Student's Solutions Manual Part One to accompany Thomas' Calculus, Eleventh Edition [11 ed.] 0321226461, 9780321226464

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STUDENT'S SOLUTIONS MANUAL PART ONE WILLIAM ARDIS JOSEPH E. BORZELLINO LINDA BUCHANAN ALEXIS T. MOGILL PATRICIA NELSON

to a c c o m p a n y

Thomas' Calculus Eleventh Edition

Weir ∙ Hass ∙ Giordano

STUDENT’S S OLUTIONS M ANUAL PART O NE A RDIS ∙ B ORZELLINO ∙ BUCHANAN ∙ M OGILL ∙ N ELSON to accompany

T HOMAS’ C ALCULUS E LEVENTH E DITION BASED

ON THE

ORIGINAL W ORK BY

George B. Thomas, Jr. Massachusetts Institute o f Technology AS

REVISED

BY

Maurice D. Weir Naval Postgraduate School

Joel Hass University o f California, Davis

Frank R. Giordano Naval Postgraduate School

Boston San Francisco New York London Toronto Sydney Tokyo Singapore Madrid Mexico City Munich Paris Cape Town Hong Kong Montreal

Reproduced by Pearson Addison-Wesley from electronic files supplied by the authors. Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN

0-321-22646-1

4 5 6 BB 08 07 06

PEARSO N

Addison Wesley

PREFACE TO THE STUDENT The Student’s Solutions Manual contains the solutions to all of the odd-numbered exercise in the 11th Edition of THOMAS’ CALCULUS by Maurice Weir, Joel Hass and Frank Giordano, excluding the Computer Algebra System (CAS) exercises. We have worked each solution to ensure that it •

conforms exactly to the methods, procedures and steps presented in the text

•

is mathematically correct

•

includes all of the steps necessary so you can follow the logical argument and algebra

•

includes a graph or figure whenever called for by the exercise, or if needed to help with the explanation

•

is formatted in an appropriate style to aid in its understanding

How to use a solution's manual

•

solve the assigned problem yourself

•

if you get stuck along the way, refer to the solution in the manual as an aid but continue to solve the problem on your own

•

if you cannot continue, reread the textbook section, or work through that section in the Student Study Guide, or consult your instructor

•

if your answer is correct by your solution procedure seems to differ from the one in the manual, and you are unsure your method is correct, consult your instructor

•

if your answer is incorrect and you cannot find your error, consult your instructor

Acknowledgments

Solutions Writers William Ardis, Collin County Community College-Preston Ridge Campus Joseph Borzellino, California Polytechnic State University Linda Buchanan, Howard College Tim Mogill Patricia Nelson, University of Wisconsin-La Crosse Accuracy Checkers Karl Kattchee, University of Wisconsin-La Crosse Marie Vanisko, California State University, Stanislaus Tom Weigleitner, VISTA Information Technologies

Thanks to Rachel Reeve, Christine O’Brien, Sheila Spinney, Elka Block, and Joe Vetere for all their guidance and help at every step.

TABLE OF CONTENTS Preliminaries 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7

Real Numbers and the Real Line 1 Lines, Circles, and Parabolas 3 Functions and Their Graphs 8 Identifying Functions; Mathematical Models 11 Combining Functions; Shifting and Scaling Graphs 13 Trigonometric Functions 19 Graphing with Calculators and Computers 22 Practice Exercises 26 Additional and Advanced Exercises 30

Limits and Continuity 33 2.1 2.2 2.3 2.4 2.5 2.6 2.7

Rates of Change and Limits 33 Calculating Limits Using the Limit Laws 36 Precise Definition of a Limit 38 One-Sided Limits and Limits at Infinity 42 Infinite Limits and Vertical Asymptotes 45 Continuity 48 Tangents and Derivatives 51 Practice Exercises 54 Additional and Advanced Exercises 56

Differentiation 61 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8

The Derivative as a Function 61 Differentiation Rules 65 The Derivative as a Rate of Change 68 Derivatives of Trigonometric Functions 71 The Chain Rule and Parametric Equations 74 Implicit Differentiation 79 Related Rates 83 Linearizations and Differentials 85 Practice Exercises 88 Additional and Advanced Exercises 94

Applications of Derivatives 97 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8

Extreme Values of Functions 97 The Mean Value Theorem 103 Monotonic Functions and the First Derivative Test 105 Concavity and Curve Sketching 111 Applied Optimization Problems 120 Indeterminate Forms and L,HopitaΓs Rule 127 Newton’s Method 129 Antiderivatives 131 Practice Exercises 134 Additional and Advanced Exercises 140

5 Integration 145 5.1 5.2 5.3 5.4 5.5 5.6

Estimating with Finite Sums 145 Sigma Notation and Limits of Finite Sums 147 The Definite Integral 149 The Fundamental Theorem of Calculus 155 Indefinite Integrals and the Substitution Rule 158 Substitution and Area Between Curves 161 Practice Exercises 168 Additional and Advanced Exercises 173

6 Applications of Definite Integrals 177 6.1 6.2 6.3 6.4 6.5 6.6 6.7

Volumes by Slicing and Rotation About an Axis 177 Volumes by Cylindrical Shells 182 Lengths of Plane Curves 185 Moments and Centers of Mass 188 Areas of Surfaces of Revolution and the Theorems of Pappus 192 Work 196 Fluid Pressures and Forces 199 Practice Exercises 201 Additional and Advanced Exercises 206

7 Transcendental Functions 209 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8

Inverse Functions and Their Derivatives 209 Natural Logarithms 211 The Exponential Function 215 ax andlog a x 218 Exponential Growth and Decay 221 Relative Rates of Growth 222 Inverse Trigonometric Functions 224 Hyperbolic Functions 231 Practice Exercises 235 Additional and Advanced Exercises 240

8 Techniques of Integration 243 8.1 8.2 8.3 8.4 8.5 8.5 8.6 8.7

Basic Integration Formulas 243 Integration by Parts 247 Integration of Rational Functions by Partial Fractions 250 Trigonometric Integrals 253 Trigonometric Substitutions 255 Integral Tables and Computer Algebra Systems 258 Numerical Integration 264 Improper Integrals 270 Practice Exercises 275 Additional and Advanced Exercises 283

9 Further Applications of Integration 287 9.1 9.2 9.3 9.4 9.5

Slope Fields and Separable Differential Equations 287 First-Order Linear Differential Equations 288 Euler’s Method 291 Graphical Solutions of Autonomous Differential Equations 292 Applications of First-Order Differential Equations 296 Practice Exercises 298 Additional and Advanced Exercises 301

10 Conic Sections and Polar Coordinates 303 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8

Conic Sections and Quadratic Equations 303 Classifying Conic Sections by Eccentricity 308 Quadratic Equations and Rotations 312 Conics and Parametric Equations; The Cycloid 314 Polar Coordinates 317 Graphing in Polar Coordinates 319 Areas and Lengths in Polar Coordinates 324 Conic Sections in Polar Coordinates 327 Practice Exercises 331 Additional and Advanced Exercises 336

11 Infinite Sequences and Series 343 11.1 Sequences 343 11.2 Infinite Series 347 11.3 The Integral Test 350 11.4 Comparison Tests 352 11.5 The Ratio and Root Tests 354 11.6 Alternating Series, Absolute and Conditional Convergence 355 11.7 Power Series 358 11.8 Taylor and Maclaurin Series 363 11.9 Convergence of Taylor Series; Error Estimates 364 11.10 Applications of Power Series 367 11.11 Fourier Series 371 Practice Exercises 373 Additional and Advanced Exercises 379

CHAPTER 1 PRELIMINARIES 1.1 REAL NUMBERS AND THE REAL LINE 1. Executing long division, ∣= 0.1, ∣= 0.2, ∣= 0.3, ∣= 0.8, ∣= 0.9 3. NT = necessarily true, NNT = Not necessarily true. Given: 2 < x < 6. a) NNT. 5 is a counter example. b) NT. 2 < x < 6 ≠ > 2 - 2 < x - 2 < 6 - 2 ≠ > 0 < x - 2 < 2 . c) NT. 2 < x < 6 ≠- 2/2 < x/2 < 6/2 => 1 < x < 3. d) NT. 2 < x < 6 ≠> 1/2 > 1/x > 1/6 ≠> 1/6 < 1/x < 1/2. e) NT. 2 < x < 6 ≠>1/2 > 1/x > 1/6 => 1/6 < 1/x < 1/2 => 6(1/6)< 6(l∕x) < 6(1/2) ≠> 1< 6/x < 3. f) NT. 2 < x < 6 => x < 6 = > ( x - 4 ) < 2 and 2 < x < 6 ≠- x > 2 ≠- —x < —2 =>—x + 4 < 2 => —(x —4) < 2. |x —4 ∣ g) NT. 2 < x < 6 => —2 > - x > —6 => —6 < —x < —2. But —2 < 2. So —6 < —x < —2 < 2 or —6 < —x < 2. h) NT. 2 < x < 6 ≠> -1 (2 ) > - l ( x ) < -1 (6 ) => - 6 < - x < - 2 5. - 2 x > 4 ≠> x < - 2 7.

5x - 3 ≤ 7 - 3x ≠> 8x < 10 ≠> x ≤ ∣

9.

2x - ∣> 7x + ⅛ => - l - 2 > 5 χ

5/4

=► | ( - ⅛ ) ≥ X or - I ≥ x

÷x

-1/3

11. I (x - 2) < I (x - 6) => 12(x - 2) < 5(x - 6) => 12x - 24 < 5x - 30 => 7x < - 6

or x < - 7

-6 /7

13. y = 3 or y = - 3 15. 2t + 5 = 4 or 2t + 5 = - 4 => 2t = - 1 or 2t = - 9 =► t = - ∣o rt = - j 7 25 17. 8 —3s = 1 or 8 —3s = — ∣ ≠> —3s = — j or —3s = - y => s = ⅛ o r s = f

19. - 2 < x < 2; solution interval (—2,2)

-2

——O----- > x 2

21. —3 ≤ t - 1 ≤ 3

-2

—→ 4

=> —2 ≤ t ≤ 4; solution interval [—2,4]

►/

23. - 4 < 3 y - 7 < 4 ≠> 3 < 3 y < l l => 1 < y < ⅛ ; solution interval (1, y )

1

— O— > y 11/3

25. —1 ≤ 5 — 1 ≤ 1 = > 0 ≤ f ≤ 2 = > 0 ≤ z < 1 0 j solution interval [0,10] 27.

2

j

x

2

=> ¾ < x < ∣; solution interval

2 ( 7 , ∣)

2 ^ x

0

10

10/35

14/35

2

2

Chapter 1 Preliminaries

29. 2s > 4 or —2s ≥ 4 => s ≥ 2 o r s ≤ —2; -2

solution intervals (—∞ , —2] U [2, ∞ )

2

31. 1 - x > 1 o r -(1 - x) > 1 => - x > 0 o r x > 2 => x < 0 or x > 2; solution intervals (—∞ , 0) U (2, ∞ )

0

2

33. ⅛ > l o r - ( ψ ) ≥ 1 => r + 1 ≥ 2 o r r + l < - 2 1

-3

≠∙ r > 1 or r ≤ —3; solution intervals (—∞ , -3 ] U [1, ∞ ) x ∣< √ 2 ≠ , - ᅟ ∕2 < 35. x2 < 2 => ∣ solution interval

37. 4 < X2 < 9 => 2 < ∣ x∣< 3 => 2 < x < 3 o r 2 < - x < 3 => 2 < x < 3 o r - 3 < x < —2; solution intervals (—3, —2) U (2,3)

—

^ →> - — o—

3 - 2

2

-o -

3

39. (x - 1)2 < 4 => ∣ x - 1∣< 2 ≠> - 2 < x - 1 < 2 ≠- - 1 < x < 3; solution interval (—1,3) x - ∣∣ < ∣ => - ∣< x - ∣< ∣ => 0 < x < l . 41. x2 - x < 0 => x2 —x ÷ ∣< ∣ => (x - I ) 2 < ∣ = > ∣ So the solution is the interval (0,1) 43. True if a ≥ 0; False if a < 0. 45. (1) ∣ a + b∣= (a + b) or ∣ a ÷ b∣= - ( a + b); both squared equal (a + b)2 (2) ab ≤ ∣ ab∣= ∣ a∣ ∣ b∣ 2 = b2 2 = a2 ; likewise, ∣ a∣= a or ∣ a∣ b∣ (3) ∣ a∣= - a , so ∣ (4) x2 ≤ y2 implies v ^ < 1^y2 θ r x ≤ y for all nonnegative real numbers x and y. Let x = ∣ a + b∣ and 2 ≤ (∣ a∣ + ∣ b∣ so that ∣ a + b∣ b∣ + ∣ b∣ . y —∣ a∣ + ∣ )2 => ∣ a + b∣≤ ∣ a∣

47. - 3 ≤ x ≤ 3 and x > —∣ => — ∣< x ≤ 3. x -l ∣ < 26 => 49. Let 6 be a real number > 0 and f(x) = 2x + 1. Suppose that ∣ x - 1 1< 6. Then ∣ < 6 => 2∣ x -1 ∣ ∣ 2x - 2 ∣ < 26 => ∣ < 26 => ∣ f(x) - f(l) ∣ < 26 (2x + 1) - 3 ∣ 51. Consider: i) a > 0; ii) a < 0; iii) a = 0. i) For a > 0, ∣ b ∣= —b. Since b = —a, a ∣= a by definition. Now, a > 0 => - a < 0. Let - a = b. By definition, ∣ ∣ - a ∣= - ( - a ) = a and ∣ a ∣= ∣ - a ∣= a. a ∣= —a. Now, a < 0 ≠> —a > 0. Let - a = b. By definition, ∣ b ∣= b and thus |—a∣= —a. So again ii) For a < 0, ∣ a ∣= ∣ ∣ -a ∣ . iii) By definition 10 ∣= 0 and since - 0 = 0, ∣ - 0 1 = 0. Thus, by i), ii), and iii) ∣ for any real number. a ∣= ∣ —a ∣ 53. a)

l=l ^

= l≠ > ∣ b ∙ l ∣= ⅛ ∣l ∣

^ ∣ i ∣= ^ b ∣∙ ∣

1

^

= ^

≠

∣1 ∣= ^

Section 1.2 Lines, Circles, and Parabolas

3

» ⅛i =k ∣ I H ÷ N =IΦ⅛ =⅛ 1.2 LINES, CIRCLES, AND PARABOLAS 1. ∆x = - 1 - (-3 ) = 2, ∆ y = - 2 - 2 = - 4 ; d = √ (∆ x ) 2 + (∆y) 2 = √ 4 + 16 = 2 √ 5 3. ∆x = - 8 . 1 - (-3.2) = -4.9, ∆y = - 2 - (-2 ) = 0; d = √ (- 4 .9 ) 2 + 02 = 4.9 5. Circle with center (0,0) and radius 1. 7. Disk (i.e., circle together with its interior points) with center (0,0) and radius ᅟ ∕3 . * = ^ = ⅛ ⅛ = 3

11∙ m

perpendicular slope = - ∣

= ^ = ⅛⅛ = θ

perpendicular slope does not exist y B ( - l, 3)

A (2 ,3)

3

*

2 -

Slope = 0

1-J______ I____ L -10 1 2

15. (a) x = 0 (b) y = - √ 2

13. (a) x = - 1 (b) y - f 17. P ( - l , 1), m = - 1 ≠> y - 1 = - l ( x - ( - l ) ) ≠> y = - χ

19. P(3,4),Q (-2,5) ^ m = ^ = ⅛ ⅛ = - ∣ ^ y - 4 = - ∣( x - 3 ) ^ y = - ∣x + ⅛ 21. m = - ∣, b = 6 => y = - ∣x + 6 23. m = 0 ,P ( -1 2 ,- 9 ) ≠> y = - 9 25. a = —1, b = 4 ≠- (0,4) and (—1,0) are on the line => m = ^ = ⅛ ⅛ = 4 => y = 4x + 4 27. P(5,-1 ), L: 2x + 5 y = 1 5 => mL = - ∣ => parallel line is y - ( - 1) = - ∣(x - 5) => y = - j x + l 29. P(4,10), L: 6x - 3y = 5 ≠∙ mL = 2 => m± = - ∣ => perpendicular line is y - 10 = - ∣(x - 4) => y = - ∣x + 12 31. x-intercept = 4, y-intercept = 3

33. x-intercept = ᅟ ∕3 , y-intercept = —√ z2

4

Chapter 1 Preliminaries

35. Ax + By = Ci «• y = - g x + ¾ and Bx - Ay = C2 «• y = ∣x - ⅞ . Since ( - g ) ( ∣) = - 1 is the product of the slopes, the lines are perpendicular. 37. New position = (xo∣ d + ∆ x, y0∣ d + ∆ y) = ( - 2 + 5,3 + (-6 )) = (3, -3 ). 39. ∆ x = 5, ∆ y = 6, B(3, -3 ). Let A = (x, y). Then Δ x = x 2 -x 1 => 5 = 3 - x =► x = —2 and ∆ y = y2 - yι ≠

6 = - 3 —y ≠

y = - 9 . Therefore, A = ( - 2 , —9).

41. C(0,2), a = 2 ≠> x2 + (y - 2)2 = 4 y

1

45. C ( —√ 3 , —2) , a = 2 => ( x + √ 3 ) 2 + (y + 2)2 = 4, X = 0 => ( 0 + √ 3 ) 2 + (y + 2)2 = 4 ≠ ( y + 2)2 = l => y + 2 = ± 1 ≠ ∙ y = - l o r y = - 3 . Also,y = 0 ≠ ( X + √ 3 } 2 ÷ ( 0 + 2)2 = 4 => (x + √ 3 ) 2 = 0 ≠

x = —χ ∕3

47. x 2 + y 2 + 4x —4y + 4 = 0 ≠

x2 + 4® + y2 - 4y = —4

≠ ∙ x 2 + 4x + 4 + y2 —4y + 4 = 4 ≠> (x + 2)2 + (y - 2)2 = 4 ≠

C = (-2 ,2 ), a = 2.

49. x 2 + y 2 —3y - 4 = 0 4 x 2 + y 2 - 3y = 4 => x2 + y 2 - 3y + ∣= ⅞ = > χ 2 + ( y - i ) 2 = ⅛ = > C = ( 0 , ∣) , a = t∙

51. x 2 + y 2 - 4x + 4y = 0 ≠ x 2 —4x + y 2 + 4y = 0 ≠ x 2 —4x + 4 + y2 + 4y + 4 = 8 = > ( x - 2 ) 2 + (y + 2)2 = 8 ≠ C(2, -2 ) , a = √ 8 .

43. C ( - l , 5), a = √ 1 0 ≠> (x + 1)2 + (y - 5)2

10

Section 1.2 Lines, Circles, and Parabolas 53 ax = — 2~a — ----2-(1— ) — 11 ^

y = (1)2 - 2(1) - 3 = - 4

≠> V = (1 ,-4 ). Ifx = 0 th e n y = - 3 . Also, y = 0 => x2 - 2x —3 = 0 =+ (x —3)(x + 1 ) = 0 =+ x = 3 o r x = —1. Axis of parabola is x = 1.

55 ∙ ax = - -2a = ----2-( — - 1 ) = 42

j j

=+ y = - ( 2 ) 2 + 4 (2 ) = 4 =+ V = (2,4). If x = 0 then y = 0. Also, y = 0 =+ - x 2 + 4x = 0 =+ -x (x - 4 ) = 0 =+ x = 4 o r x = 0. Axis of parabola is x = 2.

=+ y = - ( - 3 ) 2 - 6 (-3 ) - 5 = 4 =+ V = ( - 3 , 4). Ifx = 0 th e n y = - 5 . Also, y = 0 => - x 2 —6x —5 = 0 =+ (x + 5)(x + 1) = 0 =+ x = - 5 or x = —1. Axis of parabola is x = —3.

59.

X —

2a

-

=+ y = ∣( -

2(1/2)

l)2

-

'

+ (-l) + 4 = 2

≠> V = ( - l , j ) . Ifx = 0 then y = 4. Also, y = 0 =+ ∣x2 + x + 4 = 0 =+ x = ^1⅛VΞZ => no x intercepts. Axis of parabola is x = —1. 61. The points that lie outside the circle with center (0,0) and radius √ z7. 63. The points that lie on or inside the circle with center (1,0) and radius 2. 65. The points lying outside the circle with center (0,0) and radius 1, but inside the circle with center (0,0), and radius 2 (i.e., a washer).

5

6

Chapter 1 Preliminaries

67. x2 + y2 + 6y < 0 ≠ x2 + (y + 3)2 < 9. The interior points of the circle centered at (0, —3) with radius 3, but above the line y = -3 ∙

71. x2 + y2 ≤ 2, x ≥ 1

69. (x + 2)2 + ( y - l ) 2 < 6

73. x2 + y2 = 1 and y = 2x ≠ 1 = x2 + 4x2 = 5x2 (x = ⅛ and y = ⅛ ) o r (x = -⅛ and y= -⅛ )∙ t,iu s

'

a

(⅛ > ⅛ ) ^ ( - ^ - - ^ ) ^ ^

points of intersection.

75. y - x = 1 andy = x2 ≠ x2 - x = l => x2 - x - l = 0 ≠∙ x = ^ . If x = ^

, ⅛en y = χ + 1 = ^

.

If x = ⅛ ^ , then y = x + 1 = ^

.

Thus, A ( ⅛ ^ , 5⅛ ^ ) a n d β ( ⅛ ⅛ ) are the intersection points.

77. y = 2x2 - 1 = - x 2 ≠ 3x2 = 1 ≠ t = ∙ ^ and y = - ∣orx = - - ^ and y = - 1 . Thus, A ^ ^ , - ∣) and B ( —∙ ^ , - ∣) are the intersection points.

Section 1.2 Lines, Circles, and Parabolas 79. x2 + y 2 = 1 — (x — 1)2 + y 2 ≠> x2 = (x - 1)2 = x 2 - 2x + 1 => 0 = —2x + 1 ≠> x = ∣. Hence y2 = 1 - x 2 = ∣or y = ± δ ^ . Thus, A (l >

and B f ⅜, -

are the

intersection points.

81. (a) A ≈ (69o ,0 in), B ≈ (68o , .4 in) => m = ⅛ (b) A ≈ ( 6 8 o ,.4 i n ) ,B ≈ ( 1 0 o ,4 in ) ^

m = - ^ ∈^

≈ -2 .5 7 in , ≈ - 1 6 .1 7 in .

(c) A ≈ (10o ,4 in), B ≈ (5o ,4.6 in) ≠> m = ⅛ ^ ≈ -8 .3 7 in . 83. p = kd + 1 and p = 10.94 at d = 100 4

k= ¾ d

=

o.O994. Then p = 0.0994d + 1 is the diver's

pressure equation so that d = 50 => p = (0.0994)(50) ÷ 1 = 5.97 atmospheres. 85. C = ∣(F —32) and C = F => F = ∣F - y

=> ∣F = - ^ o r F = —40° gives the same numerical reading.

c

87. length AB = √ (5 - 1)2 + ( 5 - 2)2 = √ 1 6 + 9 = 5 length AC = √ ( 4 - 1)2 + ( - 2 - 2)2 = √ 9 + 16 = 5 length BC = √ ( 4 - 5)2 + ( - 2 - 5)2 = √ 1 + 4 9 = √ 5 0 = 5 √ 2 ≠ 5 89. Length AB = √ ( ∆ x ) 2 + (∆ y) 2 = √ l 2 + 4 2 = √ Γ 7 and length BC = √ ( ∆ x ) 2 + (∆ y) 2 = √ 4 2 + 12 = √ 1 7 . Also, slope AB = ^ and slope BC = ∣, so AB ± BC. Thus, the points are vertices of a square. The coordinate increments from the fourth vertex D(x, y) to A must equal the increments from C to B => 2 —x = ∆ x = 4 and - l - y = ∆ y = l => x = - 2 and y = - 2 . Thus D (-2 , - 2 ) is the fourth vertex. 91. Let A ( - l , 1), B (2,3), and C (2,0) denote the points. Since BC is vertical and has length ∣ BC∣= 3, let D ι( - 1 ,4) be located vertically upward from A and D 2 (-1, —2) be located vertically downward from A so that ∣ BC∣= ∣ A D ι∣= ∣ AD2∣= 3. Denote the point D3(x, y). Since the slope of AB equals the slope of CD3 we have ^ = - ∣ => 3y —9 = —x + 2 o r x + 3y = 11. Likewise, the slope of AC equals the slope of BD3 so that ^ = ∣ => 3y = 2x - 4 or 2x —3y = 4. Solving the system of equations

x + 3y = 111 we find x = 5 and y = 2 yielding the vertex D3(5,2). 2 χ -3 y = 4 J

8

Chapter 1 Preliminaries

93. 2x + ky = 3 has slope - ∣and 4x + y = 1 has slope - 4 . The lines are perpendicular when - ∣( - 4 ) = - 1 or k = —8 and parallel when —^ = —4 or k = ∣. 95. Let M(a, b) be the midpoint. Since the two triangles shown in the figure are congruent, the value a must lie midway between xι and X2, so a = ^ . Similarly, b = ^ ψ ^ .

1.3 FUNCTIONS AND THEIR GRAPHS 1. domain = (—∞ , ∞)*, range = [1, ∞ ) 3. domain = (0,00); y in range => y = ^ , t > 0 => y 2 = 7 and y > 0 => y can be any positive real number => range = (0, ∞ ). 5. 4 - z 2 = (2 -

z )(2

+ z) ≥ 0 Φ> z ∈[-2 ,2 ] = domain. Largest value is g(0) =

g (-2 ) = g(2) = √ 0 = 0 ^

y / i

= 2 and smallest value is

range = [0,2].

7. (a) Not the graph of a function of x since it fails the vertical line test. (b) Is the graph of a function of x since any vertical line intersects the graph at most once. 9. y = ^ Q ) “ 1 ≠> ^ - 1 ≥ 0 => x ≤ 1 andx > 0. So, (a) No (x > 0); (c) No; if x ≥ 1, i < 1 =≠4 - 1 < 0;

(b) No; division by 0 undefined; (d) (0, 1]

11. base = x; (height)2 + ( ∣) 2 = x 2 => height = ^ x ; area is a(x) = j (base)(height) = ∣(x) ^ x ^ = ^ x 2 ; perimeter is p(x) = x ÷ x ÷ x = 3x. 13. Let D = diagonal of a face of the cube and f = the length of an edge. Then £2 + D 2 = d 2 and (by Exercise 10) D 2 = M 2 => 3^2 = d 2 => ^ = ^

∙ The surface area is 6f 2 = y

= 2d2 and the volume is ^3 =

J

= y

Section 1.3 Functions and Their Graphs 15. The domain is ( - ∞ , ∞ ).

17. The domain is (—∞ , ∞ ).

19. The domain is (—∞ , 0) U (0, ∞ ).

21. Neither graph passes the vertical line test (b)

25.

27. (a) Line through (0, 0) and (1, 1): y = x Line through (1, 1) and (2, 0): y = —x + 2 X, 0 ≤ x < 1 f(x) = - x + 2, 1 < x ≤ 2

9

10

Chapter 1 Preliminaries

(b)

0≤ 1≤ 2≤ 3≤

2, 0, f(x) = < 2, 0,

x x x x

< < < ≤

1 2 3 4

29. (a) Line through (—1, 1) and (0, 0): y = —x Line through (0, 1) and (1, 1): y = 1 Line through (1, 1) and (3, 0): m = ^ ∣: y = - I , soy = - ∣(x - 1) + 1 = —⅜x + I x

0< x≤ 1 1< x< 3

1

- ⅛A T + 5 2

2

(b) Line through (—2, —1) and (0, 0): y = ∣x Line through (0, 2) and (1, 0): y = —2x + 2 Line through (1, —1) and (3, -1 ): y = - 1 r

f(x) =

—xA 2

-2 x + 2 -1

0

31. (a) From the graph, ^ > 1 ÷ ^ =^ x ∈(-2 ,0 ) U (4, ∞ ) Φ) l > ι + * => ∣- M

> θ

≠> x > 4 since x is positive; 0: ∣- 1 -

> 0 =>

< 0 4

=> x < —2 since x is negative; sign of (x —4)(x + 2) -2 4 Solution interval: (-2 ,0 )U (4 , ∞ )

33. (a) [xj = 0 for x ∈[0,1)

(b) M = 0 f o r x ∈( - l ,0 ]

35. For any real number x, n ≤ x ≤ n ÷ 1, where n is an integer. Now: n ≤ x ≤ n ÷ l = > —(n + 1) ≤ —x ≤ —n. By definition: ∣ ^-x] = —n and [xj = n => —[xj = —n. So ∣ ^ -x] = —[xj for all x ∈SR. 37. v = f(x) = x(14 - 2x)(22 - 2x) = 4x3 - 72x2 + 308x; 0 < x < 7. 39. (a) Because the circumference of the original circle was 8π and a piece of length x was removed. H > > ' = ⅛ = < - ⅛ ____________ (c) h = √ i t ^ (d i V w

V —

= √ 1 6 -(4 -⅛ )2 ≈ √ 1 6 -(∣ 6 -⅛ + ⅛ ) = √ ⅞ - ⅛

⅛ Γ 2h 3

7Γ1 n —

∙ V ⅛ Ξ Ξ 2π

;

2

π

—

= √ ⅛ -⅛

= ⅛

(8 π -x )2√ 1 6 π x -x 2 2

4π 2

41. A curve symmetric about the x-axis will not pass the vertical line test because the points (x, y) and (x, —y) lie on the same vertical line. The graph of the function y = f(x) = 0 is the x-axis, a horizontal line for which there is a single y-value, 0, for any x.

Section 1.4 Identifying Functions; Mathematical Models

11

1.4 IDENTIFYING FUNCTIONS; MATHEMATICAL MODELS 1. (a) linear, polynomial of degree 1, algebraic. (c) rational, algebraic.

(b) power, algebraic. (d) exponential.

3. (a) rational, algebraic. (c) trigonometric.

(b) algebraic. (d) logarithmic.

5. (a) Graph h because it is an even function and rises less rapidly than does Graph g. (b) Graph f because it is an odd function. (c) Graph g because it is an even function and rises more rapidly than does Graph h. 7. Symmetric about the origin Dec: —∞ < x < ∞ Inc: nowhere

9. Symmetric about the origin Dec: nowhere Inc: —∞ < x < 0 0< x< ∞

11. Symmetric about the y-axis Dec: —∞ < x ≤ 0 Inc: 0 < x < ∞

13. Symmetric about the origin Dec: nowhere Inc : —∞ < x < ∞

15. No symmetry Dec: 0 ≤ x < ∞ Inc: nowhere

17. Symmetric about the y-axis Dec: —∞ < x ≤ 0 Inc: 0 < x < ∞

19. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, the function is even.

12

Chapter 1 Preliminaries

21. Since f(x) = x2 + 1 = ( - x ) 2 + 1 = -f(x). The function is even. 23. Since g(x) = x3 ÷ x, g (-x ) = - x 3 - x = - ( x 3 ÷ x) = ~g(x). So the function is odd. 25. g(x) = ^ ⅛ ι = ^ j r y = g(—x). Thus the function is even. 27. h(t) = ^ ; h (-t) = ^

;

-h (t) = ⅛ . Since h(t) ≠ -h (t) and h(t) ≠ h (-t), the function is neither even nor odd.

29. h(t) = 2t + 1, h (-t) = - 2 t + 1. So h(t) ≠ h (-t). -h (t) = - 2 t - 1, so h(t) ≠ -h (t). The function is neither even nor odd. The graph supports the assumption that y is proportional to x. The constant of proportionality is estimated from the slope of the regression line, which is 0.166.

The graph supports the assumption that y is proportional to x 1∕ 2 . The constant of proportionality is estimated from the slope of the regression line, which is 2.03.

33. (a) The scatterplot of y = reaction distance versus x = speed is

Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which is approximately 1.1. (b) Calculate x' = speed squared. The scatterplot of x'versus y = braking distance is:

Section 1.5 Combining Functions; Shifting and Scaling Graphs Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which is approximately 0.059. 35. (a) y 10 9 -

β

8 7 -

∙ ∙

6 -

∙

5 4 -

∙ *

3 2 1- ∙ — θj 1

#

I 2

1 I 3 4

1 I 5 6

I 7

1 I I 8 9 10

The hypothesis is reasonable. (b) The constant of proportionality is the slope of the line ≈ ⅛ fz7p in./unit mass = 0.874 in./unit mass, (c) y(in.) = (0.87 in./unit mass) (13 unit mass) = 11.31 in. 1.5 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS 1. Df : - ∞ < x < ∞ , Dg : x ≥ 1 => Df+g = Dfg: x ≥ 1. Rf : - ∞ < y < ∞ , Rg : y ≥ 0, Rf+g : y ≥ 1, Rfg: y ≥ 0 3. Df : —∞ < x < ∞ , Dg : —∞ < x < ∞ => Df/g : —∞ < x < ∞ since g(x) ≠ 0 for any x; Dg / f : —∞ < x < ∞ since f(x) ≠ 0 for any x. Rf : y = 2, Rg : y ≥ 1, Rf/g : 0 < y ≤ 2, Rg ∕ f : y ≥ j 5. (a) (b) (c) (d) (e) (f) (g ) (h)

f(g (0)) = f(-3 ) = 2 g(f(O)) = g(5) = 22 f(g(x)) = f(x2 - 3) = x2 - 3 ÷ 5 = x2 + 2 g(f(x)) = g(x ÷ 5) = (x + 5)2 - 3 = x2 + 10x ÷ 22 f(f(-5)) = f(0) = 5 g (g (2)) = g(l) = - 2 f(f(x )) = f(χ ÷ 5) = (x ÷ 5) ÷ 5 = x ÷ 10 g(g(x)) = g(x2 - 3) = (x2 - 3)2 - 3 = x4 - 6x2 + 6

7. (a) u(v(f(x))) = u (v Q )) = u (⅛) = 4 (1) 2 - 5 = ⅛ - 5 (b) u(f(v(x))) = u (f (x2 )) = u (⅛) = 4 (⅛) - 5 = ⅛ - 5 (c) v(u(f(x))) = v (u Q )) = v (4 (∣) - 5) = (4 - 5) 2 (d) v(f(u(x))) = v(f(4x - 5)) = v ( 3 ⅛ ) = ( ⅛ )

2

(e) f(u(v(x))) = f (u (x2 )) = f (4 (x2 ) - 5) = 5⅛ (f) f(v(u(x))) = f(v(4x - 5)) = f ((4x - 5)2 ) = ⅛ 9.

(a) y = f(g(x)) (c) y = g(g(x)) (e) y = g(h(f(x)))

1

(b) y = j(g(x)) (d) y = j(j(x)) (f) y = h(j(f(x)))

g(x)

f(χ)

(f°g)(χ)

(a)

x- 7

√x

√x - 7

(b)

x+ 2

3x

3(x + 2) = 3x + 6

11.

13

14

Chapter 1 Preliminaries g(χ )

f(x)

(f ° g)(x)

(c)

x2

√ , χ —5

ᅟ ∕χ 2 —5

®

7=7

⅛

≠⅛ f =

(e)

⅛

1

+ 1

∣

(f > ⅛

x

-(χ -i)

=

x

x χ

13. (a) f(g(x)) = √ i + T = √ ψ g( f W ) = 7 ⅛ 7

(b) Domain (fog): (0, ∞ ), domain (gof): (—l,oo) (c) Range (fog): (1, ∞ ), range (gof): (0, ∞ ) 15. (a) y = - ( χ + 7)2 17. (a) Position 4 19.

23.

27.

31.

(b) y = - ( x - 4)2 (b) Position 1

(c) Position 2

(d) Position 3

Section 1.5 Combining Functions; Shifting and Scaling Graphs 35.

39.

43.

45.

49. (a) domain: [0,2]; range: [2,3]

(b) domain: [0,2]; range: [—1,0]

(d) domain: [0,2]; range: [—1,0]

15

16

Chapter 1 Preliminaries (e) domain: [—2,0]; range: [0,1]

(f) domain: [1,3]; range: [0,1]

(g) domain: [-2 ,0 ]; range: [0,1]

(h) domain: [-1 ,1 ]; range: [0,1] y 2y≈-∕U+l) + l

51. y = 3x2 - 3 53∙

y=K

1

+ p) = l + ⅛

55. y = √ 4 x + 1 57∙

y = √ 4 √ j j i = ∣√ 1 6 ≡ ^

59. y = 1 - (3X)3 = 1 - 27x3 61. Let y = - √ 2 x + 1 = f(x) and let g(x) = x1 /2 , h(x) = (x + | ) 1 /2 »i(x) = √ 2 (x + ^ ) 1 /2 , and j(x) = —£v^2 (x ÷ I / / 2 ] = ^ ) ∙ The graph of h(x) is the graph of g(x) shifted left ∣unit; the graph of i(x) is the graph of h(x) stretched vertically by a factor of √ ⅞ and the graph of j(x) = f(x) is the graph of i(x) reflected across the x-axis.

Section 1.5 Combining Functions; Shifting and Scaling Graphs 63. y = f(x) = x3 . Shift f(x) one unit right followed by a shift two units up to get g(x) = (x —1)3 + 2. y

65. Compress the graph of f(x) = £ horizontally by a factor of 2 to get g(x) = ^ . Then shift g(x) vertically down 1 unit to get h(x) = ⅛ - 1.

67. Reflect the graph of y = f(x) = ^ ∕x across the x-axis to get g(x) = - ^∕x.

17

Chapter 1 Preliminaries

18

+ f∑jι⅛=l

71. 9x2 + 25y2 = 225 => £ + £ = 1

w

75. 3(x - 1)2 + 2(y + 2)2 = 6

2

2

77. ⅛ + ⅜ = 1 has its center at (0, 0). Shiftinig 4 units left and 3 units up gives the center at (h, k) = (-4 , 3). So the equation is ^ ⅛ ^ - + ^y ^

= 1 ≠∙ ^ J A + ^y ^

= 1. Center, C, is ( - 4 , 3), and major axis, AB, is the segment

from (-8 , 3) to (0,3).

79. (a) (fg)(-x) = f(-x )g (-x ) = f(x)(-g(x)) = -(fg)(x), odd

(0 =⅞⅛=⅛ =- (0 «■““ (c) (f) (-x ) = f ^ = ^

= - ( f ) (X), odd

(d) f 2 (-x ) = f(-x )f(-x ) = f(x)f(x) = f 2 (x), even (e) g2 (-x ) = (g(-x)) 2 = (-g(x)) 2 = g2 (x), even (f) (f 0 g )(-x) = f(g(~x)) = f(-g(x)) = f(g(x)) = (f ° g)(x), even (g) (g o f)(-x) = g(f(~x)) = g(frx)) = (g ° f)(x), even (h) (f o f)(-x) = f(f(-x)) = f(f(χ)) = (f o f)(x), even (i) (g ° g)(~x) = g(g(-x)) = g(-g(x)) = -g(g(x)) = -(g ° g)(x), odd

Section 1.6 Trigonometric Functions 81. (a)

y

1.6 TRIGONOMETRIC FUNCTIONS (b) s = r0 = (10)(110o ) ( ⅛ ) ≈ ^

1. (a) s = r0 = (10) (⅞ ) = 8πm

= ψ m

3. 0 = 80o ≠> 0 = 80o ( ⅛ ) = v => s = (6) (γ ) = 8.4 in. (since the diameter = 12 in. => radius = 6 in.) 0

-π

2π 3

0

π 2

0

1

1

0

3π 4 1 72 1 “ 72

sin θ

0

cos θ

-1

2 _ 1 2

tan θ

0

√3

0

und.

-1

cot θ

und.

1

und.

0

-1

sec θ

-1

-2

1

und.

-√ 2

CSC θ

und.

2 “ 75

und.

1

7. cos x = —∣, tan x = —∣ 9. sinx = - ^ , t a n x = - √ 8 11. sin x = — ⅛ , cos x = — ⅛

√2

19

20

Chapter 1 Preliminaries 15.

13.

period = 2

period = π

period = 6 21.

23. period = j , symmetric about the origin

25. period = 4, symmetric about the y-axis

27. (a) Cos x and sec x are positive in QI and QIV and negative in QΠ and QIII. Sec x is undefined when cos x is 0. The range of sec x is (—∞ , —1] U [1, oo); the range of cos x is [—1 ,1].

Section 1.6 Trigonometric Functions (b) Sin x and esc x are positive in QI and QII and negative in QIΠ and QIV. Csc x is undefined when sin x is 0. The range of csc x is (—∞ , —1] U [1, oo); the range of sin x is [ - 1 , 1].

29. D: - ∞ < x < oo; R: y = —1,0, 1

31. cos (x — j ) = cos x cos (— j ) —sin x sin ( - ∣) = (cos x)(0) - (sin x)(—1) = sin x 33. sin (x + ∣) = sin x cos ( ∣) ÷ cos x sin Q ) = (sin x)(0) ÷ (cos x)(l) = cos x 35. cos (A —B) = cos (A + (—B)) = cos A cos (- B ) —sin A sin (—B) = cos A cos B —sin A (—sin B) = cos A cos B + sin A sin B 37. If B = A, A —B = 0 => cos (A - B) = cos 0 = 1. Also cos (A —B) = cos (A —A) = cos A cos A + sin A sin A = cos2 A ÷ sin2 A. Therefore, cos2 A ÷ sin2 A = 1. 39. cos (π ÷ x) = cos π cos x —sin π sin x = (—l)(cos x) —(0)(sin x) = —cos x 41. sin (⅞ “ x) = sin ( y ) c o s ( - x) + cos ( y ) s in (-x ) = (—l)(cos x) ÷ (0 )(sin(-x)) = —cos x 43. sin ⅛ = sin Q + f ) = sin ∣cos f + cos ∣sin f = ( ^ ) (∣) + ( ^ ) ( ^ ) = ⅛

^

45. cos ⅞ = cos ( f - 5) = cos f cos ( - ≈) - sin f sin ( - ≈) = ( ∣) ( ^ ) - ( ^ ) ( - ^ ) ≈ ⅛

2

1+cos (⅛ )

47. ∞s ∣π = -- = " ≠

49.

sin2

2+√2 4

1+

⅞= p∖ _ ×

4 •

×.

.

sin (A÷B) _ cos(A+B)

.

.

_

.

—

sin A cos B

∣cos A sin B

cos A cos B

cos A cos B

.

_

sin A cos B+cos A cos B _ CQSA CQSB ^I~COSA CQSB _ tan A÷tan B cos A cos B -sin A sin B cos A .cos B _ sιnAsinB 1 -ta n A ta n B

53. According to the figure in the text, we have the following: By the law of cosines, c2 = a2 ÷ b 2 —2ab cos θ = 12 + 12 —2 cos (A —B) = 2 —2 cos (A - B). By distance formula, c2 = (cos A —cos B)2 + (sin A —sin B)2 = cos2 A —2 cos A cos B + cos2 B + sin2 A —2 sin A sin B + sin2 B = 2 —2(cos A cos B + sin A sin B). Thus c2 = 2 —2 cos (A —B) = 2 —2(cos A cos B + sin A sin B) => cos (A —B) = cos A cos B + sin A sin B. 55. c 2 = a2 + b 2 - 2ab cos C = 22 + 32 - 2(2)(3) cos (60 o ) = 4 ÷ 9 - 12 cos (60 o ) = 13 - 12 ( ∣) = 7. Thus, c = ᅟ ∕7 ≈ 2.65.

21

22

Chapter 1 Preliminaries

57. From the figures in the text, we see that sin B = £. If C is an acute angle, then sin C = £. On the other hand, if C is obtuse (as in the figure on the right), then sin C = sin (π —C) = £. Thus, in either case, h = b sin C = c sin B => ah = ab sin C = ac sin B. By the law of cosines, cos C = a2⅛ c2 and cos B =

a2ψ c ⅛ 2 . 2

Moreover, since the sum of the

interior angles of a triangle is π, we have sin A = sin (π - (B + C)) = sin (B + C) = sin B cos C ÷ cos B sin C = (*) [ ¾ ^ ] + [ ⅛ ^ ] ( I ) = ( ⅛ ) (t f + b 2 - c 2 + c 2 - b 2 ) = ⅛ ^ ah = be sin A. Combining our results we have ah = ab sin C, ah = ac sin B, and ah = be sin A. Dividing by abc gives h^ be

sin A v

sin C c

a

sin B * b

law of sines 59. From the figure at the right and the law of cosines, b 2 = a2 ÷ 22 - 2(2a) cos B = a2 + 4 - 4a ( ∣) = a2 - 2a ÷ 4. Applying the law of sines to the figure, ^ =>

λ^

=

3^

=

=> b = ^ ∕ ∣a. Thus, combining results,

a2 - 2a + 4 = b 2 = ∣a2 => 0 = ∣a2 + 2a - 4 => 0 = a 2 ÷ 4a —8. From the quadratic formula and the fact that a > 0, we have

a =4 ±√^

4≡

=

^ ^

L464

61. A = 2, B = 2π, C = - π , D = - 1

63. A = - ∣, B = 4, C = 0, D = i

65. (a) amplitude = ∣ A ∣= 37 (c) right horizontal shift = C = 101

(b) period = ∣ B ∣= 365 (d) upward vertical shift = D = 25

1.7 GRAPHING WITH CALCULATORS AND COMPUTERS 1-3.

The most appropriate viewing window displays the maxima, minima, intercepts, and end behavior of the graphs and has little unused space.

Section 1.7 Graphing with Calculators and Computers 3. d.

1. d.

5-29.

For any display there are many appropriate display widows. The graphs given as answers in Exercises 5—30 are not unique in appearance.

5. [-2, 5] by [-15, 40]

7. [-2, 6] by [-250, 50]

9. [-4, 4] by [-5 , 5]

11. [ -2 ,6 ] b y [- 5 ,4 ]

13. [-1 ,6 ] by [-1 ,4 ]

15. [-3, 3] by [0,10]

23

24

Chapter 1 Preliminaries

17. [ - 5 ,1 ] by [-5 ,5 ]

19. [-4 , 4] by [0,3]

21. [-1 0 , 10] by [-6 , 6]

23. [-6 , 10] by [-6 , 6] 6X2 - 15x + 6 4 ? - 1C⅛

25. [-0.03, 0.03] by [-1.25, 1.25]

29. [-0.25, 0.25] by [-0.3 , 0.3]

27. [-300, 300] by [-1.25, 1.25]

Section 1.7 Graphing with Calculators and Computers

41. (a) y = 1059.14x - 2074972 (b) m = 1059.14 dollars/year, which is the yearly increase in compensation.

(d) Answers ≡ y vary slightly, y = (1059.14)(2010) - 2074972 = $53,899 43. (a) Let x represent the speed in miles per hour and d the stopping distance in feet. The quadratic regression function is d = 0.0866x2 - 1.97x + 50.1.

25

26

Chapter 1 Preliminaries

(c) From the graph in part (b), the stopping distance is about 370 feet when the vehicle is 72 mph and it is about 525 feet when the speed is 85 mph. Algebraically: dquadratic(72) = 0.0866(72)2 - 1.97(72) + 50.1 = 367.6 ft. dquadratic(85) = 0.0866(85)2 - 1.97(85) + 50.1 = 522.8 ft. (d) The linear regression function is d = 6.89x - 140.4 => diinear(72) = 6.89(72) - 140.4 = 355.7 ft and i∏ear(85) = 6.89(85) —140.4 — 445.2 ft. The linear regression line is shown on the graph in part (b). The quadratic d∣ regression curve clearly gives the better fit.

CHAPTER 1 PRACTICE EXERCISES

1. 7 + 2x ≥ 3 => 2x ≥ - 4 ≠- x ≥ - 2 3∙ ∣( x - l ) < ∣( x - 2 ) ^> 4 ( x - l ) < 5 ( x - 2 )

-1... 1 -1 -.1 ... 1 »

∂ I I I >x

0 1 2 3 4 5 6 7 8 9

= > 4 x -4 < 5 x - 10= > 6< x 5. ∣ x ÷ l ∣= 7 = > x + l = 7 o r - ( x + l) = 7=>x = 6 o rx = - 8 7. I* ~ 2 ∣> 1 ^ l ~ f < - ∣o r l - ∣> ∣= > - ∣< - ∣° r - ∣> ⅛ = > - x < - 5 o r - x > l Z Z I4 I X Z Z Z Z Z Z ≠- x > 5 o rx < - 1 9. Since the particle moved to the y-axis, - 2 + ∆x = 0 ≠> ∆x = 2. Since ∆ y = 3∆x = 6, the new coordinates are (x + ∆ x ,y + ∆y) = ( - 2 + 2,5 + 6) = (0,11). 11. The triangle ABC is neither an isosceles triangle nor is it a right triangle. The lengths of AB, BC and AC are ∕7 2 and ^/65, respectively. The slopes of AB, BC and AC are 5, —1 and ∣, respectively. χ∕S3, ᅟ 13. y = 3(x - 1) + (-6 ) ≠> y = 3x - 9 15. x = 0

Chapter 1 Practice Exercises

27

17. y = 2 19. y = -3 x + 3 21. Since 4x + 3y = 12 is equivalent to y = —∣x + 4, the slope of the given line (and hence the slope of the desired line) is - l.

y

= - ∣( x - 4 ) - 1 2 ^ y = - ∣x - f

23. Since ∣x + ∣y = 1 is equivalent to y = —∣x + 3, the slope of the given line is —∣and the slope of the perpendicular line > d ∙y = ^

+ i) +

2

^y = ^ + s

25. The area is A = π r 2 and the circumference is C = 2πr. Thus, r = ⅛ => A = π (^ -) 2 = ¢ . 27. The coordinates of a point on the parabola are (x, x2 ). The angle of inclination 0 joining this point to the origin satisfies the equation tan 0 = γ = x. Thus the point has coordinates (x, x2 ) = (tan θ, tan2 0).

33. y (-x ) = ( - x ) 2 + 1 = x2 + 1 = y(x). Even. 35. y (-x ) = 1 - cos(-x) = 1 —cos x = y(x). Even. 3’ . y(→ ) = ⅛ ⅛ ⅛

= ⅛

= - £

= -y W - θ "

39. y(—x) = -X + c o s (-x) = - x ÷ cosx. Neither even nor odd. 41. (a) The function is defined for all values of x, so the domain is ( - ∞ , ∞ ). (b) Since ∣ x∣ attains all nonnegative values, the range is [-2, ∞ ). 43. (a) Since the square root requires 16 - x2 ≥ 0, the domain is [-4, 4]. (b) For values of x in the domain, 0 ≤ 16 —x2 ≤ 16, so 0 ≤ ᅟ ∕1 6 - x2 ≤ 4. The range is [0, 4]. 45. (a) The function is defined for all values of x, so the domain is ( - ∞ , ∞ ). (b) Since 2e^x attains all positive values, the range is (-3 , ∞ ). 47. (a) The function is defined for all values of x, so the domain is ( - ∞ , ∞ ). (b) The sine function attains values from —1 to 1, so - 2 ≤ 2sin(3x + π) ≤ 2 and hence —3 ≤ 2sin(3x + π) —1 ≤ 1. The range is [—3, 1].

28

Chapter 1 Preliminaries

49. (a) The logarithm requires x —3 > 0, so the domain is (3, ∞ ). (b) The logarithm attains all real values, so the range is (—∞ , ∞ ). 51. (a) The function is defined for - 4 ≤ x ≤ 4, so the domain is [-4, 4]. (b) The function is equivalent to y = ι ∕[ x j , —4 ≤ x ≤ 4, which attains values from 0 to 2 for x in the domain. The range is [0, 2]. 53. First piece: Line through (0, 1) and (1, 0). m =

JΞ -

= -p = - l = > y = -

X+

1= 1-

X

Second piece: Line through (1, 1) and (2, 0). m = ∣ 5 ∣= ~ = - 1 => y = - ( χ - l ) + l = - x + 2 = 2 - x f(χ )= ∫1 - χ' θ ≤ x < 1 w ( 2 - X, l ≤ x ≤ 2 55. (a) (fo g )(-l) = f (g (-l)) = f ( 7 ⅛

) = fW = I =

(b) (gof)(2) = g(f(2)) = g (∣) = - ^ = = ^

1

or ^ / (

(c) (fof)(x) = f(f(x)) = f ( ∣) = τ ^ = x, x ≠ 0 (d) (g°g)W = g(gW ) = g ( - ⅛ ) ᅟ√x + 2 ∕

=

/ ÷ +2 √777j+2

=

7Γ⅛ ‰ √l+2√Γ+2

57. (a) (fog)(x) = f(g(x)) = f ( √ x + 2) = 2 - ( √ x + 2) 2 = - x , x ≥ -2 . (gof)(x) = f(g(x)) = g(2 - x2 ) = √ (2 - x2 ) + 2 = √ 4 - x (b) Domain of fog: [—2, ∞ ). Domain of gof: [—2, 2]. (c) Rangeoffog: ( - ∞ , 2]. Range of gof: [0, 2].

The graph of f2(x) = fι(∣ x∣ ) is the same as the graph of fι(x) to the right of the y-axis. The graph of f2(x) to the left of the y-axis is the reflection of y = fι(x), x ≥ 0 across the y-axis.

2

It does not change the graph.

Chapter 1 Practice Exercises

The graph of f2 (x) = f ι( ∣ x∣ ) is the same as the graph of f√x) to the right of the y-axis. The graph of f2(x) to the left of the y-axis is the reflection of y = fι(x), x ≥ 0 across the y-axis.

Whenever gι(x) is positive, the graph of y = g2 (x) = ∣ gι(x)∣ is the same as the graph of y = gι(x). When gι(x) is negative, the graph of y = g2 (x) is the reflection of the graph of y = gι(x) across the x-axis. Whenever gι(x) is positive, the graph of y = g2 (x) = ∣ gι(x)∣ is the same as the graph of y = gι(x). When gι(x) is negative, the graph of y = g2 (x) is the reflection of the graph of y = gι(x) across the x-axis.

67.

y - 4 ∙ x2

69.

period = π 73.

29

30

Chapter 1 Preliminaries

75. (a) sin B = sin ? = 7 = I ≠> b = 2 sin ? = 2 ( V I ≈ ᅟ ∕3 . By the theorem of Pythagoras, a2 + b 2 = c2 => a = V c ^ b ^ = ^ 4 - 3 = 1. (b) sinB = sin f = h = | ^

c= ⅛

= ^⅛ J = ⅛ ∙ Thus, a = √ c 2 - b

77. (a) tanB = ⅛ ^ a = ⅛

2

= ^ (^ /^ (2 )

2

= ^ = - ^ .

(b) sin A = 2 => c = ⅛

79. Let h = height of vertical pole, and let b and c denote the distances of points B and C from the base of the pole, measured along the flatground, respectively. Then, tan 50 o = £, tan 350 — ħ 5 a n d b —c = 10. Thus, h = c tan 50 o and h = b tan 35 o = (c ÷ 10) tan 35o => c tan 50o = (c ÷ 10) tan 35o => c (tan 50 o - tan 35o ) = 10 tan 35o => c = ⅛ ⅛ ^ h = ctan50 ° __ 10 tan 35 o tan 50 o tan 50 o - ta n 35 o ~

1 A QQ r n 1 U.VO ill.

81. (a)

y * sin x+cos ∣

(b) The period appears to be 4π. (c) f(x + 4π) = sin(x + 4π) + cos ( ^ ) = sin(x + 2π) + cos ( ∣+ 2π) = sin x + cos ∣ since the period of sine and cosine is 2π. Thus, f(x) has period 4π. CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES 1. (a) The given graph is reflected about the y-axis.

(c) The given graph is shifted left 1 unit, stretched vertically by a factor of 2, reflected about the x-axis, and then shifted upward 1 unit.

(b) The given graph is reflected about the x-axis.

(d) The given graph is shifted right 2 units, stretched vertically by a factor of 3, and then shifted downward 2 units.

Chapter 1 Addtional and Advanced Exercises 3. There are (infinitely) many such function pairs. For example, f(x) = 3x and g(x) = 4x satisfy f(g(x)) = f(4x) = 3(4x) = 12x = 4(3x) = g(3x) = g(f(x)). 5. If f is odd and defined at x, then f(-x ) = -fix). Thus g(-x ) = fi-x ) - 2 = -fix ) - 2 whereas -g(x) = -(fix) - 2) = -fix ) + 2. Then g cannot be odd because g(-x ) -- -g(x) =4∙ -fix ) - 2 = -fix ) + 2 =4 4 = 0, which is a contradiction. Also, g(x) is not even unless fix) = 0 for all x. On the other hand, if f is even, then g(x) = fix) - 2 is also even: g (-x ) = fi-x ) - 2 = fix) - 2 = g(x). 7. For (x, y) in the 1st quadrant, ∣ x∣ + ∣ y∣= 1 + χ 4» x + y = 1 + x » y = 1∙ For (x, y) in the 2nd quadrant, ∣ x∣ + ∣ y∣= x + 1 44 - x + y = x + 1 44 y = 2x + 1. In the 3rd quadrant, ∣ x∣ + ∣ y∣= χ + 1 cos A = ^ ⅛ - ^ = 13. By the law of cosines, b2 = a2 + c2 —2ac cos B => cos B = a - ^ c~= ∣ ∣ . Since 0 < B < π, sin B =   Λ ~ cos2 B = ^ 1 - ^ = ^

_ “ _ ^

= 1

22 + 4 2 - 3 2 _ (2)(2)(4) ~ 3√15 16

4 + 1 6 -9 16

,

15. (a) sin2 x ÷ cos2 x = 1 ≠> sin2 x = 1 —cos2 x = (1 - cos x)(l ÷ cos x) => (1 —cos x) = κ

1—cos x _ sin x

Λ 1 2 COS*X

sin x 1 + co sx

(b) Using the definition of the tangent function and the double angle formulas, we have ∙2∕x∖

twa πn 2 1( 22 )/ -

⅛ cos 2 ⅛ ⅛

-

1 -c o s (2

⅛))

ι+ ∞ √ 2

(j⅜))

≡4^∙ - 1~cosx

1 + co sx ∙

2

17. As in the proof of the law of sines of Section P.5, Exercise 57, ah = be sin A = ab sin C = ac sin B => the area of ABC = ∣(base)(height) = j ah = j be sin A = ∣ab sin C = ∣ac sin B. 19. 1. 2. 3. 4. 5. 6.

b ÷ c —(a + c) = b —a, which is positive since a < b. Thus, a ÷ c < b + c. b —c —(a —c) = b —a, which is positive since a < b. Thus, a —c < b —c. c > 0 and a < b => c —0 = c and b —a are positive => (b —a)c = be —acis positive => ac < be. a < b and c < 0 => b —a and —c are positive => (b —a )( - c) = ac —be is positive =^ be < ac. Since a > 0, a and ∣are positive => ∣> 0. Since 0 < a < b, both ∣and ∣are positive. By (3), a < b and ∣> 0 => a (∣) < b ( ∣) or 1 < ^

1 (5) < ι (⅛ ) bysince b 0

7.

⅛< b

a < b < 0 = > ∣and ∣are both negative, i.e., ^ < 0 and ∣< 0. By (4), a < b and ∣< 0 => M D

≠ I f(x) = 0. 25. y = ax 2 + bx + c = a (x 2 + ⅛x + ⅛ ) - g + c = a (x + ⅛ ) 2 - g + c (a) If a > 0 the graph is a parabola that opens upward. Increasing a causes a vertical stretching and a shift of the vertex toward the y-axis and upward. If a < 0 the graph is a parabola that opens downward. Decreasing a causes a vertical stretching and a shift of the vertex toward the y-axis and downward. (b) If a > 0 the graph is a parabola that opens upward. If also b > 0, then increasing b causes a shift of the graph downward to the left; if b < 0, then decreasing b causes a shift of the graph downward and to the right. If a < 0 the graph is a parabola that opens downward. If b > 0, increasing b shifts the graph upward to the right. If b < 0, decreasing b shifts the graph upward to the left. (c) Changing c (for fixed a and b) by ∆ c shifts the graph upward ∆ c units if ∆ c > 0, and downward —∆ c units if ∆ c < 0. 27. If m > 0, the x-intercept of y = mx + 2 must be negative. If m < 0, then the x-intercept exceeds ∣ => 0 = mx + 2 and x > ⅜ => x = —⅛ > ⅛ => 0 > m > —4. 29. (a) By Exercise #95 of Section 1.2, the coordinates of P are ( ^ , ^ ) = (∣, ∣) ∙ Thus the slope oo lf OP — Δ∑ — M2 — £ ~ ∆x — a/2 “ a * (b) The slope of AB = = - ^ . The line segments AB and OP are perpendicular when the product of their slopes is —1 = ( ∣) (—∣) = - ^ . Thus, b2 = a2 => a = b (since both are positive). Therefore, AB is perpendicular to OP when a = b.

CHAPTER 2 LIMITS AND CONTINUITY 2.1 RATES OF CHANGE AND LIMITS 1. (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approaches 1 from the left, g(x) approaches 1. There is no single number L that all the values g(x) get arbitrarily close to as x → 1. (b) 1 (c) 0 3. (a) True (d) False 5.

χ limθ

(b) True (e) False

(c) False (f) True

p does not exist because ^ = ^ = 1 if χ > 0 and ^ = ^ = - 1 if x < 0 . As x approaches 0 from the left,

i^i approaches - 1 . As x approaches 0 from the right, ^ approaches 1. There is no single number L that all the function values get arbitrarily close to as x → 0. 7. Nothing can be said about f(x) because the existence of a limit as x → xθ does not depend on how the function is defined at x0 . In order for a limit to exist, f(x) must be arbitrarily close to a single real number L when x is close enough to XQ- That is, the existence of a limit depends on the values of f(x) for x near x0 , not on the definition of f(x) at xo itself. 9. No, the definition does not require that f be defined at x = 1 in order for a limiting value to exist there. If f(l) is defined, it can be any real number, so we can conclude nothing about f(l) from lim f(x) = 5. 11. (a) f(x) = (x2 - 9)∕(x + 3) X -3.1 -6.1 f(x) X

-2 .9 -5 .9

-3.01 -6.01

-3.001 -6.001

-3.0001 -6.0001

-3.00001 -6.00001

—3.000∞l -6.000001

-2.99 -5.99

-2.999 -5.999

-2.9999 -5.9999

-2.99999 -5.99999

-2.999999 -5.999999

-5.999 -.1250156

-5.9999 -.1250015

-5.99999 -.1250001

f(x) The estimate is lim f(x) = - 6 . x —►—3

13. (a) G(x) = (x + 6)/ (x2 + 4x - 12) x -5 .9 -5.99 -.126582 -.1251564 GW

-5.999999 -.12500∞

34

Chapter 2 Limits and Continuity X G(x)

-6 .1 -.123456

(c) θ(x) — (x 2+ ^

t

12)

— (x

-6.01 -.124843

+

6χχ6 - 2 )

-

-6.001 -.124984

x -2 *^x

-6.0001 -.124998

6, a n d

χ

li∏^6

- 6 .0 0 ∞ l -.124999

x

_2

-6 -2

-6.000001 -.124999

8

0.125.

15. (a) f(x) = (x2 - l ) ∕ ( ∣ x[ - 1) X f(χ)

-1 .1 2.1

-1.01 2.01

-1.001 2.001

- 1 .0 ∞ l 2.0001

—1.00∞ l 2.00001

-1.000001 2.000001

X f(x)

-.9 1.9

-.9 9 1.99

-.9 9 9 1.999

-.9999 1.9999

-.99999 1.99999

-.999999 1.999999

f ^ ⅛ - - = x + 1, x ≥ 0 and x ≠ 1 2_ . (c) f(x) = ⅛rzγ = < - 2 )

(x + 3) = (-2 ) + 3 = 1 + 3)

liιn

[ ^ ]

(∣ x + 2∣= - ( x + 2) for x < - 2 )

(x + 3 ) ( - 1) = - ( - 2 + 3) = - 1

x → -2

19. (v a)7 0 lim ψ = ∣ = 1 → 3+ « 3 21. lim

= lim ⅛x

0 →O √ 2 0

23∙

y¾

25∙

1¾

^

= i

y

= 1

1≡ o ^

lim

I± l≡ s χ

s in χ c

’ χ → 0

χ → 0' ⅛

4A

(where0 = 3y)

^⅛ = ( A ^ ) ( A ⅛25) =1∙2=2

⅛'

-- lim s (

θ

=

^

=A

s= A

^

(where xV = v∕20)

= ⅛

=A

27∙ ⅛ 29

x →0

(b) ψ = 23 v 7 0 lim → 3- 0

=

=0 A ⅛ ( A ⅛ 5

x c o sx

I- -2 ∞ 3 -) s in x c o sx ∕

∏

χ → m0

(⅛∙1)(1) = I

( x- * 1-) ,+ lim —

' s *n

c o sx 7

χ

→ θ

sinx

= lim ( ⅛ ) ∙ lim ( ^ ) + lim ( ⅛ ) = (1)(1)+1 = 2 x →0 ∖ s⅛

31. lim t

-/

x →0

χ →0 \ — /

⅛ ^ = lim ⅛θ ^ = 1 since 0 = 1 - cos t → 0 as t → 0

> -∞ t

→o

x cosx z

0 →O

33∙ 0>। θ

⅛⅛=Jim (≡ ∙i) =I > (τ ∙⅛) =I ∙1∙ 1=5

35

⅛

x

sin

lim → 0 =

3 8

0

s in 8 x

x

_

θ ∖ sin

/

i™ ∕s i n 3 x . J _ \ _ x → 0 ' c °s3 x sin 8x7

1im f _ A _ Λ p m 3 x ∖ / . 8x A = → 0 ' cos3x ' V 3x 7 ∖ sin8x7

'

E x

→

sin

/

/ sin 3x . 1 0 ∞ s 3x sin8x

3 . ι . ι . ι 1 1 1 8

=

2

. 8x . 3 λ 3x 87

3 8

Note: In these exercises we use the result

lim ⅛ = 0 whenever 7n > 0. This result follows immediately from X→ ± ∞ x , = 0m/n = 0. Example 6 and the power rule in Theorem 8: lim = lim = f lim x → ± ∞ 'x

z

X→ ± ∞

37. (a) - 3

(b) - 3

39. (a) i

(b) i

41. (a) - ∣

(b) - ∣

43. —X ∣≤ 5vA 45.

lim

t —►00

47. (v a)7

5

≤ Λ ∣ => X —lim > OO

¾ ⅛

"*^cθs t

lim

i⅛

x→∞ 5x +7

'x'

∖ x→ ± ∞

⅛Λ ^ = 0 by the Sandwich Theorem

lim t →∞ =

lim

x→∞

→ = i5 5+ f

(b) ∣(same process as part (a))

x

∕

43

44

Chapter 2 Limits and Continuity

49. (a) v

7

51∙ (a)

lim

⅛

=

x → ∞

Xz + 3

⅝

⅛

x

lim

⅛

x → ∞

=

x

= 0

⅝

Γ ⅛

53. (va)z X→∞ lim l0x +X J +31X→∞ = lim 1⅛ (b) 0 (same process as part (a)) 5 5

∙

(a )

x ⅛

i⅛

⅛

⅛

=

(b) 0 (same process as part (a))

1+ 4

x⅛‰

⅛

= 7

^

(b) 7 (same process as part (a))

= θ

⅛

-

=

5

(b) —∣ (same process as part (a))

2√x + x 3 x -7

57. Xlim →∞

1

x

lim—∞ →

l - x (l∕5)-(1/3) _ l + χ(l∕5)-d∕3) “

χ

≡

0 0

2X5 /3 - x 1/3 + 7 x8 ∕ 5 + 3x +

61. Xlim →∞

63. Yes. If lim f(x) = L = x

a

+x → a

lim _ f(x), then lim f(x) =+ L. If lim f(x) ≠ x → a

χ → a

x → a

lim _ f(x), then lim f(x) does not exist, x → a

65. If f is an odd function of x, then f( -x ) = -f(x ). Given lim + f(x) = 3, then lim f(x) = - 3 . χ →0 X→ o 67. Yes. If lim

x → ∞

® = 2 then the ratio of the polynomials’ leading coefficients is 2, so g(x)

r

√

o

’

X

→ —∞

g(x)

lim

® = 2 as well,

69. At most 1 horizontal asymptote: If χ lim o ® = L, then the ratio of the polynomials’ leading coefficients is L, so 1™

X→ - ∞

⅛ = L as well.

g(x)

71. For any e > 0, take N = 1. Then for all x > N we have that ∣ f(x) —k ∣= ∣ k —k ∣= 0 < e. 73. I = (5,5 + ¢) ≠> 5 < x < 5 + ¢. Also, √ x - 5 < e =^ x - 5 < e2 => x < 5 + e2 . Choose 6 = e2 =>

lim

√ x —5 = 0.

x → 5÷

75. As x → 0 ' the number x is always negative. Thus, 11*∣—( - 1)∣< 6 => ∣ ⅛ + 11 < e => 0 < e which is always true independent of the value of x. Hence we can choose any 6 > 0 with - 6 < x < 0 => 77. (a)

lim

x → 400+

liπι

⅛

=

^1 ∙

Ixl = 400. Just observe that if 400 < x < 401, then [xj = 400. Thus if we choose 0 that 400 < x < 400 ÷ (t = I)

→ n0

2 -5 t

2 ’

∖

x7

= ∣ ) (3 + £) (cos ∣) = lim (3 + 20)(cos 0) = (3)(1) = 3, (0 v v x /

lim

x→ ± ∞

v

χ∕

0 → O

2.5 INFINITE LIMITS AND VERTICAL ASYMPTOTES positive positive

1.

lim ⅛ = ∞ x →0+ 3x

3.

lim -⅛ = - ∞ x →2^ x ~2

5

∙

x

7

∙

⅛

⅛

÷

⅜

= - ∞

(i⅛

9. (a)

= ∞

lim

5⅛

11. lim x →0 13.

15.

x

= ∞

(b)

⅛

lim x →0

⅛

lim. x →0

-

∞

= ∞

lim tan x = ∞ → (f)^

lim (1 + esc 0) = —∞ 0 → 0-

17. (a)

lim Λ 7 ' x →2÷ x

(b)

lim -7≡ U χ → 2- x

(c)

lim x → -2 +

x2 - 4

-

(d)

lim x → -2 “

x2 - 4

-

⅛

=

1 positive-positive

x

1≡

2-

⅛ ⅛

2

x

→m0 ÷

=

χ

Hm+ ⅛ B ⅞

=

x

!≡ 2 - ⅜ ⅛ ⅛ Γ =

⅛ ≡ 2 Γ

1

h

=

negative-negative positive-negative

- ∞

= ⅛ x

1i m 2

∣- x ≠

2

x

2

⅛

=

⅛

= b

≠

45

46

Chapter 2 Limits and Continuity

(d>⅛ ⅛⅛ 2 =⅛ ⅛ i r =⅛ ⅛x w ⅛ ⅛⅛ 2 = A ⅛u⅛n = ~∞

z

= ∑4 ’. x' ≠ 2 negative-negative

(

23. (a) ^ Um+ [ 2 - ⅛ ] = - ∞

(b)

25∙ ^

(b)

χ¾+ [⅛ + (Γ¼] =0 ° ^ x!⅛∙ [⅛+ ⅛⅛^]=0°

31.

35. y = ⅛ = x + l - ⅛

(d)

t

positive-negative

lim_ [ 2 - ⅛ ] = ∞

Section 2.5 Infinite Limits and Vertical Asymptotes 39. Here is one possibility.

41. Here is one possibility.

43. Here is one possibility.

45. Here is one possibility.

47

y

A(χ)≡⅛χ≠0 14

o

47. For every real number —B < 0, we must find a £ > 0 such that for all x, 0 < ∣ x —0∣< 6 ≠ - ^ < -B < 0 o

⅛ > B > 0 Φ>

X2

-^ < —B. Now,

< ∣ l < ^ ∙ Choose ¢ = ^ , then 0 < ∣ x∣< W < ^ j lx

=> ⅛ < —B so that lim —x⅛ = —∞ . x x→0

49. For every real number —B < 0, we must find a 0 such that for all x, 0 < ∣ x —3∣< δ => ⅛Z3)⅞ < ~B∙ Now, o ⅛

< —B < 0 O ( Γ ⅛ > B > 0 Φ> ^

< ⅛ ^

(x - 3)2 < ∣ 0

δ = ^ ∕∣, then 0 < ∣ x —31 < 5 => (^Ξ⅜∑ < - B < 0 so that χ lim3

2 (x ~ 3)2

0< ∣ x - 3∣< ^ j . Choose

= -∞ ∙

51. (a) We say that f(x) approaches infinity as x approaches xo from the left, and write χ lim_ f(x) = ∞ , if ⅞ for every positive number B, there exists a corresponding number 6 > 0 such that for all x, X0 - ⅛ < X < Xo => f(x) > B. (b) We say that f(x) approaches minus infinity as x approaches xo from the right, and write lim f(x) = —∞ , x →x o if for every positive number B (or negative number —B) there exists a corresponding number £ > 0 such that for all x, XQ < x < xo + ⅛ => f(x) < -B . (c) We say that f(x) approaches minus infinity as x approaches xo from the left, and write χ lim_ f(x) = —∞ , χo if for every positive number B (or negative number —B) there exists a corresponding number £ > 0 such that for all x, xo —(5 < x < xo 4 f(x) < -B . 53. For B > 0, ∣ —B < 0 O —7 > B D > 0 Φ> D —x < ⅛ 0 A< A ≠ - ∣< x = > 7X < —B so that lim ∣ = —∞ . B X → 0^ X 55. ForB > 0,

—⅜ < xO. Choose ^ = ⅛. Then — B Φ> 0 < x - 2 < ⅜. Choose 6 = 1. Then 2 < x < 2 + 6 ≠> 0 < x - 2 < 6 ≠ 0 < x - 2 < ⅜

=► τ ¼ > B > 0 so that lim

⅛

= ∞.

48

Chapter 2 Limits and Continuity 59. y —tan x + ^

63. y = χ 273 + ⅛

61.

2.6 CONTINUITY 1. No, discontinuous at x = 2, not defined at x = 2 3. Continuous on [ - 1, 3] 5.

7.

(a) Yes

(b) Yes,

(c) Yes

(d) Yes

(a) No

(b) No

9. f(2) = 0, since lim f(x) = -2(2) ÷4 = 0 = + x→2

x →2

lim

x → -1 +

f(x) = 0

lim f(x)

11. Nonremovable discontinuity at x = 1 because lim f(x) fails to exist ( lim f(x) = 1 and lim f(x) = 0). x→1

x→1

x

→ 1+

Removable discontinuity at x = 0 by assigning the number χ limθ f(x) = 0 to be the value of f(0) rather than f(0) = 1. 13. Discontinuous only when x —2 = 0 => x = 2 15. Discontinuous only when x2 —4x + 3 = 0 =Φ (x —3)(x - 1 ) = 0 => x = 3 o r x = l 17. Continuous everywhere. ( ∣ x — 11+ sin x defined for all x; limits exist and are equal to function values.) 19. Discontinuous only at x = 0

Section 2.6 Continuity 21. Discontinuous when 2x is an integer multiple of π, i.e., 2x = nπ, n an integer => x = γ , n an integer, but continuous at all other x. 23. Discontinuous at odd integer multiples of j , i.e., x = (2n - 1) ^, n an integer, but continuous at all other x. 25. Discontinuous when 2x + 3 < 0 o r x < - ∣ => continuous on the interval

H-∞)∙

27. Continuous everywhere: (2x — 1)1∕ 3 is defined for all x; limits exist and are equal to function values. 29. χ lim r sin (x - sin x) = sin (π - sin π) = sin (π - 0) = sin π = 0, and function continuous at x = π. 31. limι sec (y sec2 y - tan2 y — 1) = lim^ sec (y sec2 y - sec2 y) = limι sec ((y — 1) sec2 y) = sec ((1 —1) sec2 1) = sec 0 = 1,, and function continuous at y = 1. 3 3

∙

t¾

c

°

s

[ √ i 9 - ⅛

= s

2 -

cos

[√ i9 ⅛ o ] =

cos

^

=

cos

I = ^

a n d fu n c tio n

continuous at t = 0.

= δ ⅛ ¾ ^ = x + 3 , x ≠ 3 ≠> g(3) = χ lim3 (x + 3) = 6

35. g(x) = ⅛ 37

] =

= ⅛ i± i± I lħ z 2 2 — ⅛ ÷A ±1 1

(s

+

i)(s - i )

s +

ι

» »

1 → →

s

→

( ⅛ ÷ s ÷ jΛ

1

∖

S+ 1

)

3

2

39. As defined, lim f(x) = (3)2 —+1 = 8 and lim (2a)(3) = 6a. For f(x) to be continuous we must have x →3 x →3“ 6a = 8 => a = ∣. 41. The function can be extended: f(0) ≈ 2.3.

45. f(x) is continuous on [0,1] and f(0) < 0, f(l) > 0 => by the Intermediate Value Theorem f(x) takes on every value between f(0) and f(l) => the equation f(x) = 0 has at least one solution between x = 0 and x = 1.

43. The function cannot be extended to be continuous at x = 0. If f(0) = 1, it will be continuous from the right. Or if f(0) = —1, it will be continuous from the left.

49

50

Chapter 2 Limits and Continuity

47. Let f(x) = x3 - 15x + 1 which is continuous on [-4,4]. Then f(-4 ) = - 3 , f ( - l ) = 15, f(l) = -1 3 , and f(4) = 5. By the Intermediate Value Theorem, f(x) = 0 for some x in each of the intervals —4 < x < - 1 ,—l < x < 1, and 1 < x < 4. That is, x3 — 15x + 1 = 0 has three solutions in [ - 4 ,4]. Since a polynomial of degree 3 can have at most 3 solutions, these are the only solutions. 49. Answers may vary. Note that f is continuous for every value of x. (a) f(0) = 10, f(l) = 13 —8(1) ÷ 10 = 3. Since 3 < π < 10, by the Intermediate Value Theorem, there exists a c so that 0 < c < 1 and f(c) = π. (b) f(0) = 10, f(-4 ) = (-4 ) 3 - 8(-4) ÷ 10 = -22. Since -2 2 < - √ 3 < 10, by the Intermediate Value Theorem, there exists a c so that —4 < c < 0 and f(c) = - √^3. (c) f(0) = 10, f(1000) = (1000)3 - 8(1000) + 10 = 999,992,010. Since 10 < 5,000,000 < 999,992,010, by the Intermediate Value Theorem, there exists a c so that 0 < c < 1000 and f(c) = 5,000,000. 51. Answers may vary. For example, f(x) = η

^

is discontinuous at x = 2 because it is not defined there.

However, the discontinuity can be removed because f has a limit (namely 1) as x → 2. 53. (a) Suppose XQ is rational => f(xo) = 1. Choose e = ∣. For any 5 > 0 there is an irrational number x (actually f(x) —f(xo)∣ infinitely many) in the interval (XQ — f(x) = 0. Then 0 < ∣ x —xo∣< ⅛but ∣ = €, so Xlim f(x) fails to exist => f is discontinuous at x0 rational. = 1> ∣ * → Xo On the other hand, xo irrational => f(x0 ) = 0 and there is a rational number x in (xo — f(x) = 1. Again χ lknθ f(x) fails to exist => f is discontinuous at x0 irrational. That is, f is discontinuous at every point. (b) f is neither right-continuous nor left-continuous at any point XQ because in every interval (xθ —5, Xo) or (xθ, xo ÷ ¢) there exist both rational and irrational real numbers. Thus neither limits χ lim_ f(x) and χ o

lim f(x) exist by the same arguments used in part (a).

55. No. For instance, if f(x) = 0, g(x) = M , then h(x) = 0 ( [xl) = 0 is continuous at x = 0 and g(x) is not. 57. Yes, because of the Intermediate Value Theorem. If f(a) and f(b) did have different signs then f would have to equal zero at some point between a and b since f is continuous on [a, b]. 59. If f(0) = 0 or f(l) = 1, we are done (i.e., c = 0 or c = 1 in those cases). Then let f(0) = a > 0 and f(l) = b < 1 because 0 ≤ f(x) ≤ 1. Define g(x) = f(x) - x => g is continuous on [0 ,1]. Moreover, g(0) = f(0) —0 = a > 0 and g(l) = f(l) _ ι = b - 1 < 0 => by the Intermediate Value Theorem there is a number c in (0,1) such that g(c) = 0 => f(c) - c = 0 or f(c) = c. 61. By Exercises 52 in Section 2.3, we have χ limc f(x) = L hlimθ f(c ÷ h) = L. Thus, f(x) is continuous at x = c o 63. x ≈ 1.8794, -1.5321, -0.3473 65. x ≈ 1.7549 67. x ≈ 3.5156 69. x ≈ 0.7391

lim f(x) = f(c) o

x →c

lim f(c ÷ h) = f(c).

h →0

Section 2.7 Tangents and Derivatives 2.7 TANGENTS AND DERIVATIVES

1. Pi: mi = 1, P2: m2 = 5 3. P f mi = ∣, P2 : m2 = - ∣ 5. m = ∏m t4 - ( - l + h ) V ( 4 - ( - ⅛ h →0 h ιi m ^ 1 →∏±⅛±> = u m ⅛ ⅛ h )= 2 i = h h h →0 h →0 a t( - l,3 ) : y = 3 + 2 ( x - ( - l ) ) => y = 2x + 5, tangent line

7.

m =

um

2

√ ∏ h -2 √ r

=

ιi m

2 ∙ ⅛ 2 √ l+ h + 2

⅛

h h →0 h h →0 = lim _ ^ i ± 5 k ^ - — iim h →0 2 h ( √ l + h + l ) h →0

,

2

≡

—_ — i;

√ l+ h + l

at (1,2): y = 2 + l(x —1) => y = x + 1, tangent line

9.

(≤±J⅛ → J>! = l i m lim h h →0 h →0 = hlimθ (12 - 6h + h2 ) = 12;

m =

- 8 . + 1 2 h -6 h * ÷ h °

h

+

8

at (-2 , -8): y = - 8 + 12(x - (-2 )) ≠> y = 12x + 16, tangent line

11.

∏m R2i < ÷ l i - 5 = l i m (5 + 4h + h ^ 5 h h →0 h h →0 h →0 at (2,5): y - 5 = 4(x - 2), tangent line

m =

m = lim 2

h→0

s

¾ - l± ⅛

2 i l ⅛ j —h

=

5J

jjm

=

h

h→0

_

a + l 1 .2)

lim

5

∏m 5 h (2 + h ) _ h h→0

2 h f has a root between —1 and 2 by the Intermediate Value Theorem, (b), (c) root is 1.32471795724 CHAPTER 2 ADDITIONAL AND ADVANCED EXERCISES 1. (a) _X________ 0.1 Xx 0.7943

0.01 0.9550

0.001 0.9931

0.0001 0.9991

0.00001 0.9999

Apparently, lim + xx = 1 x →0

(b)

3.

v

lim_ L =

v

lirn. Lo √ K ⅛ = Lo √ K ⅞ ^ = Lo √ K ∣ = 0

The left-hand limit was needed because the function L is undefined if v > c (the rocket cannot move faster than the speed of light). 5. ∣ (t - 70) × 10^4 ∣< 0.0005 => -0.0005 < (t - 70) × 10^4 < 0.0005 10 ÷ (t - 70) × 10"4 - 10∣< 0.0005 => ∣ => - 5 < t - 7 0 < 5 => 65o < t < 7 5 o => Within5o F. 7. Show lim f(x) = lim (x2 —7) = —6 = f(l). x→1

Step 1:

x→1

∣ (x2 —7) + 6∣< e => —e < x2 —1 < e => l - e < x

2

< l + e => ^ 1 - € < x < ^ l + e.

Step 2: ∣ x - l ∣< i => — 1 - e < √ 2 x - 3 < 1 + e ≠> O ^ ± 3
L . Let e = 5 (L2 —Li). f(x) —L ι∣< ε =≠>- ε < f(x) - L 1< ε Since χ lirnθ f(x) = Li there is a ¢1 > 0 such that 0 < ∣ x - XQ∣< 61 => ∣ => - I (L2 - Li) + L 1 < f(x) < ∣(L2 - Li) + Li => 4Lι - L2 < 3f(x) < 2Lι + L 2 . Likewise,

x ⅛∏o f(x)=

L2

f(x) - L2 ∣< ε => - ε < f(x) - L2 < ε so there is a 62 such that 0 < ∣ x - XQ∣< ⅛ ≠> ∣ =► - I (L2 - Li) + L2 < f(x) < I (L2 - Li) + L2 => 2L2 + L i < 3f(x) < 4L2 - L 1 x —XQ∣< ¢: ≠> Li - 4L2 < -3f(x) < -2 L 2 —Li. If 5 = min { ⅛ ⅛} both inequalities must hold for 0 < ∣ 4L1 - L2 < 3f(x) < 2L1 + L2 1 L 1 - 4L2 < -3f(x) < -2 L 2 - L I ∫ => 5(L1 - L2 ) < 0 < Li —L2 . That is, Li —L2 < 0 and Li —L2 > 0, a contradiction. 13. (a) Since x (b) Since x

0 + , 0 < x3 < x < 1 => (x3 —x) → 0

=>

0 ^ , - l < x < x 3 < 0 => (x3 - x) → 0+ =>

(c) Since x → 0+ , 0 < x4 < x2 < 1 => (x2 —x4 ) → 0+ =>

lim + f (x3 —x) = lim f (x3 —x) =

x→-l

lim

x→ -l

⅛^ =

lim

ι

x → —1

lim + f(y) = A where y = x3 —x.

lim f (x2 —x 4 ) =

x → 0+

2 (d) Since x → 0^, —1 < x < 0 => 0 < x4 < x < 1 => (x2 - x4 ) → 0 + =>

15. Show lim f(x) =

lim f(y) = B where y = x3 —x.

lim f(y) = A where y = x2 —x4 .

y → 0+

lim f (x2 —x4 ) = A as in part (c).

x → 0+

¾ ⅛ ^ = - 2 ,x ≠ - l.

( χ2~ 1 x √ - 1 . We now prove the limit of f(x) as x → - 1 Define the continuous extension of f(x) as F(x) = < x + 1 ’ r I -2 , x = -1 exists and has the correct value. Step 1: ∣ ⅛ - ( - 2 ) ∣< ε ^ - ε < ⅛ ⅛ 3 > + 2 < ε => - ε < (x - 1) + 2 < ε, x ≠ - 1 ≠> - ε - l < x < ε - l . Step 2: ∣ x - ( - l ) ∣ - ^ < x ÷ l < ⅛ => - 6 - 1 < x < 5 - 1. Then —6 —1 = —e —1 => δ = e , o r δ - l = e - 1 => £= c Choose 6 = e. Then 0 < ∣ x —(—1)∣< ⅛ ~ - (—2) < e =>lim F(x) = —2. Since the conditions of the continuity test are met by F(x), I

∣

x → -1

then f(x) has a

continuous extension to F(x) at x = —1. x —0∣< e; i.e., choose 17. (a) Let e > 0 be given. If x is rational, then f(x) = x => ∣ f(x) —0 ∣= ∣ x —0∣< e o ∣ δ = e. Then ∣ x - 0∣< 6 => ∣ f(x) —0∣< e for x rational. If x is irrational, then f(x) = 0 => ∣ f(x) —0∣< e O 0 < e which is true no matter how close irrational x is to 0, so again we can choose £ = e. In either case, x - 0∣< ⅛ => ∣ given e > 0 there is a 6 = e > 0 such that 0 < ∣ f(x) —0∣< e. Therefore, f is continuous at x = 0. (b) Choose x = c > 0. Then within any interval (c —6, c + 6) there are both rational and irrational numbers. If c is rational, pick e — ∣. No matter how small we choose £ > 0 there is an irrational number x in (c - 6, c ÷ ⅛) => ∣ f(x) —f(c)∣= ∣ 0 - c∣= c > ∣= €. That is, f is not continuous at any rational c > 0. On the other hand, suppose c is irrational => f(c) = 0. Again pick e = ∣. No matter how small we choose 6 > 0 there is a rational number x in (c —δ^ c + ¢) with ∣ x —c∣< ∣= e o = ∣ x∣> ∣= € => f is not continuous at any irrational c > 0.

∣< x < y . Then ∣ f(x) —f(c)∣= ∣ x —0∣

If x = c < 0, repeat the argument picking e = ^1 = ^ . Therefore f fails to be continuous at any nonzero value x = c.

Chapter 2 Limits and Continuity

58

19. Yes. Let R be the radius of the equator (earth) and suppose at a fixed instant of time we label noon as the zero point, 0, on the equator => 0 ÷ πR represents the midnight point (at the same exact time). Suppose Xi is a point on the equator “just after” noon => x1 + πR is simultaneously “just after" midnight. It seems reasonable that the temperature T at a point just after noon is hotter than it would be at the diametrically opposite point just after midnight: That is, T(xι) - T(xι + πR) > 0. At exactly the same moment in time pick X2 to be a point just before midnight => x2 + πR is just before noon. Then T(x2 ) —T(x2 + πR) < 0. Assuming the temperature function T is continuous along the equator (which is reasonable), the Intermediate Value Theorem says there is a point c between 0 (noon) and πR (simultaneously midnight) such that T(c) —T(c + πR) = 0; i.e., there is always a pair of antipodal points on the earth's equator where the temperatures are the same. 1 ⅛∙V^

r (a) = lim 21. (a) At x = 0: alim → 0 ++ a→0 -

a

= lim

a→0 _ lim 1-(1 + a) -1 _ 1 2 -l-√ l+ 0 a → 0 a (-l-√ l+ a )

At x = -1 : (b) At x = 0:

lim

a → -l+

= lim r_ (a) v 7

a → 0^

lim

= a

r + (a) =

- L 4 ⅛ L -=

lim

a → - l + a7( - l - √ l + a )

lim

-A r^ ^ ¼ =

→ 0^ a ( - l + √ l+ a )

a

=

a

a → 0^

a

lim

-

lim

— 4 — = ∞v (because the

→-l

a

li

a →m0^

lim , —⅜ — → 0~ a ( - l + √ l + a) v

= a

denominator is always negative); lim + r_(a) =

→ 0^ - l + √ l + a

lim +

, 1+~J1'+ a = —∞ (because the denominator

is always positive). Therefore, lim r_ (a) does not exist, a→0

At x = -1 :

lim

r_(a) =

a→-l+

lim a→-l+

(c)

f(x )

——^ - ^ = a

lim a→-l+

—— U — = 1 - l + √ l+ a

Chapter 2 Practice Exercises 23. (a) The function f is bounded on D if f(x) ≥ M and f(x) ≤ N for all x in D. This means M ≤ f(x) ≤ N for all x in D. Choose B to be max {∣ ,∣ N∣ } . Then ∣ f(x)∣≤ B. On the other hand, if ∣ f(x)∣≤ B, then M∣ - B < f(x) ≤ B => f(x) > - B and f(x) ≤ B => f(x) is bounded on D with N = B an upper bound and M = —B a lower bound. (b) Assume f(x) ≤ N for all x and that L > N. Let e = t ^ . Since χ limθ f(x) = L there is a 6 > 0 such that 0< ∣ x - x0 ∣< ⅛ =► ∣ f(x) - L ∣< € Φ> L - e < f(x) < L + e 4> L - ⅛ ^ < f(x) < L + ⅛ * O ^ ^ < f(x)
N ≠- ^ ^ > N => N < f(x) contrary to the boundedness assumption

f(x) ≤ N. This contradiction proves L ≤ N. (c) Assume M < f(x) for all x and that L < M. Let e = ^⅛^∙ As in part (b), 0 < ∣ x —XQ ∣< 6 => L - ^ ^ < f(x) < L + ^ l ^

2

u— _ x →o “

s in ( l-c o s x ) _ 1 ∙ s in ( l-c o s x ) x ~ x → 0 l-≡ x = x

→0

sinx . x

sin x _ l + ∞ sx

3 l^

1 -c o sx x

ι . ∕0 ∖ ∖ 2∕

i

- < f(x) < ⅛ 1 + c o sx _ T + ∑ ^ -

χ

1≡

0

< M, a contradiction. s in (l-c o s x ) 1 -c o sx ’

χ

1≡

l-c o s 2 x _ θ x (l+ c o s x ) “

n

1

27. lim ⅛ ≡x > = lim ⅛sm x^ ∙ ≡x = lim ⅛sm≡x 1 ∙ lim ≡x = 1 ∙ 1 = 1. χ →0 χ →0 x →0 x →0 29. lim ⅛ ⅞ ^ = lim ⅛ ⅛ ⅛ 1 ∙ (x + 2) = lim ≡ ⅛ ⅛ > ∙ lim (x + 2) = 1 ■4 = 4 x “^ 2

x *~^ 2

x

-,- ^ 2

x

^2

ι i

χ

κ τn sin2 x 1≡ 0 x ( l÷ c o s x )

59

60

Chapter 2 Limits and Continuity

NOTES:

CHAPTER 3 DIFFERENTIATION 3.1 THE DERIVATIVE AS A FUNCTION 1. Step 1: f(x) = 4 - x2 and f(x + h) = 4 - (x + h)2 Steo 2’ f c l l l H j W _ [ 4 - ( x + h)2 ] - ( 4 - X2 ) _ (4 - x2 - 2xh *

h

*

h

-2x h - h2 _ h ~

h2 ) - 4 + x2

h

h (-2 x - h) h

= -2 x - h Step 3: f'(x) = lim (-2 x - h) = -2 x ; f , (-3 ) = 6, f'(0) = 0, f'(l) = - 2 h →0

3. Step 1: g(t) = ⅛ and g(t + h) = ^ Step 2:

_ “

h (-2 t-h ) _ (t + h)2 t2 h “

Step 3: g '« = ⅛

1

_

1

g(t + h )-g (t) _ h

(

g

O + jg g ) h

_ “

h

t2 - ( t 2 ÷ 2th + h2 ) _ - 2 t h - h 2 (t + h)2 ∙t ¼ “ (t + h)2 t2 h

_ “

-2 t-h (t + h)2 t2

≡ ⅛

= ⅛ = ⅛ ; g V D = 2’ g'(2) = - i, g' ( √ 3 ) = - ⅛

5. Step 1: p(0) = √ 3 0 and p(0 + h) = √3(⅛ + h) ς

p (g ÷ h )-p (⅛ ) _

t P

√3(⅛ + h ) - √ ⅞

h

‘ =

_

(√3⅜ + 3 ⅛ - ⅝ ⅞ ) . (√3⅜ + 3 ⅛ + ⅛ ¾ ) _

h

h

7

^

^τ⅛

= √ 5 ⅛

7. y = f(x) = 2x3 and f(x + h) = 2(x + h)3 ^ lim

6x⅛ + 6xh⅛ 2h3

=

h

h→0

h→0

— lim

h→0

f (t + h χ 2 t + l ) - t ( 2 t + 2 h + l ) ^ ( 2 t+ 2 h + i ) ( 2 t + i) I ; h 2^

⅛ = hlimθ 2ω

^

_

n

h→0

~^^^^∣ ^ 2 ^^ ÷ h ~ 2t —2ht —t — (2 t + 2 h + 1× 2 t + 1)h

^

lim

h

τ

; p'(l) = ^ , p'(3) = ∣, p' ( j ) =

= ⅛

h ⅛ ⅛ f a h + 21⅛ =

π m

h→0

9∙ s = r ( t ) = ⅛ a n d r ( t + h) = ⅛ _

(3 ⅜+ ⅞ ) . 3 ⅜ h ( √ 3 0 + 3h + √ ¾ )

3 3h = , , h ( √ 3 H 3 h + √ 3β) √30 + 3 h + √ 3 0

Step 3: p'(0) = h lim

=

(√ 3 β + 3 h + v ¾ )

h →0

^ lim θ

v

L

⅛

2⅛e±3i ⅛i ∣ j ^

= ^

χ 2 + 6 χ h + 2 h 2j = 6 χ 2

'

s

⅛

b

⅛

l

(t + h ) ( 2 t+ l) - t( 2 t + 2 h + l) (2t + 2 h ÷ l)( 2 t+ l)h

lim -------- --------- — lim --------- ---------

h → 0 h

h→0

(2 t + 2 h

÷ 1 ×2 t ÷

υ

— 1 — 1 _ “ ( 2 t+ l) ( 2 t+ l) - ( 2 t+ l) 2

11. p = f(q) = -7 ⅛ τ andf(q + h) = - 7 r ⅛ r

n

√q÷l

/

∙

≠> ⅛ = lim ⅛ w ÷ I ∑ - ⅛ i l

τ τ

√ (q + h ) + l

dq

h → 0

h

( ∖ ∕q÷ 1 - √ q + h + 1 λ

lim

= ⅛

-----------*---------

=

h⅛ 0

y q + l~ y q ÷ h + l

⅛

⅞

. ft —> Q

_

ι∙

=

h γ ∕q + h + i y q + l

( ^ ∕q + T + ^ q + h -I-1)

(q ÷ l ) - ( q + h + 1) _ ^ —> θ h ^ /q 4 - h + l^ / q + T (^ ∕q 4-14- ^ q 4- h + 1)

ι∙ _________________ —h_________________ _ 11 ∕q 4-14- ^ q 4- h 4-1) h —> 0 h —> 0 h ^ q 4- h 4-1 ^ q 4-1 ( ᅟ ᅟ ∕q 4- 1 ∖ fq 4-1 (

q 4-1 4-

q 4- 1)

2(q 4-1) ^ q 4-1

_________________—1 -^z∕q 4- h 4-1 ^ q ÷ 1 ( ^ ∕q 4 -1 4 - ^ ∕q 4 - h 4 -l)

62

Chapter 3 Differentiation

13. f(x) = x + ∣and f(x + h) = (x + h) + ^ x(x + h)2 + 9x - x2 (x ÷ h) - 9(x + h) _ x(x + h)h h(x2 + xh - 9) ⅛ ^ f 'W x(x + h)h

_

15 ⅛ = lim ’

dt =

= h¾

= ⅛

=

=

U m

1

- ^

m

= f '( - 3 ) = 0

h

h [3t 2 + 3th + h 2 - 2t - h) = h

h →O

x2 h + xh2 —9h x(x + h)h

(t3 + 3t2 h + 3th2 ÷ h 3 ) —(t2 + 2th÷h 2 ) —13 + 12

h →O

h

h →O

⅞ W

∣ ∣ m

h

3t⅛ ^ -3th 2 + h 3 —2th —h 2

U m

x3 + 2x2 h÷ xh 2 + 9 x - x 3 - x 2 h - 9 x - 9 h x(x + h)h

Kt + h)3 - ( t + h)2 l - ( t 3 - t 2 ) _

h →O

≠ ∙ S ⅛ - S ≤ = fej22i*+!!>]—L J .

(3

u

h →θ '

2 +

3 th +

h

2_

- h) '

2t

= 3t2 - 2 t ; m = ⅛ ∣ t=~ 1 = 5

17. — ! 1 2 ∕E Ξ I Ξ 2 ^ ! ^ ^

. ( ^ 3 ty Ξ Ξ ! )

h √ χ ÷ h -2 √ x -2

√ r τ w

r a t f e τ √ ^

i√ x + T ≡ ^ ^

=> f'(χ) = lim - 7 = = = - 7 = = ÷ ⅜ = — —= h → 0 √ x ÷ h - 2 √ x ~ 2 ( √ x - 2 + √ x + h - 2J

= - y = = - 7 ===⅛ ==—   = r h √ x 4-h —2 √ X - 2 ^ √ X - 2 + √ x + h - 2 J =

— ________ 8 [ ( x - 2 ) - ( x + h - 2 ) ] _______

(√ x -2 + √ x + h -2 )

=
" ∙< " ∙W

" ∣

line at (6,4) is y - 4 = - j (x - 6) => y = - ∣x + 3 + 4 => y = - ∣x + 7. 19. s = f(t) = 1 —3t2 and f(t ÷ h) = 1 —3(t + h)2 = 1 —3t2 —6th —3h2 => ⅛ = h ⅛∏0 ^ = lim

Hm ( - 6 t - 3 h ) = - 6 t ≠ ∙

h

h →0

d , l '=-1

h →0

,

=

and f(0 + h) = - s ⅛ v

√ 4 -0

,

lim

=> ⅛ = lim

√ 4 -(0 ÷ h ) ∞ h →0 2 √ 4 ≡ ⅛ -2 √ 4 ≡ ⅜ ≡ h (2 √ 4 ≡ g

lim

h →0

h →0

h √ 4 -0 √ 4 -0 -h

h √ 4 -0 √ 4 -0 -h

=

=

h → 0 2 h √ 4 -0 √ 4 -0 -h (√ 4 -0 + √ 4 -0 -h ) (4 -0 ) ^ 2 √ 4 -0 ^

23 1f , M = ^^∙

W

=

z

→x

d0∣ 0=O

(4 -0 )√ 4 -0

⅛ ≡ J ⅛ i = lim ⅛ ~⅜ i z -x

z

¾

z -x

h +

≡

= lim

h →0

≡

2

_

^

h

Ξ

2 √ 4 -⅛ > -h )

(2 √ 4 -H 2 √ 4 -0 -h j

„ ....... , lim 2 h → 0 √ 4 ^ -0 √ 4 -0 -h (√ 4 -0 + √ 4 - 0 - h ) 8

= lim

(χ + 2 )~( z + 2 )

z → x (z -x )(z + 2)(x + 2)

_

χ —z

z → x (z -x )(z + 2)(x÷2)

_ ι∙ z

—1

→ x(z + 2)(x÷2)

~1 „ (x + 2 )i

g W =

⅛

⅛^w

=6 2

21. r = f(0) = - ‰

+

z≡

x

z

-

x

-

z

→x

z

_

x

-

z≡

( z _ x )( z

_ 1 )(x _ υ -

z≡

x ( z _ x )(z

_ 1 )(x _

1}

-

z iιm χ ( z _ 1 )( x

_

1}

0 ⅛

27. Note that as x increases, the slope of the tangent line to the curve is first negative, then zero (when x = 0), then positive => the slope is always increasing which matches (b). 29. f3(x) is an oscillating function like the cosine. Everywhere that the graph of f3 has a horizontal tangent we expect f3 to be zero, and (d) matches this condition. 31. (a) f , is not defined at x = 0 ,1 ,4 . At these points, the left-hand and right-hand derivatives do not agree. For example, χ lim _ ® ^ ® = slope of line joining (- 4 ,0 ) and (0,2) = ∣but lim,+ ^ § ® = slope of

Section 3.1 The Derivative as a Function line joining (0,2) and (1, —2) = —4. Since these values are not equal, f'(0) = χ limθ ® _ ® does not exist. (b)

y,

r 33.

y' 3.2 — o

- ® £ L l __ I__ I__ I__ L_ -0,7 - 84 o - o 87 88

-3.3 - 0— 0

35. Left-hand derivative: For h < 0, f(0 -I- h) = f(h) = h2 (using y = x2 curve) => =

lim

h → o-

h

=

lim Mh

h → 0+

Then lim

h → 0^

=

lim

® ^ -®

lim h = 0;

h → o-

Right-hand derivative: For h > 0, f(0 + h) = f(h) = h (using y = x curve) => =

h

lim

h → 0+

f-f

lim 1 = 1;

h → 0+

S ± ^ f f l r/ ∣ >

lim

h → 0+

W+⅛∑W 4 ⅛e derivative f'(0) does not exist. h

37. Left-hand derivative: When h < 0,1 + h < 1 => f(l + h) = √ l + h ≠>

lim _ ^ ≡ ! > 1 _ ∕l + h + l

Right-hand derivative: When h > 0,1 + h > 1 ≠ f(l + h) = 2(1 + h) — 1 = 2h + 1 => =

lim

h → 0+

f(l+ h)-f(l) h

lim - 1 = lim 2 = 2; h h → 0+ h → 0+

Then lim

h → 0-

5 1 ± h⅛ ∑ ≡ r^

lim

h → o+

f 0 and decreasing on intervals where y' < 0 47. (a) Using the alternate formula for calculating derivatives: f'(x) = z limχ ^ 7 ® = z limχ ^ 3z _ χ3 = lim

z → X 3(z - x)

= lim ^ ⅛ z → x

± ⅛

= lim

3(z - x)

z → x

= x2

3

f'(x) = x2

v

7

(c) y' is positive for all x ≠ 0, and y' = 0 when x = 0; y' is never negative (d) y = y is increasing for all x ≠ 0 (the graph is horizontal at x = 0) because y is increasing where y' > 0; y is never decreasing 49. yJ ' = lim ® z & ! = lim ⅛ ^ = lim U ⅛ + ≤ X

→

C

X-C

X

→

X-C

C

X

→

C

The slope of the curve y = x3 at x = c is y' = slope. 51 v' = lim y ’

h→ 0

=

jim

(2 0 ^ J t)^ J 3 0 ty ^ ^

_

h

4xh + 2h 2 - i 3 h

= liV m (x ≡ +

X-C X→ C 3c2 . Notice that

∩m

h→ 0

xc + c

'

2)

= 3c2 .

3c2 ≥ 0 for all c => y = x3 never has a negative

2x2 + 4xh + 2 h 2 - 1 3 x - 1 3 h + 5 - 2 x 2 + 1 3 x -5 h

_ lim (4x + 2h —13) = 4x —13, slope at x. The slope is —1 when 4x — 13 = —1

4 4 x = 1 2 4 x = 3 => y = 2 ∙3 2 - 13∙3 + 5 = —16. Thus the tangent line is y + 16 = (—l)(x —3) 4 y = - x - 13 and the point of tangency is (3, -16). 53. No. Derivatives of functions have the intermediate value property. The function f(x) = [xj satisfies f(0) = 0 and f(l) = 1 but does not take on the value ∣anywhere in [0,1] => f does not have the intermediate value property. Thus f cannot be the derivative of any function on [0,1] => f cannot be the derivative of any function on (—∞ , ∞ ). 55. Yes; the derivative of —f is —f' so that f'(xo) exists => —f'(xo) exists as well. 57. Yes, Jimθ ^ can exist but it need not equal zero. For example, let g(t) = mt and h(t) = t. Then g(0) = h(0) = 0, but lim

t→ o h(0

= lim 7t = lim m = m, which need not be zero. t→ 0

t→ 0

Section 3.2 Differentiation Rules 59. The graphs are shown below for h = 1,0.5,0.1. The function y = ~ ^ is the derivative of the function = h limθ ^ x + ^ ^ ^ . The graphs reveal that y = √Z±Σz2Z* gets closer to y = ^

y = y∕ x so that ^

as h gets smaller and smaller.

61. Weierstrass’s nowhere differentiable continuous function.

∣ W≡ COS(JTJT) + 1:1

COS(9JTJO +

cos(9, πx)

÷(I) wι^ w j 3.2 DIFFERENTIATION RULES

1. y = - x 2 + 3 ≠

∣ = ⅛ ( - X2 ) + ⅛ (3) = - 2 X + 0 = - 2 X

3. s = 5t3 - 3t5 =>

∣ = ⅛ (5t3 ) - ⅛ (3t5 ) = 15t2 - 15t4 ≠> ^ = ^ (15t2 ) - ⅛ (15t4 ) = 30t - 60t3

5. y = J x3 - x 4

⅛ = 4x2 - 1 => 0 = 8x

7. w = 3z- 2 —Z~1=> ^ = - 6 Z^ 3 + z^ 2 = ^ + ^ ≠>

⅛ = -2

^ = 18z~4 - 2z^3 = ^ - ^

65

66

Chapter 3 Differentiation y = 6x 2 - 10x - 5x^ 2 =+ ⅛ = 12x - 10 + 10x^3 = 1 2 x - 10 + ^ ≠

9. ιι

11. Γr _- 51Sc - 2 - 55 sc - l = . > s dr - _ -

2 Sβ - 3 + Ij S5 β - 2 - —^ - 2 + 1 j j5 j ≠ . ∙ f df r - _2 S

2 5

§

= 12 - 0 - 30x^ 4 = 12 - ^

- 5 sc c - 3 - _^ 2 - ^ 5

13. (a) y = (3 - x2 ) (x3 - x + 1) ≠- y' = (3 - x2 ) ∙ ⅛ (x3 - x + 1) + (x3 - x + 1) ∙ ⅛ (3 - x 2 ) = (3 —x2 ) (3x2 - 1) + (x 3 —x + 1) (-2 x ) = - 5 x 4 + 12x2 - 2x —3 (b) y = - x 5 + 4x 3 - x2 - 3x + 3 ≠- y' = - 5 x 4 + 12x2 - 2x - 3 15. (a) y = (x 2 + l ) ( x + 5 + ∣) => y, - (x2 + 1) ∙ ⅛ (x + 5 + ∣) + (x + 5 + i ) ∙ ⅛ (x2+ 1) = (x2 + 1 ) ( 1 - x - 2 ) + (x + 5 + x - 1 ) (2x) = (x2 - 1 + 1 - x - 2 ) + (2x 2 + 10x + 2) = 3x2 + 10x + 2 - ⅛ (b) y = x 3 + 5x2 + 2x + 5 + | 4 ^

,

y

=

_

⅛ ⅛ ’ u s e t ^ e Qu o t ie n trule: u = 2x + 5 an d v = 3 x - 2 =+ u' = 2 a n d v' = 3 + (3x - 2)(2)- ( 2 x + 5)(3) _ (3x - 2)2

-

u' = 0 and

v'= (χ 2 — 1) (2x ÷ 1) + (x2 + x + 1) (2x) = 2X3 + x2 - 2x - 1 ÷ 2x3 + 2x2 + 2x = 4x 3 + 3x 2 —1 v

⅛ _ dx

vuz - u √ _ v2

o ~ 1 (4x3 + 3x2 - 1) _ (x 2 - l ) 2 (x 2 + x + l ) 2

—4χ3 - 3x2 + 1 (X 2 - 1 ) 2 (x 2 + x ∣l) i

29. y = | x 4 — | x 2 —x =^ y' = 2x3 —3x - 1 => y" = 6x2 - 3 => y'" = 12x => y(4 ) = 12 =>y^n^ = 0 for all n ≥ 5 31.

y

=⅛2 =

χ

2+

7x

33. r = O ⅛ L ≡ 35. w = ( 1^ ) (3 -

= ⅛1 Z)

= ⅛ - 1 => ⅛ = 37

d

~

^ ⅛

⅛ = 2 x -7 x ^ 2 = 2 x -⅞ = + §

-ι

2 cl

_ ^

1

2Z- 3

- 0=

2z~3

^ = 0 + 30~4 = 30~4= ∣ =+ ⅛ = -1 2 0 ~ 5 =

ι _ ^ -3 ^

=

= ( I Z^ 1 + 1) (3 -

f £ + 1 ) ( S r⅛ Λ 12q ) q J “ ~ 6

=

Z)

=

Z^ 1

- I + 3-

6

2?

Z

= Z^ 1 + 5 -

Z =+

^

^ = - z ^ 2 + 0 - 1 = —z -

2

- 1

1 .

1 q*

= ⅞

q6 - q 2 + 3q4 - 3 _ ± 2 _ ± 12φ “ 12 ⅛ 12 " -

= 2 + 14x"3 = 2 + ⅛

?

f l - 2t

, 1 _ 1 -4 4 4 M

^

& dq

=

1 α ψ. 1 α ~ 3 . Q - 5 6∏ 6∏ ^ "

=

1q ■ 6 ∏

Section 3.2 Differentiation Rules

67

39. u(0) = 5, u'(0) = - 3 , v(0) - - 1 , √(0) - 2 (a) ⅛ (uv) = u√ + v u ' ^ ⅛ (uv)∣ x=0 = u(0)√(0) + v(0)u'(0) = 5 ∙ 2+ ( -1 ) (- 3 ) = 13 d ∕u A ' '

_

A (y dx u /

v u '- u √ _

.

u √ -v u ' u2

d fu A I

A

^

dx

_

(v ∣ _ u / I x =o

(d) ⅛ (7v - 2u) = 7√ - 2u' ^

v(0)u'(0)-u(0)√(0) _

( -1 ) (- 3 ) -( 5 ) (2 ) _

u(0)v'(0)-v(0)u , (0) (u(0))2

( 5 )( 2 )- ( -1 ) ( -3 ) _ 7. (5)2 25

_

7

⅛ (7v - 2u)∣ x=0 = 7v'(0) - 2u'(0) = 7 - 2 - 2 (-3 ) = 20

41. y = x3 —4x + 1. Note that (2,1) is on the curve: 1 = 23 - 4(2) + 1 (a) Slope of the tangent at (x, y) is y' = 3x2 - 4 => slope of the tangent at (2,1) is y, (2) = 3(2)2 —4 = 8. Thus the slope of the line perpendicular to the tangent at (2,1) is —∣ => the equation of the line perpendicular to to the tangent line at (2,1) is y — 1 = —∣(x —2) or y = - ∣+ ∣. (b) The slope of the curve at x is m = 3x2 - 4 and thesmallest value for m is —4 when x = 0 and y = 1. (c) We want the slope of the curve to be 8 => y' = 8 => 3x2 - 4 = 8 => 3x2 = 12 => x2 = 4 => x = ± 2 . When x = 2, y = 1 and the tangent line has equation y —1 = 8(x —2) or y = 8x — 15; when x = —2, y = (-2 ) 3 —4 (-2 ) + 1 = 1, and the tangent line has equation y - 1 = 8(x + 2) or y = 8x + 17. 43

v = y

4x_

≠>

⅛ = dx

(x2 + 1)(4)-(4x)(2x)

(X 2 + 1 )2

4X2 +24 - 8 2X2 (X + I )

=

4 (- x 2 + l)

(X2 + I )2 *

ww h n e n χ

θ

θ

- u , y - υ a n αj y

/_ -

4 (0 + 1 ) 1

= 4, so the tangent to the curve at (0,0) is the line y = 4x. When x = 1, y = 2 => y' = 0, so the tangent to the curve at (1,2) is the line y = 2. 45. y = ax2 + bx + c passes through (0,0) => 0 = a(0) + b(0) + c =+ c = 0; y = ax2 + bx passes through (1,2) => 2 = a + b; y' = 2ax + b and since the curve is tangent to y = x at the origin, its slope is 1 at x = 0 =+ y' = 1 when x = 0 =+ 1 = 2a(0) + b =+ b = 1. Then a ÷ b = 2 =+ a = 1. In summary a = b = 1 and c = 0 so the curve is y = x2 + x. 47. (a) y = x3 - x =+ y' = 3x2 - 1. When χ = - 1 , y = 0 and y' = 2 =+ the tangent line to the curve at (—1,0) is y = 2(x + 1) or y = 2x + 2. (b)

(c)

y _ 2x

+

2J ^

χ3

~

x =

2χ ÷ 2 =+

X3

-3

X-2

= (X - 2)(X + 1)2 = 0 =+ x = 2 or x = - 1 . Since

y = 2(2) ÷ 2 = 6; the other intersection point is (2,6) 49. P(x) = an xn + an -ιx n ^ 1 4------ 1- a2 x2 + a i x + a0 =+ P'(x) = nan xn ^ 1 ÷ (n - l)a n -ιx n ^ 2 4------ F 2a2 x + a i 51. L etcbe a constant =+ ^ = 0 =+ ^ ( u ∙ c ) = u ∙ ^ + c ∙ ^ = u ∙ 0 ÷ c ^ = c ^ . Thus when one of the functions is a constant, the Product Rule is just the Constant Multiple Rule => the Constant Multiple Rule is a special case of the Product Rule. 53. (a) ⅛ (uvw) = ⅛ ((uv) ∙ w) - (uv) ⅛ + w ∙ ⅛ (uv) = uv ^ + w (u ⅛ + v ⅛) = uv ^ + wu ^ + wv ^ = uvw' + uv'w + u'vw

68

Chapter 3 Differentiation (b) ⅛ (u i U2 U3 U4 ) = ⅛ ((U1U2 U3 ) u 4 ) = (u j u 2 u3 ) ⅛ + u4 £ (u i u 2 u 3 ) =+ £ (U1U2 U3 U4 ) = U1U2 U3 ⅛ + U4 (u i u 2 ⅛ + U3 U1 ⅛ + u 3 u 2 ⅛ )

(using (a) above)

≠ ∙ ⅛ (u i u2 u 3 u4 ) = U1U2 U3 ⅛ + U1U2 U4 ⅛ + u i u3 u4 ⅛ + u 2 u 3 u 4 ⅛ =

U1U2 U3 U'4

+

U1 U2 U3 U4

+ U1U2U3 u4 + u'1 u 2 u 3 u 4

(c) Generalizing (a) and (b) above, ⅛ (up ∙ ∙un ) = U1u2 ∙ ∙ ∙un. ι u ' + uιu 2 ∙ ∙ ∙un- 2 u ' - 1 u n + . . . + u'1 u2 ∙ ∙ ∙un 55. p = ^ ⅛ — ^ . We are holding T constant, and a, b, n, R are also constant so their derivatives are zero .

dP _ dV -

(V -n b )-O -(n R T χ i) _ (V -nb) 2

V2 (0) - (an2 ) v = —32t => speed = ∣ v∣= 32t ft/sec and a = —32 ft∕sec2 (b) s = 0 => 179 - 16t2 = 0 => t = ^ ^ ≈ 3.3 sec (c) When t = γ ∕ ⅛ v = - 3 2 ^ ⅞ = -8 √ 1 7 9 ≈ -107.0 ft/sec

15. (a) at 2 and 7 seconds (c) Iv l (m∕sec)

17. (a) (c) (e) (f) (g)

(b) between 3 and 6 seconds: 3 ≤ t ≤ 6 (d)

190 ft/sec (b) 2 sec at 8 sec, 0 ft/sec (d) 10.8 sec, 90 ft/sec From t = 8 until t = 10.8 sec, a total of 2.8 sec Greatest acceleration happens 2 sec after launch From t = 2 to t = 10.8 sec; during this period, a = ^ ⅛ ⅛ z ^ ≈ “ 32 ft/sec2

19. s = 490t2 ^ v = 980t ^ a = 980 (a) Solving 160 = 490t2 =≠> t = ^ sec. The average velocity was g⅛^≤θl = 280 cm∕sec. (b) At the 160 cm mark the balls are falling at v(4∕7) = 560 cm∕sec. The acceleration at the 160 cm mark was 980 cm∕sec2 . (c) The light was flashing at a rate of ^ = 29.75 flashes per second. 21. C = position, A = velocity, and B = acceleration. Neither A nor C can be the derivative of B because B's derivative is constant. Graph C cannot be the derivative of A either, because A has some negative slopes while C has only positive values. So, C (being the derivative of neither A nor B) must be the graph of position. Curve C has both positive and negative slopes, so its derivative, the velocity, must be A and not B. That leaves B for acceleration. 23. (a) c(100) = 11,000 => cav = ⅛

= $n0

—.lx 2

(b) c(x) = 2000 + 100x => c'(x) = 100 —.2x. Marginal cost = c'(x) => the marginal cost of producing 100 machines is c'(100) = $80 (c) The cost of producing the 101s, machine is c(101) - c(100) = 100 - ⅛ = $79.90 25. b(t) = 106 + 104 t - 103 t2 ≠> b'(t) = 104 - (2) (103 t) = 103 (10 - 2t) (a) b'(0) = 104 bacteria/hr (b) b'(5) = 0 bacteria/hr (c) b'(10) = —104 bacteria/hr 27. (a)

y

= 6 ( l - ⅛ ) 2 = 6 ( l - i + ⅛ ) => $ = ⅛ - 1

(b) The largest value of ^ is 0 m/h when t = 12 and the fluid level is falling the slowest at that time. The smallest value of ^ is —1 m/h, when t = 0, and the fluid level is falling the fastest at that time.

69

70

Chapter 3 Differentiation (c) In this situation, ⅛ ≤ 0 => the graph of y is always decreasing. As ^ increases in value, the slope of the graph of y increases from - 1 to 0 over the interval 0 ≤ t ≤ 12.

29. 200 km/hr = 55 ∣m∕sec = ^ t = 25, D = ^ (25)2 = ^

m∕sec, and D = y t2 => V = y t. Thus V = ^

=> y t = ^

=> t = 25 sec. When

m

(b) (c) (d) (e)

v > 0 when 0 ≤ t < 6.25 => body moves up; v < 0 when 6.25 < t ≤ 12.5 => body moves down body changes direction at t = 6.25 sec body speeds up on (6.25,12.5] and slows down on [0,6.25) The body is moving fastest at the endpoints t = 0 and t = 12.5 when it is traveling 200 ft∕sec. It's moving slowest at t = 6.25 when the speed is 0. (f) When t = 6.25 the body is s = 625 m from the origin and farthest away.

(b) v < 0 when ^ - ^ < t < - + 3^

=> body moves left; v > 0 when 0 ≤ t
^ = —10 ÷ 3 ^ (cos x) = —10 —3 sin x 3. y = esc x - 4 √ x + 7 => ^ = -e s c x cot x - ^

+ 0 = -e s c x cot x - ^

5.y = (sec x + tan x)(sec x - tan x) => ^ = (sec x + tan x) ^ (sec x - tan x) + (sec x

- tan x) ^ (sec x+ tan x)

= (sec x + tan x) (sec x tan x - sec2 x) + (sec x - tan x) (sec x tan x + sec2 x) = (sec2 x tan x ÷ sec x tan2 x - sec3 x - sec2 x tan x) + (sec2 x tan x - sec x tan2 x + sec3 x - tan xsec2 x) = 0. ^Note also that y = sec2 x —tan 2 x = (tan2 x + 1) —tan2 x = 1 => ⅛ = 0.^

■J

7

,,

cot X

⅛

(14- cot x) ( —CSC2 x) - (cot x) ( - c s c 2 x)

(1 + c o tx ) ⅛ (cot x) - (cot x) ⅛ (1 + c o t x)

∙ y

= ----------

(l+ c o tx ) 2

------------- = ------------ '

(l+cotx> 2----- -----------

__ —CSC2 X — CSC2 X cot X + CSC2 X cot X __ —CSC2 X (1 + c o tx ) 2 (1 + c o t x) 2

= 4 sec x ÷ cot x => ^ = 4 sec x tan x —csc2 x

9. y = — ^ ÷ ^

11. y = x 2 sin x + 2x cos x - 2 sin x => ⅛ = (x2 cos x + (sin x)(2x)) + ((2x)(-sin x) + (cos x)(2)) - 2 cos x = x2 cos x + 2x sin x —2x sin x ÷ 2 cos x —2 cos x = x2 cos x 13. s = tan t —t => ⅛ ~ ⅛ (tan 0 ^ 1 - sec2 1 — 1 = tan 2 t __ 1 + esc t 1 -c s c t

^

ds __ (1 - esc t)(-c s c t cot t) - (1 + esc t)(csc t cot t) dt________________ (1 — esc t) 2

__ —CSC t cot t ÷ CSC2 t cot t —CSC t cot t —CSC2 t cot t __ —2 CSC t cot t (1 - c s c t ) 2 (1 - esc t) 2

17. r = 4 —02 sin 0 => ^ 19. r = sec 0 esc 0 + =

.

=

-

(^ 2 ⅛ (s *n ^) + (sin 0)(20)) — —(0 2 cos 0 + 20 sin 0) = —0(0 cos 0 + 2 sin 0)

^ = (sec 0)(-e s c 0 cot 0) + (esc 0)(sec 0 tan 0)

( c δ ⅛ ) (s in ? ) ( s in ? ) + (s in ? ) (e ra ? ) (c o s ? )

21∙

p=

23.

r

5

+ ⅛

=

__ sin q ÷ cos q cosq

5

,

+

ta n

(b)

≡⅛

+ ∞ ¼

=

sec2

^

c

SC2

0

q ≠- ⅞ = s ∞ 2 q

dp (cos q)(cos q - sin q) — (sin q + cos q )(-s in q) dq______________________ cos2 q

__ cos2 q —cos q sin q + sin2 q + cos q sin q __ 1 __ cos2 q cos2 q

25. (a)

=

2 SeC

n

y = esc x => y, = —esc x cot x => y" = - ((esc x) (- c s c 2 x) + (cot x ) ( - esc x cot x)) = csc3 x + esc x cot2 x = (esc x) (esc2 x + cot2 x) = (esc x)(esc2 x ÷ csc2 x - 1) = 2 csc3 x - esc x y = sec x => y, = sec x tan x 4 y" = (sec x) (sec2 x) + (tan x)(sec x tan x) = sec3 x + sec xtan 2 x = (sec x) (sec2 x ÷ tan2 x) = (sec x)(sec2 x ÷ sec2 x — 1) = 2 sec3 x —sec x

72

Chapter 3 Differentiation

27. y = sin x => y' = cos x => slope of tangent at x = —π is y '( - π ) = cos (—π) = —1; slope of tangent at x = 0 is y'(0) = cos (0) = 1; and slope of tangent at x = y is y, ( y ) = cos y = 0. The tangent at ( - π , 0) is y —0 = - l ( x + π), or y = —x —7r; the tangent at (0,0) is y —0 = l(x —0), or y = x; and the tangent at ⅛ , - l ) isy = - 1 . 29. y = sec x => y' = sec x tan x ≠> slope of tangent at x = —y is sec (—∣) tan ( - ∣) = - 2 √ z3 ; slope of tangent

y = secx

at x = ^ is sec (y ) tan Q ) = √ ^ . The tangent at the point ( - f , sec ( - f )) = ( - f , 2) is y - 2 = - 2 √ 3 (x + f ) ; the tangent at the point Q , sec ( 5 ) )

=

(j> V ^ j is y - ᅟ ∕2

= √ ^ (χ -z )∙

---- 1------ 1------------------------ 1---------- 1----> χ -π ∕2 -π ∕3 0 π∕4 π∕2

31. Yes, y = x ÷ sin x => y, = 1 + cos x; horizontal tangent occurs where 1 + cos x = 0 => cos x = —1 => x = π 33. No, y = x —cot x => y' = 1 + csc2 x; horizontal tangent occurs where 1 ÷ csc2 x = 0 => csc2 x = —1. But there are no x-values for which csc2 x = —1. 35. We want all points on the curve where the tangent line has slope 2. Thus, y = tan x 4 y' = sec2 x so that y' = 2 => sec2 x = 2 => sec x = ± ᅟ∕2 => x = ± ∣. Then the tangent line at (∣, 1) has equation y - 1 = 2 (x - J ) ; the tangent line at (—^ , - 1 ) has equation y + 1 = 2 (x + ∣) .

37. y = 4 + cot x —2 esc x 4

y' = -e s c 2 x + 2 esc x cot x = - ( ≠ ) (

-⅛f-)

(a) When x = f , then y' = - 1 ; the tangent line is y = - x + f ÷ 2. (b) To find the location of the horizontal tangent set y' = 0 => 1 - 2 cos x = 0 => x = ∣radians. When x = ∣, then y = 4 - 5/3 is the horizontal tangent. 39.

lim sin (7 - ∣) = sin ( 1 - 1 ) = sin 0 = 0

41. limθ sec [cos x + π tan ( ^ )

— 1] = sec [cos 0 + π tan ( ^ θ ) — 1] = sec [1 + π tan (^) — 1] = sec π — —1

43. lim tan (1 — ≡ i ) = tan f l - lim t →o ' , 7 t →o t /

— tan (l - 1)

0

45. s = 2 - 2 sin t => v = ^ = - 2 cos t => a = ^ = 2 sin t => j = ^ = 2 cos t. Therefore, velocity - v (^ ) 2 m/sec; speed = ∣ v ( j ) ∣= ᅟ ∕ 2 m/sec; acceleration = a ( ^ ) = ᅟ ∕2 m/sec2 ; jerk = j ( ∣) = ^/2 m∕sec 3 .

Section 3.4 Derivatives of Trigonometric Functions 47. lim f(x) = lim X→ 0

X→ 0

⅛x

i

i ) 3 x ∕

= lim 9 ( ⅛ ^ ) ( ⅛ x

→ θ ∖

3 x ∕∖

= 9 so that f is continuous at x

0 => lim f(x) = f(0) χ → 0

=> 9 = c. 49. p θ (cos x) = sin x because ⅛ (cos x) = cos x => the derivative of cos x any number of times that is a multiple of 4 is cos x. Thus, dividing 999 by 4 gives 999 = 249 - 4 + 3 ≠> ^ =

⅛

(c o s χ

)] = ⅛

(c o s χ

)

=

sin x

(cos x)

∙

closer and closer to the black curve y = cos x because ⅛ (sin x) = h limθ ⅛ ⅛ J ≡

=

cos x

η

ιe s a m e

is true as h takes on the values of —1, —0.5, —0.3 and —0.1. 53. (a)

The dashed curves of y =

are c∣ o s e r to t∣ ιe

bia c ]c c u r v e y = cos x than the corresponding dashed

curves in Exercise 51 illustrating that the centered difference quotient is a better approximation of the derivative of this function.

The dashed curves of y = ⅛ ⅛ - ⅛ —⅛

are

closer to the black curve y = —sin x than the corresponding dashed

curves in Exercise 52 illustrating that the centered difference quotient is a better approximation of the derivative of this function. 55. y = tan x => y, = sec2 x, so the smallest value y' = sec2 x takes on is y, = 1 when x = 0; y, has no maximum value since sec2 x has no largest value on ( - ∣, ∣) ; y 'i s never negative since sec2 x ≥ 1.

73

74

Chapter 3 Differentiation

57 y

=

≡2L appears to cross the y-axis at y = 1, since = 1; y = 5γ

limθ ^

5

at y = 2, since limθ 5γ

appears to cross the y-axis i

= 2; y = η ^ appears to

cross the Jy-axis at Jy = 4, since lim X

→ 0

X

= 4.

However, none of these graphs actually cross the y-axis since x = 0 is not in the domain of the functions. Also, lim ⅛x = 5, lim ⅛ x⅛ = - 3 , a ∏d lim ⅛x x→ 0

x→ 0

x→ 0

= k => the graphs of y = 5γ ^ , y = ^ lp ħ ξ and y _ sιiLkχ a ppr o a c ]1 5, —3, a nd k, respectively, as x → 0. However, the graphs do not actually cross the y-axis. 3.5 THE CHAIN RULE AND PARAMETRIC EQUATIONS

1. f(u) = 6u - 9 4 f'(u) = 6 ≠> f'(g(x)) = 6; g(x) = 5 x4 =► g'(x) = 2x3 ; therefore ∣ = f'(g(x))g'(x) = 6 - 2x3 = 12X3 3. f(u) = sin u ≠- f'(u) = cos u => f'(g(x)) = cos(3x + 1); g(x) = 3x + 1 => g'(x) = 3; therefore

= f'(g(x))g'(x)

= (cos (3x + 1))(3) = 3 cos (3x + 1) 5. f(u) = cos u => f'(u) = —sin u => f'(g(x)) = —sin (sin x); g(x) = sin x ≠> g'(x) = cos x; therefore ^ = f'(g(x))g'(x) = -(sin (sin x))cos x 7. f(u) = tan u => f'(u) = sec2 u 4 f'(g(x)) = sec2 (10x —5); g(x) = 10x - 5 4 g'(x) = 10; therefore ∣ = f'(g(x))g'(x) = (sec2 (10x —5)) (10) = 10 sec2 (10x - 5) 9. W ithu = (2x + 1),y = u5 : ⅛ = ⅛ ⅛ = 5u4 ∙ 2 = 10(2x + l) 4 11. W i t h u = ( l - s ) , y = « - ’: J - ; J 13. W ith .= (⅞ + x - l ) , y = Λ

7∙ '

J = H = ⅛

( ,

'

'( i + l + ⅛ ) = * ( τ + > - i ) , (i +

1

+⅛)

15. With u = tan x, y = sec u: ⅛ = ⅛ ^ = (sec u tan u) (sec2 x) = (sec (tan x) tan (tan x)) sec2 x 17. With u = sin x, y = u3 : ⅛ = ∣ = 3u2 cos x = 3 (sin2 x) (cos x) 19.

p

= √ 3 → = (3 - t)1∕2

⅛ = l ( 3 - t Γ 1∕ 2 4 ( 3 - t ) = - i ( 3 - t Γ 1∕ 2 = 5 ^

21. s = ⅛ sin 3t + ⅛ cos 5t =► ∣ = ⅛ cos 3t ∙ ⅛ (3t) + ⅛ (-sin 5t) ∙ £ (5t) = ^ cos 3t - ^ sin 5t - £ (cos 3t —sin 5t) 23. r = (esc ___

θ

+ cot 0 Γ 1 ^

CSC θ CSC 0 +

cot 0

⅛ = -(esc 0 + cot 0)~2 ⅛ (esc 0 + cot 0) = ⅞

⅞

⅜

= ¾ ⅛ ^ 1

Section 3.5 The Chain Rule and Parametric Equations 25. y = x2 sin4 x + x cos- 2 x ≠ ∙ ⅛ = x2 ⅛ (sin4 x) + sin4 x ∙ ⅛ (x2 ) + x ⅛ (cos- 2 x) + cos- 2 x ∙ ⅛ (x) = x2 (4 sin3 x ^ (sin x)) + 2x sin4 x + x ( - 2 cos- 3 x ∙ ⅛ (cos x)) + cos- 2 x = x2 (4 sin3 x cos x) + 2x sin4 x + x( (—2 cos- 3 x) (-sin x)) + cos- 2 x = 4X2 sin3 x cos x + 2x sin4 x + 2x sin x cos- 3 x + cos- 2 x 27.

y

= ± (3 x -2 )7

+

( 4 - ⅛ ) - 1 => ⅛ ^ ± ( 3 X - 2 ) 6 ∙⅛ ( 3 X - 2 ) + ( - 1 ) ( 4 - ⅛ ) - 2 . ⅛ ( 4 - ⅛ )

= ⅜ (3x - 2)6 ∙ 3 + (-1 ) (4 - ⅛ ) - 2 (⅛) = (3x - 2)6 -

29. y = (4x + 3)4 (x + I) - 3 =► ⅛ = (4x + 3)4 (-3)(x + I) - 4 ■⅛ (x + 1) + (x + l) - 3 (4)(4x + 3)3 ∙ ⅛ (4x + 3) = (4x + 3)4 (-3)(x + 1)- 4 (1) + (x + 1)- 3 (4)(4X + 3)3 (4) = -3(4x + 3)4 (x + I) - 4 + 16(4x + 3)3 (x + I) - 3 = W [~ 3 ∣ = sin (02 ) (-sin 20) ⅛ (20) + cos (20) (cos (02 )) ∙ ∣ (02 ) = sin (02 ) (-sin 20)(2) + (cos 20) (cos (02 )) (20) = - 2 sin (02 ) sin (20) + 20 cos (20) cos (02 )

39. y = sin2 (τrt —2) => ^ = 2 sin (πt —2) ∙ ^ sin (πt —2) = 2 sin (πt —2) - cos (πt —2) ∙ ^ (πt —2) = 2π sin (πt —2) cos (πt —2) 41. y = (1 + cos 2t)- 4 ≠

∣ = -4(1 + cos 2t)- 5 ∙ ⅛ (1 + cos 2t) = -4(1 + cos 2t)- 5 (-sin 2t) ∙ ⅛ (2t) =

* ⅛

43. y = sin (cos (2t —5)) ≠∙ ^ = cos (cos (2t - 5 ) ) - ^ cos (2t —5) = cos (cos (2t —5)) ∙ (—sin (2t - 5)) ∙ j (2t —5) = - 2 cos (cos (2t —5))(sin (2t - 5)) 45. y = [1 +

(⅛ )] 1 ⅛ J = 3 [1 + tan- (⅛ )] 2 ∙ ⅛ [1 + Un* (⅛)] = 3 [1 + Un* (⅛ )] 2 [4 tan≈ (⅛ ) ∙ J ■ (⅛)]

= 1 2 [ l+ u n * ( ⅛ ) ] 2 [tan3 (⅛ ) sec2 (⅛ ) ■⅛] = [ l+ u n * ( ⅛ ) ] 2 [tan3 (⅛ ) ≈ ≈ ( ⅛ ) ∣ 47. y = ( 1 + cos (t≈))1' 2 4

2 = l ( l + c . . ( t 1 ))^ 1' 2 . l ( l +

= - 1 ( 1 + cos(t-))-'≈ (sin(t 2 ) ) ∙ 2. = -

t

) ≈(t≈)) = l ( l + c a ( l 2 ) ) - l ' 2 ( - ⅛ ( l 1 ) . J ( 12 ∣

⅛

49. y = ( l + H 3 =+ y' = 3 ( l + ⅜)2 ( - ⅛ ) = - ⅛ (1 + J ) 2 =+ f = ( - ⅛ ) . J ( l + l ) , - ( l = ( - A) (2(1 + () ( - ⅛ ) ) + (⅛) ( ' + υ = 5 ( 1 + 1) (l +

2)

2

2

+

l) ≈ .4 ( > )

= ⅛ (i + i) + ⅛ ( ' + D = ⅜ ( ' + ⅛) « + 1 + A)

75

Chapter 3 Differentiation

76

51. y = ∣cot(3x — 1) ≠- y' = - 1 csc2 (3x - 1)(3) = - ∣csc2 (3x —1) => y" = (—5) (esc(3x - 1) ∙ ∣ csc(3x — 1)) = — 3c sc(3x

— 1 )(-csc(3x —1)cot(3x — 1) ∙ ∣(3x —1)) = 2 csc2 (3x — 1) cot(3x — 1)

≠- g'(x) = ⅛

53. g(x) = ≠

≠> g(D = 1 and g'(l) = ∣; f(u) = u5 + 1 => f'(u) = 5u4 ≠> f'(g(l)) = f , (l) = 5;

therefore, (f 0 g)'(l) = f'(g(l)) ■g'(l) = 5 ∙ ∣= ∣ ≠> g(l) = 5 and g'(l) = ∣; f(u) = cot ( ≡ ) => f'(u) = -e s c 2 ( ^ ) (⅛)

55. g(x) = 5 √ x => g'(x) = ^ _ -

(τδ) =^ f '(g( 1 )) = f '( 5 ) =

csc2

= ⅛

-

⅛

csc2

(2)

=

-

⅞ ; therefore, (f 0 g)'(l) = f'(g(l))g'(l) = - ⅛ ∙ j

π ___ 4

57. g(x) = 10x2 + x + 1 => g'(x) = 20x + 1 ^ g(0) = 1 and g'(0) = 1; f(u ) = ⅛ = ⅛ ⅛

=> f'(u) =

li ^

≡ ⅛

^

f'(g(θ)) = f 'W = °; therefore, (f 0 g)'(0) = f , (g(O))g'(O) = 0 - 1 = 0

^

59. (a) y = 2f(x) => ∣ = 2f'(x) => ∣∣^ = 2f'(2) = 2 (∣) = j (b) y = f(x) + g(x) => ∣ = f'(x ) + g'(x) ≠> ∣∣ = f'(3) + g'(3) = 2τr + 5 I x=3

(c)

y= f(x)∙ g(x)

Mt w

v - fW y - g (x )

(e)

y= f(g(x)) ^

^ ∣ = f(x)g'(x) + g(x)f'(x) ^

=⅛ ⅛ dχ

- g W f 'W - W W [g(x)]2

⅛l _ dx I x=2

^

11 ^ = f(3)g'(3) + g(3)f, (3) = 3 ∙ 5 + (-4)(2π) = 15 - 8π g(2)f, (2)-f(2)g'(2) _ g(2)li [∣

(2 )(⅜) - (8)(-3) _ 22 “

37 6

∣∣= f'(g(2))g'(2) = f'(2)(-3) = ∣( -3 ) =

∣= f'(g(x))g, (x) ^

-1

I x=2 W

(g) y =

(g(x)Γ 2

^

dx “

2

1

dx I x= 2 “

^ =

-2(g(x))- 3

g, (x)

-2(g(3)Γ 3 g'(3)

∙

^

⅛

=

J√ 8

^ ¾ ) “

=

24

^√8

-2 (-4 )^ 3

∙5 = ⅛

I x=3

(h) y = ((f(x))2 + (g(x))2 ) 172 =^ ∣ = ∣((f(x))2 + (g(x))2 ) - 1 / 2 (2f(x) ■f'(x) + 2g(x) ∙ g'(x)) ⅛ L = 5 ((f (2 ))2 + (g(2))2 )^ v 2 (2f(2)f'(2) + 2g(2)g'(2)) = ∣(82 + 22 )’ 1/2 (2 ∙ 8 ∙ ∣+ 2 - 2 - (-3 ))

^ = -

61∙ b

5 3√ 1 7

l ’! :

s

= ∞s0 ^

! = -≡ 0

‰

^

= - ≡ ( τ ) = l s o t h a t∣ = ∣ ∙ f = 1 ∙5 = 5

63. With y = x, we should get ∣ = 1 for both (a) and (b): (a) y = j + 7 => ∣ = ∣; u = 5x - 35 ≠> ∣ = 5; therefore, ∣ = ∣ ∙ ∣ = ∣∙5 = l,a s expected (b) y = l + * ^

∣= -⅛ !u = ( x - l Γ

1

≠> ∣ = -(x - 1Γ 2 (1) = ≠ ⅛ ; therefore ∣ = ∣ ∙ ∣

= ⅛ ∙ ≠ ⅛ = ≠ ⅛ ψ ∙ ( i ⅛ = 2 * α ⅛ = 1. again as expected 65. y = 2 ta n ( ≡ ) ^ (a)

I x

∣ = ( 2 sec2 ^ ) (?) = f sec2 ≡

= ∣sec2 (^) = π => slope of tangent is 2; thus, y(l) = 2 tan (∣) = 2 and y, (l) = π => tangent line is

I x=l

given by y —2 = π(x — 1) => y = πx + 2 —π (b) y, = f sec2 ( —) and the smallest value the secant function can have in —2 < x < 2 is 1 => the minimum value of y' is ^ and that occurs when f = f sec2 ( 2^ ) => 1 = sec2 ( ^ ) => ± 1 = sec ( ^ ) => χ = 0.

Section 3.5 The Chain Rule and Parametric Equations 69. x = 4 cos t, y = 2 sin t, 0 ≤ t ≤ 2π

67. x = cos 2t, y = sin 2t, 0 ≤ t ≤ π =>

cos2

2t +

sin2

2t = 1 =>

71. x = 3t, y = 9t2 , —∞ < t < ∞

x2

+

y2

= 1

=> y = x2

16 cos2 1 I 4 sin2 1 ψ 16 4

ι 1

2^ _|_ ∑1 16 r 4 ~

73. x = 2t - 5, y = 4t - 7, - ∞ < t < ∞ => x ÷ 5 = 2t => 2(x + 5) = 4t => y = 2(x + 5) - 7 => y = 2x + 3

77. x = sec2 1 — 1, y = tan t, —^ < t < ∣ => sec2 1 — 1 = tan2 1 => x = y2

79. (a) (b) (c) (d)

x= x= x= x=

a cos t, a cos t, a cos t, a cos t,

y y y y

= = = =

- a sin t, 0 ≤ a sin t, 0 ≤ t —a sin t, 0 ≤ a sin t, 0 ≤ t

t ≤ 2π ≤ 2π t ≤ 4π ≤ 4π

81. Using (-1 , - 3 ) we create the parametric equations x = —1 + at and y = —3 + bt, representing a line which goes through (-1 , —3) at t = 0. We determine a and b so that the line goes through (4, 1) when t = 1. Since 4 = —l + a = > a = 5. Since 1 = —3 ÷ b ≠> b = 4. Therefore, one possible parameterization is x = —1 + 5t, y = - 3 - 4t, 0 ≤ t ≤ 1. 83. The lower half of the parabola is given by x = y2 ÷ 1 for y ≤ 0. Substituting t for y, we obtain one possible parameterization x = t2 ÷ 1, y = t, t ≤ 0.

77

78

Chapter 3 Differentiation

85. For simplicity, we assume that x and y are linear functions of t and that the point(x, y) starts at (2, 3) for t = 0 and passes through (—1, - 1 ) at t = 1. Then x = f(t), where f(0) = 2 and f(l) = -1 . Since slope = ^ = ⅛ = —3, x = f(t) = —3t + 2 = 2 —3t. Also, y = g(t), where g(0) = 3 and g(l) = —1. Since slope = ^ = ⅛

= - 4 . y = g(t) = - 4 t + 3 = 3 - 4t.

One possible parameterization is: x = 2 —3t, y = 3 - 4t, t ≥ 0. 87. t = ≡

x = 2 c o s5 = √ 2 ,y = 2 s in 5 = v M

=> ∣∣ < ⅛

= - 2 sin t, ⅛ = 2 cos t

= —cot | = - 1 ; tangent line is y - √ ^ = - 1 (x - χ ∕2 ) or y = - x + 2 y ^ ; ^ dyz∕dt _ dx/dt

_

csc2 t _ -2 s in t

_

89. t = J => ∙ = ⅛ , y = ^

1 2 sin3 1

.

d2 y I d√l

= l.J = ⅛

_

. V

π

* J = S = ⅛

91. t = —1 ≠> χ = 5, y = 1; ⅛ = 4t, ⅛ = 4t3 ≠ or v — x —4 , ^- =

v —1 — 1 ∙ (x — 1

ur y — A

1

dt

^l

⅛

^ = g

g | ι, ι = ^

=-c o tt

= csc2 1

dx/dt

= ⅛ = t2 => ⅛ ∣^

= —sin t,

4t

2

⅛ 2I

(

S [ , = -2

= ( - 1)2 = 1; tangent line is

= 1

dx 11 =

= cos t 4

^ ∣ = - cot f = 0 ; tangent line is y = 2; ⅛ = csc2 1 ≠

= h tangent line is

= - J r 1' 1 ≠

=Φ∙ ⅛ = -^-- — —— —

93. t = y => x = cos | = 0, y = 1 + sin | = 2;

= ⅛

x

y - ∣= l . ( x - l ) o r y = x + b ⅛ = - l t -W ≠ S = S

y

⅛ = g

2

1

=

= —cot t

= -c sc 3 t 7

⅛ = ⅛

§ |

I t= 2

= -1 I t= 2

95. s = A cos (2πbt) => v = ∣ = - A sin (2πbt)(2πb) = -2πbA sin (2πbt). If we replace b with 2b to double the frequency, the velocity formula gives v = —4πbA sin (4πbt) => doubling the frequency causes the velocity to double. Also v = -2πbA sin(2πbt) => a = ^ = - 4 π 2 b2 A cos(2πbt). If we replace b with 2b in the acceleration formula, we get a = - 16π2 b 2 A cos (4πbt) => doubling the frequency causes the acceleration to quadruple. Finally, a = —4π 2 b2 A cos (2πbt) => j = ^ = 8π3 b3 A sin (2πbt). If we replace b with 2b in the jerk formula, we get j = 64π3 b3 A sin (4πbt) => doubling the frequency multiplies the jerk by a factor of 8. 97. s = (1 ÷ 4t)1∕ 2 => v = ^ = ∣(1 + 4t)" 1∕ 2 (4) = 2(1 + 4t)" 1∕ 2 => v(6) = 2(1 + 4 ∙ 6)^ 1∕ 2 = ∣m/sec; v = 2(1 + 4t)" 1∕2 => 99. v proportional to ^ =

—⅛

-

^

=

-

a

= ⅛ = - ∣∙ 2(1 + 4t)^3 ∕ 2 (4) = -4(1 + 4t)~3∕ 2 ^ a(6) = -4(1 + 4 - 6)" 3 ∕ 2 = - ⅛ m∕sec2

=> v = ^ for some constant k ≠ ∙ ^ = - ^ . Thus, a = ∣ = ^ ∙ ∣ = ^ ∙ acceleration is a constant times ⅛ so a is inversely proportional to s2 .

^ (?) ^

101. T -. 2π either f'(g(l)) = 0 or g, (l) = 0 (or both) => either the graph of f has a horizontal tangent at u = g(l), or the graph of g has a horizontal tangent at x = 1 (or both).

Section 3.6 Implicit Differentiation

79

105. As h → 0, the graph of y = ⅛ ⅛ h ⅛ approaches the graph of y = 2 cos 2x because ^limθ S ⅛ ⅛ z 2 ⅛ = A (s jn 2χ) = 2 cos 2x.

107. ⅛ = c o s t a n d ⅛ = 2 c o s 2 t ≠ ≠> 2

cos2

£ = g

1 - 1 = 0 => cos t = ± ^

y = sin 2 (5) = 1 ≠

= ¾ ^ = ⅛

^

;the n ⅛ = 0 => ⅛

11

=0

≠> t = J , j , j , j . In the 1st quadrant: t = ^ => x = sin J = ^

and

^ y , l j isthe point where the tangent line is horizontal. At the origin: x = 0 and y = 0

=> sin t = 0 => t = 0 or t = π and sin 2t = 0 => t = 0, ∣, π, y ; thus t = 0 and t = π give the tangent lines at the origin. Tangents at origin: ^

= 2 => y = 2x and ^

= - 2 => y = —2x

109. From the power rule, with y = x 1∕ 4 , we get ^ = J x 3 ∕ 4 . From the chain rule, y = √ ^ ∕x

i

≡=≠ ξ ∙ s ∣Λ =⅛ ∙ ⅛ =i--'^n ∙s-ee" ,∙

111. (a)

dg/dt

(b) f = 1.27324 sin 2t + 0.42444 sin 6t + 0.2546 sin 10t + 0.18186 sin 14t (c) The curve of y = j approximates y = ∣ the best when t is not - π , - 5 ,0 , 5, nor π.

3.6 IMPLICIT DIFFERENTIATION 1. y = X ^ 3. y = 3^

⅛ = 2χ5M = (2x)V3 =φ £ = i (2X) - 2 ∕3 ∙ 2 = ^

5. y = 7 √ x + 6 = 7(x + 6)1∕ 2 ≠> ⅛ = 5 (x + 6)~ 1∕ 2 = ⅛

7. y = (2x + 5 )- 1∕ 2 ≠

⅛ = - ∣(2x + 5)^ 3 ∕ 2 ∙ 2 = -(2 x + 5 p 3 ∕ 2

df/dt

80

Chapter 3 Differentiation

9. y = x (x2 + I ) 1/ 2 ≠- y , = x ∙ ∣(x2 + 1)

11.

= 7√ P = t 2∕7 ^

s

1^2 (2x)

+ (x2 + 1)1^2 ∙ 1 = (x 2 + 1) 1^2 (x2 + x2 + 1) = ^ X

⅛ = ∣r 5 ∕ 7

13. y = sin ((2t + 5)^ 2 ∕ 3 ) =+ ∣ = cos ((2t + 5)~2 ∕ 3 ) ∙ ( - j ) (2t + 5)^ 5 ∕ 3 ∙ 2 = - j (2t + 5)^ 5 ∕ 3 cos ((2t + 5)~ 2 ∕ 3 ) 1,

¢ )= ^ = (,-../.)^ - .

=H

W

I

- . . ∕. ) - '- H

X

-.∕.) = ^

=^

17. h(0) = 3√ 1 + co s(2 0 ) = ( 1 + cos 20)1∕ 3 =⅛> h'(0) = ∣(1 + cos 2 0 Γ 2 ∕ 3 ∙ ( - s in 20) ∙ 2 = - j (sin 20)(1 + cos 20)^ 2 ∕ 3 19. x 2 y + xy 2 = 6: Step 1: (x 2 ∣ + y ∙ 2x) + (x ∙ 2y ∣ + y 2 ∙ 1) = 0 Step 2:

x2 ∣ + 2xy ⅛ = -2 x y - y2

Step 3:

∣ (x2 + 2xy) = -2 x y - y2

S 4∙ a tteen P ^-

⅜ ~y2 x ⅛ y

^ =

dx

21. 2 x y + y2 = x + y: Step 1:

(2 -J+ ⅛ )+ ⅛ J = ∣ +S

Step 2: Step 3:

⅛ (2x + 2 y - l ) = l - 2 y

Step 4:

dy _ dx

1 - 2y 2x ÷ 2y - 1

23. x 2 (x - y)2 = x 2 - y 2 : x 2 [2(x - y) (1 - y

Step 2:

- 2 x 2 (x - y) ⅛ + 2y ⅛ = 2x - 2x2 (x - y) - 2x(x - y)2

Step 3:

I [-2 x 2 (x - y) + 2y] = 2x [1 - x(x - y) - (x - y)2 ]

Step 4:

dy _ dx _

25

i j ,

∣+ (x - y)2 (2x) = 2x - 2y ∣

Step 1:

v”

2

= — + x+l ^

2x [1 - x(x - y) - (x - y)2 ] _ -2 x 2 ( x - y ) + 2y

x [1 - x(x - y) - (x - y)2 ] _ y - x 2( x - y )

x (1 - x2 + xy - x2 + 2xy - y2 ) x2 y - x 3 + y

x - 2x3 + 3x2 y - xy2 x2 y - x3 + y

2v ^dxς = ≥ ±( xl h+ l⅛ ± __ )2

^

2

(x + l) 2

27. x = tan y ≠> 1 = (sec2 y) ⅛ ≠> ⅛ = ⅛

⅛ ⅛ dx

=

__ 1 y ( x + l) 2

= cos2 y

29. x + tan(xy) = 0 = + 1 + [sec2 (xy)] (y + x ∣) = 0 ≠ ∙ xsec 2 (xy) ^ = - 1 - y sec2 (xy) + _

—1______ y x sec2 (xy) x

—cos2 (xy) x

y x

—cos2 (xy) —y x

n 31. y s i Q ) = ι - χ y => y [ ∞ s O ) ∙ ( - D p ∙ ⅛ ] +

S [ - 7 ∞> ( 0 + ∙ ∙ G ) + * ] = - y ^ ≡

s in

Q ) ∙∣= -

χ

= - ⅛ = ,( ,^ ( ,) ( . =

⅛ -y

, a (i)l(.)..,

⅛ = ⅛

w

d

Section 3.6 Implicit Differentiation 33.

f

>∕≈ t , V = l

'«

s

l"

+ h -" -⅛ = 0 ≈

⅞ [j V

⅛

≠

S = - ^

35. sin (r0) = ∣ => [cos (r0)] (r + 0 ⅛ ) = 0 => ^ [0 cos (r0)] = - r cos (r0) ≠

= - ^

^ = ¢

^

= “ 5∙

cos (r0) ≠ 0 37. x 2 + y 2 = 1 4 A Ξ →

yV

∕∕

_

—

2x + 2 y ∕ = 0 => 2 y ∕ = - 2 x => ^ = / = - ^ ∞ w to find ⅛ . ⅛ ( / ) = ⅛ ( - ; )

y ( - i )2 + χ ∕ — _ y

39. y 2 = x2 + 2x 4 ⅛ dx2

=

√' v

=

y

2

sMi n∏ cuee √ y

=

-

_ i

2

2yy' = 2x + 2 => / = ^

⅛ d

→

y

=

i

= y- " y= J

-

y

=

3

∣i ; then y" =

y

y

(x + l ) y ' y2

3

= ^

= 43 y

^

j

^

y2 - ( χ + 1)2 y3

41∙ 2 √ y = χ - y => y

1 /2 y'

= 1- /

≠- / (y 1∕2

+

ι) = ι => ⅛ = / = ^ ⅛

= √ ^ T ^w e

can

differentiate the equation y, (y^ 1 ∕ 2 + 1) = 1 again to find y": y' (—∣y - 3 ∕ 2 y') + (y - 1 ∕ 2 + 1) y" = 0 1 f

1j

43.

dx2

2 ιy J y

y

V

y

- 3 ∕2

(y- 1∕2 + 1)

y

2y3∕ 2 (y- 1∕2 + 1 ) 3

2 (1 ÷ x ∕y ) 3

x3 + y3 = 16 => 3X2 + 3y2 y' = 0 => 3y2 y' = —3x2 => / = - ^ ; we differentiate y2 y, = —x 2 to find y": y2 y" + y' [2y - y'] = - 2 x ≠> y2 y" = - 2 x - 2y [y']2 => y" = --------/ ⅛

≠

y ,

= +

+

⅛ ∣ ,221 = = ¾ ^ =

45. y 2 + x 2 = y4 - 2 x at ( - 2 ,1 ) and ( - 2 , - 1 ) =► 2y ∣ + 2x = 4y3 ∣ - 2 => 2y ⅛ - 4y 3 ⅛ ≠∙ g P y - 4 Λ = - 2 - 2 χ ^

47. x2 + xy - y 2 = 1 4

1

S = ⅛ ⅛ ≠∙ ⅛ . , , = -

- ≡

= - 2 - 2x

∣ h ., = >

2x + y + xy' - 2yy' = 0 ≠- (x - 2y)y' = - 2 x - y => y' = ∣^ y ;

(a) the slope of the tangent line m = ∕ ∣ (2,3) = 5 => the tangent line is y — 3 = 5 (x — 2) ≠>y =

5x - 5

(b) the normal line is y —3 = — ¾(x —2) => y = —7 x + γ 49. x 2 y 2 = 9 => 2xy2 + 2x 2 yy' = 0 => x2 yy' - - x y 2 => / (a) the slope of the tangent line m = ∕ ∣ (_1 3)

=

-

-

χ5

x | H 3 ) = 3 ≠ , the tangent line is y - 3 = 3(x+ 1)

=> y = 3x + 6 (b) the normal line is y —3 = —∣(x + 1) => y = —∣x + ∣ 51. 6x 2 + 3xy + 2y2 + 17y - 6 = 0 => 12x + 3y + 3 x ∕ + 4 y ∕ + 17y' = 0 => y'(3x + 4y + 17) = - 12x — 3y (a) the slope of the tangent line m = ∕ ∣ ( _w =

3x + ^ / ( 7

1

= f => the tangent line is y —0 = ∣(x + 1)

y = vx + 7 (b) the normal line is y —0 = — g (x + 1) => y = —g x — g 53. 2xy + π sin y = 2π ≠ ∙ 2xy' + 2y + π(cos y)y' = 0 =^ y'(2x + τr cos y) = - 2 y =⅛> y' = 7 Γ ⅛ (a) the slope of the tangent line m = y '∣ ^^ =

2χ

+~π cos y

∣ ∣, ∣=

-

1 ^

**ι e tangent line is

;

81

Chapter 3 Differentiation

82

y - f = - f ( x - l ) => y = - f x + π (b) the normal line is y — ∣= ^ (x - 1) => y = ; x - ∣+ ∣ 55. y = 2 sin(πx —y) => y' = 2 [cos (πx —y)] ∙ (π —y') => y'[l + 2 cos(πx —y)] = 2τr cos (πx —y) → = 2π ∞g⅛L∑2> . 1 ÷ 2 cos (πx - y) ’

(a) the slope of the tangent line m = y'∣ (10) = τ⅛777⅛7⅛ I ’

1

-

Γ z COS (7ΓX

= 2π ≠ the tangent line is

J ' I (1 θ )

y —0 = 2π(x — 1) => y = 2πx —2π (b) the normal line is y - 0 = - ^ (x - 1) => y = - ^ ÷ ^ 57. Solving x2 ÷ xy + y2 = 7 and y = 0 => x2 = 7 => x = ± √ z7 => ( - √ ζ7,0j and ( √ z7 ,0j are the points where the curve crosses the x-axis. Now x2 + xy + y2 = 7 => 2x + y + xy, ÷ 2yy' = 0 => (x + 2y)y, = —2x —y => y' = - Γ⅛y => m = - ^ = - 2 and the slope at ( √ ^ , θ) is => the slope at ( - ᅟ ∕7 , θ j is m = m= - ^

= - 2 . Since the slope is - 2 in each case, the corresponding tangents must be parallel.

59. y4 = y2 - x2 => 4y3 y' = 2yy' —2x => 2 (2y3 - y) y, = —2x => y' = ( ^ ’Ψ ) is

x

is

r ⅛ I^ ^

= ψ ⅛

= ⅛

-J 2y3 ; the slope of the tangent line at

= -1 ; the slope of the tangent line at ( ¾ ∣)

= ⅛

— λ ∕3

I

4- 2 ^

>-*⅛,I) FT

V

3

61. y4 - 4y2 = x4 - 9x2 ≠> 4y3 y' - 8yy' = 4x3 - 18x => y' (4y3 - 8y) = 4x3 - 18x => y' = ^ = ⅛ ⅛ ⅛ = m i (-3 ,2 )ι

r7 ^

=

^3.¾

= ≡ ¾ ^ = - f : ( - 3 ,- 2 ) : m = f :(3,2): m = f :(3 ,-2 ): m = - f

m

63. x2 - 2tx + 2t2 = 4 =► 2x ⅛ - 2x - 2t ⅛ + 4 t = 0 => (2x - 2t) ⅛ = 2x - 4t => $ = ^ 2v3 - 3t2 = 4 => 6v2 & - 6t = 0 ⅜ zy Ji oy dt oι u

⅛ = dt

6y2

= ¾⅛ ;

- X ∙ thus ⅛ - ⅛^! - ⅝⅜) - J i t L ∙ t - 2 — y2(x_2t) , ι — y2 , ιnus dχ — dx/dt —

=» x2 - 2(2)x + 2(2)2 = 4 ≠> x2 - 4x + 4 = 0 ≠> (x - 2)2 = 0 =► x = 2; t = 2 =► 2y3 - 3(2)2 = 4 ≠> 2y3 = 16 ≠> y3 = 8 => y = 2; therefore ^ j = (2⅞ ÷ ¾ ) = 0 65. x + 2x3∕ 2 = t 2 + t ^ ≠

¾ √ ∏ ^ + y ( ∣) < t + ι r 1' i + 2 √ 5 + 2t ( l y - v ≈ ) ⅛ = 0 ^

¢ ^ + ^ + 2 7 5 + ( ^ ) ^ = 0

V√t+ι + ^ ^

ξ∕^⅛ 7y⅛ ^7τ ’thus

/

^

⅛ + 3x1∕ 2 ⅛ = 2 t + l ≠> (l + 3x1∕ 2 ) ⅛ = 2 t + l =s> $ = ⅛ ι y √ t + T + 2 t ^ = 4

S

= ⅛

57⅛τ 2√y ≠>

dt

-y √ y -4 y √ t+ ^ i

i

P

dt

-p = ^ τy

∖

^

= ∙ ⅛ x + 2 x » = 0 ⅛ x ( l + 2 x ∙ ∕≈ ) = 0 =+ x = 0 ;t = 0 (

=> y √ 0 ÷ 1 ÷ 2 (0 )√ y = 4 => y = 4; therefore

I

-4 √ 4 -4 (4 )√ 0 + T

=

∖

=

_5

1=0

67. (a) if f(x) = ∣x2∕ 3 —3, then f'(x) = x- 1 ^3 and f"(x) = - 1 x- 4 / 3 so the claim f"(x) = x- 1 ∕ 3 is false (b) if f(x) = ⅛ x5∕ 3 - 7, then f'(x) = ∣x2∕ 3 and f"(x) = x- 1 / 3 is true (c) f"(x) = x- 1 ∕ 3 => f'"(x) = - 5 x- 4 ∕ 3 is true (d) if f'(x) = ∣x2∕ 3 + 6, then f"(x) = x- 1 ∕ 3 is true

Section 3.7 Related Rates 69. x2 + 2xy - 3y2 = 0 ≠> 2x + 2xy' + 2y - 6yy' = 0 => y'(2x - 6y) = - 2 x - 2y => y' = ^⅞⅛ => the slope of the = 1 => the equation of the normal line at (1,1) is y - 1 = - l ( x - 1) tangent line m = y '∣ ,l n = ⅛ ⅛x ∣ (1 υ

=> y = - x + 2. To find where the normal line intersects the curve we substitute into its equation: x2 + 2x(2 —x) —3(2 - x)2 = 0 => x 2 + 4x - 2x 2 —3 (4 —4x + x 2 ) = 0 => - 4 x 2 + 16x — 12 = 0 => x2 —4x + 3 = 0 => (x —3)(x — 1) = 0 => x = 3 and y = —x + 2 = —1. Therefore, the normal to the curve at (1,1) intersects the curve at the point (3, —1). Note that it also intersects the curve at (1,1). 71. y 2 = χ => ^ = ^ . If a normal is drawn from (a, 0) to (xι, yι) on the curve its slope satisfies ^ ∈^ = —2y 1 => y 1 = —2yι(xι —a) or a = xι ÷ ∣. Since Xi ≥ 0 on the curve, we must have that a ≥ ∣. By symmetry, the two points on the parabola are (x i , λ∕x Γ ) and (xι, - χ ∕ x i ) . For the normal to be perpendicular, (⅛ ) (⅛ ) = ~1 ^

( ⅛

=

1

=^ x ι = ( a - x ι ) 2 ≠> x ι - ( x ι + ∣- x 1 ) 2 => x 1 = i a n d y 1 = ± ∣.

Therefore, Q , ± ∣) and a = ∣. 73. xy3 + x2 y = 6 ≠> x (3y 2 ⅛ ) + y 3 + x2 ⅛ + 2xy = 0 => ⅛ (3xy2 + x 2 ) = - y 3 - 2xy => ⅛ = ^ = ^

r

⅛

⅛ ⅞ ¾ J ais°, xy3 + x2 y = 6 => x (3y2 ) + y 3 ^ + x2 + y ^2x ⅛ j = 0 => ^ ( y 3 + 2xy) = -3 x y 2 - x 2

-

|

=

“ y⅛⅛y ’ t hu s ^ appears to equal ^ . The two different treatments view the graphs as functions

symmetric across the line y = x, so their slopes are reciprocals of one another at the corresponding points (a, b) and (b, a). 75. x4 ÷ 4 y 2 = 1: => y = ± i √ T ~ √ (a) y2 = ⅛

* £ = ÷ I ( 1 - XT , ' 2 ( - ⅛ 3 ) = ( 7 ⅛ ^ differentiating implicitly, we find, 4x3 ÷ 8y -4 x 3 _

< ⅛ _

⅛

DX

-4X3 8(±I √ Γ 7 )

_

= 0

±x3

(∣ -√) V 2 -

3.7 RELATED RATES 1.

A =

π r2

^

^

=

2π

r⅛

3. (a) V = πr 2 h => ^ = πr 2 ⅛

(b) V = 7rr2 h ^> ^ = 2τrrh ⅛

(c) V = 7rr2 h ≠> ⅛ = πr 2 ⅛ + 2πrh ⅛ (b) ^ = - ∣amp/sec

5. (a ) ⅛ = 1 volt/sec W

dV _ κp ∕ d l i τ ∕d R U ∕÷ U d Γ ∕ d f -

^

.

dR _ 1 ∕d V p dl d F -T k d Γ ~ κ dJ

.

dR _ 1 ∕d V V d∏ d Γ -ik d Γ ^ T d J

83

84

Chapter 3 Differentiation (d) ^ = ∣[1 — y (— 1)]

7.

=

(2) (3)

(a) s = √ x 2 + y 2 = (x 2 + y 2 ) v 2 2

(b) s = √ Ξ ^ = ( x

=

2 °hms∕sec, R is increasing

=► ⅛ = 7 ⅛ 7 ⅛

+ y2 ) ιz 2

^ ^ =

⅛ ⅛

7

+ ^⅛

⅛

7

(c) s = √ x 2 + y 2 => s2 = x 2 + y 2 => 2s ^ = 2x ^ + 2y ⅛ => 2s - 0= 2x⅛ + 2y 9.

(a) A = ∣ab sin 0 => ^ = ∣a *,

cos

(b) A = ∣ab sin 0 ≠>

^^

⅛ => ⅛ = - * ⅛

^ = ∣ ab cos θ ^ + ∣b sin θ ^

(c) A = Izab sin 0 => ⅜uι = ∣ ab cos 0 ⅛ + Iz b sin 0 ⅜ + ∣ a sin 0 ⅛ z uι uι z uι 11. Given ^ = - 2 cm∕sec, ^ = 2 cm∕sec, £ = 12 cm and w = 5 cm. (a) A = ^w => ^ = ^ ⅛ ÷ w ^ => j

= 12(2) + 5 (-2 ) = 14 cm 2 ∕sec, increasing

(b) P = 2^ + 2w => ^ = 2 ^ + 2 ^ = 2 (-2 ) + 2(2) = 0 cm∕sec, constant (c) D = √ =

W2

+ ^

W2

(

-

^√⅛ ÷u^ ~

Z2

+ ^

=> ® = 1 ( W 2 + ^ ) - V 2 ( 2 W ^

+ 2^

)

^

^ = ^ ⅛

∏ cm∕sec, decreasing

13. Given: ^ = 5 ft∕sec, the ladder is 13 ft long, and x = 12, y = 5 at the instant of time (a) Since x 2 + y 2 = 169 => ⅛ ~ ~ y ⅛ ~

-

( y ) (5) = “ 12 ft∕sec, the ladder is sliding down the wall

(b) The area of the triangle formed by the ladder and walls is A = ∣xy => ¾^ = ( 2 ) ( x ⅛ + y ⅛ ) ∙ ^ is changing at ∣[12(-12) + 5(5)] = - y (c) cos 0 = ⅞ => -si∏ 0 f = ⅛ ∙ ⅛ ^

le a r e a

= -59.5 ft2 ∕sec. f = - ⅛

∙ ⅛ = - (?) (5) = -

1ra d 7 se c

15. Let s represent the distance between the girl and the kite and x represents the horizontal distance between the girl and kite => s2 = (300)2 + x 2 => 5 = ; 5 = ^ ^ = 20 ft∕sec. 17. V = 1 7rr2 h, h = ∣(2r) = ⅜ => r = ^ (a) ‰ (b

)

r

= (τ⅛ )(

1 0 >=

⅛

=> V = ∣π ( ^ ) 2 h =

=> ⅛ = ⅜ ^ f

^

≈ 0.1H 9 m∕sec = 11.19cm∕sec

= T ^ I = 5 ⅛ = K ⅛ ) = ⅛ ≈ θ∙1492 m/sec = 14.92 cm/sec

2 19. (a) V = f y (3R - y) ^

⅛ = f [2y(3R - y) + y2 ( - l) ] ⅛ ^

⅛ = [f (6Ry - 3y2 )]^ 1 ^

^

at R = 13 and

y = 8 we have ⅛ = 1⅛ ( - 6 ) = ⅛ m/min (b) The hemisphere is on the circle r 2 + (13 - y)2 = 169 => r = √ 2 6 y —y2 m 2 1/2 ≠> - = 13 Y (26v - V2 Γ 1^2 (26 - 2 V) ^ Ʃ => - = ⅛ ≠> Z ~ 2 (c) rr = (26v - v ) y

=

2 ^

dt

2

dt

dt

∕2 6 y - y

dt

dtly=8

=

13 —s

∕^ n

√26-8 —64 V24π7

"l/min

21. I f V = l3π r 3 ,’ r = 5,and ^dt = 100πft 3 ∕min,’ then ⅛ = 4πr 2 ⅛ => *dt = 1 ft/min. ThenS = 4πr2 => fdt ’ dt dt =

87 rr dt

=

^ )(1 )

=

4θπ ft2 ∕min, the rate at which the surface area is increasing.

23. Let s represent the distance between the bicycle and balloon, h the height of the balloon and x the horizontal distance between the balloon and the bicycle. The relationship between the variables is s2 = h 2 ÷ x 2 ^ S = ; H + x ⅛ ) ≠ 5 = ⅛ (68(1) + 51(17)] = 11 ft∕sec. 25. y = QD^ 1 ≠> ^ = θ

-1

^ - Q θ ^ 2 ⅛ = ⅛ (°) - tan 0 = γ = x => sec2 0 ^ = ⅛ => ^ = cos2 0 ^ . Since ⅛ = 10 m/sec and cos2 0|I=3 = p ⅛ = 9 ⅛ = ⅛ , we have f ∣ x=3 = 1 rad/sec. 29. The distance from the origin is s = √ x 2 + y2 and we wish to find ^ ∣ (512) = 1 ( + + + ) ^ (2

χ

⅛

+ 2 y

⅛ )∣

=

(5,12)

(5)(-1) + (12)(-5) ∖ ∕25 + 144

—5 m/sec

31. Let s = 16t2 represent the distance the ball has fallen, h the distance between the ball and the ground, and I the distance between the shadow and the point directly beneath the ball. Accordingly, s + h = 50 and since the triangle LOQ and triangle PRQ are similar we have 1 =5δ¾ ^ h = 5 0 - 1 6 t 2 andI = ^ ⅛ ⅛ = ^ - 3 0 ^

f = - ^

=> ⅛

= -1 5 ∞ ft∕s e c ,

33. The volume of the ice is V = ^3 πr 3 -3 ⅛τr43 =+ ¾= 4πr 2 $dt =+ ⅜dt II r=6 = 72π ~⅛ in./min when ^dt = —10 in3 ∕min, ’the dt thickness of the ice is decreasing° at in2 ∕min,

= -y

in∕min. The surface area is S = 4πr2 => ^dt = 8πr ⅞ => dt

72π

dt I r=6

= 48π (∖ ⅛ ) 72π /

in2 ∕min.

the outer surface area of the ice is decreasing at y

=+ sec2 ^ ⅛ = - ^ ⅛ => ⅛ = ~3 ⅛ ^ ^ ^ ∙

35. When x represents the length of the shadow, then tan 0 = γ

We are given that ^ = 0.27o = ^55 rad/min. At x = 60, cos 0 = ∣ ≠> I⅛I = ∣ T ⅛ ld tl

80

I

L Γ ∣ d t

⅜

, = ⅛ ft/min ≈ 0.589ft/min ≈ 7.1 in,∕min. 16

= ⅛ δ andsecM )

37. Let x represent distance of the player from second base and s the distance to third base. Then ⅛ = —16 ft/sec (v a)7 s2 = x 2 + 8100 =+ 2s ⅛ = 2x ⅛ =+ ⅛ = *s ⅛ . When the rplayer is 30 ft from first base,’ x = 60 dt dt dt dt √ =+ s = 3O√13 and ⅛ = ^

(-1 6 ) - ^ ≈

(b) co s0 1 = y =+ - s i n 0 1 ⅛ = - ^ - ⅛ =+ ½ = ^t =

∕p∖ w

z⅛F

9°

* d t,

)

∖ v l3 ∕

∙⅛

=> ⅛ = ⅛

90 ∕x ∖ ∕dx∖ _ (s 2 ∙f) ’ ∖ s ∕ ‘ ∖ d t∕ ~

⅛

z

s

θ5

( + ‰ ) (-1 5 ) = - 1 ⅛

= (ra δ δ ) ⅛ ^

= ≡ ' ⅛ ∙ Therefore, x = 60 and s = 3 0 √ 1 3

= ⅛ r a d ∕sec; sin 02 = ^ => cos 02 ⅛ ≈ - ⅛ ∙ ⅛ =+ ½ = ^ >

Therefore, x = 60 and s = 30√T 3 => ^

ds _ ⅛ — 90 dt “ s2 si∏0ι ’ dt ^

= ⅛

∙⅛

(30∖ ∕13) (60)

-8.875 ft/sec

∕90∖ ∕dx∖ _ ∖ s2 ∕ ∖ d t∕ “

?

= ⅛

dt

s

dt

dt

s cos ^2

⅛ dt

= ∣ rad/sec.

/ 9 0 _ \ dx ∖ x2 + 810 0 ∕ dt

^

∙ J

(i) ( 9

= ( ⅛

χ

ι∙ →

0

⅛ dt

= m

(5 )

⅛ = ⅛rad/scc

3.8 LINEARIZATION AND DIFFERENTIALS 1. f(x) = x3 - 2x + 3 =+ f'(x) = 3x2 - 2 =+ L(x) = f'(2)(x - 2 ) + f(2) = 10(x - 2 ) + 7 =+ L(x) = 10x - 13 at x = 2 3. f(x) = x + i =+ f , (x) = 1 - χ - 2 +

L(x) = f(l) + f'(l)(x - 1) = 2 + 0(x - 1) = 2

5. f(x) = x2 + 2x =+ f'(x) = 2x + 2 =+ L(x) = f'(0)(x - 0 ) + f(0) = 2(x - 0) + 0 =+ L(x) = 2x at x = 0 7. f(x) = 2x2 + 4x - 3 =+ f'(x) = 4x + 4 =+ L(x) = f '( - l ) ( x + 1 ) + f ( - l ) = 0(x + 1 ) + ( - 5 ) =+ L(x) = - 5 at x = - 1

86

Chapter 3 Differentiation

9. f(x) = 3√ x = x 1∕ 3 =+ f'(χ) = (1) χ-2∕3 =φ L(X) = f , (8)(x - 8) + f(8) = ⅛ (x - 8) + 2 =+ L(x) = ⅛ x + ∣atx = 8 11. f(x) = sin x =+ f'(x) = cos x (a) L(x) = f'(0)(x - 0 ) + f(0) = l(x - 0) + 0 =+ L(x) = x at x = 0 (b) L(x) = f'(π)(x - 7Γ) + f(π) = ( - l)(x - π) + 0 =+ L(x) = π - x a tx = π

13. f(x) = sec x =+ f'(x) = sec x tan x (a) L(x) = f'(0)(x - 0 ) + f(0) = 0(x - 0) + 1 =+ L(x) = 1 at x = 0 (b) L(x) = f ( - f ) ( x + f ) + f ( - f ) = - 2 √ 3 (x + ≈) + 2 + L(x) = 2 - 2 √ 3 (x + f ) at x = —j

15. f'(x) = k (l + x)k ~1 . We have f(0) = 1 and f'(0) = k. L(x) = f(0) + f'(0)(x - 0) = 1 + k(x - 0) = 1 + kx 17. (a) (1.0002)5° = (1 + 0.0002)5° ≈ 1 + 50(0.0002) = 1 + .01 = 1.01 (b)

3√L 009

= (1 + 0.009)1∕ 3 ≈ 1 + (J ) (0.009) = 1 + 0.003 = 1.003

19. y = x3 - 3 λ ∕x = x3 - 3x1∕ 2 =+ dy == (3x2 - ∣x- 1 ∕ 2 ) dx =+ dy = ^3x2 - ^ ^ dx

21

yv

= ⅛ ι+χ2

dv Λ!1→JX≥Λ daxx = ^ 2 ⅛ daxx α y = f( 2E1+ (i+χ2)2 ) (i + x 2)≈

23. 2y3∕ 2 + xy - x = 0 =+ 3y1∕ 2 dy + y dx + x dy - dx = 0 =+ (3y1∕ 2 + x) dy = (1 —y) dx =+ dy = 5 ^ 7 dx 25. y = sin (5√ x) = sin (5x1∕ 2 ) =+ dy = (cos (5x1^2 )) (j x 1∕ 2 ) dx =+ dy = ^ ^ ^ ^ dx

27. y = 4 tan ( y ) =+ dy = 4 (sec2 ( y ) ) (χ 2 ) dx =+ dy = 4x2 sec2 ( 2θ dx

29. y = 3 esc (1 —2^∕x) = 3 esc (1 —2x1^2 ) =+ dy = 3 (—esc (1 —2x1∕ 2 )) cot (1 —2x1∕ 2 ) (—x^ 1∕ 2 )dx =+ dy = - ^ esc (1 —2^∕x) cot (1 —2 √ x ) dx 31. f(x) (a) (b) (c)

= x2 + 2x, x0 = 1, dx = 0.1 =+ f'(x) = 2x + 2 ∆ f = f(x0 + dx) - f(x0 ) = f(l.l) - f(l) = 3.41 - 3 = 0.41 df == f'(x 0 ) dx = [2(1) + 2](0.1) = 0.4 ∣ ∆ f - d f∣= ∣ 0.41 - 0.4∣= 0.01

Section 3.8 Linearizations and Differentials 33. f(x) (a) (b) (c)

= x3 - x, xo = 1, dx = 0.1 =4 f'(x) = 3x2 - 1 ∆ f = f(x0 + dx) - f(x0 ) = f(l.l) - f(l) = .231 df = f'(x 0 )dx = [3(1)2 - 1](.1) = .2 .2 3 1 - ∙2 ∣= .031 ∣ ∆ f - d f ∣= ∣

35. f(x) = x 1 , xo = 0.5, dx = 0.1 =4 f'(x) = - x 2 (a) ∆ f = f(x0 + dx) - f(x0 ) = f(.6) - f(.5) = - ∣ (b) df = f(x 0 )dx = (-4 ) (⅛ ) = - j (c) ∣ ∆ f - d f ∣= ∣ - ∣+ ∣ ∣= ⅛ 37. V = ∣πr 3 =4 dV = 4τrr2 dr 39. S = 6x2 => dS = 12x0 dx 41. V = τrr2 h, height constant =4 dV = 2πroh dr 43. Given r = 2 m, dr = .02 m (a) A = ττr2 => dA = 2πr dr = 2π(2)(.02) = .08π m2 Φ) (⅛ )(1 0 0 % ) = 2% 45. The volume of a cylinder is V = τrr2 h. When h is held fixed, we have ^ = 2τrrh, and so dV = 2πrh dr. For h = 30 in., r = 6 in., and dr = 0.5 in., the volume of the material in the shell is approximately dV = 2πrh dr = 2π(6)(30)(0.5) = 180π≈ 565.5 in3 . 47. V = πh 3 =4 dV = 3πh2 dh; recall that ΔV ≈ dV. Then ∣ Δ V ∣≤ (1%)(V) = ¾ ^ =4 ∣ 3πh2 dh∣≤ ¾ ^

=4 ∣ dV∣≤ ¾ ^

=4 ∣ dh∣≤ ⅛ h = (∣%) h. Therefore the greatest tolerated error in the measurement

of his ∣%. 49. V = 7rr2 h, h is constant =4 dV = 2-τrrh dr; recall that ΔV ≈ dV. We want ∣ Δ V ∣< - ^ V =4 ∣ dV∣< ^ ^ 4

I2τrrh dr∣≤ =⅛ =4 ∣ dr∣≤ ⅛

51. W = a + ∣= a + bg

1

= (.05%)r =4 a .05% variation in the radius can be tolerated.

=4 dW = - b g

2

dg = - ψ g =4 ^ ⅛ = 7 7 ⅛ γ = ( ⅛ ) 2 = 37.87, so a change of

gravity on the moon has about 38 times the effect that a change of the same magnitude has on Earth. 53. The error in measurement dx = (l%)(10) = 0.1 cm; V = x3 => dV = 3x2 dx = 3(10)2 (0.1) = 30 cm3 => the percentage error in the volume calculation is ( - ^ ) (100%) = 3% 55. Given D = 100 cm, dD = 1 cm, V = ∣π ( j ) 3 = ^ (102 %) =

=4 dV = ≡ D2 dD = ≈ (100)2 (l) = ^

. Then ^ (100%)

% = 3%

57. A 5% error in measuring t =4 dt = (5%)t = ^ . Then s = 16t2 =4 ds = 32t dt = 32t ( ^ ) = ⅛ = ^ = (10%)s => a 10% error in the calculation of s. 59. lim

x →0

= (⅛ ) s

87

88

Chapter 3 Differentiation

61. E(x) = f(x) - g(x) => E(x) = f(x) - m(x - a) - c. Then E(a) = 0 ≠> f(a) - m(a - a) - c = 0 =+ c = f(a). Next we calculate m: lim ^ ⅛ = 0 => lim ≤ j h ⅛ r ⅛ z ≈ = 0 => lim Γ ⅛ ⅛ 2i - ml = 0 (since c - f(a)) + f'(a) - m = 0 + as claimed.

m = f'(a). Therefore, g(x) = m(x - a) + c = f'(a)(x - a) + f(a) is the linear approximation,

63. (a) x = 1

y -χ χ z " 9

0 . 95

1. 05

1

(b) x = 1; m = 2.5, e 1 ≈ 2.7

x = —1; m = 0.3, e

v∣when m = 1.01m0 . m - - ⅛ 65. Find ∣ v∣ => ∣ =

c

=+ m ^ ∕ b ^ = m0 ≠> ^ 1 - ^ = ™ ≠ ,

=> dv = c ∙ ∣(1 - ⅛ )

O

v 2

1

- ? = ⅛ ^

( ⅛ ) d m >d m = θ∙θ l m o ^ dv = — π ⅛

®

-

γ2

m

=

__ ≡L ≤ ^ = (4 tan x) (sec2 x) - (2 sec x)(sec x tan x) = 2 sec2 x tan x

≈ 0.4

(1 ^ ^ )

= ιoom o>

m3 √ l--⅛ dv =

c2

,

Chapter 3 Practice Exercises

89

13. s = cos4 (1 - 2t) ≠> ⅛ = 4 cos3 (1 - 2t)(-sin (1 - 2t))(-2 ) = 8 cos3 (1 - 2t) sin (1 - 2t) 15. s = (sec t + tan t)5 => ^ = 5(sec t + tan t)4 (sec t tan t + sec2 1) = 5(sec t)(sec t + tan t)5 17. r = √ 2 0 sin 0 = (20 sin 0)1∕ 2 ≠> ⅛ = ∣(20 sin 0)^ 1∕ 2 (2 0 cos 0 + 2 sin 0) = ^ ^ ^ 19. r = sin √ 2 0 = sin(20) 1∕ 2 => ⅛ = cos (20)1∕ 2 ( ∣(20)^ 1∕ 2 (2)) = ^ ^

21. y = ∣x 2 esc j => ^ = ∣x 2 (- c s c 2 cot 2) ( 4 ) + (esc 2) ( 1 . 2x) = esc 2 cot 2 + χ esc 2 23. y = x - 1 / 2 sec(2x) 2 => ^ = x - 1 / 2 sec(2x) 2 tan (2x)2 (2(2x) - 2) + sec(2x) 2 ( - ∣x - 3 ∕ 2 ) = 8X1(2 sec(2x) 2 tan(2x) 2 - ∣x~ 3 ∕ 2 sec(2x) 2 = ∣x 1∕ 2 sec(2x)2 [16tan(2x) 2 —x - 2 ] or ⅛ s e c ( 2 x ) 2 [16x2 tan(2x) 2 - 1] 25. y = 5 cot x2 => ^ = 5 (—csc2 x 2 ) (2x) = —10x csc2 (x 2 ) 27. y = x2 sin2 (2x 2 ) => ⅛ = x2 (2 sin (2x 2 )) (cos (2x2 )) (4x) + sin2 (2x2 ) (2x) = 8x3 sin (2x 2 ) cos (2x 2 ) + 2x sin2 (2x 2 ) 90

— ( 4t V (t+ ι∕

e

-

• *

33.

y

∙'

2

x + l∕

ds — _ ? ( - 4 M z , at ∖ t+ U

^

dx

-

+

i ) 1 ∕ 2 ≠> ⅛ ^ i ( ι x∕ αx z ∖

= √ ⅛ I = ( l y x4 ∖

/

-3

=κ

sin 0 Λ2 cos 0 - 1 /

χ

z

- / 4t ∖ ~ 3 4 _ (t+ 1) 2 ∖ t+ ι∕

( g = j (2x + 1)1∕ 2 (2) = 3 v ^ + l 39. y = 3 (5x2 + sin 2x)“ 3 /2 ^

⅛ = 3 ( - ∣) (5x2 + sin 2x)- 5 / 2 [10x + (cos 2x)(2)] = ⅛ ^

i

⅛

∣

(ox j sin z x ι

41. xy + 2x + 3y = 1 => (xy' + y) + 2 + 3y' = 0 => xy' + 3 / = - 2 - y => y'(x + 3) = - 2 —y => y' = — ^ j 43. x3 + 4xy - 3y4 /3 = 2x => 3x2 + ^4x ^ + 4yj - 4y 1∕ 3 ^ = 2 =J> 4x ^ - 4y 1 /3 ^ = 2 - 3x 2 - 4y ⅛ (4 x -4 y V > )= 2 -3 x 2 - 4

≠

45. (xy ) 1''2 = 1 =≠∙ ∣ h, (0) = 2f(0)g(0)g'(0) + g2 (0)f'(0) = 2(1)(1) ( i ) + ( l ) 2 (-3 ) = - 2 (c) L e th (x) = -( x^) +- ι ≠> h'(x) π W = + e ιiH Λ ) g

_

(5 + 1) ( I ) - 3 ( -4 ) _ (5 + 1)2

⅛ < * ) + i) f 'w - f(2χ ) g , (χ) (g (χ )+ i)

=χ, hn 7Un )_-

( g (D + i)f , ( i ) - f ( 2i) g '( D ( g ( i) + i)

1

12

(d) Leth(x) = f(g(x)) ≠- h'(x) = f , (g(x))g'(x) ≠> h'(0) = f'(g(O))g, (O) = f'(l) ( ∣) = ( ∣) ( ∣) = ∣ (e) Leth(x) = g(f(x)) => h'(x) = g'(f(x))f'(x) ^ h'(0) = g'(f(O))f'(O) = g, (l)f'(O) = (-4 ) (-3 ) = 12 (f) Let h(x) = (x + f(x))3∕ 2 ≠> h'(x) = ∣(x + f(x))1∕ 2 (1 + f'(x)) => h'(l) = ∣(1 + f(l)) 1∕ 2 (1 + f , (l)) = 1(1+ 3)V 2(1

+

1) = ∣

(g) Leth(x) = f(x + g(x)) ^ h'(x) = f'(x + g (x))(l+ g'(x)) ^ = f ( D ( l + ∣) = ( l ) ( I ) = ∣

h'(0) = f , (g(O)) (1 + g'(0))

57. x = t2 + π => ^ = 2t; y = 3 sin 2x => ^ = 3(cos 2x)(2) = 6 cos 2x = 6 cos (2t2 + 2π) = 6 cos (2t2 ) ; thus, ^ = 6 c o s(0 )∙0 = 0 ⅛ = ∣ ∙⅛ = 6 c o s ( 2 t 2 ) -2 t => ⅛ ∣

= cos a

dw

dw

dr

∞ s (√ 8 s in (s + ⅞ )-2

; thus, ?d s = ?d r • ?d s = ----^ - l--- - ----------‘ 2 ^ 8 s ⅛ ( s + s)

(cos 0 ) ( 8 ) ( ^ }

r

-

61. y3 + y = 2cosx ⅛ 3y≈ g + g = - 2 sin x ⅛ g (3y2 + 1) = - 2 sin x ≠∙ g = ⅛ ⅛ _

-2 sin (0 ) _ n . ⅛ 3+∙ v

63. f(t)

’ *

d2 y I _ dχ l(o,i)

2t + 1

(3 y 2

_

2

+ 1 ) (- 2 cos x) - ( - 2 sin x) (βy g ) (3y2 + l) 2

(3 + 1)(-2 cos 0) - ( - 2 sin 0)(6∙0) _ (3 ÷ 1 ) 2

and f(t + h)

=

_______=2h_______ _ (2t + 2 h + l)( 2 t+ l)h ~

_ “

-2 (2 t÷ l)2

2(t+ h) + 1

_______- 2 (2t + 2 h ÷ l ) ( 2 t + l )

_ 1 2

f(t + h )- f(t) _ 2(t+h)+ι 2t+ι h ~ h χ f ∕m _ κ f(t + h ) -f (t) μ j h h → 0

— 2 t+ 1 - ( 2 t ÷ 2 h + 1) “ (2t + 2 h + l) ( 2 t+ l) h _ li -2 “ h → 0 (2 t + 2 h ÷ 1 ) ( 2 t+ 1)

=s∙ g j ^

Chapter 3 Practice Exercises

91

65. (a)

(b) v

lim f(x) = lim v 7

x→0 ~

x2 = 0 and lim f(x) =

x → 0~

= 0. Since lim f(x) = 0 = f(0) it lim - x 2 = 0 => lim f(x) v z

χ → 0+

χ → 0÷

follows that f is continuous at x = 0. (c) = lim (2x) lim f , (x) = 0 and lim f'(x) = v v 7 v χ →0 ~

'

x → 0"

x

→ o+

x→0

x→0

lim (-2x) = 0 => lim f'(x) = 0. Since this limit exists, it v 7

X → 0+

x→0

follows that f is differentiable at x = 0.

(vb)7

lim f(vx7) =

χ→l "

lim x = 1 and lim

x → l^

x → l+

f(x) =

follows that f is continuous at x = 1. (c) lim f'(x) = lim 1 = 1 and lim f , (x) = x → l^

x → l~

χ → l÷

lim (2 - x) = 1=> lim f(x) = 1.

x → l+

x→l

lim - 1 = - 1 =≠

x → l+

lim f'(x) ≠

x→l

Since lim f(x) = 1 = f(l), it x→l

lim f'(x), so lim f'(x) does

x → 1+

x→l

not exist => f is not differentiable at x = 1. 69. y = ∣÷ 2⅛4 = ∣x ÷ (2x —4)^ 1 => ⅛ = ∣- 2(2x - 4)~2 ; the slope of the tangent is - ∣ => - 1 = ∣- 2(2x - 4)" 2 => - 2 = -2(2x - 4)^ 2 => 1 =

3

⅛

=> (2x - 4)2 = 1 ≠> 4x2 - 16x + 16 = 1

=> 4X2 - 16x + 15 = 0 => (2x - 5)(2x - 3 ) = 0 ≠> x = ∣or x = ∣ => ( I ’ D

an^

(1’ ^ 4)

a re

P°in t s

on

⅛e

curve where the slope is - 1 . 71. y = 2x3 - 3X2 - 12x + 20 => ^ = 6x2 - 6x - 12; the tangent is parallel to the x-axis when ^ = 0 => 6X2 —6x — 12 = 0 => x2 - x —2 = 0 => ( x - 2)(x + 1) = 0 => x = 2 or x = —1 => (2,0) and (—1,27) are points on the curve where the tangent is parallel to the x-axis. 73. y = 2x3 - 3x2 - 12x + 20 ≠> g = 6x2 - 6x - 12 (a) The tangent is perpendicular to the line y = 1 - ^ when ^ = - ( z j τ y ) = 24; 6x2 - 6x — 12 = 24 => x2 —x —2 = 4 => x2 - x —6 = 0 => ( x - 3)(x + 2) = 0 => x = —2 o rx = 3 => (—2,16) and (3,11) are points where the tangent is perpendicular to y = 1 - ^ . (b) The tangent is parallel to the line y = √ z2 - 12x when ^ = -1 2 => 6x2 - 6x - 12 = - 1 2 => x2 - x = 0 => x(x - 1 ) = 0 => x = 0 or x = 1 => (0,20) and (1,7) are points where the tangent is parallel to y = √ 2 - 12x.

Chapter 3 Differentiation

92

75. y = tan x, — ∣< x < ∣ => ^ = sec2 x; now the slope of y = — ∣is — ∣ => the normal line is parallel to y = - 5 when ⅛ = 2. Thus, sec2 x = 2 ≠- -⅛ - - 2 => cos2 x = | => c o s χ = ^

=> x = - 1 andx = f

for —∣< x < ∣ => ( - ^ , - 1 ) and (J , 1) are points where the normal is parallel to y = —∣.

77. y = x 2 + C => ^ = 2x and y = x => ^ = 1; the parabola is tangent to y = x when 2 x = l => x = ∣ => y = ∣; thus, | = (1 ) 2 + C ≠> C = |

79. The line through (0,3) and (5, —2) has slope m = y = - x + 3 j y = 7f

3 θ£~2)

= - 1 => the line through (0,3) and (5, - 2 ) is

=> ⅛ = j ⅛ 5ι, so the curve is tangent to y = - x + 3 => ⅛ = ~

7

=> (x + 1)2 = c, x ≠ —1. Moreover, y = ^ η ∙ intersects y = —x + 3 => ^

1

= ⅛W

= —x + 3, x ≠ —1

1)2

=> c = (x + l ) ( - x + 3), x ≠ - 1 . Thus c = c => (x + = (x + l ) ( - x + 3) ≠> (x + l)[x + 1 —(—x + 3)] = 0, x ≠ —1 => (x + l)(2x - 2 ) = 0 => x = l (since x ≠ - 1 ) => c = 4. 81. x 2 + 2y2 = 9 ≠> 2x + 4y ^ = 0 => ⅛ = - ⅛ => ⅛ I VIA

VIA

VIA |

0 2)

= - 4 => the tangent line is y = 2 - ⅛(x - 1) ∙ ’

= - | x + | and the normal line is y = 2 + 4(x - 1) = 4x - 2. 83. xy + 2 x - 5 y = 2 ⅛ (χ g

+

y)

+

2 - 5 ⅛ = 0 =► g ( x - 5 ) = - y - 2

⅛ ∣= ⅛

⅛ £ ^

= 2

=> the tangent line is y = 2 + 2(x - 3) = 2x - 4 and the normal line isy = 2 + ~ ( x - 3 ) = - ∣x ÷ j . 85. x + √ ^ = 6 ^

l + ^ (x ⅛

+ y) = 0 ^

x ⅛ + y = -2 χΛy ^

∣∣^

⅛ = ∑½ ^∑ Z ^

= ⅛

=> the tangent line is y = l - ∣( x - 4 ) = - ∣x + 6 and the normal line is y = l + ∣(x - 4) = ∣x - y . 87. x3 y3 + y 2 = χ ⅛ y => [x3 ^3y2 ⅛θ + y 3 (3x2 )] + 2 y ^ = 1 + ^ => £ (3x3 y 2 + 2y - 1) = 1 - 3x2 y 3 ≠> £ = ⅛

⅜

i

=≠> 3x3 y 2 ⅛ + 2 y ^ - ^

= 1 - 3x2 y 3

=► £ ∣ ^ ^ = - ∣, but g ∣ ^ ⅛ undefined.

Therefore, the curve has slope - ∣at (1,1) but the slope is undefined at (1, —1).

X - 2 tan t, y - 2 s e c ^

t

⅛

~

~

s ιn t

x = ∣t a n f = ^ a n d y = ∣s e c f = 1 => y = ^ x + ∣J ^

_ sin = ^

= ι⅛

j

γ

,t -

j

= 2 cos3 t => ⅛ ∣ t^

= 2 cos3 ( f ) = ∣ 91. B = graph of f, A = graph of f'. Curve B cannot be the derivative of A because A has only negative slopes while some of B’s values are positive.

j

Chapter 3 Practice Exercises 93.

95. (a) 0,0

(b) largest 1700, smallest about 1400

⅛ ⅛ =⅛ ∣(⅛) ■(⅛ ] = 1 ⅛ ⅛ =⅛ (τ ■⅛ ■I) =⅛) (1) =I ιoi-

1 0 3

∙

i™ (5)-

x⅛

0

l

l≡ l'* '- - ∙^ ⅜ * ⅞ '* ' = “ ,+ 5 β→(JΓ (⅛ )

x sin x _ 2 —2 co sx -

= limJ ⅛

lim

χ→0

x

= ^limθ ^ 107. (a)

≡

1 0

∙⅛

x → 0 Ls ι n ^ J

105.

χ

s ιn

x sin x _ ι∙ 2 (l-c o s x ) ~ ⅛

■M I2√

= 4 l +

x sin x _ 2 (2 s⅛ ≈ (∣)) ^

’

ι∙

Γ

. s ixn x ∣ J

= (1)(1)(1) = 1

x

lim ⅛ ∙⅞ kcosx

X→ O

=

J i

) = l;le t0 = tanx

X /

’

θ → 0 asx→ 0 x

limn g(x) = limn ⅛t a n x v →o^

χ→O

= 1. Therefore, to make g continuous at the origin, define g(0) = 1.

S = 2πr2 + 2πrh andh constant => f = 4πr ^ + 2πh ^ = (4πr + 2πh) ∣

(b) S = 2τrr2 + 2πrh andr constant => ^ = 2πr ^ (c)

S = 2πr2 + 2πrh =► f = 4πr £ + 2π (r ^ + h ⅛) = (4πr + 2πh) £ + 2πr ⅛

(d) S constant => f = 0 ≠> 0 = (4πr + 2τrh) ⅛ + 2πr ⅛ 4

(2r + h) ⅛ = - r ⅛ ≠> δ = ⅛

I

109. A = 7τr2 => ^ = 2τrr j ; so r = 10 and ^ = - ^ m/sec ≠- ⅛ = (2π)(10) (—∣) = —4 0 m2∕sec H l∙ ⅞ = - 1 θhm∕sec, ⅞ = 0.5 ohm/sec; and ⅛ = ⅛ + ⅛ =>

R∑ f

= ⅛ ⅜ "⅛

⅜ ∙ Also,

R 1 = 75 ohms and R2 = 50 ohms => ⅛ = ⅛ + ⅛ => R = 30 ohms. Therefore, from the derivative equation, - 1 dR _ (30)2 dt ^

-1 ∕ (75)2 '

n

U

___ 1 ∕∏ c (50)2

_

-

∕J V5625

→ dR 5000/

f _ 0 M 1 f 5000-5625 _ 9(625) _ _1_ dt V 5625∙5000 / “ 50(5625) ”

50

= 0.02 ohm/sec. 113. Given ^ = 10 m/sec and ^ = 5 m/sec, let D be the distance from the origin => D2 = x2 + y2 => 2D ^ = 2x⅛ +2y⅛ ^

D f = x ∣ + y l When (x, y) = (3, -4 ), D = / 3 2

5 ^ = (5)(10) + (12)(5) => ^ = ^

+

( - 4 ) 2 = 5 and

= 22. Therefore, the particle is moving away from the origin at 22 m/sec

(because the distance D is increasing). 115. (a) From the diagram we have γ = 7 => r = ∣h. (b) V = i π r 2 h = i π ( 2 h ) 2 h =
^ = sec2 ⅛ ^ ; at point A ,x = 0 = > ⅛ = 0 = > ^ = (sec2 0) (—0.6) = —0.6. Therefore the speed of the light is 0.6 = ∣km/sec when it reaches point A.

93

94

Chapter 3 Differentiation (b)

(3/5) rad , 1 rev . 60 sec 2π rad sec min

= ~7Γ revs/min

119. (a) If f(x) = tan x and x = —∣, then f'(x) = sec2 x, f (—∣) = —1 and f , (—J) = 2. The linearization of f(x) is L(x) = 2 (x ÷ 5) + (-1 ) = 2x ÷ ^

(b)

.

If f(x) = sec x and x = —∣, then f , (x) = sec x tan x, f ( - ∣) = √ 2 a n d f , ( - f ) = - √ 2 . The linearization of f(x) is L(x) = - √ 5 (x + ^) + √ 2 = -√ 2 x + ⅛ ^ .

121. f(x) = √ x + 1 + sin x —0.5 = (x + I) 1/ 2 + sin x - 0.5 =+ f'(x) = (∣) (x + 1) 1^2 + cos x =+ L(x) = f'(0)(x - 0 ) + f(0) = 1.5(x - 0 ) + 0.5 =+ L(x) = 1.5x + 0.5, the linearization of f(x). 123. S = π r √ ,r2 + h2 , r constant =+ dS = π r ∙ ∣(r2 + h2 )^ 1^2 2h dh = ^ ™ d h . Height changes from ho to ho + dh =+dS = ⅛ ≡

125. C = 2πr =+ r = ⅛ , S = 4ττr2 = 7 , and V == = πr3 = ⅛ . It also follows that dr = ⅛ dC, dS = ^ dC and dV = ^ dC. Recall that C = 10 cm and dC = 0.4 cm. (a) dr = ^ = ⅛ cm =+ (⅛) (100%) = ( ^ ) (⅜ ) (100%) = (.04)(100%) = 4% (b) dS = f (0.4) = £ cm =+ ( f ) (100%) = (∣) ( ⅛ ) (100%) = 8% (c) dV = g (0.4) = ^ cm => ( $ 9 (100%) = ( ^ ) ( ⅛ ) (100%) = 12%

CHAPTER 3 ADDITIONAL AND ADVANCED EXERCISES 1. (a) sin 20 = 2 sin 0 cos 0 => ^ (sin 20) = ^ ( 2 sin 0 cos 0) => 2 cos 20 = 2[(sin 0)(-sin 0) + (cos 0)(cos 0)] => cos 20 = cos2 0 —sin2 0 (b) cos 20 = cos2 0 —sin2 0 => ^ (cos 20) = ^ (cos2 0 —sin2 0) => —2 sin 20 = (2 cos 0 ) ( - sin 0) —(2 sin 0)(cos 0) => sin 20 = cos 0 sin 0 + sin 0 cos 0 => sin 20 = 2 sin 0 cos 0 3. (a) f(x) = cos x => f , (x) = -sin x => f "(x) = —cos x, and g(x) = a ÷ bx + cx2 => g'(x) = b + 2cx => g"(x) = 2c; also, f(0) = g(0) ≠> cos(0) = a => a = 1; f'(0) = g'(0) => -sin(0) = b => b = 0; f"(0) = g"(0) => —cos (0) = 2c => c = —∣. Therefore, g(x) = 1 —∣x2 . (b) f(x) = sin (x ÷ a) => f , (x) = cos (x + a), and g(x) = b sin x ÷ c cos x => g'(x) = b cos x —c sin x; also, f(0) = g(0) => sin (a) = b sin (0) + c cos (0) => c = sin a; f'(0) = g'(0) => cos (a) = b cos (0) —c sin (0) => b = cos a. Therefore, g(x) = sin x cos a + cos x sin a.

Chapter 3 Additional and Advanced Exercises

95

(c) When f(x) = cos x, f"'(x) = sin x and f(4 x) = cos x; when g(x) = 1 —∣x2 , g,z x) = 0 and g(4 \x ) = 0. Thus f'"(0) = 0 = g'"(0) so the third derivatives agree at x = 0. However, the fourth derivatives do not agree since f^4  θ) = 1 but g^4  θ) = 0. In case (b), when f(x) = sin(x + a) and g(x) = sin x cos a + cos x sin a, notice that f(x) = g(x) for all x, not just x = 0. Since this is an identity, we have fW(x) = gW(x) for any x and any positive integer n. 5. If the circle (x —h)2 + (y —k)2 = a2 and y = x2 + 1 are tangent at (1,2), then the slope of this tangent is m = 2x∣ (U ) = 2 and the tangent line is y = 2x. The line containing (h, k) and (1,2) is perpendicular to y = 2x => ∣ ∑7 = —∣ => h = 5 - 2k => the location of the center is (5 - 2k, k). Also, (x —h)2 + (y - k)2 = a2 => x —h + (y - k)y' = 0 => 1 + (y')2 + (y - k)y'' = 0 => y'z = ⅛ ^ ^ ∙ At the point (1,2) we know yz = 2 from the tangent line and that y'z = 2 from the parabola. Since the second derivatives are equal at (1,2) we obtain 2 = => k = ∣. Then h = 5 - 2k = - 4 => the circle is (x + 4)2 + (y — ∣) 2 = a2 . Since (1,2) lies on the circle we have that a = 7. (a) y = uv => ⅛ = ^ v ÷ u ^ = (θ θ4u)v + u(0.05v) = 0.09uv = 0.09y => the rate of growth of the total production is 9% per year. (b ) If ^ = —0.02u and ~ = 0.03v, then ^ = (—0.02u)v + (0.03v)u = 0.01uv = 0.01y, increasing at 1% per year. 9. Answers will vary. Here is one possibility.

11. (a) s(t) = 64t — 16t2 => v(t) = ^ = 64 ~ 32t = 32(2 - t). The maximum height is reached when v(t) = 0 => t = 2 sec. The velocity when it leaves the hand is v(0) = 64 ft∕sec. (b) s(t) = 64t - 2.6t2 => v(t) = ^ = 64 - 5.2t. The maximum height is reached when v(t) = 0 => t ≈ 12.31 sec. The maximum height is about s(12.31) = 393.85 ft. 13. m(v 2 - vg) = k(xg - x2 ) ≠> m (2v ⅛) = k (-2 x ⅛) ≠> m ⅛ = k ( - ^ ) ⅞ =► m £ = - k x (∣) ⅛ . Then substituting ⅛ = v => m

= -k x , as claimed.

15. (a) To be continuous at x = π requires that X limπ _ s in x = lim (mx + b) => 0 = mπ + b => m = —- ; → X→7Γ +

COS X

{

X
m = - 1 and b = π. m, x ≥ π χ →π

17. (a) For all a, b and for all x ≠ 2, f is differentiable at x. Next, f differentiable at x = 2 => f continuous at x = 2 => χ li∏ι _ f(x) = f(2) => 2a = 4a —2b + 3 =≠> 2a —2b + 3 = 0. Also, f differentiable at x ≠ 2 f ax < 2 => f'(x) = ≤ n . In order that f , (2) exist we must have a = 2a(2) —b => a = 4a —b => 3a = b. ( 2ax - b, x > n2 Then 2a - 2b + 3 = 0 and 3a = b => a = J and b = ∣. (b) For x < 2, the graph of f is a straight line having a slope of ∣and passing through the origin; for x ≥ 2, the graph of f is a parabola. At x = 2, the value of the y-coordinate on the parabola is ∣which matches the y-coordinate of the point on the straight line at x = 2. In addition, the slope of the parabola at the match up point is ∣which is equal to the slope of the straight line. Therefore, since the graph is differentiable at the match up point, the graph is smooth there.

96

Chapter 3 Differentiation ⅛ ( f ( - χ ) ) = ⅛ H ( χ )) ≠ , f , ( - χ ) ( - D = - f z(χ ) => f z( - χ ) = f , (χ ) => f z is even.

19. fodd =► f(—x) = —f(x) ≠

21. Leth(x) = (fg)(x) = f(x)g(x) =► h'(x) = A limΛQ ^ _

∩m

*00 g(x) ~ fix ) g(xo) ÷ fix ) g(xo) ~ fixo) g(⅜ ) _

X —1 Xθ

X

Xθ

i

= lim

A —* Aθ

⅛ ± A- ⅛A ⅛ i 0

L γ χ  Γ g (x )~ g (x o )l l

∩m

X —►Xθ

^ 0

Λτ Λ

L

L

x

x

θ

JJ

∣ ∩m ^ —►Xθ

L L

) [ fix ) - fix 0 ) 1 j

χ

L

X

Xθ

JJ

= f(x0 ) χ limχo [ ^ E f 51] + g(x0 ) f'(x 0 ) = 0 ∙ χ limχo [ ® ^ ) | + g(x0 ) f , (x0 ) = g(x0 ) f'(xo). ifg is continuous at XQ∙ Therefore (fg)(x) is differentiable at XQ if f(xo) = 0, and (fg), (xo) = g(xo) f '(xo)∙ 23. If f(x) = x and g(x) = x sin (∣) , then x2 sin (∣) is differentiable at x = 0 because f'(0) = 1, f(0) = 0 and lim x sin = lim ⅛ ^ = lim ^ = 0 (so g is continuous at x = 0). In fact, from Exercise 21, →o x →o I t →∞ t 6 ' ’ ’ h'(0) = g(O)f, (O) = 0. However, for x ≠ 0, h'(x) = [χ 2 c o s (⅛)] (“ ⅛) ÷ ^χ s i∩ (^) - But x

limθ h'(x) = limθ [—cos (£) + 2x sin (£)] does not exist because cos (£) has no limit as x → 0. Therefore, the derivative is not continuous at x = 0 because it has no limit there. 25. Step 1: The formula holds for n = 2 (a single product) since y = U1U2 => ⅛ = ⅛

u2÷ u

ι ⅛ ∙

Step 2: Assume the formula holds for n = k: y = uιu 2 --∙uk => ^ = ⅛u2U3∙∙∙u k + u 1 ⅛ u 3 ∙∙∙⅛ + ...+ u 1 u 2 ∙∙∙u k.l ⅛ . If y = u 1u2 ∙ ∙ ∙ukuk+l = (uιu 2 ∙ ∙ ∙uk ) uk+1, then ∣ = ⅛ ∣^ =

(⅛

u 2u 3∙ '¾

uk+1 + uιu 2 ∙ ∙ ∙uk ⅛

+ U1 ⅞ U3 ∙∙∙Uk + - + U1U2 ∙ ∙ -Uk_, ⅛ ) Uk+l + U1U2 ∙ ∙ ∙Uk ⅛

= ⅜ u2 u3 ∙∙∙Uk+1 +U1 ⅛ U3 ∙∙∙U k+1 + ∙∙∙ + U1U2 ∙∙∙Uk- 1 ⅛ Uk+1 + U1U2 ∙∙∙1⅛ ⅛

.

Thus the original formula holds for n = (k÷ l) whenever it holds for n = k. 27. (a) T 2 = ⅛⅛ => L = § (b ) T≈ = ⅛

≠ ∙ L = 1 L ≡ > ≡ ⅛ => L ≈ 0.8156 ft

≠ T= ^ Λ

dT = ^

∙ ⅛ d L = ⅛ ⅛ dT = ^ , . , - ^ ^ , , ( 0 . 0 1 ft) ≈ 0.00613 see.

(c) Since there are 86,400 sec in a day, we have (0.00613 sec)(86,400 sec/day) ≈ 529.6 sec/day, or 8.83 min/day; the clock will lose about 8.83 min/day.

CHAPTER 4 APPLICATIONS OF DERIVATIVES 4.1 EXTREME VALUES OF FUNCTIONS 1. An absolute minimum at x = C2, an absolute maximum at x = b. Theorem 1 guarantees the existence of such extreme values because h is continuous on [a, b]. 3. No absolute minimum. An absolute maximum at x = c. Since the function’s domain is an open interval, the function does not satisfy the hypotheses of Theorem 1 and need not have absolute extreme values. 5. An absolute minimum at x = a and an absolute maximum at x = c. Note that y = g(x) is not continuous but still has extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when the hypothesis is not satisfied, absolute extrema may or may not occur. 7. Local minimum at (—1, 0), local maximum at (1, 0) 9. Maximum at (0, 5). Note that there is no minimum since the endpoint (2, 0) is excluded from the graph. 11. Graph (c), since this the only graph that has positive slope at c. 13. Graph (d), since this is the only graph representing a funtion that is differentiable at b but not at a. 15. f(x) = j x - 5 => f'(x) = ∣ => no critical points; f(-2 ) = — y , f(3) = —3 => the absolute maximum is - 3 at x = 3 and the absolute minimum is - y at x = -2

17. f(x) = x2 - 1 => f , (x) = 2x => a critical point at x = 0; f ( - l ) = 0, f(0) = - 1 , f(2) = 3 =Φ the absolute maximum is 3 at x = 2 and the absolute minimum is —1 at x = 0

98

Chapter 4 Applications of Derivatives

19. F(x) = - ^

= - χ ^ 2 => F , (x) = 2 X~3 = ⅛ , however

x = 0 is not a critical point since 0 is not in the domain; F(0.5) = —4, F(2) = —0.25 => the absolute maximum is —0.25 at x = 2 and the absolute minimum is —4 at x = 0.5

21. h(x) =

3 ∕x λ

= x 1∕ 3 => h'(x) = ∣x^ 2 ∕ 3 => a critical point

at x = 0; h ( - l ) = - 1 , h(0) = 0, h(8) = 2 => the absolute maximum is 2 at x = 8 and the absolute minimum is —1 at x = - 1 ( - ι,- i) Absmin

23. g(x) = √ T ^ x 2 = (4 =► g, (x) = I (4 -

X2 ) -

X2 ) 1∕ 2 V

2 ( - 2 X)

= ^= J

=> critical points at x = - 2 and x = 0, but not at x = 2 because 2 is not in the domain; g(—2) = 0, g(0) = 2, g(l) = ᅟ ∕ 3 => the absolute maximum is 2 at x = 0 and the absolute minimum is 0 at x = —2 25. f(0) = sin 0 => f , (0) = cos 0 => 0 = ∣is a critical point, but 0 = zy is not a critical point because 2y is not interior to the domain; f ( ^ ) = - 1 , f ( f ) = 1, f (⅞ ) = ∣ => the absolute maximum is 1 at 0 = ∣and the absolute minimum is - 1 at0 = y

27. g(x) = esc x => g'(x) = —(esc x)(cot x) => a critical point a t x = f j g ( f ) = ⅛ - g ( ! ) = 1, g ( τ ) = ⅛ ^ th e absolute maximum is ^

at x = j and x = y , and the

absolute minimum is 1 at x = j

Absmax

Absmax

1.2 1.0 0.8 0.6 0.4 0.2

c∞ * ( ∕2 , l ) - yπ = ∕3 ≤ x ≤ 2 π ∕3 ^ιr b s min _ l______ l--------- 1----0 W3 π ∕2 2ιr∕3

29. f(t) = 2 - ∣ t ∣= 2 - √ P = 2 - (t2 ) 1 /2 ^

f'(t) = - l ( t 2 ) - ‰

= - ^

= -⅛

=> a critical point at t = 0; f(—1) = 1, f(0) = 2, f(3) = —1 => the absolute maximum is 2 at t = 0 and the absolute minimum is —1 at t = 3

31. f(x) = x 4 ∕ 3 => f'(x) = ∣x 1∕ 3 => a critical point at x = 0; f(—1) = 1, f(0) = 0, f(8) = 16 =^ the absolute maximum is 16 at x = 8 and the absolute minimum is 0 at x = 0 33. g(0) = 03 ∕ 5 => g, (0) = ∣0 2 ∕ 5 => a critical point at 0 = 0; g (-3 2 ) = —8, g(0) = 0, g(l) = 1 => the absolute maximum is 1 at 0 = 1 and the absolute minimum is —8 at 0 = —32

Section 4.1 Extreme Values of Functions 35. Minimum value is 1 at x = 2.

y

-2

2

4

-2

-2,6] by [-2,4]

37. To find the exact values, note that that y' = 3x2 + 2x —8

y

= (3x —4)(x + 2), which is zero when x = - 2 or x = ∣ Local maximum at (—2, 17); local minimum at (∣, - ~ )

[-6,6] by [-5,20]

39. Minimum value is 0 when x = —1 or x = 1.

y

1

-2

[-6,6] by [-2,4]

41. The actual graph of the function has asymptotes at x = ± 1, so there are no extrema near these values. (This is an example of graρher failure.) There is a local minimum at

y

( 0 . 1)∙

[-4.7,4.7] by [-3.1,3.1]

43. Maximum value is ∣at x = 1;

y

minimum value is —j as x = - 1 .

[-5,5] by [-0.7,0.7]

6

*

99

100

Chapter 4 Applications of Derivatives

45. y' = x 2 ∕ 3 (l) + ∣x 1∕ 3 (x + 2) = ⅛⅛4

crit. pt. x= - ­ x= 0

derivative 0 undefined

extremum local max local min

value ≡ IO 1∕ 3 = 1.034 0

47∙ y, = x 2⅛^ - 2x)+ (1)√4 — χ2 —x2 + (4 —x2 ) _ ^ - x ^

4 - 2X2 ¼ -x 2

crit. pt. x = —2 x = -1 /2

derivative undefined

extremum local max

0

minimum

x = - /2 x= 2

0 undefined

maximum local min

value δ-

-

-2 2 0

x< 1 > 1

49.

crit. pt. X= 1

derivative undefined

f - 2 x - 2, t - 2 x + 6,

crit. x= x= x=

pt. -1 1 3

extremum minimum

value 2

x< 1 x> 1

derivative 0 undefined 0

extremum maximum local min maximum

53. (a) No, since f'(x) = ∣(x — 2)

1^3 ,

value 5 1 5

which is undefined at x = 2.

(b) The derivative is defined and nonzero for all x ≠ 2. Also, f(2) = 0 and f(x) > 0 for all x ≠ 2. (c) No, f(x) need not have a global maximum because its domain is all real numbers. Any restriction of f to a closed interval of the form [a, b] would have both a maximum value and minimum value on the interval. (d) The answers are the same as (a) and (b) with 2 replaced by a.

Section 4.1 Extreme Values of Functions

101

DF

(a) The construction cost is C(x) = 0.3ᅟ ∕1 6 + x2 ψ o.2(9 —x) million dollars, where 0 ≤ x ≤ 9 miles. The following is a graph of C(x).

2.65 4------- 1--------- 1--------- 1--------- 1--------- 1---------∣ --------- 1--------- 1--------- 1 0

1

2

3

4

5

6

7

8

9

χ(mBM)

Solving C'(x) = ^ = ^ - 0.2 = 0 gives x = ± ^

≈ ± 3.58 miles, but only x = 3.58 miles is a critical point is

the specified domain. Evaluating the costs at the critical and endpoints gives C(0) = $3 million, C ^ ^ ^ ≈ $2,694 million, and C(9) ≈ $2,955 million. Therefore, to minimize the cost of construction, the pipeline should be placed from the docking facility to point B, 3.58 miles along the shore from point A, and then along the shore from B to the refinery. (b) If the per mile cost of underwater construction is p, then C(x) = p ∕1 6 ÷ x2 ÷ 0.2(9 —x) and C'(x)

=

⅞ ¾

—0∙2 = 0 gives xc = ^ r ⅜ o 4 ,

w

^

m in im iz e s

the construction cost provided xc ≤ 9. The value

of p that gives xc = 9 miles is 0.218864. Consequently, if the underwater construction costs $218,864 per mile or less, then running the pipeline along a straight line directly from the docking facility to the refinery will minimize the cost of construction. In theory, p would have to be infinite to justify running the pipe directly from the docking facility to point A (i.e., for xc to be zero). For all values of p > 0.218864 there is always an xc ∈(0, 9) that will give a minimum value for C. This is proved by looking at C"(xc ) = -—

(16 + xc )

which is always positive for p > 0.

The length of pipeline is L(x) = √ T T x ^ + √ 2 5 ÷ (10 - x)2 for 0 ≤ x ≤ 10. The following is a graph of L(x).

Chapter 4 Applications of Derivatives

102

0

2

4

(

β

10

MW

Setting the derivative of L(x) equal to zero gives L'(x) = - τ = ^ √4

/

1 0 ~ x ----

√25 + (1 0 -x ) 2

+ x2

Uθ=i2— = 0. Note that ; x 2. = cos 0A and ^ r2 5+ (10- X)2 √4 + X

--- cos ⅛ , therefore, L'(x) = 0 when cos 0A = cos ⅛ , or ⅛ = ⅛ and ΔA C P is similar to ΔBDP. Use

simple proportions to determine x as follows: ∣= i ∏p ≠> x = y ≈ 2.857 miles along the coast from town A to town B. If the two towns were on opposite sides of the river, the obvious solution would be to place the pump station on a straight line (the shortest distance) between two towns, again forcing 0A = ‰∙ The shortest length of pipe is the same regardless of whether the towns are on thee same or opposite sides of the river. 59. (a) V(x) = 160x - 52x2 + 4x 3 V'(x) = 160 - 104x + 12X2 = 4(x - 2)(3x - 20) The only critical point in the interval (0, 5) is at x = 2. The maximum value of V(x) is 144 at x — 2. (b) The largest possible volume of the box is 144 cubic units, and it occurs when x = 2 units. 61. Let x represent the length of the base and √ 2 5 - x 2 the height of the triangle. The area of the triangle is represented by A(x) = ^ √ 2 5 - x2 where 0 ≤ x ≤ 5. Consequently, solving A, (x) = 0 => ^ ^ ^ i = 0 => x = ^ . Since A(0) = A(5) = 0, A(x) is maximized at x = ^ . The largest possible area is A ^ ^ ^ = y cm 2 .

63. s = - ∣gt2 + v0 t + s0 ≠ ∙ ∣ = - g t + v0 = 0 => t = ^ . Now s(t) = s0 0 t ( - f + v0 ) = 0 t = 0 or t = y∩. Thus s ( ^ = ∖

S

∕

+ Vo f ^ ^ + So = ⅛ + so > So is the maximum height over the interval 0 ≤ t ≤ ⅛ . ^

∖

g

∕

∖

e

∕

Z

e

65. Yes, since f(x) = ∣ x∣= √ x 3 = (x2 ) 1 /2 => f'(x) = 5 (x2 )

®

1 /2 (2x)

= -⅛

= A is not defined at x = 0. Thus it

is not required that f , be zero at a local extreme point since f ' may be undefined there. 67. If g(c) is a local minimum value of g, then g(x) ≥ g(c) for all x in some open interval (a, b) containing c. Since g is odd, g (-x ) = -g (x ) ≤ -g (c ) = g (-c ) for all - x in the open interval ( - b , - a ) containing - c . That is, g assumes a local maximum at the point - c . This is also clear from the graph of g because the graph of an odd function is symmetric about the origin. 69. (a) f'(x) = 3ax2 ÷ 2bx + c is a quadratic, so it can have 0, 1, or 2 zeros, which would be the critical points of f. The function f(x) = x3 —3x has two critical points at x = —1 and x = 1. The function f(x) = x3 — 1 has one critical point atx = 0. The function f(x) = x3 + x has no critical points.

Section 4.2 The Mean Value Theorem (b) The function can have either two local extreme values or no extreme values. (If there is only one critical point, the cubic function has no extreme values.) 71. Maximum value is 11 at x = 5; minimum value is 5 on the interval [—3, 2]; local maximum at (—5, 9)

[-6,6] by [0,12]

73. Maximum value is 5 on the interval [3, ∞ ) ,, minimum value is —5 on the interval ( - ∞ , -2 ].

4.2 THE MEAN VALUE THEOREM = f'(c) ≠> 3 = 2c + 2 => c = ∣.

1. Whenf(x) = x2 + 2x - 1 forO ≤ x ≤ 1, then M 3. When f(x) = x + i for ∣≤ x ≤ 2, then ⅛

^

= f'(c) => 0 = 1 - ⅜ => c = 1.

5. Does not; f(x) is not differentiable at x = 0 in (—1,8). 7. Does; f(x) is continuous for every point of [0,1] and differentiable for every point in (0,1). 9. Since f(x) is not continuous on 0 ≤ x ≤ 1, Rolle’s Theorem does not apply:

lim f(x) =

lim x = 1

≠ 0 = f(l). 11. (a) i ii

1V

;

^ T

T

;

~ S

0

4

9

18

24

*

xn

+ an.j xn' 1 + . . . + aιx + a0 , then P(rι) = P(r2 ) = 0. (b) Let ri and r2 be zeros of the polynomial P(x) = Since polynomials are everywhere continuous and differentiable, by Rolle’s Theorem P, (r) = 0 for some r between ri and r2 , where P, (x) = nx n' 1 + (n - 1) an.jXn^2 + ... + a i . 13. Since f" exists throughout [a, b] the derivative function f , is continuous there. If f , has more than one zero in [a, b], say f'(rι) = f '(r 2 ) = 0 for r i ≠ r 2 , then by Rolle's Theorem there is a c between ri and r2 such that f"(c) = 0, contrary to f" > 0 throughout [a, b]. Therefore f' has at most one zero in [a, b]. The same argument holds if f" < 0 throughout [a, b].

103

104

Chapter 4 Applications of Derivatives

15. With f(—2) = 11 > 0 and f(—1) = —1 < 0 we conclude from the Intermediate Value Theorem that f(x) = x4 ÷ 3x ÷ 1 has at least one zero between —2 and —1. Then —2 < x < —1 => —8 < x3 < —1 => —32 < 4x3 < - 4 ≠ —29 < 4x3 + 3 < - 1 => f'(x) < 0 for - 2 < x < —1 => f(x) is decreasing on [—2, —1] => f(x) = 0 has exactly one solution in the interval (—2, —1). 17. g(t) = √ t + √ t ÷ 1 - 4 => g'(t) = ^ and g(15) =

÷ ^

> 0 => g(t) is increasing for t in (0, ∞ ); g(3) = √ 3 - 2 < 0

Z15 > 0 => g(t) has exactly one zero in (0, ∞ ).

19. r(0) = 0 + sin2 ( f ) —8 => r, (0) = 1 + j sin ( f ) cos ( f ) = 1 + ∣sin ( y ) > 0 on (—∞ , ∞ ) => r(0) is increasing on (—∞ , oo); r(0) = —8 and r(8) = sin2 ( f ) > 0 => r(0) has exactly one zero in ( - ∞ , ∞ ). 21. r(0) = sec 0 — ^ ÷ 5 => r, (0) = (sec 0)(tan 0 ) + ∣ > 0 on (θ, ∣) => r(0) is increasing on (θ, ∣) ; r(0.1) ≈ —994 and r(1.5 7) ≈ 1260.5 => r(0) has exactly one zero in (θ, ∣) . 23. By Corollary 1, f , (x) = 0 for all x => f(x) = C, where C is a constant. Since f(—1) = 3 we have C = 3 => f(x) = 3 for all x. 25. g(x) = x 2 => g'(x) = 2x = f'(x) for all x. By Corollary 2, f(x) = g(x) + C. (a) f(0) = 0 => 0 = g(0) + C = 0 + C => C = 0 => f(x) = x2 => f(2) = 4 (b) f(l) = 0 => 0 = g(l) + C = l + C => C = —1 => f(x) = x2 —1 => f(2) = 3 (c) f(—2) = 3 => 3 = g (-2 ) + C ^ 3 = 4 + C ^ C = - l = > f(x) = x2 - l => f(2) = 3 27. (a) y = ⅛ + C

(b) y = ⅛ ÷ C

(c) y = ⅞ + C

29. (a) y' = - x ^ 2 = > y = ± + C

(b) y = x + ± + C

(c) y = 5x - ^ + C

31. (a) y = - ∣cos 2t + C

(b) y = 2 sin ∣÷ C

(c) y = - ∣cos 2t ÷ 2 sin ∣+ C 33. f(x) = x 2 —x + C; 0 = f(0) = 0 2 —0 ÷ C => C = 0 => f(x) = x2 - x 35. r(0) = 80 ÷ cot 0 ÷ C; 0 = r (J ) = 8 ( ∣) ÷ cot ( f ) + C => 0 = 2 π ÷ l + C

C = -2 π - 1

=> r(0) = 80 + cot 0 —2π — 1 37. v = ∣ = 9.8t + 5 => s = 4.9t2 + 5t ÷ C; at s = 10 and t = 0 we have C = 10 => s = 4.9t2 + 5t + 10 39. v = ∣ = sin(ττt) =^ s = - ∣cos(πt) ÷ C; at s = 0 and t = 0 we have C = ∣=> s = ⅛ ⅛ ) 41. a = 32 => v = 32t + Cμ at v = 20 and t = 0 we have Ci = 20 4 v = 32t ÷ 20 => s = 16t2 + 20t + C 2 ; at s = 5 and t = 0 we have C2 = 5 => s = 16t2 + 20t + 5 43. a = —4sin(2t) => v = 2cos(2t) ÷ C∏ at v = 2 and t = 0 we have Ci = 0 => v = 2cos(2t) => s = sin(2t) ÷ C2 ; at s = - 3 and t = 0 we have C2 = - 3 => s = sin(2t) - 3 45. If T(t) is the temperature of the thermometer at time t, then T(0) = —19o C and T(14) = 100o C. From the Mean Value Theorem there exists a 0 < to < 14 such that ^ y ∈y ^ = 8.5 o C/sec = T, (to), the rate at which the temperature was changing at t = t0 as measured by the rising mercury on the thermometer.

Section 4.3 Monotonic Functions and the First Derivative Test

105

47. Because its average speed was approximately 7.667 knots, and by the Mean Value Theorem, it must have been going that speed at least once during the trip. 49. Let d(t) represent the distance the automobile traveled in time t. The average speed over 0 ≤ t ≤ 2 is ^¾∈ ^ . The Mean Value Theorem says that for some 0 < ⅛ < 2, d'(t0 ) = ⅜∣} ^ ∙ The value d'(to) is the speed of the automobile at time to (which is read on the speedometer). 51. The conclusion of the Mean Value Theorem yields ∣ξ ∣= - ^ =>

C2

( ⅛ ) = a —b => c = y z ab.

53. f'(x) = [cos x sin (x + 2) + sin x cos (x + 2)] - 2 sin (x ÷ 1) cos (x + 1) = sin (x + x + 2) —sin 2(x + 1) = sin (2x ÷ 2) —sin (2x + 2) = 0. Therefore, the function has the constant value f(0) = —sin2 1 ≈ —0.7081 which explains why the graph is a horizontal line. 55. f(x) must be zero at least once between a and b by the Intermediate Value Theorem. Now suppose that f(x) is zero twice between a and b. Then by the Mean Value Theorem, f'(x) would have to be zero at least once between the two zeros of f(x), but this can’t be true since we are given that f'(x) ≠ 0 on this interval. Therefore, f(x) is zero once and only once between a and b. 57. Yes. By Corollary 2 we have f(x) = g(x) + c since f'(x) = g'(x). If the graphs start at the same point x = a, then f(a) = g(a) ≠> c = 0 ^> f(x) = g(x). 59. By the Mean Value Theorem we have ® we have f(b) —f(a) < 0 4

^

= f'(c) for some point c between a and b. Since b —a > 0 and f(b) < f(a),

f'(c) < 0.

61. f'(x) = (1 + x4 cos x)~ 1 => f"(x) = —(1 + x4 cos x)~2 (4x3 cos x - x4 sin x) = - x 3 (1 + x4 cos x)

2 (4

cos x - x sin x) < 0 for 0 ≤ x ≤ 0.1 => f'(x) is decreasing when 0 ≤ x ≤ 0.1

=> min f' ≈ 0.9999 and max f , = 1. Now we have 0.9999 ≤ ® p

≤ 1 => 0.09999 ≤ f(0.1) — 1 ≤ 0.1

=> 1.09999 ≤ f(0 .1 )≤ 1.1. 63. (a) Suppose x < 1, then by the Mean Value Theorem ⅛ Ξ 7 ^ < 0 => f(x) > f(l). Suppose x > 1, then by the Mean Value Theorem ^ ∈p

> 0 => f(x) > f(l). Therefore f(x) ≥ 1 for all x since f(l) = 1.

(b) Yes. From part (a), lim ⅛xz τ1^ ≤ θ and bm + ^ τ ≡ R ≥ θ∙ Since f'(l) exists, these two one-sided x →1 x →1 limits are equal and have the value f'(l) => f'(l) ≤ 0 and f'(l) ≥ 0 => f'(l) = 0. 4.3 MONOTONIC FUNCTIONS AND THE FIRST DERIVATIVE TEST 1. (a) f'(x) = x(x - 1) => critical points at 0 and 1 (b) f' = + + + ∣------ ∣+ + + => increasing on (—∞ , 0) and (1, ∞ ), decreasing on (0,1) 0 1 Local maximum at x = 0 and a local minimum at x = 1 (a) f'(x) = (x - l) 2 (x + 2) => critical points at - 2 and 1 (b) f' = ------ ∣+ + + ∣+ + + => increasing on (-2 ,1 ) and (1, ∞ ), decreasing on (—∞ , —2) -2 1 No local maximum and a local minimum at x = —2

106

Chapter 4 Applications of Derivatives

5. (a) f'(x) = (x — l)(x + 2)(x - 3) 4 critical points at - 2 , 1 and 3 (b) f ' = ------ ∣+ + + ∣-------- ∣+ + + =+ increasing on ( - 2 ,1 ) and (3, ∞ ), decreasing on (—∞ , —2) and (1,3) -2 1 3 (c) Local maximum at x = 1, local minima at x = —2 and x = 3 7. (a) f'(x) = x^ 1∕ 3 (x + 2) =+ critical points at - 2 and 0 (b) f , = + + + ∣------ ) ( + + + =+ increasing on ( - ∞ , - 2 ) and (0, ∞ ), decreasing on (—2,0) 0 -2 (c) Local maximum at x = —2, local minimum at x = 0 9. (a) g(t) = - t 2 - 3t + 3 =+ g, (t) = - 2 t - 3 =+ a critical point at t = —∣; g, = + + + ∣ -------, increasing on -3 /2 ( - ∞ , - ∣) , decreasing on ( - ∣, ∞ ) (b) local maximum value of g (—∣) = j a tt = - ∣ (c) absolute maximum is j at t = - ^

11. (a) h(x) = - x 3 + 2x2 =+ h'(x) = - 3 x 2 + 4x = x(4 - 3x) =+ critical points at x = 0, ∣ =+ h' = ------ ∣+ + + ∣-------- , increasing on (θ, ∣) , decreasing on (—∞ , 0) and ( ∣, ∞ ) 0 4/3 (b) local maximum value of h ( ∣) = ^ at x = | ; local minimum value of h(0) = 0 at x = 0 (c) no absolute extrema

13. (a) f(0) = 302 - 40 3 =+ f'(0) = 60 - 1202 = 60(1 - 20) =+ critical points at 0 = 0,⅜ =+ f ' = -------∣+ + + ∣ 0 1/2 increasing on (0, ∣) , decreasing on (—∞ , 0) and (- , ∞ ) (b) a local maximum is f ( j ) = ∣at 0 = ^, a local minimum is f(0) = 0 at 0 = 0 (c) no absolute extrema

Section 4.3 Monotonic Functions and the First Derivative Test

15. (a) f(r) = 3Γ3 + 16r ≠> f'(r) = 9r2 + 16 => no critical points => f' = + + + + + , increasing on ( - ∞ , ∞ ), never decreasing (b) no local extrema (c) no absolute extrema

17. (a) f(x) = x4 - 8x2 + 16 => f '(x) = 4x3 - 16x = 4x(x ÷ 2)(x - 2) => critical points at x = 0 and x = ± 2 => f ' = ------ ∣÷ + + ∣------- ∣+ + + , increasing on (-2 ,0 ) and (2, ∞ ), decreasing on ( - ∞ , - 2 ) and (0,2) -2 0 2 (b) a local maximum is f(0) = 16 at x = 0, local minima are f ( ± 2) = 0 at x = ± 2 (c) no absolute maximum; absolute minimum is 0 at x = ± 2 (d)

19. (a) H(t) = 114 - t6 => H, (t) = 6t3 —6t5 = 6t3 (l + t)(l —t) => critical points at t = 0, ± 1 => H, = + + + ∣------ ∣+ + + ∣--------, increasing on ( - ∞ , —1) and (0,1), decreasing on (—1,0) and (1, ∞ ) -1 0 1 (b) the local maxima are H(—1) = ∣at t = —1 and H(l) = ∣at t = 1, the local minimum is H(0) = 0 at t = 0 (c) absolute maximum is j at t = + 1; no absolute minimum

107

108

Chapter 4 Applications of Derivatives

1 H (t) =∣ t 4 -t*

21. (a) g(x) = x √ 8 - x2 = χ (8 - x2 ) 1/2 + g'(x) = (8 - x2 ) 1/2 + x (∣) (8 - x2 ) ~1 /2 (—2x) =

l

2 t 2 - χ ) critical points at x = ± 2 , ± 2 √ z2 => g' = ( ------ ∣+ + + ∣----- ) 2 -2 √ 2 ~2 2√2

, increasing on (—2,2), decreasing on

( - 2 √ 2 , - 2 ) and (2 ,2 √ 2 ) (b) local maxima are g(2) = 4 at x = 2 and g ( - 2 ^

= 0 at x = —2 √ ^ , local minima are g(—2) = —4 at

x = —2 and g ^2√^2^ = 0 at x = 2 √ 2 (c) absolute maximum is 4 at x = 2; absolute minimum is —4 at x = —2 (d)

23. (a) f(x) = γ - ∣ + f'(x) = ^ - ⅜ ⅛ ^

ii

= , ᅟ⅛

-

+ critical points at x = 1,3

=> f ' = + + + | ------ ) ( ------- | + + ÷ , increasing on ( - ∞ , 1) and (3, ∞ ), decreasing on (1,2) and (2,3), 1 2 3 discontinuous at x = 2 (b) a local maximum is f(l) = 2 at x = 1, a local minimum is f(3) = 6 at x = 3 (c) no absolute extrema

25. (a) f(x) = x 1∕ 3 (x + 8) = x4∕ 3 + 8x1∕ 3 4 f'(x) = ∣x 1∕ 3 + ∣x 2∕ 3 = ^ ^

=> critical points at x = 0, - 2

=> f ' = ------- ∣+ + + ) ( ÷ ÷ + , increasing on (—2,0) U (0, ∞ ), decreasing on (—∞ , —2) -2 0 (b) no local maximum, a local minimum is f(—2) = - 6 ᅟ∕2 ≈ —7.56 at x = —2 (c) no absolute maximum; absolute minimum is —6 V 2 at x = —2

Section 4.3 Monotonic Functions and the First Derivative Test (d) f(χ) 10n

f(x ) = x ,z3 (x+8)

-3

-2

-1

1

2

3

27. (a) h(x) = x 1/3 (x2 - 4) = x7 /3 - 4x 1/3 => h'(x) = j x4 /3 - ∣x x = 0, ⅛ T1

2/3

= - ^ - ^

=> critical points at

=> h, = + ÷ ÷ I ------ ) ( ------- I + ÷ ÷ , increasing on ( —∞ , 2⅛ ) and ( k - 2 ∕√ 7 » 2 ∕√ 7

∞ ) , decrea

I

(⅜0) " d (0'⅛) (b) local maximum is h ( ^

= ^ ^ ≈ 3.12 at x = ^ , the local minimum is h ^ ^

= —^ ^ ≈ —3.12

(c) no absolute extrema (d) h(x)

I

~4 ∙

h(x) = x ιz3 (x2 - 4 )

29. (a) f(x) = 2x —x2 => f'(x) = 2 —2x = 2(1 —x) => a critical point at x = 1 => f , = ÷ ÷ + ∣------ ] and f(l) = 1, 1 2 f(2) = 0 => a local maximum is 1 at x = 1, a local minimum is 0 at x = 2 (b) absolute maximum is 1 at x = 1; no absolute minimum (c)

31. (a) g(x) = x2 - 4x + 4 => g'(x) = 2x - 4 = 2(x - 2) => a critical point at x = 2 => g' = [ ------ I ÷ ÷ + and 1 2 g(l) = 1, g(2) = 0 => a local maximum is 1 at x = 1, a local minimum is g(2) = 0 at x = 2 (b) no absolute maximum; absolute minimum is 0 at x = 2

109

Chapter 4 Applications of Derivatives

110

33. (a) f(t) = 12t - 13 =4 f z(t) = 12 - 3t2 = 3(2 + 1)(2 - 1) ≠ critical points at t = ± 2 => f z = [ -------∣+ + + ∣— -3 -2 2 and f( -3 ) = - 9 , f( -2 ) = -1 6 , f(2) = 16 => local maxima are - 9 at t = - 3 and 16 at t = —2, a local minimum is —16 at t = —2 (b) absolute maximum is 16 at t = 2; no absolute minimum

35. (a) h(x) = ⅛ —2x2 + 4x ≠ ∙ hz(x) = x 2 —4x + 4 = (x - 2)2 =4 a critical point at x = 2 =4 h' = [ + + + ∣+ + + and 0 2 h(0) = 0 => no local maximum, a local minimum is 0 at x = 0 (b) no absolute maximum; absolute minimum is 0 at x = 0 (c)

37. (a) f(x) = ∣- 2 sin ( ∣) ≠- f'(x) = ∣—cos (∣) , f'(x) = 0 =4 cos ( ∣) = ∣ =4 a critical point at x = y ≠ f' = [ 0

I + + + ] andf(0) = 0, f ( ⅛ ) = ? - ᅟ ∕3 , f(2π) = π => local maxima are 0 at x = 0 a n d π 2 π ∕3 2π 3/ 3

at x = 2π, a local minimum i s ∣- √ 3 a t x = γ (b) The graph of f rises when f z > 0, falls when f z < 0, and has a local minimum value at the point where f z changes from negative to positive.

Section 4.4 Concavity and Curve Sketching

111

39. (a) f(x) = csc2 x —2 cot x ≠ f'(x) = 2(csc x )(-esc x)(cot x) - 2 (-c sc 2 x) = —2 (esc2 x) (cot x — 1) => a critical point at x = J => f' = ( --------I ÷ + ÷ ) and f (?) = 0 => no local maximum, a local minimum is 0 at x = ? π 0 π ∕4 (b) The graph of f rises when f ' > 0, falls when f' < 0, and has a local minimum value at the point where f' = 0 and the values of f , change from negative to positive. The graph of f steepens as f'(x) → ± ∞ .

41. h(0) = 3 cos ( 2f ) => h'(0) = —∣ sin 2( f ) => h' = [ ----- ] , (0,3) and (2π, —3) => a local maximum is 3 at 0 = 0, 2 0 2π a local minimum is —3 at 0 = 2π

45. (a)

47. f(x) = x3 - 3x + 2 => f'(x) = 3x2 - 3 = 3(x - l)(x ÷ 1) => f' = + + + ∣------ ∣÷ ÷ + => rising for x = c = 2 since -1 1 f'(x) > 0 for x = c = 2. 4.4 CONCAVITY AND CURVE SKETCHING

1. y = ^ - y - 2 x + j = > y , =

X2

-

X-2

= (X - 2)(X + 1) => y" = 2x - 1 = 2 (x — 0 . The graph is rising on

( - ∞ , -1 ) and (2, ∞ ), falling on (-1 ,2 ), concave up on (∣, ∞ ) and concave down on ( - ∞ , ∣) . Consequently, a local maximum is ∣at x = —1, a local minimum is —3 at x = 2, and ( j >- 1) is a point of inflection. 3. y = i (χ2 - 1)2 /3 =+ y' = (?) (I) (χ 2 - 1)

1 /3 (2O

= χ (χ2 - 1)

1 /3 ,

y' = ------ H + + +

------- ) ( + + +

=> the graph is rising on (—1,0) and (1, ∞ ), falling on (—∞ , —1) and (0,1) => a local maximum is ∣at x = 0, local minima are 0 at x = ± 1 ; y" = (x2 - 1)- 1 / 3 + (x) ( - ∣) (x2 - l) “ 4 /3 (2x) = - ^ ^ = , y" = + + + ∣ ------ ) ( ------- ) ( ------- ∣+ + + => the graph is concave up on ^ - ∞ , - ᅟ ∕3 ^ and ( √ % ∞ ^, concave down on ( - √ % √ ^ ) => points of inflection at ( ± A∕3 , ^ )

112

Chapter 4 Applications of Derivatives

5. y = x + sin 2x ≠- y' = 1 + 2 cos 2x, y' = [ ------ ∣ + + + ∣------] ≠> the graph is rising on (—∣, ∣) , falling - 2 π ∕3 - π ∕3 π ∕3 2 π ∕3 on (— ⅛ , —? ) and (? , ⅛ ) => local maxima are —¾1 + ^ at x = — ⅛ and ? + δ ^ at x = ? , local minima are — ? — ^ a t x = - ? and ⅛

at x = ⅛ ; y" = —4 sin 2x, y" = [ - 2 π ∕3

∣ + + + ∣-------- ∣+ + + ] 2 π ∕3 - π ∕2 0 π ∕2

graph is concave up on ( - f , θ ) and ( y y ) , concave down on (—y , —∣) and (θ, ∣) => points of inflection at

( - f , - * ) , ( O , O ) , a n d ( j, f ) x ∣= sin x and if x < 0, sin ∣ 7. If x ≥ 0, sin ∣ x ∣= sin (- x ) = -s in x. From the sketch the graph is rising on ( - T> - f ) » ( 0 >2) a n d ( ρ 2 π ) ’ fa l l i ∩g on ( - 2 π , - ⅜ ) , ( - ∣,θ) and ( f , γ ) ; local minima are —1 at x = ± ~ and 0 at x = 0; local maxima are 1 at x = ± ∣and 0 at x = ± 2%; concave up on (—2π, —π) and (π, 2π), and concavedown on ( - π ,0 ) and (0,π) => points of inflection are (—π, 0) and (π, 0) 9. When y = x2 - 4x + 3, then y' = 2x - 4 = 2(x - 2) and y" = 2. The curve rises on (2, ∞ ) and falls on ( - ∞ , 2). At x = 2 there is a minimum. Since y', > 0, the curve is concave up for all x.

11. When y = x3 - 3x ÷ 3, then y' = 3x2 - 3 = 3(x - l)(x ÷ 1) and y" = 6x. The curve rises on ( - ∞ , - 1 ) U (1, ∞ ) and falls on (—1,1). At x = —1 there is a local maximum and at x = 1 a local minimum. The curve is concave down on (—∞ , 0) and concave up on (0, ∞ ). There is a point of inflection at x = 0.

13. When y = - 2 x 3 + 6x 2 - 3, then y' = - 6 x 2 + 12x = -6 x (x - 2) and y" = -1 2 x + 12 = -1 2 (x - 1). The curve rises on (0,2) and falls on (—∞ , 0) and (2, ∞ ). At x = 0 there is a local minimum and at x = 2 a local maximum. The curve is concave up on (—∞ , 1) and concave down on (1, ∞ ). At x = 1 there is a point of inflection.

=> the

Section 4.4 Concavity and Curve Sketching 15. When y = (x —2)3 + 1, then y, = 3(x —2)2 and y" = 6(x - 2). The curve never falls and there are no local extrema. The curve is concave down on (—∞ , 2) and concave up on (2, ∞ ). At x = 2 there is a point of inflection.

17. When y = x4 —2x2 , then y, = 4x 3 - 4x = 4x(x + l)(x - 1) and y" = 12x2 —4 = 12 (x + ^ ^ (x — ^ ^ . The curve rises on ( - 1 ,0 ) and (1, ∞ ) and falls on (—∞ , - 1 ) and (0,1). At x = ± 1 there are local minima and at x = 0 a local maximum. The curve is concave up on

(⅛-∞) and concave down on (

√3’ Vs) ’ A, X

= 7⅛

there are points of inflection. 19. When y = 4x3 - x4 , then y' = 12x2 —4x 3 = 4x 2 (3 - x) and y" = 24x - 12X2 = 12x(2 —x). The curve rises on ( - ∞ , 3) and falls on (3, ∞ ). At x = 3 there is a local maximum, but there is no local minimum. The graph is concave up on (0, 2) and concave down on ( - ∞ , 0) and (2, ∞ ). There are inflection points at x = 0 and x = 2.

21. When y = χ 5 - 5x4 , then y' = 5x4 - 20x3 = 5x3 (x - 4) and y" = 20X3 - 60x 2 = 20x2 (x —3). The curve rises on ( - ∞ , 0) and (4, ∞ ), and falls on (0,4). There is a local maximum at x = 0, and a local minimum at x = 4. The curve is concave down on (—∞ , 3) and concave up on (3, ∞ ). At x = 3 there is a point of inflection.

23. When y = x + sin x, then y' = 1 + cos x and y" = -s in x. The curve rises on (0 ,2π). At x = 0 there is a local and absolute minimum and at x = 2π there is a local and absolute maximum. The curve is concave down on (0, π) and concave up on (π, 2π). At x = π there is a point of inflection.

25. When y = x 1∕ 5 , then y' = ∣x - 4 / 5 and y" = - ⅛ x“ 9 / 5 . The curve rises on (—∞ , ∞ ) and there are no extrema. The curve is concave up on (—∞ , 0) and concave down on (0, ∞ ). At x = 0 there is a point of inflection.

113

114

Chapter 4 Applications of Derivatives

27. When y = x 2 ≠5 , then y, = j x^ 3 ^5 and y, ' = - ∣ x^ 8 ∕ 5 . The curve is rising on (0, ∞ ) and falling on ( - ∞ , 0). At x = 0 there is a local and absolute minimum. There is no local or absolute maximum. The curve is concave down on (—∞ , 0) and (0, ∞ ). There are no points of inflection, but a cusp exists at x = 0. 29. When y = 2x - 3x2 ∕ 3 , then y' = 2 - 2x" 1 /3 and y" = ∣x~ 4 ∕ 3 . The curve is rising on (—∞ , 0) and (1, ∞ ), and falling on (0,1). There is a local maximum at x = 0 and a local minimum at x = 1. The curve is concave up on ( - ∞ , 0) and (0, ∞ ). There are no points of inflection, but a cusp exists at x = 0.

31. When y = x2 /3 ( ∣- x) = ∣x 2 /3 - x5 / 3 , then y' = ∣x^ 1∕ 3 - ∣x 2∕ 3 = ∣x^ 1∕ 3 (l - x) and y" = - ∣χ~ 4 ∕ 3 _ ^

χ

-l∕3 _ _ ∣χ -4∕3(l ψ 2χ).

The curve is rising on (0,1) and falling on ( - ∞ , 0) and (1, ∞ ). There is a local minimum at x = 0 and a local maximum at x = 1. The curve is concave up on ( - ∞ , - ∣) and concave down on ( - ∣, θ) and (0, ∞ ). There is a point of inflection at x = —∣and a cusp at x = 0. 33. When y = x ∕8 - x 2 = χ (8 - x2 ) 1 /2 , then y' = (8 - x 2 ) v 2 + (x) (1) (8 - x 2 )^ v 2 (-2 x ) = (8 y"

X2 ) ^ V 2

(8 - 2x 2 ) =

7

2 < 2 -χ χ 2 + χ )— _

^ (2 √ 2 + x ) (2 √ 2 -x )

a n d

= ( - 1 ) ( 8 - x 2 p ( - 2 x ) ( 8 - 2x2 ) + ( 8 - x 2 ∏ ( ~ 4 x )

— 2x (χ2 ~ 12) . The curve is rising on (- 2 ,2 ), and falling √ ( 8 - X2)3

on ( - 2 ∕2 ) - 2 ^ and ^ 2 ,2 χ ∕2 ^ . There are local minima x = —2 and x = 2 √ 2 , and local maxima at x = —2 ∕2 and x = 2. The curve is concave up on ( —2 ∕ 2 , 0^ and concave down on ^0 ,2 ∕2 ^ . There is a point of inflection at x = 0.

Section 4.4 Concavity and Curve Sketching 35. When y = ⅛ . then / = ⅛ 1 z ⅞ L g ∑ ≡ _ “ y

(x -3 )(x -l) (x - 2)2 _ ~

d a n α

(2x ~ + ( x ~ 2)2 - (x2 - 4x + 3)2(x - 2) _ (x -2 )4 “

_ 2 _ (x -2 )3 ∙

The curve is rising on ( - ∞ , 1) and (3, ∞ ), and falling on (1,2) and (2,3). There is a local maximum at x = 1 and a local minimum at x = 3. The curve is concave down on ( - ∞ , 2) and concave up on (2, ∞ ). There are no points of inflection because x = 2 is not in the domain.

2x, ∣ x∣> 1 and y" = -2x, ∣ x∣< 1

y

J , then

37. When y = ∣ x2 — 11=

(3

W > i . The W< i

curve rises on (—1,0) and (1, ∞ ) and falls on ( - ∞ , -1 ) and (0,1). There is a local maximum at x = 0 and local minima at x = + 1. The curve is concave up on (—∞ , —1) and (1, ∞ ), and concave down on (—1,1). There are no points of inflection because y is not differentiable at x = ± 1 (so there is no tangent line at those points).

Loc min

Locmin

Since lim y' = —∞ and lim y' = ∞ there is a χ →0- j X → O÷ cusp at x = 0. There is a local minimum at x = 0, but no local maximum. The curve is concave down on (—∞ , 0) and (0, ∞ ). There are no points of inflection. 41. y' = 2 + x - x2 = (1 + x)(2 - x), y' = ------ ∣+ + + ∣-------1 2 => rising on (-1 ,2 ), falling on ( - ∞ , —1) and (2, ∞ ) => there is a local maximum at x = 2 and a local minimum at x = - 1 ; y" = 1 - 2x, y" = + + + ∣-----1/2 => concave up on (—∞ , ∣) , concave down on (∣, ∞ )

Loc max

=> a point of inflection at x = ∣ 43. y' = x(x - 3)2 , y' = ------ ∣+ + + ∣+ + + => rising on 0 3 (0, ∞ ), falling on ( - ∞ , 0) => no local maximum, but there is a local minimum at x = 0; y" = (x - 3)2 + x(2)(x - 3) = 3(x - 3)(x - 1), y" = + + + ∣------ ∣+ + + => concave 1 3 up on (—∞ , 1) and (3, ∞ ), concave down on (1,3) => points of inflection at x = 1 and x = 3

x=0

115

116

Chapter 4 Applications of Derivatives

45. y' = x (x2 — 12) = x ^x —2√ 3 ^ (x + 2 √ ^ j ,

Locm ax

y' = ------ ∣ + + + ∣------ ∣ + + + =+ rising on -2 √ 3 ° 2√3 ^ - 2ᅟ ∕ 3 , 0^ and ^2ᅟ ∕3 , ∞ ) , falling on ^ - ∞ , - 2 √ z3^ and ^0,2 ^/5) => a local maximum at x = 0, local minima at x = ± 2 √ 3 ; y" = (1) (x2 - 12) + (x)(2x) = 3(x - 2)(x + 2), y" = + + + ∣------ ∣+ + + -2 2 => concave up on (—∞ , —2) and (2, ∞ ), concave down on (—2,2) => points of inflection at x = ± 2 47. y' = (8x - 5x2 ) (4 - x)2 = x(8 - 5x)(4 - x)2 , y' = ------ ∣+ ÷ ÷ ∣------- ∣------- => rising on (0, ∣) , 0 8/5 4 falling on (—∞ , 0) and (∣, ∞ ) =+ a local maximum at x = ∣, a local minimum at x = 0; y" = (8 - 10x)(4 - x)2 ÷ (8x - 5x2 ) (2)(4 - x )(-l) = 4(4 - x) (5X2 - 16x + 8 ), y" = + + + ∣ ------ ∣ + + + ∣------ =+ concave up 8-2√6 5

8+2√6 5

4

on ^ - ∞ , 8- - ^ ^ and (^ ± y ^ ,4 ^ , concave down on ( L L∣ ^ χ

=

J

L t∣ A^)

anj

8 ι⅛ 2 √6 a n d x =

(4, o o ) zφ points of inflection at

4

49. y' = sec2 x, y' = ( + + + ) =+ rising on ( - j , ∣) , —π ∕2 π ∕2 never falling =+ no local extrema; y" = 2(sec x)(sec x)(tan x) = 2 (sec2 x) (tan x), y" = ( ------ ∣+ + + ) =+ concave up on (θ, - π ∕2 0 π ∕2 concave down on ( - j , 0), 0 is a opoint of inflection. 51. y' = cot ∣, y' = ( + + + ∣------ ) =+ rising on (0, π), 0 τr 2π falling on (π, 2π) =+ a local maximum at 0 = π, no local minimum; y" = - 1 csc2 ∣, y" = ( ------ ) => never 0 2π concave up, concave down on (0,2π) =+ no points of inflection

Locm ax

Section 4.4 Concavity and Curve Sketching 53. y' = tan2 0 - 1 = (tan θ — l)(tan θ + 1), / = ( + + + ∣ ------ ∣+ + + ) —π ∕4 π ∕4 π ∕2 —π ∕2

=> rising on

( - ^ , - f ) and ( 5 4 ) , falling on ( - f 4 ) ≠> a local maximum at 0 = —^, a local minimum at 0 = ^ ; y" = 2 tan 0 sec2 0, y" = ( ------ ∣+ + ÷ ) 7Γ∕2 —π ∕2 0 => concave up on (θ, ^ ) , concave down on ( - f ,θ ) => a point of inflection at 0 = 0 55. y' = cos t, y, = [ + ÷ ÷ ∣------ ∣ ÷ + + ] => rising on 0 π ∕2 3 π ∕2 2π (θ, f ) and ( y , 2 π ) , falling on ( ∣, γ ) => local maxima at t=

and t = 2π, local minima at t = 0 and t = γ ;

y" = -s in t, y" = [ ------ ∣+ ÷ ÷ ] τr 2π 0 => concave up on (π, 2π), concave down on (0, π) => a point of inflection at t = π

57. y' = (x + 1) 2 ∕ 3 , y' = + + + ) ( + + + => rising on -1 (—∞ , ∞ ), never falling + no local extrema; y" = - 5 (x + 1)“ 5 / 3 , y" = + + + ) ( ------1 =► concave up on (—∞ , —1), concave down on ( - 1 , ∞ ) => a point of inflection and vertical tangent at x = —1

59. y' = x 2 ∕ 3 (x - 1), y' ---------- ) ( ------- ∣+ + + => rising on 1 0 (1, ∞ ), falling on ( - ∞ , 1) ≠> no local maximum, but a local minimum at x = 1; y" = ∣x^ 2 ∕ 3 + j x~ 5∕ 3 = I x - 5 / 3 (x + 2), y" = + + + I -------) ( + + + 0 -2 => concave up on (—∞ , —2) and (0, ∞ ), concave down on ( - 2 ,0 ) + points of inflection at x = —2 and x = 0, and a vertical tangent at x = 0 6 1 ∙ y, =

I

2x,, x > 0 ’ y / =

(—∞ , ∞ ) +

+ + +

o+

+ +

^

ris in g o n

∫ -2 , x < 0 no local extrema; y" = ∣ 2, x > 0 ’

y" = ------ ) ( + + + => concave up on (0, ∞ ), concave 0 down on (—∞ , 0) => a point of inflection at x = 0

Loc min

117

118

Chapter 4 Applications of Derivatives

63. The graph of y = f"(x) ≠ ∙ the graph of y = f(x) is concave u p o n (0 , ∞ ), concave down on (—∞ ,0 ) => a point of inflection at x = 0; the graph of y = f , (x) => y' = ÷ + + ∣------ ∣÷ + ÷ => the graph y = f(x) has both a local maximum and a local minimum

65. The graph of y = f"(x) ≠> y" = ------ ∣+ + + ∣-------=> the graph of y = f(x) has two points of inflection, the graph of y = f'(x) => y, = ------ ∣+ + + => the graph of y = f(x) has a local minimum

71. Graphs printed in color can shift during a press run, so your values may differ somewhat from those given here. (a) The body is moving away from the origin when ∣ displacement ∣ is increasing as t increases, 0 < t < 2 and 6 < t < 9.5; the body is moving toward the origin when ∣ displacement∣ is decreasing as t increases, 2 < t < 6 and 9.5 < t < 15 (b) The velocity will be zero when the slope of the tangent line for y = s(t) is horizontal. The velocity is zero when t is approximately 2, 6, or 9.5 sec. (c) The acceleration will be zero at those values of t where the curve y = s(t) has points of inflection. The acceleration is zero when t is approximately 4,7.5, or 12.5 sec. (d) The acceleration is positive when the concavity is up, 4 < t < 7.5 and 12.5 < t < 15; the acceleration is negative when the concavity is down, 0 < t < 4 and 7.5 < t < 12.5 73. The marginal cost is ^ which changes from decreasing to increasing when its derivative ^ is zero. This is a point of inflection of the cost curve and occurs when the production level x is approximately 60 thousand units. 75. When y' = (x — l) 2 (x —2), then y" = 2(x — l)(x - 2) + (x — 1)2 . The curve falls on (—∞ , 2) and rises on (2, ∞ ). At x = 2 there is a local minimum. There is no local maximum. The curve is concave upward on (—∞ , 1) and ( j , ∞ ) , and concave downward on (1, ∣) . At x = 1 or x = ∣there are inflection points.

Section 4.4 Concavity and Curve Sketching 77. The graph must be concave down for x > 0 because f"(x) = ~ ⅜ < 0∙

79. The curve will have a point of inflection at x = 1 if 1 is a solution of y" = 0; y = x3 + bx 2 + ex + d => / = 3x2 + 2bx + c => y'' = 6x + 2b and 6(1) + 2b = 0 => b = —3. 81. (a) f(x) = ax 2 + bx + c = a (x2 + ⅛x) + c = a ^x2 + £ a: + ⅛ ^ - ^ + c = a (x + ⅛ ) 2 - ⅛ ^ a parabola whose vertex is at x = — ⅛ => the coordinates of the vertex are ( - ⅛ , 2a

γ

2a ’

b2

~ 4ac ) 4a

y

(b) The second derivative, f , , (x) = 2a, describes concavity => when a > 0 the parabola is concave up and when a < 0 the parabola is concave down. 83. A quadratic curve never has an inflection point. If y = ax2 ÷ bx + c where a ≠ 0, then y' = 2ax + b and y" = 2a. Since 2a is a constant, it is not possible for y', to change signs. 85. If y = x 5 —5x4 —240, then y, = 5x3 (x —4) and y" = 20X2 (X - 3). The zeros of y' are extrema, and there is a point of inflection at x = 3.

87. If y = ∣x 5 ÷ 16x2 - 25, then y, = 4x (x3 + 8) and y" = 16 (x3 + 2 ) . The zeros of y' and y" are extrema and points of inflection, respectively.

89. The graph of f falls where f ' < 0, rises where f ' > 0, and has horizontal tangents where f ' = 0. It has local minima at points where f ' changes from negative to positive and local maxima where f , changes from positive to negative. The graph of f is concave down where f " < 0 and concave up where f " > 0. It has an inflection point each time f "changes sign, provided a tangent line exists there.

119

120

Chapter 4 Applications of Derivatives

91. (a) It appears to control the number and magnitude of the local extrema. If k < 0, there is a local maximum to the left of the origin and a local minimum to the right. The larger the magnitude of k (k < 0), the greater the magnitude of the extrema. If k > 0, the graph has only positive slopes and lies entirely in the first and third quadrants with no local extrema. The graph becomes increasingly steep and straight as k → ∞ . (b) f'(x) = 3x 2 ÷ k => the discriminant 0 2 - 4(3)(k) = - 12k is positive for k < 0, zero for k = 0, and ⅛when k < 0, one zero x = 0 when k = 0 and no real zeros

negative for k > 0; f ' has two zeros x =

when k > 0; the sign of k controls the number of local extrema. (c) As k → ∞ , f '(x) → ∞ and the graph becomes increasingly steep and straight. As k → - ∞ , the crest of the graph (local maximum) in the second quadrant becomes increasingly high and the trough (local minimum) in the fourth quadrant becomes increasingly deep.

93. (a) If y = x2 /3 (x 2 - 2 ) , then y' = ∣x - 1 / 3 (2x2 - 1) and y, ' = ∣x^ 4 ∕ 3 (10x2 ÷ 1 ). The curve rises on (~ ^

>0^

and

(^ 2 ’ o o )

and

^

s on

(~ ∞ 5

-

^ )

and ^0, ^ ^ ∙ The curve is concave up on ( - ∞ , 0) and (0, ∞ ). (b) A cusp since lim y' = oo and lim^ y' = - ∞ . 95. Yes: y = x2 + 3 sin 2x => y' = 2x + 6 cos 2x. The graph of y' is zero near —3 and this indicates a horizontal tangent near x = —3.

4.5 APPLIED OPTIMIZATION PROBLEMS 1. Let £ and w represent the length and width of the rectangle, respectively. With an area of 16 in.2 , we have that (£)(w) = 16 => w = 16£- 1 => the perimeter is P = 2£ ÷ 2w = 2£ + 32£- 1 and P'(£) = 2 - ^ = - - ^

16^ .

Solving P'(£) = 0 => 2 (r+ w -4) _ Q _^ ^ _ _ ^ 4 since £ > 0 for the length of a rectangle, £ must be 4 and w = 4 => the perimeter is 16 in., a minimum since P"(£) = ^ > 0. 3. (a) The line containing point P also contains the points (0,1) and (1,0) => the line containing P is y = 1 - x => a general point on that line is (x, 1 —x).

Section 4.5 Applied Optimization Problems

121

(b) The area A(x) = 2x(l - x), where 0 ≤ x ≤ 1. (c) When A(x) = 2x - 2x2 , then A'(x) = 0 => 2 - 4 x = 0 => x = ∣. Since A(0) = 0 and A (l) = 0, we conclude that A ( ∣) = ∣sq units is the largest area. The dimensions are 1 unit by ∣unit. 5. The volume of the box is V(x) = x(15 —2x)(8 - 2x) = 120x - 46 X2 ÷ 4 X3 , where 0 ≤ x ≤ 4. Solving V'(x) = 0 => 120 - 92x + 12X2 = 4(6 - x)(5 - 3x) = 0 => x = ∣ or 6, but 6 is not in the domain. Since V(0) = V(4) = 0, V (3) — ⅛ F ≈ 91 in3 must be the maximum volume of the box with dimensions y × y × ∣inches. 7. The area is A(x) = x(800 - 2x), where 0 ≤ x ≤ 400. Solving A'(x) = 800 - 4x = 0 => x = 200. With A(0) = A(400) = 0, the maximum area is A(200) = 80,000 m2 . The dimensions are 200 m by 400 m. 9. (a) We minimize the weight = tS where S is the surface area, and t is the thickness of the steel walls of the tank. The surace area is S = x 2 + 4xy where x is the length of a side of the square base of the tank, and y is its depth. The volume of the tank must be 500ft3 => y = ^ . Therefore, the weight of the tank is w(x) = t(x 2 + y y ). Treating the thickness as a constant gives w, (x) = t(2x - ^ ) for x.0. The critical value is at x = 10. Since w"(10) = t(2 + ^ ) > 0, there is a minimum at x = 10. Therefore, the optimum dimensions of the tank are 10 ft on the base edges and 5 ft deep. (b) Minimizing the surface area of the tank minimizes its weight for a given wall thickness. The thickness of the steel walls would likely be determined by other considerations such as structural requirements. 11. The area of the printing is (y - 4)(x - 8) = 50. Consequently, y = (y ⅛ ) ÷ 4. The area of the paper is A(x) = x (y ⅛ ÷ 4 ) , where 8 < x. Then — ( 50

A (x )-

1 n

v f

50

λ _

4(X - 8 ) 2 - 4 0 0 _ ∩

+4) - x ^ y z y ^ - —

=> the critical points are —2 and 18, but —2 is not in the domain. T husA , , (18) > 0 => a t x = 1 8 w e have a minimum. Therefore the dimensions 18 by 9 inches minimize the amount of paper. 13. The area of the triangle is A(0) = ⅛ ≡ 0 5 where 0 < 0 < π. Solving A'(0) = 0 =^ ^ ≡ ^ = 0 => 0 = ∣. Since A"(0) _ _ ab ^n g _^ A √ ( ^ 8(x 3 - 21x2 + 108x - 140) = 0 => 8(x - 2)(x - 5)(x - 14) = 0. Since 14 is not in the fomain, the possible values of x are x = 2 in. or x = 5 in. (f) The dimensions of the resulting box are 2x in., (24 —2x) in., and (18 —2x). Each of these measurements must be positive, so that gives the domain of (0, 9). 19. Let the radius of the cylinder be r cm, 0 < r < 10. Then the height is 2 √ 1 0 0 - r 2 and the volume is V(r) = 2πr 2 ᅟ ∕1 0 0 - r2 cm 3 . Then, V'(r) = 2πr2 ( ^ = = ) ( - 2 r ) + (2 π ᅠ ∕1 0 0 - r 2 )(2 r)

= ~ ^ ~ 7 ^ ⅞ ~ ^ ~ ^ 1 0 0 -^ ^

τ

he c ^

c a ^ P0 *n t

f° r θ
y∣and V'(r) < 0 for 1 0 ^ ∕∣< r < 10, the critical point corresponds to the maximum volume. The dimensions are r = 1 0 γ ∕∣≈ 8.16 cm and h = ^

≈ 11.55 cm, and the volume is y ^ ≈ 2418.40 cm3 .

21. (a) From the diagram we have 3h ÷ 2w = 108 and V = h 2 w ^ V(h) = h 2 (54 - ∣h) = 54h2 - ∣h 3 . Then V'(h) = 108h - ∣h 2 = ∣h(24 - h) = 0 => h = 0 or h = 24, but h = 0 results in no box. Since V"(h) = 108 —9h < 0 at h = 24, we have a maximum volume at h = 24 and w = 54 - ∣h = 18. (b)

(24,10368) Absmax 10000 8000 6000 (v = 54Λ2 -

4000 2000 -

OL≤ J— 5

— ∣  — ∣ —  — ∣ — 10

15

20

25

30

35

W

Section 4.5 Applied Optimization Problems

123

23. The fixed volume is V = τrr2 h + j π r 3 => h = ^ — y , where h is the height of the cylinder and r is the radius of the hemisphere. To minimize the cost we must minimize surface area of the cylinder added to twice the surface area of the hemisphere. Thus, we minimize C = 2πrh + 4τrr2 = 2ττr ( ⅛ - y ) + 4τrr2 = ^ + ∣πr 2 . Then ^ = - ^ + ∙^τrr = 0 ≠* V = ∣πr 3 ≠- r = ( ^ ) '^ 3 . From the volume equation, h = ^ - j _ -

π

0 when x > y . Thus L2 is minimized when x = y . (c) When x = y , then L ≈ 11.0 in.

.

L 35 30 25 20 15

^

5 g 7 &.. r τ¾

X

27. Note that h2 + r2 = 3 and so r = ᅟ ∕3 —h2 . Then the volume is given by V = ∣r2 h = ∣(3 - h 2 )h = πh - ∣h 3 for 0 < h < √ z3, and so ^ = π —πr 2 = π (l —r2 ). The critical point (for h > 0) occurs at h = 1. Since ^ > 0 for 0 < h < 1, and ^ < 0 for 1 < h < ᅟ ∕3 , the critical point corresponds to the maximum volume. The cone of greatest volume has radius √¾ m, height lm, and volume y m3 . 29. If f(x) = x2 + t ’ then f , (χ ) = 2x —ax- 2 and f ,'(x) = 2 + 2ax~3 . The critical points are 0 and 3y ∣, but x ≠ 0. Now f' , ( ᅟ∕J ) = 6 > 0 => atx = ᅟ∕ f there is a local minimum. However, no local maximum exists for any a. 31. (a) s(t) = -1 6 t 2 ÷ 96t + 112 => v(t) = s'(t) = -3 2 t ÷ 96. At t = 0, the velocity is v(0) = 96 ft∕sec. (b) The maximum height ocurs when v(t) = 0, when t = 3. The maximum height is s(3) = 256 ft and it occurs at t = 3 sec. (c) Note that s(t) = -1 6 t 2 + 96t + 112 = -1 6 (t + l)(t - 7), so s = 0 at t = - 1 or t = 7. Choosing the positive value of t, the velocity when s = 0 is v(7) = —128 ft∕sec.

124 33∙

Chapter 4 Applications of Derivatives b

⅛

^

h

= ^ /(8 + γ )

= 2

8 +

T andL(x) = ^ h 2 + (x + 27)2

+ (x + 27)2 when x ≥ 0. Note that L(x) is

minimized when f(x) = (8 + ^ )

2

+ (x + 27)2 is

minimized. If f , (x) = 0, then 2 ( 8 ÷ ¥ ) ( - ^ ) ÷ 2 (x ÷ 27) = 0 =Φ (x + 27) (1 - ^ ) = 0 => x = - 2 7 (not acceptable since distance is never negative or x = 12. ThenL(12) = √ 21 97 ≈ 46.87 ft. 35. (a) From the situation we have w 2 = 144 - d 2 . The stiffness of the beaħι is S = kwd3 = kd 3 (144 —d 2 ) 1^2 , where 0 ≤ d ≤ 12. Also, S'(d) = ⅛ ≡ = ^ => critical points at 0, 12, and 6 A ∙ Both d = 0 and √ 144- d2 d = 12 cause S = 0. The maximum occurs at d = 6 ^ 3 . The dimensions are 6 by 6^/3 inches. (c) 6000 5000 4000 3000 2000 1000

Both graphs indicate the same maximum value and are consistent with each other. The changing of k has no effect. 37. (a) s = 1 0 c o s ( π t) => v = —10π sin(τrt) => speed = ∣ 10π sin (πt)∣= 10π ∣ sin (πt)∣=> the maximum speed is 10π ≈ 31.42 cm/sec since the maximum value of ∣ is 1; the cart is moving the fastest at t = 0.5 sec, sin (πt)∣ 1.5 sec, 2.5 sec and 3.5 sec when ∣ is 1. At these times the distance is s = 10 cos ( ∣) = 0 cm and sin (πt)∣ a∣= 10π2 ∣ a = - 1 0 π 2 cos(πt) => ∣ cos(πt)∣=> ∣ a ∣= 0cm ∕sec 2 (b) ∣ cos (πt)∣ a ∣= 10π2 ∣ is greatest at t = 0.0 sec, 1.0 sec, 2.0 sec, 3.0 sec and 4.0 sec, and at these times the magnitude of the cart’s position is ∣ s∣= 10 cm from the rest position and the speed is 0 cm/sec. 2 2 1 /2 2 2 39. (a) s = √ (1 2 - 12t) + (8t) = ((12 - 12t) + 64t ) 2 (b) ∣ = | ((12 - 12t)

+

64t2 ) " v 2 [2(12 - 12t)(-12) + 128t] =

=> ⅛ 11=0 = ” 12 knots

and

≡ It=ι

=

$ knots

(c) The graph indicates that the ships did not see each other because s(t) > 5 for all values of t.

7

⅛

* ⅛

Section 4.5 Applied Optimization Problems (d) The graph supports the conclusions in parts (b) and (c).

(e )

t⅛

A

—y ⅛ [ Γ > '^ ^

-

-

A 5 f w - Vz2θ8 - 4 √ 13

which equals the square root of the sums of the squares of the individual speeds. 41. If v = kax - kx2 , then v' = ka - 2kx and v" = -2k, sov' = 0 => x = ∣. At x = ∣there is a maximum since v" (1)

=

“ 2k < 0. The maximum value of v is ^ .

43. The profit is p = nx - nc = n(x - c) = [a(x - c)^ 1 + b(100 - x)] (x - c) = a + b(100 - x)(x - c) = a + (be ÷ 100b)x - 100bc - bx2 . Then p'(x) = be + 100b - 2bx and p"(x) = —2b. Solving p'(x) = 0 => x = ∣+ 50. At x = ∣÷ 50 there is a maximum profit since p"(x) = —2b < 0 for all x. 45. (a) A(q) = kmq

1

+ cm + ∣q, where q > 0 => A'(q) = -km q

critical points are - ^ ⅛ ^ , 0, and ^ ψ , but only ^ ψ

2

÷ ∣= ⅜ ⅜ ~

and

^ (^ )

=

2kmq

3.

The

is in the domain. Then A" ^ ^ ^ ^ > 0 => at

q = y ⅛ there is a minimum average weekly cost. (b) A(q) = 0 => A'(q) = 0 at q = ^ / ^ as in (a). Also A"(q) = 2kmq

3

> 0 so the most economical quantity to order is

which minimizes

the average weekly cost. 47. The profit p(x) = r(x) - c(x) = 6x - (x3 - 6x2 + 15x) = - x 3 ÷ 6x2 - 9x, where x ≥ 0. Then p'(x) = - 3 X2 + 12x - 9 = —3(x —3)(x - 1) and p', (x) = —6x ÷ 12. The critical points are 1 and 3. Thus p"(l) = 6 > 0 => a t x = l there is a local minimum, and p"(3) = —6 < 0 => atx = 3 there is a local maximum. But p(3) = 0 => the best you can do is break even. 49. (a) The artisan should order px units of material in order to have enough until the next delivery. (b) The average number of units in storage until the next delivery is ψ and so the cost of storing then M ⅛ ) per day, and the total cost for x days is (^ )sx . When added to the delivery cost, the total cost for delivery and storage for each cycle is: cost per cycle = d + ^sx. (c) The average cost per day for storage and delivery of materials is: average cost per day = ⅛ h ∆ l + ^ + ^ x . To minimize the average cost per day, set the derivative equal to zero. ^ (d(x)~ 1 + ^ x ) = - d(x)^ 2 ÷ ^ = 0 ^

χ

= ÷ √ f ∙ Only the positive root makes sense in this context so that x* = ^ . To verify that x*gives a

minimum, check the second derivative ^ ( - d(x) The amount to deliver is px* = u ^

2

+ ^

125

126

Chapter 4 Applications of Derivatives (d) The line and the hyperbola intersect when ^ = ^x. Solving for x gives Xintersectio∏ =

i

y ∣∙ F° r x > 0,

Xintersection = ^ ∣ = x*. From this result, the average cost per day is minimized when the average daily cost of delivery is equal to the average daily cost of storage. 51. We have ^ = CM - M2 . Solving ^ = C - 2 M = 0=>M = j . Also, ^

= - 2 < 0 => atM = ∣there is a

maximum. 53. If x > 0, then (x — 1)2 ≥ 0 => x2 + 1 ≥ 2x => ⅛ 1 ≥ 2. In particular if a, b, c and d are positive integers,

*«■ (⅛1) (⅛ 1) (⅛j ) (⅛1) ≥ 1655. At x = c, the tangents to the curves are parallel. Justification: The vertical distance between the curves is D(x) = f(x) —g(x), so D'(x) = f '(x) —g, (x). The maximum value of D will occur at a point c where D, = 0. At such a point, f , (c) —g'(c) = 0, or f '(c) = g'(c). 57. (a) If y = cot x — χ ∕2 esc x where 0 < x < π, then y, = (esc x) ( √ ^ cot x —esc x ) . Solving y, = 0 => cos x = ^

=> x = f . For 0 < x < ∣we have y, > 0, and y' < 0 when ∣< x < π. Therefore, at x = J

there is a maximum value of y = —1.

The graph confirms the findings in (a). 59. (a) The square of the distance is D(x) = (x - ∣) 2 ÷ ( √ x + θ) 2 = x2 - 2x + ∣, so D'(x) = 2x ~ 2 and the critical point occurs at x = 1. Since D'(x) < 0 for x < 1 and Dz(x) > 0 for x > 1, the critical point corresponds to the minimum distance. The minimum distance is

∕D(1) = ^ .

A

The minimum distance is from the point (∣, 0) to the point (1, 1) on the graph of y = χ∕x, and this occurs at the value x = 1 where D(x), the distance squared, has its minimum value.

Section 4.6 Indeterminate Forms and L'HopitaΓs Rule

127

61. (a) The base radius of the cone is r = 2™„x and so the height is h = √ a 2 —r2 = y a2 — ( ¾ - i ) 2 . Therefore, V(x) = f r2 h = f ( ⅛ A ) 2 ^ √ ⅛

ψ

.

(b) To simplify the calculations, we shall consider the volume as a function of r: volume = f(r) = ∣r2 ᅟ ∕a 2 - r2 , where 0 < r < a. f z (r) = f ⅛ ( r 2 √ a 2 - r 2 ) = f r2 ZE 2a⅛ - 3r3 3 √ a 2 - r2

=

critical point occurs when r2 = y , which gives r = a √ ^ = ^ . Then

πr(2a2 - 3r2 ) . The 3ᅟ ∕a 2 —r2

_

^ f ~ ^ ∙ Using r = ^

∣ 2 _ ⅛ l

3

and h = ^ , we may now find the values of r and h

for the given values of a. When a = 4: r = ^ , h = ^ ; When a = 5: r = ⅛ , h = When a = 6: r = 2 √ 6 , h = 2 √ ⅛ When a = 8: r = ⅛ (c) Since r = ^

h= ^ ;

and h = ^ , the relationship is j = √ ⅞

4.6 INDETERMINATE FORMS AND L'HOPITAL'S RULE = ∣or l i m ^ ⅛ = ⅛

1. ΓH6pitak lim ⅜ ⅛ = ⅛ I

3. ΓHδpitaL

χ

lta

¾ ⅜ =

5. I'HSrpital: lim ⅛ x x →0 = 1¾

1 0

⅛

χ

9.

lim ≡ ∣= lim ⅛ = 4 = 1 0 →π 7r^* θ→ π ~1 ^1 ∣ im

cos x + sin x

x →π∕4

1

11.

lim ≤ ⅛ ∞ sx x →π∕4 4

13.

lim - ( x - ∣) t a n x = lim 27 x →√2 V x →τr∕2

17. lim x →0 19. x

→0

÷

x →l

√ a(a + x ) - a _

x ( c o s x - l) _ s in x -x

=

χ

_

n

a

-1

lim

→ 7Γ∕2

2

=

x →l ^

= 5

= Λ m∞ ⅛ t = 7

( l - c o s x) ∕l ÷ c o s χ  x2 1 + cos x /

lim x →0

1 2

2

- ⅛ r ~ ∞s x

χ

⅛

V22 τ+ V2 —λv ∕5 ^

2 χ * - 3 χ 3 y ⅛

lim

= ⅛

= i2 or limn ⅛ x x →0

( ∣- x) cos x + sin x ( - l ) -sin x

θ γ l ∕2

⅛

x^1

=

,

g = 5 or χ lim o ^

/ sin x ∕s i n x / 1 v x / x / 1 + cos x /

l ) = ⅛

lim ≡ f = lim ⅛ 1 ^ = 0 t →o , t →o

n

= ⅛

= lim ⅛2x = lim ≡2 x →0 x →0

7.

15. lim ⅛ x →l

1

lim o ⅞

(7 ⅛

u

f a

-

i

γ

⅛ = 1

1 ^

=

.

1

1

= ∣,w h e re a > 0 .

ι∙ -xsin X+ cos X- 1 _ um -xcos x - 2sin x _ ~ → 0 cosx —1 ~ χ - →Q —sin χ

x

ι∙ xcos x + 2sin x _ s in x → Q

x≡

0

-xsin x ÷ 3 cos x ∞ sx “

3 I ”

30

128

Chapter 4 Applications of Derivatives a ^_ 1^ 1

21. limι 23.

= l⅛

a (n

( x - ∙ᅟ ^ + x )=

lim X→ ∞

∖

v

— an limι rn

' f ^^

= an, where n is a positive integer.

fx - √ x 2 + xW i ± ^ ^ ^

lim

/

1

∖

x → ∞

∕∖ x ÷ √ x

v

2

+ x ∕

lim

J ^ - ^ =

lim

X→ ∞

X + √ X 2 +X

X→ ∞

----- Λ χ x

__ -

2 5

∙

ι∙ X→ ∞

— √ ⅛ √x 2

—1 __ __1 । ΓHopitaΓs rule ∖ √ 1+ I 2 ∖ is unnecessary )

1 +

1⅛ x→ ±∞

+

⅛ "

+2

3 4x - 1

lim

x+ z

X→ ± ∞

= 0

2 = √9 = 3

27. x lim →∞ 29 * x → lim √ 2 - ^t a n x =

∣ ∙ χ J ^

1 (∞ lsx _ xlim → √ 2 (- - 1V-cosx√ sm xJ

_ j_ s in x_

2~

1

31. Part (b) is correct because part (a) is neither in the § nor ^ form and so ΓHoρitaΓs rule may not be used. 33. If f(x) is to be continuous at x = 0, then lim f(x) = f(0) => c = f(0) = lim x →0

= lim x

→0

2⅞

3x

30x

8 1 ⅞s 3x

= lim

30

χ →o

=

10

9x

x →0

3¾ ιn 3 x

= iim

jx

x →0

9

9 c θs

ox

3χ

.

35. The graph indicates a limit near —1. The limit leads to the indeterminate form 0 : lim 2 x ~< 3 x ÷ 1 ) √ χ + 2 υ

=

lim

x→1

χ

—1Ξ1Ξ1

— 4-5 _

χ-

x→1 =

—1

lim

χ→1

1 1

37. Graphing f(x) = ^--∏M- on th window [—1, 1] by [—0.5, 1] it appears that limθf(x) = 0. However, we see that if we let u = x 6 , then lim f(x) = lim v x →0

u →0

39. (a) By similar triangles, ^

Thus ^ (vb)

1

_

“ 0™ o

= lim

u →0

2u

= lim ≡2 u →0

= ⅜. 2

= ^ where E is the point on AB such that CE ± AB :

= ∣z g y , since the coordinates of C are (cos 0, sin 0). Hence, 1 —x =

liπι (v1 - x)7 = lim

0 →O

1 ~⅞ os u lu

0 →O

0 -s ιn 0

0 ( - s i n 0) + c o s0 + 2cos0 _

∞s 0

= lim

0 →O

1-CO S0

- 0 s i n 0 + 3cos0 _

~ 0 →o ^

= lim

0 →O

0+ 3 _

o

j

g ≡ H ⅝ ÷ sm 0

sM

^

1

-

s s -θ ^

= ιi m

0 →O

. 0cQ s0÷ 2sin0 s ιn ^

Section 4.7 Newton's Method (c) We have that lim

⅛ S a - ( ι- → )

f(l - x) — (1 - cos 0)1j - lim 0 →∞ 0 →∞

=

lim (1 —cos 0 →∞

129

θ — sin 0

As 0 → ∞, (' 1 —cos 0) oscillates between 0 and 2, and so it is bounded. Since 0 limQQ K - - - σ — 1/) = 1 — 1 = 0,’ ∖ σ — sin a ,

lim (1 —cos 0)7 θ→ ∞

e 0 - sin 0

— 1 = 0. Geometrically, this means that as θ → ∞ , the distance between points P and D

approaches 0. 4.7 NEWTON'S METHOD 1. y = χ 2 + χ - 1 => y, = 2x + 1 ≠> xn+l = x n - ⅛ ζ ¾ p ; Xo = 1 => x 1 = 1 - ¾ ¾ = F

w

H

4

4

≈

3. y - x4 + x - 3 =► y' = 4x3 + 1 ≠> xn+1 = x n - ¾ ^ v — 6 *2 - 5 -

≠

= -2 4

yυ

= j

x1 = 1 - ⅛ Γ = ~ 2

-61905; x0 = - 1

x2 = - 2 - ⅛ ⅛ = - ∣≈ -1.66667

4

c

1

_

⅞

+

S^3

— 6 - J

4 _ 0 z √ X ~^ y

— 5 _ 113 _ “ 4 2000 ~

1296+750-1875 _ 6 4320+625 ~ 5 ~ 4945 ~

3

; ⅞ = 1 ≠> x 1 = 1 - ¾ 1

171 _ 5763 1 i r e ^ . , 4945 ~ 1-16542, X0 - - 1

4X

x

n~2 . V — 1 → 4χ3 ’ ^ 0 1

. _ =≠- X∏+ι — Xn

2500-113 _ 2000 “

2387 20∞ ~

V ^1

1 1

9. If XQ — h > 0 => xι — xo

if xo — h < 0 => xι — xo " ^

h

+ W

f∕ ^

f^

—h

f, ^

— h

f

h)

(2 √ h )= h ,

11. i) is equivalent to solving x3 —3x — 1 = 0. ii) is equivalent to solving x3 —3x — 1 = 0. iii) is equivalent to solving x3 —3x — 1 = 0. iv) is equivalent to solving x3 —3x — 1 = 0. All four equations are equivalent. 13. For x0 = —0.3, the procedure converges to the root —0.32218535.... (a)

Plot! ∏ φβ P¼B

_

11 - 1 - 3

∖ylB×^3+3×+l ×y2BnDer ×)

4

_

5 V _ 5 4 —^ ^ 2 — 4

ι⅛

_ 5 — 4

625-512 2000

gives x i = x0 => χ 2 = χ 0 ≠> xn = χ 0 for all n ≥ 0. That is, all of

the approximations in Newton’s method will be the root of f(x) = 0.

^h ^ ⅛

,

X1 - - 1 -

c 1l i iyn q j j

7. f(xo) = 0 and f'(xo) ≠ 0 => xn+ ι = x n — f ^

zz

_ 1 4

= f

1⅛ ⅛ = - 2 + ∣ }= - ∣ l ≈ -1.64516

x2 = - 2 -

v

1

(b)

130

Chapter 4 Applications of Derivatives

(c)

×-yi×y2÷×

-.322324152194 -.322185360292 -.322185354626 -.322185354626

(d) Values for x will vary. One possible choice is xo = 0.1. T÷×

i ×-yl×y2÷× -.329372795587 -.322200595043 -.322185354698 -.322185354626

(e) Values for x will vary. 15. f(x) = tan x —2x => f'(x) = sec2 x —2 => xn+1 = xn — ~ ⅛ 7 r r 1 ; xo = 1 => Xi = 12920445 sec

=> x2 = 1.155327774 ^

)

x i6 = x∏ = 1.165561185

17. (a) The graph of f(x) = sin 3x —0.99 + x2 in the window —2 ≤ x ≤ 2, —2 ≤ y ≤ 3 suggests three roots. However, when you zoom in on the x-axis near x = 1.2, you can see that the graph lies above the axis there. There are only two roots, one near x = —1, the other near x = 0.4. (b) f(x) = sin 3x —0.99 + x2 => f , (x) = 3 cos 3x + 2x v

x "÷'

=

x"

sin(3xn) - 0.99+x∏

- - ≡

) +

√

, .,

a n d th e

..

1 solutions

are approximately 0.35003501505249 and -1.0261731615301 19. f(x) = 2x4 - 4x 2 + 1 ≠> f'(x) = 8X3 - 8x => xn+1 = xn - ¾ ⅛ 7 ; if x0 = - 2 , then x6 = -1.30656296; if xo = —0.5, then X3 = —0.5411961; the roots are approximately ± 0.5411961 and ± 1.30656296 because f(x) is an even function. 21. From the graph we let x0 = 0.5 and f(x) = cos x - 2x

=►⅛ = - ∙- l⅛ ⅜

s

x,=.450t>3

=≠> X2 = .45018 => at x ≈ 0.45 we have cos x = 2x.

23. If f(x) = x3 + 2x - 4, then f(l) = —1 < 0 and f(2) = 8 > 0 => by the Intermediate Value Theorem the equation x 3 ÷ 2x —4 = 0 has a solution between 1 and 2. Consequently, f'(x) = 3x2 + 2 and xn+ ι = xn — ⅛ ⅛ τ ^ ∙ Thenxo = 1 => Xi = 1.2 4 1.17951.

X2 — 1.17975 ≠ X3 = 1.179509 ≠ ∙ X4 = 1.1795090 ≠

the root is approximately

Section 4.8 Antiderivatives 25. f(x) = 4x4 —4X2 => f'(x) = 16x3 —8x => Xi+ ι = xi — ^ j = xi — ^ r ⅛ ∙ Iterations are performed using the procedure in problem 13 in this section. (a) For xo = 2 or XQ = —0.8, Xi → —1 as i gets large. (b) For xo = —0.5 or x0 = 0.25, Xi → 0 as i gets large. (c) For x0 = 0.8 or x0 = 2, Xi → 1 as i gets large. (d) (If your calculator has a CAS, put it in exact mode, otherwise approximate the radicals with a decimal value.) For xo = - 3^ ∙ or x0 = - ^ , Newton’s method does not converge. The values of Xi alternate between xo = —2^

or x0 = - ^ y ∙ as i increases.

27. f(x) = (x - I) 40 4- f'(x) = 40(x - I)39 =► xn+1 = xn - ⅛

⅛

= ⅞

t l

∙ With x0 = 2, our computer

gave x8 7 = χ 88 = χ 89 = ∙ ∙ ∙ = χ 2θo = 1.11051, coming within 0.11051 of the root x = 1. 29. f(x) = x3 + 3.6x2 - 36.4 => f'(x) = 3x2 + 7.2x

xn+1 = xn -

^ ^ + 1 ^ 4 :χo=

2

χ

ι = 2.5303

≠,

=> X2 = 2.45418225 xj = 2.45238021 ≠> X4 = 2.45237921 which is 2.45 to two decimal places. Recall that 4 + x = 10 [H3 0+] ≠∙ [H3 O ] = (x) (10~4 ) = (2.45) (10" 4 ) = 0.000245 4.8 ANTIDERIVATIVES

1. (a) x2

(b)

3. (a) X- 3

(b)

5. (a)

-1 X

(b)

x3 3

(c) y - X2 + X (c) - y + X2 + 3x

3

Ξ

5

(c)

X

2 x

+

X

7. (a) √X^^

(b)

(c) 5 √ √ + 2 ^

9. (a) X2∕ 3

(b) x 1∕ 3

(c) x - 1 ∕ 3

11. (a) cos (πx)

(b) —3 cos x

(c) ⅛ + c o s ( 3 x )

13. (a) tan x

(b) 2 tan (∣)

(c) - 5 t a n (⅜ )

15. (a) —CSC X

(b) ∣esc (5x)

(c) 2 c s c ( f )

17. J (x + 1) dx = y + x + C 19. ∫ (3t2 ÷ i ) d t = t 3 + ( + C 21. ∫ (2x3 - 5x + 7) dx = ∣x4 - ∣x2 + 7x + C 23 -

/ ( ⅛-

25.

f

χ2

- ⅛)

dx

= / (x ^ 2 - x

2

x- 1 ∕ 3 dx = 2y + C = ∣x2∕ 3 + C

- j) d x = ⅛ -⅛ - iχ + C= - i - ⅛

- ∣+ C

131

132

Chapter 4 Applications of Derivatives

27. / ( y ∕x + 3√ x ) dx = J* (x1∕ 2 + x1∕ 3 ) dx = i r + τ

+ C = 5 x3^2 + 5 x4 ∕ 3 + C

29. J (8y - ^ ) dy = ∫ (8y - 2y~1∕ 4 ) dy = ⅞ - 2 ( ¢ ) + C = 4y2 - ∣y3 ∕ 4 + C 31. § 2x (1 —x- 3 ) dx = / (2x —2x- 2 ) dx = ^ —2 ( ⅛ ) + C = x2 + ∣+ C

33. ∫ ⅛ ^ dt = ∫ ( 7 + y ) dt = f ( r 1∕ 2 + r 3∕ 2 ) dt =

+ ( ^ r ) + C = 2√t -

35. f —2 cos t dt = —2 sin t + C 37. J*7 sin f d0 = -2 1 cos f + C 39. f —3 csc2 x dx = 3 cot x + C 41. J ≡ ^ o t β d 0 = - I

c sc0

+C

43. J (4 sec x tan x —2 sec2 x) dx = 4 sec x —2 tan x + C 45. f (sin 2x —csc2 x) dx = - ∣cos 2x + cot x + C 47. J ∏ ≈ ⅛ d t = f (∣+ i c o s 4 t ) d t = ∣t + i ( ^ ) + C = i + ⅛ ι + C 49. / ( 1 + tan2 0) d0 = / sec2 0 d0 = tan 0 + C 51. / cot2 x dx = / (esc2 x —1) dx = —cot x —x + C 53. / cos 0 (tan 0 + sec 0) d0 = f (sin 0 + 1) d0 = -cos 0 + 0 + C 55.

A

(< 2 ^

57. ⅛ ( I t a n (5 x 59∙ ⅛ ⅛

C)

+

-

= 4( 2 4 ^ = ( 7

X -^^^

1) + C) = 5 (sec2 (5x - 1)) (5) = sec2 (5x - 1)

+ C) = ( - l) ( - D ( x + I)’ 2 =

61. (a) Wrong:

^ ( y sin x + c ) = y sin x + y cos x = x sin x + y cos x ≠ x sin x

(b) Wrong:^ (—x cos x + C) = —cos x + x sin x ≠ x sin x (c) Right: ⅛ ( - x cos x + sin x + C) = —cos x + x sin x + cos x = x sin x 63. (a) Wrong:

+ (2 ⅛ ^ + c) = ^ ⅞ ≡

= 2(2x + 1)2 ≠ (2x + 1)2

(b) Wrong: ⅛ ((2x + 1)3 + C) = 3(2x + 1)2 (2) = 6(2x + 1)2 ≠ 3(2x + 1)2 (c) Right: ⅛ ((2 X + 1 ) 3 + C) = 6(2X + 1 ) 2

+ C

Section 4.8 Antiderivatives 65. Graph (b), because ^ = 2x => y = x 2 + C. Then y(l) = 4 4 67- ⅛ = ⅛

=

2x

x =

C = 3.

y - x 2 —7x + C; at x = 2 and y = 0 we have 0 — 22 —7(2) + C => C = 10 => y = x2 —7x + 10

^^ 7 4

? +

133

x

'

2

+

x

4

y ~ —χ - 1 + ⅛ + C; at x = 2 and y = 1 we have 1 = - 2 ^ 1 + f + C 4 ∙ C = - ∣

=> y = - χ - 1 + ⅛ - ∣o ry = - i + ⅛ - ∣ ^ ' ⅛

=

3x - 2 ∕ 3 4

y = ⅛ - + C = 9; at x = 9x 1^3 + C; at x = —1 and y = —5 we have —5 = 9(—I ) 1/ 3 + C ≠ ∙ C = 4 3

≠> y = 9χ 1∕3 _|_ 4 ^

,

⅛ = 1+

^

,

⅛

=

- π

cos t

s *n

4

s =

t

+ sin t + C; at t = 0 and s = 4 we have 4 = 0 + sin0 + C => C = 4 => s — t + sin t + 4

π0 => r = cos (τr^) + C; at r = 0 and 0 = 0 we have 0 = cos (πθ) + C => C = —1 => r = cos (π0) — 1

77. ⅛ = ⅜ sec t tan t => v = ∣sec t + C; at v = 1 and t — 0 we have 1 = ⅜ sec (0) + C => C = ⅜ => v = ∣sec t + ⅜ 79∙

4

S = 2 - 6x 4

⅛ =

2x -

I = 2x - 3x2 + Ci; at ^ = 4 and X = 0 we have 4 = 2(0) - 3(0)2 + C 1 ≠> C i = 4 3x2 + 4 => y = x 2 - x3 + 4x + C2; at y = 1 and x = 0 we have 1 = 0 2 —0 3 + 4(0) + C2 => C 2 = 1

=> y = χ 2 - χ 3 + 4x + 1 81

∙ ⅛ = $ =

2 t ^3

4

5 = -t^ 2 +

t^ 1

c

β a t^ = l a n d t - l we

≠- r = + 2t + C 2 ; at r = 1 and t = 1 we have 1 = r = I + 2t - 2

1^^1

hav

e l--(l)-2 +

to =

3χ2

- 8x + C 2 ; at ⅛ = 0 and x = 0 we have 0 =

ι ^ C ι= 2

+ 2(1) + C 2 => C 2 = —2 => r

83. § = 6 ≠> ⅛ = 6x + C β at ⅛ = - 8 and x = 0 we have - 8 = 6(0) + C i ^ 4

c

3(0)2

—t- 1

C 1 = - 8 => g

4

j = -Γ

2

+ 2

+ 2t - 2 or

= 6x - 8

- 8(0) + C 2 => C 2 = 0 => ^ = 3x2 - 8x

=> y = χ 3 —4χ 2 + C3; at y = 5 and x = 0 we have 5 = 03 —4(0)2 + C 3 +

C3 = 5 =+ y = χ 3 —4x 2 + 5

85. y ^ = -s in t + cos t =+ y", = cos t + sin t + C i; at y", = 7 and t = 0 we have 7 = cos (0) + sin (0) + Ci =+ Ci = 6 =+ y, " = cos t + sin t + 6 4 y" = sin t —cos t + 6t + C2; at y" = —1 and t = 0 we have - 1 = sin(0) —cos(O) + 6(0) + C2 =+ C2 = 0 =+ y" = sin t —cos t + 6t =+ y, = —cos t —sin t + 3t2 + C3; at y, = - 1 and t = 0 we have - 1 = —cos (0) - sin (0) + 3(0)2 + C3 =+ C 3 = 0 =+ y, = —cos t —sin t + 3t2 => y = —sin t + cos t ÷ t3 ÷ C4; at y = 0 and t = 0 we have 0 = -s in (0) + cos (0) ÷ 03 + C4 =+ C4 = —1 =+ y == -s in t + cos t + 13 — 1 87. m = y, = 3 λ ∕ x = 3x 1∕ 2 =+ y = 2x3 ∕ 2 + C; at (9,4) we have 4 = 2(9)3 ^2 ÷ C + 89.

C = —50 =+ y = 2x3 ∕ 2 —50

= 1 — ∣x 1∕ 3 =+ y == f (1 - ^ x 1∕ 3 ) dx = x —x 4 ∕ 3 + C; at (1,0.5) on the curve we have 0.5 = 1 — l 4 ∕ 3 + C =+ C = 0.5 =+ y = x - x4 /3 + ∣

91. ^ = sin x —cos x =+ y = J (sin x —cos x) dx = —cos x —sin x + C; at (—π, —1) on the curve we have —1 = —cos (—π) —sin (—π) + C =+ C = —2 =+ y == —cos x —sin x —2 93. (a) ∣ = 9.8t —3 =+ s = 4.9t2 —3t + C; (i) at s = 5 and t = 0 we have C = 5 =+ s = 4.9t2 —3t + 5; displacement = s(3) —s(l) = ((4.9)(9) - 9 + 5 ) - (4.9 - 3 + 5) = 33.2 units; (ii) at s = —2 and t = 0 we have C = - 2 => s = 4.9t2 - 3t - 2; displacement = s(3) - s(l) = ((4.9)(9) - 9 - 2 ) - (4.9 - 3 - 2) = 33.2 units;

134

Chapter 4 Applications of Derivatives (iii) at s = So and t = 0 we have C = so 4

s = 4.9t2 —3t +

SQ; displacement

= s(3) —s(l)

- ((4.9)(9) - 9 + s0 ) - (4.9 - 3 + s0 ) = 33.2 units (b) True. Given an antiderivative f(t) of the velocity function, we know that the body's position function is s = f(t) + C for some constant C. Therefore, the displacement from t = a to t = b is (f(b) + C) —(f(a) + C) = f(b) —f(a). Thus we can find the displacement from any antiderivative f as the numerical difference f(b) —f(a) without knowing the exact values of C and s. ∣ = - k t + C β at ^ = 88 and t = 0 we have Ci = 88 =+ ∣ = - k t + 88 4

95. Step 1: ⅛ = - k

s = - k ( 0 + 88t + C2; at s = 0 and t = 0 we have C2 = 0 => s = — ^ + 88t Step 2: ∣ = 0 =+ 0 = - k t + 88 ≠- t = f Step 3: 242 =

+ 88 ( ^ ) ≠- 242 = - ⅞

97. (a) v = ∫ a d t =

f

+ ⅞

=+ 242 =

=+ k = 16

(15t1∕ 2 - 3 r 1∕ 2 ) dt = lθt 3 ∕ 2 - 6t1∕ 2 + C; ⅛ (1) = 4 =+ 4 = 10(l) 3 ∕ 2 - 6(1) 1∕ 2 + C =+ C = 0

=+ v = lθt 3 ∕ 2 - 6t1∕ 2 (b) s = J v dt = ∫ ( l θ t 3 ∕ 2 - 6t1∕ 2 ) dt = 4t5 ∕ 2 - 4t3 ∕ 2 + C; s(l) = 0 = + 0 = 4(1)5 ∕ 2 - 4(1)3 ∕ 2 + C =+ C = 0 => s = 4t5∕ 2 —4t3 ∕ 2 ≡

=

a

I

=

J a dt = at + C; ^ — v0 when t = 0 => C = v0 => ∣ = at + v 0 => s = γ ÷ v0 t + Q ; s = s0

when t = 0 ≠> s0 = *

÷ vo (O) + C i => C i = s0 => s = y ÷ v0 t + s0

101. (a) ∫f ( x ) d x = 1 — √ x + Cι = —√ x + C (c) ∫ - f ( x ) dx = - (1 (e)

f

λ ∕x

[f(x) + g(x)] dx = (1 -

(b)

) + Cl = λ ∕ x + C λ ∕x

f g(x) dx =

x + 2 ÷ C ι= x + C

(d) J - g ( x ) dx = - ( x + 2) + Ci = - x + C

) + ( x + 2) + C ι = x - λ ∕ x + C

(f) ∫ [ f ( x ) - g ( x ) ] d x = ( l - √ ^ - ( x ÷ 2 ) + C 1 = - x - √ ^ + C CHAPTER 4 PRACTICE EXERCISES 1. No, since f(x) = x3 + 2x + tan x ≠> f'(x) = 3x2 + 2 + sec2 x > 0 => f(x) is always increasing on its domain 3. No absolute minimum because χ lim o (7 + x)(l 1 —3x)1∕ 3 = —∞ . Next f'(x) = (1 1 —3x)1∕ 3 —(7 + x)( 11 —3X) - 2 ∕ 3 = ^

⅛

^

=

(i∣¾⅜

=> x = 1 and x = y are critical points.

Since f' > 0 if x < 1 and f' < 0 if x > 1, f(l) = 16 is the absolute maximum. 5. Yes, because at each point of [0,1) except x = 0, the function's value is a local minimum value as well as a local maximum value. At x = 0 the function's value, 0, is not a local minimum value because each open interval around x = 0 on the x-axis contains points to the left of 0 where f equals —1. 7. No, because the interval 0 < x < 1 fails to be closed. The Extreme Value Theorem says that if the function is continuous throughout a finite closed interval a ≤ x ≤ b then the existence of absolute extrema is guaranteed on that interval.

Chapter 4 Practice Exercises 9. (a) There appear to be local minima at x = -1.75 and 1.8. Points of inflection are indicated at approximately x = 0 and x = ± 1.

(b) f'(x) = x7 - 3X5 - 5X4 ÷ 15X2 = x2 (x2 - 3) (x3 - 5). The pattern y' = indicates a local maximum at x = ᅟ∕5 and local minima at x = ± ᅟ ∕3 .

1.72

1.⅛4

l.⅛δ

1.78

X

11. (a) g(t) = sin2 t - 3t => g'(t) = 2 sin t cos t - 3 = sin (2t) - 3 => g' < 0 => g(t) is always falling and hence must decrease on every interval in its domain. (b) One, since sin2 1 - 3t —5 = 0 and sin2 1 —3t = 5 have the same solutions: f(t) = sin2 1 —3t —5 has the same derivative as g(t) in part (a) and is always decreasing with f(—3) > 0 and f(0) < 0. The Intermediate Value Theorem guarantees the continuous function f has a root in [—3 ,0]. 13. (a) f(x) = x4 ÷ 2x2 —2 => f'(x) = 4x3 + 4x. Since f(0) = - 2 < 0, f(l) = 1 > 0 and f'(x) ≥ 0 for 0 ≤ x ≤ 1, we may conclude from the Intermediate Value Theorem that f(x) has exactly one solution when 0 ≤ x ≤ 1. (b) x2 = — ∙ ^ > 0 ≠> x2 = √ 3 - 1 and x ≥ 0 ≠∙ x ≈ √.7320508076 ≈ .8555996772 15. Let V(t) represent the volume of the water in the reservoir at time t, in minutes, let V(0) = ao be the initial amount and V(1440) = ao ÷ (1400)(43,560)(7.48) gallons be the amount of water contained in the reservoir after the rain, where 24 hr = 1440 min. Assume that V(t) is continuous on [0,1440] and differentiable on (0,1440). The Mean Value Theorem says that for some to in (0,1440) we have V'(⅛) = ^ ¾ ^ _

ao ÷ (.1400)(4^560)(7.48) - ⅜

_ ^ ⅛ ⅛ ! = 316,778 gal/min. Therefore at t0 the reservoir’s volume

was increasing at a rate in excess of 225,000 gal/min. 17. No, ^ ∙ = 1 + ^ τ => ⅛ d — dx∖ x + l∕

( x+O-x(O _ (x+l)2

differs from ^ γ by the constant 1. Both functions have the same derivative

1 _ (x+ l)i

d dx ∖x + l∕

,

19. The global minimum value of ∣occurs at x = 2. 21. (a) t = 0,6,12

(b) t = 3,9

. (c) 6 < t < 12

(d) 0 < t < 6, 12 < t < 14

135

136

Chapter 4 Applications of Derivatives

27.

33. (a) y' = 16 —x2 => y' = ------ ∣+ ÷ ÷ ∣--------=> the curve is rising on (—4,4), falling on (—∞ , —4) and (4, ∞ ) -4 4 => a local maximum at x = 4 and a local minimum at x = - 4 ; y" = —2x => y" = + ÷ ÷ ∣------ => the curve 0 is concave up on (—∞ , 0), concave down on (0, ∞ ) + a point of inflection at x = 0

35. (a) y' = 6x(x ÷ l)(x —2) = 6x3 —6x 2 — 12x => y' = ------ ∣+ + + ∣-------- ∣+ + + => the graph is rising on (—1,0) - 1 0 2 and (2, ∞ ), falling on (—∞ , —1) and (0,2) => a local maximum at x = 0, local minima at x = —1 and X = 2; y" = 18x2 - 12x - 12 = 6(3x 2 - 2x - 2) = 6 ( x - ⅛ ^ ) (x =+ y" = + + + ∣ ----- ∣ + + + =+ the curve is concave up on f —∞ , 1 ^3^ j and f l-√7 3

on ( ⅛ ^ 5 ⅛ ^ ^

V

l+√7 3

=> points of inflection at x =

1≠

^

1+ V

3+

∞ ) , concave down '

Chapter 4 Practice Exercises

137

37. (a) y, = x4 - 2x2 = x2 (x2 - 2) => y' = + + + ∣ ------ ∣------- ∣+ + + => the curve is rising on ( - ∞ , - ᅟ ∕ 2 ) and k 7 -√ 2 θ √2 ( ^ 2 , ∞ ) , falling on ^ - √ 2 , V ^ ) =^

a

l° c ∏l maximum at x = —ᅟ ∕2 and a local minimum at x = √ 2 ;

y" = 4χ 3 - 4x = 4x(x - l)(x + 1) => y" = ------ ∣+ + + ∣--------∣+ + + => concave up on (-1 ,0 ) and (1, ∞ ), -1 0 1 concave down on (—∞ , —1) and (0,1) => points of inflection at x = 0 and x — ⅛ 1 (b)

39. The values of the first derivative indicate that the curve is rising on (0, ∞ ) and falling on (—∞ , 0). The slope of the curve approaches - ∞ as x → 0“ , and approaches ∞ as x → 0+ and x → 1. The curve should therefore have a cusp and local minimum at x = 0, and a vertical tangent at x = 1.

41. The values of the first derivative indicate that the curve is always rising. The slope of the curve approaches ∞ as x → 0 and as x → 1, indicating vertical tangents at both x = 0 and x = 1.

138

Chapter 4 Applications of Derivatives 45. y = 4

47.

— x3 + 2 _ y ~ 2x “ v

« x

f

I 2

r

1 x

s⅛2 x __ l i m t∞ ( × 2 ) ~ x → 0

lim → 0

57. lim

49. y = ⅛

2sin x cos x _ 2xsec 2 (x 2 ) -

sec(7x)cos(3x) = lim v 7 v 7

59. lim (esc x - cot x)7 = lim v x →0

sin(2x) _ 2xsec 2 (x 2 ) -

f √ x 2 + x + l - √ x 2 - x^ =

lim

lim

X → OO

v

v

,

A

+ 1 Λ

√x2 + x + l + vx

/ 2

x

.∙ "

____________ 2cos(2x)____________ __ 2 __ -∣ 1 Q 2x (2sec2 (x 2 )tan(x 2 )∙2x) + 2sec2 (x 2 ) 0 + 2 ∙l -

⅛ a ⅛7 x = ∣ 7

⅛ ¾ = lim

x → π∕2^ ^ t

⅛s l n≡x i = lim

x →0

x→∞ ∖

=

→ o

x → π∕2~ c o s (7 x )

x → π ∕2 -

61.

∣ sm x

= l - ^

( )

≡ ^ = 21 = 0

x→0

co sx

lim

ζ √ x 2 + x + 1 — Λ∕

x →∞

X2

—x^

7 χ 2 + χ + 1+ᅟ ∕χ

ᅟ ∕x

2

+ x + i + ∕x

2 2

- χ -x

—

-x

Notice that x = √ x ^ for x > 0 so this is equivalent to

63. (a) Maximize f(x) = λ ∕ x — ᅟ ∕ 3 6 - x = x 1∕ 2 —(36 - x)1∕ 2 where 0 ≤ x ≤ 36 => f'(x) = ∣x - 1 ∕ 2 — ∣(36 —x)^ 1∕ 2 ( - l ) = ^ ^ ^ 7

^

derivative fails to exist at 0 and 36; f(0) = —6,

and f(36) = 6 => the numbers are 0 and 36 (b) Maximize g(x) = λ ∕ x + √ z36 —x = x 1∕ 2 + (36 - x)1∕ 2 where 0 ≤ x ≤ 36 =^ g, (x) = ∣x - 1 ∕ 2 ÷ ∣(36 —x)^ 1∕ 2 ( - l ) =

2Xv⅞→

^

critical points at 0, 18 and 36; g(0) = 6,

Chapter 4 Practice Exercises

139

g(18) = 2 √ T 8 = 6 √ z2 and g(36) = 6 => the numbers are 18 and 18 y

65. A(x) = | (2x) (27 - x 2 ) for 0 ≤ x ≤ √ 2 7 => A'(x) = 3(3 + x)(3 - x) and A''(x) = -6 x . The critical points are —3 and 3, but —3 is not in the domain. Since A''(3) = - 1 8 < 0 and A ( ^ ) = 0» the maximum occurs at x = 3 => the largest area is A(3) = 54 sq units. 67. From the diagram we have ( ^ ) 2 ÷ r 2 = ( λ ∕^ ) => r 2 = ^ V=

τrr2 h

. The volume of the cylinder is

= π ( ^ - ^ ) h = ∣(12h - h 3 ) , where

0 ≤ h ≤ 2 √ 3 . Then V, (h) = ⅜ (2 ÷ h)(2 - h) => the critical points are —2 and 2, but —2 is not in the domain. At h = 2 there is a maximum since V, , (2) = - 3 π < 0. The dimensions of the largest cylinder are radius = √ z2 and height = 2. 69. The profit P = 2px ÷ py = 2px + p ( ¾ p ) , where p is the profit on grade B tires and 0 ≤ x ≤ 4. Thus P'(x) ~ (5¾ 2 (χ 2

-

lθ x ÷ 2θ) => the critical points are ^5 -

A ) , 5, and ^5 ÷ √ ⅞ j , but only (5 — A/ S ) is in

the domain. Now P, (x) > 0 for 0 < x < ^5 - χ ∕5 ^ and P'(x) < 0 for ^5 - ^ is a local maximum. Also P(0) = 8p, P ( 5 — √ 5

< x < 4 => at x = ^5 - ᅟ ∕ 5 ) there

1 lp, and P(4) = 8p => at x =

is an absolute maximum. The maximum occurs when x = ( 5 - √ 5 ) and y = 2 ( 5 - √ 5 ) , the units are hundreds of tires, i.e., x ≈ 276 tires and y ≈ 553 tires. 71. The dimensions will be x in. by 10 - 2x in. by 16 - 2x in., so V(x) = x(10 —2x)(16 — 2x) = 4x3 —52x2 ÷ 160x for 0 < x < 5. Then V, (x) = 12x2 — 104x + 160 = 4(x - 2)(3x - 20) , so the critical point in the correct domain is x = 2. This critical point corresponds to the maximum possible volume because V'(x) > 0 for 0 < x < 2 and V'(x) < 0 for 2 < x < 5. The box of largest volume has a height of 2 in. and a base measuring 6 in. by 12 in., and its volume is 144 in.3 Graphical support:

73. g(x) = 3x —x 3 + 4 => g(2) = 2 > 0 and g(3) = - 1 4 < 0 => g(x) = 0 in the interval [2,3] by the Intermediate Value Theorem. Then g'(x) = 3 — 3x2 => xn+1 = xn — ^ ⅛ ⅛ ^ ; Xo = 2 => xχ = 2.22 => x 2 = 2.196215, and so forth to x5 = 2.195823345.

140

Chapter 4 Applications of Derivatives

75. J ( x 3 + 5x - 7) dx = ⅛ + ^ - 7x + C 77. ∫ ( 3 √ t + ⅛ ) d t = ∫ ( 3 t 1∕ 2 + 4 r 2 ) d t = ^ + ⅛ + C = 2t3∕ 2 - ^ + C

79. L etu = r ÷ 5 => du = dr

81. Letu = 02 + 1 =4 du = 20d0 =4 ∣du = 0 d0

f

3 0 √ 0 2 + 1 d0 = ∫ √ u ( j du) = ∣J u 1∕ 2 d u = ∣^ )

83. Let u = 1 + x4 =4 du = 4x3 dx 4 ∫ x3 (1 +

X4 )^ V 4

+ C = u3∕ 2 + C = (02 + 1)3 / 2 + C

∣du = x3 dx

dx = J u-V 4 (I du) = 1 J u - 1∕ 4 du = | ( ⅛ ) + C = i u3 ∕ 4 4/

+

C = 1 (1 + x4 )3 /4 + C

85. Let u = ⅛ =4 du = ⅛ ds =4 10 du = ds § sec2 ⅛ ds = J (sec2 u) (10 du) = 10 J sec2 u d u = 1 0 ta n u + C = 1 0 t a n ⅛ + C 87. Let u = ᅟ ∕2 0 =4 du = ᅟ ∕2 d0 4∙ J*esc ᅟ ∕2 0 cot χ∕20

du = d0

dθ—J (esc u cot u) ( ^

du) = ^ (—esc u) + C = —∙ ^ esc √ z20+ C

89. Let u = ∣ 4 du = ∣dx 4 4 du = dx § sin2 ∣dx = J^ (sin2 u) (4 du) = J* 4 ( 1 ~ 2°s 2u) du = 2 ∫ (1 —cos 2u) du = 2 (u — 5y

a

) + C

= 2u - sin 2u + C — 2 (∣) - sin 2 (∣) + C = ∣- sin ∣+ C 91. y =

J

dx = J* (1 +

=4 C = - l 4 93. ∣= ∫

χ -2

) dx = x —x^ 1 + C = x - ( + C ;y = - 1 when x = l 4

1 —∣+ C = —1

y= χ - 1- 1

(1 5 √ t + ⅛ ) * -

f

(15t1∕ 2 + 3 r 1∕ 2 ) dt = lθt 3∕ 2 + 6t1∕ 2 + C; £ = 8 when t = 1

=4 10(1 )3∕ 2 + 6(I) 1/ 2 + C = 8 4 C = - 8 . Thus ⅛ = lθt 3∕ 2 + 6t1∕ 2 - 8 =4 r = ∫ (10t3∕ 2 + 6t1∕ 2 - 8) dt = 4t 5∕2 + 4 t 3∕2 - 8t + C; r = 0 when t = 1 =4 4(1)5∕ 2 + 4(1)3∕ 2 - 8(1) + C i = 0 4 r = 4t5 ∕ 2 + 4t3 ∕ 2 - 8t

C i = 0. Therefore,

CHAPTER 4 ADDITIONAL AND ADVANCED EXERCISES 1. If M and m are the maximum and minimum values, respectively, then m ≤ f(x) ≤ M for all x ∈I. If m = M then f is constant on I. 3. On an open interval the extreme values of a continuous function (if any) must occur at an interior critical point. On a half-open interval the extreme values of a continuous function may be at a critical point or at the closed endpoint. Extreme values occur only where f ' = 0, f ' does not exist, or at the endpoints of the interval. Thus the extreme points will not be at the ends of an open interval.

Chapter 4 Additional and Advanced Exercises 5. (a) If y' = 6(x + l)(x - 2)2 , then y' < 0 for x < - 1 and y' > 0 for x > - 1 . The sign pattern is f' = ------ ∣+ + + ∣+ + + =+ f has a local minimum at x = - 1 . Also y" = 6(x - 2)2 + 12(x + l)(x - 2) -1 2 = 6(x —2)(3x) => y', > 0 for x < 0 or x > 2, while y" < 0 for 0 < x < 2. Therefore f has points of inflection at x = 0 and x = 2. There is no local maximum. (b) If y' = 6x(x + l)(x - 2), then y' < 0 for x < - 1 and 0 < x < 2; y' > 0 for - 1 < x < 0 and x > 2. The sign sign pattern is y' = -∣+ + + ∣--------∣+ + + . Therefore f has a local maximum at x = 0 and - 1 0 2 local minima at x = —1 and x = 2. Also, y" = 18 ∣ x - ( ⅛ ^ ) ] [x - ( ⅛ ^ ) ] ’ s o ^ < θ ^o r 1~^ 3

< x< ^

and y" > 0 for all other x ≠> f has points of inflection at x = ^

.

7. If f is continuous on [a, c) and f'(x) ≤ 0 on [a, c), then by the Mean Value Theorem for all x ∈[a, c) we have ≤ 0 + f(c) —f(x) ≤ 0 =+ f(x) ≥ f(c). Also if f is continuous on (c5 b] and f'(x) ≥ 0 on (c, b], then for all x ∈(c, b] we have ^ ∈^

≥ 0 => f(x) - f(c) ≥ 0 => f(x) ≥ f(c). Therefore f(x) ≥ f(c) for all x ∈[a, b].

9. No. Corollary 1 requires that f '(x) = 0 for all x in some interval I, not f '(x) = 0 at a single point in I. 11. From (ii), f(—1) = b-⅛ ⅛ lim

x→ ± ∞

lim x→ ±∞

f(x) v 7 =

lim

x→ ±∞

=

a

θ ^

= ^ fr° m (hi), either 1=

r π ⅛ ∏ - - χ →lim r——⅛ = ± ∞ bx + c + f

x + c+ i

lim

x→ ±∞

13. The area of the ΔABC is A(x) = ∣(2) √ 1 - x

2

⅛

=

bx + f

lim o f(x) or 1 =

χ

]im oo f(x). In either case,

1=> b = 0 andc = 1. For if b = 1,’ then

bx2 + cx + 2

- ⅛ - —0 and if c —0, then

χ

lim x→ ±∞

= ( 1 - x2 ) 1+

⅛

= ± ∞ . Thus a = 1, b = 0, and c = 1.

f

’

y

where 0 ≤ x ≤ 1. Thus A'(x) = - ÷ ^ -2 => 0 and ± 1 are √ l- x

critical points. Also A ( ± 1) = 0 so A(0) = 1 is the maximum. When x = 0 the ΔABC is isosceles since AC = B C = √ 2 .

15. The time it would take the water to hit the ground from height y is y ^ , where g is the acceleration of gravity. The product of time and exit velocity (rate) yields the distance the water travels: D W = ( / I √64(h - y) = 8 y ∣(hy - y 2 ) 1 /2 , 0 ≤ y ≤ h ≠- D'(y) = - 4 ^ (hy - y2 ) - 1 / 2 (h - 2y) 4

0, ∣and h

are critical points. Now D(0) = 0, D (∣) = ^ and D(h) = 0 => the best place to drill the hole is at y = ∣. 17. The surface area of the cylinder is S = 2πr2 + 2πrh. From the diagram we have j = ⅛ ^ =+ h = - ^ rH and S(r) = 2τrr(r + h) = 2πr (r + H - r ∣) = 2π (1 - ∣) r2 + 2πHr, where 0 ≤ r ≤ R. Case 1: H < R =+ S(r) is a quadratic equation containing the origin and concave upward =+ S(r) is maximum at r = R. Case 2: H = R =+ S(r) is a linear equation containing the origin with a positive slope =+ S(r) is maximum at r = R.

141

142

Chapter 4 Applications of Derivatives

Case 3: H > R => S(r) is a quadratic equation containing the origin and concave downward. Then f = 4 π ( l - ∣) r + 2πH and ^ = 0 ≠ ∙ 4π (1 - ∣) r + 2πH = 0 ≠> r = 2(^ R ) . For simplification we let r* = 5 ^ ⅛

.

(a) If R < H < 2R, then 0 > H - 2 R => H > 2(H - R) ≠> 2^ (b) If H = 2R, then r* = ^

r

> > R which is impossible.

= R =► S(r) is maximum at r = R.

(c) IfH > 2R ,then2R + H < 2H ^

H < 2 ( H - R ) =► 2i⅛

< 1 ^

5⅛

< R ≠> r* < R. Therefore,

S(r) is a maximum at r = r* = 2αT⅛ ∙ Conclusion: If H ∈(0, R] or H = 2R, then the maximum surface area is at r = R. If H ∈(R, 2R), then r > R which is not possible. If H ∈(2R, ∞ ), then the maximum is at r = r* = 2( ^ R ) ' ⅛ ^ = lim 10 sin(5x) _ ∣ (5x) χ → 0 3 (5x) “ ≡ ¾ ⅛ P ^ = lim (b) lim sin(5x)cot(3x) v v = × lim χ —>0 sιn(3x) x→0 χ

19. (a)

(c)

» 1 =

lim

3x

x →0

∕ Λ∖ w

10 1 _ 10 3 ’i “ 3 ⅛ )θ

≡

⅛ M 3cos(3x)

= ∣ 3

⅛

lim

x→0

X™ (d)

lim

x→0

lim

(sec x - tan x) =

x → π∕2 ' U—

x™

=

2

cos(2√S)∙2 '

x-sinx _ χ 7 s

0

^ ~ x→0

ii m ⅛ x → 0 -2

i

limc o s x ⅛

=

lim

x → π∕2 - « n

x → π∕2

1-cosx _ υ w4

= ^ ^ = 0.

x

1-cosx _ n „

^x → 0 ^ ≡ ^

cosx-1 _- u

~ x →0 ~ ≡ ^

-sinx

x→0 2 ≡ ∏ S

lim

—sin x 2sinx cos^x

x →0

- _1 2

lim ⅛ ≤ ⅛ = lirn 2 x c o s Cχ2 > = lim -( 2χ2 )sin(*2) + 2cos(x2) = 2 = . u χ → 0 xs*n x x → 0 xcos x ÷s*n x x →0 -xsin x+2cos x 2 sec x fσl lim ~1 _ sec x tan x _ sec3 x+ tan2 xsec x _ 1+ 0 _ 1 2x - x≡ o 2 - 2 - 2 χ≡ 0 χ≡ 0 (h) limo ⅛ = Hmo ⅛ ≡ i = Um i ⅛ χ i l = 1±∣±4 = 3 χ →2 x 4 X→ 2 (x-2)(x + 2) X→ 2 +2 4 (fl

21. (a) The profit function is P(x) = (c —ex)x — (a + bx) = - e x 2 ÷ (c —b)x —a. P, (x) = —2ex + c —b = 0 => x = ^ . P"(x) = —2e < 0 ife > 0 so that the profit function is maximized at x = ⅛ ^ . (b) The price therefore that corresponds to a production level yeilding a maximum profit is c - e (⅛ ) = ^ d o lla rs ∙ p∣ x- ^ = 2e

(c) The weekly profit at this production level is P(x) — - e ( ⅛ ) 2 + (c —b ) ( ⅛ ) - a = ⅛ ^ - a. (d) The tax increases cost to the new profit function is F(x) = (c —ex)x — (a + bx + tx) = —ex2 + (c —b —t)x —a. Now F (x) = —2ex + c - b - t = 0 when x = 1⅛ when x =

c ~ g~t 2

units per week. Thus the price per unit is p = c - e ( c ^ ^ ^ t ) = ^ y t ! dollars. Thus, such a tax

increases the cost per unit by c y ⅛ = xo — ⅛ 23. xι = x0 — ⅛ 1 v q⅞ Qx o and -χ ⅛ with weights mo = o

In the case where x0 = ^

25. (a) a(t) = s"(t) = - k s(t) zy

zτ

= ⅞ ^ * . Since F '(x ) = —2e < 0 if e > 0, F is maximized

+t

= ⅛ ⅛

- ^ 2

= ⅛

~ and mi = 9

P ∖

Qx o i

= ∣dollars if units are priced to maximize profit. = xo( 4

/

χo

i

)

+

⅛ (q )

s o t h a t X1 i s a w e ⅛ h t e d

average of x0

∖4 ∕

q

we have x’ = a a n d x 1 = ^ ( ⅛

) + ^ Q ) = ^

( ^

+ θ

= ^ .

(k > 0) ≠> s'(t) = - k t + C b where s'(0) = 88 => C i = 88 => s, (t) = - k t + 88. So

+ 88t + C2 where s(0) = 0 4 Cj = 0 so s(t) =

ƩY

+ 8 8 t Now s(t) = 100 when

+ 88t = 100. Solving for t we obtain t = ~ ~ ^ 8f~~~^~ ∙ At such t we want s'(t) = 0, thus

Chapter 4 Additional and Advanced Exercises ~ k ^ 8 8 + √88 2

2θθk) +

so that k = ^

gg _ θ o r -

k

(8 8 ~ √ ^ Z ^

143

+ 88 = 0. In either case we obtain 882 - 200k = 0

≈ 38.72 ft∕sec2 .

(b) The initial condition that s'(0) = 44 ft/sec implies that s'(t) = - k t + 44 and s(t) = 2γ

÷ 44t where k is as above.

The car is stopped at a time t such that s'(t) = - k t + 44 = 0 => t = p At this time the car has traveled a distance s

(f)

=

τ (f)

2

÷ ^ (f)

=

¥

=

T

=

θ θ θ (^ )

=

25 feet. Thus halving the initial velocity quarters

stopping distance. 27. Yes. The curve y = x satisfies all three conditions since ^ = 1 everywhere, when x = 0, y = 0, and ^

= 0 everywhere.

29. s"(t) = a = —12 => v = s'(t) = y ÷ C . We seek Vo = s'(0) = C. We know that s(t*) = b for some t* and s is at a maximum for this t*. Since s(t) = p + C t ÷ k and s(0) = 0 we have that s(t) = y ÷ C t and also s'(t*) = 0 so that t, = (3C)1 /3 . So 4 C

=

V 3 + C (3C)

Thus v0 = s'(0) =
(3C)V 3 (C - ^ ) = b => (3C)1 / 3 ( ^ ) = b ≠> 3 1∕ 3 C4∕ 3 = f = ⅜ ¾ 3∕ 4 .

31. The graph of f(x) = ax2 + b x + c with a > 0 is a parabola opening upwards. Thus f(x) ≥ 0 for all x if f(x) = 0 for at most one real value of x. The solutions to f(x) = 0 are, by the quadratic equation 2 b ⅛ Y ⅞ — — . Thus we require (2b)2 - 4ac ≤ 0 => b2 - ac ≤ 0.

144

Chapter 4 Applications of Derivatives

NOTES:

CHAPTERS INTEGRATION 5.1 ESTIMATING WITH FINITE SUMS

Since f is increasing on [0, 1], we use left endpoints to obtain lower sums and right endpoints to obtain upper sums.

1. f(x) = x2

(a) ∆ x = i y

•

∙ 2

i(θ≈+(⅜Γ) = 1

= | and Xi = i∆ x = ∣=> a lower sum is ∑ ( ∣) ∙ ∣ i=0

(b)

∆x = ' √ l= ∣and Xi = i∆ x = ∣≠- a lower sum is ∑ ( i ) 2 ’ 4 = 4 ( ° 2 + ( j ) 2 + ( I ) 2 + ( I ) 2 ) = 3 ’ I = ⅛

(c)

∆ x = i y = ∣and x⅛ = i∆ x = ∣≠> an upper sum is ∑ ( 0 2 ' ∣~ 2 ( ( 2 ) 2 + ^ ) ~ I

(d)

∆ x = i ^ = i andx i = i∆ x = { => an upper sum is ∑ ( ∣) 2 ∙ ∣= ∣( ( J ) 2 + (∣) 2 + ( ∣) 2 +12 ) = ∣∙ (is) = 1

Since f is decreasing on [0, 1], we use left endpoints to obtain upper sums and right endpoints to obtain lower sums.

2

(a) ∆ x = ~ ^ = 2 and Xi = 1 ÷ i∆ x = 1 + 2i => a lower sum is ∑ ^ ∙ 2 = 2 (∣+ ∣) = ∣ ∣ i= ι

,

4

(b) ∆ x = ^

= 1 and χ i = 1 + i∆ x = 1 + i => a lower sum is ∑ ^ ∙ 1 = 1Q + ∣+ ∣+ ∣) = ^ i= l ' ι

(c) ∆ x = ψ

= 2 and Xi = 1 + i∆ x = 1 + 2i => an upper sum is ∑ ^ ∙ 2 = 2(1 +

1) =

∣

i=0 ' 3

(d) ∆ x = ^ p = 1 and xi = 1 + i∆ x = 1 ÷ i => an upper sum is ∑ ^ ∙ 1 = 1(1 ÷ ^ + ∣+ ∣) = ∣ i=0

,

146

Chapter 5 Integration Using 2 rectangles =+ ∆ x = 1y — if -

(1) 2

2 ^∖ 4∕

-L r ^

f3λ2 λ

U ∕

=

2 ^ ^ (f(∣) + f ( ∣))

_ 10 -_ 5.

-

32

16

Using 4 rectangles =+ ∆ x = i - ^ = ∣ ^ ≡ )+ t (i)+ r (i)+ r (I))

=Kω2+(t)2+(∣)2+ω2) =≡ 7∙ f(x) = ∣ Using 2 rectangles =+ ∆ x = ‰Δ = 2=+ 2(f(2) + f(4)) = 2(i + D = l Using 4 rectangles => ∆ x = ^

= 1

= > ι (f (D + f (i)+ f (D + f ( i) ) — 1 ( 2 , 2 , 2 , 1 ∖ 3 ^r 5 ^r 7 ^r

2λ 97

-

1488 _ 3 ∙5 ∙7 ∙ 9 -

496 _ 5 ∙7 ∙9 -

496 315

9. (a) D ≈ (0)(1) + (12)(1) + (22)(1) + (10)(1) + (5)(1) + (13)(1) + (11)(1) + (6)(1) + (2)(1) + (6)(1) = 87 inches (b) D ≈ (12)(1) + (22)(1) + (10)(1) + (5)(1) + (13)(1) + (11)(1) + (6)(1) + (2)(1) + (6)(1) + (0)(1) = 87 inches 11. (a) D ≈ (0)(10) + (44)(10) + (15)(10) + (35)(10) + (30)(10) + (44)(10) + (35)(1O) + (15)(10) + (22)(10) + (35)(10) + (44)(10) + (30)(10) = 3490 feet ≈ 0.66 miles (b) D ≈ (44)(10) + (15)(10) + (35)(10) + (30)(10) + (44)(10) + (35)(10) + (15)(10) + (22)(10) + (35)(10) + (44)(10) + (30)(10) + (35)(10) = 3840 feet ≈ 0.73 miles 13. (a) Because the acceleration is decreasing, an upper estimate is obtained using left end-points in summing acceleration ∙ ∆t∙ Thus, ∆ t = 1 and speed ≈ [32.00 + 19.41 + 11.77 + 7.14 + 4.33](1) = 74.65 ft/sec (b) Using right end-points we obtain a lower estimate: speed ≈ [19.41 + 11.77 + 7.14 + 4.33 + 2.63](1) = 45.28 ft/sec (c) Upper estimates for the speed at each second are: t 0 1 2 3 4 5 V 51.41 70.32 74.65 0 32.00 63.18 Thus, the distance fallen when t = 3 seconds is s ≈ [32.00 + 51.41 + 63.18](1) = 146.59 ft. 15. Partition [0,2] into the four subintervals [0,0.5], [0.5,1], [1,1.5], and [1.5,2]. The midpoints of these subintervals are mi = 0.25, m2 = 0.75, m3 = 1.25, and m4 = 1.75. The heights of the four approximating rectangles are f(mι) = (0.25)3 = ⅛, f(m2 ) = (0.75)3 = g , f(m3 ) = (1.25)3 = ^ , and f(m4 ) = (1.75)3 = ^ Notice that the average value is approximated by ∣[(∣) 3 (∣) + (∣) 3 (∣) + (∣) 3 (∣) + ( j ) 3 (∣) j = fl 1 length of [0,2]

approximate area under curve f(x) = x3

We use this observation in solving the next several exercises.

17. Partition [0,2] into the four subintervals [0,0.5], [0.5,1], [1,1.5], and [1.5,2]. The midpoints of the subintervals are mi = 0.25, m2 = 0.75, m3 = 1.25, and ∏14 = 1.75. The heights of the four approximating rectangles are f(mi ) = ∣+ sin2 ≡ = 1 + 1 = 1, f(m2 ) = ∣+ sin2 ⅜ = j + ∣= 1, ftm3 ) = ∣+ sin2 ⅞ = ∣+ ( - ^ ) = ∣+ ∣= 1, and f(π⅛) = ∣+ sin2 7 = ∣+ ( - ^ )

~ '■ The width °f e a c h rectangle is ∆ x = 1. Thus,

Section 5.2 Sigma Notation and Limits of Finite Sums Area ≈ (1 + 1 + 1 + 1) (∣) = 2 => average value ≈

leng⅞

2j

147

= 5 = 1∙

19. Since the leakage is increasing, an upper estimate uses right endpoints and a lower estimate uses left endpoints: (a) upper estimate = (70)(1) ÷ (97)(1) + (136)(1) + (190)(1) + (265)(1) = 758 gal, lower estimate = (50)(1) + (70)(1) + (97)(1) + (136)(1) ÷ (190)(1) = 543 gal. (b) upper estimate = (70 ÷ 97 + 136 + 190 + 265 + 369 ÷ 516 + 720) = 2363 gal, lower estimate = (50 + 70 + 97 + 136 + 190 + 265 + 369 ÷ 516) = 1693 gal. (c) worst case: 2363 ÷ 720t = 25,000 ≠> t ≈ 31.4 hrs; best case: 1693 + 720t = 25,000 => t ≈ 32.4 hrs 21. (a) The diagonal of the square has length 2, so the side length is 5/2. Area = ^ V ^ ) ~ 2 (b) Think of the octagon as a collection of 16 right triangles with a hypotenuse of length 1 and an acute angle measuring π 8,

2π _ 16 ~

Area = 16(∣) (sin ∣) (cos ∣) = 4 sin ∣= 2ᅟ ∕2 ≈ 2.828 (c) Think of the 16-gon as a collection of 32 right triangles with a hypotenuse of length 1 and an acute angle measuring 2π _ 32 ”

π 16*

Area = 32(∣) (sin ⅞) (cos ⅛) = 8 sin f = 2 √ 2 ≈ 3.061 (d) Each area is less than the area of the circle, π. As n increases, the area approaches π. 5.2 SIGMA NOTATION AND LIMITS OF FINITE SUMS 2

∑ k=l

6k _ 6(1) , k + 1 “ 1+ 1

6(2)_ — 6 . 1 2 2 + 1 ~ 23 ~

—

4

3. 52 c o s kπ = cos (lπ) + cos (2π) + cos (3π) + cos (4π) = - 1 + 1 - 1 + 1 = 0 k=l

5. ∑ ( - l ) k+1 sin £ = ( - l ) 1+1 sin f + ( - l ) 2+1 sin f + ( - l ) 3+1 sin≈ = 0 - 1 + ^ = K 1 Z Z 3 k=l

7. (a) ∑ 2k"' = 21^ 1 + 22 ^ 1 +

Z

23~I + 24^ 1 + 25 ^ 1 + 26 " 1 = 1 + 2 + 4 + 8 + 1 6 + 32

k=l

Φ) ∑ 2k = 20 + 21 + 22 + 23 + 24 + 25 = 1 + 2 + 4 + 8 + 16 + 32 k=0

(c) ∑ 2k+, = 2^ 1+1 + 20+1 + 21+1 + 22+1 + 23+1 + 24+1 = 1 + 2 + 4 + 8 + 16 + 32 k = -l

All of them represent 1 + 2 + 4 + 8 + 1 6 + 32 9. (a) Σ

⅛

= ⅛

→ ⅛

→ ⅛ + = - l + ∣- j

k=2 2

Z^

W

(C w

k=0

)

k+i

0+1

τ

1+ 1 '

V t i t = +1H + Iz lf k = -l

k+ 2

-1 + 2

0+ 2 +

2+1

1

2

τ

3

Izl2 = - 11 + 1 - 1 1+ 2

2

3

(a) and (c) are equivalent; (b) is not equivalent to the other two. 6

11. ∑ k k=l

13∙ Σ k=l

2k

15. ∑ ( - l ) k÷1 ∣ k=l

148

Chapter 5 Integration

17. (a) E 3ak = 3 E ak = 3 (-5 ) = - 1 5 k=l

k=l

(b) Σ

⅞ = i∑ b

k=l

k

= ∣(6) = l

k=l

(c) E (¾ + bk ) = E ¾ + E bk = - 5 + 6 = 1 k=l

k=l

k=l

(d) E ( ¾ ~ b k) = E ¾ k=l (e )

E

(,j k -

2ak) =

k=l

19. (a) E

= -1 1

k=l l

E

⅛

- 2 E ⅜ = 6 -2 (-5 )= 1 6

k=l

k

=

1

T

d

k=l

l

55

=

(b)

k=l (C )

f

k

3

j

≡

>

∙ ∣

k=l

∣ᅟ

k=l

6

23. E (3 k=l

25. E

k2)

×

k

2

=

10(10 + 1 )(2 (1 0 )+ 1 ) _

6

= -5 6

'

6

≡

= -7 3

k=l

(3k + 5) = E (3k2 + 5k) = 3 E k 2 + 5 E k = 3 [ ≡ ≡ + ! ) ] + 5 ( ⅛

k=l

3 8 5

6

552 = 3025

= Σ 3 - E k 2 = 3(6) - M

k=l

k

J

21. E - 2 k = - 2 E k = - 2 ( ⅛ k=l

£ k=l

k=l

k=l

k=l

'

/

= 240 /

Section 5.3 The Definite Integral

149

xi - x0 ∣= ∣ x2 - x i ∣= |1.5 - 1.2∣= 0.3, ∣ x3 - x2 ∣= |2.3 - 1.5∣= 0.8, ∣ 33. ∣ 1∙2 - 0∣= 1.2, ∣ x4 - x3 ∣= |2.6 - 2.3∣= 0.3, X5 —X4∣= ∣ and ∣ 3 —2.6∣= 0.4; the largest is ∣ ∣ P∣ ∣= 1.2. 35. f(x) = 1 —x2

Since f is decreasing on [ 0, 1] we use left endpoints to obtain upper sums, ∆ x = ^ = J and x⅛ = i∆ x = j . So an upper sum n -1

n -1 /

i=0

i=0

n—1

is ∑ < l - ⅛ ⅛ = i ∑ ( l - ( i ) ) = ⅛ Σ ( " ≈ - i 2 ) n3 n3

_ -

1 n

_ “

3

x

'

i=0

( n - l) n (2 ( n - 1) + 1) _ 6n3 “

-1 i

2n3 - 3 n 2 + n 6n3

i

i=0 2 -^ + 4

Since f is increasing on [ 0, 3] we use right endpoints to obtain upper sums, ∆ x = ^ = and Xi = i∆ x = 7. So an upper sumi s ⅛ + l ) 2 = ∑ ⅛ ) 2 + l) ≡ = a ∑ ( S + 1) i= l

= ⅛

+

18+∣ + ⅛

=

3

lim ∏ x 2 + 1)3π = Um I

n→∞

∙

i= lv

7

z

+ L ∏ = ^ (= < ^ ≡ ^ )+ 3

9β n i + 3 ⅛

=

v

i= l

2

;2

n→∞ V

1

3

+

τ h u s

^ + 3

/

= 9 + 3 = 12.

Since f is increasing on [ 0,1] we use right endpoints to obtain upper sums, ∆ x = i p = and Xi = i∆ x = ^. So an upper sum

39. f(x) = x + x2 = x(l + x)

n

n

z

-

1 / n2

n (p

+ 1Λ 2 /

ι ψ

n

⅛= ⅛ ∑ i + ⅜ ∑ i 2

i + ω

i= lx

i= l _

n

o

⅛ ∑ ⅛ + + )* = ∑

7

i= l

1 ( n(n + l)(2n + l ) _ ? y ~ 6

n

2

+ n 2n2

i= l . 2n3 + 3n2 + n 6n3

+ ⅜ ^ ∙ Thus, lim ⅛ (x i + x?)i

= ⅛

n→∞ : 1

53 THE DEFINITE INTEGRAL 3. j

7.

Γ5

(x2 - 3x) dx

2⅛

dx

f 0 (sec x) dx

I

«7 -π∕4

*2

J

2

p2

(c)

g(x)dx = 0

(b)

«7

p2

I 3f(x)dx = 3 ∣ f(x)dx = 3 (-4 ) = -1 2

«7 1

»5

J J

»7 1

»5 1

O

p5

(d)

1

f(x) dx -

p5

»72

I g(x) dx = 6 - 8 = - 2

«7 1

p5

[4f(x) - g(x)] dx = 4 I f(x) dx - I g(x) dx = 4(6) - 8 = 16 «7 1

«7 1

*7

p5

1

p2

I f(x )d x = I f (x ) d x - I f(x) dx = 6 - (-4 ) = 10

p5

p5

. [f(x) - g(x)] dx = I 1

pl p5 I g(x) dx = - J g(x) dx = - 8 «71

«71

150

Chapter 5 Integration .2

Jf

1

(c) J

1

2

f ( t ) d t = - ] f f(t)dt = - 5 UI

»4

J

p4

(d)

p3

f(z)dz = J θ f ( z ) d z - J o f(z)dz = 7 - 3 = 4

15. The area of the trapezoid is A = ∣(B + b)h = | (5 + 2)(6) = 21 ≠ ∫

2

( ∣+ 3) dx

= 21 square units

17. The area of the semicircle is A = ∣πr 2 = ∣π(3)2 =

2 7r ^

J _3 V 9 —x2 dx = ∣π square units

19. The area of the triangle on the left is A = ∣bh = ∣(2)(2) = 2. The area of the triangle on the right is A = ∣bh = ∣(1)(1) = j. Then, the total area is 2.5 => J*

2

Jf ι

UI

0 2

3

»2

/.2

f(u )d u = I f(x)dx = 5

∣ x∣dx = 2.5 square units

21. The area of the triangular peak is A = ∣bh = ∣(2)(1) = 1. The area of the rectangular base is S = ^w = (2)(1) = 2. f1 Then the total area is 3 => J (2 - ∣ x∣ ) dx = 3 square units

I

2

n2

√ 3 f(z )d z = √ 3 J [-f(x)] dx - -

2

1

f(z)dz = 5 √ 3

If f(x) dx = - 5

Section 5.3 The Definite Integral

*1/2 t2 0

J J

∕ ι ∖3

d t= ⅛ = ⅛

'2 a

a

37. JoΛ 3

»2

Jo 43. J

/»2

_

x d x = z ^ - fz ≈ ⅛ z

≈Λ =M

3

7 dx = 7(1 —3) = —14 Γ2

Γ

Jo

L

f* l

(2t - 3) dt = 2 J ι t dt - J

.i 2

(1 + i )

o

1

5x dx = 5 / x dx = 5 2$ - 2y

dz

pi

= J

2

ld z

p2

+ J

o

pi 2

J

(3x2

= 10 Γ 2

3 dt = 2 I ⅛ -

I

dz

Γ2

p2

49. J

H

3

o

J

151

+ x - 5) dx = 3 Jθ x2 dx +

= J

p2

pi 2

21

- 3(2 - 0) = 4 - 6 = - 2

l d z -

JQ x dx -

∣J

n2

p2 1

η

r

z d z ≈ l [ l - 2 ] - ∣[ ⅜2 - ⅛ 5] = - 1 - ∣(∣) = - 2

r

η

r

η

Jθ 5 dx = 3 [ f - ^ ] + [≤ - £ ] - 5[2 - 0] = (8 + 2) - 10 = 0

Chapter 5 Integration

152

51. Let ∆x = ^

= £ and let xo = 0, x1 = ∆x,

X2 = 2 ∆ x ,. . . , xn. 1 = (n — l)∆x, xn = n∆x = b. Let the ck's be the right end-points of the subintervals => ci = xι, C2 = X2, and so on. The rectangles defined have areas: f(ci ) ∆x = f(∆x) ∆x = 3(∆x)2 ∆x = 3(∆x)3 f(c2 ) ∆x = f(2∆x) ∆x = 3(2∆x)2 Δ X = 3(2)2 (∆x) 3 f(c3 ) ∆x = f(3∆x) ∆x = 3(3∆x)2 ∆x = 3(3)2 (∆x) 3 f(cn ) ∆x = f(n∆x) ∆x = 3(n∆x)2 ∆x = 3(n)2 (∆x) 3 Then Sn = ∑ f(ck) ∆x = ∑ 3k2 (∆x) 3 k=l

k=l

= 3(∆x) 3 ∑ k2 = 3 ( g ) ( ⅛ ± l⅛ ÷ ι> ) θ 3X2 dx = 53. Let ∆x = η ^ =

n

lim o ⅛ (2 + 3 + ⅛) = b3 .

and let xθ = 0, xι = ∆x,

x2 = 2 ∆ x ,. . . , xn- 1 = (n - l)∆ x, xn = n∆x = b. Let the ck's be the right end-points of the subintervals => ci = xι, c2 = x2 , and so on. The rectangles defined have areas: f(cι)∆ x = f(∆ x)∆ x = 2(∆x)(∆x) = 2(∆x)2 f(c2 ) ∆x = f(2∆x) ∆x = 2(2∆x)(∆x) = 2(2)(∆x)2 f(c3 )∆ x = f(3∆x)∆ x = 2(3∆x)(∆x) = 2(3)(∆x)2 f(cn ) ∆x = f(n∆x) ∆x = 2(n∆x)(∆x) = 2(n)(∆x)2 Then Sn = 22 ‰ ) ∆x = ∑ 2k(∆x)2 k=l

=

2(∆x) 2

=

b2 0 + ∏) v

k=l

Σ k = 2 (g ) ( 2 ^ )

57. av(f) = ( τ ⅛ ) J = -3 Γ = -2 .

£ 2x dx =

n∕

χ 2 d χ

Jo

o

lim n→∞

b2 (1 + 1) = b2 . V

n∕

( - 3 X2 - 1) dx =

- J

0

lld x

= -

3

(τ )-(

1

-

0>

Section 5.3 The Definite Integral

JQ ( t - l )

59. av(f) = ⅛ ) »3

J

2

153

dt

p3

p3

t2dt

td t

- l J 0 + U 0 ld t M (⅞ )- l(⅛ - ⅛ ) +l ( 3 - 0 ) = l ∙ 0

61. (a) av(g) = ( j ⅛ ) )

=I

(M - 1) dx

pθ pl I ( - x - 1) dx + i I (x - 1) dx pθ pθ pl

pl

= 4 J - 1χ d x 4 J_i l d x + iJo x d x ^ U o ld x = 4 ⅛ - 4 4 4 ( ° - ( - 1))+ l(τ - τ ) 4 o - ° ) _ _ 1 “ 2∙

(b) av(g) = ( 5 ⅛τ ) J = 5 J1

x d x

(c) av(g) = ( τ ⅛

-

p3 ι

(∣ x∣ - 1) dx = i J

2 J1

)

1 d x

= 2 (¥

-

p3 1

(x - 1) dx

7 ) -

2 (3 -

1

)

J√ l x l - 1) dx

= I J√ ∣ x ∣- 1) dx + 5Ji (∣x ∣- 1) dx = ± (-1 + 2) = | (see parts (a) and (b) above).

63. To find where x - x2 ≥ 0, let x - x2 = 0 => x(l - x) = 0 => x = 0 or x = 1. If 0 < x < 1, then 0 < x - x and b = 1 maximize the integral.

2

=> a = 0

65. f(x) = γ⅛2 is decreasing on [0,1] => maximum value of f occurs at 0 => max f = f(0) = 1; minimum value of f o T ⅛ dx ≤ (1 —0) max f

=> ∣ ≤J o T ⅛

equal 2.

dχ ≤ 1∙ That is, an upper bound = 1 and a lower bound = ∣.

J

pl pl θ sin (x2 ) dx ≤ (1 - 0)(1) or Jθ sin x2 dx ≤ 1 => Jθ sin x2 dx cannot >1

154

Chapter 5 Integration »b

J

a

»b

J

a

f(x) dx ≤ (b —a) max f.

f(x) dx ≥ 0.

71. sin x ≤ x for x ≥ 0 => sin x —x ≤ 0 for x ≥ 0 => J (sin x —x) dx ≤ 0 (see Exercise 70) =>

sin x dx —

x dx

θ sin x dx ≤ Jθ x dx => Jθ sin x dx ≤ ( y - y J => Jθ sin x dx ≤ ∣. Thus an upper bound is j. »b

J

a

>b

J

pb

pb

f(x) dx is a constant K. Thus I av(f) dx = I K dx ua ua

av(f) dx = (b - a)K = (b - a) ∙ a

pb

pb

Ia f(x) dx = Ia f(x) dx.

D

75. Consider the partition P that subdivides the interval [a, b] into n subintervals of width ∆ x = η p and let Ck be the right endpoint of each subinterval. So the partition is P = {a, a + ^ A a + ^ y ^ , . . . , a + - ^ ^ } and Ck = a + ⅛ z ^ ξ n n n ∕u We get the Riemann sum ^ f ( c k )∆ x = ∙^ = 1= P∣ ∙ n = c(b - a). As n → ∞ and ∣ ∣ ∣→ 0 k=l

k=l

k=l

this expression remains c(b —a). Thus, ^ c dx = c(b —a).

77. (a) U = maxi ∆x + max2 ∆x + ... + maxn ∆x where maxi = f(xι), max2 = f(x2), . . , maxn = f(xn) since f is increasing on [a, b]; L = mini ∆x + min2 ∆x + ... + minn ∆x where mini = f(⅞), min2 = f(xι), ∙ , minn = f(xn-ι) since f is increasing on [a, b]. Therefore U —L = (maxi —mini) △* ÷ (max2 —min2 ) ∆x + ... + (maxn —minn) ∆x = (f(x1 ) - f(x0 )) ∆x ÷ (f(x2 ) - f(x1 ))∆x + . . . + (f(xn) - f(xn. 1)) ∆x = (f(xn) - f(x0 )) ∆x = (f(b) - f(a)) ∆x. (b) U = maxi ∆ x i + max2 ∆ x 2 + . . . + maxn ∆ x π where maxi = f(xi ), max2 = f(x2 ) , ...» maxn = f(xn) since f is increasing on[a, b]; L = mini ∆ x i + min2 ∆ x 2 + . . . + minn ∆x n where mini = f(*oλ min2 = f(xι),. ∙ , minn = f(xn~1) since f is increasing on [a, b]. Therefore U —L = (maxi - mini) ∆ x i + (max2 - min2 ) ∆x 2 + . . . + (maxn - minn) ∆ x n = (f(xι) - f(x0 )) ∆ x i + (f(x2 ) - f(xι))∆x 2 + . . . + (f(xn) - f(xn-i)) ∆x n ≤ (f(xi ) - f(x0 )) ∆ x max + (f(x2 ) - f(xi )) ∆x max + . . . + (f(xn) - f(xn- 1)) ∆ x max. Then U - L ≤ (f(xn) - f(x0 )) ∆ x max = (f(b) - f(a)) ∆x max = ∣ f(b) - f(a)∣ ∆ x max since f(b) ≥ f(a). Thus lim (U - L) = lim (f(b) - f(a)) ∆x max = 0, since ∆ x max = ∣ ∣ . ∣ P∣ 79. (a) Partition [θ, -] into n subintervals, each of length ∆x = ^ with points x0 = 0, XI = ∆x, x2 = 2 ∆ x ,... , xn = n∆x = j . Since sin x is increasing on [0, ∣] , the upper sum U is the sum of the areas of the circumscribed rectangles of areas f(xi ) ∆x = (sin ∆x)∆x, f(x2 ) ∆x = (sin 2∆x) ∆ x ,. . . , f(xn) ∆x = (sin n∆x) ∆x. Then U = (sin ∆x + sin 2∆x + ... + sin n∆x) ∆x = ∞ s ⅜ -∞

s

(( n + D ⅜) ⅛

√≤211Z M 1+⅛)) 4n sin £

2 s in

'√2

(b)

J

0

sin x dx =

n

COS y - c o s ( ( n + ∣) ∆ x) 2 sin ^

lim

⅛ -c o s{ ≡ + i)

_

1 -c o s ξ _

n→∞

n

81. By Exercise 80, U —L = ^ ∆ x i ∙ Mi - ∑ ∆ x i ∙ mj where Mi = max{f(x) on the ith subinterval} and i=l

i=l

n

mi = min{f(x) on the ith subinterval}. Thus U - L = ∑(M i - mj)∆xi < i=l

n i=l

∙ ∆ x i provided ∆ x i < δ for each

Section 5.4 The Fundamental Theorem of Calculus n

n

i = 1 ,. . . , n. Since £ e ∙ ∆xi = e ^ ∆ x i = e(b —a) the result, U —L < e(b —a) follows, i=l

i=l

5.4 THE FUNDAMENTAL THEOREM OF CALCULUS 1. J J 2 x + 5) dx = [x2 + 5X]22 = (02 + 5(0)) - ((-2 ) 2 + 5(-2)) = 6 3

∙⅛ - ⅛ > = ⅛ - C = (⅞ - 9 - (⅞ ^ )= >

5-

χ5

∕√

7. f

* √7)

dx

= H + b " ^

x - 6 / 5 dx = [-5 x ^ 1∕ 5 ] j 2 = ( - ∣) - (-5 ) = ∣

i

9. J o sin x dx = [-cos x]θ = (-co s π) - (-co s 0) = - ( - 1 ) - (-1 ) = 2 sec2 x dx = [2 tan x]θ73 = (2 tan (≈)) - (2 tan 0) = 2 √ 3 - 0 = 2 √ 3

11. ^ 2

esc θ cot θdθ = [—esc θ f ^ = (—esc ( γ ) ) - (-esc Q )) = - √ z2 - ^ - v ^ ) = 0

13. f

15∙

£

dt =

2“

f ^

+

2 c o s 2 t) d t

= ⅛ t ÷ ⅛s i n

2 tl

θ ∕2 = ( I ( 0 ) + 4 s i n 2 ⅛ ( J^ cos t dt j = ^ (sin √ x ) = cos √ x (∣x 1∕ 2 )

155

Chapter 5 Integration

156 (b )

29∙

⅛ (X ^

Jo ≠ *

du

33∙ y =

x√

= Jo

*sin x

J

35.

=

(c o s √ * ) (⅛ ( √ x )) = (c o s √

uV 2 du =

x

) (ix

1 /2 )

= ⅛

[5 u 3 / 2 ] *0 = 5 (t4 )372 - 0 = 5 16 =► £ ( J

λA

d u ) = £ ( j t6 ) = 4t5

Γ T t 2 dt =► ^ = √ Γ + χ 2

s i n t2 d t =

J

^)

√ u d u ) = √ t 4 ( I (t4 )) = t2 (4t3 ) = 4t5

a ( J o

31. y = J

031

∙ 7 1d

0

t

γ l-1

,

2

-

J o ^ * 11 t '

dt

,∣ x ∣< 5 => ⅛ = 2

⅛ = ~ ( s i n ( √ x ) 2 ) (⅛ ( √ x )) = - ( s i n

^

/;■ ■^

(⅛ (sin x)) =

v 1 —sιn2 x 'dx

dx

/

11

y co s2 x

χ

) (1 x

1 /2 )

= - ^

(cos x) = 1≡ ⅛ = ≡ ^ = 1 since ,∣ x ∣< f

v

∣ cosx ∣

cos x

∣

2

39. x3 - 3x2 + 2x = 0 => x (x2 - 3x + 2) = 0 => x(x —2)(x — 1) = 0 =φ∙ x = 0, 1, or 2;

= ^

J

>l o

p2

(x3 —3x2 + 2x)dx - J

— x3 + x 2 j θ — ^

1

(x3 —3x2 + 2x)dx

— x3 + x 2 j J

= ( ⅛ - l 3 + l 2) - ( ⅞ - 0

3

+ 0 2)

- [ ( J - 2 > + 2 ≈ )-(⅛ -l≈ + l≈ )]= l

pθ

41. χ 1∕ 3 = 0 => x = 0; Area = — J = [- h =

4 z3 ] o-

1

( - 1 (O ) 4 ∕ 3 ) -

1x

p8 1∕ 3

dx + J

x 1∕ 3 dx

+ [ i χ 4 z 3 ]o ( - 1 ( - i ) 4 ∕ 3 ) + (3 (8 )4 ∕ 3 ) -

( I (O) 4 ∕ 3 )

— 51 “ 4

43. The area of the rectangle bounded by the lines y = 2, y = 0, x = π, and x = 0 is 2π. The area under the curve >π

J

θ (1 ÷ cos x) dx = [x ÷ sin x]θ = (π + sin π) —(0 + sin 0) = π. Therefore the area of

the shaded region is 2π —π = π.

Section 5.4 The Fundamental Theorem of Calculus 45. On [—∣, θ] : The area of the rectangle bounded by the lines y = ᅟ ∕2 , y = 0, 0 = 0, and 0 = —∣is χ ∕2 (∣) = z^

. The area between the curve y = sec 0 tan θ and y = 0 is —J*

4 sec

0 tan 0 d0 = [—sec 0]θ π ∕ 4

= (-sec 0) - (-sec ( - ∣)) = √ z2 - 1. Therefore the area of the shaded region on [ - ^ θ] is

zl ^

On [θ, ∣] : The area of the rectangle bounded by 0 = ^, 0 = 0, y = √ z2, and y = 0 is √^2 Q ) =

. The area

sec 0 tan 0 d0 = [sec 0]θ^4 = sec ∣—sec 0 = χ ∕^ — 1. Therefore the area

under the curve y = sec 0 tan 0 is j ^ of the shaded region on [θ, ∣] is z^

zl ^

÷ ( λ ∕^ - 1)

∕2 — 1) . Thus, the area of the total shaded region is —( ᅟ

+ √ 2 -l) + ( ⅛ - √ 2 + l) = ^ .

⅛

47. y = I 1t ⅛ - 3 => ⅛ = 1 and y(π)t = I ∣dt —3 = 0 - 3 = —3 => (d) is a solution to this problem. αX X J π J π 49. y = J sec t dt + 4 => ^ = sec x and y(0) = Jθ sec t dt + 4 = 0 + 4 = 4 => (b) is a solution to this problem.

51. y = j

57.

⅛ dx

=

sectdt + 3

oV = 2√x

5 2

x

~^

2

=> C = Jo

f

∣t - 1

/2

2

d t = [t1 /2 ] n L J O V

=

λ ∕x

c(ioo) - c(i) = √ io δ - √ i = $9.oo

59. (a) v = (b) a = (c) (d) (e) (f) (g)

⅛= ^ X f Wdx = f w ^ v(5) = f (5)= 2 m7sec is negative since the slope of the tangent line at t = 5 is negative

s = J^ f(x) dx = ∣(3)(3) = ∣m since the integral is the area of the triangle formed by y = f(x), the x-axis, and x = 3 t = 6 since from t = 6 to t = 9, the region lies below the x-axis At t = 4 and t = 7, since there are horizontal tangents there Toward the origin between t = 6 and t = 9 since the velocity is negative on this interval. Away from the origin between t = 0 and t = 6 since the velocity is positive there. Right or positive side, because the integral of f from 0 to 9 is positive, there being more area above the x-axis than below it.

157

Chapter 5 Integration

158

p π ∕k

61. k > 0 => one arch of y = sin kx will occur over the interval [θ, ∣] =J> the area = Jθ =

- t

c o s

(k ( 0 ) - ( - E

c o s

(°)) =

t

63. ∫^ f( t) dt = x2 - 2x + 1 => f(x) = ⅛ J j ( t ) dt = ⅛ (x2 - 2x + 1) = 2x - 2 *x+1

J 2

⅛ dt ^

f(x )

=

_

p l+ 1

rR

2 _ = - ^ ^ f'(l) = —3 ;f(l) = 2 —J

2

⅛ d t = 2 - 0 = 2i

L(x) = -3 (x - 1 ) + f(l) = -3 (x - 1) + 2 = -3 x + 5 67. (a) (b) (c) (d) (e) (f) (g)

True: since f is continuous, g is differentiable by Part 1 of the Fundamental Theorem of Calculus. True: g is continuous because it is differentiable. True, since g, (l) = f(l) = 0. False, since g"(l) = f '( l ) > 0. True, since g'(l) = 0 and g"(l) = f'(l) > 0. False: g"(x) = f , (x) > 0, so g" never changes sign. True, since g, (l) = f(l) = 0 and g'(x) = f(x) is an increasing function of x (because f'(x) > 0).

5 3 INDEFINTE INTEGRALS AND THE SUBSTITUTION RULE 1. Let u = 3x => du = 3 dx => ∣du = dx J* sin 3x dx = J* j sin u du = - ∣cos u + C = - ∣cos 3x ÷ C 3. Let u = 2t => du = 2 dt => ∣du = dt J* sec 2t tan 2t dt =

J*∣sec u tan u du =

∣sec u + C = ∣sec 2t + C

5. Let u = 7x - 2 => du = 7 dx => ∣du = dx ∫ 28(7x - 2)- 5 dx = ∫ | (28)U^ 5 du = ∫ 4u^5 du = -u ~ 4 + C = -(7 x - 2)^4 + C 7. Let u = 1 - r3 => du = - 3 r 2 dr ≠> - 3 du = 9r2 dr

f ^⅛ =f - 3 U" ∕

1 2

du = -3(2)u 1∕ 2 + C = - 6 ( 1 - r3 ) v 2 + C

9. Let u = x3 ∕ 2 — 1 => du = ∣x 1∕ 2 dx => j du = √ x dx J √ x sin2 (x3 ∕ 2 - 1) dx =

f

/

sin kx dx = [— £ cos kx] θ k

j sin2 u du = j (5 - ∣sin 2u) + C = ∣(x3 /2 - 1) - ∣sin (2x3∕ 2 - 2) + C

Section 5.5 Indefinite Integrals and the Substitution Rule Let u = cot 20 => du = —2 csc2 20 d0 => —∣du = csc2 20 d0

11. (a)

∫ csc2 20 cot 20 d0 = -

f

i u du = - j (⅛ ) + C = - £ + C = - ∣cot2 20 + C

(b) Let u = esc 20 => du = - 2 esc 20 cot 20 d0 => - ∣ du = esc 20 cot 20 d0

f csc

2

20 cot 20 d0 =

f - i u du = -

∣( f ) + C = - ⅛ + C = - ∣csc2 20 + C

13. Let u = 3 - 2s => du = —2 ds => - ∣du = ds

f

√ 3 - 2s ds = J

λ∕ u

( - I du) = - | J*u 1∕ 2 du = ( - ∣) ( j u3∕ 2 ) + C = - ∣(3 - 2s)3 ∕ 2 + C

15. Let u = 5s + 4 => du = 5 ds => ∣du = ds ∕⅛

*

√

17. L e tu = l - 0

f 0√1

( ∣d ∙ ) = J " - 1' '

⅛

4

= (⅛ ) (21 η + C = ≡ √ s Γ f 4 + C

≠> du = -2 0 d 0 => —∣du = 0 d0

2

- 0 2 d0 =

j

√ u ( - ∣du) = - i ∫ u 1∕ 4 d u = ( - ∣) (∣u 5∕ 4 ) + C = - ∣(1 - 0 2 )5 /4 + C

19. Let u = 7 —3y2 => du = —6y dy ≠- - ∣du = 3y dy / 3y√ 7 - 3y2 dy = 21. Let u = 1 + Λ∕

X

f

∕

A

U(-

=> du = ^

∣du) = - ∣J* u 1/2 du = ( - ∣) (j u3 /2 ) + C = - ∣(7 - 3y2 )3 /2 + C dx ≠> 2 du = - ^ dx

23. Let u = 3z + 4 ≠ du = 3 dz ≠- ∣du = dz J* cos (3z + 4) dz = § (cos u) (∣du) = 5 J* cos u du = ∣sin u + C = ∣sin (3z + 4) + C 25. Let u = 3x + 2 ≠ du = 3 dx ≠- j du = dx

f sec

2

J

(3x + 2) dx = J* (sec2 u) ( ∣du) = ∣ sec2 u d u = ∣tan u + C = ∣tan (3x + 2) + C

27. Let u = sin (∣) ≠ du = ∣cos (5) dx => 3 du = cos (∣) dx J* sin5 (5) cos (5) dx = J* u5 (3 du) = 3 (g u6 ) + C — ∣sin6 ( ∣) + C 2 29. Letu = ⅛ lo —O1 ≠ du = £ dr => 6 du = r dr u 5 ( 6 d u ) = 6 ∫ u 5 du = P ⅛ - l ) 5 d r√

6

(f)+

c

= (⅛ -l)δ +

c

31. Let u = x3∕ 2 + 1 ≠ du = 5 x1∕ 2 dx => 5 du = x1∕ 2 dx J* x1∕ 2 sin (x3 ∕ 2 + 1) dx =

f

(sin u) (∣du) = 5 J* sin u du = j (—cos u) + C = —∣cos (x3 ∕ 2 + 1) + C

33. Let u = sec (v + ?) => du = sec (v + 5) tan (v + ?) dv J s e c (v + ∣) tan (v + ∣) dv = J* du = u + C = sec (v ÷ f ) + C 35. Let u = cos (2t + 1) => du = —2 sin (2t ÷ 1) dt => —∣du = sin (2t + 1) dt

∕

sin (2 t+ 1) cos2 ( 2 t÷ l)

u ι

“

J

2 UƩ —

2u

r

v

“

1 ∣r 2 cos(2t+ 1) r v

159

160

Chapter 5 Integration

37. Let u = cot y => du = —esc2 y dy ≠ —du = csc2 y dy f ^ c o ty csc2 y dy = J* ^ ∕u (-du) = — J u 1∕ 2 du = - j u3 ∕ 2 + C = - j (cot y)3∕ 2 + C = - j (cot3 y) 1^2 + C 39. Let u = i - 1 = Γ 1 - 1 => du = —1^2 dt => —du = ∣dt J* ⅛ cos (∣— 1) dt = J* (cos u ) (- du) = —§ cos u du = —sin u + C = —sin (∣— 1) + C 41. Let u = sin ∣ => du = (cos ∣) (—^ ) d0 4

-d u = ^ cos ∣d0

y ^ sin ∣cos ∣dd = y - u du = —∣u 2 + C = - ∣sin2 ∣+ C 43. Let u = s3 + 2s2 —5s + 5 ≠> du = (3s2 + 4s —5) ds y ( s 3 + 2s2 - 5s + 5) (3S2 + 4s - 5) ds = y u du = ⅛ + C = ^ ± L ⅛

+

c

45. Let u = 1 + 14 => du = 4t3 dt => ∣du = t3 dt ∫t∙( l

+ t ')

,

Λ √ .∙( μ ) = l( l.∙)

+

C = ⅛ ( l+ t∙Λ c

47. Let u = x2 + 1. Then du = 2xdx and ∣ du = xdx and x2 = u —1. Thus J x 3 ᅟ ∕x ^ + T d x = J ( u - l ) ^ d u = ⅛S ( u 3 /2 - u 1 /2 )du = I [∣u5∕ 2 - ju 3 ∕ 2 ] + C = ∣ u5∕ 2 - ∣ u3 ∕ 2 + C = ⅜(x2 ÷ 1)5 /2 - ∣(x2 + i ) 3 /2 ÷ c

49. (a) Let u = tan x => du = sec2 x dx; v = u3 ≠> dv = 3u2 du => 6 dv = 18u2 du; w = 2 + v => dw = dv 1 'k √ d" √ ⅛ ⅛ √ ⅛ = < i R ∕ 1S ⅛ * = * 1+ c = - ÷ + c =

-

⅛

,

+ C = -

2 + ^ n 3χ + C

(b) Let u = tan3 x ≠ du = 3 tan2 x sec2 x dx 4 6 du = 18 tan2 x sec2 x dx; v = 2 + u ≠> dv = du J

∣ 18 tan2 x sec2 x J Y _ (2 + tan3 x)2

f J

6 du __ j (2 + u)2 - J

6dv _ v2 -

_ 5 χ Γ _ v v

_

6 ∣r _________ 6 ∣p 2+u+ b 2 + tan3 x + v

(c) Let u = 2 + tan3 x => du = 3 tan2 x sec2 x dx ≠> 6 du = 18 tan2 x sec2 x dx

∫

18 tan2 x 3sec2j 1 J V (2 + tan x ) “ "

J

∣ 6 du

6 x r

6 2 + ≡

∣∕~ , + c

51. Let u = 3(2r — 1)2 + 6 => du = 6(2r — 1)(2) dr => ⅛ du = (2r - 1) dr; v = √ u ≠> dv = ^ = i⅛ J

du ≠> ∣dv

du

c' -l ⅞ ^

Γ

i ^ = J (⅛ ) ⅛

4

)√

1“

v) (∣d ,) = 1 s ⅛ v + C = 1 » √ i i + c

= i sin √3(2r - 1)2 + 6 + C 53. Let u = 3t2 - 1 ≠- du = 6t dt ≠∙ 2 du = 12t dt s = y i2 t( 3 t 2 - l) 3 dt = y u 3 (2du) = 2 ( i u 4 ) + C = i u 4 + C = ∣(3t2 - l ) 4 + Cj s = 3 when t = l ≠> 3 = ∣(3 —1)4 + C => 3 = 8 + C ≠> C = - 5 4

s = ∣(3t2 - 1)4 - 5

55. Let u = t + ⅛ => du = dt s = y 8 sin2 (t + ⅛) dt = y 8 sin2 u du = 8 ( ∣- ∣sin 2u) + C = 4 (t + ⅞) - 2 sin (2t + ∣) + C; s = 8 whent = 0 ≠> 8 = 4 (⅞ ) - 2 sin (≈) + C ≠> C = 8 - f + l = 9 - ∣ =^ s = 4( t + ⅞) ~ 2 sin (2t + ∣) + 9 —j = 4t —2 sin (2t + ∣) + 9

Section 5.6 Substitution and Area Between Curves

161

57. Let u = 2t —∣ => du = 2 dt => —2 du = —4 dt 4 sin (2t —∣) dt = J (sin u ) ( - 2 du) = 2 cos u + Ci = 2 cos (2t — f ) + Ci;

-

⅛= f

att = 0and⅛ = 100 we have 100 = 2 cos ( - ≡) + Ci ≠> C i = 100 ≠> ⅛ = 2 c o s ( 2 t - ≡ ) + 1∞ => s = J (2cos (2t - ∣) + 100) dt = J*(cos u + 50) du = sinu + 5 0 u ÷ C 2 = sin (2t - ^ ) + 5 0 (2t - ∣) + C2 ; at t = 0 and s = 0 we have 0 = sin (—∣) + 50 ( - ∣) + C2 => C2 = 1 + 25π =Φ s = sin (2t - f ) ÷ 100t —25π + (1 + 25π) => s = sin (2t —∣) + 100t + 1 59. Let u = 2t => du = 2 dt => 3 du = 6 dt s = J* 6 sin 2t dt = f (sin u)(3 du) = —3 cos u + C = - 3 cos 2t + C; at t = 0 and s = 0 we have 0 = - 3 cos 0 + C => C = 3 => s = 3 —3 cos 2t => s (∣) = 3 —3 cos (π) = 6 m 61. All three integrations are correct. In each case, the derivative of the function on the right is the integrand on the left, and each formula has an arbitrary constant for generating the remaining antiderivatives. Moreover, sin2 x + Ci = 1 - cos2 x + Ci => C2 = 1 + Cβ also -co s 2 χ + C2 = - s y i - ∣+ C2 => C3 = C2 - ∣= Ci + ∣. 63. (a ) ( τ q )

X

v

∏∞ sin 120πt dt = 60 [-V max ( ⅛ ) cos (120πt)] J/6 ° = - ⅛ [cos 2π - cos 0]

= - ⅛ [ i- i] = o (b) Vmax = √ 2 Vmκ = √2(240) ≈ 339 volts pl/60

(C) J

/.1/60

(Vmax)2 sin2 120τrtdt = (Vmax)2 J o

o

= ⅛

[t “ ( ⅛ ) ≡ 240πt]= ⅛

2

nl/60

(⅛ M ta )d t= (⅛ j

(1 - cos 240τrt) dt

[(⅛ - ( ⅛ ) sin(4π)) - (0 - ⅛ ) sin(0))] = ⅛

5.6 SUBSTITUTION AND AREA BETWEEN CURVES 1. (a) Let u = y + 1 => du = dy; y = 0 => u = l , y = 3 => u = 4 Γ √ τ τ ι ⅛ = f y l2 * = Ci ■” ] : = ( 8 ∣ r - (I) ι ∣ ), ' , = θ

= ⅛

(b) Use the same substitution for u as in part (a); y = - 1 => u = 0, y = 0 => u = 1 £ √ y + T dy = J o 'uV≈ du = [2 w © J = ( j) (I)>C - 0 = ∣

3. (a)

Let u = tan x => du = sec2 x dx; x = 0 => u = 0, x = ∣ ≠> u = l o

tan x sec2 x d x = I u du = Jo

⅛

L2 J o

= ⅛ - - 0 = 12 2

(b) Use the same substitution as in part (a); x = —∣ => u = —1, x = 0 => u = 0

Γo tan x sec2 x d x = ΓIo u du = J-π ∕4 J-l I

Γ 21 0 = 0 —⅜ = —⅛ 2 2 L 2 J -1

5. (a) u = 1 + 14 => du = 4t3 dt => ∣du = t3 dt; t = 0 => u = 1, t = 1 => u = 2

f

t3 (1 + 14 )3 dt = Γ ∣u3 du = ⅛ 1

=

⅛“ ⅛

(b) Use the same substitution as in part (a); t = —1 => u = 2, t = l => u = 2

X 113 (1 +

14 )3 dt = X ∣u 3 du = 0

162

Chapter 5 Integration

7. (a) Letu = 4 + r2 => du = 2rdr => ∣du = r d r,r = —1 => u = 5, r = l ≠> u = 5 f 7∑τ⅛ J - l ( 4 + r2 ∕

I u - 2 du = 0

dr = 5

f J 5

2

(b) Use the same substitution as in part (a); r = 0 => u = 4, r = 1 => u = 5 ∫

0

⅛

dr

=

5

∫

4

⅛u-2 du = 5 [ - i u - 1]* = 5 ( - i ( 5 ) - 1) - 5 ( - i ( 4 ) - 1) = 1

9. (a) Let u = x2 ÷ 1 => du = 2x dx => 2 du = 4x dx; x = 0 => u = 1, x = V ^ => u = 4 0

τ⅛ τ

dx =

Ji √≡du

J

2 u ~v 2 d u =

l

[4 u l / 2 ] ι = 4 (4 )v 2 ~ 4 (O v 2 =

4

(b) Use the same substitution as in part (a); x = —√ ^ => u = 4, x = ᅟ ∕3 ≠> u = 4 I r d x = / -7= du = 0 J - √ 3 √X 2 + 1 J 4 √U

11. (a) Let u = 1 —cos 3t => du = 3 sin 3t dt => ∣du = sin 3t dt; t = 0 => u = 0, t = 5 => u = 1 —cos ? = 1 X

( l- c o s 3 t) s in 3 td t = ∕

∣udu = [⅛ (⅛ ) ] θ = ⅛(D 2 - ⅛(0)2 = ∣

(b) Use the same substitution as in part (a); t = ∣ => u = 1, t = ∣ 4 u = l - cos π = 2 £

(l-∞ s 3 t)s in 3 td t = ∕

⅜u du = [f ( < ) ] " = ∣(2)2 - ∣(1)2 = ∣

13. (a) Let u = 4 + 3 sin z => du = 3 cos z dz => ∣du = cos z dz; z = 0 => u = 4, z = 2π => u = 4 J*2π

∩4

(b) Use the same substitution as in part (a); z = —π ≠ u = 4 + 3 sin (—π) = 4, z = π => u = 4 cs V dz J -π v 4 + 3 sin Z

, °A

=

f Λ≡ (I J 4 vu '

du) = 0 '

15. Let u = t5 + 2t => du = (5t4 + 2) dt; t = 0 => u = 0, t = l => u = 3 √ t 5 + 2t (5t4 + 2 ) dt = Jθ u 1^2 du = [j u3∕ 2 ] θ = ∣(3)3 ∕ 2 - ∣(0)3 ∕ 2 = 2 √ z3

/

17. Let u = cos 20 => du = - 2 sin 20 d0 ≠- - I du = sin 20 d0; 0 = 0 ≠> u = 1,0 = ? => u = cos 2 (?) = I *π∕6

J

0

pi/2

cos- 3 20 sin 20 d0 = /

u^ 3 ( - ⅛du) = - ∣/

P 1∕ 2

Γ

/

u" 3 du = [ - j ( ⅛ ) ]

1/2 ι

=

^

- ^

= 2

19. Let u = 5 —4 cos t => du = 4 sin t dt => J du = sin t dt; t = 0 ≠ u = 5 —4 cos 0 = 1, t = π => u = 5 —4 cos π = 9 /

5 (5 - 4 cos t)1∕ 4 sin t dt = J ι 5u1^4 (∣du) = ∣/

u1^4 du = [∣( ∣u5^4 )]

1

= 95∕ 4 — 1 = 35 ∕ 2 —1

21. Let u = 4y - y2 + 4y3 + 1 => du = (4 —2y + 12y2 ) dy;y = 0 => u = l , y = l ≠ u = 4(1) —(1)2 + 4(1)3 + 1 = 8 /

(4 Y - y2 + 4 y3 + 1) - 2 ^3 (12y2 - 2y + 4) dy = /

23. Let u = 03∕ 2 ≠> du = 2 #1/2 d 0 /

3π2 √ 0

cos2 (03 /2 ) d0 = /

4

u^2∕ 3 du = [3u1^3 ] J = 3(8)1∕ 3 —3(1)1^3 = 3

2 du = √ 0 d0; 0 = 0 ≠ u = 0,0 = 3√π≡ => u = π

cos2 u (j du) = [5 (5 + 5 sin 2u)] θ = j ( f + ∣sin 2π) - 5 (0) = f

Section 5.6 Substitution and Area Between Curves 25. Let u = 4 —x 2 =+ du = —2x dx =+ — 1 du = x dx; x = —2 =>

u = 0, x = 0 =+ u = 4, x = 2 =+ u = 0

pθ ______ p2 ______ p4 A = - I x √ ,4 - x 2 dx + I x √ 4 - x2 dx = - I - ∣ u 1∕ 2 du 2 J -2 Jθ Jθ

= [1 ^ 2

= 5

- 5 3 /2

^

=

pθ

p4

p4

+ I — ⅜u 1∕ 2 du = 2 I 2 ⅜u 1∕ 2 du = I u 1∕ 2 du J4

*

Jθ

Jθ

T

27. Let u = 1 ÷ cos χ => du = —sin x dx => - d u = sin x dx; x = - π => u = 1 + cos (—π) = 0, x = 0 => u = 1 ÷ cos 0 = 2 3 (sin x) y 1 ÷ cos x dx = - J^ 3u 1∕ 2 (-d u ) = 3 ^ u 1∕ 2 du = [2u3 / 2 ] θ = 2(2)3 ∕ 2 - 2(0)3 ∕ 2 = 2 5 ∕ 2

A = - J

29. For the sketch given, a = 0, b = %; f(x) - g(x) = 1 - cos2 x = sin2 x = -1 ^ y f

A =

π

O→o1 2x)

dx =

1 Γ ( i _ cos 2x) dx = H x -

2x ;

* = U(τr - 0) - (0 - 0)] = 5

31. For the sketch given, a = - 2 , b = 2; f(x) —g(x) = 2x2 - (x4 - 2x2 ) = 4x 2 - x4 ; A = ∫J 4 x

2

-√ )d x = [⅞ -⅛

2

= ( f 4 ) - [ 4 - ( - f ) ]

= f - ^

⅜

^

= ⅛

33. For the sketch given, c = 0, d = 1; f(y) - g(y) = (12y2 - 12y3 ) - (2y2 - 2y) = 10y2 — 12y3 ÷ 2y; a

=

1 ( 1 °y 2 ^

1 2 y3

+ 2 y) d y = X

1 °y 2 d y

^ f

0

1 2 y3 d y

+ X

2y dy

= l τ y3 ] J - Cτ y4 ] J + [1 y 2 ] J

= ( ^ - 0 ) - ( 3 - 0 ) + (l-0 ) = f 2

35. We want the area between the line y = l , 0 ≤ x ≤ 2 , and the curve y = ¾-, m in u s the area of a triangle (formed by y = x and y = 1) with base 1 and height 1. Thus, A = Jθ (1 - ^ Γ

12/

4

2

3

2

dx - ^ (1)(1) = [x

-

⅛]

“ I

6

37. AREA = A l + A2 A l: For the sketch given, a = —3 and we find b by solving the equations y = χ 2 —4 and y = —x2 —2x simultaneously for x: χ 2 - 4 = - x 2 - 2x => 2x2 + 2x —4 = 0 ≠> 2(x + 2)(x - 1) => x = - 2 or x = 1 so p-2

b = - 2 : f(x) - g(x) = (x2 - 4 ) - ( - x 2 - 2x) = 2x2 + 2x - 4 => A l = J 9

⅜ + ⅜ - 4x A2:

1 -2

(2x 2 ÷ 2x - 4) dx

= ( - ^ + 4 + 8) - (- 1 8 + 9 + 12) = 9 - ^ = V;

For the sketch given, a = —2 and b = 1: f(x) —g(x) = (—x2 —2x) — (x 2 —4) = —2x 2 —2x + 4 => A2 = - X 2 (2 χ 2 + 2x - 4) dx = - [ y + x2 - 4xj

= - ( j + 1 - 4) + ( - y + 4 + 8)

= - l - l + 4 - f + 4 + 8 = 9j Therefore, AREA = A l + A 2 = ⅛ + 9 = f 39. AREA = A l + A2 + A3 A l: For the sketch given, a = —2 and b = —1: f(x) - g(x) = (—x + 2) — (4 —x2 ) = x 2 —x —2

A2:

For the sketch given, a = - 1 and b = 2: f(x) - g(x) = (4 —x2 ) - ( - x + 2) = —(x 2 —x —2) p2 Γ 12 ^ = 4 , ^ - ^ 2 ) ^ = - ( 1 - 7 - 4 , = -(5 -1 -4 ) + (-1 -1 + 2 ) = -3 + 8 -1 = ^

A3:

For the sketch given, a = 2 and b = 3: f(x) - g(x) = ( - x ÷ 2) — (4 —x2 ) = x 2 —x —2 ≠

*≈ = ∫ , ⅛ - " - ≈ ) *

= [ τ - p

⅛

= ( f - k

+

( '- i - +

* - ⅛ - 'i

163

Chapter 5 Integration

164

Therefore, AREA = A l + A 2 + A3 = ⅛ + ∣+ ( 9 - ∣- ∣) = 9 - ∣= f 41. a = - 2 , b = 2; f(x) - g( χ ) = 2 - (x 2 - 2) = 4 - x 2 ^

A = ∕j 4 - x

[4X - ⅛ ] 2 2 = ( 8 - ∣) - ( - 8 + ∣)

2)d x =

= 2 ∙(⅛ - !) = ⅛

43.

y

= 0, b = 2;

a

(8x —x4 ) dx

f(x) —g(x) = 8x —x 4 =≠> A = —

f e

-

L

ill

_

5

2

J

2 0

-

_

1? _ 5

^

-

80 - 32 _ 5

48

-

5

2 45. Limits of integration: x2 = —x2 + 4x =» 2x2 —4x = 0 => 2x(x —2) = 0 ≠ a = 0 and b = 2; f(x) —g(x) = ( - x 2 + 4x) - x 2 = - 2 X2 + 4x => A = ∫ — _

-

16 3

τ

o

∣ 2

( - 2 x 2 + 4 x ) dx = [ ^ _

-

-3 2 + 4 8 6

-

_

y

+ ⅛] θ

8 3

47. Limits of integration: x4 —4x 2 + 4 = x2 => x4 - 5x 2 + 4 = 0 => (x2 —4) (x2 - 1 ) = 0 ≠> (x + 2)(x - 2)(x + l)(x - ι l) = 0 => x = - 2 , - 1 ,1 , 2; f(x) ~ g(x) = (χ 4 —4x 2 + 4) —x 2 = x4 - 5x2 + 4 and g(x) —f(x) = x2 — (x4 —4 X2 + 4) = - x 4 + 5 X2 —4 => A = J

2

( - x 4 + 5x 2 - 4 ) d x + J * ι (x4 - 5x2 + 4)dx

+ J^ ( - x 4 + 5x 2 - 4)dx

= [-τ + y - 4x] j + [ τ - τ + 4x] ' 1 + [ τ + y - 4x]ι = ( I H + 4 ) √ ⅛ - f + 8) + ( ∣H + 4 ) √ - ∣+ ∣- 4 ) + ( - ⅛ + f - 8 ) - ( - l _ -

60 T ÷

60 _ 3 -

3 00 -1 8 0 _ ~ ~

o

8

n. ∙ /7“ Γ I τ∙ ∙ 49. Limits of integration: y = √ x = < I

v

X, X ≤ 0 , and √ rx , x ≥ 0

∕~ x = ∣÷ f 5y = x ÷ 6 or y = ∣÷ | ; for x ≤ 0: ᅟ => 5 ^ ∕→ = x + 6 => 2 5 ( - x) = x 2 + 12x + 36 => x2 + 37x + 36 = 0 => (x ÷ l)(x + 36) = 0 => x = —1, —36 (but x = —36 is not a solution); for x ≥ 0: 5 √ x = χ + 6 => 25x = x2 + 12x ÷ 36 => x 2 - 13x + 36 = 0 => (x - 4)(x - 9) = 0 => x = 4, 9; there are three intersection points and

+

j-4 )

Section 5.6 Substitution and Area Between Curves

_

Γ(x + 6)2 [ 10

— /3 6 U0

_

■ 2 / 3

γ

  3∕21 ° J

∣ [ (x + 6)2 _ [ 10

25 _ 2 . / 100 _ 10 3/ 10

2 j 3 ∕2 _ 3 ’

2 γ 3 λ

3∕2^∣4 1 J 0 L3

36 ∣∩λ

,

ψ

3/2 _

Q 3∕2

/2

(x + 6)2 ^∣9 l0 J 4 225 10

2 d 3 ∕2 3 *

∣l ∞ ᅟ_ _ 50 . 20 _ ψ 10 / ~ 10 3 -

51. Limits of integration: c = 0 and d = 3; f(y) - g(y) = 2y2 —0 = 2y 2

* = £ * « = M ^= 2 ∙9 = I 8

53. Limits of integration: 4x = y2 - 4 and 4x = 16 + y 4 y 2 - 4 = 16 + y ≠- y 2 - y - 2 0 = 0 =4 (y - 5)(y + 4) ≈ 0 =4 c = - 4 and d = 5; f(y) - g(y) = ( 1⅞ ς ) - ( ⅞ f ) = ξ ⅛ ^ => A = i J = ∣H

c5 ( - y 2 + 4

y + 20) dy

+ ⅞ + 20y] 5 4

=H - f +h ι∞ )-H M -8 o ) ≈ H -ψ + h ≡ ) = ψ 55. Limits of integration: x = —y2 and x = 2 —3y2 4 - y 2 = 2 - 3y2 =4 2y 2 - 2 = 0 4 2(y - l)(y + I) = 0 ≠> c = - I and d = I; f(y) - g(y) = (2 - 3y2 ) - ( - y 2 ) = 2 - 2y 2 = 2 (I - y 2 ) A

= 2 I,( 1~y )

2 d y = 2 [ y - ⅞] l = 2 ( l - j ) - 2 ( - l + ∣) = 4 ( j ) = f

l

57. Limits of integration: x — y2 — I and x = ∣ y∣ 0 - y - I = ∣ y∣ √l -

y2

4 4

y4 —2y 2

+ I = 4 + I = 0 (2y2 - I) (y 2 — I) = 0 4- 2y 2 - I = 0 or y 2 —I = 0

4

y2 = ∣or y2 = I 4

4

y4

2y 2

-

2y4

+ I =

y2

y

4

y2

y4

2

y2

(I -

y2 )

—3y 2

y= ± ^

or y = ± l .

Substitution shows that ⅛ ^ are not solutions =4 y = ± I ; for - I ≤ y ≤ 0, f(x) - g(x) = - y √ l - y 2 - (y2 - I) = l —y2 - y ( l - y 2 ) l ∕ 2 , and by symmetry of the graph, A

=2I 1[1 ^ y

2

^ y ( l ^ y2 ) v 2 ] d y

= 2 ∫ l (l - y 2 ) dy - 2 ∫

= 4 - C , + 2 ⅛)

jy

( l - y 2 ) l / 2 dy

2 (I —y2 )3 /2 3

0

= 2 [(0 - 0) - ( - l + ∣)] + ( ∣- 0) = 2 -l

5 3

165

166

Chapter 5 Integration

59. Limits of integration: y = —4x 2 + 4 and y = x4 — 1 =» x 4 — 1 = —4x 2 + 4 => x4 + 4x 2 - 5 = 0 => (x2 + 5) (x - l)(x + 1) = 0 => a = —l a n d b = l ; f(x) —g(x) = - 4 x 2 + 4 - x4 + 1 = —4X2 - x4 + 5 ≠- A = J* ι ( - 4 X2 —x 4 + 5) dx = [ - ^

- y + 5xj

+ 5 ) - ( f + ∣- 5 ) = 2 H - i

= ( - M

+

5 )= ^ ⅛

61. Limits of integration: χ = 4 —4y2 and x = 1 —y4 ≠- 4 —4y 2 = 1 —y 4 => y4 - 4y 2 + 3 = 0 => (y - x A ) (y +   Λ ) (y - i)(y + 1) = 0 => c = - 1 and d = 1 since x ≥ 0; f(y) —g(y) = (4 —4y2 ) — (1 —y 4 ) = 3 - 4y 2 + y 4 ≠> A = J* 1(3 - 4y 2 + y 4 ) dy

63. a = 0, b = n; f(x) —g(x) = 2 sin x —sin 2x => A = J^ (2 sin x —sin 2x) dx = [ - 2 cos x + s≤^ι] θ = [-2 (-l)

i] - ( - 2 - l + i)= 4

+

65. a = —1, b = 1; f(x) - g(x) = (1 - x2 ) - cos ( y )

67. a = —∣, b = | ; f(x) —g(x) = sec2 x - tan2 x p π ∕4

=> A = J

.4 (sec2 x —tan 2 x) dx

p it/4

= J = J - π ∕4

2 / 4 [sec

I

x - (sec2 x — 1)] dx

- * = W'Λ4,4 = 5 - ( - 5 ) = 5

69. c = 0, d = | ; f(y) —g(y) = 3 sin y^/cos y —0 = 3 sin y √ c o s y => A = 3J^

sin y ^∕cδ s y dy = - 3 [ j (cos y)3 ^2 ] θ^2

= -2 ( 0 - 1 ) = 2

Section 5.6 Substitution and Area Between Curves 71. A = A 1 + A 2 Limits of integration: x = y3 and x = y => y = y3 ≠> y3 - y = 0 => y(y - l)(y + 1) = 0 ≠> c 1 = - 1 , d 1 = 0 and c 2 = 0, d 2 = 1; f 1 (y) - gι(y) = y3 - y and f2(y) - g2(y) = y - y3 ^ by symmetry about the origin, Ai + A 2 = 2A2 ≠> A = 2 ^ (y - y3 ) dy = 2 [ ^ - ⅛] θ ^ ∖2

41

2

- A2 73. A — Ai -∣ Limits of integration: y = x and y = ⅛ => x = ^ , x ≠ 0 = > χ 3 = l = > χ = l , f 1 (x ) - g 1 (x ) = χ - 0 = X 4

A i = ∫ θ x d x = [ ^ ] θ = | ; f2 (x) - g 2 (x) = ⅜ - 0

= χ - 2 => A 2 = I x^ 2 dx = [ ^ ] ] = - i + 1 = i ; d∖ L X J 1 2 2 A — Ai 4 A2 — ∣+ ∣— 1

75. (a) The coordinates of the points of intersection of the line and parabola are c = x 2 => x = ± ^ ∕c and y — c (b) f(y) - g(y) = √ y - ( ~ √ y ) = 2 √ y => the area of the lower section is, AL = I [f(y) —g(y)] dy d0 = 2 J v ∕y dy = 2 [ j y3 ∕ 2 ] θ = ∣c 3 ∕ 2 . The area of the entire shaded region can be found by setting c = 4: ^ ^ (D 4 3 ∕ 2 = ^ = ^ . Since we want c to divide the region into subsections of equal area we have A = 2AL ^ ^ = 2 ( | c 3/ 2 ) ^ c = 42∕ 3 (c) f(x) - g(x) = c - x

2

=> AL = ^ ^ [f(x) - g(x)] dx = J* ^_(c - x2 ) dx = [ex - ^ ^

= 2 [c3 ∕ 2 - y ]

= | c3 ∕ 2 . Again, the area of the whole shaded region can be found by setting c = 4 => A = y . From the condition A = 2AL , we get ∣c3 ∕ 2 = y => c = 4 2∕ 3 as in part (b). 77. Limits of integration: y = 1 + ^ ∕x and y = - ^ => 1 +

v

6E = -2= , X ≠ 0 =>

λ ∕x

+ x = 2 => x = (2 - x)2

=> x = 4 - 4x ÷ x2 => x 2 - 5x + 4 = 0 => (x - 4)(x - 1 ) = 0 => x = 1,4 (but x = 4 does not satisfy the equation); y = ^ and y = ^ ^ ^ = I => 8 = x λ ∕ x ≠> 64 = x3 => x = 4. Therefore, AREA = Ai + A2 : fι(x) - gι(x) = (1 ÷ x 1//2 ) - 5 => Ai ≡ f (1 + x 1∕ 2 = (1 + 5 ~ I)

-

0

=

w

dx — [x + 5 x 3 ∕ 2 ⅛ ( χ ) - g 2 (x ) = 2 X ^ 1 ∕2 _ X 4 1

(4 ∙2 - ⅛ ) - ( 4 - l) = 4 - ⅛

Therefore, AREA = A 1 + A 2 = g + ⅛ = ¾ ≡ ι = g = ⅛

= ^

167

168

Chapter 5 Integration

2 2 2 2 3 3 79. Area between parabola and y = a : A — 2 Jθ (a - x ) dx = 2 [a x - ∣x ] “ — 2 (a — y j —0 = y ;

Area of triangle AOC: ∣(2a) (a2 ) = a 3 ; limit of ratio =

lim + X r = ∣which is independent of a.

a → 0+

( - I- )

81. Neither one; they are both zero. Neither integral takes into account the changes in the formulas for the region's upper and lower bounding curves at x = 0. The area of the shaded region is actually pθ

pl

A = J J —x —(x )]d x ÷

J

pθ

[x —(—x)] dx =

pl

- 2 x d x + J θ 2 x d x = 2.

J

83. Let u = 2x => du = 2 dx => ∣du = dx ; x = 1 => u = 2, x = 3 => u = 6 p3 p6 p6 J 1 ⅜ 2 s dx = J 2 ¾ f ( I d u ) = J 2 T d u = 1 ^ 2 = P(6) - F(2) 85. (a) Let u = - x => du = —dx; x = —1 => u = 1, x = 0 => u = 0 pθ

pθ

f odd => f ( -x ) = -f(x ). Then

J

f(x) dx =

J

ι

pθ

f(-u )(-d u ) = J

j

pθ

—f(u) (—du) = J

ι

pl

f(u) du = - J θ f(u) du

= -3 (b) Let u = —x => du = —dx; x = —1 => u = 1, x = 0 => u = 0 pθ

pθ

pθ

pl

feven => f(—x) = f(x). Then I f(x) dx = I f( -u ) ( - du) = — I f ( u )d u = I f(u) du = 3 J-l Jl «71 0 87. Let u = a - x => du = —dx; x = 0 => u = a, x = a => u = 0 »a

J 0

_ f(x) dx f(x)+f(a-x) -

T+ I 1 +

1 -

pθ

I

J

a

pa

f(a-u) /_ J A _ f(a-u)+f(u) V u u ' -

I J

w dx -I- I * f(a-x)dx _ Γ J o f(x)+f(a-x) + J o f(x)+f(a-x) -

o

f(a-u) du _ f(u)+f(a-u) -

Γ a f(x)+f(a-x) . J o f(x)+f(a-x) ®

pa

I J

o

f(a-x) dx f(x)+f(a-x) _

“

Jo

0X

-

l

*j 0

-

°

-

3

‘

Therefore, 21 = a => I = y 89. Let u = x + c => du = dx;x = a —c => u = a, x = b - c => u = b >b—c i—c

pb

pb

f(x + c) dx = I f(u) du = I f(x) dx «7a «7a

CHAPTER 5 PRACTICE EXERCISES 1. (a) Each time subinterval is of length ∆ t = 0.4 sec. The distance traveled over each subinterval, using the midpoint rule, is ∆ h = j (vi + vi+ ι) ∆ t, where vi is the velocity at the left endpoint and vi+1 the velocity at the right endpoint of the subinterval. We then add ∆ h to the height attained so far at the left endpoint vi to arrive at the height associated with velocity vi+1 at the right endpoint. Using this methodology we build the following table based on the figure in the text: t (sec) V (fps) h(ft)

0 0 0

t (sec) v(fps) h(ft)

6.4 50 643.2

0.4 10 2

0.8 25 9 6.8 37 660.6

1.2 55 25 7.2 25 672

1.6 100 56

2.0 190 114

7.6 12 679.4

2.4 180 188

2.8 165 257

3.2 150 320

3.6 140 378

4.0 130 432

4.4 115 481

4.8 105 525

5.2 90 564

5.6 76 592

6.0 65 620.2

8.0 0 681.8

NOTE: Your table values may vary slightly from ours depending on the v-values you read from the graph. Remember that some shifting of the graph occurs in the printing process. The total height attained is about 680 ft.

Chapter 5 Practice Exercises

169

(b) The graph is based on the table in part (a).

1 ∣ θ

10

Σ

3. (a)

¾ = 5η (-2 ) = - i1

I = 5 Σ

k=l

k=l 10

10

10

k=l

k=l

£

(d)

(F M =

£

k= 1

k=l

bk - 3

10

∑

⅜ = 25 - 3(-2 ) = 31

k=l

10

k= 1

F ∑

f

k=1

10

∑ (bt - 3 ¾ ) =

k=l

∑ (ak + bk - l ) = ∑ ak + ∑ bk -

(c)

JO

Φ) ∑

1 = - 2 ÷ 2 5 -(1)(10) = 1 3

k= 1

bk = f (1 0 )-2 5 = 0

k=1

5. Let u = 2 x - 1 4 du = 2 dx => ∣du = dx; x = 1 => u = 1, x = 5 => u = 9 J i (2x - I)" 1/ 2 dx = J u~ 1∕ 2 ( ∣du) = [u1∕ 2 ] θ = 3 - 1 = 2 7. Let u = ∣ => 2 du = dx; x = - π => u = - ^ , x = 0 => u = 0 •0

pθ

∕ 2(c ° s u)(2 du) = [2 sin u]θ π ∕ 2 = 2 sin 0 - 2 sin ( - f ) = 2(0 - ( - l ) ) = 2

cos (∣) dx = J p2

9. (a) J

p2

2f(x)dx

= ∣J

p-2

(c) J

2 3 f(x )d x =

p5

∣(12) = 4

(b) J

p5

g(x) dx = - J

(e) / _ 2 (

w Vw)

dx

j g(x)

= 5∕ ,

Area = J j

j f(x)

p2

dx - J

p5

dx = - 2 2

f (χ

(d) J

) d x + I ^ . 2^

11. x2 - 4x + 3 = 0 => (x - 3)(x - 1 ) - 0 4 (x2

p5

f(x) dx = J

j

- 4 x + 3) dx - J

(x2

x = 3 orx = l;

ι

= [(⅛ - 2(1)2 + 3(1)) - o ] - [ ( f - 2(3)2 + 3(3)) - (⅛ - 2(1)2 + 3(1))]

= (l + ι)-[θ -(i + ι)] = f 13. 5 - 5x2∕ 3 = 0 => 1 —x2∕ 3 = 0 => x = ± 1; Area = J* ι (5 —5X2 ^3 ) dx - J

(5 —5X2 ^3 ) dx

= [5x —3X5^3 ] ^ 1 - [5x —3X5∕ 3 ] J = [(5(1) - 3(1)5∕ 3 ) - (5(-1) - 3 ( - 1)5∕ 3 )] - [(5(8) - 3(8)5∕ 3 ) - (5(1) - 3(1)5 ∕ 3 )] = [2 - (-2)] - [(40 - 96) - 2] = 62

g(x)) dx = - π J

dx = ∣(6) + ∣(2) = f

—4x + 3) dx

= [y - 2x2 + 3x] θ - [ γ - 2x2 + 3xj

2 - ⅛ )* = [ ⅛ ]' = (h D -G

+ 1) =

1

17. f(x) = (1 - √ x ) 3 . g(x) = 0, a - 0, b = I => A - _£ [f ≈ ', - y ) ⅛ = ∣ ⅞

+

j '

1

= ⅛

≠ the total area is Ai ÷ A2 = f

3 3 ,y = x2 + f

∣d t ^

⅛ = 2 x + ∣ =≠> ^ = 2 - ⅛ ι y ( l ) = l + ∫ ' l d t = l a n d y ' ( l ) = 2 + l = 3

⅛ dt - 3 =+ ⅛ = ≡ 5 * dX X

; x = 5 =+ y = I ⅛ dt - 3 = - 3 U5 *

37. Let u = cos x ≠> du = —sin x dx =+ —du = sin x dx J 2(cos x)- 1 ∕ 2 sin x dx = J ' 2u - 1 ∕ 2 ( - du) = - 2 J* u^ 1^2 du = —2 ( a τ^) + C = - 4 u 1^2 + C = —4(cos x)1∕ 2 + C 39. Letu = 2 0 + 1 4- du = 2d0 =+

du = d0

∫ [20 + 1 + 2 cos(20 + 1)] d0 = ∫ ( u + 2 cos u) (∣du) = ⅛ + sin u + Ci = ≡ ≤

+ sin(20 + 1) + C i

= 02 + θ + sin (20 ÷ 1) + C, where C = Ci + ∣is still an arbitrary constant 41. ∕ ( t - 2 ) (, + ≡J,l, = ∫ ( t < - J )

i

= ∫ ( 1∙ - 4 t - = ) 4 4 - 4 ( g ) + c 4 + l + C

43. Let u = 2t3∕ 2 => du = 3 ι ∕t d t => ∣du = ᅟ ∕ t dt J ' ^ ts in (2t3 ∕ 2 )dt = ∣J ^ in u du = - ∣ cos u + C = - ∣ cos(2t3 ∕ 2 ) + C

45. J " l (3x2 - 4x + 7) dx = [x3 - 2x2 + 7x] L1 = [13 - 2(1)2 + 7(1)] - [ ( - l) 3 - 2 ( - l ) 2 + 7 (-l)] = 6 - (-1 0 ) = 16

J

2 3∙

dv = J

ι

4 V- 2 dv = [-4 v - 1]o≈ = (⅛ ) - ( ^ ) = 2

171

172

Chapter 5 Integration

X⅛=Γ ⅛=Γ t^3'2dι=[-2,-,z≈]M - ^ = ■ 51. Let u = 2x + 1 => du = 2 dx => 18 du = 36 dx; x = 0 => u = 1, x = 1 ≠- u = 3 J '1 p3 Γ τ 3

, ≡⅛=X>8-~3λ =M , =⅛]i=® - (≠>=8

—∣du = x^ 1∕ 3 dx; x = I => u = l - (⅛) 2 ^3 = I ,

53. Let u = 1 —x2∕ 3 => du = —? x^ 1∕ 3 dx 4 x = 1 =► u = 1 - 12 /3 = 0 (1

_ ”

_ , η

∙ '-

d x

-

£

( _

i λ

i

) -

[ ( _ i) ( ⅞

) ∣

[ _ ∣u η

_

-

=

( - 1 ) ( 2 ) s '≈

27 √ 3 160

55. Letu = 5r ≠> du = 5dr => ∣du = dr; r = 0 => u = 0, r = π => u = 5π ⅜π

J

_ sin≈5r Λ

57.

,-

p5π( ⅛

∖

u) (∣du) =

1 |. - « 2 . |“

=

(. -

⅛

)

-

(0 - ⅞

)

=

≈

sec2 0 d0 = [tan 0]^ 3 = tan ∣—tan 0 = ι ∕3

J

59. Let u = !O ≠∙ du = ∣ Odx ≠’> 6 du = dx; x =O’π => u = ?, x = 23π => u = ? J* 3 π

∕* π ∕2

cot2 | dx = I

p π ∕2

7Γ∕6

7Γ

(esc2 u —1) du = [6(—cotu —u)]^θ = 6 ( - c o t ∣- ∣) - 6 (—cot f —∣)

6 cot2 u du = 6 I v

7Γ∕6

'

= 6 √ 3 - 2π

61.

J

63.

Let u = sin x => du = cos x dx; x = 0 => u = 0, x = ? => u = l Jθ

sec

x tan x dx = [sec x]θπ ∕ 3 = sec 0 - sec ( - j ) = 1 - 2 = - 1

5(sin x)3 ∕ 2 cos x dx = Jθ 5u3∕ 2 du = [5 (∣) u5∕ 2 ] 0 = [2u5≠2 ] 0 = 2(1)5∕ 2- 2(0)5 ∕ 2 = 2

65. Let u = sin 3x => du = 3 cos 3x dx => ∣du = cos 3x dx; x = —∣ => u = sin ( - y ) = 1, x = ∣ ≠- u = sin ( y ) = -1 p-l p-1 J ^ 2 15 sin4 3xcos 3x dx = J ι 15u4 ( ∣du) = J i 5u4 du = [u5 ] 1 = ( - l ) 5 - (1)5 = - 2

pπ∕2

67. Let u = 1 + 3 sin2 x ≠> du = 6 sin x cos x dx => ∣du = 3 sin x cos x dx; x = 0 ≠- u = 1, x = f ≠ u = 1 + 3 sin2 ∣= 4

Γ ⅛¾≡ λ =Γ ⅛(^du)=Γ 1"~ι'≈d"=[⅛(τ)K=i"1'2i;=41/2- 1,'2=1 69. Let u = sec 0 => du = sec 0 tan 0 d0; 0 = 0 => u = sec 0 = 1, 0 = j ≠ u = sec ∣= 2 p π∕3

'π ∕3

J

o

-^ L ≤ / 2 sec 0

d0 = I

J 0

∩ π∕^

seeg tang ^ sec 0 ᅟ ∕2 sec 0

_

I

sec ⅜ tan ⅛

J o ᅟ ∕2 (sec 0)3 ∕ 2

^

=

I _ 1 J 1 χ ∕2 u3 ∕ 2

"⅛[Slι"l- ⅛ lι"^⅛ ^(^⅛ )^^ 1

d u

_

p2 1 I u - 3 ∕2 √2 J 1

d u

Chapter 5 Additional and Advanced Exercises 71. (a) av(f) =

τ

⅛

(b) av(f) =

k

_(_ k) J

j

173

(m χ + b) dx = ∣[ ⅛ + bx] ^ = ∣[ ( ≡ ^ + b (l)) - ( ≡ ⅛ ^ + b ( - l ) ) ] = ∣(2b) = b

£

(mx + b) dx = ⅛ [ ⅛ + bxj

k

k

= ⅛ [(s r

+

b (k )) -

(s ⅞ ^ +

b

( ~ k ))]

= ⅛ (2bk) = b 73. f av - ⅛

/

λ ∕a x f'(x )

dx = ⅛ [f(x)la = ⅛ [f(b) - f(a)] =

f(b ’_^ a)

so the average value of f' over [a, b] is the

slope of the secant line joining the points (a, f(a)) and (b, f(b)), which is the average rate of change of f over [a, b]. 75. We want to evaluate * 365

J

o

n365

f(x) v z dx =

365 vλ

Q

Γ

^

f*365

sin 365 ⅛ (' x - 1 0 1 )× dx +

( 0

Notice that the period of y = sin ∣^ (x - 101) is ⅜ = 365 and that we are integrating this function over an iterval of 365 *365

J

Q

Γ

I

n365

sin ⅛⅛(x — 101)× d x ÷ ⅛ doo ' 3oo

1 Q

dx is ^

∂OO

∙0+ ^

3θδ

∙ 365 = 25.

77. ^ = ᅟ ∕2 + cos3 x 79 ⅛ ,

*∙

dx

1 ∣_ f

=

dx I

J

dt ∣= —

3÷FU U

1

3+√

81. Yes. The function f, being differentiable on [a, b], is then continuous on [a, b]. The Fundamental Theorem of Calculus says that every continuous function on [a, b] is the derivative of a function on [a, b]. 83. y = J

px ι

______

√ b ^ d t = - J

px ι

_______

√ Γ + t 2 dt ^

Γ

⅛ = ⅛ - J

px 1

_______

‘

√ Γ + t 2 dt

85. We estimate the area A using midpoints of the vertical intervals, and we will estimate the width of the parking lot on each interval by averaging the widths at top and bottom. This gives the estimate A ≈ 15∙ ∕ 0 ÷ 3 6 ∣ 36 + 54 ∣ 54 + 51 ∣ 51 + 49.5 ∣ 49.5 + 54 ^ 54 + 64.4 ∣ 64.4 + 67.5 ∣ 67.5 + 4 2 λ 2

2

2

2

2

22

2

/

A ≈ 5961 ft2 . The cost is Area ∙ ($2.10/ft2 ) ≈ (5961 ft2 ) ($2.10/ft2 ) = $12,518.10 4

the job cannot be done for $11,000.

p30

87. av(I) = ⅛ J

o

(1200 - 40t) dt = ⅛ [1200t - 20t2 ]θ = ⅛ [((1200(30) - 20(30)2 ) - (1200(0) - 20(0)2 )]

= ⅛ (18,000) = 600; Average Daily Holding Cost = (600)($0.03) = $18 89. av(I) = ⅛ £ ( 4 5 0 - 0

dt = ⅛ [450t - f ] * = ⅛ [450(30) - ^

- θ] = 300; Average Daily Holding Cost

= (3OO)($O.O2) = $6 CHAPTER 5 ADDITIONAL AND ADVANCED EXERCISES pi 1. (a) Yes, because Jθ f(x) dx = ∣Jθ 7f(x) dx = ∣(7) = 1 (b) No. For example, J^ 8x dx = [4x2 ] θ = 4, but J^ √ z 8x dx = [∑  ∕ 2 [ y ^ ] = ⅛ ≠ √ 4

= ^

( l 3 ∕ 2 - O3 ∕ 2 )

174

Chapter 5 Integration px

px

px

3. y = 1 Jθ f(t) sin a(x —t) dt = ∣Jθ f(t) sin ax cos at dt —∣Jθ f(t) cos ax sin at dt =

^

cos at

Jo ®

+ 5τ

i

^t ~ 2⅛3 i Jo ®

cos at

^⅛ Jo ®

dt) + s in

= cos ax J^ f(t) cos at dt + ≡

s

ax

s in a t

dt ≠> ⅛ =

J^ f(0 sin at dt -

co s ax

s≡

^

( J*o f (t) c o s

at

dt)

J^ f(0 sin at dt)

(f(x) cos ax) + sin ax J^ f(t) sin at dt -

sγ

5

(f(x) sin ax)

pχ

px

=> ⅛ = cos ax Jθ f(t) cos at dt + sin ax Jθ f(t) sin at dt. Next, ^ = - a sin ax J^ f(t) cos at dt + (cos ax) ^

J^ f(t) cos at dt^ + a cos ax J^ f(t) sin at dt

÷ (sin ax) I ^ J o f(t) sin at dt j = —a sin ax Jθ f(t) cos at dt + (cos ax)f(x) cos ax px

px

px

+ a cos ax Jθ f(t) sin at dt ÷ (sin ax)f(x) sin ax = —a sin ax Jθ f(t) cos at dt + a cos ax Jθ f(t) sin at dt + f(x). pχ

px

Therefore, y" + a2 y = a cos ax Jθ f(t) sin at dt —a sin ax Jθ f(t) cos at dt + f(x) + a2

f s∏tax f ^ ∖

a

cos at

j t _ c^ax f f ^ s jn a

JO

Jo

>x2

j t ∣_ ^ χ j No t e also that y'(0) = y(0) = 0.

pχ2

⅛ J 0 f(t) dt = cos πx —7rx sin πx ≠ f (x2 ) (2x) = cos πx —τrx sin πx

f(t) dt = x cos πx ≠

5. (a)

f(χ 2 ) = ≡ (b)

at

/

^ ≡

.

Thus, x = 2 ≠- f(4) =

cos 2π —2π sin 2π _ 4 ~

1 4

t2 dt = [j ] θ = ∣(f(x))3 ≠ , ∣(f(x))3 = x cos πx ≠> (f(x))3 = 3x cos πx => f(x) = 3ᅟ ∕3 x c o s π x

J^

=> f(4) = 3√3(4) cos 4π = 3√ 1 2 7. J*1bf(x) dx = √ b 2 + T - √ 2 =► f(b) = ⅛ ∫ ⅛ ) dx = i (b2

9. ⅛ = 3 x 2 + 2 = > y = J* (3X2 + 2) dx =

X3

p3

pθ

J -8

p3

11. I f(x) dx = I x2∕ 3 dx + I - 4 dx =

U

x 5 /3

] -8 ÷

t

Jθ

"

4 x

lθ

= (0 - ∣(-8 ) 5 ∕ 3 ) + (-4(3) - 0) = f - 12 — 36 — 5

13. J^ g(t) dt = J^ td t + J^ sin πt dt =

⅛

=

⅛ -

2

+

E

0

π

H ) +

c o s π t

[- ;

H

c o s 2 π

-

( - ⅛

c o s

π

)]

l) " 1 /2 (2b) = ⅛

≠ f(x) = ^

π

+ 2X + C. Then (1, - 1 ) on the curve ≠> 13 + 2(1) + C = - 1 ≠> C = - 4

≠∙ y = x3 + 2x - 4

J -8

+

Chapter 5 Additional and Advanced Exercises

»• « =£( 4 ^ f,w =1(S - (f) (⅛(l)) =} - ^(- ⅜)=i +i =I 21. 8(y) = ∫ ^ s i n t2 dt => g'(y) = (sin (2 √ y ) 2 ) (⅛ ( 2 ^ ) ) - (sin ( √ y ) 2 ) (⅛ ( √ y ) ) = ^

- ^

23. Let f(x) = x5 on [0 ,1]. Partition [0,1] into n subintervals with ∆x = L ^ = £. Then ^, ^ ,... , ^ are the oo / ∙ $ right-hand endpoints of the subintervals. Since f is increasing on [0,1], U = ∑ ( ∏) (⅛) *s die u PPe r s u m f° r f(x) = x5 on [0,1] => = ∙ P

dx

lim ∑ (∏) (∏)

=

lim

∏ ( ^ ) 5 + (^ ) 5 + ∙ ∙ ∙ + (S) 5 ] =

lim

Γ15+ 25 ⅛∙ ∙ + "5 1

= [⅛] « = ⅛

25. Let y = f(x) on [0 ,1]. Partition [0,1] into n subintervals with ∆x = ^ p = ∣. Then ^, - , . . . , ~ are the right-hand endpoints of the subintervals. Since f is continuous on [0,1], ^ f Q ) ( ⅛) is a Riemann sum of y = f(x) on [0,1] =>

n

lim o ∑ f ( 0 (1) = j=ι x '

n

lim o ⅛[f (⅛) + f (1) + -

+ f (≡)] = J

o

*00 dx

27. (a) Let the polygon be inscribed in a circle of radius r. If we draw a radius from the center of the circle (and the polygon) to each vertex of the polygon, we have n isosceles triangles formed (the equal sides are equal to r, the radius of the circle) and a vertex angle of ⅛ where 0n = ^ . The area of each triangle is An = ∣r2 sin 0n => the area of the polygon is A = nAn = ~ sin 0n = 1y sin - r . (v b)z

lim

n →∞

A=

lim

~ sin ¾e =

n →∞ 2

n

lim

n →∞

⅛ sin ⅞ = 2π

n

lim n

(πr2 7)

→∞ v

(⅛)

2) = (πr v 7

lim

2π∕n → 0

(? )

= τrr2

29. (a) g(l) = ∫ 'f ( t ) dt = 0 p3 (b) g(3) = J ι f(t)dt = -1 ( 2 ) ( 1 ) = - 1 (c) g ( - l ) = ∫

f ( t ) d t = - ∫ ι f(t)dt = - ∣(π2 2 ) = - π

(d) g'(x) = f(x) = 0 => x = - 3 ,1 ,3 and the sign chart for g'(x) = f(x) is

∣+ + + ∣------ ∣+ + + . So g has a -3 1 3

relative maximum at x = 1. p -ι (e) g, ( - 1) = f(—1) = 2 is the slope and g(—1) = J f(t) dt = - π , by (c). Thus the equation is y + τr = 2(x + 1) y = 2x + 2 —π.

175

176

Chapter 5 Integration

(f) g"(x) = f'(x) = 0 at x = - 1 and g"(x) = f'(x) is negative on (—3, —1) and positive on ( - 1 , 1) so there is an inflection point for g at x = - 1 . We notice that g"(x) = f , (x) < 0 for x on (—1, 2) and g"(x) = f'(x) > 0 for x on (2, 4), even though g"(2) does not exist, g has a tangent line at x = 2, so there is an inflection point at x = 2. (g) g is continuous on [—3, 4] and so it attains its absolute maximum and minimum values on this interval. We saw in (d) that g, (x) = 0 => x = -3 ,1 ,3 . We have that g (-3 ) = ∫ ^ 3 f(t) dt = - / /( I ) dt = 4 = -2 π g(l) = ∫ ι 'f(t)d t = o 3f ( t ) d

g(3) = ∫

t= -l

»4

J ι

f(t)d t= -l + j ∙ l ∙ l = - i

Thus, the absolute minimum is —2π and the absolute maximum is 0. Thus, the range is [—2π, 0].

CHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS 6.1 VOLUMES BY SLICING AND ROTATION ABOUT AN AXIS

1. (a)

A = π(radius)2 and radius = √ 1 —χ 2 => A(x) = π (1 —x2 )

(b) A = width ∙ height, width = height = 2 √ 1 —x2 => A(x) = 4(1 —x2 ) (c) A = (side)2 and diagonal = y z2(side) => A = {¢222221£; diagonal = 2χ ∕ 1 —x2 => A(x) = 2 (1 —x2 ) (d) A = δ ^ (side)2 and side = 2 χ ∕ 1 - x2 => A(x) = A (1 ~

χ2)

3. A(x) = {⅛ 2!⅛ = β ≤ ∑ tv ⅛ L — 2χ (see Exercise 1c); a = 0, b = 4; V = J^ A(x) dx = J^ 2x dx = [x2 ] o = 16

5. A(x) = (edge)2 = [ √ T = ^ - ( - √ T ^ ^

v

= £ A«

dx = Γ 4(1 - χ2)dx = 4 [x l

= ^2χ ∕ 1 - x2 ) = 4(1 - x2 ) ; a = —1, b = 1;

τ] ^1 = 8 (1 - I) = τ

7. (a) STEP 1) A(x) = ∣(side) - (side) - (sin 5) = ∣∙ (2 ᅟ ∕sin x) ∙ ( 2 ᅟ ∕s in x) (sin ∣) = √ z3 sin x STEP 2) a = 0 ,b = π STEP 3) V = £ A (X) dx = √ 3 £ s i n x dx = [ - √ 3 cos x] θ = √3(1 + 1) = 2 √ 3 (b) STEP 1) A(x) = (side)2 = (2 √ s⅛ x) (2 ᅟ ∕s in x) = 4 sin x STEP 2) a = 0 ,b = π STEP 3) V = ∫ bA(x)dx = / J 4 sin x dx = [-4 cos x] θ = 8

9. A(y) = 5 (diameter)2 = ≡ ( √ 5 y 2 - θ j = ⅞ y 4 i

.

c = 0, d = 2; V = ∫ ct*A(y) dy = ∫θ ⅞ y4 dy 2

) ι] o = ≡(2 5e -O ) = 8π diameter o f c irc le

1— *— J— $— nr 11. (a) It follows from Cavalieri’s Principle that the volume of a column is the same as the volume of a right prism with a square base of side length s and altitude h. Thus, STEP 1) A(x) = (side length)2 = s2 ; STEP 2) a = 0, b = h; STEP 3) V = £ A (X) dx = / V dx = s2 h (b) From Cavalieri’s Principle we conclude that the volume of the column is the same as the volume of the prism described above, regardless of the number of turns => V = s2 h 13. R(x) = y = 1 - ∣ ≠> V = ∫

[R(x)]2 dx = τr∫ o (1 - f ) 2 dx = π ∫ o (1 - x + ⅛ ) dx - π [x - ⅛ + ⅛] θ

178

Chapter 6 Applications of Definite Integrals

15. R(x) = tan ( 5 y) ; u = ≡ y => du = ∣dy =≠- 4 du = π dy; y = 0 => u = 0, y = 1 => u = | ; V = J^ π[R(y)]2 dy = π j^ [tan ( f y)] 2 dy = 4 f j tan2 u du = 4 ,£

(—1 + sec2 u) du = 4[—u + tan u]θ^4

= 4 ( -^ + 1 -0 ) = 4 - π 17. R(x) = x2 => V = £ , π[R(x)]2 dx = π J^ (x2 )2 dx = ^ 1 ^

= 4 ^ 0

19. R(x) = √ 9 - x

2

=

^

=> V = ∫ ∖ [ R ( X)]2 dx = π / ^ (9 - x2 ) dx

= π ∣ 9x - ⅛] ’ = 2π [9(3) - f ] = 2 ∙ π ∙ 18 - 36π

----------

p π ∕2

21. R(x) = √ c o sx => V = Jθ

n π ∕2

π[R(x)]2 dx = π J o cosxdx

= π [sin x] [ ∕ 2 = π(l —0) = π

√ -2 —sec x tan x

23.

Γ√4

=> V = Jθ

n

π[R(x)]2 dx

= π

(ᅟ ∕2 —sec x tan x) dx

= π ∫o

(2 —2 5 /2 secxtanx + sec2 xtan 2 x^ dx I— p π ∕4

= π

p π ∕4

'

2 √ 2 J o s e c x ta n x d x + Jθ (tan x)2 sec2 x dx

= π ( j2 x ] [ ∕ 4 -2 √ 2 [s e c x ]i

sec z tan z -------u. x 7

= π [(≡ -0 )-2 √ 2 (√ 2 -

25. R(y) = √ 5 ∙ y2 => V = J

j

π[R(y)]2 dy = π J

= π[y 5 ]L1 = π [ l - ( - l ) ] = 2π

4 1 5y

dy

Section 6.1 Volumes by Slicing and Rotation About an Axis *π∕2

J

o

π[R(y)]2 dy

= ι r j β 2 sin 2y dy = π [—cos 2y] J^2 = π [ l - ( - l ) ] = 2π

2» R(y> = ⅛ = 4π

=> v = ∫ , ⅛

⅛ = ⅛ £

[⅛J 0= 4πf^ * ^^(- 1 ) l

5

⅛

⅛

= 3π

31. For the sketch given, a = —j , b = j ; R(x) = 1, r(x) = ^/cosx; V = J a π ([R(x)]2 —[r(x)]2 ) dx =

c o s x ^ ^x

/.^/2 π ^ “

= 2π J"θ7 (1 -

c o s x ^ c^x =

^π fχ “ s *n χ 3o^2 = 2π Q — 1) = π 2 —2π

π ([R(x)]2 - [r(x)]2 ) dx

33. r(x) = x and R(x) = 1 => V =

=X' π (1 - χ2) dx =π [x - τ] θ=π [(1 - i) - °] = τ

35. r(x) = x2 ÷ 1 and R(x) = x ÷ 3

^

v

=Γ , π ([R W12 - Mχ )l2) dx

=7rZι [(X + 3)2 - (χ2 + 1)2] dx = 7r J

ι

[(x2 + 6x + 9) — (x4 + 2x2 + 1)] dx

= 7r J* ι (—x4 - x2 + 6x + 8) dx __ _ [

X5

X3

I

6x 2

I oγ l

- π [~ τ ~ τ +τ +8x∣-1 = 4 ( 4 - h

f

+ 16) - ( l

+

l

+

∣- 8 ) ] = π ( - ⅛ - 3 + 2 8 - 3 + 8) . π ( ≡ )

37. r(x) = sec x and R(x) = √ z2 pπ∕4 ≠> V = J -π z 4 π([R(x)]2 - [r(x)]2 )dx = π

~

s e ° 2 X) d x

~

π

^

x

~

ta n χ

= π [ ( f - l ) - ( - f + l)]= π (π -2 )

]-√4

^

179

180

Chapter 6 Applications of Definite Integrals

39. r(y) = 1 and R(y) = 1 + y ≠ V = X 'π([R (y)] 2 -[r(y )] 2 )dy =

π

=

π

X K1

+

y>2 - 1] dy = ^ J o (1 + 2y + y2 - 1) dy

X (2 y + y 2 ) d y =

π

[y2 + ⅝] θ =

π

( 1 + 1) = τ

41. R(y) = 2 and r(y) = λ ∕y ≠> v = X 4 ^ ( t R (y)]2 - ir (y)J2 ) dy = π

X (4 - y) dy = π [4y - ^ ] θ = π(16 - 8) = 8π

43. R(y) = 2 and r(y) = 1 + χ ∕y v

^

=

X

π

(tκ (y)i2

-

Wy)i2 ) dy

X ' [4 - ( 1 + √ y ) 2 ] dy

=

π

=

π

J0 1 (4 -

1

=

π

X ' (3 -

2

3y

= 4

-5 y

2

-

√ y -y )d y

√ y -y )d y

3z2-⅛

]θ

= √ 3 - ∣- l ) = √ ⅛ 2 ) ^ ⅞ 45. (a) r(x) = √ x and R(x) = 2 => V = X = π

π

([R(x)l2 - [r(x)]2 ) dx

4x - ^1 X (4 - x) dx = π ∣

= π(16 - 8) = 8π

(b) r(y) = 0 and R(y) = y2 v

^ =

π

=

Jo

7 r (M(y)]2 -

X 2 y4 d y =

7 r[flo

My)i2 ) dy = ^

(c) r(x) = 0 and R(x) = 2 - √ x => V = =

π

X (4

-

4 lA

X (16 ~

16

([R(x)]2 - [r(x)]2 ) dx = ^X" (2 - √ x ) 2 dx

+ x ) dx = π [4x - ⅛ ^ + ^ ] θ = π (16 - y + ^ ) = ^ e

(d) r(y) = 4 - y2 and R(y) = 4 ≠> V = = π

f∖

f∖

([R(y)]2 - [r(y)]2 ) dy = π

f* [16 - ( 4 -

y2 )2 ] dy

+ 8y2 - y4 ) dy = π X (8y2 - y4 ) dy = π [f y3 - ^ ] θ = π ( f - f ) = ^

Section 6.1 Volumes by Slicing and Rotation About an Axis

181

(b) r(x) = 1 and R(x) = 2 - x2 => V = ∫ ' ι π ([R(x)]2 - [r(x)]2 ) dx = τr f ^ [(2 - x2 )2 - 1] dx = π

f ι (4 - 4x2 + x4 — 1) dx = π J* ι (3 - 4x2 + x4 ) dx = π [3x - ∣x3 + y j

= 2π (3 - ∣+ ∣)

= ⅛ (45 - 20 + 3) = ^ (c) r(x) = 1 + x2 and R(x) = 2 ≠> V = f = πf

1 (4

— 1 —2x2 —x 4 ) dx = 7rf

j

ι

π ([R(x)]2- [r(x)]2 ) dx = π f

1

[4 — (1 + x2 )2 j dx

(3 —2x2 - x4 ) dx = π ∣ 3x — j x3 — y ]

= 2π (3 —j —∣)

= ⅛ (45 - 10 - 3) = ⅛ ∕a 2 —y2 and r(y) = b —ᅟ ∕a 2 - y2 49. R(y) = b + ᅟ ≠> v = / . ^ ( ^ ( y ) ] 2 - Wy)i2 ) d y = π J → [θ , + √ ,a2 - y2 ) 2 - (b - √ a 2 —y2 ) 2j dy = π

f

a 4b

∕

a2

—y 2 d y ~ 4 b π f a √ a 2 —y2 dy

= 4bπ ∙ area of semicircle of radius a = 4bπ - -y = 2a2 bπ 2

51. (a) R(y) = √ a 2 —y 2 ≠> V = πj*

(a2 —y2 ) dy = π [a2 y - ^ j

a

= π [a2 h —a3 - —3 a)3 — ( —a3 + y ) j

= π [ a 2 h - ∣(h3 - 3 h 2 a + 3ha2 - a 3 ) - ⅛ ] = π (a 2 h - ⅛ + h2 a - ha2 ) = ⅛

→>

(b) Given ^ = 0.2 m3 ∕sec and a = 5 m, find y ∣ h=4 . From part (a), V(h) = =>

an

= 10τrh “ τrh2 =≠∙ ⅛r at

=

^ah ‘ ^at ≈ 7rh(10 - h)' Φ v at => Φl at I h=4

⅜y^ ^ = 5πh2 — 2y = 44w⅜⅛ π ( I U - 4) = (2Uπ)(o) 12Oπ m∕sec.

53. The cross section of a solid right circular cylinder with a cone removed is a disk with radius R from which a disk of radius h has been removed. Thus its area is A i = πR 2 - πh 2 = π (R2 - h 2 ) . The cross section of the hemisphere is a disk of radius √ R 2 —h2 . Therefore its area is A2 = π ( ∕R 2 - h2 ) = π (R2 —h2 ) . We can see that Ai = A2. The altitudes of both solids are R. Applying Cavalieri’s Principle we find Volume of Hemisphere = (Volume of Cylinder) —(Volume of Cone) = (πR2 ) R — ∣π (R2 ) R = ∣πR 3 . 55. R(y) = √ 2 5 6 - y 2 => V = f

i6 π[R(y)]

2

dy = π f 1J256 - y2 ) dy = τr [256y - ⅛j

= π [(256)(-7) + ⅞ - ((256)(-16) + ^ ) ] = π (⅞ + 256(16 - 7) - ⅛ ) = 1053π cm3 ≈ 3308 cm3

57. (a) R(x) = ∣ c - sin x∣ , so V = π f [R(x)]2 dx = π f g (c - sin x)2 dx = π f o (c2 - 2c sin x + sin2 x) dx = π f g (c2 - 2c sin x + 1= ≡ ⅛ ) dx = π f g (c2 + 5 - 2c sin x - s^⅛ ) dx = π [(c2 + ∣) x + 2c cos x — ^ ] θ = π [(c2 π + ∣—2c —θ) —(0 + 2c —0)] = π (c2 π + ∣- 4 c ). Let V(c) = 7r (c2 π + ∣- 4 c ) . We find the extreme values of V(c): ^ = π(2cπ —4) = 0 => c = ^ is a critical point, and V (¾) = π Q + f -

3)

= π ( j - £) = y - 4; Evaluate V at the endpoints: V(0) = y and

V(l) = π ( ∣π - 4) = y —(4 —π)π. Now we see that the function's absolute minimum value is y —4, taken on at the critical point c = y (See also the accompanying graph.) 2

(b) From the discussion in part (a) we conclude that the function's absolute maximum value is y , taken on at the endpoint c = 0.

Chapter 6 Applications of Definite Integrals

182

(c) The graph of the solid’s volume as a function of c for 0 ≤ c ≤ 1 is given at the right. As c moves away from [0,1] the volume of the solid increases without bound. If we approximate the solid as a set of solid disks, we can see that the radius of a typical disk increases without bounds as c moves away from [0 ,1].

6.2 VOLUME BY CYLINDRICAL SHELLS 1. For the sketch given, a = 0, b = 2; v = X ^ ( X ) ( ≤ ) * = £

2™

(l + j ) *

= 2 .£ (x + τ ) * =

2"

[⅞ + C

= 2π ∙ 3 = 6π 3. For the sketch given, c = 0, d = \/2 ; V = £ 2 ’ ( X ) ( S ) dy = X ^ 2 „ ∙ (y’ )dy = 2 √ ∕ y ≈ dy = 2 . [⅞]

5. For the sketch given, a = 0, b = √ ⅞ v

= f ⅛

("

) ( “

) dx = / / 2 . x ■ ( √ ^ T 1 ) dx;

∣ u = x 2 + 1 => du = 2x dx; x = 0 => u = 1, x = ᅟ ∕3 ≠> u = 4^ → V = π f i u 1∕ 2 du = π [∣u 3 ∕ 2 ] J = y (4 3 ^2 - 1 ) = ( y ) (8 - 1) = y≡ 7. a = 0 ,b = 2; V= ∫ > (X ) ⅛ = X

2πχ2

) dx = £ 2 ™ [x - ( - !)] dx

■1 dx = » £ 3x≈ dx = π [x s ]J = 8 .

9. a = 0 , b = l ; v = £ 2 » ( X ) ( X ) dx = X ' 2 .x ∣ (2 - x) - x≈J dx = 2πJ^ (2x - x2 - x 3 ) dx = 2π [x2 - y - y j = 2 π (l-i-i)= 2 √ ⅛ 3 ) = ^

= ⅞ °

11. a = 0 ,b = 1; V = £ 2 . ( X ) ( X ) dx = £ 2 .x [ √ 5 - (2x - 1)] dx = 2π f (x3 ∕ 2 - 2x2 + x) dx = 2π [∣x5 /2 - ∣x 3 + ∣x 2 l ∩ = 2 π ( l - ∣+ ∣) = 2 π ( I H ^ ) = ⅞

= 2,

=

2π

xf(x)λ = < [ 0, x = 0

/ A A ∫ x ∙ ^x , 0 < x ≤ π 13. (a) xf(x) = < x, x = 0 (

( sin x, 0 < x < rz λ = < xf(x) . ( sin x, x = 0 (b )

v

= Ja

2 π (radius) ( h e ig h t)

π

^

rz x ≠> xf(x) = sin x, 0 < x < π , _ d x

= £

2

™ ∙

f(x ) d x

a n d

x

∙ f (χ ) =

Sin X, 0 ≤ x ≤

7Γ by

part (a)

=> V = 2 π f o sin x dx = 2τr[- cos x]θ = 2 π (- cos π + cos 0) = 4π

(b )

v

2 π

= Γ =

2 4 π

∫

0'

(radius) ( h e ig h t) (y 2 -

2

y3 + y4 )

d

y =

X '

d

y =

2 4 π

2 π

∏ -

[⅞ -

y ) I1 2 ( y 2 -

y 3 )l

⅛ + ⅛] θ =

2 4 π

d

y =

(I -

2 4 π

Γ

V = ∫ ᅟ[R(y)2 - r(y)2 ]dy = ∫ o'π[y - y2 ]dy

=⅛ - C =π∣≡-∙) =l 27. (a) V = f 2π ( X ) ( ⅛ ) dy = J > y ( y - 1) dy

=2π∫ ⅛i - y) ⅛ =2»∣(- (], =2+ (!+)-(1-1)]

________

^ ΓΓ ^

r

= 2π (2 - 2 + ∣) = f (14 - 12 + 3) = ⅜

¢) v =X"2π( X ) ( ≤ ) d,1

=f ι 2πx(2 -

'

x2 - y j x) dx = 2πJ^ (2x - x2 ) dx = 2π ∣

‘

*

= 2π [(4 - ∣) - (1 - ∣)]

= 2 π [ ( V ) √ ψ ) ] = 2 π ( ∣- l ) = ⅞ (=) V = ∫ ι *2π ( “

) ( “ ,) dx = J >

= 2 , [ f x - ∣x≈ + μ η + (d) V = ∫ >

29. (a) V = / >

(⅞ - x) (2 - x) dx = 2 π ∫ ⅛

2 π [(a -S

+

- ⅛ x + x≈) dx

∣) H ≡ - ∣+ l ) ] = 2 π ( j ) = 2 ^

( X ) ( ⅛ ) dy = ∫ , 2x(y - l) two integrals are required (b) Washer: V = V i - V2 Vi = J*a π ([R1(x)]2 —[rι(x)]2 ) dx with Rι(x) = ^ i y and r 1 (x) = 0; a 1 = —2 and b 1 = 0; V2 = X

π

(1R 2W12

-

[r 2(χ )]2 ) dx with R2 (X) = y ⅛

2

and r2 (x) = ^/x; a2 = 0 and b2 = 1

≠> two integrals are required (c) Shell: V = £ 2π ( X ) ( ⅛ ) dy = £ x y ( 1≤ J t ) dy where shell height = y2 - (3y2 - 2) = 2 - 2y2 ; c = 0 and d = 1. Only one integral is required. It is, therefore preferable to use the shell method. However, whichever method you use, you will get V = π. 6.

3 LENGTHS OF PLANE CURVES

1. ⅛ = - l a n d ⅛ = 3 ^

^ (⅜ )2 + ⅛ )

2

≈ √ ( - l ) 2 + (3)2 = √ 1 0

≠> Length = ∫ ' 2 β √ i o dt = √ i o [t] L2 n = √ i δ - ( - j √ i δ ) = φ

3. f = 3t2 and ⅛ = 3t => ψ (⅛ ) 2 + ( ⅛ ∕ = ^∕(3t 2 )2 + (3t)2 = √ 9 t 4 + 9t2 = 3 t √ ^ ∏

(since t ≥ 0 on [θ, √ 3 ∣ )

u = t2 + 1 => ∣du = 3t dt; t = 0 => u = 1, t = ᅟ ∕3 => u = 4] ≠> Length = J ^ S t ^ / t ^ M dt; ∣ ^+ ∕

1

2 “ 1/2

du =

[u 3 /2 l 1 = (8 - 1) = 7

186

Chapter 6 Applications of Definite Integrals

5. ∣ = (2t + 3)V2 and⅛ = l + t => J (⅛ ) 2 + ( ⅛ ) 2 = √ (2 t + 3) + (1 + t)2 = √ t 2 + 4t + 4 = ∣ t + 2∣= t + 2 '^

Γ2

J (t + 2) dt = ^ o

13 l j

⅛ ∙ dx

=

χ A

l ∕3 _ 1 4

L

=

≈

r y

χ

χ λ

⅛

- l ∕3

+

/ ⅛ V ^⅛ ^

_

χ X

+ +

1 ≤ 16

dx

d χ

⅞

2∕3 _ 1 2

1

+ 2tj θ = y

= ^ ∖ J(x1∕ 3 + | x- 1 ∕ 3 ) 2 dx = J ι (x1∕ 3 + ∣x- 1 ∕ 3 ) dx =

[2 √ ∕ 3

+

3

χ

2 ∕3 ] 8

=

∣[ 2 χ 4∕3

ψ

χ

2 ∕3 ] «

= I [(2 - 24 + 22 ) - (2 + 1)] = ∣(32 + 4 - 3) = ⅞ 15. ∣ = √ sec 4 y - 1 => ^ ) L =

= sec4 y - 1

£ / < √ l + (sec4 y - l ) dy = f ^ s e c 2 y dy

= [tany]X 474 = 1 - (-1 ) = 2

Section 6.3 Lengths of Plane Curves 17. (a) S = 2 1 ≠

( 4 ) '= ⅛ ≈

(b)

= J . 1∖ ∕H ^ < i χ (c) L ≈ 6 .1 3

(^ )∖

2

= cos2 y

=> L = J^ ^/1 + cos2 y dy (c) L ≈ 3 .8 2

(⅛ ∖

2

= ( y + D2

≠ L = f , √ l + ( y + l ) 2 dy (c) L ≈ 9 .2 9

23∙

(a > ⅛ = ta n x => (⅛ ) = ta n 2 x ==. L = f

√ l + t an≈ x dx = f ⅛

⅛

^

*

o COSX = JIo sec x dx (c) L ≈O ∙55

25.

√2x = f 0 J i + (* )

dt, x ≥ 0 ≠

v^2 = y 1 + ( ^

=> ^ = ± 1 => y = f(x) = ± x + C where C is any

real number. 27. (a) ( ⅛ ) correspondes to ⅛ here, so take ^ as ^ . Then y = ^

+ C and since (1,1) lies on the curve, C = 0.

So y = ^ x from (1,1) to (4,2). (b) Only one. We know the derivative of the function and the value of the function at one value of x.

187

Chapter 6 Applications of Definite Integrals

188

29. (a) ⅛ = - 2 sin 2t and ⅛ = 2 cos 2t => ψ ( ⅛ ) 2 + ( ^ / = √ ( - 2 sin 2t)2 + (2 cos 2t)2 = 2 => Length = _£ Z 2 dt = [2t] θ^2 = π (b) ^ = π cos πt and ^ = —π sin πt = ^ ( ^ ) 2 + ^ ≠> Length = £ ^ π dt = [πt] ^

= χ ∕( π cos τrt)2 + (—π sin πt) 2 = π

= π

6.4 MOMENTS AND CENTERS OF MASS 1. Because the children are balanced, the moment of the system about the origin must be equal to zero: 5 ∙ 80 = x ∙ 100 => x = 4 ft, the distance of the 100-lb child from the fulcrum. 3. The center of mass of each rod is in its center (see Example 1). The rod system is equivalent to two point masses located at the centers of the rods at coordinates (^ ,θ ) and (θ, ^). Therefore x = =

i

P ⅛mi +ι∏2

= ⅛ m+m

yT

= 4⅛ andy∙, = ⅛ m =

tΓ

mi ÷m2

= ⅛ m+m

= 4⅛

(⅛.⅛) ∖4 ’ 4 / is the center of

mass location. 5. M o = f ∖

7

∙

^

= ^

+ 1 )^ = 1

0

_ α ∣9 _ 2 ~

*

⅛

6

= P

2

∙ 4 dx = [4 ⅛]

-

ya -_

2

( ι + ⅛

= 4 ∙ ∣= 8; M = / ^ 4 dx = [4 ⅛ = 4 ∙ 2 = 8 ≠> x = $ = 1

^

+

^

= [⅛ + v ] θ = ( l + ¥ ) = ^

1⅜ _ (f) _ 15 _ 5 M

) *

-

-

= f t +

9

-

3

^ ι >

=

⅛

+ f l > ( * + S - ( ⅛

+ D = ⅞ + ⅛ =

M = J*1 (1 + x ' 1 /2 ) dx = [x + 2x 1∕ 2 ] 4 = (4 + 4 ) - ( 1 + 2 ) = 5 ≠> χ = ⅛ = ^ 11∙

Mo = ∫

o

1χ ( 2

-x )d x + ∫ »1

J o

+ 1 ) ^ = [x + ⅛ ] θ

= ^ 0

2χ ∙ x

d x = ∫ o'( 2 x - x 2 )d x + ∕ Γ

Λ∙2

o il

(2 -x )d x + ∫ x d x = [ 2 x - ⅜ ] o

2χ 2

dx= [ f - f ] ^ +

Γ o l2

+

[⅜ ]ι = ( 2 - i )

+

s

F

= ^

= 2

[ ( ] ' = 0 ~ I ) + ( I ~ 1)

(f-i)= 3

≠> x = ⅛ = l

13. Since the plate is symmetric about the y-axis and its density is constant, the distribution of mass is symmetric about the y-axis and the center of mass lies on the y-axis. This means that x = 0. It remains to find y = ^ . We model the distribution of mass with vertical strips. The typical strip has center of mass: (3c , y z ) = f x, Q ^ 1 , length: 4 —x2 , width: dx, area: dA = (4 —x 2 ) dx, mass: dm = 2x —x 2 = 0 => x(2 - x) = 0 => x = 0 or x = 2. The typical vertical strip has center of mass: (x ,y ) = (x, = ^x, - y ) , length: (x - x2 ) - ( - x ) = 2x - x 2 , width: dx, area: dA = (2x - x2 ) dx, mass: dm = £ dA = 6 (2x - x2 ) dx. The moment of the strip about the x-axis is yz dm = ( - 7 ) ^ (2x —x 2 ) dx; about the y-axis it is zχ dm = x ∙ δ (2x —x2 ) dx. Thus, M x = J zy dm

= - £ (1χ2) (2χ - χ2) d χ = - f ∫ 02(2χ3 - χ4) d χ = - 2⅛ - ⅛] θ= - 1(23 - ⅜) = -1 ∙ 23 (ι - 1) = - f ; M y = J x dm = £ x - 5 (2x - x2 ) dx = 6∫θ 2 (2x2 - x3 ) = 6 [j x3 - ⅛] θ = y)

=

( ^ > ⅛ ) *s die center of mass.

19. Applying the symmetry argument analogous to the one used in Exercise 13, we find x = 0. The typical vertical strip has center of mass: (x ,y , ) = (x, ψ ) , length: cos x, width: dx, area: dA = cos x dx, mass: dm = (x,y) = (θ, f ) is the center of mass.

190

Chapter 6 Applications of Definite Integrals

21. Since the plate is symmetric about the line x = 1 and its density is constant, the distribution of mass is symmetric about this line and the center of mass lies on it. This means that x = 1. The typical vertical strip has center of mass:

length: (2x —x 2 ) — (2x2 —4x) = —3x2 ÷ 6x = 3 (2x ~ x2 ) , width: dx, area: dA = 3 (2x —x2 ) dx, mass: dm = 5 dA = 3 i (2x —x 2 ) dx. The moment about the x-axis is y dm = ∣6 (x 2 - 2x) (2x - x 2 ) dx = - 1 6 (x2 - 2x)2 dx 2

r

= — ∣δ (x4 —4x 3 + 4 X2 ) dx. Thus, M x = J y, dm = - Jθ 1 6 (x4 - 4x 3 + 4x 2 ) dx = - ∣δ J y - x 4 + ∣x 3 J θ

= -μ (⅞ -2 4 +S-2, ) = - i i ∙ 2 ' ( h l + ^ - b ' 2 , m = ∫ ,⅛ P < - > ') ∣ l∙ = 3 > [ l '- τ ] ' = 3 i ( 4 - l ) =


Therefore, y = ⅛ = ( - ⅝ ) (⅜ ) = - 5

=> (x,y) = (1, - ∣) is the center of mass. 23. Since the plate is symmetric about the line x = y and its density is constant, the distribution of mass is symmetric about this line. This means that x = y. The typical vertical strip has center of mass: (x , yz ) = (x, H ^ Ξ Ξ ^ , length: 3 — ᅟ∕9 —x2 , width: dx, area: dA = ^3 — ᅟ ∕ 9 - x2 ) dx, mass: dm = δ dA = δ ^3 - √ z9 - x2 ) dx. The moment about the x-axis is y dm = 6 ^ ± - ^ - ¾ ^ ^ ^

| [9 _ (9 _ χ 2)] d x = ^ dx. Thus, M x = £

dx =

⅛ dx = ∣[x3 ]≡ = ^ . The area

equals the area of a square with side length 3 minus one quarter the area of a disk with radius 3 => A = 32 — ^ = l ( 4 - π ) ≠> M = 6A = ^ ( 4 - π). Therefore, y = ⅛ = ( ^ ) [ ^

]

= ⅛

=> (*,y) = ( ⅛

⅛ ) ⅛^

center of mass. 25. M x = J ? dm = £ = £

∙ 6 ∙ (⅜ ) dx

ψ

(⅜ ) (χ 2 ) ( ⅜ ) d

χ

= £

⅜

d χ

=

2

£

χ

^2

dχ

= 2 [ - x - 4 ] μ 2 [ ( - ∣) - ( - i ) ] = 2 ( i ) = h M y = J zχ dm = J = /

x

x ∙ δ ∙ ( ^ ) dx

(χ 2 ) ⅛ ) d x =

2f

x d x

=

2

[f]ι

= 2 (2 - i ) = 4 - 1 = 3; M = J d m = £ ^ (⅜) dx : £ x 2 (⅜ ) dx = 2 £ dx = 2[x]? = 2(2 - 1) = 2. So ^ = ^ = 1 ^ 7 = ^

= 1 ^

( x ,y )= ( ∣, ∣) is th e

mter of mass.

27. (a) We use the shell method: V = £ 2π ( X ) ( ⅛ ) dx = f ^ x = 16πJ^ ^

[ ^ - ( - ^ ) ] dx

dx = 16πJ^ x 1∕ 2 dx = 16π [ j x3 /2 ] J = 16π (5 ∙ 8 - ∣) = y 1 (8 - 1) = ^

1

Section 6.4 Moments and Centers of Mass (b) Since the plate is symmetric about the x-axis and its density δ(x) = £ is a function of x alone, the distribution of its mass is symmetric about the x-axis. This means that y = 0. We use the vertical strip approach to find x: M y = J χ, dm = ∫ = 8 [2χV2] J =

8 (2

x∙ ^

■2 - 2) = 16; M = ∫ d

- (- ^ j

∙ δ dx = J^ x ∙ ^

∙ ∣dx = 8 f ι x -

1 /2

dx

= £ [ ^ - ( ^ ) ] - 6 dx = S ^ ( ⅛ ) ({ ) dx = » £ ^

m

= 8 [—2x^ 1∕ 2 ] * = 8[—1 —(—2)] = 8. So x = ^ = y = 2 => (x, y) = (2,0) is the center of mass.

29. The mass of a horizontal strip is dm = 6 dA = 6L dy, where L is the width of the triangle at a distance of y above its base on the x-axis as shown in the figure in the text. Also, by similar triangles we have ^ = ⅛ ^ ^

L = ⅛(h - y). Thus, M x = J y dm = ∫ 0⅛

= ⅞ (T -

T)

= ^

2

(5 H ) = ^

M

(⅛) (h - y) dy = ⅞ J j h y - y 2 ) dy = ⅞ [ ⅛ - ⅞] θ

=Jdm

√ > (⅛ )(h -y )d y = ⅛ χ h( h - y ) d y = ^

= τ (h 2 - y ) = ⅞ 1 ∙ So y = ⅛ = f ^ ) ( ⅛ ) = ∣ => the center of mass lies above the base of the n

∖

2 y

2

j

M

∖

o

y

∖ ∂bn /

3

triangle one-third of the way toward the opposite vertex. Similarly the other two sides of the triangle can be placed on the x-axis and the same results will occur. Therefore the centroid does lie at the intersection of the medians, as claimed. 31. From the symmetry about the line x = y it follows that x = y. It also follows that the line through the points (0,0) a n ^ ( i ’ 2 ) *s a m e ^ a n r ^ y = χ = j ∙ ( ∣- θ ) = j =► (χ,y) = ( M ) ∙ 33. The point of intersection of the median from the vertex (0, b) to the opposite side has coordinates (θ, ∣) ≠> y = (b - 0) ∙ i = ∣andx = ( | - 0) ∙ ∣= ∣ ≠ ∙ (χ,y) = ( f , 5) ∙

35. y = x 1∕ 2 => dy = ∣x 1^2 dx ≠ ∙ ds = √ (d x ) 2 + (dy)2 = ^ l + ⅛ d x ;

^=^^/^1^ =6X2∖A+^a*=τ[(*+ ] θ D 3/2

= ⅛ [ ( 2 ÷ 1 ) 3∕ 2 - ( 1 ) ≡ ∕ 2 ] — 2« Γ ∕9 ∖ 3 ∕ “ 3 [ U ∕

2

_

∕1 ∖ 3 ∕ 2 1 — 2 6 / 2 7 ∖ 4∕ J ~ 3 ∖ 8

1∖ _ 8^ ~

136 6

dx

191

192

Chapter 6 Applications of Definite Integrals

37. From Example 6 we have M x = Jθ a(a sin 0)(k sin 0) d0 = a2 k jθ sin2 θ dθ = =

⅛ P ^ ⅞ ^] o = ⅞

1

’M y = ∕

a ^a c o s

^ J Q ( 1 —cos 20) d0

J

0) ^0 = a2 k Q sin 0 cos 0 d0 = y [sin2 0] θ = 0; M = ∫θ ak sin 0 d0 = a k [- cos 0]J = 2ak. Therefore, x = ^ = 0 and y = ^ = ( ⅛ e ) ( ⅛ ) = ^ ^ ( ^ o

^^

s *n

is the center of mass. 39. Consider the curve as an infinite number of line segments joined together. From the derivation of arc length we have that the length of a particular segment is ds = ^ /(dx)2 ÷ (dy)2 . This implies that M xx = f δy ds,, M y = f δx ds and M = Γ 6 ds. If 6 is constant, then χ = ⅛ = J

J

y

J

’

J

M

l

=

and

ds

length

41. Since the density is constant, its value will not affect our answers, so we can set 6 = 1. A generalization of Example 6 yields Mx = J *y dm = J ^ 2

a

a2 sin 0 d0 = a2 [ - cos 0 ] ^ t"

= a2 [—cos ( f + α ) ÷ cos ( ∣—α)] = a2 (sin a + sin α) = 2a2 sin a ; M = ∫ dm = f ιr^ = a [(∣+ α ) — ( ∣—α)] = 2aα. Thus, y = ⅛ = ¾ τ

a

= ^ ~ ∙ Now

s =

a^ α

6.5 AREAS OF SURFACES OF REVOLUTION AND THE THEOREMS OF PAPPUS

=> S = 2 π j^

= sec4 x

(tan x) √ , 1 + sec4 x dx

(c) S ≈ 3 .8 4

3. ( a ) x y = l = > x = i = > ⅛ = - J => S = 2 π f ι | √ 1 + y - 4 dy (c) S ≈ 5 .0 2

j

=> ( ⅛ ) 2

a d0 = a [ ^ l °

) and a sin α = ∣

=> c = 2a sin a. Then y = ≡ f⅛ ^

a

τ)

Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus 5. (a) x1∕ 2 + y 1∕ 2 = 3 => y = (3 - x 1∕ 2 ) 2 g = 2 (3 - « '« ) ( - i " *

,

(J)

= (∣- W

',

(b)

)

______________

=> S = 2 π f l (3 —X1∕ 2 ) 2 ^ 1 + (1 —3X^ 1∕ 2 )2 dx (c) S ≈ 63.37

= tan2 y

7. (a) ⅛ = tany => ( ^ )

p√3 / py => S = 2πJθ ( jθ tan t dt ∖ pπ∕3 / py = 2π Jθ ( J tan t d t) sec y dy (c) S ≈ 2.08

9.

j= ⅛ » = ^

≡ = ⅛ s = ∫ , ⅛ ( ∕ι + ⅛ y *

[⅛]

=

0

⅛ s = ρ ∙( ∣ ) y i+ ι* = ≠ r * *

^ TTV^ J Geometry formula: base circumference = 2π(2), slant height = √ 4 2 + 22 = 2 y z5

=> Lateral surface area = ∣(4ττ) ( 2 7 5 ) = 4ττᅠ ∕5 in agreement with the integral value 11-

J = f S √ ⅛ ¾ ∕i + t i j =

1

2^ [(1 ÷ 3)

-

(∣+ 1) ] -

5

3> = f 2 > lΓ

1

1

ᅠ ∕H


du = ∣x3 dx => | du =

dx;

X = 0 = > U = l ,X = 2 = > U = y ] → s = 2 τr / ^ ’u 1/ 2 ∙ 3 d u = H i — 1 “ 3

∩25 _ ∖ 27

2 √

2 x

≠ S= ∫

ιλ _ ∕ ~

2≤,

15 ⅛ — 1 ⅛ dx

1

-

x

-

π / 1 2 5 -2 7 ∖ _ 3 ∖ 27 / “

98π 81

ι - χ →, ( ⅛ V = 0 - ⅞

√ 2X - X2

\d x ;

2

5 0'5 2

u 3 /2 ]

2x - x i

2

π √ 2 x - x √ l + ⅛ ⅛ dx

= 2πfo>dκ = 2πMo1 = 2π

1 .5

193

194

Chapter 6 Applications of Definite Integrals

∏∙ ∣ = y2 ^

⅛ ) 2 = y4 => s = ∫ ⅛ √ τ + F d y j

[u = 1 ÷ y4 => du = 4y3 dy => ∣du = y3 dy; y = 0 => u = 1, y = 1 => u = 2] → S = ∕ = l S ^ '',

>’ ■ ⅛ = √ ⅛

du

= 5 [i " 3' 2 ] : = 5 ( ^ - 1)

⅛ )2= ⅛

^

=⅛ Γ

(∣) u 1∕ 2 ( ∣du)

1 2π

^

s

2 τ

=C

= 4 √ 0" ' , √ ( 4 - 3 ) + 1 ⅛

√ ι + ⅛ ⅛

■≈ Λ

β - « ” - s « ] = - ¥ [ ( i ) 3' 3 - J 3' 2] * = - M i M

2 π

/

2πy ds

=

2π

J, y (y 3 + ⅛ )

dy

+ ⅛ j⅛

= 2 π ∫ (y, + s y^ 2 ) ⅛

[ ( M I ) - ( ⅜ - i ) ] = 2 " ( ⅛ + l ) = ⅛ < 8 ∙3 1 + 5> = ⅛

⅛ = 1 (a ≈ - χ ψ 1' 2 (-2x) = ^

=> S = 2π ^ ᅟ ∕ a2 —x2 ^ ÷ j ⅛

≠

( j ) '=

j⅛

j dx = 2π J* a √ 7a 2~ ^ 5 c^^

= 2π f a , dx = 2πa[x]

= 2πa[a - (—a)] = (2πa)(2a) = 4πa2 25. y = cosx => ⅛ = - s i n x => ^ ^

= s in 2 x => S = 2π J ^ 2(COSX) √ T

27. The area of the surface of one wok is S = f 2πx y 1 ÷ ^ ) =* ⅛ = 7 ⅛ = 2π I

*

:S = ∫ ζ

(g )’ = ⅛

2π

B

Λ

dy. Now, x2 ÷ y2 = 162 => x = 0 6

√ i 6 ^ 7 √1

+

r

2

—y2

⅛ 5 , dy = 2 . ∫ ^ √(16≈ - y≡) + y≡ dy

16 dy = 32π ∙ 9 = 288π ≈ 904.78 cm2 . The enamel needed to cover one surface of one wok is

V = S ∙ 0.5 mm = S ∙ 0.05 cm = (904.78)(0.05) cm3 = 45.24 cm3 . For 5000 woks, we need 5000 ∙ V = 5000 ∙ 45.24 cm3 = (5)(45.24)L = 226.2L => 226.2 liters of each color are needed. 29. , = √

^ pa+h

= 2 π Ja

s .

S = - l ^ ⅛ ,.,..

__

=

y (R2 —x2 ) + x2 dx = 2πR f

75

⅛

=> ( . )

,

= ⅛

s

= 2 x f √ M

⅛

Λ

⅛

⅛

pa+h

dx = 2πRh

x∣√ 1 ÷ 1 dx = 2π f 31. y = x => ^ ^ = 1 ≠> (⅛ ) = 1 => S = 2 π f ι ∣

χ(-x) ∖

∕2 dx + 2πJ^ x√*2 dx

= - 2 √ 2 π [≤] ° + 2 √ 2 π [⅛] θ = - 2 √ 2 π (θ - ∣) + 2√2π(2 - 0) = 5 √ 2 π

33. ⅛ = - sin t and ^ = cos t ≠> ^ (⅛ ) 2 + ^ ^

= √ ( - sin t)2 + (cos t)2 = 1 => S = J 2τry ds

= f ^ 2π(2 + sin t)(l) dt = 2π [2t - cos t] θπ = 2π[(4π - 1) - (0 - 1)] = 8π 2

Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus ^ ( ⅛ ) 2 + ( ⅛ ) 2 = √ 1 2 + (t + √ 2 ^ = - ^ 1 ^ 7 ^ + 3

35. ⅜ = l a n d ⅛ = t + √ 2 =^

i

S = f 2τrxds

2π (t + √ 2 ) y t 2 + 2 √ 2 t + 3 dt; [u = t2 + 2 √ 2 t + 3 => du = (2t + 2 √ 2 ) dt; t = - √ 2 => u = 1,

t = √ 2 ≠> u = 9 ∣→ p √ u du = [∣7ru3∕ 2 ] J = ⅛ (27 - 1) = ⅞ 1

37. ∣ = 2 a n d ⅛ - l

√ W

+ ( ⅛ ) 2 = √ 2 2 + 12 = √ 5 ≠> S = p π y d s = £ 2π(t + 1)√ 5 dt

= 2 π √ 5 [ ≤ + t ] θ = 3πᅠ ∕5 . Check: slant height is ᅠ ∕5 4 39. (a) An equation of the tangent line segment is (see figure) y f(mk) + f , (mk)(x —mk). When x = xk- 1 we have r i = f(mk ) + f'(m k )(xt ., - mk ) = f(∏⅛) + f'(m k ) ( - ^ ) = f(mk) - f'(m k) ^

i

Area is π(l + 2)y z5 = 3πᅠ ∕5 .

;

when x = xk we have r2 = f(mk ) + f'(m k)(xt - mk) = f(mk) + f'(m k) ^ ; ⅞-ι “* ⅞ H--------- A⅞--------- H

(b) ⅛ = (∆x k)2 + (r2 - r i )2 = (∆x k)2 + [f'(m k) ⅛ - (-f'(m k ) ⅛ t )] 2

— (∆x k)2 + [f, (mk)∆x k]2 => Lk = 1∕( ∆ x k)2 + [f'(m k)∆x k]2 , as claimed (c) From geometry it is a fact that the lateral surface area of the frustum obtained by revolving the tangent line segment about the x-axis is given by ΔS k = π(rι + r2)Lk = π[2f(mk )] y j (∆x k )2 + [f'(m k )∆x k]2 using parts (a) and (b) above. Thus, ΔS k = 2πf(mk) ^ 1 + [f'(mk)]2 ∆ x k . (d) S =

n

lim o ∑ ΔS k =

n

lim o ∑ 2πf(mk) √ 1 + [f'(m k)]2 ∆x k = J a 2πf(x) √ 1 + [f'(x)]2 dx

41. The centroid of the square is located at (2,2). The volume is V = (2π) (y) (A) = (2π)(2)(8) = 32π and the surface area is S = (2π) (y) (L) = (2π)(2) ^ 4 ^ = 32^2% (where ᅠ ∕8 is the length of a side).

43. The centroid is located at (2,0) => V = (2π) (x) (A) = (2π)(2)(π) = 4π2 45. S = 2πy L => 4πa 2 = (2πy) (πa) => y = 7 , and by symmetry x = 0 47. V = 2πyA 4

∣πab2 = (2πy) ( y ) => y = ∣ and by symmetry x = 0

49. V = 2πp A = (2π)(area of the region) ∙ (distance from the centroid to the line y = x —a). We must find the distance from (0, ^ ) to y = x —a. The line containing the centroid and perpendicular to y = x —a has slope -1 and contains the point (θ, ∣) . This line is y = - x + ∣ y The intersection of y = x - a and y = - x ÷ ^ is the point ( ⅛ y ) ⅛ r ) ∙ Thus, the distance from the centroid to the line y = x —a is ^ ∕^ ^ 4 a j ^ 3 ^

__ y ^ (4 a + 3aπ)

-^

γ

_ ( 2 fr ) ( V ^ (^ a + 3aπ)^ ( ~ )

=

^

^

^

^

51. From Example 4 and Pappus's Theorem for Volumes we have the moment about the x-axis is Mx = y M = ½) (τ) = τ-

195

196

Chapter 6 Applications of Definite Integrals

6.6 WORK 1. The force required to stretch the spring from its natural length of 2 m to a length of 5 m is F(x) = kx. The work done by F is W = J^ F(x) dx = k Jθ x dx = ∣[x2 ] o = y ∙ This work is equal to 1800 J => ∣k = 1800 ≠> k = 400N ∕m 3. We find the force constant from Hooke's law: F = kx. A force of 2 N stretches the spring to 0.02 m => 2 = k ∙ (0.02) => k = 100 ^ . The force of 4 N will stretch the rubber band y m, where F = ky => y = => y = τx⅛ => y = 0.04 m = 4 cm. The work done to stretch the rubber band 0.04 m is W = ι = 100 P x dx = 100 ⅛ 1

004

= ≡

^

kx dx

= 0∙08 J

5. (a) We find the spring's constant from Hooke's law: F = kx => k = ^ = ¾ ^ = ^

=> k = 7238 ⅛

p θ .5

(b) The work done to compress the assembly the first half inch is W = Jθ = 7238 [ y j θ = (7238) y ^ = G238∣o125)

w

x dx

995 in ∙ lb. The work done to compress the assembly the = 2 τ r [1 - (0.5)2 ] = < Z ≡ 7 5 )

second half inch is: W = f ’ °kx dx = 7238 f ‘ ° x dx = 7238 * ∕0 .5

nθ .5

kx dx = 7238 Jθ

0.5

L

J 0 .5

2

l

J

2

≈ 2714 in ∙ lb 7. The force required to haul up the rope is equal to the rope's weight, which varies steadily and is proportional to p50

n50

x, the length of the rope still hanging: F(x) = 0.624x. The work done is: W = Jθ F(x) dx = Jθ 0.624x dx = 0.624

Γ 21 50

=780 J

9. The force required to lift the cable is equal to the weight of the cable paid out: F(x) = (4.5)(180 —x) where x p !8 0

is the position of the car off the first floor. The work done is: W = Jθ = 4.5 [180x - y ] ^ = 4.5

(180 2

p !8 0

F(x) dx = 4.5 Jθ

(180 —x) dx

- ^ ) = ⅛ ψ ≤ = 72,900 ft ∙ lb

11. The force against the piston is F = pA. If V = Ax, where x is the height of the cylinder, then dV = A dx p (P 2 Λ 2 )

Fdx = ∫p A d x = J ^ p d V .

13. Let r = the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate, the amount of water in the bucket is proportional to (20 - x), the distance the bucket is being raised. The leakage rate of the water is 0.8 lb∕ft raised and the weight of the water in the bucket is F = 0.8(20 - x). So: '20

J

o

Γ

0.8(20 - x) dx = 0.8 [20x -

21

20

θ = 160 ft ∙ lb.

Section 6.6 Work

197

15. We will use the coordinate system given. (a) The typical slab between the planes at y and y + ∆y has a volume of ΔV = (10)(12) ∆ y = 120 ∆y ft3 . The force F required to lift the slab is equal to its weight: F = 62.4 ΔV = 62.4 ∙ 120 ∆y lb. The distance through which F must act is about y ft, so the work done lifting the slab is about ΔW = force × distance = 62.4 ∙ 120 ∙ y ∙ ∆y ft ∙ lb. The work it takes to lift all 20

the water is approximately W ≈ ^2 ^ ^ 0 20

= 22 62.4 ∙ 120y ∙ ∆y ft ∙ lb. This is a Riemann sum for o the function 62.4 ∙ 120y over the interval 0 ≤ y ≤ 20. The work of pumping the tank empty is the limit of these sums: f*20

r

21 20

W = J o 62.4 ∙ 120y dy = (62.4)(120) ⅞

= (62.4)(120) ( ^ ) = (62.4)(120)(200) = 1,497,600 ft - lb

(b) The time t it takes to empty the full tank with ( ^ ) - h p motor is t =

‰

= ^ ®

sec

^

= 5990.4 sec

sec

= 1.664 hr => t ≈ 1 hr and 40 min (c) Following all the steps of part (a), we find that the work it takes to lower the water level 10 ft is Γ

W = J o 62.4 ∙ 120y dy = (62.4)(120) ∣ ⅜]

to

= (62.4)(120) ( ψ ) = 374,400 ft ∙ lb and the time is t = ⅛

= 1497.6 sec = 0.416 hr ≈ 25 min (d) In a location where water weighs 62.26 | : a) W = (62.26)(24,000) = 1,494,240 ft ∙ lb. b ) t = 1^ = 5976.96 sec ≈ 1.660 hr => t ≈ 1 hr and 40 min In a location where water weighs 62.59 ^ a) W = (62.59)(24,000) = 1,502,160 ft ∙ lb b) t = ≡ ^ = 6008.64 sec ≈ 1.669 hr ≠> t ≈ 1 hr and 40.1 min 17. The slab is a disk of area πx2 = π Q ) 2 , thickness ∆y, and height below the top of the tank (10 - y). So the work to pump the oil in this slab, ΔW, is 57(10 —y ) π ( 0 2 . The work to pump all the oil to the top of the tank is w

=

∫ θ ⅛ ( 1 0 y 2 - y3 )dy = ^

[ ^ - ^ ]θ ° = ll,875π ft ∙ lb ≈ 37,306 ft ∙ lb.

19. The typical slab between the planes at y and and y + ∆y has a volume of ΔV = π(radius)2 (thickness) = π ( y ) 2 ∆y = 7Γ∙ 100 ∆y ft3 . The force F required to lift the slab is equal to its weight: F = 51.2 ΔV = 51.2 ∙ 100π ∆y lb => F = 5120π ∆y lb. The distance through which F must act is about (30 —y) ft. The work it takes to lift all the 30

30

kerosene is approximately W ≈ ^ ΔW = ]Γ 5120π(30 - y) ∆y ft ∙ lb which is a Riemann sum. The work to pump the 0

0 P30

Γ

21 3θ

tank dry is the limit of these sums: W = Jθ 5120π(30 —y) dy = 5120π ∣ 30y - ⅛ J 0

=

5120π ( y ) = (5120)(450τr)

≈ 7,238,229.48 ft ∙ lb 21. (a) Follow all the steps of Example 5 but make the substitution of 64.5 ∣ for 57 ^ . Then, w = ∫ ⅛ = 64^

(10 - y)y2 d y = ^ = 2

[ i⅛ - ^

=

- ⅛) = m

(83 ) ( ^ - 2)

ι,5π - 83 ≈ 34,582.65 ft ∙ lb

(b) Exactly as done in Example 5 but change the distance through which F acts to distance ≈ (13 —y) ft. Then W = ∫ ⅛ ( 1 3 - y ) f ¢ = ? ∣ ⅞ 4 ] ' = Φ ( 1 ⅞f - ξ ) = ( ⅞ ) (S’ ) ( τ “ ⅛ = ¾ F

198

Chapter 6 Applications of Definite Integrals = (19π) (82 ) (7)(2) ≈ 53.482.5 ft ∙ lb

23. The typical slab between the planes at y and y ÷ ∆ y has a volume of about ΔV = π(radius)2 (thickness) = π ( ^ 2 5 —y2 ) 2 ∆ y m 3 . The force F(y) required to lift this slab is equal to its weight: F(y) = 9800 ∙ Δ V = 9800π ( ι∕2 5 —y2 ) 2 ∆ y = 9800π (25 - y2 ) ∆ y N. The distance through which F(y) must act to lift the slab to the level of 4 m above the top of the reservoir is about (4 —y) m, so the work done is approximately Δ W ≈ 9800π (25 - y2 ) (4 - y) ∆ y N ∙ m. The work done lifting all the slabs from y = - 5 m to y = 0 m is 0

approximately W ≈ ^ 9800π (25 - y2 ) (4 - y) ∆ y N ∙ m. Taking the limit of these Riemann sums, we get -5

w = tf °5 9800π (25 - y 2 ) (4 - y) dy = 9 8 ∞ π £ = -9 8 0 0 π (-5 0 0 - ^

+ f

(K)0 - 25y - 4y2 + y3 ) dy = 9800π [100j∕ - f y2 - f y3 + ⅞] °

125 + ^ ) ≈ 15,073,099.75 J

∩^2 ∕∙X2 25. F = m ^ ≈ mv 3j by the chain rule => W = J* mv ^ dx = m j ι (v ⅛ )

=

m

[^ v2 (x)]

= ∣m [v2 (x2) - v2 (xι)] = ∣mv2 - 5 mv2 , as claimed. 27. 9 0 mr p h = 1⅛hr - ⅛ min ∙ ⅛ sec ≡ ∙ ⅛1 mi = 132 ft/sec; m = ≡ 32 ⅞ ’ 32 w

= (I) ( ≡

)

m = ⅛ slugs = j⅛ slugs; 124 mph = ^ ^ ^ w

s iu g s ;

≈ 181.87 ft/sec;

= (2) ( ⅛ ≡l u gs) (181.87 ft/sec)2 ≈ 64.6 ft ∙ lb

31. weight = 6 ,5 o z = ^ lb => m — j∣ ^

slugs; W = ( ∣) ( ( ^

slugs) (132ft/sec)2 ≈ 110.6ft ∙ lb

33. (a) From the diagram, 2 r (y ) = 60 - x = 60 - - ^ 5 0 ^ ( 7 ^ 3 2 5 ) for 325 ≤ y ≤ 375 ft. (b) The volume of a horizontal slice of the funnel is Δ V ≈ 7Γ[r(y)] 2 ∆ y 2 2 = π 60 — γ ∕δ θ — (y —325)

∆y

(c) The work required to lift the single slice of water is Δ W ≈ 62.4ΔV(375 - y) = 62.4(375 - y)π 60 - ^ 5 0 2 - (y - 325)2

∆ y.

The total work to pump our the funnel is W »375

J

325

2 2 62.4(375 - y)π 60 - √ 5 0 - ( y - 325)

dy

≈ 6.3358 ∙ 107 ft-lb . 35. We imagine the milkshake divided into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval [0 ,7]. The typical slab between the planes at y and y ÷ ∆ y has a volume of about Δ V = π(radius)2 (thickness) = π ( ς ⅛ ^ ) 2 ∆ y in3 . The force F(y) required to lift this slab is equal to its weight: F(y) = ∣ΔV = y ( j η ^ ) 2 Ay

oz∙

The distance through which F(y) must act to lift this slab to

the level of 1 inch above the top is about (8 - y) in. The work done lifting the slab is about

Section 6.7 Fluid Pressures and Forces ΔW = ( y ) ^ ⅛ ^ (8 - y) ∆ y in ∙ oz. The work done lifting all the slabs from y = 0 to y = 7 is 7

approximately W = ∑ 9 ⅜ (y ÷ 17.5)2 (8 —y) ∆ y in ∙ oz which is a Riemann sum. The work is the limit of 0

these sums as the norm of the partition goes to zero: W = = ⅛

£ (

2450

-

2 6 ∙2 5 y

2 7 y2

-

- y3 ) d y = 9 ⅛ [ - ⅛ -

^ 9 y3

(y + 17.5)2 (8 —y) dy - ¥

y2 +

2 4 5 °y]

Γ

11 35,780,000 i] 6 37 00 00

θ

≈ 9 ⅛ [ - y - 9 ∙ 7 3 - ^ 2 - 72 + 2450 ∙ 7] ≈ 91.32 in - oz

37. WθΓk =

•35,780,000

'35,780,000

J

J 6 6,370,000

6,370,000

= (1000) (5.975 ∙ 102 4 ) (6.672 ∙ 10~1 1 ) ( 5 ⅛

dr

= 1000 MG [ -

1 35,780,000

5.144 × 101° J

6.7 FLUID PRESSURES AND FORCES 1. To find the width of the plate at a typical depth y, we first find an equation for the line of the plate’s right-hand edge: y = x —5. If we let x denote the width of the right-hand half of the triangle at depth y, then x = 5 ÷ y and the total width is L(y) = 2x = 2(5 + y). The depth of the strip is (—y). The force exerted by the water against one side of the plate is therefore F = J* g w (-y ) ∙ L(y) dy = J* 5 62.4 ∙ (—y) ∙ 2(5 + y) dy = 124.8 ∫J^52 ( - 5 y - y 2 ) dy = 124.8 [ - j y2 - ∣y3 ] Z3 = 124.8 [ ( - ∣- 4 + ∣∙ 8) - ( - ∣- 25 + ∣∙ 125)] = (124.8) ( ψ - ψ ) = (124.8) ( 3 1∑^34) = 1684.8 lb 3. Using the coordinate system of Exercise 4, we find the equation for the line of the plate's right-hand edge is y = x - 3 => x = y + 3. Thus the total width is L(y) = 2x = 2(y + 3). The depth of the strip changes to (4 —y) ^

F= J

62A ∙ (4 - y) ∙ 2(y + 3) dy = 124.8 J / 1 2 + y - y2 ) dy

w(4 - y)L(y) dy = J

[

2

31 θ

12y + ⅜ - ⅝]

3

= (-124.8) ( - 3 6 + ∣+ 9) = (-124.8) ( - ^ ) = 2808 lb

5. Using the coordinate system of Exercise 4, we find the equation for the line of the plate's right-hand edge to be y = 2 x - 4 => x = φ and L(y) = 2x = y + 4. The depth of the strip is (1 - y). pθ 4 w(l

(a) F = J

Pθ

- y)L(y) dy = J i

= (-6 2 .4 ) [(-4)(4) - ≡ (b) F = (-6 4 .0 ) [(-4)(4) -

2

Pθ

+ y ] = (-62.4) ( - 1 6 - 24 + ⅛ ) =

f

33

of the strip is (33.5 —y) => F = Jθ w(33.5 - y)L(y) dy ∙ (33.5 - y) ∙ 63 dy = ( ⅝ ) ( 6 3 ) ( 3 3 . 5 - y) dy

= ( ⅛ ) (63) [33,5y - ⅞] θ" = ( ^ ) [(33.5)(33) - f ] (64)(63)(33)(67 - 33) (2)(12’ )

= 1309 lb

< -≈ ∙4 χ -i2 θ + 6 4 )

≡ > + ^ 1 = G6lθ)ti2°±6 4 ) ~ ιi94.7 ⅛

Using the coordinate system given in the accompanying figure, we see that the total width is L(y) = 63 and the depth

= ∫ o⅛

o

Γ

α 2

31 θ

62A ∙ (1 - y)(y + 4) dy = 62.4 J / 4 - 3y - y 2 ) dy = 62.4 [4y - ⅝ - ⅞j

y( H = 3 ft.

1 3. Suppose that h is the maximum height. Using the coordinate system given in the text, we find an equation for the line of the end plate’s right-hand edge is y = ∣x 4 x = ∣y. The total width is L(y) = 2x = ∣y and the depth of the typical horizontal strip at level y is (h —y). Then the force is F = Jθ w(h —y)L(y) dy = Fm ax , J»h

= (62.4) ( j ) [ ⅝ - ξ ] ^ = 3ᅟ ∕(

h

f

(hy - y 2 ) dy 0 (62.4) (4) (⅝ - ⅛ ) = (62.4) ( V = ( ∣h 2 ) (30) = 12h2 ≈ 12(9.288)2 ≈ 1035 ft3 . 15. The pressure at level y is p(y) = w ∙ y => the average pressure is p = ⅛ £ p(y) dy = ∣£ w ∙ y dy = ∣w [⅛] θ =

,

(κ ) ( l ) ^ T

This is the pressure at level ∣, which

is the pressure at the middle of the plate. 17. When the water reaches the top of the tank the force on the movable side is J

2

(62.4) ( 2 ^ 4 —y 2 ) (—y) dy

= (62.4)∫ ° 2 (4 - y 2 ) 1 /2 (—2y) dy = (62.4) [ j (4 - y2 ) 3 /2 ] ° 2 = (62.4) ( j ) (43 ∕ 2 ) = 332.8 ft ∙ lb. The force compressing the spring is F = 100x, so when the tank is full we have 332.8 = 100x => x ≈ 3.33 ft. Therefore the movable end does not reach the required 5 ft to allow drainage => the tank will overflow. 19. Use ac oordinate system with y = 0 at the bottom of the carton and with L(y) = 3.75 and the depth of a typical strip being J »7.75 p7 .7 5 Γ 1 7.75 w(7.75 - y)L(y) dy = ( ^ ) (3.75)£ (7.75 - y) dy 2= ( ^ ) (3.75) [7.75y - ⅜] θ o = ( W ) (3.75) e F

≈

4 ∙2 l b

21. (a) An equation of the right-hand edge is y = ∣x => x = j y and L(y) = 2x = ^ . The depth of the strip is (3 - y)

F= £

W (3

- y)L(y) dy = £ (6 2 .4 )(3 - y) ( ∣y) dy = (62.4) ∙ ( ∣) £ ( 3 y - y 2 ) dy

Chapter 6 Practice Exercises = (I) [i YM

] ’ = (I) [ ? - ? ] = »

201

( I) ( ? ) = 374∙4 lb

(b) We want to find a new water level Y such that Fγ = ^ (374.4) = 187.2 lb. The new depth of the strip is (Y —y), and Y is the new upper limit of integration. Thus, Fγ = J o w(Y - y)L(y) dy Γ ( γ - y) (I y) d y = (62∙4) ( l ) Γ ( γ y - y2 ) d y = (62.4) ( ∣) [γ . f _ ⅛] θ = (6 2 .4) (∣) ( ^ - ≤ )

6 2 ∙4

=

= (62.4) (∣) Y3 . Therefore Y3 = 5 ¾

= ^ ∣ P

≠> Y = ^

≡

= 3√ 1 T 5 ≈ 2.3811 ft. So,

ΔY = 3 —Y ≈ 3 —2.3811 ≈ 0.6189 ft ≈ 7.5 in. to the nearest half inch. (c) No, it does not matter how long the trough is. The fluid pressure and the resulting force depend only on depthe of the water. CHAPTER 6 PRACTICE EXERCISES

1. A(x) = ∣(diameter)2 = ∣( √ x - x2 ) 2 = ∣(x - 2 √ x - x2 + x4 ) ; a = 0, b = 1 => V = ^ A(x) dx = ∣J^ (x - 2x5 ∕ 2 + x4 ) dx — Z “ 4 [2

I 7

γ λ

1 _ 7∕2 ∣ ^r 5 j 0 -

π f 1 _ 4 ∣ 1  7 5/ 4^2

= 4 ⅜ (3 5 -4 0 + 1 4 )= ⅝ 3. A(x) = ∣(diameter)2 = ∣(2 sin x —2 cos x)2 = ∣∙ 4 (sin2 x —2 sin x cos x ÷ cos2 x) = π(l - sin 2x); a = ∣,b = 7 *b

J a

p 5 π ∕4

A(x) dx = π I

d π ∕4

(1 - sin 2x) dx

_ _ Γv _L cos2x] 5π∕4

= π [x + — ] ff/4

5.

A(x) = 5 (diameter)2 = ∣^2^∕x ~ ⅛ ) = j ^4x - x5∕ 2 + ⅛ ) ; a = 0, b = 4 => V = J^ A(x) dx = ? £ (4 x -

=

2Γ

x 5 /2

+ ⅛)

dx

= 5 [2 χ 2 - 7 ^

2

+ ⅛

; M

(32 - 32 ∙ μ

? ∙ 32)

( 1 - 7 ÷ 5 ) = ⅜ (35-40+ 1 4 )= >

7. (a) disk method'. V = J^ 7rR2 (x) dx = f

π

(3X4)2 dx = π J* J9X8 dx

= π [x9 ] L1 = 2π

(b) shell method'. V = f 2π ( s^e l1 ) ( j^ ! !^ dx = f 2πx (3xv4 ) dx = 2π ∙ 3 f x5 dx = 2π ∙ 36 ⅛1 = π z Ja radius/ yheight) Jo Jo L J0 Note: The lower limit of integration is 0 rather than —1. (c) shell method'. V= J > (X ) ( ⅛ ⅛ = 2 π ∫ j l - X)(3x4 ) ⅛ = 2 ∙ ⅛ √ ] ' , = ⅛ l ( h )) - ( - M ) ] = I f

202

Chapter 6 Applications of Definite Integrals (d) washer method: R(x) = 3, r(x) = 3 - 3x4 = 3 (1 - x4 ) 4 V = f π [R2 (x) - r2 (x)] dx = f = ^π J - f t

9.

-

2x4 + x8 )] dx = 9π ∕

4 1(2x

- x8 ) dx = 9π ^

-y j

π ∣ 9 - 9 (1 - x4 )2 ] dx = 18π [∣- ∣ ] = ^

= ^

l

(a) disk method: V= π ^ V x — 1) dx = JΓJ (x —1) dx —π [y —x] = π [ ( f - 5 ) - ( l- l) ] .< - 4 ) .⅛

1

(b) washer method: R(y) = 5, r(y) = y2 + 1 ≠> V = J*c π [R2 (y) - r2 (y)] dy = π ∫ = π

L (

25 -

y4

“

2y2

1)

-

dy =

7r

j

∣ 25 - (y2 + 1)2 ] dy

∫ . 2 (2 4 - y4 - 2y2 ) dy = π [24y - ⅞ - j y3 ]

2

= 2π (24 ∙ 2 - ⅛ - j ∙ 8)

= 32π (3 - ∣- ∣) = ^ (45 - 6 - 5) = ⅛ (c) disk method: R(y) = 5 - (y2 + 1) = 4 - y2 =>

=

v

π

=

L (

j

c

7 γ r 2

16 -

( y )

8 y2

d

y

=

J -2

π

(4

-

y 2

)2

d y

+ y4 ) d y

= ^ [ 1 6 y - ⅛ + ⅞ ]2 2 = 2 , ( 3 2 - ⅝

+

⅝)

= 64π (1 - ∣+ ∣) = ⅛ (15 - 10 + 3) = ⅛ Ξ 11. disk method: R(x) = tan x, a = 0, b = ^ => V = π J^ tan2 xdx = πjθ

(sec2 x — 1) dx = π[tan x —x]θ^3 = - ^ - γ —-

13. (a) disk method: V = πJ*θ (x2 —2x)* dx = π jθ (x4 - 4x3 + 4x2 ) dx = π [y - x4 + ∣x3 j θ = π ( y — 16 + y ) = ⅞ (6 - 15 + 10) = ⅞ (b) washer method: v

= f o π [ p ^ (χ 2 -

2x

+ 1 )2 ] dx = ∫ 02πdx + ∫ ᅟ (x - 1)4 dx = 2π - [π ⅛ ] θ = 2π - π ∙ j = ⅞

(c) shell method: v = £ ^ ( X ) (height) dx = 2 π f ( 2 - x) [ - (x2 - 2x)] dx = 2 ^ ( 2 - x) (2x - x2 ) dx = 2π^ (4x - 2x2 - 2x2 + x3 ) dx = 2π J o (x3 - 4x2 + 4x) dx = 2π ^ - ∣x3 + 2x2 ]

= 2π (4 - ^ + 8)

= ⅞ (36 - 32) = ⅜ (d) washer method: V = π J^ [2 - (x2 - 2X)]2 dx - π ^ 22 dx = πJ*θ ∣ 4 - 4 (x2 - 2x) + (x2 - 2x)2j dx - 8π = π

X (4

-

4χ2

+ 8x + x4 - 4x3 + 4X2 ) dx - 8π = π ^ (x4 - 4x3 + 8x + 4) dx - 8π

= π [⅛ - x4 + 4x2 + 4x] θ - 8% = π ( f - 16 + 16 + 8) - 8π = f (32 + 40) - 8π = ^

- ^

= 32Ξ

Chapter 6 Practice Exercises 15. The material removed from the sphere consists of a cylinder and two ’’caps.” From the diagram, the height of the cylinder is 2h, where h 2 + ( λ A )

= 22 , ie . h = 1. Thus

= 6π ft3 . To get the volume of a cap,

Vcy ι = ( 2 h ) π ( ^

use the disk method and x2 + y 2 = 22 : V cap = f 1 πx 2 dy = ∫ι

π

(4

-

y2 ) d y

=

π

4 y - 3] ∣ 1

= π [(8 — f) - (4 - ∣)] = y ft3 . Therefore, V rem0ved = V cyl + 2Vcap = 6π + ^ x l/2

∏∙ y = 4

L

=

∫

M

^ -

ιz 2

= H (4 + l ∙ 8 ) - ( 2 + j ) ] = l( 2

L= J

+ ^

d χ

=

∫

4

L= Γ √ ι + H

1 (χ - l ∕2

+

χ

l ∕2 )

d χ

=

4

1 [2

- 2 + χ ) dχ χ

l ∕2

+

2

χ

3 ∕2 ] J

⅛) = ⅛

+

( ∣) 2 = 1 (χ 2∕5 - 2 + χ - 2 ∕ 5 )

⅛ = i x 1≠5 - i χ-V5

^/1 + 5 (x2 ^5 —2 + x - 2 ∕ 5 ) dx

= ⅛ (1260 + 450) = ≡

∣- 2 + x)

y Γ ( χ - ⅛

= ∕ " ι (χV≈ + χ-v ≈ ) ⅛ = 1 [S

21.

∣H

- 1 χιz M

y i ( Γ + 2 + x) ⅛ = ∫

19. y = ⅛ x6 ∕ 5 - ∣x 4 ∕ 5

ft3 .

= ψ

+

L = ιf

y j ∣(x 2 ∕ 5 + 2 + x - 2 ∕ 5 ) dx — J ᅟ∣∣(x 1∕ 5 + x - 1 ∕ 5 ) 2 dx

= H ( i ' 2’ + i ' 2, ) - (1 + i) ] = I ( ψ + ⅞)

11η

= ψ

= 5 cos t —5 cos 5t + y ( j ) 2 + ^ ^

= - 5 sin t + 5 sin 5t and

= ^ / ( - 5 sin t + 5 sin 5t)2 + (5 cos t —5 cos 5t) 2 = 5ᅟ ∕ sin2 5t —2sin tsin 5t + sin2 t + cos2 t —2cos tcos 5t + cos2 5t = 5 √ 2 —2(sin tsin 5t + cos tcos 5 t) = 5^/2(1 —cos 4t) = 5 ^ 4 ( 1 ) ( 1 —cos 4t) = 10ᅟ ∕ sin2 2t = 10∣ sin 2t∣= 10sin 2t (since 0 ≤ t ≤ 5) 10sin 2tdt = [-5 cos 2t]o/2 = ( - 5 ) ( - 1 ) - (-5 )(1 ) = 10

+ Length = / ^

23. ^ = - 3 sin θ and ⅛ = 3 cos θ => J (⅛ ) 2 + ( ∣) 2 = ^ ( - 3 sin 0) 2 + (3 cos 0) 2 = √ 3 (sin 2 0 + cos 2 0) = 3 '3 π ∕2

J

0

n 3 π ∕2

3d⅛ = 3 ∫θ

,

λ d0 = 3 ( o⅜ - θ )v = ⅜

25. Intersection points: 3 —x2 = 2x 2 => 3x2 —3 = 0 => 3(x - l)(x ÷ 1) = 0 ≠> x = - l o r x = l . Symmetry suggests that x = 0. The typical vertical strip has center of mass: (x ,y ) = (x, 2 χ 2 ψ ^ ~ χ2 )^ — ( x , , length: (3 —x2 ) —2x2 = 3 ( 1 —x2 ), width: dx, area: dA = 3 (1 —x2 ) dx, and mass: dm = 6 ∙ dA = 35 (1 —x2 ) dx => the moment about the x-axis is y dm = ∣(5 (x2 + 3) (1 —x 2 ) dx = 1 6 (—x 4 - 2x2 + 3) dx => M x = ∫ y dm = p j = 2

5

[^ T “ T

+

3 x

]

1

1

=

3 6

(-

5

-

5 +

3

) = T5 ( —3 -

10

+

4 5

>= T

■ ’M

=

ι

(—x 4 —2x 2 + 3) dx

f dm = 36 ∫ ∕ 1 - x2 ) dx

203

204

Chapter 6 Applications of Definite Integrals = 30 [x - y ]

ι

= 60 (1 - ∣)

= ∣. Therefore, the centroid is (x, y) = (0, f ) .

40 ≠ ∙ y = ⅛ = ^

27. The typical vertical strip has: center of mass: (χ , ,y , ) /

= I x, ^

l

2∖

1 , length: 4 - ^ , width: dx,

area: dA = ^ 4 - ^ d x , mass: dm = the moment about the x-axis is V dm = S = J ι 2π √ 4 y —y2 y ∕ ^

* (J)’ = ⅛ d x

= ∕4 y -y 2

s

= ∫Λ √ s +> √ι + ⅛ *

= 2 √ 2 π [j (x + l) 3 ∕ 2 ] θ = 2 √ 2 π ∙ j (8 - 1) = ^

,2 ~ y ≠ ∕4 y - y2

1 + ( ⅛ V = 4y - y 2 + 4-4 y + y 2 _ 4 dy∕ 4y-y 2 4y-y 2

dy = 4π j∖ dx = 4π

35. x = ∣and y = 2t, 0 ≤ t ≤ ᅟ ∕5 => ∣ = t and ⅛ = 2 => Surface Area = J^ = 2π [j u 3 ∕ 2 ] 4 = ηf r , where u = t2 ÷ 4 4

2 π ( 2 t ) y ^ ÷ 4 dt = J 4 2πu 1∕ 2 du

du = 2t dt; t = 0 => u = 4, t =   Λ => u = 9

Chapter 6 Practice Exercises

205

37. The equipment alone: the force required to lift the equipment is equal to its weight => F1 (x) = 100 N. pb

The work done is Wι =

j* Fι(x) dx =

p40

Jθ 100 dx = [100x]θ° = 4000 J; the rope alone: the force required

to lift the rope is equal to the weight of the rope paid out at elevation x => F2 (x) = 0.8(40 - x). The work

J

'b a

/^40

F2 (x )d x = I 0

r

"1 40

L

J 0

0.8(40- x ) d x = 0.8 4 0 x - ⅛

∖

/

= 0.8 (40 2 - ^ ) = ≡ ⅛ ^ = 640 J; ∖

the total work is W = Wi + W2 = 4000 + 640 = 4640 J 39. Force constant: F = kx => 20 = k ∙ 1 => k = 20 Ib/ft; the work to stretch the spring 1 ft is W = J^ kx dx = kj^ x dx = ^20 ^ θ = 10 ft ∙ lb; the work to stretch the spring an additional foot is W = £kxdx = k £

X

dx = 2 0 [ ^ = 2 0 ( | - i ) = 2 0 ( j ) = 3 0 f t . l b

41. We imagine the water divided into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval [0,8]. The typical slab between the planes at y and y + ∆y has a volume of about ΔV = π(radius)2 (thickness) = π (I y) 2 ∆y ~ W y2 △y ^ 3 , ^ ι e ^o r c e ^(y) required to lift this slab is equal to its weight: F(y) = 62.4 Δ V _ (62.4)(25) π y2 ^ ^ ι∣ 5 The dis t a nce through which F(y)

Reservoir's Cross Section

must act to lift this slab to the level 6 ft above the top is about (6 ÷ 8 - y) ft, so the work done lifting the slab is about ΔW = ^ ®

πy2 (14 —y) ∆y ft ∙ lb. The work done

lifting all the slabs from y = 0 to y = 8 to the level 6 ft above the top is approximately W ≈ £ (6214χ25) π y2 (14 _ y) ∆y f t . lb so the work to pump the water is the limit of these Riemann sums as the norm of o the partition goes to zero: W = f ^ ^ 1 πy 2 (14 - y) dy = < ≤ ^ ≥ fθ (14y2 _ y 3) dy = (62.4) ( ^ ) [ ^ y3 - ^ ] θ = (62.4) ( ^ ) ( y ■83 - ⅞ ) ≈ 418,208.81 ft ∙ lb

43. The tank's cross section looks like the figure in Exercise 41 with right edge given by x = ⅛ y = ⅜. A typical horizontal slab has volume ΔV = π(radius)2 (thickness) = π Q ) 2 ∆y = f y2 ∆y. The force required to lift this slab is its weight: F(y) = 60 ∙ J y2 ∆y. The distance through which F(y) must act is (2 ÷ 10 —y) ft, so the work to pump the liquid is W = 60J π(12 ~ y) ^ J dy = 15π ^ — ⅛j θ — 22,500π ft ∙ lb; the time needed to empty the tank is

2⅞5^ivse c

≈ 257 sec

45. F = £ w ■ ( ⅛ ) - L(y) dy ^

F = 2 £(62.4)(2 - y)(2y) dy = 2 49.b£(2y - y2 ) dy = 249.6 [y2 - ⅝] θ

= (249.6) (4 - ∣) = (249.6) ( ∣) = 332.8 lb 47. F = £ w ∙ ( ⅛ ) ∙ L(y) dy ^

F = 6 2 .4 £ (9 - y) (2 ∙ ^ ) dy = 6 2 .4 £ (9y1∕ 2 -

3y

3∕ 2 )

dy

= 62.4 [6y3∕ 2 - 2 y s∕2] J = (62.4) (6 ∙ 8 - j . 32) = ( ψ ) (48 ∙ 5 - 64) = = 2196.48 lb 49. F = w j β (8 - y)(2)(6 - y) dy + w2 £ ( 8 - y)(2)(y + 6) dy = 2wi fθ (48 - 14y + y2 ) dy + 2w2 £ -- 2w1 ∣ 48y - 7 y 2 + ^ j

6 0

^48y + y2 - y ] ÷ 2W2 ∣

0

= 216wι + 360W2

6 (48

+ 2y - y2 ) dy

206

Chapter 6 Applications of Definite Integrals

CHAPTER 6 ADDITIONAL AND ADVANCED EXERCISES 1. V = π J^ [f(x)]2 dx = b2 —ab 4

τ ^ [f(t)]2 dt = x2 - ax for all x > a 4

τr [f(x)]2 = 2x —a ≠- f(x) - ± ^

^

√ 1 + [f'(t)] 2 dt = Cx => √ 1 + [f'(x)]2 = C =► f'(x) = √ C 2 - 1 for C ≥ 1

3. s(x) = Cx => £

≠- f(x) = Γ √ C 2 - 1 dt + k. Then f(0) = a

a = 0 + k => f(x) = Γ √ C 2 - 1 dt + a

f(x) = x √ C 2 - 1 + a,

where C ≥ 1. 5. From the symmetry of y = 1 —xn , n even, about the y-axis for —1 ≤ x ≤ 1, we have x = 0. To find y = use the vertical strips technique. The typical strip has center of mass: (x ,y ) = (x, ⅛ ^ ) , length: 1

w

—xn ,

width: dx, area: dA = (1 - xn ) dx, mass: dm = 1 ∙ dA = (1 - xn ) dx. The moment of the strip about the x-axis is y dm = ⅛ dx M x = ∫ J 1 i h ⅛ ^ dx = 2 ∫ o' ∣(1 - 2x" + x 2n) dx = [x - ⅛J⅛ + ^ r ] J _ -

1 1

____ 2___ ∣ 1 _ (n + l)(2n + 1 ) - 2(2n + 1) + (n ÷ 1) _ n+1 ' 2n+ l ~ ( n + l) ( 2 n + l) ~

2n2 + 3n + 1 —4n —2 ÷ n + 1 _ ( n + l) ( 2 n + l)

Also, M = f dA = f (1 - xn ) dx = 2 f (1 - xn ) dx = 2 Γx y=

M

=

(n + i) (2 n + i)

’T

(0 >⅛ )

= STT

2n2 ( n + l) ( 2 n + l) *

* = 2 (1 - - U ) =

is th e lo c a tio n o f th e

Therefore,

centroid. As n → ∞ , y → i so

the limiting position of the centroid is (θ, ∣) . 7. (a) Consider a single vertical strip with center of mass (x ,y ). If the plate lies to the right of the line, then the moment of this strip about the line x = b is (x —b) dm = (x —b) £ dA => the plate’s first moment about x = b is the integral ∫ (x —b)6 dA = J ⅛ dA —∫ Λ dA = M y —b the plate’s first moment about x = b is J (b —x)⅛ dA = ∫ M dA - f δx dA = b«A - M y . x 9. (a) On [0, a] a typical vertical strip has center of mass: (x ,y ) = (x, ^ - ~ - ∣^ ~ ^ ,

length: ᅟ ∕b = δ ^ᅟ ∕b

2

2

—x 2 — ᅟ∕a 2 - x2 , width: dx, area: dA = ( ᅟ ∕b

2

—x 2 — √ a 2 —x 2 ) dx, mass: dm = £ dA

—x 2 — ᅟ ∕ a 2 —x 2 ) dx. On [a, b] a typical vertical strip has center of mass: , length: χ ∕b 2 - x 2 , width: dx, area: dA = √ b 2 —x 2 dx,

(x ,y ) = (x, ^ ^ ^

mass: dm = 6 dA = 6 χ ∕b 2 —x2 dx. Thus, M x = ∫ y dm = f

0

l (ᅟ ∕b

2

= 5 ∫o K ^ “

—x2 + ᅟ∕a 2 - x 2 ^ δ ζ χ ∕b 2 - x2 - χ ∕a 2 —x2 ^ dx + J^ ∣χ ∕b 2 —x 2 6 χ ∕b 2 —x 2 dx χ2 )

“ (a2 “

χ2 )] d x

+ I J j b2 -

χ2 ) d x

= f [(b2 - a2 ) x] 0 + f [b2 X - ⅛] * = j [(b 2 _ a 2) a ] = f (ab2 - a3 ) + f ( h

3

= J^ xδ ^ᅟ∕b 2 —x 2 — ᅟ ∕a = δ f o x (b2 -

X2 ) 1^2

+

= ^ o '( b2

^ f^

- χ2 ) d x

∣^ b 3 _ ^ ) _ ( b 2a _ ^ ]

- ab2 + < ) = ⅞ - ^ = 5 f ^ ) ; M 2

- a2 ) d x +

y

= f ^ dm

—x2 j dx + J*a x x = ⅛ ∙ Then χ = h => t = (12mh)1∕ 4 . The work done is '(12m h)V 4

0

J ¾

^ lorn

= ^

j

F ω -⅛ d t = ∫

n(12m h) 1∕ 4

o

-

r

η ( 12mh)1∕ 4

t2 ∙ ⅛ d t = ⅛ [ ⅛ ] 0 ι

,

zφ

X

“ 12m +

C i;

,

= (⅛ )(1 2 m h )≡ ∕ 4 z

i‰ ^ ~ T ' 2 ᅟ ∕3m h = yv √3m h v lorn 3 3

15. The submerged triangular plate is depicted in the figure at the right. The hypotenuse of the triangle has slope —1 => y —(-2 ) = - (x - 0) => x = —(y ÷ 2) is an equation of the hypotenuse. Using a typical horizontal strip, the fluid pressure is F = / (62.4) ∙ ( ⅛ ) ∙ ( ⅛ ) dy = ∫ J 6 2 ,4 ) ( - y ) [ - ( y + 2)] dy = 62.4 ∫ J y = 62.4 [⅝ + y2 ] "θ = (62.4) [ ( - f

+ 4)

2

+ 2y) dy

- ( - ψ + 36)]

= (62.4) ( ψ - 32) = ≈ 2329.6 lb 17. (a) We establish a coordinate system as shown. Atypical horizontal strip has: center of pressure: (x ,y ) = (∣, y ) , length: L(y) = b, width: dy, area: dA = b dy, pressure: dp = u ∣ y∣ dA = ωb ∣ dy y∣ => Fx = = -

ω b

f

y dp =

[ C

F = ∫ dp =

= ^wb [⅛]

h

= -

J' h ω b

y∣ dy = -ω b J y ■ωb ∣ M

^ ) ]

L(y)dy = f hω∣ y∣ 0 -h

= -ω b [θ -⅞ ]

h

y2 dy

= ^

-ω b £

y dy

^ . Thus, y = ⅝

(b ,-h )

- y => the distance below the surface is ∣h.

208

Chapter 6 Applications of Definite Integrals (b) A typical horizontal strip has length L(y). By similar triangles from the figure at the right, ^ = z i J p => L(y) = —∣(y + a). Thus, a typical strip has center of pressure: (x ,y ) = (x ,y), length: L(y) = ^∏(y ÷ ⅛ width: dy, area: dA = —£ (y + a) dy, pressure: dp = ω ∣ y∣ dA = ω (-y ) ( - £) (y + a) dy = τ (y2 + a y) d y ^

F

« = f y ⅛,

= f _ (aa+h) y ∙ τ (y 2 + a y) d y = τ ∫ J

+h ,

(y3 + a y 2 ) d y

a4 - a(a + h)3 3

~ ⅛ β (a 4 = ^

-

(a4 + 4 a ⅛ + 6a2 h 2 + 4ah3 + h4 )) - 4 (a4 - a (a3 + 3a2 h + 3ah2+ h 3 ))]

(12a3 h + 12a2 h2 + 4ah 3 - 12a3 h - 18a2 h2- 12ah3 - 3h4 ) = ^

y∣ L(y) dy = f f _ ^ = = ^ (6a2 + 8ah + 3h2 ) ; F = J dp = J ω ∣ _ “ =

h

Γ ∕ -a3 3

⅛ ja⅛

∣ a3 λ 2y

3⅛ +⅛

∕ - ( a + h)3 3 + P - a3

+

ψ

a3

∣ a(a + h)2 l _ 2 ∕J

ω b Γ(a +

h [

- ( a 3 + 2a⅛ + ah η

=

h)3 - a 3

3

^ ^ (3a 2h

( - 6 a 2 h 2 - 8ah3 - 3h4 ) (y 2 + ay) dy = ^ [⅛ + ⅛ ]

∣ a3 - a ( a + h)2 l 2 J ψ

3a h 2 ψ

h 3)

_ 3 (2 a ⅛ + ah 2 )]

= ⅛ (6a2 h + 6ah2 + 2h 3 - 6a 2 h - 3ah2 ) = ^ (3ah2 + 2h3 ) - ⅛⅛ (3a + 2h). Thus, y = ⅞ =

(^H∏i ) (6a2 ÷ 8 a h ÷ 3 h 2 ) (⅛ )(3 a + 2 h )— =

6a2 + 8ah + 3h2 6a + 4h *

/ -1 (2 )

∕b a 2 ÷ Rah 4-3h2 λ 3T ⅛ h ( j

,

z

^

1

r

∙

the distance below the surface is

CHAPTER 7 TRANSCENDENTAL FUNCTIONS 7.1 INVERSE FUNCTIONS AND THEIR DERIVATIVES 1. Yes one-to-one, the graph passes the horizontal test. 3. Not one-to-one since (for example) the horizontal line y = 2 intersects the graph twice. 5. Yes one-to-one, the graph passes the horizontal test 7. Domain: 0 < x ≤ 1, Range: 0 ≤ y

9. Domain: —1 ≤ x ≤ 1, Range: — f ≤ y ≤ 5

11. The graph is symmetric about y = x.

(b) y = χ ∕l —X2 =χ, y2 13. Step 1: y = x2 + l = > x Step 2: y = ∖ ∕iΓ Π =

1_

=

2

χ

2 -^

χ

2 _ 1 _ y2 ^ .

χ

_ √ 1 _ y2 => y = ^ 1 - χ

= y -l= > x = ^y - 1

f-1

(x)

15. Step 1: y = x3 - 1 => x3 = y + 1 => x —( y + l ) 1∕ 3 Step 2: y = ∖ ∕x + 1 = f ^ 1 (χ ) 17. Step 1: y = (x + 1)2 =>

λ∕ y

= x + 1, since x ≥ - 1 => x =

Step 2: y = √ x - 1 = f - 1 (x) 19. Step 1: y = x5 ≠> x = y 1∕ 5 Step 2: y = 5√ x = f - 1 (x); Domain and Range of f - 1 : all reals; f (f - 1 (x)) = (x 1∕ 5 ) 5 = x and f - 1 (f(x)) = (x5 ) 1^5 = x 21. Step 1: y = x3 + 1 => x3 = y — 1 => χ = ( y - l) 1∕ 3 Step 2: y = 3√ x ^ T = f

1 (x)j

Domain and Range o f f - 1 : all reals;

χ ∕y

- 1

2

- f

1 (χ)

210

Chapter 7 Transcendental Functions f (f - 1 (x)) = ((x - 1)1 /3 ) 3 + 1 = (x - 1) + 1 = x and f - 1 (f(x)) = ((x3 + 1 ) - 1)1/3 = (x3 ) 1 /3 = x

23. Step 1: y = ⅛ =+ x2 = * =+ x = ^ = f - 1 (x)

Step 2: y = ^

Domain of f - 1 : x > 0, Range of f - 1 : y > 0; f (f - 1 (x)) = 7 ⅛ = 4 y = x and f^ 1(f(x)) = -⅛r = A W V? (τυ W

= x since x > 0

25. (a) y = 2x + 3 => 2x = y —3 =^

ω

X =

f

2 ^ H

‰

^ 1 (x ) = 2 ^ 2

= 2 ∙⅛ , = ⅛

27. (a) y = 5 —4x => 4x = 5 —y => X =

( ) d£|

^c

d x lχ = ι∕2

1 -

4

=*

f

= - 4% ¢11

" 1W

dx | x = 3

=

4 ^

4

= - i

4

29. (a) f(g(x)) = ( 3√ x ) 3 = x >g(f (χ )) = V x 3 = x (c) f'(x) = 3X2 =+ f'(l) = 3, f '( - l ) = 3; g'(x) = ∣x - 2 ∕ 3 =+ g, (l) = ∣, gz( - l ) = ∣ (d) The line y = 0 is tangent to f(x) = x3 at (0,0); the line x = 0 is tangent to g(x) = 3λ∕ x at (0,0)

31. ‰ = 3 x 2 - 6 x ^

⅛

=⅜ l x = «3)

33.

df~ 1 I d x

lx = 4

_

37

df~ 1 I dx

lx = f(2)

35. (a) y = m x = + x = ⅛ y ≠ > f (b) The graph of y

—f - 1

1 (χ)

= 1χ

(x) is a line through the origin with slope ^.

Section 7.2 Natural Logarithms 37. (a) y = x ÷ l => x = y - 1 => f - 1 (x) = x — 1 (b) y = x + b => x = y - b => f - 1 (x) = x - b (c) Their graphs will be parallel to one another and lie on opposite sides of the line y = x equidistant from that line.

39. Let xι ≠ X2 be two numbers in the domain of an increasing function f. Then, either xι < x2 or xι > x2 which implies f(xι) < f(x2 ) or f(x1) > f(x2 ), since f(x) is increasing. In either case, f(xι) ≠ f(x2 ) and f is one-to-one. Similar arguments hold if f is decreasing. 41. f(x) is increasing since x2 > Xi => 27x3 > 27x3 ; y = 27x3 => χ = ∣y 1∕ 3 ≠> f df _ 01 dx ~

γ

2

df~ 1 _ 1 I 81x2 I ∣x ι∕3 ~

. dx

_ 9X2 ∕ 3

-

1

9

_

λ

1

γ

1(

x)

= 1x

1∕ 3

j

- 2 ∕3

43. f(x) is decreasing since x2 > Xi => (1 —x2 )3 < (1 - Xi)3 ; y = (1 - x)3 => x = 1 —y 1∕ 3 => f df _ dx -

o ∕1 _

γ

  2

.

df~ 1 _ dx -

1 I - 3 ( l - x ) 2 ∣l

_ χ l∕ 3

1 (x)

= 1 —x 1/ 3 ;

- 1 _ _ 1 γ - 2 ∕3 3^3 ~ 3x

45. The function g(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if Xi ≠ x2 then f(xι) ≠ f(x2 ), so -f(x ι) ≠ -f(x2) and therefore g(xi ) ≠ g(x2). Therefore g(x) is one-to-one as well. 47. The composite is one-to-one also. The reasoning: If Xi ≠ x2 then g(xι) ≠ g(x2) because g is one-to-one. Since g(X1 ) ≠ g(x2 ), we also have f(g(xι)) ≠ f(g(x2)) because f is one-to-one. Thus, f 0 g is one-to-one because X1 ≠ X2 => f(g(xι)) ≠ f(g(x2)). 49. The first integral is the area between f(x) and the x-axis over a ≤ x ≤ b. The second integral is the area between f(x) and the y-axis for f(a) ≤ y ≤ f(b). The sum of the integrals is the area of the larger rectangle with corners at (0,0), (b, 0), (b, f(b)) and (0, f(b)) minus the area of the smaller rectangle with vertices at (0,0), (a, 0), (a, f(a)) and (0, f(a)). That is, the sum of the integrals is bf(b) - af(a).

51. (gof)(x) = x => g(f(x)) = x => g, (f(x))f, (x) = 1 7.2 NATURAL LOGARITHMS 1. (a)

In 0.75 = In ¾ = In 3 - In 4 = In 3 - In 22 = In 3 - 2 In 2

(b) In ∣= In 4 - In 9 = In 22 - In 32 = 2 In 2 - 2 In 3 (c) I n ∣= l n l - l n 2 = - l n 2

(d) In 3√ 9 = ∣In 9 = j In 32 = ∣In 3

(e)

In 3 √ 2 = In 3 + In 21∕ 2 - In 3 + j In 2

(f)

In √LL5 = ∣In 13.5 = ∣In

3. (a) In sin (9 - In (⅛ p) = In

sin θ

= ∣(In 33 - In 2) = ∣(3 In 3 - In 2)

= ln5

(b) ln(3x 2 - 9 x ) + ln (⅛ ) = l n ( 3 ⅛ ^ ) = l n ( x - 3 )

(c) | In (4t4 ) - In 2 = In √ 4 t 4 —In 2 = In 2t2 - In 2 = In ( y ) = In (t2 )

211

212

Chapter 7 Transcendental Functions

5. y = ln3x ≠> y ' = (⅛) (3) = 1 7. y = ln(t 2 ) => ⅛ = ( ⅜ ) ( 2 t) = f 9. y = In ∣= In 3x" 1 ^ 11. y = ln(0

+

l) ^

⅛ = ( ⅛ ) (-3 x " 2 ) = - 1

£ = ⅛ ) (1) = ^

13. y = lnx 3 ≠> ⅛ = (⅛) (3x2 ) = ∣ 15. y = t(ln t)2 ≠> ⅛ = (In t)2 + 2t(ln t) ∙ £ (In t) = (In t)2 + 2 ^ = (In t)2 + 2 In t 17. y = j In x - ⅛ => ⅛ = _ In t

19. yv - T 21. v∙y

. ⅛ _ =^ Λ -

_Jnx_ . √ l+lnx 7

=

t

x

3

ln

x

+ ^ .i - ^ =

x

3

⅛x

(l) ~ ^ n OO) _ 1 —In t P------- = - P-

=

)-(lnx)(∣ ) d+⅛x)(∣ (l÷lnx)2

1 ∣⅛x _ 1∏χ

χ

χ___ χ (l+lnx)2

1 x(l÷lnx)2

23. y = ln(lnx) ^ / = ( ⅛ ) (1) = ⅛ 25. y = 0[sin (In 0) + cos (In ¢)] ≠- ^ = [sin (In 0) + cos (In 0)] + 0 [cos (In 0) ■∣—sin (In 0) ∙ j] = sin (In 0) + cos (In 0) + cos (In 0) - sin (In 0) = 2 cos (In 0) tx + l1), X => vy, = —- —— + i)+χ —__ 2x(x 3x+2 27 v" = In — x√Jx+l = —l∏x — -2 In l ∏ x-— + l) 2f — + J1 — —2(x2x(x+l) χ 29 v = — ≠> ^ 1-lnt a 31. y = ln(sec(ln0)) ^ 33∙

y = l∏( 1^

)

35

∙y=£/2h ^

37∙

u = 1 andt = π =+ u = 3 =+ ∫ o⅛

dt

= £

u 0ι = In 3 - In 1 = In 3 ⅛d u = ll n ∣

43. Let u = lnx =+ du = ∣dx; x = 1 => u = 0 and x = 2 ≠> u = In 2; ∫ ⅛

dx =

∫θ ln2 2 u du = [u2 ]θ 2 = (In 2)2

Section 7.2 Natural Logarithms du = ∣dx; x = 2 =► u = In 2 and x = 4 => u = In 4;

45. Let u = In x 4 *4

J

pin 4

.

dx

_

= JIin 2 u~2 du =

2 x(ln X)22

_ _____ 1 ∣ 1 _ i n 4 ^ r ln2 -

ln 4

“I uJ ln2

L

____ 1___ ∣ 1 _ ______1___ I 1 _ ln2 2 r 1∏2 ^ 2 In 2 In 2 “

1 _ 2 In 2

1 In 4

47. Let u = 6 + 3 tan t => du = 3 sec2 1 dt; u∣ ∫6 ⅜ ⅛ d t = ∫⅛ ≈ ln ∣ 6 + 3 ta n t∣ +C + C = ln ∣ 49. Let u = cos ∣ => du = —∣sin ∣dx => —2 du = sin ∣dx; x = 0 => u = 1 and x = ∣ => Γ S

1 dx = Γ ¾

m

du = ∣ cos fj d0 => 6 du = 2 cos (3 d0; 0 = ∣ => u = ∣ and 0 = π => u = J Z Z ⅛ = 6 [In ∣ u∣ ]^

Γ 2 cot ( d0 = Γ ⅛ ⅛ d0 = 6 f ^ 3

J r , /2

53.

J

τ

sin 5

∕2

√ 1 ∕2

le tu

= ∫∑ 7 ∏ ħ ^

∕∑ ‰

1

=

U

+ ≠

L

=^

/2

= 6 fin ∖

I IJ1 ∕2

du

= ⅛

= ^ 5

dx;

J

Z

;

- In β = 6 In λ A = In 27 2

2

∕

= f ⅜ = ⅛∣ +c u ∣

2 ^ ''0 ⅛

= In 11 + ^ x j 4^ C = In (1 + ^ x ) + C 55. y = √ χ (χ + L

=

W x + 1 ))1/2 ≠ , In y = 5 In (x(x + 1)) => 2 In y = In (x) + In (x + 1) => ^- = ∣+ ⅛

^ y '= ( l ) √ ⅛ + υ β + ⅛ ) = ⅛

1/2

57.

^

= ^

=> In y = | [In t - ln(t + 1)] => ^ ⅛ = ∣( I -

7 ^^ t + τ )

2 ^

t+ T [ t ( t + l ) ]

~ t+ i)

2 √ t ( t + l ) 3 ∕2

59. y = √ 0 + 3(sin0) = (0 + 3)1∕ 2 sin0 => In y = ∣ln(0 + 3) + ln(sin 0) ^ ^

⅛

=

ln y ≈ ln ( 0 + 5 ) - l n 0 - l n ( c o s 0 ) ^

^

=> l = ( S ) ( ⅛ V =⅛ W =^

67∙

y,

-

t + i ^

y

^

[×

+

- ∣+ta∏0)

χ ⅛ τ

-

(1 4×x +

x

- ⅛ j

3 (χ+ n ]

l∏ y= j[l∏x + l n ( x - 2 ) - l n ( x 2 + l)] ^

— 1 3 ∕ ⅜ -¾ 3 V χ2 + 1

= 3t2 + 6t + 2

1 ∣ = ⅛ - ∣+ ⅛ ∣

lny = l n x + ∣l∏(x2 + l ) - ∣l n ( x + l ) ≠> £ = ; + ⅛

^

y= V ¾ f → ^

^ f

lny = ln t + l n ( t + l ) + ln(t + 2) ≠> 7 ⅛ = ∣+ ⅛ + i ⅛

∣ = t(t + l)(t+2) ({ + ⅛ + ⅛ ) = t (t + l)(t + 2) [ < t ± M ± ^ ^

63∙ y = ^

65∙

+

√ ^ + 3 (sin 0) ^5 L(x) = In 1 + 1 ∙ x =+ L(x) = x (b) Let f(x) = ln(x + 1). Since f"(x) = —7 ⅛ 1 < 0 on [0, 0.1], the graph of f is concave down on this interval and the largest error in the linear approximation will occur when x = 0.1. This error is 0.1 —ln (l.l) ≈ 0.00469 to five decimal places. (c) The approximation y = x for In (1 ÷ x) is best for smaller positive values of x; in particular for 0 ≤ x ≤ 0.1 in the graph. As x increases, so does the error x —In (1 + x). From the graph an upper bound for the error is 0.5 - ln(l + 0.5) ≈ 0.095; i.e., ∣ E(x)∣≤ 0.095 for 0 ≤ x ≤ 0.5. Note from the graph that 0.1 - In (1 + 0.1) ≈ 0.00469 estimates the error in replacing In (1 + x) by x over 0 ≤ x ≤ 0.1. This is consistent with the estimate given in part (b) above.

Section 7.3 The Exponential Function 85. y = In kx => y = In x ÷ In k; thus the graph of y = In kx is the graph of y = in x shifted vertically by In k, k > 0.

87. (a)

(b) y, ~ aSnx , Since ∣ sin x∣ cos x∣ are less than and ∣

y

or equal to 1, we have for a > 1 ⅛ ≤ y '≤ ⅛ f o r a l l x . J the graph of y looks a→+∞ more and more horizontal as a → + ∞ .

y = ln (a÷sinx)

7.3 THE EXPONENTIAL FUNCTION lnx2

1

3. (a) 2 In √ e = 2 In e 1∕ 2 = (2) Q ) In e = 1

1

,Jn x -ln v

Λ

1Π(X∕ V)

x

(b) In (In ee ) = In(e In e) = In e = 1

(c) lne^^χ2~y2^ = (—x2 —y2 ) In e = —x2 —y2 5. In y = 2t + 4 => elny = e2t+4 => y = e2t+4 7. ln (y - 4 0 ) = 5t => eln (y '

40)

= e5t =>

y

- 4 0 = e5t => y = e5 t + 4 0

9. In (y - 1) —In 2 = x ÷ In x => In (y — 1) —In 2 —In x = x ≠> In ( ⅛ ∙ ) = x => el n (⅛ i ) = ex => ⅛ i = ex => y — 1 = 2xex => y = 2xex ÷ 1 11. (a) e2k = 4 => lne 2 k = l n 4 ^ 2 k ln e = ln2 2 => 2k = 21n2 => k = ln2 (b) 100e1°k = 200 => e 10k = 2 => ln e 10k = ln2 => 10klne = ln2 => 10k = ln2 ^ k = ⅛ (c) ek ∕ 100° = a => In ek ∕ 100° = In a => y⅛θ In e = In a => γ^θ = In a => k = 1000 In a 13. (a) e - 0 ∙3t = 27 => In e -°∙3t = In 33 => (-0.3t) In e = 3 In 3 => -0.3t = 31n3 ≠> t = - 1 0 In 3 (b) ekt = ∣ => In ekt = In 2^ 1 = k t l n e = - l n 2 => t = - η^ (c) e(ln 0 ∙2 )t = 0.4 => (eln0 2 ) t = 0.4 => 0.2t = 0.4 => In 0.2t = In 0.4 => t In 0.2 = In 0.4 => t = ∣ ⅛^ 15. e v ^ = χ 2 => In e v ^ = ln x 2 => γ ∕t = 2 In x => t = 4(ln x)2 17. y = e^ 5x => y, = e^ 5x £ (-5x) => y, = —5e- 5 x 19. y = e5 ~7x => y, = e5 ^ 7x ⅛ (5 - 7x) => y' = -7 e 5 ~7x

215

216

Chapter 7 Transcendental Functions

21. y = xex —e x => y' = (ex + xex ) —e x = xex 23. y = (x 2 - 2x + 2) ex =Φ y' = (2x - 2)ex + (x2 - 2x ÷ 2) ex = x2 ex 25. y = e0 (sin θ + cos 0) => y' = e 0 (sin 0 ÷ cos 0) + e^(cos 0 - sin 0) = 2e^ cos 0 27. y = cos ( e ~02 ) => ^ = —sin ( e ^ ^ ) ⅛ ( e ^ )

29. y = In (3te t ) = In 3 + In t + In e

g(cos t+ln t) _

pl∏X

∞ s t gin t _ ^ c o s t _ ^

e

X

Z

sin et dt => y' = (sin eln x ) ∙

35. J o

(-

s *n

J

i

/ _

e

~^2 s in ( e ^ )

=

r

⅛

∙

(In x) = ^

(y~

eY s ^n x

)

=

eY c o s x

i

=> V = ⅛

⅛

39. e 2> ≈ s in (x + 3y) => ⅛ 2" = ( 1 + 3y') cos (x + 3y) ⅛ l + 3 y , = ^ ⅛ j v

^

⅛ = ec o s, + tec o st ⅛ (cos t) = (1 - 1 sin t)e co st

l^ ⅛ )= e > e o s x ^ ∕ = ⅛

∕

=

g = 1 - ( τ ⅛ ) ⅛ (1 + «’) = 1 - ⅛

37. In y = ey sin x => Q J y' = (y'e y ) (sin x) + ey cos x => ⅛

( e ~ ^ ) ) ( e ^ ) ⅛ ( - ^2 )

= In 3 ÷ In t —t => ^ = ∣- 1 = ¼ j

t

31. y = l n τ ⅛ = l n e ∙ - l n ( l + e ∙ ) = 9 - l n ( l W ) 33.

=

⅛

- 1

2e2x - cos (x + 3y) 3 cos (x + 3y)

41. f (e3x + 5 e - χ ) dx = ⅛ - 5e" x + C

43. £

3 x 2 e

dx = [ex ] ^ = e ln3 - e 1" 2 = 3 - 2 = 1

45. ∕8 e ( χ + 1 )d x = 8e(χ + 1 ) + C

47

∙ f ∣ n4e ^ 2

dx

!∏4 =

= M

2

[e(ln 9 )/2 - eθn 4 >∕2 ] = 2 (e ln3 - e ln 2 ) = 2(3 - 2) = 2

49. Let u = r 1∕ 2 =+ du = ∣r - 1 / 2 dr =+ 2 du = Γ 1^2 dr; ∫ ^

dr = ∫ e

rv2

∙ r-V 2 dr = 2 ∫ e

u

du = 2eu + C = 2er ‘/2 + C = 2e√, + C

51. Let u = —12 => du = —2t dt => —du = 2t dt; ∫ 2 t e - ,2 dt = - ∫ e u du = - e u + C = - e ~ ,2 + C 53. Let u = ∣ => du = - ^ dx => - d u = ^ dx; ∫⅛

dx = J - e

u

du = - e u + C = - e 1∕ χ + C

55. Let u = tan 0 =+ du = sec2 0 d0; 0 = 0 =+ u = 0, 0 = ∣ =+ u = l ;

J

(1 + e ta n *) sec2 θ dθ = § J sec2 θ dθ + Qeu du = [tan θ]θ^4 + [eu ] J = [tan (5) - tan (0)] + (e 1 —e 0 ) = (1 - 0 ) + ( e - 1) = e

f j

Section 7.3 The Exponential Function

217

57. Let u = sec 7rt => du = π sec πt tan πt dt ≠∙ γ = sec πt tan πt dt; f e sec(πt)s e c (π 0

ta n

(π t ) dt = ∣ f eu du = ⅛ + C = s ^

2

+C

59. Let u = ev => du = ev dv => 2 du = 2ev dv;7 v = In ?O => u = ?O, v = In f2 => u = 52; *ln(π∕2)

J

ι∏(π∕6)

2eV c o s

ev

dv

=

2

pτr∕2

Jπ∕ 6

cos

.

z

,

u du = [2 Sin u ] ∕ 6 = 2 [sin ( f ) - sin ( f )] = 2 (1 - *) = 1

61. Let u = 1 + er => du = er dr; ∫ ⅛ d r = ∫ l d u = ln ∣ u∣ + C = l n ( l + e r) + C 63. ^ = et sin (et —2) => y = J*et sin (et - 2) dt; let u = et - 2 => du = et dt => y = J*sin u du = - cos u + C = - cos (et —2) + C; y(ln 2) = 0 => - cos (eln2 - 2) + C = 0 => —cos (2 —2) + C = 0 => C = cos 0 = 1 ; thus, y = 1 —cos (et —2) x ≠> £ = -2 e * x + Q x = 0 a n d ^ = 0 ≠ > 0 = - 2 e 0 + C => C = 2; thus ⅛ = -2 e ~ x + 2 65. ⅛ ox = 2e^ αx ax ax => y = 2e- x + 2x + Ci; x = 0 and y = 1 => 1 = 2e0 + Ci 4 ∙ Q = - 1 ≠> y = 2e- x + 2x — 1 = 2 (e- x + x) - 1

67. f(x) = ex - 2x ≠> f'(x) = ex —2; f'(x) = 0 => ex = 2 => x = In 2; f(0) = 1, the absolute maximum; f(ln 2) = 2 - 2 In 2 ≈ 0.613706, the absolute minimum; f(l) = e —2 ≈ 0.71828, a relative or local maximum since f"(x) = ex is always positive. 69. f(x) = x2 In | ≠> f'(x) = 2x In j + x2 ^ τ ) ( - x - 2 ) = 2x In ∣- x = -x (2 In x + 1); f'(x) = 0 => x = 0 or In x = —∣. Since x = 0 is not in the domain of f, x = e- 1 ∕ 2 = ∙ ^ . Also, f'(x) > 0 for 0 < x < ^ and f'(x) < 0 for x > ^ . Therefore, f ( ^

= ∣In ^ ∕e = ∣In e 1^2 = ^ In e = ⅛ is the absolute maximum value

of f assumed at x = -⅛. √e 71. £ ” (e2x - ex ) dx = [⅛ - ex ] “ = ( ⅛ ^ - e1" 3 ) - ( £ - e0 ) = ( ∣- 3) - (∣- 1) = f - 2 = 2

73. L = X '√ l + ⅞ dx =^ ⅛ = ⅛i ^

y = eχ ∕ 2 + C; y(0) = 0 =>O = eo + C = > C = - l = > y = ex/2 - 1

75. (a ) ^ (x In x —x + C) = x ∙ ∣+ In x — 1 + 0 = In x (b) average value = ^⅛γ f In x dx = ^-γ [x In x - x], = ^⅛γ [(e In e —e) —(1In 1 — 1)] = ⅛ (e -e + l)= ⅛ 7 77. (a) f(x) = e* ≠> f'(x) = ex ; L(x) = f(0) + f'(0)(x - 0) => L(x) = 1 + x (b) f(0) = 1 and L(0) = 1 => error = 0; f(0.2) = e0 2 ≈ 1.22140 and L(0.2) = 1.2 ≠>error ≈ 0.02140

218

Chapter 7 Transcendental Functions y

(c) Since y" = e x > 0, the tangent line approximation always lies below the curve y = ex . Thus L(x) = x + 1 never overestimates ex .

79. f(x) = ln(x) - 1 ≠> f'(x) = ± ≠> xn + ι = x n -

l-

^

*■ => xn + ι = x n [2 - ln(x n )]. Then x i = 2

=> x2 = 2.61370564, X3 = 2.71624393 and X5 = 2.71828183, where we have used Newton’s method. 81. Note that y = In x and ey = x are the same curve; J* In x dx = area under the curve between 1 and a; J ’ In a

ey dy = area to the left of the curve between 0 and In a. The sum of these areas is equal to the area of the rectangle

θ

pa

=> J

ι

pin a

l n x d x ÷ Jθ ey dy = a ln a .

7.4 ax and loga x (b) 8l o s ^ = √ 2

1. (a) 5l°e 7 = 7

(d) log4 16 = log4 4 2 = 2 log4 4 = 2 ∙ 1 = 2

(c) 1.3l°8'≈75 = 75

(e) log3 √ 3 = log3 3 1∕2 = ∣log 3 3 = ∣∙ 1 = j = 0.5

(f) log4 (1) = log4 4 - 1 = - 1 log4 4 = - 1 ∙ 1 = - 1 3. (a) Let z = log4 x => 4 z = x ≠> 22z = x => (2z ) 2 = x ≠> 2z =

3∕ x

(b) Let z = log3 x => 3z = x => (3z ) 2 = x2 => 32z = x2 => 9z = x 2 (c) log2 (e (,n2)sinx ) = log2 2sinx = sin x S ,

' '

⅛ lΛ _ logs x

' '

⅛ 1 JL l0gx2 a

7. 3log3(7) + 9.

3

11.

y

iog3 (x2 )

In x . lnx _ In 2 , In 3

=

2⅛

=

5 e

In x . in 3 _ In 2 In x

In 3 In 2

In a _L In a — In a , In x2 _ In a In x * In x2 In x

(5) = 5iog5W => 7 + 5 inx _

3

.

1 0

iogl o (2) ^

χ

log2 x _ logs x

In x _;_ In x __ In x . In 8 _ In 2 * In 8 In 2 In x

3 In 2 __ 0 In 2

2 In x _ 9 In x

x

=

2

∕u∖ × '

=

=φ χ = 1 2 5x _ 6 ≠> x2 - 5x + 6 = 0 => (x - 2)(x - 3) = 0 => x = 2 or X = 3

= 2 x => ∕ = 2x ln 2

13. y = 5 ^ => ⅛ = 5^≈(ln 5) ( ∣s~ 1∕ 2 ) = ( ^ ) 5 ^

15. y = xπ => y' = τrx°r^ li 17. y = (cos 0 ) ^ => ⅛ = - √ 2 (cos 0)( vM

(sin 6»)

19. y = 7s κ s In 7 => ⅛ = (7secβ In 7)(ln 7)(sec 6»tan 0) = 7∞ s (ln 7)2 (sec 0 tan 0) 21. y = 2si"3* => ⅛ = (2sin3' In 2)(cos 3t)(3) = (3 cos 3t) (2sta3*) (In 2)

Section 7.4 ax and loga x 23. y = log2 50 = ⅛ f ≠ or

_ ln x y ~ ln 4

v

∣In x2 _

ψ

ln 4

In x ln 4

-

^ = (⅛ )(⅛ )(5 ) = ⅛ ∣o In x _ q In x ln 4 ~

j

κ

ln 4

/ _ ~

3 x ln 4

27. y = log2 r ∙ log4 r = ( ^ ) ( j~ ) = Q⅛ ⅛ 4) =^ ⅛ = [(S ⅛ 4 )] (^ ^n r ) (?) 29.

y

= ⅛ dx

((⅛ Γ ) = ^ x+ 1

219

x -1

= ^ ¾ M

= In ⅛ ⅛ = M x

=

r (in2χin4)

.> - M x - 1)

+

( x + l× x - l)

31. y = 0 sin(log7 0) = 0 sin (⅛ f) => ∣ = sin ( [ ^ ) + 0 [cos (∣ ^ ) ] (0⅛7) = sin(log7 0) + i⅛ cos(log 7 0) 33. y = log5 ex = ⅛⅝ = s ⅛ = > y ' = i⅛ 35.

y

⅛ ≈ [3 (l",>∕ In y = ln(x + 1)’ = x ln(x + 1) => ^ = ln(x + 1) + x ∙ ^ ^ => y' = (x + l) x [ ^ j- + ln(x + 1)]

41.

y = ( √ i) ' = (t1∕2) t = f∕≡ ^

In y = In t ^ = (∣) In t ^

1 ⅛ = (∣) (In t) + (∣) ({) = ψ + ∣

=> ^ ( ^ ) , (Ψ + I) 43.y = (sin x)x => In y = In (sin x)x = x In (sin x) => 45.

«

= In (sin x) + x ( ^ 7 ) ≠> y' = (sin x)x [In (sin x) + x cot x]

y = xlnx, x > 0 ≠> In y = (In x)2 => ^ = 2(ln x) (∣) => y' = (xl"x ) ( ⅛ ^ )

P * ≈ ⅛ + C

51. Let u = x2 =⅛ du = 2x dx ≠> ∣du = x dx; x = 1 => u = 1, x = √ z2 => u = 2; ∫ , V5 X2 dx = £ ( 1 ) 2- du = 1 [ ⅛ ] ; = ( ⅛ ) (2≈ - 2 ' ) = ⅛

53. Let u = cos t => du = —sin t dt => —du = sin t dt; t = 0 => u = 1, t = ∣ => u = 0; Γ

z2 7∞s * s in td t=

Jo

J ι

- f°7 "d u = [ L

-

ln ∕ J 1

? = (∈⅜) (70 - 7) = ∖ in / / ×

×

ln ∕

55. Letu = x2x => Inu = 2 x Inx => ⅛ ⅛ = 2 Inx + (2x) (∣) => ⅛ — 2u(lnx + 1 ) => ∣du = x2x(l + Inx)dx; x = 2 => u = 24 = 16, x = 4 => u = 48 = 65,536; f 4 x2x(l + In x) dx = ∣ f “ ’536 du = | [u]“ ’536 = ∣(65,536 - 16) = ^ f , (x) = £ => xn + ι = xn

ln(x∏)-l

zz ^ x

n+ l — X∏

[2 - ln(x n )]

xn

Then, X] = 2, X2 = 2.61370564, X3 = 2.71624393, and X5 = 2.71828183. Many other methods may be used. For example, graph y = In x — 1 and determine the zero of y.

7.5 EXPONENTIAL GROWTH AND DECAY 1. (a) y = y0 ek, ≠> O.99yo = y 0 e l °∞k ≠> k = ™ (b) 0.9 =

e

(-°∙ooool>,

≈ -0.00001

≠> (-0.00001)t = In (0.9) =► t = ⅛ ⅛

≈ 10,536 years

(c) y = y0 e ,20'≡ , >k ≈ y 0 e-°∙2 = yo (O.82) => 82% 3. ^ = —0.6y =^ y = yoβ- θ 6 yo = 100 ≠ ∙ y = 100e

061

≠ ∙ y = 100e

06

≈ 54.88 grams when t = 1 hr

5. L(x) = Loe^kx => ⅛ = L 0 e - ' 8k ≠* In ∣= —18k => k = ⅛ ≈ 0.0385 => L(x) = L 0 e^00385xj when the intensity is one-tenth of the surface value, ⅛ = L0 e^00385x => In 10 = 0.0385x => x ≈ 59.8 ft 7. y = y0 ekt and yo = 1 => y = e k, => at y = 2 and t = 0.5 we have 2 = e05k => In 2 = 0.5k => k = ^ Therefore, y =

e°n4 >,

=> y =

e 24ln4

=

4 24

= 2.81474978 ×

10'4

= In 4.

at the end of 24 hrs

9. (a) 10,000ekt => 0.2 = e t = ^

≈ 0.585 days

19. y = yoe^h = y 0 e~(k)(3/k) = yoe^3 = ^ < ⅛ = (O.O5)(yo) => after three mean lifetimes less than 5% remains 21. T - T s = (T 0 - T s )e ^ k,, T o = 90 o C, T s = 20 o C, T = 60 o C => 60 - 20 = 7 0 e- ,0k =► $ = e - ' 0k ≠> k = ⅛

≈ 0.05596

(a) 35 — 20 = 7θe^ o °5596t ≠ ∙ t ≈ 27.5 min is the total time => it will take 27.5 — 10 = 17.5 minutes longer to reach 35 o C (b) T - T s = (T 0 - T 8) e - k', T o = 90 o C, T s = - 1 5 o C => 35 + 15 = 105e-° 05596' => t ≈ 13.26 min 23. T - T s = (T 0 - T s ) e^k, ≠> 39 - T s = (46 - T s ) e^ l0k and 33 - T s = (46 - T s ) e^ 20k => ⅛ L = e - 20k = (e -' 0k ) 2 ^

∖2

/

i⅛

= (H ⅛ )

≠> (33 - T s )(46 - T s ) = (39 - T s ) 2 ^

w

1s

= e - ' 0k and

1518 - 79T s + T 2

= 1521 - 78T s + T 2 => - T s = 3 => T s = - 3 o C 25. From Example 5, the half-life of carbon-14 is 5700 yr => j c 0 = Coe-k(5700) => k = ^ => c = coe- 00001216' => (O.445)co = c 0 e -° 0∞12,6* => t = 27

⅞ ^*

≈ 0.0001216

≈ 6659 years

. From Exercise 25, k ≈ 0.0001216 for carbon-14. Thus, c = c0 e^^0 ∙0001216=t > (O.995)co = c 0 e -° 0∞1216t tl _ -

in (0.995) -0.0∞ 1216

≈ 41 years old

7. 6 RELATIVE RATES OF GROWTH (a) slower,

lim

⅛

(b) slower,

lim

⅛

v

7

’X → ∞

= e

lim

^

=

⅛ = 0 lim 6

Sandwich Theorem because ⅜ < e (c) slower,

lim X → ∞

(d) faster,

lim

(e) slower,

lim

(f) slower, v 7

X → ∞

(g) same,

lim

lim

X → ∞

ex

3x2 .+ 2 s⅛ xcosx ex

X → ∞

lim

= x

£ =

e

=

Vex

=

ex

e*

=

li

X → ∞

-

lim X → OO

lim

-⅛ = 0 χ

lim

⅛= ⅜

X → ∞

X → ∞

e ∕2

2

2

ex

e

ex

=

( ⅜ ) x = 0 since ⅜ < 1

lim

6x+2≤∞2i

< ~x for all reals and

( f ) x = ∞ since * > 1

lim =

⅛

→ OO

~

* sιn 2x

=

lim X → ∞

=

li

X → ∞

lim

⅛x = 0 =

-

= 0

X → ∞

2√xex

e

6^4⅛ 2x ex

lim

X → ∞

=

⅛ e*

Section 7.6 Relative Rates of Growth (vh)7 slower,

i

lim ⅛ X → OO e

=

223

Um ∏ ⅛ π = Um ,,nln⅛1 w e = Um ∏-⅛ᅠ τr = 0 X → ∞ ( °) X → ∞ (ln 1°)χe X → ∞ (ln 1°)e

4x = 3. (a) same, lim χ2 ⅛ lim ¾2x ^ = lim 5 = 1 v 7 χ2 x→∞ ’x → ∞ x→∞ 2 (b) faster, Um ⅛ ^ = Um (x3 - 1) = ∞

(c) same,χ Um o ^ ≡

=^

χ

s

≡

^ χ ‰

(e) slower,

^

=

Um

l

⅛

Umo o ⅛ ^ = ^

χ

^

‰

=

Um o (1 + ∏ = √ T = 1

= χ ‰

⅛i =

Um

χ

lim

1 ≈

1

^ ∙ = 0

(f) faster, lim ¾ = lim ⅛ 2x^ = lim 2 1 ^ 21 — o o v 7 X→ ∞ X→ ∞ 2 ’ X → ∞ χ2 (g) slower, lim 2⅛ - = lim ⅜ = lim ⅛ = 0 (h) same, v 7

lim ⅛ = lim 8 = 8 x→∞ x→∞ x

5. (a) same,

lim ⅛3 = lim ⅛ = ln x → ∞ ln3 x → ∞ ln χ = lim W = 1 ln χ x → ∞ (7) / ∖ r In λ∕x ζ⅛) ^n x 1 1 1∙ r (vc)7 same, Xl→ im ∞ - rl⅛ im ∞ v ÷l n x7 - = Xl→ im ∞ ⅛ 2 = ⅛ n x- = Xl→ 2

lim x→∞ (b) same, lim v7 x→∞

(vd)7 faste,r, lim X→ ∞ (ve7) faster, lim x→∞ (vf)7 same,

lim x→∞

(g) slower, (vh)7 faster, 7.

lim

ln x

=

lim X→ ∞

lnA x-

=

lim x→∞

ln x

lim

=

lim x→∞

=

^2L^

= lim X→ ∞

=

-A = lim x→∞

x= ∞

ln x

= lim lim X→ ∞ 2√x x → ∞

2

= ∞

5= 5 -⅛ - = 0

lim

lim ⅛ = lim m = lim xex = ∞ x → ∞ ln x x → ∞ (∣ x→∞ )

⅛ =

lim

X → 0 0 eχ ∕ 2

=

=

ln χ

X → OO

faster than ex/ 2 ; since for x > ee we have In x > e and eχ ∕ 2 = ∞ => ex grows 6

lim

X→ ∞

i ) x = ∞ => (lnx) x grows faster thane x ; sincex > In x for all x > 0and (⅛ lim ∏ ⅛ = v e z v X → ∞ (lnx)x / & ’

lim

X→ ∞

lim

⅛ x^

e

(⅛ )x

χ → OO ∖ lnx∕

= ∞ => xx grows faster than (In x)x . Therefore, slowest to fastest are: eχ ∕ 2 , ex , (In x)x , xx so the order is d, a, c, b 9. (a)

false;

lim

- =1

(b)

false;

lim

-⅜ ξ = ∣= 1

(c)

true; x < x ÷ 5 => ^

(d)

true; x < 2x => ^ < 1 if x > 1(or sufficiently large)

(e) true;

lim

(f) true; ^ ⅛ (g) false;

V 6/

χ

lim 4-5

lim

⅛ =

⅛= 0

= i-p lιι χ < i ψ V

X→ ∞

(h) true; ^

< 1 if x > 1 (or sufficiently large)

⅛

“>2x

=

lim

χ → OO

< ^ ± ^ < ψ

x

= i ψ i < 2 i f x > l ( o r sufficiently large)

÷∣ 4- = 1]

1= 1

= 1 + ^ < 6 if x > 1 (or sufficiently large)

lim S⅛ —L ≠ 0 => lim ⅛ = r ≠ θ∙ Then X OO 8v ∙J X OO >' - L < 1 if x is sufficiently large => L —1 < ^ < L + 1 => ^ ≤ ∣ L∣ + 1 if x is sufficiently large

11. If f(x) and g(x) grow at the same rate, then ^

Um

X→ ∞

Chapter 7 Transcendental Functions

224

≤ ∣ E ∣+

=> f = O(g). Similarly, ^

1

^

g = θΦ ∙

13. When the degree of f is less than or equal to the degree of g since = ∣(^ l e

than the degree of g, and χ limo o ^ 15.

lim χ → oo =

=

lim x→∞

ln x

= lim x → ∞ χ+1

= lim x→∞ 1

∣= 1 and lim x→∞

⅛ ±222) = ∏m x→∞

β+≡l

ln χ

1

=

X → OO ^ + ⅛ 9

lim ⅛⅛ = 0 when the degree of f is smaller x —►oc ratio of the leading coefficients) when the degrees are the same.

17.

∑±1 = y j = j Since the growth rate lim = < ∕ lim - xγ+ 1 = √ j θ and ∩m = / 6 x χ v √χ χ →∞ V χ→ ∞ χ → ∞ √χ V χ→ ∞ is transitive, we conclude that ^/1 Ox + 1 and √ x ÷ l have the same growth rate (that of ^ x ) .

19.

lim x→∞

21. (a)

xn ex

limθo ⅛7 =

χ

nx*

lim x→∞ χ

xn = o (ex ) for any non-negative integer n

l∏∏o o ⅛ = 0 ^ =

lir∏o o ⅛ ρ y

(n)

x

→n θ o

χ l^ =

00

^

^n

x

^

o

(χ l ^ ) ^o r a n ^ P o s itiy e integer n

(b) In (e 17-00 °.0 00 ) = 17,000,000 < (e 17 ×10 J = e 17 ≈ 24,154,952.75 (c) x ≈ 3.430631121 × 1015 (d) In the interval [3.41 × 101 5 ,3.45 × 1015 ] we have In x = 10 In (In x). The graphs cross at about 3.4306311 × 1015 .

2g ⅛ = lim∞ -7r lim∞ 1°g2∏ = 0 => n log2 n 6grows 23. (a) n → n→ n (log 2 n) 2 slower than n(log 2 n)2 ; lim ⅛ 5 = lim ⅛ *∙ 7

V

’

n

n3∕ 2

- → OO

n → ∞

n 1 /2

⅛ = 0 lim 7 T ∖ I 72 = i⅛ lim n → ∞ (5 ) n - 1 ' 2 In 2 ∏ - > 0 0 n 1∕ 2 => n l0g2 n grows slower than n3 ∕ 2 . Therefore, n l0g2 n

= ⅛ ln 2

grows at the slowest rate => the algorithm that takes O(n l0g2 n) steps is the most efficient in the long run. 25. It could take one million steps for a sequential search, but at most 20 steps for a binary search because 2 19 = 524,288 < 1,000,000 < 1,048,576 = 22 0 .

7.7 INVERSE TRIGONOMETRIC FUNCTIONS 1. (a)

π 4

π

3. (a)

π 3

(b)

6

(c)

(b)

π 4

(c)

π 6 π 3

5. (a)

7Γ 3

(b)

3π 4

(c)

7Γ 6

7. (a)

3π 4

(b)

π 6

(c)

2π 3

Section 7.7 Inverse Trigonometric Functions 9. (a) J

(b) - f

(c) ≡

11. (a) ⅛

(b) ∣

(c) ⅞

13. a = sin - 1 ( ⅛ ) => cos a = ∣ ∣ , tan a = ^ , sec a = ∣ ∣ , esc a = y , and cot α = y 15. a = sec- 1 ( — χ ∕5 ^ => sin α = ^ , cos a = — ^ , tan α = —2, esc a =

iy

, and cot α = — ∣

17. sin ( c o s ' 1 ^ ) = sin (* ) =

19. tan (sin’ 1 ( - 1)) = tan ( - f ) = - ^ 21. esc (sec- 1 2) + cos (tan - 1 ( - √ 3 ^ = esc (cos- 1 (5 )) + cos ( - ∣) = esc ( ∣) + cos (— ∣) = ^

+ ∣= ⅛ ^

23. sin (sin^ 1 ( - ∣) + cos^ 1 ( - ∣)) = sin ( - 5 + y ) = sin (∣) = 1 25. sec (tan

1

1 + esc

1

1) = sec (^ + sin

1

∣) = sec (J ÷ i ) = ≈ ( τ ) = - ^

27. sec’ 1 (sec ( - f )) = sec’ 1 ( ^ ) = cos’ 1 ( ^ ) = %

29. α = tan

1

∣indicates the diagram

ι 31. α = sec- 1 3y indicates the diagram

3v

33. α = sin- 1 x indicates the diagram

l< ^z zz cι

35. a = tan - 1 √ x 2 —2x indicates the diagram

=> sec (tan - 1 ∣) = sec α = ^ - -

>!

L
tan (sec- 1 3y) = tan α = χ ∕9 y 2 — 1

=> cos (sin - 1 x) = cos α = √ 1

-

x2

V1-x 2

>∕x2 -2x

xJ

=> s in (tan - 1 √ x 2 —2x]

2 -2X = sin αλ = √γ X x_1

37. α = sin- 1 ^ indicates the diagram

2y

z^ cos

(sin - 1 ⅜ ) =

cos

a = ^ 9 34y2

225

226

Chapter 7 Transcendental Functions

∣indicates the diagram

1

39. α = sec

41.

lim sin- 1 x = ⅞

43.

lim tan- 1 x = 5

45. 47.

lim sec- 1 x = 2-

x → ∞

lim csc^ 1 x =

χ → ∞

49. y' = cos

1

53. y = sec

1

55. y = esc

1

lim sin- 1 ( -χ )∕ = 0

x → ∞

(x2 )' =>

V

2x . √ 71 - χ 4

= ---- ∕τ-=.2^— , =

dx

1

(2s + 1) =>

m

∣ 2 s + l∣ √ ( 2 s + l) 2 - l

*

(x2 + 1) => ∣ =

∣ 2 s + l∣ √4s 2 + 4s

_______ 2x________ _ ∣ x2 + l ∣ ^ (x s + l) ≈ - l

∣ 2s + l ∣ √s2 + s

-2 x (x2 + l) √ x 4 + 2x2

57. yj = secX tz1 ( ∣) = cos- 1 1 at≠∙ ⅛√ =l - t72 ~

59. y = cot

1

-ᅟ ∕ t = cot

. y = ln(tan- 1 x) ≠

1 11∕ 2

∣= ⅛

. y, = esc- 1 (e, ) ≠- ⅛ = d.

65. y = s √ , 1 - s2 ⅛ cos

≠

1

11

∣=

1 + (ti∕2)'2

=

-1

2 √ t ( l+ t )

≈

≠ ----- --- 7 =

∣ e -Γ y W " l

s = s (1 - s2 ) 1^ + cos

1

s ≠

⅛ = (1 “ s2 ) 1^2 + s (∣) (1 —s2 ) 1^2 (-2 s) √ l- s

67. y = tan =

ᅟ ∕x 2 —1 + esc

1

" A

-

.

x√x2 - 1

69. y = x sin

1

∫^ ⅛

,

= tan

7 ~ '" ⅛ ---- 7= 2 √ 1- x

1

(x2 — 1)1^2 + esc

1x

1 => ⅛ = β ] ^ - i Γ ¼ ) _ 7⅛ Ξ 7vψ^ M√×2 - ι

= 0, for x > 1

x + √ 1 - x2 = x sin

- sin- 1 x + 71∙

∣∣Λ

∣ x∣ √x2 - l

1x

2

√ 1 - x2

dx = sin^1 (l)+ c

1x

+ (1 - x2 ) 1/2 ≠> ^ = sin- 1 x + x ( ^ ⅛ ) + (∣) (1 - x2 )

= sin- 1 x

1 /2 (-2 x )

Section 7.7 Inverse Trigonometric Functions 73∙

dx

X ⅛

= X p φ ^ d x = ^ ta n -1 ^ + C

75. f —7 = —= = f —7=— , where u = 5x and du = 5 dx J x√25x 2 - 2

J u√u2 - 2 s e c

= ⅛

77∙

= [4

X ^ ⅛

⅛

s in ^1

79. X 8+⅛ = ^

c

s e c

= ⅛

4

1] J =

(≡

- 1

^1 l⅛ l+

1-

s in ^1 °

c

)=

4

( i - °) = τ

β ⅛ ’ w here u = √ z2t and du = √ ^ dt; t = 0 => u = 0, t = 2 => u = 2ᅟ ∕2

X

ta n ^1

= [⅛ ∙ ⅛

81∙

+

l

⅛

=

Γ

^ (ta n '

1

⅛

^

ta n ^1 0 ) =

X_X /2 y √ ⅞ _ i = X 2 ^ 7 ⅛ ’ w h e r e u - 2y a ∏d du ^ 2 dy ;

y

4 (t a n ^ 1 1 -

t a n ^ 1 °)

= H l - °) = ⅞

= -1 =>u = - 2 , y = - ^ ≠ >

u

= [sec"1 ∣ u∣ ] Z ^ = sec~1 ∣ - √ 2 ∣- sec" 1 ∣ - 2 ∣= j - f = - ⅞

83∙

X

√ 1 - ⅜ - 1)2

’ w h e r e u = 2(r - 1) and du = 2 dr

= 1X √ ⅛

= ∣sin- 1 u + C = ∣sin~1 2(r — 1) + C 85. X yp(7∑ ^ = X 7 ⅛ , where u = x — 1 and du = dx =

87∙

ta n

⅛

^1 ^

X ( ≡ ^ j√ ⅛ ∏ ) ^ 4

+

=

c

X

=

⅛

ta n

^1 ( ⅛ )

’ w h e re

⅛

u

c

= 2x - 1 and du = 2 dx

= 5 ∙ I sec- 1 15 1+ c = 7 sec- 1 I pπ∕2

+

1+ C

pl

89’ J - π∕2⅛ ⅜ n⅛

=

2 J , 1 τ ⅛ »where u = sin 0 and du = cos 0 d0; 0 = —∣=> u = —1, 0 = ∣ => u = 1

= [2 tan- 1 u] L1 = 2 (tan- 1 1 - tan- 1 ( - l ) ) = 2 β - ( - ≈)] —π 91. f

τ ⅛ 2 . where u = ex and du = ex dx; x = 0 ≠- u = 1, x = In v√ 3 => u = x/3

⅛ ⅛2x = f

Jθ

1+ e

J 1

1+ u ’

= [tan^1 u] ∕ 93∙

a

’

= I X τ ⅛ ’ w h e re

∫ ⅛

’

= tan- 1   Λ - tan- 1 1 = ^ - ^ = ^ u

= y2

and du

= 2y dy

= | sin^1 u + C = ∣sin~1 y2 ÷ C

” ■ ∕√ 7 ⅛

^

,

7

5

f - ∣√ j - a - t ι

=

∫ 7 ΓH⅛ ⅛ > = f

τ

Γ

=

6

∕ - ∣√ 4 - ( > * + a + ι )

6

∕-

=

= ∙ 1 " ' , < ' - 2> + C

^

l

√ 2 iH Γ Π 7

=

6

ts ' n

= 6 [sin- 1 (∣) —sin~1 0] = 6 (∣—0) = π 99∙

X

y 2 -2 y

+5

=

X 4 + y2 - 2 y + l

=

X 2≈ + ( y - l ) 2 = I

ta n

1

(⅛ ^ ) + ^

* (⅛

i

)] - ι

V

= -√ 2

227

228

Chapter 7 Transcendental Functions

™- Γ ⅛ + 2

8

=

Γ

w⅛÷.>

,

=

f r a ⅛

=

8

-1

(“

lι

= 8 (tan~ 1 1 - tan" 1 0) = 8 Q - 0) = 2τr ____ dx _ Γ _____ dx _ f _____ dx (x + l)∕x 2 + 2x J (x + l) ∕x 2 + 2x + 1- 1 J (x+ l)√(x+ 1)2 - 1

103.

= Jf —7=— , where u = x ÷ 1 and du = dx 2 - 1

U√ U

=

∣ u∣ + C = sec- 1 ∣ x + 1∣ + C

sec- 1

dx = f e u du, where u = sin J = e" + C = e sin ^l χ + C

1

105.

f

107.

f ⅛ ⅛ ∙ dx = f u 2 du, where u = sin J √ 1- x2 j

=^ 109∙

x and du = ∙ 7 ---x--9

√ l-x 2

∫ ( ≡

+

⅛

c≈⅛ ⅛ dy

)

+

√ l-x 2

1

x and du =

√ 1- x2

c dy

= J⅛ ⅛

= ∫ i

du

>w h e r e

u

=

ta n

^ 1 y and du =

τ⅛

= In ∣ u∣ 4 C = In jtan- 1 y ∣ + C

111.

p2 2/ -1 ∖ pπ∕3 /— I ^ - ≡ = J 2 d = j sec2 u du, where u = sec- 1 x and du =2 —7=—∙ ; x = √ 2 => 4 u = J , x = 2 => u = ⅞ j X√X - 1 —135 — 15X2 + 675 + 3x2 = 0 => x = 3 √ z5 ; α'(x) > 0 when 0 < x < 3 ^ 5 and α'(x) < 0 for there is a maximum at 3 √ z5 ft from the front of the room

x > 31/5 4

139. Yes, sin- 1 x and —cos - 1 x differ by the constant ∣ J

141. esc’ 1 u = ∣- sec’ 1 u

143. f(x) = sec x ^

j

✓

⅛ (esc’ 1 u) = ⅛ Q - sec’ 1 u) = 0 - ^

d x

1

Since the slope of sec

= -

du

lx = f -l(b )

⅛ V

x is always positive, we the right sign by writing ^ s e c

1

,∣ u ∣> 1

|u |^ 2 _ ,

---- = sec 0, and the graph of f steepens as the values of f ' increase towards the ends of the interval. The graph of f is concave down to the left of the origin where f " < 0, and concave up to the right of the origin where f " > 0. There is an inflection point at x = 0 where f " = 0 and f ' has a local minimum value.

7.8 HYPERBOLIC FUNCTIONS 1. sinhx = - ∣ =► cosh x =

coth x = S⅛ 7= -

5’ sech x = ≡

=

tanh x

5. 2 cosh (In x) = 2 ( ⅛

λ∕

(⅛ ) 2 _ 1 = ^

= ⅛ , sech x = cosh — x = 17 ⅛ , , and csch x = 8 ’ ^ )

= e ln x + ⅛ = x +

7. cosh 5x + sinh 5x = ⅛ t ^ ψ e5> - e^i, 9. (sinh x + cosh x)4 = (^⅛ ^^ ÷ ⅛ ^ )

= ⅛∣ i = - j.

5-and csch x = ⅛ = - 5

3. c o s h x = ⅛ , x > 0 =► sinh x = √ c o sh 2 x - 1 = = ⅛ , coth x = 17 ’

= ∣, tanh x = ^

= ∕Γ + ^ 1 7 = y Γ T J = ^

= e 4 =

sinh x

_ 1 = ^ ∕ ⅛ = ⅜ , tanh x = ⅛ = ⅛ 8

1

5x

(e *)4 ~ ®4x

11. (a) sinh 2x = sinh (x + x) = sinh x cosh x + cosh x sinh x= 2sinhx cosh x (b) cosh 2x = cosh (x + χ) = cosh x cosh x + sinh x sin x= cosh2x + sinh2 x 13. y = 6 sinh j ≠ ∙ ⅛ = 6 (cosh ∣) ( ∣) = 2 cosh ∣ 15. y = 2 √ t tanh √ t = 2t1∕ 2 tanh t 1∕ 2

⅛ = [sech2 (t 1∕ 2 )] (∣t~ 1∕ 2 ) (2t1∕2)

+

(t a n h

t

ι ∕2 ) ( t - ι ∕2 )

= sech2 √ t + ^

17. y7 = ln (sin h z) dz => ⅛sinh = z⅛ 7

= cothz

19. y = (sech 0)(1 - In sech 0) => ^ = ( -

Z

¾ Γ L^) (sech 0) +

( - sech 0 tanh 0)(1 - In sech 0)

= sech 0 tanh 0 —(sech 0 tanh 0)(1 —In sech 0) = (sech 0 tanh 0)[1 —(1 —In sech 0)] = (sech 0 tanh 0)(ln sech 0) 21. y = In cosh v - ∣tanh2 v => ^ = ^

- ( ∣) (2 tanh v) (sech2 v) = tanh v - (tanh v) (sech2 v)

= (tanh v) (1 —sech2 v) = (tanh v) (tanh2 v) = tanh3 v 2 2 2 23. y = (x2 + 1) sech(In x) = (x2 + 1) ( e ∣ n, + e . ∣ nκ ) = (x + 1) ( ⅛ ) = (x + 1) ( ⅛ ) = 2 x ^

⅛ = 2

= g

232

Chapter 7 Transcendental Functions ⅛ = -∣ T l = ^ >

y = sinh - 1 √ x = sinh- 1 (x 1∕ 2 )

25.

27. y = (1 - 0) tanh - 1 0 4

31. y = cos

1

x - x sech

= ^

⅛ = (1 - 0) ( τ ⅛ ) + ( - 1 ) tanh- 1 0 = ⅛ ⅛

1 ^ ∕t

29. y = (1 —t)coth

r-

1

= (l - t)coth

1

x ^

(tV2 )

= (i -

1)

'1

dt

I ⅛ 2 ζ 1

+ ( - l) c o th

l-(t1∕2)2

[x ( i ⅛ )

⅛ = -7⅛

- tanh - 1 0

+ (1) sech^1x] = 7⅛ +

(t 1∕ 2 ) = ^

coth

√⅛ ~sech^1x

= —sech- 1 x y = csch- 1 ( ∣∕

⅛ = ------K

≡

L

_ _ ln(l)-ln(2) _

33.

[

1

sg 2

— dsιnnhn ^ (tanx) 35. vy — 1

37. (a) If y = tan (b) If y =

39. If y = 5^

sin- 1

i

coth

dχ

- ^7, r = c x^

/

_

- ; ^ ^ _^ ^sec

(sinh x) + C, then ∣ =

1

"⅛

χ

2

1

+ ∣+ C, then ^ = x coth

/ 1 ∖ 20

[

∣ secx∣∣ secx∣_

x _

- ∣ ∣ se c x ∣ ∣

|sec x |

= ⅛

= sech x, which verifies the formula

(tanh x) + C, then ^ = ^ 7 p ^ = ⅛

1x

1∏2

∖ 20

= sech x, which verifies the formula

1

= x coth

1

x, which verifies

the formula 41. J sinh 2x dx = j J sinh u du, where u = 2x and du = 2 dx _ -

cosh u I p __ cosh 2x v 2 — 2

43. f 6 cosh ( ∣—In 3) dx = 12 J* cosh u du, where u = ∣- In 3 and du = ∣dx = 12 sinh u + C = 12 sinh ( ∣- In 3) ÷ C 45. { tanh ⅞ dx = 7 f ⅛ du, where u = * and du = ∣dx J 7 J cosh u ’ 7 7 = 7 In ∣ cosh u ∣ + C i = 7 In ∣ cosh * ∣+ C 1 = 7 In ∣ ≡ ⅛ ^ ∣+ C i ≈ 7 In ∣ e χ ∕ 7 + e - χ ∕ 7 ∣- 7 In 2 + C i = 7 In |ex/7 ÷ e~x /7 | ÷ C 47. J* sech2 (x — ∣) dx = J* sech2 u du, where u = (x - ∣) and du = dx = tanh u ÷ C = tanh (x — ∣) ÷ C 4$ J*

sech √ ⅛ n h √t

fa -2 ^

s e ch

u tanh u du, where u = ᅟ ∕ t = t 1∕ 2 and du = ^

= 2(—sech u) + C = —2 sech ^ t ÷ C 5

1

∙

D

°

t h

χ

d

χ

=

f

⅛

d

χ

=

X

Γ

⅛

d

u

=

l h

| u

| ]

S

8

=

i n

ι⅛

where u = sinh x, du = cosh x dx, the lower limit is sinh (In 2) = limit is sinh (In 4) =

= - ^ - =

ι -

i n

e ln2 "

~e

111

ln2

n

i n

ι⅛

∙ 1 1 =

i n

i

*

= — ^ - = ∣and the upper

1ᅟ ∕t

Section 7.8 Hyperbolic Functions 233

f"∖ e2*+ 1) d0 =

f '"' 2e ∖ f ⅛2 ^ ))

53. JJ-^l n"4,2 2eβ cosh 0 d0 = J - l n 4 = ( ^

β

v

'

[⅛ + 01 ‘°2 J -1∏4

L2

,

- I n 2) - ( s ^ - l n 4 ) = (∣-1∏2) - ( ⅛ - l n 4 ) = ⅜ - In 2 + 2 In 2 = ⅛ + In 2

2 55. J*^ 4c0sh(tan 0) sec 0 d0 = J

_

d0 = J - l n 4

_

e -e j-e j+ e

e

_ e ~ι,

l ∞ sh

u du = [sinh u] L1 = sinh(l) —s in h ( - 1) = ( e ^ 2e 1 ) — ( L J ^ ^

θ

⅛ere u = tan 0, du = sec2 d0, the lower limit is tan ( - ∣) = —1 and the upper

w

limit is tan ( j ) = 1 57. J* cosh(in t) dt = J^ cosh u du = [sinh u] θn2 = sinh (In 2) - sinh (0) = ⅛ i - A ^ —0 = ^ y 2∙ = ∣, where u = In t, du = ∣dt, the lower limit is In 1 = 0 and the upper limit is In 2 59∙

pθ

J - ιn2c o s h 2 (1)

pθ

dx

= L

2

dx

⅛H

B

pθ

s h X + 1) dx = ∣[sinh x + x]® = U .∣ n l (∞ ln2

= | [(sinh 0 + 0 ) - (sinh( - In 2) - In 2)] = 1 [(0 + 0) - ( ⅛

^

- In 2)] = ∣ - φ

+ 1∏2

= i ( l - i + ln2) = ∣+ ∣l n 2 = ∣+ l n √ 2 61∙

sinh

1

(⅛ ) = In ( - ⅛ + √ ⅜ + l ) = In (?)

6 3 .ta n h - ^ - i) = l t o ( ⅛ ⅛ = - V

1

65. sech

67. (a)

(∣) = In (

fo

1 +

) = In 3

^ 3 /5 )9 ^

1

^ ⅛ ^ — [sinh

∣] θ ^ = sinh

1

ᅟ ∕3 - sinh0 = sinh

1

y z3

(b) sinh" 1 √ 3 = In ( √ 3 + √ 3 ∏ ) = In ( √ 3 + 2)

69. (a)

f

dx = [coth- 1 x] 5∕4 = coth- 1 2 - coth- 1 ∣

⅛

(b) coth- 1 2 - coth- 1 ∣= ∣[in 3 - In ( ^ ) ] = ∣In ∣ ,3

∕ 13 1/5 1/5

r

j

J= [ - sech

∕Λ ⅜

22

x√l-16x -1

1 2 ∕ 13

J 4 ,, < ∕4 ∕5

U] 42J' 3

.

∕ d22

2’ w h e re

u√a -u2

= - sech- 1 ∣ + sech- 1 5

(b) - sech- 1 ⅛1 + sech- 1 ∣= - In ( ⅛ = _ In (

u = 4x, du = 4 dx, a = 1

1 3 + √ 1 6 9 -1 4 4 )

+

ln

≡

)

( 5 + √ 2 Γ ∏ 6 ) = ln

+ In ( ∏ (5±3) _

≡ ) b

( ^

= b

2

- In ∣

= l n ( 2 - ∣) = l n ∣ 73. (a)

f

>c°s x 2 dx = f

>1

du = [sinh- 1 u] θ = sinh- 1 0 - sinh- 1 0 = 0, where u = sin x, du = cos x dx

(b) sinh- 1 0 - sinh- 1 0 = In (θ + √ 0 + 1) - In (θ + √ 0 + 1^ = 0

75. (a) Let E(x) = ≡ t ≤ = ®

and O(x) =

f(x )-f(-x ).

= f(x). Also, E (-x ) = t i ≡ ÷ ⅛

Then E(x) + O(x) = ≡ ^ ≤

+ ^

-1 )

= ⅛ s > = E(x) ≠> E(x) is even, and

234

Chapter 7 Transcendental Functions — _ !!x⅛ J θ — _ O(x ) =^ 0 ( x ) js o dd. Consequently, f(x) can be written as

Ξ ∣Ξ H ) )

O(—x) = M

a sum of an even and an odd function. (b) f(x) = fisld^t*) because f(x) ~ f(~x) = 0 if f is even and f(x) = f∞z≤ ∣ = (10) ( - ∣) e ^ 5 = -2e~ x/5 3. y = 1 χe4x - ⅛ e4x => ⅛ = | [x (4e4x) + e4x(l)] - ⅛ (4e4x) = xe4x + ∣e4x - ∣e4x = xe4x ∣= «

5. y = ln(sin 2 0) ^

9.

= ⅛

2 cot 0

= 8-, => ⅛ = 8-'(ln 8)(-1) = —8-(ln 8)

y

11. y = 5x3 6 => ∣ = 5(3.6)x26 = 18x26 13. y = (x + 2)x+2 ≠> In y = ln(x + 2)x+2 = (x + 2) ln(x + 2) =► £ = (x + 2) ( ⅛ ) + (1) ln(x + 2) =^ ⅛ = (x + 2)x÷2 [ln(x + 2) + 1] -1 2 -1 2 1 /2 => ⅛ = 15. y = sin √ 1 —u = sin (1 - u ) d

-u _ u /1 - u 2

-1 ∕l - u

2

19. y = ttan

1t

1 x)

=≠> y, =

-(l)ln t

_

c o s ~1 x

/2 , " - 7T~7 + l I ∣ z∣ √z2 - 1 √z2 - 1

23. y = ≡ - ( ≈ 25'

f

)

, ~u = — — √ l - u 2 √ i - ( l - u 2) ∣ U∣ √ 1 -U 2

⅛ = tan

—1______

/ 1 - x2 cos- 1 x

1 1+

1( ⅛ ) - (∣) (∣) = tan" 1 1+ τ ⅛ - ±

21. y = z sec- 1 z - √ z 2 - 1 = z sec- 1 z - (z2 - 1)1/2 ≠ =

=

’ ( / 1 - x2 )

17. y = In (cos

J O j≤ llL ≥ L

s e c

g =

~1

c

z

Λ √z2 - 1

z

+

,

j g ⅛

y = ⅜ Γ ⅛ * lny = l n ( ^ ) = l

s e c

~1

= - g

π

z

^ = z(

λ

z∣ ∣ Vz

) + (sec- 1 z) (1) - ∣(z2 - l) - 1 / 2 (2z) i/

>z > 1

= - 1 .0 < « < ≈

(2) + l n ( x ≈ + l ) - l l n ( c o s 2x> ⅛ ( = 0 + ⅛ - ( l ) ¾ ⅛

≠∙ >, = ( ⅛ + ≡ 2 *)J = ^ ( ⅛ + ' " 2>)

235

236

Chapter 7 Transcendental Functions

27. y = [ j _ -

]5

⅞

lny = 5 [ ln ( t+ l) + l n ( t - l ) - l n ( t - 2 ) - l n ( t + 3)] ≠

ι ι ∣j ^ t+ lτ t- l

s j

ι t-2

L I t+ 3 ∕

5 / Γ( t + i ) ( t - i ) ^∣ ι , _ [(t-2 )(t+ 3 )J U + l τ t - l

⅛ dt

-

j

ι t-2

L I t + 31

(1 ) (⅛ ) = √ 0 ( ^ ) + ⅛0~1∕2 ln (sin 0)

29. y = (sin 0)√* => 1∏y = √ 0 In (sin 0) ∣= (s in 0 )^ (√ 0 c o t0 + ^

i

θ ) (⅛ )

)

31. J ex sin (ex ) dx = J sin u du, where u = ex and du = ex dx = —cos u + C = —cos (ex ) + C 33. ∫ e x sec2 (ex —7) dx = ∫ sec2 u du, where u = ex —7 and du = ex dx = tan u + C = tan (ex —7) ÷ C 35.

(sec2 x) etanx dx = J eu du, where u = tan x and du = sec2 x dx

J

= eu + C = etanx + C dχ = ∣∕

37. f_j ⅛

u ^u ,

7

w

^ere u = 3x - 4, du = 3 dx; x = - 1 => u = - 7 , x = 1 => u = —1

= ∣[In ∣ u∣ ]Ξ^ = ⅛[In ∣ -1 ∣ -ln ∣ -7 ∣ ] = ⅜ [0 -ln 7 ] = - V 39∙

,a n

f

(?)

dx

= f

⅛ j⅛

dx

= ~ 3 f '^ l

d u ’ w h e re u

=

cos

( t) ’ d u = - 5 s i n (5) dx; x = 0 ≠- u = 1, x = π

=> U = ∣

= - 3 [In ∣ u∣ ] i /2 = - 3 [In ∣ ∣ ∣- In ∣ 1∣ ] = - 3 in | = In 23 = In 8 41. f H ⅛ dt = f 2 Jo

t -2 5

J-2 5 u

1 du, where u = t2 —25, du — 2t dt; t = 0 4

’

’

’

u = —25, t = 4 => u = —9 ’

= [In ∣ u∣ ] Ξ‰ = In ∣ -9 ∣ - In ∣ - 2 5 ∣= In 9 - In 25 = In ⅜ 43. J*

tan(⅛v> d v

_ J

ta n u

(ju =

J

≡

du, where u = In v and du = ∣dv

= —In ∣ cos u∣ + C = —In ∣ cos (In v)∣ +C ^ ∙

f

^

3

dχ = J u^ 3 du, where u = In x and du = ∣dx

= ⅛ + C = -j(ln x Γ 2 + C 47. J ∣csc2 (1 + In r) dr =

f

csc2 u du, where u = 1 + In r and du = ∣dr

= —cot u + C = —cot (1 ÷ In r) + C 49.

J

x3χ2 dx = ∣J* 3u du, where u = x2 and du = 2x dx - 2 ⅛

51. ∫

53 ∙

7

(

3 u

) +

c

= 2 ⅛ (

3x2

) +

c

2 dx = 3 ∫ J 1 dx = 3 [In ∣ x∣ ] J = 3 (In 7 - In 1) = 3 In 7

f G + ⅛)dx = i f (I χ + 1)dχ = H I χ2 + ln lχ l] 1= M(f + ln 4 ) - (I + ln 1)] = H + Hn 4 = H + ln √ 4 = H + ln 2

Chapter 7 Practice Exercises 55.

f

χ+ 1 2 e ( ) dx =

= —[e

57.

u] J

-fι

237

e" du, where u = —(x + 1), du = —dx; x = —2 => u = 1, x = —1 => u = 0

= —(e0 —e 1 ) = e — 1

fι

er (3er + 1)^3 ^2 dr = ∣J^ u~ 3 ∕ 2 du, where u = 3er + 1, du = 3er dr; r = 0 ≠>u = 4, r = In 5 =>u = 1 6

fι

χ (1 ÷ 7 ln χ )^ 1^3 dx = ∣J^ u - 1 ∕ 3 du, where u = 1 + 7 1 n x ,d u = ^ dx,x = 1=>u = , , , S

' =- i l"-'≈l “ =- i (∣f , ', -^ l2) =(- f) U -1) =(- i) (- ⅛) =1

59.

'

61∙

Si

1, x = e => u = 8

4 M ! =⅛(< ' '-∣ ' ) =⅛ )0 - I ) = ⅞

liτ

π ^

dv

= 5

[u 3 ] “

= ∫

Un (v + 1)]2 √ ¼ dv =

f n2 u

[(l n 4 )3

13

2

du, where u = In (v + 1), du = ^

dv;

v = 1 => u = In 2, v = 3 => u = ln4; = 5

w

2V

( 8 - 1) = 5 On 2)3

θ, du

h e r e u

=

∣d0,

d u

⅛

x

=

67.

3

[sin- 1 (5)] - 32∕2 =

r^∙^^r

^

3

=

θ=

2

d x

1 => u = 0, 0 = 8 => u = In 8

;

= ~ i => u = - ∣,x = ∣ => u = ∣

[sin- 1 ( ∣) - sin’ 1 ( - ∣)] = 3 [f - ( - f )] = 3 ( f ) = π

⅛

dt = √ 3 J * ^ ⅛

du, where u = √ 3 t, du = √ 3 dt; t = - 2 ≠- u = - 2 √ 3 , t = 2 => u = 2 ᅟ ∕3

1

= √ 3 [∣™

69.

f ^⅛ π ⅛ =f = sec

7 1

∙

r 2 ∕3 J √ 2 ∕3

1

© i ⅛

=

(2 y )√ (2 y F -ι

∣ u∣ + C = sec

1

f

2∕

dy = J ^

1

,

K

⅛ =f

(√ 5) - ≡

du

⅛

-1

’ w h e re

R

u

) ]

= ⅛

= ⅛ ⅛ - ( - 5)] = ⅞

and

du =

2

dy

∣ 2y∣ + C 3

3

4

3

dy =

Γ

2

1

du, where u = 3y, du = 3 dy; y= ^

=

7 3.

[sec- 1

J

=

Jsec^1

2-

sec- 1

5/2]

=

f

-

J

=

√-2x-x2

J √ l - ( x 2 + 2x+l)

≠> u = √ 2 , y = 5 => u = 2

⅞

/ -7 = — = dx = / -7 -—- i — — dx = / - 7=—= = ≡ dx = f >1 J √l-(x⅛l)2

9

J √l-u2

du, where u = x ÷ 1 and du = dx

= sin

75

u] 2^

1

u ÷ C = sin

∙ /-2 v2 +4v + 5 d v - 2 = 2

[tan- 1

f

1

(x ÷ 1) ÷ C

1 2 2 1 + (v 2^ 4v + 4 ) dv - 2 f 2 1+ (V + 2)

u] J = 2

(tan - 1

1-

tan - 1

dv

~

2

X T⅛

du

’

where u = v ÷ 2, du = dv; v = —2 => u = 0, v = —1 => u = 1 0) = 2 (J - 0) = ∣

238

Chapter 7 Transcendental Functions ( t+ l) √ t2 + 2 t-8

/

d t

(t+ l) √ ( t 2 + 2t+ 1) - 9

f

d t

(t+ l) √ ( t + l) 2 -3 ≈

f

d t

f

u

√u2 -3

2 d

“

where u = t + 1 and du = dt = ∣sec

∣∣ ∣ + C = ∣sec

1

∣ y ∣+ C

1

79. 3* = 2 y + , ≠- In 3* = In 2y + , => y(ln 3) = (y + 1) In 2 ≠> (In 3 - In 2)y = In 2 4

81. 9e2y = x2 ≠> e 2y = y ≠> In e 2y = In ^ ^

83. ln(y — 1) = x + Iny ≠

e'n (y

l)

4

=> y = ∣In ( ^

2y(lne) = In ^

= e*, + l n y * = ex eln y ≠

87. The limit leads to the indeterminate form 2: lim

2**"'

89. The limit leads to the indeterminate form υ§: lim

2⅛ ¾∏τ e x 1

x →0

91. The limit leads to the indeterminate form 2: 0

,1

e

x →0

t

1" ⅛-

x

indeterminate form 2 : 0

97. (a)

lim ⅛

' '

x → ∞

(b) v 7

x → oo

(v c)7

= l0g3x

lim

X→ ∞

xex

lim

⅛ →∙ = ∕⅛ A )

=

X +1

lim

x → ∞

100x

=

2

lim

X→ ∞

lim H = x→ ∞

ln 3 ∕

lim 2 ⅛

H jx → ∞

=

-x

x → ∞

lim — ½ r = x +

lim

X→ ∞

=

lim

x → ∞

lim

x → ∞

l+

⅛=

r ⅛

=2 3 ≠

x

ln 2

lim ⅛e →Q

=

lim ^ - ⅛t ^ = —oo 2 t → 0+

- =

ιn f(x )

) = e

t

lim ( M ∣= lim τ = l t / → o+ t → 0+ 1

=

lim

∏m

! s ll⅛ 2 1 j the limit leads to the

(1 + 7 ) x =

X→ ∞

'

x∕

lim

el"w = e3

x → ∞

In 2

lim

1 = 1 ≠> same rate

x → ∞

⅛ = oo => faster 100

im00 ⅛ (d) x l→ tan 71 x = ∞ => faster => same rate

(e) (f)

99. (a) (b) (c)

lim

x → ∞

⅛ xa =

e

lim

χ → oo

⅛

^

2ex

=

lim

χ → ∞

⅛ ^ = 1 => same rate 2

2

⅛ ⅛ ^ = 1 + ⅜ ≤ 2 for x sufficiently large => true (?/■ ? ) = χ 2 ⅛ 1 > M for any positive integer M whenever x > √ M => false lim

~⅛~ =

x → ∞ x + l∏x

lim

x → ∞

-1+

= In ∣ ∣ ∣= In ∣ x ∣- 1∏3

= In 10

r ? ≠ ∙ same rate

In 2

2x

β

lim ¾x ⅛ = →0 ≡ -ι

t

iim

1

x→0

93. The limit leads to the indeterminate form 2: lim ( 7 - 7 ^ = t7 ° t → 0+ V 95. Let f(x) = (1 + ∣Γ => In f(x) = ⅛ ⅛ L !1 4

⅛

lim

1

"

lim t → 0+

= In ^

y — 1 = yex => y —yex = 1 => y (1 —e x ) = 1

85. The limit leads to the indeterminate form 2: lim 1 ^ -1 = ijm 0 x →0 x χ →0

υ

(in ∣) y = In 2 => y = Λ ∣

= 1 => the same ® growth rate => false

Chapter 7 Practice Exercises

ylnx

(vd)' χ lim →∞

101.

= χ lim →∞ ⅛ lnx = 0 4

°grows slower ≠> true

≤ ∣for all x ≠- true

(e)

e

(f)

c2 ⅛s = 1 (1 + e - 2x) ≤ | (1 + 1) = 1 if x > 0 => true

df dx

ψ

i

— X→∞ lim ⅛

= eκ + 1

x = f(ln2)

1 (ex + ‰

2

_ 1 " 2 + l

_

1 3

103. y = x In 2x - x => y' = x ( ∣) + In (2x) - 1 = In 2x; solving y' = 0 => x = ∣; y' > 0 for x > ∣and y' < 0 for x < ∣ => relative minimum of - ∣a tx = ∣; f ( ⅛ ) = - ∣ and f ( ∣) = 0 => absolute minimum is —∣at x = ^ and the absolute maximum is 0 at x = ∣

105. A = J

ι

^

δ

2u du = [u2 ] J = 1, where

dx =

u = In x and du = ^ dx; x = 1 => u = 0, x = e => u = 1 107. y = h x ⅛ g = b g

109. A = xy = xe-χ2 =^ j

= g⅞

=> M

* ) ^ = ⅛

≠

⅛

= * " ∕≈

= e - ’’ + (x )(-2 x )e ^ χ2 = e^χ2 (1 - 2x2 ) . Solving ^ = 0 ≠

=> x = ∙ ^ ; ^ < 0 f o r x > - ^ and ^ > 0 f o r 0 < x < ^ x = - ^ units long by y = e - 1 ∕ 2 = ^

4- absolute maximum of ^ e - 1 ∕ 2 = ^

units high.

111. K = ln(5x) - ln(3x) = In 5 + In x - In 3 - In x = In 5 - In 3 = In j

; 1°£4 x ** log2 x

1111 5

∙ w

=

∖l n 4 ∕ ∕i∏χ∖ V "2∕

vy _— In x

=

In 2 _ In x

1∏χ In 4

κ y

_-

1

In 2 _ In 4

In 2 _ 2 In 2

~ 2I^n7x2 -_

1 2

2 - In x

=> y" = _ ∣X ~5∕ 2 (2 - In x) - I x - 5 / 2 = x“ 5 /2 ( ∣In x - 2 ); solving y' = 0 => In x = 2 => x = e2 ; y, < 0 for x > e 2 and and y' > 0 for x < e 2 ≠> a maximum of ∣; y" = 0 = > l n x = ∣ = > x = e 8 / 3 ; the curve is concave down on (θ, e8 ∕ 3 ) and concave up on (e8 ∕ 3 , ∞ ) ; so there is an inflection point at (e8 ∕ 3 , ^ ) .

1 —2x2 = 0 at

239

Chapter 7 Transcendental Functions

240

χ2

(b) y = e

χ2

=> y' = —2xe

=> y" = —2e

χ2

χ2

+ 4x 2 e

= (4X2 —2)e - x 2 ; solving y' = 0 => x = 0; y, < 0 for x > 0 and y' > 0 for x < 0 => a maximum at x = 0 of e 0 = 1; there are points of inflection at x = ± - ^ ; the curve is concave down for — ¼ < x < √2

√2

and concave

up otherwise. (c) y = (1 + x) e^x => y' = e^x - (1 + x) e~x = -x e ^ x => y" = —e -x + xe~x = (x — l)e~ x ; solving y' = 0 => —xe^x = 0 => x = 0; y' < 0 for x > 0 and y' > 0 for x < 0 => a maximum at x = 0 of (1 + 0) e0 = 1; there is a point of inflection at x = 1 and the curve is concave up for x > 1 and concave down for x < 1.

117. Since the half life is 5700 years and A(t) = Aθekt we have ^ = Aθe5700k => ∣= e5700k => In (0.5) = 5700k => k = ⅛

∙ With 10% of the original carbon-14 remaining we have 0.1A0 = A 0 e ≡

=> In (0.1) = ^

119.

0=

7r

σ

7r

_

t => t = ^ ^ ^ ^

(A ) _

c o t-ι

(5 _ .X ) 0 < x < 5 0 ≠ > ^ =

t

U oJ

w n

X3

l

30Λ

υ

x

⅛ L + . .( ” ) ,

dx

ι

+

( i√

+

, , ∕5 o - > √ 1

K 60∕

-

= 3θ [ δ ⅛

; s o lv i n g

3t>a + (so- x)a ]

=> 0.1 = e ⅛ ≡ t

≈ 18,935 years (rounded to the nearest year).

c o t-l

°

it

dx = θ ≠ '

χ2

+ (

3θ

J

—200x + 3200 = 0 => x = 100 ± 20√T7, but

100 + 2 0 √ 1 7 is not in the domain; ^ > 0 for x < 20 (5 - √ ∏ ) and ^ < 0 for 20 (5 - √ ∏ ) < x < 50 => x = 20 f 5 — ^ 1 7 ) ≈ 17.54 m maximizes θ

CHAPTER 7 ADDITIONAL AND ADVANCED EXERCISES ’b

_

b

J

A

20

o √ l-x

_

3. y = (cos

dx =

v

7

b → 1“

7

= - ∣ lim 2

χ

=

⅜χ - 1∕ 2

x → θ+

lim

l

b → 1“

- 1 b - sin - 1 0) = (sin v ,

In y = ∣In (cos λ∕x ) and lim

Λ∕ X) 1A ^

v

fsin- 1 xlj o =

lim

×

v

^m

“ 2 2

x

→

⅛ ⅛ J Λ1

χ → o+

/

0

+

(v c o s V ^ 7) 1/X v

x

=

=

e ~1/2

∏m

x → 0+

T⅛

7. A(t) = J ⅛

x

= £

(a)

lim

t →∞ lim

(b)z

t →∞

(c)

lim

'

t →o+

=

ll n

(sin - 1 b —0)7 =

v

-7 ⅛ ^= = y 2

v

x co s

v

lim

( - y + i^

d + x)] J = In 2

A(t) = lim

t →∞

A®

2, )

( l - e →) = l

= lim

t →∞

= lim+

t →0

f ( l- e - ') ( l+ e - ) ( l - e - ' ) -------

2

sin- 1 b = ⅞ 2

lim

⅛ ∕x

x → 0+

v

x

√e

dx = [ - e x]θ = 1 - e^, , V(t) = π f o e- 2x dx = [ - f e~2x] θ = f (1 - e

A

x

lim

b → l^

~ ~T

which can be interpreted as a Riemann sum with partitioning ∆ x = £ => dx

lim

b→l^

lim f (1 + e^') = π t →0+ 2

÷ ∙ ∙ ÷ ⅛)

Chapter 7 Practice Exercises

241

»■⅛=f≡jr * = ⅛ I⅛ '> = ∣⅛ ]' =∣⅛^.√⅛'i>≈∙⅛∫⅛⅛ [ (W K _ _ ]_ [ 2 In 2 J 1 - 2 In 2

=> AH A2 =

2 :1

11. In χ( χ x ) = x x In x and In (x x ) x = x In x x = x 2 In x; then, x x In x = x 2 In x => (x x — x 2 )ln x = 0 => x x = x 2 o r In x = 0. In x = 0 => x = 1; χ x = χ 2 => x In x = 2 In x ≠> x = 2. T herefore, x^χx ^ = (x x ) x w hen x = 2 or x = 1. 13. f(x) = e g W => f'(x ) = e g W g '(x), w here g'(x) =

τ⅛

≠> f'(2 ) = e 0 ( r ⅛ ) = ⅛

15. Triangle A B D is an isosceles right triangle w ith its right angle at B and an angle o f m easure J at A. W e therefore have ∣= Z D AB = Z D A E + Z C A B = tan - 1 ∣+ tan - 1 j .

17. The area o f the shaded region is

JQsin

-1

sin - 1 y dy, w hich is the sam e as the area o f the region to

x dx =

the left o f the curve y = sin x (and part o f the rectangle form ed by the coordinate axes and dashed lines y = 1, 1

x = P . The area o f the rectangle is ^ = Jθ sin sin - 1

⅞= f 2 Jo

x dx + f , Jo

/

sin x dx => Jo

y dy + Jθ

sin x dx, so w e have

sin x d x = 5 - f sin - 1 x dx. 2 Jo

19. (a) g(x) + h(x) = 0 => g(x) = - h ( x ) ; also g(x) + h(x) = 0 => g ( - x ) ÷ h ( - x ) = 0 => g(x) - h(x) = 0 => g(x) = h(x); therefore —h(x) = h(x) => h(x) = 0 => g(x) = 0 f(x) + f(-x) _ [fE(x) + fp(x)] + [⅛(~x) ÷ fp(-x)] _ fε (x) + f0 (x) + fE(x) ~ fp(x) _ f ∕ γ   . (b) 2 2 2 f(x) - f(-x) _ ⅛(x) + fo (x)] ~ [fε (-x) + fp(-x)] _ fε (x) + fp(x) ~ fε (x) + fp(x) _ f 0 ( χ ) (c) Part b => such a decom position is unique. 21

∙

m

= ∫

= ^

0

τ ⅛

= T

dx

=

2

ta n ∣

l χ

]o = f

a n d M

y = X 'τ ⅛

= tl n ( 1 +

χ2

)]o =

ln 2

^

x

= ⅛

17 = ° ⅛ sym m etry

23. A(t) = A 0 e r t j A (t) = 2A 0 ≠> 2A 0 = A 0 e π ^

25.

dx

(a) L ≈ k ( ⅛ ≤ ^ + ^

)

e

n

= 2 = > rt = l π 2 ≠ > t = ^

≠> ‘ ≈ 7 = ≡

= ⅛ = k ( ⅛ ^ - i ι ≡ ^ ) ! s o l v i n g ⅛ = 0

=> r 4 b csc 2 θ — bR 4 esc 0 co t 0 = 0 => (b esc 0) (r 4 esc 0 — R 4 cot 0) = 0; b u t b esc 0 ≠ 0 since 0 ≠ 5 => r 4 esc 0 — R 4 cot 0 = 0 => cos 0 = ^ (b) 0 = c o s^ 1 ( ∣) 4 ≈ c o s " 1 (0.48225) ≈ 6 1 0

≠> 0 = cos

1

f ^ j , the critical value o f 0

242

Chapter 7 Transcendental Functions

NOTES:

CHAPTER 8 TECHNIQUES OF INTEGRATION 8.1 BASIC INTEGRATION FORMULAS 1∙

∕^

i ⅛

[ ^

8

^

→ ⅛ = 2 √ ^ + c =

u = sin v du = cos v dv

3.

5.

Γ 1 16x dx . 0 8x2 + 2 ’

2

√ 8 ^ τ + c

→ J 3 v 4 i du = 3 ∙ ∣u 3 ∕ 2 + C = 2(sin v) 3 ∕ 2 + C

u = 8x 2 ÷ 2 du = 16x dx x = 0 => u = 2, x = 1 => u = 10

→ ∫ 2 ' 0 ⅛ = [ln ∣ u∣ ]^ o = i n l θ - l n 2 = ln 5

u = χ ∕x + 1 7∙

du

J√Γ(√⅛Γr

= ⅛

dx

= 21n∣ u ∣+ C = 2 1 n ( √ x + 1 ) + C

→ J ⅛

2du= ⅛

9.

j cot (3 — 7x) dx;

11. f e

θ

u = 3 - 7x du = - 7 dx

esc (e^ + 1) d0;

^θ d ^

csc u ÷ cot u ∣÷ C = - In ∣ csc (e 0 ÷ 1) + cot (e 0 + 1) ∣÷ C → J esc u du = - In ∣

→ ^ 3 sec u du = 3 In ∣ sec u + tan u ∣ sec ∣+ tan ∣ + C = 3 In ∣ ∣+ C

13.

15.

du =

sin u ∣+ C = — ∣In ∣ → - ∣J*cot u du = - ∣In ∣ sin (3 — 7x )∣+ C

/ u = s- 7 I esc (s - 7r) ds; . J du = ds j

csc u ÷ cot u ∣ ÷ C = —In ∣ csc (s — π) ÷ cot (s - π ) ∣+ C ^ c s c u du = - In ∣

u = x2 du = 2x dx

17.

•In 2

£2

In 2 => u = In 2

19.

21.

jta n v sec 2 v dv;

u — tan v du = sec 2 v dv

du = dx u = √ w

23.

25.

d u

= ⅛ .

x = 3u dx = 3 du

= 3 tan

1

x + C = 3 tan

1

3u + C

=

e

∣ -2 _

e

° = 2- 1= 1

Chapter 8 Techniques of Integration

244 l/6

f

2

√ l-9 x

Jo

u = 3x du = 3 dx x = 0 => u = 0, x = IO => u = I2 u = s2 = sin 1 u + C — sin du = 2s ds

dx

29.

7

31. ∕ ⅛

∕^

=

⅛

= f ∙

5

r*eπz3

1

= tan

u = In x du = ⅜ x = 1 => u = 0, x = eπz3

J

dx . x cos (In x) ’

1 s2

+ C

∞ - W + C = 6sec- ∣ 5x∣+ C

u = ex du = ex dx

35.

/2 = [∣sin 1u] Q 5(I - 0) ~ ⅛

1

u + C = tan

→f ^ u

-

^

1

=

ex + C

X

3

] J/3 sec u du = [In ∣ sec u + tan u ∣

3-

sec 0 + tan 0 ∣= In (2 + √ 3 ) - ln (l) = In (2 + √ 3 ) = In ∣ sec j ÷ tan ∣ ∣- In ∣

37.

u = x —1 du = dx x = l => u = 0, x = 2 => u = l

__ 0 Γ 2 ∕* 2 8 dx dx J l x2 - 2 x + 2 “ 0 J 1 l ÷ ( x - l ) 2

= 8 (tan

1

1 - tan

1

0) = 8 ( ∣- θ) = 2π

39

J τ ⅛

41

j l.÷ι>√,∙÷i . ° J < ⅛ l⅛

3

SX, T ⅛ = 8 ∣ W ~1 ⅛

-

= l ≠

h

■ [ “du 1 Λ ] -

/ 7 Γ ⅛ = -i " ' 1 “ +

- ∙1

1]

∙ [⅛ ^ ⅛

c

=

s i"

" « -

2 >+ c

x + l ∣+ C . → ∫ ^ ⅛ = s e c - '∣ u∣ + C = se c - 1 ∣

∣ u ∣= ∣ x + l ∣> l 43. f (sec x + cot x)2 dx = J (sec2 x + 2 sec x cot x + cot2 x) dx = f sec2 x dx + f 2 esc x dx + ∫ (esc2 x — 1) dx csc x ÷ cot x ∣—cot x —x + C = tan x —2 In ∣ 45. f esc x sin 3x dx = f (esc x)(sin 2x cos x ÷ sin x cos 2x) dx = J (esc x) (2 sin x cos2 x + sin x cos 2x) dx = ∫ ( 2 cos2 x ÷ cos 2x) dx = J [(1 + cos 2x) + cos 2x] dx = J ( l + 2 cos 2x) dx = x + sin 2x + C

∙∕⅛-dx =∕( 1 - ⅛ ) dx =x - ln ∣x +1∣+c

47

49. X√5 ⅛ f

51∙

dx =

J ⅛ ⅞ ^

dt

J v^ ( 2 x

+

⅛ f)

1

1⅞

lχ 2

+ ln

χ2 ∣

-

1∣ 1√2

= (9 + In 8) - (2 + In 1) = 7 + In 8

= ∫ [ ( 4 t - l ) + ⅛ ] d t = 2t2 - t + 2 ta n - 1 ( i ) + C

53∙ S÷^^=!77ħ¾~ S 55. X

dx =

x dx =

τ ^ = ^ i *+ 'fiτ^^ +c

∕2 ^ —(0 + 1) = √ z2 X / (s e c 2 x + sec x tan x) dx = [tan x + sec x [ ^ 4 = (1 + ᅟ

Section 8.1 Basic Integration Formulas dx = [ ⅛ τ2^ d x =x f (sec2 x - sec x tan x) dx = tan x - sec x + C 57. fτ∏⅛— = f π -^⅞⅜ J 1+ smx J (l-sm2 x) J cos x J ' 59∙ ∫ ⅛

61∙

= ∫ I = ∣(e* sin 0 —e 0 cos 0) + C, where C = y is another arbitrary constant 23. I = J e2x cos 3x dx; [u = cos 3x; du = —3 sin 3x dx, dv = e2x dx; v = ^ e 2x] => I = ∣e2x cos 3x + ∣J*e2x sin 3x dx; [u = sin 3x, du = 3 cos 3x, dv = e2x dx; v = ∣e 2x] => I = ∣e2x cos 3x + ∣θ e2x sin 3x - ∣J*e2x cos 3x dx^ = ∣e2x cos 3x + ∣e 2x sin 3x — 1 1 + C' 4

25.

j*

e

y I = ∣e2x cos 3x + ∣e2x sin 3x ÷ C' => ^ (3 sin 3x + 2 cos 3x) + C, where C = ⅛ C'

√3s+9 J

s

.

^ 3 s ÷ 9 = x2 ' → J ex ∙ ∣x dx = ∣J xex dx; [u = x, du = dx; dv = ex dx, v = ex] ; ds = ∣x dx

∣J xe x dx = j ^xex —tf ex dx^ = j (xex - ex ) + C = j ^ 3 s + 9 e v^

27. u = x, du = dx; dv = tan2 x dx, v = f tan 2 x dx = f ≡ ⅛ dx = f ¼ ⅛ J

= tan x —x; Jθ

J

x tan2 x dx = [x(tan x —x)] 0 ' —Jθ

COS z X

J

COS z X

- e ^ ^ ÷ C

a

dx = f -⅛ -----f dx J

COS z X

(tan x —x) dx = l ( ^ - j )

+

J

[ ln ∣ cosx∣ + ⅛]

π ∕3 0

Section 8.2 Integration by Parts -

”

Ξ

∙3

3

_ zΛ 4. iιnn 1 4. zd — i v l _ 111 ιn 2 3y

2

18

3

18

u = In x → J (sin u) eu du. From Exercise 21, /( s in u) eu du = eu ( ≡3L-∞su) _|_ ^ du = ∣dx dx = eu du = ∣[-x cos (In x) + x sin (In x)] + C

29. J sin (In x) dx;

31. (a) u = x, du = dx; dv = sin x dx, v = —cos x; pπ

pπ

Si = J o x sin x dx = [—x cos x]θ + Jθ cos x dx = π ÷ [sin x]θ

=

7Γ

*2π

J x sin x dx = — [—x cos x],* + J^ cos x dx

(b)

π

c^π

«

p3π

= - [—3π + [sin x]^] = 3π ∏

(c) S3 = J 2 x sin x dx = [—x cos x ]^ + J 2π cos x dx = 5π ÷ [sin x ]^ = 5π *(n+l)π

x sin x dx = ( - l ) [[-x cos x]] J [-(n + l ) π ( - l ) + n π ( - l) ] + 0 = (2n + l)π n+1

l

1,

nπ

= ( - l ) n+,

n+1

n

p in 2

p in 2

p in 2

33. V = J o 2π(ln 2 - x) ex dx = 2π In 2 JQ ex dx —2πJθ xex dx = (2π in 2) [e^ n2 - 2π ( [x e ^ n2 - f ‰

dx)

= 2π In 2 - 2π (2 In 2 - [ex]θ 2 ) = - 2 π In 2 + 2π = 2π(l - In 2)

I

f√ 2

π ∕2

f π∕ 2

35. (a) V = J o 2πx cos x dx = 2π I [x sin x] 0 z - Jθ sin x dx = 2π ( f + [cos X] Q/ 2 ) = 2π ( f + 0 - 1) = π(π - 2)

(b) V = J o 2π (∣- x) cos x dx; u = ^ —x, du = —dx; dv = cos x dx, v = sin x; V = 2π [(j —x) sin x] θ^2 + 2 π ∕

37. (a) av(y) = ⅛ ∫

sin x dx = 0 + 2 π [- cos x]θ^2 = 2π(0 ÷ 1) = 2π

(b)

2e-' cos I dt

0

≈ l [ e - . ( * t m )]J∙ (see Exercise 22) => av(y) = ⅛ (1 ~ e^2π)

39. I = J xn cos x dx; [u = xn , du = nxn => I =

xn sin

x- /

nxn ^ 1 sin

1

dx; dv = cos x dx, v = sin x]

x dx

41. I = / xn eax dx; [u = xn , du = nxn ~1 dx; dv = eax dx, v = ∣ eax ] => 1

=

χ¾ a

ax

- ? f xn ~1eax dx, a ≠ 0 aJ

'

249

Chapter 8 Techniques of Integration

250

43. J sin

1x

45. J sec

1

dx = x sin

1x

x dx = x sec

1

J

— sin y dy = x sin

1x

x —J sec y dy = x sec

1

1x

- In ∣ sec (sec

47. Yes, cos

1x

is the angle whose cosine is x which implies sin (cos

49. (a)

+ tan (sec

÷ cos (sin

1

x) + C

x —in ∣ ÷ C sec y + tan y∣

= x sec

1 x)

1x

÷ cos y + C = x sin

1 x)∣ +

C = x sec- 1 x - In x + √ x 2 - 11+ C

f sinh- 1 x dx = x sinh- 1 x —J*sinh y dy = x sinh- 1 x -

1 x)

= / 1 - x2 .

cosh y ÷ C = x sinh- 1 x —cosh (sinh- 1 x) + C;

check: d [x sinh- 1 x - cosh (sinh- 1 χ) + C] = Jsinh"1 x + ^ x -

2

- sinh (sinh- 1 x) ^ = ] dx

= sinh- 1 x dx (b) J*sinh- 1 x dx = x sinh- 1 x —J* x ( y = ^ ^ dx = x sinh- 1 x - ∣J ( l ÷ x2 )^ 1^2 2x dx = x sinh- 1 x - (1 + x2 ) 1^2 + C check: d [x sinh- 1 x - (1 + x2 ) 1^2 + c] = [sinh- 1 x + -7===^ — / x - J dx = sinh- 1 x dx

8.3 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 1∙

( ⅛ ¾

= Γ⅛ + ⅛

=> 5 x - 1 3 = A (x - 2 ) + B ( x - 3 ) = (A + B )x -(2 A + 3B)

^ 2A + 3B = 513 } ^

3. (⅛ ⅜ = r ⅜ + ( ⅛ d l u s > (x + l4)i

5∙

~ x+T

+

- B = (10 - 13) ^

=> (x

4 = A(X + 1 ) + B = AX + (A + B) ^

h

i⅛

=

1

+ i ⅛ ⅜ 6 (a f t e r

lo n δ

= -2 ^

B = -1 2 ^

B

!

4

} ^

A = l a n d B = 3;

division); ⅛ ⅜ 6 = (Γ ⅞ ⅛ ⅛ = Γ⅛ + ⅛

A = 1 7 ! th u s ,⅛ ⅛ = l + ⅛

Λ

+ BO R -

3

=^ - B = (10 + 2 ) = 1 2

+ ⅛∣

T ⅛ = τ ⅛ + TT7 => 1 = A(l + x) + B(l - x); x = 1 => A = ⅜; x = —1 ≠> B = ∣;

9∙

∕τ ⅛ n

^

+ ⅛

C = 2 jth u s ,⅛ t⅛ = ⅛ + ⅛ + ⅛

≠> 5t + 2 = A ( t - 2 ) + B ( t - 3 ) = (A + B)t + ( - 2 A - 3 B ) ≠> ^

A

= ⅛

⅛

Z + 1 = A z (z -1 ) + B (Z -1 ) + Cz2 => z + 1 = (A + C)z2 + ( -A + B)z - B

A+ C= 0 ) ≠> - A + B = l l = > B = - l ^ A -B = 1 J 7.

A ≈ 2; thus, ⅛

+ 1)2

= 7 ÷ ⅞ + ⅛Γ ^

⅛ ⅛

X+

B= 3 ^

= i∫τ ¾

∙ 7⅛ ⅛ ∫ ⅛

13.

7

^

+ l ∫⅛

= 7⅛ + ⅛

⅛

5

dx

^

]+ c = ∣M i + x ∣- i n ∣ ι-χ ∣ X+

4 = A(X - 1 ) + B(X + 6); X = 1 ≠> B = f ; x = - 6 ≠> A = ∈j = f ;

= 7 ∫ 7 ⅜ + 7J ⅛

+ 5 In ∣ = ∣In ∣ x + 6∣ x - 1∣ + C = i In ∣ (x + 6)2 (x - 1)5 ∣ +C

= 7 ⅜ + 7 ⅜τ => y - A ( y + l ) + B ( y - 3 );y = - l => B = ⅛ = ∣jy = 3 => A = | ;

Γ ^ ^ ^ H

∫ ⅛

= ∣ln5 + i l n 3 = l⅛5

y+ < H iin 5 + ^ + iX ⅛ H i⅛ ∣ y -3 ∣ + ∣i n ∣

^

Section 8.3 Integration of Rational Functions by Partial Fractions 15∙

⅛

i = f+ ⅛ + ⅛ B

l = A(t + 2 ) ( t - l ) + B t ( t - l ) + Ct(t + 2 );t = O ^

^

= l ; t = l => C = l √

p^

= - ∣∫ M

s

∫ A

+

251

A = - |; t = - 2

1 ∫A

= - 1 In ∣ t∣ + | In ∣ t + 2∣ + ∣In ∣ t - 11+ C

17∙

,⅛⅞÷ι =

(x - 2) + ⅛ ⅛⅛ (after long division); < ⅜ ⅛ = τ ¼ + o ⅛

= Ax + (A + B) => A = 3 ,A + B = 2 =+ A = 3,B = - 1 ; ∫ ' ⅛

^

3x + 2 = A(x + 1) + B

τ

= 7 > - 2 ) ⅛ + 3 ∫ 0⅛ - ∫ . ' < d ⅛ = [ ⅝ - 2 x + 3 ∣ " ∣ x + 1l+ ⅛ ] ' = ( | - 2 + 3 In 2 + ∣) - (1) = 3 In 2 - 2 19. ( ⅛ 7 = m

+ ⅛

+ (d ⅛ + ( ⅛

^

1 = A ( X + 1 )( X - 1 ) 2 + B(X - 1 ) ( X + 1 ) 2 + C ( X - 1 ) 2 + D ( X + 1 ) 2 ∙,

x = —1 =+ C = | ; x = 1 =+ D = ∣; coefficient of x3 = A + B =+ A + B = 0; constant = A —B + C + D = + A - B + C + D = l = + A - B = |; t h u s , A = i =+ B = - | ; f j ⅛ — i f dx _ 4 jx + l

21. (Γ Π )⅛ + i)

=

dx _ 1 ι n I x ÷ 1 I ______ x _ 1 f dx ∣ 1 f _ d x _ ∣1 Γ ι C 4 J x - 1 ^t ^ 4 J ( x + 1 ) 2 ^r 4 J ( x - 1 ) 2 ~ 4 111 l x - 1 I 2(x 2 - l ) ^r v

7+7 + 7 + f ^

1 = A (x2 + 1) + (Bx + C)(x + l) ;x = - l =+ A = ∣; coefficient of x2

= A + B =+ A + B = 0 =+ B = - ∣; constant = A + C =+ A + C = l =+ C = ∣; J^ (Γ ∏ ⅛ r∏ ) = l ∫ 0⅛

+ I ∕o '⅛ Γ

1 dx

= ⅛

ln

x + ∣

1∣ -

5 l n (χ 2 + 1) + I ta n - 1 x] J

= (∣In 2 - | In 2 + i tan^ 1 1) - ( ∣In 1 - ∣In 1 + ∣tan" 1 0) = ∣In 2 + ∣( f ) = ^ ± ⅜ ! ^ 23. ^

^

= ^ 1

=> y2 + 2 y + l = (Ay + B )(y 2 + l) + Cy + D

+ ( ^

= Ay3 + By2 + (A + C)y + (B + D) =+ A = 0, B = 1; A + C = 2 =+ C = 2; B + D = 1 =+ D = 0; J ⅛ ⅞ ⅛ i d y = J Λ τ d y + 2 J ^ ⅛ d y = ta n ^1 y ^ ⅛ + c

25.

2s + 2 _ (s2 + l ) ( s - 1)3 -

D As + B ∣ C I ∣ E s2 + l ψ s - 1 τ (s - 1)2 ^r (s - 1)3

9 ^∙≈> ^Γ +

= (As + B ) ( s - l) 3 + C (s 2 + l ) ( s - l) 2 + D (s 2 + l ) ( s - 1) + E (s 2 + 1) = [As4 + (- 3 A + B)s3 + (3A - 3B)s2 + ( - A + 3B)s - B] + C (s4 - 2s3 + 2s2 - 2s + 1) + D (s3 - s2 + s - 1) + E (s 2 + 1) = (A + C)s4 + (- 3 A + B - 2C + D)S3 + (3A - 3B + 2C - D + E)s2 + ( - A + 3B - 2C + D)s + ( - B + C - D + E) A + C =o , -3 A + B - 2 C + D =0 3A - 3B + 2C - D + E = 0 ►summing all equations =+ 2E = 4 =+ E = 2; = 2 - A + 3B - 2C + D -B + C -D + E = 2 summing eqs (2) and (3) =+ —2B + 2 = 0 =+ B = 1; summing eqs (3) and (4) =+ 2A + 2 = 2 =+ A = 0; C = 0 from eq (1); then —1 + 0 —D + 2 = 2 from eq (5) =+ D = —1; / ( s Π ⅜ ( ⅛ T j 5 d s = ∕ s 5 + 7 - ∕ * ( Γ = ⅛ + 2 J * ( S^ ¾ 5 = - ( s ~ 1 ) 2 + ( s ~ 1 ) 1 + tan 1 s + C

27∙

⅞ ¾ y =≡ ¾

+

(*2+¾⅛

^

203 + 502 + 80 + 4 = (A0 + B )(0 2 + 2fl + 2) + C0 + D

= A03 + (2A + B)02 + (2A + 2B + C)0 + (2B + D) + A = 2; 2A + B = 5 ≠> B = 1; 2A + 2B + C = 8 + 2B + D = 4 ≠ D = 2 √ ≡ ⅞ ⅛ ^ d β √ t s ⅛ ‰ _ -

f J

20 + 2 02 + 20 + 2

_ u σ

Γ

J

d0 , f d(0 2 + 20 + 2) _ 02 + 20 + 2 ^r J (02 + 20 + 2)i ~

J

f d(0 2 + 20 + 2) _ f _ d 0 _________1 02 + 20 + 2 02 + 20 + 2 J (0 + 1)4 + 1

C = 2;

252

Chapter 8 Techniques of Integration = ^τ⅞+2 + in (02 + 20 + 2) - tan~1 (0 + 1) + C

29.

2⅛

= 2 x + ⅛

^

= 2X + 5^

x = 1 =+ B = 1; J* - ⅛ ⅛ ^ t i

31. ⅛ ⅜

9

~

+ ⅛ Γ⅛

1

5

= *+ ⅛

=+ l = A ( x - l ) + Bx;x = 0 =+ A = - l ;

i^ i∣ = J^2x dx - J ^ + f ^ - = x2 - In ∣ x∣ + In ∣ x - 1∣ + C = x2 + In ∣ +C

(a f t e r

i

J⅛

l o n ≡ division); g ¾

1

⅜

= x+ ⅜+ ⅛

=+ 9x2 - 3x + 1 = Ax(x - 1 ) + B(x - 1 ) + Cx2 ; x = 1 =+ C = 7; x = 0 =+ B = - 1 ; A + C = 9 =+ A = 2; ¾ ⅛ 1 dx = ∫ 9 d x + 2 f ⅛ - f ⅛ + 1 ∫ ⅛ = 9 x + 2 1 n ∣ x∣ x -l∣ +C + i+ 7 1 n ∣ 33. ⅛

^

= y - 7^

J

⅛ j = f + ≡ ⅛ f =+ l ≈ A ( y 2 + l) + (By + C)y = (A + B)y2 + Cy + A

π

1

=> A = 1 ; A + B = O =+ B = - 1 ;C = O; ∫ ⅛ ⅛

dy = ∫

y

d y -∫⅛ + ∫ ⅛

= £ - In ∣ y∣ + ∣In (1 + y2 ) + C 35. ∫ ⅛

=

k

= >

l ∫ ⅛

= ∫

i

= ∣ ∙ ∣ ^

√ ⅛

¾

+ C = b (⅛ )+ C ∣

j⅛ " ¾ _„ ; ⅛⅛ y = t. cos y dy - a ) → ∫ ix ⅛ 2 = ∣J ( ⅛ - ⅛ ) ⅛ = 1 In ∣ ⅛∣ ∣+ C

3’ - ∫

=M≡+i∣+c 39.

f (x -

J

2)2 tan

(4X2

1

+

f tan~ (2x) , θx ~ J 7 x ^ 1 " θ

(2x) - 12x3 - 3x J

1)(X -2 )2

1

x

~

o f 3 Jθ

= ∣∫ t a n - 1 (2x)d(tan- 1 ( 2 x ) ) - 3 ∫ ⅛ - 6 ∫ ⅛ 5

41. ( t2 - 3 t + 2 )⅛ = U x √ ^ 5 =

c

= ^ ⅛ 7 = 2e

iτ ^ x

43. (t ≈ + 2t )⅛ = 2x + 2 l ∣∫ ⅛

^

=

√ ⅛

- ∫ ⅛

,

x Γ ^

θx

= 1⅛

^

~ 3 In ∣ x -2 ∣ + ⅛ + C

= ln ∣ ≥4∣ + C j H 4 = C e M = 3 andx = 0

x = l∏∣ t-2 ∣ -ln ∣ t- l∣ 2 ( ^ ) ∣= l n ∣ + ln 2 = ∫ ⅛

=. I l n ∣ χ + 1∣ = 1 ∫S .1 J Λ

^

In ∣ x + 1∣= In ∣ ⅛ ∣+ C;

t = 1 and x = l => In 2 = In ∣+ C => C = In 2 + In 3 = In 6 =+ In ∣ x + 11= In 6 ∣ ⅛ ∣=+ x + 1 = ^ => x = ⅛ I τ

4 5∙

V= π ∫

A

25 2 o5 y

- 1, t > 0 dx = π / “ ⅛

dx = 3π ( f™ ( - ⅛

+

1) ) d χ = [3π In ⅛

∣ ] ^ = 3π In 25

Section 8.4 Trigonometric Integrals 49. (a) ⅛ = k x ( N - x ) ^ k

-

=>

1000, t —0 and x —2 =>

250 , N — ln

= ∫ k d t => l ∫ ⅛

∫ ^

∣ W ⅛I =

4t

^

⅛

e4t

=

τ

1 10 00

+

l ⅛

In 199g ∣—C =>

≠> 499x =

e4 , (1000

51∙ < a> X'¾ π dx = X1(χ6- 4χ5 +5χ4~4χ2 +4 - ⅛ (b)

In 11000 _ x ∣— 250

π^0

- x) ≠> (499 +

e4 t ) x

+

1000 ^n (4 9 9 )

= 1000e4t ≠> x = 4≡ ⅛

=> 500 ∙ 499 + 500e4t = 1000e4t ≠> e4t = 499 =^ t = ∣In 499 ≈ 1.55 days

(b) χ = 1 N = 500 ≠> 500 = 2≡ ⅛

22

∣= kt + C;

⅛ In ⅛

= ∫k d t ≠

253

) dx = ^ ^ 7r

100% = 0.04%

(c) The area is less than 0.003

8.4 TRIGONOMETRIC INTEGRALS Γπ^ 9 Γπ ^ 9 2 4 Jθ (1 - cos2 x) sin x dx = Jθ (1 - 2cos2 x + cos4 x)sin x dx r 5 "∣π ^ pπ∣ 2 p π ∕2 pπ ^ 4 3 o = J sinxdx —J 2cos2 x sin x dx ÷ J cos4 xsin x dx = —cos x + 2 ~ — 2 ^

Cπ^ c J sin5 xdx =

π

JΓQ ^ (sin2 x) 2 sin x dx =

L

= m -(-ι + i- i) = ⅛

"

p π ∕2

p π ∕2

p π ∕2

n π ∕2

n π ∕2

3. J/ —π∕^2 cos3 x dx = J I—7Γ∕2/ n (cos2 x)cosxdx = JI—π ∕2/nx(l - sin2 x)cos x dx = J —τr∕2 c o s x d x - JI—π∕2 sin2 xcosxdx z = (1 - 1) - ( - 1 + ∣) = ∣ = [s in x -⅛ ] X

*πl2

J 0

n π ∕2

sin7 y dy = Jθ

+ 3X 0

c o s4 y

s in

y

Z

p π ∕2

2

d y

-

c o s 6 y

X

s in

d y

y

p π ∕2

(1 - cos2 y) sin y dy = Jθ

sin6 y sin y dy = Jθ

=

[^

c o s

y

+

3

⅛

y

-

3

n π ∕2

sin y dy - 3 Jθ cos2 y sin y dy

⅛

x

+ ⅛

ς

] /

=

(°) -

(~

1

+

1

-

1+ ⅜)= it

7. f o 8sin4 x dx = 8J^ ( l - y ^ ) 2 dx = 2j'0 (1 —2cos 2x + cos2 2x)dx = 2J^ dx —2 ^ cos 2x ∙ 2dx + 2 J^ l±∞⅛4χ t jx = [2x - 2sin 2x]θ + X dx + X cos4xdx = 2π + [x + ∣sin 4x] θ = 2π + π = 3π

9

∙

n π ∕4 X - π ∕4

1 6

s i

"

2 χ C 0S

= [4 C √ 4 -

2

X d x

X X

d x

= -

1 6

2

n π ∕4 X - π ∕4 ( M

X X ∕4 c

λ

v

o s 4 x d x

=

^ ) (i ^

λ

)

κ d x

=

4

p π ∕4 X - π ∕4 ( 1 ~

p π ∕4

p π ∕4

,

. .

∞Sλ2 2x)dx = 4 / ^ dx - 4 J ^ (z i±ι ≡ Λ )d x

π + π - [2x + ≡4χ] l " 4 =2π - ( f - ( - f )) = π

Chapter 8 Techniques of Integration

254 p π ∕2

11. Jθ

2 C^ A Cπ^ A C^ 9 35 sin4 xcos 3 x dx = Jθ 35 sin4 x (1 —sin2 x)cos x dx = 35 Jθ sin4 xcos x dx - 35 JQ sinθxcos x dx

Γ

∙ s

∙ 7 1 ^I2

= ∣ 3 5 ^ -3 5 ^ ]

= (7 -5 )-(0 ) = 2

β

13. f ^ 8cos3 20 sin 20 d0 = [ 8 ( - j ) ⅛ ^ ] ^ = [-cos 4 20] θz4 = (0) - (-1 ) = 1

1 5

π

y

⅛

^

d

χ

t f

17∙

Jo

19∙

J j ∕4 √ 1 ÷ tan2 x ^ = J j

=

I

s in

∙

f

A ^ ⅛ < lt = ∫

o

d ∣∣

χ

=

l∞ s t∣ dt = ∫ sec x

s in

f

d

χ

H

=

2 c o s

u

2

=

r

2

+

=

4

c o s td t - ∫π β c o s td t= [s in t]^ -[s in t]^ = l - 0 - 0 + l = 2

o

dx = j X ∣

4

sec x d x

= ll n ∣sec x + tan x ∣ ]l z4 4 = ln ( √ 2 + 1) - ln ( √ 2 - 1)

=I"(⅛ I )= Λ (1 + ^ 21. / ^ 0 √ 1 - cos 20 d0 = £ ' 2 0 √ 2 ∣ sin 0 ∣ d0 = √ 2 £ ^ 0 sin 0 d0 = √ 2 [-0cos0 + sin θ γ j 2 = √ 2(1) = √ 2

23. I „ 2 sec3 x dx; u = sec x, du = sec x tan x dx, dv = sec2 x dx, v = tan x; pθ _ pθ _ pθ J π β 2 sec3 x dx = [2 sec x tan x] _ ^ 3 - 2J ^ β sec x tan2 x dx = 2 ∙ l ∙ 0 - 2 ∙ 2 ∙ √ 3 - 2 J _ n/3 sec x (sec2 x —l)dx = 4χ ∕s - 2f

sec3 x dx + 2f

vβ

πβ

sec x dx; 2^ ^β 2 sec3 x dx = 4 ᅟ ∕3 + [21n ∣ sec x + tan x∣ ] ᅟ∣3

2j.° π ∕ 3 2sec 3 xdx = 4 √ 3 + 21n∣ l + 0∣- 2 ln ∣ 2 - √ 3 ∣= 4 √ 3 - 2 1 n ( 2 - √ 3 ) f

vβ

*π∕4

J

0

2 sec3 x dx = 2ᅟ ∕3 - In ^2 —^

.

sec4 0 d0 =

pπ∕4

~

λ

JQ

(1 + tan2 0)sec2 0 d0 =

nπ∕4

JQ

Γ

pπ∕4

a

2 0 secΛ 2 0 d0 = ∣ tanΛ tan 0 +

sec2 0 d0 + J o

a

π

∕4

π

∕2

y^ I

= (ι + 1 ) - (θ) = I *τr ∕2

j

ρ π ∕2

*π∕4

J 0

J π ∕4

∩ π ∕2

λ

λ

J= ( 0 ) - ( - 1 - ∣) = ∣ π∕4

csc4 0d0 = /

2 0 d0 = / (1 + cot2 0)csc v 7

pπ∕4

4tan 3 xdx = 4 J 0

J π ∕4

λ

p π !2

csc2 0 d 0 + ι

J π ∕4

pπ∕4

(sec2 x —l)tan x dx = 4Jθ

λ

Γ

λ

cot2 0 csc2 0 d0 = - c o t 0 - ⅛ [

3 J 7rχ4

pπ∕4

Γ

sec2 x tan x dx —4 Jθ tan x dx = p ^

2

1π∕4

—41n∣ sec x∣ J

= 2(1) - 41n√2 - 2 ∙ 0 + 41n 1 = 2 - 21n2 ∙π∕3

J

π∕6

pπ∕3

cot3 x dx = i

J π l6

pπ∕3

(csc2 x —1 )cotx dx = I γ

J π ∕6

f

pπ∕3

csc2 xcotx dx — I

J π ∕6

1 7r∕^

[ 2

cotx dx = — L

2

+ In Iesc x∣

J π ∕6

= 4 ( i - 3 ) + (l∏ ⅞ -l∏ 2 ) = ∣- l n √ 3 pθ 33. JI— % sin 3x cos 2x dx = 5I 2 ∕ —7Γ ' t

(sin x + sin 5x) dx2 L= ∣[-cos Ox - ⅛cos 5xl J —7Γ z

35. f ^ sin 3x sin 3x dx = ∣f ^ (∞s 0 —cos 6x) dx = ∣f ^ dχ - ^ f

0

= 1 ( - □1 - ∣- lO-/ l ) =0 - f

2 ᅠ

cos 6x dx = - [x ——sin 6x] ^ π = ∣+ ∣—0 = π

Section 8.5 Trigonometric Substitutions 0

255

cos 3x cos 4x dx — ∣J 0 (co s(-x) + cos 7x) dx = ∣[ -s in (- x )+ ∣sin 7x] θ = ∣(0) = 0

39. x = t2∕ 3 => t2 = x3 ; y = ^ ^- y = y ; 0 ≤ t ≤ 2 ≠> 0 ≤ x ≤ 22/ 3 ; f

22∕ 3

/ 3 ∖ /----Γ -

+

9(2 2∕ 3 ) ) 3 z 2 - l ]

= ∣ φ

Γ u = ⅛4 1

r≡(2273) /------

J

41. y = ln(sec x); y' = ss⅛71x = tan x;(y') 2 = tan2 x; Q ^ ᅟ ∕1 + tan2 x dx = J ^ = ln ( √ 2 + 1 ) - l∏(0 + 1) = ln ( √ 2 + 1)

43. V = π ∫θ sin2 x dx = π ∫θ

1

dx = ∣∫ 0 dx - f ∫θ cos 2x dx = ∣[x]J - f [sin 2x]J = f (π - 0) - f (0 - 0) = ⅞

c os2 x 2

pk+2π

9

45. (a) m2 ≠ n2 => m + n ≠ 0 and m - n ≠ 0 => J k =

5

sin(m

= ∣( i ⅛

in

n )x n

((m -

s i n (( m

= ⅛

-

-

∣ secx∣ dx = [ln∣ sec x + tan x∣ ]J z4

~ =∙¼sin(m +

nk+2π

sin mx sin nx dx = ∣J

[cos(m - n)x —cos(m + n)x]dx

n ) χ ] k + 2π

) ( k + 2 τr )) - j⅛ sin ((m + n )(k + 2 π ))) - ∣( i ⅛

n )k

s i n (( m

)- 2⅛ J

+

n )k

i n ((∞

)- 2U ⅛

-

n )k

in

((m -∏ )k) s i n (( m

)+ 2⅛ )

5⅛

i n ( ( m + n)k))

n )k

)=

÷

0

=> sin mx and sin nx are orthogonal. ρk+2π

(b) Same as part since ∣J k =

iX

+

(m

[c o s

-

n

pk+2π

cos 0 dx = π. m2 ≠ n 2 = > m ÷ n ≠ 0 and m —n ≠ 0 => J k

)x

+

c

°

s

= 2 ( ⅛

s in

((m -

n

)(k +

2 π

= 2 ( ⅛

s in

((m -

n

)M ÷ 2 ⅛

(m

n

+

)x

ld x

j

s in

((m

= 2 [ i ⅛

s i t l ((m

)) + 2 ( ⅛ +

in

n

+

)k ) "

n

)(k +

2 ⅛ )

-

(m

2 π

s in

in +

ns in

)) ~ 2 ( ⅛

s in

n

)x +

((m “

n

)k ) -

(m

n

+

((m ~

2 ⅛ )

cos mx cos nx dx

)x n

s in

] ^

)k ) “ 2 ⅛ J (("1

+

n

s in

((m

+

n

)k )

0

)k ) =

=> cos mx and cos nx are orthogonal. pk+2π

(c) Let m = n => sin mx cos nx = ∣(sin 0 + sin((m + n)x)) and j J k

pk÷2π

sin 0 dx = 0 and ∣J k

sin((m + n)x) dx = 0

=> sin mx and cos nx are orthogonal if m = n. Let m ≠ n. ∙k÷2π

J

k

f k÷2π

sin mx cos nx dx = ∣J k 1

= ~ 2 ( ⅛

c 0 s

((m ~

n

)(k +

= - 2 ( ⅛

c

((m

n

)k

°

s

-

) -

2 π

1 Γ

)) -

2 ⅛

1

ι

[sin(m - n)x + sin(m + n)x]dx = j [—^⅛ cos(m - n)x — ^L-cos(m + n)xj s

2 ⅛ )∞ c

°

s

((m

+

n

((m )k

+

n

∏

k +

) + 2 ( ⅛

c

2 π

°

s

))

((m

+

~

2 ⅛ )∞ n

)k

)

s

,1 ) k

((m “

+

2 ( ⅛

7

=

c

°

s

)

((m

+

+

2 ( ⅛ n

)k

c

) =

=> sin mx and cos nx are orthogonal. 8.5 TRIGONOMETRIC SUBSTITUTIONS 1. y = 3 t a n 0 , - f < 0 < ∣, d y = ^ , 9 + y2 = 9 ( l + tan2 0) = ^

^

7

X

M

=

≡*

(because cos 0 > 0 when —f < 0 < f ) ; ∫^ ⅛

3∙ £

5∙

⅛

= 3

∫ ^

= ∫ ^

= ln ∣ sec^ + tan0∣ + C' = l n ∣ ^

+

X∣+ C ' = l n ∣ √ 9 τ 7 + y ∣+ C

= [∣tan’ 1 1] - 2 = i tan→ 1 - 1 ta n - (-1 ) = (∣) ( 5 ) - ( ∣) ( - ≡) = ∣

f a l2-J‰ = [sin^1 U o/2 = sin^1 5 ~ sin" 1θ = f - θ = f

° s ((m 0

+

ι k÷2π 1 ,) k

)

256

Chapter 8 Techniques of Integration

7. t = 5 sin θ ,

— ∣< θ

< ∣, dt = 5 cos θ d0, χ∕25 —12 = 5 cos 0;

√ 2 5 - 12 dt = J (5 cos 0)(5 cos 0) d0 = 25 J cos2 0 d0 = 25 J

f

[sin-1

= f (0 + sin 0 cos 0) + C = f

9. x = ^ sec 0,0 < ∫ ^ ⅛

1⅛

= ∫

2

g

d0 = 25 (( + ^ ) + C

(f) + (∣) ( ^ φ ) ] + C = ⅞ sin’ 1 (∣) + ^ ≡

< ∣, dx = j sec θ tan θ d0,

θ

l + ≡

J

+ C

4x2 —49 = ᅟ ∕ 49 sec2 0 —49 = 7 tan 0;

js e c 0 d 0 = p n ∣ sec0 + tan0∣ +C = pn∣ ^

^ M

11. y = 7 sec 0,0 < 0 < ( , dy = 7 sec 0 tan 0 d0, y y2 - 49 = 7 tan 0; √ E ≡ 2 dy = ∫

f

=

7

(7 ta" ¢)(7 s ∞ ⅛

tan

⅜ d⅛

=

7 Jta n 2 0 d0 = 7 J (sec2 0 - 1) d0 = 7(tan 0 - 0) + C

[ ^ ≡ - s e c - 1 β )]+ C

13. x = sec 0,0 < 0 < ( , dx = sec 0 tan 0 d0, ᅟ ∕x 2 - l = tan 0; J

dx

_ Γ se c 0 ta n 0 d 0 _ “ J 1^0≡ Γ T -

χ2√χ2-l

C d0 _ J ^ 0 -

s m

15. x = 2 t a n 0 , - ( < 0 < ( , d x = ^ , √ x x3 dx _ λ∕ x 2 + 4

∣r ^ ÷ c

2

,

— -

AEΞ1

ι Γ, + C

+4 = ^ j

Γ (8 tan3 0) (cos 0) d0 _ o f sin3 0 d0 _ Q f (cos2 0 - 1 ) ( - sin 0) d0 . J cos2 0 ° J cos4 0 ~ ° J cos4 0 ’

∕[t = cos0] → 8 ∫ ⅛ i d t = 8 ∫ ( ⅜ - ⅛ ) d t = 8 ( - i + ⅛ ) + C = 8 ( - s e c 0 + ⅛ ^ ) + C = 8^- ^ ≡

+ ¾

f )

+

c

= 5 (X2 + 4 ) 3 7 2 - 4 √

17. w = 2 sin 0, —∣< 0 < ∣, dw = 2 cos 0 d0, √ 4 - w

/

8 dw W

2 √ 4 -

W

__ f J

8-2 cos 9 d¢ __ ι f dd _ 4 sin2 ¢-2 cosd ~ 2 J ⅛ d ~

2 -

~

∏

2

X2

+4 + C

= 2 cos 0;

z> ∣z^> _ c -

- 2 √ 4 - w 2 1 r> --------- h C

2 c 0 t w +

19. x ≈ sin 0,0 ≤ 0 ≤ ( , dx = cos 0 d0, (1 —x2 )3^2 = cos3 0; f ^

Jo

2

4χ 2

__

Γ π ∕ 3 4 sin2 < cos d d¢ __ .

(T ∑ ^ ~ Jo

---- --------- ~

4

Γ ^ 3 f 1 ~ cos2 ⅜) A A _ Λ ( ^

J0

(

∞ s2 d )

d θ

~

4

J

0

3

(c a r 2 A

1 )d ⅛

(s e c 0 -l)d 0

= 4 [ ta n 0 - 0 ] ( /3 = 4 √ 3 - ⅞ 21. x - sec 0,0 < 0 < ( , dx = sec 0 tan 0 d0, (x2 - 1)3 ^2 = tan3 0; dx (x2 - 1)3 /2

/

sec 0 tan 0 d0 tan3 0

cos 0 d0 _ sin2 0

sin 0 ÷

C

-

23. x = sin 0, —- < 0 < ^ , dx = cos 0 d0, (1 - x2 )3 ^2 = cos3 0; f

^

^

^

= f

^

^

^

=

∫ c o t ‰ s c 2 0d0 = - ^ ÷ C

= - ∣( ^ ≡ )

5

÷C

2 2 2 4 25. x = ∣ ZZtan 0, —~Z < 0 < Z dx = ∣sec ' 0 d0, ' (4x + 1) = sec 0; f ⅛ = S ^ W ? " * = 4 ∫ ∞ ≡ 2 * d » = 2(0 + sin 0 cos 0) + C = 2 tan- 1 2x + ^ ⅛

27. v = sin 0, —∣< 0 < j , dv = cos 0 d0, (1 - v2 )5^2 = cos5 0; / (T T ⅛ = ∫ ¾

^

= ∫ t a n 2 0 sec2 0 d0 = ⅛ *

+ c

= i ( - ^ )

3

+

C

y

+C

Section 8.5 Trigonometric Substitutions 1 (∣ )

29. Let e t = 3 tan 0, t = In (3 tan 0), tan f l n 4 e t2tdt _ Jo √e + 9

f ta " ( 4 / 3 ) 3 ta n 0 ∙s e c 2 0 d 0 J ta n - 1 (1∕3) tan 0∙3 sec 0

= In ( | + ∣) - In ( ⅛

31∙

p l/4

τ ⅛ ⅛ 2 du

__

J l ∕√ 3

T+ UƩ

“

‰

Γu [

=

Γπ∕4 J

π ∕6

2

f ta n ( 4 / 3 ) J ta n - 1 (1/3)

_

1 (∣ ),

≤ 0 ≤ tan

= (In IseC 0 ÷ tan 0|] tan 1^ 4/3)

0

S

dt = ^ ∣d0, √ e ^ ÷ 9 = √ , 9 tan 2 0 + 9 = 3 sec 0; L

I

tan~ (1/3)

+ θ = In 9 - In (1 + √ i 6 )

/-

√d

u

Γ1 → ∫ ∕√ 3 ⅛

d t 1]

= ⅛

iu

=

ta n

M

≤ 0 ≤ M

u

=

sec2 0


0 = ∣tan~ 1 l + C

⅜

dx; x = 3 sin 0 ,0 ≤ 0 ≤ ∣, dx = 3 cos 0 d0, ᅟ ∕ 9 —x 2 = χ ∕9 —9 sin2 0 = 3 cos 0;

3 cos 0-3 cos 0 d 0 =

dt sin t —cos t

x = 2 sec 0 ,0 < 0 < ∣ dx = 2 sec 0 tan 0 d0 ᅟ ∕ x 2 —4 = 2 tan 0

= 2 J tan2 0 d0 = 2 J ( s e c 2 0 - 1) d0 = 2(tan0 - 0) + C

39∙ (x 2 + 4) ⅛ = 3,dy = 4 A ; y = 3 ∕ ⅛ ^

dx;

3

^

c o s 20

d0 = ∣[0 + Sin 0 COS 0] J /2 = ⅞

2dz

2dz __ 1 1« U + iA -2 - √ 2 ln

∣]

^

257

258

Chapter 8 Techniques of Integration

51. ∫ s e e β d β √

⅛

√

^ √

⅛

√

lr⅛

f⅛

= ∫

τ⅛

+ J ⅛

1 + ta n

= In 11 + z ∣—In 11 - z∣ + C = In

+ c

1 -ta n

8.6 INTEGRAL TABLES AND COMPUTER ALGEBRA SYSTEMS

(We used FORMULA 13(a) with a = 1, b = 3)

3.

fi‰ =f ^

+ 2f i ⅛

=f ( ' ^ ' i ' + 2 S ( ' ^ Y ' i *

= ( D

≈

= ^

^

+

( ^

[ ⅛

3

+ 4 ∣+ C

(We used FORMULA 11 with a = 1, b = - 2 , n = 1 and a = 1, b = - 2 , n = - 1 )

5.

f x√2x -

3 dx = | ∫ ( 2 x - 3 )√ 2 x - 3 dx + ∣∫ √ 2 x - 3 dx = ∣∫ ( √ 2 x - 3 ) 3 dx + 2 ∫ ( √ 2 x ≡ 3 ) 1 dx

= ( I ) (1)

t

^

+ (1) (1)

t

^

+C = ≡

≠

∣ V

+ 1∣+ c = e - ½ ± > ι

+ c

(We used FORMULA 11 with a = 2, b = - 3 , n = 3 and a = 2, b = - 3 , n = 1)

(We used FORMULA 14 with a = - 4 , b = 9)

(We used FORMULA 13(b) with a = - 4 , b = 9)

9.

f =

x√4x - x

2

dx = J* x √ 2 ∙ 2 x - x 2 dx = M = 0 X4 ^ ∣ E r ∣= 0 (b) £ x d x = [ ⅛ ] ' = 2 - l = j ≠ ∣ E r ∣= ∫ 2χ d x - T = 0 (c)

True Value

x

l υ υ

—

f(x) = 0 => M = 0 ≠

∆x _ 3 -

1 . 12 ’

∣ Es ∣= 0

Es ∣= ∫ x d x - S = 3 - 3 = 0u (b) ∫ x d x = ∣ => ∣ 2 2 True Value

x

1U U

- U

m 1 2 2 2 1

mf(xi ) 1 5/2 3 7/2 2

f(χi ) 1 5/4 3/2 7/4 2

m 1 4 2 4 1

mf(xi ) 1 5 3 7 2

υ z o

∏∙ (a) For n = 4, ∆x = —n = ⅛41 = 4j => ∑ m f(x i) = 18 4 S = ⅛(18) = j ;

(c)

1 5/4 3/2 7/4 2

f(χ i ) 1 5/4 3/2 7/4 2

/ 0

Xi

xo xι X2 X3 X4

1 5/4 3/2 7/4 2

Section 8.7 Numerical Integration ∕ . l (x2 + l)d x I.

Xi

1

= 1ψ

(a) For n = 4, ∆ x = ⅛

= 5 = 2 =^ ⅜ = 5 5

∑ m ‰ ) = 11 => T = 1(11) = 2.75; f(x) = χ 2 + 1 ≠> f'( x ) = 2x => f"(x) = 2 => M = 2 => ∣ E r∣≤ ⅛ n (∣) 2 (2) = ⅛ or 0.08333 (b > J √

χ2

+ 1 ) dx = [ j + x ] '

1

Xo xι X2 X3 X4

-1 /2 0 1/2 1

m 1 2 2 2 1

f(xi ) 2 5/4 1 5/4 2

= (I + 1) - ( - 5 - 1) = ∣ => E r = ∫ j x

2

+

265

mf(xi ) 2 5/2 2 5/2 2

l ) d x - T = ∣- ⅛ 12

=> ∣ Eτ ∣= ∣ - ⅛ ∣≈ 0.08333 i ⅛ × l ∞ = ( f ) × l ∞ ≈ 3 % II. (a) For n = 4, ∆ x = ⅛

= 5 = ∣ => ⅜ = j;

= 1 ^

∑ mf(xi ) = 16 => S = 1 (16) = f = 2.66667; Es ∣= 0 f f(x) = 0 =^ M = 0 ≠> ∣ (b ) ∫ J ,( x 2 + l ) d x = [ τ + x ] ^ 1 = I Es ∣= ∫ j x => ∣

2

Xi xo xι X2 X3 X4

+ l ) d x - S = ∣- ∣= 0

-1 -1 /2 0 1/2 1

f(χ i ) 2 5/4 1 5/4 2

m 1 4 2 4 1

mf(xi ) 2 5 2 5 2

(c ) τ ⅛ × l θ θ = θ%

5∙ J (t3+t>dt 0 I. (a) For n = 4, ∆x =

≠> ⅜ = 5 5∑ “ % ) = 25 =^ T = 1 (25) = f ;

to tι h

f(t ) = t3 + 1 ≠> f'(t) = 3t2 + 1 ≠> f"(t) = 6t ^ M = 12 = f"(2) => ∣ E r ∣≤ ⅛ (1) 2 (12) = 1

t4

⅛^ = ¾^ = I = 1

(b) ∫ 0⅛ + t ) d t = [ l + ^ = ( z + < ) - ° = i

(°) τ ⅛ × 1 0 0 = ⅛ II. (a) For n = 4, ∆ x = ^ ∑ m f(t i ) = 36 ^ f(3)(t ) = 6 ≠> (b )

∕

0

(t3

6

5

S = 1(36) = 6: = 0 ≠> M = 0 ≠> ∣ Es ∣= 0

+ 0 dt = 6 ≠> ∣ Es ∣= ∫θ (t3 + 1) dt - S

to ti t2 t3 t4

7. ∕ ⅛ d s

ti 0 1/2 1 3/2 2

Si

(a) For n = 4, ∆x = ⅛ V m W s 3 - 12½573 Λ√ — 44,100

= .

m 1 2 2 2 1

mf(ti ) 0 5/4 4 39/4 10

=> ∣ Eτ ∣= £ ( ^ + 1 ) 0 1 - 1 - = 6 - ^ = - 1

= ∣= ∣≠> ^ = 1;

= 6 -6 = 0

I.

f(ti ) 0 5/8 2 39/8 10

^

l& rM

×100≈4% = ⅛

f(t)

ti 0 1/2 1 3/2 2

τ

≈ 0.50899; f(s) = ⅜ ^

So

⅛ 1 = _ “

1

≠>

⅜

1 / 179,573 ∖ _ 8 ∖ 44,1∞ / ~

f'(s) = - ⅛

=

f J

179,573 352,800

Si S2 S3 S4

1 5/4 3/2 7/4 2

f(ti ) 0 5/8 2 39/8 10

f(Si ) 1 16/25 4/9 16/49 1/4

m 1 4 2 4 1

m 1 2 2 2 1

mf(ti ) 0 5/2 4 39/2 10

mf(si ) 1 32/25 8/9 32/49 1/4

=> f"( s ) = ^ => M = 6 = f"(l) ^ (b ) / (β )

∣ Eτ ∣≤ ⅛ i ( i ) 2 (6) = ⅛ = 0.03125 ⅛ ds =

s^ 2 ds = [ - 1] 2 = - 1 - ( - 1) = 1 ≠> Eτ = ∫

=> ∣ Eτ ∣= 0.00899 τ ≡ ‰ × 100 = T

×

100

≈

2%

2⅜

d s - T = l - 0.50899 = -0.00899

266

Chapter 8 Techniques of Integration

II. (a) For n = 4, ∆x = ⅛⅛ = ⅛ 1 = i => ^ = + ; m M

∑

f ∕ς ∖ _

264,821

.

— 44J 1OO

I W

ς _ 1 / 264,821 λ _ 264,821 ° — 12 44,100 ) ~ 529,200

Si So Si

≈ 0.50042; f (3 s) = - ⅛ => f (4 ) (s) = ^

S2

∣( 1 ) 4 (120)

S3

=> M = 1 2 0 ^

∣ E S ∣≤ ∣ ⅛

S4

= ⅛ ≈ 0.00260

1 5/4 3/2 7/4 2

f(si ) 1 16/25 4/9 16/49 1/4

m 1 4 2 4 1

∏ r f(S i)

1 64/25 8/9 64/49 1/4

(b) Γ ⅛ ds = ∣ ≠> Es = Γ ⅛ ds - S = ∣- 0.50042 = -0.00042 => ∣ Es l = 0.00042 (0 τ J ^ ×

9.

100

= T

100

×

≈

0

∙≡

Γ sin t dt Jo

I.

(a) For n = 4, ∆x = ⅛ ∙ = V

= 5 ≠> ⅜ = f ’

∑ mf(ti ) = 2 + 2 √ 2 ≈ 4.8284 => T = f (2 + 2 √ 2 ) ≈ 1.89612; f(t) = sin t => f'(t) = cos t => f"(t) = -sin t ^ M = l =^ ∣ EI ∣ ≤ ⅛ Q ) 2 (1) = ⅛

to tι t2 t3 t4

ti 0 π∕4 π∕2 3π∕4 π

f(ti ) 0 √ 2 ∕2 1 √ 2 ∕2 0

m 1 2 2 2 1

mf(t>) 0 λ ∕2 2 λΛ 0

≈ 0.16149 (b) J^ sin t dt - [-cos t] θ = (-co s π) - (-cos 0) = 2 => ∣ E r ∣= / J sin t dt - T ≈ 2 - 1.89612 = 0.10388 (c) τ i ⅛ × 1 0 0 = ∞ × 1 0 0 ≈ 5 % II. (a) For n = 4, ∆x = ^

= ^

= ∣ ≠> ⅛ = i ;

∑ mf(ti ) = 2 + 4 √ 2 ≈ 7.6569 => S = ⅞ (2 + 4 √ 2 ) ≈ 2.00456; f ^ ( t) = -co s t => f(4 t) = sin t ^ M= 1 ^ ∣ Es ∣≤ ⅛ ( f ) 4 (1) ≈ 0.00664 (b)

n

3 TΓ∕4

π

f(ti ) 0 √ 2 ∕2 1 √ 2 ∕2 0

m 1 4 2 4 1

∏rf(ti ) 0 2√2 2 2√2 0

sin t dt =► Es = / J sin t dt - S ≈ 2 - 2.00456 = -0.00456 ≠> ∣ Es ∣≈ 0.00456

(O τ i ⅛ 11. (a)

to h t2 t3 t4

ti 0 π∕4 7Γ∕2

× 100 = ^

× 100≈0%

= 8 => ∆x = ∣ ≠> ⅜ = ⅛J

∑ mf(xi) = 1(0.0) + 2(0.12402) + 2(0.24206) + 2(0.34763) + 2(0.43301) + 2(0.48789) + 2(0.49608) + 2(0.42361) + 1(0) = 5.1086 => T = ⅛ (5.1086) = 0.31929 (b)

n

= 8 ≠> ∆x = i =+ ^ = ⅛ ;

£ mf(xi ) = 1(0.0) + 4(0.12402) + 2(0.24206) + 4(0.34763) + 2(0.43301) + 4(0.48789) + 2(0.49608) + 4(0.42361) + 1(0) = 7.8749 => S = ⅛ (7.8749) = 0.32812 (c) Let u = 1 —x2 =+ du = —2x dx =+ —∣du = x dx; x = 0 =+ u = 1, x = 1 =+ u = 0 ∫

0

x √ l - x2 dx = ∫ ° √ 5 ( - ∣du) = i £ u 1∕2 du = [1 ( ^ ) ] θ = [f u3∕ 2 ] J = 1( √ T ) 3 - i ( √ θ ) 3 = 1;

Eτ = ∫ o' x √ l^ → 2 dχ - T ≈ j - 0.31929 = 0.01404; Es = ∫θ' x √ l≡ ~ x 2 dx - S ≈ ∣- 0.32812 = 0.00521 13. (a)

n

= 8 => ∆x = f =+ ⅜ = ⅛ J

∑ mf(ti ) = 1(0.0) + 2(0.99138) + 2(1.26906) + 2(1.05961) + 2(0.75) + 2(0.48821) + 2(0.28946) + 2(0.13429) + 1(0) = 9.96402 =+ T = ⅛ (9.96402) ≈ 1.95643 (b) n = 8 => ∆x = f ≠> ^ = ¾ ; £ mf(ti ) = 1(0.0) + 4(0.99138) + 2(1.26906) + 4(1.05961) + 2(0.75) + 4(0.48821) + 2(0.28946) + 4(0.13429) + 1(0)=15.311 ≠> S ≈ ¾ (15.311) ≈ 2.00421

Section 8.7 Numerical Integration

267

(c) Let u = 2 + sin t =4 du = cos t dt; t — —^ ≠> u —2 + sin ( - ∣) = 1, t = f =4 u = 2 + sin ∣= 3

pπ∕2

L

Er

0

Z2

(5¾¾ dt -

pπ∕2

_

T ≈ 2 - 1.95643 = 0.04357; Es = ∫ ^ ⅛

dt - S

≈ 2 - 2.00421 = -0.00421 15. (a) M = 0 (see Exercise 1): Then n = 1 ≠ ∆x = 1 4

∣ E r ∣= ⅛ (l) 2 (0) = 0 < 10~4

(b) M = 0 (see Exercise 1): Then n — 2 (n must be even) =4 ∆x = ∣ =4 ∣ E l = ⅛ ( ∣) 4 (0) = 0 < 10^4 4 ' s ’ 1Ov / 17. (a) M = 2 (see Exercise 3): Then ∆ x = 2 ≠> ∣ E r ∣≤ ⅛ ( ^ ( 2 ) = ⅛ < 10~4 => n2 > ∣(104 ) =4 n > ^ ( 1 0 4 ) ≠ n > 115.4, so let n = 116 (b) M = 0 (see Exercise 3): Then n = 2 (n must be even) 4

∆ x = 1 4 l⅛l = ⅛ (l) 4 (0) = 0 < 10^4

19. (a) M = 12 (see Exercise 5): Then ∆ x = 2 - . ∣ E r ∣≤ ⅛ (^ ) 2 (12) = ⅜ < 10~4

=4 n2 > 8 (104 ) =4 n > √ 8 (1 0 4 )

=> n > 282.8, so let n = 283 (b) M = 0 (see Exercise 5): Then n = 2 (n must be even) => ∆x = 1 => ∣ Es ∣= ⅛ (l) 4 (0) = 0 < 10~4 ∣ E r ∣≤ ⅛ Q ) 2 (6) = ⅛ < 10"4 4

21. (a) M = 6 (see Exercise 7): Then ∆ x = J 4

n2 > ∣(104 ) =4 n > ^

≠> n > 70.7, so let n = 71 (b) M = 120 (see Exercise 7): Then∆x = ⅛ =4 ∣ Es ∣= ⅛ (⅛)4 (120) = ⅛ < 10^4 4 4 n > y ^ (104 ) 4

23. (a) f(x) = √ Γ ∏

Then∆x = ⅞ => ∣ EI ∣ n >

4

^

i

=4 M = - j ⅛

=4 n2 > ⅜ ( 1 0 4 ) =4 n > y ⅞ R

so let n = 76 (b) f f w (x) = - n (x + 1 F 7/2 = - ^ 7 7 = 7 4 ⅛

n4 > ^ (104 )

n > 9.04, so let n = 10 (n must be even)

4 f , (x) = I (x + I)" 1/ 2 =4 f"(x) = - ∣(x + I)’ 3 / 2 = -

4

(104 )

35 (15)

(104 )

16(180)

n >

M=

15

1 6 (√ T )

« / 3^(15) (ItH) V 16(180)

= ∣.

=4 n > 75,

⅜= ]∣. Then ∆ x = 2n ‘°

=>

n

>

,n

f i

lθ∙θ,

.

so le t

n = 12 (n must be even) 25. (a) f(x) = sin(x + 1) => f , (x) = cos(x + 1) => f"(x) = -sin (x + 1) 4 = ⅛




n> ᅟ ∕ ≡

M = 1. Then ∆ x = ^ => ∣ Eτ ∣≤ ⅛ (^ ) 2 (1)

=4 n > 81.6, soletn = 82

(b) f(3 x) = -co s(x + 1) =4 f w (x) = sin(x + 1) ≠> M = 1. Then ∆ x = ^ 4 =4 n4 > ^ ¾ j^ =4 n > y

32 f f ^

=4 n > 6.49, so let n = 8 (n must be even)

27. ∣(6.0 + 2(8.2) + 2(9.1)... + 2(12.7) + 13.0)(30) = 15,990 ft3 .

∣ Es ∣≤ ⅛ (^) 4 (1) = ⅛

< 10- 4

268

Chapter 8 Techniques of Integration

29. Use the conversion 30 mph = 44 fps (ft per sec) since time is measured in seconds. The distance traveled as the car accelerates from, say, 40 mph = 58.67 fps to 50 mph = 73.33 fps in (4.5 —3.2) = 1.3 sec is the area of the trapezoid (see figure) associated with that time interval: ∣(58.67 + 73.33)(1.3) = 85.8 ft. The total distance traveled by the Ford Mustang Cobra is the sum of all these eleven trapezoids (using y and the table below): s = (44)(1.1) + (102.67)(0.5) + (132)(0.65) + (161.33)(0.7) ÷ (190.67)(0.95) + (220)(1.2) ÷ (249.33)(1.25) + (278.67)(1.65) ÷ (308)(2.3) + (337.33)(2.8) ÷ (366.67)(5.45) = 5166.346 ft ≈ 0.9785 mi v (mph) v(fps) t (sec) ∆ t∕2

0 0 0 0

30 44 2.2 1.1

40 58.67 3.2 0.5

50 73.33 4.5 0.65

60 88 5.9 0.7

70 102.67 7.8 0.95

80 117.33 10.2 1.2

31. Using Simpson’s Rule, ∆ x = 1 => y = ∣; = 11.2

xι

Let x be the length of the tank. Then the

X2

Volume V = (Cross Sectional Area) x = 11.2x. Now 5000 lb of gasoline at 42 lb∕ft3 => V = ^ θ = 119.05ft3

X3 X4 X5

=> 119.05 = 11.2x => x ≈ 10.63 ft 33. (a) ∣ E s ∣≤ ⅛ (b) ∆ x = ∣ ^

X6

f

= ⅞j xι

≠> S = ⅛ (10.47208705) ≈ 1.37079

X2 X3 X4

0 π∕8 π ∕4 3π∕8 π ∕2

120 176 26.2 2.8

130 190.67 37.1 5.45

myi 1.5 6.4 3.6 7.6 4.0 8.4 2.1 ( f ) 4 (1) ≈ 0∙00021 m 1 4 2 4 1

f(χ i ) 1 0.974495358 0.900316316 0.784213303 0.636619772

Xi xo

110 161.33 20.6 2.3

m 1 4 2 4 2 4 1

Es ∣≤ ⅛ (Δ X4 )M J n = 4 ≠ - ∆ x = ⅛ ≤ = ∣5 ∣ f∣≤; 1 =^ M = 1 ≠> ∣

£ mf(xi ) = 10.47208705

(c >

0 1 2 3 4 5 6

xo

100 146.67 16 1.65

ys 1.5 1.6 1.8 1.9 2.0 2.1 2.1

Xi

^ myi = 33.6 =^ Cross Section Area ≈ ∣(33.6) ft2 .

90 132 12.7 1.25

mf(xii) 1 3.897981432 1.8∞632632 3.136853212 0.636619772

≈ ( τ ≡ ) × 1∞ ≈ θ∙015%

35. (a) n = 1 0 + ∆x = ⅛

= ⅞ => f

=

£ mf(xi ) = 1(0) + 2(0.09708) + 2(0.36932) + 2(0.76248) + 2(1.19513) + 2(1.57080) + 2(1.79270) + 2(1.77912) + 2(1.47727) + 2(0.87372) + 1(0) = 19.83524 4 T - ⅛ (19.83524) = 3.11571 (b) π - 3.11571 ≈ 0.02588 (c) With M = 3.11, we get ∣ Eτ ∣≤ ⅞ (⅛ ) 2 (3.11) = ⅛

(3.11) < 0.08036

37. T = y ( y o + 2yι + 2y2 + 2y3 + . . . + 2yn _i + yn ) where ∆ x = τ 1

b —a (yo + y ι + y ι + y 2 ÷ y 2 ÷ ∙ ∙ ∙ + y∏-ι +y∏-ι ÷y∏) ^

n

2

“

n

2

and f is continuous on [a, b]. So ∣ f(χι) + f(¾)

b —a ( f(⅞) + f(χι) ψ

2

∣ τ

∣ f(χ∏-ι) + f(χ∏)) ∙

,

∙

τ

2

J'

Since f is continuous on each interval ⅛ - ι , Xk], and ⅛ ^ ιH f e l is always between f(xk-ι) and f(xk), there is a point Ck in [Xk-i, Xk] with f(ck) = ⅛ ^ ⅛ i j this is a consequence of the Intermediate Value Theorem. Thus our sum is n

n

∑ ( v ) ^ ( c k ) which has the form ^ ∆xkf(ck) with ∆xk = ^ k=l

k=l

2

^o r a ^ ^∙ ^

s

*s

a

Rie m a ∩n Sum for f on [a, b].

Section 8.7 Numerical Integration

269

Exercises 39-42 were done using a graphing calculator with n = 50 39. 1.08943 41. 0.82812 43. (a) Tιo ≈ 1.983523538 Tιoo ≈ 1.999835504 (b)

Tιooo ≈ 1.999998355 n ∣ Eτ ∣= 2 - T n 0.016476462 = 1.6476462 × 10~2 10 1.64496 × 10^4 100 1.646 × 10" 6 1000

Eτ ,0n ∣≈ 10- 2 ∣ ETO ∣ (c) ∣ 2 (d) b - a = π, (∆ x ) = ⅛ , M = 1 I 17 ∣ ⅛

I z π f π2 I < 12 ∣≤

⅛

_ π3 - 12? ≤

-

ω

1

∣ ⅛

∣

45. (a) f'(x) = 2x cos(x 2 ), f"(x) = 2x ∙ (—2x)sin(x 2 ) + 2cos(x2 ) = —4x 2 sin(x 2 ) + 2cos(x 2 ) (b)

y = - 4 X2 sin(x2) + 2 cos(x2)

(c) The graph shows that 3 ≤ f"(x) ≤ 2 so ∣f , , (x)∣≤ 3 for - 1 ≤ x ≤ 1. (d) ∣ ET ∣≤ ⅛ ^ ( Δ X) 2 (3) = ⅛ (e) ForO < ∆ x < 0.1, ∣ Eτ ∣≤ ^

≤ ⅛ = 0.005 < 0.01

(f) ∏ > ≡ > ⅛ = 2 0 47. (a) Using d = ^, and A = π ( ∣) 2 = ^ yields the following areas (in square inches, rounded to the nearest tenth): 2.3, 1.6, 1.5, 2.1, 3.2, 4.8, 7.0, 9.3, 10.7, 10.7, 9.3, 6.4, 3.2 (b) If C(y) is the circumference as a function of y, then the area of a cross section is A

(y) = π ( ^

1

j

= ¾

i

, and the volume is ⅛ J o C 2 (y) dy.

(c ) f 0 A (y ) d y = ⅛ f 06 c 2 (y ) d y ≈ ⅛ ( ⅛ 5 )I 5 Λ

2

+ 2(4.5 2 + 4.4 2 + 5.12 + 6.32 + 7.82 + 9.42 + 10.82 + 11.62 + 11.62 + 10.82 + 9.0 2 ) + 6.32 ]

≈ 34.7 in3 = ⅛ £ c 2 (y) d y ≈ ⅛ ( ⅛ ) [5 ∙4 2 +

(d ) v

4 (4 ∙5 2 )

+

2 (4 ∙4 2 )

+

4 (5 ∙ 1 2 )

+

2 (6 ∙3 2 )

+

4 ( 7 ∙8 2 )

+

2 (9 ∙4 2 )

+

4

( 1 °-8 2 )

+ 2(11.6 2 ) + 4(11.6 2 ) + 2(10.8 2 ) + 4(9.0 2 ) + 6.32 ] = 34.792 in3 by Simpson’s Rule. The Simpson’s Rule estimate should be more accurate than the trapezoid estimate. The error in the Simpson’s estimate is proportional to (∆ y) 4 = 0.0625 whereas the error in the trapezoid estimate is proportional to (∆ y) 2 = 0.25, a larger number when ∆ y = 0.5 in.

270

Chapter 8 Techniques of Integration

49. (a) a = 1, e = ∣ => Length = 4 J^

^ 1 “ I cos2 1dt

= 2 [ ^ 4 - cos2 1 dt = [

f(t) dt; use the

Trapezoid Rule with n = 10 => ∆ t = 2p p π ∕2

ᅟ ∕4

= — ι OA *

z ------------------------------

■

V

”

—cos2 W

Xi

xι X2

ιo-

=

X3

lθ

X4

1dt ≈ V mf(x ) = 37 3686183

□

0 π∕20 π∕10 3π∕20 π∕5 π∕4 3π∕10 7π∕20 2π∕5 9π∕20 π∕2

Xo

✓

I

11U¾ Λ∩J

/ ∙√ U v V lU √

X5

n=0

≠ T = f (37.3686183) = ⅛ (37.3686183)

X6

= 2 934924419 ≠> Length = 2(2 934924419) ≈ 5.870 (b) ∣ f"(t)∣< 1 4 M = 1 ∣ E r ∣< ⅛ (Δt 2 M) ≤ ⅛ ! ⅛ ) 2 1 ≤ 0.0032

X7 X8 X9 X ιo

51. The length of the curve y = sin (⅛ x ) from Oto 20 is: L = f

f(χi ) 1.732050808 1.739100843 1.7594∞893 1.790560631 1.82906848 1.870828693 1.911676881 1.947791731 1.975982919 1.993872679 2

m 1 2 2 2 1 1 2 2 2 2 1

t ∕ l + f ⅛ ) dx; ⅛ = ⅛ cos (⅛ x) ≠

mf(xi ) 1.732050808 3.478201686 3.518801786 3.581121262 3.658136959 3.741657387 3.823353762 3.895583461 3.951965839 3.987745357 2

( ^

= ^ cos2 ( ^ x ) => L = J^ y Γ + ^ c o s ^ x ) dx. Using numerical integration we find L ≈ 21.07 in

= cos2 x => S = Jθ 2π(sin x) √ 1 ÷ cos2 x dx; a numerical integration gives

53. y = sin x => ^ = cos x => ^ j S ≈ 14.4

= (1 + 2 cos 2x)2 ; by symmetry of the graph we have that

55. y = x + sin 2x => ^ = 1 + 2 cos 2x => ^ j p2√3

S = 2J o

/

o

2π(x + sin 2x) √ 1 + (1 + 2 cos 2x)2 dx; a numerical integration gives S ≈ 54.9

57. A calculator or computer numerical integrator yields sin- 1 0.6 ≈ 0.643501109. 8.8 IMPROPER INTEGRALS 1

∙

Γ

3∙

∕⅛

5∙

£ =

7

∙

£

⅛

=

=

b

b

⅛

l¾ ÷

∫

£

0

b

⅛

x ^v 2 d x

⅛ = ∫-⅛ + ∫.⅛ b

b

1"

‰

=

b

1⅜ ÷

=

lta n

^'

X ]» =

M

i =

= b !im t o- M

- ι

b →m oo

b

+

fta

" '1

b

~ ^

θ) = f ~ θ = 5

!tmo÷ (2 - 2 √ b ) = 2 - 0 = 2

c

l i !no÷ f3 x ^

'

Um_ [3b1∕ 3 - 3 ( - 1)1/ 3 ] + lim + [3(1)1∕ 3 - 3c1∕ 3 ] = (0 + 3) + (3 - 0) = 6

⅛

=

b

1⅛

-

is i n ^ 1

χ

i«

=

b

¾

- (s i n ~ 1

b

“

s in

^1 °) = 1 "

0

= I

9.

f π = f Λ i- f ^ i = hm [In ∣ x — 1∣ ]^2 - lim [ln∣ x + l∣ ]^2 = lim Γhι∣ ⅛ 4∣ l 2 ,jb ljb J-∞Xz - l J-∞ x - l J-∞ x+ l b → —∞ l 1 b → —∞ L l b → —∞ L ∣ x + l∣ Jb = b ⅛ 0 (h ∣ ⅛ - l n l ⅛ l ) = l l l 3 - l n ( b J j ‰ ^ 1 ) = l n 3 - l n l = ln3

11∙

∕ 2⅛

=

b

1T

co

[2 ^ 1 ^ ^ 1 1 2 = ^ 2 ^ (21n∣ ⅛ l ∣- 2 1 n ∣ ⅛ I∣ ) = 2 1 n ( l ) - 2 1 n ( i ) = 0 + 21n2 = ln4

Section 8.8 Improper Integrals r °0

J i * = f 0 J1⅛ -∞ (x2 + 1)2 J-∞ (x2 + 1)2

= Dl→ im∞ ( - l

+ r

Γ - ⅛ . Γu = x2 + 1 Jo (χ2 + 1)2 ’ du = 2x dx

l)C J i∞m L [C- i - (-1J )] = ( - 1 + 0) + (0 + 1 ) = 0 +→

+

u = 02 + 20 15. Γ ^ 1 d⅜ 2 Jo √0 + 20 du = 2(0 + 1) d0 = √ 3 -0 = 3

17.

u = √x = ⅞ .

'∞

J dx . o (1 ÷ x)√x ’

J

b →∞

' O°

^+

£ Λ , do

⅛ T7=

d u

lim

o "i + ι

1™ f ττ~ = ^ m [ V “ lk ^ b → 0+ *'b 2 v " b → 0+ v

= 2 v"

b→∞

j

/ ⅛2 =

( v ^ - ∖ ∕b ) b → 0+ v v /

[2tan - 1 u]j o0

lim

ou + ι

i™ b

l

b →∞

(2 tan^ 1 b —2 tan~1 0) = 2z z(5) —2(0) = τr

= lim

J0

→∫ ⅛ + D =b⅛ HK+ c⅛ HK

J

=

(i + v2 ) ( i v+ tan-1 v)

h

1 + tan" 1 v ∣ ]0 = b lu∏o [In ∣ 1 + tan~1 b ∣ ⅛‰ [1∏∣ ] - In ∣ 1 + tan~ 1 0∣

b

= In (1 + f ) —ln(l ÷ 0) = In (1 4- f ) 21. f° 0ee d0 = J -∞

[0eβ - eβ l ” = (0 ∙ e0 - e0 ) -

lim l

b →-∞

= —1 —b lim

JO

7

V

[beb - e bl = - l -

lim l

b →- ∞

j

lim u

b→- ∞

(⅛ k

(ΓHopitaΓs rule for ^ form)

( ^ t)

= - 1 - 0 = -1 w

e

23. [

" ^ 5 ^

∙ ∕o

⅛

b

b →m 2- Cs i n ~ 1

29. f —7=— = J i s√s 2 - l

lim

J -1 √ W

b → 0^

j

l

f

v2

b →0 ∖

33- Γ Γ π ∕2

n

f e

f ⅛ =

→ o + Jc √×

tan0d0 =

b

⅛

M H ilΓ -, =

.

lim_ [—In ∣ cos 0∣ ]o =

b→f

47

4

lim b

⅛ τ+ 0

-t0 + ( T J

τ

= - i

s in

^1 0 = I -

0

= f

sec- 1 b = 5 - 0 = ⅞ 3

lim b → 0~

3

[ - 2 λ ∕ → ] b + lim

l

lim

v

j

-'

c

→ 0+

[2 Jx ]4 l

v

j c

2√c = 0 + 2 + 2 ∙2 - 0 = 6

c → 0+

/

=

0

→o+k2

b → l+

lim c

b

( s i n ^ 1 2) ^

«= b ^

-‰ +

lim f ⅛ l n b - ⅛ ^ = - ⅜ -

4'

→m o÷ ( τ ) = - 5 +

Jb

-'∖ F×

7

= (∣In 1 —⅛) -

Jb

lim_ ( - 2 √ ^ b ) - ( - 2 √ < i j ) + 2 √ 4 -

=

0

b

(1 —eb) = (1 —0) = 1

lim

b → -∞ '

[sec“ 1 s ] = sec- 1 2 — lim

lim b → 1+

31. f -τ∏ =

35. J o

4

= ^ H

( ¾

=

⅛

Jb

t

Γ ⅛ ln x -⅛ 1

lim

b → 0+ L 2

Jo

[ex]θ =

lim b → -∞

J ∞

25. f x ln x d x =

27

ex dx =

dx = f

J ∞

1

⅛

J b ^ ∣ ] - b∙ l ≡ ⅛ l = 0 - M D = ^

] ÷ In 1 = liin_ [—In ∣ cos b∣

b→J

]= ÷ ∞ , lim_ [—l n ∣ co sb∣

b→5

the integral diverges = f ^ ∙ Jo √x

SinceO < ⅛ < 4 = for allO < x < τr and Γ ⅛ — — Jo √x — √x — √x

271

Chapter 8 Techniques of Integration

272 , ∙n

2

„

x^ 2 e^vx dx; β = y

J= 0 ÷ e“ θ

41. Jo

-,

p l∕ln 2

→ /

Lx

1/ln2

J y2e -v dy v2 - y

= ι

J l ∕l n 2

e~y dy = j

.

[-e^ y] 1 ,ln2 =

lim

L

b → ∞

J 1 /ln 2

[ -e - b ] - [—e -1/ln2 l

lim

l

b → ∞

j

j

l

= e~1/ln2, so the integral converges.

’ ^

√ t+ ⅛ ∏ t

J∞

nce

⅛r θ ≤

t

≤

π

,0 ≤ ^

≤ ^ and f

sin t

Q

^ converges, then the original integral

converges as well by the Direct Comparison Test. 43. f τ ⅛

= f τ ⅛ + f τ ⅛ and f ⅛

J 0 1~ x

1- x

Jo

diverges =>

lim Γ∣In ∣ l z⅛ ∣ E =

=

J o 1~ x z

J ι 1 —x

b → 1-

11—x l J θ

^2

lim

⅛ In ∣ p⅛∏ - 0 = ∞ , which

b → 1- L2

I 1 —b IJ

’

γ ⅛ diverges as well.

χ ∣ 45. J j ln ∣ dx = J j ln (—x) dx + J^ In x dx; J^ In x dx =

lim + [x In x - x] ’ = [1 ∙ 0 - 1] -

lim + [b In b —b]

χ ∣ = —1 —0 = —1; J* ι l n ( - x) dx = —1 => J* 1ln ∣ dχ = ~ 2 converges.

47. J*1 τ ⅛ ; θ ≤ τ⅛ - ≤ ⅛ for 1 ≤ x < ∞ and f

ι

⅛r converges => f

η ⅛ converges by the Direct

Comparison Test. 49. £

lim √ v -l

J2

=

V→ ∞

( 1 )

hm

which diverges =>

=

A

v→∞

√ V -1

hm

— =

v→∞

∕1_ 1

∏= 1 and f √ l-0

Ji

⅜ √V

=

[2√vl * = ∞ , b→∞

l

v

j 2

diverges by the Limit Comparison Test.

poo

( - ⅛ + 5) = ∣ =^ J o ^ η

=

b

lim

=

b

hmθ [-2x

,

converges by the Direct Comparison Test.

53.

2 +∞s x

55. f Jπ

57∙

π

X —

( ^ ^"^)

=

X

—

Jπ

X

=

l*m

b → ∞

l

^

⅛

=

+ 1 conver es

δ

1 and

X °⅛ =

b

1i m

00

b

lim

^x

c o n v e r 8e s

by the Limit Comparison Test.

[ln x]b = ∞ , which diverges jπ

[ - 4 t- 1∕2 ] b =

°

b

lh∏o ( ^ + 2) = 2 ^

f ^

converges

by the Limit Comparison Test.

59, J i 7 ⅛ 0 < ^ < γ fo r χ > 1 and J^ γ diverges => f

=

+~

^ ^ fι ^

2+^0sx j χ div e r ge s by the Direct Comparison Test.

X ~ i½ ;t ⅛ ^ J4

j = . lim

dx; 0 < ∣< Lts∞* for x > π and f v

X

=> j

1∕ 2 ]

[-2e^ χ∕ 2] 1 = lim

by the Limit Comparison Test.

(-2 e -b/2 + 2e"l/2 ) = ^

ι

^

l

diverges by the Direct Comparison Test.

=> J ι e-x/2 dx converges =>

^

x

χ

converges

’

Section 8.8 Improper Integrals n °o

-^ = = 2

-∞ √ x 4 + l Γ*σo ι

J

⅛,

∙ f 0o- A4 = =

b

| = ^lim *2 ι

Γ ,φ

Jo ^ * + l , Jo

[~ ^ ] 1 -

√x + ι

lim

Γ , dx 4

Jo

i

Γ √^ J -o o ^ ^

ln 2

j ; [t = In x] → ∫ Γo ⅛ =

b

Hm+ Γ[ ^,

f ' ]1

ln 2

=

X4

I

I

Γ^

^r Jι ×2

I

converges by the Direct Comparison Test.

b

lim + ^, . τn + ⅛

(In 2)→

=> the integral converges for p < 1 and diverges for p ≥ 1

^ ) J2

x(⅛x)p

» Ct = ln χ ] → J ln2 ^ and this integral is essentially the same as in Exercise 65(a): it converges

for p > 1 and diverges for p ≤ 1

69. V = f

2πxe^x dx = 2π f xe

Jo

71. A = ∫ =

73∙

00

x

dx = 2π lim

Jo

(sec x - tan x) dx =

[-xe~ x —e~x]∩ = 2π [ lim jo

b →∞

lim

(—be -b —e - b ) — 1^∣= 2π 7

Lb→ ∞

sec x + tan x∣ sec x∣ [In ∣ ~ In ∣ ]θ =

lim

l+ ^ ( ln ∣

J

∣- l n ∣ l÷ 0 ∣ )

lim _ In 11 + sin b∣= In 2

b →5

fS ^

dx =

b

⅛ n00 [- I e ^3^l 3 =

b

⅛ nc o ( - 3 e ^3b)

-

( - 5 e ~3 3 ) = 0 + | ∙ e- 9 = j e- 9 f

oc

≈ 0.0000411 < 0.000042. Since e^x ≤ e^3x for x > 3, then J 3 e~x dx < 0.000042 and therefore

J Q e-χ2 dx can be replaced by f

e~χ2 dx without introducing an error greater than 0.000042.

(b) ∫θ 3 e- χ 2 dx= 0.88621

(b) > int((sin(t))∕t, t=0..infinity); (answer is ^)

273

274 77∙

Chapter 8 Techniques of Integration f (χ )

^

e " χ2β

= ⅛

f is increasing on (—∞ , 0]. f is decreasing on [0, ∞ ). f has a local maximum at (0, f(0)) = ^0, ^ ^ (b) Maple commands: >f: = ex p (-xA2/2)(sqrt(2*pi); >int(f,x = -1..1); >int(f, x = -2..2); >int(f, x = -3..3);

≈ 0.683 ≈ 0.954 ≈ 0.997

(c) Part (b) suggests that as n increases, the integral approaches 1. We can take ι

e-x /2

enough. This is because 0 < f(x) < w∞

p∞

'oo

J n

p -n

n

J —oo

f(x) dx as close to 1 as we want by

f(x) dx and ι f(x) dx as small as we want by choosing n large J for x > 1. (Likewise, 0 < f(x) < e for x < —1.) v2

v2

f(x) dx < / e^ dx. J e^ dx = lim / e^ dx = lim [ - 2 e " n

'∞

Un

nC

v2

C→ ∞

j

2e-n /2

As n → ∞ ,

x72

x72 ]c

= lim [ —2e“ c/2 + 2e-n /2 ] — 2e-n /2

n

C→ ∞

∏

C→ ∞

→ 0, for large enough n, f f(x) dx is as small as we want. Likewise for large enough n, J n

f J

—∞

f(x) dx is as small as we want. pb

pa

pb

79. (a) The statement is true since J ∞ f(x) dx = j and f (b)

pa

pb

7 = √X + 1

p∞

S 7 J -oo √ x 1

⅛ Γ = J Ix

√ lx

i ;

2

+ 1

√X + 1

S u 1

1 and

=

00

poo

p∞

_|x|

⅛

dx = 2

⅛

⅛

X∣ ∣ ^h I

= 2 ^ lim

poo

7T

poo

f(x) dx + / f(x) dx J b

diverges because lim

√X2 ÷ 1

6

X→ ∞

/

φ

( -

r

⅛ l

diverges; therefore, / “ ^

O

J

τ

diverges

poo

≡ τ > ; z ⅛ τ = ejx ÷⅛e

ezx ÷ 1 ’ ezx + 1

2

J —oo

÷ 5

τ 2

= x⅛ n∞ √ ^ ⅛

→ f2 ¾ ⅛ =

oq e

pb

pb

poo

poo

/

87. Γ∞

pb

pa

2

J - oo JΓ ex ÷ e

pb

f(x) dx —J* f(x) dx

f(x) dx + I f(x) dx — ι f(x) dx + I f(x) dx Ua Ua Ua

f(x) dx ÷ / f(x) dx + ι f(x) dx = I Ub U —∞ J a

= x⅛ n∞ ⅛

85. J

pa

poo

I f(x) dx + I f(x) dx = I Ua U—oo U —∞ U —∞

83.

poo

f(x) dx exists since f(x) is integrable on every interval [a, b].

= / I 81. J -oo

poo

f(x) dx + J* f(x) dx, J b f(x) dx =

⅛

Γ

2

pb

Jθ e -x dx = —2 ^lim

s⅛ x∣ f°° ∣ + ∣ c ⅞ x ∣d χ > Jθ

(—e^c + 1)' = 1

v

converges

e^x dx = 2 ^lim

x ∣d χ =

[—e^,j lθ = c →lim∞ < ex⅛ andJoI e⅛ = c →lim ∞ 1 °

x

X 4- 1

2

Γ∞s i n ⅛ Jθ

os¼ X 4^ 1

[e^x]0 = 2, so the integral converges.

dχ = 2

x + 11]θ = ∞ , which diverges => § ∞ ]sιnx^^^ [In ∣

f b^ lim ∣ ^ , 0 0 J ^ X 4- I

j χ c∣ iv e r ge s

dχ

Chapter 8 Practice Exercises CHAPTER 8 PRACTICE EXERCISES 1.

f x Z ^∑ 9dχj

u = 4 X2 - 9 du = 8x dx

3.

J x ( 2 x + I ) 1 / 2 dx;

u = 2x ÷ 1 du = 2 dx

(2x+ I)3/2 , 6

_ (2x+ I)5/2 “ 10 x⅛

→ ∣∫ ⅛

. Γ u = 9 - 4 t4 √ 9 - 4t4 ’ du = —16t3 dt

+ 4

X

u = z5∕ 3 + 1

→ 2J

s 9 2 19. f e sin (e ) cos (e") d0;

21. J > - d x = ‰

23.

+

’

27.

29.

1 f⅛ _ _ 1 , J u≡ 2u ^r

du =

r v

=

- l J e

u = cos (e^) du = - sin (e 0 ) ∙ e* d^

2 . 2 u 5∕3

_

dt 4

+

c

=

A ( z 5∕3

+

1 )5∕ 3

+

c

1 2(1 -∞ s2e)

u

d u = -ie

f -u

u

+C = - ie

1

1

u + C = sin

1

(2x) + C

u = ∣t du = ∣dt

+ C

du = - ∣u 3 + C = - ∣co s 3 (e s ) + C

x

! ⅛ = sin

cos2x

2

= In ∣ u ∣+ C = In ∣ ln v ∣+ C

1

u = 2 + tan duu — _dχ α “ x2 +l

- 4 X2 ’ du = 2 dx

+ C

0

→ - ⅛∕ ⅞ = - ⅛'» l“l + c = - ⅛⅛ ∣ 3 + 4 ∞ s >∣ +c

t

∫^

u = 2x

2dx

-9 t 2

2∕3

C = - ^ ≡

+

du = - dv

C_____ dx_____ . 25. J (x2 +l)(2+tan^1 x) ’

c

+ C

C

dv . Γ u = In V ' v ln v

,

u

2

u = cos 2x du = —2 sin 2x dx

cos2x dx; 17. J (sin 2 x )e

+

= p n ∣ u ∣+ C = i l n ( 2 5 + y 2 ) + C

→ - ⅛ f ^ = - ⅛ . 2 U 1∕2 lo J √u io

u = 1 - cos 20 du = 2 sin 20 d0

15. J r f ¾ ⅛ : Γ Λ

9 )3/2

√ u du = i ( ∫ u 3 ∕ 2 du - J u 1 ∕2 d u ) = i ( ∣u 5 ∕ 2 - ∣u 3 ∕ 2 ) + C

du = ∣z 2 ∕ 3 dz

sin 20 d^ (1 - cos 20)2

= ⅛ (4 χ 2 -

c

⅛ f ⅛ = ⅛ - 2U1∕ 2 + C = ® lo lo J √u °

t 3d t

11. f z 2 ^3 (z 5 ∕ 3 + 1 ) 2 ^3 dz;

= I ’ 5 u3/2 +

r v

du = 16x dx

f ydy . Γu = 25 + y 2 J 25 + y , du = 2y dy

du

→ I / ( ¥ )

. Γu = 8x 2 + 1

√8χ2 +1 ’

7.

I J √u

7 = 5 sin

275

276 31 j l

Chapter 8 Techniques of Integration

f ∙ J

4 dx _ 5x√25x 2 - 16

33.

35- ∫ ^ ⅛

37. J 39∙

u = ∣t du = | dt

A 9T?

∕

^

s

= ∕

1

= I sec

7

⅛ ¾

= ≡ -(⅛

= JJ⅛ ¾

⅛

→ 5 ∫ T ⅛ = 5 ta n

= l

= ∫ i7∑7i W

lm

2

1u

+ C = ∣tan

1

(f)+ C

∣ ⅜∣ +C

)+ C

-* (4 2 )+ C

⅛

= s~ - , ∣ * - >∣ +

C

41. y sin2 x dx = J* -—≡ ^ s dx = ∣—5a^ i + C

2 3 43. y s in f d0 = y (1 - cos f ) (sin f) d^;

u = cos ∣ du = - “ sin | d0

→ - 2 y ( l - u2 ) du - ⅛ - 2u + C

= | cos3 | - 2 cos | + C u = 2t du = 2 dt → ∣y tan u sec2 udu - ^ ∫ tan u du = ∣tan2 u + ∣In ∣ cos u∣ + C = ∣tan2 2t + ∣In ∣ cos 2t∣ + C

45. y tan 3 2tdt = y (tan 2t) (sec2 2t - 1) dt = ytan 2t sec2 2t dt -

f

tan2tdt;

= ∣tan2 2t - ∣In ∣ sec 2t∣ +C 47∙

∫2 ≡ ⅛ 7 = ∫ ⅛

49∙

£^ √

csc2 y

-

1 dy

csc 2x + cot 2x∣ +C = y esc 2x dx = - | In ∣ s i n y∣ = ∫ √ ° t y d y = lln ∣ l ^ = in ι - in ^ = ι∏ √ 2

dx = y j^ sin 2x dx - £ s i n 2x dx = - [ ⅛ χ ] f + [ ⅞ i ] ^ 2 51. y π √ Γ ^ 2 ^ dx = £ j s i n 2x∣ = - ( - ∣4 ) + [ F ( 4 ) ] = 2 53. y ζ 22 √ Γ → os2t dt = √ 2 ∫ ^ ∣ sin t∣ dt = 2 √ 2 £

55. y ⅛ 57∙

^

t dt = [ - 2 √ 2 cos t]

= x - y ^ = x - 2 t a n - 3 (∣) + C

2 x -l∣ £ ⅛ ⅛ dx = y [(2x + l ) + ⅛ ] d x = x + x2 + 2 h ∣ +C

59∙ y ∣ ≡ d y = y

^

⅛

√

+

= ln(y 2 + 4 ) - i t a n → ( * ) + C

2 f √ ½ d √ 4 -t2

= - √ 4 → s ÷ 2 sin -1 Q ) + C

= 2 √ 2 [0 - ( - l ) ] = 2 √ 2

Chapter 8 Practice Exercises f tan x dx J tan x + sec x

f

_

J

sin x dx __ ∖ (sin x)(l - sin x) < sin x + 1 J 1 - sin2 x

= "J¾ Γ “

+ J

dx

= ⅛

y = 5 —3x dy = —3 dx

65. J*sec (5 —3x) dx;

j sin x - 1 + cos2 x cos2 x

J

- tanx + x + C = x - t a n x + secx + C

sec y + tan y∣ = - ∣J*sec y dy = —∣In ∣ + C

→ J*sec y ∙ f - ^

= —∣In ∣ sec (5 —3x) ÷ tan(5 - 3x)∣ ÷ C ÷ C 67. f cot (∣) dx = 4 J cot Q ) d (∣) = 4 In ∣ sin Q ) ∣ u = 1 —x du = - dx

69. fn ∖ j 1 —x dx; =

2(1 _ x )5∕2 _ j

(1

_

- J l - u) λ∕u du = J*(u3 ∕ 2 - u 1∕ 2 ) du = ∣u5 ∕ 2 - ∣u3 ∕ 2 + C

x ) 3/2 + c

=

{ ⅛

_2

, ( √ H

+c

→ J ᅟ ∕ tan2 0 + 1 ∙ sec2 0 d0 = J* sec3 θ d0

71. = s^⅛ 9

+

^2 ∫

sec

^ d⅛

(FORMULA 92)

⅛ ⅛ + iln ∣ +C = sec0 + tan0∣

du 1 + u2 ’

73.

+C + ∣l n ∣ z+√Γ+7∣

^

⅛b

u = tan 0 du = sec2 0 d0

= In ∣ √ 1 + u2 + u∣ sec 0 + tan 0∣ + Ci = In ∣ + Ci

x = sin θ dx = cos θ d0

75.

x = sin θ dx = cos θ d0

→ ∫

= i0 -is in 0 c o s 0 = ⅛ 79. f ,d2* J

χ ∕x —9

cos 0 d0 in2 θ cos 0

5¾

^

= f

s in 2 θ d θ

=f ±

^

d0 = i 0 - i sin 20 + C

i

x = 3 sec 0 dx = 3 sec 0 tan 0 d0

f

l

J

3 se c 0 ta n 0 d 0 _ T ⅜ sec^9

Γ 3 se c 0 ta n 0 d 0 _

^ J

-

T ≡

-

(∖ ^^ ∩ A ∩

- J sec 0 d0

i± ⅞ ≡ ∣ + C 1 = !n ∣ x+ 4 M —In ∣ sec 0 + tan 0∣ + Ci = In 5 + √ ( 5 ) ' - l + Cι = l n ∣

81.

w = sec 0 dw = sec 0 tan 0 d0 = tan 0 —0 + C = ᅟ ∕w

2

— 1 —sec

"4 ∫ ( ^ ) ∙ 1

sec

∣ +C

0 tan0d0 = J*tan2 0d0 = J*(sec2 0 — 1) d0

w+ C

83. u = In (x + 1), du = ^ ∙ ; dv = dx, v = x; ∫ l n (x + 1) dx = x In (x ÷ 1) - J ^ ∙ dx = x In (x + 1) - J*dx + f ~ ∙ = x In (x ÷ 1) - x + In (x + 1) + Ci = (x + 1) In (x + 1) —x + Ci = (χ + 1) ln(x + 1) —(x ÷ 1) + C, where C = Ci + 1

277

278

Chapter 8 Techniques of Integration 1

85. u = tan

3x, du - 1 ^

; dv = dx, v = x;

i

∫ 1an -∙ 3x dx = x m - l 3x - ∫ ⅛ ⅛

; [ ^

7

^

→ xtan

1

3 x -j∫^

= x tan - 1 (3x) - £ In (1 + 9x2 ) + C 87. (x +

1)2

ex - ^ → ex

2 (χ + 1 ) - ^ ‰ 2

x

e

— — ♦e => ∫ ( x + l) 2 e x dx = [(x + 1)2 - 2(x + 1) + 2] e’ + C

0

89. u = cos 2x, du = - 2 sin 2x dx; dv = ex dx, v = ex ; I ~ f Q* c o s ^ x d χ — e * c o s 2x + 2 J e x sin 2x dx; u = sin 2x, du = 2 cos 2x dx; dv = ex dx, v = ex ; I = ex cos 2x + 2 ∣ ex sin 2x - 2 J*ex cos 2x dx] = ex cos 2x ÷ 2ex sin 2x —41 => I =

91∙ ∫ ½

93

= ⅛

=

∙ f ^ w

√

⅛

J ( ∣" ⅛

+

-

c θs

-

+ 2e⅛2χ

= 21n∣ x -2 ∣ -ln ∣ x -l∣ +C ( Γ w ) dx = In ∣ x∣- In ∣ x + 1∣ + ⅛

95. ∫ j ⅛ ‰ U c o s ^ y ] → √

^

=- j ∕ ^

+

+ C

l ∫ ⅛ = jin∣∑±2∣+ c

= M ⅛ f l +c ≈ - M ^ π l + c ∙ f ⅞

99

∙ ∕⅛ ⅛ v ^ = 2 ^ ( - S

^

+4

dx = J^ 7 dx - J* ^ ⅛ dx = 4 In ∣ x ∣- ∣In (x2 + 1) + 4 tan - 1 x + C

97

= ⅛1∏ ∣ ^

^

+

8(v-2) ÷ 8(v + 2)) dv = - 1 In ∣ v∣ + ⅜ In ∣ v - 2∣ + ⅛ In ∣ v + 2 ∣+ C

∣+ C

∙ ∕ ^ ⅛ 3 = 5∫⅛ H∕ I⅛ = h a ι r l t ~ ⅛ ^ ( ^ ) + ^ 5 ^ ^ - ^

101

103∙

J ⅛

⅛

dx

= ∫(

x

+ ⅛ 2 )d x = A ⅛ + 5 ⅛

= y + 5 In ∣ x + 2 ∣+ ∣In ∣ x - 1∣ + C

+ 5 ∫⅛

tan 1

~ 75+C

Chapter 8 Practice Exercises 'u = es - Γ du = es ds L ud5s = — u+ 1 J

109.

ιπ

+ J ⅛ = 1" ⅛ l + = - J ⅛

∕ τ ⅞ - ⅛ = - J¾ ⅛ ⅛ l = - √ > ^ + c (b) ∫

113-

279

7

⅛ ⅛ U y = 4 s ⅛ x ∣→

4

∫ 1¾

s i

= -

4

∞ s* + C = - i ⅛ ≡ + C = - √ l 6 ^ 7 + C

ω ∕ r ⅛ = - i ∕ ¾ ⅛ 1 = - i∣ " ∣ 4 - χ2l+c (b) ∫ ⅛ ! [ x = 2sinβj → ∫

u

⅛ g ≡

= ∕u ∙ M ( = - l ∙ ∣ ∞ sβ ∣ + C = - h ι ( 1⅞ 3 ) + C

= - I In ∣ 4 - x2 ∣ +C u = 9 - x2 du = —2x dx

US- ∫ ⅛ 1

11’ - ∫ ⅛

= i f ⅛

,

→ - 5 ∕τ

+ 5∫ ⅛

= - H " M + C = ln ⅛ + C = ln 7 ⅛ + C

= " I >- 13 - x∣ + ∣In ∣ 3 + x∣ + C = ∣In ∣ ⅛

∣+ C

119. f sin3 x cos4 x dx = f cos4 x(l —cos2 x)sin x dx = J*cos4 x sin x dx —J*cos6 x sin x dx = - ≡ ⅛ + ≡ ⅛ + C 121. f tan4 x sec2 x dx =

a

yi + C

123. J sin 50 cos 60 d0 = ⅛ f (sin (-0) + sin(110))d0 = ∣J*sin(-0)d0 + ∣J^sin(110)d0 = ∣c o s (-0) - ⅛cos 110 + C = ∣cos 0 —⅛cos 110 + C 125. J ^ 1 + cos(∣) dt = f

∕2 ∣ cos 51dt = 4 ^ 2 ∣ sin 11 + C

Es ∣< ⅛ ^ (Δx) 4 M where ∆ x = 3—1 = 2 ; f(χ) = 1 = χ 127. ∣

1

=> f'(χ) = —χ

2

=> f"(χ) = 2x

3

=> f"(x) = - 6 x

4

≠> f(4 )(x) = 24x- 5 which is decreasing on [1,3] => maximum of f(4 x) on [1,3] is f ^ ( l ) = 24 ≠- M = 24. Then ∣ Es ∣≤ 0.0001 ≠ ( ⅛ ) ( ^ ( 2 4 ) ≤ 0.0001 ^ ® (⅛) ≤ 0.0001 => ⅛ ≤ (0.0001) ® ≠> n4 ≥ 10,000 ® => n ≥ 14.37 => n ≥ 16 (n must be even) 129. ∆ χ = ⅛n

= ⅛o ^ = f0 => ⅜2 = ⅛ ; 12

∑ m f(x i ) = 12 => T = ( ⅞ ) ( 1 2 ) = π 5

0 π∕6 π∕3 π∕2 2π∕3 5π∕6 π

f(xi ) 0 1/2 3/2 2 3/2 1/2 0

m 1 2 2 2 2 2 1

mf(xi ) 0 1 3 4 3 1 0

χ> 0 π∕6 π∕3 π∕2 2π∕3 5π∕6 π

f(χi ) 0 1/2 3/2 2 3/2 1/2 0

m 1 4 2 4 2 4 1

mf(xi ) 0 2 3 8 3 2 0

Xi

7

6

Xo

xι X2 X3 X4 X5 X6

6

mf(Xi) = 18 and y = ⅞ => i=0

Xo

S = (⅞ )d 8 ) = π.

Xi X2 X3 X4 X5 X6

Chapter 8 Techniques of Integration

280

131. y„ = ⅛

= ⅛ [(-3 7 ® cos

= - ≡

s in

[3 7

Γ

(χ -

⅛

1 °D)

25] dx

+

= ⅛ [-37 ( f cos ( ⅛ (x - 101)) + 25x)] J65

cos [∣ (365 - 101)] + 25(365)) - (-3 7 ( f ) cos [ ⅛ (0 - 101)] + 25(0))]

( ⅛ (264)) + 25 + g cos ( ⅜ (-101)) = - g (cos ( ⅛ (264)) - cos ( ⅛ (-101))) + 25

≈ - g (0.16705 - 0.16705) + 25 = 25o F 133. (a) Each interval is 5 min = ⅛ hour. ⅛[2,5 + 2(2.4) + 2(2.3) + . . . + 2(2.4) + 2.3] = ^ ≈ 2.42gal (b) (60 mph) ( g hours/gal) ≈ 24.83 mi/gal 133∙

∫.⅛ ⅛

=

*3 ’ - O

1 4 3

X *√5⅛ =

+ X ⅛

= Γ

⅛

⅛

x2 e^x dx =

f

j

⅛

= ∫ ⅛

139- Γ

141.

b

θ

=

2

145∙

1

I”

2

+

∙3 b ⅛

N +

⅛

l

∫ p0o o 4 ⅛

i ∫ p jo o ⅛

=

= i

b

= 2fZ ⅛

u= du = = ^

diverges

[ h

n

b

1

-

b

l ⅛ , [l"l +

s i"^ 1

(?) = 5 - ° = 3

b l',

⅛

6

) =

l ] - l " 1+

1 = 0 -1 -(1 )= ^ 3

( - b 2 e^b - 2be'b - 2e^b ) - (-2 ) = 0 + 2 = 2

κ

7

^ ' ( τ ) ] .0 = 5

li m

b

[5

00

ta n

7

^

1

(τ )] -

5

ta n

^

dx

J

X (X2 +

.

1)2

+ lim L

2

J

⅛ 1 ∖

e

2

^

b

= ⅛ /

0)

b→ ∞

⅛

+ lim L

'

[dx =

2

2

- l ]

J

2

+ 2u - 2 In ∣ 1 + u∣ +C

^

sec2

&

_

C sec⅛⅛

= ∣y3 ∕ 2 - 4y 1∕ 2 ÷ C = j (2 - x)3 ∕ 2 - 4(2 - x)1∕ 2 ÷ C

(v ^ )3 /-----= 2 v -3 I _ 2 √ 2 - x

'“ ■ ∕+ ⅛ ÷ 2 √

δ⅜

⅛ T = t''" - , ⅛ - l ) + C

183. J 02 tan(0 3 ) d0 = ∣J*tan(0 3 )d (0 3 ) = 5 In ∣ sec 03 ∣ +C 185∙

∫^ f ⅛ f

IRQ J

td t

d z

_

f cot 0 d0 1 + sin2 0

= lJ (l + ⅛ - ⅛ )

_ 1 f d ( 9 - 4 t2 )

_

Γ cos 0 d0 . J (sin 0)(1 + sin2 0) ’

1

χ

z∣ - ⅛ - i = Zl n ∣

l n (z 2

+ 4) - l

^ =

n

c o s

^ # J0

__

Γ J

_ dx x (l + x 2 )

J

Γ dx _ Γ x dx x J xi + l

— ∣In (1 + sin2 0) + C = In ∣ sin 0∣ 191∙ J ^

^

:

[\^

=

x]

→

ta n ^1

/9 _ 4 t 2 I C

x

J

dz

J* M ⅛ L^ = In ∣ secx∣ + c = In ∣ sec v ∕y ∣+ C

l+

c

281

Chapter 8 Techniques of Integration

282

193. J ⅛

= J(-l +

τ

⅛

) d 0 - √ d 0 √ ^ + J ^

0 -2 ∣ 0 + 2∣ = -^ -ln ∣ + ln ∣ +C

f ⅛ ∣+ C = -0 + ln ∣

∕

u = sin 1 x du = √ ½ 2 √ l-x

cos (sin 1 x) dx √ 1- x2

1

→ J cos u du = sin u ÷ C = sin (sin

x) ÷ C = x + C

197. f sin ∣cos ∣dx = f ^ sin ( ∣+ ∣) dx = ∣J*sin x dx = - j cos x + C

199∙

f

⅛ ⅛ = l n ( l + e t) ÷ C

x = In y dx=⅛

201.

→ f Jo

⅛ dx = f xe~2x dx = e3x

Jo

Γ- * e^2x

lim b→∞

l

2

1 Ae - 2X 4 .

dy = e x dx ~ b ⅛ ∞ ^2e ^

4 *2b)

(θ

Γ cot v dv _ Γ cos v dv . 203. J In(sin v) J (sin v) In(sin v) ’

D “ 4

U

^n (sill v) du = cos v — sin v

→ ∫ ⅛ = ln ∣ ln (sin v )∣ u∣+ C = l n ∣ + C

205. J*eto ^ d x = J * y x d x = ∣x3 ∕ 2 + C

∞ 7∙ fτ⅛ S ⅛ >

→ - ∣∫ T ⅛

du = -5 ° s ⅛ 5 td t

209. f (27)3β+1 d0 = I J

(27)M + I d(3β +

1 )- ⅛

= - ⅛ '> " - , " + C = - S ≡ ^ '< ∞ 5 > > + C

(27)w + 1 + C = ∣( ^ ) + C

217. If u = J^ ^ 1 + (t - 1)4 dt and dv = 3(x — 1)2 dx, then du = ^/1 + (x — 1)4 dx, and v = (x — 1)3 so integration by parts =+ Jθ 3(x - 1)2 [∫θ √ 1 + (t — 1)4 dt] dx = [(x - 1)3 f

o

√ Γ ∏ t ^ l 7 dt] θ

- ∫ o'(x - 1)3 √ l + ( x - l ) 4 dx = [ - | (1 + (x - 1)4 )3 /2 ] θ - ⅛

219. u = f(x), du = f'(x) dx; dv = dx, v = x; i/ 2

f(x) dx = [x f(x )]^ - £

= ( τ - ? ) - [sin ⅛

2

/2

xf'(x) dx = [⅞ f (⅞ ) - I f ( f )] - £

= f (3b - a) - [ ( - l ) - 1] = f (3b - a) + 2

/2

COS X dx

Chapter 8 Additional and Advanced Exercises CHAPTER 8 ADDITIONAL AND ADVANCED EXERCISES 1

1. u = (sin

x)2 , du = - ψ = = y ; dv = dx, v = x;

§ (sin- 1 x)2 dx = x (sin- 1 x)2 —J ^ = u = sin- 1 x, du = =

-f ^ ^ ^

; dv = — ⅛ = , v = 2 √ T ^ χ 2 j

√ ι - x2

2

(s in

^ ;

√ ι - x2

χ ∕ 1 —x2 —f 2 dx = 2 (sin- 1 x) ᅟ ∕1 - x

~1 x )

2

- 2x ÷ C; therefore

J (sin- 1 x)2 dx = x (sin- 1 x)2 ÷ 2 (sin- 1 x) √ 1 —x2 - 2x + C 1 x,

u = sin

du == - 7=

√ 1 - x2 z

; dv = x dx, v = ⅛ ;

2 2 x Jfx s ι ∙n - ι1 x dj x = x2 sι∙∏- ι1 x J Γ 5 χ√ d⅛ q

= y sin- 1 × ~ I f sin2 0 d0 = y sin = ⅞

sin~1 x

5∙ ∫ T ⅛

+ ^

4

-⅞ n-

x

d0

= ∫^ ^ e ‰

f

dt = COS0d0 1 Γ

du

= ∣In (t —/ 1 -

f

11. 13.

d x

7+4

-

_

J(χi

t2 )

+

-

J

sin- 1

2) 2 -4x>

s in 0 - c o s 0

dX

^^ f

x ⅞ + 2 1+ I [t a n ~1 (χ +

n

lim o £ In ^ / 1 4 ^ =

→ ∫

2 In

n

⅛∕U* ψ I

1)

,

j

ta n 0 -l

I

2 (x -l)2 + lj

u = tan 0 du = sec2 θ dθ . d0 = ⅛ .

_____ du_____ ( u - l)(u 2 ÷ l )

Z

Z

I

ι ∣- i 0 + C

SCC ι∕

I

z

d x

u λ

+ tan" 1 (x - 1)] + C

x = [—costl lim -X X →∞ L

s in tdX→∞ t= lim L

J

(x2 + 2x + 2∏x 2 - 2 x + 2)

= ⅛ln /

+ c

1+ C

2x÷2 . _ _ 2 __________ 2 x - 2 x2 - 2 x +2 x2 + 2x + 2 r ( x + l ) 2 + l

lim

f_ d e _

l ^ ⅛ ∣- l t a n - 1 u + C = i l n ∣ ¾

1 2

= ^ i

X→ ∞ J -χ

x dx = ⅛ sin- 1 x - f 5 is ⅛ ^ ⅛ 2 J 2 cos v

x - ∣(f - s v ) + C = ⅛ sin" 1 x + andcosj^

c o s ⅛ d i- =

1 f u du _ 2 J u2 + 1

—|

J

+ C

→

2 J u2 + 1

1

→ f x sin

= ∫⅛ ^ d 0 = l∫(s e c 2 0 + l)d 0 = ⅛ ^ ^

t = s in 0

— 1 f _du_ _ 2 J u- 1

1

x = sin 0 dx = cos 0 d0

[ - cos x + Vcos(-x)] = lim ZJ X → 00

lim o £ In (1 + k (1)) (1) = ∫ 0'ln (l + x) dx;

v

χ =

(—cos x ÷ cos x)x = lim → ∞ 7

0= 0

θ u= 2

u du = [u In u - u] J = (2 In 2 - 2) - (In 1 - 1) = 2 In 2 - 1 = In 4 - 1

15. ^ = √ c o s2 x => 1 + ( * θ = √ 2 [sin t]o/4 = 1

= 1 + cos2x = 2cos 2 x;L = . £ 7 y 1 + ^ √ c o s2 t) dt = √ z2 f j

√ ,cos2 1dt

283

284

Chapter 8 Techniques of Integration

17. V = f b 2 π ( ^ " ) ( hs⅛ J

d x = f'2 π x y d x J

∖ radius/ y height J

a

= 6π J^ x2 √ Γ → dxj

y

J

Q

u = 1- x du = —dx χ 2 = (1 - u)2

→ —6π J ι (1 —u)2 y ∕u du = —6π f

i

(u 1∕ 2 - 2u 3 /2 + u 5 ∕ 2 ) du

= - 6 π [∣u 3 ∕ 2 - ∣u 5 ∕ 2 + j u 7∕ 2 ] ° = 6π ( ∣- ∣+ j )

= 6 √ * ^ ) = 6

19.

v

= Γ J

2π

a

( .S

7Γ ( 1⅛)

(⅛ ∣ Ω

dx

= > y

= f'2 π x e x dx J

∖ radius/ y height J

Q

= 2π [xex —ex] J = 2π

21. (a) V = ∫ ᅟ [ l - ( l n x ) 2 ]dx = π [x —x(ln x)2 ] j + 2πJ^ In x dx (FORMULA 110) = π [x - x(ln x)2 + 2(x In x - x)] * = π [ - x - x(ln x)2 + 2x In x] ® = π [—e —e + 2e —(—1)] = π (b) V = J J π (l —In x)2 dx = π J*ι [1 —2 In x ÷ (In x)2 ] dx = π [x —2(x In x —x) + x(ln x)2 ] * —2πJ^ In x dx = π [x —2(x In x —x) + x(ln x)2 - 2(x In x - x)] * = π [5x —4x In x + x(ln x)2 ] * = π [(5e —4e + e) —(5)] = π(2e —5)

23. (a)

lim x In x = 0 => lim + f(x) = 0 = f(0) => f is continuous x →0+ x →0 πx 2 (ln x)2 dx;

(b)

u = (In x)2 du = (2 In x) ⅛ dv = x2 dx v

= π

( f ) ( ln 2 ) 2

π

mo+ ⅛ (b →

(ln x)2 ] /

Γ ( f l (2

= τ

- ( ∣) b Hm+ [ ⅛ l ∏ x -


2

lim In ^ i ^

/

4 ’

2

=

b

1⅛

o

^ ≠ 1^ > lim

integral diverges if a > ∣; for a = ∣ lim b →∞ 2 X

d χ

-

b→∞

[∣In (x2 + 1) —∣In x ] a = y =

b

b

∏mθ I2f1o,1 121+1£ X

lim b2 (a ^ = ∞ if a > ∣ => the improper

1 = lim J l + ⅛ = 1 ≠" lim ⅜ In ^ ⅛ p -— In 2 1∕ 2 ^⅛2 b b2 2 b

b →∞ V

lim

b→∞

⅛ b⅛ ^
the improper integral diverges if a < ∣; in summary, the improper integral

285

Chapter 8 Techniques of Integration

286 *∞

Ji (x¾^ " ⅛) ^

x

converges only when a = ∣and has the value - ^

39. A = J ι ∣ converges if p > 1 and diverges if p < 1. Thus, p ≤ 1 for infinite area. The volume of the solid of revolution about the x-axis is V = J^ π ( - ) 2 dx = π ^

~ which converges if 2p > 1 and diverges if 2p ≤ 1. Thus we want

ρ > ∣for finite volume. In conclusion, the curve y = x^p gives infinite area and finite volume for values of p satisfying I < p ≤ 1∙ 41. e2x

(+)

cos 3x

—∣ cos 3x I = y sin 3x + ^ 43.

sin 3x

cos 3x - 1 1 => y I = ^ (3 sin 3x + 2 cos 3x) => I = ⅛ (3 sin 3x + 2 cos 3x) + C sin x

(+)

" * ^ —cos X

3 cos 3x ■

—9 sin 3x

***^ —sin x

I = —sin 3x cos x + 3 cos 3x sin x + 91 => —81 = —sin 3x cos x + 3 cos 3x sin x κ τ __ sin 3x cos x - 3 cos 3x sin x ∣p 45. e“

(+)

sinbx

=+ I = J⅛∑ (a s *n bx - b cos bx) + C 47. In (ax)

(+)

1

I = x In (ax) —J* (∣) x dx = x In (ax) - x + C 49. (a) Γ y' = -e ^ x => 2y' + 3y = 2 (-e ^ x ) + 3e^x = e“x (b) y = e^x ÷ e’ 3x/2 => y' = —e^x — ∣e~3x/2 => 2y' + 3y = 2 (-e~ x — ∣e"3x/2) + 3 (e^x + e~3x/2) = e~x (c) y = e^x + Ce~3x/2 => y' = —e^x — ∣Ce“3x/2 => 2y' + 3y = 2 (—e^x — ∣Ce-3x/2 ) + 3 (e^x + Ce-3x/2 ) = e^x 3∙

y= I £

τ

dt

^

y' = - ⅛ Γ τ d t + (D ( τ ) ≠

χ2y'

= - /3

dt

+ ex = - χ ( ∣∫ 3

+ e^= - χy + e^

d t)

=> x2 y' + xy = ex 5. y = e^x tan- 1 (2ex ) => y' = - e ^ x tan- 1 (2ex) + e -x [7 7 ⅛ ψ ] (2ex ) = - e ^ x tan- 1 (2ex) + 1⅛ => y' = - y + τ ⅛ 7.

y

__ cos x x

v

y

^

;

y' + y = r ⅛

/ __ - x sin x - cos x x2

=> xy' + y = - s i n x ; y ( f ) =

y

s^

/ __ _

y (-

sin x _ x

ln 2 )

=

e4

^ l"2*t a n ~1 (2 e ^ ln2) =

1 f cos x ∖ x ∖ x √

v

y

/ __

≡

λ

2

tan- 1

1

=

2

(?) = 5

— I => xy' = —sin x —y

- 0

9. 2^/xy ⅛ = 1 =^ 2x1∕ 2 y 1^2 dy = dx => 2y1^2 dy = x 1∕ 2 d× ^

J 2y1∕ 2 dy = J*x 1∕ 2 dx => 2 ( j y3 ∕ 2 )

= 2x1^2 + Cj => j y3 ∕ 2 - x 1∕ 2 = C, where C = ∣Ci 11. ^ = e , ^y => dy = exe^y dx => ey dy = ex dx => J*ey dy = J*ex dx => ey = ex + C ≠> e, - e *3

⅛

=

ᅟ ∕y

co s2 √ y

≠ ' d y = ( √ y cos2 ^∕y)

dx

=> ^ ^ ⅛

~

dx

^, / ^ ^ d y

=

∕

dx∙

,

= C

I∏the integral on the left-hand

side, substitute u = ^ y => du = ^ d y ≠> 2 du = -^dy, and we have f sec2 u du = J , d x = > 2 ta n u = x -∣ -C ≠∙ —x + 2 tan ^ y = C !$• λ ∕ x ⅛

=

ey+v ^ ≠ , ⅛

=

s^

=> dy = s ^

dx

=^ e ~y dy = ^ dx => J*e^y ^Y = f ^

dx

∙ I∏ the integral on the

right-hand side, substitute u = ^ ∕x => du = ^ ^ dx => 2 du = ^ dx, and we have J*e^y dy = 2 f e“ du ≠> -e ^ y = 2e“ + Ci ≠> - e ^ y = 2e√^ + C, where C = -C i ^2 ∙ dy = 2 x √ l - y2 dx => ^ d[ yi = 2xdx => f √⅛J7 = ^ 2 x d x ≠> sin

≠> y = sin(x2 + C)

1y

= x2 + C since ∣ y∣< 1

288

Chapter 9 Further Applications of Integration

19. y , = χ + y => slope of 0 for the line y = -x . For χ, y > 0, y ' = x + y => slope > 0 in Quadrant I. For χ, y < 0 , y , = x ÷ y = > slope < 0 in Quadrant III. x∣ F θ r∣ y∣> ∣ ,y > 0 , x < 0 , y ' = x + y => slope > 0 in Quadrant II above y = —x. For ∣ y∣< ∣ x∣ , y > 0 , x < 0 , y , = x ÷ y = > slope < 0 in Quadrant II below y = —x. For ∣ y∣< ∣ , x > 0 , y < 0 , y , = x + y=> slope > 0 in x∣ Quadrant IV above y = —x. For ∣ y∣> ∣ x∣ , x > 0 , y < 0 , y , = x + y=> slope < 0 in Quadrant IV below y = -x . All of the conditions are seen in slope field (d). 21. y , = — => slope = 1 on y = —x and - 1 on y = x. y , = — => slope = 0 on the y-axis, excluding (0,0), and is undefined on the x-axis. Slopes are positive for x > 0, y < 0 and x < 0, y > 0 (Quadrants II and IV), otherwise negative. Field (a) is consistent with these conditions.

25.

27.

29.

9.2 FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS 1.

x

+ ( i) y = ⅞ ,P ( x ) = i,Q ( x ) = ⅞

g+y = e ^ ⅛

= In x , x > 0 ≠> v(x ) = e∙fp w d x = elnx = J P(x ) dx = J*∣dx = In ∣ x ∣ y= ⅛ f

v (χ

)Q( χ

)

dx

= I ∕

x

(f)

dx

= I

(ex

+ C) = ⅛ ≤ ,

x

> 0

x

Section 9.2 First-Order Linear Differential Equations 3.

÷ 3y = ⅜ , x > 0 ≠> ⅛ + Q ) y = ⅛ , P(x) = 2 , Q(x ) = ⅛

xy'

f½d

x

= 3 In ∣ x∣= In x3 , x > 0 => v(x) = elnχ3 = x3

y = ⅛ f χ3 ( τ ) 5∙

x

l + 2y =

1

dx

x

- b

c

= ⅜ fsin x dx = ⅛ ( - cos x + C) = >

~”

x

,x > 0

^ ⅛ + 0 ) y = ∣- ⅛- p w = 1 - Q v(x) = e -x/2 => y = ^ J^e-x/2 ( | ex/2) dx = ex/2 y ∣dx = ex/2 ( ∣x + C) = ∣xex/2 + Cex/2

9. ⅛

-

(x )y ^ ^ ' n

≠> v(x) = 11∙

e^ 1"x

x

^

v (t)

= e1" ^

-

χ,^

χ

[(In x)2

— 1 f ≤ _ t ι rΛ _ ι ~ ( t- l) 4 I 3 ι —

∫ ( t - 1)4 [ ≡

s= ⅛

+ C] = x (In

x)2

- In x, x > 0

+ Cx

y P (t)d t = y ⅛ d t = 4 1 n ∣ t - l ∣= l n ( t - l ) 4

=> P(t) = f ⅛ . Q ( t ) = ( F ⅛ ^

= (t - l ) ^

- f ∣dx =

) —2 in x => y P(x) dx =

= I => y = x y Q ) (2 In x) dx = x

⅛ + (tT τ ) s = ( Γ ⅛ =,

P(x) ~

dt = ⅛ y ( t 2 - 1) dt

5]

I t3_________ t C ( t _ 1 )4 -Γ (t - 1 )4

3 ( t _ 1)4

13. ⅛ ÷ (cot Θ r = sec 0 => P(0) = cot 0, Q(0) = sec 0 => § P(0) d0 = J*cot 0 d0 = in ∣ sin 0∣=> v(0) = elnlsinβl

f tan 0 d0 =

= sin 0 because 0 < 0 < ∣ => r = 4 ^ J sin 0)(sec 0) d0 = ^

(in ∣ sec 0∣ + C)

= (esc 0) (in ∣ sec 0∣ + C) 15. ⅛ + 2y = 3 => P(t) = 2, Q(t) = 3 => yP(t) dt = y 2 dt = 2t => v(t) = e2' ≠> y = ⅛ y 3 e 2' dt = ⅛ (h

17∙

2,

+ c )iy (0 ) = ι ^ 3

+

c

=^ ^ f P (0)

I + (I) y = ⅜ f =* P (0) = I >Q(0) => y = ⅛ ∫ ^ ( ⅛ ^ ) d ^ l p ( ^ )

y = l - i e - 2∙

= - i ^

= l^ C

d 0 f °r 0

≠

o

d0

^

= l n 1^1 => v (0) = e 'n '" = l0 l y = ∣y s in 0 d 0 = i ( - c o s 0 + C)

= - I cos 0 + f ; y ( j ) = 1 => C = f ≠> y = - ∣cos 0 + ⅜ 19. ( χ + l ) ⅛ - 2 (x2 + x ) y = ⅛

^

⅛ -

2

[ ^ ] y = ⅛

j

^

⅛ _

2xy

Q(χ ) = ( Γ ⅛ => y P(χ ) d x = y ~ 2 x dx = - x 2 ≠- v(x) = e - χ2 ≠> y = Λ f y

=^ f R W =, c = 6 ^

2

dx

=

y=

eχ2

[i ⅛ π +

6eχ 2 -

c

] = - ⅛

+

c

≡χ2 ! y(°) = 5 ^

=, P(x) = -2 x ,

= ^ e

χ2

[(7⅞⅛] dx

- θ⅛τ + C = 5 ≠> - 1 + C = 5

⅛

1. ⅛ _ ky = 0 => P(t) = -k , Q(t) = 0 => yP(t) dt = y —k dt = - k t => v(t) = e^ kt => y = ⅛ y (e~k t ) (0) dt = ekt (0 + C) = Cek t ; y(0) = yo ≠> C = y0 => y = yoekt

23. x y ∣dx = x (In ∣ x∣ + C) = x In ∣ x∣ + Cx => (b) is correct

289

290

Chapter 9 Further Applications of Integration

25. Let y(t) = the amount of salt in the container and V(t) = the total volume of liquid in the tank at time t. Then, the departure rate is ^ (the outflow rate). 10

(a) Rate entering = ¾ j ∙ ¾⅛ =

lb∕min

(b) Volume = V(t) = 100 gal + (5t gal - 4t gal) = (100 + 1) gal (c) The volume at time t is (100 + 1) gal. The amount of salt in the tank at time t is y lbs. So the concentration at any time t is l J + t lbs∕gal. Then, the rate leaving = 1oo+1 (lbs∕gal) - 4 (gal/min) 1bs∕min

= ⅛

(d ) ⅛ = ι θ - ⅛

^

⅛ + ½

) y

= ιθ ^

p

ω = 1 o‰ ,Q (t) = ιo ^

∫P ( t) d t = ∫ ⅛

d t

= 41n(100 + t) =+ v(t) = e4ln(100+,) = (100+ t)4 =+ y = ∏ ⅛ ∫ ( 1 0 0 + 1)4 (10 dt) = ∏ δ⅛ F ( β 5 T ^ + c ) - 2(100+ t) +

π

θ⅛ ψ jy(O) = 50 ≠- 2(100+ 0) +

=* C = -(150)(100) 4 =+ y = 2(100 + t ) - ∏ ≡ ⅛

π

⅛

= 50

1

=> y = 2(100 + 1) - - ⅛ 3

(e) y(25) = 2(100 + 25) - ¾ ∞ r ≈ 188.56 lbs =+ concentration = ≤

100/

≈ 1 ^ ≈ 1.5 lb∕gal

27. Let jy be the amount of fertilizer in the tank at time t. Then rateo entering = 1min⅞ ∙ 1 ⅛ = 1 ∙⅛ and the gal min volume in the tank at time t is V(t) = 100 (gal) + [1 (gal/min) - 3 (gal∕min)]t min = (100 —2t) gal. Hence rate out = ( τ ⅛ ) 3 = ⅛ lbs∕min =+ ⅛ = (1 - ⅛ i ) lbs∕min =+ ⅛ + ( ⅛ i ) Y = 1 p

+

ω = ⅛

1

>Q ω ≈

^

= (100 - 2 t)- 3 ∕ 2 =+ y =

∫ 1

p d t 3y2

dt

= ∫ ⅛

1^

= ^

^

vω

= e '- 3ln ' 100- 20 >2

f (100 - 2t)- 3 ∕ 2 dt = (100 - 2t)" 3 ∕ 2 [ ∑

≡ ⅛

+

cl

= (100 - 2t) + C(100 - 2t)3 / 2 ; y(0) = 0 = + [100 - 2(0)] + C[100 - 2(0)]3 ∕ 2 =+ C (l∞ ) 3 ∕ 2 = - 1 ∞ =+ C = -(1 0 0 )-V 2 = - ⅛ =+ y = (100 - 2t) - 0

y '< 0 ∙l 15t y" 0 : y">0

5∙ y' = √ y , y > θ (a) There are no equilibrium values. (b ) = 2⅛ y ' = 2 ⅛ √ y = 2 y>o y,>0

7∙ y ' = (y - i)(y - 2)(y - 3) (a) y = 1 and y = 3 is an unstable equilibrium and y = 2 is a stable equilibrium value.

293

294

Chapter 9 Further Applications of Integration

(b) y" = (3y2 - 12y + 1l)(y - l)(y - 2)(y - 3) = 3(y - 1) (y - ⅛ ≠ ) (y - 2) (y - ⅛ ≠ ) (y - 3) y 0

P '< 0

P"0

p

Section 9.4 Graphical Solutions of Autonomous Differential Equations

295

Before the catastrophe, the population exhibits logistic growth and P(t) → Mo, the stable equilibrium. After the catastrophe, the population declines logistically and P(t) → Mi, the new stable equilibrium. 15. ⅛ = g - ⅛v2 , g, k, m > 0 and v(t) > 0 Equilibrium: ⅛ = g - ⅛v2 = 0 => v = ^ ≡ Concavity: ⅛ = -2(⅛ v) $ = -2 (⅛ v ) (g - ⅛v2 )

(c) vterminal = ^ ⅛

= 178.9 ∣= 122 mph

17. F = Fp - Fr ma = 50 - 5∣ v∣ ⅛ = ⅛ (5 θ -5 ∣ v∣ ) The maximum velocityj occurs when ⅛ = 0 or v = 10 ⅛ at sec 19.

L

£ + Ri = V

^

= £ - £ i= £ ( £ - i) ,V ,L ,R > 0

Equilibrium: ⅛ = ⅛ ( ^ - i ) = 0 ^ i = ⅛ Concavity: £ = - ( B ) *

=

- ( E ) 2 (Y _ i )

Phase Line:

If the switch is closed at t = 0, then i(0) = 0, and the graph of the solution looks like this:

t ►OO, it ►isteady state — R ∙ (In the steady state condition, the self-inductance acts like a simple wire connector and, as a result, the current throught the resistor can be calculated using the familiar version of Ohm’s Law.)

A.S

296

Chapter 9 Further Applications of Integration

9.5 APPLICATIONS OF FIRST ORDER DIFFERENTIAL EQUATIONS 1. Note that the total mass is 66 + 7 = 73 kg, therefore, v = voe v = 9e 3 ∙9t ∕ 73 (a) s(t) =

f 9e~

3 9t ∕ 7 3 dt

∣ - - 2∣ θe - 3 ∙9V73

+ C

y (1 —e - 3 ∙9t^7 3 ) = ^∣ Since s(0) = 0 we have C = ^∣ y and hms(t) = h m 2 ∣ ∣ 2 ≈ 168.5 The cyclist will coast about 168.5 meters. (b) 1 = 9e" 3 ∙9t∕ 73 => f = In 9 =≠> t = ¾ 9 ≈ 41.13 sec It will take about 41.13 seconds. 3. The total distance traveled = ^

≠x

¢2.75)(39.92)

_ 4 ^ ∣=> k = 22.36. Therefore, the distance traveled is given by the

function s(t) = 4.91 (1 —e - (22 -36z39∙92 )t ). The graph shows s(t) and the data points.

5. (a)

= 0.0015P(150 - P) = ^ P ( 1 5 0 - P) = ⅛P(M - P) Thus, k = 0.255 and M = 150, and P = Initial condition: P(0) = 6 => 6 = , ∣ ¾ Formula: P =

(b) 100 =

ri

⅛

≡

1+

⅛

τ

⅛

f

=

τ⅛

=> 1 + A = 25 => A = 24

,

=> 1 + 24e-°∙255t = ∣≠> 24e-°∙255t = j => e-°∙255t = ⅛ => -0.25 5t = - I n 48

^ , = r a ≈ 1^.21 weeks 125 = u ‰ => 1 + 24e-°∙255t = ∣≠> 24e-°∙255t = ⅜ =►e-°∙255t - ⅛ ≠> -0 .255t = - I n 120 ^ t= ⅛ ^ ≈ 2 1 ,2 8 It will take about 17.21 weeks to reach 100 guppies, and about 21.28 weeks to reach 125 guppies. 7. (a) Using the general solution form Example 2, part (c), ⅛ = (0.08875 × 10- 7 )(8 × 107 - y)y => y(t) = r ⅛

π

=

Apply the initial condition: y(0) = 1.6 × 107 = ¾ < => ⅛ - 1 = 4 ≠> y (l) = ⅛ (b) y(t) = 4 × 107 = γ ⅛ ‰

τ τ

^ ⅛

w

= ⅛

≈ 2.69671 x 107 kg.

=> 4e~0 ∙71t = 1 => t = - ⅛ ^ ≈ 1.95253 years.

1+y∣ = et + c , => 11 + y ∣= e , e c ' 9. (a ) ⅛ — 1 + y =^ ⅜ — (1 + y)dt => i ¾ = dt => In 11 + y∣= t + Ci => eln ∣

1 + y = ± C2et => y — Cet — 1, where C2 = ec ' and C = ± C2. Apply the initial condition: y(0) = 1 = Ce0 — 1 => C = 2 => y = 2et - 1. (b) * = θ∙5(400 - y)y => dy = 0.5(400 - y)y dt =>

(4 5 ⅛ ^ =

θ∙^ dt. U s in 8 die partial fraction decomposition in

Example 2, part (c), we obtain ⅛ θ + 4δ0ty)dy = 0.5dt ≠> θ + 4δ ^7y )dy = 200dt ≠ ∫ θ

+ 4 δ ⅛ ) d y = ∫ 2 0 0 d t ^ ln∣ y ∣- ln∣ y - 400∣= 200t + C 1 ^ ln∣ 7⅛

∣= 200t + C 1

Section 9.5 Applications of First-Order Differential Equations ^ eln l ⅛ l = => ⅛

=

e

200t+c

c e 2 °°t

=> y = ⅞ ⅞ y(0) =

2

∙ = e 200,ec ' ^ ∣ y⅛

∣= C2 e2∞, (where C2 = ec ' ) ^

7⅛

297

= ± C2 e200t

(where c = ÷ C2 ) => y = Ce200ty - 400 Ce200t 4 (1 - Ce200t)y = -4 0 0 Ce20°,

=> y = A

= ⅛

=

_ v ∞ ,

1

=

199

H

⅛

. where A = -⅛ . Apply the initial condition:

y(t) = τ ⅛

11. (a) f = kP2 ^ ∫ P ^ 2 dP = ∫ k d t ≠ - —P

1

= k t + C=>P =

δ

⅛

Initial condition: P(0) = Po=>Po = - ⅛ 4 ∙ C = ^ Solution: P =

∣ a -'ι∕p 0) = r ⅛

-

(b) There is a vertical asymptote at t = Λ 13. y = mx≠> J = m=>

xy

= 0 => y' =

*Γy

y

. So for

orthogonals: ⅛ = - ^ => ydy = —xdx ≠- ^ + γ = C χ 2 + y2 = Cl

15. kx2 + y2 = 1 => 1 - y2 = kx2 => ⅛ ≠> ⅛

H H ⅛

≠∙ y' =

^^ 1 ⅛ J 2 ^

^ = ⅛

17. y = ce

=

.

0

2 y χ 2y

—⅛ Γ ∙ ^

o

= k ,

=

(1

_

y

2) ( 2 x )

f° r t ^ι e orthogonals:

≠- ⅛ ^ d y = Xdx => In y - 4 = τ + c

x

=> ⅛ =

c

=>

e

y

^ ^ ∙^

θ

≠ e x y' = —ye x => y' = —y. So for the orthogonals: ∣ = ^ ≠ -y d y = d x ^ ⅛ = x + c => y2 - 2x + Ci 4 y = ± √ 2 x + Ci

19. 2x2 + 3y2 = 5 and y2 = x3 intersect at (1, 1). Also, 2x2 + 3y2 = 5 => 4x + 6yy' = 0 => y ' = —^ => y '(l, 1) = —∣ y2 = x3 => 2yi y 1' = 3x2 =^ Yι = j ^ => y 1'( l, 1) = ∣∙ Since y' ∙ y 1' = ( - j) (∣) = —1, the curves are orthogonal. 21. y2 = 4a2 —4ax and y2 = 4b2 + 4bx => (at intersection) 4a2 - 4ax = 4b2 + 4bx => a2 —b 2 = x(a + b) => (a + b)(a —b) = (a + b)x => x = a —b. Now, y2 = 4a2 - 4a(a —b) = 4a2 —4a2 ÷ 4ab = 4ab => y = ± 2√ab. Thus the intersections are at (a —b, ± 2 √ a b ^ . So, y2 = 4a2 —4ax => y[ = - ∣ which are equal to —- ^ ^ and ~2(

2 /b j

~

and

71

a t t ⅛e

intersections. Also, y2 = 4b2 + 4bx => y2 = ∣ which are equal to ^ = y and

298

1∙

Chapter 9 Further Applications of Integration

s2√ y

I = √y∞

=> ^ ⅛

= dx => 2tan√y = x + C => y = (tan

1(

^ ))

2

∣= sec2 x dx => ^ ^ = tan x + C => sin(y2 ) = 2tan x + Ci

3. yy , = sec(y2 )sec2 x => ^

5. y , = xey √ x ^ - 2 => e^ y dy = x √ x - 2 d x => - e ^ y = 2 ÷ C 4 e ^ y = Ξ2!i∑ 2l ⅛ + 4 ) _ ≠> _ y = ln [ ⅛

⅛

4 1 ,

c

] ^

y

= -1∏[

- ⅜ - 2> ⅝ ÷ 4 ) -

c

c

]

7. sec x dy* ÷ x cos2 y* dx = 0 =>SvC -⅛ = —⅛⅛ => tan y = —cos x —x sin x + C COS y X * 9. y ' = ∣ ≠- ye- y dy = γ => (y + 1)e- y = -In ∣ x∣ + C 11. x(x —l)dy —y dx = 0 4 x(x —l)dy = y dx => ^ = ^ ^ ≠- In y = ln(x - 1 ) - ln(x) + In Ci => In y = 1H ( ^

=> In y = ln(x — 1) —ln(x) + C

^ ) => y = ⅛ ^

Ξ

13. 2y' - y = xex/2 ≠* y ' - ∣y = ∣ex β . p(x) = - ∣, v(x) = e ^ - ^ e -χ∕2 y

∕ _ ι e -χΛ

= e- x / 2 .

_ ( e -> ^) ( ∣ ) ζe χ∕2) — s - ^ Λ ζe -χ∕2 y) = s => e- x / 2 y = j + C = > y = ex72 ^

y

15. xy' + 2y = 1 - x^ , 4 y' + (¾)y = ⅛ — ⅛ v(x) = e2 J^⅛ -- e2'n x = e'nχ2 = x2 . x2 y' + 2xy = χ - 1 => ^ ( χ 2 y) = χ - J ⅛ χ 2 y = ^ - χ f C = > y = 1 - ) + ^ 17. (1 + ex )dy + (yex + e - x )dx = 0 ≠> (1 + ex )y' + ex y = - e - x => y' = ∣ ⅛ y = v

(χ) = e

(ex

+

l)y '

^ +

=

e

ln(ex+l)

_

eχ +

(ex + l ) ( ⅛ ) y =

( ι~ ⅞ ∙

ι π

⅞ ( e x + 1) ^ ⅛[ (ex + l)y] = - e ~ x ^ (ex + l)y = e→ + C

+ c)

Chapter 9 Practice Exercises v y

_ e x+ C _ ex + l

299

e x+ C l÷e x

19. (x ÷ 3y2 ) dy + y dx = 0 => x dy ÷ y dx = - 3 y 2 dy => ^ (xy) = -3 y 2 dy => xy = —y3 + C 21. ^ = e ey

=

x y 2

=> ey dy = e (χ + 2 Mx => ey = —e (χ + 2 ) + C. We have y(0) = —2, soe

—e~(χ + 2 )

+

2e- 2

=> y = l n ( - e ^ ^ +

χ

= -e

2

+ C => C = 2e

2

and

2e^2 )

23. (x + 1)∣ + 2y = x => y' + (⅛ ^ )y = ⅛ ∙ Let v(x) = e ∫ ⅛ S oy , (x + 1)2 + (⅛ y (χ + i) 2 y = ⅛ , (

2

dx

= e21n(χ + 1 ) = eln ^x + 1 ^ = (x + 1)2 .

+ 1)2 => ⅛ [y(χ + 1)2 ] =

χ (χ

+ 1) => y(χ + 1)2 = J χ (χ + 0 *

- . y(χ + 1)2 = ^ + ^ + c => y = (x + 1)"2 (⅛ + ⅝ + c ) . We have y(0) = 1 => 1 = C. So y = ( χ + l ) - 2 (⅛

+

⅛ + 1)

^ ∙ ⅛ ÷ 3χ 2 y —x2 ∙ Let v(x) = e^ 3χ2dx = eχ3 . So eχ3y' + 3x2 eχ3y = x2 eχ3 4∙ ^ (e , 3 y^ = x2 eχ3 4 eχ3 y = ∣eχ3 + C. Wehavey(0) = - 1 =►e°3 ( - l ) = ∣e°3 + C => - 1 = ∣+ C ^ C = - f ande χ3 y = ∣eχ j - ^ y 27. x dy - (y + ^ ) d x = 0 =>

(y + y y J

=

T

= ∣- ∣e" χ3

21n(1yy + 1) = In x + C. We have y(l) = 1 => 21n(ᅟ ∕ l + 1) = in 1 + C

=> 21n2 = C = ln2 2 = ln4. So 2 1n(yy + 1) = In x + In 4 = ln(4x) => l n ( ^ + 1) = ⅛ln (4x) = ln(4x) 1/2 lβ e ι∏(√y+ι) = e in(4χ)

^ ^

+

j

=

2 χ ∕x => y = (2 λ ∕x - 1)2

29. xy' ÷ (x - 2)y = 3x3 e~x => y' + ( 2^ ) y = 3x2 e^ x . Let v(x) = e ∕( i ^ ) dx = ex -2 1 n x = ⅜y' + ⅛ ( ^ ) y =

3

=^ ⅛(y , ⅛) =

3

So

=^ y ∙ g = 3 x ÷ C W ehavey( 1 ) = 0 => 0 = 3( 1 ) ÷

c

=> c = - 3

= > y ∙^ = 3 x - 3 = > y = x2 e^ x (3x - 3) 31. To find the approximate values let yn = yn -1 + (y∏-1 ÷ cos xn _i )(0.1) with xo = 0, yo = 0, and 20 steps. Use a spreadsheet, graphing calculator, or CAS to obtain the values in the following table. X

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

X

y 0 0.1000 0.2095 0.3285 0.4568 0.5946 0.7418 0.8986 1.0649 1.2411 1.4273

1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0

33. To estimate y(3), let zn = yn -ι ÷ ( ⅛ ¾

d

y 1.6241 1.8319 2.0513 2.2832 2.5285 2.7884 3.0643 3.3579 3.6709 4.0057

) (0.05) and yn = yn -ι ÷ ∣( ⅛ f r

i

+ ⅛ ⅞ ) (0.05) with initial values

xo = θ, yo = L and 60 steps. Use a spreadsheet, graphing calculator, or CAS to obtain y(3) ≈ 0.9063. 35. Let yn = yn -ι + (

e

χn

J ι+ V ι+ s

)(d χ ) with starting values xo = 0 and yo = 2, and steps of 0.1 and —0.1. Use a spreadsheet,

programmable calculator, or CAS to generate the following graphs.

Chapter 9 Further Applications of Integration

300 (a)

i

l

l

!

[-0.2,4.5] by [-2.5,0.5]

(b) Note that we choose a small interval of x-values because the y-values decrease very rapidly and our calculator cannot handle the calculations for x ≤ —1. (This occurs because the analytic solution is y = —2 ÷ ln(2 —e - x ), which has an asymptote at x = —In 2 ≈ 0.69. Obviously, the Euler approximations are misleading for x ≤ —0.7.)

[-1,0.2] by [-10,2]

37.

X 1 1.2 1.4 1.6 y - 1 -0 .8 -0 .5 6 -0 .2 8

1.8 0.04

2.0 0.4

^ = x = > d y = x d x ≠ > y = γ + Q x = l andy = —1 = > —1 = ∣÷ C = > C = - 1 = > y(exact) — y - ∣ => y (2) = y — ∣= ∣is the exact value.

39.

x 1 1.2 1.4 1.6 1.8 2.0 y - 1 -1 .2 -0.488 -1.9046 -2.5141 -3.4192 ^ = xy => ^ = x d x ≠> ln∣ y ∣= y + C =» y = e⅛+ c = e⅛ ∙ ec = C ιe⅛ ; x = 1 and y = - 1 => —1 = C ιe ιz2 => Ci = —e 1/2y(exact) = - e 1/2 ∙ e⅛ = - e ( χ2^ 1)z2 => y(2) = - e 3z2 ≈ -4.4817 is the exact value.

41. ⅛ = y2 - 1 => y , = (y + l)(y - 1). We have y , = 0 => (y ÷ 1) = 0, (y - 1) = 0 => y = - 1 , 1. (a) Equilibrium points are —1 (stable) and 1 (unstable) (b) y ' = y2 - 1 => y ', = 2yy' => y" = 2y(y 2 - 1 ) = 2y(y ÷ l)(y - 1). So y" = 0 => y = 0, y = - 1 , y = 1.

⅛>

0

dx ⅛ < o dx?

:⅛
o

∣d x .

! ⅛ o ∙ ⅛ > o J dx? [ dx2∙ y=0

o

Chapter 9 Additional and Advanced Exercises 43. (a) Force = Mass times Acceleration (Newton's Second Law) or F = ma. Let a = ⅛ = j

,

301

⅛ = v ρ Then

ma = —mgR2 s^2 ≠> a = —gR2 s~2 => v ^ = -g R 2 s^2 ≠> v dv = —gR2 s^ 2 ds ≠ ∙ J v dv = ∫ - g R 2 s^2 ds = > ^ = ⅛ + C ι≠ > v 2 = ^

+ 2Cι = ^

=> c = v§ - 2gR 4 v2 = ^

+ vj - 2gR

(b) If vo = √ ,2gR, then v2 = ⅛

+ C. When t = 0, v = v0 and s = R => vθ = ^

≠> v = ^ ^ , since v ≥ 0 if VQ ≥ √2gR, Then ^ = i ^

=>f s ιz2ds — f ι∕2 g R 2 dt ≠- ∣s3z2 =

+C

=> λ ∕ s ds = √ 2gR 2 dt

√2gR 2 t + C, 4∙ s3z2 = ( ∣∙y ⅛ R 2 )t + C; t = 0 and s = R

≠> R3Z2 = (∣√ 2gR 2 )(0) + C ^ C = R3 β ⅛∙ s3 β = (∣√2gR 2 )t + R3z2 = ( 5 R √ ⅜ )t + R3z2

CHAPTER 9 ADDITIONAL AND ADVANCED EXERCISES 1. (a) ⅛ = kφ(c - y) => dy = -k φ (y - c)dt ≠> ^

= -k φ d t =^ f

^

= - ∫k ⅛ d t =≠> ln∣ y - c ∣= - k 0 t + Ci

=> y - c = ± ec , e^ k vt . Apply the initial condition, y(0) = yo => yo = c ÷ C => C = yo —c => y = c + (yo - c)e - k vt . (b) Steady state solution: yoo = limy(t) = lim lΓc + (y∩ v t→∞ t→∞

c )e^ z

k vt

1j = c ÷ (yo —c)(0) = c

3. The amount of CO2 in the room at time t is A(t). The rate of change in the amount of CO2, j

is the rate of internal

production (Ri) plus the inflow rate (R2) minus the outflow rate (R3). Ri = (20 ¾ ⅛ m ) (30 students) ( ^ ft3 ) (θ.O4 ≈ 1.39 ⅛ R2 = (lθθθ £ ) (0∙0004 ^

)

= 0.4 ⅛

R3 = (1o‰δ)lOOO = O . l A ⅛ f = 1.39 + 0.4 - 0.1A = 1.79 - 0.1A ≠> A' + 0.1A = 1.79. Let v(t) = e ∕ 0 1 d t . We have £ ( A e ∕ 0 1 d , ) = 1.79e∕0 1 d t => Ae0 1 t =

f 1.79e0 l t dt =

17.9e0 1 t + C. At t = 0, A = (10,000)(0.0004) = 4 ft3 CO2 => C = -13.9

=^ A = 17.9 - 13.9e-° l t . So A(60) = 17.9 - 13.9e~0 ' ≈ 17.87 ft3 of CO2 in the 10,000 ft3 room. The percent of CO2 i s S × 100 = 0.18% 5. (a) Let y be any function such that v(x)y = ^(v(x) ∙ y) = v(x)

∙ y,

+y

∙ v , (x)

f v(x)Q(x) dx + C, v(x) =

= v(x)Q(x). We have v(x) = e ∕

p dx. eJ ^ p ^ dx =>

Then v , (x) = = e~fp ∞ dx P(x) = v(x)P(x).

Thus v(x) ∙ y , ÷ y ∙ v(x) P(x) = v(x)Q(x) => y , + y P(x) = Q(x) => the given y is a solution. (b) If v and Q are continuous on [ a, b ] and x ∈(a, b), then ^ £J^ v(t)Q(t) dtj = v(x)Q(x) => J ^ v(t)Q(t) dt = J v(x)Q(x) dx. So C = yov(xo) - J v(x)Q(x) dx. From part (a), v(x)y =

f v(x)Q(x) dx + C.

Substituting for C: v(x)y = Jv(x)Q(x) dx + yov(xo) —J v(x)Q(x) dx => v(x)y = yov(xo) when x =

XQ.

302

Chapter 9 Further Applications of Integration

NOTES:

CHAPTER 10 CONIC SECTIONS AND POLAR COORDINATES 10.1 CONIC SECTIONS AND QUADRATIC EQUATIONS 2

1. x = ^ ≠ > 4 p = 8= > p = 2; focus is (2,0), directrix is x = - 2 3. y = - y = > 4 p = 6 = > p = | ; focus is (0, - ∣) , directrix is y = ∣ 5. j - ⅛

= l= > c = √ ^ + 9 = ᅟ ∕1 3 ≠* foci are ( ± Z13,0^ ; vertices are ( ± 2 ,0); asymptotes are y = ±

7. y + y2 = 1 => c = ᅟ ∕2 - 1 = 1 ≠> foci are ( ± 1, 0); vertices are ( ± v A θ)

9. y2 = 12x ≠- x = ^ ≠> 4p = 12 =≠> p = 3; focus is (3,0), directrix is x = - 3

13. y = 4x2 => y = ^

=> 4p = ∣ ≠> p = ⅛ ;

focus is (0, ⅛ ) , directrix is y = - ⅛

11. x2 = - 8 y => y = ⅛ => 4p = 8 => p = 2; focus is (0, —2), directrix is y = 2

15. x = -3 y 2 ≠ > χ = - ^

≠ > 4 p = ∣ ≠> p = ⅛ ;

focus is ( - ⅛ , 0 ) , directrix is x = ⅛

w ∣ o t

304

Chapter 10 Conic Sections and Polar Coordinates 19. 2x2 + y 2 = 2 ≠> x2 + ^ = 1

17. 16X2 + 25y2 = 400 =► ⅛ + ⅛ = 1 => c = √ a 2 - b 2 = √ 2 5 - 16 = 3

≠> c = √ a 2 - b2 = √ 2 - 1 = 1

21. 3x2 + 2y 2 = 6 ≠> ⅛ + ⅞ = 1 ≠

23. 6 X2 + 9y2 = 54 ≠> £ + ⅛ = 1 => c = ∖ ∕a2 —b 2 = √ 9 —6 = √ 3

c = √ a 2 - b 2 = √ 3 —2 = 1

25. Foci: ( ± √ 2 , θ ) , Vertices: ( ± 2,0) 4

27. x2 - y 2 = 1 ^

a = 2, c = √ 2 ≠> b 2 = a2 - c 2 = 4 - ( √ 2 ) 2 = 2 ≠> ^ + ^ = 1

c = √ a 2 + b2 = √ T ∏ = √ 2 ;

asymptotes are y = ± x

29. y2 - x

2

= 8 ^ ⅞ -⅛

= l ^ c

= √ ^ 2 + b2

= ^ 8 ÷ 8 = 4; asymptotes are y = ± x

Section 10.1 Conic Sections and Quadratic Equations 31. 8X2 - 2y2 = 16 => ⅛ - ⅛ = 1 ≠> c = √ a 2 + b2

33. 8y2 - 2x2 = 16 => ⅛ - ⅞ = 1 ^

= ^ 2 + 8 = √ 1 0 ; asymptotes are y = ± 2x

c= √ ^ + ^

= ^/2 ± 8 = ᅟ ∕ i b ; asymptotes are y = ± ∣

35. Foci: ( 0, ± ᅟ ∕2 ) , Asymptotes: y = ± x => c = ᅟ ∕2 and § = 1 =>

a

= b => c2 = a2 + b2 = 2a2 => 2 = 2a2

=> a = 1 => b = 1 => y2 - x2 = 1 37. Vertices: ( ± 3 ,0 ) , Asymptotes: y = ± ∣x = > a = 3 and ^ — ∣ => b = ∣(3) = 4 ^

⅛ -⅛ = l

39. (a) y2 = 8x ≠> 4ρ = 8 => p = 2 => directrix is x = -2 , focus is (2,0), and vertex is (0,0); therefore the new directrix is x = —1, the new focus is (3, —2), and the new vertex is (1, —2)

2

2

41. (a) ⅛ ÷ ⅛ = 1 ^ center is (0,0), vertices are (-4 ,0 ) ∕7 , 0^ and (4,0); c = ᅟ ∕a 2 - b 2 = √ 67 => foci are ^ᅟ and ( —5/7, θ) ; therefore the new center is (4,3), the new vertices are (0,3) and (8,3), and the new foci are (4 ± √ 7 , 3)

2

2

43. (a) ^ - ^ = 1 => center is (0,0), vertices are (—4,0) and (4,0), and the asymptotes are * = ± ^ or y = ± ^ ; c = ^ a 2 + b 2 = √ z25 = 5 => foci are (—5,0) and (5,0); therefore the new center is (2,0), the new vertices are (—2,0) and (6,0), the new foci are (—3,0) and (7,0), and the new asymptotes are y= ± ¾ ^

45. y2 = 4x => 4p = 4 => p = 1 => focus is (1,0), directrix is x = - 1 , and vertex is (0,0); therefore the new vertex is (-2 , -3 ), the new focus is (—1, —3), and the new directrix is x = —3; the new equation is

305

306

Chapter 10 Conic Sections and Polar Coordinates (y + 3)2 = 4 (x + 2)

47. x2 = 8y => 4p = 8 => p = 2 => focus is (0,2), directrix is y = - 2 , and vertex is (0,0); therefore the new vertex is (1, —7), the new focus is (1, —5), and the new directrix is y = —9; the new equation is ( x - l ) 2 = 8(y + 7) 49. ⅛ + ⅛ = 1 ^

center is (0,0), vertices are (0,3) and (0, -3 ); c = ∖ ∕a2 —b2 = √ ^ S M = √ ^ => foci are ^0,

A

and ^0, - √ z3^ ; therefore the new center is (-2 , -1 ), the new vertices are (-2 ,2 ) and ( -2 , -4 ), and the new foci are ^ - 2 ,- 1 ± √ 3 ) ; the new equation is ⅛±⅛ ψ ⅛+ll2

51. y + ⅛ = 1 ^

=

ι

center is (0,0), vertices are (∖ ∕3, θ) and ( - √ z3,θ) ; c = √ a 2 —b 2 = √ 3 —2 = 1 => foci are

(-1 ,0 ) and (1,0); therefore the new center is (2,3), the new vertices are ^2 ± ^ 3 ,3 ^ , and the new foci are (1,3) and (3,3); the new equation is ⅛ p ^ + ⅛ ^ - = 1 53. Z " ^

=

^ => center is (0,0), vertices are (2,0) and (-2 ,0 ); c = ∖ ∕a2 + b2 = χ ∕4 + 5 = 3 => foci are (3,0) and

(-3 ,0 ); the asymptotes are ± * = ^

=> y = ± δ^ ; therefore the new center is (2,2), the new vertices are

(4,2) and (0,2), and the new foci are (5,2) and (-1 ,2 ); the new asymptotes are y - 2 = ± ^ (^~ 2 ) ; the new equation is ^ ^ £ _ & ~J 2

=

j

55. y2 - x2 = 1 => center is (0,0), vertices are (0,1) and (0, -1 ); c = ∕ a ^0, ± ^

2

+ b2 = √ T ÷ T = ^/2 => foci are

5the asymptotes are y = ± x; therefore the new center is (—1, —1), the new vertices are (—1, 0) and

( - 1, —2), and the new foci are ( —1, —1 ± ^

; the new asymptotes are y ÷ 1 = ± (x + 1); the new equation is

(y + l)2 - ( x + l ) 2 = l 57. x2 + 4x + y2 = 12 => x2 + 4x + 4 + y2 = 12 + 4 => (x + 2)2 + y2 = 16; this is a circle: center at C (-2 ,0 ),a = 4 59. x2 ÷ 2x + 4y —3 = 0 ≠> x2 ÷ 2x ÷ 1 = —4y + 3 ÷ 1 => (x ÷ 1)2 = —4(y — 1); this is a parabola: V ( - l ,l ) , F ( - l,0 ) 61. x2 + 5y2 + 4x = 1 => x2 + 4x + 4 ÷ 5y2 = 5 =>(x + 2)2 + 5y2 = 5 => ^

÷ y2 = 1; this is an ellipse: the

center is (-2 ,0 ), the vertices are ^ - 2 ± y^5,θ) ; c = ∖ ∕a2 - b2 = ∖ ∕5 - 1 = 2 =≠> the foci are (-4 ,0 ) and (0,0)

63. x2 + 2y2 - 2x - 4y = - 1 => x2 - 2x ÷ 1 + 2 (y2 - 2y + 1) = 2 => (x - 1)2 + 2(y - 1)2 = 2 ^ ⅛ ^ ÷ (y ~ 1)2 = 1» t ^ s *s a n ellipse: the center is (1,1), the vertices are (1 ± √ 2 , 1) » c = √ a 2 - b2 = √⅛ —1 = 1 => the foci are (2,1) and (0,1) 65. x2 - y2 - 2x + 4y = 4 => x2 - 2x + 1 - (y2 - 4y + 4) = 1 => (x - 1)2 - (y - 2)2 = 1; this is a hyperbola: the center is (1,2), the vertices are (2,2) and (0,2); c = √ a 2 + b2 = ^ 1 + 1 = >∕2 => the foci are (1 ± √ ^ , 2^ the asymptotes are y —2 = ± (x — 1)

Section 10.1 Conic Sections and Quadratic Equations 67. 2x2 - y 2 + 6y = 3 => 2x2 - (y2 —6y ÷ 9) = - 6 => ^ ^

- j = 1; this is a hyperbola: the center is (0,3),

the vertices are ^0,3 ± √ z6^ ; c = √ a 2 + b2 = ^ 6 ÷ 3 — 3 => the foci are (0,6) and (0,0); the asymptotes are ⅛ =

i

⅛

^

^

i

^

x + 3

69.

71.

75. Volume of the Parabolic Solid: Vι = ^

2τrx (h - ^ x2 ) dx = 2πh f

o

^x - ^ θ dx = 2πh ^

= ≡ ^ ; Volume of the Cone: V2 = ∣π (∣) 2 h = I π (⅛ ^ h = ⅛ ^ ; therefore Vι = ∣V2 O

j

∖

Z

∕ j

∖

4τ

∕ l

Z

Z

77. A general equation of the circle is x2 + y2 + ax ÷ by + c = 0, so we will substitute the three given points into a ÷ c = -1 b + c = - 1 ►=> c = ∣and a = b = —j ; therefore this equation and solve the resulting system: 2a 4- 2b 4^ c — —8 3X2 + 3y2 —7x —7y + 4 = 0 represents the circle 79. r2 = (—2 — 1)2 + (1 —3)2 = 13 => (x + 2)2 + (y —1)2 = 13 is an equation of the circle; the distance from the center to (1.1,2.8) is ^ ( - 2 - 1.1)2 ÷ (1 - 2.8)2 = ᅟ ∕12.85 < ι ∕1 3 , the radius => the point is inside the circle 81.

= ξ ^ . The volume of the right circular cylinder formed by revolving PS about the x-axis is V2 = π ( ᅟ ∕k x ) x = πkx 2 => the volume of the solid formed by revolving Ri about the x-axis is V3 = V2 - Vι = πkx2 —z^

= z^

. Therefore the ratio of V3 to V i is 1:1.

307

308

Chapter 10 Conic Sections and Polar Coordinates

83. Let y = ^ 1 - ^ on the interval 0 ≤ x ≤ 2. The area of the inscribed rectangle is given by A(x) = 2x ^2^/ 1 - j ^ = 4 x ^ 1 - j (since the length is 2x and the height is 2y) => A'(x) = 4 √ 1 - £ - - ≠

7

. Thus A'(x) = 0 ≠> 4^/1 — ⅜ - - γ ⅛ = 0 =► 4 (1 - ⅛ ) - x2 = 0 => x2 = 2

=> x = √ ^ (only the positive square root lies in the interval). Since A(0) = A(2) = 0 we have that A ^ √ z2^ = 4 is the maximum area when the length is 2 √ z2 and the height is √ z2.

85. 9X2 - 4y 2 = 36 => y 2 = -

4

=> y = ± ∣χ ∕x 2 - 4 on the interval 2 ≤ x ≤ 4 => V =

3-

J ^4 2 r τ4 2 (x - 4 ) dx = ⅞ [⅛ - 4 x ]

2

^π

Q ∕x

2

—4^ dx

= ⅛ [(⅛ - 16) - ( I - 8)] = ⅞ (⅞ - 8) = ⅜ (56 - 24) = 24π

87. Let y = y 16 - y x2 on the interval —3 ≤ x ≤ 3. Since the plate is symmetric about the y-axis, x = 0. For a

(

∕16

ιβ χ2 ∖

I ~^ ~

/ .................

x, ^— ^ — I , length = √ 16 - y x2 , width = dx => area = dA = y 16 — y x2 dx

=Φ mass = dm = 5 dA = ¢ ^ 1 6 — y x2 dx. Moment of the strip about the x-axis: y dm = ^ — 2^— ( 6 y l 6 - y x

2

l dx = £ (8 - ∣x 2 ) dx so the moment of the plate about the x-axis is

M x = J y dm = J* 3 6 (8 — ∣x2 ) dx = 6 [8x - ^ x 3 ] θ 3 = 32(5; also the mass of the plate is M=

f 3δ↑J 16 -

y X2 dx = ^ 4 6 y^ 1 - ( ∣x ) 2 dx = 46

=> u = - 1 and x = 3 => u = 1. Hence, 46 = 126 ^ ζu ∕ 1 —u 2 + sin^ 1 u )j

^

,

f l^V1 -

f_^V1 —u

2

u 2 du = 1 2 6

du where u = ∣ => 3 du = dx; x = —3 j

∕1 - u

2

du

^ = 6π6 => y = ^ = ∣ ^ = j^ . Therefore the center of mass is(0, ^ ) .

⅜ ~ ⅛ ^ ⅛ (ΓA —⅛) = 0 => ΓA - r B = C, a constant => the points P(t) lie on a hyperbola with foci at A andB

91. PF will always equal PB because the string has constant length AB = FP + PA = AP + PB. 93. x 2 = 4py and y = p => x 2 = 4p 2 => x = ± 2p. Therefore the line y = p cuts the parabola at points (—2p, p) and (2p, p), and these points are ^ [2 p —(-2 p )] 2 ÷ (p - p)2 = 4p units apart. 10.2 CLASSIFYING CONIC SECTIONS BY ECCENTRICITY 1. 16x2 + 25y2 = 400 => ⅛ + ⅛ = 1 => c = √ a 2 - b 2 = √ 2 5 - 16 = 3 =>

e

= ξ = |; F ( ± 3 ,0 ) ;

directrices are x = 0 ± e- = ±

= ± ⅛ j

Section 10.2 Classifying Conic Sections by Eccentricity 3. 2x2 + y2 = 2 => x2 + = √ Γ T = 1

4

= 1 => c = ᅟ ∕a

e=≡ = ^

j

2

309

—b2

F ( 0 ,± l) i

directrices are y = 0 ± ∣= ± ≠γ⅛- = ± 2

5. 3χ 2 + 2y 2 = 6 => j + ^ = 1 => c = √"a2 - b 2 = √ V 2= 1

e = f = ^ ;F ( 0 ,± l) ;

directrices are y = 0 ± ∣= ± √ p r = ± 3

7. 6x2 + 9y2 = 54 => ^ ÷ ^ = 1 => c = χ ∕a 2 —b 2 = √ 9 - 6 = √ 3 => e = * = ^

; F ( ± √ 3 ,θ ) ;

directrices are x = 0 ± ∣= ± y ⅛ r = ± 3ᅟ ∕3

9. Foci: (0, ± 3 ) , e = 0.5 => c = 3 and a = ^ = ⅛ = 6 => b2 = 3 6 - 9 = 27 => ^ ÷ ⅛

=

^

11. Vertices: (0, ± 7 0 ), e = 0.1 ≠> a = 7 0 a n d c = ae = 70(0.1) = 7 => b2 = 4900 - 49 = 4851 => ⅛ ÷ 4 ⅛ 13. Focus: ^ ? θ ) , Directrix: x = ^

=> c = a e = √ 5 and ; = ^

=> ? = ^

=

^

2

= ∣( x - ^ )

15. Focus: (—4 ,0 ),Directrix: x = —16 => c = ae = 4 and ≡ = 16=> ^ = 16 ≠> ^ = 16 => e 2 = J ≠ PF = | PD ≠- √ ( x + = i

(x2

4)2

+ 32x + 256) =>

+ (y -

0)2

= ∣∣ x + 16∣≠> (x +

∣x2

y2

= 48 => ⅛ + ⅛ = 1

+

4)2

+

y

2

= 1 (χ +

16)2

≠

x2

1

= > e2 = ∣

∣ x - ^ ∣=⅛> ( x - √ 5 p y

=> J ( x - √ s ) 2 + (y - 0)2 = ^

4 - e = ^ . ThenPF = ^ P D

=> ^

=

+ 8x + 16 +

e

y2

2

= ∣. Then

310

Chapter 10 Conic Sections and Polar Coordinates

17. e = ∣ => take c = 4 and a = 5; c2 = a2 —b 2 => 16 = 25 - b 2 => b 2 = 9 => b = 3; therefore

19. One axis is from A (l, 1) to B (l, 7) and is 6 units long; the other axis is from C (3,4) to D(—1, 4) and is 4 units long. Therefore a = 3, b = 2 and the major axis is vertical. The center is the point C (l, 4) and the ellipse is given by ⅛ = 1; c 2 = a2 - b 2 = 32 - 22 = 5 + ^ => c = √ z5 ; therefore the foci are F ^1,4 ± √⅞ ) , the eccentricity is e = ^ = δ ^ , and the directrices are y = 4 ± ∣= 4 ± ^ j = 4 ± ^ .

21. The ellipse must pass through (0,0) => c = 0; the point (—1,2) lies on the ellipse ≠- —a + 2b = - 8. The ellipse is tangent to the x-axis => its center is on the y-axis, so a = 0 and b = - 4 => the equation is 4x2 + y 2 —4y = 0. Next, 4x2 + y2 - 4y + 4 = 4 => 4x2 + (y - 24)2 = 4 => x2 + ^ ^ = 1 => a = 2 and b = 1 (now using the standard symbols) => c2 = a2 - b 2 = 4 — 1 = 3 ≈> c = √ 5 => e = *j = ^ c = √ a 2 + b 2 = √ 1 + 1 = √ 2 ≠> e = ≡

23. x2 - y 2 = 1 ^ =

i

p = ᅟ∕ 2 ; asymptotes are y = ± x; F ( ± ᅟ ∕ 2 , θ) ;

directrices are x = 0 ± ∣= ± ^

2 2 25. y2 - x 2 = 8 => ^ - ^ = 1 ≠ ∙ c = ᅟ τ∕ a + b o o = χ ∕8 + 8 = 4 => e = £ = ^ = √ 2 ; asymptotes are

y = ± x; F (0, ± 4 ); directrices are y = 0 ± ∣ = ± ⅛ =

±

2

.

Section 10.2 Classifying Conic Sections by Eccentricity 2 2 27. 8x - 2y = 16 4

⅛- ⅛ = 1 ^

= ^ 2 + 8 = √Tθ ^

e

—a

=

311

c = √^2 ± b2 √ z5 ; asymptotes

"^

are y = ± 2x; F ^ ± ^/10,0^ ; directrices are x = 0 ± ∣ =

= ± 2

29. 8y2 - 2x2 - 16 => ⅛ - ⅛ = 1 => c = √ a 2 + b 2 = √ 2 + 8 = ^/10 => e = j = -s^

= 1∕ 5 ; asymptotes

are y = ± | ; F (0, ± √ 1 0 ι ; directrices are y = 0 ± ∣

31. Vertices (0, ± 1) and e = 3 => a = 1 and e = ^ = 3 => c = 3a = 3 => b 2 = c 2 —a2 = 9 — 1 = 8 => y2 — y = 1 33. Foci ( ± 3 ,0 ) and e = 3 ≠> c = 3 and e = ^ = 3 => c = 3a => a = 1 => b 2 = c 2 —a 2 = 9 — 1 = 8 => x2 — ^ = 1 35. Focus (4,0) and Directrix x = 2 => c = ae = 4 and ∣= 2 => ^ = 2 ≠> ^ = 2 = > e PF = √ 2 P D ≠> √ ( x -

4)2

+ (y -

0)2

= √2 ∣ x - 2 ∣=> (χ -

4)2

+

y2

= 2(x -

2)2

2

= 2 = > e = √ 2 . Then

≠> x 2 - 8x + 16 + y2

= 2 (x2 - 4x + 4) => - x 2 ± y 2 = —8 => ⅞ - ⅞ = 1 37. Focus (—2,0) and Directrix x = —∣ => c = ae = 2 and ∣= 5 => ⅛ ~ 2 ^

⅝

=

2 => ε 2 = 4 => e = 2. Then

PF = 2PD ≠> √ ( x ± 2)2 + (y - 0)2 = 2 ∣ x + i ∣=> (x + 2)2 + y 2 = 4 (x + ∣) 2 ≠> x2 ± 4 x + 4 ± y 2 = 4 (x 2 + x + ∣) => - 3 x 2 + y 2 = - 3 => x2 - ^ = 1 39. √ ( x - 1)2 + (y + 3)2 = 5 ∣ y - 2 ∣≠> x2 - 2x + 1 + y2 + 6y + 9 = 5 (y2 - 4y + 4) => 4x 2 - 5y 2 - 8x + 60y + 4 = 0 ≠> 4 (x2 - 2x ± 1) - 5 (y2 - 12y + 36) = - 4 + 4 - 180 => ⅛

41. To prove the reflective property for hyperbolas: ⅛ - ⅛ = 1 => a2 y2 = b 2 x 2 - a2 b 2 and ⅛ = ^ . Let P(x0 , yo) be a point of tangency (see the accompanying figure). The slope from P to F(—c, 0) is ^ ^ and from P to F 2 (c, 0) it is ^⅛^ . Let the tangent through P meet the x-axis in point A, and define the angles Z F i PA = α and ZF 2 PA = β. We will show that tan α = tan β. From the preliminary result in Exercise 22,

- ⅛

= 1

Chapter 10 Conic Sections and Polar Coordinates

312

x

(

tan α =

ob 2 _

yg A

∖

7⅛ 2⅜ T 1

x⅜ b2

=

tan β =

∖ 1+

c

+ x °b 2 c ~ y ffa2

a⅜2 +χob 2 c

=

i

x0y0 a2 + y0 a2c + xoyob2

I( ⅞ ^ ( ⅞ ^

x0yoc2 + yoa2c

_ bi y0c

jn

a s im

∩a r

m a nner,

’

= ^ . Since tan α = tan β, and a and β are acute angles, we have a — β.

,T ⅛

G⅛Λ⅛U

10.3 QUADRATIC EQUATIONS AND ROTATIONS 1.

x 2 —3xy + y2 —x = 0 => B2 - 4AC = ( - 3 ) 2 - 4(1)(1) = 5 > 0 => Hyperbola

3.

3x 2 - 7xy + √ Γ 7 y 2 = 1 => B 2 - 4AC - ( - 7 ) 2 - 4(3) √ 1 7 ≈ -0.477 < 0 => Ellipse

5. x 2 + 2xy + y2 + 2x - y + 2 = 0 ≠> B 2 - 4AC = 22 - 4(1)(1) = 0 => Parabola 7.

x2 + 4xy + 4y2 - 3x = 6 => B 2 - 4AC = 42 - 4(1)(4) = 0 ≠∙ Parabola

9.

xy + y 2 - 3x = 5 => B 2 - 4AC = 12 - 4(0)(1) = 1 > 0 ≠> Hyperbola

11. 3x2 - 5xy + 2y2 - 7x - 14y = - 1 ≠- B2 - 4AC = ( - 5 ) 2 - 4(3)(2) = 1 > 0 => Hyperbola 13. x2 - 3xy + 3y2 + 6y = 7 ≠> B 2 - 4AC = ( - 3 ) 2 - 4(1)(3) = - 3 < 0 => Ellipse 15. 6x 2 + 3xy + 2y2 + 17y + 2 = 0 ≠> B 2 - 4AC = 32 - 4(6)(2) = - 3 9 < 0 =► Ellipse 17. cot 2α = ^ ^ = 2 - θ

=> 2α = ^ ≠> α = ≡ ; therefore x = x' cos α —y' sin α,

y = x' sin α + y' cos α => x = x' ^ => ( ^

x'

^ ^ y ') ( ΔΓ

19. cot 2α = ^

'+

X

—y' ^ , y = x' ^

y ') =

ΔΓ

2

2 x '2

4

-

+ y' y^ I y'2 =

2

≠,

x '2

- y' 2 = 4 ≠> Hyperbola

= ∣ ^= = - ^ => 2α — j => α = f ; therefore x = x' cos α —y' sin α,

y = x' sin α + y' cos α ≠> x = Δ ^ => 3 ( ^ x ' - i y ' )

2

X'

- ⅜y', y = 1 x' ⅛ ∙ ^ y'

+ 2 √ 3 ( ^ x ' + ∣y ') ( i x '

+

^ ∕)

( l x'

+

+

^

y')

2

-

8

( ^ χ '- l ∕

+ 8 √ 3 ( ∣x' + ^ y ') = 0 ≠> 4 X' 2 + 16y' = 0 => Parabola

21. cot 2α = ^ ^ = ^ ∏ = 0 => 2α = j => α = ^ ; therefore x = x' cos α —y' sin α, y = x' sin α + y' cos α => x = -i^ x' - ^ y', y = ^ x' + - ^ y' =► ( ^ x ' - ^ y ' )

2

-

2

( ^ χ ' + ⅛ ' ) + ( ^ χ ' + ^ y ,)

^ ^ - ^ / )

2

= 2 ≠> y'2 = l

=> Parallel horizontal lines 23. cot 2α = ^ ^ =

v^

- ^ = 0 => 2α = ^ => α = J ; therefore x = x' cos α —y' sin α,

y = x' sin α + y' cos a => x = ^ x' Z

^

√ 2 ( ^ x' - ^ y ') 2

—$ ( ^

x' -

+ 2

^ y ') 4 $ ( ^

y', y = Z

∙z

*

y'

χ' + 2

Z

*

√ 2 ( ^ x' - ^ y ') ( ^ x' + ^ y ') χ,

+ ^ y, )

=

+

√ 2 ( ^ x'

+

^ y ') 2

0 => 2∖ ∕2 X' 2 ÷ 8 ᅟ ∕2 y ' = 0 => Parabola

Section 10.3 Quadratic Equations and Rotations 25. cot 2α = ^

313

= 2y3 = o =^ 2α = ^ => α = ∣; therefore x = x' cos α —y' sin α,

y = x' sin α + y' cos α => x = δ ^ x' - ^ y', y = δ^ x' + ^ y' => 3 ( ^ x' - ^ y ') 2 + 2 ( ^ x' - ^ y') ( ^ x '+ ^ y') + 3 ( ^ x'

+

^ y ') 2 = 1 9 => 4x'2 + 2y'2 = 19

=> Ellipse 27. cot 2α = η ^ = 4 => cos 2α = ∣(if we choose 2α in Quadrant I); thus sin a = and cos α = ^ ^ ^

= ᅟ ∕ ^ ∙ = ∙⅜ (or sin a —

/.L-cos ja _ . ∕Σ Ξ U Γ =

V

2

V

2

1

^

^ and cos a =

29. tan 2α = τ ⅛ = ∣ => 2α ≈ 26.570 => α ≈ 13.28o ≠∙ sin α ≈ 0.23, cos α ≈ 0.97; then A' ≈ 0.9, B' ≈ 0.0, C' ≈ 3.1, D' ≈ 0.7, E' ≈ -1.2, and F' = - 3 => 0.9 x'2 + 3.1 y'2 + 0.7x' - 1.2y' - 3 = 0, an ellipse 31. tan 2α = f ⅛ = ∣ => 2α ≈ 53.130 ≠> α ≈ 26.57o => sin a ≈ 0.45, cos a ≈ 0.89; then A' ≈ 0.0, B' ≈ 0.0, C' ≈ 5.0, D' ≈ 0, E' ≈ 0, and F' = - 5 => 5.0 y'2 - 5 = 0 or y' = ± 1.0, parallel lines 33. tan 2α = ^

= 5 ≠* 2α ≈ 78.69o => α ≈ 39.35o => sin α ≈ 0.63, cos a ≈ 0.77; then A' ≈ 5.0, B' ≈ 0.0,

C' ≈ -0.05, D' ≈ -5.0, E' ≈ -6.2, and F' = -1 => 5.0 x'2 - 0.05 y'2 - 5.0x' - 6.2y' - 1 = 0, a hyperbola 35. α = 90o => x = x' cos 90o —y, sin 90o = - y ' and y = x, sin 90o ÷ y' cos 90o = x' (a ) ⅛ ÷ ⅛ - 1

(b) ⅛- —p- — 1

(c) x'2 ÷ y'2 — a2

(d) y = m x = > y - m x = 0 = > D = —m and E = 1; α = 90o => D' = 1 and E' = m => my' + χ' = 0 => y' = — x' (e) y = mx + b => y - m x - b = 0 => D = - m and E = 1; Q = 90o => D' = 1, E' = m and F' = - b => my' ÷ x' - b = 0 => y' = - ⅛ x' ÷ ⅛ 37. (a) A' = cos45 o sin45° = ( ^ ) ( ^ ) = ∣, B' = 0, C' = - c o s 45o sin45o = - ∣, F' = - 1 ≠> 1 x ' 2 - 1 y'2 =

1

=^

χ '2

- y'2 =

2

(b) A' = | , C' = —| (see part (a) above), D' = E' = B' = 0, F' = —a ≠> ∣x'2 - ^ y'2 = a ≠ ∙ x'2 - y'2 = 2a 39. Yes, the graph is a hyperbola: with AC < 0 we have -4A C > 0 and B2 —4AC > 0. 41. Let a be any angle. Then A' = cos2 α + sin2 α = 1, B' = 0, C' = sin2 a + cos2 α = 1, D' = E' = 0 and F' = —a2 => x'2 ÷ y'2 = a2 . 43. (a) B2 - 4AC = 42 - 4(1)(4) = 0, so the discriminant indicates this conic is a parabola (b) The left-hand side of x2 + 4xy + 4y2 ÷ 6x + 12y + 9 = 0 factors as a perfect square: (x + 2y + 3)2 = 0 => x ÷ 2y ÷ 3 = 0 => 2y = —x —3; thus the curve is a degenerate parabola (i.e., a straight line).

314

Chapter 10 Conic Sections and Polar Coordinates

45. (a) B 2 - 4AC = 1 - 4(0)(0) = 1 =≠ hyperbola (b) xy + 2x —y = 0 => y(x - 1) = - 2 x => y = 7r7 (c) y = ⅛

-> ∣ = ⅛ 2 and we want ⅛

= -2 ,

the slope of y = —2x => —2 = —^

-4 -3 -2

2

=> ( x - l ) = 4 = > x = 3 o rx = - 1 ; x = 3 => y = - 3 => (3, - 3 ) is a point on the hyperbola where the line with slope m = - 2 is normal => the line is y + 3 = -2 (x - 3) or y = - 2 x + 3; x = —1 => y = —1 => ( - 1 , - 1 ) is a point on the hyperbola where the line with slope m = —2 is normal => the line is y + 1 = -2 (x + 1) or y = -2 x - 3 47. Assume the ellipse has been rotated to eliminate the xy-term => the new equation is A 'x'2 + C 'y'2 = 1 => the ∕⅜ ) ~ √ f e ^ 7 ⅜ σ ’ $*n c e ®2 “ ^ ^ semi-axes are y ^ and ^ / ^ => the area is π ( y ^ ) ( ᅟ = B'2 —4A'C' = —4A'C' (because B' = 0) we find that the area is v ∕ 4a 2^ -

β2

as claimed.

49. B,2 - 4A , C' = (B cos 2α ÷ (C —A) sin 2α) 2 —4 (A cos2 α ÷ B cos α sin α + C sin2 Q ) (A sin2 α —B cos α sin α ÷ C cos2 α) = B 2 COS2 2α + 2B(C - A) sin 2α cos 2α + (C - A)2 sin2 2α - 4A 2 cos2 a sin2 a + 4AB cos3 α sin α - 4AC cos4 a —4AB cos α sin3 α + 4B 2 cos2 a sin2 α - 4BC cos3 α sin α —4AC sin4 α + 4BC cos a sin3 a - 4C 2 cos2 α sin2 α = B2 COS2 2α ÷ 2BC sin 2α cos 2α - 2AB sin 2α cos 2α ÷ C 2 sin2 2α - 2AC sin2 2α + A 2 sin2 2α - 4A 2 COS2 α sin2 α + 4AB cos3 α sin α - 4AC cos4 a —4AB cos a sin3 α + B 2 sin2 2α —4BC cos3 α sin α —4AC sin4 a ÷ 4BC cos α sin3 α —4C 2 cos2 α sin2 α 2 = B + 2BC(2 sin a cos a) (cos2 a —sin2 α) —2AB(2 sin α cos α) (cos2 a —sin2 α) + C 2 (4 sin2 α cos2 α) —2AC (4 sin2 a cos2 α) ÷ A 2 (4 sin2 a cos2 α) —4A 2 cos2 α sin2 a + 4AB cos3 α sin α —4AC cos4 a —4AB cos a sin3 a —4BC cos3 α sin α —4AC sin4 α + 4BC cos a sin3 α —4C 2 cos2 α sin2 a 2 = B — 8AC sin2 a cos2 α —4AC cos4 α - 4AC sin4 a = B 2 —4AC (cos4 α + 2 sin2 α cos2 α + sin4 α) = B 2 —4AC (cos2 a + sin2 α ) 2 = B 2 - 4AC 10.4 CONICS AND PARAMETRIC EQUATIONS; THE CYCLOID 1. x = cos t, y = sin t, 0 ≤ t ≤ π =Φ cos2 1 ÷ sin2 t = 1 => x 2 + y2 = 1

3. x = 4 cos t, y = 5 sin t, 0 ≤ t ≤ π => ⅞ 2 ^ + ⅛ ⅛ = ι => ⅛ ÷ ⅛ = 1

Section 10.4 Conics and Parametric Equations; The Cycloid 7. x = - sec t, y - tan t, —f < t < ∣ => sec2 1 - tan2 1 = 1 => χ 2 —y2 = 1

9. x = t, y = √ 4 —12 , 0 ≤ t ≤ 2

11. x = —cosh t, y = sinh t, —∞ < 1 < ∞ => cosh2 1 —sinh2 t = l => x2 - y2 = l

13. Arc PF = Arc AF since each is the distance rolled and ^ = ZFCP => Arc PF = b(ZFCP); * ^ = θ => Arc AF = a0 => a0 = b(ZFCP) ≠> ZFCP = § 0; ZOCG = f - 0; ZOCG = ZOCP + ZPCE = ZOCP + ( f - α ) . Now ZOCP = π - ZFCP = π - £D0. Thus ZOD CG = 2π - ≥0 + 2≡ - α => ≡ - 0 = π - ∣0 + ∣- α => α = π - ∣0 ÷ 0 = π - ( ^ 0 ) .

Then x = OG - BG = OG - PE = (a - b) cos 0 - b cos α = (a - b) cos 0 - b cos (π - ⅛ ^ 0) = (a —b) cos 0 ÷ b cos ( 5p 0 ) . Also y = EG = CG - CE = (a - b) sin 0 —b sin α = (a - b) sin 0 —b sin (π —5p 0) = (a —b) sin 0 - b sin ( 3^ 0 ) . Therefore x = (a - b) cos 0 + b cos ( ^ 0) and y = (a - b) sin 0 - b sin ( ^ 0 ) . If b = ∣, then x = (a - ∣) cos 0 + ∣cos ( ^

^)

= ~ cos 0 + ∣cos 30 = ^ cos 0 ÷ ∣(cos 0 cos 20 - sin 0 sin 20) = y cos 0 + ∣((cos 0) (cos2 0 - sin2 0) - (sin 0)(2 sin 0 cos 0)) = ⅜ cos 0 ÷ I cos3 0 —7 cos 0 sin2 0 — ⅜ sin2 0 cos 0 = γ cos 0 ÷ ∣cos3 0 —y (cos 0) (1 —cos2 0) = a cos3 0; y = (a —∣) sin 0 —| sin ( ^ ∙ #) = ⅜ sin 0 - ∣ sin 30 = ^ sin 0 — ∣(sin 0 cos 20 + cos 0 sin 20) = ^ sin 0 - ∣((sin 0) (cos2 0 - sin2 0) + (cos 0)(2 sin 0 cos 0)) = ^ sin 0 — ∣ sin 0 cos2 0 + ∣sin3 0 - ^ cos2 0 sin 0 = j sin 0 - ^ sin 0 cos2 0 + ∣sin3 0 = j sin 0 - ^ (sin 0) (1 - sin2 0) ÷ ∣sin3 0 = a sin3 0.

315

316

Chapter 10 Conic Sections and Polar Coordinates

15. Draw line AM in the figure and note that ZAMO is a right angle since it is an inscribed angle which spans the diameter of a circle. Then AN 2 = MN 2 ÷ AM 2 . Now, OA = a, ^ = tan t, and ^ = sin t. Next MN = OP => OP 2 = AN 2 —AM 2 = a2 tan2 1 —a2 sin2 1 => OP = √ za2 tan2 1 —a2 sin2 1 ∕ sec2 1 — 1 = 3^ = (a sin t)ᅟ x = OP sin t = ~ ⅛ = a cos t

y = OP cos t = a

sin2

sin2

i

. In triangle BPO,

1 tan t and

1 => x = a sin2 1 tan t and y = a sin2 1.

17. D = y ( x - 2 ) 2 + ( y - l ) 2 => D 2 = (x - 2)2 + (y - ∣) 2 = (t - 2)2 + (t2 - ∣) 2 ^

D2 ≈ t 4 - 4 t + ⅛

=> 5⅛ P = 4t3 —4 = 0 => t = l . The second derivative is always positive for t ≠ 0 => t = 1 gives a local minimum for D2 (and hence D) which is an absolute minimum since it is the only extremum => the closest point on the parabola is (1, 1). 19. (a)

23. (a)

(c)

Section 10.5 Polar Coordinates

10.5 POLAR COORDINATES

1. a, e; b, g; c, h; d, f 3. (a) (2, f + 2nπ) and (-2 , ∣+ ( 2 n + l ) π ) , n an integer (b) (2,2nπ) and (—2, (2n + l)π), n an integer (c) (2, γ + 2nπ) and (—2, γ + (2n + l ) π ) , n an integer (d) (2, (2n ÷ l)π) and ( - 2 , 2nπ), n an integer

5. (a) (b) (c)

χ = r cos θ= 3 cos 0 = 3, y = r sin θ = 3 sin 0 = 0 => Cartesian coordinates are (3,0) x = r cos 0= - 3 cos 0 = - 3 , y = r sin θ = - 3 sin 0 = 0 => Cartesian coordinates are (—3,0) x = r cos 0= 2 cos y = - 1 , y = r sin θ = 2 sin y = ^/3 => Cartesian coordinates are ^ -1 , ᅟ ∕3 ^

(d)

∕3 ^ x = r cos 0= 2 cos y = 1, y = r sin 0 = 2 sin y = √ z3 => Cartesian coordinates are ^1, ᅟ

(e) (f)

x = r cos 0= - 3 cos π = 3, y = r sin 0 = - 3 sin π = 0 => Cartesian coordinates are (3,0) ∕3 ^ x = r cos 0= 2 cos ∣= 1, y = r sin 0 = 2 sin j = √ 3 => Cartesian coordinates are (1, ᅟ

(g) x = r cos 0 = - 3 cos 2π = - 3 , y = r sin 0 = - 3 sin 2π = 0 => Cartesian coordinates are (-3 ,0 ) (h) x = r cos 0 = - 2 cos ( - j ) = - 1 , y = r sin 0 = - 2 sin ( - ∣) = λ ∕3 => Cartesian coordinates are ( - 1 , √⅞

317

318

Chapter 10 Conic Sections and Polar Coordinates

23. r cos 0 = 2 ≠>x = 2, vertical line through (2,0) 25. r sin 0 = 0 => y = 0, the x-axis 27. r = 4 esc 0 => r = ^

=> r sin 0 = 4 => y = 4, a horizontal line through (0,4)

29. r cos 0 + r sin 0 = 1 => x + y = 1, line with slope m = - 1 and intercept b = 1 31. r 2 = l => x2 + y2 = 1, circle with center C = (0,0) and radius 1 33.

r =

sin ¢ - 2 cos θ

^

r sin 0 - 2r cos 0 = 5 => y - 2x = 5, line with slope m = 2 and intercept b = 5

35. r = cot 0 esc 0 = ( ^ f ) ( ^ )

≠ , r sin2 0 = cos 0 => r2 sin2 0 = r cos 0 ≠ ∙ y2 = x, parabola with vertex (0,0)

which opens to the right 37. r = (esc 0) e rcosβ => r sin 0 = ercos" => y = ex , graph of the natural exponential function 39. r 2 + 2r2 cos 0 sin 0 = 1 => x2 + y2 + 2xy = 1 => x2 + 2xy + y2 = 1 => (x + y)2 = 1 => x + y = ± 1 , two parallel straight lines of slope - 1 and y-intercepts b = ± 1 41. r 2 = - 4 r cos 0 ≠> χ 2 + y 2 = - 4 x ≠- x2 + 4x + y2 = 0 => x2 + 4x + 4 + y2 = 4 => (x + 2)2 + y 2 = 4, a circle with center C(—2,0) and radius 2 43.

r = 8 sin 0 => r2 = 8r sin 0 => x2 + y2 = 8y => x2 + y 2 —8y = 0 => x2 + y 2 —8y + 16 = 16 => x2 + (y —4)2 = 16, a circle with center C(0,4) and radius 4

45.

r = 2 cos 0 + 2 sin 0 => r2 = 2r cos 0 + 2r sin 0 => x2 + y2 = 2x + 2y => x2- 2x + y 2 - 2y = 0 => (x - 1)2 + (y - 1)2 = 2, a circle with center C (l, 1) and radius √ z2

47. r sin (0 + ∣) = 2 ≠

r (sin 0 cos j + cos 0 sin j ) = 2 => ^ r sin 0 + ∣r cos 0 = 2 => ^ y + ∣x = 2

∕3 y + x = 4, line with slope m = —^ => ᅟ

and intercept b = ^

49. x = 7 => r cos 0 = 7 51. x = y => rc o s 0 = r s in 0 => 0 = ∣ 53. x2 + y 2 = 4 => r 2 = 4 => r = 2 or r = - 2 55. ^ + ^ = 1 => 4x2 + 9y 2 = 36 => 4r2 cos2 0 + 9r2 sin2 0 = 36

Section 10.6 Graphing in Polar Coordinates

319

57. y 2 = 4x => r2 sin2 0 = 4r cos 0 => r sin2 θ = 4 cos θ 59. x2 + (y - 2)2 = 4 => x2 + y 2 - 4y + 4 = 4 ≠- x2 + y2 = 4y => r 2 = 4r sin 0 => r = 4 sin 0 61. (x - 3)2 + (y + 1)2 = 4 => x2 - 6x + 9 + y2 + 2y + 1 = 4 => x2 + y 2 = 6x - 2y - 6 => r 2 = 6r cos θ - 2r sin θ - 6 63. (0,0) where θ is any angle

10.6 GRAPHING IN POLAR COORDINATES 1. 1 + c o s ( - 0) = 1 + cos 0 = r => symmetric about the x-axis; 1 + cos (—0) ≠ —r and 1 + cos (π —0) ∙ ι . ι = 1 - cos 0 ≠ r => not symmetπc about the y-axis; therefore not symmetric about the origin

r= 1 + co s0

1

/

3. 1 - sin (- 0 ) = 1 + sin 0 ≠ r and 1 — sin (π —0) = 1 —sin 0 ≠ —r => not symmetric about the x-axis; 1 - sin (π - 0) = 1 - sin 0 = r => symmetric about the y-axis; therefore not symmetric about the origin

5. 2 + s in (-0 ) = 2 - sin 0 ≠ r and 2 + sin(π - 0) = 2 + sin 0 ≠ —r => not symmetric about the x-axis; 2 ÷ sin (π - 0) = 2 ÷ sin 0 = r => symmetric about the y-axis; therefore not symmetric about the origin

7. sin ( - ∣) = “ sin ( ∣) = - r => symmetric about the y-axis; sin ( ⅛ ^ ) = sin ( f ) , so the graph is symmetric about the x-axis, and hence the origin.

sin (0/2)

Chapter 10 Conic Sections and Polar Coordinates

320

9. cos (—0 = cos 0 = r 2 => (r, —0) and (—r, - 0 ) are on the graph when (r, 0) is on the graph => symmetric about the x-axis and the y-axis; therefore symmetric about the origin

11. —sin (π —0) = —sin 0 = r 2 => (r, π —0) and ( - r , π —0) are on the graph when (r, 0) is on the graph => symmetric about the y-axis and the x-axis; therefore symmetric about the origin

13. Since ( ± r, —0) are on the graph when (r, 0) is on the graph (( ± r) 2 = 4 cos 2(—0) => r 2 = 4 cos 2 0 ), the graph is symmetric about the x-axis and the y-axis => the graph is symmetric about the origin

15. Since (r, 0) on the graph => ( - r , 0) is on the graph (( ± r) 2 = —sin 20 => r 2 = - sin 2 0 ), the graph is symmetric about the origin. But - sin 2 (-0 ) = - ( - sin 20) sin 20 ≠ r 2 and - sin 2(π - 0) == - sin (2π - 20) = —sin (-2 0 ) = —(—sin 20) = sin 20 ≠ r2 => the graph is not symmetric about the x-axis; therefore the graph is not symmetric about the y-axis 17. 0 = f => r = - l => ( - l , f ) , a n d 0 = - f *

( - 1 . - 5 ) : ^ = $ = - * > 0 ; Slope = ⅛ ⅛ 1

= ⅛ ⅛ ⅛ ⅛ -s in 2 →

=> r = - l

(f)+ (-l)c o s 5

⅛ ⅜ - ( - l ) s i n *

S lo p e a t(-l,* )is

^

,

= - 1 ! sc ll °Pe

- ^ ( Ξ J )H - 1 ) COS(-J)

at

 .

π IS ( - h - 1)

1

-s in (-≈ )c o s (-f)-(-l)s in (-≈ )

19. 0 = 2 => r = l => ( l , ≈ ) 5 0 = - z

≠ r = -l

≠> ( - l , - 2 ) i 0 = ⅜ => r = - l => ( - l , ⅜ ) j 0 = -⅞

^

r= 1 ^

√ = ⅛ = α2σ cos 20;’ z 0+r cos 0 Slope = rr 'csin o s 0 - r s in 0 => Slope at (1, ∣) is Slope at (—1, —5) is

(l,-⅜ ) j 2 cos 20 sin 0+r cos θ 2 cos 20 cos 0—r sin 0 2 cos ( I ) sin ( ∣)÷ (l)c o s (∣) _ 2 cos ( f ) cos ( J ) - ( 1 ) sin ( J ) “

-1 ;

2 cos ( - ∣) sin ( - ^ ) + ( - 1 ) cos ( - ∣) Γ c 0 Γ (^ p x ^ (^ 5 ^ ^

= 1;

r2 > 4 cot 28

Section 10.6 Graphing in Polar Coordinates

29. (2, j ) is the same point as (—2, — ∣) ; r = 2 sin 2 ( - ∣) = 2 sin ( - ^ ) = - 2 => (—2, — ∣) is on the graph => (2, ¾e ) is on the graph 31. 1 + cos 0 = 1 - cos θ => cos 0 = 0 ≠> 0 = ∣, y => r = 1; points of intersection are (1, ∣) and (1, γ ) . The point of intersection (0,0) is found by graphing.

321

322

Chapter 10 Conic Sections and Polar Coordinates

33. 2 sin 0 = 2 sin 20 => sin 0 = sin 20 => sin 0 = 2 sin 0 cos 0 4 sin 0 - 2 sin 0 cos 0 = 0 => (sin 0)(1 - 2 cos 0) = 0 => sin 0 = 0 or cos 0 = ∣ => 0 = 0, π, 5 , or - I ; 0 = 0 or π =» r = 0, 0 = j => r = ι ∕ 3 , and 0 = - 5 => r = - √ z3 ; points of intersection are (0,0), f ᅟ ∕ 3 , f Y and f —  ∕3 >- f )

= 4 sin 0 => ∣= sin 0 => 0 = f , ⅜ ; points

35. ^ √ ^ j

of intersection are ( √ 2 , ≡ ) and ( √ 2 , ⅜ ) . The points ( y ^ , —∣) and I √ 2 , - y j are found by graphing.

≠> 20 = ^0 ⅞ , ψ , ψ0 37. 1 = 2 sin 20 ≠> sin 20 = ∣ 2 o ’ 0 o

0

=

⅛ ’ II ’ IT ’ IT ’ P°in t s ° f intersection are

( l > ⅞ ) - ( ⅛ ) - ( ⅛ ) < ^ ( ⅛ ) ∙ No Other points are found by graphing.

39. r2 = sin 20 and r 2 = cos 20 are generated completely for 0 ≤ 0 ≤ f ∙ Then sin 20 = cos 20 => 20 = ∣is the only solution on that interval o

0 = ∣ O r 2 = sin 2 ( ∣) = ^

O r = ± ψ j ; points of intersection are ( ± ^ , j ) . The point of intersection (0,0) is found by graphing.

41. 1 = 2 sin 20 o O 0

=

sin 20 = ∣ o 2

⅞ ’ ⅜ ’ W ’ IT

;

, ψo , ⅛o 20 = £o , ⅞ o 7

points of intersection are

( 1 >⅞ ) ’ ( 1 >⅜ ) , (1. ⅜ ) - and (1, ⅛ ) . The points of intersection (1, ⅛ ) , (1, ⅛ ) , (1, ^ ) and (1, ^ ) are found by graphing and symmetry.

43. Note that (r, 0) and ( - r , 0 + π) describe the same point in the plane. Then r = 1 —cos 0 0 —1 —cos (0 + π) = - 1 - (cos 0 cos 7r —sin 0 sin π) = - 1 ÷ cos 0 = -(1 —cos 0) = —r; therefore (r, 0) is on the graph of r = 1 - cos 0 o (—r, 0 + π) is on the graph of r = - 1 —cos 0 o the answer is (a).

Section 10.6 Graphing in Polar Coordinates

323

47.

49. (a)r2 = —4 cos θ => cos 0 = —^ ; r = 1 —cos 0 =>

r = 1 - ( - 1θ

≠> 0 = r2 - 4r + 4 => (r —2)2 = 0

=> r = 2; therefore cos 0 = - y = - 1 => 0 = π => (2, π) is a point of intersection (b) r = 0 ≠> 02 = 4 cos 0 => cos 0 = 0 => 0 = ∣, y => (θ, j ) or (θ, y ) is on the graph; r = 0 => 0 = 1 - cos 0 => cos 0 = 1 => 0 = 0 => (0,0) is on the graph. Since (0,0) = (θ, j ) for polar coordinates, the graphs intersect at the origin. 51. The maximum width of the petal of the rose which lies along the x-axis is twice the largest y value of the curve on the interval 0 ≤ 0 ≤ ∣. So we wish to maximize 2y = 2r sin 0 = 2 cos 20 sin 0 on 0 ≤ 0 ≤ ∣. Let f(0) = 2 cos 20 sin 0 = 2 (1 —2 sin2 0) (sin 0) = 2 sin 0 - 4 sin3 0 => f'(0) = 2 cos 0 —12 sin2 0 cos 0. Then f , (0) = 0 => 2 cos 0 —12 sin2 0 cos 0 = 0 => (cos 0) (1 - 6 sin2 0) = 0 => cos 0 = 0 or 1 —6 sin2 0 = 0 => 0 = ∣or sin 0 = ^ . Since we want 0 ≤ 0 ≤ ∣, we choose 0 = sin^1 ( ^ ) => f(0) = 2 sin 0 —4 sin3 0 = 2( ^

—4 ∙ ^ ^ = ^ y . We can see from the graph of r = cos 20 that a maximum does occur in the

interval 0 ≤ 0 ≤ ∣. Therefore the maximum width occurs at 0 = sin~1 ( ^ ) , a n ^ the maximum width is ⅛

324

Chapter 10 Conic Sections and Polar Coordinates

10.7 AREA AND LENGTHS IN POLAR COORDINATES *2τr

J

o

=f 0 3. A = 2

∣(4 + 2 cos 0)2 d0 =

∕*2τr

Jo ∣(16 +

f*2τr

l6 c o s 0 + 4cos 2 0)d0 = J o [8 + 8 cos 0 + 2 ( l + c 2o s 2 g )] d0

(9 + 8 cos

θ+cos 20) d0 =

Γ" ⅜cos

20 d0 *= Γ l + ≡ 2wL d0 =4 1J [0 + ⅛o∙≤] [ ∕ 4 = f 0

c ∕θ 2

2

[90 + 8 sin 0 + ∣sin 20] 2π = 18π

*zθ

2 sin 20 d0 = [ - cos 20] θz2 = 2

5. A = / ^ 1 ( 4 sin 20) d0 = ^

7. r = 2 cos 0 and r = 2 sin 0 =+ 2 cos 0 ≈ 2 sin 0 + cos 0 = sin 0 + 0 = ^ ; therefore Γπ∕4

*7Γ∕4 1

J

o

*τr∕4

∣(2 sin 0 ) 2 d0 = J o

J = [20 —sin 20] J 0

∕* π ∕4

4 (M )d ⅛ 74

= Jo

A

4 sin 2 0d0

( 2 - 2 cos 20) d0

= f —1

9. r = 2 and r = 2(1 - cos 0) =+ 2 = 2(1 —cos 0) + cos 0 ≈ 0 =+ 0 = ± ∣; therefore 'π ∕2 1

J

o

1

| [2(1 —cos ¢)]2 d0 + ∣area of the circle

*π∕2

0 + cos 0) d0 + ( ∣π) (2) JX ^ 4(1( —- 22 cos COS 0 + M ) d0 + 2π 2

0

=

4

2

1

f√ 2

=

JQ

(4 —8 cos 0 + 2 4- 2 cos 20) d0 + 2ττ

= [60 —8 sin 0 + sin 20] θ^2 ÷ 2π = 5π —8 11. r = √ z3 and r 2 = 6 cos 20 => 3 = 6 cos 20 => cos 20 = ∣ => 0 = ∣(in the 1st quadrant); we use symmetry of the graph to find the area, so

=4 Γ ^ = 2£

6 C°S 20)-∣ M

2

d0

‘(6 cos 20 - 3) d0 = 2 [3 sin 20 - 30] J /6

= 3ᅟ ∕3 —π 13. r = 1 and r = —2 cos 0 => 1 = - 2 cos 0 => cos 0 = - ∣ => 0 = y in quadrant II; therefore

a = 2fZ I κ - 2

=Γ

2π∕3

cos

^

2

-

12ι d *

U

= £ (

12(1 + cos 2 0 ) - l ] d 0 2π∕3 = Γ

= [0 + sin 20] 2π ∕3 = f + ΔΓ

4 co s2

^-

1) d ^

( l+ 2 c o s 2 0 ) d 0

Section 10.7 Area and Lengths in Polar Coordinates

θ -3 ; therefore A = j ^

15. r = 6 and r = 3 esc 0 =4 6 sin => 0 = ∣or y

sin 0 = ∣

4

_ p5π∕6

325

i

∣(62 —9 csc2 0) d0

√ > -≡ ^ < )* = ι* + i-C il = (15π - ∣√ 3 ) - (3π + ∣√ 3 ) = 12τr - 9 √ 3

17. (a) r = tan 0 and r = ( 2 )

csc

ta n

^ z^

csc

^= ( 2)

$

cos

=> sin2 0 = ( ^

cos 0 => 1 —cos2 0 = ( ^

=> cos2 0 + ( ^

cos 0 —1 = 0 => cos 0 = - √ z2 or

δ^

^

(use the quadratic formula) => 0 = J (the solution

in the first quadrant); therefore the area of Ri is Ai

= Jo

4

1 ta n 2

0 d0

= I

AO = ( ^ ) csc f = ^

ΓS ^s e c 2 r

and OB = ( ^ ) csc ^ = 1 ^

=> the area of R2 is A2 = ∣^ y J ( 2 ) ^ ( 2 ~ I - *"4)

=

= i [tan 0 - 0] ; /4 = ∣(t a n 5 - 5 ) = 5 - 5 :

1) d0

=

AB = ^

- ( ^

= ^

Z ’ therefore the area of the region shaded in the text is

2"” Z* Note: The area must be found this way since no common interval generates the region. For

example, the interval 0 ≤ 0 ≤ J generates the arc OB of r = tan 0 but does not generate the segment AB of the line r = δ ^ csc 0. Instead the interval generates the half-line from B to + ∞ on the line r = δ ^ csc 0. (b)

ta n

lim^_ =

lim

^ → π ∕2 -

=

∞

a n ^ t ^ι e

(c⅛ -⅛ ) = o s '

^

r = sec 0 as 0 →

cosσ∕

zy

hne x = 1 is r = sec 0 in polar coordinates; then Q

lim } π

y2~

( ⅛ cos '

1)

'

=

1™

^ → π ∕2 -

lirn^ (t a n $ -

sec

$)

= 0 => r = tan 0 approaches (-¾⅛) × sι∏ 0∕

=> r = sec 0 (or x = 1) is a vertical asymptote of r = tan 0. Similarly, r = —sec 0

(or x = —1) is a vertical asymptote of r = tan 0. 19. r = 02 , 0 ≤ 0 ≤ √ 5 =4 ^ = 20; therefore Length = / £ ^ ∕(0 2 )2 + (20)2 d0 = ∫ ∕ √ ⅛

4

+ ^

2

d0

0∣√ 0 2 + 4 d0 = (since 0 ≥ 0) / £ 0 √ 0 2 + 4 d0; [u = 02 + 4 =4 ∣du = 0 d0; 0 = 0 =4 u = 4, = ∫ O'Λ ∣ 0

√5

u = 9] → £

I √ ild u = i [ ju 3∕ 2 ] ° = ⅛

21. r = 1 + cos 0 => ^ = —sin 0; therefore Length =

JQ χ ∕( l + COS 0)2 + (—sin 0)2 d0

= 2 f √ 2 ^ 2 ^ 0 d 0 = 2 ∫ f √ ⅛ ⅛ ^ d 0 = 4 ∫ oπ √ ^ d 0 2 3 ∙ r = ⅛ , O ≤ 0 ≤ f => * 36 _ Γπ^ / 7 “ Jo V ( l+ c o s 0 ) 2 +

= (—

π ⅛

36 sin2 9 ( l + cos⅛)4

=

{T^

;

r -

therefore length = £

=6∫0'"(τ÷⅛) √⅛⅛S⅛ 0« =⅛√2 Γ

=^f0

sec3 ^ d0 = βjθ

72^

j C π ∕2 . . / , sin2 Jo I l+cos⅛ l V 1 ^t^ ( l+ c o s β ) 2

> O ono ≤ 0 ≤ f ) 6 X ' i ( π ⅛ ) √ l ÷1 ∙⅛

∏⅛

1i

= 4 ∫ oπ c o s (f)d 0 = 4 [ 2 s in f ] ^ = 8 ^

a σ

* " y ∙' ,- io

=6√2 Γ " ^ ⅛

sec3 u du = (use tables) 6 f [secjrta∏u] ^ ∕4 _p 1 J^

=3Γ sec u du^

,

1“ ’ 2∣ 0«

326

Chapter 10 Conic Sections and Polar Coordinates = 6 ( ^ + [1 In ∣ sec u + tan u ∣ ] θ^4 ) = 3 ^ √ 2 + In (1 + √ z2 ^

f o ψ(cos

25. r = cos 3 ∣ => ^ = —sin f cos2 ∣; therefore Length = =

∕^ T f Γ ÷ s i '^ ^

=r

3

∣) 2 + (—sin ∣cos2 ∣) 2 d0

f*'4 cos

(cos2 f) λ ∕c o s 2 ( f ) + s i n 2 (f ) d0 =

d0 = ∫ ∕

2

( f ) d0

*+1sin f] r =i +1

4 ii ⅛ s d0 = H

27. r = √ 1 + cos 20 => ^ = ∣(1 + cos 20) 1∕ 2 ( - 2 sin 20); therefore Length 1+ 2 cos 20+ cos2 20+ sin2 20 1+ cos 20

^ =^ ^ 3 ^ =1 ^ ^ = ^ ] ^

29. r = ᅟ ∕ cos 2 0 , 0 ≤ 0 ≤ ∣ => ^ = ∣(cos 20) 1∕ 2 ( - s jn 20)(2) = ^ = = Jo ' ^ (√

=f 0 ^ v ∕( l + cos 20) +

cos20^

+ (^ = )

^2πᅠ∕ cos 20) (cos 0 )^ / ^ ⅛ d0 = X

dθ =f 2 π cos

∕

sin2 20 (1 +cos 20) d0

= 2π

; therefore Surface Area

( W c o s 20) (cos 0 ) y c o s 2 0 + ⅛ d0

^

=

P π s *n 0) o ^ ~

π

λ ∕2

^ ζ ;

31. r 2 = cos 20 => r = ± √ , cos 20; use r = ᅠ ∕ cos 20 on [θ, J] => ⅛ — 5 (cos 20)^ 1^2 ( - sin 20)(2) = therefore Surface Area = 2 =

^π X

7 s'n

^ ^

=

fo

^ π [- c

^2πᅠ∕ cos 20^ (sin 0) ^ cos 20 + ^

o s

0]o ^

=

^

d0 = 4π ^

v^coΓ20(sin 0) ^ j ^ ⅛ d0

[ - δ ^ - ( - l ) ] = 2π ^2 - ^

33. Let r = f(0). Then x = f(0) cos 0 ≠> ^ = f'(0) cos 0 - f(0) sin 0 ≠> ( H ) 2 = [f'(0) cos 0 - f(0) sin 0]2 = [f'(0)]2 cos2 0 - 2f'(0) f(0) sin 0 cos 0 + [f(0)]2 sin2 0; y = f(0) sin 0 ≠> ⅛ = f'(0) sin 0 + f(0) cos 0 => (30)

=

(I)2 + ( I ) 1

⅛

1

^ ) 2

s *n

0÷ ^ ^

= [f '^ ) l 2 (∞ s2

β

y ^ F + w

cos

^]2

+ ≡

2

=

[f, (^)]2 s ' n 2 0 ÷ 2f'(0)f(0) sin 0 cos 0 + [f(0)]2 cos2 0. Therefore

0) + ≡

® = r

2

(∞ s 2 0 + ≡

^ + ⅛ )

,

0) = [f'W ] 2 + ≡

2

= r2 + (⅜ ) 2 ∙

■».

35. r = 2 K W , α ≤ S < 0 ⅛ a = 2Γ(β) => r2 + (⅛ ) , = P W

=^ f a y i ^ ) F ^ H

2

+ (2 fW

⅛ Length = ( 7 w

+

f ( ^ d0 which is twice the length of the curve r = f(0) for α ≤ 0 ≤ ^.

∣f r3 cos 0d0 5 f [a(1+ cos 0)]3(cos 0) d0 i a3 f (1+ 3 cos 0+ 3 cos2 0+ cos30) (cos 0) d0 = --- ⅛ i --------- - --------- = ----- ----- 7 ^ ----------------------------2 2 J o r d0 J o [a(l+cos0)] d0 a2 Jθ (1+2cos0 + cos2 0) d0 ∣a Γ π [cos 0+ 3 (1±≡20) ÷ 3(1 - sin2 0) (cos 0) + ( ⅛ ^ ) 21 d0 ------- ----------------------- — ---- - ----------------- J— = V(After considerable algebra using6 6 ∫ o .[ l + 2 c o s 0 + ( ^ ) ] d 0 ----

:

2

-- ⅛ f -------

the identity cos2 A =

i

+c o s2 a ) 2 '

a

Γ ^

[∣0+ 2 sin 0+∣sin20]θ

0 = 2π => u = 2a]

SCOSII, + SCOS 2⅜- 2 cos gsin2 0+ ⅛ cos40) d⅛

∫ o π (∣+2cos0÷icos20)d0

_- aJ∣1^4-∣2⅛p^jH2ΞlΞ^Ξli2!21^ΞL⅛^ π

_ i f r3 sin 0 d0 y - 1⅛ - — =

+

— a (⅜7r) _ 5 . ~ 3π ~ 6 a ,

jfv [a(1+ cos 0)] (sin 0) d0 ; [u = a(l ÷ cos 0) 3

3π

=> —∣du = sin 0 d0; 0 = 0 => u = 2a;

5 / 22a___a_____ ~∣ u3 du = ⅛ = 0. Therefore the centroid is (x, y) = (∣a, θ) 3π 3

, Λ 4 ∣ ∣ 'm ∣

Section 10.8 Conic Sections in Polar Coordinates

327

10.8 CONIC SECTIONS IN POLAR COORDINATES 1. r c o s ( 0 - j ) = 5 => r(c o s 0 cos ∣+ sin 0 sin ∣) = 5 => ^ r c o s 0 + ∣rsin 0 = 5 => ^ x + ∣y = 5 => ᅟ ∕3 x + y = 10 => y = —  ∕3 x + 10 3. r cos (0 - y ) = 3 => r (cos 0 cos y + sin 0 sin y ) = 3 => —∣r cos 0 —i y r sin 0 = 3 =>

-

l

x

~ ^ y = 3 ≠'

x

+   A y = - 6 ≠∙ y = —^ x —2 √ z3

5. r c o s ( 0 - ∣) = √ 2 ≠> r (cos 0 cos ∣+ sin 0 sin ^) = √2 ^

^ r c o s 0 + ^ r s i n 0 = √ 2 => ⅛

x

+⅛ Y

= √ 5 => x + y = 2 => y = 2 - x

7. r cos (0 - y ) = 3 ≠> r (cos 0 cos y + sin 0 sin y ) = 3 => - ∣r c o s 0 + ^ r s in 0 = 3 => ~ ∣χ + ^ y = 3 => - x +   ∕3 y = 6 => y = Ay x + 2 \/3

5 i 9. y ^ x + v ^ y = 6 ≠- ∙  ∕ 2 r cos 0 +   ∕2 r sin 0 = 6 => r ( y cos 0 + ∙ ^ sin 0j = 3 => r (cos ∣cos 0 + sin ^ sin 0)

= 3 => r cos (0 — ∣) = 3 11. y = - 5 ≠> r sin 0 = - 5 => —r sin 0 = 5 => r sin (-0 ) = 5 => rcos (∣- (—0)) = 5 => r cos (0 + ∣) = 5 13. r = 2(4) cos 0 = 8 cos 0 15. r = 2 √ 2 s in 0

328

Chapter 10 Conic Sections and Polar Coordinates

21. (x - 6)2 + y2 = 36 ≠> C = (6,0), a = 6 ≠> r = 12 cos θ is the polar equation

23. x2 + (y - 5)2 = 25 ≠- C = (0,5), a = 5 => r = 10 sin θ is the polar equation

25. x2 + 2x + y2 = 0 => (x + 1)2 + y2 = 1

27. χ 2 ⅛ y2 ⅛ y = 0 ≠ x2 + (y + ∣) 2 = ∣

4 C = (—1, 0), a = 1 ≠∙ r = —2 cos θ is the polar equation

29. e - l , x - 2 31.

e

polar equation

≠ k —2 => r - μ ∣⅛ ^ - ⅛

= 5 ,y ^ - 6

k= 6

33. e = ∣, x = l ^ k = l ^

35. e = ∣,x = - 1 0

39.

r =

25 1 0 - 5 cos 0

r

r = τ⅛

= τ⅛

= J J ⅛

= ⅛

k=10

r =

1 - (½ ) cos 0

r= ^ ¾ ^ = ⅛

=

( 1) 1 - ( ∣) cos 0

=> e = j , k = 5 => x = —5; a (1 —e2 ) = ke a

=» C = (θ, —∣) , a = ∣ ≠∙ r = —sin θ is the

[1 ~ ( I ) 2 ] = 5 = > i a = ∣= > a = ^ ^ e a = ∣

Section 10.8 Conic Sections in Polar Coordinates 41

’

r

=

400

16+ 8 sin 0

r =

1+ (⅛) sin 0

e = ∣, k = 50 => y = 50; a (1 =► a [1 -

43. r =

5—

( ∣) 2 ]

⅛

—e 2 )

r

=

1+ (∣) sin 0

= ke

= 2 5 ≠> ∣a = 25 ≠> a = ^

⅛ - Λ 2=> r = -i—⅛ -5 => e = 1,’ — 2 sin σ 1 — sin σ

k = 4 ≠∙ y = - 4

47.

49.

55.

r ≡ 1/(1 — sin β)

329

330

Chapter 10 Conic Sections and Polar Coordinates

57. (a) Perihelion = a —ae = a(l —e), Aphelion = ea + a = a(l + e) Planet Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto

Perihelion 0.3075 AU 0.7184 AU 0.9833 AU 1.3817 AU 4.9512 AU 9.0210 AU 18.2977 AU 29.8135 AU 29.6549 AU

Aphelion 0.4667 AU 0.7282 AU 1.0167 AU 1.6663 AU 5.4548 AU 10.0570 AU 20.0623 AU 30.3065 AU 49.2251 AU

59. (a) r = 4 sin 0 => r2 — 4r sin 0 => x2 + y 2 = 4y; => r cos 0 = -ᅟ ∕3

r = y ^ sec 0 => r =

=> x = √ 3 ; x = y 5 => ( √ ^ )

+ y2 = 4y

≠> y2 - 4y + 3 = 0 => (y - 3)(y - 1 ) = 0 => y = 3 or y = 1. Therefore in Cartesian coordinates, the points of intersection are ( ᅟ ∕ 3 , 3^ and ( ᅟ ∕ 3 , 1^. In polar coordinates, 4 sin 0 = √ z3 sec θ => 4 sin 0 cos θ = ᅟ ∕3 => 2 sin 0 cos 0 = - ^ => sin 20 = ^

=> 20 = j or

⅞3 ≠> 0 = 20 θ r ^3 ’0 = ⅞ ≠> r = 2’,a n d 0 = 23 0 => r = 2 √ z3 => (2, ∣) and ^ 2 ^ 3 , j ) are the points of intersection in polar coordinates. 61. r cos 0 = 4 => x = 4 => k = 4: rparabola => e = 1 => r = 1. +, 4cos zσ 63. (a) Let the ellipse be the orbit, with the Sun at one focus. Then r max = a + c and rmin = a - c => ⅛max Ξ∣⅛min r

_

(a + c ) - ( a - c ) _ (a ÷ c) + (a - c) 2a

2c _ a

c _ c

r

p

(b) Let Fι, F2 be the foci. Then PFι + PF2 = 10 where P is any point on the ellipse. If P is a vertex, then PFι = a + c and PF 2 = a - c => (a + c) ÷ (a - c) = 10 => 2a = 10 => a = 5. Since e = ςa we have 0.2 = 5⅞ => c = 1.0 => the pins should be 2 inches apart. 65. x 2 + y 2 —2ay = 0 => (r cos 0)2 + (r sin 0)2 —2ar sin 0 = 0 => r 2 cos2 0 ÷ r 2 sin2 0 - 2ar sin 0 = 0 ≠> r 2 = 2ar sin 0 => r = 2a sin 0

Chapter 10 Practice Exercises

331

67. x cos α + y sin α = p => r cos θ cos a + r sin θ sin a = p ≠> r(cos θ cos a + sin θ sin a) = p => r cos (0 —α) —p

CHAPTER 10 PRACTICE EXERCISES

1. x2 = -4 y => y = - j

=> 4p = 4 ≠- p = 1;

therefore Focus is (0, —1), Directrix is y = 1

5. 16x2 + 7y2 = 112 => y + ⅛ = 1 => c2 ≈ 16 - 7 = 9 =>

c

= 3;’e = ≡ 1 a = 4

3. y2 = 3 x = > X = ^ = > 4 p = 3 = > p = ^ therefore Focus is (∣, 0 ) , Directrix is x = —∣

7. 3x2 - y2 = 3 => x2 - ⅛ = 1 => c2 = 1 + 3 = 4 => c = 2; e =

= γ = 2; the asymptotes are

y= ±   Λx

9. x2 = —12y => —^ = y = > 4 p = 1 2 = > p = 3 =Φ focus is (0, —3), directrix is y = 3, vertex is (0,0); therefore new vertex is (2,3), new focus is (2,0), new directrix is y = 6, and the new equation is (x —2)2 = —12(y —3) a =

11. y ÷ ⅛ = 1 ^

^ a n ^ b = 3 => c = √ z25 - 9 = 4 => foci are (0, ± 4 ) , vertices are (0, ± 5 ) , center is

(0,0); therefore the new center is ( -3 , -5 ), new foci are (-3 , - 1 ) and (-3 , -9 ), new vertices are ( -3 , -1 0 ) and (-3 ,0 ), and the new equation is ^ t2 £ ψ ⅛ ^ ^ = 1 13. ⅞ ~ y = 1 =>

a =

2 λ∕2 and b = A/ 2 ≠> c = √ 8 ÷ 2 = ᅟ ∕w

=> foci are ( 0, ± A/ T OJ , vertices are

center is (0,0), and the asymptotes are y = ± 2x; therefore the new center is

, new foci are

2,2 3/2 ± A/1 0 ) , new vertices are (2 ,4 √ z2 ) and (2,0), the new asymptotes are y = 2x —4 ÷ 2 √ 2 and y = —2x ÷ 4 + 2^/2; the new equation is ^ —g ^ ---- ^ - ^ = 1

332

Chapter 10 Conic Sections and Polar Coordinates

15. x2 - 4x - 4y 2 = 0 => x 2 —4x + 4 —4y 2 = 4 => (x - 2)2 - 4y 2 — 4 => ^

2

- y 2 = 1, a hyperbola; a = 2 and

b = 1 => c = √ zΓ ÷ 4 = √ z5 ; the center is (2,0), the vertices are (0,0) and (4,0); the foci are ^2 ± √ ^ , θ ) and the asymptotes are y = ±

1y

17. y 2 —2y + 16x = —49 => y2 —2y + 1 = —16x —48 => (y — 1)2 = —16(x + 3), a parabola; the vertex is (—3,1); 4p = 16 => p = 4 => the focus is ( - 7 ,1 ) and the directrix is x = 1 19. 9X2 + 16y2 + 54x - 64y = - 1 => 9 (x2 ÷ 6x) + 16 (y2 - 4y) = - 1 => 9 (x2 ÷ 6x + 9) ÷ 16 (y2 - 4y ÷ 4) = 144 => 9(x ÷ 3)2 + 16(y - 2)2 = 144 =>

( x ^ 3)2 6

+ ⅛ y ^ = 1, an ellipse; the center is (- 3 ,2 ); a = 4 and b = 3

=> c = √ 1 6 - 9 = √^7; the foci are ( —3 ± √ z7 ,2^ ; the vertices are (1,2) and (—7,2)

21. x2 ÷ y 2 —2x —2y = 0 => x 2 —2x + 1 ÷ y2 —2y + 1 = 2 => (x — 1)2 + (y — 1)2 = 2, a circle with center(1,1) and radius = √ 2 23. B2 - 4AC = 1 - 4(1)(1) = - 3 < 0 ^

ellipse

25. B 2 - 4AC = 32 - 4(1)(2) = 1 > 0 => hyperbola 27. x2 —2xy ÷ y 2 = 0 => (χ —y)2 = 0 => x - y = 0 o r y = x, a straight line 29. B 2 - 4AC = 12 - 4(2)(2) = - 1 5 < 0 ≠> ellipse; cot 2α = V y = ⅛ x' + ⅛

' => 2 ( ^ x ' - ^ y ' )

2

= 0 => 2α = £ => α = J ; x = ^ x ' - ^ y 'and

+ ( ^ x '- ^ y ') ( ^ x ' + ^ y ') + 2 ( ^ x ' + ^ y ')

2

-1 5 = 0

=> 5x,2 + 3y'2 = 30 31. B 2 - 4 A C = ( 2 √ 3 ) 2 - 4 ( l ) ( - l ) = 16 ≠> hyperbola; cot 2α = ^ andy = ∣x' + ^ y ' ≠> ( √ 2 x ' - 1 ∕ )

2

+

2√3 ( ^ x

,

= ^

≠> 2α = f ^

α = ∣; x = ^ x ' - ∣y'

- ∣y ') ( ∣x '+ ^ y, ) - θ x ' + ^ y ') 2 = 4

≠> 2X' 2 - 2y'2 = - 4 => y'2 - x'2 = 2 33. x = ∣tan t and y = j sec t => x2 = J tan2 1

35. x = —cos t and y = cos2 t => y = (—x)2 = x2

and y 2 = ∣sec2 1 => 4x 2 = tan2 1 and 4y 2 = see2 1 ≠> 4x 2 + 1 = 4y 2 => 4y 2 - 4x2 = 1

∖

v

Chapter 10 Practice Exercises 37. 0 ≤ r ≤ 6 cos θ

39. d

41.

1

43. k

45.

i

47. r = sin 0 and r = 1 + sin 0 => sin 0 = 1 + sin 0 => 0 = 1 so no solutions exist. There are no points of intersection found by solving the system. The point of intersection (0,0) is found by graphing.

49. r = 1 + cos 0 and r = 1 —cos 0 => 1 + cos 0 = 1 - cos 0 ≠> 2 COS 0 = 0 => COS 0 = 0 => 0 = ∣, γ J 0 = j θ Γ y => r = 1. The points of intersection are (1, ∣) and (1, y ) . The point of intersection (0,0) is found by graphing.

51. r = 1 ÷ sin 0 and r = —1 + sin 0 intersect at all points of r = 1 + sin 0 because the graphs coincide. This can be seen by graphing them.

53. r = sec 0 and r = 2 sin 0 => sec 0 = 2 sin 0 => 1 = 2 sin 0 cos 0 => 1 = sin 20 => 20 = ∣ => 0 = f => r = 2 sin ∣= √ z2 => the point of intersection is ∕ 2 , 1 1 . No other points of intersection exist. (ᅟ

333

334

Chapter 10 Conic Sections and Polar Coordinates

55. r cos ( H f ) = 2 √ ^ => r (cos 0 cos ∣—sin 0 sin ∣) = 2 χ ∕3 ≠

y

5 r cos 0 —^ r s i n 0 = 2-^3 x -√ 3 ^ y-4 √ 3 ~

≠ ∙ r c o s 0 - √ z3 rsin 0 = 4 y z3 => x - √ ^ y = 4ᅟ ∕3 => y = ^ x —4

57. r = 2 sec 0 => r = ^

y

≠ r cos 0 = 2 ≠ x = 2

ι-2

59. r = - ∣esc 0 ≠> 2 r s in 0 = —2z ≠- ∙y, = - 1z

y

._ _ _ _

x

-3/2 y≡-3∕2

61. r = —4 sin 0 => r2 = —4r sin 0 => x2 + y2 ÷ 4y = 0 => x2 + (y + 2)2 = 4; circle with center (0, -2 ) and radius 2.

63. r = 2 ^ ∕2 c o s0 => r2 = 2 y ^ r c o s 0 => x2 + y2 - 2 √ 2 x = 0 => (x - y ^ j + y2 = 2; circle with center ( √ ⅞ 0 ) and radius ᅟ ∕2

y

r.-4βlnθ

x∖ ⅛ ÷2 ) i . 4

y r*2√2cosβ

Chapter 10 Practice Exercises 65. x 2 + y 2 + 5y = 0 => x2 + (y + ∣) 2 = ⅛ => C = (θ, - j ) r ∙-5 s in θ

and a = ∣; r 2 + 5r sin 0 = 0 => r = —5 sin 0

'•(-J)'-?

67. x 2 + y 2 —3x = 0 => (x - ∣) 2 + y 2 = ∣ => C = ( ∣, θ) and a = ∣; r 2 —3r cos 0 = 0 => r = 3 cos 0 r ∙3 c o s ∂ (3/2.0)

69.

r =

ι

+

cos0

J

*

=> e = 1 ≠- parabola with vertex at (1,0) 1 +cos0

73. e = 2 and r cos 0 = 2 => x = 2 is directrix => k = 2; the conic is a hyperbola; r = j ^ ~ ^

=> r =

1 + 2 cos 0

Γ

75. e = ∣and r sin 0 = 2 => y = 2 is directrix => k = 2; the conic is an ellipse; r =

1+

⅛ ^

=> r =

1

+ (⅛ p

r= — -— 2 + sin 0 77. A = 2^£ [ r 2 d0 = £ (2 - cos 0)2 d0 = ,£ (4 - 4 cos 0 + cos2 0) d⅛ = £ (4 —4 cos 0 + = Γ ( ∣- 4 c o s 0 + ⅛ * ) d 0 = [∣0 - 4 s i n 0 + ^ ] ^ «/ 0

∖Z

Z

/

LZ

∙τ

J u

1 +

^ )

d0

= ∣π 4

79. r = 1 + cos 20 and r = 1 => 1 = 1 + cos 20 => 0 = cos 20 => 20 = f => 0 = ∣; therefore pπ∕4 . λ „ pπ∕4 λ A = 4 j o | [(1 + cos 20)2 - 12 ] d0 = 2 J o ( 1 + 2 cos 20 + cos2 20 - 1) d0 = 2∫

πz4 (2 c o s2 0 o

+ ∣+ ≡ ^ ) d0 = 2 [sin 20 + ∣0 + ⅛ ^ ] " / 4 = 2 ( l + ∣+ 0 ) = 2 + ≈

81. r = —1 + cos 0 => ^ = —sin 0; Length = J^ ᅟ ∕ ( - l + cos 0)2 + (—sin 0)2 d^ = £ = £ 7

^

7

81

d0

= £

2 sin

f ⅜ = 8 sin2 (f) cos ( f ) ; r2 + (∣) 2 = [8 sin3 ( f ) ] 2 + [8 sin2 (f) cos ( f ) ] 2 = 64 sin4 ( f ) ^

L= X ^ ^ s in

= £ ' [4 -

(f)]

4 cos

85. r = χ ∕cos 20 => ^ = = X

Z

d0

= [40 -

^ζ

4

^)

6 s in

d0

= Γ

8 s in 2

(f)

d0

= JΓ

8

[k ⅛ ^ ]

d0

( f ) ] o/4 = 4 (?) - 6 sin (≡) - 0 = π - 3

; Surface Area = X ' 2π(r sin 0) ^ r 2 ! ( ^ ) 2 d0

2 π V cos 20 (sin 0) ^ cos 20 + ^

d0 = X

Z

2πᅠ ∕ cos 20 (sin 0) ^ ^ ⅛ d0 = X

Z

2π sin 0 d0

= [2 π (- cos 0)] g/ 4 = 2π (1 - ^ ) = (2 - √ 2 ) π

87.

(a) Around the x-axis: 9x2 + 4y 2 = 36 => y 2 = 9 - ∣x2 => y = ± y 9 — ∣x 2 and we use the positive root: v

=

2

π

£

⅛

9

-

4 χ2)

d χ

=

2

£

π

(9 - 1 χ 2 )

d χ

=

2 π t9 χ

-1

χ3]

0=

24π

(b) Around the y-axis: 9x2 ÷ 4y 2 = 36 => x2 = 4 — ∣y 2 => x = ± ^ 4 - ^ y2 and we use the positive root: v

≈

2

∕o

π

( ᅠ∕ 4 - b

2)

dy

= 2 ∫ o π ( 4 - ∣y 2 )d y = 2 π [ 4 y - ⅛ y 3 ] J = 16π

89.

Each portion of the wave front reflects to the other focus, and since the wave front travels at a constant speed as it expands, the different portions of the wave arrive at the second focus simultaneously, from all directions, causing a spurt at the second focus.

91.

The time for the bullet to hit the target remains constant, say t = to. Let the time it takes for sound to travel from the target to the listener be t2. Since the listener hears the sounds simultaneously, tι = to ÷ t2 where tι is the time for the sound to travel from the rifle to the listener. If v is the velocity of sound, then vtι = vto + vt2 or vtι - vt2 = vt0 . Now vti is the distance from the rifle to the listener and vt2 is the distance from the target to the listener. Therefore the difference of the distances is constant since v⅛ is constant so the listener is on a branch of a hyperbola with foci at the rifle and the target. The branch is the one with the target as focus.

93.

(a) r =

τ^

^

=> r + er cos 0 = k => √ zx2 ÷ y 2 + ex = k => √ x 2 + y2 = k - ex => x 2 + y 2

= k 2 - 2kex + e 2 x2 => χ 2 - e 2 x 2 + y2 ÷ 2kex - k 2 = 0 => (1 - e 2 ) x 2 + y 2 ÷ 2kex - k 2 = 0 (b) e = 0 => x2 + y 2 —k 2 = 0 => x2 + y 2 = k 2 => circle; 0 < e < l = > e 2 < l = > e 2 - l < 0 ≠ > B 2 - 4AC = 0 2 —4 (1 —e 2 ) (1) = 4 (e2 — 1) < 0 => ellipse; = 1 => B 2 - 4AC = 0 2 - 4(0)(1) = 0 => parabola; e > l = > e 2 > l => B 2 —4AC = 02 —4 (1 —e 2 ) (1) = 4e 2 —4 > 0 => hyperbola e

CHAPTER 10 ADDITIONAL AND ADVANCED EXERCISES 1. Directrix x = 3 and focus (4,0) => vertex is (^ , θ) ≠> p = ∣ => the equation is x — ^ =

3. x 2 = 4y => vertex is (0,0) and p = 1 => focus is (0,1); thus the distance from P(x, y) to the vertex is ^ ∕x 2 + y 2 and the distance from P to the focus is ^ x 2 ÷ ( y - 1)2 => ^ ∕x 2 + y 2 = 2 √ x 2 + (y — 1)2

Chapter 10 Additional and Advanced Exercises

337

=> x2 + y 2 = 4 [x2 + (y - 1)2 ] => x2 + y 2 = 4x 2 + 4y 2 - 8y + 4 => 3x2 + 3y2 — 8y + 4 = 0, which is a circle 5. Vertices are (0, ± 2) => a = 2; e = ≡ => 0.5 = ∣ ≠ ∙ c = 1 => foci are (0, ± 1) 7. Let the center of the hyperbola be (0, y). (a) Directrix y = —1, focus (0, —7) and e = 2 => c —∣= 6 => ∣= c —6 => a = 2c - 12. Also c = ae = 2a => a = 2(2a) - 1 2 => a = 4 ≠> c = 8; y - ( - 1 ) = ∣= ∣= 2 ≠> y = 1 => the center is (0,1); c2 = a2 + b 2 => b 2 - c 2 - a 2 = 64 - 16 = 48; therefore the equation is ^ ^

~ ⅛ —1

(b) e = 5 => c - ∣= 6 => ∣= c —6 ≠> a = 5 c - 3 0 . Also, c = ae = 5a => a = 5(5a) —30 => 24a = 30 ≠> a = ∣ => c = ^ ; y - ( - 1) = ≡ = ^ =

W

-

τl

=

T

;

= 1 ξ^ y

=

-

1 => the center is (0, —∣) ; c 2 = a2 + b 2 => b2 = c 2 —a 2

therefore the equation is ^ ^ — j⅛y = 1 or ^ ⅛ ^ — ^

= 1

9. (a) b 2 x2 + a2 y 2 = a2 b 2 ≠> ^ = - ^ ; at (x i ,y i ) the tangent line is y - yι = ( ~ ⅛ ) (χ => a2 yyι + b 2 xxχ = b 2 x 2 + a 2 y 2 = a2 b 2 => b2 xxι + a2 yyι - a 2 b 2 = 0 (b) b 2 x2 - a 2 y 2 = a2 b 2 => ⅛ = ⅛ ; at (χ ι, yι) the tangent line is y - y i = ( ^ ι ) (x - Xi) => b2 xxχ —a2 yyι = b 2 xf —a2 y 2 = a2 b 2 => b2 xxι —a 2 yyι - a2 b 2 = 0 13.

15. (9X2 + 4y 2 - 36) (4x2 + 9y 2 - 16) < 0 => 9X2 + 4y 2 - 36 ≤ 0 and 4x 2 + 9y 2 - 16 ≥ 0 or 9X2 + 4y 2 —36 ≥ 0 and 4x 2 + 9y 2 — 16 ≤ 0

17. x4 - (y 2 - 9) 2 = 0 => x 2 - (y 2 - 9) = 0 or x2 + (y2 - 9) = 0 => y 2 —x2 = 9 or x 2 + y2 = 9

χ

ι)

338

Chapter 10 Conic Sections and Polar Coordinates

19. Arc PF = Arc AF since each is the distance rolled; ZPCF = ½⅜≡ ≠> Arc PF = b(ZPCF); 0 = * ^ ≠> Arc AF = a0 => a0 = b(ZPCF) => ZPCF = (§) 0; ZOCB = ≡ - 0 and ZOCB = ZPCF - ZPCE = Z P C F -(f-α ) = ( ≈ ) 9 - (f -α ) ⅛ * - 0 = (s)< M 5 → ) *

f-s = β )^ -f +≈

=► α = π - 0 —(∣) 0 => α = π - ( s ^ ) 0 . Now x = OB + BD = OB + EP = (a + b) cos 0 + b cos α = (a + b) cos 0 + b cos (π — (i p ) 0) = (a + b) cos 0 + b cos π cos ( ( ^ ) 0) + b sin π sin ( ( ^ ) 0) = (a + b) cos 0 - b cos ((⅛ ^ ) 0) and y = PD = CB —CE = (a + b) sin 0 - b sin α = (a + b) sin 0 - b sin ((⅛ ^ ) 0) = (a + b) sin 0 —b sin π cos (( i ^ ) 0) + b cos π sin (( 1 ^ ) 0) = (a + b) sin 0 —b sin ((⅛ ^ ) 0 ); therefore x = (a + b) cos 0 —b cos ( ( a ⅞^) 0) and y = (a + b) sin 0 - b sin ( ( ^ i ) 0) ½SL= tan t 21. (a) x = e2' cos t and y = e 2t sin t => x2 + y2 = e4*cos2 1 + e4t sin2 1 = e4t. Also J = vf⅛CO => t = tan" 1 (∣) => χ 2 + y2 = e 4tan^1 (y/x) is the Cartesian equation. Since r2 = x2 + y2 and Λ

θ = tan- 1 ( 0 , the polar equation is r2 = e4β or r = e2^ for r > 0 (b) ds2 = r2 d02 + dr2 ; r = e20 => dr = 2e2* d0 ≠> ds2 = r2 d02 + (2e2* d0) 2 = (eM ) 2 d02 + 4e4β d02 ∙2π

= 5e4s d02 => ds = √ z5 e2s 2π

5 e 2*d0

= ^ (e4’ -

0

23. r = 1 + cos 0 and S = ∫ 2πp ds, where p = y = r sin 0; ds = √ r 2 d02 + dr2 = √ ,(1 + cos 0)2 d02 + sin2 0 d02 √ 1 + 2 cos 0 + cos2 0 + sin2 0 d0 = ᅟ ∕2 + 2 cos 0 d0 = ^ 4 c o s 2 ( ∣) d0 = 2 cos (∣) d0 since 0 ≤ 0 ≤ ∣. Then S = fθ 2π(r sin 0) ∙ 2 cos (∣) d0 = J^ Z 4π(l + cos 0) ∙ sin 0 cos ( ∣) d0

=f ^ 4π

t2

c o s2

( 1 )1 t2

_ ^2^21AL (

~

5

k

s in

32π∖ _ 5 / “

(1 )

c o s

(1 )

c o s

(!)1

d 0

= J Γ

l6 7 r c o s 4

( ! )

s in

(5 )

d 0

=

[Ξ^ T ^ ]

3 2 π -4 π √ 2 5

25. e = 2 and r cos 0 = 2 => x = 2 is the directrix => k = 2; the conic is a hyperbola with r = .

r = 1

(2)(2) l+ 2 c o s0

=

1 + ec0s

____4____ l÷ 2 c o s ^

27. e = ∣and r sin 0 = 2 => y = 2 is the directrix => k = 2; the conic is an ellipse with r = γ η ⅛ ^

θ z2

Chapter 10 Additional and Advanced Exercises

339

29. The length of the rope is L = 2x + 2c + y ≥ 8c. (a) The angle A (ZBED) occurs when the distance CF = £ is maximized. Now £ = ᅟ ∕x 2 —c2 + y ^ £ = √ x 2 —c2 + L - 2x —2c ^

^ = H

χ2

- c T ^ x ) - 2= ^

- 2.

-2 r ^ - - 2 = 0 => x = 2 √ x 2 - c2 Thus fα x = 0 => 2 √x -c

> x2 = 4x2 - 4C2 => 3x2 = 4c2 => ^ = ^

=

= X> 7 =Z ∙ Since 7x = sin 4z we have sin ^z = ≠> Δ = 6 0 0 ^ A = 120o

z

(b) If the ring is fixed at E (i.e., y is held constant) and E is moved to the right, for example, the rope will slip around the pegs so that BE lengthens and DE becomes shorter => BE + ED is always 2x = L —y —2c, which is constant => the point E lies on an ellipse with the pegs as foci. (c) Minimal potential energy occurs when the weight is at its lowest point => E is at the intersection of the ellipse and its minor axis. 31. If the vertex is (0,0), then the focus is (p, 0). Let P(x, y) be the present position of the comet. Then ^ (x - p)2 + y2 = 4 × 107 . Since y2 = 4px we have ^ ( x - p)2 ÷ 4px = 4 × 107 => (x —p)2 + 4ρx = 16 × 1014 . Also, x - p = 4 × 107 cos 60o = 2 × 107 => x = p ÷ 2 × 107 . Therefore (2 × 107 )2 + 4p (p ÷ 2 × 107 ) = 16 × 1014 4 4 × 1014 + 4p2 ÷ 8p × 107 = 16 × 1014 => 4p2 + 8p × 107 - 12 × 1014 = 0 ≠> p 2 + 2p × 107 - 3 × 1014 = 0 => (p ÷ 3 × 107 ) (p - 107 ) = 0 => p = —3 × 107 or p = 107 . Since p is positive we obtain p = 107 miles. 33. cot 2α = ^π-^ = 0 =>

Q

= 45o is the angle of rotation => A' = cos2 45o ÷ cos 45 o sin 45o ÷ sin2 45 o = ∣, B' = 0,

and C' = sin2 45 o - sin 450 cos 45o + cos2 45o = ∣ => ∣x'2 + ∣y'2 = 1 => b = λ ∕ ∣and a = √ 2 ≠> c2 = a2 - b2 = 2 —J = j => c = ^ . Therefore the eccentricity is e = ^ = ^ ^ = ≈ 0.82.

35. λ ∕x + ∙v ∕y = 1 => x + 2∙χ∕x y + y = 1 => 2 χ ∕x y = 1 - (χ + y) ≠> 4xy - 1 - 2(x + y) + (x + y)2 => 4xy = x2 + 2xy + y2 - 2x - 2y + 1 ≠> x2 - 2xy + y2 - 2x - 2y + 1 = 0 => B2 - 4AC = ( -2 ) 2 - 4(1)(1) = 0 => the curve is part of a parabola 37. (a) The equation of a parabola with focus (0,0) and vertex (a, 0) is r = through α =

45o

gives r -

l+ c o ⅞

2a 1 + c os ^

and rotating this parabola

^ ∙

(b) Foci at (0,0) and (2,0) => the center is (1, 0) => a = 3 and c = 1 since one vertex is at (4,0). Then e = ∣ = ∣. For ellipses with one focus at the origin and major axis along the x-axis we have r = ⅛ ~ ⅞ ) 3(1-∣

=

1 - (∣) cos 0

=

8

3 - cos 0 ’

(c) Center at (2, ∣) and focus at (0,0) => c = 2; center at (2, ∣) and vertex at (1, j ) => a = 1. Then e = ^ =

j

= 2. Also k

=

ae —∣= (1)(2) — 2

=

2 ∙

Thereforer

=

1 + e s in 0

=

1

+ 2 sin 0

=

T ÷2sIiΓ0 ∙

39. Arc PT = Arc TO since each is the same distance rolled. Now Arc PT = a(ZTAP) and Arc TO = a(ZTBO) => ZTAP = ZTBO. Since AP = a = BO we have that ΔADP is congruent to ΔBCO => CO = DP => OP is parallel to AB => ZTBO = ZTAP = 0. Then OPDC is a square => r = CD = AB —AD —CB = AB —2CB => r = 2a —2a cos 0 = 2a(l —cos 0), which is the polar equation of a cardioid.

340 41.

Chapter 10 Conic Sections and Polar Coordinates ∕3 = ≠ 2 - ≠

=> tan∕3 = tan(^ 2 - ≠ 1 ) = ⅛ ‰

1

⅛

the curves will be orthogonal when tan β is undefined, or when tan ≠ 2 = ^ => ^ = ^ . r2 = - f ( 0 ) g , (0)

^

43.

= “6a cos 30 => 30;" when3 0 = 2≡ t a n ≠ = ∣tan ^ r = 2a sin 30 => ⅛ tan ≠ = 3 7 ⅜τ = ’⅛ ⅛ j = I otan d0 j oacos30 ’

45.

tan ≠ι = ^

at 0 = f ; tan ≠ 2 = ≡ ∣= tan 0 is 1/3 at 0 = ∣; since the product of

= —cot 0 is —^

^

these slopes is —1, the tangents are perpendicular 47

r 1

_

1 l-c o s 0

________ sin 0 . r _ ( 1 - COS0)2 , 1 2

d0

⅛ d0

3 1 + COS0

3 sin 0 1 . _ ( l+ c o s 0 ) 2 ’ l - c o s 0

___ 3 _ _ l+ c o s 0

=> 1 + cos 0 = 3 —3 cos 0 => 4 cos 0 = 2 →∙ cos 0 = ∣ => 0 = ÷ f => r i = r2 = 2 => the curves intersect at the points (2, ± ? ) ; tan ≠ι = ∕

e

= - ⅛ τ ∏ is — 7- at 0 = ? ; tan ≠ 2 =

⅛

[ (1 - cos 0)2 J

∕ '

t l

°°* ∕1

= ⅛ τ ^ is

[ (1 + cos 0)2 J

∕3 at 0 = f ; therefore tan /3 is undefined at 0 = f since 1 + tan ≠ι tan ≠ 2 = 1 + ^ - ^ ^ ( ^ ) ~ θ ^ ton ≠' r jι L∣β=-

π ∕3

= ——sin (

at 0 = — ≈

⅛

=

-

√3

and tan ⅛ ∣ . _r t = ' r 4 ∖ y - π ∕6

-

sin (

^

= - v ∕3 => tan /3 is also undefined

j

=

«• n = τ ⅛ ^ ⅛ = π ^ ≡ ⅛ = τ ⅛ ⅛ ⅛ = - i⅛ ⅛ >⅛ ∞ =

ta∏≠ι = T ⅛ ⅛

^

and tan ≠ 2 = ⅛ ⅜

L (l+ c o s0 ) 2 J

= ¾ Γ

=> 1 + tan ^ ι tan ^2

[ (1 - cos 0)2 ]

= 1 + (⅛ ∏ r9 (⅛ ⅛ ⅛ ) = 1 -

1

sinθ02"g =

^ ^s

θ ^

u n c *e fined

=> the parabolas are orthogonal at each

point of intersection 51.

r = 3 sec 0 => r = ^

;^

= 4 + 4 cos 0 => 3 = 4 cos 0 + 4 cos2 0 => (2 cos 0 + 3)(2 cos 0 - 1 ) = 0

=> cos 0 = ∣or cos 0 = - ∣ => 0 = j or y (the second equation has no solutions); tan ≠ 2 = ~ z ⅛ 7 ^ = “ ⅛

HF

*S “ \/3

at

f

and

1 + tan ≠1 tan ≠ 2 = 1 + ^ ^

tan ≠ι = (- ^ )

3

j c⅞ g

= 0

53.

1 _ 1 - cos 0 ta n ≠ 2 =

ta n ≠ 2 =

1 1 - sin 0

^ --+ l

a 1 s o >t a n

θ z^

ta n

^ *s a i≡0 undefined ≠> /3 = ∣.

a

4

= ⅛ 9 . Thus at 0 = ≡ tan ≠1 = ⅛ ⅛ ⅛ = 1 - √ 2 and COS0

4 ’

≡ ⅛ M = √ 2 - 1. Then ton/3= (?)

2∙

=> 1 —cos 0 = 1 —sin 0 => cos 0 = sin 0 => 0 = ? ; tan ^ ι =

COS0 .(1 - s in 0)2 J

co s

⅛∕3 = √ 3 and tan ⅛ ∣ ⅛∣ w j = - ^

=

^ ^ =

=> 1 + tan ≠ι tan ≠ 2 = 1 + ^ - 4 ^ ( ^ )

= cot 0 is ^ at j . Then tan /3 is undefined since

v

-

^

\

s m

(≡ )

H

l + (√ 2 -l) (l-√ 2 )

V

= 7⅛ ⅛ = 2 √ 2 -2

1

J 1- ~ ⅛ -sιn 0 L(l - cos 0)2 J

=

1

- ^ s∕ — sin 0

=

f ’

Chapter 10 Additional and Advanced Exercises 55.

(a) tan α = 7 ⅛r ≠- ⅛ = ^ - => In r = - ^ - + C (by integration) ≠> r = Be’/ ( “”a) for some constant B; A =

=

e

i £ ^ β 2 g 2 9 ∕( t a n α ) j β _

^B i (tana ) e a ' l ~

l

j ^2 _

tan a [ β 2 e 2⅛∕(tana) _ β 2 e W1 ∕(u n a )j

-p (⅛ - if) since if = B2 eM2/(tana) and if = B2 e2'’1/ l< - ∣B ta < ,ie '- ∙ ∣' = (sec α) [Be^z converges by the Sandwich Theorem for sequences

lim o 2⅛2 ~ θ z^ converges (using ΓHopitaΓs rule)

7÷4 =

=

Δ LIM = nn lim N ∏+1 →∞ r n →∞

i⅛

√n^

47.

n

lim o 8 1∕ n = l => converges

49.

n

lim o (l + ^ ) n = c 7 ^

51.

n

h ιn o ^ 1 0 n =

53∙

∏⅛∞ ( n ) ^ = h ⅛

LIM n → ∞ 1+ (-)

= 0 => converges &

(Theorem 5, #3)

converges

(Theorem5,#5)

ljm o 101∕ n ∙ n 1∕ n = 1 ∙ 1 = 1 => converges

n

= f = 1 => converges

(Theorem 5, #3 and #2)

(Theorem 5, #3 and #2)

n→ ∞

55

∙ n⅛ ∞ ^

=

⅛ ⅛

= f = oo =⅛> diverges

(Theorem 5, #2)

n→ ∞

57.

n

lim o ι ^ n =

lir∏o 4 ^ ∕n = 4 ∙ 1 = 4 ≠> converges

n

59

∙ n⅛ ‰

6 1

∙

63

∙ Λ

m∞

( n ) '^ "

∙ Λ

m∞

( l ^ ) " = n⅛ ‰

ex

/

3_\ l )

6 5

n

⅛‰ ⅛

= Λ

6 7

⅛ =

n

=

m∞ e χ p

∙ n ⅛ n∞ ( ⅛

)

h

⅛‰

n

I

v

⅛∞

1

=

⅛

^

≤

p⅛ J = ∞

n

⅛‰

3 4

≡

⅛

" = Λ

eχp

(≡

P (n

m0 x 0

ln

d ⅛ ≥ 0 ^

n

⅛ ‰ (D ≈

=^

d i v e r ges

(Theorem5, #6)

(s )) = n⅛ ‰

exP

ln

(^ τ )) = Λ ,

= Λ

0 an

(Theorem 5, #2)

m∞ e χ p

(⅛ τ)

vn

=

x

m∞

ex

( j⅛ Γ * ) =

e

^ 1 =* converges

x

2

m 00 e x

lim o ⅛ = 0 => converges

P ( i ^ * ^ ^ ! ^ )

((3 iH ⅛ υ ) =

Λ

n

P (≡

ln

eχp

(I ) = e 2 /3 ≠> converges

(⅛ )) =

x

n⅛ ‰

ex

P ( - ^

= x n lh∏0 exp ( j ⅛ ) = xe0 = x, x > 0 => converges 69

∙ n ⅛ 0 0 F ⅛ = ∏ ⅛ n∞ ⅛"

7 1

∙ n⅛ ‰

ta n h n

=

θ ^

= n½ ‰ ‰ ^ =

converges

n

1Xm oo

⅛

(Theorem 5, #6) = ∏⅛‰ ⅛

= Λ

m oo 1

=

1

≠

converges

^

Section 11.1 Sequences no 7 3

75. 77

n2 sin(J)

Γ ∙ ⅛

n

81.

(n)

(c 0 s ( π ) ) ( j ) ( -⅜ + ⅞)

1∙ = n ⅛

,. " “ “

=

- c o s ( i) W

1 =

c

5

°n v e rg e s

lim o tan^^1 n = ∣ => converges

∙ n ⅛ 1oo θ ) °

79.

s *n

1∙ = n ⅛

+

⅛ = ∏ ⅛ n∞ ( θ ) "

lim

⅛ ff.

lim

( n - 7 v∏

M

∏ n → ∞U m

n→∞

2

n →∞ ∖

J

( ⅛ ) ) = 0 =≠> converges

≡ ⅛

= n→ ∞l i m

∏

-n ) =

+

f ! = ...=

n

n →∞

(n - √v n 2 - n") ( n + f

lim

n →∞ ∖

e i)

J ∖ n+√n 2 - n ∕

=

(Theorem5,#4)

li m

∏

lim

^ = 0

=> converges &

---- ‰ = — iim

n → ∞ n+√n2 -n

n →∞

---- *— 1+

Λ_ 1

= j => converges 83.

lim

; f I dx =

n → ∞ ∏J i x

lim

n →∞

⅛π =

lim

∏n → ∞ ∏

; = 0 ≠> converges &

∖

(Theorem 5,7 #1) ’

85. 1 ,1 ,2 ,4 ,8 , 16,32,... = 1, 20 , 21 , 22 , 23 , 24 ,2 5 , ... => x i = 1 andxn = 2n^2 for n ≥ 2 87. (a) f(x) = x2 —2; the sequence converges to 1.414213562 ≈ √ ^ (b) f(x) = tan(x) - 1; the sequence converges to 0.7853981635 ≈ ∣ (c) f(x) = ex ; the sequence 1,0, —1, —2, —3, - 4 , —5 ,... diverges 89. (a) If a = 2n + 1, then b = L⅜J = [ ⅛ ⅛ ≡ j = [2n2 + 2n + ∣J = 2n2 + 2n, c = ⅛ ] = [2n2 + 2n + ∣] = 2n2 + 2n + 1 and a2 + b2 = (2n + 1)2 + (2n2 + 2n)2 = 4n2 + 4n + 1 + 4n4 + 8n3 + 4n2 = 4n4 + 8n3 + 8n2 + 4n + 1 = (2n2 + 2n + 1)2 = c2 . (b)

v 7

lim

a →∞

I a2 I

⅛

= lim

,⅛ ∏ 2

a → ∞ 2n + 2n÷l

I a2 I

= 1 or lim

⅛

a →∞

= lim sin θ = a →∞

lim

θ → π ∕2

sin 0 = 1

⅛ = lim ⅛c 1 = lim ⅛c = 0 91. (a) n lim → ∞ ∏c n → ∞ cn n → ∞ cn (b) For all e > 0, there exists an N = e^θn ^ c such that n > e"θ n e ^ c => Inn > - y ^

∏c > 7 ^

⅛
In Q )

⅛= °

= ∞ lim r Vn = ∞ lim r eVn∕ 93. n lim n 1^nn → e x p (∣' lnn)n → x p (∣) = e 0 = l →∞ 95. Assume the hypotheses of the theorem and let e be a positive number. For all e there exists a Nι such that when n > Nι then ∣ ⅜ - L∣< f => - e < ¾ - L < e => L - € < ⅜, and there exists a N2 such that when n > N2 then ∣ cn - L∣< e ≠> - e < cn - L < € => cn < L + e. If n > max{N1,N2}, then L - e < an ≤ bn ≤ cn < L + e ≠ ∣ bn —L ∣< e => n lim o bn = L. 97. an + 1 ≥ a n ≠> ⅞

≡ > ⅛

^

¾ ⅛ > ¾TΓ ^

3n2 + 3 n + 4n + 4 > 3n2 + 6n + n + 2

=> 4 > 2; the steps are reversible so the sequence is nondecreasing; ^

< 3 => 3n + 1 < 3n + 3

=> 1 < 3; the steps are reversible so the sequence is bounded above by 3 99. an + ι ≤ an ≠-

2-

^ ' ≤ ?

z^

^ iF

≤ i⅛

2

4

2 ∙ 3 < n + 1 which is true for n ≥ 5; the steps are

reversible so the sequence is decreasing after a5, but it is not nondecreasing for all its terms; aι = 6, a2 = 18,

345

346

Chapter 11 Infinite Sequences and Series

a3 = 36, a4 = 54, a5 = γ

= 64.8 => the sequence is bounded from above by 64.8

101. an = 1 — ⅛ converges because ⅛ → 0 by Example 1; also it is a nondecreasing sequence bounded above by 1 103. an = ¾

i

= 1 - ⅛ and θ < ⅛ < ⅛J since £ → 0 (by Example 1) => ^ → 0, the sequence converges; also it is

a nondecreasing sequence bounded above by 1 105. an = ( ( - l ) n + 1) ( ⅛ i ) diverges because an = 0 for n odd, while for n even an = 2 (1 + £) converges to 2; it diverges by definition of divergence 107. If {an } is nonincreasing with lower bound M, then {—an } is a nondecreasing sequence with upper bound —M. By Theorem 1, {—an } converges and hence {an } converges. If {an } has no lower bound, then {—an } has no upper bound and therefore diverges. Hence, {an } also diverges. 109. an > an + ι

> W

—

—

≡ 2 √ ∏ + 1 + √ 2 n 2 + 2n > √ π + √ 2 n 2 + 2n «■ √ n + 1 > √ n v v

√∏+1

—

V

*

v

—

v

and '1'+^⅜2 ^ ≥ ^ ^ ’ t h u s ⅛ e sequence is nonincreasing and bounded below by ᅟ ∕ 2 => it converges

H l. 4 ^ = 4÷

(^ ) n

4 +

( 2 ) " s 0 an ≥ a n+ 1 o

4 + ( ∣) n ≥ 4 + Q ) n ÷ 1 o

Q ) n ≥ ( ∣) n+1 ^

1 ≥ j and

≥ 4; thus the sequence is nonincreasing and bounded below by 4 => it converges

113. Let 0 < M < 1 and let N be an integer greater than γ ⅛ . Then n > N ^

n > ⅛

=> n —nM > M

=> n > M + nM => n > M(n + 1) => ^∩- > M. 115. The sequence an = 1 + ^ - is the sequence ∣, ∣, ∣, ∣, . . . ∙ This sequence is bounded above by ∣, but it clearly does not converge, by definition of convergence. 117. Given an e > 0, by definition of convergence there corresponds an N such that for all n > N, ∣ Lχ - an ∣< e and ∣ L 2 - L i ∣= ∣ L 2 - an + an - L j ≤ ∣ L 2 - an ∣< e. Now ∣ L 2 - an ∣ + ∣ an - L 1 ∣< € + e = 2e. ∣ L 2 —L I ∣< 2∈says that the difference between two fixed values is smaller than any positive number 2e. The only nonnegative number smaller than every positive number is 0, so ∣ Lι - L 2 ∣= 0 or Li = L 2 . 119. a2k → L o

given an e > 0 there corresponds an Nι such that [2k > Nι => ∣ a2 k —L ∣< e ]. Similarly,

a2 k+ι → L o [2k + 1 > N 2 => ∣ a2k+ι - L ∣< e]. L etN = m ax{N ι,N 2 }. Thenn > N => ∣ an —L ∣< e whether n is even or odd, and hence an → L. ^ 121. ∣

- l ∣< H - 1 ^

- ⅛




( ⅛ ) " < ∣< S )

n

^ n > i¾

^ n > 6 9 2 ,8

=> N = 692; anπ = ( ∣) 1^n and lim ann = 1 '2 / n →∞ 123. (0.9)n < 10" 3 ≠> n I n (0.9) < - 3 In 10 -> n > ⅛ ^ ≈ 65.54 => N = 65; an = (⅜ ) n a n d n lim o an = 0 125. (a) f(x) = x2 - a ≠> f'(x) = 2x => xn + ι = xn - ⅛

=> xn + ι = ⅛ ⅛ ^ ~ ^ = ¾

s

=

^ 2 ^

(b) Xi = 2, x2 = 1.75, x3 = 1.732142857, x< = 1.73205081, x5 = 1.732050808; we are finding the positive number where x2 —3 = 0; that is, where x2 = 3, x > 0, or where x = √ z3 .

Section 11.2 Infinite Series

347

127. xι = 1, x 2 = 1 + co s(l) = 1.540302306, x3 = 1.540302306 + cos(l ÷ cos(l)) = 1.570791601, X4 = 1.570791601 + cos (1.570791601) = 1.570796327 = j to 9 decimal places. After a few steps, the arc (xn-ι) and line segment cos (xn-ι) are nearly the same as the quarter circle. 11.2 INFINITE SERIES 1.

(l-r)

—

=>

1 - (I)

lim s = τ-⅜ττ = 3 1 ~ (∣ n → ∞ nπ )

=> n l→ im∞ sn

3.

( n + l) ( n + 2) “

n+1

n÷2

“

1

— 2 “ 3

3/

  3

>

  2

4 ∕^

∙

^V n÷ l

n+ 2∕ “

2

n+ 2

n → oo

n

2

7. 1 — ∣+ ⅛ - ^ + . . . , the sum of this geometric series is 1 _ p ^ = 1 jjT(i j = ∣ 9. ⅞÷ ⅞ + ⅛ ÷ ∙ ∙ ∙ , ⅛ e

su m

s

°f ^

βe o m

e t r ic

series is

1^

= j

11. (5 ÷ 1) ÷ ( ∣÷ j ) ÷ ( ∣÷ ∣) + ( ∣+ ^ ) + ... , is the sum of two geometric series; the sum is ∣

5

ι

10+ 3 2

=

23 2

=

13. (1 + 1) ÷ ( ∣~ ∣) ÷ (^ ÷ ⅛ ) ÷ ( ∣~ ⅛ ) ÷ ∙ ∙ ∙ , is the sum of two geometric series; the sum is T Γ 17 I y +∣ τ τ 1 π y _ - n2

15.

(4 n -3 X 4 n + l)

+ ( ⅛

17 1

-

-

4 n -3

4 ⅛ l) =

______40n_____ (2n - l) 2 (2n + 1)2 ~ _ -

5 +∣ 6

17 6

4n + l 1

"

=^

s

n ~

(1

=+ n ⅛ n ∞

4 ⅛

5) + S"

(5

= n 1M o

9) +

(9

13) +

(1 “ ⅛

)

=

-

’’ +

(4 n - 7

4 n -3 )

1

___A__ 1 B I C I D (2n - 1) ^r (2n - 1)2 ^r (2n + 1) (2n + l) i

A(2n - l)(2n + 1)2 + B(2n + 1)2 + C(2n + l)(2n - 1)2 + D(2n - 1)2 (2n - l) 2 (2n + l) i

=► A(2n - l)(2n + 1)2 + B(2n + 1)2 + C(2n + l)(2n - 1)2 + D(2n - 1)2 = 40n ≠> A (8n3 + 4n 2 - 2n - 1) + B (4n2 + 4n + 1) + C (8n3 - 4n 2 - 2n + 1) = D (4n2 - 4n + 1) = 40n => (8A + 8C)n3 + (4A + 4B - 4C + 4D)n 2 + (-2 A + 4B - 2C - 4D)n + ( - A + B + C + D) = 40n 8A + 8 C = 0 A+ B -C + D = 0 - A + 2 B - C - 2 D = 20 —A H~ B + C + D = 0

8A + 8 C = 0 4A + 4 B - 4 C + 4 D = 0 - 2 A + 4B - 2C - 4D = 40 -A + B + C + D = 0 A+ C= 0 -A + 5 ÷ C -5 = 0

and D = —5 => -

5 Σ

=

1 9

5

∙ ⅛ =

=>

(2 n ⅛

1

(2n + 1)4

-

( 1

1

"

lim

n → ∞

⅛ )

snn =

+

9 + 9

( ⅛

-

lim

(  1 ----√7n≡ +≡1 / ) = 1

n → ∞

⅛ )

+

( ⅛

=> 4B = 20 => B = 5

k r ι => C = 0 and A = 0. Hence, ]P (2∏—ι% n + i)i I j n=l l 25 + 25

j n=l l ( 1 ~ ( 2 k w ) =+ th e s u m is n lim o 5 (1 - ⅛

(

f B+ D= 0 [ 2B - 2D = 20

~ ⅛ )

+ ∙∙∙ +

‘' ’ 1)

(2(k - 1) + 1)2 +

(2k - 1)2

(2k + 1)2

= 5

( √ ⅛

+

⅛ )

+

( ⅛

^

τ ⅛ τ )

-

1

-

√⅛ T

348 21.

Chapter 11 Infinite Sequences and Series

Sn — ( “

1 ln 3

ta 2 ) ÷

“ 1⅛ ÷

( ln 4

M ⅛ 2)

^

ln 3 ) + (ln 5 n →m ∞

Sn

=

II M ) +

,,

’ +

(111(^+ 1)

In n ) ÷

(ln (n + 2)

ln (n ÷ l)

“ h⅛

23. convergent geometric series with sum

25. convergent geometric series with sum

27.

n

⅛ n o cos(nπ) =

⅛⅜0 (—l) n ≠ 0 => diverges

n

29. convergent geometric series with sum

31. convergent geometric series with sum — ⅜γγ —2 = y - y = ∣ 1~ ( Tn) 33. difference of two geometric series with sum — ½ r ------ γpr = 3 —∣= ∣ 1- W 1^W 35∙

∏⅛n∞ ⅛

0

= ∞ ≠

∞ 37. ∑ In ⅛

=^ diverges

∞ l n (“) - In (n + 1)] ^ ) = ∑ ∣

n=l

n=l

sn - [In (1) - In (2)] + [In (2) - In (3)] + [In (3) - In (4)] + ...

+ [In (n —1) - ln(n)] + [ln(n) - ln(n + 1)] = ln(l) - In (n + 1) = - In (n + 1) => 39. convergent geometric series with sum

J^ = ^ ∣

∞ ∞ 41. ∑ ( - l ) nxn = ∑ (-x )”; a = 1, r = - x ; converges to

γτ⅛ = ⅛

for ∣ x∣< 1

n=0

n=0

43. a = 3, r =

l

≠ ; converges to — 1

λ γ⅛ = τ ^ - for - 1 < ⅛1 < 1 or —1 < x < 3

k 2 J

45. a = 1, r = 2x; converges to γ ⅛ for ∣ 2x∣< 1 or ∣ x∣< ∣ 47. a = 1, r = - ( x + l) n ; converges to r q r ^ n j = ⅛

x ÷ 1∣< 1 or - 2 < x < 0 for ∣

49. a = 1, r = sin x; converges to ∣ ∙ . 1sin χ for x ≠ (2k + 1) j , k an integer 51. O∙23 = Σ ⅛ ( ⅛ Γ = ⅛ ∏=0

53 0 7 - Γ

'

10 k ι o ∕

—

n=0

55. < ≡ - ∑ n=0

)

= i

*

1

1

/ 1 ∖, -— 2 9

^ W

( ⅛ ∣(⅞ )(⅛ )" = ⅛ 1

“ W

= ⅜ -⅛

n

⅛‰ s ∏=

-

∞> ≠^ diverges

Section 11.2 Infinite Series 57. 1.24123 = ∞ 59. (a) ∑

123 / 1 124 ∣p 100 ^r 105 103 / n=0

n

_ ~

¾ / 1 W

124 100

_ “

124 ■ 123 _ 100 ^r 105 - 102 “

124 ■ 123 _ 100 ^r 99,900 ~

123,999 _ 99,9 ∞

∞ (n+4) 1(n + 5)

n= -2

∞ Σ

^

∑ (n + 2)(n + 3) n=0

^

41,333 33,300

( n - 3 ) 1( n - 2 )

n=5

61. (a) one example i s j + ∣÷ ∣+ ⅛ ÷ . . ∙ (b) one example is —∣- ∣- ∣- ⅛ - . (c) one example i s l - ∣- ∣- ∣- ⅛ - . . . ;the series ∣÷ ∣÷ ∣÷ ∙ ∙ ∙

= k where k is any positive or

negative number. ∞

n

∞

OO

∞

( ∣) . Then ∑ an = ∑ b n = ∑ ( ∣) = 1, while £

63. Let an = b n =

n=l

n=l

n=l

n=l

∞

✓

∖

oo

( ⅛k ) = ∑ (1) diverges, v

n /

∞

n=l

OO

/

65. L eta n = ( ∣) andb n = ( ∣) . Then A = ∑ an = ∣, B = ∑ bn = 1 and £ n=l

n=l

n=l

x

OO

(⅛l ) = ∑ ×

n /

(I)

= 1 ≠ ∣.

n=l

67. Since the sum of a finite number of terms is finite, adding or subtracting a finite number of terms from a series that diverges does not change the divergence of the series. r = ∣; 2 + 2 (∣) + 2 ( ∣) 2 + ...

69. (a) ⅛

= 5 ^

j = l-r ^

o» @

= s ^

⅛ = ∣→ *

' →

^ - ⅛ ( ⅛ ) + ⅛ ( ⅛ ) 2 - ⅛ ( ⅛ ) s + --­

71. sn = 1 + 2r + r 2 + 2r3 + r4 + 2r5 + ... + r 2n + 2r2 n + 1 , n = 0 , 1 , ... ≠> sn = (1 + r2 + r 4 + ... + r 2 n ) + (2 r + 2r3 + 2r5 + ... + 2r2 n + 1 ) => = γ ⅛ >if

lf 2 1
l '∞

(-)

J

, : ) √ ,, , 3 z(In x) √ (l n x)2 - 1

= ^lim

[sec

1

∣ ]^13 = u∣

b

∏m

[sec

1b

- sec

1

dx;

Γ u = In x 1 p∞ 1 → J lIn 3 -√4u 2 — , du , -1 d u = -1 d x u

(In 3)] =

b

[cos- 1

∏m

(£) — sec- 1 (In 3)]

= cos- 1 (0) - sec- 1 (In 3) = f - sec- 1 (In 3) ≈ 1.1439 23. diverges by the nth-Term Test for divergence; n lu∏0 n sin ( ∣) = p∞ χ 25. converges by the Integral Test: J j ∣ ^ dx; =

lim (tan b →∞

1b

- tan

1 e)

n

lim

u — e* j u _ eχjχ

⅛

i

= ⅛

du =

0

⅞

i

lim [tan n →∞ l

= l≠ β

1

u]JeJ

=2 £ - tan - 1 e ≈ 0.35 r*oo

u = tan 1 x → du = 1 +dxx 2 -

27. converges by the Integral Test:

8tan 1 x τ + τ r

29. converges by the Integral Test: J

sech x dx = 2 b lim

dJ Vx

.

’

= 2 lim v(tan - 1 e b —tan - 1 e) = τr —2 tan - 1 e ≈ 0.71 b →∞ '

f s u du = [4u2 ] ^ = 4 ⅛ J π ∕4

L

Γ ⅛ ψ dx = 2 b lim

Jπ∕4

[tan- 1 e x ] ,

- ⅛) = ⅛ 16 y

4

Section 11.3 The Integral Test 31∙

f ( ⅛ b

- ⅛ )

dx

=

+ ⅞ - = a b lim

lim

b

⅛

1 00

Ia

ln

(b + 2)a ~ 1

x + ∣ =

2

∣-

ln

x + ∣

| i° ’^ — ι

4

0? =

ξ^

t ^ι e

b

Hm o In ⅞ ⅞ ^ - In ( f ) ;

series converges to In ( j ) if a = 1 and diverges to ∞ if

a > 1. If a < 1, the terms of the series eventually become negative and the Integral Test does not apply. From that point on, however, the series behaves like a negative multiple of the harmonic series, and so it diverges. 33. (a)

(b) There are (13)(365)(24)(60)(60) (109 ) seconds in 13 billion years; by part (a) sn ≤ 1 + In n where n = (13)(365)(24)(60)(60) (109 ) => sn ≤ 1 + In ((13)(365)(24)(60)(60) (109 )) = 1 + ln(13) + In (365) + In (24) + 2 ln(60) + 9 ln(10) ≈ 41.55

an is a divergent series of positive numbers, then (∣) ∑ an = 52 (⅜) also diverges and ^ < an .

35. Yes. If £ n=l

n=l

n=l

∞ There is no “smallest" divergent series of positive numbers: for any divergent series 52 a ∏of positive n=l

∞ numbers 52 (⅜) has smaller terms and still diverges, n=l

n

n

37. Let A n = 52 a k a ∏d B∏ = ∑ 2k a(2k), where {ak} is a nonincreasing sequence of positive terms converging to k=l

k=l

0. Note that {An } and {Bn } are nondecreasing sequences of positive terms. Now, Bn = 2a2 + 4a4 + 8a§ + . . . + 2n a(2∏) = 2a2 + (2a4 + 2a4) + (2aθ + 2a§ + 2a⅛ + 2a⅛) + ... + (2a(2 ∏) + 2a(2∏j + . . . + 2a(2 ∏j) ≤ 2aι + 2a2 + (2a3 + 2a4) + (2a§ + 2aβ + 2aγ + 2ag) + .. . 2n ^ 1 terms ∞ + (2a(2n-η + 2a(2n-ι+ 1 ) + . . . + 2a(2n)) = 2A(2∏) ≤ 2 ^2 ak . Therefore if 52 a k converges, k=l

then {Bn } is bounded above => 52 2ka (2k) converges. Conversely,

∞ An = aι + (a2 + a3) + (a4 + a5 + aβ + a7) + ... + an < aι + 2a2 + 4a4 + . . . + 2n a(2∏) = aι + Bn < aι + 52 2k a(2k). k=l

∞ Therefore, if 52 2k a(2 k) converges, then {An } is bounded above and hence converges, k=l

39

∙ω

Γ ^ ∙ [⅛ 'J°⅛ ]

" L " ^ ' d" = b 1⅛

[⅛ π ]L = b ⅛ ( ⅛ ) ∣ b^p+ , -5+1, p > 1 . . , r ∙ P^ => thv e improper integral converges i f p > 11 andj jdiverges ( ∞, p < 1 if p < 1. For p = 1: J^ ^ ~ = lim [In (In χ )] 2 ≈ b hm [In (hi b) - In (In 2)] = ∞ , so the improper

=

integral diverges if p = 1.

351

352

Chapter 11 Infinite Sequences and Series

∞ (b) Since the series and the integral converge or diverge together, ^

41. (a) From Fig. 11.8 in the text with f(x) = £ and ak = ^ , we have f

ι

K< ⅛ P converges

if and only if p > 1.

^ d x ≤ l ÷ ∣+ ∣+ . . . + ⅛

≤ 1 ÷ J^ f(x) dx => ln ( n ÷ 1) ≤ 1 + ^ + ^ + ... + J ≤ 1 + In n => 0 ≤ In (n ÷ 1) - In n ≤ 0 ÷ I + I + ∙ ∙ ∙ ÷ ⅛) —hi n ≤ ^ The r e f° r e the sequence { (1 + ∣+ ∣÷ ∙ ∙ ∙ ÷ ⅛) —In n} is bounded above by 1 and below by 0. ∣dx = In (n + 1) —In n b) From the graph in Fig. 11.8(a) with f(x) = ∣, ⅛ < f X'∏-t-l ' ' ' ' 0 > ⅛ - [In ( n + l ) - Inn] = (1 + 1 + 1 + ,.. + ⅛ - ⅛ (n + 1)) - (1 + ∣+ ∣+ ... + ⅛ - l n n ) .

( ' '

If we define an = 1 + ^ = j + ⅛ - In n, then 0 > a∏+ ι - a

n

+

an + ι < a n => {an } is a decreasing sequence of

nonnegative terms.

11.4 COMPARISON TESTS ∞ 1. diverges by the Limit Comparison Test (part 1) when compared with ∑ ⅛ , a divergent p-series: n=l * lim

n→∞

(-7-)

∏→ ∞

2√ n+ y∏

=

f

∩

A= 1

l

n →m ∞ ∖2 + n 1∕ 6 ∕

2

3. converges by the Direct Comparison Test; ≡ r ≤ ^ , which is the nth term of a convergent geometric series 5. diverges since n lim o j ⅛ ∙ = ∣≠ 0 7. converges by the Direct Comparison Test; ( ⅛ ) " < (⅛ ) n — ( j ) ”, the nth term of a convergent geometric series 9. diverges by the Direct Comparison Test; n > l n n => l n n > l n l n n => - < ∣ ~
n for

19. converges by the Direct Comparison Test with ^ n ≥ 2 =>

n2

(n 2

— 1) >

n3

=> n √ n ^ T >

∞ ∞ 21. converges because 52 ⅛ r = ∑ n=l

n3 ∕ 2

=≠> ^

> —^

≈ or use Limit Comparison Test with ^ .

∞ ⅛ ÷ 52 ⅛ which is the sum of two convergent series:

n=l

n=l

∞ ∞ 52 ⅛ converges by the Direct Comparison Test since ⅛ < ⅜ , and 52 ⅛ is a convergent geometric n=l

n=l

series 23. converges by the Direct Comparison Test: ^ τ⅛ ^ < 3⅛τ , which is the nth term of a convergent geometric series 25. diverges by the Limit Comparison Test (part 1) with £ , the nth term of the divergent harmonic series: 1∙ (sin n) hm A-7T∖ ∏→∞ (I)

lim

x→0

⅞x * = 1

27. converges by the Limit Comparison Test (part 1) with ⅛ , the nth term of a convergent p-series: /

lim n → ∞

10n + l

A

⅛ < ^ 2 =

Um

( ι )

⅛ ⅛ =

∏→ ∞

Um

nz ÷ 3n + 2

⅛ ⅛ =

-1

29. converges by the

lim

n → ∞ 2n + 3

Direct Comparison Test: ⅛

2

ΓQ
N, ∣ ^ - 0 ∣< 1 => —1 < ^ < 1 => an < b n . Thus, if 52 b∏ converges, then 52 a ∏converges by the Direct Comparison Test. (b) If n lim o ^ = ∞ , then there exists an integer N such that for all n > N, ^ > 1 => an > b n . Thus, if J2 bn diverges, then 52 an diverges by the Direct Comparison Test. 39.

n

lπ∏o ⅛ = ∞ =^ t bere exists an integer N such that for all n > N, ^ > 1 => an > b n . If 52 a ∏converges,

then 52 b n converges by the Direct Comparison Test

353

354

Chapter 11 Infinite Sequences and Series

11.5 THE RATIO AND ROOT TESTS

1, converges by the Ratio Test: o

∙7

⅛1 =

lim

lim

∙k - τ r =

n→ ∞

n→ ∞ a ∏

n √2

⅛ ⅛ - ∙∙⅞ 2n+1

n → ∞

n √2

L^5H J

= lim ∏→ ∞

(1 + ∣) 7 ⅛ ) = ⅜ < 1 ∖

∖2∕ 2

n∕

Λ n + 1 )!∖

3. diverges by, the Ratio Test: =

∙

n→ ∞

lim

⅜

⅛t i =

n→ ∞

lim

= n+llim

- ⅞ τ ' n→ ∞

(⅛r)

e

n!

=

¾ p i ∙ ⅛ n→ ∞

e

a ⅜^

lim

= ∞

Λ > + ∣ )' 0 l

5. converges by, the Ratio Test: =

=

∙

⅛t l =

lim n→ ∞

lim

⅛

=

^ /° ”o∖

Hm

n→ ∞

- ¾w =

⅛ ⅛ τ10n+,

n→ ∞

∏

(l + ⅛)1° (⅛ )

lim n→ ∞

'

n∕

M0∕

⅜ < 1

= (∣) n [2 + ( - l ) n ] ≤ ( ∣) n (3) which is the nth term of a convergent

7. converges by the Direct Comparison Test: ^ ~ geometric series 9. diverges; lim an =

(1 - - ) n =

lim

lim

(1 + —) n = e~3 ≈ 0.05 ≠ 0

11. converges by the Direct Comparison Test: ¾ 1 < ⅛ = ⅛ for n ≥ 2, the nth term of a convergent p-series. > ∣(⅛) for n > 2 or by the Limit Comparison Test (part 1)

13. diverges by the Direct Comparison Test: J - ⅛ = ⅛ w ith i.

15. diverges by the Direct Comparison Test: ~ 1 > - for n ≥ 3 ⅛t l = n →lim∞

17. converges by the Ratio Test: j ®

n→ ∞ a

19. converges by the Ratio Test: O

n→ ∞

21. converges byj the Ratio Test: e

n→ ∞

23. converges by the Root Test:

lim

n

n

lim

⅛⅜t l = n →lim∞

lim

⅛aa =

lim

n→ ∞

n

hm o √ ¾ ≈

⅛ (∏+1)' ⅛ ⅛ r^ ∙ (n+l)(n + 2) = 0 < 1 qΓ7⅛⅛τ⅛iπ ∙ (n+ 3)! =

3!(n+l)!3n + *

(2n + 3)!

∙ ¾ n!^ =

lim o ^ ≡ ⅛ = ∏ ⅛

n

25. converges by the Direct Comparison Test: j⅛ ⅛ = ^ ⅛

2)

lim

n→ ∞

lim

n→ ∞

⅛ =

3(n÷l)

29. diverges byj the Ratio Test: o

⅞h =

lim

n→ ∞ a

lim

⅛a =

n→ ∞ a

31. converges byj the Ratio Test: β n→ ∞ a

n

n

lim

∏→ ∞

n

n

lir∏o

< i^ ⅛ + 2 j =

s⅛

lim

n→ ∞

lim ⅛d - = n→ ∞ a

n

an

=

lim n → ^n∞∕ - n=

lim

n→ ∞

lim

2n + 1

= ∣ > 1 2

- = 0< 1

= 0< 1

(n + ιχ n

= 0< 1

an

= 0< 1

(2n + 3)(2n÷2)

which is the nth-term of a convergent p-series 27. converges by the Ratio Test:

= i3 < 1

+

2)
f'(x) = - (⅞ + ? ) < 0 => f(x) is decreasing and hence un > un+1 > 0 for n ≥ 1 and n lim o (⅛ + A) = 0 ≠> convergence; but ∑ ∣ ⅛ ∣= ∑ ∞ ∞ ~ Σ ⅛÷ Σ n=l

⅛ is the sum of a convergent and divergent series, and hence diverges

n=l

23. converges absolutely by the Ratio Test:

n

lit∏o ( ¾ i ) — n ⅛¾0

(n + im ^ . ^ (ir

25. converges absolutely by the Integral Test since I 1 (tan- 1 x) ( τ ⅛ ) dx = ∖ ι- t - χ ∕

=

b

⅛>

n=l

n=l

! ⅛

k

, b

>2 ^

( t

1 ι

"

>2 ] =

≡ [ ®

2

-

2

®

] =

lim

z

b → ∞ L

Ji

⅛

27. diverges by the nth-Term Test since n lim o ~ j = 1 ≠ 0 29. converges absolutely by the Ratio Test: ®

j

lim

(⅛ u ) =

n → ∞ ∖ u

j

lim

∕ n

n

→ ∞

∙

∞

n=l 2+⅛

= 0< 1 n+1

^+⅛ TT

an^

n=l

< ⅛ which is the nth-term of a convergent p-series

∏

33. converges absolutely since ∑ ∣ an ∣= ∑ n=l

= ∑

n=l ' v

35. converges absolutely by the Root Test:

n

'

lim

Ja±i)⅛ ±¾ ⅛ ±l"z22 >

∏m

39. converges conditionally since ^ 1 1 o

j

1

L

^

is a convergent p-series

n=l

lirn o √ i ⅛ ∣=

37. diverges by the nth-Term Test since n li+m o ∣ an ∣= =

lim n → ∞

∞

31. converges absolutely by the Direct Comparison Test since ∑ ∣ an ∣= ∑ n

= (100)n

(n + l)!

n

n

⅛ no ( ⅞ ⅛ r ) ' =

lim o ⅛ ⅛ =

( n ± l) n ^ 1 =

oo

n

lim o *a

il

⅛

n

⅛¾o ⅛ Γ

h

=

^

≠ o

ι— ^ . ^ + i^ _ √n + 1+ √n √n + 1+ √n oo

decreasing sequence of positive terms which converges to 0 => ∑

/ anc∣

ι, j is a I √n ÷ 1+ √n J

n

7 ⅛ ⅛ 7 ζ converges; but

2
52 an converges. But this is equivalent to 52 an diverges => 52 la ∏l n=l

n=l

n=l

n=l

diverges. ∞

57. (a) 52 la ∏÷ ⅛l converges by the Direct Comparison Test since ∣ an ∣ ÷ ∣ bn ∣ and hence an + b n ∣≤ ∣ n=l ∞

52 (an ÷ b n ) converges absolutely n=l ∞

∞

∞

(b) 52 ∣ bn ∣ converges => 52 n=l ∞

-

n=l

⅛ converges absolutely; since 52 an converges absolutely and n=l

∞

∞

5 2 - b n converges absolutely, we have 52 [a ∏+ ( - ⅛)]

n=l ∞

n=l

∞

∞

=

Σ2 (a ∏- ⅛ ) converges absolutely by part (a) n=l ∞

kan ∣ converges => 52Lan converges absolutely (c) Σ2 la ∏l converges => ∣ k ∣52l a ∏l = ∑2∣ n=l

n=l

n=l

n=l

59. sι = - ∣,s 2 = - i + l = ∣, b3

—

2

τ

1

4

6

8

10

12

14

16

18

20

22

≈ -0.5099,

357

358

Chapter 11 Infinite Sequences and Series

S4 = S3 + ∣≈ —0.1766, 24

sβ

=

S5 +

»7 — ⅜

26

28

30

32

34

36

38

40

42

44

≈ -0.512,

52

54

56

58

60

62

64

66

≈ -0.51106

≈ —0.312, 46

48

50

0.2 $------ 3 ------ g ------

δ

∙∙....... y - 1/2

∞

∞

∞

∞

∞

61. (a) If ∑ ∣ an ∣ converges, then ∑ an converges and ∣∑ an + ∣∑ ∣ ⅛ ∣= ∑ n=l

n=l

n=l

n=l

. . ⅛+⅛l

n=l

. u ⅜÷l¾l f a∏, if a∏ ≥ 0 converges where bn = ■ = ∣ θ jf a < θ ∙ ∞

∞

00

∞

00

(b) If ∑ ∣ an ∣ converges, then ∑ an converges and ∣∑ an - 5 ∑ l⅛l = ∑ n=l

n=l

n=l

converges where cn = ½-⅛l

=

n=l

. . 212

n=l

0, if an ≥ 0 an , if an < 0 ’

63. Here is an example figure when N = 5. Notice that U3 > u2 > Ui and u3 > U5 > U4, but un ≥ un + ι for n ≥ 5.

11.7 POWER SERIES 1.

n

lim o ⅛ ∣

i n

< 1 =>

∣

n

⅛⅜o ⅞ r < 1 => ∣ x∣< 1 => —1 < x < 1; when x = —1 we have ^2 (—l) n , a divergent n=l

∞

series; when x = 1 we have ∑ 1, a divergent series n=l

(a) the radius is Γ, the interval of convergence is —1 < x < 1 (b) the interval of absolute convergence is —1 < x < 1 (c) there are no values for which the series converges conditionally

Section 11.7 Power Series 3

∙

M

n⅛ ⅞ o


I U ∣

lim

n→∞

n

I-— xΓ'

I ( n + l) √ n ÷ 13n+1

∙^

×n

∣< 1 =>

∣

( lim ( ∕ lim 3 ^ n → ∞ n + l ∕ xy n → ∞ n + i y

< 1

=> ^ (1)(1) < 1 => ∣ x∣< 3 => - 3 < x < 3; when x = - 3 we have 52 ¾7r> a n absolutely convergent series; n=l

n

∞

'

when x = 3 we have 52 ⅛ ,

a

convergent p-series

n=l

(a) the radius is 3; the interval of convergence is —3 ≤ x ≤ 3 (b) the interval of absolute convergence is —3 ≤ x ≤ 3 (c) there are no values for which the series converges conditionally n4

im∞ (×⅛ < 1 => n l→ im∞ o xf+ n^ i xn!n < 1 => ,lx1∣ 11. n l→ im∞ n l→ ∏+∙1)× < 1 for all x (a) the radius is oo; the series converges for all x (b) the series converges absolutely for all x (c) there are no values for which the series converges conditionally 13.

lim

n→∞

∣ ⅛h ∣< 1 => I un I

lim

n→∞

∣ 7⅛πτ ’

∣ (n + l)!

< 1 => x2 lim (∑⅛τ) < 1 for all x

χ2n+ , ∣

359

n → ∞ × ∏ + l×

(a) the radius is 00; the series converges for all x (b) the series converges absolutely for all x (c) there are no values for which the series converges conditionally

360 15.

Chapter 11 Infinite Sequences and Series I⅛ t i ∣< 1 ^

lim

∣un ∣

n→∞

∣
—1 < x < 1; when x = —1 we have 52 ∞

∑

nn2⅛⅞Λ + 2n + 44

∙ ⅛ n + 3 1 < 1 ≠∙ 1lx∣ ∕~Um ᅟ l

∣√ ( n + l ) 2 + 3

n→∞

∣

V∏→ ∞

< 1 ≠- lxl < 1 ∣∣

J n21^ 3 , a conditionally convergent series; when x = 1 we have

√ ⅛ 3 ’ a divergent series

(a) the radius is 1; the interval of convergence is —1 ≤ x < 1 (b) the interval of absolute convergence is - 1 < x < 1 (c) the series converges conditionally at x = —1 17.

lim

n→∞

∣ M ∣un ∣

n→∞

lim

∣

Il sn i 1W

^

1

∙ s τ τ w I < 1 ≠> ⅛

n(x + 3)π ∣

5+

5

n→∞

lim

( ^ ) < 1 ≠> ⅛ ^ < 1

∏ /

5

0°

/

=> ∣ x + 3∣< 5 => - 5 < x + 3 < 5 => - 8 < x < 2 ; when x = - 8 we have ∑

oo

5?- = ∑ (—1)" n, a divergent

n= l

n= l

∞

∞

series; when x = 2 we have 52 y r = 52 n , a divergent series n=l

n=l

(a) the radius is 5; the interval of convergence is —8 < x < 2 (b) the interval of absolute convergence is —8 < x < 2 (c) there are no values for which the series converges conditionally 19.

lim

n→∞

M

< l^

I u∏ ∣

lim

I⅜

' ∙⅜ x

∣ 3n+1

n→∞

v

n

∣

∣< 1 => ⅜ √ lim 3 y ∏ →∞

½ < 1 ≠> ∣ χ ∣< 3

(φ j < 1 ^

n ∕

3

∞ => —3 < x < 3; when x = —3 we have 52 (—l) n χ ∕n , a divergent series; when x = 3 we have n= l

∞

^2

∕n , a divergent series

n=l

(a) the radius is 3; the interval of convergence is —3 < x < 3 (b) the interval of absolute convergence is - 3 < x < 3 (c) there are no values for which the series converges conditionally 21.

lim ∏→ ∞

H I

< l

u"

≠∙

n

lim

n→∞

I

⅛

⅜

< 1

W ∕M

(l + in ) x " ∖

' '

∕

00

^ ⅛ ) < 1 => ∣ x∣(≡) < 1 ≠- W < 1

lim (1 + 1)

y n →∞ ∖

π/ /

1

I

'

' '

n

=> —1 < x < 1; when x = - 1 we have 52 ( - l ) n (1 ÷ ⅛) »a divergent series by the nth-Term Test since n=l

n

lh∏0 (1 + ⅛)n —e ≠ 0; when x = 1 we have ∑ (1 + “) n , a divergent series

(a) the radius is 1; the interval of convergence is —1 < x < 1 (b) the interval of absolute convergence is - 1 < x < 1 (c) there are no values for which the series converges conditionally 23.

^ ⅛ ^ ∣< 1 => ∣ x∣( lim (1 + ±)n ) ( lim ∣ 1 1 ∏nχn n → oo × ∏√ y y∏→ oo v ∣ ∏0 (n + 1) < 1 => only x = 0 satisfies this inequality => e ∣ x∣ n li∣ lim

n→∞

∣ ⅛α ∣< 1 ≠> I un I

lim

n→∞

(n + 1/) < 1 'J

I

(a) the radius is 0; the series converges only for x = 0 (b) the series converges absolutely only for x = 0 (c) there are no values for which the series converges conditionally 2 5

∙ n⅛ ∞

|¾

1

1
∣

n → oo

,⅛ J

( lim

n→∞

l∏ ( n + l) ∕

< 1

- 1 < x < 1; when x = - 1 we have

∏ / 0

∞

°

1 n(ln nj2

which converges absolutely; when x = 1 we have ^2

∑

∏ +1∕

n=l

n=l

which converges

(a) the radius is 1; the interval of convergence is —1 ≤ x ≤ 1 (b) the interval of absolute convergence is - 1 ≤ x ≤ 1 (c) there are no values for which the series converges conditionally

29.

lim

n→∞

| — I < 1 => I u∏ I

∣ 7⅛ ⅛ r ∙

lim

∣( n + l) 3∕ 2

n→∞

=> ∣ 4x - 5∣< 1 => —l < 4 x - 5 < 1

< 1 => (4x —5), 2 ( lim v

(4 x -5 ) zn +1 ∣

n→∞

∏ +1∕

< 1 =^ (4x —5)7 2 < 1 v

=> 1 < x < ∣; when x = 1 we have ^2 ⅛ ~

=

Σ

n=l

⅛

which is

n=l

absolutely convergent; when x = ∣we have 22 ~ ^ ~ »a convergent p-series n=l

(a) the radius is |; the interval of convergence is 1 ≤ x ≤ ∣ (b) the interval of absolute convergence is 1 ≤ x ≤ ∣ (c) there are no values for which the series converges conditionally 31.

lim

n→∞

l ^ - l < 1 =>

I un ∣

x + ≠∙ ∣

∣y ⅛

π

lim

n→∞

( ⅛

)
1 < x < 5; when x = 1 we have 52 U)n which diverges; lπ∏o ∣ ‰ m— ∙ ( ⅛ 7 ∣< 1 => ∣

∞ when x = 5 we have ∑ ( - l ) n which also diverges; the interval of convergence is 1 < x < 5; the sum of this n=l

convergent geometric series is — 4 —r v = ⅛ ∙ If f(x) = 1 - I (x - 3) + ∣(x - 3)2 + ... + ( - ∣) n (x - 3)n + ... 1+ l 2 J , = ⅛ ^ then f (x) = —∣+ ∣(x —3) ÷ ... ÷ ( - ∣) n n(x - 3)n ^ 1 + ... is convergent when 1 < x < 5, and diverges when x = 1 or 5. The sum for f , (x) is

, the derivative of ^ -j-.

jχ ^ ^

γ^ + ... 41. (a) Differentiate the series for sin x to get cos x = 1 - ^ + ^ - ^£ + ^ - ^∣ l

v8

v2

1

=

-

fi + ⅛

n⅛ ∞

X 10

∙

∣! + ⅛ - ⅛ i + ∙ ∙ ∙ ∙

-

l(2n + 2)! ‘ ⅛

l

=

χ2

(b) sin2x = 2 x - ^ + ^

n

⅛‰

- ^

.

The senes converges for all values of x since ( (2 n + lj(2 n + 2 ) ) = ° < 1 fθΓ a l l X.

+ ^ - ⅛ 4 + ... = 2 χ - ⅛ + ⅛

(c) 2 sin Xcos X = 2 [(0 ∙ 1) + (0 ■0 + 1 - l)x + (0 ∙ ⅛ + 1 ∙ 0 + 0 ∙ 1) + ( 0 ∙l + +

4

- ^

+ ^

_

2⅜d I 2⅛^ _ 3! ' 5 !

2V 7!

I 2Ψ _ ' 9!

211 x11 11!

- ⅛

→ ...

+ (0 ∙ 0 - 1 ■i + 0 ■0 - 1 ∙ i ) x3

+ ( 0 ∙0 + l ∙ T + 0 ∙ 0 + ∣. T + 0 - 0 + l ∙ i ) x

(θ ∙ ^ + 1 -0 + 0 - ⅛ + 0 ∙ ^ + 0 - 1 + 0 - ^ + 0 - 1 )x6 + ...] ≈ 2 [ x - ^ + ^

_

43. (a)

l∙0 - 0 ∙i- 0 - l+ 0 ∙l) x

x2

5

- ...]

∣ ' ” ’

In ∣ sec x∣ + C = f tan x dx = f (x + ⅛ + ⅛ + ⅛ + ⅛⅛ + ∙ ∙ ∙ ) dx _ “

_1_ 2^. _L 2 ÷ 12 ÷

_L lZ2d I 3lx 10 l 1 45 ÷ 2520 ÷ 14,175 ÷ * * * ÷

∩ r

__ x2 ∣ x4 ∣ x6 ∣ 17x8 ∣ 31x1° I 2 ÷ 12 ÷ 45 + 2520 ÷ 14,175 ÷ * * * ’

∩

X —θ => C —0 => In ∣ SeC X∣—

converges when - ∣< x < ∣ Q PΓ2

x — ⅛ nx) — 1

sec X -

dχ

-

dx

_|_ d ± 2 √

1 17√

+ 3 + 15 +

315

∣ 62>d ∣

+ 2835 +

J

_

1 ∣γ 2 ∣ 2 √ ∣ Γ7x^ ∣ 62 √ +∣ ι+ x + 3 + + i5 ••■ 45

when —∣< x < ∣ (c) sec2 x = (sec x)(sec x) = ( l + y + ^ + ^ =

_

11 +

(I 2 (A 1 ∏ 2 + ^ 2 ∕ Λχ +T ^ 2 4 +T 4 χ 4 χ 6 ∣ 17 1 1 v2 1 2 1 62X8 Π

2 4 ∕ Aχ

+ T

1

4 . ( 6 1 , V720 7r v

+ - ) (1 + τ + ⅛ + ⅛ A 48 7r

+

A _L 6 M 48 7 20/

aχ

6

+

+ ∙∙∙)

3

rn n v p rσ p q

’ converges

Section 11.8 Taylor and Maclaurin Series ∞

45. (a) If f(x) = £

∞

an xn , then f ^ (x ) = £

n=0

n(n - l)(n - 2)∙ ∙ ∙(n - (k - 1)) an xn - k and f ^ ( 0 ) = k!a∣ t

n=k

hn xn , then bk = ^

=> ak = ^ r ; likewise if f(x) =

=> ak = bk for every nonnegative integer k

n=0 ∞

(b) If f(x) = 52 ¾ x " = θ f°r a ll x >then f ^ W = 0 for all x ≠> from part (a) that a∣ c = 0 for every n=0

nonnegative integer k ∞

47. The series ^ 2 ^ converges conditionally at the left-hand endpoint of its interval of convergence [—1,1); the n=l ∞

series 52 ω = ⅛ + ^ ( χ - 5 ) - ^ ( χ - 5 )

2

) ,

-⅛ (χ -5 )

s

7. f(x) = √ x = x1∕ 2 , f'(x) = (∣) x- 1 ∕ 2 , f"(x) = ( - ∣) x - 3 ∕ 2 , f"'(x) = ( ∣) x - 5 / 2 ; f(4) = χ ∕4 = 2, f , (4) = (5) 4 - 1 / 2 = 1 , f"(4) = ( - ∣) 4 - 3 ∕ 2 = - ⅛ ,f"'(4) = (∣) 4 - 5∕ 2 = ⅛

≠> Po(x) = 2, P 1(x) = 2 + ∣(x - 4),

P2 (x) = 2 + i (x - 4) - ⅛ (x - 4)2 , P3 (x) = 2 + ∣(x - 4) - ⅛ (x - 4)2 + ⅛ (x - 4)3

11. f(x) = (1 + x)- 1 ≠> f'(x) = -(1 + x)^2 ,f"(x) = 2(1 + x)~3 ,f"'(x) = -3!(1 + x)~4 ≠> ... f(x) = ( - l ) k k!(l + x Γ k - , j f(0) = 1, f , (0) = - 1 , f"(0) = 2, f"'(0) = - 3 ! , ... , f w (0) = ( - l ) kk! ∞

∞

= 1 - χ + χ 2 - χ 3 + . . . = 52 ( - χ )n = Σ ( - i ) n χ n

≠∙ ⅛

n=0

γ — V ιq 1□. s i n A — n≈0

(-D"*2"* 1 (2n+l)!

— V n=0

n=0

(∑ JIW ^ 2 — V (2n+l)!

~

(zlΓ≤L^2≤ll — Qγ _ 3 V (2n+l)!

~

3’

∣ 3⅛ _ 5!

n=0

15. 7 cos (—x) = 7 cos χ = 7 ∑ ⅛ ⅛ r = 7 ~ ⅛ ÷ ⅛ - ^ ÷ ∙ ∙ ∙ *

s in c e

the cosine is an even function

363

364

Chapter 11 Infinite Sequences and Series

19. f(x) = x4 - 2x3 - 5x + 4 =+ f'(x) = 4x3 - 6x2 - 5, f"(x) = 12x2 - 12x, f"'(x) = 24x - 12, fW(x) = 24 =+ f(n >(x) = 0 if n ≥ 5; f(0) = 4, f'(0) = - 5 , f"(0) = 0, f'"(0) = -1 2 , f(0) = 24, f(n >(0) = 0 if n ≥ 5 =+ x4 - 2x3 - 5x + 4 = 4 - 5x - ^ x3 + ^ x4 = x4 - 2x3 - 5x + 4 itself 21. f(x) = x3 - 2x + 4 =+ f , (x) = 3X2 - 2, f"(x) = 6x, f'"(x) = 6 +> f(x) = 24, f W(x) = 0 if n ≥ 5; f(-2 ) = 21, f , (-2 ) = -3 6 , f"(-2 ) = 50, f"'(-2 ) = -48, fW (-2) = 24, f(n )(-2 ) = 0 if n ≥ 5 =+ x4 + x2 + 1 = 21 - 36(x + 2) + f (x + 2)2 - f (x + 2)3 + ^ (x + 2)4 = 21 - 36(x + 2) + 25(x + 2)2 - 8(x + 2)3 + (x + 2)4 25. f(x) = x^ 2 =+ f , (x) = - 2 χ - 3 , f"(x) = 3! x~4 , f ,"(x) == -4 ! x~5 =+ f COS V χ + ι = Σ

( - l ) n x2n (2n)!

5. COS x = n=0

n=0

  ∞ / ∞ ∞ 1 7. e* = ∑ £ ^ x e * = x ∑ ⅛ = ∑ ⅛ = ∖ n=0

n=0

9- ∞ s x - ∑ ⅛

/

^

X4 4!

X6 6!

n=0

x + x

2

, ⅛

+

+

( -l)" (x + l)° _ (2n)!

. £

+

, 1

χ + l , (x + l) 2 _ 2! ^r 4!

(x + l) 3 6!

_ ⅛ + ...

n=0

⅛ -l+ c o s x = ⅞ - l + ∑ ⅛

n=0

_ ~

Σ

=

(2n)!

365

- i

4

+

i- ^

+

⅛ ^

+

^ -⅛

+

...

n=0

. X8 8!

X10 I 10! "r

,

_ v ⅛ (z l T ^ . (2n)! ∙∙ ~ n=2

11. cosx

=

f

⅛

^

x c o s π x

n=0

n 1j .

=

U

f

x

n=0

^

, 2 T _ 1 , cos2x - 1 1 1 v √ -l) " ( 2 x ) 2n _ - 2 τ 2 uυ⅛ A — 2 τ (2∏)! 2

1 , 1 Γ1 2 ^r 2 1

m

l

n=0

-

⅛ χ (2x)8 _ 2∙6! ^r 2∙8!

1 _ !⅛ ! 4. 2∙2! ^r 2∙4!

1

= χ2

15. T ⅛

( ι⅛ ) ~

χ2

(→ )

f

=

^

n=0

_ ,

''

-

1 ι V ^r Δ

1

(2x)2 , ( 2 x ∕ 2! ^r 4!

(-l)"(2x) 2 " _ 2∙(2n)!

1 1

(2x)t , (2x)8 6! ^r 8!

∞ 17. ⅛ = ∑ xn = 1 + x +

, V ( - 1)" 22 - ' x2 " ^r Z J (2n)!

χ2

n=0

+

χ3

∞ + ■■∙ =^ ⅛ ( ⅛ ) = ∏ ⅛ = 1 + 2x + 3x2 + ... = ^ ∏xn ~ 1

n=0

= £

(n +

j

Σ (2x)n = ∑ 2n xn+ 2 = x2 + 2x3 + 22 x4 + 23 x5 + ... n=0

∞

1 '' '

n= l

l)x n

n=0

5 < (5!) (5 × 10"4 ) 19. By the Alternating Series Estimation Theorem, the error is less than ⅛ => ∣ x∣ 5 < 600 × 10- 4 ≠> ∣ =► ∣ x∣< √ 6 × 10"2 ≈ 0.56968 x∣

21. If sin x = x and ∣ x∣< 10- 3 , then the error is less than ^

≈ 1.67 × 10^^10, by Alternating Series Estimation Theorem;

The Alternating Series Estimation Theorem says R2(x) has the same sign as - ^ . Moreover, x < sin x => 0 < sin x —x = R2(x) => x < 0 => —IO- 3 < x < 0. 23. ∣ R2(x)∣= I⅛ I < 5 ^ p ^ < 1.87 × 10 4 , where c is between 0 and x 25. ∣ R4 (x)∣< ∣ ^ x 27. ∣ R ι∣= ∣ ⅛

5∣ =

∣ ⅛ ^ ⅛ ∣< 1⅛ ⅛ ∙ ^

= (1.13) ⅛

x ∣ ⅛ < ⅛ = 111 lx l < ∙θl ∣ = (1%) W ≠

≈ 0.000294

∣ ∣ ⅛* = 1 - £ + £ - £ + ... , s1 = 1 and s2 = 1 - ⅞ ; if L is the sum of the series representing ≡ i , then by the Alternating Series Estimation Theorem, L —Si = ^ L - s2 =

⅛i

- (1 - ⅛ ) > 0. Therefore 1 - ⅛ < ⅛ i < 1

— 1 < 0 and

366

Chapter 11 Infinite Sequences and Series

(b) The graph of y = ≡

, x ≠ 0, is bounded below by the

graph of y = 1 — y and above by the graph of y = 1 as derived in part (a).

31. sin x when x = 0.1; the sum is sin (0.1) ≈ 0.099833417 33. tan

1

x when x = j ; the sum is tan

1

( j ) ≈ 0.808448

35. e 1 s i π x ≈ 0 + x + x2 + x 3 ( - l + i ) + x _ v I -- A I A

17 □

I 1 I β

1

_

_

d n 2 γ — < 1 —cos 2 xλ _ l _ b ill A — ^ ; — 2 2

=>

⅛ (s⅛ 2 χ

= 2x — ^

1 γ6

l 2

P n e

4

( - i + i)+ x

( ∣- ∣l + i ) + x

6

( l - i l + ∣) + . . .

I ~ ...

7 Y -_ l _ l f l _ 2 2

) = ⅛ (⅛ - ¾ r + ⅞ Γ - ∙∙∙) =

+ ^ p —^

5

⅛ )i 2! 2x

. W t _ ' 4!

(¾ 6 6!

∣ '

-⅛ τ + ⅛ r - ⅛

^ - ⅛2! . ⅛ 41 ! J . ⅛6 ’ .

** ’

+ ∙∙∙ => 2 s in x c o s x

+ ... = sin 2x, which checks

39. A special case of Taylor’s Theorem is f(b) = f(a) + f , (c)(b - a), where c is between a and b => f(b) —f(a) = f , (c)(b —a), the Mean Value Theorem. 41. (a) f" ≤ 0, f , (a) = 0 and x = a interior to the interval I => f(x) —f(a) = ^

(x —a)2 ≤ 0 throughout I

=> f(x) ≤ f(a) throughout I => f has a local maximum at x = a (b) similar reasoning gives f(x) - f(a) = ^ (x - a)2 ≥ 0 throughout I => f(x) ≥ f(a) throughout I => f has a local minimum at x = a 43. (a) f(x) = ( 1 + x)k => f , (x) = k(l + x)k ^ 1 => f"(x) = k(k - 1)(1 + x)k “ 2 ; f(0) = 1, f'(0) = k, and f"(0) = k(k - 1) => Q(x) = 1 + kx + ^ x2 2 ⅛ i x3 ∣ < ⅛ (b) ∣ R2 (X)∣ = ∣

x 3 ∣< ⅛ => ∣

≠> 0 < x < τ ⅛ 73 OΓ0 < x < .21544

∞ ∞ 45. If f(x) = ∑ ¾ x "> then f^k x) = ∑ n(n - l)(n - 2)∙ ∙ ∙(n - k + l)a n x n - k and fW (0) = k! ⅜ n=k

n=0

=> a^ = T p

f° r

k

a

nθnnegative integer. Therefore, the coefficients of f(x) are identical with the

corresponding coefficients in the Maclaurin series of f(x) and the statement follows. 47. (a) Suppose f(x) is a continuous periodic function with period p. Let xo be an arbitrary real number. Then f assumes a minimum m i and a maximum m2 in the interval [x0 , xo + p]; i.e., m i ≤ f(x) ≤ m 2 for all x in [xo, xo + p]. Since f is periodic it has exactly the same values on all other intervals [xo + p, xo + 2p], [xo + 2p, x0 + 3 p ],. . . , and [xo —ρ, xθ], [xo - 2p, xo —p ] ,... , and so forth. That is, for all real numbers —∞ < x < ∞ we have mi ≤ f(x) ≤ m2. Now choose M = max {∣ ,∣ m2∣ } . Then m ι∣ m2 ∣≤ M ≠> ∣ -M ≤ - ∣ m ι∣≤ m i ≤ f(x) ≤ m 2 ≤ ∣ f(x)∣≤ M for all x. (b) The dominate term in the nth order Taylor polynomial generated by cos x about x = a is ^

(x —a)n or

^ p ^ (x —a)n . In both cases, as ∣ x∣ increases the absolute value of these dominate terms tends to ∞ ,

Section 11.10 Applications of Power Series

367

causing the graph of Pn (x) to move away from cos x. 49. (a) e i,r = cos (-τr) + i sin ( - π ) = - 1 + i(0) = - 1 (b) e"'« = cos ( J ) + i sin (J) = ⅛ + ⅛ = ( ⅛ ) (1 + i) ,π ∕ 2

(c) e

= cos (—f ) + i sin ( - f ) = 0 + i(—1) = - i

51. ex = l + x + ^ + ^ + ^ + ∙.∙ => ew = l + i0 + ^ e -*

= i - i0 + ( ≠ e

i,

+ e

-

iβ

_

+

∩

+ w

+ ⅜ - ⅞ → ⅜ -∙- )

“

2i

^ ]i + l i^ ^ + -

= sin0

τ + ⅛ + ∙∙∙) (x -5 τ + ∣ 53. ex sinx = ( l + x + ⅛ + ∣ 7 - ^ + ∙∙∙) = (l)x + (l)x 2 + ( - | + ∣) x3 + ( - 1 + ∣) x4 + ( ⅛ - ⅛ + ⅛) x5 + ... = x + x2 + ∣x3 - ⅛ x 5 + . . . ; ex ∙ eix = e(1+i ^x = e x (cos x + i sin x) = ex cos x + i (ex sin x) ≠> ex sin x is the series of the imaginary part of √ 1+i )χ which we calculate next; e the imaginary part χ 2 jn a gr e e m e nt with our 0 f e (*+>) i s x + 2 x + 2 χ 3 _ £ χ 5 _ 1 χ β + . . . = x + χ 2 + 1 χ 3 _ χ χ 5 _ χ χ 6 + product calculation. The series for ex sin x converges for all values of x. 55. (a) eiβl ew ≈ = (cos 0ι + i sin 0ι)(cos 02 + i sin 02 ) = (cos 0ιcos 02 —sin 0ιsin 02 ) + i(sin 01cos 02 + sin 02 cos 0ι) = cos(0i + 02 ) + i sin(0ι + 02 ) = e i ^ + ^ (b) e’ * = cos(-0) + i sin(-0) = cos 0 - i sin 0 = (cos 0 - i sin 0) ( ^ ^ ‘^ ) = c o s ⅜ + isin 9 = ⅛ 11.10 APPLICATIONS OF POWER SERIES ι.

( i + x ) 1∕ 2 = ι + i χ + ^

⅛

^

+ ^ ^ ⅛

3. (1 - x )- 1∕ 2 = 1 - I (-X) + t ^ ⅛ 5. (l

∣) - 2 = l - 2 ( * ) + ½

+

7. (1 + χ3 )^ v 2 =

1

)< = l

+

⅛

+

÷ W

!≡ !

13. (1 - 2x)3 = 1 + 3(-2x) +

+

+

+
2a2 = 0, 6a3 = 1, 4 - 3a4 + ao = 0, 5 ∙ 4a5 + aι = 0, and in general n(n - l)a n + an _4 = 0. Since y' = b and y = a when x = 0, we have a0 = a and a i = b. Therefore a2 = 0, a3 = ^ , a4 = - ^ , a5 = - ^ , ⅜ = 0, a7 = 2 3 ⅛ 7 —κ [error∣< - ^ ≈ 0.000013 45. (a) F1 ± l l W ^ ^ ("+ D -(n -D _ 2 71. (a) tan tan^tan in - 11) -— aj +i ⅛t a n! (u n - (n + 1)) tan(ta∏-1 ( n - l ) ) - 1 + ( n + l ) ( n - 1) - n2 ⅛ n+ tan tn N N -1 1 1 1 -1 1 1 (b) y tan^ (⅛) = £ [tan~ (n + 1) —tan^ (n —1)] = (tan^ 2 —tan 0) + (tan 3 —tan^ 1) n=l n=l + (tan^1 4 - tan- 1 2) ÷ ... + (tan- 1 (N + 1) - tan- 1 (N - 1)) = tan- 1 (N ÷ 1) ÷ tan- 1 N - ∣ ∞ (c) Σ ta∏~1 (⅜) = n limo o [tan"1 (N + 1) + ta n ' 1 N - ∣] = f + f - ∣= ⅜

11.11 FOURIER SERIES 1∙ a« — A f l ⅛ = l , a k = ∣f c o s k x d x = M ^ l Z r = 0 ,b k = ; f 7ΓJ o πJo 2π J o π L k J0 Thus, the Fourier series for f(x) is 1.

J ________ I________ I________ I—

π 2

π

3π 2

2ιr

s i n h d x = j [ - sψ π L

k

J0

2

]^ =0.

371

Chapter 11 Infinite Sequences and Series

372

3. ⅜ = ⅛ ∫ 0'x ⅛ + ∫ Λ - ⅛ ) < k >2τr

=

⅛ [ ∣π 2 ÷ H ^

-

π2) -

^

p it

2

] ~ θ∙ Note, p2π

pπ

(x —2π)cos kx dx = - Jθ u cos ku du (Letu = 2τr - x). So ak = ∣ j o x cos kx dx + J p it

p2∙π

(x —2π) cos kx dx p2π

p it

Note, J j (X —2π)sin kx dx = Jθ u sin ku du (Letu = 2π —x). So ⅛ = £ Jθ x sin kx dx + J* (x —2π) sin kx dx = π X x sinkxdx = ∣[ - ∣coskx + psinkx]θ = —∣coskπ = ∣( - l ) k + 1 . Thus, the Fourier series for f(x) is ∑ ( - l ) k + l ^ p

ss.

k=l

5- a0 = ⅛ ∫ f e x dx = l ( e 2x - 1), ak = ± £ > cos kx dx = 1 [ 1⅛ ( c o s kx + k sin kx) ]θπ = j ⅛ ⅛ =

£ ^ sin kx dx = 1 [ 1⅛ (s in kx - k cos kx) ]θπ = ¾ g . ∞

9

Thus, the Fourier series for f(x) is ^ ( e 2π —1) ÷ ⅛

2π

2π

i

∑ ( f⅛ - ⅛ ⅛ )∙

2π

f H

⅜= ⅛ f f(x) dx = ⅛ f cosxdx = 0, ak = ∣ f cosxcoskxdx = < 2 jo 2 j ° J ° | ∫ 0, ll>

,

π

Gc ~ l)x ∣ sin(k + l)x 1 π 2 2 (k +i) Jo’ L s in

i[lχ

+

l s in 2 x ζ ,

k≠ 1 = 1'

k

1 Γ cos(⅛ - l)x l cos(k + l)x 1 π 4 2 ( k - l ) ^t ^ 2 ( k + l) ] 0 > -⅛ ∞

Thus, the Fourier series for f(x) is jcos x + £ keven

s

w ⅛ ) s ^n ^ ∙

2x∣ θ,

k≠ 1

o, =

k= 1

^

2k τ ∏ ),

kodd keven

k≠l k= 1

Chapter 11 Practice Exercises

p2π

373

12π

9. J o cos px dx = ⅛in px∣ θ = O if p ≠ O .

11.

f g cos px cos qx dx = f g ∣[cos (p + q)x + cos(p p 2π

Γ

Ifp = qthen Jθ cospx cos qx dx =

13.

2π

9

p2π

JQ cos2 ρxdx =

f g sin px cos qx dx = f g ∣[ sin (p + q)x + sin (p 1

q)x ]dx = ∣[ ^ ⅛ i n (p + q)x + ^ s i n (p - q)x ]θπ = 0 if p ≠ q. ∖ P

1 /

Jθ j ( l + cos 2px) dx = ∣^x ÷ ^ sin 2pxj

q)x ]dx = 4 [ ^ ∞

.

P2π

= —1[(1 — l)^pξ ÷ (1 — l ) ^ j ] = 0. Ifp = qthen

J

s

π

= π.

(ρ + q)x + ^ c o s (p - q)x ] θπ p2π

p2π

sin px cos qx dx = Jθ sin px cos px dx = Jθ ∣sin2pxdx

= -⅛ c o s2 p x ∣ o = - X ( l-1 )= 0 . 15. (a) f(x) is piecewise continuous on [0, 2π] and f , (x) = 1 for all x ≠ π => f'(x) is piecewise continuous on [0, 2π]. Then by Theorem 24, the Fourier series for f(x) converges to f(x) for all x ≠ π and converges to ∣(f(π + ) + f(π~)) = j (—π + π) = 0 at x = π. oo

(b) The Fourier series for f(x) is 52 (—l) k +

2 si ” k x .

If we differentiate this series term by term we get the series

k=l

^ ( - l ) k + 1 2 cos kx, which diverges by the nth term test for divergence for any x since lim (—l) k + 1 2 cos kx ≠ 0. k →∞

k=l

CHAPTER 11 PRACTICE EXERCISES

1. converges to 1, since n lim o an =

n

3. converges to - 1 , since n lim o an =

lim o ( 1 ÷

n

lh∏0 ( ⅛ - ) =

n

5. diverges, since {sin ψ } = { 0 ,1,0, —1 ,0 ,1 ,... }

7. converges to 0, since n lim o an11 =

n→∞

9. converges to 1, since n lim o an =

n→∞ ×

11. converges to e 5 , since n lim o ⅛ = 13. converges to 3, since lim an = ®

’

n→∞

"

lim

^ = 2

lim

(≡ ⅛ * ) =

∏

∏ ×

lim

n→∞

lim

V

n→∞

1

ψ

= θ

= 1

1

= e

n

lim

( ^ ) 1∕ n =

n→∞ × n ∕

lim

n→∞

5

by Theorem 5

⅛, n = ∣= 3 by, Theorem 5 n∕

1

∙

374

Chapter 11 Infinite Sequences and Series f —2 1 ∕

1n l/n —1) = 15. converoges to In 2, since n lim a = n lim n(2 lim 2 ∕ - 1 v → ∞ n11 →∞ ' n→∞

n

In 2)

lim

2 1≠n ln2

lim

n→o

n→∞

= 20 ∙ In 2 = In 2 17. divergo es, since n l→ imoo an = n l→ im∞

2i- s ⅛

w

= |2

= ⅛

- A

*

n!

-= (b i)÷ (k a + (k ⅛ )+ -∙ + ( ⅛

⅛ im∞ sn = n l→ im∞ × 2 3n+2 => n l→

∞ 23. ∑

e ~n = n=0 n=0

= n l→ im∞ (vn + 1/) = ∞

- ⅛

)

3n + 2 / = 2§

∞ ∑ ⅜ , a convergent geometric series with r = ∣and a = 1 => the sum-is — U ∙ = ^ ∣ 1

(e )

25. diverges, a p-series with p = ∣ 27. Since f(x) = ⅛ ≠> f'(x) = - ≡ X '

< 0 => f(x) is d1 ecreasing 4∩ —a> ι < an , ∏ an→d ∞ liymn n0 +Q

oo

an =

lim

-¼ = 0, the

00

series £ ∏=1

- V≠

converges by the Alternating Series Test. Since £ n=l

Λ diverges, the given series converges

v

conditionally. 29. The given series does not converge absolutely by the Direct Comparison Test since h ( ⅛ ) > ⅛ the nth term of a divergent series. Since f(x) =

s

^⅛ -

4

, f'( x ) = _ _

- L ,- _
an + ι < an , and n lim o an = n lh∏o ⅛ j⅛ ∣ j = 0. the given series converges conditionally by the Alternating Series Test. 31. converges absolutely by the Direct Comparison Test since ⅛ < ⅛ = ⅛ , the nth term of a convergent p-series

^∙

n

→1∞

" ^ j 1 ~ ∖ ∕n ⅛ n∞ n⅛ Γ ~ ^

~ 1

z^

converges absolutely by the Limit Comparison Test

35. converges absolutely by the Ratio Test since lim

⅛ ∙- ⅛ l=

37. converges absolutely by the Ratio Test since n ∏mo j ~ ^ ∣∙ ∣j =

n

lim

τ⅛ = 0 < l

lim o ⅛

= 0< 1

39. converges absolutely by the Limit Comparison Test since n Umo γ— ^

—γ = J

n

ljm o — —∏⅜"^2 '

√ n ( n + l ) ( n + 2)J

4 ι∙

.⅛

k

∣< > * . ⅛

∣S

P

∙5 ⅛ ∣ < >

⅛

ι

.⅛

∣ ⅛ ) ∣ x + 4∣< 3 => - 3 < x + 4 < 3 => - 7 < x < - 1 ; at x = - 7 we have £ n=l

⅛

ι

⅛e divergent n=l

harmonic series (a) the radius is 3; the interval of convergence is —7 ≤ x < —1 (b) the interval of absolute convergence i s —7 < x < - 1 (c) the series converges conditionally at x = —7 «■ . 1⅛

⅛ !
—l < 3 x - 1 < 1 =

.⅛

∣ ¾

F ∙o ⅛

∣< ι ≠∙ ∣ ⅛ -1 ∣ .⅛ , s ⅛
0 < 3 x < 2 => 0 < x < ? ; a t x = 0we have 52 ^ ⅛

1

ς

^

∣ 3 χ - ι ∣< ι

^ = 52 ⅛

-

∞ ~ Σ ⅛ ’ a nonzero constant multiple of a convergent p-series, which is absolutely convergent; at x = ∣we n=l

have 52 ^ ^ ^ ∙ = 5 2 ¼ s ~ »which converges absolutely n=l

n=l

(a) the radius is j ; the interval of convergence is 0 ≤ x ≤ ∣ (b) the interval of absolute convergence is 0 ≤ x ≤ j (c) there are no values for which the series converges conditionally 45.

lim

n → ∞

∣ ⅛t i ∣< 1 => ∣u∏ I

lim

n → ∞

∣( n + l ) n + ,

∙ ⅛n ∣< 1 => W lim ∣ (⅛ ^ )n (⅛ τ)l < ^ ≠ x ∣

∙ ' n → ∞ ∣× n + l ∕

n+ l ∕ ∣

⅛ lim

(⅛ τ)
^ ∙ 0 < 1, which holds for all x (a) the radius is ∞∙, the series converges for all x (b) the series converges absolutely for all x (c) there are no values for which the series converges conditionally 47. n lim →∞

< 1 => n l→ im∞ ∞ ∞ a n ^ Σ2 the series 52 “

(π + 2)x2n + 1 3n + ,

∕3 < x < √ ⅞ < 1 => ⅛ lim M ) < 1 => —ᅟ 3 n → ∞ V∏÷ 1 / _ ’ obtained with x = ± √ 3 , both diverge 3n _ (n + l) x 2n- '

(a) the radius is \/3 ; the interval of convergence is —ᅟ ∕3 < x < √ z3 (b) the interval of absolute convergence is —√ z3 < x < ᅟ ∕3 (c) there are no values for which the series converges conditionally

49. n l→ im∞

1 => n l→ im∞

¾

⅛

csch(n)xn

^ ∣< 1 ∣

∣ x∣ lim 1 n →∞

Ie ι _ e 2n ι ∣ e n => ∣ 1x∣ 1 n lim T ^ r ∣< 1 => “ < 1 => - e < x < e; the series 52( ± )" csch »obtained with x = ± e, →∞ ∣ n=l both diverge since lim o ( ± e)n csch n ≠ 0 (a) the radius is e; the interval of convergence is —e < x < e (b) the interval of absolute convergence is —e < x < e (c) there are no values for which the series converges conditionally 51. The given series has the form 1 —x ÷ x2 —x3 ÷ ... + (—x)n ÷ ... = ∣ τ ~ , where x = ∣; the sum is — 53. The given series has the form x - ^ + ^ - ... ÷ ( - l ) n ^ ⅛ ^ ÷ ... = sin x, where x = πi the sum is sin 7r = 0 55. The given series has the form 1 + x + ^ + ^ ÷ ... + ^ ÷ ... = e *> where x = In 2; the sum is el n ^ = 2

375

Chapter 11 Infinite Sequences and Series

376

57. Consider ∣ ⅛

as the sum of a convergent geometric series with a = 1 and r = 2x +■ ^ ∞ ∞ = 1 + (2x) + (2X)2 + (2x)3 + . . . = £ (2x)n = £ 2n xn where ∣ x∣ < ∣ 2x∣< 1 + ∣ n=0

59 M sinx = yχ UX

2 t ⅛ (2n + l)!

^

sS i Π n π1 x7

n=0

= y t d H ^ Γ X -^ (2n + l)!

n=0

_

n=0

61. ∞ s x = £ ⅛

^

cos ( x « ) = £ a g £ n=0

n=0

∞

∞

n=0

⅛ ^

= £ ⅛ ≤

n=0

∕7ΓX∖ ∏

63. ex = ∑ £ =+ c ^ ∕ 2 ) = ∑ ⅛ n=0

y (2n(+z il)!) ⅛ n=0

=

∞

⅛

ς

n=0

65. f(x) = √ 3 + x2 = (3 + x2 ) 1/2 =+ f , (x) = x (3 + x2 )- 1 / 2 =+ f"(x) = - x 2 (3 + x2 )~3 /2 + (3 + x2 ) ' 1/2 =+ f"'(x) = 3X3 (3 + x2 )~5 /2 - 3x (3 + f W

t

∕ ι  _

67. f(x) =

r

X2 ) - 3 / 2 ; f ( - l )

= 2, f , ( - l ) = - j , f " ( - l ) = - 1 + ∣= ∣,

A ∣ 2 (x + l) ∣ 3(X + 1)2 . 9(x + 1)3 5 . 3 , 3 _ 9 . 3 2 ÷ 8 ~ 3 2 => V ∙^ + X - 2 - T T Γ ÷ ~ 2 M Γ ÷ 2 -3!

÷ •”

L = (χ + 1)"1 =+ f'(χ) = - ( χ + l) - 2 =+ f"(χ) = 2(x + l) - 3 =+ f'"(x) = -6 (x + I)" 4 ; f(3) = ∣,

f , (3) = - ⅜ , f"(3) = ⅜ , f w (2) = =^ =+ ⅛ = ∣- ⅜ ( x - 3 ) + ⅜ ( x - 3 ) 2 - ⅜ ( x - 3 ) 3 + ... 69. Assume the solution has the form y = ao + aιx + a2x2 + . . . + an -ιx n => ^ = ⅛ + 2a2 x + . . ∙ + n¾xn - 1 + ... =+ ^ + y

1

+ an xn + .. .

= (aι + a0 ) + (2a2 + a j x + (3a3 + a2 )x2 + . . . + (nan + an -ι)x n - 1 + ... = 0 =+ aj + a0 = 0,2a2 + a j = 0, 3a3 + a2 = 0 and in general nan + an -ι = 0. Since y = —1 when x = 0 we have ao = —1. Therefore aι = 1, aa o2

_ ~

→ ι _21

_ 1 2

, _ → _ X „ _ ’ aa 3 — 3 2 — 3-2 ’ a 4 ~

_ → a _ _____ i _ 4 ~ 4∙3∙2 ’ ∙ ∙ ∙ ’ aa ∏ ~

=+ y = - l + χ - ∣χ 2 + ⅛ χ 3 - . . . +

1⅛

- χ n + ... = -

-⅜ -,

ς

n

_ -1 — n

l ^*p'

( - D n _ ( -D ° + l (n - l ) ∣— n!

= - e -x

71. Assume the solution has the form y = ao + aιx + a2 x2 + . . . + an -ιx n ~ 1 + an xn + ... =+ Λ = a i + 2 a 2χ + ∙ ∙ ∙ + ∏an xn ^ 1 + ■∙ ∙ =+ ^ + 2y = (aι + 2ao) + (2a2 + 2aι)x + (3a3 + 2a2 )x2 + . . . + (nan + 2an -ι)x l1-1 + ... = 0. Since y = 3 when x = 0 we have a0 = 3. Therefore aι = -2 a 0 = -2(3) = -3(2), a2 = - ∣aι = - 1 ( - 2 ∙ 3) = 3 ( ⅛ ) , a3 = - ? a2

=- H≡(¥)]=-3 (⅛)....⅛=H ) >-■=(-1) G(⅛S+))=3(4F) =+ y = 3 - 3(2x) + 3 ^ x2 - 3 ^ x3 + . . . + 3 ^ =

3

[ 1 - ( 2 X) + ^ L

- ^ '

+

... +

^ ^

+

xn + ...

...l = 3 ∑ t l ∞ J

n=0

= 3e- 2×

’

73. Assume the solution has the form y = a0 ÷ aιx + a2x2 ÷ ... + an - 1xn - 1 ÷ an xn + ... =+ ⅛ = a ι + 2 a 2χ + ■∙ ∙ + nan xn - 1 + ... =+ ^ - y = (aι - a0 ) + (2a2 - a i )x + (3a3 - a2 )x2 + . . . + (nan - an -ι)x n ~ 1 + ... = 3x =+ a i - a0 = 0, 2a2 —a 1 = 3, 3a3 - a2 = 0 and in general na∏ - an -ι = 0 for n > 2. Since y = —1 when x = 0 we have a0 = —1. Therefore = j ,a 3 = f = ⅛ ,a 4 = ⅜ = ⅛ , . . . , a n = V = ⅛ aι = - l,a 2 = ^ +> y = - l - x + ( ∣) x 2 + ⅛ x 3 + 5 ⅛ x 4 + ... + ⅛ x n + ... = 2 ( l + xZ+ ⅜x2 + ⅛ x⅜3∙J∙+Z 34-5 x4 + ...∏! + ⅛ xn +/ ...) - 3 - 3x =* 2∏!y ⅛ —3 —3x = 2ex —3x —3 n=0

Chapter 11 Practice Exercises 75. Assume the solution has the form y = ao + aιx + a2x2 + ... + an -ιx n =+ ^ = a i + 2a2 x + . . . + n a n xn ^ 1 + ... =+ ⅛ - y

1

+ an xn + ...

= (ai - a0 ) + (2a2 - a i )x + (3a3 - a2 )x2 + . . . + (nan - an - 1)xn ~ 1 + ... = x =+ a 1 - ⅜ = 0 ,2a2 - a 1 = 1, 3a3 - a2 = 0 and in general nan - an - ι = 0 for n > 2. Since y = 1 when x = 0 we have a0 = 1. Therefore aι = 1, a2 = ^ y 1 = ∣,a3 = ^ = ⅛ ,a 4 = ^ = 4^2 >■∙ ∙ >a n = ⅛ i = ⅛ ≠> y = l + x + ( 2 ) χ 2 + ⅛ χ 3 + - 2 , χ 4 + . + 2 χ n + ... ∞

= 2 ( l + x + ∣x2 + ⅛

χ3

+ 4⅛2 χ 4 + ∙∙∙ + ⅛x n + ∙ ∙ ∙ ) - l - x =

Σ ⅛ - l - x = 2ex - x - l

2

n=0

' 1 ^2

J

o

~ ~

/

e x p (-x 3 )dx = J o

1 _ 1 2 2τ 4

τ

I

Q 3

Y6

γ9

1 7M 7

ψ

Y1 2

∖

„7

J

,+ ⅜ - + . . J d x = + ∣ r - ∣

1________ 1 I 27 ∙7∙2! 2 1°∙10∙3! "r 2τ M 3 4 !

1 ∣ 15 _ 5M 9^2$

1 2

(l-x

2 16 -16∙5!

∣ ___2 1________ 1 TT^2τ τ 9 ∙29

I

„10

1/2

„13

10-3! "r

7∙2!

13∙4!

0

≈ 0.484917143

1_________i1 ι s 132 ∙2 13 15 ∙2

1

W

1

7

_

1 . 1 192 ∙2 19 ^+ ^ 212 ∙2i l

≈ 0.4872223583 7 ( * - ⅜ + ⅜ - ∙) ⅛ Σ J

⅛ ς ⅛ +⅛ l ξ ) _ _ ^ χ1→ 0 ( 2 ⅛ + f + . . ) "

81. lim e⅛ π41 = lim ^ W ^ χ →0 x →0

7 2

t2 - 2 + 2 Λ - ⅛ + ^ - . . . )

= lim ⅛ 2 ¾ — ⅛) 2

83∙ hm ( 9 J

t → 0 ×2 —2 costt

85. z ¾

M

t /

⅛

⅛

t→ 0

2t ( l - cost)

1

= lim

t → 0

2∣ 2 ( 1 - 1 + ⅛ - ⅛ + ...)

2

f ⅛ -g + ..J (t4 - ¾

^ t ^ 0

+ ∙. . )

= ⅛ 2

3

4

∙∙∙-

= -2

87. lim v( ⅛x x →0

δ

3

÷ ⅛ ÷ s), = lim χ2 x →o

2 ∣ 81x2 2 "r 40

+ ⅜ + s) = 0

=+ ⅜ + ⅜ —0 and s —∣= 0 =+ r = —3 and s = ∣ 89. (a) ∑ (sin ⅛ - sin ⅛ ) = (sin j - sin j) + (sin ∣- sin ∣) + (sin ∣- sin ∣) + ... + (sin ⅛ - sin ⅛ ) n=l 00

—cos

+ ... = £ (—l) n sin j ; f(x) = sin ∣ =+ f'(x) = —^

< 0 if x ≥ 2 =+ sin ⅛

< sin J ,a n d

n=2

∞ n

lπ∏o sin

= 0 => 52 ( ~ l) n s i n ∏c o n v e rge s by the Alternating Series Test n=2

(b) ∣ error∣< ∣ sin ^ ∣≈ 0.02381 and the sum is an underestimate because the remainder is positive 91.

lim

- ( 3 n - l) ( 3 n + 2)xn ÷ 1 I2 ∙5 ∙8 2∙4∙6∙ ∙ ∙(2n)(2n + 2)

n→ ∞ ∣

2∙4∙6∙∙∙(2n) I ‘ 2∙5∙8∙ ∙ ∙(3n - l)x n ∣
the radius of convergence is ∣

1 => ∣ ∣⅛ ∣< 1 => ∣ x∣ < i3 1x∣ 1 n lim → ∞ ∣ ∣ ∙ 2n + 2 I ∣

377

378

Chapter 11 Infinite Sequences and Series

93. ∑ 1∏(1 - ⅛) = ∑ [l n ( 1 + k) + k=2

ln

( 1 " j)]

k=2

=

∑ Pn (k + 1) - 1∏k + In(k - 1) - In k]

k=2

= [In 3 - In 2 + In 1 - In 2] + [In 4 - In 3 + In 2 - In 3] + [in 5 - In 4 + In 3 - In 4] + [In 6 - In 5 + In 4 - In 5] + ... + [In (n + 1) —In n + In (n —1) - In n] = [In 1 - In 2] + [In (n + 1) —In n] after cancellation => Σ

ln

( 1 - ⅛) =

1.4.7...(

ln

-

ln

( ⅛ ) => Σ

)(3n l)χ3°÷3 .

( 1 - ⅛) =

(3n)!

I

n

⅛ n∞

∞

1±L∑2SL L2) (3n)! A

V — 1 V y — 1 “T

γ

d⅞

~

V

i∙4>7∙ ∙(3 n -2 ) (3 n -2 )!

d χ

⅛i s

th e s u m

3n-2 _

l± Z ∑ < 3 n - 2 ) x3n-l (3∏-1)!

n=l γ

ι V^

λ

1∙4∙7∙ ∙ ∙(3n-5) λ (3n-3)!

n=l

3n-2

n=2 ∖

oo

(

ln

00

⅛ — V " L

3n

n=l _k

(⅛ ) =

(3n + l) ∣3 ∣ 1 i m ∣ * ∣n → oo (3 n+ l)(3n + 2)(3n + 3)

.

3 n 2 + 95. (a) (3n + 3)! l-4∙7 - (3 n -2 )x 3" ∣ 1 ' , n lim →∞ = ∣ x3 ∣ ∙ 0 < 1 ≠ the radius of convergence is ∞

∕U λ W

ln

1+ ∑ n=l

~ ⅞ n )"' ^ /

x3n I = xy + θ ≠ , a = 1 and b = 0

∞ ∞ 97. Yes, the series^ an bn converges as we now show. Since ∑ a ∏converges it follows that an → 0 => an < 1 n=l

n=l

∞ ∞ for n > some index N => an bn < bn for n > N => 52 a ∏bn converges by the Direct Comparison Test with 52 ⅛ n=l

∞ 99. ∑ (xn + ι - xn ) =

n lim

∞ ∑ (x k + ι - Xk) =

k=l

n

n=l

⅛∏ (xn+1 - x 1) =

lim (xn + 1 ) - x 1 => both the series and

sequence must either converge or diverge. 101. Newton’s method gives xn + ι = xn — ^ L =^L+⅛

n^ y ⅛

= ^ xn + ⅛ , a n ^ ^ the sequence {xn } has the limit L, then

≠ ∙ L = 1 and {xn } converges since ∣ ^ ^ ∣= ^ < 1

103∙ an

= i ⅛ f o r n ≥ 2 => a2 ≥ a 3 ≥ a 4 ≥ . . . , a n d s ⅛ + s ⅛ + s ⅛ + ... = i ⅛ + ⅛ + 3⅛1 + ... ∞ = 1 ^ (1 + 2 + 3 + -- -) w hich diverges so that 1 + 52 ⅛ ii ^ v e r δ e s hy the Integral Test.

1 05 ∙ a0

= ⅛ ∕ 0⅛ ⅛ = ∣X ⅛

dx

= ⅛ Γ 1 d x = I, ⅜ = l f 0⅛ s in k x d x = U 5 in k x d x = - ⅛ C

cos kx dx

=

∣ JΓCOS kx dx = 0.

= - ⅛ ( 1 - ( " 1 )k ) = { ^ ¾

keven

Thus, the Fourier series of f(x) is ∣—52 ⅛ s in ^ k odd

107∙

⅛ = ⅛ £ + - x) d * + £

(x - 2π) dx

= ⅛ [X + - χ) dχ p it

Jθ (π - u) du j = 0 where we used the

substitution u = x —π in the second integral. We haveak = ∣ Jθ (π —x) cos kx dx +

p2π

(x —2π) cos kx dx . Using

Chapter 11 Additional and Advanced Exercises p2π

the substitution u = x —π in the second integral gives § (π —u) cos ku du, =

p

(x - 2π) cos kx dx = Jθ - ( π - u) cos(ku + kπ) du

k odd

pπ

Jo ^^(7r “

u

) c o s ^ u ^u ,

k even , 0,

Thus, ak = ^

k odd k even

Now, since k is odd, letting v = π —x => I

4

π

lfΛ

-

χ

)

co s k χ d χ

= -^ X

v c

°s

k v d v

= - ^ ( - ⅛ ) = ⅛ ' k ° d d ∙ L2 + L — 1 = 0 => L = - ψ ^

≠0 9. f(x) = cos x with a = f => f ( f ) = 0.5, f' ( f ) = - ⅛ , f " ( f ) = -0 .5, f"' (5) = j ^ , f (4) ( f ) = 0.5; ∞s x = I - ^

(x - f ) - | (x - f ) 2 + ⅛ (x - f ) 3 + ...

11. ex = l + x + ^ + ^ + ... with a = 0

e

380

Chapter 11 Infinite Sequences and Series

13. f(x) = cos x with a = 22π => f(22π) = 1, f'(22π) = 0, f"(22π) = - 1 , f"'(22π) = 0, f w (22π) = 1, f(22π) = 0, f(22π) = -1 ; cos x = 1 - ∣(x - 22π)2 + ⅛ (x - 22π)4 - ⅛ (x - 22π)6 + ... 15. Yes, the sequence converges: c„ = (an + bn ) 1/n => cn = b ( ( ∣) n + 1) 1/n => lim cn = In b + lim !≤li^±H ∏

⅛⅛p (b ∕ ■

= In b + lim n→ ∞

i 7

-

s

=>

ι

lim snπ =

n→ ∞

⅛ =

,

x

τ

⅛

+ r

+ -

*

+

L

⅛

⅛ =

^

*

r

(tan- 1 n —tan-f 1 0)2= ⅞

lim

n→ ∞

®

∏ —> o o

^

r

∏—>oo

= In b + ⅛ ⅛ = In b since 0 < a < b. Thus, lim cn = eln b = b.

⅛

∙ = £

* ∞

v

19. (a) Each An+ ι fits into the corresponding upper triangular region, whose vertices are: (n, f(n) - f(n+l)), (n + l, f(n+l)) and (n, f(n)) along the line whose slope is f(n + 1 ) - f(n). All the An,s fit into the first upper triangular region whose area is 2 1 ^ ® => ^ An


th e n

Ak = ≡ ≡ ≡ ≡ ≡ ± ÷ ^ ^

g

_ £

f(x ) d x

_ £

f(x ) d x

_ ... _ £

f( x )

dx

k=l =

!U)+f(n)

g

+

f(k )

_ r

dx

f w

^

1

k=2

fn part (a). The sequence < ^

g

1

g

Ak =

k=l

f(k )

k=l

_ tu rn s ) _ £

f(x ) d x

< M ≈ ) , fto m

1

1 Ak ? is bounded above and increasing, so it converges and the limit in J

(k = l

question must exist. n

(c) Let L =

lim

52 f(k) -

n→ ∞

f l f(x) dx ~ ∣(f(l) + f(∏))

, which exists by part (b). Since f is positive and

∑ f(k) - £ f ( x ) dx = L + ∣(f(l) + M).

decreasing lim f(n) = M ≥ 0 exists. Thus lim n→ oo n —► ∞

21. (a) No, the limit does not appear to depend on the value of the constant a (b) Yes, the limit depends on the value of b (c) =

lim ∏→ ∞

23.

lim n→ ∞

i ^ ⅛ 1_

^

∣ M I u∏ I

= H

c o s (n)

∣ l∏(n + 1)

bn xn ∣

25. lim s M ⅛ i* t o ιi m (ax " + -) (, x x x →0 x →0 = limθ J ^ r 5 + 3∣^ ( ^ ⅛) χ 2 + ■••] χh ⅛

27∙

≡2x^≡x→ = -2 !

⅛ = ¾

d^

=

1

j

n→ ∞

bx∣< 1 => - ⅛ < x < i = 5 => b = ± ∣ ∣ ⅛ ⅛ ⅛ ∙ ⅛⅛∣< 1 ≠> ∣

lim

n→ ∞

lim s = e^ 1 ≈ 0.3678794412; similarly,

= - 1 =>

l-θ

+

1

,

3.+∙∙) ιs

b

b

x

finite if a —2 = 0 => a = 2;

± = -2

+ n + ⅛ => C = 2 > 1 and g

⅛ converges

5

Chapter 11 Additional and Advanced Exercises (b)

⅛

=

α

7

1

=

1

+

n +

⅛

∞ C = 1 ≤ 1 and £

=+

n

i

diverges

n=l

∞

oo

∞

29. (a) 52 a ∏= L => ⅛ ≤ an ∑ an = an L => 52 ⅛ converges by the Direct Comparison Test n=l

n=l

n= l

f _ ⅛D-A

(b) converges by the Limit Comparison Test:

∞

lim o ^ a^

=

n

n

lπn o y ^ = 1 since 52 a ∏converges and n=l

an = 0 therefore X lim → ∞ 1

31. (1 - x)

x∣< 1 =+ (⅛52 = 1 + ∑ xn where ∣

=

⅛0

-

χ

)

1

1

= £ nxn

and when x = ∣we have

n= l

n=l

4 = l + 2 ( l ) + 3 ( ∣) 2 + 4 ( l ) 3 + . . . + n ( i ) n - 1 + . . . 33. The sequence {xn } converges to f from below so en = ∣- xn > 0 for each n. By the Alternating Series Estimation Theorem en + ι ≈ j ∣ error∣< ^ (en )5 , and since the remainder is negative this is an (en )3 with ∣ overestimate => 0 < en ÷ι < ∣(en )3 . 35. (a)

⅛

π

∞

= ⅛ ( ⅛ ) = ⅛ ( l + x + x2 + x 3 + ...) = l + 2 x + 3x2 + 4x3 + ... = £ n x n - ’ n=l

(b) from part (a) we have ∑ n (∣) n ~ 1 ( ∣) = ( ∣) [∙∏ ⅛ ] = 6 ∞

(c) from part (a) we have £ npn - 1 q = ^ ^

= ^ = ^

n= l

37. (a) Rn = Coe"1* + C0 e - 2⅛ + . . . + C0 e-"⅛ = ^ ⅛

⅛

^

=> R =

n

Um50 Rn = ⅛ ⅞ = ⅛

+ R1 = e - 1 ≈ 0.36787944 and R 10 = s ¾ ⅞ ^ ≈ 0.58195028;

(b) Rn = ⅛ ⅛ P

R = _ L ≈ 0.58197671; R - R 10 ≈ 0.00002643 =+ ^ ω < 0.0001 B , ⅛ ⅛ > (1) ( ⅛ ) =+ 1 - e - ∕ 1° > iZ =+ e - n∕ ' 0 < iZ =+ —⅛ < In (⅜) =+ ⅛ > - l n (∖ ∣ ) =+ n > 6.93 =+ n = 7 lv ∖Z / z / lv ∞ 39. The convergence of 52 la ∏l implies that

lim ∣ a ∣= 0. Let N > 0 be such that ∣ an ∣< ∣ => 1 —∣ an ∣> ∣ n→∞ n N . Now ∣ ln (l + an )∣= ∣ + ∣ ≤ ∣+ ∣ ⅛∣+ ∣ ⅞∣ an ∣ an - f + ⅛ - $ + ... ∣≤ ∣ an ∣ + ... n=ι

=+ ⅛

∞ < 2∣ an ∣ . Therefore £ In (1 + an ) converges by the Direct

2 + ∣ 3 + ∣ 4 + ...= j ¾ an ∣ + ∣ an ∣ < ∣ an ∣ an ∣

∞ Comparison Test since 52 la ∏∣ converges. n=l

41. (a) s2n+1 = ⅜ + ⅝ + ⅝ + ... + ⅛ f = ⅛ + ^

+ ⅛⅛ + ... + ⅛ ≈ 2n

=

tl

(1

-

I) +

t

2 (I ~ I) +

,

*, +

t2 n

(⅛ “

2 n + l) +

2n+l “

∑

k(k+ l) +

2n+⅛

,

k=l

(b) {cn } = { ( - l ) n } =+ ∑ ∏ r converges n—1

(c) {cn } = {1,- 1 , - 1 , 1 , 1 , - 1 , - 1 , 1 , 1 , . . } =+ the series 1 - 5 - 5 + ∣+ 5 - ∣- ∣+ ... converges

381

382

Chapter 11 Infinite Sequences and Series

NOTES:

P EARSON Addison Wesley

ISBN 0-3Sl-≡Sb4b-l