Multivariable Calculus Early Transcendentals [4 ed.] 1319055923, 9781319055929

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MULTIVARIABLE

FOURTH EDITION

CALCULUS ~i:~: __

EARLY TRANSC;ENfDENTALS

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ALGEBRA Lines Slope of the line through P1

Special Factorizations x 2 - y2 = (x + y)(x - y) x 3 + y 3 = (x + y)(x 2 - xy + y2) x 3 - y 3 = (x - y)(x 2 + xy + i)

= (x1, yi) and p2 = (xz, yz): m = Y2-Y1 xz-x,

Slope-intercept equation of line with slope m and y-intercept b: Y=mx+b

Point-slope equation of line through P1 = (x 1, YI) with slope m:

Binomial Theorem (x + y)2 = x2 + 2xy + y2 (x - y)2 = x2 - 2xy + y2 (x + y) 3 = x3 + 3x 2 y + 3xy2 + y3 (x - y) 3 = x3 - 3x2 y + 3xy2 - y3

y-y1=m(x-x1)

Point-point equation ofline through P1 = (x 1, yi) and p2 = (xz, yz) : Y - YI

= m(x -

x1) where m

= Yz -

YI x2-x1 L~nes of slope m1 and m2 are parallel if and only if m 1 = m2.

Lmes of slope m I and m2 are perpendicular if and only if m1 = - ..!...

(x

m2·

n(n - I) 2 2 + y)" = x" + nx" -1 y + -x"- y 2

+ .. •+

Circles Equation of the circle with center (a, b) and radius r:

where

a}2 + (y - b)2 = r2

(x -

· t of p I p2·. Midpom

= J 1:

aX

y

2

Iim ax == oo, lim ax == 0 Iim ax == 0,

Jim Ioga X

lim ax == oo

Um

Hyperbolic Functions ex - e-x

sinhx == - - coshx ==

2 ex +e-x

1

cschx == sinhx

Iog0 x:::::

Y = COthx

----------Y = tanhx

-2 -1

-2

2

-2

----------.

-3

sinh(x

+ y) == sinh x cosh y + cosh x sinh y

cosh(x + y) == cosh x cosh y + sinh x sinh y

Inverse Hyperbolic Functions {:}

sinhy == X

==

cosh- 1 x

{:}

cosh y == x and y

V ==

tanh-l X

{:}

tanhy

y

+ JxT+T) 1 cosh- x == In(x + &=t)

sinh 2x == 2 sinh x cosh x cosh 2x == cosh2 x +sinh2 x

1

sinh- x == In(x

0

oo

y

1

y= sinhx

y == sinh-l X

-oo

y

sechx == coshx coshx cothx==sinhx

2 sinhx tanhx == coshx

:::::

=: X

tanh- I x == -In 1 c+x) 2 1-x

y

X > 1

-l

Forms Involving

55. f unsinudu=-uncosu+n f un- 1 cosudu

u2

utan- udu = /

+

2

I

I

tan- u -

u

2

Forms Involving

+ C

1 64. funsin- 1 udu=--[un+lsin- 1 u - f un+ldu], n-:p-1 n+l 1 -[un+I cos-I u + f u ~ ] , n+ I v I - u2

65. fun cos-I udu = -

n # -1

84.

fJ

a 2 + u2 du =

a2

+ u2 ,

a >

o

~Ja 2 + u2 + a: ln(u + Ja2 + u2) + C

2

85. f u Ja 2 + u2 du u

a4

8

8

= -(a 2 + 2u 2 )v a2 + u2 - -

ln(u + v a 2 + u2) + C

r I

2

Ja +u2 r;,-;; 2 +u 2 -aln /a+Jau2 +u2! +c 86. / --;-du=va

I

I

87.

90. /

9L

= b(2n2+ 3) [un(a + bu)3/2 -

92.

U+i2

~-~=

102. 103.

du =-~ln/~+a/+c uJa2 +u2 a u

I I

du u2Ja2 + u2

Ja2+u2

du

(a2

105.

Forms Involving a + bu

I I I f J

93.

94. 95.

96. 97.

98_

99.

u2 du

udu

a +bu

106.

1

= b-2 (a+bu-aln/a +bu/) +c

du

-

f I J

-b a+ u = -2b 3 [(a +bu>2-4a(a +bu) +2a 2 In/a +bu/]+ C

I__!!_!+ C

du

+ bu

l IO

./a

=-

-~J + ~+ un-1du

b(2n

I)

VQ-1-bii

/va+hu -.jal +c,

+

du=u

du =-_!_+~In/a+bu/+c u2(a+bu) au a2 u

108.

udu a I -:---~:-;,2 == :-;;---:--- + - In /a+ bu/+ C (a+ bu) b2(a + bu) b2

109.

J

ifa:,.o

ifa~o ,

,./a-0;,i - b(2n -2!_ f a(n - l)un-I 2a(n - I)

u2

Forms Involving

I

~1· 2

du

va+bi;

du _ u...;a +bu u~

- u2 , a >

2 Jiau - u du = u

J

a +budu]

2 -l~+bu = v-a ~tan -+c, -a

du =~In u(a + bu) ? a+ bu

a J2au - u2 +

o cos- 1 ( ~ ) +C

u/iau - u2 du 2 2 2u - au - 3a r,:------;;

(a - u) +c

a3

= --:---v2au -u 2 + -cos-I 6 2 a

j u(a +dubu)2 == a(a +I bu) _ a2_!_ 10 /a+bu/+c u 2 u du I ( a2 ) - 3 a +bu - - -2aln/a +bu/ (a+bu) 2 == .b a+bu

J

2un,./a-0;,i b(2n + 1)

-du=2va+bu+a u

107_

1

+c

naf un-1 Va?,--

2 == - 2 (bu - 2a).Ja"+bii + C 3b

bu

u

+ u2)3/2 = a 2 ~ + C

udu -b a+ u

I 1~ =

104 / ·

-~-+c a2u

2a)(a + bu)3/2

15b

101. /

88. / ~ =ln(u+Ja2+u2)+c a2+u2 2 u du u r;,-;; a2 2 89. = -va +u 2 - -ln(u +va2 +u2) +c 0 2 + u2 2 2

I

+ bu du= 2.2 (3bu -

100. /

Ja2+u2 Ja2+u2 r;,-;; --i--du=----+ln(u+va2 +u2)+c u u

+C

110. / ~ = c o s - 1 ('!..=.'!_)+c 2au - u2 a

lll.

J-----;c== = - - - - + du

J2au-u2

uJ2au -u 2

au

C

ESSENTIAL THEOREMS Intermediate Value Theorem If f is continuous on a closed interval [a, b] and f(a) =I= f(b), then for every value M between f (a) and f (b), there exists at least one value c E (a, b) such that f (c) = M.

The Fundamental Theorem of Calculus, Part I Assume that f is continuous on [a, b] and let F be an antiderivative off on [a, b]. Then

lb

Mean Value Theorem If f is continuous on a closed interval [a, b] and differentiable on (a, b), then there exists at least one value c E (a, b) such that

J1 (c) = f (b) - f (a) b-a

Extreme Values on a Closed Interval If f is continuous on a closed interval [a, b], then f attains both a minimum and a maximum value on [a, b]. Furthennore, if c E [a, b] and f (c) is an extreme value (min or max), then c is either a critical poi nt off in (a , b) or one of the endpoints a orb.

J(x)dx

= F(b) -

F(a)

Fundamental Theorem of Calculus, Part II Assume that f is a continuous function on [a, b]. Then the area function A(x)

A1(x)

=1x J(t) dt is an antiderivative off, that is,

=f (x)

or equivalently

-d

lx

f (t) dt =f(:c)

dx a

Furthennore, A (x) satisfies the initial condition A (a)

=0.

,

-

I

FOURTH EDITION

MULTIVARIABLE

CALCULUS EARLY TRANSCENDENTALS . .

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Jon Rogawski University of California, Los Angeles

.,, .i,, ,,,.,-

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Colin Adams

.

Robert Franzosa

Williams College

w.h.freeman . Macmillan Learning New York

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The University of Maine

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TO JULIE -Jon TO ALEXA AND COLTON -Colin TO MY FAMILY -Bob Vice President, STEM: Daryl Fox Program Director: Andy Dunaway Program Manager: Nikki Miller Dworsky Senior Marketing Manager: Nancy Bradshaw Marketing Assistant: Madeleine Inskeep Executive Development Editor: Katrina Mangold Development Editor: Tony Palermino Executive Media Editor: Catriona Kaplan Associate Editor: Andy Newton Editorial Assistant: Justin Jones Director, Content Management Enhancement: Tracey Kuehn Senior Managing Editor: Lisa Kinne Senior Content Project Manager: Kerry O'Shaughnessy Senior Workflow Project-Manager: Paul Rohloff Director of Design, Content Management: Diana Blume Design Services Manager: Natasha Wolfe Cover Design Manager: John Callahan Interior & Cover Design: Lumina Datamatics, Inc. Senior Photo Editor: Sheena Goldstein Rights and Billing Associate: Alexis Gargin Illustration Coordinator: Janice Donnola Illustrations: Network Graphics Director of Digital Production: Keri deManigold Senior Media Project Manager: Alison Lorber Media Project Manager: Hanna Squire Composition: Lumina Datamatics, Inc. Printing and Binding: LSC Communications Cover Photo: Bettmann/Getty Images

Library of Congress Control Number: 2018959767

ISBN-13: 978-1-319-05592-9 ISBN-IO: 1-319-05592-3 Copyright© 2019, 2015, 2012, 2008 by W. H. Freeman and Company All rights reserved Printed in the United States of America 1 2 3 4 5 6

23 22 21 20 19 18

W. H. Freeman and Company One New York Plaza Suite 4500 New York, NY 10004-1562 www.macmillanlearning.com

ABOUT THE AUTHORS Jon Rogawski

A

s. a successful teacher for more than 30 years, Jon Rogawski hstened and learned much from his own students. These

valuable lessons made an impact on his thinking his writing and his shaping of a calculus text. ' '

Jon Rogawski received his undergraduate and master's de-

grees in ma~ematics simultaneously from Yale University, and he earned his :hD in mathematics from Princeton University, where he studied under Robert Langlands. Before joining the Department of Mathematics at UCLA in 1986, where be was a full professor, be held teaching and visiting positions at the Institute for Advanced Study, the University of Bonn, and the University of Paris at Jussieu and Orsay. Jon's areas of interest were numlJer theory, automorpbic. forms, and harmonic analysis on ~ern.isimple groups. He pubhshed numerous research articies in leading mathematics journals, including the research monograph Automorphic Representations of Unitary Groups in Three Variables (Princeton University Press). He was the recipient of a Sloan Fellowship and an editor of the Pacific Journal of Mathematics and the Transactions of the AMS. Sadly, Jon Rogawski passed away in September 2011. Jon's commitment to presenting the beauty of calculus and the important role it plays in students' understanding of the wider world is the legacy that lives on in each new edition of Calculus.

Colin Adams olin Adams is the Thomas T. Read professor of Mathematics at Williams College, where he has taught since 1985. Colin received his undergraduate degree from MIT and his PhD from the University of Wisconsin. His research is in the area of knot theory and low-dimensional topology. He has held various grants to support his research and written numerous research articles.

C

Colin is the author or co-author of The Knot Book, How to Ace Calculus: The Streetwise Guide, How to Ace the Rest of Calculus: The Streeh,•ise Guide, Riot al the Cale Exam and Other Mathemalically Bent Stories, Why Knot?, Introduction to Topology: Pure and Applied, and 7.ombies & Calculus. He cowrote and appears in the videos '1'he Great Pi vs. e Debate" and "Derivative vs. Integral: the FinnJ Smnckdown." He is a recipient of the Haimo National Distinguished Teaching Award from the Mathematical Association of America (MAA) in 1998, an MAA Polya Lecturer for 1998-2000, a Sigma Xi Distinguished Lecturer for 2000-2002, and the recipient of the Robert Foster Cherry Teaching Award in 2003. Colin has two children and one slightly crazy dog, who is great at providing the entertainment.

Robert Franzosa obert (Bob) Franzosa is a professor of mathematics at the University of Maine where he has been on the faculty since 1983. Bob received a BS in mathematics from MIT in 1977 and a PhD in mathematics from the University of Wisconsin in 1984. His research has been in dynamical systems and in applications of topology in geographic information systems. He has been involved in mathematics education outreach in the state of Maine for most of his career. Bob is a co-author of Introduction to Topology: Pure and Applied and Algebraic Models in Our World. He was awarded the University of Maine's Presidential Outstanding Teaching award in 2003. Bob is married, has two children, three step-children, and one grandson.

SECTION 10.7 We have chosen a somewhat traditional location for the section on Taylor polynomials, placing it directly before the th secti_on on T~ylor series in Chapter 10. We feel that this placement is an improvement over the previous edition where e section .--------t was isolated in a chapter that primarily was about applications of the integral. The subject matter in the Taylor polynomials sectio wor~s well as an initial step toward the important topic of Taylor series representations of specific functions. The Taylor polynolllia~s section can serve as a follow-up to linear approximation in Section 4.1. Consequently, Taylor polynomials (except for Taylor's 4 Theorem at the end of the section, which involves integration) can be covered at any point after Section .1.

-----

Chapter 10: Infinite Series 537

11.5 Conic Sections 651 Chapter Review Exercises 665

10.1 Sequences 53 7

13.3 Arc Length and Speed 751 13.4 Curvature 760 13.5 Motion in 3-Space 772

Chapter 12: Vector Geometry 669

10.3 Convergence of Series with Positive Terms 560

13.6 Planetary Motion According to Kepler and Newton 780

12.1 Vectors in the Plane 669

Chapter Review Exercises 787

10.4 Absolute and Conditional Convergence 569

12.2 Three-Dimensional Space: Surfaces, Vectors, and Curves 680

10.5 The Ratio and Root Tests and Strategies for Choosing Tests 575

12.3 Dot Product and the Angle Between Two Vectors 690

l0, 2 Summing an Infinite Series 548

10.6 Power Series 580 10.7 Taylor Polynomials 592 10.8 Taylor Series 603 Chapter Review Exercises 615

Chapter 11: Parametric Equations, Polar Coordinates, and Conic Sections 619 11.1 Parametric Equations 619

12.4 The Cross Product 700 12.5 Planes in 3-Space 712 12.6 A Survey of Quadric Surfaces 719 12.7 Cylindrical and Spherical Coordinates 727 Chapter Review Exercises 734

Chapter 13: Calculus of Vector-Valued Functions 737

11.3 Polar Coordinates 637

13.1 Vector-Valued Functions 737

11.4 Area and Arc Length in Polar Coordinates 646

13.2 Calculus of Vector-Valued Functions 744

11.2 Arc Length and Speed 631

SECTION 14.4 The development of the concept of differentiability in Section 14.4 was rewritten to provide a clearer pathway from the basic idea of the existence of partial derivatives to the more-technical notion of differentiability. We dropped the concept of local linearity 1 - - - - - - - . . . . l introduced in previous editions because it is redundant and adds an extra layer of technical detail that can be avoided.

Chapter 14: Differentiation in Several Variables 789 14.1 Functions of Two 01 More Variables 789 14.2 Limits and Continuity in Several Variables 800 14.3 Partial Derivatives 808 .-o 14.4 Differentiability, Tangent

Planes, and Linear Approximation 818

14.5 The Gradient and Directional Derivatives 827 14.6 Multivariable Calculus Chain Rules 840 14.7 Optimization in Several Variables 850

MULTIVARIABLE CALCULUS: EARLY TRANSCENDENTALS

CONTENTS

Iii

14.8 Lagrange Multipliers: Optimizing with a Constraint 865 Chapter Review Exercises 875

16.4 Parametrized Surfaces and Surface Integrals 995

ANSWERS TO ODD-NUMBERED EXERCISES ANSl

16.5 Surface Integrals of Vector Fields 1009

REFERENCES Rl

Chapter Review Exercises 1020

Chapter 15: Multiple Integration 879 15.1 Integration in Two Variables 879 15.2 Double Integrals over More General Regions 891 15.3 Triple Integrals 905 15.4 Integration in Polar, Cylindrical, and Spherica ! Coordinates 917 15.5 Applications of Multiple Integrals 927 15.6 Change of Variables 939 Chapter Review Exercises 953

Chapter 16: Line andSurface Integrals 957 16.1 Vector Fields 95 7 16.2 Line Integrals 967 16.3 Conservative Vector Fields 983

INDEX 11

Additional content can be accessed online at www.macmillanlearning ,com/calculuset4e:

Chapti!r i 7; F;mrt~rwnt~I Theoremsof V1;,~\'l, ;. "nly~is 1023 17.1 Green's Theorem 1023 17.2 Stokes' Theorem 1037

Additional Proofs:

17.3 Divergence Theorem

L'Hopital's Rule

1049

Chapter Review Exercises 1061

ices Al

'T ;i '·1

A. The Language of Mathematics Al B. Properties of Real Numbers A7 C. Induction and the Binomial Theorem Al2

D. Additional Proofs Al6

Error Bounds for Numerical Integration Comparison Test for Improper Integrals

Additional Content: Second-Order Differential Equations Complex Numbers

, CALCULUS: EARLY TRANSCENDENTALS, FOURTH EDITION

we

On Teaching Mathematics

consider ourselves very lucky to have careers as teachers and researchers of mathematics. Through many years (over 30 each) teaching and learning mathematics we have developed many ideas about how best to present mathematical concepts and to en~age students working with and exploring them. We see teachmg mathematics as a form of storytelling, ~oth when we present in a classroom and when we write matenals f~r explo~ati~n and learning. The goal is to explain to students m a captivatmg manner, at the right pace, and in as clear a way as possible, how mathematics works and what it can do for them. We find mathematics to be intriguing and immensely beautiful. We want students to feel that way, too.

On Writing a Calculus Text It has been an exciting challenge to author the recent editions of Jon Rogawski's calculus book. We both had experience with the early editions of the text and had a lot of respect for Jon's approach to them. Jon's vision of what a calculus book could be fits very closely with our own. Jon believed that as math teachers, how we present material is as important as what we present. Although he insisted on rigor at all times, he also wanted a book that was clearly written, that could be read by a calculus student and would motivate them to engage in the material and learn more. Moreover, Jon strived to create a text in which exposition, graphics, and layout would work together to enhance all facets of a student's calculus experience. Jon paid special attention to certain aspects of the text: 1. Clear, accessible exposition that anticipates and addresses student difficulties. 2. Layout and figures that communicate the flow of ideas. 3. Highlighted features that emphasize concepts and mathematical reasoning including Conceptual Insight, Graphical Insight, Assumptions Matter, Reminder, and Historical Perspective. 4. A rich collection of examples and exercises of graduated difficulty that teach basic skills as well as problem-solving techniques, reinforce conceptual understanding, and motivate calculus through interesting applications. Each section also contains exercises that develop additional insights and challenge students to further develop their skills. vi

Our approach to writing the recent editions has been to la!( the strong foundation that Jon provided and strengthen it in tw: ways: • To fine-tune it, while keeping with the book's original philosophy, by enhancing presentations, clarifying concepts, and emphasizing major points where we felt such adjustments would benefit the reader. • To expand it slightly, both in the mathematics presented and the applications covered. The expansion in mathematics content has largely been guided by input from users and reviewers who had good suggestions for valuable additions (for example, a section on how to decide which technique to employ on an integration problem). The original editions of the text had very strong coverage of applications in physics and engineering; consequently, we have chosen to add examples that provide applications in the life and climate sciences. We hope our experience as mathematicians and teachers enables us to make positive contributions to the continued development of this calculus book. As mathematicians, we want to ensure that the theorems, proofs, arguments, and derivations are correct and are presented with an appropriate level of rigor. As teachers, we want the material to be accessible and written ~t ~e level of a student who is new to the subject matter. Workmg from the strong foundation that Jon set, we have strived to maintain the level of quality of the previous editions while making the changes that we believe will bring the book to a new level.

What's New in the Fourth Edition In_ this ~~ition we have continued the themes introduced in the thrrd ed1t10n and have implemented a number of new changes.

A Focus on Concepts We have continued to emphasize conceptual understanding over the memorization of formulas. Memorization can never be completely avoided, but it should play a minor role in the process of learning calculus. Students will remember how to a?ply a procedure or technique if they see the l . al swn of th t • th og1c progres e s eps m e proof that generates it. And they then understand the underlying concepts rather than seemg . th e topic .

PREFACE

as a black box. To further support conceptual understanding of calculus, we have added a number of new Graphical and Conceptual Insights through the book. These include insights that discuss: • The differences between the expressions "undefined" "does not exist, · " and "indeterminate" in Section 2.5 'on indeterminate forms, • How measuring angles in radians is preferred in calculus ~ver measuring in degrees because the resulting derivauve formulas are simpler (in Section 3.6 on derivative rules of trigonometric functions), • How the Fundamental Theorem of Calculus (Part II) guarantees the existence of an antiderivative for continuous functions (in Section 5.5 on the Fundamental Theorem of Calculus, Part II), • How the volume-of-revolution formulas in Section 6.3 are special cases of the main volume-by-slices approach in Section 6.2, • The relationships between a curve, parametrizations of it, and arc length computed from a parametrization (in Section 11.2 on arc length and speed), • The relationship between linear approximation in multivariable calculus (in Section 14.4) and linear approximation for a function of one variable in Section 4.1.

Simplified Derivations We simplified a number of derivations of important calculus formulas. These include: • The derivative rule for the exponential function in Section 3.2, • The formula for the area of a surface of revolution in Section 8.2, • The vector-based formulas for lines and planes in 3-space in Sections 12.2 and 12.5.

New Examples in the Life and Climate Sciences Expanding on the strong collection of applications in physics and engineering that were already in the book, we added a number of applications from other disciplines, particularly in the life and climate sciences. These include: • The rate of change of-day length in Section 3.7 • The log-wind profile in Section 3.9

vii

• A grid-connected energy system in Section 5.2 • A glacier height differential-equations model in Section 9 .1 • A predator-prey interaction in Section 11.1 • Geostrophic wind flow in Section 14.5 • Gulf Stream heat flow in Section 15.l

An Introduction to Calculus In previous editions of the text, the first mathematics material that the reader encountered was a review of precalculus. We felt that a brief introduction to calculus would be a more meaningful start to this important body of mathematics. We hope that it provides the reader with a motivating glimpse ahead and a perspective on why a review of precalculus is a beneficial way to begin.

Additional Historical Content Historical Perspectives and margin notes have been a wellreceived feature of previous editions. We added to the historical content by including a few new margin notes about past and contemporary mathematicians throughout the book. For example, we added a margin note in Section 3.1 about the contributions of Sir Isaac Newton and Gottfried Wilhelm Leibniz to the development of calculus in the seventeenth century, and a margin note in Section 12.2 about recent Field's medalist Maryam Mirzakhani.

New Examples, Figures, and Exercises Numerous examples and accompanying figures have been added to expand on the variety of applications and to clarify concepts. Figures marked with a &]I icon have been made dynamic and can be accessed via WebAssign Premium. A selection of these figures also includes brief tutorial videos explaining the concepts at work. A variety of exercises have also been added throughout the text, particularly following up on new examples in the sections. The comprehensive section exercise sets are closely coordinated with the text. These exercises vary in difficulty from routine to moderate as well as more challenging. Specialized exercises are identified by icons. For example, (¥Ii indicates problems that require the student to give a written response. There also are icons for problems that require the use of either graphingcalculator technology (GU) or more advanced software such as a computer algebra system ( CAS ).

CALCULUS: £AHii I rRANscENDENTALS, 60 Uff,Jff ED IJI ONi\lfei$ ~P 1~.eil 9,l;pce offormal precision and

. dedicated,conceptual focus, helpin~~ u~~~' ~uil,o'strong co!"putational Skills While , l l Q~}in~ally reinforci,ng the relevan~~I~;1¢&1c~l~s10,1heir future studief and the{r lives 11 a u •'JI b .. li' JI 111 • \i -o

~dJUll~

FOCUS ON CONCEPTS CONCEPTUAL INSIGHTS encourage students to

HISTORICAL PERSPECTIVES are brief vigne::t1Js that

develop a conceptual understanding of calculus by

place key discoveries and conceptual advances

explaining important ideas clearly but informally.

in their historical context. They give students a glimpse into some of the accomplishmen ts of

CONCEPTIIAl INSIGHT In our work with functions and limits so far, we have encoun. tered three expressions that are similar but have different meanings: undefined, does not exist, and Indeterminate. It is important to understand lhe meanlngs of these expressions so that you can use them correctly 10 describe functions and limits.

great mathematicians and an appreciation for their significance.

• The won! "undefined" is used for a matbematlcal expression that is not defined, such as 2/0 or In 0. • The phrase "does not exist" means J~/(x)does notexis~ that is, /(x) does

HISTORICAL PERSPECTIVE

not approach a particular numerical value as x approaches c. • The term "Indeterminate" is used when, upon substitution, a function or limit has one of the lndeterminale forms.

f

I ••

GRAPHICAL INSIGHTS enhance students' visual understanding by making the crucial connections between graphical properties and the underlying concepts.

GRAPHICAL INSIGHT The formula (sin xY = cosx seems reasonable when we compare the graphs in Figure I. The tangent lines 10 the graph of y =sinx have positive slope on the Interval ( - f, f ), and on this inierval, the derivative y' =cosx is posidve. The tangent lines have negative slope on the interval (f, *),where y' = cosx is negative. Toe tangent lines are horizontal at x = -f, f, *, where cos x = 0. y

,,

2

FIGURE 1 Tho gnphl of y =sin.rand iu derivative y' = cou.

EJ

F~-

\~~f O. Him_;· (ab'f-1 =a·'l,' -a-' +al -I =a-'(lr-l)+ (tr'" -1 ). fl'b1s shows thnr. L(a) behaves like II Jog.irithm, in the :i.cnsc tbal ln(ab) = lnu +lnb. In foci, it can be shown Ihm L(a) = lnu.J (b) Verify numcriclllly thUI L O. A,~qimc also

RICH APPLICATIONS such as

EXAMPLE 3 A Glacial Thickness Model Let p =917 k"'mJ g - 9 8 mis2 and t = 75 000 N/m2 · El • • • Th • '" Eq._(2). Use T(O) =0 ror an initial condition, and- solve for T(x). en use T(x) lo determine lhe thickness of lhe glacier I km from its terminus.

this exercise on smart phone

Solution The differential equation that we need lo solve is

' ~ ·~ T

r'!!.. = dx

growth (below) and this

75,000 (917X9.8)

example discussing glacier

lt isda ~edparable differential equation. We use lhc approximate value or 8.35 ror lhe rightha" " •• and proceed as follows:

thickness (left) reinforce

JTdT= J8.35dx

the relevance of calculus

Tenninu."

FlGURE 3 The glacier's thickness T is

modeled as a runction of distance x rrom the terminus.

I , i " =8.35x+C

to students' lives and

T(x) = ../16.7x

demonstrate the importance

+C

Since T(O) .= 0, we obtain T(x) = ./f6:fx (Figure 4). Ai a distance or I km rrom Jhe terminus, the thickness is T(IOOO) - Ji6,7liO a. 129m. '

T

of calculus in scientific

•

160

resecjrch.

120

80 40 400 FIGURE 4 T(x)

800

37, In 2009, 2012, and 2015, the number (in millions) of smart phones sold in the world was 172.4, 680.1, and 1423.9, respectively.

1200

(a) ( CAS) Let r represent time in years since 2009, and let S represent the number of smart phones sold in millions. Determine M, A, and k for a

= ./i6.')x.

= ~ k , that fits the given data points. I+Ae-' (b) What is the long-term expected maximum number of smart phones sold annually? That is, what is ,!!, X 2

x2

1- 0

=

1

•

EXAMPLE 5 Calculate lim n +Inn n---+oo n2

I

Solution Apply Theorem I, using L'Hopital's Rule in the second step: . n+Inn hm -----=-- =

!/

I

n2

n--+oo

1,

. x+Inx hm -----=-= x2

x--+oo

. 1+(1/x) 11m - -2x- -

x--+oo

=

0

I

The limit of the Balmer wavelengths bn in the next example plays a role in physics and chemistry because it determines the ionization energy of the hydrogen atom. Figure 7 plots the sequence and the graph of a function f that defines the sequence. In Figure 8, the wavelengths are shown "crowding in" toward their limiting value_ y

800 600

b3

y

= f(x)

.:::...._--====/~ ==

400 :j:___ _ _

364.5 200

.,

Limit L

.,,

0

00 UJ.J..L

=

lim Cn

n-4-00

nee,

=L

L.

EXAMPLE 8 Show that if fun lanl n-+OO

= O,

th

Jim an= O. en n-->oo

Solution We have -lanl

On :'.:: lanl r -la I= - Iim00 lanl By hypothesis, lim Ian I = 0, and thus also n ~ .n ~ n ..... oo Jude that bro On - · we can apply the Squeeze Theorem to cone n-->oo

0

EXAM PL E 9

Geometric Sequences with r < 0

lim crn n-->oo

+-I REMINDER n! (n-factorial) is the number n!

=

n(n -

l)(n - 2) ... 2. I

20

10

.

.

.. . 5

FIGURE 11

..

0,

-1 < r < 0

r

-I

5n

an= n! Both the numerator and the denominator grow without bound, so it is not clear in advance whether {an} converges. Figure 11 and Table 1 suggest that an increases initially and then tends to zero. In the next example, we verify that an = Rn/ n ! converges to zero for all R. This fact is·used in the discussion of Taylor series in Section 10.8 .

..

_._.......,_.______ n 15

10

Graph of an

1

As another application of the Squeeze Theorem, consider the sequence

=

.

#-

Therefor e.

Solution If -1 < r < 0, then 0 < lrl < 1 and lim lcrnl = O by Example 7 - Thus, n--+OO l)n c 8Jtemates heIi m er n = o b y E xample g . If r = - 1, then the sequence crn = (~;'e':n c and -c and therefore does not approach a limit. The sequence also diverges if r < - I because jcrn I grows arbitrarily large. I

For example, 4! = 4 - 3 - 2 - I = 24. By definition, O! I. y

if ·f 1

= { diverges o_

Prove that for c

= o.

=

5n

Rn EXAMPLE 10 Prove that n-+oo lim -n !

= 0 for all R .

Solution Assume fir_st that R > 0 and let M be the nonnegative integer such that

;;T .

M~R M, we write Rn /n! as a product of n factors: TABLE 1

n

I 2 3 4 10 15 20 50

5n an= n! 5 12.5 20.83 26.04 2.69 0.023 0.000039 2.92xI0- 30

Rnn! -_ (!!:_!!:_ _.. !!_) (M+I R ) (M+2 R ) · ·· (R) (R) -;; SC 7; Call this constant C .

OJ

Each factor is Jess than I.

The first M factors are greater than or equal to 1 and the la s t n - M iactors _,, are 1ess than 1. If _w_e lucmp together the first_M factors and call the product C and replace all the remammg iactors except R/n with 1, we see that '

Rn CR Os-oo ,. n. tends to zero. .,

SECTION 10.1

Sequences

543

Given a sequence {an} and a function f, we can form the new sequence {f(an)}. It is useful to know that if f is continuous and an L, then f(an) f(L). A proof is given in Appendix D.

THEOREM 4 If f is continuous and lim an= L, then lim f(an)

=f (

lim an)

= f(L)

In other words, we may pass a limit of a sequence inside a continuous function. 3n

EXAMPLE 11 Determine the limit of the sequence an=--, and then apply Theon+l rem 4 to determine the limits of the sequences {f(an)} and {g(an)}, where f(x) = ex and g(x) == x 2 .

Solution First, L

=

lim an

=

Now, with f(x) =ex, we have f(an) lim f(an) Finally, with g(x)

3n

.

3

lim - - - hm n+1 1 + n- 1

=3

3n = e = eii+I. According to Theorem 4, 0

•

= f ( lim

an )

=

lim

3n

ii+T

= e3

= x 2 , we have g(an) = a~. According to Theorem 4,

lim g(an)

= gf lim

an)

=(

lim -3n- ) n+1

2

= 32 = 9

Next, we define the concepts of a bounded sequence and a monotonic sequence, concepts of great importance for understanding convergence.

DEFINITION Bounded Sequences A sequence {an} is • Bounded from above if there is a number M such that an ::: M for all n. The number M is called an upper bound. • Bounded from below if there is a number m such that an ::: m for all n. The number m is called a lower bound. The sequence {an} is-called bounded if it is bounded from above and below. A sequence that is not bounded is called an unbounded sequence. Another y upper........_

bound

Upper M

bound

.... .... ,

..... .

L 4--- ,L -- _,.,,v........,....:.....,.._..,,.,,___--'--

'

I I

Lower

bound m

'

2

3

4

5

6

FIGURE 12 A convergent sequence is

bounded.

Thus, for instance, the sequence given by an = 3 - ¼is clearly bounded above by 3. It is also bounded below by 0, since all the terms are positive. Hence, this sequence is bounded. Upper and lower bounds are not unique. If M is an upper bound, then any number greater than M is also an upper bound, and if m is a lower bound, then any number less than mis also a lower bound (Figure 12). As we might expect, a convergent sequence {an} is necessarily bounded because the terms an get closer and closer to the limit. This fact is stated in the next theorem.

7

THEOREM 5 Convergent Sequences Are Bounded If {an} converges, then {an} is bounded.

544

CHAPTER

10

INFINITE SERIES

P ro r Le t L o th er Words, 0

I

~he Fibonacci sequence . since it is unb {Fn} diverges OUnded (F. > but the sequence d fi n - n for al/ n) _ F,.+ 1 e tned by the ratios '

an -

.

--

F,.

c

onverges. The limit is an important number k ratio (see Exercises 3n3own as the golden _ and 34).

= n~oo lim an . Then there exists N

> 0 such th at Ian - LI < I for n > N · I n

L-I M.

l+;

=n(~ -n)

. an = -

84. Use the limit definition to prove that the limit does not change if a finite number of terms are added or removed from a convergent sequence.

.... (a) Show thatifan < 2, thenan+l < 2. Conclude by induction that an< 2

fying the given inequality.

71

does not exist.

88. Let {an} be the sequence defined recursively by

In Exercises 67-70,find the limit of the sequence using L'Hopital's Rule. 2

1 76. Show that an = is decreasing. 2n+l 32 77. Show that an = n is increasing. Find an upper bound. 2 n +2

Jthe3M, ... is increasing and bounded above by limit exists and find its value.

66. an=(l+~r

(lnn) 61. an= -n-

1

87. Proceed as in Example 13 to show that the sequence

3 _4n 64. a = - - n 2+ 7 · 3n

~r

10 ::, an ::, (2 • 10") /n

85. Let bn = an+i. Use the limit definition to prove that if {an} converges, then {bn I also converges and lim a. = lim bn.

60. bn

65. an=

= (n + 10")1/ • ,

converges and lim an = L. Show that the converse is false. In other words, find a function f such that an= f(n) converges but lim f(x)

= ---'-----'5n

63. bn=--2+7, 4n

7~. an

83. Theorem 1 states that if lim f(x) = L, then the sequence a. = f (n)

59. Yn

n 3 _4n

=211" • 3

82. Use the limit definition to prove that if {an} is a convergent sequence of integers with limit L, then there exists a number M such that an = L foralln M.

,Jli+4

58. an=!!:_ 2n

en+ (-3)n

3::, an ::: (2 • 3") 1/n

48. On=

ln(n 2 - 1)

)1/3

=(2n +3")1/ n,

81. Using the limit definition, prove that if {an) converges and {bnl diverges, then (an + bnl diverges.

51. dn

4 53. an= ( 2 + nZ

73. an

g2n 46. an=_

50.

=1nsn - ln n! 52. dn =ln(n 2 + 4) -

n

-

80. Give an example of divergent sequences {an) and {bn) such that {an + bn) converges.

n!

3n2 +n + 2 zn2 _ 3

J;;r+,j-

79. Give an example of a divergent sequence {an I such that n~% Ian I converges.

= el-n bn = nl /n

42. bn

= 1.0ln

n

- - - < c. < - - -

78. Show that.an = ;/n + l - n is decreasing.

40. dn = ../n + 3 - ,Jli

n

1

For every E > 0, the interval (L - €, L + E) contains at least one element of the sequence (a.). (b) For every E > 0, the interval (L - €, L + €) contains all but at most finitely many elements of the sequence (a.).

•

(-D

l

= Jn2 + l + Jn2 + 2 + ... + Jn2 + n•

(a)

:h. Use the limit definition to prove that lim n-2 = 0

39, an= 10 +

Cn

75. Which of the following statements is equivalent to the assertion lim a. = L? Explain.

36, 1,etbn = (ff. ( ) Find a value of M such that lbn I ::: 10-s for n > M a) Use the limit definition to prove that lim b =-O · (b

1

72.

547

1

1

- < a < -• ./2n4 - n - ./2n2

for all n. ~b Sho~ that if an < 2, then an ::: an+ I· Conclude by induction that{ an} 1s mcreasmg. (c) Use (a) and (b) to conclude that L = lim an exists. Then compute L by showing that L = J2 + L .

I

l

548

CH Ap TE R lO

INFINITE SERIES

I

Further Insights and Challenges 89. Show that Jim _ H· . . n. -oo. mt:Venfythatn! > (n/2)"1 2 byobservmg that halfof the f t Of ac ors n ! are greater than or equal to n/2. 90. Letbn

= ---.!!.:.. = ! I)n ~. n k=l

2n dx < cn -+2 X n -

91. Given positive numbers 01 < b1, define two sequences recursively by On+l

1

n

(b) Show that In bn converges lo { l In x dx' and conclude that bn e- 1• lo

=~

.

bn+l = an+ bn 2

limit, denoted AGM(o1,b1), is called the arithmetic-geometric mean of 01 and b1. (e) Estimate AGM(l, ./2) to three decimal places. Geometric Arithmetic mean mean

I

I

b. + I

over the

I

· e Jim Cn· rem to deternun ..... ""

. . th nth harmonic number: e R is e -H -Inn,wher n

93.

Let On -

I

n

[" +i

J1

dx

7·

Show that an 2: Ofor n 2: · as an area . b interpreting On - an+ 1 . (b) Show that (an) is decreasing Y Jim On exi sts. · (c) Prove that n-->oo , Constant. It appears m many .1 kn0 n as Eu 1er s This limit, denoted Y, ~ "".' al is and number t1ieory, and has · Including an ·ru ys n decimal places, but It· IS · still areas of mathema11cs, 1 been calculated to mor~ than. ~~=al ~umber. The first ! 0 digits are not known whether Y is an irr y ss 0.5772156649. (a)

2

-l

x +-n

I Hint: Show that Hn 2:

(d) Prove that both (on] and {bn} converge and have the same limit. This

an+ l

n

(c) Use the Squeeze Theo

n

under y == x

I+!+-··+Hn ==l+z 3 n

(a) Show that On bn for all n (Figure 14). (b) Show that {on} is increasing and (b.} is decreasing. bn -o (c) Show that bn+I - On+l - -•.

an

c3, c4. gJes with the area (a) Calculate c1, cz, . n ofrectan (b) Use a companso that interval [n, 2n] to prove dx

I.Zn

n

(a) Show that In bn

92.

I I +·• •+2n · I +--z LelCn ==;+;+I n +

X

b•

AGM(a1,b1) FIGURE 14

10.2 Summing an Infinite Series

I

Many quantities that arise in mathematics and its applications cannot be computed exactly. We cannot write down an exact decimal expression for the number re or for values of the sine function such as sin 1. However, sometimes these quantities can be represented as infinite sums. For example, using Taylor series (Section I 0.8), we can show that I

I

I

I

I

sin I = I - - + - - - + - - - + ... 3! 5! 7! 9! II! Infinite sums of this type are called infinite series. We think of them as having been obtained by adding up all of the terms in a sequence of numbers. But what precisely does Eq. (I) mean? How do we make sense of a sum of infinitely many terms? The idea is to examine finite sums of terms at the start of the series and see how they behave. We add progressively more terms and determine whether or not the sums approach a limiting value. More specifically, for the infinite series a,

+a2 + a3 +a4 + ···+ an + •••

define the partial sums:

S2

= a1 +a2

S3 == a1 + a2 + a3

[iJ

summing an Infinite Series

SECTION 10.2

549

The idea then is to consider the sequence of values, S1, S2, S3, ... , SN,• • ·• and wheth er the limit of this sequence exists.· . For example, here are the first five partial sums of the infinite series for sm 1: S1 = 1

1 3! 1 S3 = 1 - 3!

Si

=l -

1 6

=1-

0.833

+ -1 = 1 - -1 + - 1

0.841667

- -1

0.841468

5! 1 1 S4 = 1 - - + 6 120 1

6

120

5040

1

1

1

Ss = 1 - - + - - + - - 0.8414709846 6 120 5040 362,880 Compare these values with the value obtained from a calculator: sin 1

0.8414709848079

We see that Ss differs from sin 1 by Jess than 10-9. This suggests that the partial sums converge to sin 1, and in fact, in Section 10.8 we will prove that sin 1 = lim SN

, Infinite series may begin with any value for the index. For example,

(see Example 2). It makes sense then to define the sum of an infinite series as a limit of partial sums. In general, an infinite series is an expression of the form 00

Lan = a1 + a2 + a3 + a4 + ··· n=l where {an} is any sequence. For example, When it is not necessary to specify the starting point, we write simply , Any letter may be used for the index. Thus, we may write am, ak, a;, and soon.

La•.

~·

Sequence

General term

Infinite series

1 1

3' 9' 27' ... 1 1 1 1

1'4'9' 16' ' " The Nth partial sum SN is the finite sum of the terms up to and including aN: N

SN= Lan= a1 n=l

+ a2 + a3 + ···+ aN

If the series begins at k, then SN= ak + ak+l

+ ···+ aN. 00

DEFINITION Convergence of an Infinite Series An infinite series Lan converges n=k to the sum S if the sequence of its partial sums {SN} converges to S: Jim SN= S

In this case, we write S =

Lan. 00

n=k

, If the limit does not exist, we say that the infinite series diverges. , If the limit is infinite, we • Ythat the infinite series diverges to infinity.

I

550

,

CHAPTER

10 INFINITE SERIES

. . . utin several partial sums SN. If the We can mvestigate series numencally by comp g me number S, then we hav se~uence of partial sums shows a trend of convergence next example treats a telee evidence (but not proof) that the series converges to S. aluate • scop·mg series, • where the partial sums are parttc · ul arly easy to ev ·

EXAMPLE 1

Investigate numerically:

Telescoping Seri es

1

1

1

1 +-1-+· · · 4 (5 )

L n(n + 1) = 1(2) + 2(3) + 3(4) 00

n=I

Then compute the sum of the series using the identity: 1 n(n [!lJ

TABLE 1

L

Partial

00

Sums for

n=I

1 n(n

+

N

SN

IO 50 100 200 300

0.90909 0.98039 0.990099 0.995025 0.996678

1)

1

+ 1) = ;; - ;;Ti

. . . T a bl e 1 s uggest convergence to S == 1· S o Iu t 100 The values of the partial sums listed m To prove this, we observe that because of the identity, each partial sum collapses down to just two terms: 1 1(2)

1 1

1 2

S1 = - - = - - -

Sz

=

1(~)

+ 2(~) = (

S3

=

1(~)

+

f- l) + (1- ½) =

2(~) + 3(~)

=

l -

½

(f-l) + (l-l) + (l- ¼) = l -

¼

In general, SN=

G-l) + (l-1)+-··+~-J)+ (t- N~l) 1 N+ 1

=1--In most cases (apart from telescoping series and the geometric series introduced later in this section), there is no simple formula like Eq. (2) for the partial sum SN. Therefore, we shall develop techniques for evaluating infinite series that do not rely on formulas for SN.

I

The sum S is the limit of the sequence of partial sums: S

= N--+oo Jim SN = Jim (1 N--+oo

1 - - -) N +1

=

1

It is important to keep in mind the difference between a sequence (an} and an infinite 00

series

L an. n=I

EXAMPLE 2 oo

Discuss the difference between (a,.} and

Sequences Versus Series I

Lan, where an= n(n + l)° n=I

Make sure you understand the difference between sequences and series. • With a sequence, we consider the limit of the individual terms an. • With a series, we are interested in the

sum of the terms

a 1 +az +a3 + ·· · which is defined as the limit of the sequence of partial sums.

Solution The sequence is the list of numbers J{i), verges to zero: Ji m

n--+oo

2 /3 ), 3 (~) , . . ..

This sequence con-

. I I1m - - - = O an = n--+oo n(n + l)

an,

The infinite seri~s is th~ sum of the numbers defined as the limit of the sequence of partial sums. This sum 1s not zero. In fact, the sum 1s equal to 1 by Example 1: 00

00

La,. = n=1 L n=l

1 n(n + I)

=

I I (2)

1

+ 2(3)

I + 3(4) + · · ·

=

1

•

Summing an Infinite Series

SECTION 10.2

551

The next theorem shows that infinite series may be added or subtracted like 0rdinary sums, provided that the series converge.

I:>n and L bn converge, then

THEOREM 1 linearity of Infinite Series If

~)an + bn), L)an - bn), and L can also converge, the latter for any conStaot c.

Furthennore,

~)an+ bn) =Lan+ Lhn L(an - bn) = Lan - Lhn (c any constant)

Proof These rules follow from the corresponding linearity rules for limits. For example, oo

L)an+bn)=

lim

n=l

N

~(an+bn)=

lim

n=l

=

oo

N

lim

Lan+ n=I

lim

(N

N )

Lan+ Lhn

n=l

n=l

oo

oo

Lbn =Lan+ L)n n=I

n=l

n=

I

A main goal in this chapter is to develop techniques for determining whether a series converges or diverges. It is easy to give examples of series that diverge:

• L l diverges to infinity (the partial sums increase without bound): 00

n=l

S1

= 1,

• L(-

S2

= 1 + 1 = 2,

S3

= 1 + 1 + 1 = 3,

S4

= 1 + 1 + 1 + 1 = 4,

00

on-I

diverges (the partial sums jump between 1 and 0):

n=l

S1

= 1,

S2

= 1 - 1 = 0,

S3

= 1 - 1 + 1 = 1,

S4

= 1 - 1 + 1 - 1 = 0,

Next, we study geometric series, which converge or diverge depending on the common ratio r. A geometric series with common ratio r =I- 0 is a series defined by a geometric sequence crn, where c =I- 0. If the series begins at n = 0, then -

L crn = c + er + cr 00

2

+ cr 3 + cr 4 + cr 5 + ...

n=O

For r

= ½and c = 1, we have the following series:

Figure 1 demonstrates that adding successive tenns in the series corresponds to moving stepwise from 0 to 1, where each step is a move to the right by half of the remaining distance. Thus it appears that the series converges to 1.

552

CHAPTER IO INFINITE SERIES

o----:----

I

o-----r--ro-----r-rr -----; 2

2

-

4

2

4

2

I

I+!

2+4 FIGURE 1 Partial sums of

8

I I I ½+4+g+l6

'°' ...!_2n. 00

L..,

8

o~iI.1¾ 2 4 8 I

1 J 1

1

of a geometric series: -..tial sums rn uting the pc:u u There is a simple formula for co P For the geometric series I::o crn THEOREM 2 Partial Sums of a Geometric series

n=I

withr

=f.

l, c(l _ rN+I ) SN - c + er + er

2

+ cr3 + ... + er

N_ ~ - - ~

1_ r

-

0

. 1'&or SN , multiply each side by r , . h the expression

Proof In the steps below, we start wit . s and then simplify:

take the difference between the first two hne ,

N

= c + er + er 2 + cr3 + ... + er N N+I z + er3 + ... + er + er r SN = er + Cr SN

SN - rSN SN(l - r)

= c - er N+I = c(l - rN+I)

Since r =f. 1, we may divide by (1 - r) to obtain c(l - rN+I) SN = --1---rGeometric series are important because they

Now, the partial sum formula enables us to compute the sum of the geometric series when /r/ < 1.

• arise often in applications. • can be evaluated explicitly. • are used to study other, nongeometric series (by comparison).

THEOREM 3 Sum of a Geometric Series Let c =f. O. If /r/ < 1, then 00

"

er n = c

+er + er 2 + er 3 + ... =

n=O

In words, the sum of a geometric series is the first term divided by 1 mmus the common ratio.

I

0

-C-

1-

r

If /r/ ::: 1, then the geometric series diverges.

!fr

Proof = 1, then the series certainly diverges because the artial _ ow arbitranly large. If r =f. 1, then Eq. (3) yields P sums SN - Ne gr lim SN If

=

,c(l _ rN+I)

lim ----..:. _ 1-r

e

c

-~ - -

r

1- r

lim

rN+ I

N+I /r/ < 1, then Ii m r = 0 and we obtain E (4)

lim r

N+I

·

q.

· If Ir / > l and r 4 I, then -r

does not exist and the geometric sen· ct· es 1verges.

,

r s EcTI o N

10.2

Summing an Infinite Series

553

L5-n_ 00

EXAMPLE 3 Evaluate

n=O

Solution This is a geome~~eries with common ratio r = 5-l and first tenn c = 1. By Eq. (4), ··-n

5 L n=O oo

EXAMPLE 4 faaluate

1 1 1 1 5 =1+-+ -+-+···=-=5 52 53 1 - 5-l 4

~7(-ff ={D' +7(-D' +7(-~)' +·

Solution This is a geometric series with common ratio c

= 7 ( -¾)

3

• Therefore,

r

= -¾

and first tenn

it converges to C

7(-¾)3

27

--= 3 =-1- r 1 - (- 4) 16 EXAMPLE 5 Find a fraction that has repeated decimal expansion 0.212121 ....

Solution We can write this decimal as the series /cio + You can check the result by dividing 7 by 33 on a calculator and seeing that the desired decimal expansion, {) 2121 21 ... , results.

ometric series with c = /cio and r =

i6o· Thus, it converges to lcio

c

+

1- r = 1-

+ · · · . This is a ge-

21 7 l = 99 = 33

100

EXAMPLE 6 A Probability Computation Nina and Brook are participating in an

archery competition where they take turns shooting at a target. The first one to hit the bullseye wins. Nina's success rate hitting the bullseye is 45%, while Brook's is 52%. Nina pointed out this difference, arguing that she should go first. Brook agreed to give the first tum to Nina. Should he have?

Solution We can answer this question by determining the probability that Nina wins the competition. It is done via a geometric series. Nina wins in each of the following cases.

Two events A and B are called independent occurring does not affect the probability of the other occurring. In such a case, the probability that A and B both occur is the product of the probabilities of each occurring individually. This idea applies to each case that leads to a win by Nina. For example, in the second case, the probability that Nina wins is the product of the probabilities of: Nina missing on Turn 1 (0.55), Brook missing on Turn 1 (O.48), and Nina hitting on Turn 2 (O.45). if one of them

• By hitting the bullseye on her first tum (which happens with probability 0.45), or • By having both players miss on their first tum and Nina hit on her second turn [which happens with probability (0.55)(0.48)(0.45)], or • By having both players miss on their first two turns and Nina hit on her third [which happens with probability (0.55)(0.48)(0.55)(0.48)(0.45)], and so on ... There are infinitely many different cases that result in a win for Nina, and because they are distinct from each other (that is, no two of them can occur at the same time) the probability that some one of them occurs is the sum of each of the individual probabilities. That is, the probability that Nina hits the bullseye first is: 0.45 + (0.55)(0.48)(0.45) + (0.55)2(0.48)2(0.45) + ...

This is a geometric series with c = 0.45 and r = (0.55)(0.48) = 0.264. It follows that the probability that Nina wins is 1 64 0.61. Thus, Brook would have been wise not to

~o:i

554 CHAPTER 10

INFINITE SERIES 00

2+~

"--·

EXAMPLE 7 Evaluate L-, n==O

of tWO · s as a surn Solution Write the sene . onverge: because both geometric senes c otJ 1 oo

Loo 2 + 3n = Loo 2 +;:'" o 5n

n

uuS

1

3 n

(-)

::=:

"-+.t'.-,1 5

== 2 L-, 5n

5n

·

is valid by

.-rl.i

·c series. geollletrl

5n

2.

Th

eorern 1 1

+

1- l W 5

""5 5

n==O

n==O

n=O

a series converges, sometimes

Both geometric series converge.

. tt r I{nowing th~t tion. for example, suppose 1 CONCEPTUAL INSIGHT Assumptions ~a ~e algebraic rnaruP~/~ converges. Let us say that

we can determine its sum tbrou~h si~thp r == 1/2 and c == we know that the geometric senes w1rite the sum of the series is S, and we w

1 1 !+ .. · S= 2+4+8 1 !+!+ ... ==l+S 2S = 1 + z+ 4 8

f the series is 1. . the su~ o 1· d to a divergeni: senes: Thus, 2s -- 1 + S' or S = 1. Therefore, thi roach is app ie Observe what happens when s app

s = 1+ 2+ 4 + 8+ 16 + ... 2S

= 2 + 4 + 8+ 16 + ... == S - 1

hi h . bsurd because the series diverges. This would yield 2S = S - I, or S = -: 1• w c is a we cannot employ such algebraic Thus, without the assumption that a senes converges, techniques to determine its sum. The infinite series

f

1 diverges because the Nth partial sum SN

~I

infinity. It is less clear whether the following series converges or 00

I

n

2

3

ili

=N

diverges to

verges:

5

4

L(- it+ n + 1 = 2 - 3+ 4 - 5 + 6 - .. · 1

n=l

We now introduce a usefu_l test that allows us to conclude that this series iliverges. The idea is that if the terms are not shrinking to Oin size, then the series will not converge. This is typically the first test one applies when attempting to determine whether a series diverges. The nth Term Divergence Test (also known as the Divergence Test) is often stated as follows, 00

If'\""' an converges, then lim an

= 0.

THEOREM 4 nth Term Divergence Test If lim

=

00

Lan necessarily converges. We will see n=l

I

that even though Jim n -+00

I

L ; diverges. oo

n= I

ll

. = 0, the senes

=j:.

diverges.

o '

00

then the series "

L.,i

an

n= l

n=l

In practice, we use it to prove that a given series diverges. It is important to note that it does not say that if lim an 0, then

an

Proof First, note that an Sn

oo

= Sn -

Sn- l because

= (ai + a2 + ... +an-I)+ an -_ Sn- 1 + an

If Lan converges with sum S, then n=l

lim an

=

lim (Sn - S i) n-

-

rIm

S

n -

Jim S

L a,, cannot

11-1

=S - S = 0

00

Therefore, if an does not converge to zero,

11=[

:.,r,, , ( , ~e

•

Summing an Infinite Series

SECTION 10.2

EXAMPLE 8 Prove the divergence of

L 4n+1'

Solution We have

n=l

n

oo

n

fun an= fun -

555

4n + 1

=

1

fun - - -

4 + 1/n

= -4

The nth term an does not converge to zero, so the series diverges by the nth Tenn Divergence Test (Theorem 4). • EXAMPLE 9 Determine the convergence or divergence of 00

1

n

2

3

4

I)- 1t-\+1 =2-3+4-s+ ... n=I

Solution The general term an n

n+

1

= (- l)n- t _n-

does not approach a limit. Indeed, n+l tends to 1, so the odd terms a2n+ 1 tend to I, and the even terms a2n tend to -1.

Because lim an does not exist the series diverges by the nth Term Divergence Test. '

•

The nth Term Divergence Test tells only part of the story. If an does not tend to zero, then Lan certainly diverges. But what if an does tend to zero? In this case, the series may converge or it may diverge. In other words, lim an = 0 is a necessary condition of convergence, but it is not sufficient. As we show in the next example, it is possible for a series to diverge even though its terms tend to zero. EXAMPLE 10 Sequence Tends to Zero, Yet the Series Diverges Prove the divergence of 00 1 1 1 1

JI+ ,J2+ J3+ ...

y

Partial sums, SN • •

Solution The general term 1/ Jn tends to zero. However, because each term in the partial sum SN is greater than or equal to 1/ ../N, we have 1

5 4

1

3

2

I

1

1

SN=-+-+ .. ·+JI ,J2 ../N •

..... . ...

Tenns of sequence, an

L~I ,jii diverge even though the terms an = 1/ Jn FIGURE 2 The partial sums of

1

N terms

=N(~)

l 2 3 4 5 6 7 8 9 10 11 12

tend to zero.

1

7N + 7N + " .+ 7N

This shows that SN ../N. But ../N increases without bound (Figure 2). Therefore, also increases without bound. This proves that the series diverges.

SN

10.2 SUMMARY • An infinite series is an expression 00

Z::>n =

a1

+ a2 + a3 + a4 + •••

n=I

We call an the general term of the series. An infinite series can begin at n = k for any integer k.

556

CHAPTER

10

INFINITE SERIES

an d including the Nth te....._ - ·II' fthe terrns up to . . th finite sUID o 1 The Nth partial sum 1s e N

""""' = a1 + az +

a3

+ ... +aN

SN= ~an

=

n=I . . S lim SN- If the limit exists • 1·s the limit N->OO , • By definition, the sum of an 1·nfiim·te senes nverges to the su m S · If the lirni t does . convergentorco ries is h "nfi . we say that t e I mite se . . series diverges. . hout bound, we say that not exist we say that the 1nfimte . s ·ncreases wit the ' . sums 0 fa sene 1 If the sequence of partial

series diverges to infinity.

diverges. H owe ver, a serj .

#OthW~~

• nth Term Divergence Test: If n ~ an

'

n=I

may diverge even if its general term an tends to zero. • Partial sum of a geometric series: c

+ er +

3 cr 2 + er

• Geometric series: Assume c

+ · · · + er

# O. If /rl

N -

-

c(l -

rN+I)

--'~----

1- r

< l, th en

L crn = c + er + crz + cr3 + ... = l oo

C

r

n=O

The geometric series diverges if /rl

::::

l.

triangles on the segments AD, DB, BE, EC,

Archimedes (287-212 BCE), who discovered the law of the lever, said, "Give me a place to stand on, and I can move the Earth" (quoted by Pappus of Alexandria C. 340 CE).

¼) 2 T. Then construct eight tri3 angles of total area ( ¼) T, and so on. 1n this of total area (

I

way, we obtain infinitely m any triangles that completely fill up the parabolic segment. By the formula for the sum of a geometric series, we get S= T

B

Archimedes showed that the area T' where T is the area of MBC. FIGURE 3

S of the parabolic segment is

1

Geometric series were used as early as the third century BCE by Archimedes in a brilliant argument for determining the area S of a "parabolic segment" (shaded region in Figure 3). Given two points A and C on a parabola, there is a point B between ~and C where the tangent line is parallel to AC (apparently, Archimedes was aware of the Mean Value Theorem more than 2000 years,..pefore the invention of calculus). Let T be the area of triangle MBC. Archimedes proved that if D is chosen in a similar fashion relative to AB and E is chosen relative to BC, then

¾T = area(MDB) + area(t::.BEC)

0

This construction of triangles can be continued. The next step would be to construct the four

I

I

+ -T + -16T + -.. = 4

'°' -4nI = -T34 00

T

n=O

For this and many other achievements, Archimedes is ranked together with Newton and Gauss as one of the greatest scientists of all time. The modem study of infinite series began in the seventeenth century with Newton, Leibniz, and their contemporaries. The diveroo

gence of

L 1/ n (called the harmonic series)

n=I

was known to the medieval scholar Nicole d'Oresme (1323-1382), but his proof was loSl for centuries, and the result was rediscovered on more than one occasion. It was also ~own that the sum of the reciprocal squares 2 l / n converges, and in the 1640s, me

L

n=J

Itati

p· an . ietro Mengoli put forward the challenge of finding 1·t . ...~,t s sum. Despne the efforts of the """'

b

Summing an Infinite Series

SECTION 10.2

mathematicians of the day, including Leibnit and the Bernoulli brothers Jakob and Johann, the problem resisted solution for nearly a century. In 1735, the great master Leonhard Euler (at the time, 28 years old) astonished his contemporaries by proving that

I

I

557

I

,r2

62

6

..!_ ..!..+-+· ··=-

I

12 + i2 + )2 + 42 + 52

We examine the convergence of this series in Exercises 85 and 91 in Section 10.3.

----==~~;----------------------10.2 EXERCISES preliminary Questions . ? 1. What role do partial sums play in defining the sum of an infinite senes. What is the sum of the following infinite series?

z.

l

I

I

I

I

4+8+16+32+64+···

2

I:

I lim -=0

n=I

3_Whal hap~ns if_you apply the formula for the sum of a geometric series to the followmg senes? Is the formula valid? I + 3 +3

5. Indicate whether or not the reasoning in the following statement is oo I correct: ./'ii converges because

+ 33 + 34 + ...

n-+oo

Jn

L 2. 00

6. Find an N such that SN > 25 for the series

n=l

4. Indicate whether or no1 the reasoning in the following statement is 00 I 1 ect· = 0 because -112- tends to zero. COrr ·L,n 2

°"' n=l

00

. '"'r"? 7. Does there exist an N such that SN > 25 for the senes L., · Explain. n= 1 8. Give an example of a divergent infinite series whose general term tends to zero.

Exercises l. Find a formula for the general term an (not the partial sum) of the infinite series. I I I I I 5 25 125 (a) 3+ 9+ 27 + 81 + .. · (b) - + - + - + - + ... I

()C

2

4

8

I 22 33 44 -+-------+ 1-2,J 3,2 ,J 4 . 3,2,J ...

2

(d) 12 + I

I

2

I

+ 22 + 1 + 32 + 1 + 42 + I + ...

In Exercises 9 and JO, use a computer algebra system to compute S10, S100, s500, and s1000 Jor the series. Do these values suggest convergence to the given value? 9. [ CAS) n-3 I 1 I I -4- =2.3.4 - - -4,5,6 - - +6-7-8 ---- + .. , g.9.10 10. ( CAS) n4

2. Write in summation notation: l (a) I+ 4+ 9+ 16 (c) I -

(d)

I

I

l

1

+ .. .

625

3125

L

15,625

n=l

9 +16 + 25 + 36 + ... I

I

I

00

I

4.

I l ) n+1-n+2

1 L -- as a telescoping series and find its sum. n(n - I) I 13. Calculate S3, S4, and S5 and then find the sum 1:using the 4

12. Write

n=

3

00

00

!+-+-+-+ .. , 22 32 42

(

00

In Exercises 3-6, compute the partial sums S2, S4, and S6. 3.

I

11. Calculate S3, S4, and S5 and then find the sum of the telescoping series

I

3+ 5 - 7 + ...

125

I

-=!+-+-+-+ ... 90 24 34 44

~)-1/k-l

-

identity

n=l

n2 - I

k=I

I

l

1

s. -+-+-+ ... 1-2 2-3 3.4

6·

m m ms +...

7. The series I+ + 2+ for N == 1, 2, . . . until you find an less than 0.000 I.

3

N

00

1

Z: -:iJ. j=I

converges to\· C~culate SN that approximates 4 with an error

+ .. . is known to converge to e- 1 0! I! 2! 3! (recall that 0! = 1). Calculate SN for N = 1, 2, . .. until you find an SN that approximates e- 1 with an error less than 0.001.

8. The series

1 1:-- as a telescoping series n(n + 3) 00

14. Use partial fractions to rewrite and find its sum. . f 1 15. Fmd the sumo 1-3

n= 1

I 1 + -3.5 + -5.7 + ....

.!_ _ .!_ + .!_ _

00

16. Find a formula for the partial sum SN of

the series diverges.

L(-1 )nn=l

l

and show that

7

(

558

),

CHAPTER

10

INFINITE SERIES

In r_e • C!J N

T~e Integral Test is valid for any series

L f(n), provided that for some M > 0, n=k

f is a positive, decreasing, and continuous function of x for x 2: M . The convergence of the series is determined by the convergence of

f"

f(x)dx

The infinite series oo

I

n=I

n

I>

·

f

!

J I

x

00

1 .

i

R->00 I

•

Therefore, the series "i..J -n diverges. n=l

is called the harmonic series.

1

oo

2

+ ... converge?

3

" n --+-+oz EXAMPLE 2 Does i..J (n 2 + !)2 - z2 52 1 n=' x .. d continuous for x . - . is pos1uve an Solution The function f (x) == (x2 + 1)2

creasing because J' (x) is negative:

1 - 3x

J'(x) == (x2

2

+ 1)3


oo 2u

R->oo

2

4

2(R

1 )-1

2

+ 1)

--

4

00

Thus the integral converges and therefore, " n also converges by the Integral ' ' i..J (n 2 + 1)2 n=I

I

Test.

The sum of the reciprocal powers n-P is called a p-series. As the next theorem shows, the convergence or divergence of these series is determined by the value of p. 00

THEOREM 3 Convergence of p-Series The infinite series

".2.. L, nP

converges if

11=!

p > 1 and diverges otherwise.

Proof If p .'.:: 0, then the general term n-P does not tend to zero so the series diveroes by the nth Term Divergence Test. ~f p > 0, then f(x) = x-P is ~ositive and decreasing for x ::: J, so the Integral Test applies. According to Theorem I in Section 7.?,

{"" _!_ dx =

f1

xP

I p

00

L oo

Therefore,

11=!

I

if p > I if p S I

I nP

converges for p > 1 and diverges for p S l.

•

sEcTI oN

t0.3

Convergence of Series with Positive Terms

5S3

Here are two examples of p-series: P

1

00

=-·3·

1

1

+

1

1

+ ffe + ... = oo

diverges

~_!_=!+_!_+_!_+_!_+···

p =2:

L n2

1

n=I

y

1

ffei = --YT +

;

22

32

converges

42

Another powerful method for determining convergence of positive series occurs via comparison with other series. Suppose that O an bn . Figure 4 suggests that if the larger sum bn converges, then the smaller sum Lan also converges. Similarly, if the smaller sum diverges, then the larger sum also diverges.

b2

L

i.=,1=-1--,-~~--+--'---L-n I 2 3 N

FIGURE 4

The convergence of ~,---

L bn

forces the convergence of L,. lln .

THEOREM 4 Direct Comparison Test Assume that there exists M > Osuch that O an CX)

(i) If

bn for n

M.

CX)

L bn converges, then Lan also converges.

n=I

n=I

L

00

00

(ii) If Lan diverges, then bn also diverges. n=I n=I

= 1. If L bn converges to S, 00

Agood analogy for the Direct Comparison Test, as in Figure 5, is or;e balloon containing the an term5, inside a balloon containing the bn terms. As we add air in amounts corresponding to the subsequent terms in each series, the balloon with the a. terms will always be smaller than the other since an bn . If the bigger balloon does not contain enough air to pop, then the smaller balloon does not pop either. Thus, if the larger series converges, so does the smaller series. On the other hand, if the smaller balloon contains enough air to make it pop, then the bigger balloon must also pop, implying that if the smaller series diverges, the larger series diverges as well. But in the cases that the bigger balloon pops or the smaller balloon does not pop, nothing can be said about the other balloon.

Proof We can assume, without loss of generality, that M 00

then the partial sums of

L an are bounded above by S because

n=I

n=I

00

a1 +a1+ · · · +aN

b1 +b1+ · ·· +bN

Note that the first inequality in (2) holds since an since bn 0 for all n.

}:bn n=I

0

=S

bn for all n, and the second holds

00

00

Under the assumption that L bn converges, it now follows that Lan converges n=I n=I by the Partial Sum Theorem for Positive Series (Theorem 1). This proves (i). On the 00

00

L

other hand, if Lan diverges, then bn must also diverge. Otherwise, we would have n=I n=I a contradiction to (i). 00 1 EXAMPLE 3 Does'°' r.; converge? L vn3n n=I

Solution For n

l, we have 1 1 Jn3n - 3n

-- 0, then Lan converges if and on! if'°' Y L h11 converges. • If L = oo and Lan converges, then L '°""' bn converges. • If L

= 0 and L bn converges, then '°""'

L---------------=L-::a" converges.

Convergence of Series with Positive Terms

SECTION 10.3

565

L

Proof Assume first that L is finite (possibly zero) and that bn converges. Choose a positive number R > L. Then 0 < an/bn S R for all n sufficiently large because an/ bn approaches L. Therefore, an s Rbn, The series Rbn converges because it is a con-

L st~t multiple of the convergent series L bn. Thus, L an converges by the Direct Companson Test. Next, suppose that L is nonzero (positive or infinite) and that Lan converges. Let K = lim bn/an, Then either K

= L- 1 (if L is finite) or K = 0 (if L is infinite). In

either case, K is finite and we can apply the result of the previous paragraph with th e roles of {an} and {bn} reversed to conclude that bn converges. •

L

CONCEPTUAL INSIGHT To remember the different cases of the Limit Comparison Test, you can think of it this way: If L > 0, then an ::::: Lbn for large n. In other words, the

L L

series Lan and bn are roughly multiples of each other, so one converges if and only if the other converges. If L = oo, then an is much larger than bn (for large n), so if Lan converges, bn certainly converges. Finally, if L = 0, then bn is much larger than an and the convergence of

L bn yields the convergence of L an. 2

00

EXAMPLE 6 Show that'\""

converges.

n

L, n4 -n -1

n=2

Solution If we divide the numerator and denominator by n, we can conclude that for large n, n2

n4 - n - I ::::: n2

To apply the Limit Comparison Test, we set 1

n2

an=~--- and bn = 2 n4 - n -1 n

We observe that Jim

!!II.ha

n

exists and is positive:

. an n2 n2 1 L= Jim-= lim 4 •-= Jim - - - - - = 1 bn n - n- 1 1 - n-3 - n-4 00

Since

1

L 2n

2

00

converges, our series

n~

EXAMPLE 7 Determine whether

Ln

4

n~

n

-n-l

00

1

n=3

2 .J,;2+4

I:--

-

also converges by Theorem 5.

•

converges.

Solution Apply the Limit Comparison Test with an = __l _ and bn = ~. Then +4 n . an . n . 1 L = Jim - = hm - - - = hm -;:::.=== = 1 bn +4 J1 + 4/n2 . ool. _oo Smee diverges and L > 0, the senes

Ln=3n

1 L --also diverges .

•

In the Limit Comparison Test, when attempting to find an appropriate bn to compare with an, we typically keep only the largest power of n in the numerator and denominator of an, as we did in each of the previous examples.

{I

566

CHAPTER

10

INFINITE SERIES

:-=====------

11 0 ~ -~ 3 ~S~U~M~M~A~R!_Y_ _ _ _ _ _ _ 1

-

increasing sequence. .. ·es '°"'an form an "f. • Th e partial sums SN of a pos1uve sen L., "tive series converges 1 its Partial Partial Sum Theorem for Positive Series: A posi sums SN are bounded. Otherwise, it diverges. . and continuous for x > M. Set . posiuve, ·· decreasmg, Integral Test: Assume that f 1s d ·t f( )d 00 a converges, an 1 x x di. an = f(n). If M f(x)dx converges, then L., n M

100

'°"'

1

verges, then Lan diverges.

'°"' -

. and diverges if P :'.S 1. converges if P > 1

1

00

p-Series: The series

L._,nP

L

n=I . t M > 0 such that O :'.:: On :::C: bn for all Direct Comparison Test: Assume there exis s d "f di verge< , hen '"" b an ~, L.., n n 2:: M . If L._, bn converges, then L., an converges • an I diverges. b } are positive and that th,- following Limit Comparison Test: Assume that {an) and { n limit exists:

'°'

'°'

L

On

=

Jim -b n

If L > 0, then Lan converges if and only if If L - If L

=

L bn converges.

= and L an converges, then L bn converges. = 0 and L bn converges, then Lan converges.

10.3 EXERCISES Preliminary Questions

L

an, if the partial sums SN are increasing, then (choose

n=l

the correct conclusion) (a) {an} is an increasing sequence. (b) {an} is a positive sequence.

oo

4. Which test would you use to determine whether converges? n=l 2n

5. Ralph hopes to investigate the convergence of .. 00

3 . Which test would you use to determine whether

oo

L n=l

2. What are the hypotheses of the Integral Test?

L

I

L

00

1. For the series

11

32 n- -

with

~I

L.., - . n= l

n

+ vn

-n

n

by comparing

Is Ralph on the right track?

n=I

converges?

Exercises In Exercises J-12, use the Integral Test to detennine whether the infinite series is convergent.

L

n=I

00

I

00

1.

I, choose r such that 1 < r < p. Then there exists a number M such that lan+ifanl > r for all n:::: M. Arguing as before with the inequalities reversed, we find that /aM+nl:::: rnlaMI- Since ,n tends to oo, the terms aM+n do not tend to zero, and consequently, Lan diverges. Finally, Example 4 in this section shows that both convergence and divergence are possible when p = I, so the test is inconclusive in this case.

576

CHAPTER 10

INFINITE SERIES 00 zn '\"' converges. EXAMPLE 1 Prove that L.., n!

. its JiJJJit with an - n! Solution Compute the ratio and

fn+l)'- n + 1

an+I , - ~ n!:::: -zn

- (n+ 1)!2n

l On

2 __

n.I

zn+I

Thus,

-

+ 1)! = (n +l)n!,

_ ~- Note that (n

n=I

We obtain

2

-0

an+I /- lim - - p == lim - n-.c,o n + 1 l On

Loo In.zn converges by the Ratio Test.

Since p < 1, the series

n=I co

2

EXAMPLE 2 Does'\""~ converge? L.., zn n=I n2

Solution Apply the Ratio Test with On == zn :

+

(n 0_ 2 zn On+I _ _

-;;;- I

zn+I

1

== - (n n2 2

2

+2n + 2

1) == !2 (1 +

n

+ J:nL__.. )

n

1

We obtain an+I p== Iim - / ==-1 1·im (1 an 2

I

+ -2 + 21 ) n

n

!

-2

Since p < 1, the series converges by the Ratio Test. co

n'

EXAMPLE 3 Does '\""(-ll-·- converge? L, 10OOn n=0

Solution This series diverges by the Ratio Test because p > 1:

. I

+

an+I / . (n + l)! 10OOn . n 1 p== hm - - == hm - - - - - - = hm --=oo a11 lQQQn+I n! 1000

In the next example, we demonstrate why the Ratio Test is inconclusive in the case

where p == 1. EXAMPLE 4 Ratio Test Inconclusive Show that both convergence and divergence

I> 00

are possible when p == 1 by considering

00

2

and

n=I

L

n -2.

n=I

Solution Fer an == n 2, we have

. I I-

an+I (n + . n2 + 2n + 1 2 1) == lim - - 1·1m -21)2 - = hm • ( an n = 1im 1+- +2 - l n2 n n 2 Furthermore, for bn == n- , p

p

== Jim lbn+I bn

I= 1~1Jim

an+I

1

- ---- - 1

Jim ,~, an

j

I

· T sts The Ratio and Root Tests and Strategies for Choosing e

SECTION 10.5

Thus, P = l in both cases, but, in fact,

577

Loo n diverges bYthe nth Term Divergence Test 2 .

n=I

oo . . . . .th P == 2 > I. This nz = co, and'°' n-2 converges smce tt ts a p-senes wt f-i n=l 'bl h p - I shows that both convergence and divergence are poss, e w en - · . smce lim

.11-1 ther than the ratios Our next test is based on the limit of the nth roots v l"n I ra • ·son WI'th a geo metn'c an+ I/an. Its proof, like that of the Ratio Test, ts based on a compan series (see Exercise 63).

THEOREM 2 Root Test Assume that the following limit exiS ts: L= Jim

v'faJ

(i) If L < l, then Lan converges absolutely. (ii) If L > l, then (iii) If L

Lan diverges.

= l, the test is inconclusive.

EXAMPLE 5 Does

f

n=I

(-n-)n converge? 2n +3

Solution We have L = lim

::;a;=

n

lim _ _

verges by the Root Test.

2n + 3

l

= -. Since L < 2

.

1, the senes con-

Determining Which Test to Apply We end this section with a brief review of all of the tests we have introduced for determining convergence so far and how one decides which test to apply.

Lan be given. Keep in mind that the series for which convergence or diver00

Let

n=I

00

gence is known include the geometric series

L arn, which converge for Ir I < 1, and the n=O

00

1

p-series '°' - , which converge for p > 1. n=O

nP

1. The nth Term Divergence Test Always check this test first. If lim the series diverges. But if lim an

an f= 0, then

= 0, we do not know whether the series converges or

diverges, and hence we move on to the next step.

2. Positive Series If all terms in the series are positive, try one of the following tests: (a) The Direct Comparison Test Consider whether dropping terms in the numerator or denominator gives a series that we know either converges or diverges. If a larger series converges or a smaller series diverges, then the original series does the same. For 00

example, '°' n=I

1

Jn n2 + n

converges because

(since it is a p-series with p

oo

Ln n=2

1

1

Jn n2 + n

00

1 1 < - 2 and '°' - 2 converges n n n=I

= 2 > 1). On the other hand, this does not work for

. h th . . ~1 r.: smce t en e companson senes 2n , while still converging, is

2 -..;n

n=I

r

578

CHAPTER

10

INFINITE SERIES

i

the series with L and a annot corn Pare pplY the Limit Cornp,_Jlll; · so we c often a -•so smaller than the original senes, this case, we can the Direct Comparison TeS t- In ....., in the numerator and d _;nant te,... Test as follows. sider the dou..-· . of those terms. For exani e, (b) The Limit Comparison Test series to the ratio . Pie, nominator and compare the ong faster as n mcreases. So oo ' 'n as it grows ' wt I 2 . doroiIJant over 1or r.:• n is 2

~::a1

,, L

n=2

Jet bn

=

n -...;n l

. Then n2

1

-

n

2

-I

Iirn ~ Jim an = lirn n->OO n 00 n--H,O bn n--> . . Oo . on Test apphes. Smee ' : ' I the Limit Cornpans L..:, ;;i The limit is a positive number, so n""1 · · al series. converges, so does the ongm . th resence of a factorial . ffective m e P . such (c) The Ratio Test The Ratio Test is oft~n e after cancellation. It is also effective . . . · al disappears · :h as n! smce m the ratio, the facton m the ratio, r e power n n such as zn , since 00 when there are constants to the power • 3n . . 1 if the series is then ;;·pp lying the disappears after cancellat10n. For exarnp e, n=l ~-

Z:: ',

Ratio Test yields

. I I= an+!

bm - -

n-+oo

an

3n+l

1·1m

n-+00

3

E:±IIT = n->OO lim -n + I '=•0 < 3n

1

(n)!

Therefore, the series converges. th (d) The Root Test The Root Test is often effective when there is a term of e fonn f(n)g(n)_

For example,£

:2

is a good example since applying the Root Test yields

n=l

Iim /anll/n

n-+oo

zn) 1/n

= n-+oo lim ( n2n

. 2 = hm 2 = 0 < 1 n-+OO n

Thus, the series converges.

(e) The Integral Test When the other tests fail on a positive series, consider the Integral Test. If an = f(n) is a decreasing function, then the series converges if and only if the improper integral not easily apply to

1

and Jz

00

1 x lnx dx

series does as well.

1

00

f(x)dx converges. For example, the other tests do

1 I:-n 1-1nn. However, f(x) = -n Inn ex:,

n=2

is a decreasing function

= ln(lnx) loo = 00 - Thus, the integral diverges, implying that the 2

3. Series That Are Not Positive Series (a) Alternating Series Test If the series is altema•;ng ., 1 h w u o f the1ormL..,(-lt-bn,so that O < bn+I < bn and lim bn

= 0.

Then the Altemating . n=:cI Test shows the Senes

series converges.

(b) Absolute Convergence If the series La is

t 1 . ·h is a positive series, converges using th/t no., a tematmg, then see iC~= Ian I, whICal ests 1or positiv · th · in series is absolutely convergent and ther f e senes. 1f so, e ong e ore convergent.

The Ratio and Root Tests and Strategies tor Choosing Tests

SECTION 10.5

579

10.5 SUMMARY • Ratio Test: Assume that p = lim

L - Then L - Then

On

IOn+I Iexists. On

converges absolutely if p < I.

diverges if p > I. - The test is inconclusive if p = I. On

• Root Test: Assume that L = Jim

L - Then L

- Then

On

,iiaJ exists.

converges absolutely if L < I.

diverges if L > I. - The test is inconclusive if L = I. On

10,5 EXERCISES preliminary Questions

L er". 00

. consider the geometric series 1

3. Is the Ratio Test conclusive for

1equal?

(a) In the Ratio Test, what do the t 0 (see marginal note~ In this case, there exist numbers B e S smaller than but arbitrarily close le L , and thus S contains ( - B, B) for all < B < L. It follows that S contains the open !ntc:,;:va} ( _ L, The set S _cannot contain any number x with Ix I > L, but S may c 0. Then F is differentiable On ( c _ R,c + R) .Furthermore, we can mtegrate an differentiate term by term • For x E (c - R, c + R),

= L nan(X 00

The proof of Theorem 2 is somewhat technical and is omitted. See Exercise 70 for a proof that F is continuous.

I

F'(x)

f

n=I

F(x)dx =A+

t

er-I

~ ( x -cr+1

n=O n

+l

(A any constant)

For both the derivative series and the int . also R. egra1 senes the radius of convergence is Theorem 2 is a. powerful tool for working with . po • I show how to use differentiation or antid •,,,, . . wer senes. The next two exaJIJP es 0f . b . IuerentJatJon of funct10ns to o tam power series for oth fu . power series representations er nctJons.

J

SECTION 10.6

EXAMPLE 6 Differentiating a Power Series

Power Series

585

Prove that for -1 < x < 1,

= 1+ 2x + 3x 2 + 4x 3 + 5x4 + ···

_l_ (1-x)2

Solution First, note that (1 ~x) 2 =

For

:x C~x)

I

:x, we have the following geometric series with radius of convergence R = 1:

I

I

1

1 4 3 - - = 1 + x + x 2 + x + x + ·· · 1-x By Theorem 2, we can differentiate term by term for Ix I < 1 to obtain dx

1 ( --) l -x

i11

4 3 2 = ~(1 +x +x +x +x + .. ·) dx

_l_2 = 1 + 2x + 3x 2 + 4x 3 + 5x 4 + .. ·

I

(1 - x)

EXAMP LE 7 Power Series for Arctangent -1 tan

oo

x=I:

(-ltx2n+I 2n+1

11

Prove that for -1 < x < 1, x3

x5

x1

0

=x-3+5-7+ ...

I

n=O

Solution Recall that tan- 1 xis an antiderivative of (1 + x2)- 1. We obtain a power series expansion of (1 + x2)- 1 by substituting -x 2 for x in the geometric series of Eq. (2):

111 11

I

I

1 2 4 6 --=1-x +x -x + ... 2 1 +x

This expansion is valid for lx 2 I < I-that is, for Ix I < 1. By Theorem 2, we can integrate this series term by term. The resulting expansion is also valid for Ix I < 1: tan- I x= y

y=Sso(x)

2

-1

-dx -= 1 +x 2

J

I

(l-x 2 +x 4 -x 6 + .. ·)dx

I

x3 x5 x1 =A+x--+---+· .. 3 5 7

y=tan- 1x

-2

J

111

Setting x

= 0, we obtain A= tan- 1 0 = 0. Thus, Eq. (4) is valid for -1 < x
n(X - c) 00

F(x)

==

n=O

The constant c is called the center of F(x ).

11

SECTION 10.6

Diverges

--

!

Converges absolutely lx-cl R. - F(x) may converge or diverge at the endpoints c - Rand c + R.

c+R

We set R = O if F(x) converges only for x = c and R = oo if F(x) converges for all x. • The interval of convergence of F consists of the open interval (c - R, c + R) and possibly one or both endpoints c - R and c + R. . • In many cases, the Ratio Test can be used to find the radius of convergence R• It IS necessary to check convergence at the endpoints separately. • If R > 0, then F is differentiable and has antiderivatives on (c - R, c + R). The derivative and antiderivatives can be obtained by directly differentiating and antidifferentiating, respectively, the power series for F:

I \ Possible convergence at the endpoints FIGURE 5 Interval of convergence of a

power series. -

00

F'(x)

= I:nan(X -

f

c)n-l,

n=I

'\""' 00

On

(

F(x)dx =A+ L..,-- x - c n=O n 1

)n+l

+

(A is any constant) These two power series have the same radius of convergence R. 1 • Theexpansion--

1-x

= Loo

x"isvalidforlxl < Lltcanbeusedto d' enveexpans1ons

n=O

of other related functions by substitution, integration, or differentiation.

10,6 EXERCISES Preliminary Questions

L anx" convergf; fo, x "' .5. Must it also converge for

t. Suppose that

x=4?Whataboutx = - 3?

2. Suppose that

L an(x - 6)" converges for x = 10. At which of the

3. What is the radius of convergence of F(3x) if F(x) is a power series with radius of convergence R = 12? 00

4. The power series F(x) =

n=l

points (a)-(d) must it also converge? (a) x = 8 (b) x = 11

(C)

X

=3

(d) x

=0

What is the power series expansion of F' (x) and what is its radius of convergence?

Exercises 1. Use the Ratio Test to determine the radius of convergence R of oo

n

" :... _Does it converge at the endpoints x = ± R? L, 2" oo

00

n

2. Use the Ratio Test to show that"

has radius of convergence

vn2" R = 2. Then detennine whether it converges at the endpoints R = ±2. L, n=l

3. Show that the power series (a)-(c) have the same radius of convergence, Then show that (a) diverges at both endpoints, (b) converges at one endpoint but diverges at the other, and (c) converges at both endpoints, oo n oo n oo n L, 3n

(b)

I: :3"

(c)

.., n=l

n=l

00

6. For which values of x does Ln!x" converge? n=O

L ;n

n=O

(a) " : _

Ln~3"

n=l

7. Use the Ratio Test to show that R = ,./3. n=O 00

(a)~

9n

00

S.

Show that

(b)

(x - 5)"

L n" x" diverges for all x # 0. n=O

(c)

n=l

n29n

has radius of convergence

x3n+I

00 2" 10. I:-x"

00

9. I:nx" n=O

n=l

oo

2n + I

11. "(-J)nX_ L, 2"n n=l

f 0.

(b) Choose

yn\::: n\x - y\(\x In-I + IY1•-l)

= (x _ y)(xn-1 + xn-2 y + •••+ yn-1) .

x+8

\

-R

0

I

X

R1 R

FIGURE 6 If x > 0, choose 8 > 0 less thane/Mand R1 - x .

7

) 592

CHAPTER 10 INFINITE SERIES

10. 7 Taylor Polynomials

. Of can express some fun. ctions as. polynornials Using power series, we have seen hoW we ower series for specific_ functto~s and maru infinite degree. We saw that we can take~ u·on and algebraic operations to obt . ti On mtegra , ¾i ulate them by substitution, differentta ' . . power series for other functions. . wer series for a specific given functio Next, we consider how we can obtatn a ~s special polynomial functions that tun. To do so, first we introduce Taylor polY?0 ]1)lf 'function. The Taylor polynomials I'll senes o a • • fu are out to be partial sums of the power ful tools for approximatmg nctions. 1 ·rmportant m · th err . own ng • ht smce · they are use 'als to Taylor senes . representations n the next section, we extend these Taylor polynornt of 2 functions. . F . tance f (x) sin(x ) cannot be· . diffi I tO ork with or ms 2 , In. 1cu t w · (x) - e -.x • In fact ' even simple fun ct'Ions M any functions are f tegrated using elementary functions. Nor can f-( ) _ In x can only be evaluated like J (x ) = sm · x, f (x ) = cos x, J (x) -- e.x '.andth xmust - be numerically approximated ex. actly at relatively few values of x and otherwise ~y 7 3 + 2x - 4 can be easily d'"" · 0 th th h X h1ern e o er and, polynoIDI'al s sueh as f(x) -- 3x - value of x using just multiplicati entiated and integrated. They can be evaluated at any . . . . • tu al to ask 1f there is a wa) to accurate(on and addition. Thus, given a function, 1t 1s na r . Y approximate the function'thusing• a polynomial function. . . of a func,.on •i olynornial approx1mat1on before In , ) . · k ed w1 a s1mp1e p W e 4.1, havewewor Section used the Jinearization L(x) f(a) + f (a)(x - a to approximate f(x)

=

=

near a point x

= a: J(x)

J(a)

+ J'(a)(x

- a)

We refer to L(x) as a "first-order" approximation to J(x) at x = a because f(x) and L(x) have the same value and the same first derivative at x a (Figure I):

=

L(a)

= J(a),

L'(a)

= J'(a)

A first-order approximation is useful only in a small interval around x = a. In this section, we achieve greater accuracy over larger intervals using higher-order approximations (Figure 2). These higher-order approximations will simply be polynomials with higher powers, the Taylor polynomials. Along with using Taylor polynomials to approximate functions, we will develop tools for estimating the error in the approximation. y

y

Second-order approximation,

y =f(.x)

a

a

FIGURE 1 The linear approximation L(x) is a first-order approximation to f.

FIGURE 2 A second-order approximation is more accurate over a larger interval.

In what follows, assume that f is defined on . • tives f(k) exist on I. Let a E / We sa th an open mterval / and that all deova· · Y at two function / d at x a if their derivatives up to order n at x _ s an g agree to order n - a are equal:

=

J(a)

= g(a),

J'(a)

= g'(a),

f"(a)

= g"(a),

We also say that g approximates f to order n at x

= a.

· · ·,

f(n)(a)

= g,/ .. ' '. ,

r~,~ '

g

]

.f-t .

j :-~~::;1'\

\.

= c, then the Taylor

+ J'(c)(x - c) + --(x - c)2 + •.. = '°' - - ( x 2! L.., n! J"(c)

oo J 0, then that power series is the Taylor series T(x)

_I

English mathematician Brook Taylor (1685-173!) made important contributions to calculus and physics, as well as to the theory of linear perspective used in drawing.

oo J 0: 00

f(x)

= I>n(x -

ct= ao + a1(x - c) +a2(x

-c)2 + ...

n=O

According to Theorem 2 in Section 10.6, we can compute the derivatives off by differentiating the series term by term:

=

+ a1(x J'(x) = a1 + 2a2(x J"(x) = 2a2 + 2 · 3a3(x J(x)

I .

I

10.8 Taylor Series

.-

1!

I

fork=j fork =F j

be a polynomial of degree n or less. (a) Show that if pU>(a) = 0 for j = 0, I, . . . ,n, then P(x) = 0, that is, ai = 0 for all j. Hint: Use induction, noting that if the statement is true for degree n - I, then P'(x) = 0. (b) Prove that Tn is the only polynomial of degree n or less that agrees with f at x = a to order n. Hint: If Q is another such polynomial, apply (a) to P(x) = Tn(x) - Q(x).

. . Substitute x for x in the Error Bound for I • J{i?t.a]so the fourth Maclaurin polynomial for f(sm) x -. P(x)I, noting that sin(x 2 ) dx

+ l)(x -a)H

77. Let a be any number and let

2

1/2

I

Use this to prove that Tn agrees with fat x = a to order n.

75, ~stituting x2 in P(x) == x - x3 /6, the i::~d T6 for !(x) == sin(x2) bY sfu(x) == sinx. Maclaunn polynomial for

pis

k(k-1) .. •(k- j

k!

di . ((x-a)k)I dxJ k! x=

o

that Isin(x ) - T6(x)I < (b) Show -

603

76. Prove by induction that for all k,

4

calculate [ I Q(x) dx as an approximati

find a

x

Taylor Series

ao

+ a2(x c) + 3a3(x c) + 3 · 4a4(x c)

+ a3(x -d + ... c)2 + 4a4(x - c) 3 + •.. c) 2 + 4 • 5as(x - c)3 + ...

c)2

\11

1 1

I

604 CHAPTER 10

INFINITE SERIES

In general,

j a, and in this case, the n t~ ial series breaks off at degree n. The . . & · ·t b1nom . omial series 1s an 1n,m1 e series when a bin ta positive whole number. 15 no

In general t O such that [J 0 ror a < ' d _ R is a sector as m 1gure l(A). 0-aan 0 _,_, • by the curve r = /(0) and the two rays - . . t al· we will see how m the first rd We can compute this area via a polar-coo mates m egr ' part of this section. . . the re ion into N narrow sectors of angle To derive a formula for the area, di~i~e g_ terval [a, /3] as in Figure l(B): t:;.0 = (/3 _ a)/ N corresponding to a partttwn of them 0o

=a

< 01 < 02 < . .. < 8N

= f3

y

y

0 0 =a

mi FIGURE 1 Area bounded by the curve and the two rays 0 = a and 0 = fJ. r = /(0)

y

(A) Region a

$ 0$

(B) Region divided into narrow sectors

/3

Recall that a circular sector of angle t:;.0 and radius r has area ½r 2 t:,.0 (Figure 2). If M is small, the jth narrow sector (Figure 3) is nearly a circular sector ofradius r; = J(0j ), so its area is approximately ½rJ 1::!,.0 • The total area is approximated by the sum

areaofregion

FI GURE 2 The area of a circular sector is

½r 2 t:;.0.

L 2I rJt:;.0 = 2I Lf(0; )26'.0 N

N

J=I

J=I

[I]

. I [fi This is a Riemann sum for tbe mtegral 2 }a f(0)2 d0. If f is continuou s, then the suJll approaches the integral as N -

oo, and we obtain the foll owing form ula.

Area and Arc Length in Polar Coordinates

SECTION 11.4

.

y

.

647

function then the area

THEOREM 1 Area in Polar Coordinates If f 1s a contmuous ' . a) d0-P(w1tha O· thill th e sum of the distances to two fixed points Fi and Fz is a · We assume alwa the distance F '{: that K is greater than because thee,; . 2 between the foci, segment FF- ~pfsKe reduces to the line 121 =FF 1 2, an d",t has no Points at al/ 1•, K < F1F2 •

GJ

p Fi + p Fz = K . . al f "focus") of the ellipse. Note th e pomts Fi and F2 are called the foci (plur O d obtain a circle ofrad· ~t if th e foci coincide, then Eq. (I) reduces to 2P Fi =Kan we ius }/( centered at Fi . We use the following terminology:

Th

• The midpoint of Fi Fz is the center of the ellipse. • The ~ne through the foci is the foca! axis. focal axis is the conjugate . The line through the center perpendicular to the ~y Conjugate axis B-(0,b)

-_ ___.(---====-~ -~ - ~~-t!mti"~~-:--x

Semimino, axis

Focal axis

A'_ (-a, 0) (-c, 0)

Center

B' - (0, -b)

Semimajor axis (B) Ellipse in standard position:

(A) The ellipse consists of all points P such that PF1 + PF2 = K

(¾Y (iY = +

1

FIGURE 2

The ellipse is said to be in standard position (and is called a standard ellipse) if the focal and conjugate axes are the x- and y-axes, as shown in Figure 2(B). In this case, the foci have coordinates F1 = (c, 0) and Fz = ( - c, 0) for some c > 0 . Let us prove that the equation of this ellipse has the particularly simple form

Gf+Gf =

QJ

1

where a= K/2 and b = - c2 • By the distance formula, P = (x, y) lies on the ellipse in Figure 2(B) if P F1

+ P Fz

= /(x - c) 2 + y 2 + /(x + c) 2 + y 2

= 2a

0

In the second equation, move the first term on the left over to the right and square both sides: (x + c) Strictly speaking, it is necessary to show that if p (x, y) satisfies Eq. (4), then it also satisfies Eq. (3). When we begin with Eq. (4 ) and reverse the algebraic steps, the process of taking square roots leads to the

=

relation

v'(x _ c)Z

+ y2 ±

+ c)2 + Y 2 = ± 2a

However, because a > c this equation_ ~as . unless both signs are pos,ttve. noso I u t 10ns

2

+ y2 =

4a/(x - c) 2 + y 2

2 4a - 4a/(x - c)2 + y2 + (x _ c)z + yz

= 4a 2 +

(x - c) 2 - (x + c)2

Now divide by 4, square, and simplify: 2 2 2 2 a (x - 2cx + c + y )

= a4 -

= 40 2 -

4cx

2a 2 cx + c 2x2

(az - cz)xz + azyz = a4 - a2c2 = a2(a2 - c2) xz y2 2+-2--2=1 a a -c 2 2 2 This is Eq. (2) with b = a - c as claimed.

0

,r SECTION 11.5

Conic sections

653

The ellipse intersects th . ,. . ts A A' B B' called vertices. Vertices A and A' e axes m 1our pom , , , the along the focal axis are called the focal vertices. Following common us~~• numbers a and bare referred to as the semimajor axis and the semiminor axIS (even though they are numbers rather than axes). 2

THEOREM 1 Ellipse in Standard Position Let a > b > 0, and set c = Ja . - 'i,!. The ellipse PFI + P F2 =2a with . foci. F1 = (c, 0) and r2 ,, -_ (- c, 0) has equanon

0 Furthennore, the ellipse has • semimajor axis a, semiminor axis b. • focal vertices (± a, 0), minor vertices (0, ± b).

If b > a > 0, then Eq. (5) defines an ellipse with foci (0, ±c), where c = ~ EXAMPLE 1 Find the equation of the ellipse with foci (±./TT, 0) and semimajor axis a =6. Then find the semiminor axis and sketch the graph.

Solution The foci are (±c, 0) with c = ./TT, and the semimajor axis is a = 6, so we can use the relation c = a2 - b2 to find b: b2 = a2 - c2 = 62 -

(vlli = 25

::::}

b= 5

. (x)2 + (y)2 = 1.

Thus, the semiminor axis is b =5 and the ellipse has equatton

5

6

To sketch this ellipse, plot the vertices (±6, 0) and (0, ±5) and connect them as in Figure 3. y

y

(0,5)

------------x

--e---+---t---+X

(6, 0)

(6,0)

(-6,0)

(0,-5)

(0,-5)

FIGURE 3

To write the equation of an ellipse with axes parallel to the x- and y-axes and center translated to the point C = (h, k), replace x by x - h and y by y - k in the equation (Figure 4):

2

(y _

x - h) + - b k)2 =1 (- a

EXAMPLE 2 Translating an Ellipse Find an equation of the ellipse with center at c = (6, 7), vertical focal axis, semimajor axis 5, and semiminor axis 3. Where are the foci located?

654

CHAPTER 11

y

PARA

METRIC EQUATIONS, POLAR COORDINATES, AND coNI

(x;6)2 +(Y;7)2 =1 (6, 12)

. 7) .

5

X

3

G)2+(ft= I (0, -4)

FIGURE 4 An ellipse with a vertical major axis and center at the origin, and a translation of it to an ellipse with center

C

= (6, 7).

+ (~)

===

. are located ±4 vertical

- 4 so the foci '11 ) and F2 === (6, 3)-

Furthermore, c = = "':~ - .:,.:_-6 above and below the center-that 1s, F1 - ( '

(6, 2)

-3

e~~r

- a2

(6, 3)

/

(0, 4)

d b === 5, so that a -s:: b (B· :,:;:; 3 an z 2 lo . al we have a (x) + (l) - J •· 5 Solution Since the focal axis is vertic ' h e equation 3 • \Vhe 1 . ·n would av ure 4). The ellipse centered at the ongt uation becornes . (6 7) the eq the center 1s translated to (h, k) = ' ' 2 1

3

C= (6,

C SECTIONS

. p such that the A hyperbola is the set of all pomts P to two foci F 1 and F2 is ±K:

Unj

ts

-

difference of the distances fro

tn

PFi -PF2 =± K . F b tween the foci (the hyperbola has no We assume that K is less than the distance Fi 2 . e 0 f two branches corresponding to t points if K > Fi F2 ). Note that a hyperbola conslS S the choices of sign ± (Figure 5). _ _ . f the hyperbola, the line through O As before, the midpoint of F1 F2 is th e_ cen:r h the center perpendicular to the F1 and F2 is called the focal axis, and the hne oug the points where the focal axi focal axis is called the conjugate axis. The vertic~ i:eFigure 5. The hyperbola is sai~ inters~cts the hyperbola;_ they are. labeled A and A d h erbola) when the focal and stand to be m standard position (and 1s called a . ar in ure 6. The next theorem can conjugate axes are the x- and y -axes, respectively, as g be proved in much the same way as Theorem I.

/t

Conjugate axis

Focal axis

THEOREM 2 Hyperbola in Standard Position Let a_ > 0 and b > 0, an~ set c ::::: + b 2 • The hyperbola P F 1 - P F2 = ± 2a with foci F1 = (c, 0) and F2 - (-c,O) has equation

(~f-Gf = FIGURE 5

1

A hyperbola with center (0, 0). Conjugate axis

A hyperbola has two asymptotes that we claim are the lines y = ± x. Geometrically, these are the lines through opposite comers of the rectangle whose sides pass through (± a, 0) and (0, ± b) as in Figure 6. To prove the claim, consider a point (x, y) on the hyperbola in the first quadrant. By Eq. (7), y

=

=

b -Jxz -a2 a

The following limit shows that a point (x, y) on the hyperbola approaches the line y asx--+ oo: li]J FIGURE 6 Hyperbola in standard

position.

. ( -xb) = -b. (Jxz _

hm

y -

a

a

=

= -ab

lim

(

lim (

Jxz

a2 _

-a2

-a2

x)

-x)

x 2 -a 2 +x

)

- a

2

+ x)

-a 2 +x

=0

The asymptotic behavior in the remaining quadra t • . . n s 1s s1mllar.

= ~x

s EC TIO N

11.5

Conic Sections

655

nd EXAMPLE 3 Find the foci of the hyperbola 9x 2 - 4y2 = 36. Sketch its graph a asymptotes.

y

Solution First, divide by 36 to write the equation in standard form:

---"1~+-~--~x

F2 = (-./13, O)

I

I

I

I

1 / I I

I

\ \

\

\

y2

4

9

= (vD, 0)

I F1 I

>:--3 --~ \

x2

+ b2 =

Thus, a= 2, b = 3, and c = \

Gf-Gf =

or

---=l

+9 F2

Fi =(Ji3,0),

\

2

FIGURE 7 The hyperbola 9x - 4y2

= 36.

1

foci are

= (-Jl3, 0)

T~ sketch the gr_aph, we draw the rectangle through the points (±2, Oj and (0, ±3) as in Figure 7. The diagonals of the rectangle are the asymptotes y = ± 1 x. The hyperbola passes through the vertices (±2, 0) and approaches the asymptotes. • Unlike the ellipse and hyperbola, which are defined in terms of two foci, a parabola is the set of points P equidistant from a focus F and a line V called the directrix:

y

0

PF= PV

2

Q

I I

Directrix 'D y =-c

mJ

FIGURE 8 Parabola with focus (0, c) and

directrix y = -c.

Here, when we speak of the distance from a point P to a line V, we mean the distance from P to the point Q on V closest to p, obtained by dropping a perpendicular from P to V (Figure 8). We denote this distance by PV. The line through the focus F perpendicular to V is called the axis of the parabola. The vertex is the point where the parabola intersects its axis. We say that the parabola is in standard position (and is a standard parabola) if, for some c, the focus is F = (0, c) and the directrix is y = -c, as shown in Figure 8. We prove in Exercise 75 that the vertex 2 is then located at the origin and the equation of the parabola is y = x / 4c. If c < 0, then the parabola opens downward.

THEOREM 3 Parabola in Standard Position Let F = (0, c) and directrix y = -c has equation y

c -=I=

0. The parabola with focus

[I]

l = -x2

4c

The vertex is located at the origin. The parabola opens upward if c > 0 and downward if C < 0.

EXAMPLE 4 The standard parabola with directrix y = -2 is translated so that its vertex is located at (2, 8). Find its equation, directrix, and focus.

y

Focus

.

(2, 10)

(2, 8)

Directrix y = 6 \

I

y=l.. 2\ 8X

Focus ',

'

(0, 2)

/

/

/

- - - -'---='~.....,.:":+-"- - - - x 2 Directrix y = -2

FIGURE 9 A parabola and its translate.

L

Solution By Eq. (9) with c = 2, the standard parabola with directrix y = -2 has equation y = }x 2 (Figure 9). The focus of this standard parabola is (0, c) = (0, 2), which is 2 units above the vertex (0, 0). To obtain the equation when the parabola is translated with vertex at (2, 8), we replace x by x - 2 and y by y - 8: 1 2 y - 8 = -(x - 2)

8

or

1 1 17 y=-x 2 --x+8

2

2

The vertex has moved up 8 units, so the directrix also moves up 8 units to become y The new focus is 2 units above the new vertex (2, 8), so the new focus is (2, 10).

= 6.

656

CHAPTER 11

PARAMETRI

C EQUATIONS, POLAR COORDINATES, AND CONIC

SECTIONS

Eccentricity b las are steeper. One aspe . ome hyper o t . . ctor Some ellipses are flatter than others, JUS t ass umber e called the eccen ncity. For a the shape of a conic section is measured by a n ellipse or hyperbola, distance betweeen . ---~==::..::-:;:::;~ ;.;;~on focal axis e = distance between verttces

THEOREM 4

. tandard position, For ellipses and hyperbolas ms C

e= a . circle having eccentricity 0). e < 1 (with a

.,__ REMINDER

c~r+Gr=l, c=Ja c~r-Gr = c=Ja

1. An ellipse has eccentricity O 2. A hyperbola has eccentricity e > l.

Standard ellipse:

2 -b2

Standard hyperbola:

I,

A parabola is defined to have eccentricity e

2 +b2

= 1.

. th ...; es are on the focal axi5 at (± a 0) Proof The foci are located at (± c, 0) and e ve,uC • . Therefore, 2 _ _ _....:di=·=-st=an=c.:..e..::b..::e.:..tw....,.e_e_n_fo_c-;:i--;-::-=:-::- = c = c e = distance between vertices on focal axis 2a a For an ellipse, c = a2 _ b2 and so e = c / a < 1. If the ellipse is a circle, then c :::::: o and therefore e = 0. For a hyperbola, c = a 2 + b 2 and thus e = c / a > 1. 1 How the eccentricity determines the shape of a conic is summarized in Figure IO. Consider the ratio b/a of the semiminor axis to the semimajor axis of an ellipse. The ellipse is nearly circular if b/a is close to 1, whereas it is elongated and flat if b/a is small. Now

b a

-=---=

a

g2

1-=~ a2

This shows that bf a gets smaller (and the ellipse gets flatter) as e The most round ellipse is the circle, with e = O. Similarly, for a hyperbola,

1 [Figure lO(B)].

b

=~ a The ratios ± b/ a are the slopes of the asymptotes so th [Figure IO(C)J. ' e asymptotes get steeper as e

oo

y

Circle

Parabola

!

!

1--- - - --+-- - - - - - e 0

Ellipses

(A) Eccentricity e

fi]J FIGU RE 10

Hyperbolas

(B) Ellipse flattens as e-> I.

(C) Asymptotes of the hyperbOla get steeper as e -> oo.

SECTION 11.5

conic Sections

657

1 . h w eccentricity deter-

CONCEPTUAL INSIGHT There is a more precise ~ay to ex~ am :id C have the same 2

mines the shape of a conic. We can prove that if two comes Ci C2 Changeccentricity e then there is a change of scale that makes C1 congruent to • 'rive . ' th d axes by a common pos1 mg the scale means changing the units along ex- an y·s times as large. 1 10 factor. A curve scaled by a factor of 10 has the sam_e shape but . 'd motion 0 By "congruent" we mean that after scaling it is possible to move C1 by~ ~ dir ti . . '. . ' • be di ) so that 1t lies ec Y (mvolvmg rotation and translation, but no stretching or n ng on top of C2. otrans All circles (e = 0) have the same shape because scaling by a factor' > b 01 forms a circle of radius R into a circle of radius r R. Similarly, any two para _ .as (e = 1) become congruent after suitable scaling. However, an ellipse of ecc~ntncity e = 0.5 cannot be made congruent to an ellipse of eccentricity e = 0.8 by scaling (see Exercise 76).

Directrix V x =!!

y

e

(0, b)

p

Eccentricity can be used to give a unified focus-directrix de~nitio~ of the conic sections with e > O. Given a point F (the focus), a line V (the directnx), and a number e > 0, we consider the set of all points P such that

PV

PF\

(a,0)

F=(c,0) 1 ae

//

[ill

I PF=ePV I

I

Fore = I, this is our definition of a parabola. According to the next theorem, Eq. (10) defmes a conic section of eccentricity e for all e > 0 (Figures 11 and 12).

a e

lilJ FIGURE

11 The ellipse consists of points P such that PF= ePV. Directrix V

Tl-:EOREM 5 Focus-Directrix Relationship Ellipse • If O < e < 1, then the set of points satisfying Eq. (10) is an ellipse, and xycoordinate axes can be chosen, and a, b defined, so that the ellipse has eccentricity e and is in standard position with equation

(~)2 +Gf = 1

y x=~

.fP

P

• Conversely, if a > b > 0 and c = a 2 - b2 , then the ellipse

(~)2 + Gf = 1

satisfies Eq. (10) with F = (c, 0), e = ~. and vertical directrix x = %-

Hyperbola FIGURE 12 The hyperbola consists of points P such that PF= ePV.

• If e > I, then the set of points satisfying Eq. (10) is a hyperbola, and xy-coordinate axes can be chosen, and a, b defined, so that the hyperbola has eccentricity e and is in standard position with equation

(~)2 -Gf = 1 • Conversely, if a, b > 0 and c = a2 + b2 , the hyperbola

(~f-Gf =

1

satisfies Eq. (10) with F = (c, 0), e = f., a and vertical directrix x

= !!.e ·

Proof We prove the first part of the hyperbola relationship. The remaining parts of the theorem are proven in Exercises 65, 66, and 68. We are assuming we have a set of points satisfying Eq. (10) withe > 1. Let d denote the distance between the focus and directrix. We want to set up our x- and y-axes and define a, b, c so that the focus is located at (c, 0)

658

oNIC SECTIONS PARAMETRIC EQUATIONS, POLAR COORDINATES, AND C

CH APTER 11 y

Directrix V x =

and the directrix is x that follow) is to set

= c

- - - - + - --+-_F_=....:

38. Calculate the length of the cu Figure 4. rve

polar equation r

y

31. ( GU ) Convert the equation 9 (x2

.th

WI

= a(I + cos 0), where

+ y 2) = (x 2 + y 2 _

r=(J - - - - -

Zy )2

7t

2

to pol ar coordinates, and plot it with a graphing utility.

32. Calcul ate the area of th e c ircle r 0

=

rr

3 an

d0

=

= 3 sin 0

bounded by the rays

2rr

7t

X

T·

33. Calc ul ate the area of one petal of r

= sin 40 (see Figure

I ).

FIGURE 4

= 0 1n.

Chapter Review Exercises

=

_ The graph of r e0.50 sin 0 for 0 < < . . 39 fgure5- Useacomputeralgebrasystemt - 2rr 1s shown m . t ten"th between the outer and inner loo o approximate the difference

PL

0

44. (y - 3)2

= 2x2 -

661

I

In Exercises 45-50, find the equation of the conic section indicated.

45. Ellipse with vertices (±8, 0), foci (±""3, 0)

y

46. Ellipse with foci (±8, 0), eccentricity

!

47. Hyperbola with vertices (±8, 0), asymptotes y =

±¾x

48. Hyperbola with foci (2, 0) and ( I 0, 0), eccentricity e = 4 X

SO. Parabola with vertex (4, -1), directrix x = 15

FIGURE 5

40,

51. Find the asymptotes of the hyperbola 3x2 + 6x - y2 - l Oy = l .

Show . that .r = f1(0) and r = f 2(0) defime the same curves in pol~ coordinates if ft (0) h(0 + rr). Use this to show that the followmg define the same conic section:

=-

de l-ecos0'

r=----

-de

r=---1 +ecos0

Jn Exercises 41-44, identify the conic section. Find the vertices and

foci. 41.

Gf +Gf =

42. x 2 - 2y2

I

=4 2

43. (2x+ ½y) =4-(x- y)2

49. Parabola with focus (8, 0), directrix x = -8

2

52. Show that the "conic section" with equation x - 4x +

y2 + 5 = 0

has no points.

53. Show that the relation~ = (e 2 - l)f holds on a standard ellipse or hyperbola of eccentricity e. 54. The orbit of Jupiter is an ellipse with the sun at a focus. Find the eccentricity of the orbit if the perihelion (closest distance to the sun) equals 740 x 106 km and the aphelion (farthest distance from the sun) equals 816 x 106 km.

55. Refer to Figure 25 in Section 11.5. Prove that the product of the perpendicular distances F1 R1 and F2R2 from the foci to a tangent line of an ellipse is equal to the square b2 of the sernirninor axes.

L

12 VECTORGEOMETRY V Engineers use vectors to analyze the forces on the cables in a suspension bridge, such as the Penobscot Narrows Bridge in Maine. Tension forces in the horizontal direction must balance so that there is no net force on the bridge towers. In the vertical direction the tension forces suppon the weight of th~ bridge deck. The cables must be designed so that each can suppon the combined vertical and horizontal tension forces.

ect~rs play a role in nearly all areas of mathematics and its applications. In ~hys_ical Settings, they are used to represent quantities that have both magnitude and directIOn, such as velocity and force. Newtonian mechanics, quantum physics, and special and general relativity all depend fundamentally on vectors. We could not understand electricity and magnetism without this basic mathematical concept. Computer graphics depend on vectors to help depict how light reflects off objects in a scene, and they provide a means for changing an observer's point of view. Fields such as economics and statistics use vectors to represent information in a manner that may be efficiently manipulated. . This chapter introduces the basic geometric and algebraic properties of vectors, setting the stage for the development of multi variable calculus in the chapters ahead.

12.1 Vectors in the Plane Recall that the plane is the set of points {(x, y): x, y e R} . We occasionally denote the 2 plane by R . This notation represents the idea that the plane is a "product" of two copies of the real line R, where one copy represents the points' x-coordinates and the other represents they-coordinates. (We extend this notation and idea to three-dimensional space in the next section.) A two-dimensional vector v is determined by two points in the plane: an initial point P (also called the tail or basepoint) and a terminal point Q (also called the head or tip). We write

NOTATION In this text, vectors are represented by boldface letters such as v, w,a,b,F.

I

and we draw 'll as an arrow pointing from P to Q. This vector is said to be based at P. Figure 1(A) shows the vector with initial point P = (2, 2) and terminal point Q = (7, 5). The length or magnitude ofv, denoted llvll, is the distance from P to Q. The vector v = 0 R pointing from the origin to a point R is called the position vector of R. Figure l (B) shows the position vector of the point R = (3, 5).

-

y

6 5 4 3 2

I} ;I

(A) Vectors parallel to v

(B) w is a translation ofv. FIGURE 2

/ Q-(7,5)

P= (2,2)

2345678

(A) The vector PQ

mJ

y

6 5 4

R = (3,5)

3 2 Q L......+-1-2+--+-3--+-4--+-5+ X

(B) The position vector

OR

FIGURE 1

We now introduce some vector terminology. • Two vectors v and w of nonzero length are called parallel if the lines through v and w are parallel. Parallel vectors point either in the same or in opposite directions [Figure 2(A)]. • A vector v is said to undergo a translation when it is moved to begin at a new point without changing its length or direction. The resulting vector w is called a translation of v [Figure 2(8)]. A translation w of a vector v has the same length and direction as v but a different basepoint. In almost all situations, it is convenient to treat vectors with the same length and direction as equivalent, even if they have different basepoints. With this in mind, we say that

• v and ware equivalent if w is a translation of v [Figure 3(A)]. 669

670

CHAPTER 12

VECTOR GEOMETRY

. al t £very vector can be tran . 3(B) are equiv en . s. Note that no _pair ?f. the vector_s Fi~:e 3(C)]. Therefore, lated so that its tail 1s at the ongm [Fig based at the origin. unique vector vo Every vector v is equivalent to a .

y

- ~-----x (C) Vo is the unique vect~r based at the origin and eqmvalent to v.

(B) Inequivalent vectors

(A) Vectors equivalent to v (translates of v) FIGURE 3

-

t of a vector (Figure 4). To work algebraically, we define the componen s

y Q = (a2, b2)

b = b2 - b 1 P= (a 1,b 1)

a= a2 - a 1

DEFINITION Components of a Vector The components of v = PQ, where P ==

(a1, b1) and Q

= (a 2 , b2), are the quantities

/ a = a2 - a1

b - - - - - Po=(a,b) Vo :

(x-component),

b = bz - bi

(y-component)

J

The pair of components is denoted (a, b).

I I

i,:...._---+--------+-X a

FIGURE 4 Both of the vectors v and vo have components (a,b).

The pair of components (a, b) determine the length and direction of v, but not its basepoint. Therefore, two vectors are equivalent if and only if they have the same components. Nevertheless, the standard practice is to describe a vector by its components, and thus we write

• In this text, angle brackets are used to distinguish between the vector v = (a, b) and the point P = (a,b). Some textbooks denote both v and P by(a,b). • When referring to vectors, we use the terms "length" and "magnitude" interchangeably. The term "norm" is also commonly used.

v= (a,b) Equivalent vectors have the same length and point in the same direction, but they can start at any basepoint. • When the basepoint P = (0, 0), the components of v are just the coordinates of its endpoint Q. • The length of a vector v = (a, b) in tenns of its components (by th distance fonnula; see Figure 4) is e

llvll = IIPQ/1 = Ja2 + b2 y

• The zero. vector (whose head and tail coincide) 1·8 the vector O= (0 0) of Jength zero. It 1s the only vector that lacks a direction. ' • For a vector v, the vector -vis the vector with th . . in the opposite direction. If v = (a b) th e same length as v but poinung , , en -v = (-a, -b).

P1 = (3, 7)

EXAMPLE 1 Detennine whether VJ P1 Q2

= (2,

I)

= (3, 7),

Q1

= (6, 5)

= P1Q1 and v2 -_ and

p:-7 2 Q2 are equivalent, where

p _ ( l 2-

-

'4),

Solution We can test for equivalence by compu t'rng the com FIGU RE 5

Q2

= (2, 1)

What is the magnitude of v 1?

V1=(6-3,5-7)=(3,-2),

Vz=( 2 -

ponents (Figure 5):

(-l), I - 4)

= (3, -3)

SECTION 12.1

The components of v1 and v2 are not the same, so v1 an v, = (3, -2}, its magnitude is

y

P=(l,4)

''

l \3

v = (2 -3}

,

: 2

·------- Q = (3, I)

Vo= (2, -3)

FIGURE 6 The vectors v and v0 have the same components but different basepoints.

Uv1 II= .ff1- +

Vectors in the Plane

671

d v2 are not equivalent. Since

0, and in the opposite direction if). < 0. y ')..b ------

AV

L---+---x

A.a

a+c

FIGURE 11 Vector operations using

(A) v + w = (a

components.

+ c, b + d)

(B)

Av= (Aa, Ab)

Vector Operations Using Components lfv = (a,b) and w = (c,d), then: (i) v+w=(a+c,b+d)

(ii) v - w = (a - c, b - d) (iii) AV= (Aa, Ab)

(iv)

V

+0 = 0 + V = y

We also note that if p = (ai bi) and Q _ ( b _ P---+Q . ' - a2, 2), then compo t f th are convemently computed as the d"f~ nen s o e vector 1 ,erence

---

vy V

PQ

+ W = (4,6)

= OQ -

OP= (a2,b2) - (a1,b1) = (a2 - a1,b2 - bi)

EXAMPLE 3 For v = (1 ' 4) ' w -_ (3, 2), caIculate: V

= (!, 4)

(a) V

+W

(b) 5v

Solution V

2

liII

FIGURE 12

4

+W = (1, 4) + (3, 2) = (1 + 3' 4 + 2) = (4,6) 5v = 5 (1,4) = (S, 20)

The vector sum is illustrated in Figure 12 _

SECTION 12.1

Vectors in the Plane

673

Vector operations obey the usual Jaws of algebra. THEOREM 1 Basic Properties of Vector Algebra For all vectors u, v, wand for all scalars A,

v+w=w+v u + (v + w) = (u + v) + w

Commutative Law: Associative Law: Distributive Law for Scalars:

A(v+w)=AV+AW

These properties are verified easily using components. For example, we can check that vector addition is commutative: (v1, v2) + (w1, w2) = (v 1 + w1, v2 + w2) = (w1 + v,, w2 + v2) = (w1, w2)

+ (v1, v2)

i1

Ii

Commutativity of addition of real numbers

A linear combination of vectors v and w is a vector

rv+sw where r and s are scalars. If v and w are not parallel, then every vector u in the plane can be expressed as a linear combination u = rv + sw [Figure 13(A)]. The parallelogram· P whose vertices are the origin and the terminal points of v, w and v + w is called the parallelogram spanned by v and w [Figure 13(B)]. It consists of the linear combinations rv --i- sw with O :::: r ::S 1 and O:S s :S 1.

I"

y

Jw

u = rv + s~-- --

/

..-} ,w f

/

!

I

---i-' - - ~f 0 is called the first octant. As in two dimensions, we derive the distance formula in R3 from the Pythagorean Theorem.

FIGURE 2 z

z

y X

z = 0 defines the xy-plane. FIGURE 3

Y = 0 defines the xz-plane.

x = 0 defines the yz-plane.

SECTION 12.2

Three -Dimensional Space: Surfaces, Vectors, and Curves

681

THEOREM 1 Distance Formula in R3 The distance IP - QI between the points P = (a1, b1, c1) and Q = (a2, b2, c2) is

[D

Proof First, apply the distance formula in the plane to the points P and R [Figure 4(A)]:

IP - Rl 2 = (a2 -

ad+ (b2 - bi)2

R = (a 2, b2, c 1)

la2 - ad (B)

(A)

Then observe that t:,PRQ is a right triangle [Figure 4(B)] and use the Pythagorean Theorem: IP - Ql 2 = IP - Rl 2 + IR - Ql 2 = (a2 - ad+ (b2 - b1f + (c2 - ci)2

Surfaces Surfaces in R3 will play an important role in our development of multivariable calculus. Planes are the most basic surfaces. The equation x = a ·defines a plane parallel to the yz-plane, while y =band z = c define planes parallel to the xz-plane and xy-plane, respectively. For example, the planes x = 3, y = -5, and z = 10 are illustrated in Figure 5. We explore planes further in Section 5 in this chapter. z

x=3

z = 10

y

y X

y X

FIGURE 5 Planes in 3-space.

Spheres and cylinders are some other surfaces we will encounter. The sphere of radius R with center Q = (a, b, c) consists of all points P = (x, y, z) located a distance R from Q (Figure 6). By the distance formula, the coordinates of P = (x, y, z) must satisfy

J 0, we have U[[v = (llull cos 0)ev, where ev = i!n vis the unit vector in the direction of v. Similarly, if 0 is obtuse, then U[[v is a negative multiple of ev and again u11v = (llull cos 0)ev since cos 0 < 0. Therefore, U[[v

lllJ FIGURE

11 When 0 is obtuse, U[[v and V point in opposite directions.

L

= (llull cos 0)ev =

v

llnll cos 0 - = llvll

l'

I

11

=(a+ b). (a+ b) + (a - b) · (a - b) FIGURE 8

I I

v)

llull llvll cos 0 (" • v= v llvll 2 v·v

This formula provides us with the desired expression for the projection. We present three equivalent expressions for u11v:

694

CHAPTER 12

VECTOR GEOMETRY

. of u along v is the vector r--------------:-..t. 0 The projection

Projection of u along v Assume v .,.. · u11v

(-u•V) v =(~)ev

= (S) v = fvji2

QJ

_llv_l_ l --

is called the component 01 . The scalar II v II This is sometimes denoted proJvUd compvu, . sometimes denote the scalar component of u along v and is :__ _ __ __ __

EXAMPLE 7 Find the projection ofu == (5, l,

-3) along v == (4, 4, 2).

,. rmula in Eq. (5): Solution It is convenient to use the firrst 10 U·V

-4 V • V-

= (5, 1,-3) · (4,4,2) = 20+4- 6 == lS, Ullv

= (U·V) V--(~) 36

2

+4

2

+· 22 = 36

(4 4,2) == (2,2,1)

'

V

can be written as the sum of the We show now that if v -# 0, th~n every vecto:; (see Figure 12). In fact, if we set projection u11 v and a vector u.1 v that 1s orthogonal t / u.1v FIGURE 12 Decomposition ofu as a sum u = "llv + ll.Lv of vectors parallel and orthogonal to v.

llIJ

=u -

u11v ]

then we have the following decomposition of u w1·th respe ct to v··

I

U

= Ullv + U.Lv

I

Equation (6) expresses u as a sum of vectors, one parallel to v and on_e perpendi~ular to v. We must verify, however, that u.1v is perpendicular to v. We do this by showmg that

the dot product is zero: u.1 • v = (u - u11v) · v = (u - (~) v) · v = u · v - (u · v) (v · v) v

V·V

V •V

=0

EXAMPLE 8 Find the decomposition of u = (5, 1, -3) with respect to v = (4, 4, 2). Solution In Example 7, we showed that u11v U.Lv

=U -

Ullv

= (2, 2, 1). The orthogonal vector is

= (5, 1, -3) -

(2, 2, 1)

= (3, -1, -4)

The decomposition of u with respect to v is u = (5, 1, -3) = Ullv + U.Lv

=

(2, 2, 1)

'--,-'

Projection along v

+ '(3,- .-1, -4) .,.-,

I

Orthogonal to v

The decomposition into parallel and orthogonal vectors is useful in many applica· tions, as we see in the next two examples.

EXAMPLE g Let us return to the problem posed at the start of the section (see Figure 13). We have a wind vector w = (60, 0) km/h, and the bridge is oriented 32 degrees east of north. Express w as a sum of vectors, one parallel to the bridge and one perpendicular to it. Also, compute the magnitude of the vector perpendicular to the bridge to determine the speed of the part of the wind blowing directly at the bridge. Solution To begin, note that u = (~os ~go' sin 580) is a unit vector parallel to the bridge. It is shown, but not drawn to scale, m Figure 13. The goal is to decompose was the sUJ!l FIGURE 13

ofw11u and W.Lu• ~-------__,_-

J

Dot Product and the Angle Between Two Vectors

SECTION 12.3

=

695

For w11u = (::~) u, note that u. u I since u is a unit vector. Also, w · u (60, 0) · (cos 58°, sin 58°) = 60 cos 58°. Therefore, Wllu

= 60cos58° (cos58°, sin58°)

=

(16.85, 26.96)

and then W.Lu

=w-

w11u

(43.15, -26.96)

So we have the decomposition: w = (16.85, 26.96)

+ (43.15, -26.96)

where, approximately, (16.85, 26.96) is along the bridge and (43.15, -26.96) is perpendicular to it. The magnitude of the perpendicular part of the wind, the part blowing directly at the bridge, is approximately J(43.IS)2 + (-26.96) 2 50.9 km/h.

I

1

EXAMPLE 10 What is the minimum force you must apply to pull a 20-kg wagon up a frictionless ramp inclined at an angle 0 = 15°?

Solution Let v be a vector in the direction of the ramp, and let Fg be the force on the wagon due to gravity. It has magnitude 20g newtons, or N, with g = 9.8. Referring to Figure 14, we decompose Fg as a sum

Fg =F11v+F.1v

FIGURE 14 The ai,gle between Fg and Fllv

is90° -0.

where F11v is the projection along the ramp and F.iv, called the normal force, is the force perpendicular to the ramp. The normal force F.iv is canceled by the ramp pushing back against the wagon in the opposite direction, and thus (because there is no friction) you need only pull against F11vNotice that the angle between Fg and the ramp is the complementary angle 90° - 0. Since F1 1v is parallel to the ramp, the angle between Fg and F11v is also 90° - 0, or 75°, and I\F11vll

= IIFg II cos(75°);::, 20(9.8)(0.26)

51 newtons

Since gravity pulls the wagon down the ramp with a 51-newton force, it takes a minimum force of 51 newtons to pull the wagon up the ramp.

y

bj _______ u = (a, b)

ai

GRAPHICAL INSIGHT It seems that we are using the term "component" in two ways. We say that a vector u = (a, b) has components a and b. On the other hand, u • e is called the component of u along the unit vector e. In fact, these two notions of component are the same. The components a and b are the dot products of u with the standard unit vectors:

=a u · j = (a,b) · (0, 1) = b u • i = (a,b) • (1,0)

(A)

and we have the decomposition [Figure lS(A)] u = ai + bj

But any two orthogonal unit vectors e and f give rise to a rotated coordinate system, and we see in Figure 15(8) that u = (u • e)e (B)

In other words, u • e and u · f really are the components when we express u relative to

the rotated system. FIGURE 15

+ (u • f)f

\

I I

r-·-• 696

CHAPTER 12

VECTOR GEOMETRY

)

---------------_ 12 _._3_S_U_M _M ~ AR:..=Y . ::____ _ _ _ _ • The dot product of v

____

= (a1, b1, c1) and 'W = (az, bz, cz) is v. 'W

= a1a2 + b1b2 + c1c2

• Basic Properties of the Dot Product: - Commutativity: v . w = 'W · v . (A'W) ),.(V · 'W) - Pulhng out scalars: (),.v) · 'W v · - Distributive Law: u . (v + w) u · v + u · 'W (v + w) • u v · u + 'W · u

=

= = =

-

V. V

- v. w u

V

0

= livli2 = llvll llwll cos 0,

0

where 0 is the angle between v and w that satisfies

JC.

=

Test for orthogonality: v ..l w if and only if v · 'W 0. . The angle between v and w is acute if v · 'W > 0 and obtuse if v · 'W < O. • Assume v =/. 0. Every vector u has a decomposition

[,u=

FIGURE 16

+ llJ.v

ll//v

J

where u v is parallel to v, and 01.v is -orthogonal to v (see Figure 16). The vectoru 11, 11 is called the projection of u along v. With ev = 11 11 v, the decomposition is computed

!

as follows : llllv

U·V) v = (Ullvll2 ·V) v = = (V-V

(U·V) W ev,

U·V

UJ.v

=u-

llllv

The coefficient - - is called the component of u along v·

llvll

.

component of u along v

U·V

= - - = llu II cos 0 llvll

12.3 EXERCISES

Preliminary Questions 1.

Is the dot product of two vectors a scalar or a vector?

2.

What can you say about the angle between a and b if a • b < O?

3. Which property of dot products allows us to conclude that if v is orthogonal to both u and w, then v is orthogonal to u + w? 4.

Which is the projection of v along v: (a) v or (b) ev?

Whi . . 5• Let u11v be the projection of u alon projection u along the vector !?, v._ ch of the followmg 1s the 2v and which 1s the projection of 2u along v?

( )

a

1

2ll//v

(c) 2u1 1v

(b) u11v

6. andWhich u v? of the following is equ al to cos 0, where 0 is the angle between (b) u, ev

(a) U · V

(c) e 11 • ev

iEx =";;e;;r;,c~is;.e~s- - - - - - - - - - - - - - - - - - - - - - - - - - - - - 9.

In Exercises 1-12, compute the dot product. 1.

(I , 2, 1) · (4, 3, 5)

3.

(0, I , 1) · (-7, 41, -39)

5.

(3 , I) · (4, - 7)

7.

k -j

2.

(3, -2, 2) - (I , 0, 1)

4.

(J, - 1, I)· (-2,4,-6)

6.

(¾ , ½) · (3 , ½)

8.

k-k

+ j) . (j + k) (i + j + k) . (3i + 2j (i

11.

+ 2k) . (i - 4k) ( -k) · (i - 2j + 7k)

10. (3j

5k)

12. lnExercises 13-18' determ·me w h ether th ,-tw _,1 _ e o vectors are orthogonal Ill"'' 1if not, w hether the angle between them is acute or obtuse 13. (1, 1, 1), 15 (1 2 1) .

' '

'

(I, -2, -2) (7 ,-3,-1)

14• (0,2, 4), 16. (0,2,4) ,

.

(-5,0,0) (3, 1,0)

....ii

SECTION 12.3

l8. (12, 6), (2, -4) c,ercises 19-22,jind the cosine of the angle between the vectors. In.,... 19- (0, 3, I),

(4, 0, 0)

Dot Product and the Angle Between Two Vectors

51. Fmd the angle 9 in the triangle in Figure 17 ·

y

20. (l,1,1), (2,-1,2)

21. i +j, j + 2k

697

(0, 10)

22. 3i + k, i + j + k

In exercises 23-30, find the angle between the vectors.

Z3. (2, Ji), (1 + ,./2, 1- ,./5_)

24.

zs.

26. (3, 1, 1),

(I.!, 1),

(1,0, 1)

21. (0,1.1), (1,-1,0) z9, i,

(5, J3). (""3. 2) (2, -4, 2)

28. (1, 1,-1),

3i + 2j+ k

30. i+k,

(3, 2)

L--------x

(1,-2,-1)

FIGURE 17

j-k

31. Find all values of b for which the vectors are orthogonal. (a) (b,3,2), (l,b,l) (b) (4,-2,7), (b2,b,0)

11

52. Find all three angles in the triangle in Figure I 8.

I

32. Find a vector that is orthogonal to (- 1, 2, 2) . y

33, Find two vectors that are not multiples of each other and are both orthogonal to (2, 0, - 3).

34. Find a vector that is orthogonal to v = (1, 2, 1) but not to w== (1,0,-1) .

35, Find v · e, where Jivll and vis

ff.

(6, 3)

= 3, e is a unit vector, and the angle between e

36. Assu_me that v lies in the yz-plane. Which of the following dot products is equal to zero for all choices of v? (b) V · k (a) V • (0, 2, 1) (C) V· (-3, 0, 0) (d) v · j In Exercises 37-40, simplify the expression.

37. (v - w) · v + v · w

38. (v + w) • (v + w) - 2v • w

39. (v + w) • v - (v + w) · w

40. (v + w) • v - (v - w) • w

6"'-------x

FIGURE 18

53. (a) Draw u 11 • and v 11 u for the vectors appearing as in Figure 19. (b) Which of u11v and v11u has the greater magnitude?

In Exercises 41-44, use the properties of the dot product to evaluate the expression, assuming that u · v = 2, llull = 1, and llvll = 3. 41. u • (4v)

42. (u+ v) • v

43. 2u • (3u - v)

44. (u + v) • (u - v)

FIGURE 19

45. Find the angle between v and w ifv · w = -llvll llwll.

from the component of v along u? Why or why not?

46. Find the angle between v and w ifv · w = ½llvll llwll. 47. Assume that \\vii = 3, llwl\ = 5, and the angle between v and w is 0 = j. (a) Use the relation llv + w\1 2 = (v + w) · (v + w) to show that llv +wll2 = 32 + 52 + 2v · w. (b) Find llv

+ wl\.

48. Assume that \\vii = 2, I\WII = 3, and the angle between v and w is 120°. Determine: (a) v. w

(b) 112v + wll

49. Show that if e and f are unit vectors such that

lie - fll

= f. Hint: Show that e· f = ½-

(c)

112v - 3wll

\le+ fll = ;, then

SO. Find ll2e _ 3fl\, assuming that e and f are unit vectors such that

lie + r11

= ./3/2.

54. Let u and v be two nonzero vectors. (a) Is it possible for the component of u along v to have the opposite sign

(b) What must be true of the vectors if either of these two components is O? In Exercises 55--62, find the projection ofu along v.

55.

U

= (2,5),

57.

U

= (-1,2,0),

V

= (1, 1) V

= (2,0, 1)

56.

U

= (2, -3),

58.

U

= (1, 1, 1),

V

= (1, 2)

V

= (1, 1,0)

59. u = Si+ 7j - 4k, v = k

60. u = i+ 29k, v = j

61. u = (a,b,c), v = i

62. u=(a,a,b), v=i-j

In Exercises 63 and 64, compute the component of u along v.

63.

U

= (3,2, 1),

V

= (1,0, 1)

65. Find the length of OP in Figure 20.

64. U=(3,0,9), V=(l,2,2)

• S98

CHAPTER 12

~

1illllL£1

VECTOR GEOMETRY Of carbon molecule boncteo ~ar apart from each Other I() methane roolecul~at are spaced a~e vertices of a tetrahectr as

CH4 consists

66, Find 1101-vll in Figure 20.

81. The n molecules then sit at . 22. We can model ~. four hydroge en atoms in Figure % as -I -I) and the hydrogen poss1'ble · The hydrog matt·ts center, (I ti atoms at with the carbon atom at the point 2, ~; ilie dot product to md the bond 'th the carbon ato I) and (0, 1, 1). egments from the carbon ato" w1 00)(1,1,0,) (l , 0, anytwoof , the lines .,, (0 , , ' ed between angle a form to the hydrogen atoms.

y

FIGURE 20

ln Exercises 67-72, find the decomposition a = allb respect to b.

+ a.1b

with

67. a= (1,0), b = (I, I)

68. a = (2, -3), b = (5, 0)

69. a= (4,-1,0),

70. a= (4,-1,5), b = (2, I, 1)

71. a=(x,y),

b = (0, I, I)

b=(l,-1)

72. a= (x,y,z), b= (1,1,1)

73. Let ee = (cos0,sin0). Show that ee • e,i, angles 0 and ,f,. 74.

= cos(0 -

,fl) for any two

thane molecule. FIGURE 22 A me

Let v and w be vectors in the plane.

(a) Use Theorem 2 to explain why the dot product v • w does not change if both v and w are rotated by the same angle 0. (b) Sketch the vectors e 1 = (I, 0) and ei

= (f, f

), and determine the

ei, e; obtained by rotating e1, e2 through an angle %, Verify that =e; -e;.

vectors e1 ·e2

each central atom . . appears . at ~e cen8,.,, Iron forms a crystal Iatti ce where . nd to add1t10nal rron atoms, as "'' f which correspo th lin ter of a cube, the comers O ti d the f3 between e e segin Figure 23. Use the dot produc~~ ~~jacent outer atoms. Hint: Take the roents from the central atom to . . and the comer atoms to occur at central atom to be situated at the ongm

angle

(±1, ±1, ±1).

ln Exercises 75-78, refer to Figure 21. 75. Find the angle between AB and AC. 76. Find the angle between AB and AD. -+ 77, Calculate the projection of AC along AD. -+ along AB. 78. Calculate the projection of AD

A= (0, 0, I)

FIGURE 23 An iron crystal.

D=(0, 1,0) B=(l,0,0) C=(l, 1, 0)

FIGURE 21 Unit cube in R3.

. le 6 assume that the carpenter's diagln Exercises 79-80, as m E:;mp ompute the diagonal length that anal measurements!arefixa;e J;:pnar: the result with the corresponding produces a rectangu a~ " . · split-difference approx1matwn.

d

79. x

= 234½ inches and Y = 223 in.

80. x = 87.2 cm and y = 82.7 cm

83. Let v-and w be nonzero vectors and set u = ev + ew, Use the dot product to show that the angle between u and v is equal to the angle between u and w. Explain this result geometrically with a diagram.

84. Let v, w, and a be nonzero vectors such that v . a = w · a. ls it true that v = w? Either prove this or give a counterexample. 85. In Example 9, assume that the wind is out of the north at 45 km/h. Express the corresponding wind vector as a sum of vectors, one parallel to the bridge and one perpendicular to it. Also, compute the magnitude of the ' perpendicular term to determine the speed of the part of the wind blowing directly at the bridge. 86. A plane flies with velocity v = (220, -90, JO) km/h. A wind is blow· ing o~t of the northeast with velocity w = (-30, _ 30, 0) km/h. Express the wmd ve~tor as a ~um of vectors, one parallel to the plane's velociry. one perpendicular to 1t. Is the parallel part of Uie wind blowing with N against the plane?

l

SECTION 12.3

8'/ Ca)culate the force (in newtons) rcqu·red 1 : incline (Figure 24). 10

to push a 40-kg wagon up a

oot Product and the Anale Between two Vectors

699

90. Let p and Q be antipodal (opposite) points on II sphere o~ radius r centered at the origin 11Dd let R be II third point on the sphere (Figuni 27)Prove that PR and QR are onhogonal.

10° FIGURE 24

88. A force F is applied to each of two ropes (of ne . .

. ghgible weight) atg wagon and making an angl f 350 with the horizontal (Figure 25). What is the maximum . de o . newtons) that can beFapplied without lifting the wagon ;f;i;;:;~r:uo:d~ (m

ed to opposite ends of a 40-k

fBC h

FIGURE 27

F 2

91. Prove that !Iv+ w!l 2 - Uv - w!l = 4v · w. 92. Use Exercise 91 to show that v and w are orthogonal if and only if

!Iv - w!I

93. A rhombus is a parallelogram in which all four sides have equal

FIGURE 25

89, A light beam travels . along the ray determined by a um·t vector L, strikes flat surface at pomt P, and is reflected along the ray determined by a umt vector R, where 01 = /h (Figure 26). Show that if N is the unit vector orthogonal to the surface, then R = 2(L · N)N - L

Incoming light

length. Show that the diagonals of a parallelogram are perpendicular if and only if the parallelogram is a rhombus. Hint: Take an approach similar to the solution in Example 5, and consider x · y. 94. Verify the Distributive Law:

u • (v + w)

Reflected light

-'I

= !Iv+ w!I .

= u •v + u • w

95. Verify that ()..v) • w = )..(v • w) for any scalar A.

·-·~ R

p l

FIGURE 26

I

Furthe'r Insights and Challenges 96. Prove the Law of Cosines, c2 = a2 + b2 -

2ab cos 0, .by referring to Figure 28. Hint: Consider the right triangle l:.PQR. R

98. Use (7) to prove the Triangle Inequality:

l\v + wl\ ::: l\vl\ + llw\l Hint: First use the Triangle Inequality for numbers to prove

I

I

l(v + w) · (v + w)I ::: l(v + w) · v\

I

a

I

a sin 0 :

C

99. This exercise gives another proof of the relation between the dot product and the angle 0 between two vectors v = (a 1, b1) and w = (a2,b2) in the plane. Observe that v=llvll(cos0 1,sin0 1) and w= l\wll (cos 02, sin02), with 01 and 0i as in Figure 29. Then use the addition formula for the cosine to show that

I I

0

s L..._---'------ -Ip' - " - - - ' Q b

b-acos0

v · w = l\vll llwl\ cos0

FIGURE 28

97. In this exercise, we prove the Cauchy-Schwarz inequality: If v and w are any two vectors, then lv •wl::: llvll llwll

y

y

y

w

0

= llxv + wH2 where x is a scalar value. Show that /(x) may be written /(x) = ax 2 +bx+ c, where a= llvu2, b = 2v • w, and c = llwl\2. (b) Conclude that b2 - 4ac ::: 0. Hint: Observe that /(x) 0 for all x.

(a) Let /(x)

+ l(v + w) •w\

X

FIGURE 29

700

CHAPTER

12 VECTOR GEOMETRY

100. Let v == (x, y} and Ve== (xcos0 Prove that the

+

·

Y sm 0 • -x sin 0

uidistant from two points · ts X - (x, y, z) eq th" I · /' 103. The set of all pom - ow that x lies on 1s P ane 1f ,Q in R3 is a plane (Figure 31). Sh -+

+ y cos 0)

QJ

PQ _OX= ½(uOQllz-110Pl/2)

angle between v and Ve is 0 .

101. Let vb

e a nonzero vector Th vectors i,j, k are called th d" . . e ang 1es°'• /3, Y between v and the unit of these angles are called ~~tlo~ angle~ of v (Figure 30). The cosines e irectton cosines ofv. Prove that 2 cos o, + cos 2 /3 + cos2 Y = 1

z

p

'

z

y V X

FIGURE 31

• If R IS · the ml·dpom · t of p Q then X is equidistant from P and Q "• Hint: and only if is orthogonal to P Q •

XR

y

104 Sketch the plane consisting of all points X = (x, Y, z) equidistan1 fro,;. the points p = (0, I, 0) and Q = (0, 0, 1). Use E q . (8) to show thatx lies on this plane if and only if Y = z.

X

FIGURE 30 Direction angles of v. 102. Find the direction cosines of v

=

105. Use Eq. (8) to find the equation of the plane consisting of all poin~ Q = (1, 0 , 2).

(3, 6, -2).

X = (x,y, z) equidistant from P = (2, 1, 1) and

12.4 The Cross Product Some applications of vectors require another operation called the cross product. In physics and engineering, the cross product is used to compute torque, a twisting force that causes an object to rotate. Figure 1 displays diagrams of a force F of varying strength and direction applied to a wrench to tum a bolt. The vector r is referred to as a position vector and indicates the location of the force relative to the turning axis in the bolt. The resulting twist on the bolt varies from weak to strong, and the direction of the twist is either clockwise or counterclockwise. Since the twist has magnitude and direction, it is naturally represented by a vector. The resulting twisting force is referred to as the torque on the bolt and is calculated using the cross product of rand F. F

r

Weak clockwise twist

Weak

Weak clockwise twist

counterclockwise

~. ·..

twist

· .-::: rd · t ~ f f l

' .

Weak

Weak

counterclockwise twist

counterclockwise twist

@·?T~···rtt i""f

E

F

I

)

r7.1d

r

Strong counterclockwise twist

I-~

r

Strong clockwise twist

f

F

r

:1 FIGURE 1

Unlike the dot ~roduct v_ · w (whi~h is a scalar), the cross product v x w is a vector. It is defined algebraically usmg what is known as a 3 x 3 "determinant." We incroduce

r SECTION 12.4

The Cross Product

701

2 x 2 and 3 x 3 determinants next, and then show how they are used to define the cross product. Ann x n matrix is an array consisting of n rows and n columns_ of numbers (or vectors, as we will see in the definition of cross product). The determinant of a 2 x 2 matrix is denoted and defined as follows:

\~ :1

[TI

=ad-be

Note that the determinant is the difference of the diagonal products. For example,

2 ~ IN I2 4\ = ~I

_ IV21 = 3 -4- 2. ! = 11

VI

2

The determinant of a 3 x 3 matrix is denoted and defined by a, a2

b1 b2

c,

a3

b3

C3

= a1

c2

I I I I+ I b2

c2

b3

c3 (I, !)-minor

- b1

a2

c2

a3 c3 (I, 2)-minor

Ci

a2

0

bb2 \

a3 3 (I, 3)-minor

This formula expresses the 3 x 3 determinant in terms of 2 x 2 determinants called minors. The minors are obtained by crossing out the first row and one of the three columns of the 3 x 3 matrix. For example, the minor labeled (1, 2) above is obtained as follows: b b

a2

l

a2 c21 a3 C3 ( 1,2)-minor

to obtain the (1, 2)-rninor

c2

a3 c3 Cross out row I and column 2

4 3 1 -7. 5 3

2 The theory of matrices and determinants is part of linear algebra, a subject of great importance throughout mathematics. In this section, we discuss just a few basic definitions and facts from linear algebra needed for our treatment of multivariable calculus.

EXAMPLE 1 A3 x 3 Determinant Calculate

Solution

®

@

Q)

-~

-~ =a>\~

0 -1

-~\-®\-~ -~\+®\-~ ~I

= 2(38) - 4(-7) + 3(1) = 107

Later in this section, we will see how determinants are related to area and volume. First, we introduce the cross product, which is defined as a determinant whose first row has the vector entries i,j, k. CAUTION Note in Eq. (3) that the middle

Iterm comes with a minus sign.

I

DEFINITION The Cross Product The cross product of vectors v = (v1, v2, v3) and

w = (w1, w2, w3) is the vector

The cross product differs fundamentally from the dot product in that u x v is a vector, whereas u . v is a number.

VX W

=

j

k

Vi

v2

V3

WJ

W2

W3

=

I Ii -I I.+ \vi v2 W2

V3 W3

v1

WI

v3

W3

J

W!

v21k

w2

EXAMPLE 2 Calculate v x w, where v = (-2, 1,4) and w = (3,2,5).

Solution

VXW=

-i i ;=\; :H-; :1i+1-; ;\k

=(-3)i-(-22)j+(-7)k= (-3,22,-7)

0

1'

11 702

CHAPTER 12

VECTOR GEOMETRY

u

. g of the cross product. Hov., . . tric meanin . eve F ormula (3) gives no hint of the geome •ng the nght-hand rule S t, . . t r V X W USI • U th ere 1s a simple way to visualize the vec O all lie in a plane. We s Pthat do not • ay th pose that v, w, and u are nonzero vectors . . f O is deterrruned by the . a1 {v, w, u } forms a right-handed system if the direction o v tow, your thumb po· ngh1• 1 h an d ru le: When the fingers of your rzg · h t h and curl from p· e 2) The following th nts lo h . d asu( igur . eore t e same side of the plane spanned by V an w arts are proved at the end Of lll t describes the cross product geometrically. The firS two P the section.

FIGURE 2 {v, w, u}

system.

THEOREM 1 Geometric Description of the Cross Product Given tdwotnonzero_ non. the cross pro uc v x w is th parallel vectors v and w with angle 0 between th em, e unique vector with the following three properties:

forms a right-handed

(i) v x w is orthogonal to v and w. (ii) v x w has length llvll llwll sin 0. (iii) {v, w, v x w} forms a right-handed system.

\L-

How do the three properties in Theorem I determine v x_ w? By property (i), v x \\' lies on the line orthogonal to v and w. By property (ii), v x w is one_of th ~ two vectors on this line of length II vii II wll sin 0. Finally, property (iii) tells us whic~ of these two vec. tors is v x w-namely, the vector for which {v, w, v x w} forms a nght-handed system (Figure 3).

.L--1-.,__ V

EXAMPLE 3 Let v = (2, 0, 0) and w = (0, I, I). Determine u = v x w using the ge0metric properties of the cross product rather than Eq. (3).

/,

Ii

FIGURE 3 There are two vectors orthogonal to v and w with length /Jvll 1/wll sin 0 . The right-hand rule determines which is v x w .

Solution We use Theorem I. First, by property (i), u = v x w is orthogonal to v and w. Since v lies along the x-axis, u must lie in the yz-plane (Figure 4). In other words, u = (0,b,c). But u is also orthogonal tow= (0, I, I), sou · w = b + c = 0 and thus

u

=

(0, b, -b). Next, direct computation shows that llvll = 2 and llwll = ,,/2. Furthermore, the angle between v and w is 0 = f since v • w = 0. By property (ii),

z

Hull= Jb 2

y X

FIGURE 4 The direction of u = v x w is determined by the right-hand rule. Thus, u has a positive z-component.

+ (-b) 2 =

lbl-Ji.

is equal to

IM llwll sin

i

=

2-Ji.

Th~refo_re, lbl _= 2 and b = ±2. Finally, property (iii) tells us that u points in the po;itive z-drrect10n (Figure 4). Thus, b = -2 and u = (0, -2, 2). You can verify that the formula for the cross product yields the same answer. 1 One of the most striking properties of the cross product is that it is anticommutative. Reversing the order changes the sign:

Iw

XV=

-v

X

0

w

We verify this using Eq. (3). When we interchange v and w , each o f the 2 x 2 d etemu· · . nants changes sign. For example,

~!/

FIGURE 5

=

v1w2 -

v2w1

= -(v2 w 1 _ v 1w 2) = _ ,w1

w21

Vz Anticommutativity also follows from the geometric d • . B properties (i) and (ii) in Theorem 1, v x wand w x escnpt~on of the cross product.d 1 0th and have the same length. However, v x w and w x v ~e . orthogonal to v an th v pomt m · di · by e right-hand rule, and thus v x w = -w xv (F . opposite recttons igure 5). In particular v x v - -v x v and , h ence v x v = O. V[

SECTION 12.4

The Cross Product

703

• of cross products (the proofs are The • next theorem li sts these and some further properties gtven as Exercises 53-56).

Note an important distinction between the dot product and cross product of a vector

THEOREM 2 Basic Properties of the Cross Product (i)

with itself: VXV=O V·V=\\V\\ 2

(ii)

W X V

= -V X W

V XV=

(iii) v x (iv) (AV)

W X

0

= 0 if and only if either w = Av for some scalar A or v = 0 w=

(v) (u + v)

X

V X

(AW)= A(V

w =u

u X (v + w)

X

w+V

X

X

w)

w

=u X V + u X w

. The cross product of any two of the standard basis vectors i, j, and k is equal to the third, possibly with a minus sign. More precisely (see Exercise 57), i

X

j

= k,

j Xi= -k,

j

X

k = i,

k

X

j = -i,

k

X

i= j

0

i

X

k = -j

IT]

Furthermore, we also have:

k

ixi=jxj=kxk=O FIGURE 6 Circle for computing the cross products of the basis vectors.

'2

An easy way to remember the relations (5) and (6) is to draw i, j, and k in a circle as in Figure 6. Starting at any vector, go around the circle in the clockwise direction and you obtain one of the relations (5). For example, starting at i and moving clockwise yields i x j = k. If you go around in the counterclockwise direction, you obtain the relations (6). Thus, starting at k and going counterclockwise gives the relation k x j = -i.

EXAMPLE 4 Using the ijk Relations Compute (2i + k) x (3j + 5k). Solution We use the properties of the cross product: (2i + k) x (3j + 5k)

= (2i) x (3j) + (2i) x (5k) + k x (3j) + k x (5k) = 6(i x j) + l0(i x k) + 3(k x j) + 5(k x k) = 6k- l0j - 3i + 5(0) = -3i - l0j + 6k

FIGURE 7

The cross product can be used to demonstrate a significant property of the largescale motion in the earth's atmosphere. Meteorologists study the properties and motion of small volumes (called parcels) of air to help forecast the weather and understand our climate. The rotation of the earth impacts the movement of air parcels via what is known as the Coriolis force. For a parcel of mass m with velocity v, the Coriolis force is Fc = - 2m Q x v, where Q is the angular velocity of the rotating earth (Figure 7). The vector Q is parallel to the rotation axis of the earth and has a magnitude of approximately 7.3 x 10- 5 s- 1, reflecting that the earth completes one rotation through 2rr radians in a day. The next example demonstrates that in the northern hemisphere, the impact of the Coriolis force increases with increasing latitude.

EXAMPLE 5 The Coriolis Force in Meteorology We consider three parcels of air with mass 2 kg, moving directly north at 20 mis, one at the equator, at a latitude of 45° north, and at a latitude of 80° north. The parcel velocities have the same magnitude and the same direction relative to the surface of the earth, but not the same direction relative to the earth's axis (Figure 7). In Figure 8 we illustrate that the angle that each velocity vector makes with a line parallel to the earth's axis (and therefore with Q) is equal to the latitude. Find the magnitude of the Coriolis force for each parcel.

704

CHAPTER 12

VECTOR GEOMETRY

. . at the different latitudes veloc1nes . Solution Let v0 , v45 , and vso represen _ and therefore IIFcll = 0. 0 · that R x vo 1N • At the equator, vo!IR, imply1Dg . (45°) 0.004 • t the parceI

• At latitude 45°, we have /IFell • At latitude 80°' we have

liFC II

_ miiRIIIIV45II slD - 2 . (BOo)

=

2mi1Rllllvsoll slD

.

0.0058 N. '

currents so that they te . lis force steers a1f . nd In Section 14 5 we will show how the Cono fl w directly mto them. . · ' therthan to crrculate around low-pressure systems ra forces applied to a wrench to turn EXAMPLE 6 Torque on a B~lt Figure 9 s~o;:t;~where Fis the force app~ed to the a bolt. The torque on the bolt 1s the ~ector "t th axis of the bolt to the pomt Where wrench, and r is a position vector, drrected from . ointing out of the page. Compute 1 1 the force is applied. In the figure, assume the z-ax ~ -50, 0) (both in newtons), and the torque in each case where Fi = (0, 60, 0), F2 ' ri = (0.5, 0, O), r2 = (0.3, O, 0) (both in meters).

°

y

X

~fo

Solution We have

"ti

• X

"tz FIGURE 9

I 1

= ri

x Fi

= 0.5i x

= r2 x F2 = 0.3i x

60j

= 30k N-m

(50i - 50j)

=-

I

lSk N-m

That "ti is in the positive z-direction indicates that a bolt wi~ righ~-hand thread turns upward out of the page with that combination of Fi and r1 • Snmlarly 1 the second case, the bolt turns into the page. Also note that, even though the force is greater in magnitude in the second case, the resulting torque is smaller in magnitude. This is due to the force in the second case being applied closer to the bolt and more obliquely than the force in the f"rrst case.

?

Cross Products, Area, and Volume Cross products and determinants are closely related to area and volume. Consider the parallelogram P spanned by nonzero vectors v and w with a common basepoint. In Figure IO(A), we see that P has baseb = llvll and height h = llwll sin 0, where 0 is the angle between v and w. Therefore, P has area A= bh = llvll llwll sin0 = llv x wll. z

X

(A) The area of the parallelogram p is llv x wll = llvll llwll sin 0.

X

(B) The area of the triangle Tis

y

llv x wll/2.

FIGURE 10

Notice also, as in Figure IO(B) that weals 0 kn . ' ow the are Of th · d by nonzer? vectors v and w 1s exactly half the area of a e triangle 7 spanne the parallelogram. Thus, we have the followmg:

r

I~

1,

The Cross Product

SECTION 12.4

705

Areas If 'Pis the parallelogram spanned by v and w, and 7 is the triangle spanned by v and w, then area('P) = llv x wll

and area(TI =

0

llv x wll 2

A"parallelepiped" is the solid spanned by

Itnree vectors. Each face is a parallelogram.

I If I

I

. R3

Next, consider the parallelepiped 1J spanned by three nonzero vectors u, v, w m (the three-dimensional prism in Figure 11). The base of 1J is the parallelogram spanned by v and w, so the area of the base is llv x wll. The height of 1J is h = llull · lcos01, where 0 is the angle between u and v x w. Therefore, volume of 1J = (area of base)(height) = llv x wll · llull · lcos 01 But, llv x wll llull cos 0 is equal to the dot product of v x w and u. This proves the formula volume of 1J = Ju· (v x w)I The quantity u . (v x w), called the scalar triple product, can be expressed as a determinant. Let

FIGURE 11 The volume of the parallelepiped is lu · (v x w)I.

We use the following liotation fer the determinant of the matrix whose rows are the vectors u, v, w:

Then

u • (v x w) = u •

(I

v2 W2

= u J Iv2

w2

It is awkward to write the absolute value of a determinant in the notation on the right, but we may denote it as

V31,1 -

I VJ

W3

V3 \ - u2

IVJ

W3

W3

w3

V3 \•+\VJ

WJ

w1

J

Wl

V3 I+ U3 \ VJ

WJ I

~!

= ~: WJ

W2

W3

[I]

= det (~) W

We obtain the following volume formula:

THEOREM 3 Volume via Scalar Triple Product and Determinants Let u, v, w be nonzero vectors in R3 . Then the parallelepiped 1J spanned by u, v, and w has volume

[I]

V - In (v x w)l - det (:)

EXAMPLE 7 Let v = (1,4, 5) and w = (-2, -1, 2). Calculate: (a) The area A of the parallelogram spanned by v and w (b) The volume V of the parallelepiped in Figure 12

w=(-2,-

Solution We compute the cross product and apply Theorem 3: VXW=

y X

FIGURE 12

4 51. 11 51. 11 41 2 I - _ 2 2 J+ _ 2 -l k=(l3,-12,7)

I-l

(a) The area of the parallelogram spanned by v and w is

A= llv x wll = J13 2 + (-12) 2 + 72 =

19

'I,

II

706

CHAPTER 12

VECTOR GEOMETRY

. d is the vector (b) The vertical leg of the parallelepipe

z

v=

vXw=Ak

1(6k) . (v

=

Parallelogram spanned by v and the xy-plane.

gH:: g/;+/::

J,

vxw=/:,

r

I

w,

1

I

~,k

:)I=

l 0 gram in Figure 14. EXAMPLE 8 Compute the area A of the paral e I

A

The third property in Theorem 1 is more subtle than the first two. It cannot be verified by algebra alone.

w,

w, wz O h area A= llv x wll , and thus By Eq. (7), the parallelogram spanned by v aod w as I 1v 1 w 2 - v2w1I

Solution We have/;/=

FIGURE 14

= 6(7) = 42

A of the paral1e

X

v=(l,4)

7) I

Iograrn spanned by vectors • R3 with a zero corn v In we can compute the area d w as vectors 1n Pone, ' din van d w - (w1 w 2 O) ·•1 (v1, vz} and w = (w1, wz} by regar g ·te v (vi, vz,O) an ' • • 1\ in the z-direction (Figure 13). Thus, we_ wnth z direction: . ting m e cross product v x w is a vector pom vz k = VJ

y

FIGURE 13

'

12 (0 0 6) . (13, - , x w)I = I ' '

R2

w in

6k so by Eq. (9),

= I - 101 = 10.

lj ~I

=

_ _ . 1 2 3 4

= _ JO. The area is t.L-:: absolute Value

•

Proofs of Cross-Product Properties · the first two properties · of the cross product listed in Theorem I. Let We now denve v

= (v 1, v2 , v3 ),

w

= (w1, w2, w3}

We prove that v x w is orthogonal to v by showing that v · (v x w) v (v x w)

det (:) ~"' ,~

: / - "> /:;

:,/

+ V3

= 0. By Eq. (8),

/:',

:~,

[ii]

I

Straightforward algebra (left to the reader) shows that the nght-hand side of Eq. (11) is equal to zero. This shows that v · (v x w) = 0 and thus v x w is orthogonal to v as claimed. Interchanging the roles of v and w, we conclude also that w x vis orthogonal tow, and since v x w = -w xv, it follows that v x w is orthogonal tow. This proves part (i) of Theorem I. To prove (ii), we shall use the following identity: 1/v x wl/

2

= llv/1 2 /lw/1 2 -

(v. w)2

To verify this identity, we compute II v x w /1 2 as the sum of the squares of the components ofv x w:

=

!Iv x wl/2

I

v2 w2

= (v2w3

v312 +Iv, WI

W3

- v3w2)2

I

v312 + v1

W3

+ (v1w3

wI

v212

W2

- v3w,)2

+ (v1w2

- v2w1)2

@]

On the other hand, 2 2 l/v/1 1/wl/ - (v · w)2 = (vf

+ v? + v})(wf + w 22 + w2) 3

_ (v I w I

+ V2W2 + V3W3 )2

CE] Again, algebra (left to the reader) shows that the right side of E . · bl side of Eq. (14), proving Eq. (12). q. (13) rs equal to the ng

Th e cross Product

SECTION 12.4

707

Now let 0 be the angle between v and w. By Eq. (12),

= llvu2 Uwll 2 - llv11 2 llwll 2 2 2 cos2 0) = llv11 llwll sin 0

llv x wll 2 = llv11 2llwll 2 - (v. w)2

2

2

cos 0

= llv11 2 llwll 2(1 Therefore, llv x wll = llvllllwll sin 0. Note that sin0::: 0 since, by convention, 0 lies between Oand rr . This proves (ii).

12.4 SUMMARY • Determinants of sizes 2 x 2 and 3 x 3:

I

a,, aiz\ = a11a22 - a12a21 \a21 a22 au

a21 a31

a12 a13 a22 a23 a32 a33

= a11 la22 a32

a231 a33

i

I

a12

-

la21 a31

a231+a 13 1a21 a33 a31

a221 a32

• The cross product of v = (v 1, v2, v3) and w = (w1, w2, w3} is the determinant

vx w=

i

1

Wi

2 W2

:3

W3

= \w2 v2

w,

v3 Ii - Iv, W3

I

v3 \ • + v1 W3 J Wj

v2 \ k w2

• The cross product v x w is the unique vector with the following three properties: (i) v x w is orthogonal to v and w. (ii) v x w has length llvll llwll sine (where 0 is the angle between v and w). (iii) {v, w, v x w} is a right-handed system.

• Properties of the cross product: (i) W X V = -V X W (ii) v x w = 0 if and only ifw (iii) (AV)

X

= J-.v for some scalar or v = 0

w = V X (J-.w) = J-.(v X w) X w = u X w + V X w and

(iv) (u + v)

V X

(u + w) =

V X

u+

• Cross products of standard basis vectors (Figure 15): k

FIGURE 15 Circle for computing the cross products of the basis vectors.

ixj=k, j Xi= -k, i Xi= 0,

jxk=i, k xj

j

X

= -i,

j = 0,

k Xi =j i

X

k = -j

kxk=O

• The parallelogram spanned by v and w has area \Iv x wll• The triangle spanned by v and w has area \Iv x w\l _ 2 • Cross-product identity: llv x w\1 2 = \lv\1 2 iiwil 2 - (v. w)2. • The scalar triple product is defined by u • (v x w). We have

• The parallelepiped spanned by u, v, and w has volume iu. (v x w)I.

V X

w

l

708

CHAPTER

12

VECTOR GEOMETRY

~~E~~~~======================-=-=~ -1 ~~-4 EXERCISES

-

Preliminary Questions 1.

Whatisthe(l,3)minorofthematrixl-!

!I?

;· x -::e angle between two unit vectors e and f is

3. 4. (a)

s. 6.

What is u x w, assuming that w x _u

=

I. What is the length of

7.

"

. gful and which are not? ExpJ . . are meanm a,n Which of the fo)lowmg .

(a) (u. v)

xw

(b) (u xv) · W

(c) llwll(u · v)

/

(d) llwll(u x v)

(2, 2, I)?

Find the cross product without using the formula: (4,8,2) x (4,8,2) (b) (4,8,2) x (2,4, I) What are i x j and i x k?

8.

· equal toj xi ? Which of the foJlowing vectors is

(a) ix k (b) -k

When is the cross product v x w equal to zero?

(c) i X j

~Exercises - : - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -· In Exercises 1-4, calculate the 2 x 2 determinant.

1.

3.

I! ;I

1-~ ii

2.

j_!

4.

I~

24. (v + w) x (3u + 2v)

23. (u _ 2v) x (u + 2v)

!I

25. Let v

=

(a, b, c). Calculate v x

=

251 14

z

In Exercises 5~. calculate the 3 x 3 determinant.

l1

lj

7.

i

2 -3 0

s.

2 4 -4

6.

~I

0 0 3

1-i

8.

Jg

• v x J. • and v x k .

I,

and w are vectors of length 3 in the xz-plane 26. Find v x w, w h ere v " , oriented as in Figure 16, and 0 6·

0 0 I

jl -!I y

In Exercises 9-14, calculate v x w.

9.

V

10.

V

11.

V

= (1,2, 1), = (2,0,0), = (i, 1, ½),

= (3, 1, 1) w = (-1,0, 1) w = (4,-6,3) w

X

FIGURE 16

12. v = (I, 1, 0),

w = (0, 1, 1)

In Exercises 27 and 28, refer to Figure 17.

13.

V

= (1, 2, 3),

w = (1,2,3.01)

27. Which ofu and -u is equal to v x w?

14.

V

=

(2.4, -J.25, 3),

W=

(-7.68,4,-9.6)

28. Which of the following form a right-handed system?

In Exercises 15-18, use the relations in Eqs. (5) and (6) to calculate the cross product. 1S. (i

+ j) x

(a) {v, w, u}

(b) {w, v, u}

(d) {u, v, w}

(e)

{w, v, -u}

k

16. (j - k) x (j

+

k)

17. (i - 3j + 2k) x (j - k) 18. (2i - 3j + 4k) x (i + j - 7k) In Exercises 19-24, calculate the cross product assuming that uxv=(l,1,0),

19. vxu

21. w x (u + v)

u x w = (0, 3, l),

V X W

w

= (2, -1 , 1)

20. v x (u + v) 22. (3u+4w) x w

FIGU RE 17

(c)

{v, u, w}

(f)

{v, -u, w}

SE cTI o N 12.4

The Cross Product

709

1.,et v = (3,0,0) and w = (0, I, -1). Determine u . th= v x w usmg the an e fonnula .

z9. (lletric properties of the cross product rather th

geO

•n wi,at are the possible angles 0 between tw

,.,.

_ !?

tie)( fll -

31 ,

2·

.

o urut vectors e and f if

Show that ifv and w lie in the yz-plane then v . . • • x w 1s a mu 1ltple of 1.

32, find the two unit vectors orthogonal to both (-1.2,1).

3

u

= (3, l, I) and b =

= (I, 0, 4)

e' be unit vectors in R3 such that e .l e' U th . 33' J.,et e and f th • se e geometnc properties o e cross product to compute e x (e' x e). X

34, Dete~ne the magnitude of each Coriolis force on a 1.5-kg parcel of aifwith wmd v. (a) v is 25 mis toward the east at the equator

FIGURE 19

(b) vis 25 mis toward the east at 45° N

42. Calculate the volume of the parallelepiped spanned by

(c) vis 35 mis toward the south at 30° N

(d) vis 35 mis toward the south at 60° N

U=

(2,2,1),

v=(l, 0,3),

w= (0,-4,0)

35, Determine the magnitud-o of each Coriolis force on a 2.3-kg parcel of air with wind v. (a) vis 20 mis towarcottf>

In Exercises 65-72, find an equation of the form p = f(0,t/>) in spherical coordinates for the following surfaces. 67, z = 2 66. X = 3 65. x 2 + y2 =9

2 , z =4

27, z ::: x + Y

s 2,

62. p = 1,

64. p

- 4

4

26. r 3, rr S 0

p

63.

Z
61.

22. z = r

7r

0=

y

3

r 3, 0 S z 4

,r

7f

0

c,ercises 17-26, sketch the set (described in cyli'ndnca • 1coo~d'mates). [n.,,..

,I

56. t/>=

55. p=4 0, z

733

. 5c u '·-tch the set of poinJs (dtscribtd in spherical In Exerrises J-v (b) p sint/> = R (c) P = R smtf>

32. z = 3xy

x2+l = 4

In Exercises 33-38, convert from spherical to rectangular coordinates.

33.

(3,0, i)

35. (3, rr, 0)

/

rr 5rr) 37. ( 6,6'6

3rr rr)

36. ( 5,4'1

38. (0.5, 3.7, 2)

/n Exercises 39-44, convert from rectangular to spherical coordinates. 39,

m.o,

y

./3 3 ) 40. ( 2·2·1

1)

42. (I, -1, 1)

41. (I, 1, 1) 1

J3

)

44.

43. ( 2·2·./3

-./2 -./2 ) ( 2'2'./3

In Exercises 45 and 46, convert from cylindrical to spherical coordinates.

45. (2, 0, 2)

46. (3, rr, ./3)

In Exercises 47 and 48, convert from spherical to cylindrical coordinates. 41.

48. (2,f,U

(4.o,f)

In Exercises 49-54, describe the given set in spherical coordinates. x2

+ y2 + 100 CSO. x2 + y2 + z2 = 1, z 0 51. x2 +y2 + z2 = 10, x o, 52, x2 + y2 + z2 1, x = y, 53, y2 +z2 4, X = 0

L

49.

54.

x2

z2

+ y2 = 3z2

y x

O, z 0, y

0 0

FIGURE 17

74. Let P1 =(1,-./3,5) and Pz=(-1,./3,5) in rectangular coordinates. In which quadrants do the projections of P1 and P2 onto the xy-plane lie? Find the polar angle 0 of each point. 75. Determine the spherical angles (0, cp) for the cities Helsinki, Finland (60.1 ° N, 25.0° E), and Sao Paulo, Brazil (23.52° 46.52° W).

s:

76. Find the longitude and latitude for the points on the globe with angular coordinates (0, 0 (see Exercise 71). Observe that -N pomts to _the outside of C. Fork> 0, the curve c1 defmed by r1 (s) = r(s) - kN 1s called the expansion of c(s) in the normal direction.

= llr'(s)II +kK(s).

(b) A~ P moves around ~e oval counterclockwise, 0 increases by

2rr [Figure 22(A)]. Use this and a change of variables to prove that

f

K(s)ds

IC

(cos,cs)N)

K

is an arc length parametrization of the osculating circle at r(ro).

r (so) = r2(so), 1

r'1(so) = r'2(so), r'{(so) = r'~(so)

Let r(s) be an arc length parametrization of a curve C, and let P be the terminal point of r(O). Let y(s) be the arc length parametrization of the osculating circle given in Exercise 80. Show that r(s) and y(s) agree to order 2 at s = O(in fact, the osculating circle is the unique circle that ap· proximates C to order 2 al P). 82. Let r(r) = (x(t), y(r), z(t)) be a path with curvature ,c(r) and define the scaled path r 1(t) = (>.x(t), >..y(t), >..z(r)), where>.. i= 0 is a constant. Prove that curvature varies inversely with the scale factor. That is, prove that the curvature Kl (r) of r1(1) is Kl (1) = >..- 1,c(t). This explains why the curva• ture of a circle of radius R is proportional to 1/ R (in fact, it is equal to I/ R). Hint: Use Eq. (3).

86. Follow steps (a)-(c) to prove that there is a number -r called the torsion such that dB -=-,N ds

G

dB

84. Let r(s) be an arc )P,ngth parametrization of a closed curve

(a) Show that \lr'1(s)\I

= r(ro) + ~N + ~((sinKs)T -

2 al so if

curvature

K(0)

y(s)

81. Two vector-valued functions r 1(s) and r2(s) are said to agree to order

1J se the parametrization r(0) = (f (0) cos 0 8 . 74. :Ver :c f(0) in polar coordinates has 'f( ) sm0) to show that a C"" •

711

80. Show that

d0

1]se the arc length integral to show th t ds

Curvature

79. Show that both r'(t) and r''(t) lie in the osculating plane for n vector function r(t). Hint: Differentiate r'(r) = u(r)T(r).

J"(x) •-=---dx + f'(x)2) ·

(b) . a dx (c) )'low give a proof ofEq. (6) using Eq. (l3).

13.4

(a) Show that -

ds

= T x dN ds

. and conclude that dB/ds 1s orthogonal

toT. (b) Differentiate B •B = 1 with respectto s to show thatdB/ds is orthogonal to B. (c) Conclude that dB/ds is a multiple ofN.

87. Show that if C is contained in a plane P, then B is a unit vector normal to P. Conclude that -r = 0 for a plane curve. 88.

Torsion means twisting. ls this an appropriate name for ,? Explain by interpreting -r geometrically.

= 2rr.

(c) Show that C1 has length L + 2rr k.

89. Use the identity ax (b x c) =(a• c)b - (a. b)c

to prove

y

N xB=T,

BxT=N

[ill

90. Follow steps (a)-(b) to prove dN - =-KT+,B ds (A) An oval

(B) C1is the expansion of C

in the normal direction.

FIGURE 22 As P moves around the oval, 0 increases by 2rr.

ln Exercises 85-93, we investigate the binormal vector further. 85, Let r(r) = (x(t), y(t), O). Assuming that T(t) x N(t) is nonzero, there arc two possibilities for the vector B(t). What are they? Explain.

(a) Show that dN/ds is orthogonal to N. Conclude that dN/ds lies in

the plane sparmed by T and B, and hence, dN/ds = aT scalars a, b.

+ bB for some

= 0 to show that T • dN = -N . dT and compute a. Comb . . I ds ds pute sum1ar y. Equations (15) and (17) together with dT/dt = KN are called the Frenet formulas.

(b) Use N · T

91. Show that r' x r" is a multiple of B. Conclude that

r'' 1\r' X r''I\ r'

X

B=---

1 712 CHAPTER 13

. CALCULUS OF VECTOR -VALUED FUNCTIONS

d using N == B x _ r [Eq. 06)] w· Ncan be computed Nin the followmg cases: 1th~ 93, 'Jbe vec;~ this ,nethod to fin ,'

92. Use the i

I fro curve . onnu a m the preceding problem 10 find B for the space rics. give) n by r(t) = (sin,, -cost, sint). Conclude that the space curve in

in Eq. (IS). s 2) all == 0 -(cos/,/,1 (a) r(/)- 2 -I ,)all:::: I

a pane.

(b) r(I):::: (t ,I

•

13.5 Motion in 3-Space

.ng along a path r(t). "'1. ob·ect trave1J •11a1 o~ In this section, we study the_moti~n i°~g a ~particle,_ a baseball, or comet Ba!e.e0~ ject could be a variety of thmgs, inc u . the derivauve 'ty vector JS (Figure I). Recall that the ve1oci h) _ r(t) v(t)

=r'(t) =}~

h

.

. (if it is nonzero), and its magnjh, . tion of mouon . th d •u~ d As we have seen v(t) points in the irec tion vector JS e secon denv..· ' d The acceIera "'IVe v(t) = llv(t)II is the object's spee · from r(t), we have r''(t), which we shall denote as a(t). In summary, v(t) = r'(t),

v(t)

a(t) == r" (t)

= IIV(t)II,

. d celeration vectors at t EXAMPLE 1 Calculate and plot the velocity an (Figure 2). (sin 2t, - cos 2t, ,Jt+T). Then find the speed at t - 1

= I of r(t)

Solution FIGURE 1 The trajectory of a comet is analyzed using vector calculus.

v(t) = r'(t) = (2cos2t,2sin2t, ~(t +

0-

a(t) = r''(t) = (-4sin2t,4cos2t, -i(t + The speed at t

112

),

0-

32 / ),

v(l)

(-0.83, 1 :n ,0.35)

a(l)

(-3.64, ·- L66, -0.089)

= I is J(-0.83)2 +(l.82)2 + (0.35)2

llv(l)II

2.03

I

If an object's acceleration is given, we can solve for v(t) and r(t) by integrating twice:

j

v(t) =

a(t)dt and then r(t) =

j

v(t)dt

Arbitrary constants arise in each of these antiderivatives. To determine specific functions v(t) and r(t), initial conditions need to be provided. X

FIGURE 2

EXAMPLE 2 Find r(t) if

a(t) = (cos t)i + e1j + tk,

Solution We obtain v(t)

CAUTION While the initial condition determines the constant Co that arises in the antiderivative, the constant is not necessarily equal to the value in the initial condition. To find Co, we need to find two expressions for v(O) and use them to solve forC0 .

=

j

r(O) = k

v(O) = i,

a(t)dt = (sint)i + e'j +

f

IT]

k + Co

~ith the initial condition for v, we can determine Co. Specifically, from the initial cond1t1on and from Eq. (1) we have, respectively, the following two expressions for v(0): v(O) = i and v(O) = j + Co It follows that i =j + Co, and therefore Co

=i - j. Thus,

2

. )' ,. t k . r2 ( ) = (smt1+eJ+ vt 2 +1-j=(sint+l)i+(e1-l)j+-k

2

I

Motion in 3-Space

SECTION 13.5

Now, talcing another antiderivative, we obtain 13

r(t) = / v(t) dt = (-cost + t)i + (e' - t)j +

6k +

Ct

773

0

The initial condition and Eq. (2) provide the following two expressions for r(O): r(O)

=k

and

r(O)

= -i +j + Ct

This implies that k = -i + j +Ct, and therefore c 1 = i - j + k. We now have r(t)

= (-cost+ t)i + (e 1 -

= (-cost + t +

3

t)j + ~k + i -j + k

l)i + (e' - t - I)j +

c:

•

+ 1) k

Near the surface of the earth, gravity imparts an acceleration of approximately 32 ft/s 2 in the downward direction. This means that if we have a projectile moving near the surface of the earth that has no additional means of acquiring acceleration, we know that its acceleration vector a(t) is determined by gravity. When a projectile's motion occurs within a vertical plane, we can model the motion in the xy-plane, using x for the horizontal motion and y for the vertical. In that case, we use

9 .s mls 2

a(t) = -9.8j m/s 2 or a(t) = -32j ft!s 2 . E>L~MPLE 3 A projectile is launched from the ground at angle 0 with a speed of

100 ft/s. Show that the projectile lands at a distance of 625 sin 0 cos 0 ft from the launch point. (This launch-to-landing distance is called the range of the projectile.)

on the moon, the accel:erati on due to · gravity is 1.6 rn/s 2. Du.TU1g l1is moonwalk, Apollo 14 astronaut .dJz~, Shepard used a makeshift club to hit a golf ball. It went "miles and miles and miles," he said. In Exercise 21, we explore how far it might have gone.

Solution Assume that the launch point is at the origin, and let r(t) be the position vector. F~r an initial condition, we have r(O) = 0. Furthermore, since the launch speed is 100 ft/s with an angle of 0, we have the initial condition v(O) = 100 cos 0i + 100 sin 0j (Figure 3). . We_ assume that we have an acceleration vector a(t) = -32j. We determine r(t) by

mtegratmg twice: v(t)

=

f

a(t)dt

= -32tj +

0

Co

From the initial condition and from Eq. (3), we have

y

v(O)

100 ft/sec 100 sin 0

FIGURE 3 The launch velocity is v(O)= I00cos0i+ I00sin0j.

100 sin 0j

and

v(O)

= Co

Hence, Co = 100 cos 0i + 100 sin 0j. Therefore, v(t)

0

"'--'-------i--x 100 cos 0

= 100 cos 0i +

= 100 cos 0i +

(100 sin 0 - 32t)j

Integrating again:

f

v(t)dt = (100cos0)ti + ((100sin0)t - 16t )j

+ c1

Eq. (4) implies th at r(O) = C1 • From the initial condition r(O) = Therefore,

o we

r(t) =

2

r(t)

= (100 cos 0)ti + ((100 sin 0)t -

•

8J tC _ 0 ge I •

16t 2 )j

Now, the projectile is at ground level when the y-component of r(t) . Th fore, we solve is zero. ere(l OO sin 0)t - 16t 2

=O

t(100sin0 - 16t)

=o

774

CHAPTER

13 CALCULUS OF VECTOR-VALUED FUNCTIONS

:;a~:mp~e 3 in Section 3.6, we showed th1s case, the maximum range c:c~t'!rs when (J = :rc/4. In fact, given any n! ia1 Speed, the maximum range occurs wi th 9 :rr /4; see Exercise 23.

=

. launched) and at t ::::: ~ n . ·ect:Jle was . I~ We have solutions at r = O (when the proJ The distance from th e_launch Po· ::,. (when the projectile returns to the groun~~-ated at the rime of landmg. That i:t\t !I) the landing point is the x-component of r(t) ev . ft from the launch point. ' 1'1e . . (25sin0)-625cos0sin 0 proJectile lands at (100cos0) ' velocity with respect to ti"' Of · two or·••eth. l In general, acceleration is the rate O f change . stant. By contras t , m "f th speed is con b" t' . constq"" t•. linear motion, acceleration is zero 1 e when the o ~ec s speed 1s dimensions, the acceleration can be nonzero eventhe direction of v(t) is changing, ~t. This happens when v(t) = Uv(t)II is const~t bu~ which an object travels in a circ e 11litt simplest example is uniform circular motion, ID path at constant speed (Figure 4).

FIGURE 4 In uniform circular motion, v has constant !eng~ but turns continuously. The acceleration a 1s centripetal, pointing toward the center of the circle.

. . t) and lla(t)II for the motion of a p . EXAMPLE 4 Uniform Circular Motion Fmd a( !lrt1. cle around a circle of radius R with constant speed v. · - ular path r(t) = R(cos wt, sinwr) f Solution Assume that the particle follows the crrc . Or . d f the partic1e are 0 some constant w. Then the velocity and spee

v(t) The constant w (lowercase Greek omega) is called the angular speed because the particle's angl~ along the circle changes at a rate of w radians per unit time.

I

Thus, /w/

=

v

Rw(-sinwt,coswt),

=

//v(t)II

= Rlwl

= v/ R; accordingly,

a(t)

= v'(t) =

11a(t)ll=Rw2=R

-Rw2 (coswt,sinwt),

V \ l

(

R)

v2

=R

The vector a(t) is called the centripetal acceleration: It has length v IR and points toward the center of the circle, in this case the origin [because a(t) is a negative multiple of the position vector r(t )], as in Figure 4. 1 2

Understanding the Acceleration Vector Acceleration is the rate of change of velocity, and a velocity vector provides information about the direction of motion and the speed (via its magnitude). Thus acceleration can involve change in either the direction or the magnitude of velocity. To understand how the acceleration vector a(t) encodes both types of change, we decompose a(t) into a sum of tangential and normal components. Recall the definition of unit tangent and unit normal vectors:

Tt -

N t -

v(t)

( )- /Mt)II'

Thus, v(t) When you make a left turn in an automobile at constant speed, your tangential acceleration is zero [because v'(t) = OJ and you will not be pushed back against your seat. But the car seat (via friction) pushes you to the left toward the car door, causing you to accelerate in the normal direction. Due to inertia, you feel as if you are being pushed to the right toward the passenger's seat. This force is proportional to Kv 2 , so a sharp turn (large K) or high speed (large v) produces a strong normal force.

= v(t)T(t), where v(t) = a(t)

=

dv dt

=

T'(t)

() - IIT'(t)II

llv(t)I/, so by the Scalar Product Rule,

d

dt v(t)T(t)

=

v'(t)T(t)

+ v(t)T'(t)

Furthermore, T'~t) = v(t)K(t)N(t) by Eq. (8) of Section 13.4, where ,c(t) is the curvature. Thus, we can wnte

/ a

=

aTT

+ aN N,

aT

=

v'(t),

aN

The coefficient aT(t) is called the tangential component and a (t) th

· (F'1gure 5). nent o f acce1erat10n

0

= ,c(t)v(t)2 N

e norm

al cornpO"

Motion in 3-Space

SECTIO N 13,5

ID The normal component aN is often called the centripetal acceleration. In the case of uniform circular motion it is directed toward the center of the circle.

FIGURE 5

775

Decomposition of a into tangential and normal components.

CONCEPTUAL INSIGHT The tangential component ar = v'(t) is the rate at which speed 2 v(t) changes, whereas the normal component aN = K(t)v(t) describes the change in V due to a change in direction. These interpretations become clear once we consider the

following extreme cases: • A particle travels in a straight line. Then direction does not change [K(t) = O] and a(t) = v'(t)T is parallel to the direction of motion. • A particle travels with constant speed along a curved path. Then v'(t) = 0 and the acceleration vector a(t) = K(t)v(t) 2N is normal to the direction of motion. General motion combines both tangential and normal acceleration. EXAMPLE 5 The Giant Ferris Wheel in Vienna has radius R = 30 m (Figure 6). Assume that at time t = to, a person in a seat at the bottom of the wheel has a speed of 40 m/min that is slowing at a rate of 15 m/min2 • Find the acceleration vector a for the person.

Solution At the bottom of the wheel, T = (1,0) and N = (0, 1). We are told that ar = v' = -15 at time to. The curvature of the wheel is K = 1/ R = 1/30, so the normal component is aN = Kv 2 = v2/ R = (40) 2/30 53.3. Therefore (Figure 7), a~ -15T + 53.3N = (-15,53.3) m/min The Giant Ferris Wheel in Vienna, Austria, erected in 1897 to celebrate the 50th anniversary of the coronation of Emperor Franz Joseph I.

FIGURE 6

2

The following theorem provides useful formulas for the tangential and normal components.

THEOREM 1 Tangential and Normal Components of Acceleration In the decomposition a = arT + aNN, we have

y

Ferris wheel

a-v ar=a•T=-, llv\l

aN =a· N =

-

larl 2

and arT= ( -a-v) v, V·V

aNN=a-arT=a- ( -a-v) v V·V

FIGURE 7

Proof To begin, note that T • T = 1 and N . T = 0. Thus, a· T = (arT + aNN) • T = ar a· N = (arT + aNN) • N = aN .

V

an d smce T =-,we have llvll arT=(a-T)T=

a. V) M V = (a. V) v (M

IT]

i1

I 776

CHAPTER

13 CALCULUS OF VECTOR-VALUED FUNCTIONS

and aN

-a-arT=a-(S)v N. ht triangle w1"th h ypotenuse ·des of a ng a as .

Figure 5, so by the Pythagorean Theorem, 2 aN = ...;11all2 - laTI 2 Half= laTl + laNI . .. egative, depending on whethe . - osittve or n r th Keep in mind that aN 2: 0 but aT is P urve. e object is speeding up or slowing down along th e c . e the acceleration a(t). At t = ½, deco EXAMPLE 6 For r(t) = (t 2 , 2t, Int), detemun al components, and find the curvltt, pose the acceleration vector into tangential and nonn a, ture of the path (Figure 8). W h ts T and aT- e ave Solution First we compute the tangential componen , - r"(t) = (2, 0, -t-2) 1 v(t) = r'(t) = (2t, 2, t- ), a(t) -

z y

X

At t

= ½,

N

v

a=-2T+4N

llIJ

t = r(t)

FIGURE 8

The vectors T, N, and a at

½on the curve given by

=

=

r'

a= r"

(½) = ( (½) (½)-') = (½) = (2,0, - (½)-2) = ( ,2,

2

2 , O, - 4 )

Thus,

(t 2 ,2t,Int).

(l,2,2)

-(.!.~~) - 3'3'3

(1,2,2)

V

T=M= and by Eq. (6),

(31, 32, 32) = 122) (8 4 8) aNN=a-aTT=(2,0,-4)-(-2) (3'3'3 = 3'3'-3 aT

=a• T

=

2

(2, 0, -4) ·

Next, we use Eq. (7):

Summary of steps in Example 6:

This vector has length

V

T=M aT=a-T

aN

16 64 = llaNNII =----'- + - + - = 4

9

aNN=a-aTT aN

=

llaNNII aNN

N=-aN

9

9

and thus,

N = aNN = aN

(i, 1• -i) = (~ 4

.!_ -~)

3' 3'

3

Finally, we obtain the decomposition

a=

(2, 0, -4)

= aTT + aNN =

-2T

+ 4N

Now, since aN = 4 at t = ½, curvature at t = ½divide 4 by the square of the speed. With v have v 2 = 9, and therefore K(l/2) = 4/9.

and we know aN = Kv 2 from Eq. (5), to obtain the

=

(1, 2 , 2) at t

= ½, we •

EXAMPLE 7 Nonuniform Circular Motion Figure 9 shows the acceleration vectors of three particles moving counterclockwise around a circle. In each case, state whether the particle's speed v around the circle is increasing, decreasing, or momentarily constant.

5

EcT I oN 13.5

Motion in 3-Space

777

RE g Acceleration vectors of particles flGU ·og counterclockwise (in the directio

. 0v1 f1l 'f) around a circle. of

n

(C)

(B)

(A)

Solution The rate of change of speed depends on the angle 0 between a and T: v'

=

aT

=a• T = llall IITII cos0 = llall cos(}

Here, the first equality follows from Eq. (5), the second from Eq. (6), the third from the geometric interpretation of the dot product, and the last since Tis a unit vector. • In (A), 0 is obtuse, so cos e < oand v' < 0. The particle's speed is decreasing. • In (B), 0 = so cos0 = 0 and v' = 0. The particle's speed is momentarily constant. • In (C), 0 is acute, so cos0 > Oand v' > 0. The particle's speed is increasing.

1,

13.5 SUMMARY • For an object whose path is described by a vector-valued function r(t),

v(t) = llv(t)II,

v(t) = r'(t),

a(t) = r"(t)

• The velocity vector v(t) points in the direction of motion. Its length v(t) = l\v(t)I\ is the object's speed. • The acceleration vector a is the sum of a tangential component (reflecting change in speed along the path) and a normal component (reflecting change in direction): a(t)

= llT(t)T(t) + llN(t)N(t)

Unit tangent vector Unit normal vector Tangential component

T

v(t) (t) = l\v(t)I\

Nt -

T'(t) () - I\T'(t)I\ , a•v aT = v (t) =a• T = -

(a•V) V·V

UTT= Normal component

aN

I\VII

V

= K(t)v(t)2 =

UN N

=a-

UTT

\\al\2 - \uT\2

=a-

v (a•V) V·V

13.5 EXERCISES Preliminary Questions 1. If a particle travels with constant speed, must its acceleration vector be zero? Explain. 2· For a particle in unifonn circular motion around a circle, which of the vectors v(1) or a(r) always points toward the center of the circle?

3. Two ob~ects travel to the right along the parabola y = x2 with nonzero speed. Which of the following statements must be true? (a) Their velocity vectors point in the same direction. (b) Their velocity vectors have the same length.

(c) Their acceleration vectors point in the same direction.

778

CHAPTER

13 CALCULUS OF VECTOR-VALUED FUNCTIONS

4. lJsc the deco mposition of acccl . . ponenis to explain the foll . eratJon into tangential and normal comthe acceleration and velocityo~ng staremco~ If the speed is constant. !hen cctors are onhogonal S.lfapam1 . • c e travels along a 1ocny Vectors arc (choose the slrai g h t li "~• ~en lhe acceleration and ve(a) orthogonal . correct descnpbon): (b) parallel.

--tion vector of a Panic) th f the aceel ~ u e Ira Whal is the Jeng . 0 .th constant speed 4 cm/s? vei· . I f radius 2 cm w1 •~1 around a cue e o f ircular track. I • at a cena,· C . around a n Ill 7. Two cars are racing read 110 mph, then the two cars have tho~~ bolh of their speedometers e s~ (choose one): (b) ON 6.

(a) OT

EE;x;enrc:ii;s;e;s------'------------------------------~ I. Use the table to calculate the difference quotiems r(I + h) - r(I) h -0.2 -0 I O I 0 2 h for • · • · • · · Then estimate the velociiy and speed at , = I.

=

r(0.8) r(0.9) r(l.2)

16.

18. a(t)

)\5)

5. r(0)

=

(sin0,cos0,cos30),

6. r(s)

1 - t,

t

1

= (-- , ___:!___ ). 1 +s2 1 +s 2

=

s

1

= e'j -

cos(2t)k,

t

=0

=f

=2

stant speed of v = 4 emfs (see Example 4). Draw the path, and on it, draw the acceleration vector at t = !f.

8. Sketch the path r(I) = (1 - 1 2 , 1 - t) for -2 t 2, indicating the direction of motion. Draw the velocity and acceleration vectors at t = 0 and t = 1.

= (1 2 , 1 3 ) together with the velocity and accelera-

10. (00 Toe paths r(t) = (t 2 ,t 3 ) and r1(t) = (t 4 ,t6 ) trace the same curve,.,md r 1 (1) r(l). Do .you ~xpect either the velocity vectors or the acceleration vectors of these paths at t = I to point in the same direction? Explain. Compute these vectors and draw them on a single plot of the curve.

=

In Exercises 11-14,find v(t) given a(t) and the initial velocity.

11. a(t)=(t,4),

v(0)=(¾,-2)

12. a(t)

= (e', 0, t + I),

13. a(t)

= k,

v(O)

=i

v(O)

:= (I, -3, v'2} 14. a(t)

(

1

v(0)

= cos tk,

= i,

v(0)

=

(0, 0) r(0)

=

(2, I, l)

r(O) = j 1·

-

· J,

r(O)

=i

the ground at an arc· · of 45° un. 19 A ro;ectile is Jaunche d from . · "nat iQj • P ' • tile have in order to hit the t .• , • of a 120-ni 1 · tial speed must the proJeC ower located 180 m away?

23. Show that a projectile launched at an angle 0 with initial speed Vo travels a distance ( g) sin 20 before hitting the ground. Conclude that the maximum distance (for a given vo) is attained at 0 = ¾.

v5 /

24. Show that a projectile launched at an angle 0 will hit the top of an h-meter tower located d meters away if its initial speed is

7. Find a(I) for a particle moving around a circle of radius 8 cm at a con-

9. Sketch the path r(t) tion vectors at t = I.

=

22. Golfer Judy Robinson hit a golf ball on the plane' Priplanus with 1111 initial speed of 50 mis at an angle of 40° • ~t landed exactly 2 km away. What is the acceleration due to gravity on Priplanus ?

4. r(t) 0

r(0)

s\_u

In Exercises 3-6, calculate the velocity and acceleration vectors and the speed at the time indicated.

=

2)

21. Assume that astronaut Alan S~ep~d hit his golf on the moon (ac. celeration due to gravity = 1.6 mis ) with a m;dest m., · w l speed of 35 lli/s at an angle of 30°. How far did the ball travel•

FIGURE 10

3. r(t)

(3

20. Find the initial velocity vect?r vo of~ projectile :•·.leased With initial speed JOO mis that reaches a maximum height of 300 ., ..

OL.___)r(2)

4t 2 ),

C!tyQII,/

= (t , 4) , v(O)--:. ' - ' + I) v(0) = (J,0, 1), a(t) = e, 2 t, 1 •

17. a(I) = tk,

2. Draw the vectors r(2 + h) _ r( 2 ) and r(2 + h) - r(2) the path in Figure 10 D . h for h = 0.5 for · raw v(2 ) (using a rough estimate for its length).

(t 3 ,

r

position. 15. a(t)

(1.557, 2.459, -1.970) ( 1.559, 2.634, -1. 740) ( 1.540, 2.841, -1.443) ( l .499, 3.078, -1.035) ( 1.435, 3.342, 0.428)

r(I) r(I.I)

nd (t) and v(t) given a(t) and the initial ve/o .

Jn Exercises 15-18,ft

= t 2 k,

v(O)

=i-

j

VO=

../i72. d sec 0

25. A quarterback throws a football while standing at the very center of the field on the 50-yard line. The ball leaves his hand at a height of 5 ft and has initial velocity vo = 40i + 35j + 32k ft/s. Assume an acceleration of 32 ft/s 2 due to gravity and that the i vector points down the field toward the endzone and the j vector points to the sideline. The field is 150 ft in width and 300 ft in length. (a) Determine the positi,o n function that -gives the position of the ball I seconds after it is thrown. (b) The ball is caught by a player 5 ft above the ground. Is the player in bounds or out of bounds whenjle receives the ball? Assume the player is standing vertically with both toes on the ground at the time of receptiJl)I._

26. A soccer ball is kicked from ground level with (x, y )-coordinates (85.' 20) on th~ soc_cer field shown in Figure 11 and with an initial ve· loci1?' Vo = !Oi - 5J + 25k ft/s. Assume an acceleration of 32 ft/s 2 defl0 gravity and that the goal net has a height of 8 ft and a total width of 24 ft. (a) Determine the position function that gives the position of the ball 1 seconds after it is hit. (b) Does th e ball go in the goal before hitting the ground? Explain why or why not.

SECTION 13.5 y

44.

Motion in 3-Space

779

=

Let r(t) (12,41 - 3). Find T(t) and N(t), and sh~w that the decom. 'al d normal components 1s position of a(t) mto tangenb an a(t)= (

!05 ft

2t

Ji2+4

)T+(~)N Ji2+4

d f the acceleration vector of a particle 45. Find the components ar an aN R_ cm with constant speed moving along a circular path of radrns - 100

?

:e7:~

= 5 emfs. 46. In the notation of Example 5, find the acceler~tion vector for a

vo

!05 ft 165 ft

165 ft

seated in a car at (a) the highest point of the Fems wheel and (b) points level with the center of the wheel. ,

47. Suppose that the Ferris wheel in Example 5 is rotating clockwise that the point p at angle 450 has acceleration vector a (0, -50). m/rrun . · th peed and tangenbal compointing down, as in Figure 12. Dete~ne es ponent of the acceleration of the Fems wheel.

=

FIGURE 11

z7. A constant force F = (5, 2) (in newtons) acts on a 10-kg mass. Find we p0sition o~ ~e mas,s a'. t = IO seconds if it is located at the origin at . t "'0 and has irubal ve,oc1ty vo = (2, -3) (in meters per second).

zs. Aforce F = (241 , 16 - ;t)_(i~ ~ewtons) acts on a 4-kg mass. Find the pasitio~_ofthe ~ass at t s 1f 1t 1s located at (10, 12) at t = oand has

y

Ferris wheel

cc, .

zero irobal velocity.

·

29, Aparticle follow;l J.'Ui..'l r(t) for O:S t :S T, beginning at the origin O. T

= } lo ,: (t) dt is called the average velocity vector. Suppase that v =0. Answe,_:·~d-explain the following:

'[he vector v

(a) Where is the parti 0, the level curve is an ellipse. • For c = 0, the level curve is just the point (0, 0) because x 2 + 3y2 = 0 only for (x,y) = (0,0). • There is no level curve for c < 0 because f (x, y) is never negative. The graph of f(x, y) is an elliptic paraboloid (Figure 11). As we move away from the origin, f(x, y) increases more rapidly. The graph gets steeper, and the level curves become closer together.

l 794

CHAPTER 14

DIFFERENTIATION IN SEVERAL VARIABLES

: -.REMINDER The hyperbolic paraboloid .. Figure 12 is often called a "saddle" or saddle-shaped surface."

the contour map of 'd Sketch 01 I EXAMPLE 6 Hyperbolic Parabo 2 3 g(x,Y) == x2- Y .

(xy)==c,or

Solution The level curves have equation g '

x2- 3y2 == C 3y2 == c.

2

. th hyperbola X r,,3 b • For c :/= 0 the level curve is e Jines x == ±v :,y ecause the • For c - 0' the level curve consists of the two eiiua. tion g~. ;) = 0 factors as follows: - 0 X 2 -3y

2_ ( _

-

X

+

-

. .d (f' ure 12). When you stand at the orj . The graph of g(x, y) is a hyperbohc paraboloi . .igeither direction and decreases as gin, . al ng the. x-axis g(x, y) mcreases as you move o. F rthmrmore, the graph gets steeper as Yau move along the y-axis in either drrection. u e loser together. Yau move out from the origin, so the level curves grow c ' . Sketch the graph of EXAMPLE 7 Contour Map of a Linear Function f(x,y) == 12- 2x - 3y

X

g(x,y)

C

= 30

I

g(x,y)

increasing

_,: I

"' C

increasing

= -30

g(x,~ y

decreasing

FIGURE 12 g(x, y) interval m = IO.

4 and the associated contour map with contour interval m == · ·te the equation as 2x +3y + Solution Notethatifwesetz==f(x,y),wecanwn T ;-t th Z::: · · h uation of a p1ane. o p,u e graph 12. As we discussed in Section 12.5, this 1s t e eq . , We · (f' 13) The graph mterrepts the z-••; find the intercepts of the plane with the axes 1gure · . 1? _ ~sat z = f (0, 0) = 12. To find the x-intercept, we set Y == z .== 0 to obtam ,, 2x - 3'.0)::: 0, or x = 6. Similarly, solving 12 - 3y == 0 gives the y-mtercept Y == "'· The graph 1s the plane determined by the three intercepts. In general, the level curves of a linear function f (x, Y) == qx + r :' +s are the lines with equation qx +ry + s == c. Therefore, the contour map of a linear function consists of equally spaced parallel lines. In our case, the level curves are the lines 12 - 2x - 3y = c, or 2x +3y == 12 - c (Figure 13). 1 How can we measure steepness of the graph of a function quantitatively? Let's imag. ine the surface given by z == f(x, y) as a mountain (Figure 14). We place the xy-planeat sea level, so that f (a, b) is the height (also called altitude or elevation) of the mountain above sea level at the point (a, b) in the plane. Figure 14(A) shows two points P and Qin the xy-plane, together with the points P and Qon the graph that lie above them. We define the average rate of change:

= x 2 - 3y2. Contour

/1 altitude average rate of change from P to Q = - - - 11 horizontal

where /1 altitude from P to Q.

= change in the height from Pand Q, and /1 horizontal = distance

~ONCEPTUAL INSIGHT ~e ~ill discuss the idea that rates of change depend on direc· tJon when we come to drrectJonal derivatives in Section 14.5. In single-variable calcu· !us, we measure the rate o( change by the derivative J'(a). In the multivariable case, there is n~ single rate of change because the change in f (x, y) depends on the direction: The rate 1s z_ero along~ le~el c~rve [because f(x,y) is constant along level curves], and the rate 1s nonzero m d1rectJons pointing from one level curv t th t [Fi"ute (B)]. e o e nex " 14

r

SECTION 14.1

z

C::

20

C =:

16

-

Functions of Two or More Variables

795

12

A horizontal y

I I

I

I

I

I I I

I I I

V!

20 16 == 12 == 8 == 4 ==0

C =:

D

Function does not change along the level curve

-4

y

(Interval m == 4)

Contour interval: 0.8 km Horizontal scale: 2 km ...._,

X

A '"Too' B A

400

C

Contour interval: 100 m Horizontal scale: 200 m '-----'

(A)

(B)

FIGURE 13 Graph and contour

rnapof f(x,y)

= 12-2x -3y.

Acontour map is like a topographic map that hikers would use to help understand the terrain that they encounter. They are both two-dimensional repre:;entations of the features of three-dimensional structures.

FIGURE 14

EXAMPLE 8 Average Rate of Change Depends on Direction Compute the average rate of change from A to the points B, C, and D in Figure l 4(B).

Solution The contour interval in Figure 14(B) ism = 100 m. Segments AB and AC both span two level curves, so the change in altitude is 200 m in both cases. The horizontal scale shows that AB corresponds to a horizontal change of 200 m, and AC corresponds to a horizontal change of 400 m. On the other hand, there is no change in altitude from A to D. Therefore, !),, altitude average rate of change from A to B = - - - !),, horizontal

= -200 = 1.0 200

6. altitude average rate of change from A to C = - - - !),, horizontal

= -200 = 0.5

6. altitude average rate of change from A to D = - - - 6. horizontal

=O

400

We see here explicitly that the average rate varies according to the direction.

Apath of steepest descent is the same as a path of steepest ascent traversed in the opposite direction. Water flowing down a mountain approximately follows a path of steepest descent.

When we walk up a mountain, the incline at each moment depends on the path we choose. If we walk around the mountain, our altitude does not change at all. On the other hand, at each point there is a steepest direction in which the altitude increases most rapidly. On a contour map, the steepest direction is approximately the direction that takes us to the closest point on the next highest level curve [Figure 15(A)]. We say "approximately" because the terrain may vary between level curves. A path of steepest ascent is a path that begins at a point P and, everywhere along the way, points in the steepest direction. We can approximate the path of steepest ascent by drawing a sequence of

796

CHAPTER

14

DIFFERENT

IATION IN SEVERAL VARIABLES

l curve to the next. Fio-.,_ "ble from one Ie Ve .,,....e I segments that move as directly as pDSSI • th of steepest ascent, but th SUl) shows two paths from p to Q. The solid patb is a pal curve to the next along thee clhas1i•• 1eve · from one s path 1s not, because it does not move o¾"'1 possible segment.

Approximate path of steepest ascent starting at P (B)

(A) Vectors pointing approximately in the direction of steepest ascent FIGURE 15

More Than Two Variables

x2

+ y2 +

z2 = I

z x2 + y2 + z2 =4

1

y

There are many modeling situations where it is necessary to use a function of more thilll two variables. For instance, we might want to keep track of temperature at the VarioUs points in a room using a function T(x, y, z) that depends on the three variables corre. sponding to the coordinates of each point. In making quantitative models of the economy, functions often depend on more than 100 variables. Unfortunately, it is not possible to draw the graph of a function of more than two variables. The graph of a function f(x, y, z) would consist of the set of points (x, y, z, f(x, y, z)) in four-dimensional space R 4 . However, just as we can use contour maps to visualize a three-dimensional mountain using curves on a two-dimensional plane, it is possible to draw the level surfaces of a function of three variables f(x, y,z). These are the surfaces with equation f(x, y, z) = c for different values of c. For example, the level surfaces of f(x,y,z)

x2+y2+z2=9

FIGURE 16

f(x, y, z)

The level surfaces of

= x 2 + y 2 + z 2 are spheres.

= xz + Y2 + z2

are the spheres with equation x 2 + y 2 + z 2 = c (Figure 16). In the case of a function T (x, y, z) that represents temperature of points in space, we call the level surfaces corresponding to T (x, y, z) = k the isotherms. These are the collections of points, all of which have the same temperature k. For functions of four or more variables, we can no longer visualize the graph or the level surfaces. We must rely on intuition developed through the study of functions of two and three variables.

EXAMPLE 9 Describe the level surfaces of g(x, y, z)

+

= x 2 + y2 -

z2.

Solution The level surface for c = 0 is the cone x 2 y 2 - z 2 = o. For c =J: 0, the level surfaces are the hyperboloids x 2 + y 2 - z 2 c. The hyperboloid has one sheet if c > 0 and it lies outside the cone. The hyperboloid has two sheets if c < o, one sheet lies inside the upper part of the cone, the other lies inside the lower part (Figure 17). •

=

SE cTIO N 14.1

Functions of Two or More Variables

797

Cfl -~~:·

y y

,.,~: g(x,y, z)

=c

g(x, Y, z)

=O

+ y2 _ z . 2

y

,-

. ~-- -- -

(c> 0)

FIGURE 17 Level surfaces of g(x, y, z) :::::: x2

...

·;,('

,) ----

g(x,y, z) = c (c < 0)

14.1 SUMMARY • The domain V of a function f (x 1, ... , Xn) of n variables is the set of n-tuples (a1, ... ,an) in R" for which f(a 1, ... ,an) is defined. The range off is the set of values taken on by f. • The graph of a continuous real-valued function f(x, y) is the surface in R3 consisting of the points (a, b, f(a, b)) for (a, b) in the domain V off. • A vertical trace is a curve obtained by intersecting the graph with a vertical plane x == a or y == b. • A level curve is a curve in the xy-plane defined by an equation f(x, y) == c. The level curve f(x, y) == c is the projection onto the xy-plane of the horizontal trace curve, obtained by intersecting the graph with the horizontal plane z = c. • A contour map shows the level curves f(x, y) == c for equally spaced values of c. The spacing m is called the contour interval. • When reading a contour map, keep in mind: - Your altitude does not change.when you hike along a level curve. - Your altitude increases or decreases by m (the contour interval) when you hike from one level curve to the next. • The spacing of the level curves indicates steepness: They are closer together where the graph is steeper. . the ratio . Llaltitude • The average rate of change from P to Q 1s . . Llhonzontal • A direction of steepest ascent at a point P is a direction along which f (x, y) increases most rapidly. The steepest direction is obtained (approximately) by drawing the segment from P to the nearest point on the next level curve. • Level surfaces can be used to understand a function f(x, y, z). In the case where the function represents temperature, we call the level surfaces isotherms.

14,1 EXERCISES Preliminary Questions I. What is the difference between a horizontal trace and a level curve? How are they related? 2. Describe the trace of f(x, y)

= x2 -

sin(x3y) in the xz-plane.

3, ls it possible for two different level curves of a function to intersect?

Explain_

4. Describe the contour map of f (x, y) == x with contour interval I. 5. How will the contour maps of f (x, y) == x

and g(x, y) == 2x

with contour interval I look different?

r

798

----

CHAPTER 14

DIFFERENTIATION IN SEVERAL VARIABLES

----

Ex"~~e~rc;:;i::se=s- - - - - - - - - - - - - - - - 1n Exercises /-(, at h . functi . ' eac point evaluate the function or indicate that the on is undefined there.

l.

f(x,y)=x+yx3, (1,2),(-1,6),(e,,r)

2 ·

g(x,y) = x2 -2.__ _ y2' (1,3), (3, -3), (..fi,2)

3

• h(x, Y)

4. k(x,y) 5.

Jx-y2

=, x-y = xe-Y,

h(x, y, z)

6. w(r,s,t)

(B) (A)

(1,0), (3, -3), (0, 12)

= xyz- 2,

=

(20, 2), (I, -2), (I, I)

FIGURE 19

(3, 7, -2), (3, 2, ¼), (4, -4, 0)

r-s sin/, (2,2,

. ( )--(f) with their gr~phs (A)-(F) in Figure 20. 21. Match the funcnons a

f), (,r,,r,,r), (-2,2, f)

(a) f(x,y) =/xi+ /y/ (b) J(x,y)

In Exercises 7-14, sketch the domain of the function.

7. 9.

f(x,y)= 12x-5y f(x,y)

= ln(4x 2 -

8.

f(x , y)

=

11. g(y,z) = - -2 z+y

12. f(x,y)

= sin X2:'.

= ./TR

14. f(x,y)

=

cos- 1(x

y) -1 (c) f(x,y)= l+9x2+y2 2 -O.l(x2+y2)

(d) f(x, y) = cos(y )e

-1

I 10. h(x,t) = x+t

y)

l

13. F(J,R)

- x2

=cos(x -

(e) f(x, y)

= 1+9x2 +9y 2

(f) f(x, y)

=cos(x + Y )e

2

2 -0.I(x 2+y2)

z

+ y)

In Exercises 15-18, describe the domain and range of the function.

=x,./y+zezfx

15. f(x,y,z) =xz+eY

16. f (x, y, z)

17. P(r,s,l)=J16-r 2s 2t 2

18. g(r,s) = cos-1(rs)

19. Match graphs (A) and (B) in Figure 18 with the functions: (i)

(ii)

(B)

(A)

= -x +y2 g(x,y) = X + y2 f(x,y)

z

X

(C)

(D)

z

X X

(B)

(A)

FIGURE 18

20. Match each of graphs (A) and (B) in Figure 19 with one of the following functions: (i) J(x, y) = (cosx)(cos y) (ii) g(x, y) = cos(x 2 + y2)

(E)

(F)

FIGURE 20

SECTION 14.1

., tch the functions (a}-(d) with the·

ii, ;~I(ii (J)

(d

f(1,

tr

y)::3x+4y

contour maps (A}-(D) in

=x3 _ y k(x,y) = x2 _ y

(d)

1: 18

10

0

0

-5

-5 0

5

5 FIGURE 22 Contour map with contour interval

10

5 0 -5

10

y

i

0

:c=-3

-5

'c=O

: ..

-10 -10 -5

·+-;-.5 10

(C)

0

5

10

= 12- 3x -

25, f(x,y) == x2 +4y2

= sin(x -

y)

29. Sketch contour maps of f(x, y)

-2

2

4

Exercises 41-43 refer to the map in Figure 24.

24, f(x,y)=J4-x2-y2

4y

-4

FIGURE 23

In Exercises 23-28, sketch the graph and draw several vertical and horizontal traces.

27. f(x,y)

-6

(D)

tlGI.J~E 21

23. f(x,y)

(b) from A to C.

5

(

0

5

10

I

-10 -10 -5

0

change: (a) from A to B.

(B)

% , //,

m = 6.

40. Use the contour map in Figure 23 to calculate the average rate of

-10 -10 -5

(A)

10

799

(b) g(x,y)

h(1, y) == 4x - 3y

-10 -10 -5

Functions of Two or More Variables

26. f(x,y)

= y2

28. f(x,y)

= X 2 +y2 +1

41. (a) At which of A-C is pressure increasing in the northern direction? (b) At which of A-C is pressure increasing in the westerly direction?

42. For each of A-C indicate in which of the four cardinal directions, N, S, E, or W, pressure is increasing the greatest. 43. Rank the following states in order from greatest change in pressure across the state to least: Arkansas, Colorado, North Dakota, Wisconsin.

I

= x + y with contour intervals m = 1

and 2. 30, Sketch the contour map of f(x, y)

= x 2 + y2 with level curves c = 0,

4,8, 12, 16. In Exercises 31-38, draw a contour map of f(x, y) with an appropriate

1016

contpur interval, showing at least six level curves. 31. f(x,y)

=x2 -

y

32. f(x,y)

\\ 016-"l ;

y

= 2X

~-"\ \ '0 16''). --

33. f(x,y)

= '!.

34. f(x,y)=xy

X

35. f(x , Yl = x2 + 4y2

36. f(x,y) =X +2y- J

37. f (x,y) =x2

38, f(x, y)

= 3x2 - y2

39. Find the linear function whose contour map (with contour interval m =6) is shown in Figure 22. What is the linear function if m = 3 (and the curve labeled c = 6 is relabeled c = 3)?

'1

- ~

FIGURE 24 Atmospheric pressure (in millibars) over North America on March 26, 2009.

In Exercises 44-47, let T(x, y, z) denote temperature at each point in space. Draw level surfaces (also called isotherms) corresponding to the fixed temperatures given.

44. T(x, y, z) 45. T(x,y,z)

= 2x + 3y - z, T =0, 1, 2 = x - y + 2z, T = 0, 1,2

BOO

46. 47.

CHAPTER 14

DIFFERENTIATION IN SEVERAL VARIABLES

figure 26, referto B ·iy from A to . In £,xerc1se . water dens1 e in sea . d the chang hange from A to B and 52. Fin e rate of c fto"' the averag .,, , Estimate , 53 • from A to points i 1"1 to C. f change ' , and ··· rage rate o l!j_ · tetheave · · atD . 54, Esnma ent beg1nrung th of steepest asc 55, Sketch the pa . 5 52-55•

=x2 + y2- z, T =0, 1,2 T(x,y,z) =x2 - y2 +,2, T =0, 1,2, -1, -2 T(x ,y,z)

In Exercises 48-5/, p(S, T) is seawater density (kilograms per cubic meter) as a function of salinity S (pans per thousand) and temperature T (degrees Celsius). Refer to the contour map in Figure 25.

48. Calculate the average rate of change of p with respect to T from B to A. 49. Calculate the average rate of change of p with respect to S from B

Ca)~

toe.

50. At a fixed level of salinity, is seawater density an increasing or a decreasing function of temperature?

\\~

"'~

51. Does water density appear to be more sensitive to a change in temperature at point A or point B?

~

25

A

oi-,O

\:ft±-!. - :J,',? f \.O~ oiMI \ .. \~'.'µ,

20

. terva1=20m Contour in

~, ~,,

\ .o'l-'~ \·.oi • oioO \.

,,

in 3-space be given by T(x, y, z) = x2 +y2, 56. Let temperature ding to temperatures T -:,c -2, -1,0, 1,2. 1. Draw isotherms correspon

;/

-

·n 3-space be given by T (x , i, z) = 57. Let temperature l T - O1 2 Draw isothertns corresponding to temperatures - ' ' .

7 y

VI

0 31.5

V

32.0

32.5

33.0

33.5

2km

FIGURE 26

.

I I

0

p~

om

\·

5

----

34.0

34.5

+i.9 +'·,1

58. Let temperature in 3-space be given by T(x,y,z)=x2-y2,1. Draw isotherms corresponding to temperatures T -1, 0, 1.

Salinity (ppt)

59. Let temperature in 3-space be given by T(x, y, z) = x2 - y2 _i2_ Draw isotherms corresponding to temperatures T = -2, -1,0, 1,2.

FIGURE 25 Contour map of seawater density p(S, T) (kilograms per cubic meter).

Further Insights and Challenges 60. The function f(x,t) = 1- 112e-x / 1, whose graph is shown in Figure 27, models the temperature along a metal bar after an intense burst of heat is applied at its center point. 2

(a) Sketch the vertical traces at times t = 1, 2, 3. What do these traces tell us about the way heat diffuses through the bar? (b) Sketch the vertical traces x = c for c = ±0.2, ±0.4. Describe how temperature varies in time at points near the center.

61. Let f(x,y)

=

X

vx2+y2

for(x,y) # (0,0)

Write fas a function f (r, 0) in polar coordinates, and use this to find the level curves of f.

FIGURE 27 Graphoff(x t)- 1-lf2e-x 2 /t b · · h rt] after o. ' egmnmg s o y

t

=

~ ;-;=:~=:=:--:-:: - - - - - - - - 14.2 Limits and Continuity inSeveral Variables

---

. develops limits and continuit . th This. section functions of two variables but similar d fi ~n e multi variable setting. We focus OIi or more variables. ' e mitions and results apply to functions of [}Utt

II , 11

I

·I SECTION 14.2

Limits and Continuity in Several Variables

801

Recall that on the real number line a number x is close to a if the distance l_x - a I · ' · p ( b) ·f the distance Is small. In the plane, a point (x' y) is close to another pomt a, I

=

v•(P, r) excludes P

/\;\

I I

I

p ,

•

,.f,Y

\ (

)

= J 0, there exists 8 > 0 such that if (x, y) satisfies 0 < d((x, y), (a, b)) < 8,

then 1/(x, y) - LI
0,y>0}

22. f(x,y)

+ y)

= (x -y)ex 2-Y 2

(e) {(x,y) E R 2 (f)

:

1 ::5: x ::5: 4,5 ::5: y ::5: 10}

{(x,y) E R 2 : x > 0,x 2

+ y2

::5: 10}

2

24. Show that f(x, y) = has infinitely many critical points (~s a function of two variables) and that the Second Derivative Test fa.tis for all of them. What is the minimum value off? Does f(x, y) have any local maxima? x2

25. Prove that the function f(x,y) = ½x 3 + ly l -xy satisfies f(x,y) 2'.; 0 forx 2'.; 0 andy 2:: 0. 2 (a) First, verify that the set of critical points of f is the parabola Y = x and that the Second Derivative Test fails for these points. 3 2

(b) Show that for fixed b, the function g(x) x > 0 with a critical point at x = b 1l 2 •

(c) Conclude that f(a, b) 2'.; f(b 1l 2 , b)

26.

FIGURE 21 Plot of f(x,y)

(b) {(x, y) E R

= lnx +

23. f(x, y)

= ex 2 -Y 2+4 Y

18. f(x,y)=xln(x+y)

cosx

19. f(x,y)

- y2

14. f(x, y)

y

=

f(x, b) is concave up for

= 0 for all a, b

2:: 0.

[¥1 Let f(x,y) = (x 2 + y 2 )e-x2-y2·

(a) Where does J take on its minimum value? Do not use calculus to answer this question. (b) Verify that the set of critical points off consists of the origin (0, 0) and the unit circle x 2 + y 2 = I.

In Exercises 29-32, determine the global extreme values ofthefunc· tion on the given set without using calculus.

29. f(x,y)=x+y, 30. j(x,y)

= 2x -y,

31. f(x,y)

=

32. f(x, y)

= e-x2-y2'

(x 2

0:::x:::1,

0:5:y:5:l

0::: x ::5: 1,

+ y 2 + 1)- 1, x2

+ y2

0 ::5: y ::5: 3

0::: x ::5: 3,

0::: y ::5: 5

::5: 1

A linear function f(x, y) =ax+ by+ c has no critical points. Thert· fore, the global minimum and maximum values of f(x, y) on a closed and ~o~nded d_o main must occur on the boundary of the domain. Furthemrort, 11 is not difficult to see that if the domain is a polygon, as in Figure 22· th en th e global minimum and maximum values off must occur a1 a verttt of th e polygon. In Exercises 33-36, find the global minimum and ,na:a· mum values of f(x, y) on the specified polygon, and indicate where on t/11 polygon they occur.

33 · f(x, y) 34. f(x,y)

= 2.x - 6Y + 4 on the polygon in Figure 22(A). = lly-?x +7 on the po Iygon m . Figure . 22(8).

s Ec TI o N 14.7 f(X ; y) == 12 + Sy - 20x on the poly

•

•

gOn 10 Figure 22(A). f(:x,Y) == 3x - 6y - 8 on the domain )6, Where \x\ + \y\ 3_

35·

y

Optimization in Several Variables

'

1!1l3

40. Find the maximum of f(x, y) = y2 + xy - x2 on the square domain 0 X 2, Q y 2. In Exercises 4/-49, determine the global extreme values of the junction on the given domain.

y

(4, II)

41. f(x, y) = x 3 - 2y, 0

x

42. f(x, y) = 5x - 3y, y

x - 2, y

1, 0

y

l

-x - 2, Y

3

43. /(x,y)==x 2 +2y 2 , 0~x~l,

(8,7)

45. f(x,y)=x 2 +xy2+y2 , x,y,::0, x+y~ l

(0, -4)

46. f(x,y)=x 3 +y3-3xy, 0~x~I, 0~y~l

(A) (B)

47. f(x, y)

FIGURE 22

= x2 + y2 -

2x - 4y,

x :::: 0, 0

y

3,

y :::: x

J7. Assumptions Mai~r Show that f(x y) _ d · · ' '- al • • - xy oes not have a glob al muumum or a g;ou maximum on the domain

,.

'D

50. Find the maximum volume of a box inscribed in the tetrahedron bounded by the coordinate planes and the plane

= [(x,y): 0 < x < 1,0 < y < l}

Explain why this does wt contradict Theorem 3.

1

X

38, Find _a continuous function that does not have a global maximum on

= {(x, y) : x + Y not contradict Theorem 3.

the domam_ 'D

0,x

+Y

l

+ -y + -z =l 2 3

1). Explain why this does

51. Find the volume of the largest box of the type shown in Figure 24, with one comer at the origin and the opposite comer at a point P = (x, y,z) 0\ the paraboloid ·

39. Find the maximum of f(x,y) =x + y-x 2 -y2-xy

y2

x2

on the square, 0 x 2, 0 y 2 (Figure 23). (a) First, locate the critical point of f in the square, and evaluate / at this point. (b) On the bottom edge of the square, y = 0 and f(x,0) the extreme values of f on the bottom edge.

z=l---4 9

= x -x 2. Find

withx,y,z~0

z

(c) Find the extreme values off on the remaining edges.

(d) Find the greatest among the values computed in (a), (b), and (c).

y

f(x, 2)=-2-x-x2

Edgey=2 2

-l--- -1---, y

-Edgex= 2 .- - ·., /(2, y) = -2-y-y2

Edgex=O f(0,y)=y-y2

2

-x

X

FIGURE 24

52. Find the point on the plane

Edgey=O 2 f(x, 0) =x-x 2

2

FIGURE 23 The function f(x, y) == x + Y - x - Y - xy on boundary segments of the square O :'.': x :'.': 2, O :'.': Y :'.': 2·

th

e

closest to the point P = (1, 0, 0). Hint: Minimize the square of the distance.

864

CHAPTER 14

DIFFERENTIATION IN SEVERAL VARIABLES y

53• Show that the sum of the squares of the distances from a point = (c, d) ton fixed points (a1 , bi), ... ,(an,bn) is minimized when c is . e average of the x-coordiilates a; and d is the average of the y-coordmates b; . '

y=mx+ b

54. Show that the rectangular box (including the top and bottom) with fixed volume V = 27 m 3 and smallest possible surface area is a cube (Figure 25).

L--------x ares fit minimizes the s um of the fr m the data points to the line. squares of the vertical distances o FIGURE 26 The lin~ar le'.15t-squ

z

FIGURE 25 Rectangular box with dimensions

x, y, z.

55. Consider a rectangular box B that has a bottom and sides but no top and has minimal surface area among all boxes with fixed volume 3

V=64m . (a) Do you think Bis a cube as in the solution to Exercise 54? If not, how

would its shape differ from a cube?

(b) Find the dimensions of Band compare with your response to (a).

56. Find three positive numbers that sum to 150 with the greatest possible product of the three.

. . owatts) of a laser is measured as a fu nction of cur. 61. 1:11e power (m IDIF~rd th linear least-squares fit (Exe; cise 60) for the rent (m mi111amps). m e data points. Current (milliamps)

1.0

1.1

1.2

1.3

I.4

1.5

Laser power (microwatts)

0.52

0.56

0.82

0.78

1.22

1.50

62. Let A = (a, b) be a fixed point in the plane, and let /A ( P) be_ the distance from A to the point P = (x, y). For P # A, let eAP be the umt vector pointing from A to P (Figure 27):

57. A 120-m long fence is to be cut into pieces to make three enclosures, each of which is square. How should the fence be cut up in order to minimize the total area enclosed by the fence? 58. A box with a volume of 8 m 3 is to be constructed with a gold-plated top, silver-plated bottom, and copper-plated sides. If gold plate costs $120 per square meter, silver plate costs $40 per square meter, and copper plate costs $10 per square meter, find the dimensions that will minimize the cost of the materials for the box. 59. Find the maximum volume of a cylindrical can such that the sum of its height and its circumference is 120 cm. 60. Given n data points (xi, Yi), . .. , (xn, Yn), the linear least-squares tit is the linear function

-

eAP

=

-;;;, -+

IIAPII

Show that

Note that we can derive this result without calculation: Because 'ilfA(P) points in the direction of maximal increase, it must point directly away from A at P, and because the distance f A (x , y) increases at a rate of 1 as you move away from A along the line through A and P, VJA (P) must be a unit vector.

j(x)=mx+b y

that minimizes the sum of the squares (Figure 26): n

E(m,b)

= L(Yj -

f(xj))

2

j=l

Show that the minimum value of E occurs for m and b satisfying the two equations

--t-------------.x FI_GURE 27 The distance from A to p increases most rapidly in the dIrect1on eAP·

11

n

m 1:xJ+b LXj )=1

)=1

n

= I:xm j=I

r

ltipliers· Optimizing with a Constraint Lagrange Mu ·

SECTION 14.8

865

~ ights and Challe; ; - - -- - - - - - - - - - - - - - - - - - -

f~''

. we prove that for all x this exercise,

63· ID

.

' y ::: 0:

The following problem was posed by Pierre de Fennat: Given points A= (a1,a2), B (bi,bi), and C (ci:ci) in the plane, find the point p = (x' y) that minimizes the sum of the distances

64

I " + -xfi 1 > a /3 - xy

-x

=

=

f(x,y) =AP+ BP+ CP

a I and /3 ::: I_ are numbers such that a- 1

,v~ere ve that the function

we Pro

f(x, y)

_1

+ /3 = I. To do this,

= o:-1 x" + 13-1 y/J -xy

Let e, r, g be the unit vectors pointing from P to the points A, B, C as in Figure 29. . . (a) Use Exercise 62 to show that the condition Vf(P) = Ois equivalent to

usfies f(x, y) ::: 0 for all x, y ::: 0.

=

Verify that the Second Derivative Test fails Sh h O the function g(x ) = f ( b) • · ow,. owever, that for fi~ed x' is concave up with a critical point atX === b · (c) Conclude that for all x .., 0, f (x , b) ::: f (bfi-1, b) = O. (b) b

0

e+f+g=O

s3 SbOW that the set of critical points of !( ). (9) 28). Note that this curve can also be dexs, Y_b1sdthe curve/Jy xa-1 ~Jgure . en e as x - y -I Wh t . ••e value of f(x, y) at pomts on this curve?. · a 1w

p-i

(b) Show that f(x, y) is differentiable except at ~ints A,~• C: Conclude that the minimum of f(x, y) occurs either at a pomt P sattsfymg Eq. (3) or at one of the points A, B, or C. (c) Prove that Eq. (3) holds if and only if P is the Fermat point, defined as the point p for which the angles between the segments AP, BP, C P are all 120° (Figure 29). (d) Show that the Fennat point does not exist if one o~ the ~gles in MBC · · is greater than 120°. Where does the mm1mum occur 1n this case?

y

b

,. _... ,.,,-

/

\

C

Critical points of z = f(x, y)

(A) P is the Fermat point

X

FIGURE 28 The critical points of f (x, y) a-I form acurve y x .

=

y

Constraint g(x, y) =2x+ 3y- 6 =0 /

Point on the line closest to the origin

=o: - 1x" +r I yf3 - xy

= Jx2 + yZ 00 the line 2x + 3y = 6. f(x,y)

not exist.

FIGURE 29

14.8 Lagrange Multipliers: Optimizing with a Constraint Some optimization problems involve finding the extreme values of a function f(x, y) subject to a constraint g(x, y) = 0. Suppose that we want to find the point on the line 2x + 3y = 6 closest to the origin (Figure 1). The distance from (x, y) to the origin is f(x, y) = Jx 2 + y 2, so our problem is

Minimize f(x, y) Iii) FIGURE 1 Finding the minimum of

(B) The Fermat point does

(the angles between e, f, and g are all 120°).

=

J+ x2

y2 subject to g(x, y)

= 2x + 3y - 6 = 0

We are not seeking the minimum value of f(x, y) (which is 0), but rather the minimum among all points (x, y) that lie on the line. The method of Lagrange multipliers is a general procedure for solving optimization problems with a constraint. Here is a description of the main idea.

B66

CHAPTER 14

DIFFERENTIATION IN SEVERAL VARIABLES

. Figure 2(A). We want to increas

10

GRAPHICAL INSIGHT Imagm · e standing at point Q (x y) == 0. The gradient vec1 e tr ·nt curve g , · h or the value off while remaining on the con_s ru e but we cannot move 1Il t e gradient 'i/f,Q points in the direction of maximum mcreas 'tr";nt curve. However, the gradient

ff the cons ... . t th . h direction because that would take us O hat by movmg O e ng t alon . ase J somew g points to the right, so we can still mere the constraint curve. . t the point P, where 'iif P is Orthog . ·1 arnve a . We keep moving to the nght unu we p we cannot mcrease f further by onal to the constraint curve [Figure Z(B)]. Once at s~aint curve. Thus, f (P) is a Ioca] moving either to the right or to the left along th e con maximum subject to the co?straint. to the constraint curve because it is the Now, the vector 'i/gp 1s also orthogonal alt the level curve through P. Thus gradient of g(x, y) at P and therefore is orthog~ }.. ;g P for some scalar J.. (called a La: 'i/f p and 'i/g p are parallel.In other "'.ords, Vf P max subject to the constraint 10 1 grange multiplier). Graphically, this means th at: : tangent. The same holds for a occurs at points P where the level curves of f an g local min subject to a constraint. y Tangent line at P

y

"·i'gp Q

Level curves of z = f(x, Y)

/

4 3 2

I

,11

-constraint curve g(x, y) = 0

,11

I'

(B) The local maximum off on th~ constraint curve occurs where Vfp and Vgp are parallel.

(A) f increases as we move to the

right along the constraint curve.

I

i'I I I I

llii

FIGURE 2

THEOREM 1 Lagrange Multipliers Assume that f(x, y) and g(x, y) are differentiable functions. If f (x, y) has a local minimum or a local maximum on the constraint curve g(x, y) = 0 at P = (a, b), and if 'i/gp i= 0, then there is a scalar}.. such that In Theorem 1, the assumption Vgp f. 0 guarantees (by the Implicit Function Theorem of advanced calculus) that we can parametrize the curve g(x, y) 0 near P by a path r(t) such that r(O) P and

=

=

r'(O) f. 0.

1

'i/fp =J..'vgp

1

Proof Let r(t) be a parametrization of the constraint curve g(x, y) = Onear P, chosen so that r(O) = P and r'(O) i= 0. Then f(r(O)) = f(P), and by assumption, f(r(t)) has a local min or max at t = 0. Thus, t = 0 is a critical point off (r(t)) and _dd f(r(t))I

t

t=O

= 'i/fp · r'(O) = 0

Chain Rule

This sho"."s that Vfp is orthogonal to the ~an gent vector r'(O) to the curve g(x, y) = O. The gradient 'i/gp 1s also orthogonal to r (0) [because 'i/gp is orthogonal to the level curve g(x , y) = 0 at P]. We conclude that 'i/ f p and Vgp are parallel and hence Vf is a multiple of 'i/gp as claimed. ' P 1 We refer to Eq. (I) as th_e Lagrange condition. When we write this condition in tenns of components, we obtam the Lagrange equations:

= J..gx(a,b) Jy(a,b) = >..gy(a,b) fx(a,b)

Lagrange Multipliers: Optimizing with a Constraint

SECTION 14.8

867

A · "tical · t for the optimiza. pomt P = (~, b) satisfying these equations is calle~ en porn tion problem with constraint and f(a,b) is called a cnbcal value.

EXAMPLE 1 Find the extreme values of f(x, y)

= 2x + 5y on the ellipse

Gf +Gf =

1

Solution Step I. Write out the Lagrange equations. 2 W h 2 The constraint curve is g(x, y) = O, where g(x, y) = (x/4) + (y/3) - 1. e ave

VJ= The Lagrange equations Vf p (2,5)

(2,5),

Vg=

X

(

2y)

8'9

= AVg p are

= A(i•

2 :)

A(2y) 5=9

=}

0

Step 2. Solve for A in terms of x and y. Equation (2) gives us two equations for)._; 16

A=-, X

Level curve of f(x,y) = 2.t + Sy

To justify dividing by x and y, note that x and y must be nonzero, because x = 0 or y = 0 would violate Eq. (2). Step 3. Solve for x and y using the constraint. 45 16 45 The two expressions for)._ must be equal, so we obtain~ = y or Y = x. Now 32 2 substitute this in the constraint equation and solve for x:

y

Constrain~

g(x,y)=\

0

45 2y

A=-

Gf + (45 )2 =

"--

~x

17 x2 ( 0

Thus, x -17

l!lJ FIGURE 3 The min and max occur where a level curve of J is tangent to the constraint curve g(x, y) = 0.

=±

l

/6 + 12:;4) = x2 ( 12;;4) = 1

= ±32 -, and since y = -45x ,the critical points are P = (32 , -45) 17 32 7 17 1

and Q = (- 32 _ 45)· 17' 17

Step 4. Calculate the critical values. f(P)=f ( 32 , 45) =2 (32) l7 +5 (45) =17 17 17 17

and f (Q) = -17. We conclude that the maximum of f (x, y) on the ellipse is 17 and the minimum is -17 (Figure 3).

14.7,

Assumptions Matter According to Theorem 3 in Section a continuous function on a closed, bounded domain takes on extreme values. This tells us that if the constraint curve is closed and bounded (as in the previous example, where the constraint curve is an ellipse), then every continuous function f(x, y) takes on both a minimum and a maximum value subject to the constraint. Be aware, however, that extreme values need not exist if the constraint curve is not bounded. For example, the constraint x - y = O

868

CH APTER 14

DIFFERENTIATION IN SEVERAL VARIABLES

y2 has neither a minirnu . . The function f (x' Y === x ) satisfies the constr •ni nor is a line which is unbounded. use every pomt (~, a aximum) and :nt, Yet a maximum subject to x - Y === 0 beca itive (so there 1s no m ar Ilriitjly f(a,a) = a 3 can be arbitrarily_J~ge pos . (so th ere 1s • no ffilmmum). large negative . . By investing x umts of labor f n function o.4 o.6 ilnd EXAMPLE 2 Cobb-Douglas Produc 10 roduce P(x, y)::::: 50x Y watches. Pind Y units of capital, a watch manufacturer canbpe produced on a budget of $20,000 if labor . the maximum number of watches that can . $20Operum 1• costs $100 per unit and capital coS ts . f capital is 100x + 200y · Of labor and Y umts O • · 0Ur Solution The total cost of x uruts o.4 o.6 subject to the followmg buct . . the funct10n • P(x, y) ::::: 50x Y get task is to maxmuze

y (capital)

120

constraint (Figure 4):

+ 200y _ 20,000 g(x,y)::::: looX 60 _ ---- -------[ ~ Budget_

/

constramt

i ' 80

40

Step J. Write out the Lagrange equations.

/ 120

x(labor)

Contour plot of the Cobb-Douglas production function P(x, y) = 50x 0Ay0·6. The level curves of a production function are called isoquants. FIGURE 4

Px(x,y)

= }..gx(x,y):

Py(x,y)

= }..gy(x,y):

G]

== 0

20x-0.6y°.6 = 100A 30xo.4Y-0.4::::: 200}..

Step 2. Solve for}.. in terms of x and Y· t b equal· · fior }.. that mus e · These equations yield two express10ns

l

}..=5

(y)0.6 _

-20

0

(2:)-0.4 X

Step 3. Solve for x and y using the constraint. . . 3 Multiply Eq. (5) by 5(y/x)°-4 to obtain y/x = 15/20, or y = 4X· Then substitute in Eq. (4): z

Maximum on constraint curv

f(x, Y)

x...-------fp Constrained_........-D max occurs here g(x, y) = 0 y (A)

X

(B)

FIGURE 5

IO0x + 200y

= IO0x + 200 Ux) = 20,000 =>

250x

= 20,000

2 We obtain x = 0,000 = 80 and y = ¾x = 60. The critical point is A = (80, 60). 250 Step 4. Calculate the critical values. Since P(x, y) is increasing as a function of x and y, VP points to the northeast, and it is clear that P(x, y) takes on a maximum value at A (Figure 4). The maximum is P(80, 60) = 50(80)°-4(60)°-6 = 3365.87, or roughly 3365 watches, with a cost per 20,000 watch of - - or about $5.94. 3365 GRAPHICAL INSIGHT In an ordinary optimization problem without constraint, the global maximum value is the height of the highest point on the surface z = f(x, y) [point Q in Figure 5(A)]. When a constraint is given, we restrict our attention to the curve on the surface lying above the constraint curve g(x, y) = 0 . The maximum value subject to the constraint is the height of the highest point on this curve. Figure 5(B) shows the optimization problem solved in Example 1.

The method of Lagrange multipliers is valid in any number of variables. Imagine, for instance, that we are trying to find the maximum temperature f(x, y, z) for points on a surface Sin 3-space given by g(x, y, z) = 0, as in Figure 6. This surface is a level surface for the function g, and therefore, Vgp is perpendicular to the tangent planes to this surface at every point P on the surface. Consider the level surfaces for temperature, which we have called the isotherms. They appear as surfaces in 3-space, and their intersections with s yield the level sets of temperature on S. If, as in the figure, the temperature increases as we move to the right on the surface, then it is apparent that the maximum temperature

---- =:« r

r

SECTION 14.8

Lagrange Multipliers: Oplimizinit with a Constraint

869

for the surface occurs when the last isotherm intersects the surface in just a single point and hence that isotherm is tangent to the surface. That is to say, the last isotherm and the surface share th e same tangent plane at their single pomt · of mtersec · 11·on. s However, as we know, VJPis always perpendicular to the tangent plane to the level Vurfaces for f at each point p on a level surface. So at the hottest point on the surface, gP and V/ P are both perpendicular to the same tangent plane. Hence, they must be P_ar~lel, and one must be a multiple of the other. Thus, at that point, V/P =)... Vgp. A sirru ar argument holds for the minimum temperature on the surface.

V/=

=4

/=2

-

-

(ilj 1..,Ji!E 7 As we move to the right,

FIGURE 6 As we move to the right, temperature increases, attaining a maximum on the surface of f = 4 at P.

temperature increases and then decreases.

There is one other situation to consider. Imagine that as we move left to right across our surface, temperature first increases to f = 4 and then it decreases again, as in Figure 7. There is a collection of points wit!-, foe maximal temperature of f = 4. In this case, V/ must point to the right on isot'1em1s that are to the left of f = 4 since this is the direction of increasing temperature, and Vf must point to the left on isotherms that are to the right of f = 4 since this is the direction of increasing temperature. Hence, in order for the right-pointing gradient vectors to become left-pointing gradient vectors in a continuous manner, they must be equal to Oon the f = 4 isotherm. This makes sense, since on· that isotherm there is no direction of increasing temperature. So for all of the points on the surface with maximal temperature, of which there are many, the equation Vf = J,.. Vg is satisfied, but by taking J,.. = 0. In the next example, we consider a problem in three variables.

EXAMPLE 3 Lagrange Multipliers in Three Variables Find the point on the plane

x + y + z = 11 ' ' 'm R3 . c osest to the ongm 2 4 4

Solution Our task is to minimize the distance d = Jx2 + y 2 + z2 subject to the constraint

1

+ ¼+ = 1. But, finding the minimum distance d is the same as finding the

minimum square of the distance d 2, so our problem can be stated: Minimize f(x, y, z)

=x2 + y2 + z2

subject to

g(x,y,z)

= -X2 + -Y4 + -4Z - 1 = 0

The Lagrange condition is

(111)

(2x, 2y, 2z) = J,.. -, -, --.,.-, 244 VJ '-...--' Vg

WIIIII

870

CHAPTER 14

DIFFERENTIATION IN SEVERAL VARIABLES

This yields

). == 4x == Sy :::: 8z . we obtain Substituting in the constraint equation, z 3z :::} X y z 2z ~+-=-:::: 1 2 2+4+4=2+ 4 4

2

z= 3

2 . critical point must corres~ond to the lllinini Thus, x = 2~ = and.y == z == 3• This Jane since there are pomts on the Plane U~ of J. There 1s no maxunum of Jon the P . ton the plane closest to the orj .1bat th are arbitrarily far from the origin. Hence, e pom &In~ P= (Figure 8). I

1

FIGURE 8 Point p closest to the origin on the plane.

(1, ~. D

. . be used when there is more than one c The method of Lagrange mult:1phers canul . r for each additional constraint 00• straint equation, but we must ~oth er m ect to constraints g(x, y, z) ::::: For 0¾d example, if the problem is to nunumze J(x, _Y •Z h(x, y, z) = 0, then the Lagrange condition is

ti)~~:

VJ== ).Vg+ µVh

The intersection of a sphere with a plane

Ithrough its center is called a great circle.

EXAMPLE 4 Lagrange Multiplie~s with Mujtiple2Con~tr! i~t:s a~ :~~•::;::c;~:n of~e planex+½Y+!z=0withtheurutspherex +Y +z . g gure9). . the pomt . on this. great ClfC . 1e WI"th the greatest x-coordmate. . Fmd . Our task 1s . to maxumze . • th e fu ncu·on J(x , y' z) = x subjtc' to the two conSolution

straint equations g(x,y,z) =

I

X

I

+ zY + 3z = 0,

2

2

2

h(x,y,z)=x +y + z - 1=0

The Lagrange condition is VJ =AVg+µVh (1,0,0)

=).

(I,~,~)+µ (2x,2y,2z)

Note that µ cannot be zero, since, if it were, the Lagrange condition would become (I, 0, 0) = ).(!, ½, !), and this equation is not satisfed for any value of A. Now, the Lagrange condition gives us three equations: A +2µx

= 1,

The last two equations yield A= -4µy and). = -6µz . Becauseµ -/= 0, 1·

FIGURE 9 The plane intersects the sphere in a great circle. Q is the point on this great circle with the greatest x-coordinate.

-4µy

= -6µz

=}

Now use this relation in the fast constraint equation: X

(3 )

1 + -z I = X + -I -z + -z I =0 + -y 2 3 2 2 3

13 12

X=--z

Finally, we can substitute in the second constraint equation: x2

. 637 2 t0 0 btam 144Z p

+

y2 + z2 -

1 =(-:~Zr+

GzY + z2- 1 = 0

1 or z - ±__11_ s· 13 3 - 1v'TI" mce x = -nz and y = 2z, the critical points are

v'13

18

12 )

= (--7-, 7Jff 7v'l3 '

Q=

(m -~ _2) 7 ' 7v113' 7v'l3

ctr

SECTION 14.8

Lagrange Multipliers: Optimizing with a Constraint

871

The critical point with the greatest x-coordinate [the maximum value of f(x, y, z)J is Q

.

-v'I3

Wtth x-coordinate _ _ 7

•

0.515.

14.8 SUMMARY

f

• Method of Lagrange multipliers: The local extreme values of (x_, y) subject to a constraint g(x, y) = 0 occur at points p (called critical points) satisfying the ~agrange condition 'vfp = '). 'vgp. This condition is equivalent to the Lagrange equations

= Agx(x,y), /y(x,y) = Agy(x,y) • If the constraint curve g(x, y) = Ois bounded [e.g., if g(x, Y) = 0 is a ci~cle 0 ~ ellipse1' fx(x,y)

then global minimum and maximum values of J subject to the constraint exist. • Lagrange condition for a function of three variables f(x , Y, z) subject to two constraints g(x, y, z) = O and h(x, y, z) = 0: VJ= ')..'vg

+ µ,'vh

---:-;:::;;~:-;;-;::-;:-- - - - - - - - - - - - - - - - - - - - - - - - - 14,8 EXERCISE~

·- - - - - - - - - - - - - - - - - - - - -

preliminary Questi ons

I. Suppose that the ;naxirnum of f(x, y) subject to the constraint 0 occurs at.a porn~ P = (a,b) such that 'vfp =fa o. Which of the following statements ,s i:rue. (a) 'vfp is tangent t,, g,:x , y) =Oat p . g(x,y)

=

(b) 'vfp is orthogonal'.:: g(x, y) =Oat P.

3.

On the contour map in Figure 11:

(a) Identify the points where

VJ= 7'. Vg for some scalar 7'..

(b) Identify the minimum and maximum values of f (x, Y) subject to

g(x,y)

= 0.

y

z. Figure 10 shows a. ~onstraint g(x, y) = 0 and the level curves of a

-6 -2

2

6

function f. In each case, determine whether f has a local minimum a local maximum, or neither at the labeled point. '

'vf 4 3

•'i \

g(x,y)=0

VJ

6

g(x, y) = 0

g(x,y)=0

FIGURE 10

2 -2 -6

Contour plot off(x, y) (contour interval 2) FIGURE 11

Exercises = x 2 + 2y 2

In this exercise set, use the method of Lagrange multipliers unless otherwise stated.

2. Find the extreme values of f(x, y) straint g(x, y) = 4x - 6y = 25.

I. Find the extreme values of the function f(x, y) = 2x + 4y subject to the constraint g(x, y) = x 2 + y2 - 5 = 0. (a) Show that the Lagrange equation 'vf = 7'. 'vg gives 7'.x = 1 and = 2.

(a) Show that the Lagrange equations yield 2x

,y

i= 0 and Y = 2x · (c) Use the constraint equation to determine the possible critical points

(b) Show that these equations imply 7'. (x,y).

(d) Evaluate J(x, y ) at the critical points and determine the minimum and

lllaximum values.

subject to the con-

= 4A, 4y = -67'..

(b) Show that if x = 0 or y = 0, then the Lagrange equations give x = y = 0. Since (0, 0) does not satisfy the constraint, you may assume that x and y are nonzero.

(c) Use the Lagrange equations to show that y

= -¾x.

(d) Substitute in the constraint equation to show that there is a unique critical point P.

I\ 872

CHAPTER 14

DIFFERENTIATION IN SEVERAL VARIABLES (e) Does P co fer to Fi ITCSPond to a mini . i •gure 12 to justify yo mum or maxunum value of f ? Rencrease or decrease as ur answer. Hint: Do the values of f(x y) g(x,y) == 0? (x,y) moves away from p along the line

y

y

4

-2 0 3 3 FIGURE 13 Contour map of/(x, y) == x + xy + Y aoct graph of~ . ( ) - x3 - xy + y3 == 1. e constraint g x, Y -

-4 -4

0

4

17. Find the point (a, b) on the graph of y == e' where the value ab is lht

FIGURE 12 Level curves off( 2 constraint g(x y) _ x' Y) =x + 2y2 and graph of the ' - 4X - 6Y - 25 = 0.

3

• Apply the method of La • . (x2 + I) b" grange mulllphers to the function J(x

least.

y)

=

Y su ~ect to the constraint 2 + 2 5 ' x Y = • Hint: First show that ' aI e cases x 0 and x -/- 0 separately.

y f. O· then Ire th

=

In Exercises 4-15 find th • • subiect I th . ' e ~mzmum and maximum values of the function 0 , e given constraint.

18. Find the rectangular box of maximum volume if the sum of the lenillii of the edges is 300 cm. 19. The surface area ~f a right-c~cular crne

s =71 ,J,2 +hz, and its volume 1s V == 3rrr

t

radius r and height h,

h.

ij

(a) Determine the ratio h/r for the cone with given surface area S maximum volume V. illd {b) What is the ratio h/r for a cone with given volume V and minim surface area S? um

4.

f(x,y)=2x+3y, x2+y2=4

5.

f(x,y)=x2+y2, 2x+3y=6

6.

f(x,y)=4x 2 +9y 2, xy=4

7.

f(x, y) = xy, 4x 2 +9y2

8.

f(x,y)=x 2y+x+y, xy=4

20. In Example I, we found the maximum of f(x, y) = 2x +Syon the ellipse (x/4)2 + (y /3)2 = I. Solve this problem again without using Lagrange multipliers. First, show that the ellipse is parametrized by x0 4cos_t, y = 3 sint. Then find the maximum value off (4cost,3sint) usmg smgle-vanable calculus. Is one method easier than the other?

9.

f(x,y)=x 2 +y2,

21. Find the point on the ellipse

(c) Does a cone with given volume V and maximum surface area exist?

= 32

x 4 +y4=1

10. f(x, y) = x 2y4, x 2 + 2y2 = 6

x2 +6y2 +3xy

11. f(x,y,z)=3x+2y+4z, x 2 +2y2+6z 2 =1

= 40

with the greatest x-coordinate (Figure 14).

12. f(x,y,z)=x 2 -y-z, x 2 -y2+z=0

----

13. f(x,y,z)=xy+2z, x 2 +y2+z 2 =36

14. f(x,y,z)=x 2 +y2+z 2, x+3y+2z=36

y

4

15. f(x,y,z)=xy+xz , x 2 +y2+z 2 =4

-4

16. ~Let g(x, y)

= x3 -

xy + y3

FIGURE 14 Graph of x2

+6y2 +3xy == 40 _

(a) Show that there is a unique point P = (a, b) on g(x, y) = I where

'vfp

=.l 'vgp for some scalar .l.

(b) Refer to Figure 13 to determine whether f(P) is a local minimum or

a local maximum off subject to the constraint. (c) Does Figure 13 suggest that /(P) is a global extremum subject to the constraint?

22."bUse . d . Lagrange . mult'1P1·iers to find the maximum area of a rectangle in· sen e m the ellipse (Figure 15):

x2

y2

;;i-+p- = l

I

J

r

SECTION 14.8

Lagrange Multipliers: Optimizing with a Constraint

873

32 In test, a runner starting at A must touch a point p along a riv~r and thena~~~o B in the shortest time possible (Figure 17). The;mncr should choose the point p that minimizes the total length of the pa .

(a) Define a function J(x ,y) =AP+ PB ,

FIGURE 15 Rectangle inscribed in the

, 1 find the point (xo, Yo) on the line 4x "".' . oogin,

. x 2 y2 e111pse -+- _ 1 a2 b2 - .

+9

_ . Y - 12 that 1s closest to the

,, Show that the point (xo, Yo) closest to the • • . ..,. ordinates ongm on the lme ax Ch3S CO xo

ac

where P

= (x,y)

Rephrase the runner's problem as a constrained optimization problem, assuming that the river is given by an equation g(x, y) 0. (b) Explain why the level curves of f(x , y) are ellipses. (c) Use Lagrange multipliers to justify the following state~ent: The ellipse through the point p minimizing the length of the path 1s tangent to the river. (d) Identify the point on the river in Figure 17 for which the length is minimal.

=

+ by = River

be Yo=~ a2 + b2

= a2 + b2 ,

,,, find the maximurr v,li ue of f (x y) - x• b , ""' ' Y ,or x =:: 0, y > 0 on the unex + Y = I •where" ""', 0, > 0 are constants. ,, Show that the mn ,mmn value of f(x y) _ 2 3

""'

•

:SA·

27, Find the maxim~.nn ·,alue of f(x, y) unit circle, where a, b > 0 are constants.

- x Y on

th

. . . e umtcrrcle 1s

= x• yh for x =:: o, y > 0 on the

FIGURE 17

-

zg, Find the maxim,J;n value of f (x, y, z) = x• yh z' for x, y, z =:: oon the unit sphere, where a, b, c > 0 are constants. 29, Show that the minimum distance from the origin to a point on the plane ax + by + cz = d is

ldl 30. Antonio has $5.00 to spend on a lunch consisting of hamburgers ($1.50 each) and french fries ($ 1.00 per order). Antonio's satisfaction from eating x1 hamburgers and x2 orders of french fries is measured by a function U(x1 ,x2) ,lxtxz. How much of each type of food should he purchase to maximize his satisfaction? (Assume that fractional amounts of each food can be purchased.)

=

In Exercises 33 and 34, let V be the volume of a can of radius r and height h, and let S be its surface area (including the top and bottom). 33. Find r and h that minimize S subject to the constraint V

34. Show that for both of the following two problems, P a Lagrange critical point if h = 2r:

= (r, h) is

• Minimize surface area S for fixed volume V. • Maximize volume V for fixed surface area S. Then use the contour plots in Figure 18 to explain why S has a minimum for fixed V but no maximum and, similarly, V has a maximum for fixed S but no minimum. h

31. Let Q be the point on an ellipse closest to a given point P outside the ellipse. It was known to the Greek mathematician Apollonius (third century BCE) that P Q is perpendicular to the tangent to the ellipse at Q (Figure 16). Explain in words why this conclusion is a consequence of the method of Lagrange multipliers. Hint: The circles centered at P are level curves of the function to be minimized.

= 54rr.

/

Increasing S Level curves of S

I

Increasing V

I

'----------+-r

FIGURE 18

35. Figure 19 depicts a tetrahedron whose faces lie in the coordinate

FIGURE 16

r

+ + : = I (a, b, c > 0). The a b c volume of the tetrahedron is given by V = iabc. Find the minimum value of V among all planes passing through the point P (l , I, I).

planes and in the plane with equation

=

874 CHAPTER 14

DIFFERENTIATION IN SEVERAL VARIABLES

y

C=(0,0,c)

Wall

fence y

A= (a, 0, 0)

FIGURE 20

X

FIGURE 19

W!~

36. ~e sam~ set-up as in the previous problem, find the plane that mmuruzes V 1f the plane is constrained to pass through a point P = (a,,8, y) with a,,8, y > 0. 37. Sh?w that the Lagrange equations for f(x, y) = x + y subject to the constramt g(x, y) = x + 2y = 0 have no solution. What can you conclude about the minimum and maximum values off subject tog = O? Show this directly.

38. Show that the Lagrange equations for f(x, y) = 2x + y subject to the constraint g(x, y) x 2 - y2 I have a solution but that f has no min or max on the constraint curve. Does this contradict Theorem I?

=

=

39. Let L be the minimum length of a ladder that can reach over a fence of height h to a wall located a distance b behind the wall. (a) Use Lagrange multipliers to show that L = (h 213 +b213)312 (Figure 20). Hint: Show that the problem amounts to minimizing f(x, y) = (x + b)2 + (y + h)2 subject to y/b = h/x or xy = bh. (b) Show that the value of L is also equal to the radius of the circle with center (-b, -h) that is tangent to the graph of xy = bh.

f f(x y z) = xy + xz + yz - xyz sub· al o , ' >' 0 y > 0 z > 0. !eq 40• Fl.nd the maximum v ue - 1 ,or x , - ' to the constraint x + Y + z - ' 2 2 2 b' . . . um of J(x,y,z) = x + y + z su ~ect to the l\1/ 41. Fmd the nunun I d + 2y + 3z = 6. o constraints x +Y +z = an x . f f( y z) = z subject to the two constr,:. 42. Find the maxunum O x, ' ..,,ts x2 + y2 I and x + Y + Z I. .

=

=

. 1 I I 43 Find the point lying on the intersecuon of the p ime x : zy + 4z::: 0 • 2 2 + 2 _ 9 with the greatest z-coordmate. and the sphere x +Y z -

. . off(x y z) = x + y +z subject to the two con 44. Fmd the max.11~um / 2 ! 2 + 42 2 = 9. · 2 straints x + y2 +z = 9 and 4X + 4Y

+: ;

=_I _in an ellipse. 45• The cylinder x2 + y2 = I interse_cts the plane x Find the point on such an ellipse that is farthest from th- ongm.

46. Find the minimum and maximum of f(x,y,z) 2 two constraints, 2x + z = 4 and x + y2 = I. 47. Find the minimum value of f(x, y, z) = x constraints, x + 2y + z = 3 and x - y = 4.

2

= Y + 2z subject to

+ Y2 + z2 subject to two

Further Insights and Challenges 48. Suppose that both f(x, y) and the constraint function g(x, y) are linear. Use contour maps to explain why f(x, y) does not have a maximum subject to g(x,y) = 0 unless g = af + b for some constants a,b. 49. Assumptions Matter Consider the problem of minimizing f(x,y) = x subject to g(x,y) = (x - 1)3 - y2 = 0. (a) Show, without using calculus, that the minimum occurs at P = (I, 0). (b) Show that the Lagrange condition Vf p = ). Vgp is not satisfied for any value of J... (c) Does this contradict Theorem I?

50. Marginal Utility Goods I and 2 are available at dollar prices of Pl per unit of Good I and p2 per unit of Good 2. A utility function U(x1, x2) is a function representing the utility or benefit of consuming xj units of good j. The marginal utility of the jth good is iJU /iJxj, the rate of increase in utility per unit increase in the jth good. Prove the following law of economics: Given a budget of L dollars, utility is maximized at the consumption level (a, b) where the ratio of marginal utility is equal to the ratio of prices: marginal utility of Good I marginal utility of Good 2

Ux 1 (a,b)

= C.

(a) Show that the maximum of U(x1,x2) subject to the budget constraint

is equal to c2/(4p1 p2).

(c) Prove the following interpretation: ). is the rate of increase in utility per unit increase in total budget c. 52. This exercise shows that the multiplier J.. may be interpreted as a rate of change in general. Assume that the maximum of f(x, y) subject to g(x, y) = c occurs at a point P. Then P depends on the value of c, so we may write P = (x(c), y(c)) and we have g(x(c), y(c)) = c. (a) Show that Vg(x(c),y(c)) • (x'(c),y'(c))

Hint: Differentiate the equation g(x(c), y(c)) the Chain Rule.

=1

= c with respect to c using

(b) Use the Chain Rule and the Lagrange condition VJP =). 'vgp to show that

PI

= Ux2 (a, b) = P2

51. Consider the utility function U(x1, x2) = x1x2 with budget constraint PIX!+ pzX2

(b) Calculate the value of the Lagrange multiplier). occurring in (a).

d dJ(x(c),y(c))

=).

(c) Conclude that A is the rate of increase in f per unit increase in !he "budget level" c.

,

'I I

,J.

Chapter Review Exercises

18 > 0. Show that the maxim um of I} f(x,, ... ,Xn) == x

. tX2 . . 'Xn . t to the constraints x1 + ... + bJeC Xn :: B sU ors for xi = ... = Xn == BI n. Use thi and xi 0 for . s to conclude that J - 1, ... ,n (a1a2 .. · an)l/n < a,+ ... + rorall pasitive numbers a1, ... ,a•.

-~ n

1.,et 8 > 0. Show that the maximu f 54· to ,2 + ... + x2 - B2 . r:: mo f(x, x ) ,;ect ,,, n- IS,vnB · coneI'"·, n ==x,+ .. . + Xn sUb, Ude that

- a, + ... + a~)l /2

latl + .. · + Ian I < ./n(

2

l numbers a,, ... , an, for al 55, Given constants E, E,' E2, £3, considerth S(x1 ,x2,x3) .

= x, lnx, +x2 I

Xt

+xi +x3

x, = A-

I eµ E,

for i

= l, 2, 3,

56. Boltzmann Distribution Generalize Exercise 55 to n variables: Show that there is a constant µ, such that the maximum of

S = Xt lnx1

+ · ·· + Xn lnxn

subject to the constraints xi+··•+xn=N,

= A- 1eµE,, where

occurs for :r; . emax1mumof

n X2 + X3 In X3

subject to two constramts:

Show that there is a constant µ, such that where A= N- 1(eµE, + eµE2 + eµE3 ).

875

= N,

This result lies at the heart of statistical mechanics. It is used to determine the distribution of velocities of gas molecules at temperature T; x; is the number of molecules with kinetic energy E;; µ, = -(kT)- 1, where k is Boltzmann's constant. The quantity Sis called the entropy.

cHAaPP~TEiRRRE\ Evm 1E;;~~DE~X~ER;;::-c;; ,5;: Es: - - - - - - - - - - - - - - - .

.,/;.·2 _ y2 -- -- · X -f-3 . (a) Sketch the dormJ" ef f .

(c) What are the signs of fx and fy at D?

(b) Calculate f (3, i , z...,:i f( -5 , -3).

7. Describe the level curves of: (a) f(x, y) = e4x-y

I Given f(x ,y) '

(c) Find a point sati?::'ymg f(x, y)

z.

(d) At which of the labeled points are both fx and Jy negative?

= I.

(c) f(x, y)

=

3x 2 -

4y2

(b) f(x, y)

= ln(4x -

(d) f(x,y)

= x + y2

y)

Find the domain and range of:

= - y + ..Ji=z f(x, y) = ln(4x 2 - y)

(a) f(x,y,z)

(b)

3. Sketch the graph f(x, y) and horizontal traces.

= x 2 - y + 1 and describe its vertical y

4. (CAS] Use a graphing utility to draw the graph of the func- .

X

tion cos(x 2

+ y2)el-xy in the domains [-1, l] x [-1, l], [-2,2] x [-2,2], and [-3, 3] x [-3, 3], and explain its behavior.

(B)

(A)

5. Match the functions (a)-(d) with their graphs in Figure I. (a) f(x,y)

= x2 + y

(b) f(x, y)

= x 2 + 4y2

(c) f(x, y)

= sin(4xy)e-x2 -y2

(d) f(x, y)

= sin(4x)e-x

2

-y

2

y

6. Referring to the contour map in Figure 2: (a) Estimate the average rate of change of elevation from A to B and

from A to D. (b) Estimate the directional derivative at A in the direction of v.

X

(C)

(D)

FIGURE 1

\' 876

CHAPTER 14

DIFFERENTIATION IN SEVERAL VARIABLES

. 17-20, compute fx and fy· In Exercises 18. f(x, y) = 4xy3 2 )-2x+Y 17, f( x,Y -x-Y 20. f(x,y) = ln(xl ) - sin(xy)e i- Xyl 19. f(x,Y ) _ ysin(x + z). ) for f(x,y,z Zl Calculate f xxyz R th fu • w that for any constants a,_,.,, e nction U(r 22. Fix c > 0. Sho . ti the wave equauon •X)~ . + R)sin(ax) saus es

sm(act ,.,

Contour interval = 50 m 0

W :::C

2km

FIGURE 2

8 : Match each function (aHc) with its contour graph (iHiii) in Figure 3:

(c)

2a u OX2

. f the tangent plane to the graph off( 23, Find an equauon o x,y)~ 2 + 3x 3y at P == (I, 3). xy -xy f (4 4 4)- 3 and fx(4,4) == Y ,4) = -I. D 24. Suppose that!( ' - timate /(4.1,4) and /(3.88,4.03) se~ Linear Approximauon to es .

. A proximation of J(x,y,z) = 25 Use a Linear p . al < •. 2 + 69.5. Compare with a c culator value. esumate v 7.1 + 4·9 _ 1 is tangent to the graph of z:::: /( 26. The plane z == 2x - Y X. ))

(a) f(x, y) = xy (b) f (x, y)

2

a2u

= exy

f (x, y) = sin(xy)

at p == (5, 3). nd f (5, 3). (a) Detennine f (5, 3), fx( 5, 3), a Y

y

(b) Approximate f (5.2, 2.9).

. 4 hows the contour map of a function f (x, y) together\1/j,t 27. F1gure s di ti To - , , ts (1) ,, a path r(t) in the counterclockwise rec on. ''.' ;,0 m r ' r(2), aJid r(3) are indicated on the path. Let g(t) == f(r(t)J. ,fhich of statemen~ (i)-(iv) are true? Explain. (i) g'(J) > 0. (ii) g(t) has a local minimum for some I :5 t :5 '-· (i)

(ii)

(iii)

(iii) g' (2) == 0.

(iv) g1(3)

FIGURE 3

=0.

In Exercises 9-14, evaluate the limit or state that it does not exist.

9.

11.

lim

lim

(xy + y2) xy +xy2 x2

13.

14.

lim

lim

12.

+ y2

(2x

Jim

10.

Jim

ln(3x + y) x3y2 +x2y3 x 4 + y4

+ y)e-x+y

(.r - I)(eY - !) X

FIGURE4

15. Let (xy)P

f(x,y)=

I

4 4 X +y

0

(x,y)

i

(x,y)

= (0,0)

(0,0)

Use polar coordinates to show that f(x, y) is continuous at all (x, y) if p > 2 but is discontinuous at (0, 0) if p :5 2.

16. Calculate fx(l, 3) and /y(I, 3) for f(x,y)

= J7x + y2.

28. Jason earns S(h,c) = 20h (! + 100 )1- 5 dollars per month at a used car lot, where h is the number of hours worked and c is the num· ber of cars sold. He has already worked I60 hours and sold 69 cars. Right now Jason wants to go home but wonders how much more he might earn if he stays another IO minutes with a customer who is con· sidering buying a car. Use the Linear Approximation to estimate how much extra money Jason will earn if he sells his 70th car during these 10 min.

Chapter Review Exercises

c,ercises 29-32, compute

/nv

i·9

dt

J(x,Y) == x + eY, r(t) ==

.n J(x,y,z) ,.,.

!!_ f (r(t))

= xz - i,

at the given · value oft.

(3t - 1,t2} att::2

r(t) ==

(r ' 13' 1 -

) t att

= _2

,ntatt::J

1 () 3Z. f( x, Y) -- tan- lx,rt::(cost,sint)r-rr ' - 3 1n£xercises 33-36, compute the directional d . . ,ection of v. erivative at p in the di-

33, J(x,y) == x3y4,

p

=(3, -1),

34, J(x, y, z) = zx - xy2,

p ==

35, J(x,y) == ex2+y2, 36, J(x, y, z) = sin(xy

+ z),

(,Ji ,./2,) 2 ' 2

V

'

45. Calculate az/ax, where xez + zeY

determine whether they are a local minima or a local maxima.

= (2,-1,2)

In Exercises 47-50, find the critical points of the function and analyze them using the Second Derivative Test.

= (3, -4)

47. f (x, y) == x4 - 4xy + 2y2

V

p == (0 0 0) . ' ' ' V=J+k

38, Find an equatim, cf the tangent plane at p _ (O 3 face with equation - ' , -1) to the sur-

48. f(x, y) = x 3 + 2y3

3

39, Let n # 0 be an mteger and r an arbitrary constant. Show that the tangent plane to the surface xn + yn + zn =rat p = (a, b, c) has equation

-

xy

49. f(x, y) = ex+y - xe 2Y

= sin(x + y) - 21 (x + Y2) Prove that f(x, y) = (x + 2y)exy has no critical points.

50. f(x, y) 51.

+ ez+I == xy + y -

= x4 -

2x 2 + y2- 6y. (a) Find the critical points off and use the Second Derivative Test to

46. Let f(x,y)

square.

. . . 37, Find .the unit vr.·~tcr e at p == (0 ' 0' l) pomtmg m the direction 2 a]ong which f(x , y, z) =-= xz + e-x +y increases most rapidly.

7,,x ··"

52, Find the global extrema of /(x, y) square [O, 1] x [O, 1].

53. Find the global extrema of /(x, y) {y::: 4,y::: x 2}. 54. Find the maximum of /(x, y, z) g(x,y,z) = 2x + y +4z = 1.

40. Let f(x, y) == (x - y)eX. Use the Chain Rule to calculate anau and anav (in terms of u and v), where X = u - Vandy= u + v. 41. Let f(x, y, z) == x 2 y + iz. Use the Chain Rule to calculate anas and af/8t (in terms of sand t), where

x = s + t,

y=

st,

z = 2s - t

42. Let P have spherical coordinates (p,0,¢) ==

*IP assuming that

fx(P)

= 4,

Jy(P)

Recall that x = p cos0 sin¢, y

43. Let g(u, v)

= -3,

(2, %, %). Calculate

fz(P)

=8

= p sin0 sin¢, z = P cos¢.

= f(u3 _ v3, v3 -

u3). Prove that

v2ag +u2ag ==0 au av

= x + y.

(b) Find the minimum value of/ without calculus by completing the

v == 2i + j

p == (1, l, l),

44. Let f(x, y) == g(u), where u = x 2 + y2 and g(u) is differentiable. Prove that

11 f(X, y) == xe3Y - ye3x' r(t) == (e' 1 ) J'

877

=x3 -

= 2xy -

xy -

y2 + y on the

x - yon the domain

= xyz subject to the constraint

55. Use Lagrange multipliers to find the minimum and maximum values of /(x, y) = 3x - 2y on the circle x2 + y2 = 4.

56. Find the minimum value of f(x, y) = xy subject to the constraint 5x - y == 4 in two ways: using Lagrange multipliers and setting y = 5x -4in /(x,y). 57. Find the minimum and maximum values of f(x, y) ellipse 4x 2 + 9y2 = 36.

= x2y on the

58. Find the point in the first quadrant on the curve y = x + x-1 closest to the origin.

= x +2y + 3z subject to the + y2 + z2 = 1.

59. Find the extreme values of f(x, y, z) two constraints x + y + z == I and x2

60. Find the minimum and maximum values of f (x, y, z) = x - z on the intersection of the cylinders x 2 +·i = I and x 2 + z2 = I (Figure 5).

878

CHAPTER 14

DIFFERENTIATION IN SEVERAL VARIABLES

y

FIGURE 6

FIGURE 5

61. Use Lagrange multipliers to find the dimensions of a cylindrical can with a bottom but no top, of fixed volume V with minimum surface area.

a show that the lllill!niuil\ ·

. 63. Given n nonzero numbers O'J, · · · ' n' value of

n

62. Find the dimensions of the box of maximum volume with its sides parallel to the coordinate planes that can be inscribed in the ellipsoid (Figure 6)

-b + (z)2 -c =1 (-ax)2 + (y)2

2 2 -x2a2 + .. . + xnan

f(XJ, .. . , Xn)- I I

'r'

subject to XJ

L + .. · + Xn == 1 is c, where c == ( ;=i

-1

-2 aj

)

15 MULTIPLE INTEGRATION

------------------

I

nte~als of functions of several variables called multiple integrals, are a natural extension of the single-variable integrals s~died in the first part of the text. They are · applications, · · h pu e many quantities that appear m such as vo Iumes, masses, eat flow' total charge, and net force.

icanic-rock columns making up . W . ·rs rower m yom1ng resemble the peVl "S of volume in a Riemann sum IUJII" cO resentation of the volume under the graph reP function of two vanables. As in the 3 of le-variable case, we define integrals in sing .nd (hree variables as limits of Riema r,'0"'' nn

111eVO

~ -1 Integration in Two Variables The integral of a function of two variables f(x, y), called a double integral, is denoted

5 soJ11; : . . · - - - - - - - - - - -

f fvf(x ,y) dA ~hen _f (x, ~)::: 0 on a domain Din the xy-plane, the integral represents the volume of he ~olid region between the graph of J(x, y) and the xy-plane (Figure 1). More generally, th e integral represents a signed volume, where positive contributions arise from regions above the xy-plane and negative contributions from regions below. There are many similarities between double integrals and single integrals:

z =f(x, y)

.

• Double integrals are defined as limits of Riemann sums. • Double integrals are evaluated using the Fundamental Theorem of Calculus (but we have to use it twice-see the discussion of iterated integrals below) .

,.________ .

j

y

An important difference, however, is that the domains of integration of double integrals are often more complicated. In one variable, the domain of integration is simply an interval [a,b]. In two variables, the domain Dis a plane region whose boundary can be made up of a number of different curves and segments (e.g., V in Figure 1 and R in Figure 2).

FIGURE 1 The double integral off (x, y) over the domain D yield2 the volume of the solid region between !he graph of J(x,y) and the xy-phme over D .

In this section, we focus on the simplest case where the domain is a rectangle, leaving more general domains for Section 15.2. Let R

= [a , b] x [c,d]

denote the rectangle in the plane (Figure 2) consisting of all points (x, y) such that

a :S x :Sb,

R:

Like integrals in one variable, double integrals are defined through a three-step process: subdivision, summation, and passage to the limit. Figure 3 illustrates the subdivision step which itself has three steps:

1. Subdivide [a , b] and [c, d] by choosing partitions: b

y

a = xo
1 (use the Triangle ln~uaiity), and apply the Comparison Theorem to show that l(a) converges.

(b)

(c) Show that l(a) =

fL

f (x, y) dA = F(b, d) - F(a, d) - F(b, c) + F(a, c)

1•

e-xy dy dx.

. 1•

l(a)=lna- hm T

2F

a

53. LetF(x,y)=x- 1exY.Showthat axay =yexYandusetheresultof

L

yexy dA for R = [I, 3] x [0, l].

a2F

2

54. Find a function F(x, y) satisfying ax ay = 6x y and use the result of

L

Exercise 52 to evaluate/

fa""

(d) Prove, by interchanging the order of integration, that

where n = [a,b] x [c,d].

Exercise 52 to evaluate/

A

Side of area A

6x 2y dA for R

0

X

=

y

y

4 e-x _ e-5x

y=--x

e-x _ e-ax

, though not X defined al x = 0, can be defined and made continuous at x = 0 by assigning the value f (0) =a - I. (a) Use L'Hopital's Rule to show that f(x)

e-Ty -dy

(e) Use the Comparison Theorem to show that the limit in Eq. (I)~ zero and therefore that I (a) =In a ( see Figure 19, for example). Hil'J: If a:".: 1, show that e-Ty /y::: e-T for y :".: 1, and if a < 1, show that e-TY/y :'.:: e-aT /a fora::: y::: I.

= [0, I] x [O, 4].

55. In this exercise, we use double integration to evaluate the following improper integral for a > 0 a positive constant: oo e-x -e -ax !(a)= fo - - - d x

I

2 FIGURE 19 The shaded region has area In 5.

G

SECTION 15.2

Double Integrals over More General Regions

891

~5.2 Double Integrals over More General Regions In the previous section, we restricted our attention to integrals on rectangul_ar doma(ns. Now, we shall treat the more general case of domains 'D whose boundanes are ~m~ple _closed curves (a curve is simple if it does not intersect itself and cl?sed tf tt beg_ms and ends at the same point). We assume that the boundary of 'D ts smo~th as m Figure l(A) or consists of finitely many smooth curves, joine~ together _with P~ssible comers, as in Figure l(B). A boundary curve of this typ_e 1s called ~•e~eWlse smooth. We also assume that 'D is a closed domain; that ts, 'D contams its boundary.

- -~ - - z=f(x,y)

-J(x,y) I

'

''' '' '''

'' '' '' '

:---------: ! !------..

X

v;-\ !

:-------: ' '

y X

\

--- y

'---Boundary

"'-Boundary (A) V has a smooth boundary.

:-.....__

:

(B) V has a piecewise smooth boundary, consisting of three smooth curves joined at the comers.

FIGURE 1

z=f(x,y) '' '' ''' '' ' I

''' '' '''

f(x,y)

!

~j '

'

Fortunately, we do not need to start from the beginning to define the double integral over a domain 'D of this type. Given a function f (x, y) on 'D, we choose a rectangle R = [a,b] x [c,d] containing 'D and define a new function f(x,y) that agrees with f(x,y) on 'D and is zero outside of 'D (Figure 2):

= [of(x,y)

if(x,y) E 'D if(x,y)\f'D

The double integral of f over 'D is defined as the integral of f over R: y

X

ffv f(x,y)dA = ffn f(x,y)dA FIGURE 2

ofD.

The function f is zero outside

IT] n

We say that f is integrable over 'D if the integral off over exists. The value of the integral does not depend on the particular choice of R because f is zero outside of D. This definition se~ms reasonable because the integral of f involves only the values off on D. However, f is likely to be discontinuous because its values could jump to zero beyond the boundary. Despite this possible discontinuity, the next theorem guarantees that the integral off over R exists if our original function f is continuous on D.

892

CHAPTER 15

MULTIPLE INTEGRATION

d domain V whose bound - -- - - - - - - - . us on a close ary i THEOREM 1 If f(x,y) is contJJJUO then f f(x, y) dA exists. sa simple closed piecewise smooth curve, J__1v_ __ ______

1•

In Theorem 1, we defme · continuity on 'D to mean that J IS· .,, ,. and continuous on ue,med

Isome open set containing 'D.

. al defines the signed volume b . double 1ntegr et\\, As in the previous sectton, the the graph off(x, y) and the xy-pJane._ b Riemann sums for the function j We can approximate the double_inte~O / points pin 'R,. that not belong10%a'D P) _ 1or th do t 1. . rectangle n containing 'D. Because f( m over those s"'"ple points a 1e m D·• a.,u ' any such Riemann sum reduces to a su

!fz

'D

2 •m--• •-

- 4• - • - -•

· /----+-_-"'J

I

i I

I

·r

1.5

2

S4,4

-t---'----'---'---4 X

0.5

FIGURE 3

I

ff f~(P--),, t.x·' llYi = ""£ f(Pu) t.xi t1yi i==I j==I

Sum only over points Pu that lie in 'D

f'f

. al (x + y)dA, where 'Dis the sh for the mtegr Jv . adeo . ers of the squares as sample pomts. domain in Figure 3. Use the upper-nght com . 1 1·n Figure 3 have sides oflength 6. Solution Let f(x, y) = x + y. The subrectang es . . x,, I I On! 7 f the 16 sample points he m 'D' so t..y = 2 and area t..A = 4. y o EXAMPLE 1 Compute

y

1.5

f(x,y)dA

S4,4

= ttj(Pij)t.x t..y = i(f(0.5,0.5) + f(l,0.5) + f(0 .5, 1) +/0,I) i==l j==l

+f (I.5, 1) + /(1, 1.5) + f (I.5, 1.5))

i

Domain 'D.

= (1 + 1.5 + 1.5 +2 +2.5 + 2.5 + 3)

=;

The linearity properties of the double integral carry over to general domains: li f (x, y) and g(x, y) are integrable and C is a constant, then

fl(f(x,y)+g(x,y))dA= flt(x,y)dA+ !fvg(x,y)dA f l Cf(x,y)dA =Cfl

y X

1)

FIGURE 4 The volume of the cylinder of height 1 with 'D as its base is equal to the area ofV. y

f(x,y)dA

Although we usually think of double integrals as representing volumes, it is worth noting that we can express the area of a domain D in the plane as the double integral of the constant function f (x, y) = l: area('D)

=/Iv

l dA

0

Indeed, as we see in Figure 4, the area of 'Dis equal to the volume of the vertical cylinder of height l with 'Das base. More generally, for any constant C,

!iv

C dA = C · area(D)

0

CONCEPTUAL INSIGHT Equation (3) tells us that we can approximate the area of a do· -l--'---'---'-'1-1' - -.........._x

D.X

FIGURE 5 The area of Vis approximated by the sum of the areas of the rectangles contained in V.

iv

main 'D by a Riemann sum for/

1dA. In this case, J(x, y)

= 1, and we obtain a

Riemann and adding up the areas 1J.X1 A . /J,y A • f th ose re ctangles • • sum by creating • a grid • 1o the gnd ~hat are conta1Ded ID 'D ~r th~t intersect the boundary of D (Figure 5). fbe finer the gnd, the better the approximation. The exact area is the limit as the sides of the rectangles tend to zero. ID

----

- -ar

-

Double Integrals over More General Regions

SECTION 15.2

893

· s?. We addressh . Now• how do we compute double integrals over nonrectangu1ar region this question next, in two special cases where the region lies between two graphs. In eac case, the computation involves an iterated integral.

~gions Between Two Graphs When D is a region between two graphs in the xy-plane, we can evaluate ~ouble_ i~t~grals over D as iterated integrals. Recall from Section 6.1 that V is vertically sunple if it is the region between the graphs of two continuous functions Y = g1(x) and Y = g2(x) over a fixed interval of x-values [Figure 6(A)]:

V

= {(x,y) : a::: x::: b,

g1(x)::: y::: g2(x)}

Similarly, Dis horizontally simple [Figure 6(B)] if D = {(x,y) : c::, y::, d,

h1(y)::;

y

x::: h2(y)}

y

x• hly)

d

y

L--------+x

b

a x

(A) Vertically simple region

(B) Horizontally simple region

FIGURE 6

THEOREM 2 If Dis vertically simple with description

a::: x::: b, then When you write a double integral over a vertically simple region as an iterated integral, the inner integral is an integral over the dashed segment shown in Figure 6/AJ. For a horizontally simple region, the inner integral is an integral over the dashed segment shown in Figure 6(8).

fl

D

f(x,y)dA

=

l

b 1g2(x)

a

f(x,y)dydx

81~)

If V is a horizontally simple region with description c:,sy:,sd,

then

fl

D

f(x,y)dA

=

i

C

d 1h2(y)

f(x,y)dxdy

h1(y)

Proof We sketch the proof, assuming that Dis vertically simple (the horizontally simple case is similar). Choose a rectangle R = [a,b] x [c,d] containing D, and let J be the function shown in Figure 2 that equals f on D and otherwise equals zero. We wish to

894

CHAPTER 15

MULTIPLE INTEGRATION

states that the functio the theorem n ni However, . ssible, however, to list k employ Fubini's Theorem next. . ous on 'R,. It 1~ po ontinuou Prove . b /t be conttnU f that are c s on a ti,. contmuous, ut may no tions such as . F bini's Th doth:~ Fubini' s Theorem also bolds for funC . Therefore, using u eorern (fo ·,% in R and that are zero outside the domain. t¾ second equality) we have

lb

r{ -

dA ff J(x,y)dA === f(.x,Y) - a Jlv 'R d 'd V so for fixe .x, By definition, J(x, y) is zero outsi e ' g, (x) y g2 (x). Therefore,

f

Jr,

Winifred Edgerton Merrill (1862-1951) was the first American woman to earn a Ph.D. in mathematics, awarded by Columbia University in 1886. Her thesis was a study of geometric interpretations of multiple integrals and their representation in different coordinate systems.

y='IX

y=

X

3

FIGURE 7 Domain between y

y

= 1/x.

df_(

.X,J

)dy===

1

= Jx and

J(x,y)dydx

c

j(x

,

y) is zero unless y

0J satisri

Ie1

f(.x,y)dy

g1(X)

C

· th desired equality: Substituting in Eq. (5), we obtam e

Iii

f(x,y

V

) l dA===

b lgz(x)

g1(x)

J(x,Y

) d dx

y

. . vertically simple region is similar to integratj0 Integrat10n over a honzontally or . . f the inner integral may be fu . n O over a rectangle with one difference: The hrmts ncti°ils instead of constants.

f!v

x2ydA, where Vis the region in Figure 7.

Solution Step 1. Describe 1) as a vertically simple region.

_!_~y~.Jx

lj)

I

f(Pj)

1.8

2.2

3

4

0.9

1.2

2.1

2.4

4

f(x,y)dA

'D

L,f(Pj)Area(Dj) j=I

=(1.8)(1) +(2.2)(1) +(2.1)(0.9) + (2.4)(1.2)

8.8

I

15.2 SUMMARY • We assume .that Vis · a s1mp · Ie cJosed . a closed, bounded domai·n whos e boundary 1s curve that either 1s smooth or has a finite number of comers. Th e dou ble mte · gra1 is b defime d y

!iv

f(x,y)dA =

JIn

](x,y)dA

r

tit. -

Double Integrals over More General Regions

SECTION 15.2

901

~here R is a rectangle containing 'D and j(x,y) = f(x,y) if (x,y) E 'D, ~d = 0 otherwise. The value of the integral does not depend on the choice

tf;~)

• The double integral defines the signed volume between the graph of /(x, y) and the xy-plane, where the signed volumes of regions below the xy-plane are negative.

j jv C dA = C . area('D). • If D is vertically or horizontally simple, j1f (x, y) d A can

• For any constant C'

ated integral:

be evaluated as an iter-

V

Vertically simple domain a S x Sb, g1(x) Sy :S g2(x)

b f.g2(x)

l 1 a

Horizontally simple domain c:=:y:sd, h1(y):=:x:=:h2(y)

· If f(x, y) :::: g(x, y) on 'D, then

d 1h2(Y)

c

f(x,y)dydx

g,(x)

f(x,y)dxdy

h1(y)

jfv f(x, y)dA :S jfv g(x, y)dA.

• If m is the minimum value and M the maximum value of f on 'D, then m · area('D)::::

fL

f(x, y) dA :S M · area('D)

• If z1(x, y) :::: z1(x, y) for all points in 'D, then the volume V of the solid region between the surfaces given by z = z1(x, y) and z = z1(x, y) over Dis given by

= j fv (z1(x, y) -

V

z1 (x, y)) dA

• The average value of f on 'D is

f

= _ l_

area('D)

f'flv

f(x,y)dA

= ffv f(x,y)dA ffv 1dA

• Mean Value Theorem for Double Integrals: If f(x, y) is continuous and 'Dis closed, bounded, and connected, then there exists a point P e D such that

jfv f(x, y)dA = f(P) • area(D) Equivalently, f(P) = f, where 7 is the average value off on D. • Additivity with respect to the domain: If D is a union of nonoverlapping (except possibly on their boundaries) domains D1, . .. , DN, then

!'lv{

f(x,y)dA

=

Lf'flvi_ N

f(x,y)dA

j=I

• If the domains 'D1, ... , 'D N are small and Pj is a sample point in D j, then

j1f(x,y)dA 'D

N

Lf(Pj)area('Dj) j=l

I\

902

CHAPTER 15

MULTIPLE INTEGRATION

~ 22~EX~ER;;;c=,s=Es___________

-------~

I::=~=---------------

--------

Preliminary Questions

. wing exprcssmns do not make sense?

(a)

(c)

fi x

(b)

lo' [

f(x , y)dy dx

f 1Y

(d)

lo' [

f(x,y)dydx

f(x ,y)dy dx

f(x,y)dydx

-I

2. Draw a domain in the I th 1. . simple. P ane at s neither vertically nor horizontally 3. Which of the four regio

1 1~ 0

-..fi./ 2 -x

FIGURE 21

it disk If the maximum value of f(x,y) on D •/ Un 4. LetV be the · 1s4 'ble value of f(o: ,y)dA is (choose the c _, then the largest poss, v orreq

1Ji

· F' . ns 10 igure 21 IS the domain of integration for

f(x,y)dydx ?

""

y

I. Which of the follo ,

.

answer):

I

4

(c) -

(b) 4ll'

(a) 4

]'[

Exercises y

1. . Cal~ulate the Riemann sum for f(x, y) = x - Y and the shaded domam m_Figure 22 with two choices of sample points,• and o. Which do you tbmk IS a better approximation to the integral of/ over V? Why?

y =I -x 2

y

31--f----.--l.-. 2H-=...._________..

FIGURE 24 2

3

4. Sketch the domain

FIGURE 22

and evaluate /

2. Approximate values of f(x, y) at sample points on a grid are given in Figure 23. Estimate

fL

f(x,y)dx dy for the shaded domain by comput-

ing the Riemann sum with the given sample points.

l

V: O::: x ::: I,

2

x :::

y::: 4 -

x

2

yd A as an iterated integral.

= x2y' over the

In Exercises 5-7, compute the double integral of f(x , y) I given shaded domain in Figure 25. 5,

6.

(A)

-y

y

7. (C)

(B)

y

$g_, y

I

~' ~' (A)

..

(C)

(B) I

FIGURE 25

-1.5 FIGURE 23

8. ~ ketch the domain V defined by x + y :::: 12, x pute

3. Express the domain V in Figure 24 as both a vertically simple region and a horizontally simple region, and evaluate the integral of f(x, y) = xy over V as an iterated integral in two ways.

fL

ex+y dA .

9. Integrate f(x , y) x+2

4, y

4 and com_..

=x over the region bounded by y =x2 and Y"' _

:n

oouble Integrals over More General Regions

SECTION 15.2

sketch the region D between _ 2 . I . y - X and a s1mp e region and calculate th . Y = x(l - x). Express e integral 0 f 1) ;er · f(x,y) = 2y

JO·

35

V

0

1

!]valuate/ dA, where 1) is th h I1• 2 . ,..I v x26 e s aded part inside the semicircle of radius m r gure . , calculate the double integral of f( I•· x,y)- 2 figure 27. - Y over the rhombus 'R in

=

::Sa

0

903

29. Sketch tlie domain V corresponding to

f lo

f

4

2

/4x2+5ydxdy

ft

Then change the order of integration and evaluate. 30. Change the order of integration and evaluate

f 1"

12 xcos(xy)dxdy

Explain the simplification achieved by changing the order.

y

1

31. Compute the integral of f(x , y) = (lny)- over the domain D bounded by y = e' and y = e.fi. Hint: Choose the order of integration that enables you to evaluate the integral.

4

32. Evaluate by changing the order of integration:

f4J lo

2

.fi

FIGURE 26 y

In Exercises 33-36, sketch the domain of integration. Then change the order of integration and evaluate. Explain the simplification achieved by

= ./4---.J

FIGURE

J3. Calculate the do.!ble. ,.. 1egral of f(

V=' {(x,y) : x2 +

sin y3dydx

y2 ". ~- 1 c: 0).

27 lxl + ½IYI:,: I.

changing the order.

33.

)_

x,y - x + Y over the domain

J4. Integrate f(x , y\ -~ix + y + 1)-2 . (0,0), (4,0), and (0, 8). over th e tnangle with vertices

J5. Calculatetheinte;;rnl of f(x , y) =xoverthere ion Vb by y "'x(2 - x) an:! i:eiow b . g Apply ounded above y x -_ Y(2 - y) Hmt· th d · formula to the low,,r boundary cur;e to solve ;,or Y as- a function . e qua of x. rat1c J6. Jntegratef(x ,y) = xovertheregionboundedby _ _ 2 d o· t Y-X,Y- 4x-x y = _m wo ways: as a vertically simple region and as a horizontally' simple region.

111.smx dx dy

1 0

35.

y

34.

X

j'

[1 xeY3 dydx lo y=x

37. Sketch the domain

36. 1)

Jfv e'

(4

f

[1 lo

j'

JL

2

ft

J;'3+t dx dy 23

xel dydx

r::=x 1

where O:,: x 5 2, 0 :": y

greater than I. Then compute 38. Calculate

lo

:c 2,

and x or y is

ex+y d A.

dA, where

1)

is bounded by the lines y =

y = x, x = 0, and x = I.

In Exercises 39-42, calculate the double integral of f(x, y) over the triangle indicated in Figure 28.

In Exercises 17-24, compute the double integral of f(x, y) over the domain V indicated.

y

3

17. J(x,y)=x y; O:,:x:,:5, x:,:y:,:2x+3 18. J(x,y)= -2; 0 :ox 5 3, 19. f(x, y)

= x;

0 .:C x .:c I,

20. J(x,y)=cos(2x+y);

I 5Y 5e' I

:c y 5 e' 2

½ :ox 5

½,

15 y ,:,:2x

21. J(x,y) =1ixy - x2 ; bounded below by y = x2, above by y =

2 3 4 5

2 3 4 5

(A)

(B)

y

.jx

·-+- ~.

4 -·I .· J : ± . -:k• 3 !. 2 _,___,; i l i

22. J(x, y) = sinx; bounded by x = 0, x = I, y = 0, y = cosx

I -f--~--x 2 3 4 5

23. f(x,y) = ,,x+Y; bounded by y = x - I, y = 12 - x for 2::: y,:,: 4

(C)

25'.

ff

f(x,y)dydx

J,.jy f(x,y)dxdy

1 9

27. 4 2

J. 1.jy,f(x,y)dxdy 9

26.

28.

3

4

lo';;

FIGURE 28

40. 41.

f{x,y)dydx

= ex', (A) f(x,y) = I - 2x, (B) f(x,y) = ..:;., (C)

39. f(x,y)

42. f(x, y)

y

= x + I,

(D)

2 3 4 5 (D)

24. f(x,y) = (x + y)- 1; bounded by y = x, y = I, y = e,x = 0 In Exercises 25-28, sketch the domain of integration and express as an iterated integral in the opposite order.

x+ I,

X

l1

904

CHAPTER 15

MULTIPLE INTEGRATION e height of

th "ceiling" in Figure 3J d

efin

e

ind the averag O< )' :'.: J. 55• F o1.

jjf

W f(x,

11>2(0,,j,)

.P=,J,1

0 z 2 < Rz, x 2:. 0, Y 13. Quarter circle x + y 2 0. Assume the mass density is S(x, y) = 1 for Exercise 39 and S(x, y) = xy for Exercises 40-42. Compute the given quantities. y

X

2 2 FIGURE 16 Upper half of ellipsoid x2 + y2 + (Rz/ H) = R , Z 2:: 0.

In Exercises 25-28, find the center of mass of the region with the given 111/JSs density 8.

25, Region bounded by y == 6 - x, x = O, Y = O; S(x, y) = x

26. Region bounded by y2 == x + 4 and x = O; S(x, y) = [y[

2

39. Centroid

40. Center of mass

41. Ix

42. Io

43. Calculate the moment of inertia Ix of the disk 'D defined by 2 +y2 R2 (in meters), with total mass M kilograms. How much kinetic energy (in joules) is required to rotate the disk about the x-axis with angular velocity 10 radians per second?

x

44. Calculate the moment of inertia lz of the box W == [-a,a] x [-a, a] x [O, HJ assuming that W has total mass M.

/•

938

CHAPTER 15

MULTIPLE INTEGRATION

45. Show that the

f• . . moment o merua of a sphere of radius R of lot.al mass M wi.th uniform mass dens1ty · abo ut any :ws • passmg • lhrough the cenler of the sphere · 2 M R2 N . . is 3 • ole that the mass density of the sphere 1s S = M/(jrrR3).

46: Use the result of Exercise 45 to calculate the radius of gyration of a uniform sphere of radius R about any axis through the cenler of the sphere. In_Exercises 47 and 48, prove the formula/or the right circular cylinder in Figure 18.

47. I,=

½MR 2

components in a cellain . nths) of rwo b'J. d de,.· ·'etime (m mo •oint proba I tty ensity f •1ee l 53. The ••: X and y !hat have J Unctio d m vanables n ran O ) if X 2: 0, y 2: 0, 2x + y I (48 - 2x - Y . :S 48 §2i6 o1herw1se p(x,y)::::: { 0 . . !hat bolh components function for at 1 Calculale the pro~a_b1htyNote that p(x, y) is n~nzero only Within C.S111 months withoul fading. di te axes and the )me 2x + y :::: 48 shlhe ~nded by the coor na o\\>ii , an~ bou ' Figure 20. y (months)

48

2x+y=48 36

24

l

H

12

~y

-+-+-+--~x (month:;) 12

24

FIGURE 20 FIGURE 18

49. The yo-yo in Figure 19 is made up of two disks of radius r = 3 cm and an axle of radius b = I cm. Each disk has mass M1 = 20 g, and the axle has mass M2 = 5 g. (a) Use the result of Exercise 47 to calculate the moment of inertia / of the yo-yo with respect to the axis of symmetry. Note that / is the sum of the moments of the three components of the yo-yo. (b) The yo-yo is released and falls to the end of a 100-cm string, where it spins with angular velocity w. The total mass of the yo-yo ism = 45 g, so the potential energy lost is PE = mgh = (45)(980)100 g-cm 2/s 2. Find w using the fact that the potential energy is the sum of the rotational kinetic energy and the translational kinetic energy and that the velocity v = bw because the string unravels at this rate. Axle of radius b

54. Find a constant C such that p(x,y)

ifO s x and O S y otherwise

Cxy

= {0

I- x

is a joint probability density function. Then calculate: ½; Y S ¼) (b) P(X 2: Y)

(a) P(X S

55. Find a constant C such that Cy

p(x,y) = { 0

if OS x S I and x 2 S y S x otherwise

is a joint probability density function. Then calculate the probability that Y 2: xJ/2. 56. Numbers X and Y between O and I are chosen randomly. The joint probability density is p(x, y) = I if O S x S I and OS y S I, and p(x, y) = 0 otherwise. Calculate the probability P that the product XY is at least½-

57. According to quantum mechanics, the x- and y-coordinates of a particle confined to the region n = [O, I] x [O, I] are random variables with joint probability density function 2

( )-{Csin (2,rfr)sin 2(2,rny) if(x,y) En P x' Y - 0 otherwise FIGURE 19

50. Calculate I, for the solid region W inside the hyperboloid x 2 + y2 = z2 + I between z = 0 and z = I.

The integers f and n determine the energy of the particle, and C is a constant. (a) Find the constant C.

51. Calculate P(0 S X S 2; I S Y S 2), where X and Y have joint probability density function

(b) Calculate the probability that a particle with f region [o, ¼] x [o,

x p( 'y)

={f(2xy+2x+y) ifOsxs4andOsys2 O otherwise

52. Calculate the probability that X +Y S 2 for random variables with joint probability density function as in Exercise 51.

H

= 2, n =3 lies in the

58. The wave function for the Is state of an electron in the hydrogen atom is 1/11s(P)

I = --e-P/11-0

R

Ch ange of Variables

SECTI 0 N 15,6 00

is the Bohr radius. The probab' . ility of finding th

\\'~~ W of R3 is equal to

res100

fffw .... in spherical coordinates

ll'~o,••

.

p

e e1ectron m a

Charged plate

P(x,Y,z)dV

~d-1---_ /

'

P(p)

939

= li/F1s(P)l2

integration in spherical coordinat use l . es to show th ding the e ectron at a distance greater th at the probability of ~e2 0.677. (The Bohr radius is ~~~?hr radius i_s equal to m, but this value is oot needed.)

ao =s.::

According to Coulomb's La h leetric charges of magnitude q1 andw, t e attracti ve i orce between two :qiqz/r2 (k is a constant). Let F be the ~~/::ated by a distanc7 r is f charge Q coulombs located d centimet b e on a charged particle p ~sk of radius R, with a uniform charge di~:~ ~ve the center of a circular nersquare meter (Figure 21). By symmetry Ful!o~ of density p coulombs r' acts m the vertical d' · Let 'R, be a small nclar rectan 1 f . 1rect1on. (a) • g e o size t:.r x t:.0 1 d . Show that R exerts a force on p who . ocate at distance '· se vertical component is

FIGURE 21

59·

(

kpQd

)

\ (r2 + d2)3/2 r 6.r M

(b) Explain why F is equal to the following double . t

al

60.

Let 'D be the annular region 2 -

= kpQd

(21r (R

lo lo

d

F = kpQ

rdrd0

(r 2 + d2)3/2

- 2

r

b

where b > a > o. Assume that 'D has a uniform charge distribution of p coulombs per square meter. Let F be the net force on a charged particle of charge Q coulombs located at the origin (by symmetry, F acts along the x-axis). (a) Argue as in Exercise 59 to show that

m egr , an evaluate:

F

a

_'.: < 0 < '.: ,

fb

1

,r/2

9=-rr/2 r=a

(cos0) - 2rdrd0 r

(b) Compute F.

further Insights and Challenges 61. Let 'D be the ~amain _in Figure 22. Assume that v is symmetric with

Ay=

respect to the y-axis; that 1s, both 81 (x) and 82 (x) are even functions. (a) Prove that the centroid lies on the y-axis-that is, that x = o. (b) Show that if the mass density satisfies 8(-x , y)

My =0andxcM=0.

= o(x, y),

then

y

l

b

x=a

J g2(x)

ydydx

y= g1(x)

63. Use Pappus' s Theorem in Exercise 62 to show that the torus obtained by revolving a circle of radius b centered at (0, a) about the x-axis (where b < a) has volume V = 2rr 2 ab2 . 64. Use Pappus's Theorem to compute y for the upper half of the disk x2 + y2

a2 , y

0. Hint: The disk revolved about the x-axis is a sphere.

65. Parallel-Axis Theorem Let W be a region in R3 with center of mass at the origin. Let lz be the moment of inertia of W about the z-axis, and let Ih be the moment of inertia about the vertical axis through a point P = (a, b, 0), where h = ,J a2 + b2 . By definition,

-+-----'------+-x -a

lh

a

=ffL((x -a)2 +(y-b)2)8(x,y,z)dV

Prove the Parallel-Axis Theorem:

FIGURE 22

62. Pappus's Theorem Let A be the area of the region 1) between two graphs y = 81 (x) and y = 82(x) over the interval [a, b], where 82(x) g,(x) 0. Prove Pappus's Theorem: The volume of the solid obtained by revolving 1) about the x-axis is V = 2rr Ay, where y is they-coordinate of the centroid of 1) (the average of they-coordinate). Hint: Show that

h

= lz + Mh2 •

66. Let W be a cylinder of radius 10 cm and height 20 cm, with total mass M = 500 g. Use the Parallel-Axis Theorem (Exercise 65) and the result of Exercise 47 to calculate the moment of inertia of W about an axis that is parallel to and at a distance of 30 cm from the cylinder's axis of symmetry.

15.6 Change of Variables The fonnulas for integration in polar, cylindrical, and spherical coordinates are important special cases of the general Change of Variables Fonnula for multiple integrals. In this section, we discuss the general fonnula.

940

CHAPTER 15

MULTIPLE INTEGRATION

Maps from R2 to R2 .n) to another set Y is often ca1Iec1 a A function G : X Y from a set X (the domai y and is called the image llia~ot a mapping. For x EX, the element G(x) belongs to f G We denote the image ~f-l:· 'tlie set of all ~ges_G(x) is calle~ the image or ~2 defined on a domain ?(~). In this section, we consider maps G · as our domain VariabI 1n ~l (Figure I). To prevent co~sion,_ we'll ofte~use(:• v~ (u, v)), where the co es ¾d .t, y for the range. Thus, we will wnte G(u, v) - (x ' 'Y lllponen~ x and y are functions of u and v:

J

~;1~

x=x (u , V) ,

Y = y(u, v) y

V

Domain V

-

Image 'R, = G(V)

G

FIGURE 1 G maps D to R.

L - - - - -- - x

0 ., ·Iiarw1"th 1·s the map defining nemapweare1ami • polar coordinates. . R?. . ..,, For R2.this lllap, we use variables r, 0 instead of u, v. The polar coordinates map G · ' is defined by G(r, 0)

= (r cos 0, r sin0)

EXAMPLE 1 Polar Coordinates Map Describe the image of a polar rectangle = [r1, r2] x [01, 02] under the polar coordinates map.

R

Solution Referring to Figure 2, we see that • A vertical line r = r1 (shown in red) is mapped to the set of points with radial coordinate r1 and arbitrary angle. This is the circle of radius r1. • A horizontal line 0 = 01 (dashed line in the figure) is mapped to the set of points with polar angle 01 and arbitrary r-coordinate. This is the line through the origin of angle 01.

G

--1----1---+--,

FIGURE 2 The polar coordinates map G(r,0) (rcos0,rsin0).

=

r0-plane

xy-plane

The image of n = [r1, r2] x [01, 02] under the polar coordinates map G(r, 0) is the polar rectangle in the xy-plane defined by r1 S r S r2, 01 s 0 s 0 . • 2

General mappings can be quite complicated, so it is useful to study the simplest case-linear maps-in detail. A map G(u, v) is linear if it has the form G(u, v) =(Au+ Cv, Bu+ Dv) (A, B, C, Dare constants)

We can get a clear picture of this linear map by thinking of G as a map from vectors in the uv-plane to vectors in the xy-plane. Then G has the following linearity properties (see Exercise 46): G(u1

+u2, v1 +v2) = G(u1, v1) + G(u 2, v2) G(cu, cv) = cG(u, v)

(c any constant)

I~ SECTION 15.6

Change of Variables

941

A consequence of these Vectors a and b in th properties is 1hat G maps 1he parallelogram spanned by any two G(b), as shown in Fi~\plane to 1he parallelogram spanned by the images G(a) and . . _More generally G . Jozm~g G(P) and G(Q maps the se~mentjoining ~ny two points P and Q to the segment andJ == (0, l} is ma ) (see Exei:c1se 47). The gnd generated by basis vectors i (1,0} pped to 1he gnd generated by 1he image vectors (Figure 3)

=

r = G(l , 0)

= (A, B} s = G(0, 1) = (C, D}

V

y

G

j=(O,lL ,-•

I=

(

1, 0) u

••(C, D)

1

~ x

V

Image of v-axis Y,

--

Q

---

'

(0 1)

'-

'

' q,

Image of u-axis

G

"- p u

3 A linear mapping G maps a parallelogram to a parallelogram.

l!lJ FIGURE

EXAMPLE 2 Image of a Triangle Find 1he image of the triangle T wi1h vertices (1,2), (2, 1), (3,4)under1he linear map G(u, v) = (2u -v,u + v).

Solution Because G is linear, it maps 1he segment joining two vertices of T to the segment joining the images of the two vertices. Therefore, the image of T is the triangle whose vertices are the images (Figure 4) G(3,4) = (2, 7) G(2, 1) = (3, 3), G(l, 2) = (0, 3),

•

y

V

6 (3, 4)

4

(1~2-M 2

FIGURE 4 The map G(u, v) (2u - v ,u

=

~

(0, 3)

(3, 3)

2

-

(2, 1)

2

4

2

4

+ v). To understand a nonlinear map, it is usually belp~l to dete~e the images of horizontal and vertical lines, as we did for the polar coordinate mappmg.

EXAMPLE 3 Let G(u, v)

= (uv- 1, uv) for u > 0, v > 0. Determine the images of

(a) The lines u = c and v = c -1

Find the inverse map G ·

(b) (1, 2] x [l, 2]

942

CHAPTER 15

MULTIPLE INTEGRATION

= uv -1 and y = uv. Tirns,

Solution In this map, we have x

2'. =

2

xy=U'

X

J

2

v

a oint (c, v) to a point in the xy-pJane

=

s/w·

(a) By2 the first part of Eq. (3), G maps_ line u c to the hyperbola xy :::::: c2. 1h xy =c • In other words, G maps the vertic . ta! line v c is mapped to the lllilarly, by the second part of Eq. (3), ~e h~nthzon . e through the origin of slope c28etSof · . •• pomts where y/x = c 2 , or y = c 2 x, which 1s e 1m "I: Th~ term "curvilinear rectangle" refers to a region bounded on four sides by curves as on the right in Figure 5.

I

Figure 5.

(b) The image of [1, 2] x _[l, 2] is the

that are the images of the Imes u defined by the inequalities 1

=

=

c~ztz::;::r/: ..

ctan le bounded by the four c l, = 2. By Eq. (3), this reg;:es nis y

1, u - 2,

1 -d0 4

11" 1

(b)

14

-2 rr/3 2

foy f(x )dxdy

X

11/o~ll

29. Evaluate / =

21. Prove the formula

{

,

2

3

(x +Y +z) dxdyaz.

ffl (xy + z)dV, where

xy dA as an iterated integral in the order dx dy and dydx.

[3 [1

f1

z, 2y + z = 3,

16. Let W be the region bounded by the planes y = and z = 0 for O :5 x 5 4. (a) Express the triple integral

4

f(x,y) = Jxy

11. 'D={05x:5l, l-x:5y:52-x),

15. Express

\

- ~ ~ - - - - - -- -: ---x

f(x,y)dA .

9. 'D={O:sx 54, O5y ::ox),

ffv

A where V is the shaded domain in Figure 2 .

xd ,

[1 [1 ydxdy

11 11

In

.Jv{ !

(c)

f0 "

fn

3 [

fo

lof

1°

1

r dr d0 dz ~rdzdrd0

-v9-r2

i "' 1

31. Fi nd the volume of the solid contained in the cylinder x 1 + below the surface z = (x + y )2 and above the surface z = - (x -

y)1.

ff

• vse polar coordinates to 3•· evaluate •aded region between the two . JD x dA , Where D is the 5,, c1rc 1es of . radius I in Figure 3.

Chapter Review Exercises

955

42. LetWbetheportionofthehalf-cylinderx 2 + y2 s 4.x Osuch that Os z s 3y. Use cylindrical coordinates to compute the mass of W if the mass density is p(x, y, z) = z2. 43. Use cylindrical coordinates to find the mass of a cylinder of radius 4 and height IO if the mass density at a point is equal to the square of the distance from the cylinder's central axis.

44, Find the centroid of the region W bounded, in spherical coordinates, by rp = 0 ~1 a pomt P, an outflow of gas occurs near this point. In other words, the gas is expan:e.wg around the point, as might occur when the gas is heated. When div(F) < 0 at a pom! P, the gas is compressing toward P, as might occur when the gas is cooled. When div(Y) = 0, the gas is neither compressing nor expanding near P. For example, the vector field F = (x, y, z), appearing in Figure 6(A), has div(F) :c 3 everywhere. Thinking of this as the velocity vector field for a gas, at every point, the gas is expanding. This is most obvious at the origin, but even at other points, the gas is expanding in the sense that more gas atoms are moving away from the point than are moving toward it. We say that each of these points is a source. For the vector field F = (-x, -y, -z) appearing in Figure 6(B), div(F) = -3 for all points P, and the gas is compressing at every point. We say that every point is a sink. For the vector field F = (0, 1, 0), appearing in Figure 6(C), div(F) = 0. At each point, the gas is neither expanding nor compressing. Rather, it is simply shifting in the positive y-direction. In this case, no points are sources or sinks and we say that the vector field is incompressible. For other vector fields, there can be points that are sources points that are sinks, and points that are neither. ' I

I

I y

(A)

FIGURE 6

y

(B)

y

(C)

I~

SECTION 16.1

den Thtde otber operation involving v and vector fields F = (Ft, Fi. o e curl(F) d an defmed using the cross product as follows: j

VectorFields

961

F3) is the curl of F,

k

a a a ax ay az F1 F2 F3

CUrl(F) :::: V X F ::::

:::: (aF3 _ aF2)i-(aF3 _ aF1)j+ (aF2 _ aF1)k ay

That is,

curl(F)

az

ax

az

ax

ay

= (aF3 _ aF2' aF1 _ aF3' aF2 _ aF1) ay

az

az

ax ax

ay

. Note that, in contrast to divergence, the curl of a vector field is itself a vector field.

It is straightforward to check that curl obeys the linearity rules:

= curl(F) +curl(G) curl( cF) = c curl(F) (c any constant) Calculating the Curl Calculate the curl of F= (xy, ex, Y + z). curl(F + G)

EXAMPLE 4

Solution We compute the curl as the detenninant: j

k

a a a curl(F) = ax ay az xy ex y +z

= cur!F(P)

Ii FIGURE 7 curl F(P) tells us about the rotation of the fluid.

• The term "conservative" comes from physics and the Law of Conservation of Energy (see Section 16.3). • Any letter can be used to denote a potential function. We use f. Some textbooks use V, which suggests "volt," the unit of electric potential. Others use f

a= ax' az = ay' ax= aF,

or equivalently,

aF2

aF2

11

3

y

8Ft

-

.

az.

• The radial unit vector field and the inverse-square vector field are conservative. e, = (::, ~. :} r r r

='i/r,

e, _ (.:_ !._ ~}

,2 -

,3' ,3' ,3

='i/(-,- 1),

r ={x2 + Y + z

2

2

where

~ ~EYI XEiRN Cl~ SE;S- - - - - - - - - - - - - - - - - preliminary Questions . 1· Which of the following is a unit vector fiieId m the plane?

fl Id · the plane in which 2. Sketch an example of a nonconstant vector e m each vector is parallel to (l , l) . · rthogonal to the position 3. Show that the vector field F = (-z, 0,x } is O • vector OP at each point p. Give an example of another vector field with

(a) F = (y,x}

(b)

F=(~•

Jx2x+y2)

(c) F = (x2: y2 ' x2

this property. 4. Show that f (x, y, z) = xyz is a potential function for (yz, xz,xy) and give an example of a potential function other than f ·

:7)

Exercises 1, Compute and skrtch the vector assigned to the points p Q= (-1, -1) by !he vector field F = (x2,x).

= (l

'

2) and

y

y

-__--- -_ -

2

2. Compute and shtch the vector assigned to the points p Q= (-1, -1) by the vector field F = (-y,x}. 3, Compute and sketch the vector assigned to the points p Q = (2, 1, 0) by the vector field F (xy, z2 , x ).

=

= (1 , 2) and

= (0, 1, l) and

,,., ,,,,,,.

0

X

5. F= (1,0)

6. F = (1,1)

F= yi

9. F = (O,x)

8.

11.F=(~-~ ) X +y X +y

7. F=xi

0

-2

''

\

I

\

I

0

(A)

(B)

y

y

,,,

_...,,,

_..... ___

- ' , ---" , ____ - , ' r-----/

I

\

/

I

\

2

I/ / / -

, -.....,_

-2

0

X

0

-2

""' ---""' ""'-""' ---""' ""'--

-2

2

(C)

0

2

X

2

(D)

FIGURE 10

-y - - X 12 F- - -) . - ( Jx2+y2' Jx2+yz

In Exercises 13-16, match each of the following planar vector fields with the corresponding plot in Figure JO.

-...-2

2

0

4. Compute the vector assigned to the points P = (1, 1, 0) and e, e, Q = (2, 1,2) by the vector fields e,, -, and 2 . r r 2

X

0

-2

-2 -2

In Exercises 5-12, sketch the following planar vector fields by drawing the vectors attached to points with integer coordinates in the rectangle -3 x 3, -3 y 3. Instead of drawing the vectors with their true lengths, scale them if necessary to avoid overlap.

2

_...,,..

In Exercises 17-20, match each three-dimensional vector field with the corresponding plot in Figure 11.

13. F= (2, x)

14. F = (2x +2, y}

17. F = (!, 1, 1)

18. F = (x,0,z}

15. F = (y, cos x)

16. F = (x + y,x - y)

19. F=(x,y,z}

20. F = e,

I

966

CHAPTER 16 LINE AND SURFACE INTEGRALS

9--47, find a poten/la. 1Junction for the vector field t J.. In Exercises 3 t ne does not exist. "Y inspection or show t a 0

h

(B)

39. F = (x,y)

40. F = (y,x)

41. F = (y2z,

42. F

I+ 2xyz,xy2)

= (yz,xz,y)

43. F = (ye-"Y,xf"'Y)

44. F = (2xyz,x 2z,x 2yz)

45. F=(yz 2,xz2,2xyz)

46. F

= (2xze- Oand z < 0. We are free to choose different constants C on the two halves, if desired.

tive and, for any constant C, a potential function is f(x,y,z) =x 2yz- 1+YZ + C Assumptions Matter We cannot expect the method for finding _a po~ential function to work if F does not satisfy the cross-partials condition (because m this case, no potential function exists). What goes wrong? Consider F = (y, 0). If we attempted to find a potential function, we would calculate j(x,y) f(x, y)

f =f =

ydx

= xy + g(y)

Ody= 0 + h(x)

However, there is no choice of g(y) and h(x) for which xy + g(y) = h(x). If there were, and we differentiated this equation twice, once with respect to x and once with respect toy, we would obtain the contradiction 1 = 0. The method fails in this case because F does not satisfy the cross-partials condition, and thus is not conservative.

The Vortex Field Why does Theorem 4 require that the domain is simply connected? This is an interesting question that we can answer by examining the vortex field that we introduced in the previous section, F = ( x2

-y

X

+ y1' x2 +y1

)

EXAMPLE 8 Show that the vortex field satisfies the cross-partials condition but is not conservative. Does this contradict Theorem 4?

Solution We check the cross-partials condition directly: (-x-) ax x 2 + y1 ( ay

-y ) (x1 + y2)

= (x 2 +y2) -

2

+ y2) _ y2 - x1 - (x2 + y2)2 2 2 = -(x +y2) + y(a/ay)(x + y2) _ y2 - x 1 (x2 + y2)2 - (x2 + y2)2 x(a/ax)(x (x2 + y2)2

In Example 6 in the previous section, we showed that

i

F . dr = 21r f. Ofor any

circle C centered at the origin. If F were conservative, its circulation around every closed

Conservative Vector fields

SECTION 16.3 y

---

991

. though it sariscurve would be zero by The l Thus F cannot be conservative, even fi orem . , ies the cross-partials condition. . f F does not satisfy This result does not contradict Theorem 4 because the doma1D o ) - (0 0) ~e simply connected condition of the theorem. Because F is _not defined at (~ '~-;ure its domain is 'D = {(x, y) # (0, 0)), and this domain is not sIIDply connecte I

i2/

cDNCEPTUAL INSIGHT

URE 12 The domain 'D of the vortex flG!d Fis the plane with the origin fie Th. d . .

rernoved- 1s oma.m 1s not simply connected.

REMINDER _dd tan- t == _I_ t I+ 12 1

I

. ti e on its domain Although the vortex field F 1s not conserva v . · d 'D {( ) . I connected domalD conta1De . = x, Y # (0, 0)}, it is conservative on any sIIDp Y the left half-plane m 'D. For example, on the right half-plane {(x,y): x OJ and on_ -I l We can {(x,y): x < 0), Fis conservative with potential function f(x,y)- t~ x·. . . verify that f is a potential function for F by directly computing the partial denvatlves.

aJ ax

a

-y

a/an ; = +

==

a/an ;= l+(y/x)2 -

aJ ay

-y/x 2

==

-1

a

--'--'---=- - 2 l (y/x) 2 - x

y

_1 y

1/x

+ y2

_ _!._ x 2 +y2

(x

i- 0)

(x =/:- 0)

Furthermore, on the upper half-plane and on the lower half-plane, Fis conservative with potential function g(x, y) == -tan- 1 (See Exercise 32.) . . Even if a simply connected domain 'D* is irregularly shaped, like th~ domalD ID Figure 13, we can specify a potential function for F, although the fun~tlon m~y not be expressed as simply as J or g. Nevertheless, we can define a potential function :15 follows: Fix a point (xo, yo) E D*, and for every (x, y) E 1)*, choose a path C(x,y) ID 'D* from (xo, y0) to (x, y). It can be shown that the function

f

h(x,y)=l FIGURE 13

There is a potential function for

FonV* .

y

(x,y)

Toe angle 0 is the inverse tangent of y / x .

F -dr

C(,,y)

is defmed, independent of the path chosen, and is a potential function for F on D*. GRAPHICAL INSIGHT There is an interesting geometric interpretation of the integral of the vortex field over a curve. First, we saw that in the right half-plane the function f (x, y) == tan- 1 f is a potential function for F. But tan- 1 f is just the angle 0 illustrated in Figure 14. Thus, by the Fundamental Theorem for Conservative Vector Fields, the integral of F along a curve in the right half plane is the difference between the angles at the end and the beginning of the curve; that is, the change in 0 along the curve [Figure 15(A)]. We can show that this relationship is true for any curve C in 1) = {(x, y) =/:(0, 0)}, not just those in the right half plane. Assume we have a parametrization r(t) = (x(t), y(t)) of a curve C in 'D. Consider the equation tan 0 = and differentiate implicitly with respect tot. We obtain

FIGURE 14

( sec

2

e) dedt = (-y) dx + (~) dy x 2 dt x dt

With 0 as in Figure 14, it can be shown that sec 2 0 = x ~;/ • Substituting this for sec2 0 and simplifying, we have x 2

d0

dt =

(

-y ) dx ( x ) dy x 2 + y2 dt + x 2 + y2 dt

Now, if we integrate both sides. of this equation with respect to t along C,on theet If . . side we obtam the net change m the angle 0 along C. On the right side bt · · • h - • , we o am an expression ,or t e 1me mtegral of F along C. Thus, we have net change in 0 along C =

fc F . dr

992

CHAPTER 16

LINE AND SURFACE INTEGRALS

Thi . cation of the line integ!"a) s mte1: cirele centered at the Ori _of the vortex field F explains why the integral of F aro~ d . Example 6 in the last glll cl!id oriented counterclockwise is 21r • a result we obtatne -'~ times (where n . section · d around the ongm n ts neg . · In generaJ , if a closed path r wm . s . . I5(C) and (D)] lltive 5 th if the curve winds in the clockwise direction), en [Figure .

as illustrated in Figures 15(A) and (B).

i

F-dr= 2,rn

The number n is called the winding number of the path - It plays an important role in the mathematical field of topology. y

y

0

Q

(A)

fcF • dr=02 -01

FIGURE 15

(B)

fcF · dr=02 + 2,r-01

(D) r do_es not go around the ongu;, ;o fc F • dr" O.

(C) r goes around the origin twice, so fcF · dr =4,r.

The line integral of the vortex field Fis equal to the change in 0 along the path.

16.3 SUMMARY • A vector field F on a domain V is conservative if there exists a function f such that V f =Fon V. The function f is called a potential function ofF. • A vector field F on a domain V is called path independent if for any two points P, Q EV, we have [ F · dr

le,

= [

lc2

F · dr

for any two curves C1 and C2 in V from P to Q. • The Fundamental Theorem for Conservative Vector Fields: If F

fc F · dr = f(Q) -

= V f, th&n

f(P)

for any path r from P to Q in the domain of F. This shows that conservative vector fields are path independent. In particular, if r is a closed path (P = Q), then

i

F-dr=O

• The converse is also true: On an open, connected domain, a path independent vector field is conservative. • Conservative vector fields satisfy the cross-partials condition

--=--

ax

az

CD

• Equality of the cross partial derivatives guarantees that F is conservative if the do· main V is simply connected-that is, if any loop in V can be drawn down to a point within V.

-

\ Conservative Vector Fields

SECTION 16.3

~

XERCISES

993

\

~---------------

preliJllinar~ Quest·10ns - - - - - - - - - - - - - - - - - - - - - - - - - 'file followmg statement is false.

If F .

I• fine integral of F along every curve . is a gradient vector field he ~I dded to make JI true? is, fero. Which sin 1 't n a g e word must 111 WJ!ich of the following statements are

~,;cb are troe only for conservative vector fitrule for all vector fields and f•• . . 1e ds? , 'file (me mtegral along a path fro P ·

m toQd path is ch~se~. oes not depend on which 'file (me mtegral over an oriented (b) . d curvec do is parametnze . es not depend on how C The line integral around a closed c . (c) . . urve 1s zero. d) The line mtegral changes sign if th . ( . . . e onentation is reversed The line mtegral 1s equal to the d'f' (e) • 1 ,erence of . · the iwo endpomts. a potenttal function at

(a) If F has

a potential function, then F is conservati:e.

4. Let C, 'D, and£ be the oriented curves in Figure 16, and let F = 'ii f _ 4 What are the values of be a gradient vector field such that c F · dr - ·

1

the following integrals? (a) { F . dr

(b)

y

(O The line integral is equal to the inte al of . the langenttal component ~ong the curve. gr p

(.,r;., The cross partial derivatives of the components are equal. field on an open 3· Let F be da vector .1, . , connected domain 'D •th

nnuous secon parl!a, uenvatives. Which of the t . wi con~ways troe, and which are true under addiu· hollowing statements are 1 ona ypotheses on 'D?

exercises

1

11. F = y2i + (2xy

2, LetF(x,y,z) = (x- 1z,y- 1z,ln(xy)). (a) Verify that F = v' /, where f(x, y, z) = z ln(xy).

15. F = (zsec2 x,z,y

(c) Evaluate

l l

F · dr, where r(t) =

(e', e21 , t 2 ) for 1 :5 t :5 3.

F · dr for any path C from P = ( ½, 4, 2) to Q = (2, 2, 3)

6

+ tanx)

16. F = (ex(z + 1), -cosy, ex) 17. F = (2xy

+ 5,x 2 - 4z, -4y)

18. F = (yzexY,xzexy -z,exy -

19. Evaluate

over the path r(t) = (12, sin(nt /4), / - 2') for 0 :5 t :5 2.

20. Evaluate

4. F(x,y) = (cosy,-xsiny), f(x,y) =xcosy; upper half of the unit circle centered at the origin, oriented counterclockwise

6, F(x,y,z)= ~i+j+lnxk,

f(x,y,z) =xyez;

f(x,y,z)=y+zlnx;

X

.

circle (x _ 4)2 + y2 = I in the clockwise direcl!on In Exercises 7-18, find a potential Junction for For determine that Fis not

conservative. 7.

F:::: (x,y,z)

9. F= (z, x, y)

8.

F = (y,x,z)

10. F = xj + yk

y)

fc 2xyzdx +x2 zdy +x2 ydz

In Exercises 3--0, verify that F = v' f and evaluate the line integral of F over the given path.

5, F(x,y,z) = yezi+xezj +xyezk, r(t) = (r2,t 3,t -1) for l :5 t :5 2

X

14. F=(cosz,2y,-xsinz)

13. F = (cos(xz), sin(yz), xy sin z)

contained in the region x > 0, y > 0.

f(x,y)=3x+3y2; r(t)=(t,2t- 1} on the

F. dr

12. F=(y,x,z3 )

+ e')j + ye'k

(d) In part (c), why is it necessary to specify that the path lies in the region where x and y are positive?

3. F(x,y)=(3,6y), interval I :5 t :5 4

l

FIGURE 16

I, Letf(x,y,z) = xysin(yz)andF = VJ.Evaluate F-dr, whereC is any path from (0, 0, 0) to (1, 1, n). • c

(b) Evaluate

al

(b) If Fis conservative, then the cross partial derivattves of Fare eq~ . (c) If the cross partial derivatives of Fare equal, then Fis conservauve.

£

sinxdx+zcosydy+sinydz

where C is the ellipse 4x 2 + 9y 2 = 36, oriented clockwise.

In Exercises 21-22, let F = v' f, and determine directly

1

F. dr for each

~! the two paths given, showing that they both give the sa:e answer ~ft0-ft~

'

which

2

21. f = x y - z,r1 = (t,t,0) for 0 < t < I and r2 = (t,t2,o) 0:::t:::1 - - ' for 22. f=zy+xy+xz,r1=(t,t,t) for 0:::t:51, r 2 =(t,t2,t3) for 0 :5 t :5 l

994

CHAPTER 16

LINE AND SURFACE INTEGRALS

23• A Vector field F and . shown . Fi contour Imes of a potential function for F are in igure 17 Cal I curve h . . cu ate the common value of F . dr for the s s own m figure 17 • c onented in the direction from p to Q.

1

y

r

!n

where = (x2 + y2 + z2)1/2 with distance meters ~d k == 8,99 with units N-m21c2. Calculate the work against F requrred to move~ 101 ticle of charge q =O.Ol C from (I , -5, 0) to (3, 4, 4). Nore: 'The fo aP.r. l'Ceo q is qF newtons. n

29. On the surface o~ the earth,_ the gravitational field (with z as vetti coordinate measured m meters) 1s F = (0, 0, -g). c~ (a) Find a potential function for F. (b) Beginning at res~ a ball of mass m = 2 kg moves under the .

ence of gravity (without friction) alon~ a path _from P == (3,2, 4~~nu. Q = (-21,40,50). Find the ball's velocity when 11 reaches Q. lo

I

I

I

---------x I

FIGURE 17 24 • vative.

Give a reason why th

y

30. An electron at rest at p = (5, 3, 7) moves_along a.path endin Q == (I, 1, 1) under the influence of the electnc field (m newton/ a1 coulomb) P!r F(x, y, z)

(a) Find a potential function for F.

. . e vector field F m Figure 18 is not conser-

What is the electron's speed at point Q? Use Conservation of En ergy and the value q,!m, = -1.76 x 10 11 C/kg, whern q, and m, are lh; charge and mass on the electron, respecuvely. (b)

31. LetF

, ,

I

I

I

I

,

I

I

I

I

I

I

I

I

, , ,

I

I

I

I

I

I

I

I

I

,

I

I

I

I

I

I

I

=400(x 2 +z2)- 1 (x , 0, z)

I I I I I I I

I I I I I I I

=(~.A) y2 + be the vortex field. DoterminejF-d x2+

x

y

for each of the paths in Figure 20.

X

(A)

c

r

(C)

(B)

FIGURE 18

25. Calculate the work expended when a particle is moved from o to Q along segments OP and P Q in Figure 19 in the presence of the force field 2 F = (x , y2). How much work is expended moving in a complete circuit around the square? y

R = (0, 1) >---- - ~ Q = (I, I)

I

I

32. Show that g(x, y) field.

FIGURE 19 26. Let F(x,y)

= (~. ~l )·

33. Determine

Calculate the work against F required to

move an object from (I, I) to (3,4) along any path in the first quadrant.

27. Compute the work W against the earth's gravitational field required to move a satellite of mass m = 1000 kg along any path from an orbit of altitude 4000 km to an orbit of altitude 6000 km. 28. An electric dipole with dipole moment p electric field (in newtons per coulomb)

j

FIGURE 20

= 4 x 10-5 C-m sets up an

kp (3xz,3yz,2z 2 -x 2 -y 2) F(x,y,z)=-;s

= - tan-I :: is a potential function for the vortex

whether

y

or

not

the

vector

field

F(x, y) =

{x2: y2, xi~ y2) has a potential function. 34. The vector field F(x, y)

=

domain 'D = {(x,y) # (0,0)). (a) ls 'D simply connected?

(-x-, _Y_) + y2 x2

x2+ y2

is defined 00 the

(b) Show that F satisfies the cross-partials condition. Does this guarantee

that F 1s conservative?

(c) Show that Fis conservative on 'D by finding a potential function. (d) Do these results contradict Theorem 4?

s ECT ION

~

35·

Insights and Ch~

16.4

Parametrized Surfaces and Surface Integrals

995

-------------------

suppose that F is defined on R3

. . andthatiF. tbS Cin R3 ._Prove. 1'c dr "' 0 for all closed r1 Fis path mdependent; that • , (s) . .. al d . is, ,or any tw saJlle 1rull an terrrunal points o paths c1 d . i)le , an C2 10 'D with

1 Ci

F . dr =

1

F · dr

C2

(b) Fis conservative.

l&,4 Parametrized Surfaces and Surface Integrals The bas·ic I·ctea of an mtegral · · So far, we have defined smg · 1e, appears in several gwses. double, and triple integrals and in the previous section line integrals over curves. Now• We .' • ' . consider one 1ast type of mtegral: mtegrals over surfaces. We treat sc al ar surface mtegrals in this section and vector surface integrals in the following section. JuS t as parametrized curves are a key ingredient in the discussion of line integrals, ~urface integrals require the notion of a parametrized surface-that is, a surface S, 10 3 R ' whose points are described in the fonn G(u, v)

forG(u, v) in R3, vie allow both interpretations as a point and as a vector. The intent shoufd r;a clear from the context or the notation uswt. Typically for a parametrization, we regard G(u, v) as · tson th e surrace - and -(u, aG v) and pom

aG au -(u, v) as vectors tangent to the surface av .

FIGURE 1 A parametrized surface S.

= (x(u, v) , y(u, v), z(u, v))

The variables u, v (called parameters) vary in a region V, in the uv-plane, called the parameter domain. Two parameters u and v are needed to parametrize a surface because th e surface is two-dimensional. Figure 1 shows a surface Sin R3 with parametrization G(u, v) defined for (u, v) in Din the uv-plane. V

D

(u,v)

-

G X

--t----------- u

=L + y2 = 1 is

EXAMPLE 1 Find a parametrization for the cylinder x 2 + y2 2

Solu~ion _The ~yli~der of ra~us 1 ~ith equation x conveniently parametrJ~ed m cylindrical coordmates (Figure 2). Points on the cylinder have cylindrical coordmates (1, 0, z), so we use 0 and z as parameters. We obtain G(0,z)

= (cos0,sin0,z),

0

0 < 2n,

-

G

FIGURE 2

The parametrization of a cylinder

by cylindrical coordinates amounts to

Wrapping the rectangle around the cylinder.

60

2n: 0

Parameter domain D

-oo < z
B is the assertion, "If A is true, then B is true." The contrapositive of A => B is the implication ~B => ~A, which says, "If B is false, then A is false." An implication and its contrapositive are equivalent (one is true if and only if the other is true). The converse of A => B is B => A. An implication and its converse are not necessarily equivalent. One may be true and the other false. A and B are equivalent if A => B and B => A are both true. In a proof by contradiction (in which the goal is to prove statement A), we start by assuming that A is false and show that this assumption leads to a contradiction.

A. EXERCIS ES Prelim inary Questions 1. Which is the contrapositive of A (a) B A (c) ~B

~A

3. Suppose that A B? (b) ~B · (d) ~A

B is true. Which is then automatically true, the converse or the contrapositive?

A ~B

2, Which of the choices in Question 1 is the converse of A

4. Restate as an implication: "A triangle is a polygon." B?

Exercises

1. Which is the negation of the statement, "The car and the shirt are both

blue"? Neither the car nor the shirt is blue. The car is not blue and/or the shirt is not blue. . • r tion "If the car has gas, then 2. Which is the contraposinve of the unp ica • it will run"? (a) If the car has no gas, then it will not run. (b) If the car will not run. then it has no gas.

(a) (b)

5. m and n are odd integers. 6. Either m is odd or n is odd. 1. xis a real number and y is an integer.

8. f is a linear function. In Exercises 9-14, state the contraposi·t·zve an d converse.

9. If m and n are odd integers, then mn is odd.

111 Exercises 3-8. state the negation.

10. If today is Tuesday• then we are in Belgium.

3. 'The time is 4 o'clock.

11. If today is Tuesday, then we are not in Belgium.

4. 6 1'. BC is an isosceles triangle.

AG

A P P EN D I X A

THE LANGUAGE OF MATHEMATICS

. s·

12. If x > 4, then x 2 > 16. 13. If m2 is divisible by 3, then m is divisible by 3. 14. If x 2

= 2, then x is irrational.

In Exercise /5-18, give a counterexample to show that the converse of the statement is false. IS. If m is odd, then 2m

+ I is also odd.

16. If 6.ABC is equilateral, then it is an isosceles triangle. 17. If mis divisible by 9 and 4, then mis divisible by 12. 18. If m is odd, then m3 - m is divisible by 3.

In Exercise 19-22, determine whether the converse of the statement is false. 19. If x > 4 and y > 4, then x + y > 8. 20. If x > 4, then x 2 > 16. 21. If lxl > 4, then x2 > 16.

22. If m and n are even, then mn is even.

In Exercises 23 and 24, state the contrapositive and converse (it is not necessary to know what these statements mean). 23. If f and g ate differentiable, then f g is differentiable. 24. If the force field is radial and decreases as the inverse square of the distance, then all closed orbits ate ellipses.

In Exercises 25-28, the inverse of A => B is the implication ~A => ~B. 25. Which of the following is the inverse of the implication, "If she jumped in the lake, then she got wet"? (a) If she did not get wet, then she did not jump in the lake. (b) If she did not jump in the lake, then she did not get wet. Is the inverse true?

. of these implicauon . the wverses 26. State th X is 8 rodent. . mouse, en (a) If X 1s 8 •11 miss class. (b) If you sleep late, you w~ the sun, then it's a planet. a star revolves aroun (c) If . uivalent to the converse. l=7. I . why the inverse is eq 27 EJtpalD . • rean Theorem. Is 11 true? State the inverse of the rythago " . 28. th following: If f and g ate conttnst rem I in Section 2.4 at~s e s " Does it follow logically that if 29• Theo f + is conunuou . . ? uous functions, then. g th f + g is not conunuous . d are not conunuous, en . f an g . ,,&or this fact: There 1s trad"1cuon . no smallest b roof on the fact that if r > 0, then 30 Write out a proof Y con po;itive rational number. Base your p O < r/2 < r.

.. vethat ifx + y> 2, thenx>Ior 31. Use proof by contrad1cuon to pro y > I (or both). ,I" b ontradiction to Jhow that the number is In Exercises 32-35, use proo, y c irrational. 35. -1iT ii 33. ./3 34. ~Ii

32,y½

. .

. ·angle wi"th two c,1i. a1 sides. The follow. • I · le 1s a tn · 36. An 1sosce es tnang . . gle with two e,~ual angles, then /:,. is an ing theorem holds: If 6. is a tnan · · isosceles triangle. (a) What is the hypothesis? . , ,... . . .d. a counterexample ~tJ.lt m(, ,.; pothes1s 1s necessary. (b) Show by prOVI mg . (c) What is the contrapositive? (d) What is the converse? Is it true?

37. Consider the following theorem: Let f be _a quadratic polynomial with a positive leading coefficient. Then f has a mmimum value. (a) What are the hypotheses? (b) What is the contrapositive?

(c) What is the converse? Is it true?

Further Insights and Challenges = n-/A - n l,/1.J, where

Lx J is the greatest integer

38. Let a, b, and c be the sides of a triangle and let 0 be the angle opposite c. Use the Law of Cosines (Theorem I in Section 1.4) to prove the converse of the Pythagorean Theorem.

as before and let m function.

39. Carry out the details of the following proof by contradiction that ./2 is irrational (this proof is due to R. Palais). If ,/2 is rational, then n./2 is a whole number for some whole number n. Let n be the smallest such whole number and let m = n./2 - n. (a) Prove that m < n. (b) Prove that m./2 is a whole number. Explain why (a) and (b) imply that ,/2 is irrational.

..:/A is irrational unless A is an ,th power. Hint: Let x = ,:/A. Show that if

40. Generalize the argument of Exercise 39 to prove that ,/A is irrational if A is a whole number but not a perfect square. Hint: Choose n

41. Generalize further and show that for any whole number r, the rth root

x is rational, then we may choose a smallest whole number n such that nxj is a whole number for j =I, . .. , r - I. Then consider m = nx - n[x] as before. 42. Given a finite list of prime numbers Pl,••·,PN, let M= Pl ·Pi· · · PN I. Show that M is not divisible by any of the primes Pl, . .. , PN- Use this and the fact that every number has a prime factoriza-

+

tion to prove that there exist infinitely many prime numbers. This argument was advanced by Euclid in The Elements.

~_PR_O~P ~ R~EA=-:. LN ~U= MB:=ER=S ------~~e ingenious methOd of exp , ," b ressmg f1!55ible num er using a set of . el)' 10 ,••ch symbol having a Place 1 5Ymbo1s '"" va ue and so/Ute value) emerged in I d' an ab n 1a Th · seems so simple nowadays that e idea ;'111ificance and profound im s&" . PDrtance is onger appreciated. Its simp/' .,.,, . no l .. ici.:, //es in th waY it faci/itated calculation d e . an Placed arithmetic forem_ ost amongst useful invent1_ ons: The importance of th is invent10n ,s more readily apprec. t .d . ia ed When one cons, ers that it was beyond th e two greatest men of Antiqui~• Arch.1medes and Apo/lonius.

its

- Pierre-Simon Lapla one of the great French mathem ti . ce, a c1ans of the eighteenth century

...,.. I

I

-3 - 2 - 1

FIGURE 1

0

2

3

R

The real number line.

ln th" real

l.lappmdix •-c,1;·- ·

,•

·

ua.,,i;.w.·ulne bNc

.

r::

.

•

.... ·

-~ ~ ~ r ~ . l. .u~'

ui

roctJI that a

• - ,. 1 (

,_.i

aho l:'altcU

number ~ anumber ...... ..... _ __."'"' fm.ille or infina1!C Jec1mai . , a dcci UMR IBIJ' to. l q " l ~ w V J • ol'li - \'~•Ui:lllml 3• .. maJ expansion). The~ of 1111 ml IIIJlllll,m R ,~ en the number line"' ffigu,t- I}. Thus, a rcaJ nufflbtt O u IIS,

°'

a =±.,v.o1a;a:,a..

where n is any w1,ole numbcf ml eac:h ......., o r" :a «mk nu.m~ hctw,•e~ t) ttm 9. Po - · , . ni>l\~·on IS. t'\nll~ 1)1' r example, IO;r = 3141 ' 92 R-'I th.ll u I'< r:ational • ,~ e.~r-- , • · J · .. · it, . ooe,lrn!41 repeating and is irrational if i~ u,...,.,inn is ~pt'lltilli- f\arttleffOOte, IC · 1tti a · · . t - _ _ .. , M111tlon 1~ ""1.llill P nsion is unique apart from the follov.;ni c,ceptioo: Every 'in1tc e,.,..., . . "''1 an expansion in which the digit 9 rcpais. For c..um~le. O..S = O.~W} ' ·' a t).,41), We shall take for granted that the opcnition:s of oodilioo llnJ n~uldpll1;!4tlt~n ll~ ~e: ~ned on R-that is, on the set of all decinw.s, Roughly 5pcllltlfii. lltkl1d0f'I ~ ultlpllto h lion of infinite decimals are defined in tenns of finite dccl!fillls, Ft!r J ;:; I, tic.hue th e t/l truncation of a = n.a1a2a3a4 ... to be the finite dcclmill o(d) = o .0111~ •• , u.i 1:1 1\ltt\nell by truncating at the dth place. To fonn the: sum " + b. 11&1umc lhllt bi.1th o II.flt\ /l ll.l\1 iri11· nite (possibly ending with repeated nines). This eliminates lliiY pos~lt,lo llfttbll!ult)' In l~Q expansion. Then the nth digit of a + b is equal to the t1 th digit 11 f o(d) -1- /)((/) for ti ~ul ciently large [from a certain point onward, lhc nth dlgit of"( b

TABLE 2 Order Properties If a < b and b < c, If a 0, If a < b and c < 0,

then a < c. then a+ c < b + d. then ac < be. then ac > be.

To distinguish between the conditions a band a < b, we often refer to a < b as a strict inequality. Similar conventions hold for > and ~- The rules given in Table 2 allow us to manipulate inequalities. The last order property says that an inequality reverses direction when multiplied by a negative number c. For example, - 2 < 5 but

(- 3)(-2) > (- 3)5

The algebraic and order properties of real numbers are certainly familiar. We now discuss the less familiar Least Upper Bound (LUB) Property of the real numbers. This property is one way of ex~ressing the so-called completeness of the real numbers. There are other ways of formul aung co'.'1pleteness (such as the so-called nested interval property discussed in any book on analysis) that are equivalent to the UJB Property and serve the same purpose. Completeness is used in calculus to construct rigorous proofs of basic A7

j

AB

APPENDIX B

PR OPERTIES OF REAL NUMBERS

. . h the Intermediate Value Theorem, (IVT) theorems_about continuous functions, sue as . The underlying idea is that the or the existence of extreme values on a closed mte~~- below. First, we introduce th real number line "has no holes." We elaborate on this idea e necessary definitions. b r M is called an u Suppose that S is a nonempty set of real numbers. A num e PPer bound for S if

x L

-3 -2 -1

M

0

2

3

X

FIGURE 2 M = 3 is an upper bound for the set S = (-2, !). TheLUB is L = I.

sM

for all x E S

· boonded above· A least upper bound L is an If S has an upper bound we say that S 1s upper bound for S such ' that every other upper boun d M satisfies M -> L. For example (Figure 2), • M = 3 is an upper bound for the open interval S • L = 1 is the LUB for S = (-2, l).

= (- 2, l ).

We now state the LUB Property of the real numbers.

THEOREM 1 Existence of a Least Upper Bound Let S be numbers that is bounded above. Then S has an LUB.

11.

nonempty set of real

In a similar fashion, we say that a number B is a lowet l:mi;.;i/1 for S if x B for all x ES. We say that S is bounded below if S has a lower boum:l. A i;_'1'eatest lower bound (GLB) is a lower bound M such that every other lower bound JJ. satisfies B S M. The set of real numbers also has the GLB Property: If S is a nonempty set of real numbers that is bounded below, then S has a GLB. This may be deduced inm1ediately from Theorem 1. For any nonempty set of real numbers S, let -S be the set of numbers of the form -x for x ES. Then -S has an upper bound if S has a lower bound. Consequently, -S has an LUB L by Theorem 1, and-Lis a GLB for S.

.rz I I I I

-3 -2 -1

0

I

I I I I

2

3

FIGURE 3 The rational numbers have a "hole" at the location ,./2.

X

CONCEPTUAL INSIGHT Theorem 1 may appear quite reasonable, but perhaps it is not clear why it is useful. We suggested above that the LUB Property expresses the idea that R is "complete" or "has no holes." To illustrate this idea, let's compare R to the set of rational numbers, denoted Q. Intuitively, Q is not complete because the irrational numbers are missing. For example, Q has a "hole" where the irrational number ,./2 should be located (Figure 3). This hole divides Q into two halves that are not connected to each other (the half to the left and the half to the right of ./2). Furthermore, the half on the left is bounded above but no rational number is an LlJB, and the half on the right is bounded below but no rational number is a GLB. The LUB and GLB are both equal to the irrational number ./2, which exists in only R but not Q. So unlike R, the rational numbers Q do not have the LUB property.

EXAMPLE 1 Show that 2 has a square root by applying the LUB Property to the set S = {x: x 2 < 2}

Solution First, we note that Sis bounded with the upper bound M = 2. Indeed, if x > 2, 2 then x satisfies x > 4, and hence x does not belong to S. By the LUB Property, S has a least upper bound. Call it L. We claim that L = ./2, or, equivalently, that L 2 = 2. We prove this by showing that L 2 2 and L 2 s 2. If L 2 < 2, let b = L + h, where h > 0. Then 2

= L 2 + 2Lh + h2 = L 2 + h(2L + h) IT] We can make the quantity h(2L + h ~s small as desired by choosing h > 0 small enough. b

In particular, we may choose a positive h so that h(2L + h) < 2 _ L 2. For this choice, 2 2 2 b < L + (2 - L ) = 2 by Eq. (l). Therefore, b ES. But b > L since h > 0, and thus

r APPENDIX B

PROPERTIES OF REAL

A9

NUMBERS

Lis not an . L We conclude that L2 :::. 2. upper bound for S' in contradiction to our hypothesis on .

If L2 > 2' 1et b == L - h, where h > 0. Then b2 = L2 - 2Lh + h2 = L2 - h(2L - h) Now choose h . . (2L h) < L2 - 2. Then b2 > L2 _ (L 2 positive but small enough so that O < h I deed, if x 2: b, then x2 > - == 2. But b < L, so b is a smaller lower bo~nd for :thesis that L is the LlJB- b > 2, and x does not belong to S. This contradicts our yp that L 2 2: 2, we h 2. We conclude that L2 < 2 and since we have already shown ave L == 2 as claimed. - ,

!·

i')

We . f hi h. used in the proof of now prove three important theorems, the third o w c is the LUB Property below.

THEOREM 2 Bolzano-Weierstrass Theorem Let S be a bounded, infinite set of numb Th . . . { } · S such that the hnut e:s. en there exists a sequence of distmct elements an m L == hm an exists · 1 Proof For simplicity of notation we assume that S is contained in the unit interval [O, ) ~a similar proof works in general). If ki , k2, ... , kn is a sequence of n digits (i.e., each k i is a whole number and 0 s kj s 9), let S(k1, k2, ... , kn)

be the set of x E S whose decimal expansion begins 0.k1 k2 . . . kn. The set Sis the union of the subsets S(0), S(l), ... , S(9), and since sis infinite, at least one of these subsets must be infinite. Therefore, we may choose k1 so that S(k 1) is infinite. In a similar fashion, at least one of the set S(k1, 0), S(k 2, 1), ... , S(ki, 9) must be infinite, so we may choose k2 so that S(k1,k2) is infinite. Continuing in this way, we obtain an infinite sequence {kn} such that S(k1, k2, ... , kn) is infinite for all n. We may choose a sequence of elements an E S(k1,k2, . . . ,kn) with the property that an differs froma1, ... ,an-1 for all n. Let L be the infinite decimal 0.k1k2k3 . . .. Then lim an = L since \L - an\ < 10-n for all n.

•

We use the Bolzano-Weierstrass Theorem to prove two important results about sequences {an}. Recall that an upper bound for {an} is a number M such that ai M for all j. If an upper bound exists, {an} is said to be bounded from above. Lower bounds are defined similarly and {an} is said to be bounded from below if a lower bound exists. A sequence is bounded if it is bounded from above and below. A subsequence of {an} is a sequence of elements ani, an 2 , an 3 , • •• , where n1 < n2 < n3 < • • • . Now consider a bounded sequence {an}. If infinitely many of the an are distinct, e Bolzano-Weierstrass Theorem implies that there exists a subsequence {an1, an2, · · · } th such that lim ank exists. Otherwise, infinitely many of the an must coincide and these ~00

terms form a convergent subsequence. This proves the next result.

'

THEOREM 3 Every bounded sequence has a convergent subsequence. I Section 10.1

THEOREM 4 Bounded Monotonic Sequences Converge

• If {an} is incr~asing and an SM for all n, then {an} converges and lim a11 M . · an d an 2: m for all n, then {an} converges and lim an 2: m. • If {an} is decreasmg

AlO

APPENDIX B PROPERTIES OF REAL NUMBERS

Proof Suppose that {a } is increasing and bounded above by M. The~ fan} is automatj. cally bounded below b; m = a1 since a1 :'.:: a2 :'.:: a3 .... Hence, {an} JS bounded, and by Theorem 3, we may choose a convergent subsequence an,, an2• ····Let L

= k--+oo Jim ank

Observe that an _:s L for all n. For if not, then an > L for some n and then ::: an :> L for all k such that nk ::: n. But this contradicts that ank L. Now, by defirul.ion, for any E > 0, there exists Nf > Osuch that lank - LI
Nf. If n::: nm, then anm .'.San :'.5 L, and therefore, for all n ::: nm This proves that Jim an = L, as desired. It remains to prove that L :'.S M. If L > M, let E = (L - M)/2 and choose N so that Ian - LI
N

Then an > L - E = M + E. This contradicts our assumption that M is an upper bound for {an}. Therefore, L _:s Mas claimed. I Proof of Theorem I We now use Theorem 4 to prove the LUB Property (Theorem I). As above, if x is a real number, let x(d) be the truncation of x of length d. For example,

= 1.41569, then x(3) = 1.415 We say that x is a decimal of length d if x = x(d). Any two distinct decimals of length d If x

differ by at least 1o-d. It follows that for any two real numbers A < B, there are at most finitely many decimals of length d between A and B. Now let S be a nonempty set of real numbers with an upper bound M. We shall prove that S has an LUB. Let S(d) be the set of truncations oflength d: S(d)

= {x(d) : x E S}

We claim that S(d) has a maximum element. To verify this, choose any a and x(d) > a(d), then

E

S. If x

ES

a(d) :'.5 x(d) :'.5 M

Thus, by the remark of the previous paragraph, there are at most finitely many values of x(d) in S(d) larger than a(d). The largest of these is the maximum element in S(d). For d = I, 2, .. . , choose an element xd such that Xd (d) is the maximum element in S(d). By construction, {xd(d)) is an increasing sequence (since the largest dth truncation cannot get smaller as d increases). Furthermore, xd(d) :'.5 M for all d. We now apply Theorem 4 to conclude that {xd(d)) converges to a limit L. We claim that L is the LUB of S. Observe first that Lis an upper bound for S. Indeed, if x E S, then x(d) :'.:: L for all d and thus x :'.5 L. To show that L is the LUB, suppose that M is an upper bound such that M < L. Then Xd SM for all d and hence xd(d) :'.5 M for all d. But then L

= lim xd(d) :'.5 M

This is a contradiction since M < L. Therefore, Lis the LUB of S. As mentioned above, the LUB Property is used in calculus to establish certain basic theorems about continuous functions. As an example, we prove the IVT. Another example is the theorem on the existence of extrema on a closed interval (see Appendix D).

C

- n, then + > 2n. Use this to show that if P(n) is true for n = k, then P(n) is true for n = k + 1. Conclude lhat P(n) is true for all n.

6. Use induction to prove that n ! >

zn for n2: 4.

The first few terms are 1, 1, 2, 3, 5, 8, 13, .... In Exercises 7-10, use

+F1 +···+ Fn = Fn+1 - l 8. p2I +F22 + · · · + F2n -- Fn+I Fn Rn _ Rn 1 ± Js 9· Fn ::: + - , where R± =

.Js

---z

= 7 · gk + (8k -

1)

13. Use induction to prove that s2n - 4n is divisible by 7 for all natural numbers n.

14. Use Pascal's Triangle to write out the expansions of (a+ b)6 and (a-b)4. 15. Expand(x+x-1)4. 16. What is the coefficient of x9 in (x3 + x)5?

17. Let S(n) =

= Fn-1 + Fn-2,

7· F1

+ Fn+IFn-1

12. Use induction to prove that n3 - n is divisible by 3 for all natural numbers n.

Let {Fn) be the Fibonacci sequence, defined by the recursion formula

induction to prove the identity.

= Fn+1Fn + Fn2

11. Use induction to prove that f(n) = gn - 1 is divisible by 7 for all natural numbers n. Hint: For the induction step, show that

(a) Show that P(l) is true.

Fn

•

Hint: For the induction step, show

F;+ 1 = Fn+1Fn

n

3.-+-+···+---

5. Let P(n) be the statement zn >

10. Fn+I Fn-l that

Fn+2Fn

2. 13 +23 + 33 + .. . + n3 = n2(n + 1)2 4

4. 1+ x + x

+ 12x + 8

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __

In Exercises 1--4, use the Principle of Ind . for all natural numbers n. uctio/1 to prove the formula

1 1· 2

2

t (n).

k=O k (a) Use Pascal's Triangle to compute S(n) forn=l234 (b) Prove that S(n) = 2n for all n > 1 ff . ' ' ' · evaluate at a = b = 1. - . mt. Expand (a + b)ll and

I)-1i(n). n

18. Let T(n) =

k=O

k

(a) Use Pascal's Triangle to compute T(n) for n = (b) Prove that T(n) = 0 for all n > . 1, 2, 3, 4. evaluate at a== 1, b = -1. - 1. Hint: Expand (a+ bjll and

-

D ADDITIONAL PROOFS

-----------------------------------

I Section 2.3

t were stated or used in the . . f veral theorems tha In this appendix, we provide proofs o se text. Ii f ( ) and Iim g(x) exist. Then: THEOREM 1 Basic Limit Laws Assume that x-Tc x (i) Iim (f(x) + g(x))

=lim J(x) + f_Tcg(x)

(ii) For any number k, lim kf(x) = k lim f(x) (iii) Jim f(x)g(x)

=( Jim f(x)) ( Iim g(x))

(iv) If Jim g(x)-/= 0, then lim f(x)

f( X)

]im-=::.---

Jim g(x)

g(x)

Proof Let L = Jim f(x) and M = Jim g(x). The Sum Law (i) was ;:,roved in Seck· · t fu · tion 2.9. Observe that (ii) is a special case of (iii), where g(x) = · Jc, a cm:;•,, ,an nction. Thus, it will suffice to prove the Product Law (iii). We write f(x)g(x) - LM

=f(x)(g(x) -

M)

+ M(f(x) -· !.)

and apply the Triangle Inequality to obtain lf(x)g(x) - LMI

lf(x)(g(x) - M)I + [M U V ) - - F,)l

IT]

By the limit definition, we may choose 8 > 0 so that if0 < [x - c[ < 8, then 1/(x) - LI < 1 If follows that 1/(x)[ < [LI+ 1 for O < [x - c[ < 8. Now choose any number E > 0. Applying the limit definition again, we see that by choosing a smaller 8 if necessary, we may also ensure that if 0 < [x - c[ < 8, then E

lf(x) - LI

E

and

2(IMI + l)

[g(x) - Ml

2(1LI + 1)

Using Eq. (1), we see that if 0 < Ix - c[ < 8, then lf(x)g(x) - LM[

1/(x)[ [g(x) - Ml+ [Ml 1/(x) - LI

(ILi

+ 1)

E

E

E

2(1LI + 1)

+ [Ml

E

2(1MI + 1)

0 so that if O I Y q. (2). Since g(x) approaches Mand M # 0, we m Y By choosing a srn IMl/2. Now choose any number E > O. a er oif n ecessary, we may also ensure that

W

for O< Ix - cl < 8, then \M - g(x)\ < EIM\

Then 1

s·

.

gN -

1

11

M ==

1M-g(x)1 Mg(x)

:'.S

c~

1 )

1M-g(x)1 EIMl(IMl/ 2) =E M(M/2) :'.S \M\(IMl/2)

Ince E is arbitrary, the limit in Eq. (2) is proved.

•

The following result was used in the text.

THEOREM 2 Limits Preserve Inequalities Let (a , b) be an open interval and let c E (a, b). Suppose that f(x) and g(x) are defined on (a , b), except possibly at c. Assume that for X

f(x) :'.S g(x)

E (a, b),

X

.\ I

I

#C

and that the limits lim f(x) and lim g(x) exist. Then lim f(x) ::; lim g(x)

\

ij Proof Let L = lim f(x) and M = lim g(x). To show that L '.SM, we use proof by contradiction. If L > M, let E = ! 0 so that the following two conditions are satisfied: If Ix - cl < o, then IM - g(x)I < E. If Ix - cl
g(x)

This is a contradiction since f(x) :'.S g(x). We conclude that L ::; M.

•

THEOREM 3 Limit of a Composite Function Assume that the following limits exist: L = lim g(x)

and

M

= lim

f(x)

Then Jim f(g(x)) = M.

Proof Let E > Obe given. By the limit definition, there exists 01 > ifO < Ix - LI < 01, then lf(x) - Ml
Osuch that ifO < Ix - cl< 8, then lg(x)- LI < 81,

8J

We replace x by g(x) in Eq. (3) and apply Eq. (4) to obtain: IfO < Ix - cl < 8, then lf(g(x)) - Ml < E.

•

Since Eis arbitrary this proves that Jim f(g(x)) = M. '

I Section 2.4

x--+C

THEOREM 4 Continuity of Composite Functions Let !(x) - f(g~))(be) athcompo~. . . . d f 1·s contmuous at x - g c , en F 1s 1te funct10n. If g 1s contmuous at x = c an continuous at x c. Proof By definition of continuity, and

Jim g(x) = g(c)

Jim f(x) = J(g(c))

Therefore, we may apply Theorem 3 to obtain Jim f(g(x)) = f(g(c))

•

This proves that F(x) = f(g(x)) is continuous at x = c. I Section 2 .6

THEOREM 5 Squeeze Theorem Assume that for x taining c), and

l(x) .::S f(x) .::S u(x)

f= c (in some open interval con-

Jim l(x) = Jim u(x) = L

Then Jim f(x) exists and Jim f(x) = L. Proof Let E > Obe given. We may choose 8 > 0 such that ifO < Ix - cl < 8, then ll(x) - LI < E and lu(x) - LI < E.

In principle, a different 8 may be required to obtain the two inequalities for l(x) and u(x), but we may choose the smaller of the two deltas. Thus, if O < Ix - cl < 8, we have L-

E

< l(x) < L

+E

L-

E

< u(x) < L

+E

and

Since f(x) lies between l(x) and u(x), it follows that L-

E

< l(x) .::S f(x) .::S u(x) < L

+E

and therefore lf(x) - LI < E if O < Ix - cl < 8. Since E is arbitrary, this proves that lim f(x) = L, as desired.

I Section 4.2

THEOREM 6 Existence of Extrema on a Closed Interval A continuous function f on a closed (bounded) interval/= [a,b] takes on both a minimum and a maximum value on/. Proof We prove that f takes on a maximum value in two steps (the case of a minimum is similar).

ADDITIONAL pROOFS

A19

APPENDIX D

Step 1. Prove th tf. . t paints We use a IS bounded from above. ve then there exis e a E [ proof by contradiction. If f is not bounded from abo • 3 in Appendix B, w C::aycha,b] such that f(an) 2:: n for n = 1,2, .... By Tbeorermges to a limit in [a,b]oose a subs that conve say Ii equence of elements an 1 , an2 • · · ·. 0 such that 'k-..~ ank = L. Since f is continuous, there eXIS ts 0 >

if x

E [a,b]

and Ix - LI< S, then 1/(x) - f(L)I < 1.

Therefore, if x

E [a,b]

and x

S, L

E (L -

+ S),

then /(x) < f(L)

+

0

1.

· = L. By Eq. (S), For k sufficiently large Ii . (L - S L + S) because hm f . • es m • kd~O infinity as k - ~(ank) is bounded by J(L) + 1 However, J(ank) = nk ten s d d from above 1s ank

I \

ank

This IS · a contradiction. . . . th at / 1·s not boun e Hence, ·our assumption false.

S tep 2 - Prove that/ takes on a maximum value. The range off on I = [a, b) is the set S = {f(x): x E [a,b]} . h s a least upper bound Y the previous step, S is bounded from above and therefore a th proof we 1 M by the LUB Property. Thus, f(x) :S M for all x E [a, b]. To comp_ etethe maxi~um show that f(c) = M for some c E [a,b]. This will show that f attams e

B

value Mon [a, b]. 1 d therefore, we may By definition, M - 1/n is not an upper bound for n 2: , an choose a point bn in [a, b) such that 1 M - -

::c f (bn) :'.5 M n Again by Theorem 3 in Appendix B, there exists a subsequence of elements {bn" bn 2 , ••• } in {b1, b2, ... } that converges to a limit-say, lim bnk

k--+oo

=

c

Furthermore, this limit c belongs to [a, b] because [a, b] is closed. Let E > 0. Since f is continuous, we may choose k so large that the following two conditions are satisfied: 1/(c) - f(bnk)I < E/2 and nk > 2/E. Then 1/(c) - Ml :'.5 1/(c) - f(bnk)I

+ lf(bnk) -

Ml :'.5

Thus, lf(c) - Ml is smaller than E for all positive numbers unless lf(c) - Ml= 0. Thus, f(c) = M, as desired.

I Section 5 .2

1

E

2+

llk

E.

:'.S

E

E

2+2=

E

But this is not possible •

THEOREM 7 Continuous Functions Are Integrable If f is continuous on [a, b], or if f is continuous except at finitely many jump discontinuities in [a, b], then J is integrable over [a, b].

Proof We shall make the simplifying assumption that f is differentiable and that its derivative J' is bounded. In other words, we assume that lf'(x)I ::5 K for some constarit K. This assumption is used to show that f cannot vary too much in a small interval M "f [ b ] • . ore precisely, let us pr?ve th at 1 ao,_ o 1s any closed interval contained in [a, b] and if m and Mare the mimmum and maximum values off on [ao, bo], then

\M - ml

:s

Klbo -

aol

I l

A20 A p P EN DI X D

ADDITIONAL PROOFS

y

Figure 1 illustrates the idea behind this inequality. Suppose that{~')~m and f(xz):::: =I- x2, then by the Mean ue eorem 0. Since f is continuous, there exists 8 > 0 such that if O < \x - L\ < 8, then \f(x) - f(L)\ < Since lim an

= L, there exists N > 0 such that \an \/(an) - /(L)\
N. Thus,

for n > N

It follows that lim /(an)= f (L). I Section 14.3

THEOREM 9 Clairaut's Theorem If fxy and fyx both exist and are continuous on a disk D, then fxy(a, b) = fyx(a, b) for all (a, b) E D.

1

Proof We prove that both fxy(a, b) and fyx(a, b) are equal to the limit . f(a L = lim

+ h,b + h)- f(a + h,b)- f(a,b + h) + f(a,b) h2

= f(x,b + h)- f(x,b). The numerator in the limit is equal to F(a + h) - F(a) and F'(x) = fx(x, b + h) - fx(x, b). By the MVT, there exists a 1 between a and a+ h

Let F(x)

such that

= hF'(a1) = h(fx(a1, b + h) - fx(a 1, b)) By the MVT applied to fx, there exists b1 between band b + h such that fx(a1,b + h)- fx(a1,b) = hfxy(a1,b1) F(a + h) - F(a)

A22

App END IX D

ADDITIONAL PROOFS

Thus, F(a+ h) -

F(a)==h2fxy(a1,b1)

and h 2 fxy(a1,b1) == Jim fxy(a1,b1) == fxy(a ,b) h2 • . ff since (a1,b1) approaches (a,b) as The last equality follows from the contmUity :e ument using the function F(y):::: h 0. To prove that L = fyx(a,b), repeat g d 1 . 1 of x and y reverse f(a + h, y) - f(a, y), with the roes .

L = hm

:J

I Section 14.4

• • .ty If f (x y) and fy(x, y) exist and are THEOREM 10 Criterion for Different1abl 1I x '. D . f( ) · differentiable on · continuous on an open disk D, then x, Y is

Proof Let (a, b)

E D

and set

L(x,y)=f(a,b)+fx(a,b)(x-a ) + f y\ ' ;"

'a b\fv - b)

· bl h d k where x It is convenient to switch to the vana es an , 1:!..f == f(a

+ h, b + k) -

= a + h and Y = b + k. Set

f(a, h)

Then l(x,y)

= f(a,b) + fx(a, b)h + fy{o.,h )k

and we may define the function e(h,k) = f(x,y)- l(x,y) = 1:!..f- (fx(a , b)h

+ fy(a,b)k)

To prove that f(x, y) is differentiable, we must show that . hm

e(h,k) O ---= Jh2 + k2

To do this, we write 1:!..f as a sum of two terms: l:!..f

= (f(a + h, b + k) -

f(a , b + k)) + (f(a, b + k) - f(a, b))

and apply the MVT to each term separately. We find that there exist a1 between a and a + h, and b1 between b and b + k, such that f (a+ h, b + k) - f (a, b + k) f(a, b + k) - f(a, b)

= hfx(a1, b + k)

= kfy(a, bi)

Therefore, e(h,k)

= h(fx(a1,b + k)- fx(a,b)) + k(fy(a,b1) -

/y(a,b))

and for (h, k) -1- (0, 0), e(h,k) Jh2 + k2 I

I= lh(fx(a1,b + k) - fx(a,b)) + k(fy(a,b1) - /y(a,b)) I Jh2 + k2 :s /h(fx(a1, b +k) - fx(a, b)) I+ Ik(fy(a, b1) - /y(a, b)) I 2 2 Jh

+k

= lfx(a1,b + k) -

fx(a, b)I

Jh2

+ IJy(a,b1) -

+ k2

/y(a, b)I

In the second line, we use the Triangle Inequality [see Eq. (1) in Section 1.1], and we may pass to the third line because lh/ Jh 2 + k 2I and lk/ Jh 2 + k2 / are both less than 1. Both terms in the last line tend to zero as (h, k) (0, 0) because fx and J are assumed to be continuous. This completes the proof that f(x, y) is differentiable. Y •

,,,,,------------=-~~..:.:.:===---

ANSWERS TO ODD-NUMBERED EXERCISE___ S--------

chapter 1O

· es Section 10.2 Exercis

section 10.1 Preliminary Questions

=

12 2. (c) 3 . • ~a.= 1_ 14 4 _ (b) 5_ (a) False; counterexample: an = cos rrn (b) True (c) False; counterexample: a. = (-1)"

(a) (iv)

3.

CJ

5.

= 3; C2 = a1 ~ 2; a2 = b1 = 4; b2 = CJ = I· C2 = ' 0

7. 9•

L

(b) (i)

(d) (ii)

¥

~; C3 = ~; C4 = 5; a3 = 47; a4 = 4415 6; b3 = 4; b4 = 6 l, C4 -- 12 2• CJ= !!. 6 •

= 19 !!±l an = n+s

(.·

- (-Ir' n

(b)

f~.~+i f2

Ssoo

== 0.03539816290;

,vn 2

= I +1

27 • Limit= 0 cos-

31 . •

29. 1

43.

47.

( 10 +

45.

2n -3

2

n--+oo

25

51. The sequence diverges.

2n

53 . •

"

Jim 2+7•4" 3- 4" = _!7

2 lim (lnn) = 0

n--+oo

n

(2

+ 4 ) 1/3

= 2 1/3

55• •r..T&i

59. lim

!)" = e

I

.

z

1h # O

-

)

• 2

33. S ==

n--+oo

1-n)

¥

,,6 0

23. ; 59 049

"mt'

29• S ==

· 27 . The series diverges.

35.

s == 4

7

37. S

,& +

= 13

39.

ij

9

9 1000

+•"==~=! I - TII

51. (a) Counterexample:

L 00

(

)"

½

=

I

n=l

(b) Counterexample: If an

== 1, then SN

=N

L ¼diverges 00

(c) Counterexample:

=

L cos 2rrn # I 00

(d) Counterexample:

½

·

91. (e) AGM ( I,

I

==z

n=l

65. lim (1 + n

69. lim

k

lim (-l)"+I ( n-;; I) does not exist.

47. (b) and (c)

__I _ = 0 73. 1im (2" + 3") 11• = 3 75. (b) +nS 77. Any number greater than or equal to 3 is an uppe~ bound. 79. Example: an= (-1)" 83. Example: f(x) = sm:rrx

71.

==

lim

n

5"

67.

17.

9 9 45. 0.999999., , = 1Jj + 100

=0

+ 1 ) -- In 13 57• The sequence diverges. In ( 3n+4 e"+(-J)" =0 61. Jim nsin!!. =rr

63.

5"

31. S == _I_ e-1 41. 43.

41. The sequence diverges.

n--+oo

I = ~; S4 = ~; Ss == TT;~ :;r,;'l-1

!•

n--+OO

9" lim - = 0 n! Jim 3n2tn+2 = 1 49. lim ™

lim zl/n = l

n--+oo

I ) -

;;-:n - • n

_L.. n+ I -

13. S3

I. Ss

21. lim a. == lim cos ii+T -

= 99,999

(b) M

:C, (

5 "L.., == 14; n=I

=

n-+oo

33. Limit"=' 1.61803

(-½ )") = 10

o

S1000 -

11. S3

19.

= In 3

+ ln = 2

lim

3 •S lJj• 4 -

15.S==½

Jim In ( ~f++42n )

(2n,~ 1) = 1'

35. (a) M = 999 39 . •

25.

4'

15. Diverges

17. ;-.li."!1. = 19. Diverges --+ 00 21. "foe sequence diverges. 23. iirn

536'.I

5. 9.

-

oo

11. bt "'- 2; b2 = 3; b3 = 8; b4 13. 1>) an -

-

I+~

==

144; S6 - j6ilO 6 7 S6 == t.24992 2 s - 4 · S6==, • S2 = !; 4 - 3• _ 0 39810274; 035 S10 = o.03535167962; s,~ 03.539816334; yes.

.

(c) (iii)

(d) an

205

s2 - 1- S4 ==

3

1.

n-l

0)

(b) an=

== ( - l) n+I !!'n!.;.

(c) an

section 10.1 Exercises

-J.

(a) an=

1.

"=' 1.198

n=l

53. (a) (.55)"(.48)"- 1 (.52) (b)

L~I (.55)" (.48)n-l (.52) = L~I (.55)(.52)((.~5)(.48))"-l =

I ~-a8~64 "=' 0.39

(c)

I:~ 1(.52)((.48)(.55))"- 1 =

1

~J.~64 :;::; 0.71

55. The total area is ¼-

Section 10.2 Preliminary Questions 1

The sum of an infinite series is defined as the limit of the s~qu_ence of p~al sums. If the limit of this sequence does not exist, the senes is said to diverge.

2. S = k · 'th all • I · t alid· A senes w1 3. The result is negative, so the r~su t i:n~: fo~ula is not valid positive terms cannot have a negauve su . · . senes • w1'th lrl > because a geometnc - I diverges.

4.

7.

No 5. No 6. N = l3 l t all N No, SN is increasing and converges to 1, so SN ::: or .

L 00

8.

Example:

n=l

57. (a) De-k + De-Zk + De-Jk + · · · (b) De-kt (c) t ::".

= 1°.:_~~l

+ De-2kt + De-3kt + ... = v,-k,

-¼ ln(l -

1-e-''

!})

59. The total length of the path is 2 +

../2.

63. (a) As x -,. oo the diameter of the horn goes to zero. If we can fill the horn all of the way to the end, then the paint must be capable of being spread thin enough to fit all of the way into the horn. (b) Use a volume of ,},, milliliters of paint to paint the portion of the horn between x n

an_d ~- = n + I: Overal~ we use a volume of½+¼+½+ nulhhter of pamt to pa.mt the surface.

=

ft+ ... = I

ANS1

ANS2

ANSWERS TO ODD-NUMBERED EXERCISES

Section 10.3 Preliminary Questions 1.

Section 10.4 Exercises

(b)

2 • _A function f such that an contmuous for x :::_ I.

3. 4.

5.

= f(n) must be positive, decreasing, and

Convergence of p-series or integral test Comparison Test co 1 No; L..n diverges, but since ,-• < l for n > I the Comparison n n - '

3 Converges con

11.

'°' n=I

Test tells us nothing about the convergence of

L ,:• . 00

n=I

•

3. 5

Jx= 1 (x + 1)4 x converges, so the sen es converges. oo

J25

•

JI

9

•

11. 13• 17,

.

ft x-l/J dx = oo, so the series diverges.

•

7

Id

roo

x2

dx converges, so the series converges.

. x'dx+ 1 converges, so the senes converges.

oo I d Jx=I x(x+S) x converges, so the series converges.

f2

00

converges, so the series converges. th . :C: ;;r, so e senes converges.

x(l:x)' dx I

I

n' + Sn n~•

(½

:C:

r,

so the series converges.

19. nl/,1+z• :C: (½)",sotheseriesconverges.

!

21. m!

4m

:C: 4 (

23. 0 :,: 25.

:,:

27. ~ l :,: (n+P! 29.

;!,; n , so the series converges.

:,: ;f,: for n :". ·1, so the series converges. 100 (lnn~\ :,:

33.

yr :,: ( j) n for n :". 1, so the series converges.

;;i1@- for n sufficiently large, so the series converges.

37. The series converges. 39. The series diverges. 41. The series converges. 43. The series diverges. 45. The series converges. 47. The series converges. 49. The series diverges. 51. The series converges. 53. The series diverges. 55. The series converges. 57. The series diverges. 59. The series diverges. 61. The series diverges. 63. The series converges. 65. The series diverges. 67. The series diverges. 69. The series converges. 71. The series converges. 73. The series diverges. 75. The series converges. 77. The series converges for a > I and diverges for a :C: I. 79. The series converges for p > I and diverges for p :C: I.

I: n00

5

1.0369277551

= 0.947

IS. S44

I~

100

n=l

n=I

L ;f,: = 1.6439345667 and I+ L nt(,;+ T{2

second sum is a better approximation to -

6

LTn

I)

== 1.6448848903. The

1.6449340668.

Section 10.4 Preliminary Questions I.

Example:

4.

IS - S1ool < 10-3 , and Sis larger than S100-

2. (b) 3. No

--

Sn_ 0.899782407 0.902697859 0.900744734 0.9021 !6476 0.9011 !6476

= 0.06567457397

Converges (by Limit Comparison Test) Converges (by Limit Comparison Test) Diverges (by Limit Comparison Test) Converges (by geometric series and linearity) Converges absolutely (by Integral TeS t) Converges (by Alternating Series Test) .. Converges (by Integral Test) 33. Converges cond1t.10nally

Section 10.5 Preliminary Questions (b) lei l/n lrl

2. (a) ( n~ I

r

1.

(a) lrl

3.

Yes in the frrst case, no in the second. yes in the fust case, no in the second.

(l>) Nothing

Converges absolutely 3. Converges absolutely The Ratio Test is inconclusive 7. Diverges Converges absolutely 11. Converges absolutely Diverges 15. The Ratio Test is inconclusive Converges absolutely 19. Converges absolutely

21. p = ½ < I 23. p = 21xl 25. p = lrl 27. Converges 29. Converges absolutely 31. The Ratio Test is inconclusive, so the series may converge or diverge. 33. Converges absolutely 35.

=

Jim

Jim n-p/n

=

Jim e-p(¥)

= e-{l!_"d., ¥) = e0 = I.

Therefore the root test is inconclusive. 37. Converges absolutely 39, Converges absolutely 41. Converges absolutely 43. 45. 47. 49. 51.

Converges (by geometric series and linearity) Diverges (by the Divergence Test) Converges (by the Direct Comparison Test) Diverges (by the Direct Comparison Test) Converges (by the Ratio Test)

53. Converges (by the Limit Comparison Test) 55. Diverges (by p-series) 57, Converges (by geometric series) Converges (by Limit Comparison Test) 61. Diverges (by Divergence Test) 65. (b) ,,/fir~ 2.50663

c../7 59,

n=I

91.

19. 21. 23. 25. 27. 29. 31.

1. 5. 9. 13. 17.

so the series converges.

31.

s1.

n 6 7 8 9 10

Sn I 0.875 0.912037037 0.896412037 0.904412037

Section 10.5 Exercises

'tr, so the series converges.

r,

13. S5

4.

¼) m, so the series converges.

3n .:3-n :C: 2 ( ½

n 1 2 3 4 5

17. Converges (by geometric series)

Section 10.3 Exercises 1roo

ditionally 5. Converges absolutely di · II

1: Converges absolutely 9. Converges con aona y

ennJ

n

;jn+T7I

1000 1500 2000 2500 3000

2.506837 2.506768 2.506733 2.506712 2.506698

r

\7 ANS3

ANSWERS TO OD

Section 10.6 Preliminary Qu es t·ions l

.

Yes. Tbe series must conv erge for b b (a), (c) 3. R 4 ot x

=

i.

= 4 and x = -3.

00

= Ln2xn-1; R =

F'(x)

4,

l

n=l

= 2. It does not conv

R R

3. 9,

=

3 for all thre

17, (-oo , oo) 19. (-1 I] 27. (6 , 8) 29. [ - Z -•~) 2•

33.

I

(2

+

- e, 2

l) e

2

oo

n

n=O

L

.

on the interval ( _ ! !) 3' 3

n=0

Tn(x)

'°'

n=0 00

= L3nx3n-1 n-+oo

n=l

r (n+ l)lxl'(n+l)- 3 n..!..~ nlxl 3" - 3 =

=e+

,lx-1>2

I)+~+···

e(x -

-

31. T2(x)

an

00

47. L(-1)"+ 1 (x - 5)" on the interval (4, 6)

4(x - 1) 3 + · · ·

27i

and IS - S41 ""0.000386