names different from 𝑥 and 𝑦, we adjust the notation accordingly. For example, the partial derivatives of 𝑓 (𝑢, 𝑣) are denoted by 𝑓𝑢 and 𝑓𝑣 .

Visualizing Partial Derivatives on a Graph The ordinary derivative of a one-variable function is the slope of its graph. How do we visualize the partial derivative 𝑓𝑥 (𝑎, 𝑏)? The graph of the one-variable function 𝑓 (𝑥, 𝑏) is the curve where the vertical plane 𝑦 = 𝑏 cuts the graph of 𝑓 (𝑥, 𝑦). (See Figure 14.3.) Thus, 𝑓𝑥 (𝑎, 𝑏) is the slope of the tangent line to this curve at 𝑥 = 𝑎. 𝑧

Line has slope

Line has slope

Point

(𝑎, 𝑏, 𝑓 (𝑎, 𝑏))





𝑧

𝑓𝑦 (𝑎, 𝑏)

𝑓𝑥 (𝑎, 𝑏)

Graph of

𝑓 (𝑥, 𝑏)





Graph of

𝑓 (𝑎, 𝑦)

𝑥

Point

(𝑎, 𝑏, 𝑓 (𝑎, 𝑏))

𝑦 𝑦

𝑥

Figure 14.3: The curve 𝑧 = 𝑓 (𝑥, 𝑏) on the graph of 𝑓 has slope 𝑓𝑥 (𝑎, 𝑏) at 𝑥 = 𝑎

Figure 14.4: The curve 𝑧 = 𝑓 (𝑎, 𝑦) on the graph of 𝑓 has slope 𝑓𝑦 (𝑎, 𝑏) at 𝑦 = 𝑏

Similarly, the graph of the function 𝑓 (𝑎, 𝑦) is the curve where the vertical plane 𝑥 = 𝑎 cuts the graph of 𝑓 , and the partial derivative 𝑓𝑦 (𝑎, 𝑏) is the slope of this curve at 𝑦 = 𝑏. (See Figure 14.4.) Example 3

At each point labeled on the graph of the surface 𝑧 = 𝑓 (𝑥, 𝑦) in Figure 14.5, say whether each partial derivative is positive or negative. 𝑄

𝑧

❘ ✛

𝑦 𝑃

𝑥

Figure 14.5: Decide the signs of 𝑓𝑥 and 𝑓𝑦 at 𝑃 and 𝑄

Solution

The positive 𝑥-axis points out of the page. Imagine heading off in this direction from the point marked 𝑃 ; we descend steeply. So the partial derivative with respect to 𝑥 is negative at 𝑃 , with quite a large absolute value. The same is true for the partial derivative with respect to 𝑦 at 𝑃 , since there is also a steep descent in the positive 𝑦-direction. At the point marked 𝑄, heading in the positive 𝑥-direction results in a gentle descent, whereas heading in the positive 𝑦-direction results in a gentle ascent. Thus, the partial derivative 𝑓𝑥 at 𝑄 is negative but small (that is, near zero), and the partial derivative 𝑓𝑦 is positive but small.

Estimating Partial Derivatives from a Contour Diagram The graph of a function 𝑓 (𝑥, 𝑦) often makes clear the sign of the partial derivatives. However, numerical estimates of these derivatives are more easily made from a contour diagram than a surface graph. If we move parallel to one of the axes on a contour diagram, the partial derivative is the rate of change of the value of the function on the contours. For example, if the values on the contours are increasing as we move in the positive direction, then the partial derivative must be positive.

14.1 THE PARTIAL DERIVATIVE

Example 4

743

Figure 14.6 shows the contour diagram for the temperature 𝐻(𝑥, 𝑡) (in ◦ C) in a room as a function of distance 𝑥 (in meters) from a heater and time 𝑡 (in minutes) after the heater has been turned on. What are the signs of 𝐻𝑥 (10, 20) and 𝐻𝑡 (10, 20)? Estimate these partial derivatives and explain the answers in practical terms. 𝑡 (minutes) 60 30

50 40

✻ 32

25

30

❄ ✛ 14 ✲

20

20 15 10

10 5

10 15 20 25 30

𝑥 (meters)

Figure 14.6: Temperature in a heated room: Heater at 𝑥 = 0 is turned on at 𝑡 = 0

Solution

The point (10, 20) is nearly on the 𝐻 = 25 contour. As 𝑥 increases, we move toward the 𝐻 = 20 contour, so 𝐻 is decreasing and 𝐻𝑥 (10, 20) is negative. This makes sense because the 𝐻 = 30 contour is to the left: As we move further from the heater, the temperature drops. On the other hand, as 𝑡 increases, we move toward the 𝐻 = 30 contour, so 𝐻 is increasing; as 𝑡 decreases 𝐻 decreases. Thus, 𝐻𝑡 (10, 20) is positive. This says that as time passes, the room warms up. To estimate the partial derivatives, use a difference quotient. Looking at the contour diagram, we see there is a point on the 𝐻 = 20 contour about 14 units to the right of the point (10, 20). Hence, 𝐻 decreases by 5 when 𝑥 increases by 14, so we find Rate of change of 𝐻 with respect to 𝑥 = 𝐻𝑥 (10, 20) ≈

−5 ≈ −0.36◦C/meter. 14

This means that near the point 10 m from the heater, after 20 minutes the temperature drops about 0.36, or one third, of a degree, for each meter we move away from the heater. To estimate 𝐻𝑡 (10, 20), we notice that the 𝐻 = 30 contour is about 32 units directly above the point (10, 20). So 𝐻 increases by 5 when 𝑡 increases by 32. Hence, Rate of change of 𝐻 with respect to 𝑡 = 𝐻𝑡 (10, 20) ≈

5 = 0.16◦C/minute. 32

This means that after 20 minutes the temperature is going up about 0.16, or 1/6, of a degree each minute at the point 10 m from the heater.

Using Units to Interpret Partial Derivatives The meaning of a partial derivative can often be explained using units. Example 5

Suppose that your weight 𝑤 in pounds is a function 𝑓 (𝑐, 𝑛) of the number 𝑐 of calories you consume daily and the number 𝑛 of minutes you exercise daily. Using the units for 𝑤, 𝑐 and 𝑛, interpret in everyday terms the statements 𝜕𝑤 (2000, 15) = 0.02 and 𝜕𝑐

𝜕𝑤 (2000, 15) = −0.025. 𝜕𝑛

744

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

Solution

The units of 𝜕𝑤∕𝜕𝑐 are pounds per calorie. The statement 𝜕𝑤 (2000, 15) = 0.02 𝜕𝑐 means that if you are presently consuming 2000 calories daily and exercising 15 minutes daily, you will weigh 0.02 pounds more for each extra calorie you consume daily, or about 2 pounds for each extra 100 calories per day. The units of 𝜕𝑤∕𝜕𝑛 are pounds per minute. The statement 𝜕𝑤 (2000, 15) = −0.025 𝜕𝑛 means that for the same calorie consumption and number of minutes of exercise, you will weigh 0.025 pounds less for each extra minute you exercise daily, or about 1 pound less for each extra 40 minutes per day. So if you eat an extra 100 calories each day and exercise about 80 minutes more each day, your weight should remain roughly steady.

Exercises and Problems for Section 14.1 EXERCISES 1. Given the following table of values for 𝑧 = 𝑓 (𝑥, 𝑦), estimate 𝑓𝑥 (3, 2) and 𝑓𝑦 (3, 2), assuming they exist. 𝑥∖𝑦

0

2

5

1

1

2

4

3

−1

1

2

6

−3

0

0

6. A drug is injected into a patient’s blood vessel. The function 𝑐 = 𝑓 (𝑥, 𝑡) represents the concentration of the drug at a distance 𝑥 mm in the direction of the blood flow measured from the point of injection and at time 𝑡 seconds since the injection. What are the units of the following partial derivatives? What are their practical interpretations? What do you expect their signs to be? (a)

2. Using difference quotients, estimate 𝑓𝑥 (3, 2) and 𝑓𝑦 (3, 2) for the function given by 2

𝑓 (𝑥, 𝑦) =

𝑥 . 𝑦+1

[Recall: A difference quotient is an expression of the form (𝑓 (𝑎 + ℎ, 𝑏) − 𝑓 (𝑎, 𝑏))∕ℎ.] 3. Use difference quotients with Δ𝑥 = 0.1 and Δ𝑦 = 0.1 to estimate 𝑓𝑥 (1, 3) and 𝑓𝑦 (1, 3), where 𝑓 (𝑥, 𝑦) = 𝑒−𝑥 sin 𝑦. Then give better estimates by using Δ𝑥 = 0.01 and Δ𝑦 = 0.01. 4. The price 𝑃 in dollars to purchase a used car is a function of its original cost, 𝐶, in dollars, and its age, 𝐴, in years. (a) (b) (c) (d)

What are the units of 𝜕𝑃 ∕𝜕𝐴? What is the sign of 𝜕𝑃 ∕𝜕𝐴 and why? What are the units of 𝜕𝑃 ∕𝜕𝐶? What is the sign of 𝜕𝑃 ∕𝜕𝐶 and why?

5. Your monthly car payment in dollars is 𝑃 = 𝑓 (𝑃0 , 𝑡, 𝑟), where $𝑃0 is the amount you borrowed, 𝑡 is the number of months it takes to pay off the loan, and 𝑟% is the interest rate. What are the units, the financial meaning, and the signs of 𝜕𝑃 ∕𝜕𝑡 and 𝜕𝑃 ∕𝜕𝑟?

𝜕𝑐∕𝜕𝑥

(b) 𝜕𝑐∕𝜕𝑡

7. You borrow $𝐴 at an interest rate of 𝑟% (per month) and pay it off over 𝑡 months by making monthly payments of 𝑃 = 𝑔(𝐴, 𝑟, 𝑡) dollars. In financial terms, what do the following statements tell you? (a) 𝑔(8000, 1, 24) = 376.59 𝜕𝑔 || (b) = 0.047 | 𝜕𝐴 ||(8000,1,24) 𝜕𝑔 || = 44.83 (c) | 𝜕𝑟 ||(8000,1,24)

8. The sales of a product, 𝑆 = 𝑓 (𝑝, 𝑎), are a function of the price, 𝑝, of the product (in dollars per unit) and the amount, 𝑎, spent on advertising (in thousands of dollars). (a) Do you expect 𝑓𝑝 to be positive or negative? Why? (b) Explain the meaning of the statement 𝑓𝑎 (8, 12) = 150 in terms of sales. 9. The quantity, 𝑄, of beef purchased at a store, in kilograms per week, is a function of the price of beef, 𝑏, and the price of chicken, 𝑐, both in dollars per kilogram. (a) Do you expect 𝜕𝑄∕𝜕𝑏 to be positive or negative? Explain.

14.1 THE PARTIAL DERIVATIVE

(b) Do you expect 𝜕𝑄∕𝜕𝑐 to be positive or negative? Explain. (c) Interpret the statement 𝜕𝑄∕𝜕𝑏 = −213 in terms of quantity of beef purchased.

In Exercises 16–19, determine the sign of 𝑓𝑥 and 𝑓𝑦 at the point using the contour diagram of 𝑓 in Figure 14.7. 𝑦 17

15

In Exercises 10–15, a point 𝐴 is shown on a contour diagram of a function 𝑓 (𝑥, 𝑦). (a) Evaluate 𝑓 (𝐴). (b) Is 𝑓𝑥 (𝐴) positive, negative, or zero? (c) Is 𝑓𝑦 (𝐴) positive, negative, or zero?

𝑆

19

𝑃

13 11

𝑥

𝑅

10. 𝑦

745

𝑄

11. 𝑦

13

16

10

7

5

Figure 14.7

10

𝐴

15

𝐴 20

𝑥

𝑥

12. 𝑦

16. 𝑃

17. 𝑄

18. 𝑅

19. 𝑆

13. 𝑦 20. Values of 𝑓 (𝑥, 𝑦) are in Table 14.3. Assuming they exist, decide whether you expect the following partial derivatives to be positive or negative.

76

66 82

62

𝐴 88

𝐴 5 8

94

54

𝑥

𝑥

14. 𝑦

(a)

𝑓𝑥 (−2, −1)

(b) 𝑓𝑦 (2, 1)

(c)

𝑓𝑥 (2, 1)

(d) 𝑓𝑦 (0, 3)

15. 𝑦

Table 14.3 44

23

24

25

𝐴

42 26

40 27

𝐴

38 36

𝑥

𝑥

𝑥∖𝑦

−1

1

3

5

−2

7

3

2

1

0

8

5

3

2

2

10

7

5

4

4

13

10

8

7

PROBLEMS 21. Figure 14.8 is a contour diagram for 𝑧 = 𝑓 (𝑥, 𝑦). Is 𝑓𝑥 positive or negative? Is 𝑓𝑦 positive or negative? Estimate 𝑓 (2, 1), 𝑓𝑥 (2, 1), and 𝑓𝑦 (2, 1).

𝑦 10 8

16 14 12

𝑦 6

5 4

10 8

4

−6 −2

3

6 4

2

2

2

𝑥

6

2

2

4

6

8

10

10

1

Figure 14.9

14 18

1

2

3

4

𝑥 5

Figure 14.8 22. Approximate 𝑓𝑥 (3, 5) using the contour diagram of 𝑓 (𝑥, 𝑦) in Figure 14.9.

23. When riding your bike in winter, the windchill temperature is a measure of how cold you feel as a result of the induced breeze caused by your travel. If 𝑊 represents windchill temperature (in ◦ F) that you experience, then 𝑊 = 𝑓 (𝑇 , 𝑣), where 𝑇 is the actual air temperature (in ◦ F) and 𝑣 is your speed, in meters per second. Match each of the practical interpretations below with a mathematical statement that most accurately describes it be-

746

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

low. For the remaining mathematical statement, give a practical interpretation. (i) “The faster you ride, the colder you’ll feel.” (ii) “The warmer the day, the warmer you’ll feel.” (a) 𝑓𝑇 (𝑇 , 𝑣) > 0 (b) 𝑓 (0, 𝑣) ≤ 0

30. Figure 14.10 shows contours of 𝑓 (𝑥, 𝑦) with values of 𝑓 on the contours omitted. If 𝑓𝑥 (𝑃 ) > 0, find the sign: (a)

𝑓𝑦 (𝑃 )

(b) 𝑓𝑦 (𝑄) 𝑦

(c) 𝑓𝑣 (𝑇 , 𝑣) < 0

24. People commuting to a city can choose to go either by bus or by train. The number of people who choose either method depends in part upon the price of each. Let 𝑓 (𝑃1 , 𝑃2 ) be the number of people who take the bus when 𝑃1 is the price of a bus ride and 𝑃2 is the price of a train ride. What can you say about the signs of 𝜕𝑓 ∕𝜕𝑃1 and 𝜕𝑓 ∕𝜕𝑃2 ? Explain your answers. 25. The average price of large cars getting low gas mileage (“gas guzzlers”) is 𝑥 and the average price of a gallon of gasoline is 𝑦. The number, 𝑞1 , of gas guzzlers bought in a year, depends on both 𝑥 and 𝑦, so 𝑞1 = 𝑓 (𝑥, 𝑦). Similarly, if 𝑞2 is the number of gallons of gas bought to fill gas guzzlers in a year, then 𝑞2 = 𝑔(𝑥, 𝑦). (a) What do you expect the signs of 𝜕𝑞1 ∕𝜕𝑥 and 𝜕𝑞2 ∕𝜕𝑦 to be? Explain. (b) What do you expect the signs of 𝜕𝑞1 ∕𝜕𝑦 and 𝜕𝑞2 ∕𝜕𝑥 to be? Explain.

𝑄 𝑥 −5

5 𝑃

−5

Figure 14.10 31. Figure 14.11 shows the contour diagram of 𝑔(𝑥, 𝑦). Mark the points on the contours where (a)

𝑔𝑥 = 0

(b) 𝑔𝑦 = 0

𝑦 10 60 50

30 20 10

𝑥 10

Figure 14.11 32. The surface 𝑧 = 𝑓 (𝑥, 𝑦) is shown in Figure 14.12. The points 𝐴 and 𝐵 are in the 𝑥𝑦-plane. (a) What is the sign of (ii) 𝑓𝑦 (𝐴)? (i) 𝑓𝑥 (𝐴)? (b) The point 𝑃 in the 𝑥𝑦-plane moves along a straight line from 𝐴 to 𝐵. How does the sign of 𝑓𝑥 (𝑃 ) change? How does the sign of 𝑓𝑦 (𝑃 ) change? 𝑧

𝐴 𝐵





𝑦

Time 𝑡 (months)

Conc. 𝑐 (ppm)

50 60

40

𝑥

Table 14.4

𝑓 (𝑥, 𝑦)

5

For Problems 26–28, refer to Table 12.2 on page 658 giving the temperature adjusted for wind chill, 𝐶, in ◦ F, as a function 𝑓 (𝑤, 𝑇 ) of the wind speed, 𝑤, in mph, and the temperature, 𝑇 , in ◦ F. The temperature adjusted for wind chill tells you how cold it feels, as a result of the combination of wind and temperature. 26. Estimate 𝑓𝑤 (10, 25). What does your answer mean in practical terms? 27. Estimate 𝑓𝑇 (5, 20). What does your answer mean in practical terms? 28. From Table 12.2 you can see that when the temperature is 20◦ F, the temperature adjusted for wind-chill drops by an average of about 0.8◦ F with every 1 mph increase in wind speed from 5 mph to 10 mph. Which partial derivative is this telling you about? 29. An experiment to measure the toxicity of formaldehyde yielded the data in Table 14.4. The values show the percent, 𝑃 = 𝑓 (𝑡, 𝑐), of rats surviving an exposure to formaldehyde at a concentration of 𝑐 (in parts per million, ppm) after 𝑡 months. Estimate 𝑓𝑡 (18, 6) and 𝑓𝑐 (18, 6). Interpret your answers in terms of formaldehyde toxicity.

14

16

18

20

22

24

0

100

100

100

99

97

95

2

100

99

98

97

95

92

6

96

95

93

90

86

80

15

96

93

82

70

58

36

𝑓𝑥 (𝑄)

(c)

Figure 14.12

747

14.1 THE PARTIAL DERIVATIVE

33. Figure 14.13 shows the saddle-shaped surface 𝑧 = 𝑓 (𝑥, 𝑦). (a) What is the sign of 𝑓𝑥 (0, 5)? (b) What is the sign of 𝑓𝑦 (0, 5)? 𝑧

36. Figure 14.16 shows a contour diagram for the temperature 𝑇 (in ◦ C) along a wall in a heated room as a function of distance 𝑥 along the wall and time 𝑡 in minutes. Estimate 𝜕𝑇 ∕𝜕𝑥 and 𝜕𝑇 ∕𝜕𝑡 at the given points. Give units and interpret your answers. (a)

(0, 5, 3)

𝑥 = 15, 𝑡 = 20

(b) 𝑥 = 5, 𝑡 = 12

𝑡 (minutes)



60 30

50

𝑦

40 𝑥 25

30

Figure 14.13 34. Figure 14.14 shows the graph of the function 𝑓 (𝑥, 𝑦) on the domain 0 ≤ 𝑥 ≤ 4 and 0 ≤ 𝑦 ≤ 4. Use the graph to rank the following quantities in order from smallest to largest: 𝑓𝑥 (3, 2), 𝑓𝑥 (1, 2), 𝑓𝑦 (3, 2), 𝑓𝑦 (1, 2), 0.

20

20

10

15 10

𝑥 (meters) 5

10

15

20

25

30

Figure 14.16

𝑧

𝑥

In Problems 37–39, we use Figure 14.17 to model the heat required to clear an airport of fog by heating the air. The amount of heat, 𝐻(𝑇 , 𝑤), required (in calories per cubic meter of fog) is a function of the temperature 𝑇 (in degrees Celsius) and the water content 𝑤 (in grams per cubic meter of fog). Note that Figure 14.17 is not a contour diagram, but shows cross-sections of 𝐻 with 𝑤 fixed at 0.1, 0.2, 0.3, 0.4.

𝑦

𝐻 (calories∕m3 )

Figure 14.14

600

35. Figure 14.15 shows a contour diagram for the monthly payment 𝑃 as a function of the interest rate, 𝑟%, and the amount, 𝐿, of a 5-year loan. Estimate 𝜕𝑃 ∕𝜕𝑟 and 𝜕𝑃 ∕𝜕𝐿 at the following points. In each case, give the units and the everyday meaning of your answer.

500

(a) 𝑟 = 8, 𝐿 = 4000 (c) 𝑟 = 13, 𝐿 = 7000

(b) 𝑟 = 8, 𝐿 = 6000

400 300

𝐻(𝑇 , 0.4) 𝐻(𝑇 , 0.3)

200

𝐻(𝑇 , 0.2) 100

L($)

8000

𝐻(𝑇 , 0.1)

0

7000

160

120

5000 80

2000

r(%)

1

3

5

7

40

38. Make a table of values for 𝐻(𝑇 , 𝑤) from Figure 14.17, and use it to estimate 𝐻𝑇 (𝑇 , 𝑤) for 𝑇 = 10, 20, and 30 and 𝑤 = 0.1, 0.2, and 0.3.

60

3000

30

37. Use Figure 14.17 to estimate 𝐻𝑇 (10, 0.1). Interpret the partial derivative in practical terms.

100

4000

20

Figure 14.17

140

6000

𝑇 (◦ C) 10

9

Figure 14.15

11 13 15

39. Repeat Problem 38 for 𝐻𝑤 (𝑇 , 𝑤) at 𝑇 = 10, 20, and 30 and 𝑤 = 0.1, 0.2, and 0.3. What is the practical meaning of these partial derivatives?

748

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

40. The cardiac output, represented by 𝑐, is the volume of blood flowing through a person’s heart per unit time. The systemic vascular resistance (SVR), represented by 𝑠, is the resistance to blood flowing through veins and arteries. Let 𝑝 be a person’s blood pressure. Then 𝑝 is a function of 𝑐 and 𝑠, so 𝑝 = 𝑓 (𝑐, 𝑠).

A common mistake made by medical residents is to get the patient’s blood pressure back to normal by using drugs to increase the SVR, rather than by increasing the cardiac output. On a graph of the level curves of 𝑝, put a point 𝐷 representing the patient before the heart attack, a point 𝐸 representing the patient right after the heart attack, and a third point 𝐹 representing the patient after the resident has given the drugs to increase the SVR.

(a) What does 𝜕𝑝∕𝜕𝑐 represent? Suppose now that 𝑝 = 𝑘𝑐𝑠, where 𝑘 is a constant. (b) Sketch the level curves of 𝑝. What do they represent? Label your axes. (c) For a person with a weak heart, it is desirable to have the heart pumping against less resistance, while maintaining the same blood pressure. Such a person may be given the drug nitroglycerine to decrease the SVR and the drug dopamine to increase the cardiac output. Represent this on a graph showing level curves. Put a point 𝐴 on the graph representing the person’s state before drugs are given and a point 𝐵 for after. (d) Right after a heart attack, a patient’s cardiac output drops, thereby causing the blood pressure to drop.

41. In each case, give a possible contour diagram for the function 𝑓 (𝑥, 𝑦) if (a) 𝑓𝑥 > 0 and 𝑓𝑦 > 0 (c) 𝑓𝑥 < 0 and 𝑓𝑦 > 0

(b) 𝑓𝑥 > 0 and 𝑓𝑦 < 0 (d) 𝑓𝑥 < 0 and 𝑓𝑦 < 0

In Problems 42–45, give a possible contour diagram for the function 𝑓 (𝑥, 𝑦) if 42. 𝑓𝑥 = 0, 43. 𝑓𝑦 = 0, 44. 𝑓𝑥 = 1 𝑓𝑦 ≠ 0 𝑓𝑥 ≠ 0

45. 𝑓𝑦 = −2

Strengthen Your Understanding In Problems 46–47, explain what is wrong with the statement. 46. For 𝑓 (𝑥, 𝑦), 𝜕𝑓 ∕𝜕𝑥 has the same units as 𝜕𝑓 ∕𝜕𝑦. 47. The partial derivative with respect to 𝑦 is not defined for functions such as 𝑓 (𝑥, 𝑦) = 𝑥2 + 5 that have a formula that does not contain 𝑦 explicitly. In Problems 48–49, give an example of: 48. A table of values with three rows and three columns of a linear function 𝑓 (𝑥, 𝑦) with 𝑓𝑥 < 0 and 𝑓𝑦 > 0. 49. A function 𝑓 (𝑥, 𝑦) with 𝑓𝑥 > 0 and 𝑓𝑦 < 0 everywhere.

53. If 𝑃 = 𝑓 (𝑇 , 𝑉 ) is a function expressing the pressure 𝑃 (in grams∕cm3 ) of gas in a piston in terms of the temperature 𝑇 (in degrees ◦ C) and volume 𝑉 (in cm3 ), then 𝜕𝑃 ∕𝜕𝑉 has units of grams. 54. If 𝑓𝑥 (𝑎, 𝑏) > 0, then the values of 𝑓 decrease as we move in the negative 𝑥-direction near (𝑎, 𝑏). 55. If 𝑔(𝑟, 𝑠) = 𝑟2 + 𝑠, then for fixed 𝑠, the partial derivative 𝑔𝑟 increases as 𝑟 increases. 56. Let 𝑃 = 𝑓 (𝑚, 𝑑) be the purchase price (in dollars) of a used car that has 𝑚 miles on its engine and originally cost 𝑑 dollars when new. Then 𝜕𝑃 ∕𝜕𝑚 and 𝜕𝑃 ∕𝜕𝑑 have the same sign.

Are the statements in Problems 50–60 true or false? Give reasons for your answer.

57. If 𝑓 (𝑥, 𝑦) is a function with the property that 𝑓𝑥 (𝑥, 𝑦) and 𝑓𝑦 (𝑥, 𝑦) are both constant, then 𝑓 is linear.

50. If 𝑓 (𝑥, 𝑦) is a function of two variables and 𝑓𝑥 (10, 20) is defined, then 𝑓𝑥 (10, 20) is a scalar.

58. If 𝑓 (𝑥, 𝑦) has 𝑓𝑥 (𝑎, 𝑏) = 𝑓𝑦 (𝑎, 𝑏) = 0 at the point (𝑎, 𝑏), then 𝑓 is constant everywhere.

51. If 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 , then 𝑓𝑦 (1, 1) < 0.

59. If 𝑓𝑥 = 0 and 𝑓𝑦 ≠ 0, then the contours of 𝑓 (𝑥, 𝑦) are horizontal lines.

52. If the graph of 𝑓 (𝑥, 𝑦) is a hemisphere centered at the origin, then 𝑓𝑥 (0, 0) = 𝑓𝑦 (0, 0) = 0.

14.2

60. If the contours of 𝑓 (𝑥, 𝑦) are vertical lines, then 𝑓𝑦 = 0.

COMPUTING PARTIAL DERIVATIVES ALGEBRAICALLY Since the partial derivative 𝑓𝑥 (𝑥, 𝑦) is the ordinary derivative of the function 𝑓 (𝑥, 𝑦) with 𝑦 held constant and 𝑓𝑦 (𝑥, 𝑦) is the ordinary derivative of 𝑓 (𝑥, 𝑦) with 𝑥 held constant, we can use all the differentiation formulas from one-variable calculus to find partial derivatives.

Example 1

Let 𝑓 (𝑥, 𝑦) =

𝑥2 . Find 𝑓𝑥 (3, 2) algebraically. 𝑦+1

14.2 COMPUTING PARTIAL DERIVATIVES ALGEBRAICALLY

Solution

749

We use the fact that 𝑓𝑥 (3, 2) equals the derivative of 𝑓 (𝑥, 2) at 𝑥 = 3. Since 𝑓 (𝑥, 2) =

𝑥2 𝑥2 = , 2+1 3

differentiating with respect to 𝑥, we have ( ) 2𝑥 𝜕 𝑥2 𝑓𝑥 (𝑥, 2) = = , 𝜕𝑥 3 3

and so

𝑓𝑥 (3, 2) = 2.

Example 2

Compute the partial derivatives with respect to 𝑥 and with respect to 𝑦 for the following functions. (a) 𝑓 (𝑥, 𝑦) = 𝑦2 𝑒3𝑥 (b) 𝑧 = (3𝑥𝑦 + 2𝑥)5 (c) 𝑔(𝑥, 𝑦) = 𝑒𝑥+3𝑦 sin(𝑥𝑦)

Solution

(a) This is the product of a function of 𝑥 (namely 𝑒3𝑥 ) and a function of 𝑦 (namely 𝑦2 ). When we differentiate with respect to 𝑥, we think of the function of 𝑦 as a constant, and vice versa. Thus, 𝜕 ( 3𝑥 ) 𝑒 = 3𝑦2 𝑒3𝑥 , 𝜕𝑥 𝜕 𝑓𝑦 (𝑥, 𝑦) = 𝑒3𝑥 (𝑦2 ) = 2𝑦𝑒3𝑥 . 𝜕𝑦

𝑓𝑥 (𝑥, 𝑦) = 𝑦2

(b) Here we use the chain rule: 𝜕𝑧 𝜕 = 5(3𝑥𝑦 + 2𝑥)4 (3𝑥𝑦 + 2𝑥) = 5(3𝑥𝑦 + 2𝑥)4 (3𝑦 + 2), 𝜕𝑥 𝜕𝑥 𝜕𝑧 4 𝜕 = 5(3𝑥𝑦 + 2𝑥) (3𝑥𝑦 + 2𝑥) = 5(3𝑥𝑦 + 2𝑥)4 3𝑥 = 15𝑥(3𝑥𝑦 + 2𝑥)4 . 𝜕𝑦 𝜕𝑦 (c) Since each function in the product is a function of both 𝑥 and 𝑦, we need to use the product rule for each partial derivative: ( ) 𝜕 𝜕 𝑥+3𝑦 (𝑒 ) sin(𝑥𝑦) + 𝑒𝑥+3𝑦 (sin(𝑥𝑦)) = 𝑒𝑥+3𝑦 sin(𝑥𝑦) + 𝑒𝑥+3𝑦 𝑦 cos(𝑥𝑦), 𝑔𝑥 (𝑥, 𝑦) = 𝜕𝑥 𝜕𝑥 ( ) 𝜕 𝜕 𝑥+3𝑦 (𝑒 ) sin(𝑥𝑦) + 𝑒𝑥+3𝑦 (sin(𝑥𝑦)) = 3𝑒𝑥+3𝑦 sin(𝑥𝑦) + 𝑒𝑥+3𝑦 𝑥 cos(𝑥𝑦). 𝑔𝑦 (𝑥, 𝑦) = 𝜕𝑦 𝜕𝑦

For functions of three or more variables, we find partial derivatives by the same method: Differentiate with respect to one variable, regarding the other variables as constants. For a function 𝑓 (𝑥, 𝑦, 𝑧), the partial derivative 𝑓𝑥 (𝑎, 𝑏, 𝑐) gives the rate of change of 𝑓 with respect to 𝑥 along the line 𝑦 = 𝑏, 𝑧 = 𝑐. 𝑥2 𝑦3 . 𝑧

Example 3

Find all the partial derivatives of 𝑓 (𝑥, 𝑦, 𝑧) =

Solution

To find 𝑓𝑥 (𝑥, 𝑦, 𝑧), for example, we consider 𝑦 and 𝑧 as fixed, giving 𝑓𝑥 (𝑥, 𝑦, 𝑧) =

2𝑥𝑦3 , 𝑧

and 𝑓𝑦 (𝑥, 𝑦, 𝑧) =

3𝑥2 𝑦2 , 𝑧

and 𝑓𝑧 (𝑥, 𝑦, 𝑧) = −

𝑥2 𝑦3 . 𝑧2

750

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

Interpretation of Partial Derivatives Example 4

A vibrating guitar string, originally at rest along the 𝑥-axis, is shown in Figure 14.18. Let 𝑥 be the distance in meters from the left end of the string. At time 𝑡 seconds the point 𝑥 has been displaced 𝑦 = 𝑓 (𝑥, 𝑡) meters vertically from its rest position, where 𝑦 = 𝑓 (𝑥, 𝑡) = 0.003 sin(𝜋𝑥) sin(2765𝑡). Evaluate 𝑓𝑥 (0.3, 1) and 𝑓𝑡 (0.3, 1) and explain what each means in practical terms. 𝑓 (𝑥, 2) 𝑓 (𝑥, 1)

𝑦 (meters) 0.003 0

✠ ✠ 𝑥 (meters)

1

0.5



−0.003

𝑓 (𝑥, 10) Figure 14.18: The position of a vibrating guitar string at several different times: Graph of 𝑓 (𝑥, 𝑡) for 𝑡 = 1, 2, 10.

Solution

Differentiating 𝑓 (𝑥, 𝑡) = 0.003 sin(𝜋𝑥) sin(2765𝑡) with respect to 𝑥, we have 𝑓𝑥 (𝑥, 𝑡) = 0.003𝜋 cos(𝜋𝑥) sin(2765𝑡). In particular, substituting 𝑥 = 0.3 and 𝑡 = 1 gives 𝑓𝑥 (0.3, 1) = 0.003𝜋 cos(𝜋(0.3)) sin(2765) ≈ 0.002. To see what 𝑓𝑥 (0.3, 1) means, think about the function 𝑓 (𝑥, 1). The graph of 𝑓 (𝑥, 1) in Figure 14.19 is a snapshot of the string at the time 𝑡 = 1. Thus, the derivative 𝑓𝑥 (0.3, 1) is the slope of the string at the point 𝑥 = 0.3 at the instant when 𝑡 = 1. Similarly, taking the derivative of 𝑓 (𝑥, 𝑡) = 0.003 sin(𝜋𝑥) sin(2765𝑡) with respect to 𝑡, we get 𝑓𝑡 (𝑥, 𝑡) = (0.003)(2765) sin(𝜋𝑥) cos(2765𝑡) = 8.3 sin(𝜋𝑥) cos(2765𝑡). Since 𝑓 (𝑥, 𝑡) is in meters and 𝑡 is in seconds, the derivative 𝑓𝑡 (0.3, 1) is in m/sec. Thus, substituting 𝑥 = 0.3 and 𝑡 = 1, 𝑓𝑡 (0.3, 1) = 8.3 sin(𝜋(0.3)) cos(2765(1)) ≈ 6 m/sec. 𝑦 (meters) 0.002 Slope = 𝑓𝑥 (0.3, 1) = 0.002

0.001 𝑓 (𝑥, 1) 0

0.5

1

𝑥 (meters)

Figure 14.19: Graph of 𝑓 (𝑥, 1): Snapshot of the shape of the string at 𝑡 = 1 sec

To see what 𝑓𝑡 (0.3, 1) means, think about the function 𝑓 (0.3, 𝑡). The graph of 𝑓 (0.3, 𝑡) is a position versus time graph that tracks the up-and-down movement of the point on the string where 𝑥 = 0.3.

14.2 COMPUTING PARTIAL DERIVATIVES ALGEBRAICALLY

751

(See Figure 14.20.) The derivative 𝑓𝑡 (0.3, 1) = 6 m/sec is the velocity of that point on the string at time 𝑡 = 1. The fact that 𝑓𝑡 (0.3, 1) is positive indicates that the point is moving upward when 𝑡 = 1. 𝑦 (meters)

Slope = 𝑓𝑡 (0.3, 1) = 6 m/sec

0.003 0.002 0.001 0.996

−0.001 −0.002

1

1.004

𝑡 (seconds)

Figure 14.20: Graph of 𝑓 (0.3, 𝑡): Position versus time graph of the point 𝑥 = 0.3 m from the left end of the guitar string

Exercises and Problems for Section 14.2 EXERCISES 1. (a) If 𝑓 (𝑥, 𝑦) = 2𝑥2 + 𝑥𝑦 + 𝑦2 , approximate 𝑓𝑦 (3, 2) using Δ𝑦 = 0.01. (b) Find the exact value of 𝑓𝑦 (3, 2). In Exercises 2–40, find the partial derivatives. The variables are restricted to a domain on which the function is defined. 2. 𝑓𝑥 and 𝑓𝑦 if 𝑓 (𝑥, 𝑦) = 5𝑥2 𝑦3 + 8𝑥𝑦2 − 3𝑥2 3. 𝑓𝑥 (1, 2) and 𝑓𝑦 (1, 2) if 𝑓 (𝑥, 𝑦) = 𝑥3 + 3𝑥2 𝑦 − 2𝑦2 𝜕 4. (3𝑥5 𝑦7 − 32𝑥4 𝑦3 + 5𝑥𝑦) 𝜕𝑦 𝜕𝑧 𝜕𝑧 and if 𝑧 = (𝑥2 + 𝑥 − 𝑦)7 5. 𝜕𝑥 𝜕𝑦 6. 𝑓𝑥 and 𝑓𝑦 if 𝑓 (𝑥, 𝑦) = 𝐴𝛼 𝑥𝛼+𝛽 𝑦1−𝛼−𝛽 7. 𝑓𝑥 and 𝑓𝑦 if 𝑓 (𝑥, 𝑦) = ln(𝑥0.6 𝑦0.4 ) 8. 𝑧𝑥 if 𝑧 =

3𝑥5 𝑎𝑏𝑐 1 + 2 𝑦 2𝑥 𝑎𝑦

9. 𝑧𝑥 if 𝑧 = 𝑥2 𝑦 + 2𝑥5 𝑦 11. 𝑉𝑟 if 𝑉 = 31 𝜋𝑟2 ℎ 13.

√ 𝜕 (𝑥𝑒 𝑥𝑦 ) 𝜕𝑥

14.

15. 𝐹𝑚 if 𝐹 = 𝑚𝑔 𝜕𝐴 if 𝐴 = 12 (𝑎 + 𝑏)ℎ 𝜕ℎ ( ) 1 2 𝜕 𝐵 19. 𝜕𝐵 𝑢0

17.

21. 𝐹𝑣 if 𝐹 =

𝜕 √ (𝑎 𝑥) 𝜕𝑥 ( ) 𝜕 2𝜋𝑟 12. 𝜕𝑇 𝑇 10.

𝑚𝑣2 𝑟

𝜕 sin(𝑥+𝑐𝑡) 𝑒 𝜕𝑡

𝑣2 𝑟 ( ) 𝜕 1 2 𝑚𝑣 18. 𝜕𝑚 2 ( ) 𝜕 2𝜋𝑟 20. 𝜕𝑟 𝑣 16. 𝑎𝑣 if 𝑎 =

22.

𝜕 (𝑣 + 𝑎𝑡) 𝜕𝑣0 0

23. 𝑧𝑥 if 𝑧 = sin(5𝑥3 𝑦 − 3𝑥𝑦2 ) 𝜕𝑧 || if 𝑧 = 𝑒𝑥+2𝑦 sin 𝑦 24. | 𝜕𝑦 ||(1,0.5) 25. 𝑔𝑥 if 𝑔(𝑥, 𝑦) = ln(𝑦𝑒𝑥𝑦 ) 𝜕𝑓 || if 𝑓 (𝑥, 𝑦) = 𝑥 ln(𝑦 cos 𝑥) 26. 𝜕𝑥 ||(𝜋∕3,1) 27. 𝑧𝑥 and 𝑧𝑦 for 𝑧 = 𝑥7 + 2𝑦 + 𝑥𝑦

28. 𝑓𝑥 if 𝑓 (𝑥, 𝑦) = 𝑒𝑥𝑦 (ln 𝑦) 𝐺𝑚1 𝑚2 𝜕𝐹 29. if 𝐹 = 𝜕𝑚2 𝑟2 ( ) 𝜕 1 −𝑥2 ∕𝑎2 𝑒 30. 𝜕𝑥 𝑎 ( ) 𝜕 1 −𝑥2 ∕𝑎2 31. 𝑒 𝜕𝑎 𝑎 𝜕 1 32. (𝑣 𝑡 + 𝑎𝑡2 ) 𝜕𝑡 0 2 𝜕 (sin (𝜋𝜃𝜙) + ln(𝜃 2 + 𝜙)) 33. 𝜕𝜃 ) ( 𝜕 2𝜋𝑟3∕2 34. √ 𝜕𝑀 𝐺𝑀

35. 𝑓𝑎 if 𝑓 (𝑎, 𝑏) = 𝑒𝑎 sin(𝑎 + 𝑏) √ 36. 𝐹𝐿 if 𝐹 (𝐿, 𝐾) = 3 𝐿𝐾 𝜕𝑉 𝜕𝑉 and if 𝑉 = 43 𝜋𝑟2 ℎ 37. 𝜕𝑟 𝜕ℎ 1 1 2 38. 𝑢𝐸 if 𝑢 = 𝜖0 𝐸 2 + 𝐵 2 2𝜇0 ) ( 𝜕 1 −(𝑥−𝜇)2 ∕(2𝜎 2 ) 𝑒 39. √ 𝜕𝑥 2𝜋𝜎 40.

𝜕𝑄 if 𝑄 = 𝑐(𝑎1 𝐾 𝑏1 + 𝑎2 𝐿𝑏2 )𝛾 𝜕𝐾

752

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

PROBLEMS In Problems 41–43: (a) Find 𝑓𝑥 (1, 1) and 𝑓𝑦 (1, 1). (b) Use part (a) to match 𝑓 (𝑥, 𝑦) with one of the contour diagrams (I)–(III), each shown centered at (1, 1) with the same scale in the 𝑥 and 𝑦 directions. (I)

(II)

47. Money in a bank account earns interest at a continuous rate, 𝑟. The amount of money, $𝐵, in the account depends on the amount deposited, $𝑃 , and the time, 𝑡, it has been in the bank according to the formula 𝐵 = 𝑃 𝑒𝑟𝑡 . Find 𝜕𝐵∕𝜕𝑡 and 𝜕𝐵∕𝜕𝑃 and interpret each in financial terms. 48. The acceleration 𝑔 due to gravity, at a distance 𝑟 from the center of a planet of mass 𝑚, is given by 𝐺𝑚 , 𝑟2 where 𝐺 is the universal gravitational constant. 𝑔=

(III)

(a) Find 𝜕𝑔∕𝜕𝑚 and 𝜕𝑔∕𝜕𝑟. (b) Interpret each of the partial derivatives you found in part (a) as the slope of a graph in the plane and sketch the graph. 2

41. 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 2

42. 𝑓 (𝑥, 𝑦) = 𝑒𝑥 + 𝑦2

𝑦2

43. 𝑓 (𝑥, 𝑦) = 𝑥 + 𝑒

49. The Dubois formula relates a person’s surface area, 𝑠, in m2 , to weight, 𝑤, in kg, and height, ℎ, in cm, by 𝑠 = 𝑓 (𝑤, ℎ) = 0.01𝑤0.25 ℎ0.75 .

2

2

44. (a) Let 𝑓 (𝑥, 𝑦) = 𝑥 +𝑦 . Estimate 𝑓𝑥 (2, 1) and 𝑓𝑦 (2, 1) using the contour diagram for 𝑓 in Figure 14.21. (b) Estimate 𝑓𝑥 (2, 1) and 𝑓𝑦 (2, 1) from a table of values for 𝑓 with 𝑥 = 1.9, 2, 2.1 and 𝑦 = 0.9, 1, 1.1. (c) Compare your estimates in parts (a) and (b) with the exact values of 𝑓𝑥 (2, 1) and 𝑓𝑦 (2, 1) found algebraically. 𝑦 3 2

4

2

1

𝑥 2

−1 −2

4

50. The energy, 𝐸, of a body of mass 𝑚 moving with speed 𝑣 is given by the formula ( ) 1 2 −1 . 𝐸 = 𝑚𝑐 √ 1 − 𝑣2 ∕𝑐 2 The speed, 𝑣, is nonnegative and less than the speed of light, 𝑐, which is a constant.

8

6

Find 𝑓 (65, 160), 𝑓𝑤 (65, 160), and 𝑓ℎ (65, 160). Interpret your answers in terms of surface area, height, and weight.

6

8

(a) Find 𝜕𝐸∕𝜕𝑚. What would you expect the sign of 𝜕𝐸∕𝜕𝑚 to be? Explain. (b) Find 𝜕𝐸∕𝜕𝑣. Explain what you would expect the sign of 𝜕𝐸∕𝜕𝑣 to be and why.

Figure 14.21

51. Let ℎ(𝑥, 𝑡) = 5 + cos(0.5𝑥 − 𝑡) describe a wave. The value of ℎ(𝑥, 𝑡) gives the depth of the water in cm at a distance 𝑥 meters from a fixed point and at time 𝑡 seconds. Evaluate ℎ𝑥 (2, 5) and ℎ𝑡 (2, 5) and interpret each in terms of the wave.

45. (a) Let 𝑓 (𝑤, 𝑧) = 𝑒𝑤 ln 𝑧 . Use difference quotients with ℎ = 0.01 to approximate 𝑓𝑤 (2, 2) and 𝑓𝑧 (2, 2). (b) Now evaluate 𝑓𝑤 (2, 2) and 𝑓𝑧 (2, 2) exactly.

52. A one-meter-long bar is heated unevenly, with temperature in ◦ C at a distance 𝑥 meters from one end at time 𝑡 given by

−3 −3 −2 −1

1

2

3

46. (a) The surface 𝑆 is given, for some constant 𝑎, by 2

2

𝑧 = 3𝑥 + 4𝑦 − 𝑎𝑥𝑦 Find the values of 𝑎 which ensure that 𝑆 is sloping upward when we move in the positive 𝑥-direction from the point (1, 2). (b) With the values of 𝑎 from part (a), if you move in the positive 𝑦-direction from the point (1, 2), does the surface slope up or down? Explain.

𝐻(𝑥, 𝑡) = 100𝑒−0.1𝑡 sin(𝜋𝑥)

0 ≤ 𝑥 ≤ 1.

(a) Sketch a graph of 𝐻 against 𝑥 for 𝑡 = 0 and 𝑡 = 1. (b) Calculate 𝐻𝑥 (0.2, 𝑡) and 𝐻𝑥 (0.8, 𝑡). What is the practical interpretation (in terms of temperature) of these two partial derivatives? Explain why each one has the sign it does. (c) Calculate 𝐻𝑡 (𝑥, 𝑡). What is its sign? What is its interpretation in terms of temperature?

14.3 LOCAL LINEARITY AND THE DIFFERENTIAL

53. Show that the Cobb-Douglas function 𝑄 = 𝑏𝐾 𝛼 𝐿1−𝛼 where

0 0. 60. For 𝑓 (𝑥, 𝑦), if 0.01 In Problems 61–63, give an example of: 61. A nonlinear function 𝑓 (𝑥, 𝑦) such that 𝑓𝑥 (0, 0) = 2 and 𝑓𝑦 (0, 0) = 3. 62. Functions 𝑓 (𝑥, 𝑦) and 𝑔(𝑥, 𝑦) such that 𝑓𝑥 = 𝑔𝑥 but 𝑓𝑦 ≠ 𝑔𝑦 . 63. A non-constant function 𝑓 (𝑥, 𝑦) such that 𝑓𝑥 = 0 everywhere. Are the statements in Problems 64–71 true or false? Give reasons for your answer. 64. There is a function 𝑓 (𝑥, 𝑦) with 𝑓𝑥 (𝑥, 𝑦) = 𝑦 and 𝑓𝑦 (𝑥, 𝑦) = 𝑥. 65. The function 𝑧(𝑢, 𝑣) = 𝑢 cos 𝑣 satisfies the equation

𝜕 (𝑔(𝑥)𝑓 (𝑥, 𝑦)) = 𝑔(𝑥)𝑓𝑦 (𝑥, 𝑦). 𝜕𝑦

67. The function 𝑘(𝑟, 𝑠) = 𝑟𝑠𝑒𝑠 is increasing in the 𝑠direction at the point (𝑟, 𝑠) = (−1, 2). 68. There is a function 𝑓 (𝑥, 𝑦) with 𝑓𝑥 (𝑥, 𝑦) = 𝑦2 and 𝑓𝑦 (𝑥, 𝑦) = 𝑥2 . 69. If 𝑓 (𝑥, 𝑦) has 𝑓𝑦 (𝑥, 𝑦) = 0 then 𝑓 must be a constant. 70. If 𝑓 (𝑥, 𝑦) = 𝑦𝑒𝑔(𝑥) then 𝑓𝑥 = 𝑓 . 71. If 𝑓 is a symmetric two-variable function, that is 𝑓 (𝑥, 𝑦) = 𝑓 (𝑦, 𝑥), then 𝑓𝑥 (𝑥, 𝑦) = 𝑓𝑦 (𝑥, 𝑦). 72. Which of the following functions satisfy the following equation (called Euler’s Equation):

𝜕𝑧 sin 𝑣 𝜕𝑧 − = 1. cos 𝑣 𝜕𝑢 𝑢 𝜕𝑣

14.3

𝑥𝑓𝑥 + 𝑦𝑓𝑦 = 𝑓 ? (a)

𝑥2 𝑦3

(b) 𝑥+𝑦+1 (c) 𝑥2 + 𝑦2 (d) 𝑥0.4 𝑦0.6

LOCAL LINEARITY AND THE DIFFERENTIAL In Sections 14.1 and 14.2 we studied a function of two variables by allowing one variable at a time to change. We now let both variables change at once to develop a linear approximation for functions of two variables.

Zooming In to See Local Linearity For a function of one variable, local linearity means that as we zoom in on the graph, it looks like a straight line. As we zoom in on the graph of a two-variable function, the graph usually looks like a plane, which is the graph of a linear function of two variables. (See Figure 14.22.)

754

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES





Figure 14.22: Zooming in on the graph of a function of two variables until the graph looks like a plane

Similarly, Figure 14.23 shows three successive views of the contours near a point. As we zoom in, the contours look more like equally spaced parallel lines, which are the contours of a linear function. (As we zoom in, we have to add more contours.)

Figure 14.23: Zooming in on a contour diagram until the lines look parallel and equally spaced

This effect can also be seen numerically by zooming in with tables of values. Table 14.5 shows three tables of values for 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦3 near 𝑥 = 2, 𝑦 = 1, each one a closer view than the previous one. Notice how each table looks more like the table of a linear function. Table 14.5

Zooming in on values of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦3 near (2, 1) until the table looks linear 𝑦

1 𝑥

𝑦

0

1

2

1

2

9

2

4

5

12

3

9

10

17

1.9 𝑥

𝑦

0.9

1.0

1.1

4.34

4.61

4.94

2.0

4.73

5.00

5.33

2.1

5.14

5.41

5.74

𝑥

0.99

1.00

1.01

1.99

4.93

4.96

4.99

2.00

4.97

5.00

5.03

2.01

5.01

5.04

5.07

Zooming in Algebraically: Differentiability Seeing a plane when we zoom in at a point tells us (provided the plane is not vertical) that 𝑓 (𝑥, 𝑦) is closely approximated near that point by a linear function, 𝐿(𝑥, 𝑦): 𝑓 (𝑥, 𝑦) ≈ 𝐿(𝑥, 𝑦).

The Tangent Plane The graph of the function 𝑧 = 𝐿(𝑥, 𝑦) is the tangent plane at that point. See Figure 14.24. Provided the approximation is sufficiently good, we say that 𝑓 (𝑥, 𝑦) is differentiable at the point. Section 14.8 on

14.3 LOCAL LINEARITY AND THE DIFFERENTIAL

755

Figure 14.24: The tangent plane to the surface 𝑧 = 𝑓 (𝑥, 𝑦) at the point (𝑎, 𝑏)

page 799 defines precisely what is meant by the approximation being sufficiently good. The functions we encounter are differentiable at most points in their domain. What is the equation of the tangent plane? At the point (𝑎, 𝑏), the 𝑥-slope of the graph of 𝑓 is the partial derivative 𝑓𝑥 (𝑎, 𝑏) and the 𝑦-slope is 𝑓𝑦 (𝑎, 𝑏). Thus, using the equation for a plane on page 683 of Chapter 12, we have the following result:

Tangent Plane to the Surface 𝒛 = 𝒇 (𝒙, 𝒚) at the Point (𝒂, 𝒃) Assuming 𝑓 is differentiable at (𝑎, 𝑏), the equation of the tangent plane is 𝑧 = 𝑓 (𝑎, 𝑏) + 𝑓𝑥 (𝑎, 𝑏)(𝑥 − 𝑎) + 𝑓𝑦 (𝑎, 𝑏)(𝑦 − 𝑏). Here we are thinking of 𝑎 and 𝑏 as fixed, so 𝑓 (𝑎, 𝑏), and 𝑓𝑥 (𝑎, 𝑏), and 𝑓𝑦 (𝑎, 𝑏) are constants. Thus, the right side of the equation is a linear function of 𝑥 and 𝑦. Example 1

Find the equation for the tangent plane to the surface 𝑧 = 𝑥2 + 𝑦2 at the point (3, 4).

Solution

We have 𝑓𝑥 (𝑥, 𝑦) = 2𝑥, so 𝑓𝑥 (3, 4) = 6, and 𝑓𝑦 (𝑥, 𝑦) = 2𝑦, so 𝑓𝑦 (3, 4) = 8. Also, 𝑓 (3, 4) = 32 +42 = 25. Thus, the equation for the tangent plane at (3, 4) is 𝑧 = 25 + 6(𝑥 − 3) + 8(𝑦 − 4).

Local Linearization Since the tangent plane lies close to the surface near the point at which they meet, 𝑧-values on the tangent plane are close to values of 𝑓 (𝑥, 𝑦) for points near (𝑎, 𝑏). Thus, replacing 𝑧 by 𝑓 (𝑥, 𝑦) in the equation of the tangent plane, we get the following approximation:

Tangent Plane Approximation to 𝒇 (𝒙, 𝒚) for (𝒙, 𝒚) Near the Point (𝒂, 𝒃) Provided 𝑓 is differentiable at (𝑎, 𝑏), we can approximate 𝑓 (𝑥, 𝑦): 𝑓 (𝑥, 𝑦) ≈ 𝑓 (𝑎, 𝑏) + 𝑓𝑥 (𝑎, 𝑏)(𝑥 − 𝑎) + 𝑓𝑦 (𝑎, 𝑏)(𝑦 − 𝑏). We are thinking of 𝑎 and 𝑏 as fixed, so the expression on the right side is linear in 𝑥 and 𝑦. The right side of this approximation gives the local linearization of 𝑓 near 𝑥 = 𝑎, 𝑦 = 𝑏.

756

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

Figure 14.25 shows the tangent plane approximation graphically.

Figure 14.25: Local linearization: Approximating 𝑓 (𝑥, 𝑦) by the 𝑧-value from the tangent plane

Example 2

Find the local linearization of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 at the point (3, 4). Estimate 𝑓 (2.9, 4.2) and 𝑓 (2, 2) using the linearization and compare your answers to the true values.

Solution

Let 𝑧 = 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 . In Example 1, we found the equation of the tangent plane at (3, 4) to be 𝑧 = 25 + 6(𝑥 − 3) + 8(𝑦 − 4). Therefore, for (𝑥, 𝑦) near (3, 4), we have the local linearization 𝑓 (𝑥, 𝑦) ≈ 25 + 6(𝑥 − 3) + 8(𝑦 − 4). Substituting 𝑥 = 2.9, 𝑦 = 4.2 gives 𝑓 (2.9, 4.2) ≈ 25 + 6(−0.1) + 8(0.2) = 26. This compares favorably with the true value 𝑓 (2.9, 4.2) = (2.9)2 + (4.2)2 = 26.05. However, the local linearization does not give a good approximation at points far away from (3, 4). For example, if 𝑥 = 2, 𝑦 = 2, the local linearization gives 𝑓 (2, 2) ≈ 25 + 6(−1) + 8(−2) = 3, whereas the true value of the function is 𝑓 (2, 2) = 22 + 22 = 8.

Example 3

Designing safe boilers depends on knowing how steam behaves under changes in temperature and pressure. Steam tables, such as Table 14.6, are published giving values of the function 𝑉 = 𝑓 (𝑇 , 𝑃 ) where 𝑉 is the volume (in ft3 ) of one pound of steam at a temperature 𝑇 (in ◦ F) and pressure 𝑃 (in lb/in2 ). (a) Give a linear function approximating 𝑉 = 𝑓 (𝑇 , 𝑃 ) for 𝑇 near 500◦F and 𝑃 near 24 lb/in2 . (b) Estimate the volume of a pound of steam at a temperature of 505◦ F and a pressure of 24.3 lb/in2 .

14.3 LOCAL LINEARITY AND THE DIFFERENTIAL

757

Volume (in cubic feet) of one pound of steam at various temperatures and pressures

Table 14.6

Pressure 𝑃 (lb/in2 )

Temperature 𝑇 (◦ F)

Solution

20

22

24

26

480

27.85

25.31

23.19

21.39

500

28.46

25.86

23.69

21.86

520

29.06

26.41

24.20

22.33

540

29.66

26.95

24.70

22.79

(a) We want the local linearization around the point 𝑇 = 500, 𝑃 = 24, which is 𝑓 (𝑇 , 𝑃 ) ≈ 𝑓 (500, 24) + 𝑓𝑇 (500, 24)(𝑇 − 500) + 𝑓𝑃 (500, 24)(𝑃 − 24). We read the value 𝑓 (500, 24) = 23.69 from the table. Next we approximate 𝑓𝑇 (500, 24) by a difference quotient. From the 𝑃 = 24 column, we compute the average rate of change between 𝑇 = 500 and 𝑇 = 520: 𝑓𝑇 (500, 24) ≈

𝑓 (520, 24) − 𝑓 (500, 24) 24.20 − 23.69 = = 0.0255. 520 − 500 20

Note that 𝑓𝑇 (500, 24) is positive, because steam expands when heated. Next we approximate 𝑓𝑃 (500, 24) by looking at the 𝑇 = 500 row and computing the average rate of change between 𝑃 = 24 and 𝑃 = 26: 𝑓𝑃 (500, 24) ≈

𝑓 (500, 26) − 𝑓 (500, 24) 21.86 − 23.69 = = −0.915. 26 − 24 2

Note that 𝑓𝑃 (500, 24) is negative, because increasing the pressure on steam decreases its volume. Using these approximations for the partial derivatives, we obtain the local linearization: 𝑉 = 𝑓 (𝑇 , 𝑃 ) ≈ 23.69 + 0.0255(𝑇 − 500) − 0.915(𝑃 − 24) ft3

for 𝑇 near 500 ◦ F and 𝑃 near 24 lb∕in2 .

(b) We are interested in the volume at 𝑇 = 505◦F and 𝑃 = 24.3 lb/in2 . Since these values are close to 𝑇 = 500◦F and 𝑃 = 24 lb/in2 , we use the linear relation obtained in part (a): 𝑉 ≈ 23.69 + 0.0255(505 − 500) − 0.915(24.3 − 24) = 23.54 ft3 .

Local Linearity with Three or More Variables Local linear approximations for functions of three or more variables follow the same pattern as for functions of two variables. The local linearization of 𝑓 (𝑥, 𝑦, 𝑧) at (𝑎, 𝑏, 𝑐) is given by

𝑓 (𝑥, 𝑦, 𝑧) ≈ 𝑓 (𝑎, 𝑏, 𝑐) + 𝑓𝑥 (𝑎, 𝑏, 𝑐)(𝑥 − 𝑎) + 𝑓𝑦 (𝑎, 𝑏, 𝑐)(𝑦 − 𝑏) + 𝑓𝑧 (𝑎, 𝑏, 𝑐)(𝑧 − 𝑐).

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Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

The Differential We are often interested in the change in the value of the function as we move from the point (𝑎, 𝑏) to a nearby point (𝑥, 𝑦). We rewrite the tangent plane approximation as 𝑓 (𝑥, 𝑦) − 𝑓 (𝑎, 𝑏) ≈ 𝑓𝑥 (𝑎, 𝑏) (𝑥 − 𝑎) +𝑓𝑦 (𝑎, 𝑏) (𝑦 − 𝑏) , ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏟⏟ ⏟⏟⏟ Δ𝑥

Δ𝑓

Δ𝑦

giving us a relationship between Δ𝑓 , Δ𝑥, and Δ𝑦: Δ𝑓 ≈ 𝑓𝑥 (𝑎, 𝑏)Δ𝑥 + 𝑓𝑦 (𝑎, 𝑏)Δ𝑦. If 𝑎 and 𝑏 are fixed, 𝑓𝑥 (𝑎, 𝑏)Δ𝑥 + 𝑓𝑦 (𝑎, 𝑏)Δ𝑦 is a linear function of Δ𝑥 and Δ𝑦 that can be used to estimate Δ𝑓 for small Δ𝑥 and Δ𝑦. We introduce new variables 𝑑𝑥 and 𝑑𝑦 to represent changes in 𝑥 and 𝑦.

The Differential of a Function 𝒛 = 𝒇 (𝒙, 𝒚) The differential, 𝑑𝑓 (or 𝑑𝑧), at a point (𝑎, 𝑏) is the linear function of 𝑑𝑥 and 𝑑𝑦 given by the formula 𝑑𝑓 = 𝑓𝑥 (𝑎, 𝑏) 𝑑𝑥 + 𝑓𝑦 (𝑎, 𝑏) 𝑑𝑦. The differential at a general point is often written 𝑑𝑓 = 𝑓𝑥 𝑑𝑥 + 𝑓𝑦 𝑑𝑦.

Example 4

Compute the differentials of the following functions. (a) 𝑓 (𝑥, 𝑦) = 𝑥2 𝑒5𝑦 (b) 𝑧 = 𝑥 sin(𝑥𝑦) (c) 𝑓 (𝑥, 𝑦) = 𝑥 cos(2𝑥)

Solution

(a) Since 𝑓𝑥 (𝑥, 𝑦) = 2𝑥𝑒5𝑦 and 𝑓𝑦 (𝑥, 𝑦) = 5𝑥2 𝑒5𝑦 , we have 𝑑𝑓 = 2𝑥𝑒5𝑦 𝑑𝑥 + 5𝑥2 𝑒5𝑦 𝑑𝑦. (b) Since 𝜕𝑧∕𝜕𝑥 = sin(𝑥𝑦) + 𝑥𝑦 cos(𝑥𝑦) and 𝜕𝑧∕𝜕𝑦 = 𝑥2 cos(𝑥𝑦), we have 𝑑𝑧 = (sin(𝑥𝑦) + 𝑥𝑦 cos(𝑥𝑦)) 𝑑𝑥 + 𝑥2 cos(𝑥𝑦) 𝑑𝑦. (c) Since 𝑓𝑥 (𝑥, 𝑦) = cos(2𝑥) − 2𝑥 sin(2𝑥) and 𝑓𝑦 (𝑥, 𝑦) = 0, we have 𝑑𝑓 = (cos(2𝑥) − 2𝑥 sin(2𝑥)) 𝑑𝑥 + 0 𝑑𝑦 = (cos(2𝑥) − 2𝑥 sin(2𝑥)) 𝑑𝑥.

Example 5

The density 𝜌 (in g/cm3 ) of carbon dioxide gas CO2 depends upon its temperature 𝑇 (in ◦ C) and pressure 𝑃 (in atmospheres). The ideal gas model for CO2 gives what is called the state equation: 𝜌=

0.5363𝑃 . 𝑇 + 273.15

Compute the differential 𝑑𝜌. Explain the signs of the coefficients of 𝑑𝑇 and 𝑑𝑃 . Solution

The differential for 𝜌 = 𝑓 (𝑇 , 𝑃 ) is 𝑑𝜌 = 𝑓𝑇 (𝑇 , 𝑃 ) 𝑑𝑇 + 𝑓𝑃 (𝑇 , 𝑃 )𝑑𝑃 =

0.5363 −0.5363𝑃 𝑑𝑃 . 𝑑𝑇 + 𝑇 + 273.15 (𝑇 + 273.15)2

14.3 LOCAL LINEARITY AND THE DIFFERENTIAL

759

The coefficient of 𝑑𝑇 is negative because increasing the temperature expands the gas (if the pressure is kept constant) and therefore decreases its density. The coefficient of 𝑑𝑃 is positive because increasing the pressure compresses the gas (if the temperature is kept constant) and therefore increases its density.

Where Does the Notation for the Differential Come From? We write the differential as a linear function of the new variables 𝑑𝑥 and 𝑑𝑦. You may wonder why we chose these names for our variables. The reason is historical: The people who invented calculus thought of 𝑑𝑥 and 𝑑𝑦 as “infinitesimal” changes in 𝑥 and 𝑦. The equation 𝑑𝑓 = 𝑓𝑥 𝑑𝑥 + 𝑓𝑦 𝑑𝑦 was regarded as an infinitesimal version of the local linear approximation Δ𝑓 ≈ 𝑓𝑥 Δ𝑥 + 𝑓𝑦 Δ𝑦. In spite of the problems with defining exactly what “infinitesimal” means, some mathematicians, scientists, and engineers think of the differential in terms of infinitesimals. Figure 14.26 illustrates a way of thinking about differentials that combines the definition with this informal point of view. It shows the graph of 𝑓 along with a view of the graph around the point (𝑎, 𝑏, 𝑓 (𝑎, 𝑏)) under a microscope. Since 𝑓 is locally linear at the point, the magnified view looks like the tangent plane. Under the microscope, we use a magnified coordinate system with its origin at the point (𝑎, 𝑏, 𝑓 (𝑎, 𝑏)) and with coordinates 𝑑𝑥, 𝑑𝑦, and 𝑑𝑧 along the three axes. The graph of the differential 𝑑𝑓 is the tangent plane, which has equation 𝑑𝑧 = 𝑓𝑥 (𝑎, 𝑏) 𝑑𝑥 + 𝑓𝑦 (𝑎, 𝑏) 𝑑𝑦 in the magnified coordinates. 𝑧

Plane is graph of 𝑑𝑓

𝑑𝑧

Surface is graph of 𝑓



✠ ✲ 𝑑𝑥

𝑥

𝑑𝑦

𝑦

Figure 14.26: The graph of 𝑓 , with a view through a microscope showing the tangent plane in the magnified coordinate system

Exercises and Problems for Section 14.3 EXERCISES In Exercises 1–8, find the equation of the tangent plane at the given point.

In Exercises 9–12, find the differential of the function. 9. 𝑓 (𝑥, 𝑦) = sin(𝑥𝑦)

1. 𝑧 = 𝑦𝑒𝑥∕𝑦 at the point (1, 1, 𝑒)

10. 𝑔(𝑢, 𝑣) = 𝑢2 + 𝑢𝑣

2. 𝑧 = sin(𝑥𝑦) at 𝑥 = 2, 𝑦 = 3𝜋∕4

11. 𝑧 = 𝑒−𝑥 cos 𝑦

2

2

3. 𝑧 = ln(𝑥 + 1) + 𝑦 at the point (0, 3, 9) 𝑦

12. ℎ(𝑥, 𝑡) = 𝑒−3𝑡 sin(𝑥 + 5𝑡)

2

4. 𝑧 = 𝑒 + 𝑥 + 𝑥 + 6 at the point (1, 0, 9) 5. 𝑧 = 21 (𝑥2 + 4𝑦2 ) at the point (2, 1, 4) 6. 𝑥2 + 𝑦2 − 𝑧 = 1 at the point (1, 3, 9) 7. 𝑥2 𝑦2 + 𝑧 − 40 = 0 at 𝑥 = 2, 𝑦 = 3 8. 𝑥2 𝑦 + ln(𝑥𝑦) + 𝑧 = 6 at the point (4, 0.25, 2)

In Exercises 13–16, find the differential of the function at the point. 13. 𝑔(𝑥, 𝑡) = 𝑥2 sin(2𝑡) at (2, 𝜋∕4) 14. 𝑓 (𝑥, 𝑦) = 𝑥𝑒−𝑦 at (1, 0) 15. 𝑃 (𝐿, 𝐾) = 1.01𝐿0.25 𝐾 0.75 at (100, 1) 16. 𝐹 (𝑚, 𝑟) = 𝐺𝑚∕𝑟2 at (100, 10)

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Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

In Exercises 17–20, assume points 𝑃 and 𝑄 are close. Estimate Δ𝑓 = 𝑓 (𝑄) − 𝑓 (𝑃 ) using the differential 𝑑𝑓 .

In Exercises 21–24, assume points 𝑃 and 𝑄 are close. Estimate 𝑔(𝑄).

17. 𝑑𝑓 = 10 𝑑𝑥 − 5 𝑑𝑦, 𝑃 = (200, 400), 𝑄 = (202, 405)

21. 𝑃 = (60, 80), 𝑄 = (60.5, 82), 𝑔(𝑃 ) = 100, 𝑔𝑥 (𝑃 ) = 2, 𝑔𝑦 (𝑃 ) = −3.

18. 𝑑𝑓 = 𝑦 𝑑𝑥 + 𝑥 𝑑𝑦, 𝑃 = (10, 5), 𝑄 = (9.8, 5.3) √ √ 19. 𝑑𝑓 = 6 1 + 4𝑥 + 2𝑦 𝑑𝑥 + 3 1 + 4𝑥 + 2𝑦 𝑑𝑦, 𝑃 = (1, 2), 𝑄 = (1.03, 2.05) 20. 𝑑𝑓 = (2𝑥 + 2𝑦 + 5) 𝑑𝑥 + (2𝑥 + 3) 𝑑𝑦, 𝑃 = (0, 0), 𝑄 = (0.1, −0.2)

22. 𝑃 = (−150, 200), 𝑄 = (−152, 203), 𝑔(𝑃 ) = 2500, 𝑔𝑥 (𝑃 ) = 10, 𝑔𝑦 (𝑃 ) = 20. 23. 𝑃 = (5, 8), 𝑄 = (4.97, 7.99), 𝑔(𝑃 ) = 12, 𝑔𝑥 (𝑃 ) = −0.1, 𝑔𝑦 (𝑃 ) = −0.2. 24. 𝑃 = (30, 125), 𝑄 = (25, 135), 𝑔(𝑃 ) = 840, 𝑔𝑥 (𝑃 ) = 4, 𝑔𝑦 (𝑃 ) = 1.5.

PROBLEMS 25. At a distance of 𝑥 feet from the beach, the price in dollars of a plot of land of area 𝑎 square feet is 𝑓 (𝑎, 𝑥). (a) What are the units of 𝑓𝑎 (𝑎, 𝑥)? (b) What does 𝑓𝑎 (1000, 300) = 3 mean in practical terms? (c) What are the units of 𝑓𝑥 (𝑎, 𝑥)? (d) What does 𝑓𝑥 (1000, 300) = −2 mean in practical terms? (e) Which is cheaper: 1005 square feet that are 305 feet from the beach or 998 square feet that are 295 feet from the beach? Justify your answer. 26. A student was asked to find the equation of the tangent plane to the surface 𝑧 = 𝑥3 − 𝑦2 at the point (𝑥, 𝑦) = (2, 3). The student’s answer was

30. Find an equation for the tangent plane to 𝑧 = 𝑓 (𝑥, 𝑦) at (3, −2) if the differential at (3, −2) is 𝑑𝑓 = 5𝑑𝑥 + 𝑑𝑦 and 𝑓 (3, −2) = 8. 31. Find 𝑑𝑓 at (2, −4) if the tangent plane to 𝑧 = 𝑓 (𝑥, 𝑦) at (2, −4) is 𝑧 = −3(𝑥 − 2) + 2(𝑦 + 4) + 3. 32. Give a linear function approximating 𝑧 = 𝑓 (𝑥, 𝑦) near (1, −1) using its contour diagram in Figure 14.27. 𝑦 2 −3 −2 −1 1

1234

𝑥

0

−2

−1

1

4 3 2 1 −1

2

𝑧 = 3𝑥 (𝑥 − 2) − 2𝑦(𝑦 − 3) − 1.

2

−1 −2 −3

−2

(a) At a glance, how do you know this is wrong? (b) What mistake did the student make? (c) Answer the question correctly. 27. (a) Check the local linearity of 𝑓 (𝑥, 𝑦) = 𝑒−𝑥 sin 𝑦 near 𝑥 = 1, 𝑦 = 2 by making a table of values of 𝑓 for 𝑥 = 0.9, 1.0, 1.1 and 𝑦 = 1.9, 2.0, 2.1. Express values of 𝑓 with 4 digits after the decimal point. Then make a table of values for 𝑥 = 0.99, 1.00, 1.01 and 𝑦 = 1.99, 2.00. 2.01, again showing 4 digits after the decimal point. Do both tables look nearly linear? Does the second table look more linear than the first? (b) Give the local linearization of 𝑓 (𝑥, 𝑦) = 𝑒−𝑥 sin 𝑦 at (1, 2), first using your tables, and second using the fact that 𝑓𝑥 (𝑥, 𝑦) = −𝑒−𝑥 sin 𝑦 and 𝑓𝑦 (𝑥, 𝑦) = 𝑒−𝑥 cos 𝑦. 28. Find the local linearization of the function 𝑓 (𝑥, 𝑦) = 𝑥2 𝑦 at the point (3, 1). 29. The tangent plane to 𝑧 = 𝑓 (𝑥, 𝑦) at the point (1, 2) is 𝑧 = 3𝑥 + 2𝑦 − 5. (a) Find 𝑓𝑥 (1, 2) and 𝑓𝑦 (1, 2). (b) What is 𝑓 (1, 2)? (c) Approximate 𝑓 (1.1, 1.9).

Figure 14.27 33. For the differentiable function ℎ(𝑥, 𝑦), we are told that ℎ(600, 100) = 300 and ℎ𝑥 (600, 100) = 12 and ℎ𝑦 (600, 100) = −8. Estimate ℎ(605, 98). 34. (a) Find the equation of the plane tangent to the graph of 𝑓 (𝑥, 𝑦) = 𝑥2 𝑒𝑥𝑦 at (1, 0). (b) Find the linear approximation of 𝑓 (𝑥, 𝑦) for (𝑥, 𝑦) near (1, 0). (c) Find the differential of 𝑓 at the point (1, 0). √ 35. Find the differential of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦3 at the point (1, 2). Use it to estimate 𝑓 (1.04, 1.98). 36. (a) Find the differential of 𝑔(𝑢, 𝑣) = 𝑢2 + 𝑢𝑣. (b) Use your answer to part (a) to estimate the change in 𝑔 as you move from (1, 2) to (1.2, 2.1). 37. An unevenly heated plate has temperature 𝑇 (𝑥, 𝑦) in ◦ C at the point (𝑥, 𝑦). If 𝑇 (2, 1) = 135, and 𝑇𝑥 (2, 1) = 16, and 𝑇𝑦 (2, 1) = −15, estimate the temperature at the point (2.04, 0.97). 38. A right circular cylinder has a radius of 50 cm and a height of 100 cm. Use differentials to estimate the change in volume of the cylinder if its height and radius are both increased by 1 cm.

14.3 LOCAL LINEARITY AND THE DIFFERENTIAL

761

39. Give the local linearization for the monthly car-loan payment function at each of the points investigated in Problem 35 on page 747.

quantity of a liquid to an increase in its temperature 𝑇 (in ◦ C): 𝑑𝑉 = 𝛽𝑉 𝑑𝑇 .

40. In Example 3 on page 756 we found a linear approximation for 𝑉 = 𝑓 (𝑇 , 𝑃 ) near (500, 24). Now find a linear approximation near (480, 20).

(a) Let 𝜌 be the density (in kg/m3 ) of water as a function of temperature. (For a mass 𝑚 of liquid, we have 𝜌 = 𝑚∕𝑉 .) Write an expression for 𝑑𝜌 in terms of 𝜌 and 𝑑𝑇 . (b) The graph in Figure 14.28 shows density of water as a function of temperature. Use it to estimate 𝛽 when 𝑇 = 20◦ C and when 𝑇 = 80◦ C.

41. In Example 3 on page 756 we found a linear approximation for 𝑉 = 𝑓 (𝑇 , 𝑝) near (500, 24). (a) Test the accuracy of this approximation by comparing its predicted value with the four neighboring values in the table. What do you notice? Which predicted values are accurate? Which are not? Explain your answer. (b) Suggest a linear approximation for 𝑓 (𝑇 , 𝑝) near (500, 24) that does not have the property you noticed in part (a). [Hint: Estimate the partial derivatives in a different way.] 42. In a room, the temperature is given by 𝑇 = 𝑓 (𝑥, 𝑡) degrees Celsius, where 𝑥 is the distance from a heater (in meters) and 𝑡 is the elapsed time (in minutes) since the heat has been turned on. A person standing 3 meters from the heater 5 minutes after it has been turned on observes the following: (1) The temperature is increasing by 1.2◦ C per minute, and (2) As the person walks away from the heater, the temperature decreases by 2◦ C per meter as time is held constant. Estimate how much cooler or warmer it would be 2.5 meters from the heater after 6 minutes. 43. Van der Waal’s equation relates the pressure, 𝑃 , and the volume, 𝑉 , of a fixed quantity of a gas at constant temperature 𝑇 . For 𝑎, 𝑏, 𝑛, 𝑅 constants, the equation is ( ) 𝑛2 𝑎 𝑃 + 2 (𝑉 − 𝑛𝑏) = 𝑛𝑅𝑇 . 𝑉 (a) Express 𝑃 as a function of 𝑇 and 𝑉 . (b) Write a linear approximation for the change in pressure, Δ𝑃 = 𝑃 − 𝑃0 , resulting from a change in temperature Δ𝑇 = 𝑇 − 𝑇0 and a change in pressure, Δ𝑉 = 𝑉 − 𝑉0 . 44. The gas equation for one mole of oxygen relates its pressure, 𝑃 (in atmospheres), its temperature, 𝑇 (in K), and its volume, 𝑉 (in cubic decimeters, dm3 ): 𝑇 = 16.574

1 1 − 0.52754 2 − 0.3879𝑃 + 12.187𝑉 𝑃 . 𝑉 𝑉

(a) Find the temperature 𝑇 and differential 𝑑𝑇 if the volume is 25 dm3 and the pressure is 1 atmosphere. (b) Use your answer to part (a) to estimate how much the volume would have to change if the pressure increased by 0.1 atmosphere and the temperature remained constant. 45. The coefficient, 𝛽, of thermal expansion of a liquid relates the change in the volume 𝑉 (in m3 ) of a fixed

𝜌 (kg/m3 ) 1000 990 980 970 960 𝑇 (◦ C) 0

20

40

60

80

100

Figure 14.28 46. A fluid moves through a tube of length 1 meter and radius 𝑟 = 0.005 ± 0.00025 meters under a pressure 𝑝 = 105 ±1000 pascals, at a rate 𝑣 = 0.625⋅10−9 m3 per unit time. Use differentials to estimate the maximum error in the viscosity 𝜂 given by 𝜂=

𝜋 𝑝𝑟4 . 8 𝑣

47. The period, 𝑇 , of oscillation √ in seconds of a pendulum clock is given by 𝑇 = 2𝜋 𝑙∕𝑔, where 𝑔 is the acceleration due to gravity. The length of the pendulum, 𝑙, depends on the temperature, 𝑡, according to the formula 𝑙 = 𝑙0 (1 + 𝛼(𝑡 − 𝑡0 )) where 𝑙0 is the length of the pendulum at temperature 𝑡0 and 𝛼 is a constant which characterizes the clock. The clock is set to the correct period at the temperature 𝑡0 . How many seconds a day does the clock gain or lose when the temperature is 𝑡0 + Δ𝑡? Show that this gain or loss is independent of 𝑙0 . 48. Two functions that have the same local linearization at a point have contours that are tangent at this point. (a) If 𝑓𝑥 (𝑎, 𝑏) or 𝑓𝑦 (𝑎, 𝑏) is nonzero, use the local linearization to show that an equation of the line tangent at (𝑎, 𝑏) to the contour of 𝑓 through (𝑎, 𝑏) is 𝑓𝑥 (𝑎, 𝑏)(𝑥 − 𝑎) + 𝑓𝑦 (𝑎, 𝑏)(𝑦 − 𝑏) = 0. (b) Find the slope of the tangent line if 𝑓𝑦 (𝑎, 𝑏) ≠ 0. (c) Find an equation for the line tangent to the contour of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑥𝑦 at (3, 4). In Problems 49–52, the point is on the surface in 3-space. (a) Find the differential of the equation (that is, of each side).

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Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

50. 𝑥2 + 𝑦2 + 𝑧2 + 1 = 𝑥𝑦𝑧 + 2𝑥2 + 3𝑦2 − 2𝑧2 , (1, 1, 1)

(b) Find 𝑑𝑧 at the point. (c) Find an equation of the tangent plane to the surface at the point.

51. 𝑥𝑒𝑦 + 𝑧2 + 1 = cos(𝑥 − 1) +

49. 2𝑥2 + 13 = 𝑦2 + 3𝑧2 , (2, 3, 2)

52. 𝑥𝑧2 + 𝑥𝑦 + 5 = 𝑥2 + 𝑧2 , (2, −1, 1)



𝑧2 + 3, (1, 0, 1)

Strengthen Your Understanding In Problems 53–55, explain what is wrong with the statement.

Are the statements in Problems 58–65 true or false? Give reasons for your answer.

53. An equation for the tangent plane to the surface 𝑧 = 𝑓 (𝑥, 𝑦) at the point (3, 4) is

58. The tangent plane approximation of 𝑓 (𝑥, 𝑦) = 𝑦𝑒𝑥 at the point (0, 1) is 𝑓 (𝑥, 𝑦) ≈ 𝑦.

2

59. If 𝑓 is a function with 𝑑𝑓 = 2𝑦 𝑑𝑥 + sin(𝑥𝑦) 𝑑𝑦, then 𝑓 changes by about −0.4 between the points (1, 2) and (0.9, 2.0002).

𝑧 = 𝑓 (3, 4) + 𝑓𝑥 (3, 4)𝑥 + 𝑓𝑦 (3, 4)𝑦.

54. If 𝑓𝑥 (0, 0) = 𝑔𝑥 (0, 0) and 𝑓𝑦 (0, 0) = 𝑔𝑦 (0, 0), then the surfaces 𝑧 = 𝑓 (𝑥, 𝑦) and 𝑧 = 𝑔(𝑥, 𝑦) have the same tangent planes at the point (0, 0).

60. The local linearization of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 at (1,1) gives an overestimate of the value of 𝑓 (𝑥, 𝑦) at the point (1.04, 0.95).

55. The tangent plane to the surface 𝑧 = 𝑥2 𝑦 at the point (1, 2) has equation

61. If two functions 𝑓 and 𝑔 have the same differential at the point (1, 1), then 𝑓 = 𝑔. 62. If two functions 𝑓 and 𝑔 have the same tangent plane at a point (1, 1), then 𝑓 = 𝑔.

𝑧 = 2 + 2𝑥𝑦(𝑥 − 1) + 𝑥2 (𝑦 − 2).

63. If 𝑓 (𝑥, 𝑦) is a constant function, then 𝑑𝑓 = 0. 64. If 𝑓 (𝑥, 𝑦) is a linear function, then 𝑑𝑓 is a linear function of 𝑑𝑥 and 𝑑𝑦.

In Problems 56–57, give an example of: 56. Two different functions with the same differential. 57. A surface in three space whose tangent plane at (0, 0, 3) is the plane 𝑧 = 3.

14.4

65. If you zoom close enough near a point (𝑎, 𝑏) on the contour diagram of a differentiable function, the contours are precisely parallel and exactly equally spaced.

GRADIENTS AND DIRECTIONAL DERIVATIVES IN THE PLANE

The Rate of Change in an Arbitrary Direction: The Directional Derivative The partial derivatives of a function 𝑓 tell us the rate of change of 𝑓 in the directions parallel to the coordinate axes. In this section we see how to compute the rate of change of 𝑓 in an arbitrary direction. Figure 14.29 shows the temperature, in ◦ C, at the point (𝑥, 𝑦). Estimate the average rate of change of temperature as we walk from point 𝐴 to point 𝐵. 𝑦 (m ) 300 𝐵

50



200

45

Example 1

𝐴

40

100

35

100

200

300

𝑥 (m)

Figure 14.29: Estimating rate of change on a temperature map

14.4 GRADIENTS AND DIRECTIONAL DERIVATIVES IN THE PLANE

Solution

763

At the point 𝐴 we are on the 𝐻 = 45◦ C contour. At 𝐵 we are on the 𝐻 = 50◦ C contour. The displace⃗ ment vector from 𝐴 to √ 𝐵 has 𝑥 component approximately −100𝑖 and 𝑦 component approximately 2 2 ⃗ 25𝑗 , so its length is (−100) + 25 ≈ 103. Thus, the temperature rises by 5◦ C as we move 103 meters, so the average rate of change of the temperature in that direction is about 5∕103 ≈ 0.05◦C/m. Suppose we want to compute the rate of change of a function 𝑓 (𝑥, 𝑦) at the point 𝑃 = (𝑎, 𝑏) in the direction of the unit vector 𝑢⃗ = 𝑢1 ⃗𝑖 + 𝑢2 𝑗⃗ . For ℎ > 0, consider the point 𝑄 = (𝑎 + ℎ𝑢1 , 𝑏 + ℎ𝑢2 ) whose displacement from 𝑃 is ℎ⃗𝑢 . (See Figure 14.30.) Since ‖⃗𝑢 ‖ = 1, the distance from 𝑃 to 𝑄 is ℎ. Thus, 𝑓 (𝑎 + ℎ𝑢1 , 𝑏 + ℎ𝑢2 ) − 𝑓 (𝑎, 𝑏) Change in 𝑓 Average rate of change = = . in 𝑓 from 𝑃 to 𝑄 Distance from 𝑃 to 𝑄 ℎ 𝑄 (𝑎 + ℎ𝑢1 , 𝑏 + ℎ𝑢2 ) ℎ⃗𝑢 𝑃 (𝑎, 𝑏) Figure 14.30: Displacement of ℎ⃗𝑢 from the point (𝑎, 𝑏)

Taking the limit as ℎ → 0 gives the instantaneous rate of change and the following definition:

Directional Derivative of 𝒇 at (𝒂, 𝒃) in the Direction of a Unit Vector 𝒖⃗ If 𝑢⃗ = 𝑢1 ⃗𝑖 + 𝑢2 𝑗⃗ is a unit vector, we define the directional derivative, 𝑓𝑢⃗ , by

𝑓𝑢⃗ (𝑎, 𝑏) =

Rate of change of 𝑓 in direction

𝑓 (𝑎 + ℎ𝑢1 , 𝑏 + ℎ𝑢2 ) − 𝑓 (𝑎, 𝑏) , ℎ→0 ℎ

= lim

of 𝑢⃗ at (𝑎, 𝑏) provided the limit exists. Note that the directional derivative is a scalar.

Notice that if 𝑢⃗ = ⃗𝑖 , so 𝑢1 = 1, 𝑢2 = 0, then the directional derivative is 𝑓𝑥 , since 𝑓 (𝑎 + ℎ, 𝑏) − 𝑓 (𝑎, 𝑏) = 𝑓𝑥 (𝑎, 𝑏). ℎ→0 ℎ

𝑓⃗𝑖 (𝑎, 𝑏) = lim

Similarly, if 𝑢⃗ = 𝑗⃗ then the directional derivative 𝑓𝑗⃗ = 𝑓𝑦 .

What If We Do Not Have a Unit Vector?

Example 2

We defined 𝑓𝑢⃗ for 𝑢⃗ a unit vector. If 𝑣⃗ is not a unit vector, 𝑣⃗ ≠ 0⃗ , we construct a unit vector 𝑢⃗ = 𝑣⃗ ∕‖⃗ 𝑣 ‖ in the same direction as 𝑣⃗ and define the rate of change of 𝑓 in the direction of 𝑣⃗ as 𝑓𝑢⃗ .

For each of the functions 𝑓 , 𝑔, and ℎ in Figure 14.31, decide whether the directional derivative at the indicated point is positive, negative, or zero, in the direction of the vector 𝑣⃗ = ⃗𝑖 + 2𝑗⃗ , and in the ⃗ = 2⃗𝑖 + 𝑗⃗ . direction of the vector 𝑤

764

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

𝑦

𝑦

𝑓 (𝑥, 𝑦)

𝑦

𝑔(𝑥, 𝑦)

ℎ(𝑥, 𝑦) 6

9 8

𝑣⃗ ✕✯

6

⃗ 𝑤 3

4

7 6

3 4 5

5 6 7 8 9 10

𝑣⃗

✕✯

⃗ 𝑤

𝑣⃗ ✯ ✕ 9 10

5 4 3

9 8

7

7

8 9

𝑥

8

⃗ 𝑤

6

6

7

𝑥

5

6

𝑥

⃗ = 2⃗𝑖 + 𝑗⃗ marked on each Figure 14.31: Contour diagrams of three functions with direction vectors 𝑣⃗ = ⃗𝑖 + 2𝑗⃗ and 𝑤

Solution

On the contour diagram for 𝑓 , the vector 𝑣⃗ = 𝑖⃗ + 2𝑗⃗ appears to be tangent to the contour. Thus, in this direction, the value of the function is not changing, so the directional derivative in the direction ⃗ = 2⃗𝑖 + 𝑗⃗ points from the contour marked 4 toward the contour marked of 𝑣⃗ is zero. The vector 𝑤 ⃗ 5. Thus, the values of the function are increasing and the directional derivative in the direction of 𝑤 is positive. On the contour diagram for 𝑔, the vector 𝑣⃗ = ⃗𝑖 + 2𝑗⃗ points from the contour marked 6 toward the contour marked 5, so the function is decreasing in that direction. Thus, the rate of change is ⃗ = 2⃗𝑖 + 𝑗⃗ points from the contour marked 6 toward the negative. On the other hand, the vector 𝑤 ⃗ is positive. contour marked 7, and hence the directional derivative in the direction of 𝑤 Finally, on the contour diagram for ℎ, both vectors point from the ℎ = 10 contour to the ℎ = 9 contour, so both directional derivatives are negative.

Example 3

Calculate the directional derivative of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 at (1, 0) in the direction of the vector ⃗𝑖 + 𝑗⃗ .

Solution

⃗ ⃗ First we have √ to find the unit vector in the same direction as the vector 𝑖 + 𝑗 . Since this vector has magnitude 2, the unit vector is 1 1 1 𝑢⃗ = √ (⃗𝑖 + 𝑗⃗ ) = √ ⃗𝑖 + √ 𝑗⃗ . 2 2 2

Thus,

√ √ √ √ 𝑓 (1 + ℎ∕ 2, ℎ∕ 2) − 𝑓 (1, 0) (1 + ℎ∕ 2)2 + (ℎ∕ 2)2 − 1 𝑓𝑢⃗ (1, 0) = lim = lim ℎ→0 ℎ→0 ℎ ℎ √ 2 √ √ 2ℎ + ℎ = lim = lim ( 2 + ℎ) = 2. ℎ→0 ℎ→0 ℎ

Computing Directional Derivatives from Partial Derivatives If 𝑓 is differentiable, we will now see how to use local linearity to find a formula for the directional derivative which does not involve a limit. If 𝑢⃗ is a unit vector, the definition of 𝑓𝑢⃗ says 𝑓 (𝑎 + ℎ𝑢1 , 𝑏 + ℎ𝑢2 ) − 𝑓 (𝑎, 𝑏) Δ𝑓 = lim , ℎ→0 ℎ ℎ where Δ𝑓 = 𝑓 (𝑎 + ℎ𝑢1 , 𝑏 + ℎ𝑢2 ) − 𝑓 (𝑎, 𝑏) is the change in 𝑓 . We write Δ𝑥 for the change in 𝑥, so Δ𝑥 = (𝑎 + ℎ𝑢1 ) − 𝑎 = ℎ𝑢1 ; similarly, Δ𝑦 = ℎ𝑢2 . Using local linearity, we have 𝑓𝑢⃗ (𝑎, 𝑏) = lim

ℎ→0

Δ𝑓 ≈ 𝑓𝑥 (𝑎, 𝑏)Δ𝑥 + 𝑓𝑦 (𝑎, 𝑏)Δ𝑦 = 𝑓𝑥 (𝑎, 𝑏)ℎ𝑢1 + 𝑓𝑦 (𝑎, 𝑏)ℎ𝑢2 . Thus, dividing by ℎ gives

14.4 GRADIENTS AND DIRECTIONAL DERIVATIVES IN THE PLANE

765

𝑓𝑥 (𝑎, 𝑏)ℎ𝑢1 + 𝑓𝑦 (𝑎, 𝑏)ℎ𝑢2 Δ𝑓 ≈ = 𝑓𝑥 (𝑎, 𝑏)𝑢1 + 𝑓𝑦 (𝑎, 𝑏)𝑢2 . ℎ ℎ This approximation becomes exact as ℎ → 0, so we have the following formula: 𝑓𝑢⃗ (𝑎, 𝑏) = 𝑓𝑥 (𝑎, 𝑏)𝑢1 + 𝑓𝑦 (𝑎, 𝑏)𝑢2 . Example 4

Solution

Use the preceding formula to compute the directional derivative in Example 3. Check that we get the same answer as before. 1 1 We calculate 𝑓𝑢⃗ (1, 0), where 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 and 𝑢⃗ = √ ⃗𝑖 + √ 𝑗⃗ . 2 2 The partial derivatives are 𝑓𝑥 (𝑥, 𝑦) = 2𝑥 and 𝑓𝑦 (𝑥, 𝑦) = 2𝑦. So, as before, 𝑓𝑢⃗ (1, 0) = 𝑓𝑥 (1, 0)𝑢1 + 𝑓𝑦 (1, 0)𝑢2 = (2)

(

1 √ 2

)

+ (0)

(

1 √ 2

)

=

√ 2.

The Gradient Vector Notice that the expression for 𝑓𝑢⃗ (𝑎, 𝑏) can be written as a dot product of 𝑢⃗ and a new vector: 𝑓𝑢⃗ (𝑎, 𝑏) = 𝑓𝑥 (𝑎, 𝑏)𝑢1 + 𝑓𝑦 (𝑎, 𝑏)𝑢2 = (𝑓𝑥 (𝑎, 𝑏)⃗𝑖 + 𝑓𝑦 (𝑎, 𝑏)𝑗⃗ ) ⋅ (𝑢1 ⃗𝑖 + 𝑢2 𝑗⃗ ). The new vector, 𝑓𝑥 (𝑎, 𝑏)⃗𝑖 + 𝑓𝑦 (𝑎, 𝑏)𝑗⃗ , turns out to be important. Thus, we make the following definition: The Gradient Vector of a differentiable function 𝑓 at the point (𝑎, 𝑏) is grad 𝑓 (𝑎, 𝑏) = 𝑓𝑥 (𝑎, 𝑏)⃗𝑖 + 𝑓𝑦 (𝑎, 𝑏)𝑗⃗ The formula for the directional derivative can be written in terms of the gradient as follows:

The Directional Derivative and the Gradient If 𝑓 is differentiable at (𝑎, 𝑏) and 𝑢⃗ = 𝑢1 ⃗𝑖 + 𝑢2 𝑗⃗ is a unit vector, then 𝑓𝑢⃗ (𝑎, 𝑏) = 𝑓𝑥 (𝑎, 𝑏)𝑢1 + 𝑓𝑦 (𝑎, 𝑏)𝑢2 = grad 𝑓 (𝑎, 𝑏) ⋅ 𝑢⃗ . The change in 𝑓 corresponding to a small change Δ⃗𝑟 = Δ𝑥⃗𝑖 + Δ𝑦𝑗⃗ can be estimated using the gradient: Δ𝑓 ≈ grad 𝑓 ⋅ Δ⃗𝑟 .

Example 5

Find the gradient vector of 𝑓 (𝑥, 𝑦) = 𝑥 + 𝑒𝑦 at the point (1, 1).

Solution

Using the definition, we have grad 𝑓 = 𝑓𝑥 ⃗𝑖 + 𝑓𝑦 𝑗⃗ = ⃗𝑖 + 𝑒𝑦 𝑗⃗ , so at the point (1, 1) grad 𝑓 (1, 1) = ⃗𝑖 + 𝑒𝑗⃗ .

766

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

Alternative Notation for the Gradient You can think of

𝜕𝑓 ⃗ 𝜕𝑓 ⃗ 𝑖 + 𝑗 as the result of applying the vector operator (pronounced “del”) 𝜕𝑥 𝜕𝑦 ∇=

𝜕 𝜕 ⃗ 𝑖 + 𝑗⃗ 𝜕𝑥 𝜕𝑦

to the function 𝑓 . Thus, we get the alternative notation grad 𝑓 = ∇𝑓 . If 𝑧 = 𝑓 (𝑥, 𝑦), we can write grad 𝑧 or ∇𝑧 for grad 𝑓 or for ∇𝑓 .

What Does the Gradient Tell Us? The fact that 𝑓𝑢⃗ = grad 𝑓 ⋅ 𝑢⃗ enables us to see what the gradient vector represents. Suppose 𝜃 is the angle between the vectors grad𝑓 and 𝑢⃗ . At the point (𝑎, 𝑏), we have 𝑓𝑢⃗ = grad 𝑓 ⋅ 𝑢⃗ = ‖ grad 𝑓 ‖ ‖⃗𝑢 ‖ cos 𝜃 = ‖ grad 𝑓 ‖ cos 𝜃. ⏟⏟⏟ 1

Imagine that grad 𝑓 is fixed and that 𝑢⃗ can rotate. (See Figure 14.32.) The maximum value of 𝑓𝑢⃗ occurs when cos 𝜃 = 1, so 𝜃 = 0 and 𝑢⃗ is pointing in the direction of grad 𝑓 . Then Maximum 𝑓𝑢⃗ = ‖ grad 𝑓 ‖ cos 0 = ‖ grad 𝑓 ‖.

The minimum value of 𝑓𝑢⃗ occurs when cos 𝜃 = −1, so 𝜃 = 𝜋 and 𝑢⃗ is pointing in the direction opposite to grad 𝑓 . Then Minimum 𝑓𝑢⃗ = ‖ grad 𝑓 ‖ cos 𝜋 = −‖ grad 𝑓 ‖.

When 𝜃 = 𝜋∕2 or 3𝜋∕2, so cos 𝜃 = 0, the directional derivative is zero. Zero 𝑓𝑢⃗ at 𝜃 = 𝜋∕2

Max 𝑓𝑢⃗ at 𝜃 = 0

𝑢⃗

Min 𝑓𝑢⃗ at 𝜃 = 𝜋

𝜃

grad 𝑓

Zero 𝑓𝑢⃗ at 𝜃 = −𝜋∕2

Figure 14.32: Values of the directional derivative at different angles to the gradient

Properties of the Gradient Vector We have seen that the gradient vector points in the direction of the greatest rate of change at a point and the magnitude of the gradient vector is that rate of change. Figure 14.33 shows that the gradient vector at a point is perpendicular to the contour through that point. If the contours represent equally spaced 𝑓 -values and 𝑓 is differentiable, local linearity tells us that the contours of 𝑓 around a point appear straight, parallel, and equally spaced. The greatest rate of change is obtained by moving in the direction that takes us to the next contour in the shortest possible distance; that is, perpendicular to the contour. Thus, we have the following:

14.4 GRADIENTS AND DIRECTIONAL DERIVATIVES IN THE PLANE

767

Geometric Properties of the Gradient Vector in the Plane If 𝑓 is a differentiable function at the point (𝑎, 𝑏) and grad 𝑓 (𝑎, 𝑏) ≠ 0⃗ , then: • The direction of grad 𝑓 (𝑎, 𝑏) is · Perpendicular1 to the contour of 𝑓 through (𝑎, 𝑏); · In the direction of the maximum rate of increase of 𝑓 . • The magnitude of the gradient vector, ‖ grad 𝑓 ‖, is

· The maximum rate of change of 𝑓 at that point; · Large when the contours are close together and small when they are far apart.

Change in 𝑓 is Δ𝑐 for both paths

𝑦 (m) 300



Shortest path to next contour gives greatest rate of change

❲✮

200

𝑓 (𝑥, 𝑦) = 𝑐 + Δ𝑐



Contour where

𝑓 (𝑥, 𝑦) = 𝑐

Figure 14.33: Close-up view of the contours around (𝑎, 𝑏), showing the gradient is perpendicular to the contours

Examples of Directional Derivatives and Gradient Vectors



❨ 𝐴 𝐶

✛ Contour where (𝑎, 𝑏)

45

50

100

40

35

𝑥 (m) 100 200 300 Figure 14.34: A temperature map showing directions and relative magnitudes of two gradient vectors

Example 6

Explain why the gradient vectors at points 𝐴 and 𝐶 in Figure 14.34 have the direction and the relative magnitudes they do.

Solution

The gradient vector points in the direction of greatest increase of the function. This means that in Figure 14.34, the gradient points directly toward warmer temperatures. The magnitude of the gradient vector measures the rate of change. The gradient vector at 𝐴 is longer than the gradient vector at 𝐶 because the contours are closer together at 𝐴, so the rate of change is larger. Example 2 on page 763 shows how the contour diagram can tell us the sign of the directional derivative. In the next example we compute the directional derivative in three directions, two that are close to that of the gradient vector and one that is not.

Example 7

Use the gradient to find the directional derivative of 𝑓 (𝑥, 𝑦) = 𝑥+𝑒𝑦 at the point (1, 1) in the direction of the vectors ⃗𝑖 − 𝑗⃗ , ⃗𝑖 + 2𝑗⃗ , ⃗𝑖 + 3𝑗⃗ .

Solution

In Example 5 we found grad 𝑓 (1, 1) = ⃗𝑖 + 𝑒𝑗⃗ . √ A unit vector in the direction of ⃗𝑖 − 𝑗⃗ is 𝑠⃗ = (⃗𝑖 − 𝑗⃗ )∕ 2, so ( ) ⃗𝑖 − 𝑗⃗ 1−𝑒 = √ ≈ −1.215. 𝑓𝑠⃗ (1, 1) = grad 𝑓 (1, 1) ⋅ 𝑠⃗ = (⃗𝑖 + 𝑒𝑗⃗ ) ⋅ √ 2 2 1 This

assumes that the same scale is used on both axes.

768

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

√ A unit vector in the direction of ⃗𝑖 + 2𝑗⃗ is 𝑣⃗ = (⃗𝑖 + 2𝑗⃗ )∕ 5, so ( ) ⃗𝑖 + 2𝑗⃗ 1 + 2𝑒 = √ ≈ 2.879. 𝑓𝑣⃗ (1, 1) = grad 𝑓 (1, 1) ⋅ 𝑣⃗ = (⃗𝑖 + 𝑒𝑗⃗ ) ⋅ √ 5 5 √ ⃗ = (⃗𝑖 + 3𝑗⃗ )∕ 10, so A unit vector in the direction of ⃗𝑖 + 3𝑗⃗ is 𝑤 ( ) ⃗𝑖 + 3𝑗⃗ 1 + 3𝑒 ⃗ ⃗ ⃗ = ( 𝑖 + 𝑒𝑗 ) ⋅ = √ ≈ 2.895. 𝑓𝑤⃗ (1, 1) = grad 𝑓 (1, 1) ⋅ 𝑤 √ 10 10 √ Now look back at the answers and compare with the value of ‖ grad 𝑓 ‖ = 1 + 𝑒2 ≈ 2.896. One answer is not close to this value; the other two, 𝑓𝑣⃗ = 2.879 and 𝑓𝑤⃗ = 2.895, are close but slightly smaller than ‖ grad 𝑓 ‖. Since ‖ grad 𝑓 ‖ is the maximum rate of change of 𝑓 at the point, we have for any unit vector 𝑢⃗ : 𝑓𝑢⃗ (1, 1) ≤ ‖ grad 𝑓 ‖.

with equality when 𝑢⃗ is in the direction of grad 𝑓 . Since 𝑒 ≈ 2.718, the vectors 𝑖⃗ + 2𝑗⃗ and ⃗𝑖 + 3𝑗⃗ both point roughly, but not exactly, in the direction of the gradient vector grad 𝑓 (1, 1) = ⃗𝑖 + 𝑒𝑗⃗ . Thus, the values of 𝑓𝑣⃗ and 𝑓𝑤⃗ are both close to the value of ‖ grad 𝑓 ‖. The direction of the vector ⃗𝑖 − 𝑗⃗ is not close to the direction of grad 𝑓 and the value of 𝑓𝑠⃗ is not close to the value of ‖ grad 𝑓 ‖.

Exercises and Problems for Section 14.4 Online Resource: Additional Problems for Section 14.4 EXERCISES

In Exercises 1–14, find the gradient of the function. Assume the variables are restricted to a domain on which the function is defined. 1. 𝑓 (𝑥, 𝑦) = 32 𝑥5 − 47 𝑦6

2. 𝑄 = 50𝐾 + 100𝐿

3. 𝑓 (𝑚, 𝑛) = 𝑚2 + 𝑛2 √ 5. 𝑓 (𝛼, 𝛽) = 5𝛼 2 + 𝛽

4. 𝑧 = 𝑥𝑒𝑦 6. 𝑓 (𝑟, ℎ) = 𝜋𝑟2 ℎ

7. 𝑧 = (𝑥 + 𝑦)𝑒

8. 𝑓 (𝐾, 𝐿) = 𝐾 𝐿

In Exercises 23–28, which of the following vectors gives the direction of the gradient vector at point 𝐴 on the contour diagram? The scales on the 𝑥- and 𝑦-axes are the same. ⃗𝑖 −⃗𝑖 𝑗⃗ −𝑗⃗ ⃗𝑖 + 𝑗⃗ ⃗𝑖 − 𝑗⃗ −⃗𝑖 + 2𝑗⃗ −2⃗𝑖 − 𝑗⃗ 24. 𝑦 23. 𝑦

0.3

𝑦

0.7

16

13

10

7

5 10

𝐴

15

A 20

10. 𝑓 (𝑥, 𝑦) = ln(𝑥2 + 𝑦2 )

9. 𝑓 (𝑟, 𝜃) = 𝑟 sin 𝜃

𝑥

12. 𝑧 = tan−1 (𝑥∕𝑦)

11. 𝑧 = sin(𝑥∕𝑦) 13. 𝑓 (𝛼, 𝛽) =

14. 𝑧 = 𝑥

𝑥

26. 𝑦

𝑒𝑦 𝑥+𝑦

66

76

2𝛼 + 3𝛽 2𝛼 − 3𝛽

25. 𝑦

A 5 8

3

15. 𝑓 (𝑥, 𝑦) = 𝑥 𝑦 + 7𝑥𝑦 , at (1, 2)

A 88

2

82

62

In Exercises 15–22, find the gradient at the point.

94

54

16. 𝑓 (𝑚, 𝑛) = 5𝑚2 + 3𝑛4 , at (5, 2)

𝑥

27. 𝑦

2

17. 𝑓 (𝑟, ℎ) = 2𝜋𝑟ℎ + 𝜋𝑟 , at (2, 3) 18. 𝑓 (𝑥, 𝑦) = 𝑒sin 𝑦 , at (0, 𝜋) 19. 𝑓 (𝑥, 𝑦) = sin (𝑥2 ) + cos 𝑦, at (

√ 𝜋 2

20. 𝑓 (𝑥, 𝑦) = ln(𝑥2 + 𝑥𝑦), at (4, 1) 21. 𝑓 (𝑥, 𝑦) = 1∕(𝑥2 + 𝑦2 ), at (−1, 3) √ 22. 𝑓 (𝑥, 𝑦) = tan 𝑥 + 𝑦, at (0, 1)

𝑥

28. 𝑦 23

, 0) A

44 24

25

42 26

40 27

𝐴

38 36

𝑥

𝑥

14.4 GRADIENTS AND DIRECTIONAL DERIVATIVES IN THE PLANE

In Exercises 29–34, use the contour diagram of 𝑓 in Figure 14.35 to decide if the specified directional derivative is positive, negative, or approximately zero. 𝑦 3 2

4

2

1

In Exercises 43–44, approximate the directional derivative of 𝑓 in the direction from 𝑃 to 𝑄. 43. 𝑃 = (10, 12), 𝑄 = (10.3, 12.1), 𝑓 (𝑃 ) = 50, 𝑓 (𝑄) = 52. 44. 𝑃 = (−120, 45), 𝑄 = (−122, 47), 𝑓 (𝑃 ) =200, 𝑓 (𝑄) = 205.

8

6

769

In Exercises 45–48, find the directional derivative 𝑓𝑢⃗ (1, 2) for the function 𝑓 with 𝑢⃗ = (3⃗𝑖 − 4𝑗⃗ )∕5. 𝑥

45. 𝑓 (𝑥, 𝑦) = 𝑥𝑦 + 𝑦3

2

−1 4

−2

2

2

46. 𝑓 (𝑥, 𝑦) = 3𝑥 − 4𝑦

6

47. 𝑓 (𝑥, 𝑦) = 𝑥 − 𝑦

2

49. If 𝑓 (𝑥, 𝑦) = 𝑥2 𝑦 and 𝑣⃗ = 4⃗𝑖 − 3𝑗⃗ , find the directional derivative at the point (2, 6) in the direction of 𝑣⃗ .

48. 𝑓 (𝑥, 𝑦) = sin(2𝑥 − 𝑦)

8

−3 −3

−2

−1

1

3

In Exercises 50–51, find the differential 𝑑𝑓 from the gradient.

Figure 14.35 29. At point (−2, 2), in direction ⃗𝑖 . 30. At point (0, −2), in direction 𝑗⃗ .

50. grad 𝑓 = 𝑦⃗𝑖 + 𝑥𝑗⃗

31. At point (0, −2), in direction ⃗𝑖 + 2𝑗⃗ .

51. grad 𝑓 = (2𝑥 + 3𝑒𝑦 )⃗𝑖 + 3𝑥𝑒𝑦 𝑗⃗

32. At point (0, −2), in direction ⃗𝑖 − 2𝑗⃗ . 33. At point (−1, 1), in direction ⃗𝑖 + 𝑗⃗ .

In Exercises 52–53, find grad 𝑓 from the differential.

34. At point (−1, 1), in direction −⃗𝑖 + 𝑗⃗ .

53. 𝑑𝑓 = (𝑥 + 1)𝑦𝑒𝑥 𝑑𝑥 + 𝑥𝑒𝑥 𝑑𝑦

In Exercises 35–42, use the contour diagram of 𝑓 in Figure 14.35 to find the approximate direction of the gradient vector at the given point.

In Exercises 54–55, assuming 𝑃 and 𝑄 are close, approximate 𝑓 (𝑄).

35. (−2, 0)

36. (0, −2)

39. (−2, 2)

40. (−2, −2) 41. (2, 2)

55. 𝑃 = (10, 10), 𝑄 = (10.2, 10.1), 𝑓 (𝑃 ) = 50, grad𝑓 (𝑃 ) = 0.5⃗𝑖 + 𝑗⃗ .

37. (2, 0)

52. 𝑑𝑓 = 2𝑥𝑑𝑥 + 10𝑦𝑑𝑦

38. (0, 2) 42. (2, −2)

54. 𝑃 = (100, 150), 𝑄 = (101, 153), 𝑓 (𝑃 ) = 2000, grad𝑓 (𝑃 ) = 2⃗𝑖 − 2𝑗⃗ .

56. Where is grad𝑓 longer: at a point where contour lines of 𝑓 are far apart or at a point where contour lines of 𝑓 are close together?

PROBLEMS 57. A student was asked to find the directional derivative of 𝑓 (𝑥, 𝑦) = 𝑥2 𝑒𝑦 at the point (1, 0) in the direction of 𝑣⃗ = 4⃗𝑖 + 3𝑗⃗ . The student’s answer was 8 3 𝑓𝑢⃗ (1, 0) = grad 𝑓 (1, 0) ⋅ 𝑢⃗ = ⃗𝑖 + 𝑗⃗ . 5 5 (a) At a glance, how do you know this is wrong? (b) What is the correct answer? In Problems 58–64, find the quantity. Assume that 𝑔 is a smooth function and that ∇𝑔(2, 3) = −2⃗𝑖 + 𝑗⃗ 58. 𝑔𝑦 (2.4, 3)

and

∇𝑔(2.4, 3) = 4⃗𝑖

59. 𝑔𝑥 (2, 3)

60. A vector perpendicular to the level curve of 𝑔 that passes through the point (2.4, 3)

61. A vector parallel to the level curve of 𝑔 that passes through the point (2, 3) 62. The slope of the graph of 𝑔 at the point (2.4, 3) in the direction of the vector ⃗𝑖 + 3𝑗⃗ . 63. The slope of the graph of 𝑔 at the point (2, 3) in the direction of the vector ⃗𝑖 + 3𝑗⃗ . 64. The greatest slope of the graph of 𝑔 at the point (2, 3). 65. For 𝑓 (𝑥, 𝑦) = (𝑥 + 𝑦)∕(1 + 𝑥2 ), find the directional derivative at (1, −2) in the direction of 𝑣⃗ = 3⃗𝑖 + 4𝑗⃗ . 66. For 𝑔(𝑥, 𝑦) with 𝑔(5, 10) = 100 and 𝑔𝑢⃗ (5, 10) = 0.5, where 𝑢⃗ is the unit vector in the direction of the vector ⃗𝑖 + 𝑗⃗ , estimate 𝑔(5.1, 10.1). 67. Let 𝑓 (𝑃 ) = 15 and 𝑓 (𝑄) = 20 where 𝑃 = (3, 4) and 𝑄 = (3.03, 3.96). Approximate the directional derivative of 𝑓 at 𝑃 in the direction of 𝑄.

770

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

68. (a) Give 𝑄, the point at a distance of 0.1 from 𝑃 = (4, 5) in the direction of 𝑣⃗ = −⃗𝑖 + 3𝑗⃗ . Give five decimal places in your answer. (b) Use 𝑃 and 𝑄 to approximate the directional deriva√ tive of 𝑓 (𝑥, 𝑦) = 𝑥 + 𝑦 in the direction of 𝑣⃗ . (c) Give the exact value for the directional derivative you estimated in part (b). 69. For 𝑓 (𝑥, 𝑦) = 𝑒𝑥 tan(𝑦) + 2𝑥2 𝑦, find the directional derivative at the point (0, 𝜋∕4) in the direction √ (a) ⃗𝑖 − 𝑗⃗ (b) ⃗𝑖 + 3𝑗⃗

70. Find the rate of change of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 at the point (1, 2) in the direction of the vector 𝑢⃗ = 0.6⃗𝑖 + 0.8𝑗⃗ . 71. (a) Let 𝑓 (𝑥, 𝑦) = (𝑥 + 𝑦)∕(1 + 𝑥2 ). Find the directional derivative of 𝑓 at 𝑃 = (1, −2) in the direction of: (i) 𝑣⃗ = 3⃗𝑖 − 2𝑗⃗ (ii) 𝑣⃗ = −⃗𝑖 + 4𝑗⃗

77. 𝑓⃗𝑖 (4, 1)

78. 𝑓𝑗⃗ (4, 1) √ 79. 𝑓𝑢⃗ (4, 1) where 𝑢⃗ = (⃗𝑖 − 𝑗⃗ )∕ 2 √ 80. 𝑓𝑢⃗ (4, 1) where 𝑢⃗ = (−⃗𝑖 + 𝑗⃗ )∕ 2 √ 81. 𝑓𝑢⃗ (4, 1) with 𝑢⃗ = (−2⃗𝑖 + 𝑗⃗ )∕ 5

82. The surface 𝑧 = 𝑔(𝑥, 𝑦) is in Figure 14.37. What is the sign of each of the following directional derivatives? √ (a) 𝑔𝑢⃗ (2, 5) where 𝑢⃗ = (⃗𝑖 − 𝑗⃗ )∕√2. (b) 𝑔𝑢⃗ (2, 5) where 𝑢⃗ = (⃗𝑖 + 𝑗⃗ )∕ 2. 𝑧 (2, 5, 2)

(0, 5, 4)



(0, 12, 4)

(b) What is the direction of greatest increase of 𝑓 at 𝑃 ? 𝑥

72. Let 𝑓 (5, 10) = 200 and 𝑓 (5.2, 9.9) = 197. 𝑦

(a) Approximate the directional derivative at (5, 10) in the direction from (5, 10) toward (5.2, 9.9). (b) Approximate 𝑓 (𝑄) at the point 𝑄 that is distance 0.1 from (5, 10) in the direction of (5.2, 9.9). (c) Give coordinates for the point 𝑄. 73. Let 𝑓 (100, 100) = 500 and grad𝑓 (100, 100) = 2⃗𝑖 +3𝑗⃗ . (a) Find the directional derivative of 𝑓 at the point (100, 100) in the direction ⃗𝑖 + 𝑗⃗ . (b) Use the directional derivative to approximate 𝑓 (102, 102).

Figure 14.37

83. The table gives values of a differentiable function 𝑓 (𝑥, 𝑦). At the point (1.2, 0), into which quadrant does the gradient vector of 𝑓 point? Justify your answer.

𝑦

74. Let grad𝑓 (50, 60) = 0.3⃗𝑖 + 0.5𝑗⃗ . Approximate the directional derivative of 𝑓 at the point (50, 60) in the direction of the point (49.5, 62).

𝑥

2 3

75. Let 𝑓 (𝑥, 𝑦) = 𝑥 𝑦 . At the point (−1, 2), find a vector (a) In the direction of maximum rate of change. (b) In the direction of minimum rate of change. (c) In a direction in which the rate of change is zero. 76. Let 𝑓 (𝑥, 𝑦) = 𝑒𝑥𝑦 . At the point (1, 1), find a unit vector

−1

0

1

1.0

0.7

0.1

−0.5

1.2

4.8

4.2

3.6

1.4

8.9

8.3

7.7

84. The gradient of 𝑓 at a point 𝑃 has magnitude 10 and is in the direction of 𝐴 in Figure 14.38. Find the directional derivatives of 𝑓 at 𝑃 in the six directions shown.

(a) In the direction of the steepest ascent. (b) In the direction of the steepest descent. (c) In a direction in which the rate of change is zero. For Problems 77–81 use Figure 14.36, showing level curves of 𝑓 (𝑥, 𝑦), to estimate the directional derivatives.

𝐴 𝐵 𝐹 90◦

20◦

◦ 60◦ 𝑃 70

𝑦 4

𝐶

3

𝐸

2

5 4 3 2

1

0

𝐷

Figure 14.38

1 𝑥 1

2

3

4

Figure 14.36

5

6

14.4 GRADIENTS AND DIRECTIONAL DERIVATIVES IN THE PLANE

85. Figure 14.39 represents the level curves 𝑓 (𝑥, 𝑦) = 𝑐 ; the values of 𝑓 on each curve are marked. In each of the following parts, decide whether the given quantity is positive, negative or zero. Explain your answer. (a) (b) (c) (d)

The value of ∇𝑓 ⋅ ⃗𝑖 at 𝑃 . The value of ∇𝑓 ⋅ 𝑗⃗ at 𝑃 . 𝜕𝑓 ∕𝜕𝑥 at 𝑄. 𝜕𝑓 ∕𝜕𝑦 at 𝑄. 𝑃 3 1



2 4

𝑄 𝑥

Figure 14.39

86. In Figure 14.39, which is larger: ‖∇𝑓 ‖ at 𝑃 or ‖∇𝑓 ‖ at 𝑄? Explain how you know. 87. Let 𝑃 , 𝑄, 𝑅 and 𝑆 be four distinct points in the plane. Let 𝑢⃗ be the unit vector in the direction from 𝑃 to 𝑄, 𝑣⃗ the unit vector in the direction from 𝑃 to 𝑅, and ⃗ the unit vector in the direction from 𝑃 to 𝑆. Let 𝑤 𝑓 (𝑥, 𝑦) be a linear function with 𝑓 (𝑃 ) = 10, 𝑓 (𝑄) = 7, 𝑓 (𝑅) = 15, and 𝑓 (𝑆) = 10. List the directional derivatives 𝑓𝑢⃗ (𝑃 ), 𝑓𝑣⃗ (𝑃 ), and 𝑓𝑤⃗ (𝑃 ) in increasing order. 88. Let 𝑓𝑥 (3, 1) = −5 and 𝑓𝑦 (3, 1) = 2. Find a unit vector 𝑢⃗ such that: (a) 𝑓𝑢⃗ (3, 1) > 0 (c) 𝑓𝑢⃗ (3, 1) = 0

(b) 𝑓𝑢⃗ (3, 1) < 0

89. Let 𝑓 (0, 0) = −4 and 𝑓𝑢⃗ (0, 0) = 20 for a unit vector 𝑢⃗ . Suppose that points 𝑃 and 𝑄 in Figure 14.40 are close. Find approximate values of 𝑓 (𝑃 ) and 𝑓 (𝑄). 𝑢⃗

90. 𝑥2 + 𝑦2 = 13

91. 𝑥𝑦 = 6

92. 𝑦 = 𝑥2 − 1

93. (𝑦 − 𝑥)2 + 2 = 𝑥𝑦 − 3

95. At a certain point on a heated plate, the greatest rate of temperature increase, 5◦ C per meter, is toward the northeast. If an object at this point moves directly north, at what rate is the temperature increasing? 96. An ant is at the point (1, 1, 3) on the surface of a bowl with equation 𝑧 = 𝑥2 + 2𝑦2 , where 𝑥 and 𝑦 are in cm. In what two horizontal directions can the ant move away from the point (1, 1, 3) so that its initial rate of ascent is 2 vertical cm for each horizontal cm moved? Give your answers as vectors in the plane. 2

2

97. Let 𝑇 = 𝑓 (𝑥, 𝑦) = 100𝑒−(𝑥 ∕2)−𝑦 represent the temperature, in ◦ C, at the point (𝑥, 𝑦) with 𝑥 and 𝑦 in meters. (a) Describe the contours of 𝑓 , and explain their meaning in the context of this problem. (b) Find the rate at which the temperature changes as you move away from the point (1, 1) toward the point (2, 3). Give units in your answer. (c) In what direction would you move away from (1, 1) for the temperature to increase as fast as possible? 98. You are climbing a mountain by the steepest route at a slope of 20◦ when you come upon a trail branching off at a 30◦ angle from yours. What is the angle of ascent of the branch trail? 99. You are standing at the point (1, 1, 3) on the hill whose equation is given by 𝑧 = 5𝑦 − 𝑥2 − 𝑦2 .

P

0.5⃗𝑢

In Problems 90–93, check that the point (2, 3) lies on the curve. Then, viewing the curve as a contour of 𝑓 (𝑥, 𝑦), use grad 𝑓 (2, 3) to find a vector normal to the curve at (2, 3) and an equation for the tangent line to the curve at (2, 3).

94. The temperature 𝐻 in ◦ Fahrenheit 𝑦 miles north of the Canadian border 𝑡 hours after midnight is given by 𝐻 = 30 − 0.05𝑦 − 5𝑡. A moose runs north at a speed of 20 mph. At what rate does the moose perceive the temperature to be changing?

𝑦



771

(0,0)

Q

−⃗𝑢

Figure 14.40

(a) If you choose to climb in the direction of steepest ascent, what is your initial rate of ascent relative to the horizontal distance? (b) If you decide to go straight northwest, will you be ascending or descending? At what rate? (c) If you decide to maintain your altitude, in what directions can you go?

Strengthen Your Understanding In Problems 100–102, explain what is wrong with the statement. 100. A function 𝑓 has a directional derivative given by 𝑓𝑢⃗ (0, 0) = 3⃗𝑖 + 4𝑗⃗ .

101. A function 𝑓 has gradient grad 𝑓 (0, 0) = 7. 102. The gradient vector grad 𝑓 (𝑥, 𝑦) is perpendicular to the contours of 𝑓 , and the closer together the contours for equally spaced values of 𝑓 , the shorter the gradient vector.

772

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

In Problems 103–104, give an example of: 103. A unit vector 𝑢⃗ such that 𝑓𝑢⃗ (0, 0) < 0, given that 𝑓𝑥 (0, 0) = 2 and 𝑓𝑦 (0, 0) = 3. 104. A contour diagram of a function with two points in the domain where the gradients are parallel but different lengths. Are the statements in Problems 105–116 true or false? Give reasons for your answer. 105. If the point (𝑎, 𝑏) is on the contour 𝑓 (𝑥, 𝑦) = 𝑘, then the slope of the line tangent to this contour at (𝑎, 𝑏) is 𝑓𝑦 (𝑎, 𝑏)∕𝑓𝑥 (𝑎, 𝑏). 106. The gradient vector grad 𝑓 (𝑎, 𝑏) is a vector in 3-space. 107. grad(𝑓 𝑔) = (grad 𝑓 ) ⋅ (grad 𝑔) 108. The gradient vector grad 𝑓 (𝑎, 𝑏) is tangent to the contour of 𝑓 at (𝑎, 𝑏). 109. If you know the gradient vector of 𝑓 at (𝑎, 𝑏) then you can find the directional derivative 𝑓𝑢⃗ (𝑎, 𝑏) for any unit vector 𝑢⃗ .

14.5

110. If you know the directional derivative 𝑓𝑢⃗ (𝑎, 𝑏) for all unit vectors 𝑢⃗ then you can find the gradient vector of 𝑓 at (𝑎, 𝑏). 111. The directional derivative 𝑓𝑢⃗ (𝑎, 𝑏) is parallel to 𝑢⃗ . 112. The gradient grad 𝑓 (3, 4) is perpendicular to the vector 3⃗𝑖 + 4𝑗⃗ . 113. If grad 𝑓 (1, 2) = ⃗𝑖 , then 𝑓 decreases in the −⃗𝑖 direction at (1, 2). 114. If grad 𝑓 (1, 2) = ⃗𝑖 , then 𝑓 (10, 2) > 𝑓 (1, 2). 115. At the point (3, 0), the function 𝑔(𝑥, 𝑦) = 𝑥2 + 𝑦2 has the same maximal rate of increase as that of the function ℎ(𝑥, 𝑦) = 2𝑥𝑦. 116. If 𝑓 (𝑥, 𝑦) = 𝑒𝑥+𝑦 , then the directional derivative in any direction 𝑢⃗ (with ‖⃗𝑢 √ ‖ = 1) at the point (0, 0) is always less than or equal to 2.

GRADIENTS AND DIRECTIONAL DERIVATIVES IN SPACE

The Gradient Vector and Directional Derivative of a Function of Three Variables The gradient of a function of three variables is defined in the same way as for two variables: The gradient vector of a differentiable function 𝑓 (𝑥, 𝑦, 𝑧) is ⃗. grad 𝑓 = 𝑓𝑥 ⃗𝑖 + 𝑓𝑦 𝑗⃗ + 𝑓𝑧 𝑘

As in two dimensions, directional derivatives in space give the rate of change of a function in the direction of a unit vector 𝑢⃗ . If a function 𝑓 of three variables is differentiable at the point (𝑎, 𝑏, 𝑐) ⃗ , then the directional derivative 𝑓𝑢⃗ is related to the gradient by and 𝑢⃗ = 𝑢1 ⃗𝑖 + 𝑢2 𝑗⃗ + 𝑢3 𝑘 𝑓𝑢⃗ (𝑎, 𝑏, 𝑐) = 𝑓𝑥 (𝑎, 𝑏, 𝑐)𝑢1 + 𝑓𝑦 (𝑎, 𝑏, 𝑐)𝑢2 + 𝑓𝑧 (𝑎, 𝑏, 𝑐)𝑢3 = grad 𝑓 (𝑎, 𝑏, 𝑐) ⋅ 𝑢⃗ . Since grad 𝑓 (𝑎, 𝑏, 𝑐)⋅ 𝑢⃗ = ‖ grad 𝑓 (𝑎, 𝑏, 𝑐)‖ cos 𝜃, where 𝜃 is the angle between grad 𝑓 (𝑎, 𝑏, 𝑐) and 𝑢⃗ , the value of 𝑓𝑢⃗ (𝑎, 𝑏, 𝑐) is largest when 𝜃 = 0, that is, when 𝑢⃗ is in the same direction as grad 𝑓 (𝑎, 𝑏, 𝑐). In addition, 𝑓𝑢⃗ (𝑎, 𝑏, 𝑐) = 0 when 𝜃 = 𝜋∕2, so grad 𝑓 (𝑎, 𝑏, 𝑐) is perpendicular to the level surface of 𝑓 . The properties of gradients in space are similar to those in the plane:

Properties of the Gradient Vector in Space If 𝑓 is differentiable at (𝑎, 𝑏, 𝑐) and 𝑢⃗ is a unit vector, then 𝑓𝑢⃗ (𝑎, 𝑏, 𝑐) = grad 𝑓 (𝑎, 𝑏, 𝑐) ⋅ 𝑢⃗ . If, in addition, grad 𝑓 (𝑎, 𝑏, 𝑐) ≠ 0⃗ , then • grad 𝑓 (𝑎, 𝑏, 𝑐) is perpendicular to the level surface of 𝑓 at (𝑎, 𝑏, 𝑐) • grad 𝑓 (𝑎, 𝑏, 𝑐) is in the direction of the greatest rate of increase of 𝑓 • ‖ grad 𝑓 (𝑎, 𝑏, 𝑐)‖ is the maximum rate of change of 𝑓 at (𝑎, 𝑏, 𝑐). Example 1

Find the directional derivative of 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥𝑦 + 𝑧 at the point (−1, 0, 1) in the direction of the ⃗. vector 𝑣⃗ = 2⃗𝑖 + 𝑘

14.5 GRADIENTS AND DIRECTIONAL DERIVATIVES IN SPACE

Solution

The magnitude of 𝑣⃗ is ‖⃗ 𝑣‖ =



22 + 1 =

𝑢⃗ =

773

√ 5, so a unit vector in the same direction as 𝑣⃗ is

𝑣⃗ 2 1 ⃗ = √ ⃗𝑖 + 0𝑗⃗ + √ 𝑘 . ‖⃗ 𝑣‖ 5 5

The partial derivatives of 𝑓 are 𝑓𝑥 (𝑥, 𝑦, 𝑧) = 𝑦 and 𝑓𝑦 (𝑥, 𝑦, 𝑧) = 𝑥 and 𝑓𝑧 (𝑥, 𝑦, 𝑧) = 1. Thus, 𝑓𝑢⃗ (−1, 0, 1) = 𝑓𝑥 (−1, 0, 1)𝑢1 + 𝑓𝑦 (−1, 0, 1)𝑢2 + 𝑓𝑧 (−1, 0, 1)𝑢3 ( ) ( ) 2 1 1 = (0) √ + (−1)(0) + (1) √ =√ . 5 5 5 Example 2

Let 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 and 𝑔(𝑥, 𝑦, 𝑧) = −𝑥2 − 𝑦2 − 𝑧2 . What can we say about the direction of the following vectors? (a) grad 𝑓 (0, 1, 1) (b) grad 𝑓 (1, 0, 1) (c) grad 𝑔(0, 1, 1) (d) grad 𝑔(1, 0, 1).

Solution

The cylinder 𝑥2 + 𝑦2 = 1 in Figure 14.41 is a level surface of 𝑓 and contains both the points (0, 1, 1) and (1, 0, 1). Since the value of 𝑓 does not change at all in the 𝑧-direction, all the gradient vectors are horizontal. They are perpendicular to the cylinder and point outward because the value of 𝑓 increases as we move out. Similarly, the points (0, 1, 1) and (1, 0, 1) also lie on the same level surface of 𝑔, namely 𝑔(𝑥, 𝑦, 𝑧) = −𝑥2 − 𝑦2 − 𝑧2 = −2, which is the sphere 𝑥2 + 𝑦2 + 𝑧2 = 2. Part of this level surface is shown in Figure 14.42. This time the gradient vectors point inward, since the negative signs mean that the function increases (from large negative values to small negative values) as we move inward. 𝑧

𝑥

𝑧

𝑦 𝑥

Figure 14.41: The level surface 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 = 1 with two gradient vectors

𝑦

Figure 14.42: The level surface 𝑔(𝑥, 𝑦, 𝑧) = −𝑥2 − 𝑦2 − 𝑧2 = −2 with two gradient vectors

Example 3

Consider the functions 𝑓 (𝑥, 𝑦) = 4 − 𝑥2 − 2𝑦2 and 𝑔(𝑥, 𝑦) = 4 − 𝑥2 . Calculate a vector perpendicular to each of the following: (a) The level curve of 𝑓 at the point (1, 1) (b) The surface 𝑧 = 𝑓 (𝑥, 𝑦) at the point (1, 1, 1) (c) The level curve of 𝑔 at the point (1, 1) (d) The surface 𝑧 = 𝑔(𝑥, 𝑦) at the point (1, 1, 3)

Solution

(a) The vector we want is a 2-vector in the plane. Since grad 𝑓 = −2𝑥⃗𝑖 − 4𝑦𝑗⃗ , we have grad 𝑓 (1, 1) = −2⃗𝑖 − 4𝑗⃗ . Any nonzero multiple of this vector is perpendicular to the level curve at the point (1, 1). (b) In this case we want a 3-vector in space. To find it we rewrite 𝑧 = 4 − 𝑥2 − 2𝑦2 as the level

774

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

surface of the function 𝐹 , where 𝐹 (𝑥, 𝑦, 𝑧) = 4 − 𝑥2 − 2𝑦2 − 𝑧 = 0. Then ⃗, grad 𝐹 = −2𝑥⃗𝑖 − 4𝑦𝑗⃗ − 𝑘 so ⃗, grad 𝐹 (1, 1, 1) = −2⃗𝑖 − 4𝑗⃗ − 𝑘 and grad 𝐹 (1, 1, 1) is perpendicular to the surface 𝑧 = 4 − 𝑥2 − 2𝑦2 at the point (1, 1, 1). Notice ⃗ is not the only possible answer: any multiple of this vector will do. that −2⃗𝑖 − 4𝑗⃗ − 𝑘 (c) We are looking for a 2-vector. Since grad 𝑔 = −2𝑥⃗𝑖 + 0𝑗⃗ , we have grad 𝑔(1, 1) = −2⃗𝑖 . Any multiple of this vector is perpendicular to the level curve also. (d) We are looking for a 3-vector. We rewrite 𝑧 = 4 − 𝑥2 as the level surface of the function 𝐺, where 𝐺(𝑥, 𝑦, 𝑧) = 4 − 𝑥2 − 𝑧 = 0. Then ⃗ grad 𝐺 = −2𝑥⃗𝑖 − 𝑘 So ⃗, grad 𝐺(1, 1, 3) = −2⃗𝑖 − 𝑘 and any multiple of grad 𝐺(1, 1, 3) is perpendicular to the surface 𝑧 = 4 − 𝑥2 at this point. Example 4

(a) A hiker on the surface 𝑓 (𝑥, 𝑦) = 4 − 𝑥2 − 2𝑦2 at the point (1, −1, 1) starts to climb along the path of steepest ascent. What is the relation between the vector grad 𝑓 (1, −1) and a vector tangent to the path at the point (1, −1, 1) and pointing uphill? (b) At the point (1, −1, 1) on the surface 𝑓 (𝑥, 𝑦) = 4 − 𝑥2 − 2𝑦2 , calculate a vector, 𝑛⃗ , perpendicular to the surface and a vector, 𝑇⃗ , tangent to the curve of steepest ascent.

𝑦 −5 −3 −1 0

1

2 3

𝑥



Figure 14.43: Contour diagram for 𝑧 = 𝑓 (𝑥, 𝑦) = 4 − 𝑥2 − 2𝑦2 showing direction of grad 𝑓 (1, −1)

Figure 14.44: Graph of 𝑓 (𝑥, 𝑦) = 4 − 𝑥2 − 2𝑦2 showing path of steepest ascent from the point (1, −1, 1)

14.5 GRADIENTS AND DIRECTIONAL DERIVATIVES IN SPACE

Solution

775

(a) The hiker at the point (1, −1, 1) lies directly above the point (1, −1) in the 𝑥𝑦-plane. The vector grad 𝑓 (1, −1) lies in 2-space, pointing like a compass in the direction in which 𝑓 increases most rapidly. Therefore, grad 𝑓 (1, −1) lies directly under a vector tangent to the hiker’s path at (1, −1, 1) and pointing uphill. (See Figures 14.43 and 14.44.) ⃗, (b) The surface is represented by 𝐹 (𝑥, 𝑦, 𝑧) = 4−𝑥2 −2𝑦2 −𝑧 = 0. Since grad 𝐹 = −2𝑥⃗𝑖 −4𝑦𝑗⃗ − 𝑘 a normal, 𝑛⃗ , to the surface is given by ⃗ = −2⃗𝑖 + 4𝑗⃗ − 𝑘 ⃗. 𝑛⃗ = grad 𝐹 (1, −1, 1) = −2(1)⃗𝑖 − 4(−1)𝑗⃗ − 𝑘 We take the ⃗𝑖 and 𝑗⃗ components of 𝑇⃗ to be the vector grad 𝑓 (1, −1) = −2⃗𝑖 + 4𝑗⃗ . Thus, we have that, for some 𝑎 > 0, ⃗ 𝑇⃗ = −2⃗𝑖 + 4𝑗⃗ + 𝑎𝑘 We want 𝑛⃗ ⋅ 𝑇⃗ = 0, so ⃗ ) ⋅ (−2⃗𝑖 + 4𝑗⃗ + 𝑎𝑘 ⃗ ) = 4 + 16 − 𝑎 = 0 𝑛⃗ ⋅ 𝑇⃗ = (−2⃗𝑖 + 4𝑗⃗ − 𝑘 So 𝑎 = 20 and hence ⃗. 𝑇⃗ = −2⃗𝑖 + 4𝑗⃗ + 20𝑘

Example 5

Find the equation of the tangent plane to the sphere 𝑥2 + 𝑦2 + 𝑧2 = 14 at the point (1, 2, 3).

Solution

We write the sphere as a level surface as follows: 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 + 𝑧2 = 14. We have ⃗, grad 𝑓 = 2𝑥⃗𝑖 + 2𝑦𝑗⃗ + 2𝑧𝑘 so the vector ⃗ grad 𝑓 (1, 2, 3) = 2⃗𝑖 + 4𝑗⃗ + 6𝑘 is perpendicular to the sphere at the point (1, 2, 3). Since the vector grad 𝑓 (1, 2, 3) is normal to the tangent plane, the equation of the plane is 2𝑥 + 4𝑦 + 6𝑧 = 2 ⋅ 1 + 4 ⋅ 2 + 6 ⋅ 3 = 28 or

𝑥 + 2𝑦 + 3𝑧 = 14.

We could also try to find the tangent plane to the level surface 𝑓 (𝑥, 𝑦, 𝑧) = 𝑘 by solving algebraically for 𝑧 and using the method of Section 14.3, page 755. (See Problem 47.) Solving for 𝑧 can be difficult or impossible, however, so the method of Example 5 is preferable.

Tangent Plane to a Level Surface If 𝑓 (𝑥, 𝑦, 𝑧) is differentiable at (𝑎, 𝑏, 𝑐), then an equation for the tangent plane to the level surface of 𝑓 at the point (𝑎, 𝑏, 𝑐) is 𝑓𝑥 (𝑎, 𝑏, 𝑐)(𝑥 − 𝑎) + 𝑓𝑦 (𝑎, 𝑏, 𝑐)(𝑦 − 𝑏) + 𝑓𝑧 (𝑎, 𝑏, 𝑐)(𝑧 − 𝑐) = 0.

776

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

Caution: Scale on the Axis and the Geometric Interpretation of the Gradient When we interpreted the gradient of a function geometrically (page 767), we tacitly assumed that the units and scales along the 𝑥 and 𝑦 axes were the same. If the scales are not the same, the gradient vector may not look perpendicular to the contours. Consider the function 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦 with gradient vector grad 𝑓 = 2𝑥⃗𝑖 + 𝑗⃗ . Figure 14.45 shows the gradient vector at (1, 1) using the same scales in the 𝑥 and 𝑦 directions. As expected, the gradient vector is perpendicular to the contour line. Figure 14.46 shows contours of the same function with unequal scales on the two axes. Notice that the gradient vector no longer appears perpendicular to the contour lines. Thus, we see that the geometric interpretation of the gradient vector requires that the same scale be used on both axes. 𝑦

𝑦 2

4 3

1

2 1

2

3

1

4

𝑥 1 2 3 Figure 14.45: The gradient vector with 𝑥 and 𝑦 scales equal

4

5

6

3 2 1

𝑥 1 2 3 Figure 14.46: The gradient vector with 𝑥 and 𝑦 scales unequal

Exercises and Problems for Section 14.5 Online Resource: Additional Problems for Section 14.5 EXERCISES In Exercises 1–12, find the gradient of the function. 2

1. 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥

2. 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦3 − 𝑧4 3. 𝑓 (𝑥, 𝑦, 𝑧) = 𝑒𝑥+𝑦+𝑧 4. 𝑓 (𝑥, 𝑦, 𝑧) = cos(𝑥 + 𝑦) + sin(𝑦 + 𝑧) 5. 𝑓 (𝑥, 𝑦, 𝑧) = 𝑦𝑧2 ∕(1 + 𝑥2 ) 6. 𝑓 (𝑥, 𝑦, 𝑧) = 1∕(𝑥2 + 𝑦2 + 𝑧2 ) √ 7. 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 + 𝑧2 8. 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥𝑒𝑦 sin 𝑧

9. 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥𝑦 + sin (𝑒𝑧 ) 10. 𝑓 (𝑥1 , 𝑥2 , 𝑥3 ) = 𝑥21 𝑥32 𝑥43

In Exercises 19–24, find the directional derivative using 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥𝑦 + 𝑧2 . ⃗. 19. At (1, 2, 3) in the direction of ⃗𝑖 + 𝑗⃗ + 𝑘 ⃗. ⃗ ⃗ 20. At (1, 1, 1) in the direction of 𝑖 + 2𝑗 + 3𝑘 21. As you leave the point (1, 1, 0) heading in the direction of the point (0, 1, 1). 22. As you arrive at (0, 1, 1) from the direction of (1, 1, 0). 23. At the point (2, 3, 4) in the direction of a vector making an angle of 3𝜋∕4 with grad 𝑓 (2, 3, 4). 24. At the point (2, 3, 4) in the direction of the maximum rate of change of 𝑓 .

12. 𝑓 (𝑥, 𝑦, 𝑧) = 𝑒𝑧 + 𝑦 ln(𝑥2 + 5)

In Exercises 25–30, check that the point (−1, 1, 2) lies on the given surface. Then, viewing the surface as a level surface for a function 𝑓 (𝑥, 𝑦, 𝑧), find a vector normal to the surface and an equation for the tangent plane to the surface at (−1, 1, 2).

In Exercises 13–18, find the gradient at the point.

25. 𝑥2 − 𝑦2 + 𝑧2 = 4

26. 𝑧 = 𝑥2 + 𝑦2

13. 𝑓 (𝑥, 𝑦, 𝑧) = 𝑧𝑦2 , at (1, 0, 1)

27. 𝑦2 = 𝑧2 − 3

28. 𝑥2 − 𝑥𝑦𝑧 = 3

14. 𝑓 (𝑥, 𝑦, 𝑧) = 2𝑥 + 3𝑦 + 4𝑧, at (1, 1, 1)

29. cos(𝑥 + 𝑦) = 𝑒𝑥𝑧+2

30. 𝑦 = 4∕(2𝑥 + 3𝑧)

𝑟2

11. 𝑓 (𝑝, 𝑞, 𝑟) = 𝑒𝑝 + ln 𝑞 + 𝑒 2

2

2

4

15. 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥 + 𝑦 − 𝑧 , at (3, 2, 1) 17. 𝑓 (𝑥, 𝑦, 𝑧) = sin(𝑥𝑦) + sin(𝑦𝑧), at (1, 𝜋, −1)

In Exercises 31–32, the gradient of 𝑓 and a point 𝑃 on the level surface 𝑓 (𝑥, 𝑦, 𝑧) = 0 are given. Find an equation for the tangent plane to the surface at the point 𝑃 .

18. 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥 ln(𝑦𝑧), at (2, 1, 𝑒)

⃗ , 𝑃 = (1, 2, 3) 31. grad 𝑓 = 𝑦𝑧⃗𝑖 + 𝑥𝑧𝑗⃗ + 𝑥𝑦𝑘

16. 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥𝑦𝑧, at (1, 2, 3)

⃗ , 𝑃 = (10, −10, 30) 32. grad 𝑓 = 2𝑥⃗𝑖 + 𝑧2 𝑗⃗ + 2𝑦𝑧𝑘

14.5 GRADIENTS AND DIRECTIONAL DERIVATIVES IN SPACE

777

In Exercises 33–37, find an equation of the tangent plane to the surface at the given point.

39. If 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 +3𝑥𝑦+2𝑧, find the directional deriva⃗. tive at the point (2, 0, −1) in the direction of 2⃗𝑖 +𝑗⃗ −2𝑘

33. 𝑥2 + 𝑦2 + 𝑧2 = 17 at the point (2, 3, 2)

40. (a) Let 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 − 𝑥𝑦𝑧. Find grad 𝑓 . (b) Find the equation for the tangent plane to the surface 𝑓 (𝑥, 𝑦, 𝑧) = 7 at the point (2, 3, 1).

34. 𝑥2 + 𝑦2 = 1 at the point (1, 0, 0) 35. 𝑧 = 2𝑥 + 𝑦 + 3 at the point (0, 0, 3) 36. 3𝑥2 − 4𝑥𝑦 + 𝑧2 = 0 at the point (𝑎, 𝑎, 𝑎), where 𝑎 ≠ 0 37. 𝑧 = 9∕(𝑥 + 4𝑦) at the point where 𝑥 = 1 and 𝑦 = 2 38. For 𝑓 (𝑥, 𝑦, 𝑧) = 3𝑥2 𝑦2 +2𝑦𝑧, find the directional deriva⃗ tive at the point (−1, 0, 4) in the direction of (a) ⃗𝑖 − 𝑘 ⃗ (b) −⃗𝑖 + 3𝑗⃗ + 3𝑘

41. Find the equation √ of the tangent plane at the point (3, 2, 2) to 𝑧 = 17 − 𝑥2 − 𝑦2 .

42. Find the equation of the tangent plane to 𝑧 = 8∕(𝑥𝑦) at the point (1, 2, 4). 43. Find an equation of the tangent plane and of a normal vector to the surface 𝑥 = 𝑦3 𝑧7 at the point (1, −1, −1).

PROBLEMS 44. Let 𝑓 (𝑥, 𝑦, 𝑧) represent the temperature in ◦ C at the point (𝑥, 𝑦, 𝑧) with 𝑥, 𝑦, 𝑧 in meters. Let 𝑣⃗ be your velocity in meters per second. Give units and an interpretation of each of the following quantities. (a) || grad 𝑓 || (b) grad 𝑓 ⋅ 𝑣⃗

(c)

The directional derivatives are all evaluated at the point 𝑃 and the function 𝑓 (𝑥, 𝑦) is differentiable at 𝑃 .

∇𝑓

|| grad 𝑓 ||⋅||𝑣⃗ ||

45. Consider the surface 𝑔(𝑥, 𝑦) = 4 − 𝑥2 . What is the relation between grad 𝑔(−1, −1) and a vector tangent to the path of steepest ascent at (−1, −1, 3)? Illustrate your answer with a sketch.

𝑢⃗ 1 𝑢⃗ 4

𝑃

46. Match the functions 𝑓 (𝑥, 𝑦, 𝑧) in (a)–(d) with the descriptions of their gradients in (I)–(IV). (a) 𝑥2 + 𝑦2 + 𝑧2 1 (c) 𝑥2 + 𝑦2 + 𝑧2

I II III IV

(b) 𝑥2 + 𝑦2 1 (d) 𝑥2 + 𝑦2

Points radially outward from the 𝑧-axis. Points radially inward toward the 𝑧-axis. Points radially outward from the origin. Points radially inward toward the origin.

47. Find the equation of the tangent plane at (2, 3, 1) to the surface 𝑥2 + 𝑦2 − 𝑥𝑦𝑧 = 7. Do this in two ways: (a) Viewing the surface as the level set of a function of three variables, 𝐹 (𝑥, 𝑦, 𝑧). (b) Viewing the surface as the graph of a function of two variables 𝑧 = 𝑓 (𝑥, 𝑦). 48. At what point on the surface 𝑧 = 1+𝑥2 +𝑦2 is its tangent plane parallel to the following planes? (a) 𝑧 = 5

(b) 𝑧 = 5 + 6𝑥 − 10𝑦.

49. Let 𝑔𝑥 (2, 1, 7) = 3, 𝑔𝑦 (2, 1, 7) = 10, 𝑔𝑧 (2, 1, 7) = −5. Find the equation of the tangent plane to 𝑔(𝑥, 𝑦, 𝑧) = 0 at the point (2, 1, 7). 50. The vector ∇𝑓 at point 𝑃 and four unit vectors 𝑢⃗ 1 , 𝑢⃗ 2 , 𝑢⃗ 3 , 𝑢⃗ 4 are shown in Figure 14.47. Arrange the following quantities in ascending order 𝑓𝑢⃗ 1 ,

𝑓𝑢⃗ 2 ,

𝑓𝑢⃗ 3 ,

𝑓𝑢⃗ 4 ,

the number 0.

𝑢⃗ 2 𝑢⃗ 3

Figure 14.47

51. Let 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 + 𝑧2 . At the point (1, 2, 1), find the rate of change of 𝑓 in the direction perpendicular to the plane 𝑥 + 2𝑦 + 3𝑧 = 8 and moving away from the origin. 52. Let 𝑓 (𝑥, 𝑦) = cos 𝑥 sin 𝑦 and let 𝑆 be the surface 𝑧 = 𝑓 (𝑥, 𝑦). (a) Find a normal vector to the surface 𝑆 at the point (0, 𝜋∕2, 1). (b) What is the equation of the tangent plane to the surface 𝑆 at the point (0, 𝜋∕2, 1)? 53. Let 𝑓 (𝑥, 𝑦, 𝑧) = sin(𝑥2 + 𝑦2 + 𝑧2 ). (a) Describe in words the shape of the level surfaces of 𝑓 . (b) Find grad 𝑓 . ⃗ and (c) Consider the two vectors 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 2 2 grad 𝑓 at a point (𝑥, 𝑦, 𝑧) where sin(𝑥 + 𝑦 + 𝑧2 ) ≠ 0. What is (are) the possible values(s) of the angle between these vectors?

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Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

54. Each diagram (I) – (IV) in Figure 14.48 represents the level curves of a function 𝑓 (𝑥, 𝑦). For each function 𝑓 , consider the point above 𝑃 on the surface 𝑧 = 𝑓 (𝑥, 𝑦) and choose from the lists of vectors and equations that follow:

60. (a) Where does the surface 𝑥2 + 𝑦2 − (𝑧 − 1)2 = 0 cut the 𝑥𝑦-plane? What is the shape of the curve? (b) At the points where the surface cuts the 𝑥𝑦-plane, do vectors perpendicular to the surface lie in the 𝑥𝑦-plane?

(a) A vector which could be the normal to the surface at that point; (b) An equation which could be the equation of the tangent plane to the surface at that point.

61. A unit vector is perpendicular to the surface 𝑧 = 𝑥2 −𝑦2 . At which point on the surface does this unit vector have ⃗ the largest dot product with the vector ⃗𝑖 + 2𝑗⃗ + 3𝑘?

(I)

𝑦

(II)

𝑦

4 2 3 𝑃

2 𝑃 1

4

𝑥 (III)

(a) Find the unit vectors 𝑢⃗ 1 and 𝑢⃗ 2 pointing in the direction of maximum increase of 𝐹 at the points (0, 0, 1) and (1, 1, 1) respectively. (b) Find the tangent plane to 𝑆 at the points (0, 0, 1) and (1, 1, 1). (c) Find all points on 𝑆 where a normal vector is parallel to the 𝑥𝑦-plane.

1

3

𝑦

𝑥 (IV)

4

𝑦

63. Consider the function 𝑓 (𝑥, 𝑦) = (𝑒𝑥 − 𝑥) cos 𝑦. Suppose 𝑆 is the surface 𝑧 = 𝑓 (𝑥, 𝑦).

1 3

𝑃

2 2

𝑃 1

3 4

𝑥

𝑥

Figure 14.48

Vectors ⃗ (E) 2⃗𝑖 + 2𝑗⃗ − 2𝑘 ⃗ (F) 2⃗𝑖 + 2𝑗⃗ + 2𝑘 ⃗ (G) 2⃗𝑖 − 2𝑗⃗ + 2𝑘 ⃗ (H) −2⃗𝑖 + 2𝑗⃗ + 2𝑘

62. The surface 𝑆 is represented by the equation 𝐹 = 0 where 𝐹 (𝑥, 𝑦, 𝑧) = 𝑥2 − (𝑦∕𝑧2 ).

Equations (J) 𝑥 + 𝑦 + 𝑧 = 4 (K) 2𝑥 − 2𝑦 − 2𝑧 = 2 (L) −3𝑥 − 3𝑦 + 3𝑧 = 6 𝑥 𝑦 𝑧 (M) − + − = −7 2 2 2

55. (a) What is the shape of the curve in which the following surface cuts the 𝑦𝑧-plane: 5(𝑥 − 1)2 + 2(𝑦 + 1)2 + 2(𝑧 − 3)2 = 25? (b) Does the curve in part (a) go through the origin? (c) Find an expression for a vector perpendicular to the surface at the origin. 56. Find the points on the surface 𝑦 = 4 + 𝑥2 + 𝑧2 where ⃗. the gradient is parallel to ⃗𝑖 + 𝑗⃗ + 𝑘 57. A particle moves at a speed of 3 units per second perpendicular to the surface 𝑥 = 4 + 𝑦2 + 𝑧2 from the point (9, 1, 2) toward the 𝑦𝑧-plane. (a) What is the particle’s velocity vector? (b) Where is the particle after one second? 58. For the surface 𝑧 + 7 = 2𝑥2 + 3𝑦2 , where does the tangent plane at the point (−1, 1, −2) meet the three axes? 59. Find a vector perpendicular to the surface 𝑧 = 4−𝑥2 −𝑦2 at the point above the point (1, 1, 0). (The 𝑧-axis is vertical.)

(a) Find a vector which is perpendicular to the level curve of 𝑓 through the point (2, 3) in the direction in which 𝑓 decreases most rapidly. ⃗ is a vector in 3-space (b) Suppose 𝑣⃗ = 5⃗𝑖 + 4𝑗⃗ + 𝑎𝑘 which is tangent to the surface 𝑆 at the point 𝑃 lying on the surface above (2, 3). What is 𝑎? 64. (a) Find the tangent plane to the surface 𝑥2 +𝑦2 +3𝑧2 = 4 at the point (0.6, 0.8, 1). (b) Is there a point on the surface 𝑥2 + 𝑦2 + 3𝑧2 = 4 at which the tangent plane is parallel to the plane 8𝑥 + 6𝑦 + 30𝑧 = 1? If so, find it. If not, explain why not. 65. Your house lies on the surface 𝑧 = 𝑓 (𝑥, 𝑦) = 2𝑥2 − 𝑦2 directly above the point (4, 3) in the 𝑥𝑦-plane. (a) How high above the 𝑥𝑦-plane do you live? (b) What is the slope of your lawn as you look from your house directly toward the 𝑧-axis (that is, along the vector −4⃗𝑖 − 3𝑗⃗ )? (c) When you wash your car in the driveway, on this surface above the point (4, 3), which way does the water run off? (Give your answer as a twodimensional vector.) (d) What is the equation of the tangent plane to this surface at your house? 66. (a) Sketch the contours of 𝑧 = 𝑦 − sin 𝑥 for 𝑧 = −1, 0, 1, 2. (b) A bug starts on the surface at the point (𝜋∕2, 1, 0) and walks on the surface 𝑧 = 𝑦 − sin 𝑥 in the direction parallel to the 𝑦-axis, in the direction of increasing 𝑦. Is the bug walking in a valley or on top of a ridge? Explain. (c) On the contour 𝑧 = 0 in your sketch for part (a), draw the gradients of 𝑧 at 𝑥 = 0, 𝑥 = 𝜋∕2, and 𝑥 = 𝜋.

14.5 GRADIENTS AND DIRECTIONAL DERIVATIVES IN SPACE

779

67. The function 𝑓 (𝑥, 𝑦, 𝑧) = 2𝑥 − 3𝑦 + 𝑧 + 10 gives the temperature, 𝑇 , in degrees Celsius, at the point (𝑥, 𝑦, 𝑧).

instantaneous rate of change of the temperature as you leave the point (1, 0, 0). Give units.

(a) In words, describe the isothermal surfaces. (b) Calculate 𝑓𝑧 (0, 0, 0) and interpret in terms of temperature. (c) If you are standing at the point (0, 0, 0), in what direction should you move to increase your temperature the fastest? (d) Is 𝑧 = −2𝑥 + 3𝑦 + 17 an isothermal surface? If so, what is the temperature on this isotherm?

71. A spaceship is plunging into the atmosphere of a planet. With coordinates in miles and the origin at the center of the planet, the pressure of the atmosphere at (𝑥, 𝑦, 𝑧) is

68. The concentration of salt in a fluid at (𝑥, 𝑦, 𝑧) is given by 𝐹 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦4 + 𝑥2 𝑧2 mg/cm3 . You are at the point (−1, 1, 1). (a) In which direction should you move if you want the concentration to increase the fastest? (b) You start to move in the direction you found in part (a) at a speed of 4 cm/sec. How fast is the concentration changing? 69. The temperature of a gas at the point (𝑥, 𝑦, 𝑧) is given by 𝐺(𝑥, 𝑦, 𝑧) = 𝑥2 − 5𝑥𝑦 + 𝑦2 𝑧. (a) What is the rate of change in the temperature at the ⃗? point (1, 2, 3) in the direction 𝑣⃗ = 2⃗𝑖 + 𝑗⃗ − 4𝑘 (b) What is the direction of maximum rate of change of temperature at the point (1, 2, 3)? (c) What is the maximum rate of change at the point (1, 2, 3)?

𝑃 = 5𝑒−0.1



𝑥2 +𝑦2 +𝑧2

atmospheres.

The velocity, in miles/sec, of the spaceship at (0, 0, 1) ⃗ . At (0, 0, 1), what is the rate of change is 𝑣⃗ = ⃗𝑖 − 2.5𝑘 with respect to time of the pressure on the spaceship? 72. The earth has mass 𝑀 and is located at the origin in 3-space, while the moon has mass 𝑚. Newton’s Law of Gravitation states that if the moon is located at the point (𝑥, 𝑦, 𝑧) then the attractive force exerted by the earth on the moon is given by the vector 𝑟⃗ 𝐹⃗ = −𝐺𝑀𝑚 , ‖⃗𝑟 ‖3

⃗ . Show that 𝐹⃗ = grad 𝜑, where 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 where 𝜑 is the function given by 𝜑(𝑥, 𝑦, 𝑧) =

𝐺𝑀𝑚 . ‖⃗𝑟 ‖

⃗ and 𝑎⃗ be a constant vector. For 73. Let 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 each of the quantities in (a)–(c), choose the statement in (I)–(V) that describes it. No reasons are needed.

70. The temperature at the point (𝑥, 𝑦, 𝑧) in 3-space is given, 2 2 2 in degrees Celsius, by 𝑇 (𝑥, 𝑦, 𝑧) = 𝑒−(𝑥 +𝑦 +𝑧 ) .

(a)

(a) Describe in words the shape of surfaces on which the temperature is constant. (b) Find grad 𝑇 . (c) You travel from the point (1, 0, 0) to the point (2, 1, 0) at a speed of 3 units per second. Find the

I II III IV V

grad(⃗𝑟 + 𝑎⃗ ) (b) grad(⃗𝑟 ⋅ 𝑎⃗ ) (c)

grad(⃗𝑟 × 𝑎⃗ )

Scalar, independent of 𝑎⃗ . Scalar, depends on 𝑎⃗ . Vector, independent of 𝑎⃗ . Vector, depends on 𝑎⃗ . Not defined.

Strengthen Your Understanding In Problems 74–75, explain what is wrong with the statement. 74. The gradient vector grad 𝑓 (𝑥, 𝑦) points in the direction perpendicular to the surface 𝑧 = 𝑓 (𝑥, 𝑦). 75. The tangent plane at the origin to a surface 𝑓 (𝑥, 𝑦, 𝑧) = 1 that contains the point (0, 0, 0) has equation 𝑓𝑥 (0, 0, 0)𝑥 + 𝑓𝑦 (0, 0, 0)𝑦 + 𝑓𝑧 (0, 0, 0)𝑧 + 1 = 0.

In Problems 76–78, give an example of: ⃗ is 76. A surface 𝑧 = 𝑓 (𝑥, 𝑦) such that the vector ⃗𝑖 − 2𝑗⃗ − 𝑘 normal to the tangent plane at the point where (𝑥, 𝑦) = (0, 0). ⃗. 77. A function 𝑓 (𝑥, 𝑦, 𝑧) such that grad 𝑓 = 2⃗𝑖 + 3𝑗⃗ + 4𝑘

78. Two nonparallel unit vectors 𝑢⃗ and 𝑣⃗ such that 𝑓𝑢⃗ (0, 0, 0) = 𝑓𝑣⃗ (0, 0, 0) = 0, where 𝑓 (𝑥, 𝑦, 𝑧) = 2𝑥 − 3𝑦. Are the statements in Problems 79–82 true or false? Give reasons for your answer. 79. An equation for the tangent plane to the surface 𝑧 = 𝑥2 + 𝑦3 at (1, 1) is 𝑧 = 2 + 2𝑥(𝑥 − 1) + 3𝑦2 (𝑦 − 1). 80. There is a function 𝑓 (𝑥, 𝑦) which has a tangent plane with equation 𝑧 = 0 at a point (𝑎, 𝑏). 81. There is a function with ‖ grad 𝑓 ‖ = 4 and 𝑓𝑘⃗ = 5 at some point. 82. There is a function with ‖ grad 𝑓 ‖ = 5 and 𝑓𝑘⃗ = −3 at some point.

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Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

14.6

THE CHAIN RULE

Composition of Functions of Many Variables and Rates of Change The chain rule enables us to differentiate composite functions. If we have a function of two variables 𝑧 = 𝑓 (𝑥, 𝑦) and we substitute 𝑥 = 𝑔(𝑡), 𝑦 = ℎ(𝑡) into 𝑧 = 𝑓 (𝑥, 𝑦), then we have a composite function in which 𝑧 is a function of 𝑡: 𝑧 = 𝑓 (𝑔(𝑡), ℎ(𝑡)). If, on the other hand, we substitute 𝑥 = 𝑔(𝑢, 𝑣), 𝑦 = ℎ(𝑢, 𝑣), then we have a different composite function in which 𝑧 is a function of 𝑢 and 𝑣: 𝑧 = 𝑓 (𝑔(𝑢, 𝑣), ℎ(𝑢, 𝑣)). The next example shows how to calculate the rate of change of a composite function. Example 1

Corn production, 𝐶, depends on annual rainfall, 𝑅, and average temperature, 𝑇 , so 𝐶 = 𝑓 (𝑅, 𝑇 ). Global warming predicts that both rainfall and temperature depend on time. Suppose that according to a particular model of global warming, rainfall is decreasing at 0.2 cm per year and temperature is increasing at 0.1◦C per year. Use the fact that at current levels of production, 𝑓𝑅 = 3.3 and 𝑓𝑇 = −5 to estimate the current rate of change, 𝑑𝐶∕𝑑𝑡.

Solution

By local linearity, we know that changes Δ𝑅 and Δ𝑇 generate a change, Δ𝐶, in 𝐶 given approximately by Δ𝐶 ≈ 𝑓𝑅 Δ𝑅 + 𝑓𝑇 Δ𝑇 = 3.3Δ𝑅 − 5Δ𝑇 . We want to know how Δ𝐶 depends on the time increment, Δ𝑡. A change Δ𝑡 causes changes Δ𝑅 and Δ𝑇 , which in turn cause a change Δ𝐶. The model of global warming tells us that 𝑑𝑅 = −0.2 and 𝑑𝑡

𝑑𝑇 = 0.1. 𝑑𝑡

Thus, a time increment, Δ𝑡, generates changes of Δ𝑅 and Δ𝑇 given by Δ𝑅 ≈ −0.2Δ𝑡 and Δ𝑇 ≈ 0.1Δ𝑡. Substituting for Δ𝑅 and Δ𝑇 in the expression for Δ𝐶 gives us Δ𝐶 ≈ 3.3(−0.2Δ𝑡) − 5(0.1Δ𝑡) = −1.16Δ𝑡. Thus,

Δ𝐶 ≈ −1.16 and, therefore, Δ𝑡

𝑑𝐶 ≈ −1.16. 𝑑𝑡

The relationship between Δ𝐶 and Δ𝑡, which gives the value of 𝑑𝐶∕𝑑𝑡, is an example of the chain rule. The argument in Example 1 leads to more general versions of the chain rule.

The Chain Rule for 𝒛 = 𝒇 (𝒙, 𝒚), 𝒙 = 𝒈(𝒕), 𝒚 = 𝒉(𝒕) Since 𝑧 = 𝑓 (𝑔(𝑡), ℎ(𝑡)) is a function of 𝑡, we can consider the derivative 𝑑𝑧∕𝑑𝑡. The chain rule gives 𝑑𝑧∕𝑑𝑡 in terms of the derivatives of 𝑓 , 𝑔, and ℎ. Since 𝑑𝑧∕𝑑𝑡 represents the rate of change of 𝑧 with 𝑡, we look at the change Δ𝑧 generated by a small change, Δ𝑡. We substitute the local linearizations Δ𝑥 ≈

𝑑𝑥 Δ𝑡 𝑑𝑡

and Δ𝑦 ≈

𝑑𝑦 Δ𝑡 𝑑𝑡

14.6 THE CHAIN RULE

781

into the local linearization Δ𝑧 ≈

𝜕𝑧 𝜕𝑧 Δ𝑥 + Δ𝑦, 𝜕𝑥 𝜕𝑦

yielding 𝜕𝑧 𝑑𝑥 𝜕𝑧 𝑑𝑦 Δ𝑡 + Δ𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡 ( ) 𝜕𝑧 𝑑𝑥 𝜕𝑧 𝑑𝑦 = + Δ𝑡. 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡

Δ𝑧 ≈

Thus, Δ𝑧 𝜕𝑧 𝑑𝑥 𝜕𝑧 𝑑𝑦 ≈ + . Δ𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡 Taking the limit as Δ𝑡 → 0, we get the following result.

If 𝑓 , 𝑔, and ℎ are differentiable and if 𝑧 = 𝑓 (𝑥, 𝑦), and 𝑥 = 𝑔(𝑡), and 𝑦 = ℎ(𝑡), then 𝜕𝑧 𝑑𝑥 𝜕𝑧 𝑑𝑦 𝑑𝑧 = + . 𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡

Visualizing the Chain Rule with a Diagram The diagram in Figure 14.49 provides a way of remembering the chain rule. It shows the chain of dependence: 𝑧 depends on 𝑥 and 𝑦, which in turn depend on 𝑡. Each line in the diagram is labeled with a derivative relating the variables at its ends.

𝜕𝑧 𝜕𝑥

𝑧

𝑥

𝜕𝑧 𝜕𝑦

𝑦

𝑑𝑥 𝑑𝑡

𝑑𝑦 𝑑𝑡

𝑡 Figure 14.49: Diagram for 𝑧 = 𝑓 (𝑥, 𝑦), 𝑥 = 𝑔(𝑡), 𝑦 = ℎ(𝑡). Lines represent dependence of 𝑧 on 𝑥 and 𝑦, and of 𝑥 and 𝑦 on 𝑡

The diagram keeps track of how a change in 𝑡 propagates through the chain of composed functions. There are two paths from 𝑡 to 𝑧, one through 𝑥 and one through 𝑦. For each path, we multiply together the derivatives along the path. Then, to calculate 𝑑𝑧∕𝑑𝑡, we add the contributions from the two paths. Example 2

Suppose that 𝑧 = 𝑓 (𝑥, 𝑦) = 𝑥 sin 𝑦, where 𝑥 = 𝑡2 and 𝑦 = 2𝑡 + 1. Let 𝑧 = 𝑔(𝑡). Compute 𝑔 ′ (𝑡) directly and using the chain rule.

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Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

Solution

Since 𝑧 = 𝑔(𝑡) = 𝑓 (𝑡2 , 2𝑡 + 1) = 𝑡2 sin(2𝑡 + 1), it is possible to compute 𝑔 ′ (𝑡) directly by one-variable methods: ( ) 𝑑 2 𝑑 (𝑡 ) sin(2𝑡 + 1) = 2𝑡2 cos(2𝑡 + 1) + 2𝑡 sin(2𝑡 + 1). 𝑔 ′ (𝑡) = 𝑡2 (sin(2𝑡 + 1)) + 𝑑𝑡 𝑑𝑡 The chain rule provides an alternative route to the same answer. We have 𝜕𝑧 𝑑𝑥 𝜕𝑧 𝑑𝑦 𝑑𝑧 = + = (sin 𝑦)(2𝑡) + (𝑥 cos 𝑦)(2) = 2𝑡 sin(2𝑡 + 1) + 2𝑡2 cos(2𝑡 + 1). 𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡

Example 3

The capacity, 𝐶, of a communication channel, such as a telephone line, to carry information depends on the ratio of the signal strength, 𝑆, to the noise, 𝑁. For some positive constant 𝑘, ) ( 𝑆 . 𝐶 = 𝑘 ln 1 + 𝑁 Suppose that the signal and noise are given as a function of time, 𝑡 in seconds, by 𝑆(𝑡) = 4 + cos(4𝜋𝑡)

𝑁(𝑡) = 2 + sin(2𝜋𝑡).

What is 𝑑𝐶∕𝑑𝑡 one second after transmission started? Is the capacity increasing or decreasing at that instant? Solution

By the chain rule 𝑑𝐶 𝜕𝐶 𝑑𝑆 𝜕𝐶 𝑑𝑁 = + 𝑑𝑡 𝜕𝑆 𝑑𝑡 𝜕𝑁 𝑑𝑡 ( ) 1 𝑘 𝑆 𝑘 ⋅ (−4𝜋 sin 4𝜋𝑡) + − = (2𝜋 cos 2𝜋𝑡). 1 + 𝑆∕𝑁 𝑁 1 + 𝑆∕𝑁 𝑁2 When 𝑡 = 1, the first term is zero, 𝑆(1) = 5, and 𝑁(1) = 2, so ( ) ( ) 𝑆(1) 𝑑𝐶 𝑘 5 𝑘 = − − ⋅ 2𝜋. ⋅ 2𝜋 = 2 𝑑𝑡 1 + 𝑆(1)∕𝑁(1) 1 + 5∕2 4 (𝑁(1)) Since 𝑑𝐶∕𝑑𝑡 is negative, the capacity is decreasing at time 𝑡 = 1 second.

How to Formulate a General Chain Rule A diagram can be used to write the chain rule for general compositions. To find the rate of change of one variable with respect to another in a chain of composed differentiable functions: • Draw a diagram expressing the relationship between the variables, and label each link in the diagram with the derivative relating the variables at its ends. • For each path between the two variables, multiply together the derivatives from each step along the path. • Add the contributions from each path. The diagram keeps track of all the ways in which a change in one variable can cause a change in another; the diagram generates all the terms we would get from the appropriate substitutions into the local linearizations.

14.6 THE CHAIN RULE

𝑧

𝜕𝑧 𝜕𝑥

𝑥

𝜕𝑥 𝜕𝑣

783

𝜕𝑧 𝜕𝑦

𝑦

𝜕𝑦 𝜕𝑢

𝜕𝑥 𝜕𝑢

𝜕𝑦 𝜕𝑣

𝑢

𝑣

Figure 14.50: Diagram for 𝑧 = 𝑓 (𝑥, 𝑦), 𝑥 = 𝑔(𝑢, 𝑣), 𝑦 = ℎ(𝑢, 𝑣). Lines represent dependence of 𝑧 on 𝑥 and 𝑦, and of 𝑥 and 𝑦 on 𝑢 and 𝑣

For example, we can use Figure 14.50 to find formulas for 𝜕𝑧∕𝜕𝑢 and 𝜕𝑧∕𝜕𝑣. Adding the contributions for the two paths from 𝑧 to 𝑢, we get the following results: If 𝑓 , 𝑔, ℎ are differentiable and if 𝑧 = 𝑓 (𝑥, 𝑦), with 𝑥 = 𝑔(𝑢, 𝑣) and 𝑦 = ℎ(𝑢, 𝑣), then 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦 𝜕𝑧 = + , 𝜕𝑢 𝜕𝑥 𝜕𝑢 𝜕𝑦 𝜕𝑢 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦 𝜕𝑧 = + . 𝜕𝑣 𝜕𝑥 𝜕𝑣 𝜕𝑦 𝜕𝑣

Example 4

Let 𝑤 = 𝑥2 𝑒𝑦 , 𝑥 = 4𝑢, and 𝑦 = 3𝑢2 − 2𝑣. Compute 𝜕𝑤∕𝜕𝑢 and 𝜕𝑤∕𝜕𝑣 using the chain rule.

Solution

Using the previous result, we have 𝜕𝑤 𝜕𝑤 𝜕𝑥 𝜕𝑤 𝜕𝑦 = + = 2𝑥𝑒𝑦 (4) + 𝑥2 𝑒𝑦 (6𝑢) = (8𝑥 + 6𝑥2 𝑢)𝑒𝑦 𝜕𝑢 𝜕𝑥 𝜕𝑢 𝜕𝑦 𝜕𝑢 = (32𝑢 + 96𝑢3 )𝑒3𝑢

2 −2𝑣

.

Similarly, 𝜕𝑤 𝜕𝑤 𝜕𝑥 𝜕𝑤 𝜕𝑦 = + = 2𝑥𝑒𝑦 (0) + 𝑥2 𝑒𝑦 (−2) = −2𝑥2 𝑒𝑦 𝜕𝑣 𝜕𝑥 𝜕𝑣 𝜕𝑦 𝜕𝑣 = −32𝑢2 𝑒3𝑢

Example 5

2 −2𝑣

.

A quantity 𝑧 can be expressed either as a function of 𝑥 and 𝑦, so that 𝑧 = 𝑓 (𝑥, 𝑦), or as a function of 𝑢 and 𝑣, so that 𝑧 = 𝑔(𝑢, 𝑣). The two coordinate systems are related by 𝑥 = 𝑢 + 𝑣,

𝑦 = 𝑢 − 𝑣.

(a) Use the chain rule to express 𝜕𝑧∕𝜕𝑢 and 𝜕𝑧∕𝜕𝑣 in terms of 𝜕𝑧∕𝜕𝑥 and 𝜕𝑧∕𝜕𝑦. (b) Solve the equations in part (a) for 𝜕𝑧∕𝜕𝑥 and 𝜕𝑧∕𝜕𝑦. (c) Show that the expressions we get in part (b) are the same as we get by expressing 𝑢 and 𝑣 in terms of 𝑥 and 𝑦 and using the chain rule.

784

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

Solution

(a) We have 𝜕𝑥∕𝜕𝑢 = 1 and 𝜕𝑥∕𝜕𝑣 = 1, and also 𝜕𝑦∕𝜕𝑢 = 1 and 𝜕𝑦∕𝜕𝑣 = −1. Thus, 𝜕𝑧 𝜕𝑧 𝜕𝑧 𝜕𝑧 𝜕𝑧 = (1) + (1) = + 𝜕𝑢 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 and

𝜕𝑧 𝜕𝑧 𝜕𝑧 𝜕𝑧 𝜕𝑧 = (1) + (−1) = − . 𝜕𝑣 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦

(b) Adding together the equations for 𝜕𝑧∕𝜕𝑢 and 𝜕𝑧∕𝜕𝑣, we get 𝜕𝑧 𝜕𝑧 𝜕𝑧 + =2 , 𝜕𝑢 𝜕𝑣 𝜕𝑥

𝜕𝑧 1 𝜕𝑧 1 𝜕𝑧 = + . 𝜕𝑥 2 𝜕𝑢 2 𝜕𝑣

so

Similarly, subtracting the equations for 𝜕𝑧∕𝜕𝑢 and 𝜕𝑧∕𝜕𝑣 yields 𝜕𝑧 1 𝜕𝑧 1 𝜕𝑧 = − . 𝜕𝑦 2 𝜕𝑢 2 𝜕𝑣 (c) Alternatively, we can solve the equations 𝑥 = 𝑢 + 𝑣,

𝑦 = 𝑢−𝑣

for 𝑢 and 𝑣, which yields 1 1 1 1 𝑥 + 𝑦, 𝑣 = 𝑥 − 𝑦. 2 2 2 2 Now we can think of 𝑧 as a function of 𝑢 and 𝑣, and 𝑢 and 𝑣 as functions of 𝑥 and 𝑦, and apply the chain rule again. This gives us 𝑢=

𝜕𝑧 𝜕𝑢 𝜕𝑧 𝜕𝑣 1 𝜕𝑧 1 𝜕𝑧 𝜕𝑧 = + = + 𝜕𝑥 𝜕𝑢 𝜕𝑥 𝜕𝑣 𝜕𝑥 2 𝜕𝑢 2 𝜕𝑣 and

𝜕𝑧 𝜕𝑧 𝜕𝑢 𝜕𝑧 𝜕𝑣 1 𝜕𝑧 1 𝜕𝑧 = + = − . 𝜕𝑦 𝜕𝑢 𝜕𝑦 𝜕𝑣 𝜕𝑦 2 𝜕𝑢 2 𝜕𝑣

These are the same expressions we got in part (b).

An Application to Physical Chemistry A chemist investigating the properties of a gas such as carbon dioxide may want to know how the internal energy 𝑈 of a given quantity of the gas depends on its temperature, 𝑇 , pressure, 𝑃 , and volume, 𝑉 . The three quantities 𝑇 , 𝑃 , and 𝑉 are not independent, however. For instance, according to the ideal gas law, they satisfy the equation 𝑃 𝑉 = 𝑘𝑇 where 𝑘 is a constant which depends only upon the quantity of the gas. The internal energy can then be thought of as a function of any two of the three quantities 𝑇 , 𝑃 , and 𝑉 : 𝑈 = 𝑈1 (𝑇 , 𝑃 ) = 𝑈2 (𝑇 , 𝑉 ) = 𝑈3 (𝑃 , 𝑉 ). ( ) to indicate the partial derivative of 𝑈 with respect to The chemist writes, for example, 𝜕𝑈 𝜕𝑇 𝑃 𝑇 holding 𝑃 constant, that for this computation 𝑈 is viewed as a function of 𝑇 and 𝑃 . ) ( signifying 𝜕𝑈 as Thus, we interpret 𝜕𝑇 𝑃 ( ) 𝜕𝑈1 (𝑇 , 𝑃 ) 𝜕𝑈 . = 𝜕𝑇 𝑃 𝜕𝑇 ( ) If 𝑈 is to be viewed as a function of 𝑇 and 𝑉 , the chemist writes 𝜕𝑈 for the partial derivative 𝜕𝑇 𝑉 ( ) 𝜕𝑈 (𝑇 ,𝑉 ) of 𝑈 with respect to 𝑇 holding 𝑉 constant: thus, 𝜕𝑈 = 2𝜕𝑇 . 𝜕𝑇 𝑉

785

14.6 THE CHAIN RULE

Each of the functions 𝑈1 , 𝑈2 , 𝑈3 gives rise to one of the following formulas for the differential 𝑑𝑈 : ) ) ( 𝜕𝑈 𝜕𝑈 𝑑𝑇 + 𝑑𝑃 corresponds to 𝑈1 , 𝜕𝑇 𝑃 𝜕𝑃 𝑇 ) ) ( ( 𝜕𝑈 𝜕𝑈 𝑑𝑇 + 𝑑𝑉 corresponds to 𝑈2 , 𝑑𝑈 = 𝜕𝑇 𝑉 𝜕𝑉 𝑇 ) ) ( ( 𝜕𝑈 𝜕𝑈 𝑑𝑈 = 𝑑𝑃 + 𝑑𝑉 corresponds to 𝑈3 . 𝜕𝑃 𝑉 𝜕𝑉 𝑃 All the six partial derivatives appearing in formulas for 𝑑𝑈 have physical meaning, but they are not all equally easy to measure experimentally. A relationship among the partial derivatives, usually derived from the chain rule, may make it possible to evaluate one of the partials in terms of others that are more easily measured. 𝑑𝑈 =

Example 6

Solution

(

) ( 𝜕𝑈 Suppose a gas satisfies the equation 𝑃 𝑉 = 2𝑇 and 𝑃 = 3 when 𝑉 = 4. If = 7 and 𝜕𝑃 𝑉 ) ) ) ( ( ( 𝜕𝑈 𝜕𝑈 𝜕𝑈 = 8, find the values of and . 𝜕𝑉 𝑃 𝜕𝑃 𝑇 𝜕𝑇 𝑃 ( ) ( ) 𝜕𝑈 Since we know the values of 𝜕𝑈 and , we think of 𝑈 as a function of 𝑃 and 𝑉 and use 𝜕𝑃 𝑉 𝜕𝑉 𝑃 the function 𝑈3 to write ) ) ( ( 𝜕𝑈 𝜕𝑈 𝑑𝑃 + 𝑑𝑉 𝑑𝑈 = 𝜕𝑃 𝑉 𝜕𝑉 𝑃 𝑑𝑈 = 7𝑑𝑃 + 8𝑑𝑉 . ( ) ( ) and 𝜕𝑈 , we think of 𝑈 as a function of 𝑇 and 𝑃 . Thus, we want to substitute To calculate 𝜕𝑈 𝜕𝑃 𝜕𝑇 𝑇

𝑃

for 𝑑𝑉 in terms of 𝑑𝑇 and 𝑑𝑃 . Since 𝑃 𝑉 = 2𝑇 , we have

𝑃 𝑑𝑉 + 𝑉 𝑑𝑃 = 2𝑑𝑇 , 3𝑑𝑉 + 4𝑑𝑃 = 2𝑑𝑇 . Solving gives 𝑑𝑉 = (2𝑑𝑇 − 4𝑑𝑃 )∕3, so 2𝑑𝑇 − 4𝑑𝑃 3 16 11 𝑑𝑈 = − 𝑑𝑃 + 𝑑𝑇 . 3 3

𝑑𝑈 = 7𝑑𝑃 + 8

)

(

Comparing with the formula for 𝑑𝑈 obtained from 𝑈1 , ) ) ( ( 𝜕𝑈 𝜕𝑈 𝑑𝑇 + 𝑑𝑃 , 𝑑𝑈 = 𝜕𝑇 𝑃 𝜕𝑃 𝑇 we have (

𝜕𝑈 𝜕𝑇

)

𝑃

=

16 3

and

(

𝜕𝑈 𝜕𝑃

)

𝑇

=−

11 . 3

In Example 6, we could have substituted for 𝑑𝑃 instead of 𝑑𝑉 , leading to values of ( ) and 𝜕𝑈 . See Problem 41. 𝜕𝑉

(

𝜕𝑈 𝜕𝑇

)

𝑉

𝑇 ( ) In general, if for some particular 𝑃 , 𝑉 , and 𝑇 , we can measure two of the six quantities 𝜕𝑈 , 𝜕𝑃 𝑉 ( ) ( ) ( ) ( ) ( ) 𝜕𝑈 𝜕𝑈 , 𝜕𝑈 , 𝜕𝑈 , 𝜕𝑉 , 𝜕𝑈 , then we can compute the other four using the relationship 𝜕𝑉 𝑃 𝜕𝑃 𝑇 𝜕𝑇 𝑃 𝜕𝑇 𝑉 𝑇 between 𝑑𝑃 , 𝑑𝑉 , and 𝑑𝑇 given by the gas law. General formulas for each partial derivative in terms of others can be obtained in the same way. See the following example and Problem 41.

786

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

Example 7 Solution

Express

(

𝜕𝑈 𝜕𝑇

)

in terms of 𝑃

(

𝜕𝑈 𝜕𝑇

)

𝑉

Since we are interested in the derivatives 𝑉 and use the formula 𝑑𝑈 =

(

𝜕𝑈 𝜕𝑇

)

𝜕𝑈 𝜕𝑉 ( )

and

𝑑𝑇 +

𝑉

(

)

𝜕𝑈 𝜕𝑇

𝑉

(

𝜕𝑈 𝜕𝑉

𝑇

and

)

𝜕𝑉 𝜕𝑇 ( )

and

𝑑𝑉

𝑇

(

𝜕𝑈 𝜕𝑉

)

.

𝑃

, we think of 𝑈 as a function of 𝑇 and

𝑇

corresponding to 𝑈2 .

( ) We want to find a formula for 𝜕𝑈 , which means thinking of 𝑈 as a function of 𝑇 and 𝑃 . 𝜕𝑇 𝑃 Thus, we want to substitute for 𝑑𝑉 . Since 𝑉 is a function of 𝑇 and 𝑃 , we have ) ) ( ( 𝜕𝑉 𝜕𝑉 𝑑𝑇 + 𝑑𝑃 . 𝑑𝑉 = 𝜕𝑇 𝑃 𝜕𝑃 𝑇 Substituting for 𝑑𝑉 into the formula for 𝑑𝑈 corresponding to 𝑈2 gives ) ) (( ) ) ( ( ) ( 𝜕𝑉 𝜕𝑈 𝜕𝑉 𝜕𝑈 𝑑𝑇 + 𝑑𝑇 + 𝑑𝑃 . 𝑑𝑈 = 𝜕𝑇 𝑉 𝜕𝑉 𝑇 𝜕𝑇 𝑃 𝜕𝑃 𝑇 Collecting the terms containing 𝑑𝑇 and the terms containing 𝑑𝑃 gives ) ) ( ) ) ) ( ) ( (( ( 𝜕𝑉 𝜕𝑈 𝜕𝑉 𝜕𝑈 𝜕𝑈 𝑑𝑇 + 𝑑𝑈 = + 𝑑𝑃 . 𝜕𝑇 𝑉 𝜕𝑉 𝑇 𝜕𝑇 𝑃 𝜕𝑉 𝑇 𝜕𝑃 𝑇 But we also have the formula ) ) ( ( 𝜕𝑈 𝜕𝑈 𝑑𝑈 = 𝑑𝑇 + 𝑑𝑃 𝜕𝑇 𝑃 𝜕𝑃 𝑇

corresponding to 𝑈1 .

We now have two formulas for 𝑑𝑈 in terms of 𝑑𝑇 and 𝑑𝑃 . The coefficients of 𝑑𝑇 must be identical, so we conclude ( ) ) ) ( ) ( ( 𝜕𝑈 𝜕𝑉 𝜕𝑈 𝜕𝑈 = + . 𝜕𝑇 𝑃 𝜕𝑇 𝑉 𝜕𝑉 𝑇 𝜕𝑇 𝑃 ( ) Example 7 expresses 𝜕𝑈 in terms of three other partial derivatives. Two of them, namely 𝜕𝑇 𝑃 ( ) ( ) 𝜕𝑈 1 𝜕𝑉 , the constant-volume heat capacity, and , the expansion coefficient, can be easily 𝜕𝑇 𝑉 𝑉 𝜕𝑇 𝑃 ( ) measured experimentally. The third, the internal pressure, 𝜕𝑈 , cannot be measured directly but 𝜕𝑉 𝑇 ( ) ( ) can be related to 𝜕𝑃 , which is measurable. Thus, 𝜕𝑈 can be determined indirectly using 𝜕𝑇 𝑉 𝜕𝑇 𝑃 this identity.

Exercises and Problems for Section 14.6 Online Resource: Additional Problems for Section 14.6 EXERCISES

For Exercises 1–6, find 𝑑𝑧∕𝑑𝑡 using the chain rule. Assume the variables are restricted to domains on which the functions are defined. 1. 𝑧 = 𝑥𝑦2 , 𝑥 = 𝑒−𝑡 , 𝑦 = sin 𝑡 2. 𝑧 = 𝑥 sin 𝑦 + 𝑦 sin 𝑥, 𝑥 = 𝑡2 , 𝑦 = ln 𝑡 3. 𝑧 = sin(𝑥∕𝑦), 𝑥 = 2𝑡, 𝑦 = 1 − 𝑡2 √ 4. 𝑧 = ln(𝑥2 + 𝑦2 ), 𝑥 = 1∕𝑡, 𝑦 = 𝑡

5. 𝑧 = 𝑥𝑒𝑦 , 𝑥 = 2𝑡, 𝑦 = 1 − 𝑡2

6. 𝑧 = (𝑥 + 𝑦)𝑒𝑦 , 𝑥 = 2𝑡, 𝑦 = 1 − 𝑡2

For Exercises 7–15, find 𝜕𝑧∕𝜕𝑢 and 𝜕𝑧∕𝜕𝑣. The variables are restricted to domains on which the functions are defined. 7. 8. 9. 10. 11. 12. 13. 14. 15.

𝑧 = sin(𝑥∕𝑦), 𝑥 = ln 𝑢, 𝑦 = 𝑣 𝑧 = ln(𝑥𝑦), 𝑥 = (𝑢2 + 𝑣2 )2 , 𝑦 = (𝑢3 + 𝑣3 )2 𝑧 = 𝑥𝑒𝑦 , 𝑥 = ln 𝑢, 𝑦 = 𝑣 𝑧 = (𝑥 + 𝑦)𝑒𝑦 , 𝑥 = ln 𝑢, 𝑦 = 𝑣 𝑧 = 𝑥𝑒𝑦 , 𝑥 = 𝑢2 + 𝑣2 , 𝑦 = 𝑢2 − 𝑣2 𝑧 = (𝑥 + 𝑦)𝑒𝑦 , 𝑥 = 𝑢2 + 𝑣2 , 𝑦 = 𝑢2 − 𝑣2 𝑧 = 𝑥𝑒−𝑦 + 𝑦𝑒−𝑥 , 𝑥 = 𝑢 sin 𝑣, 𝑦 = 𝑣 cos 𝑢 𝑧 = cos (𝑥2 + 𝑦2 ), 𝑥 = 𝑢 cos 𝑣, 𝑦 = 𝑢 sin 𝑣 𝑧 = tan−1 (𝑥∕𝑦), 𝑥 = 𝑢2 + 𝑣2 , 𝑦 = 𝑢2 − 𝑣2

14.6 THE CHAIN RULE

787

PROBLEMS 16. Use the chain rule to find 𝑑𝑧∕𝑑𝑡, and check the result by expressing 𝑧 as a function of 𝑡 and differentiating directly. 𝑧 = 𝑥3 𝑦2 , 𝑥 = 𝑡3 , 𝑦 = 𝑡2 17. Use the chain rule to find 𝜕𝑤∕𝜕𝜌 and 𝜕𝑤∕𝜕𝜃, given that 𝑤 = 𝑥2 + 𝑦2 − 𝑧2 , and 𝑥 = 𝜌 sin 𝜙 cos 𝜃,

𝑦 = 𝜌 sin 𝜙 sin 𝜃,

𝑧 = 𝜌 cos 𝜙.

18. Let 𝑧 = 𝑓 (𝑥, 𝑦) where 𝑥 = 𝑔(𝑡), 𝑦 = ℎ(𝑡) and 𝑓 , 𝑔, ℎ are all differentiable functions. Given the information 𝜕𝑧 || . in the table, find 𝜕𝑡 ||𝑡=1 𝑓 (3, 10) = 7

𝑓 (4, 11) = −20

𝑓𝑥 (3, 10) = 100

𝑓𝑦 (3, 10) = 0.1

𝑓𝑥 (4, 11) = 200

𝑓𝑦 (4, 11) = 0.2

𝑓 (3, 4) = −10

𝑓 (10, 11) = −1

𝑔(1) = 3

ℎ(1) = 10

𝑔 ′ (1) = 4

ℎ′ (1) = 11

21. The air pressure is decreasing at a rate of 2 pascals per kilometer in the eastward direction. In addition, the air pressure is dropping at a constant rate with respect to time everywhere. A ship sailing eastward at 10 km/hour past an island takes barometer readings and records a pressure drop of 50 pascals in 2 hours. Estimate the time rate of change of air pressure on the island. (A pascal is a unit of air pressure.) 22. A steel bar with square cross sections 5 cm by 5 cm and length 3 meters is being heated. For each dimension, the bar expands 13 ⋅ 10−6 meters for each 1◦ C rise in temperature.2 What is the rate of change in the volume of the steel bar? 23. Corn production, 𝐶, is a function of rainfall, 𝑅, and temperature, 𝑇 . (See Example 1 on page 780.) Figures 14.51 and 14.52 show how rainfall and temperature are predicted to vary with time because of global warming. Suppose we know that Δ𝐶 ≈ 3.3Δ𝑅 − 5Δ𝑇 . Use this to estimate the change in corn production between the year 2020 and the year 2021. Hence, estimate 𝑑𝐶∕𝑑𝑡 when 𝑡 = 2020. 𝑅 (in)

19. A bison is charging across the plain one morning. His path takes him to location (𝑥, 𝑦) at time 𝑡 where 𝑥 and 𝑦 are functions of 𝑡 and north is in the direction of increasing 𝑦. The temperature is always colder farther north. As time passes, the sun rises in the sky, sending out more heat, and a cold front blows in from the east. At time 𝑡 the air temperature 𝐻 near the bison is given by 𝐻 = 𝑓 (𝑥, 𝑦, 𝑡). The chain rule expresses the derivative 𝑑𝐻∕𝑑𝑡 as a sum of three terms:

15 14 13 𝑡 (years) 2020

Figure 14.51: Rainfall as a function of time

𝜕𝑓 𝑑𝑥 𝜕𝑓 𝑑𝑦 𝜕𝑓 𝑑𝐻 = + + . 𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡 𝜕𝑡 Identify the term that gives the contribution to the change in temperature experienced by the bison that is due to (a) The rising sun. (b) The coming cold front. (c) The bison’s change in latitude.

𝑇 (◦ C)

27 25 23

20. The voltage, 𝑉 (in volts), across a circuit is given by Ohm’s law: 𝑉 = 𝐼𝑅, where 𝐼 is the current (in amps) flowing through the circuit and 𝑅 is the resistance (in ohms). If we place two circuits, with resistance 𝑅1 and 𝑅2 , in parallel, then their combined resistance, 𝑅, is given by 1 1 1 = + . 𝑅 𝑅1 𝑅2 Suppose the current is 2 amps and increasing at 10−2 amp/sec and 𝑅1 is 3 ohms and increasing at 0.5 ohm/sec, while 𝑅2 is 5 ohms and decreasing at 0.1 ohm/sec. Calculate the rate at which the voltage is changing. 2 www.engineeringtoolbox.com,

2040

𝑡 (years) 2020

2040

Figure 14.52: Temperature as a function of time 24. At a point 𝑥 miles east and 𝑦 miles north of a campground, the height above sea level is 𝑓 (𝑥, 𝑦) feet. Let ∇𝑓 (10, 2) = −2⃗𝑖 + 𝑗⃗ and 𝑣⃗ = ⃗𝑖 + 3𝑗⃗ . Find the following quantities, including units, as you leave a point 10 miles east and 2 miles north of the campground. (a) The slope of the land in the direction of 𝑣⃗ .

accessed January 10th, 2016.

788

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

(b) Your vertical speed if you move in the direction of 𝑣⃗ at a speed of 2 miles per hour. (c) Your vertical speed if you move east at a speed of 2 miles per hour. 25. The function 𝑔(𝑥, 𝑦) gives the temperature, in degrees Fahrenheit, 𝑥 miles east and 𝑦 miles north of a campground. Let 𝑢⃗ be the unit vector in the direction of 𝑣⃗ = ⃗𝑖 + 3𝑗⃗ and ∇𝑔(−1, −3) = ⃗𝑖 . A camper located one mile to the west and three miles to the south of the camp starts walking back to camp in the direction 𝑢⃗ at a speed of 2.5 miles∕hr. Find the value of the following expressions, and interpret each in everyday terms for the camper. (a) 𝑔𝑢⃗ (−1, −3) (c) 2.5𝑔⃗𝑖 (−1, −3)

(b) 2.5𝑔𝑢⃗ (−1, −3)

(a) Describe precisely in words the level surfaces of 𝑓 . (b) Give a unit vector in the direction of 𝐹⃗ at the point (1, 2, 2). (c) Estimate ||𝐹⃗ || at the point (1, 2, 2). (d) Estimate 𝐹⃗ at the point (1, 2, 2). (e) The points (1, 2, 2) and (3, 0, 0) are both on the sphere 𝑥2 + 𝑦2 + 𝑧2 = 9. Estimate 𝐹⃗ at (3, 0, 0). (f) If 𝑃 and 𝑄 are any two points on the sphere 𝑥2 + 𝑦2 + 𝑧2 = 𝑘2 : (i) Compare the magnitudes of 𝐹⃗ at 𝑃 and at 𝑄. (ii) Describe the directions of 𝐹⃗ at 𝑃 and at 𝑄. 3

𝑔(𝜌)

2

26. Mina’s score on her weekly multivariable calculus quiz, 𝑆, in points, is a function of the number of hours, 𝐻, she spends studying the course materials and the number of problems, 𝑃 , she solves per week. • Her score 𝑆 goes up 2 points for each additional hour spent per week studying the course materials. • Her score 𝑆 goes up by 3 points for each additional 5 problems solved during the week. • The number of weekly hours, 𝐻, she spends studying the course materials has been decreasing at a rate of 1.5 hours per week. Mina’s weekly quiz score does not change from week to week. (a) Find the value of 𝑑𝑃 ∕𝑑𝑡, where 𝑡 is time in weeks, include units. (b) What can Mina learn from the value of the derivative in part (a)? 27. Let 𝑧 = 𝑔(𝑢, 𝑣, 𝑤) and 𝑢 = 𝑢(𝑠, 𝑡), 𝑣 = 𝑣(𝑠, 𝑡), 𝑤 = 𝑤(𝑠, 𝑡). How many terms are there in the expression for 𝜕𝑧∕𝜕𝑡? 28. Suppose 𝑤 = 𝑓 (𝑥, 𝑦, 𝑧) and that 𝑥, 𝑦, 𝑧 are functions of 𝑢 and 𝑣. Use a tree diagram to write down the chain rule formula for 𝜕𝑤∕𝜕𝑢 and 𝜕𝑤∕𝜕𝑣. 29. Suppose 𝑤 = 𝑓 (𝑥, 𝑦, 𝑧) and that 𝑥, 𝑦, 𝑧 are all functions of 𝑡. Use a tree diagram to write down the chain rule for 𝑑𝑤∕𝑑𝑡. 30. Let 𝑧 = 𝑓 (𝑡)𝑔(𝑡). Use the chain rule applied to ℎ(𝑥, 𝑦) = 𝑓 (𝑥)𝑔(𝑦) to show that 𝑑𝑧∕𝑑𝑡 = 𝑓 ′ (𝑡)𝑔(𝑡) + 𝑓 (𝑡)𝑔 ′ (𝑡). The one-variable product rule for differentiation is a special case of the two-variable chain rule. 31. Let 𝐹 (𝑢, 𝑣) be a function of two variables. Find 𝑓 ′ (𝑥) if (a) 𝑓 (𝑥) = 𝐹 (𝑥, 3) (c) 𝑓 (𝑥) = 𝐹 (𝑥, 𝑥)

32. The √ function 𝑔(𝜌) is graphed in Figure 14.53. Let 𝜌 = 𝑥2 + 𝑦2(+ 𝑧2 . Define 𝑓 ,)a function of 𝑥, 𝑦, 𝑧 by √ 𝑓 (𝑥, 𝑦, 𝑧) = 𝑔 𝑥2 + 𝑦2 + 𝑧2 . Let 𝐹⃗ = grad 𝑓 .

(b) 𝑓 (𝑥) = 𝐹 (3, 𝑥) (d) 𝑓 (𝑥) = 𝐹 (5𝑥, 𝑥2 )

1 𝜌 1

2

3

4

5

Figure 14.53 In Problems 33–34, let 𝑧 = 𝑓 (𝑥, 𝑦), 𝑥 = 𝑥(𝑢, 𝑣), 𝑦 = 𝑦(𝑢, 𝑣) and 𝑥(1, 2) = 5, 𝑦(1, 2) = 3, calculate the partial derivative in terms of some of the numbers 𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑘, 𝑝, 𝑞: 𝑓𝑥 (1, 2) = 𝑎

𝑓𝑦 (1, 2) = 𝑐

𝑥𝑢 (1, 2) = 𝑒

𝑦𝑢 (1, 2) = 𝑝

𝑓𝑥 (5, 3) = 𝑏

𝑓𝑦 (5, 3) = 𝑑

𝑥𝑣 (1, 2) = 𝑘

𝑦𝑣 (1, 2) = 𝑞

33. 𝑧𝑢 (1, 2)

34. 𝑧𝑣 (1, 2)

In Problems 35–36, let 𝑧 = 𝑓 (𝑥, 𝑦), 𝑥 = 𝑥(𝑢, 𝑣), 𝑦 = 𝑦(𝑢, 𝑣) and 𝑥(4, 5) = 2, 𝑦(4, 5) = 3. Calculate the partial derivative in terms of 𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑘, 𝑝, 𝑞, 𝑟, 𝑠, 𝑡, 𝑤: 𝑓𝑥 (4, 5) = 𝑎

𝑓𝑦 (4, 5) = 𝑐

𝑥𝑢 (4, 5) = 𝑒

𝑦𝑢 (4, 5) = 𝑝

𝑓𝑥 (2, 3) = 𝑏

𝑓𝑦 (2, 3) = 𝑑

𝑥𝑣 (4, 5) = 𝑘

𝑦𝑣 (4, 5) = 𝑞

𝑥𝑢 (2, 3) = 𝑟

𝑦𝑢 (2, 3) = 𝑠

𝑥𝑣 (2, 3) = 𝑡

𝑦𝑣 (2, 3) = 𝑤

35. 𝑧𝑢 (4, 5)

36. 𝑧𝑣 (4, 5)

For Problems 37–38, suppose that 𝑥 > 0, 𝑦 > 0 and that 𝑧 can be expressed either as a function of Cartesian coordinates (𝑥, 𝑦) or as a function of polar coordinates (𝑟, 𝜃), so that 𝑧 = 𝑓 (𝑥,√ 𝑦) = 𝑔(𝑟, 𝜃). [Recall that 𝑥 = 𝑟 cos 𝜃, 𝑦 = 𝑟 sin 𝜃, 𝑟 = 𝑥2 + 𝑦2 , and, for 𝑥 > 0, 𝑦 > 0, 𝜃 = arctan(𝑦∕𝑥).] 37. (a) Use the chain rule to find 𝜕𝑧∕𝜕𝑟 and 𝜕𝑧∕𝜕𝜃 in terms of 𝜕𝑧∕𝜕𝑥 and 𝜕𝑧∕𝜕𝑦. (b) Solve the equations you have just written down for 𝜕𝑧∕𝜕𝑥 and 𝜕𝑧∕𝜕𝑦 in terms of 𝜕𝑧∕𝜕𝑟 and 𝜕𝑧∕𝜕𝜃. (c) Show that the expressions you get in part (b) are the same as you would get by using the chain rule to find 𝜕𝑧∕𝜕𝑥 and 𝜕𝑧∕𝜕𝑦 in terms of 𝜕𝑧∕𝜕𝑟 and 𝜕𝑧∕𝜕𝜃.

14.6 THE CHAIN RULE

44. In Example 6, we calculated values of (𝜕𝑈 ∕𝜕𝑇 )𝑃 and (𝜕𝑈 ∕𝜕𝑃 )𝑇 using the relationship 𝑃 𝑉 = 2𝑇 for a specific gas. In this problem, you will derive general relationships for these two partial derivatives.

38. Show that (

𝜕𝑧 𝜕𝑥

)2

+

(

𝜕𝑧 𝜕𝑦

)2

=

(

𝜕𝑧 𝜕𝑟

)2

+

1 𝑟2

(

𝜕𝑧 𝜕𝜃

)2

.

Problems 39–44 are continuations of the physical chemistry example on page 786. ( ) as a partial derivative of one of the func39. Write 𝜕𝑈 𝜕𝑃 𝑉 tions 𝑈1 , 𝑈2 , or 𝑈3 . ( ) as a partial derivative of one of the func40. Write 𝜕𝑈 𝜕𝑃 𝑇 tions 𝑈1 , 𝑈2 , 𝑈3 . ( ) ( ) 𝜕𝑈 and 𝜕𝑉 . 41. For the gas in Example 6, find 𝜕𝑈 𝜕𝑇 𝑉

𝑇

[Hint: Use the same method as the example, but substitute for 𝑑𝑃 instead of .] /𝑑𝑉 ( ) ( ) 𝜕𝑇 𝜕𝑉 42. Show that 𝜕𝑉 =1 . 𝜕𝑇 𝑃

𝑃

43. Use Example 7 and Problem 42 to show that ( ) 𝜕𝑈 ( ( ) ) 𝜕𝑇 𝜕𝑈 𝜕𝑈 = + ( )𝑉 . 𝜕𝑉 𝜕𝑉 𝑃 𝜕𝑉 𝑇 𝜕𝑇

789

(a) Think of 𝑉 as a function of 𝑃 and 𝑇 and write an expression for 𝑑𝑉 . (b) Substitute for 𝑑𝑉 into the following formula for 𝑑𝑈 (thinking of 𝑈 as a function of 𝑃 and 𝑉 ): ( ) ) ( 𝜕𝑈 𝜕𝑈 𝑑𝑈 = 𝑑𝑃 + 𝑑𝑉 . 𝜕𝑃 𝑉 𝜕𝑉 𝑃 (c) Thinking of 𝑈 as a function of 𝑃 and 𝑇 , write an expression for 𝑑𝑈 . (d) By comparing coefficients of 𝑑𝑃 and 𝑑𝑇 in your answers to parts (b) and (c), show that ) ) ( ) ( ( 𝜕𝑈 𝜕𝑉 𝜕𝑈 = ⋅ 𝜕𝑇 𝑃 𝜕𝑉 𝑃 𝜕𝑇 𝑃 ) ) ) ( ) ( ( ( 𝜕𝑈 𝜕𝑈 𝜕𝑉 𝜕𝑈 = + ⋅ . 𝜕𝑃 𝑇 𝜕𝑃 𝑉 𝜕𝑉 𝑃 𝜕𝑃 𝑇

𝑃

Strengthen Your Understanding

In Problems 45–47, explain what is wrong with the statement. 45. If 𝑧 = 𝑓 (𝑔(𝑡), ℎ(𝑡)), then 𝑑𝑧∕𝑑𝑡 = 𝑓 (𝑔 ′ (𝑡), ℎ(𝑡)) + 𝑓 (𝑔(𝑡), ℎ′ (𝑡)). 46. If 𝐶 = 𝐶(𝑅, 𝑇 ), 𝑅 = 𝑅(𝑥, 𝑦), 𝑇 = 𝑇 (𝑥, 𝑦) and 𝑅(0, 2) = 5, 𝑇 (0, 2) = 1, then 𝐶𝑥 (0, 2) = 𝐶𝑅 (0, 2)𝑅𝑥 (0, 2) + 𝐶𝑇 (0, 2)𝑇𝑥 (0, 2). 47. If 𝑧 = 𝑓 (𝑥, 𝑦) and 𝑥 = 𝑔(𝑡), 𝑦 = ℎ(𝑡) with 𝑔(0) = 2 and ℎ(0) = 3, then 𝑑𝑧 || = 𝑓𝑥 (0, 0)𝑔 ′ (0) + 𝑓𝑦 (0, 0)ℎ′ (0). 𝑑𝑡 ||𝑡=0

In Problems 48–52, give an example of:

48. Functions 𝑥 = 𝑔(𝑡) and 𝑦 = ℎ(𝑡) such that (𝑑𝑧∕𝑑𝑡)|𝑡=0 = 9, given that 𝑧 = 𝑥2 𝑦. 49. A function 𝑧 = 𝑓 (𝑥, 𝑦) such that 𝑑𝑧∕𝑑𝑡|𝑡=0 = 10, given that 𝑥 = 𝑒2𝑡 and 𝑦 = sin 𝑡. 50. Functions 𝑧, 𝑥 and 𝑦 where you need to follow the diagram in order to answer questions about the derivative of 𝑧 with respect to the other variables. 𝑧 𝜕𝑧 𝜕𝑦

𝜕𝑧 𝜕𝑥

𝑥

𝑑𝑦 𝑑𝑡

𝑡

𝑤 𝜕𝑤 𝜕𝑣

𝜕𝑤 𝜕𝑢

𝑢 𝜕𝑢 𝜕𝑠

𝑠

𝑣 𝜕𝑢 𝜕𝑡

𝑡

𝜕𝑣 𝜕𝑠

𝑠

𝜕𝑣 𝜕𝑡

𝑡

52. Function 𝑧 = 𝑓 (𝑥, 𝑦) where 𝑥 and 𝑦 are functions of 𝜕𝑧 one variable, 𝑡, for which = 2. 𝜕𝑡 53. Let 𝑧 = 𝑔(𝑢, 𝑣) and 𝑢 = 𝑢(𝑥, 𝑦, 𝑡), 𝑣 = 𝑣(𝑥, 𝑦, 𝑡) and 𝑥 = 𝑥(𝑡), 𝑦 = 𝑦(𝑡). Then the expression for 𝑑𝑧∕𝑑𝑡 has (a) Three terms (c) Six terms (e) Nine terms

𝑦

𝑑𝑥 𝑑𝑡

51. Functions 𝑤, 𝑢 and 𝑣 where you need to follow the diagram in order to answer questions about the derivative of 𝑤 with respect to the other variables.

(b) Four terms (d) Seven terms (f) None of the above

790

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

14.7

SECOND-ORDER PARTIAL DERIVATIVES Since the partial derivatives of a function are themselves functions, we can differentiate them, giving second-order partial derivatives. A function 𝑧 = 𝑓 (𝑥, 𝑦) has two first-order partial derivatives, 𝑓𝑥 and 𝑓𝑦 , and four second-order partial derivatives.

The Second-Order Partial Derivatives of 𝒛 = 𝒇 (𝒙, 𝒚) 𝜕2 𝑧 = 𝑓𝑥𝑥 = (𝑓𝑥 )𝑥 , 𝜕𝑥2 𝜕2 𝑧 = 𝑓𝑥𝑦 = (𝑓𝑥 )𝑦 , 𝜕𝑦𝜕𝑥

𝜕2 𝑧 = 𝑓𝑦𝑥 = (𝑓𝑦 )𝑥 , 𝜕𝑥𝜕𝑦 𝜕2 𝑧 = 𝑓𝑦𝑦 = (𝑓𝑦 )𝑦 . 𝜕𝑦2

It is usual to omit the parentheses, writing 𝑓𝑥𝑦 instead of (𝑓𝑥 )𝑦 and

( ) 𝜕 𝜕𝑧 𝜕2 𝑧 instead of . 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑥

Example 1

Compute the four second-order partial derivatives of 𝑓 (𝑥, 𝑦) = 𝑥𝑦2 + 3𝑥2 𝑒𝑦 .

Solution

From 𝑓𝑥 (𝑥, 𝑦) = 𝑦2 + 6𝑥𝑒𝑦 we get 𝑓𝑥𝑥 (𝑥, 𝑦) =

𝜕 2 (𝑦 + 6𝑥𝑒𝑦 ) = 6𝑒𝑦 𝜕𝑥

𝑓𝑥𝑦 (𝑥, 𝑦) =

and

𝜕 2 (𝑦 + 6𝑥𝑒𝑦 ) = 2𝑦 + 6𝑥𝑒𝑦 . 𝜕𝑦

From 𝑓𝑦 (𝑥, 𝑦) = 2𝑥𝑦 + 3𝑥2 𝑒𝑦 we get 𝑓𝑦𝑥 (𝑥, 𝑦) =

𝜕 (2𝑥𝑦 + 3𝑥2 𝑒𝑦 ) = 2𝑦 + 6𝑥𝑒𝑦 𝜕𝑥

and

𝑓𝑦𝑦 (𝑥, 𝑦) =

𝜕 (2𝑥𝑦 + 3𝑥2 𝑒𝑦 ) = 2𝑥 + 3𝑥2 𝑒𝑦 . 𝜕𝑦

Observe that 𝑓𝑥𝑦 = 𝑓𝑦𝑥 in this example.

Example 2

Use the values of the function 𝑓 (𝑥, 𝑦) in Table 14.7 to estimate 𝑓𝑥𝑦 (1, 2) and 𝑓𝑦𝑥 (1, 2). Table 14.7

Solution

Values of 𝑓 (𝑥, 𝑦)

𝑦∖𝑥

0.9

1.0

1.1

1.8

4.72

5.83

7.06

2.0

6.48

8.00

9.60

2.2

8.62

10.65

12.88

Since 𝑓𝑥𝑦 = (𝑓𝑥 )𝑦 , we first estimate 𝑓𝑥 𝑓 (1.1, 2) − 𝑓 (1, 2) 9.60 − 8.00 = = 16.0, 0.1 0.1 𝑓 (1.1, 2.2) − 𝑓 (1, 2.2) 12.88 − 10.65 𝑓𝑥 (1, 2.2) ≈ = = 22.3. 0.1 0.1 𝑓𝑥 (1, 2) ≈

14.7 SECOND-ORDER PARTIAL DERIVATIVES

791

Thus, 𝑓𝑥𝑦 (1, 2) ≈

𝑓𝑥 (1, 2.2) − 𝑓𝑥 (1, 2) 22.3 − 16.0 = = 31.5. 0.2 0.2

Similarly, 𝑓𝑦𝑥 (1, 2) ≈

𝑓𝑦 (1.1, 2) − 𝑓𝑦 (1, 2) 0.1

( ) 𝑓 (1.1, 2.2) − 𝑓 (1.1, 2) 𝑓 (1, 2.2) − 𝑓 (1, 2) 1 ≈ − 0.1 0.2 0.2 ( ) 1 12.88 − 9.60 10.65 − 8.00 − = 31.5. = 0.1 0.2 0.2

Observe that in this example also, 𝑓𝑥𝑦 = 𝑓𝑦𝑥 .

The Mixed Partial Derivatives Are Equal It is not an accident that the estimates for 𝑓𝑥𝑦 (1, 2) and 𝑓𝑦𝑥 (1, 2) are equal in Example 2, because the same values of the function are used to calculate each one. The fact that 𝑓𝑥𝑦 = 𝑓𝑦𝑥 in Examples 1 and 2 corroborates the following general result; Problem 71 (available online) suggests why you might expect it to be true.3

Theorem 14.1: Equality of Mixed Partial Derivatives If 𝑓𝑥𝑦 and 𝑓𝑦𝑥 are continuous at (𝑎, 𝑏), an interior point of their domain, then 𝑓𝑥𝑦 (𝑎, 𝑏) = 𝑓𝑦𝑥 (𝑎, 𝑏). For most functions 𝑓 we encounter and most points (𝑎, 𝑏) in their domains, not only are 𝑓𝑥𝑦 and 𝑓𝑦𝑥 continuous at (𝑎, 𝑏), but all their higher-order partial derivatives (such as 𝑓𝑥𝑥𝑦 or 𝑓𝑥𝑦𝑦𝑦 ) exist and are continuous at (𝑎, 𝑏). In that case we say 𝑓 is smooth at (𝑎, 𝑏). We say 𝑓 is smooth on a region 𝑅 if it is smooth at every point of 𝑅.

What Do the Second-Order Partial Derivatives Tell Us? Example 3

Let us return to the guitar string of Example 4, page 750. The string is 1 meter long and at time 𝑡 seconds, the point 𝑥 meters from one end is displaced 𝑓 (𝑥, 𝑡) meters from its rest position, where 𝑓 (𝑥, 𝑡) = 0.003 sin(𝜋𝑥) sin(2765𝑡). Compute the four second-order partial derivatives of 𝑓 at the point (𝑥, 𝑡) = (0.3, 1) and describe the meaning of their signs in practical terms.

Solution

First we compute 𝑓𝑥 (𝑥, 𝑡) = 0.003𝜋 cos(𝜋𝑥) sin(2765𝑡), from which we get 𝑓𝑥𝑥 (𝑥, 𝑡) =

𝜕 (𝑓 (𝑥, 𝑡)) = −0.003𝜋 2 sin(𝜋𝑥) sin(2765𝑡), 𝜕𝑥 𝑥

𝑓𝑥𝑥 (0.3, 1) ≈ −0.01;

so

and 𝑓𝑥𝑡 (𝑥, 𝑡) =

𝜕 (𝑓 (𝑥, 𝑡)) = (0.003)(2765)𝜋 cos(𝜋𝑥) cos(2765𝑡), 𝜕𝑡 𝑥

so

𝑓𝑥𝑡 (0.3, 1) ≈ 14.

On page 750 we saw that𝑓𝑥 (𝑥, 𝑡) gives the slope of the string at any point and time. Therefore, 3 For

a proof, see M. Spivak, Calculus on Manifolds, p. 26 (New York: Benjamin, 1965).

792

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

𝑓𝑥𝑥 (𝑥, 𝑡) measures the concavity of the string. The fact that 𝑓𝑥𝑥 (0.3, 1) < 0 means the string is concave down at the point 𝑥 = 0.3 when 𝑡 = 1. (See Figure 14.54.) On the other hand, 𝑓𝑥𝑡 (𝑥, 𝑡) is the rate of change of the slope of the string with respect to time. Thus, 𝑓𝑥𝑡 (0.3, 1) > 0 means that at time 𝑡 = 1 the slope at the point 𝑥 = 0.3 is increasing. (See Figure 14.55.) The slope at 𝐵 is greater than the slope at 𝐴

The slope at 𝐵 is less than the slope at 𝐴

𝑡=1+ℎ

𝐵 𝐴

𝐵 𝑡=1

𝐴

𝑥

𝑥 0

0.3

0

1

Figure 14.54: Interpretation of 𝑓𝑥𝑥 (0.3, 1) < 0: The concavity of the string at 𝑡 = 1

0.3

1

Figure 14.55: Interpretation of 𝑓𝑥𝑡 (0.3, 1) > 0: The slope of one point on the string at two different times

Now we compute 𝑓𝑡 (𝑥, 𝑡) = (0.003)(2765) sin(𝜋𝑥) cos(2765𝑡), from which we get 𝑓𝑡𝑥 (𝑥, 𝑡) =

𝜕 (𝑓 (𝑥, 𝑡)) = (0.003)(2765)𝜋 cos(𝜋𝑥) cos(2765𝑡), 𝜕𝑥 𝑡

so 𝑓𝑡𝑥 (0.3, 1) ≈ 14

and 𝑓𝑡𝑡 (𝑥, 𝑡) =

𝜕 (𝑓 (𝑥, 𝑡)) = −(0.003)(2765)2 sin(𝜋𝑥) sin(2765𝑡), 𝜕𝑡 𝑡

so

𝑓𝑡𝑡 (0.3, 1) ≈ −7200.

On page 750 we saw that 𝑓𝑡 (𝑥, 𝑡) gives the velocity of the string at any point and time. Therefore, 𝑓𝑡𝑥 (𝑥, 𝑡) and 𝑓𝑡𝑡 (𝑥, 𝑡) will both be rates of change of velocity. That 𝑓𝑡𝑥 (0.3, 1) > 0 means that at time 𝑡 = 1 the velocities of points just to the right of 𝑥 = 0.3 are greater than the velocity at 𝑥 = 0.3. (See Figure 14.56.) That 𝑓𝑡𝑡 (0.3, 1) < 0 means that the velocity of the point 𝑥 = 0.3 is decreasing at time 𝑡 = 1. Thus, 𝑓𝑡𝑡 (0.3, 1) = −7200 m/sec2 is the acceleration of this point. (See Figure 14.57.) The velocity at 𝐵 is greater than the velocity at 𝐴

The velocity at 𝐵 is less than the velocity at 𝐴

𝐵

❘✻ ✻

✻ ✻ 𝐴

𝐵

✠ ✠

𝑡 = 1+ℎ 𝑡=1

𝐴 𝑥

0

0.3

1

Figure 14.56: Interpretation of 𝑓𝑡𝑥 (0.3, 1) > 0: The velocity of different points on the string at 𝑡 = 1

𝑥 0

0.3

1

Figure 14.57: Interpretation of 𝑓𝑡𝑡 (0.3, 1) < 0: Negative acceleration. The velocity of one point on the string at two different times

Taylor Approximations We use second derivatives to construct quadratic Taylor approximations. In Section 14.3, we saw how to approximate 𝑓 (𝑥, 𝑦) by a linear function (its local linearization). We now see how to improve this approximation of 𝑓 (𝑥, 𝑦) using a quadratic function.

Linear and Quadratic Approximations Near (0,0) For a function of one variable, local linearity tells us that the best linear approximation is the degree-1 Taylor polynomial 𝑓 (𝑥) ≈ 𝑓 (𝑎) + 𝑓 ′ (𝑎)(𝑥 − 𝑎) for 𝑥 near 𝑎.

14.7 SECOND-ORDER PARTIAL DERIVATIVES

793

A better approximation to 𝑓 (𝑥) is given by the degree-2 Taylor polynomial: 𝑓 (𝑥) ≈ 𝑓 (𝑎) + 𝑓 ′ (𝑎)(𝑥 − 𝑎) +

𝑓 ′′ (𝑎) (𝑥 − 𝑎)2 2

for 𝑥 near 𝑎.

For a function of two variables the local linearization for (𝑥, 𝑦) near (𝑎, 𝑏) is 𝑓 (𝑥, 𝑦) ≈ 𝐿(𝑥, 𝑦) = 𝑓 (𝑎, 𝑏) + 𝑓𝑥 (𝑎, 𝑏)(𝑥 − 𝑎) + 𝑓𝑦 (𝑎, 𝑏)(𝑦 − 𝑏). In the case (𝑎, 𝑏) = (0, 0), we have: Taylor Polynomial of Degree 1 Approximating 𝒇 (𝒙, 𝒚) for (𝒙, 𝒚) near (0,0) If 𝑓 has continuous first-order partial derivatives, then 𝑓 (𝑥, 𝑦) ≈ 𝐿(𝑥, 𝑦) = 𝑓 (0, 0) + 𝑓𝑥 (0, 0)𝑥 + 𝑓𝑦 (0, 0)𝑦.

We get a better approximation to 𝑓 by using a quadratic polynomial. We choose a quadratic polynomial 𝑄(𝑥, 𝑦), with the same partial derivatives as the original function 𝑓 . You can check that the following Taylor polynomial of degree 2 has this property. Taylor Polynomial of Degree 2 Approximating 𝒇 (𝒙, 𝒚) for (𝒙, 𝒚) near (0,0) If 𝑓 has continuous second-order partial derivatives, then 𝑓 (𝑥, 𝑦) ≈ 𝑄(𝑥, 𝑦) = 𝑓 (0, 0) + 𝑓𝑥 (0, 0)𝑥 + 𝑓𝑦 (0, 0)𝑦 +

Example 4

𝑓𝑦𝑦 (0, 0) 2 𝑓𝑥𝑥 (0, 0) 2 𝑥 + 𝑓𝑥𝑦 (0, 0)𝑥𝑦 + 𝑦 . 2 2

Let 𝑓 (𝑥, 𝑦) = cos(2𝑥 + 𝑦) + 3 sin(𝑥 + 𝑦) (a) Compute the linear and quadratic Taylor polynomials, 𝐿 and 𝑄, approximating 𝑓 near (0, 0). (b) Explain why the contour plots of 𝐿 and 𝑄 for −1 ≤ 𝑥 ≤ 1, −1 ≤ 𝑦 ≤ 1 look the way they do.

Solution

(a) We have 𝑓 (0, 0) = 1. The derivatives we need are as follows: 𝑓𝑥 (𝑥, 𝑦) = −2 sin(2𝑥 + 𝑦) + 3 cos(𝑥 + 𝑦) 𝑓𝑦 (𝑥, 𝑦) = − sin(2𝑥 + 𝑦) + 3 cos(𝑥 + 𝑦)

so 𝑓𝑥 (0, 0) = 3, so

𝑓𝑦 (0, 0) = 3,

𝑓𝑥𝑥 (𝑥, 𝑦) = −4 cos(2𝑥 + 𝑦) − 3 sin(𝑥 + 𝑦) so 𝑓𝑥𝑥 (0, 0) = −4, 𝑓𝑥𝑦 (𝑥, 𝑦) = −2 cos(2𝑥 + 𝑦) − 3 sin(𝑥 + 𝑦) so 𝑓𝑥𝑦 (0, 0) = −2, 𝑓𝑦𝑦 (𝑥, 𝑦) = − cos(2𝑥 + 𝑦) − 3 sin(𝑥 + 𝑦)

so 𝑓𝑦𝑦 (0, 0) = −1.

Thus, the linear approximation, 𝐿(𝑥, 𝑦), to 𝑓 (𝑥, 𝑦) at (0, 0) is given by 𝑓 (𝑥, 𝑦) ≈ 𝐿(𝑥, 𝑦) = 𝑓 (0, 0) + 𝑓𝑥 (0, 0)𝑥 + 𝑓𝑦 (0, 0)𝑦 = 1 + 3𝑥 + 3𝑦. The quadratic approximation, 𝑄(𝑥, 𝑦), to 𝑓 (𝑥, 𝑦) near (0, 0) is given by 𝑓 (𝑥, 𝑦) ≈ 𝑄(𝑥, 𝑦) = 𝑓 (0, 0) + 𝑓𝑥 (0, 0)𝑥 + 𝑓𝑦 (0, 0)𝑦 + 1 = 1 + 3𝑥 + 3𝑦 − 2𝑥2 − 2𝑥𝑦 − 𝑦2 . 2

𝑓𝑦𝑦 (0, 0) 2 𝑓𝑥𝑥 (0, 0) 2 𝑥 + 𝑓𝑥𝑦 (0, 0)𝑥𝑦 + 𝑦 2 2

794

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

Notice that the linear terms in 𝑄(𝑥, 𝑦) are the same as the linear terms in 𝐿(𝑥, 𝑦). The quadratic terms in 𝑄(𝑥, 𝑦) can be thought of as “correction terms” to the linear approximation. (b) The contour plots of 𝑓 (𝑥, 𝑦), 𝐿(𝑥, 𝑦), and 𝑄(𝑥, 𝑦) are in Figures 14.58–14.60. 𝑦

𝑦

1

2

−1

1

1

𝑥 1

2

𝑥

2

−1

1

1

0

−1

𝑦

1

−1

1

0

−1

−1

−2

Figure 14.58: Original function, 𝑓 (𝑥, 𝑦)

0

−2

−2

−1

𝑥 1

−1 Figure 14.59: Linear approximation, 𝐿(𝑥, 𝑦)

−1 Figure 14.60: Quadratic approximation, 𝑄(𝑥, 𝑦)

Notice that the contour plot of 𝑄 is more similar to the contour plot of 𝑓 than is the contour plot of 𝐿. Since 𝐿 is linear, the contour plot of 𝐿 consists of parallel, equally spaced lines. An alternative, and much quicker, way to find the Taylor polynomial in the previous example is to use the single-variable approximations. For example, since cos 𝑢 = 1 −

𝑢2 𝑢4 + +⋯ 2! 4!

and

sin 𝑣 = 𝑣 −

𝑣3 +⋯, 3!

we can substitute 𝑢 = 2𝑥 + 𝑦 and 𝑣 = 𝑥 + 𝑦 and expand. We discard terms beyond the second (since we want the quadratic polynomial), getting cos(2𝑥 + 𝑦) = 1 −

(2𝑥 + 𝑦)2 (2𝑥 + 𝑦)4 1 1 + + ⋯ ≈ 1 − (4𝑥2 + 4𝑥𝑦 + 𝑦2 ) = 1 − 2𝑥2 − 2𝑥𝑦 − 𝑦2 2! 4! 2 2

and sin(𝑥 + 𝑦) = (𝑥 + 𝑦) −

(𝑥 + 𝑦)3 + ⋯ ≈ 𝑥 + 𝑦. 3!

Combining these results, we get 1 1 cos(2𝑥 + 𝑦) + 3 sin(𝑥 + 𝑦) ≈ 1 − 2𝑥2 − 2𝑥𝑦 − 𝑦2 + 3(𝑥 + 𝑦) = 1 + 3𝑥 + 3𝑦 − 2𝑥2 − 2𝑥𝑦 − 𝑦2 . 2 2

Linear and Quadratic Approximations Near (𝒂, 𝒃) The local linearization for a function 𝑓 (𝑥, 𝑦) at a point (𝑎, 𝑏) is

Taylor Polynomial of Degree 1 Approximating 𝒇 (𝒙, 𝒚) for (𝒙, 𝒚) Near (𝒂, 𝒃) If 𝑓 has continuous first-order partial derivatives, then 𝑓 (𝑥, 𝑦) ≈ 𝐿(𝑥, 𝑦) = 𝑓 (𝑎, 𝑏) + 𝑓𝑥 (𝑎, 𝑏)(𝑥 − 𝑎) + 𝑓𝑦 (𝑎, 𝑏)(𝑦 − 𝑏).

14.7 SECOND-ORDER PARTIAL DERIVATIVES

795

This suggests that a quadratic polynomial approximation 𝑄(𝑥, 𝑦) for 𝑓 (𝑥, 𝑦) near a point (𝑎, 𝑏) should be written in terms of (𝑥−𝑎) and (𝑦−𝑏) instead of 𝑥 and 𝑦. If we require that 𝑄(𝑎, 𝑏) = 𝑓 (𝑎, 𝑏) and that the first- and second-order partial derivatives of 𝑄 and 𝑓 at (𝑎, 𝑏) be equal, then we get the following polynomial: Taylor Polynomial of Degree 2 Approximating 𝒇 (𝒙, 𝒚) for (𝒙, 𝒚) Near (𝒂, 𝒃) If 𝑓 has continuous second-order partial derivatives, then 𝑓 (𝑥, 𝑦) ≈ 𝑄(𝑥, 𝑦) = 𝑓 (𝑎, 𝑏) + 𝑓𝑥 (𝑎, 𝑏)(𝑥 − 𝑎) + 𝑓𝑦 (𝑎, 𝑏)(𝑦 − 𝑏) +

𝑓𝑦𝑦 (𝑎, 𝑏) 𝑓𝑥𝑥 (𝑎, 𝑏) (𝑥 − 𝑎)2 + 𝑓𝑥𝑦 (𝑎, 𝑏)(𝑥 − 𝑎)(𝑦 − 𝑏) + (𝑦 − 𝑏)2 . 2 2

These coefficients are derived in exactly the same way as for (𝑎, 𝑏) = (0, 0). Example 5

Find the Taylor polynomial of degree 2 at the point (1, 2) for the function 𝑓 (𝑥, 𝑦) =

Solution

Table 14.8 contains the partial derivatives and their values at the point (1, 2). Partial derivatives of 𝑓 (𝑥, 𝑦) = 1∕(𝑥𝑦)

Table 14.8 Derivative 𝑓 (𝑥, 𝑦)

1 . 𝑥𝑦

Formula

Value at (1, 2)

1∕(𝑥𝑦)

1∕2

2

Derivative

Formula

Value at (1, 2)

3

𝑓𝑥𝑥 (𝑥, 𝑦)

2∕(𝑥 𝑦) 2 2

1

𝑓𝑥 (𝑥, 𝑦)

−1∕(𝑥 𝑦)

−1∕2

𝑓𝑥𝑦 (𝑥, 𝑦)

1∕(𝑥 𝑦 )

1∕4

𝑓𝑦 (𝑥, 𝑦)

−1∕(𝑥𝑦2 )

−1∕4

𝑓𝑦𝑦 (𝑥, 𝑦)

2∕(𝑥𝑦3 )

1∕4

So, the quadratic Taylor polynomial for 𝑓 near (1, 2) is 1 ≈ 𝑄(𝑥, 𝑦) 𝑥𝑦 ( )( ) 1 1 1 1 1 1 1 (𝑦 − 2)2 = − (𝑥 − 1) − (𝑦 − 2) + (1)(𝑥 − 1)2 + (𝑥 − 1)(𝑦 − 2) + 2 2 4 2 4 2 4 1 𝑥 − 1 𝑦 − 2 (𝑥 − 1)2 (𝑥 − 1)(𝑦 − 2) (𝑦 − 2)2 − + + + . = − 2 2 4 2 4 8

Exercises and Problems for Section 14.7 Online Resource: Additional Problems for Section 14.7 EXERCISES

In Exercises 1–11, calculate all four second-order partial derivatives and check that 𝑓𝑥𝑦 = 𝑓𝑦𝑥 . Assume the variables are restricted to a domain on which the function is defined. 1. 𝑓 (𝑥, 𝑦) = (𝑥 + 𝑦)2 2

2. 𝑓 (𝑥, 𝑦) = (𝑥 + 𝑦)3 3

3. 𝑓 (𝑥, 𝑦) = 3𝑥 𝑦 + 5𝑥𝑦

2𝑥𝑦

4. 𝑓 (𝑥, 𝑦) = 𝑒

5. 𝑓 (𝑥, 𝑦) = (𝑥 + 𝑦)𝑒𝑦 7. 𝑓 (𝑥, 𝑦) = sin(𝑥∕𝑦)

6. 𝑓 (𝑥, 𝑦) = 𝑥𝑒𝑦 √ 8. 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2

9. 𝑓 (𝑥, 𝑦) = 5𝑥3 𝑦2 − 7𝑥𝑦3 + 9𝑥2 + 11 10. 𝑓 (𝑥, 𝑦) = sin(𝑥2 + 𝑦2 ) 11. 𝑓 (𝑥, 𝑦) = 3 sin 2𝑥 cos 5𝑦

796

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

In Exercises 12–19, find the quadratic Taylor polynomials about (0, 0) for the function. 12. (𝑦 − 1)(𝑥 + 1)

2

13. (𝑥 − 𝑦 + 1)

26.

5

27.

2 4

2

𝑃

𝑃 3

2 −𝑦2

14. 𝑒−2𝑥

15. 𝑒𝑥 cos 𝑦

16. 1∕(1 + 2𝑥 − 𝑦)

17. cos(𝑥 + 3𝑦)

18. sin 2𝑥 + cos 𝑦

19. ln(1 + 𝑥2 − 𝑦)

5

29. 5

1 2 3

𝑃

2

4

In Exercises 22–31, use the level curves of the function 𝑧 = 𝑓 (𝑥, 𝑦) to decide the sign (positive, negative, or zero) of each of the following partial derivatives at the point 𝑃 . Assume the 𝑥- and 𝑦-axes are in the usual positions. (b) 𝑓𝑦 (𝑃 ) (e) 𝑓𝑥𝑦 (𝑃 )

2

✛ ✛

(c) 𝑓𝑥𝑥 (𝑃 )



5 3 1

𝑃 5

1

𝑃

30.

23. 3

3

4

1

5

20. 𝑓 (𝑥, 𝑦) = ln(1 + 𝑥 − 2𝑦) √ 21. 𝑓 (𝑥, 𝑦) = 1 + 2𝑥 − 𝑦

22.

4

1

In Exercises 20–21, find the best quadratic approximation for 𝑓 (𝑥, 𝑦) for (𝑥, 𝑦) near (0, 0).

(a) 𝑓𝑥 (𝑃 ) (d) 𝑓𝑦𝑦 (𝑃 )

3

2

28.

5 4

1

4

3

𝑃

2

1

𝑃

31.

24.

✛ ✛

25. 1

2

3

4

5

1

2

𝑃

3

4

5



𝑃

1 3 5

𝑃

PROBLEMS In Problems 32–35 estimate the quantity, if possible. If it is not possible, explain why. Assume that 𝑔 is smooth and ∇𝑔(2, 3) = −7⃗𝑖 + 3𝑗⃗ ∇𝑔(2.4, 3) = −10.2⃗𝑖 + 4.2𝑗⃗

In Problems 40–44, find the linear, 𝐿(𝑥, 𝑦), and quadratic, 𝑄(𝑥, 𝑦), Taylor polynomials valid near (1, 0). Compare the values of the approximations 𝐿(0.9, 0.2) and 𝑄(0.9, 0.2) with the exact value of the function 𝑓 (0.9, 0.2). √ 41. 𝑓 (𝑥, 𝑦) = 𝑥2 𝑦 40. 𝑓 (𝑥, 𝑦) = 𝑥 + 2𝑦

32. 𝑔𝑦𝑥 (2, 3) 33. 𝑔𝑥𝑥 (2, 3) 34. 𝑔𝑦𝑦 (2, 3) 35. 𝑔𝑥𝑦 (2, 3)

42. 𝑓 (𝑥, 𝑦) = 𝑥𝑒−𝑦

In Problems 36–39 estimate the quantity, if possible. If it is not possible, explain why. Assume that ℎ is smooth and

43. 𝐹 (𝑥, 𝑦)= 𝑒𝑥 sin 𝑦 + 𝑒𝑦 sin 𝑥

∇ℎ(2.4, 3) = −20.4⃗𝑖 + 8.4𝑗⃗ ∇ℎ(2.4, 2.7) = −22.2⃗𝑖 + 9𝑗⃗ 36. ℎ𝑦𝑦 (2.4, 2.7)

37. ℎ𝑥𝑥 (2.4, 3)

38. ℎ𝑥𝑦 (2.4, 2.7)

39. ℎ𝑦𝑥 (2.4, 2.7)

44. 𝑓 (𝑥, 𝑦) = sin(𝑥 − 1) cos 𝑦 In Problems 45–46, show that the function satisfies Laplace’s equation, 𝐹𝑥𝑥 + 𝐹𝑦𝑦 = 0. 45. 𝐹 (𝑥, 𝑦) = 𝑒−𝑥 sin 𝑦 46. 𝐹 (𝑥, 𝑦) = arctan(𝑦∕𝑥)

14.7 SECOND-ORDER PARTIAL DERIVATIVES

47. If 𝑢(𝑥, 𝑡) = 𝑒𝑎𝑡 sin (𝑏𝑥) satisfies the heat equation 𝑢𝑡 = 𝑢𝑥𝑥 , find the relationship between 𝑎 and 𝑏. 48. (a) Check that 𝑢(𝑥, 𝑡) satisfies the heat equation 𝑢𝑡 = 𝑢𝑥𝑥 for 𝑡 > 0 and all 𝑥, where 2 1 𝑢(𝑥, 𝑡) = √ 𝑒−𝑥 ∕(4𝑡) 2 𝜋𝑡

Problems 53–56 give tables of values of quadratic polynomials 𝑃 (𝑥, 𝑦) = 𝑎 + 𝑏𝑥 + 𝑐𝑦 + 𝑑𝑥2 + 𝑒𝑥𝑦 + 𝑓 𝑦2 . Determine whether each of the coefficients 𝑑, 𝑒 and 𝑓 of the quadratic terms is positive, negative, or zero. 53.

(b) Graph 𝑢(𝑥, 𝑡) against 𝑥 for 𝑡 = 0.01, 0.1, 1, 10. These graphs represent the temperature in an infinitely long insulated rod that at 𝑡 = 0 is 0◦ C everywhere except at the origin 𝑥 = 0, and that is infinitely hot at 𝑡 = 0 at the origin. 49. Figure 14.61 shows a graph of 𝑧 = 𝑓 (𝑥, 𝑦). Is 𝑓𝑥𝑥 (0, 0) positive or negative? Is 𝑓𝑦𝑦 (0, 0) positive or negative? Give reasons for your answers.

797

54.

𝑥

𝑦

10

12

14

10

35

37

39

15

45

47

49

20

55

57

59

55.

𝑦

𝑦

10

12

14

10

26

36

54

15

31

41

59

20

36

46

64

10

12

14

10

13

33

61

15

28

28

36

20

93

73

61

56.

𝑥

𝑧

𝑥

10

12

14

10

90

82

74

15

75

87

99

20

10

42

74

𝑥

𝑦

𝑦 𝑥

57. You are hiking on a level trail going due east and planning to strike off cross country up the mountain to your left. The slope up to the left is too steep now and seems to be gentler the further you go along the trail, so you decide to wait before turning off.

Figure 14.61 50. If 𝑧 = 𝑓 (𝑥) + 𝑦𝑔(𝑥), what can you say about 𝑧𝑦𝑦 ? Explain your answer. 51. If 𝑧𝑥𝑦 = 4𝑦, what can you say about the value of (a) 𝑧𝑦𝑥 ? (b) 𝑧𝑥𝑦𝑥 ? (c) 𝑧𝑥𝑦𝑦 ? 52. A contour diagram for the smooth function 𝑧 = 𝑓 (𝑥, 𝑦) is in Figure 14.62. (a) Is 𝑧 an increasing or decreasing function of 𝑥? Of 𝑦? (b) Is 𝑓𝑥 positive or negative? How about 𝑓𝑦 ? (c) Is 𝑓𝑥𝑥 positive or negative? How about 𝑓𝑦𝑦 ? (d) Sketch the direction of grad 𝑓 at points 𝑃 and 𝑄. (e) Is grad 𝑓 longer at 𝑃 or at 𝑄? How do you know? 𝑦 𝑃

6 1 2 3 4

5 4 3

6

7

8

9 10

2 𝑄 𝑥 1

2

58. The weekly production, 𝑌 , in factories that manufacture a certain item is modeled as a function of the quantity of capital, 𝐾, and quantity of labor, 𝐿, at the factory. Data shows that hiring a few extra workers increases production. Moreover, for two factories with the same number of workers, hiring a few extra workers increases production more for the factory with more capital. (With more equipment, additional labor can be used more effectively.) What does this tell you about the sign of (a) 𝜕𝑌 ∕𝜕𝐿? (b) 𝜕 2 𝑌 ∕(𝜕𝐾𝜕𝐿)?

5

1

(a) Sketch a topographical contour map that illustrates this story. (b) What information does the story give about partial derivatives? Define all variables and functions that you use. (c) What partial derivative influenced your decision to wait before turning?

3

4

Figure 14.62

5

6

59. Data suggests that human surface area, 𝑆, can reasonably be modeled as a function of height, ℎ, and weight, 𝑤. In the Dubois model, we have 𝜕 2 𝑆∕𝜕𝑤2 < 0 and 𝜕 2 𝑆∕(𝜕ℎ𝜕𝑤) > 0. Two people 𝐴 and 𝐵 each gain 1 pound. Which experiences the greater increase in surface area if (a) They have the same weight but 𝐴 is taller? (b) They have the same height, but 𝐴 is heavier?

798

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

60. You plan to buy a used car. You are debating between a 5-year old car and a 10-year old car and thinking about the price. Experts report that the original price matters more when buying a 5-year old car than a 10-year old car. This suggests that we model the average market price, 𝑃 , in dollars as a function of two variables: the original price, 𝐶, in dollars, and the age of the car, 𝐴, in years. (a) Give units for the following partial derivatives and say whether you think they are positive or negative. Explain your reasoning. (i) 𝜕𝑃 ∕𝜕𝐴

(ii) 𝜕𝑃 ∕𝜕𝐶

𝑓 (𝑥, 𝑦) ≈ 𝑎 + 𝑏(𝑥 − 𝑥1 ) + 𝑐(𝑦 − 𝑦1 ) 𝑓 (𝑥, 𝑦) ≈ 𝑘 + 𝑚(𝑥 − 𝑥2 ) + 𝑛(𝑦 − 𝑦2 ). (a) What is the relationship between the values of 𝑎 and 𝑘? (b) What are the coordinates of 𝑃 ? (c) Is 𝑀 a maximum or a minimum? (d) What can you say about the sign of the constants 𝑚 and 𝑛? 𝑦

(b) Express the experts’ report in terms of partial derivatives. (c) Using a quadratic polynomial to model 𝑃 , we have 2

has coordinates (𝑥2 , 𝑦2 ). Suppose 𝑏 > 0 and 𝑐 > 0. Consider the two linear approximations to 𝑓 given by

𝑄

𝑃



✠ 𝑀

2

𝑃 = 𝑎 + 𝑏𝐶 + 𝑐𝐴 + 𝑑𝐶 + 𝑒𝐶𝐴 + 𝑓 𝐴 . 𝑥

Which term in this polynomial is most relevant to the experts’ report? 61. The tastiness, 𝑇 , of a soup depends on the volume, 𝑉 , of the soup in the pot and the quantity, 𝑆, of salt in the soup. If you have more soup, you need more salt to make it taste good. Match the three stories (a)–(c) to the three statements (I)–(III) about partial derivatives. (a) I started adding salt to the soup in the pot. At first the taste improved, but eventually the soup became too salty and continuing to add more salt made it worse. (b) The soup was too salty, so I started adding unsalted soup. This improved the taste at first, but eventually there was too much soup for the salt, and continuing to add unsalted soup just made it worse. (c) The soup was too salty, so adding more salt would have made it taste worse. I added a quart of unsalted soup instead. Now it is not salty enough, but I can improve the taste by adding salt. (I) 𝜕 2 𝑇 ∕𝜕𝑉 2 < 0 (II) 𝜕 2 𝑇 ∕𝜕𝑆 2 < 0 (III) 𝜕 2 𝑇 ∕𝜕𝑉 𝜕𝑆 > 0 62. Figure 14.63 shows the level curves of a function 𝑓 (𝑥, 𝑦) around a maximum or minimum, 𝑀. One of the points 𝑃 and 𝑄 has coordinates (𝑥1 , 𝑦1 ) and the other

Figure 14.63 63. Consider the function 𝑓 (𝑥, 𝑦) = (sin 𝑥)(sin 𝑦). (a) Find the Taylor polynomials of degree 2 for 𝑓 about the points (0, 0) and (𝜋∕2, 𝜋∕2). (b) Use the Taylor polynomials to sketch the contours of 𝑓 close to each of the points (0, 0) and (𝜋∕2, 𝜋∕2). √ 64. Let 𝑓 (𝑥, 𝑦) = 𝑥 + 2𝑦 + 1.

(a) Compute the local linearization of 𝑓 at (0, 0). (b) Compute the quadratic Taylor polynomial for 𝑓 at (0, 0). (c) Compare the values of the linear and quadratic approximations in part (a) and part (b) with the true values for 𝑓 (𝑥, 𝑦) at the points (0.1, 0.1), (−0.1, 0.1), (0.1, −0.1), (−0.1, −0.1). Which approximation gives the closest values?

65. Using a computer and your answer to Problem √ 64, draw the six contour diagrams of 𝑓 (𝑥, 𝑦) = 𝑥 + 2𝑦 + 1 and its linear and quadratic approximations, 𝐿(𝑥, 𝑦) and 𝑄(𝑥, 𝑦), in the two windows [−0.6, 0.6]×[−0.6, 0.6] and [−2, 2]×[−2, 2]. Explain the shape of the contours, their spacing, and the relationship between the contours of 𝑓 , 𝐿, and 𝑄.

Strengthen Your Understanding In Problems 66–67, explain what is wrong with the statement.

68. A function 𝑓 (𝑥, 𝑦) such that 𝑓𝑥𝑥 ≠ 0, 𝑓𝑦𝑦 ≠ 0, and 𝑓𝑥𝑦 = 0.

66. If 𝑓 (𝑥, 𝑦) ≠ 0, then the Taylor polynomial of degree 2 approximating 𝑓 (𝑥, 𝑦) near (0, 0) is also nonzero. 67. There is a function 𝑓 (𝑥, 𝑦) with partial derivatives 𝑓𝑥 = 𝑥𝑦 and 𝑓𝑦 = 𝑦2 .

69. Formulas for two different functions 𝑓 (𝑥, 𝑦) and 𝑔(𝑥, 𝑦) with the same quadratic approximation near (0, 0).

In Problems 68–70, give an example of:

70. Contour diagrams for two different functions 𝑓 (𝑥, 𝑦) and 𝑔(𝑥, 𝑦) that have the same quadratic approximations near (0, 0).

14.8 DIFFERENTIABILITY

14.8

799

DIFFERENTIABILITY In Section 14.3 we gave an informal introduction to the concept of differentiability. We called a function 𝑓 (𝑥, 𝑦) differentiable at a point (𝑎, 𝑏) if it is well approximated by a linear function near (𝑎, 𝑏). This section focuses on the precise meaning of the phrase “well approximated.” By looking at examples, we shall see that local linearity requires the existence of partial derivatives, but they do not tell the whole story. In particular, existence of partial derivatives at a point is not sufficient to guarantee local linearity at that point. We begin by discussing the relation between continuity and differentiability. As an illustration, take a sheet of paper, crumple it into a ball and smooth it out again. Wherever there is a crease it would be difficult to approximate the surface by a plane—these are points of nondifferentiability of the function giving the height of the paper above the floor. Yet the sheet of paper models a graph which is continuous—there are no breaks. As in the case of one-variable calculus, continuity does not imply differentiability. But differentiability does require continuity: there cannot be linear approximations to a surface at points where there are abrupt changes in height.

Differentiability for Functions of Two Variables For a function of two variables, as for a function of one variable, we define differentiability at a point in terms of the error and the distance from √ the point. If the point is (𝑎, 𝑏) and a nearby point is (𝑎 + ℎ, 𝑏 + 𝑘), the distance between them is ℎ2 + 𝑘2 . (See Figure 14.64.) A function 𝑓 (𝑥, 𝑦) is differentiable at the point (𝑎, 𝑏) if there is a linear function 𝐿(𝑥, 𝑦) = 𝑓 (𝑎, 𝑏) + 𝑚(𝑥 − 𝑎) + 𝑛(𝑦 − 𝑏) such that if the error 𝐸(𝑥, 𝑦) is defined by 𝑓 (𝑥, 𝑦) = 𝐿(𝑥, 𝑦) + 𝐸(𝑥, 𝑦), √ and if ℎ = 𝑥 − 𝑎, 𝑘 = 𝑦 − 𝑏, then the relative error 𝐸(𝑎 + ℎ, 𝑏 + 𝑘)∕ ℎ2 + 𝑘2 satisfies lim ℎ→0 𝑘→0

𝐸(𝑎 + ℎ, 𝑏 + 𝑘) = 0. √ ℎ 2 + 𝑘2

The function 𝑓 is differentiable on a region 𝑅 if it is differentiable at each point of 𝑅. The function 𝐿(𝑥, 𝑦) is called the local linearization of 𝑓 (𝑥, 𝑦) near (𝑎, 𝑏).

Figure 14.64: Graph of function 𝑧 = 𝑓 (𝑥, 𝑦) and its local linearization 𝑧 = 𝐿(𝑥, 𝑦) near the point (𝑎, 𝑏)

Partial Derivatives and Differentiability In the next example, we show that this definition of differentiability is consistent with our previous notion — that is, that 𝑚 = 𝑓𝑥 and 𝑛 = 𝑓𝑦 and that the graph of 𝐿(𝑥, 𝑦) is the tangent plane. Example 1

Show that if 𝑓 is a differentiable function with local linearization 𝐿(𝑥, 𝑦) = 𝑓 (𝑎, 𝑏) + 𝑚(𝑥 − 𝑎) + 𝑛(𝑦 − 𝑏), then 𝑚 = 𝑓𝑥 (𝑎, 𝑏) and 𝑛 = 𝑓𝑦 (𝑎, 𝑏).

800

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

Solution

Since 𝑓 is differentiable, we know that the relative error in 𝐿(𝑥, 𝑦) tends to 0 as we get close to (𝑎, 𝑏). Suppose ℎ > 0 and 𝑘 = 0. Then we know that 𝑓 (𝑎 + ℎ, 𝑏) − 𝐿(𝑎 + ℎ, 𝑏) 𝐸(𝑎 + ℎ, 𝑏 + 𝑘) 𝐸(𝑎 + ℎ, 𝑏) = lim = lim √ ℎ→0 ℎ→0 ℎ→0 ℎ ℎ ℎ 2 + 𝑘2 𝑓 (𝑎 + ℎ, 𝑏) − 𝑓 (𝑎, 𝑏) − 𝑚ℎ = lim ℎ→0 ℎ ) ( 𝑓 (𝑎 + ℎ, 𝑏) − 𝑓 (𝑎, 𝑏) − 𝑚 = 𝑓𝑥 (𝑎, 𝑏) − 𝑚. = lim ℎ→0 ℎ

0 = lim

A similar result holds if ℎ < 0, so we have 𝑚 = 𝑓𝑥 (𝑎, 𝑏). The result 𝑛 = 𝑓𝑦 (𝑎, 𝑏) is found in a similar manner. The previous example shows that if a function is differentiable at a point, it has partial derivatives there. Therefore, if any of the partial derivatives fail to exist, then the function cannot be differentiable. This is what happens in the following example of a cone. √

𝑥2 + 𝑦2 . Is 𝑓 differentiable at the origin?

Example 2

Consider the function 𝑓 (𝑥, 𝑦) =

Solution

√ If we zoom in on the graph of the function 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 at the origin, as shown in Figure 14.65, the sharp point remains; the graph never flattens out to look like a plane. Near its vertex, the graph does not look as if is well approximated (in any reasonable sense) by any plane. 𝑧



𝑥

𝑦 √

Figure 14.65: The function 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 is not locally linear at (0, 0): Zooming in around (0, 0) does not make the graph look like a plane

Judging from the graph of 𝑓 , we would not expect 𝑓 to be differentiable at (0, 0). Let us check this by trying to compute the partial derivatives of 𝑓 at (0, 0): √ |ℎ| 𝑓 (ℎ, 0) − 𝑓 (0, 0) ℎ2 + 0 − 0 𝑓𝑥 (0, 0) = lim = lim = lim . ℎ→0 ℎ→0 ℎ ℎ→0 ℎ ℎ

Since |ℎ|∕ℎ = ±1, depending on whether ℎ approaches 0 from the left or right, this limit does not exist and so neither does the partial derivative 𝑓𝑥 (0, 0). Thus, 𝑓 cannot be differentiable at the origin. If it were, both of the partial derivatives, 𝑓𝑥 (0, 0) and 𝑓𝑦 (0, 0), would exist. Alternatively, we could show directly that there is no linear approximation near (0, 0) that satisfies the small relative error criterion for differentiability. Any plane passing through the point (0, 0, 0) has the form 𝐿(𝑥, 𝑦) = 𝑚𝑥 + 𝑛𝑦 for some constants 𝑚 and 𝑛. If 𝐸(𝑥, 𝑦) = 𝑓 (𝑥, 𝑦) − 𝐿(𝑥, 𝑦), then √ 𝐸(𝑥, 𝑦) = 𝑥2 + 𝑦2 − 𝑚𝑥 − 𝑛𝑦. Then for 𝑓 to be differentiable at the origin, we would need to show that √ ℎ2 + 𝑘2 − 𝑚ℎ − 𝑛𝑘 = 0. lim √ ℎ→0 ℎ 2 + 𝑘2 𝑘→0

14.8 DIFFERENTIABILITY

801

Taking 𝑘 = 0 gives lim

ℎ→0

|ℎ| − 𝑚ℎ ℎ = 1 − 𝑚 lim . ℎ→0 |ℎ| |ℎ|

This limit exists only if 𝑚 = 0 for the same reason as before. But then the value of the limit is 1 and not 0 as required. Thus, we again conclude 𝑓 is not differentiable. In Example 2 the partial derivatives 𝑓𝑥 and 𝑓𝑦 did not exist at the origin and this was sufficient to establish nondifferentiability there. We might expect that if both partial derivatives do exist, then 𝑓 is differentiable. But the next example shows that this not necessarily true: the existence of both partial derivatives at a point is not sufficient to guarantee differentiability. Example 3

Consider the function 𝑓 (𝑥, 𝑦) = 𝑥1∕3 𝑦1∕3 . Show that the partial derivatives 𝑓𝑥 (0, 0) and 𝑓𝑦 (0, 0) exist, but that 𝑓 is not differentiable at (0, 0).

Solution

See Figure 14.66 for the part of the graph of 𝑧 = 𝑥1∕3 𝑦1∕3 when 𝑧 ≥ 0. We have 𝑓 (0, 0) = 0 and we compute the partial derivatives using the definition: 𝑓𝑥 (0, 0) = lim

ℎ→0

𝑓 (ℎ, 0) − 𝑓 (0, 0) 0−0 = lim = 0, ℎ→0 ℎ ℎ

and similarly 𝑓𝑦 (0, 0) = 0.

Figure 14.66: Graph of 𝑧 = 𝑥1∕3 𝑦1∕3 for 𝑧 ≥ 0

So, if there did exist a linear approximation near the origin, it would have to be 𝐿(𝑥, 𝑦) = 0. But we can show that this choice of 𝐿(𝑥, 𝑦) does not result in the small relative error that is required for differentiability. In fact, since 𝐸(𝑥, 𝑦) = 𝑓 (𝑥, 𝑦) − 𝐿(𝑥, 𝑦) = 𝑓 (𝑥, 𝑦), we need to look at the limit ℎ1∕3 𝑘1∕3 lim √ . ℎ→0 ℎ 2 + 𝑘2 𝑘→0

If this limit exists, we get the same value no matter how ℎ and 𝑘 approach 0. Suppose we take 𝑘 = ℎ > 0. Then the limit becomes ℎ2∕3 1 ℎ1∕3 ℎ1∕3 lim √ = lim √ = lim √ . ℎ→0 ℎ→0 1∕3 ℎ→0 2 2 ℎ 2 ℎ 2 ℎ +ℎ

But this limit does not exist, since small values for ℎ will make the fraction arbitrarily large. So the only possible candidate for a linear approximation at the origin does not have a sufficiently small relative error. Thus, this function is not differentiable at the origin, even though the partial derivatives 𝑓𝑥 (0, 0) and 𝑓𝑦 (0, 0) exist. Figure 14.66 confirms that near the origin the graph of 𝑧 = 𝑓 (𝑥, 𝑦) is not well approximated by any plane.

802

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

In summary, • If a function is differentiable at a point, then both partial derivatives exist there. • Having both partial derivatives at a point does not guarantee that a function is differentiable there.

Continuity and Differentiability We know that differentiable functions of one variable are continuous. Similarly, it can be shown that if a function of two variables is differentiable at a point, then the function is continuous there. In Example 3 the function 𝑓 was continuous at the point where it was not differentiable. Example 4 shows that even if the partial derivatives of a function exist at a point, the function is not necessarily continuous at that point if it is not differentiable there. Example 4

Suppose that 𝑓 is the function of two variables defined by { 𝑥𝑦 , (𝑥, 𝑦) ≠ (0, 0), 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 0, (𝑥, 𝑦) = (0, 0). Problem 27 on page 699 showed that 𝑓 (𝑥, 𝑦) is not continuous at the origin. Show that the partial derivatives 𝑓𝑥 (0, 0) and 𝑓𝑦 (0, 0) exist. Could 𝑓 be differentiable at (0, 0)?

Solution

From the definition of the partial derivative we see that ( ) 𝑓 (ℎ, 0) − 𝑓 (0, 0) 0 1 0 𝑓𝑥 (0, 0) = lim = lim ⋅ = lim = 0, 2 2 ℎ→0 ℎ→0 ℎ ℎ + 0 ℎ→0 ℎ ℎ and similarly 𝑓𝑦 (0, 0) = 0. So, the partial derivatives 𝑓𝑥 (0, 0) and 𝑓𝑦 (0, 0) exist. However, 𝑓 cannot be differentiable at the origin since it is not continuous there. In summary, • If a function is differentiable at a point, then it is continuous there. • Having both partial derivatives at a point does not guarantee that a function is continuous there.

How Do We Know If a Function Is Differentiable? Can we use partial derivatives to tell us if a function is differentiable? As we see from Examples 3 and 4, it is not enough that the partial derivatives exist. However, the following theorem gives conditions that do guarantee differentiability4:

Theorem 14.2: Continuity of Partial Derivatives Implies Differentiability If the partial derivatives, 𝑓𝑥 and 𝑓𝑦 , of a function 𝑓 exist and are continuous on a small disk centered at the point (𝑎, 𝑏), then 𝑓 is differentiable at (𝑎, 𝑏). 4 For

a proof, see M. Spivak, Calculus on Manifolds, p. 31 (New York: Benjamin, 1965).

14.8 DIFFERENTIABILITY

803

We will not prove this theorem, although it provides a criterion for differentiability which is often simpler to use than the definition. It turns out that the requirement of continuous partial derivatives is more stringent than that of differentiability, so there exist differentiable functions which do not have continuous partial derivatives. However, most functions we encounter will have continuous partial derivatives. The class of functions with continuous partial derivatives is given the name 𝐶 1 . Example 5

Show that the function 𝑓 (𝑥, 𝑦) = ln(𝑥2 + 𝑦2 ) is differentiable everywhere in its domain.

Solution

The domain of 𝑓 is all of 2-space except for the origin. We shall show that 𝑓 has continuous partial derivatives everywhere in its domain (that is, the function 𝑓 is in 𝐶 1 ). The partial derivatives are 𝑓𝑥 =

2𝑥 𝑥2 + 𝑦2

and 𝑓𝑦 =

2𝑦 . 𝑥2 + 𝑦2

Since each of 𝑓𝑥 and 𝑓𝑦 is the quotient of continuous functions, the partial derivatives are continuous everywhere except the origin (where the denominators are zero). Thus, 𝑓 is differentiable everywhere in its domain. Most functions built up from elementary functions have continuous partial derivatives, except perhaps at a few obvious points. Thus, in practice, we can often identify functions as being 𝐶 1 without explicitly computing the partial derivatives.

Exercises and Problems for Section 14.8 EXERCISES In Exercises 1–10, list the points in the 𝑥𝑦-plane, if any, at which the function 𝑧 = 𝑓 (𝑥, 𝑦) is not differentiable. √ 1. 𝑧 = − 𝑥2 + 𝑦2

2. 𝑧 =

3. 𝑧 = |𝑥| + |𝑦|

4. 𝑧 = |𝑥 + 2| − |𝑦 − 3|

PROBLEMS



(𝑥 + 1)2 + 𝑦2

In Problems 11–14, a function 𝑓 is given. (a) Use a computer to draw a contour diagram for 𝑓 . (b) Is 𝑓 differentiable at all points (𝑥, 𝑦) ≠ (0, 0)? (c) Do the partial derivatives 𝑓𝑥 and 𝑓𝑦 exist and are they continuous at all points (𝑥, 𝑦) ≠ (0, 0)? (d) Is 𝑓 differentiable at (0, 0)? (e) Do the partial derivatives 𝑓𝑥 and 𝑓𝑦 exist and are they continuous at (0, 0)? ⎧𝑥 𝑦 ⎪ + , 𝑥 ≠ 0 and 𝑦 ≠ 0, 11. 𝑓 (𝑥, 𝑦) = ⎨ 𝑦 𝑥 ⎪ 0, 𝑥 = 0 or 𝑦 = 0. ⎩ { 2𝑥𝑦 , (𝑥, 𝑦) ≠ (0, 0), 12. 𝑓 (𝑥, 𝑦) = (𝑥2 + 𝑦2 )2 (𝑥, 𝑦) = (0, 0). { 0, 2 𝑥𝑦 , (𝑥, 𝑦) ≠ (0, 0), 13. 𝑓 (𝑥, 𝑦) = 𝑥4 + 𝑦2 0, (𝑥, 𝑦) = (0, 0). { 𝑥𝑦 , (𝑥, 𝑦) ≠ (0, 0), √ 14. 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 0, (𝑥, 𝑦) = (0, 0).

2 +𝑦2 )

5. 𝑧 = 𝑒−(𝑥

6. 𝑧 = 𝑥1∕3 + 𝑦2

7. 𝑧 = |𝑥 − 3|2 + 𝑦3 8. 𝑧 = (sin 𝑥)(cos |𝑦|) √ 9. 𝑧 = 4 + (𝑥 − 1)2 + (𝑦 − 2)2 ( )2 10. 𝑧 = 1 + (𝑥 − 1)2 + (𝑦 − 2)2 15. Consider the function { 𝑓 (𝑥, 𝑦) =

𝑥𝑦2 , + 𝑦2

𝑥2

(𝑥, 𝑦) ≠ (0, 0),

0,

(𝑥, 𝑦) = (0, 0).

(a) (b) (c) (d) (e)

Use a computer to draw the contour diagram for 𝑓 . Is 𝑓 differentiable for (𝑥, 𝑦) ≠ (0, 0)? Show that 𝑓𝑥 (0, 0) and 𝑓𝑦 (0, 0) exist. Is 𝑓 differentiable at (0, 0)? Suppose 𝑥(𝑡) = 𝑎𝑡 and 𝑦(𝑡) = 𝑏𝑡, where 𝑎 and 𝑏 are constants, not both zero. If 𝑔(𝑡) = 𝑓 (𝑥(𝑡), 𝑦(𝑡)), show that 𝑎𝑏2 𝑔 ′ (0) = 2 . 𝑎 + 𝑏2 (f) Show that 𝑓𝑥 (0, 0)𝑥′ (0) + 𝑓𝑦 (0, 0)𝑦′ (0) = 0.

Does the chain rule hold for the composite function 𝑔(𝑡) at 𝑡 = 0? Explain. (g) Show that the directional derivative 𝑓𝑢⃗ (0, 0) exists for each unit vector 𝑢⃗ . Does this imply that 𝑓 is differentiable at (0, 0)?

804

Chapter 14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES

16. Consider the function 𝑓 (𝑥, 𝑦) =

√ |𝑥𝑦|.

(a) Use a computer to draw the contour diagram for 𝑓 . Does the contour diagram look like that of a plane when we zoom in on the origin? (b) Use a computer to draw the graph of 𝑓 . Does the graph look like a plane when we zoom in on the origin? (c) Is 𝑓 differentiable for (𝑥, 𝑦) ≠ (0, 0)? (d) Show that 𝑓𝑥 (0, 0) and 𝑓𝑦 (0, 0) exist. (e) Is 𝑓 differentiable at (0, 0)? [Hint: Consider the √directional derivative 𝑓𝑢⃗ (0, 0) for 𝑢⃗ = (⃗𝑖 + 𝑗⃗ )∕ 2.]

17. Consider the function { 𝑓 (𝑥, 𝑦) =

𝑥𝑦2 , + 𝑦4

𝑥2

0,

(b) Show that 𝑓𝑥 (0, 0) = 0 and 𝑓𝑦 (0, 0) = 0. (c) Are the functions 𝑓𝑥 and 𝑓𝑦 continuous at (0, 0)? (d) Is 𝑓 differentiable at (0, 0)? 𝑧

𝑦

𝑥

(𝑥, 𝑦) ≠ (0, 0), (𝑥, 𝑦) = (0, 0).

Figure 14.67: Graph of

(a) Use a computer to draw the contour diagram for 𝑓 . (b) Show that the directional derivative 𝑓𝑢⃗ (0, 0) exists for each unit vector 𝑢⃗ . (c) Is 𝑓 continuous at (0, 0)? Is 𝑓 differentiable at (0, 0)? Explain.

5 −. −. 8 65

−.

45 −.25 −.05

(a) Is 𝑓 differentiable at (0, 0)? Explain. (b) Give an example of a function 𝑓 defined on 2-space which satisfies these conditions. [Hint: The function 𝑓 does not have to be defined by a single formula valid over all of 2-space.]

𝑦

.25

18. Suppose 𝑓 (𝑥, 𝑦) is a function such that 𝑓𝑥 (0, 0) = 0 and √ 𝑓𝑦 (0, 0) = 0, and 𝑓𝑢⃗ (0, 0) = 3 for 𝑢⃗ = (⃗𝑖 + 𝑗⃗ )∕ 2.

𝑥𝑦(𝑥2 − 𝑦2 ) 𝑥2 + 𝑦2

.45

−.25

.05

.85 .65

.25

𝑥 .25

.25

−.

25

−.

25

19. Consider the following function: 𝑓 (𝑥, 𝑦) =

{

𝑥𝑦(𝑥2 − 𝑦2 ) , 𝑥2 + 𝑦2 0,

(𝑥, 𝑦) ≠ (0, 0), (𝑥, 𝑦) = (0, 0).

The graph of 𝑓 is shown in Figure 14.67, and the contour diagram of 𝑓 is shown in Figure 14.68. (a) Find 𝑓𝑥 (𝑥, 𝑦) and 𝑓𝑦 (𝑥, 𝑦) for (𝑥, 𝑦) ≠ (0, 0).

Figure 14.68: Contour diagram of 𝑥𝑦(𝑥2 − 𝑦2 ) 𝑥2 + 𝑦2

20. Suppose a function 𝑓 is differentiable at the point (𝑎, 𝑏). Show that 𝑓 is continuous at (𝑎, 𝑏).

Strengthen Your Understanding In Problems 21–22, explain what is wrong with the statement. 21. If 𝑓 (𝑥, 𝑦) is continuous at the origin, then it is differentiable at the origin. 22. If the partial derivatives 𝑓𝑥 (0, 0) and 𝑓𝑦 (0, 0) both exist, then 𝑓 (𝑥, 𝑦) is differentiable at the origin. In Problems 23–24, give an example of:

Online Resource: Review problems and Projects

23. A continuous function 𝑓 (𝑥, 𝑦) that is not differentiable at the origin. 24. A continuous function 𝑓 (𝑥, 𝑦) that is not differentiable on the line 𝑥 = 1. 25. Which of the following functions 𝑓 (𝑥, 𝑦) is differentiable at the given point? √ √ 1 − 𝑥2 − 𝑦2 at (0, 0) (b) 4 − 𝑥2 − 𝑦2 at (2, 0) (a) √ √ (c) − 𝑥2 + 2𝑦2 at (0, 0) (d) − 𝑥2 + 2𝑦2 at (2, 0)

Chapter Fifteen

OPTIMIZATION: LOCAL AND GLOBAL EXTREMA

Contents 15.1 Critical Points: Local Extrema and Saddle Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . How Do We Detect a Local Maximum or Minimum? . . . . . . . . . . . . . . . . . . . . . . . . . . . Finding and Analyzing Critical Points . . . . . . . . . . Classifying Critical Points . . . . . . . . . . . . . . . . . .

806 806 807 809

Quadratic Functions of the Form f (x, y) = ax2 + bxy + cy2 . . . . . . . . . . . . . . . . . . . 809 The Shape of the Graph of f (x, y) = ax2 + bxy + cy2 . . . . . . . . . . . . . . . . . . . 810 Classifying the Critical Points of a Function . . . . . 15.2 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . How Do We Find Global Maxima and Minima? . Maximizing Profit and Minimizing Cost . . . . . . Fitting a Line to Data: Least Squares . . . . . . . . . How Do We Know Whether a Function Has a Global Maximum or Minimum? . . . . . . . 15.3 Constrained Optimization: Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . Graphical Approach: Maximizing Production Subject to a Budget Constraint . . . . . . . . . . . . . Analytical Solution: Lagrange Multipliers . . . . . Lagrange Multipliers in General . . . . . . . . . . . . . How to Distinguish Maxima from Minima . . . . . . . Optimization with Inequality Constraints . . . . . . The Meaning of λ . . . . . . . . . . . . . . . . . . . . . . . . An Expression for λ . . . . . . . . . . . . . . . . . . . . . . The Lagrangian Function. . . . . . . . . . . . . . . . . . .

811 815 816 816 819 820 825 825 826 827 828 828 830 830 831

806

15.1

Chapter 15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA

CRITICAL POINTS: LOCAL EXTREMA AND SADDLE POINTS Functions of several variables, like functions of one variable, can have local and global extrema. (That is, local and global maxima and minima.) A function has a local extremum at a point where it takes on the largest or smallest value in a small region around the point. Global extrema are the largest or smallest values anywhere on the domain under consideration. (See Figures 15.1 and 15.2.) 𝑦 𝑧

𝑏

3

𝑦

0

11 6 1

1 −3 −1

𝑥

𝑥 𝑎

Figure 15.1: Local and global extrema for a function of two variables on 0 ≤ 𝑥 ≤ 𝑎, 0≤𝑦≤𝑏

Figure 15.2: Contour map of the function in Figure 15.1

More precisely, considering only points at which 𝑓 is defined, we say: • 𝑓 has a local maximum at the point 𝑃0 if 𝑓 (𝑃0 ) ≥ 𝑓 (𝑃 ) for all points 𝑃 near 𝑃0 . • 𝑓 has a local minimum at the point 𝑃0 if 𝑓 (𝑃0 ) ≤ 𝑓 (𝑃 ) for all points 𝑃 near 𝑃0 . For example, the function whose contour map is shown in Figure 15.2 has a local minimum value of −3 and local maximum values of 3 and 11 in the rectangle shown.

How Do We Detect a Local Maximum or Minimum? Recall that if the gradient vector of a function is defined and nonzero, then it points in a direction in which the function increases. Suppose that a function 𝑓 has a local maximum at a point 𝑃0 which is not on the boundary of the domain. If the vector grad 𝑓 (𝑃0 ) were defined and nonzero, then we could increase 𝑓 by moving in the direction of grad 𝑓 (𝑃0 ). Since 𝑓 has a local maximum at 𝑃0 , there is no direction in which 𝑓 is increasing. Thus, if grad 𝑓 (𝑃0 ) is defined, we must have grad 𝑓 (𝑃0 ) = 0⃗ . Similarly, suppose 𝑓 has a local minimum at the point 𝑃0 . If grad 𝑓 (𝑃0 ) were defined and nonzero, then we could decrease 𝑓 by moving in the direction opposite to grad 𝑓 (𝑃0 ), and so we must again have grad 𝑓 (𝑃0 ) = 0⃗ . Therefore, we make the following definition: Points where the gradient is either 0⃗ or undefined are called critical points of the function. If a function has a local maximum or minimum at a point 𝑃0 , not on the boundary of its domain, then 𝑃0 is a critical point. For a function of two variables, we can also see that the gradient vector must be zero or undefined at a local maximum by looking at its contour diagram and a plot of its gradient vectors. (See Figures 15.3 and 15.4.) Around the maximum the vectors are all pointing inward, perpendicularly to the contours. At the maximum the gradient vector must be zero or undefined. A similar argument shows that the gradient must be zero or undefined at a local minimum.

15.1 CRITICAL POINTS: LOCAL EXTREMA AND SADDLE POINTS

807

𝑦

𝑦 2 4 5

𝑥

𝑥

3 1

Figure 15.3: Contour diagram around a local maximum of a function

Figure 15.4: Gradients pointing toward the local maximum of the function in Figure 15.3

Finding and Analyzing Critical Points ⃗ = 0⃗ , which means setting all the To find critical points of 𝑓 we set grad 𝑓 = 𝑓𝑥 ⃗𝑖 + 𝑓𝑦 𝑗⃗ + 𝑓𝑧 𝑘 partial derivatives of 𝑓 equal to zero. We must also look for the points where one or more of the partial derivatives is undefined. Example 1

Find and analyze the critical points of 𝑓 (𝑥, 𝑦) = 𝑥2 − 2𝑥 + 𝑦2 − 4𝑦 + 5.

Solution

To find the critical points, we set both partial derivatives equal to zero: 𝑓𝑥 (𝑥, 𝑦) = 2𝑥 − 2 = 0 𝑓𝑦 (𝑥, 𝑦) = 2𝑦 − 4 = 0. Solving these equations gives 𝑥 = 1, 𝑦 = 2. Hence, 𝑓 has only one critical point, namely (1, 2). To see the behavior of 𝑓 near (1, 2), look at the values of the function in Table 15.1. Table 15.1

Values of 𝑓 (𝑥, 𝑦) near the point (1, 2) 𝑥

𝑦

0.8

0.9

1.0

1.1

1.2

1.8

0.08

0.05

0.04

0.05

0.08

1.9

0.05

0.02

0.01

0.02

0.05

2.0

0.04

0.01

0.00

0.01

0.04

2.1

0.05

0.02

0.01

0.02

0.05

2.2

0.08

0.05

0.04

0.05

0.08

The table suggests that the function has a local minimum value of 0 at (1, 2). We can confirm this by completing the square: 𝑓 (𝑥, 𝑦) = 𝑥2 − 2𝑥 + 𝑦2 − 4𝑦 + 5 = (𝑥 − 1)2 + (𝑦 − 2)2 . Figure 15.5 shows that the graph of 𝑓 is a paraboloid with vertex at the point (1, 2, 0). It is the same shape as the graph of 𝑧 = 𝑥2 + 𝑦2 (see Figure 12.12 on page 661), except that the vertex has been shifted to (1, 2). So the point (1, 2) is a local minimum of 𝑓 (as well as a global minimum). 𝑧

𝑥 𝑦 (1, 2, 0) Figure 15.5: The graph of 𝑓 (𝑥, 𝑦) = 𝑥2 − 2𝑥 + 𝑦2 − 4𝑦 + 5 with a local minimum at the point (1, 2)

808

Chapter 15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA

Example 2

√ Find and analyze any critical points of 𝑓 (𝑥, 𝑦) = − 𝑥2 + 𝑦2 .

Solution

We look for points where grad 𝑓 = 0⃗ or is undefined. The partial derivatives are given by 𝑥 , 𝑓𝑥 (𝑥, 𝑦) = − √ 𝑥2 + 𝑦2 𝑦 𝑓𝑦 (𝑥, 𝑦) = − √ . 𝑥2 + 𝑦2

These partial derivatives are never simultaneously zero, but they are undefined at 𝑥 = 0, 𝑦 = 0. Thus, (0, 0) is a critical point and a possible extreme point. The graph of 𝑓 (see Figure 15.6) is a cone, with vertex at (0, 0). So 𝑓 has a local and global maximum at (0, 0). 𝑧 𝑥



𝑦

Local maximum Global maximum

√ Figure 15.6: Graph of 𝑓 (𝑥, 𝑦) = − 𝑥2 + 𝑦2

Example 3

Find and analyze any critical points of 𝑔(𝑥, 𝑦) = 𝑥2 − 𝑦2 .

Solution

To find the critical points, we look for points where both partial derivatives are zero: 𝑔𝑥 (𝑥, 𝑦) = 2𝑥 = 0 𝑔𝑦 (𝑥, 𝑦) = −2𝑦 = 0. Solving gives 𝑥 = 0, 𝑦 = 0, so the origin is the only critical point. Figure 15.7 shows that near the origin 𝑔 takes on both positive and negative values. Since 𝑔(0, 0) = 0, the origin is a critical point which is neither a local maximum nor a local minimum. The graph of 𝑔 looks like a saddle. 𝑧

𝑧

𝑦 𝑥

Figure 15.7: Graph of 𝑔(𝑥, 𝑦) = 𝑥2 − 𝑦2 , showing saddle shape at the origin

𝑥

𝑦

Figure 15.8: Graph of ℎ(𝑥, 𝑦) = 𝑥2 + 𝑦2 , showing minimum at the origin

The previous examples show that critical points can occur at local maxima or minima, or at points which are neither: The functions 𝑔 and ℎ in Figures 15.7 and 15.8 both have critical points at the origin. Figure 15.9 shows level curves of 𝑔. They are hyperbolas showing both positive and negative values of 𝑔 near (0, 0). Contrast this with the level curves of ℎ near the local minimum in Figure 15.10.

15.1 CRITICAL POINTS: LOCAL EXTREMA AND SADDLE POINTS

𝑦

809

𝑦

3

3 12

−4

2

2

−2

4

1

2

4

1

8

𝑥

2

𝑥 4

0

−1

0

−1

2

−2

−2

−2

−4

−3 −3 −2 −1

1

2

6

−3 −3 −2 −1

3

Figure 15.9: Contours of 𝑔(𝑥, 𝑦) = 𝑥2 − 𝑦2 , showing a saddle shape at the origin

1

10

2

3

Figure 15.10: Contours of ℎ(𝑥, 𝑦) = 𝑥2 + 𝑦2 , showing a local minimum at the origin

Example 4

Find the local extrema of the function 𝑓 (𝑥, 𝑦) = 8𝑦3 + 12𝑥2 − 24𝑥𝑦.

Solution

We begin by looking for critical points: 𝑓𝑥 (𝑥, 𝑦) = 24𝑥 − 24𝑦, 𝑓𝑦 (𝑥, 𝑦) = 24𝑦2 − 24𝑥. Setting these expressions equal to zero gives the system of equations 𝑥 = 𝑦2 ,

𝑥 = 𝑦,

which has two solutions, (0, 0) and (1, 1). Are these local maxima, local minima or neither? Figure 15.11 shows contours of 𝑓 near the points. Notice that 𝑓 (1, 1) = −4 and the contours at nearby points have larger function values. This suggests 𝑓 has a local minimum at (1, 1). We have 𝑓 (0, 0) = 0 and the contours near (0, 0) show that 𝑓 takes both positive and negative values nearby. This suggests that (0, 0) is a critical point which is neither a local maximum nor a local minimum. 𝑦

2 1.5

−2

1

−4

0.5

3

−3 −1 1

0

−0.5 −0.5

0

𝑥

2

0.5

1

1.5

2

Figure 15.11: Contour diagram of 𝑓 (𝑥, 𝑦) = 8𝑦3 + 12𝑥2 − 24𝑥𝑦 showing critical points at (0, 0) and (1, 1)

Classifying Critical Points We can see whether a critical point of a function, 𝑓 , is a local maximum, local minimum, or neither by looking at the contour diagram. There is also an analytic method for making this distinction.

Quadratic Functions of the Form 𝒇 (𝒙, 𝒚) = 𝒂𝒙𝟐 + 𝒃𝒙𝒚 + 𝒄𝒚 𝟐 Near most critical points, a function has the same behavior as its quadratic Taylor approximation

810

Chapter 15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA

about that point. Thus, we start by investigating critical points of quadratic functions of the form 𝑓 (𝑥, 𝑦) = 𝑎𝑥2 + 𝑏𝑥𝑦 + 𝑐𝑦2 , where 𝑎, 𝑏 and 𝑐 are constants. Example 5

Find and analyze the local extrema of the function 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑥𝑦 + 𝑦2 .

Solution

To find critical points, we set 𝑓𝑥 (𝑥, 𝑦) = 2𝑥 + 𝑦 = 0, 𝑓𝑦 (𝑥, 𝑦) = 𝑥 + 2𝑦 = 0. The only critical point is (0, 0), and the value of the function there is 𝑓 (0, 0) = 0. If 𝑓 is always positive or zero near (0, 0), then (0, 0) is a local minimum; if 𝑓 is always negative or zero near (0, 0), it is a local maximum; if 𝑓 takes both positive and negative values, it is neither. The graph in Figure 15.12 suggests that (0, 0) is a local minimum. How can we be sure that (0, 0) is a local minimum? We complete the square. Writing ) ( 1 2 3 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑥𝑦 + 𝑦2 = 𝑥 + 𝑦 + 𝑦2 , 2 4 shows that 𝑓 (𝑥, 𝑦) is a sum of two nonnegative terms, so it is always greater than or equal to zero. Thus, the critical point is both a local and a global minimum. 𝑧

𝑦



𝑥

Local minimum

Figure 15.12: Graph of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑥𝑦 + 𝑦2 = (𝑥 + 21 𝑦)2 + 34 𝑦2 showing local minimum at the origin

The Shape of the Graph of 𝒇 (𝒙, 𝒚) = 𝒂𝒙𝟐 + 𝒃𝒙𝒚 + 𝒄𝒚 𝟐 In general, a function of the form 𝑓 (𝑥, 𝑦) = 𝑎𝑥2 + 𝑏𝑥𝑦 + 𝑐𝑦2 has one critical point at (0, 0). Assuming 𝑎 ≠ 0, we complete the square and write ) ] [( ( ) ] [ 𝑐 2 𝑏2 𝑏 2 𝑏 𝑐 2 2 2 − 𝑦2 𝑎𝑥 + 𝑏𝑥𝑦 + 𝑐𝑦 = 𝑎 𝑥 + 𝑥𝑦 + 𝑦 = 𝑎 𝑥 + 𝑦 + 𝑎 𝑎 2𝑎 𝑎 4𝑎2 ) ] ( [( ) 4𝑎𝑐 − 𝑏2 𝑏 2 𝑦2 . =𝑎 𝑥+ 𝑦 + 2𝑎 4𝑎2 The shape of the graph of 𝑓 depends on whether the coefficient of 𝑦2 is positive, negative, or zero. The sign of the discriminant, 𝐷 = 4𝑎𝑐 − 𝑏2 , determines the sign of the coefficient of 𝑦2 . • If 𝐷 > 0, then the expression inside the square brackets is positive or zero, so the function has a local maximum or a local minimum. ∙ If 𝑎 > 0, the function has a local minimum, since the graph is a paraboloid opening upward, like 𝑧 = 𝑥2 + 𝑦2 . (See Figure 15.13.) ∙ If 𝑎 < 0, the function has a local maximum, since the graph is a paraboloid opening downward, like 𝑧 = −𝑥2 − 𝑦2 . (See Figure 15.14.) • If 𝐷 < 0, then the function goes up in some directions and goes down in others, like 𝑧 = 𝑥2 −𝑦2 . We say the function has a saddle point, that is, a critical point at which the function value increases in some directions but decreases in others. (See Figure 15.15.) • If 𝐷 = 0, then the quadratic function is 𝑎(𝑥 + 𝑏𝑦∕2𝑎)2 , whose graph is a parabolic cylinder. (See Figure 15.16.)

15.1 CRITICAL POINTS: LOCAL EXTREMA AND SADDLE POINTS

Figure 15.13: Local minimum: 𝐷 > 0 and 𝑎 > 0

Figure 15.14: Local maximum: 𝐷 > 0 and 𝑎 < 0

Figure 15.15: Saddle point: 𝐷 0 and 𝑓𝑥𝑥 (𝑥0 , 𝑦0 ) > 0, then 𝑓 has a local minimum at (𝑥0 , 𝑦0 ). • If 𝐷 > 0 and 𝑓𝑥𝑥 (𝑥0 , 𝑦0 ) < 0, then 𝑓 has a local maximum at (𝑥0 , 𝑦0 ). • If 𝐷 < 0, then 𝑓 has a saddle point at (𝑥0 , 𝑦0 ). • If 𝐷 = 0, anything can happen: 𝑓 can have a local maximum, or a local minimum, or a saddle point, or none of these, at (𝑥0 , 𝑦0 ).

Example 6

Find the local maxima, minima, and saddle points of 𝑓 (𝑥, 𝑦) = 12 𝑥2 + 3𝑦3 + 9𝑦2 − 3𝑥𝑦 + 9𝑦 − 9𝑥.

Solution

Setting the partial derivatives of 𝑓 to zero gives 𝑓𝑥 (𝑥, 𝑦) = 𝑥 − 3𝑦 − 9 = 0, 𝑓𝑦 (𝑥, 𝑦) = 9𝑦2 + 18𝑦 − 3𝑥 + 9 = 0. 1 We assumed that 𝑎 ≠ 0. If 𝑎 = 0 and 𝑐 ≠ 0, the same argument works. If both 𝑎 = 0 and 𝑐 = 0, then 𝑓 (𝑥, 𝑦) = 𝑏𝑥𝑦, which has a saddle point.

812

Chapter 15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA

Eliminating 𝑥 gives 9𝑦2 + 9𝑦 − 18 = 0, with solutions 𝑦 = −2 and 𝑦 = 1. The corresponding values of 𝑥 are 𝑥 = 3 and 𝑥 = 12, so the critical points of 𝑓 are (3, −2) and (12, 1). The discriminant is 2 𝐷(𝑥, 𝑦) = 𝑓𝑥𝑥 𝑓𝑦𝑦 − 𝑓𝑥𝑦 = (1)(18𝑦 + 18) − (−3)2 = 18𝑦 + 9.

Since 𝐷(3, −2) = −36 + 9 < 0, we know that (3, −2) is a saddle point of 𝑓 . Since 𝐷(12, 1) = 18 + 9 > 0 and 𝑓𝑥𝑥 (12, 1) = 1 > 0, we know that (12, 1) is a local minimum of 𝑓 . The second-derivative test does not give any information if 𝐷 = 0. However, as the following example illustrates, we may still be able to classify the critical points. Example 7

Classify the critical points of 𝑓 (𝑥, 𝑦) = 𝑥4 + 𝑦4 , and 𝑔(𝑥, 𝑦) = −𝑥4 − 𝑦4 , and ℎ(𝑥, 𝑦) = 𝑥4 − 𝑦4 .

Solution

Each of these functions has a critical point at (0, 0). Since all the second partial derivatives are 0 there, each function has 𝐷 = 0. Near the origin, the graphs of 𝑓 , 𝑔 and ℎ look like the surfaces in Figures 15.13–15.15, respectively, so 𝑓 has a local minimum at (0, 0), and 𝑔 has a local maximum at (0, 0), and ℎ is saddle-shaped at (0, 0). We can get the same results algebraically. Since 𝑓 (0, 0) = 0 and 𝑓 (𝑥, 𝑦) > 0 elsewhere, 𝑓 has a local minimum at the origin. Since 𝑔(0, 0) = 0 and 𝑔(𝑥, 𝑦) < 0 elsewhere, 𝑔 has a local maximum at the origin. Lastly, ℎ is saddle-shaped at the origin since ℎ(0, 0) = 0 and, away from the origin, ℎ(𝑥, 𝑦) > 0 on the 𝑥-axis and ℎ(𝑥, 𝑦) < 0 on the 𝑦-axis.

Exercises and Problems for Section 15.1 EXERCISES 1. Figures (I)–(VI) show level curves of six functions around a critical point 𝑃 . Does each function have a local maximum, a local minimum, or a saddle point at 𝑃 ?

2. Which of the points 𝐴, 𝐵, 𝐶 in Figure 15.17 appear to be critical points? Classify those that are critical points.

(II)

(I)

𝑦 2 1 0

𝑃

𝑃

𝐺

𝐸 1

𝐹

−5 −7

−2

2

(III)

−1

−1

−2

0

(IV)

𝑥 −2 −1 1

0𝑃

2 1 1

−1

0𝑃

1 2

1

−1

𝐴

1

𝐷 𝐵

2 −1 −2

(V)

−2

(VI)

2

𝐶

1 2 −1

0

−2

Figure 15.17 𝑃

𝑃

3. Which of the points 𝐷–𝐺 in Figure 15.17 appear to be −1 12 −2

14

(a) Local maxima? (b) Local minima? (c) Saddle points?

15.1 CRITICAL POINTS: LOCAL EXTREMA AND SADDLE POINTS

4. A function 𝑓 (𝑥, 𝑦) has partial derivatives 𝑓𝑥 (1, 2) = 3, 𝑓𝑦 (1, 2) = 5. Explain how you know that 𝑓 does not have a minimum at (1, 2). 5. Assume 𝑓 (𝑥, 𝑦) has a critical point at (2, 3) with 𝑓 (2, 3) = 5. Draw possible cross-sections for 𝑥 = 2 and 𝑦 = 3, and label one value on each axis, if 𝑓 (2, 3) is:

In Exercises 14–27, find the critical points and classify them as local maxima, local minima, saddle points, or none of these. 14. 𝑓 (𝑥, 𝑦) = 𝑥2 − 2𝑥𝑦 + 3𝑦2 − 8𝑦 15. 𝑓 (𝑥, 𝑦) = 5 + 6𝑥 − 𝑥2 + 𝑥𝑦 − 𝑦2 16. 𝑓 (𝑥, 𝑦) = 𝑥2 − 𝑦2 + 4𝑥 + 2𝑦 17. 𝑓 (𝑥, 𝑦) = 400 − 3𝑥2 − 4𝑥 + 2𝑥𝑦 − 5𝑦2 + 48𝑦

(a) A local minimum (b) A local maximum (c) A saddle point

18. 𝑓 (𝑥, 𝑦) = 15 − 𝑥2 + 2𝑦2 + 6𝑥 − 8𝑦 19. 𝑓 (𝑥, 𝑦) = 𝑥2 𝑦 + 2𝑦2 − 2𝑥𝑦 + 6

In Exercises 6–13, the function has a critical point at (0, 0). What sort of critical point is it? 6. 𝑓 (𝑥, 𝑦) = 𝑥2 − cos 𝑦

7. 𝑓 (𝑥, 𝑦) = 𝑥 sin 𝑦

8. 𝑔(𝑥, 𝑦) = 𝑥4 + 𝑦3

9. 𝑓 (𝑥, 𝑦) = 𝑥6 + 𝑦6

10. 𝑘(𝑥, 𝑦) = sin 𝑥 sin 𝑦

813

20. 𝑓 (𝑥, 𝑦) = 2𝑥3 − 3𝑥2 𝑦 + 6𝑥2 − 6𝑦2 21. 𝑓 (𝑥, 𝑦) = 𝑥3 − 3𝑥 + 𝑦3 − 3𝑦 22. 𝑓 (𝑥, 𝑦) = 𝑥3 + 𝑦3 − 3𝑥2 − 3𝑦 + 10 23. 𝑓 (𝑥, 𝑦) = 𝑥3 + 𝑦3 − 6𝑦2 − 3𝑥 + 9 24. 𝑓 (𝑥, 𝑦) = (𝑥 + 𝑦)(𝑥𝑦 + 1)

11. ℎ(𝑥, 𝑦) = cos 𝑥 cos 𝑦

12. 𝑔(𝑥, 𝑦) = (𝑥 − 𝑒𝑥 )(1 − 𝑦2 ) 13. ℎ(𝑥, 𝑦) = 𝑥2 − 𝑥𝑦 + sin2 𝑦

25. 𝑓 (𝑥, 𝑦) = 8𝑥𝑦 − 14 (𝑥 + 𝑦)4 √ 26. 𝑓 (𝑥, 𝑦) = 3 𝑥2 + 𝑦2 2 +𝑦2

27. 𝑓 (𝑥, 𝑦) = 𝑒2𝑥

PROBLEMS 28. Let 𝑓 (𝑥, 𝑦) = 3𝑥2 + 𝑘𝑦2 + 9𝑥𝑦. Determine the values of 𝑘 (if any) for which the critical point at (0, 0) is: −3 −2 −1

(a) A saddle point (b) A local maximum (c) A local minimum

3 2 1

0

0 3

3

𝑆

2

29. Let 𝑓 (𝑥, 𝑦) = 𝑥 + 𝑘𝑦 − 5𝑥𝑦. Determine the values of 𝑘 (if any) for which the critical point at (0, 0) is:

𝑇

2 1

−3

(a) A saddle point (b) A local maximum (c) A local minimum

−2

−1

𝑅

−1

−2

−3

𝑄 2 3

30. Find 𝐴 and 𝐵 so that 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝐴𝑥 + 𝑦2 + 𝐵 has a local minimum value of 20 at (1, 0).

0

0 1 2 3

𝑃

−1 −2 −3

31. For 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑥𝑦 + 𝑦2 + 𝑎𝑥 + 𝑏𝑦 + 𝑐, find values of 𝑎, 𝑏, and 𝑐 giving a local minimum at (2, 5) and so that 𝑓 (2, 5) = 11. 2

Figure 15.18

2

32. (a) Find critical points for 𝑓 (𝑥, 𝑦) = 𝑒−(𝑥−𝑎) −(𝑦−𝑏) . (b) Find 𝑎 and 𝑏 such that the critical point is at (−1, 5). (c) For the values of 𝑎 and 𝑏 in part (b), is (−1, 5) a local maximum, local minimum, or a saddle point? 33. Let 𝑓 (𝑥, 𝑦) = 𝑘𝑥2 + 𝑦2 − 4𝑥𝑦. Determine the values of 𝑘 (if any) for which the critical point at (0, 0) is:

34. Decide whether you think each point is a local maximum, local minimum, saddle point, or none of these. (a)

(a) A saddle point (b) A local maximum (c) A local minimum For Problems 34–36, use the contours of 𝑓 in Figure 15.18.

𝑃

(b) 𝑄

(c) 𝑅

(d) 𝑆

35. Sketch the direction of ∇𝑓 at points surrounding each of 𝑃 , 𝑅, 𝑆, and 𝑇 . 36. At which of 𝑃 , 𝑄, 𝑅, 𝑆, or 𝑇 does ‖∇𝑓 ‖ seem largest?.

814

Chapter 15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA

For Problems 37–40, find critical points and classify them as local maxima, local minima, saddle points, or none of these. 37. 𝑓 (𝑥, 𝑦) = 𝑥3 + 𝑒−𝑦

2

38. 𝑓 (𝑥, 𝑦) = sin 𝑥 sin 𝑦 39. 𝑓 (𝑥, 𝑦) = 1 − cos 𝑥 + 𝑦2 ∕2 40. 𝑓 (𝑥, 𝑦) = 𝑒𝑥 (1 − cos 𝑦)

50. On a computer, draw contour diagrams for functions 𝑓 (𝑥, 𝑦) = 𝑘(𝑥2 + 𝑦2 ) − 2𝑥𝑦 for 𝑘 = −2, −1, 0, 1, 2. Use these figures to classify the critical point at (0, 0) for each value of 𝑘. Explain your observations using the discriminant, 𝐷. 51. The behavior of a function can be complicated near a critical point where 𝐷 = 0. Suppose that

41. At the point (1, 3), suppose that 𝑓𝑥 = 𝑓𝑦 = 0 and 𝑓𝑥𝑥 > 0, 𝑓𝑦𝑦 > 0, 𝑓𝑥𝑦 = 0. (a) What can you conclude about the behavior of the function near the point (1, 3)? (b) Sketch a possible contour diagram. 42. At the point (𝑎, 𝑏), suppose that 𝑓𝑥 = 𝑓𝑦 = 0, 𝑓𝑥𝑥 > 0, 𝑓𝑦𝑦 = 0, 𝑓𝑥𝑦 > 0. (a) What can you conclude about the shape of the graph of 𝑓 near the point (𝑎, 𝑏)? (b) Sketch a possible contour diagram. 43. Let ℎ(𝑥, 𝑦) = 𝑓 (𝑥)𝑔(𝑦) where 𝑓 (0) = 𝑔(0) = 0 and 𝑓 ′ (0) ≠ 0, 𝑔 ′ (0) ≠ 0. Show that (0, 0) is a saddle point of ℎ. 44. Let ℎ(𝑥, 𝑦) = 𝑓 (𝑥) + 𝑔(𝑦). Show that ℎ has a critical point at (𝑎, 𝑏) if 𝑓 ′ (𝑎) = 𝑔 ′ (𝑏) = 0, and, assuming 𝑓 ′′ (𝑎) ≠ 0 and 𝑔 ′′ (𝑏) ≠ 0, it is a local maximum or minimum when 𝑓 ′′ (𝑎) and 𝑔 ′′ (𝑏) have the same sign and a saddle point when they have opposite signs. 45. Let ℎ(𝑥, 𝑦) = (𝑓 (𝑥))2 + (𝑔(𝑦))2 . Show that if 𝑓 (𝑎) = 𝑔(𝑏) = 0, then (𝑎, 𝑏) is a local minimum.

𝑓 (𝑥, 𝑦) = 𝑥3 − 3𝑥𝑦2 . Show that there is one critical point at (0, 0) and that 𝐷 = 0 there. Show that the contour for 𝑓 (𝑥, 𝑦) = 0 consists of three lines intersecting at the origin and that these lines divide the plane into six regions around the origin where 𝑓 alternates from positive to negative. Sketch a contour diagram for 𝑓 near (0, 0). The graph of this function is called a monkey saddle. 52. The contour diagrams for four functions 𝑧 = 𝑓 (𝑥, 𝑦) are in (a)–(d). Each function has a critical point with 𝑧 = 0 at the origin. Graphs (I)–(IV) show the value of 𝑧 for these four functions on a small circle around the origin, expressed as function of 𝜃, the angle between the positive 𝑥-axis and a line through the origin. Match the contour diagrams (a)–(d) with the graphs (I)–(IV). Classify the critical points as local maxima, local minima or saddle points. 𝑦

(a)

46. Draw a possible contour diagram of 𝑓 such that 𝑓𝑥 (−1, 0) = 0, 𝑓𝑦 (−1, 0) < 0, 𝑓𝑥 (3, 3) > 0, 𝑓𝑦 (3, 3) > 0, and 𝑓 has a local maximum at (3, −3). 47. Draw a possible contour diagram of a function with a saddle point at (2, 1), a local minimum at (2, 4), and no other critical points. Label the contours.

𝑦

(b)

𝑥

𝑦

(c)

𝑥

𝑦

(d)

48. For constants 𝑎 and 𝑏 with 𝑎𝑏 ≠ 0 and 𝑎𝑏 ≠ 1, let 𝑥

𝑓 (𝑥, 𝑦) = 𝑎𝑥2 + 𝑏𝑦2 − 2𝑥𝑦 − 4𝑥 − 6𝑦. (a) Find the 𝑥- and 𝑦-coordinates of the critical point. Your answer will be in terms of 𝑎 and 𝑏. (b) If 𝑎 = 𝑏 = 2, is the critical point a local maximum, a local minimum, or neither? Give a reason for your answer. (c) Classify the critical point for all values of 𝑎 and 𝑏 with 𝑎𝑏 ≠ 0 and 𝑎𝑏 ≠ 1. 49. (a) Find the critical point of 𝑓 (𝑥, 𝑦) = (𝑥2 − 𝑦)(𝑥2 + 𝑦). (b) Show that at the critical point, the discriminant 𝐷 = 0, so the second-derivative test gives no information about the nature of the critical point. (c) Sketch contours near the critical point to determine whether it is a local maximum, a local minimum, a saddle point, or none of these.

(I)

𝑧

𝑥

(II)

𝑧

𝜃 𝜋 (III)

𝜃 𝜋

2𝜋

𝑧

(IV)

2𝜋

𝑧

𝜃 𝜋

2𝜋

𝜃 𝜋

2𝜋

15.2 OPTIMIZATION

815

Strengthen Your Understanding In Problems 53–55, explain what is wrong with the statement.

0.7 0.5

53. If 𝑓𝑥 = 𝑓𝑦 = 0 at (1, 3), then 𝑓 has a local maximum or local minimum at (1, 3).

0.05

0.2

54. For 𝑓 (𝑥, 𝑦), if 𝐷 = 𝑓𝑥𝑥 𝑓𝑦𝑦 − (𝑓𝑥𝑦 )2 = 0 at (𝑎, 𝑏), then (𝑎, 𝑏) is a saddle point.

0

0

0.5

0.5 0.2 0.05

0.05 0.2

0.7

55. A critical point (𝑎, 𝑏) for the function 𝑓 must be a local minimum if both cross-sections for 𝑥 = 𝑎 and 𝑦 = 𝑏 are concave up.

0.7

𝑃 0

0 0.05 0.2

In Problems 56–57, give an example of:

0.5 0.7

56. A nonlinear function having no critical points 57. A function 𝑓 (𝑥, 𝑦) with a local maximum at (2, −3, 4).

Figure 15.19 √ 63. The function 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 has a local minimum at the origin. 64. The function 𝑓 (𝑥, 𝑦) = 𝑥2 − 𝑦2 has a local minimum at the origin. 65. If 𝑓 has a local minimum at 𝑃0 then so does the function 𝑔(𝑥, 𝑦) = 𝑓 (𝑥, 𝑦) + 5. 66. If 𝑓 has a local minimum at 𝑃0 then the function 𝑔(𝑥, 𝑦) = −𝑓 (𝑥, 𝑦) has a local maximum at 𝑃0 . 67. Every function has at least one local maximum. 68. If 𝑃0 is a local maximum of 𝑓 , then 𝑓 (𝑎, 𝑏) ≤ 𝑓 (𝑃0 ) for all points (𝑎, 𝑏) in 2-space. 69. If 𝑃0 is a local maximum of 𝑓 , then 𝑃0 is also a global maximum of 𝑓 .

Are the statements in Problems 58–69 true or false? Give reasons for your answer. 58. If 𝑓𝑥 (𝑃0 ) = 𝑓𝑦 (𝑃0 ) = 0, then 𝑃0 is a critical point of 𝑓 . 59. If 𝑓𝑥 (𝑃0 ) = 𝑓𝑦 (𝑃0 ) = 0, then 𝑃0 is a local maximum or local minimum of 𝑓 . 60. If 𝑃0 is a critical point of 𝑓 , then 𝑃0 is either a local maximum or local minimum of 𝑓 . 61. If 𝑃0 is a local maximum or local minimum of 𝑓 , and not on the boundary of the domain of 𝑓 , then 𝑃0 is a critical point of 𝑓 . 62. The function whose contour diagram is shown in Figure 15.19 has a saddle point at 𝑃 .

OPTIMIZATION Suppose we want to find the highest and the lowest points in Colorado. A contour map is shown in Figure 15.20. The highest point is the top of a mountain peak (point A on the map, Mt. Elbert, 14,440 feet high). What about the lowest point? Colorado does not have large pits without drainage, like Death Valley in California. A drop of rain falling at any point in Colorado will eventually flow out of the state. If there is no local minimum inside the state, where is the lowest point? It must be on the state boundary at a point where a river is flowing out of the state (point B where the Arikaree River leaves the state, 3,315 feet high). The highest point in Colorado is a global maximum for the elevation function in Colorado and the lowest point is the global minimum. R. tte Pla N.

Longs Peak Denver

e kar Ari

R.

Grays Peak Mt. Elbert

Grand Junction

A

Arka nsas R. Pueblo

Mt. Wilson

Alamosa

. eR

B

Lamar

Pu rga toi re R.

C

. oR rad olo

R. tte Pla th u So

Sou th P latte

15.2

Figure 15.20: The highest and lowest points in the state of Colorado

816

Chapter 15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA

In general, if we are given a function 𝑓 defined on a region 𝑅, we say: • 𝑓 has a global maximum on 𝑅 at the point 𝑃0 if 𝑓 (𝑃0 ) ≥ 𝑓 (𝑃 ) for all points 𝑃 in 𝑅. • 𝑓 has a global minimum on 𝑅 at the point 𝑃0 if 𝑓 (𝑃0 ) ≤ 𝑓 (𝑃 ) for all points 𝑃 in 𝑅. The process of finding a global maximum or minimum for a function 𝑓 on a region 𝑅 is called optimization. If the region 𝑅 is not stated explicitly, we take it to be the whole 𝑥𝑦-plane unless the context of the problem suggests otherwise.

How Do We Find Global Maxima and Minima? As the Colorado example illustrates, a global extremum can occur either at a critical point inside the region or at a point on the boundary of the region. This is analogous to single-variable calculus, where a function achieves its global extrema on an interval either at a critical point inside the interval or at an endpoint of the interval. To locate global maxima and minima for a function 𝑓 on a region 𝑅: • Find the critical points of 𝑓 in the region 𝑅. • Investigate whether the critical points give global maxima or minima. • If the region 𝑅 has a boundary, investigate whether 𝑓 attains a global maximum or minimum on the boundary of 𝑅. Investigating the boundary of a region for possible maxima and minima is the topic of Section 15.1. In this section, we focus on finding global maxima and minima of functions on regions that do not include boundaries. Not all functions have a global maximum or minimum: it depends on the function and the region. First, we consider applications in which global extrema are expected from practical considerations. At the end of this section, we examine the conditions that lead to global extrema. In general, the fact that a function has a single local maximum or minimum does not guarantee that the point is the global maximum or minimum. (See Problem 38.) An exception is if the function is quadratic, in which case the local maximum or minimum is the global maximum or minimum. (See Example 1 on page 807 and Example 5 on page 810.)

Maximizing Profit and Minimizing Cost In planning production of an item, a company often chooses the combination of price and quantity that maximizes its profit. We use Profit = Revenue − Cost, and, provided the price is constant, Revenue = Price ⋅ Quantity = 𝑝𝑞. In addition, we need to know how the cost and price depend on quantity. Example 1

A company manufactures two items which are sold in two separate markets where it has a monopoly. The quantities, 𝑞1 and 𝑞2 , demanded by consumers, and the prices, 𝑝1 and 𝑝2 (in dollars), of each item are related by 𝑝1 = 600 − 0.3𝑞1 and 𝑝2 = 500 − 0.2𝑞2 . Thus, if the price for either item increases, the demand for it decreases. The company’s total production cost is given by 𝐶 = 16 + 1.2𝑞1 + 1.5𝑞2 + 0.2𝑞1 𝑞2 . To maximize its total profit, how much of each product should be produced? What is the maximum profit? 2 2 Adapted

from M. Rosser and P. Lis, Basic Mathematics for Economists, 3rd ed. (New York: Routledge, 2016), p. 351.

15.2 OPTIMIZATION

Solution

817

The total revenue, 𝑅, is the sum of the revenues, 𝑝1 𝑞1 and 𝑝2 𝑞2 , from each market. Substituting for 𝑝1 and 𝑝2 , we get 𝑅 = 𝑝1 𝑞1 + 𝑝2 𝑞2 = (600 − 0.3𝑞1 )𝑞1 + (500 − 0.2𝑞2 )𝑞2 = 600𝑞1 − 0.3𝑞12 + 500𝑞2 − 0.2𝑞22 . Thus, the total profit 𝑃 is given by 𝑃 = 𝑅 − 𝐶 = 600𝑞1 − 0.3𝑞12 + 500𝑞2 − 0.2𝑞22 − (16 + 1.2𝑞1 + 1.5𝑞2 + 0.2𝑞1 𝑞2 ) = −16 + 598.8𝑞1 − 0.3𝑞12 + 498.5𝑞2 − 0.2𝑞22 − 0.2𝑞1 𝑞2 . Since 𝑞1 and 𝑞2 cannot be negative,3 the region we consider is the first quadrant with boundary 𝑞1 = 0 and 𝑞2 = 0. To maximize 𝑃 , we look for critical points by setting the partial derivatives equal to 0: 𝜕𝑃 = 598.8 − 0.6𝑞1 − 0.2𝑞2 = 0, 𝜕𝑞1 𝜕𝑃 = 498.5 − 0.4𝑞2 − 0.2𝑞1 = 0. 𝜕𝑞2 Since grad 𝑃 is defined everywhere, the only critical points of 𝑃 are those where grad 𝑃 = 0⃗ . Thus, solving for 𝑞1 , and 𝑞2 , we find that 𝑞1 = 699.1 and 𝑞2 = 896.7. The corresponding prices are 𝑝1 = 390.27 and 𝑝2 = 320.66. To see whether or not we have found a local maximum, we compute second partial derivatives: 𝜕2 𝑃 = −0.6, 𝜕𝑞12

𝜕2 𝑃 = −0.4, 𝜕𝑞22

𝜕2 𝑃 = −0.2, 𝜕𝑞1 𝜕𝑞2

so, 𝜕2 𝑃 𝜕2 𝑃 − 𝐷= 𝜕𝑞12 𝜕𝑞22

(

𝜕2 𝑃 𝜕𝑞1 𝜕𝑞2

)2

= (−0.6)(−0.4) − (−0.2)2 = 0.2.

Therefore we have found a local maximum. The graph of 𝑃 is a paraboloid opening downward, so (699.1, 896.7) is a global maximum. This point is within the region, so points on the boundary give smaller values of 𝑃 . The company should produce 699.1 units of the first item priced at $390.27 per unit, and 896.7 units of the second item priced at $320.66 per unit. The maximum profit 𝑃 (699.1, 896.7) ≈ $433,000.

Example 2

A delivery of 480 cubic meters of gravel is to be made to a landfill. The trucker plans to purchase an open-top box in which to transport the gravel in numerous trips. The total cost to the trucker is the cost of the box plus $80 per trip. The box must have height 2 meters, but the trucker can choose the length and width. The cost of the box is $100/m2 for the ends, $50/m2 for the sides and $200/m2 for the bottom. Notice the tradeoff: A smaller box is cheaper to buy but requires more trips. What size box should the trucker buy to minimize the total cost?4 3 Restricting 4 Adapted

prices to be nonnegative further restricts the region but does not alter the solution. from Claude McMillan, Jr., Mathematical Programming, 2nd ed. (New York: Wiley, 1978), pp. 156–157.

818

Chapter 15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA

Solution

We first get an algebraic expression for the trucker’s cost. Let the length of the box be 𝑥 meters and the width be 𝑦 meters; the height is 2 meters. (See Figure 15.21.) Table 15.2 Expense





𝑦

Trucker’s itemized cost Cost in dollars

Travel: 480∕(2𝑥𝑦) at $80/trip

2m

2

𝑥

2

80 ⋅ 480∕(2𝑥𝑦)

Ends: 2 at $100/m ⋅ 2𝑦 m

400𝑦

Sides: 2 at $50/m2 ⋅ 2𝑥 m2

200𝑥

Bottom: l at $200/m2 ⋅ 𝑥𝑦 m2

200𝑥𝑦

Figure 15.21: The box for transporting gravel

The volume of the box is 2𝑥𝑦 m3 , so delivery of 480 m3 of gravel requires 480∕(2𝑥𝑦) trips. The number of trips is a whole number; however, we treat it as continuous so that we can optimize using derivatives. The trucker’s cost is itemized in Table 15.2. The problem is to minimize ( ) 96 480 + 400𝑦 + 200𝑥 + 200𝑥𝑦 = 200 + 2𝑦 + 𝑥 + 𝑥𝑦 . Total cost = 80 ⋅ 2𝑥𝑦 𝑥𝑦 The length and width of the box must be positive. Thus, the region is the first quadrant but it does not contain the boundary, 𝑥 = 0 and 𝑦 = 0. Our problem is to minimize 𝑓 (𝑥, 𝑦) =

96 + 2𝑦 + 𝑥 + 𝑥𝑦. 𝑥𝑦

The critical points of this function occur where 96 +1+𝑦=0 𝑥2 𝑦 96 + 2 + 𝑥 = 0. 𝑓𝑦 (𝑥, 𝑦) = − 𝑥𝑦2

𝑓𝑥 (𝑥, 𝑦) = −

We put the 96∕(𝑥2 𝑦) and 96∕(𝑥𝑦2 ) terms on the other side of the the equation, divide, and simplify: 96∕(𝑥2 𝑦) 1+𝑦 = 96∕(𝑥𝑦2 ) 2 + 𝑥

so

𝑦 1+𝑦 = 𝑥 2+𝑥

giving

2𝑦 = 𝑥.

Substituting 𝑥 = 2𝑦 in the equation 𝑓𝑦 (𝑥, 𝑦) = 0 gives −

96 + 2 + 2𝑦 = 0 2𝑦 ⋅ 𝑦2 𝑦4 + 𝑦3 − 24 = 0.

The only positive solution to this equation is 𝑦 = 2, so the only critical point in the region is (4, 2). To check that the critical point is a local minimum, we use the second-derivative test. Since ( )2 192 192 25 96 >0 𝐷(4, 2) = 𝑓𝑥𝑥 𝑓𝑦𝑦 − (𝑓𝑥𝑦 )2 = 3 ⋅ =9− − + 1 4 4 ⋅ 2 4 ⋅ 23 42 ⋅ 22 and 𝑓𝑥𝑥 (4, 2) > 0, the point (4, 2) is a local minimum. Since the value of 𝑓 increases without bound as 𝑥 or 𝑦 increases without bound and as 𝑥 → 0+ and 𝑦 → 0+ , it can be shown that (4, 2) is a global minimum. (See Problem 42.) Thus, the optimal box is 4 meters long and 2 meters wide. With a box of this size, the trucker would need to make 30 trips to haul all of the gravel. This large number lends some credibility to our decision to treat the number of trips as a continuous variable.

15.2 OPTIMIZATION

819

Fitting a Line to Data: Least Squares Suppose we want to fit the “best” line to some data in the plane. We measure the distance from a line to the data points by adding the squares of the vertical distances from each point to the line. The smaller this sum of squares is, the better the line fits the data. The line with the minimum sum of square distances is called the least squares line, or the regression line. If the data is nearly linear, the least squares line is a good fit; otherwise it may not be. (See Figure 15.22.)

Data almost linear: line fits well

Data not very linear: line does not fit well

Figure 15.22: Fitting lines to data points

Example 3

Find a least squares line for the following data points: (1, 1), (2, 1), and (3, 3).

Solution

Suppose the line has equation 𝑦 = 𝑏 + 𝑚𝑥. If we find 𝑏 and 𝑚 then we have found the line. So, for this problem, 𝑏 and 𝑚 are the two variables. Any values of 𝑚 and 𝑏 are possible, so this is an unconstrained problem. We want to minimize the function 𝑓 (𝑏, 𝑚) that gives the sum of the three squared vertical distances from the points to the line in Figure 15.23. 𝑦 (3, 3)

3

❄ ✻

𝑦 = 𝑏 + 𝑚𝑥

(3, 𝑏 + 3𝑚) 2 (2, 𝑏 + 2𝑚)

1

(1, 1)

❄ ✻

✻ ❄ (2, 1)

(1, 𝑏 + 𝑚)

0

𝑥 1

2

3

Figure 15.23: The least squares line minimizes the sum of the squares of these vertical distances

The vertical distance from the point (1, 1) to the line is the difference in the 𝑦-coordinates 1 − (𝑏 + 𝑚); similarly for the other points. Thus, the sum of squares is 𝑓 (𝑏, 𝑚) = (1 − (𝑏 + 𝑚))2 + (1 − (𝑏 + 2𝑚))2 + (3 − (𝑏 + 3𝑚))2 . To minimize 𝑓 we look for critical points. First we differentiate 𝑓 with respect to 𝑏: 𝜕𝑓 = −2(1 − (𝑏 + 𝑚)) − 2(1 − (𝑏 + 2𝑚)) − 2(3 − (𝑏 + 3𝑚)) 𝜕𝑏 = −2 + 2𝑏 + 2𝑚 − 2 + 2𝑏 + 4𝑚 − 6 + 2𝑏 + 6𝑚 = −10 + 6𝑏 + 12𝑚.

820

Chapter 15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA

Now we differentiate with respect to 𝑚: 𝜕𝑓 = 2(1 − (𝑏 + 𝑚))(−1) + 2(1 − (𝑏 + 2𝑚))(−2) + 2(3 − (𝑏 + 3𝑚))(−3) 𝜕𝑚 = −2 + 2𝑏 + 2𝑚 − 4 + 4𝑏 + 8𝑚 − 18 + 6𝑏 + 18𝑚 = −24 + 12𝑏 + 28𝑚. The equations

𝜕𝑓 𝜕𝑓 = 0 and = 0 give a system of two linear equations in two unknowns: 𝜕𝑏 𝜕𝑚 −10 + 6𝑏 + 12𝑚 = 0, −24 + 12𝑏 + 28𝑚 = 0.

The solution to this pair of equations is the critical point 𝑏 = −1∕3 and 𝑚 = 1. Since 𝐷 = 𝑓𝑏𝑏 𝑓𝑚𝑚 − (𝑓𝑚𝑏 )2 = (6)(28) − 122 = 24 and 𝑓𝑏𝑏 = 6 > 0, we have found a local minimum. The graph of 𝑓 (𝑏, 𝑚) is a parabola opening upward, so the local minimum is the global minimum of 𝑓 . Thus, the least squares line is 1 𝑦=𝑥− . 3 As a check, notice that the line 𝑦 = 𝑥 passes through the points (1, 1) and (3, 3). It is reasonable that introducing the point (2, 1) moves the 𝑦-intercept down from 0 to −1∕3. The general formulas for the slope and 𝑦-intercept of a least squares line are in Project 2 (available online). Many calculators have built-in formulas for 𝑏 and 𝑚, as well as for the correlation coefficient, which measures how well the data points fit the least squares line.

How Do We Know Whether a Function Has a Global Maximum or Minimum? Under what circumstances does a function of two variables have a global maximum or minimum? The next example shows that a function may have both a global maximum and a global minimum on a region, or just one, or neither. Example 4

Investigate the global maxima and minima of the following functions: (a) ℎ(𝑥, 𝑦) = 1 + 𝑥2 + 𝑦2 on the disk 𝑥2 + 𝑦2 ≤ 1. (b) 𝑓 (𝑥, 𝑦) = 𝑥2 − 2𝑥 + 𝑦2 − 4𝑦 + 5 on the 𝑥𝑦-plane. (c) 𝑔(𝑥, 𝑦) = 𝑥2 − 𝑦2 on the 𝑥𝑦-plane.

Solution

(a) The graph of ℎ(𝑥, 𝑦) = 1 + 𝑥2 + 𝑦2 is a bowl-shaped paraboloid with a global minimum of 1 at (0, 0), and a global maximum of 2 on the edge of the region, 𝑥2 + 𝑦2 = 1. (b) The graph of 𝑓 in Figure 15.5 on page 807 shows that 𝑓 has a global minimum at the point (1, 2) and no global maximum (because the value of 𝑓 increases without bound as 𝑥 → ∞, 𝑦 → ∞). (c) The graph of 𝑔 in Figure 15.7 on page 808 shows that 𝑔 has no global maximum because 𝑔(𝑥, 𝑦) → ∞ as 𝑥 → ∞ if 𝑦 is constant. Similarly, 𝑔 has no global minimum because 𝑔(𝑥, 𝑦) → −∞ as 𝑦 → ∞ if 𝑥 is constant. Sometimes a function is guaranteed to have a global maximum and minimum. For example, a continuous function, ℎ(𝑥), of one variable has a global maximum and minimum on every closed interval 𝑎 ≤ 𝑥 ≤ 𝑏. On a non-closed interval, such as 𝑎 ≤ 𝑥 < 𝑏 or 𝑎 < 𝑥 < 𝑏, or on an unbounded interval, such as 𝑎 < 𝑥 < ∞, ℎ may not have a maximum or minimum value.

15.2 OPTIMIZATION

821

What is the situation for functions of two variables? As it turns out, a similar result is true for continuous functions defined on regions which are closed and bounded, analogous to the closed and bounded interval 𝑎 ≤ 𝑥 ≤ 𝑏. In everyday language we say • A closed region is one which contains its boundary; • A bounded region is one which does not stretch to infinity in any direction. More precise definitions follow. Suppose 𝑅 is a region in 2-space. A point (𝑥0 , 𝑦0 ) is a boundary point of 𝑅 if, for every 𝑟 > 0, the circular disk with center (𝑥0 , 𝑦0 ) and radius 𝑟 contains both points which are in 𝑅 and points which are not in 𝑅. See Figure 15.24. A point (𝑥0 , 𝑦0 ) can be a boundary point of the region 𝑅 without belonging to 𝑅. The collection of all the boundary points is the boundary of 𝑅. The region 𝑅 is closed if it contains its boundary. A region 𝑅 in 2-space is bounded if the distance between every point (𝑥, 𝑦) in 𝑅 and the origin is less than some constant 𝐾. Closed and bounded regions in 3-space are defined in the same way. Example 5

√ (a) Consider the square −1 ≤ 𝑥 ≤ 1, −1 ≤ 𝑦 ≤ 1. Every point in this region is within distance 2 of the origin, so the region is bounded. The region’s boundary consists of four line segments, all of which belong to the region, so the region is closed. (b) Consider the first quadrant 𝑥 ≥ 0, 𝑦 ≥ 0. The boundary of this region consists of the origin, the positive 𝑥-axis, and the positive 𝑦-axis. All of these belong to the region, so the region is closed. However, the region is not bounded, since there is no upper bound on distances between points in the region and the origin. (c) The disk 𝑥2 + 𝑦2 < 1 is bounded, because each point in the region is within distance 1 of the origin. However, the disk is not closed, because (1, 0) is a boundary point of the region but not included in the region. (d) The half-plane 𝑦 > 0 is neither closed nor bounded. The origin is a boundary point of this region but is not included in the region. The reason that closed and bounded regions are useful is the following theorem, which is also true for functions of three or more variables:5

Theorem 15.1: Extreme Value Theorem for Multivariable Functions If 𝑓 is a continuous function on a closed and bounded region 𝑅, then 𝑓 has a global maximum at some point (𝑥0 , 𝑦0 ) in 𝑅 and a global minimum at some point (𝑥1 , 𝑦1 ) in 𝑅. If 𝑓 is not continuous or the region 𝑅 is not closed and bounded, there is no guarantee that 𝑓 achieves a global maximum or global minimum on 𝑅. In Example 4, the function 𝑔 is continuous but does not achieve a global maximum or minimum in 2-space, a region which is closed but not bounded. Example 6 illustrates what can go wrong when the region is bounded but not closed.

(𝑥0 , 𝑦0 ) 𝑅

Figure 15.24: Boundary point (𝑥0 , 𝑦0 ) of 𝑅 5 For

a proof, see Walter Rudin, Principles of Mathematical Analysis, 3rd ed. (New York: McGraw-Hill, 1976), p. 89.

822

Chapter 15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA

Example 6

Does the function 𝑓 have a global maximum or minimum on the region 𝑅 given by 0 < 𝑥2 + 𝑦2 ≤ 1?

Solution

1 𝑥2 + 𝑦2 The region 𝑅 is bounded, but it is not closed since it does not contain the boundary point (0, 0). We see from the graph of 𝑧 = 𝑓 (𝑥, 𝑦) in Figure 15.25 that 𝑓 has a global minimum on the circle 𝑥2 + 𝑦2 = 1. However, 𝑓 (𝑥, 𝑦) → ∞ as (𝑥, 𝑦) → (0, 0), so 𝑓 has no global maximum. 𝑓 (𝑥, 𝑦) =

𝑧

𝑦

𝑥 Figure 15.25: Graph showing 𝑓 (𝑥, 𝑦) =

1 𝑥2 +𝑦2

has no global maximum on 0 < 𝑥2 + 𝑦2 ≤ 1

Exercises and Problems for Section 15.2 EXERCISES 1. By looking at the weather map in Figure 12.1 on page 652, find the maximum and minimum daily high temperatures in the states of Mississippi, Alabama, Pennsylvania, New York, California, Arizona, and Massachusetts. In Exercises 2–4, estimate the position and approximate value of the global maxima and minima on the closed region shown.

0

4

3 2

4

2

3

𝑥 4

5

6

7

27

1 29

𝜋∕2

28

In Problems 5–9, without calculus, find the highest and lowest points (if they exist) on the surface. The 𝑧-axis is upward. 5. 𝑥2 + 𝑦2 + (𝑧 − 1)2 = 49 6. (𝑥 + 1)2 + (𝑦 − 3)2 + 2𝑧2 = 162

2

3

4

0

0.50 0.75

𝑥 1

0

𝜋

26

6

1

3𝜋∕2

25

2

1

2𝜋

0.50 0.25

4

0.75

9 7

21 22 23 24

−0.25

2

3

5𝜋∕2

−0.75

4

𝑦

4.

5

5 3

𝑦

3.

6

−0.50

𝑦

2.

0. 25

𝑥

5

𝜋∕2

𝜋

In Exercises 11–13, find the global maximum and minimum of the function on −1 ≤ 𝑥 ≤ 1, −1 ≤ 𝑦 ≤ 1, and say whether it occurs on the boundary of the square. [Hint: Use graphs.] 11. 𝑧 = 𝑥2 + 𝑦2 12. 𝑧 = −𝑥2 −𝑦2 13. 𝑧 = 𝑥2 − 𝑦2

7. 𝑧 = (𝑥 − 5)2 + (𝑦 − 𝜋)2 + 2𝜋

In Problems 14–21, does the function have a global maximum? A global minimum?

8. 𝑧 = 44 − 2𝑥2 − 2𝑦2

14. 𝑓 (𝑥, 𝑦) = 𝑥2 − 2𝑦2

9. 𝑥 = 4 + 𝑦2 + 2𝑧2 10. The surface 𝑧 = 27 − 𝑥2 − 𝑦2 cuts the plane 𝑧 = 2 in a curve. Without calculus, find the point on this curve with the greatest 𝑦-coordinate.

3

3

16. ℎ(𝑥, 𝑦) = 𝑥 + 𝑦 𝑥2 +𝑦2

18. 𝑓 (𝑥, 𝑦) = 𝑒

15. 𝑔(𝑥, 𝑦) = 𝑥2 𝑦2 17. 𝑓 (𝑥, 𝑦) = −2𝑥2 − 7𝑦2 19. ℎ(𝑥, 𝑦) = 1 − 𝑦2 𝑒𝑥𝑦

20. 𝑓 (𝑥, 𝑦) = 𝑥2 ∕2 + 3𝑦3 + 9𝑦2 − 3𝑥 21. 𝑔(𝑥, 𝑦) = 𝑥2 − cos(𝑥 + 𝑦)

15.2 OPTIMIZATION

823

PROBLEMS 22. (a) Compute and classify the critical points of 𝑓 (𝑥, 𝑦) = 2𝑥2 − 3𝑥𝑦 + 8𝑦2 + 𝑥 − 𝑦. (b) By completing the square, plot the contour diagram of 𝑓 and show that the local extremum found in part (a) is a global one. In Problems 23–26, find and classify the critical points of the function. Then decide whether the function has global extrema on the 𝑥𝑦-plane, and find them if they exist. 23. 𝑓 (𝑥, 𝑦) = 𝑦2 + 2𝑥𝑦 − 𝑦 − 𝑥3 − 𝑥 + 2 24. 𝑓 (𝑥, 𝑦) = 2𝑥2 − 4𝑥𝑦 + 5𝑦2 + 9𝑦 + 2 25. 𝑓 (𝑥, 𝑦) = −𝑥4 + 2𝑥𝑦2 − 2𝑦3

(a) Find the maximum profit that can be made, assuming the prices are fixed. (b) Find the rate of change of that maximum profit as 𝑝1 increases. 35. A company operates two plants which manufacture the same item and whose total cost functions are 𝐶1 = 8.5 + 0.03𝑞12

and

𝐶2 = 5.2 + 0.04𝑞22 ,

where 𝑞1 and 𝑞2 are the quantities produced by each plant. The company is a monopoly. The total quantity demanded, 𝑞 = 𝑞1 + 𝑞2 , is related to the price, 𝑝, by

2 −𝑦2

26. 𝑓 (𝑥, 𝑦) = 𝑥𝑒−𝑥

27. A closed rectangular box has volume 32 cm3 . What are the lengths of the edges giving the minimum surface area? 28. A closed rectangular box with faces parallel to the coordinate planes has one bottom corner at the origin and the opposite top corner in the first octant on the plane 3𝑥 + 2𝑦 + 𝑧 = 1. What is the maximum volume of such a box? 29. An international airline has a regulation that each passenger can carry a suitcase having the sum of its width, length and height less than or equal to 135 cm. Find the dimensions of the suitcase of maximum volume that a passenger may carry under this regulation. 30. Design a rectangular milk carton box of width 𝑤, length 𝑙, and height ℎ which holds 512 cm3 of milk. The sides of the box cost 1 cent∕cm2 and the top and bottom cost 2 cent∕cm2 . Find the dimensions of the box that minimize the total cost of materials used. 31. Find the point on the plane 3𝑥+2𝑦+𝑧 = 1 that is closest to the origin by minimizing the square of the distance. 32. What is the shortest distance from the surface 𝑥𝑦+3𝑥+ 𝑧2 = 9 to the origin? 33. For constants 𝑎, 𝑏, and 𝑐, let 𝑓 (𝑥, 𝑦) = 𝑎𝑥 + 𝑏𝑦 + 𝑐 be a linear function, and let 𝑅 be a region in the 𝑥𝑦-plane. (a) If 𝑅 is any disk, show that the maximum and minimum values of 𝑓 on 𝑅 occur on the boundary of the disk. (b) If 𝑅 is any rectangle, show that the maximum and minimum values of 𝑓 on 𝑅 occur at the corners of the rectangle. They may occur at other points of the rectangle as well. (c) Use a graph of the plane 𝑧 = 𝑓 (𝑥, 𝑦) to explain your answers in parts (a) and (b). 34. Two products are manufactured in quantities 𝑞1 and 𝑞2 and sold at prices of 𝑝1 and 𝑝2 , respectively. The cost of producing them is given by 𝐶 = 2𝑞12 + 2𝑞22 + 10. 6 Adapted

𝑝 = 60 − 0.04𝑞. How much should each plant produce in order to maximize the company’s profit?6 36. The quantity of a product demanded by consumers is a function of its price. The quantity of one product demanded may also depend on the price of other products. For example, if the only chocolate shop in town (a monopoly) sells milk and dark chocolates, the price it sets for each affects the demand of the other. The quantities demanded, 𝑞1 and 𝑞2 , of two products depend on their prices, 𝑝1 and 𝑝2 , as follows: 𝑞1 = 150 − 2𝑝1 − 𝑝2 𝑞2 = 200 − 𝑝1 − 3𝑝2 . (a) What does the fact that the coefficients of 𝑝1 and 𝑝2 are negative tell you? Give an example of two products that might be related this way. (b) If one manufacturer sells both products, how should the prices be set to generate the maximum possible revenue? What is that maximum possible revenue? 37. A company manufactures a product which requires capital and labor to produce. The quantity, 𝑄, of the product manufactured is given by the Cobb-Douglas function 𝑄 = 𝐴𝐾 𝑎 𝐿𝑏 , where 𝐾 is the quantity of capital; 𝐿 is the quantity of labor used; and 𝐴, 𝑎, and 𝑏 are positive constants with 0 < 𝑎 < 1 and 0 < 𝑏 < 1. One unit of capital costs $𝑘 and one unit of labor costs $𝓁. The price of the product is fixed at $𝑝 per unit. (a) If 𝑎 + 𝑏 < 1, how much capital and labor should the company use to maximize its profit? (b) Is there a maximum profit in the case 𝑎 + 𝑏 = 1? What about 𝑎 + 𝑏 ≥ 1? Explain.

from M. Rosser and P. Lis, Basic Mathematics for Economists, 3rd ed. (New York: Routledge, 2016), p. 354.

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Chapter 15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA

38. Let 𝑓 (𝑥, 𝑦) = 𝑥2 (𝑦 + 1)3 + 𝑦2 . Show that 𝑓 has only one critical point, namely (0, 0), and that point is a local minimum but not a global minimum. Contrast this with the case of a function with a single local minimum in one-variable calculus. 39. Find the parabola of the form 𝑦 = 𝑎𝑥2 + 𝑏 which best fits the points (1, 0), (2, 2), (3, 4) by minimizing the sum of squares, 𝑆, given by 𝑆 = (𝑎 + 𝑏)2 + (4𝑎 + 𝑏 − 2)2 + (9𝑎 + 𝑏 − 4)2 . 40. For the data points (11, 16), (12, 17), (13, 17), and (16, 20), find an expression for 𝑓 (𝑏, 𝑚), the sum of squared errors that are minimized on the least squares line 𝑦 = 𝑏 + 𝑚𝑥. (You need not do the minimization.) 41. Find the least squares line for the data points (0, 4), (1, 3), (2, 1). 42. Let 𝑓 (𝑥, 𝑦) = 80∕(𝑥𝑦) + 20𝑦 + 10𝑥 + 10𝑥𝑦 in the region 𝑅 where 𝑥, 𝑦 > 0. (a) Explain why 𝑓 (𝑥, 𝑦) > 𝑓 (2, 1) at every point in 𝑅 where (i) 𝑥 > 20 (ii) 𝑦 > 20 (iii) 𝑥 < 0.01 and 𝑦 ≤ 20 (iv) 𝑦 < 0.01 and 𝑥 ≤ 20 (b) Explain why 𝑓 must have a global minimum at a critical point in 𝑅. (c) Explain why 𝑓 must have a global minimum in 𝑅 at the point (2, 1). 43. Let 𝑓 (𝑥, 𝑦) = 2∕𝑥+3∕𝑦+4𝑥+5𝑦 in the region 𝑅 where 𝑥, 𝑦 > 0. (a) Explain why 𝑓 must have a global minimum at some point in 𝑅. (b) Find the global minimum. 44. (a) The energy, 𝐸, required to compress a gas from a fixed initial pressure 𝑃0 to a fixed final pressure 𝑃𝐹 through an intermediate pressure 𝑝 is7 ( )2 ( )2 𝑃𝐹 𝑝 𝐸= + − 1. 𝑃0 𝑝

How should 𝑝 be chosen to minimize the energy? (b) Now suppose the compression takes place in two stages with two intermediate pressures, 𝑝1 and 𝑝2 . What choices of 𝑝1 and 𝑝2 minimize the energy if 𝐸=

(

𝑝1 𝑃0

)2

+

(

𝑝2 𝑝1

)2

+

(

𝑃𝐹 𝑝2

)2

− 2?

45. The Dorfman-Steiner rule shows how a company which has a monopoly should set the price, 𝑝, of its product and how much advertising, 𝑎, it should buy. The price of advertising is 𝑝𝑎 per unit. The quantity, 𝑞, of the product sold is given by 𝑞 = 𝐾𝑝−𝐸 𝑎𝜃 , where 𝐾 > 0, 𝐸 > 1, and 0 < 𝜃 < 1 are constants. The cost to the company to make each item is 𝑐. (a) How does the quantity sold, 𝑞, change if the price, 𝑝, increases? If the quantity of advertising, 𝑎, increases? (b) Show that the partial derivatives can be written in the form 𝜕𝑞∕𝜕𝑝 = −𝐸𝑞∕𝑝 and 𝜕𝑞∕𝜕𝑎 = 𝜃𝑞∕𝑎. (c) Explain why profit, 𝜋, is given by 𝜋 = 𝑝𝑞−𝑐𝑞−𝑝𝑎𝑎. (d) If the company wants to maximize profit, what must be true of the partial derivatives, 𝜕𝜋∕𝜕𝑝 and 𝜕𝜋∕𝜕𝑎? (e) Find 𝜕𝜋∕𝜕𝑝 and 𝜕𝜋∕𝜕𝑎. (f) Use your answers to parts (d) and (e) to show that at maximum profit, 𝑝−𝑐 1 = 𝑝 𝐸

and

𝑝−𝑐 𝑎 . = 𝑝𝑎 𝜃𝑞

(g) By dividing your answers in part (f), show that at maximum profit, 𝑝𝑎 𝑎 𝜃 = . 𝑝𝑞 𝐸 This is the Dorfman-Steiner rule, that the ratio of the advertising budget to revenue does not depend on the price of advertising.

Strengthen Your Understanding In Problems 46–48, explain what is wrong with the statement.

49. A continuous function 𝑓 (𝑥, 𝑦) that has no global maximum and no global minimum on the 𝑥𝑦-plane.

46. A function having no critical points in a region 𝑅 cannot have a global maximum in the region.

50. A function 𝑓 (𝑥, 𝑦) and a region 𝑅 such that the maximum value of 𝑓 on 𝑅 is on the boundary of 𝑅.

47. No continuous function has a global minimum on an unbounded region 𝑅.

Are the statements in Problems 51–59 true or false? Give reasons for your answer.

48. If 𝑓 (𝑥, 𝑦) has a local maximum value of 1 at the origin, then the global maximum is 1. In Problems 49–50, give an example of: 7 Adapted

51. If 𝑃0 is a global maximum of 𝑓 , where 𝑓 is defined on all of 2-space, then 𝑃0 is also a local maximum of 𝑓 . 52. Every function has a global maximum.

from Aris Rutherford, Discrete Dynamic Programming, p. 35 (New York: Blaisdell, 1964).

15.3 CONSTRAINED OPTIMIZATION: LAGRANGE MULTIPLIERS

53. The region consisting of all points (𝑥, 𝑦) satisfying 𝑥2 + 𝑦2 < 1 is bounded. 54. The region consisting of all points (𝑥, 𝑦) satisfying 𝑥2 + 𝑦2 < 1 is closed. 55. The function 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 has a global minimum on the region 𝑥2 + 𝑦2 < 1. 56. The function 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 has a global maximum on the region 𝑥2 + 𝑦2 < 1.

15.3

825

57. If 𝑃 and 𝑄 are two distinct points in 2-space, and 𝑓 has a global maximum at 𝑃 , then 𝑓 cannot have a global maximum at 𝑄. 58. The function 𝑓 (𝑥, 𝑦) = sin(1 + 𝑒𝑥𝑦 ) must have a global minimum in the square region 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1. 59. If 𝑃0 is a global minimum of 𝑓 on a closed and bounded region, then 𝑃0 need not be a critical point of 𝑓 .

CONSTRAINED OPTIMIZATION: LAGRANGE MULTIPLIERS Many, perhaps most, real optimization problems are constrained by external circumstances. For example, a city wanting to build a public transportation system that will serve the greatest possible number of people has only a limited number of tax dollars it can spend on the project. In this section, we see how to find an optimum value under such constraints. In Section 15.2, we saw how to optimize a function 𝑓 (𝑥, 𝑦) on a region 𝑅. If the region 𝑅 is the entire 𝑥𝑦-plane, we have unconstrained optimization; if the region 𝑅 is not the entire 𝑥𝑦-plane, that is, if 𝑥 or 𝑦 is restricted in some way, then we have constrained optimization.

Graphical Approach: Maximizing Production Subject to a Budget Constraint Suppose we want to maximize production under a budget constraint. Suppose production, 𝑓 , is a function of two variables, 𝑥 and 𝑦, which are quantities of two raw materials, and that 𝑓 (𝑥, 𝑦) = 𝑥2∕3 𝑦1∕3 . If 𝑥 and 𝑦 are purchased at prices of 𝑝1 and 𝑝2 thousands of dollars per unit, what is the maximum production 𝑓 that can be obtained with a budget of 𝑐 thousand dollars? To maximize 𝑓 without regard to the budget, we simply increase 𝑥 and 𝑦. However, the budget constraint prevents us from increasing 𝑥 and 𝑦 beyond a certain point. Exactly how does the budget constrain us? With prices of 𝑝1 and 𝑝2 , the amount spent on 𝑥 is 𝑝1 𝑥 and the amount spent on 𝑦 is 𝑝2 𝑦, so we must have 𝑔(𝑥, 𝑦) = 𝑝1 𝑥 + 𝑝2 𝑦 ≤ 𝑐, where 𝑔(𝑥, 𝑦) is the total cost of the raw materials and 𝑐 is the budget in thousands of dollars. Let’s look at the case when 𝑝1 = 𝑝2 = 1 and 𝑐 = 3.78. Then 𝑥 + 𝑦 ≤ 3.78.

Figure 15.26 shows some contours of 𝑓 and the budget constraint represented by the line 𝑥+𝑦 = 3.78. Any point on or below the line represents a pair of values of 𝑥 and 𝑦 that we can afford. A point on the line completely exhausts the budget, while a point below the line represents values of 𝑥 and 𝑦 which can be bought without using up the budget. Any point above the line represents a pair of values that we cannot afford. To maximize 𝑓 , we find the point which lies on the level curve with the largest possible value of 𝑓 and which lies within the budget. The point must lie on the budget constraint because production is maximized when we spend all the available money. Unless we are at a point where the budget constraint is tangent to a contour of 𝑓 , we can increase 𝑓 by moving in some direction along the line representing the budget constraint in Figure 15.26. For example, if we are on the line to the left of the point of tangency, moving right on the constraint will increase 𝑓 ; if we are on the line to the right of the point of tangency, moving left will increase 𝑓 . Thus, the maximum value of 𝑓 on the budget constraint occurs at the point where the budget constraint is tangent to the contour 𝑓 = 2.

826

Chapter 15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA

𝑦 Level curves of production

✛ ✛

𝑓 =1 𝑓 =2 ✛𝑓 =3

Maximum production



Budget constraint

𝑥 + 𝑦 = 3.78

✠ 𝑃

𝑥 Figure 15.26: Optimal point, 𝑃 , where budget constraint is tangent to a level of production function

Analytical Solution: Lagrange Multipliers Figure 15.26 suggests that maximum production is achieved at the point where the budget constraint is tangent to a level curve of the production function. The method of Lagrange multipliers uses this fact in algebraic form. Figure 15.27 shows that at the optimum point, 𝑃 , the gradient of 𝑓 and the normal to the budget line 𝑔(𝑥, 𝑦) = 𝑥 + 𝑦 = 3.78 are parallel. Thus, at 𝑃 , grad 𝑓 and grad 𝑔 are parallel, so for some scalar 𝜆, called the Lagrange multiplier, grad 𝑓 = 𝜆 grad 𝑔. Calculating the gradients, we find that ( ) ( ) ( ) 2 −1∕3 1∕3 ⃗ 1 2∕3 −2∕3 ⃗ 𝑥 𝑥 𝑦 𝑖 + 𝑗 = 𝜆 ⃗𝑖 + 𝑗⃗ . 𝑦 3 3 Equating components gives 2 −1∕3 1∕3 𝑥 𝑦 =𝜆 3

and

1 2∕3 −2∕3 𝑥 𝑦 = 𝜆. 3

Eliminating 𝜆 gives 2 −1∕3 1∕3 1 2∕3 −2∕3 𝑥 𝑦 = 𝑥 𝑦 , which leads to 2𝑦 = 𝑥. 3 3 Since the constraint 𝑥 + 𝑦 = 3.78 must be satisfied, we have 𝑥 = 2.52 and 𝑦 = 1.26. Then 𝑓 (2.52, 1.26) = (2.52)2∕3(1.26)1∕3 ≈ 2. As before, we see that the maximum value of 𝑓 is approximately 2. Thus, to maximize production on a budget of $3780, we should use 2.52 units of one raw material and 1.26 units of the other. 𝑦 Level curves of production

✛ ✛

Budget constraint

𝑔(𝑥, 𝑦) = 3.78

𝑓 =1 𝑓 =2 ✛𝑓 =3

grad 𝑓 = 𝜆 grad 𝑔

✲ 𝑃





𝑥 Figure 15.27: At the point, 𝑃 , of maximum production, the vectors grad 𝑓 and grad 𝑔 are parallel

15.3 CONSTRAINED OPTIMIZATION: LAGRANGE MULTIPLIERS

827

Lagrange Multipliers in General Suppose we want to optimize an objective function 𝑓 (𝑥, 𝑦) subject to a constraint 𝑔(𝑥, 𝑦) = 𝑐. We look for extrema among the points which satisfy the constraint. We make the following definition.

Suppose 𝑃0 is a point satisfying the constraint 𝑔(𝑥, 𝑦) = 𝑐. • 𝑓 has a local maximum at 𝑃0 subject to the constraint if 𝑓 (𝑃0 ) ≥ 𝑓 (𝑃 ) for all points 𝑃 near 𝑃0 satisfying the constraint. • 𝑓 has a global maximum at 𝑃0 subject to the constraint if 𝑓 (𝑃0 ) ≥ 𝑓 (𝑃 ) for all points 𝑃 satisfying the constraint. Local and global minima are defined similarly.

As we saw in the production example, constrained extrema occur at points of tangency of contours of 𝑓 and 𝑔; they can also occur at endpoints of constraints. At a point of tangency, grad 𝑓 is perpendicular to the constraint and so parallel to grad 𝑔. At interior points on the constraint where grad 𝑓 is not perpendicular to the constraint, the value of 𝑓 can be increased or decreased by moving along the constraint. Therefore constrained extrema occur only at points where grad 𝑓 and grad 𝑔 are parallel or at endpoints of the constraint. (See Figure 15.28.) At points where the gradients are parallel, provided grad 𝑔 ≠ 0⃗ , there is a constant 𝜆 such that grad 𝑓 = 𝜆 grad 𝑔.

Optimizing 𝒇 Subject to the Constraint 𝒈 = 𝒄: If a smooth function, 𝑓 , has a maximum or minimum subject to a smooth constraint 𝑔 = 𝑐 at a point 𝑃0 , then either 𝑃0 satisfies the equations grad 𝑓 = 𝜆 grad 𝑔

and 𝑔 = 𝑐,

or 𝑃0 is an endpoint of the constraint, or grad 𝑔(𝑃0 ) = 0⃗ . To investigate whether 𝑃0 is a global maximum or minimum, compare values of 𝑓 at the points satisfying these three conditions. The number 𝜆 is called the Lagrange multiplier.

If the set of points satisfying the constraint is closed and bounded, such as a circle or line segment, then there must be a global maximum and minimum of 𝑓 subject to the constraint. If the constraint is not closed and bounded, such as a line or hyperbola, then there may or may not be a global maximum and minimum.

grad 𝑓 ✗ grad 𝑔

𝑓 =4

𝑃0

𝑓 =3

grad 𝑓

𝑓 =2



𝑔=𝑐

✗ ✎ grad 𝑔

𝑓 =1

Figure 15.28: Maximum and minimum values of 𝑓 (𝑥, 𝑦) on 𝑔(𝑥, 𝑦) = 𝑐 are at points where grad 𝑓 is parallel to grad 𝑔

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Chapter 15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA

Example 1

Find the maximum and minimum values of 𝑥 + 𝑦 on the circle 𝑥2 + 𝑦2 = 4.

Solution

The objective function is 𝑓 (𝑥, 𝑦) = 𝑥 + 𝑦, and the constraint is 𝑔(𝑥, 𝑦) = 𝑥2 + 𝑦2 = 4. Since grad 𝑓 = 𝑓𝑥 ⃗𝑖 + 𝑓𝑦 𝑗⃗ = ⃗𝑖 + 𝑗⃗ and grad 𝑔 = 𝑔𝑥 ⃗𝑖 + 𝑔𝑦 𝑗⃗ = 2𝑥⃗𝑖 + 2𝑦𝑗⃗ , the condition grad 𝑓 = 𝜆 grad 𝑔 gives 1 = 2𝜆𝑥 and 1 = 2𝜆𝑦, so 𝑥 = 𝑦. We also know that 𝑥2 + 𝑦2 = 4, √ ⃗ giving 𝑥 = 𝑦 = 2 or 𝑥 = 𝑦 = − 2. The constraint √ √ has no endpoints √ √ (it’s a circle) and grad 𝑔 ≠ 0 on the circle, so we compare at ( 2, 2) and (− 2, − 2). √ Since √ 𝑓 (𝑥, 𝑦) = 𝑥√+ 𝑦, the √values √ of 𝑓 √ maximum value of 𝑓 is 𝑓 ( 2, 2) = 2 2; the minimum value is 𝑓 (− 2, − 2) = −2 2. (See Figure 15.29.) √

𝑥2 + 𝑦2 = 4

𝑓 = −1 𝑓 = −2 √ 𝑓 = −2 2 √ √ (− 2, − 2) Minimum 𝑓



𝑦 Maximum √ √ 𝑓 ( 2, 2)

𝑥 √ 𝑓 =2 2 𝑓 =2 𝑓 =1 𝑓 =0

Figure 15.29: Maximum and minimum values of 𝑓 (𝑥, 𝑦) = 𝑥 + 𝑦 on the circle 𝑥2 + 𝑦2 = 4 are at points where contours of 𝑓 are tangent to the circle

How to Distinguish Maxima from Minima There is a second-derivative test8 for classifying the critical points of constrained optimization problems, but it is more complicated than the test in Section 15.1. However, a graph of the constraint and some contours usually shows which points are maxima, which points are minima, and which are neither.

Optimization with Inequality Constraints The production problem that we looked at first was to maximize production 𝑓 (𝑥, 𝑦) subject to a budget constraint 𝑔(𝑥, 𝑦) = 𝑝1 𝑥 + 𝑝2 𝑦 ≤ 𝑐. 8 See

J. E. Marsden and A. J. Tromba, Vector Calculus, 6th ed. (New York: W.H. Freeman, 2011), p. 220..

15.3 CONSTRAINED OPTIMIZATION: LAGRANGE MULTIPLIERS

829

Since the inputs are nonnegative, 𝑥 ≥ 0 and 𝑦 ≥ 0, we have three inequality constraints, which restrict (𝑥, 𝑦) to a region of the plane rather than to a curve in the plane. In principle, we should first check to see whether or not 𝑓 (𝑥, 𝑦) has any critical points in the interior: 𝑝1 𝑥 + 𝑝2 𝑦 < 𝑐,

𝑥>0

𝑦 > 0.

However, in the case of a budget constraint, we can see that the maximum of 𝑓 must occur when the budget is exhausted, so we look for the maximum value of 𝑓 on the boundary line: 𝑝1 𝑥 + 𝑝2 𝑦 = 𝑐,

𝑥≥0

𝑦 ≥ 0.

Strategy for Optimizing 𝑓 (𝑥, 𝑦) Subject to the Constraint 𝑔(𝑥, 𝑦) ≤ 𝑐 • Find all points in the region 𝑔(𝑥, 𝑦) < 𝑐 where grad 𝑓 is zero or undefined. • Use Lagrange multipliers to find the local extrema of 𝑓 on the boundary 𝑔(𝑥, 𝑦) = 𝑐. • Evaluate 𝑓 at the points found in the previous two steps and compare the values.

From Section 15.2 we know that if 𝑓 is continuous on a closed and bounded region, 𝑅, then 𝑓 is guaranteed to attain its global maximum and minimum values on 𝑅. Example 2

Find the maximum and minimum values of 𝑓 (𝑥, 𝑦) = (𝑥 − 1)2 + (𝑦 − 2)2 subject to the constraint 𝑥2 + 𝑦2 ≤ 45.

Solution

First, we look for all critical points of 𝑓 in the interior of the region. Setting 𝑓𝑥 (𝑥, 𝑦) = 2(𝑥 − 1) = 0 𝑓𝑦 (𝑥, 𝑦) = 2(𝑦 − 2) = 0, we find 𝑓 has exactly one critical point at 𝑥 = 1, 𝑦 = 2. Since 12 + 22 < 45, that critical point is in the interior of the region. Next, we find the local extrema of 𝑓 on the boundary curve 𝑥2 + 𝑦2 = 45. To do this, we use Lagrange multipliers with constraint 𝑔(𝑥, 𝑦) = 𝑥2 + 𝑦2 = 45. Setting grad 𝑓 = 𝜆 grad 𝑔, we get 2(𝑥 − 1) = 𝜆 ⋅ 2𝑥, 2(𝑦 − 2) = 𝜆 ⋅ 2𝑦. We can’t have 𝑥 = 0 since the first equation would become −2 = 0. Similarly, 𝑦 ≠ 0. So we can solve each equation for 𝜆 by dividing by 𝑥 and 𝑦. Setting the expressions for 𝜆 equal gives 𝑥−1 𝑦−2 = , 𝑥 𝑦 so 𝑦 = 2𝑥. Combining this with the constraint

𝑥2

+

𝑦2

= 45, we get 5𝑥2 = 45,

so 𝑥 = ±3. Since 𝑦 = 2𝑥, we have possible local extrema at 𝑥 = 3, 𝑦 = 6 and 𝑥 = −3, 𝑦 = −6.

830

Chapter 15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA

We conclude that the only candidates for the maximum and minimum values of 𝑓 in the region occur at (1, 2), (3, 6), and (−3, −6). Evaluating 𝑓 at these three points, we find 𝑓 (1, 2) = 0,

𝑓 (3, 6) = 20,

𝑓 (−3, −6) = 80.

Therefore, the minimum value of 𝑓 is 0 at (1, 2) and the maximum value is 80 at (−3, −6).

The Meaning of 𝝀 In the uses of Lagrange multipliers so far, we never found (or needed) the value of 𝜆. However, 𝜆 does have a practical interpretation. In the production example, we wanted to maximize 𝑓 (𝑥, 𝑦) = 𝑥2∕3 𝑦1∕3 subject to the constraint 𝑔(𝑥, 𝑦) = 𝑥 + 𝑦 = 3.78. We solved the equations 2 −1∕3 1∕3 𝑥 𝑦 = 𝜆, 3 1 2∕3 −2∕3 𝑥 𝑦 = 𝜆, 3 𝑥 + 𝑦 = 3.78, to get 𝑥 = 2.52, 𝑦 = 1.26 and 𝑓 (2.52, 1.26) ≈ 2. Continuing to find 𝜆 gives us 𝜆 ≈ 0.53. Now we do another, apparently unrelated, calculation. Suppose our budget is increased by one, from 3.78 to 4.78, giving a new budget constraint of 𝑥 + 𝑦 = 4.78. Then the corresponding solution is at 𝑥 = 3.19 and 𝑦 = 1.59 and the new maximum value (instead of 𝑓 = 2) is 𝑓 = (3.19)2∕3(1.59)1∕3 ≈ 2.53. Notice that the amount by which 𝑓 has increased is 0.53, the value of 𝜆. Thus, in this example, the value of 𝜆 represents the extra production achieved by increasing the budget by one—in other words, the extra “bang” you get for an extra “buck” of budget. In fact, this is true in general: • The value of 𝜆 is approximately the increase in the optimum value of 𝑓 when the budget is increased by 1 unit. More precisely: • The value of 𝜆 represents the rate of change of the optimum value of 𝑓 as the budget increases.

An Expression for 𝝀 To interpret 𝜆, we look at how the optimum value of the objective function 𝑓 changes as the value 𝑐 of the constraint function 𝑔 is varied. In general, the optimum point (𝑥0 , 𝑦0 ) depends on the constraint value 𝑐. So, provided 𝑥0 and 𝑦0 are differentiable functions of 𝑐, we can use the chain rule to differentiate the optimum value 𝑓 (𝑥0 (𝑐), 𝑦0 (𝑐)) with respect to 𝑐: 𝜕𝑓 𝑑𝑥0 𝜕𝑓 𝑑𝑦0 𝑑𝑓 = + . 𝑑𝑐 𝜕𝑥 𝑑𝑐 𝜕𝑦 𝑑𝑐 At the optimum point (𝑥0 , 𝑦0 ), we have 𝑓𝑥 = 𝜆𝑔𝑥 and 𝑓𝑦 = 𝜆𝑔𝑦 , and therefore ( ) 𝑑𝑓 𝜕𝑔 𝑑𝑥0 𝜕𝑔 𝑑𝑦0 𝑑𝑔 =𝜆 + =𝜆 . 𝑑𝑐 𝜕𝑥 𝑑𝑐 𝜕𝑦 𝑑𝑐 𝑑𝑐 But, as 𝑔(𝑥0 (𝑐), 𝑦0 (𝑐)) = 𝑐, we see that 𝑑𝑔∕𝑑𝑐 = 1, so 𝑑𝑓 ∕𝑑𝑐 = 𝜆. Thus, we have the following interpretation of the Lagrange multiplier 𝜆:

15.3 CONSTRAINED OPTIMIZATION: LAGRANGE MULTIPLIERS

831

The value of 𝜆 is the rate of change of the optimum value of 𝑓 as 𝑐 increases (where 𝑔(𝑥, 𝑦) = 𝑐). If the optimum value of 𝑓 is written as 𝑓 (𝑥0 (𝑐), 𝑦0 (𝑐)), then 𝑑 𝑓 (𝑥0 (𝑐), 𝑦0 (𝑐)) = 𝜆. 𝑑𝑐

Example 3

The quantity of goods produced according to the function 𝑓 (𝑥, 𝑦) = 𝑥2∕3 𝑦1∕3 is maximized subject to the budget constraint 𝑥 + 𝑦 ≤ 3.78. The budget is increased to allow for a small increase in production. What is the price of the product if the sale of the additional goods covers the budget increase?

Solution

We know that 𝜆 = 0.53, which tells us that 𝑑𝑓 ∕𝑑𝑐 = 0.53. The constraint corresponds to a budget of $3.78 thousand. Therefore increasing the budget by $1000 increases production by about 0.53 units. In order to make the increase in budget profitable, the extra goods produced must sell for more than $1000. Thus, if 𝑝 is the price of each unit of the good, then 0.53𝑝 is the revenue from the extra 0.53 units sold. Thus, we need 0.53𝑝 ≥ 1000 so 𝑝 ≥ 1000∕0.53 = $1890.

The Lagrangian Function Constrained optimization problems are frequently solved using a Lagrangian function, . For example, to optimize 𝑓 (𝑥, 𝑦) subject to the constraint 𝑔(𝑥, 𝑦) = 𝑐, we use the Lagrangian function (𝑥, 𝑦, 𝜆) = 𝑓 (𝑥, 𝑦) − 𝜆(𝑔(𝑥, 𝑦) − 𝑐). To see how the function  is used, compute the partial derivatives of : 𝜕𝑔 𝜕 𝜕𝑓 = −𝜆 , 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑔 𝜕 𝜕𝑓 = −𝜆 , 𝜕𝑦 𝜕𝑦 𝜕𝑦 𝜕 = −(𝑔(𝑥, 𝑦) − 𝑐). 𝜕𝜆 Notice that if (𝑥0 , 𝑦0 ) is an extreme point of 𝑓 (𝑥, 𝑦) subject to the constraint 𝑔(𝑥, 𝑦) = 𝑐 and 𝜆0 is the corresponding Lagrange multiplier, then at the point (𝑥0 , 𝑦0 , 𝜆0 ) we have 𝜕 = 0 and 𝜕𝑥

𝜕 =0 𝜕𝑦

and

𝜕 = 0. 𝜕𝜆

In other words, (𝑥0 , 𝑦0 , 𝜆0 ) is a critical point for the unconstrained Lagrangian function, (𝑥, 𝑦, 𝜆). Thus, the Lagrangian converts a constrained optimization problem to an unconstrained problem. Example 4

A company has a production function with three inputs 𝑥, 𝑦, and 𝑧 given by 𝑓 (𝑥, 𝑦, 𝑧) = 50𝑥2∕5 𝑦1∕5 𝑧1∕5 . The total budget is $24,000 and the company can buy 𝑥, 𝑦, and 𝑧 at $80, $12, and $10 per unit, respectively. What combination of inputs will maximize production?9

Solution

We need to maximize the objective function 𝑓 (𝑥, 𝑦, 𝑧) = 50𝑥2∕5 𝑦1∕5 𝑧1∕5 , subject to the constraint 𝑔(𝑥, 𝑦, 𝑧) = 80𝑥 + 12𝑦 + 10𝑧 = 24,000. 9 Adapted

from M. Rosser and P. Lis, Basic Mathematics for Economists, 3rd ed. (New York: Routledge, 2016), p. 360.

832

Chapter 15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA

The method for functions of two variables works for functions of three variables, so we construct the Lagrangian function (𝑥, 𝑦, 𝑧, 𝜆) = 50𝑥2∕5 𝑦1∕5 𝑧1∕5 − 𝜆(80𝑥 + 12𝑦 + 10𝑧 − 24,000), and solve the system of equations we get from grad  = 0⃗ : 𝜕 𝜕𝑥 𝜕 𝜕𝑦 𝜕 𝜕𝑧 𝜕 𝜕𝜆

= 20𝑥−3∕5 𝑦1∕5 𝑧1∕5 − 80𝜆 = 0, = 10𝑥2∕5 𝑦−4∕5 𝑧1∕5 − 12𝜆 = 0, = 10𝑥2∕5 𝑦1∕5 𝑧−4∕5 − 10𝜆 = 0, = −(80𝑥 + 12𝑦 + 10𝑧 − 24,000) = 0.

Simplifying this system gives 1 −3∕5 1∕5 1∕5 𝑥 𝑦 𝑧 , 4 5 𝜆 = 𝑥2∕5 𝑦−4∕5 𝑧1∕5 , 6 𝜆 = 𝑥2∕5 𝑦1∕5 𝑧−4∕5 , 80𝑥 + 12𝑦 + 10𝑧 = 24,000. 𝜆=

Eliminating 𝑧 from the first two equations gives 𝑥 = 0.3𝑦. Eliminating 𝑥 from the second and third equations gives 𝑧 = 1.2𝑦. Substituting for 𝑥 and 𝑧 into 80𝑥 + 12𝑦 + 10𝑧 = 24,000 gives 80(0.3𝑦) + 12𝑦 + 10(1.2𝑦) = 24,000, so 𝑦 = 500. Then 𝑥 = 150 and 𝑧 = 600, and 𝑓 (150, 500, 600) = 4,622 units. The graph of the constraint, 80𝑥 + 12𝑦 + 10𝑧 = 24,000, is a plane. Since the inputs 𝑥, 𝑦, 𝑧 must be nonnegative, the graph is a triangle in the first octant, with edges on the coordinate planes. On the boundary of the triangle, one (or more) of the variables 𝑥, 𝑦, 𝑧 is zero, so the function 𝑓 is zero. Thus production is maximized within the budget using 𝑥 = 150, 𝑦 = 500, and 𝑧 = 600.

Exercises and Problems for Section 15.3 Online Resource: Additional Problems for Section 15.3 EXERCISES

In Exercises 1–18, use Lagrange multipliers to find the maximum and minimum values of 𝑓 subject to the given constraint, if such values exist.

2

2. 𝑓 (𝑥, 𝑦) = 𝑥 + 3𝑦 + 2,

2

𝑥 + 𝑦 = 10

2

2

3. 𝑓 (𝑥, 𝑦) = (𝑥 − 1) + (𝑦 + 2) , 3

4. 𝑓 (𝑥, 𝑦) = 𝑥 + 𝑦,

2

𝑥2 + 𝑦2 + 𝑧2 = 1

11. 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 − 𝑦2 − 2𝑧,

𝑥2 + 𝑦2 = 𝑧

12. 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥𝑦𝑧,

𝑥2 + 𝑦2 = 1

1. 𝑓 (𝑥, 𝑦) = 𝑥 + 𝑦,

10. 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥 + 3𝑦 + 5𝑧,

2

13. 𝑓 (𝑥, 𝑦) = 𝑥2 + 2𝑦2 ,

3𝑥 + 𝑦 = 4

14. 𝑓 (𝑥, 𝑦) = 𝑥 + 3𝑦, 15. 𝑓 (𝑥, 𝑦) = 𝑥𝑦,

6. 𝑓 (𝑥, 𝑦) = 𝑥𝑦, 4𝑥2 + 𝑦2 = 8

16. 𝑓 (𝑥, 𝑦) = 𝑥3 + 𝑦,

𝑥+𝑦 ≥ 1

5𝑥 + 4𝑦 = 100

8. 𝑓 (𝑥1 , 𝑥2 ) = 𝑥1 2 + 𝑥2 2 , 𝑥1 + 𝑥2 = 1 9. 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦,

𝑥2 + 𝑦2 ≤ 2

𝑥2 + 2𝑦2 ≤ 1

5. 𝑓 (𝑥, 𝑦) = 3𝑥 − 2𝑦, 𝑥2 + 2𝑦2 = 44 7. 𝑓 (𝑥, 𝑦) = 2𝑥𝑦,

𝑥2 + 𝑦2 ≤ 4

2

𝑥 +𝑦 =5

2

𝑥2 + 𝑦2 + 4𝑧2 = 12

𝑥2 − 𝑦2 = 1

17. 𝑓 (𝑥, 𝑦) = (𝑥 + 3)2 + (𝑦 − 3)2 , 18. 𝑓 (𝑥, 𝑦) = 𝑥2 𝑦 + 3𝑦2 − 𝑦,

𝑥2 + 𝑦2 ≤ 2

𝑥2 + 𝑦2 ≤ 10

15.3 CONSTRAINED OPTIMIZATION: LAGRANGE MULTIPLIERS

19. For each point marked in Figure 15.30, decide whether: (a) The point is a local minimum, maximum, or neither for the function 𝑓 constrained by the loop. (b) The point is a global minimum, maximum, or neither subject to the constraint. 𝑓 = 10

833

In Exercises 20–23, a Cobb-Douglas production function 𝑃 (𝐾, 𝐿) and budget 𝐵(𝐾, 𝐿) are given, where 𝐾 represents capital and 𝐿 represents labor. Use Lagrange multipliers to find the values of 𝐾 and 𝐿 that maximize production given a budget constraint or minimize budget given a production constraint. Then give the value for 𝜆 and its meaning.

𝑄

Constraint

20. Maximize production: 𝑃 = 𝐾 1∕4 𝐿3∕4 Budget constraint: 𝐵 = 2𝐾 + 𝐿 = 40

𝑅



21. Maximize production: 𝑃 = 𝐾 2∕3 𝐿1∕3 Budget constraint: 𝐵 = 10𝐾 + 4𝐿 = 60

𝑃 𝑓 = 20

𝑆

22. Maximize production: 𝑃 = 𝐾 2∕5 𝐿3∕5 Budget constraint: 𝐵 = 4𝐾 + 5𝐿 = 100

𝑓 = 50 𝑓 = 30 𝑓 = 40

𝑓 = 60

23. Minimize budget: 𝐵 = 4𝐾 + 𝐿 Production constraint: 𝑃 = 𝐾 1∕2 𝐿1∕2 = 200

Figure 15.30

PROBLEMS 24. Find the maximum value of 𝑓 (𝑥, 𝑦) = 𝑥 + 𝑦 − (𝑥 − 𝑦)2 on the triangular region 𝑥 ≥ 0, 𝑦 ≥ 0, 𝑥 + 𝑦 ≤ 1. 25. For 𝑓 (𝑥, 𝑦) = 𝑥2 + 6𝑥𝑦, find the global maximum and minimum on the closed region in the first quadrant bounded by the line 𝑥 + 𝑦 = 4 and the curve 𝑥𝑦 = 3. 26. (a) Draw contours of 𝑓 (𝑥, 𝑦) = 2𝑥 + 𝑦 for 𝑧 = −7, −5, −3, −1, 1, 3, 5, 7. (b) On the same axes, graph the constraint 𝑥2 +𝑦2 = 5. (c) Use the graph to approximate the points at which 𝑓 has a maximum or a minimum value subject to the constraint 𝑥2 + 𝑦2 = 5. (d) Use Lagrange multipliers to find the maximum and minimum values of 𝑓 (𝑥, 𝑦) = 2𝑥 + 𝑦 subject to 𝑥2 + 𝑦2 = 5. 27. Let 𝑓 (𝑥, 𝑦) = 𝑥𝛼 𝑦1−𝛼 for 0 < 𝛼 < 1. Find the value of 𝛼 such that the maximum value of 𝑓 on the line 2𝑥 + 3𝑦 = 6 occurs at (1.5, 1). 28. Figure 15.31 shows contours of 𝑓 . Does 𝑓 have a maximum value subject to the constraint 𝑔(𝑥, 𝑦) = 𝑐 for 𝑥 ≥ 0, 𝑦 ≥ 0? If so, approximately where is it and what is its value? Does 𝑓 have a minimum value subject to the 𝑦 constraint? If so, approximately where and what? 𝑔(𝑥, 𝑦) = 𝑐

16 14 12



𝑓 = 100

10

𝑓 = 700 𝑓 = 600

8 6 4

☛ ✢ ✠ ❂ ✙ ✾𝑥



2 2

4

6

8

10 12 14 16

Figure 15.31

𝑓 𝑓 𝑓 𝑓

= 500 = 400 = 300 = 200

29. Each person tries to balance his or her time between leisure and work. The tradeoff is that as you work less your income falls. Therefore each person has indifference curves which connect the number of hours of leisure, 𝑙, and income, 𝑠. If, for example, you are indifferent between 0 hours of leisure and an income of $1125 a week on the one hand, and 10 hours of leisure and an income of $750 a week on the other hand, then the points 𝑙 = 0, 𝑠 = 1125, and 𝑙 = 10, 𝑠 = 750 both lie on the same indifference curve. Table 15.3 gives information on three indifference curves, I, II, and III. Table 15.3 Weekly income

Weekly leisure hours

I

II

III

I

II

III

1125

1250

1375

0

20

40

750

875

1000

10

30

50

500

625

750

20

40

60

375

500

625

30

50

70

250

375

500

50

70

90

(a) Graph the three indifference curves. (b) You have 100 hours a week available for work and leisure combined, and you earn $10/hour. Write an equation in terms of 𝑙 and 𝑠 which represents this constraint. (c) On the same axes, graph this constraint. (d) Estimate from the graph what combination of leisure hours and income you would choose under these circumstances. Give the corresponding number of hours per week you would work.

834

Chapter 15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA

30. Figure 15.32 shows ∇𝑓 for a function 𝑓 (𝑥, 𝑦) and two curves 𝑔(𝑥, 𝑦) = 1 and 𝑔(𝑥, 𝑦) = 2. Mark the following: (a) (b) (c) (d) (e)

The point(s) 𝐴 where 𝑓 has a local maximum. The point(s) 𝐵 where 𝑓 has a saddle point. The point 𝐶 where 𝑓 has a maximum on 𝑔 = 1. The point 𝐷 where 𝑓 has a minimum on 𝑔 = 1. If you used Lagrange multipliers to find 𝐶, what would the sign of 𝜆 be? Why?

33. If the right side of the constraint in Exercise 5 is changed by the small amount Δ𝑐, by approximately how much do the maximum and minimum values change? 34. If the right side of the constraint in Exercise 6 is changed by the small amount Δ𝑐, by approximately how much do the maximum and minimum values change? 35. The function 𝑃 (𝑥, 𝑦) gives the number of units produced and 𝐶(𝑥, 𝑦) gives the cost of production. (a) A company wishes to maximize production at a fixed cost of $50,000. What is the objective function 𝑓 ? What is the constraint equation? What is the meaning of 𝜆 in this situation? (b) A company wishes to minimize costs at a fixed production level of 2000 units. What is the objective function 𝑓 ? What is the constraint equation? What is the meaning of 𝜆 in this situation?



𝑔=2



𝑔=1

36. Design a closed cylindrical container which holds 100 cm3 and has the minimal possible surface area. What should its dimensions be? Figure 15.32 31. The point 𝑃 is a maximum or minimum of the function 𝑓 subject to the constraint 𝑔(𝑥, 𝑦) = 𝑥 + 𝑦 = 𝑐, with 𝑥, 𝑦 ≥ 0. For the graphs (a) and (b), does 𝑃 give a maximum or a minimum of 𝑓 ? What is the sign of 𝜆? If 𝑃 gives a maximum, where does the minimum of 𝑓 occur? If 𝑃 gives a minimum, where does the maximum of 𝑓 occur? 𝑦

(a)

✛ 𝑓 =3 ✛ 𝑓 =2 ✛𝑓 =1

✛ 𝑓 =1 ✛ 𝑓 =2 ✛𝑓 =3 ✛

𝑔(𝑥, 𝑦) = 𝑐

𝑃

𝐶 = 5𝑥2 + 2𝑥𝑦 + 3𝑦2 + 800. (a) If the production quota for the total number of items (both types combined) is 39, find the minimum production cost. (b) Estimate the additional production cost or savings if the production quota is raised to 40 or lowered to 38.

𝑦

(b)



37. A company manufactures 𝑥 units of one item and 𝑦 units of another. The total cost in dollars, 𝐶, of producing these two items is approximated by the function

𝑔(𝑥, 𝑦) = 𝑐

𝑃

𝑥

38. An international organization must decide how to spend the $2,000,000 they have been allotted for famine relief in a remote area. They expect to divide the money between buying rice at $38.5/sack and beans at $35/sack. The number, 𝑃 , of people who would be fed if they buy 𝑥 𝑥 sacks of rice and 𝑦 sacks of beans is given by

32. Figure 15.33 shows the optimal point (marked with a dot) in three optimization problems with the same constraint. Arrange the corresponding values of 𝜆 in increasing order. (Assume 𝜆 is positive.) (I)

𝑦

(II)

𝑓 =3 𝑓 =2 𝑓 =1 𝑥

𝑥𝑦 . 108

What is the maximum number of people that can be fed, and how should the organization allocate its money?

𝑦

39. The quantity, 𝑞, of a product manufactured depends on the number of workers, 𝑊 , and the amount of capital 𝑓 =2 invested, 𝐾, and is given by

𝑓 =3

(III)

𝑃 = 1.1𝑥 + 𝑦 −

❄ ✠ ✛

𝑥

𝑞 = 6𝑊 3∕4 𝐾 1∕4 . 𝑓 =1

Labor costs are $10 per worker and capital costs are $20 per unit, and the budget is $3000.

𝑦

𝑓 =3

𝑓 =2

❄ ✠ ✛

𝑥

Figure 15.33

𝑓 =1

(a) What are the optimum number of workers and the optimum number of units of capital? (b) Show that at the optimum values of 𝑊 and 𝐾, the ratio of the marginal productivity of labor (𝜕𝑞∕𝜕𝑊 ) to the marginal productivity of capital

15.3 CONSTRAINED OPTIMIZATION: LAGRANGE MULTIPLIERS

(𝜕𝑞∕𝜕𝐾) is the same as the ratio of the cost of a unit of labor to the cost of a unit of capital. (c) Recompute the optimum values of 𝑊 and 𝐾 when the budget is increased by one dollar. Check that increasing the budget by $1 allows the production of 𝜆 extra units of the good, where 𝜆 is the Lagrange multiplier. 40. A neighborhood health clinic has a budget of $600,000 per quarter. The director of the clinic wants to allocate the budget to maximize the number of patient visits, 𝑉 , which is given as a function of the number of doctors, 𝐷, and the number of nurses, 𝑁, by

835

𝐿, labor 20 𝐵 =

10, 000

800 0

𝑃 = 90

15 60 00

80 70

10 40 00

60 50 40 30 20

5

5

200 0 10

𝐾 , capital 15

20

Figure 15.34

𝑉 = 1000𝐷0.6 𝑁 0.3 . A doctor gets $40,000 per quarter; nurses get $10,000 per quarter. (a) Set up the director’s constrained optimization problem. (b) Describe, in words, the conditions which must be satisfied by 𝜕𝑉 ∕𝜕𝐷 and 𝜕𝑉 ∕𝜕𝑁 for 𝑉 to have an optimum value. (c) Solve the problem formulated in part (a). (d) Find the value of the Lagrange multiplier and interpret its meaning in this problem. (e) At the optimum point, what is the marginal cost of a patient visit (that is, the cost of an additional visit)? Will that marginal cost rise or fall with the number of visits? Why? 41. (a) In Problem 39, does the value of 𝜆 change if the budget changes from $3000 to $4000? (b) In Problem 40, does the value of 𝜆 change if the budget changes from $600,000 to $700,000? (c) What condition must a Cobb-Douglas production function, 𝑄 = 𝑐𝐾 𝑎 𝐿𝑏 , satisfy to ensure that the marginal increase of production (that is, the rate of increase of production with budget) is not affected by the size of the budget? 42. The production function 𝑃 (𝐾, 𝐿) gives the number of pairs of skis produced per week at a factory operating with 𝐾 units of capital and 𝐿 units of labor. The contour diagram for 𝑃 is in Figure 15.34; the parallel lines are budget constraints for budgets, 𝐵, in dollars. (a) On each budget constraint, mark the point that gives the maximum production. (b) Complete the table, where the budget, 𝐵, is in dollars and the maximum production is the number of pairs of skis to be produced each week. 𝐵

2000

4000

6000

8000

10000

43. A doctor wants to schedule visits for two patients who have been operated on for tumors so as to minimize the expected delay in detecting a new tumor. Visits for patients 1 and 2 are scheduled at intervals of 𝑥1 and 𝑥2 weeks, respectively. A total of 𝑚 visits per week is available for both patients combined. The recurrence rates for tumors for patients 1 and 2 are judged to be 𝑣1 and 𝑣2 tumors per week, respectively. Thus, 𝑣1 ∕(𝑣1 + 𝑣2 ) and 𝑣2 ∕(𝑣1 + 𝑣2 ) are the probabilities that patient 1 and patient 2, respectively, will have the next tumor. It is known that the expected delay in detecting a tumor for a patient checked every 𝑥 weeks is 𝑥∕2. Hence, the expected detection delay for both patients combined is given by10 𝑓 (𝑥1 , 𝑥2 ) =

𝑣2 𝑣1 𝑥 𝑥 ⋅ 1 + ⋅ 2. 𝑣1 + 𝑣2 2 𝑣1 + 𝑣2 2

Find the values of 𝑥1 and 𝑥2 in terms of 𝑣1 and 𝑣2 that minimize 𝑓 (𝑥1 , 𝑥2 ) subject to the fact that 𝑚, the number of visits per week, is fixed. 44. What is the value of the Lagrange multiplier in Problem 43? What are the units of 𝜆? What is its practical significance to the doctor? 45. Figure 15.35 shows two weightless springs with spring constants 𝑘1 and 𝑘2 attached between a ceiling and floor without tension or compression. A mass 𝑚 is placed between the springs which settle into equilibrium as in Figure 15.36. The magnitudes 𝑓1 and 𝑓2 of the forces of the springs on the mass minimize the complementary energy 𝑓2 𝑓12 + 2 2𝑘1 2𝑘2

𝑀

subject to the force balance constraint 𝑓1 + 𝑓2 = 𝑚𝑔. (c) Estimate the Lagrange multiplier 𝜆 = 𝑑𝑀∕𝑑𝐵 at a budget of $6000. Give units for the multiplier.

(a) Determine 𝑓1 and 𝑓2 by the method of Lagrange multipliers.

10 Adapted from Daniel Kent, Ross Shachter, et al., “Efficient Scheduling of Cystoscopies in Monitoring for Recurrent Bladder Cancer,” Medical Decision Making (Philadelphia: Hanley and Belfus, 1989).

836

Chapter 15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA

(b) If you are familiar with Hooke’s law, find the meaning of 𝜆.

(a) What is the practical meaning of 𝑉 (𝑝, 𝑞, 𝐼)? (b) The Lagrange multiplier 𝜆 that arises in the maximization defining 𝑉 is called the marginal utility of money. What is its practical meaning? (c) Find formulas for 𝑉 (𝑝, 𝑞, 𝐼) and 𝜆 if 𝑈 (𝑥, 𝑦) = 𝑥𝑦.

𝑘1

𝑘2

Figure 15.35

𝑉 (𝑝, 𝑞, 𝐼) is the maximum of 𝑈 (𝑥, 𝑦) subject to the constraint 𝑝𝑥 + 𝑞𝑦 = 𝐼.

m

Figure 15.36

∑3 46. (a) If 𝑖=1 𝑥𝑖 = 1, find the values of 𝑥1 , 𝑥2 , 𝑥3 making ∑3 𝑥 2 minimum. 𝑖=1 𝑖 (b) Generalize the result of part (a) to find the mini∑𝑛 ∑𝑛 mum value of 𝑖=1 𝑥𝑖 2 subject to 𝑖=1 𝑥𝑖 = 1.

47. Let 𝑓 (𝑥, 𝑦) = 𝑎𝑥2 + 𝑏𝑥𝑦 + 𝑐𝑦2 . Show that the maximum value of 𝑓 (𝑥, 𝑦) subject to the constraint 𝑥2 + 𝑦2 = 1 is equal to 𝜆, the Lagrange multiplier.

48. Find the minimum distance from the point (1, 2, 10) to the paraboloid given by the equation 𝑧 = 𝑥2 + 𝑦2 . Give a geometric justification for your answer. 49. A company produces one product from two inputs (for example, capital and labor). Its production function 𝑔(𝑥, 𝑦) gives the quantity of the product that can be produced with 𝑥 units of the first input and 𝑦 units of the second. The cost function (or expenditure function) is the three-variable function 𝐶(𝑝, 𝑞, 𝑢) where 𝑝 and 𝑞 are the unit prices of the two inputs. For fixed 𝑝, 𝑞, and 𝑢, the value 𝐶(𝑝, 𝑞, 𝑢) is the minimum of 𝑓 (𝑥, 𝑦) = 𝑝𝑥+𝑞𝑦 subject to the constraint 𝑔(𝑥, 𝑦) = 𝑢. (a) What is the practical meaning of 𝐶(𝑝, 𝑞, 𝑢)? (b) Find a formula for 𝐶(𝑝, 𝑞, 𝑢) if 𝑔(𝑥, 𝑦) = 𝑥𝑦. 50. A utility function 𝑈 (𝑥, 𝑦) for two items gives the utility (benefit) to a consumer of 𝑥 units of item 1 and 𝑦 units of item 2. The indirect utility function is the threevariable function 𝑉 (𝑝, 𝑞, 𝐼) where 𝑝 and 𝑞 are the unit prices of the two items. For fixed 𝑝, 𝑞, and 𝐼, the value

51. The function ℎ(𝑥, 𝑦) = 𝑥2 + 𝑦2 − 𝜆(2𝑥 + 4𝑦 − 15) has a minimum value 𝑚(𝜆) for each value of 𝜆. (a) Find 𝑚(𝜆). (b) For which value of 𝜆 is 𝑚(𝜆) the largest and what is that maximum value? (c) Find the minimum value of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 subject to the constraint 2𝑥+4𝑦 = 15 using the method of Lagrange multipliers and evaluate 𝜆. (d) Compare your answers to parts (b) and (c). 52. Let 𝑓 be differentiable and grad 𝑓 (2, 1) = −3⃗𝑖 + 4𝑗⃗ . You want to see if (2, 1) is a candidate for the maximum and minimum values of 𝑓 subject to a constraint satisfied by the point (2, 1). (a) Show (2, 1) is not a candidate if the constraint is 𝑥2 + 𝑦2 = 5. (b) Show (2, 1) is a candidate if the constraint is (𝑥 − 5)2 + (𝑦 + 3)2 = 25. From a sketch of the contours for 𝑓 near (2, 1) and the constraint, decide whether (2, 1) is a candidate for a maximum or minimum. (c) Do the same as part (b), but using the constraint (𝑥 + 1)2 + (𝑦 − 5)2 = 25. 53. A person’s satisfaction from consuming a quantity 𝑥1 of one item and a quantity 𝑥2 of another item is given by 𝑆 = 𝑢(𝑥1 , 𝑥2 ) = 𝑎 ln 𝑥1 + (1 − 𝑎) ln 𝑥2 , where 𝑎 is a constant, 0 < 𝑎 < 1. The prices of the two items are 𝑝1 and 𝑝2 respectively, and the budget is 𝑏. (a) Express the maximum satisfaction that can be achieved as a function of 𝑝1 , 𝑝2 , and 𝑏. (b) Find the amount of money that must be spent to achieve a particular level of satisfaction, 𝑐, as a function of 𝑝1 , 𝑝2 , and 𝑐.

Strengthen Your Understanding In Problems 54–55, explain what is wrong with the statement. 54. The function 𝑓 (𝑥, 𝑦) = 𝑥𝑦 has a maximum of 2 on the constraint 𝑥 + 𝑦 = 2. 55. If the level curves of 𝑓 (𝑥, 𝑦) and the level curves of 𝑔(𝑥, 𝑦) are not tangent at any point on the constraint 𝑔(𝑥, 𝑦) = 𝑐, 𝑥 ≥ 0, 𝑦 ≥ 0, then 𝑓 has no maximum on the constraint.

In Problems 56–60, give an example of: 56. A function 𝑓 (𝑥, 𝑦) whose maximum subject to the constraint 𝑥2 + 𝑦2 = 5 is at (3, 4). 57. A function 𝑓 (𝑥, 𝑦) to be optimized with constraint 𝑥2 + 2𝑦2 ≤ 1 such that the minimum value does not change when the constraint is changed to 𝑥2 + 𝑦2 ≤ 1 + 𝑐 for 𝑐 > 0. 58. A function 𝑓 (𝑥, 𝑦) with a minimum at (1, 1) on the constraint 𝑥 + 𝑦 = 2.

15.3 CONSTRAINED OPTIMIZATION: LAGRANGE MULTIPLIERS

837

59. A function 𝑓 (𝑥, 𝑦) that has a maximum but no minimum on the constraint 𝑥 + 𝑦 = 4.

64. If 𝑓 (𝑥, 𝑦) has a local maximum at (𝑎, 𝑏) subject to the constraint 𝑔(𝑥, 𝑦) = 𝑐, then grad𝑓 (𝑎, 𝑏) = 0⃗ .

60. A contour diagram of a function 𝑓 whose maximum value on the constraint 𝑥 + 2𝑦 = 6, 𝑥 ≥ 0, 𝑦 ≥ 0 occurs at one of the endpoints.

65. The function 𝑓 (𝑥, 𝑦) = 𝑥 + 𝑦 has no global maximum subject to the constraint 𝑥 − 𝑦 = 0.

For Problems 61–62, use Figure 15.37. The grid lines are one unit apart. 𝑦

𝑔=𝑐 𝑓= 4

66. The point (2, −1) is a local minimum of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 subject to the constraint 𝑥 + 2𝑦 = 0. 67. If grad 𝑓 (𝑎, 𝑏) and grad 𝑔(𝑎, 𝑏) point in opposite directions, then (𝑎, 𝑏) is a local minimum of 𝑓 (𝑥, 𝑦) constrained by 𝑔(𝑥, 𝑦) = 𝑐. In Problems 68–75, suppose that 𝑀 and 𝑚 are the maximum and minimum values of 𝑓 (𝑥, 𝑦) subject to the constraint 𝑔(𝑥, 𝑦) = 𝑐 and that (𝑎, 𝑏) satisfies 𝑔(𝑎, 𝑏) = 𝑐. Decide whether the statements are true or false. Give an explanation for your answer.

𝑓= 5

𝑓= 3 𝑓= 2

68. If 𝑓 (𝑎, 𝑏) = 𝑀, then 𝑓𝑥 (𝑎, 𝑏) = 𝑓𝑦 (𝑎, 𝑏) = 0.

𝑓= 1

69. If 𝑓 (𝑎, 𝑏) = 𝑀, then 𝑓 (𝑎, 𝑏) = 𝜆𝑔(𝑎, 𝑏) for some value of 𝜆.

𝑓 =0

𝑥

Figure 15.37 61. Find the maximum and minimum values of 𝑓 on 𝑔 = 𝑐. At which points do they occur? 62. Find the maximum and minimum values of 𝑓 on the triangular region below 𝑔 = 𝑐 in the first quadrant. Are the statements in Problems 63–67 true or false? Give reasons for your answer. 63. If 𝑓 (𝑥, 𝑦) has a local maximum at (𝑎, 𝑏) subject to the constraint 𝑔(𝑥, 𝑦) = 𝑐, then 𝑔(𝑎, 𝑏) = 𝑐.

Online Resource: Review problems and Projects

70. If grad 𝑓 (𝑎, 𝑏) = 𝜆 grad 𝑔(𝑎, 𝑏), then 𝑓 (𝑎, 𝑏) = 𝑀 or 𝑓 (𝑎, 𝑏) = 𝑚. 71. If 𝑓 (𝑎, 𝑏) = 𝑀 and 𝑓𝑥 (𝑎, 𝑏)∕𝑓𝑦 (𝑎, 𝑏) = 5, then 𝑔𝑥 (𝑎, 𝑏)∕𝑔𝑦 (𝑎, 𝑏) = 5. 72. If 𝑓 (𝑎, 𝑏) = 𝑚 and 𝑔𝑥 (𝑎, 𝑏) = 0, then 𝑓𝑥 (𝑎, 𝑏) = 0. 73. Increasing the value of 𝑐 increases the value of 𝑀. 74. Suppose that 𝑓 (𝑎, 𝑏) = 𝑀 and that grad 𝑓 (𝑎, 𝑏) = 3 grad 𝑔(𝑎, 𝑏). Then increasing the value of 𝑐 by 0.02 increases the value of 𝑀 by about 0.06. 75. Suppose that 𝑓 (𝑎, 𝑏) = 𝑚 and that grad 𝑓 (𝑎, 𝑏) = 3 grad 𝑔(𝑎, 𝑏). Then increasing the value of 𝑐 by 0.02 decreases the value of 𝑚 by about 0.06.

Chapter Sixteen

INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

Contents 16.1 The Definite Integral of a Function of Two Variables . . . . . . . . . . . . . . . . . . . . . . . . . . Population Density of Foxes in England . . . . . . . Definition of the Definite Integral . . . . . . . . . . . . Interpretation of the Double Integral as Volume . As Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Interpretation of the Double Integral as Average Value. . . . . . . . . . . . . . . . . . . . . . . . . . . Integral over Regions that Are Not Rectangles . . Convergence of Upper and Lower Sums to Same Limit . . . . . . . . . . . . . . . . . . . . . 16.2 Iterated Integrals . . . . . . . . . . . . . . . . . . . . . . . . The Fox Population Again: Expressing a Double Integral as an Iterated Integral. . . . . . . . The Order of Integration . . . . . . . . . . . . . . . . . . . Iterated Integrals Over Non-Rectangular Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Reversing the Order of Integration . . . . . . . . . . . . 16.3 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . 16.4 Double Integrals in Polar Coordinates . . . . . . . Integration in Polar Coordinates . . . . . . . . . . . . . What Is d A in Polar Coordinates? . . . . . . . . . . . . . 16.5 Integrals in Cylindrical and Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . Integration in Cylindrical Coordinates . . . . . . . . Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . Integration in Spherical Coordinates . . . . . . . . . . 16.6 Applications of Integration to Probability . . . . Density Functions . . . . . . . . . . . . . . . . . . . . . . . . Smoothing the Histogram .. . . . . . . . . . . . . . . . . . Joint Probability Density Functions . . . . . . . . . .

840 840 841 842 844 844 844 845 847 847 849 850 853 857 864 864 865 869 869 870 872 873 878 878 879 880

840

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

16.1

THE DEFINITE INTEGRAL OF A FUNCTION OF TWO VARIABLES The definite integral of a continuous one-variable function, 𝑓 , is a limit of Riemann sums: 𝑏

∫𝑎

𝑓 (𝑥) 𝑑𝑥 = lim

Δ𝑥→0



𝑓 (𝑥𝑖 ) Δ𝑥,

𝑖

where 𝑥𝑖 is a point in the 𝑖th subdivision of the interval [𝑎, 𝑏]. In this section we extend this definition to functions of two variables. We start by considering how to estimate total population from a twovariable population density.

Population Density of Foxes in England The fox population in parts of England can be important to public health officials because animals can spread diseases, such as rabies. Biologists use a contour diagram to display the fox population density, 𝐷; see Figure 16.1, where 𝐷 is in foxes per square kilometer.1 The bold contour is the coastline, which may be thought of as the 𝐷 = 0 contour; clearly the density is zero outside it. We can think of 𝐷 as a function of position, 𝐷 = 𝑓 (𝑥, 𝑦) where 𝑥 and 𝑦 are in kilometers from the southwest corner of the map. kilometers north 150

1.5

125 0.5

North

1

100 2

1.5

75

1.5

25

2

0.5

50

2

1

1.5

1 0.5

30

60

90

120

150

180

kilometers east

Figure 16.1: Population density of foxes in southwestern England

Example 1

Estimate the total fox population in the region represented by the map in Figure 16.1.

Solution

We subdivide the map into the rectangles shown in Figure 16.1 and estimate the population in each rectangle. For simplicity, we use the population density at the northeast corner of each rectangle. For example, in the bottom left rectangle, the density is 0 at the northeast corner; in the next rectangle to the east (right), the density in the northeast corner is 1. Continuing in this way, we get the estimates in Table 16.1. To estimate the population in a rectangle, we multiply the density by the area of the rectangle, 30 ⋅ 25 = 750 km2 . Adding the results, we obtain Estimate of population = (0.2 + 0.7 + 1.2 + 1.2 + 0.1 + 1.6 + 0.5 + 1.4 + 1.1 + 1.6 + 1.5 + 1.8 + 1.5 + 1.3 + 1.1 + 2.0 + 1.4 + 1.0 + 1.0 + 0.6 + 1.2)750 = 18,000 foxes. 1 From

“On the spatial spread of rabies among foxes”, Murray, J. D. et al, Proc. R. Soc. Lond. B, 229: 111–150, 1986.

16.1 THE DEFINITE INTEGRAL OF A FUNCTION OF TWO VARIABLES

841

Taking the upper and lower bounds for the population density on each rectangle enables us to find upper and lower estimates for the population. Using the same rectangles, the upper estimate is approximately 35,000 and the lower estimate is 4,000. There is a wide discrepancy between the upper and lower estimates; we could make them closer by taking finer subdivisions. Table 16.1

Estimates of population density (northeast corner) 0.0

0.0

0.2

0.7

1.2

1.2

0.0

0.0

0.0

0.0

0.1

1.6

0.0

0.0

0.5

1.4

1.1

1.6

0.0

0.0

1.5

1.8

1.5

1.3

0.0

1.1

2.0

1.4

1.0

0.0

0.0

1.0

0.6

1.2

0.0

0.0

Definition of the Definite Integral The sums used to approximate the fox population are Riemann sums. We now define the definite integral for a function 𝑓 of two variables on a rectangular region. Given a continuous function 𝑓 (𝑥, 𝑦) defined on a region 𝑎 ≤ 𝑥 ≤ 𝑏 and 𝑐 ≤ 𝑦 ≤ 𝑑, we subdivide each of the intervals 𝑎 ≤ 𝑥 ≤ 𝑏 and 𝑐 ≤ 𝑦 ≤ 𝑑 into 𝑛 and 𝑚 equal subintervals respectively, giving 𝑛𝑚 subrectangles. (See Figure 16.2.) 𝑦 𝑑 = 𝑦𝑚

𝑦3 𝑦2 𝑦1 𝑐 = 𝑦0 𝑥 𝑎 = 𝑥0 𝑥1 𝑥2 𝑥3

𝑏 = 𝑥𝑛

Figure 16.2: Subdivision of a rectangle into 𝑛𝑚 subrectangles

The area of each subrectangle is Δ𝐴 = Δ𝑥 Δ𝑦, where Δ𝑥 = (𝑏 − 𝑎)∕𝑛 is the width of each subdivision on the 𝑥-axis, and Δ𝑦 = (𝑑 − 𝑐)∕𝑚 is the width of each subdivision on the 𝑦-axis. To compute the Riemann sum, we multiply the area of each subrectangle by the value of the function at a point in the rectangle and add the resulting numbers. Choosing the maximum value, 𝑀𝑖𝑗 , of the ∑ function on each rectangle and adding for all 𝑖, 𝑗 gives the upper sum, 𝑖,𝑗 𝑀𝑖𝑗 Δ𝑥Δ𝑦. ∑ The lower sum, 𝑖,𝑗 𝐿𝑖𝑗 Δ𝑥Δ𝑦, is obtained by taking the minimum value on each rectangle. If (𝑢𝑖𝑗 , 𝑣𝑖𝑗 ) is any point in the 𝑖𝑗-th subrectangle, any other Riemann sum satisfies ∑ ∑ ∑ 𝑀𝑖𝑗 Δ𝑥Δ𝑦. 𝐿𝑖𝑗 Δ𝑥Δ𝑦 ≤ 𝑓 (𝑢𝑖𝑗 , 𝑣𝑖𝑗 ) Δ𝑥 Δ𝑦 ≤ 𝑖,𝑗

𝑖,𝑗

𝑖,𝑗

We define the definite integral by taking the limit as the numbers of subdivisions, 𝑛 and 𝑚, tend to infinity. By comparing upper and lower sums, as we did for the fox population, it can be shown that the limit exists when the function, 𝑓 , is continuous. We get the same limit by letting Δ𝑥 and Δ𝑦 tend to 0. Thus, we have the following definition:

842

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

Suppose the function 𝑓 is continuous on 𝑅, the rectangle 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑐 ≤ 𝑦 ≤ 𝑑. If (𝑢𝑖𝑗 , 𝑣𝑖𝑗 ) is any point in the 𝑖𝑗-th subrectangle, we define the definite integral of 𝑓 over 𝑅 ∫𝑅

𝑓 𝑑𝐴 =

lim

Δ𝑥,Δ𝑦→0



𝑓 (𝑢𝑖𝑗 , 𝑣𝑖𝑗 )Δ𝑥Δ𝑦.

𝑖,𝑗

Such an integral is called a double integral. The case when 𝑅 is not rectangular is considered on page 844. Sometimes we think of 𝑑𝐴 as being the area of an infinitesimal rectangle of length 𝑑𝑥 and height 𝑑𝑦, so that 𝑑𝐴 = 𝑑𝑥 𝑑𝑦. Then we use the notation2 𝑓 𝑑𝐴 = 𝑓 (𝑥, 𝑦) 𝑑𝑥 𝑑𝑦. ∫𝑅 ∫𝑅 For this definition, we used a particular type of Riemann sum with equal-sized rectangular subdivisions. In a general Riemann sum, the subdivisions do not all have to be the same size.

Interpretation of the Double Integral as Volume Just as the definite integral of a positive one-variable function can be interpreted as an area, so the double integral of a positive two-variable function can be interpreted as a volume. In the one-variable case we visualize the Riemann sums as the total area of rectangles above the subdivisions. In the two-variable case we get solid bars instead of rectangles. As the number of subdivisions grows, the tops of the bars approximate the surface better, and the volume of the bars gets closer to the volume under the graph of the function. (See Figure 16.3.) 𝑧

𝑧

𝑥

𝑥 𝑦

𝑦

Figure 16.3: Approximating volume under a graph with finer and finer Riemann sums

Thus, we have the following result: If 𝑥, 𝑦, 𝑧 represent length and 𝑓 is positive, then Volume under graph of 𝑓 above region 𝑅

Example 2

=

∫𝑅

𝑓 𝑑𝐴.

Let 𝑅 be the rectangle 0 ≤ 𝑥 ≤ 1 and 0 ≤ 𝑦 ≤ 1. Use Riemann sums to make upper and lower 2 2 estimates of the volume of the region above 𝑅 and under the graph of 𝑧 = 𝑒−(𝑥 +𝑦 ) . 2 Another

common notation for the double integral is ∫ ∫𝑅 𝑓 𝑑𝐴.

16.1 THE DEFINITE INTEGRAL OF A FUNCTION OF TWO VARIABLES

Solution

843

If 𝑅 is the rectangle 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1, the volume we want is given by Volume =

2 +𝑦2 )

∫𝑅

𝑒−(𝑥

𝑑𝐴.

We divide 𝑅 into 16 subrectangles by dividing each edge into four parts. Figure 16.4 shows that 2 2 𝑓 (𝑥, 𝑦) = 𝑒−(𝑥 +𝑦 ) decreases as we move away from the origin. Thus, to get an upper sum we evaluate 𝑓 on each subrectangle at the corner nearest the origin. For example, in the rectangle 0 ≤ 𝑥 ≤ 0.25, 0 ≤ 𝑦 ≤ 0.25, we evaluate 𝑓 at (0, 0). Using Table 16.2, we find that 𝑧

𝑦

𝑥 2 +𝑦2 )

Figure 16.4: Graph of 𝑒−(𝑥

above the rectangle 𝑅

Upper sum = (1 + 0.9394 + 0.7788 + 0.5698 + 0.9394 + 0.8825 + 0.7316 + 0.5353 + 0.7788 + 0.7316 + 0.6065 + 0.4437 + 0.5698 + 0.5353 + 0.4437 + 0.3247)(0.0625) = 0.68. To get a lower sum, we evaluate 𝑓 at the opposite corner of each rectangle because the surface slopes down in both the 𝑥 and 𝑦 directions. This yields a lower sum of 0.44. Thus, 0.44 ≤

2 +𝑦2 )

∫𝑅

𝑒−(𝑥

𝑑𝐴 ≤ 0.68.

To get a better approximation of the volume under the graph, we use more subdivisions. See Table 16.3. Table 16.2

2 +𝑦2 )

Values of 𝑓 (𝑥, 𝑦) = 𝑒−(𝑥

on the rectangle 𝑅 𝑦

𝑥

Table 16.3

0.0

0.25

0.50

0.75

1.00

0.0

1

0.9394

0.7788

0.5698

0.3679

0.25

0.9394

0.8825

0.7316

0.5353

0.3456

0.50

0.7788

0.7316

0.6065

0.4437

0.2865

0.75

0.5698

0.5353

0.4437

0.3247

0.2096

1.00

0.3679

0.3456

0.2865

0.2096

0.1353

2 +𝑦2 )

Riemann sum approximations to ∫𝑅 𝑒−(𝑥

𝑑𝐴

Number of subdivisions in 𝑥 and 𝑦 directions 8

16

32

64

Upper

0.6168

0.5873

0.5725

0.5651

Lower

0.4989

0.5283

0.5430

0.5504

844

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

The exact value of the double integral, 0.5577 …, is trapped between the lower and upper sums. Notice that the lower sum increases and the upper sum decreases as the number of subdivisions increases. However, even with 64 subdivisions, the lower and upper sums agree with the exact value of the integral only in the first decimal place.

Interpretation of the Double Integral as Area In the special case that 𝑓 (𝑥, 𝑦) = 1 for all points (𝑥, 𝑦) in the region 𝑅, each term in the Riemann sum is of the form 1 ⋅ Δ𝐴 = Δ𝐴 and the double integral gives the area of the region 𝑅: Area(𝑅) =

∫𝑅

1 𝑑𝐴 =

∫𝑅

𝑑𝐴

Interpretation of the Double Integral as Average Value As in the one-variable case, the definite integral can be used to define the average value of a function: Average value of 𝑓 on the region 𝑅

=

1 𝑓 𝑑𝐴 Area of 𝑅 ∫𝑅

We can rewrite this as Average value × Area of 𝑅 =

∫𝑅

𝑓 𝑑𝐴.

If we interpret the integral as the volume under the graph of 𝑓 , then we can think of the average value of 𝑓 as the height of the box with the same volume that is on the same base. (See Figure 16.5.) Imagine that the volume under the graph is made out of wax. If the wax melted within the perimeter of 𝑅, then it would end up box-shaped with height equal to the average value of 𝑓 . 𝑧







Base of the box is the rectangle 𝑅

Average value of 𝑓

𝑥 𝑦

Figure 16.5: Volume and average value

Integral over Regions that Are Not Rectangles We defined the definite integral ∫𝑅 𝑓 (𝑥, 𝑦) 𝑑𝐴, for a rectangular region 𝑅. Now we extend the definition to regions of other shapes, including triangles, circles, and regions bounded by the graphs of piecewise continuous functions. To approximate the definite integral over a region, 𝑅, which is not rectangular, we use a grid of rectangles approximating the region. We obtain this grid by enclosing 𝑅 in a large rectangle and subdividing that rectangle; we consider just the subrectangles which are inside 𝑅. As before, we pick a point (𝑢𝑖𝑗 , 𝑣𝑖𝑗 ) in each subrectangle and form a Riemann sum ∑ 𝑓 (𝑢𝑖𝑗 , 𝑣𝑖𝑗 )Δ𝑥Δ𝑦. 𝑖,𝑗

This time, however, the sum is over only those subrectangles within 𝑅. For example, in the case of the fox population we can use the rectangles which are entirely on land. As the subdivisions become

16.1 THE DEFINITE INTEGRAL OF A FUNCTION OF TWO VARIABLES

845

finer, the grid approximates the region 𝑅 more closely. For a function, 𝑓 , which is continuous on 𝑅, we define the definite integral as follows: ∑ 𝑓 (𝑢𝑖𝑗 , 𝑣𝑖𝑗 )Δ𝑥Δ𝑦 𝑓 𝑑𝐴 = lim ∫𝑅 Δ𝑥,Δ𝑦→0 𝑖,𝑗 where the Riemann sum is taken over the subrectangles inside 𝑅. You may wonder why we can leave out the rectangles which cover the edge of 𝑅—if we included them, might we get a different value for the integral? The answer is that for any region that we are likely to meet, the area of the subrectangles covering the edge tends to 0 as the grid becomes finer. Therefore, omitting these rectangles does not affect the limit.

Convergence of Upper and Lower Sums to Same Limit We have said that if 𝑓 is continuous on the rectangle 𝑅, then the difference between upper and lower sums for 𝑓 converges to 0 as Δ𝑥 and Δ𝑦 approach 0. In the following example, we show this in a particular case. The ideas in this example can be used in a general proof. Example 3

Let 𝑓 (𝑥, 𝑦) = 𝑥2 𝑦 and let 𝑅 be the rectangle 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1. Show that the difference between upper and lower Riemann sums for 𝑓 on 𝑅 converges to 0, as Δ𝑥 and Δ𝑦 approach 0.

Solution

The difference between the sums is ∑ ∑ ∑ 𝑀𝑖𝑗 Δ𝑥Δ𝑦 − 𝐿𝑖𝑗 Δ𝑥Δ𝑦 = (𝑀𝑖𝑗 − 𝐿𝑖𝑗 ) Δ𝑥Δ𝑦, where 𝑀𝑖𝑗 and 𝐿𝑖𝑗 are the maximum and minimum of 𝑓 on the 𝑖𝑗-th subrectangle. Since 𝑓 increases in both the 𝑥 and 𝑦 directions, 𝑀𝑖𝑗 occurs at the corner of the subrectangle farthest from the origin and 𝐿𝑖𝑗 at the closest. Moreover, since the slopes in the 𝑥 and 𝑦 directions don’t decrease as 𝑥 and 𝑦 increase, the difference 𝑀𝑖𝑗 −𝐿𝑖𝑗 is largest in the subrectangle 𝑅𝑛𝑚 which is farthest from the origin. Thus, ∑ ∑ (𝑀𝑖𝑗 − 𝐿𝑖𝑗 ) Δ𝑥Δ𝑦 ≤ (𝑀𝑛𝑚 − 𝐿𝑛𝑚 ) Δ𝑥Δ𝑦 = (𝑀𝑛𝑚 − 𝐿𝑛𝑚 )Area(𝑅).

Thus, the difference converges to 0 as long as (𝑀𝑛𝑚 − 𝐿𝑛𝑚 ) does. The maximum 𝑀𝑛𝑚 of 𝑓 on the 𝑛𝑚-th subrectangle occurs at (1, 1), the subrectangle’s top right corner, and the minimum 𝐿𝑛𝑚 occurs at the opposite corner, (1 − 1∕𝑛, 1 − 1∕𝑚). Substituting into 𝑓 (𝑥, 𝑦) = 𝑥2 𝑦 gives ) ( ) ( 2 1 2 1 1 1 1 2 = − 2+ − + 1− . 𝑀𝑛𝑚 − 𝐿𝑛𝑚 = (1)2 (1) − 1 − 𝑛 𝑚 𝑛 𝑛 𝑚 𝑛𝑚 𝑛2 𝑚 The right-hand side converges to 0 as 𝑛, 𝑚 → ∞, that is, as Δ𝑥, Δ𝑦 → 0.

Exercises and Problems for Section 16.1 Online Resource: Additional Problems for Section 16.1 EXERCISES

1. Table 16.4 gives values of the function 𝑓 (𝑥, 𝑦), which is increasing in 𝑥 and decreasing in 𝑦 on the region 𝑅 ∶ 0 ≤ 𝑥 ≤ 6, 0 ≤ 𝑦 ≤ 1. Make the best possible upper and lower estimates of ∫𝑅 𝑓 (𝑥, 𝑦) 𝑑𝐴. Table 16.4

2. Values of 𝑓 (𝑥, 𝑦) are in Table 16.5. Let 𝑅 be the rectangle 1 ≤ 𝑥 ≤ 1.2, 2 ≤ 𝑦 ≤ 2.4. Find Riemann sums which are reasonable over and underestimates for ∫𝑅 𝑓 (𝑥, 𝑦) 𝑑𝐴 with Δ𝑥 = 0.1 and Δ𝑦 = 0.2. Table 16.5

𝑥

𝑦

𝑥

0

3

6

0

5

7

10

0.5

4

5

7

1

3

4

6

𝑦

1.0

1.1

1.2

2.0

5

7

10

2.2

4

6

8

2.4

3

5

4

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

3. Figure 16.6 shows contours of 𝑔(𝑥, 𝑦) on the region 𝑅, with 5 ≤ 𝑥 ≤ 11 and 4 ≤ 𝑦 ≤ 10. Using Δ𝑥 = Δ𝑦 = 2, find an overestimate and an underestimate for ∫𝑅 𝑔(𝑥, 𝑦) 𝑑𝐴. 𝑦

5. Figure 16.8 shows a contour plot of population density, people per square kilometer, in a rectangle of land 3 km by 2 km. Estimate the population in the region represented by Figure 16.8.

10

𝑦 5

8

2 400

4 3

6

600

400

2 1

9

11 1000

Figure 16.6 4. Figure 16.7 shows contours of 𝑓 (𝑥, 𝑦) on the rectangle 𝑅 with 0 ≤ 𝑥 ≤ 30 and 0 ≤ 𝑦 ≤ 15. Using Δ𝑥 = 10 and Δ𝑦 = 5, find an overestimate and an underestimate for ∫𝑅 𝑓 (𝑥, 𝑦) 𝑑𝐴. 𝑦

15

10

0 60 00 8

𝑥 7

5

40 60 0 0

20 0

1

4

𝑥 1

2

3

Figure 16.8 In Exercises 6–7, for 𝑥 and 𝑦 in meters and 𝑅 a region on the 𝑥𝑦-plane, what does the integral represent? Give units.

8

10 6

5 2

6.

4

𝑥 10

20

7.

PROBLEMS In Problems 8–14, decide (without calculation) whether the integrals are positive, negative, or zero. Let 𝐷 be the region inside the unit circle centered at the origin, let 𝑅 be the right half of 𝐷, and let 𝐵 be the bottom half of 𝐷. 8. ∫𝐷 1 𝑑𝐴

𝜎(𝑥, 𝑦) 𝑑𝐴, where 𝜎(𝑥, 𝑦) is bacteria population, in ∫𝑅 thousands per m2 .

30

Figure 16.7

9. ∫𝑅 5𝑥 𝑑𝐴

10. ∫𝐵 5𝑥 𝑑𝐴

11. ∫𝐷 (𝑦3 + 𝑦5 ) 𝑑𝐴

12. ∫𝐵 (𝑦3 + 𝑦5 ) 𝑑𝐴

13. ∫𝐷 (𝑦 − 𝑦3 ) 𝑑𝐴

1 ℎ(𝑥, 𝑦) 𝑑𝐴, where ℎ(𝑥, 𝑦) is the height of Area of 𝑅 ∫𝑅 a tent, in meters.

16. Table 16.6 gives values of 𝑓 (𝑥, 𝑦), the number of milligrams of mosquito larvae per square meter in a swamp. If 𝑥 and 𝑦 are in meters and 𝑅 is the rectangle 0 ≤ 𝑥 ≤ 8, 0 ≤ 𝑦 ≤ 6, estimate ∫𝑅 𝑓 (𝑥, 𝑦) 𝑑𝐴. Give units and interpret your answer. Table 16.6 𝑥

3

14. ∫𝐵 (𝑦 − 𝑦 ) 𝑑𝐴

𝑦

15. Figure 16.9 shows contours of 𝑓 (𝑥, 𝑦). Let 𝑅 be the square −0.5 ≤ 𝑥 ≤ 1, −0.5 ≤ 𝑦 ≤ 1. Is the integral ∫𝑅 𝑓 𝑑𝐴 positive or negative? Explain your reasoning. 𝑦

0

4

8

0

1

3

6

3

2

5

9

6

4

9

15

17. Table 16.7 gives values of 𝑓 (𝑥, 𝑦), the depth of volcanic ash, in meters, after an eruption. If 𝑥 and 𝑦 are in kilometers and 𝑅 is the rectangle 0 ≤ 𝑥 ≤ 100, 0 ≤ 𝑦 ≤ 100, estimate the volume of volcanic ash in 𝑅 in km3 . Table 16.7

2.0 1.5 1.0

𝑥

0.5 𝑦 −1

2

0

200

846

3

−0.5 −1.0 −1.0 −0.5

4

𝑥 0 1

2

0

0.5

Figure 16.9

1.0

1.5

2.0

0

50

100

0

0.82

0.56

0.43

50

0.63

0.45

0.3

100

0.55

0.44

0.26

18. Table 16.8 gives the density of cacti, 𝑓 (𝑥, 𝑦), in a desert region, in thousands of cacti per km2 . If 𝑥 and 𝑦 are in kilometers and 𝑅 is the square 0 ≤ 𝑥 ≤ 30, 0 ≤ 𝑦 ≤ 30, estimate the number of cacti in the region 𝑅.

16.2 ITERATED INTEGRALS

Table 16.8

and green), make a rough estimate of how many cubic miles of rain fell on the state during this time.

𝑥

𝑦

847

0

10

20

30

0

8.5

8.2

7.9

8.1

10

9.5

10.6

10.5

10.1

20

9.3

10.5

10.4

9.5

30

8.3

8.6

9.3

9.1

19. Use four subrectangles to approximate the volume of the object whose base is the region 0 ≤ 𝑥 ≤ 4 and 0 ≤ 𝑦 ≤ 6, and whose height is given by 𝑓 (𝑥, 𝑦) = 𝑥+𝑦. Find an overestimate and an underestimate and average the two. 20. Figure 16.10 shows the rainfall, in inches, in Tennessee on May 1–2, 2010.3 Using three contours (red, yellow,

Figure 16.10

Strengthen Your Understanding In Problems 21–22, explain what is wrong with the statement.

27. If 𝑅 is the rectangle 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1 then ∫𝑅 𝑒𝑥𝑦 𝑑𝐴 > 3.

21. For all 𝑓 , the integral ∫𝑅 𝑓 (𝑥, 𝑦) 𝑑𝐴 gives the volume of the solid under the graph of 𝑓 over the region 𝑅.

28. If 𝑅 is the rectangle 0 ≤ 𝑥 ≤ 2, 0 ≤ 𝑦 ≤ 3 and 𝑆 is the rectangle −2 ≤ 𝑥 ≤ 0, −3 ≤ 𝑦 ≤ 0, then ∫𝑅 𝑓 𝑑𝐴 = − ∫𝑆 𝑓 𝑑𝐴.

22. If 𝑅 is a region in the third quadrant where 𝑥 < 0, 𝑦 < 0, then ∫𝑅 𝑓 (𝑥, 𝑦) 𝑑𝐴 is negative. In Problems 23–24, give an example of:

29. Let 𝜌(𝑥, 𝑦) be the population density of a city, in people per km2 . If 𝑅 is a region in the city, then ∫𝑅 𝜌 𝑑𝐴 gives the total number of people in the region 𝑅.

23. A function 𝑓 (𝑥, 𝑦) and rectangle 𝑅 such that the Riemann sums obtained using the lower left-hand corner of each subrectangle are an overestimate.

30. If ∫𝑅 𝑓 𝑑𝐴 = 0, then 𝑓 (𝑥, 𝑦) = 0 at all points of 𝑅.

24. A function 𝑓 (𝑥, 𝑦) whose average value over the square 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1 is negative.

32. If 𝑓 and 𝑔 are two functions continuous on a region 𝑅, then ∫𝑅 𝑓 ⋅ 𝑔 𝑑𝐴 = ∫𝑅 𝑓 𝑑𝐴 ⋅ ∫𝑅 𝑔 𝑑𝐴.

Are the statements in Problems 25–34 true or false? Give reasons for your answer.

33. If 𝑅 is the rectangle 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 2 and 𝑆 is the square 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1, then ∫𝑅 𝑓 𝑑𝐴 = 2 ∫𝑆 𝑓 𝑑𝐴.

25. The double integral ∫𝑅 𝑓 𝑑𝐴 is always positive. 26. If 𝑓 (𝑥, 𝑦) = 𝑘 for all points (𝑥, 𝑦) in a region 𝑅 then ∫𝑅 𝑓 𝑑𝐴 = 𝑘 ⋅ Area(𝑅).

16.2

31. If 𝑔(𝑥, 𝑦) = 𝑘𝑓 (𝑥, 𝑦), where 𝑘 is constant, then ∫𝑅 𝑔 𝑑𝐴 = 𝑘 ∫𝑅 𝑓 𝑑𝐴.

34. If 𝑅 is the rectangle 2 ≤ 𝑥 ≤ 4, 5 ≤ 𝑦 ≤ 9, 𝑓 (𝑥, 𝑦) = 2𝑥 and 𝑔(𝑥, 𝑦) = 𝑥 + 𝑦, then the average value of 𝑓 on 𝑅 is less than the average value of 𝑔 on 𝑅.

ITERATED INTEGRALS In Section 16.1 we approximated double integrals using Riemann sums. In this section we see how to compute double integrals exactly using one-variable integrals.

The Fox Population Again: Expressing a Double Integral as an Iterated Integral To estimate the fox population, we computed a sum of the form ∑ Total population ≈ 𝑓 (𝑢𝑖𝑗 , 𝑣𝑖𝑗 )Δ𝑥 Δ𝑦, 𝑖,𝑗

where 1 ≤ 𝑖 ≤ 𝑛 and 1 ≤ 𝑗 ≤ 𝑚 and the values 𝑓 (𝑢𝑖𝑗 , 𝑣𝑖𝑗 ) can be arranged as in Table 16.9. 3 www.srh.noaa.gov/images/ohx/rainfall/TN_May2010_rainfall_map.png,

accessed June 13, 2016.

848

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

Table 16.9

Estimates for fox population densities for 𝑛 = 𝑚 = 6 0.0

0.0

0.2

0.7

1.2

1.2

0.0

0.0

0.0

0.0

0.1

1.6

0.0

0.0

0.5

1.4

1.1

1.6

0.0

0.0

1.5

1.8

1.5

1.3

0.0

1.1

2.0

1.4

1.0

0.0

0.0

1.0

0.6

1.2

0.0

0.0

For any values of 𝑛 and 𝑚, we can either add across the rows first or add down the columns first. If we add rows first, we can write the sum in the form ( 𝑛 ) 𝑚 ∑ ∑ Total population ≈ 𝑓 (𝑢𝑖𝑗 , 𝑣𝑖𝑗 )Δ𝑥 Δ𝑦. 𝑗=1

The inner sum,

𝑛 ∑

𝑖=1

180

𝑓 (𝑢𝑖𝑗 , 𝑣𝑖𝑗 ) Δ𝑥, approximates the integral ∫0

𝑓 (𝑥, 𝑣𝑖𝑗 ) 𝑑𝑥. Thus, we have

𝑖=1

𝑚 ∑

Total population ≈

𝑗=1

(

180

∫0

𝑓 (𝑥, 𝑣𝑖𝑗 ) 𝑑𝑥

)

Δ𝑦. 180

The outer Riemann sum approximates another integral, this time with integrand ∫0 𝑓 (𝑥, 𝑦) 𝑑𝑥, which is a function of 𝑦. Thus, we can write the total population in terms of nested, or iterated, onevariable integrals: ( ) 150

Total population =

∫0

180

𝑓 (𝑥, 𝑦) 𝑑𝑥

∫0

𝑑𝑦.

Since the total population is represented by ∫𝑅 𝑓 𝑑𝐴, this suggests the method of computing double integrals in the following theorem:4

Theorem 16.1: Writing a Double Integral as an Iterated Integral If 𝑅 is the rectangle 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑐 ≤ 𝑦 ≤ 𝑑 and 𝑓 is a continuous function on 𝑅, then the integral of 𝑓 over 𝑅 exists and is equal to the iterated integral 𝑦=𝑑

∫𝑅 𝑦=𝑑

The expression ∫𝑦=𝑐

(

𝑓 𝑑𝐴 =

∫𝑦=𝑐

(

𝑥=𝑏

∫𝑥=𝑎

𝑓 (𝑥, 𝑦) 𝑑𝑥

)

𝑑𝑦.

) 𝑥=𝑏 𝑑 𝑏 ∫𝑥=𝑎 𝑓 (𝑥, 𝑦) 𝑑𝑥 𝑑𝑦 can be written ∫𝑐 ∫𝑎 𝑓 (𝑥, 𝑦) 𝑑𝑥 𝑑𝑦.

To evaluate the iterated integral, first perform the inside integral with respect to 𝑥, holding 𝑦 constant; then integrate the result with respect to 𝑦. Example 1

A building is 8 meters wide and 16 meters long. It has a flat roof that is 12 meters high at one corner and 10 meters high at each of the adjacent corners. What is the volume of the building?

Solution

If we put the high corner on the 𝑧-axis, the long side along the 𝑦-axis, and the short side along the 𝑥-axis, as in Figure 16.11, then the roof is a plane with 𝑧-intercept 12, and 𝑥 slope (−2)∕8 = −1∕4, and 𝑦 slope (−2)∕16 = −1∕8. Hence, the equation of the roof is 𝑧 = 12 − 14 𝑥 − 81 𝑦. 4 For

a proof, see M. Spivak, Calculus on Manifolds, pp. 53 and 58 (New York: Benjamin, 1965).

16.2 ITERATED INTEGRALS

849

The volume is given by the double integral Volume =

∫𝑅

(12 − 14 𝑥 − 81 𝑦) 𝑑𝐴,

where 𝑅 is the rectangle 0 ≤ 𝑥 ≤ 8, 0 ≤ 𝑦 ≤ 16. Setting up an iterated integral, we get 16

Volume =

∫0

8

∫0

(12 − 14 𝑥 − 18 𝑦) 𝑑𝑥 𝑑𝑦.

The inside integral is ( ) |𝑥=8 (12 − 41 𝑥 − 18 𝑦) 𝑑𝑥 = 12𝑥 − 81 𝑥2 − 81 𝑥𝑦 || = 88 − 𝑦. ∫0 |𝑥=0 8

Then the outside integral gives

16

Volume =

∫0

|16 (88 − 𝑦) 𝑑𝑦 = (88𝑦 − 21 𝑦2 )|| = 1280. |0

The volume of the building is 1280 cubic meters.

𝑧 (m)

✻ ✻ 12

10





❄✛ 16



𝑦 (m)

8

𝑥 (m)



10





Figure 16.11: A slant-roofed building

Figure 16.12: Cross-section of a building 8

Notice that the inner integral ∫0 (12 − 14 𝑥 − 81 𝑦) 𝑑𝑥 in Example 1 gives the area of the cross section of the building perpendicular to the 𝑦-axis in Figure 16.12. 16 8 The iterated integral ∫0 ∫0 (12 − 14 𝑥 − 81 𝑦) 𝑑𝑥𝑑𝑦 thus calculates the volume by adding the volumes of thin cross-sectional slabs.

The Order of Integration In computing the fox population, we could have chosen to add columns (fixed 𝑥) first, instead of the rows. This leads to an iterated integral where 𝑥 is constant in the inner integral instead of 𝑦. Thus, ) 𝑏( 𝑑 𝑓 (𝑥, 𝑦) 𝑑𝐴 = 𝑓 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥 ∫𝑅 ∫𝑎 ∫𝑐 where 𝑅 is the rectangle 𝑎 ≤ 𝑥 ≤ 𝑏 and 𝑐 ≤ 𝑦 ≤ 𝑑. For any function we are likely to meet, it does not matter in which order we integrate over a rectangular region 𝑅; we get the same value for the double integral either way. 𝑑

∫𝑅

𝑓 𝑑𝐴 =

∫𝑐

(

∫𝑎

𝑏

) 𝑓 (𝑥, 𝑦) 𝑑𝑥 𝑑𝑦 =

∫𝑎

𝑏(

∫𝑐

𝑑

) 𝑓 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥

850

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

Example 2

Compute the volume of Example 1 as an iterated integral by integrating with respect to 𝑦 first.

Solution

Rewriting the integral, we have ( 8

Volume =

∫0

16

∫0

(12 −

)

1 𝑥 − 81 𝑦) 𝑑𝑦 4

8

𝑑𝑥 =

∫0

(

(12𝑦 −

|8 (176 − 4𝑥) 𝑑𝑥 = (176𝑥 − 2𝑥2 )|| = 1280 meter3 . ∫0 |0 8

=

𝑦=16 1 1 2 || 𝑥𝑦 − 𝑦 ) | 4 16 |𝑦=0

)

𝑑𝑥

Iterated Integrals Over Non-Rectangular Regions Example 3

The density at the point (𝑥, 𝑦) of a triangular metal plate, as shown in Figure 16.13, is 𝛿(𝑥, 𝑦). Express its mass as an iterated integral. 𝑦 2 𝑦 = 2 − 2𝑥

𝑥 1 Figure 16.13: A triangular metal plate with density 𝛿(𝑥, 𝑦) at the point (𝑥, 𝑦)

Solution

Approximate the triangular region using a grid of small rectangles of sides Δ𝑥 and Δ𝑦. The mass of one rectangle is given by Mass of rectangle ≈ Density ⋅ Area ≈ 𝛿(𝑥, 𝑦)Δ𝑥Δ𝑦. Summing over all rectangles gives a Riemann sum which approximates the double integral: Mass =

∫𝑅

𝛿(𝑥, 𝑦) 𝑑𝐴,

where 𝑅 is the triangle. We want to compute this integral using an iterated integral. Think about how the iterated integral over the rectangle 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑐 ≤ 𝑦 ≤ 𝑑 works: 𝑏

𝑑

∫𝑎 ∫𝑐

𝑓 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥.

The inside integral with respect to 𝑦 is along vertical strips which begin at the horizontal line 𝑦 = 𝑐 and end at the line 𝑦 = 𝑑. There is one such strip for each 𝑥 between 𝑥 = 𝑎 and 𝑥 = 𝑏. (See Figure 16.14.) For the triangular region in Figure 16.13, the idea is the same. The only difference is that the individual vertical strips no longer all go from 𝑦 = 𝑐 to 𝑦 = 𝑑. The vertical strip that starts at the point (𝑥, 0) ends at the point (𝑥, 2 − 2𝑥), because the top edge of the triangle is the line 𝑦 = 2 − 2𝑥. See Figure 16.15. On this vertical strip, 𝑦 goes from 0 to 2 − 2𝑥. Hence, the inside integral is 2−2𝑥

∫0

𝛿(𝑥, 𝑦) 𝑑𝑦.

16.2 ITERATED INTEGRALS

𝑦

𝑦

𝑦

2

𝑑

851

2 (𝑥, 2 − 2𝑥)

(0, 𝑦)

(1 − 21 𝑦, 𝑦)

𝑐 𝑥

𝑥 𝑎

𝑥

𝑏 Figure 16.14: Integrating over a rectangle using vertical strips

(𝑥, 0)

𝑥 1 Figure 16.16: Integrating over a triangle using horizontal strips

1

Figure 16.15: Integrating over a triangle using vertical strips

Finally, since there is a vertical strip for each 𝑥 between 0 and 1, the outside integral goes from 𝑥 = 0 to 𝑥 = 1. Thus, the iterated integral we want is 1

Mass =

2−2𝑥

𝛿(𝑥, 𝑦) 𝑑𝑦 𝑑𝑥.

∫0 ∫0

We could have chosen to integrate in the opposite order, keeping 𝑦 fixed in the inner integral instead of 𝑥. The limits are formed by looking at horizontal strips instead of vertical ones, and expressing the 𝑥-values at the end points in terms of 𝑦. See Figure 16.16. To find the right endpoint of the strip, we use the equation of the top edge of the triangle in the form 𝑥 = 1 − 12 𝑦. Thus, a

horizontal strip goes from 𝑥 = 0 to 𝑥 = 1 − 21 𝑦. Since there is a strip for every 𝑦 from 0 to 2, the iterated integral is 2

Mass =

∫0 ∫0

1− 21 𝑦

𝛿(𝑥, 𝑦) 𝑑𝑥 𝑑𝑦.

Limits on Iterated Integrals • The limits on the outer integral must be constants. • The limits on the inner integral can involve only the variable in the outer integral. For example, if the inner integral is with respect to 𝑥, its limits can be functions of 𝑦.

Example 4

Find the mass 𝑀 of a metal plate 𝑅 bounded by 𝑦 = 𝑥 and 𝑦 = 𝑥2 , with density given by 𝛿(𝑥, 𝑦) = 1 + 𝑥𝑦 kg∕meter2 . (See Figure 16.17.) 𝑦 (meters)

(1, 1)

𝑦=𝑥 𝑦 = 𝑥2

𝑥 (meters) Figure 16.17: A metal plate with density 𝛿(𝑥, 𝑦)

852

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

Solution

The mass is given by 𝑀=

∫𝑅

𝛿(𝑥, 𝑦) 𝑑𝐴.

We integrate along vertical strips first; this means we do the 𝑦 integral first, which goes from the bottom boundary 𝑦 = 𝑥2 to the top boundary 𝑦 = 𝑥. The left edge of the region is at 𝑥 = 0 and the right edge is at the intersection point of 𝑦 = 𝑥 and 𝑦 = 𝑥2 , which is (1, 1). Thus, the 𝑥-coordinate of the vertical strips can vary from 𝑥 = 0 to 𝑥 = 1, and so the mass is given by 1

𝑀=

1

𝑥

∫0 ∫𝑥2

𝛿(𝑥, 𝑦) 𝑑𝑦 𝑑𝑥 =

𝑥

∫0 ∫𝑥2

(1 + 𝑥𝑦) 𝑑𝑦 𝑑𝑥.

Calculating the inner integral first gives 1

𝑀=

𝑥

(1 + 𝑥𝑦) 𝑑𝑦 𝑑𝑥 =

1(

𝑦+𝑥

𝑦2 2

)

|𝑦=𝑥 | | 2 𝑑𝑥 |𝑦=𝑥

∫0 ∫𝑥2 ∫0 ) ( 2 ) 1 1( 𝑥 𝑥3 𝑥4 𝑥6 || 5 𝑥3 𝑥5 − 𝑑𝑥 = − + − = 0.208 kg. 𝑥 − 𝑥2 + = = ∫0 2 2 2 3 8 12 ||0 24

Example 5

A semicircular city of radius 3 km borders the ocean on the straight side. Find the average distance from points in the city to the ocean.

Solution

Think of the ocean as everything below the 𝑥-axis in the 𝑥𝑦-plane and think of the city as the upper half of the circular disk of radius 3 bounded by 𝑥2 + 𝑦2 = 9. (See Figure 16.18.) 𝑦

𝑥2 + 𝑦2 = 9

3 (𝑥,

√ 9 − 𝑥2 )

√ (− 9 − 𝑦2 , 𝑦)

√ ( 9 − 𝑦2 , 𝑦) 𝑥

−3

(𝑥, 0)

3

Figure 16.18: The city by the ocean showing a typical vertical strip and a typical horizontal strip

The distance from any point (𝑥, 𝑦) in the city to the ocean is the vertical distance to the 𝑥-axis, namely 𝑦. Thus, we want to compute Average distance =

1 𝑦 𝑑𝐴, Area(𝑅) ∫𝑅

where 𝑅 is the region between the upper half of the circle 𝑥2 + 𝑦2 = 9 and the 𝑥-axis. The area of 𝑅 is 𝜋32 ∕2 = 9𝜋∕2. To compute the integral, let’s take the inner integral with respect to 𝑦. A vertical strip goes from the 𝑥-axis, namely 𝑦 = 0, to √ the semicircle. The upper limit must be expressed in terms of 𝑥, so we 2 2 solve 𝑥 + 𝑦 = 9 to get 𝑦 = 9 − 𝑥2 . Since there is a strip for every 𝑥 from −3 to 3, the integral is: √



⎞ 3 ⎛ 2 |𝑦= 9−𝑥2 ⎞ ⎜ ⎟ 𝑑𝑥 ⎜𝑦 | ⎟ 𝑦 𝑑𝐴 = 𝑦 𝑑𝑦 𝑑𝑥 = ∫−3 ⎜∫0 ∫𝑅 ∫−3 ⎜ 2 ||𝑦=0 ⎟ ⎟ ⎝ ⎠ ⎝ ⎠ ( ) 3 3 𝑥3 || 1 1 1 (9 − 𝑥2 ) 𝑑𝑥 = 9𝑥 − = (18 − (−18)) = 18. = | ∫−3 2 2 3 |−3 2 3⎛

9−𝑥2

16.2 ITERATED INTEGRALS

853

Therefore, the average distance is 18∕(9𝜋∕2) = 4∕𝜋 = 1.273 km. What if we choose the inner integral with respect to 𝑥? Then we get the limits by looking at √ 2 2 2 horizontal strips, not vertical, and we √ solve 𝑥 + 𝑦 = 9 for 𝑥 in terms of 𝑦. We get 𝑥 = − 9 − 𝑦 at the left end of the strip and 𝑥 = 9 − 𝑦2 at the right. There is a strip for every 𝑦 from 0 to 3, so 3

∫𝑅

𝑦 𝑑𝐴 =

∫0

(



9−𝑦2

∫−√9−𝑦2

𝑦 𝑑𝑥

)

𝑑𝑦 =

∫0



2 3 √ |𝑥= 9−𝑦 ⎞ ⎟ 𝑑𝑦 = ⎜𝑦𝑥| √ 2𝑦 9 − 𝑦2 𝑑𝑦 | ∫ ⎜ |𝑥=− 9−𝑦2 ⎟ 0 ⎠ ⎝

3⎛

|3 2 2 = − (9 − 𝑦2 )3∕2 || = − (0 − 27) = 18. 3 3 |0

We get the same result as before. The average distance to the ocean is (2∕(9𝜋))18 = 4∕𝜋 = 1.273 km. In the examples so far, a region was given and the problem was to determine the limits for an iterated integral. Sometimes the limits are known and we want to determine the region. 6

2

√ 𝑥 𝑦3 + 1 𝑑𝑦 𝑑𝑥.

Example 6

Sketch the region of integration for the iterated integral

Solution

The inner integral is with respect to 𝑦, so we imagine the region built of vertical strips. The bottom of each strip is on the line 𝑦 = 𝑥∕3, and the top is on the horizontal line 𝑦 = 2. Since the limits of the outer integral are 0 and 6, the whole region is contained between the vertical lines 𝑥 = 0 and 𝑥 = 6. Notice that the lines 𝑦 = 2 and 𝑦 = 𝑥∕3 meet where 𝑥 = 6. See Figure 16.19. 𝑦

∫0 ∫𝑥∕3

𝑦=2 (6, 2)

2 𝑦 = 𝑥∕3

𝑥 6 Figure 16.19: The region of integration for Example 6, showing the vertical strip

Reversing the Order of Integration It is sometimes helpful to reverse the order of integration in an iterated integral. An integral which is difficult or impossible with the integration in one order can be quite straightforward in the other. The next example is such a case. 6

2

√ 𝑥 𝑦3 + 1 𝑑𝑦 𝑑𝑥 using the region sketched in Figure 16.19.

Example 7

Evaluate

Solution

√ Since 𝑦3 + 1 has no elementary antiderivative, we cannot calculate the inner integral symbolically. We try reversing the order of integration. From Figure 16.19, we see that horizontal strips go from 𝑥 = 0 to 𝑥 = 3𝑦 and that there is a strip for every 𝑦 from 0 to 2. Thus, when we change the order of integration we get

∫0 ∫𝑥∕3

6

2

∫0 ∫𝑥∕3

√ 𝑥 𝑦3 + 1 𝑑𝑦 𝑑𝑥 =

2

∫0 ∫0

3𝑦

√ 𝑥 𝑦3 + 1 𝑑𝑥 𝑑𝑦.

Now we can at least do the inner integral because we know the antiderivative of 𝑥. What about the

854

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

outer integral? 2

3𝑦

∫0 ∫0

√ 𝑥 𝑦3 + 1 𝑑𝑥 𝑑𝑦 =

∫0

2( 2√ 𝑥

𝑦3

2

) 𝑥=3𝑦 2 | 9𝑦2 3 (𝑦 + 1)1∕2 𝑑𝑦 + 1 || 𝑑𝑦 = ∫0 2 |𝑥=0

|2 = (𝑦3 + 1)3∕2 || = 27 − 1 = 26. |0

Thus, reversing the order of integration made the integral in the previous problem much easier. Notice that to reverse the order it is essential first to sketch the region over which the integration is being performed.

Exercises and Problems for Section 16.2 Online Resource: Additional Problems for Section 16.2 EXERCISES

In Exercises 1–4, sketch the region of integration. 𝜋

1.

𝑦 sin 𝑥 𝑑𝑦 𝑑𝑥

∫0 ∫0 2

3.

1

𝑥

2.

∫0 ∫𝑦2

𝑦2

1

𝑦2 𝑥 𝑑𝑥 𝑑𝑦

∫0 ∫0

𝑦

4.

𝑥𝑦 𝑑𝑥 𝑑𝑦

In Exercises 21–26, write ∫𝑅 𝑓 𝑑𝐴 as an iterated integral for the shaded region 𝑅. 21.

22.

𝑦

12

cos 𝜋𝑥

𝑦 𝑑𝑦 𝑑𝑥

∫0 ∫𝑥−2

1 𝑥

0 1

For Exercises 5–12, evaluate the integral. 5.

3

(4𝑥 + 3𝑦) 𝑑𝑥 𝑑𝑦

∫0 ∫0 3

7.

∫0 ∫0 1

9.

3

11.

1

𝑦𝑒𝑥𝑦 𝑑𝑥 𝑑𝑦

10.

12.

6 𝑥 1

−1

2

3

−2

𝑦

𝑥

𝑦 𝑑𝑥 𝑑𝑦

∫0 ∫0

𝑦

25.

sin 𝑥

3

5

2

3

𝑦

26.

3

𝑥 𝑑𝑦 𝑑𝑥

∫0

∫0

𝑦

24.

1

𝑥2 𝑦 𝑑𝑦 𝑑𝑥

𝜋∕2

sin 𝑥 𝑑𝑥 𝑑𝑦

4

𝑦

23.

2

2

𝑦

∫0 ∫0

(𝑥2 + 𝑦2 ) 𝑑𝑦 𝑑𝑥

∫0 ∫0

1

∫0 ∫0

3

2

2

8.

3

𝑥

∫0 ∫0

6𝑥𝑦 𝑑𝑦 𝑑𝑥

2

4

6.

4

𝑦

2

2

2 1 1

For Exercises 13–20, sketch the region of integration and evaluate the integral. 3

13.

∫1 ∫0 5

15.

2

𝑒𝑥+𝑦 𝑑𝑦 𝑑𝑥

14.

sin 𝑥 𝑑𝑦 𝑑𝑥

16.

3𝑦

𝑥𝑦 𝑑𝑥 𝑑𝑦

∫1 ∫𝑦

18.

∫0 ∫𝑥

2

2

19.

∫0 ∫0

2𝑥

3

𝑥𝑒𝑥 𝑑𝑦 𝑑𝑥

1

20.

∫0 ∫1

1

4

For Exercises 27–28, write ∫𝑅 𝑓 𝑑𝐴 as an iterated integral in two different ways for the shaded region 𝑅.

𝑥2 𝑦3 𝑑𝑥 𝑑𝑦

27.

𝑦



𝑥

30𝑥 𝑑𝑦 𝑑𝑥

𝑦

1+𝑥2

𝑥 √ 𝑑𝑦 𝑑𝑥 𝑦

𝑦

28.

2

2

1

1 𝑥

0 2

𝑥

0

3

𝑒𝑥 𝑑𝑦 𝑑𝑥

∫1 ∫√𝑦 1

1

𝑥

∫0 ∫0 4

2𝑥

∫1 ∫𝑥 2

17.

4

𝑥

0

1

2

𝑥

0 1

2

16.2 ITERATED INTEGRALS

855

For Exercises 29–33, evaluate the integral.

31. ∫𝑅 (5𝑥2 + 1) sin 3𝑦 𝑑𝐴, where 𝑅 is the rectangle −1 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 𝜋∕3.

√ 29. ∫𝑅 𝑥 + 𝑦 𝑑𝐴, where 𝑅 is the rectangle 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 2.

32. ∫𝑅 𝑥𝑦 𝑑𝐴, where 𝑅 is the triangle 𝑥+𝑦 ≤ 1, 𝑥 ≥ 0, 𝑦 ≥ 0.

30. The integral in Exercise 29 using the other order of integration.

33. ∫𝑅 (2𝑥 + 3𝑦)2 𝑑𝐴, where 𝑅 is the triangle with vertices at (−1, 0), (0, 1), and (1, 0).

PROBLEMS In Problems 34–37, integrate 𝑓 (𝑥, 𝑦) = 𝑥𝑦 over the region 𝑅. 𝑦 35. 34. 𝑦 2

1

47.

∫0 ∫𝑒𝑦

1

8

49. 𝑅 𝑅 𝑥

2

∫0 ∫ √ 3 𝑦

48.

1 𝑑𝑥 𝑑𝑦 1 + 𝑥4

50.

5

I. −1 𝑦

37. (1, 1)

1

V.

𝑅

𝑅 𝑥

𝑥

2

1

cos(𝑦2 ) 𝑑𝑦 𝑑𝑥

∫0 ∫𝑥 1

𝑥 2

𝑒2𝑦−𝑦 𝑑𝑦 𝑑𝑥

∫0 ∫0

10

5

𝑥𝑦2 𝑑𝑥 𝑑𝑦

∫0 ∫0 10

III. 36. 𝑦

1

𝑥 𝑑𝑥 𝑑𝑦 ln 𝑥

51. Each of the integrals (I)–(VI) takes one of two distinct values. Without evaluating, group them by value.

𝑥

1

𝑒

2

II.

∫0

5

10

10

𝑥𝑦2 𝑑𝑥 𝑑𝑦 IV.

5

∫0

∫0 5

𝑢𝑣2 𝑑𝑢 𝑑𝑣

∫0 ∫0

𝑥𝑦2 𝑑𝑦 𝑑𝑥

∫0 ∫0

5

∫0

10

VI.

∫0 ∫0

𝑥𝑦2 𝑑𝑦 𝑑𝑥

10

𝑢𝑣2 𝑑𝑣 𝑑𝑢

52. Find the volume under the graph of the function 𝑓 (𝑥, 𝑦) = 6𝑥2 𝑦 over the region shown in Figure 16.20. 𝑦

38. (a) Use four subrectangles to approximate the volume of the object whose base is the region 0 ≤ 𝑥 ≤ 4 and 0 ≤ 𝑦 ≤ 6, and whose height is given by 𝑓 (𝑥, 𝑦) = 𝑥𝑦. Find an overestimate and an underestimate and average the two. (b) Integrate to find the exact volume of the threedimensional object described in part (a).

8

For Problems 39–42, sketch the region of integration then rewrite the integral with the order of integration reversed.

0

3

39.

∫0 ∫0

𝑓 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥

𝑓 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥

∫−3 ∫0 2

42.

2−𝑦

𝑓 (𝑥, 𝑦) 𝑑𝑥 𝑑𝑦

∫0 ∫𝑦−2

1

1

∫0 ∫𝑦 1

45.

𝑥 1

2

𝑒𝑥 𝑑𝑥 𝑑𝑦

1√

∫0 ∫√𝑦

2

3

4

54. Compute the integral ∫ ∫𝑅

In Problems 43–50, evaluate the integral by reversing the order of integration. 43.

2

53. (a) Find the volume below the surface 𝑧 = 𝑥2 + 𝑦2 and above the 𝑥𝑦-plane for −1 ≤ 𝑥 ≤ 1,−1 ≤ 𝑦 ≤ 1. (b) Find the volume above the surface 𝑧 = 𝑥2 + 𝑦2 and below the plane 𝑧 = 2 for −1 ≤ 𝑥 ≤ 1, −1 ≤ 𝑦 ≤ 1.

9−𝑥2

3

41.

𝑓 (𝑥, 𝑦) 𝑑𝑥 𝑑𝑦

√ 4−𝑥2

2

40.

4

Figure 16.20

6

∫0 ∫2𝑦

6

2 + 𝑥3 𝑑𝑥 𝑑𝑦

1

1

44.

∫0 ∫𝑦 3

46.

sin (𝑥2 ) 𝑑𝑥 𝑑𝑦

9

∫0 ∫𝑦2

𝑦 sin(𝑥2 ) 𝑑𝑥 𝑑𝑦

(2𝑥2 + 𝑦) 𝑑𝐴,

where 𝑅 is the triangular region with vertices at (0, 1), (−2, 3) and (2, 3). 55. (a) Sketch the region in the 𝑥𝑦-plane bounded by the 𝑥-axis, 𝑦 = 𝑥, and 𝑥 + 𝑦 = 1. (b) Express the integral of 𝑓 (𝑥, 𝑦) over this region in terms of iterated integrals in two ways. (In one, use 𝑑𝑥 𝑑𝑦; in the other, use 𝑑𝑦 𝑑𝑥.) (c) Using one of your answers to part (b), evaluate the integral exactly with 𝑓 (𝑥, 𝑦) = 𝑥.

856

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES 2

56. Let 𝑓 (𝑥, 𝑦) = 𝑥2 𝑒𝑥 and let 𝑅 be the triangle bounded by the lines 𝑥 = 3, 𝑥 = 𝑦∕2, and 𝑦 = 𝑥 in the 𝑥𝑦-plane.

In Problems 62–63 set up, but do not evaluate, an iterated integral for the volume of the solid.

(a) Express ∫𝑅 𝑓 𝑑𝐴 as a double integral in two different ways. (b) Evaluate one of them.

62. Under the graph of 𝑓 (𝑥, 𝑦) = 25 − 𝑥2 − 𝑦2 and above the 𝑥𝑦-plane.

2

57. Find the average value of 𝑓 (𝑥, 𝑦) = 𝑥 + 4𝑦 on the rectangle 0 ≤ 𝑥 ≤ 3 and 0 ≤ 𝑦 ≤ 6. 58. Find the average value of 𝑓 (𝑥, 𝑦) = 𝑥𝑦2 on the rectangle 0 ≤ 𝑥 ≤ 4, 0 ≤ 𝑦 ≤ 3. 59. Figure 16.21 shows two metal plates carrying electrical charges. The charge density (in coulombs per square meter) of each at the point (𝑥, 𝑦) is 𝜎(𝑥, 𝑦) = 6𝑥 + 6 for 𝑥, 𝑦 in meters. (a) Without calculation, decide which plate carries a greater total charge, and explain your reasoning. (b) Find the total charge on both plates, and compare to your answer from part (a). 𝑦

64. A solid with flat base in the 𝑥𝑦-plane is bounded by the vertical planes 𝑦 = 0 and 𝑦 − 𝑥 = 4, and the slanted plane 2𝑥 + 𝑦 + 𝑧 = 4. (a) Draw the base of the solid. (b) Set up, but do not evaluate, an iterated integral for the volume of the solid. In Problems 65–69, find the volume of the solid region. 65. Under the graph of 𝑓 (𝑥, 𝑦) = 𝑥𝑦 and above the square 0 ≤ 𝑥 ≤ 2, 0 ≤ 𝑦 ≤ 2 in the 𝑥𝑦-plane. 66. Under the graph of 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2 and above the triangle 0 ≤ 𝑦 ≤ 𝑥, 0 ≤ 𝑥 ≤ 1.

𝑦

1

63. Below the graph of 𝑓 (𝑥, 𝑦) = 25 − 𝑥2 − 𝑦2 and above the plane 𝑧 = 16.

1

67. Under the graph of 𝑓 (𝑥, 𝑦) = 𝑥+𝑦 and above the region 𝑦2 ≤ 𝑥, 0 ≤ 𝑥 ≤ 9, 𝑦 ≥ 0. Plate 1

68. Under the graph of 2𝑥 + 𝑦 + 𝑧 = 4 in the first octant.

Plate 2

𝑥

𝑥

1

1

Figure 16.21 2

60. The population density in people per km for the trapezoid-shaped town in Figure 16.22 for 𝑥, 𝑦 in kilometers is 𝛿(𝑥, 𝑦) = 100𝑥 + 200𝑦. Find the town’s population.

69. The solid region 𝑅 bounded by the coordinate planes and the graph of 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 1. Assume 𝑎, 𝑏, and 𝑐 > 0. 70. If 𝑅 is the region 𝑥 + 𝑦 ≥ 𝑎, 𝑥2 + 𝑦2 ≤ 𝑎2 , with 𝑎 > 0, evaluate the integral ∫𝑅

𝑥𝑦 𝑑𝐴.

𝑦

71. The region 𝑊 lies below the surface 𝑓 (𝑥, 𝑦) = 2 2 2𝑒−(𝑥−1) −𝑦 and above the disk 𝑥2 + 𝑦2 ≤ 4 in the 𝑥𝑦plane.

6

3

𝑥 0

6

Figure 16.22 61. The quarter-disk-shaped metal plate in Figure 16.23 has radius 3 and density 𝜎(𝑥, 𝑦) = 2𝑦 gm/cm2 , with 𝑥, 𝑦 in cm. Find the mass of the plate.

(a) Describe in words the contours of 𝑓 , using 𝑓 (𝑥, 𝑦) = 1 as an example. (b) Write an integral giving the area of the crosssection of 𝑊 in the plane 𝑥 = 1. (c) Write an iterated double integral giving the volume of 𝑊 . 72. Find the average distance to the 𝑥-axis for points in the region in the first quadrant bounded by the 𝑥-axis and the graph of 𝑦 = 𝑥 − 𝑥2 . 73. Give the contour diagram of a function 𝑓 whose average value on the square 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1 is

𝑦 3

(a) Greater than the average of the values of 𝑓 at the four corners of the square. (b) Less than the average of the values of 𝑓 at the four corners of the square. 𝑥 3

Figure 16.23

74. The function 𝑓 (𝑥, 𝑦) = 𝑎𝑥 + 𝑏𝑦 has an average value of 20 on the rectangle 0 ≤ 𝑥 ≤ 2, 0 ≤ 𝑦 ≤ 3. (a) What can you say about the constants 𝑎 and 𝑏?

16.3 TRIPLE INTEGRALS

(b) Find two different choices for 𝑓 that have average value 20 on the rectangle, and give their contour diagrams on the rectangle. 75. The function 𝑓 (𝑥, 𝑦) = 𝑎𝑥2 + 𝑏𝑥𝑦 + 𝑐𝑦2 has an average value of 20 on the square 0 ≤ 𝑥 ≤ 2, 0 ≤ 𝑦 ≤ 2.

857

(a) What can you say about the constants 𝑎, 𝑏, and 𝑐? (b) Find two different choices for 𝑓 that have average value 20 on the square, and give their contour diagrams on the square.

Strengthen Your Understanding In Problems 76–77, explain what is wrong with the statement. 1

𝑥

1

𝑦

1

𝑦

76. ∫0 ∫0 𝑓 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥 = ∫0 ∫0 𝑓 (𝑥, 𝑦) 𝑑𝑥 𝑑𝑦 𝑦

1

77. ∫0 ∫0 𝑥𝑦 𝑑𝑥 𝑑𝑦 = ∫0 ∫0 𝑥𝑦 𝑑𝑦 𝑑𝑥 In Problems 78–80, give an example of:

83. The region of integration of the iterated integral 2

𝑥3

∫1 ∫𝑥2 𝑓 𝑑𝑦 𝑑𝑥 lies completely in the first quadrant (that is, 𝑥 ≥ 0, 𝑦 ≥ 0). 84. If the limits 𝑎, 𝑏, 𝑐 and 𝑑 in the iterated integral 𝑏 𝑑 ∫𝑎 ∫𝑐 𝑓 𝑑𝑦 𝑑𝑥 are all positive, then the value of 𝑏 𝑑 ∫𝑎 ∫𝑐 𝑓 𝑑𝑦 𝑑𝑥 is also positive. 𝑏

78. An iterated double integral, with limits of integration, giving the volume of a cylinder standing vertically with a circular base in the 𝑥𝑦-plane. 79. A nonconstant function, 𝑓 , whose integral is 4 over the triangular region with vertices (0, 0), (1, 0), (1, 1). 80. A double integral representing the volume of a triangular prism of base area 6. Are the statements in Problems 81–88 true or false? Give reasons for your answer. 1

86. If 𝑅 is the region inside a circle of√ radius 𝑎, centered at 𝑎

the origin, then ∫𝑅 𝑓 𝑑𝐴 = ∫−𝑎 ∫0

𝑎2 −𝑥2

𝑓 𝑑𝑦 𝑑𝑥.

87. If 𝑓 (𝑥, 𝑦) = 𝑔(𝑥) ⋅ ℎ(𝑦), where 𝑔 and ℎ are singlevariable functions, then ( 𝑏 ) ( 𝑑 ) 𝑏 𝑑 𝑓 𝑑𝑦 𝑑𝑥 = 𝑔(𝑥) 𝑑𝑥 ⋅ ℎ(𝑦) 𝑑𝑦 . ∫𝑎 ∫𝑐 ∫𝑎 ∫ 𝑐

12

81. The iterated integral ∫0 ∫5 𝑓 𝑑𝑥 𝑑𝑦 is computed over the rectangle 0 ≤ 𝑥 ≤ 1, 5 ≤ 𝑦 ≤ 12. 82. If 𝑅 is the region inside the triangle with vertices (0, 0), (1, 1) and (0, 2), then the double integral ∫𝑅 𝑓 𝑑𝐴 can be evaluated by an iterated integral of the form 2 1 ∫0 ∫0 𝑓 𝑑𝑥 𝑑𝑦.

16.3

1

85. If 𝑓 (𝑥, 𝑦) is a function of 𝑦 only, then ∫𝑎 ∫0 𝑓 𝑑𝑥 𝑑𝑦 = 𝑏 ∫𝑎 𝑓 𝑑𝑦.

88. If 𝑓 (𝑥, 𝑦) = 𝑔(𝑥) + ℎ(𝑦), where 𝑔 and ℎ are singlevariable functions, then ) ( 𝑏 ) ( 𝑑 𝑑 𝑏 ℎ(𝑦) 𝑑𝑦 . 𝑓 𝑑𝑥 𝑑𝑦 = 𝑔(𝑥) 𝑑𝑥 + ∫𝑎 ∫𝑐 ∫𝑎 ∫𝑐

TRIPLE INTEGRALS A continuous function of three variables can be integrated over a solid region 𝑊 in 3-space in the same way as a function of two variables is integrated over a flat region in 2-space. Again, we start with a Riemann sum. First we subdivide 𝑊 into smaller regions, then we multiply the volume of each region by a value of the function in that region, and then we add the results. For example, if 𝑊 is the box 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑐 ≤ 𝑦 ≤ 𝑑, 𝑝 ≤ 𝑧 ≤ 𝑞, then we subdivide each side into 𝑛, 𝑚, and 𝑙 pieces, thereby chopping 𝑊 into 𝑛𝑚𝑙 smaller boxes, as shown in Figure 16.24.

Figure 16.24: Subdividing a three-dimensional box

The volume of each smaller box is Δ𝑉 = Δ𝑥Δ𝑦Δ𝑧,

858

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

where Δ𝑥 = (𝑏 − 𝑎)∕𝑛, and Δ𝑦 = (𝑑 − 𝑐)∕𝑚, and Δ𝑧 = (𝑞 − 𝑝)∕𝑙. Using this subdivision, we pick a point (𝑢𝑖𝑗𝑘 , 𝑣𝑖𝑗𝑘 , 𝑤𝑖𝑗𝑘 ) in the 𝑖𝑗𝑘-th small box and construct a Riemann sum ∑ 𝑓 (𝑢𝑖𝑗𝑘 , 𝑣𝑖𝑗𝑘 , 𝑤𝑖𝑗𝑘 ) Δ𝑉 . 𝑖,𝑗,𝑘

If 𝑓 is continuous, as Δ𝑥, Δ𝑦, and Δ𝑧 approach 0, this Riemann sum approaches the definite integral, ∫𝑊

𝑓 𝑑𝑉 , called a triple integral, which is defined as

∫𝑊

𝑓 𝑑𝑉 =

lim

Δ𝑥,Δ𝑦,Δ𝑧→0



𝑓 (𝑢𝑖𝑗𝑘 , 𝑣𝑖𝑗𝑘 , 𝑤𝑖𝑗𝑘 ) Δ𝑥 Δ𝑦 Δ𝑧.

𝑖,𝑗,𝑘

As in the case of a double integral, we can evaluate this integral as an iterated integral: Triple integral as an iterated integral ) ) 𝑞( 𝑑( 𝑏 𝑓 𝑑𝑉 = 𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑥 𝑑𝑦 𝑑𝑧, ∫𝑝 ∫𝑊 ∫𝑐 ∫𝑎 where 𝑦 and 𝑧 are treated as constants in the innermost (𝑑𝑥) integral, and 𝑧 is treated as a constant in the middle (𝑑𝑦) integral. Other orders of integration are possible.

Example 1

A cube 𝐶 has sides of length 4 cm and is made of a material of variable density. If one corner is at the origin and the adjacent corners are on the positive 𝑥, 𝑦, and 𝑧 axes, then the density at the point (𝑥, 𝑦, 𝑧) is 𝛿(𝑥, 𝑦, 𝑧) = 1 + 𝑥𝑦𝑧 gm/cm3 . Find the mass of the cube.

Solution

Consider a small piece Δ𝑉 of the cube, small enough so that the density remains close to constant over the piece. Then Mass of small piece = Density ⋅ Volume ≈ 𝛿(𝑥, 𝑦, 𝑧) Δ𝑉 . To get the total mass, we add the masses of the small pieces and take the limit as Δ𝑉 → 0. Thus, the mass is the triple integral 4

𝑀=

∫𝐶

𝛿 𝑑𝑉 = 4

=

∫0 ∫0

4

∫0 ∫0 ∫0

4

4

4

(1 + 𝑥𝑦𝑧) 𝑑𝑥 𝑑𝑦 𝑑𝑧 =

4

(4 + 8𝑦𝑧) 𝑑𝑦 𝑑𝑧 =

∫0

4(

∫0 𝑦=4 )|

∫0

(

)|𝑥=4 1 𝑥 + 𝑥2 𝑦𝑧 || 𝑑𝑦 𝑑𝑧 2 |𝑥=0 4

(16 + 64𝑧) 𝑑𝑧 = 576 gm. 4𝑦 + 4𝑦2 𝑧 || 𝑑𝑧 = ∫0 |𝑦=0

Example 2

Express the volume of the building described in Example 1 on page 848 as a triple integral.

Solution

The building is given by 0 ≤ 𝑥 ≤ 8, 0 ≤ 𝑦 ≤ 16, and 0 ≤ 𝑧 ≤ 12 − 𝑥∕4 − 𝑦∕8. (See Figure 16.25.) To find its volume, divide it into small cubes of volume Δ𝑉 = Δ𝑥 Δ𝑦 Δ𝑧 and add. First, make a vertical stack of cubes above the point (𝑥, 𝑦, 0). This stack goes from 𝑧 = 0 to 𝑧 = 12 − 𝑥∕4 − 𝑦∕8, so

Volume of vertical stack ≈

∑ 𝑧

Δ𝑉 =

∑ 𝑧

Δ𝑥 Δ𝑦 Δ𝑧 =

( ∑ 𝑧

)

Δ𝑧

Δ𝑥 Δ𝑦.

859

16.3 TRIPLE INTEGRALS

Next, line up these stacks parallel to the 𝑦-axis to form a slice from 𝑦 = 0 to 𝑦 = 16. So ) ( ∑∑ Δ𝑧 Δ𝑦 Δ𝑥. Volume of slice ≈ 𝑦

𝑧

Finally, line up the slices along the 𝑥-axis from 𝑥 = 0 to 𝑥 = 8 and add up their volumes, to get ∑∑∑ Δ𝑧 Δ𝑦 Δ𝑥. Volume of building ≈ 𝑥

𝑦

𝑧

Thus, in the limit, 8

Volume of building =

16

∫0 ∫0

12−𝑥∕4−𝑦∕8

1 𝑑𝑧 𝑑𝑦 𝑑𝑥.

∫0 𝑧

𝑧 (0, 0, 12) (0, 0, 12) (𝑥, 𝑦, 12 −

1 𝑥 4





1 𝑦) 8

(0, 16, 10) (8, 16, 8)

(8, 0, 10) (8, 0, 8)

(8, 0, 10)



Δ𝑧





16

(𝑥, 𝑦, 0)

8

8



𝑥

Δ𝑥



(8, 0, 0)

𝑦



(0, 16, 0) 𝑦

16

(8, 16, 0)

Δ𝑦

𝑥



Figure 16.25: Volume of building (shown to left) divided into blocks and slabs for a triple integral

The preceding examples show that the triple integral has interpretations similar to the double integral: • If 𝜌(𝑥, 𝑦, 𝑧) is density, then •

∫𝑊

∫𝑊

𝜌 𝑑𝑉 is the total quantity in the solid region 𝑊 .

1 𝑑𝑉 is the volume of the solid region 𝑊 .



𝑥2 + 𝑦2 and

Example 3

Set up an iterated integral to compute the mass of the solid cone bounded by 𝑧 = 𝑧 = 3, if the density is given by 𝛿(𝑥, 𝑦, 𝑧) = 𝑧.

Solution

We break the cone in Figure 16.26 into small cubes of volume Δ𝑉 = Δ𝑥 Δ𝑦 Δ𝑧, on which the density is approximately constant, and approximate the mass of each cube by 𝛿(𝑥, 𝑦, 𝑧)√ Δ𝑥 Δ𝑦 Δ𝑧. Stacking the cubes vertically above the point (𝑥, 𝑦, 0), starting on the cone at height 𝑧 = 𝑥2 + 𝑦2 and going up to 𝑧 = 3, tells us that the inner integral is 3





3

𝛿(𝑥, 𝑦, 𝑧) 𝑑𝑧 = 𝑥2 +𝑦2



𝑧 𝑑𝑧. √

𝑥2 +𝑦2

860

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

√ There is a stack for every point in the 𝑥𝑦-plane in the shadow of the cone. The cone 𝑧 = 𝑥2 + 𝑦2 intersects the horizontal plane 𝑧 = 3 in the circle 𝑥2 + 𝑦2 = 9, so there is a stack for all (𝑥, √ 𝑦) in the 2 + 𝑦2 ≤ 9. Lining up the stacks parallel to the 𝑦-axis gives a slice from 𝑦 = − 9 − 𝑥2 to region 𝑥 √ 𝑦 = 9 − 𝑥2 , for each fixed value of 𝑥. Thus, the limits on the middle integral are √

∫−



9−𝑥2

9−𝑥2

3

∫√𝑥2 +𝑦2

𝑧 𝑑𝑧 𝑑𝑦.

Finally, there is a slice for each 𝑥 between −3 and 3, so the integral we want is 3

Mass =



9−𝑥2

∫−3 ∫−√9−𝑥2

3



𝑧 𝑑𝑧 𝑑𝑦 𝑑𝑥. √

𝑥2 +𝑦2

Notice that setting up the limits on the two outer integrals was just like setting up the limits for a double integral over the region 𝑥2 + 𝑦2 ≤ 9.

√ Figure 16.26: The cone 𝑧 = 𝑥2 + 𝑦2 with its shadow on the 𝑥𝑦-plane

As the previous example illustrates, for a region 𝑊 contained between two surfaces, the innermost limits correspond to these surfaces. The middle and outer limits ensure that we integrate over the “shadow” of 𝑊 in the 𝑥𝑦-plane.

Limits on Triple Integrals ∙ The limits for the outer integral are constants. ∙ The limits for the middle integral can involve only one variable (that in the outer integral). ∙ The limits for the inner integral can involve two variables (those on the two outer integrals).

Exercises and Problems for Section 16.3 Online Resource: Additional Problems for Section 16.3 EXERCISES

In Exercises 1–4, find the triple integrals of the function over the region 𝑊 .

3. 𝑓 (𝑥, 𝑦, 𝑧) = sin 𝑥 cos(𝑦 + 𝑧), 𝑊 is the cube 0 ≤ 𝑥 ≤ 𝜋, 0 ≤ 𝑦 ≤ 𝜋, 0 ≤ 𝑧 ≤ 𝜋.

1. 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 + 5𝑦2 − 𝑧, 𝑊 is the rectangular box 0 ≤ 𝑥 ≤ 2, −1 ≤ 𝑦 ≤ 1, 2 ≤ 𝑧 ≤ 3.

4. 𝑓 (𝑥, 𝑦, 𝑧) = 𝑒−𝑥−𝑦−𝑧 , 𝑊 is the rectangular box with corners at (0, 0, 0), (𝑎, 0, 0), (0, 𝑏, 0), and (0, 0, 𝑐).

2. ℎ(𝑥, 𝑦, 𝑧) = 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧, 𝑊 is the rectangular box 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1, 0 ≤ 𝑧 ≤ 2.

16.3 TRIPLE INTEGRALS

Sketch the region of integration in Exercises 5–13. 1

5.

√ 1−𝑧2

√ 1−𝑥2

∫−1 ∫−√1−𝑥2 ∫0 1

10.

1

∫−1 ∫0 ∫−√1−𝑧2 1

9.

1

√ 1−𝑥2

∫0 ∫−1 ∫−√1−𝑥2 1

8.

1

√ 1−𝑧2

∫0 ∫−1 ∫0 1

7.

√ 1−𝑥2

∫0 ∫−1 ∫0 1

6.

1

√ 1−𝑧2

1

11.

∫0 ∫0

𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑧 𝑑𝑥 𝑑𝑦

1

12. 𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑦 𝑑𝑧 𝑑𝑥 13.

√ 1−𝑥2 −𝑧2

∫0 ∫−√1−𝑧2 ∫0

∫−√1−𝑥2 −𝑦2 √ 1−𝑧2

√ 1−𝑦2 −𝑧2

√ 1−𝑧2

∫0 ∫0

√ 1−𝑥2 −𝑧2

∫−√1−𝑥2 −𝑧2

𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑧 𝑑𝑥 𝑑𝑦

𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑥 𝑑𝑦 𝑑𝑧

𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑦 𝑑𝑥 𝑑𝑧

In Exercises 14–15, for 𝑥, 𝑦 and 𝑧 in meters, what does the integral over the solid region 𝐸 represent? Give units.

𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑦 𝑑𝑧 𝑑𝑥

√ 1−𝑥2 −𝑧2

√ 1−𝑥2 −𝑦2

∫0 ∫−√1−𝑧2 ∫−√1−𝑦2 −𝑧2 1

𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑧 𝑑𝑥 𝑑𝑦

√ 1−𝑦2

861

𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑦 𝑑𝑧 𝑑𝑥

14.

𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑦 𝑑𝑥 𝑑𝑧

15.

∫𝐸 ∫𝐸

1 𝑑𝑉 𝛿(𝑥, 𝑦, 𝑧) 𝑑𝑉 , where 𝛿(𝑥, 𝑦, 𝑧) is density, in kg/m3 .

PROBLEMS In Problems 16–20, decide whether the integrals are positive, negative, or zero. Let 𝑆 be the solid sphere 𝑥2 + 𝑦2 + 𝑧2 ≤ 1, and 𝑇 be the top half of this sphere (with 𝑧 ≥ 0), and 𝐵 be the bottom half (with 𝑧 ≤ 0), and 𝑅 be the right half of the sphere (with 𝑥 ≥ 0), and 𝐿 be the left half (with 𝑥 ≤ 0). 16.

19.

∫𝑇 ∫𝑇

𝑒𝑧 𝑑𝑉

17.

∫𝐵

sin 𝑧 𝑑𝑉

𝑒𝑧 𝑑𝑉

20.

18.

∫𝑅

∫𝑆

sin 𝑧 𝑑𝑉

sin 𝑧 𝑑𝑉

√ Let 𝑊 be the solid cone bounded by 𝑧 = 𝑥2 + 𝑦2 and 𝑧 = 2. For Problems 21–29, decide (without calculating its value) whether the integral is positive, negative, or zero. 21. ∫𝑊 𝑦 𝑑𝑉

22. ∫𝑊 𝑥 𝑑𝑉

23. ∫𝑊 𝑧 𝑑𝑉

24. ∫𝑊 𝑥𝑦 𝑑𝑉

25. ∫𝑊 𝑥𝑦𝑧 𝑑𝑉 √ 27. ∫𝑊 𝑥2 + 𝑦2 𝑑𝑉 √ 29. ∫𝑊 (𝑧 − 𝑥2 + 𝑦2 ) 𝑑𝑉

26. ∫𝑊 (𝑧 − 2) 𝑑𝑉 28. ∫𝑊 𝑒−𝑥𝑦𝑧 𝑑𝑉

In Problems 30–34, let 𝑊 be the solid cylinder bounded by 𝑥2 + 𝑦2 = 1, 𝑧 = 0, and 𝑧 = 2. Decide (without calculating its value) whether the integral is positive, negative, or zero. 30. ∫𝑊 𝑥 𝑑𝑉

31. ∫𝑊 𝑧 𝑑𝑉

32. ∫𝑊 (𝑥2 + 𝑦2 − 2) 𝑑𝑉

33. ∫𝑊 (𝑧 − 1) 𝑑𝑉

36. Find the volume of the region bounded by 𝑧 = 𝑥2 , 0 ≤ 𝑥 ≤ 5, and the planes 𝑦 = 0, 𝑦 = 3, and 𝑧 = 0. 37. Find the volume of the region in the first octant bounded by the coordinate planes and the surface 𝑥 + 𝑦 + 𝑧 = 2. 38. A trough with triangular cross-section lies along the 𝑥axis for 0 ≤ 𝑥 ≤ 10. The slanted sides are given by 𝑧 = 𝑦 and 𝑧 = −𝑦 for 0 ≤ 𝑧 ≤ 1 and the ends by 𝑥 = 0 and 𝑥 = 10, where 𝑥, 𝑦, 𝑧 are in meters. The trough contains a sludge whose density at the point (𝑥, 𝑦, 𝑧) is 𝛿 = 𝑒−3𝑥 kg per m3 . (a) Express the total mass of sludge in the trough in terms of triple integrals. (b) Find the mass. 39. Find the volume of the region bounded by 𝑧 = 𝑥+𝑦, 𝑧 = 10, and the planes 𝑥 = 0, 𝑦 = 0. In Problems 40–45, write a triple integral, including limits of integration, that gives the specified volume. 40. Between 𝑧 = 𝑥 + 𝑦 and 𝑧 = 1 + 2𝑥 + 2𝑦 and above 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 2. 41. Between the paraboloid 𝑧 = 𝑥2 + 𝑦2 and the sphere 𝑥2 + 𝑦2 + 𝑧2 = 4 and above the disk 𝑥2 + 𝑦2 ≤ 1. 42. Between 2𝑥 + 2𝑦 + 𝑧 = 6 and 3𝑥 + 4𝑦 + 𝑧 = 6 and above 𝑥 + 𝑦 ≤ 1, 𝑥 ≥ 0, 𝑦 ≥ 0. 43. Under the sphere 𝑥2 + 𝑦2 + 𝑧2 = 9 and above the region between 𝑦 = 𝑥 and 𝑦 = 2𝑥 − 2 in the 𝑥𝑦-plane in the first quadrant.

34. ∫𝑊 𝑒−𝑦 𝑑𝑉

44. Between the top portion of the sphere 𝑥2 + 𝑦2 + 𝑧2 = 9 and the plane 𝑧 = 2.

35. Find the volume of the region bounded by the planes 𝑧 = 3𝑦, 𝑧 = 𝑦, 𝑦 = 1, 𝑥 = 1, and 𝑥 = 2.

45. Under the sphere 𝑥2 + 𝑦2 + 𝑧2 = 4 and above the region 𝑥2 + 𝑦2 ≤ 4, 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 2 in the 𝑥𝑦-plane.

862

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

In Problems 46–49, write limits of integration for the integral ∫𝑊 𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑉 where 𝑊 is the quarter or half sphere or cylinder shown. 𝑧 𝑟

46.

𝑧

47.

2 2

1 𝑥 𝑥

𝑦

𝑟

1 𝑧

48.

Figure 16.27

𝑧

49.

𝑟

𝑥

𝑦

𝑟

𝑟

58. (a) What is the equation of the plane passing through the points (1, 0, 0), (0, 1, 0), and (0, 0, 1)? (b) Find the volume of the region bounded by this plane and the planes 𝑥 = 0, 𝑦 = 0, and 𝑧 = 0.

𝑦 𝑟 𝑥

𝑟 𝑟 𝑦

50. Find the volume of the region between the plane 𝑧 = 𝑥 and the surface 𝑧 = 𝑥2 , and the planes 𝑦 = 0, and 𝑦 = 3. 51. Find the volume of the region bounded by 𝑧 = 𝑥 + 𝑦, 0 ≤ 𝑥 ≤ 5, 0 ≤ 𝑦 ≤ 5, and the planes 𝑥 = 0, 𝑦 = 0, and 𝑧 = 0.

Problems 59–61 refer to Figure 16.28, which shows triangular portions of the planes 2𝑥+4𝑦+𝑧 = 4, 3𝑥−2𝑦 = 0, 𝑧 = 2, and the three coordinate planes 𝑥 = 0, 𝑦 = 0, and 𝑧 = 0. For each solid region 𝐸, write down an iterated integral for the triple integral ∫𝐸 𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑉 .

52. Find the volume of the pyramid with base in the plane 𝑧 = −6 and sides formed by the three planes 𝑦 = 0 and 𝑦 − 𝑥 = 4 and 2𝑥 + 𝑦 + 𝑧 = 4. 53. Find the volume between the planes 𝑧 = 1 + 𝑥 + 𝑦 and 𝑥 + 𝑦 + 𝑧 = 1 and above the triangle 𝑥 + 𝑦 ≤ 1, 𝑥 ≥ 0, 𝑦 ≥ 0 in the 𝑥𝑦-plane. 54. Find the mass of a triangular-shaped solid bounded by the planes 𝑧 = 1 + 𝑥, 𝑧 = 1 − 𝑥, 𝑧 = 0, and with 0 ≤ 𝑦 ≤ 3. The density is 𝛿 = 10 − 𝑧 gm/cm3 , and 𝑥, 𝑦, 𝑧 are in cm. 55. Find the mass of the solid bounded by the 𝑥𝑦-plane, 𝑦𝑧plane, 𝑥𝑧-plane, and the plane (𝑥∕3)+(𝑦∕2)+(𝑧∕6) = 1, if the density of the solid is given by 𝛿(𝑥, 𝑦, 𝑧) = 𝑥 + 𝑦. 56. Find the mass of the pyramid with base in the plane 𝑧 = −6 and sides formed by the three planes 𝑦 = 0 and 𝑦 − 𝑥 = 4 and 2𝑥 + 𝑦 + 𝑧 = 4, if the density of the solid is given by 𝛿(𝑥, 𝑦, 𝑧) = 𝑦. 57. Let 𝐸 be the solid pyramid bounded by the planes 𝑥 + 𝑧 = 6, 𝑥 − 𝑧 = 0, 𝑦 + 𝑧 = 6, 𝑦 − 𝑧 = 0, and above the plane 𝑧 = 0 (see Figure 16.27). The density at any point in the pyramid is given by 𝛿(𝑥, 𝑦, 𝑧) = 𝑧 grams per cm3 , where 𝑥, 𝑦, and 𝑧 are measured in cm. (a) Explain in practical terms what the triple integral ∫𝐸 𝑧 𝑑𝑉 represents. (b) In evaluating the integral from part (a), how many separate triple integrals would be required if we chose to integrate in the 𝑧-direction first? (c) Evaluate the triple integral from part (a) by integrating in a well-chosen order.

Figure 16.28 59. 𝐸 is the region bounded by 𝑦 = 0, 𝑧 = 0, 3𝑥 − 2𝑦 = 0, and 2𝑥 + 4𝑦 + 𝑧 = 4. 60. 𝐸 is the region bounded by 𝑥 = 0, 𝑦 = 0, 𝑧 = 0, 𝑧 = 2, and 2𝑥 + 4𝑦 + 𝑧 = 4. 61. 𝐸 is the region bounded by 𝑥 = 0, 𝑧 = 0, 3𝑥 − 2𝑦 = 0, and 2𝑥 + 4𝑦 + 𝑧 = 4. 62. Figure 16.29 shows part of a spherical ball of radius 5 cm. Write an iterated triple integral which represents the volume of this region.

✻ 2 cm

❄ Figure 16.29

16.3 TRIPLE INTEGRALS

63. A solid region 𝐷 is a half cylinder of radius 1 lying horizontally with its rectangular base in the 𝑥𝑦-plane and its axis along the 𝑦-axis from 𝑦 = 0 to 𝑦 = 10. (The region is above the 𝑥𝑦-plane.) (a) What is the equation of the curved surface of this half cylinder? (b) Write the limits of integration of the integral ∫𝐷 𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑉 in Cartesian coordinates. 64. Set up, but do not evaluate, an iterated integral for the volume of the solid formed by the intersections of the cylinders 𝑥2 + 𝑧2 = 1 and 𝑦2 + 𝑧2 = 1.

68. 𝑑𝑧 𝑑𝑦 𝑑𝑥

69. 𝑑𝑧 𝑑𝑥 𝑑𝑦

70. 𝑑𝑦 𝑑𝑧 𝑑𝑥

71. 𝑑𝑦 𝑑𝑥 𝑑𝑧

863

Problems 72–74 refer to Figure 16.32, which shows 𝐸, the region in the first octant bounded by the plane 𝑥 + 𝑦 = 2 and the parabolic cylinder 𝑧 = 4 − 𝑥2 . For the given order of integration, write an iterated integral, or sum of integrals, equivalent to the triple integral ∫𝐸 𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑉 .

Problems 65–67 refer to Figure 16.30, which shows 𝐸, the region in the first octant bounded by the parabolic cylinder 𝑧 = 6𝑦2 and the elliptical cylinder 𝑥2 + 3𝑦2 = 12. For the given order of integration, write an iterated integral equivalent to the triple integral ∫𝐸 𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑉 .

Figure 16.32 72. 𝑑𝑧 𝑑𝑦 𝑑𝑥

Figure 16.30 65. 𝑑𝑧 𝑑𝑥 𝑑𝑦

66. 𝑑𝑥 𝑑𝑧 𝑑𝑦

67. 𝑑𝑦 𝑑𝑧 𝑑𝑥

Problems 68–71 refer to Figure 16.31, which shows 𝐸, the region in the first octant bounded by the planes 𝑧 = 5 and 5𝑥 + 3𝑧 = 15 and the elliptical cylinder 4𝑥2 + 9𝑦2 = 36. For the given order of integration, write an iterated integral equivalent to the triple integral ∫𝐸 𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑉 .

73. 𝑑𝑦 𝑑𝑧 𝑑𝑥

74. 𝑑𝑦 𝑑𝑥 𝑑𝑧

Problems 75–76 concern the center of mass, the point at which the mass of a solid body in motion can be considered to be concentrated. If the object has density 𝜌(𝑥, 𝑦, 𝑧) at the point (𝑥, 𝑦, 𝑧) and occupies a region 𝑊 , then the coordinates (𝑥, ̄ 𝑦, ̄ 𝑧) ̄ of the center of mass are given by

𝑥̄ =

1 𝑥𝜌 𝑑𝑉 𝑚 ∫𝑊

𝑦̄ =

1 𝑦𝜌 𝑑𝑉 𝑚 ∫𝑊

𝑧̄ =

1 𝑧𝜌 𝑑𝑉 𝑚 ∫𝑊

where 𝑚 = ∫𝑊 𝜌 𝑑𝑉 is the total mass of the body. 75. A solid is bounded below by the square 𝑧 = 0, 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1 and above by the surface 𝑧 = 𝑥 + 𝑦 + 1. Find the total mass and the coordinates of the center of mass if the density is 1 gm/cm3 and 𝑥, 𝑦, 𝑧 are measured in centimeters.

Figure 16.31

76. Find the center of mass of the tetrahedron that is bounded by the 𝑥𝑦, 𝑦𝑧, 𝑥𝑧 planes and the plane 𝑥 + 2𝑦 + 3𝑧 = 1. Assume the density is 1 gm/cm3 and 𝑥, 𝑦, 𝑧 are in centimeters.

Strengthen Your Understanding In Problems 77–78, explain what is wrong with the statement. 77. Let 𝑆 be the solid sphere 𝑥2 + 𝑦2 + 𝑧2 ≤ 1 and let 𝑈 be the upper half of 𝑆 where 𝑧 ≥ 0. Then ∫𝑆 𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑉 = 2 ∫𝑈 𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑉 .

In Problems 79–80, give an example of: 79. A function 𝑓 such that ∫𝑅 𝑓 𝑑𝑉 = 7, where 𝑅 is the cylinder 𝑥2 + 𝑦2 ≤ 4, 0 ≤ 𝑧 ≤ 3.

80. A nonconstant function 𝑓 (𝑥, 𝑦, 𝑧) such that if 𝐵 is the region enclosed by the sphere of radius 1 centered at the 1 𝑥 𝑦 1 1 𝑥 78. ∫0 ∫0 ∫0 𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑧 𝑑𝑦 𝑑𝑥 = ∫0 ∫𝑦 ∫0 𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑧 𝑑𝑥 𝑑𝑦 origin, the integral ∫𝐵 𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑥 𝑑𝑦 𝑑𝑧 is zero.

864

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES 1

Are the statements in Problems 81–90 true or false? Give reasons for your answer. 81. If 𝜌(𝑥, 𝑦, 𝑧) is mass density of a material in 3-space, then ∫𝑊 𝜌(𝑥, 𝑦, 𝑧) 𝑑𝑉 gives the volume of the solid region 𝑊 .

1

√ 1−𝑧 √ − 1−𝑧

∫0 ∫0 ∫

𝑓 𝑑𝑥 𝑑𝑦 𝑑𝑧 are equal.

86. If 𝑊 is a rectangular solid in 3-space, then ∫𝑊 𝑓 𝑑𝑉 = 𝑏 𝑑 𝑘 ∫𝑎 ∫𝑐 ∫𝑒 𝑓 𝑑𝑧 𝑑𝑦 𝑑𝑥, where 𝑎, 𝑏, 𝑐, 𝑑, 𝑒, and 𝑘 are constants.

82. The region of integration of the triple iterated inte1 1 𝑥 gral ∫0 ∫0 ∫0 𝑓 𝑑𝑧 𝑑𝑦 𝑑𝑥 lies above a square in the 𝑥𝑦plane and below a plane.

87. If 𝑊 is the unit cube 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1, 0 ≤ 𝑧 ≤ 1 and ∫𝑊 𝑓 𝑑𝑉 = 0, then 𝑓 = 0 everywhere in the unit cube.

83. If 𝑊 is the unit ball 𝑥2 + 𝑦2 + 𝑧√2 ≤ 1 then an iterated √

88. If 𝑓 > 𝑔 at all points in the solid region 𝑊 , then ∫𝑊 𝑓 𝑑𝑉 > ∫𝑊 𝑔 𝑑𝑉 .

integral over 𝑊 is

1 ∫0

84. The iterated integrals 1 ∫0

1−𝑧 ∫0

1−𝑦−𝑧 ∫0

∫0

1−𝑥2

1 ∫0

∫0

1−𝑥 ∫0

1−𝑥2 −𝑦2

1−𝑥−𝑦 ∫0

𝑓 𝑑𝑧 𝑑𝑦 𝑑𝑥 and

1

1

1−𝑥2

89. If 𝑊1 and 𝑊2 are solid regions with volume(𝑊1 ) > volume(𝑊2 ) then ∫𝑊 𝑓 𝑑𝑉 > ∫𝑊 𝑓 𝑑𝑉 . 1

𝑓 𝑑𝑥 𝑑𝑦 𝑑𝑧 are equal.

85. The iterated integrals ∫−1 ∫0 ∫0

16.4

𝑓 𝑑𝑧 𝑑𝑦 𝑑𝑥.

𝑓 𝑑𝑧 𝑑𝑦 𝑑𝑥 and

2

90. Both double and triple integrals can be used to compute volume.

DOUBLE INTEGRALS IN POLAR COORDINATES

Integration in Polar Coordinates We started this chapter by putting a rectangular grid on the fox population density map, to estimate the total population using a Riemann sum. However, sometimes a polar grid is more appropriate. Example 1

A biologist studying insect populations around a circular lake divides the area into the polar sectors of radii 2, 3, and 4 km in Figure 16.33. The approximate population density in each sector is shown in millions per square km. Estimate the total insect population around the lake. Shore of the lake

13 10

17 10

2

Lake

4

20

3

17

14 8

Figure 16.33: An insect-infested lake showing the insect population density by sector

Solution

To get the estimate, we multiply the population density in each sector by the area of that sector. Unlike the rectangles in a rectangular grid, the sectors in this grid do not all have the same area. The inner sectors have area 1 5𝜋 (𝜋32 − 𝜋22 ) = ≈ 3.93 km2 , 4 4 and the outer sectors have area 7𝜋 1 (𝜋42 − 𝜋32 ) = ≈ 5.50 km2 , 4 4 so we estimate Population ≈ (20)(3.93) + (17)(3.93) + (14)(3.93) + (17)(3.93) +(13)(5.50) + (10)(5.50) + (8)(5.50) + (10)(5.50) = 492.74 million insects.

16.4 DOUBLE INTEGRALS IN POLAR COORDINATES

865

What Is 𝒅𝑨 in Polar Coordinates? The previous example used a polar grid rather than a rectangular grid. A rectangular grid is constructed from vertical and horizontal lines of the form 𝑥 = 𝑘 (a constant) and 𝑦 = 𝑙 (another constant). In polar coordinates, 𝑟 = 𝑘 gives a circle of radius 𝑘 centered at the origin and 𝜃 = 𝑙 gives a ray emanating from the origin (at angle 𝑙 with the 𝑥-axis). A polar grid is built out of these circles and rays. Suppose we want to integrate 𝑓 (𝑟, 𝜃) over the region 𝑅 in Figure 16.34. 𝑦

𝑦

𝜃𝑛 = 𝛽 𝑟𝑚 = 𝑏 𝑅



Δ𝑟

𝑟0 = 𝑎







Δ𝐴



𝑟Δ𝜃 ❯

Δ𝜃



𝜃0 = 𝛼 𝑟



Arc of circle of radius 𝑟

𝜃

𝑥 Figure 16.34: Dividing up a region using a polar grid

𝑥

Figure 16.35: Calculating area Δ𝐴 in polar coordinates

Choosing (𝑟𝑖𝑗 , 𝜃𝑖𝑗 ) in the 𝑖𝑗-th bent rectangle in Figure 16.34 gives a Riemann sum: ∑ 𝑓 (𝑟𝑖𝑗 , 𝜃𝑖𝑗 ) Δ𝐴. 𝑖,𝑗

To calculate the area Δ𝐴, look at Figure 16.35. If Δ𝑟 and Δ𝜃 are small, the shaded region is approximately a rectangle with sides 𝑟 Δ𝜃 and Δ𝑟, so Δ 𝐴 ≈ 𝑟Δ𝜃Δ𝑟. Thus, the Riemann sum is approximately ∑

𝑓 (𝑟𝑖𝑗 , 𝜃𝑖𝑗 ) 𝑟𝑖𝑗 Δ𝜃 Δ𝑟.

𝑖,𝑗

If we take the limit as Δ𝑟 and Δ𝜃 approach 0, we obtain 𝛽

∫𝑅

𝑓 𝑑𝐴 =

𝑏

∫𝛼 ∫𝑎

𝑓 (𝑟, 𝜃) 𝑟 𝑑𝑟 𝑑𝜃.

When computing integrals in polar coordinates, use 𝑥 = 𝑟 cos 𝜃, 𝑦 = 𝑟 sin 𝜃, 𝑥2 + 𝑦2 = 𝑟2 . Put 𝑑𝐴 = 𝑟 𝑑𝑟 𝑑𝜃 or 𝑑𝐴 = 𝑟 𝑑𝜃 𝑑𝑟.

Example 2

Compute the integral of 𝑓 (𝑥, 𝑦) = 1∕(𝑥2 + 𝑦2 )3∕2 over the region 𝑅 shown in Figure 16.36. 𝑦

𝑅

𝜋 4

𝑥 1

2

Figure 16.36: Integrate 𝑓 over the polar region

866

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

Solution

The √ region 𝑅 is described by the inequalities 1 ≤ 𝑟 ≤ 2, 0 ≤ 𝜃 ≤ 𝜋∕4. In polar coordinates, 𝑟 = 𝑥2 + 𝑦2 , so we can write 𝑓 as 1 1 1 = = 3. (𝑥2 + 𝑦2 )3∕2 (𝑟2 )3∕2 𝑟

𝑓 (𝑥, 𝑦) = Then

2

𝜋∕4

∫𝑅

𝑓 𝑑𝐴 =

∫0

∫1 𝜋∕4

=

Example 3

∫0

1 𝑟 𝑑𝑟 𝑑𝜃 = ∫0 𝑟3

𝜋∕4

(

2

∫1

𝑟

−2

𝑑𝑟

)

𝑑𝜃

𝜋∕4 𝜋 1 1 |𝑟=2 𝑑𝜃 = . 𝑑𝜃 = − || ∫0 𝑟 |𝑟=1 2 8

For each region in Figure 16.37, decide whether to integrate using polar or Cartesian coordinates. On the basis of its shape, write an iterated integral of an arbitrary function 𝑓 (𝑥, 𝑦) over the region. 𝑦

(a)

(b)

2

𝑦

𝑦

(d)

3

2

2

1 𝑥 𝑥 1 −1

𝑦

(c)

3

−3

1

3

1 𝑥

3 −1

−3

2 𝑥 −2

−1

Figure 16.37

Solution

(a) Since this is a rectangular region, Cartesian coordinates are likely to be a better choice. The rectangle is described by the inequalities 1 ≤ 𝑥 ≤ 3 and −1 ≤ 𝑦 ≤ 2, so the integral is 2

3

∫−1 ∫1

𝑓 (𝑥, 𝑦) 𝑑𝑥 𝑑𝑦.

(b) A circle is best described in polar coordinates. The radius is 3, so 𝑟 goes from 0 to 3, and to describe the whole circle, 𝜃 goes from 0 to 2𝜋. The integral is 2𝜋

3

𝑓 (𝑟 cos 𝜃, 𝑟 sin 𝜃) 𝑟 𝑑𝑟 𝑑𝜃.

∫0

∫0

(c) The bottom boundary of this trapezoid is the line 𝑦 = (𝑥∕2) − 1 and the top is the line 𝑦 = 3, so we use Cartesian coordinates. If we integrate with respect to 𝑦 first, the lower limit of the integral is (𝑥∕2) − 1 and the upper limit is 3. The 𝑥 limits are 𝑥 = 0 to 𝑥 = 2. So the integral is 3

2

∫0 ∫(𝑥∕2)−1

𝑓 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥.

(d) This is another polar region: it is a piece of a ring in which 𝑟 goes from 1 to 2. Since it is in the second quadrant, 𝜃 goes from 𝜋∕2 to 𝜋. The integral is 𝜋

∫𝜋∕2 ∫1

2

𝑓 (𝑟 cos 𝜃, 𝑟 sin 𝜃) 𝑟 𝑑𝑟 𝑑𝜃.

16.4 DOUBLE INTEGRALS IN POLAR COORDINATES

867

Exercises and Problems for Section 16.4 Online Resource: Additional Problems for Section 16.4 EXERCISES

For the regions 𝑅 in Exercises 1–4, write ∫𝑅 𝑓 𝑑𝐴 as an iterated integral in polar coordinates. 𝑦

1.

𝑦

3.

𝑦

8. 𝑥

5

3

1 𝑥 2

Sketch the region of integration in Exercises 9–15. 𝑦

4

9.

2

−1

𝑓 (𝑟, 𝜃) 𝑟 𝑑𝑟 𝑑𝜃

∫𝜋∕2 ∫0 2𝜋

11.

𝑥 1

∫0

2

In Exercises 5–8, choose rectangular or polar coordinates to set up an iterated integral of an arbitrary function 𝑓 (𝑥, 𝑦) over the region. 𝑦 6. 5. 𝑦

1

𝑓 (𝑟, 𝜃) 𝑟 𝑑𝑟 𝑑𝜃

3𝜋∕2

∫3 ∫3𝜋∕4

𝑓 (𝑟, 𝜃) 𝑟 𝑑𝜃 𝑑𝑟

2∕ sin 𝜃

𝜋∕2

15.

𝑓 (𝑟, 𝜃) 𝑟 𝑑𝑟 𝑑𝜃 1∕ cos 𝜃

∫0

∫0 4

14.

5

4

∫𝜋∕6 ∫0 𝜋∕4

13.

𝑓 (𝑟, 𝜃) 𝑟 𝑑𝑟 𝑑𝜃

∫1 𝜋∕3

12.

𝑓 (𝑟, 𝜃) 𝑟 𝑑𝜃 𝑑𝑟

1

𝜋

10.

𝑥 −2

𝜋∕2

∫0 ∫−𝜋∕2

1

−1

4

𝑥 √ 2 √ − 2

4.

2

2

−4

√ − 2

0.5

𝑦

−2

√ 𝑦 2

2.

0.5

𝑥

7. −4 −2

𝑓 (𝑟, 𝜃) 𝑟 𝑑𝑟 𝑑𝜃

∫𝜋∕4 ∫0

𝑥

2

−5

5

𝑥 1

−5

5

PROBLEMS 𝑦

In Problems 16–18, evaluate the integral. √ 16. ∫𝑅 𝑥2 + 𝑦2 𝑑𝑥𝑑𝑦 where 𝑅 is 4 ≤ 𝑥2 + 𝑦2 ≤ 9.

✛ 𝑅1

18. ∫𝑅 (𝑥2 − 𝑦2 ) 𝑑𝐴, where 𝑅 is the first quadrant region between the circles of radius 1 and radius 2.

0

21.

∫−1 ∫−√1−𝑥2 √

∫0

𝑥 𝑑𝑦 𝑑𝑥

√ 4−𝑦2

2

∫𝑦

20.

∫0

𝑥

∫−𝑥

𝑑𝑦 𝑑𝑥

𝑥𝑦 𝑑𝑥 𝑑𝑦

Problems 22–26 concern Figure 16.38, which shows regions 𝑅1 , 𝑅2 , and 𝑅3 contained in the semicircle 𝑥2 + 𝑦2 = 4 with 𝑦 ≥ 0.

3𝑥 + 𝑦 = 0

−1

Convert the integrals in Problems 19–21 to polar coordinates and evaluate. 19.

𝑅2



𝑥 −2

√ 6



𝑅3

17. ∫𝑅 sin(𝑥2 + 𝑦2 ) 𝑑𝐴, where 𝑅 is the disk of radius 2 centered at the origin.

√ 1−𝑥2

𝑥 2 + 𝑦2 = 4

2

2

Figure 16.38 22. In Cartesian coordinates, write ∫𝑅 2𝑦 𝑑𝐴 as an iterated 1 integral in two different ways and then evaluate it. 23. In Cartesian coordinates, write ∫𝑅 2𝑦 𝑑𝐴 as an iterated 2 integral in two different ways. 24. Evaluate ∫𝑅 (𝑥2 + 𝑦2 ) 𝑑𝐴. 3

25. Evaluate ∫𝑅 12𝑦 𝑑𝐴, where 𝑅 is the region formed by combining the regions 𝑅1 and 𝑅2 .

868

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

26. Evaluate ∫𝑆 𝑥 𝑑𝐴, where 𝑆 is the region formed by combining the regions 𝑅2 and 𝑅3 . 3

1

27. Consider the integral ∫0 ∫𝑥∕3 𝑓 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥. (a) Sketch the region 𝑅 over which the integration is being performed. (b) Rewrite the integral with the order of integration reversed. (c) Rewrite the integral in polar coordinates. 𝜋∕2

4∕ sin 𝜃

28. Describe the region of integration for ∫𝜋∕4 ∫1∕ sin 𝜃 𝑓 (𝑟, 𝜃)𝑟 𝑑𝑟 𝑑𝜃.

(b) The population density decreases the farther you live from the shoreline of the bay; it also decreases the farther you live from the ocean. Which of the following functions best describes this situation? (i) 𝛿(𝑟, 𝜃) = (4 − 𝑟)(2 + cos 𝜃) (ii) 𝛿(𝑟, 𝜃) = (4 − 𝑟)(2 + sin 𝜃) (iii) 𝛿(𝑟, 𝜃) = (𝑟 + 4)(2 + cos 𝜃) (c) Estimate the population using your answers to parts (a) and (b). 𝑦 (km)

29. Evaluate the integral by converting it into Cartesian coordinates: ∫0

4

2∕ cos 𝜃

𝜋∕6

∫0

𝑟 𝑑𝑟 𝑑𝜃.

Ocean City

1

30. (a) Sketch the region of integration of 1

∫0

√ 4−𝑥2

∫√1−𝑥2

2

𝑥 𝑑𝑦 𝑑𝑥 +

∫1 ∫0

√ 4−𝑥2

𝑥 (km)

■ Bay

𝑥 𝑑𝑦 𝑑𝑥

(b) Evaluate the quantity in part (a).

Figure 16.39

31. Find the volume of the region between the graph of 𝑓 (𝑥, 𝑦) = 25 − 𝑥2 − 𝑦2 and the 𝑥𝑦 plane. 32. Find the volume of an √ ice cream cone bounded by the hemisphere 𝑧 = 8 − 𝑥2 − 𝑦2 and the cone 𝑧 = √ 𝑥2 + 𝑦2 .

33. (a) For 𝑎 > 0, find the volume under the graph of 2 2 𝑧 = 𝑒−(𝑥 +𝑦 ) above the disk 𝑥2 + 𝑦2 ≤ 𝑎2 . (b) What happens to the volume as 𝑎 → ∞?

34. A circular metal disk of radius 3 lies in the 𝑥𝑦-plane with its center at the origin. At a distance 𝑟 from the ori1 . gin, the density of the metal per unit area is 𝛿 = 2 𝑟 +1 (a) Write a double integral giving the total mass of the disk. Include limits of integration. (b) Evaluate the integral. 35. A city surrounds a bay as shown in Figure 16.39. The population density of the city (in thousands of people per square km) is 𝛿(𝑟, 𝜃), where 𝑟 and 𝜃 are polar coordinates and distances are in km. (a) Set up an iterated integral in polar coordinates giving the total population of the city.

36. A disk of radius 5 cm has density 10 gm/cm2 at its center and density 0 at its edge, and its density is a linear function of the distance from the center. Find the mass of the disk. 37. Electric charge is distributed over the 𝑥𝑦-plane, with density inversely proportional to the distance from the origin. Show that the total charge inside a circle of radius 𝑅 centered at the origin is proportional to 𝑅. What is the constant of proportionality? 38. (a) Graph 𝑟 = 1∕(2 cos 𝜃) for −𝜋∕2 ≤ 𝜃 ≤ 𝜋∕2 and 𝑟 = 1. (b) Write an iterated integral representing the area inside the curve 𝑟 = 1 and to the right of 𝑟 = 1∕(2 cos 𝜃). Evaluate the integral. 39. (a) Sketch the circles 𝑟 = 2 cos 𝜃 for −𝜋∕2 ≤ 𝜃 ≤ 𝜋∕2 and 𝑟 = 1. (b) Write an iterated integral representing the area inside the circle 𝑟 = 2 cos 𝜃 and outside the circle 𝑟 = 1. Evaluate the integral.

Strengthen Your Understanding In Problems 40–44, explain what is wrong with the statement. 40. If 𝑅 is the region bounded by 𝑥 = 1, 𝑦 = 0, 𝑦 = 𝑥, then 𝜋∕4 1 in polar coordinates ∫𝑅 𝑥 𝑑𝐴 = ∫0 ∫0 𝑟2 cos 𝜃 𝑑𝑟 𝑑𝜃. 41. If 𝑅 is the region 𝑥2 + 𝑦2 ≤ 4, then ∫𝑅 (𝑥2 + 𝑦2 ) 𝑑𝐴 = 2𝜋 2 ∫0 ∫0 𝑟2 𝑑𝑟 𝑑𝜃. √ 1 1 𝜋∕2 1 42. ∫0 ∫0 𝑥2 + 𝑦2 𝑑𝑦 𝑑𝑥 = ∫0 ∫0 𝑟2 𝑑𝑟 𝑑𝜃

2



1

𝜋

43. ∫1 ∫0

4−𝑥2

𝜋∕2

1 𝑑𝑦 𝑑𝑥 = ∫0 𝜋

2

∫1 𝑟 𝑑𝑟 𝑑𝜃

1

44. ∫0 ∫0 𝑟 𝑑𝑟 𝑑𝜃 = ∫0 ∫0 𝑟 𝑑𝑟 𝑑𝜃 In Problems 45–48, give an example of: 45. A region 𝑅 of integration in the first quadrant which suggests the use of polar coordinates. 46. An integrand 𝑓 (𝑥, 𝑦) that suggests the use of polar coordinates.

869

16.5 INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES

47. A function 𝑓 (𝑥, 𝑦) such that ∫𝑅 𝑓 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥 in polar coordinates has an integrand without a factor of 𝑟.

reasons for your answer.

48. A region 𝑅 such that ∫𝑅 𝑓 (𝑥, 𝑦) 𝑑𝐴 must be broken into two integrals in Cartesian coordinates, but only needs one integral in polar coordinates.

50. The integral ∫0 ∫0 𝑑𝑟 𝑑𝜃 gives the area of the unit circle.

2𝜋

7

1

(a)

∫−1 ∫−√1−𝑥2 2𝜋

(c)

𝑟 𝑑𝑟 𝑑𝜃

∫0 ∫0

∫−1 ∫−√1−𝑥2 2𝜋

(d)

2𝜋

1

(e)

1

𝑑𝑦 𝑑𝑥 (b)

1

∫0

∫0

√ 1−𝑥2

1

𝑟 𝑑𝜃 𝑑𝑟

(f)

52. Let 𝑅 be the region inside the semicircle 𝑥2 + 𝑦2 = 9 𝜋 3 with 𝑦 ≥ 0. Then ∫𝑅 (𝑥 + 𝑦) 𝑑𝐴 = ∫0 ∫0 𝑟 𝑑𝑟 𝑑𝜃

𝑥 𝑑𝑦 𝑑𝑥

𝜋

∫0 ∫0

𝜋∕2

1

𝑑𝑟 𝑑𝜃 𝑑𝜃 𝑑𝑟

𝑟 𝑑𝑟 𝑑𝜃 gives the area of the 54. The integral ∫0 ∫0 region 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 𝑥. 2𝜋

∫0 𝑟2 cos 𝜃 𝑑𝑟 𝑑𝜃

1∕ cos 𝜃

𝜋∕4

2𝜋

1

55. The integral ∫0 ∫0 𝑟3 𝑑𝑟 𝑑𝜃 gives the area of the unit circle.

Are the statements in Problems 50–55 true or false? Give

16.5

1

53. The integrals ∫0 ∫0 𝑟2 cos 𝜃 𝑑𝑟 𝑑𝜃 and 2 ∫0 are equal.

1

∫0

∫0

𝜋∕4

51. The quantity 8 ∫5 ∫0 𝑟 𝑑𝜃 𝑑𝑟 gives the area of a ring with radius between 5 and 7.

49. Which of the following integrals give the area of the unit circle? √ 1−𝑥2

1

INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES Some double integrals are easier to evaluate in polar, rather than Cartesian, coordinates. Similarly, some triple integrals are easier in non-Cartesian coordinates.

Cylindrical Coordinates The cylindrical coordinates of a point (𝑥, 𝑦, 𝑧) in 3-space are obtained by representing the 𝑥 and 𝑦 coordinates in polar coordinates and letting the 𝑧-coordinate be the 𝑧-coordinate of the Cartesian coordinate system. (See Figure 16.40.)

Relation Between Cartesian and Cylindrical Coordinates Each point in 3-space is represented using 0 ≤ 𝑟 < ∞, 0 ≤ 𝜃 ≤ 2𝜋, −∞ < 𝑧 < ∞. 𝑥 = 𝑟 cos 𝜃, 𝑦 = 𝑟 sin 𝜃, 𝑧 = 𝑧. As with polar coordinates in the plane, note that 𝑥2 + 𝑦2 = 𝑟2 .

𝑧 𝑃 = (𝑟, 𝜃, 𝑧) 𝑧 𝜃

𝑥

𝑦 𝑟

(𝑟, 𝜃, 0) Figure 16.40: Cylindrical coordinates: (𝑟, 𝜃, 𝑧)

870

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

A useful way to visualize cylindrical coordinates is to sketch the surfaces obtained by setting one of the coordinates equal to a constant. See Figures 16.41–16.43. 𝑟=1

𝑧

✠ 𝑟=2



𝑥

𝑦

Figure 16.41: The surfaces 𝑟 = 1 and 𝑟=2

Figure 16.42: The surfaces 𝜃 = 𝜋∕4 and 𝜃 = 3𝜋∕4

Figure 16.43: The surfaces 𝑧 = −1 and 𝑧=3

Setting 𝑟 = 𝑐 (where 𝑐 is constant) gives a cylinder around the 𝑧-axis whose radius is 𝑐. Setting 𝜃 = 𝑐 gives a half-plane perpendicular to the 𝑥𝑦 plane, with one edge along the 𝑧-axis, making an angle 𝑐 with the 𝑥-axis. Setting 𝑧 = 𝑐 gives a horizontal plane |𝑐| units from the 𝑥𝑦-plane. We call these fundamental surfaces. The regions that can most easily be described in cylindrical coordinates are those regions whose boundaries are such fundamental surfaces. (For example, vertical cylinders, or wedge-shaped parts of vertical cylinders.) Example 1

Describe in cylindrical coordinates a wedge of cheese cut from a cylinder 4 cm high and 6 cm in radius; this wedge subtends an angle of 𝜋∕6 at the center. (See Figure 16.44.)

Solution

The wedge is described by the inequalities 0 ≤ 𝑟 ≤ 6, and 0 ≤ 𝑧 ≤ 4, and 0 ≤ 𝜃 ≤ 𝜋∕6. 𝑧





6 cm



𝜋 6

✛ 4 cm



𝑦 𝑥 Figure 16.44: A wedge of cheese

Integration in Cylindrical Coordinates To integrate a double integral ∫𝑅 𝑓 𝑑𝐴 in polar coordinates, we had to express the area element 𝑑𝐴 in terms of polar coordinates: 𝑑𝐴 = 𝑟 𝑑𝑟 𝑑𝜃. To evaluate a triple integral ∫𝑊 𝑓 𝑑𝑉 in cylindrical coordinates, we need to express the volume element 𝑑𝑉 in cylindrical coordinates. In Figure 16.45, consider the volume element Δ𝑉 bounded by fundamental surfaces. The area of the base is Δ𝐴 ≈ 𝑟 Δ𝑟 Δ𝜃. Since the height is Δ𝑧, the volume element is given approximately by Δ𝑉 ≈ 𝑟 Δ𝑟 Δ𝜃 Δ𝑧.

16.5 INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES

871

When computing integrals in cylindrical coordinates, put 𝑑𝑉 = 𝑟 𝑑𝑟 𝑑𝜃 𝑑𝑧. Other orders of integration are also possible. 𝑧

𝑟Δ𝜃 ✛ ✛ Δ𝑟 ✻ ✛ Δ𝑧







Δ𝑉

✻ 𝑧



𝑦



𝑟



Δ𝜃

𝑥 Figure 16.45: Volume element in cylindrical coordinates

Example 2

Find the mass of the wedge of cheese in Example 1, if its density is 1.2 grams/cm3 .

Solution

If the wedge is 𝑊 , its mass is ∫𝑊

1.2 𝑑𝑉 .

In cylindrical coordinates this integral is 4

6

𝜋∕6

∫0 ∫0

∫0

4

𝜋∕6

1.2 𝑟 𝑑𝑟 𝑑𝜃 𝑑𝑧 =

4 𝜋∕6 |6 0.6𝑟2 || 𝑑𝜃 𝑑𝑧 = 21.6 𝑑𝜃 𝑑𝑧 ∫0 ∫0 |0

∫0 ∫0 ( ) 𝜋 4 = 45.239 grams. = 21.6 6

Example 3

A water tank in the shape of a hemisphere has radius 𝑎; its base is its plane face. Find the volume, 𝑉 , of water in the tank as a function of ℎ, the depth of the water.

Solution

In Cartesian coordinates, a sphere of radius 𝑎 has the equation 𝑥2 + 𝑦2 + 𝑧2 = 𝑎2 . (See Figure 16.46.) In cylindrical coordinates, 𝑟2 = 𝑥2 + 𝑦2 , so this becomes 𝑟2 + 𝑧2 = 𝑎2 . Thus, if we √want to describe the amount of water in the tank in cylindrical coordinates, we let 𝑟 go from 0 to 𝑎2 − 𝑧2 , we let 𝜃 go from 0 to 2𝜋, and we let 𝑧 go from 0 to ℎ, giving Volume of water

2𝜋

=

∫𝑊

1 𝑑𝑉 =



√ 𝑎2 −𝑧2

2𝜋

𝑟 𝑑𝑟 𝑑𝑧 𝑑𝜃 =

√ ℎ 2 |𝑟= 𝑎2 −𝑧2 𝑟

| 𝑑𝑧 𝑑𝜃 ∫0 2 ||𝑟=0 ( ) 𝑧=ℎ 2𝜋 2𝜋 ℎ 1 1 2 𝑧3 || 2 2 = (𝑎 − 𝑧 ) 𝑑𝑧 𝑑𝜃 = 𝑎 𝑧− 𝑑𝜃 ∫0 ∫0 2 ∫0 2 3 ||𝑧=0 ( ) ( ) 2𝜋 1 ℎ3 ℎ3 𝑎2 ℎ − 𝑑𝜃 = 𝜋 𝑎2 ℎ − . = ∫0 2 3 3 ∫0

∫0 ∫0

∫0

872

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

𝑧 𝑟2 + 𝑧2 = 𝑎2

𝑟







✻ ✛

𝑥 Figure 16.46: Hemispherical water tank with radius 𝑎 and water of depth ℎ

Spherical Coordinates In Figure 16.47, the point 𝑃 has coordinates (𝑥, 𝑦, 𝑧) in√ the Cartesian coordinate system. We define spherical coordinates 𝜌, 𝜙, and 𝜃 for 𝑃 as follows: 𝜌 = 𝑥2 + 𝑦2 + 𝑧2 is the distance of 𝑃 from the origin; 𝜙 is the angle between the positive 𝑧-axis and the line through the origin and the point 𝑃 ; and 𝜃 is the same as in cylindrical coordinates.

Figure 16.47: Spherical coordinates: (𝜌, 𝜙, 𝜃)

In cylindrical coordinates, 𝑥 = 𝑟 cos 𝜃,

and 𝑦 = 𝑟 sin 𝜃,

and 𝑧 = 𝑧.

From Figure 16.47 we have 𝑧 = 𝜌 cos 𝜙 and 𝑟 = 𝜌 sin 𝜙, giving the following relationship:

Relation Between Cartesian and Spherical Coordinates Each point in 3-space is represented using 0 ≤ 𝜌 < ∞, 0 ≤ 𝜙 ≤ 𝜋, and 0 ≤ 𝜃 ≤ 2𝜋. 𝑥 = 𝜌 sin 𝜙 cos 𝜃 𝑦 = 𝜌 sin 𝜙 sin 𝜃 𝑧 = 𝜌 cos 𝜙. Also, 𝜌2 = 𝑥2 + 𝑦2 + 𝑧2 . This system of coordinates is useful when there is spherical symmetry with respect to the origin, either in the region of integration or in the integrand. The fundamental surfaces in spherical coordinates are 𝜌 = 𝑘 (a constant), which is a sphere of radius 𝑘 centered at the origin, 𝜃 = 𝑘 (a constant), which is the half-plane with its edge along the 𝑧-axis, and 𝜙 = 𝑘 (a constant), which is a cone if 𝑘 ≠ 𝜋∕2 and the 𝑥𝑦-plane if 𝑘 = 𝜋∕2. (See Figures 16.48–16.50.)

16.5 INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES

Figure 16.49: The surfaces 𝜃 = 𝜋∕4 and 𝜃 = 3𝜋∕4

Figure 16.48: The surfaces 𝜌 = 1 and 𝜌 = 2

873

Figure 16.50: The surfaces 𝜙 = 𝜋∕6 and 𝜙 = 2𝜋∕3

Integration in Spherical Coordinates To use spherical coordinates in triple integrals we need to express the volume element, 𝑑𝑉 , in spherical coordinates. From Figure 16.51, we see that the volume element can be approximated by a box with curved edges. One edge has length Δ𝜌. The edge parallel to the 𝑥𝑦-plane is an arc of a circle made from rotating the cylindrical radius 𝑟 (= 𝜌 sin 𝜙) through an angle Δ𝜃, and so has length 𝜌 sin 𝜙 Δ𝜃. The remaining edge comes from rotating the radius 𝜌 through an angle Δ𝜙, and so has length 𝜌 Δ𝜙. Therefore, Δ𝑉 ≈ Δ𝜌(𝜌 Δ𝜙)(𝜌 sin 𝜙 Δ𝜃) = 𝜌2 sin 𝜙 Δ𝜌 Δ𝜙 Δ𝜃. 𝜌 sin 𝜙 Δ𝜃

𝑧

❄ ✠

𝜌 Δ𝜙

Δ𝜃

✛ ❘ 𝜌 sin 𝜙







Δ𝜌

𝜌



𝜙

𝑦 Δ𝜃

𝜃 𝑥

Figure 16.51: Volume element in spherical coordinates

Thus:

When computing integrals in spherical coordinates, put 𝑑𝑉 = 𝜌2 sin 𝜙 𝑑𝜌 𝑑𝜙 𝑑𝜃. Other orders of integration are also possible.

Example 4

Use spherical coordinates to derive the formula for the volume of a ball of radius 𝑎.

Solution

In spherical coordinates, a ball of radius 𝑎 is described by the inequalities 0 ≤ 𝜌 ≤ 𝑎, 0 ≤ 𝜃 ≤ 2𝜋, and 0 ≤ 𝜙 ≤ 𝜋. Note that 𝜃 goes from 0 to 2𝜋, whereas 𝜙 goes from 0 to 𝜋. We find the volume by integrating the constant density function 1 over the ball: 2𝜋

Volume =

∫𝑅

1 𝑑𝑉 = =

∫0

𝜋

∫0 ∫0

1 3 𝑎 3 ∫0

2𝜋

2𝜋

𝑎

𝜌2 sin 𝜙 𝑑𝜌 𝑑𝜙 𝑑𝜃 =

∫0

𝜋

1 3 𝑎 sin 𝜙 𝑑𝜙 𝑑𝜃 ∫0 3

2𝜋 |𝜋 2 4𝜋𝑎3 . − cos 𝜙|| 𝑑𝜃 = 𝑎3 𝑑𝜃 = 3 ∫0 3 |0

874

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

Example 5

Find the magnitude of the gravitational force exerted by a solid hemisphere of radius 𝑎 and constant density 𝛿 on a unit mass located at the center of the base of the hemisphere.

Solution

Assume the base of the hemisphere rests on the 𝑥𝑦-plane with center at the origin. (See Figure 16.52.) Newton’s law of gravitation says that the force between two masses 𝑚1 and 𝑚2 at a distance 𝑟 apart is 𝐹 = 𝐺𝑚1 𝑚2 ∕𝑟2 , where 𝐺 is the gravitational constant. In this example, symmetry shows that the net component of the force on the particle at the origin due to the hemisphere is in the 𝑧 direction only. Any force in the 𝑥 or 𝑦 direction from some part of the hemisphere is canceled by the force from another part of the hemisphere directly opposite the first. To compute the net 𝑧-component of the gravitational force, we imagine a small piece of the hemisphere with volume Δ𝑉 , located at spherical coordinates (𝜌, 𝜃, 𝜙). This piece has mass 𝛿Δ𝑉 and exerts a force of magnitude 𝐹 on the unit mass at the origin. The 𝑧-component of this force is given by its projection onto the 𝑧-axis, which can be seen from the figure to be 𝐹 cos 𝜙. The distance from the mass 𝛿Δ𝑉 to the unit mass at the origin is the spherical coordinate 𝜌. Therefore, the 𝑧-component of the force due to the small piece Δ𝑉 is 𝑧-component of force

=

𝐺(𝛿Δ𝑉 )(1) cos 𝜙. 𝜌2

Adding the contributions of the small pieces, we get a vertical force with magnitude 2𝜋

𝐹 =

∫0

𝜋∕2

∫0 2𝜋

=

2𝜋

=

∫0

𝐺𝛿 𝜌2

)

(cos 𝜙)𝜌2 sin 𝜙 𝑑𝜌 𝑑𝜙 𝑑𝜃 = 2𝜋

𝜋∕2

∫0

∫0

∫0

𝑎(

𝐺𝛿𝑎 cos 𝜙 sin 𝜙 𝑑𝜙 𝑑𝜃 =

∫0

( ) 1 𝑑𝜃 = 𝐺𝛿𝑎𝜋. 𝐺𝛿𝑎 2

2𝜋

𝜋∕2

|𝜌=𝑎 𝐺𝛿(cos 𝜙 sin 𝜙)𝜌|| 𝑑𝜙𝑑𝜃 |𝜌=0

∫0 ∫0 ) 𝜙=𝜋∕2 ( (cos 𝜙)2 || 𝑑𝜃 𝐺𝛿𝑎 − | 2 |𝜙=0

The integral in this example is improper because the region of integration contains the origin, where the force is undefined. However, it can be shown that the result is nevertheless correct. 𝑧



Δ𝑉

✲𝜙 ✛

of force

✒✛ Unit mass

Force, 𝐹 , due to mass 𝛿𝑑𝑉

𝑦 𝑎



𝑧-component

𝑥

Figure 16.52: Gravitational force of hemisphere on mass at origin

Exercises and Problems for Section 16.5 Online Resource: Additional Problems for Section 16.5 EXERCISES

1. Match the equations in (a)–(f) with one of the surfaces in (I)–(VII). (a) 𝑥 = 5 (d) 𝑧 = 1

(b) 𝑥2 + 𝑧2 = 7 (c) 𝜌 = 5 (e) 𝑟 = 3 (f) 𝜃 = 2𝜋

(I) (II) (III) (IV) (V) (VI) (VII)

Cylinder, centered on 𝑥-axis. Cylinder, centered on 𝑦-axis. Cylinder, centered on 𝑧-axis. Plane, perpendicular to the 𝑥-axis. Plane, perpendicular to the 𝑦-axis. Plane, perpendicular to the 𝑧-axis. Sphere.

16.5 INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES

In Exercises 2–7, find an equation for the surface.

14.

875

15.

2. The vertical plane 𝑦 = 𝑥 in cylindrical coordinates. 3. The top half of the sphere 𝑥2 +𝑦2 +𝑧2 = 1 in cylindrical coordinates. √ 4. The cone 𝑧 = 𝑥2 + 𝑦2 in cylindrical coordinates. √ 5. The cone 𝑧 = 𝑥2 + 𝑦2 in spherical coordinates. 6. The plane 𝑧 = 10 in spherical coordinates. 7. The plane 𝑧 = 4 in spherical coordinates. In Exercises 8–9, evaluate the triple integrals in cylindrical coordinates over the region 𝑊 . 2

16. A piece of a sphere; angle at the center is 𝜋∕3.

2

8. 𝑓 (𝑥, 𝑦, 𝑧) = sin(𝑥 + 𝑦 ), 𝑊 is the solid cylinder with height 4 and with base of radius 1 centered on the 𝑧 axis at 𝑧 = −1. 9. 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2 + 𝑧2 , 𝑊 is the region 0 ≤ 𝑟 ≤ 4, 𝜋∕4 ≤ 𝜃 ≤ 3𝜋∕4, −1 ≤ 𝑧 ≤ 1. In Exercises 10–11, evaluate the triple integrals in spherical coordinates. 10. 𝑓 (𝜌, 𝜃, 𝜙) = sin 𝜙, over the region 0 ≤ 𝜃 ≤ 2𝜋, 0 ≤ 𝜙 ≤ 𝜋∕4, 1 ≤ 𝜌 ≤ 2. 11. 𝑓 (𝑥, 𝑦, 𝑧) = 1∕(𝑥2 + 𝑦2 + 𝑧2 )1∕2 over the bottom half of the sphere of radius 5 centered at the origin.

17.

For Exercises 12–18, choose coordinates and set up a triple integral, including limits of integration, for a density function 𝑓 over the region. 12. 13. 18.

PROBLEMS In Problems 19–21, if 𝑊 is the region in Figure 16.53, what are the limits of integration?

?

19.

𝑧 4

21.

𝑦 𝑥

Figure 16.53: Cone with flat top, symmetric about 𝑧-axis

𝑓 (𝑟, 𝜃, 𝑧)𝑟 𝑑𝑧 𝑑𝑟 𝑑𝜃

?

?

𝑔(𝜌, 𝜙, 𝜃)𝜌2 sin 𝜙 𝑑𝜌 𝑑𝜙 𝑑𝜃

∫? ∫ ? ∫ ? ?

(2, 0, 4)

?

∫? ∫? ∫? ?

20.

?

?

∫? ∫? ∫?

?

ℎ(𝑥, 𝑦, 𝑧) 𝑑𝑧 𝑑𝑦 𝑑𝑥

22. Write a triple integral in cylindrical coordinates giving the volume of a sphere of radius 𝐾 centered at the origin. Use the order 𝑑𝑧 𝑑𝑟 𝑑𝜃. 23. Write a triple integral in spherical coordinates giving the volume of a sphere of radius 𝐾 centered at the origin. Use the order 𝑑𝜃 𝑑𝜌 𝑑𝜙.

876

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

In Problems 24–26, for the regions 𝑊 shown, write the limits of integration for ∫𝑊 𝑑𝑉 in the following coordinates: (a)

(b) Cylindrical

Cartesian

(c) Spherical

In Problems 29–34, write a triple integral including limits of integration that gives the specified volume. 29. Under 𝜌 = 3 and above 𝜙 = 𝜋∕3. 30. Under 𝜌 = 3 and above 𝑧 = 𝑟.

24.

31. The region between 𝑧 = 5 and 𝑧 = 10, with 2 ≤ 𝑥2 + 𝑦2 ≤ 3 and 0 ≤ 𝜃 ≤ 𝜋. √ 32. Between the cone 𝑧 = 𝑥2 + 𝑦2 and the first quadrant of the 𝑥𝑦-plane, with 𝑥2 + 𝑦2 ≤ 7.

𝑧 1

𝑦

1

33. The cap of the solid sphere 𝑥2 + 𝑦2 + 𝑧2 ≤ 10 cut off by the plane 𝑧 = 1.

𝑥 One-eighth sphere

−1

25.

35. (a) Write an integral (including limits of integration) representing √ the volume of the region inside the cone 𝑧 = 3(𝑥2 + 𝑦2 ) and below the plane 𝑧 = 1. (b) Evaluate the integral. √ 36. Find the volume between the cone 𝑧 = 𝑥2 + 𝑦2 and the plane 𝑧 = 10 + 𝑥 above the disk 𝑥2 + 𝑦2 ≤ 1. √ 37. Find the volume between the cone 𝑥 = 𝑦2 + 𝑧2 and the sphere 𝑥2 + 𝑦2 + 𝑧2 = 4.

𝑧 1

𝑦 𝑥

Cone, topped by sphere of radius 1 centered at origin, 90◦ at vertex

26.

34. Below the cone 𝑧 = 𝑟, above the 𝑥𝑦-plane, and inside the sphere 𝑥2 + 𝑦2 + 𝑧2 = 8.

38. The sphere of radius 2 centered at the origin is sliced horizontally at 𝑧 = 1. What is the volume of the cap above the plane 𝑧 = 1? 39. Suppose 𝑊 is the region outside the cylinder 𝑥2 + 𝑦2 = 1 and inside the sphere 𝑥2 + 𝑦2 + 𝑧2 = 2. Calculate

𝑧

∫𝑊 √ 1∕ 2

(𝑥2 + 𝑦2 ) 𝑑𝑉 .

40. Write and evaluate a triple integral representing the volume of a slice of the cylindrical cake of height 2 and radius 5 between the planes 𝜃 = 𝜋∕6 and 𝜃 = 𝜋∕3. 𝑦

𝑥 Cone, flat on top, 𝜋∕2 at vertex

41. Write a triple integral representing the volume of the cone in Figure 16.54 and evaluate it.



(a) Spherical coordinates. (b) Cylindrical coordinates. Write your answer as the difference of two integrals.



28. Write a triple integral representing the volume of the region between spheres of radius 1 and 2, both centered at the origin. Include limits of integration but do not evaluate. Use:

m

(a) Cylindrical coordinates (b) Spherical coordinates

5c

27. Write a triple√integral representing the volume above the cone 𝑧 = 𝑥2 + 𝑦2 and below the sphere of radius 2 centered at the origin. Include limits of integration but do not evaluate. Use:

✛ 5∕√2 cm ✲ Figure 16.54

42. Find the average distance from the origin of (a) The points in the interval |𝑥| ≤ 12. (b) The points in the plane in the disc 𝑟 ≤ 12. (c) The points in space in the ball 𝜌 ≤ 12.

16.5 INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES

In Problems 43–44, without performing the integration, decide whether the integral is positive, negative, or zero. 43. 𝑊1 is the unit ball, 𝑥2 + 𝑦2 + 𝑧2 ≤ 1. (a) ∫𝑊 sin 𝜙 𝑑𝑉

(b) ∫𝑊 cos 𝜙 𝑑𝑉

1

44. 𝑊2 is 0 ≤ 𝑧 ≤ ball.



1

1 − 𝑥2 − 𝑦2 , the top half of the unit

(a) ∫𝑊 (𝑧2 − 𝑧) 𝑑𝑉 2

(b) ∫𝑊 (−𝑥𝑧) 𝑑𝑉 2

45. The insulation surrounding a pipe of length 𝑙 is the region between two cylinders with the same axis. The inner cylinder has radius 𝑎, the outer radius of the pipe, and the insulation has thickness ℎ. Write a triple integral, including limits of integration, giving the volume of the insulation. Evaluate the integral. 46. Assume 𝑝, 𝑞, 𝑟 are positive constants. Find the volume contained between the coordinate planes and the plane 𝑥 𝑦 𝑧 + + = 1. 𝑝 𝑞 𝑟 47. A cone stands with its flat base on a table. The cone’s circular base has radius 𝑎; the vertex (tip) is at a height of ℎ above the center of the base. Write a triple integral, including limits of integration, representing the volume of the cone. Evaluate the integral. 48. A half-melon is approximated by the region between two concentric spheres, one of radius 𝑎 and the other of radius 𝑏, with 0 < 𝑎 < 𝑏. Write a triple integral, including limits of integration, giving the volume of the half-melon. Evaluate the integral. 49. A bead is made by drilling a cylindrical hole of radius 1 mm through a sphere of radius 5 mm. See Figure 16.55. (a) Set up a triple integral in cylindrical coordinates representing the volume of the bead. (b) Evaluate the integral.

877

50. A pile of hay is in the region 0 ≤ 𝑧 ≤ 2 − 𝑥2 − 𝑦2 , where 𝑥, 𝑦, 𝑧 are in meters. At height 𝑧, the density of the hay is 𝛿 = (2 − 𝑧) kg/m3 . (a) Write an integral representing the mass of hay in the pile. (b) Evaluate the integral. 51. Find the mass 𝑀 of the solid region 𝑊 given in spherical coordinates by 0 ≤ 𝜌 ≤ 3, 0 ≤ 𝜃 < 2𝜋, 0 ≤ 𝜙 ≤ 𝜋∕4. The density, 𝛿(𝑃 ), at any point 𝑃 is given by the distance of 𝑃 from the origin. 52. Write an integral representing the mass of a sphere of radius 3 if the density of the sphere at any point is twice the distance of that point from the center of the sphere. 53. A sphere has density at each point proportional to the square of the distance of the point from the 𝑧-axis. The density is 2 gm∕cm3 at a distance of 2 cm from the axis. What is the mass of the sphere if it is centered at the origin and has radius 3 cm? 54. The density of a solid sphere at any point is proportional to the square of the distance of the point to the center of the sphere. What is the ratio of the mass of a sphere of radius 1 to a sphere of radius 2? 55. A spherical shell centered at the origin has an inner radius of 6 cm and an outer radius of 7 cm. The density, 𝛿, of the material increases linearly with the distance from the center. At the inner surface, 𝛿 = 9 gm/cm3 ; at the outer surface, 𝛿 = 11 gm/cm3 . (a) Using spherical coordinates, write the density, 𝛿, as a function of radius, 𝜌. (b) Write an integral giving the mass of the shell. (c) Find the mass of the shell. 56. (a) Write an iterated integral which represents the mass of a solid ball of radius 𝑎. The density at each point in the ball is 𝑘 times the distance from that point to a fixed plane passing through the center of the ball. (b) Evaluate the integral. 57. In the region under 𝑧 = 4 − 𝑥2 − 𝑦2 and above the 𝑥𝑦plane the density of a gas is 𝛿 = 𝑒−𝑥−𝑦 gm/cm3 , where 𝑥, 𝑦, 𝑧 are in cm. Write an integral, with limits of integration, representing the mass of the gas.

1 mm

❄ ✲ ✛ ✛

✲ 5 mm

58. The density, 𝛿, of the cylinder 𝑥2 + 𝑦2 ≤ 4, 0 ≤ 𝑧 ≤ 3 varies with the distance, 𝑟, from the 𝑧-axis: 𝛿 = 1 + 𝑟 gm∕cm3 . Find the mass of the cylinder if 𝑥, 𝑦, 𝑧 are in cm.

Figure 16.55

59. The density of material at a point in a solid cylinder is proportional to the distance of the point from the 𝑧-axis. What is the ratio of the mass of the cylinder 𝑥2 +𝑦2 ≤ 1, 0 ≤ 𝑧 ≤ 2 to the mass of the cylinder 𝑥2 + 𝑦2 ≤ 9, 0 ≤ 𝑧 ≤ 2?

878

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

60. Electric charge is distributed throughout 3-space, with density proportional to the distance from the 𝑥𝑦-plane. Show that the total charge inside a cylinder of radius 𝑅 and height ℎ, sitting on the 𝑥𝑦-plane and centered along the 𝑧-axis, is proportional to 𝑅2 ℎ2 . 61. Electric charge is distributed throughout 3-space with density inversely proportional to the distance from the origin. Show that the total charge inside a sphere of radius 𝑅 is proportional to 𝑅2 . For Problems 62–65, use the definition of center of mass given on page 863. Assume 𝑥, 𝑦, 𝑧 are in cm. 62. Let 𝐶 be a solid cone with both height√and radius 1 and contained between the surfaces 𝑧 = 𝑥2 + 𝑦2 and

𝑧 = 1. If 𝐶 has constant mass density of 1 gm/cm3 , find the 𝑧-coordinate of 𝐶’s center of mass. 63. The density of the cone 𝐶 in Problem 62 is given by 𝛿(𝑧) = 𝑧2 gm/cm3 . Find (a) The mass of 𝐶. (b) The 𝑧-coordinate of 𝐶’s center of mass. 64. For 𝑎 > 0, consider the family of solids bounded below by the paraboloid 𝑧 = 𝑎(𝑥2 + 𝑦2 ) and above by the plane 𝑧 = 1. If the solids all have constant mass density 1 gm/cm3 , show that the 𝑧-coordinate of the center of mass is 2∕3 and so independent of the parameter 𝑎. 65. Find the location of the center of mass of a hemisphere of radius 𝑎 and density 𝑏 gm/cm3 .

Strengthen Your Understanding In Problems 66–68, explain what is wrong with the statement. 2𝜋

1

𝜋

69. An integral in spherical coordinates that gives the volume of a hemisphere.

66. The integral

70. An integral for which it is more convenient to use spherical coordinates than to use Cartesian coordinates.

67. Changing the order of integration gives

71. Which of the following integrals give the volume of the unit sphere?

1 𝑑𝜌 𝑑𝜙 𝑑𝜃 gives the volume ∫0 ∫0 ∫0 inside the sphere of radius 1. 2𝜋

∫0

∫0

𝜌2 sin 𝜙 𝑑𝜌 𝑑𝜙 𝑑𝜃

∫0 2∕ cos 𝜙

=

2𝜋

2∕ cos 𝜙

𝜋∕4

∫0

2𝜋

𝜋∕4

∫0

∫0

𝜌2 sin 𝜙 𝑑𝜃 𝑑𝜙 𝑑𝜌.

(a) (b)

68. The volume of a cylinder of height and radius 1 is ∫0

1

1

∫0 ∫0

1 𝑑𝑧 𝑑𝑟 𝑑𝜃.

In Problems 69–70, give an example of:

16.6

𝜋

∫0

∫0 ∫0 𝜋

(c) 2𝜋

∫0

2𝜋

∫0 ∫0 𝜋

(d)

∫0 ∫0 𝜋

(e)

∫0 ∫0

2𝜋

2𝜋

2𝜋

2𝜋

1

∫0

1 𝑑𝜌 𝑑𝜃 𝑑𝜙

1

∫0 ∫0 ∫0 ∫0

1 𝑑𝜌 𝑑𝜃 𝑑𝜙 1

𝜌2 sin 𝜙 𝑑𝜌 𝑑𝜃 𝑑𝜙 1

𝜌2 sin 𝜙 𝑑𝜌 𝑑𝜙 𝑑𝜃 1

𝜌 𝑑𝜌 𝑑𝜙 𝑑𝜃

APPLICATIONS OF INTEGRATION TO PROBABILITY To represent how a quantity such as height or weight is distributed throughout a population, we use a density function. To study two or more quantities at the same time and see how they are related, we use a multivariable density function.

Density Functions Distribution of Weight and Height in Expectant Mothers Table 16.10 shows the distribution of weight and height in a survey of expectant mothers. The histogram in Figure 16.56 is constructed so that the volume of each bar represents the percentage in the corresponding weight and height range. For example, the bar representing the mothers who weighed 60–70 kg and were 160–165 cm tall has base of area 10 kg ⋅ 5 cm = 50 kg cm. The volume of this bar is 12%, so its height is 12%∕50 kg cm = 0.24%∕ kg cm. Notice that the units on the vertical axis are % per kg cm, so the volume of a bar is a %. The total volume is 100% = 1.

16.6 APPLICATIONS OF INTEGRATION TO PROBABILITY

Table 16.10

879

Distribution of weight and height in a survey of expectant mothers, in % 45-50 kg

50-60 kg

60-70 kg

70-80 kg

80-105 kg

Totals by height

150-155 cm

2

4

4

2

1

13

155-160 cm

0

12

8

2

1

23

160-165 cm

1

7

12

4

3

27

165-170 cm

0

8

12

6

2

28

170-180 cm

0

1

3

4

1

9

Totals by weight

3

32

39

18

8

100

0.25% 0.20% 0.15% 0.10%

45

percent per kg cm

0.05% 65 kg

85 105 150

155

160

165

170

175

180

cm

Figure 16.56: Histogram representing the data in Table 16.10

Example 1

Find the percentage of mothers in the survey with height between 170 and 180 cm.

Solution

We add the percentages across the row corresponding to the 170–180 cm height range; this is equivalent to adding the volumes of the corresponding rectangular solids in the histogram. Percentage of mothers = 0 + 1 + 3 + 4 + 1 = 9%.

Smoothing the Histogram If we group the data using narrower weight and height groups (and a larger sample), we can draw a smoother histogram and get finer estimates. In the limit, we replace the histogram with a smooth surface, in such a way that the volume under the surface above a rectangle is the percentage of mothers in that rectangle. We define a density function, 𝑝(𝑤, ℎ), to be the function whose graph is the smooth surface. It has the property that Fraction of sample with weight between 𝑎 and 𝑏 and height between 𝑐 and 𝑑

=

Volume under graph of 𝑝 over the rectangle

𝑏

=

∫𝑎 ∫𝑐

𝑑

𝑝(𝑤, ℎ) 𝑑ℎ 𝑑𝑤.

𝑎 ≤ 𝑤 ≤ 𝑏, 𝑐 ≤ ℎ ≤ 𝑑

This density also gives the probability that a mother is in these height and weight groups.

880

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

Joint Probability Density Functions We generalize this idea to represent any two characteristics, 𝑥 and 𝑦, distributed throughout a population.

A function 𝑝(𝑥, 𝑦) is called a joint probability density function, or pdf, for 𝑥 and 𝑦 if Probability that member of

Volume under graph of 𝑝 =

population has 𝑥 between 𝑎 and 𝑏 and 𝑦 between 𝑐 and 𝑑 where



above the rectangle 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑐 ≤ 𝑦 ≤ 𝑑

𝑏

=

∫𝑎 ∫𝑐

𝑑

𝑝(𝑥, 𝑦) 𝑑𝑦 𝑑𝑥,



∫−∞ ∫−∞

𝑝(𝑥, 𝑦) 𝑑𝑦 𝑑𝑥 = 1

and 𝑝(𝑥, 𝑦) ≥ 0 for all 𝑥 and 𝑦.

The probability that 𝑥 falls in an interval of width Δ𝑥 around 𝑥0 and 𝑦 falls in an interval of width Δ𝑦 around 𝑦0 is approximately 𝑝(𝑥0 , 𝑦0 )Δ𝑥Δ𝑦.

A joint density function need not be continuous, as in Example 2. In addition, as in Example 4, the integrals involved may be improper and must be computed by methods similar to those used for improper one-variable integrals. Example 2

Let 𝑝(𝑥, 𝑦) be defined on the square 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1 by 𝑝(𝑥, 𝑦) = 𝑥 + 𝑦; let 𝑝(𝑥, 𝑦) = 0 if (𝑥, 𝑦) is outside this square. Check that 𝑝 is a joint density function. In terms of the distribution of 𝑥 and 𝑦 in the population, what does it mean that 𝑝(𝑥, 𝑦) = 0 outside the square?

Solution

First, we have 𝑝(𝑥, 𝑦) ≥ 0 for all 𝑥 and 𝑦. To check that 𝑝 is a joint density function, we show that the total volume under the graph is 1: ∞



∫−∞ ∫−∞

1

1

𝑝(𝑥, 𝑦) 𝑑𝑦 𝑑𝑥 =

(𝑥 + 𝑦) 𝑑𝑦 𝑑𝑥 ∫0 ∫0 ) 1 ( 2 ) 1 ) 1( 1( 𝑦2 || 𝑥 𝑥 || 1 𝑑𝑥 = + 𝑥𝑦 + = 𝑥 + 𝑑𝑥 = = 1. ∫0 ∫0 2 ||0 2 2 2 ||0

The fact that 𝑝(𝑥, 𝑦) = 0 outside the square means that the variables 𝑥 and 𝑦 never take values outside the interval [0, 1]; that is, the value of 𝑥 and 𝑦 for any individual in the population is always between 0 and 1.

Example 3

Two variables 𝑥 and 𝑦 are distributed in a population according to the density function of Example 2. Find the fraction of the population with 𝑥 ≤ 1∕2, the fraction with 𝑦 ≤ 1∕2, and the fraction with both 𝑥 ≤ 1∕2 and 𝑦 ≤ 1∕2.

Solution

The fraction with 𝑥 ≤ 1∕2 is the volume under the graph to the left of the line 𝑥 = 1∕2: ) 1 ) 1∕2 1 1∕2 ( 1∕2 ( 𝑦2 || 1 𝑑𝑥 𝑥𝑦 + 𝑥+ (𝑥 + 𝑦) 𝑑𝑦 𝑑𝑥 = 𝑑𝑥 = | ∫0 ∫0 ∫0 ∫0 2 |0 2 ( 2 ) 1∕2 𝑥 𝑥 || 1 1 3 = + = + = . 2 2 ||0 8 4 8

Since the function and the regions of integration are symmetric in 𝑥 and 𝑦, the fraction with 𝑦 ≤ 1∕2

16.6 APPLICATIONS OF INTEGRATION TO PROBABILITY

881

is also 3∕8. Finally, the fraction with both 𝑥 ≤ 1∕2 and 𝑦 ≤ 1∕2 is 1∕2

1∕2

∫0

∫0

) 1∕2 ) 1∕2 ( 𝑦2 || 1 1 𝑥 + 𝑑𝑥 𝑥𝑦 + (𝑥 + 𝑦) 𝑑𝑦 𝑑𝑥 = 𝑑𝑥 = ∫0 ∫0 2 ||0 2 8 ( ) 1∕2 1 2 1 || 1 1 1 = 𝑥 + 𝑥 | = + = . 4 8 |0 16 16 8 1∕2 (

Recall that a one-variable density function 𝑝(𝑥) is a function such that 𝑝(𝑥) ≥ 0 for all 𝑥, and ∞ ∫−∞ 𝑝(𝑥) 𝑑𝑥 = 1. Example 4

Let 𝑝1 and 𝑝2 be one-variable density functions for 𝑥 and 𝑦, respectively. Check that 𝑝(𝑥, 𝑦) = 𝑝1 (𝑥)𝑝2 (𝑦) is a joint density function.

Solution

Since both 𝑝1 and 𝑝2 are density functions, they are nonnegative everywhere. Thus, their product 𝑝1 (𝑥)𝑝2 (𝑥) = 𝑝(𝑥, 𝑦) is nonnegative everywhere. Now we must check that the volume under the ∞ ∞ graph of 𝑝 is 1. Since ∫−∞ 𝑝2 (𝑦) 𝑑𝑦 = 1 and ∫−∞ 𝑝1 (𝑥) 𝑑𝑥 = 1, we have ∞



∫−∞ ∫−∞



𝑝(𝑥, 𝑦) 𝑑𝑦 𝑑𝑥 =



∫−∞ ∫−∞



𝑝1 (𝑥)𝑝2 (𝑦) 𝑑𝑦 𝑑𝑥 =



=

Example 5

∫−∞

∫−∞

𝑝1 (𝑥)

(



∫−∞

) 𝑝2 (𝑦) 𝑑𝑦 𝑑𝑥



𝑝1 (𝑥)(1) 𝑑𝑥 =

∫−∞

𝑝1 (𝑥) 𝑑𝑥 = 1.

A machine in a factory is set to produce components 10 cm long and 5 cm in diameter. In fact, there is a slight variation from one component to the next. A component is usable if its length and diameter deviate from the correct values by less than 0.1 cm. With the length, 𝑥, in cm and the diameter, 𝑦, in cm, the probability density function is √ 50 2 −100(𝑥−10)2 −50(𝑦−5)2 𝑝(𝑥, 𝑦) = 𝑒 𝑒 . 𝜋 What is the probability that a component is usable? (See Figure 16.57.)

Figure 16.57: The density function 𝑝(𝑥, 𝑦) =

Solution

√ 50 2 −100(𝑥−10)2 −50(𝑦−5)2 𝑒 𝑒 𝜋

We know that

Probability that 𝑥 and 𝑦 satisfy 𝑥0 − Δ𝑥 ≤ 𝑥 ≤ 𝑥0 + Δ𝑥 𝑦0 − Δ𝑦 ≤ 𝑦 ≤ 𝑦0 + Δ𝑦

√ 50 2 𝑦0 +Δ𝑦 𝑥0 +Δ𝑥 −100(𝑥−10)2 −50(𝑦−5)2 𝑒 𝑒 𝑑𝑥 𝑑𝑦. = 𝜋 ∫𝑦0 −Δ𝑦 ∫𝑥0 −Δ𝑥

882

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

Thus, Probability that component is usable

√ 50 2 5.1 10.1 −100(𝑥−10)2 −50(𝑦−5)2 𝑒 𝑒 𝑑𝑥 𝑑𝑦. = 𝜋 ∫4.9 ∫9.9

The double integral must be evaluated numerically. This yields √ 50 2 Probability that (0.02556) = 0.57530. = 𝜋 component is usable Thus, there is a 57.530% chance that the component is usable.

Exercises and Problems for Section 16.6 EXERCISES In Exercises 1–6, check whether 𝑝 is a joint density function. Assume 𝑝(𝑥, 𝑦) = 0 outside the region 𝑅. 1. 𝑝(𝑥, 𝑦) = 1∕2, where 𝑅 is 4 ≤ 𝑥 ≤ 5, −2 ≤ 𝑦 ≤ 0 2. 𝑝(𝑥, 𝑦) = 1, where 𝑅 is 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 2

In Exercises 11–14, a joint probability density function is given by 𝑝(𝑥, 𝑦) = 0.005𝑥 + 0.025𝑦 in 𝑅, the rectangle 0 ≤ 𝑥 ≤ 10, 0 ≤ 𝑦 ≤ 2, and 𝑝(𝑥, 𝑦) = 0 else. Find the probability that a point (𝑥, 𝑦) satisfies the given conditions. 11. 𝑥 ≤ 4

12. 𝑦 ≥ 1

4. 𝑝(𝑥, 𝑦) = 6(𝑦 − 𝑥), where 𝑅 is 0 ≤ 𝑥 ≤ 𝑦 ≤ 2

13. 𝑥 ≤ 4 and 𝑦 ≥ 1

14. 𝑥 ≥ 5 and 𝑦 ≥ 1

5. 𝑝(𝑥, 𝑦) = (2∕𝜋)(1 − 𝑥2 − 𝑦2 ), where 𝑅 is 𝑥2 + 𝑦2 ≤ 1

In Exercises 15–22, let 𝑝 be the joint density function such that 𝑝(𝑥, 𝑦) = 𝑥𝑦 in 𝑅, the rectangle 0 ≤ 𝑥 ≤ 2, 0 ≤ 𝑦 ≤ 1, and 𝑝(𝑥, 𝑦) = 0 outside 𝑅. Find the fraction of the population satisfying the given constraints.

3. 𝑝(𝑥, 𝑦) = 𝑥 + 𝑦, where 𝑅 is −1 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1

6. 𝑝(𝑥, 𝑦) = 𝑥𝑦𝑒−𝑥−𝑦 , where 𝑅 is 𝑥 ≥ 0, 𝑦 ≥ 0 In Exercises 7–10, a joint probability density function is given by 𝑝(𝑥, 𝑦) = 𝑥𝑦∕4 in 𝑅, the rectangle 0 ≤ 𝑥 ≤ 2, 0 ≤ 𝑦 ≤ 2, and 𝑝(𝑥, 𝑦) = 0 else. Find the probability that a point (𝑥, 𝑦) satisfies the given conditions. 7. 𝑥 ≤ 1 and 𝑦 ≤ 1 9. 𝑥 ≥ 1 and 𝑦 ≤ 1

8. 𝑥 ≥ 1 and 𝑦 ≥ 1 10. 1∕3 ≤ 𝑥 ≤ 1

15. 𝑥 ≥ 3

16. 𝑥 = 1

17. 𝑥 + 𝑦 ≤ 3

18. −1 ≤ 𝑥 ≤ 1

19. 𝑥 ≥ 𝑦

20. 𝑥 + 𝑦 ≤ 1

21. 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1∕2 22. Within a distance 1 from the origin

PROBLEMS 23. Let 𝑥 and 𝑦 have joint density function { 2 (𝑥 + 2𝑦) for 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1, 3 𝑝(𝑥, 𝑦) = 0 otherwise.

𝑦 𝑦=𝑥 𝑦=

1

Find the probability that (a) 𝑥 > 1∕3.

(b) 𝑥 < (1∕3) + 𝑦.

24. The joint density function for 𝑥, 𝑦 is given by { 𝑘𝑥𝑦 for 0 ≤ 𝑥 ≤ 𝑦 ≤ 1, 𝑓 (𝑥, 𝑦) = 0 otherwise. (a) Determine the value of 𝑘. (b) Find the probability that (𝑥, 𝑦) lies in the shaded region in Figure 16.58.

𝑥

0 1

Figure 16.58



𝑥

16.6 APPLICATIONS OF INTEGRATION TO PROBABILITY

25. A joint density function is given by { 𝑘𝑥2 for 0 ≤ 𝑥 ≤ 2 and 0 ≤ 𝑦 ≤ 1, 𝑓 (𝑥, 𝑦) = 0 otherwise. (a) Find the value of the constant 𝑘. (b) Find the probability that (𝑥, 𝑦) satisfies 𝑥 + 𝑦 ≤ 2. (c) Find the probability that (𝑥, 𝑦) satisfies 𝑥 ≤ 1 and 𝑦 ≤ 1∕2. 26. A point is chosen at random from the region 𝑆 in the 𝑥𝑦-plane containing all points (𝑥, 𝑦) such that −1 ≤ 𝑥 ≤ 1, −2 ≤ 𝑦 ≤ 2 and 𝑥 − 𝑦 ≥ 0 (“at random” means that the density function is constant on 𝑆). (a) Determine the joint density function for 𝑥 and 𝑦. (b) If 𝑇 is a subset of 𝑆 with area 𝛼, then find the probability that a point (𝑥, 𝑦) is in 𝑇 . 27. A probability density function on a square has constant values in different triangular regions as shown in Figure 16.59. Find the probability that (a) 𝑥 ≥ 2 (b) 𝑦 ≥ 𝑥 (c) 𝑦 ≥ 𝑥 and 𝑥 ≥ 2

883

29. The probability that a radioactive substance will decay at time 𝑡 is modeled by the density function 𝑝(𝑡) = 𝜆𝑒−𝜆𝑡 for 𝑡 ≥ 0, and 𝑝(𝑡) = 0 for 𝑡 < 0. The positive constant 𝜆 depends on the material, and is called the decay rate. (a) Check that 𝑝 is a density function. (b) Two materials with decay rates 𝜆 and 𝜇 decay independently of each other; their joint density function is the product of the individual density functions. Write the joint density function for the probability that the first material decays at time 𝑡 and the second at time 𝑠. (c) Find the probability that the first substance decays before the second. 30. Figure 16.60 represents a baseball field, with the bases at (1, 0), (1, 1), (0, 1), and home plate at (0, 0). The outer bound of the outfield is a piece of a circle about the origin with radius 4. When a ball is hit by a batter we record the spot on the field where the ball is caught. Let 𝑝(𝑟, 𝜃) be a function in the plane that gives the density of the distribution of such spots. Write an expression that represents the probability that a hit is caught in (a) The right field (region 𝑅). (b) The center field (region 𝐶).

𝑦 (m) 4

𝑦 0.12

0.16

4

3 0.06

0.08

0.01

0.04

2 𝐶

1 0.01

0.02

𝑥 (m) 1

2

3

4

Figure 16.59: Probability density on a square (per m2 )

1 𝜋 6

𝑅 𝜋 6𝜋 6

𝑥 1

4

Figure 16.60 28. A health insurance company wants to know what proportion of its policies are going to cost the company a lot of money because the insured people are over 65 and sick. In order to compute this proportion, the company defines a disability index, 𝑥, with 0 ≤ 𝑥 ≤ 1, where 𝑥 = 0 represents perfect health and 𝑥 = 1 represents total disability. In addition, the company uses a density function, 𝑓 (𝑥, 𝑦), defined in such a way that the quantity 𝑓 (𝑥, 𝑦) Δ𝑥 Δ𝑦 approximates the fraction of the population with disability index between 𝑥 and 𝑥+Δ𝑥, and aged between 𝑦 and 𝑦+Δ𝑦. The company knows from experience that a policy no longer covers its costs if the insured person is over 65 and has a disability index exceeding 0.8. Write an expression for the fraction of the company’s policies held by people meeting these criteria.

31. Two independent random numbers 𝑥 and 𝑦 between 0 and 1 have joint density function { 1 if 0 ≤ 𝑥, 𝑦 ≤ 1 𝑝(𝑥, 𝑦) = 0 otherwise. This problem concerns the average 𝑧 = (𝑥+𝑦)∕2, which has a one-variable probability density function of its own. (a) Find 𝐹 (𝑡), the probability that 𝑧 ≤ 𝑡. Treat separately the cases 𝑡 ≤ 0, 0 < 𝑡 ≤ 1∕2, 1∕2 < 𝑡 ≤ 1, 1 < 𝑡. Note that 𝐹 (𝑡) is the cumulative distribution function of 𝑧. (b) Find and graph the probability density function of 𝑧. (c) Are 𝑥 and 𝑦 more likely to be near 0, 1∕2, or 1? What about 𝑧?

884

Chapter 16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES

Strengthen Your Understanding

32. If 𝑝1 (𝑥, 𝑦) and 𝑝2 (𝑥, 𝑦) are joint density functions, then 𝑝1 (𝑥, 𝑦) + 𝑝2 (𝑥, 𝑦) is a joint density function.

35. A one-variable function 𝑔(𝑦) such that 𝑓 is a joint density function: { 𝑔(𝑦) for 0 ≤ 𝑥 ≤ 2 and 0 ≤ 𝑦 ≤ 1, 𝑓 (𝑥, 𝑦) = 0 otherwise

33. If 𝑝(𝑤, ℎ) is the probability density function of the weight and height of mothers discussed in Section 16.6, then the probability that a mother weighs 60 kg and has a height of 170 cm is 𝑝(60, 170).

For Problems 36–39, let 𝑝(𝑥, 𝑦) be a joint density function for 𝑥 and 𝑦. Are the following statements true or false?

In Problems 32–33, explain what is wrong with the statement.

In Problems 34–35, give an example of: 34. Values for 𝑎, 𝑏, 𝑐 and 𝑑 such that 𝑓 is a joint density function: { 1 for 𝑎 ≤ 𝑥 ≤ 𝑏 and 𝑐 ≤ 𝑦 ≤ 𝑑, 𝑓 (𝑥, 𝑦) = 0 otherwise

Online Resource: Review problems and Projects



𝑏

36.

∫𝑎 ∫−∞

𝑝(𝑥, 𝑦) 𝑑𝑦 𝑑𝑥 is the probability that 𝑎 ≤ 𝑥 ≤ 𝑏.

37. 0 ≤ 𝑝(𝑥, 𝑦) ≤ 1 for all 𝑥. 𝑏

38.

∫𝑎

𝑝(𝑥, 𝑦) 𝑑𝑥 is the probability that 𝑎 ≤ 𝑥 ≤ 𝑏. ∞

39.



∫−∞ ∫−∞

𝑝(𝑥, 𝑦) 𝑑𝑦 𝑑𝑥 = 1.

Chapter Seventeen

PARAMETERIZATION AND VECTOR FIELDS

Contents 17.1 Parameterized Curves . . . . . . . . . . . . . . . . . . . . . Parametric Equations in Three Dimensions. . . . . Using Position Vectors to Write Parameterized Curves as Vector-Valued Functions . . . . . . . . . . Parametric Equation of a Line . . . . . . . . . . . . . . . . Intersection of Curves and Surfaces. . . . . . . . . . . 17.2 Motion, Velocity, and Acceleration. . . . . . . . . . . The Velocity Vector . . . . . . . . . . . . . . . . . . . . . . . Computing the Velocity . . . . . . . . . . . . . . . . . . . . The Components of the Velocity Vector . . . . . . . . . Velocity Vectors and Tangent Lines . . . . . . . . . . . . The Acceleration Vector. . . . . . . . . . . . . . . . . . . . Components of the Acceleration Vector . . . . . . . . . Motion in a Circle and Along a Line . . . . . . . . . . The Length of a Curve . . . . . . . . . . . . . . . . . . . . . 17.3 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Introduction to Vector Fields . . . . . . . . . . . . . . . . Velocity Vector Fields . . . . . . . . . . . . . . . . . . . . . Force Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . Definition of a Vector Field . . . . . . . . . . . . . . . . . Visualizing a Vector Field Given by a Formula . . . . Finding a Formula for a Vector Field . . . . . . . . . . . Gradient Vector Fields . . . . . . . . . . . . . . . . . . . . . 17.4 The Flow of a Vector Field . . . . . . . . . . . . . . . . . How Do We Find a Flow Line? . . . . . . . . . . . . . . Approximating Flow Lines Numerically . . . . . . .

886 886 888 889 890 896 896 896 897 898 899 899 899 901 905 905 905 906 906 907 908 909 913 913 916

886

Chapter 17 PARAMETERIZATION AND VECTOR FIELDS

17.1

PARAMETERIZED CURVES A curve in the plane may be parameterized by a pair of equations of the form 𝑥 = 𝑓 (𝑡), 𝑦 = 𝑔(𝑡). As the parameter 𝑡 changes, the point (𝑥, 𝑦) traces out the curve. In this section we find parametric equations for curves in three dimensions, and we see how to write parametric equations using position vectors.

Parametric Equations in Three Dimensions We describe motion in the plane by giving parametric equations for 𝑥 and 𝑦 in terms of 𝑡. To describe a motion in 3-space parametrically, we need a third equation giving 𝑧 in terms of 𝑡. Example 1

Find parametric equations for the curve 𝑦 = 𝑥2 in the 𝑥𝑦-plane.

Solution

A possible parameterization in two dimensions is 𝑥 = 𝑡, 𝑦 = 𝑡2 . Since the curve is in the 𝑥𝑦-plane, the 𝑧-coordinate is zero, so a parameterization in three dimensions is 𝑥 = 𝑡,

Example 2

𝑦 = 𝑡2 ,

𝑧 = 0.

Find parametric equations for a particle that starts at (0, 3, 0) and moves around a circle as shown in Figure 17.1. 𝑧

𝑦



Start

𝑥

Figure 17.1: Circle of radius 3 in the 𝑦𝑧-plane, centered at origin

Solution

Since the motion is in the 𝑦𝑧-plane, we have 𝑥 = 0 at all times 𝑡. Looking at the 𝑦𝑧-plane from the positive 𝑥-direction we see motion around a circle of radius 3 in the clockwise direction. Thus, 𝑥 = 0,

Example 3

𝑦 = 3 cos 𝑡,

Describe in words the motion given parametrically by 𝑥 = cos 𝑡,

Solution

𝑧 = −3 sin 𝑡.

𝑦 = sin 𝑡,

𝑧 = 𝑡.

The particle’s 𝑥- and 𝑦-coordinates give circular motion in the 𝑥𝑦-plane, while the 𝑧-coordinate increases steadily. Thus, the particle traces out a rising spiral, like a coiled spring. (See Figure 17.2.) This curve is called a helix.

17.1 PARAMETERIZED CURVES

887

𝑧

𝑦

𝑥 Figure 17.2: The helix 𝑥 = cos 𝑡, 𝑦 = sin 𝑡, 𝑧 = 𝑡

Example 4

⃗ and through the point Find parametric equations for the line parallel to the vector 2⃗𝑖 + 3𝑗⃗ + 4𝑘 (1, 5, 7).

Solution

Let’s imagine a particle at the point (1, 5, 7) at time 𝑡 = 0 and moving through a displacement of ⃗ for each unit of time, 𝑡. When 𝑡 = 0, 𝑥 = 1 and 𝑥 increases by 2 units for every unit 2⃗𝑖 + 3𝑗⃗ + 4𝑘 of time. Thus, at time 𝑡, the 𝑥-coordinate of the particle is given by 𝑥 = 1 + 2𝑡. Similarly, the 𝑦-coordinate starts at 𝑦 = 5 and increases at a rate of 3 units for every unit of time. The 𝑧-coordinate starts at 𝑦 = 7 and increases by 4 units for every unit of time. Thus, the parametric equations of the line are 𝑥 = 1 + 2𝑡, 𝑦 = 5 + 3𝑡, 𝑧 = 7 + 4𝑡.

We can generalize the previous example as follows: Parametric Equations of a Line through the point (𝑥0 , 𝑦0 , 𝑧0 ) and parallel to the vector 𝑎⃗𝑖 + ⃗ are 𝑏𝑗⃗ + 𝑐 𝑘 𝑥 = 𝑥0 + 𝑎𝑡, 𝑦 = 𝑦0 + 𝑏𝑡, 𝑧 = 𝑧0 + 𝑐𝑡. Notice that the coordinates 𝑥, 𝑦, and 𝑧 are linear functions of the parameter 𝑡. Example 5

(a) Describe in words the curve given by the parametric equations 𝑥 = 3 + 𝑡, 𝑦 = 2𝑡, 𝑧 = 1 − 𝑡. (b) Find parametric equations for the line through the points (1, 2, −1) and (3, 3, 4).

Solution

⃗ (a) The curve is a line through the point (3, 0, 1) and parallel to the vector ⃗𝑖 + 2𝑗⃗ − 𝑘. (b) The line is parallel to the vector between the points 𝑃 = (1, 2, −1) and 𝑄 = (3, 3, 4). ⃗ = 2⃗𝑖 + 𝑗⃗ + 5𝑘 ⃗. 𝑃⃖⃖⃖⃖⃖𝑄⃗ = (3 − 1)⃗𝑖 + (3 − 2)𝑗⃗ + (4 − (−1))𝑘 Thus, using the point 𝑃 , the parametric equations are 𝑥 = 1 + 2𝑡,

𝑦 = 2 + 𝑡,

𝑧 = −1 + 5𝑡.

Using the point 𝑄 gives the equations 𝑥 = 3 + 2𝑡, 𝑦 = 3 + 𝑡, 𝑧 = 4 + 5𝑡, which represent the same line. The point where 𝑡 = 0 in the second equations is given by 𝑡 = 1 in the first equations.

888

Chapter 17 PARAMETERIZATION AND VECTOR FIELDS

Using Position Vectors to Write Parameterized Curves as Vector-Valued Functions A point in the plane with coordinates (𝑥, 𝑦) can be represented by the position vector 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ ⃗ . (See Figure 17.4.) in Figure 17.3. Similarly, in 3-space we write 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 𝑧

𝑦

(𝑥, 𝑦, 𝑧)

(𝑥, 𝑦)

𝑟⃗ ⃗ 𝑧𝑘

𝑦𝑗⃗

𝑟⃗

𝑥⃗𝑖 𝑥

𝑦

𝑥 𝑦𝑗⃗

𝑥⃗𝑖

Figure 17.4: Position vector 𝑟⃗ for the point (𝑥, 𝑦, 𝑧)

Figure 17.3: Position vector 𝑟⃗ for the point (𝑥, 𝑦)

We can write the parametric equations 𝑥 = 𝑓 (𝑡), 𝑦 = 𝑔(𝑡), 𝑧 = ℎ(𝑡) as a single vector equation ⃗ 𝑟⃗ (𝑡) = 𝑓 (𝑡)⃗𝑖 + 𝑔(𝑡)𝑗⃗ + ℎ(𝑡)𝑘 called a parameterization. As the parameter 𝑡 varies, the point with position vector 𝑟⃗ (𝑡) traces out a curve in 3-space. For example, the circular motion in the plane 𝑥 = cos 𝑡, 𝑦 = sin 𝑡

can be written as

𝑟⃗ = (cos 𝑡)⃗𝑖 + (sin 𝑡)𝑗⃗

and the helix in 3-space ⃗. 𝑥 = cos 𝑡, 𝑦 = sin 𝑡, 𝑧 = 𝑡 can be written as 𝑟⃗ = (cos 𝑡)⃗𝑖 + (sin 𝑡)𝑗⃗ + 𝑡𝑘 See Figure 17.5.

𝑧 𝐶 𝑟⃗ (𝑏) 𝑟⃗ (𝑎)

𝑎

𝑟⃗ (𝑡)

𝑏 𝑦 𝑥

Figure 17.5: The parameterization sends the interval, 𝑎 ≤ 𝑡 ≤ 𝑏, to the curve, 𝐶, in 3-space 1 2

Example 6

Use vectors to give a parameterization for the circle of radius

Solution

The circle of radius 1 centered at the origin is parameterized by the vector-valued function 𝑟⃗ 1 (𝑡) = cos 𝑡⃗𝑖 + sin 𝑡𝑗⃗ ,

centered at the point (−1, 2).

0 ≤ 𝑡 ≤ 2𝜋.

The point (−1, 2) has position vector 𝑟⃗ 0 = −⃗𝑖 + 2𝑗⃗ . The position vector, 𝑟⃗ (𝑡), of a point on the circle of radius 12 centered at (−1, 2) is found by adding 12 𝑟⃗ 1 to 𝑟⃗ 0 . (See Figures 17.6 and 17.7.) Thus, 𝑟⃗ (𝑡) = 𝑟⃗ 0 + 12 𝑟⃗ 1 (𝑡) = −⃗𝑖 + 2𝑗⃗ + 21 (cos 𝑡⃗𝑖 + sin 𝑡𝑗⃗ ) = (−1 +

1 2

cos 𝑡)⃗𝑖 + (2 +

or, equivalently, 𝑥 = −1 + 12 cos 𝑡,

𝑦=2+

1 2

sin 𝑡,

0 ≤ 𝑡 ≤ 2𝜋.

1 2

sin 𝑡)𝑗⃗ ,

17.1 PARAMETERIZED CURVES

889

𝑦 2.5 𝑟⃗1 (𝑡) = cos 𝑡⃗𝑖 + sin 𝑡𝑗⃗

1 𝑟⃗ 2 1

𝑦

2

1 1.5

𝑟⃗

❄ −1

1

𝑟⃗ 0

𝑥 1

0.5 𝑥 −1

−1

Figure 17.6: The circle 𝑥2 + 𝑦2 = 1 parameterized by 𝑟⃗ 1 (𝑡) = cos 𝑡⃗𝑖 + sin 𝑡𝑗⃗

Figure 17.7: The circle of radius

1 2

and center

(−1, 2) parameterized by 𝑟⃗ (𝑡) = 𝑟⃗ 0 + 21 𝑟⃗ 1 (𝑡)

Parametric Equation of a Line Consider a straight line in the direction of a vector 𝑣⃗ passing through the point (𝑥0 , 𝑦0 , 𝑧0 ) with position vector 𝑟⃗ 0 . We start at 𝑟⃗ 0 and move up and down the line, adding different multiples of 𝑣⃗ to 𝑟⃗ 0 . (See Figure 17.8.) 𝑡 = −1 𝑟⃗ 0

𝑡=0

𝑡=2

𝑡=1

𝑡=3

𝑣⃗ 𝑟⃗

Figure 17.8: The line 𝑟⃗ (𝑡) = 𝑟⃗ 0 + 𝑡𝑣⃗

In this way, every point on the line can be written as 𝑟⃗ 0 + 𝑡⃗ 𝑣 , which yields the following:

Parametric Equation of a Line in Vector Form ⃗ in the direction of the The line through the point with position vector 𝑟⃗0 = 𝑥0 ⃗𝑖 + 𝑦0 𝑗⃗ + 𝑧0 𝑘 ⃗ has parametric equation vector 𝑣⃗ = 𝑎⃗𝑖 + 𝑏𝑗⃗ + 𝑐 𝑘 𝑟⃗ (𝑡) = 𝑟⃗ 0 + 𝑡⃗ 𝑣.

Example 7

(a) Find parametric equations for the line passing through the points (2, −1, 3) and (−1, 5, 4). (b) Represent the line segment from (2, −1, 3) to (−1, 5, 4) parametrically.

Solution

⃗ (a) The line passes through (2, −1, 3) and is parallel to the displacement vector 𝑣⃗ = −3⃗𝑖 + 6𝑗⃗ + 𝑘 from (2, −1, 3) to (−1, 5, 4). Thus, the parametric equation is ⃗ + 𝑡(−3⃗𝑖 + 6𝑗⃗ + 𝑘 ⃗ ). 𝑟⃗ (𝑡) = 2⃗𝑖 − 𝑗⃗ + 3𝑘 (b) In the parameterization in part (a), 𝑡 = 0 corresponds to the point (2, −1, 3) and 𝑡 = 1 corresponds to the point (−1, 5, 4). So the parameterization of the segment is ⃗ + 𝑡(−3⃗𝑖 + 6𝑗⃗ + 𝑘 ⃗ ), 𝑟⃗ (𝑡) = 2⃗𝑖 − 𝑗⃗ + 3𝑘

0 ≤ 𝑡 ≤ 1.

890

Chapter 17 PARAMETERIZATION AND VECTOR FIELDS

Intersection of Curves and Surfaces Parametric equations for a curve enable us to find where a curve intersects a surface. Example 8

Find the points at which the line 𝑥 = 𝑡, 𝑦 = 2𝑡, 𝑧 = 1 + 𝑡 pierces the sphere of radius 10 centered at the origin.

Solution

The equation for the sphere of radius 10 and centered at the origin is 𝑥2 + 𝑦2 + 𝑧2 = 100. To find the intersection points of the line and the sphere, substitute the parametric equations of the line into the equation of the sphere, giving 𝑡2 + 4𝑡2 + (1 + 𝑡)2 = 100, so 6𝑡2 + 2𝑡 − 99 = 0, which has the two solutions at approximately 𝑡 = −4.23 and 𝑡 = 3.90. Using the parametric equation for the line, (𝑥, 𝑦, 𝑧) = (𝑡, 2𝑡, 1 + 𝑡), we see that the line cuts the sphere at the two points (𝑥, 𝑦, 𝑧) = (−4.23, 2(−4.23), 1 + (−4.23)) = (−4.23, −8.46, −3.23), and (𝑥, 𝑦, 𝑧) = (3.90, 2(3.90), 1 + 3.90) = (3.90, 7.80, 4.90). We can also use parametric equations to find the intersection of two curves.

Example 9

⃗ and 𝑟⃗ 2 (𝑡) = Two particles move through space, with equations 𝑟⃗ 1 (𝑡) = 𝑡⃗𝑖 + (1 + 2𝑡)𝑗⃗ + (3 − 2𝑡)𝑘 ⃗ . Do the particles ever collide? Do their paths cross? (−2 − 2𝑡)⃗𝑖 + (1 − 2𝑡)𝑗⃗ + (1 + 𝑡)𝑘

Solution

To see if the particles collide, we must find out if they pass through the same point at the same time 𝑡. So we must find a solution to the vector equation 𝑟⃗ 1 (𝑡) = 𝑟⃗ 2 (𝑡), which is the same as finding a common solution to the three scalar equations 𝑡 = −2 − 2𝑡,

1 + 2𝑡 = 1 − 2𝑡,

3 − 2𝑡 = 1 + 𝑡.

Separately, the solutions are 𝑡 = −2∕3, 𝑡 = 0, and 𝑡 = 2∕3, so there is no common solution, and the particles don’t collide. To see if their paths cross, we find out if they pass through the same point at two possibly different times, 𝑡1 and 𝑡2 . So we solve the equations 𝑡1 = −2 − 2𝑡2 ,

1 + 2𝑡1 = 1 − 2𝑡2 ,

3 − 2𝑡1 = 1 + 𝑡2 .

We solve the first two equations simultaneously and get 𝑡1 = 2, 𝑡2 = −2. Since these values also satisfy the third equation, the paths cross. The position of the first particle at time 𝑡 = 2 is the same as the position of the second particle at time 𝑡 = −2, namely the point (2, 5, −1). Example 10

Are the lines 𝑥 = −1 + 𝑡, 𝑦 = 1 + 2𝑡, 𝑧 = 5 − 𝑡 and 𝑥 = 2 + 2𝑡, 𝑦 = 4 + 𝑡, 𝑧 = 3 + 𝑡 parallel? Do they intersect?

Solution

In vector form the lines are parameterized by ⃗ + 𝑡(⃗𝑖 + 2𝑗⃗ − 𝑘 ⃗) 𝑟⃗ = −⃗𝑖 + 𝑗⃗ + 5𝑘 ⃗ ⃗) ⃗ ⃗ ⃗ ⃗ 𝑟⃗ = 2𝑖 + 4𝑗 + 3𝑘 + 𝑡(2𝑖 + 𝑗 + 𝑘

17.1 PARAMETERIZED CURVES

891

⃗ and 2⃗𝑖 + 𝑗⃗ + 𝑘 ⃗ are not multiples of each other, so the lines are Their direction vectors ⃗𝑖 + 2𝑗⃗ − 𝑘 not parallel. To find out if they intersect, we see if they pass through the same point at two possibly different times, 𝑡1 and 𝑡2 : −1 + 𝑡1 = 2 + 2𝑡2 ,

1 + 2𝑡1 = 4 + 𝑡2 ,

5 − 𝑡1 = 3 + 𝑡2 .

The first two equations give 𝑡1 = 1, 𝑡2 = −1. Since these values do not satisfy the third equation, the paths do not cross, and so the lines do not intersect. The next example shows how to tell if two different parameterizations give the same line. Example 11

Show that the following two lines are the same: ⃗ + 𝑡(3⃗𝑖 + 6𝑗⃗ − 3𝑘 ⃗) 𝑟⃗ = −⃗𝑖 − 𝑗⃗ + 𝑘 ⃗ + 𝑡(−⃗𝑖 − 2𝑗⃗ + 𝑘 ⃗) 𝑟⃗ = ⃗𝑖 + 3𝑗⃗ − 𝑘 ⃗ and −⃗𝑖 − 2𝑗⃗ + 𝑘 ⃗ , are multiples of each other, The direction vectors of the two lines, 3⃗𝑖 + 6𝑗⃗ − 3𝑘 so the lines are parallel. To see if they are the same, we pick a point on the first line and see if it is on ⃗. the second line. For example, the point on the first line with 𝑡 = 0 has position vector −⃗𝑖 − 𝑗⃗ + 𝑘 Solving ⃗ + 𝑡(−⃗𝑖 − 2𝑗⃗ + 𝑘 ⃗ ) = −⃗𝑖 − 𝑗⃗ + 𝑘 ⃗, ⃗𝑖 + 3𝑗⃗ − 𝑘

Solution

we get 𝑡 = 2, so the two lines have a point in common. Thus, they are the same line, parameterized in two different ways.

Exercises and Problems for Section 17.1 Online Resource: Additional Problems for Section 17.1 EXERCISES

In Exercises 1–6, find a parameterization for the curve. 𝑦 𝑦 2. 1.

In Exercises 7–17, find parametric equations for the line. ⃗ and through 7. The line in the direction of the vector ⃗𝑖 − 𝑘 the point (0, 1, 0).

2

1

⃗ and 8. The line in the direction of the vector ⃗𝑖 + 2𝑗⃗ − 𝑘 through the point (3, 0, −4).

𝑥 −1

1 −1

𝑥 2

−2 𝑦

3.

𝑦

4.

⃗ and 11. The line in the direction of the vector 3⃗𝑖 − 3𝑗⃗ + 𝑘 through the point (1, 2, 3).

1

2 1

𝑥 −1

𝑥 1

2

5.

1

3

𝑦 3

2

2

1

⃗ and 12. The line in the direction of the vector 2⃗𝑖 + 2𝑗⃗ − 3𝑘 through the point (−3, 4, −2). 13. The line through (−3, −2, 1) and (−1, −3, −1).

𝑦

6.

9. The line parallel to the 𝑧-axis passing through the point (1, 0, 0). ⃗ and 10. The line in the direction of the vector 5𝑗⃗ + 2𝑘 through the point (5, −1, 1).

Segment of parabola

14. The line through the points (1, 5, 2) and (5, 0, −1). 15. The line through the points (2, 3, −1) and (5, 2, 0). 16. The line through (3, −2, 2) and intersecting the 𝑦-axis at 𝑦 = 2. 𝑥

1

1 𝑥 1

2

2

17. The line intersecting the 𝑥-axis at 𝑥 = 3 and the 𝑧-axis at 𝑧 = −5.

892

Chapter 17 PARAMETERIZATION AND VECTOR FIELDS

In Exercises 18–34, find a parameterization for the curve. 18. A line segment between (2, 1, 3) and (4, 3, 2). 19. A circle of radius 3 centered on the 𝑧-axis and lying in the plane 𝑧 = 5. 20. A line perpendicular to the plane 𝑧 = 2𝑥 − 3𝑦 + 7 and through the point (1, 1, 6).

33. The ellipse of major diameter 6 along the 𝑥-axis and minor diameter 4 along the 𝑦-axis, centered at the origin. 34. The ellipse of major diameter 3 parallel to the 𝑥-axis and minor diameter 2 parallel to the 𝑧-axis, centered at (0, 1, −2).

21. The circle of radius 2 in the 𝑥𝑦-plane, centered at the origin, clockwise.

In Exercises 35–42, find a parametric equation for the curve segment.

22. The circle of radius 2 parallel to the 𝑥𝑦-plane, centered at the point (0, 0, 1), and traversed counterclockwise when viewed from below.

35. Line from (−1, 2, −3) to (2, 2, 2).

23. The circle of radius 2 in the 𝑥𝑧-plane, centered at the origin. 24. The circle of radius 3 parallel to the 𝑥𝑦-plane, centered at the point (0, 0, 2). 25. The circle of radius 3 in the 𝑦𝑧-plane, centered at the point (0, 0, 2). 26. The circle of radius 5 parallel to the 𝑦𝑧-plane, centered at the point (−1, 0, −2). 27. The curve 𝑥 = 𝑦2 in the 𝑥𝑦-plane.

36. Line from 𝑃0 = (−1, −3) to 𝑃1 = (5, 2). 37. Line from 𝑃0 = (1, −3, 2) to 𝑃1 = (4, 1, −3). 38. Semicircle from (0, 0, 5) to (0, 0, −5) in the 𝑦𝑧-plane with 𝑦 ≥ 0. 39. Semicircle from (1, 0, 0) to (−1, 0, 0) in the 𝑥𝑦-plane with 𝑦 ≥ 0. √ 40. Graph of 𝑦 = 𝑥 from (1, 1) to (16, 4).

41. Arc of a circle of radius 5 from 𝑃 = (0, 0) to 𝑄 = (10, 0). 42. Quarter-ellipse from (4, 0, 3) to (0, −3, 3) in the plane 𝑧 = 3.

28. The curve 𝑦 = 𝑥3 in the 𝑥𝑦-plane. 29. The curve 𝑥 = −3𝑧2 in the 𝑥𝑧-plane. 30. The √ curve in which the plane 𝑧 = 2 cuts the surface 𝑧 = 𝑥2 + 𝑦2 .

31. The curve 𝑦 = 4 − 5𝑥4 through the point (0, 4, 4), parallel to the 𝑥𝑦-plane. 32. The ellipse of major diameter 5 parallel to the 𝑦-axis and minor diameter 2 parallel to the 𝑧-axis, centered at (0, 1, −2).

In Exercises 43–46, find parametric equations for a helix satisfying the given conditions. 43. Centered on the 𝑧-axis, with radius 10. 44. Centered on the 𝑥-axis, with radius 5. 45. Centered on the 𝑦-axis, with radius 2. 46. Centered on the vertical line passing through (3, 5, 0), with radius 1.

PROBLEMS In Problems 47–51, parameterize the line through 𝑃 = (2, 5) and 𝑄 = (12, 9) so that the points 𝑃 and 𝑄 correspond to the given parameter values. 47. 𝑡 = 0 and 1

48. 𝑡 = 0 and 5

49. 𝑡 = 20 and 30

50. 𝑡 = 10 and 11

51. 𝑡 = 0 and −1 52. At the point where 𝑡 = −1, find an equation for the plane perpendicular to the line 𝑥 = 5 − 3𝑡,

𝑦 = 5𝑡 − 7,

𝑧 = 6. 𝑡

53. Determine whether the following line is parallel to the plane 2𝑥 − 3𝑦 + 5𝑧 = 5: 𝑥 = 5 + 7𝑡,

𝑦 = 4 + 3𝑡,

𝑧 = −3 − 2𝑡.

54. Show that the equations 𝑥 = 3 + 𝑡, 𝑦 = 2𝑡, 𝑧 = 1 − 𝑡 satisfy the equations 𝑥 + 𝑦 + 3𝑧 = 6 and 𝑥 − 𝑦 − 𝑧 = 2. What does this tell you about the curve parameterized by these equations? 55. (a) Explain why the line of intersection of two planes must be parallel to the cross product of a normal vector to the first plane and a normal vector to the second. (b) Find a vector parallel to the line of intersection of the two planes 𝑥 + 2𝑦 − 3𝑧 = 7 and 3𝑥 − 𝑦 + 𝑧 = 0. (c) Find parametric equations for the line in part (b). 56. Find an equation for the plane containing the point (2, 3, 4) and the line 𝑥 = 1 + 2𝑡, 𝑦 = 3 − 𝑡, 𝑧 = 4 + 𝑡. 57. (a) Find an equation for the line perpendicular to the plane 2𝑥 − 3𝑦 = 𝑧 and through the point (1, 3, 7). (b) Where does the line cut the plane? (c) What is the distance between the point (1, 3, 7) and the plane?

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17.1 PARAMETERIZED CURVES

58. Consider two points 𝑃0 and 𝑃1 in 3-space. (a) Show that the line segment from 𝑃0 to 𝑃1 can be parameterized by ⃖⃖⃖⃖⃖⃖⃗0 + 𝑡𝑂𝑃 ⃖⃖⃖⃖⃖⃖⃗1 , 𝑟⃗ (𝑡) = (1 − 𝑡)𝑂𝑃

0 ≤ 𝑡 ≤ 1.

(b) What is represented by the parametric equation ⃖⃖⃖⃖⃖⃖⃗0 + (1 − 𝑡)𝑂𝑃 ⃖⃖⃖⃖⃖⃖⃗1 , 𝑟⃗ (𝑡) = 𝑡𝑂𝑃

0 ≤ 𝑡 ≤ 1?

59. (a) Find a vector parallel to the line of intersection of the planes 2𝑥 − 𝑦 − 3𝑧 = 0 and 𝑥 + 𝑦 + 𝑧 = 1. (b) Show that the point (1, −1, 1) lies on both planes. (c) Find parametric equations for the line of intersection. 60. Find the intersection of the line 𝑥 = 5 + 7𝑡, 𝑦 = 4 + 3𝑡, 𝑧 = −3 − 2𝑡 and the plane 2𝑥 − 3𝑦 + 5𝑧 = −7. In Problems 61–64, are the lines 𝐿1 and 𝐿2 the same line? 61. 𝐿1 : 𝑥 = 5 + 𝑡, 𝑦 = 3 − 2𝑡, 𝑧 = 5𝑡 𝐿2 : 𝑥 = 5 + 2𝑡, 𝑦 = 3 − 4𝑡, 𝑧 = 10𝑡 62. 𝐿1 : 𝑥 = 2 + 3𝑡, 𝑦 = 1 + 4𝑡, 𝑧 = 6 − 𝑡 𝐿2 : 𝑥 = 2 + 6𝑡, 𝑦 = 4 + 3𝑡, 𝑧 = 3 − 2𝑡 63. 𝐿1 : 𝑥 = 2 + 3𝑡, 𝑦 = 1 + 4𝑡, 𝑧 = 6 − 𝑡 𝐿2 : 𝑥 = 5 + 6𝑡, 𝑦 = 5 + 8𝑡, 𝑧 = 5 − 2𝑡

72. The line perpendicular to the surface 𝑧 = 𝑥2 + 𝑦2 at the point (1, 2, 5). 73. The line through the point (−4, 2, 3) and parallel to a line in the 𝑦𝑧-plane which makes a 45◦ angle with the positive 𝑦-axis and the positive 𝑧-axis. 74. Is the point (−3, −4, 2) visible from the point (4, 5, 0) if there is an opaque ball of radius 1 centered at the origin? 75. Two particles are traveling through space. At time 𝑡 the first particle is at the point (−1 + 𝑡, 4 − 𝑡, −1 + 2𝑡) and the second particle is at (−7 + 2𝑡, −6 + 2𝑡, −1 + 𝑡). (a) Describe the two paths in words. (b) Do the two particles collide? If so, when and where? (c) Do the paths of the two particles cross? If so, where? 76. Match the parameterizations with their graphs in Figure 17.9. (a) (b) (c) (d)

𝑥 = 2 cos 4𝜋𝑡, 𝑦 = 2 sin 4𝜋𝑡, 𝑧 = 𝑡 𝑥 = 2 cos 4𝜋𝑡, 𝑦 = sin 4𝜋𝑡, 𝑧 = 𝑡 𝑥 = 0.5𝑡 cos 4𝜋𝑡, 𝑦 = 0.5𝑡 sin 4𝜋𝑡, 𝑧 = 𝑡 𝑥 = 2 cos 4𝜋𝑡, 𝑦 = 2 sin 4𝜋𝑡, 𝑧 = 0.5𝑡3

(I)

64. 𝐿1 : 𝑥 = 1 + 2𝑡, 𝑦 = 1 − 3𝑡, 𝑧 = 1 + 𝑡 𝐿2 : 𝑥 = 1 − 4𝑡, 𝑦 = 6𝑡, 𝑧 = 4 − 2𝑡

(II)

𝑧

𝑧

In Problems 65–67 two parameterized lines are given. Are they the same line? 𝑦

⃗ 65. 𝑟⃗ 1 (𝑡) = (5 − 3𝑡)⃗𝑖 + 2𝑡𝑗⃗ + (7 + 𝑡)𝑘 ⃗ ⃗ ⃗ 𝑟⃗ 2 (𝑡) = (5 − 6𝑡)𝑖 + 4𝑡𝑗 + (7 + 3𝑡)𝑘

𝑦

⃗ 66. 𝑟⃗ 1 (𝑡) = (5 − 3𝑡)⃗𝑖 + (1 + 𝑡)𝑗⃗ + 2𝑡𝑘 ⃗ 𝑟⃗ 2 (𝑡) = (2 + 6𝑡)⃗𝑖 + (2 − 2𝑡)𝑗⃗ + (2 − 4𝑡)𝑘 ⃗ 67. 𝑟⃗ 1 (𝑡) = (5 − 3𝑡)⃗𝑖 + (1 + 𝑡)𝑗⃗ + 2𝑡𝑘 ⃗ ⃗ ⃗ 𝑟⃗ 2 (𝑡) = (2 + 6𝑡)𝑖 + (2 − 2𝑡)𝑗 + (3 − 4𝑡)𝑘

𝑥 𝑥 (III)

(IV)

𝑧

𝑧

68. If it exists, find the value of 𝑐 for which the lines 𝑙(𝑡) = (𝑐 + 𝑡, 1 + 𝑡, 5 + 𝑡) and 𝑚(𝑠) = (𝑠, 1 − 𝑠, 3 + 𝑠) intersect. 𝑦

⃗) 69. (a) Where does the line 𝑟⃗ = 2⃗𝑖 + 5𝑗⃗ + 𝑡(3⃗𝑖 + 𝑗⃗ + 2𝑘 cut the plane 𝑥 + 𝑦 + 𝑧 = 1? (b) Find a vector perpendicular to the line and lying in the plane. (c) Find an equation for the line that passes through the point of intersection of the line and plane, is perpendicular to the line, and lies in the plane.

𝑦

𝑥 𝑥

Figure 17.9

In Problems 70–73, find parametric equations for the line. 70. The line of intersection of the planes 𝑥 − 𝑦 + 𝑧 = 3 and 2𝑥 + 𝑦 − 𝑧 = 5. 71. The line of intersection of the planes 𝑥 + 𝑦 + 𝑧 = 3 and 𝑥 − 𝑦 + 2𝑧 = 2.

In Problems 77–80 find 𝑐 so that one revolution about the 𝑧axis of the helix gives an increase of Δ𝑧 in the 𝑧-coordinate. 77. 𝑥 = 2 cos 𝑡, 𝑦 = 2 sin 𝑡, 𝑧 = 𝑐𝑡, Δ𝑧 = 15

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Chapter 17 PARAMETERIZATION AND VECTOR FIELDS

78. 𝑥 = 2 cos 𝑡, 𝑦 = 2 sin 𝑡, 𝑧 = 𝑐𝑡, Δ𝑧 = 50 79. 𝑥 = 2 cos 3𝑡, 𝑦 = 2 sin 3𝑡, 𝑧 = 𝑐𝑡, Δ𝑧 = 10 80. 𝑥 = 2 cos 𝜋𝑡, 𝑦 = 2 sin 𝜋𝑡, 𝑧 = 𝑐𝑡, Δ𝑧 = 20 81. For 𝑡 > 0, a particle moves along the curve 𝑥 = 𝑎 + 𝑏 sin 𝑘𝑡, 𝑦 = 𝑎 + 𝑏 cos 𝑘𝑡, where 𝑎, 𝑏, 𝑘 are positive constants. (a) Describe the motion in words. (b) What is the effect on the curve of the following changes? (i) Increasing 𝑏 (ii) Increasing 𝑎 (iii) Increasing 𝑘 (iv) Setting 𝑎 and 𝑏 equal 82. In the Atlantic Ocean off the coast of Newfoundland, Canada, the temperature and salinity (saltiness) vary throughout the year. Figure 17.10 shows a parametric curve giving the average temperature, 𝑇 (in ◦ C) and salinity (in grams of salt per kg of water) for 𝑡 in months, with 𝑡 = 1 corresponding to mid-January.1 (a) (b) (c) (d)

Why does the parameterized curve form a loop? When is the water temperature highest? When is the water saltiest? Estimate 𝑑𝑇 ∕𝑑𝑡 at 𝑡 = 6, and give the units. What is the meaning of your answer for seawater? salinity (gm/kg)

𝑡=3



36.1

✻ 35.9 𝑡 = 1 35.7

𝑇 (◦ C) 14

16

18

20

22

Figure 17.10

83. A light shines on the helix of Example 3 on page 886 from far down each axis. Sketch the shadow the helix casts on each of the coordinate planes: 𝑥𝑦, 𝑥𝑧, and 𝑦𝑧. 84. The paraboloid 𝑧 = 𝑥2 +𝑦2 and the plane 𝑧 = 2𝑥+4𝑦+4 intersect in a curve in 3-space. (a) Show that the shadow of the intersection in the 𝑥𝑦plane is a circle and find its center and radius. (b) Parameterize the circle in the 𝑥𝑦-plane. (c) Parameterize the intersection of the paraboloid and the plane in 3-space. 85. For a positive constant 𝑎 and 𝑡 ≥ 0, the parametric equations I-V represent the curves described in (a)-(e). Match each description (a)-(e) with its parametric equations and write an equation involving only 𝑥 and 𝑦 for the curve. (a) Line through the origin. 1 Based

(b) (c) (d) (e)

Line not through the origin. Hyperbola opening along 𝑥-axis. Circle traversed clockwise. Circle traversed counterclockwise.

I. 𝑥 = 𝑎 sin 𝑡, 𝑦 = 𝑎 cos 𝑡 II. 𝑥 = 𝑎 sin 𝑡, 𝑦 = 𝑎 sin 𝑡 III. 𝑥 = 𝑎 cos 𝑡, 𝑦 = 𝑎 sin 𝑡 IV. 𝑥 = 𝑎 cos2 𝑡, 𝑦 = 𝑎 sin2 𝑡 V. 𝑥 = 𝑎∕ cos 𝑡, 𝑦 = 𝑎 tan 𝑡 86. (a) Find a parametric equation for the line through the ⃗. point (2, 1, 3) and in the direction of 𝑎⃗𝑖 + 𝑏𝑗⃗ + 𝑐 𝑘 (b) Find conditions on 𝑎, 𝑏, 𝑐 so that the line you found in part (a) goes through the origin. Give a reason for your answer. 87. Consider the line 𝑥 = 5 − 2𝑡, 𝑦 = 3 + 7𝑡, 𝑧 = 4𝑡 and the plane 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑. All the following questions have many possible answers. Find values of 𝑎, 𝑏, 𝑐, 𝑑 such that: (a) The plane is perpendicular to the line. (b) The plane is perpendicular to the line and through the point (5, 3, 0). (c) The line lies in the plane. 88. Explain the significance of the constants 𝛼 > 0 and 𝛽 > 0 in the family of helices given by 𝑟⃗ = 𝛼 cos 𝑡⃗𝑖 + ⃗. 𝛼 sin 𝑡𝑗⃗ + 𝛽𝑡𝑘 89. Find parametric equations of the line passing through the points (1, 2, 3), (3, 5, 7) and calculate the shortest distance from the line to the origin. 90. Show that for a fixed value of 𝜃, the line parameterized by 𝑥 = cos 𝜃 + 𝑡 sin 𝜃, 𝑦 = sin 𝜃 − 𝑡 cos 𝜃 and 𝑧 = 𝑡 lies on the graph of the hyperboloid 𝑥2 + 𝑦2 = 𝑧2 + 1. ⃗ 91. A line has equation 𝑟⃗ = 𝑎⃗ +𝑡𝑏⃗ where 𝑟⃗ = 𝑥⃗𝑖 +𝑦𝑗⃗ +𝑧𝑘 and 𝑎⃗ and 𝑏⃗ are constant vectors such that 𝑎⃗ ≠ 0⃗ , 𝑏⃗ ≠ 0⃗ , 𝑏⃗ not parallel or perpendicular to 𝑎⃗ . For each of the planes (a)–(c), pick the equation (i)–(ix) which represents it. Explain your choice. (a) A plane perpendicular to the line and through the origin. (b) A plane perpendicular to the line and not through the origin. (c) A plane containing the line. (i) 𝑎⃗ ⋅ 𝑟⃗ = ||𝑏⃗ || (ii) 𝑏⃗ ⋅ 𝑟⃗ = ||𝑎⃗ || (iii) 𝑎⃗ ⋅ 𝑟⃗ = 𝑏⃗ ⋅ 𝑟⃗ (iv) (𝑎⃗ × 𝑏⃗ ) ⋅ (⃗𝑟 − 𝑎⃗ ) = 0 (v) 𝑟⃗ − 𝑎⃗ = 𝑏⃗ (vi) 𝑎⃗ ⋅ 𝑟⃗ = 0 ⃗ (vii) 𝑏 ⋅ 𝑟⃗ = 0 (viii) 𝑎⃗ + 𝑟⃗ = 𝑏⃗ ⃗ ⃗ (ix) (𝑎⃗ ×𝑏 )⋅(⃗𝑟 −𝑏 ) = ||𝑎⃗ ||

92. (a) Find a parametric equation for the line through the point (1, 5, 2) and in the direction of the vector ⃗. 2⃗𝑖 + 3𝑗⃗ − 𝑘 (b) By minimizing the square of the distance from a point on the line to the origin, find the exact point on the line which is closest to the origin.

on http://www.vub.ac.be. Accessed November, 2011.

17.1 PARAMETERIZED CURVES

93. A plane from Denver, Colorado, (altitude 1650 meters) flies to Bismark, North Dakota (altitude 550 meters). It travels at 650 km/hour at a constant height of 8000 meters above the line joining Denver and Bismark. Bismark is about 850 km in the direction 60◦ north of east from Denver. Find parametric equations describing the plane’s motion. Assume the origin is at sea level beneath Denver, that the 𝑥-axis points east and the 𝑦-axis points north, and that the earth is flat. Measure distances in kilometers and time in hours.

(iii) (𝑥(𝑡) + 1, 𝑦(𝑡)) 𝑦

(A)

(i) 𝑃1 and 𝑃2 ?

(𝑥(𝑡) + 1, 𝑦(𝑡) + 1) 𝑦

(B)

𝑥

(C)

𝑦

94. The vector 𝑛⃗ is perpendicular to the plane 𝑃1 . The vector 𝑣⃗ is parallel to the line 𝐿. (a) If 𝑛⃗ ⋅ 𝑣⃗ = 0, what does this tell you about the directions of 𝑃1 and 𝐿? (Are they parallel? Perpendicular? Or is it impossible to tell?) (b) Suppose 𝑛⃗ × 𝑣⃗ ≠ 0⃗ . The plane 𝑃2 has normal 𝑛⃗ × 𝑣⃗ . What can you say about the directions of

(iv)

𝑥

(D)

𝑦

𝑥

(E)

𝑦

895

𝑥

(F)

𝑦

𝑥

𝑥

(ii) 𝐿 and 𝑃2 ?

95. Figure 17.11 shows the parametric curve 𝑥 = 𝑥(𝑡), 𝑦 = 𝑦(𝑡) for 𝑎 ≤ 𝑡 ≤ 𝑏.

(G)

𝑦

(H)

𝑦

𝑦 𝑥

𝑥

𝑥

Figure 17.11 (a) Match a graph to each of the parametric curves given, for the same 𝑡 values, by (i) (−𝑥(𝑡), −𝑦(𝑡))

(ii) (−𝑥(𝑡), 𝑦(𝑡))

(b) Which of the following could be the formulas for the functions 𝑥(𝑡), 𝑦(𝑡)? (i) 𝑥 = 10 cos 𝑡 𝑦 = 10 sin 𝑡 (ii) 𝑥 = (10 + 8𝑡) cos 𝑡 𝑦 = (10 + 8𝑡) sin 𝑡 2 2 (iii) 𝑥 = 𝑒𝑡 ∕200 cos 𝑡 𝑦 = 𝑒𝑡 ∕200 sin 𝑡 (iv) 𝑥 = (10 − 8𝑡) cos 𝑡 𝑦 = (10 − 8𝑡) sin 𝑡 (v) 𝑥 = 10 cos(𝑡2 + 𝑡) 𝑦 = 10 sin(𝑡2 + 𝑡)

Strengthen Your Understanding In Problems 96–97, explain what is wrong with the statement.

Are the statements in Problems 101–112 true or false? Give reasons for your answer.

96. The curve parameterized by 𝑟⃗ 1 (𝑡) = 𝑟⃗ (𝑡 − 2), defined for all 𝑡, is a shift in the ⃗𝑖 -direction of the curve parameterized by 𝑟⃗ (𝑡).

101. The parametric curve 𝑥 = 3𝑡 + 2, 𝑦 = −2𝑡 for 0 ≤ 𝑡 ≤ 5 passes through the origin.

⃗ 97. All points of the curve 𝑟⃗ (𝑡) = 𝑅 cos 𝑡⃗𝑖 + 𝑅 sin 𝑡𝑗⃗ + 𝑡𝑘 are the same distance, 𝑅, from the origin. In Problems 98–100, give an example of: 98. Parameterizations of two different circles that have the same center and equal radii.

102. The parametric curve 𝑥 = 𝑡2 , 𝑦 = 𝑡4 for 0 ≤ 𝑡 ≤ 1 is a parabola. 103. A parametric curve 𝑥 = 𝑔(𝑡), 𝑦 = ℎ(𝑡) for 𝑎 ≤ 𝑡 ≤ 𝑏 is always the graph of a function 𝑦 = 𝑓 (𝑥). 104. The parametric curve 𝑥 = (3𝑡 + 2)2 , 𝑦 = (3𝑡 + 2)2 − 1 for 0 ≤ 𝑡 ≤ 3 is a line.

99. Parameterizations of two different lines that intersect at the point (1, 2, 3).

105. The parametric curve 𝑥 = − sin 𝑡, 𝑦 = − cos 𝑡 for 0 ≤ 𝑡 ≤ 2𝜋 traces out a unit circle counterclockwise as 𝑡 increases.

100. A parameterization of the line 𝑥 = 𝑡, 𝑦 = 2𝑡, 𝑧 = 3 + 4𝑡 that is not given by linear functions.

106. A parameterization of the graph of 𝑦 = ln 𝑥 for 𝑥 > 0 is given by 𝑥 = 𝑒𝑡 , 𝑦 = 𝑡 for −∞ < 𝑡 < ∞.

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Chapter 17 PARAMETERIZATION AND VECTOR FIELDS

107. Both 𝑥 = −𝑡 + 1, 𝑦 = 2𝑡 and 𝑥 = 2𝑠, 𝑦 = −4𝑠 + 2 describe the same line. 108. The line of intersection of the two planes 𝑧 = 𝑥 + 𝑦 and 𝑧 = 1 − 𝑥 − 𝑦 can be parameterized by 𝑥 = 𝑡, 𝑦 = 1 − 𝑡, 𝑧 = 12 . 2 109. The two lines given by 𝑥 = 𝑡, 𝑦 = 2 + 𝑡, 𝑧 = 3 + 𝑡 and 𝑥 = 2𝑠, 𝑦 = 1 − 𝑠, 𝑧 = 𝑠 do not intersect.

17.2

110. The line parameterized by 𝑥 = 1, 𝑦 = 2𝑡, 𝑧 = 3 + 𝑡 is parallel to the 𝑥-axis. 111. The equation 𝑟⃗ (𝑡) = 3𝑡⃗𝑖 + (6𝑡 + 1)𝑗⃗ parameterizes a line. 112. The lines parameterized by 𝑟⃗ 1 (𝑡) = 𝑡⃗𝑖 + (−2𝑡 + 1)𝑗⃗ and 𝑟⃗ 2 (𝑡) = (2𝑡 + 5)⃗𝑖 + (−𝑡)𝑗⃗ are parallel.

MOTION, VELOCITY, AND ACCELERATION In this section we see how to find the vector quantities of velocity and acceleration from a parametric equation for the motion of an object.

The Velocity Vector The velocity of a moving particle can be represented by a vector with the following properties:

The velocity vector of a moving object is a vector 𝑣⃗ such that: ∙ The magnitude of 𝑣⃗ is the speed of the object. ∙ The direction of 𝑣⃗ is the direction of motion. Thus, the speed of the object is ‖⃗ 𝑣 ‖ and the velocity vector is tangent to the object’s path. Example 1

A child is sitting on a Ferris wheel of diameter 10 meters, making one revolution every 2 minutes. Find the speed of the child and draw velocity vectors at two different times.

Solution

The child moves at a constant speed around a circle of radius 5 meters, completing one revolution every 2 minutes. One revolution around a circle of radius 5 is a distance of 10𝜋, so the child’s speed is 10𝜋∕2 = 5𝜋 ≈ 15.7 m/min. Hence, the magnitude of the velocity vector is 15.7 m/min. The direction of motion is tangent to the circle, and hence perpendicular to the radius at that point. Figure 17.12 shows the direction of the vector at two different times. Velocity

15.7 m/min

5m

5m

Velocity

15.7 m/min

Figure 17.12: Velocity vectors of a child on a Ferris wheel (note that vectors would be in opposite direction if viewed from the other side)

Computing the Velocity We find the velocity, as in one-variable calculus, by taking a limit. If the position vector of the particle is 𝑟⃗ (𝑡) at time 𝑡, then the displacement vector between its positions at times 𝑡 and 𝑡 + Δ𝑡 is Δ⃗𝑟 = 𝑟⃗ (𝑡 + Δ𝑡) − 𝑟⃗ (𝑡). (See Figure 17.13.) Over this interval, Average velocity =

Δ⃗𝑟 . Δ𝑡

In the limit as Δ𝑡 goes to zero we have the instantaneous velocity at time 𝑡:

17.2 MOTION, VELOCITY, AND ACCELERATION

897

The velocity vector, 𝑣⃗ (𝑡), of a moving object with position vector 𝑟⃗ (𝑡) at time 𝑡 is 𝑣⃗ (𝑡) = lim

Δ𝑡→0

𝑟⃗ (𝑡 + Δ𝑡) − 𝑟⃗ (𝑡) Δ⃗𝑟 = lim , Δ𝑡→0 Δ𝑡 Δ𝑡

whenever the limit exists. We use the notation 𝑣⃗ =

𝑑⃗𝑟 = 𝑟⃗ ′ (𝑡). 𝑑𝑡

Notice that the direction of the velocity vector 𝑟⃗ ′ (𝑡) in Figure 17.13 is approximated by the direction of the vector Δ⃗𝑟 and that the approximation gets better as Δ𝑡 → 0. Δ⃗𝑟 = 𝑟⃗ (𝑡 + Δ𝑡) − 𝑟⃗ (𝑡)



𝑟⃗ (𝑡)

𝑟⃗ ′ (𝑡) 𝑟⃗ (𝑡 + Δ𝑡) Figure 17.13: The change, Δ⃗𝑟 , in the position vector for a particle moving on a curve and the velocity vector 𝑣⃗ = 𝑟⃗ ′ (𝑡)

The Components of the Velocity Vector If we represent a curve parametrically by 𝑥 = 𝑓 (𝑡), 𝑦 = 𝑔(𝑡), 𝑧 = ℎ(𝑡), then we can write its position ⃗ . Now we can compute the velocity vector: vector as: 𝑟⃗ (𝑡) = 𝑓 (𝑡)⃗𝑖 + 𝑔(𝑡)𝑗⃗ + ℎ(𝑡)𝑘 𝑟⃗ (𝑡 + Δ𝑡) − 𝑟⃗ (𝑡) Δ𝑡 ⃗ ) − (𝑓 (𝑡)⃗𝑖 + 𝑔(𝑡)𝑗⃗ + ℎ(𝑡)𝑘 ⃗) (𝑓 (𝑡 + Δ𝑡)⃗𝑖 + 𝑔(𝑡 + Δ𝑡)𝑗⃗ + ℎ(𝑡 + Δ𝑡)𝑘 = lim Δ𝑡→0 Δ𝑡 ( ) 𝑓 (𝑡 + Δ𝑡) − 𝑓 (𝑡) ⃗ 𝑔(𝑡 + Δ𝑡) − 𝑔(𝑡) ⃗ ℎ(𝑡 + Δ𝑡) − ℎ(𝑡) ⃗ = lim 𝑖 + 𝑗 + 𝑘 Δ𝑡→0 Δ𝑡 Δ𝑡 Δ𝑡 ′ ⃗ ′ ′ ⃗ = 𝑓 (𝑡)𝑖 + 𝑔 (𝑡)𝑗⃗ + ℎ (𝑡)𝑘 𝑑𝑥 ⃗ 𝑑𝑦 ⃗ 𝑑𝑧 ⃗ = 𝑖 + 𝑗 + 𝑘. 𝑑𝑡 𝑑𝑡 𝑑𝑡

𝑣⃗ (𝑡) = lim

Δ𝑡→0

Thus, we have the following result: The components of the velocity vector of a particle moving in space with position vector ⃗ at time 𝑡 are given by 𝑟⃗ (𝑡) = 𝑓 (𝑡)⃗𝑖 + 𝑔(𝑡)𝑗⃗ + ℎ(𝑡)𝑘 ⃗ = 𝑣⃗ (𝑡) = 𝑓 ′ (𝑡)⃗𝑖 + 𝑔 ′ (𝑡)𝑗⃗ + ℎ′ (𝑡)𝑘

𝑑𝑥 ⃗ 𝑑𝑦 ⃗ 𝑑𝑧 ⃗ 𝑖 + 𝑗 + 𝑘. 𝑑𝑡 𝑑𝑡 𝑑𝑡

Example 2

Find the components of the velocity vector for the child on the Ferris wheel in Example 1 using a coordinate system which has its origin at the center of the Ferris wheel and which makes the rotation counterclockwise.

Solution

The Ferris wheel has radius 5 meters and completes 1 revolution counterclockwise every 2 minutes. The motion is parameterized by an equation of the form 𝑟⃗ (𝑡) = 5 cos(𝜔𝑡)⃗𝑖 + 5 sin(𝜔𝑡)𝑗⃗ ,

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Chapter 17 PARAMETERIZATION AND VECTOR FIELDS

where 𝜔 is chosen to make the period 2 minutes. Since the period of cos(𝜔𝑡) and sin(𝜔𝑡) is 2𝜋∕𝜔, we must have 2𝜋 = 2, so 𝜔 = 𝜋. 𝜔 Thus, the motion is described by the equation 𝑟⃗ (𝑡) = 5 cos(𝜋𝑡)⃗𝑖 + 5 sin(𝜋𝑡)𝑗⃗ , where 𝑡 is in minutes. The velocity is given by 𝑣⃗ =

𝑑𝑥 ⃗ 𝑑𝑦 ⃗ 𝑖 + 𝑗 = −5𝜋 sin(𝜋𝑡)⃗𝑖 + 5𝜋 cos(𝜋𝑡)𝑗⃗ . 𝑑𝑡 𝑑𝑡

To check, we calculate the magnitude of 𝑣⃗ , √ √ ‖⃗ 𝑣 ‖ = (−5𝜋)2 sin2 (𝜋𝑡) + (5𝜋)2 cos2 (𝜋𝑡) = 5𝜋 sin2 (𝜋𝑡) + cos2 (𝜋𝑡) = 5𝜋 ≈ 15.7,

which agrees with the speed we calculated in Example 1. To see that the direction is correct, we must show that the vector 𝑣⃗ at any time 𝑡 is perpendicular to the position vector of the child at time 𝑡. To do this, we compute the dot product of 𝑣⃗ and 𝑟⃗ : 𝑣⃗ ⋅ 𝑟⃗ = (−5𝜋 sin(𝜋𝑡)⃗𝑖 + 5𝜋 cos(𝜋𝑡)𝑗⃗ ) ⋅ (5 cos(𝜋𝑡)⃗𝑖 + 5 sin(𝜋𝑡)𝑗⃗ ) = −25𝜋 sin(𝜋𝑡) cos(𝜋𝑡) + 25𝜋 cos(𝜋𝑡) sin(𝜋𝑡) = 0. So the velocity vector, 𝑣⃗ , is perpendicular to 𝑟⃗ and hence tangent to the circle. The direction is counterclockwise, since in the first quadrant, 𝑥 is decreasing while 𝑦 is increasing. (See Figure 17.14.) 𝑦

𝑣⃗ = −5𝜋 sin(𝜋𝑡)⃗𝑖 + 5𝜋 cos(𝜋𝑡)𝑗⃗ 𝑟⃗ = 5 cos(𝜋𝑡)⃗𝑖 + 5 sin(𝜋𝑡)𝑗⃗ 𝑥

Figure 17.14: Velocity and radius vector of motion around a circle

Velocity Vectors and Tangent Lines Since the velocity vector is tangent to the path of motion, it can be used to find parametric equations for the tangent line, if there is one. Example 3

Find the tangent line at the point (1, 1, 2) to the curve defined by the parametric equation ⃗. 𝑟⃗ (𝑡) = 𝑡2 ⃗𝑖 + 𝑡3 𝑗⃗ + 2𝑡𝑘

Solution

⃗ . The velocity At time 𝑡 = 1 the particle is at the point (1, 1, 2) with position vector 𝑟⃗ 0 = ⃗𝑖 + 𝑗⃗ + 2𝑘 ′ 2 ′ ⃗. ⃗ ⃗ ⃗ vector at time 𝑡 is 𝑟⃗ (𝑡) = 2𝑡𝑖 + 3𝑡 𝑗 + 2𝑘 , so at time 𝑡 = 1 the velocity is 𝑣⃗ = 𝑟⃗ (1) = 2⃗𝑖 + 3𝑗⃗ + 2𝑘 The tangent line passes through (1, 1, 2) in the direction of 𝑣⃗ , so it has the parametric equation ⃗ ) + 𝑡(2⃗𝑖 + 3𝑗⃗ + 2𝑘 ⃗ ). 𝑟⃗ (𝑡) = 𝑟⃗ 0 + 𝑡⃗ 𝑣 = (⃗𝑖 + 𝑗⃗ + 2𝑘

17.2 MOTION, VELOCITY, AND ACCELERATION

899

The Acceleration Vector Just as the velocity of a particle moving in 2-space or 3-space is a vector quantity, so is the rate of change of the velocity of the particle, namely its acceleration. Figure 17.15 shows a particle at time 𝑡 with velocity vector 𝑣⃗ (𝑡) and then a little later at time 𝑡 + Δ𝑡. The vector Δ⃗ 𝑣 = 𝑣⃗ (𝑡 + Δ𝑡) − 𝑣⃗ (𝑡) is the change in velocity and points approximately in the direction of the acceleration. So, Average acceleration =

Δ⃗ 𝑣 . Δ𝑡

In the limit as Δ𝑡 → 0, we have the instantaneous acceleration at time 𝑡:

The acceleration vector of an object moving with velocity 𝑣⃗ (𝑡) at time 𝑡 is 𝑣⃗ (𝑡 + Δ𝑡) − 𝑣⃗ (𝑡) Δ⃗ 𝑣 = lim , Δ𝑡→0 Δ𝑡→0 Δ𝑡 Δ𝑡 2 𝑑 𝑟⃗ 𝑑 𝑣⃗ = if the limit exists. We use the notation 𝑎⃗ = = 𝑟⃗ ′′ (𝑡). 𝑑𝑡 𝑑𝑡2 𝑎⃗ (𝑡) = lim

𝑣⃗ (𝑡)

𝑣⃗ (𝑡) 𝑣⃗ (𝑡 + Δ𝑡) 𝑣⃗ (𝑡 + Δ𝑡)

Δ𝑣⃗ = 𝑣⃗ (𝑡 + Δ𝑡) − 𝑣⃗ (𝑡)

Figure 17.15: Computing the difference between two velocity vectors

Components of the Acceleration Vector If we represent a curve in space parametrically by 𝑥 = 𝑓 (𝑡), 𝑦 = 𝑔(𝑡), 𝑧 = ℎ(𝑡), we can express the acceleration in components. The velocity vector 𝑣⃗ (𝑡) is given by ⃗. 𝑣⃗ (𝑡) = 𝑓 ′ (𝑡)⃗𝑖 + 𝑔 ′ (𝑡)𝑗⃗ + ℎ′ (𝑡)𝑘 From the definition of the acceleration vector, we have 𝑎⃗ (𝑡) = lim

Δ𝑡→0

𝑣⃗ (𝑡 + Δ𝑡) − 𝑣⃗ (𝑡) 𝑑 𝑣⃗ = . Δ𝑡 𝑑𝑡

Using the same method to compute 𝑑 𝑣⃗ ∕𝑑𝑡 as we used to compute 𝑑⃗𝑟 ∕𝑑𝑡 on page 897, we obtain

The components of the acceleration vector, 𝑎⃗ (𝑡), at time 𝑡 of a particle moving in space with ⃗ at time 𝑡 are given by position vector 𝑟⃗ (𝑡) = 𝑓 (𝑡)⃗𝑖 + 𝑔(𝑡)𝑗⃗ + ℎ(𝑡)𝑘 ⃗ = 𝑎⃗ (𝑡) = 𝑓 ′′ (𝑡)⃗𝑖 + 𝑔 ′′ (𝑡)𝑗⃗ + ℎ′′ (𝑡)𝑘

𝑑2𝑥 ⃗ 𝑑2𝑦 ⃗ 𝑑2𝑧 ⃗ 𝑘. 𝑖 + 𝑗 + 𝑑𝑡2 𝑑𝑡2 𝑑𝑡2

Motion in a Circle and Along a Line We now consider the velocity and acceleration vectors for two basic motions: uniform motion around a circle, and motion along a straight line.

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Chapter 17 PARAMETERIZATION AND VECTOR FIELDS

Example 4

Find the acceleration vector for the child on the Ferris wheel in Examples 1 and 2.

Solution

The child’s position vector is given by 𝑟⃗ (𝑡) = 5 cos(𝜋𝑡)⃗𝑖 + 5 sin(𝜋𝑡)𝑗⃗ . In Example 2 we saw that the velocity vector is 𝑑𝑥 ⃗ 𝑑𝑦 ⃗ 𝑣⃗ (𝑡) = 𝑖 + 𝑗 = −5𝜋 sin(𝜋𝑡)⃗𝑖 + 5𝜋 cos(𝜋𝑡)𝑗⃗ . 𝑑𝑡 𝑑𝑡 Thus, the acceleration vector is 𝑎⃗ (𝑡) =

𝑑2𝑦 𝑑2𝑥 ⃗ 𝑖 + 𝑗⃗ = −(5𝜋) ⋅ 𝜋 cos(𝜋𝑡)⃗𝑖 − (5𝜋) ⋅ 𝜋 sin(𝜋𝑡)𝑗⃗ 𝑑𝑡2 𝑑𝑡2 = −5𝜋 2 cos(𝜋𝑡)⃗𝑖 − 5𝜋 2 sin(𝜋𝑡)𝑗⃗ .

Notice that 𝑎⃗ (𝑡) = −𝜋 2 𝑟⃗ (𝑡). Thus, the acceleration vector is a multiple of 𝑟⃗ (𝑡) and points toward the origin. The motion of the child on the Ferris wheel is an example of uniform circular motion, whose properties follow. (See Problem 45.) Uniform Circular Motion: For a particle whose motion is described by 𝑟⃗ (𝑡) = 𝑅 cos(𝜔𝑡)⃗𝑖 + 𝑅 sin(𝜔𝑡)𝑗⃗ • Motion is in a circle of radius 𝑅 with period 2𝜋∕|𝜔|. • Velocity, 𝑣⃗ , is tangent to the circle and speed is constant ‖⃗ 𝑣 ‖ = |𝜔|𝑅. • Acceleration, 𝑎⃗ , points toward the center of the circle with ‖𝑎⃗ ‖ = ‖⃗ 𝑣 ‖2 ∕𝑅.

In uniform circular motion, the acceleration vector is perpendicular to the velocity vector, 𝑣⃗ , because 𝑣⃗ does not change in magnitude, only in direction. There is no acceleration in the direction of 𝑣⃗ . We now look at straight-line motion in which the velocity vector always has the same direction but its magnitude changes. In straight-line motion, the acceleration vector points in the same direction as the velocity vector if the speed is increasing and in the opposite direction to the velocity vector if the speed is decreasing. Example 5

Consider the motion given by the vector equation ⃗ ). 𝑟⃗ (𝑡) = 2⃗𝑖 + 6𝑗⃗ + (𝑡3 + 𝑡)(4⃗𝑖 + 3𝑗⃗ + 𝑘 ⃗ and relate the Show that this is straight-line motion in the direction of the vector 4⃗𝑖 + 3𝑗⃗ + 𝑘 acceleration vector to the velocity vector.

Solution

The velocity vector is ⃗ ). 𝑣⃗ = (3𝑡2 + 1)(4⃗𝑖 + 3𝑗⃗ + 𝑘 Since (3𝑡2 + 1) is a positive scalar, the velocity vector 𝑣⃗ always points in the direction of the vector ⃗ . In addition, 4⃗𝑖 + 3𝑗⃗ + 𝑘 √ √ Speed = ‖⃗ 𝑣 ‖ = (3𝑡2 + 1) 42 + 32 + 12 = 26(3𝑡2 + 1). Notice that the speed is decreasing until 𝑡 = 0, then starts increasing. The acceleration vector is ⃗ ). 𝑎⃗ = 6𝑡(4⃗𝑖 + 3𝑗⃗ + 𝑘

17.2 MOTION, VELOCITY, AND ACCELERATION

901

⃗ , which is the For 𝑡 > 0, the acceleration vector points in the same direction as 4⃗𝑖 + 3𝑗⃗ + 𝑘 same direction as 𝑣⃗ . This makes sense because the object is speeding up. For 𝑡 < 0, the acceleration ⃗ ) points in the opposite direction to 𝑣⃗ because the object is slowing down. vector 6𝑡(4⃗𝑖 + 3𝑗⃗ + 𝑘

Motion in a Straight Line: For a particle whose motion is described by 𝑟⃗ (𝑡) = 𝑟⃗ 0 + 𝑓 (𝑡)𝑣⃗ • Motion is along a straight line through the point with position vector 𝑟⃗ 0 parallel to 𝑣⃗ . • Velocity, 𝑣⃗ , and acceleration, 𝑎⃗ , are parallel to the line. If 𝑓 (𝑡) = 𝑡, then we have 𝑟⃗ (𝑡) = 𝑟⃗ 0 + 𝑡𝑣, the equation of a line obtained on page 889.

The Length of a Curve The speed of a particle is the magnitude of its velocity vector: √ ( )2 ( 𝑑𝑦 )2 ( )2 𝑑𝑥 𝑑𝑧 + + . Speed = ‖⃗ 𝑣‖ = 𝑑𝑡 𝑑𝑡 𝑑𝑡

As in one dimension, we can find the distance traveled by a particle along a curve by integrating its speed. Thus, 𝑏

Distance traveled =

‖⃗ 𝑣 (𝑡)‖ 𝑑𝑡. ∫𝑎 If the particle never stops or reverses its direction as it moves along the curve, the distance it travels will be the same as the length of the curve. This suggests the following formula, which is justified in Problem 71 (available online): If the curve 𝐶 is given parametrically for 𝑎 ≤ 𝑡 ≤ 𝑏 by smooth functions and if the velocity vector 𝑣⃗ is not 0⃗ for 𝑎 < 𝑡 < 𝑏, then 𝑏

Length of 𝐶 =

Example 6

‖⃗ 𝑣 ‖𝑑𝑡.

Find the circumference of the ellipse given by the parametric equations 𝑥 = 2 cos 𝑡,

Solution

∫𝑎

𝑦 = sin 𝑡,

0 ≤ 𝑡 ≤ 2𝜋.

The circumference of this curve is given by an integral which must be calculated numerically:

2𝜋

Circumference =

∫0



(

𝑑𝑥 𝑑𝑡

)2

+

(

𝑑𝑦 𝑑𝑡

)2

2𝜋

𝑑𝑡 =

∫0 2𝜋

=

√ (−2 sin 𝑡)2 + (cos 𝑡)2 𝑑𝑡 √ 4 sin2 𝑡 + cos2 𝑡 𝑑𝑡 = 9.69.

∫0 Since the ellipse is inscribed in a circle of radius 2 and circumscribes a circle of radius 1, we would expect the length of the ellipse to be between 2𝜋(2) ≈ 12.57 and 2𝜋(1) ≈ 6.28, so the value of 9.69 is reasonable.

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Chapter 17 PARAMETERIZATION AND VECTOR FIELDS

Exercises and Problems for Section 17.2 Online Resource: Additional Problems for Section 17.2 EXERCISES

In Exercises 1–6, find the velocity and acceleration vectors.

14. 𝑥 = cos 3𝑡, 𝑦 = sin 5𝑡 for 0 ≤ 𝑡 ≤ 2𝜋. 15. 𝑥 = cos(𝑒𝑡 ), 𝑦 = sin(𝑒𝑡 ) for 0 ≤ 𝑡 ≤ 1. Check by calculating the length by another method.

1. 𝑥 = 2 + 3𝑡, 𝑦 = 4 + 𝑡, 𝑧 = 1 − 𝑡 2. 𝑥 = 2 + 3𝑡2 , 𝑦 = 4 + 𝑡2 , 𝑧 = 1 − 𝑡2

⃗ for 1 ≤ 𝑡 ≤ 2. 16. 𝑟⃗ (𝑡) = 2𝑡⃗𝑖 + ln 𝑡𝑗⃗ + 𝑡2 𝑘

3. 𝑥 = 𝑡, 𝑦 = 𝑡2 , 𝑧 = 𝑡3 4. 𝑥 = 𝑡, 𝑦 = 𝑡3 − 𝑡

In Exercises 17–18, find the velocity and acceleration vectors of the uniform circular motion and check that they are perpendicular. Check that the speed and magnitude of the acceleration are constant.

5. 𝑥 = 3 cos 𝑡, 𝑦 = 4 sin 𝑡 6. 𝑥 = 3 cos (𝑡2 ), 𝑦 = 3 sin (𝑡2 ), 𝑧 = 𝑡2 In Exercises 7–12, find the velocity 𝑣⃗ (𝑡) and speed ‖𝑣⃗ (𝑡)‖. Find any times at which the particle stops. 7. 𝑥 = 𝑡, 𝑦 = 𝑡2 , 𝑧 = 𝑡3 8. 𝑥 = cos 3𝑡, 𝑦 = sin 5𝑡

18. 𝑥 = 2𝜋, 𝑦 = 2 sin(3𝑡), 𝑧 = 2 cos(3𝑡) In Exercises 19–20, find the velocity and acceleration vectors of the straight-line motion. Check that the acceleration vector points in the same direction as the velocity vector if the speed is increasing and in the opposite direction if the speed is decreasing.

9. 𝑥 = 3𝑡2 , 𝑦 = 𝑡3 + 1 10. 𝑥 = (𝑡 − 1)2 , 𝑦 = 2, 𝑧 = 2𝑡3 − 3𝑡2 11. 𝑥 = 3 sin(𝑡2 ) − 1, 𝑦 = 3 cos(𝑡2 ) 12. 𝑥 = 3 sin2 𝑡, 𝑦 = cos 𝑡 − 1,

17. 𝑥 = 3 cos(2𝜋𝑡), 𝑦 = 3 sin(2𝜋𝑡), 𝑧 = 0

𝑧 = 𝑡2

19. 𝑥 = 2 + 𝑡2 , 𝑦 = 3 − 2𝑡2 , 𝑧 = 5 − 𝑡2

In Exercises 13–16, find the length of the curve.

20. 𝑥 = −2𝑡3 − 3𝑡 + 1, 𝑦 = 4𝑡3 + 6𝑡 − 5, 𝑧 = 6𝑡3 + 9𝑡 − 2

13. 𝑥 = 3 + 5𝑡, 𝑦 = 1 + 4𝑡, 𝑧 = 3 − 𝑡 for 1 ≤ 𝑡 ≤ 2. Check by calculating the length by another method.

21. Find parametric equations for the tangent line at 𝑡 = 2 for Exercise 10.

PROBLEMS 22. A particle passes through the point 𝑃 = (5, 4, −2) at time 𝑡 = 4, moving with constant velocity 𝑣⃗ = ⃗ . Find a parametric equation for its mo2⃗𝑖 − 3𝑗⃗ + 𝑘 tion. In Problems 23–24, find all values of 𝑡 for which the particle is moving parallel to the 𝑥-axis and to the 𝑦-axis. Determine the end behavior and graph the particle’s path. 23. 𝑥 = 𝑡2 − 6𝑡,

𝑦 = 𝑡3 − 3𝑡

24. 𝑥 = 𝑡3 − 12𝑡,

𝑦 = 𝑡2 + 10𝑡

25. The table gives 𝑥 and 𝑦 coordinates of a particle in the plane at time 𝑡. Assuming that the particle moves smoothly and that the points given show all the major features of the motion, estimate the following quantities: (a) The velocity vector and speed at time 𝑡 = 2. (b) Any times when the particle is moving parallel to the 𝑦-axis. (c) Any times when the particle has come to a stop. 𝑡

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

𝑥

1

4

6

7

6

3

2

3

5

𝑦

3

2

3

5

8

10

11

10

9

26. A particle starts at the point 𝑃 = (3, 2, −5) and moves along a straight line toward 𝑄 = (5, 7, −2) at a speed of 5 cm/sec. Let 𝑥, 𝑦, 𝑧 be measured in centimeters. (a) Find the particle’s velocity vector. (b) Find parametric equations for the particle’s motion.

27. A particle moves at a constant speed along a line from the point 𝑃 = (2, −1, 5) at time 𝑡 = 0 to the point 𝑄 = (5, 3, −1). Find parametric equations for the particle’s motion if: (a) The particle takes 5 seconds to move from 𝑃 to 𝑄. (b) The speed of the particle is 5 units per second.

28. A particle travels along the line 𝑥 = 1 + 𝑡, 𝑦 = 5 + 2𝑡, 𝑧 = −7 + 𝑡, where 𝑡 is in seconds and 𝑥, 𝑦, 𝑧 are in meters. (a) When and where does the particle hit the plane 𝑥 + 𝑦 + 𝑧 = 1? (b) How fast is the particle going when it hits the plane? Give units.

17.2 MOTION, VELOCITY, AND ACCELERATION

29. A stone is thrown from a rooftop at time 𝑡 = 0 seconds. Its position at time 𝑡 is given by ⃗. 𝑟⃗ (𝑡) = 10𝑡⃗𝑖 − 5𝑡𝑗⃗ + (6.4 − 4.9𝑡2 )𝑘 The origin is at the base of the building, which is standing on flat ground. Distance is measured in meters. The ⃗ points up. vector ⃗𝑖 points east, 𝑗⃗ points north, and 𝑘 (a) How high is the rooftop above the ground? (b) At what time does the stone hit the ground? (c) How fast is the stone moving when it hits the ground? (d) Where does the stone hit the ground? (e) What is the stone’s acceleration when it hits the ground? 30. A child wanders slowly down a circular staircase from the top of a tower. With 𝑥, 𝑦, 𝑧 in feet and the origin at the base of the tower, her position 𝑡 minutes from the start is given by 𝑥 = 10 cos 𝑡, (a) (b) (c) (d)

𝑦 = 10 sin 𝑡,

𝑧 = 90 − 5𝑡.

How tall is the tower? When does the child reach the bottom? What is her speed at time 𝑡? What is her acceleration at time 𝑡?

31. The origin is on flat ground and the 𝑧-axis points upward. For time 0 ≤ 𝑡 ≤ 10 in seconds and distance in centimeters, a particle moves along a path given by ⃗. 𝑟⃗ = 2𝑡⃗𝑖 + 3𝑡𝑗⃗ + (100 − (𝑡 − 5)2 )𝑘 (a) When is the particle at the highest point? What is that point? (b) When in the interval 0 ≤ 𝑡 ≤ 10 is the particle moving fastest? What is its speed at that moment? (c) When in the interval 0 ≤ 𝑡 ≤ 10 is the particle moving slowest? What is its speed at that moment? 32. The function 𝑤 = 𝑓 (𝑥, 𝑦, 𝑧) has grad 𝑓 (7, 2, 5) = 4⃗𝑖 − ⃗ . A particle moves along the curve 𝑟⃗ (𝑡), arriving 3𝑗⃗ + 𝑘 ⃗ when at the point (7, 2, 5) with velocity 2⃗𝑖 + 3𝑗⃗ + 6𝑘 𝑡 = 0. Find the rate of change of 𝑤 with respect to time at 𝑡 = 0. 33. Suppose 𝑥 measures horizontal distance in meters, and 𝑦 measures distance above the ground in meters. At time 𝑡 = 0 in seconds, a projectile starts from a point ℎ meters above the origin with speed 𝑣 meters/sec at an angle 𝜃 to the horizontal. Its path is given by 𝑥 = (𝑣 cos 𝜃)𝑡,

1 𝑦 = ℎ + (𝑣 sin 𝜃)𝑡 − 𝑔𝑡2 . 2

Using this information about a general projectile, analyze the motion of a ball which travels along the path 𝑥 = 20𝑡,

𝑦 = 2 + 25𝑡 − 4.9𝑡2 .

(a) When does the ball hit the ground?

903

(b) Where does the ball hit the ground? (c) At what height above the ground does the ball start? (d) What is the value of 𝑔, the acceleration due to gravity? (e) What are the values of 𝑣 and 𝜃? 34. A particle is moving on a path in the 𝑥𝑧-plane given by 𝑥 = 20𝑡, 𝑧 = 5𝑡 − 0.5𝑡2 , where 𝑧 is the height of the particle above the ground in meters, 𝑥 is the horizontal distance in meters, and 𝑡 is time in seconds. (a) What is the equation of the path in terms of 𝑥 and 𝑧 only? (b) When is the particle at ground level? (c) What is the velocity of the particle at time 𝑡? (d) What is the speed of the particle at time 𝑡? (e) Is the speed ever 0? (f) When is the particle at the highest point? 35. The base of a 20-meter tower is at the origin; the base of a 20-meter tree is at (0, 20, 0). The ground is flat and the 𝑧-axis points upward. The following parametric equations describe the motion of six projectiles each launched at time 𝑡 = 0 in seconds. (I) (II) (III) (IV) (V) (VI)

⃗ 𝑟⃗ (𝑡) = (20 + 𝑡2 )𝑘 ⃗ 𝑟⃗ (𝑡) = 2𝑡2 𝑗⃗ + 2𝑡2 𝑘 ⃗ 𝑟⃗ (𝑡) = 20⃗𝑖 + 20𝑗⃗ + (20 − 𝑡2 )𝑘 ⃗ 𝑟⃗ (𝑡) = 2𝑡𝑗⃗ + (20 − 𝑡2 )𝑘 ⃗ 𝑟⃗ (𝑡) = (20 − 2𝑡)⃗𝑖 + 2𝑡𝑗⃗ + (20 − 𝑡)𝑘 ⃗ 𝑟⃗ (𝑡) = 𝑡⃗𝑖 + 𝑡𝑗⃗ + 𝑡𝑘

(a) Which projectile is launched from the top of the tower and goes downward? When and where does it hit the ground? (b) Which projectile hits the top of the tree? When? From where is it launched? (c) Which projectile is not launched from somewhere on the tower and hits the tree? Where and when does it hit the tree? 36. A particle moves on a circle of radius 5 cm, centered at the origin, in the 𝑥𝑦-plane (𝑥 and 𝑦 measured in centimeters). It starts at the point (0, 5) and moves counterclockwise, going once around the circle in 8 seconds. (a) Write a parameterization for the particle’s motion. (b) What is the particle’s speed? Give units. 37. A particle moves along a curve with velocity vector 𝑣⃗ (𝑡) = − sin 𝑡⃗𝑖 + cos 𝑡𝑗⃗ . At time 𝑡 = 0 the particle is at (2, 3). (a) Find the displacement vector for the particle from time 𝑡 = 0 to 𝑡 = 𝜋. (b) Find the position of the particle at time 𝑡 = 𝜋. (c) Find the distance traveled by the particle from time 𝑡 = 0 to time 𝑡 = 𝜋.

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Chapter 17 PARAMETERIZATION AND VECTOR FIELDS

38. Determine the position vector 𝑟⃗ (𝑡) for a rocket which is launched from the origin at time 𝑡 = 0 seconds, reaches its highest point of (𝑥, 𝑦, 𝑧) = (1000, 3000, 10,000), where 𝑥, 𝑦, 𝑧 are in meters, and after the launch is subject only to the acceleration due to gravity, 9.8 m/sec2 .

44. An ant crawls along the radius from the center to the edge of a circular disk of radius 1 meter, moving at a constant rate of 1 cm/sec. Meanwhile, the disk is turning counterclockwise about its center at 1 revolution per second.

39. Emily is standing on the outer edge of a merry-goround, 10 meters from the center. The merry-go-round completes one full revolution every 20 seconds. As Emily passes over a point 𝑃 on the ground, she drops a ball from 3 meters above the ground.

(a) Parameterize the motion of the ant. (b) Find the velocity and speed of the ant. (c) Determine the acceleration and magnitude of the acceleration of the ant.

(a) How fast is Emily going? (b) How far from 𝑃 does the ball hit the ground? (The acceleration due to gravity is 9.8 m/sec2 .) (c) How far from Emily does the ball hit the ground? 40. A point 𝑃 moves in a circle of radius 𝑎. Show that 𝑟⃗ (𝑡), the position vector of 𝑃 , and its velocity vector 𝑟⃗ ′ (𝑡) are perpendicular. 41. A wheel of radius 1 meter rests on the 𝑥-axis with its center on the 𝑦-axis. There is a spot on the rim at the point (1, 1). See Figure 17.16. At time 𝑡 = 0 the wheel starts rolling on the 𝑥-axis in the direction shown at a rate of 1 radian per second. (a) Find parametric equations describing the motion of the center of the wheel. (b) Find parametric equations describing the motion of the spot on the rim. Plot its path. 𝑦 2

(a) Show that the particle moves on a circle and find the radius, direction, and period. (b) Determine the velocity vector of the particle and its direction and speed. (c) What are the direction and magnitude of the acceleration vector of the particle? 46. You bicycle along a straight flat road with a safety light attached to one foot. Your bike moves at a speed of 25 km/hr and your foot moves in a circle of radius 20 cm centered 30 cm above the ground, making one revolution per second. (a) Find parametric equations for 𝑥 and 𝑦 which describe the path traced out by the light, where 𝑦 is distance (in cm) above the ground. (b) Sketch the light’s path. (c) How fast (in revolutions/sec) would your foot have to be rotating if an observer standing at the side of the road sees the light moving backward? 47. How do the motions of objects 𝐴 and 𝐵 differ, if 𝐴 has position vector 𝑟⃗ 𝐴 (𝑡) and 𝐵 has position vector 𝑟⃗ 𝐵 (𝑡) = 𝑟⃗ 𝐴 (2𝑡) for 𝑡 ≥ 0. Illustrate your answer with 𝑟⃗ 𝐴 (𝑡) = 𝑡⃗𝑖 + 𝑡2 𝑗⃗ .

Spot

1

45. The motion of a particle is given by 𝑟⃗ (𝑡) = 𝑅 cos(𝜔𝑡)⃗𝑖 + 𝑅 sin(𝜔𝑡)𝑗⃗ , with 𝑅 > 0, 𝜔 > 0.

Figure 17.16

48. At time 𝑡 = 0 an object is moving with velocity vector 𝑣⃗ = 2⃗𝑖 + 𝑗⃗ and acceleration vector 𝑎⃗ = ⃗𝑖 + 𝑗⃗ . Can it be in uniform circular motion about some point in the plane?

⃗ represents the 42. Suppose 𝑟⃗ (𝑡) = cos 𝑡 ⃗𝑖 + sin 𝑡 𝑗⃗ + 2𝑡 𝑘 position of a particle on a helix, where 𝑧 is the height of the particle above the ground.

49. Figure 17.17 shows the velocity and acceleration vectors of an object in uniform circular motion about a point in the plane at a particular moment. Is it moving round the circle in the clockwise or counterclockwise direction?

𝑥 −1

0

1

(a) Is the particle ever moving downward? When? (b) When does the particle reach a point 10 units above the ground? (c) What is the velocity of the particle when it is 10 units above the ground? (d) When it is 10 units above the ground, the particle leaves the helix and moves along the tangent. Find parametric equations for this tangent line. ⃗ is 43. Show that the helix 𝑟⃗ = 𝛼 cos 𝑡⃗𝑖 + 𝛼 sin 𝑡𝑗⃗ + 𝛽𝑡𝑘 parameterized with constant speed.

𝑎⃗

𝑣⃗

Figure 17.17

17.3 VECTOR FIELDS

50. Let 𝑣⃗ (𝑡) be the velocity of a particle moving in the plane. Let 𝑠(𝑡) be the magnitude of 𝑣⃗ and let 𝜃(𝑡) be the angle of 𝑣⃗ (𝑡) with the positive 𝑥-axis at time 𝑡, so that 𝑣⃗ = 𝑠 cos 𝜃 ⃗𝑖 + 𝑠 sin 𝜃 𝑗⃗ . Let 𝑇⃗ be the unit vector in the direction of 𝑣⃗ , and ⃗ ⃗ × 𝑣⃗ , perlet 𝑁 be the unit vector in the direction of 𝑘 pendicular to 𝑣⃗ . Show that the acceleration 𝑎⃗ (𝑡) is given

905

by 𝑑𝑠 ⃗ 𝑑𝜃 ⃗ 𝑇 +𝑠 𝑁 . 𝑑𝑡 𝑑𝑡 This shows how to separate the acceleration into the 𝑑𝑠 ⃗ 𝑇 , due to changing speed sum of one component, 𝑑𝑡 𝑑𝜃 ⃗ and a perpendicular component, 𝑠 𝑁 , due to chang𝑑𝑡 ing direction of the motion. 𝑎⃗ =

Strengthen Your Understanding In Problems 51–53, explain what is wrong with the statement. 51. When a particle moves around a circle its velocity and acceleration are always orthogonal. 52. A particle with position 𝑟⃗ (𝑡) at time 𝑡 has acceleration equal to 3 m/sec2 at time 𝑡 = 0. 53. A parameterized curve 𝑟⃗ (𝑡), 𝐴 ≤ 𝑡 ≤ 𝐵, has length 𝐵 − 𝐴. In Problems 54–55, give an example of: 54. A function 𝑟⃗ (𝑡) such that the particle with position 𝑟⃗ (𝑡) at time 𝑡 has velocity 𝑣⃗ = ⃗𝑖 + 2𝑗⃗ and acceleration ⃗ at 𝑡 = 0. 𝑎⃗ = 4⃗𝑖 + 6𝑘 55. An interval 𝑎 ≤ 𝑡 ≤ 𝑏 corresponding to a piece of the ⃗ of length 10. helix 𝑟⃗ (𝑡) = cos 𝑡⃗𝑖 + sin 𝑡𝑗⃗ + 𝑡𝑘 Are the statements in Problems 56–63 true or false? Give reasons for your answer. 56. A particle whose motion in the plane is given by 𝑟⃗ (𝑡) = 𝑡2 ⃗𝑖 +(1−𝑡)𝑗⃗ has the same velocity at 𝑡 = 1 and 𝑡 = −1. 57. A particle whose motion in the plane is given by 𝑟⃗ (𝑡) = 𝑡2 ⃗𝑖 + (1 − 𝑡)𝑗⃗ has the same speed at 𝑡 = 1 and 𝑡 = −1.

17.3

58. If a particle is moving along a parameterized curve 𝑟⃗ (𝑡) then the acceleration vector at any point is always perpendicular to the velocity vector at that point. 59. If a particle is moving along a parameterized curve 𝑟⃗ (𝑡) then the acceleration vector at a point cannot be parallel to the velocity vector at that point. 60. If 𝑟⃗ (𝑡) for 𝑎 ≤ 𝑡 ≤ 𝑏 is a parameterized curve, then 𝑟⃗ (−𝑡) for 𝑎 ≤ 𝑡 ≤ 𝑏 is the same curve traced backward. 61. If 𝑟⃗ (𝑡) for 𝑎 ≤ 𝑡 ≤ 𝑏 is a parameterized curve 𝐶 and the speed ||𝑣⃗ (𝑡)|| = 1, then the length of 𝐶 is 𝑏 − 𝑎. ⃗, 62. If a particle moves with motion 𝑟⃗ (𝑡) = 3𝑡⃗𝑖 + 2𝑡𝑗⃗ + 𝑡𝑘 then the particle stops at the origin.

63. If a particle moves with constant speed, the path of the particle must be a line. For Problems 64–67, decide if the statement is true or false for all smooth parameterized curves 𝑟⃗ (𝑡) and all values of 𝑡 for which 𝑟⃗ ′ (𝑡) ≠ 0⃗ . 64. The vector 𝑟⃗ ′ (𝑡) is tangent to the curve at the point with position vector 𝑟⃗ (𝑡). 65. 𝑟⃗ ′ (𝑡) × 𝑟⃗ (𝑡) = 0⃗ 66. 𝑟⃗ ′ (𝑡) ⋅ 𝑟⃗ (𝑡) = 0 67. 𝑟⃗ ′′ (𝑡) = −𝜔2 𝑟⃗ (𝑡)

VECTOR FIELDS

Introduction to Vector Fields A vector field is a function that assigns a vector to each point in the plane or in 3-space. One example of a vector field is the gradient of a function 𝑓 (𝑥, 𝑦); at each point (𝑥, 𝑦) the vector grad 𝑓 (𝑥, 𝑦) points in the direction of maximum rate of increase of 𝑓 . In this section we look at other vector fields representing velocities and forces.

Velocity Vector Fields Figure 17.18 shows the flow of a part of the Gulf Stream, a current in the Atlantic Ocean.2 It is an example of a velocity vector field: each vector shows the velocity of the current at that point. The current is fastest where the velocity vectors are longest in the middle of the stream. Beside the stream are eddies where the water flows round and round in circles. 2 Based

on data supplied by Avijit Gangopadhyay of the Jet Propulsion Laboratory.

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Chapter 17 PARAMETERIZATION AND VECTOR FIELDS

km 600

500

400

300

200

100

0

0

100

200

300

400

500

600

km

Figure 17.18: The velocity vector field of the Gulf Stream

Force Fields Another physical quantity represented by a vector is force. When we experience a force, sometimes it results from direct contact with the object that supplies the force (for example, a push). Many forces, however, can be felt at all points in space. For example, the earth exerts a gravitational pull on all other masses. Such forces can be represented by vector fields. Figure 17.19 shows the gravitational force exerted by the earth on a mass of one kilogram at different points in space. This is a sketch of the vector field in 3-space. You can see that the vectors all point toward the earth (which is not shown in the diagram) and that the vectors farther from the earth are smaller in magnitude.

Figure 17.19: The gravitational field of the earth

Definition of a Vector Field Now that you have seen some examples of vector fields, we give a more formal definition.

17.3 VECTOR FIELDS

907

A vector field in 2-space is a function 𝐹⃗ (𝑥, 𝑦) whose value at a point (𝑥, 𝑦) is a 2-dimensional vector. Similarly, a vector field in 3-space is a function 𝐹⃗ (𝑥, 𝑦, 𝑧) whose values are 3dimensional vectors. Notice the arrow over the function, 𝐹⃗ , indicating that its value is a vector, not a scalar. We often represent the point (𝑥, 𝑦) or (𝑥, 𝑦, 𝑧) by its position vector 𝑟⃗ and write the vector field as 𝐹⃗ (⃗𝑟 ).

Visualizing a Vector Field Given by a Formula Since a vector field is a function that assigns a vector to each point, a vector field can often be given by a formula. Example 1

Sketch the vector field in 2-space given by 𝐹⃗ (𝑥, 𝑦) = −𝑦⃗𝑖 + 𝑥𝑗⃗ .

Solution

Table 17.1 shows the value of the vector field at a few points. Notice that each value is a vector. To plot the vector field, we plot 𝐹⃗ (𝑥, 𝑦) with its tail at (𝑥, 𝑦). (See Figure 17.20.) Table 17.1

Values of 𝐹⃗ (𝑥, 𝑦) = −𝑦⃗𝑖 + 𝑥𝑗⃗ 𝑦 −1

0

1

0

⃗𝑖 − 𝑗⃗ ⃗𝑖

−𝑗⃗ 0⃗

−⃗𝑖 − 𝑗⃗ −⃗𝑖

1

⃗𝑖 + 𝑗⃗

𝑗⃗

−⃗𝑖 + 𝑗⃗

−1 𝑥

Now we look at the formula. The magnitude of the vector at (𝑥, 𝑦) is the distance from (𝑥, 𝑦) to the origin since √ ‖𝐹⃗ (𝑥, 𝑦)‖ = ‖ − 𝑦⃗𝑖 + 𝑥𝑗⃗ ‖ = 𝑥2 + 𝑦2 .

Therefore, all the vectors at a fixed distance from the origin (that is, on a circle centered at the origin) have the same magnitude. The magnitude gets larger as we move farther from the origin. What about the direction? Figure 17.20 suggests that at each point (𝑥, 𝑦) the vector 𝐹⃗ (𝑥, 𝑦) is perpendicular to the position vector 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ . We confirm this using the dot product: 𝑟⃗ ⋅ 𝐹⃗ (𝑥, 𝑦) = (𝑥⃗𝑖 + 𝑦𝑗⃗ ) ⋅ (−𝑦⃗𝑖 + 𝑥𝑗⃗ ) = 0.

Thus, the vectors are tangent to circles centered at the origin and get longer as we go out. In Figure 17.21, the vectors have been scaled so that they do not obscure each other. 𝑦

𝑦

1 −1

𝑥

𝑥

1 −1

Figure 17.20: The value 𝐹⃗ (𝑥, 𝑦) is placed at the point (𝑥, 𝑦)

Figure 17.21: The vector field 𝐹⃗ (𝑥, 𝑦) = −𝑦⃗𝑖 + 𝑥𝑗⃗ , vectors scaled smaller to fit in diagram

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Chapter 17 PARAMETERIZATION AND VECTOR FIELDS

𝐹⃗ (𝑥, 𝑦) = 𝑥𝑗⃗

(b) 𝐺⃗ (𝑥, 𝑦) = 𝑥⃗𝑖 .

Example 2

Sketch the vector fields in 2-space given by (a)

Solution

(a) The vector 𝑥𝑗⃗ is parallel to the 𝑦-direction, pointing up when 𝑥 is positive and down when 𝑥 is negative. Also, the larger |𝑥| is, the longer the vector. The vectors in the field are constant along vertical lines since the vector field does not depend on 𝑦. (See Figure 17.22.) 𝑦

𝑦

𝑥

Figure 17.22: The vector field 𝐹⃗ (𝑥, 𝑦) = 𝑥𝑗⃗

𝑥

Figure 17.23: The vector field 𝐹⃗ (𝑥, 𝑦) = 𝑥⃗𝑖

(b) This is similar to the previous example, except that the vector 𝑥⃗𝑖 is parallel to the 𝑥-direction, pointing to the right when 𝑥 is positive and to the left when 𝑥 is negative. Again, the larger |𝑥| is the longer the vector, and the vectors are constant along vertical lines, since the vector field does not depend on 𝑦. (See Figure 17.23.)

Example 3

⃗. Describe the vector field in 3-space given by 𝐹⃗ (⃗𝑟 ) = 𝑟⃗ , where 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘

Solution

The notation 𝐹⃗ (⃗𝑟 ) = 𝑟⃗ means that the value of 𝐹⃗ at the point (𝑥, 𝑦, 𝑧) with position vector 𝑟⃗ is the vector 𝑟⃗ with its tail at (𝑥, 𝑦, 𝑧). Thus, the vector field points outward. See Figure 17.24. Note that the lengths of the vectors have been scaled down so as to fit into the diagram.

𝑚 𝐹⃗

Figure 17.24: The vector field 𝐹⃗ (⃗𝑟 ) = 𝑟⃗

𝑀 Figure 17.25: Force exerted on mass 𝑚 by mass 𝑀

Finding a Formula for a Vector Field Example 4

Newton’s Law of Gravitation states that the magnitude of the gravitational force exerted by an object of mass 𝑀 on an object of mass 𝑚 is proportional to 𝑀 and 𝑚 and inversely proportional to the square of the distance between them. The direction of the force is from 𝑚 to 𝑀 along the line connecting them. (See Figure 17.25.) Find a formula for the vector field 𝐹⃗ (⃗𝑟 ) that represents the gravitational force, assuming 𝑀 is located at the origin and 𝑚 is located at the point with position vector 𝑟⃗ .

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17.3 VECTOR FIELDS

Solution

Since the mass 𝑚 is located at 𝑟⃗ , Newton’s law says that the magnitude of the force is given by ‖𝐹⃗ (⃗𝑟 )‖ =

𝐺𝑀𝑚 , ‖⃗𝑟 ‖2

where 𝐺 is the universal gravitational constant. A unit vector in the direction of the force is −⃗𝑟 ∕‖⃗𝑟 ‖, where the negative sign indicates that the direction of force is toward the origin (gravity is attractive). By taking the product of the magnitude of the force and a unit vector in the direction of the force, we obtain an expression for the force vector field: ( ) 𝐺𝑀𝑚 𝑟⃗ −𝐺𝑀𝑚⃗𝑟 ⃗ 𝐹 (⃗𝑟 ) = − = . 2 3 ‖⃗𝑟 ‖ ‖⃗𝑟 ‖ ‖⃗𝑟 ‖

We have already seen a picture of this vector field in Figure 17.19.

Gradient Vector Fields The gradient of a scalar function 𝑓 is a function that assigns a vector to each point, and is therefore a vector field. It is called the gradient field of 𝑓 . Many vector fields in physics are gradient fields. Example 5

Sketch the gradient field of the functions in Figures 17.26–17.28. 𝑦

𝑦

𝑦

−1

5

9

3

1

𝑥

5

𝑥

𝑥

1

3

−3

Figure 17.26: The contour map of 𝑓 (𝑥, 𝑦) = 𝑥2 + 2𝑦2

Solution

−3

1

7

Figure 17.27: The contour map of 𝑔(𝑥, 𝑦) = 5 − 𝑥2 − 2𝑦2

Figure 17.28: The contour map of ℎ(𝑥, 𝑦) = 𝑥 + 2𝑦 + 3

See Figures 17.29–17.31. For a function 𝑓 (𝑥, 𝑦), the gradient vector of 𝑓 at a point is perpendicular to the contours in the direction of increasing 𝑓 and its magnitude is the rate of change in that direction. The rate of change is large when the contours are close together and small when they are far apart. Notice that in Figure 17.29 the vectors all point outward, away from the local minimum of 𝑓 , and in Figure 17.30 the vectors of grad 𝑔 all point inward, toward the local maximum of 𝑔. Since ℎ is a linear function, its gradient is constant, so grad ℎ in Figure 17.31 is a constant vector field. 𝑦

𝑦

𝑥

Figure 17.29: grad 𝑓

𝑦

𝑥

Figure 17.30: grad 𝑔

𝑥

Figure 17.31: grad ℎ

910

Chapter 17 PARAMETERIZATION AND VECTOR FIELDS

Exercises and Problems for Section 17.3 EXERCISES 13. 𝐹⃗ (𝑥, 𝑦) = −𝑦𝑗⃗ 15. 𝐹⃗ (⃗𝑟 ) = 𝑟⃗ ∕‖⃗𝑟 ‖

For Exercises 1–6, find formulas for the vector fields. (There are many possible answers.) 1.

𝑦

2.

17. 𝐹⃗ = 𝑦⃗𝑖 − 𝑥𝑗⃗

𝑦

𝑥

14. 𝐹⃗ (⃗𝑟 ) = 2⃗𝑟 16. 𝐹⃗ (⃗𝑟 ) = −⃗𝑟 ∕‖⃗𝑟 ‖3

𝑥

18. 𝐹⃗ (𝑥, 𝑦) = 2𝑥⃗𝑖 + 𝑥𝑗⃗

19. 𝐹⃗ (𝑥, 𝑦) = (𝑥 + 𝑦)⃗𝑖 + (𝑥 − 𝑦)𝑗⃗ ⃗ in the tables with vector 20. Match vector fields 𝐴⃗ –𝐷 fields (I)–(IV) in Figure 17.32. ⃗ Vector field 𝐁

⃗ Vector field 𝐀 𝑦∖𝑥

−1

1

−1

⃗𝑖 + 𝑗⃗ ⃗𝑖 + 𝑗⃗

⃗𝑖 + 𝑗⃗ ⃗𝑖 + 𝑗⃗

1 3.

𝑦

4.

𝑦∖𝑥

−1

1

−1

−⃗𝑖 − 𝑗⃗ ⃗𝑖 + 𝑗⃗

−⃗𝑖 − 𝑗⃗ ⃗𝑖 + 𝑗⃗

1

𝑦

⃗ Vector field 𝐂

𝑥

𝑥

𝑦∖𝑥

−1

1

𝑦∖𝑥

−1

1

−1

−2⃗𝑖 + 𝑗⃗ −2⃗𝑖 + 𝑗⃗

2⃗𝑖 + 𝑗⃗ 2⃗𝑖 + 𝑗⃗

−1

⃗𝑖 + 𝑗⃗ −⃗𝑖 + 𝑗⃗

−⃗𝑖 − 𝑗⃗ ⃗𝑖 − 𝑗⃗

1

5.

𝑦

6.

𝑥

In Exercises 7–10, assume 𝑥, 𝑦 > 0 and decide if (a) The vector field is parallel to the 𝑥-axis, parallel to the 𝑦-axis, or neither. (b) As 𝑥 increases, the length increases, decreases, or neither. (c) As 𝑦 increases, the length increases, decreases, or neither. Assume 𝑥, 𝑦 > 0.

9. 𝐹⃗ = (𝑥 + 𝑒1−𝑦 )⃗𝑖

(I)

(II)

(III)

(IV)

8. 𝐹⃗ = 𝑦⃗𝑖 + 𝑗⃗ 10. grad(𝑥4 + 𝑒3𝑦 )

Figure 17.32 21. For each description of a vector field in (a)-(d), choose one or more of the vector fields I-IX. (a) Pointing radially outward, increasing in length away from the origin. (b) Pointing in a circular direction around the origin, remaining the same length. (c) Pointing towards the origin, increasing in length farther from the origin. (d) Pointing clockwise around the origin. 𝑥⃗𝑖 + 𝑦𝑗⃗ √ 𝑥2 + 𝑦2 IV. −⃗𝑟

I. Sketch the vector fields in Exercises 11–19 in the 𝑥𝑦-plane.

VII. 𝑦⃗𝑖 + 𝑥𝑗⃗ 11. 𝐹⃗ (𝑥, 𝑦) = 2⃗𝑖 + 3𝑗⃗

1

𝑦

𝑥

7. 𝐹⃗ = 𝑥𝑗⃗

⃗ Vector field 𝐃

12. 𝐹⃗ (𝑥, 𝑦) = 𝑦⃗𝑖

−𝑦⃗𝑖 + 𝑥𝑗⃗ √ 𝑥2 + 𝑦2 V. −𝑦⃗𝑖 + 𝑥𝑗⃗ 𝑟⃗ VIII. ||⃗𝑟 ||3 II.

III. 𝑟⃗ VI. 𝑦⃗𝑖 − 𝑥𝑗⃗ 𝑟⃗ IX. − ||⃗𝑟 ||3

17.3 VECTOR FIELDS

22. Each vector field in Figures (I)–(IV) represents the force on a particle at different points in space as a result of another particle at the origin. Match up the vector fields with the descriptions below. (a) A repulsive force whose magnitude decreases as distance increases, such as between electric charges of the same sign. (b) A repulsive force whose magnitude increases as distance increases. (c) An attractive force whose magnitude decreases as distance increases, such as gravity. (d) An attractive force whose magnitude increases as distance increases.

(I)

(II)

(III)

(IV)

911

PROBLEMS In Problems 23–27, give an example of a vector field 𝐹⃗ (𝑥, 𝑦) in 2-space with the stated properties. 23. 𝐹⃗ is constant 24. 𝐹⃗ has a constant direction but ‖𝐹⃗ ‖ is not constant 25. ‖𝐹⃗ ‖ is constant but 𝐹⃗ is not constant

Problems 29–30 concern the vector fields 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ , ⃗ = 𝑥⃗𝑖 − 𝑦𝑗⃗ . 𝐺⃗ = −𝑦⃗𝑖 + 𝑥𝑗⃗ , and 𝐻 ⃗ with their sketches in (I)–(III). 29. Match 𝐹⃗ , 𝐺⃗ , 𝐻 𝑦

(I)

26. Neither ‖𝐹⃗ ‖ nor the direction of 𝐹⃗ is constant 27. 𝐹⃗ is perpendicular to 𝐺⃗ = (𝑥+𝑦)⃗𝑖 +(1+𝑦2 )𝑗⃗ at every point

𝑥

28. Match the level curves in (I)–(IV) with the gradient fields in (A)–(D). All figures use the same square window. (I)

𝑥

𝑦

(III)

(II) 3

2 1 0

−2

−3

𝑥

1 2

−2

−3

−1

0

−1 3

(III)

(IV) 0

0

1

0. 5

0

−0 .5 0

0. 5

−0 .5

(A)

𝑦

(II)

0. 5 −0 −1 . 5

−0 .5 0. 5

30. Match the vector fields with their sketches, (I)–(IV). ⃗ (c) 𝐺⃗ + 𝐻 ⃗ (d) −𝐹⃗ +𝐺⃗ (a) 𝐹⃗ + 𝐺⃗ (b) 𝐹⃗ + 𝐻 (I)

𝑦

(II)

𝑦

(B)

𝑥

(III) (C)

𝑦

𝑥

(IV)

𝑦

(D)

𝑥

𝑥

912

Chapter 17 PARAMETERIZATION AND VECTOR FIELDS

31. Match vector fields (a)–(f) with their graphs (I)–(VI). −𝑦⃗𝑖 + 𝑥𝑗⃗ (c) 𝑦𝑗⃗ ⃗ (e) 2𝑘

(b) 𝑥⃗𝑖 ⃗ (d) 𝑧𝑘

(a)

(f) 𝑟⃗

(a) If 𝐴 > 0, 𝐾 > 0, explain why the following expression could represent the velocity field for the combined flow of the river and the fountain:

𝑧

(I)

𝑧

(II)

𝑥

𝑦

𝑧

(III)

𝑥

𝑦

𝑥

(b) What is the significance of the constants 𝐴 and 𝐾? (c) Using a computer, sketch the vector field 𝑣⃗ for 𝐾 = 1 and 𝐴 = 1 and 𝐴 = 2, and for 𝐴 = 0.2, 𝐾 = 2. 39. Figures 17.33 and 17.34 show the gradient of the functions 𝑧 = 𝑓 (𝑥, 𝑦) and 𝑧 = 𝑔(𝑥, 𝑦).

𝑧

(IV)

𝑧

(V)

𝑣⃗ = 𝐴⃗𝑖 + 𝐾(𝑥2 + 𝑦2 )−1 (𝑥⃗𝑖 + 𝑦𝑗⃗ ).

𝑥

𝑦

38. In the middle of a wide, steadily flowing river there is a fountain that spouts water horizontally in all directions. The river flows in the ⃗𝑖 -direction in the 𝑥𝑦-plane and the fountain is at the origin.

𝑦

(a) For each function, draw a rough sketch of the level curves, showing possible 𝑧-values. (b) The 𝑥𝑧-plane cuts each of the surfaces 𝑧 = 𝑓 (𝑥, 𝑦) and 𝑧 = 𝑔(𝑥, 𝑦) in a curve. Sketch each of these curves, making clear how they are similar and how they are different from one another.

𝑧

(VI)

𝑦

𝑥

𝑦

𝑥

𝑦

𝑦 𝑥

In Problems 32–34, write formulas for vector fields with the given properties.

Figure 17.33: Gradient of 𝑧 = 𝑓 (𝑥, 𝑦)

𝑥

Figure 17.34: Gradient of 𝑧 = 𝑔(𝑥, 𝑦)

32. All vectors are parallel to the 𝑥-axis; all vectors on a vertical line have the same magnitude. 33. All vectors point toward the origin and have constant length. 34. All vectors are of unit length and perpendicular to the position vector at that point. ⃗ . Find a 35. (a) Let 𝐹⃗ = 𝑥⃗𝑖 + (𝑥 + 𝑦)𝑗⃗ + (𝑥 − 𝑦 + 𝑧)𝑘 ⃗ point at which 𝐹 is parallel to 𝑙, the line 𝑥 = 5 + 𝑡, 𝑦 = 6 − 2𝑡, 𝑧 = 7 − 3𝑡. (b) Find a point at which 𝐹⃗ and 𝑙 are perpendicular. (c) Give an equation for and describe in words the set of all points at which 𝐹⃗ and 𝑙 are perpendicular. In Problems 36–37, let 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ and 𝐺⃗ = −𝑦⃗𝑖 + 𝑥𝑗⃗ .

40. Let 𝐹⃗ = 𝑢⃗𝑖 + 𝑣𝑗⃗ be a vector field in 2-space with magnitude 𝐹 = ‖𝐹⃗ ‖.

(a) Let 𝑇⃗ = (1∕𝐹 )𝐹⃗ . Show that 𝑇⃗ is the unit vector in the direction of 𝐹⃗ . See Figure 17.35. ⃗ = (1∕𝐹 )(𝑘 ⃗ × 𝐹⃗ ) = (1∕𝐹 )(−𝑣⃗𝑖 + 𝑢𝑗⃗ ). (b) Let 𝑁 ⃗ Show that 𝑁 is the unit vector pointing to the left ⃗ See Figure 17.35. of and at right angles to 𝐹. 𝑦

𝐹⃗ ⃗ 𝑁 𝑇⃗

⃗ = 𝑎𝐹⃗ + 𝐺⃗ if: 36. Sketch the vector field 𝐿 (a)

𝑎=0

(b) 𝑎 > 0

(c)

𝑎 0

(c)

𝑥

Figure 17.35 𝑏 0 and 1 ∫ 𝐹⃗ ⋅ 𝑑⃗𝑟 < 0. 𝐶2

Are the statements in Problems 65–67 true or false? Explain why or give a counterexample. 65. ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 is a vector. 66. Suppose 𝐶1 is the unit square joining the points (0, 0), (1, 0), (1, 1), (0, 1) oriented clockwise and 𝐶2 is the same square but traversed twice in the opposite direction. If ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 3, then ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = −6. 1

18.2

2

67. The line integral of 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ = 𝑟⃗ along the semicircle 𝑥2 + 𝑦2 = 1, 𝑦 ≥ 0, oriented counterclockwise, is zero. Are the statements in Problems 68–74 true or false? Give reasons for your answer. 68. The line integral ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 is a scalar. 69. If 𝐶1 and 𝐶2 are oriented curves and 𝐶1 is longer than 𝐶2 , then ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 > ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 . 1

2

70. If 𝐶 is an oriented curve and ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 0, then 𝐹⃗ = 0⃗ . 71. If 𝐹⃗ = ⃗𝑖 is a vector field in 2-space, then ∫𝐶 𝐹⃗ ⋅𝑑⃗𝑟 > 0, where 𝐶 is the oriented line from (0, 0) to (1, 0). 72. If 𝐹⃗ = ⃗𝑖 is a vector field in 2-space, then ∫𝐶 𝐹⃗ ⋅𝑑⃗𝑟 > 0, where 𝐶 is the oriented line from (0, 0) to (0, 1). 73. If 𝐶1 is the upper semicircle 𝑥2 + 𝑦2 = 1, 𝑦 ≥ 0 and 𝐶2 is the lower semicircle 𝑥2 +𝑦2 = 1, 𝑦 ≤ 0, both oriented counterclockwise, then for any vector field 𝐹⃗ , we have ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = − ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 . 1

2

74. The work done by the force 𝐹⃗ = −𝑦⃗𝑖 +𝑥𝑗⃗ on a particle moving clockwise around the boundary of the square −1 ≤ 𝑥 ≤ 1, −1 ≤ 𝑦 ≤ 1 is positive.

COMPUTING LINE INTEGRALS OVER PARAMETERIZED CURVES The goal of this section is to show how to use a parameterization of a curve to convert a line integral into an ordinary one-variable integral.

932

Chapter 18 LINE INTEGRALS

Using a Parameterization to Evaluate a Line Integral Recall the definition of the line integral, ∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 =

lim ‖Δ⃗𝑟 𝑖 ‖→0



𝐹⃗ (⃗𝑟 𝑖 ) ⋅ Δ⃗𝑟 𝑖 ,

where the 𝑟⃗ 𝑖 are the position vectors of points subdividing the curve into short pieces. Now suppose we have a smooth parameterization, 𝑟⃗ (𝑡), of 𝐶 for 𝑎 ≤ 𝑡 ≤ 𝑏, so that 𝑟⃗ (𝑎) is the position vector of the starting point of the curve and 𝑟⃗ (𝑏) is the position vector of the end. Then we can divide 𝐶 into 𝑛 pieces by dividing the interval 𝑎 ≤ 𝑡 ≤ 𝑏 into 𝑛 pieces, each of size Δ𝑡 = (𝑏−𝑎)∕𝑛. See Figures 18.19 and 18.20. At each point 𝑟⃗ 𝑖 = 𝑟⃗ (𝑡𝑖 ) we want to compute 𝐹⃗ (⃗𝑟 𝑖 ) ⋅ Δ⃗𝑟 𝑖 .

✛Δ𝑡✲ 𝑡𝑖

𝑡𝑖+1



𝑟⃗ (𝑡𝑛−1 )

𝑟⃗ (𝑡𝑖 ) ✸

𝑡 𝑡𝑛−1 𝑡𝑛 = 𝑏

𝑟⃗ (𝑡1 )

.

𝑟⃗ (𝑡𝑛 ) = 𝑟⃗ (𝑏)



...

𝑡0 = 𝑎 𝑡1



...

𝑟⃗ (𝑡𝑖+1 )

Δ⃗𝑟 𝑖 = 𝑟⃗ (𝑡𝑖+1 ) − 𝑟⃗ (𝑡𝑖 ) 𝑟⃗ (𝑡0 ) = 𝑟⃗ (𝑎)

Figure 18.20: Corresponding subdivision of the parameterized path 𝐶

Figure 18.19: Subdivision of the interval 𝑎 ≤ 𝑡 ≤ 𝑏

Since 𝑡𝑖+1 = 𝑡𝑖 + Δ𝑡, the displacement vectors Δ𝑟⃗𝑖 are given by Δ𝑟⃗𝑖 = 𝑟⃗ (𝑡𝑖+1 ) − 𝑟⃗ (𝑡𝑖 ) = 𝑟⃗ (𝑡𝑖 + Δ𝑡) − 𝑟⃗ (𝑡𝑖 ) 𝑟⃗ (𝑡𝑖 + Δ𝑡) − 𝑟⃗ (𝑡𝑖 ) ⋅ Δ𝑡 = Δ𝑡 ′ ≈ 𝑟⃗ (𝑡𝑖 )Δ𝑡, where we use the facts that Δ𝑡 is small and that 𝑟⃗ (𝑡) is differentiable to obtain the last approximation. Therefore, ∑ ∑ 𝐹⃗ ⋅ 𝑑⃗𝑟 ≈ 𝐹⃗ (⃗𝑟 (𝑡𝑖 )) ⋅ 𝑟⃗ ′ (𝑡𝑖 ) Δ𝑡. 𝐹⃗ (⃗𝑟 𝑖 ) ⋅ Δ⃗𝑟 𝑖 ≈ ∫𝐶 Notice that 𝐹⃗ (⃗𝑟 (𝑡𝑖 )) ⋅ 𝑟⃗ ′ (𝑡𝑖 ) is the value at 𝑡𝑖 of a one-variable function of 𝑡, so this last sum is really a one-variable Riemann sum. In the limit as Δ𝑡 → 0, we get a definite integral: 𝑏

lim

Δ𝑡→0



𝐹⃗ (⃗𝑟 (𝑡𝑖 )) ⋅ 𝑟⃗ ′ (𝑡𝑖 ) Δ𝑡 =

∫𝑎

𝐹⃗ (⃗𝑟 (𝑡)) ⋅ 𝑟⃗ ′ (𝑡) 𝑑𝑡.

Thus, we have the following result: If 𝑟⃗ (𝑡), for 𝑎 ≤ 𝑡 ≤ 𝑏, is a smooth parameterization of an oriented curve 𝐶 and 𝐹⃗ is a vector field which is continuous on 𝐶, then 𝑏

∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫𝑎

𝐹⃗ (⃗𝑟 (𝑡)) ⋅ 𝑟⃗ ′ (𝑡) 𝑑𝑡.

In words: To compute the line integral of 𝐹⃗ over 𝐶, take the dot product of 𝐹⃗ evaluated on 𝐶 with the velocity vector, 𝑟⃗ ′ (𝑡), of the parameterization of 𝐶, then integrate along the curve.

18.2 COMPUTING LINE INTEGRALS OVER PARAMETERIZED CURVES

933

Even though we assumed that 𝐶 is smooth, we can use the same formula to compute line integrals over curves which are piecewise smooth, such as the boundary of a rectangle. If 𝐶 is piecewise smooth, we apply the formula to each one of the smooth pieces and add the results by applying property 4 on page 928. Example 1

Compute ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 where 𝐹⃗ = (𝑥 + 𝑦)⃗𝑖 + 𝑦𝑗⃗ and 𝐶 is the quarter unit circle, oriented counterclockwise as shown in Figure 18.21. 𝑦

𝐶 𝑥 Figure 18.21: The vector field 𝐹⃗ = (𝑥 + 𝑦)⃗𝑖 + 𝑦𝑗⃗ and the quarter circle 𝐶

Solution

Since most of the vectors in 𝐹⃗ along 𝐶 point generally in a direction opposite to the orientation of 𝐶, we expect our answer to be negative. The first step is to parameterize 𝐶 by 𝑟⃗ (𝑡) = 𝑥(𝑡)⃗𝑖 + 𝑦(𝑡)𝑗⃗ = cos 𝑡⃗𝑖 + sin 𝑡𝑗⃗ ,

0≤𝑡≤

𝜋 . 2

Substituting the parameterization into 𝐹⃗ , we get 𝐹⃗ (𝑥(𝑡), 𝑦(𝑡)) = (cos 𝑡 + sin 𝑡)⃗𝑖 + sin 𝑡𝑗⃗ . The vector 𝑟⃗ ′ (𝑡) = 𝑥′ (𝑡)⃗𝑖 + 𝑦′ (𝑡)𝑗⃗ = − sin 𝑡⃗𝑖 + cos 𝑡𝑗⃗ . Then 𝜋∕2

∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 =

((cos 𝑡 + sin 𝑡)⃗𝑖 + sin 𝑡𝑗⃗ ) ⋅ (− sin 𝑡⃗𝑖 + cos 𝑡𝑗⃗ )𝑑𝑡

∫0 𝜋∕2

=

(− cos 𝑡 sin 𝑡 − sin2 𝑡 + sin 𝑡 cos 𝑡)𝑑𝑡

∫0 𝜋∕2

=

∫0

− sin2 𝑡 𝑑𝑡 = −

𝜋 ≈ −0.7854. 4

So the answer is negative, as expected.

Example 2

Consider the vector field 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ . (a) Suppose 𝐶1 is the line segment joining (1, 0) to (0, 2) and 𝐶2 is a part of a parabola with its vertex at (0, 2), joining the same points in the same order. (See Figure 18.22.) Verify that ∫𝐶1

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫ 𝐶2

𝐹⃗ ⋅ 𝑑⃗𝑟 .

(b) If 𝐶 is the triangle shown in Figure 18.23, show that ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 0. 𝑦

𝑦

2 1

2

𝐶2

1

𝐶1

𝐶 1

Figure 18.22

1

𝑥 Figure 18.23

𝑥

934

Chapter 18 LINE INTEGRALS

Solution

(a) We parameterize 𝐶1 by 𝑟⃗ (𝑡) = (1 − 𝑡)⃗𝑖 + 2𝑡𝑗⃗ with 0 ≤ 𝑡 ≤ 1. Then 𝑟⃗ ′ (𝑡) = −⃗𝑖 + 2𝑗⃗ , so 1

∫𝐶1

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫0

1

𝐹⃗ (1 − 𝑡, 2𝑡) ⋅ (−⃗𝑖 + 2𝑗⃗ ) 𝑑𝑡 =

((1 − 𝑡)⃗𝑖 + 2𝑡𝑗⃗ ) ⋅ (−⃗𝑖 + 2𝑗⃗ ) 𝑑𝑡

∫0 1

=

(5𝑡 − 1) 𝑑𝑡 =

∫0

3 . 2

To parameterize 𝐶2 , we use the fact that it is part of a parabola with vertex at (0, 2), so its equation is of the form 𝑦 = −𝑘𝑥2 + 2 for some 𝑘. Since the parabola crosses the 𝑥-axis at (1, 0), we find that 𝑘 = 2 and 𝑦 = −2𝑥2 + 2. Therefore, we use the parameterization 𝑟⃗ (𝑡) = 𝑡⃗𝑖 + (−2𝑡2 + 2)𝑗⃗ with 0 ≤ 𝑡 ≤ 1, which has 𝑟⃗ ′ = ⃗𝑖 − 4𝑡𝑗⃗ . This traces out 𝐶2 in reverse, since 𝑡 = 0 gives (0, 2), and 𝑡 = 1 gives (1, 0). Thus, we make 𝑡 = 0 the upper limit of integration and 𝑡 = 1 the lower limit: 0

∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 = 2

∫1

𝐹⃗ (𝑡, −2𝑡2 + 2) ⋅ (⃗𝑖 − 4𝑡𝑗⃗ ) 𝑑𝑡 = −

1

∫0

(𝑡⃗𝑖 + (−2𝑡2 + 2)𝑗⃗ ) ⋅ (⃗𝑖 − 4𝑡𝑗⃗ ) 𝑑𝑡

1

=−

∫0

(8𝑡3 − 7𝑡) 𝑑𝑡 =

3 . 2

So the line integrals of 𝐹⃗ along 𝐶1 and 𝐶2 have the same value. (b) We break ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 into three pieces, one of which we have already computed (namely, the piece connecting (1, 0) to (0, 2), where the line integral has value 3∕2). The piece running from (0, 2) to (0, 0) can be parameterized by 𝑟⃗ (𝑡) = (2 − 𝑡)𝑗⃗ with 0 ≤ 𝑡 ≤ 2. The piece running from (0, 0) to (1, 0) can be parameterized by 𝑟⃗ (𝑡) = 𝑡⃗𝑖 with 0 ≤ 𝑡 ≤ 1. Then 2

1

3 𝐹⃗ ⋅ 𝑑⃗𝑟 = + 𝐹⃗ (0, 2 − 𝑡) ⋅ (−𝑗⃗ ) 𝑑𝑡 + 𝐹⃗ (𝑡, 0) ⋅ ⃗𝑖 𝑑𝑡 ∫𝐶 ∫0 2 ∫0 2

1

=

3 + (2 − 𝑡)𝑗⃗ ⋅ (−𝑗⃗ ) 𝑑𝑡 + 𝑡⃗𝑖 ⋅ ⃗𝑖 𝑑𝑡 ∫0 2 ∫0

=

1 3 3 + (𝑡 − 2) 𝑑𝑡 + 𝑡 𝑑𝑡 = + (−2) + = 0. ∫0 2 ∫0 2 2

2

Example 3

1

Let 𝐶 be the closed curve consisting of the upper half-circle of radius 1 and the line forming its diameter along the 𝑥-axis, oriented counterclockwise. (See Figure 18.24.) Find ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 where 𝐹⃗ (𝑥, 𝑦) = −𝑦⃗𝑖 + 𝑥𝑗⃗ . 𝑦 1

𝐶1

𝐶2 −1

𝑥 1

Figure 18.24: The curve 𝐶 = 𝐶1 + 𝐶2 for Example 3

Solution

We write 𝐶 = 𝐶1 + 𝐶2 where 𝐶1 is the half-circle and 𝐶2 is the line, and compute ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 and 1 ∫ 𝐹⃗ ⋅ 𝑑⃗𝑟 separately. We parameterize 𝐶1 by 𝑟⃗ (𝑡) = cos 𝑡⃗𝑖 + sin 𝑡𝑗⃗ , with 0 ≤ 𝑡 ≤ 𝜋. Then 𝐶2

𝜋

∫ 𝐶1

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫0

(− sin 𝑡⃗𝑖 + cos 𝑡𝑗⃗ ) ⋅ (− sin 𝑡⃗𝑖 + cos 𝑡𝑗⃗ ) 𝑑𝑡

18.2 COMPUTING LINE INTEGRALS OVER PARAMETERIZED CURVES 𝜋

=

935

𝜋

(sin2 𝑡 + cos2 𝑡) 𝑑𝑡 =

∫0

∫0

1 𝑑𝑡 = 𝜋.

For 𝐶2 , we have ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 0, since the vector field 𝐹⃗ has no ⃗𝑖 component along the 𝑥-axis 2 (where 𝑦 = 0) and is therefore perpendicular to 𝐶2 at all points. Finally, we can write ∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫ 𝐶1

𝐹⃗ ⋅ 𝑑⃗𝑟 +

∫ 𝐶2

𝐹⃗ ⋅ 𝑑⃗𝑟 = 𝜋 + 0 = 𝜋.

It is no accident that the result for ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 is the same as the length of the curve 𝐶1 . See 1 Problem 56 on page 931 and Problem 41 on page 938. The next example illustrates the computation of a line integral over a path in 3-space. Example 4

⃗ and is subject to a force A particle travels along the helix 𝐶 given by 𝑟⃗ (𝑡) = cos 𝑡⃗𝑖 + sin 𝑡𝑗⃗ + 2𝑡𝑘 ⃗ ⃗ ⃗ ⃗ 𝐹 = 𝑥𝑖 + 𝑧𝑗 − 𝑥𝑦𝑘 . Find the total work done on the particle by the force for 0 ≤ 𝑡 ≤ 3𝜋.

Solution

The work done is given by a line integral, which we evaluate using the given parameterization: 3𝜋

Work done =

∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫0

𝐹⃗ (⃗𝑟 (𝑡)) ⋅ 𝑟⃗ ′ (𝑡) 𝑑𝑡

3𝜋

=

⃗ ) ⋅ (− sin 𝑡⃗𝑖 + cos 𝑡𝑗⃗ + 2𝑘 ⃗ ) 𝑑𝑡 (cos 𝑡⃗𝑖 + 2𝑡𝑗⃗ − cos 𝑡 sin 𝑡𝑘

∫0 3𝜋

=

(− cos 𝑡 sin 𝑡 + 2𝑡 cos 𝑡 − 2 cos 𝑡 sin 𝑡) 𝑑𝑡

∫0 3𝜋

=

∫0

(−3 cos 𝑡 sin 𝑡 + 2𝑡 cos 𝑡) 𝑑𝑡 = −4.

The Differential Notation ∫𝑪 𝑷 𝒅𝒙 + 𝑸 𝒅𝒚 + 𝑹𝒅𝒛 There is an alternative notation for line integrals that is often useful. For the vector field 𝐹⃗ = ⃗ and an oriented curve 𝐶, if we write 𝑑⃗𝑟 = 𝑑𝑥⃗𝑖 + 𝑑𝑦𝑗⃗ + 𝑑𝑧𝑘 ⃗ 𝑃 (𝑥, 𝑦, 𝑧)⃗𝑖 + 𝑄(𝑥, 𝑦, 𝑧)𝑗⃗ + 𝑅(𝑥, 𝑦, 𝑧)𝑘 we have 𝑃 (𝑥, 𝑦, 𝑧)𝑑𝑥 + 𝑄(𝑥, 𝑦, 𝑧)𝑑𝑦 + 𝑅(𝑥, 𝑦, 𝑧)𝑑𝑧. 𝐹⃗ ⋅ 𝑑⃗𝑟 = ∫𝐶 ∫𝐶 𝑥𝑦 𝑑𝑥 − 𝑦2 𝑑𝑦 where 𝐶 is the line segment from (0, 0) to (2, 6).

Example 5

Evaluate

Solution

We parameterize 𝐶 by 𝑥 = 𝑡, 𝑦 = 3𝑡, 0 ≤ 𝑡 ≤ 2. Thus, 𝑑𝑥 = 𝑑𝑡, 𝑑𝑦 = 3𝑑𝑡, so

∫𝐶

2

∫𝐶

𝑥𝑦 𝑑𝑥 − 𝑦2 𝑑𝑦 =

∫0

2

𝑡(3𝑡)𝑑𝑡 − (3𝑡)2 (3𝑑𝑡) =

∫0

(−24𝑡2 ) 𝑑𝑡 = −64.

Line integrals can be expressed either using vectors or using differentials. If the independent variables are distances, then visualizing a line integral in terms of dot products can be useful. However, if the independent variables are, for example, temperature and volume, then the dot product does not have physical meaning, so differentials are more natural.

936

Chapter 18 LINE INTEGRALS

Independence of Parameterization Since there are many different ways of parameterizing a given oriented curve, you may be wondering what happens to the value of a given line integral if you choose another parameterization. The answer is that the choice of parameterization makes no difference. Since we initially defined the line integral without reference to any particular parameterization, this is exactly as we would expect. Example 6

Consider the oriented path which is a straight-line segment 𝐿 running from (0, 0) to (1, 1). Calculate the line integral of the vector field 𝐹⃗ = (3𝑥 − 𝑦)⃗𝑖 + 𝑥𝑗⃗ along 𝐿 using each of the parameterizations (a) 𝐴(𝑡) = (𝑡, 𝑡),

Solution

(b) 𝐷(𝑡) = (𝑒𝑡 − 1, 𝑒𝑡 − 1),

0 ≤ 𝑡 ≤ 1,

0 ≤ 𝑡 ≤ ln 2.

The line 𝐿 has equation 𝑦 = 𝑥. Both 𝐴(𝑡) and 𝐷(𝑡) give a parameterization of 𝐿: each has both coordinates equal and each begins at (0,0) and ends at (1,1). Now let’s calculate the line integral of the vector field 𝐹⃗ = (3𝑥 − 𝑦)⃗𝑖 + 𝑥𝑗⃗ using each parameterization. (a) Using 𝐴(𝑡), we get 1

∫𝐿

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫0

1

1

((3𝑡 − 𝑡)⃗𝑖 + 𝑡𝑗⃗ ) ⋅ (⃗𝑖 + 𝑗⃗ ) 𝑑𝑡 =

∫0

3𝑡 𝑑𝑡 =

(b) Using 𝐷(𝑡), we get ln 2

∫𝐿

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫0 ln 2

=

((

) ) 3(𝑒𝑡 − 1) − (𝑒𝑡 − 1) ⃗𝑖 + (𝑒𝑡 − 1)𝑗⃗ ⋅ (𝑒𝑡 ⃗𝑖 + 𝑒𝑡 𝑗⃗ ) 𝑑𝑡 2𝑡

𝑡

3(𝑒 − 𝑒 ) 𝑑𝑡 = 3

∫0

3𝑡2 || 3 = . 2 ||0 2

(

𝑒2𝑡 − 𝑒𝑡 2

)

|ln 2 3 | = . | 2 |0

The fact that both answers are the same illustrates that the value of a line integral is independent of the parameterization of the path. Problems 59–61 (available online) give another way of seeing this.

Exercises and Problems for Section 18.2 Online Resource: Additional Problems for Section 18.2 EXERCISES

𝑏

In Exercises 1–3, write ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 in the form ∫𝑎 𝑔(𝑡) 𝑑𝑡. (Give a formula for 𝑔 and numbers for 𝑎 and 𝑏. You do not need to evaluate the integral.) 1. 𝐹⃗ = 𝑦⃗𝑖 + 𝑥𝑗⃗ and 𝐶 is the semicircle from (0, 1) to (0, −1) with 𝑥 > 0. ⃗ and 𝐶 is the line from (0, 1, 0) to (2, 3, 2). 2. 𝐹⃗ = 𝑥⃗𝑖 +𝑧2 𝑘 ⃗ and 𝐶 is the unit 3. 𝐹⃗ = (cos 𝑥)⃗𝑖 + (cos 𝑦)𝑗⃗ + (cos 𝑧)𝑘 circle in the plane 𝑧 = 10, centered on the 𝑧-axis and oriented counterclockwise when viewed from above. In Exercises 4–8, find the line integral. 4.

(3⃗𝑖 + (𝑦 + 5)𝑗⃗ ) ⋅ 𝑑⃗𝑟 where 𝐶 is the line from (0, 0) ∫𝐶 to (0, 3).

5.

(2𝑥⃗𝑖 + 3𝑦𝑗⃗ ) ⋅ 𝑑⃗𝑟 where 𝐶 is the line from (1, 0, 0)

∫𝐶 to (5, 0, 0). 6. ∫𝐶 (2𝑦2 ⃗𝑖 + 𝑥𝑗⃗ ) ⋅ 𝑑⃗𝑟 where 𝐶 is the line segment from (3, 1) to (0, 0).

7. ∫𝐶 (𝑥⃗𝑖 + 𝑦𝑗⃗ ) ⋅ 𝑑⃗𝑟 where 𝐶 is the semicircle with center at (2, 0) and going from (3, 0) to (1, 0) in the region 𝑦 > 0. 8. Find ∫𝐶 ((𝑥2 + 𝑦)⃗𝑖 + 𝑦3 𝑗⃗ ) ⋅ 𝑑⃗𝑟 where 𝐶 consists of the three line segments from (4, 0, 0) to (4, 3, 0) to (0, 3, 0) to (0, 3, 5). In Exercises 9–23, find ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 for the given 𝐹⃗ and 𝐶. 9. 𝐹⃗ = 2⃗𝑖 + 𝑗⃗ ; 𝐶 is the 𝑥-axis from 𝑥 = 10 to 𝑥 = 7. 10. 𝐹⃗ = 3𝑗⃗ − ⃗𝑖 ; 𝐶 is the line 𝑦 = 𝑥 from (1, 1) to (5, 5). 11. 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ and 𝐶 is the line from (0, 0) to (3, 3). 12. 𝐹⃗ = 𝑦⃗𝑖 − 𝑥𝑗⃗ and 𝐶 is the right-hand side of the unit circle, starting at (0, 1). 13. 𝐹⃗ = 𝑥2 ⃗𝑖 + 𝑦2 𝑗⃗ and 𝐶 is the line from the point (1, 2) to the point (3, 4). 14. 𝐹⃗ = −𝑦 sin 𝑥⃗𝑖 + cos 𝑥𝑗⃗ and 𝐶 is the parabola 𝑦 = 𝑥2 between (0, 0) and (2, 4). 15. 𝐹⃗ = 𝑦3 ⃗𝑖 + 𝑥2 𝑗⃗ and 𝐶 is the line from (0, 0) to (3, 2).

18.2 COMPUTING LINE INTEGRALS OVER PARAMETERIZED CURVES

16. 𝐹⃗ = 2𝑦⃗𝑖 − (sin 𝑦)𝑗⃗ counterclockwise around the unit circle 𝐶 starting at the point (1, 0). 17. 𝐹⃗ = ln 𝑦⃗𝑖 + ln 𝑥𝑗⃗ and 𝐶 is the curve given parametrically by (2𝑡, 𝑡3 ), for 2 ≤ 𝑡 ≤ 4. ⃗ , and 𝐶 is the line 𝑥 = 𝑦 = 𝑧 from 18. 𝐹⃗ = 𝑥⃗𝑖 + 6𝑗⃗ − 𝑘 (0, 0, 0) to (2, 2, 2). 19. 𝐹⃗ = (2𝑥−𝑦+4)⃗𝑖 +(5𝑦+3𝑥−6)𝑗⃗ and 𝐶 is the triangle with vertices (0, 0), (3, 0), (3, 2) traversed counterclockwise. ⃗ for ⃗ and 𝐶 is 𝑟⃗ = 𝑡⃗𝑖 + 𝑡2 𝑗⃗ + 𝑡3 𝑘 20. 𝐹⃗ = 𝑥⃗𝑖 + 2𝑧𝑦𝑗⃗ + 𝑥𝑘 1 ≤ 𝑡 ≤ 2. ⃗ and 𝐶 is the line from the origin 21. 𝐹⃗ = 𝑥3 ⃗𝑖 + 𝑦2 𝑗⃗ + 𝑧𝑘 to the point (2, 3, 4). ⃗ and 𝐶 is the helix 𝑥 = cos 𝑡, 𝑦 = 22. 𝐹⃗ = −𝑦⃗𝑖 + 𝑥𝑗⃗ + 5𝑘 sin 𝑡, 𝑧 = 𝑡, for 0 ≤ 𝑡 ≤ 4𝜋. ⃗ and 𝐶 is the circle of radius 23. 𝐹⃗ = 𝑒𝑦 ⃗𝑖 + ln(𝑥2 + 1)𝑗⃗ + 𝑘 2 centered at the origin in the 𝑦𝑧-plane in Figure 18.25. 𝑧

𝑦

■ 𝑥

Start

Figure 18.25

937

24. Every line integral can be written in both vector notation and differential notation. For example, ∫𝐶

(2𝑥⃗𝑖 + (𝑥 + 𝑦)𝑗⃗ ) ⋅ 𝑑⃗𝑟 =

∫𝐶

2𝑥 𝑑𝑥 + (𝑥 + 𝑦) 𝑑𝑦.

(a) Express ∫𝐶 3 𝑑𝑥 + 𝑥𝑦 𝑑𝑦 in vector notation. (b) Express ∫𝐶 (100 cos 𝑥⃗𝑖 + 𝑒𝑦 sin 𝑥𝑗⃗ ) ⋅ 𝑑⃗𝑟 in differential notation. In Exercises 25–26, express the line integral ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 in differential notation. 25. 𝐹⃗ = 3𝑥⃗𝑖 − 𝑦 sin 𝑥𝑗⃗ ⃗ 26. 𝐹⃗ = 𝑦2 ⃗𝑖 + 𝑧2 𝑗⃗ + (𝑥2 − 5)𝑘 In Exercises 27–28, find 𝐹⃗ so that the line integral equals ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 . 27. ∫𝐶 (𝑥 + 2𝑦)𝑑𝑥 + 𝑥2 𝑦 𝑑𝑦 28. ∫𝐶 𝑒−3𝑦 𝑑𝑥 − 𝑦𝑧(sin 𝑥) 𝑑𝑦 + (𝑦 + 𝑧) 𝑑𝑧 Evaluate the line integrals in Exercises 29–34. 29. ∫𝐶 𝑦 𝑑𝑥 + 𝑥 𝑑𝑦 where 𝐶 is the parameterized path 𝑥 = 𝑡2 , 𝑦 = 𝑡3 , 1 ≤ 𝑡 ≤ 5. 30. ∫𝐶 𝑑𝑥 + 𝑦 𝑑𝑦 + 𝑧 𝑑𝑧 where 𝐶 is one turn of the helix 𝑥 = cos 𝑡, 𝑦 = sin 𝑡, 𝑧 = 3𝑡, 0 ≤ 𝑡 ≤ 2𝜋. 31. ∫𝐶 3𝑦 𝑑𝑥 + 4𝑥 𝑑𝑦 where 𝐶 is the straight-line path from (1, 3) to (5, 9). 32. ∫𝐶 𝑥 𝑑𝑥 + 𝑧 𝑑𝑦 − 𝑦 𝑑𝑧 where 𝐶 is the circle of radius 3 in the 𝑦𝑧-plane centered at the origin, oriented counterclockwise when viewed from the positive 𝑦-axis. 33. ∫𝐶 (𝑥 + 𝑦) 𝑑𝑥 + 𝑥2 𝑑𝑦 where 𝐶 is the path 𝑥 = 𝑡2 , 𝑦 = 𝑡, 0 ≤ 𝑡 ≤ 10. 34. ∫𝐶 𝑥 𝑑𝑦 where 𝐶 is the quarter circle centered at the origin going counterclockwise from (2, 0) to (0, 2).

PROBLEMS 35. Evaluate the line integral of 𝐹⃗ = (3𝑥 − 𝑦)⃗𝑖 + 𝑥𝑗⃗ over two different paths from (0, 0) to (1, 1). (a) The path (𝑡, 𝑡2 ), with 0 ≤ 𝑡 ≤ 1 (b) The path (𝑡2 , 𝑡), with 0 ≤ 𝑡 ≤ 1 36. Curves 𝐶1 and 𝐶2 are parameterized as follows:

(b) 𝐶(𝑡) =

(

𝑡2 − 1 𝑡2 − 1 , 3 3

for − 1 ≤ 𝑡 ≤ 1 3𝜋 𝜋 𝐶2 is (𝑥(𝑡), 𝑦(𝑡)) = (cos 𝑡, sin 𝑡) for ≤ 𝑡 ≤ . 2 2

37. Calculate the line integral of 𝐹⃗ = (3𝑥 − 𝑦)⃗𝑖 + 𝑥𝑗⃗ along the line segment 𝐿 from (0, 0) to (1, 1) using each of the parameterizations (a) 𝐵(𝑡) = (2𝑡, 2𝑡),

0 ≤ 𝑡 ≤ 1∕2

,

1≤𝑡≤2

In Problems 38–39 evaluate the line integral using a shortcut available in two special situations. • If 𝐹⃗ is perpendicular to 𝐶 at every point of 𝐶, then

𝐶1 is (𝑥(𝑡), 𝑦(𝑡)) = (0, 𝑡)

(a) Sketch 𝐶1 and 𝐶2 with arrows showing their orientation. (b) Suppose 𝐹⃗ = (𝑥+3𝑦)⃗𝑖 +𝑦𝑗⃗ . Calculate ∫𝐶 𝐹⃗ ⋅𝑑⃗𝑟 , where 𝐶 is the curve given by 𝐶 = 𝐶1 + 𝐶2 .

)

∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 = 0.

• If 𝐹⃗ is tangent to 𝐶 at every point of 𝐶 and has constant magnitude on 𝐶, then

∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 = ±‖𝐹⃗ ‖ × Length of 𝐶.

Choose the + sign if 𝐹⃗ points in the direction of integration and choose the − sign if 𝐹⃗ points in the direction opposite to the direction of integration.

938

Chapter 18 LINE INTEGRALS

38. ∫𝐶 (𝑥⃗𝑖 + 𝑦𝑗⃗ ) ⋅ 𝑑⃗𝑟 , where 𝐶 is the circle of radius 10 centered at the origin, oriented counterclockwise. 39. ∫𝐶 (−𝑦⃗𝑖 + 𝑥𝑗⃗ ) ⋅ 𝑑⃗𝑟 , where 𝐶 is the circle of radius 10 centered at the origin, oriented counterclockwise. 40. Justify the following without parameterizing the paths. (a) ∫𝐶 (𝑥2 + cos 𝑦) 𝑑𝑦 = 0 where 𝐶 is the straight line path from (10, 5) to (20, 5). (b) ∫𝐶 (𝑥2 + cos 𝑦) 𝑑𝑥 = 0 where 𝐶 is the straight line path from (10, 5) to (10, 50). 41. Let 𝐹⃗ = −𝑦⃗𝑖 + 𝑥𝑗⃗ and let 𝐶 be the unit circle oriented counterclockwise. (a) Show that 𝐹⃗ has a constant magnitude of 1 on 𝐶. (b) Show that 𝐹⃗ is always tangent to the circle 𝐶. (c) Show that ∫ 𝐹⃗ ⋅ 𝑑⃗𝑟 = Length of 𝐶. 𝐶

42. A spiral staircase in a building is in the shape of a helix of radius 5 meters. Between two floors of the building, the stairs make one full revolution and climb by 4 meters. A person carries a bag of groceries up two floors. The combined mass of the person and the groceries is 𝑚 = 70 kg and the gravitational force is 𝑚𝑔 downward, where 𝑔 = 9.8 m∕sec2 is the acceleration due to gravity. Calculate the work done by the person against gravity. ⃗ for 0 ≤ 𝑡 ≤ 1, we know 43. If 𝐶 is 𝑟⃗ = (𝑡 + 1)⃗𝑖 + 2𝑡𝑗⃗ + 3𝑡𝑘 ∫𝐶 𝐹⃗ (⃗𝑟 ) ⋅ 𝑑⃗𝑟 = 5. Find the value of the integrals: 0 ⃗ ) ⋅ (⃗𝑖 + 2𝑗⃗ + 3𝑘 ⃗ ) 𝑑𝑡 (a) ∫1 𝐹⃗ ((𝑡 + 1)⃗𝑖 + 2𝑡𝑗⃗ + 3𝑡𝑘 1 ⃗ )⋅(2𝑡⃗𝑖 +4𝑡𝑗⃗ +6𝑡𝑘 ⃗ ) 𝑑𝑡 (b) ∫0 𝐹⃗ ((𝑡2 +1)⃗𝑖 +2𝑡2 𝑗⃗ +3𝑡2 𝑘 1 2 2⃗ 2⃗ ⃗ ⃗ ) 𝑑𝑡 ⃗ ⃗ ⃗ (c) ∫ 𝐹 ((𝑡 +1)𝑖 +2𝑡 𝑗 +3𝑡 𝑘 )⋅(2𝑡𝑖 +4𝑡𝑗 +6𝑡𝑘 −1

Strengthen Your Understanding In Problems 44–45, explain what is wrong with the statement. 44. For the vector field 𝐹⃗ = 𝑥⃗𝑖 − 𝑦𝑗⃗ and oriented path 𝐶 parameterized by 𝑥 = cos 𝑡, 𝑦 = sin 𝑡, 0 ≤ 𝑡 ≤ 𝜋∕2, we have 𝜋∕2

∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫0

(cos 𝑡⃗𝑖 − sin 𝑡𝑗⃗ ) ⋅ (cos 𝑡⃗𝑖 + sin 𝑡𝑗⃗ ) 𝑑𝑡.

45. If ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 0, then 𝐹⃗ is perpendicular to 𝐶 at every point on 𝐶.

46. A vector field 𝐹⃗ such that, for the parameterized path 𝑟⃗ (𝑡) = 3 cos 𝑡⃗𝑖 + 3 sin 𝑡𝑗⃗ , −𝜋∕2 ≤ 𝑡 ≤ 𝜋∕2, the integral ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 can be computed geometrically, without using the parameterization. 47. A parameterized path 𝐶 such that, for the vector field 𝐹⃗ (𝑥, 𝑦) = sin 𝑦⃗𝑖 , the integral ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 is nonzero and can be computed geometrically, without using the parameterization. Are the statements in Problems 48–56 true or false? Give reasons for your answer. 48. If 𝐶1 and 𝐶2 are oriented curves with 𝐶2 beginning where 𝐶1 ends, then ∫𝐶 +𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 > ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 . 2

1

49. The line integral ∫𝐶 4⃗𝑖 ⋅ 𝑑⃗𝑟 over the curve 𝐶 parameterized by 𝑟⃗ (𝑡) = 𝑡⃗𝑖 + 𝑡2 𝑗⃗ , for 0 ≤ 𝑡 ≤ 2, is positive. 50. If 𝐶1 is the curve parameterized by 𝑟⃗ 1 (𝑡) = cos 𝑡⃗𝑖 + sin 𝑡𝑗⃗ , with 0 ≤ 𝑡 ≤ 𝜋, and 𝐶2 is the curve parameterized by 𝑟⃗ 2 (𝑡) = cos 𝑡⃗𝑖 − sin 𝑡𝑗⃗ , 0 ≤ 𝑡 ≤ 𝜋, then for any vector field 𝐹⃗ we have ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 . 1

2

51. If 𝐶1 is the curve parameterized by 𝑟⃗ 1 (𝑡) = cos 𝑡⃗𝑖 + sin 𝑡𝑗⃗ , with 0 ≤ 𝑡 ≤ 𝜋, and 𝐶2 is the curve parameterized by 𝑟⃗ 2 (𝑡) = cos(2𝑡)⃗𝑖 + sin(2𝑡)𝑗⃗ , 0 ≤ 𝑡 ≤ 𝜋2 , then for any vector field 𝐹⃗ we have ∫ 𝐹⃗ ⋅ 𝑑⃗𝑟 = ∫ 𝐹⃗ ⋅ 𝑑⃗𝑟 . 𝐶1

53. If 𝐶1 is the line segment from (0, 0) to (1, 0) and 𝐶2 is the line segment from (0, 0) to (2, 0), then for any vector field 𝐹⃗ , we have ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 2 ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 . 2

1

54. If 𝐶 is a circle of radius 𝑎, centered at the origin and oriented counterclockwise, then ∫𝐶 (2𝑥⃗𝑖 + 𝑦𝑗⃗ ) ⋅ 𝑑⃗𝑟 = 0. 55. If 𝐶 is a circle of radius 𝑎, centered at the origin and oriented counterclockwise, then ∫𝐶 (2𝑦⃗𝑖 + 𝑥𝑗⃗ ) ⋅ 𝑑⃗𝑟 = 0.

In Problems 46–47, give an example of:

1

52. If 𝐶 is the curve parameterized by 𝑟⃗ (𝑡), for 𝑎 ≤ 𝑡 ≤ 𝑏 with 𝑟⃗ (𝑎) = 𝑟⃗ (𝑏), then ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 0 for any vector field 𝐹⃗ . (Note that 𝐶 starts and ends at the same place.)

𝐶2

56. If 𝐶1 is the curve parameterized by 𝑟⃗ 1 (𝑡) = 𝑡⃗𝑖 + 𝑡2 𝑗⃗ , with 0 ≤ 𝑡 ≤ 2, and 𝐶2 is the curve parameterized by 𝑟⃗ 2 (𝑡) = (2 − 𝑡)⃗𝑖 + (2 − 𝑡)2 𝑗⃗ , 0 ≤ 𝑡 ≤ 2, then for any vector field 𝐹⃗ we have ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = − ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 . 1

2

57. If 𝐶1 is the path parameterized by 𝑟⃗ 1 (𝑡) = (𝑡, 𝑡), 0 ≤ 𝑡 ≤ 1, and if 𝐶2 is the path parameterized by 𝑟⃗ 2 (𝑡) = (1 − 𝑡, 1 − 𝑡), 0 ≤ 𝑡 ≤ 1, and if 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ , which of the following is true? (a) ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 > ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 1 2 (b) ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 < ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 1 2 (c) ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 1

2

58. If 𝐶1 is the path parameterized by 𝑟⃗ 1 (𝑡) = (𝑡, 𝑡), for 0 ≤ 𝑡 ≤ 1, and if 𝐶2 is the path parameterized by 𝑟⃗ 2 (𝑡) = (sin 𝑡, sin 𝑡), for 0 ≤ 𝑡 ≤ 1, and if 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ , which of the following is true? (a) ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 > ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 1 2 (b) ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 < ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 1 2 (c) ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 1

2

18.3 GRADIENT FIELDS AND PATH-INDEPENDENT FIELDS

18.3

939

GRADIENT FIELDS AND PATH-INDEPENDENT FIELDS For a function, 𝑓 , of one variable, the Fundamental Theorem of Calculus tells us that the definite integral of a rate of change, 𝑓 ′ , gives the total change in 𝑓 : 𝑏

∫𝑎

𝑓 ′ (𝑡) 𝑑𝑡 = 𝑓 (𝑏) − 𝑓 (𝑎).

What about functions of two or more variables? The quantity that describes the rate of change is the gradient vector field. If we know the gradient of a function 𝑓 , can we compute the total change in 𝑓 between two points? The answer is yes, using a line integral.

Finding the Total Change in 𝒇 from grad 𝒇 : The Fundamental Theorem To find the change in 𝑓 between two points 𝑃 and 𝑄, we choose a smooth path 𝐶 from 𝑃 to 𝑄, then divide the path into many small pieces. See Figure 18.26. First we estimate the change in 𝑓 as we move through a displacement Δ⃗𝑟 𝑖 from 𝑟⃗ 𝑖 to 𝑟⃗ 𝑖+1 . Suppose 𝑢⃗ is a unit vector in the direction of Δ⃗𝑟 𝑖 . Then the change in 𝑓 is given by 𝑓 (⃗𝑟 𝑖+1 ) − 𝑓 (⃗𝑟 𝑖 ) ≈ Rate of change of 𝑓 × Distance moved in direction of 𝑢⃗ = 𝑓𝑢⃗ (⃗𝑟 𝑖 )‖Δ⃗𝑟 𝑖 ‖ = grad 𝑓 ⋅ 𝑢⃗ ‖Δ⃗𝑟 𝑖 ‖ = grad 𝑓 ⋅ Δ⃗𝑟 𝑖 .

since Δ⃗𝑟 𝑖 = ‖Δ⃗𝑟 𝑖 ‖⃗𝑢

Therefore, summing over all pieces of the path, the total change in 𝑓 is given by Total change = 𝑓 (𝑄) − 𝑓 (𝑃 ) ≈

𝑛−1 ∑

grad 𝑓 (⃗𝑟 𝑖 ) ⋅ Δ⃗𝑟 𝑖 .

𝑖=0

In the limit as ‖Δ𝑟⃗𝑖 ‖ approaches zero, this suggests the following result:

Theorem 18.1: The Fundamental Theorem of Calculus for Line Integrals Suppose 𝐶 is a piecewise smooth oriented path with starting point 𝑃 and ending point 𝑄. If 𝑓 is a function whose gradient is continuous on the path 𝐶, then ∫𝐶

grad 𝑓 ⋅ 𝑑⃗𝑟 = 𝑓 (𝑄) − 𝑓 (𝑃 ).

Notice that there are many different paths from 𝑃 to 𝑄. (See Figure 18.27.) However, the value of the line integral ∫𝐶 grad 𝑓 ⋅ 𝑑⃗𝑟 depends only on the endpoints of 𝐶; it does not depend on where 𝐶 goes in between. Problem 88 (available online) shows how the Fundamental Theorem for Line Integrals can be derived from the one-variable Fundamental Theorem of Calculus. Δ⃗𝑟 𝑖 = 𝑟⃗ 𝑖+1 − 𝑟⃗ 𝑖 𝑄 ... ❄ ✸ 𝑟⃗ 𝑖+1

𝑟⃗ 𝑛−1

𝑟⃗ 𝑛

𝑄

...

𝑟⃗ 𝑖

𝑃

𝑟⃗ 1

𝑃

𝑟⃗ 0 Figure 18.26: Subdivision of the path from 𝑃 to 𝑄. We estimate the change in 𝑓 along Δ⃗𝑟 𝑖

Figure 18.27: There are many different paths from 𝑃 to 𝑄: all give the same value of ∫𝐶 grad 𝑓 ⋅ 𝑑⃗𝑟

940

Chapter 18 LINE INTEGRALS

Example 1

Suppose that grad 𝑓 is everywhere perpendicular to the curve joining 𝑃 and 𝑄 shown in Figure 18.28. (a) Explain why you expect the path joining 𝑃 and 𝑄 to be a contour. (b) Using a line integral, show that 𝑓 (𝑃 ) = 𝑓 (𝑄).

𝑃

𝑄

Figure 18.28: The gradient vector field of the function 𝑓

Solution

(a) The gradient of 𝑓 is everywhere perpendicular to the path from 𝑃 to 𝑄, as you expect along a contour. (b) Consider the path from 𝑃 to 𝑄 shown in Figure 18.28 and evaluate the line integral ∫𝐶

grad 𝑓 ⋅ 𝑑⃗𝑟 = 𝑓 (𝑄) − 𝑓 (𝑃 ).

Since grad 𝑓 is everywhere perpendicular to the path, the line integral is 0. Thus, 𝑓 (𝑄) = 𝑓 (𝑃 ).

Example 2

Consider the vector field 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ . In Example 2 on page 933 we calculated ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 and 1 ∫ 𝐹⃗ ⋅ 𝑑⃗𝑟 over the oriented curves shown in Figure 18.29 and found they were the same. Find a 𝐶2

scalar function 𝑓 with grad 𝑓 = 𝐹⃗ . Hence, find an easy way to calculate the line integrals, and explain why we could have expected them to be the same. 𝑦 2 1

𝐶2 𝐶1 1

𝑥

Figure 18.29: Find the line integral of 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ over the curves 𝐶1 and 𝐶2

Solution

We want to find a function 𝑓 (𝑥, 𝑦) for which 𝑓𝑥 = 𝑥 and 𝑓𝑦 = 𝑦. One possibility for 𝑓 is 𝑓 (𝑥, 𝑦) =

𝑥2 𝑦2 + . 2 2

You can check that grad 𝑓 = 𝑥⃗𝑖 + 𝑦𝑗⃗ . Now we can use the Fundamental Theorem to compute the line integral. Since 𝐹⃗ = grad 𝑓 we have ∫ 𝐶1

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫𝐶1

grad 𝑓 ⋅ 𝑑⃗𝑟 = 𝑓 (0, 2) − 𝑓 (1, 0) =

3 . 2

Notice that the calculation looks exactly the same for 𝐶2 . Since the value of the integral depends only on the values of 𝑓 at the endpoints, it is the same no matter what path we choose.

18.3 GRADIENT FIELDS AND PATH-INDEPENDENT FIELDS

941

Path-Independent, or Conservative, Vector Fields In the previous example, the line integral was independent of the path taken between the two (fixed) endpoints. We give vector fields whose line integrals have this property a special name. A vector field 𝐹⃗ is said to be path-independent, or conservative, if for any two points 𝑃 and 𝑄, the line integral ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 has the same value along any piecewise smooth path 𝐶 from 𝑃 to 𝑄 lying in the domain of 𝐹⃗ . If, on the other hand, the line integral ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 does depend on the path 𝐶 joining 𝑃 to 𝑄, then 𝐹⃗ is said to be a path-dependent vector field. Now suppose that 𝐹⃗ is any continuous gradient field, so 𝐹⃗ = grad 𝑓 . If 𝐶 is a path from 𝑃 to 𝑄, the Fundamental Theorem for Line Integrals tells us that ∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 = 𝑓 (𝑄) − 𝑓 (𝑃 ).

Since the right-hand side of this equation does not depend on the path, but only on the endpoints of the path, the vector field 𝐹⃗ is path-independent. Thus, we have the following important result: If 𝐹⃗ is a continuous gradient vector field, then 𝐹⃗ is pathindependent.

Why Do We Care About Path-Independent, or Conservative, Vector Fields? Many of the fundamental vector fields of nature are path-independent—for example, the gravitational field and the electric field of particles at rest. The fact that the gravitational field is path-independent means that the work done by gravity when an object moves depends only on the starting and ending points and not on the path taken. For example, the work done by gravity (computed by the line integral) on a bicycle being carried to a sixth floor apartment is the same whether it is carried up the stairs in a zig-zag path or taken straight up in an elevator. When a vector field is path-independent, we can define the potential energy of a body. When the body moves to another position, the potential energy changes by an amount equal to the work done by the vector field, which depends only on the starting and ending positions. If the work done had not been path-independent, the potential energy would depend both on the body’s current position and on how it got there, making it impossible to define a useful potential energy. Project 1 (available online) explains why path-independent force vector fields are also called conservative vector fields: When a particle moves under the influence of a conservative vector field, the total energy of the particle is conserved. It turns out that the force field is obtained from the gradient of the potential energy function.

Path-Independent Fields and Gradient Fields We have seen that every gradient field is path-independent. What about the converse? That is, given a path-independent vector field 𝐹⃗ , can we find a function 𝑓 such that 𝐹⃗ = grad 𝑓 ? The answer is yes, provided that 𝐹⃗ is continuous.

⃖⃖⃗ How to Construct 𝒇 from 𝑭 First, notice that there are many different choices for 𝑓 , since we can add a constant to 𝑓 without changing grad 𝑓 . If we pick a fixed starting point 𝑃 , then by adding or subtracting a constant to 𝑓 , we can ensure that 𝑓 (𝑃 ) = 0. For any other point 𝑄, we define 𝑓 (𝑄) by the formula 𝑓 (𝑄) =

∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 ,

where 𝐶 is any path from 𝑃 to 𝑄.

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Chapter 18 LINE INTEGRALS

Since 𝐹⃗ is path-independent, it does not matter which path we choose from 𝑃 to 𝑄. On the other hand, if 𝐹⃗ is not path-independent, then different choices might give different values for 𝑓 (𝑄), so 𝑓 would not be a function (a function has to have a single value at each point). We still have to show that the gradient of the function 𝑓 really is 𝐹⃗ ; we do this on page 943. However, by constructing a function 𝑓 in this manner, we have the following result:

Theorem 18.2: Path-independent Fields Are Gradient Fields If 𝐹⃗ is a continuous path-independent vector field on an open region 𝑅, then 𝐹⃗ = grad 𝑓 for some 𝑓 defined on 𝑅.

Combining Theorems 18.1 and 18.2, we have

A continuous vector field 𝐹⃗ defined on an open region is path-independent if and only if 𝐹⃗ is a gradient vector field.

The function 𝑓 is sufficiently important that it is given a special name:

If a vector field 𝐹⃗ is of the form 𝐹⃗ = grad 𝑓 for some scalar function 𝑓 , then 𝑓 is called a potential function for the vector field 𝐹⃗ .

Warning Physicists use the convention that a function 𝜙 is a potential function for a vector field 𝐹⃗ if 𝐹⃗ = − grad 𝜙. See Problem 89 (available online). Example 3

Show that the vector field 𝐹⃗ (𝑥, 𝑦) = 𝑦 cos 𝑥⃗𝑖 + (sin 𝑥 + 𝑦)𝑗⃗ is path-independent.

Solution

Let’s suppose 𝐹⃗ does have a potential function 𝑓 , so that 𝐹⃗ = grad 𝑓 . This means 𝜕𝑓 = 𝑦 cos 𝑥 𝜕𝑥

and

𝜕𝑓 = sin 𝑥 + 𝑦. 𝜕𝑦

Integrating the expression for 𝜕𝑓 ∕𝜕𝑥 with respect to 𝑥 shows that 𝑓 (𝑥, 𝑦) = 𝑦 sin 𝑥 + 𝐶(𝑦)

where 𝐶(𝑦) is a function of 𝑦 only.

The constant of integration here is an arbitrary function 𝐶(𝑦) of 𝑦, since 𝜕(𝐶(𝑦))∕𝜕𝑥 = 0. Differentiating this expression for 𝑓 (𝑥, 𝑦) with respect to 𝑦 and using 𝜕𝑓 ∕𝜕𝑦 = sin 𝑥 + 𝑦 gives 𝜕𝑓 = sin 𝑥 + 𝐶 ′ (𝑦) = sin 𝑥 + 𝑦. 𝜕𝑦 Thus, we must have 𝐶 ′ (𝑦) = 𝑦, so 𝑔(𝑦) = 𝑦2 ∕2 + 𝐴, where 𝐴 is some constant. Thus, 𝑦2 +𝐴 2 is a potential function for 𝐹⃗ . Therefore, 𝐹⃗ is path-independent. 𝑓 (𝑥, 𝑦) = 𝑦 sin 𝑥 +

18.3 GRADIENT FIELDS AND PATH-INDEPENDENT FIELDS

Example 4

943

The gravitational field, 𝐹⃗ , of an object of mass 𝑀 is given by 𝐺𝑀 𝑟⃗ . 𝐹⃗ = − 𝑟3 Show that 𝐹⃗ is a gradient field by finding 𝑓 , a potential function for 𝐹⃗ .

Solution

The vector 𝐹⃗ points directly in toward the origin. If 𝐹⃗ = grad 𝑓 , then 𝐹⃗ must be perpendicular to the level surfaces of 𝑓 , so the level surfaces of 𝑓 must be spheres. Also, if grad 𝑓 = 𝐹⃗ , then ‖ grad 𝑓 ‖ = ‖𝐹⃗ ‖ = 𝐺𝑀∕𝑟2 is the rate of change of 𝑓 in the direction toward the origin. Now, differentiating with respect to 𝑟 gives the rate of change in a radially outward direction. Thus, if we write 𝑤 = 𝑓 (𝑥, 𝑦, 𝑧), we have ( ) ) ( 𝑑𝑤 𝐺𝑀 𝑑 1 1 = − 2 = 𝐺𝑀 − 2 = 𝐺𝑀 . 𝑑𝑟 𝑑𝑟 𝑟 𝑟 𝑟 So for the potential function, let’s try 𝑤=

𝐺𝑀 or 𝑓 (𝑥, 𝑦, 𝑧) = √ . 𝑥2 + 𝑦2 + 𝑧2

𝐺𝑀 𝑟

We check that 𝑓 is the potential function by calculating 𝑓𝑥 = 𝑓𝑦 = 𝑓𝑧 = So

𝐺𝑀 𝜕 −𝐺𝑀𝑥 = , √ 2 𝜕𝑥 𝑥2 + 𝑦2 + 𝑧2 (𝑥 + 𝑦2 + 𝑧2 )3∕2 −𝐺𝑀𝑦 𝐺𝑀 𝜕 = , √ 𝜕𝑦 𝑥2 + 𝑦2 + 𝑧2 (𝑥2 + 𝑦2 + 𝑧2 )3∕2 𝐺𝑀 𝜕 −𝐺𝑀𝑧 . = √ 2 𝜕𝑧 𝑥2 + 𝑦2 + 𝑧2 (𝑥 + 𝑦2 + 𝑧2 )3∕2

⃗ = grad 𝑓 = 𝑓𝑥 ⃗𝑖 + 𝑓𝑦 𝑗⃗ + 𝑓𝑧 𝑘

(𝑥2

−𝐺𝑀 ⃗ ) = −𝐺𝑀 𝑟⃗ = 𝐹⃗ . (𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 2 2 3∕2 +𝑦 +𝑧 ) 𝑟3

Our computations show that 𝐹⃗ is a gradient field and that 𝑓 = 𝐺𝑀∕𝑟 is a potential function for 𝐹⃗ . Path-independent vector fields are rare, but often important. Section 18.4 gives a method for determining whether a vector field has the property.

⃖⃖⃗ Why Path-Independent Vector Fields Are Gradient Fields: Showing grad 𝒇 = 𝑭 Suppose 𝐹⃗ is a path-independent vector field. On page 941 we defined the function 𝑓 , which we hope will satisfy grad 𝑓 = 𝐹⃗ , as follows: 𝑓 (𝑥0 , 𝑦0 ) =

∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 ,

where 𝐶 is a path from a fixed starting point 𝑃 to a point 𝑄 = (𝑥0 , 𝑦0 ). This integral has the same value for any path from 𝑃 to 𝑄 because 𝐹⃗ is path-independent. Now we show why grad 𝑓 = 𝐹⃗ . We consider vector fields in 2-space; the argument in 3-space is essentially the same. First, we write the line integral in terms of the components 𝐹⃗ (𝑥, 𝑦) = 𝐹1 (𝑥, 𝑦)⃗𝑖 + 𝐹2 (𝑥, 𝑦)𝑗⃗ and the components 𝑑⃗𝑟 = 𝑑𝑥⃗𝑖 + 𝑑𝑦𝑗⃗ : 𝑓 (𝑥0 , 𝑦0 ) =

∫𝐶

𝐹1 (𝑥, 𝑦) 𝑑𝑥 + 𝐹2 (𝑥, 𝑦) 𝑑𝑦.

We want to compute the partial derivatives of 𝑓 , that is, the rate of change of 𝑓 at (𝑥0 , 𝑦0 ) parallel to the axes. To do this easily, we choose a path which reaches the point (𝑥0 , 𝑦0 ) on a horizontal or vertical line segment. Let 𝐶 ′ be a path from 𝑃 which stops short of 𝑄 at a fixed point (𝑎, 𝑏) and let

944

Chapter 18 LINE INTEGRALS

(𝑎, 𝑦0 ) 𝐿𝑥 ✲ 𝐿𝑦



𝐶′

𝑄 = (𝑥0 , 𝑦0 ) 𝐾𝑦

𝑄 = (𝑥0 , 𝑦0 )

(𝑎, 𝑏)

𝐶

(𝑎, 𝑏) ✲ ✻ 𝐾𝑥 (𝑥0 , 𝑏)







𝑃

𝑃

Figure 18.31: The path 𝐶 ′ + 𝐾𝑥 + 𝐾𝑦 is used to show 𝑓𝑦 = 𝐹2

Figure 18.30: The path 𝐶 ′ + 𝐿𝑦 + 𝐿𝑥 is used to show 𝑓𝑥 = 𝐹1

𝐿𝑥 and 𝐿𝑦 be the paths shown in Figure 18.30. Then we can split the line integral into three pieces. Since 𝑑⃗𝑟 = 𝑗⃗ 𝑑𝑦 on 𝐿𝑦 and 𝑑⃗𝑟 = ⃗𝑖 𝑑𝑥 on 𝐿𝑥 , we have: 𝑓 (𝑥0 , 𝑦0 ) =

∫𝐶 ′

𝐹⃗ ⋅𝑑⃗𝑟 +

∫𝐿

𝑦

𝐹⃗ ⋅𝑑⃗𝑟 +

∫𝐿

𝐹⃗ ⋅𝑑⃗𝑟 =

𝑥

∫𝐶 ′

𝐹⃗ ⋅𝑑⃗𝑟 +

𝑦0

∫𝑏

𝑥0

𝐹2 (𝑎, 𝑦) 𝑑𝑦+

∫𝑎

𝐹1 (𝑥, 𝑦0 ) 𝑑𝑥.

The first two integrals do not involve 𝑥0 . Thinking of 𝑥0 as a variable and differentiating with respect to it gives 𝑦

0 𝜕 𝜕 𝜕 𝐹2 (𝑎, 𝑦)𝑑𝑦 + 𝐹⃗ ⋅ 𝑑⃗𝑟 + 𝜕𝑥0 ∫𝐶 ′ 𝜕𝑥0 ∫𝑏 𝜕𝑥0 ∫𝑎 = 0 + 0 + 𝐹1 (𝑥0 , 𝑦0 ) = 𝐹1 (𝑥0 , 𝑦0 ),

𝑓𝑥0 (𝑥0 , 𝑦0 ) =

𝑥0

𝐹1 (𝑥, 𝑦0 )𝑑𝑥

and thus 𝑓𝑥 (𝑥, 𝑦) = 𝐹1 (𝑥, 𝑦). A similar calculation for 𝑦 using the path from 𝑃 to 𝑄 shown in Figure 18.31 gives 𝑓𝑦0 (𝑥0 , 𝑦0 ) = 𝐹2 (𝑥0 , 𝑦0 ). Therefore, as we claimed, grad 𝑓 = 𝑓𝑥 ⃗𝑖 + 𝑓𝑦 𝑗⃗ = 𝐹1 ⃗𝑖 + 𝐹2 𝑗⃗ = 𝐹⃗ .

Exercises and Problems for Section 18.3 Online Resource: Additional Problems for Section 18.3 EXERCISES

1. If 𝐹⃗ = grad(𝑥2 + 𝑦4 ), find ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 where 𝐶 is the quarter of the circle 𝑥2 + 𝑦2 = 4 in the first quadrant, oriented counterclockwise. 2. If 𝐹⃗ = grad(sin(𝑥𝑦) + 𝑒𝑧 ), find ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 where 𝐶 consists of√a line from (0, 0, 0) to (0, 0, 1)√followed by a √ line to (0, 2, 3), followed by a line to ( 2, 5, 2). In Exercises 3–6, let 𝐶 be the curve consisting of a square of side 2, centered at the origin with sides on the lines 𝑥 = ±1, 𝑦 = ±1 and traversed counterclockwise. What is the sign of the line integrals of the vector fields around the curve 𝐶? Indicate whether each vector field is path-independent. 𝑦 𝑦 3. 4. 𝑥 𝑥

5.

𝑦

6.

𝑥

𝑦

𝑥

In Exercises 7–12, does the vector field appear to be pathindependent (conservative)? 8. 7.

18.3 GRADIENT FIELDS AND PATH-INDEPENDENT FIELDS

9.

10.

11.

12.

945

17. 𝐹⃗ = 3𝑥2 ⃗𝑖 + 4𝑦3 𝑗⃗ around the top of the unit circle from (1, 0) to (−1, 0). 18. 𝐹⃗ = (𝑥 + 2)⃗𝑖 + (2𝑦 + 3)𝑗⃗ and 𝐶 is the line from (1, 0) to (3, 1). 19. 𝐹⃗ = 2 sin(2𝑥 + 𝑦)⃗𝑖 + sin(2𝑥 + 𝑦)𝑗⃗ along the path consisting of a line from (𝜋, 0) to (2, 5) followed by a line to (5𝜋, 0) followed by a quarter circle to (0, 5𝜋).

13. Find 𝑓 if grad 𝑓 = 2𝑥𝑦⃗𝑖 + 𝑥2 𝑗⃗ . 14. Find 𝑓 if grad 𝑓 = 2𝑥𝑦⃗𝑖 + (𝑥2 + 8𝑦3 )𝑗⃗ . 15. Find 𝑓 if grad 𝑓 = (𝑦𝑧𝑒𝑥𝑦𝑧 + 𝑧2 cos(𝑥𝑧2 ))⃗𝑖 + 𝑥𝑧𝑒𝑥𝑦𝑧 𝑗⃗ + ⃗. (𝑥𝑦𝑒𝑥𝑦𝑧 + 2𝑥𝑧 cos(𝑥𝑧2 ))𝑘 2 16. Let 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥 + 2𝑦3 + 3𝑧4 and 𝐹⃗ = grad 𝑓 . Find ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 where 𝐶 consists of four line segments from (4, 0, 0) to (4, 3, 0) to (0, 3, 0) to (0, 3, 5) to (0, 0, 5). In Exercises 17–25, use the Fundamental Theorem of Line Integrals to calculate ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 exactly.

⃗ and 𝐶 is the line from 20. 𝐹⃗ = 2𝑥⃗𝑖 − 4𝑦𝑗⃗ + (2𝑧 − 3)𝑘 (1, 1, 1) to (2, 3, −1). 21. 𝐹⃗ = 𝑥2∕3 ⃗𝑖 + 𝑒7𝑦 𝑗⃗ and 𝐶 is the unit circle oriented clockwise. 22. 𝐹⃗ = 𝑥2∕3 ⃗𝑖 + 𝑒7𝑦 𝑗⃗ and 𝐶 is the quarter of the unit circle in the first quadrant, traced counterclockwise from (1, 0) to (0, 1). ⃗ along the curve con23. 𝐹⃗ = 𝑦𝑒𝑥𝑦 ⃗𝑖 + 𝑥𝑒𝑥𝑦 𝑗⃗ + (cos 𝑧)𝑘 sisting of a line from (0, 0, 𝜋) to (1, 1, 𝜋) followed by the parabola 𝑧 = 𝜋𝑥2 in the plane 𝑦 = 1 to the point (3, 1, 9𝜋). 24. 𝐹⃗ = 𝑦 sin(𝑥𝑦)⃗𝑖 + 𝑥 sin(𝑥𝑦)𝑗⃗ and 𝐶 is the parabola 𝑦 = 2𝑥2 from (1, 2) to (3, 18). 2 2 2 2 2 2 ⃗ and 25. 𝐹⃗ = 2𝑥𝑦2 𝑧𝑒𝑥 𝑦 𝑧 ⃗𝑖 + 2𝑥2 𝑦𝑧𝑒𝑥 𝑦 𝑧 𝑗⃗ + 𝑥2 𝑦2 𝑒𝑥 𝑦 𝑧 𝑘 𝐶 is the circle of radius 1 in the plane 𝑧 = 1, centered on the 𝑧-axis, starting at (1, 0, 1) and oriented counterclockwise viewed from above.

PROBLEMS 26. Let 𝑣⃗ = grad(𝑥2 + 𝑦2 ). Consider the path 𝐶 which is a line between any two of the following points: (0, 0); (5, 0); (−5, 0); (0, 6); (0, −6); (5, 4); (−3, −5). Suppose you want to choose the path 𝐶 in order to maximize ∫𝐶 𝑣⃗ ⋅ 𝑑⃗𝑟 . What point should be the start of 𝐶? What point should be the end of 𝐶? Explain your answer. 27. Let 𝐹⃗ = grad(2𝑥2 + 3𝑦2 ). Which one of the three paths 𝑃 𝑄, 𝑄𝑅, and 𝑅𝑆 in Figure 18.32 should you choose as 𝐶 in order to maximize ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 ?

29. The vector field 𝐹⃗ (𝑥, 𝑦) = 𝑥⃗𝑖 +𝑦𝑗⃗ is path-independent. Compute geometrically the line integrals over the three paths 𝐴, 𝐵, and 𝐶 shown in Figure 18.33 from (1, 0) to (0, 1) and check that they are equal. Here 𝐴 is a portion of a circle, 𝐵 is a line, and 𝐶 consists of two line segments meeting at a right angle. 𝑦 (0, 1) 𝐴

𝑦 𝑅 𝐶

4

𝐵 (1, 0)

3

✠𝑥

𝑄

2

𝑆

Figure 18.33 1 𝑥

𝑃 1

2

3

4

Figure 18.32 28. Compute

(

) ⃗ ⋅𝑑⃗𝑟 where cos(𝑥𝑦)𝑒sin(𝑥𝑦) (𝑦⃗𝑖 + 𝑥𝑗⃗ ) + 𝑘

∫𝐶 𝐶 is the line from (𝜋, 2, 5) to (0.5, 𝜋, 7).

30. The vector field 𝐹⃗ (𝑥, 𝑦) = 𝑥⃗𝑖 +𝑦𝑗⃗ is path-independent. Compute algebraically the line integrals over the three paths 𝐴, 𝐵, and 𝐶 shown in Figure 18.34 from (0, 0) to (1, 1) and check that they are equal. Here 𝐴 is a line segment, 𝐵 is part of the graph of 𝑓 (𝑥) = 𝑥2 , and 𝐶 consists of two line segments meeting at a right angle.

946

Chapter 18 LINE INTEGRALS 𝑦 (1, 1)

𝐴

2 2 2 ⃗ , find 40. If 𝐹⃗ (𝑥, 𝑦, 𝑧) = 2𝑥𝑒𝑥 +𝑦𝑧 ⃗𝑖 + 𝑧𝑒𝑥 +𝑦𝑧 𝑗⃗ + 𝑦𝑒𝑥 +𝑦𝑧 𝑘 ⃗ exactly the line integral of 𝐹 along the curve consisting of the two half circles in the plane 𝑧 = 0 in Figure 18.35.

𝐶

𝑦

𝐵 3 𝑥

𝑥

Figure 18.34 Figure 18.35 In Problems 31–34, decide whether the vector field could be a gradient vector field. Justify your answer. 31. 𝐹⃗ (𝑥, 𝑦) = 𝑥⃗𝑖 32. 𝐺⃗ (𝑥, 𝑦) = (𝑥2 − 𝑦2 )⃗𝑖 − 2𝑥𝑦𝑗⃗ ⃗ 33. 𝐹⃗ (⃗𝑟 ) = 𝑟⃗ ∕||⃗𝑟 ||3 , where 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 𝑦 −𝑧 𝑥 ⃗ ⃗𝑖 + √ 34. 𝐹⃗ (𝑥, 𝑦, 𝑧) = √ 𝑗⃗ + √ 𝑘 2 2 2 2 2 𝑥 +𝑧 𝑥 +𝑧 𝑥 + 𝑧2

35. Find a potential function for 𝐹⃗ = 5𝑦⃗𝑖 + (5𝑥 + 𝑦)𝑗⃗ . Use it to evaluate the integral ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 on the path 𝐶 if (a) 𝐶 runs from (10, 0) to (0, −10) along the circle of radius 10 centered at the origin. (b) 𝐶 runs from (10, 0) to (0, −10) along a straight line.

36. Suppose 𝐶 is a path that begins and ends at the same point 𝑃 = (15, 20). What, if anything, can you say about ∫𝐶 (𝑝(𝑥, 𝑦)⃗𝑖 + 𝑞(𝑥, 𝑦)𝑗⃗ ) ⋅ 𝑑⃗𝑟 ? (a) With no assumptions about 𝑝 and 𝑞. (b) If 𝑝(𝑥, 𝑦)⃗𝑖 + 𝑞(𝑥, 𝑦)𝑗⃗ has a potential function. 37. Let 𝐹⃗ = −𝑦⃗𝑖 + 𝑥𝑗⃗ and let 𝐶 be the circle of radius 5 centered at the origin, oriented counterclockwise. (a) Evaluate ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 . (b) Give a potential function for 𝐹⃗ or explain why there are none.

2 2 41. Let grad 𝑓 = 2𝑥𝑒𝑥 sin 𝑦⃗𝑖 +𝑒𝑥 cos 𝑦𝑗⃗ . Find the change in 𝑓 between (0, 0) and (1, 𝜋∕2):

(a) By computing a line integral. (b) By computing 𝑓 . 42. Let 𝐶 be the quarter of the unit circle centered at the origin, traversed counterclockwise starting on the negative 𝑥-axis. Find the exact values of (a)

∫𝐶

(2𝜋𝑥⃗𝑖 + 𝑦2 𝑗⃗ ) ⋅ 𝑑⃗𝑟 (b)

∫𝐶

(−2𝑦⃗𝑖 + 𝑥𝑗⃗ ) ⋅ 𝑑⃗𝑟

For the vector fields in Problems 43–46, find the line integral along the curve 𝐶 from the origin along the 𝑥-axis to the point (3, 0) and then counterclockwise around √ the circumfer√ ence of the circle 𝑥2 + 𝑦2 = 9 to the point (3∕ 2, 3∕ 2). 43. 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ ⃗ = −𝑦⃗𝑖 + 𝑥𝑗⃗ 44. 𝐻 45. 𝐹⃗ = 𝑦(𝑥 + 1)−1 ⃗𝑖 + ln(𝑥 + 1)𝑗⃗ 46. 𝐺⃗ = (𝑦𝑒𝑥𝑦 + cos(𝑥 + 𝑦))⃗𝑖 + (𝑥𝑒𝑥𝑦 + cos(𝑥 + 𝑦))𝑗⃗ 47. Let 𝐶 be the helix 𝑥 = cos 𝑡, 𝑦 = sin 𝑡, 𝑧 = 𝑡 for 0 ≤ 𝑡 ≤ 1.25𝜋. Find ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 exactly for 2 2 2 ⃗. 𝐹⃗ = 𝑦𝑧2 𝑒𝑥𝑦𝑧 ⃗𝑖 + 𝑥𝑧2 𝑒𝑥𝑦𝑧 𝑗⃗ + 2𝑥𝑦𝑧𝑒𝑥𝑦𝑧 𝑘

38. Let 𝐹⃗ = 𝑦⃗𝑖 . (a) Evaluate ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 if 𝐶1 is the straight line path 1 from (0, 0) to (1, 0). (b) Evaluate ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 if 𝐶2 is the path along three 2 edges of a square, from (0, 0) to (0, 1) to (1, 1) to (1, 0). (c) Does 𝐹⃗ have a potential function? Either give one or explain why there are none. 39. If 𝑑𝑓 = 𝑝 𝑑𝑥 + 𝑞 𝑑𝑦 for smooth 𝑓 , explain why 𝜕𝑝 𝜕𝑞 = . 𝜕𝑦 𝜕𝑥

⃗ and 𝐺⃗ = (2𝑥 + 𝑦)⃗𝑖 + 48. Let 𝐹⃗ = 2𝑥⃗𝑖 + 2𝑦𝑗⃗ + 2𝑧𝑘 ⃗ . Let 𝐶 be the line from the origin to the 2𝑦𝑗⃗ + 2𝑧𝑘 point (1, 5, 9). Find ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 and use the result to find ∫𝐶 𝐺⃗ ⋅ 𝑑⃗𝑟 . 49. (a) If 𝐹⃗ = 𝑦𝑒𝑥 𝑖⃗ + 𝑒𝑥 𝑗⃗ , explain how the Fundamental Theorem of Calculus for Line Integrals enables you to calculate ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 where 𝐶 is any curve going from the point (1, 2) to the point (3, 7). Explain why it does not matter how the curve goes. (b) If 𝐶 is the line from the point (1, 2) to the point (3, 7), calculate the line integral in part (a) without using the Fundamental Theorem.

18.3 GRADIENT FIELDS AND PATH-INDEPENDENT FIELDS 𝑦

50. Calculate the line integral ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 exactly, where 𝐶 is the curve from 𝑃 to 𝑄 in Figure 18.36 and (𝑦) ( ) (𝑦) ( ) 𝑥 ⃗𝑖 − cos 𝑥 cos sin 𝑗⃗ . 𝐹⃗ = sin 2 2 2 2

9

1

The curves 𝑃 𝑅, 𝑅𝑆 and 𝑆𝑄 are trigonometric functions of period 2𝜋 and amplitude 1. 𝑦

𝑅( 3𝜋 , 2

, 3𝜋 )𝑃 (− 3𝜋 2 2

947

5 3

7

𝑄 𝑥

3𝜋 ) 2

𝑃

𝑥

Figure 18.38 (− 3𝜋 , − 3𝜋 )𝑄 2 2

, − 3𝜋 ) 𝑆( 3𝜋 2 2

Figure 18.36 51. The domain of 𝑓 (𝑥, 𝑦) is the 𝑥𝑦-plane; values of 𝑓 are in Table 18.1. Find ∫𝐶 grad 𝑓 ⋅ 𝑑⃗𝑟 , where 𝐶 is (a) A line from (0, 2) to (3, 4). (b) A circle of radius 1 centered at (1, 2) traversed counterclockwise. Table 18.1 𝑦

⧵𝑥

0

1

2

3

4

0

53

57

59

58

56

1

56

58

59

59

57

2

57

58

59

60

59

3

59

60

61

62

61

4

62

63

65

66

69

54. Consider the line integrals, ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 , for 𝑖 = 1, 2, 3, 4, 𝑖 where 𝐶𝑖 is the path from 𝑃𝑖 to 𝑄𝑖 shown in Figure 18.39 and 𝐹⃗ = grad 𝑓 . Level curves of 𝑓 are also shown in Figure 18.39. (a) Which of the line integral(s) is (are) zero? (b) Arrange the four line integrals in ascending order (from least to greatest). (c) Two of the nonzero line integrals have equal and opposite values. Which are they? Which is negative and which is positive?

𝑄1

52. Figure 18.37 shows the vector field 𝐹⃗ (𝑥, 𝑦) = 𝑥𝑗⃗ . (a) Find paths 𝐶1 , 𝐶2 , and 𝐶3 from 𝑃 to 𝑄 such that ∫𝐶1

𝐹⃗ ⋅ 𝑑⃗𝑟 = 0,

∫𝐶2

𝐹⃗ ⋅ 𝑑⃗𝑟 > 0,

∫𝐶3

𝑃3

𝑄4



Increasing values of 𝑓

𝑃4 𝑃1

𝑃2

𝐹⃗ ⋅ 𝑑⃗𝑟 < 0.

𝑄3 𝑄2

Figure 18.39

(b) Is 𝐹⃗ a gradient field? Explain. 𝑦

55. Consider the vector field 𝐹⃗ shown in Figure 18.40. 𝑦

𝑥 𝑄

𝑃

𝐶

Figure 18.37 53. (a) Figure 18.38 shows level curves of 𝑓 (𝑥, 𝑦). Sketch a vector at 𝑃 in the direction of grad 𝑓 . (b) Is the length of grad 𝑓 at 𝑃 longer, shorter, or the same length as the length of grad 𝑓 at 𝑄? (c) If 𝐶 is a curve going from 𝑃 to 𝑄, find ∫𝐶 grad 𝑓 ⋅ 𝑑⃗𝑟 .

𝑥

Figure 18.40

⃗ positive, negative, or zero? (a) Is ∫𝐶 𝐹⃗ ⋅ 𝑑𝑟

948

Chapter 18 LINE INTEGRALS

(b) From your answer to part (a), can you determine whether or not 𝐹⃗ = grad 𝑓 for some function 𝑓 ? (c) Which of the following formulas best fits 𝐹⃗ ? 𝑦 𝑥 ⃗ 𝑖 + 2 𝑗⃗ , 𝐹⃗1 = 2 𝑥 + 𝑦2 𝑥 + 𝑦2 𝐹⃗2 = −𝑦⃗𝑖 + 𝑥𝑗⃗ , −𝑦 𝑥 ⃗𝑖 + 𝐹⃗3 = 2 𝑗⃗ . (𝑥 + 𝑦2 )2 (𝑥2 + 𝑦2 )2 56. If 𝐹⃗ is a path-independent vector field, with (1,0) (1,1) ∫(0,0) 𝐹⃗ ⋅ 𝑑⃗𝑟 = 5.1 and ∫(1,0) 𝐹⃗ ⋅ 𝑑⃗𝑟 = 3.2 and (1,1) ∫ 𝐹⃗ ⋅ 𝑑⃗𝑟 = 4.7, find (0,1)

(0,0)

∫(0,1)

𝐹⃗ ⋅ 𝑑⃗𝑟 .

57. The path 𝐶 is a line segment of length 10 in the plane starting at (2, 1). For 𝑓 (𝑥, 𝑦) = 3𝑥 + 4𝑦, consider ∫𝐶

grad 𝑓 ⋅ 𝑑⃗𝑟 .

(a) Where should the other end of the line segment 𝐶 be placed to maximize the value of the integral? (b) What is the maximum value of the integral?

⃗ and 𝑎⃗ = 𝑎1 ⃗𝑖 + 𝑎2 𝑗⃗ + 𝑎3 𝑘 ⃗,a 58. Let 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 constant vector. (a) Find grad(⃗𝑟 ⋅ 𝑎⃗ ). (b) Let 𝐶 be a path from the origin to the point with position vector 𝑟⃗0 . Find ∫𝐶 grad(⃗𝑟 ⋅ 𝑎⃗ ) ⋅ 𝑑⃗𝑟 . (c) If ||⃗𝑟 0 || = 10, what is the maximum possible value of ∫𝐶 grad(⃗𝑟 ⋅ 𝑎⃗ ) ⋅ 𝑑⃗𝑟 ? Explain. 59. The force exerted by gravity on a refrigerator of mass ⃗. 𝑚 is 𝐹⃗ = −𝑚𝑔 𝑘 (a) Find the work done against this force in moving from the point (1, 0, 0) to the point (1, 0, 2𝜋) along the curve 𝑥 = cos 𝑡, 𝑦 = sin 𝑡, 𝑧 = 𝑡 by calculating a line integral. (b) Is 𝐹⃗ conservative (that is, path independent)? Give a reason for your answer. 60. A particle subject to a force 𝐹⃗ (𝑥, 𝑦) = 𝑦⃗𝑖 − 𝑥𝑗⃗ moves clockwise along the arc of the unit circle, centered at the origin, that begins at (−1, 0) and ends at (0, 1). (a) Find the work done by 𝐹⃗ . Explain the sign of your answer. (b) Is 𝐹⃗ path-independent? Explain.

Strengthen Your Understanding In Problems 61–63, explain what is wrong with the statement.

Are the statements in Problems 70–80 true or false? Give reasons for your answer.

61. If 𝐹⃗ is a gradient field and 𝐶 is an oriented path from point 𝑃 to point 𝑄, then ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 𝐹⃗ (𝑄) − 𝐹⃗ (𝑃 ).

70. The fact that the line integral of a vector field 𝐹⃗ is zero around the unit circle 𝑥2 + 𝑦2 = 1 means that 𝐹⃗ must be a gradient vector field. 71. If 𝐶 is the line segment that starts at (0, 0) and ends at (𝑎, 𝑏) then ∫𝐶 (𝑥⃗𝑖 + 𝑦𝑗⃗ ) ⋅ 𝑑⃗𝑟 = 12 (𝑎2 + 𝑏2 ).

62. Given any vector field 𝐹⃗ and a point 𝑃 , the function 𝑓 (𝑄) = ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 , where 𝐶 is a path from 𝑃 to 𝑄, is a potential function for 𝐹⃗ . 63. If a vector field 𝐹⃗ is not a gradient vector field, then ∫ 𝐹⃗ ⋅ 𝑑⃗𝑟 can’t be evaluated. 𝐶

In Problems 64–65, give an example of: 64. A vector field 𝐹⃗ such that ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 100, for every oriented path 𝐶 from (0, 0) to (1, 2). 65. A path-independent vector field. In Problems 66–69, each of the statements is false. Explain why or give a counterexample. 66. If ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 0 for one particular closed path 𝐶, then 𝐹⃗ is path-independent. 67. ∫ 𝐹⃗ ⋅ 𝑑⃗𝑟 is the total change in 𝐹⃗ along 𝐶. 𝐶

68. If the vector fields 𝐹⃗ and 𝐺⃗ have ∫𝐶 𝐹⃗ ⋅𝑑⃗𝑟 = ∫𝐶 𝐺⃗ ⋅𝑑⃗𝑟 for a particular path 𝐶, then 𝐹⃗ = 𝐺⃗ . 69. If the total change of a function 𝑓 along a curve 𝐶 is zero, then 𝐶 must be a contour of 𝑓 .

72. The circulation of any vector field 𝐹⃗ around any closed curve 𝐶 is zero. 73. If 𝐹⃗ = grad 𝑓 , then 𝐹⃗ is path-independent. 74. If 𝐹⃗ is path-independent, then ∫𝐶 𝐹⃗ ⋅𝑑⃗𝑟 = ∫𝐶 𝐹⃗ ⋅𝑑⃗𝑟 , 1 2 where 𝐶1 and 𝐶2 are any paths. 75. The line integral ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 is the total change of 𝐹⃗ along 𝐶. 76. If 𝐹⃗ is path-independent, then there is a potential function for 𝐹⃗ . 77. If 𝑓 (𝑥, 𝑦) = 𝑒cos(𝑥𝑦) , and 𝐶1 is the upper semicircle 𝑥2 +𝑦2 = 1 from (−1, 0) to (1, 0), and 𝐶2 is the line from (−1, 0) to (1, 0), then ∫𝐶 grad 𝑓 ⋅ 𝑑⃗𝑟 = ∫𝐶 grad 𝑓 ⋅ 𝑑⃗𝑟 . 1

2

78. If 𝐹⃗ is path-independent, and 𝐶 is any closed curve, then ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 0. 79. The vector field 𝐹⃗ (𝑥, 𝑦) = 𝑦2 ⃗𝑖 + 𝑘𝑗⃗ , where 𝑘 is constant, is a gradient field. 80. If ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 0, where 𝐶 is any circle of the form 𝑥2 + 𝑦2 = 𝑎2 , then 𝐹⃗ is path-independent.

18.4 PATH-DEPENDENT VECTOR FIELDS AND GREEN’S THEOREM

18.4

949

PATH-DEPENDENT VECTOR FIELDS AND GREEN’S THEOREM Suppose we are given a vector field but are not told whether it is path-independent. How can we tell if it has a potential function, that is, if it is a gradient field?

How to Tell If a Vector Field Is Path-Dependent Using Line Integrals One way to decide if a vector field is path-dependent is to find two paths with the same endpoints such that the line integrals of the vector field along the two paths have different values. Example 1

Is the vector field 𝐺⃗ shown in Figure 18.41 path-independent? At any point 𝐺⃗ has magnitude equal to the distance from the origin and direction perpendicular to the line joining the point to the origin. 𝑦 𝑄 𝐶1 𝐶2 𝑥 𝑃

𝐶2

Figure 18.41: Is this vector field path-independent?

Solution

We choose 𝑃 = (1, 0) and 𝑄 = (0, 1) and two paths between them: 𝐶1 , a quarter circle of radius 1, and 𝐶2 , formed by parts of the 𝑥- and 𝑦-axes. (See Figure 18.41.) Along 𝐶1 , the line integral ∫𝐶 𝐺⃗ ⋅ 𝑑⃗𝑟 > 0, since 𝐺⃗ points in the direction of the curve. 1 Along 𝐶2 , however, we have ∫ 𝐺⃗ ⋅ 𝑑⃗𝑟 = 0, since 𝐺⃗ is perpendicular to 𝐶2 everywhere. 𝐶2

Thus, 𝐺⃗ is not path-independent.

Path-Dependent Fields and Circulation Notice that the vector field in the previous example has nonzero circulation around the origin. What can we say about the circulation of a general path-independent vector field 𝐹⃗ around a closed curve, 𝐶? Suppose 𝐶 is a simple closed curve, that is, a closed curve that does not cross itself. If 𝑃 and 𝑄 are any two points on the path, then we can think of 𝐶 (oriented as shown in Figure 18.42) as made up of the path 𝐶1 followed by −𝐶2 . Since 𝐹⃗ is path-independent, we know that ∫ 𝐶1

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫ 𝐶2

𝐹⃗ ⋅ 𝑑⃗𝑟 .

Thus, we see that the circulation around 𝐶 is zero: ∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫𝐶1

𝐹⃗ ⋅ 𝑑⃗𝑟 +

∫−𝐶2

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫ 𝐶1

𝐹⃗ ⋅ 𝑑⃗𝑟 −

∫𝐶2

𝐹⃗ ⋅ 𝑑⃗𝑟 = 0.

If the closed curve 𝐶 does cross itself, we break it into simple closed curves as shown in Figure 18.43 and apply the same argument to each one. Now suppose we know that the line integral around any closed curve is zero. For any two points, 𝑃 and 𝑄, with two paths, 𝐶1 and 𝐶2 , between them, create a closed curve, 𝐶, as in Figure 18.42. Since the circulation around this closed curve, 𝐶, is zero, the line integrals along the two paths, 𝐶1 and 𝐶2 , are equal.1 Thus, 𝐹⃗ is path-independent. 1A

similar argument is used in Problems 57 and 58 on page 931.

950

Chapter 18 LINE INTEGRALS

𝑄 𝐶1 𝐶

𝐶

𝐶2

𝑃

Figure 18.42: A simple closed curve 𝐶 broken into two pieces, 𝐶1 and 𝐶2

Figure 18.43: A curve 𝐶 which crosses itself can be broken into simple closed curves

Thus, we have the following result:

A vector field is path-independent if and only if

∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 = 0 for every closed curve 𝐶.

Hence, to see if a field is path-dependent, we look for a closed path with nonzero circulation. For instance, the vector field in Example 1 has nonzero circulation around a circle around the origin, showing it is path-dependent.

How to Tell If a Vector Field Is Path-Dependent Algebraically: The Curl Example 2

Does the vector field 𝐹⃗ = 2𝑥𝑦⃗𝑖 + 𝑥𝑦𝑗⃗ have a potential function? If so, find it.

Solution

Let’s suppose 𝐹⃗ does have a potential function, 𝑓 , so 𝐹⃗ = grad 𝑓 . This means that 𝜕𝑓 = 2𝑥𝑦 and 𝜕𝑥

𝜕𝑓 = 𝑥𝑦. 𝜕𝑦

Integrating the expression for 𝜕𝑓 ∕𝜕𝑥 shows that we must have 𝑓 (𝑥, 𝑦) = 𝑥2 𝑦 + 𝐶(𝑦)

where 𝐶(𝑦) is a function of 𝑦.

Differentiating this expression for 𝑓 (𝑥, 𝑦) with respect to 𝑦 and using the fact that 𝜕𝑓 ∕𝜕𝑦 = 𝑥𝑦, we get 𝜕𝑓 = 𝑥2 + 𝐶 ′ (𝑦) = 𝑥𝑦. 𝜕𝑦 Thus, we must have 𝐶 ′ (𝑦) = 𝑥𝑦 − 𝑥2 . But this expression for 𝐶 ′ (𝑦) is impossible because 𝐶 ′ (𝑦) is a function of 𝑦 alone. This argument shows that there is no potential function for the vector field 𝐹⃗ . Is there an easier way to see that a vector field has no potential function, other than by trying to find the potential function and failing? The answer is yes. First we look at a 2-dimensional vector field 𝐹⃗ = 𝐹1 ⃗𝑖 + 𝐹2 𝑗⃗ . If 𝐹⃗ is a gradient field, then there is a potential function 𝑓 such that 𝜕𝑓 𝜕𝑓 ⃗𝑖 + 𝑗⃗ . 𝐹⃗ = 𝐹1 ⃗𝑖 + 𝐹2 𝑗⃗ = 𝜕𝑥 𝜕𝑦 Thus, 𝐹1 =

𝜕𝑓 𝜕𝑥

and 𝐹2 =

𝜕𝑓 . 𝜕𝑦

18.4 PATH-DEPENDENT VECTOR FIELDS AND GREEN’S THEOREM

951

Let us assume that 𝑓 has continuous second partial derivatives. Then, by the equality of mixed partial derivatives, 𝜕𝐹2 𝜕𝐹1 𝜕2 𝑓 𝜕2 𝑓 = = = . 𝜕𝑦 𝜕𝑦𝜕𝑥 𝜕𝑥𝜕𝑦 𝜕𝑥 Thus, we have the following result: If 𝐹⃗ (𝑥, 𝑦) = 𝐹1 ⃗𝑖 + 𝐹2 𝑗⃗ is a gradient vector field with continuous partial derivatives, then 𝜕𝐹2 𝜕𝐹1 − = 0. 𝜕𝑥 𝜕𝑦 If 𝐹⃗ (𝑥, 𝑦) = 𝐹1 ⃗𝑖 + 𝐹2 𝑗⃗ is an arbitrary vector field, then we define the 2-dimensional or scalar curl of the vector field 𝐹⃗ to be 𝜕𝐹2 𝜕𝐹1 − . 𝜕𝑥 𝜕𝑦 Notice that we now know that if 𝐹⃗ is a gradient field, then its curl is 0. We do not (yet) know whether the converse is true. (That is: If the curl is 0, does 𝐹⃗ have to be a gradient field?) However, the curl already enables us to show that a vector field is not a gradient field. Example 3

Show that 𝐹⃗ = 2𝑥𝑦⃗𝑖 + 𝑥𝑦𝑗⃗ cannot be a gradient vector field.

Solution

We have 𝐹1 = 2𝑥𝑦 and 𝐹2 = 𝑥𝑦. Since 𝜕𝐹1 ∕𝜕𝑦 = 2𝑥 and 𝜕𝐹2 ∕𝜕𝑥 = 𝑦, in this case 𝜕𝐹2 ∕𝜕𝑥 − 𝜕𝐹1 ∕𝜕𝑦 ≠ 0 so 𝐹⃗ cannot be a gradient field.

Green’s Theorem We now have two ways of seeing that a vector field 𝐹⃗ in the plane is path-dependent. We can evaluate ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 for some closed curve and find it is not zero, or we can show that 𝜕𝐹2 ∕𝜕𝑥 − 𝜕𝐹1 ∕𝜕𝑦 ≠ 0. It’s natural to think that 𝜕𝐹2 𝜕𝐹1 − 𝐹⃗ ⋅ 𝑑⃗𝑟 and ∫𝐶 𝜕𝑥 𝜕𝑦 might be related. The relation is called Green’s Theorem.

Theorem 18.3: Green’s Theorem Suppose 𝐶 is a piecewise smooth simple closed curve that is the boundary of a region 𝑅 in the plane and oriented so that the region is on the left as we move around the curve. See Figure 18.44. Suppose 𝐹⃗ = 𝐹1 ⃗𝑖 + 𝐹2 𝑗⃗ is a smooth vector field on a region2 containing 𝑅 and 𝐶. Then ) ( 𝜕𝐹2 𝜕𝐹1 ⃗ − 𝑑𝑥 𝑑𝑦. 𝐹 ⋅ 𝑑⃗𝑟 = ∫𝐶 ∫𝑅 𝜕𝑥 𝜕𝑦 The online supplement at www.wiley.com/college/hughes-hallett contains a proof of Green’s Theorem with different, but equivalent, conditions on the region 𝑅. 2 The

region is an open region containing 𝑅 and 𝐶.

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Chapter 18 LINE INTEGRALS

𝑦 𝐶

𝑅

𝑥 Figure 18.44: Boundary 𝐶 oriented with 𝑅 on the left

We first prove Green’s Theorem in the case where the region 𝑅 is the rectangle 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑐 ≤ 𝑦 ≤ 𝑑. Figure 18.45 shows the boundary of 𝑅 divided into four curves. On 𝐶1 , where 𝑦 = 𝑐 and 𝑑𝑦 = 0, we have 𝑑⃗𝑟 = 𝑑𝑥⃗𝑖 and thus 𝑏

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫𝐶1

∫𝑎

𝐹1 (𝑥, 𝑐) 𝑑𝑥.

Similarly, on 𝐶3 where 𝑦 = 𝑑 we have 𝑎

∫𝐶3

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫𝑏

𝑏

𝐹1 (𝑥, 𝑑) 𝑑𝑥 = −

∫𝑎

𝐹1 (𝑥, 𝑑) 𝑑𝑥.

Hence 𝑏

∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 = 1 +𝐶3

∫𝑎

𝑏

𝐹1 (𝑥, 𝑐) 𝑑𝑥 −

∫𝑎

𝑏

𝐹1 (𝑥, 𝑑) 𝑑𝑥 = −

(𝐹1 (𝑥, 𝑑) − 𝐹1 (𝑥, 𝑐)) 𝑑𝑥.

∫𝑎

By the Fundamental Theorem of Calculus, 𝑑

𝐹1 (𝑥, 𝑑) − 𝐹1 (𝑥, 𝑐) =

∫𝑐

𝜕𝐹1 𝑑𝑦 𝜕𝑦

and therefore 𝑏

∫𝐶1 +𝐶3

𝐹⃗ ⋅ 𝑑⃗𝑟 = −

∫𝑎 ∫𝑐

𝑑

𝑏 𝜕𝐹 𝑑 𝜕𝐹1 1 𝑑𝑦 𝑑𝑥 = − 𝑑𝑥 𝑑𝑦. ∫𝑐 ∫𝑎 𝜕𝑦 𝜕𝑦

Along the curve 𝐶2 , where 𝑥 = 𝑏, and the curve 𝐶4 , where 𝑥 = 𝑎, we get, by a similar argument, 𝑑

∫𝐶2 +𝐶4

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫𝑐

𝑑

(𝐹2 (𝑏, 𝑦) − 𝐹2 (𝑎, 𝑦)) 𝑑𝑦 =

∫𝑐 ∫𝑎

𝑏

𝜕𝐹2 𝑑𝑥 𝑑𝑦. 𝜕𝑥

Adding the line integrals over 𝐶1 + 𝐶3 and 𝐶2 + 𝐶4 , we get ) ( 𝜕𝐹2 𝜕𝐹1 ⃗ − 𝑑𝑥 𝑑𝑦. 𝐹 ⋅ 𝑑⃗𝑟 = ∫𝑅 𝜕𝑥 ∫𝐶 𝜕𝑦 If 𝑅 is not a rectangle, we subdivide it into small rectangular pieces as shown in Figure 18.46. The contribution to the integral of the non-rectangular pieces can be made as small as we like by making the subdivision fine enough. The double integrals over each piece add up to the double integral over the whole region 𝑅. Figure 18.47 shows how the circulations around adjacent pieces cancel along the common edge, so the circulations around all the pieces add up to the circulation around the boundary 𝐶. Since Green’s Theorem holds for the rectangular pieces, it holds for the whole region 𝑅.

18.4 PATH-DEPENDENT VECTOR FIELDS AND GREEN’S THEOREM

𝑦

953

𝐶3

𝑑 𝐶4

𝐶2

𝑐 𝐶1

𝐶

✣ ✕ 𝑅

𝑥 𝑏 Figure 18.45: The boundary of a rectangle broken into 𝐶1 , 𝐶2 , 𝐶3 , 𝐶4 𝑎

Figure 18.47: Two adjacent small closed curves

Figure 18.46: Region 𝑅 bounded by a closed curve 𝐶 and split into many small regions, Δ𝑅

(

) 𝑦2 ⃗𝑖 + 𝑥𝑗⃗ ⋅ 𝑑⃗𝑟 where 𝐶 is the counterclockwise path around

Example 4

Use Green’s Theorem to evaluate

Solution

We have 𝐹1 = 𝑦2 and 𝐹2 = 𝑥. By Green’s Theorem,

∫𝐶 the perimeter of the rectangle 0 ≤ 𝑥 ≤ 2, 0 ≤ 𝑦 ≤ 3.

∫𝐶

(

) 𝑦2 ⃗𝑖 + 𝑥𝑗⃗ ⋅ 𝑑⃗𝑟 =

∫𝑅

(

𝜕𝐹2 𝜕𝐹1 − 𝜕𝑥 𝜕𝑦

)

3

𝑑𝑥 𝑑𝑦 =

∫0 ∫0

2

(1 − 2𝑦) 𝑑𝑥 𝑑𝑦 = −12.

The Curl Test for Vector Fields in the Plane We already know that if 𝐹⃗ = 𝐹1 ⃗𝑖 + 𝐹2 𝑗⃗ is a gradient field with continuous partial derivatives, then 𝜕𝐹2 𝜕𝐹1 − = 0. 𝜕𝑥 𝜕𝑦 Now we show that the converse is true if the domain of 𝐹⃗ has no holes in it. This means that we assume that 𝜕𝐹2 𝜕𝐹1 − =0 𝜕𝑥 𝜕𝑦 and show that 𝐹⃗ is path-independent. If 𝐶 is any oriented simple closed curve in the domain of 𝐹⃗ and 𝑅 is the region inside 𝐶, then ) ( 𝜕𝐹2 𝜕𝐹1 − 𝑑𝑥 𝑑𝑦 = 0 ∫𝑅 𝜕𝑥 𝜕𝑦 since the integrand is identically 0. Therefore, by Green’s Theorem, ) ( 𝜕𝐹2 𝜕𝐹1 − 𝑑𝑥𝑑𝑦 = 0. 𝐹⃗ ⋅ 𝑑⃗𝑟 = ∫𝐶 ∫𝑅 𝜕𝑥 𝜕𝑦 Thus, 𝐹⃗ is path-independent and therefore a gradient field. This argument is valid for every closed curve, 𝐶, provided the region 𝑅 is entirely in the domain of 𝐹⃗ . Thus, we have the following result:

The Curl Test for Vector Fields in 2-Space Suppose 𝐹⃗ = 𝐹1 ⃗𝑖 + 𝐹2 𝑗⃗ is a vector field with continuous partial derivatives such that • The domain of 𝐹⃗ has the property that every closed curve in it encircles a region that lies entirely within the domain. In particular, the domain of 𝐹⃗ has no holes. 𝜕𝐹2 𝜕𝐹1 − = 0. • 𝜕𝑥 𝜕𝑦 Then 𝐹⃗ is path-independent, so 𝐹⃗ is a gradient field and has a potential function.

954

Chapter 18 LINE INTEGRALS

Why Are Holes in the Domain of the Vector Field Important? The reason for assuming that the domain of the vector field 𝐹⃗ has no holes is to ensure that the region 𝑅 inside 𝐶 is actually contained in the domain of 𝐹⃗ . Otherwise, we cannot apply Green’s Theorem. The next two examples show that if 𝜕𝐹2 ∕𝜕𝑥 − 𝜕𝐹1 ∕𝜕𝑦 = 0 but the domain of 𝐹⃗ contains a hole, then 𝐹⃗ can either be path-independent or be path-dependent.

Example 5

Let 𝐹⃗ be the vector field given by 𝐹⃗ (𝑥, 𝑦) =

−𝑦⃗𝑖 + 𝑥𝑗⃗ . 𝑥2 + 𝑦2

𝜕𝐹2 𝜕𝐹1 − . Does the curl test imply that 𝐹⃗ is path-independent? 𝜕𝑥 𝜕𝑦 (b) Calculate 𝐹⃗ ⋅ 𝑑⃗𝑟 , where 𝐶 is the unit circle centered at the origin and oriented counterclock∫𝐶 wise. Is 𝐹⃗ a path-independent vector field? (c) Explain why the answers to parts (a) and (b) do not contradict Green’s Theorem. (a) Calculate

Solution

(a) Taking partial derivatives, we have 𝜕𝐹2 𝜕 = 𝜕𝑥 𝜕𝑥

(

𝑥 𝑥2 + 𝑦2

)

=

𝑦2 − 𝑥2 1 𝑥 ⋅ 2𝑥 − = . 𝑥2 + 𝑦2 (𝑥2 + 𝑦2 )2 (𝑥2 + 𝑦2 )2

𝜕𝐹1 𝜕 = 𝜕𝑦 𝜕𝑦

(

−𝑦 2 𝑥 + 𝑦2

)

=

𝑦 ⋅ 2𝑦 𝑦2 − 𝑥2 −1 + = . 𝑥2 + 𝑦2 (𝑥2 + 𝑦2 )2 (𝑥2 + 𝑦2 )2

Similarly,

Thus, 𝜕𝐹2 𝜕𝐹1 − = 0. 𝜕𝑥 𝜕𝑦 Since 𝐹⃗ is undefined at the origin, the domain of 𝐹⃗ contains a hole. Therefore, the curl test does not apply. (b) See Figure 18.49. On the unit circle, 𝐹⃗ is tangent to the circle and ||𝐹⃗ || = 1. Thus,3 ∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 = ||𝐹⃗ || ⋅ Length of curve = 1 ⋅ 2𝜋 = 2𝜋.

Since the line integral around the closed curve 𝐶 is nonzero, 𝐹⃗ is not path-independent. We observe that 𝐹⃗ = grad(arctan(𝑦∕𝑥)) and arctan(𝑦∕𝑥) is 𝜃 from polar coordinates, for −𝜋∕2 < 𝜃 < 𝜋∕2. The fact that 𝜃 increases by 2𝜋 each time we wind once around the origin counterclockwise explains why 𝐹⃗ is not path-independent. (c) The domain of 𝐹⃗ is the “punctured plane,” as shown in Figure 18.48. Since 𝐹⃗ is not defined at the origin, which is inside 𝐶, Green’s Theorem does not apply. In this case

2𝜋 = 3 See

Problem 56 on page 931.

∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 ≠

∫𝑅

(

𝜕𝐹2 𝜕𝐹1 − 𝜕𝑥 𝜕𝑦

)

𝑑𝑥𝑑𝑦 = 0.

18.4 PATH-DEPENDENT VECTOR FIELDS AND GREEN’S THEOREM

𝑦

955

𝑦 𝐶 𝑅 𝑥

𝑥 1

Figure 18.48: The domain of 𝐹⃗ (𝑥, 𝑦) = the plane minus the origin

−𝑦⃗𝑖 +𝑥𝑗⃗ 𝑥2 +𝑦2

is

Figure 18.49: The region 𝑅 is not contained in the ⃗ +𝑥𝑗⃗ domain of 𝐹⃗ (𝑥, 𝑦) = −𝑦𝑥2𝑖 +𝑦 2

Although the vector field 𝐹⃗ in the last example was not defined at the origin, this by itself does not prevent the vector field from being path-independent, as we see in the following example.

Example 6

Consider the vector field 𝐹⃗ given by 𝐹⃗ (𝑥, 𝑦) =

𝑥⃗𝑖 + 𝑦𝑗⃗ . 𝑥2 + 𝑦2

𝜕𝐹2 𝜕𝐹1 − . Does the curl test imply that 𝐹⃗ is path-independent? 𝜕𝑥 𝜕𝑦 (b) Explain how we know that 𝐹⃗ ⋅ 𝑑⃗𝑟 = 0, where 𝐶 is the unit circle centered at the origin and ∫𝐶 oriented counterclockwise. Does this imply that 𝐹⃗ is path-independent? (c) Check that 𝑓 (𝑥, 𝑦) = 12 ln(𝑥2 + 𝑦2 ) is a potential function for 𝐹⃗ . Does this imply that 𝐹⃗ is path-independent? (a) Calculate

Solution

(a) Taking partial derivatives, we have 𝜕𝐹2 𝜕 = 𝜕𝑥 𝜕𝑥

(

𝑦 2 𝑥 + 𝑦2

)

−2𝑥𝑦 = , 2 (𝑥 + 𝑦2 )2

and

𝜕𝐹1 𝜕 = 𝜕𝑦 𝜕𝑦

(

𝑥 2 𝑥 + 𝑦2

)

=

−2𝑥𝑦 . + 𝑦2 )2

(𝑥2

Therefore, 𝜕𝐹2 𝜕𝐹1 − = 0. 𝜕𝑥 𝜕𝑦 This does not imply that 𝐹⃗ is path-independent: The domain of 𝐹⃗ contains a hole since 𝐹⃗ is undefined at the origin. Thus, the curl test does not apply. (b) Since 𝐹⃗ (𝑥, 𝑦) = 𝑥⃗𝑖 + 𝑦𝑗⃗ = 𝑟⃗ on the unit circle 𝐶, the field 𝐹⃗ is everywhere perpendicular to 𝐶. Thus ∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 = 0.

The fact that ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 0 when 𝐶 is the unit circle does not imply that 𝐹⃗ is path-independent. To be sure that 𝐹⃗ is path-independent, we would have to show that ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 0 for every closed curve 𝐶 in the domain of 𝐹⃗ , not just the unit circle. (c) To check that grad 𝑓 = 𝐹⃗ , we differentiate 𝑓 : 𝑓𝑥 =

1 2𝑥 𝑥 1 𝜕 ln(𝑥2 + 𝑦2 ) = = 2 , 2 2 2 𝜕𝑥 2𝑥 +𝑦 𝑥 + 𝑦2

956

Chapter 18 LINE INTEGRALS

and 𝑓𝑦 =

𝑦 1 𝜕 1 2𝑦 ln(𝑥2 + 𝑦2 ) = = , 2 2 2 2 𝜕𝑦 2𝑥 +𝑦 𝑥 + 𝑦2

so that grad 𝑓 =

𝑥⃗𝑖 + 𝑦𝑗⃗ = 𝐹⃗ . 𝑥2 + 𝑦2

Thus, 𝐹⃗ is a gradient field and therefore is path-independent—even though 𝐹⃗ is undefined at the origin.

The Curl Test for Vector Fields in 3-Space The curl test is a convenient way of deciding whether a 2-dimensional vector field is path-independent. Fortunately, there is an analogous test for 3-dimensional vector fields, although we cannot justify it until Chapter 20. ⃗ is a vector field on 3-space we define a new vector field, If 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝐹1 ⃗𝑖 + 𝐹2 𝑗⃗ + 𝐹3 𝑘 ⃗ curl 𝐹 , on 3-space by ) ) ) ( ( ( 𝜕𝐹3 𝜕𝐹2 𝜕𝐹1 𝜕𝐹3 𝜕𝐹2 𝜕𝐹1 ⃗. ⃗𝑖 + − − − 𝑘 curl 𝐹⃗ = 𝑗⃗ + 𝜕𝑦 𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑦 The vector field curl 𝐹⃗ can be used to determine whether the vector field 𝐹⃗ is path-independent.

The Curl Test for Vector Fields in 3-Space Suppose 𝐹⃗ is a vector field on 3-space with continuous partial derivatives such that • The domain of 𝐹⃗ has the property that every closed curve in it can be contracted to a point in a smooth way, staying at all times within the domain. • curl 𝐹⃗ = 0⃗ . Then 𝐹⃗ is path-independent, so 𝐹⃗ is a gradient field and has a potential function.

For the 2-dimensional curl test, the domain of 𝐹⃗ must have no holes. This meant that if 𝐹⃗ was defined on a simple closed curve 𝐶, then it was also defined at all points inside 𝐶. One way to test for holes is to try to “lasso” them with a closed curve. If every closed curve in the domain can be pulled to a point without hitting a hole, that is, without straying outside the domain, then the domain has no holes. In 3-space, we need the same condition to be satisfied: we must be able to pull every closed curve to a point, like a lasso, without straying outside the domain. Example 7

Decide if the following vector fields are path-independent and whether or not the curl test applies. ⃗ −𝑦⃗𝑖 + 𝑥𝑗⃗ 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 ⃗ (b) 𝐺⃗ = + 𝑧2 𝑘 (a) 𝐹⃗ = 2 2 2 3∕2 (𝑥 + 𝑦 + 𝑧 ) 𝑥2 + 𝑦2

Solution

(a) Suppose 𝑓 = −(𝑥2 + 𝑦2 + 𝑧2 )−1∕2 . Then 𝑓𝑥 = 𝑥(𝑥2 + 𝑦2 + 𝑧2 )−3∕2 and 𝑓𝑦 and 𝑓𝑧 are similar, so grad 𝑓 = 𝐹⃗ . Thus, 𝐹⃗ is a gradient field and therefore path-independent. Calculations show curl 𝐹⃗ = 0⃗ . The domain of 𝐹⃗ is all of 3-space minus the origin, and any closed curve in the domain can be pulled to a point without leaving the domain. Thus, the curl test applies. (b) Let 𝐶 be the circle 𝑥2 +𝑦2 = 1, 𝑧 = 0 traversed counterclockwise when viewed from the positive

18.4 PATH-DEPENDENT VECTOR FIELDS AND GREEN’S THEOREM

957

𝑧-axis. Since 𝑧 = 0 on the curve 𝐶, the vector field 𝐺⃗ reduces to the vector field in Example 5 and is everywhere tangent to 𝐶 and of magnitude 1, so ∫𝐶

𝐺⃗ ⋅ 𝑑⃗𝑟 = ‖𝐺⃗ ‖ ⋅ Length of curve = 1 ⋅ 2𝜋 = 2𝜋.

Since the line integral around this closed curve is nonzero, 𝐺⃗ is path-dependent. Computations show curl 𝐺⃗ = 0⃗ . However, the domain of 𝐺⃗ is all of 3-space minus the 𝑧-axis, and it does not satisfy the curl test domain criterion. For example, the circle, 𝐶, is lassoed around the 𝑧-axis, and cannot be pulled to a point without hitting the 𝑧-axis. Thus, the curl test does not apply.

Exercises and Problems for Section 18.4 EXERCISES In Exercises 1–10, decide if the given vector field is the gradient of a function 𝑓 . If so, find 𝑓 . If not, explain why not.

11. 𝐹⃗ = 𝑦⃗𝑖 − 𝑥𝑗⃗ around the unit circle. 12. 𝐹⃗ = 𝑥𝑦𝑗⃗ around the square 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1.

1. 𝑦⃗𝑖 − 𝑥𝑗⃗ 2. 2𝑥𝑦⃗𝑖 + 𝑥2 𝑗⃗

13. 𝐹⃗ = (2𝑥2 +3𝑦)⃗𝑖 +(2𝑥+3𝑦2 )𝑗⃗ around the triangle with vertices (2, 0), (0, 3), (−2, 0).

3. 𝑦⃗𝑖 + 𝑦𝑗⃗ 4. 2𝑥𝑦⃗𝑖 + 2𝑥𝑦𝑗⃗

14. 𝐹⃗ = 3𝑦⃗𝑖 + 𝑥𝑦𝑗⃗ around the unit circle. ( ) 15. Use Green’s Theorem to evaluate ∫𝐶 𝑦2 ⃗𝑖 + 𝑥𝑗⃗ ⋅

5. (𝑥2 + 𝑦2 )⃗𝑖 + 2𝑥𝑦𝑗⃗ 6. (2𝑥𝑦3 + 𝑦)⃗𝑖 + (3𝑥2 𝑦2 + 𝑥)𝑗⃗

𝑑⃗𝑟 where 𝐶 is the counterclockwise path around the perimeter of the rectangle 0 ≤ 𝑥 ≤ 2, 0 ≤ 𝑦 ≤ 3.

⃗ ⃗𝑖 𝑗⃗ 𝑘 + + 𝑥 𝑦 𝑧 ⃗ ⃗ ⃗𝑖 𝑗 𝑘 + + 8. 𝑥 𝑦 𝑥𝑦 7.

16. If 𝐶 goes counterclockwise around the perimeter of the rectangle 𝑅 with vertices (10, 10), (30, 10), (30, 20), and (10, 20), use Green’s theorem to evaluate

⃗ 9. 2𝑥 cos(𝑥2 + 𝑧2 )⃗𝑖 + sin(𝑥2 + 𝑧2 )𝑗⃗ + 2𝑧 cos(𝑥2 + 𝑧2 )𝑘 𝑦 𝑥 ⃗𝑖 − 10. 2 𝑗⃗ 𝑥 + 𝑦2 𝑥2 + 𝑦2 In Exercises 11–14, use Green’s Theorem to calculate the circulation of 𝐹⃗ around the curve, oriented counterclockwise.

∫𝐶

−𝑦 𝑑𝑥 + 𝑥 𝑑𝑦.

17. Calculate ∫𝐶 ((3𝑥 + 5𝑦)⃗𝑖 + (2𝑥 + 7𝑦)𝑗⃗ ) ⋅ 𝑑⃗𝑟 where 𝐶 is the circular path with center (𝑎, 𝑏) and radius 𝑚, oriented counterclockwise. Use Green’s Theorem.

PROBLEMS 2 18. Find the line integral of 𝐹⃗ = 𝑒𝑥 ⃗𝑖 + (7𝑥 + 1)𝑗⃗ around the closed curve 𝐶 consisting of the two line segments and the circular arc in Figure 18.50.

𝑦 5

19. (a) Sketch 𝐹⃗ = 𝑦⃗𝑖 and determine the sign of the circulation of 𝐹⃗ around the unit circle centered at the origin and oriented counterclockwise. (b) Use Green’s Theorem to compute the circulation in part (a) exactly.

𝐶

20. Let 𝐹⃗ = (sin 𝑥)⃗𝑖 + (𝑥 + 𝑦)𝑗⃗ . Find the line integral of 𝐹⃗ around the perimeter of the rectangle with corners (3, 0), (3, 5), (−1, 5), (−1, 0), traversed in that order. 𝑥 5

Figure 18.50

((sin(𝑥2 ) cos 𝑦)⃗𝑖 + (sin(𝑦2 ) + 𝑒𝑥 )𝑗⃗ ) ⋅ 𝑑⃗𝑟 where ∫𝐶 𝐶 is the square of side 1 in the first quadrant of the 𝑥𝑦plane, with one vertex at the origin and sides along the axes, and oriented counterclockwise when viewed from above.

21. Find

958

Chapter 18 LINE INTEGRALS

In Problems 22–23 find the line integral of 𝐹⃗ around the closed curve in Figure 18.51. The arc is part of a circle. 𝑦

45◦

𝑥 3

27. Let 𝐶1 be the curve consisting of the circle of radius 2, centered at the origin and oriented counterclockwise, and 𝐶2 be the curve consisting of the line segment from (0, 0) to (1, 1) followed by the line segment from (1, 1) to (3, 1). Let 𝐹⃗ = 2𝑥𝑦2 ⃗𝑖 + (2𝑦𝑥2 + 2𝑦)𝑗⃗ and let 𝐺⃗ = (𝑦 + 𝑥)⃗𝑖 + (𝑦 − 𝑥)𝑗⃗ . Compute the following line integrals. (a)

∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟

(b) ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟

(c)

∫𝐶 𝐺⃗ ⋅ 𝑑⃗𝑟

(d) ∫𝐶 𝐺⃗ ⋅ 𝑑⃗𝑟

1

2

1

2

28. Prove that Green’s theorem is true when the integrand of the line integral has a potential function.

Figure 18.51

29. Consider the following parametric equations: 22. 𝐹⃗ = (𝑥 − 𝑦)⃗𝑖 + 𝑥𝑗⃗ 23. 𝐹⃗ = (𝑥 + 𝑦)⃗𝑖 + sin 𝑦𝑗⃗ 24. Let 𝐹⃗ = 2𝑥𝑒𝑦 ⃗𝑖 + 𝑥2 𝑒𝑦 𝑗⃗ and 𝐺⃗ = (𝑥 − 𝑦)⃗𝑖 + (𝑥 + 𝑦)𝑗⃗ . Let 𝐶 be the line from (0, 0) to (2, 4). Find exactly: (a) ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟

(b) ∫𝐶 𝐺⃗ ⋅ 𝑑⃗𝑟

25. Let 𝐹⃗ = 𝑦⃗𝑖 +𝑥𝑗⃗ and 𝐺⃗ = 3𝑦⃗𝑖 −3𝑥𝑗⃗ . In Figure 18.52, the curve 𝐶2 is the semicircle centered at the origin from (−1, 1) to (1, −1) and 𝐶1 is the line segment from (−1, 1) to (1, −1), and 𝐶 = 𝐶2 − 𝐶1 . Find the following line integrals: (a) (c) (e) (g)

𝐹⃗ ⋅ 𝑑⃗𝑟

(b)

𝐹⃗ ⋅ 𝑑⃗𝑟

(d)

𝐺⃗ ⋅ 𝑑⃗𝑟

(f)

∫𝐶1 ∫𝐶2 ∫𝐶 ∫𝐶

∫𝐶 ∫𝐶2 ∫𝐶1

𝐹⃗ ⋅ 𝑑⃗𝑟 𝐺⃗ ⋅ 𝑑⃗𝑟 𝐺⃗ ⋅ 𝑑⃗𝑟

⃗ ,0 ≤ 𝑡 ≤ 2 𝐶1 ∶ 𝑟⃗ (𝑡) = 𝑡 cos(2𝜋𝑡) ⃗𝑖 + 𝑡 sin(2𝜋𝑡) 𝑘 ⃗ ,0 ≤ 𝑡 ≤ 2 𝐶2 ∶ 𝑟⃗ (𝑡) = 𝑡 cos(2𝜋𝑡) ⃗𝑖 + 𝑡 𝑗⃗ + 𝑡 sin(2𝜋𝑡) 𝑘 (a) Describe, in words, the motion of a particle moving along each of the paths. (b) Evaluate ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 , for the vector field 𝐹⃗ = 2 ⃗. 𝑦𝑧 ⃗𝑖 + 𝑧(𝑥 + 1) 𝑗⃗ + (𝑥𝑦 + 𝑦 + 1) 𝑘 (c) Find a nonzero vector field 𝐺⃗ such that: ∫𝐶1

𝐺⃗ ⋅ 𝑑⃗𝑟 =

∫𝐶2

𝐺⃗ ⋅ 𝑑⃗𝑟 .

Explain how you reasoned to find 𝐺⃗ . (d) Find two different, nonzero vector fields 𝐻⃗ 1 , 𝐻⃗2 such that: ∫𝐶1

𝐻⃗ 1 ⋅ 𝑑⃗𝑟 =

∫𝐶1

𝐻⃗2 ⋅ 𝑑⃗𝑟 .

Explain how you reasoned to find the two fields. 30. Show that the line integral of 𝐹⃗ = 𝑥𝑗⃗ around a closed curve in the 𝑥𝑦-plane, oriented as in Green’s Theorem, measures the area of the region enclosed by the curve.

(𝐹⃗ + 𝐺⃗ ) ⋅ 𝑑⃗𝑟

In Problems 31–33, use the result of Problem 30 to calculate the area of the region within the parameterized curves. In each case, sketch the curve.

𝑦

31. The ellipse 𝑥2 ∕𝑎2 + 𝑦2 ∕𝑏2 = 1 parameterized by 𝑥 = 𝑎 cos 𝑡, 𝑦 = 𝑏 sin 𝑡, for 0 ≤ 𝑡 ≤ 2𝜋.

𝐶1 𝑥

𝐶2

Figure 18.52 ) ( 26. Calculate ∫𝐶 (𝑥2 − 𝑦)⃗𝑖 + (𝑦2 + 𝑥)𝑗⃗ ⋅ 𝑑⃗𝑟 if: (a) 𝐶 is the circle (𝑥 − 5)2 + (𝑦 − 4)2 = 9 oriented counterclockwise. (b) 𝐶 is the circle (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑅2 in the 𝑥𝑦-plane oriented counterclockwise.

32. The hypocycloid 𝑥2∕3 + 𝑦2∕3 = 𝑎2∕3 parameterized by 𝑥 = 𝑎 cos3 𝑡, 𝑦 = 𝑎 sin3 𝑡, 0 ≤ 𝑡 ≤ 2 𝜋. 33. The folium of Descartes, 𝑥3 + 𝑦3 = 3𝑥𝑦, parameterized 3𝑡 3𝑡2 by 𝑥 = ,𝑦= , for 0 ≤ 𝑡 < ∞. 3 1+𝑡 1 + 𝑡3 34. The vector field 𝐹⃗ is defined on the disk 𝐷 of radius 5 centered at the origin in the plane: 𝐹⃗ = (−𝑦3 + 𝑦 sin (𝑥𝑦))⃗𝑖 + (4𝑥(1 − 𝑦2 ) + 𝑥 sin (𝑥𝑦))𝑗⃗ . Consider the line integral ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 , where 𝐶 is some closed curve contained in 𝐷. For which 𝐶 is the value of this integral the largest? [Hint: Assume 𝐶 is a closed curve, made up of smooth pieces and never crossing itself, and oriented counterclockwise.]

18.4 PATH-DEPENDENT VECTOR FIELDS AND GREEN’S THEOREM

35. Example 1 on page 949 showed that the vector field in Figure 18.53 could not be a gradient field by showing that it is not path-independent. Here is another way to see the same thing. Suppose that the vector field were the gradient of a function 𝑓 . Draw and label a diagram showing what the contours of 𝑓 would have to look like, and explain why it would not be possible for 𝑓 to have a single value at any given point. 𝑦

959

39. Arrange the line integrals 𝐿1 , 𝐿2 , 𝐿3 in ascending order, where 𝐿𝑖 =

∫𝐶𝑖

(−𝑥2 𝑦⃗𝑖 + (𝑥𝑦2 − 𝑥)𝑗⃗ ) ⋅ 𝑑⃗𝑟 .

The points 𝐴, 𝐵, 𝐷 lie on the unit circle and 𝐶𝑖 is one of the curves shown in Figure 18.54. 𝐶1 : Line segment 𝐴 to 𝐵 𝐶2 : Line segment 𝐴 to 𝐷 followed by line segment 𝐷 to 𝐵 𝐶3 : Semicircle 𝐴𝐷𝐵 𝑦 𝐷 𝐶3

𝐶3

𝑥 𝐶2

𝐶2 𝐵

𝐴 −1

Figure 18.53

𝐶1

𝐶1

𝑥

1

Figure 18.54 40. For all 𝑥, 𝑦, let 𝐹⃗ = 𝐹1 (𝑥, 𝑦)⃗𝑖 + 𝐹2 (𝑥, 𝑦)𝑗⃗ satisfy

36. Repeat Problem 35 for the vector field in Problem 52 on page 947.

𝜕𝐹2 𝜕𝐹1 − = 3. 𝜕𝑥 𝜕𝑦

37. (a) By finding potential functions, show that each of ⃗ is a gradient field on the vector fields 𝐹⃗ , 𝐺⃗ , 𝐻 some domain (not necessarily the whole plane). ⃗ around the unit (b) Find the line integrals of 𝐹⃗ , 𝐺⃗ , 𝐻 circle in the 𝑥𝑦-plane, centered at the origin, and traversed counterclockwise. (c) For which of the three vector fields can Green’s Theorem be used to calculate the line integral in part (b)? Why or why not?

(a) Calculate ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 where 𝐶1 is the unit circle in 1 the 𝑥𝑦-plane centered at the origin, oriented counterclockwise. (b) Calculate ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 where 𝐶2 is the boundary of 2 the rectangle of 4 ≤ 𝑥 ≤ 7, 5 ≤ 𝑦 ≤ 7, oriented counterclockwise. (c) Let 𝐶3 be the circle of radius 7 centered at the point (10, 2); let 𝐶4 be the circle of radius 8 centered at the origin; let 𝐶5 be the square of side 14 centered at (7, 7) with sides parallel to the axes; 𝐶3 , 𝐶4 , 𝐶5 are all oriented counterclockwise. Arrange the integrals ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 , ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 , ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 in in3 4 5 creasing order.

𝑦⃗𝑖 − 𝑥𝑗⃗ ⃗ 𝑥⃗𝑖 + 𝑦𝑗⃗ 𝐹⃗ = 𝑦⃗𝑖 + 𝑥𝑗⃗ , 𝐺⃗ = 2 ,𝐻 = 2 2 𝑥 +𝑦 (𝑥 + 𝑦2 )1∕2 38. (a) For which of the following can you use Green’s Theorem to evaluate the integral? Explain. (𝑥2 + 𝑦2 ) 𝑑𝑥 + (𝑥2 + 𝑦2 ) 𝑑𝑦 where 𝐶 is the ∫𝐶 curve defined by 𝑦 = 𝑥, 𝑦 = 𝑥2 , 0 ≤ 𝑥 ≤ 1 with counterclockwise orientation. 1 1 𝑑𝑥 − √ 𝑑𝑦 where 𝐶 is II ∫𝐶 √𝑥2 + 𝑦2 𝑥2 + 𝑦2 the unit circle centered at the origin, oriented counterclockwise. I

III

𝐹⃗ ⋅ 𝑑⃗𝑟 where 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ where 𝐶 is ∫𝐶 the line segment from the origin to (1, 1).

(b) Use Green’s Theorem to evaluate the integrals in part (a) that can be done that way.

2 41. Let 𝐹⃗ = (3𝑥2 𝑦 + 𝑦3 + 𝑒𝑥 )⃗𝑖 + (𝑒𝑦 + 12𝑥)𝑗⃗ . Consider the line integral of 𝐹⃗ around the circle of radius 𝑎, centered at the origin and traversed counterclockwise.

(a) Find the line integral for 𝑎 = 1. (b) For which value of 𝑎 is the line integral a maximum? Explain. 42. Let −𝑦⃗𝑖 + 𝑥𝑗⃗ 𝐹⃗ (𝑥, 𝑦) = 𝑥2 + 𝑦2 and let oriented curves 𝐶1 and 𝐶2 be as in Figure 18.55. The curve 𝐶2 is an arc of the unit circle centered at the origin. Show that (a) The curl of 𝐹⃗ is zero. (b) ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 . 1

2

960

Chapter 18 LINE INTEGRALS

(c) ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 𝜃, the angle at the origin subtended 1 by the oriented curve 𝐶1 .

43. The electric field 𝐸⃗ , at the point with position vector 𝑟⃗ in 3-space, due to a charge 𝑞 at the origin is given by

𝑦

𝐸⃗ (⃗𝑟 ) = 𝑞

𝐵

𝑟⃗ . ||⃗𝑟 ||3

(a) Compute curl 𝐸⃗ . Is 𝐸⃗ a path-independent vector field? Explain. (b) Find a potential function 𝜑 for 𝐸⃗ , if possible.

𝐶1 𝐷 𝐶2 𝜃

𝐴 𝐶

𝑥

Figure 18.55

Strengthen Your Understanding In Problems 44–45, explain what is wrong with the statement. 44. If ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 0 for a specific closed path 𝐶, then 𝐹⃗ must be path-independent. 45. Let 𝐹⃗ = 𝐹1 (𝑥, 𝑦)⃗𝑖 + 𝐹2 (𝑥, 𝑦)𝑗⃗ with 𝜕𝐹2 𝜕𝐹1 − =3 𝜕𝑥 𝜕𝑦 and let 𝐶 be the path consisting of line segments from (0, 0) to (1, 1) to (2, 0). Then

∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 = 3.

48. A vector field that is not a gradient field. Are the statements in Problems 49–56 true or false? Give reasons for your answer. 49. If 𝑓 (𝑥) and 𝑔(𝑦) are continuous one-variable functions, then the vector field 𝐹⃗ = 𝑓 (𝑥)⃗𝑖 + 𝑔(𝑦)𝑗⃗ is pathindependent. 50. If 𝐹⃗ = grad 𝑓 , and 𝐶 is the perimeter of a square of side length 𝑎 oriented counterclockwise and surrounding the region 𝑅, then ∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫𝑅

𝑓 𝑑𝐴.

51. If 𝐹⃗ and 𝐺⃗ are both path-independent vector fields, then 𝐹⃗ + 𝐺⃗ is path-independent. In Problems 46–48, give an example of: 46. A function 𝑄(𝑥, 𝑦) such that 𝐹⃗ = 𝑥𝑦⃗𝑖 + 𝑄(𝑥, 𝑦)𝑗⃗ is a gradient field. 47. Two oriented curves, 𝐶1 and 𝐶2 , from (1, 0) to (0, 1) such that if −𝑦⃗𝑖 + 𝑥𝑗⃗ , 𝐹⃗ (𝑥, 𝑦) = 𝑥2 + 𝑦2 then ∫𝐶1

𝐹⃗ ⋅ 𝑑⃗𝑟 ≠

∫𝐶2

𝐹⃗ ⋅ 𝑑⃗𝑟 .

[Note that the scalar curl of 𝐹⃗ is 0 where 𝐹⃗ is defined.]

Online Resource: Review problems and Projects

52. If 𝐹⃗ and 𝐺⃗ are both path-dependent vector fields, then 𝐹⃗ + 𝐺⃗ is path-dependent. 53. The vector field 𝐹⃗ (⃗𝑟 ) = 𝑟⃗ in 3-space is pathindependent. 54. A constant vector field 𝐹⃗ independent.

= 𝑎⃗𝑖 + 𝑏𝑗⃗ is path-

55. If 𝐹⃗ is path-independent and 𝑘 is a constant, then the vector field 𝑘𝐹⃗ is path-independent. 56. If 𝐹⃗ is path-independent and ℎ(𝑥, 𝑦) is a scalar function, then the vector field ℎ(𝑥, 𝑦)𝐹⃗ is pathindependent.

Chapter Nineteen

FLUX INTEGRALS AND DIVERGENCE

Contents 19.1 The Idea of a Flux Integral . . . . . . . . . . . . . . . . Flow Through a Surface . . . . . . . . . . . . . . . . . . . Orientation of a Surface . . . . . . . . . . . . . . . . . . . . The Area Vector . . . . . . . . . . . . . . . . . . . . . . . . . .

962 962 962 963

The Flux of a Constant Vector Field Through a Flat Surface . . . . . . . . . . . . . . . . . . . 963

The Flux Integral . . . . . . . . . . . . . . . . . . . . . . . . . Flux and Fluid Flow . . . . . . . . . . . . . . . . . . . . . . LI I Calculating Flux Integrals Using d A = n dA . . . Notes on Orientation . . . . . . . . . . . . . . . . . . . . . . 19.2 Flux Integrals for Graphs, Cylinders, and Spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Flux of a Vector Field Through the Graph of z = f (x, y) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . The Area Vector of a Coordinate Patch . . . . . . . . . Surface Area of a Graph . . . . . . . . . . . . . . . . . . . Flux of a Vector Field Through a Cylindrical Surface . . . . . . . . . . . . . . . . . . . . . . Flux of a Vector Field Through a Spherical Surface . . . . . . . . . . . . . . . . . . . . . . . 19.3 The Divergence of a Vector Field . . . . . . . . . . . . Definition of Divergence . . . . . . . . . . . . . . . . . . . Why Do the Two Definitions of Divergence Give the Same Result? . . . . . . . . . . . . . . . . . . . . Divergence-Free Vector Fields . . . . . . . . . . . . . . Magnetic Fields . . . . . . . . . . . . . . . . . . . . . . . . . 19.4 The Divergence Theorem . . . . . . . . . . . . . . . . . . The Boundary of a Solid Region . . . . . . . . . . . . . Calculating the Flux from the Flux Density . . . . The Divergence Theorem and Divergence-Free Vector Fields. . . . . . . . . . . . . . Electric Fields . . . . . . . . . . . . . . . . . . . . . . . . . .

964 965 967 968 973 973 973 975 975 977 982 982 985 986 987 991 991 991 993 994

962

Chapter 19 FLUX INTEGRALS AND DIVERGENCE

19.1

THE IDEA OF A FLUX INTEGRAL

Flow Through a Surface Imagine water flowing through a fishing net stretched across a stream. Suppose we want to measure the flow rate of water through the net, that is, the volume of fluid that passes through the surface per unit time. Example 1

A flat square surface of area 𝐴, in m2 , is immersed in a fluid. The fluid flows with constant velocity 𝑣⃗ , in m/sec, perpendicular to the square. Write an expression for the rate of flow in m3 /sec. ✛

𝐴 𝑣⃗

Figure 19.1: Fluid flowing perpendicular to a surface

Solution

In one second a given particle of water moves a distance of ‖⃗ 𝑣 ‖ in the direction perpendicular to the square. Thus, the entire body of water moving through the square in one second is a box of length ‖⃗ 𝑣 ‖ and cross-sectional area 𝐴. So the box has volume ‖⃗ 𝑣 ‖𝐴 m3 , and Flow rate = ‖⃗ 𝑣 ‖𝐴 m3 /sec.

This flow rate is called the flux of the fluid through the surface. We can also compute the flux of vector fields, such as electric and magnetic fields, where no flow is actually taking place. If the vector field is constant and perpendicular to the surface, and if the surface is flat, as in Example 1, the flux is obtained by multiplying the speed by the area. Next we find the flux of a constant vector field through a flat surface that is not perpendicular to the vector field, using a dot product. In general, we break a surface into small pieces which are approximately flat and where the vector field is approximately constant, leading to a flux integral.

Orientation of a Surface Before computing the flux of a vector field through a surface, we need to decide which direction of flow through the surface is the positive direction; this is described as choosing an orientation.1 At each point on a smooth surface there are two unit normals, one in each direction. Choosing an orientation means picking one of these normals at every point of the surface in a continuous way. The unit normal vector in the direction of the orientation is denoted by 𝑛⃗ . For a closed surface (that is, the boundary of a solid region), we choose the outward orientation unless otherwise specified. We say the flux through a piece of surface is positive if the flow is in the direction of the orientation and negative if it is in the opposite direction. (See Figure 19.2.) 1 Although

we will not study them, there are a few surfaces for which this cannot be done. See page 968.

19.1 THE IDEA OF A FLUX INTEGRAL

963

𝐴⃗

Negative flow

𝑛⃗ 𝑆 Direction of orientation: Positive flow

Figure 19.2: An oriented surface showing directions of positive and negative flow

Figure 19.3: Area vector 𝐴⃗ = 𝑛⃗ 𝐴 of flat surface with area 𝐴 and orientation 𝑛⃗

The Area Vector The flux through a flat surface depends both on the area of the surface and its orientation. Thus, it is useful to represent its area by a vector as shown in Figure 19.3.

The area vector of a flat, oriented surface is a vector 𝐴⃗ such that • The magnitude of 𝐴⃗ is the area of the surface. • The direction of 𝐴⃗ is the direction of the orientation vector 𝑛⃗ .

The Flux of a Constant Vector Field Through a Flat Surface Suppose the velocity vector field, 𝑣⃗ , of a fluid is constant and 𝐴⃗ is the area vector of a flat surface. The flux through this surface is the volume of fluid that flows through in one unit of time. The skewed box in Figure 19.4 has cross-sectional area ‖𝐴⃗ ‖ and height ‖⃗ 𝑣 ‖ cos 𝜃, so its volume is ( ) ‖⃗ 𝑣 ‖ cos 𝜃 ‖𝐴⃗ ‖ = 𝑣⃗ ⋅ 𝐴⃗ . Thus, we have the following result: If 𝑣⃗ is constant and 𝐴⃗ is the area vector of a flat surface, then Flux through surface = 𝑣⃗ ⋅ 𝐴⃗ .

𝑣⃗

𝐴⃗ 𝜃

✻ ‖𝑣⃗ ‖ cos 𝜃

❄ Figure 19.4: Flux of 𝑣⃗ through a surface with area vector 𝐴⃗ is the volume of this skewed box

Example 2

Water is flowing down a cylindrical pipe 2 cm in radius with a velocity of 3 cm/sec. Find the flux of the velocity vector field through the ellipse-shaped region shown in Figure 19.5. The normal to the ellipse makes an angle of 𝜃 with the direction of flow and the area of the ellipse is 4𝜋∕(cos 𝜃) cm2 .

964

Chapter 19 FLUX INTEGRALS AND DIVERGENCE





2 cm



𝜃



𝐴⃗ (where ‖𝐴⃗ ‖ = 4𝜋∕ (cos 𝜃) cm2 )

𝑣⃗ ( where ‖𝑣⃗ ‖ = 3 cm/sec)

Figure 19.5: Flux through ellipse-shaped region across a cylindrical pipe

Solution

There are two ways to approach this problem. One is to use the formula we just derived, which gives Flux through ellipse = 𝑣⃗ ⋅ 𝐴⃗ = ‖⃗ 𝑣 ‖‖𝐴⃗ ‖ cos 𝜃 = 3(Area of ellipse) cos 𝜃 ) ( 4𝜋 cos 𝜃 = 12𝜋 cm3 ∕sec. =3 cos 𝜃 The second way is to notice that the flux through the ellipse is equal to the flux through the circle perpendicular to the pipe in Figure 19.5. Since the flux is the rate at which water is flowing down the pipe, we have Flux through circle =

Velocity of water

×

Area of circle

) ( cm (𝜋22 cm2 ) = 12𝜋 cm3 ∕sec. = 3 sec

The Flux Integral If the vector field, 𝐹⃗ , is not constant or the surface, 𝑆, is not flat, we divide the surface into a patchwork of small, almost flat pieces. (See Figure 19.6.) For a particular patch with area Δ𝐴, we pick a unit orientation vector 𝑛⃗ at a point on the patch and define the area vector of the patch, Δ𝐴⃗ , as Δ𝐴⃗ = 𝑛⃗ Δ𝐴. (See Figure 19.7.) If the patches are small enough, we can assume that 𝐹⃗ is approximately constant on each piece. Then we know that Flux through patch ≈ 𝐹⃗ ⋅ Δ𝐴⃗ , so, adding the fluxes through all the small pieces, we have Flux through whole surface ≈



𝐹⃗ ⋅ Δ𝐴⃗ ,

As each patch becomes smaller and ‖Δ𝐴⃗ ‖ → 0, the approximation gets better and we get ∑ Flux through 𝑆 = lim 𝐹⃗ ⋅ Δ𝐴⃗ . ‖Δ𝐴⃗ ‖→0

Thus, provided the limit exists, we make the following definition:

The flux integral of the vector field 𝐹⃗ through the oriented surface 𝑆 is ∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

lim ‖Δ𝐴⃗ ‖→0



𝐹⃗ ⋅ Δ𝐴⃗ .

If 𝑆 is a closed surface oriented outward, we describe the flux through 𝑆 as the flux out of 𝑆.

965

19.1 THE IDEA OF A FLUX INTEGRAL

𝑆 Area Δ𝐴

𝑃

❘❄

Δ𝐴⃗

𝑛⃗ 𝑛⃗

Δ𝐴⃗ Vector field

𝐹⃗ (𝑥, 𝑦, 𝑧)

Figure 19.6: Surface 𝑆 divided into small, almost flat pieces, showing a typical orientation vector 𝑛⃗ and area vector Δ𝐴⃗

Figure 19.7: Flux of a vector field through a curved surface 𝑆

In computing a flux integral, we have to divide the surface up in a reasonable way, or the limit might not exist. In practice this problem seldom arises; however, one way to avoid it is to define flux integrals by the method used to compute them shown in Section 21.3.

Flux and Fluid Flow If 𝑣⃗ is the velocity vector field of a fluid, we have Rate fluid flows through surface 𝑆

Flux of 𝑣⃗ through 𝑆

=

=

∫𝑆

𝑣⃗ ⋅ 𝑑 𝐴⃗

The rate of fluid flow is measured in units of volume per unit time. Example 3

Find the flux of the vector field 𝐵⃗ (𝑥, 𝑦, 𝑧) shown in Figure 19.8 through the square 𝑆 of side 2 shown in Figure 19.9, oriented in the 𝑗⃗ direction, where −𝑦⃗𝑖 + 𝑥𝑗⃗ . 𝐵⃗ (𝑥, 𝑦, 𝑧) = 2 𝑥 + 𝑦2

𝑧

𝑧 𝑆

𝑆

Δ𝐴⃗

2







2

Δ𝑧

𝑦 𝑥 Figure 19.8: The vector field 𝐵⃗ in planes 𝑧 = 0, 𝑧 = 1, 𝑧 = 2, where −𝑦⃗𝑖 + 𝑥𝑗⃗ 𝐵⃗ (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦2

Solution

3

𝐵⃗

𝑥





1

Δ𝑥

𝑦 1

3 Figure 19.10: A small patch of surface with area ‖Δ𝐴⃗ ‖ = Δ𝑥Δ𝑧

Figure 19.9: Flux of 𝐵⃗ through the square 𝑆 of side 2 in 𝑥𝑦-plane and oriented in 𝑗⃗ direction

Consider a small rectangular patch with area vector Δ𝐴⃗ in 𝑆, with sides Δ𝑥 and Δ𝑧 so that ‖Δ𝐴⃗ ‖ = Δ𝑥 Δ𝑧. Since Δ𝐴⃗ points in the 𝑗⃗ direction, we have Δ𝐴⃗ = 𝑗⃗ Δ𝑥Δ𝑧. (See Figure 19.10.) At the point (𝑥, 0, 𝑧) in 𝑆, substituting 𝑦 = 0 into 𝐵⃗ gives 𝐵⃗ (𝑥, 0, 𝑧) = (1∕𝑥)𝑗⃗ . Thus, we have Flux through small patch ≈ 𝐵⃗ ⋅ Δ𝐴⃗ =

(

) 1⃗ 1 𝑗 ⋅ (𝑗⃗ Δ𝑥Δ𝑧) = Δ𝑥 Δ𝑧. 𝑥 𝑥

966

Chapter 19 FLUX INTEGRALS AND DIVERGENCE

Therefore, Flux through surface =

∫𝑆

𝐵⃗ ⋅ 𝑑 𝐴⃗ =

lim ‖Δ𝐴⃗ ‖→0



∑1 Δ𝑥 Δ𝑧. 𝑥 Δ𝑥 → 0

𝐵⃗ ⋅ Δ𝐴⃗ = lim

Δ𝑧 → 0

1 𝑥

This last expression is a Riemann sum for the double integral ∫𝑅 𝑥 ≤ 3, 0 ≤ 𝑧 ≤ 2. Thus, 2

Flux through surface =

∫𝑆

𝐵⃗ ⋅ 𝑑 𝐴⃗ =

𝑑𝐴, where 𝑅 is the square 1 ≤ 3

1 1 𝑑𝐴 = 𝑑𝑥 𝑑𝑧 = 2 ln 3. ∫0 ∫1 𝑥 ∫𝑅 𝑥

The result is positive since the vector field is passing through the surface in the positive direction.

Example 4

Each of the vector fields in Figure 19.11 consists entirely of vectors parallel to the 𝑥𝑦-plane, and is constant in the 𝑧 direction (that is, the vector field looks the same in any plane parallel to the 𝑥𝑦plane). For each one, say whether you expect the flux through a closed surface surrounding the origin to be positive, negative, or zero. In part (a) the surface is a closed cube with faces perpendicular to the axes; in parts (b) and (c) the surface is a closed cylinder. In each case we choose the outward orientation. (See Figure 19.12.) (a)

𝑦

𝑦

(b)

𝑦

(c)

𝑥

𝑥

𝑥

Figure 19.11: Flux of a vector field through the closed surfaces whose cross-sections are shown in the 𝑥𝑦-plane 𝑧

𝑧

𝑦 𝑥

𝑛⃗

𝑥

𝑛⃗ 𝑦

Figure 19.12: The closed cube and closed cylinder, both oriented outward

Solution

(a) Since the vector field appears to be parallel to the faces of the cube which are perpendicular to the 𝑦- and 𝑧-axes, we expect the flux through these faces to be zero. The fluxes through the two faces perpendicular to the 𝑥-axis appear to be equal in magnitude and opposite in sign, so we expect the net flux to be zero. (b) Since the top and bottom of the cylinder are parallel to the flow, the flux through them is zero. On the curved surface of the cylinder, 𝑣⃗ and Δ𝐴⃗ appear to be everywhere parallel and in the same direction, so we expect each term 𝑣⃗ ⋅ Δ𝐴⃗ to be positive, and therefore the flux integral ∫𝑆 𝑣⃗ ⋅ 𝑑 𝐴⃗ to be positive. (c) As in part (b), the flux through the top and bottom of the cylinder is zero. In this case 𝑣⃗ and Δ𝐴⃗ are not parallel on the round surface of the cylinder, but since the fluid appears to be flowing inward as well as swirling, we expect each term 𝑣⃗ ⋅ Δ𝐴⃗ to be negative, and therefore the flux integral to be negative.

19.1 THE IDEA OF A FLUX INTEGRAL

967

⃖⃖⃗ = 𝒏 ⃖⃗ 𝒅𝑨 Calculating Flux Integrals Using 𝒅𝑨 For a small patch of surface Δ𝑆 with unit normal 𝑛⃗ and area Δ𝐴, the area vector is Δ𝐴⃗ = 𝑛⃗ Δ𝐴. The next example shows how we can use this relationship to compute a flux integral. Example 5

An electric charge 𝑞 is placed at the origin in 3-space. The resulting electric field 𝐸⃗ (⃗𝑟 ) at the point with position vector 𝑟⃗ is given by 𝑟⃗ 𝐸⃗ (⃗𝑟 ) = 𝑞 , ‖⃗𝑟 ‖3

𝑟⃗ ≠ 0⃗ .

Find the flux of 𝐸⃗ out of the sphere of radius 𝑅 centered at the origin. (See Figure 19.13.) 𝑆



Δ𝐴⃗

Solution

Figure 19.13: Flux of 𝐸⃗ = 𝑞⃗𝑟 ∕‖⃗𝑟 ‖3 through the surface of a sphere of radius 𝑅 centered at the origin

This vector field points radially outward from the origin in the same direction as 𝑛⃗ . Thus, since 𝑛⃗ is a unit vector, 𝐸⃗ ⋅ Δ𝐴⃗ = 𝐸⃗ ⋅ 𝑛⃗ Δ𝐴 = ‖𝐸⃗ ‖ Δ𝐴. On the sphere, ‖𝐸⃗ ‖ = 𝑞∕𝑅2 , so ∫𝑆

𝐸⃗ ⋅ 𝑑 𝐴⃗ =

lim

‖Δ𝐴⃗ ‖→0



∑ 𝑞 ∑ 𝑞 Δ𝐴. Δ𝐴 = lim 2 2 Δ𝐴→0 𝑅 𝑅 Δ𝐴→0

𝐸⃗ ⋅ Δ𝐴⃗ = lim

The last sum approximates the surface area of the sphere. In the limit as the subdivisions get finer we have ∑ lim Δ𝐴 = Surface area of sphere. Δ𝐴→0

Thus, the flux is given by

∑ 𝑞 𝑞 𝑞 Δ𝐴 = 2 ⋅ (Surface area of sphere) = 2 (4𝜋𝑅2 ) = 4𝜋𝑞. 𝐸⃗ ⋅ 𝑑 𝐴⃗ = 2 lim ∫𝑆 𝑅 Δ𝐴→0 𝑅 𝑅 This result is known as Gauss’s law. To compute a flux with an integral instead of Riemann sums, we often write 𝑑 𝐴⃗ = 𝑛⃗ 𝑑𝐴, as in the next example. Example 6

Let 𝑆 be the surface of the cube bounded by the six planes 𝑥 = ±1, 𝑦 = ±1, and 𝑧 = ±1. Compute the flux of the electric field 𝐸⃗ of the previous example outward through 𝑆.

Solution

It is enough to compute the flux of 𝐸⃗ through a single face, say the top face 𝑆1 defined by 𝑧 = 1, where −1 ≤ 𝑥 ≤ 1 and −1 ≤ 𝑦 ≤ 1. By symmetry, the flux of 𝐸⃗ through the other five faces of 𝑆 must be the same. ⃗ 𝑑𝑥 𝑑𝑦 and On the top face, 𝑆1 , we have 𝑑 𝐴⃗ = 𝑛⃗ 𝑑𝐴 = 𝑘 ⃗ 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑘 . 𝐸⃗ (𝑥, 𝑦, 1) = 𝑞 2 2 (𝑥 + 𝑦 + 1)3∕2

968

Chapter 19 FLUX INTEGRALS AND DIVERGENCE

The corresponding flux integral is given by 1

∫𝑆1

𝐸⃗ ⋅ 𝑑 𝐴⃗ = 𝑞

1

∫−1 ∫−1

1 1 ⃗ 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑘 1 ⃗ 𝑑𝑥 𝑑𝑦 = 𝑞 ⋅𝑘 𝑑𝑥 𝑑𝑦. 2 2 3∕2 2 2 3∕2 ∫ ∫ (𝑥 + 𝑦 + 1) −1 −1 (𝑥 + 𝑦 + 1)

Computing this integral numerically shows that Flux through top face =

∫𝑆1

𝐸⃗ ⋅ 𝑑 𝐴⃗ ≈ 2.0944𝑞.

Thus, Total flux of 𝐸⃗ out of cube =

∫𝑆

𝐸⃗ ⋅ 𝑑 𝐴⃗ ≈ 6(2.0944𝑞) = 12.5664𝑞.

Example 5 on page 967 showed that the flux of 𝐸⃗ through a sphere of radius 𝑅 centered at the origin is 4𝜋𝑞. Since 4𝜋 ≈ 12.5664, Example 6 suggests that Total flux of 𝐸⃗ out of cube = 4𝜋𝑞. By computing the flux integral in Example 6 exactly, it is possible to verify that the flux of 𝐸⃗ through the cube and the sphere are exactly equal. When we encounter the Divergence Theorem in Chapter 20 we will see why this is so.

Notes on Orientation Two difficulties can occur in choosing an orientation. The first is that if the surface is not smooth, it may not have a normal vector at every point. For example, a cube does not have a normal vector along its edges. When we have a surface, such as a cube, which is made of a finite number of smooth pieces, we choose an orientation for each piece separately. The best way to do this is usually clear. For example, on the cube we choose the outward orientation on each face. (See Figure 19.14.)

𝑃

✠ 𝑛⃗

Figure 19.14: The orientation vector field 𝑛⃗ on the cube surface determined by the choice of unit normal vector at the point 𝑃

Figure 19.15: The Möbius strip is an example of a non-orientable surface

The second difficulty is that there are some surfaces which cannot be oriented at all, such as the Möbius strip in Figure 19.15.

Exercises and Problems for Section 19.1 Online Resource: Additional Problems for Section 19.1 EXERCISES

In Exercises 1–4, find the area vector of the oriented flat surface.

2. The circular disc of radius 5 in the 𝑥𝑦-plane, oriented upward.

1. The triangle with vertices (0, 0, 0), (0, 2, 0), (0, 0, 3) oriented in the negative 𝑥 direction.

3. 𝑦 = 10, 0 ≤ 𝑥 ≤ 5, 0 ≤ 𝑧 ≤ 3, oriented away from the 𝑥𝑧-plane. 4. 𝑦 = −10, 0 ≤ 𝑥 ≤ 5, 0 ≤ 𝑧 ≤ 3, oriented away from the 𝑥𝑧-plane.

969

19.1 THE IDEA OF A FLUX INTEGRAL

5. Find an oriented flat surface with area vector 150𝑗⃗ .

14.

(0, 0, 2)

(2, 2, 2) (2, 0, 2)

(3, 2, 4)

𝑦

(3, 0, 0)

𝑥 𝑧

15.

⃗ ) ⋅ 𝑑 𝐴⃗ , where 𝑆 is the disk of ra(2⃗𝑖 + 3𝑘 ∫𝑆 dius 4 perpendicular to the 𝑥-axis, centered at (5, 0, 0) and oriented

(0, 2, 3)

(0, 0, 3)

(2, 0, 4)

(3, 2, 0) 𝑥

𝑧

16. (0, 0, 4)

(b) Away from the origin.

(3, 0, 4)

𝑦

7. Compute

(a) Toward the origin.

𝑧

(0, 2, 2)

6. Let 𝑆 be the disk of radius 3 perpendicular the the 𝑦axis, centered at (0, 6, 0) and oriented away from the origin. Is (𝑥⃗𝑖 + 𝑦𝑗⃗ ) ⋅ 𝑑 𝐴⃗ a vector or a scalar? ∫𝑆 ⃗ ) ⋅ 𝑑 𝐴⃗ , where 𝑆 is the square of (4⃗𝑖 + 5𝑘 ∫𝑆 side length 3 perpendicular to the 𝑧-axis, centered at (0, 0, −2) and oriented

𝑧

13.

8. Compute

(a) Toward the origin.

𝑧

𝑧

(b)

𝑦

𝑥

𝑥 (2, 2, 0)

(2, 0, 0)

𝑥

For Exercises 17–20 find the flux of the constant vector field ⃗ through the given surface. 𝑣⃗ = ⃗𝑖 − 𝑗⃗ + 3𝑘 17. A disk of radius 2 in the 𝑥𝑦-plane oriented upward. 18. A triangular plate of area 4 in the 𝑦𝑧-plane oriented in the positive 𝑥-direction. 19. A square plate of area 4 in the 𝑦𝑧-plane oriented in the positive 𝑥-direction. 20. The triangular plate with vertices (1, 0, 0), (0, 1, 0), (0, 0, 1), oriented away from the origin.

𝑦

𝑥 𝑧

(c)

𝑦 (2, 2, 0)

(b) Away from the origin.

In Exercises 9–12, for each of the surfaces in (a)–(e), say whether the flux of 𝐹⃗ through the surface is positive, negative, or zero. The normal vector shows the orientation. (a)

(0, 2, 0) 𝑦

⃗ = 2⃗𝑖 + 3𝑗⃗ + 5𝑘 ⃗ In Exercises 21–23, find the flux of 𝐻 through the surface 𝑆.

𝑧

(d)

21. 𝑆 is the cylinder 𝑥2 + 𝑦2 = 1, closed at the ends by the planes 𝑧 = 0 and 𝑧 = 1 and oriented outward.

𝑦 𝑥 𝑦 𝑥

23. 𝑆 is the disk of radius 1 in the plane 𝑥 + 𝑦 + 𝑧 = 1 oriented in upward.

𝑧

(e)

22. 𝑆 is the disk of radius 1 in the plane 𝑥 = 2 oriented in the positive 𝑥-direction.

Find the flux of the vector fields in Exercises 24–26 out of the closed box 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 2, 0 ≤ 𝑧 ≤ 3. 𝑦

⃗ 24. 𝐹⃗ = 3⃗𝑖 + 2𝑗⃗ + 𝑘

25. 𝐺⃗ = 𝑥⃗𝑖

𝑥

⃗ = 𝑧𝑥𝑘 ⃗ 26. 𝐻 ⃗. 9. 𝐹⃗ (𝑥, 𝑦, 𝑧) = ⃗𝑖 + 2𝑗⃗ + 𝑘 10. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑧⃗𝑖 . ⃗. 11. 𝐹⃗ (𝑥, 𝑦, 𝑧) = −𝑧⃗𝑖 + 𝑥𝑘 12. 𝐹⃗ (⃗𝑟 ) = 𝑟⃗ . ⃗ In Exercises 13–16, compute the flux of 𝑣⃗ = ⃗𝑖 + 2𝑗⃗ − 3𝑘 through the rectangular region with the orientation shown.

In Exercises 27–30, calculate the flux integral. (𝑥⃗𝑖 +4𝑗⃗ )⋅𝑑 𝐴⃗ where 𝑆 is the disk of radius 5 perpen∫𝑆 dicular to the 𝑥-axis, centered at (3, 0, 0) and oriented toward the origin. 28. ∫𝑆 𝑟⃗ ⋅ 𝑑 𝐴⃗ where 𝑆 is the sphere of radius 3 centered at the origin. ⃗ ) ⋅ 𝑑 𝐴⃗ where 𝑆 is a disk of 29. ∫𝑆 (sin 𝑥 ⃗𝑖 + (𝑦2 + 𝑧2 )𝑗⃗ + 𝑦2 𝑘 radius 𝜋 in the plane 𝑥 = 3𝜋∕2, oriented in the positive 𝑥-direction. ⃗ ) ⋅ 𝑑 𝐴⃗ where 𝑆 is a disk of radius 3 in 30. ∫𝑆 (5⃗𝑖 + 5𝑗⃗ + 5𝑘 the plane 𝑥 + 𝑦 + 𝑧 = 1, oriented upward. 27.

970

Chapter 19 FLUX INTEGRALS AND DIVERGENCE

In Exercises 31–34, calculate the flux integral using a shortcut arising from two special cases: • If 𝐹⃗ is tangent at every point of 𝑆, then ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 0. • If 𝐹⃗ is perpendicular at every point of 𝑆 and has constant magnitude on 𝑆, then ∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ = ±‖𝐹⃗ ‖ ⋅ Area of 𝑆.

Choose the positive sign if 𝐹⃗ points in the same direction as the orientation of 𝑆; choose the negative sign if 𝐹⃗ points in the direction opposite the orientation of 𝑆. 31. ∫𝑆 (𝑥⃗𝑖 + 𝑦𝑗⃗ ) ⋅ 𝑑 𝐴⃗ , where 𝑆 is the cylinder of radius 10, centered on the 𝑧-axis between 𝑧 = 0 and 𝑧 = 10 and oriented away from the 𝑧-axis. 32. ∫𝑆 (−𝑦⃗𝑖 + 𝑥𝑗⃗ ) ⋅ 𝑑 𝐴⃗ , where 𝑆 is the cylinder of radius 10, centered on the 𝑧-axis between 𝑧 = 0 and 𝑧 = 10 and oriented away from the 𝑧-axis. ⃗ ) ⋅ 𝑑 𝐴⃗ , where 𝑆 is the sphere of radius 33. ∫𝑆 (𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 20, centered at the origin and oriented outward. 34. ∫𝑆 (−𝑦⃗𝑖 + 𝑥𝑗⃗ ) ⋅ 𝑑 𝐴⃗ , where 𝑆 is the sphere of radius 20, centered at the origin and oriented outward. In Exercises 35–57, calculate the flux of the vector field through the surface. 35. 𝐹⃗ = 2⃗𝚤 + 3⃗𝚥 through the square of side 𝜋 in the 𝑥𝑦plane, oriented upward. 36. 𝐹⃗ = 2⃗𝚤 + 3⃗𝚥 through the unit disk in the 𝑦𝑧-plane, centered at the origin and oriented in the positive 𝑥direction. ⃗ through the square of side 1.6 cen37. 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 tered at (2, 5, 8), parallel to the 𝑥𝑧-plane and oriented away from the origin. √ ⃗ through a square of side 14 in a horizontal 38. 𝐹⃗ = 𝑧𝑘 plane 2 units below the 𝑥𝑦-plane and oriented downward. 39. 𝐹⃗ = −𝑦⃗𝑖 + 𝑥𝑗⃗ and 𝑆 is the square plate in the 𝑦𝑧plane with corners at (0, 1, 1), (0, −1, 1), (0, 1, −1), and (0, −1, −1), oriented in the positive 𝑥-direction. ⃗ and 𝑆 is a disk of radius 2 in the 𝑦𝑧40. 𝐹⃗ = 7⃗𝑖 + 6𝑗⃗ + 5𝑘 plane, centered at the origin and oriented in the positive 𝑥-direction.

⃗ and 𝑆 is a square of side 2 in the 41. 𝐹⃗ = 𝑥⃗𝑖 + 2𝑦𝑗⃗ + 3𝑧𝑘 plane 𝑦 = 3, oriented in the positive 𝑦-direction. ⃗ and 𝑆 is a sphere of radius 𝜋 centered 42. 𝐹⃗ = 7⃗𝑖 +6𝑗⃗ +5𝑘 at the origin. 43. 𝐹⃗ = −5⃗𝑟 through the sphere of radius 2 centered at the origin. ⃗ and 𝑆 is the rectangle 𝑧 = 4, 44. 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + (𝑧2 + 3)𝑘 0 ≤ 𝑥 ≤ 2, 0 ≤ 𝑦 ≤ 3, oriented in the positive 𝑧direction. 45. 𝐹⃗ = 6⃗𝑖 + 7𝑗⃗ through a triangle of area 10 in the plane 𝑥 + 𝑦 = 5, oriented in the positive 𝑥-direction. ⃗ , through the square of side 4 in 46. 𝐹⃗ = 6⃗𝑖 + 𝑥2 𝑗⃗ − 𝑘 the plane 𝑦 = 3, centered on the 𝑦-axis, with sides parallel to the 𝑥 and 𝑧 axes, and oriented in the positive 𝑦-direction. ⃗ through the rectangle 47. 𝐹⃗ = (𝑥+3)⃗𝑖 +(𝑦+5)𝑗⃗ +(𝑧+7)𝑘 𝑥 = 4, 0 ≤ 𝑦 ≤ 2, 0 ≤ 𝑧 ≤ 3, oriented in the positive 𝑥-direction. 48. 𝐹⃗ = 7⃗𝑟 through the sphere of radius 3 centered at the origin. 49. 𝐹⃗ = −3⃗𝑟 through the sphere of radius 2 centered at the origin. ⃗ through the rectangle 𝑥 = 4, 50. 𝐹⃗ = 2𝑧⃗𝑖 + 𝑥𝑗⃗ + 𝑥𝑘 0 ≤ 𝑦 ≤ 2, 0 ≤ 𝑧 ≤ 3, oriented in the positive 𝑥direction. 51. 𝐹⃗ = ⃗𝑖 + 2𝑗⃗ through a square of side 2 lying in the plane 𝑥 + 𝑦 + 𝑧 = 1, oriented away from the origin. ⃗ through the disk of radius 3 in the 52. 𝐹⃗ = (𝑥2 + 𝑦2 )𝑘 𝑥𝑦-plane, centered at the origin and oriented upward. ⃗ through the disk 𝑥2 + 𝑦2 ≤ 9 ori53. 𝐹⃗ = cos(𝑥2 + 𝑦2 )𝑘 ented upward in the plane 𝑧 = 1. 2 2 54. 𝐹⃗ = 𝑒𝑦 +𝑧 ⃗𝑖 through the disk of radius 2 in the 𝑦𝑧plane, centered at the origin and oriented in the positive 𝑥-direction. 55. 𝐹⃗ = −𝑦⃗𝑖 + 𝑥𝑗⃗ through the disk in the 𝑥𝑦-plane with radius 2, oriented upward and centered at the origin.

56. 𝐹⃗ = 𝑟⃗ through the disk of radius 2 parallel to the 𝑥𝑦plane oriented upward and centered at (0, 0, 2). 57. 𝐹⃗ = (2 − 𝑥)⃗𝑖 through the cube whose vertices include the points (0, 0, 0), (3, 0, 0), (0, 3, 0), (0, 0, 3), and oriented outward.

PROBLEMS 58. Let 𝐵 be the surface of a box centered at the origin, with edges parallel to the axes and in the planes 𝑥 = ±1, 𝑦 = ±1, 𝑧 = ±1, and let 𝑆 be the sphere of radius 1 centered at origin. (a) Indicate whether the following flux integrals are positive, negative, or zero. No reasons needed. (b) ∫ 𝑦⃗𝑖 ⋅ 𝑑 𝐴⃗ (a) ∫ 𝑥⃗𝑖 ⋅ 𝑑 𝐴⃗ 𝐵

𝐵

(c)

∫𝑆 |𝑥|⃗𝑖 ⋅ 𝑑 𝐴⃗

(d) ∫𝑆 (𝑦 − 𝑥)⃗𝑖 ⋅ 𝑑 𝐴⃗

(b) Explain with reasons how you know which flux integral is greater:

∫𝑆

𝑥⃗𝑖 ⋅ 𝑑 𝐴⃗

or

∫𝐵

𝑥⃗𝑖 ⋅ 𝑑 𝐴⃗ ?

19.1 THE IDEA OF A FLUX INTEGRAL

971

59. Suppose that 𝐸⃗ is a uniform electric field on 3-space, ⃗ , for all points (𝑥, 𝑦, 𝑧), so 𝐸⃗ (𝑥, 𝑦, 𝑧) = 𝑎⃗𝚤 + 𝑏⃗𝚥 + 𝑐 𝑘 where 𝑎, 𝑏, 𝑐 are constants. Show, with the aid of symmetry, that the flux of 𝐸⃗ through each of the following closed surfaces 𝑆 is zero:

64. Find the flux of 𝐹⃗ = 𝑟⃗ ∕||𝑟||2 out of the sphere of radius 𝑅 centered at the origin. 65. Consider the flux of the vector field 𝐹⃗ = 𝑟⃗ ∕||⃗𝑟 ||𝑝 for 𝑝 ≥ 0 out of the sphere of radius 2 centered at the origin.

(a) 𝑆 is the cube bounded by the planes 𝑥 = ±1, 𝑦 = ±1, and 𝑧 = ±1. (b) 𝑆 is the sphere 𝑥2 + 𝑦2 + 𝑧2 = 1. (c) 𝑆 is the cylinder bounded by 𝑥2 + 𝑦2 = 1, 𝑧 = 0, and 𝑧 = 2.

66. Let 𝑆 be the cube with side length 2, faces parallel to the coordinate planes, and centered at the origin.

60. Water is flowing down a cylindrical pipe of radius 2 cm; its speed is (3 − (3∕4)𝑟2 ) cm/sec at a distance 𝑟 cm from the center of the pipe. Find the flux through the circular cross section of the pipe, oriented so that the flux is positive. 61. (a) What do you think will be the electric flux through the cylindrical surface that is placed as shown in the constant electric field in Figure 19.16? Why? (b) What if the cylinder is placed upright, as shown in Figure 19.17? Explain.

Figure 19.16

Figure 19.17

(a) For what value of 𝑝 is the flux a maximum? (b) What is that maximum value?

(a) Calculate the total flux of the constant vector field ⃗ out of 𝑆 by computing the flux 𝑣⃗ = −⃗𝑖 + 2𝑗⃗ + 𝑘 through each face separately. (b) Calculate the flux out of 𝑆 for any constant vector ⃗. field 𝑣⃗ = 𝑎⃗𝑖 + 𝑏𝑗⃗ + 𝑐 𝑘 (c) Explain why the answers to parts (a) and (b) make sense. 67. Let 𝑆 be the tetrahedron with vertices at the origin and at (1, 0, 0), (0, 1, 0) and (0, 0, 1). (a) Calculate the total flux of the constant vector field ⃗ out of 𝑆 by computing the flux 𝑣⃗ = −⃗𝑖 + 2𝑗⃗ + 𝑘 through each face separately. (b) Calculate the flux out of 𝑆 in part (a) for any constant vector field 𝑣⃗ . (c) Explain why the answers to parts (a) and (b) make sense. 68. Let 𝑃 (𝑥, 𝑦, 𝑧) be the pressure at the point (𝑥, 𝑦, 𝑧) in a ⃗ . Let 𝑆 be the surfluid. Let 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑃 (𝑥, 𝑦, 𝑧)𝑘 face of a body submerged in the fluid. If 𝑆 is oriented inward, show that ∫𝑆 𝐹⃗ ⋅𝑑 𝐴⃗ is the buoyant force on the body, that is, the force upward on the body due to the pressure of the fluid surrounding it. [Hint: 𝐹⃗ ⋅ 𝑑 𝐴⃗ = ⃗ ⋅ 𝑑 𝐴⃗ = (𝑃 (𝑥, 𝑦, 𝑧) 𝑑 𝐴⃗ ) ⋅ 𝑘 ⃗ .] 𝑃 (𝑥, 𝑦, 𝑧)𝑘 69. A region of 3-space has a temperature which varies from point to point. Let 𝑇 (𝑥, 𝑦, 𝑧) be the temperature at a point (𝑥, 𝑦, 𝑧). Newton’s law of cooling says that grad 𝑇 is proportional to the heat flow vector field, 𝐹⃗ , where 𝐹⃗ points in the direction in which heat is flowing and has magnitude equal to the rate of flow of heat. (a) Suppose 𝐹⃗ = 𝑘 grad 𝑇 for some constant 𝑘. What is the sign of 𝑘? (b) Explain why this form of Newton’s law of cooling makes sense. (c) Let 𝑊 be a region of space bounded by the surface 𝑆. Explain why Rate of heat

62. Let 𝑆 be part of a cylinder centered on the 𝑦-axis. Ex⃗ have plain why the three vectors fields 𝐹⃗ , 𝐺⃗ , and 𝐻 the same flux through 𝑆. Do not compute the flux. ⃗ 𝐹⃗ = 𝑥⃗𝑖 + 2𝑦𝑧𝑘 ⃗ 𝐺⃗ = 𝑥⃗𝑖 + 𝑦 sin 𝑥𝑗⃗ + 2𝑦𝑧𝑘 ⃗ = 𝑥⃗𝑖 + cos(𝑥2 + 𝑧)𝑗⃗ + 2𝑦𝑧𝑘 ⃗ 𝐻 63. Find the flux of 𝐹⃗ = 𝑟⃗ ∕‖⃗𝑟 ‖3 out of the sphere of radius 𝑅 centered at the origin.

loss from 𝑊

=𝑘

∫𝑆

(grad 𝑇 ) ⋅ 𝑑 𝐴⃗ .

70. The 𝑧-axis carries a constant electric charge density of 𝜆 units of charge per unit length, with 𝜆 > 0. The resulting electric field is 𝐸⃗ . (a) Sketch the electric field, 𝐸⃗ , in the 𝑥𝑦-plane, given 𝑥⃗𝑖 + 𝑦𝑗⃗ . 𝐸⃗ (𝑥, 𝑦, 𝑧) = 2𝜆 2 𝑥 + 𝑦2

972

Chapter 19 FLUX INTEGRALS AND DIVERGENCE

(b) Compute the flux of 𝐸⃗ outward through the cylinder 𝑥2 + 𝑦2 = 𝑅2 , for 0 ≤ 𝑧 ≤ ℎ. 71. An infinitely long straight wire lying along the 𝑧-axis ⃗ direction. carries an electric current 𝐼 flowing in the 𝑘 Ampère’s Law in magnetostatics says that the current gives rise to a magnetic field 𝐵⃗ given by 𝐼 −𝑦⃗𝑖 + 𝑥𝑗⃗ . 𝐵⃗ (𝑥, 𝑦, 𝑧) = 2𝜋 𝑥2 + 𝑦2

(a) Sketch the field 𝐵⃗ in the 𝑥𝑦-plane. (b) Let 𝑆1 be the disk with center at (0, 0, ℎ), radius 𝑎, ⃗ diand parallel to the 𝑥𝑦-plane, oriented in the 𝑘 ⃗ rection. What is the flux of 𝐵 through 𝑆1 ? Does your answer seem reasonable? (c) Let 𝑆2 be the rectangle given by 𝑥 = 0, 𝑎 ≤ 𝑦 ≤ 𝑏, 0 ≤ 𝑧 ≤ ℎ, and oriented in the −⃗𝑖 direction. What is the flux of 𝐵⃗ through 𝑆2 ? Does your answer seem reasonable?

Strengthen Your Understanding In Problems 72–73, explain what is wrong with the statement. 72. For a certain vector field 𝐹⃗ and oriented surface 𝑆, we ⃗. have ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 2⃗𝑖 − 3𝑗⃗ + 𝑘 73. If 𝑆 is a region in the 𝑥𝑦-plane oriented upwards then ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ > 0.

85. If ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ > ∫𝑆 𝐺⃗ ⋅ 𝑑 𝐴⃗ then ||𝐹⃗ || > ||𝐺⃗ || at all points on the surface 𝑆.

86. For each of the surfaces in (a)–(e), pick the vector field 𝐹⃗ 1 , 𝐹⃗ 2 , 𝐹⃗ 3 , 𝐹⃗ 4 , 𝐹⃗ 5 , with the largest flux through the surface. The surfaces are all squares of the same size. Note that the orientation is shown.

In Problems 74–75, give an example of: 74. A nonzero vector field 𝐹⃗ such that ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 0, where 𝑆 is the triangular surface with corners (1, 0, 0), (0, 1, 0), (0, 0, 1), oriented away from the origin.

⃗ 𝐹⃗ 1 = 2⃗𝑖 − 3𝑗⃗ − 4𝑘 ⃗ ⃗ 𝐹 2 = ⃗𝑖 − 2𝑗⃗ + 7𝑘 ⃗ 𝐹⃗ 3 = −7⃗𝑖 + 5𝑗⃗ + 6𝑘

75. A nonconstant vector field 𝐹⃗ (𝑥, 𝑦, 𝑧) and an oriented surface 𝑆 such that ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 1.

⃗ 𝐹⃗ 4 = −11⃗𝑖 + 4𝑗⃗ − 5𝑘 ⃗ 𝐹⃗ 5 = −5⃗𝑖 + 3𝑗⃗ + 5𝑘

Are the statements in Problems 76–85 true or false? Give reasons for your answer.

𝑧

(a)

𝑧

(b)

𝑦

𝑥

76. The value of a flux integral is a scalar. 77. The area vector 𝐴⃗ of a flat, oriented surface is parallel to the surface. 78. If 𝑆 is the unit sphere centered at the origin, oriented outward and the flux integral ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ is zero, then 𝐹⃗ = 0⃗ . 79. The flux of the vector field 𝐹⃗ = ⃗𝑖 through the plane 𝑥 = 0, with 0 ≤ 𝑦 ≤ 1, 0 ≤ 𝑧 ≤ 1, oriented in the ⃗𝑖 direction is positive. 80. If 𝑆 is the unit sphere centered at the origin, oriented ⃗ = 𝑟⃗ , then the flux outward and 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 integral ∫ 𝐹⃗ ⋅ 𝑑 𝐴⃗ is positive.

𝑦

𝑥

𝑧

(c)

𝑦 𝑥 𝑦 𝑥

𝑆

81. If 𝑆 is the cube bounded by the six planes 𝑥 = ±1, 𝑦 = ⃗ , then ±1, 𝑧 = ±1, oriented outward, and 𝐹⃗ = 𝑘 ∫ 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 0.

𝑧

(e)

𝑆

82. If 𝑆 is an oriented surface in 3-space, and −𝑆 is the same surface, but with the opposite orientation, then ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ = − ∫−𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ . 83. If 𝑆1 is a rectangle with area 1 and 𝑆2 is a rectangle with area 2, then 2∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ = ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ . 1

2

84. If 𝐹⃗ = 2𝐺⃗ , then ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 2 ∫𝑆 𝐺⃗ ⋅ 𝑑 𝐴⃗ .

𝑧

(d)

𝑦 𝑥

19.2 FLUX INTEGRALS FOR GRAPHS, CYLINDERS, AND SPHERES

19.2

973

FLUX INTEGRALS FOR GRAPHS, CYLINDERS, AND SPHERES In Section 19.1 we computed flux integrals in certain simple cases. In this section we see how to compute flux through surfaces that are graphs of functions, through cylinders, and through spheres.

Flux of a Vector Field Through the Graph of 𝒛 = 𝒇 (𝒙, 𝒚) Suppose 𝑆 is the graph of the differentiable function 𝑧 = 𝑓 (𝑥, 𝑦), oriented upward, and that 𝐹⃗ is a smooth vector field. In Section 19.1 we subdivided the surface into small pieces with area vector Δ𝐴⃗ and defined the flux of 𝐹⃗ through 𝑆 as follows: ∑ 𝐹⃗ ⋅ 𝑑 𝐴⃗ = lim 𝐹⃗ ⋅ Δ𝐴⃗ . ∫𝑆 ‖Δ𝐴⃗ ‖→0 How do we divide 𝑆 into small pieces? One way is to use the cross sections of 𝑓 with 𝑥 or 𝑦 constant and take the patches in a wire frame representation of the surface. So we must calculate the area vector of one of these patches, which is approximately a parallelogram.

The Area Vector of a Coordinate Patch ⃗ has magAccording to the geometric definition of the cross product on page 729, the vector 𝑣⃗ × 𝑤 ⃗ and direction perpendicular to this nitude equal to the area of the parallelogram formed by 𝑣⃗ and 𝑤 parallelogram and determined by the right-hand rule. Thus, we have ⃗. Area vector of parallelogram = 𝐴⃗ = 𝑣⃗ × 𝑤 𝑟⃗ 𝑦

Coordinate patch

𝑟⃗ 𝑥

𝑧

✠ ✠

𝑣⃗ 𝑦

𝑣⃗ 𝑥 Δ𝑦

❘ Δ𝑥

Δ𝑥

𝑥 Figure 19.18: Surface showing coordinate patch and tangent vectors 𝑟⃗ 𝑥 and 𝑟⃗ 𝑦

Δ𝑦 Figure 19.19: Parallelogram-shaped patch in the tangent plane to the surface

Consider the patch of surface above the rectangular region with sides Δ𝑥 and Δ𝑦 in the 𝑥𝑦plane shown in Figure 19.18. We approximate the area vector, Δ𝐴⃗ , of this patch by the area vector of the corresponding patch on the tangent plane to the surface. See Figure 19.19. This patch is the parallelogram determined by the vectors 𝑣⃗ 𝑥 and 𝑣⃗ 𝑦 , so its area vector is given by Δ𝐴⃗ ≈ 𝑣⃗ 𝑥 × 𝑣⃗ 𝑦 . ⃗. To find 𝑣⃗ 𝑥 and 𝑣⃗ 𝑦 , notice that a point on the surface has position vector 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑓 (𝑥, 𝑦)𝑘 Thus, a cross section of 𝑆 with 𝑦 constant has tangent vector 𝜕⃗𝑟 ⃗, = ⃗𝑖 + 𝑓𝑥 𝑘 𝜕𝑥 and a cross section with 𝑥 constant has tangent vector 𝑟⃗ 𝑥 =

𝑟⃗ 𝑦 =

𝜕⃗𝑟 ⃗. = 𝑗⃗ + 𝑓𝑦 𝑘 𝜕𝑦

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Chapter 19 FLUX INTEGRALS AND DIVERGENCE

The vectors 𝑟⃗ 𝑥 and 𝑣⃗ 𝑥 are parallel because they are both tangent to the surface and parallel to the 𝑥𝑧-plane. Since the 𝑥-component of 𝑟⃗ 𝑥 is ⃗𝑖 and the 𝑥-component of 𝑣⃗ 𝑥 is (Δ𝑥)⃗𝑖 , we have 𝑣⃗ 𝑥 = (Δ𝑥)⃗𝑟 𝑥 . Similarly, we have 𝑣⃗ 𝑦 = (Δ𝑦)⃗𝑟 𝑦 . So the upward-pointing area vector of the parallelogram is ( ) ) ( ⃗ Δ𝑥 Δ𝑦. Δ𝐴⃗ ≈ 𝑣⃗ 𝑥 × 𝑣⃗ 𝑦 = 𝑟⃗ 𝑥 × 𝑟⃗ 𝑦 Δ𝑥 Δ𝑦 = −𝑓𝑥 ⃗𝑖 − 𝑓𝑦 𝑗⃗ + 𝑘 This is our approximation for the area vector Δ𝐴⃗ on the surface. Replacing Δ𝐴⃗ , Δ𝑥, and Δ𝑦 by 𝑑 𝐴⃗ , 𝑑𝑥 and 𝑑𝑦, we write ( ) ⃗ 𝑑𝑥 𝑑𝑦. 𝑑 𝐴⃗ = −𝑓𝑥 ⃗𝑖 − 𝑓𝑦 𝑗⃗ + 𝑘

The Flux of 𝑭⃗ Through a Surface Given by a Graph of 𝒛 = 𝒇 (𝒙, 𝒚) Suppose the surface 𝑆 is the part of the graph of 𝑧 = 𝑓 (𝑥, 𝑦) above2 a region 𝑅 in the 𝑥𝑦-plane, and suppose 𝑆 is oriented upward. The flux of 𝐹⃗ through 𝑆 is ∫𝑆

Example 1

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

∫𝑅

( ) ⃗ 𝑑𝑥 𝑑𝑦. 𝐹⃗ (𝑥, 𝑦, 𝑓 (𝑥, 𝑦)) ⋅ −𝑓𝑥 ⃗𝑖 − 𝑓𝑦 𝑗⃗ + 𝑘

⃗ and 𝑆 is the rectangular plate with corners (0, 0, 0), Compute ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ where 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑧𝑘 (1, 0, 0), (0, 1, 3), (1, 1, 3), oriented upward. See Figure 19.20. (0, 1, 3)

𝑧

(1, 1, 3)

𝑦

(1, 0, 0) 𝑥 ⃗ on the rectangular surface 𝑆 Figure 19.20: The vector field 𝐹⃗ = 𝑧𝑘

Solution

We find the equation for the plane 𝑆 in the form 𝑧 = 𝑓 (𝑥, 𝑦). Since 𝑓 is linear, with 𝑥-slope equal to 0 and 𝑦-slope equal to 3, and 𝑓 (0, 0) = 0, we have 𝑧 = 𝑓 (𝑥, 𝑦) = 0 + 0𝑥 + 3𝑦 = 3𝑦. Thus, we have ⃗ ) 𝑑𝑥 𝑑𝑦 = (0⃗𝑖 − 3𝑗⃗ + 𝑘 ⃗ ) 𝑑𝑥 𝑑𝑦 = (−3𝑗⃗ + 𝑘 ⃗ ) 𝑑𝑥 𝑑𝑦. 𝑑 𝐴⃗ = (−𝑓𝑥 ⃗𝑖 − 𝑓𝑦 𝑗⃗ + 𝑘 The flux integral is therefore 1

∫𝑆

2 The

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

∫0 ∫0

1

1

⃗ ⋅ (−3𝑗⃗ + 𝑘 ⃗ ) 𝑑𝑥 𝑑𝑦 = 3𝑦𝑘

formula is also correct when the graph is below the region 𝑅.

∫0 ∫0

1

3𝑦 𝑑𝑥 𝑑𝑦 = 1.5.

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Surface Area of a Graph Since the magnitude of an area vector is area, we can find area of a surface by integrating the magnitude ‖𝑑 𝐴⃗ ‖. If a surface is the graph of a function 𝑧 = 𝑓 (𝑥, 𝑦), we have √ ⃗ ‖ 𝑑𝑥 𝑑𝑦 = (𝑓𝑥 )2 + (𝑓𝑦 )2 + 1 𝑑𝑥 𝑑𝑦. ‖𝑑 𝐴⃗ ‖ = ‖ − 𝑓𝑥 ⃗𝑖 − 𝑓𝑦 𝑗⃗ + 𝑘 Thus we have the following result:

Suppose a surface 𝑆 is the part of the graph 𝑧 = 𝑓 (𝑥, 𝑦) where (𝑥, 𝑦) is in a region 𝑅 in the 𝑥𝑦-plane. Then √ Area of 𝑆 = (𝑓𝑥 )2 + (𝑓𝑦 )2 + 1 𝑑𝑥 𝑑𝑦. ∫𝑅

Example 2

Find the area of the surface 𝑧 = 𝑓 (𝑥, 𝑦) where 0 ≤ 𝑥 ≤ 4, 0 ≤ 𝑦 ≤ 5, when: (a) 𝑓 (𝑥, 𝑦) = 2𝑥 + 3𝑦 + 4 (b) 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2

Solution

(a) Since 𝑓𝑥 = 2 and 𝑓𝑦 = 3, we have 5

Area =

∫0 ∫0

4√

√ 22 + 32 + 1 𝑑𝑥 𝑑𝑦 = 20 14.

(b) Since 𝑓𝑥 = 2𝑥 and 𝑓𝑦 = 2𝑦. we have 5

Area =

∫0 ∫0

4√

4𝑥2 + 4𝑦2 + 1 𝑑𝑥 𝑑𝑦 = 140.089.

Surface area integrals can often only be evaluated numerically.

Flux of a Vector Field Through a Cylindrical Surface Consider the cylinder of radius 𝑅 centered on the 𝑧-axis illustrated in Figure 19.21 and oriented away from the 𝑧-axis. The coordinate patch in Figure 19.22 has surface area given by Δ𝐴 ≈ 𝑅 Δ𝜃 Δ𝑧. 𝑧 𝑧 Coordinate patch



Δ𝜃 𝑅



✛ 𝑅

✛ 𝑟⃗ 𝑧

𝜃 ✻ 𝑅Δ Δ𝑧



❄ 𝑛⃗

𝑟⃗ 𝜃

𝑦

𝑆 𝑛⃗ 𝑥

𝑦 Figure 19.21: Outward-oriented cylinder

𝜃

𝑥 Figure 19.22: Coordinate patch with area Δ𝐴⃗ on surface of a cylinder

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Chapter 19 FLUX INTEGRALS AND DIVERGENCE

The outward unit normal 𝑛⃗ points in the direction of 𝑥⃗𝑖 + 𝑦𝑗⃗ , so 𝑛⃗ =

𝑅 cos 𝜃⃗𝑖 + 𝑅 sin 𝜃 𝑗⃗ 𝑥⃗𝑖 + 𝑦𝑗⃗ = = cos 𝜃⃗𝑖 + sin 𝜃 𝑗⃗ . 𝑅 ‖𝑥⃗𝑖 + 𝑦𝑗⃗ ‖

Therefore, the area vector of the coordinate patch is approximated by ( ) Δ𝐴⃗ = 𝑛⃗ Δ𝐴 ≈ cos 𝜃⃗𝑖 + sin 𝜃 𝑗⃗ 𝑅 Δ𝑧 Δ𝜃. Replacing Δ𝐴⃗ , Δ𝑧, and Δ𝜃 by 𝑑 𝐴⃗ , 𝑑𝑧, and 𝑑𝜃, we write ( ) 𝑑 𝐴⃗ = cos 𝜃⃗𝑖 + sin 𝜃 𝑗⃗ 𝑅 𝑑𝑧 𝑑𝜃. This gives the following result:

The Flux of a Vector Field Through a Cylinder The flux of 𝐹⃗ through the cylindrical surface 𝑆, of radius 𝑅 and oriented away from the 𝑧-axis, is given by ∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

∫𝑇

( ) 𝐹⃗ (𝑅, 𝜃, 𝑧) ⋅ cos 𝜃⃗𝑖 + sin 𝜃 𝑗⃗ 𝑅 𝑑𝑧 𝑑𝜃,

where 𝑇 is the 𝜃𝑧-region corresponding to 𝑆.

Example 3

Compute ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ where 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑦𝑗⃗ and 𝑆 is the part of the cylinder of radius 2 centered on the 𝑧-axis with 𝑥 ≥ 0, 𝑦 ≥ 0, and 0 ≤ 𝑧 ≤ 3. The surface is oriented toward the 𝑧-axis. 𝑧

𝑥 𝑦

Figure 19.23: The vector field 𝐹⃗ = 𝑦𝑗⃗ on the surface 𝑆

Solution

In cylindrical coordinates, we have 𝑅 = 2 and 𝐹⃗ = 𝑦𝑗⃗ = 2 sin 𝜃 𝑗⃗ . Since the orientation of 𝑆 is toward the 𝑧-axis, the flux through 𝑆 is given by 3

𝜋∕2

∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ = −

∫𝑇

2 sin 𝜃 𝑗⃗ ⋅ (cos 𝜃⃗𝑖 + sin 𝜃 𝑗⃗ )2 𝑑𝑧 𝑑𝜃 = −4

∫0

∫0

sin2 𝜃 𝑑𝑧 𝑑𝜃 = −3𝜋.

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Flux of a Vector Field Through a Spherical Surface Consider the piece of the sphere of radius 𝑅 centered at the origin, oriented outward, as illustrated in Figure 19.24. The coordinate patch in Figure 19.24 has surface area given by Δ𝐴 ≈ 𝑅2 sin 𝜙 Δ𝜙 Δ𝜃. ⃗ , so The outward unit normal 𝑛⃗ points in the direction of 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 𝑛⃗ =

𝑟⃗ ⃗ = sin 𝜙 cos 𝜃⃗𝑖 + sin 𝜙 sin 𝜃 𝑗⃗ + cos 𝜙𝑘. ‖⃗𝑟 ‖

Therefore, the area vector of the coordinate patch is approximated by ( ) 𝑟⃗ ⃗ 𝑅2 sin 𝜙 Δ𝜙 Δ𝜃. Δ𝐴⃗ ≈ 𝑛⃗ Δ𝐴 = Δ𝐴 = sin 𝜙 cos 𝜃⃗𝑖 + sin 𝜙 sin 𝜃 𝑗⃗ + cos 𝜙𝑘 ‖⃗𝑟 ‖ Replacing Δ𝐴⃗ , Δ𝜙, and Δ𝜃 by 𝑑 𝐴⃗ , 𝑑𝜙, and 𝑑𝜃, we write ( ) 𝑟⃗ ⃗ 𝑅2 sin 𝜙 𝑑𝜙 𝑑𝜃. 𝑑𝐴 = sin 𝜙 cos 𝜃⃗𝑖 + sin 𝜙 sin 𝜃 𝑗⃗ + cos 𝜙𝑘 𝑑 𝐴⃗ = ‖⃗𝑟 ‖

Thus, we obtain the following result:

The Flux of a Vector Field Through a Sphere The flux of 𝐹⃗ through the spherical surface 𝑆, with radius 𝑅 and oriented away from the origin, is given by ∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ = =

∫𝑆 ∫𝑇

𝑟⃗ 𝑑𝐴 𝐹⃗ ⋅ ‖⃗𝑟 ‖

( ) ⃗ 𝑅2 sin 𝜙 𝑑𝜙 𝑑𝜃, 𝐹⃗ (𝑅, 𝜃, 𝜙) ⋅ sin 𝜙 cos 𝜃⃗𝑖 + sin 𝜙 sin 𝜃 𝑗⃗ + cos 𝜙𝑘

where 𝑇 is the 𝜃𝜙-region corresponding to 𝑆.

𝑛⃗ 𝑧

(𝑅, 𝜃, 𝜙)

✻ 𝑅

✠⑥ ✛𝑅Δ𝜙 Coordinate patch ❘ ✛ 𝑅 sin 𝜙Δ𝜃 ❃ ✮

𝜙 Δ𝜙

𝑦

❄ Δ𝜃 𝜃

𝑥

Figure 19.24: Coordinate patch with area Δ𝐴⃗ on surface of a sphere

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Chapter 19 FLUX INTEGRALS AND DIVERGENCE

Example 4

⃗ through 𝑆, the upper hemisphere of radius 2 centered at the origin, oriented Find the flux of 𝐹⃗ = 𝑧𝑘 outward.

Solution

The hemisphere 𝑆 is parameterized by spherical coordinates 𝜃 and 𝜙, with 0 ≤ 𝜃 ≤ 2𝜋 and 0 ≤ 𝜙 ≤ ⃗ = 2 cos 𝜙𝑘 ⃗ , the flux is 𝜋∕2. Since 𝑅 = 2 and 𝐹⃗ = 𝑧𝑘 ∫𝑆

Example 5

⃗ ⋅ (sin 𝜙 cos 𝜃⃗𝑖 + sin 𝜙 sin 𝜃 𝑗⃗ + cos 𝜙𝑘 ⃗ )4 sin 𝜙 𝑑𝜙 𝑑𝜃 2 cos 𝜙𝑘 ) ( ( ) 𝜋∕2 2𝜋 − cos3 𝜙 ||𝜋∕2 16𝜋 2 = . 8 sin 𝜙 cos 𝜙 𝑑𝜙 𝑑𝜃 = 2𝜋 8 = | ∫0 ∫0 3 3 |𝜙=0

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

∫𝑆

The magnetic field 𝐵⃗ due to an ideal magnetic dipole, 𝜇⃗ , located at the origin is a multiple of 3(𝜇⃗ ⋅ 𝑟⃗ )⃗𝑟 𝜇⃗ + . 𝐵⃗ (⃗𝑟 ) = − 3 ‖⃗𝑟 ‖ ‖⃗𝑟 ‖5

Figure 19.25 shows a sketch of 𝐵⃗ in the plane 𝑧 = 0 for the dipole 𝜇⃗ = ⃗𝑖 . Notice that 𝐵⃗ is similar to the magnetic field of a bar magnet with its north pole at the tip of the vector ⃗𝑖 and its south pole at the tail of the vector ⃗𝑖 . Compute the flux of 𝐵⃗ outward through the sphere 𝑆 with center at the origin and radius 𝑅. 𝑦

𝑥

Figure 19.25: The magnetic field of a dipole, ⃗𝑖 , at the origin: 𝐵⃗ = −

Solution

⃗𝑖 3(⃗𝑖 ⋅ 𝑟⃗ )⃗𝑟 + 3 ‖⃗𝑟 ‖ ‖⃗𝑟 ‖5

Since ⃗𝑖 ⋅ 𝑟⃗ = 𝑥 and ‖⃗𝑟 ‖ = 𝑅 on the sphere of radius 𝑅, we have ( ) ( ) ⃗𝑖 ⋅ 𝑟⃗ )‖⃗𝑟 ‖2 ⃗𝑖 ⋅ 𝑟⃗ )⃗𝑟 ⃗𝑖 ⃗𝑖 ⋅ 𝑟⃗ 3( 3( 𝑟 ⃗ − ⋅ − 𝑑𝐴 𝑑𝐴 = + + 𝐵⃗ ⋅ 𝑑 𝐴⃗ = ∫𝑆 ∫𝑆 ∫𝑆 ‖⃗𝑟 ‖3 ‖⃗𝑟 ‖5 ‖⃗𝑟 ‖ ‖⃗𝑟 ‖4 ‖⃗𝑟 ‖6 =

2⃗𝑖 ⋅ 𝑟⃗ 2𝑥 2 𝑥 𝑑𝐴. 𝑑𝐴 = 𝑑𝐴 = 4 ∫𝑆 ‖⃗𝑟 ‖4 ∫𝑆 ‖⃗𝑟 ‖4 𝑅 ∫𝑆

But the sphere 𝑆 is centered at the origin. Thus, the contribution to the integral from each positive 𝑥-value is canceled by the contribution from the corresponding negative 𝑥-value; so ∫𝑆 𝑥 𝑑𝐴 = 0. Therefore, 2 𝑥 𝑑𝐴 = 0. 𝐵⃗ ⋅ 𝑑 𝐴⃗ = ∫𝑆 𝑅4 ∫𝑆

19.2 FLUX INTEGRALS FOR GRAPHS, CYLINDERS, AND SPHERES

979

Exercises and Problems for Section 19.2 EXERCISES In Exercises 1–4, find the area vector 𝑑 𝐴⃗ for the surface 𝑧 = 𝑓 (𝑥, 𝑦), oriented upward. 1. 𝑓 (𝑥, 𝑦) = 3𝑥 − 5𝑦

2. 𝑓 (𝑥, 𝑦) = 8𝑥 + 7𝑦

3. 𝑓 (𝑥, 𝑦) = 2𝑥2 − 3𝑦2

4. 𝑓 (𝑥, 𝑦) = 𝑥𝑦 + 𝑦2

⃗ 13. 𝐹⃗ (𝑥, 𝑦, 𝑧) = ⃗𝑖 + 2𝑗⃗ + 3𝑘 𝑆: radius 10, 𝑥 ≥ 0, 𝑦 ≥ 0, 0 ≤ 𝑧 ≤ 5 ⃗ 14. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥⃗𝑖 + 2𝑦𝑗⃗ + 3𝑧𝑘 𝑆: radius 10, 0 ≤ 𝑧 ≤ 5 ⃗ 15. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑧2 ⃗𝑖 + 𝑒𝑥 𝑗⃗ + 𝑘 𝑆: radius 6, inside sphere of radius 10

In Exercises 5–8, write an iterated integral for the flux of 𝐹⃗ through the surface 𝑆, which is the part of the graph of 𝑧 = 𝑓 (𝑥, 𝑦) corresponding to the region 𝑅, oriented upward. Do not evaluate the integral. ⃗ 5. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 10⃗𝑖 + 20𝑗⃗ + 30𝑘 𝑓 (𝑥, 𝑦) = 2𝑥 − 3𝑦 𝑅: −2 ≤ 𝑥 ≤ 3, 0 ≤ 𝑦 ≤ 5 ⃗ ⃗ 6. 𝐹 (𝑥, 𝑦, 𝑧) = 𝑧⃗𝑖 + 𝑥𝑗⃗ + 𝑦𝑘 𝑓 (𝑥, 𝑦) = 50 − 4𝑥 + 10𝑦 𝑅: 0 ≤ 𝑥 ≤ 4, 0 ≤ 𝑦 ≤ 8 ⃗ 7. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑦𝑧⃗𝑖 + 𝑥𝑦𝑗⃗ + 𝑥𝑦𝑘 𝑓 (𝑥, 𝑦) = cos 𝑥 + sin 2𝑦 𝑅: Triangle with vertices (0, 0), (0, 5), (5, 0) 8. 𝐹⃗ (𝑥, 𝑦, 𝑧) = cos(𝑥 + 2𝑦)𝑗⃗ 𝑓 (𝑥, 𝑦) = 𝑥𝑒3𝑦 𝑅: Quarter disk of radius 5 centered at the origin, in quadrant I In Exercises 9–12, compute the flux of 𝐹⃗ through the surface 𝑆, which is the part of the graph of 𝑧 = 𝑓 (𝑥, 𝑦) corresponding to region 𝑅, oriented upward.

⃗ 16. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥2 𝑦𝑧𝑗⃗ + 𝑧3 𝑘 𝑆: radius 2, between the 𝑥𝑦-plane and the paraboloid 𝑧 = 𝑥2 + 𝑦2 In Exercises 17–20, compute the flux of 𝐹⃗ through the cylindrical surface 𝑆 centered on the 𝑧-axis, oriented away from the 𝑧-axis. ⃗ 17. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑧𝑗⃗ + 6𝑥𝑘 𝑆: radius 5, 𝑦 ≥ 0, 0 ≤ 𝑧 ≤ 20 ⃗ 18. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑦⃗𝑖 + 𝑥𝑧𝑘 𝑆: radius 10, 𝑥 ≥ 0, 𝑦 ≥ 0, 0 ≤ 𝑧 ≤ 3 ⃗ 19. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥𝑦𝑧𝑗⃗ + 𝑥𝑒𝑧 𝑘 𝑆: radius 2, 0 ≤ 𝑦 ≤ 𝑥, 0 ≤ 𝑧 ≤ 10 20. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥𝑦⃗𝑖 + 2𝑧𝑗⃗ 𝑆: radius 1, 𝑥 ≥ 0, 0 ≤ 𝑦 ≤ 1∕2, 0 ≤ 𝑧 ≤ 2 In Exercises 21–24, write an iterated integral for the flux of 𝐹⃗ through the spherical surface 𝑆 centered at the origin, oriented away from the origin. Do not evaluate the integral. ⃗ 21. 𝐹⃗ (𝑥, 𝑦, 𝑧) = ⃗𝑖 + 2𝑗⃗ + 3𝑘

⃗ 9. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 3⃗𝑖 − 2𝑗⃗ + 6𝑘 𝑓 (𝑥, 𝑦) = 4𝑥 − 2𝑦 𝑅: 0 ≤ 𝑥 ≤ 5, 0 ≤ 𝑦 ≤ 10 ⃗ 10. 𝐹⃗ (𝑥, 𝑦, 𝑧) = ⃗𝑖 − 2𝑗⃗ + 𝑧𝑘 𝑓 (𝑥, 𝑦) = 𝑥𝑦 𝑅: 0 ≤ 𝑥 ≤ 10, 0 ≤ 𝑦 ≤ 10 ⃗ ⃗ 11. 𝐹 (𝑥, 𝑦, 𝑧) = cos 𝑦⃗𝑖 + 𝑧𝑗⃗ + 𝑘 2 𝑓 (𝑥, 𝑦) = 𝑥 + 2𝑦 𝑅: 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1 ⃗ 12. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥⃗𝑖 + 𝑧𝑘 𝑓 (𝑥, 𝑦) = 𝑥 + 𝑦 + 2 𝑅: Triangle with vertices (−1, 0), (1, 0), (0, 1) In Exercises 13–16, write an iterated integral for the flux of 𝐹⃗ through the cylindrical surface 𝑆 centered on the 𝑧axis, oriented away from the 𝑧-axis. Do not evaluate the integral.

𝑆: radius 10, 𝑧 ≥ 0 ⃗ 22. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥⃗𝑖 + 2𝑦𝑗⃗ + 3𝑧𝑘 𝑆: radius 5, entire sphere 23. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑧2 ⃗𝑖 𝑆: radius 2, 𝑥 ≥ 0 ⃗ 24. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑒𝑥 𝑘 𝑆: radius 3, 𝑦 ≥ 0, 𝑧 ≤ 0 In Exercises 25–27, compute the flux of 𝐹⃗ through the spherical surface 𝑆 centered at the origin, oriented away from the origin. 25. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑧⃗𝑖 𝑆: radius 20, 𝑥 ≥ 0, 𝑦 ≥ 0, 𝑧 ≥ 0 ⃗ 26. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑦⃗𝑖 − 𝑥𝑗⃗ + 𝑧𝑘 𝑆: radius 4, entire sphere

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Chapter 19 FLUX INTEGRALS AND DIVERGENCE

27. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥⃗𝑖 + 𝑦𝑗⃗ 𝑆: radius 1, above the cone 𝜙 = 𝜋∕4.

𝑧

28.

𝑧

29. (0, 0, 4)

(2, 0, 4)

⃗ through the In Exercises 28–29, compute the flux of 𝑣⃗ = 𝑧𝑘 rectangular region with the orientation shown.

(0, 2, 3)

(0, 0, 3)

(0, 2, 0) 𝑦

𝑦 (2, 2, 0)

𝑥 (2, 2, 0)

(2, 0, 0)

𝑥

PROBLEMS In Problems 30–46 compute the flux of the vector field 𝐹⃗ through the surface 𝑆. ⃗ and 𝑆 is the portion of the plane 𝑥 + 𝑦 + 𝑧 = 1 30. 𝐹⃗ = 𝑧𝑘 that lies in the first octant, oriented upward. ⃗ and 𝑆 is the part of the plane 31. 𝐹⃗ = (𝑥 − 𝑦)⃗𝑖 + 𝑧𝑗⃗ + 3𝑥𝑘 𝑧 = 𝑥 + 𝑦 above the rectangle 0 ≤ 𝑥 ≤ 2, 0 ≤ 𝑦 ≤ 3, oriented upward. ⃗ and 𝑆 is the part of the surface 32. 𝐹⃗ = 2𝑥𝑗⃗ + 𝑦𝑘 𝑧 = −𝑦 + 1 above the square 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1, oriented upward. ⃗ and 𝑆 is the part of the surface 33. 𝐹⃗ = −𝑦𝑗⃗ + 𝑧𝑘 𝑧 = 𝑦2 + 5 over the rectangle −2 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1, oriented upward.

⃗ and 𝑆 is the cone 𝑧 = 43. 𝐹⃗ = −𝑥𝑧⃗𝑖 −𝑦𝑧𝑗⃗ +𝑧2 𝑘 for 0 ≤ 𝑧 ≤ 6, oriented upward.

2 ⃗ and 𝑆 is the cone 𝑧 = 44. √ 𝐹⃗ = 𝑦𝑧4 ⃗𝑖 − 𝑥𝑧4 𝑗⃗ + 𝑒𝑧 𝑘 2 2 𝑥 + 𝑦 for 1 ≤ 𝑧 ≤ 2, oriented upward.

⃗ and 𝑆 is the surface 𝑦 = 𝑥2 + 𝑧2 , 45. 𝐹⃗ = 𝑦⃗𝑖 + 𝑗⃗ − 𝑥𝑧𝑘 with 𝑥2 + 𝑧2 ≤ 1, oriented in the positive 𝑦-direction. ⃗ and 𝑆 is the oriented triangular 46. 𝐹⃗ = 𝑥2 ⃗𝑖 + 𝑦2 𝑗⃗ + 𝑧2 𝑘 surface shown in Figure 19.26. 𝑧 1 𝑆

⃗ and 𝑆 is the part of 34. 𝐹⃗ = ln(𝑥2 )⃗𝑖 + 𝑒𝑥 𝑗⃗ + cos(1 − 𝑧)𝑘 the surface 𝑧 = −𝑦 + 1 above the square 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1, oriented upward. ⃗ and 𝑆 is a closed cylinder of radius 35. 𝐹⃗ = 5⃗𝑖 + 7𝑗⃗ + 𝑧𝑘 3 centered on the 𝑧-axis, with −2 ≤ 𝑧 ≤ 2, and oriented outward. ⃗ and 𝑆 is a closed cylinder of radius 36. 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 2 centered on the 𝑦-axis, with −3 ≤ 𝑦 ≤ 3, and oriented outward. ⃗ and 𝑆 is the part of the surface 37. 𝐹⃗ = 3𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 𝑧 = −2𝑥 − 4𝑦 + 1, oriented upward, with (𝑥, 𝑦) in the triangle 𝑅 with vertices (0, 0), (0, 2), (1, 0). 38. 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ and 𝑆 is the part of the surface 𝑧 = 25 − (𝑥2 + 𝑦2 ) above the disk of radius 5 centered at the origin, oriented upward. ⃗ and 𝑆 is as in Exercise 38. 39. 𝐹⃗ = cos(𝑥2 + 𝑦2 )𝑘 40. 𝐹⃗ = 𝑟⃗ and 𝑆 is the part of the plane 𝑥 + 𝑦 + 𝑧 = 1 above the rectangle 0 ≤ 𝑥 ≤ 2, 0 ≤ 𝑦 ≤ 3, oriented downward. 41. 𝐹⃗ = 𝑟⃗ and 𝑆 is the part of the surface 𝑧 = 𝑥2 + 𝑦2 above the disk 𝑥2 + 𝑦2 ≤ 1, oriented downward. ⃗ and 𝑆 is the hemisphere 42. 𝐹⃗ = 𝑥𝑧⃗𝑖 + 𝑦𝑘 𝑥2 + 𝑦2 + 𝑧2 = 9, 𝑧 ≥ 0, oriented upward.

√ 𝑥2 + 𝑦2

𝑦 1

1 𝑥

Figure 19.26

In Problems 47–50 find the area of the surface 𝑧 = 𝑓 (𝑥, 𝑦) over the region 𝑅 in the 𝑥𝑦-plane. 47. 𝑓 (𝑥, 𝑦) = 50 + 5𝑥 − 𝑦, 𝑅: −5 ≤ 𝑥 ≤ 5, 0 ≤ 𝑦 ≤ 10 48. 𝑓 (𝑥, 𝑦) = 50 + 5𝑥 − 𝑦, 𝑅: circle of radius 3 centered at the origin 49. 𝑓 (𝑥, 𝑦) = 𝑥𝑒𝑦 , 𝑅: 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1 50. 𝑓 (𝑥, 𝑦) = (sin 𝑥)(sin 𝑦), 𝑅: 0 ≤ 𝑥 ≤ 𝜋∕2, 0 ≤ 𝑦 ≤ 𝜋∕2 51. Let 𝑆 be the hemisphere 𝑥2 + 𝑦2 + 𝑧2 = 𝑎2 of radius 𝑎, where 𝑧 ≥ 0. (a) Express the surface area of 𝑆 as an integral in Cartesian coordinates.

19.2 FLUX INTEGRALS FOR GRAPHS, CYLINDERS, AND SPHERES

(b) Change variables to express the area integral in polar coordinates. (c) Find the area of 𝑆. In Problems 52–53, compute the flux of 𝐹⃗ through the cylindrical surface in Figure 19.27, oriented away from the 𝑧-axis. 𝑧

6

981

𝐹⃗ through this rectangle, 𝑆, oriented upward. In each of the following cases, how does the flux change? (a) The rectangle is twice as wide in the 𝑥-direction, with new corners at the origin, (2, 0, 0), (2, 1, 3), (0, 1, 3). (b) The rectangle is moved so that its corners are at (1, 0, 0), (2, 0, 0), (2, 1, 3), (1, 1, 3). (c) The orientation is changed to downward. (d) The rectangle is tripled in size, so that its new corners are at the origin, (3, 0, 0), (3, 3, 9), (0, 3, 9). (0, 1, 3)

𝑧 𝑆

(1, 1, 3) 𝑥1

1

𝑦

Figure 19.27 𝑦

52. 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ ⃗ 53. 𝐹⃗ = 𝑥𝑧⃗𝑖 + 𝑦𝑧𝑗⃗ + 𝑧3 𝑘 In Problems 54–57, compute the flux of 𝐹⃗ through the spherical surface, 𝑆. 54. 55. 56. 57. 58.

59.

60.

61.

62.

63.

⃗ and 𝑆 is the upper hemisphere of radius 2 𝐹⃗ = 𝑧𝑘 centered at the origin, oriented outward. ⃗ and 𝑆 is the spherical cap given by 𝐹⃗ = 𝑦⃗𝑖 − 𝑥𝑗⃗ + 𝑧𝑘 𝑥2 + 𝑦2 + 𝑧2 = 1, 𝑧 ≥ 0, oriented upward. ⃗ and 𝑆 is the upper hemisphere of the sphere 𝐹⃗ = 𝑧2 𝑘 𝑥2 + 𝑦2 + 𝑧2 = 25, oriented away from the origin. ⃗ and 𝑆 is the surface of the sphere 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 𝑥2 + 𝑦2 + 𝑧2 = 𝑎2 , oriented outward. ⃗ over the quarter Compute the flux of 𝐹⃗ = 𝑥⃗𝑖 +𝑦𝑗⃗ +𝑧𝑘 cylinder 𝑆 given by 𝑥2 + 𝑦2 = 1, 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1, 0 ≤ 𝑧 ≤ 1, oriented outward. ⃗ through the surCompute the flux of 𝐹⃗ = 𝑥⃗𝑖 + 𝑗⃗ + 𝑘 face 𝑆 given by 𝑥 = sin 𝑦 sin 𝑧, with 0 ≤ 𝑦 ≤ 𝜋∕2, 0 ≤ 𝑧 ≤ 𝜋∕2, oriented in the direction of increasing 𝑥. ⃗ through Compute the flux of 𝐹⃗ = (𝑥 + 𝑧)⃗𝑖 + 𝑗⃗ + 𝑧𝑘 the surface 𝑆 given by 𝑦 = 𝑥2 + 𝑧2 , with 0 ≤ 𝑦 ≤ 1, 𝑥 ≥ 0, 𝑧 ≥ 0, oriented toward the 𝑥𝑧-plane. ⃗ . Calculate the Let 𝐹⃗ = (𝑥𝑧𝑒𝑦𝑧 )⃗𝑖 + 𝑥𝑧𝑗⃗ + (5 + 𝑥2 + 𝑦2 )𝑘 flux of 𝐹⃗ through the disk 𝑥2 + 𝑦2 ≤ 1 in the 𝑥𝑦-plane, oriented upward. ⃗. ⃗ = (𝑒𝑥𝑦 +3𝑧+5)⃗𝑖 +(𝑒𝑥𝑦 +5𝑧+3)𝑗⃗ +(3𝑧+𝑒𝑥𝑦 )𝑘 Let 𝐻 ⃗ Calculate the flux of 𝐻 through the square of side 2 with one vertex at the origin, one edge along the positive 𝑦-axis, one edge in the 𝑥𝑧-plane with 𝑥 > 0, 𝑧 > 0, ⃗. and the normal 𝑛⃗ = ⃗𝑖 − 𝑘 The vector field, 𝐹⃗ , in Figure 19.28 depends only on 𝑧; ⃗ , where 𝑔 is an increasing that is, it is of the form 𝑔(𝑧)𝑘 ⃗ function. The integral ∫ 𝐹 ⋅ 𝑑 𝐴⃗ represents the flux of 𝑆

(1, 0, 0) 𝑥

Figure 19.28 64. Electric charge is distributed in space with density (in coulomb/m3 ) given in spherical coordinates by { 𝛿0 (a constant) 𝜌 ≤ 𝑎 𝛿(𝜌, 𝜙, 𝜃) = 0 𝜌 > 𝑎. (a) Describe the charge distribution in words. (b) Find the electric field 𝐸⃗ due to 𝛿. Assume that 𝐸⃗ can be written in spherical coordinates as 𝐸⃗ = 𝐸(𝜌)⃗ 𝑒 𝜌 , where 𝑒⃗ 𝜌 is the unit outward normal to the sphere of radius 𝜌. In addition, 𝐸⃗ satisfies Gauss’s Law for any simple closed surface 𝑆 enclosing a volume 𝑊 : ∫𝑆

𝐸⃗ ⋅ 𝑑 𝐴⃗ = 𝑘

∫𝑊

𝛿 𝑑𝑉 ,

𝑘 a constant.

65. Electric charge is distributed in space with density (in coulomb/m3 ) given in cylindrical coordinates by { 𝛿0 (a constant) if 𝑟 ≤ 𝑎 𝛿(𝑟, 𝜃, 𝑧) = 0 if 𝑟 > 𝑎 (a) Describe the charge distribution in words. (b) Find the electric field 𝐸⃗ due to 𝛿. Assume that 𝐸⃗ can be written in cylindrical coordinates as 𝐸⃗ = 𝐸(𝑟)⃗ 𝑒 𝑟 , where 𝑒⃗ 𝑟 is the unit outward vector to the cylinder of radius 𝑟, and that 𝐸⃗ satisfies Gauss’s Law (see Problem 64).

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Chapter 19 FLUX INTEGRALS AND DIVERGENCE

Strengthen Your Understanding In Problems 66–67, explain what is wrong with the statement.

69. An oriented surface 𝑆 on the cylinder of radius 10 centered on the 𝑧-axis such that ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 600, where 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ .

66. Flux outward through the cone, given in cylindrical coordinates ( by 𝑧 = 𝑟, can be ) computed using the formula 𝑑 𝐴⃗ = cos 𝜃⃗𝑖 + sin 𝜃 𝑗⃗ 𝑅 𝑑𝑧 𝑑𝜃.

Are the statements in Problems 70–72 true or false? Give reasons for your answer.

67. For the surface 𝑧 = 𝑓 (𝑥, 𝑦) oriented upward, the formula

70. If 𝑆 is the part of the graph of 𝑧 = 𝑓 (𝑥, 𝑦) above 𝑎 ≤ 𝑥√≤ 𝑏, 𝑐 ≤ 𝑦 ≤ 𝑑, then 𝑆 has surface area 𝑏 𝑑 ∫𝑎 ∫𝑐 (𝑓𝑥 )2 + (𝑓𝑦 )2 + 1 𝑑𝑥 𝑑𝑦.

( ) ⃗ 𝑑𝑥 𝑑𝑦 𝑑 𝐴⃗ = 𝑛⃗ 𝑑𝐴 = −𝑓𝑥 ⃗𝑖 − 𝑓𝑦 𝑗⃗ + 𝑘

71. If 𝐴⃗ (𝑥, 𝑦) is the area vector for 𝑧 = 𝑓 (𝑥, 𝑦) oriented upward and 𝐵⃗ (𝑥, 𝑦) is the area vector for 𝑧 = −𝑓 (𝑥, 𝑦) oriented upward, then 𝐴⃗ (𝑥, 𝑦) = −𝐵⃗ (𝑥, 𝑦).

⃗ and 𝑑𝐴 = 𝑑𝑥 𝑑𝑦. gives 𝑛⃗ = −𝑓𝑥 ⃗𝑖 − 𝑓𝑦 𝑗⃗ + 𝑘 In Problems 68–69, give an example of:

72. If 𝑆 is the sphere 𝑥2 + 𝑦2 + 𝑧2 = 1 oriented outward and ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 0, then 𝐹⃗ (𝑥, 𝑦, 𝑧) is perpendicular to ⃗ at every point of 𝑆. 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘

68. A function 𝑓 (𝑥, 𝑦) such that, for the surface 𝑧 = 𝑓 (𝑥, 𝑦) ⃗ ) 𝑑𝑥 𝑑𝑦. oriented upwards, we have 𝑑 𝐴⃗ = (⃗𝑖 + 𝑗⃗ + 𝑘

19.3

THE DIVERGENCE OF A VECTOR FIELD Imagine that the vector fields in Figures 19.29 and 19.30 are velocity vector fields describing the flow of a fluid.3 Figure 19.29 suggests outflow from the origin; for example, it could represent the expanding cloud of matter in the big-bang theory of the origin of the universe. We say that the origin is a source. Figure 19.30 suggests flow into the origin; in this case we say that the origin is a sink. In this section we use the flux out of a closed surface surrounding a point to measure the outflow per unit volume there, also called the divergence, or flux density. 𝑦

𝑦

𝑥

Definition of Divergence

Figure 19.29: Vector field showing a source

𝑥

Figure 19.30: Vector field showing a sink

To measure the outflow per unit volume of a vector field at a point, we calculate the flux out of a small sphere centered at the point, divide by the volume enclosed by the sphere, then take the limit of this flux-to-volume ratio as the sphere contracts around the point.

Geometric Definition of Divergence The divergence, or flux density, of a smooth vector field 𝐹⃗ , written div𝑭⃗ , is a scalar-valued function defined by ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ div 𝐹⃗ (𝑥, 𝑦, 𝑧) = lim . Volume→0 Volume of 𝑆 Here 𝑆 is a sphere centered at (𝑥, 𝑦, 𝑧), oriented outward, that contracts down to (𝑥, 𝑦, 𝑧) in the limit. The limit can be computed using other shapes as well, such as the cubes in Example 2. 3 Although

not all vector fields represent physically realistic fluid flows, it is useful to think of them in this way.

19.3 THE DIVERGENCE OF A VECTOR FIELD

983

In Cartesian coordinates, the divergence can also be calculated using the following formula. We show these definitions are equivalent later in the section.

Cartesian Coordinate Definition of Divergence ⃗ , then If 𝐹⃗ = 𝐹1 ⃗𝑖 + 𝐹2 𝑗⃗ + 𝐹3 𝑘 div 𝐹⃗ =

𝜕𝐹1 𝜕𝐹2 𝜕𝐹3 + + . 𝜕𝑥 𝜕𝑦 𝜕𝑧

The dot product formula gives an easy way to remember the Cartesian coordinate definition, 𝜕 ⃗ 𝜕 𝜕 ⃗ and suggests another common notation for div 𝐹⃗ , namely ∇ ⋅ 𝐹⃗ . Using ∇ = 𝑖 + 𝑗⃗ + 𝑘 , 𝜕𝑥 𝜕𝑦 𝜕𝑧 we can write ( ) 𝜕𝐹1 𝜕𝐹2 𝜕𝐹3 𝜕 ⃗ 𝜕 ⃗ 𝜕 ⃗ ⃗ ⃗ ⃗)= + + . div 𝐹 = ∇⋅ 𝐹 = 𝑖 + 𝑗 + 𝑘 ⋅ (𝐹1 ⃗𝑖 + 𝐹2 𝑗⃗ + 𝐹3 𝑘 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧

Example 1

Calculate the divergence of 𝐹⃗ (⃗𝑟 ) = 𝑟⃗ at the origin (a) Using the geometric definition. (b) Using the Cartesian coordinate definition.

Solution

(a) Using the method of Example 5 on page 967, we can calculate the flux of 𝐹⃗ out of the sphere of radius 𝑎, centered at the origin; it is 4𝜋𝑎3 . So we have Flux 4𝜋𝑎3 = lim 4 = lim 3 = 3. 𝑎→0 Volume 𝑎→0 𝑎→0 𝜋𝑎3 3

div 𝐹⃗ (0, 0, 0) = lim

⃗ , so (b) In Cartesian coordinates, 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 div 𝐹⃗ =

𝜕 𝜕 𝜕 (𝑥) + (𝑦) + (𝑧) = 1 + 1 + 1 = 3. 𝜕𝑥 𝜕𝑦 𝜕𝑧

The next example shows that the divergence can be negative if there is net inflow to a point. Example 2

(a) Using the geometric definition, find the divergence of 𝑣⃗ = −𝑥⃗𝑖 at: (i) (0, 0, 0) (ii) (2, 2, 0). (b) Confirm that the coordinate definition gives the same results.

Solution

(a) (i) The vector field 𝑣⃗ = −𝑥⃗𝑖 is parallel to the 𝑥-axis and is shown in the 𝑥𝑦-plane in Figure 19.31. To compute the flux density at (0, 0, 0), we use a cube 𝑆1 , centered at the origin with edges parallel to the axes, of length 2𝑐. Then the flux through the faces perpendicular to the 𝑦- and 𝑧-axes is zero (because the vector field is parallel to these faces). On the faces perpendicular to the 𝑥-axis, the vector field and the outward normal are parallel but point in opposite directions. On the face at 𝑥 = 𝑐, where 𝑣⃗ = −𝑐 ⃗𝑖 and Δ𝐴⃗ = ‖𝐴⃗ ‖⃗𝑖 , we have 𝑣⃗ ⋅ Δ𝐴⃗ = −𝑐 ‖Δ𝐴⃗ ‖.

On the face at 𝑥 = −𝑐, where 𝑣⃗ = 𝑐 ⃗𝑖 and Δ𝐴⃗ = −‖𝐴⃗ ‖⃗𝑖 , the dot product is still negative: 𝑣⃗ ⋅ Δ𝐴⃗ = −𝑐 ‖Δ𝐴⃗ ‖.

984

Chapter 19 FLUX INTEGRALS AND DIVERGENCE

Therefore, the flux through the cube is given by ∫𝑆1

𝑣⃗ ⋅ 𝑑 𝐴⃗ =

∫Face 𝑥=−𝑐

𝑣⃗ ⋅ 𝑑 𝐴⃗ +

∫Face 𝑥=𝑐

𝑣⃗ ⋅ 𝑑 𝐴⃗

= −𝑐 ⋅ Area of one face + (−𝑐) ⋅ Area of other face = −2𝑐(2𝑐)2 = −8𝑐 3 . Thus, 𝑣⃗ ⋅ 𝑑 𝐴⃗ ( ) ∫𝑆 −8𝑐 3 = lim = −1. div 𝑣⃗ (0, 0, 0) = lim Volume→0 Volume of cube 𝑐→0 (2𝑐)3 Since the vector field points inward toward the 𝑦𝑧-plane, it makes sense that the divergence is negative at the origin. (ii) Take 𝑆2 to be a cube as before, but centered this time at the point (2, 2, 0). See Figure 19.31. As before, the flux through the faces perpendicular to the 𝑦- and 𝑧-axes is zero. On the face at 𝑥 = 2 + 𝑐, 𝑣⃗ ⋅ Δ𝐴⃗ = −(2 + 𝑐) ‖Δ𝐴⃗ ‖. On the face at 𝑥 = 2 − 𝑐 with outward normal, the dot product is positive, and 𝑣⃗ ⋅ Δ𝐴⃗ = (2 − 𝑐) ‖Δ𝐴⃗ ‖.

Therefore, the flux through the cube is given by ∫𝑆

2

𝑣⃗ ⋅ 𝑑 𝐴⃗ =

∫Face 𝑥=2−𝑐

𝑣⃗ ⋅ 𝑑 𝐴⃗ +

∫Face 𝑥=2+𝑐

𝑣⃗ ⋅ 𝑑 𝐴⃗

= (2 − 𝑐) ⋅ Area of one face − (2 + 𝑐) ⋅ Area of other face = −2𝑐(2𝑐)2 = −8𝑐 3 . Then, as before, div 𝑣⃗ (2, 2, 0) =

∫𝑆 𝑣⃗ ⋅ 𝑑 𝐴⃗

lim

Volume→0

Volume of cube

= lim

𝑐→0

(

−8𝑐 3 (2𝑐)3

)

= −1.

Although the vector field is flowing away from the point (2, 2, 0) on the left, this outflow is smaller in magnitude than the inflow on the right, so the net outflow is negative. ⃗ , the formula gives (b) Since 𝑣⃗ = −𝑥⃗𝑖 + 0𝑗⃗ + 0𝑘 div 𝑣⃗ =

𝜕 𝜕 𝜕 (−𝑥) + (0) + (0) = −1 + 0 + 0 = −1. 𝜕𝑥 𝜕𝑦 𝜕𝑧 (2, 2, 0)

𝑦

✠ 𝑆2

𝑆1

𝑥

✒ (0, 0, 0)

Figure 19.31: Vector field 𝑣⃗ = −𝑥⃗𝑖 in the 𝑥𝑦-plane

19.3 THE DIVERGENCE OF A VECTOR FIELD

985

Why Do the Two Definitions of Divergence Give the Same Result? The geometric definition defines div 𝐹⃗ as the flux density of 𝐹⃗ . To see why the coordinate definition is also the flux density, imagine computing the flux out of a small box-shaped surface 𝑆 at (𝑥0 , 𝑦0 , 𝑧0 ), with sides of length Δ𝑥, Δ𝑦, and Δ𝑧 parallel to the axes. On 𝑆1 (the back face of the box shown in Figure 19.32, where 𝑥 = 𝑥0 ), the outward normal is in the negative 𝑥-direction, so 𝑑 𝐴⃗ = −𝑑𝑦 𝑑𝑧 ⃗𝑖 . Assuming 𝐹⃗ is approximately constant on 𝑆1 , we have ∫𝑆1

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

∫𝑆1

𝐹⃗ ⋅ (−⃗𝑖 ) 𝑑𝑦 𝑑𝑧 ≈ −𝐹1 (𝑥0 , 𝑦0 , 𝑧0 )

∫𝑆1

𝑑𝑦 𝑑𝑧

= −𝐹1 (𝑥0 , 𝑦0 , 𝑧0 ) ⋅ Area of 𝑆1 = −𝐹1 (𝑥0 , 𝑦0 , 𝑧0 ) Δ𝑦 Δ𝑧. On 𝑆2 , the face where 𝑥 = 𝑥0 + Δ𝑥, the outward normal points in the positive 𝑥-direction, so 𝑑 𝐴⃗ = 𝑑𝑦 𝑑𝑧 ⃗𝑖 . Therefore, ∫𝑆2

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

∫𝑆2

𝐹⃗ ⋅ ⃗𝑖 𝑑𝑦 𝑑𝑧 ≈ 𝐹1 (𝑥0 + Δ𝑥, 𝑦0 , 𝑧0 )

∫𝑆2

𝑑𝑦 𝑑𝑧

= 𝐹1 (𝑥0 + Δ𝑥, 𝑦0 , 𝑧0 ) ⋅ Area of 𝑆2 = 𝐹1 (𝑥0 + Δ𝑥, 𝑦0 , 𝑧0 ) Δ𝑦 Δ𝑧. 𝑆1 (Back) Δ𝑦

𝑧



𝑆4 (Back)

𝑆6 𝑆 (𝑥0 , 𝑦0 , 𝑧0 ) 3 𝑆2 Δ𝑥

✠ Δ𝑧 𝑦



𝑆5 (Bottom)

𝑥

Figure 19.32: Box used to find div 𝐹⃗ at (𝑥0 , 𝑦0 , 𝑧0 )

Thus, ∫𝑆1

𝐹⃗ ⋅ 𝑑 𝐴⃗ +

∫𝑆2

𝐹⃗ ⋅ 𝑑 𝐴⃗ ≈ 𝐹1 (𝑥0 + Δ𝑥, 𝑦0 , 𝑧0 )Δ𝑦Δ𝑧 − 𝐹1 (𝑥0 , 𝑦0 , 𝑧0 )Δ𝑦Δ𝑧 𝐹1 (𝑥0 + Δ𝑥, 𝑦0 , 𝑧0 ) − 𝐹1 (𝑥0 , 𝑦0 , 𝑧0 ) Δ𝑥Δ𝑦Δ𝑧 Δ𝑥 𝜕𝐹1 ≈ Δ𝑥Δ𝑦Δ𝑧. 𝜕𝑥

=

By an analogous argument, the contribution to the flux from 𝑆3 and 𝑆4 (the surfaces perpendicular to the 𝑦-axis) is approximately 𝜕𝐹2 Δ𝑥 Δ𝑦 Δ𝑧, 𝜕𝑦 and the contribution to the flux from 𝑆5 and 𝑆6 is approximately 𝜕𝐹3 Δ𝑥 Δ𝑦 Δ𝑧. 𝜕𝑧 Thus, adding these contributions, we have Total flux through 𝑆 ≈

𝜕𝐹 𝜕𝐹1 𝜕𝐹 Δ𝑥 Δ𝑦 Δ𝑧 + 2 Δ𝑥 Δ𝑦 Δ𝑧 + 3 Δ𝑥 Δ𝑦 Δ𝑧. 𝜕𝑥 𝜕𝑦 𝜕𝑧

986

Chapter 19 FLUX INTEGRALS AND DIVERGENCE

Since the volume of the box is Δ𝑥 Δ𝑦 Δ𝑧, the flux density is 𝜕𝐹 𝜕𝐹 𝜕𝐹1 Δ𝑥Δ𝑦Δ𝑧 + 2 Δ𝑥Δ𝑦Δ𝑧 + 3 Δ𝑥Δ𝑦Δ𝑧 Total flux through 𝑆 𝜕𝑥 𝜕𝑦 𝜕𝑧 ≈ Volume of box Δ𝑥Δ𝑦Δ𝑧 𝜕𝐹1 𝜕𝐹2 𝜕𝐹3 = + + . 𝜕𝑥 𝜕𝑦 𝜕𝑧

Divergence-Free Vector Fields A vector field 𝐹⃗ is said to be divergence free or solenoidal if div𝐹⃗ = 0 everywhere that 𝐹⃗ is defined. Example 3

Figure 19.33 shows, for three values of the constant 𝑝, the vector field 𝐸⃗ =

𝑟⃗ ‖⃗𝑟 ‖𝑝

⃗ , 𝑟⃗ ≠ 0⃗ . 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘

(a) Find a formula for div 𝐸⃗ . (b) Is there a value of 𝑝 for which 𝐸⃗ is divergence-free? If so, find it. 𝑦

𝑦

𝑥

𝑥

𝑝=0

Solution

𝑦

𝑝=1

𝑥

𝑝=3

Figure 19.33: The vector field 𝐸⃗ (⃗𝑟 ) = 𝑟⃗ ∕‖⃗𝑟 ‖𝑝 for 𝑝 = 0, 1, and 3 in the 𝑥𝑦-plane

(a) The components of 𝐸⃗ are 𝐸⃗ =

(𝑥2

𝑦 𝑥 𝑧 ⃗. ⃗𝑖 + 𝑘 𝑗⃗ + 2 2 𝑝∕2 2 2 + +𝑧 ) (𝑥 + 𝑦 + 𝑧2 )𝑝∕2 (𝑥 + 𝑦2 + 𝑧2 )𝑝∕2 𝑦2

We compute the partial derivatives ( ) 𝑝𝑥2 𝑥 1 𝜕 = 2 − 2 2 2 2 𝑝∕2 2 2 𝑝∕2 2 𝜕𝑥 (𝑥 + 𝑦 + 𝑧 ) (𝑥 + 𝑦 + 𝑧 ) (𝑥 + 𝑦 + 𝑧2 )(𝑝∕2)+1 ( ) 𝑝𝑦2 𝑦 1 𝜕 = − 𝜕𝑦 (𝑥2 + 𝑦2 + 𝑧2 )𝑝∕2 (𝑥2 + 𝑦2 + 𝑧2 )𝑝∕2 (𝑥2 + 𝑦2 + 𝑧2 )(𝑝∕2)+1 ( ) 𝑝𝑧2 𝑧 1 𝜕 = − . 𝜕𝑧 (𝑥2 + 𝑦2 + 𝑧2 )𝑝∕2 (𝑥2 + 𝑦2 + 𝑧2 )𝑝∕2 (𝑥2 + 𝑦2 + 𝑧2 )(𝑝∕2)+1 So 𝑝(𝑥2 + 𝑦2 + 𝑧2 ) 3 − (𝑥2 + 𝑦2 + 𝑧2 )𝑝∕2 (𝑥2 + 𝑦2 + 𝑧2 )(𝑝∕2)+1 3−𝑝 3−𝑝 = . = 2 2 2 𝑝∕2 (𝑥 + 𝑦 + 𝑧 ) ‖⃗𝑟 ‖𝑝

div 𝐸⃗ =

(b) The divergence is zero when 𝑝 = 3, so 𝐹⃗ (⃗𝑟 ) = 𝑟⃗ ∕‖⃗𝑟 ‖3 is a divergence-free vector field. Notice that the divergence is zero even though the vectors point outward from the origin.

987

19.3 THE DIVERGENCE OF A VECTOR FIELD

Magnetic Fields An important class of divergence-free vector fields is the magnetic fields. One of Maxwell’s Laws of Electromagnetism is that the magnetic field 𝐵⃗ satisfies div 𝐵⃗ = 0. Example 4

An infinitesimal current loop, similar to that shown in Figure 19.34, is called a magnetic dipole. Its magnitude is described by a constant vector 𝜇⃗ , called the dipole moment. The magnetic field due to a magnetic dipole with moment 𝜇⃗ is a multiple of

Show that div 𝐵⃗ = 0.

𝜇⃗ 3(𝜇⃗ ⋅ 𝑟⃗ )⃗𝑟 𝐵⃗ = − + , 3 ‖⃗𝑟 ‖ ‖⃗𝑟 ‖5

𝑟⃗ ≠ 0⃗ .

𝜇⃗ Area = 𝑎



Figure 19.34: A current loop

Solution

To show that div 𝐵⃗ = 0 we can use the following version of the product rule for the divergence: if 𝑔 is a scalar function and 𝐹⃗ is a vector field, then div(𝑔 𝐹⃗ ) = (grad 𝑔) ⋅ 𝐹⃗ + 𝑔 div 𝐹⃗ . (See Problem 37 on page 990.) Thus, since 𝜇⃗ is constant and div 𝜇⃗ = 0, we have ( ) ( ) ( ) ) ( 𝜇⃗ 1 1 1 div = div ⋅ 𝜇⃗ + 0 𝜇⃗ = grad ‖⃗𝑟 ‖3 ‖⃗𝑟 ‖3 ‖⃗𝑟 ‖3 ‖⃗𝑟 ‖3 and

div

(

(𝜇⃗ ⋅ 𝑟⃗ )⃗𝑟 ‖⃗𝑟 ‖5

)

(

𝑟⃗ = div 𝜇⃗ ⋅ 𝑟⃗ ||⃗𝑟 ||5

)

𝑟⃗ = grad(𝜇⃗ ⋅ 𝑟⃗ ) ⋅ + (𝜇⃗ ⋅ 𝑟⃗ ) div ‖⃗𝑟 ‖5

(

𝑟⃗ ‖⃗𝑟 ‖5

)

.

From Problems 83 and 84 of Section 14.5 (available online) and Example 3 on page 986, we have ) ) ( ( −3⃗𝑟 −2 1 𝑟⃗ = = grad , grad( 𝜇 ⃗ ⋅ 𝑟 ⃗ ) = 𝜇 ⃗ , div . ‖⃗𝑟 ‖3 ‖⃗𝑟 ‖5 ‖⃗𝑟 ‖5 ‖⃗𝑟 ‖5 Putting these results together gives ) ) ( ( 𝑟⃗ 𝑟⃗ 1 ⃗ ⋅ 𝜇⃗ + 3 grad(𝜇⃗ ⋅ 𝑟⃗ ) ⋅ + 3(𝜇⃗ ⋅ 𝑟⃗ ) div div 𝐵 = − grad ‖⃗𝑟 ‖3 ‖⃗𝑟 ‖5 ‖⃗𝑟 ‖5 3⃗𝑟 ⋅ 𝜇⃗ 3𝜇⃗ ⋅ 𝑟⃗ 6𝜇⃗ ⋅ 𝑟⃗ = + − 5 5 ‖⃗𝑟 ‖ ‖⃗𝑟 ‖ ‖⃗𝑟 ‖5 = 0.

988

Chapter 19 FLUX INTEGRALS AND DIVERGENCE

Exercises and Problems for Section 19.3 Online Resource: Additional Problems for Section 19.3 EXERCISES

Are the quantities in Exercises 1–2 vectors or scalars? Calculate them. ( ) ⃗ 1. div (𝑥2 + 𝑦)⃗𝑖 + (𝑥𝑦𝑒𝑧 )𝑗⃗ − ln(𝑥2 + 𝑦2 )𝑘 (

2 2 ⃗ 2. div (2 sin(𝑥𝑦) + tan 𝑧)⃗𝑖 + (tan 𝑦)𝑗⃗ + (𝑒𝑥 +𝑦 )𝑘

)

3. Which of the following two vector fields, sketched in the 𝑥𝑦-plane, appears to have the greater divergence at the origin? The scales are the same on each. 𝑦

(I)

(II)

𝑦

) ( ( ) ⃗ 9. 𝐹⃗ = ln 𝑥2 + 1 ⃗𝑖 + (cos 𝑦) 𝑗⃗ + (𝑥𝑦𝑒𝑧 ) 𝑘

10. 𝐹⃗ (⃗𝑟 ) = 𝑎⃗ × 𝑟⃗

−𝑦⃗𝑖 + 𝑥𝑗⃗ 11. 𝐹⃗ (𝑥, 𝑦) = 𝑥2 + 𝑦2 12. 𝐹⃗ (⃗𝑟 ) =

,

𝑟⃗ ≠ 𝑟⃗ 0

13. For each of the following vector fields, sketched in the 𝑥𝑦-plane, decide if the divergence is positive, zero, or negative at the indicated point. (a)

𝑥

𝑟⃗ − 𝑟⃗ 0 ‖⃗𝑟 − 𝑟⃗ 0 ‖

𝑦

𝑦

(b)

𝑥 𝑥

In Exercises 4–12, find the divergence of the vector field. ⃗ .) (Note: 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 4. 𝐹⃗ (𝑥, 𝑦) = −𝑦⃗𝑖 + 𝑥𝑗⃗ 5. 𝐹⃗ (𝑥, 𝑦) = −𝑥⃗𝑖 + 𝑦𝑗⃗

(c)

𝑥

𝑦

𝑥

⃗ 6. 𝐹⃗ (𝑥, 𝑦, 𝑧) = (−𝑥 + 𝑦)⃗𝑖 + (𝑦 + 𝑧)𝑗⃗ + (−𝑧 + 𝑥)𝑘 2 2 7. 𝐹⃗ (𝑥, 𝑦) = (𝑥 − 𝑦 )⃗𝑖 + 2𝑥𝑦𝑗⃗ ⃗) 8. 𝐹⃗ (𝑥, 𝑦, 𝑧) = 3𝑥2 ⃗𝑖 − sin(𝑥𝑧)(⃗𝑖 + 𝑘

PROBLEMS 14. Draw two vector fields that have positive divergence everywhere. 15. Draw two vector fields that have negative divergence everywhere. 16. Draw two vector fields that have zero divergence everywhere. 17. A small sphere of radius 0.1 surrounds the point (2, 3, −1). The flux of a vector field 𝐺⃗ into this sphere is 0.00004𝜋. Estimate div 𝐺⃗ at the point (2, 3, −1). 18. A smooth vector field 𝐹⃗ has div 𝐹⃗ (1, 2, 3) = 5. Estimate the flux of 𝐹⃗ out of a small sphere of radius 0.01 centered at the point (1, 2, 3). 19. Let 𝐹⃗ be a vector field with div 𝐹⃗ = 𝑥2 + 𝑦2 − 𝑧. (a) Estimate ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ where 𝑆 is (i) A sphere of radius 0.1 centered at (2, 0, 0). (ii) A box of side 0.2 with edges parallel to the axes and centered at (0, 0, 10).

(b) The point (2, 0, 0) is called a source for the vector field 𝐹⃗ ; the point (0, 0, 10) is called a sink. Explain the reason for these names using your answer to part (a). 20. The flux of 𝐹⃗ out of a small sphere of radius 0.1 centered at (4, 5, 2) is 0.0125. Estimate: (a) div 𝐹⃗ at (4, 5, 2) (b) The flux of 𝐹⃗ out of a sphere of radius 0.2 centered at (4, 5, 2). ⃗ through 21. (a) Find the flux of 𝐹⃗ = 2𝑥⃗𝑖 − 3𝑦𝑗⃗ + 5𝑧𝑘 a cube with four of its corners at the points (𝑎, 𝑏, 𝑐), (𝑎 + 𝑤, 𝑏, 𝑐), (𝑎, 𝑏 + 𝑤, 𝑐), (𝑎, 𝑏, 𝑐 + 𝑤) and edge length 𝑤. See Figure 19.35. (b) Use the geometric definition and part (a) to find div 𝐹⃗ at the point (𝑎, 𝑏, 𝑐). (c) Find div 𝐹⃗ using partial derivatives.

19.3 THE DIVERGENCE OF A VECTOR FIELD 𝑆1 (Back)

𝑧 (𝑎, 𝑏, 𝑐 + 𝑤)

(𝑎, 𝑏, 𝑐)

𝑆4 (Back)







𝑆6

✲ 𝑆3

𝑦

989

30. A certain distribution of electric charge produces the electric field shown in Figure 19.36. Where are the charges that produced this electric field concentrated? Which concentrations are positive and which are negative? 𝑦

𝑆2

(𝑎 + 𝑤, 𝑏, 𝑐)■

𝑆5 (Bottom)

𝑥 𝑥

Figure 19.35

−4 −3 −2 −1

1

2

⃗ . Use the 22. Suppose 𝐹⃗ = (3𝑥 + 2)⃗𝑖 + 4𝑥𝑗⃗ + (5𝑥 + 1)𝑘 method of Exercise 21 to find div 𝐹⃗ at the point (𝑎, 𝑏, 𝑐) by two different methods. 23. Use the geometric definition of divergence to find div 𝑣⃗ at the origin, where 𝑣⃗ = −2⃗𝑟 . Check that you get the same result using the definition in Cartesian coordinates. 24. (a) Let 𝑓 (𝑥, 𝑦) = 𝑎𝑥𝑦 + 𝑎𝑥2 𝑦 + 𝑦3 . Find div grad 𝑓 . (b) Choose 𝑎 so that div grad 𝑓 = 0 for all 𝑥, 𝑦. 25. Let 𝐹⃗ = (9𝑎2 𝑥 + 10𝑎𝑦2 )⃗𝑖 + (10𝑧3 − 6𝑎𝑦)𝑗⃗ − (3𝑧 + ⃗ . Find the value(s) of 𝑎 making div 𝐹⃗ 10𝑥2 + 10𝑦2 )𝑘 (a) 0

(b) A minimum

26. Let 𝐹⃗ (⃗𝑟 ) = 𝑟⃗ ∕‖⃗𝑟 ‖3 (in 3-space), 𝑟⃗ ≠ 0⃗ .

(a) Calculate div 𝐹⃗ . (b) Sketch 𝐹⃗ . Does 𝐹⃗ appear to have nonzero divergence? Does this agree with your answer to part (a)?

27. The vector field 𝐹⃗ (⃗𝑟 ) = 𝑟⃗ ∕||⃗𝑟 ||3 is not defined at the origin. Nevertheless, we can attempt to use the flux definition to compute div 𝐹⃗ at the origin. What is the result? ⃗. 28. Let 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑧𝑘 (a) Calculate div 𝐹⃗ . (b) Sketch 𝐹⃗ . Does 𝐹⃗ appear to have nonzero divergence? Does this agree with your answer to part (a)? 29. The divergence of a magnetic vector field 𝐵⃗ must be zero everywhere. Which of the following vector fields cannot be a magnetic vector field? (a) (b) (c)

⃗ 𝐵⃗ (𝑥, 𝑦, 𝑧) = −𝑦⃗𝑖 + 𝑥𝑗⃗ + (𝑥 + 𝑦)𝑘 ⃗ ⃗ ⃗ ⃗ 𝐵 (𝑥, 𝑦, 𝑧) = −𝑧𝑖 + 𝑦𝑗 + 𝑥𝑘 𝐵⃗ (𝑥, 𝑦, 𝑧) = (𝑥2 − 𝑦2 − 𝑥)⃗𝑖 + (𝑦 − 2𝑥𝑦)𝑗⃗

Problems 30–31 involve electric fields. Electric charge produces a vector field 𝐸⃗ , called the electric field, which represents the force on a unit positive charge placed at the point. Two positive or two negative charges repel one another, whereas two charges of opposite sign attract one another. The divergence of 𝐸⃗ is proportional to the density of the electric charge (that is, the charge per unit volume), with a positive constant of proportionality.

Figure 19.36 31. The electric field at the point 𝑟⃗ as a result of a point charge at the origin is 𝐸⃗ (⃗𝑟 ) = 𝑘⃗𝑟 ∕‖⃗𝑟 ‖3 .

(a) Calculate div 𝐸⃗ for 𝑟⃗ ≠ 0⃗ . (b) Calculate the limit suggested by the geometric definition of div 𝐸⃗ at the point (0, 0, 0). (c) Explain what your answers mean in terms of charge density.

32. Due to roadwork ahead, the traffic on a highway slows linearly from 55 miles/hour to 15 miles/hour over a 2000-foot stretch of road, then crawls along at 15 miles/hour for 5000 feet, then speeds back up linearly to 55 miles/hour in the next 1000 feet, after which it moves steadily at 55 miles/hour. (a) Sketch a velocity vector field for the traffic flow. (b) Write a formula for the velocity vector field 𝑣⃗ (miles/hour) as a function of the distance 𝑥 feet from the initial point of slowdown. (Take the direction of motion to be ⃗𝑖 and consider the various sections of the road separately.) (c) Compute div 𝑣⃗ at 𝑥 = 1000, 5000, 7500, 10,000. Be sure to include the proper units. 33. The velocity field 𝑣⃗ in Problem 32 does not give a complete description of the traffic flow, for it takes no account of the spacing between vehicles. Let 𝜌 be the density (cars/mile) of highway, where we assume that 𝜌 depends only on 𝑥. (a) Using your highway experience, arrange in ascending order: 𝜌(0), 𝜌(1000), 𝜌(5000). (b) What are the units and interpretation of the vector field 𝜌𝑣⃗ ? (c) Would you expect 𝜌𝑣⃗ to be constant? Why? What does this mean for div(𝜌𝑣⃗ )? (d) Determine 𝜌(𝑥) if 𝜌(0) = 75 cars/mile and 𝜌𝑣⃗ is constant. (e) If the highway has two lanes, find the approximate number of feet between cars at 𝑥 = 0, 1000, and 5000.

990

Chapter 19 FLUX INTEGRALS AND DIVERGENCE

⃗ , an arbitrary function 𝑓 (𝑥, 𝑦, 𝑧), 34. For 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 and an arbitrary vector field 𝐹⃗ (𝑥, 𝑦, 𝑧), which of the following is a vector field and which is a constant vector field? (a) grad 𝑓 (b) (div 𝐹⃗ )⃗𝑖 (c) (div 𝑟⃗ )⃗𝑖 (d) (div ⃗𝑖 )𝐹⃗

(e)

grad(div 𝐹⃗ )

⃗ and 𝑐⃗ = 𝑐1 ⃗𝑖 + 𝑐2 𝑗⃗ + 𝑐3 𝑘 ⃗,a 35. Let 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 constant vector; let 𝑆 be a sphere of radius 𝑅 centered at the origin. Find (a) div(⃗𝑟 × 𝑐⃗ )

(b) ∫𝑆 (⃗𝑟 × 𝑐⃗ ) ⋅ 𝑑 𝐴⃗

36. Show that if 𝑎⃗ is a constant vector and 𝑓 (𝑥, 𝑦, 𝑧) is a function, then div(𝑓 𝑎⃗ ) = (grad 𝑓 ) ⋅ 𝑎⃗ . 37. Show that if 𝑔(𝑥, 𝑦, 𝑧) is a scalar-valued function and 𝐹⃗ (𝑥, 𝑦, 𝑧) is a vector field, then div(𝑔 𝐹⃗ ) = (grad 𝑔) ⋅ 𝐹⃗ + 𝑔 div 𝐹⃗ . 38. If 𝑓 (𝑥, 𝑦, 𝑧) and 𝑔(𝑥, 𝑦, 𝑧) are functions with continuous second partial derivatives, show that div(grad 𝑓 × grad 𝑔) = 0. In Problems 39–41, use Problems 37 and 38 to find the divergence of the vector field. The vectors 𝑎⃗ and 𝑏⃗ are constant. 39. 𝐹⃗ =

1 𝑎⃗ × 𝑟⃗ ‖⃗𝑟 ‖𝑝

1 40. 𝐵⃗ = 𝑎 𝑟⃗ 𝑥

41. 𝐺⃗ = (𝑏⃗ ⋅ 𝑟⃗ )𝑎⃗ × 𝑟⃗

42. A vector field, 𝑣⃗ , in the plane is a point source at the origin if its direction is away from the origin at every point, its magnitude depends only on the distance from the origin, and its divergence is zero away from the origin. (a) Explain why a point ( source at ) the origin must be of the form 𝑣⃗ = 𝑓 (𝑥2 + 𝑦2 ) (𝑥⃗𝑖 + 𝑦𝑗⃗ ) for some positive function 𝑓 . (b) Show that 𝑣⃗ = 𝐾(𝑥2 + 𝑦2 )−1 (𝑥⃗𝑖 + 𝑦𝑗⃗ ) is a point source at the origin if 𝐾 > 0. (c) What is the magnitude ‖𝑣⃗ ‖ of the source in part (b) as a function of the distance from its center? (d) Sketch the vector field 𝑣⃗ = (𝑥2 + 𝑦2 )−1 (𝑥⃗𝑖 + 𝑦𝑗⃗ ). (e) Show that 𝜙 = 𝐾2 log(𝑥2 + 𝑦2 ) is a potential function for the source in part (b). 43. A vector field, 𝑣⃗ , in the plane is a point sink at the origin if its direction is toward the origin at every point, its magnitude depends only on the distance from the origin, and its divergence is zero away from the origin. (a) Explain why a point sink at)the origin must be of ( the form 𝑣⃗ = 𝑓 (𝑥2 + 𝑦2 ) (𝑥⃗𝑖 + 𝑦𝑗⃗ ) for some negative function 𝑓 . (b) Show that 𝑣⃗ = 𝐾(𝑥2 + 𝑦2 )−1 (𝑥⃗𝑖 + 𝑦𝑗⃗ ) is a point sink at the origin if 𝐾 < 0. (c) Determine the magnitude ‖𝑣⃗ ‖ of the sink in part (b) as a function of the distance from its center. (d) Sketch 𝑣⃗ = −(𝑥2 + 𝑦2 )−1 (𝑥⃗𝑖 + 𝑦𝑗⃗ ). (e) Show that 𝜙 = 𝐾2 log(𝑥2 + 𝑦2 ) is a potential function for the sink in part (b).

Strengthen Your Understanding In Problems 44–46, explain what is wrong with the statement.

53. If 𝐹⃗ is a vector field in 3-space, then div𝐹⃗ is also a vector field.

44. div(2𝑥⃗𝑖 ) = 2⃗𝑖 . ⃗ we have 45. For 𝐹⃗ (𝑥, 𝑦, 𝑧) = (𝑥2 + 𝑦)⃗𝑖 + (2𝑦 + 𝑧)𝑗⃗ − 𝑧2 𝑘 ⃗. div 𝐹⃗ = 2𝑥⃗𝑖 + 2𝑗⃗ − 2𝑧𝑘 46. The divergence of 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 + 𝑦𝑧 is given by div 𝑓 (𝑥, 𝑦, 𝑧) = 2𝑥 + 𝑧 + 𝑦.

⃗ has zero 54. A constant vector field 𝐹⃗ = 𝑎⃗𝑖 + 𝑏𝑗⃗ + 𝑐 𝑘 divergence.

In Problems 47–49, give an example of: 47. A vector field 𝐹⃗ (𝑥, 𝑦, 𝑧) whose divergence is a nonzero constant. 48. A nonzero vector field 𝐹⃗ (𝑥, 𝑦, 𝑧) whose divergence is zero. 49. A vector field that is not divergence free. Are the statements in Problems 50–62 true or false? Give reasons for your answer. 50. div(𝐹⃗ + 𝐺⃗ ) = div 𝐹⃗ + div 𝐺⃗ 51. grad(𝐹⃗ ⋅ 𝐺⃗ ) = 𝐹⃗ (div 𝐺⃗ ) + (div 𝐹⃗ )𝐺⃗ 52. div 𝐹⃗ is a scalar whose value can vary from point to point.

55. If a vector field 𝐹⃗ in 3-space has zero divergence then ⃗ where 𝑎, 𝑏 and 𝑐 are constants. 𝐹⃗ = 𝑎⃗𝑖 + 𝑏𝑗⃗ + 𝑐 𝑘 56. If 𝐹⃗ is a vector field in 3-space, and 𝑓 is a scalar function, then div(𝑓 𝐹⃗ ) = 𝑓 div𝐹⃗ . 57. If 𝐹⃗ is a vector field in 3-space, and 𝐹⃗ = grad 𝑓 , then div 𝐹⃗ = 0. 58. If 𝐹⃗ is a vector field in 3-space, then grad(div 𝐹⃗ ) = 0⃗ . 59. The field 𝐹⃗ (⃗𝑟 ) = 𝑟⃗ is divergence free. 60. If 𝑓 (𝑥, 𝑦, 𝑧) is any given continuous scalar function, then there is at least one vector field 𝐹⃗ such that div𝐹⃗ = 𝑓 . 61. If 𝐹⃗ and 𝐺⃗ are vector fields satisfying div𝐹⃗ = div𝐺⃗ then 𝐹⃗ = 𝐺⃗ . 62. There exist a scalar function 𝑓 and a vector field 𝐹⃗ satisfying div(grad 𝑓 ) = grad(div 𝐹⃗ ).

19.4 THE DIVERGENCE THEOREM

19.4

991

THE DIVERGENCE THEOREM The Divergence Theorem is a multivariable analogue of the Fundamental Theorem of Calculus; it says that the integral of the flux density over a solid region equals the flux integral through the boundary of the region.

The Boundary of a Solid Region The boundary, 𝑆, of a solid region, 𝑊 , may be thought of as the skin between the interior of the region and the space around it. For example, the boundary of a solid ball is a spherical surface, the boundary of a solid cube is its six faces, and the boundary of a solid cylinder is a tube sealed at both ends by disks. (See Figure 19.37). A surface which is the boundary of a solid region is called a closed surface. 𝑊 = Solid cube 𝑆 = 6 square faces

𝑊 = Ball 𝑆 = Sphere

𝑊 =Solid cylinder 𝑆 = Tube and two disks

Figure 19.37: Several solid regions and their boundaries

Calculating the Flux from the Flux Density Consider a solid region 𝑊 in 3-space whose boundary is the closed surface 𝑆. There are two ways to find the total flux of a vector field 𝐹⃗ out of 𝑊 . One is to calculate the flux of 𝐹⃗ through 𝑆: Flux out of 𝑊 =

∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ .

Another way is to use div 𝐹⃗ , which gives the flux density at any point in 𝑊 . We subdivide 𝑊 into small boxes, as shown in Figure 19.38. Then, for a small box of volume Δ𝑉 , Flux out of box ≈ Flux density ⋅ Volume = div 𝐹⃗ Δ𝑉 . What happens when we add the fluxes out of all the boxes? Consider two adjacent boxes, as shown in Figure 19.39. The flux through the shared wall is counted twice, once out of the box on each side. When we add the fluxes, these two contributions cancel, so we get the flux out of the solid region formed by joining the two boxes. Continuing in this way, we find that Fluxes through inner wall cancel

✠ ✛

Δ𝑉

Figure 19.38: Subdivision of region into small boxes

Figure 19.39: Adding the flux out of adjacent boxes

992

Chapter 19 FLUX INTEGRALS AND DIVERGENCE

Flux out of 𝑊 =



Flux out of small boxes ≈



div 𝐹⃗ Δ𝑉 .

We have approximated the flux by a Riemann sum. As the subdivision gets finer, the sum approaches an integral, so Flux out of 𝑊 = div 𝐹⃗ 𝑑𝑉 . ∫𝑊 We have calculated the flux in two ways, as a flux integral and as a volume integral. Therefore, these two integrals must be equal. This result holds even if 𝑊 is not a rectangular solid. Thus, we have the following result.4

Theorem 19.1: The Divergence Theorem If 𝑊 is a solid region whose boundary 𝑆 is a piecewise smooth surface, and if 𝐹⃗ is a smooth vector field on a solid region5 containing 𝑊 and 𝑆, then ∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

∫𝑊

div 𝐹⃗ 𝑑𝑉 ,

where 𝑆 is given the outward orientation.

Example 1

Use the Divergence Theorem to calculate the flux of the vector field 𝐹⃗ (⃗𝑟 ) = 𝑟⃗ through the sphere of radius 𝑎 centered at the origin.

Solution

In Example 5 on page 967 we computed the flux using the definition of a flux integral, giving ∫𝑆

𝑟⃗ ⋅ 𝑑 𝐴⃗ = 4𝜋𝑎3 .

⃗ ) = 3 and the Divergence Theorem: Now we use div 𝐹⃗ = div(𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 ∫𝑆

Example 2

𝑟⃗ ⋅ 𝑑 𝐴⃗ =

∫𝑊

div 𝐹⃗ 𝑑𝑉 =

∫𝑊

4 3 𝑑𝑉 = 3 ⋅ 𝜋𝑎3 = 4𝜋𝑎3 . 3

Use the Divergence Theorem to calculate the flux of the vector field

⃗ 𝐹⃗ (𝑥, 𝑦, 𝑧) = (𝑥2 + 𝑦2 )⃗𝑖 + (𝑦2 + 𝑧2 )𝑗⃗ + (𝑥2 + 𝑧2 )𝑘

through the cube in Figure 19.40. 4 A proof of the Divergence Theorem using the coordinate definition of the divergence can be found in the online supplement at www.wiley.com/college/hughes-hallett. 5 The region containing 𝑊 and 𝑆 is open.

19.4 THE DIVERGENCE THEOREM

993

𝑧 1

𝑥

1

1

𝑦

Figure 19.40

Solution

The divergence of 𝐹⃗ is div 𝐹⃗ = 2𝑥 + 2𝑦 + 2𝑧. Since div 𝐹⃗ is positive everywhere in the first quadrant, the flux through 𝑆 is positive. By the Divergence Theorem, |1 (𝑥2 + 2𝑥(𝑦 + 𝑧))|| 𝑑𝑦 𝑑𝑧 ∫0 ∫0 ∫0 ∫0 ∫0 |0 1 1 1 1 | = 1 + 2(𝑦 + 𝑧) 𝑑𝑦 𝑑𝑧 = (𝑦 + 𝑦2 + 2𝑦𝑧)|| 𝑑𝑧 ∫0 ∫0 ∫0 |0 1 1 | = (2 + 2𝑧) 𝑑𝑧 = (2𝑧 + 𝑧2 )|| = 3. ∫0 |0 1

∫𝑆

1

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

1

1

1

2(𝑥 + 𝑦 + 𝑧) 𝑑𝑥 𝑑𝑦 𝑑𝑧 =

The Divergence Theorem and Divergence-Free Vector Fields An important application of the Divergence Theorem is the study of divergence-free vector fields. Example 3

In Example 3 on page 986 we saw that the following vector field is divergence free: 𝐹⃗ (⃗𝑟 ) =

𝑟⃗ , ‖⃗𝑟 ‖3

𝑟⃗ ≠ 0⃗ .

Calculate ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ , using the Divergence Theorem if possible, for the following surfaces: (a) 𝑆1 is the sphere of radius 𝑎 centered at the origin. (b) 𝑆2 is the sphere of radius 𝑎 centered at the point (2𝑎, 0, 0). Solution

(a) We cannot use the Divergence Theorem directly because 𝐹⃗ is not defined everywhere inside the sphere (it is not defined at the origin). Since 𝐹⃗ points outward everywhere on 𝑆1 , the flux out of 𝑆1 is positive. On 𝑆1 , 𝑎 𝐹⃗ ⋅ 𝑑 𝐴⃗ = ‖𝐹⃗ ‖𝑑𝐴 = 𝑑𝐴, 𝑎3 so 1 1 1 𝑑𝐴 = (Area of 𝑆1 ) = 4𝜋𝑎2 = 4𝜋. 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 2 2 ∫𝑆1 ∫ 𝑎 𝑆1 𝑎 𝑎2 Notice that the flux is not zero, although div 𝐹⃗ is zero everywhere it is defined. (b) Let 𝑊 be the solid region enclosed by 𝑆2 . Since div 𝐹⃗ = 0 everywhere in 𝑊 , we can use the Divergence Theorem in this case, giving ∫𝑆2

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

∫𝑊

div 𝐹⃗ 𝑑𝑉 =

∫𝑊

0 𝑑𝑉 = 0.

994

Chapter 19 FLUX INTEGRALS AND DIVERGENCE

The Divergence Theorem applies to any solid region 𝑊 and its boundary 𝑆, even in cases where the boundary consists of two or more surfaces. For example, if 𝑊 is the solid region between the sphere 𝑆1 of radius 1 and the sphere 𝑆2 of radius 2, both centered at the same point, then the boundary of 𝑊 consists of both 𝑆1 and 𝑆2 . The Divergence Theorem requires the outward orientation, which on 𝑆2 points away from the center and on 𝑆1 points toward the center. (See Figure 19.41.) 𝑆2



𝑊

✒ 𝑆1 Figure 19.41: Cutaway view of the region 𝑊 between two spheres, showing orientation vectors

Example 4

Let 𝑆1 be the sphere of radius 1 centered at the origin and let 𝑆2 be the ellipsoid 𝑥2 + 𝑦2 + 4𝑧2 = 16, both oriented outward. For 𝑟⃗ 𝐹⃗ (⃗𝑟 ) = , 𝑟⃗ ≠ 0⃗ , ‖⃗𝑟 ‖3 show that

∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

1

Solution

∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ .

2

The ellipsoid contains the sphere; let 𝑊 be the solid region between them. Since 𝑊 does not contain the origin, div 𝐹⃗ is defined and equal to zero everywhere in 𝑊 . Thus, if 𝑆 is the boundary of 𝑊 , then div 𝐹⃗ 𝑑𝑉 = 0. 𝐹⃗ ⋅ 𝑑 𝐴⃗ = ∫𝑆 ∫𝑊 But 𝑆 consists of 𝑆2 oriented outward and 𝑆1 oriented inward, so 0=

∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ −

2

∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ ,

1

and thus ∫𝑆2

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

∫𝑆1

𝐹⃗ ⋅ 𝑑 𝐴⃗ .

In Example 3 we showed that ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 4𝜋, so ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 4𝜋 also. Note that it would have 1 2 been more difficult to compute the integral over the ellipsoid directly.

Electric Fields The electric field produced by a positive point charge 𝑞 placed at the origin is 𝑟⃗ 𝐸⃗ = 𝑞 . ‖⃗𝑟 ‖3

Using Example 3, we see that the flux of the electric field through any sphere centered at the origin is 4𝜋𝑞. In fact, using the idea of Example 4, we can show that the flux of 𝐸⃗ through any closed

19.4 THE DIVERGENCE THEOREM

995

surface containing the origin is 4𝜋𝑞. See Problems 37 and 38 on page 997. This is a special case of Gauss’s Law, which states that the flux of an electric field through any closed surface is proportional to the total charge enclosed by the surface. Carl Friedrich Gauss (1777–1855) also discovered the Divergence Theorem, which is sometimes called Gauss’s Theorem.

Exercises and Problems for Section 19.4 EXERCISES For Exercises 1–5, compute the flux integral ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ in two ways, if possible, directly and using the Divergence Theorem. In each case, 𝑆 is closed and oriented outward. 1. 𝐹⃗ (⃗𝑟 ) = 𝑟⃗ and 𝑆 is the cube enclosing the volume 0 ≤ 𝑥 ≤ 2, 0 ≤ 𝑦 ≤ 2, and 0 ≤ 𝑧 ≤ 2. 2. 𝐹⃗ = 𝑦𝑗⃗ and 𝑆 is a closed vertical cylinder of height 2, with its base a circle of radius 1 on the 𝑥𝑦-plane, centered at the origin. ⃗ and 𝑆 is the surface of the box 3. 𝐹⃗ = 𝑥2 ⃗𝑖 + 2𝑦2 𝑗⃗ + 3𝑧2 𝑘 with faces 𝑥 = 1, 𝑥 = 2, 𝑦 = 0, 𝑦 = 1, 𝑧 = 0, 𝑧 = 1. ⃗ and 𝑆 is the 4. 𝐹⃗ = (𝑧2 + 𝑥)⃗𝑖 + (𝑥2 + 𝑦)𝑗⃗ + (𝑦2 + 𝑧)𝑘 closed cylinder 𝑥2 + 𝑧2 = 1, with 0 ≤ 𝑦 ≤ 1, oriented outward.

⃗ and 𝑆 is a square pyramid with height 5. 𝐹⃗ = −𝑧⃗𝑖 + 𝑥𝑘 3 and base on the 𝑥𝑦-plane of side length 1. In Exercises 6–12, find the flux of the vector field out of the closed box 0 ≤ 𝑥 ≤ 2, 0 ≤ 𝑦 ≤ 3, 0 ≤ 𝑧 ≤ 4. ⃗ 6. 𝐹⃗ = 4⃗𝑖 + 7𝑗⃗ − 𝑘 ⃗ 7. 𝐺⃗ = 𝑦⃗𝑖 + 𝑧𝑘 ⃗ = 𝑥𝑦⃗𝑖 + 𝑧𝑗⃗ + 𝑦𝑘 ⃗ 8. 𝐻 ⃗ 9. 𝐽⃗ = 𝑥𝑦2 𝑗⃗ + 𝑥𝑘 ⃗ = 𝑒𝑧 ⃗𝑖 + sin(𝑥𝑦)𝑘 ⃗ 10. 𝑁 ⃗ = (3𝑥 + 4𝑦)⃗𝑖 + (4𝑦 + 5𝑧)𝑗⃗ + (5𝑧 + 3𝑥)𝑘 ⃗ 11. 𝑀 ⃗ where div 𝑀 ⃗ = 𝑥𝑦 + 5 12. 𝑀

PROBLEMS ⃗ out of a sphere of 13. Find the flux of 𝐹⃗ = 𝑧⃗𝑖 + 𝑦𝑗⃗ + 𝑥𝑘 radius 3 centered at the origin. ⃗ out of a sphere 14. Find the flux of 𝐹⃗ = 𝑥𝑦⃗𝑖 + 𝑦𝑧𝑗⃗ + 𝑧𝑥𝑘 of radius 1 centered at the origin. ⃗ through the 15. Find the flux of 𝐹⃗ = 𝑥3 ⃗𝑖 + 𝑦3 𝑗⃗ + 𝑧3 𝑘 closed surface bounding the solid region 𝑥2 + 𝑦2 ≤ 4, 0 ≤ 𝑧 ≤ 5, oriented outward. 16. The region 𝑊 lies between the spheres 𝑥2 + 𝑦2√+ 𝑧2 = 4 and 𝑥2 + 𝑦2 + 𝑧2 = 9 and within the cone 𝑧 = 𝑥2 + 𝑦2 with 𝑧 ≥ 0; its boundary is the closed surface, 𝑆, ori⃗ ented outward. Find the flux of 𝐹⃗ = 𝑥3 ⃗𝑖 + 𝑦3 𝑗⃗ + 𝑧3 𝑘 out of 𝑆. ⃗, 17. For 𝐹⃗ = (2𝑥 + sin 𝑧)⃗𝑖 + (𝑥𝑧 − 𝑦)𝑗⃗ + (𝑒𝑥 + 2𝑧)𝑘 ⃗ find the flux of 𝐹 out of the closed silo-shaped region 2 2 within the √cylinder 𝑥 + 𝑦 = 1, below the hemisphere 𝑧 = 1 + 1 − 𝑥2 − 𝑦2 , and above the 𝑥𝑦-plane.

18. Find the flux of 𝐹⃗ through the closed cylinder of radius 2, centered on the 𝑧-axis, with 3 ≤ 𝑧 ≤ 7, if ⃗. 𝐹⃗ = (𝑥 + 3𝑒𝑦𝑧 )⃗𝑖 + (ln(𝑥2 𝑧2 + 1) + 𝑦)𝑗⃗ + 𝑧𝑘 2 2 19. Find the flux of 𝐹⃗ = 𝑒𝑦 𝑧 ⃗𝑖 + (tan(0.001𝑥2 𝑧2 ) + 𝑦2 )𝑗⃗ + ⃗ out of the closed box 0 ≤ 𝑥 ≤ 5, (ln(1 + 𝑥2 𝑦2 ) + 𝑧2 )𝑘 0 ≤ 𝑦 ≤ 4, 0 ≤ 𝑧 ≤ 3. 2 ⃗ ⃗ 20. Find the flux √ of 𝐹 = 𝑥 𝑖⃗ + 𝑧𝑗⃗ + 𝑦𝑘 out of the closed cone 𝑥 = 𝑦2 + 𝑧2 , with 0 ≤ 𝑥 ≤ 1.

21. Suppose 𝐹⃗ is a vector field with div 𝐹⃗ = 10. Find the flux of 𝐹⃗ out of a cylinder of height 𝑎 and radius 𝑎, centered on the 𝑧-axis and with base in the 𝑥𝑦-plane.

⃗ , and let 22. Let 𝐹⃗ = (5𝑥 + 7𝑦)⃗𝑖 + (7𝑦 + 9𝑧)𝑗⃗ + (9𝑧 + 11𝑥)𝑘 𝑄𝑖 be the flux of 𝐹⃗ through the surfaces 𝑆𝑖 for 𝑖 =1–4. Arrange 𝑄𝑖 in ascending order, where (a) 𝑆1 is the sphere of radius 2 centered at the origin (b) 𝑆2 is the cube of side 2 centered at the origin and with sides parallel to the axes (c) 𝑆3 is the sphere of radius 1 centered at the origin (d) 𝑆4 is a pyramid with all four corners lying on 𝑆3 23. A cone has its tip at the point (0, 0, 5), and its base is the disk 𝐷, 𝑥2 + 𝑦2 ≤ 1, in the 𝑥𝑦-plane. The surface of the cone is the curved and slanted face, 𝑆, oriented upward, and the flat base, 𝐷, oriented downward. The ⃗ flux of the constant vector field 𝐹⃗ = 𝑎⃗𝑖 + 𝑏𝑗⃗ + 𝑐 𝑘 through 𝑆 is given by

∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ = 3.22.

Is it possible to calculate ∫𝐷 𝐹⃗ ⋅ 𝑑 𝐴⃗ ? If so, give the answer. If not, explain what additional information you would need to be able to make this calculation. 24. If 𝑉 is a volume surrounded by a closed surface 𝑆, show that 31 ∫𝑆 𝑟⃗ ⋅ 𝑑 𝐴⃗ = 𝑉 . 25. A vector field 𝐹⃗ satisfies div𝐹⃗ = 0 everywhere. Show that ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 0 for every closed surface 𝑆.

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Chapter 19 FLUX INTEGRALS AND DIVERGENCE

26. Let 𝑆 be the cube in the first quadrant with side 2, one corner at the origin and edges parallel to the axes. Let 2⃗

𝐹⃗ 1 = (𝑥𝑦 + 3𝑥𝑧 )⃗𝑖 + (3𝑥 𝑦 + 2𝑦𝑧 )𝑗⃗ + 3𝑧𝑦 𝑘 ⃗ 𝐹⃗ 2 = (𝑥𝑦2 + 5𝑒𝑦𝑧 )⃗𝑖 + (𝑦𝑧2 + 7 sin(𝑥𝑧))𝑗⃗ + (𝑥2 𝑧 + cos(𝑥𝑦))𝑘 ) ) ) ( ( ( 3 𝑦 𝑧3 ⃗ 𝑥3 ⃗ 𝑘. 𝐹⃗ 3 = 𝑥𝑧2 + 𝑖 + 𝑦𝑧2 + 𝑗⃗ + 𝑧𝑦2 + 3 3 3 2

2

2

2

Arrange the flux integrals of 𝐹⃗ 1 , 𝐹⃗ 2 , 𝐹⃗ 3 out of 𝑆 in increasing order. 27. Let div 𝐹⃗ = 2(6 − 𝑥) and 0 ≤ 𝑎, 𝑏, 𝑐 ≤ 10. (a) Find the flux of 𝐹⃗ out of the rectangular box given by 0 ≤ 𝑥 ≤ 𝑎, 0 ≤ 𝑦 ≤ 𝑏, 0 ≤ 𝑧 ≤ 𝑐. (b) For what values of 𝑎, 𝑏, 𝑐 is the flux largest? What is that largest flux? 28. (a) Find div(⃗𝑟 ∕||⃗𝑟 ||2 ) where 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ for 𝑟⃗ ≠ 0⃗ . (b) Can you use the Divergence Theorem to compute the flux of 𝑟⃗ ∕||⃗𝑟 ||2 out of a closed cylinder of radius 1, length 2, centered at the origin, and with its axis along the 𝑧-axis? (c) Compute the flux of 𝑟⃗ ∕||⃗𝑟 ||2 out of the cylinder in part (b). (d) Find the flux of 𝑟⃗ ∕||⃗𝑟 ||2 out of a closed cylinder of radius 2, length 2, centered at the origin, and with its axis along the 𝑧-axis. ⃗ and let 𝐹⃗ be the vector field 29. Let 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 given by 𝑟⃗ . 𝐹⃗ = ||⃗𝑟 ||3 (a) Find the flux of 𝐹⃗ out of the sphere 𝑥2 +𝑦2 +𝑧2 = 1 oriented outward. (b) Calculate div 𝐹⃗ . Show your work and simplify your answer completely. (c) Use your answers to parts (a) and (b) to calculate the flux out of a box of side 10 centered at the origin and with sides parallel to the coordinate planes. (The box is also oriented outward.)

In Problems 30–31, find the flux of 𝐹⃗ = 𝑟⃗ ∕||⃗𝑟 ||3 through the surface. [Hint: Use the method of Problem 29.] 30. 𝑆 is the ellipsoid 𝑥2 + 2𝑦2 + 3𝑧2 = 6.

31. 𝑆 is the closed cylinder 𝑦2 + 𝑧2 = 4, −2 ≤ 𝑥 ≤ 2. 32. (a) Let div 𝐹⃗ = 𝑥2 + 𝑦2 + 𝑧2 + 3. Calculate ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ 1 where 𝑆1 is the sphere of radius 1 centered at the origin. (b) Let 𝑆2 be the sphere of radius 2 centered at the origin; let 𝑆3 be the sphere of radius 3 centered at the origin; let 𝑆4 be the box of side 6 centered at the origin with edges parallel to the axes. Without calculating them, arrange the following integrals in increasing order: ∫𝑆2

𝐹⃗ ⋅ 𝑑 𝐴⃗ ,

∫𝑆3

𝐹⃗ ⋅ 𝑑 𝐴⃗ ,

∫𝑆4

𝐹⃗ ⋅ 𝑑 𝐴⃗ .

33. Suppose div 𝐹⃗ = 𝑥𝑦𝑧2 . (a) Find div 𝐹⃗ at the point (1, 2, 1). [Note: You are given div 𝐹⃗ , not 𝐹⃗ .] (b) Using your answer to part (a), but no other information about the vector field 𝐹⃗ , estimate the flux out of a small box of side 0.2 centered at the point (1, 2, 1) and with edges parallel to the axes. (c) Without computing the vector field 𝐹⃗ , calculate the exact flux out of the box. 34. Suppose div 𝐹⃗ = 𝑥2 +𝑦2 +3. Find a surface 𝑆 such that ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ is negative, or explain why no such surface exists. 35. As a result of radioactive decay, heat is generated uniformly throughout the interior of the earth at a rate of 30 watts per cubic kilometer. (A watt is a rate of heat production.) The heat then flows to the earth’s surface where it is lost to space. Let 𝐹⃗ (𝑥, 𝑦, 𝑧) denote the rate of flow of heat measured in watts per square kilometer. By definition, the flux of 𝐹⃗ across a surface is the quantity of heat flowing through the surface per unit of time. (a) What is the value of div 𝐹⃗ ? Include units. (b) Assume the heat flows outward symmetrically. ⃗ and Verify that 𝐹⃗ = 𝛼⃗𝑟 , where 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 𝛼 is a suitable constant, satisfies the given conditions. Find 𝛼. (c) Let 𝑇 (𝑥, 𝑦, 𝑧) denote the temperature inside the earth. Heat flows according to the equation 𝐹⃗ = −𝑘 grad 𝑇 , where 𝑘 is a constant. Explain why this makes sense physically. (d) If 𝑇 is in ◦ C, then 𝑘 = 30,000 watts/km◦ C. Assuming the earth is a sphere with radius 6400 km and surface temperature 20◦ C, what is the temperature at the center? 36. If a surface 𝑆 is submerged in an incompressible fluid, a force 𝐹⃗ is exerted on one side of the surface by the pressure in the fluid. If the 𝑧-axis is vertical, with the positive direction upward and the fluid level at 𝑧 = 0, then the component of force in the direction of a unit vector 𝑢⃗ is given by the following: 𝐹⃗ ⋅ 𝑢⃗ = −

∫𝑆

𝑧𝛿𝑔⃗𝑢 ⋅ 𝑑 𝐴⃗ ,

where 𝛿 is the density of the fluid (mass/volume), 𝑔 is the acceleration due to gravity, and the surface is oriented away from the side on which the force is exerted. In this problem we consider a totally submerged closed surface enclosing a volume 𝑉 . We are interested in the force of the liquid on the external surface, so 𝑆 is oriented inward. Use the Divergence Theorem to show that: (a) The force in the ⃗𝑖 and 𝑗⃗ directions is zero. ⃗ direction is 𝛿𝑔𝑉 , the weight of (b) The force in the 𝑘 the volume of fluid with the same volume as 𝑉 . This is Archimedes’ Principle.

19.4 THE DIVERGENCE THEOREM

37. According to Coulomb’s Law, the electrostatic field 𝐸⃗ at the point 𝑟⃗ due to a charge 𝑞 at the origin is given by 𝑟⃗ . 𝐸⃗ (⃗𝑟 ) = 𝑞 ‖⃗𝑟 ‖3

(a) Compute div 𝐸⃗ . (b) Let 𝑆𝑎 be the sphere of radius 𝑎 centered at the origin and oriented outward. Show that the flux of 𝐸⃗ through 𝑆𝑎 is 4𝜋𝑞. (c) Could you have used the Divergence Theorem in part (b)? Explain why or why not. (d) Let 𝑆 be an arbitrary, closed, outward-oriented surface surrounding the origin. Show that the flux of 𝐸⃗ through 𝑆 is again 4𝜋𝑞. [Hint: Apply the Diver-

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gence Theorem to the solid region lying between a small sphere 𝑆𝑎 and the surface 𝑆.] 38. According to Coulomb’s Law, the electric field 𝐸⃗ at the point 𝑟⃗ due to a charge 𝑞 at the point 𝑟⃗ 0 is given by 𝐸⃗ (⃗𝑟 ) = 𝑞

(⃗𝑟 − 𝑟⃗ 0 )

.

‖⃗𝑟 − 𝑟⃗ 0 ‖3

Suppose 𝑆 is a closed, outward-oriented surface and that 𝑟⃗ 0 does not lie on 𝑆. Use Problem 37 to show that { 4𝜋𝑞 if 𝑞 lies inside 𝑆, 𝐸⃗ ⋅ 𝑑 𝐴⃗ = ∫𝑆 0 if 𝑞 lies outside 𝑆.

Strengthen Your Understanding In Problems 39–40, explain what is wrong with the statement.

Are the statements in Problems 48–55 true or false? Give reasons for your answer.

39. The flux integral ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ can be evaluated using the Divergence Theorem, where 𝐹⃗ = 2𝑥⃗𝑖 − 3𝑗⃗ and 𝑆 is the triangular surface with corners (1, 0, 0), (0, 1, 0), (0, 0, 1) oriented away from the origin.

48. ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ = div 𝐹⃗ .

40. If 𝑆 is the boundary of a solid region 𝑊 , where 𝑆 is oriented outward, and 𝐹⃗ is a vector field, then ∫𝑆

div 𝐹⃗ 𝑑 𝐴⃗ =

∫𝑊

𝐹⃗ 𝑑𝑉 .

In Problems 41–42, give an example of:

49. If 𝐹⃗ is a divergence-free vector field in 3-space and 𝑆 is a closed surface oriented inward, then ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 0. 50. If 𝐹⃗ is a vector field in 3-space satisfying div 𝐹⃗ = 1, and 𝑆 is a closed surface oriented outward, then ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ is equal to the volume enclosed by 𝑆. 51. Let 𝑊 be the solid region between the sphere 𝑆1 of radius 1 and 𝑆2 of radius 2, both centered at the origin and oriented outward. If 𝐹⃗ is a vector field in 3-space, then ∫𝑊 div 𝐹⃗ 𝑑𝑉 = ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ − ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ . 2

41. A surface 𝑆 that is the boundary of a solid region such ⃗. that ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 0 if 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑦⃗𝑖 + 𝑥𝑧𝑗⃗ + 𝑦2 𝑘 42. A vector field 𝐹⃗ such that the flux of 𝐹⃗ out of a sphere of radius 1 centered at the origin is 3. Are the statements in Problems 43–47 true or false? The smooth vector field 𝐹⃗ is defined everywhere in 3-space and has constant divergence equal to 4. 43. The field 𝐹⃗ has a net inflow per unit volume at the point (−3, 4, 0). 44. The vector field 𝐹⃗ could be 𝐹⃗ = 𝑥⃗𝑖 + (3𝑦)𝑗⃗ + (𝑦 − ⃗. 5𝑥)𝑘 45. The vector field 𝐹⃗ could be a constant field. 46. The flux of 𝐹⃗ through a circle of radius 5 lying anywhere on the 𝑥𝑦-plane and oriented upward is 4(𝜋52 ). 47. The flux of 𝐹⃗ through a closed cylinder of radius 1 centered along the 𝑦-axis, 0 ≤ 𝑦 ≤ 3 and oriented outward is 4(3𝜋).

Online Resource: Review problems and Projects

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52. Let 𝑆1 be the square 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1, 𝑧 = 0 oriented downward and let 𝑆2 be the square 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1, 𝑧 = 1 oriented upward. If 𝐹⃗ is a vector field, then ∫𝑊 div 𝐹⃗ 𝑑𝑉 = ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ + ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ , where 2 1 𝑊 is the solid cube 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1, 0 ≤ 𝑧 ≤ 1. 53. Let 𝑆1 be the square 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1, 𝑧 = 0 oriented downward and let 𝑆2 be the square 0 ≤ 𝑥 ≤ 1, 0 ≤ ⃗ , then 𝑦 ≤ 1, 𝑧 = 1 oriented upward. If 𝐹⃗ = cos(𝑥𝑦𝑧)𝑘 ∫𝑊 div 𝐹⃗ 𝑑𝑉 = ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ + ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ , where 𝑊 2 1 is the solid cube 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1, 0 ≤ 𝑧 ≤ 1. 54. If 𝑆 is a sphere of radius 1, centered at the origin, oriented outward, and 𝐹⃗ is a vector field satisfying ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 0, then div 𝐹⃗ = 0 at all points inside 𝑆. 55. Let 𝑆ℎ be the surface consisting of a cylinder of height ℎ, closed at the top. The curved sides are 𝑥2 + 𝑦2 = 1, for 0 ≤ 𝑧 ≤ ℎ, and the top 𝑥2 + 𝑦2 ≤ 1, for 𝑧 = ℎ, oriented outward. If 𝐹⃗ is divergence free, then ∫𝑆 𝐹⃗ ⋅𝑑 𝐴⃗ ℎ is independent of the height ℎ.

Chapter Twenty

THE CURL AND STOKES’ THEOREM

Contents 20.1 The Curl of a Vector Field . . . . . . . . . . . . . . . . Circulation Density . . . . . . . . . . . . . . . . . . . . . . Definition of the Curl . . . . . . . . . . . . . . . . . . . . Why Do the Two Definitions of Curl Give the Same Result? . . . . . . . . . . . . . . . . . . . Curl-Free Vector Fields . . . . . . . . . . . . . . . . . . . 20.2 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . The Boundary of a Surface . . . . . . . . . . . . . . . . Calculating the Circulation from the Circulation Density . . . . . . . . . . . . . . . . . . . . . Curl-Free Vector Fields . . . . . . . . . . . . . . . . . . . Curl Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.3 The Three Fundamental Theorems . . . . . . . . . The Gradient and the Curl . . . . . . . . . . . . . . . . . The Curl and the Divergence . . . . . . . . . . . . . .

1000 1000 1001 1004 1004 1008 1008 1008 1010 1011 1015 1015 1016

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20.1

Chapter 20 THE CURL AND STOKES’ THEOREM

THE CURL OF A VECTOR FIELD The divergence is a scalar derivative which measures the outflow of a vector field per unit volume. Now we introduce a vector derivative, the curl, which measures the circulation of a vector field. Imagine holding the paddle-wheel in Figure 20.1 in the flow shown by Figure 20.2. The speed at which the paddle-wheel spins measures the strength of circulation. Notice that the angular velocity depends on the direction in which the stick is pointing. If the stick is pointing horizontally the paddlewheel does not spin; if the stick is vertical, the paddle wheel spins.

Figure 20.1: A device for measuring circulation

Figure 20.2: A vector field (in the planes 𝑧 = 1, 𝑧 = 2, 𝑧 = 3) with circulation about the 𝑧-axis

Circulation Density We measure the strength of the circulation using a closed curve. Suppose 𝐶 is a circle with center 𝑃 = (𝑥, 𝑦, 𝑧) in the plane perpendicular to 𝑛⃗ , traversed in the direction determined from 𝑛⃗ by the right-hand rule. (See Figures 20.3 and 20.4.)

𝑛⃗ 𝑃 𝐶

Figure 20.3: Direction of 𝐶 relates to direction of 𝑛⃗ by the right-hand rule

Figure 20.4: When the thumb points in the direction of 𝑛⃗ , the fingers curl in the forward direction around 𝐶

We make the following definition:

The circulation density of a smooth vector field 𝐹⃗ at (𝑥, 𝑦, 𝑧) around the direction of the unit vector 𝑛⃗ is defined, provided the limit exists, to be 𝐹⃗ ⋅ 𝑑⃗𝑟 ∫𝐶 Circulation around 𝐶 ⃗ circ𝑛⃗ 𝐹 (𝑥, 𝑦, 𝑧) = lim = lim , Area→0 Area→0 Area inside 𝐶 Area inside 𝐶 The circle 𝐶 is in the plane perpendicular to 𝑛⃗ and oriented by the right-hand rule.

20.1 THE CURL OF A VECTOR FIELD

1001

We can use other closed curves for 𝐶, such as rectangles, that lie in a plane perpendicular to 𝑛⃗ and include the point (𝑥, 𝑦, 𝑧). The circulation density determines the angular velocity of the paddle-wheel in Figure 20.1 provided you could make one sufficiently small and light and insert it without disturbing the flow. Example 1

Consider the vector field 𝐹⃗ in Figure 20.2. Suppose that 𝐹⃗ is parallel to the 𝑥𝑦-plane and that at a distance 𝑟 from the 𝑧-axis it has magnitude 2𝑟. Calculate circ𝑛⃗ 𝐹⃗ at the origin for ⃗ ⃗ (a) 𝑛⃗ = 𝑘 (b) 𝑛⃗ = −𝑘 (c) 𝑛⃗ = ⃗𝑖 .

Solution

(a) Take a circle 𝐶 of radius 𝑎 in the 𝑥𝑦-plane, centered at the origin, traversed in a direction deter⃗ by the right-hand rule. Then, since 𝐹⃗ is tangent to 𝐶 everywhere and points in mined from 𝑘 the forward direction around 𝐶, we have Circulation around 𝐶 =

∫𝐶

Thus, the circulation density is

𝐹⃗ ⋅ 𝑑⃗𝑟 = ‖𝐹⃗ ‖ ⋅ Circumference of 𝐶 = 2𝑎(2𝜋𝑎) = 4𝜋𝑎2 .

4𝜋𝑎2 Circulation around 𝐶 circ𝑘⃗ 𝐹⃗ = lim = lim = 4. 𝑎→0 𝜋𝑎2 𝑎→0 Area inside 𝐶 ⃗ the circle is traversed in the opposite direction, so the line integral changes sign. (b) If 𝑛⃗ = −𝑘 Thus, circ−𝑘⃗ 𝐹⃗ = −4. (c) The circulation around ⃗𝑖 is calculated using circles in the 𝑦𝑧-plane. Since 𝐹⃗ is everywhere perpendicular to such a circle 𝐶, 𝐹⃗ ⋅ 𝑑⃗𝑟 = 0. ∫𝐶 Thus, we have circ⃗𝑖 𝐹⃗ = lim

𝑎→0

∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 𝜋𝑎2

= lim

𝑎→0

0 = 0. 𝜋𝑎2

Definition of the Curl Example 1 shows that the circulation density of a vector field can be positive, negative, or zero, depending on the direction. We assume that there is one direction in which the circulation density is greatest and define a single vector quantity that incorporates all these different circulation densities. We give two definitions, one geometric and one algebraic, which turn out to lead to the same result.

Geometric Definition of Curl The curl of a smooth vector field 𝐹⃗ , written curl 𝐹⃗ , is the vector field with the following properties: • The direction of curl 𝐹⃗ (𝑥, 𝑦, 𝑧) is the direction 𝑛⃗ for which circ𝑛⃗ 𝐹⃗ (𝑥, 𝑦, 𝑧) is the greatest. • The magnitude of curl 𝐹⃗ (𝑥, 𝑦, 𝑧) is the circulation density of 𝐹⃗ around that direction. If the circulation density is zero around every direction, then we define the curl to be 0⃗ .

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Chapter 20 THE CURL AND STOKES’ THEOREM

Cartesian Coordinate Definition of Curl ⃗ , then If 𝐹⃗ = 𝐹1 ⃗𝑖 + 𝐹2 𝑗⃗ + 𝐹3 𝑘 ) ) ) ( ( ( 𝜕𝐹3 𝜕𝐹2 𝜕𝐹1 𝜕𝐹3 𝜕𝐹2 𝜕𝐹1 ⃗. ⃗𝑖 + − − − 𝑘 curl 𝐹⃗ = 𝑗⃗ + 𝜕𝑦 𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑦

The cross-product formula gives an easy way to remember the Cartesian coordinate definition 𝜕 ⃗ 𝜕 𝜕 ⃗ and suggests another common notation for curl 𝐹⃗ , namely ∇ × 𝐹⃗ . Using ∇ = 𝑖 + 𝑗⃗ + 𝑘 , 𝜕𝑥 𝜕𝑦 𝜕𝑧 we can write | ⃗𝑖 𝑗⃗ 𝑘 ⃗ || | | 𝜕 𝜕 𝜕 || curl 𝐹⃗ = ∇ × 𝐹⃗ = || 𝜕𝑥 𝜕𝑦 |. 𝜕𝑧 | | |𝐹 𝐹 𝐹 | | 1 2 3 | Example 2

For each field in Figure 20.5, use the sketch and the geometric definition to decide whether the curl at the origin appears to point up or down, or to be the zero vector. Then check your answer using the coordinate definition of curl and the formulas in the caption of Figure 20.5. Note that the vector ⃗ -components and are independent of 𝑧. fields have no 𝑘 (a)

𝑦

(b)

𝑥

𝑦

(c)

𝑦

𝑥

𝑥

Figure 20.5: Sketches in the 𝑥𝑦-plane of (a) 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ (b) 𝐹⃗ = 𝑦⃗𝑖 − 𝑥𝑗⃗ (c) 𝐹⃗ = −(𝑦 + 1)⃗𝑖

Solution

(a) This vector field shows no rotation, and the circulation around any circle in the 𝑥𝑦-plane centered ⃗ is zero. The at the origin appears to be zero, so we suspect that the circulation density around 𝑘 coordinate definition of curl gives ( ) ) ) ( ( 𝜕𝑦 𝜕𝑥 ⃗ 𝜕(0) 𝜕𝑦 ⃗ 𝜕𝑥 𝜕(0) ⃗ curl 𝐹⃗ = − − − 𝑘 = 0⃗ . 𝑖 + 𝑗 + 𝜕𝑦 𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑦 (b) This vector field appears to be rotating around the 𝑧-axis. By the right-hand rule, the circulation ⃗ is negative, so we expect the 𝑧-component of the curl to point down. The density around 𝑘 coordinate definition gives ( ) ) ) ( ( 𝜕𝑦 𝜕(0) ⃗ 𝜕(0) 𝜕(−𝑥) ⃗ 𝜕(−𝑥) 𝜕𝑦 ⃗ ⃗ ⃗. curl 𝐹 = − − − 𝑘 = −2𝑘 𝑖 + 𝑗 + 𝜕𝑦 𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑦 (c) At first glance, you might expect this vector field to have zero curl, as all the vectors are parallel to the 𝑥-axis. However, if you find the circulation around the curve 𝐶 in Figure 20.6, the sides contribute nothing (they are perpendicular to the vector field), the bottom contributes a negative quantity (the curve is in the opposite direction to the vector field), and the top contributes a larger positive quantity (the curve is in the same direction as the vector field and the magnitude of the

20.1 THE CURL OF A VECTOR FIELD

1003

vector field is larger at the top than at the bottom). Thus, the circulation around 𝐶 is positive and hence we expect the curl to be nonzero and point up. The coordinate definition gives ( ) ) ) ( ( 𝜕(−(𝑦 + 1)) 𝜕(0) ⃗ 𝜕(0) 𝜕(0) ⃗ 𝜕(0) 𝜕(−(𝑦 + 1)) ⃗ ⃗. curl 𝐹⃗ = − − − 𝑘 =𝑘 𝑖 + 𝑗 + 𝜕𝑦 𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑦 Another way to see that the curl is nonzero in this case is to imagine the vector field representing the velocity of moving water. A boat sitting in the water tends to rotate, as the water moves faster on one side than the other.

𝑦 𝜔 ⃗

𝐶

𝑟⃗ 𝑃 𝑥 Figure 20.6: Rectangular curve in 𝑥𝑦-plane

𝑣⃗

Figure 20.7: Rotating flywheel

Example 3

A flywheel is rotating with angular velocity 𝜔 ⃗ and the velocity of a point 𝑃 with position vector 𝑟⃗ is given by 𝑣⃗ = 𝜔 ⃗ × 𝑟⃗ . (See Figure 20.7.) Calculate curl 𝑣⃗ .

Solution

⃗ , using the determinant notation introduced in Section 13.4, we have If 𝜔 ⃗ = 𝜔1 ⃗𝑖 + 𝜔2 𝑗⃗ + 𝜔3 𝑘 | ⃗ ⃗ ⃗ | | 𝑖 𝑗 𝑘 | | | ⃗. 𝑣⃗ = 𝜔 ⃗ × 𝑟⃗ = || 𝜔1 𝜔2 𝜔3 || = (𝜔2 𝑧 − 𝜔3 𝑦)⃗𝑖 + (𝜔3 𝑥 − 𝜔1 𝑧)𝑗⃗ + (𝜔1 𝑦 − 𝜔2 𝑥)𝑘 | | | 𝑥 𝑦 𝑧 | | |

The curl formula can also be written using a determinant:

| | ⃗ ⃗𝑖 𝑗⃗ 𝑘 | | | | 𝜕 𝜕 𝜕 | curl 𝑣⃗ = || | 𝜕𝑥 𝜕𝑦 𝜕𝑧 | | | 𝜔 𝑧 − 𝜔 𝑦 𝜔 𝑥 − 𝜔 𝑧 𝜔 𝑦 − 𝜔 𝑥| | 2 3 3 1 1 2 | ( ) ) ( 𝜕 𝜕 𝜕 𝜕 = (𝜔 𝑦 − 𝜔2 𝑥) − (𝜔3 𝑥 − 𝜔1 𝑧) ⃗𝑖 + (𝜔 𝑧 − 𝜔3 𝑦) − (𝜔 𝑦 − 𝜔2 𝑥) 𝑗⃗ 𝜕𝑦 1 𝜕𝑧 𝜕𝑧 2 𝜕𝑥 1 ) ( 𝜕 𝜕 ⃗ (𝜔3 𝑥 − 𝜔1 𝑧) − (𝜔2 𝑧 − 𝜔3 𝑦) 𝑘 + 𝜕𝑥 𝜕𝑦 ⃗ = 2𝜔 = 2𝜔1 ⃗𝑖 + 2𝜔2 𝑗⃗ + 2𝜔3 𝑘 ⃗. Thus, as we would expect, curl 𝑣⃗ is parallel to the axis of rotation of the flywheel (namely, the direction of 𝜔 ⃗ ) and the magnitude of curl 𝑣⃗ is larger the faster the flywheel is rotating (that is, the larger the magnitude of 𝜔 ⃗ ).

1004

Chapter 20 THE CURL AND STOKES’ THEOREM

Why Do the Two Definitions of Curl Give the Same Result? Using Green’s Theorem in Cartesian coordinates, we can show that for curl 𝐹⃗ defined in Cartesian coordinates

curl 𝐹⃗ ⋅ 𝑛⃗ = circ𝑛⃗ 𝐹⃗ .

This shows that curl 𝐹⃗ defined in Cartesian coordinates satisfies the geometric definition, since the left-hand side takes its maximum value when 𝑛⃗ points in the same direction as curl 𝐹⃗ , and in that case its value is ‖ curl 𝐹⃗ ‖. The following example justifies this formula in a specific case. Example 4

Use the definition of curl in Cartesian coordinates and Green’s Theorem to show that ( ) ⃗ = circ⃗ 𝐹⃗ . curl 𝐹⃗ ⋅ 𝑘 𝑘

Solution

Using the definition of curl in Cartesian coordinates, the left-hand side of the formula is (

curl 𝐹⃗

)

⃗ = ⋅𝑘

𝜕𝐹2 𝜕𝐹1 − . 𝜕𝑥 𝜕𝑦

⃗ is calculated using circles Now let’s look at the right-hand side. The circulation density around 𝑘 ⃗ ⃗ ⃗ perpendicular to 𝑘 ; hence, the 𝑘 -component of 𝐹 does not contribute to it; that is, the circulation ⃗ is the same as the circulation density of 𝐹1 ⃗𝑖 + 𝐹2 𝑗⃗ around 𝑘 ⃗ . But in any density of 𝐹⃗ around 𝑘 ⃗ plane perpendicular to 𝑘 , 𝑧 is constant, so in that plane 𝐹1 and 𝐹2 are functions of 𝑥 and 𝑦 alone. Thus, 𝐹1 ⃗𝑖 + 𝐹2 𝑗⃗ can be thought of as a two-dimensional vector field on the horizontal plane through the point (𝑥, 𝑦, 𝑧) where the circulation density is being calculated. Let 𝐶 be a circle in this plane, with radius 𝑎 and centered at (𝑥, 𝑦, 𝑧), and let 𝑅 be the region enclosed by 𝐶. Green’s Theorem says that ) ( 𝜕𝐹2 𝜕𝐹1 ⃗ ⃗ − 𝑑𝐴. (𝐹 𝑖 + 𝐹2 𝑗 ) ⋅ 𝑑⃗𝑟 = ∫𝐶 1 ∫𝑅 𝜕𝑥 𝜕𝑦 When the circle is small, 𝜕𝐹2 ∕𝜕𝑥 − 𝜕𝐹1 ∕𝜕𝑦 is approximately constant on 𝑅, so ) ( ) ( ) ( 𝜕𝐹2 𝜕𝐹1 𝜕𝐹2 𝜕𝐹1 𝜕𝐹2 𝜕𝐹1 − 𝑑𝐴 ≈ − ⋅ Area of 𝑅 = − 𝜋𝑎2 . ∫𝑅 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 Thus, taking a limit as the radius of the circle goes to zero, we have ) ( 𝜕𝐹2 𝜕𝐹1 ⃗ ⃗ − 𝑑𝐴 (𝐹1 𝑖 + 𝐹2 𝑗 ) ⋅ 𝑑⃗𝑟 ∫𝑅 𝜕𝑥 ∫ 𝜕𝑦 𝜕𝐹2 𝜕𝐹1 − . circ𝑘⃗ 𝐹⃗ (𝑥, 𝑦, 𝑧) = lim 𝐶 = lim = 2 2 𝑎→0 𝑎→0 𝜕𝑥 𝜕𝑦 𝜋𝑎 𝜋𝑎

Curl-Free Vector Fields A vector field is said to be curl free or irrotational if curl 𝐹⃗ = 0⃗ everywhere that 𝐹⃗ is defined.

20.1 THE CURL OF A VECTOR FIELD

Example 5

1005

Figure 20.8 shows the vector field 𝐵⃗ for three values of the constant 𝑝, where 𝐵⃗ is defined on 3-space by −𝑦⃗𝑖 + 𝑥𝑗⃗ 𝐵⃗ = 2 . (𝑥 + 𝑦2 )𝑝∕2 (a) Find a formula for curl 𝐵⃗ . (b) Is there a value of 𝑝 for which 𝐵⃗ is curl free? If so, find it.

Figure 20.8: The vector field 𝐵⃗ (⃗𝑟 ) = (−𝑦⃗𝑖 + 𝑥𝑗⃗ )∕(𝑥2 + 𝑦2 )𝑝∕2 for 𝑝 = 0, 2, and 4

Solution

(a) We can use the following version of the product rule for curl. If 𝜙 is a scalar function and 𝐹⃗ is a vector field, then curl(𝜙𝐹⃗ ) = 𝜙 curl 𝐹⃗ + (grad 𝜙) × 𝐹⃗ . (See Problem 32 on page 1007.) We write 𝐵⃗ = 𝜙𝐹⃗ =

(𝑥2

1 (−𝑦⃗𝑖 + 𝑥𝑗⃗ ). Then + 𝑦2 )𝑝∕2

⃗ curl 𝐹⃗ = curl(−𝑦⃗𝑖 + 𝑥𝑗⃗ ) = 2𝑘 ) ( −𝑝 1 = 2 (𝑥⃗𝑖 + 𝑦𝑗⃗ ). grad 𝜙 = grad (𝑥2 + 𝑦2 )𝑝∕2 (𝑥 + 𝑦2 )(𝑝∕2)+1 Thus, we have ) ( 1 1 × (−𝑦⃗𝑖 + 𝑥𝑗⃗ ) curl(−𝑦⃗𝑖 + 𝑥𝑗⃗ ) + grad (𝑥2 + 𝑦2 )𝑝∕2 (𝑥2 + 𝑦2 )𝑝∕2 −𝑝 1 ⃗ + = 2𝑘 (𝑥⃗𝑖 + 𝑦𝑗⃗ ) × (−𝑦⃗𝑖 + 𝑥𝑗⃗ ) 2 2 𝑝∕2 2 (𝑥 + 𝑦 ) (𝑥 + 𝑦2 )(𝑝∕2)+1 −𝑝 1 ⃗ + ⃗ 2𝑘 (𝑥2 + 𝑦2 )𝑘 = (𝑥2 + 𝑦2 )𝑝∕2 (𝑥2 + 𝑦2 )(𝑝∕2)+1 2−𝑝 ⃗. = 2 𝑘 (𝑥 + 𝑦2 )𝑝∕2

curl 𝐵⃗ =

(b) The curl is zero when 𝑝 = 2. Thus, when 𝑝 = 2 the vector field is curl free: −𝑦⃗𝑖 + 𝑥𝑗⃗ 𝐵⃗ = . 𝑥2 + 𝑦2

1006

Chapter 20 THE CURL AND STOKES’ THEOREM

Exercises and Problems for Section 20.1 Online Resource: Additional Problems for Section 20.1 EXERCISES

In Exercises 1–5, is the quantity a vector or a scalar? Calculate it. ⃗) 1. curl(𝑧⃗𝑖 − 𝑥𝑗⃗ + 𝑦𝑘 ⃗) 2. circ⃗𝑖 (𝑧⃗𝑖 − 𝑥𝑗⃗ + 𝑦𝑘

In Exercises 14–17, does the vector field appear to have nonzero curl at the origin? The vector field is shown in the 𝑥𝑦-plane; it has no 𝑧-component and is independent of 𝑧. 𝑦 𝑦 14. 15.

⃗) 3. curl(−2𝑧⃗𝑖 − 𝑧𝑗⃗ + 𝑥𝑦𝑘 ⃗) 4. circ𝑘⃗ (−2𝑧⃗𝑖 − 𝑧𝑗⃗ + 𝑥𝑦𝑘 𝑥

⃗) 5. curl(𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘

𝑥

In Exercises 6–13, compute the curl of the vector field. ⃗ 6. 𝐹⃗ = 3𝑥⃗𝑖 − 5𝑧𝑗⃗ + 𝑦𝑘 2 2 ⃗ ⃗ 7. 𝐹 = (𝑥 − 𝑦 )𝑖 + 2𝑥𝑦𝑗⃗ ⃗ 8. 𝐹⃗ = (−𝑥 + 𝑦)⃗𝑖 + (𝑦 + 𝑧)𝑗⃗ + (−𝑧 + 𝑥)𝑘 ⃗ ⃗ ⃗ ⃗ 9. 𝐹 = 2𝑦𝑧𝑖 + 3𝑥𝑧𝑗 + 7𝑥𝑦𝑘 ⃗ 10. 𝐹⃗ = 𝑥2 ⃗𝑖 + 𝑦3 𝑗⃗ + 𝑧4 𝑘 2 𝑥 ⃗ 11. 𝐹⃗ = 𝑒 ⃗𝑖 + cos 𝑦𝑗⃗ + 𝑒𝑧 𝑘 2 ⃗ ⃗ ⃗ 12. 𝐹 = (𝑥 + 𝑦𝑧)𝑖 + (𝑦 + 𝑥𝑧𝑦)𝑗⃗ + (𝑧𝑥3 𝑦2 + 𝑥7 𝑦6 )𝑘 13. 𝐹⃗ (⃗𝑟 ) = 𝑟⃗ ∕‖⃗𝑟 ‖

16.

𝑦

17.

𝑦

𝑥

𝑥

PROBLEMS

18. Let 𝐹⃗ be the vector field in Figure 20.2 on page 1000. It is rotating counterclockwise around the 𝑧-axis when viewed from above. At a distance 𝑟 from the 𝑧-axis, 𝐹⃗ has magnitude 2𝑟. (a) Find a formula for 𝐹⃗ . (b) Find curl 𝐹⃗ using the coordinate definition and relate your answer to circulation density. 19. Use the geometric definition to find the curl of the vector field 𝐹⃗ (⃗𝑟 ) = 𝑟⃗ . Check your answer using the coordinate definition. 20. A smooth vector field 𝐺⃗ has curl 𝐺⃗ (0, 0, 0) = 2⃗𝑖 − ⃗ . Estimate the circulation around a circle of ra3𝑗⃗ + 5𝑘 dius 0.01 centered at the origin in each of the following planes: (a) 𝑥𝑦-plane, oriented counterclockwise when viewed from the positive 𝑧-axis. (b) 𝑦𝑧-plane, oriented counterclockwise when viewed from the positive 𝑥-axis. (c) 𝑥𝑧-plane, oriented counterclockwise when viewed from the positive 𝑦-axis. 21. Three small circles, 𝐶1 , 𝐶2 , and 𝐶3 , each with radius 0.1 and centered at the origin, are in the 𝑥𝑦-, 𝑦𝑧-, and 𝑥𝑧-planes, respectively. The circles are oriented counterclockwise when viewed from the positive 𝑧-, 𝑥-, and 𝑦-axes, respectively. A vector field, 𝐹⃗ , has circulation around 𝐶1 of 0.02𝜋, around 𝐶2 of 0.5𝜋, and around 𝐶3 of 3𝜋. Estimate curl 𝐹⃗ at the origin.

22. Using your answers to Exercises 10–11, make a conjecture about a particular form of the vector field 𝐹⃗ ≠ 0⃗ that has curl 𝐹⃗ = 0⃗ . What form? Show why your conjecture is true. 23. (a) Find curl 𝐺⃗ if 𝐺⃗ = (𝑎𝑦3 + 𝑏𝑒𝑧 )⃗𝑖 + (𝑐𝑧 + 𝑑𝑥2 )𝑗⃗ + ⃗ and 𝑎, 𝑏, 𝑐, 𝑑, 𝑒, 𝑓 are constants. (𝑒 sin 𝑥 + 𝑓 𝑦)𝑘 (b) If curl 𝐺⃗ is everywhere parallel to the 𝑦𝑧-plane, what can you say about the constants 𝑎–𝑓 ? (c) If curl 𝐺⃗ is everywhere parallel to the 𝑧-axis, what can you say about the constants 𝑎–𝑓 ? 24. Figure 20.9 gives a sketch of the velocity vector field 𝐹⃗ = 𝑦⃗𝑖 + 𝑥𝑗⃗ in the 𝑥𝑦-plane. (a) What is the direction of rotation of a thin twig placed at the origin along the 𝑥-axis? (b) What is the direction of rotation of a thin twig placed at the origin along the 𝑦-axis? (c) Compute curl 𝐹⃗ . 𝑦

𝑥

Figure 20.9

20.1 THE CURL OF A VECTOR FIELD

25. A tornado is formed when a tube of air circling a horizontal axis is tilted up vertically by the updraft from a thunderstorm. If 𝑡 is time, this process can be modeled by the wind velocity field ⃗ ) × 𝑟⃗ and 0 ≤ 𝑡 ≤ 𝜋 . 𝐹⃗ (𝑡, 𝑥, 𝑦, 𝑧) = (cos 𝑡𝑗⃗ + sin 𝑡𝑘 2 Determine the direction of curl 𝐹⃗ : (a) At 𝑡 = 0 (c) For 0 < 𝑡 < 𝜋∕2

(b) At 𝑡 = 𝜋∕2

26. A large fire becomes a fire-storm when the nearby air acquires a circulatory motion. The associated updraft has the effect of bringing more air to the fire, causing it to burn faster. Records show that a fire-storm developed during the Chicago Fire of 1871 and during the Second World War bombing of Hamburg, Germany, but there was no fire-storm during the Great Fire of London in 1666. Explain how a fire-storm could be identified using the curl of a vector field. 27. A vortex that rotates at constant angular velocity 𝜔 about the 𝑧-axis has velocity vector field 𝑣⃗ = 𝜔(−𝑦⃗𝑖 + 𝑥𝑗⃗ ). (a) Sketch the vector field with 𝜔 = 1 and the vector field with 𝜔 = −1. (b) Determine the speed ‖𝑣⃗ ‖ of the vortex as a function of the distance from its center.

1007

(c) Compute div 𝑣⃗ and curl 𝑣⃗ . (d) Compute the circulation of 𝑣⃗ counterclockwise about the circle of radius 𝑅 in the 𝑥𝑦-plane, centered at the origin. 28. A central vector field is one of the form 𝐹⃗ = 𝑓 (𝑟)⃗𝑟 where 𝑓 is any function of 𝑟 = ‖⃗𝑟 ‖. Show that any central vector field is irrotational. 29. Show that curl (𝐹⃗ + 𝐶⃗ ) = curl 𝐹⃗ for a constant vector field 𝐶⃗ . 30. If 𝐹⃗ is any vector field whose components have continuous second partial derivatives, show div curl 𝐹⃗ = 0. 31. We have seen that the Fundamental Theorem of Calculus for Line Integrals implies ∫𝐶 grad 𝑓 ⋅𝑑⃗𝑟 = 0 for any smooth closed path 𝐶 and any smooth function 𝑓 . (a) Use the geometric definition of curl to deduce that curl grad 𝑓 = 0⃗ . (b) Show that curl grad 𝑓 = 0⃗ using the coordinate definition. 32. Show that curl (𝜙𝐹⃗ ) = 𝜙 curl 𝐹⃗ + (grad 𝜙) × 𝐹⃗ for a scalar function 𝜙 and a vector field 𝐹⃗ . 33. Show that if 𝐹⃗ = 𝑓 grad 𝑔 for some scalar functions 𝑓 and 𝑔, then curl 𝐹⃗ is everywhere perpendicular to 𝐹⃗ .

Strengthen Your Understanding In Problems 34–35, explain what is wrong with the statement. 34. A vector field 𝐹⃗ has curl given by curl 𝐹⃗ = 2𝑥 − 3𝑦. 35. If all the vectors of a vector field 𝐹⃗ are parallel, then curl 𝐹⃗ = 0⃗ .

44. curl(𝑓 𝐺⃗ ) = (grad 𝑓 ) × 𝐺⃗ + 𝑓 (curl 𝐺⃗ ). 45. For any vector field 𝐹⃗ , the curl of 𝐹⃗ is perpendicular at every point to 𝐹⃗ . 46. If 𝐹⃗ is as shown in Figure 20.10, then curl 𝐹⃗ ⋅ 𝑗⃗ > 0.

In Problems 36–37, give an example of: 36. A vector field 𝐹⃗ (𝑥, 𝑦, 𝑧) such that curl 𝐹⃗ = 0⃗ . 37. A vector field 𝐹⃗ (𝑥, 𝑦, 𝑧) such that curl 𝐹⃗ = 𝑗⃗ . In Problems 38–46, is the statement true or false? Assume 𝐹⃗ and 𝐺⃗ are smooth vector fields and 𝑓 is a smooth function on 3-space. Explain. 38. The circulation density, circ𝑛⃗ 𝐹⃗ (𝑥, 𝑦, 𝑧), is a scalar. 39. curl grad 𝑓 = 0. 40. If 𝐹⃗ is a vector field with div𝐹⃗ = 0 and curl𝐹⃗ = 0⃗ , then 𝐹⃗ = 0⃗ . 41. If 𝐹⃗ and 𝐺⃗ are vector fields, then curl(𝐹⃗ + 𝐺⃗ ) = curl𝐹⃗ + curl𝐺⃗ . 42. If 𝐹⃗ and 𝐺⃗ are vector fields, then curl(𝐹⃗ ⋅ 𝐺⃗ ) = curl𝐹⃗ ⋅ curl𝐺⃗ . 43. If 𝐹⃗ and 𝐺⃗ are vector fields, then curl(𝐹⃗ × 𝐺⃗ ) = (curl𝐹⃗ ) × (curl𝐺⃗ ).

Figure 20.10

47. Of the following vector fields, which ones have a curl which is parallel to one of the axes? Which axis? ⃗ (b) 𝑦⃗𝑖 +𝑧𝑗⃗ +𝑥𝑘 ⃗ (c) 𝑦⃗𝑖 −𝑥𝑗⃗ +𝑧𝑘 ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ (d) 𝑥𝑖 +𝑧𝑗 −𝑦𝑘 (e) 𝑧𝑖 +𝑥𝑗 +𝑦𝑘

(a)

⃗ −𝑧⃗𝑖 +𝑦𝑗⃗ +𝑥𝑘

1008

20.2

Chapter 20 THE CURL AND STOKES’ THEOREM

STOKES’ THEOREM The Divergence Theorem says that the integral of the flux density over a solid region is equal to the flux through the surface bounding the region. Similarly, Stokes’ Theorem says that the integral of the circulation density over a surface is equal to the circulation around the boundary of the surface.

The Boundary of a Surface The boundary of a surface 𝑆 is the curve or curves running around the edge of 𝑆 (like the hem around the edge of a piece of cloth). An orientation of 𝑆 determines an orientation for its boundary, 𝐶, as follows. Pick a positive normal vector 𝑛⃗ on 𝑆, near 𝐶, and use the right-hand rule to determine a direction of travel around 𝑛⃗ . This in turn determines a direction of travel around the boundary 𝐶. See Figure 20.11. Another way of describing the orientation on 𝐶 is that someone walking along 𝐶 in the forward direction, body upright in the direction of the positive normal on 𝑆, would have the surface on their left. Notice that the boundary can consist of two or more curves, as the surface on the right in Figure 20.11 shows. 𝐶

𝑛⃗

𝑛⃗

𝑆

𝐶

𝐶

𝑆 𝑛⃗ Figure 20.11: Two oriented surfaces and their boundaries

Calculating the Circulation from the Circulation Density Consider a closed, oriented curve 𝐶 in 3-space. We can find the circulation of a vector field 𝐹⃗ around 𝐶 by calculating the line integral: Circulation

=

around 𝐶

∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 .

If 𝐶 is the boundary of an oriented surface 𝑆, there is another way to calculate the circulation using curl 𝐹⃗ . We subdivide 𝑆 into pieces as shown on the surface on the left in Figure 20.11. If 𝑛⃗ is a positive unit normal vector to a piece of surface with area Δ𝐴, then Δ𝐴⃗ = 𝑛⃗ Δ𝐴. In addition, circ𝑛⃗ 𝐹⃗ is the circulation density of 𝐹⃗ around 𝑛⃗ , so ) ( Circulation of 𝐹⃗ around ≈ circ𝑛⃗ 𝐹⃗ Δ𝐴 = ((curl 𝐹⃗ ) ⋅ 𝑛⃗ )Δ𝐴 = (curl𝐹⃗ ) ⋅ Δ𝐴⃗ . boundary of the piece Next we add up the circulations around all the small pieces. The line integral along the common edge of a pair of adjacent pieces appears with opposite sign in each piece, so it cancels out. (See Figure 20.12.) When we add up all the pieces the internal edges cancel and we are left with the circulation around 𝐶, the boundary of the entire surface. Thus, Circulation around 𝐶

=



Circulation around boundary of pieces





Taking the limit as Δ𝐴 → 0, we get Circulation around 𝐶

=

∫𝑆

curl 𝐹⃗ ⋅ 𝑑 𝐴⃗ .

curl 𝐹⃗ ⋅ Δ𝐴⃗ .

20.2 STOKES’ THEOREM

1009

Figure 20.12: Two adjacent pieces of the surface

We have expressed the circulation as a line integral around 𝐶 and as a flux integral over 𝑆; thus, the two integrals must be equal. Hence we have1

Theorem 20.1: Stokes’ Theorem If 𝑆 is a smooth oriented surface with piecewise smooth, oriented boundary 𝐶, and if 𝐹⃗ is a smooth vector field on a solid region2 containing 𝑆 and 𝐶, then ∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫𝑆

curl 𝐹⃗ ⋅ 𝑑 𝐴⃗ .

The orientation of 𝐶 is determined from the orientation of 𝑆 according to the right-hand rule.

Example 1

Let 𝐹⃗ (𝑥, 𝑦, 𝑧) = −2𝑦⃗𝑖 + 2𝑥𝑗⃗ . Use Stokes’ Theorem to find ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 , where 𝐶 is a circle. (a) Parallel to the 𝑦𝑧-plane, of radius 𝑎, centered at a point on the 𝑥-axis, with either orientation. (b) Parallel to the 𝑥𝑦-plane, of radius 𝑎, centered at a point on the 𝑧-axis, oriented counterclockwise as viewed from a point on the 𝑧-axis above the circle.

Solution

⃗ . Figure 20.13 shows sketches of 𝐹⃗ and curl 𝐹⃗ . We have curl 𝐹⃗ = 4𝑘 (a) Let 𝑆 be the disk enclosed by 𝐶. Since 𝑆 lies in a vertical plane and curl 𝐹⃗ points vertically everywhere, the flux of curl 𝐹⃗ through 𝑆 is zero. Hence, by Stokes’ Theorem, ∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫𝑆

curl 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 0.

It makes sense that the line integral is zero. If 𝐶 is parallel to the 𝑦𝑧-plane (even if it is not lying in the plane), the symmetry of the vector field means that the line integral of 𝐹⃗ over the top half of the circle cancels the line integral over the bottom half.

Figure 20.13: The vector fields 𝐹⃗ and curl 𝐹⃗ (in the planes 𝑧 = −1, 𝑧 = 0, 𝑧 = 1) 1 A proof of Stokes’ Theorem using the coordinate definition of curl can be found in the online supplement at www.wiley.com/college/hughes-hallett. 2 The region containing 𝑆 and 𝐶 is open.

1010

Chapter 20 THE CURL AND STOKES’ THEOREM

(b) Let 𝑆 be the horizontal disk enclosed by 𝐶. Since curl 𝐹⃗ is a constant vector field pointing in ⃗ , we have, by Stokes’ Theorem, the direction of 𝑘 ∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫𝑆

curl 𝐹⃗ ⋅ 𝑑 𝐴⃗ = ‖ curl 𝐹⃗ ‖ ⋅ Area of 𝑆 = 4𝜋𝑎2 .

Since 𝐹⃗ is circling around the 𝑧-axis in the same direction as 𝐶, we expect the line integral to be positive. In fact, in Example 1 on page 1001, we computed this line integral directly.

Curl-Free Vector Fields Stokes’ Theorem applies to any oriented surface 𝑆 and its boundary 𝐶, even in cases where the boundary consists of two or more curves. This is useful in studying curl-free vector fields. Example 2

⃗ direction. The induced magnetic field 𝐵⃗ (𝑥, 𝑦, 𝑧) is A current 𝐼 flows along the 𝑧-axis in the 𝑘 ) ( ⃗𝑖 + 𝑥𝑗⃗ −𝑦 2𝐼 , 𝐵⃗ (𝑥, 𝑦, 𝑧) = 𝑐 𝑥2 + 𝑦2 where 𝑐 is the speed of light. Example 5 on page 1005 shows that curl 𝐵⃗ = 0⃗ . (a) Compute the circulation of 𝐵⃗ around the circle 𝐶1 in the 𝑥𝑦-plane of radius 𝑎, centered at the origin, and oriented counterclockwise when viewed from above. (b) Use part (a) and Stokes’ Theorem to compute ∫𝐶 𝐵⃗ ⋅ 𝑑⃗𝑟 , where 𝐶2 is the ellipse 𝑥2 + 9𝑦2 = 9 2 in the plane 𝑧 = 2, oriented counterclockwise when viewed from above.

Solution

(a) On the circle 𝐶1 , we have ‖𝐵⃗ ‖ = 2𝐼∕(𝑐𝑎). Since 𝐵⃗ is tangent to 𝐶1 everywhere and points in the forward direction around 𝐶1 , ∫𝐶

4𝜋𝐼 2𝐼 ⋅ 2𝜋𝑎 = . 𝐵⃗ ⋅ 𝑑⃗𝑟 = ‖𝐵⃗ ‖ ⋅ Length of 𝐶1 = 𝑐𝑎 𝑐 1

(b) We cannot use Stokes’ Theorem on the elliptical disk bounded by 𝐶2 in the plane 𝑧 = 2 because curl 𝐵⃗ is not defined at (0, 0, 2). Instead, we will use the theorem on a conical surface connecting 𝐶1 and 𝐶2 . Let 𝑆 be the conical surface extending from 𝐶1 to 𝐶2 in Figure 20.14. The boundary of this surface has two pieces, −𝐶2 and 𝐶1 . The orientation of 𝐶1 leads to the outward normal on 𝑆, which forces us to choose the clockwise orientation on 𝐶2 . By Stokes’ Theorem, ∫𝑆

curl 𝐵⃗ ⋅ 𝑑 𝐴⃗ =

Since curl 𝐵⃗ = 0⃗ , we have

∫−𝐶 ∫𝑆

𝐵⃗ ⋅ 𝑑⃗𝑟 + 2

∫𝐶

𝐵⃗ ⋅ 𝑑⃗𝑟 = − 1

∫𝐶

𝐵⃗ ⋅ 𝑑⃗𝑟 + 2

∫𝐶

𝐵⃗ ⋅ 𝑑⃗𝑟 . 1

curl 𝐵⃗ ⋅ 𝑑 𝐴⃗ = 0, so the two line integrals must be equal:

∫ 𝐶2

𝐵⃗ ⋅ 𝑑⃗𝑟 =

∫ 𝐶1

4𝜋𝐼 . 𝐵⃗ ⋅ 𝑑⃗𝑟 = 𝑐

1011

20.2 STOKES’ THEOREM

𝑧

−𝐶2

𝑛⃗2

𝑛⃗1 𝑛⃗

𝑆2 𝑆

𝑥

𝐶

𝑆1

𝐶1 𝑦

Figure 20.15: The flux of a curl is the same through the two surfaces 𝑆1 and 𝑆2 if they determine the same orientation on the boundary, 𝐶

Figure 20.14: Surface joining 𝐶1 to 𝐶2 , oriented to satisfy the conditions of Stokes’ Theorem

Curl Fields A vector field 𝐹⃗ is called a curl field if 𝐹⃗ = curl 𝐺⃗ for some vector field 𝐺⃗ . Recall that if 𝐹⃗ = grad𝑓 , then 𝑓 is called a potential function. By analogy, if a vector field 𝐹⃗ = curl 𝐺⃗ , then 𝐺⃗ is called a vector potential for 𝐹⃗ . The following example shows that the flux of a curl field through a surface depends only on the boundary of the surface. This is analogous to the fact that the line integral of a gradient field depends only on the endpoints of the path. Example 3

Suppose 𝐹⃗ = curl 𝐺⃗ , and that 𝑆1 and 𝑆2 are two oriented surfaces with the same boundary 𝐶. Show that, if 𝑆1 and 𝑆2 determine the same orientation on 𝐶 (as in Figure 20.15), then ∫𝑆1

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

𝐹⃗ ⋅ 𝑑 𝐴⃗ .

∫𝑆2

If 𝑆1 and 𝑆2 determine opposite orientations on 𝐶, then ∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ = −

1

Solution

∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ .

2

If 𝑆1 and 𝑆2 determine the same orientation on 𝐶, then since 𝐹⃗ = curl 𝐺⃗ , by Stokes’ Theorem we have curl 𝐺⃗ ⋅ 𝑑 𝐴⃗ = 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 𝐺⃗ ⋅ 𝑑⃗𝑟 ∫𝑆1 ∫𝐶 ∫𝑆1 and ∫𝑆2

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

∫𝑆2

curl 𝐺⃗ ⋅ 𝑑 𝐴⃗ =

∫𝐶

𝐺⃗ ⋅ 𝑑⃗𝑟 .

In each case the line integral on the right must be computed using the orientation determined by the surface. Thus, the two flux integrals of 𝐹⃗ are the same if the orientations are the same and they are opposite if the orientations are opposite.

Exercises and Problems for Section 20.2 EXERCISES ⃗ , find the circulation of 𝐹⃗ around 𝐶, a 1. If curl 𝐹⃗ = 𝑘 circle of radius 1, centered at the origin, with (a) 𝐶 in the 𝑥𝑦-plane, oriented counterclockwise when viewed from above.

(b) 𝐶 in the 𝑦𝑧-plane, oriented counterclockwise when viewed from the positive 𝑥-axis.

1012

Chapter 20 THE CURL AND STOKES’ THEOREM

In Exercises 2–3, for the circle 𝐶 and curl 𝐹⃗ as described, is the circulation of 𝐹⃗ around 𝐶 positive, negative, or zero? 2. 𝐶 is in the 𝑥𝑦-plane oriented counterclockwise when viewed from above, and curl 𝐹⃗ points parallel to 𝑗⃗ everywhere. 3. 𝐶 is in the 𝑦𝑧-plane oriented clockwise when viewed from the positive 𝑥-axis, and curl 𝐹⃗ points parallel to and in the direction of −⃗𝑖 . In Exercises 4–8, calculate the circulation, ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 , in two ways, directly and using Stokes’ Theorem. ⃗ and 𝐶 is the upper half of 4. 𝐹⃗ = (𝑥 + 𝑧)⃗𝑖 + 𝑥𝑗⃗ + 𝑦𝑘 the circle 𝑥2 + 𝑧2 = 9 in the plane 𝑦 = 0, together with the 𝑥-axis from (3, 0, 0) to (−3, 0, 0), traversed counterclockwise when viewed from the positive 𝑦-axis. 5. 𝐹⃗ = 𝑦⃗𝑖 − 𝑥𝑗⃗ and 𝐶 is the boundary of 𝑆, the part of the surface 𝑧 = 4 − 𝑥2 − 𝑦2 above the 𝑥𝑦-plane, oriented upward. 6. 𝐹⃗ = (𝑥−𝑦+𝑧)(⃗𝑖 +𝑗⃗ ) and 𝐶 is the triangle with vertices (0, 0, 0), (5, 0, 0), (5, 5, 0), traversed in that order. ⃗ and 𝐶 is the boundary of 𝑆, 7. 𝐹⃗ = 𝑥𝑦⃗𝑖 + 𝑦𝑧𝑗⃗ + 𝑥𝑧𝑘 2 the surface 𝑧 = 1 − 𝑥 for 0 ≤ 𝑥 ≤ 1 and −2 ≤ 𝑦 ≤ 2, oriented upward. Sketch 𝑆 and 𝐶. ⃗ and 𝐶 is the boundary of 𝑆, the 8. 𝐹⃗ = 𝑦⃗𝑖 + 𝑧𝑗⃗ + 𝑥𝑘 paraboloid 𝑧 = 1 − (𝑥2 + 𝑦2 ), 𝑧 ≥ 0 oriented upward. [Hint: Use polar coordinates.] In Exercises 9–12, use Stokes’ Theorem to calculate the integral.

⃗ and 𝐶 is the unit 9. ∫𝐶 𝐹⃗ ⋅𝑑⃗𝑟 where 𝐹⃗ = 𝑥2 ⃗𝑖 +𝑦2 𝑗⃗ +𝑧2 𝑘 circle in the 𝑥𝑧-plane, oriented counterclockwise when viewed from the positive 𝑦-axis. ⃗ 10. ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 where 𝐹⃗ = (𝑦 − 𝑥)⃗𝑖 + (𝑧 − 𝑦)𝑗⃗ + (𝑥 − 𝑧)𝑘 2 2 and 𝐶 is the circle 𝑥 + 𝑦 = 5 in the 𝑥𝑦-plane, oriented counterclockwise when viewed from above. ⃗ 11. ∫𝑆 curl 𝐹⃗ ⋅ 𝑑 𝐴⃗ where 𝐹⃗ = −𝑦⃗𝑖 + 𝑥𝑗⃗ + (𝑥𝑦 + cos 𝑧)𝑘 and 𝑆 is the disk 𝑥2 + 𝑦2 ≤ 9, oriented upward in the 𝑥𝑦-plane. ⃗ and 𝑆 12. ∫𝑆 curl 𝐹⃗ ⋅ 𝑑 𝐴⃗ where 𝐹⃗ = (𝑥 + 7)𝑗⃗ + 𝑒𝑥+𝑦+𝑧 𝑘 is the rectangle 0 ≤ 𝑥 ≤ 3, 0 ≤ 𝑦 ≤ 2 , 𝑧 = 0, oriented upward. 13. Let 𝐹⃗ = 𝑦⃗𝑖 − 𝑥𝑗⃗ and let 𝐶 be the unit circle in the 𝑥𝑦-plane centered at the origin and oriented counterclockwise when viewed from above. (a) (b) (c) (d)

Calculate ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 by parameterizing the circle. Calculate curl 𝐹⃗ . Calculate ∫𝐶 𝐹⃗ ⋅𝑑⃗𝑟 using your result from part (b). What theorem did you use in part (c)?

⃗ , find curl 𝐹⃗ . 14. (a) If 𝐹⃗ = (cos 𝑥)⃗𝑖 + 𝑒𝑦 𝑗⃗ + (𝑥 − 𝑦 − 𝑧)𝑘 (b) Find ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 where 𝐶 is the circle of radius 3 in the plane 𝑥+𝑦+𝑧 = 1, centered at (1, 0, 0) oriented counterclockwise when viewed from above. 15. Can you use Stokes’ Theorem to compute the line inte⃗ ) ⋅ 𝑑⃗𝑟 where 𝐶 is the straight gral ∫𝐶 (2𝑥⃗𝑖 + 2𝑦𝑗⃗ + 2𝑧𝑘 line from the point (1, 2, 3) to the point (4, 5, 6)? Why or why not?

PROBLEMS 16. At all points in 3-space curl 𝐹⃗ points in the direction ⃗ . Let 𝐶 be a circle in the 𝑦𝑧-plane, oriented of ⃗𝑖 − 𝑗⃗ − 𝑘 clockwise when viewed from the positive 𝑥-axis. Is the circulation of 𝐹⃗ around 𝐶 positive, zero, or negative? ⃗ , find ∫ 𝐹⃗ ⋅ 𝑑⃗𝑟 , where 𝐶 17. If curl 𝐹⃗ = (𝑥2 + 𝑧2 )𝑗⃗ + 5𝑘 𝐶 is a circle of radius 3, centered at the origin, with (a) 𝐶 in the 𝑥𝑦-plane, oriented counterclockwise when viewed from above. (b) 𝐶 in the 𝑥𝑧-plane, oriented counterclockwise when viewed from the positive 𝑦-axis. ⃗ ). 18. (a) Find curl(𝑦⃗𝑖 + 𝑧𝑗⃗ + 𝑥𝑘 ⃗ ) ⋅ 𝑑⃗𝑟 where 𝐶 is the bound(b) Find ∫𝐶 (𝑦⃗𝑖 + 𝑧𝑗⃗ + 𝑥𝑘 ary of the triangle with vertices (2, 0, 0), (0, 3, 0), (−2, 0, 0), traversed in that order. ⃗ . Find curl 𝐹⃗ . 19. (a) Let 𝐹⃗ = 𝑦⃗𝑖 + 𝑧𝑗⃗ + 𝑥𝑘 ⃗ (b) Calculate ∫𝐶 𝐹 ⋅ 𝑑⃗𝑟 where 𝐶 is (i) A circle of radius 2 centered at (1, 1, 3) in the plane 𝑧 = 3, oriented counterclockwise when viewed from above. (ii) The triangle obtained by tracing out the path (2, 0, 0) to (2, 0, 5) to (2, 3, 5) to (2, 0, 0).

⃗ ). 20. (a) Find curl(𝑧⃗𝑖 + 𝑥𝑗⃗ + 𝑦𝑘 ⃗ ) ⋅ 𝑑⃗𝑟 where 𝐶 is a square (b) Find ∫𝐶 (𝑧⃗𝑖 + 𝑥𝑗⃗ + 𝑦𝑘 of side 2 lying in the plane 𝑥 + 𝑦 + 𝑧 = 5, oriented counterclockwise when viewed from the origin. In Problems 21–26, find ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 where 𝐶 is a circle of radius 2 in the plane 𝑥 + 𝑦 + 𝑧 = 3, centered at (1, 1, 1) and oriented clockwise when viewed from the origin. ⃗ 21. 𝐹⃗ = ⃗𝑖 + 𝑗⃗ + 3𝑘 ⃗ 22. 𝐹⃗ = −𝑦⃗𝑖 + 𝑥𝑗⃗ + 𝑧𝑘 ⃗ 23. 𝐹⃗ = 𝑦⃗𝑖 − 𝑥𝑗⃗ + (𝑦 − 𝑥)𝑘 ⃗ 24. 𝐹⃗ = (2𝑦 + 𝑒𝑥 )⃗𝑖 + ((sin 𝑦) − 𝑥)𝑗⃗ + (2𝑦 − 𝑥 + cos 𝑧2 )𝑘 ⃗ 25. 𝐹⃗ = −𝑧𝑗⃗ + 𝑦𝑘 ⃗ 26. 𝐹⃗ = (𝑧 − 𝑦)⃗𝑖 + (𝑥 − 𝑧)𝑗⃗ + (𝑦 − 𝑥)𝑘 27. For positive constants 𝑎, 𝑏, and 𝑐, let 𝑓 (𝑥, 𝑦, 𝑧) = ln(1 + 𝑎𝑥2 + 𝑏𝑦2 + 𝑐𝑧2 ). (a) What is the domain of 𝑓 ?

20.2 STOKES’ THEOREM

(b) Find grad𝑓 . (c) Find curl(grad𝑓 ). (d) Find ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 where 𝐶 is the helix 𝑥 = cos 𝑡, 𝑦 = sin 𝑡, 𝑧 = 𝑡 for 0 ≤ 𝑡 ≤ 13𝜋∕2 and ⃗ 2𝑥⃗𝑖 + 4𝑦𝑗⃗ + 6𝑧𝑘 𝐹⃗ = . 1 + 𝑥2 + 2𝑦2 + 3𝑧2 28. Figure 20.16 shows an open cylindrical can, 𝑆, standing on the 𝑥𝑦-plane. (𝑆 has a bottom and sides, but no top.) (a) Give equation(s) for the rim, 𝐶. (b) If 𝑆 is oriented outward and downward, find ⃗ ) ⋅ 𝑑 𝐴⃗ . ∫𝑆 curl(−𝑦⃗𝑖 + 𝑥𝑗⃗ + 𝑧𝑘

1013

33. In the region between the circles 𝐶1 ∶ 𝑥2 + 𝑦2 = 4 and 𝐶2 ∶ 𝑥2 + 𝑦2 = 25 in the 𝑥𝑦-plane, the vector field 𝐹⃗ ⃗ . If 𝐶1 and 𝐶2 are both oriented counhas curl 𝐹⃗ = 3𝑘 terclockwise when viewed from above, find the value of 𝐹⃗ ⋅ 𝑑⃗𝑟 − 𝐹⃗ ⋅ 𝑑⃗𝑟 . ∫𝐶1 ∫𝐶2 ⃗ and let 𝐶1 and 𝐶2 be 34. Let curl 𝐹⃗ = 3𝑥⃗𝑖 + 3𝑦𝑗⃗ − 6𝑧𝑘 the closed curves in the planes 𝑧 = 0 and 𝑧 = 5 in Figure 20.18. Find ∫𝐶1

𝐹⃗ ⋅ 𝑑⃗𝑟 + 𝐶2

𝑧

∫𝐶2

𝐹⃗ ⋅ 𝑑⃗𝑟 .

𝑧

✻ 2

𝐶 𝑥 2 + 𝑦2 = 9

✲ 5

𝑥 2 + 𝑦2 = 4 𝑦 𝑥



𝑥 𝐶1 𝑦

Figure 20.16

Figure 20.18 ⃗ )⋅𝑑⃗𝑟 , where 𝐶 is a circle of 29. Evaluate ∫𝐶 (−𝑧⃗𝑖 +𝑦𝑗⃗ +𝑥𝑘 radius 2, parallel to the 𝑥𝑧-plane and around the 𝑦-axis with the orientation shown in Figure 20.17. 𝑧 𝐶



𝑥

✛2

𝑦

Figure 20.17 ⃗ around 30. Evaluate the circulation of 𝐺⃗ = 𝑥𝑦⃗𝑖 +𝑧𝑗⃗ +3𝑦𝑘 a square of side 6, centered at the origin, lying in the 𝑦𝑧plane, and oriented counterclockwise viewed from the positive 𝑥-axis. 31. Find the flux of 𝐹⃗ = curl((𝑥3 + cos(𝑧2 ))⃗𝑖 + (𝑥 + ⃗ ) through the upper half of the sin(𝑦2 ))𝑗⃗ + (𝑦2 sin(𝑥2 ))𝑘 sphere of radius 2, with center at the origin and oriented upward. 32. Suppose that 𝐶 is a closed curve in the 𝑥𝑦-plane, oriented counterclockwise when viewed from above. Show that 12 ∫𝐶 (−𝑦⃗𝑖 + 𝑥𝑗⃗ ) ⋅ 𝑑⃗𝑟 equals the area of the region 𝑅 in the 𝑥𝑦-plane enclosed by 𝐶.

3 ⃗ ). 35. (a) Find curl(𝑥3 ⃗𝑖 + sin(𝑦3 )𝑗⃗ + 𝑒𝑧 𝑘 (b) What does your answer to part (a) tell you about 3 ⃗ )⋅𝑑⃗𝑟 where 𝐶 is the circle ∫𝐶 (𝑥3 ⃗𝑖 +sin(𝑦3 )𝑗⃗ +𝑒𝑧 𝑘 (𝑥 − 10)2 + (𝑦 − 20)2 = 1 in the 𝑥𝑦-plane, oriented clockwise? (c) If 𝐶 is any closed curve, what can you say about 3 ⃗ ) ⋅ 𝑑⃗𝑟 ? ∫𝐶 (𝑥3 ⃗𝑖 + sin(𝑦3 )𝑗⃗ + 𝑒𝑧 𝑘

36. For 𝐶, the intersection of the cylinder 𝑥2 + 𝑦2 = 9 and the plane 𝑧 = −2 − 𝑥 + 2𝑦 oriented counterclockwise when viewed from above, use Stokes’ Theorem to find ) ( 𝑧2 ⃗ ⋅ 𝑑⃗𝑟 . (𝑥2 − 3𝑦2 )⃗𝑖 + ( + 𝑦)𝑗⃗ + (𝑥 + 2𝑧2 )𝑘 ∫𝐶 2 37. Let 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝐹1 (𝑥, 𝑦)⃗𝑖 + 𝐹2 (𝑥, 𝑦)𝑗⃗ , where 𝐹1 and 𝐹2 are continuously differentiable for all 𝑥, 𝑦. (a) Describe in words how 𝐹⃗ varies through space. (b) Find an expression for curl 𝐹⃗ in terms of 𝐹1 and 𝐹2 . (c) Let 𝐶 be a closed curve in the 𝑥𝑦-plane, oriented counterclockwise when viewed from above, and let 𝑆 be the region enclosed by 𝐶. Use your answer to part (b) to simplify the statement of Stokes’ Theorem for this 𝐹⃗ and 𝐶. (d) The result in part (c) is usually known by another name. What is it?

1014

Chapter 20 THE CURL AND STOKES’ THEOREM

38. Water in a bathtub has velocity vector field near the drain given, for 𝑥, 𝑦, 𝑧 in cm, by 𝑦 + 𝑥𝑧 𝑦𝑧 − 𝑥 1 ⃗ ⃗𝑖 − 𝐹⃗ = − 2 𝑘 cm/sec. 𝑗⃗ − 2 (𝑧 + 1)2 (𝑧2 + 1)2 𝑧 +1 (a) Rewriting 𝐹⃗ as follows, describe in words how the water is moving:

(c) Find the divergence of 𝐹⃗ . (d) Find the flux of the water through the hemisphere of radius 1, centered at the origin, lying below the 𝑥𝑦-plane and oriented downward. (e) Find ∫𝐶 𝐺⃗ ⋅ 𝑑⃗𝑟 where 𝐶 is the edge of the drain, oriented clockwise when viewed from above, and where

⃗ −𝑦⃗𝑖 + 𝑥𝑗⃗ −𝑧(𝑥⃗𝑖 + 𝑦𝑗⃗ ) 𝑘 . 𝐹⃗ = 2 + − 2 2 2 2 (𝑧 + 1) (𝑧 + 1) 𝑧 +1 (b) The drain in the bathtub is a disk in the 𝑥𝑦-plane with center at the origin and radius 1 cm. Find the rate at which the water is leaving the bathtub. (That is, find the rate at which water is flowing through the disk.) Give units for your answer.

1 𝐺⃗ = 2

(

𝑦 ⃗ 𝑥2 + 𝑦2 ⃗ 𝑥 ⃗ 𝑘 𝑖 − 2 𝑗 − 2 2 𝑧 +1 𝑧 +1 (𝑧 + 1)2

)

.

(f) Calculate curl 𝐺⃗ . (g) Explain why your answers to parts (d) and (e) are equal.

Strengthen Your Understanding In Problems 39–40, explain what is wrong with the statement. 39. The line integral ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 can be evaluated using ⃗ and 𝐶 Stokes’ Theorem, where 𝐹⃗ = 2𝑥⃗𝑖 − 3𝑗⃗ + 𝑘 is an oriented curve from (0, 0, 0) to (3, 4, 5). 40. If 𝑆 is the unit circular disc 𝑥2 + 𝑦2 ≤ 1, 𝑧 = 0, in the 𝑥𝑦-plane, oriented downward, 𝐶 is the unit circle in the 𝑥𝑦-plane oriented counterclockwise, and 𝐹⃗ is a vector field, then ∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫𝑆

curl 𝐹⃗ ⋅ 𝑑 𝐴⃗ .

In Problems 41–42, give an example of:

45. Let 𝑆 be the cylinder 𝑥2 + 𝑧2 = 1, 0 ≤ 𝑦 ≤ 2, oriented with inward-pointing normal. Then the boundary of 𝑆 consists of two circles 𝐶1 (𝑥2 + 𝑧2 = 1, 𝑦 = 0) and 𝐶2 (𝑥2 + 𝑧2 = 1, 𝑦 = 2), both oriented clockwise when viewed from the positive 𝑦-axis. 46. If 𝐶 is the boundary of an oriented surface 𝑆, oriented by the right-hand rule, then ∫𝐶 curl 𝐹⃗ ⋅𝑑⃗𝑟 = ∫𝑆 𝐹⃗ ⋅𝑑 𝐴⃗ . 47. Let 𝑆1 be the disk 𝑥2 + 𝑦2 ≤ 1, 𝑧 = 0 and let 𝑆2 be the upper unit hemisphere 𝑥2 + 𝑦2 + 𝑧2 = 1, 𝑧 ≥ 0, both oriented upward. If 𝐹⃗ is a vector field then ∫𝑆 curl𝐹⃗ ⋅ 𝑑 𝐴⃗ = ∫𝑆 curl𝐹⃗ ⋅ 𝑑 𝐴⃗ . 1

2

48. Let 𝑆 be the closed unit sphere 𝑥2 +𝑦2 +𝑧2 = 1, oriented outward. If 𝐹⃗ is a vector field, then ∫𝑆 curl𝐹⃗ ⋅𝑑 𝐴⃗ = 0.

41. An oriented closed curve 𝐶 such that ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 0, ⃗. where 𝐹⃗ (𝑥, 𝑦, 𝑧) = 𝑥⃗𝑖 + 𝑦2 𝑗⃗ + 𝑧3 𝑘

49. If 𝐹⃗ and 𝐺⃗ are vector fields satisfying curl𝐹⃗ = curl𝐺⃗ , then ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = ∫𝐶 𝐺⃗ ⋅ 𝑑⃗𝑟 , where 𝐶 is any oriented circle in 3-space.

42. A surface 𝑆, oriented appropriately to use Stokes’ Theorem, which has as its boundary the circle 𝐶 of radius 1 centered at the origin, lying in the 𝑥𝑦-plane, and oriented counterclockwise when viewed from above.

50. If 𝐹⃗ is a vector field satisfying curl𝐹⃗ = 0⃗ , then ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 0, where 𝐶 is any oriented path around a rectangle in 3-space.

In Problems 43–51, is the statement true or false? Give a reason for your answer.

51. Let 𝑆 be an oriented surface, with oriented boundary 𝐶, and suppose that 𝐹⃗ is a vector field such that ∫ 𝐹⃗ ⋅ 𝑑⃗𝑟 = 0. Then curl 𝐹⃗ = 0⃗ everywhere on 𝑆. 𝐶

43. If curl 𝐹⃗ is everywhere perpendicular to the 𝑧-axis, and 𝐶 is a circle in the 𝑥𝑦-plane, then the circulation of 𝐹⃗ around 𝐶 is zero. 44. If 𝑆 is the upper unit hemisphere 𝑥2 + 𝑦2 + 𝑧2 = 1, 𝑧 ≥ 0, oriented upward, then the boundary of 𝑆 used in Stokes’ Theorem is the circle 𝑥2 + 𝑦2 = 1, 𝑧 = 0, with orientation counterclockwise when viewed from the positive 𝑧-axis.

52. The circle 𝐶 has radius 3 and lies in a plane through the ⃗. origin. Let 𝐹⃗ = (2𝑧 + 3𝑦)⃗𝑖 + (𝑥 − 𝑧)𝑗⃗ + (6𝑦 − 7𝑥)𝑘 What is the equation of the plane and what is the orientation of the circle that make the circulation, ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 , a maximum? [Note: You should specify the orientation of the circle by saying that it is clockwise or counterclockwise when viewed from the positive or negative 𝑥- or 𝑦- or 𝑧-axis.]

20.3 THE THREE FUNDAMENTAL THEOREMS

20.3

1015

THE THREE FUNDAMENTAL THEOREMS We have now seen three multivariable versions of the Fundamental Theorem of Calculus. In this section we will examine some consequences of these theorems. Fundamental Theorem of Calculus for Line Integrals ∫𝐶

grad𝑓 ⋅ 𝑑⃗𝑟 = 𝑓 (𝑄) − 𝑓 (𝑃 ).

Stokes’ Theorem ∫𝑆

curl 𝐹⃗ ⋅ 𝑑 𝐴⃗ =

∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 .

Divergence Theorem ∫𝑊

div𝐹⃗ 𝑑𝑉 =

∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ .

Notice that, in each case, the region of integration on the right is the boundary of the region on the left (except that for the first theorem we simply evaluate 𝑓 at the boundary points); the integrand on the left is a sort of derivative of the integrand on the right; see Figure 20.19. Boundary of region 𝑊 is surface 𝑆 The boundary of the curve 𝐶 consists of the points 𝑃 and 𝑄

𝑆 𝑄

Boundary of surface 𝑆 is curve 𝐶

𝐶

𝐶 𝑊 𝑆

𝑃

Figure 20.19: Regions and their boundaries for the three fundamental theorems

The Gradient and the Curl Suppose that 𝐹⃗ is a smooth gradient field, so 𝐹⃗ = grad 𝑓 for some function 𝑓 . Using the Fundamental Theorem for Line Integrals, we saw in Chapter 18 that ∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 = 0

for any closed curve 𝐶. Thus, for any unit vector 𝑛⃗ 𝐹⃗ ⋅ 𝑑⃗𝑟 ∫𝐶 0 = lim = 0, circ𝑛⃗ 𝐹⃗ = lim Area→0 Area of 𝐶 Area→0 Area where the limit is taken over circles 𝐶 in a plane perpendicular to 𝑛⃗ , and oriented by the right-hand rule. Thus, the circulation density of 𝐹⃗ is zero in every direction, so curl 𝐹⃗ = 0⃗ , that is, curl grad 𝑓 = 0⃗ . (This formula can also be verified using the coordinate definition of curl. See Problem 31 on page 1007.)

1016

Chapter 20 THE CURL AND STOKES’ THEOREM

Is the converse true? Is any vector field whose curl is zero a gradient field? Suppose that curl 𝐹⃗ = 0⃗ and let us consider the line integral ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 for a closed curve 𝐶 contained in the domain of 𝐹⃗ . If 𝐶 is the boundary curve of an oriented surface 𝑆 that lies wholly in the domain of curl 𝐹⃗ , then Stokes’ Theorem asserts that ∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫𝑆

curl 𝐹⃗ ⋅ 𝑑 𝐴⃗ =

∫𝑆

0⃗ ⋅ 𝑑 𝐴⃗ = 0.

If we knew that ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 0 for every closed curve 𝐶, then 𝐹⃗ would be path-independent, and hence a gradient field. Thus, we need to know whether every closed curve in the domain of 𝐹⃗ is the boundary of an oriented surface contained in the domain. It can be quite difficult to determine if a given curve is the boundary of a surface (suppose, for example, that the curve is knotted in a complicated way). However, if the curve can be contracted smoothly to a point, remaining all the time in the domain of 𝐹⃗ , then it is the boundary of a surface, namely, the surface it sweeps through as it contracts. Thus, we have proved the test for a gradient field that we stated in Chapter 18.

The Curl Test for Vector Fields in 3-Space Suppose 𝐹⃗ is a smooth vector field on 3-space such that • The domain of 𝐹⃗ has the property that every closed curve in it can be contracted to a point in a smooth way, staying at all times within the domain. • curl 𝐹⃗ = 0⃗ . Then 𝐹⃗ is path-independent, and thus is a gradient field.

Example 7 on page 956 shows how the curl test is applied.

The Curl and the Divergence In this section we will use the second two fundamental theorems to get a test for a vector field to be a curl field, that is, a field of the form 𝐹⃗ = curl 𝐺⃗ for some 𝐺⃗ . Example 1

Suppose that 𝐹⃗ is a smooth curl field. Use Stokes’ Theorem to show that for any closed surface, 𝑆, contained in the domain of 𝐹⃗ , 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 0. ∫𝑆

Solution

Suppose 𝐹⃗ = curl 𝐺⃗ . Draw a closed curve 𝐶 on the surface 𝑆, thus dividing 𝑆 into two surfaces 𝑆1 and 𝑆2 as shown in Figure 20.20. Pick the orientation for 𝐶 corresponding to 𝑆1 ; then the orientation of 𝐶 corresponding to 𝑆2 is the opposite. Thus, using Stokes’ Theorem,

∫𝑆1

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

∫𝑆1

curl 𝐺⃗ ⋅ 𝑑 𝐴⃗ =

∫𝐶

𝐺⃗ ⋅ 𝑑⃗𝑟 = −

∫𝑆2

curl 𝐺⃗ ⋅ 𝑑 𝐴⃗ = −

Thus, for any closed surface 𝑆, we have

∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

∫𝑆1

𝐹⃗ ⋅ 𝑑 𝐴⃗ +

∫𝑆2

𝐹⃗ ⋅ 𝑑 𝐴⃗ = 0.

∫𝑆2

𝐹⃗ ⋅ 𝑑 𝐴⃗ .

20.3 THE THREE FUNDAMENTAL THEOREMS

1017

Orientation of 𝐶 corresponding to 𝑆1

𝐶

✠ 𝑆 2

𝑆1



𝑛⃗ 1

𝑛⃗ 2

𝑆

Orientation of 𝐶 corresponding to 𝑆2

Figure 20.20: The closed surface 𝑆 divided into two surfaces 𝑆1 and 𝑆2

Thus, if 𝐹⃗ = curl 𝐺⃗ , we use the result of Example 1 to see that

div 𝐹⃗ =

lim

Volume→0

∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗

Volume enclosed by 𝑆

=

lim

Volume→0

0 = 0, Volume

where the limit is taken over spheres 𝑆 contracting down to a point. So we conclude that: div curl 𝐺⃗ = 0. (This formula can also be verified using coordinates. See Problem 30 on page 1007.) Is every vector field whose divergence is zero a curl field? It turns out that we have the following analogue of the curl test, though we will not prove it.

The Divergence Test for Vector Fields in 3-Space Suppose 𝐹⃗ is a smooth vector field on 3-space such that • The domain of 𝐹⃗ has the property that every closed surface in it is the boundary of a solid region completely contained in the domain. • div 𝐹⃗ = 0. Then 𝐹⃗ is a curl field.

Example 2

𝑟⃗ 2𝐼 Consider the vector fields 𝐸⃗ = 𝑞 and 𝐵⃗ = 𝑐 ‖⃗𝑟 ‖3

(

−𝑦⃗𝑖 + 𝑥𝑗⃗ 𝑥2 + 𝑦2

)

.

(a) Calculate div 𝐸⃗ and div 𝐵⃗ . (b) Do 𝐸⃗ and 𝐵⃗ satisfy the divergence test? (c) Is either 𝐸⃗ or 𝐵⃗ a curl field? Solution

(a) Example 3 on page 986 shows that div 𝐸⃗ = 0. The following calculation shows div 𝐵⃗ = 0 also:

div 𝐵⃗ =

2𝐼 𝑐

(

𝜕 𝜕𝑥

(

−𝑦 𝑥2 + 𝑦2

)

+

𝜕 𝜕𝑦

(

𝑥 𝑥2 + 𝑦2

)

+

𝜕 (0) 𝜕𝑧

)

1018

Chapter 20 THE CURL AND STOKES’ THEOREM

2𝐼 = 𝑐

(

2𝑥𝑦 −2𝑦𝑥 + 2 2 2 2 (𝑥 + 𝑦 ) (𝑥 + 𝑦2 )2

)

= 0.

(b) The domain of 𝐸⃗ is 3-space minus the origin, so a region is contained in the domain if it misses the origin. Thus, the surface of a sphere centered at the origin is contained in the domain of 𝐸, but the solid ball inside is not. Hence, 𝐸⃗ does not satisfy the divergence test. The domain of 𝐵⃗ is 3-space minus the 𝑧-axis, so a region is contained in the domain if it avoids the 𝑧-axis. If 𝑆 is a surface bounding a solid region 𝑊 , then the 𝑧-axis cannot pierce 𝑊 without piercing 𝑆 as well. Hence, if 𝑆 avoids the 𝑧-axis, so does 𝑊 . Thus, 𝐵⃗ satisfies the divergence test. (c) In Example 3 on page 993 we computed the flux of 𝑟⃗ ∕‖⃗𝑟 ‖3 through a sphere centered at the origin, and found it was 4𝜋, so the flux of 𝐸⃗ through this sphere is 4𝜋𝑞. Thus, 𝐸⃗ cannot be a curl field, because by Example 1, the flux of a curl field through a closed surface is zero. On the other hand, 𝐵⃗ satisfies the divergence test, so it must be a curl field. In fact, Problem 26 shows that ( ) −𝐼 ⃗ . 𝐵⃗ = curl ln(𝑥2 + 𝑦2 )𝑘 𝑐

Exercises and Problems for Section 20.3 Online Resource: Additional Problems for Section 20.3 EXERCISES

In Exercises 1–6, is the vector field a gradient field? 1. 2. 3. 4. 5. 6.

𝐹⃗ 𝐹⃗ 𝐹⃗ 𝐹⃗ 𝐺⃗ 𝐹⃗

⃗ = 2𝑥⃗𝑖 + 𝑧𝑗⃗ + 𝑦𝑘 ⃗ ⃗ ⃗ = 𝑦𝑖 + 𝑧𝑗 + 𝑥𝑘 ⃗ = (𝑦 + 2𝑧)⃗𝑖 + (𝑥 + 𝑧)𝑗⃗ + (2𝑥 + 𝑦)𝑘 ⃗ = (𝑦 − 2𝑧)⃗𝑖 + (𝑥 − 𝑧)𝑗⃗ + (2𝑥 − 𝑦)𝑘 = −𝑦⃗𝑖 + 𝑥𝑗⃗ ⃗ = 𝑦𝑧⃗𝑖 + (𝑥𝑧 + 𝑧2 )𝑗⃗ + (𝑥𝑦 + 2𝑦𝑧)𝑘

In Exercises 7–12, is the vector field a curl field? 7. 8. 9. 10. 11. 12. 13.

𝐹⃗ 𝐹⃗ 𝐹⃗ 𝐹⃗ 𝐹⃗ 𝐹⃗

⃗ = 𝑧⃗𝑖 + 𝑥𝑗⃗ + 𝑦𝑘 ⃗ = 𝑧⃗𝑖 + 𝑦𝑗⃗ + 𝑥𝑘 ⃗ = 2𝑥⃗𝑖 − 𝑦𝑗⃗ − 𝑧𝑘 ⃗ ⃗ = (𝑥 + 𝑦)𝑖 + (𝑦 + 𝑧)𝑗⃗ + (𝑥 + 𝑧)𝑘 2 ⃗ ⃗ ⃗ = (−𝑥𝑦)𝑖 + (2𝑦𝑧)𝑗 + (𝑦𝑧 − 𝑧 ))𝑘 ⃗ = (𝑥𝑦)⃗𝑖 + (𝑥𝑦)𝑗⃗ + (𝑥𝑦)𝑘 Let 𝐹⃗ be a vector field defined everywhere except the 𝑧-axis and with curl 𝐹⃗ = 0 at all points of its domain. Determine whether Stokes’ Theorem implies that ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 0 for a circle 𝐶 of radius 1, where (a) 𝐶 is parallel to the 𝑥𝑦-plane with center (0, 0, 1).

(b) 𝐶 is parallel to the 𝑦𝑧-plane with center (1, 0, 0). 14. Let 𝐹⃗ be a vector field defined everywhere except the origin and with div 𝐹⃗ = 0 at all points of its domain. Determine whether the Divergence Theorem implies that ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 0 for a sphere 𝑆 of radius 1, where (a) 𝑆 is centered at (0, 1, 1). (b) 𝑆 is centered at (0.5, 0, 0). In Exercises 15–18, can the curl test and the divergence test be applied to a vector field whose domain is the given region? 15. All points (𝑥, 𝑦, 𝑧) such that 𝑧 > 0. 16. All points (𝑥, 𝑦, 𝑧) not on the 𝑦-axis. 17. All points (𝑥, 𝑦, 𝑧) not on the positive 𝑧-axis. 18. All points (𝑥, 𝑦, 𝑧) except the 𝑥-axis with 0 ≤ 𝑥 ≤ 1.

PROBLEMS ⃗ , for some constant 𝑏. Show that the fol19. Let 𝐵⃗ = 𝑏𝑘 lowing are all possible vector potentials for 𝐵⃗ : (a) 𝐴⃗ = −𝑏𝑦⃗𝑖

(b) 𝐴⃗ = 𝑏𝑥𝑗⃗

(c)

𝐴⃗ = 12 𝐵⃗ × 𝑟⃗ .

⃗. 20. Find a vector field 𝐹⃗ such that curl 𝐹⃗ = 2⃗𝑖 − 3𝑗⃗ + 4𝑘 [Hint: Try 𝐹⃗ = 𝑣⃗ × 𝑟⃗ for some vector 𝑣⃗ .]

20.3 THE THREE FUNDAMENTAL THEOREMS

21. Find a vector potential for the constant vector field 𝐵⃗ whose value at every point is 𝑏⃗ . 22. Express (3𝑥+2𝑦)⃗𝑖 +(4𝑥+9𝑦)𝑗⃗ as the sum of a curl-free vector field and a divergence-free vector field. In Problems 23–24, does a vector potential exist for the vector field given? If so, find one. ⃗ 23. 𝐺⃗ = 𝑥2 ⃗𝑖 + 𝑦2 𝑗⃗ + 𝑧2 𝑘 ⃗ 24. 𝐹⃗ = 2𝑥⃗𝑖 + (3𝑦 − 𝑧2 )𝑗⃗ + (𝑥 − 5𝑧)𝑘 25. An electric charge 𝑞 at the origin produces an electric field 𝐸⃗ = 𝑞⃗𝑟 ∕‖⃗𝑟 ‖3 . (a) Does curl 𝐸⃗ = 0⃗ ? (b) Does 𝐸⃗ satisfy the curl test? (c) Is 𝐸⃗ a gradient field?

−𝐼 ⃗ is a vector potential 26. Show that 𝐴⃗ = ln(𝑥2 + 𝑦2 )𝑘 𝑐 for ) ( ⃗𝑖 + 𝑥𝑗⃗ −𝑦 2𝐼 . 𝐵⃗ = 𝑐 𝑥2 + 𝑦2 27. Suppose 𝑐 is the speed of light. A thin wire along the 𝑧-axis carrying a current 𝐼 produces a magnetic field ) ( 2𝐼 −𝑦⃗𝑖 + 𝑥𝑗⃗ ⃗ . 𝐵 = 𝑐 𝑥2 + 𝑦2

1019

(a) Does curl 𝐵⃗ = 0⃗ ? (b) Does 𝐵⃗ satisfy the curl test? (c) Is 𝐵⃗ a gradient field? 28. Use Stokes’ Theorem to show that if 𝑢(𝑥, 𝑦) and 𝑣(𝑥, 𝑦) are two functions of 𝑥 and 𝑦 and 𝐶 is a closed curve in the 𝑥𝑦-plane oriented counterclockwise, then

∫𝐶

(𝑢⃗𝑖 + 𝑣𝑗⃗ ) ⋅ 𝑑⃗𝑟 =

∫𝑅

(

𝜕𝑣 𝜕𝑢 − 𝜕𝑥 𝜕𝑦

)

𝑑𝑥𝑑𝑦

where 𝑅 is the region in the 𝑥𝑦-plane enclosed by 𝐶. This is Green’s Theorem. 29. Suppose that 𝐴⃗ is a vector potential for 𝐵⃗ . (a) Show that 𝐴⃗ + grad 𝜓 is also a vector potential for 𝐵⃗ , for any function 𝜓 with continuous secondorder partial derivatives. (The vector potentials 𝐴⃗ and 𝐴⃗ +grad 𝜓 are called gauge equivalent and the transformation, for any 𝜓, from 𝐴⃗ to 𝐴⃗ + grad 𝜓 is called a gauge transformation.) (b) What is the divergence of 𝐴⃗ +grad 𝜓? How should 𝜓 be chosen such that 𝐴⃗ + grad 𝜓 has zero divergence? (If div 𝐴⃗ = 0, the magnetic vector potential 𝐴⃗ is said to be in Coulomb gauge.)

Strengthen Your Understanding In Problems 30–31, explain what is wrong with the statement. 30. The curl of a vector field 𝐹⃗ is given by curl 𝐹⃗ = 𝑥⃗𝑖 . 31. For a certain vector field 𝐹⃗ , we have curl div 𝐹⃗ = 𝑦⃗𝑖 . In Problems 32–33, give an example of: 32. A vector field 𝐹⃗ that is not the curl of another vector field. 33. A function 𝑓 such that div grad 𝑓 ≠ 0. In Problems 34–37, is the statement true or false? Give a reason for your answer. 34. There exists a vector field 𝐹⃗ with curl𝐹⃗ = ⃗𝑖 . 35. There exists a vector field 𝐹⃗ (whose components have continuous second partial derivatives) satisfying

Online Resource: Review problems and Projects

curl𝐹⃗ = 𝑥⃗𝑖 . 36. Let 𝑆 be an oriented surface, with oriented boundary 𝐶, and suppose that 𝐹⃗ is a vector field such that ∫𝑆 curl𝐹⃗ ⋅ 𝑑 𝐴⃗ = 0. Then 𝐹⃗ is a gradient field. 37. If 𝐹⃗ is a gradient field, then ∫𝑆 curl 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 0, for any smooth oriented surface, 𝑆, in 3-space. 38. Let 𝑓 (𝑥, 𝑦, 𝑧) be a scalar function with continuous second partial derivatives. Let 𝐹⃗ (𝑥, 𝑦, 𝑧) be a vector field with continuous second partial derivatives. Which of the following quantities are identically zero? (a)

curl grad 𝑓

(b) 𝐹⃗ × curl 𝐹⃗

(c)

grad div 𝐹⃗

(d) div curl 𝐹⃗

(e)

div grad 𝑓

Chapter Twentyone

PARAMETERS, COORDINATES, AND INTEGRALS

Contents 21.1 Coordinates and Parameterized Surfaces . . . . 1022 Polar Coordinates as Families of Parameterized Curves . . . . . . . . . . . . . . . . . . . . 1022 Cylindrical and Spherical Coordinates . . . . . . . . . 1022 General Parameterizations . . . . . . . . . . . . . . . . . 1023 How Do We Parameterize a Surface? . . . . . . . . 1023 Using Position Vectors . . . . . . . . . . . . . . . . . . . . 1024 Parameterizing a Surface of the Form z = f (x, y) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1024 Parameterizing Planes . . . . . . . . . . . . . . . . . . . . 1024 Parameterizations Using Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 1025 Parameterizing a Sphere Using Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . 1026

Parameterizing Surfaces of Revolution . . . . . . . Parameter Curves . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Change of Coordinates in a Multiple Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Polar Change of Coordinates Revisited . . . . . . . General Change of Coordinates . . . . . . . . . . . . Change of Coordinates in Triple Integrals . . . . . 21.3 Flux Integrals over Parameterized Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Area of a Parameterized Surface . . . . . . . . . . . .

1028 1029 1033 1033 1034 1036 1038 1039

1022

21.1

Chapter 21 PARAMETERS, COORDINATES, AND INTEGRALS

COORDINATES AND PARAMETERIZED SURFACES In Chapter 17 we parameterized curves in 2- and 3-space, and in Chapter 16 we used polar, cylindrical, and spherical coordinates to simplify iterated integrals. We now take a second look at parameterizations and coordinate systems, and see that they are the same thing in different disguises: functions from one space to another. We have already seen this with parameterized curves, which we view as a function from an interval 𝑎 ≤ 𝑡 ≤ 𝑏 to a curve in 𝑥𝑦𝑧-space. See Figure 21.1. 𝑧 𝐶 𝑟⃗ (𝑏) 𝑟⃗ (𝑎) 𝑎

𝑟⃗ (𝑡)

𝑏

𝑦 𝑥 Figure 21.1: The parameterization is a function from the interval, 𝑎 ≤ 𝑡 ≤ 𝑏, to 3-space, whose image is the curve, 𝐶

Polar, Cylindrical, and Spherical Coordinates Revisited The equations for polar coordinates, 𝑥 = 𝑟 cos 𝜃 𝑦 = 𝑟 sin 𝜃, can also be viewed as defining a function from the 𝑟𝜃-plane into the 𝑥𝑦-plane. This function transforms the rectangle on the left of Figure 21.2 into the quarter disk on the right. We need two parameters to describe this disk because it is a two-dimensional object.

Polar Coordinates as Families of Parameterized Curves Polar coordinates give two families of parameterized curves, which form the polar coordinate grid. The lines 𝑟 = Constant in the 𝑟𝜃-plane correspond to circles in the 𝑥𝑦-plane, each circle parameterized by 𝜃; the lines 𝜃 = Constant correspond to rays in the 𝑥𝑦-plane, each ray parameterized by 𝑟.

Cylindrical and Spherical Coordinates Similarly, cylindrical and spherical coordinates may be viewed as functions from 3-space to 3-space. Cylindrical coordinates take rectangular boxes in 𝑟𝜃𝑧-space and map them to cylindrical regions in 𝑥𝑦𝑧-space; spherical coordinates take rectangular boxes in 𝜌𝜙𝜃-space and map them to spherical regions in 𝑥𝑦𝑧-space. 𝜃 𝜋∕2

𝑟=1

𝑟=2

𝑦

𝑟=3

𝜃 = 3𝜋∕8

✲ 3𝜋∕8

𝜃 = 3𝜋∕8

3

𝜋∕4

𝜃 = 𝜋∕4

2

𝜋∕8

𝜃 = 𝜋∕8

1

𝜃 = 𝜋∕4

𝜃 = 𝜋∕8

𝑥

𝑟 1

2

3

4

1

2

Figure 21.2: A grid in the 𝑟𝜃-plane and the corresponding curved grid in the 𝑥𝑦-plane

3

4

21.1 COORDINATES AND PARAMETERIZED SURFACES

1023

General Parameterizations In general, a parameterization or coordinate system provides a way of representing a curved object by means of a simple region in the parameter space (an interval, rectangle, or rectangular box), along with a function mapping that region into the curved object. In the next section, we use this idea to parameterize curved surfaces in 3-space.

How Do We Parameterize a Surface? In Section 17.1 we parameterized a circle in 2-space using the equations 𝑥 = cos 𝑡,

𝑦 = sin 𝑡.

In 3-space, the same circle in the 𝑥𝑦-plane has parametric equations 𝑥 = cos 𝑡,

𝑦 = sin 𝑡,

𝑧 = 0.

We add the equation 𝑧 = 0 to specify that the circle is in the 𝑥𝑦-plane. If we wanted a circle in the plane 𝑧 = 3, we would use the equations 𝑥 = cos 𝑡,

𝑦 = sin 𝑡,

𝑧 = 3.

Suppose now we let 𝑧 vary freely, as well as 𝑡. We get circles in every horizontal plane, forming a cylinder as in the left of Figure 21.3. Thus, we need two parameters, 𝑡 and 𝑧, to parameterize the cylinder. 𝜃 = 𝜋∕3 𝜃 = 𝜋∕6

𝑧

𝑧

𝜃 = 𝜋∕2

𝜃 = 2𝜋∕3

𝜃=0

❘ ❘

𝑧=3 𝑧=2 𝑧=1 𝑧=0





𝑦

𝑥

𝑦

𝑥



𝑧 = −3

Figure 21.3: The cylinder 𝑥 = cos 𝑡, 𝑦 = sin 𝑡, 𝑧 = 𝑧

We can contrast curves and surfaces. A curve, though it may live in two or three dimensions, is itself one-dimensional; if we move along it we can only move backward and forward in one direction. Thus, it only requires one parameter to trace out a curve. A surface is 2-dimensional; at any given point there are two independent directions we can move. For example, on the cylinder we can move vertically, or we can circle around the 𝑧-axis horizontally. So we need two parameters to describe it. We can think of the parameters as map coordinates, like longitude and latitude on the surface of the earth. Just as polar coordinates give a polar grid on a circular region, so the parameters for a surface give a grid on the surface. See Figure 21.3 on the right. In the case of the cylinder our parameters are 𝑡 and 𝑧, so 𝑥 = cos 𝑡,

𝑦 = sin 𝑡,

𝑧 = 𝑧,

0 ≤ 𝑡 < 2𝜋,

−∞ < 𝑧 < ∞.

The last equation, 𝑧 = 𝑧, looks strange, but it reminds us that we are in three dimensions, not two, and that the 𝑧-coordinate on our surface is allowed to vary freely. In general, we express the coordinates, (𝑥, 𝑦, 𝑧), of a point on a surface 𝑆 in terms of two parameters, 𝑠 and 𝑡: 𝑥 = 𝑓1 (𝑠, 𝑡), 𝑦 = 𝑓2 (𝑠, 𝑡), 𝑧 = 𝑓3 (𝑠, 𝑡).

1024

Chapter 21 PARAMETERS, COORDINATES, AND INTEGRALS

As the values of 𝑠 and 𝑡 vary, the corresponding point (𝑥, 𝑦, 𝑧) sweeps out the surface, 𝑆. (See Figure 21.4.) The function which sends the point (𝑠, 𝑡) to the point (𝑥, 𝑦, 𝑧) is called the parameterization of the surface. 𝑧 𝑆

✻ 𝑅



(𝑠, 𝑡)

𝑥 (

(𝑥, 𝑦, 𝑧) = ) 𝑓1 (𝑠, 𝑡), 𝑓2 (𝑠, 𝑡), 𝑓3 (𝑠, 𝑡)

𝑦

Figure 21.4: The parameterization sends each point (𝑠, 𝑡) in the parameter region, 𝑅, to a point (𝑥, 𝑦, 𝑧) = (𝑓1 (𝑠, 𝑡), 𝑓2 (𝑠, 𝑡), 𝑓3 (𝑠, 𝑡)) in the surface, 𝑆

Using Position Vectors ⃗ to combine the three parametric equations for a We can use the position vector 𝑟⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 surface into a single vector equation. For example, the parameterization of the cylinder 𝑥 = cos 𝑡, 𝑦 = sin 𝑡, 𝑧 = 𝑧 can be written as ⃗ 𝑟⃗ (𝑡, 𝑧) = cos 𝑡⃗𝑖 + sin 𝑡𝑗⃗ + 𝑧𝑘 0 ≤ 𝑡 < 2𝜋, −∞ < 𝑧 < ∞. For a general parameterized surface 𝑆, we write ⃗. 𝑟⃗ (𝑠, 𝑡) = 𝑓1 (𝑠, 𝑡)⃗𝑖 + 𝑓2 (𝑠, 𝑡)𝑗⃗ + 𝑓3 (𝑠, 𝑡)𝑘

Parameterizing a Surface of the Form 𝒛 = 𝒇 (𝒙, 𝒚) The graph of a function 𝑧 = 𝑓 (𝑥, 𝑦) can be given parametrically simply by letting the parameters 𝑠 and 𝑡 be 𝑥 and 𝑦: 𝑥 = 𝑠, 𝑦 = 𝑡, 𝑧 = 𝑓 (𝑠, 𝑡). Example 1

Give a parametric description of the lower hemisphere of the sphere 𝑥2 + 𝑦2 + 𝑧2 = 1.

Solution

√ The surface is the graph of the function 𝑧 = − 1 −√ 𝑥2 − 𝑦2 over the region 𝑥2 + 𝑦2 ≤ 1 in the plane. Then parametric equations are 𝑥 = 𝑠, 𝑦 = 𝑡, 𝑧 = − 1 − 𝑠2 − 𝑡2 , where the parameters 𝑠 and 𝑡 vary inside and on the unit circle. In practice we often think of 𝑥 and 𝑦 as parameters rather than introduce new parameters 𝑠 and 𝑡. Thus, we may write 𝑥 = 𝑥, 𝑦 = 𝑦, 𝑧 = 𝑓 (𝑥, 𝑦).

Parameterizing Planes Consider a plane containing two nonparallel vectors 𝑣⃗ 1 and 𝑣⃗ 2 and a point 𝑃0 with position vector 𝑟⃗ 0 . We can get to any point on the plane by starting at 𝑃0 and moving parallel to 𝑣⃗ 1 or 𝑣⃗ 2 , adding multiples of them to 𝑟⃗ 0 . (See Figure 21.5.)

𝑣⃗2 𝑃0

𝑣⃗1

Figure 21.5: The plane 𝑟⃗ (𝑠, 𝑡) = 𝑟⃗ 0 + 𝑠𝑣⃗ 1 + 𝑡𝑣⃗ 2 and some points corresponding to various choices of 𝑠 and 𝑡

21.1 COORDINATES AND PARAMETERIZED SURFACES

1025

Since 𝑠𝑣⃗ 1 is parallel to 𝑣⃗ 1 and 𝑡⃗ 𝑣 2 is parallel to 𝑣⃗ 2 , we have the following result:

Parameterizing a Plane The plane through the point with position vector 𝑟⃗ 0 and containing the two nonparallel vectors 𝑣⃗ 1 and 𝑣⃗ 2 has parameterization 𝑟⃗ (𝑠, 𝑡) = 𝑟⃗ 0 + 𝑠𝑣⃗ 1 + 𝑡⃗ 𝑣 2.

⃗ , and 𝑣⃗ 1 = 𝑎1 ⃗𝑖 + 𝑎2 𝑗⃗ + 𝑎3 𝑘 ⃗ , and 𝑣⃗ 2 = 𝑏1 ⃗𝑖 + 𝑏2 𝑗⃗ + 𝑏3 𝑘 ⃗ , then the If 𝑟⃗ 0 = 𝑥0 ⃗𝑖 + 𝑦0 𝑗⃗ + 𝑧0 𝑘 parameterization of the plane can be expressed with the parametric equations 𝑥 = 𝑥0 + 𝑠𝑎1 + 𝑡𝑏1 ,

𝑦 = 𝑦0 + 𝑠𝑎2 + 𝑡𝑏2 ,

𝑧 = 𝑧0 + 𝑠𝑎3 + 𝑡𝑏3 .

Notice that the parameterization of the plane expresses the coordinates 𝑥, 𝑦, and 𝑧 as linear functions of the parameters 𝑠 and 𝑡. Example 2

Write a parameterization for the plane through the point (2, −1, 3) and containing the vectors 𝑣⃗ 1 = ⃗ and 𝑣⃗ 2 = ⃗𝑖 − 4𝑗⃗ + 5𝑘 ⃗. 2⃗𝑖 + 3𝑗⃗ − 𝑘

Solution

A possible parameterization is ⃗ + 𝑠(2⃗𝑖 + 3𝑗⃗ − 𝑘 ⃗ ) + 𝑡(⃗𝑖 − 4𝑗⃗ + 5𝑘 ⃗) 𝑣 2 = 2⃗𝑖 − 𝑗⃗ + 3𝑘 𝑟⃗ (𝑠, 𝑡) = 𝑟⃗ 0 + 𝑠𝑣⃗ 1 + 𝑡⃗ ⃗, = (2 + 2𝑠 + 𝑡)⃗𝑖 + (−1 + 3𝑠 − 4𝑡)𝑗⃗ + (3 − 𝑠 + 5𝑡)𝑘 or equivalently, 𝑥 = 2 + 2𝑠 + 𝑡,

𝑦 = −1 + 3𝑠 − 4𝑡,

𝑧 = 3 − 𝑠 + 5𝑡.

Parameterizations Using Spherical Coordinates Recall the spherical coordinates 𝜌, 𝜙, and 𝜃 introduced on page 872 of Chapter 16. On a sphere of radius 𝜌 = 𝑎 we can use 𝜙 and 𝜃 as coordinates, similar to latitude and longitude on the surface of the earth. (See Figure 21.6.) The latitude, however, is measured from the equator, whereas 𝜙 is measured from the north pole. If the positive 𝑥-axis passes through the Greenwich meridian, the longitude and 𝜃 are equal for 0 ≤ 𝜃 ≤ 𝜋. 𝑧

𝜙 𝑦 𝜃 𝑥

Figure 21.6: Parameterizing the sphere by 𝜙 and 𝜃

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Chapter 21 PARAMETERS, COORDINATES, AND INTEGRALS

Example 3

You are at a point on a sphere with 𝜙 = 3𝜋∕4. Are you in the northern or southern hemisphere? If 𝜙 decreases, do you move closer to or farther from the equator?

Solution

The equator has 𝜙 = 𝜋∕2. Since 3𝜋∕4 > 𝜋∕2, you are in the southern hemisphere. If 𝜙 decreases, you move closer to the equator.

Example 4

On a sphere, you are standing at a point with coordinates 𝜃0 and 𝜙0 . Your antipodal point is the point on the other side of the sphere on a line through you and the center. What are the 𝜃, 𝜙 coordinates of your antipodal point?

Solution

Figure 21.7 shows that the coordinates are 𝜃 = 𝜃0 + 𝜋 if 𝜃0 < 𝜋 or 𝜃 = 𝜃0 − 𝜋 if 𝜋 ≤ 𝜃0 ≤ 2𝜋, and 𝜙 = 𝜋 − 𝜙0 . Notice that if you are on the equator, then so is your antipodal point. 𝑦

𝜋 + 𝜃0

Point (above 𝑥𝑦-plane)

𝜃0

𝑧

𝜙0

𝜋 − 𝜙0

𝑥

Point

𝑥𝑦-plane seen edge-on

View from above

Side view

Antipodal Point (below 𝑥𝑦-plane)

Antipodal Point

Figure 21.7: Two views of the 𝑥𝑦𝑧-coordinate system showing coordinates of antipodal points

Parameterizing a Sphere Using Spherical Coordinates The sphere with radius 1 centered at the origin is parameterized by 𝑥 = sin 𝜙 cos 𝜃,

𝑦 = sin 𝜙 sin 𝜃,

𝑧 = cos 𝜙,

where 0 ≤ 𝜃 ≤ 2𝜋 and 0 ≤ 𝜙 ≤ 𝜋. (See Figure 21.8.) 𝑧

𝜙

Ra diu s

=

1



sin

𝜃

sin

𝑦 𝜙

✛ sin 𝜙 sin 𝜃





𝑥



𝜙 ✛ cos 𝜃

cos 𝜙

Figure 21.8: The relationship between 𝑥, 𝑦, 𝑧 and 𝜙, 𝜃 on a sphere of radius 1

21.1 COORDINATES AND PARAMETERIZED SURFACES

1027

We can also write these equations in vector form:

⃗. 𝑟⃗ (𝜃, 𝜙) = sin 𝜙 cos 𝜃 ⃗𝑖 + sin 𝜙 sin 𝜃 𝑗⃗ + cos 𝜙 𝑘

Since 𝑥2 + 𝑦2 + 𝑧2 = sin2 𝜙(cos2 𝜃 + sin2 𝜃) + cos2 𝜙 = sin2 𝜙 + cos2 𝜙 = 1, this verifies that the point with position vector 𝑟⃗ (𝜃, 𝜙) does lie on the sphere of radius 1. Notice that the 𝑧-coordinate depends only on the parameter 𝜙. Geometrically, this means that all points on the same latitude have the same 𝑧-coordinate. Example 5

Find parametric equations for the following spheres: (a) Center at the origin and radius 2. (b) Center at the point with Cartesian coordinates (2, −1, 3) and radius 2.

Solution

(a) We must scale the distance from the origin by 2. Thus, we have 𝑥 = 2 sin 𝜙 cos 𝜃,

𝑦 = 2 sin 𝜙 sin 𝜃,

𝑧 = 2 cos 𝜙,

where 0 ≤ 𝜃 ≤ 2𝜋 and 0 ≤ 𝜙 ≤ 𝜋. In vector form, this is written ⃗. 𝑟⃗ (𝜃, 𝜙) = 2 sin 𝜙 cos 𝜃⃗𝑖 + 2 sin 𝜙 sin 𝜃 𝑗⃗ + 2 cos 𝜙𝑘 (b) To shift the center of the sphere from the origin to the point (2, −1, 3), we add the vector parameterization we found in part (a) to the position vector of (2, −1, 3). (See Figure 21.9.) This gives ⃗ + (2 sin 𝜙 cos 𝜃⃗𝑖 + 2 sin 𝜙 sin 𝜃 𝑗⃗ + 2 cos 𝜙𝑘 ⃗) 𝑟⃗ (𝜃, 𝜙) = 2⃗𝑖 − 𝑗⃗ + 3𝑘 ⃗, = (2 + 2 sin 𝜙 cos 𝜃)⃗𝑖 + (−1 + 2 sin 𝜙 sin 𝜃)𝑗⃗ + (3 + 2 cos 𝜙)𝑘 where 0 ≤ 𝜃 ≤ 2𝜋 and 0 ≤ 𝜙 ≤ 𝜋. Alternatively, 𝑥 = 2 + 2 sin 𝜙 cos 𝜃,

𝑦 = −1 + 2 sin 𝜙 sin 𝜃,

𝑧 = 3 + 2 cos 𝜙.

𝑧

✒ 2 sin 𝜙 cos 𝜃⃗𝑖 + 2 sin 𝜙 sin 𝜃 𝑗⃗ + ⃗ 2 cos 𝜙𝑘



⃗ 2⃗𝑖 − 𝑗⃗ + 3𝑘 𝑦

𝑥 Figure 21.9: Sphere with center at the point (2, −1, 3) and radius 2

Note that the same point can have more than one value for 𝜃 or 𝜙. For example, points with 𝜃 = 0 also have 𝜃 = 2𝜋, unless we restrict 𝜃 to the range 0 ≤ 𝜃 < 2𝜋. Also, the north pole, at 𝜙 = 0, and the south pole, at 𝜙 = 𝜋, can have any value of 𝜃.

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Chapter 21 PARAMETERS, COORDINATES, AND INTEGRALS

Parameterizing Surfaces of Revolution Many surfaces have an axis of rotational symmetry and circular cross sections perpendicular to that axis. These surfaces are referred to as surfaces of revolution. Example 6

Find a parameterization of the cone whose base is the circle 𝑥2 + 𝑦2 = 𝑎2 in the 𝑥𝑦-plane and whose vertex is at height ℎ above the 𝑥𝑦-plane. (See Figure 21.10.) 𝑧





𝑟⃗ 1





✻ 𝑟 ⃗ 𝑧𝑘



𝑧

❄❄

𝑎

𝑎 𝜃 𝑟⃗0 𝑦

𝑥

Figure 21.10: The cone whose base is the circle 𝑥2 + 𝑦2 = 𝑎2 in the 𝑥𝑦-plane and whose vertex is at the point (0, 0, ℎ) and the vertical cross section through the cone

Solution

We use cylindrical coordinates, 𝑟, 𝜃, 𝑧. (See Figure 21.10.) In the 𝑥𝑦-plane, the radius vector, 𝑟⃗ 0 , from the 𝑧-axis to a point on the cone in the 𝑥𝑦-plane is 𝑟⃗ 0 = 𝑎 cos 𝜃⃗𝑖 + 𝑎 sin 𝜃 𝑗⃗ . Above the 𝑥𝑦-plane, the radius of the circular cross section, 𝑟, decreases linearly from 𝑟 = 𝑎 when 𝑧 = 0 to 𝑟 = 0 when 𝑧 = ℎ. From the similar triangles in Figure 21.10, 𝑎 𝑟 = . ℎ ℎ−𝑧 Solving for 𝑟, we have ) ( 𝑧 𝑎. 𝑟= 1− ℎ The horizontal radius vector, 𝑟⃗ 1 , at height 𝑧 has components similar to 𝑟⃗ 0 , but with 𝑎 replaced by 𝑟: ) ( ) ( 𝑧 𝑧 𝑎 cos 𝜃⃗𝑖 + 1 − 𝑎 sin 𝜃 𝑗⃗ . 𝑟⃗ 1 = 𝑟 cos 𝜃⃗𝑖 + 𝑟 sin 𝜃 𝑗⃗ = 1 − ℎ ℎ As 𝜃 goes from 0 to 2𝜋, the vector 𝑟⃗1 traces out the horizontal circle in Figure 21.10. We get the ⃗ , so position vector, 𝑟⃗ , of a point on the cone by adding the vector 𝑧𝑘 ) ( ) ( ⃗, ⃗ = 𝑎 1 − 𝑧 cos 𝜃⃗𝑖 + 𝑎 1 − 𝑧 sin 𝜃 𝑗⃗ + 𝑧𝑘 𝑟⃗ = 𝑟⃗ 1 + 𝑧𝑘 ℎ ℎ These equations can be written as ) ( 𝑧 𝑎 cos 𝜃, 𝑥= 1− ℎ The parameters are 𝜃 and 𝑧.

for 0 ≤ 𝑧 ≤ ℎ and 0 ≤ 𝜃 ≤ 2𝜋.

( ) 𝑧 𝑦= 1− 𝑎 sin 𝜃, ℎ

𝑧 = 𝑧.

21.1 COORDINATES AND PARAMETERIZED SURFACES

Example 7

1029

Consider the bell of a trumpet. A model for the radius 𝑧 = 𝑓 (𝑥) of the horn (in cm) at a distance 𝑥 cm from the large open end is given by the function 𝑓 (𝑥) =

6 . (𝑥 + 1)0.7

The bell is obtained by rotating the graph of 𝑓 about the 𝑥-axis. Find a parameterization for the first 24 cm of the bell. (See Figure 21.11.) 𝑧

𝑥

𝑦

Figure 21.11: The bell of a trumpet obtained by rotating the graph of 𝑧 = 𝑓 (𝑥) about the 𝑥-axis

Solution

At distance 𝑥 from the large open end of the horn, the cross section parallel to the 𝑦𝑧-plane is a circle of radius 𝑓 (𝑥), with center on the 𝑥-axis. Such a circle can be parameterized by 𝑦 = 𝑓 (𝑥) cos 𝜃, 𝑧 = 𝑓 (𝑥) sin 𝜃. Thus, we have the parameterization ( ) ( ) 6 6 𝑥 = 𝑥, 𝑦 = cos 𝜃, 𝑧 = sin 𝜃, 0 ≤ 𝑥 ≤ 24, 0 ≤ 𝜃 ≤ 2𝜋. (𝑥 + 1)0.7 (𝑥 + 1)0.7 The parameters are 𝑥 and 𝜃.

Parameter Curves On a parameterized surface, the curve obtained by setting one of the parameters equal to a constant and letting the other vary is called a parameter curve. If the surface is parameterized by ⃗, 𝑟⃗ (𝑠, 𝑡) = 𝑓1 (𝑠, 𝑡)⃗𝑖 + 𝑓2 (𝑠, 𝑡)𝑗⃗ + 𝑓3 (𝑠, 𝑡)𝑘 there are two families of parameter curves on the surface, one family with 𝑡 constant and the other with 𝑠 constant. Example 8

Consider the vertical cylinder 𝑥 = cos 𝑡,

𝑦 = sin 𝑡,

𝑧 = 𝑧.

(a) Describe the two parameter curves through the point (0, 1, 1). (b) Describe the family of parameter curves with 𝑡 constant and the family with 𝑧 constant. Solution

(a) Since the point (0, 1, 1) corresponds to the parameter values 𝑡 = 𝜋∕2 and 𝑧 = 1, there are two parameter curves, one with 𝑡 = 𝜋∕2 and the other with 𝑧 = 1. The parameter curve with 𝑡 = 𝜋∕2 has the parametric equations ( ) ( ) 𝜋 𝜋 𝑥 = cos = 0, 𝑦 = sin = 1, 𝑧 = 𝑧, 2 2 with parameter 𝑧. This is a line through the point (0, 1, 1) parallel to the 𝑧-axis. The parameter curve with 𝑧 = 1 has the parametric equations 𝑥 = cos 𝑡,

𝑦 = sin 𝑡,

𝑧 = 1,

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Chapter 21 PARAMETERS, COORDINATES, AND INTEGRALS

with parameter 𝑡. This is a unit circle parallel to and one unit above the 𝑥𝑦-plane centered on the 𝑧-axis. (b) First, fix 𝑡 = 𝑡0 for 𝑡 and let 𝑧 vary. The curves parameterized by 𝑧 have equations 𝑥 = cos 𝑡0 ,

𝑦 = sin 𝑡0 ,

𝑧 = 𝑧.

These are vertical lines on the cylinder parallel to the 𝑧-axis. (See Figure 21.12.) The other family is obtained by fixing 𝑧 = 𝑧0 and varying 𝑡. Curves in this family are parameterized by 𝑡 and have equations 𝑥 = cos 𝑡,

𝑦 = sin 𝑡,

𝑧 = 𝑧0 .

They are circles of radius 1 parallel to the 𝑥𝑦-plane centered on the 𝑧-axis. (See Figure 21.13.) 𝑧

𝑧

𝑦

𝑦 𝑥

𝑥

Figure 21.12: The family of parameter curves with 𝑡 = 𝑡0 for the cylinder 𝑥 = cos 𝑡, 𝑦 = sin 𝑡, 𝑧 = 𝑧

Example 9

Figure 21.13: The family of parameter curves with 𝑧 = 𝑧0 for the cylinder 𝑥 = cos 𝑡, 𝑦 = sin 𝑡, 𝑧 = 𝑧

Describe the families of parameter curves with 𝜃 = 𝜃0 and 𝜙 = 𝜙0 for the sphere 𝑥 = sin 𝜙 cos 𝜃,

𝑦 = sin 𝜙 sin 𝜃,

𝑧 = cos 𝜙,

where 0 ≤ 𝜃 ≤ 2𝜋, 0 ≤ 𝜙 ≤ 𝜋. Solution

Since 𝜙 measures latitude, the family with 𝜙 constant consists of the circles of constant latitude. (See Figure 21.14.) Similarly, the family with 𝜃 constant consists of the meridians (semicircles) running between the north and south poles. (See Figure 21.15.) 𝑧

𝑧

𝑦

𝑦 𝑥

𝑥

Figure 21.14: The family of parameter curves with 𝜙 = 𝜙0 for the sphere parameterized by (𝜃, 𝜙)

Figure 21.15: The family of parameter curves with 𝜃 = 𝜃0 for the sphere parameterized by (𝜃, 𝜙)

21.1 COORDINATES AND PARAMETERIZED SURFACES

1031

We have seen parameter curves before on pages 661-663 of Section 12.2: The cross sections with 𝑥 = 𝑎 or 𝑦 = 𝑏 on a surface 𝑧 = 𝑓 (𝑥, 𝑦) are examples of parameter curves. So are the grid lines on a computer sketch of a surface. The small regions shaped like parallelograms surrounded by nearby pairs of parameter curves are called parameter rectangles. See Figure 21.16.

Parameter curve: with 𝑦 = 𝑏





Parameter curve: with 𝑥 = 𝑎

Figure 21.16: Parameter curves 𝑥 = 𝑎 or 𝑦 = 𝑏 on a surface 𝑧 = 𝑓 (𝑥, 𝑦); the darker region is a parameter rectangle

Exercises and Problems for Section 21.1 EXERCISES In Exercises 1–4 decide if the parameterization describes a curve or a surface. 1. 2. 3. 4.

⃗ 𝑟⃗ (𝑠) = 𝑠⃗𝑖 + (3 − 𝑠)𝑗⃗ + 𝑠2 𝑘 𝑟⃗ (𝑠, 𝑡) = (𝑠 + 𝑡)⃗𝑖 + (3 − 𝑠)𝑗⃗ ⃗ 𝑟⃗ (𝑠, 𝑡) = cos 𝑠 ⃗𝑖 + sin 𝑠 𝑗⃗ + 𝑡2 𝑘 ⃗ 𝑟⃗ (𝑠) = cos 𝑠 ⃗𝑖 + sin 𝑠 𝑗⃗ + 𝑠2 𝑘

Describe in words the objects parameterized by the equations in Exercises 5–8. (Note: 𝑟 and 𝜃 are cylindrical coordinates.) 5. 𝑥 = 𝑟 cos 𝜃 0≤𝑟≤5 6. 𝑥 = 5 cos 𝜃 0 ≤ 𝜃 ≤ 2𝜋

𝑦 = 𝑟 sin 𝜃 0 ≤ 𝜃 ≤ 2𝜋 𝑦 = 5 sin 𝜃 0≤𝑧≤7

𝑧=7 𝑧=𝑧

7. 𝑥 = 5 cos 𝜃 0 ≤ 𝜃 ≤ 2𝜋

𝑦 = 5 sin 𝜃

8. 𝑥 = 𝑟 cos 𝜃 0≤𝑟≤5

𝑦 = 𝑟 sin 𝜃 0 ≤ 𝜃 ≤ 2𝜋

𝑧 = 5𝜃 𝑧=𝑟

In Exercises 9–12, for a sphere parameterized using the spherical coordinates 𝜃 and 𝜙, describe in words the part of the sphere given by the restrictions. 9. 0 ≤ 𝜃 < 2𝜋,

0 ≤ 𝜙 ≤ 𝜋∕2

10. 𝜋 ≤ 𝜃 < 2𝜋,

0≤𝜙≤𝜋 0≤𝜙≤𝜋

11. 𝜋∕4 ≤ 𝜃 < 𝜋∕3, 12. 0 ≤ 𝜃 ≤ 𝜋,

𝜋∕4 ≤ 𝜙 < 𝜋∕3

PROBLEMS In Problems 13–16, give parametric equations for the plane through the point with position vector 𝑟⃗ 0 and containing the vectors 𝑣⃗ 1 and 𝑣⃗ 2 .

In Problems 19–20, give two nonparallel vectors and the coordinates of a point in the plane with given parametric equations

⃗ 13. 𝑟⃗ 0 = ⃗𝑖 , 𝑣⃗ 1 = 𝑗⃗ , 𝑣⃗ 2 = 𝑘

19. 𝑥 = 2𝑠 + 3𝑡,

⃗ , 𝑣⃗ 2 = ⃗𝑖 14. 𝑟⃗ 0 = 𝑗⃗ , 𝑣⃗ 1 = 𝑘

20. 𝑥 = 2 + 𝑠 + 𝑡,

⃗ , 𝑣⃗ 2 = ⃗𝑖 + 𝑘 ⃗ 15. 𝑟⃗ 0 = ⃗𝑖 + 𝑗⃗ , 𝑣⃗ 1 = 𝑗⃗ + 𝑘

In Problems 21–22, parameterize the plane through the point with the given normal vector.

⃗ , 𝑣⃗ 1 = ⃗𝑖 − 𝑘 ⃗ , 𝑣⃗ 2 = −𝑗⃗ + 𝑘 ⃗ 16. 𝑟⃗ 0 = ⃗𝑖 + 𝑗⃗ + 𝑘 In Problems 17–18, parameterize the plane that contains the three points. 17. (0, 0, 0), (1, 2, 3), (2, 1, 0) 18. (1, 2, 3), (2, 5, 8), (5, 2, 0)

𝑦 = 𝑠 − 5𝑡, 𝑦 = 𝑠 − 𝑡,

𝑧 = −𝑠 + 2𝑡 𝑧 = −1 + 𝑠 + 𝑡

⃗ 21. (3, 5, 7), ⃗𝑖 + 𝑗⃗ + 𝑘 ⃗ 22. (5, 1, 4), ⃗𝑖 + 2𝑗⃗ + 3𝑘 ⃗ 23. Does the plane 𝑟⃗ (𝑠, 𝑡) = (2 + 𝑠)⃗𝑖 + (3 + 𝑠 + 𝑡)𝑗⃗ + 4𝑡𝑘 contain the following points? (a) (4, 8, 12) (b) (1, 2, 3)

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Chapter 21 PARAMETERS, COORDINATES, AND INTEGRALS

(b) Find the volume of the post.

24. Are the following two planes parallel? 𝑥 = 2 + 𝑠 + 𝑡,

𝑦 = 4 + 𝑠 − 𝑡,

𝑥 = 2 + 𝑠 + 2𝑡,

𝑦 = 𝑡,

𝑧 = 1 + 2𝑠,

and

✛4′′✲

𝑧 = 𝑠 − 𝑡.



𝑦 = 𝑡,

𝑧 = 1 for −∞ < 𝑠 < ∞, −∞ < 𝑡
1, the value of the 𝑠, 𝑡23. If the Jacobian || | | 𝜕(𝑠, 𝑡) | integral is greater than the original 𝑥, 𝑦-integral. 24. The Jacobian cannot be negative.

1038

21.3

Chapter 21 PARAMETERS, COORDINATES, AND INTEGRALS

FLUX INTEGRALS OVER PARAMETERIZED SURFACES Most of the flux integrals we are likely to encounter can be computed using the methods of Sections 19.1 and 19.2. In this section, we briefly consider the general case: how to compute the flux of a smooth vector field 𝐹⃗ through a smooth oriented surface, 𝑆, parameterized by 𝑟⃗ = 𝑟⃗ (𝑠, 𝑡), for (𝑠, 𝑡) in some region 𝑅 of the parameter space. The method is similar to the one used for graphs in Section 19.2. We consider a parameter rectangle on the surface 𝑆 corresponding to a rectangular region with sides Δ𝑠 and Δ𝑡 in the parameter space. (See Figure 21.21.) Parameter rectangle

𝑆



𝑡 𝑅





■ 𝜕⃗𝑟 𝜕𝑠

𝜕⃗𝑟 𝜕𝑡

Δ𝑡 ✻

Δ𝑠

Δ𝑡

❄✛ ✲ Δ𝑠

𝑠 Figure 21.21: Parameter rectangle on the surface 𝑆 corresponding to a small rectangular region in the parameter space, 𝑅

If Δ𝑠 and Δ𝑡 are small, the area vector, Δ𝐴⃗ , of the patch is approximately the area vector of the parallelogram defined by the vectors 𝑟⃗ (𝑠 + Δ𝑠, 𝑡) − 𝑟⃗ (𝑠, 𝑡) ≈

𝜕⃗𝑟 Δ𝑠, 𝜕𝑠

and

𝑟⃗ (𝑠, 𝑡 + Δ𝑡) − 𝑟⃗ (𝑠, 𝑡) ≈

𝜕⃗𝑟 Δ𝑡. 𝜕𝑡

Thus, 𝜕⃗𝑟 𝜕⃗𝑟 × Δ𝑠 Δ𝑡. 𝜕𝑠 𝜕𝑡 We assume that the vector 𝜕⃗𝑟 ∕𝜕𝑠 × 𝜕⃗𝑟 ∕𝜕𝑡 is never zero and points in the direction of the unit normal orientation vector 𝑛⃗ . If the vector 𝜕⃗𝑟 ∕𝜕𝑠 × 𝜕⃗𝑟 ∕𝜕𝑡 points in the opposite direction to 𝑛⃗ , we reverse the order of the cross product. Replacing Δ𝐴⃗ , Δ𝑠, and Δ𝑡 by 𝑑 𝐴⃗ , 𝑑𝑠, and 𝑑𝑡, we write ( ) 𝜕⃗𝑟 𝜕⃗𝑟 ⃗ 𝑑𝐴 = × 𝑑𝑠 𝑑𝑡. 𝜕𝑠 𝜕𝑡 Δ𝐴⃗ ≈

The Flux of a Vector Field Through a Parameterized Surface The flux of a smooth vector field 𝐹⃗ through a smooth oriented surface 𝑆 parameterized by 𝑟⃗ = 𝑟⃗ (𝑠, 𝑡), where (𝑠, 𝑡) varies in a parameter region 𝑅, is given by ) ( 𝜕⃗𝑟 𝜕⃗𝑟 × 𝑑𝑠 𝑑𝑡. 𝐹⃗ ⋅ 𝑑 𝐴⃗ = 𝐹⃗ (⃗𝑟 (𝑠, 𝑡)) ⋅ ∫𝑆 ∫𝑅 𝜕𝑠 𝜕𝑡 We choose the parameterization so that 𝜕⃗𝑟 ∕𝜕𝑠×𝜕⃗𝑟 ∕𝜕𝑡 is never zero and points in the direction of 𝑛⃗ everywhere.

21.3 FLUX INTEGRALS OVER PARAMETERIZED SURFACES

1039

Example 1

Find the flux of the vector field 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ through the surface 𝑆, oriented downward and given by 𝑥 = 2𝑠, 𝑦 = 𝑠 + 𝑡, 𝑧 = 1 + 𝑠 − 𝑡, where 0 ≤ 𝑠 ≤ 1, 0 ≤ 𝑡 ≤ 1.

Solution

Since 𝑆 is parameterized by ⃗, 𝑟⃗ (𝑠, 𝑡) = 2𝑠⃗𝑖 + (𝑠 + 𝑡)𝑗⃗ + (1 + 𝑠 − 𝑡)𝑘 we have 𝜕⃗𝑟 ⃗ = 2⃗𝑖 + 𝑗⃗ + 𝑘 𝜕𝑠

and

𝜕𝑟 ⃗ ⃗ = 𝑗 −𝑘, 𝜕𝑡

so |⃗ ⃗ ⃗ | |𝑖 𝑗 𝑘 | | | 𝜕⃗𝑟 𝜕⃗𝑟 ⃗. × = || 2 1 1 || = −2⃗𝑖 + 2𝑗⃗ + 2𝑘 𝜕𝑠 𝜕𝑡 | | | 0 1 −1 | | |

⃗ points upward, we use 2⃗𝑖 − 2𝑗⃗ − 2𝑘 ⃗ for downward orientation. Since the vector −2⃗𝑖 + 2𝑗⃗ + 2𝑘 Thus, the flux integral is given by 1

∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

1

⃗ ) 𝑑𝑠 𝑑𝑡 (2𝑠⃗𝑖 + (𝑠 + 𝑡)𝑗⃗ ) ⋅ (2⃗𝑖 − 2𝑗⃗ − 2𝑘

∫0 ∫0 1

1

1

1

=

(4𝑠 − 2𝑠 − 2𝑡) 𝑑𝑠𝑑𝑡 = (2𝑠 − 2𝑡) 𝑑𝑠 𝑑𝑡 ∫0 ∫0 ∫0 ∫0 ) ( 1 1 |1 |𝑠=1 𝑑𝑡 = 𝑠2 − 2𝑠𝑡|| (1 − 2𝑡) 𝑑𝑡 = 𝑡 − 𝑡2 || = 0. = ∫0 ∫0 |0 |𝑠=0

Area of a Parameterized Surface The area Δ𝐴 of a small parameter rectangle is the magnitude of its area vector Δ𝐴⃗ . Therefore, Area of 𝑆 =



Δ𝐴 =



‖Δ𝐴⃗ ‖ ≈

∑ ‖ 𝜕⃗𝑟 𝜕⃗𝑟 ‖ ‖ ‖ ‖ 𝜕𝑠 × 𝜕𝑡 ‖ Δ𝑠 Δ𝑡. ‖ ‖

Taking the limit as the area of the parameter rectangles tends to zero, we are led to the following expression for the area of 𝑆.

The Area of a Parameterized Surface The area of a surface 𝑆 which is parameterized by 𝑟⃗ = 𝑟⃗ (𝑠, 𝑡), where (𝑠, 𝑡) varies in a parameter region 𝑅, is given by ‖ 𝜕⃗𝑟 𝜕⃗𝑟 ‖ ‖ 𝑑𝑠 𝑑𝑡. ‖ × 𝑑𝐴 = ‖ ∫𝑆 ∫𝑅 ‖ 𝜕𝑠 𝜕𝑡 ‖ ‖

1040

Chapter 21 PARAMETERS, COORDINATES, AND INTEGRALS

Example 2

Compute the surface area of a sphere of radius 𝑎.

Solution

We take the sphere 𝑆 of radius 𝑎 centered at the origin and parameterize it with the spherical coordinates 𝜙 and 𝜃. The parameterization is 𝑥 = 𝑎 sin 𝜙 cos 𝜃,

𝑦 = 𝑎 sin 𝜙 sin 𝜃,

𝑧 = 𝑎 cos 𝜙,

for 0 ≤ 𝜃 ≤ 2𝜋,

0 ≤ 𝜙 ≤ 𝜋.

We compute 𝜕⃗𝑟 𝜕⃗𝑟 ⃗ ) × (−𝑎 sin 𝜙 sin 𝜃⃗𝑖 + 𝑎 sin 𝜙 cos 𝜃 𝑗⃗ ) × = (𝑎 cos 𝜙 cos 𝜃⃗𝑖 + 𝑎 cos 𝜙 sin 𝜃 𝑗⃗ − 𝑎 sin 𝜙𝑘 𝜕𝜙 𝜕𝜃 ⃗) = 𝑎2 (sin2 𝜙 cos 𝜃⃗𝑖 + sin2 𝜙 sin 𝜃 𝑗⃗ + sin 𝜙 cos 𝜙𝑘 and so ‖ 𝜕⃗𝑟 𝜕⃗𝑟 ‖ 2 ‖ ‖ ‖ 𝜕𝜙 × 𝜕𝜃 ‖ = 𝑎 sin 𝜙. ‖ ‖ Thus, we see that the surface area of the sphere 𝑆 is given by Surface area =

∫𝑆

𝑑𝐴 =

𝜋 2𝜋 ‖ 𝜕⃗𝑟 𝜕⃗𝑟 ‖ ‖ 𝑑𝜙𝑑𝜃 = ‖ × 𝑎2 sin 𝜙 𝑑𝜃 𝑑𝜙 = 4𝜋𝑎2 . ‖ ∫𝜙=0 ∫𝜃=0 ∫𝑅 ‖ ‖ 𝜕𝜙 𝜕𝜃 ‖

Exercises and Problems for Section 21.3 EXERCISES In Exercises 1–4 compute 𝑑 𝐴⃗ for the given parameterization for one of the two orientations. 1. 2. 3. 4.

𝑥 = 𝑠 + 𝑡, 𝑦 = 𝑠 − 𝑡, 𝑧 = 𝑠𝑡 𝑥 = sin 𝑡, 𝑦 = cos 𝑡, 𝑧 = 𝑠 + 𝑡 𝑥 = 𝑒𝑠 , 𝑦 = cos 𝑡, 𝑧 = sin 𝑡 𝑥 = 0, 𝑦 = 𝑢 + 𝑣, 𝑧 = 𝑢 − 𝑣

√ ⃗ and 𝑆 is the cone 𝑥2 + 𝑦2 = 𝑧, with 9. 𝐹⃗ = 𝑥2 𝑦2 𝑧𝑘 0 ≤ 𝑧 ≤ 𝑅, oriented downward. Parameterize the cone using cylindrical coordinates. (See Figure 21.22.) 𝑧 (0, 0, 𝑅)

In Exercises 5–9 compute the flux of the vector field 𝐹⃗ through the parameterized surface 𝑆.

𝑟⃗ 𝑟

⃗ and 𝑆 is oriented upward and given, for 5. 𝐹⃗ = 𝑧𝑘 0 ≤ 𝑠 ≤ 1, 0 ≤ 𝑡 ≤ 1, by 𝑥 = 𝑠 + 𝑡,

𝑟⃗ 𝜃 𝑛⃗ 𝑦

𝑧 = 𝑠2 + 𝑡2 .

𝑦 = 𝑠 − 𝑡,

6. 𝐹⃗ = 𝑥⃗𝑖 + 𝑦𝑗⃗ and 𝑆 is oriented downward and given, for 0 ≤ 𝑠 ≤ 1, 0 ≤ 𝑡 ≤ 1, by 𝑥 = 2𝑠,

𝑦 = 𝑠 + 𝑡,

𝑧 = 1 + 𝑠 − 𝑡.

𝑥

Figure 21.22

7. 𝐹⃗ = 𝑥⃗𝑖 through the surface 𝑆 oriented downward and parameterized for 0 ≤ 𝑠 ≤ 4, 0 ≤ 𝑡 ≤ 𝜋∕6 by 𝑥 = 𝑒𝑠 ,

𝑦 = cos(3𝑡),

𝑧 = 6𝑠.

8. 𝐹⃗ = 𝑦⃗𝑖 + 𝑥𝑗⃗ and 𝑆 is oriented away from the 𝑧-axis and given, for 0 ≤ 𝑠 ≤ 𝜋, 0 ≤ 𝑡 ≤ 1, by 𝑥 = 3 sin 𝑠,

𝑦 = 3 cos 𝑠,

𝑧 = 𝑡 + 1.

In Exercises 10–11, find the surface area. 10. A cylinder of radius 𝑎 and length 𝐿. 11. The region 𝑆 in the plane 𝑧 = 3𝑥 + 2𝑦 such that 0 ≤ 𝑥 ≤ 10 and 0 ≤ 𝑦 ≤ 20.

21.3 FLUX INTEGRALS OVER PARAMETERIZED SURFACES

1041

PROBLEMS 12. Compute the flux of the vector field 𝐹⃗ = (𝑥 + 𝑧)⃗𝑖 + ⃗ through the surface 𝑆 given by 𝑦 = 𝑥2 + 𝑧2 , 𝑗⃗ + 𝑧𝑘 1∕4 ≤ 𝑥2 + 𝑧2 ≤ 1 oriented away from the 𝑦-axis. 13. Find the area of the ellipse 𝑆 on the plane 2𝑥+𝑦+𝑧 = 2 cut out by the circular cylinder 𝑥2 + 𝑦2 = 2𝑥. (See Figure 21.23.) 𝑧

𝑆

𝑦

(b) Express the surface area of the cone as an integral. (c) Use a numerical method to find the surface area of the cone with vertex 𝑃 = (2, 0, 1). As we remarked in Section 19.1, the limit defining a flux integral might not exist if we subdivide the surface in the wrong way. One way to get around this is to take the formula for a flux integral over a parameterized surface that we have developed in this section and to use it as the definition of the flux integral. In Problems 17–20 we explore how this works. 17. Use a parameterization to verify the formula for a flux integral over a surface graph on page 974. 18. Use a parameterization to verify the formula for a flux integral over a cylindrical surface on page 976.

𝑥

Figure 21.23 14. Consider the surface 𝑆 formed by rotating the graph of 𝑦 = 𝑓 (𝑥) around the 𝑥-axis between 𝑥 = 𝑎 and 𝑥 = 𝑏. Assume that 𝑓 (𝑥) ≥ 0 for 𝑎 ≤√𝑥 ≤ 𝑏. Show that the 𝑏 surface area of 𝑆 is 2𝜋 ∫𝑎 𝑓 (𝑥) 1 + 𝑓 ′ (𝑥)2 𝑑𝑥.

15. A rectangular channel of width 𝑤 and depth ℎ meters lies in the 𝑗⃗ direction. At a point 𝑑1 meters from one side and 𝑑2 meters from the other side, the velocity vector of fluid in the channel is 𝑣⃗ = 𝑘𝑑1 𝑑2 𝑗⃗ meters/sec. Find the flux through a rectangle stretching the full width and depth of the channel, and perpendicular to the flow.

16. The base of a cone is the unit circle centered at the origin in the 𝑥𝑦-plane and vertex 𝑃 = (𝑎, 𝑏, 𝑐), where 𝑐 > 0. (a) Parameterize the cone.

19. Use a parameterization to verify the formula for a flux integral over a spherical surface on page 977. 20. One problem with defining the flux integral using a parameterization is that the integral appears to depend on the choice of parameterization. However, the flux through a surface ought not to depend on how the surface is parameterized. Suppose that the surface 𝑆 has two parameterizations, 𝑟⃗ = 𝑟⃗ (𝑠, 𝑡) for (𝑠, 𝑡) in the region 𝑅 of 𝑠𝑡-space, and also 𝑟⃗ = 𝑟(𝑢, 𝑣) for (𝑢, 𝑣) in the region 𝑇 in 𝑢𝑣-space, and suppose that the two parameterizations are related by the change of coordinates 𝑢 = 𝑢(𝑠, 𝑡)

𝑣 = 𝑣(𝑠, 𝑡).

Suppose that the Jacobian determinant 𝜕(𝑢, 𝑣)∕𝜕(𝑠, 𝑡) is positive at every point (𝑠, 𝑡) in 𝑅. Use the change of coordinates formula for double integrals on page 1035 to show that computing the flux integral using either parameterization gives the same result.

Strengthen Your Understanding In Problems 21–22, explain what is wrong with the statement. 21. The area of the surface parameterized by 𝑥 = 𝑠, 𝑦 = 𝑡, 𝑧 = 𝑓 (𝑠, 𝑡) above the square 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1 is given by the integral 1

Area =

∫0 ∫0

24. A vector field 𝐹⃗ such that ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ > 0, where 𝑆 is ⃗ , 0 ≤ 𝑠 ≤ 1, the surface 𝑟⃗ = (𝑠 − 𝑡)⃗𝑖 + 𝑡2 𝑗⃗ + (𝑠 + 𝑡)𝑘 𝜕⃗𝑟 𝜕⃗𝑟 × . 0 ≤ 𝑡 ≤ 1, oriented in the direction of 𝜕𝑠 𝜕𝑡 Are the statements in Problems 25–27 true or false? Give reasons for your answer.

1

𝑓 (𝑠, 𝑡) 𝑑𝑠 𝑑𝑡.

22. The surface 𝑆 parameterized by 𝑥 = 𝑓 (𝑠, 𝑡), 𝑦 = 𝑔(𝑠, 𝑡), 𝑧 = ℎ(𝑠, 𝑡), where 0 ≤ 𝑠 ≤ 2, 0 ≤ 𝑡 ≤ 3, has area 6. In Problems 23–24, give an example of: 23. A parameterization 𝑟⃗ = 𝑟⃗ (𝑠, 𝑡) of the 𝑥𝑦-plane such that 𝑑𝐴 = 2 𝑑𝑠 𝑑𝑡.

25. If 𝑟⃗ (𝑠, 𝑡), 0 ≤ 𝑠 ≤ 1, 0 ≤ 𝑡 ≤ 1 is an oriented parameterized surface 𝑆, and 𝐹⃗ is a vector field that is everywhere tangent to 𝑆, then the flux of 𝐹⃗ through 𝑆 is zero. 26. For any parameterization of the surface 𝑥2 −𝑦2 +𝑧2 = 6, ⃗ )𝑑𝑥 𝑑𝑦. 𝑑 𝐴⃗ at (1, 2, 3) is a multiple of (2⃗𝑖 − 4𝑗⃗ + 6𝑘 27. If you parameterize the plane 3𝑥 + 4𝑦 + 5𝑧 = 7, then there is a constant 𝑐 such that, at any point (𝑥, 𝑦, 𝑧), ⃗ )𝑑𝑥 𝑑𝑦. 𝑑 𝐴⃗ = 𝑐(3⃗𝑖 + 4𝑗⃗ + 5𝑘

1042

Chapter 21 PARAMETERS, COORDINATES, AND INTEGRALS

28. Let 𝑆 be the hemisphere 𝑥2 + 𝑦2 + 𝑧2 = 1 with 𝑥 ≤ 0, oriented away from the origin. Which of the following integrals represents the flux of 𝐹⃗ (𝑥, 𝑦, 𝑧) through 𝑆? 𝜕⃗𝑟 𝐹⃗ (𝑥, 𝑦, 𝑧(𝑥, 𝑦)) ⋅ × 𝜕𝑥 𝜕⃗𝑟 (b) × 𝐹⃗ (𝑥, 𝑦, 𝑧(𝑥, 𝑦)) ⋅ ∫𝑅 𝜕𝑦 (a)

∫𝑅

𝜕⃗𝑟 𝑑𝑥 𝑑𝑦 𝜕𝑦 𝜕⃗𝑟 𝑑𝑦 𝑑𝑥 𝜕𝑥

Online Resource: Review problems and Projects

𝜕⃗𝑟 𝐹⃗ (𝑥, 𝑦(𝑥, 𝑧), 𝑧) ⋅ × 𝜕𝑥 𝜕⃗𝑟 𝐹⃗ (𝑥, 𝑦(𝑥, 𝑧), 𝑧) ⋅ (d) × ∫𝑅 𝜕𝑧 𝜕⃗𝑟 𝐹⃗ (𝑥(𝑦, 𝑧), 𝑦, 𝑧) ⋅ (e) × ∫𝑅 𝜕𝑦 𝜕⃗𝑟 𝐹⃗ (𝑥(𝑦, 𝑧), 𝑦, 𝑧) ⋅ (f) × ∫𝑅 𝜕𝑧 (c)

∫𝑅

𝜕⃗𝑟 𝑑𝑥 𝑑𝑧 𝜕𝑧 𝜕⃗𝑟 𝑑𝑧 𝑑𝑥 𝜕𝑥 𝜕⃗𝑟 𝑑𝑦 𝑑𝑧 𝜕𝑧 𝜕⃗𝑟 𝑑𝑧 𝑑𝑦 𝜕𝑦

Ready Reference Ready Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1044

1044

READY REFERENCE

READY REFERENCE 𝑃

A Library of Functions Linear functions (p. 4) have the form 𝑦 = 𝑓 (𝑥) = 𝑏 + 𝑚𝑥, where 𝑚 is the slope, or rate of change of 𝑦 with respect to 𝑥 (p. 5) and 𝑏 is the vertical intercept, or value of 𝑦 when 𝑥 is zero (p. 5). The slope is 𝑚=

1 0.8 (0.95)𝑡

0.6

(0.9)𝑡

0.4

𝑓 (𝑥2 ) − 𝑓 (𝑥1 ) Δ𝑦 Rise = = Run Δ𝑥 𝑥2 − 𝑥1

(p. 5). 0.2

(0.5)𝑡

(0.8)𝑡

(0.1)𝑡

𝑦 = −𝑥

𝑦=𝑥

𝑦 = −0.5𝑥

𝑡

0

𝑦 = −2𝑥 𝑦 𝑦 = 2𝑥 𝑦 = 0.5𝑥

2

4

6

8

10

12

Figure R.4: Exponential decay: 𝑃 = 𝑎𝑡 , for 0 < 𝑎 < 1 (p. 17)

𝑥

Common Logarithm and Natural Logarithm log10 𝑥 = log 𝑥 = power of 10 that gives 𝑥 (p. 32) log10 𝑥 = 𝑐 means 10𝑐 = 𝑥 (p. 32) ln 𝑥 = power of 𝑒 that gives 𝑥 (p. 32) ln 𝑥 = 𝑐 means 𝑒𝑐 = 𝑥 (p. 32) log 𝑥 and ln 𝑥 are not defined if 𝑥 is negative or 0 (p. 32).

Figure R.1: The family 𝑦 = 𝑚𝑥 (with 𝑏 = 0) (p. 6) 𝑦

𝑦=𝑥 𝑦 = −1 + 𝑥 𝑥 𝑦=2+𝑥 𝑦=1+𝑥

Figure R.2: The family 𝑦 = 𝑏 + 𝑥 (with 𝑚 = 1) (p. 6) Exponential functions have the form 𝑃 = 𝑃0 𝑎𝑡 (p. 15) or 𝑃 = 𝑃0 𝑒𝑘𝑡 (p. 17), where 𝑃0 is the initial quantity (p. 15), 𝑎 is the growth (decay) factor per unit time (p. 15), |𝑘| is the continuous growth (decay) rate (pp. 17, 587), and 𝑟 = |𝑎−1| is the growth (decay) rate per unit time (p. 15). Suppose 𝑃0 > 0. If 𝑎 > 1 or 𝑘 > 0, we have exponential growth; if 0 < 𝑎 < 1 or 𝑘 < 0, we have exponential decay (p. 16). The doubling time (for growth) is the time required for 𝑃 to double (p. 16). The half-life (for decay) is the time required for 𝑃 to be reduced by a factor of one half (p. 16). The continuous growth rate 𝑘 = ln(1 + 𝑟) is slightly less than, but very close to, 𝑟, provided 𝑟 is small (p. 588).

Properties of Logarithms (p. 33) 4. log (10𝑥 ) = 𝑥

1. log(𝐴𝐵) = log 𝐴 + log 𝐵 ( ) 2. log 𝐵𝐴 = log 𝐴 − log 𝐵

5. 10log 𝑥 = 𝑥

6. log 1 = 0 3. log (𝐴𝑝 ) = 𝑝 log 𝐴 The natural logarithm satisfies properties 1, 2, and 3, and ln 𝑒𝑥 = 𝑥, 𝑒ln 𝑥 = 𝑥, ln 1 = 0 (p. 33). Trigonometric Functions The sine and cosine are defined 𝑡 = in Figure R.5 (see also p. 40). The tangent is tan 𝑡 = sin cos 𝑡 slope of the line through the origin (0, 0) and 𝑃 if cos 𝑡 ≠ 0 (p. 43). The period of sin and cos is 2𝜋 (p. 40), the period of tan is 𝜋 (p. 44).

(0, 1)

𝑃



𝑥 = cos 𝑡 𝑦 = sin 𝑡

𝑦 𝑡

❄ ✛𝑥✲

(1, 0)

𝑃 40

10𝑡

5𝑡 3𝑡

2𝑡

30

Figure R.5: The definitions of sin 𝑡 and cos 𝑡 (p. 40)

(1.5)𝑡

20 10

𝑡 1

2

3

4

5

6

7

Figure R.3: Exponential growth: 𝑃 = 𝑎𝑡 , for 𝑎 > 1 (p. 17)

A sinusoidal function (p. 41) has the form 𝑦 = 𝐶 + 𝐴 sin(𝐵(𝑡 + ℎ)) or 𝑦 = 𝐶 + 𝐴 cos(𝐵(𝑡 + ℎ)).

READY REFERENCE

The amplitude is |𝐴|, (p. 41), and the period is 2𝜋∕|𝐵| (p. 41). cos 𝑡

1

Graphs for zero and negative powers: 𝑦

𝑦



sin 𝑡

Amplitude = 1

𝑦=

❄ 𝑡 −3𝜋

−𝜋

−2𝜋

𝜋

−1

2𝜋

1045

1 = 𝑥−2 𝑥2

3𝜋

𝑥 𝑦=

✛Period = 2𝜋✲

𝑥 1 = 𝑥−1 𝑥

𝑦

Figure R.6: Graphs of cos 𝑡 and sin 𝑡 (p. 41) 𝑦 = 𝑥0

Trigonometric Identities sin2 𝑥 + cos2 𝑥 = 1

𝑥

sin(2𝑥) = 2 sin 𝑥 cos 𝑥 cos(2𝑥) = cos 𝑥 − sin2 𝑥 = 2 cos2 𝑥 − 1 = 1 − 2 sin2 𝑥 2

cos(𝑎 + 𝑏) = cos 𝑎 cos 𝑏 − sin 𝑎 sin 𝑏

Figure R.10: Comparison of zero and negative powers of 𝑥

sin(𝑎 + 𝑏) = sin 𝑎 cos 𝑏 + cos 𝑎 sin 𝑏 Inverse Trigonometric Functions: arcsin 𝑦 = 𝑥 means sin 𝑥 = 𝑦 with −(𝜋∕2) ≤ 𝑥 ≤ (𝜋∕2) (p. 44), arccos 𝑦 = 𝑥 means cos 𝑥 = 𝑦 with 0 ≤ 𝑥 ≤ 𝜋 (Problem 74, p. 49), arctan 𝑦 = 𝑥 means tan 𝑥 = 𝑦 with −(𝜋∕2) < 𝑥 < (𝜋∕2) (p. 45). The domain of arcsin and arccos is [−1, 1] (p. 44), the domain of arctan is all numbers (p. 45). Power Functions have the form 𝑓 (𝑥) = 𝑘𝑥𝑝 (p. 49). Graphs for positive powers: 𝑦

𝑦

𝑥5 𝑥3

10

𝑓 (𝑥) = 𝑎𝑛 𝑥𝑛 + 𝑎𝑛−1 𝑥𝑛−1 + ⋯ + 𝑎1 𝑥 + 𝑎0 , 𝑎𝑛 ≠ 0 (p. 51). The degree is 𝑛 (p. 51) and the leading coefficient is 𝑎𝑛 (p. 51).

𝑥4 𝑥2

10

5 −2

Polynomials have the form

𝑥

−1

5 𝑥

1

2

Quadratic

𝑥

−5

−3 −2 −1

−10

1

2

Cubic

(𝑛 = 2)

(𝑛 = 3)

3

−5

Figure R.7: Odd integer powers of 𝑥: “Seat” shaped for 𝑘 > 1 (p. 50)

Figure R.8: Even integer ⋃ powers of 𝑥: -shaped (p. 50) Quartic

Quintic

(𝑛 = 4)

𝑦 𝑥3 𝑥2 𝑥3∕2

2

(𝑛 = 5)

Figure R.11: Graphs of typical polynomials of degree 𝑛 (p. 51)

𝑥

𝑥1∕2 𝑥1∕3 1

𝑝(𝑥) , where 𝑞(𝑥) 𝑝 and 𝑞 are polynomials (p. 53). There is usually a vertical asymptote at 𝑥 = 𝑎 if 𝑞(𝑎) = 0 and a horizontal asymptote at 𝑦 = 𝐿 if lim𝑥→∞ 𝑓 (𝑥) = 𝐿 or lim𝑥→−∞ 𝑓 (𝑥) = 𝐿 (p. 53). Rational Functions have the form 𝑓 (𝑥) =

Hyperbolic Functions: 𝑥

0 1

2

Figure R.9: Comparison of some fractional powers of 𝑥

cosh 𝑥 =

𝑒𝑥 + 𝑒−𝑥 2

sinh 𝑥 =

𝑒𝑥 − 𝑒−𝑥 2

(p. 174).

1046

READY REFERENCE

Relative Growth Rates of Functions

Operations on Functions

Power Functions As 𝑥 → ∞, higher powers of 𝑥 dominate, as 𝑥 → 0, smaller powers dominate.

Shifts, Stretches, and Composition Multiplying by a constant, 𝑐, stretches (if 𝑐 > 1) or shrinks (if 0 < 𝑐 < 1) the graph vertically. A negative sign (if 𝑐 < 0) reflects the graph about the 𝑥-axis, in addition to shrinking or stretching (p. 24). Replacing 𝑦 by (𝑦 − 𝑘) moves a graph up by 𝑘 (down if 𝑘 is negative) (p. 24). Replacing 𝑥 by (𝑥 − ℎ) moves a graph to the right by ℎ (to the left if ℎ is negative) (p. 24). The composite of 𝑓 and 𝑔 is the function 𝑓 (𝑔(𝑥)); 𝑓 is the outside function, 𝑔 the inside function (p. 24).

𝑥5

𝑥4 𝑥3 𝑥2

1

✛ ✛ ✛ ✛ ✛ ✛

15 10

𝑥1.5 𝑥

5

𝑥 1

2

3

4

𝑥 𝑥1.5 𝑥2 𝑥3 𝑥4 𝑥5 𝑥

1

Figure R.12: For large 𝑥: Figure R.13: For 0 ≤ 𝑥 ≤ 1: Large powers of 𝑥 dominate Small powers of 𝑥 dominate

Power Functions Versus Exponential Functions Every exponential growth function eventually dominates every power function (p. 50).

Symmetry We say 𝑓 is an even function if 𝑓 (−𝑥) = 𝑓 (𝑥) (p. 25) and 𝑓 is an odd function if 𝑓 (−𝑥) = −𝑓 (𝑥) (p. 25). Inverse Functions A function 𝑓 has an inverse if (and only if) its graph intersects any horizontal line at most once (p. 27). If 𝑓 has an inverse, it is written 𝑓 −1 , and 𝑓 −1 (𝑥) = 𝑦 means 𝑓 (𝑦) = 𝑥 (p. 27). Provided the 𝑥 and 𝑦 scales are equal, the graph of 𝑓 −1 is the reflection of the graph of 𝑓 about the line 𝑦 = 𝑥 (p. 28). Proportionality We say 𝑦 is proportional to 𝑥 if 𝑦 = 𝑘𝑥 for 𝑘 a nonzero constant. We say 𝑦 is inversely proportional to 𝑥 if 𝑦 = 𝑘(1∕𝑥) (p. 6).

1.1𝑥 1029

Limits and Continuity

𝑥10

𝑥 250

500

750

Figure R.14: Exponential function eventually dominates power function

Power Functions Versus Logarithm Functions The power function 𝑥𝑝 dominates 𝐴 log 𝑥 for large 𝑥 for all values of 𝑝 > 0 and 𝐴 > 0. 𝑥1∕3

1000

100 log 𝑥

log 𝑥 𝑥1∕3

1

𝑥

−1

109

100

Figure R.15: Comparison of 𝑥1∕3 and log 𝑥

𝑥

Figure R.16: Comparison of 𝑥1∕3 and 100 log 𝑥

Idea of Limit (p. 60) If there is a number 𝐿 such that 𝑓 (𝑥) is as close to 𝐿 as we please whenever 𝑥 is sufficiently close to 𝑐 (but 𝑥 ≠ 𝑐), then lim𝑥→𝑐 𝑓 (𝑥) = 𝐿. Definition of Limit()Section 1.10 online) If there is a number 𝐿 such that for any 𝜖 > 0, there exists a 𝛿 > 0 such that if |𝑥 − 𝑐| < 𝛿 and 𝑥 ≠ 𝑐, then |𝑓 (𝑥) − 𝐿| < 𝜖, then lim𝑥→𝑐 𝑓 (𝑥) = 𝐿. One-sided Limits (p. 67) If 𝑓 (𝑥) approaches 𝐿 as 𝑥 approaches 𝑐 through values greater than 𝑐, then lim𝑥→𝑐 + 𝑓 (𝑥) = 𝐿. If 𝑓 (𝑥) approaches 𝐿 as 𝑥 approaches 𝑐 through values less than 𝑐, then lim𝑥→𝑐 − 𝑓 (𝑥) = 𝐿. Limits at Infinity (p. 68) If 𝑓 (𝑥) gets as close to 𝐿 as we please when 𝑥 gets sufficiently large, then lim𝑥→∞ 𝑓 (𝑥) = 𝐿. Similarly, if 𝑓 (𝑥) approaches 𝐿 as 𝑥 gets more and more negative, then lim𝑥→−∞ 𝑓 (𝑥) = 𝐿. Limits of Continuous Functions (p. 62) If a function 𝑓 (𝑥) is continuous at 𝑥 = 𝑐, the limit is the value of 𝑓 (𝑥) there: lim 𝑓 (𝑥) = 𝑓 (𝑐). Thus, to find limits for a continuous 𝑥→𝑐 function: Substitute 𝑐. Theorem: Properties of Limits (p. 71) Assuming all the limits on the right-hand side exist: ( ) 1. If 𝑏 is a constant, then lim (𝑏𝑓 (𝑥)) = 𝑏 lim 𝑓 (𝑥) . 𝑥→𝑐

Numerical comparisons of growth rates: Table R.1 Comparison of 𝑥0.001 and 1000 log 𝑥 𝑥

𝑥0.001

1000 log 𝑥

Table R.2 Comparison of 𝑥100 and 1.01𝑥 𝑥

𝑥100

1.01𝑥

𝑥→𝑐

2. lim (𝑓 (𝑥) + 𝑔(𝑥)) = lim 𝑓 (𝑥) + lim 𝑔(𝑥). 𝑥→𝑐 ) ( 𝑥→𝑐 ) ( 𝑥→𝑐 3. lim (𝑓 (𝑥)𝑔(𝑥)) = lim 𝑓 (𝑥) lim 𝑔(𝑥) . 𝑥→𝑐

𝑥→𝑐

𝑥→𝑐

lim𝑥→𝑐 𝑓 (𝑥) 𝑓 (𝑥) = , provided lim𝑥→𝑐 𝑔(𝑥) ≠ 0. 4. lim 𝑥→𝑐 𝑔(𝑥) lim𝑥→𝑐 𝑔(𝑥) 5. For any constant 𝑘, lim 𝑘 = 𝑘. 𝑥→𝑐

5000

10

105

5⋅

106

104

400

10

1.6 ⋅ 1043

6. lim 𝑥 = 𝑐.

106000

106

6 ⋅ 106

105

10500

1.4 ⋅ 10432

107000

107

7 ⋅ 106

106

10600

2.4 ⋅ 104321

Idea of Continuity (p. 58) A function is continuous on an interval if its graph has no breaks, jumps, or holes in that interval.

𝑥→𝑐

READY REFERENCE 𝑓 (𝑥)

Definition of Continuity (p. 60) The function 𝑓 is continuous at 𝑥 = 𝑐 if 𝑓 is defined at 𝑥 = 𝑐 and lim 𝑓 (𝑥) = 𝑓 (𝑐).

𝐴

𝑥→𝑐



The function is continuous on an interval if it is continuous at every point in the interval. Theorem: Continuity of Sums, Products, Quotients (p. 71) Suppose that 𝑓 and 𝑔 are continuous on an interval and that 𝑏 is a constant. Then, on that same interval, the following functions are also continuous: 𝑏𝑓 (𝑥), 𝑓 (𝑥) + 𝑔(𝑥), 𝑓 (𝑥)𝑔(𝑥). Further, 𝑓 (𝑥)∕𝑔(𝑥) is continuous provided 𝑔(𝑥) ≠ 0 on the interval. Theorem: Continuity of Composite Functions (p. 71) Suppose 𝑓 and 𝑔 are continuous and 𝑓 (𝑔(𝑥)) is defined on an interval. Then on that interval 𝑓 (𝑔(𝑥)) is continuous. Intermediate Value Theorem (p. 61) Suppose 𝑓 is continuous on a closed interval [𝑎, 𝑏]. If 𝑘 is any number between 𝑓 (𝑎) and 𝑓 (𝑏), then there is at least one number 𝑐 in [𝑎, 𝑏] such that 𝑓 (𝑐) = 𝑘. The Extreme Value Theorem (p. 203) If 𝑓 is continuous on the interval [𝑎, 𝑏], then 𝑓 has a global maximum and a global minimum on that interval. The Squeeze Theorem (p. 79) If 𝑏(𝑥) ≤ 𝑓 (𝑥) ≤ 𝑎(𝑥) for all 𝑥 close to 𝑥 = 𝑐 except possibly at 𝑥 = 𝑐, and lim 𝑏(𝑥) = 𝐿 = lim 𝑎(𝑥), then lim 𝑓 (𝑥) = 𝐿. 𝑥→𝑐

𝑥→𝑐

Slope = Derivative = 𝑓 ′ (𝑎)

𝑥

𝑎

Figure R.18: Visualizing the instantaneous rate of change of 𝑓 (p. 93) Units of 𝑓 (𝑥) (p. 109). If The units of 𝑓 ′ (𝑥) are: Units of 𝑥 𝑓 > 0 on an interval, then 𝑓 is increasing over that interval (p. 101). If 𝑓 ′ < 0 on an interval, then 𝑓 is decreasing over that interval (p. 101). If 𝑓 ′′ > 0 on an interval, then 𝑓 is concave up over that interval (p. 115). If 𝑓 ′′ < 0 on an interval, then 𝑓 is concave down over that interval (p. 115). The tangent line at (𝑎, 𝑓 (𝑎)) is the graph of 𝑦 = 𝑓 (𝑎) + 𝑓 ′ (𝑎)(𝑥−𝑎) (p. 178). The tangent line approximation says that for values of 𝑥 near 𝑎, 𝑓 (𝑥) ≈ 𝑓 (𝑎) + 𝑓 ′ (𝑎)(𝑥 − 𝑎). The expression 𝑓 (𝑎) + 𝑓 ′ (𝑎)(𝑥 − 𝑎) is called the local linearization of 𝑓 near 𝑥 = 𝑎 (p. 178). ′

Derivatives of elementary functions 𝑑 𝑛 (𝑥 ) = 𝑛𝑥𝑛−1 (p. 132) 𝑑𝑥

𝑑 𝑥 (𝑒 ) = 𝑒𝑥 (p. 141) 𝑑𝑥

𝑥→𝑐

𝑑 𝑥 (𝑎 ) = (ln 𝑎)𝑎𝑥 (p. 142) 𝑑𝑥

The Derivative The slope of the secant line of 𝑓 (𝑥) over an interval [𝑎, 𝑏] gives: Average rate of change

=

of 𝑓 over [𝑎, 𝑏]

𝑓 (𝑏) − 𝑓 (𝑎) 𝑏−𝑎

Slope = Average rate of change

=

1047

𝑓 (𝑥)

𝑓 (𝑎+ℎ)−𝑓 (𝑎) ℎ

𝐵 𝑓✻ (𝑎 + ℎ) − 𝑓 (𝑎)



𝐴





𝑎

(p. 91).





𝑥

𝑎+ℎ

Figure R.17: Visualizing the average rate of change of 𝑓 (p. 93) The derivative of 𝑓 at 𝑎 is the slope of the line tangent to the graph of 𝑓 at the point (𝑎, 𝑓 (𝑎)): 𝑓 ′ (𝑎) = lim ℎ→0

𝑓 (𝑎 + ℎ) − 𝑓 (𝑎) , ℎ

and gives the instantaneous rate of change of 𝑓 at 𝑎 (p. 92). The function 𝑓 is differentiable at 𝑎 if this limit exists (p. 92). The second derivative of 𝑓 , denoted 𝑓 ′′ , is the derivative of 𝑓 ′ (p. 115).

𝑑 (sin 𝑥) = cos 𝑥 𝑑𝑥

𝑑 1 (ln 𝑥) = (p. 165) 𝑑𝑥 𝑥

𝑑 (cos 𝑥) = − sin 𝑥 (p. 160) 𝑑𝑥

𝑑 1 (arctan 𝑥) = (p. 166) 𝑑𝑥 1 + 𝑥2 1 𝑑 (arcsin 𝑥) = √ (p. 166) 𝑑𝑥 1 − 𝑥2 Derivatives of sums, differences, and constant multiples 𝑑 [𝑓 (𝑥) ± 𝑔(𝑥)] = 𝑓 ′ (𝑥) ± 𝑔 ′ (𝑥) (p. 131) 𝑑𝑥 𝑑 [𝑐𝑓 (𝑥)] = 𝑐𝑓 ′ (𝑥) (p. 130) 𝑑𝑥 Product and quotient rules (𝑓 𝑔)′ = 𝑓 ′ 𝑔 + 𝑓 𝑔 ′ (p. 146) ( )′ 𝑓 𝑓 ′𝑔 − 𝑓 𝑔′ (p. 147) = 𝑔 𝑔2 Chain rule 𝑑 𝑓 (𝑔(𝑥)) = 𝑓 ′ (𝑔(𝑥)) ⋅ 𝑔 ′ (𝑥) (p. 152) 𝑑𝑥

1048

READY REFERENCE

Derivative of an inverse function (p. 166). If 𝑓 has a differentiable inverse, 𝑓 −1 , then 𝑑 −1 1 . (𝑓 (𝑥)) = ′ −1 𝑑𝑥 𝑓 (𝑓 (𝑥)) Implicit differentiation (p. 171) If 𝑦 is implicitly defined as a function of 𝑥 by an equation, then, to find 𝑑𝑦∕𝑑𝑥, differentiate the equation (remembering to apply the chain rule).

Theorems About Derivatives Theorem: Local Extrema and Critical Points (pp. 194, 198) Suppose 𝑓 is defined on an interval and has a local maximum or minimum at the point 𝑥 = 𝑎, which is not an endpoint of the interval. If 𝑓 is differentiable at 𝑥 = 𝑎, then 𝑓 ′ (𝑎) = 0. The Mean Value Theorem (p. 186) If 𝑓 is continuous on [𝑎, 𝑏] and differentiable on (𝑎, 𝑏), then there exists a number 𝑐, with 𝑎 < 𝑐 < 𝑏, such that

Applications of the Derivative

𝑓 ′ (𝑐) =

A function 𝑓 has a local maximum at 𝑝 if 𝑓 (𝑝) is greater than or equal to the values of 𝑓 at points near 𝑝, and a local minimum at 𝑝 if 𝑓 (𝑝) is less than or equal to the values of 𝑓 at points near 𝑝 (p. 193). It has a global maximum at 𝑝 if 𝑓 (𝑝) is greater than or equal to the value of 𝑓 at any point in the interval, and a global minimum at 𝑝 if 𝑓 (𝑝) is less than or equal to the value of 𝑓 at any point in the interval (p. 203). A critical point of a function 𝑓 (𝑥) is a point 𝑝 in the domain of 𝑓 where 𝑓 ′ (𝑝) = 0 or 𝑓 ′ (𝑝) is undefined (p. 193). Theorem: Local maxima and minima which do not occur at endpoints of the domain occur at critical points (pp. 194, 198). The First-Derivative Test for Local Maxima and Minima (p. 194): • If 𝑓 ′ changes from negative to positive at 𝑝, then 𝑓 has a local minimum at 𝑝. ′

• If 𝑓 changes from positive to negative at 𝑝, then 𝑓 has a local maximum at 𝑝. The Second-Derivative Test for Local Maxima and Minima (p. 196): • If 𝑓 ′ (𝑝) = 0 and 𝑓 ′′ (𝑝) > 0 then 𝑓 has a local minimum at 𝑝. • If 𝑓 ′ (𝑝) = 0 and 𝑓 ′′ (𝑝) < 0 then 𝑓 has a local maximum at 𝑝. • If 𝑓 ′ (𝑝) = 0 and 𝑓 ′′ (𝑝) = 0 nothing.

then

𝑓 (𝑏) − 𝑓 (𝑎) . 𝑏−𝑎

The Increasing Function Theorem (p. 187) Suppose that 𝑓 is continuous on [𝑎, 𝑏] and differentiable on (𝑎, 𝑏). • If 𝑓 ′ (𝑥) > 0 on (𝑎, 𝑏), then 𝑓 is increasing on [𝑎, 𝑏]. • If 𝑓 ′ (𝑥) ≥ 0 on (𝑎, 𝑏), then 𝑓 is nondecreasing on [𝑎, 𝑏]. The Constant Function Theorem (p. 187) Suppose that 𝑓 is continuous on [𝑎, 𝑏] and differentiable on (𝑎, 𝑏). If 𝑓 ′ (𝑥) = 0 on (𝑎, 𝑏), then 𝑓 is constant on [𝑎, 𝑏]. The Racetrack Principle (p. 188) Suppose that 𝑔 and ℎ are continuous on [𝑎, 𝑏] and differentiable on (𝑎, 𝑏), and that 𝑔 ′ (𝑥) ≤ ℎ′ (𝑥) for 𝑎 < 𝑥 < 𝑏. • If 𝑔(𝑎) = ℎ(𝑎), then 𝑔(𝑥) ≤ ℎ(𝑥) for 𝑎 ≤ 𝑥 ≤ 𝑏. • If 𝑔(𝑏) = ℎ(𝑏), then 𝑔(𝑥) ≥ ℎ(𝑥) for 𝑎 ≤ 𝑥 ≤ 𝑏. Theorem: Differentiability and Local Linearity (p. 179) Suppose 𝑓 is differentiable at 𝑥 = 𝑎 and 𝐸(𝑥) is the error in the tangent line approximation, that is: 𝐸(𝑥) = 𝐸(𝑥) 𝑓 (𝑥) − 𝑓 (𝑎) − 𝑓 ′ (𝑎)(𝑥 − 𝑎). Then lim = 0. 𝑥→𝑎 𝑥 − 𝑎 Theorem: A Differentiable Function Is Continuous (p. 125) If 𝑓 (𝑥) is differentiable at a point 𝑥 = 𝑎, then 𝑓 (𝑥) is continuous at 𝑥 = 𝑎.

the test tells us

To find the global maximum and minimum of a function on an interval we compare values of 𝑓 at all critical points in the interval and at the endpoints of the interval (or lim 𝑓 (𝑥) if the interval is unbounded) (p. 204). 𝑥→±∞

An inflection point of 𝑓 is a point at which the graph of 𝑓 changes concavity (p. 196); 𝑓 ′′ is zero or undefined at an inflection point (p. 196). L’Hopital’s rule (p. 252) If 𝑓 and 𝑔 are continuous, 𝑓 (𝑎) = 𝑔(𝑎) = 0, and 𝑔 ′ (𝑎) ≠ 0, then

The Definite Integral The definite integral of 𝒇 from 𝑎 to 𝑏 (p. 284), denoted 𝑏 ∫𝑎 𝑓 (𝑥) 𝑑𝑥, is the limit of the left and the right sums as the width of the rectangles is shrunk to 0, where

Left-hand sum =

𝑛−1 ∑

𝑓 (𝑥𝑖 )Δ𝑥

(p. 277)

𝑖=0

= 𝑓 (𝑥0 )Δ𝑥 + 𝑓 (𝑥1 )Δ𝑥 + ⋯ + 𝑓 (𝑥𝑛−1 )Δ𝑥

𝑓 (𝑥) 𝑓 ′ (𝑎) = ′ . lim 𝑥→𝑎 𝑔(𝑥) 𝑔 (𝑎) Parametric equations (p. 259) If a curve is given by the parametric equations 𝑥 = 𝑓 (𝑡), 𝑦 = 𝑔(𝑡), the slope of the curve as a function of 𝑡 is 𝑑𝑦∕𝑑𝑥 = (𝑑𝑦∕𝑑𝑡)∕(𝑑𝑥∕𝑑𝑡).

Right-hand sum =

𝑛 ∑

𝑓 (𝑥𝑖 )Δ𝑥

(p. 283)

𝑖=1

= 𝑓 (𝑥1 )Δ𝑥 + 𝑓 (𝑥2 )Δ𝑥 + ⋯ + 𝑓 (𝑥𝑛 )Δ𝑥

READY REFERENCE 𝑏

✻ ✛

Area = 𝑓 (𝑥0 )Δ𝑥

∫𝑎 𝑐

∫𝑎



∫𝑎

∫𝑐

𝑔(𝑥) 𝑑𝑥

∫𝑎

𝑏

𝑓 (𝑥) 𝑑𝑥 =

𝑏

𝑓 (𝑥)

𝑓 (𝑥) 𝑑𝑥 ±

𝑏

𝑓 (𝑥) 𝑑𝑥 +

∫𝑎

𝑓 (𝑥0 )

𝑏

𝑏

(𝑓 (𝑥) ± 𝑔(𝑥)) 𝑑𝑥 =

1049

∫𝑎

𝑓 (𝑥) 𝑑𝑥

𝑏

𝑐𝑓 (𝑥) 𝑑𝑥 = 𝑐

∫𝑎

𝑓 (𝑥) 𝑑𝑥

𝑥

𝑎 = 𝑥0

𝑥1

𝑥2 ⋅ ⋅ ⋅

𝑥𝑛 = 𝑏

Antiderivatives

Figure R.19: Left-hand sum (p. 277)

✻✛

An antiderivative of a function 𝑓 (𝑥) is a function 𝐹 (𝑥) such that 𝐹 ′ (𝑥) = 𝑓 (𝑥) (p. 316). There are infinitely many antiderivatives of 𝑓 since 𝐹 (𝑥) + 𝐶 is an antiderivative of 𝑓 for any constant 𝐶, provided 𝐹 ′ (𝑥) = 𝑓 (𝑥) (p. 316). The indefinite integral of 𝑓 is the family of antiderivatives ∫ 𝑓 (𝑥) 𝑑𝑥 = 𝐹 (𝑥) + 𝐶 (p. 322).

Area = 𝑓 (𝑥1 )Δ𝑥

𝑓 (𝑥1 )

𝑓 (𝑥)



Construction Theorem (Second Fundamental Theorem of Calculus) If 𝑓 is a continuous function on an interval and 𝑥 𝑎 is any number in that interval, then 𝐹 (𝑥) = ∫𝑎 𝑓 (𝑡) 𝑑𝑡 is an antiderivative of 𝑓 (p. 336).

𝑥

𝑎 = 𝑥0

𝑥1

𝑥2 ⋅ ⋅ ⋅

𝑥𝑛 = 𝑏

Figure R.20: Right-hand sum (p. 277)

Properties of Indefinite Integrals (p. 325) 𝑏

If 𝑓 is nonnegative, ∫𝑎 𝑓 (𝑥) 𝑑𝑥 represents the area under the curve between 𝑥 = 𝑎 and 𝑥 = 𝑏 (p. 285). If 𝑓 has any 𝑏 sign, ∫𝑎 𝑓 (𝑥) 𝑑𝑥 is the sum of the areas above the 𝑥-axis, counted positively, and the areas below the 𝑥-axis, counted negatively (p. 286). If 𝐹 ′ (𝑡) is the rate of change of some 𝑏 quantity 𝐹 (𝑡), then ∫𝑎 𝐹 ′ (𝑡) 𝑑𝑡 is the total change in 𝐹 (𝑡) between 𝑡 = 𝑎 and 𝑡 = 𝑏 (p. 294). The average value of 𝑓 𝑏 1 𝑓 (𝑥) 𝑑𝑥 (p. 308). on the interval [𝑎, 𝑏] is given by 𝑏 − 𝑎 ∫𝑎



(𝑓 (𝑥) ± 𝑔(𝑥)) 𝑑𝑥 =



𝑏



❄𝑥

𝑎



𝑏

Figure R.21: The definite integral 𝑏 ∫𝑎 𝑓 (𝑥) 𝑑𝑥 (p. 285)

𝑏−𝑎

𝑏

The units of ∫𝑎 𝑓 (𝑥) 𝑑𝑥 are Units of 𝑓 (𝑥) × Units of 𝑥 (p. 292). The Fundamental Theorem of Calculus If 𝑓 is continuous on [𝑎, 𝑏] and 𝑓 (𝑥) = 𝐹 ′ (𝑥), then 𝑏

∫𝑎

𝑓 (𝑥) 𝑑𝑥 = 𝐹 (𝑏) − 𝐹 (𝑎)

(p. 293).

Properties of Definite Integrals (pp. 302, 304) If 𝑎, 𝑏, and 𝑐 are any numbers and 𝑓 , 𝑔 are continuous functions, then 𝑎

∫𝑏

𝑏

𝑓 (𝑥) 𝑑𝑥 = −

∫𝑎

𝑓 (𝑥) 𝑑𝑥



✲𝑏

Figure R.22: Area and average value (p. 309)

𝑔(𝑥) 𝑑𝑥

𝑓 (𝑥) 𝑑𝑥

𝑥𝑛+1 + 𝐶, 𝑛+1

𝑛 ≠ −1 (p. 323)

1 𝑑𝑥 = ln |𝑥| + 𝐶 (p. 324) ∫ 𝑥

✻ Average value of 𝑓 𝑥





𝑘 𝑑𝑥 = 𝑘𝑥 + 𝐶 (p. 323)

𝑥𝑛 𝑑𝑥 =

Area under curve = Area of rectangle



𝑎

𝑐𝑓 (𝑥) 𝑑𝑥 = 𝑐

𝑓 (𝑥)

𝑓 (𝑥)

𝑓 (𝑥) 𝑑𝑥 ±

Some antiderivatives: ∫

Area = ∫𝑎 𝑓 (𝑥) 𝑑𝑥







𝑒𝑥 𝑑𝑥 = 𝑒𝑥 + 𝐶 (p. 324)

cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝐶 (p. 324) sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝐶 (p. 324)

𝑑𝑥 = arctan 𝑥 + 𝐶 (p. 372) ∫ 1 + 𝑥2 𝑑𝑥 = arcsin 𝑥 + 𝐶 (p. 369) ∫ √1 − 𝑥2

Substitution (p. 342) For integrals of the form ∫ 𝑓 (𝑔(𝑥))𝑔 ′ (𝑥) 𝑑𝑥, let 𝑤 = 𝑔(𝑥). Choose 𝑤, find 𝑑𝑤∕𝑑𝑥 and substitute for 𝑥 and 𝑑𝑥. Convert limits of integration for definite integrals.

1050

READY REFERENCE

Integration by Parts (p. 353) Used mainly for products; also for integrating ln 𝑥, arctan 𝑥, arcsin 𝑥. 𝑢𝑣′ 𝑑𝑥 = 𝑢𝑣 − 𝑢′ 𝑣 𝑑𝑥 ∫ ∫ Partial Fractions (p. 366) To integrate a rational function, 𝑃 (𝑥)∕𝑄(𝑥), express as a sum of a polynomial and terms of the form 𝐴∕(𝑥 − 𝑐)𝑛 and (𝐴𝑥 + 𝐵)∕𝑞(𝑥), where 𝑞(𝑥) is an unfactorable quadratic. Trigonometric Substitutions (p. 369) To simplify √ 2 − 𝑎2 , try 𝑥 = 𝑎 sin 𝜃 (p. 370). To simplify 𝑎2 + 𝑥2 𝑥√ or 𝑎2 + 𝑥2 , try 𝑥 = 𝑎 tan 𝜃 (p. 372). Numerical Approximations for Definite Integrals (p. 376) Riemann sums (left, right, midpoint) (p. 377), trapezoid rule (p. 378), Simpson’s rule (p. 381) Approximation Errors (p. 379) 𝑓 concave up: midpoint underestimates, trapezoid overestimates (p. 379) 𝑓 concave down: trapezoid underestimates, midpoint overestimates (p. 379) Evaluating Improper Integrals (p. 385) • Infinite limit of integration: 𝑏 lim𝑏→∞ ∫𝑎 𝑓 (𝑥) 𝑑𝑥 (p. 385)



∫𝑎 𝑓 (𝑥) 𝑑𝑥

=

• For 𝑎 < 𝑏, if integrand is unbounded at 𝑥 = 𝑏, then: 𝑏 𝑐 ∫𝑎 𝑓 (𝑥) 𝑑𝑥 = lim𝑐→𝑏− ∫𝑎 𝑓 (𝑥) 𝑑𝑥 (p. 389)

To find the center of mass of two- and three-dimensional objects, use the formula separately on each coordinate (p. 435). Applications to Physics Work done = Force × Distance (p. 440) Pressure = Density × 𝑔 × Depth (p. 444) Force = Pressure × Area (p. 444) Applications to Economics Present and future value of income stream, 𝑃 (𝑡) (p. 450); consumer and producer surplus (p. 453). 𝑀

Present value =

𝑃 (𝑡)𝑒−𝑟𝑡 𝑑𝑡 (p. 450)

∫0 𝑀

Future value =

𝑃 (𝑡)𝑒𝑟(𝑀−𝑡) 𝑑𝑡 (p. 450)

∫0

Applications to Probability Given a density function 𝑝(𝑥), the fraction of the population for which 𝑥 is between 𝑎 and 𝑏 is the area under the graph of 𝑝 between 𝑎 and 𝑏 (p. 459). The cumulative distribution function 𝑃 (𝑡) is the fraction having values of 𝑥 below 𝑡 (p. 460). The median is the value 𝑇 such that half the population has values of 𝑥 less than or equal to 𝑇 (p. 466). The mean (p. 468) is defined by ∞

Mean value =

Testing Improper Integrals for Convergence by Comparison (p. 395):

∫−∞

𝑥𝑝(𝑥)𝑑𝑥.



• If 0 ≤ 𝑓 (𝑥) ≤ 𝑔(𝑥) and ∫𝑎 𝑔(𝑥) 𝑑𝑥 converges, then ∞ ∫𝑎 𝑓 (𝑥) 𝑑𝑥 converges

Polar Coordinates 𝑦

• If 0 ≤ 𝑔(𝑥) ≤ 𝑓 (𝑥) and ∞ ∫𝑎 𝑓 (𝑥) 𝑑𝑥 diverges

∞ ∫𝑎

𝑔(𝑥) 𝑑𝑥 diverges, then

𝑃

(𝑥, 𝑦)



Applications of Integration

𝑟 𝑦

Total quantities can be approximated by slicing them into small pieces and summing the pieces. The limit of this sum is a definite integral which gives the exact total quantity. Applications to Geometry To calculate the volume of a solid, slice the volume into pieces whose volumes you can estimate (pp. 402, 410). Use this method to calculate volumes of revolution (p. 411) and volumes of solids with known cross sectional area (p. 412). Curve 𝑓 (𝑥) from 𝑥 = 𝑎 to 𝑥 = 𝑏 has 𝑏

Arc length =

∫𝑎

√ 1 + (𝑓 ′ (𝑥))2 𝑑𝑥

(p. 414).

Mass and Center of Mass from Density, 𝛿 𝑏

Total mass =

𝛿(𝑥) 𝑑𝑥 (p. 430)

∫𝑎 𝑏

Center of mass =

∫𝑎 𝑥𝛿(𝑥) 𝑑𝑥 𝑏

∫𝑎 𝛿(𝑥) 𝑑𝑥

(p. 434)

𝜃



𝑥

❄𝑥 ✲

Figure R.23: Cartesian and polar coordinates for the point 𝑃 The polar coordinates (p. 421) of a point are related to its Cartesian coordinates by ∙ 𝑥 = 𝑟 cos 𝜃 and 𝑦 = 𝑟 sin 𝜃 √ 𝑦 ∙ 𝑟 = 𝑥2 + 𝑦2 and tan 𝜃 = , 𝑥 ≠ 0 𝑥 For a constant 𝑎, the equation 𝑟 = 𝑎 gives a circle of radius 𝑎, and the equation 𝜃 = 𝑎 gives a ray from the origin making an angle of 𝜃 with the positive 𝑥-axis. The equation 𝑟 = 𝜃 gives an Archimedean spiral (p. 423).

READY REFERENCE

Area in Polar Coordinates (p. 426) For a curve 𝑟 = 𝑓 (𝜃), with 𝛼 ≤ 𝜃 ≤ 𝛽, which does not cross itself, 𝛽

Area of region enclosed =

1 𝑓 (𝜃)2 𝑑𝜃. 2 ∫𝛼

Slope and Arclength in Polar Coordinates (p. 427) For a curve 𝑟 = 𝑓 (𝜃), we can express 𝑥 and 𝑦 in terms of 𝜃 as a parameter, giving 𝑥 = 𝑟 cos 𝜃 = 𝑓 (𝜃) cos 𝜃

𝑦 = 𝑟 sin 𝜃 = 𝑓 (𝜃) sin 𝜃.

and

Then

𝑑𝑦∕𝑑𝜃 𝑑𝑦 = 𝑑𝑥 𝑑𝑥∕𝑑𝜃

Slope = and 𝛽

Arc length =

∫𝛼

√ (

𝑑𝑥 𝑑𝜃

)2

+

(

𝑑𝑦 𝑑𝜃

)2

𝑑𝜃.

Alternatively (p. 429),

Convergence Tests Theorem: Convergence Properties of Series (p. 490) ∑ ∑ 1. If 𝑎𝑛 and 𝑏𝑛 converge and if 𝑘 is a constant, then ∑ ∑ ∑ • (𝑎𝑛 + 𝑏𝑛 ) converges to 𝑎𝑛 + 𝑏𝑛 . ∑ ∑ • 𝑘𝑎𝑛 converges to 𝑘 𝑎𝑛 .

2. Changing a finite number of terms in a series does not change whether or not it converges, although it may change the value of its sum if it does converge. ∑ 3. If lim 𝑎𝑛 ≠ 0 or lim 𝑎𝑛 does not exist, then 𝑎𝑛 di𝑛→∞ 𝑛→∞ verges. ∑ ∑ 4. If 𝑎𝑛 diverges, then 𝑘𝑎𝑛 diverges if 𝑘 ≠ 0.

Theorem: Integral Test (p. 492) Suppose 𝑐 ≥ 0 and 𝑓 (𝑥) is a decreasing positive function, defined for all 𝑥 ≥ 𝑐, with 𝑎𝑛 = 𝑓 (𝑛) for all 𝑛.

𝛽

Arc length =

∫𝛼

√ (𝑓 ′ (𝜃))2 + (𝑓 (𝜃))2 𝑑𝜃.

1051



• If

𝑓 (𝑥) 𝑑𝑥 converges, then

∫𝑐



𝑎𝑛 converges.



Sequences and Series

• If

A sequence 𝑠1 , 𝑠2 , 𝑠3 , … , 𝑠𝑛 , … has a limit 𝐿, written lim 𝑠𝑛 = 𝐿, if we can make 𝑠𝑛 as close to 𝐿 as we please by 𝑛→∞ choosing a sufficiently large 𝑛. The sequence converges if a limit exists, diverges if no limit exists (see p. 476). Limits of sequences satisfy the same properties as limits of functions stated in Theorem 1.2 (p. 71) and lim 𝑥𝑛 = 0 if |𝑥| < 1

𝑛→∞

lim 1∕𝑛 = 0

𝑛→∞

(p. 476)

A sequence 𝑠𝑛 is bounded if there are constants 𝐾 and 𝑀 such that 𝐾 ≤ 𝑠𝑛 ≤ 𝑀 for all 𝑛 (p. 477). A convergent sequence is bounded. A sequence is monotone if it is either increasing, that is 𝑠𝑛 < 𝑠𝑛+1 for all 𝑛, or decreasing, that is 𝑠𝑛 > 𝑠𝑛+1 for all 𝑛 (p. 477). Theorem: Convergence of a Monotone, Bounded Sequence (p. 477): If a sequence 𝑠𝑛 is bounded and monotone, it converges. ∑ A series is an infinite sum 𝑎𝑛 = 𝑎1 + 𝑎2 + ⋯. The 𝑛th partial sum is 𝑆𝑛 = 𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛 (p. 488). If ∑ 𝑆 = lim 𝑆𝑛 exists, then the series 𝑎𝑛 converges, and its 𝑛→∞ sum is 𝑆 (p. 488). If a series does not converge, we say that it diverges (p. 488). The sum of a finite geometric series is (p. 482): 𝑎(1 − 𝑥𝑛 ) , 𝑥 ≠ 1. 1−𝑥 The sum of an infinite geometric series is (p. 483): 𝑎 𝑎 + 𝑎𝑥 + 𝑎𝑥2 + ⋯ + 𝑎𝑥𝑛 + ⋯ = , |𝑥| < 1. 1−𝑥 ∑ The 𝑝-series 1∕𝑛𝑝 converges if 𝑝 > 1 and diverges if 𝑝 ≤ 1 ∑ (p. 493). The harmonic series 1∕𝑛 diverges (p. 491), ∑ the alternating harmonic series (−1)𝑛−1 (1∕𝑛) converges (p. 500). An alternating series can be absolutely or conditionally convergent (p. 500). 𝑎 + 𝑎𝑥 + 𝑎𝑥2 + ⋯ + 𝑎𝑥𝑛−1 =

∫𝑐

𝑓 (𝑥) 𝑑𝑥 diverges, then



𝑎𝑛 diverges.

Theorem: Comparison Test (p. 494) Suppose 0 ≤ 𝑎𝑛 ≤ 𝑏𝑛 for all 𝑛. ∑ ∑ • If 𝑏𝑛 converges, then 𝑎𝑛 converges. ∑ ∑ • If 𝑎𝑛 diverges, then 𝑏𝑛 diverges. Theorem: Limit Comparison Test (p. 496) Suppose 𝑎𝑛 > 0 and 𝑏𝑛 > 0 for all 𝑛. If lim

𝑛→∞

𝑎𝑛 =𝑐 𝑏𝑛

where 𝑐 > 0,

∑ ∑ then the two series 𝑎𝑛 and 𝑏𝑛 either both converge or both diverge. Theorem: Convergence of Absolute Values Implies ∑ Convergence (p. 497): If |𝑎𝑛 | converges, then so does ∑ 𝑎𝑛 . ∑ ∑ ∑ We say 𝑎𝑛 is absolutely convergent if 𝑎𝑛 and |𝑎𝑛 | ∑ both converge and conditionally convergent if 𝑎𝑛 con∑ verges but |𝑎𝑛 | diverges (p. 500). ∑ Theorem: The Ratio Test (p. 498) For a series 𝑎𝑛 , suppose the sequence of ratios |𝑎𝑛+1 |∕|𝑎𝑛 | has a limit: lim

𝑛→∞

• If 𝐿 < 1, then



|𝑎𝑛+1 | = 𝐿. |𝑎𝑛 |

𝑎𝑛 converges.

• If 𝐿 > 1, or if 𝐿 is infinite, then



𝑎𝑛 diverges.

• If 𝐿 = 1, the test does not tell us anything about the ∑ convergence of 𝑎𝑛 .

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READY REFERENCE

Theorem: Alternating Series Test (p. 499) A series of the form ∞ ∑ (−1)𝑛−1 𝑎𝑛 = 𝑎1 − 𝑎2 + 𝑎3 − 𝑎4 + ⋯ + (−1)𝑛−1 𝑎𝑛 + ⋯ 𝑛=1

converges if 0 < 𝑎𝑛+1 < 𝑎𝑛 for all 𝑛 and lim𝑛→∞ 𝑎𝑛 = 0. Theorem: Error Bounds for Alternating Series 𝑛 ∑ (−1)𝑖−1 𝑎𝑖 be the 𝑛th partial sum of (p. 500) Let 𝑆𝑛 = 𝑖=1

an alternating series and let 𝑆 = lim 𝑆𝑛 . Suppose that 0 < 𝑛→∞ 𝑎𝑛+1 < 𝑎𝑛 for all 𝑛 and lim𝑛→∞ 𝑎𝑛 = 0. Then ||𝑆 − 𝑆𝑛 || < 𝑎𝑛+1 . Absolutely and conditionally convergent series behave differently when their terms are rearranged (p. 501) • Absolutely convergent series: Rearranging terms does not change the sum.

The Maclaurin polynomial of degree 𝑛 is the Taylor polynomial of degree 𝑛 near 𝑥 = 0 (p. 517). The Taylor series approximating 𝑓 (𝑥) for 𝑥 near 𝑎 is: 𝑓 (𝑥) = 𝑓 (𝑎) + 𝑓 ′ (𝑎)(𝑥 − 𝑎) + +

𝑃 (𝑥) = 𝐶0 + 𝐶1 (𝑥 − 𝑎) + 𝐶2 (𝑥 − 𝑎)2 + ⋯

|𝐸𝑛 (𝑥)| = |𝑓 (𝑥) − 𝑃𝑛 (𝑥)| ≤

𝑥3 𝑥5 𝑥7 𝑥9 + − + −⋯ 3! 5! 7! 9! 𝑥2 𝑥4 𝑥6 𝑥8 cos 𝑥 = 1 − + − + −⋯ 2! 4! 6! 8! 𝑥2 𝑥3 𝑥4 𝑒𝑥 = 1 + 𝑥 + + + +⋯ 2! 3! 4! sin 𝑥 = 𝑥 −

𝑛=0

𝑛=0

𝑎𝑛 = 𝐶𝑛 (𝑥 − 𝑎)𝑛 .

Taylor Series for ln 𝑥 about 𝑥 = 1 converges for 0 < 𝑥 ≤ 2 (p. 526): (𝑥 − 1) −

𝑛→∞

• If lim |𝑎𝑛+1 |∕|𝑎𝑛 | = 0, then 𝑅 = ∞.

1 + 𝑝𝑥 +

𝑛→∞

𝑛→∞

nonzero, then 𝑅 = 1∕𝐾.

Theorem: A Power Series is its Own Taylor Series (p. 527) If a power series about 𝑥 = 𝑎 converges to 𝑓 (𝑥) for |𝑥 − 𝑎| < 𝑅, then the power series is the Taylor series for 𝑓 (𝑥) about 𝑥 = 𝑎.

Approximations

(𝑥 − 1)2 (𝑥 − 1)3 (𝑥 − 1)4 + − +⋯ 2 3 4

The Binomial Series for (1 + 𝑥)𝑝 converges for −1 < 𝑥 < 1 (p. 525):

• If lim |𝑎𝑛+1 |∕|𝑎𝑛 | is infinite, then 𝑅 = 0.

• If lim |𝑎𝑛+1 |∕|𝑎𝑛 | = 𝐾|𝑥 − 𝑎|, where 𝐾 is finite and

𝑀 |𝑥 − 𝑎|𝑛+1 , (𝑛 + 1)!

where max ||𝑓 (𝑛+1) || ≤ 𝑀 on the interval between 𝑎 and 𝑥. Taylor Series for sin 𝑥, cos 𝑥, 𝑒𝑥 (p. 524):

+𝐶𝑛 (𝑥 − 𝑎)𝑛 + ⋯ ∞ ∑ 𝐶𝑛 (𝑥 − 𝑎)𝑛 (p. 504). = The radius of convergence is 0 if the series converges only for 𝑥 = 𝑎, ∞ if it converges for all 𝑥, and the positive number 𝑅 if it converges for |𝑥 − 𝑎| < 𝑅 and diverges for |𝑥 − 𝑎| > 𝑅 (p. 506). The interval of convergence is the interval between 𝑎 − 𝑅 and 𝑎 + 𝑅, including any endpoint where the series converges (p. 506). Theorem: Method for Computing Radius of Convergence (p. 507) To calculate the radius of convergence, ∞ ∑ 𝐶𝑛 (𝑥 − 𝑎)𝑛 , use the ratio test with 𝑅, for the power series

(p. 525)

The Maclaurin series is the Taylor series approximating 𝑓 (𝑥) near 𝑥 = 0 (p. 524). Theorem: The Lagrange Error Bound for 𝑃𝑛 (𝑥) (p. 540) Suppose 𝑓 and all its derivatives are continuous. If 𝑃𝑛 (𝑥) is the 𝑛th Taylor polynomial for to 𝑓 (𝑥) about 𝑎, then

• Conditionally convergent series: Rearranging terms can change the sum to any number.

Power Series

𝑓 (𝑛) (𝑎) (𝑥 − 𝑎)𝑛 + ⋯ 𝑛!

𝑓 ′′ (𝑎) (𝑥 − 𝑎)2 + ⋯ 2!

𝑝(𝑝 − 1) 2 𝑝(𝑝 − 1)(𝑝 − 2) 3 𝑥 + 𝑥 +⋯ 2! 3!

The Fourier Series of 𝑓 (𝑥) is (p. 550) 𝑓 (𝑥) = 𝑎0 + 𝑎1 cos 𝑥 + 𝑎2 cos 2𝑥 + 𝑎3 cos 3𝑥 + ⋯ +𝑏1 sin 𝑥 + 𝑏2 sin 2𝑥 + 𝑏3 sin 3𝑥 + ⋯ , where 𝜋

The Taylor polynomial of degree 𝑛 approximating 𝑓 (𝑥) for 𝑥 near 𝑎 is:

𝑎0 =

1 𝑓 (𝑥) 𝑑𝑥, 2𝜋 ∫−𝜋

𝑓 ′′ (𝑎) 𝑃𝑛 (𝑥) = 𝑓 (𝑎) + 𝑓 (𝑎)(𝑥 − 𝑎) + (𝑥 − 𝑎)2 + ⋯ ⋯ 2! 𝑓 (𝑛) (𝑎) (𝑥 − 𝑎)𝑛 (p. 517) + 𝑛!

𝑎𝑘 =

1 𝑓 (𝑥) cos(𝑘𝑥) 𝑑𝑥 𝜋 ∫−𝜋

𝑏𝑘 =

1 𝑓 (𝑥) sin(𝑘𝑥) 𝑑𝑥 𝜋 ∫−𝜋



𝜋

for 𝑘 a positive integer,

𝜋

for 𝑘 a positive integer.

READY REFERENCE

Differential Equations A differential equation for the function 𝑦(𝑥) is an equation involving 𝑥, 𝑦 and the derivatives of 𝑦 (p. 562). The order of a differential equation is the order of the highest-order derivative appearing in the equation (p. 563). A solution to a differential equation is any function 𝑦 that satisfies the equation (p. 563). The general solution to a differential equation is the family of functions that satisfies the equation (p. 563). An initial value problem is a differential equation together with an initial condition; a solution to an initial value problem is called a particular solution (p. 563). An equilibrium solution to a differential equation is a particular solution where 𝑦 is constant and 𝑑𝑦∕𝑑𝑥 = 0 (p. 592). First-order equations: methods of solution. A slope field corresponding to a differential equation is a plot in the 𝑥𝑦plane of small line segments with slope given by the differential equation (p. 567). Euler’s method approximates the solution of an initial value problem with a string of small line segments (p. 575). Differential equations of the form 𝑑𝑦∕𝑑𝑥 = 𝑔(𝑥)𝑓 (𝑦) can be solved analytically by separation of variables (p. 580). First-order equations: applications. The differential equation for exponential growth and decay is of the form 𝑑𝑃 = 𝑘𝑃 . 𝑑𝑡 The solution is of the form 𝑃 = 𝑃0 𝑒𝑘𝑡 , where 𝑃0 is the initial value of 𝑃 , and positive 𝑘 represents growth while negative 𝑘 represents decay (p. 587). Applications of growth and decay include continuously compounded interest (p. 587), population growth (p. 606), and Newton’s law of heating and cooling (p. 590). The logistic equation for population growth is of the form ) ( 𝑑𝑃 𝑃 , = 𝑘𝑃 1 − 𝑑𝑡 𝐿 where 𝐿 is the carrying capacity of the population (p. 607). The solution to the logistic equation is of the form 𝐿 − 𝑃0 𝐿 𝑃 = , where 𝐴 = (p. 608). 1 + 𝐴𝑒−𝑘𝑡 𝑃0 Systems of differential equations. Two interacting populations, 𝑤 and 𝑟, can be modeled by two equations 𝑑𝑤 = 𝑎𝑤 − 𝑐𝑤𝑟 𝑑𝑡

and

𝑑𝑟 = −𝑏𝑟 + 𝑘𝑤𝑟 𝑑𝑡

(p. 620).

Solutions can be visualized as trajectories in the 𝑤𝑟-phase plane (p. 620). An equilibrium point is one at which 𝑑𝑤∕𝑑𝑡 = 0 and 𝑑𝑟∕𝑑𝑡 = 0 (p. 621). A nullcline is a curve along which 𝑑𝑤∕𝑑𝑡 = 0 or 𝑑𝑟∕𝑑𝑡 = 0 (p. 627).

1053

Functions of three variables can be represented by the family of level surfaces 𝑓 (𝑥, 𝑦, 𝑧) = 𝑐 for various values of the constant 𝑐 (p. 689). A linear function 𝑓 (𝑥, 𝑦) has equation 𝑓 (𝑥, 𝑦) = 𝑧0 + 𝑚(𝑥 − 𝑥0 ) + 𝑛(𝑦 − 𝑦0 ) (p. 683) = 𝑐 + 𝑚𝑥 + 𝑛𝑦, where 𝑐 = 𝑧0 − 𝑚𝑥0 − 𝑛𝑦0 . Its graph is a plane with slope 𝑚 in the 𝑥-direction, slope 𝑛 in the 𝑦-direction, through (𝑥0 , 𝑦0 , 𝑧0 ) (p. 683). Its table of values has linear rows (of same slope) and linear columns (of same slope) (p. 684). Its contour diagram is equally spaced parallel straight lines (p. 684). The limit of 𝑓 at the point (𝑎, 𝑏), written lim(𝑥,𝑦)→(𝑎,𝑏) 𝑓 (𝑥, 𝑦), is the number 𝐿, if one exists, such that 𝑓 (𝑥, 𝑦) is as close to 𝐿 as we please whenever the distance from the point (𝑥, 𝑦) to the point (𝑎, 𝑏) is sufficiently small, but not zero. (p. 697). A function 𝑓 is continuous at the point (𝑎, 𝑏) if lim(𝑥,𝑦)→(𝑎,𝑏) 𝑓 (𝑥, 𝑦) = 𝑓 (𝑎, 𝑏). A function is continuous on a region 𝑅 if it is continuous at each point of 𝑅 (p. 697).

Vectors A vector 𝑣⃗ has magnitude (denoted ‖𝑣⃗ ‖) and direction. Examples are displacement vectors (p. 702), velocity and acceleration vectors (pp. 711, 712). and force (p. 712). We can add vectors, and multiply a vector by a scalar (p. 703). ⃗ , are parallel if one is a Two non-zero vectors, 𝑣⃗ and 𝑤 scalar multiple of the other (p. 704). A unit vector has magnitude 1. The vectors ⃗𝑖 , 𝑗⃗ , and ⃗𝑘 are unit vectors in the directions of the coordinate axes. A unit vector in the direction of any nonzero vector 𝑣⃗ is 𝑢⃗ = 𝑣⃗ ∕‖𝑣⃗ ‖ (p. 708). We resolve 𝑣⃗ into components by ⃗ (p. 705). writing 𝑣⃗ = 𝑣1 ⃗𝑖 + 𝑣2 𝑗⃗ + 𝑣3 𝑘 ⃗ and 𝑤 ⃗ ⃗ = 𝑤1 ⃗𝑖 + 𝑤2 𝑗⃗ + 𝑤3 𝑘 If 𝑣⃗ = 𝑣1 ⃗𝑖 + 𝑣2 𝑗⃗ + 𝑣3 𝑘 then √ ‖𝑣⃗ ‖ = 𝑣21 + 𝑣22 + 𝑣23 (p. 706) ⃗ (p. 707), ⃗ = (𝑣1 + 𝑤1 )⃗𝑖 + (𝑣2 + 𝑤2 )𝑗⃗ + (𝑣3 + 𝑤3 )𝑘 𝑣⃗ + 𝑤 ⃗ (p. 707). 𝜆𝑣⃗ = 𝜆𝑣1 ⃗𝑖 + 𝜆𝑣2 𝑗⃗ + 𝜆𝑣3 𝑘

The displacement vector from 𝑃1 = (𝑥1 , 𝑦1 , 𝑧1 ) to 𝑃2 = (𝑥2 , 𝑦2 , 𝑧2 ) is ⃗ ⃖⃖⃖⃖⃖⃖⃖ ⃗ ⃗ ⃗ 𝑃 1 𝑃2 = (𝑥2 − 𝑥1 )𝑖 + (𝑦2 − 𝑦1 )𝑗 + (𝑧2 − 𝑧1 )𝑘

(p. 706).

⃖⃖⃖⃖⃖⃗ (p. 706). A vector The position vector of 𝑃 = (𝑥, 𝑦, 𝑧) is 𝑂𝑃 in 𝑛 dimensions is a string of numbers 𝑣⃗ = (𝑣1 , 𝑣2 , … , 𝑣𝑛 ) (p. 714).

Multivariable Functions Points in 3-space are represented by a system of Cartesian coordinates √ (p. 654). The distance between (𝑥, 𝑦, 𝑧) and (𝑎, 𝑏, 𝑐) is (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 + (𝑧 − 𝑐)2 (p. 656). Functions of two variables can be represented by graphs (p. 661), contour diagrams (p. 668), cross-sections (p. 663), and tables (p. 653).

Dot Product (Scalar Product) (p. 718). ⃗ = ‖𝑣⃗ ‖‖𝑤 ⃗ ‖ cos 𝜃 where Geometric definition: 𝑣⃗ ⋅ 𝑤 ⃗ and 0 ≤ 𝜃 ≤ 𝜋. 𝜃 is the angle between 𝑣⃗ and 𝑤 ⃗ = 𝑣1 𝑤1 + 𝑣2 𝑤2 + 𝑣3 𝑤3 . Algebraic definition: 𝑣⃗ ⋅ 𝑤

1054

READY REFERENCE

⃗ are perpendicular if and Two nonzero vectors 𝑣⃗ and 𝑤 ⃗ = 0 (p. 720). Magnitude and dot product only if 𝑣⃗ ⋅ 𝑤 are related by 𝑣⃗ ⋅ 𝑣⃗ = ‖𝑣⃗ ‖2 (p. 720). If 𝑢⃗ = (𝑢1 , … , 𝑢𝑛 ) and 𝑣⃗ = (𝑣1 , … , 𝑣𝑛 ) then the dot product of 𝑢⃗ and 𝑣⃗ is 𝑢⃗ ⋅ 𝑣⃗ = 𝑢1 𝑣1 + … + 𝑢𝑛 𝑣𝑛 (p. 722). The equation of the plane with normal vector 𝑛⃗ = ⃗ and containing the point 𝑃0 = (𝑥0 , 𝑦0 , 𝑧0 ) is 𝑎⃗𝑖 + 𝑏𝑗⃗ + 𝑐 𝑘 𝑛⃗ ⋅ (⃗𝑟 − 𝑟⃗ 0 ) = 𝑎(𝑥 − 𝑥0 ) + 𝑏(𝑦 − 𝑦0 ) + 𝑐(𝑧 − 𝑧0 ) = 0 or 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑, where 𝑑 = 𝑎𝑥0 + 𝑏𝑦0 + 𝑐𝑧0 (p. 721). If 𝑣⃗ parallel and 𝑣⃗ perp are components of 𝑣⃗ which are parallel and perpendicular, respectively, to a unit vector 𝑢⃗ , then 𝑣⃗ parallel = (𝑣⃗ ⋅ 𝑢⃗ )⃗𝑢 and 𝑣⃗ perp = 𝑣⃗ − 𝑣⃗ parallel (p. 723). The work, 𝑊 , done by a force 𝐹⃗ acting on an object through a displacement 𝑑⃗ is 𝑊 = 𝐹⃗ ⋅ 𝑑⃗ (p. 724). Cross Product (Vector Product) (p. 729, 729) Geometric definition ( ) Area of parallelogram ⃗ = 𝑣⃗ × 𝑤 𝑛⃗ ⃗ with edges 𝑣⃗ and 𝑤 ⃗ ‖ sin 𝜃)⃗ = (‖𝑣⃗ ‖‖𝑤 𝑛,

⃗ and 𝑛⃗ where 0 ≤ 𝜃 ≤ 𝜋 is the angle between 𝑣⃗ and 𝑤 ⃗ pointing is the unit vector perpendicular to 𝑣⃗ and 𝑤 in the direction given by the right-hand rule. Algebraic definition ⃗ = (𝑣2 𝑤3 − 𝑣3 𝑤2 )⃗𝑖 + (𝑣3 𝑤1 − 𝑣1 𝑤3 )𝑗⃗ 𝑣⃗ × 𝑤 ⃗ +(𝑣1 𝑤2 − 𝑣2 𝑤1 )𝑘 ⃗,𝑤 ⃗. ⃗ = 𝑤1 ⃗𝑖 + 𝑤2 𝑗⃗ + 𝑤3 𝑘 𝑣⃗ = 𝑣1 ⃗𝑖 + 𝑣2 𝑗⃗ + 𝑣3 𝑘 To find the equation of a plane through three points that do not lie on a line, determine two vectors in the plane and then find a normal vector using the cross product (p. 731). The area of a parallelogram with edges 𝑣⃗ and ⃗ is ‖𝑣⃗ × 𝑤 ⃗ ‖. The volume of a parallelepiped with edges 𝑤 | | 𝑎⃗ , 𝑏⃗ , 𝑐⃗ is |(𝑏⃗ × 𝑐⃗ ) ⋅ 𝑎⃗ | (p. 733). | | The angular velocity (p. 733) of a flywheel can be represented by a vector 𝜔 ⃗ whose direction is parallel to the axis of rotation and magnitude is the angular speed of rotation. The velocity vector 𝑣⃗ of a point 𝑃 on the flywheel is 𝑣⃗ = 𝜔 ⃗ × 𝑟⃗ where 𝑟⃗ is a vector from the axis of rotation to 𝑃.

Differentiation of Multivariable Functions Partial derivatives of 𝑓 (p. 741). Rate of change of 𝑓 with respect to 𝑥 𝑓𝑥 (𝑎, 𝑏) = at the point (𝑎, 𝑏) 𝑓 (𝑎 + ℎ, 𝑏) − 𝑓 (𝑎, 𝑏) = lim , ℎ→0 ℎ 𝑓𝑦 (𝑎, 𝑏) = Rate of change of 𝑓 with respect to 𝑦 at the point (𝑎, 𝑏) 𝑓 (𝑎, 𝑏 + ℎ) − 𝑓 (𝑎, 𝑏) = lim . ℎ→0 ℎ

On the graph of 𝑓 , the partial derivatives 𝑓𝑥 (𝑎, 𝑏) and 𝑓𝑦 (𝑎, 𝑏) give the slope in the 𝑥 and 𝑦 directions, respectively (p. 742). The tangent plane to 𝑧 = 𝑓 (𝑥, 𝑦) at (𝑎, 𝑏) is 𝑧 = 𝑓 (𝑎, 𝑏) + 𝑓𝑥 (𝑎, 𝑏)(𝑥 − 𝑎) + 𝑓𝑦 (𝑎, 𝑏)(𝑦 − 𝑏)

(p. 755).

Partial derivatives can be estimated from a contour diagram or table of values using difference quotients (p. 742), and can be computed algebraically using the same rules of differentiation as for one-variable calculus (p. 748). Partial derivatives for functions of three or more variables are defined and computed in the same way (p 749). The gradient vector grad 𝑓 of 𝑓 is grad 𝑓 (𝑎, 𝑏) = 𝑓𝑥 (𝑎, 𝑏)⃗𝑖 + 𝑓𝑦 (𝑎, 𝑏)𝑗⃗ (2 variables) (p. 765) or ⃗ grad 𝑓 (𝑎, 𝑏, 𝑐) = 𝑓𝑥 (𝑎, 𝑏, 𝑐)⃗𝑖 + 𝑓𝑦 (𝑎, 𝑏, 𝑐)𝑗⃗ + 𝑓𝑧 (𝑎, 𝑏, 𝑐)𝑘 (3 variables) (p. 772). The gradient vector at 𝑃 : Points in the direction of increasing 𝑓 ; is perpendicular to the level curve or level surface of 𝑓 through 𝑃 ; and has magnitude ‖ grad 𝑓 ‖ equal to the maximum rate of change of 𝑓 at 𝑃 (pp. 767, 772). The magnitude is large when the level curves or surfaces are close together and small when they are far apart. The directional derivative of 𝑓 at 𝑃 in the direction of a unit vector 𝑢⃗ is (pp. 763, 765) Rate of change 𝑓𝑢⃗ (𝑃 ) = of 𝑓 in direction = grad 𝑓 (𝑃 ) ⋅ 𝑢⃗ of 𝑢⃗ at 𝑃 The tangent plane approximation to 𝑓 (𝑥, 𝑦) for (𝑥, 𝑦) near the point (𝑎, 𝑏) is 𝑓 (𝑥, 𝑦) ≈ 𝑓 (𝑎, 𝑏) + 𝑓𝑥 (𝑎, 𝑏)(𝑥 − 𝑎) + 𝑓𝑦 (𝑎, 𝑏)(𝑦 − 𝑏). The right-hand side is the local linearization (p. 755). The differential of 𝑧 = 𝑓 (𝑥, 𝑦) at (𝑎, 𝑏) is the linear function of 𝑑𝑥 and 𝑑𝑦 𝑑𝑓 = 𝑓𝑥 (𝑎, 𝑏) 𝑑𝑥 + 𝑓𝑦 (𝑎, 𝑏) 𝑑𝑦

(p. 758).

Local linearity with three or more variables follows the same pattern as for functions of two variables (p. 757). The tangent plane to a level surface of a function of three-variables 𝑓 at (𝑎, 𝑏, 𝑐) is (p. 775) 𝑓𝑥 (𝑎, 𝑏, 𝑐)(𝑥 − 𝑎) + 𝑓𝑦 (𝑎, 𝑏, 𝑐)(𝑦 − 𝑏) + 𝑓𝑧 (𝑎, 𝑏, 𝑐)(𝑧 − 𝑐) = 0. The Chain Rule for the partial derivative of one variable with respect to another in a chain of composed functions (p. 782): • Draw a diagram expressing the relationship between the variables, and label each link in the diagram with the derivative relating the variables at its ends. • For each path between the two variables, multiply together the derivatives from each step along the path.

READY REFERENCE

• Add the contributions from each path. If 𝑧 = 𝑓 (𝑥, 𝑦), and 𝑥 = 𝑔(𝑡), and 𝑦 = ℎ(𝑡), then 𝜕𝑧 𝑑𝑥 𝜕𝑧 𝑑𝑦 𝑑𝑧 = + 𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡

(p. 781).

If 𝑧 = 𝑓 (𝑥, 𝑦), with 𝑥 = 𝑔(𝑢, 𝑣) and 𝑦 = ℎ(𝑢, 𝑣), then 𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦 = + , 𝜕𝑢 𝜕𝑥 𝜕𝑢 𝜕𝑦 𝜕𝑢 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦 𝜕𝑧 = + (p. 783). 𝜕𝑣 𝜕𝑥 𝜕𝑣 𝜕𝑦 𝜕𝑣 Second-order partial derivatives (p. 790) 𝜕2 𝑧 = 𝑓𝑥𝑥 = (𝑓𝑥 )𝑥 , 𝜕𝑥2 𝜕2 𝑧 = 𝑓𝑥𝑦 = (𝑓𝑥 )𝑦 , 𝜕𝑦𝜕𝑥

𝜕2 𝑧 = 𝑓𝑦𝑥 = (𝑓𝑦 )𝑥 , 𝜕𝑥𝜕𝑦 𝜕2 𝑧 = 𝑓𝑦𝑦 = (𝑓𝑦 )𝑦 . 𝜕𝑦2

1055

Optimization A function 𝑓 has a local maximum at the point 𝑃0 if 𝑓 (𝑃0 ) ≥ 𝑓 (𝑃 ) for all points 𝑃 near 𝑃0 , and a local minimum at the point 𝑃0 if 𝑓 (𝑃0 ) ≤ 𝑓 (𝑃 ) for all points 𝑃 near 𝑃0 (p. 806). A critical point of a function 𝑓 is a point where grad 𝑓 is either 0⃗ or undefined. If 𝑓 has a local maximum or minimum at a point 𝑃0 , not on the boundary of its domain, then 𝑃0 is a critical point (p. 806). A quadratic function 𝑓 (𝑥, 𝑦) = 𝑎𝑥2 + 𝑏𝑥𝑦 + 𝑐𝑧2 generally has one critical point, which can be a local maximum, a local minimum, or a saddle point (p. 809). Second derivative test for functions of two variables (p. 811). Suppose grad 𝑓 (𝑥0 , 𝑦0 ) = 0⃗ . Let 𝐷 = 𝑓𝑥𝑥 (𝑥0 , 𝑦0 )𝑓𝑦𝑦 (𝑥0 , 𝑦0 ) − (𝑓𝑥𝑦 (𝑥0 , 𝑦0 ))2 . • If 𝐷 > 0 and 𝑓𝑥𝑥 (𝑥0 , 𝑦0 ) > 0, then 𝑓 has a local minimum at (𝑥0 , 𝑦0 ). • If 𝐷 > 0 and 𝑓𝑥𝑥 (𝑥0 , 𝑦0 ) < 0, then 𝑓 has a local maximum at (𝑥0 , 𝑦0 ). • If 𝐷 < 0, then 𝑓 has a saddle point at (𝑥0 , 𝑦0 ). • If 𝐷 = 0, anything can happen.

Theorem: Equality of Mixed Partial Derivatives. If 𝑓𝑥𝑦 and 𝑓𝑦𝑥 are continuous at (𝑎, 𝑏), an interior point of their domain, then 𝑓𝑥𝑦 (𝑎, 𝑏) = 𝑓𝑦𝑥 (𝑎, 𝑏) (p. 791). Taylor Polynomial of Degree 1 Approximating 𝒇 (𝒙, 𝒚) for (𝒙, 𝒚) near (𝒂, 𝒃) (p. 794) 𝑓 (𝑥, 𝑦) ≈ 𝐿(𝑥, 𝑦) = 𝑓 (𝑎, 𝑏)+𝑓𝑥 (𝑎, 𝑏)(𝑥−𝑎)+𝑓𝑦 (𝑎, 𝑏)(𝑦−𝑏). Taylor Polynomial of Degree 2 (p. 795) 𝑓 (𝑥, 𝑦) ≈ 𝑄(𝑥, 𝑦) = 𝑓 (𝑎, 𝑏) + 𝑓𝑥 (𝑎, 𝑏)(𝑥 − 𝑎) + 𝑓𝑦 (𝑎, 𝑏)(𝑦 − 𝑏) 𝑓𝑥𝑥 (𝑎, 𝑏) (𝑥 − 𝑎)2 + 𝑓𝑥𝑦 (𝑎, 𝑏)(𝑥 − 𝑎)(𝑦 − 𝑏) 2 𝑓𝑦𝑦 (𝑎, 𝑏) (𝑦 − 𝑏)2 . + 2

+

Definition of Differentiability (p. 799). A function 𝑓 (𝑥, 𝑦) is differentiable at the point (𝑎, 𝑏) if there is a linear function 𝐿(𝑥, 𝑦) = 𝑓 (𝑎, 𝑏) + 𝑚(𝑥 − 𝑎) + 𝑛(𝑦 − 𝑏) such that if the error 𝐸(𝑥, 𝑦) is defined by 𝑓 (𝑥, 𝑦) = 𝐿(𝑥, 𝑦) + 𝐸(𝑥, 𝑦), and if ℎ = 𝑥 −√𝑎, 𝑘 = 𝑦 − 𝑏, then the relative error 𝐸(𝑎 + ℎ, 𝑏 + 𝑘)∕ ℎ2 + 𝑘2 satisfies lim ℎ→0 𝑘→0

𝐸(𝑎 + ℎ, 𝑏 + 𝑘) = 0. √ ℎ2 + 𝑘2

Theorem: Continuity of Partial Derivatives Implies Differentiability (p. 802). If the partial derivatives, 𝑓𝑥 and 𝑓𝑦 , of a function 𝑓 exist and are continuous on a small disk centered at the point (𝑎, 𝑏), then 𝑓 is differentiable at (𝑎, 𝑏).

Unconstrained optimization A function 𝑓 defined on a region 𝑅 has a global maximum on 𝑅 at the point 𝑃0 if 𝑓 (𝑃0 ) ≥ 𝑓 (𝑃 ) for all points 𝑃 in 𝑅, and a global minimum on 𝑅 at the point 𝑃0 if 𝑓 (𝑃0 ) ≤ 𝑓 (𝑃 ) for all points 𝑃 in 𝑅 (p. 816). For an unconstrained optimization problem, find the critical points and investigate whether the critical points give global maxima or minima (p. 816). A closed region is one which contains its boundary; a bounded region is one which does not stretch to infinity in any direction (p. 821). Extreme Value Theorem for Multivariable Functions. If 𝑓 is a continuous function on a closed and bounded region 𝑅, then 𝑓 has a global maximum at some point (𝑥0 , 𝑦0 ) in 𝑅 and a global minimum at some point (𝑥1 , 𝑦1 ) in 𝑅 (p. 821). Constrained optimization Suppose 𝑃0 is a point satisfying the constraint 𝑔(𝑥, 𝑦) = 𝑐. A function 𝑓 has a local maximum at 𝑃0 subject to the constraint if 𝑓 (𝑃0 ) ≥ 𝑓 (𝑃 ) for all points 𝑃 near 𝑃0 satisfying the constraint (p. 827). It has a global maximum at 𝑃0 subject to the constraint if 𝑓 (𝑃0 ) ≥ 𝑓 (𝑃 ) for all points 𝑃 satisfying the constraint (p. 827). Local and global minima are defined similarly (p. 827). A local maximum or minimum of 𝑓 (𝑥, 𝑦) subject to a constraint 𝑔(𝑥, 𝑦) = 𝑐 occurs at a point where the constraint is tangent to a level curve of 𝑓 , and thus where grad 𝑔 is parallel to grad 𝑓 (p. 827). To optimize 𝑓 subject to the constraint 𝑔 = 𝑐 (p. 827), find the points satisfying the equations grad 𝑓 = 𝜆 grad 𝑔

and

𝑔 = 𝑐.

Then compare values of 𝑓 at these points, at points on the constraint where grad 𝑔 = 0⃗ , and at the endpoints of the constraint. The number 𝜆 is called the Lagrange multiplier.

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To optimize 𝑓 subject to the constraint 𝑔 ≤ 𝑐 (p. 829), find all points in the interior 𝑔(𝑥, 𝑦) < 𝑐 where grad 𝑓 is zero or undefined; then use Lagrange multipliers to find the local extrema of 𝑓 on the boundary 𝑔(𝑥, 𝑦) = 𝑐. Evaluate 𝑓 at the points found and compare the values. The value of 𝜆 is the rate of change of the optimum value of 𝑓 as 𝑐 increases (where 𝑔(𝑥, 𝑦) = 𝑐) (p. 831). The Lagrangian function (𝑥, 𝑦, 𝜆) = 𝑓 (𝑥, 𝑦) − 𝜆(𝑔(𝑥, 𝑦) − 𝑐) can be used to convert a constrained optimization problem for 𝑓 subject the constraint 𝑔 = 𝑐 into an unconstrained problem for  (p. 831).

Multivariable Integration The definite integral of 𝑓 , a continuous function of two variables, over 𝑅, the rectangle 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑐 ≤ 𝑦 ≤ 𝑑, is called a double integral, and is a limit of Riemann sums ∫𝑅

𝑓 𝑑𝐴 =

lim

Δ𝑥,Δ𝑦→0



𝑓 (𝑢𝑖𝑗 , 𝑣𝑖𝑗 )Δ𝑥Δ𝑦

(p. 842).

𝑖,𝑗

The Riemann sum is constructed by subdividing 𝑅 into subrectangles of width Δ𝑥 and height Δ𝑦, and choosing a point (𝑢𝑖𝑗 , 𝑣𝑖𝑗 ) in the 𝑖𝑗-th rectangle. A triple integral of 𝑓 , a continuous function of three variables, over 𝑊 , the box 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑐 ≤ 𝑦 ≤ 𝑑, 𝑝 ≤ 𝑧 ≤ 𝑞 in 3-space, is defined in a similar way using three-variable Riemann sums (p. 858). Interpretations If 𝑓 (𝑥, 𝑦) is positive, ∫𝑅 𝑓 𝑑𝐴 is the volume under graph of 𝑓 above the region 𝑅 (p. 842). If 𝑓 (𝑥, 𝑦) = 1 for all 𝑥 and 𝑦, then the area of 𝑅 is ∫𝑅 1 𝑑𝐴 = ∫𝑅 𝑑𝐴 (p. 844). If 𝑓 (𝑥, 𝑦) is a density, then ∫𝑅 𝑓 𝑑𝐴 is the total quantity in the region 𝑅 (p. 840). The average value of 𝑓 (𝑥, 𝑦) on the region 𝑅 1 is ∫ 𝑓 𝑑𝐴 (p. 844). In probability, if 𝑝(𝑥, 𝑦) is a Area of 𝑅 𝑅 𝑏 𝑑 joint density function then ∫𝑎 ∫𝑐 𝑝(𝑥, 𝑦) 𝑑𝑦 𝑑𝑥 is the fraction of population with 𝑎 ≤ 𝑥 ≤ 𝑏 and 𝑐 ≤ 𝑦 ≤ 𝑑 (p. 880). Iterated integrals Double and triple integrals can be written as iterated integrals 𝑑

∫𝑅

𝑓 𝑑𝐴 =

∫𝑊

𝑓 (𝑥, 𝑦) 𝑑𝑥𝑑𝑦 (p. 848)

∫𝑐 ∫𝑎 𝑞

𝑓 𝑑𝑉 =

𝑏

𝑑

∫𝑝 ∫𝑐 ∫𝑎

𝑏

𝑓 (𝑥, 𝑦, 𝑧) 𝑑𝑥 𝑑𝑦 𝑑𝑧 (p. 858)

Other orders of integration are possible. For iterated integrals over non-rectangular regions (p. 850), limits on outer integral are constants and limits on inner integrals involve only the variables in the integrals further out (pp. 851, 860). Integrals in other coordinate systems When computing double integrals in polar coordinates, put 𝑑𝐴 = 𝑟 𝑑𝑟 𝑑𝜃 or 𝑑𝐴 = 𝑟 𝑑𝜃 𝑑𝑟 (p. 864). Cylindrical coordinates are given by 𝑥 = 𝑟 cos 𝜃, 𝑦 = 𝑟 sin 𝜃, 𝑧 = 𝑧, for 0 ≤ 𝑟 < ∞, 0 ≤ 𝜃 ≤ 2𝜋, −∞ < 𝑧 < ∞ (p. 869). Spherical coordinates are given by 𝑥 = 𝜌 sin 𝜙 cos 𝜃, 𝑦 = 𝜌 sin 𝜙 sin 𝜃, 𝑧 = 𝜌 cos 𝜙, for 0 ≤ 𝜌 < ∞, 0 ≤ 𝜙 ≤ 𝜋, 0 ≤ 𝜃 ≤ 2𝜋

(p. 872). When computing triple integrals in cylindrical or spherical coordinates, put 𝑑𝑉 = 𝑟 𝑑𝑟 𝑑𝜃 𝑑𝑧 for cylindrical coordinates (p. 871), 𝑑𝑉 = 𝜌2 sin 𝜙 𝑑𝜌 𝑑𝜙 𝑑𝜃 for spherical coordinates (p. 873). Other orders of integration are also possible. For a change of variables 𝑥 = 𝑥(𝑠, 𝑡), 𝑦 = 𝑦(𝑠, 𝑡), the Jacobian is | | | 𝜕𝑥 𝜕(𝑥, 𝑦) 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 || 𝜕𝑠 = ⋅ − ⋅ =| 𝜕(𝑠, 𝑡) 𝜕𝑠 𝜕𝑡 𝜕𝑡 𝜕𝑠 || | 𝜕𝑦 | 𝜕𝑠 |

𝜕𝑥 𝜕𝑡

𝜕𝑦 𝜕𝑡

| | | | | | | | | | |

(p. 1034).

To convert an integral from 𝑥, 𝑦 to 𝑠, 𝑡 coordinates (p. 1035): Substitute for 𝑥 and 𝑦 in terms of 𝑠 and 𝑡, change the 𝑥𝑦 region 𝑅 into an 𝑠𝑡 region 𝑇 , and change the area element by | | making the substitution 𝑑𝑥𝑑𝑦 = | 𝜕(𝑥,𝑦) | 𝑑𝑠𝑑𝑡. For triple in| 𝜕(𝑠,𝑡) | tegrals, there is a similar formula (p. 1036).

Parameterizations and Vector Fields Parameterized curves The motion of a particle is described by parametric equations 𝑥 = 𝑓 (𝑡), 𝑦 = 𝑔(𝑡) (2-space) or 𝑥 = 𝑓 (𝑡), 𝑦 = 𝑔(𝑡), 𝑧 = ℎ(𝑡) (3-space). The path of the particle is a parameterized curve (p. 886). Parameterizations are also written in vector ⃗ (p. 888). For a curve form 𝑟⃗ (𝑡) = 𝑓 (𝑡)⃗𝑖 + 𝑔(𝑡)𝑗⃗ + ℎ(𝑡)𝑘 segment we restrict the parameter to to a closed interval 𝑎 ≤ 𝑡 ≤ 𝑏 (p. 889). Parametric equations for the graph of 𝑦 = 𝑓 (𝑥) are 𝑥 = 𝑡, 𝑦 = 𝑓 (𝑡). Parametric equations for a line through (𝑥0 , 𝑦0 ) in the direction of 𝑣⃗ = 𝑎⃗𝑖 + 𝑏𝑗⃗ are 𝑥 = 𝑥0 + 𝑎𝑡, 𝑦 = 𝑦0 + 𝑏𝑡. In 3-space, the line through (𝑥0 , 𝑦0 , 𝑧0 ) in the direction of 𝑣⃗ = ⃗ is 𝑥 = 𝑥0 +𝑎𝑡, 𝑦 = 𝑦0 +𝑏𝑡, 𝑧 = 𝑧0 +𝑐𝑡 (p. 887). In 𝑎⃗𝑖 +𝑏𝑗⃗ +𝑐 𝑘 vector form, the equation for a line is 𝑟⃗ (𝑡) = 𝑟⃗ 0 + 𝑡𝑣⃗ , where ⃗ (p. 889). 𝑟⃗ 0 = 𝑥0 ⃗𝑖 + 𝑦0 𝑗⃗ + 𝑧0 𝑘 Parametric equations for a circle of radius 𝑅 in the plane, centered at the origin are 𝑥 = 𝑅 cos 𝑡 , 𝑦 = 𝑅 sin 𝑡 (counterclockwise), 𝑥 = 𝑅 cos 𝑡, 𝑦 = −𝑅 sin 𝑡 (clockwise). To find the intersection points of a curve 𝑟⃗ (𝑡) = ⃗ with a surface 𝐹 (𝑥, 𝑦, 𝑧) = 𝑐, solve 𝑓 (𝑡)⃗𝑖 + 𝑔(𝑡)𝑗⃗ + ℎ(𝑡)𝑘 𝐹 (𝑓 (𝑡), 𝑔(𝑡), ℎ(𝑡)) = 𝑐 for 𝑡 (p. 890). To find the intersection points of two curves 𝑟⃗ 1 (𝑡) and 𝑟⃗ 2 (𝑡), solve 𝑟⃗ 1 (𝑡1 ) = 𝑟⃗ 2 (𝑡2 ) for 𝑡1 and 𝑡2 (p. 890). The length of a curve segment 𝐶 given parametrically 𝑏 for 𝑎 ≤ 𝑡 ≤ 𝑏 with velocity vector 𝑣⃗ is ∫𝑎 ‖𝑣⃗ ‖𝑑𝑡 if 𝑣⃗ ≠ 0⃗ for 𝑎 < 𝑡 < 𝑏 (p. 901). The velocity and acceleration of a moving object with position vector 𝑟⃗ (𝑡) at time 𝑡 are Δ⃗𝑟 Δ𝑡 Δ𝑣⃗ 𝑎⃗ (𝑡) = lim Δ𝑡→0 Δ𝑡

𝑣⃗ (𝑡) = lim

Δ𝑡→0

We write 𝑣⃗ =

(p. 897) (p. 899)

𝑑⃗𝑟 𝑑 𝑣⃗ 𝑑 2 𝑟⃗ = 𝑟⃗ ′ (𝑡) and 𝑎⃗ = = 2 = 𝑟⃗ ′′ (𝑡). 𝑑𝑡 𝑑𝑡 𝑑𝑡

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The components of the velocity and acceleration vectors are 𝑑𝑥 ⃗ 𝑑𝑦 ⃗ 𝑑𝑧 ⃗ 𝑖 + 𝑗 + 𝑣⃗ (𝑡) = 𝑘 (p. 897) 𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑2𝑦 𝑑2𝑧 ⃗ 𝑑2𝑥 (p. 899) 𝑎⃗ (𝑡) = 2 ⃗𝑖 + 2 𝑗⃗ + 2 𝑘 𝑑𝑡 𝑑𝑡 𝑑𝑡 √ The speed is ‖𝑣⃗ ‖ = (𝑑𝑥∕𝑑𝑡)2 + (𝑑𝑦∕𝑑𝑡)2 + (𝑑𝑧∕𝑑𝑡)2 (p. 901). Analogous formulas for velocity, speed, and acceleration hold in 2-space. Uniform Circular Motion (p. 900) For a particle 𝑟⃗ (𝑡) = 𝑅 cos(𝜔𝑡)⃗𝑖 + 𝑅 sin(𝜔𝑡)𝑗⃗ : motion is in a circle of radius 𝑅 with period 2𝜋∕𝜔; velocity, 𝑣⃗ , is tangent to the circle and speed is constant ‖𝑣⃗ ‖ = 𝜔𝑅; acceleration, 𝑎⃗ , points toward the center of the circle with ‖𝑎⃗ ‖ = ‖𝑣⃗ ‖2 ∕𝑅. Motion in a Straight Line (p. 901) For a particle 𝑟⃗ (𝑡) = 𝑟⃗ 0 + 𝑓 (𝑡)𝑣⃗ 0 : Motion is along a straight line through the point with position vector 𝑟⃗ 0 parallel to 𝑣⃗ 0 ; velocity, 𝑣⃗ , and acceleration, 𝑎⃗ , are parallel to the line. Vector fields A vector field in 2-space is a function 𝐹⃗ (𝑥, 𝑦) whose value at a point (𝑥, 𝑦) is a 2-dimensional vector (p. 905). Similarly, a vector field in 3-space is a function 𝐹⃗ (𝑥, 𝑦, 𝑧) whose values are 3-dimensional vectors (p. 905). Examples are the gradient of a differentiable function 𝑓 , the velocity field of a fluid flow, and force fields (p. 905). A flow line of a vector field 𝑣⃗ = 𝐹⃗ (⃗𝑟 ) is a path 𝑟⃗ (𝑡) whose velocity vector equals 𝑣⃗ , thus 𝑟⃗ ′ (𝑡) = 𝑣⃗ = 𝐹⃗ (⃗𝑟 (𝑡)) (p. 914). The flow of a vector field is the family of all of its flow line (p. 914). Flow lines can be approximated numerically using Euler’s method (p. 916). Parameterized surfaces We parameterize a surface with two parameters, 𝑥 = 𝑓1 (𝑠, 𝑡), 𝑦 = 𝑓2 (𝑠, 𝑡), 𝑧 = 𝑓3 (𝑠, 𝑡) (p. 1024). We also use ⃗ the vector form 𝑟⃗ (𝑠, 𝑡) = 𝑓1 (𝑠, 𝑡)⃗𝑖 + 𝑓2 (𝑠, 𝑡)𝑗⃗ + 𝑓3 (𝑠, 𝑡)𝑘 (p. 1024). Parametric equations for the graph of 𝑧 = 𝑓 (𝑥, 𝑦) are 𝑥 = 𝑠, 𝑦 = 𝑡, and 𝑧 = 𝑓 (𝑠, 𝑡) (p. 1024). Parametric equation for a plane through the point with position vector 𝑟⃗ 0 and containing the two nonparallel vectors 𝑣⃗ 1 and 𝑣⃗ 2 is 𝑟⃗ (𝑠, 𝑡) = 𝑟⃗ 0 + 𝑠𝑣⃗ 1 + 𝑡𝑣⃗ 2 (p. 1025). Parametric equation for a sphere of radius 𝑅 centered at the ori⃗, gin is 𝑟⃗ (𝜃, 𝜙) = 𝑅 sin 𝜙 cos 𝜃 ⃗𝑖 + 𝑅 sin 𝜙 sin 𝜃 𝑗⃗ + cos 𝜙 𝑘 0 ≤ 𝜃 ≤ 2𝜋, 0 ≤ 𝜙 ≤ 𝜋 (p. 1025). Parametric equation for a cylinder of radius 𝑅 along the 𝑧-axis is 𝑟⃗ (𝜃, 𝑧) = ⃗ , 0 ≤ 𝜃 ≤ 2𝜋, −∞ < 𝑧 < ∞ 𝑅 cos 𝜃⃗𝑖 + 𝑅 sin 𝜃 𝑗⃗ + 𝑧𝑘 (p. 1023). A parameter curve is the curve obtained by holding one of the parameters constant and letting the other vary (p. 1029).

Line Integrals

𝑛−1

𝐹⃗ ⋅ 𝑑⃗𝑟 =

𝑏

∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 =

lim ‖Δ⃗𝑟 𝑖 ‖→0



𝐹⃗ (⃗𝑟 𝑖 ) ⋅ Δ⃗𝑟 𝑖 ,

𝑖=0

where the direction of Δ⃗𝑟 𝑖 is the direction of the orientation (p. 923).

∫𝑎

𝐹⃗ (⃗𝑟 (𝑡)) ⋅ 𝑟⃗ ′ (𝑡) 𝑑𝑡 (p. 932).

Fundamental Theorem for Line Integrals (p. 939): Suppose 𝐶 is a piecewise smooth oriented path with starting point 𝑃 and endpoint 𝑄. If 𝑓 is a function whose gradient is continuous on the path 𝐶, then

∫𝐶

grad 𝑓 ⋅ 𝑑⃗𝑟 = 𝑓 (𝑄) − 𝑓 (𝑃 ).

Path-independent fields and gradient fields A vector field 𝐹⃗ is said to be path-independent, or conservative, if for any two points 𝑃 and 𝑄, the line integral ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 has the same value along any piecewise smooth path 𝐶 from 𝑃 to 𝑄 lying in the domain of 𝐹⃗ (p. 941). A gradient field is a vector field of the form 𝐹⃗ = grad 𝑓 for some scalar function 𝑓 , and 𝑓 is called a potential function for the vector field 𝐹⃗ (p. 942). A vector field 𝐹⃗ is path-independent if and only if 𝐹⃗ is a gradient vector field (p. 942). A vector field 𝐹⃗ is path-independent if and only if ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 0 for every closed curve 𝐶 (p. 950). If 𝐹⃗ is 𝜕𝐹 𝜕𝐹 a gradient field, then 𝜕𝑥2 − 𝜕𝑦1 = 0 (p. 951). The quantity 𝜕𝐹2 𝜕𝑥



𝜕𝐹1 𝜕𝑦

is called the 2-dimensional or scalar curl of 𝐹⃗ .

Green’s Theorem (p. 951): Suppose 𝐶 is a piecewise smooth simple closed curve that is the boundary of an open region 𝑅 in the plane and oriented so that the region is on the left as we move around the curve. Suppose 𝐹⃗ = 𝐹1 ⃗𝑖 + 𝐹2 𝑗⃗ is a smooth vector field defined at every point of the region 𝑅 and boundary 𝐶. Then

∫𝐶

The line integral of a vector field 𝐹⃗ along an oriented curve 𝐶 (p. 922) is

∫𝐶

The line integral measures the extent to which 𝐶 is going with 𝐹⃗ or against it (p. 924). For oriented curves 𝐶, 𝐶1 , and 𝐶2 , ∫−𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = − ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 , where −𝐶 is the curve 𝐶 parameterized in the opposite direction, and ∫𝐶 +𝐶 𝐹⃗ ⋅𝑑⃗𝑟 = ∫𝐶 𝐹⃗ ⋅𝑑⃗𝑟 +∫𝐶 𝐹⃗ ⋅𝑑⃗𝑟 , where 𝐶1 +𝐶2 is the 1 2 1 2 curve obtained by joining the endpoint of 𝐶1 to the starting point of 𝐶2 (p. 928). The work done by a force 𝐹⃗ along a curve 𝐶 is ∫𝐶 𝐹⃗ ⋅𝑑⃗𝑟 (p. 925). The circulation of 𝐹⃗ around an oriented closed curve is ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 (p. 927). Given a parameterization of 𝐶, 𝑟⃗ (𝑡), for 𝑎 ≤ 𝑡 ≤ 𝑏, the line integral can be calculated as

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫𝑅

(

𝜕𝐹2 𝜕𝐹1 − 𝜕𝑥 𝜕𝑦

)

𝑑𝑥 𝑑𝑦.

Curl test for vector fields in 2-space: If

𝜕𝐹2 𝜕𝑥



𝜕𝐹1 𝜕𝑦

=

0 and the domain of 𝐹⃗ has no holes, then 𝐹⃗ is pathindependent, and hence a gradient field (p. 953). The condition that the domain have no holes is important. It is not always true that if the scalar curl of 𝐹⃗ is zero then 𝐹⃗ is a gradient field (p. 954).

1058

READY REFERENCE

Surface Integrals A surface is oriented if a unit normal vector 𝑛⃗ has been chosen at every point on it in a continuous way (p. 962). For a closed surface, we usually choose the outward orientation (p. 962). The area vector of a flat, oriented surface is a vector 𝐴⃗ whose magnitude is the area of the surface, and whose direction is the direction of the orientation vector 𝑛⃗ (p. 963). If 𝑣⃗ is the velocity vector of a constant fluid flow and 𝐴⃗ is the area vector of a flat surface, then the total flow through the surface in units of volume per unit time is called the flux of 𝑣⃗ through the surface and is given by 𝑣⃗ ⋅ 𝐴⃗ (p. 963). The surface integral or flux integral of the vector field 𝐹⃗ through the oriented surface 𝑆 is ∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

lim ‖Δ𝐴⃗ ‖→0



𝐹⃗ ⋅ Δ𝐴⃗ ,

where the direction of Δ𝐴⃗ is the direction of the orientation (p. 964). If 𝑣⃗ is a variable vector field and then ∫𝑆 𝑣⃗ ⋅ 𝑑 𝐴⃗ is the flux through the surface 𝑆 (p. 965). Simple flux integrals can be calculated by putting 𝑑 𝐴⃗ = 𝑛⃗ 𝑑𝐴 and using geometry or converting to a double integral (p. 967). The flux through a graph of 𝑧 = 𝑓 (𝑥, 𝑦) above a region 𝑅 in the 𝑥𝑦-plane, oriented upward, is ∫𝑅

) ( ⃗ 𝑑𝑥 𝑑𝑦 𝐹⃗ (𝑥, 𝑦, 𝑓 (𝑥, 𝑦)) ⋅ −𝑓𝑥 ⃗𝑖 − 𝑓𝑦 𝑗⃗ + 𝑘

∫𝑅



(𝑓𝑥 )2 + (𝑓𝑦 )2 + 1 𝑑𝑥 𝑑𝑦

(p. 975).

The flux through a cylindrical surface 𝑆 of radius 𝑅 and oriented away from the 𝑧-axis is ∫𝑇

( ) 𝐹⃗ (𝑅, 𝜃, 𝑧) ⋅ cos 𝜃⃗𝑖 + sin 𝜃 𝑗⃗ 𝑅 𝑑𝑧 𝑑𝜃

‖ 𝜕⃗𝑟 𝜕⃗𝑟 ‖ ‖ ‖ × ‖ 𝑑𝑠 𝑑𝑡 ‖ ∫𝑅 ‖ 𝜕𝑠 𝜕𝑡 ‖ ‖ ‖

(p. 1039).

Divergence and Curl Divergence

Definition of Divergence (p. 982). Geometric definition: The divergence of 𝐹⃗ is div 𝐹⃗ (𝑥, 𝑦, 𝑧) =

lim

∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗

Volume→0

Volume of 𝑆

.

Here 𝑆 is a sphere centered at (𝑥, 𝑦, 𝑧), oriented outwards, that contracts down to (𝑥, 𝑦, 𝑧) in the limit. Cartesian coordinate definition: If 𝐹⃗ = 𝐹1 ⃗𝑖 +𝐹2 𝑗⃗ + ⃗ , then 𝐹3 𝑘 div 𝐹⃗ =

𝜕𝐹1 𝜕𝐹2 𝜕𝐹3 + + . 𝜕𝑥 𝜕𝑦 𝜕𝑧

The divergence can be thought of as the outflow per unit volume of the vector field. A vector field 𝐹⃗ is said to be divergence free or solenoidal if div𝐹⃗ = 0 everywhere that 𝐹⃗ is defined. Magnetic fields are divergence free (p. 986). The Divergence Theorem (p. 992). If 𝑊 is a solid region whose boundary 𝑆 is a piecewise smooth surface, and if 𝐹⃗ is a smooth vector field which is defined everywhere in 𝑊 and on 𝑆, then

(p. 976),

where 𝑇 is the 𝜃𝑧-region corresponding to 𝑆. The flux through a spherical surface 𝑆 of radius 𝑅 and oriented away from the origin is ( ) ⃗ ∫𝑇 𝐹⃗ (𝑅, 𝜃, 𝜙) ⋅ sin 𝜙 cos 𝜃⃗𝑖 + sin 𝜙 sin 𝜃 𝑗⃗ + cos 𝜙𝑘 𝑅2 sin 𝜙 𝑑𝜙 𝑑𝜃,

∫𝑆

𝑑𝐴 =

(p. 974).

The area of the part of the graph of 𝑧 = 𝑓 (𝑥, 𝑦) above a region 𝑅 in the 𝑥𝑦-plane is Area of 𝑆 =

The area of a parameterized surface 𝑆, parameterized by 𝑟⃗ = 𝑟⃗ (𝑠, 𝑡), where (𝑠, 𝑡) varies in a parameter region 𝑅, is

(p. 977)

where 𝑇 is the 𝜃𝜙-region corresponding to 𝑆. The flux through a parameterized surface 𝑆, parameterized by 𝑟⃗ = 𝑟⃗ (𝑠, 𝑡), where (𝑠, 𝑡) varies in a parameter region 𝑅, is ) ( 𝜕⃗𝑟 𝜕⃗𝑟 ⃗ 𝑑𝑠 𝑑𝑡 (p. 1038). 𝐹 (⃗𝑟 (𝑠, 𝑡)) ⋅ × ∫𝑅 𝜕𝑠 𝜕𝑡 We choose the parameterization so that 𝜕⃗𝑟 ∕𝜕𝑠 × 𝜕⃗𝑟 ∕𝜕𝑡 is never zero and points in the direction of 𝑛⃗ everywhere.

∫𝑆

𝐹⃗ ⋅ 𝑑 𝐴⃗ =

∫𝑊

div 𝐹⃗ 𝑑𝑉 ,

where 𝑆 is given the outward orientation. In words, the Divergence Theorem says that the total flux out of a closed surface is the integral of the flux density over the volume it encloses. Curl The circulation density of a smooth vector field 𝐹⃗ at (𝑥, 𝑦, 𝑧) around the direction of the unit vector 𝑛⃗ is defined to be Circulation around 𝐶 circ𝑛⃗ 𝐹⃗ (𝑥, 𝑦, 𝑧) = lim Area→0 Area inside 𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 ∫𝐶 = lim Area→0 Area inside 𝐶

(p. 1000).

Circulation density is calculated using the right-hand rule (p. 1000).

READY REFERENCE

Definition of curl (p. 1001). Geometric definition The curl of 𝐹⃗ , written curl 𝐹⃗ , is the vector field with the following properties • The direction of curl 𝐹⃗ (𝑥, 𝑦, 𝑧) is the direction 𝑛⃗ for which circ𝑛⃗ (𝑥, 𝑦, 𝑧) is greatest. • The magnitude of curl 𝐹⃗ (𝑥, 𝑦, 𝑧) is the circulation density of 𝐹⃗ around that direction. Cartesian coordinate definition If 𝐹⃗ = 𝐹1 ⃗𝑖 +𝐹2 𝑗⃗ + ⃗ , then 𝐹3 𝑘 ( ) ( ) 𝜕𝐹3 𝜕𝐹2 𝜕𝐹1 𝜕𝐹3 ⃗𝑖 + curl 𝐹⃗ = − − 𝑗⃗ 𝜕𝑦 𝜕𝑧 𝜕𝑧 𝜕𝑥 ( ) 𝜕𝐹2 𝜕𝐹1 ⃗. + − 𝑘 𝜕𝑥 𝜕𝑦

Curl and circulation density are related by circ𝑛⃗ 𝐹⃗ = curl 𝐹⃗ ⋅ 𝑛⃗ (p. 1004). A vector field is said to be curl free or irrotational if curl 𝐹⃗ = 0⃗ everywhere that 𝐹⃗ is defined (p. 1004). Given an oriented surface 𝑆 with a boundary curve 𝐶 we use the right-hand rule to determine the orientation of 𝐶 (p. 1008).

1059

Stokes’ Theorem (p. 1009). If 𝑆 is a smooth oriented surface with piecewise smooth, oriented boundary 𝐶, and if 𝐹⃗ is a smooth vector field which is defined on 𝑆 and 𝐶, then ∫𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 =

∫𝑆

curl 𝐹⃗ ⋅ 𝑑 𝐴⃗ .

Stokes’ Theorem says that the total circulation around 𝐶 is the integral over 𝑆 of the circulation density. A curl field is a vector field 𝐹⃗ that can be written as 𝐹⃗ = curl 𝐺⃗ for some vector field 𝐺⃗ , called a vector potential for 𝐹⃗ (p. 1011). Relation between divergence, gradient, and curl The curl and gradient are related by curl grad 𝑓 = 0 (p. 1015). Divergence and curl are related by div curl 𝐹⃗ = 0 (p. 1017). The curl test for vector fields in 3-space (p. 1016) Suppose that curl 𝐹⃗ = 0⃗ , and that the domain of 𝐹⃗ has the property that every closed curve in it can be contracted to a point in a smooth way, staying at all times within the domain. Then 𝐹⃗ is path-independent, so 𝐹⃗ is a gradient field and has a potential function. The divergence test for vector fields in 3-space (p. 1017) Suppose that div 𝐹⃗ = 0, and that the domain of 𝐹⃗ has the property that every closed surface in it is the boundary of a solid region completely contained in the domain. Then 𝐹⃗ is a curl field.

1061

ANSWERS TO ODD NUMBERED PROBLEMS 𝑝

Section 1.1

45 Distance from

25 20 15 10 5

Kalamazoo

1 Pop 7 million in 2015

155 120

5 𝑦 = (1∕2)𝑥 + 2 7 𝑦 = 2𝑥 + 2 9 Slope:−12∕7 Vertical intercept: 2∕7

arrive in start in Chicago Kalamazoo

11 Slope: 2 Vertical intercept: −2∕3 13 (a) (b) (c) (d) (e) (f)

(V) (VI) (I) (IV) (III) (II)

15 We use the point-slope form, getting 𝑦 − 𝑐 = 𝑚(𝑥 − 𝑎). Alternatively, we can substitute the point (𝑎, 𝑐) into the equation 𝑦 = 𝑏 + 𝑚𝑥 and solve for 𝑏: 𝑐 = 𝑚𝑎 + 𝑏 𝑏 = 𝑐 − 𝑚𝑎. Substituting our value of 𝑏 into 𝑦 = 𝑏 + 𝑚𝑥 gives: 𝑦 = 𝑐 − 𝑚𝑎 + 𝑚𝑥 = 𝑐 + 𝑚(𝑥 − 𝑎). 17 𝑦 =

− 51 𝑥

+

7 5

19 Parallel: 𝑦 = 𝑚(𝑥 − 𝑎) + 𝑏 Perpendicular: 𝑦 = (−1∕𝑚)(𝑥 − 𝑎) + 𝑏

23 Domain: 0 ≤ 𝑥 ≤ 5 Range: 0 ≤ 𝑦 ≤ 4

27 𝑉 = 𝑘𝑟3 29 𝑆 = 𝑘ℎ2

53 (a) 𝐶 = 4.16 + 0.12𝑤 (b) 0.12 $/gal (c) $4.16 55 (a) $0.025/cubic foot (b) 𝑐 = 15 + 0.025𝑤 𝑐 = cost of water 𝑤 = cubic feet of water (c) 3400 cubic feet (i) 𝑓 (2001) = 272

35 𝑓 (12) = 7.049 37 𝑓 (63) = 𝑓 (12) 39

67 (a) 𝐶1 = 40 + 0.15𝑚 𝐶2 = 50 + 0.10𝑚 (b) 𝐶 (cost in dollars)

150 100 50 0

driving speed

83 False; 𝑦 = 𝑥 + 1 at points (1, 2) and (2, 3) 85 (b), (c)

Section 1.2 1 Concave up 3 Neither 5 5; 7% 7 3.2; 3% (continuous) 9 𝑃 = 15(1.2840)𝑡 ; growth 11 𝑃 = 𝑃0 (1.2214)𝑡 ; growth 13 (a) 1.5 (b) 50% 15 (a) 𝑃 = 1000 + 50𝑡 (b) 𝑃 = 1000(1.05)𝑡 17 (a) (b) (c) (d)

𝐷 to 𝐸, 𝐻 to 𝐼 𝐴 to 𝐵, 𝐸 to 𝐹 𝐶 to 𝐷, 𝐺 to 𝐻 𝐵 to 𝐶, 𝐹 to 𝐺

21 Table D

65 (a) 7.094 meters (b) 1958, 1883

𝑎 (years)

77 𝑦 = 2𝑥 + 3

19 (a) ℎ(𝑥) = 31 − 3𝑥 (b) 𝑔(𝑥) = 36(1.5)𝑥

63 (a) 2007–2009 (b) 2012–2014

(5, 6)

75 𝑦 = 0.5 − 3𝑥 is decreasing

81 True

61 (a) −0.01 (b) −0.2

31 𝑁 = 𝑘∕𝑙2 33 𝑉 (thousand dollars)

(d) (120, 14): equilibrium price.

79 False

About 10 days About 0.775 days per day Flower later when snow cover lasts longer 𝑦 = 41.25 + 0.775𝑥. Other answers are possible

80 120 160 200

73 𝑥 = 3 not a function of 𝑥

51 (a) (b) (c) (d)

59 𝑄(𝑚) = 𝑇 + 𝐿 + 𝑃 𝑚 𝑇 = fuel for take-off 𝐿 = fuel for landing 𝑃 = fuel per mile in the air 𝑚 = length of the trip (miles)

25 Domain: all 𝑥 Range: 0 < 𝑦 ≤ 1∕2

𝑞 40

49 1939: more addl grass/addl rainfall

(ii) 𝑓 (2014) = 525 (b) (𝑓 (2014) − 𝑓 (2001))∕(2014 − 2001) = 19.46 billionaires/yr (c) 𝑓 (𝑡) = 19.46𝑡 − 38,667.5

21 Domain: 1 ≤ 𝑥 ≤ 5 Range: 1 ≤ 𝑦 ≤ 6

𝑠(𝑞) = 11 + 𝑞∕40

Time arrive in Detroit

47 (a) About 10.4 kg∕hectare per mm (b) Additional 1 mm annual rainfall corresponds to additional 10.4 kg grass per hectare (c) 𝑄 = −440 + 10.4𝑟

57 (a)

𝑑(𝑞) = 20 − 𝑞∕20 (120, 14)

𝐶1 (𝑚) = 40 + 0.15𝑚 𝐶2 (𝑚) = 50 + 0.10𝑚 𝑚 (miles) 200 400 600 800

23 𝑓 (𝑠) = 2(1.1)𝑠 𝑔(𝑠) = 3(1.05)𝑠 ℎ(𝑠) = (1.03)𝑠 25 (a) 𝑔(𝑥) (b) ℎ(𝑥) (c) 𝑓 (𝑥) 27 𝑦 = 3(2𝑥 ) 29 𝑦 = 2(3𝑥 ) ( )𝑥 31 𝑦 = 4 12

33 (a) 𝐻

20 15 time

41 distance from exit

time

(c) For distances less than 200 miles, 𝐶1 is cheaper. For distances more than 200 miles, 𝐶2 is cheaper. 69 (a) Higher price: customers want less but owner wants to sell more (b) (60, 18): No (120, 12): Yes (c) Shaded region is all possible quantities of cakes owner is willing to sell and customers willing to buy

10 5 𝑌 2010 2011 2012 2013 2014 2015

(b) No (c) No

1062

Section 1.3

35 (a) revenue

1

3

𝑦

(a)

𝑦

(a)

4

4 𝑥 −2

𝑥 −2

advertising

2

2 −4

−4

(b) temperature

𝑦

(b)

𝑦

(b)

4

4 𝑥 −2

𝑥 time

−2

39 𝑑 = 670(1.096)

2 −4

37 17.882 mn barrels/day ℎ∕1000

−4



= 670(1.00009124)

41 (a) II (b) 18.9% growth; 10.9% loss; 20.2% loss

4

4

45 (a) 2𝑃0 , 4𝑃0 , 8𝑃0 (b) 𝑡∕50; 𝑃 = 𝑃0 2𝑡∕50

−2

2

𝑥 𝑥

−2

47 (b) 80.731% Users in 2006, thousands 81% per year 124.8% per year Less

53 (a) 923 trillion BTUs, 1163 trillion BTUs (b) 𝑦

𝑥

𝑥

𝑥

−2

2

𝑦

(e)

𝑦

(e)

4

4

𝑎, 𝑐, 𝑝 𝑎, 𝑑, 𝑞 𝑐=𝑝 𝑎 and 𝑏 are reciprocals 𝑝 and 𝑞 are reciprocals

2

−4

−4

2014

(c) 2013, 268 trillion BTUs

57 𝑦 = 𝑒

4

−2

−0.25𝑥

𝑦

(d)

4

2000 1750 1500 1250 1000 750

55 (a) (b) (c) (d)

−4

𝑦

(d)

2012

2

−4

51 (a) False (b) 2011

2010

𝑦

(c)

𝑦

(c)

43 (a) 𝑆(𝑡) = 219(1.05946)𝑡 (b) 5.946%

49 (a) (b) (c) (d)

2

𝑥

𝑥 −2

−2

2

2

concave up

59 (2, 2𝑒), (3, 3𝑒) not on the graph

−4

−4

61 𝑞 = 2.2(0.97)𝑡 63 𝑦 = 𝑒−𝑥 − 5 65 True

𝑦

(f)

67 True

𝑦

(f)

4

4

69 True 71 True

𝑥 −2

−4

2

𝑥 −2

−4

2

1063 5

49 𝑓 (𝑥) = 𝑥3 𝑔(𝑥) = 𝑥 + 1 √ 51 𝑓 (𝑥) = 𝑥 𝑔(𝑥) = 𝑥2 + 4

𝑦 4 𝑝(𝑡) 𝑡 −4

4

6

(b) 210.391 microgm∕cubic m 43 (a) 𝐴 = 100𝑒−0.17𝑡 (b) 𝑡 ≈ 4 hours

53 2(𝑦 − 1)3 − (𝑦 − 1)2 57 (a) −1 (b)

𝐴 (mg) 100

𝑦

−4

4 𝑓 (𝑥)

7

4

−4

𝑓 −1 (𝑥)

6

𝑡 (hours)

−4

𝑡

4

6 𝑤(𝑡)

59 Not invertible

−4

61 Not invertible 67 𝑔(𝑓 (𝑡)) ft3

4 2 (𝑥 + 1)2 𝑥2 + 1 𝑡2 (𝑡 + 1)

= 𝑓 −1 (𝐶) 71 (a) 𝑞 = 𝐶−100 2 (b) Number of articles produced at given cost

11 (a) (b) (c) (d) (e)

𝑒 𝑒2 2 𝑒𝑥 𝑒2𝑥 𝑒𝑡 𝑡2

73 (a) 𝑓 (15) ≈ 48 (b) Yes (c) 𝑓 −1 (120) ≈ 35 Rock is 35 millions yrs old at depth of 120 meters

13 (a) (b) (c) (d) (e)

𝑡2 + 2𝑡 + 2 𝑡4 + 2𝑡2 + 2 5 2𝑡2 + 2 𝑡4 + 2𝑡2 + 2

9 (a) (b) (c) (d) (e)

50

𝑥 −4

69 𝑓 −1 (30) min

75 Shift left

55 2054 57 It is a fake

91 False

23 odd

93 True

25 Odd

95 Impossible

27 Not invertible

97 Impossible

29 not invertible 31 Invertible

Section 1.4

35 (a) 𝑦 = 2𝑥2 + 1

𝑦 = 2𝑥2 + 1

8 6

𝑦 = 𝑥2

4 2

51 2023

81 𝑓 (𝑥) = 𝑥2 + 2

89 False

21 Even

49 (a) 𝐵(𝑡) = 𝐵0 𝑒0.067𝑡 (b) 𝑃 (𝑡) = 𝑃0 𝑒0.033𝑡 (c) 𝑡 = 20.387; in 2000

79 𝑓 −1 (𝑥) = 𝑥

87 True

19 Neither

47 7.925 hours

53 (a) 0.00664 (b) 𝑡 = 22.277; April 11, 2032

85 True

17 2𝑧ℎ − ℎ2

10 mg 18% 3.04 mg 11.60 hours

77 𝑓 (𝑔(𝑥)) ≠ 𝑥

83 𝑓 (𝑥) = 1.5𝑥, 𝑔(𝑥) = 1.5𝑥 + 3

15 2𝑧 + 1

(c) 𝑡 = 4.077 hours 45 (a) (b) (c) (d)

59 (a) 63.096 million (b) 3.162 61 𝐴: 𝑒𝑥 𝐵: 𝑥2 𝐶: 𝑥1∕2 𝐷: ln 𝑥 63 (a) Increases (b) Moves to the left 65 (a) No effect (b) Moves toward the origin 67 Moves toward the origin

1 1∕2

69 Function even

5𝐴2

71 Only for 𝑥 > 1

3

5 −1 + ln 𝐴 + ln 𝐵

73 Correct property is ln(𝐴𝐵) = ln 𝐴 + ln 𝐵

7 (log 11)∕(log 3) = 2.2

75 𝑦 = 0.7 log 𝑥

9 (log(2∕5))∕(log 1.04) = −23.4

77 ln(𝑥 − 3)

11 1.68

79 False

13 6.212

81 False

15 0.26 (b) 𝑦 = 2(𝑥2 + 1) (c) No

17 1 19 (log 𝑎)∕(log 𝑏)

37 𝑦 = −(𝑥 + 1)2 + 3

21 (log 𝑄 − log 𝑄0 )∕(𝑛 log 𝑎)

39 Can’t be done

23 ln(𝑎∕𝑏)

41 About 60

Section 1.5 1 Negative 0 Undefined

25 𝑃 = 15𝑒0.4055𝑡

43 0.4

27 𝑃 = 174𝑒−0.1054𝑡

45 −0.9

29 𝑝−1 (𝑡) ≈ 58.708 log 𝑡

47

3

𝑔(𝑓 (𝑥)) 𝑥

−3

3

31 𝑓 −1 (𝑡) = 𝑒𝑡−1 33 0.0693 35 13,500 bacteria 37 (a) 2024 (b) 336.49 million people 39 19.807 mn barrels per day

−3

41 (a) 𝑄0 (1.0033)𝑥



1064 3 Positive Positive Positive

31 (cos−1 (0.5) − 1)∕2 ≈ 0.0236

63 Depth = 7 + 1.5 sin(𝜋𝑡∕3)

33 (arctan(0.5) − 1)∕2 = −0.268

65 (a) 𝑓 (𝑡) = 10 sin (𝜋𝑡∕14) (b)𝑃 (watts)

35 One year

10

37 0.3 seconds 39 (a) 4.76◦ (b) 7.13◦ (c) 2.86◦



41 (a) ℎ(𝑡) (b) 𝑓 (𝑡) (c) 𝑔(𝑡)

𝑡 (hrs after 6 am) 7

14

𝑦

43

(c) 1 pm; 10 watts (d) 𝑔(𝑡) = 10 sin (𝜋𝑡∕9)

2𝜋 𝑘

5 Positive Positive Positive

−2𝜋

2𝜋

𝑥

𝑘

−𝑘 −2𝜋

45

69 (a) (b) (c) (d)

𝑦 2𝜋



𝑘 𝑥 −2𝜋

−𝑘

67 (a) 𝑓 (𝑡) = −0.5 + sin 𝑡 𝑔(𝑡) = 1.5 + sin 𝑡 ℎ(𝑡) = −1.5 + sin 𝑡 𝑘(𝑡) = 0.5 + sin 𝑡 (b) 𝑔(𝑡) = 1 + 𝑘(𝑡)

𝑘

2𝜋

Average depth of water 𝐴 = 7.5 𝐵 = 0.507 The time of a high tide

71 (a) About 10 ppm (b) Estimating coefficients 𝑦 = (1∕6) + 381 (c) Period: about 12 months; amplitude: about 3.5 ppm 𝑦 = 3.5 sin(𝜋𝑡∕6) (d) ℎ(𝑡) = 3.5 sin(𝜋𝑡∕6) + (𝑡∕6) + 381

−2𝜋

7 Positive Negative Negative

ppm

47

𝑦 2𝜋 −2𝜋



𝑘

2𝜋

−𝑘

𝑥

𝑘

ℎ(𝑡)

395 390 385 380 375

𝑡 (months since Dec 2005) 12 24

36 48 60

−2𝜋 𝑦

49

2𝜋 9 Negative Positive Negative

𝑘 𝑥 −2𝜋

−𝑘

𝑘

2𝜋

73 (a) 0.4 and 2.7 (b) arcsin(0.4) ≈ 0.4 𝜋 − arcsin(0.4) ≈ 2.7 (c) −0.4 and −2.7. (d) −0.4 ≈ − arcsin(0.4) −2.7 ≈ arcsin(0.4) − 𝜋 75 𝑓 has period 2𝜋, but 𝑔 has different period 77 Period is less than 2𝜋

−2𝜋



79 sin(2𝜋𝑥∕23) 81 True

51 12◦ C

83 False

53 7.5 months

85 True

3

55 275 thousand ft ∕sec

87 False

57 1 month

13 2; 0.1

59 If 𝑓 (𝑥) = sin 𝑥 and 𝑔(𝑥) = 𝑥2 then sin 𝑥2 = 𝑓 (𝑔(𝑥)) sin2 𝑥 = 𝑔(𝑓 (𝑥)) sin(sin 𝑥) = 𝑓 (𝑓 (𝑥))

15 𝑓 (𝑥) = 5 cos(𝑥∕3)

61 (a)

11 8𝜋; 3

17 𝑓 (𝑥) = −8 cos(𝑥∕10) 19 𝑓 (𝑥) = 2 cos(5𝑥) 21 𝑓 (𝑥) = 3 sin(𝜋𝑥∕9) 23 𝑓 (𝑥) = 3 + 3 sin ((𝜋∕4)𝑥) 25 0.588

89 False 91 False 93 False 95 True 97 True

𝑃

900 800 700

Section 1.6 1 As 𝑥 → ∞, 𝑦 → ∞ As 𝑥 → −∞, 𝑦 → −∞

100 Jan. 1

𝑡 (months) Jul. 1

Jan. 1

27 −0.259 29 0.727

(b) 𝑃 = 800 − 100 cos(𝜋𝑡∕6)

3 𝑓 (𝑥) → −∞ as 𝑥 → +∞ 𝑓 (𝑥) → −∞ as 𝑥 → −∞ 5 𝑓 (𝑥) → +∞ as 𝑥 → +∞ 𝑓 (𝑥) → +∞ as 𝑥 → −∞

1065 7 𝑓 (𝑥) → 3 as 𝑥 → +∞ 𝑓 (𝑥) → 3 as 𝑥 → −∞ 9 𝑓 (𝑥) → 0 as 𝑥 → +∞ 𝑓 (𝑥) → 0 as 𝑥 → −∞ 11 0.2𝑥5 13 1.05𝑥 15 25 − 40𝑥2 + 𝑥3 + 3𝑥5 17 (I) (II) (III) (IV) (V)

(a) (a) (a) (a) (a)

3 4 4 5 5

(b) (b) (b) (b) (b)

51 (a) 1.3 m2 (b) 86.8 kg (c) ℎ = 112.6𝑠4∕3

35 (a) Continuous (b) Not continuous (c) Not continuous

53 (a) 𝑅 = 𝑘𝑟4 (𝑘 is a constant) (b) 𝑅 = 4.938𝑟4 (c) 3086.42 cm3 /sec

37 (a) No (b) Yes

55 (a) (b) (c) (d)

Negative Positive Negative Negative Positive

0 𝑡 = 2𝑣0 ∕𝑔 𝑡 = 𝑣0 ∕𝑔 (𝑣0 )2 ∕(2𝑔)



57 Yes

𝑔(𝑡)

59 No

19 𝑦 = − 12 (𝑥 + 2)2 (𝑥 − 2)

61 No

21 𝑓 (𝑥) = 𝑘𝑥(𝑥 + 3)(𝑥 − 4) (𝑘 < 0)

63 Horizontal: 𝑦 = 1; Vertical: 𝑥 = −2, 𝑥 = 2

23 𝑓 (𝑥) = 𝑘(𝑥 + 2)(𝑥 − 2)2 (𝑥 − 5) (𝑘 < 0)

65 (a) −∞, −∞ (b) 3∕2, 3∕2 (c) 0, +∞

25 𝑟

67 (3∕𝑥) + 6∕(𝑥 − 2) Horizontal asymptote: 𝑥-axis Vertical asymptote: 𝑥 = 0 and 𝑥 = 2

27 (III), (IV) 29 (II), (III), (IV)

𝑡

39 Not continuous 41 Does not exist 43 2

33 IV

69 ℎ(𝑡) = 𝑎𝑏𝑡 𝑔(𝑡) = 𝑘𝑡3 𝑓 (𝑡) = 𝑐𝑡2

35 I, II, V

71 𝑦 = 𝑥

49 𝑘 = 𝜋 − 12

37 V

73 𝑦 = 100𝑒−0.2𝑧 .

51 No value

39 II

75 May tend to negative infinity

53 𝑘 = 𝜋∕6

41 Logarithmic

77 Only for 𝑥 > 1

43 Rational

79 𝑓 (𝑥) = 𝑥∕(𝑥2 + 1) crosses its horizontal asymptote 𝑦 = 0

55 (a) 𝑥 = 1, 3, 5 (b) 𝑥 = 3 { 1.2𝑡 57 𝑄 = 0.6𝑒0.001 𝑒−.002𝑡 59 𝑥 = 5; 𝑔(5) = 6

31 (II), (IV)

45 1, 2, 3, 4, or 5 roots

45 1 47 0.693

81 𝑓 (𝑥) = 3𝑥∕(𝑥 − 10) 83 𝑓 (𝑥) = 1∕(𝑥 + 7𝜋)

(a) 5 roots

𝑥

85 𝑓 (𝑥) = (𝑥 − 1)∕(𝑥 − 2) 87 True

𝑥

91 (a) (b) (c) (d)

𝑥

(d) 2 roots

𝑥

(e) 1 root

𝑥

69 (a) No (b) 𝑘 = 3; other answers possible

5 (a) 𝑥 = −2, 3 (b) lim 𝑓 (𝑥) = −3; lim 𝑓 (𝑥) does not exist 𝑥→−2

9 (b) 3 11 Yes 47 3 zeros: 𝑥 ≈ −1, 𝑥 ≈ 3, 𝑥 > 10

13 Yes

49 ℎ = 𝑉 ∕𝑥2

17 (a) Continuous (b) Not continuous

15 No

23 4



25 6 27 5∕9 29 𝑘 = 2

ℎ = 𝑉 ∕𝑥2

31 𝑘 = 5∕3

𝑥

71 𝑒

1 2 Does not exist 1 0 2

7 1

33 𝑘 = 4

61 𝑡 = ±3; 𝑞(3) = −3, 𝑞(−3) = 3

67 Three zeros: one between 5 and 10, one between 10 and 12, the third either less than 5 or greater than 12

1 (a) 𝑥 = −1, 1 (b) −3 < 𝑥 < −1, −1 < 𝑥 < 1, 1 < 𝑥 < 3 3 (a) (b) (c) (d) (e) (f)

0.5 < 𝑡

65 No. 𝑓 (0) ≠ limit at 0

(I) (III) (IV) (II)

Section 1.7

(c) 3 roots

0 ≤ 𝑡 ≤ 0.5

63 No. Limit does not exist at 0

89 True. (b) 4 roots

7

5

𝑥→3

73 (a) 𝑥 = 1∕(𝑛𝜋), 𝑛 = 1, 2, 3, … (b) 𝑥 = 2∕(𝑛𝜋), 𝑛 = 1, 5, 9, … (c) 𝑥 = 2∕(𝑛𝜋), 𝑛 = 3, 7, 11, … 75 𝑓 (𝑥) = 5 for some 𝑥, not necessarily for 𝑥 = 1 { 1 𝑥 ≥ 15 77 𝑓 (𝑥) = −1 𝑥 < 15 { 1 𝑥≤2 79 𝑓 (𝑥) = 𝑥 𝑥>2 { 1 𝑥≤3 81 False, 𝑓 (𝑥) = 2 𝑥>3 83 False

Section 1.8 1 (a) (b) (c) (d) (e) (f)

1 0 0 1 Does not exist 1

1066 3 (a) (b) (c) (d) (e) (f) (g) (h)

4 2 −4 2 4 0 Does not exist 2

39 𝑒

3 Does not exist

41 (a)

5 −1∕9

10

7 −1∕11 9 −4

5

11 6∕7 13 6 15 5∕6

𝑥

5 (a) 3 (b) Does not exist or −∞

−10 −5

5

10

17 −1∕9

7 16 9 (a) (b) (c) (d)

19 −1∕16 8 6 15 4

21 1∕6

(b) Yes

11 Other answers are possible

43 −1

23 4

45 0

25 (a) Yes (b) 1

47 ∞

𝑥

49 0

27 0

51 5∕7

29 4∕5

53 (a)

31 ∞

9

33 28 35 10 37 4 39 5

2

13 Other answers are possible

41 6

𝑥 3

𝑦

5

43 7 or −7 45 Any 𝑘 47 𝑘 ≤ 5

3

(b) No

𝑥

49 𝑘 ≥ 0

55 Other answers are possible

51 5 53 1∕3

𝑦

55 2∕5

4

15 Other answers are possible

59 0 61 0

𝑥 −5

5 −3

5 𝑥 3

(i) 50 mg (ii) 150 mg (b) 𝑡 = 1, 2, 3, 4 sec

57 (a)

17 ∞

69 1 71 Does not exist 73 𝑐 = ±∞, 𝐿 = 0 75 The limit of a function does not depend on the value of a function at a point

lim 𝑓 (𝑥) = −∞;

𝑥→−∞

67 𝑓 (𝑥) = (𝑥 + 3)(𝑥 − 1)∕(𝑥 − 1)

83 False

lim 𝑓 (𝑥) = −∞

69 True

lim 𝑓 (𝑥) = −∞;

71 True

lim 𝑓 (𝑥) = ∞

73 True

23 ∞ 25 0 27 0

𝑥→∞

𝑥→−∞ 𝑥→∞

33

67 0

𝑥−5 (𝑥 + 5)(𝑥 − 5) (b) 1∕10 1 nor − 210 exist. (c) Neither lim 𝑥 −25 𝑥→5 𝑥 − 5 65 𝑃 (𝑥)∕𝑄(𝑥) is not necessarily defined at all 𝑥

61 (a)

21 0

31

65 There are many possibilities, one being 𝑔(𝑥) = (𝑥 ln 𝑥 − ln 𝑥)∕(𝑥 − 1); (1, 0)

59 2

19 ∞

29

63 There are many possibilities, one being 𝑔(𝑥) = (𝑥3 − 3𝑥2 + 𝑥 − 3)∕(𝑥 − 3); (3, 10)

75 True

lim 𝑓 (𝑥) = 0;

𝑥→−∞

77 False

lim 𝑓 (𝑥) = 0

𝑥→+∞

79 True

35 Yes; not on any interval containing 0 = 37 lim𝑥→2+ 𝑓 (𝑥) lim𝑥→2 𝑓 (𝑥) = 0

lim𝑥→2− 𝑓 (𝑥)

𝑓 (𝑥) 𝑥 1 2 3 4

83 1 85 3

1

0

81 False =

87 (a) (b) (c) (d)

79 False 81 False

Section 1.10 1 (a) (b) (c) (d) (e)

No No Yes Yes Yes

3 0.1, 0.05, 0.01, 0.005, 0.001, 0.0005 Follows Does not follow (although true) Follows Does not follow

Section 1.9 1 3

77 True

5 0.025 7 0.02 11 0.45, 0.0447, 0.00447 13 𝛿 = 1 − 𝑒−0.1 15 𝛿 = 1∕3 17 (b) 0

1067 (c)

100

0.01

𝑓 (𝑥)

50

1

2 3

−50

−0.01 −0.015

0.015

93 US: 156 volts max, 60 cycles/sec Eur: 339 volts max, 50 cycles/sec

𝑥

0

47 (a) (b) (c) (d)

35 False 37 False

(i) 𝑉 = 3𝜋𝑟2 (ii) 𝑉 = 𝜋𝑟2 ℎ (i) 𝑉

95 (a)

4

(b)

−100

(d) −0.015 < 𝑥 < 0.015, −0.01 < 𝑦 < 0.01

91 One hour

−5 −1 Does not exist Does not exist

𝑟

49 Yes

39 False

(ii)

51 No

Chapter 1 Review

55 1.098

1 𝑦 = 2𝑥 − 10

57 (b) 1

−3𝑥2

+3 √ 5 𝑦 = 2 + 9 − (𝑥 + 1)2

3 𝑦=

59 0

𝑘

61 −5∕3

7 𝑦 = −5𝑥∕(𝑥 − 2) 9

65 (a) (b) (c) (d)

heart rate

1 −1 Does not exist −5

𝑦

97 (a)

4

67 10

3

69 −1

time administration of drug

11 (a) (b) (c) (d) (e) (f) (g)

𝑉

53 0

2

71 9 73 (b) (c) (d) (e) (f)

[0, 7] [−2, 5] 5 (1, 7) Concave up 1 No

1

200 bushels 80 lbs About 0 ≤ 𝑌 ≤ 550 Decreasing Concave down

75 (a) (b)

𝑥 −1 0 99 (a) (b) (c) (d)

(i) 𝑞 = 320 − (2∕5)𝑝 (ii) 𝑝 = 800 − (5∕2)𝑞

𝑝 (dollars)

1

2

3

5

III IV I II

13 (a) Min gross income for $100,000 loan (b) Size of loan for income of $75,000

800

101 20

15 𝑡 ≈ 35.003

550

103 Velocity: Not continuous Distance: Continuous

17 𝑡 = log(12.01∕5.02)∕ log(1.04∕1.03) 90.283

4

= velocity

19 𝑃 = 5.23𝑒−1.6904𝑡 21 𝑓 (𝑥) = 𝑥3 𝑔(𝑥) = ln 𝑥

𝑞 (items) 100 320

23 Amplitude: 2 Period: 2𝜋∕5 25 (a) 𝑓 (𝑥) → ∞ as 𝑥 → ∞; 𝑓 (𝑥) → −∞ as as 𝑥 → −∞ (b) 𝑓 (𝑥) → −∞ as as 𝑥 → ±∞ (c) 𝑓 (𝑥) → 0 as 𝑥 → ±∞ (d) 𝑓 (𝑥) → 6 as 𝑥 → ±∞ √ 27 0.25 𝑥

77 (a)

29 𝑦 = 𝑒0.4621𝑥 or 𝑦 = (1.5874)𝑥

83 Parabola opening downward

31 𝑦 = −𝑘(𝑥2 + 5𝑥) (𝑘 > 0)

85 10 hours

time

(i) Attractive force, pulling atoms together (ii) Repulsive force, pushing atoms apart (b) Yes

distance

79 2024 81 (a) Increased: 2011; decreased: 2012 (b) False (c) True

time

105 (a) 5 (b) 5 (c) 5

87 About 14.21 years

33 𝑧 = 1 − cos 𝜃

89 (a) 81% (b) 32.9 hours (c) 𝑃

35 𝑥 = 𝑘(𝑦2 − 4𝑦) (𝑘 > 0)

𝑦

107 (a)

2

37 𝑦 = −(𝑥 + 5)(𝑥 + 1)(𝑥 − 3) 39 Simplest is 𝑦 = 1 − 𝑒−𝑥 41 𝑓 (𝑥) = sin (2(𝜋∕5)𝑥)

𝑦=5 𝑃0 𝑥

43 Not continuous 45 lim𝑥→3+ 𝑓 (𝑥) = 54, lim𝑥→3− 𝑓 (𝑥) = −54, lim𝑥→3 𝑓 (𝑥) does not exist

−2

2

𝑡 510

32.9

𝑥=3

1068

Section 2.2

(b) 64

(b) No

141 5 sin 𝑥 − 20 sin3 𝑥 + 16 sin5 𝑥

109 False 111 Cannot be determined

1 12 3 (a) 70 $/kg; 50 $/kg (b) About 60 $/kg

Section 2.1

113 (a) 3.5 (b) Yes √ 115 1∕(2 𝑥)

1 265∕3 km/hr 3 −3 angstroms∕sec

5 (b) 0.24 (c) 0.22

117 48∕7

5 2 meters∕sec

7 negative

121 11

7 23.605 𝜇m∕sec

123 25

9 (a) (i) 6.3 m/sec (ii) 6.03 m/sec (iii) 6.003 m/sec (b) 6 m/sec

125 The limit appears to be 1 127 (a) 1 (b) lim

𝑥→0

1 does not exist 𝑥

13

−1.01 −0.099

distance

9 𝑓 ′ (2) ≈ 9.89 11 (a) 𝑥 = 2 and 𝑥 = 4 (b) 𝑔(4) (c) 𝑔 ′ (4)

time

13

15 27

𝑦 𝐹 𝐴

𝐸

𝐶

17 1.9 19 1

0.01

21

𝑥 4𝜋

2𝜋 3𝜋

−1

0.099

(d) −0.099 < 𝑥 < 0.099, −1.01 < 𝑦 < −0.99

𝑦 = sin 𝑥 𝜋

11 (a) (i) −1.00801 m/sec (ii) −0.8504 m/sec (iii) −0.834 m/sec (b) −0.83 m/sec

129 (b) −1 (c) −0.99

131 (b) 0 (c)

𝑦 1

𝑥

𝐷 𝐵

Slope

−3

−1

0

1∕2

1

2

Point

𝐹

𝐶

𝐸

𝐴

𝐵

𝐷 15 16 million km2 covered by ice in Feb 1983

−0.01 −0.0033

0.0033

19 About 41 21 (a) (b) (c) (d)

25 𝑣𝑎𝑣𝑔 = 2.5 ft/sec 𝑣(0.2) = 4.5 ft/sec

(d) −0.0033 < 𝑥 < 0.0033, −0.01 < 𝑦 < 0.01 133 (b) 3 (c)

17 14 million km2 covered by ice on Feb 1, 2015

23 From smallest to largest: 0, slope at 𝐶, slope at 𝐵 slope of 𝐴𝐵, 1, slope at 𝐴

27

𝑓 (𝑡)

𝑓 (4) 𝑓 (2) − 𝑓 (1) (𝑓 (2) − 𝑓 (1))∕(2 − 1) 𝑓 ′ (1)

23 (4, 25); (4.2, 25.3); (3.9, 24.85)

3.01

25 𝑦

𝑦 = 𝑓 (𝑥) 2.99 −0.047

0.047



(d) −0.047 < 𝑥 < 0.047, 2.99 < 𝑦 < 3.01 135 (b) 2 (c)

29 (a) (b) (c) (d) (e)

0.0049

36 feet 34 ft/sec 18 ft/sec 75 ft, 0 ft/sec

𝑦

80 60 40 20

(d) −0.0049 < 𝑥 < 0.0049, 1.99 < 𝑦 < 2.01

(f) 𝑡 = 1.6 sec 33 6

𝑥 3

𝑥4

(e) Slope =

❄ ❄ 𝑥 + ✛(d) ℎ✲ ℎ

𝑥

𝑡 2

31 3

2𝑥3

✻ ✻𝑓 (𝑥 + ℎ) − 𝑓 (𝑥) (c) ❄ 𝑓 (𝑥+ℎ)−𝑓 (𝑥) ℎ

(a) 𝑓 (𝑥)

(1.6, 75.1)

1

137 (a) 𝑓 (𝑥) = −(𝑥 − 1)2 (𝑥 − 3)3 (b)

1



(b) 𝑓 (𝑥 + ℎ)

2.01

1.99 −0.0049

𝑡

139 (a) 𝑥 − + −𝑥+𝑥2 +2𝑥3 −5𝑥4 +2𝑥5 +4𝑥6 −4𝑥7 +𝑥8

35 Expand and simplify first

3

27 (a) (𝑓 (𝑏) − 𝑓 (𝑎))∕(𝑏 − 𝑎) (b) Slopes same (c) Yes 29 𝑔 ′ (−4) = 5 31 𝑦 = 7𝑥 − 9 33 𝑓 ′ (2) ≈ 6.77 35 𝑓 ′ (1) ≈ −1.558 𝑓 ′ ( 𝜋4 ) ≈ −1

37 𝑓 (𝑡) = 𝑡2

37 300 million people; 2.867 million people∕yr

39 False

39 6

41 True

41 −4

43 True

45 −6

𝑥

1069 47 −1

15

33 (a) (b) (c) (d)

4

49 1∕4

𝑓 (𝑥)

51 3

2

53 1∕2

𝑥3 𝑥4 𝑥5 𝑥3

35

55 1

1 𝑥

𝑥

57 12

−2

2

−1

59 3 61 −1∕4

𝑥

67 Derivative at a point is slope of tangent line

−2 −1 −2

69 𝑓 (𝑥) = 2𝑥 + 1

𝑓 ′ (𝑥)

𝑓 ′ (𝑥)

2

65 𝑦 = −2𝑥 + 3

1

37

2

−4

71 True

2

−2

4

63 𝑦 = 12𝑥 + 16

1

1

𝑥 𝑓 ′ (𝑥)

2

73 (a) 17

Section 2.3

10 8 6 4 2

1 (a) About 3 (b) Positive: 0 < 𝑥 < 4 Negative: 4 < 𝑥 < 12 3 Between −9 and −6 Between 6 and 9 5

39

𝑓 (𝑥)

𝑥 𝑥

−3 −2 −1

4

−4

1

2

10 8 6 4 2

𝑥 4 −4

3

41

𝑓 ′ (𝑥) 𝑥 −1

−3 −2 −1

4

𝑥 −4

1

2

19

4

1

2 3

𝑓 ′ (𝑥) 𝑥

7

𝑓 ′ (𝑥)

43

𝑓 ′ (𝑥)

3 𝑓 (𝑥) 𝑥

−1

𝑥

1

−4

45 (V) 47 (II)

9

49 (a) Graph II (b) Graph I (c) Graph III

15

51 current

𝑥 −3 − 1

3

time

2

𝑓 ′ (𝑥)

1 11

𝑥

4

1

𝑥 −4

57

𝑦

21 −2∕𝑥3 23 −1∕(𝑥 + 1)2

4

25 2𝑥 + 2 27 −1∕(2𝑥3∕2 )

−4

29

𝑥 13

𝑥

4

1

−1

2

3

4

Other answers possible

𝑥 −4

4

−4

Other answers possible 31 𝑓 ′ (𝑥) positive: 4 ≤ 𝑥 ≤ 8 𝑓 ′ (𝑥) negative: 0 ≤ 𝑥 ≤ 3 𝑓 ′ (𝑥) greatest: at 𝑥 ≈ 8

63 counterexample: 𝑓 (𝑥) = 1, 𝑔(𝑥) = 2 65 𝑓 (𝑥) = 𝑏 + 2𝑥 67 True 69 False

1070

Section 2.4 1 (a) Costs $1300 for 200 gallons (b) Costs about $6 for 201st gallon 3 (a) Positive (b) ◦ F/min 5 (a) Quarts; dollars. (b) Quarts; dollars/quart 7 (a) acre-feet per week (b) water storage increases; recent rainfall increases inflow 9 Dollars/percent; positive 11 Ft/sec; negative 13 Dollars per percent; positive 15 Joule/cm; positive

49 (a) Cell divisions per hour per 10−4 M of concentration (b) 10−4 M (c) 0.02 cell divisions per hour 51 (a) 0.0729 people/sec (b) 14 seconds

15 (a) Positive; negative (b) Neither; positive (c) Number of cars increasing at about 600,000 million cars per year in 2005 17 (a)velocity (m/s)

53 Number of people 65.5–66.5 inches Units: People per inch 𝑃 ′ (66) between 17 and 34 million people/in 𝑃 ′ (𝑥) is never negative

35 30 25 20 15 10 5 0

55 No units are given 57 Minutes/page 61 True 63 False 65 (b), (d)

(b) 3.6

17 Dollars/year 19 (a) ◦ 𝐶∕km (b) Air temp decreases about 6.5◦ for a 1 km increase in altitude 21 (a) Δ𝑅 ≈ 3Δ𝑆 (b) 0.6 (c) 13.6

time (sec)

2 4 6 8 101214161820

19

Section 2.5

𝑦 4

1 (a) Increasing, concave up (b) Decreasing, concave down 3 𝐵

𝑥

5 (a)

𝑓 (𝑥)

−4

23 About 1.445 billion people in 2020; growing at about 8.64 million people per year

4 −4

25 (a) 𝑓 (b) 𝑔 and ℎ

𝑥 𝑦

21

27 (b) Pounds∕(Calories∕day) 29 𝑓 (points) = Revenue 𝑓 ′ (4.3) ≈ 55 million dollars/point 31 (a) In 2008 pop Mexico increase 1.35 m people/yr (b) Pop 113.2 m in 2009 (c) 113.2 m: about 0.74 yrs for pop increase of 1 m 33 (a) (b) (c) (d)

(b)

𝑓 (𝑥)

23 (c)

𝑦

𝑓 (𝑥) 𝑥

𝑥 −4

(d)

𝑓 (𝑥) 𝑥

(b) 𝑑𝑃 ∕𝑑𝑡 < 0, 𝑑 2 𝑃 ∕𝑑𝑡2 > 0 (but 𝑑𝑃 ∕𝑑𝑡 is close to zero) 27 (a) utility

7 height quantity

41 Barrels/year; negative 43 (a) Gal/minute (b) (i) 0 (ii) Negative (iii) 0

47 (a) mm (of rain) (b) mm (of leaf width) per mm (of rain) (c) 4.36 mm

4

25 (a) 𝑑𝑃 ∕𝑑𝑡 > 0, 𝑑 2 𝑃 ∕𝑑𝑡2 > 0

39 mpg/mph

45 (a) Mn km2 /day Growing 73,000 km2 /day (b) 0.365 mn km2 Grew by 0.365 mn km2

4

𝑥

Depth 3 ft at 𝑡 = 5 hrs Depth increases 0.7 ft/hr Time 7 hrs when depth 5 ft Depth at 5 ft increases 1 ft in 1.2 hrs

35 (a) The weight of the object, in Newtons, at a distance of 80 kilometers above the surface of the earth; positive (b) For an object at a distance of 80 km from the surface of the earth, 𝑓 ′ (80) represents the approximate number of Newtons by which the weight of an object will change for each additional kilometer that the object moves away from the earth’s surface; negative (c) The distance of the object from the surface of the earth, in kilometers, that would be necessary for the object to weigh 200 Newtons; positive (d) For an object that weighs 200 Newtons, (𝑓 −1 )′ (200) represents the approximate amount by which the object’s distance from the earth’s surface must change in order for its weight to increase by 1 Newton; negative

𝑥 −4

(b) Derivative of utility is positive 2nd derivative of utility is negative time

31 No 33 No

9 𝑓 ′ (𝑥) < 0 𝑓 ′′ (𝑥) = 0 11

𝑓 ′ (𝑥)

0

13 𝑓 ′ (𝑥) < 0 𝑓 ′′ (𝑥) < 0

35 Yes 37 (a) (b) (c) (d) (e)

𝑡3 , 𝑡4 , 𝑡5 𝑡2 , 𝑡3 𝑡1 , 𝑡2 , 𝑡5 𝑡1 , 𝑡4 , 𝑡5 𝑡3 , 𝑡4

1071

Chapter 2 Review

39 10

8 6 4 2

23 −1∕𝑥2

1 72∕7 = 10.286 cm∕sec

25 (a)

(i) 𝐶 and 𝐷

3 (ln 3)∕2 = 0.549 mm∕sec 5 0 mm∕sec

𝑥

−2 −4

2

4

6

8 10 12

7 0 mm∕sec 9 distance

(ii) 𝐵 and 𝐶 (b) 𝐴 and 𝐵, 𝐶 and 𝐷

41 22 only possible value 43 (a) All positive 𝑥 (b) All positive 𝑥 (c) All positive 𝑥 47 𝑓 (𝑥) = 𝑥

27 10𝑥 29 6𝑥 31 2𝑎

2

time

49 True 11 (a)

51 True

√ 35 −1∕(2( 𝑎)3 )

𝑦 𝑦 = 𝑥 sin 𝑥

Section 2.6 1 (a) 𝑥 = 1 (b) 𝑥 = 1, 2, 3

8

37 From smallest to largest: 0, Vel at 𝐶, Vel at 𝐵, Av vel between 𝐴 and 𝐵, 1, Vel at 𝐴

4

39 (a) 𝐴, 𝐶, 𝐹 , and 𝐻. (b) Acceleration 0

3 No

𝑥

5 Yes

−10

7 Yes

−5

5

10

−4

9 No

41 (a) 𝑓 ′ (𝑡) > 0: depth increasing 𝑓 ′ (𝑡) < 0: depth decreasing (b) Depth increasing 20 cm/min (c) 12 meters/hr 43 (a) Thousand $ per $ per gallon Rate change revenue with price/gal (b) $ per gal/thousand $ Rate change price/gal with revenue

11 (a)

𝑓 (𝑥) (b) Seven (c) Increasing at 𝑥 = 1 Decreasing at 𝑥 = 4 (d) 6 ≤ 𝑥 ≤ 8 (e) 𝑥 = −9

𝑥

33 −2∕𝑎3

47 Concave up 49 (a)

6

13

2

𝑓 ′ (𝑥)

(b)

𝑥

5

Student B’s

✛answer = slope

Student C’s answer =slope of this line

of this line

❘✛

𝑓 (𝑥)

Student A’s answer =slope of this line

4

𝑥 15

2

1

15 (a) 𝐵 = 𝐵0 at 𝑟 = 𝑟0 𝐵0 is max

−1

𝑥 2

3

𝑥

4

1

2

3

4

5

6

𝑓 ′ (𝑥)

𝐵 17

𝐵0

𝑓 ′ (𝑥)

𝑟

𝑥

𝑟0 (b) Yes (c) No 17 (a) Yes (b) Yes

19

𝑦

19 (a) No; yes. (b) 𝑓 ′ (𝑥) not differentiable at 𝑥 = 0; 𝑓 ′′ (𝑥) neither differentiable nor continuous at 𝑥 = 0

2 1

−1

𝑥

21 Other cases are possible 23 𝑓 (𝑥) = |𝑥 − 2|

25 𝑓 (𝑥) = (𝑥2 − 1)∕(𝑥2 − 4)

27 True; 𝑓 (𝑥) = 𝑥2 { 29 True; 𝑓 (𝑥) = 31 (a) (b) (c) (d)

(b) Student C’s 𝑓 (𝑥 + ℎ) − 𝑓 (𝑥 − ℎ) (c) 𝑓 ′ (𝑥) = 2ℎ 51 𝑥1 = 0.9, 𝑥2 = 1, 𝑥3 = 1.1 𝑦1 = 2.8, 𝑦2 = 3, 𝑦3 = 3.2

5 4 3

−1 2 21

4

6

𝑦

55 (a) Dose for 140 lbs is 120 mg Dose increases by 3mg/lb (b) About 135 mg

1 𝑥≥0

57 𝑃 ′ (𝑡) = 0.008𝑃 (𝑡)

−1 𝑥 < 0

Not a counterexample Counterexample Not a counterexample Not a counterexample

53 (a) Negative (b) 𝑑𝑤∕𝑑𝑡 = 0 (c) |𝑑𝑤∕𝑑𝑡| increases; 𝑑𝑤∕𝑑𝑡 decreases

𝑥 −4

4

59 (a) 𝑓 ′ (0.6) ≈ 0.5 𝑓 ′ (0.5) ≈ 2 (b) 𝑓 ′′ (0.6) ≈ −15 (c) Maximum: near 𝑥 = 0.8 minimum: near 𝑥 = 0.3

1072 𝑦

√ 61 (a) At (0, √ 19): slope = 0 At ( 19, 0): slope is undefined (b) slope ≈ √ 1∕2 15): slope ≈ 1∕2 (c) At (−2, √ At (−2, √− 15): slope ≈ −1∕2 At (2, 15): slope ≈ −1∕2

3 2 1

𝑔 ′′ (𝑥)

59 (a) 𝑓 ′ (𝑥) = 7𝑥6 ; 𝑓 ′′ (𝑥) = 42𝑥5 ; 𝑓 ′′′ (𝑥) = 210𝑥4 (b) 𝑛 = 8.

𝑔(𝑥)

✠ 𝑥

−2

2

67 −2∕3𝑧3 (power rule) 69 𝑦 = 2𝑥 − 1

−6 𝑎=2

71 𝑦 = −9.333 + 6.333𝑥 73 (a) 𝑓 ′ (1) < 𝑓 ′ (0) < 𝑓 ′ (−1) < 𝑓 ′ (4) (b) 𝑓 ′ (1) = −1, 𝑓 ′ (0) = 2, 𝑓 ′ (−1) = 11, 𝑓 ′ (4) = 26

4000 𝑦 3

6

𝑔 ′′ (𝑥) 3 2 1

9 12 15 18 21 24

April July Oct Jan April July Oct Jan April

−2

75 For 𝑥 < 0 or 2 < 𝑥 < 3

𝑔(𝑥)



𝑥 2

−2

(b) Max of 4500 on June 1st (c) Min of 3500 on Feb 1st (d) Growing fastest: April 1st Decreasing fastest: July 15 and Dec 15 (e) About 400 deer/month

−4 −6 𝑎=3

65 (a) Concave down 73 (a) 4𝑥(𝑥2 + 1), 6𝑥(𝑥2 + 1)2 , 8𝑥(𝑥2 + 1)3 (b) 2𝑛𝑥(𝑥2 + 1)𝑛−1

Section 3.1

temperature

200◦ C

3 Do not apply 5

𝑦′

=

𝜋𝑥𝜋−1

(power rule)

10

7 11𝑥

9 −12𝑥−13

20◦ C

time

11 −3𝑥−7∕4 ∕4 13 3𝑥−1∕4 ∕4 15 3𝑡2 − 6𝑡 + 8 17 −5𝑡−6 19 −(7∕2)𝑟−9∕2

(b) 120◦ < 𝑇 < 140◦ (c) 135◦ < 𝑇 < 140◦ (d) 45 < 𝑡 < 50

21 𝑥−3∕4 ∕4 23 −(3∕2)𝑥−5∕2

67 (a) 𝑓 ′ (0) = 1.00000 𝑓 ′ (0.3) = 1.04534 𝑓 ′ (0.7) = 1.25521 𝑓 ′ (1) = 1.54314 (b) They are about the same

25 6𝑥1∕2 − 52 𝑥−1∕2

69 0, because 𝑓 (𝑥) constant 71 (a) −2𝑎∕𝑒 (b)

𝑎𝑥2

𝑔 ′′ (𝑥)

+ 4𝑎2 𝑥2 ∕𝑒

𝑎𝑥2

𝑥 2

−4

79 (a) 0 (b) −6 (c) (IV) 81 (a) 0 (b) 6 (c) (I) 83 (a) 12.5 liters (b) −6.25 liters per atmosphere (c) 1 atmosphere 85 (a) (b) (c) (d) (e)

15.2 m/sec 5.4 m/sec −9.8 m/sec2 34.9 m 5.2 sec

87 (a) 5159𝑣−0.33 (b) 11030 m3 ∕sec per km3 89 Approximately 10% 91 Approximately 20% 93 (a) 0.0467𝑥−1∕3 (b) 0.015 mm per meter (c) 0.09 mm √ 95 (a) 𝑑𝑇 ∕𝑑𝑙 = 𝜋∕ 𝑔𝑙 (b) Positive, so period increases as length increases 99 𝑛 = 4, 𝑎 = 3∕32 101 (a) 𝑓 ′ (𝑥) = 𝑔 ′ (𝑥) = 3𝑥2 + 6𝑥 − 2 (b) One is a vertical shift of the other, so they have the same slopes everywhere (c) Any vertical shift has the same derivative 103 𝑥2 + 𝑎 105 𝑓 (𝑥) = 𝑥2 , 𝑔(𝑥) = 3𝑥

27 17 + 12𝑥−1∕2

107 𝑓 (𝑥) = 3𝑥2

29 20𝑥3 − 2∕𝑥3

109 False

31 −12𝑥3 − 12𝑥2 − 6

111 True

33 6𝑡 − 6∕𝑡3∕2 + 2∕𝑡3

113 𝑦 = 2𝑥 and 𝑦 = −6𝑥

37 (1∕2)𝜃 −1∕2 + 𝜃 −2

3 2 1 𝑔(𝑥)

−2

77 (a) 0 (b) 5040

35 3𝑡1∕2 + 2𝑡

𝑦

−2

𝑦

39 (𝑧2 − 1)∕3𝑧2 √ 41 1∕(2 𝜃) + 1∕(2𝜃 3∕2 )

(−2, 12)

45 𝑎∕𝑐

𝑦 = −6𝑥

43

3𝑥2 ∕𝑎

+ 2𝑎𝑥∕𝑏 − 𝑐

47 3𝑎𝑏 √ 49 𝑏∕(2 𝑡) 51 4.44

55 −3.98 57 −4.15

𝑦 = 𝑥2 − 2𝑥 + 4 𝑦 = 2𝑥 (2, 4) 𝑥

2

53 −3.85

−6 𝑎=1

(−1∕4)𝑥−3∕2 + (3∕8)𝑥−5∕2 −

65 6𝑥 (power rule)

−4 5000

=

63 Rules of this section do not apply

−2

63 (a) Period 12 months

= 61 𝑑 2 𝑤∕𝑑𝑥2 (3∕4)𝑥−5∕2 ; 𝑑 3 𝑤∕𝑑𝑥3 −7∕2 (15∕8)𝑥

115 (a) 𝑥 < 1∕2 (b) 𝑥 > 1∕4

1073 ( ) 𝑒𝑥 𝑒𝑥 𝑑 𝑒𝑥 = − 2 𝑑𝑥 ( 𝑥 ) 𝑥 𝑥 𝑒𝑥 𝑑 𝑒𝑥 2𝑒𝑥 = − 3 𝑑𝑥 ( 𝑥2 ) 𝑥2 𝑥 𝑒𝑥 3𝑒𝑥 𝑑 𝑒𝑥 = 3 − 4 3 𝑑𝑥 ( 𝑥 ) 𝑥 𝑥 𝑒𝑥 𝑑 𝑒𝑥 𝑛𝑒𝑥 = 𝑛 − 𝑛+1 (b) 𝑑𝑥 𝑥𝑛 𝑥 𝑥 83 (a) 𝑔 ′ (𝑎) = −𝑓 ′ (𝑎)∕(𝑓 (𝑎))2 < 0 since 𝑓 ′ (𝑎) > 0 (b) If 𝑔 ′ (𝑎) = −𝑓 ′ (𝑎)∕(𝑓 (𝑎))2 = 0, so must 𝑓 ′ (𝑎) (c) Not if 𝑓 (𝑎) = 0, too ( 1 1 + + 87 𝑓 ′ (𝑥) = 𝑓 (𝑥) 𝑥 − 𝑟2 )𝑥 − 𝑟1 1 ⋯+ 𝑥 − 𝑟𝑛

9 (1 − 𝑥)∕𝑒𝑥

(c) 𝑥 < 0 or 𝑥 > 0 117 𝑛 = 1∕13

11 (1 − (𝑡 + 1) ln 2)∕2𝑡

119 𝑎 = 2, 𝑏 = 2

13 6∕(5𝑟 + 2)2

121 Infinitely many, each of the form 𝑓 (𝑥) = 𝐶

15 1∕(5𝑡 + 2)2

81 (a)

17 2𝑒𝑡 + 2𝑡𝑒𝑡 + 1∕(2𝑡3∕2 )

Section 3.2

19 2𝑦 − 6, 𝑦 ≠ 0 √ 𝑧(3 − 𝑧−2 )∕2 21

1 2𝑒𝑥 + 2𝑥 3 5 ln(𝑎)𝑎5𝑥

23 2𝑟(𝑟 + 1)∕(2𝑟 + 1)2

5 10𝑥 + (ln 2)2𝑥

25 17𝑒𝑥 (1 − ln 2)∕2𝑥

7 4(ln 10)10𝑥 − 3𝑥2

27 1, 𝑥 ≠ −1

9 ((ln 3)3𝑥 )∕3 − (33𝑥−3∕2 )∕2

29 (−4 − 6𝑥)(6𝑥𝑒 − 3𝜋) + (2 − 4𝑥 − 3𝑥2 )(6𝑒𝑥𝑒−1 )

2 𝑥

11 (ln 4) 4

31 (a) 4 (b) Does not exist (c) −4

13 5 ⋅ 5𝑡 ln 5 + 6 ⋅ 6𝑡 ln 6 15 𝑒𝑥𝑒−1 17 (ln 𝜋)𝜋 𝑥

33 (a) −2 (b) Does not exist (c) 0

19 (ln 𝑘)𝑘𝑥 21 𝑒𝑡+2

Section 3.4

35 Approx 0.4

1 99(𝑥 + 1)98

25 2 + 1∕(3𝑥4∕3 ) + 3𝑥 ln 3

37 Approx 0.7

27 𝑓 ′′ (𝑡) = (ln 5)2 5𝑡+1 ; 𝑓 ′′′ (𝑡) = (ln 5)3 5𝑡+1

39 Approx −21.2

3 56𝑥(4𝑥2 + 1)6 √ 5 𝑒𝑥 ∕(2 𝑒𝑥 + 1)

𝑥

23 𝑎 ln 𝑎 + 𝑎𝑥

𝑎−1

29 2𝑥 + (ln 2)2𝑥

45 𝑦 = 5𝑥

33 𝑒𝑥+5

47 𝑦 = 7𝑥 − 5

35 Rules do not apply

13 𝜋𝑒𝜋𝑥

49 4𝑥 (ln 4 ⋅ 𝑓 (𝑥) + ln 4 ⋅ 𝑔(𝑥) + 𝑓 ′ (𝑥) + 𝑔 ′ (𝑥))

37 Rules do not apply

+ 𝑓 (𝑥)𝑔 ′ (𝑥)ℎ(𝑥) 51 (𝑓 ′ (𝑥)𝑔(𝑥)ℎ(𝑥) 𝑓 (𝑥)𝑔(𝑥)ℎ′ (𝑥))∕(ℎ(𝑥))2

39 Rules do not apply 41 16.021; pop incr ≈ 16 animals/yr when 𝑡 = 5 43 1013.454 mb; increasing at 2.261 mb/hr ) ( 45 (a) 17.9 ⋅ ln 1.025 1.025𝑡 m. tonnes/yr (b) 0.580 m. tonnes/year (c) 2.9 m. tonnes (d) Smaller 47 (a) 1.247 million people per year (b) 2.813 million people per year US, since 𝑑𝑈 ∕𝑑𝑡 > 𝑑𝑀∕𝑑𝑡 at 𝑡 = 0 49 𝑔(𝑥) = 𝑥2 ∕2 + 𝑥 + 1

𝑒𝑥 1 2 𝑥 2

𝑥 51

7 5(𝑤4 − 2𝑤)4 (4𝑤3 − 2) √ 9 2𝑟3 ∕ 𝑟4 + 1 [ ] 11 𝑒2𝑥 2𝑥2 + 2𝑥 + (ln 5 + 2)5𝑥

43 𝑥 > −2

31 Rules do not apply

𝑓 ′ (𝑥)

41 𝑓 ′ (𝑥) = 2𝑒2𝑥

=0

+𝑥+1

15 −200𝑥𝑒−𝑥 −

53 (a) 19 (b) −11 55 𝑓 (𝑥) = 𝑥10 𝑒𝑥 57 (a) 𝑓 (140) = 15,000: If the cost $140 per board then 15,000 skateboards are sold 𝑓 ′ (140) = −100: Every dollar of increase from $140 will decrease the total sales by about 100 boards (b) 𝑑𝑅∕𝑑𝑝|𝑝=140 = 1000 (c) Positive Increase by about $1000 59 (a) 𝑔(𝑣) = 1∕𝑓 (𝑣) 𝑔(80) = 20 km∕liter 𝑔 ′ (80) = −(1∕5) km∕liter for each 1 km∕hr increase in speed (b) ℎ(𝑣) = 𝑣 ⋅ 𝑓 (𝑣) ℎ(80) = 4 liters∕hr ℎ′ (80) = 0.09 liters∕hr for each 1 km∕hr inc. in speed 61 (a) 𝑃 (𝑡) = 𝐴(𝑡)∕𝑁(𝑡) (b) Purchase more shares to offset loss (c) Purchase 28,750 additional shares

53 𝑓 (𝑥) = 𝑒𝑥 55 False 59 (a) 𝑓 ′ (0) = −1 (b) 𝑦 = −𝑥 (c) 𝑦 = 𝑥

63 0 67 Product rule gives two terms

63 0 < 𝑎 < 1 and 𝐶 < 0

69 Write 𝑓 (𝑥) = 𝑒𝑥 (𝑥 + 1) and apply the product rule

65 𝑎 > 1 and 𝐶 < 0

71 𝑓 (𝑥) = 𝑥2 = 𝑥 ⋅ 𝑥 73 True

Section 3.3 1 5𝑥4 + 10𝑥 𝑥

3 𝑒 (𝑥 + 1) √ √ 5 2𝑥 ∕(2 𝑥) + 𝑥(ln 2)2𝑥 1

7 3𝑥 [(ln 3)(𝑥2 √ − 𝑥2 )+ (2𝑥 − 1∕(2 𝑥))]

75 (a) (b) (c) (d)

19 𝑒5−2𝑡 (1 − 2𝑡) 21 (2𝑡 − 𝑐𝑡2 )𝑒−𝑐𝑡 √ √ 23 (𝑒 𝑠 )∕(2 𝑠) √ 25 3𝑠2 ∕(2 𝑠3 + 1) √ √ 27 (𝑒−𝑧 )∕(2 𝑧) − 𝑧𝑒−𝑧 29 5 ⋅ ln 2 ⋅ 25𝑡−3

𝑦

5

31 −(ln 10)(10 2 − 2 )∕2 √ 33 (1 − 2𝑧 ln 2)∕(2𝑧+1 𝑧) √ 𝑥 + 3(𝑥2 + 6𝑥 − 9) 35 √ 2 𝑥2 + 9(𝑥 + 3)2

37 −(3𝑒3𝑥 + 2𝑥)∕(𝑒3𝑥 + 𝑥2 )2 39 −1.5𝑥2 (𝑥3 + 1)−1.5 41 (2𝑡 + 3)(1 − 𝑒−2𝑡 ) + (𝑡2 + 3𝑡)(2𝑒−2𝑡 )

43 30𝑒5𝑥 − 2𝑥𝑒−𝑥

2

2 2𝑤𝑒𝑤 (5𝑤2

+ 8) √ 2 47 −3𝑡𝑒−3𝑡 ∕ 𝑒−3𝑡2 + 5

45

(𝑦2 ) +𝑦2 ]

49 2𝑦𝑒[𝑒

51 6𝑎𝑥(𝑎𝑥2 + 𝑏)2

65 0

61 𝑒

17 (ln 𝜋)𝜋

2

(𝑥+2)

Not a counterexample Not a counterexample Not a counterexample Counterexample

77 False, choose 𝑔(2) = −1∕0.7 and ℎ(2) = 2 79 True

53 𝑎𝑒−𝑏𝑥 − 𝑎𝑏𝑥𝑒−𝑏𝑥 −𝑐𝑥

55 𝑎𝑏𝑐𝑒−𝑐𝑥 𝑒−𝑏𝑒 ( )2 ( 3 )( ) 3𝑥 − 2 6𝑥3 + 15𝑥 − 2 57 6𝑥 𝑥2 + 5

59 (a) −2 (b) Chain rule does not apply (c) 2 61 (a) 1 (b) 1 (c) 1 63 0 65 1∕2 67 𝑦 = 4451.66𝑥 − 3560.81

1074 √ √ 69 −1∕ 2 < 𝑥 < 1∕ 2

71 𝑓 ′ (𝑥) = [(2𝑥 + 1)9 (3𝑥 − 1)6 ] × (102𝑥 + 1) 𝑓 ′′ (𝑥) = [9(2𝑥 + 1)8 (2)(3𝑥 − 1)6 + (2𝑥 + 1)9 (6)(3𝑥 − 1)5 (3)] × (102𝑥 + 1) + (2𝑥 + 1)9 (3𝑥 − 1)6 (102) 73 (a) (b) (c) (d) (e)

31 −3𝑒−3𝜃 ∕ cos2 (𝑒−3𝜃 )

√ 5 1∕ 1 − (𝑥 + 1)2

33 −2𝑒2𝑥 sin(𝑒2𝑥 )

9 2

29 5 sin4 𝜃 cos 𝜃

𝐻(4) = 1 𝐻 ′ (4) = 30 𝐻(4) = 4 𝐻 ′ (4) = 56 𝐻 ′ (4) = −1

7 (6𝑥 + 15)∕(𝑥2 + 5𝑥 + 3)

11 𝑒−𝑥 ∕(1 − 𝑒−𝑥 )

35 − sin 𝛼 + 3 cos 𝛼 2

3

37 3𝜃 cos 𝜃 − 𝜃 sin 𝜃

13 𝑒𝑥 ∕(𝑒𝑥 + 1)

39 cos(cos 𝑥 + sin 𝑥) ⋅ (cos 𝑥 − sin 𝑥)

15 𝑎𝑒𝑎𝑥 ∕ (𝑒𝑎𝑥 + 𝑏) 17 3𝑤2 ln(10𝑤) + 𝑤2

41 (−𝑡 sin 𝑡 − 3 cos 𝑡)∕𝑡4 √ 1 − cos 𝑥(1 − cos 𝑥 − sin 𝑥) 43 √ 2 1 − sin 𝑥(1 − cos 𝑥)2

19 𝑒 21 1∕𝑡 23 −1∕(1 + (2 − 𝑥)2 ) 2

45 (6 sin 𝑥 cos 𝑥)∕(cos2 𝑥 + 1)2

75 (a) 𝑔 ′ (2) = 42 (b) ℎ′ (2) = −8

25 𝑒arctan(3𝑡 ) (6𝑡)∕(1 + 9𝑡4 )

47 −𝑎𝑏 sin(𝑏𝑡 + 𝑐)

27

49 2𝑥3 𝑒5𝑥 cos(2𝑥) 3𝑥2 𝑒5𝑥 sin(2𝑥)

77 (a) 13,394 fish (b) 8037 fish/month 79 𝑓 (3) = 2.51 billion dollars; 𝑓 ′ (3) = 0.301 billion dollars of 4th quarter sales per year 81 (a) $5000; 2% compounded continuously (b) 𝑓 (10) = 6107.01 dollars; 𝑓 ′ (10) = 122.14 dollars per year

5𝑥3 𝑒5𝑥 sin(2𝑥)

+

+

51 𝑓 ′′ (𝜃) = −(𝜃 cos 𝜃 + 2 sin 𝜃); 𝑓 ′′′ (𝜃) = 𝜃 sin 𝜃 − 3 cos 𝜃 55 𝑦 = 3𝑥 + 1 57

= − cos 𝑥

83 𝑏 = 1∕40 and 𝑎 = 169.36

59 (a) 0 (b) sin2 𝑥 + cos2 𝑥 = 1, so 𝑓 ′ (𝑥) = 0

85 (a) 𝑃 (1 + 𝑟∕100)𝑡 ln(1 + 𝑟∕100) (b) 𝑃 𝑡(1 + 𝑟∕100)𝑡−1 ∕100

61 (a) 𝑣(𝑡) = 2𝜋 cos(2𝜋𝑡) (b) 𝑦

𝑡 1

2

3

91 Increasing 𝑥

𝑥

𝑣

4

93 𝑔 (𝑥) = 5𝑒 (𝑒 + 2)

2𝜋

95 Not necessarily equal to 0 97 𝑓 (𝑥) =

(𝑥2

+

101 False; 𝑓 (𝑥) =

𝑣 = 2𝜋 cos 2𝜋𝑡 𝑡

1)2

99 False; 𝑓 (𝑥) = 6, 𝑔(𝑥) = 10 𝑒−𝑥 , 𝑔(𝑥)

=

𝑥2

1

2

3

−2𝜋

111 (a) 𝑔 ′ (1) = 3∕4 (b) ℎ′ (1) = 3∕2

63 (a) 𝑡 = (𝜋∕2)(𝑚∕𝑘)1∕2 ; 𝑡 = 0; 𝑡 = (3𝜋∕2)(𝑚∕𝑘)1∕2 (b) 𝑇 = 2𝜋(𝑚∕𝑘)1∕2 √ (c) 𝑑𝑇 ∕𝑑𝑚 = 𝜋∕ 𝑘𝑚; Positive sign means an increase in mass causes the period to increase

115 𝑓 ′′ (𝑥)(𝑔(𝑥))−1 − 2𝑓 ′ (𝑥)(𝑔(𝑥))−2 𝑔 ′ (𝑥) + 2𝑓 (𝑥)(𝑔(𝑥))−3 (𝑔 ′ (𝑥))2 − 𝑓 (𝑥)(𝑔(𝑥))−2 𝑔 ′′ (𝑥)

65 (a) 0 ≤ 𝑡 ≤ 2 (b) No, not at 𝑡 = 2

109 (a) 4 (b) 2 (c) 1∕2

67 CO2 increasing at 1.999 ppm/month on Dec 1, 2008

Section 3.5 3 cos2 𝜃 − sin2 𝜃 = cos 2𝜃

37 3𝑤−1∕2 − 2𝑤−3 + 5∕𝑤 √ 39 −(𝑥 + 1)∕( 1 − (𝑥 + 1)2 ) 41 1∕(1 + 2𝑢 + 2𝑢2 ) 43 −1 < 𝑥 < 1 𝑑 (log 𝑥) 𝑑𝑥

=

1 (ln 10)𝑥

49 (a) 𝑓 ′ (𝑥) = 32.7∕𝑥 (b) 0.01635 mm of leaf width per mm of rain (c) 2.4525 mm 51 (a) 𝑓 ′ (𝑥) = 0 (b) 𝑓 is a constant function 53 (a) (b) (c) (d) (e)

10𝑥4 + 9𝑥2 + 1 𝑓 ′ > 0 so 𝑓 increasing everywhere 6 20 1∕20

55 Any 𝑥 with 25 < 𝑥 < 50 57 Any 𝑥 with 75 < 𝑥 < 100 59 2.8

103 Zero; decreasing 105 Positive; positive

33 −1∕𝑧(ln 𝑧)2

45

𝑦 = 15 + sin 2𝜋𝑡

89 Decreasing ′

√ 31 −𝑥∕ 1 − 𝑥2

47 𝑔(5000) = 32.189 years 𝑔 ′ (5000) = 0.004 years per dollar

16 15 14

87 (a) 𝑓 (𝑡): linear 𝑔(𝑡): exponential ℎ(𝑡): quadratic polynomial (b) 1.3 ppm/yr; 1.451 ppm/yr; 2.133 ppm/yr (c) Linear < Exp < Quad (d) No; Exponential eventually largest

29 𝑘

35 (cos 𝑥 − sin 𝑥)∕(sin 𝑥 + cos 𝑥)

53 Decreasing, concave up 𝑑 50 𝑦∕𝑑𝑥50

(ln 2)𝑧(ln 2−1)

69 CO2 decreasing at 1.666 ppm/month on June 1, 2008

61 1.4 63 −0.12 65 1∕5 67 (a) (b) (c) (d)

Pop is 319 m in 2014 2014 Pop incr by 2.44 m∕yr 0.41 yr/million

69 1∕3 71 (a) 1 (b) 1 (c) 1 73 (𝑓 −1 )′ does not change sign 75 𝑤′ (𝑥) = 4𝑥3 ∕(1 + 𝑥4 ) 77 Formula for (𝑓 −1 )′ incorrect

7 −8 sin(2𝑡)

71 (a) 𝑑 ′ (𝑡) = −2𝑒−𝑡 sin 𝑡 (b) 𝑡 = 0, 𝜋, 2𝜋, 3𝜋, … (c) 0; the wave stops

9 3𝜋 sin(𝜋𝑥)

75 Cannot use product rule

81 𝑓 (𝑥) = 𝑥

5 3 cos(3𝑥)

79 𝑦 = 𝑐 ln 𝑥 for constant 𝑐.

11 3𝜋 cos(𝜋𝑡)(2 + sin(𝜋𝑡))2

77 sin 𝑥

83 False; 𝑓 (𝑥) = 𝑥3

13 𝑒𝑡 cos(𝑒𝑡 )

79 True

85 (b)

15 (cos 𝑦)𝑒sin 𝑦

81 (sin 𝑥 + 𝑥 cos 𝑥)𝑥 sin 𝑥

17 3 cos(3𝜃)𝑒sin(3𝜃)

83 2 sin(𝑥4 ) + 8𝑥4 cos(𝑥4 ) √ 85 𝑘 = 7.46, (3𝜋∕4, 1∕ 2)

19

2𝑥∕ cos2 (𝑥2 )

21 4 cos(8𝑥)(3 + sin(8𝑥))−0.5 23 cos 𝑥∕ cos2 (sin 𝑥)

Section 3.6

Section 3.7 1 𝑑𝑦∕𝑑𝑥 = −𝑥∕𝑦 3 𝑑𝑦∕𝑑𝑥 = (𝑦2 − 𝑦 − 2𝑥)∕(𝑥 − 3𝑦2 − 2𝑥𝑦)

25 2 sin(3𝑥) + 6𝑥 cos(3𝑥)

1 2𝑡∕(𝑡2 + 1)

5 −(1 + 𝑦)∕(1 + 𝑥)

27 𝑒−2𝑥 [cos 𝑥 − 2 sin 𝑥]

3 10𝑥∕(5𝑥2 + 3)

7 𝑑𝑦∕𝑑𝑥 = (𝑦 − 2𝑥𝑦3 )∕(3𝑥2 𝑦2 − 𝑥)

1075 √ 9 𝑑𝑦∕𝑑𝑥 = − 𝑦∕𝑥

(b) 𝐴 = 6.325 (stretch factor) 𝑐 = 0.458 (horizontal shift)

11 −3𝑥∕2𝑦 13 −𝑦∕(2𝑥) 15 𝑑𝑦∕𝑑𝑥 = (2 − 𝑦 cos(𝑥𝑦))∕(𝑥 cos(𝑥𝑦)) 17 (𝑦2 + 𝑥4 𝑦4 − 2𝑥𝑦)∕(𝑥2 − 2𝑥𝑦 − 2𝑥5 𝑦3 ) 19 (𝑎 − 𝑥)∕𝑦 21 (𝑦 + 𝑏 sin(𝑏𝑥))∕ (𝑎 cos(𝑎𝑦) − 𝑥)

(b) 𝑦 = 3𝑥 − 3 17 𝑎 = 1; 𝑓 (𝑎) = 1 Underestimate 𝑓 (1.2) ≈ 1.4

33 (a) 0 (b) Positive for 𝑥 > 0 Negative for 𝑥 < 0 Zero for 𝑥 = 0 (c) Increasing everywhere (d) 1, −1

19 0.1 21 (a) 2.5 (b) 7.25 (c) Underestimate 23 331.3 + 0.606𝑇 m∕sec

23 Slope is infinite

𝑦

25 −23∕9

1

27 𝑦 = 𝑒2 𝑥 29 𝑦 = 𝑥∕𝑎

𝑥

31 (a) (4 − 2𝑥)∕(2𝑦 + 7) (b) Horizontal if 𝑥 = 2, Vertical if 𝑦 = −7∕2

−5

5 −1

33 (a) (1, −1) and (1, 0). (b) 𝑑𝑦∕𝑑𝑥 = −(𝑦 + 2𝑥)∕(𝑥 + 2𝑦) (c) 1 at (1, −1) and −2 at (1, 0)

(e) Yes; derivative positive everywhere 37 tanh 𝑥 → 1 as 𝑥 → ∞ 39 𝑘 = 0

35 (a) 16,398 m (b) 16,398 + 682(𝜃 − 20) m (c) True: 17,070 m Approx: 17,080 m

45 False

Section 3.9 √

43 𝑑𝑦∕𝑑𝑥 = (𝑥2 − 4)∕(𝑦 − 2)

1

45 True

37 (a) 1492 m (b) 1492 + 143(𝜃 − 20) m (c) True: 1638 m Approx: 1635 m

1 + 𝑥 ≈ 1 + 𝑥∕2

3 1∕𝑥 ≈ 2 − 𝑥

√ √ 47 (−1∕3, 2 2∕3); (7∕3, 4 2∕3)

2

5 𝑒𝑥 ≈ 2𝑒𝑥 − 𝑒

39 𝑓 (𝑥) ≈ 1 + 𝑘𝑥 𝑒0.3 ≈ 1.3

9 |Error| < 0.2 Overestimate, 𝑥 > 0 Underestimate, 𝑥 < 0

2 1 𝑥 −1

12 3 4

5

41 𝐸(𝑥) = 𝑥4 − (1 + 4(𝑥 − 1)) 𝑘 = 6; 𝑓 ′′ (1) = 12 𝐸(𝑥) ≈ 6(𝑥 − 1)2

11 (a) 𝐿(𝑥) = 1 + 𝑥 (b) Positive for 𝑥 ≠ 0 (c) 0.718

43 𝐸(𝑥) = 𝑒𝑥 − (1 + 𝑥) 𝑘 = 1∕2; 𝑓 ′′ (0) = 1 𝐸(𝑥) ≈ (1∕2)𝑥2

−1 −2

Section 3.8

𝑓 (𝑥) = 𝑒𝑥

1 3 cosh(3𝑧 + 5) 3 2 cosh 𝑡 ⋅ sinh 𝑡

−1

5 3𝑡2 sinh 𝑡 + 𝑡3 cosh 𝑡





𝐸(1)

✻ ✻𝑓 (1)

45 𝐸(𝑥) = ln 𝑥 − (𝑥 − 1) 𝑘 = −1∕2; 𝑓 ′′ (1) = −1 𝐸(𝑥) ≈ −(1∕2)(𝑥 − 1)2



𝑦

𝐿(1) 1 𝐿(𝑥) = 1 + 𝑥 ❄❄ 𝑥 1

47 (c) 0 53 Only near 𝑥 = 0 55 𝑓 (𝑥) = 𝑥3 + 1, 𝑔(𝑥) = 𝑥4 + 1 57 𝑓 (𝑥) = |𝑥 + 1|

7 18∕ cosh2 (12 + 18𝑥) 9 tanh(1 + 𝜃)

(d) 𝐸(1) larger (e) 0.005

11 0 15 (𝑡2 + 1)∕2𝑡 19 sinh(2𝑥) = 2 sinh 𝑥 cosh 𝑥 23 0

13 (a) 99.5 + 0.2(𝑥 − 50) (b) 52.5 (c) Approx straight near 𝑥 = 50

Section 3.10 1 False 3 False

15 (a) and (c)

25 1

5 True

27 |𝑘| ≤ 3

𝑦

31 (a)

7 No; no

𝑦

29 (a) 0.54 𝑇 ∕𝑤

𝑦 = 2𝑒𝑥 + 5𝑒−𝑥

40

𝑓 (𝑥) = 3 True value Approximation

𝑥3

− 3𝑥2

9 No; no

+ 3𝑥 + 1



11 𝑓 ′ (𝑐) = −0.5, 𝑓 ′ (𝑥1 ) > −0.5, 𝑓 ′ (𝑥2 ) < −0.5 Error

𝑦 = 3𝑥 − 3

20

2 (0.5, 6.3) −3

10.034 mn users/yr 5.7 mn users/yr 𝑡 = 1.6; mid 2011 Rates of change constant

33 𝑓 (1 + Δ𝑥) ≥ 𝑓 (1) + 𝑓 ′ (1)Δ𝑥

43 True

41 Need to use implicit differentiation

27 (a) (b) (c) (d)

31 (b) 1% increase

41 True

39 (a) −(1∕2𝑃 )𝑓 (1 − 𝑓 2 )

23.681𝐿−0.421 23.681 days per cm 4.736 days 40.9 + 23.681(𝐿 − 1) days

29 (a) 9.14, 4.21, 1.51 (b) −134.16𝑠−2.118 (c) −0.0338 attacks per bug per hour per number of bugs (d) 𝑎 ≈ 1.51 − 0.0338(𝑠 − 50) attacks per bug per hour (e) Linear: −0.169, Power function: −0.166

35 𝑓 ′ (𝑥) = sinh 𝑥

35 (a) 𝑦 − 3 = −4(𝑥 − 4)∕3 and 𝑦 + 3 = 4(𝑥 − 4)∕3 (b) 𝑦 = 3𝑥∕4 and 𝑦 = −3𝑥∕4 (c) (0, 0)

25 (a) (b) (c) (d)

3

𝑥

13 6 distinct zeros 19 Racetrack 21 Constant Function 23 21 ≤ 𝑓 (2) ≤ 25 31 Not continuous

𝑥 −3

33 1 ≤ 𝑥 ≤ 2

1076 35 Possible answer: 𝑓 (𝑥) = |𝑥| 37 Possible{ answer 𝑥2 𝑓 (𝑥) = 1∕2

𝑓 (𝑥)

5

if 0 ≤ 𝑥 < 1

3

𝑥

−1

Critical point Not local max or min

if 𝑥 = 1

39 False Local min

41 False

𝑥 83 4 √ √ 5 Critical points: 𝑥 = 0, 𝑥 = − √ 6, 𝑥 = √ 6 Inflection points: 𝑥 = 0, 𝑥 = − 3, 𝑥 = 3

85 𝑥 ≈ 1, −0.4, or 2.4 87 𝑦 = −4𝑥 + 6

Chapter 3 Review

91 Approx 1.9

3 (𝑡2 + 2𝑡 + 2)∕(𝑡 + 1)2 7 𝑥2 ln 𝑥 9 (cos 𝜃)𝑒sin 𝜃 11 − tan(𝑤 − 1)

95 (a) (b) (c) (d) (e) (f)

13 𝑘𝑥𝑘−1 + 𝑘𝑥 ln 𝑘 15 3 sin2 𝜃 cos 𝜃 17 6 tan(2 + 3𝛼) cos−2 (2 + 3𝛼) 19 (−𝑒−𝑡 − 1)∕(𝑒−𝑡 − 𝑡) 21 1∕ sin2 𝜃 − 2𝜃 cos 𝜃∕ sin3 𝜃

31

33 𝑒(tan 2 + tan 𝑟)𝑒−1 ∕ cos2 𝑟 35 2𝑒2𝑥 sin(3𝑥) (sin(3𝑥) + 3 cos(3𝑥)) ) ( 37 2sin 𝑥 (ln 2) cos2 𝑥 − sin 𝑥

107 (a) 9.801 million (b) 0.0196 million/year 109 (a) 𝑣(𝑡) = 10𝑒𝑡∕2 (b) 𝑣(𝑡) = 𝑠(𝑡)∕2

39 𝑒𝜃−1

41 (−𝑐𝑎𝑡2 + 2𝑎𝑡 − 𝑏𝑐)𝑒−𝑐𝑡 43 (ln 5)5𝑥 47 −2∕(𝑥2 + 4) 2 3

49 20𝑤∕(𝑎 − 𝑤 )

113 (a) (b) (c) (d)

51 𝑎𝑒𝑎𝑢 ∕(𝑎2 + 𝑏2 ) 53 ln 𝑥∕(1 + ln 𝑥)2 √ √ 𝑒𝑡 + 1∕(2 𝑒𝑡 + 1)

55 𝑒𝑡 cos

61 3𝑥2 + 3𝑥 ln 3 63 − 52

sin(5𝜃) √ cos(5𝜃)

0.33268∕𝑘 years 30.24 years 30.24 − 2749.42(𝑘 − 0.011) years Exponential: 33.3 years Approx 32.99 years

+ 12 sin(6𝜃) cos(6𝜃)

√ 5 −3∕2 𝑧 2

75 𝑑𝑦∕𝑑𝑥 = 0 77 𝑑𝑦∕𝑑𝑥 = −3 ′

5 −1∕2 𝑧 2

5 −3∕2 𝑧 2

+

1 local max

81 𝑥 = −1 and 𝑥 = 5

𝑓 ′ (𝑥)





local max



local min

𝑥

✻ neither max nor min

inflection points of 𝑓𝑓 ′′ (𝑥)

☛ ❯

Section 4.1

2

79 𝑓 (𝑡) = 6𝑡 − 8𝑡 + 3 𝑓 ′′ (𝑡) = 12𝑡 − 8

✻ local max 37



local min



71 −16 − 12𝑥 + 48𝑥2 − 32𝑥3 + 28𝑥6 − 9𝑥8 + 20𝑥9 +

35

125 (a) 0 (b) 0 (c) ln(1 − 1∕𝑡) + ln(𝑡∕(𝑡 − 1)) = ln 1

69 𝑓 ′ (𝑡) = 4(sin(2𝑡) − cos(3𝑡))3 (2 cos(2𝑡) + 3 sin(3𝑡))

5 (5𝑧)−1∕2 2

27 (a) Increasing for 0 < 𝑥 < 4 Decreasing for 𝑥 < 0 and 𝑥 > 4 (b) Local max: 𝑓 (4) Local min: 𝑓 (0) √ 29 (a) ± 𝑎∕3 (b) 𝑎 = 12

123 (a) 1 (b) 1 (c) sin(arcsin 𝑥) = 𝑥

67 𝑘𝑒𝑘𝑡 (sin 𝑎𝑡 + cos 𝑏𝑡)+ 𝑒𝑘𝑡 (𝑎 cos 𝑎𝑡 − 𝑏 sin 𝑏𝑡)

73 𝑓 ′ (𝑧) =

21 (a) Critical point 𝑥 ≈ 0; Inflection points between −1 and 0 and between 0 and 1 (b) Critical point at 𝑥 = 0, √ Inflection points at 𝑥 = ±1∕ 2

33 (a) 𝑓 ′ (𝑥) = 0 has no solutions (b) 𝑓 ′ (𝑥) > 0, 𝑓 ′′ (𝑥) < 0

121 (𝑓 ∕𝑔)′ ∕(𝑓 ∕𝑔) = (𝑓 ′ ∕𝑓 ) − (𝑔 ′ ∕𝑔)

65 4𝑠3 − 1

+ 2𝑛𝜋,

+ 2𝑛𝜋

31 (a) 𝑥 = 𝑏 (b) Local minimum

119 (a) 𝑦(𝑛) = (−1)𝑛+1 (𝑛 − 1)!𝑥−𝑛 (b) 𝑦(𝑛) = 𝑥𝑒𝑥 + 𝑛𝑒𝑥

𝑧

𝜋 3

23 𝑓 ′ (2) < 𝑓 ′ (6)

115 1∕6

57 18𝑥2 + 8𝑥 − 2

min: 𝑥 =

5𝜋 3

25 (a) Increasing for 𝑥 > 0 Decreasing for 𝑥 < 0 (b) Local and global min: 𝑓 (0)

111 (a) 𝑣 = −2𝜋𝜔𝑦0 sin(2𝜋𝜔𝑡) 𝑎 = −4𝜋 2 𝜔2 𝑦0 cos(2𝜋𝜔𝑡) (b) Amplitudes: different (𝑦0 , 2𝜋𝜔𝑦0 , 4𝜋 2 𝜔2 𝑦0 ) Periods = 1∕𝜔

45 (2𝑎𝑏𝑟 − 𝑎𝑟4 )∕(𝑏 + 𝑟3 )2

59 (2 ln 3)𝑧 + (ln 4)𝑒

19 Crit pts: max: 𝑥 =

105 (a) 2, 1, −3 (b) 𝑓 (2.1) ≈ 0.7, overestimate 𝑓 (1.98) ≈ 1.06, underestimate Second better

cos 𝛼∕ cos2 (sin 𝛼)

2

17 Critical point: 𝑥 = 1∕3, local maximum

103 1

(𝑒𝜃 − 𝑒−𝜃 )

13 Critical points: 𝑥 = 0, 𝑥 = 4 𝑥 = 0: local minimum 𝑥 = 4: local maximum 15 Critical point: 𝑥 = 0 𝑥 = 0: local maximum

99 Not perpendicular; 𝑥 ≈ 1.3 101 0

27 2−4𝑧 [−4 ln(2) sin(𝜋𝑧) + 𝜋 cos(𝜋𝑧)] 𝑒tan(sin 𝛼)

11 Critical points: 𝑥 = 0 and 𝑥 = 2 Extrema: 𝑓 (2) local minimum 𝑓 (0) not a local extremum.

𝑦 = 20𝑥 − 48 𝑦 = 11𝑥∕9 − 16∕9 𝑦 = −4𝑥 + 20 𝑦 = −24𝑥 + 57 𝑦 = 8.06𝑥 − 15.84 𝑦 = −0.94𝑥 + 6.27

97 1.909 radians (109.4◦ ) or 1.231 radians (70.5◦ )

23 −(2𝑤 ln 2 + 𝑒𝑤 )∕(2𝑤 + 𝑒𝑤 )2 √ √ 25 1∕( sin(2𝑧) cos3 (2𝑧)) 𝜃 +𝑒−𝜃 )

9 Critical points: 𝑥 = 0 and 𝑥 = 1 Extrema: 𝑓 (1) local minimum 𝑓 (0) not a local extremum

93 (a) 𝐻 ′ (2) = 11 (b) 𝐻 ′ (2) = −1∕4 (c) 𝐻 ′ (2) = 𝑟′ (1) ⋅ 3 (we don’t know 𝑟′ (1)) (d) 𝐻 ′ (2) = −3

5 −8∕(4 + 𝑡)2

29 𝑒(𝑒

7 Critical point: 𝑥 = 3∕5 No inflection points

89 Approx 1

1 200𝑡(𝑡2 + 1)99

39 (a) 𝐷, 𝐺 (b) 𝐷, 𝐹 (c) One; Zero

𝑥

1077 41

𝑦′′ = 0

𝑦′′ = 0

37 (a) Ordering: 𝑎∕𝑞 Storage: 𝑏𝑞 √ (b) 𝑎∕𝑏

81 Impossible

𝑦′ < 0 everywhere

𝑦

83 Impossible

𝑦′′ = 0

85 True

39 (a) 0 ≤ 𝑦 ≤ 𝑎

87 False 89 Yes 91 Not enough information. 𝑦′′ > 0

𝑦′′ < 0

𝑦′′ > 0

𝑦′′ < 0

93 (a) 𝑥 = ±2 √ (b) 𝑥 = 0, ± 5

rate (gm/sec) Max rate

95 (a) No critical points (b) Decreasing everywhere

𝑦 = 𝑓 (𝑥) 𝑥 𝑥1

𝑥2

𝑥3

97 Impossible

Section 4.2 𝑦

1 43

Global and local max

8 6 4 2

𝑦 𝑦 = 𝑓 (𝑥)

𝑎∕2

Local min

𝑥

𝑦′ > 0 𝑦′′ > 0

2

3

4

5

43 𝑥 = 𝐿∕2

𝑦

3 (a)

45 (−1, −1∕2); (1, 1∕2) 47 (b) Yes, at 𝑥 = 0 (c) Max: 𝑥 = −2, Min: 𝑥 = 2 (d) 5 > 𝑔(0) > 𝑔(2)

𝑥

𝑥

4

𝑥1 (b) 𝑥 = 4, 𝑦 = 57 45 (a) Local maxima at 𝑥 ≈ 1, 𝑥 ≈ 8, Local minimum at 𝑥 ≈ 4 (b) Local maxima at 𝑥 ≈ 2.5, 𝑥 ≈ 9.5 Local minimum at 𝑥 ≈ 6.5,

5 Max: 9 at 𝑥 = −3; Min: −16 at 𝑥 = −2

47depth of water

9 Max: 8 at 𝑥 = 4; Min: −1 at 𝑥 = −1, 1



time

55 𝑓 (𝑥) = 1 − 𝑥 √ √ 2≤𝑥≤ 5 57

7 Max: 2 at 𝑥 = 1; Min: −2 at 𝑥 = −1, 8

59 True 61 True

11 (a) 𝑓 (1) local minimum; 𝑓 (0), 𝑓 (2) local maxima (b) 𝑓 (1) global minimum 𝑓 (2) global maximum

63 True

13 (a) 𝑓 (2𝜋∕3) local maximum 𝑓 (0) and 𝑓 (𝜋) local minima (b) 𝑓 (2𝜋∕3) global maximum 𝑓 (0) global minimum

69 True

65 False 67 True 71 (a) 𝑥 = 𝑎∕2 (b) 𝑥 = 𝑎∕2

15 Global min = 2 at 𝑥 = 1 No global max

51 (a) Positive

19 Global max = 2 at 𝑡 = 0, ±2𝜋, … Global min = −2 at 𝑡 = ±𝜋, ±3𝜋, …

55 (a) 0 (b) Two (c) Local maximum

49 (a) 𝑓 is always increasing; 𝑔 is always decreasing (b) 5 ) (∑𝑛 51 𝑥 = 𝑖=1 𝑎𝑖 ∕𝑛

53 On 1 ≤ 𝑥 ≤ 2, global minimum at 𝑥 = 1

49 𝑥 = 0, local min if 𝑚 even, inflect if 𝑚 odd; 𝑥 = 𝑚∕(𝑚 + 𝑛), local max; 𝑥 = 1, local min if 𝑛 even, inflect if 𝑛 odd 53 𝑎 = −1∕3

(b) 𝑦 = 𝑎∕2 41 𝑟 = 3𝐵∕(2𝐴)

57 50

time at which water reaches widest part of urn

Section 4.3

17 Global min = 1 at 𝑥 = 1 No global max

1 2500 3 9000 5 Square of side 16 cm 7 𝑥 = ℎ = 81∕3 cm

21 Global min −4∕27 at 𝑥 = ln(2∕3)

9 𝑟 = (4∕𝜋)1∕3 cm, ℎ = 2(4∕𝜋)1∕3 cm

23 0.91 < 𝑦 ≤ 1.00 25 0 ≤ 𝑦 ≤ 2𝜋

11 11.270

59 𝐵 = 𝑓 , 𝐴 = 𝑓 ′ , 𝐶 = 𝑓 ′′

27 0 ≤ 𝑦 < 1.61

13 (±1, 0) , (±1, 1∕2)

61 III even; I, II odd I is 𝑓 ′′ , II is 𝑓 , III is 𝑓 ′

29 Yes, at 𝐺; No

15 (a) 1∕(2𝑒) (b) (ln 2) + 1

63 Consider 𝑓 (𝑥) = 𝑥3

31 𝑥 = −𝑏∕2𝑎, Max if 𝑎 < 0, min if 𝑎 > 0

65 Consider 𝑓 (𝑥) = 𝑥4

33 𝑇 = 𝑆∕2

19 1∕2

67 𝑓 (𝑥) = 𝑥

35 𝑡 = 2∕𝑏

21 (a) 𝑥𝑦 + 𝜋𝑦2 ∕8 (b) 2𝑥 + 𝑦 + 𝜋𝑦∕2 (c) 𝑥 = 100∕(4 + 𝜋), 𝑦 = 200∕(4 + 𝜋)

69 𝑓 (𝑥) = cos 𝑥 or 𝑓 (𝑥) = sin 𝑥 71 True

17 When the rectangle is a square

𝑦

23 (a) 𝑥𝑦 + 𝜋𝑦2 ∕4 + 𝜋𝑥2 ∕4 (b) 𝜋𝑥 + 𝜋𝑦 (c) 𝑥 = 0, 𝑦 = 100∕𝜋; 𝑥 = 100∕𝜋, 𝑦 = 0

73 False 75 False

25 0, 10

77 True 79 True

𝑦 (gm)

Global and local min

1 𝑦′ = 2 𝑦′′ = 0

𝑎

𝑡

√ √ 27 10 + 5 3, 10 5

1078 √ 29 (1∕2, 1∕ 2) √ 31 (−2.5, 2.5); Minimum distance is 2.958

5 (a)

15 (a) large 𝑎

33 Radius 3.854 cm Height 7.708 cm Volume 359.721 cm3





small 𝑎

large 𝑎





𝑥 𝑥

35 13.13 mi from first smokestack √ 37 Min 𝑣 = 2𝑘; no max

(b) Nonzero critical point moves down to the left (c) 𝑥 = 0, 2∕𝑎

39 Minimum: 0.148𝑚𝑔 newtons Maximum: 1.0𝑚𝑔 newtons

41 Maximum revenue = $27,225 Minimum = $0 43 (a) (b) (c) (d)

(d) 𝑥 = 𝑏∕3, 𝑏

small 𝑎

(b)

small 𝑏



7 𝐴 has 𝑎 = 1, 𝐵 has 𝑎 = 2, 𝐶 has 𝑎 = 5 9 (a) Larger |𝐴|, steeper (b) Shifted horizontally by 𝐵 Left for 𝐵 > 0; right for 𝐵 < 0 Vertical asymptote 𝑥 = −𝐵 (c) 𝑦 𝐴 = 2, 𝐵 = 5 𝐴 = 20, 𝐵 = 0 10 ✛ 𝐴 = 2, 𝐵 = 0

𝑞∕𝑟 months (𝑟𝑎∕𝑞) + 𝑟𝑏 dollars 𝐶 =√ (𝑟𝑎∕𝑞) + 𝑟𝑏 + 𝑘𝑞∕2 dollars 𝑞 = 2𝑟𝑎∕𝑘

45 0.8 miles from Town 1 47 65.1 meters 49 (a) 𝑥 = 𝑒 (b) 𝑛 = 3 (c) 31∕3 > 𝜋 1∕𝜋

−10

53 (a) 10 (b) 9

✠ 𝑥 (c) 𝑎 moves critical points to the right; 𝑏 moves one critical point left, one up, one right √ (d) 𝑥 = 𝑎, 𝑎 ± 𝑏

17 𝐶 has 𝑎 = 1, 𝐵 has 𝑎 = 2, 𝐴 has 𝑎 = 3

𝑥 10

51 Max slope = 1∕(3𝑒) at 𝑥 = 1∕3

19 (a)

𝑦 𝑏=1

0.5

−10

✠ 𝑏=2 ✠ ✠

55 (a) drag

(thousands of pounds)

11 (a)



3 2 1

Induced✲ Drag

200



❄ ✛

large 𝑎

(b)

Parasite Drag



400

600

(b) (1∕𝑏, 1∕𝑏𝑒)



speed (mph)

𝑥 𝑥

13 (a) large 𝑎

25 (a) 𝑥 = −1∕𝑏 (b) Local minimum





27 𝑘 > 0 29 (a) 𝑥 = 𝑒𝑎 (b) 𝑎 = −1:

𝑥



𝑥

1

small 𝑎

(b) Critical point moves right (c) 𝑥 = 𝑎

10

(b) −2 < 𝑎 < 2

large 𝑎

small 𝑎

−10 −100

(c) 𝑎 moves critical point right; 𝑏 moves critical point up (d) 𝑥 = 𝑎

Section 4.4 ❘

𝑎=1 𝑎 = 20

100

small 𝑏

59 Max at 𝑥 = 10

1 (a)

23 (a)



61 Optimum may occur at an endpoint

𝑥

(b)

large 𝑏

large 𝑎



𝑏=4

𝑥

large 𝑏

57 𝑦 = 𝑚𝑔∕𝑘

small 𝑎

𝑥 3

21 Two critical points: 𝑥 = 𝑏, 𝑥 = 𝑏∕(𝑛 + 1)

(b) 160 mph or 320 mph; no; yes (c) 220 mph

3 (a)

𝑏=3

1

small 𝑎

Total Drag

large 𝑏

1

2

3

1

2

3

−1

❄ ❄

𝑎 = 1:

𝑥

(b) 2 critical √ points move closer to origin (c) 𝑥 = ± 1∕(3𝑎)

small 𝑏



𝑥

(c) 𝑎 moves one critical point up, does not move the other; 𝑏 moves one critical point up to right, moves the other right

1 𝑥 −1

1079 𝑦 (c) Max at (𝑒𝑎−1 , 𝑒𝑎−1 ) for any 𝑎

(c)

33 𝑓 ′′ (𝑥) = 𝑎2 𝑒−𝑎𝑥 + 𝑏2 𝑒𝑏𝑥 ; always positive

𝑊

𝐴 = 50, 𝑏 = 2, 𝑐 = 5

𝐴 = −𝐵 = 1 𝑥

35 𝑓 ′ (𝑥) = 𝑎∕𝑥 − 𝑏; positive for 𝑥 < 𝑎∕𝑏, negative for 𝑥 > 𝑎∕𝑏

𝐴 = 50, 𝑏 = 2, 𝑐 = 1

37 𝑓 ′′ (𝑥) = 12𝑎𝑥2 − 2𝑏, graph 𝑓 ′′ is upward parabola with neg vert intercept

𝐴 = 20, 𝑏 = 2, 𝑐 = 1 𝑡

39 (a) 𝐴: value for large negative 𝑥 𝐵: value for large positive 𝑥 (c) (I) (8, −4)

𝑦

(d) Yes 69 Two critical points only if 𝑎 and 𝑏 have the same sign.

𝐴=2 𝐵=1

(II) (−2, 5) (III) (7, 0)

𝑥

73 One possibility: 𝑔(𝑥) = 𝑎𝑥3 + 𝑏𝑥2 , 𝑎, 𝑏 ≠ 0

(IV) (2, 6). 41 (a) Local max: 𝑥 = 1∕𝑏 No local minima Inflection point: 𝑥 = 2∕𝑏 (b) Varying 𝑎 stretches or flattens the graph vertically. Incr 𝑏 shifts critical, inflection points to left; lowers max (c) Varying 𝑎 𝑦 4 𝑏𝑒 3 𝑏𝑒 2 𝑏𝑒 1 𝑏𝑒

71 𝑓 (𝑥) = 𝑥3 + 𝑥

75 (b), (c)

Section 4.5

𝑦

1 $5000, $2.40, $4

𝐴=2 𝐵 = −1 𝑥

𝑎=4

3 Below 4500 5 About $1.1 m, 70, $1.2 m 7 𝐶(𝑞) = 500+6𝑞, 𝑅(𝑞) = 12𝑞, 𝜋(𝑞) = 6𝑞−500

𝑎=3

9 𝐶(𝑞) = 5000 + 15𝑞, 𝑅(𝑞) = 60𝑞, 𝜋(𝑞) = 45𝑞 − 5000

𝑎=2

11 𝜋(𝑞) = 490𝑞 − 𝑞 2 − 150 Max at 𝑞 = 245

𝑎=1

𝑦

13 $0.20/item

𝑥 1

15 (a) Increase production (b) 𝑞 = 8000

2 𝑥 𝐴 = −2 𝐵 = −1

Varying 𝑏

𝑦

17 Above 2000 19 Greater than 100 21 (a) $0 (b) $96.56 (c) Raise the price by $5 23 (a)number of bees

2100 𝑏=1

𝑥 𝐴 = −2 𝐵=1

𝑏=2 1

𝑥

𝑏=3

2

(b)

47 𝑦 = 5(1 − 𝑒−𝑏𝑥 ) 2 ∕2

51 𝑦 = (2 + 2𝑒)∕(1 + 𝑒1−𝑡 ) 53 𝑦 = −2𝑥4 + 4𝑥2 √ 55 𝑦 = ( 2∕𝜋) cos(𝜋𝑡2 )



𝑥

59 (a) For 𝑎 ≤ 0, all 𝑏; 𝑥 = 𝑏 ± (b) 𝑎 = 1∕5, 𝑏 = 3

−𝑎

𝑥

Slope=

𝐶(𝑞0 ) 𝑞0

= 𝑎(𝑞0 ) 𝑞

27 (a)

𝑥 2 + 𝑦2 = 9 𝑥 2 + 𝑦2 = 4 𝑥 2 + 𝑦2 = 1

(2𝑎, −𝑏∕4)

𝑦 𝐴=𝐵=1

𝐶(𝑞)

𝑞0

𝑎 2𝑎

57 𝑦 = 2𝑥𝑒1−𝑥∕3

(i) 𝑁 ′ (𝑥) = 20 (ii) 𝑁(𝑥)∕𝑥 = (100∕𝑥) + 20



𝑈



𝑥 (acres) 100

25 (a) $ Slope= 𝑀𝐶

63 (a) Intercept: 𝑥 = 𝑎 Asymptotes: 𝑥 = 0, 𝑈 =0 (b) Local min: (2𝑎, −𝑏∕4) Local max: none (c)

45 𝐴𝑥3 + 𝐶𝑥, Two

61 (a)

100 50

(b) U-shaped (c) Incr (𝐴 > 0) or Decr (𝐴 < 0) (d) Max: 𝐴 < 0, 𝐵 < 0 Min: 𝐴 > 0, 𝐵 > 0

49 𝑦 = 𝑒−(𝑥−2)

𝑁(𝑥) = 100 + 20𝑥

65 (a) +∞ (b) 𝑟 = (2𝐴∕𝐵)1∕6 ; local minimum (c) 𝑟 = (2𝐴∕𝐵)1∕6 𝑏 𝐴𝑒−𝑒 ,

67 (a) Vertical intercept: 𝑊 = Horizontal asymptote 𝑊 = 𝐴 (b) No critical points, inflection point at 𝑡 = 𝑏∕𝑐, 𝑊 = 𝐴𝑒−1

𝑦

❘4 ❘ ❘

𝑥+𝑦=4 𝑥 4

1080 79 𝑒−𝜆

(c) 0.4 radians/min

(c) 8

𝑦

29 (a)

𝐶=4

4

2



𝐶=3



𝐶=2



𝑄=

81 5

55 (a) ℎ(𝑡) = 300 − 30𝑡 0 ≤ 𝑡 ≤ 10 (b) 𝜃 = arctan((200 − 30𝑡)∕150) 𝑑𝜃∕𝑑𝑡 ) )( ( =

83 2

57 𝑑𝐷∕𝑑𝑡 = 2 ⋅ 𝑑𝑅∕𝑑𝑡

93 𝑓 (𝑥) = 𝑥

59 𝑦 = 𝑓 (𝑥) = 2𝑥 + 1 and 𝑥 = 𝑔(𝑡) = 5𝑡

95 (b)

2

𝑥1∕2 𝑦1∕2 𝑥

2

53 1∕3

4

150 − 15 1502 +(200−30𝑡)2 (c) When elevator is at level of observer

61 True √ (c) 2 2

31 Cannot determine whether local or global, max or min 33 2 or 15 35 (e) 37 (a) (b) (c) (d) (e) (f)

Positive Cost = 𝑤𝑥, Revenue = 𝑝𝑓 (𝑥) 𝑤∕𝑝 Negative 1∕(𝑝𝑓 ′′ (𝑥∗ )), negative Less

63 (c) 65 (1∕𝑉 )𝑑𝑉 ∕𝑑𝑡 67 (a) (b) (c) (d) (e) (g)

900 34 ships End of battle 30 ships 𝑦′ (𝑡) = 𝑘𝑥; negative 𝑘 = 2, 68 ships/hour

Section 4.7 1 Yes 3 Yes

Section 4.6 1 −0.32◦ C∕min; −0.262◦ C∕min 3 (a) 0 (b) −4.5 (c) −8 √ 5 (a) 𝜋∕ 9.8𝑙 (b) Decr

5 No 7 No 9 Yes 11 No

85 −25∕2 87 1 89 −2 91 Limit of derivative does not exist

Section 4.8 1 The particle moves on straight lines from (0, 1) to (1, 0) to (0, −1) to (−1, 0) and back to (0, 1) 3 Two diamonds meeting at (1, 0) 5 𝑥 = 3 cos 𝑡, 𝑦 = −3 sin 𝑡, 0 ≤ 𝑡 ≤ 2𝜋 7 𝑥 = 2 + 5 cos 𝑡, 𝑦 = 1 + 5 sin 𝑡, 0 ≤ 𝑡 ≤ 2𝜋 9 𝑥 = 𝑡, 𝑦 = −4𝑡 + 7 11 𝑥 = −3 cos 𝑡, 𝑦 = −7 sin 𝑡, 0 ≤ 𝑡 ≤ 2𝜋 13 Clockwise for all 𝑡 √ √ 15 Clockwise: − 1∕3 < 𝑡 < 1∕3 √ √ Counterclockwise: 𝑡 < − 1∕3 or 𝑡 > 1∕3

17 Clockwise: 2𝑘𝜋 < 𝑡 < (2𝑘 + 1)𝜋 Counterclockwise: (2𝑘 − 1)𝜋 < 𝑡 < 2𝑘𝜋

13 1∕4

19 Line segment: 𝑦+𝑥 =4,1 ≤ 𝑥 ≤ 3

15 1.5

21 Line 𝑦 = (𝑥 − 13)∕3, left to right

17 0

23 Parabola 𝑦 = 𝑥2 − 8𝑥 + 13, left to right

19 0

25 Circle 𝑥2 + 𝑦2 = 9, counterclockwise

21 (1∕3)𝑎−2∕3

27 (𝑦 − 4) = (4∕11)(𝑥 − 6)

11 −81∕𝑅2

23 −0.7

29 𝑦 = −(4∕3)𝑥

13 0.211 units/sec

25 10∕9

31 Speed = 2|𝑡|, stops at 𝑡 = 0

15 24 meters3 ∕yr

27 0.75

17 0 atmospheres/hour

29 Does not exist

33 Speed = ((2𝑡 − 4)2 + (3𝑡2 − 12)2 )1∕2 , Stops at 𝑡 = 2

19 −1.440 atmospheres/hour

31 Does not exist

35

21 30 cm2 /min

33 7∕9

23 −7 cm2 /sec

35 0

25 Moving closer at 4 mph

37 0

27 (a) 𝐶𝐷 − 𝐷2 (b) 𝐷 < 𝐶

41 𝑥0.2

7

−1◦ 𝐶

minute

9 (a) 5644 ft (b) −124 ft∕min

(i) 5𝑟 $/yr (ii) 5𝑟𝑒0.02𝑟 $/yr (b) 24.916 $/yr

29 (a)

31 5.655 ft3 ∕sec 33 180𝜋 = 565.487 cm3 /hr 35 2∕9 ohms/min 37 𝑔; 𝑔𝑒−𝑘∕𝑚 ; acceleration decreases from 𝑔 at 𝑡=0 39 (a) −92.8𝑉 −2.4 atm/cm3 (b) Decreasing at 0.0529 atm/min

39 0.1𝑥7

𝑥 1

𝑓 (𝑡)

𝑡

0 8

16 24 32

43 −2

𝑦

45 Positive 47 Negative

𝑔(𝑡)

1

49 0 51 0 53 none, no 55 ∞ − ∞, yes

𝑡

0 8

16 24 32

57 ∞0 , yes 59 0

41 (a) 94.248 m2 ∕min (b) 0.0000267 m/min

61 0

43 (a) 80𝜋 = 251.327 sec (b) 𝑉 = 3𝜋ℎ3 ∕25 cm3 (c) 0.0207 cm/sec

65 0

45 8∕𝜋 meters/min

69 −2

47 0.253 meters/second

71 0

49 Between 12◦ F/min and 20◦ F/min √ 51 (a) 𝑧 = 0.25 + 𝑥2 (b) 0.693 km/min

73 Does not exist

63 0 67 1∕2

75 𝑒 77 𝑒𝑘𝑡

37 (a) The part of the line with 𝑥 < 10 and 𝑦 < 0 (b) The line segment between (10, 0) and (11, 2) 39 (a) Both parameterize line 𝑦 = 3𝑥 − 2 (b) Slope = 3 𝑦-intercept = −2 41 (a) 𝑎 = 𝑏 = 0, 𝑘 = 5 or −5 (b) 𝑎 = 0, 𝑏 = 5, 𝑘 = 5 or √ −5 √ (c) 𝑎 = 10, 𝑏 = −10, 𝑘 = 200 or − 200

43 A straight line through the point (3, 5)

1081 45 (a) 𝑑𝑦∕𝑑𝑥 = 4𝑒𝑡 (b) 𝑦 = 2𝑥2 (c) 𝑑𝑦∕𝑑𝑥 = 4𝑥

(e) 𝑓 increasing: 𝑥 < 1 and 𝑥 > 2 Decreasing: 1 < 𝑥 < 2 𝑓 concave up: 𝑥 > 3∕2 𝑓 concave down: 𝑥 < 3∕2

(e) 𝑓 increasing: for 𝑥 < 0 and 𝑥 > 2 𝑓 decreasing: for 0 < 𝑥 < 2 𝑓 concave up: for 𝑥 > 1 𝑓 concave down: for 𝑥 < 1

47 (a) (11, −2) (b) √ 1∕12 145 (c)

49 (a) Yes, 𝑡 = 1, (𝑥, 𝑦) = (−2, −1) (b) Yes, 𝑡 = −1, (𝑥, 𝑦) = (2, 3) (c) Never

𝑓 (𝑥) = 2𝑥3 − 9𝑥2 + 12𝑥 + 1

6 5

| incr. | decreasing | incr. | | concave down | concave up |

51 (a) Increasing (b) 𝑃 to 𝑄 (c) Concave down

−1

53 (a) No (b) 𝑘 = 1 (c) Particle 𝐵

1

3 𝑥 1



3 55 (a) 𝑦 + 12 = − 3 (𝑥 − 𝜋) (b) 𝑡 = 𝜋 (c) 0.291, concave up

9 lim𝑥→−∞ 𝑓 (𝑥) = −∞ lim𝑥→∞ 𝑓 (𝑥) = 0

√ 57 (a) 𝑡 = 𝜋∕4; at that time, speed = 9∕2 (b) Yes, when 𝑡 = 𝜋∕2 or 𝑡 = 3𝜋∕2 (c) Concave down everywhere 59 Line traversed in the wrong direction 61 𝑥 = 2 cos 𝑡, 𝑦 = 2 sin 𝑡, 0 ≤ 𝑡 ≤

𝑦 1 𝑥 1

69

𝑦 1

𝑥 −1

(a) 𝑓 ′ (𝑥) = (1 − 𝑥)𝑒−𝑥 𝑓 ′′ (𝑥) = (𝑥 − 2)𝑒−𝑥 (b) Only critical point is at 𝑥 = 1 (c) Inflection point: 𝑓 (2) = 2∕𝑒2 (d) Global max: 𝑓 (1) = 1∕𝑒 Local and global min: none (e) 𝑓 increasing: 𝑥 < 1 𝑓 decreasing: 𝑥 > 1 𝑓 concave up: 𝑥 > 2 𝑓 concave down: 𝑥 < 2

−4

5 (a) 𝑓 ′ (𝑥) = − 𝑒−𝑥 sin 𝑥 + 𝑒−𝑥 cos 𝑥 𝑓 ′′ (𝑥) = −2𝑒−𝑥 cos 𝑥 (b) Critical points: 𝑥 = 𝜋∕4 and 5𝜋∕4 (c) Inflection points: 𝑥 = 𝜋∕2 and 3𝜋∕2 (d) Endpoints: 𝑓 (0) = 0 and 𝑓 (2𝜋) = 0 Global max: √ 𝑓 (𝜋∕4) = 𝑒−𝜋∕4 ( 2∕2) Global min: √ 𝑓 (5𝜋∕4) = −𝑒−5𝜋∕4 ( 2∕2) (e) 𝑓 increasing: 0 < 𝑥 < 𝜋∕4 and 5𝜋∕4 < 𝑥 < 2𝜋 𝑓 decreasing: 𝜋∕4 < 𝑥 < 5𝜋∕4 𝑓 concave down: for 0 ≤ 𝑥 < 𝜋∕2 and 3𝜋∕2 < 𝑥 ≤ 2𝜋 𝑓 concave up: for 𝜋∕2 < 𝑥 < 3𝜋∕2

𝜋 2

63 False 67

1

𝑓 (𝑥) = 𝑥𝑒−𝑥 𝑥 1

13 Max: 22.141 at 𝑥 = 𝜋; Min: 2 at 𝑥 = 0 15 Max: 1 at 𝑥 = 0; Min: Approx 0 at 𝑥 = 10 17 Global max = 1 at 𝑡 = 0 No global min 19 Local max: 𝑓 (−3) = 12 Local min: 𝑓 (1) = −20 Inflection pt: 𝑥 = −1 Global max and min: none

Chapter 4 Review 50 40 30 20 10

Local and global max Local max

Local max Local min Local and global min

𝑥

2

3

4

5

𝑓 (𝑥) = 𝑥3 + 3𝑥2 − 9𝑥 − 15

12

𝑓 (𝑥)

1

2

11 D

−1

1

2

6

3 (a) 𝑓 ′ (𝑥) = 3𝑥(𝑥 − 2) 𝑓 ′′ (𝑥) = 6(𝑥 − 1) (b) 𝑥 = 0 𝑥=2 (c) Inflection point: 𝑥 = 1 (d) Endpoints: 𝑓 (−1) = −4 and 𝑓 (3) = 0 Critical points: 𝑓 (0) = 0 and 𝑓 (2) = −4 Global max: 𝑓 (0) = 0 and 𝑓 (3) = 0 Global min: 𝑓 (−1) = −4 and 𝑓 (2) = −4

𝜋 2

| conc. down | | incr. |

3𝜋 2

𝜋 concave up decreasing

|

𝑥 2𝜋

| conc. down | |

increasing

7 lim𝑥→∞ 𝑓 (𝑥) = ∞ lim𝑥→−∞ 𝑓 (𝑥) = −∞ (a) 𝑓 ′ (𝑥) = 6(𝑥 − 2)(𝑥 − 1) 𝑓 ′′ (𝑥) = 6(2𝑥 − 3) (b) 𝑥 = 1 and 𝑥 = 2 (c) 𝑥 = 3∕2 (d) Critical points: 𝑓 (1) = 6, 𝑓 (2) = 5 Local max: 𝑓 (1) = 6 Local min: 𝑓 (2) = 5 Global max and min: none

𝑥

−3

1

−20

21 Global and local min: 𝑥 = 2 Global and local max: none Inflection pts: none

𝑓 (𝑥) = 𝑥 − ln 𝑥 𝑥 2

1082 23 Local max: 𝑓 (−2∕5) Global and local min: 𝑓 (0) √ Inflection pts: 𝑥 = (−2 ± 2)∕5 Global max: none

67 ( Max area )= 1( ± √1 , 0 ; ± √1 , 2

2

69 (0.59, 0.35) √ 71 𝑟 = 2𝐴∕(4 + 𝜋); ℎ=

𝐴 2





4+𝜋 √ 2𝐴



𝜋 4

1 √ 2



73 40 feet by 80 feet

95 𝑑𝑀∕𝑑𝑡 = 𝐾(1 − 1∕(1 + 𝑟))𝑑𝑟∕𝑑𝑡

)

2 97 Height: 0.3∕(𝜋ℎ √ ) meters/hour Radius: (0.3∕ 3)∕(𝜋ℎ2 ) meters/hour

99 (a) Increasing (b) Not changing (c) Decreasing

√ 2𝐴 √ 4+𝜋

101 398.103 mph 103 (a) Decreases (b) −0.25 cm3 /min

75 (a) 𝜋(𝑞) max when 𝑅(𝑞) > 𝐶(𝑞) and 𝑅 and 𝑄 are farthest apart

𝑦

105 (a)

2 𝑓 (𝑥) = 𝑥2 𝑒5𝑥

$

𝐶(𝑞)

𝑥

𝑅(𝑞)

−0.4

𝑥 1∕𝑎



25 Max: 𝑟 = 23 𝑎; Min: 𝑟 = 0, 𝑎 27 𝐴 has 𝑎 = 1, 𝐵 has 𝑎 = 2, 𝐶 has 𝑎 = 3 √ 29 (a) Critical points: 𝑥 = 0, 𝑥 = 𝑎, √ 𝑥=− 𝑎 √ Inflection points: 𝑥 = 𝑎∕3, √ 𝑥 = − 𝑎∕3 (b) 𝑎 = 4, √ √𝑏 = 21 (c) 𝑥 = 4∕3, 𝑥 = − 4∕3

31 Domain: All real numbers except 𝑥 = 𝑏; Critical points: 𝑥 = 0, 𝑥 = 2𝑏

−2 maximum 𝜋(𝑞)

𝑎 = 0.5 𝑦

𝑞 𝑞0

2

(b) 𝐶 ′ (𝑞0 ) = 𝑅′ (𝑞0 ) = 𝑝 (c) $

𝑥 1∕𝑎

𝐶 ′ (𝑞) −2

33 18 35 (a) 𝐶(𝑞) = 3 + 0.4𝑞 m (b) 𝑅(𝑞) = 0.5𝑞 m (c) 𝜋(𝑞) = 0.1𝑞 − 3 m

𝑎=1

𝑝 𝑦 2

37 2.5 gm∕hr 39 Line 𝑦 = −2𝑥 + 9, right to left 41 (a) (b) (c) (d)

𝑞 𝑞0

𝑥3 𝑥1 , 𝑥 5 𝑥2 0

𝑥 1∕𝑎

77 About 331 meters, Maintain course

43

𝑓 ′ (𝑥) 𝑓 has a local min.

𝑓 has crit. pt. Neither max or min

79 (a) 𝑔(𝑒) is a global maximum; there is no minimum (b) There are exactly two solutions (c) 𝑥 = 5 and 𝑥 ≈ 1.75 81 (a)

−2 𝑎=2 (b) Same graph all 𝑎

depth of water

𝑑𝑦∕𝑑𝑥 𝑑(𝑡)

45 𝑦 = 𝑥3 − 6𝑥2 + 9𝑥 + 5 slope =

47 𝑦 = 3𝑥−1∕2 ln 𝑥

𝐾

(1−𝑥2 )∕2

49 𝑦 = 2𝑥𝑒 √ √ 8∕3 cm, ℎ = 2∕3 cm √ 53 𝑟 = ℎ = 8∕3𝜋 cm

𝑡

51 𝑥 =

55 −4.81 ≤ 𝑓 (𝑥) ≤ 1.82

(b)

depth of water

𝑑(𝑡)

57 Max: 𝑚 = −5𝑘∕(12𝑗) 59 (a) 𝑡 = 0 and 𝑡 = 2∕𝑏 (b) 𝑏 = 0.4; 𝑎 = 3.547 (c) Local min: 𝑡 = 0; Local max: 𝑡 = 5 61

𝑎 = −2 𝑎=0 𝑎=2 𝑎=4 𝑎=6

✲ ✲ ✲ ✲ ✲

𝑥 (c) 𝑑𝑦∕𝑑𝑥 simplifies to √ 107 (a) 1∕ 𝑥2 − 1

109 (a) 𝑑𝑦∕𝑑𝑥 =

𝑥∕(2𝑥(1 + 𝑥))

tan(𝑥∕2) √ 2 (1−cos 𝑥)∕(1+cos 𝑥)

(b) 𝑑𝑦∕𝑑𝑥 = 1∕2 on 0 < 𝑥 < 𝜋 𝑑𝑦∕𝑑𝑥 = −1∕2 on 𝜋 < 𝑥 < 2𝜋

𝑦

𝑡 0.5

83 −0.0545 m3 ∕min

✛ 𝑦 = − 1 𝑥3 2



85 0∕0, yes 87 −1∕2

𝑥 −4𝜋

−2𝜋

2𝜋

89 −1∕6 63 1.688 mm/sec away from lens

91 1∕𝜋 ≈ 0.32𝜇m/day

65 Minimum: −2 amps Maximum: 2 amps

93 (a) Angular velocity (b) 𝑣 = 𝑎(𝑑𝜃∕𝑑𝑡)

−0.5

4𝜋

1083

Section 5.1 1 (a) (b) (c) (d) (e)

Section 5.2

Left sum Upper estimate 6 Δ𝑡 = 2 Upper estimate ≈ 24.4

3 (a) 54 meters; upper (b) 30 meters; lower

27 14,052.5

Right Upper 3 2

29 2.545

3 (a) Left; smaller (b) 0, 2, 6, 1∕3

35 (a) −4 (b) 0 (c) 8

1 (a) (b) (c) (d)

31 1.106 33 21.527

5 (a) 224

37 (a) (b) (c) (d)

5 Between 140 and 150 meters 7 (a) 𝑏 = 15 (b) 75 feet

32 24

9 (a) 𝑎 = 20, 𝑏 = 90, 𝑐 = 12 (b) 660 feet 11 (a) 408 (b) 390

39 −16

𝑓 (𝑡)

16

41 22/3

8

15 Lower est = 46 m Upper est = 118 m Average = 82 m

43 0.80

𝑡

13 lower: 235.2 meters; upper: 295.0 meters

2

4

6

45 (a) 0.808 (b) 0.681

8

47 Left-hand sum = 792; Right-hand sum = 2832.

(b) 96

49 Left-hand sum = 50.5; Right-hand sum = 18.625. √ 51 Left-hand sum√= (𝜋∕4)( 2 + 1); Right-hand sum = (𝜋∕4)( 2 + 1). √ √ 53 Left-hand √= 1 + 2 + 3; Right-hand √ sum sum = 2 + 3 + 2.

17 Change: 0 cm, no change in position Total distance: 12 cm

32

19 Change: 15 cm to the left Total distance: 15 cm

24

21 (a) 120, 240, 180 ft (b) 180 ft 23 (a) (b) (c) (d)

8

2; 15, 17, 19, 21, 23; 10, 13, 18, 20, 30 122; 162 4; 15, 19, 23; 10, 18, 30 112; 192

55 (a)

𝑡 2

4

6

sin 𝑥

8 𝑥 −𝜋

𝜋

(c) About 200 (b)

sin 𝑥

32

27 1∕5

𝑥 −𝜋

24

29 0.0157 31 (a)

𝑓 (𝑡)

16

25 (a) Lower estimate = 5.25 mi Upper estimate = 5.75 mi (b) Lower estimate = 11.5 mi Upper estimate = 14.5 mi (c) Every 30 seconds

1 2𝜋 2𝜋 − 1∕2 𝜋 − 3∕2

16

𝑣 (ft/sec) 88

𝑓 (𝑡)

𝜋

(c) 0

8

57 (a)

1

𝑓 (𝑥)

𝑡 2

4

6

8

𝑡 (sec) 2.2

𝑥

(b) 96.8 ft 33 About 65 km from home 3 hours About 90 km

32

35 (a) Up from 0 to 90 seconds; down from 100 to 190 seconds (b) 870 ft

16

2

(b) 1 (c) 1; same value 59 2∕125

24

3

61 ∫0 2𝑡 𝑑𝑡

𝑓 (𝑡)

1∕2

(7𝑡2 + 3) 𝑑𝑡 √ 4 65 ∫1 𝑡2 + 𝑡 𝑑𝑡 4√ 67 ∫0 𝑡 𝑑𝑡 = 16∕3

63 ∫0

8

37 (a) 𝐴: 8 hrs 𝐵: 4 hrs (b) 100 km/hr (c) 𝐴: 400 km 𝐵: 100 km 39 (a)

1

(d) About 136

𝑡 2

4

6

8

2

69 ∫1 (8𝑡 − 8) 𝑑𝑡 = 4 7 205.5

50

71 𝑎 = 1

11 About 350

73 𝑎 = 1, 𝑏 = 5

13 28.5

75 Too many terms in sum

15 1.4936

77 𝑓 (𝑥) = −1, [𝑎, 𝑏] = [0, 1]

17 0.3103

79 False

19 (a) 160 (b) 219 (c) −18.5

81 False 85

47 True

21 (a) 80 (b) 123 (c) 88

49 False

23 IV: 93.47

51 About 0.0635 miles or 335 feet

25 7.667

𝑣(𝑡) 𝑡 5 (b) 125 feet (c) 4 times as far 41 20 minutes 43 Only true if car accelerates at a constant rate 45 𝑓 (𝑥) = 𝑥2 , [𝑎, 𝑏] = [−2, −1]

83 True

1 𝑓 (𝑥) 𝑥 10

1084 87

39 (a) 0.375 thousand/hour (b) 1.75 thousands

(e) III (f) I

𝑦

41 (a) 22◦ C (b) 183◦ C (c) Smaller

45 Amount oil pumped from well from 0 to 𝑡0

𝑥 𝑎

𝑏

47 (a) (b) (c) (d)

Boys: black curve; girls: colored curve About 43 cm Boys: about 23 cm; girls: about 18 cm About 13 cm taller

49 (a) 𝑆(0) = 182,566 acre-feet; 𝑆(3) 171,770 acre-feet (b) Max in April; min in November (c) Between June and July

89 (c) 𝑛 ≥ 105

43 (a) Less (b) 0.856 =

45 Lower 12, upper 15 47 Lower 12, upper 20 49

𝑓 (𝑥)



51 −4.3

Section 5.3 3 Foot-pounds

61 (III), (II), (I) √ √ 𝑑 ( 𝑥) ≠ 𝑥 63 𝑑𝑥

3

9 ∫1 2𝑡 𝑑𝑡 = 8 5

11 ∫1 1∕𝑡 𝑑𝑡 = ln 5

51

(miles/hour)

𝑥

50

15 (a) 3𝑥2 + 1 (b) 10

𝑎

17 (a) sin 𝑡 cos 𝑡 (b) (i) ≈ 0.056 1 (sin2 (0.4) 2

𝑡 (hours)

− sin2 (0.2))

1

2

3

4

19 75 21 (a) Removal rate 500 kg∕day on day 12 (b) Days, days, kilograms (c) 4000 kg removed between day 5 and day 15 23 (a) (b)

𝑓 (𝑥)

65 velocity

7 ln(4) ⋅ 4𝑡 𝑑𝑡 = 336

2 ∫0

𝑅(𝑡) 𝑑𝑡

67 0.732 gals, 1.032 gals; Better estimates possible

1

1

2

55 (a)

61 Not enough information

7 2

63 39

9 2

65

5 ∫0

𝑓 (𝑡) 𝑑𝑡 25 (a) (b) 177.270 billion barrels (c) Lower estimate of year’s oil consumption 27 (a) Emissions 2000–2012, m. tons CO2 equiv (b) 1912.7 m. tons CO2 equiv (Possible answers from 1898 to 1928) 29 13.295 billion tons 12 ∫0 𝑓 (𝑡) 𝑑𝑡

𝑥 𝑎

1

21 ∫0 𝑥2 𝑑𝑥

67 𝑓 (𝑥) could have both positive and negative values

3

23 ∫2 cos(𝑥) 𝑑𝑥

69 Time should be in days

25 (a) 4 (b) 16 (c) 12 27

71 𝑓 (𝑥) = 2 − 𝑥

3 ∫0 𝑓 (𝑥) 𝑑𝑥 = 1 1 ∫ 𝑓 (𝑥) 𝑑𝑥 2 −1

73 True 3

+ ∫1 𝑓 (𝑥) 𝑑𝑥

75 False 77 False 79 False

35 45.8◦ C.

31 (a)

83 True

39 𝑓 (3) − 𝑓 (2), (𝑓 (4) − 𝑓 (2))∕2, 𝑓 (4) − 𝑓 (3)

𝑏

19 0.172

33 8.9064 kwh 37 (a) Rate is 50 ng/ml per hour after 1 hour (b) Concentration is 730 mg/ml after 3 hours

𝐹 (𝑏) − 𝐹 (𝑎)

15 0.083

29 (a) −2 (b) 6 (c) 1

31 About $11, 600

✻ ❄

𝐹 (𝑥)

13 4.389 17 7.799

(c) lower estimate: 2.81 upper estimate: 3.38

(i) 1/4 (ii) 1/4

59 12

11 (a) 8.5 (b) 1.7

𝑡

Negative Negative Positive Negative

57 (a) 0.1574 (b) 0.9759

5 2𝑐1 + 12𝑐22

(2, 1) 𝑅(𝑡)

𝑏

(iii) 0 (b) Not true

3 9

(1, 1.6)

53 (a) (b) (c) (d)

Section 5.4 1 10

(0, 2) 2

81 False

(i) 14 (ii) −7 (b) 𝑎 = 9, 𝑏 = 11 𝑏

33 (a) ∫𝑎 𝑥 𝑑𝑥 = (𝑏2 − 𝑎2 )∕2 (b)

5

(i) ∫2 𝑥 𝑑𝑥 = 21∕2 8

41 7.1024 < 𝑓 (2) < 7.1724

(ii) ∫−3 𝑥 𝑑𝑥 = 55∕2

43 (a) (b) (c) (d)

(iii) ∫1 5𝑥 𝑑𝑥 = 20

V IV III II

𝑏

59 (II), (III), (I)

7 Change in salinity; gm/liter

(ii)

𝑥 𝑎

4

57 ∫0 𝑟(𝑡) 𝑑𝑡 < 4𝑟(4)

5 Change in velocity; km/hr

13



55 0.4; cost increases by 0.4 million/inch after 24 in fallen

1 Dollars

3 ∫2

𝑓 (𝑏) − 𝑓 (𝑎)

53 5; costs $5 mil more to plow 24 in than 15 in

3

35 10 37 42

85 True 87 False 89 (a) Does not follow (b) Follows (c) Follows

Chapter 5 Review 1 (a) Lower estimate = 122 ft Upper estimate = 298 ft

1085 (b)

𝑣 80

(b)

60

(c) (d) (e) (f)

40 20 𝑡 2

4

6

3 20 5 396 7 (a) Upper estimate = 34.16 m/sec Lower estimate = 27.96 m/sec (b) 31.06 m/sec; It is too high 1

9 ∫−2 (12𝑡3 − 15𝑡2 + 5) 𝑑𝑡 = −75 11 36

negative: from 𝑡 = 40 to just before 𝑡 = 60 About 500 feet at 𝑡 = 42 seconds Just before 𝑡 = 60 Just after 𝑡 = 40 A catastrophe Total area under curve is positive About 280 feet

15 13.457

20,000

63 Underestimate ∑𝑛 5 6 67 (a) 𝑖=1 𝑖 ∕𝑛 (b) (2𝑛4 + 6𝑛3 + 5𝑛2 − 1)∕12𝑛4 (c) 1∕6 ∑𝑛−1 2 3 69 (a) 𝑖=0 (𝑛 + 𝑖) ∕𝑛 (b) 7∕3 + 1∕(6𝑛2 ) − 3∕(2𝑛) (c) 7/3 (d) 7/3

10,000

21 0.399 miles

3 1, 0, −1∕2, 0, 1

23 (a) Emissions 1970–2000, m. metric tons (b) 772.8 million metric tons

5

29 𝑥1 local min; 𝑥2 inflection pt; 𝑥3 local max

𝑥 𝑥1

𝑥2

𝑥3

𝐹 (0) = 0 31 𝑥1 local max; 𝑥2 inflection pt; 𝑥3 local min

1 𝑥

2

1

5

(b) (1∕5)(∫0 𝑓 (𝑥) 𝑑𝑥 − ∫2 𝑓 (𝑥) 𝑑𝑥) 29 (a) 𝑓 (1), 𝑓 (2) (b) 2, 2.31, 2.80, 2.77

𝑥2 𝑥 𝑥3

𝑥1

7

31 (a) 𝐹 (0) = 0 (b) 𝐹 increases (c) 𝐹 (1) ≈ 0.7468 𝐹 (2) ≈ 0.8821 𝐹 (3) ≈ 0.8862

𝐹 (0) = 1

1

𝐹 (𝑥)

𝐹 (0) = 0 𝑥 1

33 4 35 3 37 3

9

41 (a) 318,000 megawatts; megawatts (b) 11.5% (c) 488,730 megawatts

𝐹 (0) = 1

1

Odd integrand; areas cancel 0.4045 −0.4049 No. Different sized rectangles

1

𝑥 11

𝐹 (0) = 1 𝑥

2

5

(b) ∫0 𝑓 (𝑥) 𝑑𝑥 − ∫2 𝑓 (𝑥) 𝑑𝑥 1 2

𝐹 (0) = 0

35 (b) Maximum in July 2016 Minimum in Jan 2017 (c) Increasing fastest in May 2016 Decreasing fastest in Oct 2016 37 Statement has 𝑓 (𝑥) and 𝐹 (𝑥) reversed 39 𝑓 (𝑥) = 1 − 𝑥

51 2.5; ice 2.5 in thick at 4 am 5

1 2 3 4 5 67

1

1

53 (a) 2 ∫0 𝑓 (𝑥) 𝑑𝑥

3 2 1

𝑥 𝐹 (0) = 0

−1

47 30∕7 49 V < IV < II < III < I I, II, III positive IV, V negative

33 (a) 𝑓 (3) = 1; 𝑓 (7) = 0 (b) 𝑥 = 0, 5.5, 7 (c) 𝑦

711,556.784

0.5𝑇 𝑇 𝑟(𝑡) 𝑑𝑡 > ∫0.5𝑇 𝑟(𝑡) 𝑑𝑡 ∫0 𝑇ℎ 0.5𝑇 ∫0 𝑟(𝑡) 𝑑𝑡 < ∫0 𝑟(𝑡) 𝑑𝑡

5

27 (a) −1 < 𝑥 < 1 (b) 𝑥 < 0

𝐹 (0) = 1

5

27 (a) (1∕5) ∫0 𝑓 (𝑥) 𝑑𝑥

(c) ∫−2 𝑓 (𝑥) 𝑑𝑥 −

𝑥 10 20 30 40 50

𝐹 (𝑥)

1 (a) Increasing (b) Concave up

45

25 (a) 0; 3000; 12,000; 21,000; 27,000; 30,000 (b) 𝑦

30,000

25 (a) Overestimate = 7 tons Underestimate = 5 tons (b) Overestimate = 74 tons Underestimate = 59 tons (c) Every 2 days

(8, 0) 𝑥 8

(6, −4)

61 Exact

19 (a) 260 ft (b) Every 0.5 sec

43

4

Section 6.1

1

17 ∫−1 |𝑥| 𝑑𝑥 = 1

39 (a) (b) (c) (d)

𝐹 (𝑥)

(0, 1)

59 (𝑎) < (𝑐) < (𝑏) < (𝑑)

71 (a) cos(𝑎𝑐)∕𝑐 − cos(𝑏𝑐)∕𝑐 (b) − cos(𝑐𝑥)∕𝑐

13 1.142

(2, 3)

2

∫−2 𝑓 (𝑥) 𝑑𝑥

13 62 15 (a) 𝐹 (2) = 6 (b) 𝐹 (100) = 𝐹 (0)

41 True 43 Maximum = 6.17 at 𝑥 = 1.77 45

17 (a) −16 (b) 84

55 About 554 feet

19 52.545

57 (a) Positive: 0 ≤ 𝑡 < 40 and just before 𝑡 = 60;

21 82, 107, 119 23 (0, 1); (2, 3); (6, −4); (8, 0)

𝑡1 −2

𝑡2

𝑡3

𝑡4

𝑡5

1086

Section 6.2

(b) 2∕𝜋

29 (a) 80 ft/sec (b) 640 ft

1 ∫ 𝑞(𝑥) 𝑑𝑥 = 𝑝(𝑥) + 𝐶

91 5.500 ft

3 (II) and (IV) are antiderivatives

93 𝐶(𝑥) = 5𝑥2 + 4000𝑥 + 1,000,000 riyals

5 (III), (IV) and (V) are antiderivatives 7 5𝑡 ∕2 9

+

33 10 ft; 4 sec

95 26

2

𝑡3 ∕3

31 128 ft/sec2

𝑡2 ∕2

11 ln |𝑧|

97 (II)

35 Positive, zero, negative, positive, zero

99 None

37 If 𝑦 = cos(𝑡2 ), 𝑑𝑦∕𝑑𝑡 ≠ − sin(𝑡2 )

101 All 𝑛 ≠ −1

39 𝑑𝑦∕𝑑𝑥 = 0

103 𝑓 (𝑥) = −1

41 𝑑𝑦∕𝑑𝑥 = −5 sin (5𝑥)

105 True

43 True

107 False

45 True

109 True

47 False

23 𝑥4 ∕4 − 𝑥2 ∕2 + 𝐶

111 False

49 True

25 2𝑡3∕2 ∕3 + 𝐶

113 80

51 True

27 𝑧2 ∕2 + 𝑒𝑧 + 𝐶

117 5 to 1

13 sin 𝑡 15

𝑦5 ∕5

+ ln |𝑦|

17 − cos 𝑡

19 5𝑥2 ∕2 − 2𝑥3∕2 ∕3 21

𝑡4 ∕4



𝑡3 ∕6

− 𝑡2 ∕2

29 𝑥4 − 7𝑥 + 𝐶

53 (a) 𝑡 = 𝑠∕( 12 𝑣max ) 55 (a) First second: −𝑔∕2 Second: −3𝑔∕2 Third: −5𝑔∕2 Fourth: −7𝑔∕2 (b) Galileo seems to have been correct

Section 6.3

31 2𝑡 − cos 𝑡 + 𝐶

3 𝑦 = 𝑥2 + 𝐶

33 7 tan 𝑡 + 𝐶

5 𝑦 = 𝑥4 ∕4 + 𝑥5 + 𝐶

35 𝐹 (𝑥) = 𝑥2 (only possibility)

7 𝑦 = sin 𝑥 + 𝐶

Section 6.4

9 𝑦 = 𝑥3 + 5

37 𝐹 (𝑥) = 2𝑥 + 2𝑥2 + (5∕3)𝑥3 (only possibility)

11 𝑦 = 𝑒𝑥 + 6

3 (a) Si(4) ≈ 1.76 Si(5) ≈ 1.55 (b) (sin 𝑥)∕𝑥 is negative on that interval

39 𝐹 (𝑥) = 𝑥3 ∕3 (only possibility)

13 80.624 ft∕sec downward

41 𝐹 (𝑥) = − cos 𝑥 + 1 (only possibility)

17 (a) 𝑅(𝑝) = 25𝑝 − 𝑝2 (b) Increasing for 𝑝 < 12.5 Decreasing for 𝑝 > 12.5

15 10 sec

43 2𝑡2 + ln |𝑡| + 𝐶

𝑥

5 𝑓 (𝑥) = 5 + ∫1 (sin 𝑡)∕𝑡 𝑑𝑡 7

19 (a) 𝑎(𝑡) = −9.8 m/sec2 𝑣(𝑡) = −9.8𝑡 + 40 m/sec ℎ(𝑡) = −4.9𝑡2 + 40𝑡 + 25 m (b) 106.633 m; 4.082 sec (c) 8.747 sec

45 7𝑒𝑥 + 𝐶 𝑥

47 4𝑒 + 3 cos 𝑥 + 𝐶 49 𝑥3 ∕3 + 3𝑥2 + 9𝑥 + 𝐶 2 51 3 ln |𝑡| + + 𝐶 𝑡 53 𝑡6 ∕6 + 𝑡4 ∕4

21 (a) (b) (c) (d) (e)

55 𝑥 + ln |𝑥| + 𝐶 57 ln 3 ≈ 1.0986

𝑥

9

160

61 609∕4 − 39𝜋 ≈ 29.728

𝐴

63 ln 2 + 3∕2 ≈ 2.193

𝑥

11 cos(𝑥2 )

𝑣(𝑡) Highest point

65 1

5

67 False

Ground

𝑡 (sec) 10

−160 (f) Height = 400 feet (g) 𝑣(𝑡) = −32𝑡 + 160 Height = 400 feet

73 True 75 False

23 5∕6 miles

77 1∕2

25 −33.56 ft/sec2

79 22 81

𝑒2

27 (a) 𝑦 = − cos 𝑥 + 2𝑥 + 𝐶

+ 4 ln 2 − 7

25 (a) (b) (c) (d)

𝑦 15

87 2 + 𝜋 3 ∕6

𝐶 = 20

sin 𝑥 − 𝑥(𝑥 − 𝜋) 𝐴



𝑥 𝜋 2

𝐵

𝜋

𝐶 = 15 −2𝜋 𝐶 = 10 𝐶=5

5 −𝜋 −5

𝜋

−10 −15

89 (a) 0

0 −2 1≤𝑥≤6 8

𝐶=0 𝐶 = −1.99 27 (a) 𝐹 ′ (𝑥) = 1∕(ln 𝑥) (b) Increasing, concave down 𝐶 = −5

10

Height =

𝑦

17 Concave up 𝑥 < 0, concave down 𝑥 > 0

23 500

83 1∕2 85 𝑏 = 7

13 (1 + 𝑥)200 15 arctan(𝑥2 ) 19 𝐹 (0) = 0 𝐹 (0.5) = 0.041 𝐹 (1) = 0.310 𝐹 (1.5) = 0.778 𝐹 (2) = 0.805 𝐹 (2.5) = 0.431 √ 21 Max at 𝑥 = 𝜋; √ 𝐹 ( 𝜋) = 0.895

69 True 71 False

𝐹 (𝑥)

32 ft∕sec2 Constant rate of change 5 sec 10 sec

velocity (ft/sec)

59 2𝑒 − 2 ≈ 3.437

𝐹 (𝑥)

(b) 𝑦 = − cos 𝑥 + 2𝑥 − 1.99

𝐶𝑥= −10 2𝜋 𝐶 = −15 𝐶 = −20

29 −3.905 31 −1.4; exact 33 −8, an overestimate; or −6, an underestimate 35 cos(sin2 𝑡)(cos 𝑡) 37 4𝑥𝑒𝑥

4

√ 39 𝑒−𝑥 ∕ 𝜋𝑥 6

41 3𝑥2 𝑒−𝑥 − 𝑒−𝑥

2

1087 43 (a) (b) (c) (d)

−5 0 12 −5

51 (a) 𝑥 = −1, 𝑥 = 1, 𝑥 = 3 (b) Local min at 𝑥 = −1, 𝑥 = 3; local max at 𝑥 = 1 (c)

95 (a) −(1∕3) cos(3𝑥), −(1∕4) cos(4𝑥), −(1∕3) cos(3𝑥 + 5) (b) −(1∕𝑎) cos(𝑎𝑥 + 𝑏) 97 (a) (1∕2)(ln |𝑥 − 3| − ln |𝑥 − 1|), (1∕3)(ln |𝑥 − 4| − ln |𝑥 − 1|), (1∕4)(ln |𝑥 + 3| − ln |𝑥 − 1|) (b) (1∕(𝑏 − 𝑎))(ln |𝑥 − 𝑏| − ln |𝑥 − 𝑎|)

45 𝐹 (𝑥) is non-decreasing; min at 𝑥 = −2 47 𝐹 (𝑥) =

𝑥 ∫0 𝑡2 𝑑𝑡

𝑥

49 True

−2

51 True

2

4 𝐹 (𝑥)

53 True 55 Increasing; concave up 57 (a) 𝑅(0) = 0, 𝑅 is an odd function. (b) Increasing everywhere (c) Concave up for 𝑥 > 0, concave down for 𝑥 < 0. (d) 𝑅(𝑥)

Section 7.1 𝑦

53 (a)

1 (a) (ln 2)∕2 (b) (ln 2)∕2

2

3 (1∕3)𝑒3𝑥 + 𝐶

𝑥

35

2 4 6

−10

7 −0.5 cos(2𝑥) + 𝐶

−2

𝑥

9 cos(3 − 𝑡) + 𝐶

10

11 (𝑟 + 1)4 ∕4 + 𝐶

−35

(b) 𝑓 (0) = 𝑓 (7) = −1 (c) 0

(e) 1∕2

13 (1∕18)(1 + 2𝑥3 )3 + 𝐶 15 (1∕6)(𝑥2 + 3)3 + 𝐶

55 36

Chapter 6 Review 3

5 −𝑒−𝑥 + 𝐶

3

3

19 (1∕3)𝑒𝑥 +1 + 𝐶 √ 21 −2 4 − 𝑥 + 𝐶

61 𝑒 − 1 − sin 1

𝐹 (𝑥)

2

17 (1∕5)𝑦5 + (1∕2)𝑦4 + (1∕3)𝑦3 + 𝐶

57 1∕30 √ 2−1 59

23 −10𝑒−0.1𝑡+4 + 𝐶

63 𝑐 = 3

25 −(1∕8)(cos 𝜃 + 5)8 + 𝐶

65 𝑐 = 6

1

27 (1∕7) sin7 𝜃 + 𝐶

67 0.4

𝑥 1

2

3

73 2𝑒−𝑥

9 𝑥4 ∕4 − 2𝑥 + 𝐶

33 𝑡 + 2 ln |𝑡| − 1∕𝑡 + 𝐶 √ √ 35 (1∕ 2) arctan( 2𝑥) + 𝐶 √ 37 2 sin 𝑥 + 𝐶 √ 39 2 𝑥 + 𝑒𝑥 + 𝐶

4

75 (a) 𝑓 (𝑡) = 𝑄 − (b) 𝑄∕2

7 𝑥4 ∕4 + 𝐶

𝑄 𝑡 𝐴

77 (a) 𝑡 = 2, 𝑡 = 5 (b) 𝑓 (2) ≈ 55, 𝑓 (5) ≈ 40 (c) −10

11 −4∕𝑡 + 𝐶 13

31 (1∕3)(ln 𝑧)3 + 𝐶

71 3𝑥2 sin(𝑥6 )

5 (a) About −19 (b) About 6

8𝑤3∕2 ∕3

29 (1∕35) sin7 5𝜃 + 𝐶

69 16∕3

+𝐶

41 (1∕2) ln(𝑥2 + 2𝑥 + 19) + 𝐶

43 ln(𝑒𝑥 + 𝑒−𝑥 ) + 𝐶

15 sin 𝜃 + 𝐶

79 (a) 14,000 revs/min2 (b) 180 revolutions

17 𝑥2 ∕2 + 2𝑥1∕2 + 𝐶

81 19.55 ft∕sec2

47 (1∕2) sinh(2𝑤 + 1) + 𝐶

19 3 sin 𝑡 + 2𝑡3∕2 + 𝐶

83 (b) Highest pt: 𝑡 = 2.5 sec Hits ground: 𝑡 = 5 sec (c) Left sum: 136 ft (an overest.) Right sum: 56 ft (an underest.) (d) 100 ft

49

21 tan 𝑥 + 𝐶 23 (1∕2)𝑥2 + 𝑥 + ln |𝑥| + 𝐶

25 2𝑥 ∕ ln 2 + 𝐶

27 2𝑒𝑥 − 8 sin 𝑥 + 𝐶 29 4

85 𝑉 (𝑟) =

31 𝑥4 ∕4 + +2𝑥3 − 4𝑥 + 4.

45 (1∕3) cosh 3𝑡 + 𝐶

33 𝑒 + 3

53 sin 𝑥2 + 𝐶 55 (1∕5) cos(2 − 5𝑥) + 𝐶 57 (1∕2) ln(𝑥2 + 1) + 𝐶

(4∕3)𝜋𝑟3

59 0 61 1 − (1∕𝑒)

𝐻 (height)

63 2𝑒(𝑒 − 1)

35 sin 𝑥 + 4

65 2(sin 2 − sin 1)

39 4𝑥2 + ln |𝑥| + 𝐶

67 40

41 −3 cos 𝑝 + 𝐶 43

10𝑒𝑡

69 ln 3

+ 15

71 14∕3

45 2𝑧 − cos 𝑧 + 6

73 (2∕5)(𝑦 + 1)5∕2 − (2∕3)(𝑦 + 1)3∕2 + 𝐶

47 − ln 𝑥 49 Acceleration is zero at points 𝐴 and 𝐶 height

cosh3 𝑥 + 𝐶

51 (𝜋∕4)𝑡4 + 2𝑡2 + 𝐶

87

𝑥

1 3

𝑡 (time) 𝑡1 2𝑡1

𝑡3

75 (2∕5)(𝑡 + 1)5∕2 − (2∕3)(𝑡 + 1)3∕2 + 𝐶 77 (2∕7)(𝑥 − 2)7∕2 + (8∕5)(𝑥 − 2)5∕2 + (8∕3)(𝑥 − 2)3∕2 + 𝐶 79 (2∕3)(𝑡 + 1)3∕2 − 2(𝑡 + 1)1∕2 + 𝐶

𝐵 89 Increasing; concave down

𝐴

time

𝐶 𝐷

91 (a) Zeros: 𝑥 = 0, 5; Critical points: 𝑥 = 3 (b) Zeros: 𝑥 = 1; Critical points: 𝑥 = 0, 5 93 (a) Δ𝑥 = (𝑏 − 𝑎)∕𝑛 𝑥𝑖 = 𝑎 + 𝑖(𝑏 − 𝑎)∕𝑛 (b) (𝑏4 − 𝑎4 )∕4

81 ln(1 + 𝑒𝑡 ) + 𝐶 83 𝑡 = 𝑠2 85 Let 𝑡 = 𝜋 − 𝑥 87 1∕𝑤2 89 𝑒11 − 𝑒7

1088 91 3

161 ∫ sin(𝑥3 − 3𝑥)(𝑥2 − 1) 𝑑𝑥

93 218∕3

163 False

81 Li(𝑥) ln 𝑥 − 𝑥 + 𝐶

95 Substitute 𝑤 = sin 𝑥

165 (𝑚𝑔∕𝑘)𝑡 − (𝑚2 𝑔∕𝑘2 )(1 − 𝑒−𝑘𝑡∕𝑚 ) + ℎ0

83 Write arctan 𝑥 = (1)(arctan 𝑥) 85 ∫ 𝜃2 sin 𝜃 𝑑𝜃

97 Substitute 𝑤 = sin 𝑥, 𝑤 = arcsin 𝑥 √ 99 Substitute 𝑤 = 𝑥 + 1, 𝑤 = 1 + 𝑥

101 𝑒

𝑔(𝑥)

79 ℎ(𝑥) − 𝑔(𝑥) + 𝐶

87 ∫ 𝑒𝑥 sin 𝑥 𝑑𝑥

Section 7.2

+𝐶

1 (a) 𝑥3 𝑒𝑥 ∕3 − (1∕3) ∫ 𝑥3 𝑒𝑥 𝑑𝑥 (b) 𝑥2 𝑒𝑥 − 2 ∫ 𝑥𝑒𝑥 𝑑𝑥

103 2(1 + 𝑔(𝑥))3∕2 ∕3 + 𝐶 105 𝑤 = sin 𝑡, 𝑘 = 1, 𝑛 = −1

3 −𝑡 cos 𝑡 + sin 𝑡 + 𝐶

107 𝑘 = 0.3, 𝑛 = −1∕2, 𝑤0 = 12, 𝑤1 = 132 109 𝑘 = 0.5, 𝑛 = 7, 𝑤0 = 0.5, 𝑤1 = 1

5

1 5𝑡 𝑡𝑒 5

1 5𝑡 𝑒 25



+𝐶

7 −10𝑝𝑒(−0.1)𝑝 − 100𝑒(−0.1)𝑝 + 𝐶

111 𝑤 = sin 𝜙, 𝑘 = 1

9 (1∕2)𝑥2 ln 𝑥 − (1∕4)𝑥2 + 𝐶

113 𝑤 = 𝑧3 , 𝑘 = 1∕3 115 𝑎 = 3, 𝑏 = 11, 𝐴 = 0.5, 𝑤 = 2𝑡 − 3 √ 117 (a) 2 𝑥 + 𝐶 √ (b) 2√𝑥 + 1 + 𝐶 √ (c) 2 𝑥 − 2 ln( 𝑥 + 1) + 𝐶

127 (1∕2) ln 3

5 No formula 7 V-26; 𝑎 = 4, 𝑏 = −1 or 𝑎 = −1, 𝑏 = 4 9 VI-31, then VI-29; 𝑎 = 3

25 (2∕3)𝑥

3∕2

ln 𝑥 − (4∕9)𝑥

3∕2

+𝐶

27 𝑟2 [(ln 𝑟)2 − ln 𝑟 + (1∕2)]∕2 + 𝐶 29 𝑧 arctan 7𝑧 −

𝑓 (𝑥) =

1 1+𝑥

𝑥 2 129 (a) 10 (b) 5 131 (a) 0 (b) 2∕3 133 (a) (b) (c) (d)

1 14

ln(1 + 49𝑧2 ) + 𝐶

(sin2 𝜃)∕2 + 𝐶 −(cos2 𝜃)∕2 + 𝐶 −(cos 2𝜃)∕4 + 𝐶 Functions differ by a constant

11 No formula 13 V-25; 𝑎 = 3, 𝑏 = 4, 𝑐 = −2 15 (1∕6)𝑥6 ln 𝑥 − (1∕36)𝑥6 + 𝐶 17 −(1∕5)𝑥3 cos 5𝑥 + (3∕25)𝑥2 sin 5𝑥 + (6∕125)𝑥 cos 5𝑥 − (6∕625) sin 5𝑥 + 𝐶 19 (1∕7)𝑥7 + (5∕2)𝑥4 + 25𝑥 + 𝐶 21 −(1∕4) sin3 𝑥 cos 𝑥 − (3∕8) sin 𝑥 cos 𝑥 + (3∕8)𝑥 + 𝐶

31 (1∕2)𝑥2 𝑒𝑥 − (1∕2)𝑒𝑥 + 𝐶

23 ((1∕3)𝑥2 − (2∕9)𝑥 + 2∕27)𝑒3𝑥 + 𝐶

33 𝑥 cosh 𝑥 − sinh 𝑥 + 𝐶 √ √ √ 35 2 𝑥𝑒 𝑥 − 2𝑒 𝑥 + 𝐶

25 ((1∕3)𝑥4 − (4∕9)𝑥3 + (4∕9)𝑥2 − (8∕27)𝑥 + (8∕81))𝑒3𝑥 + 𝐶 √ √ 27 (1∕ 3) arctan(𝑦∕ 3) + 𝐶

2

0.54931

1 No formula

15 𝑡(ln 𝑡)2 − 2𝑡 ln 𝑡 + 2𝑡 + 𝐶

23 −2𝑡(5 − 𝑡)1∕2 − (4∕3)(5 − 𝑡)3∕2 − 14(5 − 𝑡)1∕2 + 𝐶

125 2𝑉0 ∕𝜔

Section 7.3 3 IV-17; 𝑛 = 4

21 −𝑥−1 ln 𝑥 − 𝑥−1 + 𝐶

123 𝑒3 − 𝑒2 − 𝑒 + 1

93 (a) −(1∕𝑎)𝑇 𝑒−𝑎𝑇 + (1∕𝑎2 )(1 − 𝑒−𝑎𝑇 ) (b) lim𝑇 →∞ 𝐸 = 1∕𝑎2

13 −(1∕2) sin 𝜃 cos 𝜃 + 𝜃∕2 + 𝐶

19 −(𝜃 + 1) cos(𝜃 + 1) + sin(𝜃 + 1) + 𝐶

121 2 cosh 1 − 2

91 Approximately 77

11 (1∕6)𝑞 6 ln 5𝑞 − (1∕36)𝑞 6 + 𝐶

17 (2∕3)𝑦(𝑦 + 3)3∕2 − (4∕15)(𝑦 + 3)5∕2 + 𝐶

119 (1∕2)(𝑒4 − 1)

89 True

2

37 (1∕18)(2𝑥 + 1)3 (3 ln(2𝑥 + 1) − 1) + 𝐶

39 5 ln 5 − 4 ≈ 4.047 41 −11𝑒

−10

+ 1 ≈ 0.9995

43 𝜋∕4 − (1∕2) ln 2 ≈ 0.439 45 𝜋∕2 − 1 ≈ 0.571

135 𝑤 = cos 𝑥 𝑘 = −1, 𝑎 = 1, 𝑏 = −1 or 𝑘 = 1, 𝑎 = −1, 𝑏 = 1

47 (a) (b) (c) (d) (e) (f) (g)

Parts Substitution Substitution Substitution Substitution Parts Parts

29 (1∕4) arcsin(4𝑥∕5) + 𝐶

31 (5∕16) sin 3𝜃 sin 5𝜃 + (3∕16) cos 3𝜃 cos 5𝜃 + 𝐶 | sin 𝑥+1 | 1 sin 𝑥 33 2 2 + 14 ln | sin |+𝐶 cos 𝑥 | 𝑥−1 | 5𝑥 35 (1∕34)𝑒 (5 sin 3𝑥 − 3 cos 3𝑥) + 𝐶 37 −(1∕2)𝑦2 cos 2𝑦 + (1∕2)𝑦 sin 2𝑦 + (1∕4) cos 2𝑦 + 𝐶

39 41

1 2 (tan 7𝑥∕cos2 7𝑥) + 21 tan 7𝑥 21 − 2 tan1 2𝜃 + 𝐶 1 (ln |𝑥 + 1| − ln |𝑥 + 3|) + 𝐶 2

+𝐶

137 𝑤 = ln(𝑥2 + 1), 𝑘 = 1∕2, 𝑎 = ln 5, 𝑏 = ln 26

49 𝑤 = 5 − 3𝑥, 𝑘 = −2∕3

43

139 13

51 𝑤 = ln 𝑥, 𝑘 = 3

141 0.661

53 2 arctan 2 − (ln 5)∕2

45 −(1∕3)(ln |𝑧| − ln |𝑧 − 3|) + 𝐶

143 One possible answer √ 𝑔(𝑥) = 𝑒 𝑥 𝑓 (𝑥) = 0.5 cos 𝑥,

55 2 ln 2 − 1

49 −(1∕3) sin2 𝑥 cos 𝑥 − (2∕3) cos 𝑥 + 𝐶

57 𝜋

51

147 114.414 million

59 Integration by parts gives: (1∕2) sin 𝜃 cos 𝜃 + (1∕2)𝜃 + 𝐶 The identity for cos2 𝜃 gives: (1∕2)𝜃 + (1∕4) sin 2𝜃 + 𝐶

72

149 (a) ∫0 𝑓 (𝑡) 𝑑𝑡 3

(b) ∫0 𝑓 (24𝑇 ) 24 𝑑𝑇 151 (a) 50 thousand liters/minute 15.06 thousand liters/minute (b) 1747 thousand liters 153 (a) 𝐸(𝑡) = 1.4𝑒0.07𝑡 (b) 0.2(𝑒7 − 1) ≈ 219 million megawatt-hours (c) 1972 (d) Graph 𝐸(𝑡) and estimate 𝑡 such that 𝐸(𝑡) = 219 155 (a) First case: 19,923 Second case: 1.99 billion (b) In both cases, 6.47 yrs (c) 3.5 yrs 157 Integrand needs extra factor of 𝑓 ′ (𝑥) 159 Change limits of integration for substitution in definite integral

61 (1∕2)𝑒𝜃 (sin 𝜃 + cos 𝜃) + 𝐶 63

(1∕2)𝜃𝑒𝜃 (sin 𝜃

+ cos 𝜃) − (1∕2)𝑒 sin 𝜃 + 𝐶 𝜃

47 arctan(𝑧 + 2) + 𝐶 1 5

cosh5 𝑥 −

1 3

cosh3 𝑥 + 𝐶

3

53 −(1∕9)(cos 𝑎𝜃) + (1∕15)(cos5 𝑎𝜃) + 𝐶 55

1 (1 3





2 ) 2

57 9∕8 − 6 ln 2 = −3.034 59 1∕2 61 𝜋∕12

65 𝑥𝑓 ′ (𝑥) − 𝑓 (𝑥) + 𝐶

63 0.5398

67 Integrate by parts choosing 𝑢 = 𝑥𝑛 , 𝑣′ = cos 𝑎𝑥

65 𝑘 = 0.5, 𝑤 = 2𝑥 + 1, 𝑛 = 3

69 Integrate by parts choosing 𝑢 = cos𝑛−1 𝑥, 𝑣′ = cos 𝑥

69 Form (iii) with 𝑎 = 1, 𝑏 = −5, 𝑐 = 6, 𝑛 = 7

71 125

75 (a) 𝑟′ (𝑡) = 𝑡∕(𝑡2 + 1)3∕2 > 0; 𝑟(𝑡) increases from 0 to 1 (b) Increasing, concave √ up (c) 𝑣(𝑡) = 𝑡 − ln |𝑡 + 𝑡2 + 1|

73 𝑓 (𝑥) = 𝑥3 , 𝑔(𝑥) = (1∕3) cos 𝑥 75 ℎ(𝑥) − 0.5𝑔(𝑥) + 𝐶 77 (a) −𝑎2 𝑒−𝑎 − 2𝑎𝑒−𝑎 − 2𝑒−𝑎 + 2 (b) Increasing (c) Concave up

67 Form (i) with 𝑎 = −1∕6, 𝑏 = −1∕4, 𝑐 = 5 71 𝑎 = 5, 𝑏 = 4, 𝜆 = 2

77 If 𝑎 = 3, denominator factors; answer involves ln, not arctan

1089 79 ∫ sin 𝑥 cos 𝑥 𝑑𝑥 = (sin2 𝑥)∕2 + 𝐶 √ 81 ∫ 1∕ 2𝑥 − 𝑥2 𝑑𝑥 83 True

85 False 87 (a) 0 √ (b) 𝑉0 ∕ 2 (c) 156 volts

Section 7.4

81 Use 𝑥 = 2 tan 𝜃

(b) Overestimate

83 ∫ 1∕(𝑥3 − 𝑥) 𝑑𝑥 √ 85 ∫ (1∕ 9 − 4𝑥2 ) 𝑑𝑥 87 True

89 (c), (b) 91 (a) (𝑘∕(𝑏 − 𝑎)) ln |(2𝑏 − 𝑎)∕𝑏| (b) 𝑇 → ∞

𝑥 𝑎

𝑏

Section 7.5 1 (a) Underestimate

(c) Underestimate

1 (1∕6)∕(𝑥) + (5∕6)∕(6 + 𝑥) 3 1∕(𝑤 − 1) − 1∕𝑤 − 1∕𝑤2 − 1∕𝑤3 5 −2∕𝑦 + 1∕(𝑦 − 2) + 1∕(𝑦 + 2) 7 1∕(2(𝑠 − 1)) − 1∕(2(𝑠 + 1)) − 1∕(𝑠2 + 1) 9 −2 ln |5 − 𝑥| + 2 ln |5 + 𝑥| + 𝐶 1 ln ||𝑦2 + 1|| + 𝐶 2 13 ln |𝑠| − ln |𝑠 + 2| + 𝐶 11 ln |𝑦 − 1| + arctan 𝑦 −

15 ln |𝑥 − 2| + ln |𝑥 + 1| + ln |𝑥 − 3| + 𝐾 17 2 ln |𝑥 − 5| − ln ||𝑥2 + 1|| + 𝐾 19 𝑥3 ∕3 + ln |𝑥 + 1| − ln |𝑥 + 2| + 𝐶

𝑥 𝑎

𝑥 𝑎

𝑏

(b) Overestimate

𝑏

(d) Overestimate

21 arcsin(𝑥 − 2) + 𝐶

23 (a) Yes; 𝑥 = 3 sin 𝜃 (b) No 25 𝑤 = 3𝑥2 − 𝑥 − 2, 𝑘 = 2 27 𝑎 = −5∕3, 𝑏 = 4∕5, 𝑐 = −2∕15, 𝑑 = −3∕5 29 (a) 2 ln |𝑥| + ln |𝑥 + 3| + 𝐶

𝑥 𝑎

𝑏

𝑥 𝑎

𝑏

𝑎

𝑏

31 𝑥 = (4 tan 𝜃) − 3 33 𝑤 = (𝑥 + 1)2

(c) Overestimate

35 𝑤 = 1 − (𝑧 − 1)2 37 𝑤 = (𝜃 − 2)2 − 4 39 ln |𝑥 + 2| − ln |𝑥 + 3| + 𝐶

𝑥

41 − ln |𝑥 − 1| + 2 ln |𝑥 − 2| + 𝐶

43 (ln |𝑥 + 1| − ln |𝑥 + 4|)∕3 + 𝐶

45 −4 ln |𝑥 − 1| + 7 ln |𝑥 − 2| + 𝐶 47 ln |𝑥| − (1∕2) ln ||𝑥2 + 1|| + arctan 𝑥 + 𝐾 49 𝑦 − 5 arctan(𝑦∕5) + 𝐶 51 2 ln |𝑠 + 2| − ln ||𝑠2 + 1|| + 4 arctan 𝑠 + 𝐾 53 (1∕3)(ln |𝑒𝑥 − 1| − ln |𝑒𝑥 + 2|) + 𝐶 √ 55 −𝑥 9 − 𝑥2 ∕2 + (9∕2) arcsin(𝑥∕3) + 𝐶 √ √ 57 √ (5∕2) ln |(5− 25 − 9𝑥2 )∕(5+ 25 − 9𝑥2 )|+ 25 − 9𝑥2 + 𝐶 √ √ 59 (1∕2) ln |(1− 1 + 16𝑥2 )∕(1+ 1 + 16𝑥2 )|+ 𝐶 √ 61 (1∕25)(𝑥∕ 25 + 4𝑥2 ) + 𝐶 √ 63 (1∕27)( 12 ln |(3𝑥 + 1 + 9𝑥2 )(3𝑥 − √ √ 1 + 9𝑥2 )| + 3𝑥∕ 1 + 9𝑥2 ) + 𝐶 √ 65 −4 4 − 𝑥2 + 13 (4 − 𝑥2 )3∕2 + 𝐶 67 2 ln 2 − 𝜋∕4

√ 69 8(2∕3 − 5∕(6 2)) √ √ 71 (1∕6) ln((3 − 2 2)(7 + 4 3)) 73 ln |𝑥 + 1| + ln |𝑥 − 1| + 𝐶 √ 75 (b) − 5 − 𝑦2 ∕(5𝑦) + 𝐶

𝑥 𝑎

𝑏

5 (a) Underestimate

(d) Underestimate

𝑎

𝑥

𝑥 𝑎

𝑏

(b) Overestimate

𝑎

𝑡(𝑝) = 𝑏 ln (𝑝(1 − 𝑎))∕(𝑎(1 − 𝑝)) 𝑎 = 0.01 𝑏 = 0.218 1.478

𝑏 𝑥

𝑥 𝑎

𝑏

3 (a) Underestimate

(c) Overestimate

𝑎

77 (a) (𝑎 ln |𝑥 − 𝑎| − 𝑏 ln |𝑥 − 𝑏|)∕(𝑎 − 𝑏) + 𝐶 (b) ln |𝑥 − 𝑎| − 𝑎∕(𝑥 − 𝑎) + 𝐶 79 (a) (b) (c) (d)

𝑏

𝑏 𝑥

𝑥 𝑎

𝑏

1090 (d) Underestimate

𝑎

𝑏 𝑥

𝑎

𝑏 𝑥

29 MID: over; TRAP: under

27 Does not converge

31 TRAP: over; MID: under

29 0.01317

33 (a) 𝑓 (𝑥) (b) 𝑓 (𝑥)

31 1∕ ln 3

35 (a) Increasing; concave down

35 Does not converge

37 (a) 1 (d) Ratio of errors: LEFT = 1.78, RIGHT = 2.18, TRAP = 3.96, MID = 3.93, SIMP = 15.91

37 Converges to 1/4

33 1∕3

39 Does not converge 41 (a) 0.421 (b) 0.500 43 The area is infinite 45 1∕2 47 2∕27

39 (a) 0 (c) MID(𝑛) = 0 for all 𝑛

49 9 × 109 joules √ 𝑏𝜋 51

41 Large |𝑓 ′′ | gives large error

43 445 lbs. of fertilizer

53 𝜋 4 ∕120

49 Midpoint rule exact for linear functions 7 (a) (b) (c) (d)

27 135 81 67.5

51 If 𝑓 concave down, TRAP(𝑛) ≤ MID(𝑛) 53 𝑓 (𝑥) = 1 − 𝑥2

61 True

57 False

63 False

59 False

65 False

61 True

67 True

63 False

𝑓 (𝑥) = 𝑥2 + 1

69 False.

65 False

71 (a) Γ(1) = 1 Γ(2) = 1 (c) Γ(𝑛) = (𝑛 − 1)!

67 True Area shaded = LEFT(2)

𝑥 2

69 (a) (b) (c) (d) (e)

II III V V III

Section 7.7 1 Converges; behaves like 1∕𝑥2

4

3 Diverges; behaves like 1∕𝑥

Section 7.6 𝑓 (𝑥) = 𝑥2 + 1

1 (a)

57 𝑓 (𝑥) = 1∕𝑥 59 𝑓 (𝑥) = 1∕(𝑥 − 3)2

55 True

9 (a) LEFT(2)= 12; RIGHT(2)= 44 (b) LEFT(2) underestimate; RIGHT(2) overestimate

55 1∕4

5 Diverges; behaves like 1∕𝑥

𝑦

7 Converges; behaves like 5∕𝑥3 9 Converges; behaves like 1∕𝑥2

Area shaded = RIGHT(2)

11 Inconclusive 13 Converges

𝑥

𝑥 2

1

4

15 Does not converge 17 Converges 19 Does not converge

11 (a) (b) (c) (d)

𝜋∕2 𝜋∕2 𝜋∕2 √ 2𝜋∕2

(b)

𝑦

21 Converges 23 Converges 25 Converges 27 Converges

13 18.1

𝑥

15 4.863

1

17 I, IV 19 0.1649 21 LEFT(6) = 31 RIGHT(6) = 39 TRAP(6) = 35 23 (a) 12 (b) SIMP(2) = 12 SIMP(4) = 12 SIMP(100) = 12 25 (a) RIGHT = 0.368, LEFT = 1, TRAP = 2 1 0.684, MID =0.882, ∫0 𝑒−𝑥 ∕2 𝑑𝑥 = 0.856 (b) LEFT = 0.368, RIGHT = 1, TRAP = 2 0 0.684, MID =0.882, ∫−1 𝑒−𝑥 ∕2 𝑑𝑥 = 0.856 27 (a) TRAP(5) = 0.3846 (b) Concave down (c) 0.3863

3 (a) 0.9596, 0.9995, 0.99999996 (b) 1.0 5 Diverges 7 −1 9 Converges to 1∕2 11 1 13 ln 2 15 Converges to 1/16

29 Does not converge 31 (a) (b) (c) (d) (e) (f) (g)

Converges Either Converges Either Converges Diverges Converges

33 Converges for 𝑝 > 1 Diverges for 𝑝 ≤ 1 ∞

2

35 (a) ∫3 𝑒−𝑥 𝑑𝑥 ≤ 𝑒−9 ∕3 (b)

2 ∞ ∫𝑛 𝑒−𝑥

𝑑𝑥 ≤ (1∕𝑛)𝑒−𝑛

2

17 4

37 Cannot compare, first integrand sometimes larger than second

19 Does not converge

39 Comparison test not applicable

21 𝜋∕4

41 𝑓 (𝑥) = 3∕(2𝑥2 + 1)

23 Diverges

43 True

25 Does not converge

45 True

1091 √ 93 (2∕3)𝑥3∕2 + 2 𝑥 + 𝐶 √ 95 (4∕3)( 𝑥 + 1)3∕2 + 𝐶

47 False

Chapter 7 Review

187 (a) (𝑘∕𝐿) ln 3 (b) 𝑇 → ∞ 189 (a) 0.5 ml (b) 99.95%

97 −1∕(4(𝑧2 − 5)2 ) + 𝐶

1 (𝑡 + 1)3 ∕3

99 (1 +

3 5𝑥 ∕ ln 5

101 𝑒𝑥

5 3𝑤2 ∕2 + 7𝑤 + 𝐶

tan 𝑥)4 ∕4

2 +𝑥

+𝐶

191 (a) 360 degree-days (b)𝐻 , ◦ C 𝑡 = 𝑇 = 18

+𝐶

3

𝑡, days

109

1 3

15 −𝑒−𝑧 + 𝐶

111

49932 16

17 𝑥2 ∕2 + ln |𝑥| − 𝑥−1 + 𝐶

113 201,760

19 (1∕2)𝑒𝑡 + 𝐶

115 3∕2 + ln 2

21 ((1∕2)𝑥2 − (1∕2)𝑥 + 1∕4)𝑒2𝑥 + 𝐶

117 𝑒 − 2 ≈ 0.71828

23 (1∕2)𝑦2 ln 𝑦 − (1∕4)𝑦2 + 𝐶

119 −11𝑒−10 + 1

13 2𝑥5∕2 ∕5 + 3𝑥5∕3 ∕5 + 𝐶

2

sinh3 𝑥 + 𝐶

18 (c) 𝑇2 = 36

121 3(𝑒 − 𝑒)

25 𝑥(ln 𝑥)2 − 2𝑥 ln 𝑥 + 2𝑥 + 𝐶

123 0

27 (1∕3) sin3 𝜃 + 𝐶

125

29 (1∕2)𝑢2 + 3𝑢 + 3 ln |𝑢| − 1∕𝑢 + 𝐶

127 (ln |𝑥| − ln |𝐿 − 𝑥|)∕𝐿 + 𝐶

31 tan 𝑧 + 𝐶

33 (1∕12)𝑡12 − (10∕11)𝑡11 + 𝐶

35 (1∕3)(ln 𝑥)3 + 𝐶 37

39 (1∕2)𝑒𝑡

+

2 +1

+ ln |𝑥| + 𝐶

+𝐶

2

ln |𝑥 − 1| −

1 2

ln |𝑥 + 1| + 𝐶

129 arcsin(𝑥∕5) + 𝐶

131 (1∕3) arcsin(3𝑥) + 𝐶 133 2 ln |𝑥 − 1| − ln |𝑥 + 1| − ln |𝑥| + 𝐾 135 arctan(𝑥 + 1) + 𝐶

137 (1∕𝑏) arcsin(𝑏𝑥∕𝑎) + 𝐶 𝑥

𝑥

139 (1∕2)(ln |𝑒 − 1| − ln |𝑒 + 1|) + 𝐶

41 (1∕10) sin (5𝜃)+𝐶 (other forms of answer are possible)

141 Does not converge

43 arctan 𝑧 + 𝐶

143 6

45 −(1∕8) cos4 2𝜃 + 𝐶

145 Does not converge

47 (−1∕4) cos4 𝑧 + (1∕6) cos6 𝑧 + 𝐶

147 Does not converge

49 (2∕3)(1 + sin 𝜃)3∕2 + 𝐶 51

𝑡3 𝑒𝑡



3𝑡2 𝑒𝑡

+ 6𝑡𝑒𝑡

𝑦 𝐻 = 𝑔(𝑡) 𝑡 = 𝑇2 30 20 𝑆 = Area = 360 𝐻 = 𝐻min 10 𝑥 18 36

2

1 2

− 6𝑒𝑡

+𝐶

𝐻 = 𝐻min = 10

10

107 (𝑥 + sin 𝑥)4 ∕4 + 𝐶

11 (−1∕2) cos 2𝜃 + 𝐶

𝑥2

𝑆 = Area = 360

5

105 sin (2𝜃)∕6 − sin (2𝜃)∕10 + 𝐶

9 (1∕5)𝑒5𝑧 + 𝐶

(1∕3)𝑥3

𝐻 = 𝑓 (𝑡) = 30

30

103 −(2∕9)(2 + 3 cos 𝑥)3∕2 + 𝐶

7 − cos 𝑡 + 𝐶

149 Converges to 𝜋∕8 √ 151 Converges to 2 15

193 (a) (ln 𝑥)2 ∕2, (ln 𝑥)3 ∕3, (ln 𝑥)4 ∕4 (b) (ln 𝑥)𝑛+1 ∕(𝑛 + 1) 195 (a) (−9 cos 𝑥 + cos(3𝑥))∕12 (b) (3 sin 𝑥 − sin(3𝑥))∕4 197 (a) 𝑥 + 𝑥∕(2(1 + 𝑥2 )) − (3∕2) arctan 𝑥 (b) 1 − (𝑥2 ∕(1 + 𝑥2 )2 ) − 1∕(1 + 𝑥2 )

Section 8.1 ∑

1 (a) 2𝑥 Δ𝑥 (b) 9 ∑√ 𝑦 Δ𝑦 3 (a) (b) 18

53 (3𝑧 + 5)4 ∕12 + 𝐶

153 Does not converge

5 15

55 arctan(sin 𝑤) + 𝐶

155 Converges to value between 0 and 𝜋∕2

7 15/2

57 − cos(ln 𝑥) + 𝐶

157 Does not converge

59 𝑦 − (1∕2)𝑒

−2𝑦

+𝐶

61 ln | ln 𝑥| + 𝐶 √ 63 sin 𝑥2 + 1 + 𝐶

65 𝑢𝑒𝑘𝑢 ∕𝑘 − 𝑒𝑘𝑢 ∕𝑘2 + 𝐶 √ √ 67 (1∕ 2)𝑒 2+3 + 𝐶

69 (𝑢3 ln 𝑢)∕3 − 𝑢3 ∕9 + 𝐶

71 − cos(2𝑥)∕(4 sin2 (2𝑥)) | (cos(2𝑥)−1) | + 81 ln | (cos(2𝑥)+1) |+𝐶 | | 73 −𝑦2 cos(𝑐𝑦)∕𝑐 + 2𝑦 sin(𝑐𝑦)∕𝑐 2 + 2 cos(𝑐𝑦)∕𝑐 3 + 𝐶 75

1 5𝑥 𝑒 (5 cos(3𝑥) 34

+ 3 sin(3𝑥)) + 𝐶

√ √ 77 ( 3∕4)(2𝑥 √ 1 + 4𝑥2 + ln |2𝑥 + 1 + 4𝑥2 |) + 𝐶

79 𝑥2 ∕2 − 3𝑥 − ln |𝑥 + 1| + 8 ln |𝑥 + 2| + 𝐶

81 (1∕𝑏) (ln |𝑥| − ln |𝑥 + 𝑏∕𝑎|) + 𝐶 83

(𝑥3 ∕27)

+ 2𝑥 − (9∕𝑥) + 𝐶

85 −(1∕ ln 10)101−𝑥 + 𝐶 ) ( 87 (1∕2)𝑣2 − 1∕4 arcsin 𝑣+ √ (1∕4)𝑣 1 − 𝑣2 + 𝐾

89 𝑧3 ∕3 + 5𝑧2 ∕2 + 25𝑧 + 125 ln |𝑧 − 5| + 𝐶 91 (1∕3) ln | sin(3𝜃)| + 𝐶

159 𝑒9 ∕3 − 𝑒6 ∕2 + 1∕6 √ 161 Area = 2 2

163 𝑤 = 3𝑥5 + 2, 𝑝 = 1∕2, 𝑘 = 1∕5

9 (5∕2)𝜋 11 1/6 13 (a) I:𝐶; II:𝐵; III:𝐴; IV:𝐷 1 2 (b) II:∫0 𝑥 𝑑𝑥 + ∫1 (2 − 𝑥) 𝑑𝑥 2

III:∫1 (2𝑥 − 2) 𝑑𝑥

165 𝐴 = 2∕5, 𝐵 = −3∕5 167

15

0.5𝑢𝑒𝑢 𝑑𝑢; ∫ 𝑘 = 0.5, 𝑢 = −𝑥2

17

6 ∫0 7 ∫0 2

𝜋(𝑥2 ∕9) 𝑑𝑥 = 8𝜋 cm3 √ 20 72 − 𝑦2 𝑑𝑦 = 245𝜋 m3

169 Substitute 𝑤 = 𝑥 into second integral

19 ∫0 (2 − 𝑦)2 𝑑𝑦 = 8∕3 m3

171 Substitute 𝑤 = 1 − 𝑥2 , 𝑤 = ln 𝑥

21 ∫0 6 𝑑𝑥 = 90 cm3

173 𝑤 = 2𝑥

23 Semicircle 𝑟 = 9

175 5∕6

25 Triangle; 𝑏, ℎ = 5, 7

177 (a) 3.5 (b) 35

27 (a) ∫0 (3𝑥 − 𝑥2 ) 𝑑𝑥 = 4.5 9 √ (b) ∫0 ( 𝑦 − (𝑦∕3)) 𝑑𝑦 = 4.5

2

179 11∕9 181 Wrong; improper integral treated as proper integral, integral diverges 183 RT(5) < RT(10) < TR(10) < Exact < MID(10) < LF(10) < LF(5) 185 (a) 4 places: 2 seconds 8 places: ≈ 6 hours 12 places: ≈ 6 years 20 places: ≈ 600 million years (b) 4 places: 2 seconds 8 places: ≈ 3 minutes 12 places: ≈ 6 hours 20 places: ≈ 6 years

15

3

2

6

29 (a) ∫0 𝑥2 𝑑𝑥 + ∫2 (6 − 𝑥) 𝑑𝑥 = 2.667 + 8 = 10.667 √ 4 (b) ∫0 ((6 − 𝑦) − 𝑦) 𝑑𝑦 = 10.667 31 Hemisphere, 𝑟 = 12 33 Sphere, 𝑟 = 8 35 Hemisphere 𝑟 = 2 37 (a) 0.2. ( ) (b) 0.2𝜋 16 − ℎ2 39 36𝜋 = 113.097 m3

41 (a) 3Δ𝑥; 4 ∫0 3 𝑑𝑥 = 12 cm3

1092

✛2 cm

Radius =

1 + 𝑥3

Section √8.3

𝑦



Δ𝑥

✛ ✻

1 (−1∕2, 3∕2) √ 3 (3, − 3) √ 5 ( 2, 𝜋∕4) √ 7 (2 2, −𝜋∕6)

(𝑦 = −1)





𝑥

3 cm

9 (b)



𝑦 2

43 𝑉 = (𝜋 2 ∕2) ≈ 4.935

1 𝑥

(b) 8(1 − ℎ∕3)Δℎ; 3 ∫0 8(1 − ℎ∕3) 𝑑ℎ = 12 cm3

𝑦 Radius =

45 4𝜋∕5

1400(160 − ℎ) 𝑑ℎ = 1.785 ⋅ 107 m3

47 8∕15 √ 3∕8 49

45 36𝜋 = 113.097 m3 47 III 49 IV 51 Change to

10 ∫−10

( ) 𝜋 102 − 𝑥2 𝑑𝑥

53 Region between positive 𝑥-axis, positive 𝑦axis, and 𝑦 = 1 − 𝑥 55 False

2 1

51 𝑉 ≈ 42.42

𝑥

53 𝑉 = (𝑒2 − 1)∕2 ) 4 ( 55 𝑉 = ∫0 𝜋 3𝑦2 ∕4 − 12𝑦 + 36 𝑑𝑦 = 64𝜋 ) 1 ( 57 𝑉 = ∫0 𝜋 (2 − 𝑦)4 − 𝑦4 𝑑𝑦 = 6𝜋 59 (a) 16𝜋∕3 (b) 1.48

61 𝑉 = 32𝜋𝑎3 ∕105 √ 63 40, 000𝐿𝐻 3∕2 ∕(3 𝑎)

57 True

65 3509 cubic inches √ 4 67 ∫0 1 + (4 − 2𝑥)2 𝑑𝑥

59 True

Section 8.2 3 ∫0

1 (a) 4𝜋 (b) 36𝜋

𝑥2

𝑑𝑥

9 ∫0

69 (a) (𝜋∕2)(𝑎2 − 𝑥2 ) Δ𝑥 𝑎 (b) (𝜋∕2) ∫−𝑎 (𝑎2 − 𝑥2 ) 𝑑𝑥 (c) 2𝜋𝑎3 ∕3 √ 71 64 3∕15

𝜋𝑦 𝑑𝑦 3 (a) (b) 40.5𝜋

73 𝑘 = 6

5 𝜋∕5

75 122.782 m

7 256𝜋∕15

77 80

9 𝜋(𝑒2 − 𝑒−2 )∕2

79 (a) 3𝜋 (c) 2

11 𝜋∕2 13 2𝜋∕15 6

15 𝑉 = ∫0 𝜋𝑦2 ∕9 𝑑𝑦 = 8𝜋 ) ln 2 ( 17 𝑉 = ∫0 𝜋 4 − 𝑒2𝑦 𝑑𝑦 = 4𝜋 ln 2 − 3𝜋∕2

2

(c) Cartesian: √ ( √ 3∕4, 1∕4); (− 3∕4, 1∕4) or polar: 𝑟 = 1∕2, 𝜃 = 𝜋∕6 or 5𝜋∕6 (d) 𝑦

𝑥



150

43 ∫0

❄✛ ✛ ✻𝑤



Δℎ



✛2 cm

1

−2

sin 𝑥



4 cm

−2 −1 −1

𝑧

−2 −1 −1

1

2

−2 11 Looks the same 13 Rotated by 90◦ clockwise 15 𝜋∕4 ≤ 𝜃 ≤ 5𝜋∕4; 0 ≤ 𝜃 ≤ 𝜋∕4 and 5𝜋∕4 ≤ 𝜃 ≤ 2𝜋 √ √ 8 ≤ 𝑟 ≤ 18 and 𝜋∕4 ≤ 𝜃 ≤ 𝜋∕2 17 19 0 ≤ 𝜃 ≤ 𝜋∕2 and 1 ≤ 𝑟 ≤ 2∕ cos 𝜃 21 −1 √ 2(𝑒𝜋 − 𝑒𝜋∕2 ) 23

𝑦

25

1

𝑥 −1

81 (a) Change in 𝑥 is 27 Change in 𝑦 is 54 (b) (20, 65) (c) 61 83 𝑉 = 2𝜋

1 −1

27 4𝜋 3 29 (a) 𝑟 = 1∕ cos 𝜃; 𝑟 = 2 𝜋∕3 (b) 21 ∫−𝜋∕3 (22 − (1∕ cos 𝜃)2 ) 𝑑𝜃 √ (c) (4𝜋∕3) − 3

19 2.958

85 2595 cubic feet

21 2.302

87 𝑦 = 0.8𝑥5∕4 √ 31 2𝜋𝑎 ( )2 3 √ 89 ∫1 1 + 4𝑥3 − 24𝑥2 + 36𝑥 + 3 𝑑𝑥 33 (5𝜋∕6) + 7 3∕8 √ ( ) 2 5 35 (a) 𝑦 91 ∫−5 1 + (𝑒𝑥∕𝑎 − 𝑒−𝑥∕𝑎 )∕2 𝑑𝑥 𝑟 = 2 sin 𝜃 √ 𝜋∕4 2 93 Change to ∫0 1 + cos 𝑥 𝑑𝑥 2 𝜃 = 𝜋∕4 √ 1 95 Change to ∫0 (−2𝜋 sin(2𝜋𝑡))2 + (2𝜋 cos(2𝜋𝑡))2 𝑑𝑡 𝑟 = 2 cos 𝜃 1

23 𝜋 √ 41 25

27 ≈ 24.6 5

29 𝑉 = ∫0 𝜋((5𝑥)2 − (𝑥2 )2 ) 𝑑𝑥 5

31 𝑉 = ∫0 𝜋((4 + 5𝑥)2 − (4 + 𝑥2 )2 ) 𝑑𝑥 2

33 𝑉 = ∫0 [𝜋(9 − 𝑦3 )2 − 𝜋(9 − 4𝑦)2 ] 𝑑𝑦 9

35 𝑉 = ∫0 [𝜋(2 + 13 𝑥)2 − 𝜋22 ] 𝑑𝑥 37 3.820

97 Region bounded by 𝑦 = 2𝑥, 𝑥-axis, 0 ≤ 𝑥 ≤ 1

𝑥

99 𝑓 (𝑥) = 𝑥2

39 (a) 8𝑎∕3 (b) 32𝑎2 𝜋∕5

101 False

41 𝑉 = (16∕7)𝜋 ≈ 7.18

105 (c) 𝑓 (𝑥) =

1

103 False √

3𝑥

(b) (𝜋∕2) − 1

2

1093 37 (a)

39 Mass < 27𝜋 gm

𝑦 1.5

41 𝛿(𝑥) = 𝑥

√ 𝑟= 2

43 False

𝑥 −2

2 𝑟2 = 4 cos 2𝜃

45 False 47 False

23 (a) $5716.59 per year (b) $74,081.82

49 False

√ (b) 2 3 − 2𝜋∕3

27 3.641%∕ yr

1 30 ft-lb

29 9.519%∕ yr

3 1.333 ft-lb

31 (a) Option 1 (b) Option 1: $10.929 million; Option 2: $10.530 million

39 Horiz: (±1.633, ±2.309); (0, 0) Vert: (±2.309, ±1.633); (0, 0)

5 27/2 joules

41 (a) 𝑦 = (2∕𝜋)𝑥 + (2∕𝜋) (b) 𝑦 = 1

9 20 ft-lb

7 1.176 ⋅ 107 lb 10

43 21.256

11 1.489 ⋅ 10

47 2.828

13 3437.5 ft-lbs

49 Points are in quadrant IV

15 180,000 ft-lb

joules

33 In 10 years ∑𝑛−1 (2000 − 100𝑡𝑖 )𝑒−0.03𝑡𝑖 Δ𝑡 35 (a) 𝑖=0 𝑀

(b) ∫0 𝑒−0.03𝑡 (2000 − 100𝑡) 𝑑𝑡 (c) After 20 years $16,534.63

17 3,088,800 ft-lbs

𝑦

51

25 Accept installments

Section 8.5

−1.5

(b) 15 years 21 (a) Lump sum better at 6% Continuous payments better at 3% (b) Interest rates remain high, above 5.3% per year

39 (a) price

19 2,822,909.50 ft-lbs

𝑟 = 3 sin 2𝜃 𝑥

21 661,619.41 ft-lb 23 (a) 62.4𝜋(8𝑦 − 6𝑦2 + 𝑦3 )Δ𝑦 1 (b) ∫0 62.4𝜋(8𝑦 − 6𝑦2 + 𝑦3 ) 𝑑𝑦 = 140.4𝜋 25 354,673 ft-lbs

Supply

27 1,170,000 lbs

𝑞∗

29 4,992,000 lbs 53 𝑟 = 100 55 𝑟 = 1 + cos 𝜃

Section 8.4 1 1 − 𝑒−10 gm 𝑁

∑ (2 + 6𝑥𝑖 )Δ𝑥

3 (a)

𝑖=1

(b) 16 grams ∑𝑁 [600 + 𝑖=1 √ 300 sin(4 𝑥𝑖 + 0.15)](20∕𝑁) (c) ≈ 11513

5 (b)

7 2 cm to right of origin

31 (a) 1.76 ⋅ 106 nt/m2 180 (b) 1.96 ⋅ 107 ∫0 ℎ 𝑑ℎ = 3.2 ⋅ 1011 nt 33 (a) 780,000 lb/ft2 About 5400 lb/in2 √ 3 (b) 124.8 ∫−3 (12,500 − ℎ) 9 − ℎ2 𝑑ℎ 7 = 2.2 ⋅ 10 lb 7000

35 ∫0 37 (a) (b) (c) (d) (e)

𝑑ℎ∕𝑣(ℎ) seconds

14 + 0.2ℎ kg (137.2 + 1.96ℎ) Δℎ joules 6m 858.48 joules 1387.68 joules

𝐷(𝑞)

(b) price

Supply

∶ 𝑆(𝑞)

𝑝∗ 𝑞∗

Demand ∶ quantity

𝐷(𝑞)

(c) price

41 Not all of rope raised 20 m

9 45

Supply

43 196 joules

13 1 gm

45 𝐴 tall, thin; 𝐵 short, fat

𝑏 = ∫𝑎 ∑𝑁 𝑖=1

𝛿(𝑥)[𝑓 (𝑥) − 𝑔(𝑥)]𝑑𝑥 ) ( 17 (a) 2𝜋𝑟𝑖 50∕(1 + 𝑟𝑖 ) Δ𝑟 (b) 3.14 ⋅ 106 kg (c) 5009 meters

19 (a) (b) (c) (d)

Demand ∶ quantity

39 Potential √ = 2𝜋𝜎( 𝑅2 + 𝑎2 − 𝑅)

11 186,925 15 𝑀

∶ 𝑆(𝑞)

𝑝∗

𝐶(𝑠) = 10 + 2𝑠; 30 g/m3 157.080 g; overestimate (10 + 2𝑠)𝜋Δ𝑠 1,884.956 g

21 Total mass = 12; 𝑥̄ = 2.06 23 (a) 𝑥̄ = (3∕4) ((2 + 𝑘)∕(3 + 𝑘)) 25 𝑥̄ = 4∕(3𝜋), 𝑦̄ = 0 27 (a) 10∕3 gm (b) 𝑥 = 3∕5 cm; 𝑦 = 3∕8 cm 29 1.25 cm from center of base 31 (a) 16000𝛿∕3 gm (b) 2.5 cm above center of base 60

33 ∫0 (1∕144)𝑔(𝑡) 𝑑𝑡 ft3 35 𝑀𝑈 ≈ 6.50 × 1027 g 𝑀𝐿 ≈ 5.46 × 1027 g ( ) ( ) 10 10 37 ∫0 𝑥3 𝑑𝑥 ∕ ∫0 𝑥2 𝑑𝑥

47 True

𝑞∗

49 True 51 True ( ) (𝑎 + 𝑙1 )(𝑎 + 𝑙2 ) 𝐺𝑀1 𝑀2 ln 53 𝑙1 𝑙2 𝑎(𝑎 + 𝑙1 + 𝑙2 )

∶ 𝑆(𝑞)

𝑝∗ Demand ∶ quantity

𝐷(𝑞)

(d) price

55 60 joules

Section 8.6 1 𝐶(1 +

0.02)20

Supply

dollars

∶ 𝑆(𝑞)

𝑝∗

3 𝐶∕𝑒0.02(10) dollars

𝑞∗

15

5 ∫0 𝐶 𝑒0.02(15−𝑡) 𝑑𝑡 dollars

Demand ∶ quantity

𝐷(𝑞)

7 ln(25,000∕𝐶)∕30 per year 9 $22,658.65, $27,675.34

(e) price

11 (a) Future value = $6389.06 Present value = $864.66 (b) 17.92 years Supply

13 3.466%∕ yr 15 $4000∕ yr; $21,034.18 17 $1000∕ yr; $24,591.23 19 (a) $300 million

∶ 𝑆(𝑞)

𝑝∗ 𝑞∗

Demand ∶ quantity

𝐷(𝑞)

1094 1

11 For small Δ𝑥 around 70, fraction of families with incomes in that interval about 0.05Δ𝑥

(f) price

13 𝐴 is 300 kelvins; 𝐵 is 500 kelvins Supply

∶ 𝑆(𝑞)

𝑝∗ Demand ∶ quantity

𝑞∗

𝐷(𝑞)

0.5

15 (a) About 0.9 m–1.1 m

𝑟

19 (a) Cumulative distribution increasing (b) Vertical 0.2, horizontal 2

1

21 (a) (b) (c) (d)

43 Smaller with 4%

23 (b) About 3∕4

45 Dollars

25 Prob 0.02Δ𝑥 in interval around 1 27

𝑡

0

1

2

3

4

$

3855

4241

4665

5131

5644

Section 8.7 1 (a)-(II), (b)-(I), (c)-(III) 3

% of population per dollar of income

22.1% 33.0% 30.1% 𝐶(ℎ) = 1 − 𝑒−0.4ℎ

∞ ∫−∞

2

3

(b) Mean: 1.5 Bohr radii Median: 1.33 Bohr radii Most likely: 1 Bohr radius

41 (a) Less (b) Cannot tell (c) Less

49 Assuming 10%/yr and 𝑡 in yrs from now:

𝑃 (𝑟)

27 Median cannot be 1 since all of the area is to the left of 𝑥 = 1

𝑝(𝑡) 𝑑𝑡 ≠ 1

⎧ 0 ⎪ 29 𝑝(𝑥) = ⎨ 1 ⎪ ⎩ 0

29 𝑃 (𝑥) grows without bound as 𝑥 → ∞, instead of approaching 1 31 Cumulative distribution increasing { 0, 𝑡 < 0 33 𝑃 (𝑡) = 𝑡, 0 ≤ 𝑡 ≤ 1 1, 𝑡 > 1 35 𝑃 (𝑥) = (𝑥 − 3)∕4, 3 ≤ 𝑥 ≤ 7

for

𝑥1

31 True

33 False

Chapter 8 Review

37 False

1

Section 8.8

income

5 2.65 m 7 (a) 𝐴 = 0.015, 𝐵 = 0.005 (b) 33.33 (c) 37.5 (d) fraction of population

% of population having at least this income

1 0.8 0.6 0.4 0.2 0

income

5 pdf; 1∕4

𝑥 (test score) 25 50 75 100

𝑃 (𝑥) 1 9 20% 11 22%

𝑥

13 Mean 2∕3; Median 2 −

1 4

15 (a) 0.684 ∶ 1 (b) 1.6 hours (c) 1.682 hours



2 = 0.586

17 (a) 𝑃 (𝑡) = Fraction of population who survive up to 𝑡 years after treatment (b) 𝑆(𝑡) = 𝑒−𝐶𝑡 (c) 0.178

7 cdf; 1

𝑃 (𝑥) 1∕5

19 (a) 𝑝(𝑥) =

−1 1 √ 𝑒 2 15 2𝜋

(

) 𝑥−100 2 15

(b) 6.7% of the population

𝑥

21 (c) 𝜇 represents the mean of the distribution, while 𝜎 is the standard deviation 23 (b)

5

25 (a) 𝑝(𝑟) = 4𝑟2 𝑒−2𝑟

9 512𝜋∕15 = 107.233 9√ 9 − 𝑦 𝑑𝑦 11 (a) 12𝜋 ∫0 (b) 216𝜋 2

13 𝑉 = ∫0 𝜋(𝑦3 )2 𝑑𝑦 ] 2[ 15 𝑉 = ∫0 𝜋(10 − 4𝑦)2 − 𝜋(2)2 𝑑𝑦 ] 2[ 17 𝑉 = ∫0 𝜋(4𝑦 + 3)2 − 𝜋(𝑦3 + 3)2 𝑑𝑥 23 45.230

1

25 4.785 27 15.865

𝑝(𝑥)

0.5

1 6

1

29 ∫0 𝑓 (𝑥) 𝑑𝑥 < 1∕2

𝑝(𝑟) 𝑟

𝑥 2

7 3𝜋∕2 = 4.712

19 𝑉 = 𝜋 𝑎 √ 21 2 ∫−𝑎 1 + 𝑏2 𝑥2 ∕(𝑎2 (𝑎2 − 𝑥2 )) 𝑑𝑥

9 cdf; 1∕3 1 3

√ 𝑟 3 2 ∫−𝑟 𝑟2 − 𝑥2 𝑑𝑥 = 𝜋𝑟2 ∑ √ 5 (a) 2 9 − 𝑦 Δ𝑦 (b) 36

4

1

2

3

1

31 ∫0 𝑓 −1 (𝑥) 𝑑𝑥 > 1∕2 √ √ 1 1 + (𝑓 ′ (𝑥))2 𝑑𝑥 > 2 33 ∫0

1095 35

Section 9.1

𝑦 4 √ 𝑦 = 2 4 − 𝑥2

𝑥 2 𝑦 2

√ 𝑦 = 4 − 𝑥2 𝑥 2

−2

√ 𝑦 = − 4 − 𝑥2

25

37 𝑉 = ∫0 𝜋((8 − 𝑦∕5)2 − (8 − 39 (a) 𝑉 = 5𝜋∕6 (b) 𝑉 = 4𝜋∕5



𝑦)2 ) 𝑑𝑦

41 736𝜋∕3 = 770.737 43 2𝜋∕3 45 𝜋 2 ∕2 − 2𝜋∕3 = 2.840 47 𝜋∕24 49 Area under graph of 𝑓 from 𝑥 = 0 to 𝑥 = 3 51 𝑒 − 𝑒−1 53 (a) 𝑎 = 𝑏∕𝑙 (b) (1∕3)𝜋𝑏2 𝑙 55 Volume = 6𝜋 2 57 𝑦 = (1∕3)𝑒3𝑡 from 𝑡 = 3 to 𝑡 = 8 59 (𝑥 − 𝑎)2 + 𝑦2 = 𝑎2 √ 61 (𝜋∕3 + 3∕2)𝑎2 ; 61%

21 Not geometric

3 2∕3, 4∕5, 6∕7, 8∕9, 10∕11

23 Not geometric

5 1, −1∕2, 1∕4, −1∕8, 1∕16

25 10 terms; 0.222

7 𝑠𝑛 = 2𝑛+1

27 14 terms; 15.999

9 𝑠 𝑛 = 𝑛2 + 1

29 −486

11 𝑠𝑛 = 𝑛∕(2𝑛 + 1)

31 1∕(1 − 5𝑥), −1∕5 < 𝑥 < 1∕5

13 (a) (b) (c) (d)

(IV) (III) (II) (I)

35 1∕(1 − 𝑧∕2), −2 < 𝑧 < 2

15 (a) (b) (c) (d) (e)

(II) (III) (I) (IV) (V)

43 Does not converge

17 Converges to 0

45 Does not converge

19 Converges to 0

47 32 tons

21 Converges to 0 25 Converges to 1∕2

49 (a) 0.232323 … = 0.23 + 0.23(0.01) + 0.23(0.01)2 + ⋯ (b) 0.23∕(1 − 0.01) = (23)∕(99)

27 Diverges

51 (a)

23 Converges to 0

33 13,27,50,85 35 𝑠𝑛 = 𝑠𝑛−1 + 2, 𝑠1 = 1 37 𝑠𝑛 = 2𝑠𝑛−1 − 1, 𝑠1 = 3 39 𝑠𝑛 = 𝑠𝑛−1 + 𝑛, 𝑠1 = 1 45 45 47 2 49 0.878, 0.540, 0.071, −0.416, −0.801, −0.990 51 0, 6, −6, 6, −6, 6, … and 3, 0, 2, −2, 2, … 53 1.5, 2, 3, 4, 5, 6, 7 … and 1.75, 2.17, 3, 4, 5, 6, … 55 Converges to 0.5671

65 1∕4

59 (a) 2, 4, 2𝑛 (b) 33 generations; overlap

71 518,363 ft-lbs 73 (a) Force on dam ∑ 1000(62.4ℎ𝑖 )Δℎ ≈ 𝑁−1 𝑖=0 50 ∫0

= 1000(62.4ℎ) 𝑑ℎ = 78,000,000 pounds

(b)

87 (a) (b) 𝜋ℎ∕𝑎 (c) 𝑑ℎ∕𝑑𝑡 = −𝑘 (d) ℎ0 ∕𝑘

55 (a) 𝑃𝑛 = 250(0.04) + 250(0.04)2 + 250(0.04)3 + ⋯ + 250(0.04)𝑛−1 (b) 𝑃𝑛 = 250 ⋅ 0.04(1 − (0.04)𝑛−1 )∕(1 − 0.04) (c) lim 𝑃𝑛 ≈ 10.42 𝑛→∞

Difference between them is 250 mg 57

𝑞 (quantity, mg)

𝑄0 = 250 125

𝑄1 𝑄2 𝑄3 𝑄4 𝑄5

1

𝑃1 𝑃2 𝑃3 𝑃4 𝑃5 𝑡 (time, days) 2 3 4 5 6

59 $400 million

71 False

61 (a) $1000 (b) When the interest rate is 5%, the present value equals the principal (c) The principal (d) Because the present value is more than the principal

77 False 79 (b)

(b) 1.042 ⋅ 1012 cubic feet

𝜋ℎ2 ∕(2𝑎)

53 (a) 17.9(1.025) + 17.9(1.025)2 + … + 17.9(1.025)𝑛 ; 733.9(1.025𝑛 − 1) (b) 2040 (c) Right Riemann sum with 𝑛 subdivisions

67 𝑠𝑛 = 2 − 1∕𝑛

75 False

85 (a) 𝜋𝑃 𝑅4 ∕(8𝜂𝑙)

(i) $16.43 million

(ii) $24.01 million (b) $16.87 million

69 𝑠𝑛 = 𝑛

77 33 billion m3 /year )2 √ ∑𝑁 ( 5 Δℎ 79 (a) 𝑖=1 𝜋 (3.5 ⋅ 10 )∕ ℎ + 600 𝑅

2 41 8∕(1 √ √ − (𝑥 − 5)) − 6 < 𝑥 < −2 and 2 < 𝑥 < 6

65 Converges to 3∕7

73 True

83 ∫0 2𝜋𝑁𝑟𝑆(𝑟)𝑑𝑟

39 3 + 𝑥∕(1 − 𝑥), −1 < 𝑥 < 1

61 (a) 𝑘 rows; 𝑎𝑛 = 𝑛 (b) 𝑇𝑛 = 𝑇𝑛−1 + 𝑛, 𝑛 > 1; 𝑇1 = 1

75 0.366(𝑘 + 1.077)𝑔𝜋 joules

81 92% of pop younger than 70

37 𝑦∕(1 + 𝑦), −1 < 𝑦 < 1

31 1, 5, 7, 17, 31, 65

57 Converges to 1

69 1000 ft-lb

33 1∕𝑥, 0 < 𝑥 < 4

29 1, 3, 6, 10, 15, 21

63 2𝜋𝑎

67 4400 ft-lb

19 Geometric; diverges

1 3, 5, 9, 17, 33

Section 9.2 1 Series 3 Sequence 5 Series 7 Series 9 Yes, 𝑎 = 2, ratio = 1∕2

65 |𝑥| ≥ 1

67 3 + 6 + 12 + 24 + ⋯

69 16∕3 + 8∕3 + 4∕3 + 2∕3 71 𝑎𝑛 = 2𝑟𝑛 , 𝑏𝑛 = 3𝑟𝑛 , |𝑟| < 1 73 (c)

Section 9.3

89 The thin spherical shell

11 No

91 (c) 𝜋 2 𝑎3 ∕8 √ √√ 93 (a) 12 𝑡 1 + 4𝑡 + 14 arcsinh (2 𝑡) (b) 𝑡

13 No

3 1∕2, 2∕3, 3∕4, 4∕5, 5∕6

15 Yes, 𝑎 = 1, ratio = −𝑦2

5 Diverges

17 No

7 Converges

1 1, 3, 6, 10, 15

1096 11 𝑓 (𝑥) = (−1)𝑥 ∕𝑥 undefined

59 lim𝑛→∞ 𝑎𝑛 ≠ 0

27 −3 < 𝑥 < 3

13 Diverges

61 Does not converge

29 −2 < 𝑥 < 2

15 Diverges

63 Converges to approximately 0.3679

31 −∞ < 𝑥 < ∞

17 Converges

65 Diverges

19 Converges

67 Converges

21 Converges

69 Diverges

23 Diverges

71 Converges

33 −1∕5 ≤ 𝑥 < 1∕5 ∑∞ 𝑛 35 𝑛=0 (−2𝑧) , −1∕2 < 𝑧 < 1∕2 ∑∞ 𝑛 37 𝑛=0 3(𝑧∕2) , −2 < 𝑧 < 2

25 Converges

73 Converges

27 Diverges

75 Diverges

29 Diverges

77 Converges

31 Converges

79 Diverges

41 (a) (b) (c) (d)

35 (a) ln(𝑛 + 1) (b) Diverges

81 Diverges

43 5 ≤ 𝑅 ≤ 7

83 Diverges

37 (b) 𝑆3 = 1 − 1∕4; 𝑆10 = 1 − 1∕11; 𝑆𝑛 = 1 − 1∕(𝑛 + 1)

45 𝑘 = 0.1, 𝐶0 = 3, 𝐶1 = 1, 𝐶2 = 4, 𝐶3 = 1

85 Diverges

49 (a) Odd; 𝑔(0) = 0 (c) sin 𝑥

45 Terms approaching zero does not guarantee convergence

89 Converges

47 Converges for 𝑘 < −1 and diverges for 𝑘 ≥ −1 ∑∞ 49 𝑛=1 1∕𝑛

51 True

53 False 55 True 57 False 63 (a) (b) (c) (d)

39 (a) 5.833 (b) −4 < 𝑧 < 10

87 Diverges

51 Limit may be > 2 ∑∞ 𝑛 53 𝑛=0 (𝑥 − 2) ∕𝑛 ∑∞ 𝑛 55 𝑛=1 (1∕𝑛)(𝑥 − 10)

91 Converges 93 (a) Converges (b) Converges

57 False

95 (a) Converges (b) Diverges

59 False

97 (a) Diverges (b) Diverges

63 False

61 True 65 True

99 Unaffected

1.596 3.09 1.635; 3.13 0.05; 0.01

67 True

101 Affected 103 Converges for 𝑎 > 2 and diverges for 𝑎 ≤ 2 105 Converges for 𝑎 > 1 and diverges for 𝑎 ≤ 1

67 (d)

107 99 or more terms 109 2 or more terms

Section 9.4 ∑ 5 Behaves like

7 Behaves like

9 Converges

∞ 𝑛=1 ∑∞ 𝑛=1

113 (−1)2𝑛 = 1 so series is not alternating 1∕𝑛; Diverges 1∕𝑛4 ; Converges

11 Converges 13 Converges 15 Converges 17 Converges 19 Converges 21 Diverges 23 Converges 25 Converges 27 Diverges 29 Diverges 31 Diverges 33 Alternating

False True False Cannot determine

115 Need 1∕𝑛3∕2 ≤ 1∕𝑛2 for all 𝑛 to make comparison ∑ 117 (−1)𝑛 𝑛

69 True 71 (a) All real numbers (b) 𝐽 (0) = 1 (c) 𝑆4 (𝑥) = 1 − (d) 0.765 (e) 0.765

+

𝑥4 64



𝑥6 2304

+

𝑥8 147,456

Chapter 9 Review 1 3(211 − 1)∕210

119 False

3 619.235

121 True 123 False

5 𝑏5 (1 − 𝑏6 )∕(1 − 𝑏) when 𝑏 ≠ 1; 6 when 𝑏 = 1

125 True

7 (317 − 1)∕(2 ⋅ 320 )

127 False

9 36, 48, 52, 53.333; 𝑆𝑛 = 36(1 − (1∕3)𝑛 )∕(1 − 1∕3); 𝑆 = 54

129 False 133 False

11 −810, −270, −630, −390; 𝑆𝑛 = −810(1 − (−2∕3)𝑛 )∕(1 + 2∕3); 𝑆 = −486

135 (b)

13 Converges 4∕7

139 Converges

15 Diverges

131 True

17 Converges

Section 9.5

19 Converges

35 Not alternating

1 Yes

21 Converges

37 Converges

3 No

23 Diverges

39 Converges

5 1 ⋅ 3 ⋅ 5 ⋯ (2𝑛 − 1)𝑥𝑛 ∕(2𝑛 ⋅ 𝑛!); 𝑛 ≥ 1

25 Converges

41 Converges

7 (−1)𝑘 (𝑥 − 1)2𝑘 ∕(2𝑘)!; 𝑘 ≥ 0

27 Divergent

9 (𝑥 −

𝑎)𝑛 ∕(2𝑛−1

⋅ 𝑛!); 𝑛 ≥ 1

29 Divergent

43 (a) No (b) Converges (c) Converges

11 1

33 Converges

13 1

35 Converges

45 Conditionally convergent

15 2

37 Converges

47 Divergent

17 2

39 Converges

49 Conditionally convergent

19 1

41 Diverges

51 Conditionally convergent

21 1∕4

43 Diverges

53 Terms not positive

23 1

45 Convergent

55 Limit of ratios is 1

25 (a) 𝑅 = 1 (b) −1 < 𝑥 ≤ 1

47 Converges

57 𝑎𝑛+1 > 𝑎𝑛 or lim𝑛→∞ 𝑎𝑛 ≠ 0

𝑥2 4

49 Converges

1097 51 Diverges

17 (a) (b) (c) (d)

53 Converges 55 Diverges 57 Diverges

19 (−1)𝑛 𝑥𝑛 ; 𝑛 ≥ 0

2 −1 −2∕3 12

21 (−1)𝑛−1 𝑥𝑛 ∕𝑛; 𝑛 ≥ 1 23 (−1)𝑘 𝑥2𝑘+1 ∕(2𝑘 + 1); 𝑘 ≥ 0

59 ∞

19 𝑃3 (𝑥) = 1 − 5𝑥 + 7𝑥2 + 𝑥3 𝑓 (𝑥) = 𝑃3 (𝑥)

61 1

21 1 − 𝑥∕3 + 5𝑥2 ∕7 + 8𝑥3

63 −3 < 𝑥 < 7

23 −3 + 5𝑥 − 𝑥2 − 𝑥4 ∕24 + 𝑥5 ∕30

65 −∞ < 𝑥 < ∞ 69 9, 15, 23

25 (a) Positive (b) 𝑃1 (𝑥) = ln 4 + (1∕2)𝑥 (c) 𝑃2 (𝑥) = ln 4 + (1∕2)𝑥 − (1∕8)𝑥2 𝑃3 (𝑥) = ln 4+(1∕2)𝑥−(1∕8)𝑥2 +(1∕24)𝑥3

71 Converges if 𝑟 > 1, diverges if 0 < 𝑟 ≤ 1

27 𝑎 > 0, 𝑏 < 0, 𝑐 < 0

73 (a) All real numbers (b) Even

29 𝑎 < 0, 𝑏 < 0, 𝑐 > 0

67 𝑠𝑛 = 2𝑛 + 3

31

75 240 million tons

=3

(i) 𝐵0 4𝑛 (ii) 𝐵0 2𝑛∕10 (iii) (21.9 )𝑛 (b) 10.490 hours

83 (a)

85 5∕6 ∑ ∑ 87 (3∕2)𝑛 , (1∕2)𝑛 , other answers possible

89 Converges

91 Not enough information 93 Converges 95 (a) 𝐶 0

𝐶1 𝐶2 𝐶3 (b) 1∕3 + 2∕9 + 4∕27 + ⋯ + (1∕3)(2∕3)𝑛−1 (c) 1

43 (a) 1 + 3𝑥 + 2𝑥2 (b) 1 + 3𝑥 + 7𝑥2 (c) No

3 𝑃2 (𝑥) = 1 + (1∕2)𝑥 − (1∕8)𝑥2 𝑃3 (𝑥) = 1 + (1∕2)𝑥 − (1∕8)𝑥2 + (1∕16)𝑥3 𝑃4 (𝑥) = 1 + (1∕2)𝑥 − (1∕8)𝑥2 + (1∕16)𝑥3 − (5∕128)𝑥4 5 𝑃2 (𝑥) = 1 − 𝑥2 ∕2! 𝑃4 (𝑥) = 1 − 𝑥2 ∕2! + 𝑥4 ∕4! 𝑃6 (𝑥) = 1 − 𝑥2 ∕2! + 𝑥4 ∕4! − 𝑥6 ∕6! 7 𝑃3 (𝑥) = 𝑃4 (𝑥) = 𝑥 − (1∕3)𝑥3 9 𝑃2 (𝑥) = 1 − (1∕2)𝑥 + (3∕8)𝑥2 𝑃3 (𝑥) = 1 − (1∕2)𝑥 + (3∕8)𝑥2 − (5∕16)𝑥3 𝑃4 (𝑥) = 1 − (1∕2)𝑥 + (3∕8)𝑥2 − (5∕16)𝑥3 + (35∕128)𝑥4

(1∕2)𝑥4

45 (a) 𝑃4 (𝑥) = 1 + + (b) If we substitute 𝑥2 for 𝑥 in the Taylor polynomial for 𝑒𝑥 of degree 2, we will get 2 𝑃4 (𝑥), the Taylor polynomial for 𝑒𝑥 of degree 4 (c) 𝑃20 (𝑥) = 1 + 𝑥2 ∕1! + 𝑥4 ∕2! + ⋯ + 𝑥20 ∕10! (d) 𝑒−2𝑥 ≈ 1 − 2𝑥 + 2𝑥2 − (4∕3)𝑥3 + (2∕3)𝑥4 − (4∕15)𝑥5 47 (a) Infinitely many; 3 (b) That near 𝑥 = 0 Taylor poly only accurate near 0

𝑥2 8 𝑃4 (𝑥) = 31 [1 − 𝑥−2 3 2 3 + (𝑥−2) − (𝑥−2) 32 33 (𝑥−2)4

11 𝑃3 (𝑥) = 1 −

+

34 √

𝑥 2





𝑥3 16

]

) ( 15 𝑃3 (𝑥) = 22 [−1 + 𝑥 + 𝜋4 ( )2 ( )3 + 21 𝑥 + 𝜋4 − 16 𝑥 + 𝜋4 ]

61 False 63 False 65 False

51 𝑒−0.1

57 1 61 Only converges for −1 < 𝑥 < 1 63 𝑓 (𝑥) = cos 𝑥 65 False 67 True 69 True 71 True

Section 10.3 3 cos (𝜃 2 ) = 1 − 𝜃 4 ∕2! + 𝜃 8 ∕4! − 𝜃 12 ∕6! + ⋯

9 𝜙3 cos(𝜙2 ) = 𝜙3 − 𝜙7 ∕2! + 𝜙11 ∕4! − 𝜙15 ∕6! + ⋯

Section 10.2 1 𝑓 (𝑥) = 1 + 3𝑥∕2 + 3𝑥2 ∕8 − 𝑥3 ∕16 + ⋯ 3 𝑓 (𝑥) = −𝑥 + 𝑥3 ∕3! − 𝑥5 ∕5! + 𝑥7 ∕7! + ⋯ 5 𝑓 (𝑥) = 1 + 𝑥 + 𝑥2 + 𝑥3 + ⋯ 𝑦2 ∕9

9 ln 5 + (2∕5)𝑥 − (2∕25)𝑥2 + (8∕475)𝑥3 √ √ 11 cos 𝜃√= 2∕2 − ( 2∕2) (𝜃 √ − 𝜋∕4) −( 2∕4) (𝜃 − 𝜋∕4)2 +( 2∕12) (𝜃 − 𝜋∕4)3 −⋯ √ √ 13 sin 𝜃 √ = − 2∕2 + ( 2∕2) (𝜃 + 𝜋∕4) + (√2∕4) (𝜃 + 𝜋∕4)2 − ( 2∕12) (𝜃 + 𝜋∕4)3 + ⋯

15 1∕𝑥 = 1 − (𝑥 − 1) + (𝑥 − 1)2 − (𝑥 − 1)3 + ⋯ 1 𝑥

49 𝑒3

5 arcsin 𝑥 = 𝑥 + (1∕6)𝑥3 + (3∕40)𝑥5 + (5∕112)𝑥7 + ⋯ √ 7 1∕ 1 − 𝑧2 = 1 + (1∕2)𝑧2 + (3∕8)𝑧4 + (5∕16)𝑧6 + ⋯

59 False

17

47 ln(3∕2)

1 𝑒−𝑥 = 1 − 𝑥 + 𝑥2 ∕2! − 𝑥3 ∕3! + ⋯

57 𝑓 (𝑥) = 1 + 𝑥; 𝑔(𝑥) = 1 + 𝑥 + 𝑥2

7 𝑓 (𝑦) = 1 − 𝑦∕3 − − 5𝑦3 ∕81 − ⋯

43 𝑒2 45 4∕3

55 216

𝑥2

55 𝑓 (𝑥) = sin 𝑥

1 𝑃3 (𝑥) = 1 + 𝑥 + 𝑥2 + 𝑥3 𝑃5 (𝑥) = 1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4 + 𝑥5 𝑃7 (𝑥) = 1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4 + 𝑥5 + 𝑥6 + 𝑥7

= 360

33 𝑓 (𝑛) (0) > 0 for all 𝑛

53 4∕5

53 𝑓 ′ (0) = 1

Section 10.1

2 𝑑 (𝑥2 𝑒𝑥 )|𝑥=0 = 0 𝑑𝑥 6 2 𝑑 (𝑥2 𝑒𝑥 )|𝑥=0 = 6!2 𝑑𝑥6

41 1

49 (b) 0, 0.2

97 12

31

39 (a) −1 < 𝑥 < 1 (b) 1

41 (a) 𝑓 s are Figure 10.9 𝑔s are Figure 10.8 (b) 𝐴 = (0, 1) 𝐵 = (0, 2) (c) 𝑓1 = III, 𝑓2 = I, 𝑓3 = II 𝑔1 = III, 𝑔2 = II, 𝑔3 = I

81 £250

29 1

35 𝑓 (𝑛) (0) = 0 for even 𝑛

39 (a) 1∕2 (b) 1∕6

79 (a) $124.50 (b) $125

27 (a) 0 (b) Increasing (c) Concave down

37 −1 < 𝑥 < 1

35 −2.616

77 $926.40

13

𝑓 ′′′ (1)

25 (−1)𝑘 𝑥4𝑘+2 ∕(2𝑘)!; 𝑘 ≥ 0

= −1 − (𝑥 + 1) − (𝑥 + 1)2 − (𝑥 + 1)3 − ⋯

11 1 + 𝑡2 ∕2! + 𝑡4 ∕4! + 𝑡6 ∕6! + ⋯ 13 1 + 3𝑥 + 3𝑥2 + 𝑥3 0 ⋅ 𝑥𝑛 for 𝑛 ≥ 4 15 1 + (1∕2)𝑦2 + (3∕8)𝑦4 + ⋯ + ((1∕2)(3∕2) ⋯ (1∕2 + 𝑛 − 1)𝑦2𝑛 )∕𝑛! + ⋯; 𝑛≥1 17 ln(3∕2) √ √ 𝑇 + ℎ = 𝑇 (1 + (1∕2)(ℎ∕𝑇 ) 19 − (1∕8)(ℎ∕𝑇 )2 + (1∕16)(ℎ∕𝑇 )3 + ⋯) 21 1∕(𝑎 + 𝑟)2 = (1∕𝑎2 )(1 − 2(𝑟∕𝑎) + 3(𝑟∕𝑎)2 − 4(𝑟∕𝑎)3 + ⋯) √ 23 𝑎∕ 𝑎2 + 𝑥2 = 1−(1∕2)(𝑥∕𝑎)2 +(3∕8)(𝑥∕𝑎)4 − (5∕16)(𝑥∕𝑎)6 + ⋯ 25 𝑒𝑡 cos 𝑡 = 1 + 𝑡 − 𝑡3 ∕3 − 𝑡4 ∕6 + ⋯ 27 1 + 𝑡 + 𝑡2 ∕2 + 𝑡3 ∕3 29 (a) 1∕2−(𝑥−1)∕4+(𝑥−1)2 ∕8−(𝑥−1)3 ∕16+ ⋯ + (−1)𝑛 (𝑥 − 1)𝑛 ∕2𝑛+1 + ⋯

1098 (b) (𝑥 − 1)2 ∕2 − (𝑥 − 1)3 ∕4 + (𝑥 − 1)4 ∕8 − (𝑥 − 1)5 ∕16 + ⋯ + (−1)𝑛 (𝑥 − 1)𝑛+2 ∕2𝑛+1 + ⋯ ∑ 2𝑚+1 ∕(2𝑚 + 1)! 31 sinh 2𝑥 = ∞ 𝑚=0 (2𝑥) ∑ cosh 2𝑥 = ∞ (2𝑥)2𝑚 ∕(2𝑚)! 𝑚=0 33 (a) 2𝑥 + 𝑥3 ∕3 + 𝑥5 ∕60 + ⋯ (b) 𝑃3 (𝑥) = 2𝑥 + 𝑥3 ∕3

35 1∕(1 − 𝑥)2

(i) 4, 0.2 (ii) 1 (b) −4 ≤ 𝑥 ≤ 4

23 (a)

25 Four decimal places: 𝑛 = 7 Six decimal places:𝑛 = 9

√ 37 1 + sin 𝜃 ≤ 𝑒𝜃 ≤ 1∕ 1 − 2𝜃 39 (a) (b) (c) (d)

15 𝐹4 (𝑥) = 1∕2 − (2∕𝜋) sin(2𝜋𝑥) − (1∕𝜋) sin(4𝜋𝑥) − (2∕3𝜋) sin(6𝜋𝑥) − (1∕2𝜋) sin(8𝜋𝑥)

(b) |𝐸2 | ≤ 0.17

21 (a) Underestimate (b) 1

29 (a) (b) (c) (d)

I IV III II

41 𝑓 (𝑥) = (1∕3)𝑥3 − (1∕42)𝑥7 + (1∕1320)𝑥11 − (1∕75,600)𝑥15 + ⋯ 43 1

𝐹4 (𝑥)

𝜋 ≈ 2.67 𝜋 ≈ 2.33 |𝐸𝑛 | ≤ 0.78 Derivatives unbounded near 𝑥 = 1

𝑥 1

31 (a) (1∕2)𝑛+1 ∕(𝑛 + 1)! (b) (1∕2)𝑛+1 ∕(𝑛 + 1)! (c) Equal

17 52.93

33 |𝑓 (𝑎) − 𝑃𝑛 (𝑎)| = 0

45 (a) 𝑓 (𝑥) = 1 + (𝑎 − 𝑏)𝑥 + (𝑏2 − 𝑎𝑏)𝑥2 + ⋯ (b) 𝑎 = 1∕2, 𝑏 = −1∕2

23 (a) (b) (c) (d)

35 𝑓 (𝑥) = sin 𝑥

37 𝑒𝑥 on [−1, 1]

41 True

47 (a) If 𝑀 >> 𝑚, then 𝜇 ≈ 𝑚𝑀∕𝑀 = 𝑚 (b) 𝜇 = 𝑚[1 − 𝑚∕𝑀 + (𝑚∕𝑀)2 − (𝑚∕𝑀)3 + ⋯] (c) −0.0545%

Section 10.5 1 Not a Fourier series 3 Fourier series

49 (a) 𝜃 ′′ + (𝑔∕𝑙)𝜃 = 0 (b) 0.76%

53 (a) Set 𝑑𝑉 ∕𝑑𝑟 = 0, solve for 𝑟. Check for max or min. (b) 𝑉 (𝑟) = −𝑉0 + 72𝑉0 𝑟−2 ⋅ (𝑟 − 𝑟0 )2 (1∕2) 0 +⋯ (d) 𝐹 = 0 when 𝑟 − 𝑟0

1

1

−𝜋

𝑥 −3𝜋 −2𝜋 −𝜋 − 11 55

𝑥 𝜋

𝐹1 (𝑥) = 𝐹2 (𝑥) =

4 𝜋

sin 𝑥

𝑥3 3!

−𝜋

35 Any odd function with period 2𝜋

𝑥

37 (b)

−1

+⋯

63 |𝑥|

𝐹3 (𝑥) =

4 𝜋

65 True

sin 𝑥 +

4 3𝜋

Chapter 10 Review

sin 3𝑥

67 True 7 99.942% of the total energy

69 (c)

Section 10.4 1 |𝐸3 | ≤ 0.00000460, 𝐸3 = 0.00000425 3 |𝐸3 | ≤ 0.000338, 𝐸3 = 0.000336

9 𝐻𝑛 (𝑥) =∑ 𝜋∕4 + 𝑛𝑖=1 ((−1)𝑖+1 sin(𝑖𝑥))∕𝑖 + ∑[𝑛∕2] (−2∕((2𝑖 − 1)2 𝜋)) cos((2𝑖 − 1)𝑥), 𝑖=1 where [𝑛∕2] denotes the biggest integer smaller than or equal to 𝑛∕2

7 |𝐸3 | ≤ 16.5, 𝐸3 = 0.224

𝜋

9 1∕5! 11 1∕5!

✛ ℎ(𝑥) ✛ 𝐻3 (𝑥) ✛ 𝐻2 (𝑥) ✛ 𝐻 (𝑥) 1

13 1∕50

0.03 𝐸1

𝑥 𝑥

−0.5

−𝜋

0.5

19 (a) Overestimate: 0 10; 𝑃 tends to 10

5 (VI)

21 Slope field (b)

𝑦

9 (a) Not a solution (b) Solution (c) Not a solution 11 (a) (I), (III) (b) (IV) (c) (II), (IV) 15 No 17 (a) Decreasing (b) 7.75 21 𝑃 = 15∕𝑡

(4, 1) (0, 0)

5 Possible curves:

3 (II) 7 (V)

𝑥

0

49 False

55 True

43 (a) 𝑉 (𝑥) ≈ 𝑉 (0) + 𝑉 ′′ (0)𝑥2 ∕2 𝑉 ′′ (0) > 0 (b) Force ≈ −𝑉 ′′ (0)𝑥 𝑉 ′′ (0) > 0 Toward origin 𝑥2

𝑦

41 𝑑𝑦∕𝑑𝑥 = 𝑥∕𝑦 and 𝑦 = 100 when 𝑥 = 0

𝑥

23 (a) (b) (c) (d) (e)

(II) (I) (V) (III) (IV)

25 (a) (b) (c) (d) (e) (f)

IV I III V II VI

1100 𝑦 27 (a) 𝑦′ = 0.05𝑦(10 − 𝑦).

3 Separable; 1∕(𝑒𝑦 − 1)𝑑𝑦 = 𝑥 𝑑𝑥

29 (d) 𝑦′ = 0.05𝑥(5 − 𝑥).

5 Separable; (1∕𝑦)𝑑𝑦 = 1∕(1 − 𝑥)𝑑𝑥

31 Graph of 𝑦 = 𝑥 not tangent to slope field at (1, 1)

9 𝑄 = 50𝑒(1∕5)𝑡

7 𝑃 = 20𝑒0.02𝑡

33 Slopes of 𝑦′ = 𝑦 in quadrant II are positive

11 𝑚 = 5𝑒3𝑡−3

35 𝑑𝑦∕𝑑𝑥 = 𝑦

13 𝑧 = 5𝑒5𝑡−5

37

𝑦

𝑥

15 𝑢 = 1∕(1 − (1∕2)𝑡) 17 𝑦 = 10𝑒−𝑥∕3 19 𝑃 = 104𝑒𝑡 − 4 21 𝑄 = 400 − 350𝑒0.3𝑡

𝑥

23 𝑅 = 1 − 0.9𝑒1−𝑦 25 𝑦 = (1∕3)(3 + 𝑡)

(c) 𝑦2 − 𝑥2 = 2𝐶

27 𝑦 = 3𝑥5 29 𝑧 = 5𝑒𝑡+𝑡 39 True

57 Impossible to separate the variables in 𝑑𝑦∕𝑑𝑥 = 𝑥 + 𝑦

3 ∕3

47 True

31 𝑤 = −1∕(ln | cos 𝜓| − 1∕2) √ 3 6𝑥2 + 𝐵 33 (a) 𝑦 = √ √ 3 3 (b) √ 𝑦 = 6𝑥2 + 1; 𝑦 = 6𝑥2 + 8; 𝑦 = 3 6𝑥2 + 27

49 True

𝑦

41 False 43 False 45 False

59 𝑒−𝑦 𝑑𝑦 = 𝑒𝑥 𝑑𝑥 61 𝑓 (𝑥) = cos 𝑥 63 𝑑𝑦∕𝑑𝑥 = 𝑥∕𝑦 65 True 67 False 69 False

51 True 53 False

4

Section 11.5

Section 11.3 −4

5 1.97 7 (b) 𝑦(𝑥) =



0.25

True answer

(III) (IV) (I) (II)

3 (a) 𝑦 = 2: stable; 𝑦 = −1: unstable (b) 𝑦

35 (a) 𝑦(𝑡) = 100 − 𝐴𝑒−𝑡 (b) 𝑦

110



0.16

Approximation

𝑥

𝑥

25

1

𝑡

(b) At 𝑥 = 1, 𝑦 ≈ 0.16 (c) Underestimate

(c) 𝑦(𝑡) = 100 − 75𝑒−𝑡 𝑦(𝑡) = 100 + 10𝑒−𝑡 (d) 𝑦(𝑡) = 100 − 75𝑒−𝑡

11 (a) 𝑦 = 0.667 (b) 𝑦 = 0.710 13 Error in ten-step is half error ins five-step; 𝑦 = 0.753

37 (a)

5

𝑦

𝑦

𝛼

(0, 1)

15 (a) Δ𝑥 = 0.5, 𝑦(1) ≈ 1.5 Δ𝑥 = 0.25, 𝑦 ≈ 1.75 (b) 𝑦 = 𝑥2 + 1, so 𝑦(1) = 2 (c) Yes 17 (a)

4

𝑥4 ∕4

𝑦

9 (a)

1 (a) (b) (c) (d)

𝑥

3 1548, 1591.5917, 1630.860

(0, 0)

𝑥

𝑡

(0, −1)

(i) 𝐵 ≈ 1050 (ii) 𝐵 ≈ 1050.63 (iii) 𝐵 ≈ 1050.94 (b) One end asymptotic to 𝑦 = 0, other end unbounded (c) 𝑦 = −1∕(𝑥 + 𝐶), 𝑥 ≠ −𝐶

21 Overestimate if 𝑥(0) < 0 23 𝑑𝑦∕𝑑𝑥 = 𝑦, 𝑦(0) = 1

7 (a) 𝐻 = 200 − 180𝑒−𝑘𝑡 (b) 𝑘 ≈ 0.027 (if 𝑡 is in minutes) 9 (a) Positive (b) Approximately 10 minutes

25 True

39 𝑄 = 𝐴𝑒𝑡∕𝑘

11 Decreasing

27 Error decreases by factor of approx 1∕2

41 𝑄 = 𝑏 − 𝐴𝑒−𝑡

13 10,250

29 We would need 𝑛 = 300

43 𝑅 = −(𝑏∕𝑎) + 𝐴𝑒𝑎𝑡 ( ) 45 𝑦 = −1∕ 𝑘(𝑡 + 𝑡3 ∕3) + 𝐶

15 Decrease

Section 11.4 1 (a) Yes (d) No (g) No (j) Yes

(b) No (e) Yes (h) Yes (k) Yes

(c) Yes (f) Yes (i) No (l) No

47 𝐿 = 𝑏

2 + 𝐴𝑒𝑘((1∕2)𝑥 +𝑎𝑥)

49 𝑥 = 𝑒𝐴𝑡 −𝑡

51 𝑦 = 2(2−𝑒 ) 53 (a) and (b)

19 Longest: Lake Superior Shortest: Lake Erie The ratio is about 75 21 Is a solution 23 (a) 𝑦 = 0, 𝑦 = 4, 𝑦 = −2 (b) 𝑦 = 0 stable, 𝑦 = 4 and 𝑦 = −2 unstable

1101 𝑦 13 𝑑𝐴∕𝑑𝑡 = −0.17𝐴 + 130; 𝐴 = 764.7 − 764.7𝑒−0.17𝑡 ; 625 mg

7

𝑑𝑄∕𝑑𝑡

15 (a) 𝑑𝑃 ∕𝑑𝑡 = 0.079 − 0.000857𝑡 (b) 𝑃 = 0.079𝑡 − 0.000857𝑡2 ∕2 + 7.052 (c) 9.435 bn

𝑥

800

17 (a) Mt. Whitney: 17.50 inches Mt. Everest: 10.23 inches (b) 18,661.5 feet

25 (a) 𝑁 = amount of nicotine in mg at time 𝑡; 𝑡 = number of hours since smoking a cigarette (b) 𝑑𝑁∕𝑑𝑡 = −0.347𝑁 (c) 𝑁 = 0.4𝑒−0.347𝑡 27 (a) 𝐺 = size in acres of Grinnell Glacier in year 𝑡; 𝑡 = number of years since 2007 (b) 𝑑𝐺∕𝑑𝑡 = −0.043𝐺 (c) 𝐺 = 142𝑒−0.043𝑡

9 (a) 𝑃 = 0, 𝑃 = 2000 (b) 𝑑𝑃 ∕𝑑𝑡 positive; 𝑃 increasing

19 (b) 𝑑𝑄∕𝑑𝑡 = −0.347𝑄 + 2.5 (c) 𝑄 = 7.2 mg 21 𝑃 = 𝐴𝑉

11 (a) 𝑘 = 0.035; gives relative growth rate when 𝑃 is small 𝐿 = 6000; gives limiting value on the size of 𝑃 (b) 𝑃 = 3000

𝑘

23 5.7 days, 126.32 liters 25 (a) 𝑑𝑝∕𝑑𝑡 = −𝑘(𝑝 − 𝑝∗ ) (b) 𝑝 = 𝑝∗ + (𝑝0 − 𝑝∗ )𝑒−𝑘𝑡 (c) 𝑝

13 𝑃 = 2800∕(1 + 𝐴𝑒−0.05𝑡 ) 15 𝑃 = 4000∕(1 + 𝐴𝑒−0.68𝑡 )

𝑝0 > 𝑝∗

𝑡

𝑑𝑃 ∕𝑑𝑡 = 0.01𝑟𝑃 − 𝐻 𝑃 = 300 million fish 𝑃 = 100𝐻∕𝑟 No

21 𝑃 = 8500∕(1 + 16𝑒−0.8𝑡 ) 23 (a) Between 80,000 and 85,000 cases (b) About 6500 cases/week

(d) As 𝑡 → ∞, 𝑝 → 𝑝∗

25 (a) 539,226 cases (b) 47,182 cases/week

27 (a) 𝑄 = (𝑟∕𝛼)(1 − 𝑒−𝛼𝑡 ) 𝑄∞ = 𝑟∕𝛼

35 (a) 𝑑𝑄∕𝑑𝑡 = −0.5365𝑄 𝑄 = 𝑄0 𝑒−0.5365𝑡 (b) 4 mg

27 (a) 30,000 cases (b) 6000 cases/week 29 22 days

𝑄

31 (a) 𝑃 = 5000∕(1 + 499𝑒−1.78𝑡 ) (b) 𝑦

𝑟 𝛼

37 (a) 𝑘 ≈ 0.000121 (b) 779.4 years

𝑄=

𝑟 (1 − 𝑒−𝛼𝑡 ) 𝛼

5000 𝑡

39 (a) 𝑑𝑇 ∕𝑑𝑡 = −𝑘(𝑇 − 68) (b) 𝑇 = 68 + 22.3𝑒−0.06𝑡 ; 3:45 am.

2500 (b) Doubling 𝑟 doubles 𝑄∞ ; Altering 𝑟 does not alter the time it takes to reach (1∕2)𝑄∞ (c) Both 𝑄∞ and the time to reach (1∕2)𝑄0 are halved by doubling 𝛼

41 (a) 𝑑𝐷∕𝑑𝑡 = 𝑘𝐷 𝑑𝑄∕𝑑𝑡 = 𝑘𝑄 𝑄 = 0, stable 𝑄 = 𝐶𝑒𝑘𝑡 𝑘 ≈ −0.182 𝑄(12) ≈ 1.126 mg

29 𝑡 = 133.601 min √ 31 𝑦 = 𝐴 𝑥

45 𝑑𝑦∕𝑑𝑥 ≠ 0 when 𝑦 = 2 49 𝑑𝑄∕𝑑𝑡 = 𝑄 − 500

(c) 𝑡 ≈ 3.5; 𝑃 ≈ 2500

35 Units of time are not consistent √ 37 𝑑𝑄∕𝑑𝑡 = 0.2 3 𝑄 + 100 ∑59 39 (a) 𝐵60 = 𝑘=0 100𝑒(0.02)𝑘∕12 (b) 𝐵60 = 6304.998 (c) Smaller because deposits are made later

Section 11.6 (III) (V) (I) (II) (IV)

Section 11.7

3 𝑑𝐵∕𝑑𝑡 = 0.037𝐵 + 5000

1 (b) 1

5 𝑑𝐵∕𝑑𝑡 = −0.065𝐵 − 50,000

41 (a) (b) (c) (d)

1261 bn barrels About 28 bn barrels 1434 bn barrels Very close

49

(b) (a)

11 𝑑𝐷∕𝑑𝑡 = −0.75(𝐷 − 4) Equilibrium = 4 g/cm2

39 (a) 2017 (b) 2025

47 𝑑𝑃 ∕𝑑𝑡 = 0.2𝑃 (1 − 𝑃 ∕150)

2500 1250

37 𝑑𝑁∕ 𝑑𝑡 = 𝑁(0.155 − 0.001𝑁); 155 million

45 Inflection point should be at 𝑃 = 50

(c)

9 𝑑𝑎∕𝑑𝑡 = −𝑘𝑎∕𝑚 𝑎 = 𝑔𝑒−𝑘𝑡∕𝑚

35 (a) 𝐿 = 12,000 (b) 360

43 𝑃 = 0 also equilibrium

𝑦

3

7 (a) 𝐻 ′ = 𝑘(𝐻 − 50); 𝐻(0) = 90 (b) 𝐻(𝑡) = 50 + 40𝑒−0.05754𝑡 (c) 24 minutes

𝑡 3.5

33 (a) 2500 fish (b) 2230 fish (c) 1500 fish

33 (a) (0, 𝑎) √ (b) 𝑑𝑦∕𝑑𝑥 = −𝑦∕ 𝑎2 − 𝑦2

47 Surrounding air is 350◦ F not 40◦ F

1 (a) (b) (c) (d) (e)

19 𝑘 = 0.3, 𝐿 = 100, 𝐴 = 1∕3, 𝑃 = 100∕(1 + 𝑒−0.3𝑡 ∕3), 𝑡 = − ln(3)∕0.3

𝑝0 < 𝑝∗

33 (a) 𝑑𝑄∕𝑑𝑡 = −𝑘𝑄 (b) Half-life is (ln 2)∕𝑘 (c) Decreasing

43 (a) (b) (c) (d) (e)

17 𝑘 = 10, 𝐿 = 2, 𝐴 = 3, 𝑃 = 2∕(1 + 3𝑒−10𝑡 ), 𝑡 = ln(3)∕10

𝑝∗

29 (a) 𝑑𝐻∕𝑑𝑡 = −𝑘(𝐻 − 22) (b) 𝐻 = 22 − 19𝑒−𝑘𝑡 31 (a) (b) (d) (e)

𝑄

𝑑𝑃 ∕𝑑𝑡

𝑥 𝑃

𝐷

5

𝑑𝐴∕𝑑𝑡

20

4 51 True

𝑡

5000

𝐴

53 (a) 𝐿 ≈ 13 (b) 𝑘 ≈ 0.30

1102

Section 11.9

55 𝑑𝐼∕𝑑𝑡 = 0.001𝐼(1 − 10(𝐼∕𝑀)); 𝑀∕10

15 Horizontal nullclines; 𝑦 = 0, 𝑦 = 1 − 2𝑥 Vertical nullclines; 𝑥 = 0, 𝑦 = (1∕2)(2 − 𝑥) Equilibrium points; (0, 0), (0, 1), (2, 0)

1 (4, 10)

𝑦

3

𝐼 𝑀 10

15 10 𝑡

5

1

III

𝑥 2

4

6 I

Section 11.8 1 𝑥 and 𝑦 increase, about same rate 3 𝑥 decreases quickly while 𝑦 increases more slowly

7 (a) 𝑑𝑥∕𝑑𝑡 > 0 and 𝑑𝑦∕𝑑𝑡 > 0 (b) 𝑑𝑥∕𝑑𝑡 < 0 and 𝑑𝑦∕𝑑𝑡 = 0 (c) 𝑑𝑥∕𝑑𝑡 = 0 and 𝑑𝑦∕𝑑𝑡 > 0

5 (0, 0) and (5, 3)

9

0.5

𝑦 1

9 (a) Both 𝑥 and 𝑦 are decreasing (b) 𝑥 is increasing; 𝑦 is decreasing

4 2

✻ ✒ ✲

15 1

𝑥

17 Helpful 19 (a) Susceptibles: 1950,1850, 1550, 1000, 750, 550, 350, 250, 200, 200, 200 Infecteds: 20, 80, 240, 460, 500, 460, 320, 180, 100, 40, 20 (b) Day 22, 500 (c) 1800, 200

5

10

3

25 𝑟 = 𝑟0 𝑒−𝑡 , 𝑤 = 𝑤0 𝑒𝑡

5



𝑃0 𝑃2

𝑃0 𝑃2

𝑦

13 Vertical nullclines: 𝑥 = 0, 𝑥 + 𝑦 = 2 Horizontal nullclines: 𝑦 = 0, 𝑥 + 𝑦 = 1 Equilibrium points: (0, 0), (0, 1), (2, 0)

Robins

𝑡 𝑃0 𝑃2

𝑥 2

𝑥

Worms





0.5

15

17 𝑑𝑥∕𝑑𝑡 = 0 when 𝑥 = 0 or 𝑥 + 𝑦∕3 = 1 𝑑𝑦∕𝑑𝑡 = 0 when 𝑦 = 0 or 𝑦 + 𝑥∕2 = 1 Equilibrium points: (0, 0), (0, 1), (1, 0), (4∕5, 3∕5)

27 Worms decrease, robins increase. Long run: populations oscillate

1.5 1



𝑦

11

21 (a) 𝑎 is negative; 𝑏 is negative; 𝑐 is positive (b) 𝑎 = −𝑐

population





11 𝑃3 13 (0, 0), (3, 0), (0, 6), (4.5, 1.5)

2

𝑦 6

7 (0, 0) and (−5, 3)

29

II

5 Tends towards point (4, 10)

𝑃0



𝑦 + 𝑥∕2 = 1 𝑑𝑦∕𝑑𝑡 = 0

𝑥+𝑦 =2 𝑑𝑥∕𝑑𝑡 = 0

2



1 𝑥+𝑦 =1 𝑑𝑦∕𝑑𝑡 = 0

✠ 𝑥 2

1



𝑦 3

𝑥

37 (10, 30)

1

39 (a) 𝑑𝑦∕𝑑𝑥 = 𝑥∕5𝑦 (b) US victory (c) No

2 (II)

𝑦 2

(I)

( 45 , 35 )

1

43 𝑑𝑥∕𝑑𝑡, 𝑑𝑦∕𝑑𝑡 can both be positive

(IV)

(III)

45 𝑑𝑥∕𝑑𝑡 = 0.5𝑥; 𝑑𝑦∕𝑑𝑡 = −0.1𝑦 + 0.3𝑥𝑦 47 𝐷1 is flu and 𝐷2 is HIV

( 45 , 35 )

1

35 (a) Predator-prey (b) 𝑥, 𝑦 tend to ≈ 1

41 (a) 𝑥2 − 5𝑦2 = 604.75 (b) About 25,000 troops

𝑥 + 𝑦∕3 = 1 𝑑𝑥∕𝑑𝑡 = 0

𝑦

31 Robins increase; Worms constant, then decrease; Both oscillate in long run 33 (a) Symbiosis (b) Both → ∞ or both → 0

3

(III)

𝑥

(II)

1

1

49 False

2

51 𝑎 = 0.00001, 𝑘 = 0.06 53 (a) 𝑤 = 0, 𝑟 = 0 and 𝑤 = 𝑏∕𝑘, 𝑟 = 𝑎∕𝑐 (b) Worm equilibrium unchanged, robin equilibrium reduced

(I)

𝑥 1

2

19 (a) 𝑑𝑆∕𝑑𝑡 = 0 where 𝑆 = 0 or 𝐼 = 0 𝑑𝐼 ∕𝑑𝑡 = 0 where 𝐼 = 0 or 𝑆 = 192

1103 (b) Where 𝑆 > 192, 𝑑𝑆∕𝑑𝑡 < 0 and 𝑑𝐼∕𝑑𝑡 > 0 Where 𝑆 < 192, 𝑑𝑆∕𝑑𝑡 < 0 and 𝑑𝐼∕𝑑𝑡 < 0 (c) 𝐼

13 𝑦(𝑡) = 𝐶1 cos (𝑡∕3) + 𝐶2 sin (𝑡∕3) 15 (a) 𝑑 2 𝑠∕𝑑𝑡2 = −2.5𝑠 (b) 𝑠(𝑡) = 5 cos(1.581𝑡) − 6.325 cos(1.581𝑡) (c) −8.06 17 Middle point, moving down (i) 𝑦 = 6 sinh 𝑤𝑡∕ sinh 𝑤 (ii) 𝑦 = 6 cosh 𝑤𝑡∕ cosh 𝑤

19 (b)

𝑆 192 21 (a) In the absence of the other, each company grows exponentially The two companies restrain each other’s growth if they are both present (b) (0, 0) and (1, 2) (c) In the long run, one of the companies will go out of business

𝐵 4 3 2

23 (a) (b) (c) (d)

Spring (iii) Spring (iv) Spring (iv) Spring (i)

2

3

4

29 𝐶 = 1∕20 farads

5 𝑦(𝑥) = 40 + 𝐴𝑒0.2𝑥

31 Initial displacement is 3

7 𝐻 = 𝐴𝑒0.5𝑡 − 20 9 𝑃 = 100𝐴𝑒0.4𝑡 ∕(1 + 𝐴𝑒0.4𝑡 )

9 𝑝(𝑡) = 𝐶1 𝑒−𝑡∕2 cos 𝐶2 𝑒−𝑡∕2 sin



3 𝑡+ 2



3 𝑡 2 −2𝑡

17 𝑦(𝑡) = −2𝑒−3𝑡 + 3𝑒−2𝑡

1

19 𝑦(𝑡) = 51 𝑒4𝑡 + 45 𝑒−𝑡

𝑥 1

2

21 𝑦(𝑡) = 54 𝑒−𝑡 − 14 𝑒−5𝑡

3

15

1 𝑦 2

− 4 ln |𝑦| = 3 ln |𝑥| − 𝑥

+

7 2

− 4 ln 5

17 𝑧(𝑡) = 1∕(1 − 0.9𝑒𝑡 ) 19 20 ln |𝑦| − 𝑦 = 100 ln |𝑥| − 𝑥 + 20 ln 20 − 19

21 𝑦 = ln(𝑒𝑥 + 𝑒 − 1)

23 𝑧 = sin(−𝑒cos 𝜃 + 𝜋∕6 + 𝑒) +

13 𝑦(𝑡) = 𝐶1 𝑒𝑡∕3 + 𝐶2 𝑒−𝑡∕3 √ √ 15 𝑦(𝑡) = 𝑒−𝑡 (𝐴 sin 2𝑡 + 𝐵 cos 2𝑡)

2

11 𝑃 = (40000∕3)(𝑒0.03𝑡 − 1) √ 3 13 𝑦 = 33 − 6 cos 𝑥

25 𝑦 = − ln( ln32 sin2 𝑡 cos 𝑡

11 𝑦(𝑡) = 𝐴 + 𝐵𝑒

3

57 Curve shows 𝑠′ (0) < 0 59 𝑑 2 𝑠∕𝑑𝑡2 + 2𝑑𝑠∕𝑑𝑡 + 𝑠 = 0

3 (a) (I) (b) (IV) (c) (III)

7 𝑧(𝑡) = 𝐶1 𝑒−𝑡∕2 + 𝐶2 𝑒−3𝑡∕2

𝑦

55 As 𝑡 → ∞, 𝑄(𝑡) → 0

27 (a) 𝐴 sin(𝜔𝑡 + 𝜙) = (𝐴 sin 𝜙) cos 𝜔𝑡 + (𝐴 cos 𝜙) sin 𝜔𝑡

5 𝑠(𝑡) = √ √ 𝐶1 cos 7𝑡 + 𝐶2 sin 7𝑡

25 𝑑𝑥∕𝑑𝑡 ≠ 0

53 (a) 𝑄(𝑡) = 2𝑡𝑒−𝑡∕2 (b) 𝑄(𝑡) = (2 + 𝑡)𝑒−𝑡∕2

1 (I) satisfies equation (d) (II) satisfies equation (c)

3 𝑦(𝑡) = 𝐶1 𝑒−2𝑡 cos 𝑡 + 𝐶2 𝑒−2𝑡 sin 𝑡

23 (a) 𝑑𝑃1 ∕𝑑𝑡 = 0 where 𝑃 = 0 or 𝑃1 + 3𝑃2 = 13 𝑑𝑃2 ∕𝑑𝑡 = 0 where 𝑃 = 0 or 𝑃2 + 0.4𝑃1 = 6

47 𝑧(𝑡) = 3𝑒−2𝑡 49 (a) 𝑑 2 𝑦∕𝑑𝑡2 − 𝑦 = 0 (b) 𝑦 = 𝐶1 𝑒𝑡 + 𝐶2 𝑒−𝑡 , so 𝑥 = 𝐶2 𝑒−𝑡 − 𝐶1 𝑒𝑡

Chapter 11 Review

1 𝑦(𝑡) = 𝐶1 𝑒−𝑡 + 𝐶2 𝑒−3𝑡

𝐴

43 (a) 𝑎 > 447.216 (b) 𝑎 = 447.216 (c) 0 < 𝑎 < 447.216

25 (a) Starts higher; period same (b) Period increases

Section 11.11

1

27

21 (a) 𝑥(𝑡) = 5 cos 4𝑡 𝐴 = 5, period = 𝜋∕2 (b) 𝑥(𝑡) = √− cos(𝑡∕5) + 10 sin(𝑡∕5) 𝐴 = 101, period = 10𝜋

33 𝑦′′ = 4𝑦

1

(b) 𝑠(𝑡) = −4𝑒−2𝑡 + 3𝑒−5𝑡 (c) 𝑠 = −1.58; 𝑠 = 0 (d) 𝑡 ≥ 1.843

11 𝐴 ≈ −1.705 √ 𝛼=± 5

2 ln 2 3

cos 𝑡 −

2 ln 2 3

+ 1)∕ ln 2

27 (a) 𝑦 = 3 and 𝑦 = −2 (b) 𝑦 = 3 unstable; 𝑦 = −2 stable 29 No 31 (a) 𝑦(1) ≈ 3.689 (b) Overestimate

23 𝑦(𝑡) = 2𝑒−3𝑡 sin 𝑡 29 False 31 False

Section√ 11.10 1 𝐴=

1 −2𝑡 𝑒 1−𝑒

27 𝑝(𝑡) =

20𝑒(𝜋∕2)−𝑡

29 (a) (b) (c) (d)

32 + 42 = 5

3 𝜔 = 2, 𝐴 = 13, 𝜓 = 𝑡𝑎𝑛−1 12∕5

+

−𝑒 −3𝑡 𝑒 1−𝑒

𝑦

sin 𝑡

(IV) (II) (I) (III)

5

31 (iii)

𝑠

5 (a)

25 𝑦(𝑡) =

33 (iv)

10

𝑥

35 (ii)

𝑠 = 4 cos 𝑡 + 3 sin 𝑡

5 𝑡

−2𝜋

2𝜋 −5 −10

(c) 𝐴 = 5, 𝜙 = 0.93

5

37 Overdamped: 𝑐 < 2 Critically damped: 𝑐 = 2 Underdamped: 𝑐 > 2 39 Overdamped if: √ 𝑏>2 5 Critically √ damped if: 𝑏=2 5 Underdamped √ if: 0 0 21 Positive, Negative, 10, 2, −4 23 (i)(c); (ii)(a) 25 (a) Both negative (b) Both negative 27 𝑓𝑇 (5, 20) ≈ 1.2◦ F∕◦ F

𝑥

0 2

13 (a) 𝑓 (𝐴) = 88 (b) Negative (c) Negative

17 𝑓𝑥 > 0, 𝑓𝑦 < 0

𝑥

− −2 1

(b)

9 (a) Negative (b) Positive

15 (a) 𝑓 (𝐴) = 40 (b) Negative (c) Positive

2

5

4

3

3

4 5

4 ⃗ −4⃗𝑖 − 11𝑗⃗ − 17𝑘 ⃗ 3.64⃗𝑖 + 2.43𝑗⃗ − 2.43𝑘 79.0◦ 0.784. ⃗ (many answers possible) 2⃗𝑖 − 2𝑗⃗ + 𝑘 ⃗. −4⃗𝑖 − 11𝑗⃗ − 17𝑘 √ ⃗ )∕ 6 25 ±(−⃗𝑖 + 𝑗⃗ − 2𝑘

0.2

0

23 (a) (b) (c) (d) (e) (f) (g)

0.1

1

⃗ (multiples of) 21 −5⃗𝑖 + 3𝑗⃗ + 𝑘

𝑥 10

1 𝑓𝑥 (3, 2) ≈ −2∕5; 𝑓𝑦 (3, 2) ≈ 3∕5

17 0 19 0⃗

39 (a) (b) (c) (d) (e) (f)

59 0, 0⃗

𝑥 10

53 (a) 1.5 (b) 𝑦 = 1

55 9𝑥 − 16𝑦 + 12𝑧 = 5 0.23 ⃗ 57 7.0710⃗𝑖 + 2.5882𝑗⃗ + 6.580𝑘

60

30 20 10

(c)

𝑦 5 4

3

2 1

57 (a) ((𝑢2 𝑣3 − 𝑢3 𝑣2 )2 + (𝑢3 𝑣1 − 𝑢1 𝑣3 )2 + (𝑢1 𝑣2 − 𝑢2 𝑣1 )2 )1∕2 (b) ||𝑢1 𝑣2 − 𝑢2 𝑣1 || (c) 𝑚 = (𝑢2 𝑣3 − 𝑢3 𝑣2 )∕(𝑢2 𝑣1 − 𝑢1 𝑣2 ), 𝑛 = (𝑢3 𝑣1 − 𝑢1 𝑣3 )∕(𝑢2 𝑣1 − 𝑢1 𝑣2 )

50

40

𝑄

55 4𝜋⃗𝑖

0 1 − 2 −

𝑥 −4

−5

⃗ 47 (a) 4𝑘 (b) 3𝑗⃗ (c) 2⃗𝑖

−3

⃗ 45 ⃗𝑖 − 3𝑗⃗ − 5𝑘

1112 (d)

19 2𝐵∕𝑢0

𝑦

− −1 2

31 𝑑𝑓 = −3𝑑𝑥 + 2𝑑𝑦 at (2, −4)

23 (15𝑥2 𝑦 − 3𝑦2 ) cos(5𝑥3 𝑦 − 3𝑥𝑦2 )

33 376

25 𝑦

35 𝑑𝑓 = 13 𝑑𝑥 + 2𝑑𝑦 𝑓 (1.04, 1.98) ≈ 2.973

6

𝑦−1

27 𝑧𝑥 = 7𝑥 + 𝑦𝑥 𝑧𝑦 = 2𝑦 ln 2 + 𝑥𝑦 ln 𝑥

𝑥

0

5

3

4

2

1

− − 5 −3 4

(c) 2.1

21 2𝑚𝑣∕𝑟

29 𝐺𝑚1

37 136.09◦ C

∕𝑟2

31 −𝑒−𝑥

2 ∕𝑎2

(𝑎2 − 2𝑥2 )∕𝑎4 1 ⋅ 2𝜃 33 cos(𝜋𝜃𝜙) ⋅ 𝜋𝜙 + 2 𝜃 +𝜙 𝑎 𝑎 35 𝑓𝑎 = 𝑒 sin(𝑎 + 𝑏) + 𝑒 cos(𝑎 + 𝑏)

43 There are many possibilities.

37 𝜕𝑉 ∕𝜕𝑟 = 83 𝜋𝑟ℎ, 𝜕𝑉 ∕𝜕ℎ = 43 𝜋𝑟2

𝑦

39 −(𝑥 − 𝜇)𝑒−(𝑥−𝜇) 8 42

−6

6

√ ∕( 2𝜋𝜎 3 )

41 (a) 𝑓𝑥 (1, 1) = 2; 𝑓𝑦 (1, 1) = 2 (b) (II)

−2 −8 0

2 ∕(2𝜎 2 )

𝑥

43 (a) 𝑓𝑥 (1, 1) = 2; 𝑓𝑦 (1, 1) = 2𝑒 (b) (I)

49 1.277

45 There are many possibilities.

𝑦 2 −4

1

−2

−2

𝑥

−1 0 −1 2

−2

1

2

(c) 𝑧 = 2 + 23 (𝑥 − 2) − 12 (𝑦 − 3)

51 (a) 𝑒𝑦 𝑑𝑥+𝑥𝑒𝑦 𝑑𝑦+2𝑧 𝑑𝑧 = − sin(𝑥−1) 𝑑𝑥+ 𝑧 √ 𝑑𝑧 𝑧2 +3

(b) 𝑑𝑧 = − 23 𝑑𝑥 −

0.005

m2 /kg,

0.006

55 No such points exist

57 sphere of radius 3 centered at the origin

57 (0, 1), (0, −1), (−2, 1) and (−2, −1)

59 True

59 𝜕𝑓 ∕𝜕𝑥 or 𝜕𝑓 ∕𝜕𝑦?

61 False

61 𝑓 (𝑥, 𝑦) = 2𝑥 + 3𝑦 + 𝑥2

63 True

63 𝑓 (𝑥, 𝑦) = 𝑦2 + 1

65 False

65 True

𝑥4 )⃗𝑖 − ( 24 𝑦5 )𝑗⃗ 1 ( 15 2 7

71 False

3 2𝑚⃗𝑖 + 2𝑛𝑗⃗ ( )

49 𝑓 (𝑥, 𝑦) = 4𝑥 − 𝑦

3 𝑧 = 6𝑦 − 9

51 False

5 𝑧 = −4 + 2𝑥 + 4𝑦

53 False

7 𝑧 = −36𝑥 − 24𝑦 + 148

55 True

9 𝑑𝑓 = 𝑦 cos(𝑥𝑦) 𝑑𝑥 + 𝑥 cos(𝑥𝑦) 𝑑𝑦

57 True

11 𝑑𝑧 = −𝑒−𝑥 cos(𝑦)𝑑𝑥 − 𝑒−𝑥 sin(𝑦)𝑑𝑦

59 True

13 𝑑𝑔 = 4 𝑑𝑥

5 𝜕𝑧∕𝜕𝑥 = 𝜕𝑧∕𝜕𝑦 =

14𝑥+7 (𝑥2 +𝑥−𝑦)− 6 −7(𝑥2 + 𝑥

7 𝑓𝑥 = 0.6∕𝑥 𝑓𝑦 = 0.4∕𝑦

15 𝑑𝑃 ≈ 2.395 𝑑𝐾 + 0.008 𝑑𝐿

9 2𝑥𝑦 + 10𝑥4 𝑦 11 𝑉𝑟 = 23 𝜋𝑟ℎ √ √ 13 𝑒 𝑥𝑦 (1 + 𝑥𝑦∕2) 15 𝑔

17 (𝑎 + 𝑏)∕2

− 𝑦)6

Section 14.4

69 False

1 𝑧 = 𝑒𝑥

3 𝑓𝑥 (1, 2) = 15, 𝑓𝑦 (1, 2) = −5

53 𝑧 = 𝑓 (3, 4) + 𝑓𝑥 (3, 4)(𝑥 − 3) + 𝑓𝑦 (3, 4)(𝑦 − 4) 55 Equation not linear

47 𝑓𝑦 = 0

1 (a) 7.01 (b) 7

𝑑𝑦

51 ℎ𝑥 (2, 5) ≈ −0.38 cm/meter ℎ𝑡 (2, 5) ≈ 0.76 cm/second

Section 14.3

Section 14.2

2 3

(c) 𝑧 = 1 − 32 (𝑥 − 1) − 23 (𝑦 − 0)

m2 /cm

67 False

4

45 (a) 𝑑𝜌 = −𝛽𝜌 𝑑𝑇 (b) 0.00015, 𝛽 ≈ 0.0005

49 (a) 4𝑥 𝑑𝑥 = 2𝑦 𝑑𝑦 + 6𝑧 𝑑𝑧 (b) 𝑑𝑧 = 32 𝑑𝑥 − 12 𝑑𝑦

47 (a) 𝑃 𝑟𝑒𝑟𝑡 (b) 𝑒𝑟𝑡 m2 ,

43 (a) 𝑛𝑅𝑇 ∕(𝑉 − 𝑛𝑏) − 𝑛2 𝑎∕𝑉 2 (b) Δ𝑃 ≈ (𝑛𝑅∕(𝑉0 − 𝑛𝑏))Δ𝑇 + (2𝑛2 𝑎∕𝑉03 − 𝑛𝑅𝑇0 ∕((𝑉0 − 𝑛𝑏)2 ))Δ𝑉

47 −43200Δ𝑡 Slow if Δ𝑡 > 0; fast if Δ𝑡 < 0

45 (a) 𝑓𝑤 (2, 2) ≈ 2.78 𝑓𝑧 (2, 2) ≈ 4.01 (b) 𝑓𝑤 (2, 2) ≈ 2.773 𝑓𝑧 (2, 2) = 4

−4

39 𝑃 (𝑟, 𝐿) ≈ 80 + 2.5(𝑟 − 8) + 0.02(𝐿 − 4000) 𝑃 (𝑟, 𝐿) ≈ 120 + 3.33(𝑟 − 8) + 0.02(𝐿 − 6000) 𝑃 (𝑟, 𝐿) ≈ 160 + 3.33(𝑟 − 13) + 0.02(𝐿 − 7000)

5

√ 5𝛼

5𝛼 2 +𝛽

15 60⃗𝑖 + 85𝑗⃗

19 0.99

19 (𝜋∕2)1∕2 ⃗𝑖

21 95

21

23 12.005

23 𝑖⃗ 25 𝑖⃗ + 𝑗⃗

27 (b) 𝑓 (𝑥, 𝑦) ≈ 0.3345 − 0.33(𝑥 − 1) − 0.15(𝑦 − 2) (c) 𝑓 (𝑥, 𝑦) ≈ 0.3345 − 0.3345(𝑥 − 1) − 0.1531(𝑦 − 2) 29 (a) 𝑓𝑥 (1, 2) = 3; 𝑓𝑦 (1, 2) = 2 (b) 2

2

√1

5𝛼 2 +𝛽

)

𝑗⃗

9 sin 𝜃 ⃗𝑖 + 𝑟 cos 𝜃 𝑗⃗ 1 𝑥 𝑥 𝑥 11 ∇𝑧 = cos ( )𝑖⃗ − 2 cos ( )𝑗⃗ 𝑦 𝑦 𝑦 𝑦 ) ( ) ( −12𝛽 12𝛼 ⃗𝑖 + 𝑗⃗ 13 2 2 (2𝛼 − 3𝛽) (2𝛼 − 3𝛽) 17 10𝜋⃗𝑖 + 4𝜋 𝑗⃗

Dollars∕Square foot Larger plots at same distance $3∕ft2 more Dollars∕Foot Farther from beach but same area $2∕ft less (e) 998 ft2

(

7 ∇𝑧 = 𝑒𝑦 ⃗𝑖 + 𝑒𝑦 (1 + 𝑥 + 𝑦)𝑗⃗

17 −5

25 (a) (b) (c) (d)

⃗𝑖 +

1 (2⃗𝑖 100

− 6𝑗⃗ )

27 ⃗𝑖 − 𝑗⃗ 29 Negative 31 Negative 33 Approximately zero 35 −⃗𝑖 37 𝑖⃗ 39 −⃗𝑖 + 𝑗⃗ 41 ⃗𝑖 + 𝑗⃗

1113 43 6.325

37 𝑥 + 4𝑦 + 10𝑧 = 18

129 (a)

45 −46∕5

39 10∕3

47 22∕5

41 2𝑧 + 3𝑥 + 2𝑦 = 17

𝑧

49 84∕5

43 𝑥 + 3𝑦 + 7𝑧 = −9 ⃗ ⃗𝑖 + 3𝑗⃗ + 7𝑘

51 (2𝑥 + 3𝑒𝑦 )𝑑𝑥 + 3𝑥𝑒𝑦 𝑑𝑦

45 grad 𝑔(−1, −1) lies directly under path of steepest descent

53 (𝑥 + 1)𝑦𝑒𝑥 ⃗𝑖 + 𝑥𝑒𝑥 𝑗⃗ 55 50.2

𝑦

𝑥

57 (a) Should be number (b) 11∕5 59 −2 61 ⃗𝑖 + 2𝑗⃗ or any multiple

49 3𝑥 + 10𝑦 − 5𝑧 + 19 = 0 √ 51 16∕ 14

𝑦

(b)

63 0.316 65 1

𝑧 = 16 𝑧=9 𝑧=4 𝑧=1 𝑥 𝑧=0 𝑧=1 𝑧=4

4

67 100 69 (a) (b) 71 (a) (b) (c) 73 (a) (b) 75 (a) (b) (c)

√ − 2∕2 √ 3 + 1∕2 √ 2∕√13 1∕ 17 ⃗𝑖 + 1 𝑗⃗ 2 √ 5∕ 2 510 −16⃗𝑖 + 12𝑗⃗ 16⃗𝑖 − 12𝑗⃗ 12⃗𝑖 + 16𝑗⃗ ; answers may vary

77 1.7; closer estimate is 1.35 79 2.5; better estimate is 1.8 81 −0.9; better estimate is −1.8 83 Fourth quadrant 85 (a) (b) (c) (d)

Negative Positive Positive Negative

−2

131 (a) Circles centered at 𝑃 (b) away from 𝑃 (c) 1 √ √ √ √ 133 (3 5 − 2 2)⃗𝑖 + (4 2 − 3 5)𝑗⃗ √ 135 4 2, 6⃗𝑖 + 2𝑗⃗ √ 139 (a) 𝑚2 + 𝑛2 √ (b) (𝐶2 − 𝐶1 )∕ 𝑚2 + 𝑛2

91 3⃗𝑖 + 2𝑗⃗ ; 3(𝑥 − 2) + 2(𝑦 − 3) = 0

145 True

93 −5⃗𝑖 ; 𝑥 = 2 √ 95 5∕ 2

147 False

101 grad 𝑓 (0, 0) is vector, not scalar 103 −⃗𝑖 105 False 107 False 109 True 111 False 113 True 115 True 117 (a) Perpendicular to contour of 𝑓 at 𝑃 (b) Maximum directional derivative of 𝑓 at 𝑃 (c) Directional derivative 𝑓𝑢⃗ (𝑃 )

143 True

⃗) + 𝑗⃗ + 𝑘

−2𝑥𝑦𝑧2

𝑧2

2𝑦𝑧 ⃗ ⃗𝑖 + 𝑗⃗ + 𝑘 (1 + 𝑥2 )2 1 + 𝑥2 1 + 𝑥2 √ ⃗ )∕ 𝑥2 + 𝑦2 + 𝑧2 7 (𝑥⃗𝑖 + 𝑦𝑗⃗ + 𝑧𝑘 ⃗ 9 𝑦⃗𝑖 + 𝑥𝑗⃗ + 𝑒𝑧 cos (𝑒𝑧 ) 𝑘

2 ⃗ 11 𝑒𝑝 ⃗𝑖 + (1∕𝑞)𝑗⃗ + 2𝑟𝑒𝑟 𝑘

13 0⃗ ⃗ 15 6⃗𝑖 + 4𝑗⃗ − 4𝑘 ⃗ 17 −𝜋⃗𝑖 − 𝜋 𝑘 √ 19 9∕ 3 √ 21 −1∕ 2 √ 23 − 77∕2

121 356.5

⃗ ; 2(𝑦 − 1) − 4(𝑧 − 2) = 0 27 2𝑗⃗ − 4𝑘

127 Yes

67 (a) Parallel planes: 2𝑥 − 3𝑦 + 𝑧 = 𝑇 − 10 (b) 𝑓𝑧 (0, 0, 0) = 1, temp increases 1◦ C per unit in 𝑧-direction ⃗ (c) 2⃗𝑖 − 3𝑗⃗ + 𝑘 (d) Yes; 27◦ C √ 69 (a) −25∕ 21 ⃗ (b) √ −8⃗𝑖 + 7𝑗⃗ + 4𝑘 129 (c)

77 𝑓 (𝑥, 𝑦, 𝑧) = 2𝑥 + 3𝑦 + 4𝑧 + 100

𝑒𝑥 𝑒𝑦 𝑒𝑧 (⃗𝑖

⃗; 25 −2⃗𝑖 − 2𝑗⃗ + 4𝑘 −2(𝑥 + 1) − 2(𝑦 − 1) + 4(𝑧 − 2) = 0

125 Yes

23 −9.2 −16⃗𝑖 + 6𝑗⃗ 16𝑥 − 6𝑦 − 𝑧 = 23

75 𝑓𝑥 (0, 0, 0)𝑥 + 𝑓𝑦 (0, 0, 0)𝑦 + 𝑓𝑧 (0, 0, 0)𝑧 = 0

1 2𝑥⃗𝑖

5

65 (a) (b) (c) (d)

73 (a) is (V); (b) is (IV); (c) is (V)

Section 14.5 3

63 (a) 6.33⃗𝑖 + 0.76𝑗⃗ (b) −34.69

71 1.131 atm/sec

119 19.612 123 (a) −3.268 (b) −4.919

55 (a) Circle: (𝑦 + 1)2 + (𝑧 − 3)2 = 10 (b) Yes ⃗ (c) Multiples of −10⃗𝑖 + 4𝑗⃗ − 12𝑘 √ ⃗ ⃗ ⃗ 57 (a) (−3𝑖 + 6𝑗 + 12𝑘 )∕ 21 (b) (8.345, 2.309, 4.619) 61 (−1∕6, 1∕3, −1∕12)

(c) 𝑗⃗

89 𝑓 (𝑃 ) ≈ 6, 𝑓 (𝑄) ≈ −24

97 (a) ellipses centered at (0, 0) (b) decreasing at 49.9◦ C per meter (c) −⃗𝑖 − 2𝑗⃗ . Other answers possible √ 13 meters ascended/horizontal meter 99 (a) (b) 3.54 meters ascended/horizontal meter (c) 𝑢⃗ = 3⃗𝑖 + 2𝑗⃗ ; 𝑢⃗ = −3⃗𝑖 − 2𝑗⃗

53 (a) Spheres centered at the origin (b) 2𝑥 sin(𝑥2 + 𝑦2 + 𝑧2 )⃗𝑖 + 2𝑦 sin(𝑥2 + 𝑦2 + ⃗ 𝑧2 )𝑗⃗ + 2𝑧 sin(𝑥2 + 𝑦2 + 𝑧2 )𝑘 (c) 0, 180◦

59 Any multiple of 2⃗𝑖 + 2𝑗⃗ + ⃗𝑘

⃗ ⃗ 141 (a) 𝑎 √𝑖 + 2𝑏𝑗 (2 + 𝑎)∕(2 − 𝑎) (b) √ (c) (2 − 𝑎)∕(2 + 𝑎)

87 𝑓𝑢⃗ (𝑃 ) < 𝑓𝑤⃗ (𝑃 ) < 𝑓𝑣⃗ (𝑃 )

47 (a) (𝑥 − 2) + 4(𝑦 − 3) − 6(𝑧 − 1) = 0 (b) 𝑧 = 1 + (1∕6)(𝑥 − 2) + (2∕3)(𝑦 − 3)

⃗ ; −2(𝑥 + 1) + (𝑧 − 2) = 0 29 −2⃗𝑖 + 𝑘 31 6(𝑥 − 1) + 3(𝑦 − 2) + 2(𝑧 − 1) = 0 33 2𝑥 + 3𝑦 + 2𝑧 = 17 35 𝑧 = 2𝑥 + 𝑦 + 3

79 False 81 False

Section 14.6 𝑑𝑧 = 𝑒−𝑡 sin(𝑡)(2 cos 𝑡 − sin 𝑡) 𝑑𝑡 ) ( 2 3 2 cos 2𝑡2 1+𝑡2 2 1

1−𝑡

2

(1−𝑡 )

5 2𝑒1−𝑡 (1 − 2𝑡2 ) ( ) 1 ln 𝑢 𝜕𝑧 = cos 7 𝜕𝑢 𝑣𝑢 𝑣 ) ( 𝜕𝑧 ln 𝑢 ln 𝑢 = − 2 cos 𝜕𝑣 𝑣 𝑣 𝜕𝑧 𝑒𝑣 9 = 𝜕𝑢 𝑢 𝜕𝑧 = 𝑒𝑣 ln 𝑢 𝜕𝑣 2 2 𝜕𝑧 = 2𝑢𝑒(𝑢 −𝑣 ) (1 + 𝑢2 + 𝑣2 ) 11 𝜕𝑢 2 2 𝜕𝑧 = 2𝑣𝑒(𝑢 −𝑣 ) (1 − 𝑢2 − 𝑣2 ) 𝜕𝑣 𝜕𝑧 = 13 𝜕𝑢 (𝑒−𝑣 cos 𝑢 − 𝑣(cos 𝑢)𝑒−𝑢 sin 𝑣 ) sin 𝑣 − (−𝑢(sin 𝑣)𝑒−𝑣 cos 𝑢 + 𝑒−𝑢 sin 𝑣 )𝑣 sin 𝑢

1114 𝜕𝑧 = 𝜕𝑣 (𝑒−𝑣 cos 𝑢 − 𝑣(cos 𝑢)𝑒−𝑢 sin 𝑣 )𝑢 cos 𝑣 + (−𝑢(sin 𝑣)𝑒−𝑣 cos 𝑢 + 𝑒−𝑢 sin 𝑣 ) cos 𝑢 𝜕𝑧 −2𝑢𝑣2 15 = 4 𝜕𝑢 𝑢 + 𝑣4 2𝑣𝑢2 𝜕𝑧 = 4 𝜕𝑣 𝑢 + 𝑣4 17 −2𝜌 cos 2𝜙, 0

𝑓𝑦𝑥 = 30𝑥2 𝑦 − 21𝑦2 𝑓𝑦𝑦 = 10𝑥3 − 42𝑥𝑦

(c) (III) 63 (a) 𝑥𝑦 1 − 12 (𝑥 − (b)

11 𝑓𝑥𝑥 = −12 sin 2𝑥 cos 5𝑦 𝑓𝑥𝑦 = −30 cos 2𝑥 sin 5𝑦 𝑓𝑦𝑥 = −30 cos 2𝑥 sin 5𝑦 𝑓𝑦𝑦 = −75 sin 2𝑥 cos 5𝑦 15 𝑄(𝑥, 𝑦) = 1 + 𝑥 + 𝑥2 ∕2 − 𝑦2 ∕2 19 𝑄(𝑥, 𝑦) = −𝑦 + 𝑥2 − 𝑦2 ∕2

21 −5 pascal/hour

21 1 + 𝑥 − 𝑦∕2 − 𝑥2 ∕2 + 𝑥𝑦∕2 − 𝑦2 ∕8

23 −0.6

23 (a) (b) (c) (d) (e)

Negative Zero Negative Zero Zero

25 (a) (b) (c) (d) (e)

Positive Zero Positive Zero Zero

27 (a) (b) (c) (d) (e)

Zero Negative Zero Negative Zero

29 (a) (b) (c) (d) (e)

Positive Positive Zero Zero Zero

31 (a) (b) (c) (d) (e)

Positive Negative Negative Negative Positive

17 𝑄(𝑥, 𝑦) = 1 − 𝑥2 ∕2 − 3𝑥𝑦 − (9∕2)𝑦2

√ 10 = 0.316 ◦ F/mile 25 (a) 1∕ √ (b) 2.5∕ 10 = 0.791◦ F∕hr (c) 2.5 ◦ F∕hr 27 Three 𝑑𝑤 𝑑𝑡

=

𝜕𝑤 𝑑𝑥 𝜕𝑥 𝑑𝑡

+

𝜕𝑤 𝑑𝑦 𝜕𝑦 𝑑𝑡

+

𝜕𝑤 𝑑𝑧 𝜕𝑧 𝑑𝑡

𝐹𝑢 (𝑥, 3) 𝐹𝑣 (3, 𝑥) 𝐹𝑢 (𝑥, 𝑥) + 𝐹𝑣 (𝑥, 𝑥) 𝐹𝑢 (5𝑥, 𝑥2 )(5) + 𝐹𝑣 (5𝑥, 𝑥2 )(2𝑥)

31 (a) (b) (c) (d)

33 𝑏 ⋅ 𝑒 + 𝑑 ⋅ 𝑝 35 𝑏 ⋅ 𝑒 + 𝑑 ⋅ 𝑝 𝜕𝑧 𝜕𝑧 𝜕𝑧 = cos 𝜃 + sin 𝜃 37 (a) 𝜕𝑟 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑧 𝜕𝑧 = 𝑟(cos 𝜃 − sin 𝜃 ) 𝜕𝜃 𝜕𝑦 𝜕𝑥 𝜕𝑧 cos 𝜃 𝜕𝑧 𝜕𝑧 = sin 𝜃 + (b) 𝜕𝑦 𝜕𝑟 𝑟 𝜕𝜃 𝜕𝑧 sin 𝜃 𝜕𝑧 𝜕𝑧 = cos 𝜃 − 𝜕𝑥 𝜕𝑟 𝑟 𝜕𝜃 𝜕𝑈3 ) 𝜕𝑃 𝑉 𝜕𝑈 ( 𝜕𝑇 )𝑉 = 7∕2 ( 𝜕𝑈 ) = 11∕4 𝜕𝑉 𝑇

39 ( 41

− 12 (𝑦 − 𝜋2 )2

𝑦

13 𝑄(𝑥, 𝑦) = 1 + 2𝑥 − 2𝑦 + 𝑥2 − 2𝑥𝑦 + 𝑦2

19 (a) 𝜕𝑓 ∕𝜕𝑡 (b) (𝜕𝑓 ∕𝜕𝑥)(𝑑𝑥∕𝑑𝑡) (c) (𝜕𝑓 ∕𝜕𝑦)(𝑑𝑦∕𝑑𝑡)

29

𝜋 2 ) 2

𝑥

0

0 𝑦

𝜋 2

𝑥

𝜋 2

65 𝑓 (𝑥, 𝑦):

𝑦 0.6

33 −8 𝑓𝑥 (𝑔(𝑡), ℎ(𝑡))𝑔 ′ (𝑡)

45 𝑑𝑧∕𝑑𝑡 = 𝑓𝑦 (𝑔(𝑡), ℎ(𝑡))ℎ′ (𝑡)

+

35 3

1.4

37 Not possible

47 𝑑𝑧∕𝑑𝑡|𝑡=0 = 𝑓𝑥 (2, 3)𝑔 ′ (0) + 𝑓𝑦 (2, 3)ℎ′ (0)

39 6

49 𝑓 (𝑥, 𝑦) = 4𝑥 + 2𝑦

41 𝐿(𝑥, 𝑦) = 𝑦 𝑄(𝑥, 𝑦) = 𝑦 + 2(𝑥 − 1)𝑦 𝐿(0.9, 0.2) = 0.2 𝑄(0.9, 0.2) = 0.16 𝑓 (0.9, 0.2) = 0.162

2𝑠2

+ 𝑡 and 𝑣 = 51 𝑤 = 𝑢𝑣, 𝑢 = answers are possible

𝑒𝑠𝑡 ,

53 (c) 𝑏

57 ∫0 𝐹𝑢 (𝑥, 𝑦) 𝑑𝑦

many other

1.1

−0.6

47 𝑎 = −𝑏2

−0.6

49 Positive, negative

Section 14.7

51 (a) 𝑧𝑦𝑥 = 4𝑦 (b) 𝑧𝑥𝑦𝑥 = 0 (c) 𝑧𝑥𝑦𝑦 = 4

1 𝑓𝑥𝑥 = 2 𝑓𝑦𝑦 = 2 𝑓𝑦𝑥 = 2 𝑓𝑥𝑦 = 2

𝑥 0.6

0.8

𝐿(𝑥, 𝑦):

53 𝑑 = 𝑒 = 𝑓 = 0 55 𝑑 = 0, 𝑒 > 0, 𝑓 < 0

3 𝑓𝑥𝑥 = 6𝑦 𝑓𝑥𝑦 = 6𝑥 + 15𝑦2 𝑓𝑦𝑥 = 6𝑥 + 15𝑦2 𝑓𝑦𝑦 = 30𝑥𝑦

𝑦

9 𝑓𝑥𝑥 = 30𝑥𝑦2 + 18 𝑓𝑥𝑦 = 30𝑥2 𝑦 − 21𝑦2

0.6

0 101

1.7

Trail

1.4

1000 1.1

0 98

5 𝑓𝑥𝑥 = 0 𝑓𝑦𝑥 = 𝑒𝑦 = 𝑓𝑥𝑦 𝑓𝑦𝑦 = 𝑒𝑦 (𝑥 + 2 + 𝑦) ( ( )) ( ) 1 7 𝑓𝑥𝑥 = − sin 𝑥𝑦 2 ( ( )) ( 𝑦 ) ( ) 𝑥 1 −𝑥 𝑓𝑥𝑦 = − sin 𝑦 2 𝑦 ( ( )) ( ) 𝑦 −1 + cos 𝑥𝑦 = 𝑓𝑦𝑥 2 𝑦 ( ( )) ( )2 −𝑥 𝑓𝑦𝑦 = − sin 𝑥𝑦 2 ( ( )) ( ) 𝑦 𝑥 2𝑥 + cos 𝑦 3

𝑦

Elevation in meters

10 20

57 (a)

990

(b) 𝜕ℎ∕𝜕𝑥 = 0, 𝜕ℎ∕𝜕𝑦 > 0, (𝜕2 ℎ)∕(𝜕𝑥𝜕𝑦) < 0 (c) (𝜕2 ℎ)∕(𝜕𝑥𝜕𝑦) 59 (a) 𝐴 (b) 𝐵 61 (a) (II) (b) (I)

−0.6

0.8

−0.6

𝑄(𝑥, 𝑦):

𝑥 0.6

1115 𝑦 69 𝑓 (𝑥, 𝑦) = 2𝑥 + 𝑦2 , 𝑔(𝑥, 𝑦) = 2𝑥 + 𝑦2 + 𝑥3

0.6

Section 14.8

1.4

1 (0, 0) 3 𝑥-axis and 𝑦-axis

1.1

−0.6

𝑥 0.6

0.8

5 None 7 None 9 (1, 2) 11 (a)

25 (a) (b) (c) (d)

𝑦

−2

𝑥

−4 2 −3

4

3

2

4

2.6 2.3

1 Vector; 3𝑒−1 ⃗𝑖 − 12 𝑒−1 𝑗⃗

−2

2.0 1.7 1.4 1.1

𝑥

0.8

Chapter 14 Review

−3 −4

2

2

(b) (c) (d) (e)

⃗ 3 Vector;−(sin 𝑥)𝑒𝑦 ⃗𝑖 + (cos 𝑥)𝑒𝑦 𝑗⃗ + 𝑘

No No No Exist, not continuous

13 (a)

5 𝑓𝑥 = 2𝑥𝑦 + 3𝑥2 − 7𝑦6 𝑓𝑦 = 𝑥2 − 42𝑥𝑦5 √ 7 𝜋∕ 𝑙𝑔

𝑦 2

9 𝑓𝑥 =

.4

.4

−2

−.2 −.4

−.2

270𝑥3 𝑦7 − 168𝑥2 𝑦6 − 15𝑥𝑦2 + 16𝑦 (15𝑥𝑦 − 8)2 √ 19 𝜋𝑥𝑦∕ 2𝜋𝑥𝑦𝑤 − 13𝑥7 𝑦3 𝑣 )−9∕2 ( 𝑤−1 21 27 𝑥2 𝑦𝑤−𝑥𝑦3 𝑤7 ( 2 ) 3 7 3 6

−.2

−2

(b) (c) (d) (e)

𝑥

Yes Yes No Exist, not continuous

𝑥 𝑦+6𝑥𝑦 𝑤 −7𝑥𝑦 𝑤 (𝑤−1)2

√ 23 −1∕(4𝜋𝐿 𝐿𝐶)

𝑦

15 (a)

25 𝑢𝑥𝑥 = 𝑒𝑥 sin 𝑦,

2

2 −.8 −.6 −.4 −.2

−2

𝑄(𝑥, 𝑦):

.8

29 2𝑥⃗𝑖 + (2𝑦 + 3𝑦2 )𝑗⃗

.6 .4

⃗ )∕(𝑥𝑦𝑧) 31 −((1∕𝑥)⃗𝑖 + (1∕𝑦)𝑗⃗ + (1∕𝑧)𝑘

.2

−.2 −.4 −.6 −.8

2

.2 .4 .6

33 ∇𝑧 = 2𝑥 cos (𝑥2 + 𝑦2 )⃗𝑖 + 2𝑦 cos (𝑥2 + 𝑦2 )𝑗⃗ ( ) ⃗ 35 cos(𝑥2 + 𝑦2 + 𝑧2 ) 2𝑥⃗𝑖 + 2𝑦𝑗⃗ + 2𝑧𝑘 37 − (𝑡

.8

𝑦 −2

2

17 (a)

45 0

2

.2

−.4

2

−.4 −.2

√ 47 2∕ 3

.4

⃗ 49 5⃗𝑖 + 4𝑗⃗ + 3𝑘

.4 .2

51 −4𝑥 − 3𝑦 + 4𝑧 = 9

𝑥 −2

−.4

.2 .4 −2

.2

67 None since 𝑓𝑥𝑦 ≠ 𝑓𝑦𝑥

−.24 −. −. 2

−2

(2𝑡−2) ⃗ √ 𝑗 𝑠

43 −1

𝑦

𝑥

+

⃗ 41 2⃗𝑖 + 𝑘

2 −.

−2

1.4 1.1 0.8

2 −2𝑡+4) ⃗𝑖 √ (2𝑠 𝑠)

39 𝑦[cos(𝑥𝑦) − sin(𝑥𝑦)]⃗𝑖 + 𝑥[cos(𝑥𝑦) − sin(𝑥𝑦)]𝑗⃗

(b) Yes (d) No (f) No

0.8 1.1 1.4

𝑢𝑦𝑦 = −𝑒𝑥 sin 𝑦

27 𝑓𝑥𝑥𝑦 = 𝑓𝑦𝑥𝑥 = 2 cos(𝑥 − 2𝑦)

𝑥 −2

𝑥4 −𝑥2 𝑦2 (𝑥2 +𝑦2 )2

17

−2

2

3.8 3.5 3.2 2.9 2.6 2.3 2.0 1.7 1.4 1.1 0.8

−.2 −.4

𝑦

−. 2 4

4 −.

, 𝑓𝑦 =

13 𝑓𝑁 = 𝑐𝛼𝑁 𝛼−1 𝑉 𝛽 (√ )) ( √ 15 𝑥∕ 2 𝜔𝑥 cos2 𝜔𝑥

𝑥 𝐿(𝑥, 𝑦):

2𝑥𝑦3 (𝑥2 +𝑦2 )2

11 𝜕𝑓 ∕𝜕𝑝 = (1∕𝑞)𝑒𝑝∕𝑞 𝜕𝑓 ∕𝜕𝑞 = −(𝑝∕𝑞 2 )𝑒𝑝∕𝑞

.2

.4 .2

.4 .2

.2

−2

Differentiable Not differentiable Not differentiable Differentiable

34

3

𝑦

−2

34

−4 −3

−3 −4

𝑓 (𝑥, 𝑦):

2

2 −0.6

(c) No, no 19 (a) 𝑓𝑥 (𝑥, 𝑦) = (𝑥4 𝑦 + 4𝑥2 𝑦3 − 𝑦5 )∕(𝑥2 + 𝑦2 )2 𝑓𝑦 (𝑥, 𝑦) = (𝑥5 − 4𝑥3 𝑦2 − 𝑥𝑦4 )∕(𝑥2 + 𝑦2 )2 (c) Yes (d) Yes √ 21 Counterexample: 𝑥2 + 𝑦2 √ 23 𝑓 (𝑥, 𝑦) = 𝑥2 + 𝑦2

.4

2

53 𝑥 + 𝑦 + 𝑧 = 3 55 cos 𝑡 sin(cos 𝑡) − sin2 𝑡 cos(cos 𝑡) 57 100𝑡3 59 3∕𝑡 + 2𝑡∕(𝑡2 + 1) 61 𝑄(𝑥, 𝑦) = 2 + 6𝑥 + 𝑦 + 6𝑥2 + 3𝑥𝑦

1116 𝑧

63 𝑄(𝑥, 𝑦) = 1 + (𝑥 − 3) − 21 (𝑦 − 5)

Cross-section for 𝑦

− 21 (𝑥 − 3)2 + 21 (𝑥 − 3)(𝑦 − 5) − 18 (𝑦 − 5)2

65 (a) 2𝑥 − 4𝑦 + 𝑎𝑧 = 𝑎 − 2 (b) 𝑎 = 2 67 (a) (b) (c) (d)

(c) None 31 𝑎 = −9, 𝑏 = −12, 𝑐 = 50 33 (a) 𝑘 < 4 (b) None (c) 𝑘 ≥ 4

5

𝑄, 𝑅 𝑄, 𝑃 𝑃 , 𝑄, 𝑅, 𝑆 None

𝑥

35

2

71 𝐹 = 684 newtons, 𝜕𝐹 ∕𝜕𝑚 = 9.77 newtons/kg, 𝜕𝐹 ∕𝜕𝑟 = −0.000214 newtons/meter

(b)

−3 −2 −1 321 0

5

=2 𝑦

𝑃,𝑆 𝑅, 𝑆 𝑃 , 𝑄, 𝑅, 𝑆 None

0

3 ✒ ✒ 2 ❑ 𝑆 1❑ 𝑇 ❖ ✗ ✿ −3 −2 −1 −1 −2 ✒ ③ ■ ✙ ❘ ☛𝑅❲ 𝑄

𝑧 Cross-section for 𝑥

73 (a) 20 hours per day (b) 18.615 hours per day 75 (a) (b) (c) (d)

=3

−3

2

3

3 ❘◆ ✌✠ ✲ ✛ ✯ ❨

0

0 1 23 −1 −2 −3

𝑃

77 0.3

𝑧

79 0.8

Cross-section for 𝑦

81 0 √ 83 (a) −5 2∕2 (b) 4⃗𝑖 + 𝑗⃗

5

37 Saddle point: (0, 0).

85 (a) 98.387 ft/mile (b) 295.161 ft/hour

𝑥

87 (a) −4𝑒−81 ◦ C∕meter −81 ◦ C∕sec (b) −40𝑒 √ 932𝑒−81 ◦ C∕meter (c)

2 (c)

89 −2𝑥⃗𝑖 − 2𝑦𝑗⃗

𝑧 Cross-section for 𝑥

91 Yes 95 (a) (b) (c) (d)

=3

𝐹𝑢 (𝑥, 𝑦, 3) 𝐹𝑤 (3, 𝑦, 𝑥) 𝐹𝑢 (𝑥, 𝑦, 𝑥) + 𝐹𝑤 (𝑥, 𝑦, 𝑥) 𝐹𝑢 (𝑥, 𝑦, 𝑥𝑦) + 𝑦𝐹𝑤 (𝑥, 𝑦, 𝑥𝑦)

=2

39 Critical points: (0, 0), (±𝜋, 0), (±2𝜋, 0), (±3𝜋, 0), ⋯ Local minima: (0, 0), (±2𝜋, 0), ±4𝜋, 0), ⋯ Saddle points: (±𝜋, 0), (±3𝜋, 0), (±5𝜋, 0), ⋯ 41 (a) (1, 3) is a local minimum (b)

0

𝑦 5

0 12

𝑦

97 𝑑𝑃 ≈ 47.6 𝑑𝐿 + 17.8 𝑑𝐾 101 (a) −3⃗𝑖 + 4𝑗⃗ − ⃗𝑘 (b) −3⃗𝑖 + 4𝑗⃗

64

3

32 16 4

3

1✠

𝑧

105 15◦ C/minute

𝑥

Cross-section for 𝑦

107 Approx 7.5 at (1.94, 1.08)

=3

1

109 𝑥 − 𝑦 111 (a) Negative, positive, Up if positive, down if negative (b) 𝜋 < 𝑡 < 2𝜋 (c) 0 < 𝑥 < 3𝜋∕2 and 0 < 𝑡 < 𝜋∕2 or 3𝜋∕2 < 𝑡 < 5𝜋∕2

5 𝑥 2

113 (a) 𝐴0 + 𝐴1 + 2𝐴2 + 𝐴3 + 2𝐴4 + 4𝐴5 + (𝐴1 + 2𝐴3 +2𝐴4 )(𝑥−1)+(𝐴2 +𝐴4 +4𝐴5 )(𝑦−2), 1 + 𝐵1 𝑡, 2 + 𝐶1 𝑡 (b) 𝐴1 𝐵1 + 2𝐴3 𝐵1 + 2𝐴4 𝐵1 + 𝐴2 𝐶1 + 𝐴4 𝐶1 + 4𝐴5 𝐶1 115

𝜕𝑤 𝜕𝑥 𝜕𝑤 𝜕𝑧

𝜕𝑥 𝑑𝑢 𝜕𝑢 𝑑𝑡 𝑑𝑧 𝑑𝑡

+

𝜕𝑤 𝜕𝑦 𝑑𝑢 𝜕𝑦 𝜕𝑢 𝑑𝑡

+

𝜕𝑤 𝜕𝑥 𝑑𝑣 𝜕𝑥 𝜕𝑣 𝑑𝑡

+

𝜕𝑤 𝜕𝑦 𝑑𝑣 𝜕𝑦 𝜕𝑣 𝑑𝑡

+

7 Saddle point 9 Local minimum 13 Local minimum 17 Local max: (1, 5)

1 (I) and (V) Local maximum, (II) and (VI) Local minimum, (III) and (IV) Saddle point

19 Saddle point: (0, 0) Saddle point: (2, 0) Local min: (1, 0.25)

3 (a) None (b) 𝐸, 𝐺 (c) 𝐷, 𝐹

21 Saddle pts: (1, −1), (−1, 1) Local max: (−1, −1) Local min: (1, 1)

𝑧

5 (a)

Cross-section for 𝑥

=2

6

11 Local maximum 15 Local max: (4, 2)

Section 15.1

𝑦

47

23 Local max: (−1, 0) Saddle pts: (1, 0), (−1, 4) Local min: (1, 4)

5 4 −4 −3 −2 −1

3 2 1

0

−1 −1

1

25 Saddle point: (0, 0) Local max: (1, 1), (−1, −1)

5

27 Local min: (0, 0)

𝑦 3

29 (a) All values of 𝑘 (b) None

1 2 34

−1

49 (a) (0, 0) (b) 𝐷 = −24𝑥2 (c) Saddle point

2

𝑥 3

4

5

6

1117 𝑦 𝑓 >0

𝑓 0 𝑥

−2

𝑓 1 41 (a)

(i)

2𝜋

3 ∫0 (− sin 𝑡 cos(cos 𝑡) + cos 𝑡 cos(sin 𝑡))𝑑𝑡

𝑦

5 24

43 9/2 45 47

7 −4

3 √ 2

ln( √3 + 1) 2

2 𝑒(1.25𝜋) ∕2

−1

49 (a) 7𝑒3 − 2𝑒 (b) 7𝑒3 − 2𝑒

9 −6

𝑥

41 (a) 𝑒 (b) 𝑒

11 9 13 82∕3

51 (a) 9 (b) 0

15 12

53 (a)

𝑦

17 116.28 (ii)

19 12

𝑦

9 5 7 3

21 21

𝑥

25 ∫𝐶 3𝑥𝑑𝑥 − 𝑦 sin 𝑥𝑑𝑦

𝑥

𝑄

1

23 0

𝑃

27 (𝑥 + 2𝑦)⃗𝑖 + 𝑥2 𝑦𝑗⃗ 29 3124 31 144 33 77,000/3

(iii)

35 (a) 11∕6 (b) 7∕6

𝑦

37 (a) 3∕2 (b) 3∕2

(b) Shorter (c) 6

39 200𝜋

𝑥

(iv)

𝑦

𝑥

43 (a) −5 (b) 5 (c) 0

55 (a) Positive (b) Not gradient (c) 𝐹⃗2

45 𝐹 could point with 𝐶 at some points and against 𝐶 at others

57 (a) (8, 9) (b) 50

47 𝑦 = 𝜋∕2, 𝑥 = 𝑡, 0 ≤ 𝑡 ≤ 3, ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 = 3 49 True

59 (a) 2𝜋𝑚𝑔 (b) Yes

51 True

61 𝑓 (𝑄) − 𝑓 (𝑃 ) where 𝐹⃗ = grad 𝑓

53 False

63 Methods other than Theorem 18.1 can be used

55 False

65 Gradient of any function

57 (a)

71 True

Section 18.3

73 True 75 False

1 12

77 True

3 Negative, not path-independent

79 False

43 Positive

5 Negative, not path-independent

45 0

7 Path-independent

83 If 𝐴′ (𝑥) = 𝑎(𝑥), then 𝑓 (𝑥, 𝑦) = 𝐴(𝑥) is potential function 𝑥 + 𝑥2 ∕2 + 𝑥3 ∕3 + 𝐶, any 𝐶

(b) (i), (iii)

47 11

9 Path-independent

49 6

11 Path-independent

51 13

13 𝑓 (𝑥, 𝑦) =

53 −1.2 ⋅ 107 meter2 ∕sec

15 𝑓 (𝑥, 𝑦, 𝑧) = 𝑒𝑥𝑦𝑧 + sin(𝑥𝑧2 ) + 𝐶 𝐶 = constant

55 14𝜋; tangent to 𝐶 Same direction, ||𝐹⃗ || = 7 −14𝜋; tangent to 𝐶 Opposite direction, ||𝐹⃗ || = 7

59 −2.5 ⋅ 10−5 𝐺𝑀𝑚 61 If ∫ 𝐹⃗ ⋅ 𝑑⃗𝑟 < 0, then ∫ 𝐶

−𝐶

𝐹⃗ ⋅ 𝑑⃗𝑟 > 0

𝑥2 𝑦

17 −2

23 𝑒3 − 1 25 0

65 False

31 Yes

67 True

33 Yes.

69 False

35 5𝑥𝑦 + 𝑦2 ∕2

75 − ∫𝐶 𝐸⃗ ⋅ 𝑑⃗𝑟 77 Spheres centered at origin

89 (a) Increases

21 0

27 𝑃 𝑄

73 False

87 (a) 𝐹⃗ − grad 𝜙 = −(𝑥 + 2𝑦) grad ℎ (b) −50

19 2

63 𝐹⃗ = ⃗𝑖 − 𝑗⃗

71 True

+𝐾

85 (a) 𝐹⃗ − grad 𝜙 = −𝑦 grad ℎ (b) 30

(a) 50 (b) 50

Section 18.4 1 No 3 No 5 𝑓 (𝑥, 𝑦) = 𝑥3 ∕3 + 𝑥𝑦2 + 𝐶 7 Yes, 𝑓 = ln 𝐴|𝑥𝑦𝑧| where 𝐴 > 0 9 No 11 −2𝜋

37 (a) 50𝜋 (b) No, integral over closed path not zero

13 −6

39 Use 𝑓𝑥𝑦 = 𝑓𝑦𝑥

17 −3𝜋𝑚2

15 −12

1126 19 (a)

𝑦

1 𝑥 −1 −1

1

37 (a) Possible answers are: 𝐹⃗ = grad(𝑥𝑦) 𝐺⃗ = grad(arctan(𝑥∕𝑦)), 𝑦 ≠ 0 ) ( ⃗ = grad (𝑥2 + 𝑦2 )1∕2 , (𝑥, 𝑦) ≠ (0, 0) 𝐻 (b) 0, −2𝜋, 0 ⃗ ; holes in domain (c) Does not apply to 𝐺⃗ , 𝐻

𝑦

47

(0, 1)

23 −9𝜋∕8

0 10 −8𝜋 7

𝐶

𝑥

(0, −1)

45 Green’s Theorem does not apply; Line integral depends on 𝐹⃗

21 𝑒 − cos 1

27 (a) (b) (c) (d)

(0, 1)

41 (a) 21𝜋∕2 (b) 2

(b) −𝜋

0 0 0 −6𝜋 −6𝜋 0 −6𝜋

𝑦

39 𝐿1 < 𝐿2 < 𝐿3

43 (a) 0⃗ (b) 𝑞∕||⃗𝑟 ||

25 (a) (b) (c) (d) (e) (f) (g)

47

𝐶1

𝑥 (1, 0)

49 (a) Closed curve oriented counterclockwise (b) Closed curve oriented clockwise with 𝑦 > 0 or Closed curve oriented counterclockwise if 𝑦 < 0 (Other answers are possible)

𝑦

51 (a)

𝐶2

23.3 23

1 𝑄

𝑃

22.7 𝑥

1

49 True

29 (b) 0 (c) 𝐺⃗ = ∇(𝑥𝑦𝑧 + 𝑧𝑦 + 𝑧) (d) 𝐻⃗1 = ∇(𝑦𝑥2 ), 𝐻⃗2 = ∇(𝑦(𝑥 + 𝑧))

51 True 53 True 55 True

31 𝜋𝑎𝑏

Chapter 18 Review

𝑦

1 Positive

𝑏

3 (a) Zero (b) 𝐶1 , 𝐶3 : Zero 𝐶2 : Negative 𝐶4 : Positive (c) Zero

𝑥 𝑎

2

(b) Longer (c) −0.3 53 (a) 𝜋∕2 (b) 0 57 (a) 𝜔 = 3000 rad/hr 𝐾 = 3 ⋅ 107 m2 ⋅rad/hr (b) Inside tornado:

𝑦

5 Scalar; 12 7 −58

𝑥

9 50 11 18

33 3∕2

13 1372 15 Not path-independent

𝑦

17 Path-independent

2

19 Path-independent

View from great distance:

21 Path-independent 23 350

𝑦

25 27𝜋∕2 27 45

𝑥

35

29 0, 100

𝑥

31 (ii), (iv)

2

33 (a) 24 (b) 12 (c) −12

𝑦

37 18 39 −36 41 (a) 0 (b) 24

𝑥

43 (a) (b) (c) (d)

0 6 75𝜋∕2 14

45 (a) 9∕2 (b) −9∕2

(c) 𝑟 < 100 m, circulation is 2𝜔𝜋𝑟2 𝑟 ≥ 100 m, circulation is 2𝐾𝜋 √ 59 (a) −(𝜋∕2)(−6𝑎2 + 𝑎4 ), 𝑎 = 3 2 2 (b) Integrand√ 3 − 𝑥 − 𝑦 positive inside disk of radius 3, negative outside 61 18𝑎 + 18𝑏 + 36𝑐 + (81𝑑∕2), −18𝑎 − 18𝑏 − 36𝑐 − (81𝑑∕2), curves go in opposite directions

1127

Section 19.1

77 False

5 0

1 −3⃗𝑖 3 15𝑗⃗

79 True

7 4𝑥

5 Rectangle in 𝑥𝑧 plane with area 150, oriented pos 𝑦 direction

83 False

11 0

85 False

7 (a) 45 (b) −45

87 (a) 0 (b) 0

13 (a) Positive (b) Zero (c) Negative

9 2𝑥∕(𝑥2 + 1) − sin 𝑦 + 𝑥𝑦𝑒𝑧

81 True

9 (a) (b) (c) (d) (e)

Positive Positive Negative Negative Positive

11 (a) (b) (c) (d) (e)

Negative Positive Negative Negative Zero

𝑦

15

Section 19.2 (

) ⃗ 𝑑𝑥 𝑑𝑦 −3⃗𝑖 + 5𝑗⃗ + 𝑘 ( ) ⃗ 𝑑𝑥 𝑑𝑦 3 −4𝑥⃗𝑖 + 6𝑦𝑗⃗ + 𝑘 1

5

3 5 ∫−2 ∫0 5

𝑥

70 𝑑𝑦 𝑑𝑥

5−𝑥

7 ∫0 ∫0

(𝑦𝑧 sin 𝑥 − 2𝑥𝑦 cos 2𝑦 + 𝑥𝑦) 𝑑𝑦 𝑑𝑥

9 −500

13 −12

𝑦

11 −5∕3 − sin 1 = −2.508

15 4

5

𝜋∕2

∫0 10 (cos 𝜃 + 2 sin 𝜃) 𝑑𝑧 𝑑𝜃 ) 2𝜋 8 ( 15 ∫0 ∫−8 6𝑧2 cos 𝜃 + 6 sin 𝜃𝑒6 cos 𝜃 𝑑𝑧 𝑑𝜃 13 ∫0

17 12𝜋 19 4 21 0

17 2000 √ 19 100 2∕3

√ 23 10𝜋∕ 3 25 6

2𝜋

𝜋∕2

𝜋∕2

𝜋

27 −75𝜋

21 ∫0 ∫0 100 (sin 𝜙 cos 𝜃 + 2 sin 𝜙 sin 𝜃 + 3 cos 𝜙) sin 𝜙 𝑑𝜙 𝑑𝜃

29 −𝜋 3

23 ∫−𝜋∕2 ∫0 16 cos2 𝜙 sin2 𝜙 cos 𝜃 𝑑𝜙 𝑑𝜃

31 2000𝜋

25 8000∕3 √ 27 (8 − 5 2)𝜋∕6 = 0.486

33 32000𝜋 35 0

29 6

37 12.8

31 6

39 Zero

33 18

41 24

35 36𝜋

43 −160𝜋 √ 45 130∕ 2

37 7∕3 39 𝜋 sin 25

47 42

41 𝜋∕2

49 −96𝜋 √ 51 4 3

43 1296𝜋 45 𝜋

53 𝜋 sin 9

√ 47 100 27

55 0

49 2.228

57 −27

√ 51 (a) ∫𝑅 𝑎∕ 𝑎2 − 𝑥2 − 𝑦2 𝑑𝑥 𝑑𝑦 √ 2𝜋 𝑎 (b) ∫0 ∫0 𝑎𝑟∕ 𝑎2 − 𝑟2 𝑑𝑟 𝑑𝜃. 2 (c) 2𝜋𝑎

61 (a) Zero (b) Zero 63 4𝜋

53 36𝜋

65 (a) 0 (b) 32𝜋

55 2𝜋∕3 57 4𝜋𝑎3

67 (a) Zero (b) Zero

59 −1

𝑦

71 (a)

61 11𝜋∕2

4 2 𝑥 −4 −2 −2

2

4

−4 (b) 0 (c) 𝐼ℎ ln |𝑏∕𝑎|∕2𝜋

73 Sign of ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ depends on both 𝐹⃗ and 𝑆 ⃗ 75 𝐹⃗ = 𝑧𝑘 𝑆: 0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 1, 𝑧 = 1, oriented upwards

65 (a) Constant inside cylinder radius 𝑎 ⎧1 𝑒𝑟 if 𝑟 ≤ 𝑎 ⎪ 𝑘𝛿0 𝑟⃗ (b) 𝐸⃗ = ⎨ 21 𝑎2 ⎪ 2 𝑘𝛿0 𝑒⃗ 𝑟 if 𝑟 > 𝑎 𝑟 ⎩ ) √ ( ⃗ ∕ 𝑓2 + 𝑓2 + 1 67 𝑛⃗ = −𝑓𝑥 ⃗𝑖 − 𝑓𝑦 𝑗⃗ + 𝑘 𝑥 𝑦 √ 𝑑𝐴 = 𝑓𝑥2 + 𝑓𝑦2 + 1 𝑑𝑥 𝑑𝑦

69 𝑟 = 10, 0 ≤ 𝜃 ≤ 2, 0 ≤ 𝑧 ≤ 3, oriented outwards 71 False

Section 19.3 1 2𝑥 + 𝑥𝑒𝑧 3 (I)

𝑥

17 −0.030 (i) 0.016𝜋∕3 (ii) −0.08 (b) Flux positive at (2, 0, 0) and negative at (0, 0, 10)

19 (a)

21 (a) 4𝑤3 (b) 4 (c) 4 23 div 𝑣⃗ = −6 25 (a) −1∕3, 1 (b) 1∕3 27 Undefined 29 (b) 31 (a) 0 (b) Undefined 33 (a) 𝜌(0) < 𝜌(1000) < 𝜌(5000) (b) cars/hour (d) 𝜌(𝑥) = 4125∕(55 − 𝑥∕50) if 0 ≤ 𝑥 < 2000 𝜌(𝑥) = 4125∕15 = 275 if 2000 ≤ 𝑥 < 7000 4125 𝜌(𝑥) = (15 + (𝑥 − 7000)∕25) if 7000 ≤ 𝑥 < 8000 𝜌(𝑥) = 4125∕55 = 75 if 𝑥 ≥ 8000 (e) 139 ft. at 𝑥 = 0 89 ft. at 𝑥 = 1000 38 ft. at 𝑥 = 5000 35 (a) 0 (b) 0 39 0 41 𝑏⃗ ⋅ (⃗ 𝑎 × 𝑟⃗ ) 43 (d)

1128 45 div 𝐹⃗ = 2𝑥 + 2 − 2𝑧 ⃗ 47 𝐹⃗ (𝑥, 𝑦, 𝑧) = 2𝑥⃗𝑖 + 3𝑦𝑗⃗ + 4𝑧𝑘

23 𝑏 > 0

19 0

25 𝑎 > 0

⃗ 21 50⃗𝑖 + 300𝑗⃗ + 2𝑘

49 𝐹⃗ (𝑥, 𝑦) = 2𝑥⃗𝑖

27 (a) 3/2 (b) 3/2

⃗ 23 (a) (𝑓 −𝑐)⃗𝑖 +(𝑏𝑒𝑧 −𝑒 cos 𝑥)𝑗⃗ +(2𝑑𝑥−3𝑎𝑦2 )𝑘 (b) 𝑓 = 𝑐 (c) 𝑓 = 𝑐, 𝑏 = 𝑒 = 0

51 False 53 False 55 False 57 False 59 False 61 False

Section 19.4 1 24 3 8 5 Zero 7 24 9 72

29 32𝜋 31 162 33 120𝜋 35 Flux through 𝑆1 = Flux through 𝑆2 < Flux through 𝑆3 < Flux through 𝑆4

𝑦

39 (a) Negative (b) Positive (c) Zero 41 (a) Flux = 𝑐 3 (b) div 𝐹⃗ = 1 (c) div 𝐹⃗ = 1

13 36𝜋 15 620𝜋

45 −2𝜋

17 5𝜋

47 42𝜋 ⃗ )∕(16𝜋) 49 −7(⃗𝑖 + 𝑘

19 420 21 10𝜋𝑎3 23 Yes; −3.22 25 ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ = ∫𝑊 div 𝐹⃗ 𝑑𝑉 = 0 27 (a) 𝑐𝑏(12𝑎 − 𝑎2 ) (b) 6, 10, 10; 3600 29 (a) 4𝜋 (b) 0 (c) 4𝜋 31 4𝜋 33 (a) 2 (b) 0.016 (c) 0.016053 ⋯ 35 (a) 30 watts/km3 (b) 𝛼 = 10 watts/km3 (d) 6847◦ C 37 (a) 0 (c) No

27 (a) 𝑤 = 1

37 𝑎 = 6 Cannot say anything about 𝑏 and 𝑐

43 (a) 2𝑐 3 (b) 2 (c) 2

11 288

25 (a) Horizontal (b) Vertical (c) Parallel to the 𝑦𝑧-plane, making angle 𝑡 with horizontal

𝑥

𝑤 = −1

51 (a) div 𝐹⃗ = 0 (b) 12

𝑦

53 (20,000𝜋∕3) − 128 55 (a) 35𝜋 (b) 105𝜋 (c) 70𝜋 57 (a) (b) (c) (d) (e) 61 (a)

0, except at origin 4𝜋 0 4𝜋 4𝜋 if origin inside 0 if origin outside

𝑥

(i) Total charge inside 𝑊 (ii) Total current out of 𝑆 √ (b) |𝜔| ⋅ 𝑥2 + 𝑦2 (c) div 𝑣⃗ = 0 ⃗ curl 𝑣⃗ = 2𝜔𝑘 (d) 2𝜋𝜔𝑅2

63 (b) ‖⃗ 𝑣 ‖ = 𝐾∕𝑟2 (c) Flux = 4𝜋𝐾∕3 (d) Zero 65 (a)

𝑧

39 𝑆 not the boundary of a solid region

35 Counterexample: 𝐹⃗ = 𝑦⃗𝑖

41 Any sphere

37 𝐹⃗ = 𝑧⃗𝑖

43 False

39 True.

−𝑥

45 False.

𝑦

47 True.

41 True 43 False

49 True

45 False

51 True

47 (a)-𝑧, (c)-𝑦, (d)-𝑥

53 True

Section 20.2

55 True

Chapter 19 Review 1 Scalar; −100𝜋 3 0 5 0 7 75𝜋

(b) 32𝜋∕5

Section 20.1 ⃗ 1 Vector; ⃗𝑖 + 𝑗⃗ − 𝑘 3 Vector; (𝑥 + 1)⃗𝑖 − (𝑦 + 2)𝑗⃗

11 −12

5 Vector; 0⃗ ⃗ 7 4𝑦𝑘

13 0

⃗ 9 4𝑥⃗𝑖 − 5𝑦𝑗⃗ + 𝑧𝑘

9 80

17 3(8 + sin 4)

11 0⃗ 13 0⃗

19 20

15 Zero curl

21 2𝜋

17 Nonzero curl

15 Zero

1 (a) 𝜋 (b) 0 3 Positive 5 −8𝜋 7 −2 9 0 11 18𝜋 13 (a) (b) (c) (d)

−2𝜋 ⃗ −2𝑘 −2𝜋 Stokes’ Theorem

15 No 17 (a) 45𝜋 (b) 81𝜋∕2

1129 ⃗ 19 (a) −⃗𝑖 − 𝑗⃗ − 𝑘 (b) (i) −4𝜋 (ii) 15∕2

9 Defined; vector 11 Nonzero curl 13 div 𝐹⃗ = 𝑦 + 𝑧 + 𝑥 ⃗ curl 𝐹⃗ = −𝑦⃗𝑖 − 𝑧𝑗⃗ − 𝑥𝑘 𝐹⃗ not solenoidal; not irrotational

21 0 23 0 √ 25 8𝜋∕ 3

27 (a) All 3-space ⃗





𝑗 +2𝑐𝑧𝑘 (b) 2𝑎𝑥𝑖 +2𝑏𝑦 1+𝑎𝑥2 +𝑏𝑦2 +𝑐𝑧2 (c) 0 (d) ln(3 + 507𝜋 2 ∕4) − ln(2)

29 −8𝜋 31 4𝜋

15 div 𝐹⃗ = 0 curl 𝐹⃗ = (2𝑦 − cos(𝑥 + 𝑧))⃗𝑖 ( ( ) ) ⃗ − 2𝑥 − 𝑒𝑦+𝑧 𝑗⃗ + cos(𝑥 + 𝑧) − 𝑒𝑦+𝑧 𝑘 ⃗ 𝐹 is solenoidal; not irrotational 17 (a) 18𝜋 (b) 18𝜋

33 63𝜋 35 (a) 0⃗ (b) 0 (c) 0

19 20𝜋

37 (a) Parallel to 𝑥𝑦-plane; same in all horizontal planes ⃗ (b) (𝜕𝐹2 ∕𝜕𝑥 − 𝜕𝐹1 ∕𝜕𝑦)𝑘 (d) Green’s Theorem

25 (a) (b) (c) (d) (e) (f) (g) (h)

39 𝐶 not the boundary of a surface 41 Any oriented circle 43 True 45 False 47 True

21 0 23 (a) is (I); (b) is (I); (c) is (V); (d) is (III); (e) is (IV); (f) is (V)

1 Yes 3 Yes

31 0

51 False

Section 20.3

−1125𝜋 Not defined 0 Not defined 384𝜋∕5 353∕4 Not defined 54

27 (a) 23 √ (b) Approx 0.0003𝜋∕ 3

29 (a) (b) (c) (d) (e) (f)

49 True

div 𝐹⃗ < 0, div 𝐺⃗ < 0 curl 𝐹⃗ = 0, curl 𝐺⃗ = 0 Yes Yes No No

5 No

33 4𝜋

7 Yes

35 210

9 Yes

37 12𝜋∕5

11 Yes

39 150𝜋

13 (a) No (b) Yes

41 𝑒 − 1

15 Curl yes; Divergence yes 17 Curl yes; Divergence yes 21 (1∕2)𝑏⃗ × 𝑟⃗ 23 No 25 (a) Yes (b) Yes (c) Yes 27 (a) Yes (b) No (c) No 29 (b) ∇2 𝜓 = − div 𝐴⃗ 31 Curl of scalar function not defined 33 𝑓 (𝑥, 𝑦, 𝑧) = 𝑥2 35 False 37 True 39 (a) curl 𝐸⃗ = 0⃗ (b) 3-space minus a point if 𝑝 > 0 3-space if 𝑝 ≤ 0. (c) Satisfies test for all 𝑝. 𝜙(𝑟) = 𝑟2−𝑝 if 𝑝 ≠ 2. 𝜙(𝑟) = ln 𝑟 if 𝑝 = 2.

Chapter 20 Review 1 ⃗𝑖 − 𝑗⃗ − ⃗𝑘 3 (𝑐), (𝑑), (𝑓 ) 5 𝐶2 , 𝐶3 , 𝐶4 , 𝐶6

𝑦 = 1+𝑠 𝑧= 𝑠+𝑡

7 Defined; scalar

43 (a) 18 (b) 81∕2 45 (a) No (b) No 47 (a) (b) (c) (d) (e)

2𝜋 0⃗ except on 𝑧-axis No Yes; 0 No

49 (a) (b) (c) (d)

50 30𝜋 8 −18𝜋

⃗, 17 𝑟⃗ (𝑠, 𝑡) = (𝑠 + 2𝑡)⃗𝑖 + (2𝑠 + 𝑡)𝑗⃗ + 3𝑠𝑘 other answers possible ⃗ , 3⃗𝑖 − 5𝑗⃗ + 2𝑘 ⃗ 19 (0, 0, 0), 2⃗𝑖 + 𝑗⃗ − 𝑘 ⃗, 21 𝑟⃗ (𝑠, 𝑡) = (3 + 𝑠 + 𝑡)⃗𝑖 + (5 − 𝑠)𝑗⃗ + (7 − 𝑡)𝑘 other answers possible 23 (a) Yes (b) No 25 𝑠 = 𝑠0 : lines parallel to 𝑦-axis with 𝑧 = 1 𝑡 = 𝑡0 : lines parallel to 𝑥-axis with 𝑧 = 1 27 𝑠 = 𝑠0 : parabolas in planes parallel to 𝑦𝑧plane 𝑡 = 𝑡0 : parabolas in planes parallel to 𝑥𝑧-plane 29 𝑠 = 4, 𝑡 = 2 (𝑥, 𝑦, 𝑧) = (𝑥0 + 10, 𝑦0 − 4, 𝑧0 + 18) 31 Horizontal circle ( ( ) ) 33 (a) 𝑥 = cos 𝜋3 𝑡 + 3 cos 𝜃 ( ( ) ) 𝑦 = cos 𝜋3 𝑡 + 3 sin 𝜃 𝑧=𝑡 0 ≤ 𝜃 ≤ 2𝜋, 0 ≤ 𝑡 ≤ 48 (b) 456𝜋 in3 35 If 𝜃 < 𝜋, then (𝜃 + 𝜋, 𝜋∕4) If 𝜃 ≥ 𝜋, then (𝜃 − 𝜋, 𝜋∕4) 37 𝑥 = 𝑟 cos 𝜃, 0≤𝑟≤𝑎 𝑦 = 𝑟 sin 𝜃, 0 ≤ 𝜃 ≤ 2𝜋 𝑧 = (1 − 𝑟∕𝑎) ℎ 39 (a) −𝑥 + 𝑦 + 𝑧 = 1, 0 ≤ 𝑥 ≤ 2, −1 ≤ 𝑦 − 𝑧 ≤ 1 (b) 𝑧

41 (a) 𝑧 = (𝑥2 ∕2) + (𝑦2 ∕2) 0 ≤ 𝑥+𝑦 ≤ 2 0 ≤ 𝑥−𝑦 ≤ 2 (b)

1 Curve 3 Surface 5 Horizontal disk of radius 5 in plane 𝑧 = 7 7 Helix radius 5 about 𝑧-axis 9 Top hemisphere 11 Vertical segment 13 𝑥 = 1 𝑦=𝑠 𝑧=𝑡 15 𝑥 = 1 + 𝑡

𝑧

𝑥

51 (a) (4(𝑝 + 𝑞)𝜋𝑅3 ∕3) (b) (4(𝑝 + 𝑞)𝜋𝑅3 ∕3)

Section 21.1

𝑦

𝑥

𝑦

43 Radius: 𝑅 sin 𝜙 45 𝑥 + 𝑦 − 𝑧 − 3 = 0 47 True 49 True 51 True 53 False

Section 21.2 1 1 3 𝑒2𝑠

1130 5 𝑎 = 1∕10, 𝑏 = 1

(b)

𝑧

19





√ 2∕2 + 𝑖 2∕2

7 𝑎 = 1∕50, 𝑏 = 1∕10

21

9 3

23 −250 √ 3 25 2𝑖 4 √ √ 27 (1∕ 2) cos(−𝜋∕12) + (𝑖∕ 2) sin(−𝜋∕12)

11 𝜌2 sin 𝜙 13 13.5 15 72 ∞

2𝑡−𝑥

2

17 (a) (1∕(2𝜋𝜎 2 ) ∫−∞ ∫−∞ 𝑒−(𝑥 +𝑦 √ 2 2 𝑡 (b) (1∕( 𝜋𝜎)) ∫−∞ 𝑒−𝑢 ∕𝜎 𝑑𝑢 √ −𝑡2 ∕𝜎 2 (c) (1∕( 𝜋𝜎))𝑒

2 )∕(2𝜎 2 )

√ (d) Normal, mean 0, standard deviation 𝜎∕ 2

19 𝑅 does not correspond to 𝑇 21 𝑥 = 2𝑠, 𝑦 = 3𝑡 23 False

Section 21.3

29 −𝑖, −1, 𝑖, 1 𝑖−36 = 1, 𝑖−41 = −𝑖

𝑥

𝑑𝑦 𝑑𝑥

𝑦

19

+

1 (a) 𝑦 ≤ 30 (b) two zeros 3 −1.05

⃗ ) 𝑑𝑠 𝑑𝑡 3 −𝑒𝑠 (cos 𝑡 𝑗⃗ + sin 𝑡 𝑘

5 2.5

5 4/3

7 𝑥 = −1.1

9 −𝜋𝑅7 ∕28 √ 11 200 14 √ 13 6𝜋

15 𝑘ℎ𝑤3 ∕6 meter3 ∕sec. 21 Integral gives volume

𝑏2 )

Appendix A

⃗ ) 𝑑𝑠 𝑑𝑡 1 ((𝑠 + 𝑡)⃗𝑖 − (𝑠 − 𝑡)𝑗⃗ − 2𝑘

7 6(𝑒4 − 1)

9 0.45 11 1.3 13 (a) 𝑥 = −1.15 (b) 𝑥 = 1, 𝑥 = 1.41, and 𝑥 = −1.41 15 (a) 𝑥 ≈ 0.7 (b) 𝑥 ≈ 0.4

25 True

17 (a) 4 zeros (b) [0.65, 0.66], [0.72, 0.73], [1.43, 1.44], [1.7, 1.71]

27 False

19 (b) 𝑥 ≈ 5.573

23 𝑟⃗ (𝑠, 𝑡) = 2𝑠⃗𝑖 + 𝑡𝑗⃗

Chapter 21 Review 1 Cone, height 7 and radius 14 3 𝑎 = 1∕15, 𝑏 = 1∕15 5 −1∕2 7 0 9 𝑥 = 2 + 5 sin 𝜙 cos 𝜃 𝑦 = −1 + 5 sin 𝜙 sin 𝜃 𝑧 = 3 + 5 cos 𝜙 11 (a) Cylinder (b) Helices 13 𝑥 = 𝑎 sin 𝜙 cos 𝜃 0 ≤ 𝜙 ≤ 𝜋 𝑦 = 𝑏 sin 𝜙 sin 𝜃 0 ≤ 𝜃 ≤ 2𝜋 𝑧 = 𝑐 cos 𝜙 15 (a) 𝑥2 + 𝑦2 = 9, 𝑥 ≥ 0, 1 ≤ 𝑧 ≤ 2.

21 Bounded −5 ≤ 𝑓 (𝑥) ≤ 4 23 Not bounded

Appendix B 1 2𝑒𝑖𝜋∕2 √ 3 2𝑒𝑖𝜋∕4

5 0𝑒𝑖𝜃 , for any 𝜃. √ 10𝑒𝑖(arctan(−3)+𝜋) 7

9 −3 − 4𝑖

11 −5 + 12𝑖 13 1∕4 − 9𝑖∕8 √ 15 −1∕2 + 𝑖 3∕2

17 −125𝑖

31 𝐴1 = 1 + 𝑖 𝐴2 = 1 − 𝑖 37 True

17 0 2𝜋𝑐(𝑎2

3∕2 + 𝑖∕2

39 False 41 True

Appendix C 1 (a) 𝑓 ′ (𝑥) = 3𝑥2 + 6𝑥 + 3 (b) At most one (c) [0, 1] (d) 𝑥 ≈ 0.913 √ 4 100 ≈ 3.162 3

5 𝑥 ≈ 0.511

7 𝑥 ≈ 1.310 9 𝑥 ≈ 1.763 11 𝑥 ≈ 0.682328

Appendix D 1 3, 0 radians 3 2, 3𝜋∕4 radians 5 7𝑗⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ 7 ‖3 √𝑖 + 4𝑗 ‖ = ‖ − 5𝑖 ‖ = ‖5𝑗 ‖, ‖𝑖 + 𝑗 ‖ = ‖ 2𝑗⃗ ‖ √ 9 5𝑗⃗ and −6𝑗⃗ ; 2𝑗⃗ and −6𝑗⃗

11 (a) (−3∕5)⃗𝑖 + (4∕5)𝑗⃗ (b) (3∕5)⃗𝑖 + (−4∕5)𝑗⃗ 13 8⃗𝑖 − 6𝑗⃗ 15 ⃗𝑖 + 2𝑗⃗ 17 Equal 19 Equal

√ 2, ⃗𝑖 − 𝑗⃗ √ √ 23 Pos: (1∕ √ 2)⃗𝑖 + (1∕ √ 2)𝑗⃗ Vel: (−1∕ 2)⃗𝑖 + (1∕ 2)𝑗⃗ Speed: 1 21 ⃗𝑖 + 𝑗⃗ ,

INDEX 𝑆-𝐼-𝑅 model, 616 SARS, online threshold value, 618, online Δ𝑡, 277 Δ, Delta, notation, 5, 144 ∇𝑓 , gradient, 766 ∇⋅ 𝐹⃗ , divergence, 983 𝛾, Euler’s constant, online ∫ , 322 ∫𝐶 𝐹⃗ ⋅ 𝑑⃗𝑟 , 922 𝑏 ∫𝑎 , 284 𝜇, mean, 469 ∇ × 𝐹⃗ , curl, 1002 𝜕∕𝜕𝑥, partial derivative, 741 𝜌, 𝜃, 𝜙, spherical coordinates, 872 𝜎, standard deviation, 469 ∫𝑆 𝐹⃗ ⋅ 𝑑 𝐴⃗ , 964 ∑ , Sigma, notation, 283 →, tends to, 53 ⃗ , 705 ⃗𝑖 , 𝑗⃗ , 𝑘 ⃗ , cross product, 729 𝑣⃗ × 𝑤 ⃗ , dot product, 718 𝑣⃗ ⋅ 𝑤 𝑑∕𝑑𝑥 notation, 108 𝑒, the number, 17 as limit, 255, online property of, 32 𝑓 ′ (𝑥) notation, 100 𝑔, 330, 441 𝑖, the number, online 𝑛 factorial, 475 𝑝-series, 493 𝑟, 𝜃, 𝑧, cylindrical coordinates, 869 𝑟, 𝜃, polar coordinates, 420 ∫𝐶

𝑃 𝑑𝑥 + 𝑄 𝑑𝑦 + 𝑅𝑑𝑧, 935

absolute growth rate, 586 absolute value function, 124 absolutely convergent, 500, 1051 acceleration, 332, 899–901 average, 118 instantaneous, 119 straight line motion, 901 uniform circular motion, 900 vector, 712, 899–901, online components of, 899 limit definition, 899 acceleration due to gravity, 330 accuracy, online to 𝑝 decimal places, online adding sines and cosines, 637

addition of vectors, online components, 707, 1053 geometric view, 702 properties, 713 air pressure, 183 alternating series, 499 error bounds for, 500, 543 test, 499, 1052 Ampere’s law, 972, online amplitude, 41, 637 analytical solution of differential equation, 580 angular frequency, 553 antiderivative, 316, 1049 computing, 317, 323 constructing with definite integral, 335 Construction Theorem for, 336, 1049 differential equations and, 329 family of, 316 from graph, 317 Fundamental Theorem and, 317 most general, 322 of 0, 322 of 1∕𝑥, 324, 1049 of sin 𝑥, cos 𝑥, 1049 of sin 𝑥, cos 𝑥, 324 of 𝑒𝑥 , 324, 1049 2 of 𝑒−𝑥 , 316, 336 of 𝑥𝑛 , 323, 1049 of constant function, 323 of rational functions, 369 of same function, 322 of sin𝑚 𝑥 cos𝑛 𝑥, 361 properties of, 325 reduction formulas, 358 𝑥𝑛 cos 𝑎𝑥, 358 𝑥𝑛 sin 𝑎𝑥, 358 𝑥𝑛 𝑒𝑥 , 358 cos𝑛 𝑥, 358 tables of, 360 visualizing, 316 apple orchard yield, online approximation bisection, online error in Taylor polynomial, 539, 540, 1052 Euler’s method, 576 finite difference, online Fourier polynomial, 547

For “online” material, please see www.wiley.com/college/hughes-hallett.

1131

global vs. local, 549 linear, 178, 514, 753–757, 1047 Newton’s method, online of definite integral, 376 Picard’s method, online quadratic, 515, 792–795 tangent line, 178, 514, 1047 tangent plane, 755 arc length, 413, 901, 1050, online circle, 40 in polar coordinates, 427 of hanging cable, online of parametric curve, 414 arccosh, online arccosine function, 49, 1045 Archimedean spiral, 423 Archimedes’ Principle, 418, 996 arcsine function, 44, 1045 derivative of, 166 domain and range, 44, 1045 graph of, 45 arcsinh, online arctangent function, 45, 1045 derivative of, 166 domain and range, 45, 1045 graph, 45 Taylor series, 531 area between curves, 305 definite integral and, 285, 295, 402 double integral for, 844 finding by slicing, 402 of parallelogram, 732 of parameterized surface, 1039 polar coordinates, 425 area vector, 732, 736, 963 of parallelogram, 973 Aristotle (384-322 BC), 332 arithmetic mean, 222 Aswan Dam, 448 asymptote horizontal, 18, 53, 69, 1045 of rational functions, 53, 1045 vertical, 33, 53, 1045 average cost, 241 average rate of change, 91, 1047 average value of function, 308, 1049 two-variable, 844, 1056 average velocity, 84, online slope and, 87

1132

INDEX

axes coordinate, 654 right-handed, 654 bakery, 12 bank deposits as series, 484 Barbra Streisand, 486 barometric pressure, 604 base of exponential function, 15 of logarithm function, 32, 1044 basketball, online Bay of Fundy, 161 beef consumption, 659 behaves like principle, 395 bell-shaped curve, 224, 312, 468 bending of light, 537 Bernoulli, Johann (1667-1748), 605 Bessel functions, online binomial series, 525, 1052 bioavailability, 359 biodiesel fuel, 11, 22 birthdays, online bisection, method of, online bladder, 321 body mass index, BMI, 653 Bohr radius, 472 Bohr, Niels (1885-1962), 472 Boltzmann constant, online bonds, online trading at a discount, 487 trading at a premium, 487 bound, 540 above, 307 below, 307 best possible, 207, online greatest lower, online least upper, online lower, 207, online upper, 207, online boundary of region, 821 of solid region, 991 of surface, 1008 point, 755, 821 boundary-value problem, 636 bounded region, 821, 1055 bounded sequence, 477 Boyle’s Law, online Brahe, Tycho (1546–1601), online bridge design, online caffeine, 37, 604 calculators round-off error, 141, online Cantor Set, online carbon dioxide level, 351 carbon-14, 16, 596, online

cardiac output, 748 cardioid, 428, 429 carrying capacity, 56, 118, 607, 1053 Cartesian coordinates, 420 conversion to cylindrical, 869 spherical, 872 three-dimensional, 654 catalog of surfaces, 691 catenary, 174, 177, online Cauchy, Augustin (1789-1857) formal definition of limit, 60 cdf, 461 center of mass, 432, 878, 1050 continuous mass density, 434 moment, 432 of point masses, 433 triple integral for, 863 central vector field, online cephalexin, antibiotic, online chain rule, 152, 780–784 and units, 152 application to chemistry, 784 applications, 164 diagram for, 781 integration and, 342 change of coordinates, 1033–1036 channel capacity, 782 chaos, online characteristic equation, 642 critically damped case, 643 overdamped case, 642 underdamped case, 644 Chernobyl accident, 21 chikungunya, 613 chlorofluorocarbons, 114, 123 circle parameterization of, 259 circulation, 927 density, 1000, 1058 path-dependent field and, 950 closed curve, 927 closed curves as trajectories in phase plane, 621 closed form of series, 481 closed interval, 2 closed region, 821, 1055 closed surface, 962, 991 CO2 in pond water, 295 coal production in US, 299 coaxial cable, online Cobb-Douglas function, 241, 675, 753, online contour diagram of, 674 formula for, 675

returns to scale, 680 code-red computer virus, 614 common ratio, 482 comparison test for improper integrals, 395 for series, 494, 1051 competition, 624 Competitive Exclusion Principle of, 626 completing the square, 362, 373, 810 complex number algebra of, online complex plane, online conjugates, online definition, online differential equations and, 641 imaginary part, online polar representation, online powers of, online real part, online roots of, online complex plane, online polar coordinates, online compliance, 114 components of vector, 262, 705, 707, online composite functions, 24, 780 derivative of, 151 compound interest, 450, 598 continuous, 587 Rule of Seventy, online compressibility index, online concavity, 14 downward, 14, 116, 1047 error in definite integrals and, 379 Euler’s method and, 576 inflection point and, 196 of parametric curve, 266 power functions and, 134 second derivative and, 116, 192 upward, 14, 116, 1047 concentration, 601 conditionally convergent, 500, 1051 cone, 691 parameterization of, 1028 conjugates, online conservation of energy, online conservative force, online conservative vector field, 941 consols, online Constant Function Theorem, 187, 190, 1048 constant of proportionality, 6 constrained optimization, 825–830, 1055 graphical approach, 826

INDEX

inequality constraint, 829 Lagrange multiplier, 827 Lagrangian function, 831 constraint, 827 Construction Theorem for Antiderivatives, 336 consumer surplus definite integral for, 454 definition, 453 consumption vector, 715 continuity, 3, 58 at a point, 58, 60 definition, 60 definition of, 1047 differentiability and, 125, 802, 1048 extrema and, 203 graphically, 58 limits and, 58 numerically, 58 of composite functions, 71 of function, 59, 697 of sums, products, quotients of functions, 71, 1047 of vector field, 923 on an interval, 58, 60, 1046, 1047 continuous rate, 17 continuous variable, 2, 4 contour diagram, 668–675 algebraic formula, 671 Cobb-Douglas, 674 critical point, 808 local maximum, 806 density and, 840 linear function, 684 partial derivative and, 742 reading, 652 saddle, 673 table and, 673 contour line, 669 convergence of 𝑝-series, 493 of alternating series, 499 of geometric series, 483 of improper integrals, 385, 390 of partial sums, 488, 1051 of power series, 504 of sequence, 476, 1051 bounded, monotone, 477 of Taylor series, 526, 542 of upper and lower sums, 845 radius of, 507 rearranging terms, 501 coordinate plane, 655 axis, 654 coordinates

Cartesian, 420 Cartesian, three-space, 654 cylindrical, 869–870 polar, 420 relation between Cartesian and polar, 421 space-time, 715 spherical, 872–874 corn production, 670, 780 Coroner’s Rule of Thumb, 250 correlation coefficient, 820 cosh 𝑥, 174, 1045 Taylor polynomial, 538 cosine function, 40 addition formula, online derivative of, 158, 160 graph of, 41, 1045 Taylor polynomial approximation, 518 Taylor series, 524 convergence, 542 cost average, 241 fixed, 11, 233, 328 marginal, 110, 235, 328 definition of, 235 total, 233 variable, 11 cost function, 233 Coulomb’s law, online coupon, online credit multiplier, online critical point, 193, 318, 806, 1048 classifying, 809, 811 contour diagram and, 808 critical value, 193 discriminant and, 811 extrema and, 194, 806, 1048 how to find, 807 local maximum contour diagram and, 806 graph of, 808 local minimum graph of, 807 second derivative test and, 809, 811 cross product, 728–733 components of, online definition, 729, online determinant and, 733 diagram of, 731 equation of plane and, 731 properties, 731 cross-section, 412 cross-section of functions, 662–663 cubic polynomial, 51, 1045 cumulative distribution function, 460

1133

probability and, 464 properties of, 461 curl, 1000–1002 alternative notation, 1002 Cartesian coordinates, 1059 definition Cartesian coordinate, 1001 geometric, 1001 device for measuring, 1000 divergence and, 1016 field, 1011, 1059 divergence test for, 1017 formula for, 1001, 1059 gradient and, 1015 scalar, 951 test for gradient field, 1016, 1059 three-space, 956 two-space, 953 curl free, 1004, 1010 curve closed, 927 indifference, 833 integral, 914 length of, 414, 901, online level, 669 graph and, 671 oriented, 922 parameter, 1029 parameterization, 264, 886 piecewise smooth, 923 curve fitting, 819 cylinder parabolic, 664, 691 parameterization of, 1023 cylindrical coordinates, 869–870 conversion to Cartesian, 869 integration in, 870 volume and, 871 volume element, 871 damped spring equation, 640, 641 damping, 641, 642 coefficient, 641 critical, 643 over, 642 term, 641 under, 644 daylight hours as a function of latitude, online Madrid, online decreasing, 5 decreasing function, 6 derivative of, 101, 115, 192, 1047 integral of, 277 Decreasing Function Theorem, 190 definite integral, 283, 1048

1134

INDEX ∞

2

∫−∞ 𝑒−𝑥 𝑑𝑥, online odd and even functions, 306 symmetry and, 306 as antiderivative, 336 as area, 285, 295, 402, 1049 in polar coordinates, 425 as area above axis minus area below axis, 286 as average, 308 as total change, 294, 1049 by substitution, 346 changing limits of, 346 comparison of, 307 definition of, 284 for arc length, 414 in polar coordinates, 427 of parametric curve, 414 for average value, 308 for center of mass, 432 for consumer surplus, 454 for cumulative distribution function, 460 for electric potential, 449 for force, 444 for force of gravity, 449, online for fraction of population, 459 for kinetic energy, online for mass, 430 for mean, 468 for present/future value, 452 for probability, 464 for producer surplus, 454 for surface area, online for volume, 403 for volume of revolution, 411 for work, 441 Fundamental Theorem of Calculus, 293, 297, 325, 1049 interpretation, 285, 294 limits of integration, 302 notation for, 292 of density, 430 of density function, 459 of rate of change, 294 of sums and multiples, 304 one variable, 840 properties of, 302, 1049 units, 292 definite integral, double, 840–853 change of coordinates, 1035 change of variables, 1056 definition, 841, 842, 1056 for area, 844 for average value, 844 for fraction of population, 880 for probability, 880 for surface area, 1039

for volume, 842 of joint density function, 880 polar coordinates, 864–867 definite integral, triple, 857–860 change of coordinates, 1036 cylindrical coordinates, 869– 872 for center of mass, 863 for electrical energy, online for moment of inertia, online spherical coordinates, 872–874 degree homogeneous function, online degree of a polynomial, 51 degree-days, online Delta, Δ, notation, 5, 144 demand curve, 12, 453 density, 429 circulation, 1000 of earth, 439 of water, 444 probability, 459 slicing and, 430 density function, 459, 1050 definite integral and, 840 flux, 982 joint, 880, 1056 probability and, 464 properties of, 459, 880 two-variable, 847, 878, 880 probability and, 880 dependent variable, 2, 4, 652 depreciation, 10 derivative, 1047 𝑒𝑥 , 1047 𝑛𝑡ℎ , 137 of inverse function, 166 approximation of, online as a function, 99 definition of, 100 finding algebraically, 103 finding graphically, 100 finding numerically, 101 formulas for, 103 at a point, 91 chain rule, 152, 1047 critical points, 193, 1048 definition of, 92, 1047 differentiability and, 123 directional, 762–765, 772 estimating graphically, 99 estimating numerically, 94 graphical interpretation, 101 higher, 115 higher-order partial, 790 inflection point, 1048 interpretation, 91, 109

Leibniz notation for, 108 local maxima/minima, 194 test for, 194 of 𝑎𝑥 , 142 of 𝑒𝑥 , 141 of composite functions, 151, 1047 of constant functions, 103 of constant multiples of a function, 130, 1047 of cosine, 159, 1047 of exponential functions, 140, 141, 165, 1047 of hyperbolic functions, 176 of implicit functions, 171 of integral, online of inverse trigonometric functions, 166, 1047 of linear functions, 103 of ln 𝑥, 165, 1047 of polynomials, 134 of positive integer powers, 131 of power functions, 132, 164, 1047 of products of functions, 145, 1047 of quotients of functions, 146, 1047 of sine, 160, 161, 1047 of sums and differences of functions, 130, 1047 of tangent function, 160, online ordinary, 740 partial, 740–744 power rule, 104 product rule, 146, 1047 quotient rule, 147, 1047 second derivative, 115, 1047 concavity and, 116 test, 196 second-order partial, 790 slope of curve, 93 in polar coordinates, 427 slope of tangent line, 93 units of, 109, 1047 visualizing, 92, 567, 1047 derivative function, 100 Descartes, René (1596-1650), 332 determinant, online area and, 732 cross product, online Jacobian, 1034 volume and, 732 diagram, for chain rule, 781 difference quotient, 5, 91 partial derivative and, 741 differentiability, 754, 799

INDEX

continuity and, 125, 802, 1048 determining from graph, 123 local linearity and, 179, 181, 1048 partial derivatives and, 799, 802 differentiable, 92, 100, 123 everywhere, 100 differential, 758–759 computing, 758 local linearity, and, 758 notation, 759 differential equations, 562, 1053 𝐿𝑅𝐶 circuits, 649 𝑆-𝐼-𝑅 model, 616 SARS, online account balance, 587 analytical solution, 580 antiderivatives and, 329 arbitrary constants, 564 compartmental analysis, 601 concentration, 601 damped spring, 641 damped spring equation, 640 damping, 642, 643, 645 decay, 586 electric circuit, 640 equilibrium solution, 592, 1053 Euler’s method, 575, 916 existence of solutions, 569 exponential decay, 581 exponential growth, 580 SARS, online exponential growth and decay, 1053 first-order, 563 flow of vector field, 914 general solution, 563, 1053 exponential equation, 581, 587 growth, 586 guess-and-check spring equation, 634 Hooke’s Law, 634 implicit functions and, 569 initial conditions, 329, 563, 1053 initial value problem, 329, 563, 635, 1053 linear second-order, 641 logistic, 607, 609, 1053 Lotka-Volterra, 619 net worth of a company, 599 Newton’s Law of Cooling (Heating), 590 nullclines, 626, 627 numerical solutions of firstorder, 575 oil prices, 606

order of, 1053 particular solution, 329, 563, 1053 pendulum, 639 phase plane, 617 Picard’s method, online pollution, 588 population growth, 611 predator-prey model, 620 second-order, 564, 632 characteristic equation, 642 separable, 583 separation of variables, 580, 600 slope field, 567, 1053 solution(s) to, 563, 1053 spring equation, 634 systems of, 616, 1053 thickness of ice, 597 uniqueness of solutions, 569 differential notation for line integral, 935 differentiation implicit, 171 of series, 531 convergence, 531 radius of convergence, 531 dipole, 535, 978, 987, online direction cosine, 707, online direction field, online direction of vector, online directional derivative, 762–765 definition, 763 examples, 767 from contour diagram, 762 gradient vector and, 765 partial derivatives and, 764 three-variable, 772 directly proportional, 6 disability index, 883 discrete variable, 2, 4 discriminant, 810, 811 disease, 616 disease incidence, 554 displacement, 432 displacement vector, 702–706, online direction of, 702 magnitude of, 702 distance estimating from velocity, 272 on parametric curve, 414 visualizing on velocity graph, 273 distance formula in three-space, 656 in two-space, 656 distribution function, 457 cumulative, 460

1135

probability and, 464 distribution of resources, online divergence, 982–983 alternative notation, 983 curl and, 1016 definition Cartesian coordinate, 983 geometric, 982 free, 1058 of geometric series, 483 of harmonic series, 490 of improper integrals, 386, 390 of partial sums, 488, 1051 of sequence, 476, 1051 test for curl field, 1017, 1059 with spherical symmetry, online Divergence Theorem, 991–994, 1015 divergence-free, 986, 993 diverging to infinity, 63, 387 domain, 2 restricting, 3 dominance l’Hopital’s rule and, 254 dominates, 50 Dorfman-Steiner rule, 824 dot product, 718–720 definition, 718 equation of plane and, 721 line integral and, 922 properties, 719 work and, 724 double-angle formulas, 351, 358, 1045, online double factorial, online doubling time, 16, 587, 1044 drug desensitization, online drug dosage as series, 481, 486 Dubois formula, 56, 247, 752 Ebbinghaus model, 605 economy of scale, 233 Einstein, Albert (1879-1955), 537 Theory of Relativity, 157 electric charge, online electric circuits, 640, 649, online resonance, 537 electric dipole, 535 electric field, 989, online electric potential, online definite integral for, 449 electron wave function, online elementary functions, 342 ellipsoid, 691 ellipsoid, volume of, 1036 end behavior of a function, 53 endocrinologist, 301 energy conservation of, online

1136

INDEX

electrical, 352 potential, 941 wind, online energy spectrum, 552 musical instruments, 553, 557 energy theorem, 551 entropy function, online Envelope Theorem, online epidemic, 616, 618, online equation graphing in polar coordinates, 422 equations of motion, 330 equilibrium stable, online equilibrium point, 627 predator-prey model, 621 equilibrium price/quantity, 453 equilibrium solution of differential equation, 563 of differential equation, 592, 598, 1053 stable, 592 unstable, 592 equipotential surface, online erf(𝑥), 340, 359, online error, online alternating series bounds, 500 Taylor polynomial, 543 concavity and, 379 Euler’s method, 577 in approximating definite integral, 1050 in approximating derivative, online in linear approximation, 178 in numerical integration, 378, 379 left and right rules, 379 Riemann sum approximation, 277 round-off, 98, 141, online tangent line approximation, 179, 180 estimation of, 179 Taylor polynomial, 539, 540 trapezoid vs. midpoint rule, 380 error bound, 540 Lagrange, 540 error function, erf(𝑥), 340, 359, online escape velocity, 606 Ethanol fuel, 37 Euler’s Constant, online Euler’s formula, 645, online Euler’s method, 575, 1053 accuracy of, 577 error, 577

for flow lines, 916 Euler’s theorem, 753 Euler, Leonhard (1707-1783), 575, online Eulerian logarithmic integral, 359 even function, 25, 1046 definite integral of, 306 expansions, Taylor, 524 explicit function, 171 exponential decay, 15, 17, 1044 differential equation for, 586 half-life, 16, 587, 1044 radioactive decay, 16 exponential function, 13, 15, 1044 𝑦-intercept, 17 population growth, 13 as solution to differential equation, 580 base 𝑒, 17 base of, 15 compared to power functions, 50 concavity, 14 derivative of, 140, 165 domain, 15 formula for, 15, 17, 1044 Padé approximant to, 537 table of values, 14 Taylor polynomial approximation, 518 Taylor series, 524 convergence, 545 exponential growth, 14, 15, 17, 1044 annual, 588, 1044 continuous, 588, 1044 differential equation for, 586 doubling time, 16, 587, 1044 growth factor, 14, 16 rate absolute, 586 annual, 588, 1044 continuous, 17, 588, 1044 continuous vs. annual, 588 relative, 586 SARS, online exponential growth and decay differential equation, 1053 extrapolation, 4, 611 extrema, 203, 806 continuity and, 203 critical point and, 194, 1048 global, 821 local, 193 Extreme Value Theorem, 203, 821, 1047 extremum on closed bounded region, 821

factorial, 517 double, online factoring, online family of functions, 5, 224 antiderivatives, 316 exponential, 16 with limit, 226 Gompertz growth, 233 linear, 5 logistic, 228 motion due to gravity, 226 normal density, 224 power, 49 sinusoidal, 41 surge, 231 family of level surfaces, 692 Fermat’s Principle, 222 Fibonacci sequence, 479 finite difference approximation, online firebreaks, online first derivative inflection point and, 197 first-derivative test, 194, 1048 First Fundamental Theorem of Calculus, 293 fixed cost, 11, 233, 328 flow fluid, 905 flux and, 965 through surface, 963 flow line definition, 913 Euler’s method, 916 numerical solution, 916 fluid flow and flux, 965 flux orientation and, 962 through cylinder, 976 through function graph, 974 through parameterized surface, 1038 through sphere, 977 flux density, 982 flux diagram, online flux integral, 962–968 definition, 964 Divergence Theorem and, 991 independent of parameterization, 1041 fog clearance, 747 foot-pound, 444 force, 333, 444, 1050 between atoms, online between molecules, online conservative, online definite integral for, 444

INDEX

from pressure, 444 gravitational, 712 of gravity definite integral for, 449, online spring, 440 units, 440 vector, 712 force of gravity, 441 Four, Rule of, 2 Fourier coefficients formulas for period 2𝜋, 547 period 𝑏, 553 justification of, 555, 558, 559 Fourier polynomial, 547 square wave, 548 Fourier series, 546, 1052 period 2𝜋, 550 period 𝑏, 553 square wave, 550 Fourier, Joseph (1768–1830), 546 fox population, 840, 847 fraction of population from cumulative distribution function, 461 from density function, 459, 880 frequency angular, 553 fuel consumption, 223 function, 2 absolute value, 124 average value of, 308, 1049 bounded, online Cobb-Douglas, 241, 675, 753 composite, 24, 780 concave (up or down), 14, 1047 continuous, 695, 697 at a point, 697 cost, 233 cross-section of, 662–663 cumulative distribution, 460 decreasing, 5, 6, 1047 density, 459 two-variable, 880 differentiable two-variable, 799–803 differential of, 758 discontinuous, 695 distribution, 457 drug buildup, 18 elementary, 342 energy theorem, 552 even, 25 exponential, 13, 15, 1044 family of, 226 fixing one variable, 663 Fourier series for, 549

gamma, online global approximation, 549 global behavior of, 50 graph of, 2 hyperbolic, 1045 increasing, 4, 6, 1047 input, 2 inverse, 26 inverse trigonometric, 44 invertible, 26 joint cost, online Lagrangian, 831 limit of, 697 linear, 4, 5, 664, 682–685, 690, 1044 local approximation, 549 local behavior of, 50 logarithm, 32 logistic, 609 monotonic, 6 notation, 652 objective, 827 odd, 25 output, 2 periodic, 40, 546 piecewise linear, 124 polynomial, 51, 1045 potential, 942, 1057 power, 49 probability density, 459, 464, 880 profit, 234 pulse train, 550 quadratic, 690, 809 graph of, 810 rational, 53, 1045 reflection across 𝑥-axis, 23 representation, 2 revenue, 234 shift and stretch, 23 sinusoidal, 41, 1044 smooth, 791 surge, 231 table, 2 Taylor series for, 524 three-variable, 689 level surface of, 689 surface, 692 trigonometric, 39 cosine, 40 sine, 40 tangent, 43 two-variable, 652 algebraic formula, 653 contour diagram of, 668 graph of, 660–663 surface, 692

1137

unbounded, online utility, 836 zeros of, 51, online fundamental harmonic, 550 Fundamental Theorem of Calculus, 293, 297, 317, 325, 336, 939, 1049 for Line Integrals, 939, 1015 line integral, 1057 future value, 1050 annual compounding, 450 continuous compounding, 450 definite integral for, 452 definition, 450 of income stream, 451 G. H. Hardy, 1847 − 1947, online Galilei, Galileo (1564-1642), 13, 596, online Dialogues Concerning Two New Sciences, 332 gamma function, online gauge equivalent, 1019 Gause yeast population model, 631, online Gauss’s law, 967, online Gauss’s Theorem, 995 Gauss, Carl Friedrich (1777–1855), 995 general solution of differential equation, 329 general term, 524 general term of a series, 488 genetics, online geometric mean, 222 geometric series, 480, 482, 1051 as binomial series, 525 common ratio, 482 convergence of, 483 divergence of, 483 finite, sum of, 482, 1051 infinite, 482 infinite, sum of, 483, 1051 partial sum of, 483 geometry definite integral and, 410 Gini’s index of inequality, online global behavior of a function, 50, 52 global extremum, 806 closed bounded region, 821 definition, 816 how to find, 816, 820 global maxima/minima, 203, 806, 1048 global warming, 787 Golden Gate Bridge, online golden ratio, 479 Gompertz equation, online

1138

INDEX

Gompertz growth, 233 gradient field curl and, 1015 curl test for, 1016 line integral of, 939 path-independence and, 942 gradient vector, 765 alternative notation, 766 examples, 767 field, 909 geometric properties, 766, 772 three-variable, 772 two-variable, 765 Grand Canyon flooding, online Grand Coulee Dam, 409 graph circular symmetry of, 662 in polar coordinates, 422 in three-space, 655 partial derivative and, 742 plane, 683 two-variable function, 683 gravitational constant, 909, online gravitational field, online picture of, 906 gravity acceleration due to, 139, 184, 330, 332, 334, 606, online force of, 247, 449, online Gravity, Law of, 908 Great Lakes, 588, 594 Great Pyramid of Egypt, 405, 443 greatest lower bound, online Green’s Theorem, 951, 1019 Greenland Ice Sheet, 112 Gregory, James (1638–1675), 531 Grinnell Glacier, 595 growth continuous, 17 growth factor, 14, 16 growth of children, 301 growth rate absolute, 586 annual, 588 continuous, 588 relative, 586 guess-and-check, 342 spring equation, 634 guitar string, vibrating, 750 Gulf Stream, 905, 913 Half Dome Yosemite National Park, 695 half-life, 16, 587, 1044 hanging cable, 566 arc length, online harmonic series, 490 harmonics, 550

heat equation, online heated metal plate, 740 heater in room, 743, online height velocity graph, 301 helix, 886 higher derivative, 115 higher-order partial derivative, 790 histogram, 457, 458, 879 homogeneous function, online Hooke’s Law, 440, 634 Hooke, Robert (1635–1703), 634 Hoover Dam, 445 l’Hopital, Marquis de (1661–1704), 605, online l’Hopital’s rule, 252, 1048 horizontal asymptote, 18, 53 horizontal line, 5 horizontal line test, 27 Hubbert, M. King (1903–1989), 607 hybrid cars, online hydroelectric power, 11, 22 hydrogen, reduced mass of, 537 hyperbola, 176 hyperbolic cosine, 174, 1045 identities, 175 sine, 174, 1045 tangent, 176 hyperbolic functions derivative of, 176 hyperbolic tangent inverse, online hyperboloid of one sheet, 691 of two sheets, 691 hyperinflation, 38 ideal gas equation, 758 identities, 1045 imaginary numbers, online imaginary part of complex number, online implicit differentiation, 171, 1048 implicit function as solution to differential equation, 569 derivative of, 171 improper integral, 385, 1050 comparing, 394, 395 with 1∕𝑥𝑝 , 396 with 𝑒−𝑎𝑥 , 396 comparison test, 395, 1050 convergence/divergence, 385, 390 energy and, 388 infinite integrand, 389 income stream, 451 increasing function, 4, 6, 187

derivative of, 101, 115, 192, 1047 integral of, 277 Increasing Function Theorem, 187, 1048 indefinite integral, 322, 1049 computing, 323 properties of, 325, 1049 visualizing, 316 independent variable, 2, 4, 652 index in power series, 524 index in Taylor series, 524 indifference curve, 680, 833 inertia, moment of, online triple integral for, online inertia, principle of, 333 infinite series geometric, 482 sum of, 480 inflection point, 196, 318, 1048 first derivative and, 197 initial condition, 329 initial value problem, 329, 563, 636, 1053 mass/spring system, 635 input into a function, 2 instantaneous velocity, 87 growth rate, 588 rate of change, 91, 741, 1047 speed, 262 velocity, 86, 896 integral definite one variable, 840 definite vs. indefinite, 323 definite see definite integral, 283 double limits of, 851 equation, online improper see improper integral, 385 indefinite see indefinite integral, 322 iterated, 847, 848 triple limits of, 860 integral test for series, 492 integrand, 284 infinite, 389 sums and multiples, 304 integration applications, 1050 Cartesian coordinates, 848, 858, 1056 cylindrical coordinates, 870 iterated, 847

INDEX

limits of, 284, 302, 851, 860 non-rectangular region, 844, 850–853 numerical methods, 376 error, 378 LEFT(𝑛), 377 left-hand sum, 284 MID(𝑛), 377 midpoint rule, 376 Riemann sums, 284, 377 RIGHT(𝑛), 377 right-hand sum, 284 SIMP(𝑛), 381 TRAP(𝑛), 378 of series, 531 order of, 849, 853 polar coordinates, 864–867 reduction formulas, 358, 360 spherical coordinates, 873 techniques by parts, 353, 354, 1050 completing the square, 362, 373 factoring, 362 guess-and-check, 342 long division, 362 partial fractions, 366, 1050 reduction formulas, 360 sine substitution, 369 substitution, 342, 343, 346, 1049 tangent substitution, 372 trigonometric substitutions, 369 using table, 360 intercept factored form and, 51 vertical, 4, 5, 1044 interest, 598 compound, 450 continuously compounded, 587 Rule of Seventy, online interior point, 755 Intermediate Value Theorem, 60, 1047 intersection of curve and surface, 890 of two curves, 890 interval notation, 2 interval of convergence, 505, 506, 1052 inverse hyperbolic tangent function, online inverse function, 26 definition, 27

derivative of, 166 domain and range, 26 for exponential function, 35 formulas for, 27 graphs of, 27 inverse square law, online inverse trigonometric functions, 44 arccosine, 49, 1045 arcsine, 44, 1045 arctangent, 45, 1045 derivative of, 166 inversely proportional, 6 invertible function, 26 irrotational vector field, 1004, 1059 island species, online isotherms, 652 isotopes, 37 iterated integral, 847, 848 double integral and, 848 double integral, and, 1056 limits of, 851 non-rectangular region, 850– 853 numerical view, 847 triple integral and, 858 triple integral, and, 1056 iteration, online Iwo Jima, Battle of, 625 Jacobian, 1034, 1056 joint cost function, online joint density function, 880, 1056 joule, 440 Keeling Curve, 157 Kepler’s Laws, 713, online Kepler, Johann (1571–1630), online kinetic energy, online definite integral for, online l’Hopital’s rule, 252, 253, 1048 0 ⋅ ∞, 254 and 𝑒, 255 limits involving ∞, 253 l’Hopital, Marquis de (1661–1704), 605, online Lagrange error bound, 540, 1052 Lagrange multiplier, 827 constrained optimization, 827 meaning of, 830–831 Lagrange, Joseph-Louis (1736– 1813), 540, 825 Lagrangian function, 831 Lake Mead, 409 Lambert function, online Lanchester, F.W. (1868–1946) differential equations, 625 square law, 625

1139

Laplace equation, online lapse rate, 111 Law of Cosines, 719, 727 law Ampere’s, online Kepler’s, online leading coefficient, 51 leaf decomposition, 603 least squares, 819, online least upper bound, online left rule, LEFT(𝑛), 377 error in approximation, 380 LEFT(𝑛), 377 left-hand limit, 67 left-hand sum, 1048 Leibniz, Gottfried Wilhelm (16461716), 108, 584 lemniscate, 429 Lennard-Jones model, 232 Leonardo di Pisa, 479 level curve, 669 graph and, 671 set, 669, 671, 689 surface, 689, 692 tangent plane to, 775 light reflection, 222 light, bending of, 537 limaçon, 424 limit, 697, 1046 𝜖, 𝛿 definition, 1046 at infinity, 68, 1046 continuity and, 58 continuous function, 62 definite integral and, 283 definition, 60, 1046 evaluating limits using a new variable, 78 instantaneous acceleration and, 119 instantaneous velocity and, 86 left-hand, 67 local linearity and, 252 meaning of, 60 of (1 + ℎ)1∕ℎ , 141, online of 𝑥𝑛 ∕𝑛, 543 of (sin ℎ)/ℎ, 159 of a constant function, 71 of a constant multiple, 71 of a product, 71 of a quotient, 71 of a sum, 71 of improper integrals, 385 of sequence, 476 of sum, 131 one-sided, 67, 1046

1140

INDEX

properties of, 71, 1046 quotients of functions, 75 right-hand, 67 round-off error and, 141 rules, 70 two-sided, 67 limit comparison test, 496, 1051 limits of the type 0∕0, 76 limits of integration, 284, 302 substitution and, 346 line best fitting, online contour, 669 equation of, 5, 1044 least squares, 819 parametric equation, 261 parametric equation for, 887, 889, 1056 regression, 819 tangent, 93 line integral, 922–928 circulation, 927 computing, 931–936 conversion to one-variable integral, 932 definition, 922 differential notation for, 935 for electric potential, online for work, 925 Fundamental Theorem of Calculus for, 939, 1015, 1057 justification, online independent of parameterization, 936 meaning of, 924 of gradient field, 939 properties, 928 simple case, 931 using parameterization, 932 linear approximation, 178, 514, 755, 1047 linear function, 4, 5, 664, 682–685, 690, 1044 contour diagram of, 684 derivative of, 103 equation for, 683 intercept of, 5, 1044 numerical view, 683 slope of, 5, 1044 table of, 5, 684 two-variable, 682 linearization, local, 753–757 differential, and, 758 from table, 756 three-variable or more, 757 two-variable function, 753, 755 linearization, 178

Lissajous figure, 266, online Liu Hui, online loading curve, 221 local behavior of a function, 50, 52 local extrema, 193, 806 how to find, 806, 811 local linearity, 179, 252, 1048 differentiability and, 179 local linearization, 178, 1047 local maxima/minima, 193, 806, 1048 tests for, 193 logarithm function, 32 derivative of, 165 domain, 32, 1044 graph, 33 intercept, 32 Taylor polynomial approximation, 520 Taylor series, 527 vertical asymptote, 33 logarithms, 32 base 𝑒 (natural), 32 base ten (common), 32, 1044 definition of, 32 properties, 33 rules of manipulation, 33, 1044 solving equations, 33 logistic equation, 607 logistic function, 609 logistic model, 117, 228, 607, 1053, online analytic solution to, 609 carrying capacity, 118, 607 equation, 612 peak oil, 606 qualitative solution to, 607 SARS, online spread of information, online long division, 362, 368 Lotka-Volterra equations, 619 lottery winnings, 455 lower bound, 207, online lung, 114 Machin’s formula for 𝜋, online Maclaurin polynomial 𝑛th , 517 first, 514 second, 516 Maclaurin series, 524 magnetic field, 987, online magnitude of vector, online Magnus force, 735 marginal, 233, 235 costs/revenues, 110, 235, 328 marginal utility, 836 Marquis de l’Hopital (1661–1704), 605, online

mass center of, 432 from density, 430 of earth, 439 relativistic, 538 vs weight, 444 Mass Action, Law of, online mass vs weight, 441 Massachusetts Turnpike, 430 Mauna Loa Observatory, 48, 157 maxima/minima concavity and, 195 global, 203, 1048 continuity and, 203 on (𝑎, 𝑏), 204 on [𝑎, 𝑏], 204 local, 193, 238, 1048 first derivative test, 1048 first-derivative test, 194 second derivative test, 195, 1048 Maxwell distribution, online mean, 467 arithmetic vs. geometric, 222 definite integral for, 468 of normal distribution, 469 Mean Value Inequality, 190 Mean Value Theorem, 186, 1048, online measles, 554 median, 466 metal plate, heated, 740 meteorology, 111 Michelson-Morley experiment, 538 MID(𝑛), 377 midpoint rule, MID(𝑛), 376 error in approximation, 379, 381 mixed partials, 791 Mobius strip, 968 modeling, 212 and differential equations, 562, 597 compartmental analysis, 601 competitive exclusion, 626 epidemic, 616 growth and decay, 586 logistic peak oil, 606 optimization, 214 predator-prey, 620 US population, 611 vibrating spring, 633, 640 modeling with random numbers, online moment, 432 moment of inertia triple integral for, online

INDEX

monkey saddle, 681, 814 monotone sequence, 477 monotonic function, 6, 115, 123, 192 Montgolfier brothers, online morphine, 604 motion 𝑥𝑦-plane, 263 damped simple harmonic, 645 parametric equations, 259, 886 position vectors and, online simple harmonic, 636 straight line, 901 undamped, 645 uniform circular, 900 Mt. Shasta, online murder, 590, 596 natural logarithm, 32 derivative of, 165 graph of, 33 Nelson, Admiral Horatio, online newton, 440, 444 Newton’s law of gravity, online Newton’s method, online chaos, online failure, online initial estimate, online Newton, Isaac (1642–1727) Law of Gravity, 713, 908, online Newton, Isaac (1642-1727) Law of Cooling (Heating), 157, 590 Law of Gravity, 50 laws of motion, 332 First Law, 333 Second Law, 333, 600, 634, 641 Nicotine, 595 Noise levels, online nondecreasing function, 187 nondifferentiability, examples, 124 normal distribution, 224, 312, 468, 469 standard, 469 normal line, online normal vector, 721 to curve, 766 to plane, 721 nullclines, 626, 627 numerical methods accuracy, online bisection, online decimal answer, online differential equations, 575 error, online left and right rules, 379 trapezoid and midpoint rules, 380

Euler’s method, 575 for flow lines, 916 finding derivative, 101 integration, 376, 1050 error, 378, 379 left rule, 377 midpoint rule, 377 right rule, 377 Simpson’s rule, 381 trapezoid rule, 378 iterative, online Newton’s method, online Picard’s method, online objective function, 827 odd function, 25, 1046 definite integral of, 306 oil prices, 606 oil production worldwide, 615 Olympic pole vault, 4 one-sided limit, 67 open interval, 2 optimization, 203, 212 constrained, 242, 825–830 maximizing/minimizing averages, 217 unconstrained, 815–822 optimal values, 203 orientation of curve, 922 of surface, 962 origin, 654 orthogonal surfaces, online oscillations, 633 of spring, 634 damped, 641 output of a function, 2 ozone, 123 ozone depleting gas index, 114 Padé approximants, 537 parabola, 51 parabolic cylinder, 664, 691 paraboloid, 661 parallel lines, 7 parallelepiped, volume of, 733 parallelogram area of, 732 parameter, 6, 224 change of, 260 curve, 1029 rectangle, 1031 parameterization, 1031 length of a curve and, 414 line in three-space, of, 1056 line integral and, 932, 936 of circle, 259

1141

of cone, 1028 of curve, 264, 886 changing, 260 of cylinder, 1023 of graph of function, 265 of helix, 886 of line, 261, 887, 889 of plane, 1057, 1025 of sphere, 1025 of surface, 1023–1031 of surface of revolution, 1028 of torus, online speed and, 262 surface cylindrical coordinates, 1028 tangent lines and, 264 using position vector, 888, 1024 velocity, 262 components, 262 parametric curve, 886 concavity of, 266 slope of, 266 parametric equations, 259, 886 parasite drag, 223 Pareto’s Law, online partial derivative, 740–744 alternative notation, 741 computing algebraically, 748 graphically, 743 contour diagram and, 742 definition, 741 difference quotient and, 741 differentiability and, 799, 802 directional derivatives and, 764 graph and, 742 higher-order, 790 interpretation of, 743 rate of change and, 740–741 second-order, 790 units of, 743 partial fractions, 366, 1050 method of, 1050 partial sum, 488, 1051 of geometric series, 483 particular solution, 329 parts integration by, 353, 1050 pascal, 444 path-dependent vector field circulation and, 950 definition, 941 path-independent vector field, 941– 944 definition of, 1057 definition, 941 gradient field and, 942, 943

1142

INDEX

pdf, 459 peak oil, 606 US, 610 pendulum, 139, 247, 639, 761 differential equation for, 523 period, 41, 636, 1045 angular frequency and, 553 periodic function, 40, 546 permittivity, online perpendicular lines, 7 perpendicular vector, 719 pH, 168 phase difference, 41 phase plane, 617, 620 equilibrium point, 627 nullclines, 627 trajectories, 618, 621, 629 phase shift (angle), 41, 637 vs. horizontal translation, 638 phyllotaxis, 479 Picard’s method, online piecewise linear function, 125 piecewise smooth curve, 923 Pisa, Tower of, 333 Planck, Max (1858-1947) radiation law, 399, online plane, 682, 691 contour diagram of, 684 coordinate, 655 equation for, 683, 721, 731 parameterization of, 1057 parameterization of, 1025 points on, 683 tangent, 754 planimeter, online plutonium, 37 point boundary, 755, 821 interior, 755 sink, 990 source, 990 Poiseuille’s Law, 56, online polar coordinates, 420 Archimedean spiral, 423 area, 425 area element, 865 cardioid, 428 circle, 422 cylindrical, 869–870 graphing equations, 422 integration in, 864–867 lemniscate, 429 limaçon, 424 negative 𝑟, 423 roses, 423 slope, 427 spherical, 872–874 polynomial, 51, 1045

cubic, 51, 1045 degree of, 51 derivative of, 134 double zero, 52 factoring, online Fourier, 547 intercepts, 51 leading coefficient, 54 quadratic, 51, 1045 quartic, 51, 1045 quintic, 51, 1045 zeros, 52 population of Burkina Faso, 13, 33 of China, 38 of Hungary, online of India, 38, 595 of Jamaica, online of Mexico, 144, 311 of US, 36, 144, 157, 611, online of world, 21, 38, 142, 156, 247, 297, 352, 479, , online 604 population growth, 611 equilibrium population, 621 exponential, 13 logistic, 611 predator-prey model, 619 population vector, 715 position vector, 706, online motion and, online notation for vector field, 907 parameterization, 888 parameterization with, 1024 positive flow, 962 potential electric line integral for, online energy, 941 function, 942, 1057 vector, 1011, 1059 potential energy, 538 power function, 49, 1045 compared to exponential functions, 50 concavity of, 134 derivative of, 131, 164 formula for, 49 long-term behavior, 1046 power rule, 104, 132 power series, 504, 1052 convergence of, 504 interval of convergence, 506, 1052 radius of convergence, 506 ratio test, 507, 1052 Taylor, 524 predator-prey model, 620

equilibrium values, 620 prednisone, immunosuppressant, online present value, 486, 1050 annual compounding, 450 continuous compounding, 450 definite integral for, 452 definition, 450 of income stream, 451 pressure, 444, 1050 air, 183 barometric, 604 definite integral of, 444 units, 444 price vector, 715 principal, online Principle of Competitive Exclusion, 626 probability, 464, 878 cumulative distribution function, 464 definite integral for, 464 density function and, 464, 880 double integral for, 880 histogram, 457, 879 probability density function, 459 producer surplus, 453 definite integral for, 454 product rule, 146 three dimensional, online production and price of raw materials, online production function Cobb-Douglas, 241, 675 general formula, 675 profit, 234 maximizing, 816 maximizing, 237, 238 projectile motion of, 185 projection of vector on line, 723 projection, stereographic, online properties of addition and scalar multiplication, 713 continuous functions, 71 cross product, 731 dot product, 719 gradient vector, 766 line integral, 928 proportional directly, 6 inversely, 6 pulmonologist, 114 pulse train, 550, 558

INDEX

Pyramid of Egypt, 405, 443 Pythagoras’ theorem, 656 Quabbin Reservoir, 321 quadratic approximation, 515, 792– 795 quadratic formula, online quadratic function, 690, 809 discriminant of, 810 graph of, 810 quadratic polynomial, 51, 1045 second derivative test and, 809 quarantine, online quartic polynomial, 51, 1045 quintic polynomial, 51, 1045 quotient rule, 147 Racetrack Principle, 188, 189, 1048 radians, 39 conversion to degrees, 40 vs degrees for calculus, 94 radius of convergence, 506 and ratio test, 507 range, 2 Rankine model of tornado, online rate of change, 243, 740–741, 762 absolute, online average, 91, 1047 instantaneous, 91, 1047 related, 244 relative, online rate of substitution economic and technical, online ratio test finding radius of convergence, 507 power series, 507, 1052 series of constants, 498, 1051 rational functions, 53, 1045 asymptotes, 53, 1045 end behavior, 53 intercepts, 54 real part of complex number, online rectangle parameter, 1031 recurrence relations, 476 reduction formulas 𝑝(𝑥)cos 𝑥, 360 𝑝(𝑥)sin 𝑥, 360 𝑝(𝑥)𝑒𝑥 , 360 𝑥𝑛 ln 𝑥, 360 cos𝑛 𝑥, 360 sin𝑛 𝑥, 360 reflection coefficient, online reflection of light, 222 reflection, Law of, 222 region bounded, 821, 1055

closed, 821, 1055 regression line, 819 related rate, 243, 244 relative growth rate, 586 relative rate of change, online relativistic mass, 538 Relativity, Theory of, 157, 538 rent controls, 456 repeating decimal, 486 resistance, wind, online resonance, 537 returns to scale, 680 revenue marginal, 235 definition of, 236 total, 234 revenue function, 234 Richardson arms race model, 631, online Riemann sum, 284, 287 area, 402 in polar coordinates, 425 density, 430 for arc length, 413 for center of mass, 432 for consumer surplus, 454 for force, 444 for mass, 430 for mean, 467 for present/future value, 452 for producer surplus, 454 for volume of revolution, 411 for work, 440 slicing and, 403 three-variable, 857 two-variable, 841 right rule, RIGHT(𝑛), 377 error in approximation, 380 RIGHT(𝑛), 377, 384 right-hand limit, 67 right-hand rule, 729, 731, 1000 right-hand sum, 1048 right-handed axes, 654 rise, 5, 1044 Rolle’s Theorem, 187, online root mean square, online root test, online roots, online by bisection method, online by factoring, online by Newton’s method, online by numerical methods, online by zooming, online roses, 423 round-off error, 141 Rule of Four, 2 Rule of Seventy, online

1143

run, 5, 1044 saddle, 663, 808 monkey, 814 saddle point, 810 Saint Louis arch, 177 sample a signal, 479 SARS, online satellite power, online saturation level, 18, 226 scalar, 702, online scalar curl, online scalar multiplication, online definition, 703 properties, 713 vectors components, 707 scalar product, 718 second derivative, 115 concavity and, 116, 192 inflection point and, 196 interpretation, 117 maxima/minima, 195 test, 195, 1048 second derivative test, 809, 811 Second Fundamental Theorem of Calculus, 336 second-order differential equations, 632 second-order partial derivative, 790 interpretation of, 791 sensitivity of a drug, 210 separable differential equation, 583 separation of variables, 580, 600, 1053 exponential growth equation, 580 justification, 583 sequence, 474, 476 bounded, 477 bounded, monotone convergence of, 477 Calkin-Wilf-Newman, 480 convergence of, 476, 1051 defined recursively, 475 divergence of, 476, 1051 Fibonacci, 479 general term, 474 limit of, 476 monotone, 477 of partial sums, 488 smoothing, 479 series, 480 𝑝-series, 493 general term, 524 alternating, 499 harmonic, 500 alternating series test, 499

1144

INDEX

binomial, 525, 1052 closed form, 481 comparison test, 494, 1051 convergence, 488 absolute vs conditional, 500, 1051 convergence properties, 490, 1051 divergence, 488 Fourier, 546, 549, 1052 period 𝑏, 553 general term, 488 geometric, 480, 482, 1051 as binomial series, 525 harmonic, 490 alternating, 500 infinite, 480 integral test, 492, 1051 limit comparison test, 496, 1051 of constants, 1051 partial sum of, 488, 1051 power, 504, 1052 ratio test, 498, 1051 root test, online sum of, 480 Taylor, 524, 1052 terms in, 480 Si(𝑥), 338 ∑ Sigma, , notation, 283 sigmoid curve, 607 signal, 479 SIMP(𝑛), 381 simple harmonic motion, 636 Simpson’s rule approximating by quadratic, online Simpson’s rule, SIMP(𝑛), 381 sine function, 40 addition formula, online derivative of, 158, 160 graph of, 41, 1045 Taylor polynomial approximation, 517 Taylor series, 524 convergence, 545 sine-integral, Si(𝑥), 338, 339, online sinh 𝑥, 174, 1045 Taylor polynomial, 538 sink, 982, 988, 990 sinusoidal functions, 41, 1044 slicing, 403, 430 slope, 5 of curve, 86, 92 of line, 5, 1044 of parametric curve, 266 parallel lines, 7 perpendicular lines, 7 polar coordinates, 427

units of, 5 velocity and, 87 slope field, 567, 1053, online smooth curve, 923 function, 791 smooth a sequence, 479 Snowy tree cricket, 2 solar panels, 12 solar photovoltaic installations, 595 solar power, 47 solenoidal vector field, 986, 1058 solid angle, online solid of revolution, 411 surface area, online solution curve, 568 source, 982, 988, 990 Soviet-German pipeline, online space-time coordinates, 715 species on islands, online speed, 84, 414, 901, online instantaneous, 262 parameterized curve, 263 velocity and, 711 vs. velocity, 84 speed of sound, 183 sphere equation for, 657 parameterization of, 1025 surface area of, 1040 spherical coordinates, 872–874 conversion to Cartesian, 872 integration in, 873 parameterizing a sphere, 1026 volume element, 873 spinning baseball, 735 spiral, Archimedean, 423 spring, 163, 633, 835 spring constant, 634 square wave, 547 Fourier polynomial, 548 Fourier series, 550 Srinivasa Ramanujan, 1887 − 1920, online stable equilibrium, 592, online standard deviation of normal distribution, 469 state equation, 758 Statue of Liberty, 222 stereographic projection, online Stokes’ Theorem, 1008–1011, 1015 streamline, 914 Streisand, Barbra, 486 strontium-90, 22, online substitution, 1049 into definite integrals, 346 into indefinite integrals, 343

into Taylor series, 530 subtraction of vectors components, 707, 1053 geometric view, 703 sum left-hand, 277 of infinite series, 480 overestimate vs. underestimate, 277 Riemann, 284, 287 right-hand, 277 visualizing left- and right-hand, 277, 278 sum formulas for sin/cos, 1045 summation notation, 283 superposition, 641 supply curve, 12, 453 surface of revolution parametric equations, 1028 boundary of, 1008 catalog of, 691 closed, 962, 991 cylindrical, 691 level, 689 nonorientable, 968 orientation of, 962 parameterization, 1023–1031 saddle-shaped, 663, 808 three-variable function, 692 two-variable function, 692 surface area, 1039, online of sphere, 1040 of surface of revolution, 1041 surge function, 206, 231 surplus, consumer and producer, 453 survival time, 471 symbiosis, 619, 624 symmetry, 25 definite integral and, 306 systems of differential equations, 616, 1053 table, 2 contour diagram and, 673 exponential function, and, 14 linear function and, 684 linear function, and, 5 reading, 653 two-variable function, 653 table of integrals, 360 tangent approximation, 756 plane, 754 tangent field, online tangent function, 43 asymptotes, 44 derivative of, 160, online

INDEX

domain, 44 graph, 44 period, 44 tangent line, 93, 1047 approximation, 178, 514, 1047 error, 178–180, 185, 1048 for parametric curves, 264, 898 slope of, 93 velocity vector and, 898 tangent plane to level surface, 775 tangential surfaces, online tanh 𝑥, 176 tax cut, 487 Taylor expansions, 524 Taylor polynomial for ln 𝑥 near 𝑥 = 1, 520 about 𝑥 = 0, 517 about 𝑥 = 𝑎, 520 accuracy, 521 alternating series error bound, 543 degree 𝑛, 517, 1052 degree one, 514, 793, 794 degree two, 516, 793, 795 error, 539 bound, 540, 1052 for 1∕(1 − 𝑥), 519 for cos 𝑥, 518 for cosh 𝑥, 538 for sinh 𝑥, 538 for sin 𝑥, 517 for 𝑒𝑥 , 518 inequalities, online linear approximation and, 180 Taylor series, 524, 1052 (1 + 𝑥)𝑝 , 525 cos 𝑥, 524, 1052 sin 𝑥, 524, 1052 𝑒𝑥 , 524, 1052 2 𝑒−𝑥 , 530 general term, 524 about 𝑥 = 0, 524 about 𝑥 = 𝑎, 525 arctan 𝑥, 531 comparing, 534 condition for, 524 convergence of, 526, 541 𝑒𝑥 , 545 cos 𝑥, 541, 542 ln(1 + 𝑥), 527 sin 𝑥, 545 differentiation of, 531 convergence, 531 radius of convergence, 531 general, 524 integration of, 531

ln(1 + 𝑥), 527 manipulations of, 530 multiplying, 532 substitution, 530, 532 terminal velocity, 113, 226, 601 terms in series, 480 theorem Antiderivative Construction, 336 Constant Function, 187 Divergence, 992, 1015, 1058 Euler’s, 753 Extreme Value, 203, 821 Fundamental, for Line Integrals, 939, 1015 Fundamental, of Calculus, 293, 336 Green’s, 951, 1019 Increasing Function, 187 Lagrange Error Bound, 540 Mean Value, 186 Pythagoras’, 656 Stokes’, 1008, 1009, 1015 Theory of Relativity, 157, 538 thermal conductivity, online Three Gorges Dam, 448 threshold value, 618, online tin pest, online Titanic, 445, 448 topographical map, 669 tornado model, online Torricelli’s Law, online torus, online total cost, 233 total quantity from density, 430 total revenue, 234 total utility, 121 tractrix, 606 Trafalgar, Battle of, online trajectories phase plane, 618, 621, 629 transmission coefficient, online TRAP(𝑛), 378, 384 trapezoid rule, TRAP(𝑛), 378 error in approximation, 379, 381 trapezoid, area of, 378 triangular wave, 547 trigonometric functions, 39 amplitude, 41, 637, 1045 cosine, 40 derivatives of, 158, online inverse, 44 period, 41, 1045 phase difference, 41 phase shift, 637 sine, 40

1145

superposition of, 637 tangent, 43 trigonometric identity, 40, 1045 addition formula, online double angle, 351, 358, online trigonometric substitutions, 369 method of, 369 truss, online two-sided limit, 67 unbounded function, online unconstrained optimization, 815–822 unit circle, 39, 41, 1044 unit vector, 708 units force, 440, 444 pressure, 444 work, 440 universal gravitational constant, 909 unstable equilibrium, 592 upper bound, 207, online urologist, 321 utility, 680 utility function, 836 Van der Waal’s equation, 173, 538, 761 variable continuous, 2 dependent, 2, 4, 652 discrete, 2 discrete vs. continuous, 4 independent, 2, 4, 652 variable cost, 11 vector, 702, online 𝑛-dimensional, 714 acceleration, 712, 899, online addition, 702, 707, online area, 732, 736, 963 components, 705, 707, online consumption, 715 cross product, 728–733 direction, online displacement, 702–706, online dot product, 718–720 force, 712 geometric definition, 702 gradient, 765 magnitude, 706, 720, 1053, online normal, 721 notation, 705 orthogonal, 719 parallel, 704 perpendicular, 719 population, 715 position, 706, online potential, 1059

1146

INDEX

price, 715 projection on line, 723 scalar multiplication, 703, online subtraction, 703, 707 unit, 708, online velocity, 711, 896–898, online zero, 706 vector area of parallelogram, 973 vector field, 905–910 central, online conservative, 941 continuous, 923 curl of, 1000–1002 curl-free, 1004, 1010, 1059 definition, 907 definition of, 1057 divergence free, 986, 1058 divergence of, 982–983 divergence-free, 993 electric, 989, 994, online flow, 913, 914, 1057 flow line definition of, 1057 flow line of, 914 force, 906 gradient, 909, 939 gravitational, 906, 908, online integral curve, 914 irrotational, 1004, 1059 magnetic, 987, online path-independent, 941–944 solenoidal, 986, 1058 streamline, 914 velocity, 905 writing with position vector, 907

vector potential, 1011 gauge equivalence, 1019 velocity 𝑥𝑦-plane, 263 as derivative, 92 average, 84, online escape, 606 instantaneous, 86, 119, 896 of a falling body, 600 parameterized curve, 263 slope and, 87 speed, 414 speed and, 711, 901 terminal, 113, 601 vector, 262, 711, 896–898, 1057, online components of, 897 geometric definition, 896 limit definition, 897 tangent line and, 898 vector field, 905 visualizing from distance graph, 86 vs. speed, 84 Verhulst, P. F. (1804–1849), 607 Vermeer (1632–1675), 38 vertical asymptote, 33, 53 vibrating guitar string, 750 voltage, 47, online volume definite integral for, 403 double integral for, 842 element, 857 cylindrical, 870 spherical, 873 finding by slicing, 403 from cross-sections, 412

of ellipsoid, 1036 of parallelepiped, 732 of revolution, 411 surface area, online vortex, free, online vorticity, online warfarin, anticoagulant, 595 water clock, online water, density of, 444 wave, 667 wave-guide, online weather map, 652 weight, 441 weight vs mass, 444 wind chill, 658, 678, 746 wind power, 22 wind resistance, online winning probability, online work, 439, 724, 1050 definite integral for, 441 definition, 440, 724, 925 dot product and, 724, 925 line integral for, 925 units, 440 Yosemite National Park Half Dome, 695 Zeno, 84 zero vector, 703 components of, 706 zeros, online zeros of a function, 51 continuity and, 60 double, 52 multiplicity 𝑚, online zooming, online

Differentiation Formulas 1. (𝑓 (𝑥) ± 𝑔(𝑥))′ = 𝑓 ′ (𝑥) ± 𝑔 ′ (𝑥)

′ 2. ( = 𝑘𝑓 ′ (𝑥) (𝑘𝑓 (𝑥))) ′ 𝑓 ′ (𝑥)𝑔(𝑥) − 𝑓 (𝑥)𝑔 ′ (𝑥) 𝑓 (𝑥) = 4. 𝑔(𝑥) (𝑔(𝑥))2 𝑑 𝑛 (𝑥 ) = 𝑛𝑥𝑛−1 6. 𝑑𝑥 1 𝑑 (ln 𝑥) = 9. 𝑑𝑥 𝑥 𝑑 1 12. (tan 𝑥) = 𝑑𝑥 cos2 𝑥 1 𝑑 (arctan 𝑥) = 14. 𝑑𝑥 1 + 𝑥2

3. (𝑓 (𝑥)𝑔(𝑥))′ = 𝑓 ′ (𝑥)𝑔(𝑥) + 𝑓 (𝑥)𝑔 ′ (𝑥) 5. (𝑓 (𝑔(𝑥)))′ = 𝑓 ′ (𝑔(𝑥)) ⋅ 𝑔 ′ (𝑥) 𝑑 𝑥 (𝑒 ) = 𝑒𝑥 8. 𝑑𝑥 𝑑 (sin 𝑥) = cos 𝑥 11. 10. 𝑑𝑥 1 𝑑 (arcsin 𝑥) = √ 13. 𝑑𝑥 1 − 𝑥2 7.

𝑑 𝑥 (𝑎 ) = 𝑎𝑥 ln 𝑎 (𝑎 > 0) 𝑑𝑥 𝑑 (cos 𝑥) = − sin 𝑥 𝑑𝑥

A Short Table of Indefinite Integrals I. Basic Functions 1.



𝑥𝑛 𝑑𝑥 =

1 𝑛+1 𝑥 + 𝐶, 𝑛+1

1 𝑑𝑥 = ln |𝑥| + 𝐶 ∫ 𝑥 1 𝑥 3. 𝑎𝑥 𝑑𝑥 = 𝑎 + 𝐶, ∫ ln 𝑎

𝑛 ≠ −1

2.

4.



5. 6.

𝑎>0

7.

ln 𝑥 𝑑𝑥 = 𝑥 ln 𝑥 − 𝑥 + 𝐶

∫ ∫ ∫

sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝐶 cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝐶 tan 𝑥 𝑑𝑥 = − ln | cos 𝑥| + 𝐶

II. Products of 𝒆𝒙 , cos 𝒙, and sin 𝒙 8. 9. 10. 11. 12.

∫ ∫ ∫ ∫ ∫

1 𝑒𝑎𝑥 [𝑎 sin(𝑏𝑥) − 𝑏 cos(𝑏𝑥)] + 𝐶 + 𝑏2 1 𝑒𝑎𝑥 cos(𝑏𝑥) 𝑑𝑥 = 2 𝑒𝑎𝑥 [𝑎 cos(𝑏𝑥) + 𝑏 sin(𝑏𝑥)] + 𝐶 𝑎 + 𝑏2 1 [𝑎 cos(𝑎𝑥) sin(𝑏𝑥) − 𝑏 sin(𝑎𝑥) cos(𝑏𝑥)] + 𝐶, 𝑎 ≠ 𝑏 sin(𝑎𝑥) sin(𝑏𝑥) 𝑑𝑥 = 𝑏2 − 𝑎2 1 [𝑏 cos(𝑎𝑥) sin(𝑏𝑥) − 𝑎 sin(𝑎𝑥) cos(𝑏𝑥)] + 𝐶, 𝑎 ≠ 𝑏 cos(𝑎𝑥) cos(𝑏𝑥) 𝑑𝑥 = 2 𝑏 − 𝑎2 1 sin(𝑎𝑥) cos(𝑏𝑥) 𝑑𝑥 = [𝑏 sin(𝑎𝑥) sin(𝑏𝑥) + 𝑎 cos(𝑎𝑥) cos(𝑏𝑥)] + 𝐶, 𝑎 ≠ 𝑏 𝑏2 − 𝑎2 𝑒𝑎𝑥 sin(𝑏𝑥) 𝑑𝑥 =

𝑎2

III. Product of Polynomial 𝒑(𝒙) with ln 𝒙, 𝒆𝒙 , cos 𝒙, sin 𝒙 13. 14.

∫ ∫

𝑥𝑛 ln 𝑥 𝑑𝑥 =

1 𝑛+1 1 𝑥 ln 𝑥 − 𝑥𝑛+1 + 𝐶, 𝑛+1 (𝑛 + 1)2

𝑛 ≠ −1

1 1 𝑝(𝑥)𝑒𝑎𝑥 − 𝑝′ (𝑥)𝑒𝑎𝑥 𝑑𝑥 𝑎 𝑎∫ 1 1 1 = 𝑝(𝑥)𝑒𝑎𝑥 − 2 𝑝′ (𝑥)𝑒𝑎𝑥 + 3 𝑝′′ (𝑥)𝑒𝑎𝑥 − ⋯ 𝑎 𝑎 𝑎 (+ − + − …) (signs alternate)

𝑝(𝑥)𝑒𝑎𝑥 𝑑𝑥 =

1 1 𝑝(𝑥) sin 𝑎𝑥 𝑑𝑥 = − 𝑝(𝑥) cos 𝑎𝑥 + 𝑝′ (𝑥) cos 𝑎𝑥 𝑑𝑥 𝑎 𝑎∫ 1 1 1 = − 𝑝(𝑥) cos 𝑎𝑥 + 2 𝑝′ (𝑥) sin 𝑎𝑥 + 3 𝑝′′ (𝑥) cos 𝑎𝑥 − ⋯ 𝑎 𝑎 𝑎 (− + + − − + + …) (signs alternate in pairs after first term) 1 1 16. 𝑝(𝑥) cos 𝑎𝑥 𝑑𝑥 = 𝑝(𝑥) sin 𝑎𝑥 − 𝑝′ (𝑥) sin 𝑎𝑥 𝑑𝑥 ∫ 𝑎 𝑎∫ 1 1 1 = 𝑝(𝑥) sin 𝑎𝑥 + 𝑝′ (𝑥) cos 𝑎𝑥 − 𝑝′′ (𝑥) sin 𝑎𝑥 − ⋯ 𝑎 𝑎2 𝑎3 (+ + − − + + − − …) (signs alternate in pairs) 15.



IV. Integer Powers of sin 𝒙 and cos 𝒙 17. 18. 19. 20. 21. 22. 23.

∫ ∫ ∫ ∫ ∫ ∫

1 𝑛−1 sin𝑛 𝑥 𝑑𝑥 = − sin𝑛−1 𝑥 cos 𝑥 + sin𝑛−2 𝑥 𝑑𝑥, 𝑛 positive 𝑛 𝑛 ∫ 1 𝑛−1 cos𝑛 𝑥 𝑑𝑥 = cos𝑛−1 𝑥 sin 𝑥 + cos𝑛−2 𝑥 𝑑𝑥, 𝑛 positive 𝑛 𝑛 ∫ −1 cos 𝑥 1 𝑚−2 1 𝑑𝑥 = + 𝑑𝑥, 𝑚 ≠ 1, 𝑚 positive 𝑚 − 1 sin𝑚−1 𝑥 𝑚 − 1 ∫ sin𝑚−2 𝑥 sin𝑚 𝑥 1 | (cos 𝑥) − 1 || 1 𝑑𝑥 = ln || +𝐶 sin 𝑥 2 | (cos 𝑥) + 1 || 1 1 sin 𝑥 𝑚−2 1 𝑑𝑥 = + 𝑑𝑥, 𝑚 ≠ 1, 𝑚 positive cos𝑚 𝑥 𝑚 − 1 cos𝑚−1 𝑥 𝑚 − 1 ∫ cos𝑚−2 𝑥 1 | (sin 𝑥) + 1 || 1 𝑑𝑥 = ln || +𝐶 cos 𝑥 2 | (sin 𝑥) − 1 ||

sin𝑚 𝑥 cos𝑛 𝑥 𝑑𝑥: If 𝑚 is odd, let 𝑤 = cos 𝑥. If 𝑛 is odd, let 𝑤 = sin 𝑥. If both 𝑚 and 𝑛 are even and positive, ∫ convert all to sin 𝑥 or all to cos 𝑥 (using sin2 𝑥 + cos2 𝑥 = 1), and use IV-17 or IV-18. If 𝑚 and 𝑛 are even and one of them is negative, convert to whichever function is in the denominator and use IV-19 or IV-21. If both 𝑚 and 𝑛 are even and negative, substitute 𝑤 = tan 𝑥, which converts the integrand to a rational function that can be integrated by the method of partial fractions.

V. Quadratic in the Denominator 𝑥 1 1 𝑑𝑥 = arctan + 𝐶, 𝑎 ≠ 0 ∫ 𝑥2 + 𝑎2 𝑎 𝑎 𝑏𝑥 + 𝑐 𝑥 𝑏 𝑐 25. 𝑑𝑥 = ln |𝑥2 + 𝑎2 | + arctan + 𝐶, 𝑎 ≠ 0 ∫ 𝑥2 + 𝑎2 2 𝑎 𝑎 1 1 26. 𝑑𝑥 = (ln |𝑥 − 𝑎| − ln |𝑥 − 𝑏|) + 𝐶, 𝑎 ≠ 𝑏 ∫ (𝑥 − 𝑎)(𝑥 − 𝑏) 𝑎−𝑏 𝑐𝑥 + 𝑑 1 27. 𝑑𝑥 = [(𝑎𝑐 + 𝑑) ln |𝑥 − 𝑎| − (𝑏𝑐 + 𝑑) ln |𝑥 − 𝑏|] + 𝐶, ∫ (𝑥 − 𝑎)(𝑥 − 𝑏) 𝑎−𝑏 24.

VI. Integrands Involving

√ √ √ 𝒂𝟐 + 𝒙𝟐 , 𝒂𝟐 − 𝒙𝟐 , 𝒙𝟐 − 𝒂𝟐 ,

𝒂>𝟎

1 𝑥 𝑑𝑥 = arcsin + 𝐶 ∫ √ 2 𝑎 2 𝑎 −𝑥 √ 1 | | 29. 𝑑𝑥 = ln |𝑥 + 𝑥2 ± 𝑎2 | + 𝐶 √ | | ∫ 2 2 𝑥 ±𝑎 ( ) √ √ 1 1 𝑥 𝑎2 ± 𝑥2 + 𝑎2 𝑎2 ± 𝑥2 𝑑𝑥 = 𝑑𝑥 + 𝐶 30. ∫ ∫ √𝑎2 ± 𝑥2 2 ( ) √ √ 1 1 2 𝑥 𝑥2 − 𝑎2 − 𝑎 𝑥2 − 𝑎2 𝑑𝑥 = 31. 𝑑𝑥 + 𝐶 √ ∫ ∫ 2 𝑥2 − 𝑎2 28.

𝑎≠𝑏

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