thomas calculus 12th solutions

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INSTRUCTOR’S SOLUTIONS MANUAL SINGLE VARIABLE Collin County Community College

WILLIAM ARDIS

THOMAS’ CALCULUS TWELFTH EDITION BASED ON THE ORIGINAL WORK BY

George B. Thomas, Jr. Massachusetts Institute of Technology

AS

REVISED BY

Maurice D. Weir Naval Postgraduate School

Joel Hass

University of California, Davis

608070 _ISM_ThomasCalc_WeirHass_ttl.qxd:harsh_569709_ttl

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3:11 PM

Page 2

This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.

The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Pearson Addison-Wesley from electronic files supplied by the author. Copyright © 2010, 2005, 2001 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-13: 978-0-321-60807-9 ISBN-10: 0-321-60807-0 1 2 3 4 5 6 BB 12 11 10 09

PREFACE TO THE INSTRUCTOR This Instructor's Solutions Manual contains the solutions to every exercise in the 12th Edition of THOMAS' CALCULUS by Maurice Weir and Joel Hass, including the Computer Algebra System (CAS) exercises. The corresponding Student's Solutions Manual omits the solutions to the even-numbered exercises as well as the solutions to the CAS exercises (because the CAS command templates would give them all away). In addition to including the solutions to all of the new exercises in this edition of Thomas, we have carefully revised or rewritten every solution which appeared in previous solutions manuals to ensure that each solution ì conforms exactly to the methods, procedures and steps presented in the text ì is mathematically correct ì includes all of the steps necessary so a typical calculus student can follow the logical argument and algebra ì includes a graph or figure whenever called for by the exercise, or if needed to help with the explanation ì is formatted in an appropriate style to aid in its understanding Every CAS exercise is solved in both the MAPLE and MATHEMATICA computer algebra systems. A template showing an example of the CAS commands needed to execute the solution is provided for each exercise type. Similar exercises within the text grouping require a change only in the input function or other numerical input parameters associated with the problem (such as the interval endpoints or the number of iterations). For more information about other resources available with Thomas' Calculus, visit http://pearsonhighered.com.

TABLE OF CONTENTS 1 Functions 1 1.1 1.2 1.3 1.4

Functions and Their Graphs 1 Combining Functions; Shifting and Scaling Graphs 8 Trigonometric Functions 19 Graphing with Calculators and Computers 26 Practice Exercises 30 Additional and Advanced Exercises 38

2 Limits and Continuity 43 2.1 2.2 2.3 2.4 2.5 2.6

Rates of Change and Tangents to Curves 43 Limit of a Function and Limit Laws 46 The Precise Definition of a Limit 55 One-Sided Limits 63 Continuity 67 Limits Involving Infinity; Asymptotes of Graphs 73 Practice Exercises 82 Additional and Advanced Exercises 86

3 Differentiation 93 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9

Tangents and the Derivative at a Point 93 The Derivative as a Function 99 Differentiation Rules 109 The Derivative as a Rate of Change 114 Derivatives of Trigonometric Functions 120 The Chain Rule 127 Implicit Differentiation 135 Related Rates 142 Linearizations and Differentials 146 Practice Exercises 151 Additional and Advanced Exercises 162

4 Applications of Derivatives 167 4.1 4.2 4.3 4.4 4.5 4.6 4.7

Extreme Values of Functions 167 The Mean Value Theorem 179 Monotonic Functions and the First Derivative Test 188 Concavity and Curve Sketching 196 Applied Optimization 216 Newton's Method 229 Antiderivatives 233 Practice Exercises 239 Additional and Advanced Exercises 251

5 Integration 257 5.1 5.2 5.3 5.4 5.5 5.6

Area and Estimating with Finite Sums 257 Sigma Notation and Limits of Finite Sums 262 The Definite Integral 268 The Fundamental Theorem of Calculus 282 Indefinite Integrals and the Substitution Rule 290 Substitution and Area Between Curves 296 Practice Exercises 310 Additional and Advanced Exercises 320

6 Applications of Definite Integrals 327 6.1 6.2 6.3 6.4 6.5 6.6

Volumes Using Cross-Sections 327 Volumes Using Cylindrical Shells 337 Arc Lengths 347 Areas of Surfaces of Revolution 353 Work and Fluid Forces 358 Moments and Centers of Mass 365 Practice Exercises 376 Additional and Advanced Exercises 384

7 Transcendental Functions 389 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8

Inverse Functions and Their Derivatives 389 Natural Logarithms 396 Exponential Functions 403 Exponential Change and Separable Differential Equations 414 ^ Indeterminate Forms and L'Hopital's Rule 418 Inverse Trigonometric Functions 425 Hyperbolic Functions 436 Relative Rates of Growth 443 Practice Exercises 447 Additional and Advanced Exercises 458

8 Techniques of Integration 461 8.1 8.2 8.3 8.4 8.5 8.6 8.7

Integration by Parts 461 Trigonometric Integrals 471 Trigonometric Substitutions 478 Integration of Rational Functions by Partial Fractions 484 Integral Tables and Computer Algebra Systems 491 Numerical Integration 502 Improper Integrals 510 Practice Exercises 518 Additional and Advanced Exercises 528

9 First-Order Differential Equations 537 9.1 9.2 9.3 9.4 9.5

Solutions, Slope Fields and Euler's Method 537 First-Order Linear Equations 543 Applications 546 Graphical Solutions of Autonomous Equations 549 Systems of Equations and Phase Planes 557 Practice Exercises 562 Additional and Advanced Exercises 567

10 Infinite Sequences and Series 569 10.1 Sequences 569 10.2 Infinite Series 577 10.3 The Integral Test 583 10.4 Comparison Tests 590 10.5 The Ratio and Root Tests 597 10.6 Alternating Series, Absolute and Conditional Convergence 602 10.7 Power Series 608 10.8 Taylor and Maclaurin Series 617 10.9 Convergence of Taylor Series 621 10.10 The Binomial Series and Applications of Taylor Series 627 Practice Exercises 634 Additional and Advanced Exercises 642

TABLE OF CONTENTS 10 Infinite Sequences and Series 569 10.1 Sequences 569 10.2 Infinite Series 577 10.3 The Integral Test 583 10.4 Comparison Tests 590 10.5 The Ratio and Root Tests 597 10.6 Alternating Series, Absolute and Conditional Convergence 602 10.7 Power Series 608 10.8 Taylor and Maclaurin Series 617 10.9 Convergence of Taylor Series 621 10.10 The Binomial Series and Applications of Taylor Series 627 Practice Exercises 634 Additional and Advanced Exercises 642

11 Parametric Equations and Polar Coordinates 647 11.1 11.2 11.3 11.4 11.5 11.6 11.7

Parametrizations of Plane Curves 647 Calculus with Parametric Curves 654 Polar Coordinates 662 Graphing in Polar Coordinates 667 Areas and Lengths in Polar Coordinates 674 Conic Sections 679 Conics in Polar Coordinates 689 Practice Exercises 699 Additional and Advanced Exercises 709

12 Vectors and the Geometry of Space 715 12.1 12.2 12.3 12.4 12.5 12.6

Three-Dimensional Coordinate Systems 715 Vectors 718 The Dot Product 723 The Cross Product 728 Lines and Planes in Space 734 Cylinders and Quadric Surfaces 741 Practice Exercises 746 Additional Exercises 754

13 Vector-Valued Functions and Motion in Space 759 13.1 13.2 13.3 13.4 13.5 13.6

Curves in Space and Their Tangents 759 Integrals of Vector Functions; Projectile Motion 764 Arc Length in Space 770 Curvature and Normal Vectors of a Curve 773 Tangential and Normal Components of Acceleration 778 Velocity and Acceleration in Polar Coordinates 784 Practice Exercises 785 Additional Exercises 791

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

14 Partial Derivatives 795 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8 14.9 14.10

Functions of Several Variables 795 Limits and Continuity in Higher Dimensions 804 Partial Derivatives 810 The Chain Rule 816 Directional Derivatives and Gradient Vectors 824 Tangent Planes and Differentials 829 Extreme Values and Saddle Points 836 Lagrange Multipliers 849 Taylor's Formula for Two Variables 857 Partial Derivatives with Constrained Variables 859 Practice Exercises 862 Additional Exercises 876

15 Multiple Integrals 881 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8

Double and Iterated Integrals over Rectangles 881 Double Integrals over General Regions 882 Area by Double Integration 896 Double Integrals in Polar Form 900 Triple Integrals in Rectangular Coordinates 904 Moments and Centers of Mass 909 Triple Integrals in Cylindrical and Spherical Coordinates 914 Substitutions in Multiple Integrals 922 Practice Exercises 927 Additional Exercises 933

16 Integration in Vector Fields 939 16.1 16.2 16.3 16.4 16.5 16.6 16.7 16.8

Line Integrals 939 Vector Fields and Line Integrals; Work, Circulation, and Flux 944 Path Independence, Potential Functions, and Conservative Fields 952 Green's Theorem in the Plane 957 Surfaces and Area 963 Surface Integrals 972 Stokes's Theorem 980 The Divergence Theorem and a Unified Theory 984 Practice Exercises 989 Additional Exercises 997

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

CHAPTER 1 FUNCTIONS 1.1 FUNCTIONS AND THEIR GRAPHS 1. domain œ (
_ß _); range œ [1ß _)

2. domain œ [0ß _); range œ (
_ß 1]

3. domain œ Ò
2ß _); y in range and y œ È5x  10   ! Ê y can be any positive real number Ê range œ Ò!ß _). 4. domain œ (
_ß 0Ó  Ò3, _); y in range and y œ Èx2
3x   ! Ê y can be any positive real number Ê range œ Ò!ß _). 5. domain œ (
_ß 3Ñ  Ð3, _); y in range and y œ Ê3
t!Ê

4 3
t

4 3
t,

now if t  3 Ê 3
t  ! Ê

4 3
t

 !, or if t  3

 ! Ê y can be any nonzero real number Ê range œ Ð
_ß 0Ñ  Ð!ß _).

6. domain œ (
_ß
%Ñ  Ð
4, 4Ñ  Ð4, _); y in range and y œ 2


%  t  4 Ê
16 Ÿ t
16  ! Ê nonzero real number Ê range œ Ð
_ß

#
"'
18 Ó

Ÿ

2 t2
16

2 t2
16 ,

2 t2
16

now if t 
% Ê t2
16  ! Ê 2

 !, or if t  % Ê t
16  ! Ê

2 t2
16

 !, or if

 ! Ê y can be any

 Ð!ß _).

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. 8. (a) Not the graph of a function of x since it fails the vertical line test. (b) Not the graph of a function of x since it fails the vertical line test. #

9. base œ x; (height)#  ˆ #x ‰ œ x# Ê height œ

È3 #

x; area is a(x) œ

" #

(base)(height) œ

" #

(x) Š

È3 # x‹

œ

È3 4

x# ;

perimeter is p(x) œ x  x  x œ 3x. 10. s œ side length Ê s#  s# œ d# Ê s œ

d È2

; and area is a œ s# Ê a œ

" #

d#

11. Let D œ diagonal length of a face of the cube and j œ the length of an edge. Then j#  D# œ d# and D# œ 2j# Ê 3j# œ d# Ê j œ

d È3

. The surface area is 6j# œ

6d# 3

12. The coordinates of P are ˆxß Èx‰ so the slope of the line joining P to the origin is m œ ˆx, Èx‰ œ ˆ m"# ,

#

œ 2d# and the volume is j$ œ Š d3 ‹ Èx x

œ

" Èx

$Î#

œ

(x  0). Thus,

"‰ m .

13. 2x  4y œ 5 Ê y œ
"# x  54 ; L œ ÈÐx
0Ñ2  Ðy
0Ñ2 œ Éx2  Ð
"# x  54 Ñ2 œ Éx2  4" x2
54 x  œ É 54 x2
54 x 

25 16

œ É 20x

2


20x  25 16

œ

È20x2
20x  25 4

14. y œ Èx
3 Ê y2  3 œ x; L œ ÈÐx
4Ñ2  Ðy
0Ñ2 œ ÈÐy2  3
4Ñ2  y2 œ ÈÐy2
1Ñ2  y2 œ Èy4
2y2  1  y2 œ Èy4
y2  1

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

d$ 3È 3

25 16

.

2

Chapter 1 Functions

15. The domain is a
_ß _b.

16. The domain is a
_ß _b.

17. The domain is a
_ß _b.

18. The domain is Ð
_ß !Ó.

19. The domain is a
_ß !b  a!ß _b.

20. The domain is a
_ß !b  a!ß _b.

21. The domain is a
_ß
5b  Ð
5ß
3Ó  Ò3, 5Ñ  a5, _b 22. The range is Ò2, 3Ñ. 23. Neither graph passes the vertical line test (a)

(b)

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 1.1 Functions and Their Graphs 24. Neither graph passes the vertical line test (a)

(b)

Ú xyœ" Þ Ú yœ1
x Þ or or kx  yk œ 1 Í Û Í Û ß ß Ü x  y œ
" à Ü y œ
"
x à 25.

x y

0 0

1 1

27. Faxb œ œ

2 0

26.

x y

0 1

1 0

2 0

" , x0 28. Gaxb œ œ x x, 0 Ÿ x

4
x2 , x Ÿ 1 x2  2x, x  1

29. (a) Line through a!ß !b and a"ß "b: y œ x; Line through a"ß "b and a#ß !b: y œ
x  2 x, 0 Ÿ x Ÿ 1 f(x) œ œ
x  2, 1  x Ÿ 2 Ú Ý 2, ! Ÿ x  " Ý !ß " Ÿ x  # (b) f(x) œ Û Ý Ý 2ß # Ÿ x  $ Ü !ß $ Ÿ x Ÿ % 30. (a) Line through a!ß 2b and a#ß !b: y œ
x  2 " Line through a2ß "b and a&ß !b: m œ !&

# œ
x  #, 0  x Ÿ # f(x) œ œ "
$ x  &$ , #  x Ÿ &

f(x) œ œ

œ
"$ , so y œ
"$ ax
2b  " œ
"$ x 


$
! !
Ð
"Ñ œ
"
$
% #
! œ #

(b) Line through a
"ß !b and a!ß
$b: m œ Line through a!ß $b and a#ß
"b: m œ


" $

& $


$, so y œ
$x
$ œ
#, so y œ
#x  $


$x
$,
"  x Ÿ !
#x  $, !  x Ÿ #

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

3

4

Chapter 1 Functions

31. (a) Line through a
"ß "b and a!ß !b: y œ
x Line through a!ß "b and a"ß "b: y œ " Line through a"ß "b and a$ß !b: m œ !
" $
" œ Ú
x
" Ÿ x  ! " !xŸ" f(x) œ Û Ü
"# x  $# "x$


" #

(b) Line through a
2ß
1b and a0ß 0b: y œ 12 x Line through a0ß 2b and a1ß 0b: y œ
2x  2 Line through a1ß
1b and a3ß
1b: y œ
1 32. (a) Line through ˆ T# ß !‰ and aTß "b: m œ faxb œ


(b)

"
! T
aTÎ#b

œ
"# , so y œ
"# ax
"b  " œ
"# x 

Ú

1 2x

faxb œ Û
2x  2 Ü
1

$ #


2 Ÿ x Ÿ 0 0xŸ1 1xŸ3

œ T# , so y œ T# ˆx
T# ‰  0 œ T# x
"

!, 0 Ÿ x Ÿ T# # T T x
", #  x Ÿ T

Ú A, Ý Ý Ý
Aß faxb œ Û Aß Ý Ý Ý Ü
Aß

! Ÿ x  T# T # Ÿx T T Ÿ x  $#T $T # Ÿ x Ÿ #T

33. (a) ÚxÛ œ 0 for x − [0ß 1)

(b) ÜxÝ œ 0 for x − (
1ß 0]

34. ÚxÛ œ ÜxÝ only when x is an integer. 35. For any real number x, n Ÿ x Ÿ n  ", where n is an integer. Now: n Ÿ x Ÿ n  " Ê
Ðn  "Ñ Ÿ
x Ÿ
n. By definition: Ü
xÝ œ
n and ÚxÛ œ n Ê
ÚxÛ œ
n. So Ü
xÝ œ
ÚxÛ for all x − d . 36. To find f(x) you delete the decimal or fractional portion of x, leaving only the integer part.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 1.1 Functions and Their Graphs 37. Symmetric about the origin Dec:
_  x  _ Inc: nowhere

38. Symmetric about the y-axis Dec:
_  x  ! Inc: !  x  _

39. Symmetric about the origin Dec: nowhere Inc:
_  x  ! !x_

40. Symmetric about the y-axis Dec: !  x  _ Inc:
_  x  !

41. Symmetric about the y-axis Dec:
_  x Ÿ ! Inc: !  x  _

42. No symmetry Dec:
_  x Ÿ ! Inc: nowhere

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5

6

Chapter 1 Functions

43. Symmetric about the origin Dec: nowhere Inc:
_  x  _

44. No symmetry Dec: ! Ÿ x  _ Inc: nowhere

45. No symmetry Dec: ! Ÿ x  _ Inc: nowhere

46. Symmetric about the y-axis Dec:
_  x Ÿ ! Inc: !  x  _

47. 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. 48. faxb œ x
& œ

" x&

and fa
xb œ a
xb
& œ

" a
x b&

œ
ˆ x"& ‰ œ
faxb. Thus the function is odd.

49. Since faxb œ x#  " œ a
xb#  " œ
faxb. The function is even. 50. Since Òfaxb œ x#  xÓ Á Òfa
xb œ a
xb#
xÓ and Òfaxb œ x#  xÓ Á Ò
faxb œ
axb#
xÓ the function is neither even nor odd. 51. Since gaxb œ x$  x, ga
xb œ
x$
x œ
ax$  xb œ
gaxb. So the function is odd. 52. gaxb œ x%  $x#
" œ a
xb%  $a
xb#
" œ ga
xbß thus the function is even. 53. gaxb œ

" x#
"

54. gaxb œ

x x#
" ;

55. hatb œ

" t
";

œ

" a
xb#
"

œ ga
xb. Thus the function is even.

ga
xb œ
x#x
" œ
gaxb. So the function is odd.

ha
tb œ

"
t
" ;


h at b œ

" "
t.

Since hatb Á
hatb and hatb Á ha
tb, the function is neither even nor odd.

56. Since l t$ | œ l a
tb$ |, hatb œ ha
tb and the function is even.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 1.1 Functions and Their Graphs 57. hatb œ 2t  ", ha
tb œ
2t  ". So hatb Á ha
tb.
hatb œ
2t
", so hatb Á
hatb. The function is neither even nor odd. 58. hatb œ 2l t l  " and ha
tb œ 2l
t l  " œ 2l t l  ". So hatb œ ha
tb and the function is even. 59. s œ kt Ê 25 œ kÐ75Ñ Ê k œ

" 3

Ê s œ 3" t; 60 œ 3" t Ê t œ 180

60. K œ c v# Ê 12960 œ ca18b2 Ê c œ 40 Ê K œ 40v# ; K œ 40a10b# œ 4000 joules 61. r œ 62. P œ

k s

Ê6œ

k v

k 4

Ê k œ 24 Ê r œ

Ê 14.7 œ

k 1000

24 s ;

10 œ

24 s

Ê k œ 14700 Ê P œ

Êsœ

14700 v ;

12 5

23.4 œ

14700 v

Êvœ

24500 39

¸ 628.2 in3

63. v œ f(x) œ xÐ"%
2xÑÐ22
2xÑ œ %x$
72x#  $!)x; !  x  7Þ 64. (a) Let h œ height of the triangle. Since the triangle is isosceles, AB #  AB # œ 2# Ê AB œ È2Þ So, #

h#  "# œ ŠÈ2‹ Ê h œ " Ê B is at a!ß "b Ê slope of AB œ
" Ê The equation of AB is y œ f(x) œ
x  "; x − Ò!ß "Ó. (b) AÐxÑ œ 2x y œ 2xÐ
x  "Ñ œ
2x#  #x; x − Ò!ß "Ó. 65. (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. 66. (a) Graph f because it is linear. (b) Graph g because it contains a!ß "b. (c) Graph h because it is a nonlinear odd function. x #

67. (a) From the graph, (b)

x #

1

x  0:

x #

x  0:

x 2

4 x


1


Ê 4 x

x #



1

4 x

Ê x − (
2ß 0)  (%ß _)


1
4x  0 # 2x
8 0 Ê x
2x

0 Ê

(x
4)(x2) #x

0

(x
4)(x2) #x

0

Ê x  4 since x is positive;
1


4 x

0 Ê

x#
2x
8 2x

0 Ê

Ê x 
2 since x is negative; sign of (x
4)(x  2)  
ïïïïïðïïïïïðïïïïî
2 % Solution interval: (
#ß 0)  (%ß _)

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

7

8

Chapter 1 Functions 3 2 x
1  x  1 3 2 x
1  x  1

68. (a) From the graph, (b) Case x 
1:

Ê x − (
_ß
5)  (
1ß 1) Ê

3(x1) x
1

2

Ê 3x  3  2x
2 Ê x 
5. Thus, x − (
_ß
5) solves the inequality. Case
1  x  1:

3 x
1



2 x 1

Ê

3(x1) x
1

2

Ê 3x  3  2x
2 Ê x 
5 which is true if x 
1. Thus, x − (
1ß 1) solves the inequality. 3 Case 1  x: x
1  x2 1 Ê 3x  3  2x
2 Ê x 
5 which is never true if 1  x, so no solution here. In conclusion, x − (
_ß
5)  (
1ß 1). 69. A curve symmetric about the x-axis will not pass the vertical line test because the points ax, yb and ax,
yb lie on the same vertical line. The graph of the function y œ faxb œ ! is the x-axis, a horizontal line for which there is a single y-value, !, for any x. 70. price œ 40  5x, quantity œ 300
25x Ê Raxb œ a40  5xba300
25xb 71. x2  x2 œ h2 Ê x œ

h È2

œ

È2 h 2 ;

cost œ 5a2xb  10h Ê Cahb œ 10Š

È2 h 2 ‹

 10h œ 5hŠÈ2  2‹

72. (a) Note that 2 mi = 10,560 ft, so there are È800#  x# feet of river cable at $180 per foot and a10,560
xb feet of land cable at $100 per foot. The cost is Caxb œ 180È800#  x#  100a10,560
xb. (b) Ca!b œ $"ß #!!ß !!! Ca&!!b ¸ $"ß "(&ß )"# Ca"!!!b ¸ $"ß ")'ß &"# Ca"&!!b ¸ $"ß #"#ß !!! Ca#!!!b ¸ $"ß #%$ß ($# Ca#&!!b ¸ $"ß #()ß %(* Ca$!!!b ¸ $"ß $"%ß )(! Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from the point P. 1.2 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS 1. Df :
_  x  _, Dg : x   1 Ê Df

g

œ Dfg : x   1. Rf :
_  y  _, Rg : y   0, Rf g : y   1, Rfg : y   0

2. Df : x  1   0 Ê x  
1, Dg : x
1   0 Ê x   1. Therefore Df Rf œ Rg : y   0, Rf g : y   È2, Rfg : y   0

g

œ Dfg : x   1.

3. Df :
_  x  _, Dg :
_  x  _, DfÎg :
_  x  _, DgÎf :
_  x  _, Rf : y œ 2, Rg : y   1, RfÎg : 0  y Ÿ 2, RgÎf : "# Ÿ y  _ 4. Df :
_  x  _, Dg : x   0 , DfÎg : x   0, DgÎf : x   0; Rf : y œ 1, Rg : y   1, RfÎg : 0  y Ÿ 1, RgÎf : 1 Ÿ y  _ 5. (a) 2 (d) (x  5)#
3 œ x#  10x  22 (g) x  10

(b) 22 (e) 5 (h) (x#
3)#
3 œ x%
6x#  6

(c) x#  2 (f)
2

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 1.2 Combining Functions; Shifting and Scaling Graphs 6. (a)
"3 (d)

(b) 2

" x

(c)

(e) 0

(g) x
2

(h)

(f)

" " x 1 1

œ

x x

"

# 1

x" x#

œ

" x 1 3 4


1œ


x x1

7. af‰g‰hbaxb œ fagahaxbbb œ faga4
xbb œ fa3a4
xbb œ fa12
3xb œ a12
3xb  1 œ 13
3x 8. af‰g‰hbaxb œ fagahaxbbb œ fagax2 bb œ fa2ax2 b
1b œ fa2x2
1b œ 3a2x2
1b  4 œ 6x2 1 9. af‰g‰hbaxb œ fagahaxbbb œ fˆgˆ 1x ‰‰ œ fŠ 1 1 % ‹ œ fˆ 1 x 4x ‰ œ É 1 x 4x  " œ É 15x4x" x

2

10. af‰g‰hbaxb œ fagahaxbbb œ fŠgŠÈ2
x‹‹ œ f

ŠÈ2
x‹ 2

ŠÈ2
x‹

 œ fˆ $
x ‰ œ 1 2
x

2 x $ x 2 3
$2 xx

8
3x 7
2x

œ

11. (a) af‰gbaxb (d) a j‰jbaxb

(b) a j‰gbaxb (e) ag‰h‰f baxb

(c) ag‰gbaxb (f) ah‰j‰f baxb

12. (a) af‰jbaxb (d) af‰f baxb

(b) ag‰hbaxb (e) a j‰g‰f baxb

(c) ah‰hbaxb (f) ag‰f‰hbaxb

g(x)

f(x)

(f ‰ g)(x)

(a)

x
7

Èx

Èx
7

(b)

x2

3x

3(x  2) œ 3x  6

(c)

x#

Èx
5

Èx#
5

(d)

x x
1

x x
1

" x
1 " x

1

13.

(e) (f)

" x

gaxb
" g ax b

œ

x x
(x
1)

œx

x

" x

x

" lx
"l .

14. (a) af‰gbaxb œ lgaxbl œ (b) af‰gbaxb œ

x x 1 x x 1
1

x x"

œ

Ê"


" g ax b

œ

x x"

Ê"


x x"

œ

" g ax b

Ê

" x"

œ

" gaxb ß so

gaxb œ x  ".

(c) Since af‰gbaxb œ Ègaxb œ lxl, gaxb œ x . (d) Since af‰gbaxb œ fˆÈx‰ œ l x l, faxb œ x# . (Note that the domain of the composite is Ò!ß _Ñ.) #

The completed table is shown. Note that the absolute value sign in part (d) is optional. gaxb faxb af‰gbaxb " " lxl x
" lx
"l x" x# Èx

x
" x

Èx #

x

15. (a) faga
1bb œ fa1b œ 1 (d) gaga2bb œ ga0b œ 0

x x"

lxl lxl (b) gafa0bb œ ga
2b œ 2 (e) gafa
2bb œ ga1b œ
1

(c) fafa
1bb œ fa0b œ
2 (f) faga1bb œ fa
1b œ 0

16. (a) faga0bb œ fa
1b œ 2
a
1b œ 3, where ga0b œ 0
1 œ
1 (b) gafa3bb œ ga
1b œ
a
1b œ 1, where fa3b œ 2
3 œ
1 (c) gaga
1bb œ ga1b œ 1
1 œ 0, where ga
1b œ
a
1b œ 1

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

9

10

Chapter 1 Functions (d) fafa2bb œ fa0b œ 2
0 œ 2, where fa2b œ 2
2 œ 0 (e) gafa0bb œ ga2b œ 2
1 œ 1, where fa0b œ 2
0 œ 2 (f) fˆgˆ "# ‰‰ œ fˆ
#" ‰ œ 2
ˆ
#" ‰ œ 5# , where gˆ "# ‰ œ "#
1 œ
"#

17. (a) af‰gbaxb œ fagaxbb œ É 1x  1 œ É 1 x x ag‰f baxb œ gafaxbb œ

1 Èx  1

(b) Domain af‰gb: Ð
_,
1Ó  Ð0, _Ñ, domain ag‰f b: Ð
1, _Ñ (c) Range af‰gb: Ð1, _Ñ, range ag‰f b: Ð0, _Ñ 18. (a) af‰gbaxb œ fagaxbb œ 1
2Èx  x ag‰f baxb œ gafaxbb œ 1
kxk (b) Domain af‰gb: Ò0, _Ñ, domain ag‰f b: Ð
_, _Ñ (c) Range af‰gb: Ð0, _Ñ, range ag‰f b: Ð
_, 1Ó 19. af‰gbaxb œ x Ê fagaxbb œ x Ê

g ax b g ax b
2

œ x Ê gaxb œ agaxb
2bx œ x † gaxb
2x

Ê gaxb
x † gaxb œ
2x Ê gaxb œ
1 2x
x œ

2x x
1

20. af‰gbaxb œ x  2 Ê fagaxbb œ x  2 Ê 2agaxbb3
4 œ x  2 Ê agaxbb3 œ 21. (a) y œ
(x  7)#

(b) y œ
(x
4)#

22. (a) y œ x#  3

(b) y œ x#
5

x6 2

3 x6 Ê gaxb œ É 2

23. (a) Position 4

(b) Position 1

(c) Position 2

(d) Position 3

24. (a) y œ
(x
1)#  4

(b) y œ
(x  2)#  3

(c) y œ
(x  4)#
1

(d) y œ
(x
2)#

25.

26.

27.

28.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 1.2 Combining Functions; Shifting and Scaling Graphs 29.

30.

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

11

12

Chapter 1 Functions

41.

42.

43.

44.

45.

46.

47.

48.

49.

50.

51.

52.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 1.2 Combining Functions; Shifting and Scaling Graphs 53.

54.

55. (a) domain: [0ß 2]; range: [#ß $]

(b) domain: [0ß 2]; range: [
1ß 0]

(c) domain: [0ß 2]; range: [0ß 2]

(d) domain: [0ß 2]; range: [
1ß 0]

(e) domain: [
2ß 0]; range: [!ß 1]

(f) domain: [1ß 3]; range: [!ß "]

(g) domain: [
2ß 0]; range: [!ß "]

(h) domain: [
1ß 1]; range: [!ß "]

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

13

14

Chapter 1 Functions

56. (a) domain: [0ß 4]; range: [
3ß 0]

(b) domain: [
4ß 0]; range: [!ß $]

(c) domain: [
4ß 0]; range: [!ß $]

(d) domain: [
4ß 0]; range: ["ß %]

(e) domain: [#ß 4]; range: [
3ß 0]

(f) domain: [
2ß 2]; range: [
3ß 0]

(g) domain: ["ß 5]; range: [
3ß 0]

(h) domain: [0ß 4]; range: [0ß 3]

58. y œ a2xb#
1 œ %x#
1

57. y œ 3x#
3 59. y œ "# ˆ" 

"‰ x#

œ

" #



" #x#

60. y œ 1 

" axÎ$b#

œ1

61. y œ È%x  1

62. y œ 3Èx  1

# 63. y œ É%
ˆ x# ‰ œ "# È16
x#

64. y œ "$ È%
x#

65. y œ "
a3xb$ œ "
27x$

66. y œ "
ˆ x# ‰ œ "


$

* x#

x$ )

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 1.2 Combining Functions; Shifting and Scaling Graphs 67. Let y œ
È#x  " œ faxb and let gaxb œ x"Î# , "Î# "Î# haxb œ ˆx  " ‰ , iaxb œ È#ˆx  " ‰ , and #

#

"Î# jaxb œ
’È#ˆx  "# ‰ “ œ faBb. The graph of

haxb is the graph of gaxb shifted left

" #

unit; the

graph of iaxb is the graph of haxb stretched vertically by a factor of È#; and the graph of jaxb œ faxb is the graph of iaxb reflected across the x-axis. 68. Let y œ È"


x #

œ faxbÞ Let gaxb œ a
xb"Î# ,

haxb œ a
x  #b"Î# , and iaxb œ œ È"


x #

" È # a
x

 #b"Î#

œ faxbÞ The graph of gaxb is the

graph of y œ Èx reflected across the x-axis. The graph of haxb is the graph of gaxb shifted right two units. And the graph of iaxb is the graph of haxb compressed vertically by a factor of È#. 69. y œ faxb œ x$ . Shift faxb one unit right followed by a shift two units up to get gaxb œ ax
"b3  #.

70. y œ a"
Bb$  # œ
Òax
"b$  a
#bÓ œ faxb. Let gaxb œ x$ , haxb œ ax
"b$ , iaxb œ ax
"b$  a
#b, and jaxb œ
Òax
"b$  a
#bÓ. The graph of haxb is the graph of gaxb shifted right one unit; the graph of iaxb is the graph of haxb shifted down two units; and the graph of faxb is the graph of iaxb reflected across the x-axis.

71. Compress the graph of faxb œ of 2 to get gaxb œ unit to get haxb œ

" #x . Then " #x
".

" x

horizontally by a factor

shift gaxb vertically down 1

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

15

16

Chapter 1 Functions

72. Let faxb œ œ

"

#

ŠxÎÈ#‹

" x#

and gaxb œ

"œ

# x#

"œ

" # ’Š"ÎÈ#‹B“

" # Š B# ‹

"

 "Þ Since

È# ¸ "Þ%, we see that the graph of faxb stretched horizontally by a factor of 1.4 and shifted up 1 unit is the graph of gaxb.

$ 73. Reflect the graph of y œ faxb œ È x across the x-axis $ to get gaxb œ
Èx.

74. y œ faxb œ a
#xb#Î$ œ Òa
"ba#bxÓ#Î$ œ a
"b#Î$ a#xb#Î$ œ a#xb#Î$ . So the graph of faxb is the graph of gaxb œ x#Î$ compressed horizontally by a factor of 2.

75.

76.

77. *x#  #&y# œ ##& Ê

x# &#



y# $#

œ"

78. "'x#  (y# œ ""# Ê

x# # È Š (‹



y# %#

œ"

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 1.2 Combining Functions; Shifting and Scaling Graphs 79. $x#  ay
#b# œ $ Ê

x# "#



a y
#b # #

ŠÈ$‹

80. ax  "b#  #y# œ % Ê

œ"

Ê

83.

x# "'

#

ŠÈ#‹



y# *




y
a
#b‘# #

ŠÈ$‹



# # 82. 'ˆx  $# ‰  *ˆy
"# ‰ œ &%

81. $ax
"b#  #ay  #b# œ ' ax
" b #


x
a
"b‘# ##

#

œ"

Ê

’x
ˆ
$# ‰“ $#



ˆy
"# ‰# #

ŠÈ'‹

œ"

œ " has its center at a!ß !b. Shiftinig 4 units

left and 3 units up gives the center at ah, kb œ a
%ß $b. #
x
a
4b‘#  ay
3#3b œ " 4# a y
$b # œ ". Center, C, is a
%ß 3#

So the equation is Ê

ax  % b # 4#



$b, and

major axis, AB, is the segment from a
)ß $b to a!ß $b.

84. The ellipse

x# %

y# #&



œ " has center ah, kb œ a!ß !b.

Shifting the ellipse 3 units right and 2 units down produces an ellipse with center at ah, kb œ a$ß
#b and an equation

ax
3 b# %




y
a
#b‘# #&

œ ". Center,

C, is a3ß
#b, and AB, the segment from a$ß $b to a$ß
(b is the major axis.

85. (a) (fg)(
x) œ f(
x)g(
x) œ f(x)(
g(x)) œ
(fg)(x), odd (b) Š gf ‹ (
x) œ

f(
x) g(
x)

œ

f(x)
g(x)

œ
Š gf ‹ (x), odd

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

y# # È Š #‹

œ"

17

18

Chapter 1 Functions (c) ˆ gf ‰ (
x) œ (d) (e) (f) (g) (h) (i)

g(
x) f(
x)

œ


g(x) f(x)

œ
ˆ gf ‰ (x), odd

f # (
x) œ f(
x)f(
x) œ f(x)f(x) œ f # (x), even g# (
x) œ (g(
x))# œ (
g(x))# œ g# (x), even (f ‰ g)(
x) œ f(g(
x)) œ f(
g(x)) œ f(g(x)) œ (f ‰ g)(x), even (g ‰ f)(
x) œ g(f(
x)) œ g(f(x)) œ (g ‰ f)(x), even (f ‰ f)(
x) œ f(f(
x)) œ f(f(x)) œ (f ‰ f)(x), even (g ‰ g)(
x) œ g(g(
x)) œ g(
g(x)) œ
g(g(x)) œ
(g ‰ g)(x), odd

86. Yes, f(x) œ 0 is both even and odd since f(
x) œ 0 œ f(x) and f(
x) œ 0 œ
f(x). 87. (a)

(b)

(c)

(d)

88.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 1.3 Trigonometric Functions 1.3 TRIGONOMETRIC FUNCTIONS 1. (a) s œ r) œ (10) ˆ 451 ‰ œ 81 m radians and

51 4

1 ‰ 3. ) œ 80° Ê ) œ 80° ˆ 180° œ

41 9

2. ) œ

s r

œ

101 8

œ

51 4

1 ‰ (b) s œ r) œ (10)(110°) ˆ 180° œ


1

)


231

0

1 #

s r

œ

30 50

31 4 " È2
È" 2

sin )

0

cos )


1

tan )

0

È3

0

und.


"

und.

" È3

und.

0


1

und.


È 2


1


#

und.


È23

sec ) csc )

0

"

"

0

" und.

7. cos x œ
45 , tan x œ
34 9. sin x œ


È8 3

, tan x œ
È8

"

6.

È2


3#1

)


1'

sin )

"

cos )

!

" #

tan )

und.


È 3

cot )

!


È"3

sec )

und.

#

csc )

"


È23

8. sin x œ

2 È5

10. sin x œ

12 13

13.

14.

period œ 1


13

È
#3

12. cos x œ


, cos x œ

" È2

&1 ' " # È
#3


È"3

"


È"3


È 3

"


È 3

2 È3

È2


È23


#

È2

#


"# È3 #

" È5

, tan x œ
12 5 È3 #

, tan x œ

" È3

period œ 41 16.

period œ 2

m

‰ ¸ 34° œ 0.6 rad or 0.6 ˆ 180° 1

11. sin x œ
È"5 , cos x œ
È25

15.

551 9

Ê s œ (6) ˆ 491 ‰ œ 8.4 in. (since the diameter œ 12 in. Ê radius œ 6 in.)

È
#3
"#

cot )

œ

ˆ 180° ‰ œ 225° 1

4. d œ 1 meter Ê r œ 50 cm Ê ) œ 5.

1101 18

period œ 4

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

1 % " È2

19

20

Chapter 1 Functions

17.

18.

period œ 6

period œ 1

19.

20.

period œ 21

period œ 21

21.

22.

period œ 21

period œ 21

23. period œ 1# , symmetric about the origin

24. period œ 1, symmetric about the origin

25. period œ 4, symmetric about the s-axis

26. period œ 41, symmetric about the origin

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 1.3 Trigonometric Functions 27. (a) Cos x and sec x are positive for x in the interval ˆ
12 , 12 ‰; and cos x and sec x are negative for x in the intervals ˆ
321 ,
12 ‰ and ˆ 12 , 321 ‰. Sec x is undefined when cos x is 0. The range of sec x is (
_ß
1]  ["ß _); the range of cos x is [
"ß 1]. (b) Sin x and csc x are positive for x in the intervals ˆ
321 ,
1‰ and a!, 1b; and sin x and csc x are negative for x in the intervals a
1, !b and ˆ1, 321 ‰. Csc x is undefined when sin x is 0. The range of csc x is (
_ß
1]  [1ß _); the range of sin x is [
"ß "].

28. Since cot x œ

" tan x

, cot x is undefined when tan x œ 0

and is zero when tan x is undefined. As tan x approaches zero through positive values, cot x approaches infinity. Also, cot x approaches negative infinity as tan x approaches zero through negative values.

29. D:
_  x  _; R: y œ
1, 0, 1

30. D:
_  x  _; R: y œ
1, 0, 1

31. cos ˆx
1# ‰ œ cos x cos ˆ
1# ‰
sin x sin ˆ
1# ‰ œ (cos x)(0)
(sin x)(
1) œ sin x 32. cos ˆx  1# ‰ œ cos x cos ˆ 1# ‰
sin x sin ˆ 1# ‰ œ (cos x)(0)
(sin x)(1) œ
sin x 33. sin ˆx  1# ‰ œ sin x cos ˆ 1# ‰  cos x sin ˆ 1# ‰ œ (sin x)(0)  (cos x)(1) œ cos x 34. sin ˆx
1# ‰ œ sin x cos ˆ
1# ‰  cos x sin ˆ
1# ‰ œ (sin x)(0)  (cos x)(
1) œ
cos x 35. cos (A
B) œ cos (A  (
B)) œ cos A cos (
B)
sin A sin (
B) œ cos A cos B
sin A (
sin B) œ cos A cos B  sin A sin B 36. sin (A
B) œ sin (A  (
B)) œ sin A cos (
B)  cos A sin (
B) œ sin A cos B  cos A (
sin B) œ sin A cos B
cos A sin B 37. If B œ A, A
B œ 0 Ê cos (A
B) œ cos 0 œ 1. Also cos (A
B) œ cos (A
A) œ cos A cos A  sin A sin A œ cos# A  sin# A. Therefore, cos# A  sin# A œ 1.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

21

22

Chapter 1 Functions

38. If B œ 21, then cos (A  21) œ cos A cos 21
sin A sin 21 œ (cos A)(1)
(sin A)(0) œ cos A and sin (A  21) œ sin A cos 21  cos A sin 21 œ (sin A)(1)  (cos A)(0) œ sin A. The result agrees with the fact that the cosine and sine functions have period 21. 39. cos (1  x) œ cos 1 cos B
sin 1 sin x œ (
1)(cos x)
(0)(sin x) œ
cos x 40. sin (21
x) œ sin 21 cos (
x)  cos (21) sin (
x) œ (0)(cos (
x))  (1)(sin (
x)) œ
sin x 41. sin ˆ 3#1
x‰ œ sin ˆ 3#1 ‰ cos (
x)  cos ˆ 3#1 ‰ sin (
x) œ (
1)(cos x)  (0)(sin (
x)) œ
cos x 42. cos ˆ 3#1  x‰ œ cos ˆ 3#1 ‰ cos x
sin ˆ 3#1 ‰ sin x œ (0)(cos x)
(
1)(sin x) œ sin x œ sin ˆ 14  13 ‰ œ sin

44. cos

111 1#

45. cos

1 12

œ cos ˆ 13
14 ‰ œ cos

46. sin

51 1#

œ sin ˆ 231
14 ‰ œ sin ˆ 231 ‰ cos ˆ
14 ‰  cos ˆ 231 ‰ sin ˆ
14 ‰ œ Š

21 ‰ 3

È2

1 8

œ

1
cos ˆ 281 ‰ #

œ

1
#

49. sin#

1 1#

œ

1  cos ˆ 211# ‰ #

œ

1 #

3 4

Ê sin ) œ „

52. sin2 ) œ cos2 ) Ê

sin2 ) cos2 )

1 4

œ cos

47. cos#

51. sin2 ) œ

cos

#

È3 #

È3 2

œ

1 3

 cos

cos

21 3

1 4

sin


sin

1 4

cos ˆ
14 ‰
sin

1 3

1 3

È2 È3 # ‹Š # ‹

71 1#

œ cos ˆ 14 

1 4

È2 ˆ"‰ # ‹ #

43. sin

œŠ

sin 1 3

21 3

œŠ

Š

È2 ˆ "‰ # ‹
#

sin ˆ
14 ‰ œ ˆ "# ‰ Š


Š

È2 # ‹

œ

È2 È3 # ‹Š # ‹


Š

È3 È2 # ‹Š # ‹

2
È2 4

48. cos#

51 1#

œ

1‰ 1
cos ˆ 10 1# #

œ

2  È3 4

50. sin#

31 8

œ

1  cos ˆ 681 ‰ #

Ê tan2 ) œ 1 Ê tan ) œ „ 1 Ê ) œ 14 ,

31 51 71 4 , 4 , 4

cos2 ) cos2 )

œ


È3 È2 # ‹ Š
# ‹

œ

Ê ) œ 13 ,

È 6
È 2 4 È 2
È 6 4 1
È 3 2È 2

œ

 ˆ
"# ‰ Š
œ

œ

1
Š

È3 ‹ #

#

1  Š #

È2 ‹ #

È2 # ‹

œ

œ

œ

1
È 3 2È 2

2  È3 4

2
È2 4

21 41 51 3 , 3 , 3

53. sin 2)
cos ) œ 0 Ê 2sin ) cos )
cos ) œ 0 Ê cos )a2sin )
1b œ 0 Ê cos ) œ 0 or 2sin )
1 œ 0 Ê cos ) œ 0 or sin ) œ "# Ê ) œ 12 , 321 , or ) œ 16 , 561 Ê ) œ 16 , 12 , 561 , 321 54. cos 2)  cos ) œ 0 Ê 2cos2 )
1  cos ) œ 0 Ê 2cos2 )  cos )
1 œ 0 Ê acos )  1ba2cos )
1b œ 0 Ê cos )  1 œ 0 or 2cos )
1 œ 0 Ê cos ) œ
1 or cos ) œ "# Ê ) œ 1 or ) œ 13 , 531 Ê ) œ 13 , 1, 531 55. tan (A  B) œ

sin (A
B) cos (A
B)

œ

sin A cos B
cos A cos B cos A cos Bsin A sin B

œ

sin A cos B cos A sin B cos A cos B
cos A cos B sin A sin B cos A cos B  cos A cos B cos A cos B

œ

tan A
tan B 1tan A tan B

56. tan (A
B) œ

sin (AB) cos (AB)

œ

sin A cos Bcos A cos B cos A cos B
sin A sin B

œ

sin A cos B cos A sin B cos A cos B  cos A cos B sin A sin B cos A cos B
cos A cos B cos A cos B

œ

tan Atan B 1
tan A tan B

57. According to the figure in the text, we have the following: By the law of cosines, c# œ a#  b#
2ab cos ) œ 1#  1#
2 cos (A
B) œ 2
2 cos (A
B). By distance formula, c# œ (cos A
cos B)#  (sin A
sin B)# œ cos# A
2 cos A cos B  cos# B  sin# A
2 sin A sin B  sin# B œ 2
2(cos A cos B  sin A sin B). Thus c# œ 2
2 cos (A
B) œ 2
2(cos A cos B  sin A sin B) Ê cos (A
B) œ cos A cos B  sin A sin B.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 1.3 Trigonometric Functions 58. (a) cosaA
Bb œ cos A cos B  sin A sin B sin ) œ cosˆ 1#
)‰ and cos ) œ sinˆ 1#
)‰ Let ) œ A  B

sinaA  Bb œ cos’ 1#
aA  Bb“ œ cos’ˆ 1#
A‰
B“ œ cos ˆ 1#
A‰ cos B  sin ˆ 1#
A‰ sin B œ sin A cos B  cos A sin B (b) cosaA
Bb œ cos A cos B  sin A sin B cosaA
a
Bbb œ cos A cos a
Bb  sin A sin a
Bb Ê cosaA  Bb œ cos A cos a
Bb  sin A sin a
Bb œ cos A cos B  sin A a
sin Bb œ cos A cos B
sin A sin B Because the cosine function is even and the sine functions is odd. 59. c# œ a#  b#
2ab cos C œ 2#  3#
2(2)(3) cos (60°) œ 4  9
12 cos (60°) œ 13
12 ˆ "# ‰ œ 7. Thus, c œ È7 ¸ 2.65. 60. c# œ a#  b#
2ab cos C œ 2#  3#
2(2)(3) cos (40°) œ 13
12 cos (40°). Thus, c œ È13
12 cos 40° ¸ 1.951. 61. From the figures in the text, we see that sin B œ hc . If C is an acute angle, then sin C œ hb . On the other hand, if C is obtuse (as in the figure on the right), then sin C œ sin (1
C) œ hb . Thus, in either case, h œ b sin C œ c sin B Ê ah œ ab sin C œ ac sin B. a #
b # c # 2ab

By the law of cosines, cos C œ

and cos B œ

a #
c #  b # . 2ac

Moreover, since the sum of the

interior angles of a triangle is 1, we have sin A œ sin (1
(B  C)) œ sin (B  C) œ sin B cos C  cos B sin C #

#

#

#

#

#

b c c b ˆ h ‰ h ‰ œ ˆ hc ‰ ’ a
2ab a2a#  b#
c#  c#
b# b œ “  ’ a
2ac “ b œ ˆ 2abc

ah bc

Ê ah œ bc sin A.

Combining our results we have ah œ ab sin C, ah œ ac sin B, and ah œ bc sin A. Dividing by abc gives h sin A sin C sin B bc œ ðóóóóóóóñóóóóóóóò a œ c œ b . law of sines 62. By the law of sines,

sin A #

œ

sin B 3

œ

È3/2 c .

By Exercise 61 we know that c œ È7. Thus sin B œ

3È 3 2È 7

¶ 0.982.

63. From the figure at the right and the law of cosines, b# œ a#  2#
2(2a) cos B œ a#  4
4a ˆ "# ‰ œ a#
2a  4. Applying the law of sines to the figure, Ê

È2/2 a

œ

È3/2 b

sin A a

œ

sin B b

Ê b œ É 3# a. Thus, combining results,

a#
2a  4 œ b# œ

3 #

a# Ê 0 œ

" #

a#  2a
4

Ê 0 œ a#  4a
8. From the quadratic formula and the fact that a  0, we have aœ

4
È4# 4(1)(8) #

œ

4 È 3 4 #

¶ 1.464.

64. (a) The graphs of y œ sin x and y œ x nearly coincide when x is near the origin (when the calculator is in radians mode). (b) In degree mode, when x is near zero degrees the sine of x is much closer to zero than x itself. The curves look like intersecting straight lines near the origin when the calculator is in degree mode.

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23

24

Chapter 1 Functions

65. A œ 2, B œ 21, C œ
1, D œ
1

66. A œ "# , B œ 2, C œ 1, D œ

" #

67. A œ
12 , B œ 4, C œ 0, D œ

68. A œ

L 21 ,

" 1

B œ L, C œ 0, D œ 0

69-72. Example CAS commands: Maple f := x -> A*sin((2*Pi/B)*(x-C))+D1; A:=3; C:=0; D1:=0; f_list := [seq( f(x), B=[1,3,2*Pi,5*Pi] )]; plot( f_list, x=-4*Pi..4*Pi, scaling=constrained, color=[red,blue,green,cyan], linestyle=[1,3,4,7], legend=["B=1","B=3","B=2*Pi","B=3*Pi"], title="#69 (Section 1.3)" ); Mathematica Clear[a, b, c, d, f, x] f[x_]:=a Sin[21/b (x
c)] + d Plot[f[x]/.{a Ä 3, b Ä 1, c Ä 0, d Ä 0}, {x,
41, 41 }]

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Section 1.3 Trigonometric Functions 69. (a) The graph stretches horizontally.

(b) The period remains the same: period œ l B l. The graph has a horizontal shift of

" #

period.

70. (a) The graph is shifted right C units.

(b) The graph is shifted left C units. (c) A shift of „ one period will produce no apparent shift. l C l œ ' 71. (a) The graph shifts upwards l D lunits for D  ! (b) The graph shifts down l D lunits for D  !Þ

72. (a) The graph stretches l A l units.

(b) For A  !, the graph is inverted.

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25

26

Chapter 1 Functions

1.4 GRAPHING WITH CALCULATORS AND COMPUTERS 1-4.

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

1. d.

2. c.

3. d.

4. b.

5-30.

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

5. Ò
2ß 5Ó by Ò
15ß 40Ó

6. Ò
4ß 4Ó by Ò
4ß 4Ó

7. Ò
2ß 6Ó by Ò
250ß 50Ó

8. Ò
1ß 5Ó by Ò
5ß 30Ó

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 1.4 Graphing with Calculators and Computers 9. Ò
4ß 4Ó by Ò
5ß 5Ó

10. Ò
2ß 2Ó by Ò
2ß 8Ó

11. Ò
2ß 6Ó by Ò
5ß 4Ó

12. Ò
4ß 4Ó by Ò
8ß 8Ó

13. Ò
1ß 6Ó by Ò
1ß 4Ó

14. Ò
1ß 6Ó by Ò
1ß 5Ó

15. Ò
3ß 3Ó by Ò0ß 10Ó

16. Ò
1ß 2Ó by Ò0ß 1Ó

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27

28

Chapter 1 Functions

17. Ò
5ß 1Ó by Ò
5ß 5Ó

18. Ò
5ß 1Ó by Ò
2ß 4Ó

19. Ò
4ß 4Ó by Ò0ß 3Ó

20. Ò
5ß 5Ó by Ò
2ß 2Ó

21. Ò
"!ß "!Ó by Ò
'ß 'Ó

22. Ò
&ß &Ó by Ò
#ß #Ó

23. Ò
'ß "!Ó by Ò
'ß 'Ó

24. Ò
$ß &Ó by Ò
#ß "!Ó

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 1.4 Graphing with Calculators and Computers 25. Ò
0Þ03ß 0Þ03Ó by Ò
1Þ25ß 1Þ25Ó

26. Ò
0Þ1ß 0Þ1Ó by Ò
3ß 3Ó

27. Ò
300ß 300Ó by Ò
1Þ25ß 1Þ25Ó

28. Ò
50ß 50Ó by Ò
0Þ1ß 0Þ1Ó

29. Ò
0Þ25ß 0Þ25Ó by Ò
0Þ3ß 0Þ3Ó

30. Ò
0Þ15ß 0Þ15Ó by Ò
0Þ02ß 0Þ05Ó

31. x#  #x œ %  %y
y# Ê y œ # „ È
x#
#x  ). The lower half is produced by graphing y œ #
È
x#
#x  ).

32. y#
"'x# œ " Ê y œ „ È"  "'x# . The upper branch is produced by graphing y œ È"  "'x# .

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29

30

Chapter 1 Functions

33.

34.

35.

36.

37.

38Þ

39.

40.

CHAPTER 1 PRACTICE EXERCISES 1. The area is A œ 1 r# and the circumference is C œ #1 r. Thus, r œ 2. The surface area is S œ %1 r# Ê r œ ˆ %S1 ‰ surface area gives S œ %1 r# œ %1 ˆ $%V1 ‰

"Î#

#Î$

C #1

#

Ê A œ 1ˆ #C1 ‰ œ

C# %1 .

$ $V . The volume is V œ %$ 1 r$ Ê r œ É %1 . Substitution into the formula for

.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Chapter 1 Practice Exercises 3. The coordinates of a point on the parabola are axß x# b. The angle of inclination ) joining this point to the origin satisfies the equation tan ) œ 4. tan ) œ

rise run

œ

h &!!

x# x

œ x. Thus the point has coordinates axß x# b œ atan )ß tan# )b.

Ê h œ &!! tan ) ft.

5.

6.

Symmetric about the origin.

Symmetric about the y-axis.

7.

8.

Neither

Symmetric about the y-axis.

9. ya
xb œ a
xb#  " œ x#  " œ yaxb. Even. 10. ya
xb œ a
xb&
a
xb$
a
xb œ
x&  x$  x œ
yaxb. Odd. 11. ya
xb œ "
cosa
xb œ "
cos x œ yaxb. Even. 12. ya
xb œ seca
xb tana
xb œ 13. ya
xb œ

a
xb% " a
xb$
#a
xb

œ

x% "
x$ #x

sina
xb cos# a
xb

œ


sin x cos# x

œ
sec x tan x œ
yaxb. Odd.

%

" œ
xx$
# x œ
yaxb. Odd.

14. ya
xb œ a
xb
sina
xb œ a
xb  sin x œ
ax
sin xb œ
yaxb. Odd. 15. ya
xb œ
x  cosa
xb œ
x  cos x. Neither even nor odd. 16. ya
xb œ a
xbcosa
xb œ
x cos x œ
yaxb. Odd. 17. Since f and g are odd Ê fa
xb œ
faxb and ga
xb œ
gaxb. (a) af † gba
xb œ fa
xbga
xb œ Ò
faxbÓÒ
gaxbÓ œ faxbgaxb œ af † gbaxb Ê f † g is even (b) f 3 a
xb œ fa
xbfa
xbfa
xb œ Ò
faxbÓÒ
faxbÓÒ
faxbÓ œ
faxb † faxb † faxb œ
f 3 axb Ê f 3 is odd. (c) fasina
xbb œ fa
sinaxbb œ
fasinaxbb Ê fasinaxbb is odd. (d) gaseca
xbb œ gasecaxbb Ê gasecaxbb is even. (e) lga
xbl œ l
gaxbl œ lgaxbl Ê lgl is evenÞ

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31

32

Chapter 1 Functions

18. Let faa
xb œ faa  xb and define gaxb œ fax  ab. Then ga
xb œ faa
xb  ab œ faa
xb œ faa  xb œ fax  ab œ gaxb Ê gaxb œ fax  ab is even. 19. (a) The function is defined for all values of x, so the domain is a
_ß _b. (b) Since l x l attains all nonnegative values, the range is Ò
#ß _Ñ. 20. (a) Since the square root requires "
x   !, the domain is Ð
_ß "Ó. (b) Since È"
x attains all nonnegative values, the range is Ò
#ß _Ñ. 21. (a) Since the square root requires "'
x#   !, the domain is Ò
%ß %Ó. (b) For values of x in the domain, ! Ÿ "'
x# Ÿ "', so ! Ÿ È"'
x# Ÿ %. The range is Ò!ß %Ó. 22. (a) The function is defined for all values of x, so the domain is a
_ß _b. (b) Since $#
x attains all positive values, the range is a"ß _b. 23. (a) The function is defined for all values of x, so the domain is a
_ß _b. (b) Since #e
x attains all positive values, the range is a
$ß _b. 24. (a) The function is equivalent to y œ tan #x, so we require #x Á

k1 #

for odd integers k. The domain is given by x Á

k1 %

for

odd integers k. (b) Since the tangent function attains all values, the range is a
_ß _b. 25. (a) The function is defined for all values of x, so the domain is a
_ß _b. (b) The sine function attains values from
" to ", so
# Ÿ #sina$x  1b Ÿ # and hence
$ Ÿ #sina$x  1b
" Ÿ ". The range is Ò
3ß 1Ó. 26. (a) The function is defined for all values of x, so the domain is a
_ß _b. & (b) The function is equivalent to y œ È x# , which attains all nonnegative values. The range is Ò!ß _Ñ. 27. (a) The logarithm requires x
$  !, so the domain is a$ß _b. (b) The logarithm attains all real values, so the range is a
_ß _b. 28. (a) The function is defined for all values of x, so the domain is a
_ß _b. (b) The cube root attains all real values, so the range is a
_ß _b. 29. (a) (b) (c) (d)

Increasing because volume increases as radius increases Neither, since the greatest integer function is composed of horizontal (constant) line segments Decreasing because as the height increases, the atmospheric pressure decreases. Increasing because the kinetic (motion) energy increases as the particles velocity increases.

30. (a) Increasing on Ò2, _Ñ (c) Increasing on a
_, _b

(b) Increasing on Ò
1, _Ñ (d) Increasing on Ò "# , _Ñ

31. (a) The function is defined for
% Ÿ x Ÿ %, so the domain is Ò
%ß %Ó. (b) The function is equivalent to y œ Èl x l,
% Ÿ x Ÿ %, which attains values from ! to # for x in the domain. The range is Ò!ß #Ó.

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Chapter 1 Practice Exercises

33

32. (a) The function is defined for
# Ÿ x Ÿ #, so the domain is Ò
#ß #Ó. (b) The range is Ò
"ß "Ó. 33. First piece: Line through a!ß "b and a"ß !b. m œ Second piece: Line through a"ß "b and a#ß !b. m faxb œ œ

"
x, ! Ÿ x  " #
x, " Ÿ x Ÿ #

34. First piece: Line through a!ß !b and a2ß 5b. m œ Second piece: Line through a2ß 5b and a4ß !b. m faxb œ


10


5 2 x, 5x 2 ,

!
"
" "
! œ " œ "
" œ !#

" œ "


" Ê y œ
x  " œ "
x œ
" Ê y œ
ax
"b  " œ
x  # œ #
x

5
! 5 5 2
! œ 2 Ê y œ 2x !
5
5 œ 4
2 œ 2 œ
52 Ê

y œ
52 ax
2b  5 œ
52 x  10 œ 10


!Ÿx2 (Note: x œ 2 can be included on either piece.) 2ŸxŸ4

35. (a) af‰gba
"b œ faga
"bb œ fŠ È
""  # ‹ œ fa"b œ (b) ag‰f ba#b œ gafa#bb œ gˆ "2 ‰ œ (c) af‰f baxb œ fafaxbb œ fˆ "x ‰ œ

" É "#  #

" "Îx

œ

" È#Þ&

" "

œ"

or É &#

œ x, x Á !

(d) ag‰gbaxb œ gagaxbb œ gŠ Èx" # ‹ œ

"

" É Èx # #

œ

% x# È É "  #È x  #

$ 36. (a) af‰gba
"b œ faga
"bb œ fˆÈ
"  "‰ œ fa!b œ #
! œ # $ (b) ag‰f ba#b œ faga#bb œ ga#
#b œ ga!b œ È !"œ"

(c) af‰f baxb œ fafaxbb œ fa#
xb œ #
a#
xb œ x $ $ $ È (d) ag‰gbaxb œ gagaxbb œ gˆÈ x  "‰ œ É x""

#

37. (a) af‰gbaxb œ fagaxbb œ fˆÈx  #‰ œ #
ˆÈx  #‰ œ
x, x  
#. ag‰f baxb œ fagaxbb œ ga#
x# b œ Èa#
x# b  # œ È%
x# (b) Domain of f‰g: Ò
#ß _ÑÞ Domain of g‰f: Ò
#ß #ÓÞ

(c) Range of f‰g: Ð
_ß #ÓÞ Range of g‰f: Ò!ß #ÓÞ

% 38. (a) af‰gbaxb œ fagaxbb œ fŠÈ"
x‹ œ ÉÈ"
x œ È "
x.

ag‰f baxb œ fagaxbb œ gˆÈx‰ œ É"
Èx (b) Domain of f‰g: Ð
_ß "ÓÞ Domain of g‰f: Ò!ß "ÓÞ 39.

y œ faxb

(c) Range of f‰g: Ò!ß _ÑÞ Range of g‰f: Ò!ß "ÓÞ y œ af‰f baxb

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5x 2

34

Chapter 1 Functions

40.

41.

42.

The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x   0 across the y-axis. 43.

It does not change the graph.

44.

Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is the same as the graph of y œ g" (x). When g" (x) is negative, the graph of y œ g# (x) is the reflection of the graph of y œ g" (x) across the x-axis.

Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is the same as the graph of y œ g" (x). When g" (x) is negative, the graph of y œ g# (x) is the reflection of the graph of y œ g" (x) across the x-axis.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Chapter 1 Practice Exercises 45.

46.

The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x   0 across the y-axis.

Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is the same as the graph of y œ g" (x). When g" (x) is negative, the graph of y œ g# (x) is the reflection of the graph of y œ g" (x) across the x-axis. 47.

48.

The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x   0 across the y-axis. 49. (a) y œ gax
3b  (c) y œ ga
xb (e) y œ 5 † gaxb

" #

The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x   0 across the y-axis. (b) y œ gˆx  3# ‰
2 (d) y œ
gaxb (f) y œ ga5xb

50. (a) Shift the graph of f right 5 units (b) Horizontally compress the graph of f by a factor of 4 (c) Horizontally compress the graph of f by a factor of 3 and a then reflect the graph about the y-axis (d) Horizontally compress the graph of f by a factor of 2 and then shift the graph left "# unit. (e) Horizontally stretch the graph of f by a factor of 3 and then shift the graph down 4 units. (f) Vertically stretch the graph of f by a factor of 3, then reflect the graph about the x-axis, and finally shift the graph up "4 unit. 51. Reflection of the grpah of y œ Èx about the x-axis followed by a horizontal compression by a factor of 1 2 then a shift left 2 units.

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35

36

Chapter 1 Functions

52. Reflect the graph of y œ x about the x-axis, followed by a vertical compression of the graph by a factor of 3, then shift the graph up 1 unit.

53. Vertical compression of the graph of y œ

1 x2

by a

factor of 2, then shift the graph up 1 unit.

54. Reflect the graph of y œ x1Î3 about the y-axis, then compress the graph horizontally by a factor of 5.

55.

56.

period œ 1 57.

period œ 41 58.

period œ 2

period œ 4

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Chapter 1 Practice Exercises 59.

60.

period œ 21

period œ 21 1 3

61. (a) sin B œ sin

œ

b c

œ

b #

Ê b œ 2 sin

1 3

œ 2Š

È3 # ‹

œ È3. By the theorem of Pythagoras,

a#  b# œ c# Ê a œ Èc#
b# œ È4
3 œ 1. 1 3

(b) sin B œ sin

œ

b c

œ

2 c

Ê cœ

2 sin 13

œ È23 œ Š ‹ #

4 È3

#

. Thus, a œ Èc#
b# œ ÊŠ È43 ‹
(2)# œ É 34 œ

62. (a) sin A œ

a c

Ê a œ c sin A

(b) tan A œ

a b

Ê a œ b tan A

63. (a) tan B œ

b a

Ê aœ

(b) sin A œ

a c

Ê cœ

64. (a) sin A œ

a c

(c) sin A œ

a c

b tan B

œ

a sin A

È c #
b # c

65. Let h œ 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° œ hc , tan 35° œ hb , and b
c œ 10. Thus, h œ c tan 50° and h œ b tan 35° œ (c  10) tan 35° Ê c tan 50° œ (c  10) tan 35° Ê c (tan 50°
tan 35°) œ 10 tan 35° tan 35° Ê c œ tan10 50°
tan 35° Ê h œ c tan 50° œ

10 tan 35° tan 50° tan 50°
tan 35°

¸ 16.98 m.

66. Let h œ height of balloon above ground. From the figure at the right, tan 40° œ ha , tan 70° œ hb , and a  b œ 2. Thus, h œ b tan 70° Ê h œ (2
a) tan 70° and h œ a tan 40° Ê (2
a) tan 70° œ a tan 40° Ê a(tan 40°  tan 70°) 70° œ 2 tan 70° Ê a œ tan 240°tantan 70° Ê h œ a tan 40° œ

2 tan 70° tan 40° tan 40°tan 70°

¸ 1.3 km.

67. (a)

(b) The period appears to be 41.

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2 È3

.

37

38

Chapter 1 Functions (c) f(x  41) œ sin (x  41)  cos ˆ x#41 ‰ œ sin (x  21)  cos ˆ x#  21‰ œ sin x  cos since the period of sine and cosine is 21. Thus, f(x) has period 41.

x #

68. (a)

(b) D œ (
_ß 0)  (!ß _); R œ [
1ß 1] (c) f is not periodic. For suppose f has period p. Then f ˆ #"1  kp‰ œ f ˆ #"1 ‰ œ sin 21 œ 0 for all integers k. Choose k so large that

" #1

 kp 

" 1

Ê 0

" (1/21)kp

 1. But then

f ˆ #"1  kp‰ œ sin Š (1/#1")kp ‹  0 which is a contradiction. Thus f has no period, as claimed. CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES 1. There are (infinitely) many such function pairs. For example, f(x) œ 3x and g(x) œ 4x satisfy f(g(x)) œ f(4x) œ 3(4x) œ 12x œ 4(3x) œ g(3x) œ g(f(x)). 2. Yes, there are many such function pairs. For example, if g(x) œ (2x  3)$ and f(x) œ x"Î$ , then (f ‰ g)(x) œ f(g(x)) œ f a(2x  3)$ b œ a(2x  3)$ b

"Î$

œ 2x  3.

3. If f is odd and defined at x, then f(
x) œ
f(x). Thus g(
x) œ f(
x)
2 œ
f(x)
2 whereas
g(x) œ
(f(x)
2) œ
f(x)  2. Then g cannot be odd because g(
x) œ
g(x) Ê
f(x)
2 œ
f(x)  2 Ê 4 œ 0, which is a contradiction. Also, g(x) is not even unless f(x) œ 0 for all x. On the other hand, if f is even, then g(x) œ f(x)
2 is also even: g(
x) œ f(
x)
2 œ f(x)
2 œ g(x). 4. If g is odd and g(0) is defined, then g(0) œ g(
0) œ
g(0). Therefore, 2g(0) œ 0 Ê g(0) œ 0. 5. For (xß y) in the 1st quadrant, kxk  kyk œ 1  x Í x  y œ 1  x Í y œ 1. For (xß y) in the 2nd quadrant, kxk  kyk œ x  1 Í
x  y œ x  1 Í y œ 2x  1. In the 3rd quadrant, kxk  kyk œ x  1 Í
x
y œ x  1 Í y œ
2x
1. In the 4th quadrant, kxk  kyk œ x  1 Í x  (
y) œ x  1 Í y œ
1. The graph is given at the right.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Chapter 1 Additional and Advanced Exercises 6. We use reasoning similar to Exercise 5. (1) 1st quadrant: y  kyk œ x  kxk Í 2y œ 2x Í y œ x. (2) 2nd quadrant: y  kyk œ x  kxk Í 2y œ x  (
x) œ 0 Í y œ 0. (3) 3rd quadrant: y  kyk œ x  kxk Í y  (
y) œ x  (
x) Í 0 œ 0 Ê all points in the 3rd quadrant satisfy the equation. (4) 4th quadrant: y  kyk œ x  kxk Í y  (
y) œ 2x Í 0 œ x. Combining these results we have the graph given at the right: 7. (a) sin# x  cos# x œ 1 Ê sin# x œ 1
cos# x œ (1
cos x)(1  cos x) Ê (1
cos x) œ Ê

1
cos x sin x

œ

sin# x 1cos x

sin x 1cos x

(b) Using the definition of the tangent function and the double angle formulas, we have #

tan ˆ x# ‰ œ

sin# ˆ x# ‰ cos# ˆ #x ‰

œ

"

x ‹‹ cos Š2 Š #

#

"
cos Š2 Š #x ‹‹ #

œ

1
cos x 1cos x

.

8. The angles labeled # in the accompanying figure are equal since both angles subtend arc CD. Similarly, the two angles labeled ! are equal since they both subtend arc AB. Thus, triangles AED and BEC are similar which )
b implies a
b c œ 2a cos a c Ê (a
c)(a  c) œ b(2a cos )
b) Ê a#
c# œ 2ab cos )
b# Ê c# œ a#  b#
2ab cos ).

9. As in the proof of the law of sines of Section 1.3, Exercise 61, ah œ bc sin A œ ab sin C œ ac sin B Ê the area of ABC œ "# (base)(height) œ "# ah œ "# bc sin A œ "# ab sin C œ "# ac sin B. 10. As in Section 1.3, Exercise 61, (Area of ABC)# œ œ

" 4

(base)# (height)# œ

" 4

a# h # œ

" 4

a# b# sin# C

a# b# a"
cos# Cb . By the law of cosines, c# œ a#  b#
2ab cos C Ê cos C œ

Thus, (area of ABC)# œ œ

" 4

" 16

" 4

a# b# a"
cos# Cb œ #

Š4a# b#
aa#  b#
c# b ‹ œ

" 16

" 4

a# b# Œ"
Š a

#

 b#
c# ‹ #ab

#

œ

a# b# 4

a#  b#
c# 2ab

Š"


.

aa #  b #
c # b 4a# b#

#

‹

ca2ab  aa#  b#
c# bb a2ab
aa#  b#
c# bbd

" ca(a  b)#
c# b ac#
(a
b)# bd œ 16 c((a  b)  c)((a  b)
c)(c  (a
b))(c
(a
b))d a  b  c
a  b  c a
b  c a  b
c œ ˆ # ‰ ˆ # ‰ ˆ # ‰ ˆ # ‰‘ œ s(s
a)(s
b)(s
c), where s œ a#bc .

œ

" 16

Therefore, the area of ABC equals Ès(s
a)(s
b)(s
c) . 11. If f is even and odd, then f(
x) œ
f(x) and f(
x) œ f(x) Ê f(x) œ
f(x) for all x in the domain of f. Thus 2f(x) œ 0 Ê f(x) œ 0.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

39

40

Chapter 1 Functions f(
x)  f(
(
x)) œ f(x) #f(
x) œ E(x) Ê E # even function. Define O(x) œ f(x)
E(x) œ f(x)
f(x) #f(
x) œ f(x)
#f(
x) . Then O(
x) œ f(
x)
#f(
(
x)) œ f(
x)#
f(x) œ
Š f(x)
#f(
x) ‹ œ
O(x) Ê O is an odd function

12. (a) As suggested, let E(x) œ

f(x)  f(
x) #

Ê E(
x) œ

is an

Ê f(x) œ E(x)  O(x) is the sum of an even and an odd function. (b) Part (a) shows that f(x) œ E(x)  O(x) is the sum of an even and an odd function. If also f(x) œ E" (x)  O" (x), where E" is even and O" is odd, then f(x)
f(x) œ 0 œ aE" (x)  O" (x)b
(E(x)  O(x)). Thus, E(x)
E" (x) œ O" (x)
O(x) for all x in the domain of f (which is the same as the domain of E
E" and O
O" ). Now (E
E" )(
x) œ E(
x)
E" (
x) œ E(x)
E" (x) (since E and E" are even) œ (E
E" )(x) Ê E
E" is even. Likewise, (O"
O)(
x) œ O" (
x)
O(
x) œ
O" (x)
(
O(x)) (since O and O" are odd) œ
(O" (x)
O(x)) œ
(O"
O)(x) Ê O"
O is odd. Therefore, E
E" and O"
O are both even and odd so they must be zero at each x in the domain of f by Exercise 11. That is, E" œ E and O" œ O, so the decomposition of f found in part (a) is unique. 13. y œ ax#  bx  c œ a Šx#  ba x 

b# 4a# ‹




b# 4a

 c œ a ˆx 

b ‰# 2a




b# 4a

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. 14. (a) If a  0, the graph rises to the right of the vertical line x œ
b and falls to the left. If a  0, the graph falls to the right of the line x œ
b and rises to the left. If a œ 0, the graph reduces to the horizontal line y œ c. As kak increases, the slope at any given point x œ x! increases in magnitude and the graph becomes steeper. As kak decreases, the slope at x! decreases in magnitude and the graph rises or falls more gradually. (b) Increasing b shifts the graph to the left; decreasing b shifts it to the right. (c) Increasing c shifts the graph upward; decreasing c shifts it downward. 15. Each of the triangles pictured has the same base b œ v?t œ v(1 sec). Moreover, the height of each triangle is the same value h. Thus "# (base)(height) œ

" #

bh

œ A" œ A# œ A$ œ á . In conclusion, the object sweeps out equal areas in each one second interval.

16. (a) Using the midpoint formula, the coordinates of P are ˆ a# 0 ß b# 0 ‰ œ ˆ #a ß b# ‰ . Thus the slope of OP œ

?y ?x

œ

b/2 a/2

œ

b a

.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Chapter 1 Additional and Advanced Exercises (b) The slope of AB œ

b
0 0
a

41

œ
ba . The line segments AB and OP are perpendicular when the product #

of their slopes is
" œ ˆ ba ‰ ˆ
ba ‰ œ
ba# . Thus, b# œ a# Ê a œ b (since both are positive). Therefore, AB is perpendicular to OP when a œ b. 17. From the figure we see that 0 Ÿ ) Ÿ cos ) œ

and AB œ AD œ 1. From trigonometry we have the following: sin ) œ sin ) cos ) .

œ CD, and tan ) œ œ We can see that: w " area ˜AEB  area sector DB  area ˜ADC Ê # aAEbaEBb  "# aADb2 )  "# aADbaCDb AE AB

œ AE, tan ) œ

1 2

CD AD

EB AE

Ê "# sin ) cos )  "# a"b2 )  "# a"batan )b Ê "# sin ) cos )  "# ) 

" sin ) # cos )

18. af‰gbaxb œ fagaxbb œ aacx  db  b œ acx  ad  b and ag‰f baxb œ gafaxbb œ caax  bb  d œ acx  cb  d Thus af‰gbaxb œ ag‰f baxb Ê acx  ad  b œ acx  bc  d Ê ad  b œ bc  d. Note that fadb œ ad  b and gabb œ cb  d, thus af‰gbaxb œ ag‰f baxb if fadb œ gabb.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

EB AB

œ EB,

42

Chapter 1 Functions

NOTES:

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

CHAPTER 2 LIMITS AND CONTINUITY 2.1 RATES OF CHANGE AND TANGENTS TO CURVES 1. (a)

?f ?x

œ

f(3)
f(2) 3
#

2. (a)

?g ?x

œ

g(1)
g(
1) 1
(
1)

3. (a)

?h ?t

œ

h ˆ 341 ‰
h ˆ 14 ‰ 1 31 4
4

œ

?g ?t

œ

g(1)
g(0) 1
0

(2
1)
(2  1) 1
0

4. (a) 5.

?R ?)

œ

R(2)
R(0) 2
0

6.

?P ?)

œ

P(2)
P(1) 2
1

7. (a)

?y ?x

œ

?y ?x

œ

?y ?x

œ

?y ?x

œ

?y ?x

œ

?y ?x

œ

?y ?x

œ

œ œ

28
9 1

œ

œ

œ

1
1 2

œ 19

(b)

?f ?x

œ

f(1)
f(
") 1
(
1)

œ

2
0 #

œ1

œ0

(b)

?g ?x

œ

g(0)
g(
2) 0
(
2)

œ

0
4 #

œ
2

(b)

?h ?t

œ

h ˆ 1# ‰
h ˆ 16 ‰ 1
1 # 6

œ

?g ?t

œ

g(1)
g(
1) 1
(
1)

œ


1
1 1 #

È 8 1
È 1 #

œ
14 œ
12

3
" #

œ

(8
16  10)
("
%  &) 1

ˆa2  h b2
3 ‰
ˆ 2 2
3 ‰ h

œ

(b)

0
È3 1 3

œ


3 È 3 1

(2
1)
(2
") #1

œ0

œ1 œ2
2œ0

4  4h  h2
3
1 h

œ

4h  h2 h

œ 4  h. As h Ä 0, 4  h Ä 4 Ê at Pa2, 1b the slope is 4.

(b) y
1 œ 4ax
2b Ê y
1 œ 4x
8 Ê y œ 4x
7 8. (a)

ˆ 5
a1  h b 2 ‰
ˆ 5
1 2 ‰ h

œ

5
1
2h
h2
4 h

œ


2h
h2 h

œ
2
h. As h Ä 0,
2
h Ä
2 Ê at Pa1, 4b the

slope is
2. (b) y
4 œ a
2bax
1b Ê y
4 œ
2x  2 Ê y œ
2x  6 9. (a)

ˆa2  h b2
2 a 2  h b
3 ‰
ˆ 2 2
2 a 2 b
3 ‰ h

œ

4  4h  h2
4
2h
3
a
3b h

œ

2h  h2 h

œ 2  h. As h Ä 0, 2  h Ä 2 Ê at

Pa2,
3b the slope is 2. (b) y
a
3b œ 2ax
2b Ê y  3 œ 2x
4 Ê y œ 2x
7. 10. (a)

ˆa1  h b2
4 a 1  h b ‰
ˆ 1 2
4 a 1 b ‰ h

œ

1  2h  h2
4
4h
a
3b h

œ

h2
2h h

œ h
2. As h Ä 0, h
2 Ä
2 Ê at

Pa1,
3b the slope is
2. (b) y
a
3b œ a
2bax
1b Ê y  3 œ
2x  2 Ê y œ
2x
1. 11. (a)

a2  h b 3
2 3 h

œ

8  12h  4h2  h3
8 h

œ

12h  4h2  h3 h

œ 12  4h  h2 . As h Ä 0, 12  4h  h2 Ä 12, Ê at

Pa2, 8b the slope is 12. (b) y
8 œ 12ax
2b Ê y
8 œ 12x
24 Ê y œ 12x
16. 12. (a)

2
a1  h b3
ˆ 2
1 3 ‰ h

œ

2
1
3h
3h2
h3
1 h

œ


3h
3h2
h3 h

œ
3
3h
h2 . As h Ä 0,
3
3h
h2 Ä
3, Ê at

Pa1, 1b the slope is
3. (b) y
1 œ a
3bax
1b Ê y
1 œ
3x  3 Ê y œ
3x  4. 13. (a)

a1  hb3
12a1  hb
ˆ13
12a"b‰ h 2

œ

1  3h  3h2  h3
12
12h
a
11b h

œ


9h  3h2  h3 h

œ
9  3h  h2 . As h Ä 0,


9  3h  h Ä
9 Ê at Pa1,
11b the slope is
9. (b) y
a
11b œ a
9bax
1b Ê y  11 œ
9x  9 Ê y œ
9x
2.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

44

Chapter 2 Limits and Continuity

14. (a)

?y ?x

œ

a2  h b 3
3 a 2  h b 2  4
ˆ 2 3
3 a 2 b 2  4 ‰ h 2

8  12h  6h2  h3
12
12h
3h2  4
0 h

œ

œ

3h2  h3 h

œ 3h  h2 . As h Ä 0,

3h  h Ä 0 Ê at Pa2, 0b the slope is 0. (b) y
0 œ 0ax
2b Ê y œ 0. 15. (a)

?p ?t

Slope of PQ œ

Q

650
225 20
10 650
375 20
14 650
475 20
16.5 650
550 20
18

Q" (10ß 225) Q# (14ß 375) Q$ (16.5ß 475) Q% (18ß 550)

œ 42.5 m/sec œ 45.83 m/sec œ 50.00 m/sec œ 50.00 m/sec

(b) At t œ 20, the sportscar was traveling approximately 50 m/sec or 180 km/h. 16. (a)

Slope of PQ œ

Q Q" (5ß 20) Q# (7ß 39) Q$ (8.5ß 58) Q% (9.5ß 72)

80
20 10
5 80
39 10
7 80
58 10
8.5 80
72 10
9.5

?p ?t

œ 12 m/sec œ 13.7 m/sec œ 14.7 m/sec œ 16 m/sec

(b) Approximately 16 m/sec 17. (a)

(b)

?p ?t

œ

174
62 2004
2002

œ

112 #

œ 56 thousand dollars per year

(c) The average rate of change from 2001 to 2002 is ??pt œ

62
27 20022
2001 œ 35 thousand dollars per year. 111
62 The average rate of change from 2002 to 2003 is ??pt œ 2003
2002 œ 49 thousand dollars per year. So, the rate at which profits were changing in 2002 is approximatley "# a35  49b œ 42 thousand dollars

18. (a) F(x) œ (x  2)/(x
2) x 1.2 F(x)
4.0 ?F ?x ?F ?x ?F ?x

œ

?g ?x ?g ?x

œ

œ œ

1.1
3.4

1.01
3.04

1.001
3.004

1.0001
3.0004

1
3


4.0
(
3) œ
5.0; 1.2
1
3.04
(
3) œ
4.04; 1.01
1
3.!!!%
(
3) œ
4.!!!%; 1.0001
1

?F ?x ?F ?x

œ œ


3.4
(
3) œ
4.4; 1.1
1
3.004
(
3) œ
4.!!%; 1.001
1

È g(2)
g(1) œ #2

1" ¸ 0.414213 2
1 È1  h
" g(1  h)
g(1) (1  h)
1 œ h

?g ?x

œ

g(1.5)
g(1) 1.5
1

(b) The rate of change of F(x) at x œ 1 is
4. 19. (a)

œ

œ

È1.5
" 0.5

¸ 0.449489

(b) g(x) œ Èx 1h È1  h ŠÈ1  h
1‹ /h

1.1 1.04880

1.01 1.004987

1.001 1.0004998

1.0001 1.0000499

1.00001 1.000005

1.000001 1.0000005

0.4880

0.4987

0.4998

0.499

0.5

0.5

(c) The rate of change of g(x) at x œ 1 is 0.5.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

per year.

Section 2.1 Rates of Change and Tangents to Curves (d) The calculator gives lim hÄ! 20. (a) i) ii) (b)

f(3)
f(2) 3
2

œ

f(T)
f(2) T
#

œ

"
" 3 #

1 " " T
# T
#

T f(T) af(T)
f(2)b/aT
2b

"

œ

6

1

È1  h
" h

œ "# .

œ
"6


#TT T
#

2 #T

œ

45

œ

2
T #T(T
2)

2.1 0.476190
0.2381

œ

2
T
#T(2
T)

2.01 0.497512
0.2488

œ
#"T , T Á 2

2.001 0.499750
0.2500

2.0001 0.4999750
0.2500

2.00001 0.499997
0.2500

2.000001 0.499999
0.2500

(c) The table indicates the rate of change is
0.25 at t œ 2. " ‰ (d) lim ˆ
#T œ
4" TÄ#

NOTE: Answers will vary in Exercises 21 and 22. s 15
0 ˜s 20
15 10 ˜s 30
20 21. (a) Ò0, 1Ó: ˜ ˜t œ 1
0 œ 15 mphà Ò1, 2.5Ó: ˜t œ 2.5
1 œ 3 mphà Ò2.5, 3.5Ó: ˜t œ 3.5
2.5 œ 10 mph (b) At Pˆ "# , 7.5‰: Since the portion of the graph from t œ 0 to t œ 1 is nearly linear, the instantaneous rate of change

will be almost the same as the average rate of change, thus the instantaneous speed at t œ

" #

is

15
7.5 1
0.5

œ 15 mi/hr.

At Pa2, 20b: Since the portion of the graph from t œ 2 to t œ 2.5 is nearly linear, the instantaneous rate of change will
20 be nearly the same as the average rate of change, thus v œ 20 2.5
2 œ 0 mi/hr. For values of t less than 2, we have Slope of PQ œ

Q

?s ?t

15
20 1
2 œ 5 mi/hr 19
20 1.5
2 œ 2 mi/hr 19.9
20 1.9
2 œ 1 mi/hr

Q" (1ß 15) Q# (1.5ß 19) Q$ (1.9ß 19.9)

Thus, it appears that the instantaneous speed at t œ 2 is 0 mi/hr. At Pa3, 22b: s Q Slope of PQ œ ? ?t 35
22 4
3 30
22 3.5
3 23
22 3.1
3

Q" (4ß 35) Q# (3.5ß 30) Q$ (3.1ß 23)

Slope of PQ œ

Q

œ 13 mi/hr

Q" (2ß 20)

œ 16 mi/hr

Q# (2.5ß 20)

œ 10 mi/hr

Q$ (2.9ß 21.6)

20
22 2
3 œ 2 mi/hr 20
22 2.5
3 œ 4 mi/hr 21.6
22 2.9
3 œ 4 mi/hr

Thus, it appears that the instantaneous speed at t œ 3 is about 7 mi/hr. (c) It appears that the curve is increasing the fastest at t œ 3.5. Thus for Pa3.5, 30b s Q Slope of PQ œ ? Q Slope of PQ œ ?t 35
30 4
3.5 œ 10 mi/hr 34
30 3.75
3.5 œ 16 mi/hr 32
30 3.6
3.5 œ 20 mi/hr

Q" (4ß 35) Q# (3.75ß 34) Q$ (3.6ß 32)

˜A ˜t

œ

10
15 3
0

(b) At Pa1, 14b: Q Q" (2ß 12.2) Q# (1.5ß 13.2) Q$ (1.1ß 13.85)

¸
1.67

gal day à

Ò0, 5Ó:

Q# (3.25ß 25) Q$ (3.4ß 28)

Slope of PQ œ

˜A ˜t

?A ?t

12.2
14 2
1 œ
1.8 gal/day 13.2
14 1.5
1 œ
1.6 gal/day 13.85
14 1.1
1 œ
1.5 gal/day

œ

3.9
15 5
0

¸
2.2

gal day à Ò7,

10Ó:

?s ?t

22
30 3
3.5 œ 16 mi/hr 25
30 3.25
3.5 œ 20 mi/hr 28
30 3.4
3.5 œ 20 mi/hr

Q" (3ß 22)

Thus, it appears that the instantaneous speed at t œ 3.5 is about 20 mi/hr. 22. (a) Ò0, 3Ó:

?s ?t

˜A ˜t

Q Q" (0ß 15) Q# (0.5ß 14.6) Q$ (0.9ß 14.86)

œ

0
1.4 10
7

¸
0.5

Slope of PQ œ

gal day ?A ?t

15
14 0
1 œ
1 gal/day 14.6
14 0.5
1 œ
1.2 gal/day 14.86
14 0.9
1 œ
1.4 gal/day

Thus, it appears that the instantaneous rate of consumption at t œ 1 is about
1.45 gal/day. At Pa4, 6b:

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

46

Chapter 2 Limits and Continuity Q Q" (5ß 3.9) Q# (4.5ß 4.8) Q$ (4.1ß 5.7)

Slope of PQ œ 3.9
6 5
4 4.8
6 4.5
4 5.7
6 4.1
4

?A ?t

Q

œ
2.1 gal/day

Q" (3ß 10)

œ
2.4 gal/day

Q# (3.5ß 7.8)

œ
3 gal/day

Q$ (3.9ß 6.3)

Slope of PQ œ

10
6 3
4 œ
4 gal/day 7.8
6 3.5
4 œ
3.6 gal/day 6.3
6 3.9
4 œ
3 gal/day

Thus, it appears that the instantaneous rate of consumption at t œ 1 is
3 gal/day. At Pa8, 1b: Q Slope of PQ œ ??At Q Slope of PQ œ Q" (9ß 0.5) Q# (8.5ß 0.7) Q$ (8.1ß 0.95)

0.5
1 9
8 œ
0.5 gal/day 0.7
1 8.5
8 œ
0.6 gal/day 0.95
1 8.1
8 œ
0.5 gal/day

Q" (7ß 1.4)

4.8
7.8 4.5
3.5 œ
3 gal/day 6
7.8 4
3.5 œ
3.6 gal/day 7.4
7.8 3.6
3.5 œ
4 gal/day

Q" (2.5ß 11.2)

Q# (7.5ß 1.3) Q$ (7.9ß 1.04)

?A ?t

?A ?t

1.4
1 7
8 œ
0.6 gal/day 1.3
1 7.5
8 œ
0.6 gal/day 1.04
1 7.9
8 œ
0.6 gal/day

Thus, it appears that the instantaneous rate of consumption at t œ 1 is
0.55 gal/day. (c) It appears that the curve (the consumption) is decreasing the fastest at t œ 3.5. Thus for Pa3.5, 7.8b s Q Slope of PQ œ ??At Q Slope of PQ œ ? ?t Q" (4.5ß 4.8) Q# (4ß 6) Q$ (3.6ß 7.4)

Q# (3ß 10) Q$ (3.4ß 8.2)

Thus, it appears that the rate of consumption at t œ 3.5 is about
4 gal/day.

11.2
7.8 2.5
3.5 œ
3.4 gal/day 10
7.8 3
3.5 œ
4.4 gal/day 8.2
7.8 3.4
3.5 œ
4 gal/day

2.2 LIMIT OF A FUNCTION AND LIMIT LAWS 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 Ä 1. (b) 1 (c) 0 (d) 0.5 2. (a) 0 (b)
1 (c) Does not exist. As t approaches 0 from the left, f(t) approaches
1. As t approaches 0 from the right, f(t) approaches 1. There is no single number L that f(t) gets arbitrarily close to as t Ä 0. (d)
1 3. (a) True (d) False (g) True

(b) True (e) False

(c) False (f) True

4. (a) False (d) True

(b) False (e) True

(c) True

5.

x

lim x Ä 0 kx k x kx k

does not exist because

x kx k

œ

x x

œ 1 if x  0 and

approaches
1. As x approaches 0 from the right,

x kx k

x kxk

œ

x
x

œ
1 if x  0. As x approaches 0 from the left,

approaches 1. There is no single number L that all the

function values get arbitrarily close to as x Ä 0. 6. As x approaches 1 from the left, the values of

" x
1

become increasingly large and negative. As x approaches 1

from the right, the values become increasingly large and positive. There is no one number L that all the function values get arbitrarily close to as x Ä 1, so lim x
" 1 does not exist. xÄ1

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 2.2 Limit of a Function and Limit Laws

47

7. Nothing can be said about f(x) because the existence of a limit as x Ä x! does not depend on how the function is defined at x! . 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 x! . That is, the existence of a limit depends on the values of f(x) for x near x! , not on the definition of f(x) at x! itself. 8. Nothing can be said. In order for lim f(x) to exist, f(x) must close to a single value for x near 0 regardless of the xÄ0

value f(0) itself. 9. No, the definition does not require that f be defined at x œ 1 in order for a limiting value to exist there. If f(1) is defined, it can be any real number, so we can conclude nothing about f(1) from lim f(x) œ 5. xÄ1

10. No, because the existence of a limit depends on the values of f(x) when x is near 1, not on f(1) itself. If lim f(x) exists, its value may be some number other than f(1) œ 5. We can conclude nothing about lim f(x), xÄ1

xÄ1

whether it exists or what its value is if it does exist, from knowing the value of f(1) alone. 11.

lim (2x  5) œ 2(
7)  5 œ
14  5 œ
9

x Ä
(

12. lim a
x#  5x
2b œ
(2)#  5(2)
2 œ
4  10
2 œ 4 xÄ#

13. lim 8(t
5)(t
7) œ 8(6
5)(6
7) œ
8 tÄ'

14.

lim ax$
2x#  4x  8b œ (
2)$
2(
2)#  4(
2)  8 œ
8
8
8  8 œ
16

x Ä
#

15. lim

x3

œ

x Ä # x6

17.

23 26

16. lim# 3s(2s
1) œ 3 ˆ 23 ‰
2 ˆ 23 ‰
1‘ œ 2 ˆ 43
1‰ œ

5 8

sÄ

$

lim 3(2x
1)# œ 3(2(
1)
1)# œ 3(
3)# œ 27

x Ä
"

y2

18. lim

# y Ä # y  5y  6

19.

œ

œ

22 (2)#  5(#)  6

œ

4 4  10  6

œ

4 #0

œ

" 5

%

lim (5
y)%Î$ œ [5
(
3)]%Î$ œ (8)%Î$ œ ˆ(8)"Î$ ‰ œ 2% œ 16

y Ä
$

20. lim (2z
8)"Î$ œ (2(0)
8)"Î$ œ (
8)"Î$ œ
2 zÄ!

21. lim

3

22. lim

È5h  4
2 h

h Ä ! È3h  1  1

hÄ0

œ

5 È4  2

23. lim

œ

x
5

# x Ä & x
25

24.

œ

3 È3(0)  1  1

œ lim

hÄ0

œ

3 È1  1

È5h  4
2 h

†

œ

3 2

È5h  4  2 È5h  4  2

œ lim

a5h  4b
4

h Ä 0 hŠÈ5h  4  2‹

œ lim

5h

h Ä 0 hŠÈ5h  4  2‹

œ lim

5 4 x
5

œ lim

x3

x Ä & (x  5)(x
5)

lim # x Ä
$ x  4x  3

œ lim

œ lim

x3

1

x Ä & x5

x Ä
$ (x  3)(x  1)

œ lim

œ

" 55

œ

" 10

1

œ

"
3  1

x Ä
$ x  1

5

h Ä 0 È5h  4  2

œ
"2

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

2 3

48 25.

Chapter 2 Limits and Continuity x#  3x
"0 x5

lim

x Ä
&

(x  5)(x
2) x5

œ lim

x Ä
&

(x
5)(x
2) x
2

26. lim

x#
7x  "0 x
#

œ lim

27. lim

t#  t
2 t#
1

t Ä " (t
1)(t  1)

xÄ#

tÄ"

t#  3t  2 # t Ä
" t
t
2

29.

lim $ # x Ä
# x  2x

lim


2x
4

5y$  8y#

1 x
1 x x Ä 1
1

32. lim

xÄ0

33. lim

1 xc1

u%
"

y# (5y  8)

œ lim

uÄ1

œ lim

4x
x# x Ä % 2
Èx

œ lim

36. lim

lim

x Ä
"

œ

(v
2) av#  2v  4b

Èx
3

x Ä * ˆÈ x
3 ‰ ˆ È x  3 ‰

x
1 x Ä 1 Èx  3
2

37. lim

œ

Èx#  12
4 x
2 xÄ2

x Ä
"

œ lim

xÄ2

(x
2)(x  2)

x2 x Ä
2 È x #  5
3

lim

x Ä
2

8
16

1 ‰ x
1

œ lim

xÄ1

œ lim

uÄ1

au#  "b (u  1) u#  u  1

"

œ lim

(x  2)(x
2)

444 (4)(8)

œ

x
1

x2

œ
2

4 3

12 32

œ

3 8

xÄ%

œ lim ŠÈx  3  #‹ œ È4  2 œ 4 xÄ1

ax #  8 b
*

œ lim

x Ä
1 (x  1) ŠÈx#  8  $‹


2 33

œ

(x
2) ŠÈx#  12  4‹

x Ä 2 Èx#  12  4

œ

2
1

œ lim x ˆ2  Èx‰ œ 4(2  2) œ 16

(x
1) ˆÈx  3  #‰ (x  3)
4 xÄ1

x Ä
1 È x #  )  $

œ

œ

" 6

œ lim

(x  1) ŠÈx#  8  $‹

œ
"3

ax#  12b
16

œ lim

x Ä 2 (x
2) ŠÈx#  12  4‹

œ

4 È16  4

ax  2b ŠÈx#  5  3‹

È x#  5  3 x
2 x Ä
2

œ lim

" È9  3

œ

x Ä * Èx  3

(1  1)(1  1) 111

œ

# v Ä # (v  2) av  4b

x ˆ2  È x ‰ ˆ 2
È x ‰ 2
Èx xÄ%

œ lim

œ

v#  2v  4

œ lim

œ lim

œ lim

2

x Ä 1 ax
1bax  1b

x Ä
2 ŠÈx#  5
3‹ ŠÈx#  5  3‹

ax  2b ŠÈx#  5  3‹

œ
#"

œ lim
x1 œ
1

ŠÈx#  12
4‹ ŠÈx#  12  4‹

x Ä 2 (x
2) ŠÈx#  12  4‹

œ

œ

ŠÈx#  8
$‹ ŠÈx#  8  $‹

lim

(x  1)(x
1) lim x Ä
1 (x  1) ŠÈx#  )  $‹

lim

5y  8

(x
1) ˆÈx  3  2‰ È ˆ x  3
#‰ ˆ È x  3  #‰ xÄ1

39. lim

40.

œ
21

œ lim

È x#  8
3 x1

œ lim


2 4

xÄ1

# v Ä # (v
2)(v  2) av  4b

x(4
x) x Ä % 2
Èx

œ


2

œ
13

1 œ lim Š ax
12x bax  1b † x ‹ œ lim

au#  "b (u  1)(u
1) au#  u  1b (u
1)

œ lim

Èx
3 x
9

xÄ*

x

xÄ1

œ lim

35. lim

b 1b b ax c 1b c 1bax b 1b

ax

3 #


1  2
1
2

# y Ä ! 3y
16

xÄ1

œ

œ

# x Ä
# x

œ lim ˆ 1
x x † ax

œ lim

t2

t Ä
" t
2

œ lim

1cx x

12 11

œ

œ lim

œ lim

x Ä 1 x
1

% v Ä # v
16

t2


2(x  2)

v$
8

34. lim

xÄ#

œ lim

œ lim

$ u Ä 1 u
1

38.

t Ä
" (t
2)(t  1)

# # y Ä ! y a3y
16b

 x b1 1 x

œ lim (x
5) œ 2
5 œ
3

t Ä " t1

(t  2)(t  1)

œ lim

x Ä
&

œ lim

# x Ä
# x (x  2)

% # y Ä 0 3y
16y

31. lim

(t  2)(t
1)

œ lim

28.

30. lim

xÄ#

œ lim (x
2) œ
&
# œ
7

œ

œ

œ lim

" 2

ax  2b ŠÈx#  5  3‹

x Ä
2

È9  3
4

ax #  5 b
9

œ
23

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 2.2 Limit of a Function and Limit Laws 41.

2
È x#
5 x3 x Ä
3

lim

œ

Š2
Èx#
5‹ Š2  Èx#
5‹

œ lim

(x  3) Š2  Èx#
5‹

x Ä
3

9
x#

lim

œ

x Ä
3 (x  3) Š2  Èx#
5‹

4
x x Ä 4 5
È x#  9

œ lim

x Ä 4 Š5
Èx#  9‹ Š5  Èx#  9‹

a4
xb Š5  Èx#  9‹

xÄ4

x Ä
3 (x  3) Š2  Èx#
5‹

a4
xb Š5  Èx#  9‹

œ lim

42. lim

16
x#

x Ä
3 (x  3) Š2  Èx#
5‹

(3
x)(3  x)

lim

œ lim

(4
x)(4  x)

xÄ4

œ lim

3
x

x Ä
3 2  È x #
5

œ

6 2  È4

œ

3 2

a4
xb Š5  Èx#  9‹ 25
ax#  9b

xÄ4

a4
xb Š5  Èx#  9‹

œ lim

4
ax #
5 b

œ lim

œ lim

xÄ4

5  È x#  9 4x

œ

5  È25 8

œ

5 4

2

43. lim a2sin x
1b œ 2sin 0
1 œ 0
1 œ
1

44. lim sin2 x œ Š lim sin x‹ œ asin 0b2 œ 02 œ 0

45. lim sec x œ

46. lim tan x œ

xÄ0

xÄ0

47. lim

xÄ0

1  x  sin x 3cos x

1

lim x Ä 0 cos x œ

œ

1  0  sin 0 3cos 0

1 cos 0

œ

œ

1 1

xÄ0

œ1

100 3

œ

xÄ0

xÄ0

sin x

lim x Ä 0 cos x

œ

sin 0 cos 0

œ

0 1

œ0

1 3

48. lim ax2
1ba2
cos xb œ a02
1ba2
cos 0b œ a
1ba2
1b œ a
1ba1b œ
1 xÄ0

49. x Ä lim
1Èx  4 cosax  1b œ x Ä lim
1Èx  4 † x Ä lim
1cosax  1b œ È
1  4 † cos 0 œ È4
1 † 1 œ È4
1 50. lim È7  sec2 x œ É lim a7  sec2 xb œ É7  lim sec2 x œ È7  sec2 0 œ É7  a1b2 œ 2È2 xÄ0

xÄ0

xÄ0

51. (a) quotient rule (c) sum and constant multiple rules

(b) difference and power rules

52. (a) quotient rule (c) difference and constant multiple rules

(b) power and product rules

53. (a) xlim f(x) g(x) œ ’xlim f(x)“ ’ x lim g(x)“ œ (5)(
2) œ
10 Äc Äc Äc (b) xlim 2f(x) g(x) œ 2 ’xlim f(x)“ ’ xlim g(x)“ œ 2(5)(
2) œ
20 Äc Äc Äc (c) xlim [f(x)  3g(x)] œ xlim f(x)  3 xlim g(x) œ 5  3(
2) œ
1 Äc Äc Äc lim f(x) f(x) 5 5 xÄc (d) xlim œ lim f(x)
lim g(x) œ 5
(
2) œ 7 Ä c f(x)
g(x) x

54. (a) (b) (c) (d) 55. (a) (b)

Äc

x

Äc

lim [g(x)  3] œ lim g(x)  lim 3 œ
$  $ œ !

xÄ%

xÄ%

xÄ%

lim xf(x) œ lim x † lim f(x) œ (4)(0) œ 0

xÄ%

xÄ%

xÄ%

#

lim [g(x)] œ ’ lim g(x)“ œ [
3]# œ 9

xÄ%

#

g(x) x Ä % f(x)
1

lim

xÄ%

œ

Ä%

lim g(x)

x

lim f(x)
lim 1

xÄ%

xÄ%

œ


3 0
1

œ3

lim [f(x)  g(x)] œ lim f(x)  lim g(x) œ 7  (
3) œ 4

xÄb

xÄb

xÄb

lim f(x) † g(x) œ ’ lim f(x)“ ’ lim g(x)“ œ (7)(
3) œ
21

xÄb

xÄb

xÄb

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

49

50

Chapter 2 Limits and Continuity lim 4g(x) œ ’ lim 4“ ’ lim g(x)“ œ (4)(
3) œ
12

(c)

xÄb

xÄb

xÄb

xÄb

xÄb

7
3

œ
37

lim [p(x)  r(x)  s(x)] œ lim p(x)  lim r(x)  lim s(x) œ 4  0  (
3) œ 1

56. (a)

x Ä
#

x Ä
#

x Ä
#

x Ä
#

lim p(x) † r(x) † s(x) œ ’ lim p(x)“ ’ lim r(x)“ ’ lim s(x)“ œ (4)(0)(
3) œ 0

(b)

x Ä
#

x Ä
#

x Ä
#

x Ä
#

lim [
4p(x)  5r(x)]/s(x) œ ’
4 lim p(x)  5 lim r(x)“ ‚ lim s(x) œ [
4(4)  5(0)]/
3 œ

(c)

x Ä
#

57. lim

hÄ!

x Ä
#

(1  h)#
1# h

œ lim

hÄ!

(
2  h)#
(
2)# h hÄ!

58. lim 59. lim

hÄ!

ˆ #"
h ‰
ˆ "# ‰ h hÄ! È7  h
È7 h hÄ!

61. lim

1  2h  h#
1 h

œ lim

hÄ!

4
4hh#
4 h hÄ!

œ lim

[3(2  h)
4]
[3(2)
4] h

60. lim

œ

xÄb

lim f(x)/g(x) œ lim f(x)/ lim g(x) œ

(d)

œ lim

2 2
h
"

œ lim

h(2  h) h

œ lim

hÄ!

hÄ!

œ lim (h
4) œ
4 hÄ!


2
(
2  h)
2h(
#  h)

hÄ!

h ŠÈ7  h  È7‹


h

œ lim

œ
"4

h Ä ! h(4
2h)

œ lim

(7  h)
7

h Ä ! h ŠÈ7  h  È7‹

œ lim

h

h Ä ! h ŠÈ7hÈ7‹

œ lim

È3(0  h)  1
È3(0)  1 h hÄ!

œ lim

3

h Ä ! È3h  1  1

œ

œ lim

ŠÈ3h  1
"‹ ŠÈ3h  1  "‹ h ŠÈ3h  1 "‹

hÄ!

(3h  1)
"

œ lim

h Ä ! h ŠÈ3h  1  1 ‹

œ lim

3h

h Ä ! h ŠÈ3h  1  "‹

3 #

63. lim È5
2x# œ È5
2(0)# œ È5 and lim È5
x# œ È5
(0)# œ È5; by the sandwich theorem, xÄ!

xÄ!

lim f(x) œ È5

xÄ!

64. lim a2
x# b œ 2
0 œ 2 and lim 2 cos x œ 2(1) œ 2; by the sandwich theorem, lim g(x) œ 2 xÄ!

65. (a)

xÄ!

lim Š1


xÄ!

x# 6‹

œ1


0 6

xÄ!

œ 1 and lim 1 œ 1; by the sandwich theorem, lim

(b) For x Á 0, y œ (x sin x)/(2
2 cos x) lies between the other two graphs in the figure, and the graphs converge as x Ä 0.

66. (a)

lim Š "#


xÄ!

lim

xÄ!

1
cos x x#

x# 24 ‹

œ lim

1

xÄ! #


lim

x#

x Ä ! #4

œ

" #

x sin x

x Ä ! 2
2 cos x

xÄ!


0œ

" #

and lim

"

xÄ! #

œ1

œ "# ; by the sandwich theorem,

œ "# .

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

"

h Ä ! È 7 h  È 7

" #È 7

62. lim

"6 3

œ lim (2  h) œ 2

h(h
4) h

ŠÈ7  h
È7‹ ŠÈ7  h  È7‹

hÄ!

x Ä
#

œ3

œ lim


2h

hÄ!

œ lim

3h

hÄ! h

x Ä
#

Section 2.2 Limit of a Function and Limit Laws (b) For all x Á 0, the graph of f(x) œ (1
cos x)/x# lies between the line y œ "# and the parabola yœ

" #


x# /24, and the graphs converge as x Ä 0.

67. (a) f(x) œ ax#
*b/(x  3) x
3.1 f(x)
6.1
2.9
5.9

x f(x)


3.01
6.01


3.001
6.001


3.0001
6.0001


3.00001
6.00001


3.000001
6.000001


2.99
5.99


2.999
5.999


2.9999
5.9999


2.99999
5.99999


2.999999
5.999999

The estimate is lim f(x) œ
6. x Ä
$

(b)

(c) f(x) œ

x#
9 x3

œ

(x  3)(x
3) x3

œ x
3 if x Á
3, and lim (x
3) œ
3
3 œ
6. x Ä
$

68. (a) g(x) œ ax#
#b/ Šx
È2‹ x g(x)

1.4 2.81421

1.41 2.82421

1.414 2.82821

1.4142 2.828413

1.41421 2.828423

1.414213 2.828426

(b)

(c) g(x) œ

x#
2 x
È2

œ

Šx  È2‹ Šx
È2‹ Šx
È2‹

œ x  È2 if x Á È2, and

69. (a) G(x) œ (x  6)/ ax#  4x
12b x
5.9
5.99 G(x)
.126582
.1251564 x G(x)


6.1
.123456


6.01
.124843


5.999
.1250156
6.001
.124984

lim

x Ä È#


5.9999
.1250015
6.0001
.124998

Šx  È2‹ œ È2  È2 œ 2È2.


5.99999
.1250001
6.00001
.124999


5.999999
.1250000
6.000001
.124999

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

51

52

Chapter 2 Limits and Continuity (b)

(c) G(x) œ

x6 ax#  4x
12b

œ

x6 (x  6)(x
2)

œ

" x
#

70. (a) h(x) œ ax#
2x
3b / ax#
4x  3b x 2.9 2.99 h(x) 2.052631 2.005025 x h(x)

3.1 1.952380

3.01 1.995024

"

if x Á
6, and lim

x Ä
' x
2

œ

"
'
2

œ
"8 œ
0.125.

2.999 2.000500

2.9999 2.000050

2.99999 2.000005

2.999999 2.0000005

3.001 1.999500

3.0001 1.999950

3.00001 1.999995

3.000001 1.999999

(b)

(c) h(x) œ

x#
2x
3 x#
4x  3

œ

(x
3)(x  1) (x
3)(x
1)

œ

x1 x
1

71. (a) f(x) œ ax#
1b / akxk
1b x
1.1
1.01 f(x) 2.1 2.01
.9 1.9

x f(x)


.99 1.99

if x Á 3, and lim

x1

x Ä $ x
1

œ

31 3
1

œ

4 #

œ 2.


1.001 2.001


1.0001 2.0001


1.00001 2.00001


1.000001 2.000001


.999 1.999


.9999 1.9999


.99999 1.99999


.999999 1.999999

(b)

(c) f(x) œ

x#
" kx k
1

(x  1)(x
1)


1 œ  (x  x1)(x
1)
(x  1)

œ x  1, x   0 and x Á 1 , and lim (1
x) œ 1
(
1) œ 2. x Ä
1 œ 1
x, x  0 and x Á
1

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 2.2 Limit of a Function and Limit Laws 72. (a) F(x) œ ax#  3x  2b / a2
kxkb x
2.1
2.01 F(x)
1.1
1.01
1.9
.9

x F(x)


1.99
.99


2.001
1.001


2.0001
1.0001


2.00001
1.00001


2.000001
1.000001


1.999
.999


1.9999
.9999


1.99999
.99999


1.999999
.999999

(b)

(c) F(x) œ

x#  3x  2 2
kx k

(x  2)(x  1)

œ  (x  2)(x#
x") 2x

73. (a) g()) œ (sin ))/) ) .1 g()) .998334

, x 0 , and lim (x  1) œ
2  1 œ
1. x Ä
# œ x  1, x  0 and x Á
2

.01 .999983

.001 .999999

.0001 .999999

.00001 .999999

.000001 .999999


.1 .998334


.01 .999983


.001 .999999


.0001 .999999


.00001 .999999


.000001 .999999

74. (a) G(t) œ (1
cos t)/t# t .1 G(t) .499583

.01 .499995

.001 .499999

.0001 .5

.00001 .5

.000001 .5


.1 .499583


.01 .499995


.001 .499999


.0001 .5


.00001 .5


.000001 .5

) g()) lim g()) œ 1

)Ä!

(b)

t G(t)

lim G(t) œ 0.5

tÄ!

(b)

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

53

54

Chapter 2 Limits and Continuity

75. xlim f(x) exists at those points c where xlim x% œ xlim x# . Thus, c% œ c# Ê c# a1
c# b œ 0 Ê c œ 0, 1, or
1. Äc Äc Äc Moreover, lim f(x) œ lim x# œ 0 and lim f(x) œ lim f(x) œ 1. xÄ!

xÄ!

x Ä
1

xÄ1

76. Nothing can be concluded about the values of f, g, and h at x œ 2. Yes, f(2) could be 0. Since the conditions of the sandwich theorem are satisfied, lim f(x) œ
5 Á 0. xÄ#

lim f(x)
lim 5 % xÄ% œ xÄlim x
lim 2 œ

f(x)
5 x Ä % x
2

77. 1 œ lim

xÄ%

lim f(x)
5

xÄ%

%
#

xÄ%

xÄ%

78. (a) 1 œ lim

f(x) x#

lim f(x) lim f(x) œ xÄlim# x# œ xÄ %# Ê

(b) 1 œ lim

f(x) x#

œ ’ lim

x Ä
# x Ä
#

xÄ

#

x Ä
#

79. (a) 0 œ 3 † 0 œ ’ lim

xÄ#

Ê lim f(x)
5 œ 2(1) Ê lim f(x) œ 2  5 œ 7. xÄ%

lim f(x) œ 4.

x Ä
#

f(x) lim x" “ x “ ’x Ä
#

œ ’ lim

x Ä
#

f(x) ˆ " ‰ x “
#

Ê

lim

x Ä
#

f(x) x

œ
2.

f(x)
5 x
# “ ’xlim Ä#


5 (x
2)“ œ lim ’Š f(x) x
# ‹ (x
2)“ œ lim [f(x)
5] œ lim f(x)
5

f(x)
5 x
# “ ’xlim Ä#

(x
2)“ Ê lim f(x) œ 5 as in part (a).

xÄ#

Ê lim f(x) œ 5.

xÄ#

xÄ#

xÄ#

(b) 0 œ 4 † 0 œ ’ lim

xÄ#

80. (a) 0 œ 1 † 0 œ ’ lim

f(x) # “ ’ lim xÄ! x xÄ!

(b) 0 œ 1 † 0 œ 81. (a)

lim x sin

xÄ!

(b)
1 Ÿ sin

82. (a)

" x

’ lim f(x) # “ ’ lim xÄ! x xÄ!

" x

xÄ#

#

x“ œ ’ lim

f(x)

# xÄ! x

x“ œ

lim ’ f(x) x# xÄ!

# “ ’ lim x# “ œ lim ’ f(x) x# † x “ œ lim f(x). That is, lim f(x) œ 0.

xÄ!

† x“ œ

xÄ!

lim f(x) . xÄ! x

That is,

xÄ!

lim f(x) xÄ! x

œ 0.

œ0

Ÿ 1 for x Á 0:

x  0 Ê
x Ÿ x sin

" x

Ÿ x Ê lim x sin

" x

œ 0 by the sandwich theorem;

x  0 Ê
x   x sin

" x

  x Ê lim x sin

" x

œ 0 by the sandwich theorem.

xÄ! xÄ!

lim x# cos ˆ x"$ ‰ œ 0

xÄ!

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

xÄ!

Section 2.3 The Precise Definition of a Limit (b)
1 Ÿ cos ˆ x"$ ‰ Ÿ 1 for x Á 0 Ê
x# Ÿ x# cos ˆ x"$ ‰ Ÿ x# Ê lim x# cos ˆ x"$ ‰ œ 0 by the sandwich xÄ!

theorem since lim x# œ 0. xÄ!

83-88. Example CAS commands: Maple: f := x -> (x^4
16)/(x
2); x0 := 2; plot( f(x), x œ x0-1..x0+1, color œ black, title œ "Section 2.2, #83(a)" ); limit( f(x), x œ x0 ); In Exercise 85, note that the standard cube root, x^(1/3), is not defined for x (surd(x+1, 3)
1)/x. Mathematica: (assigned function and values for x0 and h may vary) Clear[f, x] f[x_]:=(x3
x2
5x
3)/(x  1)2 x0=
1; h = 0.1; Plot[f[x],{x, x0
h, x0  h}] Limit[f[x], x Ä x0] 2.3 THE PRECISE DEFINITION OF A LIMIT 1. Step 1: Step 2:

kx
5k  $ Ê
$  x
5  $ Ê
$  5  x  $  5 $  5 œ 7 Ê $ œ 2, or
$  5 œ 1 Ê $ œ 4. The value of $ which assures kx
5k  $ Ê 1  x  7 is the smaller value, $ œ 2.

Step 1: Step 2:

kx
2k  $ Ê
$  x
2  $ Ê
$  #  x  $  2
$  2 œ 1 Ê $ œ 1, or $  2 œ 7 Ê $ œ 5. The value of $ which assures kx
2k  $ Ê 1  x  7 is the smaller value, $ œ 1.

Step 1: Step 2:

kx
(
3)k  $ Ê
$  x  $  $ Ê
$
3  x  $
3
$
3 œ
7# Ê $ œ "# , or $
$ œ
"# Ê $ œ 5# .

2.

3.

The value of $ which assures kx
(
3)k  $ Ê
7#  x 
"# is the smaller value, $ œ "# . 4.

Step 1:

¸x
ˆ
3# ‰¸  $ Ê
$  x 

3 #

 $ Ê
$


3 #

x$


3 #

Step 2:

œ
7# Ê $ œ #, or $
3# œ
"# Ê $ œ 1. The value of $ which assures ¸x
ˆ
3# ‰¸  $ Ê
7#  x 
"# is the smaller value, $ œ ".

Step 1:

¸x
"# ¸  $ Ê
$  x



$


3 #

5. " #

 $ Ê
$ 

" #

x$

" #

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

55

56

Chapter 2 Limits and Continuity Step 2:

" or $  #" œ 47 Ê $ œ 14 . "¸ 4 ¸ The value of $ which assures x
#  $ Ê 9  x 


$ 

" #

œ

4 9

Ê $œ

" 18 ,

4 7

is the smaller value, $ œ

" 18 .

6.

Step 1: Step 2:

kx
3k  $ Ê
$  x
3  $ Ê
$  3  x  $  3
$  $ œ 2.7591 Ê $ œ 0.2409, or $  $ œ 3.2391 Ê $ œ 0.2391. The value of $ which assures kx
3k  $ Ê 2.7591  x  3.2391 is the smaller value, $ œ 0.2391.

7. Step 1: Step 2:

kx
5k  $ Ê
$  x
5  $ Ê
$  5  x  $  5 From the graph,
$  5 œ 4.9 Ê $ œ 0.1, or $  5 œ 5.1 Ê $ œ 0.1; thus $ œ 0.1 in either case.

8. Step 1: Step 2:

kx
(
3)k  $ Ê
$  x  3  $ Ê
$
3  x  $
3 From the graph,
$
3 œ
3.1 Ê $ œ 0.1, or $
3 œ
2.9 Ê $ œ 0.1; thus $ œ 0.1.

9. Step 1: Step 2:

kx
1k  $ Ê
$  x
1  $ Ê
$  1  x  $  1 9 7 From the graph,
$  1 œ 16 Ê $ œ 16 , or $  1 œ 25 16 Ê $ œ

10. Step 1: Step 2:

kx
3k  $ Ê
$  x
3  $ Ê
$  3  x  $  3 From the graph,
$  3 œ 2.61 Ê $ œ 0.39, or $  3 œ 3.41 Ê $ œ 0.41; thus $ œ 0.39.

11. Step 1:

kx
2k  $ Ê
$  x
2  $ Ê
$  2  x  $  2 From the graph,
$  2 œ È3 Ê $ œ 2
È3 ¸ 0.2679, or $  2 œ È5 Ê $ œ È5
2 ¸ 0.2361; thus $ œ È5
2.

Step 2:

12. Step 1: Step 2:

9 16 ;

thus $ œ

kx
(
1)k  $ Ê
$  x  1  $ Ê
$
1  x  $
1 From the graph,
$
1 œ
thus $ œ

È5
2 # .

È5 #

Ê $œ

È5
2 #

¸ 0.1180, or $
1 œ


13. Step 1: Step 2:

kx
(
1)k  $ Ê
$  x  1  $ Ê
$
1  x  $
1 7 16 From the graph,
$
1 œ
16 9 Ê $ œ 9 ¸ 0.77, or $
1 œ
25 Ê

14. Step 1:

¸x
"# ¸  $ Ê
$  x


Step 2:

7 16 .

From the graph,
$  thus $ œ 0.00248.

" #

œ

" #  1 2.01

$ Ê
$  Ê $œ

1 2




" #

x$

" #.01

" #

¸ 0.00248, or $ 

" #

œ

È3 #

9 25

Ê $œ

2
È3 #

œ 0.36; thus $ œ

1 1.99

Ê $œ

1 1.99




¸ 0.1340;

9 25

" #

œ 0.36.

¸ 0.00251;

15. Step 1: Step 2:

k(x  1)
5k  0.01 Ê kx
4k  0.01 Ê
0.01  x
4  0.01 Ê 3.99  x  4.01 kx
4k  $ Ê
$  x
4  $ Ê
$  4  x  $  4 Ê $ œ 0.01.

16. Step 1:

k(2x
2)
(
6)k  0.02 Ê k2x  4k  0.02 Ê
0.02  2x  4  0.02 Ê
4.02  2x 
3.98 Ê
2.01  x 
1.99 kx
(
2)k  $ Ê
$  x  2  $ Ê
$
2  x  $
2 Ê $ œ 0.01.

Step 2: 17. Step 1: Step 2:

¹Èx  1
"¹  0.1 Ê
0.1  Èx  1
"  0.1 Ê 0.9  Èx  1  1.1 Ê 0.81  x  1  1.21 Ê
0.19  x  0.21 kx
0k  $ Ê
$  x  $ . Then,
$ œ
!Þ"* Ê $ œ !Þ"* or $ œ !Þ#"; thus, $ œ 0.19.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 2.3 The Precise Definition of a Limit 18. Step 1: Step 2:

¸Èx
"# ¸  0.1 Ê
0.1  Èx
"#  0.1 Ê 0.4  Èx  0.6 Ê 0.16  x  0.36 ¸x
"4 ¸  $ Ê
$  x
4"  $ Ê
$  4"  x  $  4" . Then,
$ 

19. Step 1: Step 2:

20. Step 1: Step 2:

21. Step 1: Step 2:

22. Step 1: Step 2:

57

" 4

œ 0.16 Ê $ œ 0.09 or $ 

" 4

œ 0.36 Ê $ œ 0.11; thus $ œ 0.09.

¹È19
x
$¹  " Ê
"  È19
x
$  1 Ê 2  È19
x  % Ê 4  19
x  16 Ê
%  x
19 
16 Ê 15  x  3 or 3  x  15 kx
10k  $ Ê
$  x
10  $ Ê
$  10  x  $  10. Then
$  10 œ 3 Ê $ œ 7, or $  10 œ 15 Ê $ œ 5; thus $ œ 5. ¹Èx
7
4¹  1 Ê
"  Èx
7
%  1 Ê 3  Èx
7  5 Ê 9  x
7  25 Ê 16  x  32 kx
23k  $ Ê
$  x
23  $ Ê
$  23  x  $  23. Then
$  23 œ 16 Ê $ œ 7, or $  23 œ 32 Ê $ œ 9; thus $ œ 7. ¸ "x
4" ¸  0.05 Ê
0.05 

" x




" 4

 0.05 Ê 0.2 

" x

 0.3 Ê

kx
4k  $ Ê
$  x
4  $ Ê
$  4  x  $  4. 2 2 Then
$  % œ 10 3 or $ œ 3 , or $  4 œ 5 or $ œ 1; thus $ œ 3 .

10 #

x

10 3

or

10 3

 x  5.

kx#
3k  !.1 Ê
0.1  x#
3  0.1 Ê 2.9  x#  3.1 Ê È2.9  x  È3.1 ¹x
È3¹  $ Ê
$  x
È3  $ Ê
$  È3  x  $  È3. Then
$  È3 œ È2.9 Ê $ œ È3
È2.9 ¸ 0.0291, or $  È3 œ È3.1 Ê $ œ È3.1
È3 ¸ 0.0286; thus $ œ 0.0286.

23. Step 1: Step 2:

kx#
4k  0.5 Ê
0.5  x#
4  0.5 Ê 3.5  x#  4.5 Ê È3.5  kxk  È4.5 Ê
È4.5  x 
È3.5, for x near
2. kx
(
2)k  $ Ê
$  x  2  $ Ê
$
#  x  $
2. Then
$
# œ
È4.5 Ê $ œ È4.5
# ¸ 0.1213, or $
# œ
È3.5 Ê $ œ #
È3.5 ¸ 0.1292; thus $ œ È4.5
2 ¸ 0.12.

24. Step 1: Step 2:

25. Step 1: Step 2:

¸ "x
(
1)¸  0.1 Ê
0.1 

" x

 1  0.1 Ê
11 10 

" x

9 10 10 10 
10 Ê
10 11  x 
9 or
9  x 
11 .

kx
(
1)k  $ Ê
$  x  1  $ Ê
$
"  x  $
". " 10 " Then
$
" œ
10 9 Ê $ œ 9 , or $
" œ
11 Ê $ œ 11 ; thus $ œ

" 11 .

kax#
5b
11k  " Ê kx#
16k  1 Ê
"  x#
16  1 Ê 15  x#  17 Ê È15  x  È17. kx
4k  $ Ê
$  x
4  $ Ê
$  %  x  $  %. Then
$  % œ È15 Ê $ œ %
È15 ¸ 0.1270, or $  % œ È17 Ê $ œ È17
% ¸ 0.1231; thus $ œ È17
4 ¸ 0.12.

26. Step 1: Step 2:

27. Step 1: Step 2:

¸ 120 ¸ x
5  " Ê
" 

120 x


&1 Ê 4

120 x

6 Ê

" 4



x 120



" 6

Ê 30  x  20 or 20  x  30.

kx
24k  $ Ê
$  x
24  $ Ê
$  24  x  $  24. Then
$  24 œ 20 Ê $ œ 4, or $  24 œ 30 Ê $ œ 6; thus Ê $ œ 4. kmx
2mk  0.03 Ê
0.03  mx
2m  0.03 Ê
0.03  2m  mx  0.03  2m Ê 0.03 2
0.03 m x2 m . kx
2k  $ Ê
$  x
2  $ Ê
$  2  x  $  2. 0.03 0.03 Then
$  2 œ 2
0.03 m Ê $ œ m , or $  2 œ #  m Ê $ œ

0.03 m .

In either case, $ œ

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

0.03 m .

58

Chapter 2 Limits and Continuity kmx
3mk  c Ê
c  mx
3m  c Ê
c  3m  mx  c  3m Ê 3


28. Step 1:

kx
3k  $ Ê
$  x
3  $ Ê
$  $  B  $  $. Then
$  $ œ $
mc Ê $ œ mc , or $  $ œ $  mc Ê $ œ

Step 2:

¸(mx  b)
ˆ m#  b‰¸  - Ê
c  mx
m#  c Ê
c  ¸x
"# ¸  $ Ê
$  x
"#  $ Ê
$  "#  x  $  "# .

29. Step 1: Step 2:

Then
$ 

" #

œ

" #




c m

Ê $œ

c m,

or $ 

" #

œ

" #



c m

c m. m #

Ê $œ

c m

 x 3

In either case, $ œ

c m.

c m. " #




In either case, $ œ

c m.

 mx  c 

m #

Ê

c m

c m

x

" #



c m.

k(mx  b)
(m  b)k  0.05 Ê
0.05  mx
m  0.05 Ê
0.05  m  mx  0.05  m 0.05 Ê 1
0.05 m x" m .

30. Step 1:

kx
1k  $ Ê
$  x
1  $ Ê
$  "  x  $  ". 0.05 0.05 Then
$  " œ "
0.05 m Ê $ œ m , or $  " œ "  m Ê $ œ

Step 2:

0.05 m .

In either case, $ œ

0.05 m .

31. lim (3
2x) œ 3
2(3) œ
3 xÄ3

ka3
2xb
(
3)k  0.02 Ê
0.02  6
2x  0.02 Ê
6.02 
2x 
5.98 Ê 3.01  x  2.99 or 2.99  x  3.01. 0  k x
3k  $ Ê
$  x
3  $ Ê
$  $  x  $  $ . Then
$  $ œ 2.99 Ê $ œ 0.01, or $  $ œ 3.01 Ê $ œ 0.01; thus $ œ 0.01.

Step 1: Step 2:

32.

lim (
3x
#) œ (
3)(
1)
2 œ 1

x Ä
1

k(
3x
2)
1k  0.03 Ê
0.03 
3x
3  0.03 Ê 0.01  x  1 
0.01 Ê
1.01  x 
0.99. kx
(
1)k  $ Ê
$  x  1  $ Ê
$
"  x  $
1. Then
$
" œ
1.01 Ê $ œ 0.01, or $
" œ
0.99 Ê $ œ 0.01; thus $ œ 0.01.

Step 1: Step 2:

33. lim

x#
4

x Ä # x
#

34.

35.

œ lim

xÄ#

#

(x  2)(x
2) (x
2)

œ lim (x  2) œ #  # œ 4, x Á 2 xÄ#

(x  2)(x
2) (x
2)

Step 1:

¹Š xx

24 ‹

Step 2:

Ê 1.95  x  2.05, x Á 2. kx
2k  $ Ê
$  x
2  $ Ê
$  2  x  $  2. Then
$  2 œ 1.95 Ê $ œ 0.05, or $  2 œ 2.05 Ê $ œ 0.05; thus $ œ 0.05.

lim

x Ä
&

x#  6x  5 x5


4¹  0.05 Ê
0.05 

œ lim

x Ä
&

(x  5)(x  1) (x  5)


%  0.05 Ê 3.95  x  2  4.05, x Á 2

œ lim (x  1) œ
4, x Á
5. x Ä
&

(x  5)(x  ") (x  5)

Step 1:

# ¹Š x x 6x5 5 ‹

Step 2:

Ê
5.05  x 
4.95, x Á
5. kx
(
5)k  $ Ê
$  x  5  $ Ê
$
&  x  $
&. Then
$
& œ
5.05 Ê $ œ 0.05, or $
& œ
4.95 Ê $ œ 0.05; thus $ œ 0.05.


(
4)¹  0.05 Ê
0.05 

 4  0.05 Ê
4.05  x  1 
3.95, x Á
5

lim È1
5x œ È1
5(
3) œ È16 œ 4

x Ä
$

Step 1:

¹È1
5x
4¹  0.5 Ê
0.5  È1
5x
4  0.5 Ê 3.5  È1
5x  4.5 Ê 12.25  1
5x  20.25

Step 2:

Ê 11.25 
5x  19.25 Ê
3.85  x 
2.25. kx
(
3)k  $ Ê
$  x  3  $ Ê
$
$  x  $
$. Then
$
$ œ
3.85 Ê $ œ 0.85, or $
$ œ
2.25 Ê 0.75; thus $ œ 0.75.

36. lim

4

xÄ# x

Step 1:

œ

4 #

œ2

¸ 4x
2¸  0.4 Ê
0.4 

4 x


2  0.4 Ê 1.6 

4 x

 2.4 Ê

10 16



x 4



10 24

Ê

10 4

x

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

10 6

or

5 3

 x  25 .

Section 2.3 The Precise Definition of a Limit Step 2:

59

kx
2k  $ Ê
$  x
2  $ Ê
$  #  x  $  #. Then
$  # œ 53 Ê $ œ "3 , or $  # œ #5 Ê $ œ "# ; thus $ œ 3" .

37. Step 1: Step 2:

k(9
x)
5k  % Ê
%  4
x  % Ê
%
4 
x  %
4 Ê %  %  x  4
% Ê %
%  x  4  %. kx
4k  $ Ê
$  x
4  $ Ê
$  %  x  $  %. Then
$  4 œ
%  4 Ê $ œ %, or $  % œ %  % Ê $ œ %. Thus choose $ œ %.

38. Step 1:

k(3x
7)
2k  % Ê
%  3x
9  % Ê 9
%  3x  *  % Ê 3


Step 2:

39. Step 1: Step 2:

40. Step 1: Step 2:

41. Step 1: Step 2:

42. Step 1: Step 2:

43. Step 1: Step 2:

 x  3  3% .

kx
3k  $ Ê
$  x
3  $ Ê
$  3  x  $  3. Then
$  3 œ $
3% Ê $ œ 3% , or $  3 œ 3  3% Ê $ œ 3% . Thus choose $ œ 3% . ¹Èx
5
2¹  % Ê
%  Èx
5
2  % Ê 2
%  Èx
5  2  % Ê (2
%)#  x
5  (2  %)# Ê (2
%)#  &  x  (2  %)#  5. kx
9k  $ Ê
$  x
9  $ Ê
$  9  x  $  9. Then
$  * œ %#
%%  * Ê $ œ %%
%# , or $  * œ %#  %%  * Ê $ œ %%  %# . Thus choose the smaller distance, $ œ %%
%# . ¹È4
x
2¹  % Ê
%  È4
x
2  % Ê 2
%  È4
x  2  % Ê (2
%)#  %
x  (2  %)# Ê
(2  %)#  x
4 
(2
%)# Ê
(2  %)#  %  x 
(2
%)#  %. k x
0k  $ Ê
$  x  $ . Then
$ œ
(2  %)#  4 œ
%#
%% Ê $ œ %%  %# , or $ œ
(2
%)#  4 œ 4%
%# . Thus choose the smaller distance, $ œ 4%
%# . For x Á 1, kx#
1k  % Ê
%  x#
"  % Ê "
%  x#  "  % Ê È1
%  kxk  È1  % Ê È"
%  x  È1  % near B œ ". kx
1k  $ Ê
$  x
1  $ Ê
$  "  x  $  ". Then
$  " œ È1
% Ê $ œ "
È1
%, or $  1 œ È"  % Ê $ œ È"  %
1. Choose $ œ min š"
È1
%ß È1  %
"›, that is, the smaller of the two distances. For x Á
2, kx#
4k  % Ê
%  x#
4  % Ê 4
%  x#  4  % Ê È4
%  kxk  È4  % Ê
È4  %  x 
È4
% near B œ
2. kx
(
2)k  $ Ê
$  x  2  $ Ê
$
2  x  $
2. Then
$
2 œ
È%  % Ê $ œ È%  %
#, or $
# œ
È%
% Ê $ œ #
È%
%. Choose $ œ min šÈ%  %
#ß #
È%
%› . ¸ "x
1¸  % Ê
% 

" x


"% Ê "
% 

" x

"% Ê

" 1%

% "%,

" 1
%.

x

kx
1k  $ Ê
$  x
1  $ Ê "
$  x  "  $ . Then "
$ œ " " % Ê $ œ "
" " % œ " % % , or "  $ œ "
" % Ê $ œ Choose $ œ

44. Step 1:

% 3

" "
%


"œ

% "
%.

the smaller of the two distances.

¸ x"#
"3 ¸  % Ê
% 

" x#




" 3

% Ê

" 3


% 

" x#



" 3

% Ê

1
3% 3



" x#



1  $% 3

3 È $. Ê É 1 3 $%  kxk  É "
3 $% , or É " 3 $%  x  É "
$ % for x near

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Ê

3 "


%$ x# 

3 " 

%$ 60

Chapter 2 Limits and Continuity Step 2:

¹x
È3¹  $ Ê
$  x
È3  $ Ê È3
$  x  È3  $ . Then È3
$ œ É " 3 $% Ê $ œ È3
É " 3 $% , or È3  $ œ É "
3 $% Ê $ œ É "
3 $%
È3. Choose $ œ min šÈ3
É " 3 $% ß É "
3 $%
È3›.

45. Step 1: Step 2:

46. Step 1: Step 2:

47. Step 1:

#

¹Š xx 
3* ‹
(
6)¹  % Ê
%  (x
3)  6  %, x Á
3 Ê
%  x  3  % Ê
%
$  x  %
$. kx
(
3)k  $ Ê
$  x  3  $ Ê
$
$  x  $
3. Then
$
$ œ
%
$ Ê $ œ %, or $
$ œ %
$ Ê $ œ %. Choose $ œ %. #

¹Š xx

11 ‹
2¹  % Ê
%  (x  1)
2  %, x Á 1 Ê "
%  x  "  %. kx
1k  $ Ê
$  x
1  $ Ê "
$  x  "  $ . Then "
$ œ "
% Ê $ œ %, or "  $ œ "  % Ê $ œ %. Choose $ œ %. x  1: l(4
2x)
2l  % Ê !  2
2x  % since x  1Þ Thus, 1


% #

 x  !;

x   1: l(6x
4)
2l  % Ê ! Ÿ 6x
6  % since x   1. Thus, " Ÿ x  1  6% . Step 2:

48. Step 1: Step 2:

kx
1k  $ Ê
$  x
1  $ Ê "
$  x  1  $ . Then 1
$ œ "
#% Ê $ œ #% , or "  $ œ 1  6% Ê $ œ 6% . Choose $ œ 6% . x  !: k2x
0k  % Ê
%  2x  ! Ê
#%  x  0; x   0: ¸ x#
!¸  % Ê ! Ÿ x  #%.

k x
0k  $ Ê
$  x  $ . Then
$ œ
#% Ê $ œ #% , or $ œ #% Ê $ œ #%. Choose $ œ #% .

49. By the figure,
x Ÿ x sin

" x

Ÿ x for all x  0 and
x   x sin

then by the sandwich theorem, in either case, lim x sin xÄ!

50. By the figure,
x# Ÿ x# sin

" x

" x

" x

  x for x  0. Since lim (
x) œ lim x œ 0, xÄ!

œ 0.

xÄ!

Ÿ x# for all x except possibly at x œ 0. Since lim a
x# b œ lim x# œ 0, then

by the sandwich theorem, lim x# sin xÄ!

" x

xÄ!

œ 0.

xÄ!

51. As x approaches the value 0, the values of g(x) approach k. Thus for every number %  0, there exists a $  ! such that !  kx
0k  $ Ê kg(x)
kk  %. 52. Write x œ h  c. Then !  lx
cl  $ Í
$  x
c  $ , x Á c Í
$  ah  cb
c  $ , h  c Á c Í
$  h  $ , h Á ! Í !  lh
!l  $ . Thus, limfaxb œ L Í for any %  !, there exists $  ! such that lfaxb
Ll  % whenever !  lx
cl  $ x Äc

Í lfah  cb
Ll  % whenever !  lh
!l  $ Í lim fah  cb œ L. hÄ!

53. Let f(x) œ x# . The function values do get closer to
1 as x approaches 0, but lim f(x) œ 0, not
1. The function f(x) œ x# never gets arbitrarily close to
1 for x near 0.

xÄ!

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 2.3 The Precise Definition of a Limit

61

54. Let f(x) œ sin x, L œ "# , and x! œ 0. There exists a value of x (namely, x œ 16 ) for which ¸sin x
"# ¸  % for any given

%  0. However, lim sin x œ 0, not "# . The wrong statement does not require x to be arbitrarily close to x! . As another\ xÄ!

example, let g(x) œ sin "x , L œ #" , and x! œ 0. We can choose infinitely many values of x near 0 such that sin you can see from the accompanying figure. However, lim sin xÄ!

" x

" x

œ

" #

as

fails to exist. The wrong statement does not require all

values of x arbitrarily close to x! œ 0 to lie within %  0 of L œ "# . Again you can see from the figure that there are also infinitely many values of x near 0 such that sin "x œ 0. If we choose %  ¸sin "x
#" ¸  % for all values of x sufficiently near x! œ 0.

#

55. kA
*k Ÿ 0.01 Ê
0.01 Ÿ 1 ˆ x# ‰
9 Ÿ 0.01 Ê 8.99 Ÿ Ê

2É 8.99 1

ŸxŸ

2É 9.01 1

1 x# 4

" 4

we cannot satisfy the inequality

Ÿ 9.01 Ê

4 1

(8.99) Ÿ x# Ÿ

4 1

(9.01)

or 3.384 Ÿ x Ÿ 3.387. To be safe, the left endpoint was rounded up and the right

endpoint was rounded down. 56. V œ RI Ê (120)(10) 51

V R

ŸRŸ

œ I Ê ¸ VR
5¸ Ÿ 0.1 Ê
0.1 Ÿ (120)(10) 49

120 R


5 Ÿ 0.1 Ê 4.9 Ÿ

120 R

Ÿ 5.1 Ê

10 49

 

R 1#0

 

10 51

Ê

Ê 23.53 Ÿ R Ÿ 24.48.

To be safe, the left endpoint was rounded up and the right endpoint was rounded down. 57. (a)
$  x
1  0 Ê "
$  x  1 Ê f(x) œ x. Then kf(x)
2k œ kx
2k œ 2
x  2
1 œ 1. That is, kf(x)
2k   1   "# no matter how small $ is taken when "
$  x  1 Ê lim f(x) Á 2. xÄ1

(b) 0  x
1  $ Ê "  x  "  $ Ê f(x) œ x  1. Then kf(x)
1k œ k(x  1)
1k œ kxk œ x  1. That is, kf(x)
1k   1 no matter how small $ is taken when "  x  "  $ Ê lim f(x) Á 1. xÄ1

(c)
$  x
1  ! Ê "
$  x  1 Ê f(x) œ x. Then kf(x)
1.5k œ kx
1.5k œ 1.5
x  1.5
1 œ 0.5. Also, !  x
1  $ Ê 1  x  "  $ Ê f(x) œ x  1. Then kf(x)
1.5k œ k(x  1)
1.5k œ kx
0.5k œ x
0.5  "
0.5 œ 0.5. Thus, no matter how small $ is taken, there exists a value of x such that
$  x
1  $ but kf(x)
1.5k   "# Ê lim f(x) Á 1.5. xÄ1

58. (a) For 2  x  2  $ Ê h(x) œ 2 Ê kh(x)
4k œ 2. Thus for %  2, kh(x)
4k   % whenever 2  x  2  $ no matter how small we choose $  0 Ê lim h(x) Á 4. xÄ#

(b) For 2  x  2  $ Ê h(x) œ 2 Ê kh(x)
3k œ 1. Thus for %  1, kh(x)
3k   % whenever 2  x  2  $ no matter how small we choose $  0 Ê lim h(x) Á 3. xÄ#

(c) For 2
$  x  2 Ê h(x) œ x# so kh(x)
2k œ kx#
2k . No matter how small $  0 is chosen, x# is close to 4 when x is near 2 and to the left on the real line Ê kx#
2k will be close to 2. Thus if %  1, kh(x)
2k   % whenever 2
$  x  2 no mater how small we choose $  0 Ê lim h(x) Á 2. xÄ#

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

62

Chapter 2 Limits and Continuity

59. (a) For 3
$  x  3 Ê f(x)  4.8 Ê kf(x)
4k   0.8. Thus for %  0.8, kf(x)
4k   % whenever 3
$  x  3 no matter how small we choose $  0 Ê lim f(x) Á 4. xÄ$

(b) For 3  x  3  $ Ê f(x)  3 Ê kf(x)
4.8k   1.8. Thus for %  1.8, kf(x)
4.8k   % whenever 3  x  3  $ no matter how small we choose $  0 Ê lim f(x) Á 4.8. xÄ$

(c) For 3
$  x  3 Ê f(x)  4.8 Ê kf(x)
3k   1.8. Again, for %  1.8, kf(x)
3k   % whenever $
$  x  3 no matter how small we choose $  0 Ê lim f(x) Á 3. xÄ$

60. (a) No matter how small we choose $  0, for x near
1 satisfying
"
$  x 
"  $ , the values of g(x) are near 1 Ê kg(x)
2k is near 1. Then, for % œ "# we have kg(x)
2k   "# for some x satisfying
"
$  x 
"  $ , or !  kx  1k  $ Ê

lim g(x) Á 2.

x Ä
1

(b) Yes, lim g(x) œ 1 because from the graph we can find a $  ! such that kg(x)
1k  % if !  kx
(
1)k  $ . x Ä
1

61-66. Example CAS commands (values of del may vary for a specified eps): Maple: f := x -> (x^4-81)/(x-3);x0 := 3; plot( f(x), x=x0-1..x0+1, color=black, # (a) title="Section 2.3, #61(a)" ); L := limit( f(x), x=x0 ); # (b) epsilon := 0.2; # (c) plot( [f(x),L-epsilon,L+epsilon], x=x0-0.01..x0+0.01, color=black, linestyle=[1,3,3], title="Section 2.3, #61(c)" ); q := fsolve( abs( f(x)-L ) = epsilon, x=x0-1..x0+1 ); # (d) delta := abs(x0-q); plot( [f(x),L-epsilon,L+epsilon], x=x0-delta..x0+delta, color=black, title="Section 2.3, #61(d)" ); for eps in [0.1, 0.005, 0.001 ] do # (e) q := fsolve( abs( f(x)-L ) = eps, x=x0-1..x0+1 ); delta := abs(x0-q); head := sprintf("Section 2.3, #61(e)\n epsilon = %5f, delta = %5f\n", eps, delta ); print(plot( [f(x),L-eps,L+eps], x=x0-delta..x0+delta, color=black, linestyle=[1,3,3], title=head )); end do: Mathematica (assigned function and values for x0, eps and del may vary): Clear[f, x] y1: œ L
eps; y2: œ L  eps; x0 œ 1; f[x_]: œ (3x2
(7x  1)Sqrt[x]  5)/(x
1) Plot[f[x], {x, x0
0.2, x0  0.2}] L: œ Limit[f[x], x Ä x0] eps œ 0.1; del œ 0.2; Plot[{f[x], y1, y2},{x, x0
del, x0  del}, PlotRange Ä {L
2eps, L  2eps}]

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 2.4 One-Sided Limits 2.4 ONE-SIDED LIMITS 1. (a) True (e) True (i) False

(b) True (f) True (j) False

(c) False (g) False (k) True

(d) True (h) False (l) False

2. (a) True (e) True (i) True

(b) False (f) True (j) False

(c) False (g) True (k) True

(d) True (h) True

3. (a)

lim f(x) œ

x Ä #b

2 #

 " œ #, lim c f(x) œ $
# œ " xÄ#

(b) No, lim f(x) does not exist because lim b f(x) Á lim c f(x) xÄ# xÄ# xÄ# (c) lim c f(x) œ 4#  1 œ 3, lim b f(x) œ 4#  " œ $ xÄ%

xÄ%

(d) Yes, lim f(x) œ 3 because 3 œ lim c f(x) œ lim b f(x) xÄ% xÄ% xÄ% 4. (a)

lim f(x) œ

x Ä #b

2 #

œ 1, lim c f(x) œ $
# œ ", f(2) œ 2 xÄ#

(b) Yes, lim f(x) œ 1 because " œ lim b f(x) œ lim c f(x) xÄ# xÄ# xÄ# (c) lim c f(x) œ 3
(
1) œ 4, lim b f(x) œ 3
(
1) œ 4 x Ä
"

x Ä
"

(d) Yes, lim f(x) œ 4 because 4 œ x Ä
"

lim

x Ä
"c

f(x) œ

lim

x Ä
"b

f(x)

5. (a) No, lim b f(x) does not exist since sin ˆ "x ‰ does not approach any single value as x approaches 0 xÄ! (b) lim c f(x) œ lim c 0 œ 0 xÄ!

(c)

xÄ!

lim f(x) does not exist because lim b f(x) does not exist xÄ! xÄ!

6. (a) Yes, lim b g(x) œ 0 by the sandwich theorem since
Èx Ÿ g(x) Ÿ Èx when x  0 xÄ! (b) No, lim c g(x) does not exist since Èx is not defined for x  0 xÄ!

(c) No, lim g(x) does not exist since lim c g(x) does not exist xÄ! xÄ! 7. (aÑ

lim f(x) œ " œ lim b f(x) xÄ1 (c) Yes, lim f(x) œ 1 since the right-hand and left-hand (b)

x Ä 1c

xÄ1

limits exist and equal 1

8. (a)

(b)

lim f(x) œ 0 œ lim c f(x) xÄ1

x Ä 1b

(c) Yes, lim f(x) œ 0 since the right-hand and left-hand xÄ1

limits exist and equal 0

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

63

64

Chapter 2 Limits and Continuity

9. (a) domain: 0 Ÿ x Ÿ 2 range: 0  y Ÿ 1 and y œ 2 (b) xlim f(x) exists for c belonging to Äc (0ß 1)  ("ß #) (c) x œ 2 (d) x œ 0

10. (a) domain:
_  x  _ range:
" Ÿ y Ÿ 1 (b) xlim f(x) exists for c belonging to Äc (
_ß
1)  (
"ß ")  ("ß _) (c) none (d) none

11.

x Ä
!Þ&c

lim

13.

x Ä
#b

14.

x Ä 1c

15.

h Ä !b

lim

2 0.5  2 È3 É 3/2 É xx  É

1 œ
0.5  1 œ 1/2 œ

lim

x Ä 1b

" 1 È0 œ ! É "1
É xx
# œ # œ

5‰ ˆ x x 1 ‰ ˆ 2x ˆ
2 ‰ 2(
2)  5 ˆ1‰ x#  x œ
2  1 Š (
2)#  (
2) ‹ œ (2) 2 œ 1

lim

ˆ x " 1 ‰ ˆ x x 6 ‰ ˆ 3
7 x ‰ œ ˆ 1 " 1 ‰ ˆ 1 1 6 ‰ ˆ 3
7 1 ‰ œ ˆ "# ‰ ˆ 71 ‰ ˆ 27 ‰ œ 1

lim

Èh#  4h  5
È5 h

œ lim b hÄ! 16.

12.

lim

h Ä !c

(b)

18. (a)

(b)

ah#  4h  5b
5

h ŠÈh#  4h  5  È5‹

È6
È5h#  11h  6 h

œ lim c hÄ! 17. (a)

œ lim b Š hÄ!

x Ä
#c

lim

x Ä 1b

lim

x Ä 1c

œ lim c Š hÄ!

6
a5h#  11h  6b

lim

lim

œ lim b hÄ!

h ŠÈ6  È5h#  11h  6‹

x Ä
#b

kx 2 k x 2

(x  3)

œ

kx  2 k x2

œ

lim

lim

x Ä
#c

È2x (x
1) kx
1 k

È2x (x
1) kx
1 k

œ lim b xÄ1


h(5h  11)

h ŠÈ6  È5h#  11h  6‹

(x  3)

x Ä
#b

œ

œ

04 È5  È5

œ

2 È5

È5h#  11h  6 È6
È5h#  11h  6 È ‹ Š È66  ‹ h  È5h#  11h  6

x Ä
#b

lim

h(h  4)

h ŠÈh#  4h  5  È5‹

œ lim c hÄ!

œ

(x  3)

Èh#  4h  5
È5 È # 4h  5  È5 ‹ Š Èhh#  ‹ h  4h  5  È5

(x2) (x#)


(0  11) È6  È6

11 œ
2È 6

akx  2k œ ax  2b for x 
2b

(x  3) œ a(
2)  3b œ 1

(x  3) ’
(x(x#2) ) “

lim

œ

x Ä
#c

akx  2k œ
(x  2) for x 
2b

(x  3)(
1) œ
(
2  3) œ
1

È2x (x
1) (x
1)

akx
1k œ x
1 for x  1b

œ lim b È2x œ È2 xÄ1

œ lim c xÄ1

È2x (x
1)
(x
1)

akx
1k œ
(x
1) for x  1b

œ lim c
È2x œ
È2 xÄ1

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 2.4 One-Sided Limits 19. (a)

) Ä $b

20. (a)

t Ä %b

Ú) Û )

lim

œ1

3 3

lim at
ÚtÛb œ 4
4 œ 0

21. lim

sin È2) È 2)

22. lim

sin kt t

23. lim

sin 3y 4y

)Ä!

tÄ!

yÄ!

24.

œ

tan 2x x xÄ!

25. lim

2t

27. lim

xÄ!

)Ä!

œ lim c ˆ "3 † hÄ! sin 2x ‰ ˆ cos 2x x xÄ!

œ lim

œ 2 lim

t

sin t t Ä ! ˆ cos t ‰

x csc 2x cos 5x

œ

3h ‰ sin 3h

x  x cos x

2 3

lim at
ÚtÛb œ 4
3 œ 1

" ‰ cos 5x

xÄ!

œ

" 3

Œ

œ

" lim

)Ä!c

(where ) œ kt) (where ) œ 3y)

3 4

" 3

œ

sin ) )

"

‹ Š lim x Ä ! cos 2x xÄ!

œ Š #" lim

†1œ

2 sin 2x #x ‹

t

Ä!

" 3

(where ) œ 3h)

œ1†2œ2

t

"

‹ Š lim cos 5x ‹ x Ä ! sin 2x xÄ!

6x# cos x

œ lim ˆ sin xxcos x 

œk†1œk

tÄ!

x Ä ! sin x sin 2x

x Ä ! sin x cos x

t Ä %c

œ

œ 2 Š lim cos t‹ Œ lim" sin t  œ 2 † " † " œ 2

t cos t sin t

tÄ!

xÄ!

(b)

Ú) Û )

lim

3 sin ) 4 )lim Ä! )

œ

œ Š lim

sin 2x

x Ä ! x cos 2x

œ 2 lim

xÄ!

)Ä!

" " sin 3h 3 h lim Ä !c ˆ 3h ‰

œ

œ lim

œ lim ˆ sinx2x †

sin ) )

œ k lim

sin 3y 3 4 ylim Ä ! 3y

28. lim 6x# (cot x)(csc 2x) œ lim 29. lim

k sin ) )

œ lim

3 sin 3y " 4 ylim 3y Ä!

œ

h

t Ä ! tan t

k sin kt kt

tÄ!

) Ä $c

(where x œ È2))

œ1

sin x x

xÄ!

œ lim

lim h Ä !c sin 3h

26. lim

œ lim

(b)

2x

œ lim ˆ3 cos x †

x sin x

xÄ!

x cos x ‰ sin x cos x

œ lim ˆ sinx x † xÄ!

†

œ ˆ #" † 1‰ (1) œ

2x ‰ sin 2x

" ‰ cos x

" #

œ3†"†1œ3

 lim

x

x Ä ! sin x

œ lim Š sin" x ‹ † lim ˆ cos" x ‰  lim Š sin" x ‹ œ (1)(1)  1 œ 2 xÄ!

30. lim

xÄ!

xÄ!

x

x#
x  sin x #x

1
cos ) ) Ä ! sin 2)

31. lim

œ lim

)Ä!

)Ä!

34. lim

sin (sin h) sin h sin )

) Ä ! sin 2)

36. lim

sin 5x

x Ä ! sin 4x

œ

œ lim

)Ä!

œ lim

)Ä!

sin ) )

sin ) )

)Ä!

 "# (1) œ 0

1
cos2 ) a2sin ) cos )ba1  cos )b

œ lim

)Ä!

sin2 ) a2sin ) cos )ba1  cos )b

œ lim

xÄ!

xa1 c cos xb 9x2 sin2 3x 9x2

œ lim

1 c cos x 9x

2 x Ä ! ˆ sin3x3x ‰

œ

" lim ˆ 1 9 x

Ä!

cos x ‰ x

2 lim ˆ sin3x3x ‰

xÄ!

œ

" 9 a0 b 12

œ0

œ 1 since ) œ sin h Ä 0 as h Ä 0 2) ‰ #)

5x œ lim ˆ sin sin 4x †

4x 5x

)Ä!

œ lim

" #

œ 1 since ) œ 1
cos t Ä 0 as t Ä 0

sin ) œ lim ˆ sin 2) †

xÄ!

 "# ˆ sinx x ‰‰ œ 0


œ0

0 a2ba2b

xa1
cos xb sin2 3x xÄ!

sin(1
cos t) 1
cos t

x

a1
cos )ba1  cos )b a2sin ) cos )ba1  cos )b

œ lim

33. lim

35. lim

" #

xÄ!

œ lim

x
x cos x sin2 3x xÄ!

hÄ!

œ lim ˆ #x


sin ) a2cos )ba1  cos )b

32. lim

tÄ!

xÄ!

œ

" # )lim Ä!

† 54 ‰ œ

ˆ sin) ) †

5 4 xlim Ä!

2) ‰ sin 2)

ˆ sin5x5x †

œ

" #

4x ‰ sin 4x

†1†1œ œ

5 4

" #

†1†1œ

5 4

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

65

66

Chapter 2 Limits and Continuity

37. lim ) cos ) œ 0 † 1 œ 0 )Ä!

38. lim sin ) cot 2) œ lim sin ) )Ä!

)Ä!

tan 3x

39. lim

x Ä ! sin 8x

œ

3 8 xlim Ä!

40. lim

yÄ!

œ

sin 3x œ lim ˆ cos 3x †

xÄ!

" ‰ sin 8x

œ lim sin ) )Ä!

œ lim

xÄ!

3 8

†1†1†1œ

sin 3y sin 4y cos 5y y cos 4y sin 5y

yÄ!

œ lim

) cot 4) 2 2 ) Ä ! sin ) cot 2)

42. lim

)Ä!

sin ) cos ) 3) 2 ) cos sin 3)

œ lim

)Ä!

2 lim 4) cos2 4) cos ) ) Ä ! cos 2) sin 4)

œ lim

cos 4) sin 4) 2 2) 2 sin ) cos sin2 2)

" sin 8x

cos 2)

) Ä ! 2 cos )

†

8x 3x

œ

1 2

† 83 ‰

3 8

yÄ!

sin ) sin 3)

2 ) Ä ! ) cos ) cos 3)

)

œ lim

sin 4y cos 5y 3†4†5y œ lim Š siny3y ‹ Š cos 4y ‹ Š sin 5y ‹ Š 3†4†5y ‹

cos 5y ˆ 3†4 ‰ lim Š sin3y3y ‹ Š sin4y4y ‹ Š sin5y5y ‹ Š cos 4y ‹ 5 yÄ!

tan ) 2 ) Ä ! ) cot 3)

cos 2) 2sin ) cos )

sin 3x œ lim ˆ cos 3x †

ˆ cos"3x ‰ ˆ sin3x3x ‰ ˆ sin8x8x ‰ œ

sin 3y cot 5y y cot 4y

41. lim

œ

cos 2) sin 2)

œ1†1†1†1†

œ

12 5

œ lim ˆ sin) ) ‰ˆ sin3)3) ‰ˆ cos ) 3cos 3) ‰ œ a1ba1bˆ 13†1 ‰ œ 3 )Ä!

) cos 4) sin2 2) 2 2 ) Ä ! sin ) cos 2) sin 4)

œ lim

12 5

) cos 4) a2sin ) cos )b2 2 2 ) Ä ! sin ) cos 2) sin 4)

œ lim

) cos 4) ˆ4sin2 ) cos2 )‰ 2 2 ) Ä ! sin ) cos 2) sin 4)

œ lim

4) cos ) 4) cos ) œ lim ˆ sin4)4) ‰Š coscos ‹ œ lim Š sin14) ‹Š coscos ‹ œ ˆ 11 ‰Š 11†12 ‹ œ 1 2 2) 2 2) 2

)Ä!

2

)Ä!

2

4)

43. Yes. If lim b f(x) œ L œ lim c f(x), then xlim f(x) œ L. If lim b f(x) Á lim c f(x), then xlim f(x) does not exist. Äa Äa xÄa xÄa xÄa xÄa 44. Since xlim f(x) œ L if and only if lim b f(x) œ L and lim c f(x) œ L, then xlim f(x) can be found by calculating Äc Äc xÄc xÄc lim b f(x). xÄc

45. If f is an odd function of x, then f(
x) œ
f(x). Given lim b f(x) œ 3, then lim c f(x) œ
$. xÄ! xÄ! 46. If f is an even function of x, then f(
x) œ f(x). Given lim c f(x) œ 7 then xÄ#

can be said about

lim

x Ä
#c

lim

x Ä
#b

f(x) œ 7. However, nothing

f(x) because we don't know lim b f(x). xÄ#

47. I œ (5ß 5  $ ) Ê 5  x  &  $ . Also, Èx
5  % Ê x
5  %# Ê x  &  %# . Choose $ œ %# Ê lim Èx
5 œ 0. x Ä &b

48. I œ (%
$ ß %) Ê %
$  x  4. Also, È%
x  % Ê %
x  %# Ê x  %
%# . Choose $ œ %# Ê lim È%
x œ 0. x Ä %c

49. As x Ä 0
the number x is always negative. Thus, ¹ kxxk
(
1)¹  % Ê ¸
xx  1¸  % Ê 0  % which is always true independent of the value of x. Hence we can choose any $  0 with
$  x  ! Ê

2 ¸ x
2 ¸ 50. Since x Ä # we have x  2 and kx
2k œ x
2. Then, ¹ kxx

2 k
" ¹ œ x
2
"  % Ê 0  %

which is always true so long as x  #. Hence we can choose any $  !, and thus #  x  #  $ 2 Ê ¹ kxx

2k
"¹  % . Thus,

x
2

lim x Ä
#b kx
2k

x

lim x Ä ! c kx k

œ 1.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

œ
1.

Section 2.5 Continuity 51. (a) (b)

lim

x Ä %!!b

67

ÚxÛ œ 400. Just observe that if 400  x  401, then ÚxÛ œ 400. Thus if we choose $ œ ", we have for any

number %  ! that 400  x  400  $ Ê lÚxÛ
400l œ l400
400l œ !  %. lim c ÚxÛ œ 399. Just observe that if 399  x  400 then ÚxÛ œ 399. Thus if we choose $ œ ", we have for any

x Ä %!!

number %  ! that 400
$  x  400 Ê lÚxÛ
399l œ l399
399l œ !  %. (c) Since lim b ÚxÛ Á lim c ÚxÛ we conclude that lim ÚxÛ does not exist. x Ä %!!

x Ä %!!

52. (a)

x Ä %!!

lim f(x) œ lim b Èx œ È0 œ 0; ¸Èx
0¸  % Ê
%  Èx  % Ê !  x  %# for x positive. Choose $ œ %# xÄ! Ê lim b f(x) œ 0.

x Ä !b

xÄ!

(b)

lim f(x) œ lim c x# sin ˆ x" ‰ œ 0 by the sandwich theorem since
x# Ÿ x# sin ˆ x" ‰ Ÿ x# for all x Á 0. x Ä !c xÄ! Since kx#
0k œ k
x#
0k œ x#  % whenever kxk  È%, we choose $ œ È% and obtain ¸x# sin ˆ "x ‰
0¸  %

if
$  x  0. (c) The function f has limit 0 at x! œ 0 since both the right-hand and left-hand limits exist and equal 0. 2.5 CONTINUITY 1. No, discontinuous at x œ 2, not defined at x œ 2 2. No, discontinuous at x œ 3, " œ lim c g(x) Á g(3) œ 1.5 xÄ$ 3. Continuous on [
1ß 3] 4. No, discontinuous at x œ 1, 1.5 œ lim c k(x) Á lim b k(x) œ ! xÄ" xÄ" 5. (a) Yes

(b) Yes,

(c) Yes

(d) Yes

6. (a) Yes, f(1) œ 1

lim

x Ä
"b

f(x) œ 0

(b) Yes, lim f(x) œ 2 xÄ1

(c) No

(d) No

7. (a) No

(b) No

8. [
"ß !)  (!ß ")  ("ß #)  (#ß $) 9. f(2) œ 0, since lim c f(x) œ
2(2)  4 œ 0 œ lim b f(x) xÄ# xÄ# 10. f(1) should be changed to 2 œ lim f(x) xÄ1

11. Nonremovable discontinuity at x œ 1 because lim f(x) fails to exist ( limc f(x) œ 1 and lim b f(x) œ 0). xÄ" xÄ1 xÄ"

Removable discontinuity at x œ 0 by assigning the number lim f(x) œ 0 to be the value of f(0) rather than f(0) œ 1. xÄ!

12. Nonremovable discontinuity at x œ 1 because lim f(x) fails to exist ( lim c f(x) œ 2 and lim b f(x) œ 1). xÄ" xÄ1 xÄ" Removable discontinuity at x œ 2 by assigning the number lim f(x) œ 1 to be the value of f(2) rather than f(2) œ 2. xÄ#

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

68

Chapter 2 Limits and Continuity

13. Discontinuous only when x
2 œ 0 Ê x œ 2

14. Discontinuous only when (x  2)# œ 0 Ê x œ
2

15. Discontinuous only when x#
%x  $ œ ! Ê (x
3)(x
1) œ 0 Ê x œ 3 or x œ 1 16. Discontinuous only when x#
3x
10 œ 0 Ê (x
5)(x  2) œ 0 Ê x œ 5 or x œ
2 17. Continuous everywhere. ( kx
1k  sin x defined for all x; limits exist and are equal to function values.) 18. Continuous everywhere. ( kxk  " Á 0 for all x; limits exist and are equal to function values.) 19. Discontinuous only at x œ 0 20. Discontinuous at odd integer multiples of 1# , i.e., x = (2n
") 1# , n an integer, but continuous at all other x. 21. Discontinuous when 2x is an integer multiple of 1, i.e., 2x œ n1, n an integer Ê x œ

n1 # ,

n an integer, but

continuous at all other x. 22. Discontinuous when

1x #

is an odd integer multiple of 1# , i.e.,

1x #

œ (2n
1) 1# , n an integer Ê x œ 2n
1, n an

integer (i.e., x is an odd integer). Continuous everywhere else. 23. Discontinuous at odd integer multiples of 1# , i.e., x = (2n
1) 1# , n an integer, but continuous at all other x. 24. Continuous everywhere since x%  1   1 and
" Ÿ sin x Ÿ 1 Ê 0 Ÿ sin# x Ÿ 1 Ê 1  sin# x   1; limits exist and are equal to the function values. 25. Discontinuous when 2x  3  0 or x 
3# Ê continuous on the interval

3# ß _‰ . 26. Discontinuous when 3x
1  0 or x 

" 3

Ê continuous on the interval
3" ß _‰ .

27. Continuous everywhere: (2x
1)"Î$ is defined for all x; limits exist and are equal to function values. 28. Continuous everywhere: (2
x)"Î& is defined for all x; limits exist and are equal to function values. 29. Continuous everywhere since lim xÄ3 30. Discontinuous at x œ
2 since

x2
x
6 x
3

œ lim xÄ3

ax
3bax  2b x
3

œ lim ax  2b œ 5 œ ga3b xÄ3

lim faxb does not exist while fa
2b œ 4. x Ä
2

31. xlim sin (x
sin x) œ sin (1
sin 1) œ sin (1
0) œ sin 1 œ 0, and function continuous at x œ 1. Ä1 32. lim sin ˆ 1# cos (tan t)‰ œ sin ˆ 1# cos (tan (0))‰ œ sin ˆ 1# cos (0)‰ œ sin ˆ 1# ‰ œ 1, and function continuous at t œ !. tÄ!

33. lim sec ay sec# y
tan# y
1b œ lim sec ay sec# y
sec# yb œ lim sec a(y
1) sec# yb œ sec a("
") sec# 1b yÄ1

yÄ1

yÄ1

œ sec 0 œ 1, and function continuous at y œ ". 34. lim tan
14 cos ˆsin x"Î$ ‰‘ œ tan
14 cos (sin(0))‘ œ tan ˆ 14 cos (0)‰ œ tan ˆ 14 ‰ œ 1, and function continuous at x œ !. xÄ!

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 2.5 Continuity 35. lim cos ’ È19
13 sec 2t “ œ cos ’ È19
13 sec 0 “ œ cos tÄ!

1 È16

œ cos

1 4

œ

È2 # ,

and function continuous at t œ !.

36. lim1 Écsc# x  5È3 tan x œ Écsc# ˆ 16 ‰  5È3 tan ˆ 16 ‰ œ Ê4  5È3 Š È"3 ‹ œ È9 œ 3, and function continuous at xÄ

'

x œ 1' . 37. g(x) œ

x#
9 x
3

(x  3)(x
3) (x
3)

œ

38. h(t) œ

t#  3t
10 t
#

39. f(s) œ

s$
" s#
1

40. g(x) œ

œ

œ

œ x  3, x Á 3 Ê g(3) œ lim (x  3) œ 6 xÄ$

(t  5)(t
2) t
#

as#  s  1b (s
1) (s  1)(s
1)

x#
16 x#
3x
4

œ

œ t  5, t Á # Ê h(2) œ lim (t  5) œ 7 tÄ#

s#  s  " s1 ,

œ

(x  4)(x
4) (x
4)(x  1)

œ

x4 x1

s Á 1 Ê f(1) œ lim Š s sÄ1

#

s1 s1 ‹

4‰ , x Á 4 Ê g(4) œ lim ˆ xx  1 œ

xÄ%

œ

3 #

8 5

41. As defined, lim c f(x) œ (3)#
1 œ 8 and lim b (2a)(3) œ 6a. For f(x) to be continuous we must have xÄ$ xÄ$ 6a œ 8 Ê a œ 43 .

42. As defined,

lim

x Ä
#c

g(x) œ
2 and

4b œ
2 Ê b œ
"# .

lim

x Ä
#b

g(x) œ b(
2)# œ 4b. For g(x) to be continuous we must have

43. As defined, lim c f(x) œ 12 and lim b f(x) œ a# a2b
2a œ 2a#
2a. For f(x) to be continuous we must have xÄ# xÄ# 12 œ 2a#
2a Ê a œ 3 or a œ
2. 44. As defined, lim c g(x) œ
b b1

xÄ0

0
b b1

œ


b b1

œ b Ê b œ 0 or b œ
2.

45. As defined,

lim

x Ä
1 c

f(x) œ
2 and

and lim b g(x) œ a0b2  b œ b. For g(x) to be continuous we must have xÄ0

lim

x Ä
1 b

f(x) œ aa
1b  b œ
a  b, and

lim f(x) œ aa1b  b œ a  b and

x Ä 1c

lim f(x) œ 3. For f(x) to be continuous we must have
2 œ
a  b and a  b œ 3 Ê a œ

x Ä 1b

5 #

and b œ "# .

46. As defined, lim c g(x) œ aa0b  2b œ 2b and lim b g(x) œ a0b2  3a
b œ 3a
b, and xÄ0 xÄ0 lim g(x) œ a2b2  3a
b œ 4  3a
b and lim b g(x) œ 3a2b
5 œ 1. For g(x) to be continuous we must xÄ0

x Ä 2c

have 2b œ 3a
b and 4  3a
b œ 1 Ê a œ
3# and b œ
3# .

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

69

70

Chapter 2 Limits and Continuity

47. The function can be extended: f(0) ¸ 2.3.

48. The function cannot be extended to be continuous at x œ 0. If f(0) ¸ 2.3, it will be continuous from the right. Or if f(0) ¸
2.3, it will be continuous from the left.

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

50. The function can be extended: f(0) ¸ 7.39.

51. f(x) is continuous on [!ß "] and f(0)  0, f(1)  0 Ê by the Intermediate Value Theorem f(x) takes on every value between f(0) and f(1) Ê the equation f(x) œ 0 has at least one solution between x œ 0 and x œ 1.

52. cos x œ x Ê (cos x)
x œ 0. If x œ
1# , cos ˆ
1# ‰
ˆ
1# ‰  0. If x œ 1# , cos ˆ 1# ‰
for some x between


1 #

and

1 #

1 #

 0. Thus cos x
x œ 0

according to the Intermediate Value Theorem, since the function cos x
x is continuous.

53. Let f(x) œ x$
15x  1, which is continuous on [
4ß 4]. Then f(
4) œ
3, f(
1) œ 15, f(1) œ
13, and f(4) œ 5. By the Intermediate Value Theorem, f(x) œ 0 for some x in each of the intervals
%  x 
1,
"  x  1, and "  x  4. That is, x$
15x  1 œ 0 has three solutions in [
%ß 4]. Since a polynomial of degree 3 can have at most 3 solutions, these are the only solutions. 54. Without loss of generality, assume that a  b. Then F(x) œ (x
a)# (x
b)#  x is continuous for all values of x, so it is continuous on the interval [aß b]. Moreover F(a) œ a and F(b) œ b. By the Intermediate Value Theorem, since a  a # b  b, there is a number c between a and b such that F(x) œ a # b .

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 2.5 Continuity 55. Answers may vary. Note that f is continuous for every value of x. (a) f(0) œ 10, f(1) œ 1$
8(1)  10 œ 3. Since $  1  10, by the Intermediate Value Theorem, there exists a c so that !  c  1 and f(c) œ 1. (b) f(0) œ 10, f(
4) œ (
4)$
8(
4)  10 œ
22. Since
22 
È3  10, by the Intermediate Value Theorem, there exists a c so that
4  c  0 and f(c) œ
È3. (c) f(0) œ 10, f(1000) œ (1000)$
8(1000)  10 œ 999,992,010. Since 10  5,000,000  999,992,010, by the Intermediate Value Theorem, there exists a c so that !  c  1000 and f(c) œ 5,000,000.

56. All five statements ask for the same information because of the intermediate value property of continuous functions. (a) A root of f(x) œ x$
3x
1 is a point c where f(c) œ 0. (b) The points where y œ x$ crosses y œ 3x  1 have the same y-coordinate, or y œ x$ œ 3x  1 Ê f(x) œ x$
3x
1 œ 0. (c) x$
3x œ 1 Ê x$
3x
1 œ 0. The solutions to the equation are the roots of f(x) œ x$
3x
1. (d) The points where y œ x$
3x crosses y œ 1 have common y-coordinates, or y œ x$
3x œ 1 Ê f(x) œ x$
3x
1 œ !. (e) The solutions of x$
3x
1 œ 0 are those points where f(x) œ x$
3x
1 has value 0. 57. Answers may vary. For example, f(x) œ

sin (x
2) x
2

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

However, the discontinuity can be removed because f has a limit (namely 1) as x Ä 2. 58. Answers may vary. For example, g(x) œ

" x1

has a discontinuity at x œ
1 because lim g(x) does not exist. x Ä
"

Š lim c g(x) œ
_ and lim b g(x) œ _.‹ x Ä
" x Ä
" 59. (a) Suppose x! is rational Ê f(x! ) œ 1. Choose % œ "# . For any $  0 there is an irrational number x (actually infinitely many) in the interval (x!
$ ß x!  $ ) Ê f(x) œ 0. Then 0  kx
x! k  $ but kf(x)
f(x! )k œ 1  "# œ %, so x lim f(x) fails to exist Ê f is discontinuous at x! rational. Äx !

On the other hand, x! irrational Ê f(x! ) œ 0 and there is a rational number x in (x!
$ ß x!  $ ) Ê f(x) œ 1. Again x lim f(x) fails to exist Ê f is discontinuous at x! irrational. That is, f is discontinuous at Äx !

every point. (b) f is neither right-continuous nor left-continuous at any point x! because in every interval (x!
$ ß x! ) or (x! ß x!  $ ) there exist both rational and irrational real numbers. Thus neither limits lim f(x) and x Ä x!

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

x Ä x
!

60. Yes. Both f(x) œ x and g(x) œ x
g ˆ "# ‰

œ0 Ê

f(x) g(x)

" #

are continuous on [!ß "]. However

is discontinuous at x œ

f(x) g(x)

is undefined at x œ

" #

since

" #.

61. No. For instance, if f(x) œ 0, g(x) œ ÜxÝ, then h(x) œ 0 aÜxÝb œ 0 is continuous at x œ 0 and g(x) is not. 62. Let f(x) œ œ

" x
1

" (x  1)
1

œ

and g(x) œ x  1. Both functions are continuous at x œ 0. The composition f ‰ g œ f(g(x)) " x

is discontinuous at x œ 0, since it is not defined there. Theorem 10 requires that f(x) be

continuous at g(0), which is not the case here since g(0) œ 1 and f is undefined at 1. 63. 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 [aß b].

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71

72

Chapter 2 Limits and Continuity

64. Let f(x) be the new position of point x and let d(x) œ f(x)
x. The displacement function d is negative if x is the left-hand point of the rubber band and positive if x is the right-hand point of the rubber band. By the Intermediate Value Theorem, d(x) œ 0 for some point in between. That is, f(x) œ x for some point x, which is then in its original position. 65. If f(0) œ 0 or f(1) œ 1, we are done (i.e., c œ 0 or c œ 1 in those cases). Then let f(0) œ a  0 and f(1) œ b  1 because 0 Ÿ f(x) Ÿ 1. Define g(x) œ f(x)
x Ê g is continuous on [0ß 1]. Moreover, g(0) œ f(0)
0 œ a  0 and g(1) œ f(1)
1 œ b
1  0 Ê by the Intermediate Value Theorem there is a number c in (!ß ") such that g(c) œ 0 Ê f(c)
c œ 0 or f(c) œ c. 66. Let % œ

kf(c)k #

 0. Since f is continuous at x œ c there is a $  0 such that kx
ck  $ Ê kf(x)
f(c)k  %

Ê f(c)
%  f(x)  f(c)  %. If f(c)  0, then % œ "# f(c) Ê " #

" #

If f(c)  0, then % œ
f(c) Ê

f(c)  f(x)  3 #

3 #

f(c)  f(x) 

f(c) Ê f(x)  0 on the interval (c
$ ß c  $ ). " #

f(c) Ê f(x)  0 on the interval (c
$ ß c  $ ).

67. By Exercises 52 in Section 2.3, we have xlim faxb œ L Í lim fac  hb œ L. Äc hÄ0

Thus, faxb is continuous at x œ c Í xlim faxb œ facb Í lim fac  hb œ facb. Äc hÄ0

68. By Exercise 67, it suffices to show that lim sinac  hb œ sin c and lim cosac  hb œ cos c. hÄ0

hÄ0

Now lim sinac  hb œ lim
asin cbacos hb  acos cbasin hb‘ œ asin cbŠ lim cos h‹  acos cbŠ lim sin h‹ hÄ0

hÄ0

hÄ0

hÄ0

By Example 11 Section 2.2, lim cos h œ " and lim sin h œ !. So lim sinac  hb œ sin c and thus faxb œ sin x is hÄ0

continuous at x œ c. Similarly,

hÄ0

hÄ0

lim cosac  hb œ lim
acos cbacos hb
asin cbasin hb‘ œ acos cbŠ lim cos h‹
asin cbŠ lim sin h‹ œ cos c.

hÄ0

hÄ0

Thus, gaxb œ cos x is continuous at x œ c.

hÄ0

69. x ¸ 1.8794,
1.5321,
0.3473

70. x ¸ 1.4516,
0.8547, 0.4030

71. x ¸ 1.7549

72. x ¸ 1.5596

73. x ¸ 3.5156

74. x ¸
3.9058, 3.8392, 0.0667

75. x ¸ 0.7391

76. x ¸
1.8955, 0, 1.8955

hÄ0

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 2.6 Limits Involving Infinity; Asymptotes of Graphs 2.6 LIMITS INVOLVING INFINITY; ASMYPTOTES OF GRAPHS 1. (a) (c) (e) (g)

lim f(x) œ 0

(b)

xÄ2

lim

x Ä
3 c

f(x) œ 2

lim

x Ä
3 b

f(x) œ
2

(d) lim f(x) œ does not exist xÄ3

lim b f(x) œ
1

(f)

lim f(x) œ does not exist

(h) x lim f(x) œ 1 Ä_

xÄ0 xÄ0

lim f(x) œ  _

x Ä 0c

(i) x Ä lim f(x) œ 0
_ 2. (a) (c)

lim f(x) œ 2

(b)

lim f(x) œ 1

(d) lim f(x) œ does not exist

xÄ4

x Ä 2c

xÄ2

lim f(x) œ  _ x Ä
3 b (g) lim f(x) œ  _ (e)

x Ä
3

(i)

lim f(x) œ
3

x Ä 2b

lim f(x) œ
_

(f)

x Ä
3 c

lim

(h)

x Ä 0b

(k) x lim f(x) œ 0 Ä_

lim f(x) œ  _

lim f(x) œ does not exist

(j)

x Ä 0c

xÄ0

(l) x Ä lim f(x) œ
1
_

Note: In these exercises we use the result

"

lim mÎn xÄ „_ x

Theorem 8 and the power rule in Theorem 1:

lim

xÄ „_

œ 0 whenever

ˆ xm"În ‰ œ

lim

(b)
3

4. (a) 1

(b) 1

5. (a)

" #

(b)

" #

6. (a)

" 8

(b)

" 8

7. (a)
53

(b)

10.
3") Ÿ 11.

lim

tÄ_

12. r Ä lim_

 0. This result follows immediately from ˆ x" ‰mÎn œ Š

"

lim ‹ xÄ „_ x

mÎn

(b)
53

3 4

9.
"x Ÿ

m n

xÄ „_

3. (a)
3

8. (a)

f(x) œ  _

sin 2x x

Ÿ

" x

cos ) 3)

Ÿ

" 3)

2
t  sin t t  cos t

Ê x lim Ä_ Ê

13. (a) x lim Ä_

2x  3 5x  7

lim

) Ä
_

œ lim

2 t

tÄ_

r  sin r 2r  7
5 sin r

sin 2x x

œrÄ lim_

œ x lim Ä_

œ 0 by the Sandwich Theorem

cos ) 3)

œ 0 by the Sandwich Theorem


1  ˆ sint t ‰ 1  ˆ cost t ‰

œ

1  ˆ sinr r ‰ 2  7r
5 ˆ sinr r ‰ 2  3x 5  7x

3 4

œ

2 5

0
10 10

œ
1

œrÄ lim_

10 20
0

œ

(b)

" #

2 5

(same process as part (a))

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

œ 0mÎn œ 0.

73

74

Chapter 2 Limits and Continuity 2  Š x7$ ‹

$

2x  7 14. (a) x lim œ x lim Ä _ x$
x#  x  7 Ä_ (b) 2 (same process as part (a)) " x

 x"#

15. (a) x lim Ä_

x1 x#  3

œ x lim Ä_

1  x3#

16. (a) x lim Ä_

3x  7 x#
2

œ x lim Ä_

1
x2#

17. (a) x lim Ä_

7x$ x$
3x#  6x

18. (a) x lim Ä_

x$

20. (a) x lim Ä_ 9 #

2x%

 x7#

œ x lim Ä_

œ x lim Ä_

10x&  x%  31 x'

19. (a) x lim Ä_

(b)

"
4x  1

3 x

œ2

œ0

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

œ0

(b) 0 (same process as part (a)) œ(

7 1
3x  x6# " x$ 4  x"$ x#

1


œ x lim Ä_

2

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

œ!

 x"#  x31' 1

10 x

œ x lim Ä_

9x%  x  5x#
x  6

1
"x  x"#  x7$

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

œ0

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

9  x"$

5 x#


x"$  x6%

œ

9 #

(same process as part (a))
2x$
2x  3 3x$  3x#
5x

21. (a) x lim Ä_


2
x2#  x3$

œ x lim Ä_

œ
23

3  3x
x5#

(b)
23 (same process as part (a))
x % x%
7x$  7x#  9

22. (a) x lim Ä_


" 1
7x  x7#  x9%

œ x lim Ä_

œ
1

(b)
1 (same process as part (a)) 8


8


3

3


3 x2 x2 É 8x 23. x lim œ Êx lim œ É 82
00 œ È4 œ 2 2x2  x œ x lim Ä_ Ä _ Ê 2  1x Ä _ 2  1x 2

24. x Ä lim Šx x
1‹
_ 8x2
3 2

1 Î3

œxÄ lim
_Œ

5

1

"  1x
x12 8



x

3 x2

1 Î3



œ Œx Ä lim
_

5

1

"  1x
x12 8


3 x2

5


x

1 Î3



œ ˆ " 8
0
0 0 ‰

1 Î3

œ ˆ "8 ‰

1 Î3

œ

x2 x2 ‰ œ_ 25. x Ä lim lim œ Œx Ä lim œ ˆ 01
_ Š 1
x ‹ œ x Ä 0
_ x2
7x
_Π1
7x 
_ 1
7x  3

1




5

1




5

5

x x x2 x2 É x
5x œ x lim 26. x lim œ Êx lim œ É 1 0
0
0 0 œ È0 œ 0 Ä _ x3  x
2 Ä _Ê 1  x12
x23 Ä _ 1  x12
x23 2

27. x lim Ä_

2Èx  xc" 3x
7

29. x Ä lim
_

30. x lim Ä_

œ x lim Ä_

3 x
È 5 x È 3 xÈ 5 x È

x "x % x #
x $

Š

œxÄ lim
_

œ x lim Ä_

2 ‹  Š x"# ‹ x"Î# 3
7x

œ0

1
xÐ"Î&Ñ Ð"Î$Ñ 1  xÐ"Î&Ñ Ð"Î$Ñ

x  x"# 1
x"

œxÄ lim
_

28. x lim Ä_ " ‹ 1
Š #Î"& x

" ‹ 1  Š #Î"&

2  Èx 2
Èx

œ x lim Ä_

2 ‹" x"Î# 2 Š "Î# ‹
1 x

Š

œ
1

œ1

x

œ_

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

" #

Section 2.6 Limits Involving Infinity; Asymptotes of Graphs 2x&Î$
x"Î$  7 x)Î&  3x  Èx

31. x lim Ä_

3 x
5x  3 È 2x  x#Î$
4

32. x Ä lim
_ 33. x lim Ä_

È x2  1 x1

34. x Ä lim
_

œ x lim Ä_ œxÄ lim
_

œ x lim Ä_

È x2  1 x1

1

" x#Î$

2

È x 2  1 ÎÈ x 2 a x  1 b ÎÈ x 2

œxÄ lim
_

"  7 x"*Î"& x)Î& 3 "  x$Î& x""Î"!

2x"Î"&



5  3x " x"Î$

È

3 œ
5# È x


4x

œ x lim Ä_

È x 2  1 ÎÈ x 2 a x  1 bÎ È x 2

œ_

È a x 2  1 bÎx 2 ax  1bÎx

œ x lim Ä_

È 1  1 Îx 2 a1  1 Îx b

È a x 2  1 bÎ x 2

È1  0 a1  0 b

œ

È 1  1 Îx 2

œxÄ lim œ x lim œ
_ ax  1bÎa
xb Ä _ a
1
1 Î x b

œ1

È1  0 a
1
0b

a x
3 bÎ x a x
3 bÎ x a1
3 Îx b x
3 35. x lim œ x lim œ x lim œ x lim œ Ä _ È4x2  25 Ä _ È4x2  25ÎÈx2 Ä _ Èa4x2  25bÎx2 Ä _ È4  25Îx2 ˆ4
3x3 ‰ÎÈx6

4
3x 36. x Ä lim œxÄ lim œxÄ lim
_ Èx6  9
_ Èx6  9ÎÈx6
_ 3

"

œ_

37.

lim x Ä !b 3x

39.

lim x Ä #c x
2

41.

lim x Ä
)b x8

3

2x

4

43. lim

# x Ä ( (x
7)

œ
_ œ
_ œ_

lim "Î$ x Ä !b 3x

46. (a)

lim "Î& x Ä !b x 4

47. lim

#Î& xÄ! x

49.

51. 52.

lim

x Ä ˆ 1# ‰

œ lim

4

# x Ä ! ax"Î& b

œ x lim Ä_

ˆ
4 Îx 3  3‰ È 1  9 Îx 6

Š positive positive ‹

40.

lim x Ä $b x
3

Š negative positive ‹

42.

lim x Ä
&c 2x10

positive Š positive ‹

44. lim

œ_

"

3x


"

# x Ä ! x (x1)

2

(b)

lim "Î$ x Ä !c 3x

(b)

lim "Î& x Ä !c x

48. lim

"

#Î$ xÄ! x

tan x œ _

50.

œ3

œ_

positive Š negative ‹

2

a0  3 b È1  0

" #

positive Š negative ‹

lim x Ä !c 2x

5

œ

œ

œ
_

38.

œ_

2

ˆ4
3x3 ‰Îˆ
x3 ‰ Èax6  9bÎx6

positive Š positive ‹

œ_

2

45. (a)

a1
0 b È4  0

2

œ
1

œ_

Š negative negative ‹

œ
_

negative Š positive †positive ‹

œ
_ œ
_

œ lim

"

# x Ä ! ax"Î$ b

œ_

lim
sec x œ _

x Ä ˆ #1 ‰

lim (1  csc )) œ
_

)Ä!

lim (2
cot )) œ
_ and lim c (2
cot )) œ _, so the limit does not exist )Ä!

) Ä !b

"

œ lim b xÄ#

" (x2)(x
2)

œ_

Š positive"†positive ‹

"

œ lim c xÄ#

" (x2)(x
2)

œ
_

Š positive†"negative ‹

53. (a)

lim # x Ä # b x
4

(b)

lim # x Ä # c x
4

(c)

lim # x Ä
#b x
4

(d)

lim # x Ä
#c x
4

"

œ

lim x Ä
#b (x2)(x
2)

"

œ
_

Š positive†"negative ‹

"

œ

lim x Ä
#c (x2)(x
2)

"

œ_

Š negative"†negative ‹

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

75

76

Chapter 2 Limits and Continuity

54. (a)

lim # x Ä "b x
1

(b)

lim # x Ä "c x
1

(c)

lim # x Ä
"b x
1

(d)

lim # x Ä
"c x
1

x

œ lim b xÄ"

x (x1)(x
1)

œ_

positive Š positive †positive ‹

x

œ lim c xÄ"

x (x1)(x
1)

œ
_

positive Š positive †negative ‹

x

œ

lim x Ä
"b (x1)(x
1)

x

œ_

negative Š positive †negative ‹

x

œ

lim x Ä
"c (x1)(x
1)

x

œ
_

negative Š negative †negative ‹

55. (a)

lim x Ä !b #

x#




" x

œ 0  lim b xÄ!

"
x

œ
_

" Š negative ‹

(b)

lim x Ä !c #

x#




" x

œ 0  lim c xÄ!

"
x

œ_

" Š positive ‹

(c)

lim # x Ä $È2

(d)

lim x Ä
1 #

56. (a)

x#

x#

lim x Ä
#b

" x





" x

œ

x#
1 2x  4 x#
1

2#Î$ #

œ

" #

(d)

lim x Ä !c 2x  4

œ

(b) (c) (d) (e) 58. (a)

x#
3x  2 x$
2x#

lim b

x#
3x  2 x$
2x#

lim

xÄ#

#

x
3x  2 x$
2x# x
3x  2 x$
2x#

lim

œ lim c xÄ#

lim x Ä
#b

(c)

x Ä 0c

(d)

x Ä "b

(e)

lim x Ä !b x(x  #)

x#
1 2x  4

lim x Ä
#c

œ
_

positive Š negative ‹

œ0

(x
2)(x
1) x# (x
2)

œ
_

(x
2)(x
1) x# (x
2)

œ lim b xÄ#

(x
2)(x
1) x# (x
2)

œ lim c xÄ#

œ lim

œ lim

(x
2)(x
1) x# (x
2)

œ
_

xÄ!

lim

x#
3x  2 x$
4x

lim

x#
3x  2 x$
4x x
"

2†0 #4

(x
2)(x
1) x# (x
2)

xÄ#

œ lim b xÄ#

x#
3x  2 x$
4x

(b)

œ

œ lim

x#
3x  2 x$
4x

lim

x Ä #b

and

œ lim b xÄ#

x
3x  2 x$
2x#

#

xÄ!

œ lim b xÄ!

#

lim

x Ä #c

(x  1)(x
1) 2x  4

(b)


" 4

lim b

xÄ#

3 #

Š positive positive ‹

œ lim b xÄ"

xÄ!

œ 2
"Î$
2
"Î$ œ 0

œ_

lim x Ä "b 2x  4

57. (a)

" #"Î$


ˆ
"1 ‰ œ

(c)

x#
1




œ

xÄ#

(x
2)(x
") x(x
#)(x  2)

(x
2)(x
")

œ lim c xÄ! œ lim b xÄ"

(x
2)(x
") x(x
#)(x  2) (x
2)(x
") x(x
#)(x  2)

œ

x
1 x#

œ

" 4

,xÁ2

x
1 x#

œ

" 4

,xÁ2

x
1 x#

œ

" 4

,xÁ2 †negative Š negative positive†negative ‹

œ lim b xÄ#

lim x Ä
#b x(x
#)(x  2)

†negative Š negative positive†negative ‹

(x
1) x(x  #)

œ

(x
1)

lim x Ä
#b x(x  #)

œ lim c xÄ! œ lim b xÄ"

œ_

(x
1) x(x  #)

œ

negative Š positive †positive ‹

x
"

negative Š negative †positive ‹

œ_

œ

" 8

œ_

(x
1) x(x  #)

œ
_

lim x Ä !c x(x  #)

" #(4)

0 (1)(3)

negative Š negative †positive ‹ negative Š negative †positive ‹

œ0

so the function has no limit as x Ä 0. lim
2


59. (a)

t Ä !b

60. (a)

t Ä !b

61. (a)

x Ä !b

(c)

x Ä "b

3 ‘ t"Î$

œ
_

" lim
t$Î&  7‘ œ _

lim


2


lim




lim

"  ’ x#Î$

2 “ (x
1)#Î$

œ_

lim

"  ’ x#Î$

2 “ (x
1)#Î$

œ_

(b)

t Ä !c

(b)

t Ä !c

lim

"  ’ x#Î$

2 “ (x
1)#Î$

œ_

(b)

x Ä !c

lim

"  ’ x#Î$

2 “ (x
1)#Î$

œ_

(d)

x Ä "c

" t$Î&

3 ‘ t"Î$

œ_

 7‘ œ
_

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 2.6 Limits Involving Infinity; Asymptotes of Graphs lim

"
’ x"Î$

1 “ (x
1)%Î$

œ_

(b)

x Ä !c

lim

"
’ x"Î$

1 “ (x
1)%Î$

œ
_

(d)

x Ä "c

62. (a)

x Ä !b

(c)

x Ä "b

lim

"
’ x"Î$

1 “ (x
1)%Î$

œ
_

lim

"
’ x"Î$

1 “ (x
1)%Î$

œ
_

63. y œ

" x
1

64. y œ

" x1

65. y œ

" #x  4

66. y œ


3 x
3

67. y œ

x3 x2

68. y œ

2x x1

œ1

" x#

69. Here is one possibility.

œ#


2 x1

70. Here is one possibility.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

77

78

Chapter 2 Limits and Continuity

71. Here is one possibility.

72. Here is one possibility.

73. Here is one possibility.

74. Here is one possibility.

75. Here is one possibility.

76. Here is one possibility.

77. Yes. If x lim Ä_

f(x) g(x)

œ 2 then the ratio of the polynomials' leading coefficients is 2, so x Ä lim
_

f(x) g(x)

œ 2 as well.

78. Yes, it can have a horizontal or oblique asymptote. 79. At most 1 horizontal asymptote: If x lim Ä_ f(x) lim x Ä
_ g(x)

f(x) g(x)

œ L, then the ratio of the polynomials' leading coefficients is L, so

œ L as well.

Èx  9  Èx  4 80. x lim Š Èx  9
Èx  4‹ œ x lim ’Èx  9
Èx  4“ † ’ Èx  9  Èx  4 “ œ x lim Ä_ Ä_ Ä_ 5 Èx 5 0 œ x lim œ x lim œ 11 œ 0 9 4 Ä _ Èx  9  Èx  4 Ä_

ax  9 b
a x  4 b Èx  9  Èx  4

É1  x  É1  x

È 2 È 2 81. x lim Š Èx2  25
Èx2
"‹ œ x lim ’Èx2  25
Èx2
"“ † ’ Èx2  25  Èx2
" “ œ x lim Ä_ Ä_ Ä_ x  25  x
"

œ x lim Ä_

26 Èx2  25  Èx2
"

œ x lim Ä_

26 x

É1  x252  É1
x12

œ

0 11

œ0

È 2 ˆx  3 ‰
ˆ x ‰
x 82. x Ä lim lim lim Š Èx2  3  x‹ œ x Ä ’Èx2  3  x“ † ’ Èxx2  33
“œxÄ
_
_
_ Èx2  3
x x 3 È x2
3x 3 œxÄ lim œ lim œ lim œ 1 0 1 œ 0 2 È 3 x
_ x Ä
_ É1  2
È x Ä
_ É1  32  1 x 3
x x x x2 2

2

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

ˆx2  25‰
ˆx2
"‰ Èx2  25  Èx2
"

Section 2.6 Limits Involving Infinity; Asymptotes of Graphs È

ˆ4x2 ‰
ˆ4x2  3x
2‰


4x  3x
2 83. x Ä lim lim lim Š 2x  È4x2  3x
2‹ œ x Ä ’2x  È4x2  3x
2“ † ’ 2x “œxÄ
_
_
_ 2x
È4x2  3x
2 2x
È4x2  3x
2 c3x b 2 c3x b 2 3
x2 Èx2
3x  2 cx œxÄ lim œxÄ lim œxÄ lim œxÄ lim
_ 2x
È4x2  3x
2
_ È2x
É4  3
22
_ 2x
É4  3
22
_
2
É4  3
22 x x x cx x x x x2 œ
32

02 œ
43 2

È 2 84. x lim Š È9x2
x
3x‹ œ œ x lim ’È9x2
x
3x“ † ’ È9x2
x  3x “ œ x lim Ä_ Ä_ Ä_ 9x
x  3x

œ x lim Ä_


x È9x2
x  3x

œ x lim Ä_


xx 2 É 9x2 x


xx2  3x x


1 É9
"x  3

œ x lim Ä_

œ


1 33

ˆ9x2
x‰
ˆ9x2 ‰ È9x2
x  3x

œ
"6

È 2 È 2 85. x lim Š Èx2  3x
Èx2
2x‹ œ x lim ’ Èx2  3x
Èx2
2x“ † ’ Èx2  3x  Èx2
2x “ œ x lim Ä_ Ä_ Ä_ x  3x  x
2x 5x 5 5 5 œ x lim œ x lim œ 11 œ # È 2 3 2 Ä_ È 2 Ä_ x  3x 

x
2x

É1  x  É1
x

Èx#  x
Èx#
x œ lim ’Èx#  x
Èx#
x“ † ’ Èx#  x  Èx#
x “ œ lim 86. x lim È x#  x  È x#
x Ä_ xÄ_ xÄ_ 2x 2 2 œ x lim œ x lim œ 11 œ 1 È # " " Ä_ È # Ä_ x x

ˆx2  3x‰
ˆx2
2x‰ Èx2  3x  Èx2
2x

x
x

ax #  x b
a x #
x b È x#  x  È x#
x

É1  x  É1
x

87. For any %  0, take N œ 1. Then for all x  N we have that kf(x)
kk œ kk
kk œ 0  %. 88. For any %  0, take N œ 1. Then for all y 
N we have that kf(x)
kk œ kk
kk œ 0  %.
" x#

89. For every real number
B  0, we must find a $  0 such that for all x, 0  kx
0k  $ Ê


" x#

Ê


B  ! Í
" x#

" x#

#

B0 Í x 

" B

" ÈB

Í kxk 

. Choose $ œ

" ÈB

, then 0  kxk  $ Ê kxk 

xÄ!

 B  ! Í lxl  B" . Choose $ œ B" . Then !  kx
0k  $ Ê lxl 

" B

Ê

" lx l

" lx l


2 (x
3)#


B  ! Í

2 (x
3)#

$ œ É B2 , then 0  kx
3k  $ Ê

B0 Í
2 (x
3)#

(x
3) 2

#



" B

Í (x
3)# 


B  0 so that lim


2

# x Ä $ (x
3)

2 B

 B. Now,

x Ä ! lx l
2 (x
3)#

Now,

#

 B  ! Í (x  5) 

Ê kx  5k 

" ÈB

Ê

" (x  5)#

" B

Í kx  5k 

 B so that lim

"

" ÈB

# x Ä
& (x  5)

. Choose $ œ

œ _.


B.

Í !  kB
$k  É B2 . Choose

œ
_.

92. For every real number B  0, we must find a $  0 such that for all x, 0  kx
(
5)k  $ Ê 1 (x  5)#

"

 B so that lim

91. For every real number
B  0, we must find a $  0 such that for all x, 0  kx
3k  $ Ê Now,

" ÈB


B so that lim
x"# œ
_.

90. For every real number B  0, we must find a $  0 such that for all x, !  kx
0k  $ Ê " lx l


B. Now,

" ÈB

1 (x  5)#

 B.

. Then 0  kx
(
5)k  $

œ _.

93. (a) We say that f(x) approaches infinity as x approaches x! from the left, and write lim c f(x) œ _, if x Ä x! for every positive number B, there exists a corresponding number $  0 such that for all x, x!
$  x  x! Ê f(x)  B. (b) We say that f(x) approaches minus infinity as x approaches x! from the right, and write lim b f(x) œ
_, x Ä x!

if for every positive number B (or negative number
B) there exists a corresponding number $  0 such that for all x, x!  x  x!  $ Ê f(x) 
B.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

79

80

Chapter 2 Limits and Continuity (c) We say that f(x) approaches minus infinity as x approaches x! from the left, and write lim f(x) œ
_, x Ä x!

if for every positive number B (or negative number
B) there exists a corresponding number $  0 such that for all x, x!
$  x  x! Ê f(x) 
B. 94. For B  0,

" x

 B  0 Í x  B" . Choose $ œ B" . Then !  x  $ Ê 0  x 

95. For B  0,

" x


B  0 Í
x"  B  0 Í
x 

Ê
B"  x Ê 96. For B  !,

" x
#

" x


B so that lim c xÄ!

" x

" B

Ê

" x
#

" x
#

Ê 99. y œ

101. y œ

" x
#

 B  ! so that lim b xÄ#

œx1

x#
% x
"

 B so that lim b xÄ!

" x

œ _.

" B

Í x
2 
B" Í x  2
B" . Choose $ œ B" . Then " x
#


B  0 so that lim c xÄ#

" x
#

œ
_.

 B Í !  x
2  B" . Choose $ œ B" . Then #  x  #  $ Ê !  x
#  $ Ê !  x
2 

" 1
x#

" x
"

œx"


$ x
"

œ _.

 B Í 1
x# 

" #B . Then "
$  x  " Ê " 1
x#  B for !  x  1 and x# x
"

" x

œ
_.


B Í
x
" #  B Í
(x
2) 

98. For B  0 and !  x  1, $

Ê

Í
B"  x. Choose $ œ B" . Then
$  x  !

2
$  x  2 Ê
$  x
2  ! Ê
B"  x
2  0 Ê 97. For B  0,

" B

" B

Í ("
x)("  x)  B" . Now


$  x
1  0 Ê "
x  $  x near 1 Ê

"

lim # x Ä "c "
x

" #B

1x #

 1 since x  1. Choose Ê ("
x)("  x)  B" ˆ 1 # x ‰  B"

œ _.

100. y œ

x#  " x
1

œx"

102. y œ

x2
" #x  %

œ #" x
" 

# x
1

$ #x  %

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

" B

Section 2.6 Limits Involving Infinity; Asymptotes of Graphs 103. y œ

x#
1 x

105. y œ

x È 4
x#

œx


107. y œ x#Î$ 

" x

104. y œ

x$  1 x#

106. y œ


" È 4
x#

œx

" x#

108. y œ sin ˆ x# 1 1 ‰

" x"Î$

109. (a) y Ä _ (see accompanying graph) (b) y Ä _ (see accompanying graph) (c) cusps at x œ „ 1 (see accompanying graph)

110. (a) y Ä 0 and a cusp at x œ 0 (see the accompanying graph) (b) y Ä 32 (see accompanying graph) (c) a vertical asymptote at x œ 1 and contains the point Š
1,

3 ‹ 3 2È 4

(see accompanying graph)

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

81

82

Chapter 2 Limits and Continuity

CHAPTER 2 PRACTICE EXERCISES 1. At x œ
1: Ê

f(x) œ

lim

x Ä
"c

lim

x Ä
"b

f(x) œ 1

lim f(x) œ 1 œ f(
1)

x Ä
1

Ê f is continuous at x œ
1. At x œ 0: lim c f(x) œ lim b f(x) œ 0 Ê lim f(x) œ 0. xÄ!

xÄ!

xÄ!

But f(0) œ 1 Á lim f(x) xÄ!

Ê f is discontinuous at x œ 0. If we define fa!b œ !, then the discontinuity at x œ ! is removable. At x œ 1: lim c f(x) œ
1 and lim f(x) œ 1 xÄ"

Ê lim f(x) does not exist

xÄ"

xÄ1

Ê f is discontinuous at x œ 1. 2. At x œ
1: Ê

f(x) œ 0 and

lim

x Ä
"


lim

x Ä
"

f(x) œ
1

lim f(x) does not exist

x Ä
"

Ê f is discontinuous at x œ
1. At x œ 0: lim
f(x) œ
_ and lim f(x) œ _ xÄ!

Ê lim f(x) does not exist

xÄ!

xÄ!

Ê f is discontinuous at x œ 0. At x œ 1: lim
f(x) œ lim f(x) œ 1 Ê lim f(x) œ 1. xÄ"

xÄ1

xÄ"

But f(1) œ 0 Á lim f(x) xÄ1

Ê f is discontinuous at x œ 1. If we define fa"b œ ", then the discontinuity at x œ " is removable. 3. (a) (b) (c) (d) (e) (f)

lim a3fatbb œ 3 lim fatb œ 3(
7) œ
21

t Ä t!

t Ä t!

#

lim afatbb# œ Š lim fatb‹ œ a
(b# œ 49

t Ä t!

t Ä t!

lim afatb † gatbb œ lim fatb † lim gatb œ (
7)(0) œ 0

t Ä t!

t Ä t!

lim fatb t Ä t! g(t)
7

Ät

t Ä t!

lim fatb

œ

Ät

t

œ

!

lim agatb
7b

t

t

!

Ät

lim fatb

Ät

t

!

Ät

lim gatb
lim 7 t

!

!

œ


7 0
7

œ1

lim cos agatbb œ cos Š lim gatb‹ œ cos ! œ 1

t Ä t!

t Ä t!

lim kfatbk œ ¹ lim fatb¹ œ k
7k œ 7

t Ä t!

t Ä t!

(g) lim afatb  gatbb œ lim fatb
lim gatb œ
7  0 œ
7 t Ä t!

(h)

4. (a) (b) (c) (d)

t Ä t!

lim Š " ‹ t Ä t! fatb

œ

" lim fatb

t

Ät

t Ä t!

"
7

œ

!

œ
71

lim
g(x) œ
lim g(x) œ
È2

xÄ!

xÄ!

lim ag(x) † f(x)b œ lim g(x) † lim f(x) œ ŠÈ2‹ ˆ "# ‰ œ

xÄ!

xÄ!

xÄ!

lim af(x)  g(x)b œ lim f(x)  lim g(x) œ

xÄ!

"

lim x Ä ! f(x)

œ

" lim f(x)

xÄ!

xÄ!

œ

" " #

œ2

xÄ!

" #

È2 #

 È2

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Chapter 2 Practice Exercises (e) (f)

" #

lim ax  f(x)b œ lim x  lim f(x) œ 0 

xÄ!

xÄ!

f(x)†cos x x
1 xÄ!

lim

xÄ!

lim f(x)† lim cos x

œ

xÄ!

xÄ!

lim x
lim 1

xÄ!

xÄ!

œ

ˆ "# ‰ (1) 0
1

œ

83

" #

œ
#"

5. Since lim x œ 0 we must have that lim (4
g(x)) œ 0. Otherwise, if lim (%
g(x)) is a finite positive xÄ!

xÄ!

xÄ!

’ 4
xg(x) “

’ 4
xg(x) “

œ
_ and lim b œ _ so the limit could not equal 1 as xÄ! x Ä 0. Similar reasoning holds if lim (4
g(x)) is a finite negative number. We conclude that lim g(x) œ 4. number, we would have lim c xÄ!

xÄ!

6. 2 œ lim

x Ä
%

xÄ!

’x lim g(x)“ œ lim x † lim xÄ!

x Ä
%

’ lim g(x)“ œ
4 lim xÄ!

x Ä
%

(since lim g(x) is a constant) Ê lim g(x) œ xÄ!

xÄ!

2
%

x Ä
%

œ
#" .

’ lim g(x)“ œ
4 lim g(x) xÄ!

xÄ!

7. (a) xlim faxb œ xlim x"Î$ œ c"Î$ œ facb for every real number c Ê f is continuous on a
_ß _b. Äc Äc (b) xlim gaxb œ xlim x$Î% œ c$Î% œ gacb for every nonnegative real number c Ê g is continuous on Ò!ß _Ñ. Äc Äc " c#Î$ " c"Î'

(c) xlim haxb œ xlim x
#Î$ œ Äc Äc (d) xlim kaxb œ xlim x
"Î' œ Äc Äc

œ hacb for every nonzero real number c Ê h is continuous on a
_ß !b and a
_ß _b. œ kacb for every positive real number c Ê k is continuous on a!ß _b

8. (a) - ˆˆn
"# ‰1ß ˆn  "# ‰1‰, where I œ the set of all integers. n−I (b) - an1ß an  1b1b, where I œ the set of all integers. n−I (c) a
_ß 1b  a1ß _b (d) a
_ß !b  a!ß _b 9.

(a)

x#
4x  4 x
2

lim x Ä !c x(x  7) (b) 10. (a)

#

x#  x

œ lim

1 x# (x  1)

x#  x

x #
a# x %
a%

œ xlim Äa

13. lim

(x  h)#
x# h

œ lim

(x  h)#
x# h xÄ!

œ lim

xÄ!

x

œ _ and lim b xÄ!

"
Èx

ax #
a # b ax #  a # b a x #
a # b

hÄ!

œ lim

xÄ!

1 x# (x  1)

2
(2  x) 2x(#  x)

œ lim

#

x x

"

"

œ

œ

œ

0 2(9)

œ0

, x Á 0 and x Á
1.

œ _.

, x Á 0 and x Á
1. The limit does not

1

lim # x Ä
"b x (x  1)

"

# x Ä 0 x (x  1)

& % $ x Ä ! x  2x  x

# x Ä
" x (x  1)

" x #  a#

œ xlim Äa

œ lim

œ _ Ê lim

œ lim

x
2

x Ä # x(x  7)

x1

x Ä 1 1  Èx

ax#  2hx  h# b
x# h xÄ!

, x Á 2, and lim

# x Ä ! x (x  1)(x  1)

œ lim

ax#  2hx  h# b
x# h

x
2

œ lim

œ
_ and

x Ä 1 ˆ1
È x ‰ ˆ 1  È x ‰

, x Á 2; the limit does not exist because

x Ä # x(x  7)

x(x  1)

"

12. xlim Äa

"
" x #

x(x  1)

lim # x Ä
"c x (x  1)

œ lim

#

œ lim

$ # x Ä
" x ax  2x  1b

1
Èx 1
x

15. lim

(x
2)(x
2)

œ lim

11. lim

14. lim

œ
_

$ # x Ä ! x ax  2x  1b

lim & % $ x Ä
" x  2x  x

hÄ!

x
2 x(x  7)

x
2

x Ä ! x(x  7)

x Ä # x(x  7)(x
#)

œ lim

lim & % $ x Ä ! x  2x  x

xÄ1

œ lim

œ _ and lim b xÄ!

x
4x  4

exist because

(x
2)(x
2)

x Ä ! x(x  7)(x
2)

lim $ # x Ä # x  5x
14x

Now lim c xÄ! (b)

œ lim

lim $ # x Ä ! x  5x
14x

œ _.

" #

" #a #

œ lim (2x  h) œ 2x hÄ!

œ lim (2x  h) œ h xÄ!


"

x Ä ! 4  #x

œ
"4

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

84

Chapter 2 Limits and Continuity (#  x)$
8 x

16. lim

xÄ!

ax$  6x#  12x  8b
8 x

œ lim

xÄ!

ˆx1Î3
1‰ x1Î3
1 œ lim ˆÈx
1‰ È x
1 xÄ1 xÄ1 œ 1 1 1 1 1 œ 23

17. lim

18.

tan 2x

œ lim

sin 2x

x Ä ! cos 2x

lim csc x œ limc x
1

20.

x Ä 1c

21.

xÄ1

22.

xÄ1

ax
1bˆÈx  1‰

œ lim

2Î3 1Î3 x Ä 1 ax
1bax  x 1b

Èx  1

œ lim

2Î3 1Î3 x Ä 1 x  x 1

1 sin x

†

1x ‰ˆ 1x ‰ˆ 2x ‰ œ lim ˆ sin2x2x ‰ˆ cos cos 2x sin 1x 1x œ 1 † 1 † 1 †

cos 1x sin 1x

xÄ!

2 1

œ

2 1

œ_

lim sin ˆ x2  sin x‰ œ sin ˆ 12  sin 1‰ œ sin ˆ 12 ‰ œ 1

lim cos2 ax
tan xb œ cos2 a1
tan 1b œ cos2 a1b œ a
1b2 œ 1

23. lim xÄ0 24. lim xÄ0

8x 3sin x
x

œ lim xÄ0

cos 2x
1 sin x

2x
1 œ lim ˆ cossin † x xÄ0

8 3 sinx x
1

œ

8 3 a1 b
1

œ4

cos 2x  1 ‰ cos 2x  1

œ lim xÄ0

"Î$

lim [4 g(x)]"Î$ œ 2 Ê ’ lim b 4 g(x)“ x Ä !b xÄ! lim

x Ä È&

27. lim

xÄ1

28.

ˆx2Î3  x1Î3 1‰ˆÈx  1‰ ˆÈx  1‰ax2Î3  x1Î3 1b

ˆx1Î3
4‰ˆx1Î3  4‰ ˆx1Î3
4‰ˆx1Î3  4‰ ˆx2Î3  4x1Î3 16‰ˆÈx  )‰ x2Î3
16 œ lim œ lim † ˆÈx  )‰ax2Î3  4x1Î3 16b È È Èx
8 x
8 x
8 x Ä 64 x Ä 64 x Ä 64 ˆx1Î3  4‰ˆÈx  )‰ ax
64bˆx1Î3  4‰ˆÈx  )‰ 4  4ba8  8b 8 œ lim ax
64bax2Î3  4x1Î3 16b œ lim x2Î3  4x1Î3 16 œ a16  16  16 œ 3 x Ä 64 x Ä 64

x Ä ! tan 1x

26.

xÄ!

lim

19. lim

25.

†

œ lim ax#  6x  12b œ 12

" x  g(x)

3x#  1 g(x)

xÄ1

5
x#

œ

(x  g(x)) œ

lim

x Ä È&

œ2 Ê " #

œ lim xÄ0


sin2 2x sin xacos 2x  1b

œ lim xÄ0


4sin x cos2 x cos 2x  1

œ0 Ê

Ê È5  lim

x Ä È5

g(x) œ

" #

Ê

lim

x Ä È5

g(x) œ

" #


È5

xÄ1

lim g(x) œ _ since lim a5
x# b œ 1

x Ä
#

x Ä
#

lim f(x) œ lim c x Ä
"c x Ä
"

lim x Ä
"c

x ax #
1 b x#
1

œ

lim

x Ä
"c

x ax #
1 b kx #
1 k

x œ
1, and

#

lim f(x) œ lim b xkaxx#

11k b œ lim b x Ä
"b x Ä
" x Ä
" œ lim (
x) œ
(
1) œ 1. Since

x ax #
1 b
a x #
"b

x Ä
1

lim f(x) Á lim b f(x) x Ä
"c x Ä
" Ê

lim f(x) does not exist, the function f cannot be

x Ä
1

extended to a continuous function at x œ
1. At x œ 1:

lim f(x) œ lim c

x Ä "c

xÄ"

x ax #
1 b kx #
1 k

x ax #
1 b kx #
1 k

œ lim c xÄ"

x ax #
1 b x#
"

x ax #
1 b
ax #
1 b

œ lim c (
x) œ
1, and xÄ"

lim f(x) œ lim b œ lim b œ lim b x œ 1. Again lim f(x) does not exist so f xÄ1 xÄ" xÄ" xÄ1 cannot be extended to a continuous function at x œ 1 either.

x Ä "b

œ


4a0ba1b2 11

lim 4 g(x) œ 8, since 2$ œ 8. Then lim b g(x) œ 2. xÄ!

x Ä !b

œ _ Ê lim g(x) œ 0 since lim a3x#  1b œ 4

lim x Ä
# Èg(x)

29. At x œ
1:

œ2 Ê

cos2 2x
1 sin xacos 2x  1b

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

œ0

Chapter 2 Practice Exercises 30. The discontinuity at x œ 0 of f(x) œ sin ˆ "x ‰ is nonremovable because lim sin xÄ!

" x

does not exist.

31. Yes, f does have a continuous extension to a œ 1: " define f(1) œ lim xx

È œ 43 . % x xÄ1

32. Yes, g does have a continuous extension to a œ 1# : ) 5 g ˆ 1# ‰ œ lim1 45)cos
#1 œ
4 . )Ä #

33. From the graph we see that lim h(t) Á lim
h(t) tÄ! tÄ! so h cannot be extended to a continuous function at a œ 0.

34. From the graph we see that lim c k(x) Á lim b k(x) xÄ! xÄ! so k cannot be extended to a continuous function at a œ 0.

35. (a) f(
1) œ
1 and f(2) œ 5 Ê f has a root between
1 and 2 by the Intermediate Value Theorem. (b), (c) root is 1.32471795724 36. (a) f(
2) œ
2 and f(0) œ 2 Ê f has a root between
2 and 0 by the Intermediate Value Theorem. (b), (c) root is
1.76929235424 # $

#x  $ x 37. x lim œ x lim œ Ä _ &x  ( Ä _ &  (x

x 39. x Ä lim
_

#


%x  ) $x $

#! &!

ˆ"
œxÄ lim
_ $x

œ

#

# &

% $x#

#

#x  $ 38. x Ä lim œxÄ lim
_ &x#  (
_ & 



) ‰ $x$

$ x# ( x#

œ

#! &!

œ!
!!œ!

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

œ

# &

85

86

Chapter 2 Limits and Continuity "

" x# 40. x lim œ x lim œ Ä _ x #
(x  " Ä _ "
(x  x"#

! "
!!

œ!

#

%

x
(x x
( 41. x Ä lim œxÄ lim œ
_
_ x  1
_ "  "x

$

x x x" 42. x lim œxÄ lim œ_ Ä _ "#x$  "#)
_ "#  "#) x$

sin x " sin x 43. x lim Ÿ x lim œ ! since int x Ä _ as x Ä _ Êx lim œ !. Ä _ gx h Ä _ gx h Ä _ gx h

44.

lim

)Ä_

45. x lim Ä_

cos )
" )

Ÿ lim

#

) Ä _)

x  sin x  #Èx x  sin x

#Î$

œ ! Ê lim

)Ä_

œ x lim Ä_

cos )
" )

"  sinx x  È#x "  sinx x

"

œ

&Î$

x x " x 46. x lim œ x lim #x
œ Ä _ x#Î$  cos# x Ä _Œ "  cos#Î$

œ !.

"!! "!

"! "!

œ"

œ"

x

47. (a) y œ (b) y œ

x2  4 x
3

is undefined at x œ 3: lim c xx
34 œ
_ and lim b xx
34 œ  _, thus x œ 3 is a vertical asymptote. xÄ3 xÄ3 2

x2
x
2 x2
2x  1

2

is undefined at x œ 1: lim c xÄ1

x2
x
2 x2
2x  1

œ
_ and lim b xÄ1

x2
x
2 x2
2x  1

œ
_, thus x œ 1 is a vertical

asymptote. (c) y œ

x2  x
' x2  2x
8

is undefined at x œ 2 and
4: lim

x x
' 2

œ lim b x Ä
%

lim 2 x Ä
%b x  2x
8 48. (a) y œ

1
x2 1
x2 x2  " : x lim Ä _ x2  "

x3 x4

x2  x
'

2 x Ä 2 x  2x
8

x3

œ lim

x Ä 2 x4

œ 56 ; lim c x Ä
%

x2  x
' x2  2x
8

œ lim c x Ä
%

x3 x4

œ_

œ
_. Thus x œ
4 is a vertical asymptote. 1


1

x2 œ x lim Ä _ 1

1 x2

œ

1


1


1 1

1
x œ
1 and x Ä lim œ lim x2 œ
_ x2  " x Ä
_ 1  x12

œ

10 È1  0

2


1 1

œ
1, thus y œ
1 is a

horizontal asymptote. (b) y œ

Èx  4 Èx  4 Èx  4 : x lim Ä _ Èx  4

(c) y œ

È x2  4 È x2  4 : x lim x x Ä_

œx Ä lim
_

É1  x42
1

œ

œ x lim Ä_

È1  0
1

1  È4x É1  B4

œ x lim Ä_ œ

1
1

É1  x42 1

œ

œ 1 , thus y œ 1 is a horizontal asymptote.

È1  0 1

œ 1 and x lim Ä
_

thus y œ

1

2

" 3

œx Ä lim
_

É1  x42

È

9

" 3

x xx

2

is a horizontal asymptote.

0.1 0.7943

œx Ä lim
_

É1  x42 x

cx

0.01 0.9550

0.001 0.9931

0.0001 0.9991

1

9

x 9 0 " x2 É 9x and x Ä lim lim œ É 19  21 œ 0 œ 3,
_ x Ä
_ Ê 9  x12

CHAPTER 2 ADDITIONAL AND ADVANCED EXERCISES 1. (a)

x x2

œ
1, thus y œ 1 and y œ
1 are horizontal asymptotes.

x 9 x 9 0 x2 (d) y œ É 9x lim É 9x lim œ É 91  21: 21 œ 0 œ xÄ_ x Ä _ Ê 9  x12 2

È x2  4 x

0.00001 0.9999

Apparently, lim b xx œ 1 xÄ!

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Chapter 2 Additional and Advanced Exercises 87 (b)

2. (a)

x ˆ "x ‰"ÎÐln xÑ Apparently,

10

100

1000

0.3679

0.3679

0.3679

"ÎÐln xÑ lim ˆ " ‰ xÄ_ x

œ 0.3678 œ

" e

(b)

3.

lim L œ lim c L! É"
v Ä cc vÄc

v# c#

lim v œ L! É1
vÄcc # œ L! É1
#

c# c#

œ0

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). 4. (a) ¹

Èx #


1¹  0.2 Ê
0.2 

Èx #


1  0.2 Ê 0.8 

Èx #

 1.2 Ê 1.6  Èx  2.4 Ê 2.56  x  5.76.

(b) ¹

Èx #


1¹  0.1 Ê
0.1 

Èx #


1  0.1 Ê 0.9 

Èx #

 1.1 Ê 1.8  Èx  2.2 Ê 3.24  x  4.84.

5. k10  (t
70) ‚ 10
%
10k  0.0005 Ê k(t
70) ‚ 10
% k  0.0005 Ê
0.0005  (t
70) ‚ 10
%  0.0005 Ê
5  t
70  5 Ê 65°  t  75° Ê Within 5° F. 6. We want to know in what interval to hold values of h to make V satisfy the inequality lV
"!!!l œ l$'1h
"!!!l Ÿ "!. To find out, we solve the inequality: **! l$'1h
"!!!l Ÿ "! Ê
"! Ÿ $'1h
"!!! Ÿ "! Ê **! Ÿ $'1h Ÿ "!"! Ê $' 1 Ÿ hŸ

"!"! $'1

Ê )Þ) Ÿ h Ÿ )Þ*. where 8.8 was rounded up, to be safe, and 8.9 was rounded down, to be safe. The interval in which we should hold h is about )Þ*
)Þ) œ !Þ" cm wide (1 mm). With stripes 1 mm wide, we can expect to measure a liter of water with an accuracy of 1%, which is more than enough accuracy for cooking.

7. Show lim f(x) œ lim ax#
7b œ
' œ f(1). xÄ1

xÄ1

Step 1: kax
7b  6k  % Ê
%  x#
1  % Ê 1
%  x#  1  % Ê È1
%  x  È1  %. #

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

88

Chapter 2 Limits and Continuity Step 2: kx
1k  $ Ê
$  x
1  $ Ê
$  "  x  $  ". Then
$  " œ È1
% or $  " œ È1  %. Choose $ œ min š1
È1
%ß È1  %
1› , then 0  kx
1k  $ Ê kax#
(b
6k  % and lim f(x) œ
6. By the continuity test, f(x) is continuous at x œ 1. xÄ1

8. Show lim" g(x) œ lim" xÄ

xÄ

%

" 2x

œ 2 œ g ˆ 4" ‰ .

%

Step 1: ¸ #"x
2¸  % Ê
%  #"x
#  % Ê #
%  #"x  #  % Ê Step 2: ¸B
"4 ¸  $ Ê
$  x
4"  $ Ê
$  4"  x  $  4" . Then
$  Choose $ œ

" 4

œ

" 4  #%

% 4(#%)

Ê $œ

" 4




" 4  #%

œ

% 4(2  %)

, or $ 

" 4

œ

, the smaller of the two values. Then 0  ¸x

By the continuity test, g(x) is continuous at x œ

" 4

" 4
#% Ê
4" ¸  $

" 4
#%

x

" 4
#% ¸ #"x


" 4#%

.

" 4

% 4(2
%)

$œ




œ

Ê

2¸  % and lim"

.

xÄ

%

" #x

œ 2.

.

9. Show lim h(x) œ lim È2x
3 œ " œ h(2). xÄ#

xÄ#

Step 1: ¹È2x
3
1¹  % Ê
%  È2x
3
"  % Ê "
%  È2x
3  "  % Ê

(1
%)#  $ #

x

("  %)#  3 . #

Step 2: kx
2k  $ Ê
$  x
2  $ or
$  #  x  $  #. ("
% )#  $ Ê $œ # ("  % Ñ #  $ ("  %Ñ#
"
#œ # #

Then
$  # œ

#


Ê $œ

œ%

# ("
%)#  $ œ "
(1#
%) # #  %# . Choose $ œ %

œ%



%# #,

%# #

, or $  # œ

("  %)#  $ #

the smaller of the two values . Then,

!  kx
2k  $ Ê ¹È2x
3
"¹  %, so lim È2x
3 œ 1. By the continuity test, h(x) is continuous at x œ 2. xÄ#

10. Show lim F(x) œ lim È9
x œ # œ F(5). xÄ&

xÄ&

Step 1: ¹È9
x
2¹  % Ê
%  È9
x
#  % Ê 9
(2
%)#  x  *
(#  %)# . Step 2: 0  kx
5k  $ Ê
$  x
&  $ Ê
$  &  x  $  &. Then
$  & œ *
(#  %)# Ê $ œ (#  %)#
% œ %#  #%, or $  & œ *
(#
%)# Ê $ œ %
(#
%)# œ %#
#%. Choose $ œ %#
#%, the smaller of the two values. Then, !  kx
5k  $ Ê ¹È9
x
#¹  %, so lim È9
x œ #. By the continuity test, F(x) is continuous at x œ 5.

xÄ&

11. Suppose L" and L# are two different limits. Without loss of generality assume L#  L" . Let % œ

" 3

(L#
L" ).

Since x lim f(x) œ L" there is a $"  0 such that 0  kx
x! k  $" Ê kf(x)
L" k  % Ê
%  f(x)
L"  % Äx !

Ê
"3 (L#
L" )  L"  f(x) 

" 3

(L#
L" )  L" Ê 4L"
L#  3f(x)  2L"  L# . Likewise, x lim f(x) œ L# Ä x! so there is a $# such that 0  kx
x! k  $# Ê kf(x)
L# k  % Ê
%  f(x)
L#  % Ê
"3 (L#
L" )  L#  f(x)  3" (L#
L" )  L# Ê 2L#  L"  3f(x)  4L#
L" Ê L"
4L# 
3f(x) 
2L#
L" . If $ œ min e$" ß $# f both inequalities must hold for 0  kx
x! k  $ : 4L"
L#  3f(x)  2L"  L# Ê 5(L"
L# )  0  L"
L# . That is, L"
L#  0 and L"
L#  0, L"
%L# 
3f(x) 
2L#
L"  a contradiction. 12. Suppose xlim f(x) œ L. If k œ !, then xlim kf(x) œ xlim 0 œ ! œ ! † xlim f(x) and we are done. Äc Äc Äc Äc % If k Á 0, then given any %  !, there is a $  ! so that !  lx
cl  $ Ê lfaxb
Ll  l5l Ê lkllfaxb
Ll  % Ê lkafaxb
Lb|  % Ê lakfaxbb
akLbl  %. Thus, xlim kf(x) œ kL œ kŠxlim f(x)‹. Äc Äc

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Chapter 2 Additional and Advanced Exercises 89 13. (a) Since x Ä 0 , 0  x$  x  1 Ê ax$
xb Ä 0
Ê

lim f ax$
xb œ lim c f(y) œ B where y œ x$
x. yÄ!

x Ä !b

(b) Since x Ä 0
,
1  x  x$  0 Ê ax$
xb Ä 0 Ê

(c) Since x Ä 0 , 0  x%  x#  1 Ê ax#
x% b Ä 0 Ê

lim f ax$
xb œ lim b f(y) œ A where y œ x$
x. yÄ!

x Ä !c

lim f ax#
x% b œ lim b f(y) œ A where y œ x#
x% . yÄ!

x Ä !b

(d) Since x Ä 0
,
1  x  0 Ê !  x%  x#  1 Ê ax#
x% b Ä 0 Ê

lim f ax#
x% b œ A as in part (c).

x Ä !b

14. (a) True, because if xlim (f(x)  g(x)) exists then xlim (f(x)  g(x))
xlim f(x) œ xlim [(f(x)  g(x))
f(x)] Äa Äa Äa Äa œ xlim g(x) exists, contrary to assumption. Äa " x

(b) False; for example take f(x) œ

and g(x) œ
x" . Then neither lim f(x) nor lim g(x) exists, but xÄ!

lim (f(x)  g(x)) œ lim ˆ "x
x" ‰ œ lim 0 œ 0 exists.

xÄ!

xÄ!

xÄ!

xÄ!

(c) True, because g(x) œ kxk is continuous Ê g(f(x)) œ kf(x)k is continuous (it is the composite of continuous functions).
1, x Ÿ 0 Ê f(x) is discontinuous at x œ 0. However kf(x)k œ 1 is (d) False; for example let f(x) œ œ 1, x  0 continuous at x œ 0. x#
"

15. Show lim f(x) œ lim

x Ä
1 x  1

x Ä
1

(x  1)(x
") (x  1)

œ lim

x Ä
1

Define the continuous extension of f(x) as F(x) œ œ

œ
#, x Á
1.

x#
1 x1 ,


2

x Á
" . We now prove the limit of f(x) as x Ä
1 , x œ
1

exists and has the correct value. #

Step 1: ¹ xx 
1"
(
#)¹  % Ê
% 

(x  1)(x
") (x  1)

 #  % Ê
%  (x
1)  #  %, x Á
" Ê
%
"  x  %
".

Step 2: kx
(
1)k  $ Ê
$  x  1  $ Ê
$
"  x  $
". Then
$
" œ
%
" Ê $ œ %, or $
" œ %
" Ê $ œ %. Choose $ œ %. Then !  kx
(
1)k  $ #

Ê ¹ xx 
1"
a
#b¹  % Ê

lim F(x) œ
2. Since the conditions of the continuity test are met by F(x), then f(x) has a

x Ä
1

continuous extension to F(x) at x œ
1. 16. Show lim g(x) œ lim xÄ$

xÄ$

x#
2x
3 2x
6

œ lim

xÄ$

(x
3)(x  ") 2(x
3)

œ #, x Á 3. #

Define the continuous extension of g(x) as G(x) œ œ

x
2x
3 2x
6 ,

2

xÁ3 . We now prove the limit of g(x) as , xœ3

x Ä 3 exists and has the correct value. Step 1: ¹ x

#


2x
3 #x
6


2¹  % Ê
% 

(x
3)(x  ") 2(x
3)


#  % Ê
% 

x" #


#  % , x Á $ Ê $
#%  x  $  #% .

Step 2: kx
3k  $ Ê
$  x
3  $ Ê $
$  x  $  $. Then, $
$ œ $
#% Ê $ œ #%, or $  $ œ $  #% Ê $ œ #%. Choose $ œ #%. Then !  kx
3k  $ Ê ¹x

#


2x
3 2x
6


2¹  % Ê lim

xÄ$

(x
3)(x  ") #(x
3)

œ 2. Since the conditions of the continuity test hold for G(x),

g(x) can be continuously extended to G(x) at B œ 3. 17. (a) Let %  ! be given. If x is rational, then f(x) œ x Ê kf(x)
0k œ kx
0k  % Í kx
0k  %; i.e., choose $ œ %. Then kx
0k  $ Ê kf(x)
0k  % for x rational. If x is irrational, then f(x) œ 0 Ê kf(x)
0k  % Í !  % which is true no matter how close irrational x is to 0, so again we can choose $ œ %. In either case, given %  ! there is a $ œ %  ! such that !  kx
0k  $ Ê kf(x)
0k  %. Therefore, f is continuous at x œ 0. (b) Choose x œ c  !. Then within any interval (c
$ ß c  $ ) there are both rational and irrational numbers. If c is rational, pick % œ #c . No matter how small we choose $  ! there is an irrational number x in (c
$ ß c  $ ) Ê kf(x)
f(c)k œ k0
ck œ c 

c #

œ %. That is, f is not continuous at any rational c  0. On

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

90

Chapter 2 Limits and Continuity the other hand, suppose c is irrational Ê f(c) œ 0. Again pick % œ #c . No matter how small we choose $  ! there is a rational number x in (c
$ ß c  $ ) with kx
ck  œ kxk 

c #

œ% Í

œ % Ê f is not continuous at any irrational c  0.

If x œ c  0, repeat the argument picking % œ nonzero value x œ c. 18. (a) Let c œ

c #

kc k #

œ


c # .

x

c #

Then kf(x)
f(c)k œ kx
0k

3c #.

Therefore f fails to be continuous at any

m n

be a rational number in [0ß 1] reduced to lowest terms Ê f(c) œ "n . Pick % œ

" #n

œ %. Therefore f is discontinuous at x œ c, a rational number.

" #n .

No matter how small $  ! is taken, there is an irrational number x in the interval (c
$ ß c  $ ) Ê kf(x)
f(c)k œ ¸0
"n ¸ œ

" n



(b) Now suppose c is an irrational number Ê f(c) œ 0. Let %  0 be given. Notice that number reduced to lowest terms with denominator 2 and belonging to [0ß 1]; denominator 3 belonging to [0ß 1];

" 4

and

[0ß 1]; etc. In general, choose N so that

" N

3 4

with denominator 4 in [0ß 1];

" 3

and

" 2 3 5, 5, 5

2 3

and

" #

is the only rational

the only rationals with 4 5

with denominator 5 in

 % Ê there exist only finitely many rationals in [!ß "] having

denominator Ÿ N, say r" , r# , á , rp . Let $ œ min ekc
ri k : i œ 1ß á ß pf . Then the interval (c
$ ß c  $ ) contains no rational numbers with denominator Ÿ N. Thus, 0  kx
ck  $ Ê kf(x)
f(c)k œ kf(x)
0k œ kf(x)k Ÿ N"  % Ê f is continuous at x œ c irrational. (c) The graph looks like the markings on a typical ruler when the points (xß f(x)) on the graph of f(x) are connected to the x-axis with vertical lines.

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 Ê 0  1R represents the midnight point (at the same exact time). Suppose x" is a point on the equator “just after" noon Ê x"  1R 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(x" )
T(x"  1R)  0. At exactly the same moment in time pick x# to be a point just before midnight Ê x#  1R is just before noon. Then T(x# )
T(x#  1R)  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 1R (simultaneously midnight) such that T(c)
T(c  1R) œ 0; i.e., there is always a pair of antipodal points on the earth's equator where the temperatures are the same. #

#

# # " 20. xlim f(x)g(x) œ xlim af(x)  g(x)b‹
Šxlim af(x)
g(x)b‹ “ ’af(x)  g(x)b
af(x)
g(x)b “ œ "% ’Šxlim Äc Äc % Äc Äc œ "% ˆ$#
a
"b# ‰ œ #.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Chapter 2 Additional and Advanced Exercises 91 21. (a) At x œ 0: lim r (a) œ lim aÄ!

œ lim

1
("  a)

aÄ!

a Ä ! a ˆ
"
È1  a‰

At x œ
1: (b) At x œ 0:

lim

a Ä
"b

œ

r (a) œ


"  È1  a a
1
"
È1  0

œ lim c aÄ!

1
("  a) a ˆ
"  È1  a‰

" #

œ

aÄ!

1
(1  a)

lim

a Ä
"b a ˆ
1
È1  a‰

lim r
(a) œ lim c aÄ!

a Ä !c

È1  a

œ lim Š
"  a


"
È1  a a

œ lim c aÄ!


a

œ lim

a Ä
1 a ˆ
"
È1  a‰ È1  a

œ lim c Š
"
a aÄ!


a a ˆ
1  È 1  a ‰


"
È1  a

‹ Š
"
È1  a ‹

œ lim c aÄ!

œ


"
"
È0

œ1


"  È1  a

‹ Š
"  È1  a ‹


" œ _ (because the
"  È1  a
" œ
_ (because the
"  È1  a

denominator is always negative); lim b r
(a) œ lim b aÄ! aÄ! is always positive). Therefore, lim r
(a) does not exist.

denominator

aÄ!

At x œ
1:

lim

a Ä
"b

r
(a) œ

lim

a Ä
"b


1
È 1  a a

œ

lim


"

a Ä
1b
"  È1  a

œ1

(c)

(d)

22. f(x) œ x  2 cos x Ê f(0) œ 0  2 cos 0 œ 2  0 and f(
1) œ
1  2 cos (
1) œ
1
#  0. Since f(x) is continuous on [
1ß !], by the Intermediate Value Theorem, f(x) must take on every value between [
1
#ß #]. Thus there is some number c in [
1ß !] such that f(c) œ 0; i.e., c is a solution to x  2 cos x œ 0. 23. (a) The function f is bounded on D if f(x)   M and f(x) Ÿ N for all x in D. This means M Ÿ f(x) Ÿ N for all x in D. Choose B to be max ekMk ß kNkf . Then kf(x)k Ÿ B. On the other hand, if kf(x)k Ÿ B, then
B Ÿ f(x) Ÿ B Ê f(x)  
B and f(x) Ÿ B Ê f(x) is bounded on D with N œ B an upper bound and M œ
B a lower bound. (b) Assume f(x) Ÿ N for all x and that L  N. Let % œ L
# N . Since x lim f(x) œ L there is a $  ! such that Äx !

0  kx
x! k  $ Ê kf(x)
Lk  % Í L
%  f(x)  L  % Í L
Í

LN #

 f(x) 

3L
N # .

But L  N Ê

LN #

L
N #

 f(x)  L 

L
N #

 N Ê N  f(x) contrary to the boundedness assumption

f(x) Ÿ N. This contradiction proves L Ÿ N.

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92

Chapter 2 Limits and Continuity (c) Assume M Ÿ f(x) for all x and that L  M. Let % œ Ê L


M
L #

 f(x)  L 

M
L #

3L
M #

Í

 f(x)

M
L # . As in part (b), 0  kx
L  M  M, a contradiction. #

24. (a) If a   b, then a
b   0 Ê ka
bk œ a
b Ê max Öaß b× œ

ab #



ka
b k #

If a Ÿ b, then a
b Ÿ 0 Ê ka
bk œ
(a
b) œ b
a Ê max Öaß b× œ œ

2b #

xÄ0

ab #

sina"
cos xb x

œ lim

lim b sinsinÈxx œ

xÄ0

†

sin x

xÄ0

sinasin xb x

xÄ0

sinax#  xb x xÄ0

œ lim

29. lim

sinax#
%b x
2

œ lim

30. lim

sinˆÈx
$‰ x
9

28. lim

xÄ2

xÄ9

sin x "  cos x

xÄ0

œ lim

ka
b k #

.

sina"
cos xb "
cos x

lim b sinB x †

xÄ0

27. lim




œ lim

xÄ0 x

26.

ab a
b 2a #  # œ # œ a. ka
b k ab œ a # b  b
# a #  #

œ

œ b.

(b) Let min Öaß b× œ 25. lim œ

x! k  $

sinasin xb sin x

†

"
cos x x

Èx sin Èx

†

†

œ lim

sin x x

x Èx

xÄ0

sinasin xb sin x

sinax#
%b x#
%

† ax  2b œ lim

xÄ9

xÄ0

sinˆÈx
$‰ Èx
$

† lim

"
cos# x

x Ä 0 xa"  cos xb

" Èx  $

† lim

sin x

xÄ0 x

œ " † lim

sin# x

x Ä 0 xa"  cos xb

œ " † " œ ".

sinax#  xb # x Ä 0 x x

† lim ax  "b œ " † " œ "

sinax#
%b x#
%

† lim ax  2b œ " † % œ %

xÄ2

†

sina"
cos xb "
cos x

œ " † lim b sin"Èx † lim b Èx œ " † ! † ! œ !. x Ä 0 Š Èx ‹ x Ä 0

† ax  "b œ lim

œ lim

œ lim

œ " † ˆ #! ‰ œ !.

sinax#  xb # x Ä 0 x x

xÄ2

"  cos x "  cos x

†

œ lim

xÄ9

xÄ0

xÄ2

sinˆÈx
$‰ Èx
$

† lim

"

x Ä 9 Èx  $

œ"†

" '

œ

" '

31. Since the highest power of x in the numerator is 1 more than the highest power of x in the denominator, there is an oblique asymptote. y œ 32. As x Ä „ _,

2x3Î2  2x
3 Èx  1 1 x

œ 2x


3 Èx  1 ,

thus the oblique asymptote is y œ 2x.

Ä 0 Ê sinˆ 1x ‰ Ä 0 Ê 1  sinˆ 1x ‰ Ä 1, thus as x Ä „ _, y œ x  x sinˆ 1x ‰ œ xˆ1  sinˆ 1x ‰‰ Ä x;

thus the oblique asymptote is y œ x. 33. As x Ä „ _, x2  1 Ä x2 Ê Èx2  1 Ä Èx2 ; as x Ä
_, Èx2 œ
x, and as x Ä  _, Èx2 œ x; thus the oblique asymptotes are y œ x and y œ
x. 34. As x Ä „ _, x  2 Ä x Ê Èx2  2x œ Èxax  2b Ä Èx2 ; as x Ä
_, Èx2 œ
x, and as x Ä  _, Èx2 œ x; asymptotes are y œ x and y œ
x.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

CHAPTER 3 DIFFERENTIATION 3.1 TANGENTS AND THE DERIVATIVE AT A POINT 1. P" : m" œ 1, P# : m# œ 5

2. P" : m" œ
2, P# : m# œ 0

3. P" : m" œ 5# , P# : m# œ
"#

4. P" : m" œ 3, P# : m# œ
3

5. m œ lim

hÄ!

c4
(
"  h)# d
a4
(
1)# b h


a1
2h  h# b1 h hÄ!

œ lim

œ lim

hÄ!

h(#
h) h

œ 2;

at (
"ß $): y œ $  #(x
(
1)) Ê y œ 2x  5, tangent line

6. m œ lim

hÄ!

c(1  h
1)#  1d
c("
")#  1d h

h#

œ lim

hÄ! h

œ lim h œ 0; at ("ß "): y œ 1  0(x
1) Ê y œ 1, hÄ!

tangent line

È 2È 1  h
2È 1 œ lim 2 1 h h
2 h hÄ! hÄ! 4(1  h)
4 œ lim œ lim È1 2h  1 h Ä ! 2h ŠÈ1  h  1‹ hÄ!

7. m œ lim

†

2È 1  h  2 2È 1  h  #

œ 1;

at ("ß #): y œ 2  1(x
1) Ê y œ x  1, tangent line

8. m œ lim

hÄ!

"

( 1
h)#


( "")#

h


a
2h  h# b # h Ä ! h(
1  h)

œ lim

1
(
1  h)# # h Ä ! h(
1h) 2
h lim # œ 2; h Ä ! (
1  h)

œ lim œ

at (
"ß "): y œ 1  2(x
(
1)) Ê y œ 2x  3, tangent line

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

94

Chapter 3 Differentiation (
2  h)$
(
2)$ h

9. m œ lim

hÄ!


8  12h
6h#  h$  8 h

œ lim

hÄ!

œ lim a12
6h  h# b œ 12; hÄ!

at (
2ß
8): y œ
8  12(x
(
2)) Ê y œ 12x  16, tangent line

10. m œ lim

(

h

hÄ!

œ

"2 8(
8)

hÄ!

hÄ!

at ˆ
#ß
"8 ‰ : y œ
8"
Ê yœ
11. m œ lim

hÄ!

x


" #,

12
6h  h# 8(
2  h)$

œ lim

3 œ
16 ;

3 16


8
(
#  h)$
8h(
#  h)$

œ lim


a12h
6h#  h$ b
8h(
#  h)$

œ lim

hÄ!

" " #
h)$
( #)$

3 16 (x


(
2))

tangent line

c(2  h)#  1d
5 h

œ lim

hÄ!

a5  4h  h# b
5 h

hÄ!

at (2ß 5): y
5 œ 4(x
2), tangent line 12. m œ lim

hÄ!

c("  h)
2(1  h)# d
(
1) h

œ lim

hÄ!

h(4  h) h

œ lim

a1  h
2
4h
2h# b  1 h

3 (3

h h)
2


3

h

hÄ!

œ lim

hÄ!

(3  h)
3(h  1) h(h  1)

h Ä ! h(h  1)

at ($ß $): y
3 œ
2(x
3), tangent line 14. m œ lim

hÄ!

8 (2

h)#


2

h

hÄ!

(2  h)$
8 h hÄ!

œ lim

a8  12h  6h#  h$ b
8 h hÄ!

œ lim

16. m œ lim

hÄ!

c(1  h)$  3(1  h)d
4 h

hÄ!

at ("ß %): y
4 œ 6(t
1), tangent line 17. m œ lim

hÄ!

È4  h
2 h

œ lim

hÄ!

œ "4 ; at (%ß #): y
2 œ 18. m œ lim

hÄ!

œ

" È9  3

È(8  h)  1
3 h

" 4

È4  h
2 h

†

hÄ!

h a12  6h  h# b h hÄ!

a1  3h  3h#  h$  3  3hb
4 h

œ lim

œ lim

œ lim

at (2ß )): y
8 œ 12(t
2), tangent line

È4  h  2 È4  h  2

œ
3;

œ
2;

8
2 a4  4h  h# b h(2  h)# hÄ!

8
2(2  h)# # h Ä ! h(2  h)

œ lim

at (2ß 2): y
2 œ
2(x
2) 15. m œ lim


2h

œ lim

h(
3
2h) h

œ lim

at ("ß
"): y  1 œ
3(x
1), tangent line 13. m œ lim

œ %;

œ lim


2h(4  h) h(2  h)#

œ


8 4

œ
2;

œ 12;

œ lim

hÄ!

(4  h)
4

h Ä ! h ŠÈ4  h  #‹

h a6  3h  h# b h

œ lim

œ 6;

h

h Ä ! h ŠÈ4  h  #‹

œ

" È4  #

(x
4), tangent line

œ lim

hÄ!

È9  h
3 h

œ 6" ; at (8ß 3): y
3 œ

19. At x œ
1, y œ 5 Ê m œ lim

hÄ!

" 6

†

È9  h  3 È9  h  3

œ lim

(9  h)
9

h Ä ! h ŠÈ9  h  3‹

œ lim

h

h Ä ! h ŠÈ9  h  3‹

(x
8), tangent line

5(
"  h)#
5 h

œ lim

hÄ!

5 a1
2h  h# b
5 h

œ lim

hÄ!

5h(
2  h) h

œ
10, slope

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 3.1 Tangents and the Derivative at a Point c1
(2  h)# d
(
3) h

20. At x œ 2, y œ
3 Ê m œ lim

hÄ!

" #

21. At x œ 3, y œ

Ê m œ lim

"

h)
1

(3


#"

h
1 h 1

22. At x œ 0, y œ
1 Ê m œ lim

hÄ!

hÄ!

2
(2  h) 2h(2  h)

œ lim

h

hÄ!

œ lim

hÄ!


(
1) h

a1
4
4h
h# b  3 h

hÄ!

hÄ!


h

œ lim

œ lim


h(4  h) h

œ
4, slope

œ
"4 , slope

h Ä ! 2h(2  h)

(h
1)  (h  ") h(h  1)

œ lim

œ lim

2h

h Ä ! h(h  1)

œ 2, slope

c(x  h)#  4(x  h)
1d
ax#  4x
1b h hÄ! a2xh  h#  4hb lim œ lim (2x  h  4) œ 2x h hÄ! hÄ!

23. At a horizontal tangent the slope m œ 0 Ê 0 œ m œ lim ax#  2xh  h#  4x  4h
1b
ax#  4x
1b h hÄ!

œ lim

œ

 4;

2x  4 œ 0 Ê x œ
2. Then f(
2) œ 4
8
1 œ
5 Ê (
2ß
5) is the point on the graph where there is a horizontal tangent. c(x  h)$
3(x  h)d
ax$
3xb h

24. 0 œ m œ lim

hÄ!

œ lim

hÄ!

3x# h  3xh#  h$
3h h

ax$  3x# h  3xh#  h$
3x
3hb
ax$
3xb h

œ lim

hÄ!

œ lim a3x#  3xh  h#
3b œ 3x#
3; 3x#
3 œ 0 Ê x œ
1 or x œ 1. Then hÄ!

f(
1) œ 2 and f(1) œ
2 Ê (
"ß 2) and ("ß
2) are the points on the graph where a horizontal tangent exists. 25.
1 œ m œ lim

"

h)
1

(x


x
" 1

h

hÄ!

œ lim

hÄ!

(x
1)
(x  h
1) h(x
1)(x  h
1)


h

œ lim

h Ä ! h(x
1)(x  h
1)

œ
(x
" 1)#

Ê (x
1)# œ 1 Ê x#
2x œ 0 Ê x(x
2) œ 0 Ê x œ 0 or x œ 2. If x œ 0, then y œ
1 and m œ
1 Ê y œ
1
(x
0) œ
(x  1). If x œ 2, then y œ 1 and m œ
1 Ê y œ 1
(x
2) œ
(x
3). 26.

" 4

Èx  h
Èx

œ m œ lim œ lim

h

y œ 2  "4 (x
4) œ

hÄ!

f(2  h)
f(2) h

x 4

Èx  h
Èx h

hÄ!

h Ä ! h ŠÈx  h  Èx‹

27. lim

œ lim

h

hÄ!

œ

" #È x

. Thus,

" 4

œ

†

Èx  h  Èx Èx  h  Èx

" #Èx

(x  h)
x

œ lim

h Ä ! h ŠÈx  h  Èx‹

Ê Èx œ 2 Ê x œ 4 Ê y œ 2. The tangent line is

 1.

œ lim

hÄ!

a100
4.9(#  h)# b
a100
4.9(2)# b h


4.9 a4  4h  h# b  4.9(4) h

œ lim

hÄ!

œ lim (
19.6
4.9h) œ
19.6. The minus sign indicates the object is falling downward at a speed of 19.6 m/sec. hÄ!

28. lim

hÄ!

f(10  h)
f(10) h

hÄ!

1(3  h)#
1(3)# h hÄ!

f(3  h)
f(3) h hÄ!

œ lim

f(2  h)
f(2) h hÄ!

œ lim

29. lim

30. lim

3(10  h)#
3(10)# h

œ lim

hÄ!

41 3

3 a20h  h# b h

œ lim

hÄ!

œ 60 ft/sec.

1 c9  6h h#
9d h hÄ!

œ lim

(2  h)$
431 (2)$ h

œ lim

41 3

hÄ!

31. At ax0 , mx0  bb the slope of the tangent line is lim

hÄ!

œ lim 1(6  h) œ 61 hÄ!

c12h  6h#  h$ d h

œ lim

hÄ!

amax0  hb  bb
am x0  bb ax 0  h b
x 0

41 3

c12  6h  h# d œ 161

œ lim

hÄ!

mh h

The equation of the tangent line is y
am x0  bb œ max
x0 b Ê y œ mx  b. 32. At x œ 4, y œ

1 È4 È

œ

" #

œ lim – 22h
È44hh † hÄ!

and m œ lim

hÄ!

2  È4  h 2  È4  h —

È4

1 h

h


"#

œ lim – hÄ!

È4

1 h

h


"#

†

2È 4  h 2È 4  h —

œ lim m œ m. hÄ!

È

œ lim Š 22h
È44hh ‹ hÄ!

hb
h œ lim
È 4
a4  È œ lim
È 2h 4  hŠ2  4  h‹  2h 4  hŠ2  È4  h‹  hÄ! hÄ!

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

95

96

Chapter 3 Differentiation
1 1 œ lim
È œ
È 1 È œ
16 2 4  hŠ2  È4  h‹  2 4Š2  4‹ hÄ!

f(0  h)
f(0) h hÄ!

33. Slope at origin œ lim

h# sin ˆ "h ‰ h hÄ!

œ lim

œ lim h sin ˆ "h ‰ œ 0 Ê yes, f(x) does have a tangent at hÄ!

the origin with slope 0. g(0  h)
g(0) h

34. lim

hÄ!

œ lim

hÄ!

h sin ˆ "h ‰ h

œ lim sin h" . Since lim sin hÄ!

hÄ!

" h

does not exist, f(x) has no tangent at

the origin. 35.

lim

h Ä !c

f(0  h)
f(0) h

lim f(0  h)h
f(0) hÄ! 36.

œ lim c hÄ!


1
0 h

œ _, and lim b hÄ!

f(0  h)
f(0) h

1
0 h

œ lim b hÄ!

œ _ Ê yes, the graph of f has a vertical tangent at the origin.

œ _, and lim b U(0  h)h
U(0) œ lim b hÄ! hÄ! does not have a vertical tangent at (!ß ") because the limit does not exist. lim

h Ä !c

œ _. Therefore,

U(0  h)
U(0) h

œ lim c hÄ!

0
1 h

1
1 h

œ 0 Ê no, the graph of f

37. (a) The graph appears to have a cusp at x œ 0.

(b)

lim c

hÄ!

f(0  h)
f(0) h

œ lim c hÄ!

h#Î&
0 h

œ lim c hÄ!

" h$Î&

œ
_ and lim b hÄ!

" h$Î&

œ _ Ê limit does not exist

Ê the graph of y œ x#Î& does not have a vertical tangent at x œ 0.

38. (a) The graph appears to have a cusp at x œ 0.

(b)

lim

h Ä !c

f(0  h)
f(0) h

œ lim c hÄ!

h%Î&
0 h

œ lim c hÄ!

" h"Î&

œ
_ and lim b hÄ!

" h"Î&

œ _ Ê limit does not exist

Ê y œ x%Î& does not have a vertical tangent at x œ 0.

39. (a) The graph appears to have a vertical tangent at x œ !.

(b)

lim

hÄ!

f(0  h)
f(0) h

œ lim

hÄ!

h"Î&
0 h

œ lim

"

%Î& hÄ! h

œ _ Ê y œ x"Î& has a vertical tangent at x œ 0.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 3.1 Tangents and the Derivative at a Point 40. (a) The graph appears to have a vertical tangent at x œ 0.

(b)

lim

hÄ!

f(0  h)
f(0) h

œ lim

hÄ!

h$Î&
0 h

"

œ lim

#Î& hÄ! h

œ _ Ê the graph of y œ x$Î& has a vertical tangent at x œ 0.

41. (a) The graph appears to have a cusp at x œ 0.

(b)

lim c

hÄ!

f(0  h)
f(0) h

œ lim c hÄ!

4h#Î&
2h h

œ lim c hÄ!

4 h$Î&


2 œ
_ and lim b hÄ!

4 h$Î&


#œ_

Ê limit does not exist Ê the graph of y œ 4x#Î&
2x does not have a vertical tangent at x œ 0.

42. (a) The graph appears to have a cusp at x œ 0.

(b)

lim

hÄ!

f(0  h)
f(0) h

œ lim

hÄ!

h&Î$
5h#Î$ h

œ lim h#Î$
hÄ!

5 h"Î$

œ 0
lim

y œ x&Î$
5x#Î$ does not have a vertical tangent at x œ !.

5

"Î$ hÄ! h

does not exist Ê the graph of

43. (a) The graph appears to have a vertical tangent at x œ 1 and a cusp at x œ 0.

(b) x œ 1:

lim

hÄ!

(1  h)#Î$
(1  h
1)"Î$
" h

œ lim

hÄ!

(1  h)#Î$
h"Î$
" h

œ
_

Ê y œ x#Î$
(x
1)"Î$ has a vertical tangent at x œ 1;

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

97

98

Chapter 3 Differentiation x œ 0:

lim

hÄ!

f(0  h)
f(0) h

œ lim

hÄ!

h#Î$
(h
1)"Î$
(
1)"Î$ h

" œ lim ’ h"Î$


hÄ!

(h
")"Î$ h

 h" “

does not exist Ê y œ x#Î$
(x
1)"Î$ does not have a vertical tangent at x œ 0. 44. (a) The graph appears to have vertical tangents at x œ 0 and x œ 1.

(b) x œ 0:

lim

hÄ!

f(0  h)
f(0) h

œ lim

hÄ!

h"Î$  (h
1)"Î$
(
")"Î$ h

œ _ Ê y œ x"Î$  (x
1)"Î$ has a

vertical tangent at x œ 0;

x œ 1:

lim

hÄ!

f(1  h)
f(1) h

œ lim

hÄ!

(1  h)"Î$  ("  h
1)"Î$
1 h

œ _ Ê y œ x"Î$  (x
1)"Î$ has a

vertical tangent at x œ ".

45. (a) The graph appears to have a vertical tangent at x œ 0.

(b)

lim b

hÄ!

f(0  h)
f(0) h

œ lim b xÄ!

Èh
0 h

œ lim

"

h Ä ! Èh


È kh k
0

f(0  h)
f(0) h

œ lim c œ lim c h hÄ! hÄ! Ê y has a vertical tangent at x œ 0. lim

h Ä !c

œ _;
È kh k
kh k

œ lim c hÄ!

" È kh k

œ_

46. (a) The graph appears to have a cusp at x œ 4.

(b)

lim b

f(4  h)
f(4) h

œ lim b hÄ!

Èk4
(4  h)k
0 h

lim

f(4  h)
f(4) h

œ lim c

Èk4
(4  h)k h

hÄ!

h Ä !c

hÄ!

œ lim b hÄ!

œ lim c hÄ!

È kh k h

È kh k
lhl

œ lim b hÄ!

œ lim c hÄ!

" Èh


" È kh k

œ _;

œ
_

Ê y œ È%
x does not have a vertical tangent at x œ 4. 47-50. Example CAS commands: Maple: f := x -> x^3 + 2*x;x0 := 0; plot( f(x), x=x0-1/2..x0+3, color=black, title="Section 3.1, #47(a)" ); q := unapply( (f(x0+h)-f(x0))/h, h );

# part (a) # part (b)

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 3.2 The Derivative as a Function L := limit( q(h), h=0 ); # part (c) sec_lines := seq( f(x0)+q(h)*(x-x0), h=1..3 ); # part (d) tan_line := f(x0) + L*(x-x0); plot( [f(x),tan_line,sec_lines], x=x0-1/2..x0+3, color=black, linestyle=[1,2,5,6,7], title="Section 3.1, #47(d)", legend=["y=f(x)","Tangent line at x=0","Secant line (h=1)", "Secant line (h=2)","Secant line (h=3)"] ); Mathematica: (function and value for x0 may change) Clear[f, m, x, h] x0 œ p; f[x_]: œ Cos[x]  4Sin[2x] Plot[f[x], {x, x0
1, x0  3}] dq[h_]: œ (f[x0+h]
f[x0])/h m œ Limit[dq[h], h Ä 0] ytan: œ f[x0]  m(x
x0) y1: œ f[x0]  dq[1](x
x0) y2: œ f[x0]  dq[2](x
x0) y3: œ f[x0]  dq[3](x
x0) Plot[{f[x], ytan, y1, y2, y3}, {x, x0
1, x0  3}] 3.2 THE DERIVATIVE AS A FUNCTION 1. Step 1: f(x) œ 4
x# and f(x  h) œ 4
(x  h)# f(x  h)
f(x) h

Step 2:

œ

c4
(x  h)# d
a4
x# b h

œ

a4
x#
2xh
h# b
4  x# h

œ


2xh
h# h

œ

h(
2x
h) h

œ
2x
h Step 3: f w (x) œ lim (
2x
h) œ
2x; f w (
$) œ 6, f w (0) œ 0, f w (1) œ
2 hÄ!

2. F(x) œ (x
1)#  1 and F(x  h) œ (x  h
1)#  " Ê Fw (x) œ lim œ lim

hÄ!

hÄ!

ax#  2xh  h#
2x
2h  1  1b
ax#
2x  1  1b h w

w

œ lim

w

œ 2(x
1); F (
1) œ
4, F (0) œ
2, F (2) œ 2 3. Step 1: g(t) œ

" t#

and g(t  h) œ "

Step 2:

"

#
# g(t  h)
g(t) œ (t h)h t h
2t
h)
2t
h œ h( (t  h)# t# h œ (t  h)# t#

Step 3: gw (t) œ lim


2t
h

# # h Ä ! (t  h) t

4. k(z) œ

1
z #z

and k(z  h) œ

œ œ


" 2z#

œ

Œ


2t t# †t#

1
(z  h) 2(z  h)


("
z)(z  h) lim (1
z
h)z #(z  h)zh hÄ!

2xh  h#
2h h

œ lim (2x  h
2) hÄ!

" (t  h)#

œ

œ

hÄ!

c(x  h
1)#  1d
c(x
1)#  1d h

t#
(t h)# (t h)# †t#


h

œ


2 t$

t#
at#  2th  h# b (t  h)# †t# †h

œ

œ


2th
h# (t  h)# t# h

2 ; gw (
1) œ 2, gw (2) œ
"4 , gw ŠÈ3‹ œ
3È 3

Ê kw (z) œ lim

hÄ!

Š

"

(z
h) " z #(z
h)
#z ‹

#
z
h  z#  zh lim z
z
zh 2(z  h)zh hÄ!

h

œ lim


h

h Ä ! 2(z  h)zh

œ lim


"

h Ä ! #(z  h)z

; kw (
") œ
"# , kw (1) œ
"# , kw ŠÈ2‹ œ
4"

5. Step 1: p()) œ È3) and p()  h) œ È3()  h)

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

99

100

Chapter 3 Differentiation

Step 2:

p()  h)
p()) h

œ

œ

È3()  h)
È3) h

3h h ŠÈ3)  3h  È3)‹

Step 3: pw ()) œ lim

œ

ŠÈ3)  3h
È3)‹ h

œ

3 È3)  3h  È3)

3

œ

h Ä ! È3)  3h  È3)

†

œ

3 È 3)  È 3)

3 2È 3 )

ŠÈ3)  3h  È3)‹ ŠÈ3)  3h  È3)‹

; pw (1) œ

œ

(3)  3h)
3) h ŠÈ3)  3h  È3)‹

, pw (3) œ "# , pw ˆ 32 ‰ œ

3 2È 3

3 #È2

È2s  2h  1
È2s  1 h hÄ!

6. r(s) œ È2s  1 and r(s  h) œ È2(s  h)  1 Ê rw (s) œ lim œ lim

ŠÈ2s  h  1
È2s  1‹ h

hÄ!

œ lim

ŠÈ2s  2h  1  È2s  1‹

†

œ lim

2h

œ

" È2s  1

; rw (0) œ 1, rw (1) œ

2

" È3

, rw ˆ #" ‰ œ

hÄ!

6x# h  6xh#  2h$ h

hÄ!

9. s œ r(t) œ œ lim

t 2t1

Š

dr ds

œ

2 2È2s  1

2 ax$  3x# h  3xh#  h$ b
2x$ h hÄ!

2(x  h)$
2x$ h hÄ!

œ lim

œ lim

œ lim a6x#  6xh  2h# b œ 6x# hÄ!

œ lim

th 2(th)1

and r(t  h) œ

(t b h)(2t b 1) c t(2t b 2h b 1) ‹ (2t b 2h b 1)(2t b 1)

h

hÄ!

œ lim

œ

2t#  t  2ht  h
2t#
2ht
t (2t  2h  1)(2t  1)h hÄ! " " (2t  1)(2t  1) œ (2t  1)#

dv dt

œ lim

hÄ!

œ lim

10.

2 È2s  1  È2s  1

3 2 2 3 2 2 3 2 ˆas  hb3
2as  hb2  3‰
ˆs3
2s2  3‰ œ lim s  3s h  3sh  h
2s
h 4sh
h  3
s  2s
3 h hÄ! hÄ! 2 2 3 2 hˆ3s2  3sh  h2
4s
h‰ lim 3s h  3sh hh
4sh
h œ lim œ lim a3s2  3sh  h2
4s
hb œ 3s2
2s h hÄ! hÄ! hÄ!

8. r œ s3
2s2  3 Ê œ

dy dx

h a6x#  6xh  2h# b h

œ lim

œ

" È2

7. y œ f(x) œ 2x$ and f(x  h) œ 2(x  h)$ Ê œ lim

h Ä ! h ŠÈ2s  2h  1  È2s  1‹

h Ä ! È2s  2h  1  È2s  1

h Ä ! h ŠÈ2s  2h  1  È2s  1‹

(2s  2h  1)
(2s  1)

œ lim

ŠÈ2s  2h  1  È2s  1‹

’(t  h)


hÄ!

ht#  h# t  h h(t  h)t hÄ!

œ lim

Œ

" “
ˆt
" ‰ t h h

œ lim

11. p œ f(q) œ

t

" Èq  1

h

hÄ!

(t  h)(2t  1)
t(2t  2h  1) (2t  2h  1)(2t  1)h h

h Ä ! (2t  2h  1)(2t  1)h

œ lim

hÄ!

t#  ht  1 h Ä ! (t  h)t

and f(q  h) œ œ lim

œ

h


t

" " t h h

t#  1 t#

" È(q  h)  1

Ê

h Ä ! (2t  2h  1)(2t  1)

Š

œ lim

h(t

h)t
t (t (t h)t

dp dq

h)

‹

h

hÄ!

œ1

"

œ lim

" t#

œ lim

Š È(q

hÄ!

"

h)

1

‹
Š Èq"

1

‹

h

Èq  1
Èq  h  1

h Ä ! hÈ q  h  1 È q  1

h

hÄ!

t ‰ Š 2(t bt bh)hb 1 ‹
ˆ 2t b 1

œ lim

ds dt

œ lim

œ lim

Èq b 1 c Èq b h b 1 Èq b h b 1 Èq b 1


Ê

œ

ˆÈ q  1
È q  h  1 ‰ ˆ È q  1  È q  h  1 ‰ 1)
(q  h  1) † ˆÈq  1  Èq  h  1‰ œ lim hÈq  h  1(qÈq  1 ˆÈ q  1  È q  h  1 ‰ h Ä ! h Èq  h  1 Èq  1 hÄ!
h
" lim œ lim Èq  h  1 Èq  1 ˆÈq  1  Èq  h  1‰ h Ä ! h È q  h  1 È q  1 ˆÈ q  1  È q  h  1 ‰ hÄ!
"
" œ È q  1 È q  1 ˆÈ q  1  È q  1 ‰ 2(q  1) Èq  1

dz dw

œ lim

œ lim œ

12.

Š È3(w " h)
2
h

hÄ!

œ lim

hÄ!

"

È3w
2 ‹

ŠÈ3w
2
È3w  3h
2‹ hÈ3w  3h
2 È3w
2

œ lim

œ lim

È3w
2
È3w  3h
2

h Ä ! hÈ3w  3h
2 È3w
2

†

ŠÈ3w
2È3w3h
2‹ ŠÈ3w
2  È3w  3h
2‹


3

h Ä ! È3w  3h
2 È3w
2 ŠÈ3w
2  È3w  3h
2‹

œ

œ

œ lim

(3w
2)
(3w  3h
2)

h Ä ! hÈ3w  3h
2 È3w
2 ŠÈ3w
2  È3w  3h
2‹


3 È3w
2 È3w
2 ŠÈ3w
2  È3w
2‹


3 2(3w
2) È3w
2

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 3.2 The Derivative as a Function 13. f(x) œ x  œ

x(x  h)#  9x
x# (x  h)
9(x  h) x(x  h)h

œ

h(x#  xh
9) x(x  h)h

" #x

14. k(x) œ w

hÄ!

œ œ 16.

dy dx

œ

œ

x#
9 x#

œ

9 9 (x b h) “
’x  x “

’(x  h) 

h

x$  2x# h  xh#  9x
x$
x# h
9x
9h x(x  h)h

œ

x#  xh
9 h Ä ! x(x  h)

and k(x  h) œ

x# h  xh#
9h x(x  h)h

œ

œ1


9 x#

; m œ f w (
3) œ 0

Š # "x h
k(x  h)
k(x) œ lim h h hÄ! hÄ!
h
"
" lim œ lim (2  x)(#  x  h) œ (2  x)# ; h Ä ! h(2  x)(2  x  h) hÄ! " 2  (x  h)

Ê kw (x) œ lim

$ # # $ # # $ # c(t  h)$
(t  h)# d
at$
t# b œ lim at  3t h  3th  h b
h at  2th  h b
t  t h hÄ! hÄ! # # # # $ # lim 3t h  3th hh
2th
h œ lim h a3t  3th h h
2t
hb œ lim a3t#  3th  hÄ! hÄ! hÄ! ¸ 3t#
2t; m œ ds œ 5 dt tœ
"

#

" ‹ x

œ lim

ax b hb b 3 1 c ax b hb

œ lim

b3
1x c x

h

hÄ!

œ lim

17. f(x) œ

4h

b h b 3ba1 c xb c ax b 3ba1 c x c hb a1 c x c hba1 c xb h

œ lim

8 ŠÈx
2
Èx  h
2‹ hÈ x  h
2 È x
2

4

†

8 È(x  h)
2

ŠÈx
2  Èx  h
2‹

œ


8h hÈx  h
2 Èx
2 ŠÈx
2  Èx  h
2‹

œ


8 Èx
2 Èx
2 ŠÈx
2  Èx
2‹

œ

œ

œ

h#
2t
hb

x  h  3
x2
xh
3x
x
3  x2  3x  xh  3h h a1
x
h b a 1
x b

4 ; dy ¹ a1
xb2 dx xœ
2

f(x  h)
f(x) h

Ê

ŠÈx
2  Èx  h
2‹

œ lim

hÄ!

h Ä ! a1
x
hba1
xb

and f(x  h) œ

8 Èx
2

ax

œ lim

hÄ!

h Ä ! ha1
x
hba1
xb

œ

f(x  h)
f(x) h

Ê

9 (x  h)

; f w (x) œ lim

" 16

k (2) œ
ds dt

x#  xh
9 x(x  h)

œ

(#  x)
(2  x  h) h(2  x)(2  x  h)

œ lim

15.

and f(x  h) œ (x  h) 

9 x

œ

œ

4 a3 b 2

4 9

È(x b h) c 2
Èx c 2 8

œ

8

h

8[(x
2)
(x  h
2)] hÈx  h
2 Èx
2 ŠÈx
2  Èx  h
2‹
8

Ê f w (x) œ lim

h Ä ! Èx  h
2 Èx
2 ŠÈx
2  Èx  h
2‹


4 (x
2)Èx
2

; m œ f w (6) œ


4 4È 4

œ
"# Ê the equation of the tangent

line at (6ß 4) is y
4 œ
"# (x
6) Ê y œ
"# x  $  % Ê y œ
"# x  (. ˆ1  È4
(z  h)‰
Š1  È4
z‹

18. gw (z) œ lim

h

hÄ!

œ

h

hÄ!

(4
z
h)
(4
z) lim h Ä ! h ŠÈ4
z
h  È4
z‹
" œ
"# 2È 4
3
"# z  $#  # Ê w

ŠÈ4
z
h
È4
z‹

œ lim

œ


h lim h Ä ! h ŠÈ4
z
h  È4
z‹

†

ŠÈ4
z
h  È4
z‹ ŠÈ4
z
h  È4
z‹
"

œ lim

h Ä ! ŠÈ4
z
h  È4
z‹

œ


" 2È 4
z

m œ gw (3) œ

Ê the equation of the tangent line at ($ß #) is w
2 œ
"# (z
3)

Êwœ

œ
"# z  (# .

19. s œ f(t) œ 1
3t# and f(t  h) œ 1
3(t  h)# œ 1
3t#
6th
3h# Ê a1
3t#
6th
3h# b
a1
3t# b h hÄ!

œ lim

20. y œ f(x) œ "
œ lim

"
" x

21. r œ f()) œ hÄ!

h

h

hÄ!

œ lim

x

" x

2 È4
)

œ lim (
6t
3h) œ
6t Ê hÄ!

and f(x  h) œ 1
œ lim

h

h Ä ! x(x  h)h

œ

œ lim

Ê

dy dx "

h Ä ! x(x  h)

œ

œ lim

" x#

" 3

Ê

È œ

dy dx ¹x= 3

2 È4
()  h)

Ê

dr d)

œ lim

hÄ!

f(t  h)
f(t) h

œ6

f(x  h)
f(x) h hÄ!

œ lim

œ lim

Š1


x

" ‹
Š1
" ‹ h x h

hÄ!

f()  h)
f()) œ lim h hÄ! hÄ! È È 2È4
)
#È%
)
h Š2 %
)  2 4
)
h‹ lim † È Š2 4
)  #È4
)
h‹ h Ä ! hÈ 4
) È 4
)
h

and f()  h) œ

2È 4
)
2È 4
)
h hÈ 4
) È 4
)
h

" xh

ds ¸ dt t=c"

ds dt

È4 c ) c h
È4 c ) 2

2

h

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

;

101

102

Chapter 3 Differentiation 4(%
))
4(%
)
h)

œ lim

h Ä ! 2hÈ4
) È4
)
h ŠÈ4
)  È4
)
h‹

œ

2 (4
)) Š2È4
)‹

œ

" (4
))È4
)

Ê

dr ¸ d) )œ!

œ lim

2

h Ä ! È4
) È4
)
h ŠÈ4
)  È%
)
h‹

œ

" 8

22. w œ f(z) œ z  Èz and f(z  h) œ (z  h)  Èz  h Ê œ lim

Šz  h  Èz  h‹
ˆz  Èz‰

hÄ!

h

œ 1  lim

(z  h)
z

h Ä ! h ŠÈz  h  Èz‹

h  Èz  h
Èz h hÄ!

œ lim

œ 1  lim

"

h Ä ! Èz  h  Èz

"

dw dz

œ lim

hÄ!

œ lim –1  hÄ!

œ"

" 2È z

Ê

f(z  h)
f(z) h Èz  h
Èz h

dw ¸ dz zœ4

œ

†

ŠÈz  h  Èz‹ ŠÈz  h  Èz‹ —

5 4

"

fazb
faxb a x  #b
a z  # b x
z
" z #
x # 23. f w axb œ zlim œ zlim œ zlim œ zlim œ zlim œ Äx z
x Ä x z
x Ä x az
xbaz  #bax  #b Ä x az
xbaz  #bax  #b Ä x az  #bax  #b ˆz2
3z  4‰
ˆx2
3x  4‰


" ax  #b #

fazb
faxb z
3z
x  3x z
x
3z  3x 24. f w axb œ zlim œ zlim œ zlim œ zlim z
x z
x z
x Ä x z
x Äx Äx Äx az
xbaz  xb
3‘ az
xbaz  xb
3az
xb az  xb
3‘ œ 2x
3 œ zlim œ zlim œ zlim z
x z
x Äx Äx Äx z

2

2

2

2

x

gazb
gaxb z a x
"b
x a z
" b
z  x
" zc"
x " 25. gw axb œ zlim œ zlim œ zlim œ zlim œ zlim œ Äx z
x Äx z
x Ä x az
xbaz
"bax
"b Ä x az
xbaz
"bax
"b Ä x az
"bax
"b g az b
g a x b 26. gw axb œ zlim œ zlim Äx z
x Äx

ˆ"  Èz‰
ˆ"  Èx‰ z
x

œ zlim Äx

Èz
Èx z
x

†

Èz  Èx Èz  Èx


" a x
"b #

z
x " œ zlim œ zlim œ Ä x az
x bˆÈ z  È x ‰ Ä x Èz  Èx

" #È x

27. Note that as x increases, the slope of the tangent line to the curve is first negative, then zero (when x œ 0), then positive Ê the slope is always increasing which matches (b). 28. Note that the slope of the tangent line is never negative. For x negative, f#w (x) is positive but decreasing as x increases. When x œ 0, the slope of the tangent line to x is 0. For x  0, f#w (x) is positive and increasing. This graph matches (a). 29. f$ (x) is an oscillating function like the cosine. Everywhere that the graph of f$ has a horizontal tangent we expect f$w to be zero, and (d) matches this condition. 30. The graph matches with (c). 31. (a) f w is not defined at x œ 0, 1, 4. At these points, the left-hand and right-hand derivatives do not agree. For example, lim c xÄ!

f(x)
f(0) x
0

œ slope of line joining (
%ß 0) and (!ß #) œ

" #

but lim b xÄ!

line joining (0ß 2) and ("ß
2) œ
4. Since these values are not equal, f w (0) œ

f(x)
f(0) x
0

f(0) lim f(x)x

0 xÄ!

(b)

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

œ slope of

does not exist.

Section 3.2 The Derivative as a Function 32. (a)

103

(b) Shift the graph in (a) down 3 units

33.

(b) The fastest is between the 20th and 30th days; slowest is between the 40th and 50th days.

34. (a)

35. Answers may vary. In each case, draw a tangent line and estimate its slope. ‰F (a) i) slope ¸ 1.54 Ê dT ii) slope ¸ 2.86 Ê dt ¸ 1.54 hr iii) slope ¸ 0 Ê

dT dt

¸ 0‰ hrF

iv) slope ¸
3.75

dT ‰F dt ¸ 2.86 hr ‰F Ê dT dt ¸
3.75 hr

(b) The tangent with the steepest positive slope appears to occur at t œ 6 Ê 12 p.m. and slope ¸ 7.27 Ê The tangent with the steepest negative slope appears to occur at t œ 12 Ê 6 p.m. and ‰F slope ¸
8.00 Ê dT dt ¸
8.00 hr (c)

36. Answers may vary. In each case, draw a tangent line and estimate the slope. lb (a) i) slope ¸
20.83 Ê dW ii) slope ¸
35.00 Ê dt ¸
20.83 month iii) slope ¸
6.25 Ê

dW dt

dW dt

lb ¸
35.00 month

lb ¸
6.25 month

(b) The tangentwith the steepest positive slope appears to occur at t œ 2.7 months. and slope ¸ 7.27 lb Ê dW dt ¸
53.13 month

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

dT dt

¸ 7.27‰ hrF .

104

Chapter 3 Differentiation

(c)

37. Left-hand derivative: For h  0, f(0  h) œ f(h) œ h# (using y œ x# curve) Ê œ lim c hÄ!

h#
0 h

œ lim c h œ 0; hÄ!

Right-hand derivative: For h  0, f(0  h) œ f(h) œ h (using y œ x curve) Ê œ lim b hÄ! Then lim c hÄ!

h
0 h

lim

h Ä !c

œ lim b 1 œ 1; hÄ!

f(0  h)
f(0) h

Á lim b hÄ!

f(0  h)
f(0) h

lim

h Ä !b

œ lim c 0 œ 0; hÄ!

f(1  h)
f(1) h

lim

h Ä !c

Right-hand derivative: When h  !, 1  h  1 Ê f(1  h) œ 2(1  h) œ 2  2h Ê

Then lim c hÄ!

(2  2h)
2 h

œ lim b hÄ!

f(1  h)
f(1) h

2h h

hÄ!

È1  h
" h

œ lim c

lim

h Ä !b

ŠÈ1  h
"‹ h

hÄ!

†

ŠÈ1  h  "‹ ŠÈ1  h  1‹

œ lim c hÄ!

lim

h Ä !c

Then lim c hÄ!

(2h  1)
" h f(1  h)
f(1) h

40. Left-hand derivative:

lim

h Ä !c

Right-hand derivative: œ lim b hÄ! Then lim c hÄ!


h h(1  h)

œ lim b 2 œ 2; hÄ!

f(1  h)
f(") h

lim b

hÄ!

œ lim b hÄ!

f(1  h)
f(1) h

f(1  h)
f(1) h

Á lim b hÄ!

(1  h)
" h ŠÈ1  h  "‹

œ lim c hÄ!

Á lim b hÄ!

" È1  h  1

lim

h Ä !b

Ê the derivative f w (1) does not exist. (1  h)
" h

œ lim c hÄ!

f(1  h)
f(") h
" 1h

f(1  h)
f(1) h

f("  h)
f(1) h

Right-hand derivative: When h  0, 1  h  1 Ê f(1  h) œ 2(1  h)
1 œ 2h  1 Ê œ lim b hÄ!

œ lim c 1 œ 1; hÄ!

Š 1 " h
"‹

œ lim b hÄ!

h

œ lim b hÄ!

Š

1
(1 h) 1 h ‹

h

œ
1; f(1  h)
f(1) h

Ê the derivative f w (1) does not exist.

41. f is not continuous at x œ 0 since lim faxb œ does not exist and fa0b œ
1 xÄ!

42. Left-hand derivative: Right-hand derivative: Then lim c hÄ!

g(h)
g(0) h

lim

h Ä !c

g(h)
g(0) h

lim

h Ä !b

œ lim b hÄ!

2
2 h

Ê the derivative f w (1) does not exist.

39. Left-hand derivative: When h  0, 1  h  1 Ê f(1  h) œ È1  h Ê œ lim c

œ lim c hÄ!

œ lim b 2 œ 2; hÄ!

f(1  h)
f(1) h

Á lim b hÄ!

f(0  h)
f(0) h

Ê the derivative f w (0) does not exist.

38. Left-hand derivative: When h  !, 1  h  1 Ê f(1  h) œ 2 Ê

œ lim b hÄ!

f(0  h)
f(0) h

œ lim c hÄ!

g(h)
g(0) h

œ lim b hÄ!

g(h)
g(0) h

h1Î3
0 h h

2Î3

œ lim c hÄ!


0 h

1 h2Î3

œ lim b hÄ!

œ +_;

1 h1Î3

œ +_;

œ  _ Ê the derivative gw (0) does not exist.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

œ #" ;

f("h)
f(1) h

Section 3.2 The Derivative as a Function 43. (a) The function is differentiable on its domain
$ Ÿ x Ÿ 2 (it is smooth) (b) none (c) none 44. (a) The function is differentiable on its domain
# Ÿ x Ÿ 3 (it is smooth) (b) none (c) none 45. (a) The function is differentiable on
$ Ÿ x  0 and !  x Ÿ 3 (b) none (c) The function is neither continuous nor differentiable at x œ 0 since lim c f(x) Á lim b f(x) xÄ! xÄ! 46. (a) f is differentiable on
# Ÿ x 
1,
"  x  0, 0  x  2, and 2  x Ÿ 3 (b) f is continuous but not differentiable at x œ
1: lim f(x) œ 0 exists but there is a corner at x œ
1 since x Ä
1

œ
3 and lim b f(
"  h)h
f(
1) œ 3 Ê f w (
1) does not exist hÄ! hÄ! (c) f is neither continuous nor differentiable at x œ 0 and x œ 2: at x œ 0, lim c f(x) œ 3 but lim b f(x) œ 0 Ê lim f(x) does not exist; lim c

f(
1  h)
f(
") h

xÄ!

xÄ0

xÄ!

at x œ 2, lim f(x) exists but lim f(x) Á f(2) xÄ#

xÄ#

47. (a) f is differentiable on
" Ÿ x  0 and 0  x Ÿ 2 (b) f is continuous but not differentiable at x œ 0: lim f(x) œ 0 exists but there is a cusp at x œ 0, so f(0  h)
f(0) h hÄ!

f w (0) œ lim

xÄ!

does not exist

(c) none 48. (a) f is differentiable on
$ Ÿ x 
2,
2  x  2, and 2  x Ÿ 3 (b) f is continuous but not differentiable at x œ
2 and x œ 2: there are corners at those points (c) none 49. (a) f w (x) œ lim

hÄ!

f(x  h)
f(x) h

œ lim

hÄ!


(x  h)#
a
x# b h

œ lim

hÄ!


x#
2xh
h#  x# h

œ lim (
2x
h) œ
2x hÄ!

(b)

(c) yw œ
2x is positive for x  0, yw is zero when x œ 0, yw is negative when x  0 (d) y œ
x# is increasing for
_  x  0 and decreasing for !  x  _; the function is increasing on intervals where yw  0 and decreasing on intervals where yw  0 f(x  h)
f(x) h hÄ!

50. (a) f w (x) œ lim

œ lim

hÄ!

Š xc"
h
h

1 x ‹

œ lim

hÄ!


x  (x  h) x(x  h)h

œ lim

"

h Ä ! x(x  h)

œ

" x#

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

105

106

Chapter 3 Differentiation

(b)

(c) yw is positive for all x Á 0, yw is never 0, yw is never negative (d) y œ
"x is increasing for
_  x  0 and !  x  _ 51. (a) Using the alternate formula for calculating derivatives: f w (x) œ zlim Äx œ

$ $ lim z
x z Ä x 3(z
x)

œ

az#  zx  x# b lim (z
x)3(z
x) zÄx

œ

# # lim z  zx3  x zÄx

f(z)
f(x) z
x

#

w

$

Š z3


œ zlim Äx

œ x Ê f (x) œ x

x$ 3 ‹

z
x

#

(b)

(c) yw is positive for all x Á 0, and yw œ 0 when x œ 0; yw is never negative (d) y œ

x$ 3

is increasing for all x Á 0 (the graph is horizontal at x œ 0) because y is increasing where yw  0; y is

never decreasing

52. (a) Using the alternate form for calculating derivatives: f w (x) œ zlim Äx œ

% % lim z
x z Ä x 4(z
x)

œ

$  xz#  x# z x$ b lim (z
x) az 4(z
x) zÄx

œ

f(z)
f(x) z
x

$ # # $ lim z  xz 4 x z  x zÄx

œ zlim Äx $

Œ

z% 4




x% 4


z
x

w

œ x Ê f (x) œ x$

(b)

(c) yw is positive for x  0, yw is zero for x œ 0, yw is negative for x  0 (d) y œ

x% 4

is increasing on 0  x  _ and decreasing on
_  x  0

# # # a2(x  h)#
13(x  h)  5b
a2x#
13x  5b œ lim 2x  4xh  2h
13x
h13h  5
2x  13x
5 h hÄ! hÄ! # lim 4xh  2hh
13h œ lim (4x  2h
13) œ 4x
13, slope at x. The slope is
1 when hÄ! hÄ!

53. yw œ lim œ

4x
13 œ
"

Ê 4x œ 12 Ê x œ 3 Ê y œ 2 † 3#
13 † 3  5 œ
16. Thus the tangent line is y  16 œ (
1)(x
3) Ê y œ
x
"$ and the point of tangency is (3ß
16). 54. For the curve y œ Èx, we have yw œ lim

hÄ!

œ lim

"

h Ä ! Èx  h  Èx

œ

" #Èx

ŠÈx  h
Èx‹ h

†

ŠÈx  h  Èx‹ ŠÈx  h  Èx‹

œ lim

(x  h)
x

h Ä ! ŠÈx  h  Èx‹ h

. Suppose ˆ+ß Èa‰ is the point of tangency of such a line and (
"ß !) is the point

on the line where it crosses the x-axis. Then the slope of the line is

Èa
0 a
(
1)

œ

Èa a1

which must also equal

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 3.2 The Derivative as a Function " ; 2È a

using the derivative formula at x œ a Ê

exist: its point of tangency is ("ß "), its slope is

Èa a1

œ

" #È a

œ

" Ê 2a œ a  1 Ê a œ 1. #Èa " # ; and an equation of the line is

Thus such a line does y
1œ

" #

(x
1)

Ê y œ "# x  "# . 55. Yes; the derivative of
f is
f w so that f w (x! ) exists Ê
f w (x! ) exists as well. 56. Yes; the derivative of 3g is 3gw so that gw (7) exists Ê 3gw (7) exists as well. 57. Yes, lim

g(t)

t Ä ! h(t)

can exist but it need not equal zero. For example, let g(t) œ mt and h(t) œ t. Then g(0) œ h(0)

œ 0, but lim

g(t)

t Ä ! h(t)

œ lim

tÄ!

mt t

œ lim m œ m, which need not be zero. tÄ!

58. (a) Suppose kf(x)k Ÿ x# for
" Ÿ x Ÿ 1. Then kf(0)k Ÿ 0# Ê f(0) œ 0. Then f w (0) œ lim œ lim

hÄ!

f(h)
0 h

œ lim

hÄ!

f(h) h .

For khk Ÿ 1,
h# Ÿ f(h) Ÿ h# Ê
h Ÿ

hÄ!

f(h) h

f(0  h)
f(0) h

Ÿ h Ê f w (0) œ lim

hÄ!

f(h) h

œ0

by the Sandwich Theorem for limits. (b) Note that for x Á 0, kf(x)k œ ¸x# sin "x ¸ œ kx# k ksin xk Ÿ kx# k † 1 œ x# (since
" Ÿ sin x Ÿ 1). By part (a), f is differentiable at x œ 0 and f w (0) œ 0.

59. The graphs are shown below for h œ 1, 0.5, 0.1. The function y œ y œ Èx so that

" #È x

œ lim

hÄ!

Èx  h
Èx h

" 2È x

. The graphs reveal that y œ

is the derivative of the function

Èx  h
Èx h

gets closer to y œ

" #È x

as h gets smaller and smaller.

60. The graphs are shown below for h œ 2, 1, 0.5. The function y œ 3x# is the derivative of the function y œ x$ so that 3x# œ lim

hÄ!

(xh)$
x$ h

. The graphs reveal that y œ

(xh)$
x$ h

gets closer to y œ 3x# as h

gets smaller and smaller.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

107

108

Chapter 3 Differentiation

61. The graphs are the same. So we know that for f(x) œ kxk , we have f w (x) œ

kx k x

.

62. Weierstrass's nowhere differentiable continuous function.

63-68. Example CAS commands: Maple: f := x -> x^3 + x^2 - x; x0 := 1; plot( f(x), x=x0-5..x0+2, color=black, title="Section 3.2, #63(a)" ); q := unapply( (f(x+h)-f(x))/h, (x,h) ); # (b) L := limit( q(x,h), h=0 ); # (c) m := eval( L, x=x0 ); tan_line := f(x0) + m*(x-x0); plot( [f(x),tan_line], x=x0-2..x0+3, color=black, linestyle=[1,7], title="Section 3.2 #63(d)", legend=["y=f(x)","Tangent line at x=1"] ); Xvals := sort( [ x0+2^(-k) $ k=0..5, x0-2^(-k) $ k=0..5 ] ): # (e) Yvals := map( f, Xvals ): evalf[4](< convert(Xvals,Matrix) , convert(Yvals,Matrix) >); plot( L, x=x0-5..x0+3, color=black, title="Section 3.2 #63(f)" ); Mathematica: (functions and x0 may vary) (see section 2.5 re. RealOnly ): 1-x; a := 0; b := 1; N :=[ 4, 10, 20, 50 ]; P := [seq( RiemannSum( f(x), x=a..b, partition=n, method=random, output=plot ), n=N )]: display( P, insequence=true ); 95-98. Example CAS commands: Maple: with( Student[Calculus1] ); f := x -> sin(x); a := 0; b := Pi; plot( f(x), x=a..b, title="#95(a) (Section 5.3)" ); N := [ 100, 200, 1000 ]; # (b) for n in N do Xlist := [ a+1.*(b-a)/n*i $ i=0..n ]; Ylist := map( f, Xlist ); end do: for n in N do # (c) Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist)); end do; avg := FunctionAverage( f(x), x=a..b, output=value ); evalf( avg ); FunctionAverage(f(x),x=a..b,output=plot); # (d) fsolve( f(x)=avg, x=0.5 ); fsolve( f(x)=avg, x=2.5 ); fsolve( f(x)=Avg[1000], x=0.5 ); fsolve( f(x)=Avg[1000], x=2.5 ); 89-98. Example CAS commands: Mathematica: (assigned function and values for a, b, and n may vary) Sums of rectangles evaluated at left-hand endpoints can be represented and evaluated by this set of commands Clear[x, f, a, b, n] {a, b}={0, 1}; n =10; dx = (b
a)/n; f = Sin[x]2 ; xvals =Table[N[x], {x, a, b
dx, dx}]; yvals = f /.x Ä xvals; boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals, xvals  dx, yvals}]; Plot[f, {x, a, b}, Epilog Ä boxes]; Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N Sums of rectangles evaluated at right-hand endpoints can be represented and evaluated by this set of commands. Clear[x, f, a, b, n] {a, b}={0, 1}; n =10; dx = (b
a)/n; f = Sin[x]2 ;

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

281

282

Chapter 5 Integration

xvals =Table[N[x], {x, a  dx, b, dx}]; yvals = f /.x Ä xvals; boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals
dx,xvals, yvals}]; Plot[f, {x, a, b}, Epilog Ä boxes]; Sum[yvals[[i]] dx, {i, 1,Length[yvals]}]//N Sums of rectangles evaluated at midpoints can be represented and evaluated by this set of commands. Clear[x, f, a, b, n] {a, b}={0, 1}; n =10; dx = (b
a)/n; f = Sin[x]2 ; xvals =Table[N[x], {x, a  dx/2, b
dx/2, dx}]; yvals = f /.x Ä xvals; boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals
dx/2, xvals  dx/2, yvals}]; Plot[f, {x, a, b},Epilog Ä boxes]; Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N 5.4 THE FUNDAMENTAL THEOREM OF CALCULUS 1.

'c (2x  5) dx œ cx#  5xd
#! œ a0#  5(0)b
a(
2)#  5(
2)b œ 6

2.

'c ˆ5
x# ‰ dx œ ’5x
x4 “ %

0

2

4

3.

#


$

3

'

2

0

xax
3b dx œ

'

2

0

5.

3

'c ax2
2x  3b dx œ ’ x3

3

1

'

4

0

Š3x


x$ 4‹

#

dx œ ’ 3x#


'c ax$
2x  3b dx œ ’ x4 2

6.

%

2

7.

'

8.

' '

1Î3

9.

'

1

10.

'

31Î4

11.

'

1Î3

12.

1

% x% 16 “ !

1

3

œ Š a23b


œ

3 a2 b 2 2 ‹

133 4

3


Š a03b


3


1 #

#
#

3 a0 b 2 2 ‹

œ
10 3

œ Š a13b
a1b#  3a1b‹
Š (
31)
(
1)#  3(
1)‹ œ

œ Š 3(4) #



x#  3x“

2 3x2 2 “!

(
3)# 4 ‹

4% 16 ‹

3

#


Š 3(0) #


(0)% 16 ‹

œ8 %

œ Š 24
2#  3(2)‹
Š (
42)
(
2)#  3(
2)‹ œ 12 %

"

$

ˆx#  Èx‰ dx œ ’ x3  32 x$Î# “ œ ˆ "3  32 ‰
0 œ 1

32

1

$#

x
'Î& dx œ 
5x
"Î& ‘ " œ ˆ
#5 ‰
(
5) œ

0

1Î$

2 sec# x dx œ [2 tan x]!

5 #

œ ˆ2 tan ˆ 13 ‰‰
(2 tan 0) œ 2È3
0 œ 2È3

a1  cos xb dx œ [x  sin x]1! œ a1  sin 1b
a0  sin 0b œ 1

1Î4

0


x#  3x“


Š5(
3)


!

0

0

4# 4‹

ax2
3xb dx œ ’ x3


1

4.

œ Š5(4)


$1Î%

csc ) cot ) d) œ [
csc )]1Î% œ ˆ
csc ˆ 341 ‰‰
ˆ
csc ˆ 14 ‰‰ œ
È2
Š
È2‹ œ 0 1Î$

4 sec u tan u du œ [4 sec u]!

œ 4 sec ˆ 13 ‰
4 sec 0 œ 4(2)
4(1) œ 4

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

20 3

Section 5.4 The Fundamental Theorem of Calculus 13.

14.

'

0

"  cos 2t # 1Î2

'

dt œ

0

ˆ" 

1Î2 #

2t 'c ÎÎ "
cos dt œ ' # 1 3

1Î3

1 3

1Î3 #

œ ˆ "# ˆ 13 ‰


'

1Î4

15.

'

1Î6

16.

0

0

" 4

" #

cos 2t‰ dt œ  "# t 

ˆ"


" #

'

1Î4

0

sin 2t‘ 1Î# œ ˆ "# (0) 

cos 2t‰ dt œ  "# t


sin 2 ˆ 13 ‰‰
ˆ #" ˆ
13 ‰


tan# x dx œ

!

" 4

" 4

1Î6

0

sin 2 ˆ
13 ‰‰ œ

1 6




" 4

sin 231 

1 6

" 4



sin ˆ
321 ‰ œ

1 3

œ ˆtan ˆ 14 ‰
14 ‰
atan a0b
0b œ 1


asec2 x  2sec x tan x  tan2 xbdx œ

'

1Î6

0




'

1Î8

1 Î8

sin 2x dx œ ’
"# cos 2x“

!

0

œ ˆ
"# cos 2ˆ 18 ‰‰
ˆ
"# cos 2a0b‰ œ

a2sec2 x  2sec x tan x
1bdx 1 6


2

2
È 2 4

1

#

1 3

1 3

œ Š4 tan ˆ
14 ‰


1 ˆ
14 ‰ ‹


Š4 tan ˆ 13 ‰


1 ˆ
13 ‰ ‹

œ (4(
1)  4)
Š4 Š
È3‹  3‹ œ 4È3
3

19.

'"
" (r  1)# dr œ '"
" ar#  2r  1b dr œ ’ r3  r#  r“
" œ Š (
31)

20.

'
È (t  1) at#  4b dt œ '
È at$  t#  4t  4b dt œ ’ t4  t3  2t#  4t“ÈÈ$

$

"

È3

È3

3

œ

%

ŠÈ3‹ 4

$



ŠÈ3‹

 2 ŠÈ3‹  4È3


3

'È" Š u#

22.

'cc y y
2y dy œ 'cc ay2
2y
2 b dy œ ’ y3

(




2

3

'

È2

1

$

 (
1)#  (
1)‹
Š 13  1#  1‹ œ
38

$




%

#

21.

1

$

3

%

23.

È3 4

'c
Î Î% ˆ4 sec# t  t1 ‰ dt œ '
Î Î% a4 sec# t  1t
# b dt œ 4 tan t
1t ‘

11Î%Î$ 1

18.

sin 2 ˆ 1# ‰‰ œ
14

1 4

œ [2 tan x  2sec x
x]1!Î6 œ ˆ2 tanˆ 16 ‰  2secˆ 16 ‰
ˆ 16 ‰‰
a2 tan 0  2sec 0
0b œ 2È3
17.

" 4

sin 2t‘
1Î$

1 Î4

'

sin 2(0)‰
ˆ "# ˆ 1# ‰ 

1Î$

" 4

asec# x
1bdx œ [tan x
x]!

asec x  tan xb2 dx œ

" 4

" u& ‹

du œ

'È" Š u#

(

2

)

u
u
& ‹ du œ ’ 16 

1

5

3

3

3

s#  È s s#

ds œ

'

1

È2

Š
È3‹ 4

$



Š
È3‹ 3

$ #

 2 Š
È3‹  4 Š
È3‹ œ 10È3 )

" " 4u% “È#

)

1 œ Š 16 

c1

" 4(1)% ‹




ŠÈ2‹ 16



"

œ Š ac31b  ac21b ‹
Š ac33b  ac23b ‹ œ c3 3

 2y
1 “

ˆ1  s
$Î# ‰ ds œ ’s


È# 2 “ Ès "

œ È 2


3

2 É È2 


Š1


2 È1 ‹

%

4 ŠÈ2‹

œ
4 3

22 3

œ È2
2$Î%  1

4 œ È2
È 81

24.

'

8

ˆx1Î3  1‰ˆ2
x2Î3 ‰ x1Î3 1

dx œ

'

8

2x1Î3
x  2
x2Î3 x1Î3 1

dx œ

'

1

8

ˆ2
x2Î3  2x
1Î3
x1Î3 ‰ dx œ

2x
35 x5Î3  3x2Î3
34 x4Î3 ‘3 œ Š2a8b
35 a8b5Î3  3a8b2Î3
34 a8b4Î3 ‹
Š2a1b
35 a1b5Î3  3a1b2Î3
34 a1b4Î3 ‹ 1 œ
137 20 25.

'

1

sin 2x 1Î2 2 sin x

dx œ

'

1

2 sin x cos x 1Î2 2 sin x

dx œ

'

1

1Î2

1

cos x dx œ ’sin x“

1Î2

œ asin a1bb
ˆsin ˆ 12 ‰‰ œ
1

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

283

284 26.

Chapter 5 Integration

'

1Î3

0

'

œ

'

acos x  sec xb2 dx œ 1Î3

0

ˆ 12 cos 2x 

5 2

1Î3

0

'

acos2 x  2  sec2 xbdx œ

1Î3

0

ˆ cos 2x2  1  2  sec2 x‰dx

1Î3

 sec2 x‰dx œ ’ 14 sin 2x  52 x  tan x“

!

œ ˆ 14 sin 2ˆ 13 ‰  52 ˆ 13 ‰  tanˆ 13 ‰‰
ˆ 14 sin 2a0b  52 a0b  tana0b‰ œ

'c% kxk dx œ ' !% kxk dx  '! 4

27.

28.

'

1

" ! #

acos x  kcos xk b dx œ 1 #

œ sin

29. (a)

!

d dx

30. (a)

'

(b)

d dx

31. (a)

'

(b)

d dt

32. (a)

'

'

1Î#

!

" # (cos

' !% x dx  '!

x  cos x) dx 

4

#

x dx œ ’
x# “

'

1

" 1Î# #

!

#


%

%

#

 ’ x# “ œ Š
0#  !

(cos x
cos x) dx œ

'

1Î#

!

(
4)# # ‹

Èx

cos t dt œ [sin t]! œ sin Èx
sin 0 œ sin Èx Ê

Èx

'

Œ

!

sin x

d dx

Œ

'

Èx

!

cos t dt
œ

d ˆÈ ‰‰ cos t dt
œ ˆcos Èx‰ ˆ dx x œ ˆcos Èx ‰ ˆ "# x
"Î# ‰ œ

3t# dt œ ct$ d "

sin x

1

'

Πt%

sin x

1

Œ'

t%

!

tan )

!

d d)

œ sin$ x
1 Ê

d dx

Œ

'

sin x

1

3t# dt
œ

d dx

d dx

ˆsin Èx‰ œ cos Èx ˆ "# x
"Î# ‰

cos Èx 2È x

asin$ x
1b œ 3 sin# x cos x

'

t%

!

t%

u"Î# du œ  23 u$Î# ‘ ! œ

2 3

at% b

$Î#


0œ

2 ' 3 t

Ê

d dt

Œ'

Œ

'

t%

!

Èu du
œ

ˆ 23 t' ‰ œ 4t&

d dt

Èu du
œ Èt% ˆ dtd at% b‰ œ t# a4t$ b œ 4t&

) sec# y dy œ [tan y]tan œ tan (tan ))
0 œ tan (tan )) Ê !

'

Œ

!

tan )

d d)

tan )

!

sec# y dy
œ

d d)

(tan (tan )))

sec# y dy
œ asec# (tan ))b ˆ dd) (tan ))‰ œ asec# (tan ))b sec# )

33. y œ

'

35. y œ

'È sin t# dt œ
'

x

!

È1  t# dt Ê

'

x2

2

sin t3 dt Ê

'

2

x

34. y œ

sin t# dt Ê

œx†

dy dx

'

x2

dy dx

œ

d dx Œ

2

'

1

x

" t

dt Ê

dy dx

œ

" x

,x0

# d ˆÈ ‰‰ œ
Šsin ˆÈx‰ ‹ ˆ dx x œ
(sin x) ˆ "# x
"Î# ‰ œ
2sinÈxx

sin t3 dt
 1 †

'

x2

2

sin t3 dt œ x † sin ax# b

3 d # dx ax b



'

2

x2

sin t3 dt

sin t3 dt x

#

#

dy dx

œ È1  x#

x2

'c t t 4 dt
' 1

Èx

!

x

œ 2x# sin x6  37. y œ

dy dx

!

36. y œ x

œ 16

1Î#

œ asec# (tan ))b sec# ) (b)

0# #‹

cos x dx œ [sin x]!

d 3t# dt
œ a3 sin# xb ˆ dx (sin x)‰ œ 3 sin# x cos x

Èu du œ

!

#

 Š 4#


cos Èx 2È x

œ (b)

kxk dx œ


9È 3 8




sin 0 œ 1

Èx

'

4

51 6

3

t# t#  4

dt Ê

x# x#  4




x# x#  4

œ0

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 5.4 The Fundamental Theorem of Calculus

'

38. y œ Œ

x

0

39. y œ

'

40. y œ

'

sin x

dt È1
t#

!

tan x

dt 1  t#

!

, kxk  Ê

'

3

10

at3  "b dt
Ê

dy dx

1 #

dy dx

Ê

œ 3Œ

dy dx

0

x

" È1
sin# x

œ

'

10

d at3  "b dt
dx Œ

d ˆ dx (sin x)‰ œ

0

x

10

10

'

at3  "b dt
œ 3ax3  "b Œ

" Ècos# x

(cos x) œ

cos x kcos xk

œ

cos x cos x

" d ‰ ˆ dx œ ˆ 1  tan (tan x)‰ œ ˆ sec"# x ‰ asec# xb œ 1 #x

'
$
# a
x#
2xbdx  '
#! a
x#
2xbdx
'!# a
x#
2xbdx $

œ
’
x3
x# “


#
$

$

 ’
x3
x# “

!

$


#

$


’
x3
x# “

#

!

$

œ
ŠŠ
(
32)
(
2)# ‹
Š
(
33)
(
3)# ‹‹ $

 ŠŠ
03
0# ‹
Š
(
32)
(
2)# ‹‹ $

$

$


ŠŠ
23
2# ‹
Š
03
0# ‹‹ œ

28 3

42. 3x#
3 œ 0 Ê x# œ 1 Ê x œ „ 1; because of symmetry about the y-axis, Area œ 2 Œ


'!" a3x#
3bdx  '"# a3x#
3bdx


"

#

2 Š
cx$
3xd !  cx$
3xd " ‹ œ 2 c
aa1$
3(1)b
a0$
3(0)bb  aa2$
3(2)b
a1$
3(1)bd œ 2(6) œ 12

43. x$
3x#  2x œ 0 Ê x ax#
3x  2b œ 0 Ê x(x
2)(x
1) œ 0 Ê x œ 0, 1, or 2; Area œ

'!" ax$
3x#  2xbdx
'"# ax$
3x#  2xbdx "

%

%

œ ’ x4
x$  x# “
’ x4
x$  x# “ !

œ

% Š 14

$

#


1 1 ‹
%

% Š 04

$

# "

#


0 0 ‹ %


’Š 24
2$  2# ‹
Š 14
1$  1# ‹“ œ

10

at3  "b dt


œ 1 since kxk 

41.
x#
2x œ 0 Ê
x(x  2) œ 0 Ê x œ 0 or x œ
2; Area œ


0

x

" #

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

1 #

285

286

Chapter 5 Integration

44. x"Î$
x œ 0 Ê x"Î$ ˆ1
x#Î$ ‰ œ 0 Ê x"Î$ œ 0 or 1
x#Î$ œ 0 Ê x œ 0 or 1 œ x#Î$ Ê x œ 0 or 1 œ x# Ê x œ 0 or „ 1; Area œ
œ

'c! ˆx"Î$
x‰dx  '!" ˆx"Î$
x‰dx
'") ˆx"Î$
x‰dx "

%Î$


’ 34

x




! x# # “
"

œ
’Š 34 (0)%Î$


 ’ 34 x%Î$


0# #‹

" x# # “!


’ 43 x%Î$
(
1)# # ‹“


Š 34 (
1)%Î$


 ’Š 34 (1)%Î$


1# #‹


Š 34 (0)%Î$


0# # ‹“


’Š 34 (8)%Î$


8# #‹


Š 34 (1)%Î$


1# # ‹“

œ

" 4

" 4




ˆ
2!


$ 4

 #" ‰ œ

) x# # “"

83 4

45. The area of the rectangle bounded by the lines y œ 2, y œ 0, x œ 1, and x œ 0 is 21. The area under the curve y œ 1  cos x on [0ß 1] is

'!

1

(1  cos x) dx œ [x  sin x]!1 œ (1  sin 1)
(0  sin 0) œ 1. Therefore the area of

the shaded region is 21
1 œ 1. 46. The area of the rectangle bounded by the lines x œ 16 , x œ " #

51 6 ,

y œ sin

ˆ 561
16 ‰ œ 13 . The area under the curve y œ sin x on  16 ß 561 ‘ is

œ ˆ
cos

51 ‰ 6

È3 # ‹


ˆ
cos 16 ‰ œ
Š


È3 #



'

1 6

œ

51Î6

1Î6

" #

œ sin

51 6

, and y œ 0 is &1Î'

sin x dx œ [
cos x]1Î'

œ È3. Therefore the area of the shaded region is È3
13 .

47. On 
14 ß 0‘ : The area of the rectangle bounded by the lines y œ È2, y œ 0, ) œ 0, and ) œ
14 is È2 ˆ 14 ‰ œ

1È2 4

. The area between the curve y œ sec ) tan ) and y œ 0 is


'c

!

1Î4

sec ) tan ) d) œ [
sec )]!
1Î%

œ (
sec 0)
ˆ
sec ˆ
14 ‰‰ œ È2
1. Therefore the area of the shaded region on 
14 ß !‘ is

1È2 4

On 0ß 14 ‘ : The area of the rectangle bounded by ) œ 14 , ) œ 0, y œ È2, and y œ 0 is È2 ˆ 14 ‰ œ under the curve y œ sec ) tan ) is of the shaded region on !ß 14 ‘ is È

'

1Î4

!

1È2 4

1Î%

sec ) tan ) d) œ [sec )]!

œ sec

1 4

 Š È 2
1‹ .

1È2 4

. The area


sec 0 œ È2
1. Therefore the area


ŠÈ2
1‹ . Thus, the area of the total shaded region is

È

1È2 #

Š 1 4 2  È2
1‹  Š 1 4 2
È2  1‹ œ

.

48. The area of the rectangle bounded by the lines y œ 2, y œ 0, t œ
14 , and t œ 1 is 2 ˆ1
ˆ
14 ‰‰ œ 2  area under the curve y œ sec# t on 
14 ß !‘ is under the curve y œ 1
t# on [!ß "] is

'c

!

1Î4

"

$

œ

dt
3 œ 0
3 œ
3 Ê (d) is a solution to this problem.

dy dx

œ

50. y œ

'c sec t dt  4

Ê

dy dx

œ sec x and y(
1) œ

51. y œ

'

sec t dt  4 Ê

dy dx

œ sec x and y(0) œ

x

1

!

x

and y(1) œ

'

" t

dt
3 Ê

" x

1

1

Thus, the total

. Therefore the area of the shaded region is ˆ2  1# ‰


'

" 1 t

$

5 3

49. y œ

x

$

!

2 3

'

!

. The

! ˆ 1‰ sec# t dt œ [tan t]
1Î% œ tan 0
tan
4 œ 1. The area

'! a1
t# b dt œ ’t
t3 “ " œ Š1
13 ‹
Š0
03 ‹ œ 32 .

area under the curves on 
14 ß "‘ is 1 

1 #

'cc sec t dt  4 œ 0  4 œ 4 1

1

!

5 3

œ

" 3



1 #

.

Ê (c) is a solution to this problem.

sec t dt  4 œ 0  4 œ 4 Ê (b) is a solution to this problem.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 5.4 The Fundamental Theorem of Calculus 52. y œ

'

53. y œ

'

x

" " t x

#

55. Area œ

dt
3 Ê

dy dx

œ

" x

and y(1) œ

'

"

" t

"

dt
3 œ 0
3 œ
3 Ê (a) is a solution to this problem.

sec t dt  3

'c ÎÎ

b 2 b 2

54. y œ

$

4h ˆ b# ‰ 3b#

œ ˆ bh #



ˆ
bh # 

bh ‰ 6



Œh ˆ
# ‰
b

bh ‰ 6

œ bh


x

"

È1  t# dt
2

bÎ2

4h ˆ
#b ‰ 3b#

œ

bh 3

'

4hx$ 3b# “ bÎ2

ˆh
ˆ 4h ‰ # ‰ dx œ ’hx
b# x

œ Œhˆ #b ‰


$

2 3


bh

56. k  0 Ê one arch of y œ sin kx will occur over the interval 0ß 1k ‘ Ê the area œ œ
"k cos ˆk ˆ 1k ‰‰
ˆ
k" cos (0)‰ œ 57.

dc dx

œ

58. r œ

" #È x

'

$

!

œ

Š2


" #

x
"Î# Ê c œ

'

dx œ 2

'

2 (x  1)# ‹

x

!

$

!

'

1Îk

!

sin kx dx œ 


" k

œ t"Î# ‘ 0 œ Èx; c(100)
c(1) œ È100
È1 œ $9.00 x

" (x  1)# ‹

$

dx œ 2 x
ˆ x
11 ‰‘ ! œ 2 ’Š3 

" (3  1) ‹


Š0 

" (0  1) ‹“

œ 2 3 "4
1‘ œ 2 ˆ2 4" ‰ œ 4.5 or $4500 59. (a) t œ 0 Ê T œ 85
3È25
0 œ 70‰ F; t œ 16 Ê T œ 85
3È25
16 œ 76‰ F; t œ 25 Ê T œ 85
3È25
25 œ 85‰ F (b) average temperatuve œ œ

1 25 Š85a25b

1 25
0

 2a25
25b

' 25 Š85
3È25
t‹ dt œ 251 ’85t  2a25
tb3Î2 “ 25 !

0

3 Î2

‹


1 25 Š85a0b

 2a25
0b

3Î2

‰

‹ œ 75 F

3 60. (a) t œ 0 Ê H œ È0  1  5a0b1Î3 œ 1 ft; t œ 4 Ê H œ È4  1  5a4b1Î3 œ È5  5È 4 ¸ 10.17 ft; 1 Î 3 t œ 8 Ê H œ È8  1  5a8b œ 13 ft

(b) average height œ œ

61.

'

62.

'

x

1

!

x

1 2 8 Š 3 a8

 1b

3 Î2

1 8
0



' 8 ŠÈt  1  5 t1Î3 ‹ dt œ 18 ’ 23 at  1b3Î2  154 t4Î3 “ 8

15 4

!

0

a8b

4Î3

‹


f(t) dt œ x#
2x  1 Ê f(x) œ f(t) dt œ x cos 1x Ê f(x) œ

63. f(x) œ 2


'#

x

"

9 1t

d dx

1 2 8 Š 3 a0

'

d dx

'

!

1

x

x

 1b

3Î2

f(t) dt œ



d dx

4Î3 15 ‹ 4 a0b

œ

29 3

¸ 9.67 ft

ax#
2x  1b œ 2x
2

f(t) dt œ cos 1x
1x sin 1x Ê f(4) œ cos 1(4)
1(4) sin 1(4) œ 1

dt Ê f w (x) œ
1  (x9  1) œ


9 x 2

L(x) œ
3(x
1)  f(1) œ
3(x
1)  2 œ
3x  5

Ê f w (1) œ
3; f(1) œ 2


'#" " 1 9 t dt œ 2
0 œ 2;

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

1 Îk

cos kx‘ !

2 k

"
"Î# dt # t

Š1


287

288

Chapter 5 Integration

64. g(x) œ 3 

'

1

x#

sec (t
1) dt Ê gw (x) œ asec ax#
1bb (2x) œ 2x sec ax#
1b Ê gw (
1) œ 2(
1) sec a(
1)#
1b #

a
"b " œ
2; g(
1) œ 3  ' sec (t
1) dt œ 3  ' sec (t
1) dt œ 3  0 œ 3; L(x) œ
2(x
(
1))  g(
1) 1

1

œ
2(x  1)  3 œ
2x  1 65. (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 gw (1) œ f(1) œ 0. False, since gww (1) œ f w (1)  0. True, since gw (1) œ 0 and gww (1) œ f w (1)  0. False: gww (x) œ f w (x)  0, so gww never changes sign. True, since gw (1) œ f(1) œ 0 and gw (x) œ f(x) is an increasing function of x (because f w (x)  0).

66. Let a œ x0  x1  x2 â  xn œ b be any partition of Òa, bÓ and let F be any antiderivative of f. n

(a) !Faxi b
Faxi
1 b‘ iœ1

œ Fax1 b
Fax0 b‘  Fax2 b
Fax1 b‘  Fax3 b
Fax2 b‘  â  Faxn
1 b
Faxn
2 b‘  Faxn b
Faxn
1 b‘ œ
Fax0 b  Fax1 b
Fax1 b  Fax2 b
Fax2 b  â  Faxn
1 b
Faxn
1 b  Faxn b œ Faxn b
Fax0 b œ Fabb
Faab (b) Since F is any antiderivative of f on Òa, bÓ Ê F is differentiable on Òa, bÓ Ê F is continuous on Òa, bÓ. Consider any subinterval Òxi
1 , xi Ó in Òa, bÓ, then by the Mean Value Theorem there is at least one number ci in Ðxi
1 , xi Ñ such that n

Faxi b
Faxi
1 b‘ œ Fw aci baxi
xi
1 b œ faci baxi
xi
1 b œ faci b?xi . Thus Fabb
Faab œ !Faxi b
Faxi
1 b‘ iœ1

n

œ !faci b?xi . iœ1

n

(c) Taking the limit of Fabb
Faab œ !faci b?xi we obtain Ê Fabb
Faab œ 'a faxb dx

iœ1

lim aFabb
Faabb œ

mPmÄ0

n

lim Œ!faci b?xi


mPmÄ0

iœ1

b

67-70. Example CAS commands: Maple: with( plots ); f := x -> x^3-4*x^2+3*x; a := 0; b := 4; F := unapply( int(f(t),t=a..x), x ); # (a) p1 := plot( [f(x),F(x)], x=a..b, legend=["y = f(x)","y = F(x)"], title="#67(a) (Section 5.4)" ): p1; dF := D(F); # (b) q1 := solve( dF(x)=0, x ); pts1 := [ seq( [x,f(x)], x=remove(has,evalf([q1]),I) ) ]; p2 := plot( pts1, style=point, color=blue, symbolsize=18, symbol=diamond, legend="(x,f(x)) where F '(x)=0" ): display( [p1,p2], title="81(b) (Section 5.4)" ); incr := solve( dF(x)>0, x ); # (c) decr := solve( dF(x) x^2; f := x -> sqrt(1-x^2); F := unapply( int( f(t), t=a..u(x) ), x ); dF := D(F); # (b) cp := solve( dF(x)=0, x ); solve( dF(x)>0, x ); solve( dF(x)29. :

V œ 'a 1R# (x) dx œ 'c1 1 a3x% b dx œ 1 'c1 9x) dx b

1

1

#

"

œ 1 cx* d " œ 21

(b) =2/66 7/>29. : shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x a3x% b dx œ 21 † 3'0 x& dx œ 21 † 3 ’ x6 “ œ 1 b

1

1

'

!

Note: The lower limit of integration is 0 rather than
1. (c) =2/66 7/>29. : shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ 21'c1 (1
x) a3x% b dx œ 21 ’ 3x5
b

"

1

&

(d) A+=2/< 7/>29. :

" x' 2 “ "

œ 21 ˆ 35
"# ‰
ˆ
35
"# ‰‘ œ

121 5

R(x) œ 3, r(x) œ 3
3x% œ 3 a1
x% b Ê V œ 'a 1 cR# (x)
r# (x)d dx œ 'c1 1 ’9
9 a1
x% b “ dx b

1

œ 91 'c1 c1
a1
2x%  x) bd dx œ 91 'c1 a2x%
x) b dx œ 91 ’ 2x5
1

1

8. (a) A+=2/< 7/>29. : R(x) œ

, r(x) œ

4 x$

" #

&

" x* 9 “ "

b

21†13 5

œ

261 5

2

(b) =2/66 7/>29. :

V œ 21'1 x ˆ x4$
"# ‰ dx œ 21 ’
4x"
2

(c) =2/66 7/>29. :

" #

# x# 4 “"



16 5

 4" ‰ œ

1 20

(
2
10  64  5) œ

b

2


x

(d) A+=2/< 7/>29. :

# x# 4 “"

571 #0

œ 21 ˆ
4#
1‰
ˆ
4
4" ‰‘ œ 21 ˆ 54 ‰ œ

shell ‰ shell V œ 21'a ˆ radius Š height ‹ dx œ 21'1 (2
x) ˆ x4$
"# ‰ dx œ 21'1 ˆ x8$
4 x

œ 181  25
"9 ‘ œ

# # # & Ê V œ 'a 1cR# (x)
r# (x)d dx œ '1 1 ’ˆ x4$ ‰
ˆ "# ‰ “ dx œ 1 
16
x4 ‘ " 5 x

"‰ " ˆ 16 " ‰‘ œ 1 ˆ
10 œ 1 ˆ 5†16
32
#

5
4

œ 21 ’
x4# 

#

2

4 x#

51 #


1  x# ‰ dx

œ 21 (
1  2
2  1)
ˆ
4  4
1  4" ‰‘ œ

31 #

V œ 'a 1cR# (x)
r# (x)d dx b

# œ 1 '1 ’ˆ 7# ‰
ˆ4
2

dx

œ

491 4


161'1 a1
2x$  x' b dx

œ

491 4


161 ’x  x#


œ

491 4 491 4 491 4


161 ˆ2  4"
5†"3# ‰
ˆ1  1
5" ‰‘ "
161 ˆ 4"
160  5" ‰

œ œ 9.

4 ‰# x$ “

2




161 160

# x & 5 “"

(40
1  32) œ

491 4




711 10

œ

1031 20

(a) .3=5 7/>29. :

# V œ 1 '1 ŠÈx
1‹ dx œ 1'1 (x
1) dx œ 1 ’ x#
x“

#

5

5

‰ ˆ" ‰‘ œ 1 ˆ 24 ‰ œ 1 ˆ 25 #
5
#
1 #
4 œ 81

& "

(b) A+=2/< 7/>29. :

R(y) œ 5, r(y) œ y#  1 Ê V œ 'c 1 cR# (y)
r# (y)d dy œ 1 'c2 ’25
ay#  1b “ dy d

2

œ 1'c2 a25
y%
2y#
1b dy œ 1 'c2 a24
y%
2y# b dy œ 1 ’24y
2

2

#

y& 5


32 y$ “

# #

œ 21 ˆ24 † 2


Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

32 5




2 3

† 8‰

378

Chapter 6 Applications of Definite Integrals œ 321 ˆ3


2 5


"3 ‰ œ

321 15

(45
6
5) œ

10881 15

(c) .3=5 7/>29. : R(y) œ 5
ay#  1b œ 4
y#

Ê V œ 'c 1R# (y) dy œ 'c2 1 a4
y# b dy d

2

#

œ 1 'c2 a16
8y#  y% b dy 2

œ 1 ’16y


8y$ 3

œ 641 ˆ1


2 3



# y& 5 “ #

 "5 ‰ œ

œ 21 ˆ32


641 15

64 3



(15
10  3) œ

32 ‰ 5 5121 15

10. (a) =2/66 7/>29. :

shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y Šy
d

4

œ 21'0 Šy#
4

œ

21 1#

† 64 œ

y$ 4‹

$

dy œ 21 ’ y3


321 3

%

y% 16 “ !

y# 4‹

dy

œ 21 ˆ 64 3


64 ‰ 4

(b) =2/66 7/>29. : shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x ˆ2Èx
x‰ dx œ 21'0 ˆ2x$Î#
x# ‰ dx œ 21 ’ 45 x&Î#
b

4

œ 21 ˆ 45 † 32


64 ‰ 3

œ

4

1281 15

% x$ 3 “!

(c) =2/66 7/>29. :

shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21(4
x) ˆ2Èx
x‰ dx œ 21'0 ˆ8x"Î#
4x
2x$Î#  x# ‰ dx b

4

$Î# œ 21 ’ 16
2x#
54 x&Î#  3 x

œ 641 ˆ1
45 ‰ œ

641 5

4

% x$ 3 “!

œ 21 ˆ 16 3 † 8
32


(d) =2/66 7/>29. :

shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(4
y) Šy
d

4

œ 21'0 Š4y
2y#  4

y$ 4‹

dy œ 21 ’2y#
32 y$ 

y# 4‹

% y% 16 “ !

4 5

† 32 

64 ‰ 3

œ 641 ˆ 34
1


dy œ 21'0 Š4y
y#
y#  4

œ 21 ˆ32


2 3

4 5

y$ 4‹

 32 ‰

dy

† 64  16‰ œ 321 ˆ2


8 3

 1‰ œ

321 3

11. .3=5 7/>29. : R(x) œ tan x, a œ 0, b œ

1 3

Ê V œ 1 '0 tan# x dx œ 1'0 asec# x
1b dx œ 1[tan x
x]! 1Î3

12. .3=5 7/>29. :

1Î3

1Î$

V œ 1'0 (2
sin x)# dx œ 1 '0 a4
4 sin x  sin# xb dx œ 1'0 ˆ4
4 sin x  1

1

œ 1 4x  4 cos x 

x #




sin 2x ‘ 1 4 !

1

œ 1 ˆ41
4 

1 #


0‰
(0  4  0
0)‘ œ

œ

1cos 2x ‰ dx # 9 1 1 ˆ #
8‰ œ 1#

1 Š3È31‹ 3

(91
16)

13. (a) .3=5 7/>29. :

V œ 1'0 ax#
2xb dx œ 1'0 ax%
4x$  4x# b dx œ 1 ’ x5
x%  34 x$ “ œ 1 ˆ 32 5
16  2

œ

161 15

2

#

(6
15  10) œ

#

&

!

161 15

32 ‰ 3

(b) A+=2/< 7/>29. :

V œ '0 1’1#
ax#
2x  "b “ dx œ '0 1 dx
'0 1 ax
"b% dx œ #1
’1 2

2

#

2

(c) =2/66 7/>29. :

# ax"b& & “!

œ #1
1 †

shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ 21'0 (2
x) c
ax#
2xbd dx œ 21'0 (2
x) a2x
x# b dx b

2

2

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

# &

œ

)1 &

Chapter 6 Practice Exercises œ 21'0 a4x
2x#
2x#  x$ b dx œ 21'0 ax$
4x#  4xb dx œ 21 ’ x4
34 x$  2x# “ œ 21 ˆ4
2

œ

21 3

2

#

%

81 3

(36
32) œ

32 3

!

 8‰

(d) A+=2/< 7/>29. :

V œ 1 '0 c2
ax#
2xbd dx
1'0 2# dx œ 1'0 ’4
4 ax#
2xb  ax#
2xb “ dx
81 2

2

#

2

#

œ 1'0 a4
4x#  8x  x%
4x$  4x# b dx
81 œ 1'0 ax%
4x$  8x  4b dx
81 2

2

#

&

‰ œ 1 ’ x5
x%  4x#  4x“
81 œ 1 ˆ 32 5
16  16  8
81 œ !

1 5

(32  40)
81 œ

721 5




401 5

œ

321 5

14. .3=5 7/>29. :

V œ 21'0 4 tan# x dx œ 81'0 asec# x
1b dx œ 81[tan x
x]! 1Î4

1Î4

1Î%

œ 21(4
1)

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#  ŠÈ$‹ œ ## , i.e. h œ ". Thus #

Vcyl œ a#hb1ŠÈ$‹ œ '1 ft$ . To get the volume of a cap, use the disk method and x#  y# œ ## : Vcap œ '" 1x# dy 2

œ '" 1a%
y# bdy œ 1’%y
2

œ 1ˆ8
83 ‰
ˆ%
3" ‰‘ œ

# y3 3 “"

&1 3

Vremoved œ Vcyl  #Vcap œ '1 

ft$ . Therefore, "!1 3

œ

#)1 3

ft$ .

16. We rotate the region enclosed by the curve y œ É12 ˆ1


4x# ‰ 121

and the x-axis around the x-axis. To find the

11Î2

volume we use the .3=5 method: V œ 'a 1R# (x) dx œ 'c11Î2 1 ŠÉ12 ˆ1
b

œ 121'c11Î2 Š1
11Î2

4x# 121 ‹

œ 1321 ˆ1
"3 ‰ œ 17. y œ x"Î#


x$Î# 3

Ê

dx œ 121 ’x


2641 3

dy dx

œ

""Î# 4x$ 363 “ ""Î#

ˆ4 

2 3

18. x œ y#Î$ Ê

" #

#

x"Î#
"# x"Î# Ê Š dy dx ‹ œ

œ '1

8

† 8‰
ˆ2  23 ‰‘ œ

È9y#Î$
4 3y"Î$

œ

" 4

" 3

5 12

12 Š1


4x# 121 ‹

dx

#

4

ˆ2 

14 ‰ 3

œ

4y #Î$ 9

" #

ˆx"Î#  x"Î# ‰ dx œ

" #

2x"Î#  23 x$Î# ‘ % "

10 3 dx Ê L œ '1 Ê1  Š dy ‹ dy œ '1 É1  #

8

8

4 9y#Î$

dy

'18 È9y#Î$  4 ˆy"Î$ ‰ dy; u œ 9y#Î$  4 Ê du œ 6y"Î$ dy; y œ 1 Ê u œ 13,

y œ 8 Ê u œ 40d Ä L œ 19. y œ

" #

#

dy œ

11Î2

ˆ x"
2  x‰ Ê L œ ' É1  4" ˆ x"
2  x‰ dx 1 4

y"Î$ Ê Š dx dy ‹ œ

2 3

11Î2

$

4

dx dy

dx œ 1 '

œ 881 ¸ 276 in$

4

" #

#

4 ‰ 11 ˆ 4 ‰ ˆ 11 ‰ “ œ 1321 ’1
ˆ 363 œ 241 ’ 11 Š 4 ‹“ 2
363 #

# Ê L œ '1 É 4" ˆ x"  2  x‰ dx œ '1 É 4" ax"Î#  x"Î# b dx œ '1

œ

4x# ‰ 121 ‹

x'Î&
58 x%Î& Ê

dy dx

" 18

œ

'1340 u"Î# du œ 18"  32 u$Î# ‘ %! œ #"7 40$Î#
13$Î# ‘ ¸ 7.634 "$ " #

#

dy x"Î&
"# x"Î& Ê Š dx ‹ œ

" 4

ˆx#Î&
2  x#Î& ‰

# Ê L œ '1 É1  "4 ax#Î&
2  x#Î& b dx Ê L œ '1 É 4" ax#Î&  2  x#Î& b dx œ ' É 4" ax"Î&  x"Î& b dx 32

32

32

1

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

379

380

Chapter 6 Applications of Definite Integrals œ '1

32

œ

" 48

20. x œ

" #

ˆx"Î&  x"Î& ‰ dx œ

(1260  450) œ

" 1#

y$ 

" y

Ê

" % œ '1 É 16 y  2

" #

œ

1710 48

" 5 # 6 285 8

$#

x'Î&  45 x%Î& ‘ " œ

" y#

#

dx dy

œ

" 4



" y%

dy œ '1 ÊŠ 4" y# 

y#


2

8 œ ˆ 12
"# ‰
ˆ 1"#
1‰ œ

7 1#



" #

21. S œ 'a 21y Ê1  Š dy dx ‹ dx;

dy dx

œ

#

b

" 16

Ê Š dx dy ‹ œ

œ

" y# ‹

#

" #

ˆ 65 † 2' 

y%


" #



" y%

5 4

† 2% ‰
ˆ 56  54 ‰‘ œ

" #

ˆ 315 6 

" % Ê L œ '1 Ê1  Š 16 y
2

dy œ '1 Š 4" y#  2

" y# ‹

dy œ ’ 1"# y$
y" “

" #



75 ‰ 4

" y% ‹

dy

# "

13 12 #

" È2x
1

Ê Š dy dx ‹ œ

" #x
1

Ê S œ '0 21È2x  1 É1  3

" #x
1

dx


2 È ' Èx  1 dx œ 2È21  2 (x  1)$Î# ‘ $ œ 2È21 † 2 (8
1) œ œ 21'0 È2x  1 É 2x 2x
1 dx œ 2 21 0 3 3 ! 3

3

22. S œ 'a 21y Ê1  Š dy dx ‹ dx; #

b

œ

1 6

dy dx

% ' œ x# Ê Š dy dx ‹ œ x Ê S œ 0 21 † #

1

x$ 3

È1  x% dx œ

1 6

281È2 3

'01 È1  x% a4x$ b dx

'01 È1  x% d a1  x% b œ 16 ’ 32 a1  x% b$Î# “ " œ 19 ’2È2
1“ !

23. S œ 'c 21x Ê1  Š dx dy ‹ dy; #

d

dx dy

ˆ "# ‰ (4  2y) È4y  y#

œ

œ

2y È4y  y#

#

Ê 1  Š dx dy ‹ œ

4y  y#
4  4y
y# 4y  y#

œ

4 4y  y#

Ê S œ '1 21 È4y
y# É 4y 4 y# dy œ 41'1 dx œ 41 2

2

24. S œ 'c 21x Ê1  Š dx dy ‹ dy; #

d

œ 1'2 È4y  1 dy œ 6

1 4

dx dy

œ

1 2È y

#

Ê 1  Š dx dy ‹ œ 1 

 32 (4y  1)$Î# ‘ ' œ #

1 6

(125
27) œ

1 6

" 4y

œ

4y
1 4y

(98) œ

Ê S œ '2 21Èy † 6

È4y
1 È4y

dy

491 3

25. The equipment alone: the force required to lift the equipment is equal to its weight Ê F" (x) œ 100 N.

The work done is W" œ 'a F" (x) dx œ '0 100 dx œ [100x]%! ! œ 4000 J; the rope alone: the force required b

40

to lift the rope is equal to the weight of the rope paid out at elevation x Ê F# (x) œ 0.8(40
x). The work done is W# œ 'a F# (x) dx œ '0 0.8(40
x) dx œ 0.8 ’40x
b

40

the total work is W œ W"  W# œ 4000  640 œ 4640 J

%! x# # “!

œ 0.8 Š40#


40# # ‹

œ

(0.8)(1600) #

œ 640 J;

26. The force required to lift the water is equal to the water's weight, which varies steadily from 8 † 800 lb to 8 † 400 lb over the 4750 ft elevation. When the truck is x ft off the base of Mt. Washington, the water weight is x‰ F(x) œ 8 † 800 † ˆ 2†24750 œ (6400) ˆ1
†4750

œ '0

4750

6400 ˆ1


x ‰ 9500

dx œ 6400 ’x


œ 22,800,000 ft † lb

x ‰ 9500

lb. The work done is W œ 'a F(x) dx

%(&! x# 2†9500 “ !

b

œ 6400 Š4750


4750# 4†4750 ‹

œ ˆ 34 ‰ (6400)(4750)

27. Force constant: F œ kx Ê 20 œ k † 1 Ê k œ 20 lb/ft; the work to stretch the spring 1 ft is W œ '0 kx dx œ k'0 x dx œ ’20 x# “ œ 10 ft † lb; the work to stretch the spring an additional foot is 1

1

#

"

! #

W œ '1 kx dx œ k '1 x dx œ 20 ’ x# “ œ 20 ˆ 4#
"# ‰ œ 20 ˆ 3# ‰ œ 30 ft † lb 2

2

#

"

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Chapter 6 Practice Exercises 28. Force constant: F œ kx Ê 200 œ k(0.8) Ê k œ 250 N/m; the 300 N force stretches the spring x œ œ

300 250

381

F k

œ 1.2 m; the work required to stretch the spring that far is then W œ '0 F(x) dx œ '0 250x dx 1Þ2

1Þ2

œ [125x# ]!"Þ# œ 125(1.2)# œ 180 J 29. We imagine the water divided into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval [0ß 8]. The typical slab between the planes at y and y  ?y has a volume of about ?V œ 1(radius)# (thickness) # œ 1 ˆ 54 y‰ ?y œ 25161 y# ?y ft$ . The force F(y) required to lift this slab is equal to its weight: F(y) œ 62.4 ?V œ

(62.4)(25) 16

1y# ?y lb. The distance through which F(y)

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 ?W œ

(62.4)(25) 16

1y# (14
y) ?y ft † lb. The work done

lifting all the slabs from y œ 0 to y œ 8 to the level 6 ft above the top is approximately 8

W¸! !

(62.4)(25) 16

1y# (14
y) ?y ft † lb so the work to pump the water is the limit of these Riemann sums as the norm of

the partition goes to zero: W œ '0

8

$ œ (62.4) ˆ 25161 ‰ Š 14 3 †8


8% 4‹

(62.4)(25) (16)

(62.4)(25)1 16

1y# (14
y) dy œ

'08 a14y#
y$ b dy œ (62.4) ˆ 25161 ‰ ’ 143 y$
y4 “ ) %

!

¸ 418,208.81 ft † lb

30. The same as in Exercise 29, but change the distance through which F(y) must act to (8
y) rather than (6  8
y). Also change the upper limit of integration from 8 to 5. The integral is:W œ '0

5

œ (62.4) ˆ 25161 ‰'0 a8y#
y$ b dy œ (62.4) ˆ 25161 ‰ ’ 83 y$
5

& y% 4 “!

(62.4)(25)1 16

y# (8
y) dy

œ (62.4) ˆ 25161 ‰ Š 83 † 5$


5% 4‹

31. The tank's cross section looks like the figure in Exercise 29 with right edge given by x œ #

slab has volume ?V œ 1(radius)# (thickness) œ 1 ˆ #y ‰ ?y œ F(y) œ 60 †

1 4

y œ y# . A typical horizontal

y# ?y. The force required to lift thisslab is its weight:

y# ?y. The distance through which F(y) must act is (2  10
y) ft, so the work to pump the liquid is

W œ 60'0 1a12
ybŠ y4 ‹dy œ 151 ’ 12y 3
10

22,5001 ft†lb 275 ft†lb/sec

1 4

5 10

¸ 54,241.56 ft † lb

#

$

"! y% 4 “!

œ 22,5001 ft † lb; the time needed to empty the tank is

¸ 257 sec

32. A typical horizontal slab has volume about ?V œ (20)(2x)?y œ (20) ˆ2È16
y# ‰ ?y and the force required to lift this slab is its weight F(y) œ (57)(20) ˆ2È16
y# ‰ ?y. The distance through which F(y) must act is (6  4
y) ft, so the work to pump the olive oil from the half-full tank is W œ 57'c4 (10
y)(20) ˆ2È16
y# ‰ dy 0

"Î# œ 2880 'c4 10È16
y# dy  1140'c4 a16
y# b (
2y) dy 0

0

œ 22,800 † (area of a quarter circle having radius 4)  23 (1140) ’a16
y# b œ 335,153.25 ft † lb

$Î# !

“

%

œ (22,800)(41)  48,640

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

382

Chapter 6 Applications of Definite Integrals

33. Intersection points: 3
x# œ 2x# Ê 3x#
3 œ 0 Ê 3(x
1)(x  1) œ 0 Ê x œ
1 or x œ 1. Symmetry suggests that x œ 0. The typical @/3-+6 strip has # # # center of mass: (µ x ßµ y ) œ Šxß 2x
a3  x b ‹ œ Šxß x
3 ‹ , #

#

#

#

#

length: a3
x b
2x œ 3 a1
x b, width: dx, area: dA œ 3 a1
x# b dx, and mass: dm œ $ † dA œ 3$ a1
x# b dx Ê the moment about the x-axis is µ y dm œ œ

3 #

$ ax#  3b a1
x# b dx œ

3 #

&

2x$ 3

$ ’
x5


" x$ 3 “ "

œ 3$ ’x


 3x“

" "

3 #

$ a
x%
2x#  3b dx Ê Mx œ ' µ y dm œ

œ 3$ ˆ
5"


2 3

3$ 15

 3‰ œ

œ 6$ ˆ1
"3 ‰ œ 4$ Ê y œ

Mx M

œ

(
3
10  45) œ

32$ 5 †4 $

œ

8 5

32$ 5

3 #

$ 'c1 a
x%
2x#  3b dx "

; M œ ' dm œ 3$ 'c1 a1
x# b dx "

. Therefore, the centroid is (xß y) œ ˆ!ß 85 ‰ .

34. Symmetry suggests that x œ 0. The typical @/3-+6 # strip has center of mass: (µ x ßµ y ) œ Šxß x# ‹ , length: x# , width: dx, area: dA œ x# dx, mass: dm œ $ † dA œ $ x# dx Ê the moment about the x-axis is µ y dm œ #$ x# † x# dx œ

$ #

x% dx Ê Mx œ ' µ y dm œ

$ #

'c22 x% dx œ 10$ cx& d ##

35. The typical @/3-+6 strip has: center of mass: (µ x ßµ y ) œ Œxß

#

4
x4 #


, length: 4


area: dA œ Š4
œ $ Š4


x# 4‹

x# 4 ‹dx,

width: dx,

mass: dm œ $ † dA

dx Ê the moment about the x-axis is #

Š4
x4 ‹

µ y dm œ $ †

x# 4,

#

Š4


x# 4‹

dx œ

$ #

Š16


moment about the y-axis is µ x dm œ $ Š4
œ

$ 2

’16x


% x& 5†16 “ !

$ #

œ

64


64 ‘ 5

œ

x% 16 ‹ x# 4‹

œ

16†$ †3 32†$

œ

3 2

and y œ

Mx M

œ

dx. Thus, Mx œ ' µ y dm œ

4

4

My M

x$ 4‹

† x dx œ $ Š4x


; My œ ' µ x dm œ $ '0 Š4x


128$ 5

œ $ (32
16) œ 16$ ; M œ ' dm œ $ '0 Š4
Ê xœ

dx; the

128†$ †3 5†32†$

x# 4‹

œ

dx œ $ ’4x


12 5

% x$ 12 “ !

x$ 4‹

dx œ $ ’2x#


œ $ ˆ16


64 ‰ 1#

œ

$ #

'04 Š16
16x ‹ dx %

% x% 16 “ !

32$ 3

‰ . Therefore, the centroid is (xß y) œ ˆ 3# ß 12 5 .

36. A typical 29+6 strip has: # center of mass: (µ x ßµ y ) œ Š y
# 2y ß y‹ , length: 2y
y# , width: dy, area: dA œ a2y
y# b dy, mass: dm œ $ † dA œ $ a2y
y# b dy; the moment about the x-axis is µ y dm œ $ † y † a2y
y# b dy œ $ a2y#
y$ b ; the moment # about the y-axis is µ x dm œ $ † ay
2yb † a2y
y# b dy #

œ a4y
y b dy Ê Mx œ ' µ y dm œ $ '0 a2y#
y$ b dy $ #

#

œ $ ’ 23 y$
œ

$ #

yœ

ˆ 43†8 Mx M

2

%




œ

# y% 4 “!

32 ‰ 5

4†$ †3 3 † 4 †$

œ

œ $ ˆ 23 † 8
32$ 15

16 ‰ 4

œ $ ˆ 16 3


16 ‰ 4

œ

$ †16 12

œ

4$ 3

; My œ ' µ x dm œ

$ #

'02 a4y#
y% b dy œ #$ ’ 34 y$
y5 “ #

$ ; M œ ' dm œ $ '0 a2y
y# b dy œ $ ’y#
y3 “ œ $ ˆ4
83 ‰ œ

2

# !

&

!

4$ 3

Ê xœ

œ 1. Therefore, the centroid is (xß y) œ ˆ 58 ß 1‰ .

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

My M

œ

$ †32†3 15†$ †4

œ

8 5

and

Chapter 6 Practice Exercises

383

37. A typical horizontal strip has: center of mass: (µ x ßµ y ) #

œ Š y
#2y ß y‹ , length: 2y
y# , width: dy, area: dA œ a2y
y# b dy, mass: dm œ $ † dA œ (1  y) a2y
y# b dy Ê the moment about the x-axis is µ y dm œ y(1  y) a2y
y# b dy # œ a2y  2y$
y$
y% b dy œ a2y#  y$
y% b dy; the moment about the y-axis is # µ x dm œ Š y
2y ‹ (1  y) a2y
y# b dy œ " a4y#
y% b (1  y) dy œ #

Ê Mx œ ' µ y dm œ '0 a2y#  y$
y% b dy œ ’ 23 y$  2

œ

16 60

œ

" #

" #

#

(20  15
24) œ $

Š 4†32  2%


2& 5




(11) œ

4 15

2' 6‹

2

44 40

œ

11 10

y$ 3




; My œ ' µ x dm œ '0

2

œ 4 ˆ 43  2


œ '0 a2y  y#
y$ b dy œ ’y#  ‰ ˆ 38 ‰ œ œ ˆ 44 15

44 15

y% 4

4 5

œ ˆ4 

8 3




16 ‰ 4

œ ˆ 16 3 




16 4

32 ‰ 5

œ 16 ˆ "3 

a4y#  4y$
y%
y& b dy œ


86 ‰ œ 4 ˆ2
45 ‰ œ

# y% 4 “!




" #

# y& 5 “!

a4y#  4y$
y%
y& b dy

œ

8 3

24 5

" #

" 4


25 ‰

’ 43 y$  y%


y& 5




; M œ ' dm œ '0 (1  y) a2y
y# b dy

# y' 6 “!

2

Ê xœ

My M

‰ ˆ 83 ‰ œ œ ˆ 24 5

9 5

and y œ

Mx M

‰ . Therefore, the center of mass is (xß y) œ ˆ 95 ß 11 10 .

3 3 38. A typical vertical strip has: center of mass: (µ x ßµ y ) œ ˆxß 2x3$Î# ‰ , length: x$Î# , width: dx, area: dA œ x$Î# dx, µ 3 3 3 9$ mass: dm œ $ † dA œ $ † dx Ê the moment about the x-axis is y dm œ †$ dx œ dx; the moment about x$Î#

3 the y-axis is µ x dm œ x † $ x$Î# dx œ

(a) Mx œ $ '1

9

M œ $ '1

9

(b) Mx œ '1

9

" #

3 x$Î# x #

9$ #

ˆ x9$ ‰ dx œ

3$ x"Î#

#x$Î#

# *

20$ 9

’
x# “ œ "

3 ‰ ; My œ $ '1 x ˆ x$Î# dx œ 3$ 2x"Î# ‘ " œ 12$ ;

*

*

9 #

*

9

dx œ
6$ x"Î# ‘ " œ 4$ Ê x œ

ˆ x9$ ‰ dx œ

2x$

x$Î#

dx.

My M

œ

12$ 4$

œ 3 and y œ

Mx M

œ

ˆ 209$ ‰ 4$

œ

5 9

* 3 ‰ 3 ‰ 
x" ‘ * œ 4; My œ ' x# ˆ $Î# dx œ 2x$Î# ‘ " œ 52; M œ '1 x ˆ x$Î# dx " x 1

œ 6 x"Î# ‘ " œ 12 Ê x œ

9

My M

œ

13 3

and y œ

Mx M

9

œ

" 3

strip 39. F œ 'a W † Š depth ‹ † L(y) dy Ê F œ 2 '0 (62.4)(2
y)(2y) dy œ 249.6'0 a2y
y# b dy œ 249.6 ’y#
b

2

2

# y$ 3 “!

œ (249.6) ˆ4
83 ‰ œ (249.6) ˆ 43 ‰ œ 332.8 lb strip 40. F œ 'a W † Š depth ‹ † L(y) dy Ê F œ '0 75 ˆ 56
y‰ (2y  4) dy œ 75'0 ˆ 53 y  5Î6

b

5Î6

10 3


2y#
4y‰ dy

7 7 # 2 $ ‘ &Î' 50 ‰ 25 ‰ 125 ‰‘ #‰ ˆ 18 œ 75 '0 ˆ 10 dy œ 75  10
ˆ 67 ‰ ˆ 36
ˆ 32 ‰ ˆ 216 3
3 y
2y 3 y
6 y
3 y ! œ (75) 5Î6

œ (75) ˆ 25 9


175 216




250 ‰ 3†#16

‰ œ ˆ 9†75 #16 (25 † 216
175 † 9
250 † 3) œ

strip 41. F œ 'a W † Š depth ‹ † L(y) dy Ê F œ 62.4'0 (9
y) Š2 † b

4

%

œ 62.4 6y$Î#
25 y&Î# ‘ ! œ (62.4) ˆ6 † 8


2 5

Èy 2 ‹

(75)(3075) 9†#16

¸ 118.63 lb.

dy œ 62.4'0 ˆ9y"Î#
3y$Î# ‰ dy 4

‰ (48 † 5
64) œ † 32‰ œ ˆ 62.4 5

(62.4)(176) 5

œ 2196.48 lb

strip 42. Place the origin at the bottom of the tank. Then F œ '0 W † Š depth ‹ † L(y) dy, h œ the height of the mercury column, h

strip depth œ h
y, L(y) œ 1 Ê F œ '0 849(h
y) " dy œ (849)'0 (h
y) dy œ 849’hy
h

œ

849 # 2 h .

Now solve

849 # 2 h

h

h

y# # “!

œ 849 Šh#


h# #‹

œ 40000 to get h ¸ 9.707 ft. The volume of the mercury is s2 h œ 12 † 9.707 œ 9.707 ft$ Þ

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

384

Chapter 6 Applications of Definite Integrals

CHAPTER 6 ADDITIONAL AND ADVANCED EXERCISES 1. V œ 1 'a cf(x)d# dx œ b#
ab Ê 1'a cf(t)d# dt œ x#
ax for all x  a Ê 1 [f(x)]# œ 2x
a Ê f(x) œ É 2x1 a b

x

2. V œ 1 '0 [f(x)]# dx œ a#  a Ê 1 '0 [f(t)]# dt œ x#  x for all x  a Ê 1[f(x)]# œ 2x  1 Ê f(x) œ É 2x1
1 a

x

3. s(x) œ Cx Ê '0 È1  [f w (t)]# dt œ Cx Ê È1  [f w (x)]# œ C Ê f w (x) œ ÈC#
1 for C   1 x

Ê f(x) œ '0 ÈC#
1 dt  k. Then f(0) œ a Ê a œ 0  k Ê f(x) œ '0 ÈC#
1 dt  a Ê f(x) œ xÈC#
1  a, x

x

where C   1. 4. (a) The graph of f(x) œ sin x traces out a path from (!ß !) to (!ß sin !) whose length is L œ '0 È1  cos# ) d). !

The line segment from (0ß 0) to (!ß sin !) has length È(!
0)#  (sin !
0)# œ È!#  sin# !. Since the shortest distance between two points is the length of the straight line segment joining them, we have ! immediately that ' È1  cos# ) d)  È!#  sin# ! if 0  ! Ÿ 1 . #

0

(b) In general, if y œ f(x) is continuously differentiable and f(0) œ 0, then '0 È1  [f w (t)]# dt  È!#  f # (!) !

for !  0. 5. We can find the centroid and then use Pappus' Theorem to calculate the volume. faxb œ x, gaxb œ x2 , faxb œ gaxb 1 Ê x œ x2 Ê x2
x œ 0 Ê x œ 0, x œ 1; $ œ 1; M œ '0 cx
x2 ddx œ  "# x2
13 x3 ‘0 œ ˆ "#
13 ‰
0 œ 1

1 6

xœ

1 1 Î6

'01 xcx
x2 ddx œ 6'01 cx2
x3 ddx œ 6 31 x3
41 x4 ‘10 œ 6ˆ 31
41 ‰
0 œ 21

yœ

1 1 Î6

'01 12 ’x2
ax2 b2 “dx œ 3'01 cx2
x4 ddx œ 3 13 x3
15 x5 ‘10 œ 3ˆ 13
15 ‰
0 œ 25 Ê The centroid is ˆ 12 , 25 ‰.

3 is the distance from ˆ 12 , 25 ‰ to the axis of rotation, y œ x. To calculate this distance we must find the point on y œ x that also lies on the line perpendicular to y œ x that passes through ˆ 12 , 25 ‰. The equation of this line is y
25 œ
1ˆx
12 ‰ Êxyœ

9 10 .

9 3 œ Ɉ 20


The point of intersection of the lines x  y œ

1 ‰2 2

9  ˆ 20


2 ‰2 5

œ

1 . 10È2

9 and y œ x is ˆ 20 ,

9 10

Thus V œ 21Š 101È2 ‹ˆ 16 ‰ œ

9 ‰ 20 .

Thus,

1 . 30È2

6. Since the slice is made at an angle of 45‰ , the volume of the wedge is half the volume of the cylinder of radius height 1. Thus, V œ

" ˆ " ‰2 # ’1 # a1b

“œ

" #

and

1 8.

7. y œ 2Èx Ê ds œ É "x  1 dx Ê A œ '0 2Èx É "x  1 dx œ 3

4 3

(1  x)$Î# ‘ $ œ !

28 3

8. This surface is a triangle having a base of 21a and a height of 21ak. Therefore the surface area is " # # # (21a)(21ak) œ 21 a k. d# x dt#

9. F œ ma œ t# Ê

œaœ

t# m

Ê vœ

x œ 0 when t œ 0 Ê C" œ 0 Ê x œ W œ ' F dx œ '0

Ð12mhÑ"Î%

œ

(12mh)$Î# 18m

œ

F(t) †

12mh†È12mh 18m

œ

2h 3

dx dt

dx t$ dt œ 3m  C; v œ 0 when t œ 0 Ê t% "Î% . 12m . Then x œ h Ê t œ (12mh)

dt œ '0

Ð12mhÑ"Î%

† 2È3mh œ

4h 3

t# †

$

t 3m

dt œ

" 3m

'

’ t6 “

Ð12mh)"Î%

0

Cœ0 Ê

dx dt

œ

t$ 3m

The work done is

" ‰ œ ˆ 18m (12mh)'Î%

È3mh

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Ê xœ

t% 12m

 C" ;

Chapter 6 Additional and Advanced Exercises 10. Converting to pounds and feet, 2 lb/in œ

†

2 lb 1 in

œ 24 lb/ft. Thus, F œ 24x Ê W œ '0

1Î2

12 in 1 ft

24x dx

"Î# " " ‰ œ c12x# d ! œ 3 ft † lb. Since W œ "# mv!#
"# mv"# , where W œ 3 ft † lb, m œ ˆ 10 lb‰ ˆ 3# ft/sec # " œ 320 slugs, and v" œ 0 ft/sec, we have 3 œ ˆ #" ‰ ˆ 3#"0 v#! ‰ Ê v!# œ 3 † 640. For the projectile height, s œ
16t#  v! t (since s œ 0 at t œ 0) Ê ds dt œ v œ
32t  v! . At the top of the ball's path, v œ 0 Ê #

and the height is s œ
16 ˆ 3v#! ‰  v! ˆ 3v#! ‰ œ

v!# 64

3†640 64

œ

385

tœ

v! 3#

œ 30 ft.

11. From the symmetry of y œ 1
xn , n even, about the y-axis for
1 Ÿ x Ÿ 1, we have x œ 0. To find y œ MMx , we n use the vertical strips technique. The typical strip has center of mass: (µ x ßµ y ) œ ˆxß 1 2 x ‰ , length: 1
xn , width: dx, area: dA œ a1
xn b dx, mass: dm œ 1 † dA œ a1
xn b dx. The moment of the strip about the 1 1 n # n # " nb1 2n b 1 x-axis is µ y dm œ a1  x b dx Ê M œ ' a1  x b dx œ 2' " a1
2xn  x2n b dx œ x
2x  x ‘ #

œ1


2 n
1



" #n
1

œ

x c1 # (n
1)(2n
1)  2(2n
")
(n
1) (n
1)(#n
1)

œ

0 # 2n#
3n
1  4n  2
n
1 (n
1)(#n
1)

Also, M œ 'c1 dA œ 'c1 a1
xn b dx œ 2 '0 a1
xn b dx œ 2 x
1

yœ

Mx M

œ

1

#

†

2n (n
1)(2n
1)

1

(n
1) 2n

œ

xn b 1 ‘ " n
1 !

n
1

œ

2n# (n
1)(#n
1)

œ 2 ˆ1


" ‰ n
1

#n
1 !

. œ

2n n
1.

Therefore,

Ê ˆ!ß #n n
1 ‰ is the location of the centroid. As n Ä _, y Ä

n 2n
1

the limiting position of the centroid is ˆ!ß

" #

so

"‰ # .

12. Align the telephone pole along the x-axis as shown in the accompanying figure. The slope of the top length of pole is 9 ‰ ˆ 14.5 " 81  81 œ 8"1 † 40 † (14.5
9) œ 815.5 †40 40 11 ‰ y œ 891  8111†80 x œ 8"1 ˆ9  80 x is an

œ

11 81†80 .

Thus,

equation of the

line representing the top of the pole. Then, My œ 'a x † 1y# dx œ 1 '0 x  8"1 ˆ9  b

40

11 80

# x‰‘ dx

b 11 ‰# '040 x ˆ9  80 x dx; M œ 'a 1y# dx 40 40 11 ‰‘# ‰# dx. œ 1 '0  8"1 ˆ9  80 x dx œ 64"1 '0 ˆ9  11 80 x

œ

" 641

Thus, x œ

My M

¸

129,700 5623.3

¸ 23.06 (using a calculator to compute

the integrals). By symmetry about the x-axis, y œ 0 so the center of mass is about 23 ft from the top of the pole. 13. (a) Consider a single vertical strip with center of mass (µ x ßµ y ). If the plate lies to the right of the line, then µ µ
b) $ dA Ê the plate's first moment the moment of this strip about the line x œ b is (x
b) dm œ (x about x œ b is the integral ' (x
b)$ dA œ ' $ x dA
' $ b dA œ My
b$ A. (b) If the plate lies to the left of the line, the moment of a vertical strip about the line x œ b is ab
µ x b dm œ ab
µ x b $ dA Ê the plate's first moment about x œ b is ' (b
x)$ dA œ ' b$ dA
' $ x dA œ b$ A
My . 14. (a) By symmetry of the plate about the x-axis, y œ 0. A typical vertical strip has center of mass: (µ x ßµ y ) œ (xß 0), length: 4Èax, width: dx, area: 4Èax dx, mass: dm œ $ dA œ kx † 4Èax dx, for some a proportionality constant k. The moment of the strip about the y-axis is M œ ' µ x dm œ ' 4kx# Èax dx y

œ 4kÈa'0 x a

&Î#

dx œ

4kÈa  27

x

(Î# ‘ a

0

œ 4ka

"Î#

† a 2 7

(Î#

œ

œ 4kÈa'0 x$Î# dx œ 4kÈa  25 x&Î# ‘ 0 œ 4ka"Î# † 25 a&Î# œ a

Ê (xß y) œ

a

ˆ 5a ‰ 7 ß0

0

8ka% 7

. Also, M œ ' dm œ '0 4kxÈax dx

8ka$ 5

. Thus, x œ

a

My M

œ

8ka% 7

†

5 8ka$

œ

5 7

is the center of mass. y#

# #
a (b) A typical horizontal strip has center of mass: (µ x ßµ y ) œ Œ 4a # ß y
œ Š y
8a4a ß y‹ , length: a


width: dy, area: Ša
œ 'c2a y kyk Ša
2a

a

y# 4a ‹

y# 4a ‹

dy, mass: dm œ $ dA œ kyk Ša


dy œ 'c2a
y# Ša
0

y# 4a ‹

y# 4a ‹

dy  '0 y# Ša
2a

dy. Thus, Mx œ ' µ y dm y# 4a ‹

dy

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

y# 4a

,

386

Chapter 6 Applications of Definite Integrals œ 'c2a Š
ay#  0

%

32a& 20a

œ
8a3 

dy  '0 Šay#
2a

y% 4a ‹ 8a% 3






y% 4a ‹

œ 0; My œ ' µ x dm œ 'c2a Š y 2a

32a& #0a

'c2a2a kyk ay#  4a# b Š 4a 4a y ‹ dy œ 32a" #

! y& #0a “ #a

dy œ ’
3a y$  #


4a# 8a ‹

#

" 8a

œ

" 3 #a #

'c02a a
16a% y  y& b dy  32a" '02a a16a% y
y& b dy œ 3#"a

œ

" 32a#

’8a% † 4a#


" 32a#

2a

œ

#

#

64a' 6 “

œ

" 16a#

#

’
8a% y#  32a' 3 ‹

Š32a'


œ

#a

œ2†

#

16a% 4 ‹

#

Š2a † 4a


" #a

œ

! y' 6 “ #a



1 3#a#

’8a% y#


† 32 a32a' b œ

4 3

#a y' 6 “!

a% ;

'c2a2a kyk a4a#
y# b dy

" 4a

%

" 4a

dy

" 16a#

'c02a a
4a# y  y$ b dy  4a" '02a a4a# y
y$ b dy œ 4a" ’
2a# y#  y4 “ !

" 4a

yœ

c2a

’8a% † 4a#


M œ ' dm œ 'c2a kyk Š 4a 4ay ‹ dy œ

y# 4a ‹

kyk Ša


' kyk a16a%
y% b dy

#

#



#a y& #0a “ !

2a

œ

64a' 6 “

 ’ 3a y$


%

%

$

a8a
4a b œ 2a . Therefore, x œ



" 4a

’2a# y#


œ ˆ 34 a% ‰ ˆ 2a"$ ‰ œ

My M

#a y% 4 “!

2a 3

and

œ 0 is the center of mass.

Mx M

15. (a) On [0ß a] a typical @/3-+6 strip has center of mass: (µ x ßµ y ) œ Šx,

È b#  x #
È a#  x# ‹, #

length: Èb#
x#
Èa#
x# , width: dx, area: dA œ ŠÈb#
x#
Èa#
x# ‹ dx, mass: dm œ $ dA œ $ ŠÈb#
x#
Èa#
x# ‹ dx. On [aß b] a typical @/3-+6 strip has center of mass: È # # (µ x ßµ y ) œ Šxß b # x ‹ , length: Èb#
x# , width: dx, area: dA œ Èb#
x# dx,

mass: dm œ $ dA œ $ Èb#
x# dx. Thus, Mx œ ' µ y dm œ '0

a

" #

ŠÈb#
x#  Èa#
x# ‹ $ ŠÈb#
x#
Èa#
x# ‹ dx  'a

b

" #

Èb#
x# $ Èb#
x# dx

œ

$ #

'0a cab#
x# b
aa#
x# bd dx  #$ 'ab ab#
x# b dx œ #$ '0a ab#
a# b dx  #$ 'ab ab#
x# b dx

œ

$ #

cab#
a# b xd !  #$ ’b# x


œ

$ #

aab#
a$ b  #$ Š 23 b$
ab# 

a

b

x$ 3 “a

œ a$ 3‹

$ #

b$ 3‹

cab#
a# b ad  #$ ’Šb$


œ

$ b$ 3




$ a$ 3

œ $ Šb

$

 a$ 3 ‹;

a$ 3 ‹“


Š b# a


My œ ' µ x dm

œ '0 x$ ŠÈb#
x#
Èa#
x# ‹ dx  'a x$ Èb#
x# dx a

b

œ $ '0 x ab#
x# b a

œ

$ #

”

2 ab #  x # b 3

$Î#

"Î#

dx
$ '0 x aa#
x# b a

a

$ 2 aa • #”

#

 x# b 3

$Î#

# $Î#

#

œ
’ab
a b

# $Î#


ab b

a

dx  $ 'a x ab#
x# b

$ 2 ab •
#”

b

#

!

0

$ 3

"Î#

# $Î#

$ 3

“  ’0
aa b

 x# b 3

"Î#

$Î#

• a

$ 3

“
’0
ab#
a# b #

#

$Î#

We calculate the mass geometrically: M œ $ A œ $ Š 14b ‹
$ Š 14a ‹ œ œ

$ ab $  a $ b 3

yœ (b) lim

œ

Mx M 4

b Ä a 31

†

4 $1 ab#  a# b #

#

4 aa
ab
b b 31(a
b)

Ša

#


ab
b a
b

#

œ

4 31

$

$

a Š bb#  a# ‹ œ

dx

b

4 (b  a) aa#
ab
b# b 31 (b  a)(b
a)

“œ

$1 4

$ b$ 3




$ a$ 3

œ

$ ab $  a $ b 3

ab#
a# b . Thus, x œ

œ Mx ; My M

œ

4 aa#
ab
b# b 31(a
b)

2a 1

2a ‰ Ê (xß y) œ ˆ 2a 1 ß 1 is the limiting

; likewise

.

‹ œ ˆ 341 ‰ Š a

#


a#
a# ‹ a
a

#

œ ˆ 341 ‰ Š 3a 2a ‹ œ

position of the centroid as b Ä a. This is the centroid of a circle of radius a (and we note the two circles coincide when b œ a).

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Chapter 6 Additional and Advanced Exercises 16. Since the area of the traingle is 36, the diagram may be labeled as shown at the right. The centroid of the triangle is ˆ 3a , 24 ‰ a . The shaded portion is 144
36 œ 108. Write ax, yb for the centroid of the remaining region. The centroid of the whole square is obviously a6, 6b. Think of the square as a sheet of uniform density, so that the centroid of the square is the average of the centroids of the two regions, weighted by area: 'œ

$'ˆ 3a ‰
"!)axb "%%

and ' œ

‰ $'ˆ 24 a
"!)ayb "%%

which we solve to get x œ )


a *

and y œ

)a a  " b . a

Set

x œ 7 in. (Given). It follows that a œ *, whence y œ œ

7 "*

'% *

in. The distances of the centroid ax, yb from the other sides are easily computed. (Note that if we set y œ 7 in.

above, we will find x œ 7 "* .) 17. The submerged triangular plate is depicted in the figure at the right. The hypotenuse of the triangle has slope
1 Ê y
(
2) œ
(x
0) Ê x œ
(y  2) is an equation of the hypotenuse. Using a typical horizontal strip, the fluid strip strip pressure is F œ ' (62.4) † Š depth ‹ † Š length ‹ dy

c2

c2

œ 'c6 (62.4)(
y)[
(y  2)] dy œ 62.4 'c6 ay#  2yb dy $

œ 62.4 ’ y3  y# “

#

‰‘ œ (62.4) ˆ
83  4‰
ˆ
216 3  36

'

‰ œ (62.4) ˆ 208 3
32 œ

(62.4)(112) 3

¸ 2329.6 lb

18. Consider a rectangular plate of length j and width w. The length is parallel with the surface of the fluid of weight density =. The force on one side of the plate is F œ ='cw (
y)(j) dy œ
=j ’ y# “ 0

#

! w

œ

=jw# #

. The

average force on one side of the plate is Fav œ œ

= w

#

’
y# “

! w

œ

=w #

. Therefore the force

= w

'c0w (
y)dy

=jw# #

‰ œ ˆ =w # (jw) œ (the average pressure up and down) † (the area of the plate).

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

387

388

Chapter 6 Applications of Definite Integrals

NOTES:

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

CHAPTER 7 TRANSCENDENTAL FUNCTIONS 7.1 INVERSE FUNCTIONS AND THEIR DERIVATIVES 1. Yes one-to-one, the graph passes the horizontal line test. 2. Not one-to-one, the graph fails the horizontal line test. 3. Not one-to-one since (for example) the horizontal line y œ 2 intersects the graph twice. 4. Not one-to-one, the graph fails the horizontal line test. 5. Yes one-to-one, the graph passes the horizontal line test 6. Yes one-to-one, the graph passes the horizontal line test 7. Not one-to-one since the horizontal line y œ 3 intersects the graph an infinite number of times. 8. Yes one-to-one, the graph passes the horizontal line test 9. Yes one-to-one, the graph passes the horizontal line test 10. Not one-to-one since (for example) the horizontal line y œ 1 intersects the graph twice. 11. Domain: 0
x Ÿ 1, Range: 0 Ÿ y

13. Domain: 1 Ÿ x Ÿ 1, Range:  1# Ÿ y Ÿ

12. Domain: x
1, Range: y  0

1 #

14. Domain: _
x
_, Range:  1#
y Ÿ

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

1 #

390

Chapter 7 Transcendental Functions

15. Domain: 0 Ÿ x Ÿ 6, Range: 0 Ÿ y Ÿ 3

16. Domain: 2 Ÿ x Ÿ 1, Range:  1 Ÿ y
3

17. The graph is symmetric about y œ x.

(b) y œ È1  x# Ê y# œ 1  x# Ê x# œ 1  y# Ê x œ È1  y# Ê y œ È1  x# œ f
" (x) 18. The graph is symmetric about y œ x.

yœ

" x

Ê xœ

" y

Ê yœ

" x

œ f
" (x)

19. Step 1: y œ x#  1 Ê x# œ y  1 Ê x œ Èy  1 Step 2: y œ Èx  1 œ f
" (x) 20. Step 1: y œ x# Ê x œ Èy, since x Ÿ !. Step 2: y œ Èx œ f
" (x) 21. Step 1: y œ x$  1 Ê x$ œ y  1 Ê x œ (y  1)"Î$ Step 2: y œ $Èx  1 œ f
" (x) 22. Step 1: y œ x#  2x  1 Ê y œ (x  1)# Ê Èy œ x  1, since x   1 Ê x œ 1  Èy Step 2: y œ 1  Èx œ f
" (x) 23. Step 1: y œ (x  1)# Ê Èy œ x  1, since x   1 Ê x œ Èy  1 Step 2: y œ Èx  1 œ f
" (x) 24. Step 1: y œ x#Î$ Ê x œ y$Î# Step 2: y œ x$Î# œ f
" (x)

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 7.1 Inverse Functions and Their Derivatives 25. Step 1: y œ x& Ê x œ y"Î& 5 Step 2: y œ È x œ f
" (x); Domain and Range of f
" : all reals; &

f af
" (x)b œ ˆx"Î& ‰ œ x and f
" (f(x)) œ ax& b

"Î&

œx

"Î%

œx

26. Step 1: y œ x% Ê x œ y"Î% Step 2: y œ %Èx œ f
" (x); Domain of f
" : x   0, Range of f
" : y   0; %

f af
" (x)b œ ˆx"Î% ‰ œ x and f
" (f(x)) œ ax% b

27. Step 1: y œ x$  1 Ê x$ œ y  1 Ê x œ (y  1)"Î$ Step 2: y œ $Èx  1 œ f
" (x); Domain and Range of f
" : all reals; $

f af
" (x)b œ ˆ(x  1)"Î$ ‰  1 œ (x  1)  1 œ x and f
" (f(x)) œ aax$  1b  1b 28. Step 1: y œ

" #

x

7 #

Ê
"

" #

xœy

7 #

Step 2: y œ

" x#

Ê x# œ

" Èx

œ f
" (x)

" y

Ê xœ

7 #

30. Step 1: y œ

" x$

" x"Î$
"

Step 2: y œ Domain of f

f af
" axbb œ

" y

Ê xœ

Step 2: y

32. Step 1: y œ

œ

" Š "x ‹

œ x since x  0

" y"Î$

: x Á 0, Range of f
" : y Á 0;

" $ ax "Î$ b

œ

" x "

œ x and f
" afaxbb œ ˆ x"$ ‰

x3 x
2 Ê yax  2b 3
1 œ 2x axb; x
1 œ f
"

f af
" axbb œ

" É x"#

3 "
" œÉ (x); x œ f

31. Step 1: y œ Domain of f

œx

" Èy

Šx‹

Ê x$ œ

"Î$

œ x and f
" (f(x)) œ 2 ˆ "# x  7# ‰  7 œ (x  7)  7 œ x

Domain of f
" : x  0, Range of f
" : y  0; f af
" (x)b œ "" # œ "" œ x and f
" (f(x)) œ Š Èx ‹

œ ax$ b

Ê x œ 2y  7

Step 2: y œ 2x  7 œ f (x); Domain and Range of f
" : all reals; f af
" (x)b œ "# (2x  7)  7# œ ˆx  7# ‰  29. Step 1: y œ

"Î$


"Î$

œ ˆ x" ‰


"

œx

œ x  3 Ê x y  2y œ x  3 Ê x y  x œ 2y  3 Ê x œ

: x Á 1, Range of f
" : y Á 2;

b3‰ ˆ 2x xc1  3 b3‰ ˆ 2x xc1
2 Èx Èx
3

œ

a2x  3b  3ax
1b a2x  3b
2ax
1b

œ

5x 5

œ x and f
" afaxbb œ

3 2ˆ xx b c2‰  3 3 ˆ xx b c2‰
1

œ

2y  3 y
1

2 ax  3 b  3 a x
2 b ax  3 b
ax
2 b

œ

5x 5

œx

Ê yˆÈx  3‰ œ Èx Ê yÈx  3y œ Èx Ê yÈx  Èx œ 3y Ê x œ Š y 3y
1‹

2

2

‰ œ f
1 a xb ; Step 2: y œ ˆ x 3x
1

Domain of f
" : Ð_, 0Ó  a1, _b, Range of f
" : Ò0, 9Ñ  a9, _b; f af
" axbb œ f


"

2 Ɉ x 3x c1‰ 2

Ɉ x 3x c1‰
3

afaxbb œ


; If x  1 or x Ÿ 0 Ê

Èx

3Š È x c 3 ‹

Èx

Š Èx c 3 ‹
1 

3x x
1

 0Ê

2 Ɉ x 3x c1‰ 2

Ɉ x 3x c1‰
3

œ

3x xc1 3x xc1
3

œ

3x 3x
3ax
1b

œ

2

œ

9x ˆÈ x
ˆÈ x
3 ‰‰ 2

œ

9x 9

œx

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

3x 3

œ x and

391

392

Chapter 7 Transcendental Functions

33. Step 1: y œ x2  2x, x Ÿ 1 Ê y  1 œ ax  1b2 , x Ÿ 1 Ê Èy  1 œ x  1, x Ÿ 1 Ê x œ 1  Èy  1 Step 2: y œ 1  Èx  1 œ f
1 axb; Domain of f
" : Ò1, _Ñ, Range of f
" : Ð_, 1Ó; 2

f af
" axbb œ Š1  Èx  1‹  2Š1  Èx  1‹ œ 1  2Èx  1  x  1  2  2Èx  1 œ x and f
" afaxbb œ 1  Èax2  2xb  1, x Ÿ 1 œ 1  Éax  1b2 , x Ÿ 1 œ 1  lx  1l œ 1  a1  xb œ x 34. Step 1: y œ a2x3  1b 3 x Step 2: y œ É

5


1 2

1 Î5

Ê y5 œ 2x3  1 Ê y5  1 œ 2x3 Ê

y5
1 2

3 y œ x3 Ê x œ É

5


1 2

œ f
1 ax b ;

Domain of f
" : a_, _b, Range of f
" : a_, _b; 3 x f af
" axbb œ Œ2ŠÉ

5


1 2 ‹

3

1 Î5

 1

œ Š 2Š x

5


1 2 ‹

 1‹

1 Î5

œ aax5  1b  1b

1 Î5

œ ax5 b

1 Î5

œ x and

5

1Î5 ’a2x3 1b “
1

3

f
" afaxbb œ Ê

2

3 a2x σ

35. (a) y œ 2x  3 Ê 2x œ y  3 Ê x œ y#  3# Ê f
" (x) œ (c)

df ¸ dx xœ 1

36. (a) y œ

" 5

œ 2,

df c" dx ¹ xœ1

x7 Ê

" 5

œ

(c)

œ

" df c" 5 , dx ¹ xœ$%Î&

(c)

œ 4,

df c" dx ¹ xœ3

38. (a) y œ 2x# Ê x# œ Ê xœ (c)

df ¸ dx xœ&

" È2

" #

3



3 #

" #


"

(b)

(x) œ 5x  35

œ5

œ

(b) 5 4



x 4

" 4

(b)

y

Èy Ê f

3 2x œÉ 2 œ x

(b) x #

37. (a) y œ 5  4x Ê 4x œ 5  y Ê x œ 54  y4 Ê f
" (x) œ df ¸ dx xœ1Î#

 1 b
1 2

xœy7

Ê x œ 5y  35 Ê f df ¸ dx xœ 1

3


"

(x) œ

È x#

œ 4xk xœ5 œ 20,

df c" dx ¹ xœ&0

œ

" #È 2

x
"Î# ¹

xœ50

œ

" #0

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 7.1 Inverse Functions and Their Derivatives $

3 3 39. (a) f(g(x)) œ ˆÈ x‰ œ x, g(f(x)) œ È x3 œ x

w

#

w

(b)

w

(c) f (x) œ 3x Ê f (1) œ 3, f (1) œ 3; gw (x) œ 3" x
#Î$ Ê gw (1) œ 3" , gw (1) œ

" 3

(d) The line y œ 0 is tangent to f(x) œ x$ at (!ß !); the line x œ 0 is tangent to g(x) œ $Èx at (0ß 0)

40. (a) h(k(x)) œ

" 4

ˆ(4x)"Î$ ‰$ œ x,

k(h(x)) œ Š4 † (c) hw (x) œ w

k (x) œ

x$ 4‹

"Î$

(b)

œx

3x# w w 4 Ê h (2) œ 3, h (2) 4
#Î$ Ê kw (2) œ "3 , 3 (4x)

œ 3; kw (2) œ

(d) The line y œ 0 is tangent to h(x) œ

x$ 4

" 3

at (!ß !);

the line x œ 0 is tangent to k(x) œ (4x)"Î$ at (!ß !) œ 3x#  6x Ê

41.

df dx

43.

df " dx ¹ x œ 4

df c" dx ¹ x œ f(3)

df " dx ¹ x œ f(2)

œ

45. (a) y œ mx Ê x œ

" m

œ

(b) The graph of y œ f 46. y œ mx  b Ê x œ

y m

"

df dx

º

œ

xœ2

"

df dx

œ

º

xœ3

" ˆ 3" ‰

œ3

y Ê f
" (x) œ
"

" 9

œ

" m

œ 2x  4 Ê

42.

df dx

44.

dg " dx ¹x œ 0

b m

dg " dx ¹ x œ f(0)

œ

"

dg dx

º

œ

xœ0

"

df dx

º

œ

xœ5

œ

" 6

" 2

x

(x) is a line through the origin with slope



œ

df " dx ¹ x œ f(5)

Ê f
" (x) œ

" m

x

b m;

" m.

the graph of f
" (x) is a line with slope

" m

and y-intercept  mb .

47. (a) y œ x  1 Ê x œ y  1 Ê f
" (x) œ x  1 (b) y œ x  b Ê x œ y  b Ê f
" (x) œ x  b (c) Their graphs will be parallel to one another and lie on opposite sides of the line y œ x equidistant from that line.

48. (a) y œ x  1 Ê x œ y  1 Ê f
" (x) œ 1  x; the lines intersect at a right angle (b) y œ x  b Ê x œ y  b Ê f
" (x) œ b  x; the lines intersect at a right angle (c) Such a function is its own inverse.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

393

394

Chapter 7 Transcendental Functions

49. Let x" Á x# be two numbers in the domain of an increasing function f. Then, either x"
x# or x"  x# which implies f(x" )
f(x# ) or f(x" )  f(x# ), since f(x) is increasing. In either case, f(x" ) Á f(x# ) and f is one-to-one. Similar arguments hold if f is decreasing. 50. f(x) is increasing since x#  x" Ê

" 3

x# 



5 6

" 3

x"  56 ;

df dx

œ

" 3

51. f(x) is increasing since x#  x" Ê 27x$#  27x"$ ; y œ 27x$ Ê x œ df dx

œ 81x# Ê

df " dx

œ

" ¸ 81x# 13 x"Î$

œ

" 9x#Î$

œ

" 9

df c" dx

Ê " 3

œ

df dx

œ 24x# Ê

dx

œ

" ¸
24x# 12 Ð1 xÑ"Î$

œ

œ3

y"Î$ Ê f
" (x) œ

" 3

x"Î$ ;

x
#Î$

52. f(x) is decreasing since x#  x" Ê 1  8x$#
1  8x"$ ; y œ 1  8x$ Ê x œ df c"

" ˆ "3 ‰


" 6("
x)#Î$

" #

(1  y)"Î$ Ê f
" (x) œ

" #

(1  x)"Î$ ;

œ  "6 (1  x)
#Î$

53. f(x) is decreasing since x#  x" Ê (1  x# )$
(1  x" )$ ; y œ (1  x)$ Ê x œ 1  y"Î$ Ê f
" (x) œ 1  x"Î$ ; df dx

œ 3(1  x)# Ê

df c" dx

œ

"
3(1
x)# ¹ 1cx"Î$ &Î$

54. f(x) is increasing since x#  x" Ê x# df dx

œ

5 3

x#Î$ Ê

df c" dx

œ

5 3

" ¹ x#Î$ x$Î&

œ


" 3x#Î$

œ  "3 x
#Î$

&Î$

 x" ; y œ x&Î$ Ê x œ y$Î& Ê f
" (x) œ x$Î& ; œ

3 5x#Î&

œ

3 5

x
#Î&

55. The function g(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x" Á x# then f(x" ) Á f(x# ), so f(x" ) Á f(x# ) and therefore g(x" ) Á g(x# ). Therefore g(x) is one-to-one as well. 56. The function h(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x" Á x# then f(x" ) Á f(x# ), so " " f(x" ) Á f(x# ) , and therefore h(x" ) Á h(x# ). 57. The composite is one-to-one also. The reasoning: If x" Á x# then g(x" ) Á g(x# ) because g is one-to-one. Since g(x" ) Á g(x# ), we also have f(g(x" )) Á f(g(x# )) because f is one-to-one. Thus, f ‰ g is one-to-one because x" Á x# Ê f(g(x" )) Á f(g(x# )). 58. Yes, g must be one-to-one. If g were not one-to-one, there would exist numbers x" Á x# in the domain of g with g(x" ) œ g(x# ). For these numbers we would also have f(g(x" )) œ f(g(x# )), contradicting the assumption that f ‰ g is one-to-one. 59. (g ‰ f)(x) œ x Ê g(f(x)) œ x Ê gw (f(x))f w (x) œ 1 60. W(a) œ 'f(a) 1 ’af
" (y)b  a# “ dy œ 0 œ 'a 21x[f(a)  f(x)] dx œ S(a); Ww (t) œ 1’af
" (f(t))b  a# “ f w (t) f(a)

a

#

#

œ 1 at#  a# b f w (t); also S(t) œ 21f(t)'a x dx  21'a xf(x) dx œ c1f(t)t#  1f(t)a# d  21'a xf(x) dx Ê Sw (t) t

t

t

œ 1t# f w (t)  21tf(t)  1a# f w (t)  21tf(t) œ 1 at#  a# b f w (t) Ê Ww (t) œ Sw (t). Therefore, W(t) œ S(t) for all t − [aß b]. 61-68. Example CAS commands: Maple: with( plots );#63 f := x -> sqrt(3*x-2); domain := 2/3 .. 4; x0 := 3; Df := D(f);

# (a)

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 7.1 Inverse Functions and Their Derivatives

395

plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[1,3], legend=["y=f(x)","y=f '(x)"], title="#61(a) (Section 7.1)" ); q1 := solve( y=f(x), x ); # (b) g := unapply( q1, y ); m1 := Df(x0); # (c) t1 := f(x0)+m1*(x-x0); y=t1; m2 := 1/Df(x0); # (d) t2 := g(f(x0)) + m2*(x-f(x0)); y=t2; domaing := map(f,domain); # (e) p1 := plot( [f(x),x], x=domain, color=[pink,green], linestyle=[1,9], thickness=[3,0] ): p2 := plot( g(x), x=domaing, color=cyan, linestyle=3, thickness=4 ): p3 := plot( t1, x=x0-1..x0+1, color=red, linestyle=4, thickness=0 ): p4 := plot( t2, x=f(x0)-1..f(x0)+1, color=blue, linestyle=7, thickness=1 ): p5 := plot( [ [x0,f(x0)], [f(x0),x0] ], color=green ): display( [p1,p2,p3,p4,p5], scaling=constrained, title="#63(e) (Section 7.1)" ); Mathematica: (assigned function and values for a, b, and x0 may vary) If a function requires the odd root of a negative number, begin by loading the RealOnly package that allows Mathematica to do this. See section 2.5 for details. (x  1) œ '0 tx e t dt œ lim c
tx e t d b0  x '0 tx 1 e t dt œ lim ˆ
beb  0x e! ‰  x>(x) œ x>(x) bÄ_

(c) >(n  1) œ n>(n) œ n!: n œ 0: >(0  1) œ >(1) œ 0!; n œ k: Assume >(k  1) œ k! n œ k  1: >(k  1  1) œ (k  1) >(k  1) œ (k  1)k! œ (k  1)! Thus, >(n  1) œ n>(n) œ n! for every positive integer n.

x

bÄ_

for some k  0; from part (b) induction hypothesis definition of factorial

x n n 32. (a) >(x) ¸ ˆ xe ‰ É 2x1 and n>(n) œ n! Ê n! ¸ n ˆ ne ‰ É 2n1 œ ˆ ne ‰ È2n1

(b)

533

n 10 20 30 40 50 60

ˆ ne ‰n È2n1 3598695.619 2.4227868 ‚ 10") 2.6451710 ‚ 10$# 8.1421726 ‚ 10%( 3.0363446 ‚ 10'% 8.3094383 ‚ 10)"

calculator 3628800 2.432902 ‚ 10") 2.652528 ‚ 10$# 8.1591528 ‚ 10%( 3.0414093 ‚ 10'% 8.3209871 ‚ 10)"

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

534

Chapter 8 Techniques of Integration

(c)

ˆ ne ‰n È2n1 3598695.619

n 10 ÐÑ

cos 3x

2e2x

Ð
Ñ

1 3 sin

4e2x

ÐÑ


19 cos 3x

33. e2x

Iœ

e2x 3

sin 3x 

2e2x 9

13 9

Iœ

e2x 9

(3 sin 3x  2 cos 3x) Ê I œ

e2x 13

(3 sin 3x  2 cos 3x)  C

sin 4x

3e3x

Ð
Ñ


4" cos 4x

9e3x

ÐÑ

"
16 sin 4x

3x

I œ
e4 cos 4x  35.

calculator 3628800

3x

cos 3x
49 I Ê

ÐÑ

34. e3x

ˆ ne ‰n È2n1 e1Î12n 3628810.051

3e3x 16

sin 4x


I Ê

9 16

sin 3x

ÐÑ

sin x

3 cos 3x

Ð
Ñ


cos x


9 sin 3x

ÐÑ


sin x

Iœ

25 16

e3x 16

(3 sin 4x
4 cos 4x) Ê I œ

e3x 25

(3 sin 4x
4 cos 4x)  C

I œ
sin 3x cos x  3 cos 3x sin x  9I Ê
8I œ
sin 3x cos x  3 cos 3x sin x Ê I œ sin 3x cos x83 cos 3x sin x  C 36.

cos 5x

ÐÑ

sin 4x


sin 5x

Ð
Ñ


"4 cos 4x


25cos 5x

ÐÑ

"
16 sin 4

I œ
"4 cos 5x cos 4x
Ê Iœ

" 9

sin 5x sin 4x 

9 I Ê
16 I œ
"4 cos 5x cos 4x


5 16

sin 5x sin 4x

sin bx

aeax

Ð
Ñ


"b cos bx

a# eax

ÐÑ


b"# sin bx

ax

I œ
eb cos bx  Ê Iœ

25 16

(4 cos 5x cos 4x  5 sin 5x sin 4x)  C ÐÑ

37. eax

5 16

eax a#
b#

aeax b#

sin bx


a# b#

I Ê Ša

#


b# b# ‹ I

œ

eax b#

(a sin bx
b cos bx)

(a sin bx
b cos bx)  C

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Chapter 8 Additional and Advanced Exercises ÐÑ

38. eax

cos bx

aeax

Ð
Ñ

" b

a# eax

ÐÑ


b"# cos bx

Iœ

eax b

sin bx 

Ê Iœ

" x

cos bx


a# b#

I Ê Ša

#


b# b# ‹ I

eax b#

œ

(a cos bx  b sin bx)

(a cos bx  b sin bx)  C

eax a#
b#

39. ln (ax)

aeax b#

sin bx

ÐÑ

1

Ð
Ñ

x

I œ x ln (ax)
' ˆ "x ‰ x dx œ x ln (ax)
x  C 40. ln (ax) " x

Iœ

" 3

ÐÑ

x#

Ð
Ñ

" 3

x$

x$ ln (ax)
' ˆ "x ‰ Š x3 ‹ dx œ $

41.

' 1 dxsin x œ '

42.

' 1
sin dxx
cos x œ '

œ'

2 dz ‹ 1 z# 1  Š 2z # ‹ 1 z

Š

2 dz (1  z)#

" 3

x$ ln (ax)
9" x$  C

œ

2 1z

œ'

2 dz ‹ 1 z# 2z 1
z# 1
Š
‹ 1 z# 1 z#

Š

Cœ

2 1  tan ˆ #x ‰

2 dz 1
z#
2z
1  z#

C

œ'

œ ln k1  zk  C

dz 1
z

œ ln ¸tan ˆ x# ‰  1¸  C 43.

'01Î2 1
dxsin x œ '01

44.

'11ÎÎ32

45.

) '01Î2 2
dcos '1 ) œ 0

œ

46.

dx 1  cos x

1 3È 3

'12Î12Î3

œ

" È2

1

Š

2 dz (1
z)#

œ '1ÎÈ3

2 dz ‹ 1 z# 1
z# 1Š ‹ 1 z#

Š

2 dz ‹ 1 z# 1
z# 2
Š ‹ 1 z#

cos ) d) sin ) cos )
sin )

È$ z# “ 4 "

' sin t dt cos t œ ' œ

'11ÎÈ3

œ '0

1

œ '0

1

"

œ

1
2 z ‘ ! œ
(1
2) œ 1

dz z#

" œ

1z ‘ "ÎÈ$ œ È3
1

2 dz 2
2z#
1  z#

œ '0

1

œ

2 dz z#
3

2 È3

’tan"

z È3 “

" !

œ

2 È3

tan"

" È3

È 31 9

œ ’ "# ln z


47.

œ

2 dz ‹ 1 z# 2z 1
Š ‹ 1 z#

Š

È3

œ '1

È3

z# ‹ Š 2 dz# ‹ 1
z# 1
z

Š1 Ô 2z Š1

z# ‹

× 2z #
Š 1
z# ‹Ø Õ Š1
z# ‹

œ '1

œ Š "# ln È3
43 ‹
ˆ0
4" ‰ œ

2 dz ‹ 1 z# 2z 1
z#  Š ‹ 1 z# 1 z#

Š

œ'

2 dz 2z  1
z#

œ'

2 a1  z# b dz 2z  2z$
2z
2z$

ln 3 4

1  z# 2z

œ

" 4

œ

" È2

1 2 ln ¹ zz
¹C
1
È2

(ln 3
2) œ

" #

dz

" #




2 dz (z
1)#  2

È3

œ '1

Šln È3
1‹

È

tan ˆ t ‰
1  È2

ln º tan ˆ #t ‰
1
È2 º  C #

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

535

536 48.

Chapter 8 Techniques of Integration

t dt ' 1cos cos 'Š t œ

œ'

49.

a1  z# b dz a1
z # b z #

1 z# ‹ Š 2 dz# ‹ 1
z# 1
z # 1  Š 1 z# ‹ 1
z

œ'

dz z # a1
z # b

œ'
'

' sec ) d) œ ' cosd) ) œ ' Š

2 dz ‹ 1 z# # Š 1
z# ‹ 1 z

2 a1  z# b dz a1
z # b #  a 1
z # b a 1  z # b dz 1
z#

œ'

dz z#

œ'

2 dz 1  z#

œ'

œ ln k1  zk
ln k1
zk  C œ ln »

50.

' csc ) d) œ ' sind)) œ ' Š

2 dz ‹ 1 z# 2z Š ‹ 1 z#

1
tan Š )# ‹


2'

1  tan Š )# ‹ »

œ'

dz z

œ'

dz z#
1

2 dz (1
z)(1  z)

2 a1  z# b dz a1
z # b a 1
z #  1
z # b

œ
"z
2 tan" z  C œ
cot ˆ #t ‰
t  C œ'

dz 1
z

'

dz 1z

C

œ ln kzk  C œ ln ¸tan #) ¸  C

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

CHAPTER 9 FIRST-ORDER DIFFERENTIAL EQUATIONS 9.1 SOLUTIONS, SLOPE FIELDS AND EULER'S METHOD 1. y w œ x
y Ê slope of 0 for the line y œ x. For x, y  0, y w œ x
y Ê slope  0 in Quadrant I. For x, y  0, y w œ x
y Ê slope  0 in Quadrant III. For kyk  kxk, y  0, x  0, y w œ x
y Ê slope  0 in Quadrant II above y œ x. For kyk  kxk, y  0, x  0, y w œ x
y Ê slope  0 in Quadrant II below y œ x. For kyk  kxk, x  0, y  0, y w œ x
y Ê slope  0 in Quadrant IV above y œ x. For kyk  kxk, x  0, y  0, y w œ x
y Ê slope  0 in Quadrant IV below y œ x. All of the conditions are seen in slope field (d). 2. y w œ y
1 Ê slope is constant for a given value of y, slope is 0 for y œ 1, slope is positive for y  1 and negative for y  1. These characteristics are evident in slope field (c).

3. y w œ  xy Ê slope œ 1 on y œ x and 1 on y œ x. y w œ  xy Ê slope œ 0 on the y-axis, excluding a0, 0b, 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.

4. y w œ y2  x2 Ê slope is 0 for y œ x and for y œ x. For kyk  kxk slope is positive and for kyk  kxk slope is negative. Field (b) has these characteristics.

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

538

Chapter 9 First-Order Differential Equations

5.

6.

7. y œ 1
'1 at  yatbbdt Ê x

8. y œ '1 1t dt Ê x

dy dx

dy dx

œ x  yaxb; ya1b œ 1
'1 at  yatbbdt œ 1; 1

œ x1 ; ya1b œ '1 1t dt œ 0; 1

9. y œ 2  '0 a1
yatbbsin t dt Ê x

dy dx

dy dx

dy dx

œ x  y, ya1b œ 1

œ x1 , ya1b œ 0

œ a1
yaxbbsin x; ya0b œ 2  '0 a1
yatbbsin t dt œ 2; 0

dy dx

œ a1
ybsin x,

ya0b œ 2 10. y œ 1
'0 yatb dt Ê x

dy dx

œ yaxb; ya0b œ 1
'0 yatb dt œ 1; 0


1 ‰ # (.5)

11. y" œ y!
Š1 

y! x! ‹

dx œ 1
ˆ1 

y# œ y "
Š 1 

y" x" ‹

dx œ 0.25
ˆ1 

y$ œ y #
Š 1 

y# x# ‹

dx œ 0.3
ˆ1 

dy dx


ˆ x" ‰ y œ 1 Ê P(x) œ

Ê yœ

" x

" x

' x † 1 dx œ x" Š x#

Ê y(3.5) œ

#

3.5 #



4 3.5

œ

4.25 7

, Q(x) œ 1 Ê

œ y, ya0b œ 1

œ 0.25,


0.25 ‰ (.5) 2.5

0.3 ‰ 3 (.5)

dy dx

œ 0.3,

œ 0.75;

' P(x) dx œ ' x" dx œ ln kxk œ ln x, x  0 Ê v(x) œ eln x œ x


C‹ ; x œ 2, y œ 1 Ê 1 œ 1


C 2

Ê C œ 4 Ê y œ

x #



4 x

¸ 0.6071

12. y" œ y!
x! (1  y! ) dx œ 0
1(1  0)(.2) œ .2, y# œ y"
x" (1  y" ) dx œ .2
1.2(1  .2)(.2) œ .392, y$ œ y#
x# (1  y# ) dx œ .392
1.4(1  .392)(.2) œ .5622; dy 1
y

œ x dx Ê  ln k1  yk œ

x# #


C; x œ 1, y œ 0 Ê  ln 1 œ

" #

#


C Ê C œ  #" Ê ln k1  yk œ  x#


" #

#

Ê y œ 1  ea1
x bÎ2 Ê y(1.6) ¸ .5416 13. y" œ y!
(2x! y!
2y! ) dx œ 3
[2(0)(3)
2(3)](.2) œ 4.2, y# œ y"
(2x" y"
2y" ) dx œ 4.2
[2(.2)(4.2)
2(4.2)](.2) œ 6.216, y$ œ y#
(2x# y#
2y# ) dx œ 6.216
[2(.4)(6.216)
2(6.216)](.2) œ 9.6969; dy dx

œ 2y(x
1) Ê

dy y

œ 2(x
1) dx Ê ln kyk œ (x
1)#
C; x œ 0, y œ 3 Ê ln 3 œ 1
C Ê C œ ln 3  1

Ê ln y œ (x
1)#
ln 3  1 Ê y œ eÐx1Ñ ln 3
1 œ eln 3 ex 2x œ 3exÐx2Ñ Ê y(.6) ¸ 14.2765 #

#

14. y" œ y!
y#! (1
2x! ) dx œ 1
1# [1
2(1)](.5) œ .5, y# œ y"
y#" (1
2x" ) dx œ .5
(.5)# [1
2(.5)](.5) œ .5, y$ œ y#
y## (1
2x# ) dx œ .5
(.5)# [1
2(0)](.5) œ .625; dy y#

œ (1
2x) dx Ê  y" œ x
x#
C; x œ 1, y œ 1 Ê 1 œ 1
(1)#
C Ê C œ 1 Ê

Ê yœ

" 1
x
x#

Ê y(.5) œ

" 1
.5
(.5)#

œ4

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

" y

œ 1  x  x#

Section 9.1 Solutins, Slope Fields and Euler's Method

539

#

15. y" œ y!
2x! ex! dx œ 2
2(0)(.1) œ 2, # # y# œ y"
2x" ex" dx œ 2
2(.1) eÞ1 (.1) œ 2.0202, # # y$ œ y#
2x# ex# dx œ 2.0202
2(.2) eÞ2 (.1) œ 2.0618, # # # # dy œ 2xex dx Ê y œ ex
C; y(0) œ 2 Ê 2 œ 1
C Ê C œ 1 Ê y œ ex
1 Ê y(.3) œ eÞ3
1 ¸ 2.0942 16. y" œ y!
ay! ex! b dx œ 2
a2 † e0 b (.5) œ 3, y2 œ y1
ay1 ex1 b dx œ 3
a3 † e0.5 b (.5) œ 5.47308, y3 œ y2
ay2 ex2 b dx œ 5.47308
a5.47308 † e1.0 b (.5) œ 12.9118, dy dx

œ y ex Ê

Ê y œ 2ee 17. y" y# y$ y% y& dy y

x

œ ex dx Ê lnlyl œ ex
C; x œ 0, y œ 2 Ê ln 2 œ 1
C Ê C œ ln 2  1 Ê lnlyl œ ex
ln 2  1

dy y


"

Ê ya1.5b œ 2ee

1.5


"

¸ 65.0292

œ 1
1(.2) œ 1.2, œ 1.2
(1.2)(.2) œ 1.44, œ 1.44
(1.44)(.2) œ 1.728, œ 1.728
(1.728)(.2) œ 2.0736, œ 2.0736
(2.0736)(.2) œ 2.48832; œ dx Ê ln y œ x
C" Ê y œ Cex ; y(0) œ 1 Ê 1 œ Ce! Ê C œ 1 Ê y œ ex Ê y(1) œ e ¸ 2.7183

18. y" œ 2
ˆ 21 ‰ (.2) œ 2.4, ‰ y# œ 2.4
ˆ 2.4 1.2 (.2) œ 2.8, ‰ y$ œ 2.8
ˆ 2.8 1.4 (.2) œ 3.2, 3.2 y% œ 3.2
ˆ 1.6 ‰ (.2) œ 3.6, ‰ y& œ 3.6
ˆ 3.6 1.8 (.2) œ 4; dy y

œ

dx x

Ê ln y œ ln x
C Ê y œ kx; y(1) œ 2 Ê 2 œ k Ê y œ 2x Ê y(2) œ 4 #

19. y" œ 1
’ (
È1) “ (.5) œ .5, 1 #


.5) y# œ .5
’ (È “ (.5) œ .39794, 1.5 #

y$ œ .39794
’ (
.39794) “ (.5) œ .34195, È2 #

y% œ .34195
’ (
.34195) È2.5 “ (.5) œ .30497, y& œ .27812, y' œ .25745, y( œ .24088, y) œ .2272; dy " dx È y# œ Èx Ê  y œ 2 x
C; y(1) œ 1 Ê 1 œ 2
C Ê C œ 1 Ê y œ

" 1
#È x

Ê y(5) œ

" 1
#È 5

20. y" œ 1
a0 † sin 1b ˆ "3 ‰ œ 1, y# œ 1
ˆ "3 † sin 1‰ ˆ 3" ‰ œ 1.09350, y$ œ 1.09350
ˆ 23 † sin 1.09350‰ ˆ 3" ‰ œ 1.29089, y% œ 1.29089
ˆ 33 † sin 1.29089‰ ˆ 3" ‰ œ 1.61125, y& œ 1.61125
ˆ 43 † sin 1.61125‰ ˆ 3" ‰ œ 2.05533,

y' œ 2.05533
ˆ 53 † sin 2.05533‰ ˆ 3" ‰ œ 2.54694; y w œ x sin y Ê csc y dy œ x dx Ê lnlcsc y
cot yl œ 12 x2
C Ê csc y
cot y œ e
2 x C œ Ce
2 x 1 2

Ê

1  cos y sin y

1 2

2 2 2 œ Ce
2 x Ê cotˆ 2y ‰ œ Ce
2 x ; ya0b œ 1 Ê cotˆ 21 ‰ œ Ce0 œ C Ê cotˆ 2y ‰ œ cotˆ 21 ‰e
2 x 1

1

2 Ê y œ 2 cot
1 Šcotˆ 12 ‰e
2 x ‹, ya2b œ 2 cot
1 ˆcotˆ 12 ‰e
2 ‰ œ 2.65591 1

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

1

¸ .2880

540

Chapter 9 First-Order Differential Equations

21. y œ 1  x
a1
x0
y0 bex
x0 Ê yax0 b œ 1  x0
a1
x0
y0 bex0
x0 œ 1  x0
a1
x0
y0 ba1b œ y0 dy dx

œ 1
a1
x0
y0 bex
x0 Ê y œ 1  x
a1
x0
y0 bex
x0 œ

dy dx

xÊ

dy dx

œ x
y

22. y w œ faxb, yax0 b œ y0 Ê y œ 'x fatbdt
C, yax0 b œ 'x fatbdt
C œ C Ê C œ y0 Ê y œ 'x fatbdt
y0 x

x0

x

0

0

0

23-34. Example CAS commands: Maple: ode := diff( y(x), x ) = y(x); icA := [0, 1]; icB := [0, 2]; icC := [0,-1]; DEplot( ode, y(x), x=0..2, [icA,icB,icC], arrows=slim, linecolor=blue, title="#23 (Section 9.1)" ); Mathematica: To plot vector fields, you must begin by loading a graphics package. x*y+y*z; g1 := (x,y,z) -> x^2+y^2-2; g2 := (x,y,z) -> x^2+z^2-2; h := unapply( f(x,y,z)-lambda[1]*g1(x,y,z)-lambda[2]*g2(x,y,z), (x,y,z,lambda[1],lambda[2]) ); hx := diff( h(x,y,z,lambda[1],lambda[2]), x ); hy := diff( h(x,y,z,lambda[1],lambda[2]), y ); hz := diff( h(x,y,z,lambda[1],lambda[2]), z ); hl1 := diff( h(x,y,z,lambda[1],lambda[2]), lambda[1] ); hl2 := diff( h(x,y,z,lambda[1],lambda[2]), lambda[2] ); sys := { hx=0, hy=0, hz=0, hl1=0, hl2=0 }; q1 := solve( sys, {x,y,z,lambda[1],lambda[2]} ); q2 := map(allvalues,{q1}); for p in q2 do eval( [x,y,z,f(x,y,z)], p ); ``=evalf(eval( [x,y,z,f(x,y,z)], p )); end do;

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

# (a) #(b)

# (c) # (d)

is

œ1

Section 14.9 Taylor's Formula for Two Variables Mathematica: (assigned functions will vary) Clear[x, y, z, lambda1, lambda2] f[x_,y_,z_]:= x y  y z g1[x_,y_,z_]:= x2  y2
2 g2[x_,y_,z_]:= x2  z2
2 h = f[x, y, z]
lambda1 g1[x, y, z]
lambda2 g2[x, y, z]; hx= D[h, x]; hy= D[h, y]; hz= D[h,z]; hL1=D[h, lambda1]; hL2= D[h, lambda2]; critical=Solve[{hx==0, hy==0, hz==0, hL1==0, hL2==0, g1[x,y,z]==0, g2[x,y,z]==0}, {x, y, z, lambda1, lambda2}]//N {{x, y, z}, f[x, y, z]}/.critical 14.9 TAYLOR'S FORMULA FOR TWO VARIABLES 1. f(xß y) œ xey Ê fx œ ey , fy œ xey , fxx œ 0, fxy œ ey , fyy œ xey Ê f(xß y) ¸ f(0ß 0)  xfx (0ß 0)  yfy (0ß 0)  "# cx# fxx (0ß 0)  2xyfxy (0ß 0)  y# fyy (0ß 0)d œ 0  x † 1  y † 0  "# ax# † 0  2xy † 1  y# † 0b œ x  xy quadratic approximation;

fxxx œ 0, fxxy œ 0, fxyy œ ey , fyyy œ xey Ê f(xß y) ¸ quadratic  "6 cx$ fxxx (!ß !)  3x# yfxxy (0ß 0)  3xy# fxyy (!ß !)  y$ fyyy (0ß 0)d

œ x  xy  "6 ax$ † 0  3x# y † 0  3xy# † 1  y$ † 0b œ x  xy  "# xy# , cubic approximation

2. f(xß y) œ ex cos y Ê fx œ ex cos y, fy œ
ex sin y, fxx œ ex cos y, fxy œ
ex sin y, fyy œ
ex cos y Ê f(xß y) ¸ f(0ß 0)  xfx (0ß 0)  yfy (!ß 0)  "# cx# fxx (!ß !)  2xyfxy (!ß !)  y# fyy (0ß 0)d

œ 1  x † 1  y † 0  "# cx# † 1  2xy † 0  y# † (
1)d œ 1  x  "# ax#
y# b , quadratic approximation;

fxxx œ ex cos y, fxxy œ
ex sin y, fxyy œ
ex cos y, fyyy œ ex sin y Ê f(xß y) ¸ quadratic  "6 cx$ fxxx (0ß 0)  3x# yfxxy (!ß 0)  3xy# fxyy (0ß 0)  y$ fyyy (0ß 0)d œ 1  x  "# ax#
y# b  6" cx$ † 1  3x# y † 0  3xy# † (
1)  y$ † 0d œ 1  x  "# ax#
y# b  6" ax$
3xy# b , cubic approximation

3. f(xß y) œ y sin x Ê fx œ y cos x, fy œ sin x, fxx œ
y sin x, fxy œ cos x, fyy œ 0 Ê f(xß y) ¸ f(0ß 0)  xfx (0ß 0)  yfy (!ß 0)  "# cx# fxx (0ß 0)  2xyfxy (0ß 0)  y# fyy (0ß 0)d œ 0  x † 0  y † 0  "# ax# † 0  2xy † 1  y# † 0b œ xy, quadratic approximation;

fxxx œ
y cos x, fxxy œ
sin x, fxyy œ 0, fyyy œ 0 Ê f(xß y) ¸ quadratic  "6 cx$ fxxx (0ß 0)  3x# yfxxy (!ß 0)  3xy# fxyy (0ß 0)  y$ fyyy (0ß 0)d œ xy  "6 ax$ † 0  3x# y † 0  3xy# † 0  y$ † 0b œ xy, cubic approximation

4. f(xß y) œ sin x cos y Ê fx œ cos x cos y, fy œ
sin x sin y, fxx œ
sin x cos y, fxy œ
cos x sin y, fyy œ
sin x cos y Ê f(xß y) ¸ f(0ß 0)  xfx (0ß 0)  yfy (0ß 0)  "# cx# fxx (0ß 0)  2xyfxy (0ß 0)  y# fyy (0ß 0)d œ 0  x † 1  y † 0  "# ax# † 0  2xy † 0  y# † 0b œ x, quadratic approximation;

fxxx œ
cos x cos y, fxxy œ sin x sin y, fxyy œ
cos x cos y, fyyy œ sin x sin y Ê f(xß y) ¸ quadratic  "6 cx$ fxxx (0ß 0)  3x# yfxxy (!ß 0)  3xy# fxyy (0ß 0)  y$ fyyy (0ß 0)d

œ x  "6 cx$ † (
1)  3x# y † 0  3xy# † (
1)  y$ † 0d œ x
6" ax$  3xy# b, cubic approximation

5. f(xß y) œ ex ln (1  y) Ê fx œ ex ln (1  y), fy œ

ex 1
y

, fxx œ ex ln (1  y), fxy œ

ex 1
y

Ê f(xß y) ¸ f(0ß 0)  xfx (0ß 0)  yfy (0ß 0)  "# cx# fxx (0ß 0)  2xyfxy (0ß 0)  y# fyy (0ß 0)d

œ 0  x † 0  y † 1  "# cx# † 0  2xy † 1  y# † (
1)d œ y  "# a2xy
y# b , quadratic approximation; fxxx œ ex ln (1  y), fxxy œ

ex 1
y

x

, fxyy œ
(1
e y)# , fyyy œ

x

, fyy œ
(1
e y)#

2ex (1
y)$

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

857

858

Chapter 14 Partial Derivatives Ê f(xß y) ¸ quadratic  "6 cx$ fxxx (0ß 0)  3x# yfxxy (!ß 0)  3xy# fxyy (0ß 0)  y$ fyyy (0ß 0)d

œ y  "2 a2xy
y# b  6" cx$ † 0  3x# y † 1  3xy# † (
1)  y$ † 2d

œ y  "# a2xy
y# b  6" a3x# y
3xy#  2y$ b , cubic approximation 4 2 (2x
y
1)# , fxy œ (2x
y
1)# , " # # fyy œ (2x
" y
1)# Ê f(xß y) ¸ f(0ß 0)  xfx (0ß 0)  yfy (0ß 0)  # cx fxx (0ß 0)  2xyfxy (0ß 0)  y fyy (0ß 0)d œ 0  x † 2  y † 1  "# cx# † (
4)  2xy † (
2)  y# † (
1)d œ 2x  y  "# a
4x#
4xy
y# b œ (2x  y)
"# (2x  y)# , quadratic approximation; fxxx œ (2x
16y
1)$ , fxxy œ (2x
8y
1)$ , fxyy œ (2x
4y
1)$ , fyyy œ (2x
2y
1)$ Ê f(xß y) ¸ quadratic  "6 cx$ fxxx (0ß 0)  3x# yfxxy (!ß 0)  3xy# fxyy (0ß 0)  y$ fyyy (0ß 0)d œ (2x  y)
"# (2x  y)#  6" ax$ † 16  3x# y † 8  3xy# † 4  y$ † 2b œ (2x  y)
"# (2x  y)#  3" a8x$  12x# y  6xy#  y# b œ (2x  y)
"# (2x  y)#  3" (2x  y)$ , cubic approximation

6. f(xß y) œ ln (2x  y  1) Ê fx œ

2 2x
y
1

, fy œ

" #x
y
1

, fxx œ

7. f(xß y) œ sin ax#  y# b Ê fx œ 2x cos ax#  y# b , fy œ 2y cos ax#  y# b , fxx œ 2 cos ax#  y# b
4x# sin ax#  y# b , fxy œ
4xy sin ax#  y# b , fyy œ 2 cos ax#  y# b
4y# sin ax#  y# b Ê f(xß y) ¸ f(0ß 0)  xfx (0ß 0)  yfy (0ß 0)  "# cx# fxx (0ß 0)  2xyfxy (0ß 0)  y# fyy (0ß 0)d œ 0  x † 0  y † 0  "# ax# † 2  2xy † 0  y# † 2b œ x#  y# , quadratic approximation;

fxxx œ
12x sin ax#  y# b
8x$ cos ax#  y# b , fxxy œ
4y sin ax#  y# b
8x# y cos ax#  y# b , fxyy œ
4x sin ax#  y# b
8xy# cos ax#  y# b , fyyy œ
12y sin ax#  y# b
8y$ cos ax#  y# b Ê f(xß y) ¸ quadratic  "6 cx$ fxxx (0ß 0)  3x# yfxxy (!ß 0)  3xy# fxyy (0ß 0)  y$ fyyy (0ß 0)d œ x#  y#  "6 ax$ † 0  3x# y † 0  3xy# † 0  y$ † 0b œ x#  y# , cubic approximation

8. f(xß y) œ cos ax#  y# b Ê fx œ
2x sin ax#  y# b , fy œ
2y sin ax#  y# b , fxx œ
2 sin ax#  y# b
4x# cos ax#  y# b , fxy œ
4xy cos ax#  y# b , fyy œ
2 sin ax#  y# b
4y# cos ax#  y# b Ê f(xß y) ¸ f(0ß 0)  xfx (0ß 0)  yfy (0ß 0)  "# cx# fxx (0ß 0)  2xyfxy (0ß 0)  y# fyy (0ß 0)d œ 1  x † 0  y † 0  "# cx# † 0  2xy † 0  y# † 0d œ 1, quadratic approximation;

fxxx œ
12x cos ax#  y# b  8x$ sin ax#  y# b , fxxy œ
4y cos ax#  y# b  8x# y sin ax#  y# b , fxyy œ
4x cos ax#  y# b  8xy# sin ax#  y# b , fyyy œ
12y cos ax#  y# b  8y$ sin ax#  y# b Ê f(xß y) ¸ quadratic  "6 cx$ fxxx (0ß 0)  3x# yfxxy (!ß 0)  3xy# fxyy (0ß 0)  y$ fyyy (0ß 0)d œ 1  "6 ax$ † 0  3x# y † 0  3xy# † 0  y$ † 0b œ 1, cubic approximation

9. f(xß y) œ

" 1xy

Ê fx œ

" (1  x  y)#

œ fy , fxx œ

Ê f(xß y) ¸ f(0ß 0)  xfx (0ß 0)  yfy (0ß 0) 

2 (1  x  y)$ " # # cx fxx (0ß 0)

œ fxy œ fyy  2xyfxy (0ß 0)  y# fyy (0ß 0)d

œ 1  x † 1  y † 1  "# ax# † 2  2xy † 2  y# † 2b œ 1  (x  y)  ax#  2xy  y# b

œ 1  (x  y)  (x  y)# , quadratic approximation; fxxx œ Ê f(xß y) ¸ quadratic  "6 cx$ fxxx (0ß 0)  3x# yfxxy (!ß 0) 

6 œ fxxy œ fxyy œ fyyy (1  x  y)% # 3xy fxyy (0ß 0)  y$ fyyy (0ß 0)d $

œ 1  (x  y)  (x  y)#  "6 ax$ † 6  3x# y † 6  3xy# † 6  y † 6b

œ 1  (x  y)  (x  y)#  ax$  3x# y  3xy#  y$ b œ 1  (x  y)  (x  y)#  (x  y)$ , cubic approximation 10. f(xß y) œ fxy œ

" 1  x  y
xy

1 ("  x  y
xy)#

Ê fx œ

, fyy œ

1y (1  x  y
xy)#

, fy œ

1x ("  x  y
xy)#

, fxx œ

2(1  y)# (1  x  y
xy)$

,

2("  x)# (1  x  y
xy)$

Ê f(xß y) ¸ f(0ß 0)  xfx (0ß 0)  yfy (0ß 0)  "# cx# fxx (0ß 0)  2xyfxy (0ß 0)  y# fyy (0ß 0)d

œ 1  x † 1  y † 1  "# ax# † 2  2xy † 1  y# † 2b œ 1  x  y  x#  xy  y# , quadratic approximation;

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Section 14.10 Partial Derivatives with Constrained Variables fxxx œ

6(1  y)$ (1  x  y
xy)%

, fxxy œ

[4(1  x  y
xy)
6(1  y)(1  x)](1  y) (1  x  y
xy)%

,

$

[4(1  x  y
xy)
6(1  x)(1  y)](1  x)  x) , fyyy œ (1 6(1 (1  x  y
xy)% x  y
xy)% Ê f(xß y) ¸ quadratic  "6 cx$ fxxx (0ß 0)  3x# yfxxy (!ß 0)  3xy# fxyy (0ß 0) œ 1  x  y  x#  xy  y#  "6 ax$ † 6  3x# y † 2  3xy# † 2  y$ † 6b # # $ # # $

fxyy œ

 y$ fyyy (0ß 0)d

œ 1  x  y  x  xy  y  x  x y  xy  y , cubic approximation 11. f(xß y) œ cos x cos y Ê fx œ
sin x cos y, fy œ
cos x sin y, fxx œ
cos x cos y, fxy œ sin x sin y, fyy œ
cos x cos y Ê f(xß y) ¸ f(0ß 0)  xfx (0ß 0)  yfy (0ß 0)  "# cx# fxx (0ß 0)  2xyfxy (0ß 0)  y# fyy (0ß 0)d œ 1  x † 0  y † 0  "# cx# † (
1)  2xy † 0  y# † (
1)d œ 1


x# #




y# #

, quadratic approximation. Since all partial

derivatives of f are products of sines and cosines, the absolute value of these derivatives is less than or equal to 1 Ê E(xß y) Ÿ "6 c(0.1)$  3(0.1)$  3(0.1)$  0.1)$ d Ÿ 0.00134. 12. f(xß y) œ ex sin y Ê fx œ ex sin y, fy œ ex cos y, fxx œ ex sin y, fxy œ ex cos y, fyy œ
ex sin y Ê f(xß y) ¸ f(0ß 0)  xfx (0ß 0)  yfy (0ß 0)  "# cx# fxx (0ß 0)  2xyfxy (0ß 0)  y# fyy (0ß 0)d

œ 0  x † 0  y † 1  "# ax# † 0  2xy † 1  y# † 0b œ y  xy , quadratic approximation. Now, fxxx œ ex sin y,

fxxy œ ex cos y, fxyy œ
ex sin y, and fyyy œ
ex cos y. Since kxk Ÿ 0.1, kex sin yk Ÿ ke0Þ1 sin 0.1k ¸ 0.11 and kex cos yk Ÿ ke0Þ1 cos 0.1k ¸ 1.11. Therefore, E(xß y) Ÿ "6 c(0.11)(0.1)$  3(1.11)(0.1)$  3(0.11)(0.1)$  (1.11)(0.1)$ d Ÿ 0.000814. 14.10 PARTIAL DERIVATIVES WITH CONSTRAINED VARIABLES 1. w œ x#  y#  z# and z œ x#  y# : Î x œ x(yß z) Ñ y yœy Ä w Ê Š ``wy ‹ œ (a) Œ  Ä z z Ï zœz Ò œ 2x `` xy  2y Ê 0 œ 2x `` xy  2y Ê

`x `y

œ

" #y

`x `z

œ

1 2x

`w `x `x `z

`w `x `x `z

œ 2x `` yx  2y `` yy



`w `y `y `z



`w `z `x `z `z ; `z

œ 0 and

`z `z

œ 2x `` xz  2y `` yz



`w `y `y `z



`w `z `y `z `z ; `z

œ 0 and

`z `z

œ 2x `` xz  2y `` yz

Ê ˆ ``wz ‰y œ (2x) ˆ #"x ‰  (2y)(0)  (2z)(1) œ 1  2z

2. w œ x#  y
z  sin t and x  y œ t: Î xœx Ñ ÎxÑ Ð yœy Ó y Ä Ð Ä w Ê Š ``wy ‹ œ (a) Ó zœz xz ÏzÒ Ït œ x  yÒ ß

`t `y

`z `y

œ 0 and

" Ê ˆ ``wz ‰x œ (2x)(0)  (2y) Š 2y ‹  (2z)(1) œ 1  2z

Î x œ x(yß z) Ñ y yœy Ä w Ê ˆ ``wz ‰y œ (c) Œ  Ä z Ï zœz Ò Ê 1 œ 2x `` xz Ê

`w `z `z `z `y ; `y



z

xœx Ñ x y œ y(xß z) Ä w Ê ˆ ``wz ‰x œ (b) Œ  Ä z Ï zœz Ò `y `z

`w `y `y `y



œ
xy Ê Š ``wy ‹ œ (2x) ˆ
xy ‰  (2y)(1)  (2z)(0) œ
2y  2y œ 0

Î

Ê 1 œ 2y `` yz Ê

`w `x `x `y

`w `x `x `y



`w `y `y `y



`w `z `z `y



`w `t `x `t `y ; `y

œ 0,

`z `y

œ 0, and

œ 1 Ê Š ``wy ‹ œ (2x)(0)  (1)(1)  (
1)(0)  (cos t)(1) œ 1  cos t œ 1  cos (x  y) xßt

Îx œ t
yÑ ÎyÑ Ð yœy Ó z Ä Ð Ä w Ê Š ``wy ‹ œ (b) Ó z z œ zt ÏtÒ Ï tœt Ò ß

Ê

`x `y

œ

`t `y




`y `y

`w `x `x `y



`w `y `y `y



`w `z `z `y



`w `t `z `t `y ; `y

œ 0 and

`t `y

œ0

œ
1 Ê Š ``wy ‹ œ (2x)(
1)  (1)(1)  (
1)(0)  (cos t)(0) œ 1
2at
yb œ 1  2y
2t zßt

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

859

860

Chapter 14 Partial Derivatives

Î xœx Ñ ÎxÑ Ð yœy Ó y Ä Ð Ä w Ê ˆ ``wz ‰x y œ (c) Ó œ z z ÏzÒ Ït œ x  yÒ ß

`w `x `x `z



`w `y `y `z



`w `z `z `z



`w `t `x `t `z ; `z

œ 0 and

`y `z

œ0



`w `z `z `z



`w `t `y `t `z ; `z

œ 0 and

`t `z

œ0



`w `z `z `t



`w `t `x `t `t ; `t

œ 0 and

`z `t

œ0

`w `z `z `t



`w `t `y `t `t ; `t

œ 0 and

`z `t

œ0

Ê ˆ ``wz ‰x y œ (2x)(0)  (1)(0)  (
1)(1)  (cos t)(0) œ
1 ß

Îx œ t
yÑ ÎyÑ Ð yœy Ó z Ä Ð Ä w Ê ˆ ``wz ‰y t œ (d) Ó zœz ÏtÒ Ï tœt Ò ß

`w `x `x `z



`w `y `y `z

Ê ˆ ``wz ‰y t œ (2x)(0)  (1)(0)  (
1)(1)  (cos t)(0) œ
1 ß

Î xœx Ñ ÎxÑ Ð y œ t
xÓ z Ä Ð Ä w Ê ˆ ``wt ‰x z œ (e) Ó zœz ÏtÒ Ï tœt Ò ß

`w `x `x `t



`w `y `y `t

Ê ˆ ``wt ‰x z œ (2x)(0)  (1)(1)  (
1)(0)  (cos t)(1) œ 1  cos t ß

Îx œ t
yÑ ÎyÑ Ð yœy Ó z Ä Ð Ä w Ê ˆ ``wt ‰y z œ (f) Ó zœz ÏtÒ Ï tœt Ò ß

`w `x `x `t



`w `y `y `t



Ê ˆ ``wt ‰y z œ (2x)(1)  (1)(0)  (
1)(0)  (cos t)(1) œ cos t  2x œ cos t  2(t
y) ß

3. U œ f(Pß Vß T) and PV œ nRT Î PœP Ñ P VœV Ä U Ê ˆ ``UP ‰V œ (a) Œ  Ä V PV ÏT œ Ò

`U `P `P `P

`U `V `V `P



`U `T `T `P



œ

`U `P

V ‰ ‰ ˆ ` U ‰ ˆ nR  ˆ `` U V (0)  ` T

nR

œ

`U `P

V ‰  ˆ ``UT ‰ ˆ nR

nRT ÎP œ V Ñ V Ä U Ê ˆ ``UT ‰V œ (b) Œ  Ä VœV T Ï TœT Ò ‰ `U œ ˆ ``UP ‰ ˆ nR V  `T

`U `P `P `T

4. w œ x#  y#  z# and y sin z  z sin x œ 0 Î xœx Ñ x yœy Ä w Ê ˆ ``wx ‰y œ (a) Œ  Ä y Ï z œ z(xß y) Ò (y cos z) Ê

`z `x

 (sin x)

ˆ ``wx ‰ yk (0ß1ß1)

`z `x

 z cos x œ 0 Ê

`z `x

`x `z

`w `x `x `x

œ

`U `V `V `T





`w `y `y `x

z cos x y cos z
sin x . #

`U `T `T `T

‰ ˆ `U ‰ œ ˆ ``UP ‰ ˆ nR V  ` V (0) 

`w `z `y `z `x ; `x



œ 0 and œ

1 1

œ1

œ (2x)

`x `z

 (2y)(0)  (2z)(1)

At (0ß 1ß 1),

`z `x

`U `T

œ (2x)(1)  (2y)(0)  (2z)(1)k Ð0ß1ß1Ñ œ 21

Î x œ x(yß z) Ñ y yœy Ä w Ê ˆ ``wz ‰y œ (b) Œ  Ä z Ï zœz Ò œ (2x)



`y `z  y x) `` xz œ

 2z. Now (sin z)

Ê y cos z  sin x  (z cos

`w `x `x `z



`w `y `y `z

cos z  sin x  (z cos x) 0 Ê

`x `z

œ

y cos z  sin x . z cos x

`w `z `z `z



`x `z

`y `z œ 0 (!ß "ß 1), `` xz œ (11)(1)0

œ 0 and

At

œ

" 1

Ê ˆ ``wz ‰Ck (!,"ß1Ñ œ 2(0) ˆ 1" ‰  21 œ 21 5. w œ x# y#  yz
z$ and x#  y#  z# œ 6 Î xœx Ñ x yœy (a) Œ  Ä Ä w Ê Š ``wy ‹ œ y x Ï z œ z(xß y) Ò

`w `x `x `y



`w `y `y `y



`w `z `z `y

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Section 14.10 Partial Derivatives with Constrained Variables œ a2xy# b (0)  a2x# y  zb (1)  ay
3z# b `x `y

`z `y

œ 0 Ê 2y  (2z)

`z `y

œ0 Ê

`z `y

`z `y

`x `y

`w `x `x `y

`x `y

 a2x# y  zb (1)  ay
3z# b (0) œ a2x# yb

œ 0 Ê (2x)

`x `y

`x `y

 2y œ 0 Ê

`w `y `y `y





`x `y

Now (2x) `z `y

œ
yz . At (wß xß yß z) œ (4ß 2ß 1ß
1),

œ c(2)(2)# (1)  (
1)d  c1
3(
1)# d (1) œ 5 Î x œ x(yß z) Ñ y yœy (b) Œ  Ä Ä w Ê Š ``wy ‹ œ z z Ï zœz Ò œ a2xy# b

`z `y .

œ 2x# y  z  ay
3z# b

 2y  (2z)

`z `y

œ 0 and

œ
"1 œ 1 Ê Š ``wy ‹ ¹

x (4ß2ß1ßc1)

`w `z `z `y

 2x# y  z. Now (2x)

œ
yx . At (wß xß yß z) œ (4ß 2ß 1ß
1),

`x `y

`x `y

 2y  (2z)

œ
"2 Ê Š ``wy ‹ ¹ z

œ (2)(2)(1) ˆ
"# ‰  (2)(2)# (1)  (
1) œ 5

`z `y

œ 0 and

(4ß2ß1ßc1)

#

6. y œ uv Ê 1 œ v œv

`u `y

 u Š
uv

`u `y

u

`u `y ‹

`v `y ;

œ Šv

#

x œ u#  v# and u v

#

‹

`u `y

Ê

`u `y

`x `y

œ

œ 0 Ê 0 œ 2u

`u `y

 2v

`v `y

At (uß v) œ ŠÈ2ß 1‹ ,

v v#  u# .

`v `y

Ê `u `y

œ ˆ
uv ‰ "

œ

#

1#  ŠÈ2‹

`u `y

Ê 1

œ
1

Ê Š `` uy ‹ œ
1 x

r x œ r cos ) 7. Œ  Ä Œ Ê ˆ ``xr ‰) œ cos ); x#  y# œ r# Ê 2x  2y ) y œ r sin )  Ê ``xr œ xr Ê ˆ ``xr ‰ œ È #x #

`y `x

œ 2r

`r `x

and

`y `x

8. If x, y, and z are independent, then ˆ ``wx ‰y z œ ß

`w `x `x `x



`w `y `y `x

`w `z `z `x





`w `t `t `x

œ (2x)(1)  (
2y)(0)  (4)(0)  (1) ˆ ``xt ‰ œ 2x  ``xt . Thus x  2z  t œ 25 Ê 1  0  Ê ˆ ``wx ‰ œ 2x
1. On the other hand, if x, y, and t are independent, then ˆ ``wx ‰ yßz

œ

Ê 1

`r `x

x
y

y

`w `x `x `x

œ 0 Ê 2x œ 2r

`t `x

œ0 Ê

`t `x

œ
1

yßt

 ``wy `` yx  ``wz `` xz  ``wt ``xt œ (2x)(1)  (
2y)(0)  4 `` xz  (1)(0) œ 2 `` xz  0 œ 0 Ê `` xz œ
#" Ê ˆ ``wx ‰yßt œ 2x  4 ˆ
#" ‰ œ 2x
2.

9. If x is a differentiable function of y and z, then f(xß yß z) œ 0 Ê

`f `x `x `x



2x  4

`f `y `y `x



`z `x .

`f `z `z `x

Thus, x  2z  t œ 25

œ0 Ê

`f `x



`f `y `y `x

œ0

Ê Š `` xy ‹ œ
`` f/f/`` yz . Similarly, if y is a differentiable function of x and z, Š `` yz ‹ œ
`` f/f/`` xz and if z is a z

x

differentiable function of x and y, ˆ `` xz ‰y œ
`` f/f/`` xy . Then Š `` xy ‹ Š `` yz ‹ ˆ `` xz ‰y z

œ Š


` f/` y ˆ ` f/` z ‰ ` f/` x ` f/` z ‹
` f/` x Š
` f/` y ‹

10. z œ z  f(u) and u œ xy Ê œ x ˆ1  y

df ‰ du


y ˆx

df ‰ du

`z `x

x

œ
1.

œ1

df ` u du ` x

œ1y

df du ;

also

`z `y

œ0

df ` u du ` y

œx

df du

so that x

`z `x


y

œ 0 and

`x `y

œ0 Ê

`g `y



`z `y

œx

11. If x and y are independent, then g(xß yß z) œ 0 Ê

`g `x `x `y



`g `y `y `y



`g `z `z `y

`y Ê Š `` yz ‹ œ
`` g/ g/` z , as claimed. x

12. Let x and y be independent. Then f(xß yß zß w) œ 0, g(xß yß zß w) œ 0 and Ê `` xf `` xx  `` yf `` yx  `` zf `` xz  ``wf ``wx œ `` xf  `` zf `` xz  ``wf ``wx `g `x `g `y `g `z `g `w `g `g `z `g `w `x `x  `y `x  `z `x  `w `x œ `x  `z `x  `w `x œ 0

`y `x

œ0

œ 0 and imply

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

`g `z `z `y

œ0

861

862

Chapter 14 Partial Derivatives



`f `z `g `z

`z `x `z `x

 

`f `w `g `w

`w `x `w `x

œ
œ


`f `x `g `x

Ê ˆ `` xz ‰y œ

c ``xf » c `g »

`x `f `z `g `z

`f `w `g » `w `f `w `g » `w

œ

`x `y œ 0 `g `g `z `y  `z `y

Likewise, f(xß yß zß w) œ 0, g(xß yß zß w) œ 0 and œ



`f `y `f `z `g `z

 `z `y `z `y

`f `z `z `y

 

`f `w `g `w

 `w `y `w `y

`f `w `w `y

œ 0 and (similarly)

œ
`` yf œ


`g `y

Ê Š ``wy ‹ œ x

`g `g `f `w
`x `w `g `f `f `g `z `w  `z `w

 ``xf

`f `z » `g `z `f `z » `g `z

c ``yf c `` gy » `f `w `g » `w

œ

œ


`f `x `f `z

`g `w `g `w

`g `x `f `g `w `z

 ``wf



`f `x `f `y `f `z `x `y  `y `y  `z `y `g `w ` w ` y œ 0 imply

Ê 

`g `g `f `y
`z `y `g `f `f `g `z `w  `z `w

 `` fz

œ


`f `z `f `z

`g `y `g `w

`g `z `f `g `w `z

 ``yf 

, as claimed. 

`f `w `w `y

, as claimed.

CHAPTER 14 PRACTICE EXERCISES 1. Domain: All points in the xy-plane Range: z   0 Level curves are ellipses with major axis along the y-axis and minor axis along the x-axis.

2. Domain: All points in the xy-plane Range: 0  z  _ Level curves are the straight lines x  y œ ln z with slope
1, and z  0.

3. Domain: All (xß y) such that x Á 0 and y Á 0 Range: z Á 0 Level curves are hyperbolas with the x- and y-axes as asymptotes.

4. Domain: All (xß y) so that x#
y   0 Range: z   0 Level curves are the parabolas y œ x#
c, c   0.

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Chapter 14 Practice Exercises 5. Domain: All points (xß yß z) in space Range: All real numbers Level surfaces are paraboloids of revolution with the z-axis as axis.

6. Domain: All points (xß yß z) in space Range: Nonnegative real numbers Level surfaces are ellipsoids with center (0ß 0ß 0).

7. Domain: All (xß yß z) such that (xß yß z) Á (0ß !ß 0) Range: Positive real numbers Level surfaces are spheres with center (0ß 0ß 0) and radius r  0.

8. Domain: All points (xß yß z) in space Range: (0ß 1] Level surfaces are spheres with center (0ß 0ß 0) and radius r  0.

9.

lim

Ðxß yÑ Ä Ð1ß ln 2Ñ

ey cos x œ eln 2 cos 1 œ (2)(
1) œ
2 2
y

10.

lim Ðxß yÑ Ä Ð0ß 0Ñ x
cos y

11.

lim # # Ðxß yÑ Ä Ð1ß 1Ñ x  y xÁ „y

12.

13.

14.

xy

x$ y$  1 Ðxß yÑ Ä Ð1ß 1Ñ xy  1

lim

lim

P Ä Ð1ß
1ß eÑ

lim

œ

œ

2
0 0
cos 0

œ2 xy

lim Ðxß yÑ Ä Ð1ß 1Ñ (x  y)(x
y) xÁ „y

œ

œ

(xy  1) ax# y#
xy
1b xy  1 Ðx ß y Ñ Ä Ð 1 ß 1Ñ

lim

1

lim Ðxß yÑ Ä Ð1ß 1Ñ x
y

œ

lim

œ

Ðxß yÑ Ä Ð1ß 1Ñ

" 1
1

œ

" #

ax# y#  xy  1b œ 1# † 1#  1 † 1  1 œ 3

ln kx  y  zk œ ln k1  (
1)  ek œ ln e œ 1

P Ä Ð1 ß
1 ß
1 Ñ

tan" (x  y  z) œ tan" (1  (
1)  (
1)) œ tan" (
1) œ
14

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

863

864

Chapter 14 Partial Derivatives

15. Let y œ kx# , k Á 1. Then

kx#

œ

y

lim # Ðxß yÑ Ä Ð0ß 0Ñ x  y y Á x#

lim # # axß kx# b Ä Ð0ß 0Ñ x  kx

œ

k 1  k#

which gives different limits for

œ

1
k# k

which gives different limits for

different values of k Ê the limit does not exist. 16. Let y œ kx, k Á 0. Then

x#
y# xy Ðxß yÑ Ä Ð0ß 0Ñ

lim

x#
(kx)# x(kx) (xß kxÑ Ä Ð0ß 0Ñ

œ

lim

xy Á 0

different values of k Ê the limit does not exist. 17. Let y œ kx. Then

x#  y#

œ

lim # # Ðxß yÑ Ä Ð0ß 0Ñ x
y

x#  k# x# x #
k# x#

1  k# 1
k#

œ

which gives different limits for different values

of k Ê the limit does not exist so f(0ß 0) cannot be defined in a way that makes f continuous at the origin. 18. Along the x-axis, y œ 0 and

sin (x  y) kxk
kyk

lim

Ðxß yÑ Ä Ð0ß 0Ñ

œ lim

sin x k xk

xÄ0

œœ

1, x  0 , so the limit fails to exist
", x  0

Ê f is not continuous at (0ß 0). 19.

`g `r

œ cos )  sin ),

20.

`f `x

œ

21.

`f ` R"

" #

Š x# 2x
y# ‹ 

œ
R"# , "

`f ` R#

`g `)

œ
r sin )  r cos )

y ‹ x# y # 1
ˆx‰

Š

œ

œ
R"# ,

x#

`f ` R$

#

x
y#




x#

y
y#

œ

xy x#
y#

`f `y

,

œ

" #

Š x# 2y
y# ‹ 

Š 1x ‹ y #

1
ˆx‰

œ

y x#
y#



x x#
y#

œ

x
y x#
y#

œ
R"# $

22. hx (xß yß z) œ 21 cos (21x  y
3z), hy (xß yß z) œ cos (21x  y
3z), hz (xß yß z) œ
3 cos (21x  y
3z) 23.

`P `n

œ

RT V

,

`P `R

œ

nT V

`P `T

,

œ

nR V

,

`P `V

œ
nRT V#

24. fr (rß jß Tß w) œ
2r"# j É 1Tw , fj (rß jß Tß w) œ
#r"j# É

25.

œ

" 4rj

É T1"w œ

`g `x

œ

" y

,

`g `y

" 4rjT

œ1


, fT (rß jß Tß w) œ ˆ #"rj ‰ Š È"1w ‹ Š 2È" T ‹

T 1w

É 1Tw , fw (rß jß Tß w) œ ˆ #"rj ‰ É T1 ˆ
"# w$Î# ‰ œ
4r"jw É 1Tw Ê

x y#

` #g ` x#

œ 0,

` #g ` y#

œ

2x y$

,

` #g ` y` x

œ

` #g ` x` y

œ
y"#

26. gx (xß y) œ ex  y cos x, gy (xß y) œ sin x Ê gxx (xß y) œ ex
y sin x, gyy (xß y) œ 0, gxy (xß y) œ gyx (xß y) œ cos x 27.

`f `x

œ 1  y
15x# 

2x x#
1

,

`f `y

œx Ê

` #f ` x#

œ
30x 

22x# ax#
1b#

,

` #f ` y#

œ 0,

` #f ` y` x

œ

` #f ` x` y

œ1

28. fx (xß y) œ
3y, fy (xß y) œ 2y
3x
sin y  7ey Ê fxx (xß y) œ 0, fyy (xß y) œ 2
cos y  7ey , fxy (xß y) œ fyx (xß y) œ
3 29.

`w `x

Ê Ê 30.

`w `x

Ê Ê

œ y cos (xy  1),

`w `y

œ x cos (xy  1),

dx dt

œ et ,

dy dt

dw t ˆ " ‰ dt œ [y cos (xy  1)]e  [x cos (xy  1)] t
1 ; dw ¸ ˆ " ‰ dt tœ0 œ 0 † 1  [1 † (
1)] 0
1 œ
1

œ ey ,

`w `y

œ xey  sin z,

dw y "Î#  axey  dt œ e t dw ¸ dt tœ1 œ 1 † 1  (2 † 1

`w `z

œ y cos z  sin z,

sin zb ˆ1 

"‰ t

dx dt

œ

" t
1

t œ 0 Ê x œ 1 and y œ 0

œ t"Î# ,

dy dt

œ 1  "t ,

dz dt

œ1

 (y cos z  sin z)1; t œ 1 Ê x œ 2, y œ 0, and z œ 1


0)(2)  (0  0)1 œ 5

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Chapter 14 Practice Exercises 31.

`w `x

Ê Ê

33.

`w `u `w `v

œ

`x `u `x `v

œ

ˆ 1
x x#

`f `x

œ y  z,

`f `y

œ x  z,

œ

Ê

`w `x

dw dx dw dx

œ ˆ 1
x x#


`x `r

`x `s

œ 1,

œ cos s,




" ‰ u x#
1 a2e cos vb ; u œ v œ 0 Ê " ‰ `w ¸ u x#
1 a
2e sin vb Ê ` v Ð0ß0Ñ œ `f `z

œ y  x,

dx dt

df dt œ
(y  z)(sin t)  (x  z)(cos df ¸ dt tœ1 œ
(sin 1  cos 2)(sin 1) 

Ê 34.

œ
cos (2x
y),

`y `r

`y `s

œ s,

œr

`w ` r œ [2 cos (2x
y)](1)  [
cos (2x
y)](s); r œ 1 and s œ 0 Ê x œ 1 and y œ 0 `w ¸ `w ` r Ð1ß0Ñ œ (2 cos 21)
(cos 21 )(0) œ 2; ` s œ [2 cos (2x
y)](cos s)  [
cos (2x
y)](r) `w ¸ ` s Ð1ß0Ñ œ (2 cos 21)(cos 0)
(cos 21)(1) œ 2
1

Ê

32.

`w `y

œ 2 cos (2x
y),

œ

dw ` s ds ` x

œ (5)

dw ds

and

`w `y

œ

dw ` s ds ` y

œ
sin t,

dy dt

`w ¸ ` u Ð0ß0Ñ

xœ2 Ê ˆ 52




œ cos t,

dz dt

"‰ 5 (0)

œ ˆ 52
5" ‰ (2) œ

1
y cos xy 2y  x cos xy

œ
2 sin 2t

(cos 1  cos 2)(cos 1)
2(sin 1  cos 1)(sin 2)

œ (1)

dw ds

œ

`w `x

Ê

dw ds


5

`w `y

œ5

œ

dy dx ¹ Ð0ß1Ñ

1
" 2

dw ds


5

dw ds

dy dx ¹ Ð0ßln 2Ñ

1  y cos xy œ
FFxy œ
2y  x cos xy

dy dx

xby


e œ
FFxy œ
2y 2x
exby

ß

1‰ 4

#

œ


i


j Ê f increases most rapidly in the direction u œ


È2 #

i


Ê f increases most rapidly in the direction u œ
u œ
È12 i 

1 È2

i


1 È2

" È2

œ

È2 #

and decreases most

38. ™ f œ 2xec2y i
2x# ec2y j Ê ™ f k Ð1ß0Ñ œ #i
#j Ê k ™ f k œ È2#  (
2)# œ 2È2; u œ 1 È2

#

œ
"# i
"# j Ê k ™ f k œ Ɉ
"# ‰  ˆ
"# ‰ œ

È2 # j È È È È rapidly in the direction
u œ #2 i  #2 j ; (Du f)P! œ k ™ f k œ #2 and (Dcu f)P! œ
#2 ; 7 u" œ kvvk œ È33i #

4j4# œ 53 i  54 j Ê (Du" f)P! œ ™ f † u" œ ˆ
"# ‰ ˆ 35 ‰  ˆ
"# ‰ ˆ 45 ‰ œ
10 ™f k™f k

È2 #

dy dx

2 œ
2 ln0
2
2 œ
(ln 2  1)

37. ™ f œ (
sin x cos y)i
(cos x sin y)j Ê ™ f k ˆ 14 È2 #

œ0

œ
1

36. F(xß y) œ 2xy  ex
y
2 Ê Fx œ 2y  ex
y and Fy œ 2x  ex
y Ê

uœ

;

t)
2(y  x)(sin 2t); t œ 1 Ê x œ cos 1, y œ sin 1, and z œ cos 2

Ê at (xß y) œ (!ß 1) we have

Ê at (xß y) œ (!ß ln 2) we have

2 5

œ0

35. F(xß y) œ 1
x
y#
sin xy Ê Fx œ
1
y cos xy and Fy œ
2y
x cos xy Ê œ

865

™f k™f k

œ

1 È2

i


1 È2

j

j and decreases most rapidly in the direction

j ; (Du f)P! œ k ™ f k œ 2È2 and (Dcu f)P! œ
2È2 ; u" œ

v kv k

œ

i
j È 1#
1#

œ

1 È2

i

1 È2

j

Ê (Du" f)P! œ ™ f † u" œ (2) Š È"2 ‹  (
2) Š È"2 ‹ œ 0 2 3 6 39. ™ f œ Š 2x
3y
6z ‹ i  Š 2x
3y
6z ‹ j  Š 2x
3y
6z ‹ k Ê ™ f k Ð 1ß 1ß1Ñ œ 2i  3j  6k ;

uœ

™f k™f k

œ

2i
3j
6k È 2#
3#
6#

œ

2 7

i  73 j  76 k Ê f increases most rapidly in the direction u œ

2 7

i  37 j  67 k and

decreases most rapidly in the direction
u œ
27 i
37 j
67 k ; (Du f)P! œ k ™ f k œ 7, (Du f)P! œ
7; u" œ

v kv k

œ

2 7

i  37 j  67 k Ê (Du" f)P! œ (Du f)P! œ 7

40. ™ f œ (2x  3y)i  (3x  2)j  (1
2z)k Ê ™ f k Ð0ß0ß0Ñ œ 2j  k ; u œ rapidly in the direction u œ

2 È5

j

" È5

™f k™f k

œ

2 È5

j

" È5

k Ê f increases most

k and decreases most rapidly in the direction
u œ
È25 j


(Du f)P! œ k ™ f k œ È5 and (Du f)P! œ
È5 ; u" œ

v kvk

œ

i
j
k È 1#
1#
1#

Ê (Du" f)P! œ ™ f † u" œ (0) Š È"3 ‹  (2) Š È"3 ‹  (1) Š È"3 ‹ œ

3 È3

œ

" È3

i

" È3

j

" È3

k

œ È3

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

" È5

k;

;

866

Chapter 14 Partial Derivatives

41. r œ (cos 3t)i  (sin 3t)j  3tk Ê v(t) œ (
3 sin 3t)i  (3 cos 3t)j  3k Ê v ˆ 13 ‰ œ
3j  3k Ê u œ
È"2 j 

" È2

k ; f(xß yß z) œ xyz Ê ™ f œ yzi  xzj  xyk ; t œ

Ê ™ f k Ð 1ß0ß1Ñ œ
1j Ê ™ f † u œ (
1j) † Š
È"2 j 

" È2

k‹ œ

1 3

yields the point on the helix (
1ß 0ß 1)

1 È2

42. f(xß yß z) œ xyz Ê ™ f œ yzi  xzj  xyk ; at (1ß 1ß 1) we get ™ f œ i  j  k Ê the maximum value of Du f k œ k ™ f k œ È3 Ð1ß1ß1Ñ

43. (a) Let ™ f œ ai  bj at (1ß 2). The direction toward (2ß 2) is determined by v" œ (2
1)i  (2
2)j œ i œ u so that ™ f † u œ 2 Ê a œ 2. The direction toward (1ß 1) is determined by v# œ (1
1)i  (1
2)j œ
j œ u so that ™ f † u œ
2 Ê
b œ
2 Ê b œ 2. Therefore ™ f œ 2i  2j ; fx a1, 2b œ fy a1, 2b œ 2. (b) The direction toward (4ß 6) is determined by v$ œ (4
1)i  (6
2)j œ 3i  4j Ê u œ 35 i  45 j Ê ™f†uœ

14 5

.

44. (a) True

(b) False

(c) True

(d) True

45. ™ f œ 2xi  j  2zk Ê ™ f k Ð0ß 1ß 1Ñ œ j
2k , ™ f k Ð0ß0ß0Ñ œ j , ™ f k Ð0ß 1ß1Ñ œ j  2k

46. ™ f œ 2yj  2zk Ê ™ f k Ð2ß2ß0Ñ œ 4j , ™ f k Ð2ß 2ß0Ñ œ
4j , ™ f k Ð2ß0ß2Ñ œ 4k , ™ f k Ð2ß0ß 2Ñ œ
4k

47. ™ f œ 2xi
j
5k Ê ™ f k Ð2ß 1ß1Ñ œ 4i
j
5k Ê Tangent Plane: 4(x
2)
(y  1)
5(z
1) œ 0 Ê 4x
y
5z œ 4; Normal Line: x œ 2  4t, y œ
1
t, z œ 1
5t 48. ™ f œ 2xi  2yj  k Ê ™ f k Ð1ß1ß2Ñ œ 2i  2j  k Ê Tangent Plane: 2(x
1)  2(y
1)  (z
2) œ 0 Ê 2x  2y  z
6 œ 0; Normal Line: x œ 1  2t, y œ 1  2t, z œ 2  t 49.

`z `x

œ

2x x#
y#

Ê

`z ¸ ` x Ð0ß1ß0Ñ

œ 0 and

`z `y

œ

2y x#
y#

Ê

`z ` y ¹ Ð0ß1ß0Ñ

œ 2; thus the tangent plane is

2(y
1)
(z
0) œ 0 or 2y
z
2 œ 0

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Chapter 14 Practice Exercises 50.

`z `x

œ
2x ax#  y# b

#

`z ¸ ` x ˆ1ß1ß 12 ‰

Ê

œ
#" and

`z `y

œ
2y ax#  y# b

#

Ê

`z ` y ¹ ˆ1ß1ß 1 ‰ 2

867

œ
"# ; thus the tangent

plane is
"# (x
1)
"# (y
1)
ˆz
"# ‰ œ 0 or x  y  2z
3 œ 0 51. ™ f œ (
cos x)i  j Ê ™ f k Ð1ß1Ñ œ i  j Ê the tangent line is (x
1)  (y
1) œ 0 Ê x  y œ 1  1; the normal line is y
1 œ 1(x
1) Ê y œ x
1  1

52. ™ f œ
xi  yj Ê ™ f k Ð1ß2Ñ œ
i  2j Ê the tangent line is
(x
1)  2(y
2) œ 0 Ê y œ

" #

x  #3 ; the normal

line is y
2 œ
2(x
1) Ê y œ
2x  4

53. Let f(xß yß z) œ x#  2y  2z
4 and g(xß yß z) œ y
1. Then ™ f œ 2xi  2j  2kk a1 1 12 b œ 2i  2j  2k â â â i j kâ â â and ™ g œ j Ê ™ f ‚ ™ g œ â 2 2 2 â œ
2i  2k Ê the line is x œ 1
2t, y œ 1, z œ "#  2t â â â0 " 0â ß ß

54. Let f(xß yß z) œ x  y#  z
2 and g(xß yß z) œ y
1. Then ™ f œ i  2yj  kk a 12 1 12 b œ i  2j  k and â â â i j kâ â â ™ g œ j Ê ™ f ‚ ™ g œ â 1 2 1 â œ
i  k Ê the line is x œ "#
t, y œ 1, z œ "#  t â â â0 " 0â ß ß

55. f ˆ 14 ß 14 ‰ œ

" #

, fx ˆ 14 ß 14 ‰ œ cos x cos yk Ð1Î4ß1Î4Ñ œ

Ê L(xß y) œ

" #

 "# ˆx
14 ‰
"# ˆy
14 ‰ œ

" #

" # " #

, fy ˆ 14 ß 14 ‰ œ
sin x sin yk Ð1Î4ß1Î4Ñ œ
"#

 x
"# y; fxx (xß y) œ
sin x cos y, fyy (xß y) œ
sin x cos y, and

fxy (xß y) œ
cos x sin y. Thus an upper bound for E depends on the bound M used for kfxx k , kfxy k , and kfyy k . With M œ

È2 #

we have kE(xß y)k Ÿ

with M œ 1, kE(xß y)k Ÿ

" #

" #

Š

È2 ˆ¸ # ‹ x

#
14 ¸  ¸y
14 ¸‰ Ÿ

# (1) ˆ¸x
14 ¸  ¸y
14 ¸‰ œ

" #

È2 4

(0.2)# Ÿ 0.0142;

(0.2)# œ 0.02.

56. f(1ß 1) œ 0, fx (1ß 1) œ yk Ð1ß1Ñ œ 1, fy (1ß 1) œ x
6yk Ð1ß1Ñ œ
5 Ê L(xß y) œ (x
1)
5(y
1) œ x
5y  4; fxx (xß y) œ 0, fyy (xß y) œ
6, and fxy (xß y) œ 1 Ê maximum of kfxx k , kfyy k , and kfxy k is 6 Ê M œ 6 Ê kE(xß y)k Ÿ

" #

(6) akx
1k  ky
1kb# œ

" #

(6)(0.1  0.2)# œ 0.27

57. f(1ß 0ß 0) œ 0, fx (1ß 0ß 0) œ y
3zk Ð1ß0ß0Ñ œ 0, fy (1ß 0ß 0) œ x  2zk Ð1ß0ß0Ñ œ 1, fz (1ß 0ß 0) œ 2y
3xk Ð1ß0ß0Ñ œ
3 Ê L(xß yß z) œ 0(x
1)  (y
0)
3(z
0) œ y
3z; f(1ß 1ß 0) œ 1, fx (1ß 1ß 0) œ 1, fy (1ß 1ß 0) œ 1, fz ("ß "ß !) œ
1 Ê L(xß yß z) œ 1  (x
1)  (y
1)
1(z
0) œ x  y
z
1 58. f ˆ0ß !ß 14 ‰ œ 1, fx ˆ!ß 0ß 14 ‰ œ
È2 sin x sin (y  z)¹

ˆ0ß0ß 1 ‰

œ 0, fy ˆ!ß 0ß 14 ‰ œ È2 cos x cos (y  z)¹

4

fz ˆ!ß 0ß 14 ‰ œ È2 cos x cos (y  z)¹

ˆ0ß0ß 1 ‰

È2 #

œ 1 Ê L(xß yß z) œ 1  1(y
0)  1 ˆz
14 ‰ œ 1  y  z


Ê L(xß yß z) œ

È2 È2 È2 ˆ1 1 ‰ ˆ1 1 ‰ # , fy 4 ß 4 ß 0 œ # , fz 4 ß 4 ß 0 œ # È È È È È  #2 ˆy
14 ‰  #2 (z
0) œ #2
#2 x  #2

, fx ˆ 14 ß 14 ß 0‰ œ
È2 #




È2 #

ˆx
14 ‰

œ 1,

4

4

f ˆ 14 ß 14 ß 0‰ œ

ˆ0ß0ß 1 ‰

y

È2 #

z

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

1 4

;

868

Chapter 14 Partial Derivatives

59. V œ 1r# h Ê dV œ 21rh dr  1r# dh Ê dVk Ð1Þ5ß5280Ñ œ 21(1.5)(5280) dr  1(1.5)# dh œ 15,8401 dr  2.251 dh. You should be more careful with the diameter since it has a greater effect on dV. 60. df œ (2x
y) dx  (
x  2y) dy Ê df k Ð1ß2Ñ œ 3 dy Ê f is more sensitive to changes in y; in fact, near the point (1ß 2) a change in x does not change f. 61. dI œ

" R

dV


V R#

" 100

dR Ê dI¸ Ð24ß100Ñ œ

dV


24 100#

dR Ê dI¸ dVœ
1ßdRœ
20 œ
0.01  (480)(.0001) œ 0.038,

" ‰ 20 ‰ or increases by 0.038 amps; % change in V œ (100) ˆ
24 ¸
4.17%; % change in R œ ˆ
100 (100) œ
20%;

Iœ

24 100

œ 0.24 Ê estimated % change in I œ

dI I

‚ 100 œ

0.038 0.24

‚ 100 ¸ 15.83% Ê more sensitive to voltage change.

62. A œ 1ab Ê dA œ 1b da  1a db Ê dAk Ð10ß16Ñ œ 161 da  101 db; da œ „ 0.1 and db œ „ 0.1 ¸ ¸ 2.61 ¸ Ê dA œ „ 261(0.1) œ „ 2.61 and A œ 1(10)(16) œ 1601 Ê ¸ dA A ‚ 100 œ 1601 ‚ 100 ¸ 1.625% 63. (a) y œ uv Ê dy œ v du  u dv; percentage change in u Ÿ 2% Ê kduk Ÿ 0.02, and percentage change in v Ÿ 3% Ê kdvk Ÿ 0.03;

dy y

Ÿ 2%  3% œ 5% (b) z œ u  v Ê dzz œ

œ

v du
u dv uv

du
dv u
v

œ

œ

du u
v

Ê ¸ dzz ‚ 100¸ Ÿ ¸ du u ‚ 100  64. C œ Ê

dv v



du u



dv u
v

Þ

Þ

Þ

Þ

Þ

Þ

Ÿ

¸ du Ê ¹ dy y ‚ 100¹ œ u ‚ 100  du u



dv v

dv v

¸ ¸ dv ¸ ‚ 100¸ Ÿ ¸ du u ‚ 100  v ‚ 100

(since u  0, v  0)

‚ 100¸ œ ¹ dy y ‚ 100¹

(0.425)(7) 7 71.84w0 425 h0 725 Ê Cw œ 71.84w1 425 h0 725 2.975 5.075 dC œ 71.84w 1 425 h0 725 dw  71.84w0 425 h1 725 Þ

dv v

Þ

and Ch œ

(0.725)(7) 71.84w0 425 h1 725 Þ

Þ

dh; thus when w œ 70 and h œ 180 we have

dCk Ð70ß180Ñ ¸
(0.00000225) dw
(0.00000149) dh Ê 1 kg error in weight has more effect 65. fx (xß y) œ 2x
y  2 œ 0 and fy (xß y) œ
x  2y  2 œ 0 Ê x œ
2 and y œ
2 Ê (
2ß
2) is the critical point; # œ 3  0 and fxx  0 Ê local minimum value fxx (
2ß
2) œ 2, fyy (
#ß
2) œ 2, fxy (
#ß
2) œ
1 Ê fxx fyy
fxy of f(
#ß
2) œ
8 66. fx (xß y) œ 10x  4y  4 œ 0 and fy (xß y) œ 4x
4y
4 œ 0 Ê x œ 0 and y œ
1 Ê (0ß
1) is the critical point; # œ
56  0 Ê saddle point with f(0ß
1) œ 2 fxx (0ß
1) œ 10, fyy (0ß
1) œ
4, fxy (0ß
1) œ 4 Ê fxx fyy
fxy 67. fx (xß y) œ 6x#  3y œ 0 and fy (xß y) œ 3x  6y# œ 0 Ê y œ
2x# and 3x  6 a4x% b œ 0 Ê x a1  8x$ b œ 0 Ê x œ 0 and y œ 0, or x œ
"# and y œ
"# Ê the critical points are (0ß 0) and ˆ
"# ß
"# ‰ . For (!ß !):

# fxx (!ß !) œ 12xk Ð0ß0Ñ œ 0, fyy (!ß !) œ 12yk Ð0ß0Ñ œ 0, fxy (!ß 0) œ 3 Ê fxx fyy
fxy œ
9  0 Ê saddle point with # f(0ß 0) œ 0. For ˆ
"# ß
"# ‰: fxx œ
6, fyy œ
6, fxy œ 3 Ê fxx fyy
fxy œ 27  0 and fxx  0 Ê local maximum " " " value of f ˆ
# ß
# ‰ œ 4

68. fx (xß y) œ 3x#
3y œ 0 and fy (xß y) œ 3y#
3x œ 0 Ê y œ x# and x%
x œ 0 Ê x ax$
1b œ 0 Ê the critical points are (0ß 0) and (1ß 1) . For (!ß !): fxx (!ß !) œ 6xk Ð0ß0Ñ œ 0, fyy (!ß !) œ 6yk Ð0ß0Ñ œ 0, fxy (!ß 0) œ
3 # Ê fxx fyy
fxy œ
9  0 Ê saddle point with f(0ß 0) œ 15. For (1ß 1): fxx (1ß 1) œ 6, fyy (1ß 1) œ 6, fxy (1ß 1) œ
3 # Ê fxx fyy
fxy œ 27  0 and fxx  0 Ê local minimum value of f(1ß 1) œ 14

69. fx (xß y) œ 3x#  6x œ 0 and fy (xß y) œ 3y#
6y œ 0 Ê x(x  2) œ 0 and y(y
2) œ 0 Ê x œ 0 or x œ
2 and y œ 0 or y œ 2 Ê the critical points are (0ß 0), (0ß 2), (
2ß 0), and (
2ß 2) . For (!ß !): fxx (!ß !) œ 6x  6k Ð0ß0Ñ # œ 6, fyy (!ß !) œ 6y
6k Ð0ß0Ñ œ
6, fxy (!ß 0) œ 0 Ê fxx fyy
fxy œ
36  0 Ê saddle point with f(0ß 0) œ 0. For # (0ß 2): fxx (!ß 2) œ 6, fyy (0ß #) œ 6, fxy (!ß 2) œ 0 Ê fxx fyy
fxy œ 36  0 and fxx  0 Ê local minimum value of

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Chapter 14 Practice Exercises

869

# f(!ß 2) œ
4. For (
#ß 0): fxx (
2ß 0) œ
6, fyy (
#ß 0) œ
6, fxy (
2ß 0) œ 0 Ê fxx fyy
fxy œ 36  0 and fxx  0

Ê local maximum value of f(
2ß 0) œ 4. For (
2ß 2): fxx (
2ß 2) œ
6, fyy (
2ß 2) œ 6, fxy (
2ß 2) œ 0 # Ê fxx fyy
fxy œ
36  0 Ê saddle point with f(
2ß 2) œ 0. 70. fx (xß y) œ 4x$
16x œ 0 Ê 4x ax#
4b œ 0 Ê x œ 0, 2,
2; fy (xß y) œ 6y
6 œ 0 Ê y œ 1. Therefore the critical points are (0ß 1), (2ß 1), and (
2ß 1). For (!ß 1): fxx (!ß 1) œ 12x#
16k Ð0ß1Ñ œ
16, fyy (!ß 1) œ 6, fxy (!ß 1) œ 0 # Ê fxx fyy
fxy œ
96  0 Ê saddle point with f(0ß 1) œ
3. For (2ß 1): fxx (2ß 1) œ 32, fyy (2ß 1) œ 6, # fxy (2ß 1) œ 0 Ê fxx fyy
fxy œ 192  0 and fxx  0 Ê local minimum value of f(2ß 1) œ
19. For (
#ß 1): # fxx (
2ß 1) œ 32, fyy (
#ß 1) œ 6, fxy (
2ß 1) œ 0 Ê fxx fyy
fxy œ 192  0 and fxx  0 Ê local minimum value of

f(
2ß 1) œ
19. 71. (i)

On OA, f(xß y) œ f(0ß y) œ y#  3y for 0 Ÿ y Ÿ 4 Ê f w (!ß y) œ 2y  3 œ 0 Ê y œ
3# . But ˆ!ß
3# ‰

is not in the region. Endpoints: f(0ß 0) œ 0 and f(0ß 4) œ 28. (ii) On AB, f(xß y) œ f(xß
x  4) œ x#
10x  28 for 0 Ÿ x Ÿ 4 Ê f w (xß
x  4) œ 2x
10 œ 0 Ê x œ 5, y œ
1. But (5ß
1) is not in the region. Endpoints: f(4ß 0) œ 4 and f(!ß 4) œ 28. (iii) On OB, f(xß y) œ f(xß 0) œ x#
3x for 0 Ÿ x Ÿ 4 Ê f w (xß 0) œ 2x
3 Ê x œ critical point with f ˆ 3# ß !‰ œ
94 .

3 #

and y œ 0 Ê ˆ 3# ß 0‰ is a

Endpoints: f(0ß 0) œ 0 and f(%ß 0) œ 4. (iv) For the interior of the triangular region, fx (xß y) œ 2x  y
3 œ 0 and fy (xß y) œ x  2y  3 œ 0 Ê x œ 3 and y œ
3. But (3ß
3) is not in the region. Therefore the absolute maximum is 28 at (0ß 4) and the absolute minimum is
94 at ˆ 3# ß !‰ .

On OA, f(xß y) œ f(0ß y) œ
y#  4y  1 for 0 Ÿ y Ÿ 2 Ê f w (!ß y) œ
2y  4 œ 0 Ê y œ 2 and x œ 0. But (0ß 2) is not in the interior of OA. Endpoints: f(0ß 0) œ 1 and f(0ß 2) œ 5. (ii) On AB, f(xß y) œ f(xß 2) œ x#
2x  5 for 0 Ÿ x Ÿ 4 Ê f w (xß 2) œ 2x
2 œ 0 Ê x œ 1 and y œ 2 Ê (1ß 2) is an interior critical point of AB with f(1ß 2) œ 4. Endpoints: f(4ß 2) œ 13 and f(!ß 2) œ 5. (iii) On BC, f(xß y) œ f(4ß y) œ
y#  4y  9 for 0 Ÿ y Ÿ 2 Ê f w (4ß y) œ
2y  4 œ 0 Ê y œ # and x œ 4. But (4ß 2) is not in the interior of BC. Endpoints: f(4ß 0) œ 9 and f(%ß 2) œ 13. (iv) On OC, f(xß y) œ f(xß 0) œ x#
2x  1 for 0 Ÿ x Ÿ 4 Ê f w (xß 0) œ 2x
2 œ 0 Ê x œ 1 and y œ 0 Ê (1ß 0) is an interior critical point of OC with f(1ß 0) œ 0. Endpoints: f(0ß 0) œ 1 and f(4ß 0) œ 9. (v) For the interior of the rectangular region, fx (xß y) œ 2x
2 œ 0 and fy (xß y) œ
2y  4 œ 0 Ê x œ 1 and y œ 2. But (1ß 2) is not in the interior of the region. Therefore the absolute maximum is 13 at (4ß 2) and the absolute minimum is 0 at (1ß 0).

72. (i)

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

870 73. (i)

Chapter 14 Partial Derivatives On AB, f(xß y) œ f(
2ß y) œ y#
y
4 for
2 Ÿ y Ÿ 2 Ê f w (
2ß y) œ 2y
1 Ê y œ "# and x œ
2 Ê ˆ
2ß "# ‰ is an interior critical point in AB

with f ˆ
2ß "# ‰ œ
17 4 . Endpoints: f(
2ß
2) œ 2 and

f(2ß 2) œ
2. On BC, f(xß y) œ f(xß 2) œ
2 for
2 Ÿ x Ÿ 2 Ê f w (xß 2) œ 0 Ê no critical points in the interior of BC. Endpoints: f(
2ß 2) œ
2 and f(2ß 2) œ
2. (iii) On CD, f(xß y) œ f(2ß y) œ y#
5y  4 for
2 Ÿ y Ÿ 2 Ê f w (2ß y) œ 2y
5 œ 0 Ê y œ 5# and x œ 2. But ˆ#ß 5# ‰ is not in the region. (ii)

Endpoints: f(2ß
2) œ 18 and f(2ß 2) œ
2. (iv) On AD, f(xß y) œ f(xß
2) œ 4x  10 for
2 Ÿ x Ÿ 2 Ê f w (xß
2) œ 4 Ê no critical points in the interior of AD. Endpoints: f(
2ß
2) œ 2 and f(2ß
2) œ 18. (v) For the interior of the square, fx (xß y) œ
y  2 œ 0 and fy (xß y) œ 2y
x
3 œ 0 Ê y œ 2 and x œ 1 Ê (1ß 2) is an interior critical point of the square with f(1ß 2) œ
2. Therefore the absolute maximum "‰ ˆ is 18 at (2ß
2) and the absolute minimum is
17 4 at
#ß # . On OA, f(xß y) œ f(0ß y) œ 2y
y# for 0 Ÿ y Ÿ 2 Ê f w (!ß y) œ 2
2y œ 0 Ê y œ 1 and x œ 0 Ê (!ß 1) is an interior critical point of OA with f(0ß 1) œ 1. Endpoints: f(0ß 0) œ 0 and f(0ß 2) œ 0. (ii) On AB, f(xß y) œ f(xß 2) œ 2x
x# for 0 Ÿ x Ÿ 2 Ê f w (xß 2) œ 2
2x œ 0 Ê x œ 1 and y œ 2 Ê (1ß 2) is an interior critical point of AB with f(1ß 2) œ 1. Endpoints: f(0ß 2) œ 0 and f(2ß 2) œ 0. (iii) On BC, f(xß y) œ f(2ß y) œ 2y
y# for 0 Ÿ y Ÿ 2 Ê f w (2ß y) œ 2
2y œ 0 Ê y œ 1 and x œ 2 Ê (2ß 1) is an interior critical point of BC with f(2ß 1) œ 1. Endpoints: f(2ß 0) œ 0 and f(2ß 2) œ 0. (iv) On OC, f(xß y) œ f(xß 0) œ 2x
x# for 0 Ÿ x Ÿ 2 Ê f w (xß 0) œ 2
2x œ 0 Ê x œ 1 and y œ 0 Ê (1ß 0) is an interior critical point of OC with f(1ß 0) œ 1. Endpoints: f(0ß 0) œ 0 and f(0ß 2) œ 0. (v) For the interior of the rectangular region, fx (xß y) œ 2
2x œ 0 and fy (xß y) œ 2
2y œ 0 Ê x œ 1 and y œ 1 Ê (1ß 1) is an interior critical point of the square with f(1ß 1) œ 2. Therefore the absolute maximum is 2 at (1ß 1) and the absolute minimum is 0 at the four corners (0ß 0), (0ß 2), (2ß 2), and (2ß 0).

74. (i)

On AB, f(xß y) œ f(xß x  2) œ
2x  4 for
2 Ÿ x Ÿ 2 Ê f w (xß x  2) œ
2 œ 0 Ê no critical points in the interior of AB. Endpoints: f(
2ß 0) œ 8 and f(2ß 4) œ 0. (ii) On BC, f(xß y) œ f(2ß y) œ
y#  4y for 0 Ÿ y Ÿ 4 Ê f w (2ß y) œ
2y  4 œ 0 Ê y œ 2 and x œ 2 Ê (2ß 2) is an interior critical point of BC with f(2ß 2) œ 4. Endpoints: f(2ß 0) œ 0 and f(2ß 4) œ 0. (iii) On AC, f(xß y) œ f(xß 0) œ x#
2x for
2 Ÿ x Ÿ 2 Ê f w (xß 0) œ 2x
2 Ê x œ 1 and y œ 0 Ê (1ß 0) is an interior critical point of AC with f(1ß 0) œ
1. Endpoints: f(
2ß 0) œ 8 and f(2ß 0) œ 0. (iv) For the interior of the triangular region, fx (xß y) œ 2x
2 œ 0 and fy (xß y) œ
2y  4 œ 0 Ê x œ 1 and y œ 2 Ê (1ß 2) is an interior critical point of the region with f(1ß 2) œ 3. Therefore the absolute maximum is 8 at (
2ß 0) and the absolute minimum is
1 at (1ß 0).

75. (i)

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Chapter 14 Practice Exercises 76. (i)

(ii)

871

On AB, faxß yb œ faxß xb œ 4x#
2x%  16 for
2 Ÿ x Ÿ 2 Ê f w axß xb œ 8x
8x$ œ 0 Ê x œ 0 and y œ 0, or x œ 1 and y œ 1, or x œ
1 and y œ
1 Ê a0ß 0b, a1ß 1b, a
1ß
1b are all interior points of AB with fa0ß 0b œ 16, fa1ß 1b œ 18, and fa
1ß
1b œ 18. Endpoints: fa
2ß
2b œ 0 and fa2ß 2b œ 0. On BC, faxß yb œ fa2ß yb œ 8y
y% for
2 Ÿ y Ÿ 2 3 Ê f w a2ß yb œ 8
4y$ œ 0 Ê y œ È 2 and x œ 2 3 Ê Š2ß È 2‹ is an interior critical point of BC with 3 3 f Š2ß È 2‹ œ 6 È 2. Endpoints: fa2ß
2b œ
32 and fa2ß 2b œ 0.

3 (iii) On AC, faxß yb œ faxß
2b œ
8x
x% for
2 Ÿ x Ÿ 2 Ê f w axß
2b œ
8
4x$ œ 0 Ê x œ È
2 and y œ
2 3 3 3 Ê ŠÈ
2ß
2‹ is an interior critical point of AC with f ŠÈ
2ß
2‹ œ 6 È 2. Endpoints:

fa
2ß
2b œ 0 and fa2ß
2b œ
32. (iv) For the interior of the triangular region, fx axß yb œ 4y
4x$ œ 0 and fy axß yb œ 4x
4y$ œ 0 Ê x œ 0 and y œ 0, or x œ 1 and y œ 1 or x œ
1 and y œ
1. But neither of the points a0ß 0b and a1ß 1b, or a
1ß
1b are interior to the region. Therefore the absolute maximum is 18 at (1ß 1) and (
1ß
1), and the absolute minimum is
32 at a2ß
2b. On AB, f(xß y) œ f(
1ß y) œ y$
3y#  2 for
1 Ÿ y Ÿ 1 Ê f w (
1ß y) œ 3y#
6y œ 0 Ê y œ 0 and x œ
1, or y œ 2 and x œ
1 Ê (
1ß 0) is an interior critical point of AB with f(
1ß 0) œ 2; (
1ß 2) is outside the boundary. Endpoints: f(
1ß
1) œ
2 and f(
1ß 1) œ 0. (ii) On BC, f(xß y) œ f(xß 1) œ x$  3x#
2 for
1 Ÿ x Ÿ 1 Ê f w (xß 1) œ 3x#  6x œ 0 Ê x œ 0 and y œ 1, or x œ
2 and y œ 1 Ê (0ß 1) is an interior critical point of BC with f(!ß 1) œ
2; (
2ß 1) is outside the boundary. Endpoints: f(
"ß 1) œ 0 and f("ß 1) œ 2. (iii) On CD, f(xß y) œ f("ß y) œ y$
3y#  4 for
1 Ÿ y Ÿ 1 Ê f w ("ß y) œ 3y#
6y œ 0 Ê y œ 0 and x œ 1, or y œ 2 and x œ 1 Ê ("ß 0) is an interior critical point of CD with f("ß 0) œ 4; (1ß 2) is outside the boundary. Endpoints: f(1ß 1) œ 2 and f("ß
1) œ 0. (iv) On AD, f(xß y) œ f(xß
1) œ x$  3x#
4 for
1 Ÿ x Ÿ 1 Ê f w (xß
1) œ 3x#  6x œ 0 Ê x œ 0 and y œ
1, or x œ
2 and y œ
1 Ê (0ß
1) is an interior point of AD with f(0ß
1) œ
4; (
#ß
1) is outside the boundary. Endpoints: f(
1ß
1) œ
2 and f("ß
1) œ 0. (v) For the interior of the square, fx (xß y) œ 3x#  6x œ 0 and fy (xß y) œ 3y#
6y œ 0 Ê x œ 0 or x œ
2, and y œ 0 or y œ 2 Ê (0ß 0) is an interior critical point of the square region with f(!ß 0) œ 0; the points (0ß 2), (
2ß 0), and (
2ß 2) are outside the region. Therefore the absolute maximum is 4 at (1ß 0) and the absolute minimum is
4 at (0ß
1).

77. (i)

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

872

Chapter 14 Partial Derivatives

On AB, f(xß y) œ f(
1ß y) œ y$
3y for
1 Ÿ y Ÿ 1 Ê f w (
1ß y) œ 3y#
3 œ 0 Ê y œ „ 1 and x œ
1 yielding the corner points (
1ß
1) and (
1ß 1) with f(
1ß
1) œ 2 and f(
1ß 1) œ
2. (ii) On BC, f(xß y) œ f(xß 1) œ x$  3x  2 for
1 Ÿ x Ÿ 1 Ê f w (xß 1) œ 3x#  3 œ 0 Ê no solution. Endpoints: f(
"ß 1) œ
2 and f("ß 1) œ 6. (iii) On CD, f(xß y) œ f("ß y) œ y$  3y  2 for
1 Ÿ y Ÿ 1 Ê f w ("ß y) œ 3y#  3 œ 0 Ê no solution. Endpoints: f(1ß 1) œ 6 and f("ß
1) œ
2. (iv) On AD, f(xß y) œ f(xß
1) œ x$
3x for
1 Ÿ x Ÿ 1 Ê f w (xß
1) œ 3x#
3 œ 0 Ê x œ „ 1 and y œ
1 yielding the corner points (
1ß
1) and (1ß
1) with f(
1ß
1) œ 2 and f(1ß
1) œ
2 (v) For the interior of the square, fx (xß y) œ 3x#  3y œ 0 and fy (xß y) œ 3y#  3x œ 0 Ê y œ
x# and x%  x œ 0 Ê x œ 0 or x œ
1 Ê y œ 0 or y œ
1 Ê (!ß 0) is an interior critical point of the square region with f(0ß 0) œ 1; (
1ß
1) is on the boundary. Therefore the absolute maximum is 6 at ("ß 1) and the absolute minimum is
2 at (1ß
1) and (
1ß 1).

78. (i)

79. ™ f œ 3x# i  2yj and ™ g œ 2xi  2yj so that ™ f œ - ™ g Ê 3x# i  2yj œ -(2xi  2yj) Ê 3x# œ 2x- and 2y œ 2y- Ê - œ 1 or y œ 0. CASE 1: - œ 1 Ê 3x# œ 2x Ê x œ 0 or x œ 23 ; x œ 0 Ê y œ „ 1 yielding the points (0ß 1) and (!ß
1); x œ Ê yœ „

È5 3

yielding the points Š 32 ß

È5 3 ‹

and Š 32 ß


2 3

È5 3 ‹.

CASE 2: y œ 0 Ê x#
1 œ 0 Ê x œ „ 1 yielding the points (1ß 0) and (
1ß 0). Evaluations give f a!ß „ 1b œ 1, f Š 23 ß „

È5 3 ‹

œ

23 27

, f("ß 0) œ 1, and f(
"ß 0) œ
1. Therefore the absolute

maximum is 1 at a!ß „ 1b and (1ß 0), and the absolute minimum is
1 at (
"ß !). 80. ™ f œ yi  xj and ™ g œ 2xi  2yj so that ™ f œ - ™ g Ê yi  xj œ -(2xi  2yj) Ê y œ 2-x and xy œ 2-y Ê x œ 2-(2-x) œ 4-# x Ê x œ 0 or 4-# œ 1. CASE 1: x œ 0 Ê y œ 0 but (0ß 0) does not lie on the circle, so no solution. CASE 2: 4-# œ 1 Ê - œ "# or - œ
"# . For - œ "# , y œ x Ê 1 œ x#  y# œ 2x# Ê x œ C œ „ È"2 yielding the points Š È"2 ß È"2 ‹ and Š
È"2 ,
È"2 ‹ . For - œ
#" , y œ
x Ê 1 œ x#  y# œ 2x# Ê x œ „

" È2

and

y œ
x yielding the points Š
È"2 ß È"2 ‹ and Š È"2 ,
È"2 ‹ . Evaluations give the absolute maximum value f Š È"2 ß È"2 ‹ œ f Š
È"2 ß
È"2 ‹ œ

" #

and the absolute minimum

value f Š
È"2 ß È"2 ‹ œ f Š È"2 ß
È"2 ‹ œ
#" . 81. (i) f(xß y) œ x#  3y#  2y on x#  y# œ 1 Ê ™ f œ 2xi  (6y  2)j and ™ g œ 2xi  2yj so that ™ f œ - ™ g Ê 2xi  (6y  2)j œ -(2xi  2yj) Ê 2x œ 2x- and 6y  2 œ 2y- Ê - œ 1 or x œ 0. CASE 1: - œ 1 Ê 6y  2 œ 2y Ê y œ
"# and x œ „

È3 #

yielding the points Š „

È3 #

ß
#" ‹ .

CASE 2: x œ 0 Ê y# œ 1 Ê y œ „ 1 yielding the points a!ß „ 1b . Evaluations give f Š „

È3 #

ß
"# ‹ œ

" #

, f(0ß 1) œ 5, and f(!ß
1) œ 1. Therefore

" #

and 5 are the extreme

values on the boundary of the disk. (ii) For the interior of the disk, fx (xß y) œ 2x œ 0 and fy (xß y) œ 6y  2 œ 0 Ê x œ 0 and y œ
"3 Ê ˆ!ß
13 ‰ is an interior critical point with f ˆ!ß
3" ‰ œ
3" . Therefore the absolute maximum of f on the disk is 5 at (0ß 1) and the absolute minimum of f on the disk is
"3 at ˆ!ß
3" ‰ .

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Chapter 14 Practice Exercises

873

82. (i) f(xß y) œ x#  y#
3x
xy on x#  y# œ 9 Ê ™ f œ (2x
3
y)i  (2y
x)j and ™ g œ 2xi  2yj so that ™ f œ - ™ g Ê (2x
3
y)i  (2y
x)j œ -(2xi  2yj) Ê 2x
3
y œ 2x- and 2y
x œ 2yÊ 2x("
-)
y œ 3 and
x  2y(1
-) œ 0 Ê 1
- œ

x 2y

x and (2x) Š 2y ‹
y œ 3 Ê x#
y# œ 3y

Ê x# œ y#  3y. Thus, 9 œ x#  y# œ y#  3y  y# Ê 2y#  3y
9 œ 0 Ê (2y
3)(y  3) œ 0 Ê y œ
3, 3# . For y œ
3, x#  y# œ 9 Ê x œ 0 yielding the point (0ß
3). For y œ 3# , x#  y# œ 9 Ê x# 

9 4

œ 9 Ê x# œ

Ê xœ „

27 4

È

¸ 20.691, and f Š 3 # 3 , 3# ‹ œ 9


27È3 4

3È 3 #

È

. Evaluations give f(0ß
3) œ 9, f Š
3 # 3 ß 3# ‹ œ 9 

27È3 4

¸
2.691.

(ii) For the interior of the disk, fx (xß y) œ 2x
3
y œ 0 and fy (xß y) œ 2y
x œ 0 Ê x œ 2 and y œ 1 Ê (2ß 1) is an interior critical point of the disk with f(2ß 1) œ
3. Therefore, the absolute maximum of f on the disk is 9 

27È3 4

È

at Š
3 # 3 ß 3# ‹ and the absolute minimum of f on the disk is
3 at (2ß 1).

83. ™ f œ i
j  k and ™ g œ 2xi  2yj  2zk so that ™ f œ - ™ g Ê i
j  k œ -(2xi  2yj  2zk) Ê 1 œ 2x-,
1 œ 2y-, 1 œ 2z- Ê x œ
y œ z œ -" . Thus x#  y#  z# œ 1 Ê 3x# œ 1 Ê x œ „ È"3 yielding the points Š È"3 ß
È"3 ,

" È3 ‹

and Š
È"3 ,

f Š È"3 ß
È"3 ß È"3 ‹ œ

3 È3

" È3

,
È"3 ‹ . Evaluations give the absolute maximum value of

œ È3 and the absolute minimum value of f Š
È"3 ß È"3 ß
È"3 ‹ œ
È3.

84. Let f(xß yß z) œ x#  y#  z# be the square of the distance to the origin and g(xß yß z) œ x#
zy
4. Then ™ f œ 2xi  2yj  2zk and ™ g œ 2xi
zj
yk so that ™ f œ - ™ g Ê 2x œ 2-x, 2y œ
-z, and 2z œ
-y Ê x œ 0 or - œ 1. CASE 1: x œ 0 Ê zy œ
4 Ê z œ
4y and y œ
4z Ê 2 Š
y4 ‹ œ
-y and 2 ˆ
4z ‰ œ
-z Ê 8 -

8 -

œ y# and

œ z# Ê y# œ z# Ê y œ „ z. But y œ x Ê z# œ
4 leads to no solution, so y œ
z Ê z# œ 4

Ê z œ „ 2 yielding the points (0ß
2ß 2) and (0ß 2ß
2). CASE 2: - œ 1 Ê 2z œ
y and 2y œ
z Ê 2y œ
ˆ
y# ‰ Ê 4y œ y Ê y œ 0 Ê z œ 0 Ê x#
4 œ 0 Ê

x œ „ 2 yielding the points (
2ß 0ß 0) and (2ß !ß 0). Evaluations give f(0ß
2ß 2) œ f(0ß 2ß
2) œ 8 and f(
2ß 0ß 0) œ f(2ß !ß 0) œ 4. Thus the points (
2ß 0ß 0) and (2ß !ß 0) on the surface are closest to the origin.

85. The cost is f(xß yß z) œ 2axy  2bxz  2cyz subject to the constraint xyz œ V. Then ™ f œ - ™ g Ê 2ay  2bz œ -yz, 2ax  2cz œ -xz, and 2bx  2cy œ -xy Ê 2axy  2bxz œ -xyz, 2axy  2cyz œ -xyz, and 2bxz  2cyz œ -xyz Ê 2axy  2bxz œ 2axy  2cyz Ê y œ ˆ bc ‰ x. Also 2axy  2bxz œ 2bxz  2cyz Ê z œ ˆ ca ‰ x. Then x ˆ bc x‰ ˆ ca x‰ œ V Ê x$ œ #

Height œ z œ ˆ ac ‰ Š cabV ‹

"Î$

#

c# V ab

œ Š abcV ‹

#

Ê width œ x œ Š cabV ‹

"Î$

"Î$

#

, Depth œ y œ ˆ bc ‰ Š cabV ‹

"Î$

#

œ Š bacV ‹

"Î$

, and

.

86. The volume of the pyramid in the first octant formed by the plane is V(aß bß c) œ

" 3

ˆ "# ab‰ c œ

" 6

abc. The point

(2ß 1ß 2) on the plane Ê  "b  2c œ 1. We want to minimize V subject to the constraint 2bc  ac  2ab œ abc. ac ab Thus, ™ V œ bc 6 i  6 j  6 k and ™ g œ (c  2b
bc)i  (2c  2a
ac)j  (2b  a
ab)k so that ™ V œ ac ab abc Ê bc 6 œ -(c  2b
bc), 6 œ -(2c  2a
ac), and 6 œ -(2b  a
ab) Ê 6 œ -(ac  2ab
abc), abc abc 6 œ -(2bc  2ab
abc), and 6 œ -(2bc  ac
abc) Ê -ac œ 2-bc and 2-ab œ 2-bc. Now - Á 0 since 2 a

a Á 0, b Á 0, and c Á 0 Ê ac œ 2bc and ab œ bc Ê a œ 2b œ c. Substituting into the constraint equation gives y 2 2 2 x z a  a  a œ 1 Ê a œ 6 Ê b œ 3 and c œ 6. Therefore the desired plane is 6  3  6 œ 1 or x  2y  z œ 6.

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™g

874

Chapter 14 Partial Derivatives

87. ™ f œ (y  z)i  xj  xk , ™ g œ 2xi  2yj , and ™ h œ zi  xk so that ™ f œ - ™ g  . ™ h Ê (y  z)i  xj  xk œ -(2xi  2yj)  .(zi  xk) Ê y  z œ 2-x  .z, x œ 2-y, x œ .x Ê x œ 0 or . œ 1. CASE 1: x œ 0 which is impossible since xz œ 1. CASE 2: . œ 1 Ê y  z œ 2-x  z Ê y œ 2-x and x œ 2-y Ê y œ (2-)(2-y) Ê y œ 0 or 4-# œ 1. If y œ 0, then x# œ 1 Ê x œ „ 1 so with xz œ 1 we obtain the points (1ß 0ß 1) and (
1ß 0ß
1). If 4-# œ 1, then - œ „ "# . For - œ
"# , y œ
x so x#  y# œ 1 Ê x# œ "# Ê xœ „

" È2

with xz œ 1 Ê z œ „ È2, and we obtain the points Š È"2 ß
È"2 ß È2‹ and

Š
È"2 ß È"2 ß
È2‹ . For - œ

" #

, y œ x Ê x# œ

" #

Ê xœ „

" È2

with xz œ 1 Ê z œ „ È2,

and we obtain the points Š È"2 ß È"2 , È2‹ and Š
È"2 ß
È"2 ß
È2‹ . Evaluations give f(1ß 0ß 1) œ 1, f(
1ß 0ß
1) œ 1, f Š È"2 ß
È"2 ß È2‹ œ f Š È"2 ß È"2 ß È2‹ œ

3 #

, and f Š
È"2 ß
È"2 ß
È2‹ œ

3 #

" #

, f Š
È"2 ß È"2 ,
È2‹ œ 3 #

. Therefore the absolute maximum is

Š È"2 ß È"2 ß È2‹ and Š
È"2 ß
È"2 ß
È2‹ , and the absolute minimum is

" #

" #

,

at

at Š
È"2 ß È"2 ß
È2‹ and

Š È"2 ß
È"2 ß È2‹ . 88. Let f(xß yß z) œ x#  y#  z# be the square of the distance to the origin. Then ™ f œ 2xi  2yj  2zk , ™ g œ i  j  k , and ™ h œ 4xi  4yj
2zk so that ™ f œ - ™ g  . ™ h Ê 2x œ -  4x., 2y œ -  4y., and 2z œ -
2z. Ê - œ 2x(1
2.) œ 2y(1
2.) œ 2z(1  2.) Ê x œ y or . œ "# . CASE 1: x œ y Ê z# œ 4x# Ê z œ „ 2x so that x  y  z œ 1 Ê x  x  2x œ 1 or x  x
2x œ 1 (impossible) Ê x œ "4 Ê y œ "4 and z œ "# yielding the point ˆ "4 ß "4 ß "# ‰ . CASE 2: . œ

" #

Ê - œ 0 Ê 0 œ 2z(1  1) Ê z œ 0 so that 2x#  2y# œ 0 Ê x œ y œ 0. But the origin

(!ß 0ß 0) fails to satisfy the first constraint x  y  z œ 1. Therefore, the point ˆ "4 ß 4" ß "# ‰ on the curve of intersection is closest to the origin. 89. (a) y, z are independent with w œ x# eyz and z œ x#
y# Ê œ a2xeyz b

`x `y

`w `y

`w `x `w `y `w `z `x `y  `y `y  `z `y œ 2x `` xy
2y Ê `` xy œ yx

œ

 azx# eyz b (1)  ayx# eyz b (0); z œ x#
y# Ê 0

; therefore,

Š ``wy ‹ œ a2xeyz b ˆ xy ‰  zx# eyz œ a2y  zx# b eyz z

(b) z, x are independent with w œ x# eyz and z œ x#
y# Ê œ a2xeyz b (0)  azx# eyz b

`y `z

`w `z

œ

`w `x `x `z



 ayx# eyz b (1); z œ x#
y# Ê 1 œ 0
2y

1 ˆ ``wz ‰ œ azx# eyz b Š
2y ‹  yx# eyz œ x# eyz Šy
x

`w `y `w `z `y `z  `z `z `y `y " ` z Ê ` z œ
#y

; therefore,

z 2y ‹

(c) z, y are independent with w œ x# eyz and z œ x#
y# Ê

`w `z

œ

œ a2xeyz b `` xz  azx# eyz b (0)  ayx# eyz b (1); z œ x#
y# Ê 1 1 ‰ ˆ ``wz ‰ œ a2xeyz b ˆ 2x  yx# eyz œ a1  x# yb eyz

`w `x `w `y `w `z `x `z  `y `z  `z `z œ 2x `` xz
0 Ê `` xz œ #"x

; therefore,

y

90. (a) T, P are independent with U œ f(Pß Vß T) and PV œ nRT Ê ``UT œ ``UP `` TP  ‰ ˆ ``VT ‰  ˆ ``UT ‰ (1); PV œ nRT Ê P ``VT œ nR Ê ``VT œ œ ˆ ``UP ‰ (0)  ˆ `` U V ˆ ``UT ‰ œ ˆ `` U ‰ ˆ nR ‰ V P  P

`U `T

`U `V `U `T `V `T  `T `T nR P ; therefore,

`U `P `U `V (b) V, T are independent with U œ f(Pß Vß T) and PV œ nRT Ê `` U V œ `P `V  `V `V  U‰ œ ˆ ``UP ‰ ˆ ``VP ‰  ˆ `` V (1)  ˆ ``UT ‰ (0); PV œ nRT Ê V ``VP  P œ (nR) ˆ ``VT ‰ œ 0 Ê

ˆ `` U ‰ V T

œ

ˆ ``UP ‰ ˆ
VP ‰



`U `V

`U `T `T `V `P P `V œ
V

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

; therefore,

Chapter 14 Practice Exercises

875

91. Note that x œ r cos ) and y œ r sin ) Ê r œ Èx#  y# and ) œ tan" ˆ yx ‰ . Thus, `w `x

œ

`w `r `r `x



`w `) `) `x

œ ˆ ``wr ‰ Š Èx#x
y# ‹  ˆ ``w) ‰ Š x#
yy# ‹ œ (cos ))

`w `r


ˆ sinr ) ‰

`w `)

`w `y

œ

`w `r `r `y



`w `) ` ) )y

œ ˆ ``wr ‰ Š Èx#y
y# ‹  ˆ ``w) ‰ Š x#
x y# ‹ œ (sin ))

`w `r

 ˆ cosr ) ‰

`w `)

92. zx œ fu 93.

`u `y

`v `x

 fv

œ afu  afv , and zy œ fu

`u `x œ a " `w " `w a `x œ b `y

œ b and

Ê 94.

`u `x

`w `x

œ

and

`w `z

œ

2 r
s

œ

Ê

œ

2 x#
y#
2z

and

`w `s

œ

`w dw ` x œ du b ``wx œ a

" (r
s)#

`w `x `x `s

`w `y `y `s



Solving this system yields Ê ae cos vb

`u `y

`u `x


ae sin vb

u

`v `y

dw ` u du ` y

œ

2(r
s) 2 ar#
2rs
s# b

œ

" r
s

œ

`w `x `x `r

`g `)

œ

Ê

`f `x `x `) ` #g ` )#



`f `y `y `)

œ (
r sin )) Š `` x) 

`y `) ‹



" r
s






rs (r
s)#

`w `y

`w `z `z `r

dw du

œ œ

Ê

`f `x

`v `y

" r
s



œ

dw du

2(r  s) #(r
s)#

and

" `w b `y

œ

rs (r
s)#

œ

dw du

,

 ’ (r
" s)# “ (2s) œ

2r
2s (r
s)#

2 r
s


y œ 0 Ê aeu sin vb

`u `x

 aeu cos vb

`v `x

œ 0.

Similarly, e cos v
x œ 0 u

`u `y

œ eu cos v. Therefore Š `` ux i 

 (r cos ))

œ

rs (r
s)#

 ’ (r
" s)# “ (2r) œ

`v u ` x œ 1; e sin v `v u sin v. ` x œ
e

 aeu cos vb `u `y

`v `y

j‹ † Š `` vx i 

œ 1. Solving this

`v `y

j‹

u

cos vb jd œ 0 Ê the vectors are orthogonal Ê the angle

`f `x

 (r cos )) Š ``x`fy

sin vb i  ae

` #f ` y ` y` x ` ) ‹

" `w a `x

2y x#
y#
2z

œ 0 and e sin v
y œ 0 Ê aeu sin vb u

`x `)

`w `y `y `r



,

œb

u

u

#

œ

cos v and

œ eu sin v and

œ (
r sin )) Š `` xf#

and

aeu sin vb

`u `y

œ (
r sin ))

dw du

`w `z `z `s

œ cae cos vb i  ae sin vb jd † ca
e between the vectors is the constant 1# . 96.

œ bfu
bfv

œ

`u `x
u

œe

u

second system yields



`v `y

 fv

`w `y

œa

`w `r

Ê

95. eu cos v
x œ 0 Ê aeu cos vb u

`u `x `w `y

2(r
s) (r
s)#
(r  s)#
4rs -

œ

2x x#
y#
2z

Ê

`u `y

;

`f `y


(r cos ))


(r cos ))  (r cos )) Š `` x) 

`y `) ‹

#

`x `)



` #f ` y ` y# ` ) ‹


(r sin ))

`f `y


(r sin ))

œ (
r sin )  r cos ))(
r sin )  r cos ))
(r cos )  r sin )) œ (
2)(
2)
(0  2) œ 4
2 œ 2 at (rß )) œ ˆ2ß 1# ‰ . 97. (y  z)#  (z
x)# œ 16 Ê ™ f œ
2(z
x)i  2(y  z)j  2(y  2z
x)k ; if the normal line is parallel to the yz-plane, then x is constant Ê `` xf œ 0 Ê
2(z
x) œ 0 Ê z œ x Ê (y  z)#  (z
z)# œ 16 Ê y  z œ „ 4. Let x œ t Ê z œ t Ê y œ
t „ 4. Therefore the points are (tß
t „ 4ß t), t a real number.

98. Let f(xß yß z) œ xy  yz  zx
x
z# œ 0. If the tangent plane is to be parallel to the xy-plane, then ™ f is perpendicular to the xy-plane Ê ™ f † i œ 0 and ™ f † j œ 0. Now ™ f œ (y  z
1)i  (x  z)j  (y  x
2z)k so that ™ f † i œ y  z
1 œ 0 Ê y  z œ 1 Ê y œ 1
z, and ™ f † j œ x  z œ 0 Ê x œ
z. Then
z(1
z)  ("
z)z  z(
z)
(
z)
z# œ 0 Ê z
2z# œ 0 Ê z œ "# or z œ 0. Now z œ "# Ê x œ
"# and y œ Ê ˆ
"# ß "# ß "# ‰ is one desired point; z œ 0 Ê x œ 0 and y œ 1 Ê (0ß 1ß 0) is a second desired point. 99. ™ f œ -(xi  yj  zk) Ê

`f `x

œ -x Ê f(xß yß z) œ

" #

-x#  g(yß z) for some function g Ê -y œ

`f `y

œ

`g `y

-y#  h(z) for some function h Ê -z œ `` zf œ `` gz œ hw (z) Ê h(z) œ #" -z#  C for some arbitrary constant C Ê g(yß z) œ "# -y#  ˆ "# -z#  C‰ Ê f(xß yß z) œ "# -x#  "# -y#  "# -z#  C Ê f(0ß 0ß a) œ "# -a#  C Ê g(yß z) œ

" #

and f(0ß 0ß
a) œ

" #

-(
a)#  C Ê f(0ß 0ß a) œ f(0ß 0ß
a) for any constant a, as claimed.

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

" #

876

Chapter 14 Partial Derivatives ß

f(0
su" ß 0
su# ß 0
su$ )f(0ß 0ß 0)

œ lim sÄ0

‰ 100. ˆ df ds u (0 0 0) ß ß

,s0

s

És# u#"
s# u##
s# u#$  0

œ lim sÄ0

s

,s0

sÉu#"
u##
u#$

œ lim œ lim kuk œ 1; s sÄ0 sÄ0 however, ™ f œ Èx#
xy#
z# i  Èx#
yy#
z# j  Èx#
zy#
z# k fails to exist at the origin (0ß 0ß 0) 101. Let f(xß yß z) œ xy  z
2 Ê ™ f œ yi  xj  k . At (1ß 1ß 1), we have ™ f œ i  j  k Ê the normal line is x œ 1  t, y œ 1  t, z œ 1  t, so at t œ
1 Ê x œ 0, y œ 0, z œ 0 and the normal line passes through the origin. 102. (b) f(xß yß z) œ x#
y#  z# œ 4 Ê ™ f œ 2xi
2yj  2zk Ê at (2ß
3ß 3) the gradient is ™ f œ 4i  6j  6k which is normal to the surface (c) Tangent plane: 4x  6y  6z œ 8 or 2x  3y  3z œ 4 Normal line: x œ 2  4t, y œ
3  6t, z œ 3  6t

CHAPTER 14 ADDITIONAL AND ADVANCED EXERCISES fx (0ß h)  fx (0ß 0) h

1. By definition, fxy (!ß 0) œ lim

hÄ0

so we need to calculate the first partial derivatives in the

numerator. For (xß y) Á (0ß 0) we calculate fx (xß y) by applying the differentiation rules to the formula for

fy (xß y) œ

00 h

œ lim

hÄ0

2.

`w `x

x$  xy# x#
y#

x# y  y$ x#
y#

ax#
y# b (2x)  ax#  y# b (2x) ax #
y # b #

4x# y$ Ê fx (0ß h) ax #
y # b# f(0ß0) For (xß y) œ (0ß 0) we apply the definition: fx (!ß 0) œ lim f(hß 0)  œ lim 0 h 0 œ h hÄ0 hÄ0 f (hß 0)  fy (!ß 0) fxy (0ß 0) œ lim hh 0 œ
1. Similarly, fyx (0ß 0) œ lim y , so for (xß y) Á h hÄ0 hÄ0

f(xß y): fx (xß y) œ




 (xy)

4x$ y# ax#
y# b#

Ê fy (hß 0) œ

h$ h#

x# y  y $ x#
y#

œ



$

œ
hh# œ
h. 0. Then by definition (0ß 0) we have

œ h; for (xß y) œ (0ß 0) we obtain fy (0ß 0) œ lim h0 h

œ 0. Then by definition fyx (0ß 0) œ lim

hÄ0

œ 1  ex cos y Ê w œ x  ex cos y  g(y);

`w `y

hÄ0

f(0ß h)  f(!ß 0) h

œ 1. Note that fxy (0ß 0) Á fyx (0ß 0) in this case.

œ
ex sin y  gw (y) œ 2y
ex sin y Ê gw (y) œ 2y

Ê g(y) œ y#  C; w œ ln 2 when x œ ln 2 and y œ 0 Ê ln 2 œ ln 2  eln 2 cos 0  0#  C Ê 0 œ 2  C Ê C œ
2. Thus, w œ x  ex cos y  g(y) œ x  ex cos y  y#
2. 3. Substitution of u  u(x) and v œ v(x) in g(uß v) gives g(u(x)ß v(x)) which is a function of the independent variable x. Then, g(uß v) œ 'u f(t) dt Ê v

œ Š
``u

#

#

fzz œ Š ddr#f ‹ ˆ ``zr ‰  ` #r ` y#

œ

` g du ` u dx



` g dv ` v dx

œ Š ``u

#

df ` # r dr ` x#

'uv f(t) dt‹ dxdu  Š ``v 'uv f(t) dt‹ dxdv

'vu f(t) dt‹ dudx  Š ``v 'uv f(t) dt‹ dvdx œ
f(u(x)) dudx  f(v(x)) dvdx œ f(v(x)) dvdx
f(u(x)) dudx

4. Applying the chain rules, fx œ

Ê

dg dx

œ

#

df ` r dr ` z#

x#
z# 3 ˆÈx#
y#
z# ‰

df ` r dr ` x

#

Ê fxx œ Š ddr#f ‹ ˆ ``xr ‰ 

. Moreover,

; and

`r `z

œ

`r `x

œ

x È x #
y #
z#

z È x #
y #
z#

Ê

` #r ` z#

Ê

œ

#

` r ` x#

œ

#

#

. Similarly, fyy œ Š ddr#f ‹ Š ``yr ‹  #

#

y
z 3 ˆÈx#
y#
z# ‰

x#
y# 3 ˆÈ x #
y #
z # ‰

;

`r `y

œ

y È x #
y #
z#

. Next, fxx  fyy  fzz œ 0

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

df ` # r dr ` y#

and

Chapter 14 Additional and Advanced Exercises #

#

df dr

d dr

x#

(f w ) œ ˆ
2r ‰ f w , where f w œ

œ Cr# Ê f(r) œ
Cr  b œ


y #
z# ‰

Ê

df dr

y#

d# f

x #
y #
z# ‰

x#
y#

#

‰  Š ddr#f ‹ Š x#
yz #
z# ‹  ˆ df dr Œ ˆÈ Ê

y #
z#

#

‰ Ê Š ddr#f ‹ Š x#
xy#
z# ‹  ˆ df dr Œ ˆÈ

3

df f

3

877

x #
z#

  Š dr# ‹ Š x#
y#
z# ‹  ˆ dr ‰ Œ ˆÈx#
y#
z# ‰3  d# f dr#

œ0 Ê

df

 Š Èx#
2y#
z# ‹

d# f dr#

œ0 Ê

df dr



2 df r dr

œ0

œ
2 rdr Ê ln f w œ
2 ln r  ln C Ê f w œ Cr# , or

w

w

 b for some constants a and b (setting a œ
C)

a r

5. (a) Let u œ tx, v œ ty, and w œ f(uß v) œ f(u(tß x)ß v(tß y)) œ f(txß ty) œ tn f(xß y), where t, x, and y are independent variables. Then ntnc1 f(xß y) œ ``wt œ ``wu ``ut  ``wv ``vt œ x ``wu  y ``wv . Now, `w `w `u `w `v `w `w ˆ `w ‰ ˆ `w ‰ ˆ " ‰ ˆ ``wx ‰ . Likewise, ` x œ ` u ` x  ` v ` x œ ` u (t)  ` v (0) œ t ` u Ê ` u œ t `w `y

œ

`w `u `u `y



ntnc1 f(xß y) œ x `w `x

Ê

œ

`f `x

`w `v `v `y `w `u

`w `v

y

`w `y

and

œ ˆ ``wu ‰ (0)  ˆ ``wv ‰ (t) Ê œ

`f `x

Ê nf(xß y) œ x

Also from part (a), œ

` `y

œ t#

ˆt

`w ‰ `v

` #w ` v` u

œt

` #w ` x#

` `x

œ

x

y

ˆ ``wx ‰

` `x

`w ‰ `u

t

œ

`f `x

`w `v . ` w `v ` v` u ` t

` w `u ` u# ` t

` #w ` u ` u` v ` y

Ê ˆ t"# ‰

`w `u

#

` #w ` x#

œ ˆ "t ‰ Š ``wy ‹ . Therefore,

œ ˆ xt ‰ ˆ ``wx ‰  ˆ yt ‰ Š ``wy ‹. When t œ 1, u œ x, v œ y, and w œ f(xß y)

(b) From part (a), ntnc1 f(xß y) œ x n(n
1)tnc2 f(xß y) œ x

`w `v

y

` #w ` v ` v# ` y

` #w ` u#

ˆt

` #w ` v#

œ t#

, ˆ t"# ‰

` #w ` y#

`f `y

, as claimed.

Differentiating with respect to t again we obtain

#

œ

y

œ

` #w ` u ` u` v ` t

œt

, and

` #w ` v#

` #w ` v ` v# ` t

y

` #w ` u ` u# ` x

t

` #w ` y` x

` `y

œ

, and ˆ t"# ‰

œ x#

` #w ` u#

 2xy

` #w ` v ` v` u ` x

œ

# t# `` uw#

ˆ ``wx ‰ œ

` `y

ˆt

` #w ` y` x

œ

`w ‰ `u

,

` #w ` u` v

` #w ` y#

œt

œ

 y# ` `y

` #w ` v#

.

Š ``wy ‹

` #w ` u ` u# ` y

t

` #w ` v ` v` u ` y

` #w ` v` u

‰ Š ``y`wx ‹  Š yt# ‹ Š `` yw# ‹ for t Á 0. When t œ 1, w œ f(xß y) and Ê n(n
1)tnc2 f(xß y) œ Š xt# ‹ Š `` xw# ‹  ˆ 2xy t# #

#

#

#

#

#

#

#

we have n(n
1)f(xß y) œ x# Š `` xf# ‹  2xy Š ``x`fy ‹  y# Š `` yf# ‹ as claimed. 6. (a) lim

rÄ0

sin 6r 6r

œ lim

tÄ0

sin t t

œ 1, where t œ 6r

f(0
hß 0)  f(0ß 0) h hÄ0 36 sin 6h lim œ0 12 hÄ0

(b) fr (0ß 0) œ lim œ

f(rß )
h)  f(rß )) h hÄ0

(c) f) (rß )) œ lim

ˆ sin6h6h ‰ 1 h hÄ0

œ lim

œ lim

hÄ0

œ lim

hÄ0

6 cos 6h  6 12h

(applying l'Hopital's rule twice) s ˆ sin6r6r ‰  ˆ sin6r6r ‰ h hÄ0

œ lim

œ lim

0

hÄ0 h

7. (a) r œ xi  yj  zk Ê r œ krk œ Èx#  y#  z# and ™ r œ (b) rn œ ˆÈx#  y#  z# ‰

sin 6h  6h 6h#

œ0

x È x #
y #
z#

i

y È x #
y #
z#

j

z È x #
y #
z#

kœ

r r

n

ÐnÎ2Ñ

1

(d) dr œ dxi  dyj  dzk Ê r † dr œ x dx  y dy  z dz, and dr œ rx dx  ry dy  rz dz œ

x r

Ê ™ arn b œ nx ax#  y#  z# b (c) Let n œ 2 in part (b). Then

" #

ÐnÎ2Ñ 1

ÐnÎ2Ñ

i  ny ax#  y#  z# b j  nz ax#  y#  z# b k œ nrn 2 r # ™ ar# b œ r Ê ™ ˆ "# r# ‰ œ r Ê r# œ #" ax#  y#  z# b is the function. 1

dx 

y r

dy 

z r

dz

Ê r dr œ x dx  y dy  z dz œ r † dr (e) A œ ai  bj  ck Ê A † r œ ax  by  cz Ê ™ (A † r) œ ai  bj  ck œ A 8. f(g(t)ß h(t)) œ c Ê 0 œ

df dt

œ

` f dx ` x dt



` f dy ` y dt

œ Š `` xf i 

`f `y

j‹ † Š dx dt i 

dy dt

j‹ , where

dx dt

i

dy dt

j is the tangent vector

Ê ™ f is orthogonal to the tangent vector 9. f(xß yß z) œ xz#
yz  cos xy
1 Ê ™ f œ az#
y sin xyb i  (
z
x sin xy)j  (2xz
y)k Ê ™ f(0ß 0ß 1) œ i
j Ê the tangent plane is x
y œ 0; r œ (ln t)i  (t ln t)j  tk Ê rw œ ˆ "t ‰ i  (ln t  1)j  k ; x œ y œ 0, z œ 1 Ê t œ 1 Ê rw (1) œ i  j  k . Since (i  j  k) † (i
j) œ rw (1) † ™ f œ 0, r is parallel to the plane, and r(1) œ 0i  0j  k Ê r is contained in the plane.

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

878

Chapter 14 Partial Derivatives

10. Let f(xß yß z) œ x$  y$  z$
xyz Ê ™ f œ a3x#
yzb i  a3y#
xzb j  a3z#
xyb k Ê ™ f(0ß
1ß 1) œ i  3j  3k $

Ê the tangent plane is x  3y  3z œ 0; r œ Š t4
2‹ i  ˆ 4t
3‰ j  (cos (t
2)) k #

Ê rw œ Š 3t4 ‹ i
ˆ t4# ‰ j
(sin (t
2)) k ; x œ 0, y œ
1, z œ 1 Ê t œ 2 Ê rw (2) œ 3i
j . Since rw (2) † ™ f œ 0 Ê r is parallel to the plane, and r(2) œ
i  k Ê r is contained in the plane. 11.

`z `x

œ 3x#
9y œ 0 and

`z `y

œ 3y#
9x œ 0 Ê y œ

" 3

#

x# and 3 ˆ "3 x# ‰
9x œ 0 Ê

" 3

x%
9x œ 0

Ê x ax$
27b œ 0 Ê x œ 0 or x œ 3. Now x œ 0 Ê y œ 0 or (!ß 0) and x œ 3 Ê y œ 3 or (3ß 3). Next ` #z ` x#

œ 6x,

` #z ` y#

œ 6y, and

and for (3ß 3),

` #z ` #z ` x# ` y#

` #z ` x` y #

` #z ` #z ` x# ` y#

œ
9. For (!ß 0), #

` #z ` x#


Š ``x`zy ‹ œ 243  0 and

#

#


Š ``x`zy ‹ œ
81 Ê no extremum (a saddle point),

œ 18  0 Ê a local minimum.

12. f(xß y) œ 6xyeÐ2x
3yÑ Ê fx (xß y) œ 6y(1
2x)eÐ2x
3yÑ œ 0 and fy (xß y) œ 6x(1
3y)eÐ2x
3yÑ œ 0 Ê x œ 0 and y œ 0, or x œ "# and y œ 3" . The value f(0ß 0) œ 0 is on the boundary, and f ˆ "# ß "3 ‰ œ e"2 . On the positive y-axis,

f(0ß y) œ 0, and on the positive x-axis, f(xß 0) œ 0. As x Ä _ or y Ä _ we see that f(xß y) Ä 0. Thus the absolute maximum of f in the closed first quadrant is e"2 at the point ˆ #" ß 3" ‰ .

13. Let f(xß yß z) œ P! (x! ß y! ß y! ) is

y# x# a#  b# !‰ ˆ 2x a# x

 

z# c#
1 !‰ ˆ 2y b# y

Ê ™fœ  ˆ 2zc#! ‰ z

2y 2x a# i  b# j  # 2y#! ! œ 2x a#  b# 

#

2z c# k Ê an equation of the plane tangent 2z#! ˆ x! ‰ ˆ y! ‰ ˆ z! ‰ c# œ 2 or a# x  b# y  c# z œ 1.

#

at the point

#

The intercepts of the plane are Š xa! ß 0ß 0‹ , Š0ß by! ß 0‹ and Š!ß !ß zc! ‹ . The volume of the tetrahedron formed by the #

#

#

plane and the coordinate planes is V œ ˆ "3 ‰ ˆ #" ‰ Š xa! ‹ Š by! ‹ Š cz! ‹ Ê we need to maximize V(xß yß z) œ subject to the constraint f(xß yß z) œ #

" and ’
(abc) 6 “ Š xyz# ‹ œ

2z c#

x# a#



y# b#



#

z# c#

" œ 1. Thus, ’
(abc) 6 “ Š x# yz ‹ œ

2x a#

(abc)# 6

#

" -, ’
(abc) 6 “ Š xy# z ‹ œ

(xyz)"

2y b#

-,

-. Multiply the first equation by a# yz, the second by b# xz, and the third by c# xy. Then equate

the first and second Ê a# y# œ b# x# Ê y œ substitute into f(xß yß z) œ 0 Ê x œ

a È3

b a

x, x  0; equate the first and third Ê a# z# œ c# x# Ê z œ ca x, x  0;

Ê yœ

Ê zœ

b È3

c È3

Ê Vœ

È3 #

abc.

14. 2(x
u) œ
-, 2(y
v) œ -,
2(x
u) œ ., and
2(y
v) œ
2.v Ê x
u œ v
y, x
u œ
.# , and y
v œ .v Ê x
u œ
.v œ
.# Ê v œ

" #

or . œ 0.

CASE 1: . œ 0 Ê x œ u, y œ v, and - œ 0; then y œ x  1 Ê v œ u  1 and v# œ u Ê v œ v#  1 1 „ È1  4 Ê # " " " " # v œ # and u œ v Ê u œ 4 ; x
4 œ #
Ê y œ 78 . Then f ˆ
8" ß 87 ß "4 ß "# ‰ œ ˆ
8"

Ê v#
v  1 œ 0 Ê v œ

CASE 2:

no real solution. " 4 œ # 2 ˆ 38 ‰

y and y œ x  1 Ê x
# #
"4 ‰  ˆ 78
#" ‰ œ

Ê 2x œ
4" Ê x œ
8" Ê the minimum distance is 38 È2.


x


" #

(Notice that f has no maximum value.) 15. Let (x! ß y! ) be any point in R. We must show lim

Ðhß kÑ Ä Ð0ß 0Ñ

lim

Ðxß yÑ Ä Ðx! ß y! Ñ

f(xß y) œ f(x! ß y! ) or, equivalently that

kf(x!  hß y!  k)
f(x! ß y! )k œ 0. Consider f(x!  hß y!  k)
f(x! ß y! )

œ [f(x!  hß y!  k)
f(x! ß y!  k)]  [f(x! ß y!  k)
f(x! ß y! )]. Let F(x) œ f(xß y!  k) and apply the Mean Value Theorem: there exists 0 with x!  0  x!  h such that Fw (0 )h œ F(x!  h)
F(x! ) Ê hfx (0ß y!  k) œ f(x!  hß y!  k)
f(x! ß y!  k). Similarly, k fy (x! ß () œ f(x! ß y!  k)
f(x! ß y! ) for some ( with y!  (  y!  k. Then kf(x!  hß y!  k)
f(x! ß y! )k Ÿ khfx (0ß y!  k)k  kkfy (x! ß ()k . If M, N are positive real numbers such that kfx k Ÿ M and kfy k Ÿ N for all (xß y) in the xy-plane, then kf(x!  hß y!  k)
f(x! ß y! )k Ÿ M khk  N kkk . As (hß k) Ä 0, kf(x!  hß y!  k)
f(x! ß y! )k Ä 0 Ê lim kf(x!  hß y!  k)
f(x! ß y! )k Ðhß kÑ Ä Ð0ß 0Ñ

œ 0 Ê f is continuous at (x! ß y! ).

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Chapter 14 Additional and Advanced Exercises 16. At extreme values, ™ f and v œ

dr dt

df dt

are orthogonal because

œ ™f†

879

œ 0 by the First Derivative Theorem for

dr dt

Local Extreme Values. 17.

`f `x

œ 0 Ê f(xß y) œ h(y) is a function of y only. Also,

Moreover,

`f `y

œ

`g `x

`g `y

œ

`f `x

œ 0 Ê g(xß y) œ k(x) is a function of x only.

Ê hw (y) œ kw (x) for all x and y. This can happen only if hw (y) œ kw (x) œ c is a constant.

Integration gives h(y) œ cy  c" and k(x) œ cx  c# , where c" and c# are constants. Therefore f(xß y) œ cy  c" and g(xß y) œ cx  c# . Then f(1ß 2) œ g(1ß 2) œ 5 Ê 5 œ 2c  c" œ c  c# , and f(0ß 0) œ 4 Ê c" œ 4 Ê c œ Ê c# œ

9 #

. Thus, f(xß y) œ

" #

y  4 and g(xß y) œ

" #

" #

x  #9 .

18. Let g(xß y) œ Du f(xß y) œ fx (xß y)a  fy (xß y)b. Then Du g(xß y) œ gx (xß y)a  gy (xß y)b œ fxx (xß y)a#  fyx (xß y)ab  fxy (xß y)ba  fyy (xß y)b# œ fxx (xß y)a#  2fxy (xß y)ab  fyy (xß y)b# . 19. Since the particle is heat-seeking, at each point (xß y) it moves in the direction of maximal temperature increase, that is in the direction of ™ T(xß y) œ aec2y sin xb i  a2ec2y cos xb j . Since ™ T(xß y) is parallel to 2ec2y cos x ec2y sin x œ È œ
2 ln #2

the particle's velocity vector, it is tangent to the path y œ f(x) of the particle Ê f w (x) œ

2 cot x.

Integration gives f(x) œ 2 ln ksin xk  C and f ˆ 14 ‰ œ 0 Ê 0 œ 2 ln ¸sin 14 ¸  C Ê C

œ ln Š È22 ‹

#

œ ln 2. Therefore, the path of the particle is the graph of y œ 2 ln ksin xk  ln 2. 20. The line of travel is x œ t, y œ t, z œ 30
5t, and the bullet hits the surface z œ 2x#  3y# when 30
5t œ 2t#  3t# Ê t#  t
6 œ 0 Ê (t  3)(t
2) œ 0 Ê t œ 2 (since t  0). Thus the bullet hits the surface at the point (2ß 2ß 20). Now, the vector 4xi  6yj
k is normal to the surface at any (xß yß z), so that n œ 8i  12j
k is normal to the surface at (2ß 2ß 20). If v œ i  j
5k , then the velocity of the particle †25 ‰ after the ricochet is w œ v
2 projn v œ v
Š 2knvk†#n ‹ n œ v
ˆ 2209 n œ (i  j
5k)
ˆ 400 209 i 

œ
191 209 i


391 209

j


995 209

600 209

j


50 209

k‰

k.

21. (a) k is a vector normal to z œ 10
x#
y# at the point (!ß 0ß 10). So directions tangential to S at (!ß 0ß 10) will be unit vectors u œ ai  bj . Also, ™ T(xß yß z) œ (2xy  4) i  ax#  2yz  14b j  ay#  1b k Ê ™ T(!ß 0ß 10) œ 4i  14j  k . We seek the unit vector u œ ai  bj such that Du T(0ß 0ß 10) œ (4i  14j  k) † (ai  bj) œ (4i  14j) † (ai  bj) is a maximum. The maximum will occur when ai  bj has the same direction as 4i  14j , or u œ È"53 (2i  7j). (b) A vector normal to S at (1ß 1ß 8) is n œ 2i  2j  k . Now, ™ T(1ß 1ß 8) œ 6i  31j  2k and we seek the unit vector u such that Du T(1ß 1ß 8) œ ™ T † u has its largest value. Now write ™ T œ v  w , where v is parallel to ™ T and w is orthogonal to ™ T. Then Du T œ ™ T † u œ (v  w) † u œ v † u  w † u œ w † u. Thus Du T(1ß 1ß 8) is a maximum when u has the same direction as w . Now, w œ ™ T
Š ™knTk#†n ‹ n 62
2 ‰ œ (6i  31j  2k)
ˆ 124
(2i  2j  k) œ ˆ6

4
1

œ
98 9 i

127 9

j


58 9

k Ê uœ

w kwk

152 ‰ i 9

 ˆ31


152 ‰ j 9

 ˆ2


76 ‰ 9 k

" œ
È29,097 (98i
127j  58k).

22. Suppose the surface (boundary) of the mineral deposit is the graph of z œ f(xß y) (where the z-axis points up into the air). Then
`` xf i
`` yf j  k is an outer normal to the mineral deposit at (xß y) and `` xf i  `` yf j points in the direction of steepest ascent of the mineral deposit. This is in the direction of the vector

`f `x

i

`f `y

j at (0ß 0) (the location of the 1st borehole)

that the geologists should drill their fourth borehole. To approximate this vector we use the fact that (0ß 0ß
1000), (0ß 100ß
950), and (100ß !ß
1025) lie on the graph of z œ f(xß y). The plane containing these three points is a good â â j k â â i â â "00 50 â approximation to the tangent plane to z œ f(xß y) at the point (0ß 0ß 0). A normal to this plane is â 0 â â
25 â â "00 0 Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

880

Chapter 14 Partial Derivatives œ
2500i  5000j
10,000k, or
i  2j
4k. So at (0ß 0) the vector

geologists should drill their fourth borehole in the direction of

" È5

`f `x

`f `y

i

j is approximately
i  2j . Thus the

(
i  2j) from the first borehole.

23. w œ ert sin 1x Ê wt œ rert sin 1x and wx œ 1ert cos 1x Ê wxx œ
1# ert sin 1x; wxx œ positive constant determined by the material of the rod Ê
1# ert sin 1x œ

" c#

" c#

wt , where c# is the

arert sin 1xb

# #

Ê ar  c# 1# b ert sin 1x œ 0 Ê r œ
c# 1# Ê w œ ec 1 t sin 1x 24. w œ ert sin kx Ê wt œ rert sin kx and wx œ kert cos kx Ê wxx œ
k# ert sin kx; wxx œ Ê
k# ert sin kx œ

" c#

" c#

wt # #

arert sin kxb Ê ar  c# k# b ert sin kx œ 0 Ê r œ
c# k# Ê w œ ec k t sin kx. # #

Now, w(Lß t) œ 0 Ê ec k t sin kL œ 0 Ê kL œ n1 for n an integer Ê k œ

n1 L

# # # # Ê w œ ec n 1 tÎL sin ˆ nL1 x‰ .

# # # # As t Ä _, w Ä 0 since ¸sin ˆ nL1 x‰¸ Ÿ 1 and ec n 1 tÎL Ä 0.

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

CHAPTER 15 MULTIPLE INTEGRALS 15.1 DOUBLE AND ITERATED INTEGRALS OVER RECTANGLES 1.

'12 '04 2xy dy dx œ '12 cx y# d 40 dx œ '12 16x dx œ c8 x# d 21

2.

'02 'c11 ax
yb dy dx œ '02
xy
12 y# ‘ "
"

3.

'c01 'c11 (x  y  1) dx dy œ 'c01 ’ x2

4.

'01 '01 Š1
x
2 y ‹ dx dy œ '01 ’x
x6

5.

'03 '02 a4
y# b dy dx œ '03 ’4y
y3 “ # dx œ '03 163 dx œ
163 x‘30

6.

'03 'c02 ax# y
2xyb dy dx œ '03 ’ x 2y

7.

'01 '01 1
yx y dx dy œ '01 clnl1  x yld"0 dy œ '01 lnl1  yldy œ cy lnl1  yl
y  lnl1  yld 10 œ 2 ln 2
1

8.

'14 '04 ˆ 2x  Èy‰ dx dy œ '14
41 x2  xÈy‘ !4 dy œ '14 ˆ4  4 y1/2 ‰dy œ
4y  38 y3/2 ‘41

9.

'0ln 2 '1ln 5 e2x
y dy dx œ '0ln 2 ce2x
y dln" 5 dx œ '0ln 2 a5e2x
e2x
1 b dx œ
52 e2x
"# e2x
1 ‘0ln 2

10.

'01 '12 x y ex dy dx œ '01
"# x y2 ex ‘2" dx œ '01 32 x ex dx œ
32 x ex
32 ex ‘10

11.

'c21 '01Î2 y sin x dx dy œ 'c21 c
y cos xd10 Î2 dy œ 'c21 y dy œ
"# y2 ‘2 1 œ 32

12.

'121 '01 asin x  cos yb dx dy œ '121 c
cos x  x cos yd01 dy œ '121 a2  1 cos yb dy œ c2y  1 sin yd121

13.

' ' a6 y#
2 xbdA œ ' ' a6 y#
2 xb dy dx œ ' c2 y3
2 x yd20 dx œ ' a16
4 xb dx œ c16 x
2 x2 d10 œ 14 0 0 0 0

#

#

dx œ '0 2x dx œ c x# d 0 œ 4 2

"

 yx  x“

#

3




2

dy œ 'c1 (2y  2) dy œ cy#  2yd
" œ 1 0


" "

x y# 2 “0 dy

œ '0 Š 65
1

$

!

# #

1

œ 24

!


xy# “


#

!

y# 2 ‹dy

œ ’ 56 y


œ

2 3

œ 16

dx œ '0 a4x
2x# b dx œ ’2x#
3

2

1

y3 6 “0

œ

1

3

2x$ 3 “!

œ0

œ

92 3

œ 32 a5
eb

3 2

œ 21

1

R

14.

'' R

Èx y2 dA

œ '0

4

'12 Èy x dy dx œ '04 ’
Èyx “2 dx œ '04 "# x1Î2 dx œ
31 x3Î2 ‘40 2

1

8 3

' ' x y cos y dA œ ' ' x y cos y dy dx œ ' cx y sin y  x cos yd10 dx œ ' a
2xb dx œ c
x2 dc1 1 œ 0 c1 0 c1 c1 1

15.

œ

1

1

1

R

' ' y sinax  yb dA œ ' ' y sinax  yb dy dx œ ' c
y cosax  yb  sinax  ybd10 dx c1 0 c1 0

16.

1

0

R

œ 'c1 asinax  1b
1 cosax  1b
sin xbdx œ c
cosax  1b
1 sinax  1b  cos xdc0 1 œ 4 0

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

882

Chapter 15 Multiple Integrals

' ' ex  y dA œ ' ' ex  y dy dx œ ' c
ex  y dln0 2 dx œ ' a
ex  ln 2  ex b dx œ c
ex  ln 2  ex dln0 2 œ 0 0 0 0 ln 2

17.

ln 2

ln 2

ln 2

R

' ' x y ex y2 dA œ ' ' x y ex y2 dy dx œ ' ’ "# ex y2 “ dx œ ' ˆ "# ex
"# ‰ dx œ
"# ex
"# x‘20 œ "# ae2
3b 0 0 0 0 2

18.

1

1

2

'' R

20.

'' R

2

0

R

19.

" #

x y3 x2
1 dA

œ '0

y x2 y2
1 dA

1

'02 xx
y 1 dy dx œ '01 ’ 4axx y
1b “2 dx œ '01 x 4
x 1 dx œ c2 lnlx2  1ld10 3

4

2

2

2

0

œ '0

1

œ 2 ln 2

'01 ax yby
1 dx dy œ '01 ctan1 ax ybd10 dy œ '01 tan1 y dy œ
y tan1 y
"# lnl1  y2 l‘10 2

21.

'12 '12

22.

'01 '01 y cos xy dx dy œ '01 csin xyd 1! dy œ '01 sin 1y dy œ

1" cos 1y‘ "! œ
1" (
1
1) œ 12

1 xy

dy dx œ '1

2

" x

(ln 2
ln 1) dx œ (ln 2) '1

2

" x

œ

1 4


"# ln 2

dx œ (ln 2)#

" " 23. V œ ' ' fax, yb dA œ 'c1 'c1 ax2  y2 b dy dx œ 'c1
x2 y  31 y3 ‘ 1 dx œ 'c1 ˆ2 x2  32 ‰ dx œ
32 x3  32 x‘ 1 œ 1

1

1

1

R

24. V œ ' ' fax, yb dA œ '0

2

R

œ

8 3

'02 a16
x2
y2 b dy dx œ '02
16 y
x2 y
13 y3 ‘20 dx œ '02 ˆ 883
2 x2 ‰ dx œ
883 x
23 x3 ‘20

160 3

25Þ V œ ' ' fax, yb dA œ '0

'01 a2
x
yb dy dx œ '01
2 y
x y
"# y2 ‘ "! dx œ '01 ˆ 32
x‰ dx œ
32 x
"# x2 ‘ "! œ 1

26Þ V œ ' ' fax, yb dA œ '0

'02 y2 dy dx œ '04 ’ y4 “2 dx œ '04 1 dx œ cxd40 œ 4

1

R

4

R

27Þ V œ ' ' fax, yb dA œ '0

2

0

1Î2

R

'01Î4 2 sin x cos y dy dx œ '01Î2 c2 sin x sin yd01Î4 dx œ '01Î2 ŠÈ2 sin x‹ dx œ ’
È2 cos x“1Î2 0

œ È2 28. V œ ' ' fax, yb dA œ '0

1

R

'02 a4
y2 b dy dx œ '01
4 y
13 y3 ‘20 dx œ '01 ˆ 163 ‰ dx œ
163 x‘ "! œ 163

15.2 DOUBLE INTEGRALS OVER GENERAL REGIONS 1.

2.

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Section 15.2 Double Integrals Over General Regions 3.

4.

5.

6.

7.

8.

9. (a)

'!# 'x8 dy dx 3

(b)

'!8 '0y

10. (a)

'!3 '02x dy dx

(b)

'!6 'y3Î2 dx dy

11. (a)

'!3 'x3x dy dx

(b)

'!9 'yÈÎ3y dx dy

12. (a)

'!# '1e dy dx

(b)

'1e 'ln2 y dx dy

13. (a)

'!9 '0

2

x

1Î3

dx dy

2

Èx

(b)

dy dx

'0 'y dx dy 3

9 2

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

883

884

Chapter 15 Multiple Integrals

14. (a)

'!1Î4 'tan1 x dy dx

(b)

15. (a)

'01 '0tan

16. (a)

dx dy

'!ln 3 'e1c dy dx x

'1Î3 'ln y dx dy 1

(b)

c1 y

ln 3

'!1 '01 dy dx  '1e 'ln1 x dy dx '01 '0e dx dy y

(b)

17. (a) (b)

18. (a)

'!1 'x3  2x dy dx

'01 '0y dx dy  '13 '0a3  ybÎ2 dx dy

'21 'xx
2 dy dx 2

'0 'Èy dx dy  '13 'yÈy2 dx dy 1

(b)

19.

Èy

'01 '0x (x sin y) dy dx œ '01 c
x cos yd x! dx 1 1 œ '0 (x
x cos x) dx œ ’ x2
(cos x  x sin x)“ #

œ

1# #

!

2

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Section 15.2 Double Integrals Over General Regions 20.

'01 '0sin x y dy dx œ '01 ’ y2 “ sin x dx œ '01 "# sin# x dx #

!

œ

21.

" 4

'01 (1
cos 2x) dx œ "4
x
"2 sin 2x‘ !1 œ 14

'1ln 8 '0ln yexby dx dy œ '1ln 8 cexbyd !ln y dy œ '1ln 8 ayey
eyb dy œ c(y
1)ey
ey d 1ln 8 œ 8(ln 8
1)
8  e œ 8 ln 8
16  e

'12 'yy

#

22.

dx dy œ '1 ay#
yb dy œ ’ y3
2

$

œ ˆ 83
2‰
ˆ "3
"# ‰ œ

7 3




œ

3 #

5 6

'01 '0y 3y$ exy dx dy œ '01 c3y# exy d 0y #

23.

# y# # “"

#

dy

œ '0 Š3y# ey
3y# ‹ dy œ ’ey
y$ “ œ e
2 1

$

"

$

!

24.

Èx

'14 '0 œ

3 #

3 #

eyÎÈx dy dx œ

'14
32 Èx eyÎÈx ‘ 0Èx dx

% (e
1) '1 Èx dx œ
23 (e
1) ˆ 32 ‰ x$Î# ‘ " œ 7(e
1) 4

25.

'12 'x2x

26.

'01 '01cx ax#  y# b dy dx œ '01 ’x# y  y3 “ "

x y

dy dx œ '1 cx ln yd x2x dx œ (ln 2) '1 x dx œ 2

2

$

x

0

$

œ ’ x3
27.

x% 4




" (1x)% 1# “ !

œ ˆ "3


" 4

#

œ '0 Š "#
u  1

u# #


vÈ u “

ln 2

dx œ '0 ’x# (1
x)  1


0‰
ˆ0
0


'01 '01cu ˆv
Èu‰ dv du œ '01 ’ v2

3 #

"

u

0


u"Î#  u$Î# ‹ du œ ’ u2


" ‰ 1#

œ

(1x)$ 3 “

dx œ '0 ’x#
x$  1

(1x)$ 3 “

dx

" 6

du œ '0 ’ 12u#
u
Èu(1
u)“ du 1

u# #



u$ 6

#

"


32 u$Î#  25 u&Î# “ œ !

" #




" #



" 6




2 3



Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

2 5

œ
#" 

2 5

" œ
10

885

886 28.

Chapter 15 Multiple Integrals

'12 '0ln t es ln t ds dt œ '12 ces ln td 0ln t dt œ '12 (t ln t
ln t) dt œ ’ t2

#

œ (2 ln 2
1
2 ln 2  2)
ˆ
"4  1‰ œ 29.

" 4

ln t


t# 4


t ln t  t“

# "

'c02 'vcv 2 dp dv œ 2'c02 cpd vv dv œ 2'c02
2v dv œ
2 cv# d c2 œ 8 0

30.

È1cs

'01 '0

#

È1cs

8t dt ds œ '0 c4t# d 0 1

œ '0 4 a1
s# b ds œ 4 ’s
1

31.

#

ds

" s$ 3 “!

œ

8 3

'c11ÎÎ33 '0sec t 3 cos t du dt œ ' 11ÎÎ33 c(3 cos t)ud 0sec t 1Î3

œ 'c1Î3 3 dt œ 21

32.

'03Î2 '14 2u 4 v 2u dv du œ '03Î2
2u v 4 ‘ 14 2u du 3Î2 $Î2 œ '0 a3
2ub du œ c3u
u# d ! œ 92 #

33.

'24 '0Ð4

y)Î2

34.

' 02 '0x
2 dy dx

dx dy

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Section 15.2 Double Integrals Over General Regions 35.

'01 'xx dy dx

36.

'01 '1cy1cydx dy

37.

'1e 'ln1ydx dy

38.

'12 '0ln x dy dx

39.

'09 '0

40.

'04 '0

#

È

1 2

È9cy

È4cx

16x dx dy

y dy dx

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

887

888

Chapter 15 Multiple Integrals È1cx

41.

'c11 '0

42.

'c22 '0

43.

'01 'ee x y dx dy

44.

'01Î2 '0sin

45.

'1e 'ln3 x ax  ybdy dx

46.

'01Î3 'tan 3x Èx y dy dx

È4cy

#

3y dy dx

#

6x dx dy

y

c1 y

x y2 dx dy

3

È

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Section 15.2 Double Integrals Over General Regions 47.

48.

'01 'x1 siny y dy dx œ '01 '0y siny y dx dy œ '01 sin y dy œ 2

'02 'x2 2y# sin xy dy dx œ '02 '0y2y# sin xy dx dy 2 2 œ '0 c
2y cos xyd 0y dy œ '0 a
2y cos y#  2yb dy #

œ c
sin y#  y# d ! œ 4
sin 4

49.

'01 'y1 x# exy dx dy œ '01 '0x x# exy dy dx œ '01 cxexyd 0x dx œ '0 axex
xb dx œ ’ "2 ex
1

È4cy

'02 '04cx 4xey dy dx œ '04 '0 #

50.

2y

œ '0 ’ #x(4ey) “ 4

51.

'02

# 2y

Èln 3 Èln 3

'y/2

Èln 3

œ '0

52.

" x# # “!

#

#

È4cy

0

dy œ '0

4 2y e

Èln 3

ex dx dy œ '0 #

#

#

Èln 3

$

dy dx œ '0

1

'03y

#

e2 #

dx dy 2y

%

dy œ ’ e4 “ œ

#

2xex dx œ cex d 0

'03 'È1xÎ3 ey

xe2y 4 y

œ

!

'02x ex

#

e)  " 4

dy dx

œ eln 3
1 œ 2

$

ey dx dy

œ '0 3y# ey dy œ cey d ! œ e
1 1

53.

$

$

"

'01Î16 'y1Î2 cos a161x& b dx dy œ '01Î2 '0x "Î%

%

cos a161x& b dy dx

161x b œ '0 x% cos a161x& b dx œ ’ sin a80 “ 1 1Î2

&

"Î# !

œ

" 801

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

889

890 54.

Chapter 15 Multiple Integrals

'08 'È2x $

œ '0

2

55.

dy dx œ '0

2

"

y %
1 y$ y %
1

dy œ

" 4

'0y y "
1 dx dy $

%

#

cln ay%  1bd ! œ

ln 17 4

' ' ay
2x# b dA R

xb1

œ 'c1 'cxc1 ay
2x# b dy dx  '0 0

1

'x1cc1x ay
2x# b dy dx

x " 1
x œ 'c1
"2 y#
2x# y‘
x
1 dx  '0
2" y#
2x# y‘ x
1 dx 0

1

œ 'c1
"# (x  1)#
2x# (x  1)
"# (
x
1)#  2x# (
x
1)‘dx 0

 '0
"# (1
x)#
2x# (1
x)
"# (x
1)#  2x# (x
1)‘ dx 1

œ
4 'c1 ax$  x# b dx  4 '0 ax$
x# b dx 0

1

%

œ
4 ’ x4 

56.

0

x$ 3 “ c1

" x$ 3 “!

%

 4 ’ x4


%

œ 4 ’ (41) 

(1)$ 3 “

3  4 ˆ 4"
3" ‰ œ 8 ˆ 12


4 ‰ 12

8 œ
12 œ
32

' ' xy dA œ ' ' xy dy dx  ' ' xy dy dx 0 x 2Î3 x 2Î3

R

2x

2Î3

1

2x 2 œ '0
"2 xy# ‘ x dx  '2Î3
2" xy# ‘ x 1

x

2 x

dx

œ '0 ˆ2x$
"# x$ ‰ dx  '2Î3
"# x(2
x)#
"# x$ ‘ dx 2Î3

1

œ '0

2Î3

3 #

x$ dx  '2Î3 a2x
x# b dx 1

2Î3 " 2‰ 8 ‰‘ ‰ ˆ
4 ˆ 2 ‰ ˆ 27 œ
38 x% ‘ 0 
x#
23 x$ ‘ #Î$ œ ˆ 38 ‰ ˆ 16 œ 81  1
3
9
3

57. V œ '0

1

'x2cx ax#  y# b dy dx œ '01 ’x# y  y3 “ 2cx dx œ '01 ’2x#
7x3 $

$

x

œ ˆ 23


7 12




2cx#

58. V œ 'c2 'x 1

œ ˆ 23


" 5

" ‰ 12


ˆ0
0


4cx#

1

œ

x# dy dx œ 'c2 cx# yd x 32 5




16 ‰ 4



(2x)$ 3 “

È 4 cx

40 œ ˆ 60


2

7x% 12




13 81 " (2x)% 12 “ !

12 60




15 ‰ 60


ˆ
320 60 

384 60




240 ‰ 60

œ

189 60

œ

63 20

4cx (x  4) dy dx œ 'c4 cxy  4yd 3x dx œ 'c4 cx a4
x# b  4 a4
x# b
3x#
12xd dx 1

1

#

#

(3
y) dy dx œ '0 ’3y
2

œ ’ 32 xÈ4
x#  6 sin" ˆ x# ‰
2x  61. V œ '0

$

œ

"

"

'0

16 ‰ 81

dx œ ’ 2x3


1

1

2


ˆ 36 81


dx œ 'c2 a2x#
x%
x$ b dx œ
23 x$
15 x&
14 x% ‘ #

œ 'c4 a
x$
7x#
8x  16b dx œ

41 x%
37 x$
4x#  16x‘ % œ ˆ
4"


60. V œ '0

27 81

4 3

2cx#

1


4" ‰
ˆ
16 3 

59. V œ 'c4 '3x

16 ‰ 12



6 81

È 4c x

y# 2 “0

# x$ 6 “!

#

7 3

‰  12‰
ˆ 64 3
64 œ

dx œ '0 ’3È4
x#
Š 4#x ‹“ dx 2

œ 6 ˆ 1# ‰
4 

#

8 6

œ 31


16 6

œ

918 3

'03 a4
y# b dx dy œ '02 c4x
y# xd $! dy œ '02 a12
3y# b dy œ c12y
y$ d !# œ 24
8 œ 16

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

157 3




" 4

œ

625 12

Section 15.2 Double Integrals Over General Regions 62. V œ '0

2

'04cx

#

2

œ
8x
43 x$  63. V œ '0

2

4cx#

a4
x#
yb dy dx œ '0 ’a4
x# b y
" 10

#

x& ‘ ! œ 16




32 3

32 10

œ

y# 2 “!

480320
96 30

œ

" #

a4
x# b dx œ '0 Š8
4x#  2

#

!

xb1

1

1Îx

2

66. V œ 4 '0

1Î3

'x1cc1x (3
3x) dy dx œ 6 'c01 a1
x# b dx  6 '01 (1
x)# dx œ 4  2 œ 6

2

2

" x


ˆ
1
x" ‰‘dx œ 2 '1 ˆ1  x" ‰ dx œ 2 cx  ln xd #" 2

'0sec x a1  y# b dy dx œ 4 '01Î3 ’y  y3 “ sec x dx œ 4 '01Î3 Šsec x  sec3 x ‹ dx $

$

0

1Î$

c7 ln ksec x  tan xk  sec x tan xd !

œ

’7 ln Š2  È3‹  2È3“

2 3

67.

68.

'1_ 'ec1 x"y dy dx œ '1_ ’ lnx y “ " x

$

$

ec x

_

dx œ '1
ˆ x$x ‰ dx œ
lim

bÄ_

1/ ˆ1cx ‰ È1cx 1 70. 'c1 'c1/È1cx (2y  1) dy dx œ 'c1 cy#  ydº 1

1/

# 1Î#

#

#

c1/ a1c

x# b1Î#

œ 4 lim c csin" b
0d œ 21 bÄ1 71.

dx

128 15

65. V œ '1 'c1Îx (x  1) dy dx œ '1 cxy  yd 1Î1xÎx dx œ '1
1  œ 2(1  ln 2)

69.

x% #‹

%

0

2 3

2

# 2 x '02cx a12
3y# b dy dx œ '02 c12y
y$ d # dx œ '0 c24
12x
(2
x)$ d dx œ ’24x
6x#  (24x) “ œ 20 !

64. V œ 'c1 'cxc1 (3
3x) dy dx  '0

œ

dx œ '0

_ _ 'c_ ' _ ax
1b"ay
1b -dx dy œ 2 '0_ Š y 2
1 ‹ Š #

#

#

lim

bÄ_

dx œ 'c1 È 2


x" ‘ b œ
lim 1

1

1 x #

bÄ_

ˆ "b
1‰ œ 1

dx œ 4 lim c csin" xd ! bÄ1 b

tan" b
tan" 0‹ dy œ 21 lim

bÄ_

'0b y "
1 dy #

œ 21 Š lim tan" b
tan" 0‹ œ (21) ˆ 1# ‰ œ 1# bÄ_

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

891

892 72.

Chapter 15 Multiple Integrals

'0_ '0_ xecÐx

2yÑ

_

_

c
xe
x
e
x d b0 dy œ '0 e
2y lim

bÄ_

œ '0 ec2y dy œ 73.

_

dx dy œ '0 e
2y lim " # b lim Ä_

a
be
b
e
b  1b dy

bÄ_

a
ec2b  1b œ

" #

' ' f(xß y) dA ¸ "4 f ˆ
"# ß 0‰  8" f(0ß 0)  8" f ˆ "4 ß 0‰ œ "4 ˆ
"# ‰  8" ˆ0  "4 ‰ œ
323 R

74.

' ' f(xß y) dA ¸ "4 ’f ˆ 47 ß 114 ‰  f ˆ 94 ß 114 ‰  f ˆ 74 ß 134 ‰  f ˆ 94 ß 134 ‰“ œ R

75. The ray ) œ

1 6

" 16

(29  31  33  35) œ

128 16

œ8

meets the circle x#  y# œ 4 at the point ŠÈ3ß 1‹ Ê the ray is represented by the line y œ

È

È

È

$Î# 3 4cx 3 x# b ' ' f(xß y) dA œ ' ' È È4
x# dy dx œ ' ’a4
x# b
Èx3 È4
x# “ dx œ ”4x
x3$  a4È 0 xÎ 3 0 3 3 • #

R

76.

'2_ '02 ax xb "(y1) #

bÄ_ bÄ_

77. V œ '0

1

0

cln (x
1)
ln xd 2b œ 6 lim

lim

bÄ_

_

dx œ 6 '2

bÄ_

dx x(x1)

[ln (b
1)
ln b
ln 1  ln 2]

$

x

7x 3

œ ˆ 23


" ‰ 1#

7 12




$



(2x)$ 3 “

$

dx œ ’ 2x3



ˆ0
0


16 ‰ 12

œ

œ '0

 '2

œ

2 tan

ˆ1 1" ‰ y 1
y# dy 1  1 ˆ 21 ‰ ln 5  "

21

ˆ2  1 1
y # "

"

21
2 tan

y‰

21


2


%

7x 12




" (2x)% 12 “ !

4 3

'02 atan" 1x
tan" xb dx œ '02 'x1x 1
"y

œ 2 tan

3 ‰ x # x

'x2cx ax#  y# b dy dx œ '01 ’x# y  y3 “ 2cx dx

œ '0 ’2x#


2

2

"Î$

0

ln ˆ1
"b ‰  ln 2“ œ 6 ln 2

1

78.

_

1) ' ˆ x#3x  dy dx œ '2 ’ 3(y ax# xb “ dx œ 2

'2b ˆ x" 1
"x ‰ dx œ 6

œ 6 lim

œ 6 ’ lim

_

#Î$

È3

dy dx œ '0

2

#

'yyÎ1

" 1
y #

dx dy  '2

21

# # " ‰
dy œ ˆ 12" y 1 cln a1  y bd !  2 tan " 21

" 21

ln a1  41# b
2 tan" 2  #

ln a1  41 b 

" #1

" #1

'y2Î1 1
"y

#

dx dy 21

ln a1  y# b‘ 2

ln 5

ln 5 #

79. To maximize the integral, we want the domain to include all points where the integrand is positive and to exclude all points where the integrand is negative. These criteria are met by the points (xß y) such that 4
x#
2y#   0 or x#  2y# Ÿ 4, which is the ellipse x#  2y# œ 4 together with its interior. 80. To minimize the integral, we want the domain to include all points where the integrand is negative and to exclude all points where the integrand is positive. These criteria are met by the points (xß y) such that x#  y#
9 Ÿ 0 or x#  y# Ÿ 9, which is the closed disk of radius 3 centered at the origin. 81. No, it is not possible. By Fubini's theorem, the two orders of integration must give the same result.

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

x È3

œ

. Thus,

20È3 9

Section 15.2 Double Integrals Over General Regions 82. One way would be to partition R into two triangles with the line y œ 1. The integral of f over R could then be written as a sum of integrals that could be evaluated by integrating first with respect to x and then with respect to y:

' ' f(xß y) dA R

œ '0

1

'22ccÐ2yyÎ2Ñ f(xß y) dx dy  '12 'y2c1ÐyÎ2Ñ f(xß y) dx dy.

Partitioning R with the line x œ 1 would let us write the integral of f over R as a sum of iterated integrals with order dy dx. 83.

' bb ' bb e

x# y#

dx dy œ '

b ' e b b

b

#

y#

e

x#

dx dy œ ' b e b

#

y#

Œ' b e b

x#

dx dy œ Œ' b e b

x#

dx Œ' b e b

y#

dy

#

# # # œ Œ'cb ecx dx œ Œ2 '0 ecx dx œ 4 Œ'0 ecx dx ; taking limits as b Ä _ gives the stated result.

b

84.

'01 '03 (yx1)

dy dx œ '0

3

#

œ

b

#Î$

" 3 b lim Ä 1c

'0

b

dy (y1)#Î$



'01 (yx1)

dx dy œ '0

3

#

#Î$

" 3

b

'b

3

lim

b Ä 1b

dy (y1)#Î$

œ

" (y1)#Î$

lim

b Ä 1c

$

"

’ x3 “ dy œ !

" 3

'03 (ydy1)

#Î$


(y
1)"Î$ ‘ b  lim
(y
1)"Î$ ‘ 3 0 b b Ä 1b

3 3 œ ’ lim c (b
1)"Î$
(
1)"Î$ “
’ lim b (b
1)"Î$
(2)"Î$ “ œ (0  1)
Š0
È 2‹ œ 1  È 2 bÄ1 bÄ1

85-88. Example CAS commands: Maple: f := (x,y) -> 1/x/y; q1 := Int( Int( f(x,y), y=1..x ), x=1..3 ); evalf( q1 ); value( q1 ); evalf( value(q1) ); 89-94. Example CAS commands: Maple: f := (x,y) -> exp(x^2); c,d := 0,1; g1 := y ->2*y; g2 := y -> 4; q5 := Int( Int( f(x,y), x=g1(y)..g2(y) ), y=c..d ); value( q5 ); plot3d( 0, x=g1(y)..g2(y), y=c..d, color=pink, style=patchnogrid, axes=boxed, orientation=[-90,0], scaling=constrained, title="#89 (Section 15.2)" ); r5 := Int( Int( f(x,y), y=0..x/2 ), x=0..2 ) + Int( Int( f(x,y), y=0..1 ), x=2..4 ); value( r5); value( q5-r5 ); 85-94. Example CAS commands: Mathematica: (functions and bounds will vary) You can integrate using the built-in integral signs or with the command Integrate. In the Integrate command, the integration begins with the variable on the right. (In this case, y going from 1 to x).

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

893

894

Chapter 15 Multiple Integrals

Clear[x, y, f] f[x_, y_]:= 1 / (x y) Integrate[f[x, y], {x, 1, 3}, {y, 1, x}] To reverse the order of integration, it is best to first plot the region over which the integration extends. This can be done with ImplicitPlot and all bounds involving both x and y can be plotted. A graphics package must be loaded. Remember to use the double equal sign for the equations of the bounding curves. Clear[x, y, f] ˆ "# ‰ œ '0 t
"Î# e
t dt œ '0 ay# b 24. Q œ '0

" #

#

È1 # ‹

c)
cos )d #!1 œ

'cÈhh 'cÈhhccxx 'xhby

œ È1, where y œ Èt

21kR$ 3

#

#

#

#

dz dy dx

ah  r# b r dr d) œ '0 ’ hr2  r4 “ 21

#

%

Èh !

d) œ '0

21

h# 4

d) œ

h# 1 #

.

Since the top of the bowl has area 101, then we calibrate the bowl by comparing it to a right circular cylinder whose cross sectional area is 101 from z œ 0 to z œ 10. If such a cylinder contains to a depth w then we have 101w œ rain, w œ 3 and h œ È60.

h# 1 #

Ê wœ

h# 20

h# 1 #

cubic inches of water . So for 1 inch of rain, w œ 1 and h œ È20; for 3 inches of

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Chapter 15 Additional and Advanced Exercises 26. (a) An equation for the satellite dish in standard position is z œ "# x#
"# y# . Since the axis is tilted 30°, a unit vector v œ 0i
aj
bk normal to the plane of the È3 # È
#3

water level satisfies b œ v † k œ cos ˆ 16 ‰ œ Ê a œ È1  b# œ  "# Ê v œ  "# j Ê  "# (y  1)
Ê zœ

" È3

È3 #

y
Š "# 

k

ˆz  "# ‰ œ 0 " È3 ‹

is an equation of the plane of the water level. Therefore

' Èx byby c È dz dy dx, where R is the interior of the ellipse

the volume of water is V œ ' '

1

1 2

3

#

1 2

R

1 2

1

3

#

È 3  Ê 3
4 Š È 3
1‹ 2

x#
y# 

È

2

and " œ

y1


2 È3


Ê 43 3

œ 0. When x œ 0, then y œ ! or y œ " , where ! œ

2 È3

Ê Vœ

3

#

(b) x œ 0 Ê z œ

" #

y# and

!

œ y; y œ 1 Ê

dz dy

"Î#

Š yb1c È3 cy ‹

' 'c "


4 Š È
1‹ 2

2 3

#

2

3

"Î#

Š yb1c È3 cy# ‹ 2 3

2

yb

' Èb 1

1 2

x#

1 2

1 2

4

2

#

c È3 1

1 dz dx dy

y#

œ 1 Ê the tangent line has slope 1 or a 45° slant

dz dy

Ê at 45° and thereafter, the dish will not hold water. 27. The cylinder is given by x#
y# œ 1 from z œ 1 to _ Ê œ '0

21

_

'0 '1 1

z ar#  z# b&Î#

'0 '0 ' 21

dz r dr d) œ a lim Ä_

1

' ' ' z ar#
z# b
&Î# dV D

a

rz 1 ar#  z# b&Î#

dz dr d)

œ a lim Ä_

'021 '01 ’ˆ "3 ‰

œ a lim Ä_

'021 ’ 3" ar#
a# b
"Î#  3" ar#
1b
"Î# “ " d) œ a lim ' 21 ’ 3" a1
a# b
"Î#  3" ˆ2
"Î# ‰  3" aa# b
"Î#
3" “ d) Ä_ 0

œ a lim 21 ’ 3" a1
a# b Ä_

a

r “ ar#  z# b$Î# 1


"Î#

 3" Š

'021 '01 ’ˆ 3" ‰

dr d) œ a lim Ä_

È2 # ‹


ˆ "3 ‰

r ar#  a# b$Î#

!

r “ ar#  1b$Î#

dr d)

È2 # “.

 3" ˆ "a ‰
3" “ œ 21 ’ 3"  ˆ 3" ‰

28. Let's see? The length of the "unit" line segment is: L œ 2'0 dx œ 2. 1

The area of the unit circle is: A œ 4'0

1

È1 c x

'0

The volume of the unit sphere is: V œ 8'0

1

2

dy dx œ 1.

È1 c x

'0

2

È1 c x c y

'0

2

2

dz dy dx œ 43 1.

Therefore, the hypervolume of the unit 4-sphere should be: Vhyper œ 16'0

1

È1cx

'0

2

È1cx cy

'0

2

2

È1cx cy cz

'0

2

2

2

dw dz dy dx.

Mathematica is able to handle this integral, but we'll use the brute force approach. Vhyper œ 16'0

1

È1cx

œ 16'0

'0

œ 16'0

'0

œ 16'0

'0

1

1

1

È1cx È1cx

È1cx

'0

2

2

2

2

È1cx cy

'0

È1cx cy

'0

2

2

È1cx cy cz

'0

2

2

2

dw dz dy dx œ 16'0

1

2

z2 È 1
x2
y2 É 1
1 c x2 c y2

dz dy dx œ –

È1cx

'0

dz œ

a1  x2  y2 b'1/2 È1  cos2 ) sin ) d) dy dx œ 16'0 0

1 4 a1

1

2

È1cx cy

'0

2

2

È 1
x 2
y 2
z2

z È1
x2
y2 œ cos ) È1  x2  y2 sin

È1cx

'0

2

) d)

—

a1  x2  y2 b'1/2 sin2 ) d) dy dx 0

1

1
x3 $

‘ dx œ 83 1' a1  x2 b3/2 dx œ ” 0 1

0 x œ cos ) œ  83 1'1/2 sin4 ) d) dx œ sin ) d) •

2 ) ‰2 œ  83 1'1/2 ˆ 1
cos d) œ  23 1'1/2 a1  2 cos 2)
cos2 2)bd) œ  23 1'1/2 ˆ #3  2 cos 2)
2 0

dz dy dx

3/2  x2  y2 b dy dx œ 41'0 ŠÈ1  x2  x2 È1  x2  3" a1  x2 b ‹ dx

œ 41'0 È1  x2
a1  x2 b  1

2

0

0

cos 4) ‰ d) 2

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

œ

12 2

937

938

Chapter 15 Multiple Integrals

NOTES:

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

CHAPTER 16 INTEGRATION IN VECTOR FIELDS 16.1 LINE INTEGRALS 1. r œ ti
a"  tbj Ê x œ t and y œ 1  t Ê y œ 1  x Ê (c) 2. r œ i
j
tk Ê x œ 1, y œ 1, and z œ t Ê (e) 3. r œ a2 cos tbi
a2 sin tbj Ê x œ 2 cos t and y œ 2 sin t Ê x#
y# œ 4 Ê (g) 4. r œ ti Ê x œ t, y œ 0, and z œ 0 Ê (a) 5. r œ ti
tj
tk Ê x œ t, y œ t, and z œ t Ê (d) 6. r œ tj
a2  2tbk Ê y œ t and z œ 2  2t Ê z œ 2  2y Ê (b) 7. r œ at#  1b j
2tk Ê y œ t#  1 and z œ 2t Ê y œ

z# 4

 1 Ê (f)

8. r œ a2 cos tbi
a2 sin tbk Ê x œ 2 cos t and z œ 2 sin t Ê x#
z# œ 4 Ê (h) 9. ratb œ ti
a1  tbj , 0 Ÿ t Ÿ 1 Ê

œ i  j Ê ¸ ddtr ¸ œ È2 j ; x œ t and y œ 1  t Ê x
y œ t
("  t) œ 1

dr dt

Ê 'C faxß yß zb ds œ '0 fatß 1  tß 0b ¸ ddtr ¸ dt œ '0 (1) ŠÈ2‹ dt œ ’È2 t“ œ È2 1

"

1

!

10. r(t) œ ti
(1  t)j
k , 0 Ÿ t Ÿ 1 Ê œ t  (1  t)
1  2 œ 2t  2 Ê

dr dt

œ i  j Ê ¸ ddtr ¸ œ È2; x œ t, y œ 1  t, and z œ 1 Ê x  y
z  2

'C f(xß yß z) ds œ '01 (2t  2) È2 dt œ È2 ct#  2td "! œ È2

11. r(t) œ 2ti
tj
(2  2t)k , 0 Ÿ t Ÿ 1 Ê

dr dt

œ 2i
j  2k Ê ¸ ddtr ¸ œ È4
1
4 œ 3; xy
y
z

œ (2t)t
t
(2  2t) Ê 'C f(xß yß z) ds œ '0 a2t#  t
2b 3 dt œ 3
23 t$  "# t#
2t‘ ! œ 3 ˆ 23  1

12. r(t) œ (4 cos t)i
(4 sin t)j
3tk , 21 Ÿ t Ÿ 21 Ê Ê ¸ ddtr ¸ œ È16 sin# 1 œ c20td #
# 1 œ 801

"

dr dt

dr dt

Ê ¸ ddtr ¸ œ È1
9
4 œ È14 ; x
y
z œ (1  t)
(2  3t)
(3  2t) œ 6  6t Ê

œ  i  3 j  2k

'C f(xß yß z) ds

œ '0 (6  6t) È14 dt œ 6È14 ’t  t2 “ œ Š6È14‹ ˆ "# ‰ œ 3È14 " !

14. r(t) œ ti
tj
tk , 1 Ÿ t Ÿ _ Ê

_

dr dt

13 #

'C f(xß yß z) ds œ 'c2211 (4)(5) dt

13. r(t) œ (i
2j
3k)
t(i  3j  2k) œ (1  t)i
(2  3t)j
(3  2t)k , 0 Ÿ t Ÿ 1 Ê

#


2‰ œ

œ (4 sin t)i
(4 cos t)j
3k

t
16 cos# t
9 œ 5; Èx#
y# œ È16 cos# t
16 sin# t œ 4 Ê

1

" #

œ i
j
k Ê ¸ ddtr ¸ œ È3 ;

È3 x #  y#  z#

_ Ê 'C f(xß yß z) ds œ '1 Š 3t#3 ‹ È3 dt œ
 1t ‘ " œ lim ˆ b"
1‰ œ 1

œ

È3 t#  t#  t#

œ

È3 3t#

È

bÄ_

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

940

Chapter 16 Integration in Vector Fields

15. C" : r(t) œ ti
t# j , 0 Ÿ t Ÿ 1 Ê

œ i
2tj Ê ¸ ddtr ¸ œ È1
4t# ; x
Èy  z# œ t
Èt#  0 œ t
ktk œ 2t

dr dt

$Î# since t   0 Ê 'C f(xß yß z) ds œ '0 2tÈ1
4t# dt œ ’ "6 a"
4t# b “ œ "

1

!

"

C# : r(t) œ i
j
tk, 0 Ÿ t Ÿ 1 Ê

dr dt

1

"

#

"

5 6

#

16. C" : r(t) œ tk , 0 Ÿ t Ÿ 1 Ê

dr dt

(5)$Î# 

" 6

" 6

œ

Š5È5  1‹ ;

œ k Ê ¸ ddtr ¸ œ 1; x
Èy  z# œ 1
È1  t# œ 2  t#

Ê 'C f(xß yß z) ds œ '0 a2  t# b (1) dt œ
2t  "3 t$ ‘ ! œ 2 

œ 'C f(xß yß z) ds
'C f(xß yß z) ds œ

" 6

È5


" 3

œ

5 3

; therefore 'C f(xß yß z) ds

3 #

œ k Ê ¸ ddtr ¸ œ 1; x
Èy  z# œ 0
È0  t# œ t#

Ê 'C f(xß yß z) ds œ '0 at# b (1) dt œ ’ t3 “ œ  3" ; 1

"

$

!

"

C# : r(t) œ tj
k, 0 Ÿ t Ÿ 1 Ê

œ j Ê ¸ ddtr ¸ œ 1; x
Èy  z# œ 0
Èt  1 œ Èt  1

dr dt

" Ê 'C f(xß yß z) ds œ '0 ˆÈt  1‰ (1) dt œ
23 t$Î#  t‘ ! œ 1

#

C$ : r(t) œ ti
j
k , 0 Ÿ t Ÿ 1 Ê

dr dt #

œ  "6 17. r(t) œ ti
tj
tk , 0  a Ÿ t Ÿ b Ê Ê

" !

$

dr dt

 1 œ  3" ;

œ i Ê ¸ ddtr ¸ œ 1; x
Èy  z# œ t
È1  1 œ t

Ê 'C f(xß yß z) ds œ '0 (t)(1) dt œ ’ t2 “ œ 1

2 3

" #

Ê

'C f(xß yß z) ds œ 'C

"

œ i
j
k Ê ¸ ddtr ¸ œ È3 ;

f ds
'C f ds
'C f ds œ  3"
ˆ 3" ‰
#

xyz x #  y #  z#

œ

'C f(xß yß z) ds œ 'ab ˆ 1t ‰ È3 dt œ ’È3 ln ktk “ b œ È3 ln ˆ ba ‰ , since 0  a Ÿ b

$

ttt t#  t#  t#

œ

" #

1 t

a

18. r(t) œ aa cos tb j
aa sin tb k , 0 Ÿ t Ÿ 21 Ê

dr dt

œ (a sin t) j
(a cos t) k Ê ¸ ddtr ¸ œ Èa# sin# t
a# cos# t œ kak ;

21 1  kak sin t, 0 Ÿ t Ÿ 1 Èx#
z# œ È0
a# sin# t œ œ Ê 'C f(xß yß z) ds œ '0  kak# sin t dt
'1 kak# sin t dt kak sin t, 1 Ÿ t Ÿ 21

1

#1

œ ca# cos td !  ca# cos td 1 œ ca# (1)  a# d  ca#  a# (1)d œ 4a# Ê 'C x ds œ '0 t

È5 2

4

È5 2 dt

È5 2

'04 t dt œ ’ È45 t2 “ 4 œ 4È5

19. (a) ratb œ ti
"# tj , 0 Ÿ t Ÿ 4 Ê

dr dt

œ i
"# j Ê ¸ ddtr ¸ œ

(b) ratb œ ti
t j , 0 Ÿ t Ÿ 2 Ê

dr dt

œ i
2tj Ê ¸ ddtr ¸ œ È1
4t2 Ê 'C x ds œ '0 t È1
4t2 dt

2

3Î2 2

1 œ ’ 12 a1
4t2 b

“ œ !

œ

2

17È17
" 12

20. (a) ratb œ ti
4tj , 0 Ÿ t Ÿ 1 Ê

dr dt

œ i
4j Ê ¸ ddtr ¸ œ È17 Ê 'C Èx
2y ds œ '0 Èt
2a4tb È17 dt 1

œ È17'0 È9t dt œ 3È17'0 Èt dt œ ’2È17 t2Î3 “ œ 2È17 1

1

1

!

(b) C" : ratb œ ti , 0 Ÿ t Ÿ 1 Ê

'C Èx
2y ds œ 'C

1

œ i Ê ¸ ddtr ¸ œ 1; C2 : ratb œ i
tj, 0 Ÿ t Ÿ 1 Ê

dr dt

C2

œ '0 Èt dt
'0 È1
2t dt œ 1

2

21. ratb œ 4ti  3tj , 1 Ÿ t Ÿ 2 Ê

dr dt 2

16t œ 15'c1 t e16t dt œ ’ 15 “ 32 e 2

2

dr dt

œ j Ê ¸ ddtr ¸ œ 1

Èx
2y ds
' Èx
2y ds œ ' Èt
2a0b dt
' È1
2atb dt 1

2

c1


23 t2Î3 ‘ 1 !

2

0

0

2


’ 13 a1
2tb2Î3 “ œ !

2 3




È Š5 3 5

 31 ‹ œ

5È 5  1 3

œ 4i  3j Ê ¸ ddtr ¸ œ 5 Ê 'C y ex ds œ 'c1 a3tb ea4tb † 5dt 2

64 œ  15 32 e


15 16 32 e

œ

15 16 32 ae

2

2

 e64 b

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

!

Section 16.1 Line Integrals 22. ratb œ acos tbi
asin tbj , 0 Ÿ t Ÿ 21 Ê

941

œ asin tbi
acos tbj Ê ¸ ddtr ¸ œ Èsin2 t
cos2 t œ 1 Ê 'C ax  y
3b ds

dr dt

œ '0 acos t  sin t
3b † 1 dt œ csin t
cos t
3td 201 œ 61 21

23. ratb œ t2 i
t3 j , 1 Ÿ t Ÿ 2 Ê œ '1

2

œ '1Î2

œ 2ti
3t2 j Ê ¸ ddtr ¸ œ Éa2tb2
a3t2 b2 œ tÈ4
9t2 Ê 'C

3Î2 1 a4
9t2 b “ œ † tÈ4
9t2 dt œ '1 t È4
9t2 dt œ ’ 27

ˆt2 ‰2

2

2

at3 b4Î3

ŸtŸ1Ê

1 2

dr dt

1

25. C" : ratb œ ti
t2 j , 0 Ÿ t Ÿ 1 Ê Ê

dr dt

ds

œ 3t2 i
4t3 j Ê ¸ ddtr ¸ œ Éa3t2 b2
a4t3 b2 œ t2 È9
16t2 Ê 'C

1 a9
16t2 b † t2 È9
16t2 dt œ '1Î2 t È9
16t2 dt œ ’ 48

Èt4 t3

x2 y4Î3

80È10
13È13 27

1

24. ratb œ t3 i
t4 j , 1

dr dt

3 Î2

1

“

1Î2

œ

Èy x

ds

125
13È13 48

œ i
2tj Ê ¸ ddtr ¸ œ È1
4t2 ; C2 : ratb œ a1  tbi
a1  tbj, 0 Ÿ t Ÿ 1

dr dt

œ i  j Ê ¸ ddtr ¸ œ È2 Ê 'C ˆx
Èy‰ds œ 'C ˆx
Èy‰ds
'C ˆx
Èy‰ds 1

2

œ '0 Št
Èt2 ‹È1
4t2 dt
'0 Ša1  tb
È1  t‹ È2dt œ '0 2tÈ1
4t2 dt
'0 Š1  t
È1  t‹ È2dt 1

1

œ ’ 16 a1
4t2 b

1

3 Î2 1

1

0

0

Î “
È2’t  "# t2  23 a1  tb3 2 “ œ

26. C" : ratb œ ti , 0 Ÿ t Ÿ 1 Ê

5È 5
1 6




7È 2 6

1

œ

5È 5  7È 2
1 6

œ i Ê ¸ ddtr ¸ œ 1; C2 : ratb œ i
tj, 0 Ÿ t Ÿ 1 Ê ddtr œ j Ê ¸ ddtr ¸ œ 1; C3 : ratb œ a1  tbi
j, 0 Ÿ t Ÿ 1 Ê ddtr œ i Ê ¸ ddtr ¸ œ 1; C4 : ratb œ a1  tbj, 0 Ÿ t Ÿ 1 Ê ddtr œ j Ê ¸ ddtr ¸ œ 1; Ê 'C œ '0

1

1 x2  y2  1 ds dt t2  1


'0

œ ctan
1 td 0
1

1

œ 'C

1 x2  y2  1 ds

1

dt t2  2

dr dt


'0

1

x# #

2

dt a1
tb2  2

1 t
1 È2 ’tan Š È2 ‹“

27. r(x) œ xi
yj œ xi



'C

1 0





'0

1 x2  y2  1 ds

1


'C

œ '0 (2x)È1
x# dx œ ’ 23 a1
x# b 2

4

1 x2  y2  1 ds

1 0


ctan
1 a1  tbd 0 œ 1

1 2




2 1
1 È2 tan Š È2 ‹

#

“ œ !


'C

dr ¸ œ i
xj Ê ¸ dx œ È1
x# ; f(xß y) œ f Šxß x# ‹ œ

dr dx

$Î# #

1 x2  y2  1 ds

dt a1
tb2  1

1
1 t
1 È2 ’tan Š È2 ‹“

j, 0 Ÿ x Ÿ 2 Ê

3

2 3

ˆ5$Î#  1‰ œ

#

Š x# ‹

10È5
2 3

28. r(t) œ a1  tbi
#1 a1  tb2 j, 0 Ÿ t Ÿ 1 Ê ¸ ddtr ¸ œ É1
a1  tb# ; f(xß y) œ f Ša1  tbß #1 a1  tb2 ‹ œ Ê

'C f ds œ '01 a1
tb 

œ 0  ˆ "# 

4 1 4 a1
tb #

É1  a1
tb

" ‰ #0

œ

œ 2x Ê 'C f ds

x$

É1
a1  tb# dt œ ' Ša1  tb
14 a1  tb4 ‹ dt œ ’ "# a1  tb2  0 1

a1
tb  14 a1
tb4 É1  a1
tb#

1 20 a1

 tb5 “

" !

11 #0

29. r(t) œ (2 cos t) i
(2 sin t) j , 0 Ÿ t Ÿ

1 #

Ê

dr dt

œ (2 sin t) i
(2 cos t) j Ê ¸ ddtr ¸ œ 2; f(xß y) œ f(2 cos tß 2 sin t)

œ 2 cos t
2 sin t Ê 'C f ds œ '0 (2 cos t
2 sin t)(2) dt œ c4 sin t  4 cos td ! 1Î2

30. r(t) œ (2 sin t) i
(2 cos t) j , 0 Ÿ t Ÿ œ 4 sin# t  2 cos t Ê

1Î#

1 4

Ê

dr dt

œ 4  (4) œ 8

œ (2 cos t) i
(2 sin t) j Ê ¸ ddtr ¸ œ 2; f(xß y) œ f(2 sin tß 2 cos t)

'C f ds œ '01Î4 a4 sin# t  2 cos t b (2) dt œ c4t  2 sin 2t  4 sin td 01Î% œ 1  2Š1
È2‹

31. y œ x2 , 0 Ÿ x Ÿ 2 Ê ratb œ ti
t2 j , 0 Ÿ t Ÿ 2 Ê

dr dt

œ i
2tj Ê ¸ ddtr ¸ œ È1
4t2 Ê A œ 'C fax, yb ds

3 Î2 œ 'C ˆx
Èy‰ds œ '0 Št
Èt2 ‹È1
4t2 dt œ '0 2tÈ1
4t2 dt œ ’ 16 a1
4t2 b “ œ 2

2

2 0

17È17
1 6

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

942

Chapter 16 Integration in Vector Fields

32. 2x
3y œ 6, 0 Ÿ x Ÿ 6 Ê ratb œ ti
ˆ2  23 t‰j , 0 Ÿ t Ÿ 6 Ê œ 'C a4
3x
2ybds œ '0 ˆ4
3t
2ˆ2  23 t‰‰ 6

33. r(t) œ at#  1b j
2tk , 0 Ÿ t Ÿ 1 Ê

dr dt

È13 3

dt œ

È13 3

dr dt

È13 3

œ i  23 j Ê ¸ ddtr ¸ œ

Ê A œ 'C fax, yb ds

'06 ˆ8
35 t‰dt œ È313
8t
65 t2 ‘ 60 œ 26È13

œ 2tj
2k Ê ¸ ddtr ¸ œ 2Èt#
1; M œ 'C $ (xß yß z) ds œ '0 $ (t) Š2Èt#
1‹ dt 1

3/2 œ '0 ˆ 3# t‰ Š2Èt#
1‹ dt œ ’at#
1b “ œ 2$Î#  1 œ 2È2  1 "

1

!

34. r(t) œ at#  1b j
2tk , 1 Ÿ t Ÿ 1 Ê ddtr œ 2tj
2k Ê ¸ dr ¸ œ 2Èt#
1; M œ ' $ (xß yß z) ds dt

C

œ 'c1 1

ˆ15Èat#

 1b
2‰ Š2Èt#
1‹ dt

œ 'c1 30 at#
1b dt œ ’30 Š t3
t‹“ 1

$

"
"

œ 60 ˆ 3"
1‰ œ 80;

Mxz œ 'C y$ (xß yß z) ds œ 'c1 at#  1b c30 at#
1bd dt 1

œ 'c1 30 at%  1b dt œ ’30 Š t5  t‹“ 1

&

œ 48 Ê y œ

Mxz M

"


"

œ 60 ˆ 5"  1‰

48 œ  80 œ  53 ; Myz œ 'C x$ (xß yß z) ds œ 'C 0 $ ds œ 0 Ê x œ 0; z œ 0 by symmetry (since $ is

independent of z) Ê (xß yß z) œ ˆ!ß  35 ß 0‰ 35. r(t) œ È2t i
È2t j
a4  t# b k , 0 Ÿ t Ÿ 1 Ê

dr dt

œ È2i
È2j  2tk Ê ¸ ddtr ¸ œ È2
2
4t# œ 2È1
t# ;

(a) M œ 'C $ ds œ '0 (3t) Š2È1
t# ‹ dt œ ’2 a1
t# b 1

$Î# "

“ œ 2 ˆ2$Î#  1‰ œ 4È2  2 !

(b) M œ 'C $ ds œ '0 a1b Š2È1
t# ‹ dt œ ’tÈ1
t#
ln Št
È1
t# ‹“ œ ’È2
ln Š1
È2‹“  a0
ln 1b "

1

!

œ È2
ln Š1
È2‹ 36. r(t) œ ti
2tj
23 t$Î# k , 0 Ÿ t Ÿ 2 Ê

dr dt

œ i
2j
t"Î# k Ê ¸ ddtr ¸ œ È1
4
t œ È5
t;

# M œ 'C $ ds œ '0 ˆ3È5
t‰ ˆÈ5
t‰ dt œ '0 3(5
t) dt œ
32 (5
t)# ‘ ! œ 2

2

3 #

a7#  5# b œ

Myz œ 'C x$ ds œ '0 t[3(5
t)] dt œ '0 a15t
3t# b dt œ
"25 t#
t$ ‘ ! œ 30
8 œ 38; 2

2

2

2

# œ '0 ˆ10t$Î#
2t&Î# ‰ dt œ
4t&Î#
47 t(Î# ‘ ! œ 4(2)&Î#
47 (2)(Î# œ 16È2
2

œ

38 36

œ

19 18

,yœ

Mxz M

œ

76 36

œ

19 9

, and z œ

(24) œ 36;

#

Mxz œ 'C y$ ds œ '0 2t[3(5
t)] dt œ 2 '0 a15t
3t# b dt œ 76; Mxy œ 'C z$ ds œ '0 2

3 #

Mxy M

œ

144È2 7†36

37. Let x œ a cos t and y œ a sin t, 0 Ÿ t Ÿ 21. Then

dx dt

œ

4 7

32 7

È2 œ

dz dt

œ0

2 $Î# [3(5 3 t

144 7


t)] dt

È2 Ê x œ

Myz M

È2

œ a sin t,

dy dt

œ a cos t,

‰
Š dy ˆ dz ‰ dt œ a dt; Iz œ ' ax#
y# b $ ds œ ' aa# sin# t
a# cos# tb a$ dt Ê Êˆ dx dt dt ‹
dt C 0 #

#

21

#

œ '0 a$ $ dt œ 21$ a$ . 21

38. r(t) œ tj
(2  2t)k , 0 Ÿ t Ÿ 1 Ê

dr dt

œ j  2k Ê ¸ ddtr ¸ œ È5; M œ 'C $ ds œ '0 $ È5 dt œ $ È5; 1

" Ix œ 'C ay#
z# b $ ds œ '0 ct#
(2  2t)# d $ È5 dt œ '0 a5t#  8t
4b $ È5 dt œ $ È5
53 t$  4t#
4t‘ ! œ 1

1

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

5 3

$ È5 ;

Section 16.1 Line Integrals " Iy œ 'C ax#
z# b $ ds œ '0 c0#
(2  2t)# d $ È5 dt œ '0 a4t#  8t
4b $ È5 dt œ $ È5
43 t$  4t#
4t‘ ! œ 1

1

Iz œ 'C ax#
y# b $ ds œ '0 a0#
t# b $ È5 dt œ $ È5 ’ t3 “ œ 1

"

$

!

39. r(t) œ (cos t)i
(sin t)j
tk , 0 Ÿ t Ÿ 21 Ê

" 3

4 3

$ È5 ;

$ È5

œ ( sin t)i
(cos t)j
k Ê ¸ ddtr ¸ œ Èsin# t
cos# t
1 œ È2;

dr dt

(a) Iz œ 'C ax#
y# b $ ds œ '0 acos# t
sin# tb $ È2 dt œ 21$ È2 21

(b) Iz œ 'C ax#
y# b $ ds œ '0 $ È2 dt œ 41$ È2 41

40. r(t) œ (t cos t)i
(t sin t)j


2È2 $Î# k, 3 t

0ŸtŸ1 Ê

dr dt

œ (cos t  t sin t)i
(sin t
t cos t)j
È2t k

" Ê ¸ ddtr ¸ œ È(t
1)# œ t
1 for 0 Ÿ t Ÿ 1; M œ 'C $ ds œ '0 (t
1) dt œ
"2 (t
1)# ‘ ! œ 1

Mxy œ 'C z$ ds œ '

È Š 2 3 2 t$Î# ‹ (t 0

œ

2È 2 3

ˆ 27
52 ‰ œ

1

2È 2 3

ˆ 24 ‰ 35 œ


1) dt œ

16È2 35

Ê zœ

2È 2 3

'0 ˆt&Î#
t$Î# ‰ dt œ

Mxy M

œ Š 1635 2 ‹ ˆ 23 ‰ œ

È

32È2 105

œ '0 at# cos# t
t# sin# tb (t
1) dt œ '0 at$
t# b dt œ ’ t4
t3 “ œ 1

2È 2 3

1

1

%

"

$

!

" 4

" #

a2#  1# b œ

3 #

;


27 t(Î#
25 t&Î# ‘ " !

; Iz œ 'C ax#
y# b $ ds




" 3

œ

7 12

41. $ (xß yß z) œ 2  z and r(t) œ (cos t)j
(sin t)k , 0 Ÿ t Ÿ 1 Ê M œ 21  2 as found in Example 3 of the text; also ¸ ddtr ¸ œ 1; Ix œ 'C ay#
z# b $ ds œ '0 acos# t
sin# tb (2  sin t) dt œ '0 (2  sin t) dt œ 21  2 1

42. r(t) œ ti


2È2 $Î# j 3 t




t# #

k, 0 Ÿ t Ÿ 2 Ê

1

dr dt

œ i
È2 t"Î# j
tk Ê ¸ ddtr ¸ œ È1
2t
t# œ È(1
t)# œ 1
t for

0 Ÿ t Ÿ 2; M œ 'C $ ds œ '0 ˆ t"1 ‰ (1
t) dt œ '0 dt œ 2; Myz œ 'C x$ ds œ '0 t ˆ t"1 ‰ (1
t) dt œ ’ t2 “ œ 2; 2

Mxz œ 'C y$ ds œ '

2È2 $Î# 3 t 0

yœ

Mxz M

œ

16 15

2

, and z œ

Mxy M

œ

2

dt œ # 3

# È ’ 4152 t&Î# “ !

œ

2

2

$

œ '0 ˆt#
89 t$ ‰ dt œ ’ t3
29 t% “ œ $

; Mxy œ 'C z$ ds œ '0

2 # t

#

dt œ

#

$ # ’ t6 “ !

; Ix œ 'C ay#
z# b $ ds œ '0 ˆ 98 t$
"4 t% ‰ dt œ ’ 92 t%


Iy œ 'C ax#
z# b $ ds œ '0 ˆt#
4" t% ‰ dt œ ’ t3
2

32 15

2

# !

8 3




32 9

œ

# t& 20 “ !

œ

8 3




32 20

œ

64 15

œ

# t& 20 “ !

œ

; Iz œ 'C ax#
y# b $ ds

56 9

43-46. Example CAS commands: Maple: f := (x,y,z) -> sqrt(1+30*x^2+10*y); g := t -> t; h := t -> t^2; k := t -> 3*t^2; a,b := 0,2; ds := ( D(g)^2 + D(h)^2 + D(k)^2 )^(1/2): 'ds' = ds(t)*'dt'; F := f(g,h,k): 'F(t)' = F(t); Int( f, s=C..NULL ) = Int( simplify(F(t)*ds(t)), t=a..b ); `` = value(rhs(%));

# !

% 3

# (a) # (b) # (c)

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Ê xœ

Myz M




œ

32 9

32 20

œ 1, 232 45

;

943

944

Chapter 16 Integration in Vector Fields

Mathematica: (functions and domains may vary) Clear[x, y, z, r, t, f] f[x_,y_,z_]:= Sqrt[1
30x2
10y] {a,b}= {0, 2}; x[t_]:= t y[t_]:= t2 z[t_]:= 3t2 r[t_]:= {x[t], y[t], z[t]} v[t_]:= D[r[t], t] mag[vector_]:=Sqrt[vector.vector] Integrate[f[x[t],y[t],z[t]] mag[v[t]], {t, a, b}] N[%] 16.2 VECTOR FIELDS, WORK, CIRCULATION, AND FLUX 1. f(xß yß z) œ ax#
y#
z# b `f `y

#

#


"Î#

#
$Î#

œ  y ax
y
z b

and

`f `y

œ

y x #  y#  z#

and

`f `z

" #

2. f(xß yß z) œ ln Èx#
y#
z# œ similarly,

`f `x

Ê

`f `z

#


$Î#

#
$Î#

#

œ z ax
y
z b

ln ax#
y#
z# b Ê

œ

3. g(xß yß z) œ ez  ln ax#
y# b Ê

œ  #" ax#
y#
z# b

z x #  y #  z#

`g `x

Ê ™fœ

œ  x# 2x  y# ,

`g `y

`f `x

(2x) œ x ax#
y#
z# b

Ê ™fœ

œ

" #


$Î#

; similarly,


xi
yj
zk ax#  y#  z# b$Î#

Š x# y"# z# ‹ (2x) œ

x x#  y#  z#

;

xi  yj  zk x #  y#  z#

œ  x# 2y  y# and

`g `z

œ ez

z Ê ™ g œ Š x#
2xy# ‹ i  Š x# 2y  y# ‹ j
e k

`g `x

4. g(xß yß z) œ xy
yz
xz Ê

œ y
z,

`g `y

œ x
z, and

`g `z

œ y
x Ê ™ g œ (y
z)i
(B
z)j
(x
y)k

5. kFk inversely proportional to the square of the distance from (xß y) to the origin Ê È(M(xß y))#
(N(xß y))# œ

k x#  y#

y
x È x #  y# i  È x#  y# j Then M(xß y) œ Èx
#ax and N(xß y) œ Èx
#ay  y#  y# ky k
kx a œ x#  y# Ê F œ # # $Î# i  # # $Î# j , for any constant ax  y b ax  y b

, k  0; F points toward the origin Ê F is in the direction of n œ

Ê F œ an , for some constant a  0. Ê È(M(xß y))#
(N(xß y))# œ a Ê

k0

6. Given x#
y# œ a#
b# , let x œ Èa#
b# cos t and y œ Èa#
b# sin t. Then r œ ŠÈa#
b# cos t‹ i  ŠÈa#
b# sin t‹ j traces the circle in a clockwise direction as t goes from 0 to 21 Ê v œ ŠÈa#
b# sin t‹ i  ŠÈa#
b# cos t‹ j is tangent to the circle in a clockwise direction. Thus, let F œ v Ê F œ yi  xj and F(0ß 0) œ 0 . 7. Substitute the parametric representations for r(t) œ x(t)i
y(t)j
z(t)k representing each path into the vector field F , and calculate 'C F †

dr dt

.

(a) F œ 3ti
2tj
4tk and

dr dt

œi
j
k Ê F†

(b) F œ 3t# i
2tj
4t% k and œ

7 3


2œ

dr dt

dr dt

œ 9t Ê

œ i
2tj
4t$ k Ê F †

dr dt

'01 9t dt œ 9#

œ 7t#
16t( Ê

'01 a7t#
16t( b dt œ
37 t$
2t) ‘ "!

13 3

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Section 16.2 Vector Fields, Work, Circulation, and Flux (c) r" œ ti
tj and r# œ i
j
tk ; F" œ 3ti
2tj and F# œ 3i
2j
4tk and

œ k Ê F# †

d r# dt

d r# dt

d r" dt

œ i
j Ê F" †

'01 4t dt œ 2

œ 4t Ê

Ê

d r" dt 5 #

'01 5t dt œ #5 ;

œ 5t Ê


2œ

9 #

8. Substitute the parametric representation for r(t) œ x(t)i
y(t)j
z(t)k representing each path into the vector field F, and calculate 'C F †

dr dt

.

" ‰ (a) F œ ˆ t#  1 j and

dr dt

œi
j
kÊF†

" ‰ (b) F œ ˆ t#  1 j and

dr dt

œ i
2tj
4t$ k Ê F †

dr dt

" t#  1

œ

dr dt

" ‰ (c) r" œ ti
tj and r# œ i
j
tk ; F" œ ˆ t#  1 j

Ê F# †

d r# dt

œ 0 Ê '0

1

" t#  1

dt œ

Ê '0

1

œ

2t t#  1 and ddtr"

" t#  1

Ê '0

1

"

dt œ ctan
" td ! œ 2t t#  1

1 4 "

dt œ cln at#
1bd ! œ ln 2

œ i
j Ê F" †

d r" dt

œ

" t#  1

; F# œ

" #

j and

d r# dt

œk

1 4

9. Substitute the parametric representation for r(t) œ x(t)i
y(t)j
z(t)k representing each path into the vector field F, and calculate 'C F †

dr dt

.

'01 ˆ2Èt  2t‰ dt œ
43 t$Î#  t# ‘ "! œ "3 1 " F œ t# i  2tj
tk and ddtr œ i
2tj
4t$ k Ê F † ddtr œ 4t%  3t# Ê '0 a4t%  3t# b dt œ
45 t&  t$ ‘ ! œ  "5 1 r" œ ti
tj and r# œ i
j
tk ; F" œ 2tj
Èt k and ddtr œ i
j Ê F" † ddtr œ 2t Ê '0 2t dt œ 1; 1 F# œ Èti  2j
k and ddtr œ k Ê F# † ddtr œ 1 Ê '0 dt œ 1 Ê 1
1 œ 0

(a) F œ Èti  2tj
Ètk and (b) (c)

dr dt

œi
j
k Ê F†

œ 2Èt  2t Ê

dr dt

"

#

"

#

10. Substitute the parametric representation for r(t) œ x(t)i
y(t)j
z(t)k representing each path into the vector field F, and calculate 'C F †

dr dt

. œ 3t# Ê '0 3t# dt œ 1 1

(a) F œ t# i
t# j
t# k and

dr dt

œi
j
k Ê F†

(b) F œ t$ i  t' j
t& k and

dr dt

œ i
2tj
4t$ k Ê F †

%

œ ’ t4


t) 4

"


94 t* “ œ !

dr dt

œ t$
2t(
4t) Ê '0 at$
2t(
4t) b dt 1

17 18

(c) r" œ ti
tj and r# œ i
j
tk ; F" œ t# i and F# œ i
tj
tk and

dr dt

d r# dt

œ k Ê F# †

d r# dt

d r" dt

œ i
j Ê F" †

œ t Ê '0 t dt œ 1

" #

Ê

d r" dt " 3

œ t# Ê '0 t# dt œ




1

" #

œ

" 3

;

5 6

11. Substitute the parametric representation for r(t) œ x(t)i
y(t)j
z(t)k representing each path into the vector field F, and calculate 'C F †

dr dt

.

(a) F œ a3t#  3tb i
3tj
k and

†

(b) F œ a3t#  3tb i
3t% j
k

Ê F†

Ê

dr dt œ i
j
k Ê F and ddtr œ i
2tj
4t$ k

dr dt

œ 3t#
1 Ê

œ 6t&
4t$
3t#  3t

'0 a6t&
4t$
3t#  3tb dt œ
t'
t%
t$  3# t# ‘ "! œ 3# 1

(c) r" œ ti
tj and r# œ i
j
tk ; F" œ a3t#  3tb i
k and Ê

dr dt

'01 a3t#
1b dt œ ct$
td "! œ 2

d r" dt

œ i
j Ê F" †

d r" dt

œ 3t#  3t

œ k Ê F# †

d r# dt

œ1 Ê

'0 a3t#  3tb dt œ
t$  32 t# ‘ "! œ  "# ; F# œ 3tj
k and ddtr 1

Ê  "#
1 œ

#

'01 dt œ 1

1 2

12. Substitute the parametric representation for r(t) œ x(t)i
y(t)j
z(t)k representing each path into the vector field F, and calculate 'C F †

dr dt

.

(a) F œ 2ti
2tj
2tk and

dr dt

œi
j
k Ê F†

dr dt

œ 6t Ê

'01 6t dt œ c3t# d "! œ 3

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

945

946

Chapter 16 Integration in Vector Fields

(b) F œ at#
t% b i
at%
tb j
at
t# b k and Ê '0 a6t&
5t%
3t# b dt œ ct'
t&
1

dr dt œ i " t$ d ! œ 3


2tj
4t$ k Ê F †

(c) r" œ ti
tj and r# œ i
j
tk ; F" œ ti
tj
2tk and F# œ (1
t)i
(t
1)j
2k and

d r# dt

œ k Ê F# †

d r# dt

dr dt

œ 6t&
5t%
3t#

œ i
j Ê F" †

dr" dt

œ 2t Ê '0 2t dt œ "; 1

d r" dt

œ 2 Ê '0 2 dt œ 2 Ê "
2 œ 3 1

13. x œ t, y œ 2t
1, 0 Ÿ t Ÿ 3 Ê dx œ dt Ê 'C ax  yb dx œ '0 at  a2t
1bb dt œ '0 at  1b dt œ
 "# t2  t‘ ! œ  15 2 3

14. x œ t, y œ t2 , 1 Ÿ t Ÿ 2 Ê dy œ 2t dt Ê 'C

x y

dy œ '1

2

t t2 a2tb dt

3

3

œ '1 2 dt œ c2td21 œ 2 2

15. C1 : x œ t, y œ 0, 0 Ÿ t Ÿ 3 Ê dy œ 0; C2 : x œ 3, y œ t, 0 Ÿ t Ÿ 3 Ê dy œ dt Ê 'C ax2
y2 b dy

œ 'C ax2
y2 b dx
'C ax2
y2 b dx œ '0 at2
02 b † 0
'0 a32
t2 b dt œ '0 a9
t2 bdt œ
9t
13 t3 ‘ ! œ 36 3

1

3

3

3

2

16. C1 : x œ t, y œ 3t, 0 Ÿ t Ÿ 1 Ê dx œ dt; C2 : x œ 1  t, y œ 3, 0 Ÿ t Ÿ 1 Ê dx œ dt; C3 : x œ 0, y œ 3  t, 0 Ÿ t Ÿ 3 Ê dx œ 0 Ê 'C Èx
y dx œ 'C Èx
y dx
'C Èx
y dx
'C Èx
y dx 1

2

3

œ '0 Èt
3t dt
'0 Èa1  tb
3 a1bdt
'0 È0
a3  tb † 0 œ '0 2Èt dt  '0 È4  t dt 1

1

3

1

œ
43 t2Î3 ‘ !
’ 23 a4  tb2Î3 “ œ 1

4 3

!


Š2È3 

16 3 ‹

1

1

œ 2È3  4

17. ratb œ ti  j
t2 k , 0 Ÿ t Ÿ 1 Ê dx œ dt, dy œ 0, dz œ 2t dt (a) (b) (c)

'C ax
y  zb dx œ '01 at  1  t2 b dt œ
12 t2  t  13 t3 ‘ 1! œ  56 'C ax
y  zb dy œ '01 at  1  t2 b † 0 œ 0

'C ax
y  zb dz œ '01 at  1  t2 b 2t dt œ '01 a2t2  2t  2t3 b dt œ

1

œ
23 t3  t2  12 t4 ‘ ! œ  56

18. ratb œ acos tbi
asin tbj  acos tbk , 0 Ÿ t Ÿ 1 Ê dx œ sin t dt, dy œ cos t dt, dz œ sin t dt (a)

'C x z dx œ '01 acos tb acos tbasin tbdt œ '01 cos2 t sin tdt œ ’ 13 acos tb3 “ 1 œ 23

(b)

'C x z dy œ '01 acos tb acos tbacos tbdt œ '01 cos3 t dt œ '01 a1  sin2 tb cos t dt œ ’ 13 asin tb3  sin t“ 1 œ 0

(c)

!

'C x y z dz œ '0 acos tbasin tb acos tbasin tbdt œ '0 1 1 1 œ  18 '0 a1  cos 4tb dt œ
 18 t
32 sin 4t‘ ! œ  18 1

1

cos t sin t dt œ 2

2

 14

'0

1

19. r œ ti
t# j
tk , 0 Ÿ t Ÿ 1, and F œ xyi
yj  yzk Ê F œ t$ i
t# j  t$ k and Ê F†

dr dt

œ 2t$ Ê work œ '0 2t$ dt œ 1

sin 2t dt œ

dr dt

2

 41

œ i
2tj
k

" #

20. r œ (cos t)i
(sin t)j
6t k , 0 Ÿ t Ÿ 21, and F œ 2yi
3xj
(x
y)k Ê F œ (2 sin t)i
(3 cos t)j
(cos t
sin t)k and œ 3 cos# t  2sin2 t
œ
32 t


3 4

" 6

sin 2t  t


cos t
sin 2t 2




" 6

" 6

dr dt

œ ( sin t)i
(cos t)j
6" k Ê F †

sin t Ê work œ '0 ˆ3 cos# t  2 sin2 t


sin t 

" 6

cos

#1 t‘ !

21

" 6

cos t


" 6

dr dt

sin t‰ dt

œ1

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

'

1

!

1
cos 4t 2 0

dt

Section 16.2 Vector Fields, Work, Circulation, and Flux 21. r œ (sin t)i
(cos t)j
tk , 0 Ÿ t Ÿ 21, and F œ zi
xj
yk Ê F œ ti
(sin t)j
(cos t)k and dr dt

œ (cos t)i  (sin t)j
k Ê F †

œ
cos t
t sin t 

t 2




sin 2t 4

dr dt

œ t cos t  sin# t
cos t Ê work œ '0 at cos t  sin# t
cos tb dt 21

#1


sin t‘ ! œ 1

22. r œ (sin t)i
(cos t)j
6t k , 0 Ÿ t Ÿ 21, and F œ 6zi
y# j
12xk Ê F œ ti
acos# tbj
(12 sin t)k and dr dt

œ (cos t)i  (sin t)j
6" k Ê F †

dr dt

œ t cos t  sin t cos# t
2 sin t

Ê work œ '0 at cos t  sin t cos# t
2 sin tb dt œ
cos t
t sin t
21

1 3

#1

cos$ t  2 cos t‘ ! œ 0

23. x œ t and y œ x# œ t# Ê r œ ti
t# j , 1 Ÿ t Ÿ 2, and F œ xyi
(x
y)j Ê F œ t$ i
at
t# b j and dr dt

œ i
2tj Ê F †

dr dt

œ t$
a2t#
2t$ b œ 3t$
2t# Ê 'C xy dx
(x
y) dy œ 'C F †

#

œ
34 t%
32 t$ ‘
" œ ˆ12


16 ‰ 3

 ˆ 34  23 ‰ œ

45 4




18 3

œ

dr dt

dt œ 'c" a3t$
2t# b dt #

69 4

24. Along (0ß 0) to (1ß 0): r œ ti , 0 Ÿ t Ÿ 1, and F œ (x  y)i
(x
y)j Ê F œ ti
tj and

dr dt

œi Ê F†

dr dt

œ t;

Along (1ß 0) to (0ß 1): r œ (1  t)i
tj , 0 Ÿ t Ÿ 1, and F œ (x  y)i
(x
y)j Ê F œ (1  2t)i
j and dr dr dt œ i
j Ê F † dt œ 2t; Along (0ß 1) to (0ß 0): r œ (1  t)j , 0 Ÿ t Ÿ 1, and F œ (x  y)i
(x
y)j Ê F œ (t  1)i
(1  t)j and dr dt

œ j Ê F †

dr dt

œ t  1 Ê 'C (x  y) dx
(x
y) dy œ '0 t dt
'0 2t dt
'0 (t  1) dt œ '0 (4t  1) dt 1

1

1

1

dr dy

œ 2yi
j and F †

"

œ c2t#  td ! œ 2  1 œ 1 25. r œ xi
yj œ y# i
yj , 2   y   1, and F œ x# i  yj œ y% i  yj Ê Ê

dr dy

œ 2y&  y

4‰ 3 63 39 'C F † T ds œ '2c1 F † dydr dy œ '2c1 a2y&  yb dy œ
3" y'  "# y# ‘
" œ ˆ 3"  #" ‰  ˆ 64 3  # œ #  3 œ # #

26. r œ (cos t)i
(sin t)j , 0 Ÿ t Ÿ ÊF†

dr dt

1 #

, and F œ yi  xj Ê F œ (sin t)i  (cos t)j and

œ  sin# t  cos# t œ 1 Ê

'C F † dr œ '0

1Î2

dr dt

œ ( sin t)i
(cos t)j

(1) dt œ  1#

27. r œ (i
j)
t(i
2j) œ (1
t)i
(1
2t)j , 0 Ÿ t Ÿ 1, and F œ xyi
(y  x)j Ê F œ a1
3t
2t# b i
tj and dr dt

œ i
2j Ê F †

dr dt

œ 1
5t
2t# Ê work œ 'C F †

dr dt

dt œ '0 a1
5t
2t# b dt œ
t
25 t#
23 t$ ‘ ! œ 1

"

28. r œ (2 cos t)i
(2 sin t)j , 0 Ÿ t Ÿ 21, and F œ ™ f œ 2(x
y)i
2(x
y)j Ê F œ 4(cos t
sin t)i
4(cos t
sin t)j and ddtr œ (2 sin t)i
(2 cos t)j Ê F †

25 6

dr dt

œ 8 asin t cos t
sin# tb
8 acos# t
cos t sin tb œ 8 acos# t  sin# tb œ 8 cos 2t Ê work œ 'C ™ f † dr œ 'C F †

dr dt

dt œ '0 8 cos 2t dt œ c4 sin 2td #!1 œ 0 21

29. (a) r œ (cos t)i
(sin t)j , 0 Ÿ t Ÿ 21, F" œ xi
yj , and F# œ yi
xj Ê F" œ (cos t)i
(sin t)j , and F# œ ( sin t)i
(cos t)j Ê F" †

dr dt

dr dt

œ ( sin t)i
(cos t)j ,

œ 0 and F# †

dr dt

œ sin# t
cos# t œ 1

Ê Circ" œ '0 0 dt œ 0 and Circ# œ '0 dt œ 21; n œ (cos t)i
(sin t)j Ê F" † n œ cos# t
sin# t œ 1 and 21

21

F# † n œ 0 Ê Flux" œ '0 dt œ 21 and Flux# œ '0 0 dt œ 0 21

21

(b) r œ (cos t)i
(4 sin t)j , 0 Ÿ t Ÿ 21 Ê F# œ (4 sin t)i
(cos t)j Ê F" †

dr dt

dr dt

œ ( sin t)i
(4 cos t)j , F" œ (cos t)i
(4 sin t)j , and

œ 15 sin t cos t and F# †

dr dt

œ 4 Ê Circ" œ '0 15 sin t cos t dt 21

œ
"25 sin# t‘ ! œ 0 and Circ# œ '0 4 dt œ 81; n œ Š È417 cos t‹ i
Š È"17 sin t‹ j Ê F" † n #1

21

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

947

948

Chapter 16 Integration in Vector Fields œ

4 È17

cos# t


sin# t and F# † n œ  È1517 sin t cos t Ê Flux" œ '0 (F" † n) kvk dt œ '0 Š È417 ‹ È17 dt 21

4 È17

21

# ‘ œ 81 and Flux# œ '0 (F# † n) kvk dt œ '0 Š È1517 sin t cos t‹ È17 dt œ
 15 2 sin t ! œ 0 21

21

#1

30. r œ (a cos t)i
(a sin t)j , 0 Ÿ t Ÿ 21, F" œ 2xi  3yj , and F# œ 2xi
(x  y)j Ê

œ (a sin t)i
(a cos t)j ,

dr dt

F" œ (2a cos t)i  (3a sin t)j , and F# œ (2a cos t)i
(a cos t  a sin t)j Ê n kvk œ (a cos t)i
(a sin t)j , F" † n kvk œ 2a# cos# t  3a# sin# t, and F# † n kvk œ 2a# cos# t
a# sin t cos t  a# sin# t Ê Flux" œ '0 a2a# cos# t  3a# sin# tb dt œ 2a#
2t
21

sin 2t ‘ #1 4 !

Flux# œ '0 a2a# cos# t  a# sin t cos t  a# sin# tb dt œ 2a#
2t
21

31. F" œ (a cos t)i
(a sin t)j ,

d r" dt

sin 2t ‘ #1 4 !

œ 1a# , and

a# #

#1

 3a#
2t 

œ (a sin t)i
(a cos t)j Ê F" †

sin 2t ‘ #1 4 ! d r" dt




csin# td !  a#
2t 

sin 2t ‘ #1 4 !

œ 1a#

œ 0 Ê Circ" œ 0; M" œ a cos t,

N" œ a sin t, dx œ a sin t dt, dy œ a cos t dt Ê Flux" œ 'C M" dy  N" dx œ '0 aa# cos# t
a# sin# tb dt œ '0 a# dt œ a# 1;

1

1

F # œ ti ,

d r# dt

œ i Ê F# †

d r# dt

œ t Ê Circ# œ 'ca t dt œ 0; M# œ t, N# œ 0, dx œ dt, dy œ 0 Ê Flux# a

œ 'C M# dy  N# dx œ 'ca 0 dt œ 0; therefore, Circ œ Circ"
Circ# œ 0 and Flux œ Flux"
Flux# œ a# 1 a

32. F" œ aa# cos# tb i
aa# sin# tb j ,

d r" dt

œ (a sin t)i
(a cos t)j Ê F" †

d r" dt

œ a$ sin t cos# t
a$ cos t sin# t

Ê Circ" œ '0 aa$ sin t cos# t
a$ cos t sin# tb dt œ  2a3 ; M" œ a# cos# t, N" œ a# sin# t, dy œ a cos t dt, 1

$

dx œ a sin t dt Ê Flux" œ 'C M" dy  N" dx œ '0 aa$ cos$ t
a$ sin$ tb dt œ 1

F # œ t# i ,

d r# dt

œ i Ê F# †

d r# dt

œ t# Ê Circ# œ 'ca t# dt œ a

2a$ 3

4 3

a$ ;

; M# œ t# , N# œ 0, dy œ 0, dx œ dt

Ê Flux# œ 'C M# dy  N# dx œ 0; therefore, Circ œ Circ"
Circ# œ 0 and Flux œ Flux"
Flux# œ 33. F" œ (a sin t)i
(a cos t)j ,

d r" dt

œ (a sin t)i
(a cos t)j Ê F" †

d r" dt

4 3

a$

œ a# sin# t
a# cos# t œ a#

Ê Circ" œ '0 a# dt œ a# 1 ; M" œ a sin t, N" œ a cos t, dx œ a sin t dt, dy œ a cos t dt 1

Ê Flux" œ 'C M" dy  N" dx œ '0 aa# sin t cos t
a# sin t cos tb dt œ 0; F# œ tj , 1

dr# dt

œ i Ê F# †

d r# dt

œ0

Ê Circ# œ 0; M# œ 0, N# œ t, dx œ dt, dy œ 0 Ê Flux# œ 'C M# dy  N# dx œ 'ca t dt œ 0; therefore, a

Circ œ Circ"
Circ# œ a# 1 and Flux œ Flux"
Flux# œ 0 34. F" œ aa# sin# tb i
aa# cos# tb j ,

d r" dt

œ (a sin t)i
(a cos t)j Ê F" †

Ê Circ" œ '0 aa$ sin$ t
a$ cos$ tb dt œ 1

4 3

d r" dt

œ a$ sin$ t
a$ cos$ t

a$ ; M" œ a# sin# t, N" œ a# cos# t, dy œ a cos t dt, dx œ a sin t dt

Ê Flux" œ 'C M" dy  N" dx œ '0 aa$ cos t sin# t
a$ sin t cos# tb dt œ 1

2 3

a$ ; F# œ t# j ,

d r# dt

œ i Ê F# †

d r# dt

œ0

Ê Circ# œ 0; M# œ 0, N# œ t# , dy œ 0, dx œ dt Ê Flux# œ 'C M# dy  N# dx œ 'ca t# dt œ  23 a$ ; therefore, a

Circ œ Circ"
Circ# œ

4 3

a$ and Flux œ Flux"
Flux# œ 0

35. (a) r œ (cos t)i
(sin t)j , 0 Ÿ t Ÿ 1, and F œ (x
y)i  ax#
y# b j Ê F œ (cos t
sin t)i  acos# t
sin# tb j Ê F †

dr dt

(b) r œ (1  2t)i , 0 Ÿ t Ÿ 1, and F œ (x
y)i F†

dr dt

œ 4t  2 Ê 'C F † T ds œ '0 (4t  1

œ (sin t)i
(cos t)j and

œ  sin t cos t  sin# t  cos t Ê 'C F † T ds

œ '0 a sin t cos t  sin# t  cos tb dt œ
 2" sin# t  1

dr dt

sin 2t 1 ‘1 4  sin t ! œ  #  ax#
y# b j Ê ddtr œ 2i and F œ (1 " 2) dt œ c2t#  2td ! œ 0 t #




 2t)i  (1  2t)# j Ê

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Section 16.2 Vector Fields, Work, Circulation, and Flux (c) r" œ (1  t)i  tj , 0 Ÿ t Ÿ 1, and F œ (x
y)i  ax#
y# b j Ê Ê F†

d r" dt

œ (2t  1)
a1  2t
2t# b œ 2t# Ê Flow" œ 'C F †

d r" dt

"

#

#

0 Ÿ t Ÿ 1, and F œ (x
y)i  ax
y b j Ê œ i  a2t#  2t
1b j Ê F † "

œ
t#  23 t$ ‘ ! œ

" 3

d r# dt

œ i  j and F œ (1  2t)i  a1  2t
2t# b j

d r" dt

œ '0 2t# dt œ 1

#

2 3

; r# œ ti
(t  1)j ,

#

œ i
j and F œ i  at
t  2t
1b j

œ 1  a2t#  2t
1b œ 2t  2t# Ê Flow# œ 'C F †

dr # dt

949

#

Ê Flow œ Flow"
Flow# œ

2 3

" 3




dr # dt

œ '0 a2t  2t# b dt 1

œ1

36. From (1ß 0) to (0ß 1): r" œ (1  t)i
tj , 0 Ÿ t Ÿ 1, and F œ (x
y)i  ax#
y# b j Ê

d r" dt

œ i
j ,

F œ i  a1  2t
2t# b j , and n" kv" k œ i
j Ê F † n" kv" k œ 2t  2t# Ê Flux" œ '0 a2t  2t# b dt 1

"

œ
t#  23 t$ ‘ ! œ

" 3

;

From (0ß 1) to (1ß 0): r# œ ti
(1  t)j , 0 Ÿ t Ÿ 1, and F œ (x
y)i  ax#
y# b j Ê

d r# dt

œ i  j ,

#

F œ (1  2t)i  a1  2t
2t b j , and n# kv# k œ i
j Ê F † n# kv# k œ (2t  1)
a1
2t  2t# b œ 2
4t  2t# Ê Flux# œ '0 a2
4t  2t# b dt œ
2t
2t#  23 t$ ‘ ! œ  23 ; 1

"

From (1ß 0) to (1ß 0): r$ œ (1
2t)i , 0 Ÿ t Ÿ 1, and F œ (x
y)i  ax#
y# b j Ê #

d r$ dt

œ 2i ,

#

F œ (1
2t)i  a1  4t
4t b j , and n$ kv$ k œ 2j Ê F † n$ kv$ k œ 2 a1  4t
4t b Ê Flux$ œ 2 '0 a1  4t
4t# b dt œ 2
t  2t#
43 t$ ‘ ! œ 1

"

37. (a) y œ 2x, 0 Ÿ x Ÿ 2 Ê ratb œ ti
2tj , 0 Ÿ t Ÿ 2 Ê œ 4t2
8t2 œ 12t2 Ê Flow œ 'C F †

dr dt

2 3




2 3

œ

œ Ša2tb2 i
2atba2tbj‹ † ai
2jb

2

dr dt

œ Šat2 b i
2atbat2 bj‹ † ai
2tjb

dr dt

œ i
2tj Ê F †

2

dt œ '0 5t4 dt œ ct5 d ! œ 32 2

dr dt

2

œ Šˆ "# t3 ‰ i
2atbˆ "# t3 ‰j‹ † ai
3t2 jb œ 14 t6
32 t6 œ 74 t6 Ê Flow œ 'C F † 2

dr dt

dr dt

œ i
3t2 j

dt œ '0 74 t6 dt œ
14 t7 ‘ ! 2

2

œ 32 38. (a) C1 : ratb œ a1  tbi
j , 0 Ÿ t Ÿ 2 Ê

C4 : ratb œ i
at  1bj , 0 Ÿ t Ÿ 2 Ê Ê Flow œ 'C F †

dr dt

dt œ 'C F † 1

dr dt

œ i Ê F †

dr dt

C2 : ratb œ i
a1  tbj , 0 Ÿ t Ÿ 2 Ê C3 : ratb œ at  1bi  j , 0 Ÿ t Ÿ 2 Ê

dr dt

dr dt dr dt

dr dt

œ j Ê F †

œiÊF† œjÊF†

dt
'C F †

dr dt

2

dr dt dr dt

œ aa1bi
aa1  tb
2a1bbjb † aib œ 1; dr dt

œ aa1  tbi
aa1b
2a1  tbbjb † ajb œ 2t  1;

œ aa1bi
aat  1b
2a1bbjb † aib œ 1; œ aat  1bi
aa1b
2at  1bbjb † ajb œ 2t  1;

dt
'C F † 3

dr dt

dt
'C F † 4

dr dt

dt

œ '0 a1b dt
'0 a2t  1b dt
'0 a1b dt
'0 a2t  1b dt œ ctd 2!
ct2  td !
ctd !2
ct2  td ! 2

2

2

2

2

œ 2
2  2
2 œ 0 (b) x2
y2 œ 4 Ê ratb œ a2cos tbi
a2sin tbj , 0 Ÿ t Ÿ 21 Ê ÊF†

dr dt

dr dt

2

œ a2sin tbi
a2cos tbj

œ aa2sin tbi
a2cos t
2a2sin tbbjb † aa2sin tbi
a2cos tbjb œ 4sin2 t
4cos2 t
8sin t cos t

œ 4cos 2t
4sin 2t Ê Flow œ 'C F †

dr dt

(c) answers will vary, one possible path is: C1 : ratb œ ti , 0 Ÿ t Ÿ 1 Ê ddtr œ i Ê F † C2 : ratb œ a1  tbi
tj , 0 Ÿ t Ÿ 1 Ê C3 : ratb œ a1  tbj , 0 Ÿ t Ÿ 1 Ê

dr dt

" 3

2

(c) answers will vary, one possible path is y œ 12 x3 , 0 Ÿ x Ÿ 2 Ê ratb œ ti
"# t3 j , 0 Ÿ t Ÿ 2 Ê ÊF†



dr dt

œ i
2tj Ê F †

dr dt

" 3

dt œ '0 12t2 dt œ c4t3 d ! œ 32

dr dt

(b) y œ x2 , 0 Ÿ x Ÿ 2 Ê ratb œ ti
t2 j , 0 Ÿ t Ÿ 2 Ê œ t4
4t4 œ 5t4 Ê Flow œ 'C F †

Ê Flux œ Flux"
Flux#
Flux$ œ

2 3

dr dt

dt œ '0 a4cos 2t
4sin 2tb dt œ c2sin 2t  2cos 2td 2!1 œ 0 21

dr dt

œ aa0bi
at
2a1bbjb † aib œ 0;

œ i
j Ê F †

œ j Ê F †

dr dt

dr dt

œ ati
aa1  tb
2tbjb † ai
jb œ 1;

œ aa1  tbi
a0
2a1  tbbjb † ajb œ 2t  1;

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

950

Chapter 16 Integration in Vector Fields Ê Flow œ 'C F †

dt œ 'C F †

dr dt

1

dr dt

dt
'C F † 2

dr dt

dt
'C F † 3

dr dt

dt œ '0 a0b dt
'0 a1b dt
'0 a2t  1b dt 1

1

1

1

œ 0
ctd 1!
ct2  td ! œ 1
a1b œ 0 39. F œ  Èx#y y# i


j on x#
y# œ 4;

x È x#  y#

at (2ß 0), F œ j ; at (0ß 2), F œ i ; at (2ß 0), È F œ j ; at (!ß 2), F œ i ; at ŠÈ2ß È2‹ , F œ  #3 i
#" j ; at ŠÈ2ß È2‹ , F œ Fœ

È3 #

È3 #

i
#" j ; at ŠÈ2ß È2‹ ,

i  #" j ; at ŠÈ2ß È2‹ , F œ

È3 #

i  #" j

40. F œ xi
yj on x#
y# œ 1; at (1ß 0), F œ i ; at (1ß 0), F œ i ; at (0ß 1), F œ j ; at (0ß 1), F œ j ; at Š "# ß at Š "# ß

È3 # ‹,

at Š "# ß 

È3 # ‹,

at Š "# ß 

È3 # ‹,

Fœ

" #

F œ  "# i
Fœ

È3 # ‹,

" #

i

i
È3 #

È3 #

È3 #

j;

j;

j;

F œ  "# i 

È3 #

j.

41. (a) G œ P(xß y)i
Q(xß y)j is to have a magnitude Èa#
b# and to be tangent to x#
y# œ a#
b# in a counterclockwise direction. Thus x#
y# œ a#
b# Ê 2x
2yyw œ 0 Ê yw œ  xy is the slope of the tangent line at any point on the circle Ê yw œ  ba at (aß b). Let v œ bi
aj Ê kvk œ Èa#
b# , with v in a counterclockwise direction and tangent to the circle. Then let P(xß y) œ y and Q(xß y) œ x Ê G œ yi
xj Ê for (aß b) on x#
y# œ a#
b# we have G œ bi
aj and kGk œ Èa#
b# . (b) G œ ˆÈx#
y# ‰ F œ ŠÈa#
b# ‹ F . 42. (a) From Exercise 41, part a, yi
xj is a vector tangent to the circle and pointing in a counterclockwise direction Ê yi  xj is a vector tangent to the circle pointing in a clockwise direction Ê G œ Èyxi #
xjy# is a unit vector tangent to the circle and pointing in a clockwise direction. (b) G œ F 43. The slope of the line through (xß y) and the origin is pointing away from the origin Ê F œ 

xi  yj È x#  y#

y x

Ê v œ xi
yj is a vector parallel to that line and

is the unit vector pointing toward the origin.

44. (a) From Exercise 43,  Èxxi #yjy# is a unit vector through (xß y) pointing toward the origin and we want kFk to have magnitude Èx#
y# Ê F œ Èx#
y# Š Èxxi #yjy# ‹ œ xi  yj . (b) We want kFk œ

C È x#  y#

where C Á 0 is a constant Ê F œ

C È x#  y#

yj Š Èxxi #yjy# ‹ œ C Š xx#i   y# ‹.

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Section 16.2 Vector Fields, Work, Circulation, and Flux

951

45. Yes. The work and area have the same numerical value because work œ 'C F † dr œ 'C yi † dr œ 'b [f(t)i] †
i
a

df dt

j‘ dt

[On the path, y equals f(t)]

œ 'a f(t) dt œ Area under the curve b

46. r œ xi
yj œ xi
f(x)j Ê from the origin Ê F †

'C

Ê

dr dx

dr dx

œ

F † T ds œ 'C F †

[because f(t)  0]

œ i
f w (x)j ; F œ k†y†f (x) È x#  y# w




kx È x#  y#

dx œ 'a k b

dr dx

d dx

k È x#  y#

œ

(xi
yj) has constant magnitude k and points away

kx  k†f(x)†f (x) Èx#  [f(x)]# w

œk

d dx

Èx#
[f(x)]# , by the chain rule

Èx#
[f(x)]# dx œ k
Èx#
[f(x)]# ‘ b a

œ k ˆÈb#
[f(b)]#  Èa#
[f(a)]# ‰ , as claimed. 47. F œ 4t$ i
8t# j
2k and 48. F œ 12t# j
9t# k and

dr dt

œ i
2tj Ê F †

dr dt

œ 3j
4k Ê F †

49. F œ (cos t  sin t)i
(cos t)k and

dr dt

dr dt

œ 12t$ Ê Flow œ '0 12t$ dt œ c3t% d ! œ 48 2

œ 72t# Ê Flow œ '0 72t# dt œ c24t$ d ! œ 24 1

œ ( sin t)i
(cos t)k Ê F †

dr dt

#

dr dt

"

œ  sin t cos t
1

Ê Flow œ '0 ( sin t cos t
1) dt œ
2" cos# t
t‘ ! œ ˆ #"
1‰  ˆ #"
0‰ œ 1 1

1

50. F œ (2 sin t)i  (2 cos t)j
2k and

dr dt

œ (2 sin t)i
(2 cos t)j
2k Ê F †

dr dt

œ 4 sin# t  4 cos# t
4 œ 0

Ê Flow œ 0 1 #

51. C" : r œ (cos t)i
(sin t)j
tk , 0 Ÿ t Ÿ Ê F†

dr dt

Ê F œ (2 cos t)i
2tj
(2 sin t)k and

1Î2

C# : r œ j


1 #

1Î#

( sin 2t
2t cos t
2 sin t) dt œ
2" cos 2t
2t sin t
2 cos t  2 cos t‘ !

(1  t)k , 0 Ÿ t Ÿ 1 Ê F œ 1(1  t)j
2k and

Ê Flow# œ '0 1 dt œ 1

c1td "!

Ê Flow$ œ '0 2t dt œ 1

œx

dx dt


y

dy dt


z

" ct# d ! dz dt

dr dt

œ  1# k Ê F †

dr dt

œ 1
1;

œ 1

œ 1 ;

C$ : r œ ti
(1  t)j , 0 Ÿ t Ÿ 1 Ê F œ 2ti
2(1  t)k and

dr dt

œ ( sin t)i
(cos t)j
k

œ 2 cos t sin t
2t cos t
2 sin t œ  sin 2t
2t cos t
2 sin t

Ê Flow" œ '0

52. F †

dr dt

dr dt

œij Ê F†

dr dt

œ 2t

œ 1 Ê Circulation œ (1
1)  1
1 œ 0

œ

` f dx ` x dt




` f dy ` y dt

by the chain rule Ê Circulation œ 'C F †

dr dt




` f dz ` z dt

dt œ 'a

, where f(xß yß z) œ b

d dt afaratbbb

" #

ax#
y#
x# b Ê F †

dr dt

œ

d dt afaratbbb

dt œ farabbb  faraabb. Since C is an entire ellipse,

rabb œ raab, thus the Circulation œ 0. 53. Let x œ t be the parameter Ê y œ x# œ t# and z œ x œ t Ê r œ ti
t# j
tk , 0 Ÿ t Ÿ 1 from (0ß 0ß 0) to (1ß 1ß 1) Ê œ

dr dt

œ i
2tj
k and F œ xyi
yj  yzk œ t$ i
t# j  t$ k Ê F †

œ t$
2t$  t$ œ 2t$ Ê Flow œ '0 2t$ dt 1

" #

54. (a) F œ ™ axy# z$ b Ê F † œ 'a

(b)

dr dt

b

d dt afaratbbb

dr dt

œ

` f dx ` x dt




` f dy ` y dt




` z dz ` z dt

œ

df dt

, where f(xß yß z) œ xy# z$ Ê )C F †

dr dt

dt œ farabbb  faraabb œ 0 since C is an entire ellipse.

Ð2ß1ß 1Ñ

'C F † ddtr œ 'Ð1ß1ß1Ñ

d dt

Ð#ß"ß
"Ñ

axy# z$ b dt œ cxy# z$ d Ð"ß"ß"Ñ œ (2)(1)# (1)$  (1)(1)# (1)$ œ 2  1 œ 3

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

dt

952

Chapter 16 Integration in Vector Fields

55-60. Example CAS commands: Maple: with( LinearAlgebra );#55 F := r -> < r[1]*r[2]^6 | 3*r[1]*(r[1]*r[2]^5+2) >; r := t -> < 2*cos(t) | sin(t) >; a,b := 0,2*Pi; dr := map(diff,r(t),t); # (a) F(r(t)); # (b) q1 := simplify( F(r(t)) . dr ) assuming t::real; # (c) q2 := Int( q1, t=a..b ); value( q2 ); Mathematica: (functions and bounds will vary): Exercises 55 and 56 use vectors in 2 dimensions Clear[x, y, t, f, r, v] f[x_, y_]:= {x y6 , 3x (x y5
2)} {a, b}={0, 21}; x[t_]:= 2 Cos[t] y[t_]:= Sin[t] r[t_]:={x[t], y[t]} v[t_]:= r'[t] integrand= f[x[t], y[t]] . v[t] //Simplify Integrate[integrand,{t, a, b}] N[%] If the integration takes too long or cannot be done, use NIntegrate to integrate numerically. This is suggested for exercises 57 - 60 that use vectors in 3 dimensions. Be certain to leave spaces between variables to be multiplied. Clear[x, y, z, t, f, r, v] f[x_, y_, z_]:= {y
y z Cos[x y z], x2
x z Cos[x y z], z
x y Cos[x y z]} {a, b}={0, 21}; x[t_]:= 2 Cos[t] y[t_]:= 3 Sin[t] z[t_]:= 1 r[t_]:={x[t], y[t], z[t]} v[t_]:= r'[t] integrand= f[x[t], y[t],z[t]] . v[t] //Simplify NIntegrate[integrand,{t, a, b}] 16.3 PATH INDEPENDENCE, POTENTIAL FUNCTIONS, AND CONSERVATIVE FIELDS 1.

`P `y

œxœ

`N `z

2.

`P `y

œ x cos z œ

3.

`P `y

œ 1 Á 1 œ

5.

`N `x

œ0Á1œ

6.

`P `y

œ0œ

`N `z

,

,

`M `z `N `z

œyœ ,

`N `z

`M `y `M `z

`M `z

`P `x

,

`N `x

`M `y

œzœ

œ y cos z œ

`P `x

,

`N `x

Ê Conservative œ sin z œ

Ê Not Conservative

`M `y

4.

Ê Conservative `N `x

œ 1 Á 1 œ

`M `y

Ê Not Conservative

Ê Not Conservative œ0œ

`P `x

,

`N `x

œ ex sin y œ

`M `y

Ê Conservative

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Section 16.3 Path Independence, Potential Functions, and Conservative Fields 7.

`f `x

`f `z

Ê 8.

`f `x

`f `y

œ 2x Ê f(xß yß z) œ x#
g(yß z) Ê

`f `z

œ xe `f `x


h(z) Ê

`f `z

œ 2xe

y2z

`f `y œ y2z

`f `y

œ y sin z Ê f(xß yß z) œ xy sin z
g(yß z) Ê `f `z

œ

Ê f(xß yß z) œ

z y #  z#

" #

œ

y 1  x# y# `g `y

œ

z È 1
y # z#

Ê

`f `z

œ

y È 1
y # z#

`g `y

œx
z Ê

œ z Ê g(yß z) œ zy
h(z)

w

`g `y

`g `y

œ xey2z Ê

œ 0 Ê f(xß yß z)

w

Ê h (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ xey2z
C œ x sin z


`g `y

`g `y

œ x sin z Ê

œ 0 Ê g(yß z) œ h(z)

w

`f `x

ln ay#
z# b
g(xß y) Ê " #

`g `x œ #

œ

ln x
sec# (x
y) Ê g(xß y)

ln ay#
z b
(x ln x  x)
tan (x
y)
h(y)

y)


Ê f(xß yß z) œ tan
" (xy)
g(yß z) Ê

Ê


h(z)

œ xy cos z
h (z) œ xy cos z Ê h (z) œ 0 Ê h(z) œ C Ê f(xß yß z)

Ê `` yf œ y# y z#
sec# (x
y)
hw (y) œ sec# (x
œ "# ln ay#
z# b
(x ln x  x)
tan (x
y)
C `f `x

3y# #


2z#
C

w

œ (x ln x  x)
tan (x
y)
h(y) Ê f(xß yß z) œ

12.

`g `y

xey2z



h (z) œ 2xe

œ xy sin z
C `f `z

œx



h(z) Ê f(xß yß z) œ x#


œ x
y
h (z) œ x
y Ê h (z) œ 0 Ê h(z) œ C Ê f(xß yß z)

w

Ê f(xß yß z) œ xy sin z
h(z) Ê

11.

`f `y

3y #

#

3y# #

w

œ ey2z Ê f(xß yß z) œ xey2z
g(yß z) Ê y2z

10.

œ 3y Ê g(yß z) œ

œ y
z Ê f(xß yß z) œ (y
z)x
g(yß z) Ê

œ (y
z)x
zy
C `f `x

`g `y

œ hw (z) œ 4z Ê h(z) œ 2z#
C Ê f(xß yß z) œ x#


Ê f(xß yß z) œ (y
z)x
zy
h(z) Ê

9.

œ

953

`f `y

y y#  z#

œ

Ê hw (y) œ 0 Ê h(y) œ C Ê f(xß yß z)




x 1  x# y#

`g `y

œ

x 1  x# y#




z È1
y# z#

Ê g(yß z) œ sin
" (yz)
h(z) Ê f(xß yß z) œ tan
" (xy)
sin
" (yz)
h(z)
hw (z) œ

y È 1
y # z#

" z




Ê hw (z) œ

" z

Ê h(z) œ ln kzk
C

Ê f(xß yß z) œ tan
" (xy)
sin
" (yz)
ln kzk
C 13. Let F(xß yß z) œ 2xi
2yj
2zk Ê exact; Ê

`f `x

`f `z

`P `y

`N `z

`M `P `N `M `z œ 0 œ `x , `x œ 0 œ `y `g `f # ` y œ ` y œ 2y Ê g(yß z) œ y


œ0œ

#

œ 2x Ê f(xß yß z) œ x
g(yß z) Ê

œ f(2ß 3ß 6)  f(!ß !ß !) œ 2#
3#
(6)# œ 49

exact;

`f `x

`N `z

œxœ

œ yz Ê f(xß yß z) œ xyz
g(yß z) Ê

œ xyz
h(z) Ê Ê

`P `y

Ð3ß5ß0Ñ

'Ð1ß1ß2Ñ

`f `z

w

,

`f `y

`M `z

œyœ

œ xz


`g `y

`P `x

,

`N `x

œzœ

œ xz Ê

`g `y

h(z) Ê f(xß yß z) œ x#
y# œ h(z)

'Ð0Ð2ß0ß3ß0ßÑ 6Ñ 2x dx
2y dy
2z dz

œ hw (z) œ 2z Ê h(z) œ z#
C Ê f(xß yß z) œ x#
y#
z#
C Ê

14. Let F(xß yß z) œ yzi
xzj
xyk Ê

Ê M dx
N dy
P dz is

,

`M `y

Ê M dx
N dy
P dz is

œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z)

w

œ xy
h (z) œ xy Ê h (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ xyz
C

yz dx
xz dy
xy dz œ f(3ß 5ß 0)  f(1ß 1ß 2) œ 0  2 œ 2

15. Let F(xß yß z) œ 2xyi
ax#  z# b j  2yzk Ê Ê M dx
N dy
P dz is exact;

`f `x

`P `y

œ 2z œ

`N `z

,

`M `z

œ0œ

`P `x

œ 2xy Ê f(xß yß z) œ x# y
g(yß z) Ê

Ê g(yß z) œ yz#
h(z) Ê f(xß yß z) œ x# y  yz#
h(z) Ê

`f `z

,

`N `x

`f `y w

œ 2x œ

œ x#


`g `y

`M `y

œ x#  z# Ê

`g `y

œ z#

œ 2yz
h (z) œ 2yz Ê hw (z) œ 0 Ê h(z) œ C

Ê f(xß yß z) œ x# y  yz#
C Ê 'Ð0ß0ß0Ñ 2xy dx
ax#  z# b dy  2yz dz œ f("ß #ß $)  f(!ß !ß !) œ 2  2(3)# œ 16 Ð1ß2ß3Ñ

16. Let F(xß yß z) œ 2xi  y# j  ˆ 1 4 z# ‰ k Ê Ê M dx
N dy
P dz is exact;

`f `x

`P `y

œ0œ

`N `z

,

`M `z

œ0œ

`P `x

,

`N `x

œ 2x Ê f(xß yß z) œ x#
g(yß z) Ê

œ0œ `f `y

œ

`M `y

`g `y

$

œ y# Ê g(yß z) œ  y3
h(z)

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

954

Chapter 16 Integration in Vector Fields Ê f(xß yß z) œ x# 

y$ 3

`f `z


h(z) Ê

œ hw (z) œ  1 4 z# Ê h(z) œ 4 tan
" z
C Ê f(xß yß z)

œ x# 

y$ 3

 4 tan
" z
C Ê 'Ð0ß0ß0Ñ 2x dx  y# dy 

œ ˆ9 

27 3

 4 † 14 ‰  (!  !  0) œ 1

Ð3ß3ß1Ñ

17. Let F(xß yß z) œ (sin y cos x)i
(cos y sin x)j
k Ê Ê M dx
N dy
P dz is exact; `g `y

œ cos y sin x Ê

`f `x

4 1  z#

`P `y

dz œ f(3ß 3ß 1)  f(!ß !ß !)

œ0œ

`N `z

`M `z

,

`P `x

œ0œ

,

`N `x

œ cos y cos x œ `f `y

œ sin y cos x Ê f(xß yß z) œ sin y sin x
g(yß z) Ê `f `z

œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ sin y sin x
h(z) Ê

`M `y

œ cos y sin x


`g `y

œ hw (z) œ 1 Ê h(z) œ z
C

Ê f(xß yß z) œ sin y sin x
z
C Ê 'Ð1ß0ß0Ñ sin y cos x dx
cos y sin x dy
dz œ f(0ß 1ß 1)  f(1ß !ß !) Ð0ß1ß1Ñ

œ (0
1)  (0
0) œ 1 18. Let F(xß yß z) œ (2 cos y)i
Š "y  2x sin y‹ j
ˆ "z ‰ k Ê Ê M dx
N dy
P dz is exact; " y

œ

`g `y

 2x sin y Ê

" y

œ

`f `x

`P `y

`N `z

œ0œ

`M `z

,

œ0œ

`P `x

œ 2 cos y Ê f(xß yß z) œ 2x cos y
g(yß z) Ê

, `f `y

`N `x

œ 2 sin y œ

œ 2x sin y
`f `z

Ê g(yß z) œ ln kyk
h(z) Ê f(xß yß z) œ 2x cos y
ln kyk
h(z) Ê

`M `y

`g `y

œ hw (z) œ

" z

Ê h(z) œ ln kzk
C Ê f(xß yß z) œ 2x cos y
ln kyk
ln kzk
C

Ê 'Ð0ß2ß1Ñ

Ð1ß1Î2ß2Ñ

2 cos y dx
Š "y  2x sin y‹ dy


œ ˆ2 † 0
ln

1 #

" z

dz œ f ˆ1ß 1# ß 2‰  f(!ß #ß ")


ln 2‰  (0 † cos 2
ln 2
ln 1) œ ln #

`P `y

19. Let F(xß yß z) œ 3x# i
Š zy ‹ j
(2z ln y)k Ê Ê M dx
N dy
P dz is exact;

`f `x

œ

2z y

1 # `N `z

œ

`M `z

,

œ0œ

`P `x

`f `y

œ 3x# Ê f(xß yß z) œ x$
g(yß z) Ê

Ê f(xß yß z) œ x$
z# ln y
h(z) Ê œ x$
z# ln y
C Ê 'Ð1ß1ß1Ñ 3x# dx
Ð1ß2ß3Ñ

`N `x

,

œ0œ

œ

`g `y

œ

`M `y z# y

Ê g(yß z) œ z# ln y
h(z)

`f `z

œ 2z ln y
hw (z) œ 2z ln y Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z)

z# y

dy
2z ln y dz œ f(1ß 2ß 3)  f("ß "ß ")

œ (1
9 ln 2
C)  (1
0
C) œ 9 ln 2 #

`P `y

20. Let F(xß yß z) œ (2x ln y  yz)i
Š xy  xz‹ j  (xy)k Ê Ê M dx
N dy
P dz is exact; x# y

œ

 xz Ê

`g `y

`f `x

œ x œ

`N `z

,

`M `z

œ y œ

`P `x

,

`N `x

œ 2x ln y  yz Ê f(xß yß z) œ x# ln y  xyz
g(yß z) Ê `f `z

œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ x# ln y  xyz
h(z) Ê

œ

2x y

`f `y

œ

zœ x# y

`M `y

 xz


`g `y

œ xy
hw (z) œ xy Ê hw (z) œ 0

Ê h(z) œ C Ê f(xß yß z) œ x# ln y  xyz
C Ê 'Ð1ß2ß1Ñ (2x ln y  yz) dx
Š xy  xz‹ dy  xy dz Ð2ß1ß1Ñ

#

œ f(2ß 1ß 1)  f("ß 2ß 1) œ (4 ln 1  2
C)  (ln 2  2
C) œ  ln 2 21. Let F(xß yß z) œ Š "y ‹ i
Š 1z 

x y# ‹ j

Ê M dx
N dy
P dz is exact; Ê

`g `y

œ

" z

Ê g(yß z) œ

Ê f(xß yß z) œ

x y




y z

y z

 ˆ zy# ‰ k Ê

`f `x

œ

" y

Ð2ß2ß2Ñ

" y

œ  z"# œ

Ê f(xß yß z) œ


h(z) Ê f(xß yß z) œ


C Ê 'Ð1ß1ß1Ñ

`P `y

x y

dx
Š 1z 




y z

x y# ‹

x y

`N `z

`M `z

,

œ0œ `f `y  zy#


g(yß z) Ê `f `z


h(z) Ê dy 

y z#

œ

`P `x

,

`N `x

œ  y1# œ

œ  yx#


`g `y

œ

" z



Ê `f `x

œ

`P `y

2xi  2yj  2zk x #  y #  z#

œ  4yz œ 3%

2x x #  y #  z#

`N `z

,

`M `z

Šand let 3# œ x#
y#
z# Ê œ  4xz œ 3% #

`P `x #

,

`N `x

œ  4xy œ 3% #

`3 `x

dz œ f(2ß 2ß 2)  f("ß 1ß 1) œ ˆ 2#


`M `y

Ê f(xß yß z) œ ln ax
y
z b
g(yß z) Ê

œ

x 3

,

`3 `y

œ

y 3

,

`3 `z

œ 3z ‹

Ê M dx
N dy
P dz is exact; `f `y

œ

2y x #  y #  z#

x y#


hw (z) œ  zy# Ê hw (z) œ 0 Ê h(z) œ C

œ0 22. Let F(xß yß z) œ

`M `y




`g `y

œ

2y x #  y #  z#

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

2 #


C‰  ˆ "1


" 1


C‰

Section 16.3 Path Independence, Potential Functions, and Conservative Fields Ê œ

`g `y œ 0 2z x #  y #  z#

`f `z

Ê g(yß z) œ h(z) Ê f(xß yß z) œ ln ax#
y#
z# b
h(z) Ê

Ð2ß2ß2Ñ

Ê 'Ð 1ß 1ß 1Ñ

œ

955


hw (z)

2z x #  y#  z#

Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ ln ax#
y#
z# b
C 2x dx  2y dy  2z dz x #  y #  z#

œ f(2ß 2ß 2)  f("ß 1ß 1) œ ln 12  ln 3 œ ln 4

23. r œ (i
j
k)
t(i
2j  2k) œ (1
t)i
(1
2t)j
(1  2t)k, 0 Ÿ t Ÿ 1 Ê dx œ dt, dy œ 2 dt, dz œ 2 dt Ð2ß3ß 1Ñ

Ê 'Ð1ß1ß1Ñ y dx
x dy
4 dz œ '0 (2t
1) dt
(t
1)(2 dt)
4(2) dt œ '0 (4t  5) dt œ c2t#  5td ! œ 3 1

1

24. r œ t(3j
4k), 0 Ÿ t Ÿ 1 Ê dx œ 0, dy œ 3 dt, dz œ 4 dt Ê

' 000304

Ð ß ß Ñ

Ð ß ß Ñ

"

#

x# dx
yz dy
Š y# ‹ dz

œ '0 a12t# b (3 dt)
Š 9t# ‹ (4 dt) œ '0 54t# dt œ c18t# d ! œ 18 1

25.

`P `y

1

#

œ0œ

`N `z

,

`M `z

œ 2z œ

`P `x

,

`N `x

,

`M `z

"

`M `y

œ0œ

Ê M dx
N dy
P dz is exact Ê F is conservative

Ê path independence 26.

`P `y

œ  ˆÈ

yz x #  y#  z# ‰

œ

$

`N `z

œ  ˆÈ

xz $ x #  y#  z# ‰

œ

`P `x

,

`N `x

œ  ˆÈ

xy x #  y#  z# ‰

$

œ

`M `y

Ê M dx
N dy
P dz is exact Ê F is conservative Ê path independence 27.

`P `y `f `x

œ0œ œ

2x y

`N `z

,

œ0œ

Ê f(xß y) œ

Ê f(xß y) œ 28.

`M `z

x# y



" y

`N `z

,

`M `z

#

x y

`P `x

`N `x

,

œ  2x y# œ

œ  xy#
gw (y) œ


C Ê F œ ™ Šx `P `x

`N `x

#

œ cos z œ

`f `x

œ ex ln y Ê f(xß yß z) œ ex ln y
g(yß z) Ê

,

œ

ex y

1
x# y#

" y#

Ê gw (y) œ

Ê g(y) œ  "y
C


1 y ‹

`P `y

œ0œ

Ê F is conservative Ê there exists an f so that F œ ™ f;

#

`f `y


g(y) Ê

`M `y

œ

`M `y

Ê F is conservative Ê there exists an f so that F œ ™ f; `f `y

œ

ex y




œ y sin z
h(z) Ê f(xß yß z) œ e ln y
y sin z
h(z) Ê x

`g ex `y œ y `f `z œ y x

`g `y


sin z Ê

œ sin z Ê g(yß z)

w

cos z
h (z) œ y cos z Ê hw (z) œ 0

Ê h(z) œ C Ê f(xß yß z) œ ex ln y
y sin z
C Ê F œ ™ ae ln y
y sin zb 29.

`P `y `f `x

œ0œ

`N `z

,

`M `z

#

`P `x

œ x
y Ê f(xß yß z) œ

Ê f(xß yß z) œ œ

œ0œ

" 3

x$
xy


" $ 3 x
xy " $ z 3 y
ze

(a) work œ 'A F † B

dr dt

, " 3 " 3

`N `x

œ1œ

`M `y

Ê F is conservative Ê there exists an f so that F œ ™ f; `f `y

$

x
xy
g(yß z) Ê


y$
h(z) Ê

œx


`g `y

œ y#
x Ê

`f `z

`g `y z

œ y# Ê g(yß z) œ

" 3

y$
h(z)

œ hw (z) œ zez Ê h(z) œ zez  e
C Ê f(xß yß z)  ez
C Ê F œ ™ ˆ "3 x$
xy
3" y$
zez  ez ‰

dt œ 'A F † dr œ
3" x$
xy
3" y$
zez  ez ‘ Ð"ß!ß!Ñ œ ˆ 3"
0
0
e  e‰  ˆ 3"
0
0  1‰ B

Ð"ß!ß"Ñ

œ1

(b) work œ 'A F † dr œ
"3 x$
xy
3" y$
zez  ez ‘ Ð"ß!ß!Ñ œ 1 B

Ð"ß!ß"Ñ

(c) work œ 'A F † dr œ
"3 x$
xy
3" y$
zez  ez ‘ Ð"ß!ß!Ñ œ 1 B

Ð"ß!ß"Ñ

Note: Since F is conservative, 'A F † dr is independent of the path from (1ß 0ß 0) to (1ß 0ß 1). B

30.

`P `y

œ xeyz
xyzeyz
cos y œ

that F œ ™ f;

`f `x

œe

yz

`N `z

,

`M `z

œ yeyz œ

`P `x

,

`N `x

œ zeyz œ

Ê f(xß yß z) œ xe
g(yß z) Ê yz

`f `y

`M `y

œ xze


Ê g(yß z) œ z sin y
h(z) Ê f(xß yß z) œ xe
z sin y
h(z) Ê yz

Ê F is conservative Ê there exists an f so

yz

`f `z

`g `y

œ xzeyz
z cos y Ê w

`g `y

œ z cos y

œ xye
sin y
h (z) œ xyeyz
sin y yz

Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ xeyz
z sin y
C Ê F œ ™ axeyz
z sin yb

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

956

Chapter 16 Integration in Vector Fields

(a) work œ 'A F † dr œ cxeyz
z sin yd Ð"ß!ß"Ñ B

Ð"ß1Î#ß!Ñ

œ (1
0)  (1
0) œ 0

(b) work œ 'A F † dr œ cxeyz
z sin yd Ð"ß!ß"Ñ B

Ð"ß1Î#ß!Ñ

(c) work œ 'A F † dr œ cxeyz
z sin yd Ð"ß!ß"Ñ B

Ð"ß1Î#ß!Ñ

œ0 œ0

Note: Since F is conservative, 'A F † dr is independent of the path from (1ß 0ß 1) to ˆ1ß 1# ß 0‰ . B

31. (a) F œ ™ ax$ y# b Ê F œ 3x# y# i
2x$ yj ; let C" be the path from (1ß 1) to (0ß 0) Ê x œ t  1 and y œ t
1, 0 Ÿ t Ÿ 1 Ê F œ 3(t  1)# (t
1)# i
2(t  1)$ (t
1)j œ 3(t  1)% i  2(t  1)% j and r" œ (t  1)i
(t
1)j Ê dr" œ dt i  dt j Ê

'C

"

F † dr" œ '0 c3(t  1)%
2(t  1)% d dt 1

1 " œ '0 5(t  1)% dt œ c(t  1)& d ! œ 1; let C# be the path from (0ß 0) to (1ß 1) Ê x œ t and y œ t, 1 0 Ÿ t Ÿ 1 Ê F œ 3t% i
2t% j and r# œ ti
tj Ê dr# œ dt i
dt j Ê 'C F † dr# œ '0 a3t%
2t% b dt 1 œ '0 5t% dt œ 1

Ê 'C F † dr œ 'C F † dr"
'C "

#

#

F † dr# œ 2 Ð1ß1Ñ

(b) Since f(xß y) œ x$ y# is a potential function for F, 'Ð 1ß1Ñ F † dr œ f(1ß 1)  f(1ß 1) œ 2 32.

`P `y `f `x

œ0œ

`N `z

,

`M `z

œ0œ

`P `x

,

`N `x

œ 2x sin y œ

#

œ 2x cos y Ê f(xß yß z) œ x cos y
g(yß z) Ê #

Ê f(xß yß z) œ x cos y
h(z) Ê (a) (b) (c) (d)

`M `y

`f `z

Ê F is conservative Ê there exists an f so that F œ ™ f; `f `y

œ x# sin y


w

`g `y

œ x# sin y Ê

`g `y

œ 0 Ê g(yß z) œ h(z)

#

œ h (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ x cos y
C Ê F œ ™ ax# cos yb

'C 2x cos y dx  x# sin y dy œ cx# cos yd Ð!ß"Ñ Ð"ß!Ñ œ 0  1 œ 1 'C 2x cos y dx  x# sin y dy œ cx# cos yd Ð"ß!Ñ Ð
"ß1Ñ œ 1  (1) œ 2 'C 2x cos y dx  x# sin y dy œ cx# cos yd Ð"ß!Ñ Ð
"ß!Ñ œ 1  1 œ 0 'C 2x cos y dx  x# sin y dy œ cx# cos yd Ð"ß!Ñ Ð"ß!Ñ œ 1  1 œ 0

33. (a) If the differential form is exact, then all x, and

`N `x

œ

`M `y

`P `y

œ

`N `z

Ê 2ay œ cy for all y Ê 2a œ c,

`M `z

œ

`P `x

Ê 2cx œ 2cx for

Ê by œ 2ay for all y Ê b œ 2a and c œ 2a

(b) F œ ™ f Ê the differential form with a œ 1 in part (a) is exact Ê b œ 2 and c œ 2 34. F œ ™ f Ê g(xß yß z) œ 'Ð0ß0ß0Ñ F † dr œ 'Ð0ß0ß0Ñ ™ f † dr œ f(xß yß z)  f(0ß 0ß 0) Ê ÐxßyßzÑ

`g `z

œ

`f `z

ÐxßyßzÑ

`g `x

œ

`f `x

 0,

`g `y

œ

`f `y

 0, and

 0 Ê ™ g œ ™ f œ F, as claimed

35. The path will not matter; the work along any path will be the same because the field is conservative. 36. The field is not conservative, for otherwise the work would be the same along C" and C# . 37. Let the coordinates of points A and B be axA , yA , zA b and axB , yB , zB b, respectively. The force F œ ai
bj
ck is conservative because all the partial derivatives of M, N, and P are zero. Therefore, the potential function is fax, y, zb œ ax
by
cz
C, and the work done by the force in moving a particle along any path from A to B is faBb  faAb œ f axB , yB , zB b  faxA , yA , zA b œ aaxB
byB
czB
Cb  aaxA
byA
czA
Cb Ä œ aaxB  xA b
bayB  yA b
cazB  zA b œ F † BA

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Section 16.4 Green's Theorem in the Plane 38. (a) Let GmM œ C Ê F œ C ’ `P `y

Ê

œ


3yzC ax#  y#  z# b&Î#

`f `x

œ

xC ax#  y#  z# b$Î# yC Ê `` gy œ ax#  y#  z# b$Î#

some f; œ

œ

`N `z

,

x ax#  y#  z# b$Î# `M `z

œ

i


y ax#  y#  z# b$Î#


3xzC ax#  y#  z# b&Î#

Ê f(xß yß z) œ 

œ

,

C ax#  y#  z# b"Î#

0 Ê g(yß z) œ h(z) Ê

Ê h(z) œ C" Ê f(xß yß z) œ 

`P `x

C ax#  y#  z# b"Î#

`f `z

œ

j


`N `x

œ

z ax#  y#  z# b$Î#
3xyC ax#  y#  z# b&Î#


g(yß z) Ê

`f `y

k“ `M `y

œ

œ


hw (z) œ

zC ax#  y#  z# b$Î#

Ê F œ ™ f for

yC ax#  y#  z# b$Î#


C" . Let C" œ 0 Ê f(xß yß z) œ




`g `y

zC ax#  y#  z# b$Î# GmM ax#  y#  z# b"Î#

is a potential

function for F. (b) If s is the distance of (xß yß z) from the origin, then s œ Èx#
y#
z# . The work done by the gravitational field F is work œ 'P F † dr œ ’ Èx#GmM “  y #  z# P#

T#

"

T"

œ

GmM

s#



GmM

s"

œ GmM Š s"# 

"

s" ‹ ,

as claimed.

16.4 GREEN'S THEOREM IN THE PLANE 1. M œ y œ a sin t, N œ x œ a cos t, dx œ a sin t dt, dy œ a cos t dt Ê `N `y

`M `x

œ 0,

`M `y

œ 1,

`N `x

œ 1, and

œ 0;

Equation (3):

)C M dy  N dx œ '021 [(a sin t)(a cos t)  (a cos t)(a sin t)] dt œ '021 0 dt œ 0;

' ' Š ``Mx
``Ny ‹ dx dy œ ' ' 0 dx dy œ 0, Flux R

R

Equation (4):

)C M dx
N dy œ '021 [(a sin t)(a sin t)  (a cos t)(a cos t)] dt œ '021 a# dt œ 21a# ; Èa c x

' ' Š ``Nx  ``My ‹ dx dy œ ' ' ca cc a

R

#

œ 2a

ˆ 1#




1‰ #

#

#

2 dy dx œ 'ca 4Èa#  x# dx œ 4 ’ x2 Èa#  x#
a

sin
" xa “

a

ca

#

œ 2a 1, Circulation

2. M œ y œ a sin t, N œ 0, dx œ a sin t dt, dy œ a cos t dt Ê Equation (3):

a# #

)C M dy  N dx œ '0

21

`M `x

œ 0,

`M `y

œ 1,

`N `x

œ 0, and

`N `y

œ 0;

#1 a# sin t cos t dt œ a#
2" sin# t‘ ! œ 0; ' ' 0 dx dy œ 0, Flux

R

21 #1 Equation (4): )C M dx
N dy œ '0 aa# sin# tb dt œ a#
2t  sin4 2t ‘ ! œ 1a# ; ' ' Š ``Nx  ``My ‹ dx dy

œ ' ' 1 dx dy œ '0

21

R

'0

a

r dr d) œ '0  21

R

a# #

d) œ 1a# , Circulation

3. M œ 2x œ 2a cos t, N œ 3y œ 3a sin t, dx œ a sin t dt, dy œ a cos t dt Ê `N `y

`M `x

œ 2,

`M `y

`N `x

œ 0,

œ 0, and

œ 3;

Equation (3):

)C M dy  N dx œ '021 [(2a cos t)(a cos t)
(3a sin t)(a sin t)] dt

œ '0 a2a# cos# t  3a# sin# tb dt œ 2a#
2t
21

sin 2t ‘ #1 4 !

 3a#
2t 

sin 2t ‘ #1 4 !

œ 21a#  31a# œ 1a# ;

' ' Š ``Mx
``Ny ‹ œ ' ' 1 dx dy œ ' ' r dr d) œ '  a## d) œ 1a# , Flux 0 0 0 21

R

a

21

R

Equation (4):

)C M dx
N dy œ '021 [(2a cos t)(a sin t)
(3a sin t)(a cos t)] dt

#1 œ '0 a2a# sin t cos t  3a# sin t cos tb dt œ 5a#
12 sin# t‘ ! œ 0; ' ' 0 dx dy œ 0, Circulation 21

R

4. M œ x# y œ a$ cos# t, N œ xy# œ a$ cos t sin# t, dx œ a sin t dt, dy œ a cos t dt Ê ``Mx œ 2xy, ``My œ x2 , ``Nx œ y# , and ``Ny œ 2xy; Equation (3):

)C M dy  N dx œ '021 aa% cos$ t sin t
a% cos t sin$ tb œ ’ a4

%

cos% t


a% 4

sin% t“

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

#1 !

œ 0;

957

958

Chapter 16 Integration in Vector Fields

' ' Š ``Mx
``Ny ‹ dx dy œ ' ' (2xy
2xy) dx dy œ 0, Flux R

R

21 21 Equation (4): )C M dx
N dy œ '0 aa% cos# t sin# t
a% cos# t sin# tb dt œ '0 a2a% cos# t sin# tb dt 21 41 %1 œ '0 "# a% sin# 2t dt œ a4 '0 sin# u du œ a4
u2  sin42u ‘ ! œ 1#a ; ' ' Š ``Nx  ``My ‹ dx dy œ ' ' ay#
x# b dx dy %

%

R

21 a 21 œ '0 '0 r# † r dr d) œ '0 a4

%

`M `x

5. M œ x  y, N œ y  x Ê Circ œ ' '

%

d) œ

1 a% #

, Circulation

œ 1,

`M `y

œ 1,

`N `x

`N `y

œ 1,

R

œ 1 Ê Flux œ ' ' 2 dx dy œ '0

1

R

'01 2 dx dy œ 2;

[1  (1)] dx dy œ 0

R `M `x

6. M œ x#
4y, N œ x
y# Ê

`M `y

œ 2x,

œ 4,

`N `x

œ 1,

`N `y

œ 2y Ê Flux œ ' ' (2x
2y) dx dy R

1 1 1 1 " " œ '0 '0 (2x
2y) dx dy œ '0 cx#
2xyd ! dy œ '0 (1
2y) dy œ cy
y# d ! œ 2; Circ œ ' '

œ '0

1

'01 3 dx dy œ 3 `M `x

7. M œ y#  x# , N œ x#
y# Ê œ '0

3

œ 2x,

'0 (2x
2y) dy dx œ '0 a2x x

3

#

`M `y

œ 2y,

`N `x

#


x b dx œ


" 3

œ 2x, $ x$ ‘ !

`N `y

œ 2y Ê Flux œ ' ' (2x
2y) dx dy R

œ 9; Circ œ ' ' (2x  2y) dx dy R

3 x 3 œ '0 '0 (2x  2y) dy dx œ '0 x# dx œ 9

8. M œ x
y, N œ  ax#
y# b Ê

`M `x

`M `y

œ 1,

œ 1,

`N `x

œ 2x,

1 x 1 œ '0 '0 (1  2y) dy dx œ '0 ax  x# b dx œ "6 ; Circ œ ' '

œ '0 a2x  xb dx œ  1

#

R

œ '0 œ '0

1

Èx

'x

2

`M `x

`M `y

œ y,

œ x
2y,

`N `x

œ 1,

`N `y

œ 1 Ê Flux œ ' ' ay
a1bb dy dx R

' ' a1  ax
2ybb dy dx ay  1b dy dx œ '0 ˆ "# x  Èx  "# x4
x# ‰ dx œ  11 60 ; Circ œ

2

7 a1  x  2yb dy dx œ '0 ˆÈx  x3Î2  x  x#
x3
x4 ‰ dx œ  60

Circ œ ' ' R

1

`M `x

œ 1,

`M `y

œ 3,

`N `x

œ 2,

`N `y

œ 1 Ê Flux œ ' ' a1
a1bb dy dx œ 0 R

È2 È2 2 x Î2 a2  3b dy dx œ 'cÈ2 '
È 2 c x Î2 a1b dy dx œ  È22 'cÈ2 È2  x2 dx œ  1È2 Èa

2b

a

11. M œ x3 y2 , N œ "# x4 y Ê 2

R

1 x (2x  1) dx dy œ '0 '0 (2x  1) dy dx

1

10. M œ x
3y, N œ 2x  y Ê

œ '0

œ 2y Ê Flux œ ' ' (1  2y) dx dy

R

Èx

'x

`N `y

7 6

9. M œ xy
y2 , N œ x  y Ê 1

(1  4) dx dy

R

`M `x

œ 3x2 y2 ,

2b

`M `y

œ 2x3 y,

`N `x

œ 2x3 y,

`N `y

œ "# x4 Ê Flux œ ' ' ˆ3x2 y2
"# x4 ‰ dy dx

'xx
x ˆ3x2 y2
"# x4 ‰ dy dx œ '02 ˆ3x5  72 x6
3x7  x8 ‰ dx œ 649 ; Circ œ ' ' 2

R

a2x3 y  2x3 yb dy dx œ 0

R

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

Section 16.4 Green's Theorem in the Plane 12. M œ

x 1  y2 ,

È1
y2

œ 'c1 '
È1
y2 1

`M `x

N œ tan
1 y Ê 2 1  y2

1 `M 1  y2 , ` y

œ


2x y , `N a 1  y 2 b2 ` x

œ

œ 0,

`N `y

œ

Ê Flux œ ' ' Š 1 1 y2


1 1  y2

R

1 1  y2 ‹

dx dy

dx dy œ 'c1 4 1 1
y2y dx œ 41È2  41 ; Circ œ ' ' Š0  Š a1
2xy2yb2 ‹‹ dy dx È

1

2

R

È1
y2

y œ 'c1 '
È1
y2 Š a1 2x ‹ dy dx œ 'c1 a0b dx œ 0  y 2 b2 1

1

`M `x

13. M œ x
ex sin y, N œ x
ex cos y Ê

Ècos 2)

1Î4

Ê Flux œ ' ' dx dy œ 'c1Î4 '0 R

œ 1
ex sin y, Î

`M `y

œ ex cos y,

1 4

1Î%

Ècos 2)

1Î4

R

R

y x

, N œ ln ax#
y# b Ê

Ê Flux œ ' ' Š x#
yy#
R

Circ œ ' ' Š x# 2x  y# 

x x#  y# ‹

R

`M `x

15. M œ xy, N œ y# Ê œ '0 Š 3x#  1

#

3x% # ‹

2y x#  y# ‹

dx œ

`M `x

dx dy œ '0

1

dx dy œ '0

1

`M `y

œ y,


y x#  y#

œ

Ê Flux œ ' ' (x sin y) dx dy œ '0 R

œ 0,

'0

1Î2

1Î2

R

`M `x

, N œ ex
tan
" y Ê

R

`N `y

" 1  y#




1 3cx œ 'c1 'x b 1 %

#

Ê

œ '0

'0

x$

2y x#  y#

'x

#

`N `x

,

`M `y

2xy dy dx œ ' #

1

2 0 3

x

œ 6xy# ,

`N `x

"!

2 33

20. M œ 4x  2y, N œ 2x  4y Ê

dx œ `M `y

œ 2,

x dy dx œ '0 ax#
x$ b dx œ  1"# `N `y

œ cos y,

œ x sin y #

#

œ 3y 

" 1  y#

,

`N `y

œ

" 1  y#

`N `x

œ

ex y

#1 !

'0aÐ1

cos )Ñ

(3r sin )) r dr d)

œ 4a$  a4a$ b œ 0

Ê Circ œ ' ' ’ ey  Š1
x

R

ex y ‹“

dx dy œ ' ' (1) dx dy R

œ 8xy# Ê work œ )C 2xy$ dx
4x# y# dy œ ' ' a8xy#  6xy# b dx dy R

`N `x

œ 2 Ê work œ )C (4x  2y) dx
(2x  4y) dy

œ ' ' [2  (2)] dx dy œ 4 ' ' dx dy œ 4(Area of the circle) œ 4(1 † 4) œ 161 R

#

'01Î2 2 cos y dx dy œ '01Î2 1 cos y dy œ c1 sin yd 1Î# œ1 !

R

ex y

'xx 3y dy dx

1

21

œ1


œ

1 1  dy dx œ  'c1 ca3  x# b  ax%
1bd dx œ 'c1 ax%
x#  2b dx œ  44 15

19. M œ 2xy$ , N œ 4x# y# Ê 1

`M `y

`N `y

(x sin y) dx dy œ '0 Š 18 sin y‹ dy œ  18 ;

$

ex y

,

R

x

1Î2

œ '0 a$ (1
cos ))$ (sin )) d) œ ’ a4 (1
cos ))% “ 18. M œ y
ex ln y, N œ

2x x#  y#

1

dx dy œ ' ' 3y dx dy œ '0

" 1  y# ‹

21

œ

œ 2y Ê Flux œ ' ' (y
2y) dy dx œ '0

œ  cos y,

Circ œ ' ' [cos y  ( cos y)] dx dy œ '0

Ê Flux œ ' ' Š3y 

`N `x

" #

#

`M `y

œ 0,

,

Î

1 4

#

R

`M `x

x x#  y#

;

r dr d) œ ' 1Î4 ˆ "# cos 2)‰ d) œ

'12 ˆ r sinr ) ‰ r dr d) œ '01 sin ) d) œ 2;

; Circ œ ' ' x dy dx œ '0

" 5

1Î2

x 1  y#

œ

1

16. M œ  sin y, N œ x cos y Ê

17. M œ 3xy 

`M `y

" #

œ ex sin y

'12 ˆ r cosr ) ‰ r dr d) œ '01 cos ) d) œ 0

`N `x

œ x,

,

`N `y

œ 1
ex cos y,

r dr d) œ ' 1Î4 ˆ "# cos 2)‰ d) œ
4" sin 2)‘
1Î% œ

Circ œ ' ' a1
ex cos y  ex cos yb dx dy œ ' ' dx dy œ 'c1Î4 '0 14. M œ tan
"

`N `x

R

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

959

960

Chapter 16 Integration in Vector Fields `M `y

21. M œ y# , N œ x# Ê œ '0

1

1cx

'0

œ 2y,

œ 2x Ê )C y# dx
x# dy œ ' ' (2x  2y) dy dx

`N `x

R

(2x  2y) dy dx œ '0 a3x
4x  1b dx œ cx
2x#  xd ! œ 1
2  1 œ 0 1

22. M œ 3y, N œ 2x Ê

`M `y

œ 3,

#

`N `x

œ 2 Ê )C 3y dx
2x dy œ ' ' a2  3b dx dy œ '0

`M `y

œ 6,

1

R

1 œ '0 sin x dx œ 2

23. M œ 6y
x, N œ y
2x Ê

"

$

'0sin x a1bdy dx

œ 2 Ê )C (6y
x) dx
(y
2x) dy œ ' ' (2  6) dy dx

`N `x

R

œ 4(Area of the circle) œ 161 24. M œ 2x
y# , N œ 2xy
3y Ê

`M `y

œ 2y,

`N `x

œ 2y Ê

)C a2x
y# b dx
(2xy
3y) dy œ ' ' (2y  2y) dx dy œ 0 R

25. M œ x œ a cos t, N œ y œ a sin t Ê dx œ a sin t dt, dy œ a cos t dt Ê Area œ œ

'0

21

" #

'0

21

" #

aa# cos# t
a# sin# tb dt œ

'021 aab cos# t
ab sin# tb dt œ "# '021 ab dt œ 1ab

" #

)C

" #

)C x dy  y dx

x dy  y dx

a# dt œ 1a#

26. M œ x œ a cos t, N œ y œ b sin t Ê dx œ a sin t dt, dy œ b cos t dt Ê Area œ œ

" #

)C x dy  y dx 41 3 ' œ "# '0 a3 sin# t cos# tb acos# t
sin# tb dt œ "# '0 a3 sin# t cos# tb dt œ 38 '0 sin# 2t dt œ 16 sin# u du 0

27. M œ x œ cos$ t, N œ y œ sin$ t Ê dx œ 3 cos# t sin t dt, dy œ 3 sin# t cos t dt Ê Area œ 21

œ

3 16


u2 

21

sin 2u ‘ %1 4 !

œ

3 8

21

" #

1

28. C1 : M œ x œ t, N œ y œ 0 Ê dx œ dt, dy œ 0; C2 : M œ x œ a21  tb  sina21  tb œ 21  t
sin t, N œ y œ 1  cosa21  tb œ 1  cos t Ê dx œ acos t  1b dt, dy œ sin t dt Ê Area œ œ

" #

" #

)C x dy  y dx œ "# )C

"

x dy  y dx


" #

)C

2

x dy  y dx

'021 a0bdt
"# '021 ca21  t
sin tbasin tb  a1  cos tb acos t  1bd dt œ  "# '021 a2 cos t
t sin t  2  21 sin tb dt

œ  12 c3 sin t  t cos t  2t  21 cos td20 1 œ 31 29. (a) M œ f(x), N œ g(y) Ê (b) M œ ky, N œ hx Ê

`M `y

`M `y

œ 0,

œ k,

`N `x

`N `x

œ 0 Ê )C f(x) dx
g(y) dy œ ' ' Š ``Nx  R

œh

`M `y ‹

dx dy œ ' ' 0 dx dy œ 0 R

Ê )C ky dx
hx dy œ ' ' Š ``Nx  ``My ‹ dx dy

œ ' ' (h  k) dx dy œ (h  k)(Area of the region)

R

R

30. M œ xy# , N œ x# y
2x Ê

`M `y

œ 2xy,

`N `x

œ 2xy
2 Ê )C xy# dx
ax# y
2xb dy œ ' ' Š ``Nx 

œ ' ' (2xy
2  2xy) dx dy œ 2 ' ' dx dy œ 2 times the area of the square R

R

R

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

`M `y ‹

dx dy

Section 16.4 Green's Theorem in the Plane

961

31. The integral is 0 for any simple closed plane curve C. The reasoning: By the tangential form of Green's Theorem, with M œ 4x$ y and N œ x% , )C 4x$ y dx
x% dy œ ' ' ’ ``x ax% b 

` `y

R

œ ' ' ðóóñóóò a4x$  4x$ b dx dy œ 0.

a4x$ yb“ dx dy

R

0 32. The integral is 0 for any simple closed curve C. The reasoning: By the normal form of Green's theorem, with ` ` $ $ M œ x$ and N œ y$ , )C y$ dy
x$ dx œ ' ' ”ðñò ` x ay b  ï ` y ax b • dx dy œ 0.

R

0 `M `x

33. Let M œ x and N œ 0 Ê

œ 1 and

`N `y

œ0 Ê

0

)C M dy  N dx œ ' ' Š ``Mx
``Ny ‹ dx dy

œ ' ' (1
0) dx dy Ê Area of R œ ' ' dx dy œ )C x dy; similarly, M œ y and N œ 0 Ê R

`N `x

R

œ 0 Ê )C M dx
N dy œ ' ' Š ``Nx
R

œ ' ' dx dy œ Area of R

Ê )C x dy

R

`M `y ‹

`M `y

œ 1 and

dy dx Ê )C y dx œ ' ' (0  1) dy dx Ê  )C y dx R

R

34.

'ab f(x) dx œ Area of R œ )C y dx, from Exercise 33

35. Let $ (xß y) œ 1 Ê x œ

My M

' ' x $ (xßy) dA

œ 'R '

$ (xßy) dA

' ' x dA

œ 'R '

R

dA

' ' x dA

œ

Ê Ax œ ' ' x dA œ ' ' (x
0) dx dy

R

A

R

R

R

œ )C x#

dy, Ax œ ' ' x dA œ ' ' (0
x) dx dy œ  ) xy dx, and Ax œ ' ' x dA œ ' ' ˆ 23 x
"3 x‰ dx dy

œ)

#

#

" C 3

R

R

" 3

" #

x dy  xy dx Ê

)C x

C

dy œ )C xy dx œ

#

" 3

)C x

dy  xy dx œ Ax

36. If $ (xß y) œ 1, then Iy œ ' ' x# $ (xß y) dA œ ' ' x# dA œ ' ' ax#
0b dy dx œ R

R

R

R

#

R

" 3

)C

x$ dy,

' ' x# dA œ ' ' a0
x# b dy dx œ  ) x# y dx, and ' ' x# dA œ ' ' ˆ 34 x#
4" x# ‰ dy dx C R

R

œ)

" C 4

37. M œ 38. M œ

`f `y

" 4

ellipse

$

" 4

#

x dy  x y dx œ , N œ  `` xf Ê

`M `y

" 4

œ

x# y
"3 y$ , N œ x Ê " 4

)C x ` #f ` y#

`M `y

,

R

$

#

dy  x y dx Ê

`N `x

œ

1 4

" 3

œ  `` xf# Ê )C #

x#
y# ,

`N `x

)C x

`f `y

R

$

dy œ  )C x# y dx œ

dx 

`f `x

œ 1 Ê Curl œ

" 4

)C

dy œ ' ' Š `` xf#  #

R

`N `x



`M `y

x$ dy  x# y dx œ Iy

` #f ` y# ‹

dx dy œ 0 for such curves C

œ 1  ˆ "4 x#
y# ‰  0 in the interior of the

x#
y# œ 1 Ê work œ 'C F † dr œ ' ' ˆ1  4" x#  y# ‰ dx dy will be maximized on the region R

R œ {(xß y) | curl F}   0 or over the region enclosed by 1 œ 2y 39. (a) ™ f œ Š x# 2x  y# ‹ i
Š x#  y# ‹ j Ê M œ

2x x#  y#

,Nœ

" 4

x#
y#

2y x#  y#

; since M, N are discontinuous at (0ß 0), we

compute 'C ™ f † n ds directly since Green's Theorem does not apply. Let x œ a cos t, y œ a sin t Ê dx œ a sin t dt, dy œ a cos t dt, M œ

2 a

cos t, N œ

2 a

sin t, 0 Ÿ t Ÿ 21, so 'C ™ f † n ds œ 'C M dy  N dx

œ '0
ˆ 2a cos t‰aa cos tb  ˆ 2a sin t‰aa sin tb ‘dt œ '0 2acos2 t
sin2 tbdt œ 41. Note that this holds for any 21

21

Copyright © 2010 Pearson Education Inc. Publishing as Addison-Wesley.

962

Chapter 16 Integration in Vector Fields a  0, so 'C ™ f † n ds œ 41 for any circle C centered at a0, 0b traversed counterclockwise and 'C ™ f † n ds œ 41 if C is traversed clockwise.

(b) If K does not enclose the point (0ß 0) we may apply Green's Theorem: 'C ™ f † n ds œ 'C M dy  N dx œ ' ' Š ``Mx
R

`N `y ‹

dx dy œ ' ' Š ax2  y2 b2
2 ˆy 2
x 2 ‰

R

2 ˆx 2
y 2 ‰ ‹ ax2  y2 b2

dx dy œ ' ' 0 dx dy œ 0. If K does enclose the point R

(0ß 0) we proceed as follows: Choose a small enough so that the circle C centered at (0ß 0) of radius a lies entirely within K. Green's Theorem applies to the region R that lies between K and C. Thus, as before, 0 œ ' ' Š ``Mx
R

`N `y ‹

dx dy

œ 'K M dy  N dx
'C M dy  N dx where K is traversed counterclockwise and C is traversed clockwise.

Hence by part (a) 0 œ ’ ' M dy  N dx “  41 Ê 41 œ K

'K ™ f † n ds œ œ 0

'K M dy  N dx

œ 'K ™ f † n ds. We have shown:

if (0ß 0) lies inside K if (0ß 0) lies outside K

41

40. Assume a particle has a closed trajectory in R and let C" be the path Ê C" encloses a simply connected region R" Ê C" is a simple closed curve. Then the flux over R" is )C F † n ds œ 0, since the velocity vectors F are "

tangent to C" . But 0 œ )C F † n ds œ )C M dy  N dx œ ' ' Š ``Mx
"

"

R"

`N `y ‹

dx dy Ê Mx
Ny œ 0, which is a

contradiction. Therefore, C" cannot be a closed trajectory. 41.

'gg yy

#Ð Ñ

"Ð Ñ

`N `x

dx dy œ N(g# (y)ß y)  N(g" (y)ß y) Ê

'cd 'gg yy ˆ ``Nx dx‰ dy œ 'cd [N(g# (y)ß y)  N(g" (y)ß y)] dy #Ð Ñ

"Ð Ñ

œ 'c N(g# (y)ß y) dy  'c N(g" (y)ß y) dy œ 'c N(g# (y)ß y) dy
'd N(g" (y)ß y) dy œ 'C N dy
'C N dy d

œ )C dy

d

Ê

)C N dy œ ' ' R

d

c

#

`N `x

"

dx dy

42. The curl of a conservative two-dimensional field is zero. The reasoning: A two-dimensional field F œ Mi
Nj can be considered to be the restriction to the xy-plane of a three-dimensional field whose k component is zero, and whose i and j components are independent of z. For such a field to be conservative, we must have `N `M `N `M ` x œ ` y by the component test in Section 16.3 Ê curl F œ ` x  ` y œ 0. 43-46. Example CAS commands: Maple: with( plots );#43 M := (x,y) -> 2*x-y; N := (x,y) -> x+3*y; C := x^2 + 4*y^2 = 4; implicitplot( C, x=-2..2, y=-2..2, scaling=constrained, title="#43(a) (Section 16.4)" ); curlF_k := D[1](N) - D[2](M): # (b) 'curlF_k' = curlF_k(x,y); top,bot := solve( C, y ); # (c) left,right := -2, 2; q1 := Int( Int( curlF_k(x,y), y=bot..top ), x=left..right ); value( q1 ); Mathematica: (functions and bounds will vary) The ImplicitPlot command will be useful for 43 and 44, but is not needed for 43 and 44. In 44, the equation of the line from (0, 4) to (2, 0) must be determined first.

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Section 16.5 Surfaces and Area

963

Clear[x, y, f]