INSTRUCTOR’S SOLUTIONS MANUAL PART ONE ARDIS • BORZELLINO • BUCHANAN • MOGILL • NELSON to accompany
THOMAS’ CALCULUS EL...
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INSTRUCTOR’S SOLUTIONS MANUAL PART ONE ARDIS • BORZELLINO • BUCHANAN • MOGILL • NELSON to accompany
THOMAS’ CALCULUS ELEVENTH 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
Frank R. Giordano Naval Postgraduate School
weir_22646_SSM_TTL_CPY.qxd
9/27/04
9:59 AM
Page 2
Reproduced by Pearson Addison-Wesley from electronic files supplied by the authors. Copyright © 2005 Pearson Education, Inc. Publishing as Pearson Addison-Wesley, 75 Arlington Street, Boston, MA 02116. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN
0-321-22653-4
1 2 3 4 5 6 BB 07 06 05 04
PREFACE TO THE INSTRUCTOR This Instructor's Solutions Manual contains the solutions to every exercise in the 11th Edition of THOMAS' CALCULUS by Maurice Weir, Joel Hass and Frank Giordano, 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).
Acknowledgments Solutions Writers William Ardis, Collin County Community College-Preston Ridge Campus Joseph Borzellino, California Polytechnic State University Linda Buchanana, Howard College Tim Mogill Patricia Nelson, University of Wisconsin-La Crosse Accuracy Checkers Karl Kattchee, University of Wisconsin-La Crosse Marie Vanisko, California State University, Stanislaus Tom Weigleitner, VISTA Information Technologies
Thanks to Rachel Reeve, Christine O'Brien, Sheila Spinney, Elka Block, and Joe Vetere for all their guidance and help at every step.
CHAPTER 1 PRELIMINARIES 1.1 REAL NUMBERS AND THE REAL LINE 1. Executing long division,
" 9
2. Executing long division,
" 11
œ 0.1,
2 9
œ 0.2,
œ 0.09,
2 11
3 9
œ 0.3,
œ 0.18,
3 11
8 9
œ 0.8,
œ 0.27,
9 11
9 9
œ 0.9
œ 0.81,
11 11
œ 0.99
3. NT = necessarily true, NNT = Not necessarily true. Given: 2 < x < 6. a) NNT. 5 is a counter example. b) NT. 2 < x < 6 Ê 2 2 < x 2 < 6 2 Ê 0 < x 2 < 2. c) NT. 2 < x < 6 Ê 2/2 < x/2 < 6/2 Ê 1 < x < 3. d) NT. 2 < x < 6 Ê 1/2 > 1/x > 1/6 Ê 1/6 < 1/x < 1/2. e) NT. 2 < x < 6 Ê 1/2 > 1/x > 1/6 Ê 1/6 < 1/x < 1/2 Ê 6(1/6) < 6(1/x) < 6(1/2) Ê 1 < 6/x < 3. f) NT. 2 < x < 6 Ê x < 6 Ê (x 4) < 2 and 2 < x < 6 Ê x > 2 Ê x < 2 Ê x + 4 < 2 Ê (x 4) < 2. The pair of inequalities (x 4) < 2 and (x 4) < 2 Ê | x 4 | < 2. g) NT. 2 < x < 6 Ê 2 > x > 6 Ê 6 < x < 2. But 2 < 2. So 6 < x < 2 < 2 or 6 < x < 2. h) NT. 2 < x < 6 Ê 1(2) > 1(x) < 1(6) Ê 6 < x < 2 4. NT = necessarily true, NNT = Not necessarily true. Given: 1 < y 5 < 1. a) NT. 1 < y 5 < 1 Ê 1 + 5 < y 5 + 5 < 1 + 5 Ê 4 < y < 6. b) NNT. y = 5 is a counter example. (Actually, never true given that 4 y 6) c) NT. From a), 1 < y 5 < 1, Ê 4 < y < 6 Ê y > 4. d) NT. From a), 1 < y 5 < 1, Ê 4 < y < 6 Ê y < 6. e) NT. 1 < y 5 < 1 Ê 1 + 1 < y 5 + 1 < 1 + 1 Ê 0 < y 4 < 2. f) NT. 1 < y 5 < 1 Ê (1/2)(1 + 5) < (1/2)(y 5 + 5) < (1/2)(1 + 5) Ê 2 < y/2 < 3. g) NT. From a), 4 < y < 6 Ê 1/4 > 1/y > 1/6 Ê 1/6 < 1/y < 1/4. h) NT. 1 < y 5 < 1 Ê y 5 > 1 Ê y > 4 Ê y < 4 Ê y + 5 < 1 Ê (y 5) < 1. Also, 1 < y 5 < 1 Ê y 5 < 1. The pair of inequalities (y 5) < 1 and (y 5) < 1 Ê | y 5 | < 1. 5. 2x 4 Ê x 2 6. 8 3x 5 Ê 3x 3 Ê x Ÿ 1 7. 5x $ Ÿ ( 3x Ê 8x Ÿ 10 Ê x Ÿ
ïïïïïïïïïñqqqqqqqqp x 1 5 4
8. 3(2 x) 2(3 x) Ê 6 3x 6 2x Ê 0 5x Ê 0 x 9. 2x
10.
" #
Ê
" 5
6 x 4
7x
ˆ
10 ‰ 6
3x4 2
7 6
Ê "#
x or
" 3
7 6
ïïïïïïïïïðqqqqqqqqp x 0
5x
x
Ê 12 2x 12x 16
Ê 28 14x Ê 2 x
qqqqqqqqqðïïïïïïïïî x 2
2 11.
Chapter 1 Preliminaries 4 5
" 3
(x 2)
(x 6) Ê 12(x 2) 5(x 6)
Ê 12x 24 5x 30 Ê 7x 6 or x 67 12. x2 5 Ÿ
123x 4
Ê (4x 20) Ÿ 24 6x
Ê 44 Ÿ 10x Ê 22 5 Ÿ x
qqqqqqqqqñïïïïïïïïî x 22/5
13. y œ 3 or y œ 3 14. y 3 œ 7 or y 3 œ 7 Ê y œ 10 or y œ 4 15. 2t 5 œ 4 or 2t & œ 4 Ê 2t œ 1 or 2t œ 9 Ê t œ "# or t œ 9# 16. 1 t œ 1 or 1 t œ 1 Ê t œ ! or t œ 2 Ê t œ 0 or t œ 2 17. 8 3s œ 18.
s #
9 2
or 8 3s œ #9 Ê 3s œ 7# or 3s œ 25 # Ê sœ
1 œ 1 or
s #
1 œ 1 Ê
s #
œ 2 or
s #
7 6
or s œ
25 6
œ ! Ê s œ 4 or s œ 0
19. 2 x 2; solution interval (2ß 2) 20. 2 Ÿ x Ÿ 2; solution interval [2ß 2]
qqqqñïïïïïïïïñqqqqp x 2 2
21. 3 Ÿ t 1 Ÿ 3 Ê 2 Ÿ t Ÿ 4; solution interval [2ß 4] 22. 1 t 2 1 Ê 3 t 1; solution interval (3ß 1)
qqqqðïïïïïïïïðqqqqp t 3 1
23. % 3y 7 4 Ê 3 3y 11 Ê 1 y solution interval ˆ1ß
11 3
;
11 ‰ 3
24. 1 2y 5 " Ê 6 2y 4 Ê 3 y 2; solution interval (3ß 2) 25. 1 Ÿ
z 5
1Ÿ1 Ê 0Ÿ
z 5
qqqqðïïïïïïïïðqqqqp y 3 2
Ÿ 2 Ê 0 Ÿ z Ÿ 10;
solution interval [0ß 10] 26. 2 Ÿ
1 Ÿ 2 Ê 1 Ÿ solution interval 23 ß 2‘ 3z #
27. "# 3 Ê
2 7
28. 3
" x
x 2 x
2 5
" #
2 7
Ÿ 3 Ê 32 Ÿ z Ÿ 2; qqqqñïïïïïïïïñqqqqp z 2 2/3
Ê 7# x" 5# Ê
7 #
" x
5 #
; solution interval ˆ 27 ß 25 ‰
43 Ê 1
Ê 2x
3z #
Ê
2 7
2 x
( Ê 1
x #
" 7
x 2; solution interval ˆ 27 ß 2‰
qqqqðïïïïïïïïðqqqqp x 2 2/7
Section 1.1 Real Numbers and the Real Line 29. 2s 4 or 2s 4 Ê s 2 or s Ÿ 2; solution intervals (_ß 2] [2ß _) 30. s 3
" #
or (s 3)
" #
Ê s 5# or s
7 #
Ê s 5# or s Ÿ 7# ; solution intervals ˆ_ß 7# ‘ 5# ß _‰
ïïïïïïñqqqqqqñïïïïïïî s 7/2 5/2
31. 1 x 1 or (" x) 1 Ê x 0 or x 2 Ê x 0 or x 2; solution intervals (_ß !) (2ß _) 32. 2 3x 5 or (2 3x) 5 Ê 3x 3 or 3x 7 Ê x 1 or x 73 ; solution intervals (_ß 1) ˆ 73 ß _‰ 33.
r" #
ïïïïïïðqqqqqqðïïïïïïî x 1 7/3
1 or ˆ r# 1 ‰ 1 Ê r 1 2 or r 1 Ÿ 2
Ê r 1 or r Ÿ 3; solution intervals (_ß 3] [1ß _) 34.
3r 5
"
Ê
or ˆ 3r5 "‰
2 5
or 3r5 53 Ê r 37 or r 1 solution intervals (_ß ") ˆ 73 ß _‰ 3r 5
2 5 7 5
ïïïïïïðqqqqqqðïïïïïïî r 1 7/3
35. x# # Ê kxk È2 Ê È2 x È2 ; solution interval ŠÈ2ß È2‹
qqqqqqðïïïïïïðqqqqqqp x È# È #
36. 4 Ÿ x# Ê 2 Ÿ kxk Ê x 2 or x Ÿ 2; solution interval (_ß 2] [2ß _)
ïïïïïïñqqqqqqñïïïïïïî r 2 2
37. 4 x# 9 Ê 2 kxk 3 Ê 2 x 3 or 2 x 3 Ê 2 x 3 or 3 x 2; solution intervals (3ß 2) (2ß 3) 38.
" 9
x#
Ê
x
" #
" 3
kxk
" #
Ê
" 3
x
or #" x 3" ; solution intervals ˆ "# ß 3" ‰ ˆ 3" ß #" ‰ Ê
" 3
" 4
" #
or
" 3
x
39. (x 1)# 4 Ê kx 1k 2 Ê 2 x 1 2 Ê 1 x 3; solution interval ("ß $)
qqqqðïïïïðqqqqðïïïïðqqqp x 3 2 2 3 " #
qqqqðïïïïðqqqqðïïïïðqqqp x 1/2 1/3 1/3 1/2
qqqqqqðïïïïïïïïðqqqqp x 1 3
40. (x 3)# # Ê kx 3k È2 Ê È2 x 3 È2 or 3 È2 x 3 È2 ; solution interval Š3 È2ß 3 È2‹
qqqqqqðïïïïïïïïðqqqqp x 3 È # 3 È #
3
4
Chapter 1 Preliminaries
41. x# x 0 Ê x# x +
1 4
(x^4 16)/(x 2); x0 := 2; plot( f(x), x œ x0-1..x0+1, color œ black, title œ "Section 2.1, #41(a)" ); limit( f(x), x œ x0 ); In Exercise 43, note that the standard cube root, x^(1/3), is not defined for x (surd(x+1, 3) 1)/x. Mathematica: (assigned function and values for x0 and h may vary) Clear[f, x] f[x_]:=(x3 x2 5x 3)/(x 1)2 x0= 1; h = 0.1; Plot[f[x],{x, x0 h, x0 h}] Limit[f[x], x Ä x0]
73
74
Chapter 2 Limits and Continuity
2.2 CALCULATING LIMITS USING THE LIMIT LAWS 1. 3. 4. 5.
lim (2x 5) œ 2(7) 5 œ 14 5 œ 9
lim ax# 5x 2b œ (2)# 5(2) 2 œ 4 10 2 œ 4 lim ax$ 2x# 4x 8b œ (2)$ 2(2)# 4(2) 8 œ 8 8 8 8 œ 16
x Ä c#
lim 8(t 5)(t 7) œ 8(6 5)(6 7) œ 8 x3
œ
9.
lim y Ä c& 5 y
y#
23 26
# y Ä # y 5y 6
œ
5 8
(5)# 5 (5)
œ
y2
10. lim
13.
6.
tÄ'
lim x Ä # x6
12.
lim (10 3x) œ 10 3(12) œ 10 36 œ 26
x Ä 1#
xÄ#
7.
11.
2.
x Ä c(
œ
8. œ
25 10
œ
22 (2)# 5(#) 6
lim# 3s(2s 1) œ 3 ˆ 23 ‰ 2 ˆ 23 ‰ 1‘ œ 2 ˆ 43 1‰ œ
sÄ
$
4
lim x Ä & x7
œ
4 57
œ
4 #
œ 2
5 #
œ
4 4 10 6
œ
4 #0
œ
" 5
lim 3(2x 1)# œ 3(2(1) 1)# œ 3(3)# œ 27
x Ä c"
lim (x 3)"*)% œ (4 3)"*)% œ (1)"*)% œ 1
x Ä c%
%
lim (5 y)%Î$ œ [5 (3)]%Î$ œ (8)%Î$ œ ˆ(8)"Î$ ‰ œ 2% œ 16
y Ä c$
14. lim (2z 8)"Î$ œ (2(0) 8)"Î$ œ (8)"Î$ œ 2 zÄ!
15. lim
3
œ
3 È3(0) 1 1
œ
3 È1 1
œ
3 2
16. lim
5
œ
5 È5(0) 4 2
œ
5 È4 #
œ
5 4
17. lim
È3h 1 " h
h Ä ! È3h 1 1 h Ä ! È5h 4 2
hÄ0
œ
3 È" "
œ
È5h 4 2 h hÄ0 5 È4 2
19. lim
œ
x5
# x Ä & x 25
20. 21.
œ lim
a3h "b 1
È5h 4 2 h hÄ0
†
È5h 4 2 È5h 4 2
œ lim
a5h 4b 4
h Ä 0 hŠÈ3h 1 "‹
œ lim
3h
œ lim
5h
h Ä 0 hŠÈ3h 1 "‹
h Ä 0 hŠÈ5h 4 2‹
h Ä 0 hŠÈ5h 4 2‹
5 4 x5
œ lim
x Ä & (x 5)(x 5)
x3
lim
È3h 1 1 È3h 1 1
œ lim
lim # x Ä c$ x 4x 3 x Ä c&
†
hÄ0
œ lim
3
h Ä 0 È3h1"
$ #
18. lim œ
È3h 1 " h
œ lim
x# 3x "0 x5
œ lim
œ lim
x3
x Ä c$ (x 3)(x 1)
œ lim
x Ä c&
1
x Ä & x5
(x 5)(x 2) x5
œ
œ lim
" 55
œ
" 10
1
œ
" 3 1
x Ä c$ x 1
œ "2
œ lim (x 2) œ & # œ 7 x Ä c&
œ lim
5
h Ä 0 È5h 4 2
2 3
Section 2.2 Calculating Limits Using the Limit Laws (x 5)(x 2) x2
22. lim
x# 7x "0 x#
œ lim
23. lim
t# t 2 t# 1
t Ä " (t 1)(t 1)
xÄ#
tÄ"
t# 3t 2
lim # t Ä c" t t 2
25.
lim $ # x Ä c# x 2x
2x 4
5y$ 8y#
u% "
$ u Ä 1 u 1
œ lim
y# (5y 8)
œ lim
œ lim
4x x#
œ lim
x Ä 1 Èx 3 2
lim
x Ä c"
œ lim
xÄ%
œ
x Ä c"
35.
x2 x Ä c2 È x # 5 3
œ lim
œ
lim
"
œ lim
x ˆ2 È x ‰ ˆ 2 È x ‰ 2 Èx
œ lim
xÄ1
x1
œ lim
x2
x Ä 2 Èx# 12 4
œ lim
x Ä c2
(x 3) Š2 Èx# 5‹
9 x#
œ
lim
12 32
œ
3 8
" 6
œ lim x ˆ2 Èx‰ œ 4(2 2) œ 16 xÄ%
(x 1) ˆÈx 3 #‰ (x 3) 4
2 33
œ lim ŠÈx 3 #‹ xÄ1
œ "3
ax# 12b 16 x Ä 2 (x 2) ŠÈx# 12 4‹
œ lim 4 È16 4
œ
œ lim
" 2
ax 2b ŠÈx# 5 3‹ ax # 5 b 9
x Ä c2
œ
Š2 Èx# 5‹ Š2 Èx# 5‹
x Ä c3
x Ä c3 (x 3) Š2 Èx# 5‹
È x# 5 3 x2
œ
œ
4 3
ax # 8 b * x Ä c1 (x 1) ŠÈx# 8 $‹
ax 2b ŠÈx# 5 3‹
œ lim
444 (4)(8)
œ
œ
œ lim
œ
œ
(1 1)(1 1) 111
" È9 3
œ
x Ä * Èx 3
x Ä c2 ŠÈx# 5 3‹ ŠÈx# 5 3‹
(x 2)(x 2)
œ
v# 2v 4 (v 2) av# 4b vÄ#
(x 2) ŠÈx# 12 4‹
œ lim
œ #"
œ lim
ŠÈx# 12 4‹ ŠÈx# 12 4‹
xÄ2
œ 13
au# "b (u 1) u# u 1
x Ä c1 È x # ) $
ax 2b ŠÈx# 5 3‹
2 È x# 5 x3 x Ä c3
lim
uÄ1
œ lim
(x 2)(x 2)
lim
x Ä c2
8 16
(x 1) ŠÈx# 8 $‹
x Ä 2 (x 2) ŠÈx# 12 4‹
lim
œ
ŠÈx# 8 $‹ ŠÈx# 8 $‹
lim
(x 1)(x 1)
Èx# 12 4 x2 xÄ2
œ
5y 8
œ È4 2 œ 4
33. lim
34.
œ 21
x Ä 1 ˆÈ x 3 # ‰ ˆ È x 3 # ‰
x Ä c1 (x 1) ŠÈx# ) $‹
œ lim
2 4
(x 1) ˆÈx 3 2‰
œ lim
È x# 8 3 x1
œ lim
x(4 x)
x Ä % 2 Èx
x1
31. lim
Èx 3
x Ä * ˆÈ x 3 ‰ ˆ È x 3 ‰
x Ä % 2 Èx
œ
2
œ lim
(v 2) av# 2v 4b (v 2)(v 2) av# 4b vÄ#
3 #
1 2 1 2
# y Ä ! 3y 16
au# "b (u 1)(u 1) au# u 1b (u 1)
œ
œ
# x Ä c# x
# # y Ä ! y a3y 16b
Èx 3 x9
t2
t Ä c" t 2
œ lim
uÄ1
12 11
œ
œ lim
2(x 2)
œ lim
30. lim
t2
œ lim
v$ 8 % 16 v vÄ#
xÄ*
32.
t Ä c" (t 2)(t 1)
œ lim
28. lim
29. lim
xÄ#
t Ä " t1
(t 2)(t 1)
œ lim
œ lim (x 5) œ 2 5 œ 3
œ lim
# x Ä c# x (x 2)
% # y Ä 0 3y 16y
27. lim
(t 2)(t 1)
œ lim
24.
26. lim
xÄ#
È9 3 4
œ 23
œ lim
(3 x)(3 x)
4 ax # 5 b
x Ä c3 (x 3) Š2 Èx# 5‹
x Ä c3 (x 3) Š2 Èx# 5‹
œ lim
3x
x Ä c3 2 È x # 5
œ
6 2 È4
œ
3 2
75
76
Chapter 2 Limits and Continuity 4x x Ä 4 5 È x# 9
œ lim
a4 xb Š5 Èx# 9‹
œ lim
36. lim
x Ä 4 Š5 Èx# 9‹ Š5 Èx# 9‹
a4 xb Š5 Èx# 9‹
xÄ4
16 x#
œ lim
(4 x)(4 x)
xÄ4
25 ax# 9b
xÄ4
a4 xb Š5 Èx# 9‹
œ lim
a4 xb Š5 Èx# 9‹
5 È x# 9 4x
œ lim
xÄ4
œ
5 È25 8
œ
5 4
37. (a) quotient rule (b) difference and power rules (c) sum and constant multiple rules 38. (a) quotient rule (b) power and product rules (c) difference and constant multiple rules 39. (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
40. (a) (b) (c) (d) 41. (a) (b) (c) (d) 42. (a) (b) (c)
Äc
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
x
Ä%
x
Ä%
œ
3 01
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
lim 4g(x) œ ’ lim 4“ ’ lim g(x)“ œ (4)(3) œ 12
xÄb
xÄb
xÄb
lim f(x)/g(x) œ lim f(x)/ lim g(x) œ
xÄb
xÄb
xÄb
7 3
œ 73
lim [p(x) r(x) s(x)] œ lim p(x) lim r(x) lim s(x) œ 4 0 (3) œ 1
x Ä c#
x Ä c#
x Ä c#
x Ä c#
lim p(x) † r(x) † s(x) œ ’ lim p(x)“ ’ lim r(x)“ ’ lim s(x)“ œ (4)(0)(3) œ 0
x Ä c#
x Ä c#
(1 h)# 1# h hÄ!
œ lim
hÄ!
(2 h)# (2)# h
45. lim
[3(2 h) 4] [3(2) 4] h
hÄ!
x Ä c#
x Ä c#
44. lim
hÄ!
x Ä c#
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
x Ä c#
43. lim
" ‰ ˆ #" h ‰ ˆ # h hÄ!
46. lim
x
1 2h h# 1 h
œ lim
hÄ!
œ lim
hÄ!
44hh# 4 h
œ lim
hÄ!
œ lim
3h
hÄ! h
2 2 h "
2h
œ lim
x Ä c#
h(2 h) h
hÄ!
x Ä c#
œ lim (2 h) œ 2
h(h 4) h
hÄ!
œ lim (h 4) œ 4 hÄ!
œ3
œ lim
hÄ!
2 (2 h) 2h(# h)
œ lim
h
h Ä ! h(4 2h)
œ "4
"6 3
Section 2.2 Calculating Limits Using the Limit Laws È7 h È7 h hÄ!
47. lim
œ lim
ŠÈ7 h È7‹ ŠÈ7 h È7‹
œ lim
h ŠÈ7 h È7‹
hÄ!
h
h Ä ! h ŠÈ7hÈ7‹
h Ä ! È 7 h È 7
È3(0 h) 1 È3(0) 1 h hÄ!
48. lim
œ lim
"
œ lim
3h
h Ä ! h ŠÈ3h 1 "‹
œ lim
œ
œ lim
" #È 7
ŠÈ3h 1 "‹ ŠÈ3h 1 "‹ h ŠÈ3h 1 "‹
hÄ!
œ lim
œ
3
h Ä ! È3h 1 1
(7 h) 7
h Ä ! h ŠÈ7 h È7‹
(3h 1) "
œ lim
h Ä ! h ŠÈ3h 1 1 ‹
3 #
49. 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!
50. 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!
51. (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.
52. (a)
lim Š "#
xÄ!
lim
xÄ!
1cos x x#
x# 24 ‹
œ lim
1
xÄ! #
lim
x#
x Ä ! #4
œ
" #
x sin x
x Ä ! 22 cos x
xÄ!
0œ
" #
and lim
"
xÄ! #
œ1
œ "# ; by the sandwich theorem,
œ "# .
(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.
53. 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 Äc Äc Äc Ê c œ 0, 1, or 1. Moreover, lim f(x) œ lim x# œ 0 and lim f(x) œ lim f(x) œ 1. xÄ!
xÄ!
x Ä c1
xÄ1
54. 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#
55. 1 œ lim
xÄ%
f(x)5 x 2
lim f(x) lim 5 lim f(x) 5 xÄ% xÄ% œ xÄ% Ê lim f(x) 5 œ 2(1) Ê lim f(x) œ 2 5 œ 7. lim x lim 2 œ %# x
Ä%
x
Ä%
xÄ%
xÄ%
77
78
Chapter 2 Limits and Continuity
56. (a) 1 œ lim
f(x) x#
lim f(x) # œ xÄlim x# œ
(b) 1 œ lim
f(x) x#
œ ’ lim
x Ä c# x Ä c#
xÄ#
57. (a) 0 œ 3 † 0 œ ’ lim
xÄ#
x Ä c#
lim f(x)
xÄ#
%
Ê
lim f(x) œ 4.
x Ä c#
f(x) lim x" “ x “ ’x Ä c#
œ ’ lim
x Ä c#
f(x) ˆ " ‰ x “ #
Ê
lim
x Ä c#
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#
58. (a) 0 œ 1 † 0 œ ’ lim
f(x) # “ ’ lim xÄ! x xÄ!
(b) 0 œ 1 † 0 œ 59. (a)
lim x sin
xÄ!
(b) 1 Ÿ sin
60. (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!
(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 theorem since lim x# œ 0.
xÄ!
xÄ!
2.3 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.
xÄ!
Section 2.3 Precise Definition of a Limit 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# .
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
Step 2:
$
Step 1:
¸x "# ¸ $ Ê $ x
Step 2:
$
œ
3 #
$ Ê $ " #
3 #
x$
3 #
Ê $ œ #, or $ œ Ê $ œ 1. The value of $ which assures ¸x ˆ 3# ‰¸ $ Ê 7# x "# is the smaller value, $ œ ". 3 #
7 #
3 #
5. " #
$ Ê $
" #
x$
" 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:
9 16 ;
thus $ œ
7 16 .
79
80
Chapter 2 Limits and Continuity
12. Step 1: Step 2:
kx (1)k $ Ê $ x 1 $ Ê $ 1 x $ 1 From the graph, $ 1 œ thus $ œ
È 5 2 # .
È5 #
Ê $œ
È 5 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:
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 È 3 #
¸ 0.1340;
œ 0.36; thus $ œ
1 1.99
Ê $œ
1 1.99
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: 18. 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. ¸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" B $ 4" . Then, $
19. Step 1: Step 2:
20. Step 1: Step 2:
21. Step 1: Step 2:
22. Step 1: Step 2:
" 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.
Section 2.3 Precise Definition of a Limit 23. Step 1: Step 2:
81
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
11 1 0.1 Ê 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:
28. Step 1: Step 2:
29. Step 1: Step 2:
¸ 120 ¸ x 5 " Ê "
Step 2:
&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 $ # x $ #. 0.03 0.03 Then $ # œ # 0.03 m Ê $ œ m , or $ # œ # m Ê $ œ
0.03 m .
In either case, $ œ
kmx 3mk c Ê c mx 3m c Ê c 3m mx c 3m Ê 3 kx 3k $ Ê $ x 3 $ Ê $ $ B $ $. Then $ $ œ $ mc Ê $ œ mc , or $ $ œ $ mc Ê $ œ ¸(mx b) ˆ m# b‰¸ - Ê c mx m# c Ê c ¸x "# ¸ $ Ê $ x "# $ Ê $ "# x $ "# . Then $
30. Step 1:
120 x
" #
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.
In either case, $ œ
c m.
m #
Ê
c m
c m. " #
mx c
0.03 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 . kx 1k $ Ê $ x 1 $ Ê $ " x $ ". 0.05 0.05 Then $ " œ " 0.05 m Ê $ œ m , or $ " œ " m Ê $ œ
0.05 m .
In either case, $ œ
0.05 m .
31. lim (3 2x) œ 3 2(3) œ 3 xÄ3
Step 1: Step 2:
32.
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.
lim (3x #) œ (3)(1) 2 œ 1
x Ä c1
Step 1: Step 2:
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.
82
Chapter 2 Limits and Continuity Then $ " œ 1.01 Ê $ œ 0.01, or $ " œ 0.99 Ê $ œ 0.01; thus $ œ 0.01.
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 ‹ 4¹ 0.05 Ê 0.05
Step 2:
Ê 1.95 x 2.05, x Á 2. kx 2k $ Ê $ x 2 $ Ê $ # x $ 2. Then $ # œ 1.95 Ê $ œ 0.05, or $ # œ 2.05 Ê $ œ 0.05; thus $ œ 0.05.
lim
x Ä c&
x# 6x 5 x5
œ lim
x Ä c&
(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 c&
(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 c$
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
œ
4 #
œ2
Step 1:
¸ 4x
2¸ 0.4 Ê 0.4
Step 2:
kx 2k $ Ê $ x 2 $ Ê $ # x $ #. Then $ # œ 53 Ê $ œ "3 , or $ # œ 5# Ê $ œ "# ; thus $ œ 3" .
4 x
2 0.4 Ê 1.6
4 x
2.4 Ê
10 16
x 4
10 24
Ê
10 4
x
10 6
or
5 3
x 25 .
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:
% 3
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 # % Ê # % Èx 5 # % Ê (# %)# x 5 (# %)# Ê (# %)# & x (# %)# 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 # % Ê # % È4 x # % Ê (# %)# % x (# %)# Ê (# %)# x 4 (# %)# Ê (# %)# % x (# %)# %. kx 0k $ Ê $ x $ . Then $ œ (# %)# 4 œ %# %% Ê $ œ %% %# , or $ œ (# %)# 4 œ 4% %# . Thus choose the smaller distance, $ œ 4% %# .
Section 2.3 Precise Definition of a Limit 41. Step 1: Step 2:
83
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.
42. Step 1: Step 2:
43. Step 1: Step 2:
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 %
"% 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:
" x
" "%
"œ
% "%.
the smaller of the two distances.
¸ x"# "3 ¸ % Ê %
" x#
" 3
% Ê
" 3
%
" x#
" 3
% Ê
1 3% 3
" x#
1 $% 3
Ê
3 "
%$ x#
3 "
%$3 È $. Ê É 1 3 $% kxk É " 3 $% , or É " 3 $% x É "$ % for x near
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 xx11 ‹ 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 #%.
kx 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
" x
x for x 0. Since lim (x) œ lim x œ 0, xÄ!
xÄ!
84
Chapter 2 Limits and Continuity then by the sandwich theorem, in either case, lim x sin xÄ!
50. By the figure, x# Ÿ x# sin
" x
" x
œ 0.
Ÿ x# for all x except possibly at x œ 0. Since lim ax# 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 $ xÄc
Í lfah cb Ll % whenever ! lh !l $ Í limfah 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 xÄ!
function f(x) œ x# never gets arbitrarily close to 1 for x near 0.
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 xÄ!
x! . As another 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
" x
œ
" #
as you can see from the accompanying figure. However, lim sin xÄ!
" x
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 % 4" we cannot satisfy the inequality ¸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
1 x# 4
Ÿ 9.01 Ê
4 1
(8.99) Ÿ x# Ÿ
4 1
(9.01)
É 9.01 Ê 2É 8.99 1 ŸxŸ2 1 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
Section 2.3 Precise Definition of a Limit
85
(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 f(x) œ 2 Ê kf(x) 4k œ 2. Thus for % 2, kf(x) 4k % whenever 2 x 2 $ no matter how small we choose $ 0 Ê lim f(x) Á 4. xÄ#
(b) For 2 x 2 $ Ê f(x) œ 2 Ê kf(x) 3k œ 1. Thus for % 1, kf(x) 3k % whenever 2 x 2 $ no matter how small we choose $ 0 Ê lim f(x) Á 3. xÄ#
(c) For 2 $ x 2 Ê f(x) œ x# so kf(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, kf(x) 2k % whenever 2 $ x 2 no mater how small we choose $ 0 Ê lim f(x) Á 2. xÄ#
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 c1
(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 c1
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 ));
86
Chapter 2 Limits and Continuity 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}]
2.4 ONE-SIDED LIMITS AND LIMITS AT INFINITY 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 #
2 #
" œ #, lim f(x) œ $ # œ " xÄ#
(b) No, lim f(x) does not exist because lim f(x) Á lim f(x) xÄ# xÄ# xÄ# (c) lim f(x) œ 4# 1 œ 3, lim f(x) œ 4# " œ $ xÄ%
xÄ%
(d) Yes, lim f(x) œ 3 because 3 œ lim f(x) œ lim f(x) xÄ% xÄ% xÄ% 4. (a)
lim f(x) œ
x Ä #
2 #
œ 1, lim f(x) œ $ # œ ", f(2) œ 2 xÄ#
(b) Yes, lim f(x) œ 1 because " œ lim f(x) œ lim f(x) xÄ# xÄ# xÄ# (c) lim f(x) œ 3 (1) œ 4, lim f(x) œ 3 (1) œ 4 x Ä c"
x Ä c"
(d) Yes, lim f(x) œ 4 because 4 œ x Ä c"
lim
x Ä c"
f(x) œ
lim
x Ä c"
f(x)
5. (a) No, lim f(x) does not exist since sin ˆ "x ‰ does not approach any single value as x approaches 0 xÄ! (b) lim f(x) œ lim 0 œ 0 xÄ!
(c)
xÄ!
lim f(x) does not exist because lim f(x) does not exist xÄ! xÄ!
6. (a) Yes, lim g(x) œ 0 by the sandwich theorem since Èx Ÿ g(x) Ÿ Èx when x 0 xÄ! (b) No, lim g(x) does not exist since Èx is not defined for x 0 xÄ!
(c) No, lim g(x) does not exist since lim g(x) does not exist xÄ! xÄ!
Section 2.4 One-Sided Limits and Limits at Infinity 7. (aÑ
lim f(x) œ " œ lim f(x) xÄ1 (c) Yes, lim f(x) œ 1 since the right-hand and left-hand (b)
x Ä 1
xÄ1
limits exist and equal 1
8. (a)
(b)
lim f(x) œ 0 œ lim f(x) xÄ1
x Ä 1
(c) Yes, lim f(x) œ 0 since the right-hand and left-hand xÄ1
limits exist and equal 0
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 c!Þ&
lim
13.
x Ä c#
14.
x Ä 1
15.
lim
lim
lim
hÄ!
2 0.5 2 È3 É 3/2 É xx É 1 œ 0.5 1 œ 1/2 œ
12.
lim
x Ä 1
" 1 È0 œ ! É "1 É xx # œ # œ
5‰ ˆ x x 1 ‰ ˆ 2x ˆ 2 ‰ 2(2) 5 ˆ"‰ x# x œ # " Š (#)# (2) ‹ œ (2) # œ 1
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 Èh# 4h 5 È5 h
œ lim hÄ!
œ lim Š hÄ!
ah# 4h 5b 5 h ŠÈh# 4h 5 È5‹
Èh# 4h 5 È5 È # 4h 5 È5 ‹ Š Èhh# ‹ h 4h 5 È5
œ lim hÄ!
h(h 4) h ŠÈh# 4h 5 È5‹
œ
04 È5 È5
œ
2 È5
87
88 16.
Chapter 2 Limits and Continuity lim
h Ä !
È6 È5h# 11h 6 h
6 a5h# 11h 6b
œ lim hÄ! 17. (a)
19. (a)
) Ä $
20. (a)
t Ä %
(x2) (x#)
Ú) Û )
lim
akx 2k œ x 2 for x 2b
(x 3) ’ (x(x#2) ) “
lim
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 È2x œ È2 xÄ1
œ lim xÄ1
È2x (x 1) (x 1)
akx 1k œ (x 1) for x 1b
œ
œ1
3 3
lim at ÚtÛb œ 4 4 œ 0 sin È2) È 2)
22. lim
sin kt t
23. lim
sin 3y 4y
)Ä!
tÄ!
yÄ!
œ
26. lim
2t
t Ä ! tan t
27. lim
xÄ!
)Ä!
3 sin 3y " 4 ylim 3y Ä!
sin 2x ‰ ˆ cos 2x x
œ lim
xÄ!
œ 2 lim
t
sin t t Ä ! ˆ cos t ‰
x csc 2x cos 5x
œ
œ
œ lim
tÄ!
" ‰ cos 5x
29. lim
x x cos x
) Ä $
(b)
t Ä %
œ lim ˆ sin xxcos x
30. lim
xÄ!
x
x# x sin x #x
xÄ!
xÄ!
œ lim ˆ #x xÄ!
œ
2 3
lim at ÚtÛb œ 4 3 œ 1
t cos t sin t
œk†1œk
œ
" 3
Œ
)
œ
" limc
Ä!
(where ) œ 3y)
œ
sin ) )
"
‹ Š lim x Ä ! cos 2x xÄ!
" 3
†1œ
2 sin 2x #x ‹
" 3
(where ) œ 3h)
œ1†2œ2
œ 2 Š lim cos t‹ Œ lim" sin t œ 2 † " † " œ 2 tÄ!
œ Š #" lim
t
Ä!
t
"
‹ Š lim cos 5x ‹ x Ä ! sin 2x xÄ! 2x
œ lim ˆ3 cos x † xÄ!
x cos x ‰ sin x cos x
xÄ!
" #
(where ) œ kt)
3 4
œ lim ˆ sinx x † xÄ!
x sin x
†
" ‰ cos x
œ lim Š sin" x ‹ † lim ˆ cos" x ‰ lim Š sin" x ‹ œ (1)(1) 1 œ 2 xÄ!
Ú) Û )
lim
3 sin ) 4 )lim Ä! )
œ Š lim
sin 2x
6x# cos x sin x sin 2x xÄ!
x Ä ! sin x cos x
œ
" " sin 3h 3 h lim Ä ! ˆ 3h ‰
28. lim 6x# (cot x)(csc 2x) œ lim xÄ!
)Ä!
x Ä ! x cos 2x
œ 2 lim
sin ) )
œ k lim
sin 3y 3 4 ylim Ä ! 3y
3h ‰ sin 3h
œ lim ˆ sinx2x † xÄ!
k sin ) )
œ lim
œ lim ˆ "3 † hÄ!
h
tan 2x x
k sin kt kt
(b)
(where x œ È2))
œ1
sin x x
xÄ!
tÄ!
lim h Ä ! sin 3h
xÄ!
œ lim
œ lim
25. lim
œ 211 È6
œ lim È2x œ È2 xÄ1
21. lim
24.
lim
(0 11) È6 È6
œ
(x 3) œ (2) 3 œ 1
lim
x Ä c#
x Ä c#
œ lim xÄ1
È2x (x 1) kx 1 k
lim
x Ä 1
œ
h(5h 11)
h ŠÈ6 È5h# 11h 6‹
(x 3)
lim
x Ä c#
œ
È2x (x 1) kx 1 k
lim
(b)
kx 2 k x2
(x 3)
lim
x Ä 1
œ
œ
x Ä c#
18. (a)
kx 2 k x 2
(x 3)
lim
È5h# 11h 6 È6 È5h# 11h 6 È ‹ Š È66 ‹ h È5h# 11h 6
œ lim hÄ!
h ŠÈ6 È5h# 11h 6‹
x Ä c#
(b)
œ lim Š hÄ!
x
"# ˆ sinx x ‰‰ œ 0
" #
"# (1) œ 0
œ ˆ #" † 1‰ (1) œ
2x ‰ sin 2x
" #
œ3†"†1œ3
lim
x
x Ä ! sin x
Section 2.4 One-Sided Limits and Limits at Infinity 31. lim
sin(1 cos t) 1cos t
32. lim
sin (sin h) sin h
tÄ!
hÄ!
sin )
33. lim
) Ä ! sin 2)
34. lim
sin 5x
35. lim
tan 3x
œ
3 8 xlim Ä!
36. lim
yÄ!
œ
)Ä!
sin ) )
sin ) )
œ lim
)Ä!
œ 1 since ) œ 1 cos t Ä 0 as t Ä 0
œ 1 since ) œ sin h Ä 0 as h Ä 0
sin ) œ lim ˆ sin 2) †
2) ‰ #)
5x œ lim ˆ sin sin 4x †
4x 5x
sin 3x œ lim ˆ cos 3x †
" ‰ sin 8x
)Ä!
x Ä ! sin 4x
x Ä ! sin 8x
œ lim
xÄ!
xÄ!
" # )lim Ä!
œ
† 54 ‰ œ
œ lim
yÄ!
2) ‰ sin 2)
ˆ sin5x5x †
5 4 xlim Ä!
sin 3x œ lim ˆ cos 3x † 3 8
†1†1†1œ
4x ‰ sin 4x
†
8x 3x
†1†1œ œ
5 4
3 8
"
lim
xÄ „_
12 5
œ
12 5
œ 0 whenever
m n
0. This result follows immediately from
ˆ xm"În ‰ œ
lim
xÄ „_
37. (a) 3
(b) 3
38. (a) 1
(b) 1
39. (a)
" #
(b)
" #
40. (a)
" 8
(b)
" 8
41. (a) 53
45.
lim
tÄ_
46. r Ä lim_
ˆ x" ‰mÎn œ Š
(b) sin 2x x
Ÿ
" x
cos ) 3)
Ÿ
" 3)
2 t sin t t cos t
Ê x lim Ä_ Ê
47. (a) x lim Ä_
lim
) Ä c_
œ lim
2 t
tÄ_
r sin r 2r 7 5 sin r
2x 3 5x 7
$
sin 2x x
œrÄ lim_
œ x lim Ä_
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
2x 7 48. (a) x lim œ x lim Ä _ x$ x# x 7 Ä_ (b) 2 (same process as part (a))
3 4
œ 0 by the Sandwich Theorem
cos ) 3)
œ
010 10
œ 1
œrÄ lim_
2 5
2 Š x7$ ‹
1 "x x"# x7$
10 200
œ
(b)
œ2
" #
2 5
"
lim ‹ xÄ „_ x
(b) 53
3 4
44. 3") Ÿ
5 4
† 83 ‰
œ1†1†1†1†
lim mÎn xÄ „_ x
Example 6 and the power rule in Theorem 8:
43. "x Ÿ
†1†1œ
yÄ!
cos 5y ˆ 3†4 ‰ lim Š sin3y3y ‹ Š sin4y4y ‹ Š sin5y5y ‹ Š cos 4y ‹ 5 yÄ!
42. (a)
" #
sin 4y cos 5y 3†4†5y œ lim Š siny3y ‹ Š cos 4y ‹ Š sin 5y ‹ Š 3†4†5y ‹
sin 3y sin 4y cos 5y y cos 4y sin 5y
Note: In these exercises we use the result
" #
œ
" sin 8x
xÄ!
ˆ cos"3x ‰ ˆ sin3x3x ‰ ˆ sin8x8x ‰ œ
sin 3y cot 5y y cot 4y
ˆ sin) ) †
(same process as part (a))
mÎn
œ 0mÎn œ 0.
89
90
Chapter 2 Limits and Continuity " x
x"#
49. (a) x lim Ä_
x1 x# 3
œ x lim Ä_
1 x3#
50. (a) x lim Ä_
3x 7 x# 2
œ x lim Ä_
1 x2#
51. (a) x lim Ä_
7x$ x$ 3x# 6x
52. (a) x lim Ä_
" x$ 4x 1
&
3 x
x7#
œ x lim Ä_
(b)
9 #
2x%
œ0
(b) 0 (same process as part (a))
" x$
œ x lim Ä_
10 x
œ x lim Ä_
œ(
x"# x31' 1
2
(b) 7 (same process as part (a))
œ!
1 x4# x"$
%
9x% x 5x# x 6
(b) 0 (same process as part (a))
7 1 3x x6#
10x x 31 53. (a) x lim œ x lim x' Ä_ Ä_ (b) 0 (same process as part (a))
54. (a) x lim Ä_
œ0
(b) 0 (same process as part (a))
œ0
9 x"$
5 x#
x"$ x6%
œ
9 #
(same process as part (a))
55. (a) x lim Ä_
2x$ 2x 3 3x$ 3x# 5x
œ x lim Ä_
2 x2# x3$ 3 3x x5#
œ 23
(b) 23 (same process as part (a)) %
x 56. (a) x lim œ x lim Ä _ x% 7x$ 7x# 9 Ä_ (b) 1 (same process as part (a))
57. x lim Ä_
2Èx x" 3x 7
59. x Ä lim c_
œ x lim Ä_
$ & È xÈ x $ & È È x x
2 ‹ Š x"# ‹ x"Î# 3 7x
œxÄ lim c_
60. x lim Ä_
x" x% x# x$
61. x lim Ä_
2x&Î$ x"Î$ 7 x)Î& 3x Èx
62. x Ä lim c_
Š
œ x lim Ä_
$ È x 5x 3 2x x#Î$ 4
" 1 7x x7# x9%
œ0
1 xÐ"Î&Ñ Ð"Î$Ñ 1 xÐ"Î&Ñ Ð"Î$Ñ
x x"# 1 x"
œ x lim Ä_ œxÄ lim c_
œ 1
58. x lim Ä_
œxÄ lim c_
" ‹ x#Î"& " 1 Š #Î"& ‹ x
1Š
2 Èx 2 Èx
œ x lim Ä_
2 ‹" x"Î# 2 Š "Î# ‹ 1 x
Š
œ 1
œ1
œ_ " 7 )Î& x"*Î"& x " 3 ""Î"! x$Î& x
2x"Î"& 1
" x#Î$
2
5 3x " x"Î$
4x
œ_
œ 5#
63. Yes. If lim f(x) œ L œ lim f(x), then xlim f(x) œ L. If lim f(x) Á lim f(x), then xlim f(x) does not exist. Äa Äa xÄa xÄa xÄa xÄa 64. Since xlim f(x) œ L if and only if lim f(x) œ L and lim f(x) œ L, then xlim f(x) can be found by calculating Äc Äc xÄc xÄc lim f(x). xÄc
65. If f is an odd function of x, then f(x) œ f(x). Given lim f(x) œ 3, then lim f(x) œ $. xÄ! xÄ! 66. If f is an even function of x, then f(x) œ f(x). Given lim f(x) œ 7 then lim f(x) œ 7. However, nothing xÄ# x Ä c# can be said about
lim
x Ä c#
f(x) because we don't know lim f(x). xÄ#
Section 2.4 One-Sided Limits and Limits at Infinity 67. 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 c_
f(x) g(x)
91
œ 2 as well.
68. Yes, it can have a horizontal or oblique asymptote. 69. At most 1 horizontal asymptote: If x lim Ä_ f(x) lim x Ä c_ g(x)
f(x) g(x)
œ L, then the ratio of the polynomials' leading coefficients is L, so
œ L as well.
Èx# x Èx# x œ lim ’Èx# x Èx# x“ † ’ Èx# x Èx# x “ œ lim 70. x lim È x# x È x# x Ä_ xÄ_ xÄ_ 2x 2 2 œ x lim œ lim œ œ 1 È # 1 1 " " Ä_ È # xÄ_ x x
x x
ax # x b a x # x b È x# x È x# x
É1 x É1 x
71. For any % 0, take N œ 1. Then for all x N we have that kf(x) kk œ kk kk œ 0 %. 72. For any % 0, take N œ 1. Then for all y N we have that kf(x) kk œ kk kk œ 0 %. 73. 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 &
74. I œ (% $ ß %) Ê % $ x 4. Also, È% x % Ê % x %# Ê x % %# . Choose $ œ %# Ê lim È% x œ 0. x Ä %
75. 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
x
lim x Ä ! kx k
œ 1.
2 ¸ x 2 ¸ 76. 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,
77. (a) (b)
lim
x Ä %!!
x 2
lim x Ä c# kx2k
œ 1.
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 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 ÚxÛ Á lim ÚxÛ we conclude that lim ÚxÛ does not exist. x Ä %!!
x Ä %!!
78. (a)
x Ä %!!
lim f(x) œ lim Èx œ È0 œ 0; ¸Èx 0¸ % Ê % Èx % Ê ! x %# for x positive. Choose $ œ %# xÄ! Ê lim f(x) œ 0.
x Ä !
xÄ!
(b)
lim f(x) œ lim 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 ! xÄ! Since kx# 0k œ kx# 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. 79.
lim
xÄ „_
x sin
" x
" )Ä0 )
œ lim
sin ) œ 1, ˆ) œ x" ‰
80.
lim
cos
" x
x Ä c_ 1 x"
œ lim )Ä!
cos ) 1)
œ
" 1
œ 1, ˆ) œ x" ‰
92
Chapter 2 Limits and Continuity 3x 4
81.
lim x Ä „ _ 2x 5
82.
xÄ_
83. 84.
œ
lim
3 4x
5 x Ä „ _ 2 x
œ lim
3 4t
t Ä 0 2 5t
œ
3 #
, ˆt œ "x ‰
"Îx lim ˆ "x ‰ œ lim zz œ 1, ˆz œ x" ‰ zÄ!
ˆ3 2x ‰ ˆcos "x ‰ œ lim (3 2))(cos )) œ (3)(1) œ 3, ˆ) œ x" ‰
lim
xÄ „_
)Ä0
lim ˆ x3# cos x" ‰ ˆ1 sin x" ‰ œ lim a3)# cos )b (1 sin )) œ (0 1)(1 0) œ 1, ˆ) œ x" ‰ )Ä!
xÄ_
2.5 INFINITE LIMITS AND VERTICAL ASYMPTOTES "
œ_
1.
lim x Ä ! 3x
3.
lim x Ä # x 2
5.
lim x Ä c) x8
7.
lim # x Ä ( (x7)
3
2x
4
œ _ œ _ œ_
lim "Î$ x Ä ! 3x
10. (a)
lim "Î& x Ä ! x 4
11. lim
#Î& xÄ! x
13.
15. 16.
lim
x Ä ˆ 1# ‰
œ lim
4
# x Ä ! ax"Î& b
œ_
Š positive positive ‹
lim x Ä ! 2x
positive Š negative ‹
4.
lim x Ä $ x 3
Š negative positive ‹
6.
lim x Ä c& 2x10
3x
œ_
positive Š positive ‹
8.
lim # x Ä ! x (x1)
œ _
(b)
lim "Î$ x Ä ! 3x
(b)
lim "Î& x Ä ! x
œ_
2
positive Š negative ‹
2.
œ_
2
9. (a)
œ _
Š positive positive ‹
œ_
5
"
"
2
2
12. lim
"
#Î$ xÄ! x
tan x œ _
14.
lim
x Ä ˆ #1 ‰
Š negative negative ‹ negative Š positive †positive ‹
œ _ œ _
œ lim
"
# x Ä ! ax"Î$ b
sec x œ _
lim (1 csc )) œ _
) Ä !
lim (2 cot )) œ _ and lim (2 cot )) œ _, so the limit does not exist )Ä!
) Ä !
"
œ lim xÄ#
" (x2)(x2)
œ_
Š positive"†positive ‹
"
œ lim xÄ#
" (x2)(x2)
œ _
Š positive†"negative ‹
17. (a)
lim # x Ä # x 4
(b)
lim # x Ä # x 4
(c)
lim # x Ä c# x 4
(d)
lim # x Ä c# x 4
"
œ
lim x Ä c# (x2)(x2)
"
œ _
Š positive†"negative ‹
"
œ
lim x Ä c# (x2)(x2)
"
œ_
Š negative"†negative ‹
x
œ lim xÄ"
x (x1)(x1)
œ_
positive Š positive †positive ‹
x
œ lim xÄ"
x (x1)(x1)
œ _
positive Š positive †negative ‹
18. (a)
lim # x Ä " x 1
(b)
lim # x Ä " x 1
(c)
lim # x Ä c" x 1
x
œ
x
lim x Ä c" (x1)(x1)
œ_
negative Š positive †negative ‹
œ_
Section 2.5 Infinite Limits and Vertical Asymptotes (d)
œ
x
lim # x Ä "c x 1
x
lim x Ä "c (x1)(x1)
œ _
negative Š negative †negative ‹
19. (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) (d)
20. (a) (c) (d)
xÄ
$ È
x# x Ä 1 #
lim x Ä #b lim b
xÄ"
" x
x# 1 2x 4
x# 3x 2 x$ 2x#
lim
#
x 3x 2 x$ 2x# #
x 3x 2 x$ 2x# #
x 3x 2 x$ 2x#
lim
xÄ!
(c)
lim c
x# 3x 2 x$ 4x
lim b
x# 3x 2 x$ 4x
(e)
xÄ"
x"
lim x Ä !b x(x #) and
œ lim c xÄ#
œ
2†0 #4
(b)
x# 1
lim x Ä #c 2x 4
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)
2)(x 1) lim (x x# (x 2) xÄ# 2)(x 1) lim (x x# (x 2) xÄ!
œ lim b xÄ#
x# 3x 2 x$ 4x
x Ä #b xÄ0
(x 1)(x 1) 2x 4
œ lim b xÄ#
œ
(b)
(d)
lim
3 #
Š positive positive ‹
œ lim b xÄ!
œ
x# 3x 2 x$ 4x
lim
x Ä #b
œ 2"Î$ 2"Î$ œ 0
" 4
œ
lim
xÄ#
" #"Î$
ˆ "1 ‰ œ
œ lim b xÄ"
x# 3x 2 x$ 2x#
lim
" #
œ_
lim
x Ä #c
2#Î$ #
œ œ
x# 1 2x 4
x 1
x Ä #b
22. (a)
lim x Ä !c 2x 4
(b)
(e)
" x
#
x Ä !b
(d)
2
lim
21. (a)
(c)
x# #
lim
œ
œ lim c xÄ#
œ lim
xÄ#
(x 2)(x ") x(x #)(x 2)
œ
" 4
,xÁ2
x1 x#
œ
" 4
,xÁ2
x1 x#
œ
" 4
,xÁ2 †negative Š negative positive†negative ‹
œ lim b xÄ#
(x 2)(x ")
œ lim b xÄ"
x1 x#
œ _
lim x Ä #b x(x #)(x 2)
œ lim c xÄ!
†negative Š negative positive†negative ‹
(x 2)(x ") x(x #)(x 2) (x 2)(x ") x(x #)(x 2)
œ
(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
(c) Using a symmetric difference quotient, the horse's speed is approximately
?F ?t
œ
%# &* $$
œ
# #'
¸ !Þ!(( furlongs/sec.
(d) The horse is running the tastest during the last furlong (between the 9th and 10th furlong markers). This furlong takes only 11 seconds to run, which is the least amount of time for a furlong. (e) The horse accelerates the fastest during the first furlong (between markers 0 and 1). 3.4 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS 1. y œ 10x 3 cos x Ê
dy dx
œ 10 3
œ
3 x#
3. y œ csc x 4Èx 7 Ê
dy dx
2. y œ
3 x
5 sin x Ê
4. y œ x# cot x
" x#
dy dx
Ê
dy dx
5
(cos x) œ 10 3 sin x
(sin x) œ
3 x#
œ csc x cot x
œ x#
œ x# csc# x 2x cot x
d dx
d dx
d dx
5 cos x
4 #È x
0 œ csc x cot x ax# b
2 x$
œ (sec x tan x)
d dx
(cot x) cot x †
d dx
2 Èx
œ x# csc# x (cot x)(2x)
2 x$
2 x$
5. y œ (sec x tan x)(sec x tan x) Ê #
dy dx
(sec x tan x) (sec x tan x) #
d dx
(sec x tan x)
œ (sec x tan x) asec x tan x sec xb (sec x tan x) asec x tan x sec xb œ asec# x tan x sec x tan# x sec$ x sec# x tan xb asec# x tan x sec x tan# x sec$ x tan x sec# xb œ 0. ŠNote also that y œ sec# x tan# x œ atan# x 1b tan# x œ 1 Ê
dy dx
œ 0.‹
142
Chapter 3 Differentiation
6. y œ (sin x cos x) sec x Ê
œ (sin x cos x)
dy dx
d dx
(sec x) sec x
œ (sin x cos x)(sec x tan x) (sec x)(cos x sin x) œ œ
sin# x cos x sin x cos# x cos x sin x cos# x
œ
" cos# x
d dx (sin x cos x) (sin x cos x) sin x x sin x cos cos cos# x x
œ sec# x
ŠNote also that y œ sin x sec x cos x sec x œ tan x 1 Ê 7. y œ œ
(1 cot x)
œ
dy dx
csc# x csc# x cot x csc# x cot x (1 cot x)#
8. y œ œ
Ê
cot x 1 cot x
Ê
cos x 1 sin x
sin x sin# x cos# x (1 sin x)#
9. y œ
4 cos x
" tan x
10. y œ
cos x x
x cos x
œ
(cot x) (cot x) (1 cot x)#
œ
(1 sin x)
œ
dy dx
d dx
œ
(1 cot x) acsc# xb (cot x) acsc# xb (1 cot x)#
d (cos x) (cos x) dx (1 sin x) b (cos x) acos xb œ (1 sin x) a(1sin xsin (1 sin x)# x)# (1 sin x) sin x 1 " (1 sin x)# œ (1 sin x)# œ 1 sin x
œ
dy dx
(1 cot x)
œ sec# x.‹
csc# x (1 cot x)#
d dx
œ 4 sec x cot x Ê Ê
d dx
dy dx
œ 4 sec x tan x csc# x
dy dx
x(sin x) (cos x)(1) x#
11. y œ x# sin x 2x cos x 2 sin x Ê #
(cos x)(1) x(sin x) cos# x
œ
x sin x cos x x#
cos x x sin x cos# x
dy dx
œ ax# cos x (sin x)(2x)b a(2x)(sin x) (cos x)(2)b 2 cos x
dy dx
œ ax# (sin x) (cos x)(2x)b a2x cos x (sin x)(2)b 2(sin x)
œ x cos x 2x sin x 2x sin x 2 cos x 2 cos x œ x# cos x 12. y œ x# cos x 2x sin x 2 cos x Ê #
œ x sin x 2x cos x 2x cos x 2 sin x 2 sin x œ x# sin x 13. s œ tan t t Ê
ds dt
œ
14. s œ t# sec t 1 Ê 15. s œ œ 16. s œ œ
1 csc t 1 csc t
Ê
ds dt
œ
d dt
(tan t) 1 œ sec# t 1 œ tan# t
ds dt
œ 2t
d dt
(sec t) œ 2t sec t tan t
(1 csc t)(csc t cot t) (" csc t)(csc t cot t) (1 csc t)#
csc t cot t csc# t cot t csc t cot t csc# t cot t (1 csc t)# sin t 1 cos t " cos t 1
Ê
ds dt
œ
17. r œ 4 )# sin ) Ê
œ
2 csc t cot t (1 csc t)#
(1 cos t)(cos t) (sin t)(sin t) (1 cos t)#
dr d)
18. r œ ) sin ) cos ) Ê
œ ˆ) # dr d)
d d)
œ
cos t cos# t sin# t (1 cos t)#
œ
cos t " (1 cos t)#
œ 1 "cos t
(sin )) (sin ))(2))‰ œ a)# cos ) 2) sin )b œ )() cos ) # sin ))
œ () cos ) (sin ))(1)) sin ) œ ) cos )
dr 19. r œ sec ) csc ) Ê d) œ (sec ))(csc ) cot )) (csc ))(sec ) tan )) " " " " cos ) sin ) ‰ # # œ ˆ cos ) ‰ ˆ sin ) ‰ ˆ sin ) ‰ ˆ sin" ) ‰ ˆ cos" ) ‰ ˆ cos ) œ sin# ) cos# ) œ sec ) csc )
20. r œ (1 sec )) sin ) Ê 21. p œ &
" cot q
dr d)
œ (" sec )) cos ) (sin ))(sec ) tan )) œ (cos ) ") tan# ) œ cos ) sec# )
œ 5 tan q Ê
22. p œ (1 csc q) cos q Ê
dp dq
dp dq
œ sec# q
œ (1 csc q)(sin q) (cos q)(csc q cot q) œ (sin q 1) cot# q œ sin q csc# q
Section 3.4 Derivatives of Trigonometric Functions 23. p œ œ
sin q cos q cos q
Ê
dp dq
œ
(cos q)(cos q sin q) (sin q cos q)(sin q) cos# q
cos# q cos q sin q sin# q cos q sin q cos# q
24. p œ
tan q 1 tan q
Ê
dp dq
œ
œ
" cos# q
œ sec# q
(1 tan q) asec# qb (tan q) asec# qb (1 tan q)#
œ
sec# q tan q sec# q tan q sec# q (1 tan q)#
œ
sec# q (1 tan q)#
25. (a) y œ csc x Ê yw œ csc x cot x Ê yww œ a(csc x) acsc# xb (cot x)(csc x cot x)b œ csc$ x csc x cot# x œ (csc x) acsc# x cot# xb œ (csc x) acsc# x csc# x 1b œ 2 csc$ x csc x (b) y œ sec x Ê yw œ sec x tan x Ê yww œ (sec x) asec# xb (tan x)(sec x tan x) œ sec$ x sec x tan# x œ (sec x) asec# x tan# xb œ (sec x) asec# x sec# x 1b œ 2 sec$ x sec x 26. (a) y œ 2 sin x Ê yw œ 2 cos x Ê yww œ 2(sin x) œ 2 sin x Ê ywww œ 2 cos x Ê yÐ%Ñ œ 2 sin x (b) y œ 9 cos x Ê yw œ 9 sin x Ê yww œ 9 cos x Ê ywww œ 9(sin x) œ 9 sin x Ê yÐ%Ñ œ 9 cos x 27. y œ sin x Ê yw œ cos x Ê slope of tangent at x œ 1 is yw (1) œ cos (1) œ "; slope of tangent at x œ 0 is yw (0) œ cos (0) œ 1; and slope of tangent at x œ 3#1 is yw ˆ 3#1 ‰ œ cos 3#1
œ 0. The tangent at (1ß !) is y 0 œ 1(x 1), or y œ x 1; the tangent at (0ß 0) is y 0 œ 1(x 0), or y œ x; and the tangent at ˆ 31 ‰ # ß 1 is y œ 1.
28. y œ tan x Ê yw œ sec# x Ê slope of tangent at x œ 13 is sec# ˆ 13 ‰ œ 4; slope of tangent at x œ 0 is sec# (0) œ 1; and slope of tangent at x œ
1 3
is sec# ˆ 13 ‰ œ 4. The tangent
at ˆ 13 ß tanˆ 13 ‰‰ œ Š 13 ß È3‹ is y È3 œ 4ˆx 13 ‰ ; the tangent at (0ß 0) is y œ x; and the tangent at ˆ 13 ß tan ˆ 13 ‰‰ œ Š 13 ß È3‹ is y È3 œ 4 ˆx 13 ‰ . 29. y œ sec x Ê yw œ sec x tan x Ê slope of tangent at x œ 13 is sec ˆ 13 ‰ tan ˆ 13 ‰ œ 2È3 ; slope of tangent is sec ˆ 14 ‰ tan ˆ 14 ‰ œ È2 . The tangent at the point ˆ 1 ß sec ˆ 1 ‰‰ œ ˆ 1 ß #‰ is y 2 œ #È3 ˆx 1 ‰ ; at x œ
1 4
3
3
3
3
the tangent at the point ˆ 14 ß sec ˆ 14 ‰‰ œ Š 14 ß È2‹ is y È2 œ È2 ˆx 14 ‰ .
30. y œ 1 cos x Ê yw œ sin x Ê slope of tangent at È
x œ 13 is sin ˆ 13 ‰ œ #3 ; slope of tangent at x œ ‰ œ 1. The tangent at the point is sin ˆ 31 # ˆ 13 ß " cos ˆ 13 ‰‰ œ ˆ 13 ß 3# ‰ È
is y 3# œ #3 ˆx 13 ‰ ; the tangent at the point ˆ 3#1 ß " cos ˆ 3#1 ‰‰ œ ˆ 3#1 ß 1‰ is y 1 œ x 3#1
31 #
143
144
Chapter 3 Differentiation
31. Yes, y œ x sin x Ê yw œ " cos x; horizontal tangent occurs where 1 cos x œ 0 Ê cos x œ 1 Ê xœ1 32. No, y œ 2x sin x Ê yw œ 2 cos x; horizontal tangent occurs where 2 cos x œ 0 Ê cos x œ #. But there are no x-values for which cos x œ #. 33. No, y œ x cot x Ê yw œ 1 csc# x; horizontal tangent occurs where 1 csc# x œ 0 Ê csc# x œ 1. But there are no x-values for which csc# x œ 1. 34. Yes, y œ x 2 cos x Ê yw œ 1 2 sin x; horizontal tangent occurs where 1 2 sin x œ 0 Ê 1 œ 2 sin x Ê "# œ sin x Ê x œ 16 or x œ 561 35. We want all points on the curve where the tangent line has slope 2. Thus, y œ tan x Ê yw œ sec# x so that yw œ 2 Ê sec# x œ 2 Ê sec x œ „ È2 Ê x œ „ 14 . Then the tangent line at ˆ 14 ß "‰ has equation y 1 œ 2 ˆx 14 ‰ ; the tangent line at ˆ 14 ß "‰ has equation y 1 œ 2 ˆx 14 ‰ .
36. We want all points on the curve y œ cot x where the tangent line has slope 1. Thus y œ cot x Ê yw œ csc# x so that yw œ 1 Ê csc# x œ 1 Ê csc# x œ 1 Ê csc x œ „ 1 Ê x œ 1# . The tangent line at ˆ 1# ß !‰ is y œ x 12 .
2 cos x ‰ 37. y œ 4 cot x 2 csc x Ê yw œ csc# x 2 csc x cot x œ ˆ sin" x ‰ ˆ 1 sin x
(a) When x œ 1# , then yw œ 1; the tangent line is y œ x w
1 #
2.
(b) To find the location of the horizontal tangent set y œ 0 Ê 1 2 cos x œ 0 Ê x œ then y œ % È3 is the horizontal tangent. 38. y œ 1 È2 csc x cot x Ê yw œ È2 csc x cot x csc# x œ ˆ sin" x ‰ Š
1 3
È2 cos x 1 ‹ sin x
(a) If x œ 14 , then yw œ 4; the tangent line is y œ 4x 1 4. (b) To find the location of the horizontal tangent set yw œ 0 Ê È2 cos x 1 œ 0 Ê x œ xœ
31 4 ,
radians. When x œ 13 ,
31 4
radians. When
then y œ 2 is the horizontal tangent.
39. lim sin ˆ "x #" ‰ œ sin ˆ #" #" ‰ œ sin 0 œ 0 xÄ2
40. lim1 È1 cos (1 csc x) œ É1 cos ˆ1 csc 16 ‰ œ È1 cos (1 † 2) œ È2 xÄ
6
1 ‰ ‘ < ˆ 1 ‰ ‘ < ˆ1‰ ‘ 41. lim sec x • 3 x • 2 2*x; plot(f(x), x=1..2); diff(f(x),x); fp := unapply (ww ,x); L:=x -> f(a) fp(a)*(x a); plot({f(x), L(x)}, x=1..2); err:=x -> abs(f(x) L(x)); plot(err(x), x=1..2, title = #absolute error function#); err(1); Mathematica: (function, x1, x2, and a may vary): Clear[f, x] {x1, x2} = {1, 2}; a = 1; f[x_]:=x3 x2 2x Plot[f[x], {x, x1, x2}] lin[x_]=f[a] f'[a](x a) Plot[{f[x], lin[x]}, {x, x1, x2}] err[x_]=Abs[f[x] lin[x]]
180
Chapter 3 Differentiation
Plot[err[x], {x, x1,x 2}] err//N After reviewing the error function, plot the error function and epsilon for differing values of epsilon (eps) and delta (del) eps = 0.5; del = 0.4 Plot[{err[x], eps},{x, a del, a del}] CHAPTER 3 PRACTICE EXERCISES 1. y œ x& 0.125x# 0.25x Ê 2. y œ 3 0.7x$ 0.3x( Ê
4. y œ x( È7x
" 1 1
Ê
œ 5x% 0.25x 0.25
œ 2.1x# 2.1x'
dy dx
3. y œ x$ 3 ax# 1# b Ê
dy dx
dy dx
œ 3x# 3(2x 0) œ 3x# 6x œ 3x(x 2)
dy dx
œ 7x' È7
5. y œ (x 1)# ax# 2xb Ê
dy dx
œ (x 1)# (2x 2) ax# 2xb (2(x 1)) œ 2(x 1) c(x 1)# x(x 2)d
dy dx
œ (2x 5)(1)(4 x)# (1) (4 x)" (2) œ (4 x)# c(2x 5) 2(4 x)d
#
œ 2(x 1) a2x 4x 1b 6. y œ (2x 5)(4 x)" Ê œ 3(4 x)
#
$
7. y œ a)# sec ) 1b Ê 8. y œ Š1
csc ) #
)# 4‹
9. s œ
Èt 1 Èt
Ê
ds dt
œ
10. s œ
" Èt 1
Ê
ds dt
œ
#
Ê
ˆ1 Èt‰†
" sin# x
2 sin x
œ 2 Š1
"
"
Èt Èt Š #Èt ‹
#
ˆÈ t 1 ‰ dy dx
#
"
Èt ‹
#
œ
ds dt
)# ˆ csc ) cot ) 4‹ #
ˆ1 Èt‰ Èt 2Èt ˆ1 Èt‰
#
œ
#) ‰ œ Š1
csc ) #
)# 4 ‹ (csc
) cot ) ))
" # #Èt ˆ1 Èt‰
" # 2 È t ˆÈ t 1 ‰
dy dx
œ (2 csc x)(csc x cot x) 2( csc x cot x) œ (2 csc x cot x)(1 csc x)
œ 4 cos$ (1 2t)(sin (1 2t))(2) œ 8 cos$ (1 2t) sin (1 2t)
œ 3 cot# ˆ 2t ‰ ˆcsc# ˆ 2t ‰‰ ˆ t#2 ‰ œ
15. s œ (sec t tan t)& Ê
ds dt
16. s œ csc& a1 t 3t# b Ê &
csc ) #
œ (4 tan x) asec# xb (2 sec x)(sec x tan x) œ 2 sec# x tan x
œ csc# x 2 csc x Ê ds dt
œ
#
ˆÈt 1‰ (0) 1 Š
13. s œ cos% (1 2t) Ê 14. s œ cot$ ˆ 2t ‰ Ê
dy d)
ˆ1 Èt‰
11. y œ 2 tan# x sec# x Ê 12. y œ
#
œ 3 a)# sec ) 1b (2) sec ) tan ))
dy d)
6 t#
cot# ˆ 2t ‰ csc# ˆ 2t ‰
œ 5(sec t tan t)% asec t tan t sec# tb œ 5(sec t)(sec t tan t)& ds dt
œ 5 csc% a1 t 3t# b acsc a1 t 3t# b cot a1 t 3t# bb (1 6t)
œ 5(6t 1) csc a1 t 3t# b cot a1 t 3t# b 17. r œ È2) sin ) œ (2) sin ))"Î# Ê
dr d)
œ
" #
(2) sin ))"Î# (#) cos ) 2 sin )) œ
) cos ) sin ) È2) sin )
Chapter 3 Practice Exercises 18. r œ 2)Ècos ) œ 2) (cos ))"Î# Ê
œ 2) ˆ "# ‰ (cos ))"Î# (sin )) 2(cos ))"Î# œ
dr d)
) sin ) Ècos )
181
2Ècos )
2 cos ) ) sin ) Ècos )
œ
19. r œ sin È2) œ sin (2))"Î# Ê 20. r œ sin Š) È) 1‹ Ê
cos È2) È 2)
œ cos (2))"Î# ˆ "# (2))"Î# (2)‰ œ
œ cos Š) È) 1‹ Š1
" ‹ 2È ) 1
2È) " 1 #È ) "
œ
œ
" #
22. y œ 2Èx sin Èx Ê
dy dx
" 2 œ 2Èx ˆcos Èx‰ Š 2È ‹ ˆsin Èx‰ Š 2È ‹ œ cos Èx x x
x# csc
Ê
2 x
x# ˆcsc
2 x
cot x2 ‰ ˆ x#2 ‰ ˆcsc x2 ‰ ˆ "# † 2x‰ œ csc
cos Š) È) 1‹
dy dx
21. y œ
" #
dr d)
dr d)
2 x
cot
2 x
x csc
2 x
sin Èx Èx
dy "Î# sec (2x)# tan (2x)# (2(2x) † 2) sec (2x)# ˆ "# x$Î# ‰ dx œ x 8x"Î# sec (2x)# tan (2x)# "# x$Î# sec (2x)# œ "# x"Î# sec (2x)# c16 tan (2x)# x# d or #x"$Î# seca#xb2 16x# tana2xb#
23. y œ x"Î# sec (2x)# Ê œ
"‘
24. y œ Èx csc (x 1)$ œ x"Î# csc (x 1)$ Ê
dy dx
œ x"Î# acsc (x 1)$ cot (x 1)$ b a3(x 1)# b csc (x 1)$ ˆ "# x"Î# ‰
œ 3Èx (x 1)# csc (x 1)$ cot (x 1)$ or
" csc(x #È x
csc (x 1)$ 2È x
œ
" #
Èx csc (x 1)$ x" 6(x 1)# cot (x 1)$ ‘
1)$ c1 6x(x 1)# cot (x 1)$ d
25. y œ 5 cot x# Ê 26. y œ x# cot 5x Ê
dy dx
œ 5 acsc# x# b (2x) œ 10x csc# ax# b œ x# acsc# 5xb (5) (cot 5x)(2x) œ 5x# csc# 5x 2x cot 5x
dy dx
27. y œ x# sin# a2x# b Ê
dy dx
œ x# a2 sin a2x# bb acos a2x# bb (4x) sin# a2x# b (2x) œ 8x$ sin a2x# b cos a2x# b 2x sin# a2x# b
28. y œ x# sin# ax$ b Ê
dy dx
œ x# a2 sin ax$ bb acos ax$ bb a3x# b sin# ax$ b a2x$ b œ 6 sin ax$ b cos ax$ b 2x$ sin# ax$ b
29. s œ ˆ t 4t 1 ‰ 30. s œ
#
" 15(15t 1)$
Èx
Ê
œ 2 ˆ t 4t 1 ‰
$
(4t)(1) Š (t 1)(4) ‹ œ 2 ˆ t 4t 1 ‰ (t 1)#
" œ 15 (15t 1)$ Ê
#
31. y œ Š x 1 ‹ Ê 2È x
ds dt
#
dy dx
32. y œ Š 2Èx 1 ‹ Ê
dy dx
(x 1)#
2È x
œ 2 Š 2È x 1 ‹
# "Î# 33. y œ É x x # x œ ˆ1 "x ‰ Ê
dy dx
œ
34. y œ 4xÉx Èx œ 4x ˆx x"Î# ‰ "Î# œ ˆx Èx‰ ’2x Š1
" ‹ #È x
"Î#
" #
4 (t 1)#
" œ 15 (3)(15t 1)% (15) œ
ds dt
" (x 1) Š #È ‹ ˆÈx‰ (1) x
Èx
œ 2 Šx 1‹ †
$
œ
(x 1) 2x (x 1)$
ˆ2Èx 1‰ Š È" ‹ ˆ2Èx‰ Š È" ‹ x x ˆ2 È x 1 ‰
#
ˆ1 "x ‰"Î# ˆ x"# ‰ œ
Ê
dy dx
œ
œ "
"Î#
3 (15t 1)%
1x (x 1)$
#x # É 1
œ 4x ˆ "# ‰ ˆx x"Î# ‰
œ (t 8t $1)
4Èx Š È"x ‹
ˆ2 È x 1 ‰$
œ
4 ˆ2 È x 1 ‰$
" x
ˆ1 "# x"Î# ‰ ˆx x"Î# ‰"Î# (4)
"Î# ˆ2x Èx 4x 4Èx‰ œ 4 ˆx Èx‰“ œ ˆx Èx‰
6x 5Èx É x Èx
182
Chapter 3 Differentiation #
35. r œ ˆ cossin) ) 1 ‰ Ê œ 2 ˆ cossin) ) " ‰ Š cos
#
#
1 ‰ 36. r œ ˆ 1sin )cos Ê )
œ
2(sin ) ") (1 cos ))$
)) (sin ))(sin )) œ 2 ˆ cossin) ) 1 ‰ ’ (cos ) 1)(cos “ (cos ) 1)#
dr d)
) cos ) sin# ) ‹ (cos ) ")#
œ
(2 sin )) (1 cos )) (cos ) 1)$
1 ‰ (1 cos ))(cos )) (sin ) ")(sin )) œ 2 ˆ 1sin )cos “ ) ’ (1 cos ))#
dr d)
2(sin ) 1)(cos ) sin ) 1) (1 c os ))$
acos ) cos# ) sin# ) sin )b œ
37. y œ (2x 1) È2x 1 œ (2x 1)$Î# Ê
œ
dy dx
3 #
(2x 1)"Î# (2) œ 3È2x 1
38. y œ 20(3x 4)"Î% (3x 4)"Î& œ 20(3x 4)"Î#! Ê 39. y œ 3 a5x# sin 2xb 40. y œ a3 cos$ 3xb
2 sin ) (cos ) ")#
œ
$Î#
"Î$
Ê
Ê
" ‰ œ 20 ˆ 20 (3x 4)"*Î#! (3) œ
œ 3 ˆ 3# ‰ a5x# sin 2xb
dy dx
œ "3 a3 cos$ 3xb
dy dx
dy dx
%Î$
&Î#
[10x (cos 2x)(2)] œ
a3 cos# 3xb (sin 3x)(3) œ
3 (3x 4)"*Î#!
9(5x cos 2x) a5x# sin 2xb&Î#
3 cos# 3x sin 3x a3 cos$ 3xb%Î$
2 41. xy 2x 3y œ 1 Ê axyw yb 2 3yw œ 0 Ê xyw 3yw œ 2 y Ê yw (x 3) œ 2 y Ê yw œ yx
3
42. x# xy y# 5x œ 2 Ê 2x Šx œ 5 2x y Ê
œ
dy dx
dy dx
dy dx
dy dx
ˆ4x 4y"Î$ ‰ œ 2 3x# 4y Ê
44. 5x%Î& 10y'Î& œ 15 Ê 4x"Î& 12y"Î& " #
45. (xy)"Î# œ 1 Ê
(xy)"Î# Šx
46. x# y# œ 1 Ê x# Š2y 47. y# œ
x x 1
Ê 2y
x‰ 48. y# œ ˆ 11
x
"Î#
dy dx
dy dx ‹
œ
dp dq
œ
dp dq
dy dx
2y
dy dx
œ 5 2x y Ê
dy dx
(x 2y)
" x 1x
Ê 4y$ dp dq
dy dx
œ 2 Ê 4x
œ
dy dx
œ 0 Ê 12y"Î& dy dx
dy dx
4y"Î$
œ 4x"Î& Ê
œ x"Î# y"Î# Ê
œ 2xy# Ê
dy dx
œ yx
Ê
dy dx
œ
6q œ 0 Ê 3p#
dp dq
dy dx
dy dx
œ 2 3x# 4y
dy dx
dy dx
" œ "3 x"Î& y"Î& œ 3(xy) "Î&
œ x" y Ê
dy dx
œ yx
" #y(x 1)#
dy dx
œ
œ
(1 x)(1) (1 x)Ð") (" x)#
4 Šp q
dy dx
2 3x# 4y 4x 4y"Î$
dy dx
y‹ œ 0 Ê x"Î# y"Î#
Ê
dy dx
dp dq ‹
" 2y$ (1 x)#
4q
dp dq
œ 6q 4p Ê
dp dq
a3p# 4qb œ 6q 4p
6q 4p 3p# 4q
50. q œ a5p# 2pb Ê
&œ! Ê x
4y‹ 4y"Î$
y# (2x) œ 0 Ê 2x# y
49. p$ 4pq 3q# œ 2 Ê 3p# Ê
dy dx
(x 1)(1) (x)(1) (x 1)#
Ê y% œ
dy dx
5 2x y x 2y
43. x$ 4xy 3y%Î$ œ 2x Ê 3x# Š4x Ê
y‹ 2y
$Î#
#
Ê 1 œ 3# a5p# 2pb &Î#
œ a5p3(5p 2p1)b
&Î#
Š10p
dp dq
2
dp dq ‹
Ê 23 a5p# 2pb
&Î#
œ
dp dq
(10p 2)
Chapter 3 Practice Exercises dr ‰ 51. r cos 2s sin# s œ 1 Ê r(sin 2s)(2) (cos 2s) ˆ ds 2 sin s cos s œ 0 Ê
Ê
dr ds
2r sin 2s sin 2s cos 2s
œ
œ
(2r 1)(sin 2s) cos 2s
52. 2rs r s s# œ 3 Ê 2 ˆr s 53. (a) x$ y$ œ 1 Ê 3x# 3y# Ê
d# y dx#
(b) y# œ 1 Ê
d# y dx#
x# ‹ y#
2xy# a2yx# b Š
œ
y%
Ê 2y
2 x
2xy x# Š
œ
œ
dy dx
y# x%
" ‹ yx#
œ
54. (a) x# y# œ 1 Ê 2x 2y (b)
dy dx
œ
x y
d# y dx#
Ê
œ
y(1) x y#
dr ds
d# y dx#
œ
#
œ xy# Ê
dy dx 2x% y
y%
œ
" yx#
œ
dy dx
1 2s œ 0 Ê
(2s 1) œ 1 2s 2r Ê y# (2x) ax# b Š2y
dr ds
" 2s 2r 2s 1
œ
dy dx ‹
y%
2xy$ 2x% y&
Ê
dy dx
œ ayx# b
"
Ê
dy dx
œ
d# y dx#
œ ayx# b
#
’y(2x) x#
dy dx “
2xy# 1 y$ x%
œ 0 Ê 2y
dy dx dy dx
dr ds
2xy#
Ê
2 x#
œ0 Ê
œ
(cos 2s) œ 2r sin 2s 2 sin s cos s
œ (2r 1)(tan 2s)
dr ‰ ds
dy dx
dr ds
y x Š xy ‹
œ
œ
y#
dy dx
œ 2x Ê
y# x# y$
œ
" y$
x y
(since y# x# œ 1)
55. (a) Let h(x) œ 6f(x) g(x) Ê hw (x) œ 6f w (x) gw (x) Ê hw (1) œ 6f w (1) gw (1) œ 6 ˆ "# ‰ a%b œ (
(b) Let h(x) œ f(x)g# (x) Ê hw (x) œ f(x) a#g(x)b gw (x) g# (x)f w (x) Ê hw (0) œ #f(0)g(0)gw (0) g# (0)f w (0) œ #(1)(1) ˆ "# ‰ (1)# ($) œ # (c) Let h(x) œ œ
f(x) g(x) 1
(& 1) ˆ "# ‰ 3 a%b (& 1)#
(g(x) 1)f (x) f(x)g (x) (g(x) 1)
Ê hw (x) œ œ
w
w
#
Ê hw (1) œ
(g(1) ")f (1) f(1)g (1) (g(1) 1) w
w
#
& "#
(d) Let h(x) œ f(g(x)) Ê hw (x) œ f w (g(x))gw (x) Ê hw (0) œ f w (g(0))gw (0) œ f w (1) ˆ "# ‰ œ ˆ "# ‰ ˆ "# ‰ œ
" %
(e) Let h(x) œ g(f(x)) Ê hw (x) œ gw (f(x))f w (x) Ê hw (0) œ gw (f(0))f w (0) œ gw (1)f w (0) œ a%b ($) œ "# (f) Let h(x) œ (x f(x))$Î# Ê hw (x) œ 3# (x f(x))"Î# a1 f w (x)b Ê hw (1) œ 3# (1 f(1))"Î# a1 f w (1)b œ 3# (1 3)"Î# ˆ1 "# ‰ œ *# (g) Let h(x) œ f(x g(x)) Ê hw (x) œ f w (x g(x)) a1 gw (x)b Ê hw (0) œ f w (g(0)) a1 gw (0)b œ f w (1) ˆ1 "# ‰ œ ˆ "# ‰ ˆ $# ‰ œ
%$56. (a) Let h(x) œ Èx f(x) Ê hw (x) œ Èx f w (x) f(x) † (b) Let h(x) œ (f(x))"Î# Ê hw (x) œ
" #
" #È x
(f(x))"Î# af w (x)b Ê hw (0) œ
(c) Let h(x) œ f ˆÈx‰ Ê hw (x) œ f w ˆÈx‰ †
" #È x
" œ 5" (3) ˆ #" ‰ #È 1 " "Î# (2) œ 3" # (9)
Ê hw (1) œ È1 f w (1) f(1) † " #
(f(0))"Î# f w (0) œ
Ê hw (1) œ f w ŠÈ1‹ †
" #È 1 w
œ
" 5
†
" #
œ
œ 13 10
" 10
(d) Let h(x) œ f(1 5 tan x) Ê hw (x) œ f w (1 5 tan x) a5 sec# xb Ê h (0) œ f w (1 5 tan 0) a5 sec# 0b œ f w (1)(5) œ "5 (5) œ 1 (2 cos x)f (x) f(x)(sin x) f(0)(0) Ê hw (0) œ (2 1)f(2(0) œ 3(9 2) œ (2 cos x)
1) h(x) œ 10 sin ˆ 1#x ‰ f # (x) Ê hw (x) œ 10 sin ˆ 1#x ‰ a2f(x)f w (x)b f # (x) ˆ10 cos ˆ 1#x ‰‰ ˆ 1# ‰ hw (1) œ 10 sin ˆ 1# ‰ a2f(1)f w (1)b f # (1) ˆ10 cos ˆ 1# ‰‰ ˆ 1# ‰ œ 20(3) ˆ "5 ‰ ! œ 12
(e) Let h(x) œ (f) Let Ê
57. x œ t# 1 Ê dy dt
œ
dy dx
†
dx dt
f(x) 2 cos x
dx dt
Ê hw (x) œ
"Î$
œ 2 au# 2ub
"Î$
Ê
dt du
œ
5; thus
" 3
dy dt ¹ t=0
au# 2ub
ds du
œ
ds dt
w
#
œ 2t; y œ 3 sin 2x Ê
œ 6 cos a2t# b † 2t Ê
58. t œ au# 2ub
w
†
dy dx
32
œ 3(cos 2x)(2) œ 6 cos 2x œ 6 cos a2t# 21b œ 6 cos a2t# b ; thus,
œ 6 cos (0) † 0 œ 0
#Î$
dt du
#
(2u 2) œ
2 # 3 au "Î$
œ ’2 au# 2ub
2ub
#Î$
(u 1); s œ t# 5t Ê
5“ ˆ 32 ‰ au#
2ub
#Î$
(u 1)
ds dt
œ 2t 5
183
184
Chapter 3 Differentiation Ê
ds ¸ du u=2
œ ’2 a2# 2(2)b
59. r œ 8 sin ˆs 16 ‰ Ê œ
; thus,
2É8 sin ˆs 16 ‰
Ê
dw ¸ ds s=0
œ
#Î$
(2 1) œ 2 ˆ2 † 8"Î$ 5‰ ˆ8#Î$ ‰ œ 2(2 † 2 5) ˆ 4" ‰ œ
†
dw dr
œ
2É8 sin ˆ 16 ‰ d ) ‰‰ dt
d) dt
œ
dr ds
cos ŠÉ8 sin ˆs 16 ‰ 2‹ # É8 sinˆ s 16 ‰
(cos 0)(8) Š
È3 ‹
2È4
œ0 Ê
d) dt
#
and
61. y$ y œ 2 cos x Ê 3y# 2 sin (0) 3 1
œ 0;
d# y dx# ¹ (0ß1)
Ê
œ
d# y dx#
d# y dx#
œ
"3 8#Î$
œ
" 3
œ 2 sin x Ê a3y#
" 3
1b
x#Î$ 3" y#Î$ dy ˆ #Î$ ‰ ˆ 23 dx ‹ y # #Î$
ax 2 3
dy dx
a3y# 1b (2 cos x) (2 sin x) Š6y
ˆx#Î$ ‰ Š 23 y"Î$
4
(3 1)(2 cos 0) (2 sin 0)(6†0) (3 1)#
62. x"Î$ y"Î$ œ 4 Ê Ê
œ
dy dx
œ
" œ cos ˆÈr 2‰ Š #È ‹ r
† 8 cos ˆs 16 ‰‘
(2)t 1) œ )# Ê
dy dx
d) dt
œ
) # 2)t 1
; r œ a)# 7b
0 and )# t ) œ 1 Ê ) œ 1 so that œ
ˆ 6" ‰ (1)
œ
"Î$
d) ¸ dt t=0, )=1
œ
1 1
œ 1
" 6
a3y# 1b œ 2 sin x Ê
dy dx
œ
2 sin x 3y# 1
Ê
dy dx ¹ (0ß1)
dy dx ‹
#
œ #"
dy dx
x"Î$ ‰
b
dw dr
9 #
œ È3
#Î$ #Î$ dr " # (2)) œ 32 ) a)# 7b ; now t œ d) œ 3 a) 7b dr ¸ 2 dr ¸ dr ¸ " #Î$ œ 6 Ê dt t=0 œ d) t=0 † ddt) ¸ t=0 d) )=1 œ 3 (1 7)
Ê
œ
œ
dw ds
cos ŠÉ8 sin ˆ 16 ‰ 2‹†8 cos ˆ 16 ‰
60. )# t ) œ 1 Ê ˆ)# t ˆ2)
œ
5“ ˆ 23 ‰ a2# 2(2)b
œ 8 cos ˆs 16 ‰ ; w œ sin ˆÈr 2‰ Ê
dr ds
cos É8 sin ˆs 16 ‰ 2
"Î$
œ0 Ê Ê
dy dx
#Î$
œ yx#Î$ Ê
d# y dx# ¹ (8ß8)
œ
dy dx ¹ (8ß8)
œ 1;
dy dx
œ
y#Î$ x#Î$
ˆ8#Î$ ‰ 23 †8"Î$ †(1)‘ ˆ8#Î$ ‰ ˆ 23 †8"Î$ ‰ 8%Î$
" 6
" " f(t h) f(t) 2t 1 (2t 2h 1) " " œ #(th)1h #t1 œ (2t 2t 1 and f(t h) œ #(t h) 1 Ê h
2h 1)(2t 1)h f(t h) f(t) 2h 2 w œ lim (2t 2h 21)(#t 1) (2t 2h 1)(2t 1)h œ (2t 2h 1)(2t 1) Ê f (t) œ hlim h Ä! hÄ! # (2t 1)#
63. f(t) œ œ œ
g(x h) g(x) h g(x h) g(x) lim œ lim h hÄ! hÄ!
64. g(x) œ 2x# 1 and g(x h) œ 2(x h)# 1 œ 2x# 4xh 2h# 1 Ê œ
a2x# 4xh 2h# 1b a2x# 1b h
œ
4xh 2h# h
œ 4x 2h Ê gw (x) œ
œ 4x
(4x 2h)
65. (a)
lim f(x) œ lim c x# œ 0 and lim b f(x) œ lim b x# œ 0 Ê lim f(x) œ 0. Since lim f(x) œ 0 œ f(0) it xÄ! xÄ! xÄ! xÄ! xÄ! follows that f is continuous at x œ 0. (c) lim c f w (x) œ lim c (2x) œ 0 and lim b f w (x) œ lim b (2x) œ 0 Ê lim f w (x) œ 0. Since this limit exists, it (b)
x Ä !c
xÄ!
xÄ!
xÄ!
follows that f is differentiable at x œ 0.
xÄ!
xÄ!
Chapter 3 Practice Exercises
185
66. (a)
lim f(x) œ lim c x œ 0 and lim b f(x) œ lim b tan x œ 0 Ê lim f(x) œ 0. Since lim f(x) œ 0 œ f(0), it xÄ! xÄ! xÄ! xÄ! xÄ! follows that f is continuous at x œ 0. (c) lim c f w (x) œ lim c 1 œ 1 and lim b f w (x) œ lim b sec# x œ 1 Ê lim f w (x) œ 1. Since this limit exists it (b)
x Ä !c
xÄ!
xÄ!
xÄ!
xÄ!
xÄ!
follows that f is differentiable at x œ 0. 67. (a)
lim f(x) œ lim c x œ 1 and lim b f(x) œ lim b (2 x) œ 1 Ê lim f(x) œ 1. Since lim f(x) œ 1 œ f(1), it xÄ" xÄ" xÄ" xÄ" xÄ" follows that f is continuous at x œ 1. (c) lim c f w (x) œ lim c 1 œ 1 and lim b f w (x) œ lim b 1 œ 1 Ê lim c f w (x) Á lim b f w (x), so lim f w (x) does (b)
x Ä "c
xÄ"
xÄ"
xÄ"
not exist Ê f is not differentiable at x œ 1.
xÄ"
xÄ"
xÄ1
xÄ"
lim f(x) œ lim c sin 2x œ 0 and lim b f(x) œ lim b mx œ 0 Ê lim f(x) œ 0, independent of m; since xÄ! xÄ! xÄ! xÄ! f(0) œ 0 œ lim f(x) it follows that f is continuous at x œ 0 for all values of m.
68. (a)
x Ä !c
xÄ!
lim f w (x) œ lim c (sin 2x)w œ lim c 2 cos 2x œ 2 and lim b f w (x) œ lim b (mx)w œ lim b m œ m Ê f is x Ä !c xÄ! xÄ! xÄ! xÄ! xÄ! differentiable at x œ 0 provided that lim c f w (x) œ lim b f w (x) Ê m œ 2.
(b)
xÄ!
69. y œ œ
" #
x #
" #x 4
œ
2(2x 4)
" # x #
(2x 4)" Ê
dy dx
œ
" #
xÄ!
2(2x 4)# ; the slope of the tangent is 3# Ê 3#
Ê 2 œ 2(2x 4)# Ê 1 œ
" (2x 4)#
Ê 4x# 16x 15 œ 0 Ê (2x 5)(2x 3) œ 0 Ê x œ
Ê (2x 4)# œ 1 Ê 4x# 16x 16 œ 1
5 #
or x œ
3 #
Ê ˆ 5# ß 94 ‰ and ˆ 3# ß "4 ‰ are points on the
curve where the slope is . 3 #
70. y œ x
" 2x
Ê xœ „
Ê " #
dy dx
œ1
2 (2x)#
Ê ˆ "# ß "# ‰ and ˆ
71. y œ 2x$ 3x# 12x 20 Ê #
#
" " #x# ; the slope of the tangent is 3 Ê 3 œ 1 #x# " "‰ # ß # are points on the curve where the slope is 3.
œ1
dy dx
Ê 2œ
œ 6x# 6x 12; the tangent is parallel to the x-axis when
dy dx
" #x #
Ê x# œ
" 4
œ0
Ê 6x 6x 12 œ 0 Ê x x 2 œ 0 Ê (x 2)(x 1) œ 0 Ê x œ 2 or x œ 1 Ê (#ß !) and ("ß #7) are points on the curve where the tangent is parallel to the x-axis. 72. y œ x$ Ê
dy dx
œ 3x# Ê
dy dx ¹ (2ß8)
œ 12; an equation of the tangent line at (#ß )) is y 8 œ 12(x 2)
Ê y œ 12x 16; x-intercept: 0 œ 12x 16 Ê x œ 43 Ê ˆ 43 ß !‰ ; y-intercept: y œ 12(0) 16 œ 16 Ê (0ß 16)
186
Chapter 3 Differentiation
73. y œ 2x$ 3x# 12x 20 Ê
dy dx
œ 6x# 6x 12
(a) The tangent is perpendicular to the line y œ 1
x 24
when
dy dx
œ Š ˆ" " ‰ ‹ œ 24; 6x# 6x 12 œ 24 #4
Ê x# x 2 œ 4 Ê x# x 6 œ 0 Ê (x 3)(x 2) œ 0 Ê x œ 2 or x œ 3 Ê (#ß 16) and ($ß 11) are x points where the tangent is perpendicular to y œ 1 24 . dy È (b) The tangent is parallel to the line y œ 2 12x when dx œ 12 Ê 6x# 6x 12 œ 12 Ê x# x œ 0 Ê x(x 1) œ 0 Ê x œ 0 or x œ 1 Ê (!ß 20) and ("ß () are points where the tangent is parallel to y œ È2 12x. 74. y œ
1 sin x x
Ê
Since m" œ
x(1 cos x) (1 sin x)(1) x#
dy dx
œ
" m#
the tangents intersect at right angles.
75. y œ tan x, 1# x
1 #
Ê
dy dx
Ê m" œ
dy dx ¹ x=1
œ
1 # 1#
œ 1 and m# œ
dy 1# dx ¹ x=c1 1#
œ 1.
œ sec# x; now the slope
of y œ x# is "# Ê the normal line is parallel to y œ x# when #
Ê cos x œ
dy dx
" #
œ 2. Thus, sec# x œ 2 Ê
Ê cos x œ
for 1# x
1 #
„" È2
Ê xœ
1 4
" cos# x
œ2
and x œ
1 4
Ê ˆ 14 ß 1‰ and ˆ 14 ß "‰ are points
where the normal is parallel to y œ x# .
76. y œ 1 cos x Ê
dy dx
œ sin x Ê
dy dx ¹ ˆ 1 ß1‰
œ 1
2
Ê the tangent at ˆ 1# ß 1‰ is the line y 1 œ ˆx 1# ‰ Ê y œ x 1# 1; the normal at ˆ 1# ß 1‰ is
y 1 œ (1) ˆx 1# ‰ Ê y œ x
77. y œ x# C Ê thus,
" #
œ
ˆ "# ‰#
78. y œ x$ Ê
dy dx
dy dx
1 #
1
œ 2x and y œ x Ê
C Ê Cœ œ 3x# Ê
dy dx
œ 1; the parabola is tangent to y œ x when 2x œ 1 Ê x œ
" #
Ê yœ
" #
;
" 4
dy dx ¹ x=a
œ 3a# Ê the tangent line at aaß a$ b is y a$ œ 3a# (x a). The tangent line
intersects y œ x$ when x$ a$ œ 3a# (x a) Ê (x a) ax# xa a# b œ 3a# (x a) Ê (x a) ax# xa 2a# b œ 0 Ê (x a)# (x 2a) œ 0 Ê x œ a or x œ 2a. Now
dy dx ¹ x=c2a
œ 3(2a)# œ 12a# œ 4 a3a# b, so the slope at
x œ 2a is 4 times as large as the slope at aaß a$ b where x œ a. 79. The line through (!ß $) and (5ß 2) has slope m œ y œ x 3; y œ
c x 1
Ê
dy dx
œ
c (x 1)# ,
3 (2) 05
œ 1 Ê the line through (!ß $) and (&ß 2) is
so the curve is tangent to y œ x 3 Ê
Ê (x 1)# œ c, x Á 1. Moreover, y œ
c x 1
intersects y œ x 3 Ê #
c x 1
dy dx
œ 1 œ
c (x 1)#
œ x 3, x Á 1
Ê c œ (x 1)(x 3), x Á 1. Thus c œ c Ê (x 1) œ (x 1)(x 3) Ê (x 1)[x 1 (x 3)] œ !, x Á 1 Ê (x 1)(2x 2) œ 0 Ê x œ 1 (since x Á 1) Ê c œ 4.
Chapter 3 Practice Exercises 80. Let Šbß „ Èa# b# ‹ be a point on the circle x# y# œ a# . Then x# y# œ a# Ê 2x 2y Ê
dy dx ¹ x=b
œ
b „È a # b #
y Š „ Èa# b# ‹ œ
Ê normal line through Šbß „ Èa# b# ‹ has slope „È a # b # b
(x b) Ê y … Èa# b# œ
„È a # b # b
„È a # b # b
œ0 Ê
dy dx
dy dx
œ xy
Ê normal line is
x … Èa# b# Ê y œ „
È a# b # b
x
which passes through the origin. 81. x# 2y# œ 9 Ê 2x 4y œ "4 x
9 4
5 #
x œ 2y Ê
dy dx
œ "4 Ê the tangent line is y œ 2 "4 (x 1)
dy dx ¹ (1ß2)
and the normal line is y œ 2 4(x 1) œ 4x 2.
82. x$ y# œ 2 Ê 3x# 2y œ 3# x
œ0 Ê
dy dx
œ0 Ê
dy dx
œ
dy dx
3x# 2y
Ê
dy dx ¹ (1ß1)
and the normal line is y œ 1 23 (x 1) œ
83. xy 2x 5y œ 2 Ê Šx
y‹ 2 5
dy dx
œ0 Ê
dy dx
(x 5) œ y 2 Ê
Ê the tangent line is y œ 2 2(x 3) œ 2x 4 and the normal line is y œ 2 84. (y x)# œ 2x 4 Ê 2(y x) Š dy dx 1‹ œ 2 Ê (y x) Ê the tangent line is y œ 2 34 (x 6) œ 85. x Èxy œ 6 Ê 1
" #Èxy
dy dx
Šx
3 4
x
dy dx
œ 1 (y x) Ê
y 2 x 5
dy dx
œ
Ê
1 #
(x 3) œ "# x 7# .
dy dx
œ
1 yx yx
3 2
x"Î# 3y"Î#
y œ 4 "4 (x 1) œ 4" x
dy dx
œ2
dy dx ¹ (6ß2)
œ
3 4
dy dx
y œ 2Èxy Ê
dy dx
2Èxy y x
œ
Ê
dy dx
œ
x"Î# 2y"Î#
Ê
dy dx ¹ (1ß4)
œ
dy dx ¹ (4ß1) 4 5
x
5 4
11 5
.
œ "4 Ê the tangent line is
and the normal line is y œ 4 4(x 1) œ 4x.
17 4
87. x$ y$ y# œ x y Ê ’x$ Š3y# Ê
œ0 Ê
dy dx
dy dx ¹ (3ß2)
Ê
Ê the tangent line is y œ 1 54 (x 4) = 54 x 6 and the normal line is y œ " 45 (x 4) œ 86. x$Î# 2y$Î# œ 17 Ê
(x 1)
and the normal line is y œ 2 43 (x 6) œ 43 x 10.
5 #
y‹ œ 0 Ê x
3 #
x "3 .
2 3
dy dx
œ #3 Ê the tangent line is y œ 1
dy dx ‹
y$ a3x# b“ 2y
a3x$ y# 2y 1b œ 1 3x# y$ Ê
dy dx
œ
dy dx
œ1
1 3x# y$ 3x$ y# 2y 1
Ê
dy dx
Ê 3x$ y#
dy dx ¹ (1ß1)
dy dx
2y
œ 24 , but
dy dx
dy dx ¹ (1ß1) is
dy dx
œ " 3x# y$
undefined.
Therefore, the curve has slope "# at ("ß ") but the slope is undefined at ("ß 1). 88. y œ sin (x sin x) Ê
dy dx
œ [cos (x sin x)](1 cos x); y œ 0 Ê sin (x sin x) œ 0 Ê x sin x œ k1,
k œ 2, 1, 0, 1, 2 (for our interval) Ê cos (x sin x) œ cos (k1) œ „ 1. Therefore,
dy dx
œ 0 and y œ 0 when
1 cos x œ 0 and x œ k1. For #1 Ÿ x Ÿ 21, these equations hold when k œ 2, 0, and 2 (since cos (1) œ cos 1 œ 1). Thus the curve has horizontal tangents at the x-axis for the x-values 21, 0, and 21 (which are even integer multiples of 1) Ê the curve has an infinite number of horizontal tangents. 89. x œ
" #
tan t, y œ " #
Ê xœ
$
œ 2 cos 90. x œ "
tan
ˆ 13 ‰
" t#
" #
sec t Ê
1 3
œ
œ
" 4
È3 #
,yœ"
3 t
dy dx
œ
dy/dt dx/dt
œ
" #
sec
1 3
and y œ
Ê
dy dx
œ
dy/dt dx/dt
œ
" #
sec t tan t " # # sec t
œ
tan t sec t
œ sin t Ê
œ1 Ê yœ
È3 #
x 4" ;
Š t3# ‹ Š t2$ ‹
œ 32 t Ê
d# y dx
dy dx ¹ tœ2
#
dy dx ¹ tœ1Î3
œ
w
dy /dt dx/dt
1 3
œ sin
œ
" #
cos t sec t #
œ
È3 #
;tœ
1 3
œ 2 cos$ t Ê
d y¸ dx tœ1Î3 #
œ 3# (2) œ 3; t œ 2 Ê x œ 1
#
" ##
œ
5 4
and
187
188
Chapter 3 Differentiation
yœ1
3 #
œ "# Ê y œ 3x
"3 4
d# y dx
;
#
œ
w
œ
dy /dt dx/dt
ˆ 3 ‰ #
Š t2 ‹
œ
3 $ 4 t
#
Ê
œ
d y dx ¹ tœ2 #
$
3 4
(2)$ œ 6
91. B œ graph of f, A œ graph of f w . Curve B cannot be the derivative of A because A has only negative slopes while some of B's values are positive. 92. A œ graph of f, B œ graph of f w . Curve A cannot be the derivative of B because B has only negative slopes while A has positive values for x 0. 93.
94.
95. (a) 0, 0
(b) largest 1700, smallest about 1400
96. rabbits/day and foxes/day sin x
97. lim
# x Ä ! 2x x
98. lim
3x tan 7x #x
99. lim
sin r
xÄ!
r Ä ! tan 2r
100.
sin 7x ‰ 2x cos 7x
xÄ!
2r tan 2r
103.
104. 105.
)Ä!
xÄ!
4 tan# ) tan ) 1 tan# ) &
œ
lim c ) Ä ˆ1‰
lim
tan x x
œ lim
)Ä!
2 sin# ˆ #) ‰ )# )Ä!
œ lim ˆ cos" x †
tan ) )
xÄ!
sin (sin )) sin )
Š" tan5# ) ‹
Š5 cot7 ) cot8# ) ‹
œ
œ
‹œ
3 #
ˆ1 † 1 † 27 ‰ œ 2
" #
. Let x œ sin ). Then x Ä 0 as ) Ä 0
(4 0 0) (1 0)
(0 2) (5 0 0)
œ lim
x sin x # x x Ä ! 2 ˆ2 sin ˆ # ‰‰
œ4
œ 52 †
x x
œ lim ’ sin## ˆ# x ‰ † xÄ!
#
sin x x “
(1)(1)(1) œ 1
œ lim ’
sin x ‰ x
" ˆ 27 ‰
†
œ ˆ "# ‰ (1) ˆ 1" ‰ œ
cos 2r
Š4 tan" ) tan"# ) ‹
Š cot"# ) 2‹
œ lim b )Ä!
œ lim
sin 7x 7x
œ1
sin x x
x sin x œ lim 2(1xsincosx x) x Ä ! 2 2 cos x xÄ! ˆ x# ‰ ˆx‰ œ lim ’ sin ˆ x ‰ † sin #ˆ x ‰ † sinx x “ œ xÄ! # #
xÄ!
xÄ!
sin 2r r Ä ! ˆ 2r ‰
lim
1cos ) )# )Ä!
lim Š cos"7x †
† "# ‰ œ ˆ "# ‰ (1) lim
2
lim
3 #
)Ä!
œ lim
1 2 cot# ) 5 cot# ) 7 cot ) 8
lim b
œ
)Ä!
2
102.
œ (1) ˆ "1 ‰ œ 1
(sin )) ˆ sin ) ‰ œ lim Š sinsin œ lim ) ‹ )
sin (sin )) sin )
)Ä!
lim c ) Ä ˆ1‰
œ lim ˆ 3x 2x
rÄ!
Ê lim
101.
" (#x 1) “
œ lim ˆ sinr r †
sin (sin )) )
lim
)Ä!
xÄ!
œ lim ’ˆ sinx x ‰ †
)Ä!
sin ˆ #) ‰ ˆ #) ‰
†
sin ˆ #) ‰ ˆ #) ‰
† "# “ œ (1)(1) ˆ "# ‰ œ
" #
œ 1; let ) œ tan x Ê ) Ä 0 as x Ä 0 Ê lim g(x) œ lim xÄ!
œ 1. Therefore, to make g continuous at the origin, define g(0) œ 1.
xÄ!
tan (tan x) tan x
Chapter 3 Practice Exercises 106.
lim f(x) œ lim
xÄ!
(tan x) œ lim ’ tantan † x
tan (tan x)
x Ä ! sin (sin x)
sin x sin (sin x)
xÄ!
#105); let ) œ sin x Ê ) Ä 0 as x Ä 0 Ê
(b) S œ 21r# 21rh and r constant Ê (c) S œ 21r# 21rh Ê (d) S constant Ê
dh dt
(b) r constant Ê
dr dt
109. A œ 1r# Ê 110. V œ s$ Ê 111.
dR" dt
dV dt
œ0 Ê
dh ‰ ˆr dr dt h dt È r# h #
dr dt
œ 3s# †
ds dt
œ 1 ohm/sec,
dR# dt
ds dt
œ
œ
(using the result of
œ 1. Therefore, to make f
œ (41r 21h)
dr dt
Ê (2r
dh dt
dr dt
dr dt 21r h) dr dt œ r
dh dt dh dt
Ê
dr dt
œ
r dh 2r h dt
;
1Èr# h#
dr 1 r# Èr# h# “ dt
; so r œ 10 and Ê
dr dt
lim ) ) Ä ! sin )
œ (41r 21h)
21r
dr dt
sin x
x Ä ! sin (sin x)
œ ’1Èr# h#
dr dt
1 r# dr Èr# h# “ dt
1rh dh Èr# h# dt
œ
dS dt
dr ‰ dt
1Èr# h#
1r# dr dt È r# h #
œ
dS dt
œ0 Ê
œ 21 r
dA dt
dr dt
œ ’1Èr# h#
dS dt
(c) In general,
œ 1r †
dS dt
(a) h constant Ê
œ 21r dh dt #1 ˆr dh h dt
œ 0 Ê 0 œ (41r 21h)
dS dt
108. S œ 1rÈr# h# Ê
œ 41r
dS dt
œ 41r dr dt 21 h
dS dt dS dt
œ 1 † lim
sin x lim x Ä ! sin (sin x)
continuous at the origin, define f(0) œ 1. 107. (a) S œ 21r# 21rh and h constant Ê
" cos x “
†
dr dt
" dV 3s# dt
dh 1rh Èr# h# dt
œ 12 m/sec Ê
; so s œ 20 and
œ 0.5 ohm/sec; and
" R
œ
" R"
" R#
dV dt
dA dt
œ (21)(10) ˆ 12 ‰ œ 40 m# /sec
œ 1200 cm$ /min Ê " dR R# dt
Ê
œ
" dR" R"# dt
ds dt
œ
" dR# R## dt
" 3(20)#
(1200) œ 1 cm/min
. Also,
" " R" œ 75 ohms and R# œ 50 ohms Ê R" œ 75 50 Ê R œ 30 ohms. Therefore, from the derivative 9(625) " dR " " " " " ˆ ‰ Ê dR ˆ 50005625 ‰ (30)# dt œ (75)# (1) (50)# (0.5) œ 5625 5000 dt œ (900) 5625†5000 œ 50(5625) œ 50
equation,
œ 0.02 ohm/sec. 112.
dR dt
œ 3 ohms/sec and
X œ 20 ohms Ê
dZ dt
dX dt
œ 2 ohms/sec; Z œ ÈR# X# Ê
œ
(10)(3) (20)(2) È10# 20#
113. Given
dx dt
œ 10 m/sec and
œ 2x
dx dt
2y
&
dD dt
dy dt
Ê D
" È5
œ
dX R dR dt X dt È R # X#
so that R œ 10 ohms and
¸ 0.45 ohm/sec.
œ 5 m/sec, let D be the distance from the origin Ê D# œ x# y# Ê 2D
dy dt
dD dt
œ
dZ dt
œx
œ (5)(10) (12)(5) Ê
dD dt
y
dx dt
œ
110 5
dy dt
dD dt
. When (xß y) œ ($ß %), D œ É$# a%b# œ & and
œ 22. Therefore, the particle is moving away from the origin at 22 m/sec
(because the distance D is increasing). 114. Let D be the distance from the origin. We are given that œ x# ˆ x
$Î# ‰#
œ x# x$ Ê 2D
œ 2x
dD dt
3x#
dx dt
dx dt
dD dt
œ 11 units/sec. Then D# œ x# y#
œ x(2 3x)
dx dt
and substitution in the derivative equation gives (2)(6)(11) œ (3)(2 9) 115. (a) From the diagram we have (b) V œ
" 3
1 r# h œ
" 3
#
10 h
1 ˆ 25 h‰ h œ
œ
4 r 41 h$ 75
116. From the sketch in the text, s œ r) Ê Ê
ds dt
œr
d) dt
œ (1.2)
d) dt
. Therefore,
Ê rœ Ê
dV dt
2 5
œ
; x œ 3 Ê D œ È 3# 3$ œ 6 dx dt
Ê
dx dt
œ 4 units/sec.
h. 41h# dh 25 dt
ds d) dr dt œ r dt ) dt . ds dt œ 6 ft/sec and r
œ 5 and h œ 6 Ê
dh dt
125 œ 144 1 ft/min.
Also r œ 1.2 is constant Ê
dr dt
œ0
, so
dV dt
œ 1.2 ft Ê
d) dt
œ 5 rad/sec
189
190
Chapter 3 Differentiation
117. (a) From the sketch in the text,
d) dt
point A, x œ 0 Ê ) œ 0 Ê
œ 0.6 rad/sec and x œ tan ). Also x œ tan ) Ê dx dt
dx dt
œ sec# )
d) dt ;
at
#
œ asec 0b (0.6) œ 0.6. Therefore the speed of the light is 0.6 œ
3 5
km/sec
when it reaches point A. (3/5) rad sec
(b)
†
1 rev 21 rad
118. From the figure,
a r
†
60 sec min
œ
b BC
œ
18 1
Ê
a r
revs/min œ
b Èb# r#
. We are given
that r is constant. Differentiation gives, " r
†
da dt
‰ ŠÈb# r# ‹ ˆ db dt (b) Š È
œ
b#
b œ 2r and Ê œ
da dt
db dt
r#
b ‰ ‹ ˆ db dt b# r#
. Then,
œ 0.3r
Ô È(2r)# r# (0.3r) (2r) É2r(#0.3r)# × (2r) r Ù œ rÖ (2r)# r# Õ Ø
È3r# (0.3r) 4r# (0.3r)
È3r#
3r
œ
a3r# b (0.3r) a4r# b (0.3r) 3 È 3 r#
œ
0.3r 3È 3
œ
r 10È3
m/sec. Since
da dt
is positive,
the distance OA is increasing when OB œ 2r, and B is moving toward O at the rate of 0.3r m/sec. 119. (a) If f(x) œ tan x and x œ 14 , then f w (x) œ sec# x, f ˆ 14 ‰ œ 1 and f w ˆ 14 ‰ œ 2. The linearization of f(x) is L(x) œ 2 ˆx 14 ‰ (1) œ 2x
1 2 #
.
(b) If f(x) œ sec x and x œ 14 , then f w (x) œ sec x tan x, f ˆ 1 ‰ œ È2 and f w ˆ 1 ‰ œ È2. The linearization 4
4
of f(x) is L(x) œ È2 ˆx 14 ‰ È2 œ È2x
120. f(x) œ
" 1 tan x
È2(% 1) . 4
Ê f w (x) œ
sec# x (1 tan x)#
. The linearization at x œ 0 is L(x) œ f w (0)(x 0) f(0) œ 1 x.
121. f(x) œ Èx 1 sin x 0.5 œ (x 1)"Î# sin x 0.5 Ê f w (x) œ ˆ "# ‰ (x 1)"Î# cos x
Ê L(x) œ f w (0)(x 0) f(0) œ 1.5(x 0) 0.5 Ê L(x) œ 1.5x 0.5, the linearization of f(x).
122. f(x) œ œ
2 1 x
2 (1 x)#
È1 x 3.1 œ 2(1 x)" (1 x)"Î# 3.1 Ê f w (x) œ 2(1 x)# (1) "# (1 x)"Î#
" 2È 1 x
Ê L(x) œ f w (0)(x 0) f(0) œ 2.5x 0.1, the linearization of f(x).
123. S œ 1 rÈr# h# , r constant Ê dS œ 1 r † "# ar# h# b Ê dS œ
1 r h! adhb Ér# h#!
"Î#
#h dh œ
1rh Èr# h# dh.
Height changes from h! to h! dh
Chapter 3 Additional and Advanced Exercises 124. (a) S œ 6r# Ê dS œ 12r dr. We want kdSk Ÿ (2%) S Ê k12r drk Ÿ
12r# 100
Ê kdrk Ÿ
r 100
191
. The measurement of the
edge r must have an error less than 1%. #
3r dr ‰ (b) When V œ r$ , then dV œ 3r# dr. The accuracy of the volume is ˆ dV V (100%) œ Š r$ ‹ (100%) r ‰ œ ˆ 3r ‰ (dr)(100%) œ ˆ 3r ‰ ˆ 100 (100%) œ 3%
125. C œ 21r Ê r œ dV œ
#
C 21
, S œ 41 r # œ
C# 1
, and V œ
4 3
1 r$ œ
C$ 61 #
. It also follows that dr œ
" #1
dC, dS œ
2C 1
dC and
dC. Recall that C œ 10 cm and dC œ 0.4 cm. 0.2 ˆ drr ‰ (100%) œ ˆ 0.2 ‰ ˆ 2101 ‰ (100%) œ (.04)(100%) œ 4% (a) dr œ 0.4 21 œ 1 cm Ê 1 8 1 ‰ ˆ dS ‰ ˆ 8 ‰ ˆ 100 (b) dS œ 20 (100%) œ 8% 1 (0.4) œ 1 cm Ê S (100%) œ 1 C 21 #
10# 21 #
#
(0.4) œ
20 1#
‰ ˆ 20 ‰ 61 cm Ê ˆ dV V (100%) œ 1# Š 1000 ‹ (100%) œ 12%
126. Similar triangles yield
35 h
œ
(c) dV œ
Ê dh œ 120a# da œ
15 6 120 a#
Ê h œ 14 ft. The same triangles imply that 20h a œ 6a Ê h œ 120a" 6 " ‰ 2 ‰ ˆ „ 1"# ‰ œ ˆ "#! ‰ˆ „ "# da œ ˆ 120 œ „ 45 ¸ „ .0444 ft œ „ 0.53 inches. a# "
CHAPTER 3 ADDITIONAL AND ADVANCED EXERCISES 1. (a) sin 2) œ 2 sin ) cos ) Ê #
#
d d)
Ê cos 2) œ cos ) sin ) (b) cos 2) œ cos# ) sin# ) Ê
(sin 2)) œ d d)
d d)
(cos 2)) œ
(2 sin ) cos )) Ê 2 cos 2) œ 2[(sin ))(sin )) (cos ))(cos ))] d d)
acos# ) sin# )b Ê 2 sin 2) œ (2 cos ))(sin )) (2 sin ))(cos ))
Ê sin 2) œ cos ) sin ) sin ) cos ) Ê sin 2) œ 2 sin ) cos ) 2. The derivative of sin (x a) œ sin x cos a cos x sin a with respect to x is cos (x a) œ cos x cos a sin x sin a, which is also an identity. This principle does not apply to the equation x# 2x 8 œ 0, since x# 2x 8 œ 0 is not an identity: it holds for 2 values of x (2 and 4), but not for all x. 3. (a) f(x) œ cos x Ê f w (x) œ sin x Ê f ww (x) œ cos x, and g(x) œ a bx cx# Ê gw (x) œ b 2cx Ê gww (x) œ 2c; also, f(0) œ g(0) Ê cos (0) œ a Ê a œ 1; f w (0) œ gw (0) Ê sin (0) œ b Ê b œ 0; f ww (0) œ gww (0) Ê cos (0) œ 2c Ê c œ "# . Therefore, g(x) œ 1 "# x# . (b) f(x) œ sin (x a) Ê f w (x) œ cos (x a), and g(x) œ b sin x c cos x Ê gw (x) œ b cos x c sin x; also, f(0) œ g(0) Ê sin (a) œ b sin (0) c cos (0) Ê c œ sin a; f w (0) œ gw (0) Ê cos (a) œ b cos (0) c sin (0) Ê b œ cos a. Therefore, g(x) œ sin x cos a cos x sin a. (c) When f(x) œ cos x, f www (x) œ sin x and f Ð%Ñ (x) œ cos x; when g(x) œ 1 "# x# , gwww (x) œ 0 and gÐ%Ñ (x) œ 0. Thus f www (0) œ 0 œ gwww (0) so the third derivatives agree at x œ 0. However, the fourth derivatives do not agree since f Ð%Ñ (0) œ 1 but gÐ%Ñ (0) œ 0. In case (b), when f(x) œ sin (x a) and g(x) œ sin x cos a cos x sin a, notice that f(x) œ g(x) for all x, not just x œ 0. Since this is an identity, we have f ÐnÑ (x) œ gÐnÑ (x) for any x and any positive integer n.
4. (a) y œ sin x Ê yw œ cos x Ê yww œ sin x Ê yww y œ sin x sin x œ 0; y œ cos x Ê yw œ sin x Ê yww œ cos x Ê yww y œ cos x cos x œ 0; y œ a cos x b sin x Ê yw œ a sin x b cos x Ê yww œ a cos x b sin x Ê yww y œ (a cos x b sin x) (a cos x b sin x) œ 0 (b) y œ sin (2x) Ê yw œ 2 cos (2x) Ê yww œ 4 sin (2x) Ê yww 4y œ 4 sin (2x) 4 sin (2x) œ 0. Similarly, y œ cos (2x) and y œ a cos (2x) b sin (2x) satisfy the differential equation yw w 4y œ 0. In general, y œ cos (mx), y œ sin (mx) and y œ a cos (mx) b sin (mx) satisfy the differential equation yww m# y œ 0.
192
Chapter 3 Differentiation
5. If the circle (x h)# (y k)# œ a# and y œ x# 1 are tangent at ("ß #), then the slope of this tangent is m œ 2xk (1 2) œ 2 and the tangent line is y œ 2x. The line containing (hß k) and ("ß #) is perpendicular to ß
y œ 2x Ê
k2 h1
œ "# Ê h œ 5 2k Ê the location of the center is (5 2kß k). Also, (x h)# (y k)# œ a#
Ê x h (y k)yw œ 0 Ê 1 ayw b# (y k)yw w œ 0 Ê yww œ w
1 ay b ky w
#
. At the point ("ß #) we know
ww
y œ 2 from the tangent line and that y œ 2 from the parabola. Since the second derivatives are equal at ("ß #) we obtain 2 œ
1 (2) k#
#
Ê kœ
9 #
# . Then h œ 5 2k œ 4 Ê the circle is (x 4)# ˆy 9# ‰ œ a# . Since ("ß #)
lies on the circle we have that a œ
5È 5 2
.
6. The total revenue is the number of people times the price of the fare: r(x) œ xp œ x ˆ3
x ‰# , where 40 x ‰ ˆ x ‰ 40 3 40
" ‰ dr x ‰# x ‰ˆ dr ‘ 0 Ÿ x Ÿ 60. The marginal revenue is dx œ ˆ3 40 2x ˆ3 40 40 Ê dx œ ˆ3 2x 40 x x dr œ 3 ˆ3 40 ‰ ˆ1 40 ‰ . Then dx œ 0 Ê x œ 40 (since x œ 120 does not belong to the domain). When 40 people
are on the bus the marginal revenue is zero and the fare is p(40) œ ˆ3 7. (a) y œ uv Ê
dy dt
œ
du dt
x ‰# 40 ¹ x=40
œ $4.00.
v u dv dt œ (0.04u)v u(0.05v) œ 0.09uv œ 0.09y Ê the rate of growth of the total production is
9% per year. (b) If
œ 0.02u and
du dt
dv dt
œ 0.03v, then
dy dt
œ (0.02u)v (0.03v)u œ 0.01uv œ 0.01y, increasing at 1% per
year. 8. When x# y# œ 225, then yw œ xy . The tangent line to the balloon at (12ß 9) is y 9 œ Ê yœ
4 3
4 3
(x 12)
x 25. The top of the gondola is 15 8
œ 23 ft below the center of the balloon. The intersection of y œ 23 and y œ 43 x 25 is at the far right edge of the gondola Ê 23 œ Ê xœ
3 #
4 3
x 25
. Thus the gondola is 2x œ 3 ft wide.
9. Answers will vary. Here is one possibility.
10. s(t) œ 10 cos ˆt 14 ‰ Ê v(t) œ 10 (a) s(0) œ 10 cos ˆ 14 ‰ œ È
ds dt
œ 10 sin ˆt 14 ‰ Ê a(t) œ
dv dt
œ
d# s dt#
œ 10 cos ˆt 14 ‰
2
(b) Left: 10, Right: 10 (c) Solving 10 cos ˆt 14 ‰ œ 10 Ê cos ˆt 14 ‰ œ 1 Ê t œ 341 when the particle is farthest to the left. Solving 10 cos ˆt 14 ‰ œ 10 Ê cos ˆt 14 ‰ œ 1 Ê t œ 14 , but t 0 Ê t œ 21 41 œ 741 when the particle is farthest to the right. Thus, v ˆ 341 ‰ œ 0, v ˆ 741 ‰ œ 0, a ˆ 341 ‰ œ 10, and a ˆ 741 ‰ œ 10. (d) Solving 10 cos ˆt 14 ‰ œ 0 Ê t œ
1 4
Ê v ˆ 14 ‰ œ 10, ¸v ˆ 14 ‰¸ œ 10 and a ˆ 14 ‰ œ !.
Chapter 3 Additional and Advanced Exercises 11. (a) s(t) œ 64t 16t# Ê v(t) œ
ds dt
193
œ 64 32t œ 32(2 t). The maximum height is reached when v(t) œ 0
Ê t œ 2 sec. The velocity when it leaves the hand is v(0) œ 64 ft/sec. (b) s(t) œ 64t 2.6t# Ê v(t) œ ds dt œ 64 5.2t. The maximum height is reached when v(t) œ 0 Ê t ¸ 12.31 sec. The maximum height is about s(12.31) œ 393.85 ft. 12. s" œ 3t$ 12t# 18t 5 and s# œ t$ 9t# 12t Ê v" œ 9t# 24t 18 and v# œ 3t# 18t 12; v" œ v# Ê 9t# 24t 18 œ 3t# 18t 12 Ê 2t# 7t 5 œ 0 Ê (t 1)(2t 5) œ 0 Ê t œ 1 sec and t œ 2.5 sec. 13. m av# v#! b œ k ax#! x# b Ê m ˆ2v substituting
dx dt
œv Ê m
dv dt
dv ‰ dt
œ k ˆ2x
dx ‰ dt
Ê m
dv dt
2x ‰ œ k ˆ 2v
dx dt
Ê m
dv dt
œ kx ˆ "v ‰
dx dt
œ 2At B Ê v ˆ t" # t# ‰ œ 2A ˆ t" # t# ‰ B œ A at" t# b B is the
instantaneous velocity at the midpoint. The average velocity over the time interval is vav œ œ
Bt# Cb aAt#" t# t"
. Then
œ kx, as claimed.
14. (a) x œ At# Bt C on ct" ß t# d Ê v œ aAt##
dx dt
Bt" Cb
œ
at# t" b cA at# t" b Bd t# t" #
?x ?t
œ A at# t" b B.
(b) On the graph of the parabola x œ At Bt C, the slope of the curve at the midpoint of the interval ct" ß t# d is the same as the average slope of the curve over the interval. 15. (a) To be continuous at x œ 1 requires that lim c sin x œ lim b (mx b) Ê 0 œ m1 b Ê m œ 1b ; xÄ1 xÄ1 (b) If yw œ œ
cos x, x 1 is differentiable at x œ 1, then lim c cos x œ m Ê m œ 1 and b œ 1. xÄ1 m, x 1
16. faxb is continuous at ! because lim
xÄ!
œ
x ‰ ˆ 1 cos x ‰ lim ˆ 1 xcos # 1 cos x xÄ!
œ
" cos x x
œ ! œ fa!b. f w (0) œ lim
f(x) f(0) x0
ˆ 1 "cos x ‰
w
# lim ˆ sinx x ‰ xÄ!
xÄ!
œ
" #
œ lim
xÄ!
1 cos x 0 x
x
. Therefore f (0) exists with value
" #
.
17. (a) For all a, b and for all x Á 2, f is differentiable at x. Next, f differentiable at x œ 2 Ê f continuous at x œ 2 Ê lim c f(x) œ f(2) Ê 2a œ 4a 2b 3 Ê 2a 2b 3 œ 0. Also, f differentiable at x Á 2 xÄ2
Ê f w (x) œ œ
a, x 2 . In order that f w (2) exist we must have a œ 2a(2) b Ê a œ 4a b Ê 3a œ b. 2ax b, x 2
Then 2a 2b 3 œ 0 and 3a œ b Ê a œ
3 4
and b œ
9 4
.
(b) For x #, the graph of f is a straight line having a slope of
$ %
and passing through the origin; for x #, the graph of f
is a parabola. At x œ #, the value of the y-coordinate on the parabola is
$ #
which matches the y-coordinate of the point
on the straight line at x œ #. In addition, the slope of the parabola at the match up point is
$ %
which is equal to the
slope of the straight line. Therefore, since the graph is differentiable at the match up point, the graph is smooth there. 18. (a) For any a, b and for any x Á 1, g is differentiable at x. Next, g differentiable at x œ 1 Ê g continuous at x œ 1 Ê lim b g(x) œ g(1) Ê a 1 2b œ a b Ê b œ 1. Also, g differentiable at x Á 1 x Ä "
Ê gw (x) œ œ
a, x 1 . In order that gw (1) exist we must have a œ 3a(1)# 1 Ê a œ 3a 1 3ax# 1, x 1
Ê a œ "# . (b) For x Ÿ ", the graph of f is a straight line having a slope of
" #
and a y-intercept of ". For x ", the graph of f is
a parabola. At x œ ", the value of the y-coordinate on the parabola is
$ #
which matches the y-coordinate of the point
on the straight line at x œ ". In addition, the slope of the parabola at the match up point is "# which is equal to the slope of the straight line. Therefore, since the graph is differentiable at the match up point, the graph is smooth there. 19. f odd Ê f(x) œ f(x) Ê
d dx
(f(x)) œ
d dx
(f(x)) Ê f w (x)(1) œ f w (x) Ê f w (x) œ f w (x) Ê f w is even.
194
Chapter 3 Differentiation
20. f even Ê f(x) œ f(x) Ê
d dx
(f(x)) œ
d dx
(f(x)) Ê f w (x)(1) œ f w (x) Ê f w (x) œ f w (x) Ê f w is odd.
21. Let h(x) œ (fg)(x) œ f(x) g(x) Ê hw (x) œ x lim Äx œ x lim Äx œ
f(x) g(x) f(x) g(x! ) f(x) g(x! ) f(x! ) g(x! ) x x!
!
g(x! ) f(x! ) x lim ’ g(x)x x! “ Ä x!
!
h(x) h(x! ) x x!
œ x lim Äx
!
f(x) g(x) f(x! ) g(x! ) x x!
f(x! ) !) œ x lim ’f(x) ’ g(x)x xg(x ““ x lim ’g(x! ) ’ f(x)x x! ““ Äx Äx ! !
w
g(x! ) f (x! ) œ 0 †
g(x! ) lim ’ g(x)x x! “ x Ä x!
!
w
g(x! ) f (x! ) œ g(x! ) f w (x! ), if g is
continuous at x! . Therefore (fg)(x) is differentiable at x! if f(x! ) œ 0, and (fg)w (x! ) œ g(x! ) f w (x! ). 22. From Exercise 21 we have that fg is differentiable at 0 if f is differentiable at 0, f(0) œ 0 and g is continuous at 0. (a) If f(x) œ sin x and g(x) œ kxk , then kxk sin x is differentiable because f w (0) œ cos (0) œ 1, f(0) œ sin (0) œ 0 and g(x) œ kxk is continuous at x œ 0. (b) If f(x) œ sin x and g(x) œ x#Î$ , then x#Î$ sin x is differentiable because f w (0) œ cos (0) œ 1, f(0) œ sin (0) œ 0 and g(x) œ x#Î$ is continuous at x œ 0. (c) If f(x) œ 1 cos x and g(x) œ $Èx, then $Èx (1 cos x) is differentiable because f w (0) œ sin (0) œ 0, f(0) œ 1 cos (0) œ 0 and g(x) œ x"Î$ is continuous at x œ 0. (d) If f(x) œ x and g(x) œ x sin ˆ "x ‰ , then x# sin ˆ x" ‰ is differentiable because f w (0) œ 1, f(0) œ 0 and sin ˆ "x ‰
lim x sin ˆ "x ‰ œ lim
xÄ!
" x
xÄ!
œ lim
tÄ_
sin t t
œ 0 (so g is continuous at x œ 0).
23. If f(x) œ x and g(x) œ x sin ˆ "x ‰ , then x# sin ˆ x" ‰ is differentiable at x œ 0 because f w (0) œ 1, f(0) œ 0 and lim x sin ˆ "x ‰ œ lim
xÄ!
sin ˆ "x ‰ " x
xÄ!
œ lim
tÄ_
sin t t
œ 0 (so g is continuous at x œ 0). In fact, from Exercise 21,
h (0) œ g(0) f (0) œ 0. However, for x Á 0, hw (x) œ x# cos ˆ "x ‰‘ ˆ x"# ‰ 2x sin ˆ x" ‰ . But lim hw (x) œ lim cos ˆ "x ‰ 2x sin ˆ x" ‰‘ does not exist because cos ˆ x" ‰ has no limit as x Ä 0. Therefore, w
w
xÄ!
xÄ!
the derivative is not continuous at x œ 0 because it has no limit there. 24. From the given conditions we have f(x h) œ f(x) f(h), f(h) 1 œ hg(h) and lim g(h) œ 1. Therefore, hÄ!
w
f (x) œ
lim f(x h)h f(x) hÄ! w
œ
lim f(x) f(h)h f(x) hÄ!
œ
lim f(x) ’ f(h)h 1 “ hÄ!
œ f(x) ’ lim g(h)“ œ f(x) † 1 œ f(x)
Ê f w (x) œ f(x) and f axbexists at every value of x.
hÄ!
25. Step 1: The formula holds for n œ 2 (a single product) since y œ u" u# Ê
dy dx
œ
du" dx
u# u"
du# dx
.
Step 2: Assume the formula holds for n œ k: y œ u" u# âuk Ê
du# duk dx u$ âuk á u" u# âuk-1 dx d(u" u# âuk ) If y œ u" u# âuk ukb1 œ au" u# âuk b ukb1 , then dy ukb1 u" u# âuk dudxkb1 dx œ dx dukb1 du# duk ‰ " œ ˆ du dx u# u$ âuk u" dx u$ âuk â u" u# âukc1 dx ukb1 u" u# âuk dx dukb1 du# duk " œ du dx u# u$ âukb1 u" dx u$ â ukb1 â u" u# âukc1 dx ukb1 u" u# âuk dx . dy dx
œ
du" dx
u# u$ âuk u"
.
Thus the original formula holds for n œ (k1) whenever it holds for n œ k. 26. Recall ˆ mk ‰ œ œ
m! m! m! m! ˆm‰ ˆm‰ ˆ m ‰ k! (m k)! . Then 1 œ 1! (m 1)! œ m and k k 1 œ k! (m k)! (k 1)! (m k 1)! m! (k 1) m! (m k) (m 1)! ˆm 1‰ œ (k m!1)!(m(m 1)k)! œ (k 1)! ((m (k 1)! (m k)!
1) (k 1))! œ k 1 . Now, we prove
Leibniz's rule by mathematical induction. Step 1: If n œ 1, then
d(uv) dv du dx œ u dx v dx . Assume that the statement is true for n œ k, that is: " # k k# k" d (uv) du d u dv dk v ˆk‰ d u d v ˆ k ‰ du d v dxk œ dxk v k dxk" dx 2 dxk# dx# á k 1 dv dxk" u dxk . kb" k k" k (uv) d dk u dv dk" u d# v ddxk"u v ddxuk dv ‘ If n œ k 1, then d dx(uv) œ dx Š d dx k" k ‹ œ dx ’k dxk dx k dxk" dx# “ k
Step 2:
k
Chapter 3 Additional and Advanced Exercises ’ˆ k2 ‰ du dx
dk" u d# v dxk" dx#
ˆ k2 ‰
kb"
dk# u d$ v dxk# dx$ “
á ’ˆ k k 1 ‰
k"
d# u dk" v dx# dxk"
dv d u‘ d u d u dv dxk u dxk" œ dxk" v (k 1) dxk dx k kb" k" du d v d v d u ˆ k k 1 ‰ ˆ kk ‰‘ dx dxk u dxk" œ dxk" v (k k kb" dv d v ˆ k k 1 ‰ du dx dxk u dxk" . k
ˆ k1 ‰
k
1)
ˆ kk 1 ‰
du dk u dx dxk
v“
k" # ˆ k2 ‰‘ ddxk"u ddxv# á dk u dv dk" u d# v ˆ k 2 1 ‰ dx k" dxk dx dx#
á
Therefore the formula (c) holds for n œ (k 1) whenever it holds for n œ k. 27. (a) T# œ (b) T# œ
41 # L g #
41 L g
ÊLœ
T# g 41 #
ÊTœ
#1 È L; Èg
ÊLœ
a1 sec# ba32.2 ft/sec# b 41 #
dT œ
#1 Èg
†
" dL #È L
Ê L ¸ 0.8156 ft œ
1 ÈLg dL;
dT œ
1 Èa!Þ)"&' ftba32.2 ft/sec# b a!Þ!"
ftb ¸ 0.00613 sec.
(c) Since there are 86,400 sec in a day, we have a0.00613 secba86,400 sec/dayb ¸ 529.6 sec/day, or 8.83 min/day; the clock will lose about 8.83 min/day. 28. v œ s$ Ê
dv dt
# œ $s# ds dt œ ka's b Ê
ds dt
œ #k. If s! œ the initial length of the cube's side, then s" œ s! #k
Ê #k œ s! s" . Let t œ the time it will take the ice cube to melt. Now, t œ œ
" "Î$ " ˆ $% ‰
¸ "" hr.
s! #k
œ
s! s ! s "
œ
av! b"Î$ "Î$ av! b ˆ $% v! ‰ "Î$
195
196
Chapter 3 Differentiation
NOTES:
CHAPTER 4 APPLICATIONS OF DERIVATIVES 4.1 EXTREME VALUES OF FUNCTIONS 1. An absolute minimum at x œ c# , an absolute maximum at x œ b. Theorem 1 guarantees the existence of such extreme values because h is continuous on [aß b]. 2. An absolute minimum at x œ b, an absolute maximum at x œ c. Theorem 1 guarantees the existence of such extreme values because f is continuous on [aß b]. 3. No absolute minimum. An absolute maximum at x œ c. Since the function's domain is an open interval, the function does not satisfy the hypotheses of Theorem 1 and need not have absolute extreme values. 4. No absolute extrema. The function is neither continuous nor defined on a closed interval, so it need not fulfill the conclusions of Theorem 1. 5. An absolute minimum at x œ a and an absolute maximum at x œ c. Note that y œ g(x) is not continuous but still has extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when the hypothesis is not satisfied, absolute extrema may or may not occur. 6. Absolute minimum at x œ c and an absolute maximum at x œ a. Note that y œ g(x) is not continuous but still has absolute extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when the hypothesis is not satisfied, absolute extrema may or may not occur. 7. Local minimum at a"ß !b, local maximum at a"ß !b 8. Minima at a#ß !b and a#ß !b, maximum at a!ß #b 9. Maximum at a!ß &b. Note that there is no minimum since the endpoint a#ß !b is excluded from the graph. 10. Local maximum at a$ß !b, local minimum at a#ß !b, maximum at a"ß #b, minimum at a!ß "b 11. Graph (c), since this the only graph that has positive slope at c. 12. Graph (b), since this is the only graph that represents a differentiable function at a and b and has negative slope at c. 13. Graph (d), since this is the only graph representing a funtion that is differentiable at b but not at a. 14. Graph (a), since this is the only graph that represents a function that is not differentiable at a or b.
198
Chapter 4 Applications of Derivatives
15. f(x) œ
2 3
x 5 Ê f w (x) œ
f(2) œ
19 3 ,
2 3
Ê no critical points;
f(3) œ 3 Ê the absolute maximum
is 3 at x œ 3 and the absolute minimum is 19 3 at x œ 2
16. f(x) œ x 4 Ê f w (x) œ 1 Ê no critical points; f(4) œ 0, f(1) œ 5 Ê the absolute maximum is 0 at x œ 4 and the absolute minimum is 5 at x œ "
17. f(x) œ x# 1 Ê f w (x) œ 2x Ê a critical point at x œ 0; f(1) œ 0, f(0) œ 1, f(2) œ 3 Ê the absolute maximum is 3 at x œ 2 and the absolute minimum is 1 at x œ 0
18. f(x) œ % x# Ê f w (x) œ 2x Ê a critical point at x œ 0; f(3) œ 5, f(0) œ 4, f(1) œ 3 Ê the absolute maximum is 4 at x œ 0 and the absolute minimum is 5 at x œ 3
19. F(x) œ x"# œ x# Ê Fw (x) œ 2x$ œ
2 x$
, however
x œ 0 is not a critical point since 0 is not in the domain; F(0.5) œ 4, F(2) œ 0.25 Ê the absolute maximum is 0.25 at x œ 2 and the absolute minimum is 4 at x œ 0.5
Section 4.1 Extreme Values of Functions 20. F(x) œ "x œ x" Ê Fw (x) œ x# œ
" x#
, however
x œ 0 is not a critical point since 0 is not in the domain; F(2) œ "# , F(1) œ 1 Ê the absolute maximum is 1 at x œ 1 and the absolute minimum is
21. h(x) œ $Èx œ x"Î$ Ê hw (x) œ
" 3
" #
at x œ 2
x#Î$ Ê a critical point
at x œ 0; h(1) œ 1, h(0) œ 0, h(8) œ 2 Ê the absolute maximum is 2 at x œ 8 and the absolute minimum is 1 at x œ 1
22. h(x) œ 3x#Î$ Ê hw (x) œ #x"Î$ Ê a critical point at x œ 0; h(1) œ 3, h(0) œ 0, h(1) œ 3 Ê the absolute maximum is 0 at x œ 0 and the absolute minimum is 3 at x œ 1 and at x œ 1
23. g(x) œ È4 x# œ a4 x# b Ê gw (x) œ
" #
a4 x# b
"Î#
"Î#
(2x) œ
x È 4 x#
Ê critical points at x œ 2 and x œ 0, but not at x œ 2 because 2 is not in the domain; g(2) œ 0, g(0) œ 2, g(1) œ È3 Ê the absolute maximum is 2 at x œ 0 and the absolute minimum is 0 at x œ 2 24. g(x) œ È5 x# œ a& x# b a5 x# b (2x) x "‰ w ˆ Ê g (x) œ # œ È # Ê critical points at x œ È5 "Î#
"Î#
&x
and x œ 0, but not at x œ È5 because È5 is not in the domain; f ŠÈ5‹ œ 0, f(0) œ È5 Ê the absolute maximum is 0 at x œ È5 and the absolute minimum is È5 at x œ 0
25. f()) œ sin ) Ê f w ()) œ cos ) Ê ) œ 1 #
1 #
is a critical point,
but ) œ is not a critical point because #1 is not interior the domain; f ˆ #1 ‰ œ 1, f ˆ 1# ‰ œ 1, f ˆ 561 ‰ œ "# Ê the absolute maximum is 1 at ) œ 1# and the absolute minimum is 1 at ) œ #1
to
199
200
Chapter 4 Applications of Derivatives
26. f()) œ tan ) Ê f w ()) œ sec# ) Ê f has no critical points in 1‰ ˆ 1 3 ß 4 . The extreme values therefore occur at the ‰ œ È3 and f ˆ 14 ‰ œ 1 Ê the absolute endpoints: f ˆ 1 3 maximum is 1 at ) œ 14 and the absolute minimum is È3 at ) œ 1 3
27. g(x) œ csc x Ê gw (x) œ (csc x)(cot x) Ê a critical point at x œ 1# ; g ˆ 13 ‰ œ È23 , g ˆ 1# ‰ œ 1, g ˆ 231 ‰ œ È23 Ê the absolute maximum is
at x œ
2 È3
absolute minimum is 1 at x œ
1 3
and x œ
21 3 ,
and the
1 #
28. g(x) œ sec x Ê gw (x) œ (sec x)(tan x) Ê a critical point at x œ 0; g ˆ 13 ‰ œ 2, g(0) œ 1, g ˆ 16 ‰ œ È23 Ê the absolute maximum is 2 at x œ 13 and the absolute minimum is 1 at x œ 0
29. f(t) œ 2 ktk œ # Èt# œ # at# b Ê f w (t) œ "# at# b
"Î#
"Î#
(2t) œ Èt # œ kttk t
Ê a critical point at t œ 0; f(1) œ 1, f(0) œ 2, f(3) œ 1 Ê the absolute maximum is 2 at t œ 0 and the absolute minimum is 1 at t œ 3
30. f(t) œ kt 5k œ È(t 5)# œ a(t 5)# b œ
" #
a(t 5)# b
"Î#
(2(t 5)) œ
t5 È(t 5)#
"Î#
œ
Ê f w (t) t5 kt 5 k
Ê a critical point at t œ 5; f(4) œ 1, f(5) œ 0, f(7) œ 2 Ê the absolute maximum is 2 at t œ 7 and the absolute minimum is 0 at t œ 5 31. f(x) œ x%Î$ Ê f w (x) œ
4 3
x"Î$ Ê a critical point at x œ 0; f(1) œ 1, f(0) œ 0, f(8) œ 16 Ê the absolute
maximum is 16 at x œ 8 and the absolute minimum is 0 at x œ 0 32. f(x) œ x&Î$ Ê f w (x) œ
5 3
x#Î$ Ê a critical point at x œ 0; f(1) œ 1, f(0) œ 0, f(8) œ 32 Ê the absolute
maximum is 32 at x œ 8 and the absolute minimum is 1 at x œ 1 33. g()) œ )$Î& Ê gw ()) œ
3 5
)#Î& Ê a critical point at ) œ 0; g(32) œ 8, g(0) œ 0, g(1) œ 1 Ê the absolute
maximum is 1 at ) œ 1 and the absolute minimum is 8 at ) œ 32
Section 4.1 Extreme Values of Functions 34. h()) œ 3)#Î$ Ê hw ()) œ 2)"Î$ Ê a critical point at ) œ 0; h(27) œ 27, h(0) œ 0, h(8) œ 12 Ê the absolute maximum is 27 at ) œ 27 and the absolute minimum is 0 at ) œ 0 35. Minimum value is 1 at x œ #.
36. To find the exact values, note that yw œ $x# #, which is zero when x œ „ É #$ . Local maximum at ŠÉ #$ ß %
%È ' * ‹
¸ a!Þ)"'ß &Þ!)*b; local
minimum at ŠÉ #$ ß %
%È ' * ‹
¸ a!Þ)"'ß #Þ*""b
37. To find the exact values, note that that yw œ $x# #x ) œ a$x %bax #b, which is zero when x œ # or x œ %$ . ‰ Local maximum at a#ß "(b; local minimum at ˆ %$ ß %" #(
38. Note that yw œ $x# 'x $ œ $ax "b# , which is zero at x œ ". The graph shows that the function assumes lower values to the left and higher values to the right of this point, so the function has no local or global extreme values.
39. Minimum value is 0 when x œ " or x œ ".
201
202
Chapter 4 Applications of Derivatives
40. The minimum value is 1 at x œ !.
41. The actual graph of the function has asymptotes at x œ „ ", so there are no extrema near these values. (This is an example of grapher failure.) There is a local minimum at a!ß "b.
42. Maximum value is 2 at x œ "; minimum value is 0 at x œ " and x œ $.
" # at x œ "à "# as x œ ".
43. Maximum value is minimum value is
" # at x œ 0à "# as x œ 2.
44. Maximum value is minimum value is
Section 4.1 Extreme Values of Functions 45. yw œ x#Î$ a"b #$ x"Î$ ax #b œ crit. pt. x œ %& xœ!
derivative ! undefined
&x % $ x $È
extremum local max local min
46. yw œ x#Î$ a#xb #$ x"Î$ ax# %b œ crit. pt. x œ " xœ! xœ"
derivative ! undefined !
" a #xb a"bÈ% #È % x # x# a% x# b % #x # œÈ È % x# % x#
crit. pt. x œ # x œ È # x œ È# xœ#
)x# ) $ x $È
extremum minimum local max minimum
47. yw œ x œ
value "# "Î$ œ "Þ!$% #& "! 0
derivative undefined ! ! undefined
value $ 0 $
x#
extremum local max minimum maximum local min
value ! # # !
48. yw œ x# #È$" x a 1b #xÈ$ x œ
x# a%xba$ xb #È $ x
crit. pt. xœ0 x œ "# & xœ$
_5x# "#x #È $ x
derivative ! ! undefined
#, 49. yw œ œ ", crit. pt. xœ"
œ
extremum minimum local max minimum
value ! "%% "Î# ¸ %Þ%'# "#& "& !
extremum minimum
value #
x" x"
derivative undefined
203
204
Chapter 4 Applications of Derivatives
", x ! 50. yw œ œ # #x, x ! crit. pt. xœ! xœ"
51. yw œ œ
derivative undefined !
2x 2, 2x 6,
crit. pt. x œ 1 xœ1 xœ3
extremum local min local max
value $ %
x1 x1
derivative ! undefined !
extremum maximum local min maximum
value 5 1 5
"% x# "# x "& % , xŸ" x$ 'x# )x, x" w w # if x ", and limc f a" hb œ ". Also, f axb œ $x "#x ) if x ", and
52. We begin by determining whether f w axb is defined at x œ ", where faxb œ œ Clearly, f w axb œ "# x
" #
hÄ!
limb f w a" hb œ ". Since f is continuous at x œ ", we have that f w a"b œ ". Thus,
hÄ!
f w axb œ œ
"# x "# , $x "#x ) , #
Note that "# x But #
#È $ $
crit. pt. x œ " x ¸ $Þ"&&
" #
xŸ" x"
œ ! when x œ ", and $x# "#x ) œ ! when x œ
¸ !Þ)%& ", so the critical points occur at x œ " and x œ derivative ! !
extremum local max local min
È "# „ È"## %a$ba)b œ "# „' %) #a$b È # # $ $ ¸ $Þ"&&.
œ#„
#È$ $ .
value 4 ¸ $Þ!(*
53. (a) No, since f w axb œ #$ ax #b"Î$ , which is undefined at x œ #.
(b) The derivative is defined and nonzero for all x Á #. Also, fa#b œ ! and faxb ! for all x Á #. (c) No, faxb need not have a global maximum because its domain is all real numbers. Any restriction of f to a closed interval of the form Òa, bÓ would have both a maximum value and minimum value on the interval. (d) The answers are the same as (a) and (b) with 2 replaced by a. x$ *x, x Ÿ $ or ! Ÿ x $ $x$ *, x $ or ! x $ . Therefore, f w axb œ œ . $ x *x, $ x ! or x $ $x$ *, $ x ! or x $ (a) No, since the left- and right-hand derivatives at x œ !, are * and *, respectively. (b) No, since the left- and right-hand derivatives at x œ $, are ") and "), respectively.
54. Note that faxb œ œ
Section 4.1 Extreme Values of Functions
205
(c) No, since the left- and right-hand derivatives at x œ $, are ") and "), respectively. (d) The critical points occur when f w axb œ ! (at x œ „ È$) and when f w axb is undefined (at x œ ! and x œ „ $). The minimum value is ! at x œ $, at x œ !, and at x œ $; local maxima occur at ŠÈ$ß 'È$‹ and ŠÈ$ß 'È$‹. 55.
(a) The construction cost is Caxb œ !Þ$È"' x# !Þ#a* xb million dollars, where ! Ÿ x Ÿ * miles. The following is a graph of Caxb.
Solving Cw axb œ
!Þ$x È"' x#
!Þ# œ ! gives x œ „
)È & &
¸ „ $Þ&) miles, but only x œ $Þ&) miles is a critical point is È
the specified domain. Evaluating the costs at the critical and endpoints gives Ca!b œ $3 million, CŠ ) & & ‹ ¸ $2.694 million, and Ca*b ¸ $2.955 million. Thereform, to minimize the cost of construction, the pipeline should be placed from the docking facility to point B, 3.58 miles along the shore from point A, and then along the shore from B to the refinery. (b) If the per mile cost of underwater construction is p, then Caxb œ pÈ"' x# !Þ#a* xb and Cw axb œ È !Þ$x x# !Þ# œ ! gives xc œ Èp#!Þ) , which minimizes the construction cost provided xc Ÿ *. The value !Þ!% "'
of p that gives xc œ * miles is !Þ#"))'%. Consequently, if the underwater construction costs $218,864 per mile or less, then running the pipeline along a straight line directly from the docking facility to the refinery will minimize the cost of construction. In theory, p would have to infinite to justify running the pipe directly from the docking facility to point A (i.e., for xc to be zero). For all values of p !Þ#"))'% there is always an xc − Ð!ß *Ñ that will give a minimum value for C. This is proved by looking at Cww axc b œ
"'p a"' x#c b$Î#
which is always positive for p !.
56. There are two options to consider. The first is to build a new road straight from village A to Village B. THe second is to build a new highway segment from village A to the Old Road, reconstruct a segment of Old Road, and build a new highway segment from Old Road to village B, as shown in the figure. The cost of the first option is C" œ !Þ&a"&!b million dollars œ 75 million dollars.
206
Chapter 4 Applications of Derivatives
The construction cost for the second option is C# axb œ !Þ&Š#È#&!! x# ‹ !Þ$a"&! #xb million dollars for ! Ÿ x Ÿ (& miles. The following is a graph of C# axb.
Solving Cw# axb œ
x È#&!! x#
!Þ' œ ! give x œ „ $(Þ& miles, but only x œ $(Þ& miles is in the specified domain. In
summary, C" œ $75 million, C# a!b œ $95 million, C# a$(Þ&b œ $85 million, and C# a(&b œ $90.139 million. Consequently, a new road straight from village A to village B is the least expensive option. 57.
The length of pipeline is Laxb œ È% x# É#& a"! xb# for ! Ÿ x Ÿ "!. The following is a graph of Laxb.
Setting the derivative of Laxb equal to zero gives Lw axb œ "! x É#& a"! xb#
x È % x#
a"! xb É#& a"! xb#
œ !. Note that
x È % x#
œ cos )A and
œ cos )B , therefore, Lw axb œ ! when cos )A œ cos )B , or )A œ )B and ˜ACP is similar to ˜BDP. Use
simple proportions to determine x as follows:
x 2
œ
"!x &
Êxœ
#! (
¸ #Þ)&( miles along the coast from town A to town B.
If the two towns were on opposite sides of the river, the obvious solution would be to place the pump station on a straight line (the shortest distance) between two towns, again forcing )A œ )B . The shortest length of pipe is the same regardless of whether the towns are on thee same or opposite sides of the river.
Section 4.1 Extreme Values of Functions
207
58.
(a) The length of guy wire is Laxb œ È*!! x# É#&!! a"&! xb# for ! Ÿ x Ÿ "&!. The following is a graph of Laxb.
Setting Lw axb equal to zero gives Lw axb œ a"&! xb É#&!! a"&! xb#
x È*!! x#
a"&! xb É#&!! a"&! xb#
œ !. Note that
x È*!! x#
œ cos )A and
œ cos )B . Therefore, Lw axb œ ! when cos )A œ cos )B , or )A œ )B and ˜ACE is similar to ˜ABD.
Use simple proportions to determine x:
x $!
œ
"&! x &!
Êxœ
##& %
œ &'Þ#& feet.
(b) If the heights of the towers are hB and hC , and the horizontal distance between them is s, then Laxb œ Éh#C x# Éh#B as xb# and Lw axb œ as x b É h B as x b #
x Éh#C x#
as x b É h B as x b #
. However,
x Éh#C x#
œ cos )G and
œ cos )B . Therefore, Lw axb œ ! when cos )C œ cos )B , or )C œ )B and ˜ACE is similar to ˜ABD.
Simple proportions can again be used to determine the optimum x: hxc œ
sx hB
Ê x œ Š hB hc hc ‹s.
59. (a) Vaxb œ "'!x # %x$ Vw axb œ "'! "!%x "#x# œ %ax #ba$x #!b The only critical point in the interval a!ß &b is at x œ #. The maximum value of Vaxb is 144 at x œ #. (b) The largest possible volume of the box is 144 cubic units, and it occurs when x œ # units. 60. (a) Pw axb œ # #!!x# The only critical point in the interval a!ß _b is at x œ "!. The minimum value of Paxb is %! at x œ "!. (b) The smallest possible perimeter of the rectangel is 40 units and it occurs at x œ "! units which makes the rectangle a 10 by 10 square. 61. Let x represent the length of the base and È#& x# the height of the triangle. The area of the triangle is represented by # Aaxb œ x È#& x# where ! Ÿ x Ÿ &. Consequently, solving Aw axb œ ! Ê #& #x œ ! Ê x œ & . Since #È#& x#
#
Aa!b œ Aa&b œ !, Aaxb is maximized at x œ
& È# .
The largest possible area is AŠ È ‹ œ
È#
#& %
cm# .
208
Chapter 4 Applications of Derivatives
62. (a) From the diagram the perimeter P œ #x #1r œ %!! Ê x œ #!! 1r. The area A is 2rx Ê Aarb œ %!!r #1r# where ! Ÿ r Ÿ #!! 1 . (b) Aw arb œ %!! %1r so the only critical point is r œ
"!! 1 .
Since Aarb œ ! if r œ ! and x œ #!! 1r œ !, the values r œ "!! 1 ¸ 31.83 m and x œ "!! m maximize the area over the interval ! Ÿ r Ÿ 63. s œ "# gt# v! t s! Ê
ds dt
#!! 1 .
œ gt v! œ ! Ê t œ
2
Thus sŠ vg! ‹ œ "# gŠ vg! ‹ v0 Š vg! ‹ s0 œ 64.
Now satb œ s0 Í tˆ gt2 v0 ‰ œ 0 Í t œ 0 or t œ
s0 s0 is the maximum height over the interval 0 Ÿ t Ÿ
œ ! Ê tan t œ " Ê t œ never negative) Ê the peak current is #È# amps. dI dt
œ #sin t #cos t, solving
v!2 2g
v! g.
dI dt
65. Yes, since f(x) œ kxk œ Èx# œ ax# b
"Î#
Ê f w (x) œ
" #
ax# b
1 %
2v0 g . 2v0 g .
n1 where n is a nonnegative integer (in this exercise t is
"Î#
(2x) œ
x ax# b"Î#
œ
x kx k
is not defined at x œ 0. Thus it
is not required that f w be zero at a local extreme point since f w may be undefined there. 66. If f(c) is a local maximum value of f, then f(x) Ÿ f(c) for all x in some open interval (aß b) containing c. Since f is even, f(x) œ f(x) Ÿ f(c) œ f(c) for all x in the open interval (bß a) containing c. That is, f assumes a local maximum at the point c. This is also clear from the graph of f because the graph of an even function is symmetric about the y-axis. 67. If g(c) is a local minimum value of g, then g(x) g(c) for all x in some open interval (aß b) containing c. Since g is odd, g(x) œ g(x) Ÿ g(c) œ g(c) for all x in the open interval (bß a) containing c. That is, g assumes a local maximum at the point c. This is also clear from the graph of g because the graph of an odd function is symmetric about the origin. 68. If there are no boundary points or critical points the function will have no extreme values in its domain. Such functions do indeed exist, for example f(x) œ x for _ x _. (Any other linear function f(x) œ mx b with m Á 0 will do as well.) 69. (a) f w axb œ $ax# #bx c is a quadratic, so it can have 0, 1, or 2 zeros, which would be the critical points of f. The function faxb œ x$ $x has two critical points at x œ " and x œ ". The function faxb œ x$ " has one critical point at x œ !Þ The function faxb œ x$ x has no critical points.
(b) The function can have either two local extreme values or no extreme values. (If there is only one critical point, the cubic function has no extreme values.)
Section 4.1 Extreme Values of Functions
209
70. (a)
fa!b œ ! is not a local extreme value because in any open interval containing x œ !, there are infinitely many points where faxb œ " and where faxb œ ". (b) One possible answer, on the interval Ò!ß "Ó: " a" xbcos "x , !Ÿx" faxb œ œ !, x œ " This function has no local extreme value at x œ ". Note that it is continuous on Ò!ß "Ó. 71. Maximum value is 11 at x œ &; minimum value is 5 on the interval Ò$ß #Ó; local maximum at a&ß *b
72. Maximum value is 4 on the interval Ò&ß (Ó; minimum value is % on the interval Ò#ß "Ó.
73. Maximum value is & on the interval Ò$ß _Ñ; minimum value is & on the interval Ð_ß #Ó.
210
Chapter 4 Applications of Derivatives
74. Minimum value is 4 on the interval Ò"ß $Ó
75-80. Example CAS commands: Maple: with(student): f := x -> x^4 - 8*x^2 + 4*x + 2; domain := x=-20/25..64/25; plot( f(x), domain, color=black, title="Section 4.1 #75(a)" ); Df := D(f); plot( Df(x), domain, color=black, title="Section 4.1 # 75(b)" ) StatPt := fsolve( Df(x)=0, domain ) SingPt := NULL; EndPt := op(rhs(domain)); Pts :=evalf([EndPt,StatPt,SingPt]); Values := [seq( f(x), x=Pts )]; Maximum value is 2.7608 and occurs at x=2.56 (right endpoint). %
Minimum value $ is -6.2680 and occurs at x=1.86081 (singular point). Mathematica: (functions may vary) (see section 2.5 re. RealsOnly ): 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 ); 89-92. Example CAS commands: Maple: with( Student[Calculus1] ); f := x -> sin(x); a := 0; b := Pi; plot( f(x), x=a..b, title="#23(a) (Section 5.1)" ); 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 ); 83-92. 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]
Section 5.4 The Fundamental Theorem of Calculus {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 ; 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
'
4
0
Š3x
x$ 4‹
'c ax$ 2x 3b dx œ ’ x4
%
'
6.
'
7.
'
1
# #
Š5(3)
4% 16 ‹
#
Š 3(0) #
(3)# 4 ‹
(0)% 16 ‹
5
ˆx# Èx‰ dx œ ’ x3 23 x$Î# “ œ ˆ "3 23 ‰ 0 œ 1 &
x$Î# dx œ 25 x&Î# ‘ ! œ
32
1
(5)&Î# 0 œ 2(5)$Î# œ 10È5
$#
x'Î& dx œ 5x"Î& ‘ " œ ˆ #5 ‰ (5) œ
'cc x2 1
2
2 5
#
dx œ
5 #
'cc 2x# dx œ c2x" d " ˆ 2 ‰ ˆ 2 ‰ # œ 1 # œ 1 1
2
133 4
œ8 %
%
"
$
œ
œ Š 24 2# 3(2)‹ Š (42) (2)# 3(2)‹ œ 12
!
0
0
#
4# 4‹
œ Š 3(4) #
x# 3x“
2
5.
8.
% x% 16 “ !
#
dx œ ’ 3x#
2
4.
œ Š5(4)
315
316
Chapter 5 Integration
'
1
9.
'
1
10.
'
1Î3
11.
'
51Î6
12.
'
31Î4
13.
'
1Î3
14.
15.
'
0
0
sin x dx œ [cos x]1! œ (cos 1) (cos 0) œ (1) (1) œ 2 (1 cos x) dx œ [x sin x]1! œ (1 sin 1) (0 sin 0) œ 1
0
1Î$
1Î6
1Î4
0
œ ˆ2 tan ˆ 13 ‰‰ (2 tan 0) œ 2È3 0 œ 2È3
2 sec# x dx œ [2 tan x]!
&1Î' csc# x dx œ [cot x]1Î' œ ˆcot ˆ 561 ‰‰ ˆcot ˆ 16 ‰‰ œ ŠÈ3‹ ŠÈ3‹ œ 2È3
$1Î%
csc ) cot ) d) œ [csc )]1Î% œ ˆcsc ˆ 341 ‰‰ ˆcsc ˆ 14 ‰‰ œ È2 ŠÈ2‹ œ 0 1Î$
4 sec u tan u du œ [4 sec u]!
0
" cos 2t # 1Î2
dt œ
'
0
ˆ"
1Î2 #
" #
œ 4 sec ˆ 13 ‰ 4 sec 0 œ 4(2) 4(1) œ 4
cos 2t‰ dt œ "# t
" 4
!
sin 2t‘ 1Î# œ ˆ "# (0)
" 4
sin 2(0)‰ ˆ "# ˆ 1# ‰
" 4
sin 2 ˆ 1# ‰‰
œ 14 Î 1Î$ 2t 'c ÎÎ " cos dt œ ' ˆ "# "# cos 2t‰ dt œ "# t 4" sin 2t‘ 1Î$ # Î 1 3
16.
1 3
1 3
1 3
œ ˆ "# ˆ 13 ‰
" 4
sin 2 ˆ 13 ‰‰ ˆ #" ˆ 13 ‰
" 4
sin 2 ˆ 13 ‰‰ œ
1 6
" 4
sin 231
17.
'c Î#Î# a8y# sin yb dy œ ’ 8y3
18.
'cÎ Î% ˆ4 sec# t t1 ‰ dt œ 'Î Î% a4 sec# t 1t# b dt œ 4 tan t 1t ‘ 11Î%Î$
1
$
1
1
1Î#
cos y“
1Î#
œŒ
8 ˆ 1# ‰ 3
$
8 ˆ 1# ‰ 3
cos 1# Œ
1 6 $
œ Š4 tan ˆ
cos ˆ 1# ‰ œ
1‰ 4
Š4 tan ˆ 13 ‰
1 ˆ 13 ‰ ‹
È3 4
21 $ 3
œ (4(1) 4) Š4 ŠÈ3‹ 3‹ œ 4È3 3
'"" (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
%
$
$
(1)# (1)‹ Š 13 1# 1‹ œ 38
$
3
%
ŠÈ3‹
$
ŠÈ3‹
4
3
21.
'È" Š u#
" u& ‹
22.
' " ˆ v"
"‰ v%
23.
'
(
2
1
1 3
1 3
1 ˆ 14 ‰ ‹
19.
È2
sin ˆ 321 ‰ œ
1
#
1 3
1Î2
" 4
$
s# È s s#
(
2
dv œ
ds œ
'
œ È2 %È8 1
1
È 3 ‹
#
Š 2 ŠÈ3‹ 4È3
" du œ 'È Š u#
%
&
u ‹ du œ
u) ’ 16
4
$
" " 4u% “È#
ŠÈ3‹
È2
ˆ1 s$Î# ‰ ds œ ’s
#
)
œ
1) Š 16
$
È# 2 “ Ès "
#
2 ŠÈ3‹ 4 ŠÈ3‹ œ 10È3
3
" ' " av$ v% b dv œ 2v1 3v" ‘ ""Î# œ Š 2(1) 1Î2
$
#
œ È 2
" 4(1)% ‹
" 3(1)$ ‹
2 É È2
ŠÈ2‹ 16
" % 4 ŠÈ2‹
" $ 3 ˆ "# ‰
œ 34
Œ
" # 2 ˆ "# ‰
Š1
2 È1 ‹
œ È2 2$Î% 1
œ 56
Section 5.4 The Fundamental Theorem of Calculus 24.
'
4
1 Èu Èu 9
du œ
'
4
9
ˆu"Î# 1‰ du œ 2Èu u‘ % œ Š2È4 4‹ Š2È9 9‹ œ 3 *
'c% kxk dx œ '%! kxk dx '! 4
25.
4
kxk dx œ
'%! x dx '!
4
#
x dx œ ’ x# “
œ 16 26.
'
1
" ! #
acos x kcos xk b dx œ 1 #
œ sin
27. (a)
!
d dx
28. (a)
'
(b)
d dx
29. (a)
'
(b)
d dt
30. (a)
'
!
" # (cos
x cos x) dx
'
1
" 1Î# #
#
’ x# “ œ Š 0# !
(cos x cos x) dx œ
'
1Î#
!
(4)# # ‹
#
Š 4#
Èx
cos t dt œ [sin t]! œ sin Èx sin 0 œ sin Èx Ê
Èx
'
Œ
!
sin x
1
d dx
Œ
'
Èx
!
cos t dt œ
1Î#
d dx
ˆsin Èx‰ œ cos Èx ˆ "# x"Î# ‰
!
sin x
Èu du œ
Œ'
t%
!
!
d dx
Œ
'
sin x
1
3t# dt œ
d dx
asin$ x 1b œ 3 sin# x cos x
d 3t# dt œ a3 sin# xb ˆ dx (sin x)‰ œ 3 sin# x cos x
1
tan )
œ sin$ x 1 Ê
sin x
'
cos Èx 2È x
d ˆÈ ‰‰ cos t dt œ ˆcos Èx‰ ˆ dx x œ ˆcos Èx ‰ ˆ "# x"Î# ‰ œ
3t# dt œ ct$ d "
Œ t%
'
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
d dt
ˆ 23 t' ‰ œ 4t&
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 !
d d)
tan )
!
sec# y dy œ
d d)
(tan (tan )))
œ asec# (tan ))b sec# ) (b)
d d)
31. y œ
'
33. y œ
'
34. y œ
'
35. y œ
'
36. y œ
'
!
'
Œ x
!
Èx
sec# y dy œ asec# (tan ))b ˆ dd) (tan ))‰ œ asec# (tan ))b sec# )
x#
sin t# dt Ê
sin x
dt È1 t# tan x
dt 1 t#
, kxk Ê
dy dx
dy dx
dy dx
cos Èt dt Ê
!
!
tan )
È1 t# dt Ê
!
!
0# #‹
cos x dx œ [sin x]!
cos Èx 2È x
œ (b)
1Î#
%
#
%
sin 0 œ 1
Èx
'
'
!
32. y œ
'
1
x
" t
#
dt Ê
d ˆÈ ‰‰ œ Šsin ˆÈx‰ ‹ ˆ dx x œ (sin x) ˆ "# x"Î# ‰ œ
dy dx
1 #
œ È1 x#
dy dx
œ
" x
,x0
sin x 2È x
d œ Šcos Èx# ‹ ˆ dx ax# b‰ œ 2x cos kxk
Ê
dy dx
œ
" È1 sin# x
d ˆ dx (sin x)‰ œ
" Ècos# x
(cos x) œ
" d ‰ ˆ dx œ ˆ 1 tan (tan x)‰ œ ˆ sec"# x ‰ asec# xb œ 1 #x
cos x kcos xk
œ
cos x cos x
œ 1 since kxk
1 #
317
318
Chapter 5 Integration
37. x# 2x œ 0 Ê x(x 2) œ 0 Ê x œ 0 or x œ 2; Area œ
'$# ax# 2xbdx '#! ax# 2xbdx '!# ax# 2xbdx $
œ ’ x3 x# “ œ ŠŠ
(2)$ 3
# $
$
’ x3 x# “ #
(2) ‹ Š
!
$
#
(3)$ 3
’ x3 x# “
#
!
#
(3) ‹‹
$
ŠŠ 03 0# ‹ Š (32) (2)# ‹‹ $
$
$
ŠŠ 23 2# ‹ Š 03 0# ‹‹ œ
28 3
38. 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
39. 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
" #
40. x$ 4x œ 0 Ê x ax# 4b œ 0 Ê x(x 2)(x 2) œ 0 Ê x œ 0, 2, or 2. Area œ œ
% ’ x4
#
2x “ %
! #
% ’ x4
'c! ax$ 4xbdx '!# ax$ 4xbdx 2
#
#
%
2x “ œ Š 04 2(0)# ‹ !
Š (42) 2(2)# ‹ ’Š 24 2(2)# ‹ Š 04 2(0)# ‹“ œ 8 %
41. x"Î$ œ 0 Ê x œ 0; Area œ
%
'c"! x"Î$ dx '!) x"Î$ dx
! ) œ 34 x%Î$ ‘ " 34 x%Î$ ‘ ! œ ˆ 34 (0)%Î$ ‰ ˆ 34 (1)%Î$ ‰ ˆ 34 (8)%Î$ ‰ ˆ 34 (0)%Î$ ‰
œ
51 4
Section 5.4 The Fundamental Theorem of Calculus 42. 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
'c"! ˆx"Î$ x‰dx '!" ˆx"Î$ x‰dx '") ˆx"Î$ x‰dx
œ ’ 34 x%Î$
! 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
ˆ2!
$ 4
#" ‰ œ
) x# # “"
83 4
43. 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. 44. 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 ˆcos
51 ‰ 6
È3 # ‹
ˆcos 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 .
45. 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! Î sec ) tan ) d) œ [sec )]!1Î% 1 4
œ (sec 0) ˆsec ˆ 14 ‰‰ œ È2 1. Therefore the area of the shaded region on 14 ß !‘ is
1È2 4
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‹ . . 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
.
46. 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 Î sec# t dt œ [tan t]! 1Î% œ tan 0 tan ˆ 14 ‰ œ 1. 1 4
'! a1 t# b dt œ ’t t3 “ " œ Š1 13 ‹ Š0 03 ‹ œ 32 .
area under the curves on 14 ß "‘ is 1
"
$
2 3
œ
dt 3 œ 0 3 œ 3 Ê (d) is a solution to this problem.
œ
48. y œ
'c sec t dt 4
Ê
dy dx
œ sec x and y(1) œ
49. y œ
'! sec t dt 4
Ê
dy dx
œ sec x and y(0) œ
x
1
x
and y(1) œ
'
" t
dy dx
" x
1
1
Thus, the total
. Therefore the area of the shaded region is ˆ2 1# ‰
dt 3 Ê
'cc sec t dt 4 œ 0 4 œ 4 1
1
'!! sec t dt 4 œ 0 4 œ 4
. The
The area
5 3
'
" 1 t
$
!
47. y œ
x
$
1 #
5 3
œ
" 3
1 #
.
Ê (c) is a solution to this problem.
Ê (b) is a solution to this problem.
319
320
Chapter 5 Integration
50. y œ
'"
51. y œ
'
53. s œ
'
x
" t
54. v œ
'
'c ÎÎ
b 2 b 2
$
œ ˆ bh #
ˆ bh #
!
bh ‰ 6
Š2
b Œh ˆ # ‰
2 (x 1)# ‹
bh ‰ 6
dx œ 2
'
$
!
œ
bh 3
t
t!
È1 t# dt 2 g(x) dx v!
$
2 3
bh
" (x 1)# ‹
Š1
"
bÎ2
4h ˆ #b ‰ 3b#
œ bh
x
Ê (a) is a solution to this problem.
4hx$ 3b# “ bÎ2
ˆh ˆ 4h ‰ # ‰ dx œ ’hx b# x 4h ˆ b# ‰ 3b#
$
'"" "t dt 3 œ 0 3 œ 3
f(x) dx s!
œ Œhˆ #b ‰
'
and y(1) œ
'
55. Area œ
56. r œ
" x
52. y œ
#
t!
œ
dy dx
sec t dt 3
x
t
dt 3 Ê
$
dx œ 2 x ˆ x11 ‰‘ ! œ 2 ’Š3
" (3 1) ‹
Š0
" (0 1) ‹“
œ 2 3 "4 1‘ œ 2 ˆ2 4" ‰ œ 4.5 or $4500 57.
dc dx
œ
" #È x
œ
" #
x"Î# Ê c œ
'
x
!
" "Î# dt # t
œ t"Î# ‘ 0 œ Èx x
c(100) c(1) œ È100 È1 œ $9.00 58. By Exercise 57, c(400) c(100) œ È400 È100 œ 20 10 œ $10.00 59. (a) v œ (b) a œ (c) s œ (d) (e) (f) (g)
ds dt df dt
'
!
œ
d dt
'
t
!
f(x) dx œ f(t) Ê v(5) œ f(5) œ 2 m/sec
is negative since the slope of the tangent line at t œ 5 is negative 3
f(x) dx œ
" #
(3)(3) œ
9 #
m since the integral is the area of the triangle formed by y œ f(x), the x-axis,
and x œ 3 t œ 6 since from t œ 6 to t œ 9, the region lies below the x-axis At t œ 4 and t œ 7, since there are horizontal tangents there Toward the origin between t œ 6 and t œ 9 since the velocity is negative on this interval. Away from the origin between t œ 0 and t œ 6 since the velocity is positive there. Right or positive side, because the integral of f from 0 to 9 is positive, there being more area above the x-axis than below it.
60. (a) v œ (b) a œ
dg dt df dt
œ
d dt
'
!
t
g(x) dx œ g(t) Ê v(3) œ g(3) œ 0 m/sec.
is positive, since the slope of the tangent line at t œ 3 is positive
(c) At t œ 3, the particle's position is
'
!
$
g(x) dx œ
" #
(3)(6) œ 9
(d) The particle passes through the origin at t œ 6 because s(6) œ
'
!
'
g(x) dx œ 0
(e) At t œ 7, since there is a horizontal tangent there (f) The particle starts at the origin and moves away to the left for 0 t 3. It moves back toward the origin for 3 t 6, passes through the origin at t œ 6, and moves away to the right for t 6. (g) Right side, since its position at t œ 9 is positive, there being more area above the x-axis than below it at t œ *.
Section 5.4 The Fundamental Theorem of Calculus 61. 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 62. lim x"$ xÄ!
63.
'
64.
'
x
1
x
!
'
!
x
t%
t# dt "
œ lim
t# ! t% " dt
'x
x$
xÄ!
x "
#
9 1t
!
sin kx dx œ
" k
1 Îk
cos kx‘ !
2 k
x# %
xÄ!
f(t) dt œ x cos 1x Ê f(x) œ
'
1Îk
œ lim x$x#" œ lim $ax%" "b œ _.
f(t) dt œ x# 2x 1 Ê f(x) œ
65. f(x) œ 2
'
321
d dx
xÄ!
'
d dx
'
!
1
x
x
f(t) dt œ
d dx
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
Ê f w (1) œ 3; f(1) œ 2
'
#
" "
9 1t
dt œ 2 0 œ 2;
L(x) œ 3(x 1) f(1) œ 3(x 1) 2 œ 3x 5 66.
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 67. (a) (b) (c) (d) (e) (f) (g)
True: since f is continuous, g is differentiable by Part 1 of the Fundamental Theorem of Calculus. True: g is continuous because it is differentiable. True, since 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).
68. (a) True: by Part 1 of the Fundamental Theorem of Calculus, hw (x) œ f(x). Since f is differentiable for all x, h has a second derivative for all x. (b) True: they are continuous because they are differentiable. (c) True, since hw (1) œ f(1) œ 0. (d) True, since hw (1) œ 0 and hww (1) œ f w (1) 0. (e) False, since hww (1) œ f w (1) 0. (f) False, since hww (x) œ f w (x) 0 never changes sign. (g) True, since hw (1) œ f(1) œ 0 and hw (x) œ f(x) is a decreasing function of x (because f w (x) 0). 69.
70. The limit is 3x#
322
Chapter 5 Integration
71-74. Example CAS commands: Maple: p:=x^2*cos(x); 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="#71(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="71(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) x^3/3-x^2/2-2*x+1/3; g := x -> x-1; plot( [f(x),g(x)], x=-5..5, legend=["y = f(x)","y = g(x)"], title="#91(a) (Section 5.6)" ); q1 := [ -5, -2, 1, 4 ]; # (b) q2 := [seq( fsolve( f(x)=g(x), x=q1[i]..q1[i+1] ), i=1..nops(q1)-1 )]; for i from 1 to nops(q2)-1 do # (c) area[i] := int( abs(f(x)-g(x)),x=q2[i]..q2[i+1] ); end do; add( area[i], i=1..nops(q2)-1 ); # (d) Mathematica: (assigned functions may vary) Clear[x, f, g] f[x_] = x2 Cos[x] g[x_] = x3 x Plot[{f[x], g[x]}, {x, 2, 2}] After examining the plots, the initial guesses for FindRoot can be determined. pts = x/.Map[FindRoot[f[x]==g[x],{x, #}]&, {1, 0, 1}] i1=NIntegrate[f[x] g[x], {x, pts[[1]], pts[[2]]}] i2=NIntegrate[f[x] g[x], {x, pts[[2]], pts[[3]]}] i1 i2
343
344
Chapter 5 Integration
CHAPTER 5 PRACTICE EXERCISES 1. (a) Each time subinterval is of length ?t œ 0.4 sec. The distance traveled over each subinterval, using the midpoint rule, is ?h œ "# avi vib1 b ?t, where vi is the velocity at the left endpoint and vib1 the velocity at
the right endpoint of the subinterval. We then add ?h to the height attained so far at the left endpoint vi to arrive at the height associated with velocity vib1 at the right endpoint. Using this methodology we build the following table based on the figure in the text: t (sec) 0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 4.4 4.8 5.2 5.6 6.0 v (fps) 0 10 25 55 100 190 180 165 150 140 130 115 105 90 76 65 h (ft) 0 2 9 25 56 114 188 257 320 378 432 481 525 564 592 620.2 t (sec) v (fps) h (ft)
6.4 50 643.2
6.8 37 660.6
7.2 25 672
7.6 12 679.4
8.0 0 681.8
NOTE: Your table values may vary slightly from ours depending on the v-values you read from the graph. Remember that some shifting of the graph occurs in the printing process. The total height attained is about 680 ft. (b) The graph is based on the table in part (a).
2. (a) Each time subinterval is of length ?t œ 1 sec. The distance traveled over each subinterval, using the midpoint rule, is ?s œ "# avi vib1 b ?t, where vi is the velocity at the left, and vib1 the velocity at the
right, endpoint of the subinterval. We then add ?s to the distance attained so far at the left endpoint vi to arrive at the distance associated with velocity vib1 at the right endpoint. Using this methodology we build the table given below based on the figure in the text, obtaining approximately 26 m for the total distance traveled: t (sec) 0 1 2 3 4 5 6 7 8 9 10 v (m/sec) 0 0.5 1.2 2 3.4 4.5 4.8 4.5 3.5 2 0 s (m) 0 0.25 1.1 2.7 5.4 9.35 14 18.65 22.65 25.4 26.4
(b) The graph shows the distance traveled by the moving body as a function of time for 0 Ÿ t Ÿ 10.
3. (a) (c)
10
!
kœ1 10
ak 4
œ
" 4
10
! ak œ
kœ1
" 4
(2) œ #"
(b)
10
10
10
kœ1
kœ1
kœ1
10
10
10
kœ1
kœ1
kœ1
! (bk 3ak ) œ ! bk 3 ! ak œ 25 3(2) œ 31
! (ak bk 1) œ ! ak ! bk ! " œ 2 25 (1)(10) œ 13
kœ1
Chapter 5 Practice Exercises 10
10
kœ1
kœ1
! ˆ 5 bk ‰ œ ! #
(d)
20
5 #
10
! bk œ kœ1
20
kœ1 20
(b)
kœ1
! ˆ" #
(c)
kœ1 20
2bk ‰ 7
20
œ !
kœ1 20
" #
20
! bk œ
2 7
kœ1 20
" #
20
20
20
kœ1
kœ1
kœ1
! (ak bk ) œ ! ak ! bk œ 0 7 œ 7
(20) 27 (7) œ 8
! aak 2b œ ! ak ! 2 œ 0 2(20) œ 40
(d)
kœ1
kœ1
kœ1 " #
5. Let u œ 2x 1 Ê du œ 2 dx Ê 5
1
'
(2x 1)"Î# dx œ
9
1
3
1
x ax# 1b
7. Let u œ
"Î$
'
dx œ
8
0
du œ dx; x œ 1 Ê u œ 1, x œ 5 Ê u œ 9 *
u"Î# ˆ "# du‰ œ sqrt(1-x^2);a := -1; b := 1; N := [2, 4, 8 ]; for n in N do xx := [seq( a+i*(b-a)/n, i=0..n )]; pts := [seq([x,f(x)],x=xx)]; L := simplify(add( distance(pts[i+1],pts[i]), i=1..n )); T := sprintf("#31(a) (Section 6.3)\nn=%3d L=%8.5f\n", n, L ); P[n] := plot( [f(x),pts], x=a..b, title=T ): end do: display( [seq(P[n],n=N)], insequence=true, scaling=constrained ); L := ArcLength( f(x), x=a..b, output=integral ): L = evalf( L ); 37-40. Example CAS commands: Maple: with( plots ); with( student ); x := t -> t^3/3; y := t -> t^2/2; a := 0;
21
# (b) # (a)
# (c)
383
384
Chapter 6 Applications of Definite Integrals b := 1; N := [2, 4, 8 ]; for n in N do tt := [seq( a+i*(b-a)/n, i=0..n )]; pts := [seq([x(t),y(t)],t=tt)]; L := simplify(add( student[distance](pts[i+1],pts[i]), i=1..n )); T := sprintf("#37(a) (Section 6.3)\nn=%3d L=%8.5f\n", n, L ); P[n] := plot( [[x(t),y(t),t=a..b],pts], title=T ): end do: display( [seq(P[n],n=N)], insequence=true ); ds := t ->sqrt( simplify(D(x)(t)^2 + D(y)(t)^2) ): L := Int( ds(t), t=a..b ): L = evalf(L);
# (b) # (a)
# (c)
31-40. Example CAS commands: Mathematica: (assigned function and values for a, b, and n may vary) Clear[x, f] {a, b} = {1, 1}; f[x_] = Sqrt[1 x2 ] p1 = Plot[f[x], {x, a, b}] n = 8; pts = Table[{xn, f[xn]}, {xn, a, b, (b a)/n}]/ / N Show[{p1,Graphics[{Line[pts]}]}] Sum[ Sqrt[ (pts[[i 1, 1]] pts[[i, 1]])2 (pts[[i 1, 2]] pts[[i, 2]])2 ], {i, 1, n}] NIntegrate[ Sqrt[ 1 f'[x]2 ],{x, a, b}] 6.4 MOMENTS AND CENTERS OF MASS 1. Because the children are balanced, the moment of the system about the origin must be equal to zero: 5 † 80 œ x † 100 Ê x œ 4 ft, the distance of the 100-lb child from the fulcrum. 2. Suppose the log has length 2a. Align the log along the x-axis so the 100-lb end is placed at x œ a and the 200-lb end at x œ a. Then the center of mass x satisfies x œ at a distance a or
2 3
a 3
œ
2a 3
œ
" 3
(2a) which is
" 3
100(a) 200(a) 300
Ê x œ 3a . That is, x is located
of the length of the log from the 200-lb (heavier) end (see figure)
of the way from the lighter end toward the heavier end. " 3
(2a)
èëëéëëê 100 lbs. ñïïïïïïïïïïïïïïñïïïïñïïïïïïñ a 200 lbs a x œ a/3 ! 3. The center of mass of each rod is in its center (see Example 1). The rod system is equivalent to two point m masses located at the centers of the rods at coordinates ˆ L# ß !‰ and ˆ0ß L# ‰. Therefore x œ y m
œ
x" m" x# m# m" m#
œ
L # †m0
mm
œ
L 4
and y œ
mx m
œ
y" m# y# m# m" m#
œ
0 L2 †m mm
œ
L 4
Ê
ˆ L4 ß L4 ‰
is the center of
mass location. 4. Let the rods have lengths x œ L and y œ 2L. The center of mass of each rod is in its center (see Example 1). The rod system is equivalent to two point masses located at the centers of the rods at coordinates ˆ L# ß !‰ and (!ß L). Therefore x œ
L # †m0†2m
m2m
œ
L 6
and y œ
!†mL†2m m2m
œ
2L 3
‰ Ê ˆ L6 ß 2L 3 is the center of mass location.
Section 6.4 Moments and Centers of Mass 5. M! œ '0 x † 4 dx œ ’4 x# “ œ 4 † 2
#
#
!
6. M! œ '1 x † 4 dx œ ’4 x# “ œ 3
$
#
"
4 #
œ3
9 #
Ê xœ
ˆ 15 ‰ 2 ˆ 92 ‰
œ
M! M
x# 3‹
œ
8. M! œ '0 x ˆ2 x4 ‰ dx œ '0 Š2x 4
4
M œ '0 ˆ2 x4 ‰ dx œ ’2x 4
9. M! œ '1 x Š1 4
" Èx ‹
2
% x# 8 “!
15 9
$ x$ 9 “!
#
dx œ ’ x# œ
x# 4‹
dx œ ’x#
œ8
16 8
œ ˆ 92
% x$ 12 “ !
œ
27 ‰ 9
œ ˆ16
64 ‰ 12
œ
œ
œ6 Ê xœ
4
#
M! M
% 2x$Î# 3 “"
32 3 †6
œ ˆ8
16 ‰ 3
œ 3(4 1) œ 9; M œ 3'1Î4 ˆx$Î# x&Î# ‰ dx œ 3 x"Î#2 1
2 ‘" 3x$Î# "Î%
M! M
œ
1
2
œ 3; M œ '0 (2 x) dx '1 x dx œ ’2x 1
2
" x# # “!
#
# # ’ x# “ "
œ ˆ2
12. M! œ '0 x(x 1) dx '1 2x dx œ '0 ax# xb dx '1 2x dx œ ’ x3 œ3
2
2 3
œ
32 3 ;
ˆ 73 ‰ 6 5
œ
2 ‘" x"Î# "Î%
œ
15 #
14 3
1
1
2
2
œ
23 6 ;
Ê xœ
M! M
‰ ˆ 72 ‰ œ œ ˆ 23 6
4528 6
œ
œ
73 6
;
73 30
œ 3 ’(2 2) Š2 † 16 ‰‘ 3
$
"‰ #
" x$ 3 “!
" #
2 ˆ "# ‰ ‹“
œ 3 ˆ2
14 ‰ 3
" x# 2 “!
ˆ 4#
#
$
’ x3 “ œ ˆ1 "3 ‰ ˆ 83 "3 ‰ "
"‰ #
œ3 Ê xœ
#
"
M! M
œ1
# cx# d " œ ˆ "3 2" ‰ (4 1)
M œ '0 (x 1) dx '1 2 dx œ ’ x# x“ c2xd #" œ ˆ "# 1‰ (4 #) œ 2
5 6
$ x# 6 “!
9 20
2
1
œ 16 †
œ 3 ˆ2 32 ‰ ˆ4
11. M! œ '0 x(2 x) dx '1 x † x dx œ '0 a2x x# b dx '1 x# dx œ ’ 2x# 9 3
16 3
ˆ "# 32 ‰ œ
1
1
œ2
16 9
10. M! œ '1Î4 x † 3 ˆx$Î# x&Î# ‰ dx œ 3'1Î4 ˆx"Î# x$Î# ‰ dx œ 3 2x"Î# œ
16 8
3
œ 16
4
1
œ
M! M
M œ '0 ˆ1 3x ‰ dx œ ’x
15 # ;
% M œ '1 ˆ1 x"Î# ‰ dx œ x 2x"Î# ‘ " œ (4 4) (1 2) œ 5 Ê x œ
M! M
œ1
5 3
dx œ '1 ˆx x"Î# ‰ dx œ ’ x#
œ 6 14 œ 20 Ê x œ
M! M
3
3
œ
9 6
œ 8; M œ '0 4 dx œ [4x]#! œ 4 † 2 œ 8 Ê x œ
(9 1) œ 16; M œ '1 4 dx œ [4x]$" œ 12 4 œ 8 Ê x œ
7. M! œ '0 x ˆ1 x3 ‰ dx œ '0 Šx 3
4 #
385
!
3 #
œ
7 #
23 21
13. Since the plate is symmetric about the y-axis and its density is constant, the distribution of mass is symmetric about the y-axis and the center of mass lies on the y-axis. This means that x œ 0. It remains to find y œ MMx . We model the distribution of mass with @/3-+6 strips. The typical strip has center of mass: # (µ x ßµ y ) œ Šxß x 4 ‹ , length: 4 x# , width: dx, area: #
#
dA œ a4 x b dx, mass: dm œ $ dA œ $ a4 x# b dx. The moment of the strip about the x-axis is # µ C dm œ Š x #4 ‹ $ a4 x# b dx œ #$ a16 x% b dx. The moment of the plate about the x-axis is Mx œ ' µ C dm œ 'c2 #$ a16 x% b dx œ 2
$ #
’16x
# x& 5 “ #
plate is M œ ' $ a4 x# b dx œ $ ’4x ‰ mass is the point (xß y) œ ˆ!ß 12 5 .
œ
$ #
’Š16 † 2
# x$ 3 “ #
2& 5‹
œ 2$ ˆ8 83 ‰ œ
32$ 3 .
2& 5 ‹“
œ
$ †2 #
ˆ32
Therefore y œ
Mx M
œ
Š16 † 2
32 ‰ 5
$ Š 128 5 ‹
Š 323$ ‹
œ
œ
128$ 5 .
12 5 .
The mass of the
The plate's center of
386
Chapter 6 Applications of Definite Integrals
14. Applying the symmetry argument analogous to the one in Exercise 13, we find x œ 0. To find y œ MMx , we use the @/3-+6 strips technique. The typical strip has center of # mass: (µ x ßµ y ) œ Šxß 25 x ‹ , length: 25 x# , width: dx, #
#
area: dA œ a25 x bdx, mass: dm œ $ dA œ $ a25 x# b dx. The moment of the strip about the x-axis is # µ y dm œ Š 25 # x ‹ $ a25 x# b dx œ
œ 'c5 #$ a25 x# b dx œ 5
#
œ $ † 625 ˆ5 œ 2$ Š5$
10 3
5$ 3‹
$ #
'c55
a25 x# b dx. The moment of the plate about the x-axis is Mx œ ' µ y dm #
$ #
$ #
a625 50x# x% b dx œ
’625x
50 3
x$
& x& 5 “ &
œ 2 † #$ Š625 † 5
50 3
† 5$
1‰ œ $ † 625 † ˆ 38 ‰ . The mass of the plate is M œ ' dm œ 'c5 $ a25 x# b dx œ $ ’25x 5
œ
4 3
$ † 5$ . Therefore y œ
Mx M
œ
$ †5% † ˆ 83 ‰ $ †5$ †ˆ 43 ‰
5& 5‹ & x$ 3 “ &
œ 10. The plate's center of mass is the point (xß y) œ (!ß 10).
15. Intersection points: x x# œ x Ê 2x x# œ 0 Ê x(2 x) œ 0 Ê x œ 0 or x œ 2. The typical @/3-+6 # strip has center of mass: (µ x ßµ y ) œ Šxß ax x b (x) ‹ #
œ Šxß
x# # ‹,
#
length: ax x b (x) œ 2x x# , width: dx,
area: dA œ a2x x# b dx, mass: dm œ $ dA œ $ a2x x# b dx. The moment of the strip about the x-axis is # µ y dm œ Š x# ‹ $ a2x x# b dx; about the y-axis it is µ x dm œ x † $ a2x x# b dx. Thus, Mx œ ' µ y dm œ '0 ˆ #$ x# ‰ a2x x# b dx œ #$ '0 a2x$ x% b dx œ #$ ’ x# 2
2
%
# x& 5 “!
œ #$ Š2$
œ 45$ ; My œ ' µ x dm œ '0 x † $ a2x x# b dx œ $ '0 a2x# x$ b œ $ ’ 23 x$ 2
2
M œ ' dm œ '0 $ a2x x# b dx œ $ '0 a2x x# b dx œ $ ’x# 2
2
œ ˆ 43$ ‰ ˆ 43$ ‰ œ 1 and y œ
Mx M
# x$ 3 “!
# x% 4 “!
2& 5‹
œ #$ † 2$ ˆ1 45 ‰
œ $ Š2 †
œ $ ˆ4 83 ‰ œ
4$ 3
2$ 3
2% 4‹
œ
. Therefore, x œ
$ †2% 1#
œ
My M
œ ˆ 45$ ‰ ˆ 43$ ‰ œ 35 Ê (xß y) œ ˆ1ß 35 ‰ is the center of mass.
16. Intersection points: x# 3 œ 2x# Ê 3x# 3 œ 0 Ê 3(x 1)(x 1) œ 0 Ê x œ 1 or x œ 1. Applying the symmetry argument analogous to the one in Exercise 13, we find x œ 0. The typical @/3-+6 strip has center of mass: # # # (µ x ßµ y ) œ Šxß 2x ax 3b ‹ œ Šxß x 3 ‹ , #
#
#
#
#
length: 2x ax 3b œ 3 a1 x b, width: dx, area: dA œ 3 a1 x# b dx, mass: dm œ $ dA œ 3$ a1 x# b dx. The moment of the strip about the x-axis is µ y dm œ 3 $ ax# 3b a1 x# b dx œ 3 $ ax% 3x# x# 3b dx œ #
œ
3 #
$ 'c1 ax% 2x# 3b dx œ 1
#
3 #
&
$ ’ x5
2x$ 3
M œ ' dm œ 3$ 'c1 a1 x# b dx œ 3$ ’x 1
Ê (xß y) œ ˆ0ß 85 ‰ is the center of mass.
3x“ " x$ 3 “ "
" "
œ
3 #
3 #
$ ax% 2x# 3b dx; Mx œ ' µ y dm
† $ † 2 ˆ 5"
2 3
45 ‰ 3‰ œ 3$ ˆ 310 œ 325$ ; 15
œ 3$ † 2 ˆ1 3" ‰ œ 4$ . Therefore, y œ
Mx M
œ 5$††$32†4 œ 58
4$ 3
;
Section 6.4 Moments and Centers of Mass
387
17. The typical 29+6 strip has center of mass: $ (µ x ßµ y ) œ Š y y ß y‹ , length: y y$ , width: dy, #
area: dA œ ay y$ b dy, mass: dm œ $ dA œ $ ay y$ b dy. The moment of the strip about the y-axis is $ # µ x dm œ $ Š y y ‹ ay y$ b dy œ $ ay y$ b dy #
œ
$ #
#
#
%
'
ay 2y y b dy; the moment about the x-axis is
1 $ µ y dm œ $ y ay y$ b dy œ $ ay# y% b dy. Thus, Mx œ ' µ y dm œ $ '0 ay# y% b dy œ $ ’ y3
My œ ' µ x dm œ
$ #
'01 ay# 2y% y' b dy œ #$ ’ y3
$
œ $ '0 ay y$ b dy œ $ ’ y# 1
œ
#
" y% 4 “!
2y& 5
œ $ ˆ "# 4" ‰ œ
$ 4
" y( 7 “!
œ
$ #
ˆ "3
. Therefore, x œ
2 5
7" ‰ œ
$ #
œ $ ˆ "3 "5 ‰ œ
15 ‰ ˆ 35 3†42 œ 5†7
4$ ‰ ˆ 4 ‰ œ ˆ 105 $ œ
My M
" y& 5 “!
16 105
2$ 15
;
4$ 105
; M œ ' dm
Mx M
2$ ‰ ˆ 4 ‰ œ ˆ 15 $
and y œ
16 8 ‰ Ê (xß y) œ ˆ 105 ß 15 is the center of mass.
8 15
18. Intersection points: y œ y# y Ê y# 2y œ 0 Ê y(y 2) œ 0 Ê y œ 0 or y œ 2. The typical 29+6 strip has center of mass: # # (µ x ßµ y ) œ Š ay yby ß y‹ œ Š y ß y‹ , #
2
#
length: y ay yb œ 2y y# , width: dy, area: dA œ a2y y# b dy, mass: dm œ $ dA œ $ a2y y# b dy. The moment about the y-axis is µ x dm œ #$ † y# a2y y# b dy 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 dy. Thus, Mx œ ' µ y dm œ '0 $ a2y# y$ b dy œ $ ’ 2y3 2
œ '0
2
$ #
a2y$ y% b dy œ
œ $ ’y#
# y$ 3 “!
$
$ #
%
’ y2
œ $ ˆ4 83 ‰ œ
# y& 5 “!
4$ 3
œ
$ #
ˆ8
# y% 4 “!
16$ 1#
ˆ 405 32 ‰ œ
4$ 5
; M œ ' dm œ '0 $ a2y y# b dy
œ
$ #
My M
œ ˆ 45$ ‰ ˆ 43$ ‰ œ
32 ‰ 5
. Therefore, x œ
œ
(4 3) œ
4$ 3
; My œ ' µ x dm
16 ‰ 4
œ $ ˆ 16 3
2
3 5
and y œ
Mx M
œ ˆ 43$ ‰ ˆ 43$ ‰ œ 1
Ê (xß y) œ ˆ 35 ß "‰ is the center of mass. 19. Applying the symmetry argument analogous to the one used in Exercise 13, we find x œ 0. The typical @/3-+6 strip has center of mass: (µ x ßµ y ) œ ˆxß cos# x ‰ , length: cos x, width: dx,
area: dA œ cos x dx, mass: dm œ $ dA œ $ cos x dx. The moment of the strip about the x-axis is µ y dm œ $ † cos# x † cos x dx 2x ‰ œ #$ cos# x dx œ #$ ˆ 1 cos dx œ 4$ (1 cos 2x) dx; thus, # 1Î2
Mx œ ' µ y dm œ 'c1Î2 4$ (1 cos 2x) dx œ 1Î#
œ $ [sin x]1Î# œ 2$ . Therefore, y œ
Mx M
œ
$ 4
x $1 4 †# $
œ
sin 2x ‘ 1Î# # 1Î# 1 8
œ
$ 4
area: dA œ sec# x dx, mass: dm œ $ dA œ $ sec# x dx. The # moment about the x-axis is µ y dm œ Š sec x ‹ a$ sec# xb dx œ
$ #
sec% x dx. Mx œ 'c1Î4 µ y dm œ
#
$ #
'11ÎÎ44 sec% x dx
$1 4
1Î2
; M œ ' dm œ $ '1Î2 cos x dx
Ê (xß y) œ ˆ!ß 18 ‰ is the center of mass.
20. Applying the symmetry argument analogous to the one used in Exercise 13, we find x œ 0. The typical vertical strip has # center of mass: (µ x ßµ y ) œ Šxß sec# x ‹ , length: sec# x, width: dx,
1Î4
ˆ 1# 0‰ ˆ 1# ‰‘ œ
388
Chapter 6 Applications of Definite Integrals œ
$ #
'c11ÎÎ44 atan# x 1b asec# xb dx œ #$ '11ÎÎ44 (tan x)# asec# xb dx #$ '11ÎÎ44 sec# x dx œ 2$ ’ (tan3x) “ 1Î4
œ
$ 2
3" ˆ 3" ‰‘ #$ [1 (1)] œ
$
Therefore, y œ
Mx M
œ ˆ 43$ ‰ ˆ 2"$ ‰ œ
2 3
$ 3
$ œ
4$ 3
1Î%
1 Î4
1Î4
#$ [tan x]1Î%
; M œ ' dm œ $ 'c1Î4 sec# x dx œ $ [tan x]1Î4 œ $ [1 (1)] œ 2$ . 1Î4
Ê (xß y) œ ˆ!ß 32 ‰ is the center of mass.
21. Since the plate is symmetric about the line x œ 1 and its density is constant, the distribution of mass is symmetric about this line and the center of mass lies on it. This means that x œ 1. The typical @/3-+6 strip has center of mass: # # # (µ x ßµ y ) œ Šxß a2x x ba2x 4xb ‹ œ Šxß x 2x ‹ , #
#
#
#
#
length: a2x x b a2x 4xb œ 3x 6x œ 3 a2x x# b , width: dx, area: dA œ 3 a2x x# b dx, mass: dm œ $ dA œ 3$ a2x x# b dx. The moment about the x-axis is # µ y dm œ 3 $ ax# 2xb a2x x# b dx œ 3 $ ax# 2xb dx #
#
œ 3# $ ax% 4x$ 4x# b dx. Thus, Mx œ ' µ y dm œ '0
2
œ $ 3 2
& Š 25
%
2
4 3
$
%
†2 ‹œ $†2 3 #
œ '0 3$ a2x x# b dx œ 3$ ’x# 2
# x$ 3 “!
ˆ 25
1
2‰ 3
3 2
&
$ ax% 4x$ 4x# b dx œ 32 $ ’ x5 x% 34 x$ “ %
œ $ †2 3 #
10 ‰ ˆ 6 15 15
œ 3$ ˆ4 83 ‰ œ 4$ . Therefore, y œ
Mx M
œ
8$ 5
; M œ ' dm
# !
œ ˆ 85$ ‰ ˆ 4"$ ‰ œ 25
Ê (xß y) œ ˆ1ß 25 ‰ is the center of mass. 22. (a) Since the plate is symmetric about the line x œ y and its density is constant, the distribution of mass is symmetric about this line. This means that x œ y. The typical @/3-+6 strip has center of mass: È # (µ x ßµ y ) œ Šxß 9 x ‹ , length: È9 x# , width: dx, #
area: dA œ È9 x# dx, mass: dm œ $ dA œ $ È9 x# dx. The moment about the x-axis is È # µ y dm œ $ Š 9# x ‹ È9 x# dx œ
$ #
a9 x# b dx. Thus, Mx œ ' µ y dm œ '0
3
$ #
a9 x# b dx œ
$ #
’9x
$ x$ 3 “!
(27 9) œ 9$ ; M œ ' dm œ ' $ dA œ $ ' dA œ $ (Area of a quarter of a circle of radius 3) œ $ ˆ 941 ‰ œ 4 ‰ Therefore, y œ MMx œ (9$ ) ˆ 91$ œ 14 Ê (xß y) œ ˆ 14 ß 14 ‰ is the center of mass. œ
$ #
(b) Applying the symmetry argument analogous to the one used in Exercise 13, we find that x œ 0. The typical vertical strip has the same parameters as in part (a). 3 Thus, M œ ' µ y dm œ ' $ a9 x# b dx x
œ #'0
3
$ #
c3 #
a9 x# b dx œ 2(9$ ) œ 18$ ;
M œ ' dm œ ' $ dA œ $ ' dA œ $ (Area of a semi-circle of radius 3) œ $ ˆ 921 ‰ œ 91$ 2 . Therefore, y œ 4 as in part (a) Ê (xß y) œ ˆ0ß 1 ‰ is the center of mass.
Mx M
2 ‰ œ (18$ ) ˆ 91$ œ
4 1
, the same y
91$ 4
.
Section 6.4 Moments and Centers of Mass
389
23. Since the plate is symmetric about the line x œ y and its density is constant, the distribution of mass is symmetric about this line. This means that x œ y. The typical @/3-+6 strip has È # center of mass: (µ x ßµ y ) œ Šxß 3 9 x ‹ , #
length: 3 È9 x# , width: dx, area: dA œ Š3 È9 x# ‹ dx, mass: dm œ $ dA œ $ Š3 È9 x# ‹ dx. The moment about the x-axis is µ y dm œ $
Š3 È9 x# ‹ Š3 È9 x# ‹ #
dx œ
$ #
c9 a9 x# bd dx œ
$ x# #
dx. Thus, Mx œ '0
3
$ x# #
dx œ
$ 6
$
cx$ d ! œ #
9$ #
equals the area of a square with side length 3 minus one quarter the area of a disk with radius 3 Ê A œ 3 œ
9 4
9$ 4
(4 1) Ê M œ $ A œ
(4 1). Therefore, y œ
Mx M
œ ˆ 9#$ ‰ ’ 9$(44 1) “ œ
2 41
. The area 19 4
Ê (xß y) œ ˆ 4 2 1 ß 4 2 1 ‰ is the
center of mass. 24. Applying the symmetry argument analogous to the one used in Exercise 13, we find that y œ 0. The typical @/3-+6 strip has center of mass: (µ x ßµ y ) œ Œxß " x$
length:
ˆ x"$ ‰ œ
2 x$
" x$
x"$ #
œ (xß 0),
, width: dx, area: dA œ
2 x$
dx,
2$ x$
mass: dm œ $ dA œ dx. The moment about the y-axis is a µ x dm œ x † 2x$$ dx œ 2x$# dx. Thus, My œ ' µ x dm œ '1 2x$# dx a œ 2$ x" ‘ " œ 2$ ˆ "a 1‰ œ
xœ
My M
œ
’ 2$(aa 1) “
25. Mx œ ' µ y dm œ '1
2
# ’ $ aa#a 1b “
Š x2# ‹ #
2$ (a1) a
œ
2a a1
; M œ ' dm œ '1
a
Ê (xß y) œ
2$ x$
ˆ a 2a ‰ 1ß 0 .
dx œ $ x"# ‘ " œ $ ˆ a"# 1‰ œ a
$ aa# 1b a#
. Therefore,
Also, a lim x œ 2. Ä_
† $ † ˆ x2# ‰ dx
œ '1 ˆ x"# ‰ ax# b ˆ x2# ‰ dx œ '1 2
2
2 x#
dx œ 2'1 x# dx 2
#
œ 2 cx" d " œ 2 ˆ "# ‰ (1)‘ œ 2 ˆ "# ‰ œ 1;
My œ ' µ x dm œ '1 x † $ † ˆ x2# ‰ dx 2
œ '1 x ax# b ˆ x2# ‰ dx œ 2'1 x dx œ 2 ’ x# “ 2
2
#
# "
œ 2 ˆ2 "# ‰ œ 4 1 œ 3; M œ ' dm œ '1 $ ˆ x2# ‰ dx œ '1 x# ˆ x2# ‰ dx œ 2'1 dx œ 2[x]"# œ 2(2 1) œ 2. So xœ
My M
œ
3 #
and y œ
Mx M
œ
" #
2
2
2
Ê (xß y) œ ˆ 3# ß "# ‰ is the center of mass.
26. We use the @/3-+6 strip approach: 1 # M œ'µ y dm œ ' ax x b ax x# b † $ dx x
œ
0
" #
#
'0 ax# x% b † 12x dx 1
œ 6'0 ax$ x& b dx œ 6 ’ x4 1
œ 6 ˆ "4 6" ‰ œ
%
6 4
1œ
" #
" x' 6 “!
;
My œ ' µ x dm œ '0 x ax x# b † $ dx œ '0 ax# x$ b † 12x dx œ 12'0 ax$ x% b dx œ 12 ’ x4 1
1
1
%
" x& 5 “!
œ 12 ˆ "4 5" ‰
390
Chapter 6 Applications of Definite Integrals œ
12 #0
xœ
œ
; M œ ' dm œ ' ax x# b † $ dx œ 12'0 ax# x$ b dx œ 12 ’ x3 1
3 5
1
$
0
My M
œ
3 5
and y œ
Mx M
œ
" #
" x% 4 “!
œ 12 ˆ "3 4" ‰ œ
12 12
œ 1. So
Ê ˆ 35 ß "# ‰ is the center of mass.
shell ‰ shell 27. (a) We use the shell method: V œ 'a 21 ˆ radius Š height ‹ dx œ '1 21x ’ È4x Š È4x ‹“ dx b
œ 161'1
4
x Èx
4
% dx œ 161'1 x"Î# dx œ 161 32 x$Î# ‘ " œ 161 ˆ 32 † 8 32 ‰ œ 4
(b) Since the plate is symmetric about the x-axis and its density $ (x) œ
" x
321 3
(8 1) œ
2241 3
is a function of x alone, the
distribution of its mass is symmetric about the x-axis. This means that y œ 0. We use the vertical strip 4 4 4 approach to find x: My œ ' µ x dm œ '1 x † ’ È4x Š È4x ‹“ † $ dx œ '1 x † È8x † x" dx œ 8'1 x"Î# dx 4 œ 8 2x"Î# ‘ " œ 8(2 † 2 2) œ 16; M œ ' dm œ '1 ’ È4x Š È ‹“ † $ dx œ 8'1 Š È"x ‹ ˆ "x ‰ dx œ 8'1 x$Î# dx x %
4
%
œ 8 2x"Î# ‘ " œ 8[1 (2)] œ 8. So x œ
My M
4
œ
4
œ 2 Ê (xß y) œ (2ß 0) is the center of mass.
16 8
(c)
28. (a) We use the disk method: V œ 'a 1R# (x) dx œ '1 1 ˆ x4# ‰ dx œ 41'1 x# dx œ 41 x" ‘ " b
4
4
%
‘ œ 41 " 4 (1) œ 1[1 4] œ 31
(b) We model the distribution of mass with vertical strips: Mx œ ' µ y dm œ '1
4
2 œ 2'1 x$Î# dx œ 2 ’ È x dm œ '1 x † “ œ 2[1 (2)] œ 2; My œ ' µ x %
4
$Î#
4
"
%
2‘ œ 2 ’ 2x3 “ œ 2 16 3 3 œ "
œ 2(4 2) œ 4. So x œ
My M
28 3
œ
; M œ ' dm œ '1
4
ˆ 28 ‰ 3 4
œ
7 3
and y œ
2 x
Mx M
† $ dx œ 2'1
4
œ
2 4
œ
" #
Èx x
2 x
ˆ 2x ‰ 2
† ˆ 2x ‰ † $ dx œ '1
4
2 x#
† Èx dx
† $ dx œ 2'1 x"Î# dx 4
dx œ 2'1 x"Î# dx œ 2 2x"Î# ‘ " %
4
Ê (xß y) œ ˆ 73 ß "# ‰ is the center of mass.
(c)
29. The mass of a horizontal strip is dm œ $ dA œ $ L dy, where L is the width of the triangle at a distance of y above its base on the x-axis as shown in the figure in the text. Also, by similar triangles we have Ê Lœ
b h
(h y). Thus, Mx œ ' µ y dm œ '0 $ y ˆ bh ‰ (h y) dy œ h
œ
$b h
Š h#
$
h$ 3‹
œ
$b h
#
h# 2‹
Šh
œ $ bh# ˆ "# 3" ‰ œ œ
$ bh 2
. So y œ
Mx M
$ bh# 6
œ
$b h
# ˆ 2 ‰ Š $bh 6 ‹ $ bh
œ
h 3
œ
hy h
'0h ahy y# b dy œ $hb ’ hy#
; M œ ' dm œ '0 $ ˆ hb ‰ (h y) dy œ h
L b
$b h
#
h
y$ 3 “!
'0h ah yb dy œ $hb ’hy y2 “ h
Ê the center of mass lies above the base of the
#
!
Section 6.4 Moments and Centers of Mass triangle one-third of the way toward the opposite vertex. Similarly the other two sides of the triangle can be placed on the x-axis and the same results will occur. Therefore the centroid does lie at the intersection of the medians, as claimed. 30. From the symmetry about the y-axis it follows that x œ 0. It also follows that the line through the points (!ß !) and (!ß $) is a median Ê y œ "3 (3 0) œ 1 Ê (xß y) œ (!ß ").
31. From the symmetry about the line x œ y it follows that x œ y. It also follows that the line through the points (!ß !) and ˆ "# ß "# ‰ is a median Ê y œ x œ 23 † ˆ "# 0‰ œ 3" Ê (xß y) œ ˆ "3 ß 3" ‰ . 32. From the symmetry about the line x œ y it follows that x œ y. It also follows that the line through the point (!ß !) and ˆ #a ß #a ‰ is a median Ê y œ x œ 32 ˆ #a 0‰ œ 3" a Ê (xß y) œ ˆ 3a ß 3a ‰ . 33. The point of intersection of the median from the vertex (0ß b) to the opposite side has coordinates ˆ!ß #a ‰ Ê y œ (b 0) † "3 œ b3 and x œ ˆ #a !‰ † 32 œ 3a Ê (xß y) œ ˆ 3a ß b3 ‰ .
34. From the symmetry about the line x œ
a #
it follows that
xœ It also follows that the line through the points a ˆ # ß !‰ and ˆ #a ß b‰ is a median Ê y œ "3 (b 0) œ b3 a #.
Ê (xß y) œ ˆ #a ß b3 ‰ .
35. y œ x"Î# Ê dy œ
" #
x"Î# dx
Ê ds œ È(dx)# (dy)# œ É1 Mx œ $ '0 Èx É1 2
œ $ '0 Éx 2
dx œ
" ‰$Î# 4
œ
2$ 3
œ
2$ ˆ 9 ‰$Î# 3 ’ 4
’ˆ2
" 4
ˆ 4" ‰
" 4x
" 4x
dx
2$ 3
$Î# ’ˆx 4" ‰ “
dx ;
# !
ˆ 4" ‰$Î# “ $Î#
“œ
2$ 3
"‰ ˆ 27 8 8 œ
13$ 6
391
392
Chapter 6 Applications of Definite Integrals
36. y œ x$ Ê dy œ 3x# dx Ê dx œ É(dx)# a3x# dxb# œ È1 9x% dx; Mx œ $ '0 x$ È1 9x% dx; 1
" 36
[u œ 1 9x% Ê du œ 36x$ dx Ê
du œ x$ dx;
x œ 0 Ê u œ 1, x œ 1 Ê u œ 10] Ä Mx œ $ '1
10
" 36
u"Î# du œ
$ 36
23 u$Î# ‘ "! œ "
$ 54
ˆ10$Î# 1‰
37. From Example 6 we have Mx œ '0 a(a sin ))(k sin )) d) œ a# k'0 sin# ) d) œ 1
œ
a# k #
)
sin 2) ‘ 1 # !
œ
a# k 1 #
1
'01 (1 cos 2)) d)
; My œ '0 a(a cos ))(k sin )) d) œ a# k '0 sin ) cos ) d) œ 1
1
M œ '0 ak sin ) d) œ ak[ cos )]1! œ 2ak. Therefore, x œ 1
a# k #
My M
œ 0 and y œ
Mx M
a# k #
1
csin# )d ! œ 0;
#
" ‰ œ Š a 2k1 ‹ ˆ 2ak œ
a1 4
Ê ˆ!ß a41 ‰
is the center of mass. 38. Mx œ ' µ y dm œ '0 (a sin )) † $ † a d) 1
œ '0 aa# sin )b a1 k kcos )kb d) 1
œ a# '0 (sin ))(1 k cos )) d) 1Î2
a# '1Î2 (sin ))(1 k cos )) d) 1
œ a# '0 sin ) d) a# k'0 sin ) cos ) d) a# '1Î2 sin ) d) a# k '1Î2 sin ) cos ) d) 1Î2
1Î2
1Î#
#
1
a# k ’ sin# ) “
œ a# [ cos )]!
1Î# !
1
#
a# [ cos )]11Î# a# k ’ sin# ) “
1 1Î#
œ a [0 (1)] a k ˆ "# 0‰ a# [(1) 0] a# k ˆ0 "# ‰ œ a# #
#
a# k #
a#
œ 2a# a# k œ a# (2 k); 1 1 M œ'µ x dm œ ' (a cos )) † $ † a d) œ ' aa# cos )b a1 k kcos )kb d) y
0
0
œa
'0
œ a#
'01Î2 cos ) d) a# k '
#
1Î2
#
(cos ))(1 k cos )) d) a 1Î2
0
#
œ a [sin
1Î# ) ]!
œ a# (1 0)
a# k #
a# k #
a# k #
)
'1Î2 (cos ))(1 k cos )) d) 1
2) ‰ 2) ‰ ˆ 1 cos d) a# '1Î2 cos ) d) a# k'1Î2 ˆ 1 cos d) # #
sin 2) ‘ 1Î# # !
1
a# [sin )]11Î#
1
a# k #
ˆ 1# 0‰ (! 0)‘ a# (0 1)
)
a# k #
sin 2) ‘ 1 # 1Î#
(1 0) ˆ 1# 0‰‘ œ a#
a# k 1 4
a#
M œ '0 $ † a d) œ a'0 (1 k kcos )k) d) œ a '0 (1 k cos )) d) a'1Î2 (1 k cos )) d) 1
1
1Î#
œ a[) k sin )]! œ
a1 #
1Î2
a# k 1 4
œ 0;
1
a[) k sin )]11Î# œ a ˆ 1# k‰ 0‘ a (1 0) ˆ 1# k‰‘
ak a ˆ 1# k‰ œ a1 2ak œ a(1 2k). So x œ
My M
œ 0 and y œ
Mx M
œ
a# (2 k) a(1 #k)
œ
a(2 k) 1 #k
ka ‰ Ê ˆ0ß 2a 1 #k is the center of mass.
39. Consider the curve as an infinite number of line segments joined together. From the derivation of arc length we have that the length of a particular segment is ds œ È(dx)# (dy)# . This implies that Mx œ ' $ y ds, My œ ' $ x ds and M œ ' $ ds. If $ is constant, then x œ yœ
Mx M
' y ds
œ ' ds œ
' y ds length
My M
' x ds
œ ' ds œ
' x ds length
and
.
40. Applying the symmetry argument analogous to the one used in Exercise 13, we find that x œ 0. The typical
Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus x#
a vertical strip has center of mass: (µ x ßµ y ) œ Œxß 2 4p , length: a
mass: dm œ $ dA œ $ Ša œ
$ #
2
%
&
#
œ $ ’ax 8a$ Èpa 3
2
Èpa œ 2 † $ ’ax È Mx M
œŠ
Èpa
2
Èpa
x$ 12p “ !
œ
3 5
Èpa
8a# $Èpa 5
2$ paÈpa 12p ‹
œ 2$ Š2aÈpa
8a# $ Èpa 3 ‹ Š 8a$È 5 pa ‹
2
x& 80p# “ 0
œ 2 † #$ ’a# x
"6 ‰ ‰ œ 2a# $ Èpa ˆ 8080 œ 2a# $ Èpa ˆ 64 80 œ
x$ 12p “ c2 pa
. So y œ
c2Èpa
#
16 ‰ 80
width: dx, area: dA œ Ša
dx. Thus, Mx œ ' µ y dm œ 'c2Èpa "# Ša
#Èpa x x 'c22ÈÈpapa Ša# 16p ‹ dx œ #$ ’a# x 80p “
œ 2a# $ Èpa ˆ1
œ
x# 4p ‹
x# 4p ,
x# 4p ‹ Ša
x# 4p ‹ $
œ 4a$ Èpa ˆ1
dx,
dx 2& p# a# Èpa ‹ 80p#
œ $ Š2a# Èpa
; M œ ' dm œ $
x# 4p ‹
2
Èpa
'
c2Èpa
4 ‰ 12
Ša
x# 4p ‹
dx
œ 4a$ Èpa ˆ 121#4 ‰
a, as claimed.
41. Since the density is constant, its value will not affect our answers, so we can set $ œ ". 1Î2 ! A generalization of Example 6 yields M œ ' µ y dm œ ' a# sin ) d) œ a# [ cos )]1Î2 ! 1Î2 !
1Î# !
x
1Î2 !
œ a# cos ˆ 1# !‰ cos ˆ 1# !‰‘ œ a# (sin ! sin !) œ 2a# sin !; M œ ' dm œ '1Î# ! a d) œ a[)]11ÎÎ22 !! œ a ˆ 1# !‰ ˆ 1# !‰‘ œ 2a!. Thus, y œ Ê c œ 2a sin !. Then y œ
a(2a sin !) 2a!
œ
ac s ,
Mx M
œ
2a# sin ! 2a!
œ
a sin ! !
lim
! Ä !
(b)
sin ! ! cos ! ! ! cos !
! f(!)
¸
d h
œ
a sin ! ! sin ! ! cos ! ! ! cos ! .
a(sin ! ! cos !) a(! ! cos !)
œ
œ a cos ! d Ê d œ
0.4 0.664879
0.6 0.662615
0.8 0.659389
1.0 0.655145
6.5 AREAS OF SURFACES OF REVOLUTION AND THE THEOREMS OF PAPPUS 1. (a)
dy dx
#
% œ sec# x Ê Š dy dx ‹ œ sec x
Ê S œ 21'0
1Î4
(c) S ¸ 3.84
(tan x) È1 sec% x dx
(b)
a(sin ! ! cos !) . !
The graphs below suggest that
2 3.
0.2 0.666222
c #
as claimed.
42. (a) First, we note that y œ (distance from origin to AB) d Ê Moreover, h œ a a cos ! Ê
. Now s œ a(2!) and a sin ! œ
393
394 2. (a)
Chapter 6 Applications of Definite Integrals dy dx
#
(b)
œ 2x Ê Š dy dx ‹ œ 4x
Ê S œ 21'0 x# È1 4x# dx 2
(c) S ¸ 53.23
3. (a) xy œ 1 Ê x œ Ê S œ 21'1
2
" y
" y
Ê
dx dy
#
œ y"# Ê Š dx dy ‹ œ
" y%
(b)
È1 y% dy
(c) S ¸ 5.02
4. (a)
dx dy
#
# œ cos y Ê Š dx dy ‹ œ cos y
(b)
Ê S œ 21'0 (sin y) È1 cos# y dy 1
(c) S ¸ 14.42
# 5. (a) x"Î# y"Î# œ 3 Ê y œ ˆ3 x"Î# ‰ "Î# ‰ ˆ ˆ Ê dy "# x"Î# ‰ dx œ 2 3 x #
"Î# ‰ ˆ Ê Š dy dx ‹ œ 1 3x
#
# # Ê S œ 21'1 ˆ3 x"Î# ‰ É1 a1 3x"Î# b dx 4
(c) S ¸ 63.37
(b)
Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus dx dy
6. (a)
#
"Î# ‰ ˆ œ 1 y"Î# Ê Š dx dy ‹ œ 1 y
#
395
(b)
# Ê S œ 21 '1 ˆy 2Èy‰ É1 a1 y"Î# b dx 2
(c) S ¸ 51.33
dx dy
7. (a)
#
(b)
# œ tan y Ê Š dx dy ‹ œ tan y
Ê S œ 21'0 Š'0 tan t dt‹ È1 tan# y dy 1Î3
y
œ 21'0 Š'0 tan t dt‹ sec y dy 1Î3
y
(c) S ¸ 2.08
dy dx
8. (a)
#
(b)
# œ Èx# 1 Ê Š dy dx ‹ œ x 1
È5
Ê S œ 21'1 Š'1 Èt# 1 dt‹ È1 ax# 1b dx
È5
x
œ 21'1 Š'1 Èt# 1 dt‹ x dx x
(c) S ¸ 8.55
9. y œ œ
x #
1È5 #
Ê
dy dx
' ˆ x ‰ É1 œ "# ; S œ 'a 21y Ê1 Š dy dx ‹ dx Ê S œ 0 21 #
x #
4
" 4
dx œ
1È5 #
'04 x dx
%
# ’ x# “ œ 41È5; Geometry formula: base circumference œ 21(2), slant height œ È4# 2# œ 2È5
!
Ê Lateral surface area œ
10. y œ
#
b
Ê x œ 2y Ê
dx dy
" #
(41) Š2È5‹ œ 41È5 in agreement with the integral value
# È È ' È # ' œ 2; S œ 'c 21x Ê1 Š dx dy ‹ dy œ 0 21 † 2y 1 2 dy œ 41 5 0 y dy œ 21 5 cy d ! #
d
2
2
#
œ 21È5 † 4 œ 81È5; Geometry formula: base circumference œ 21(4), slant height œ È4# 2# œ 2È5 Ê Lateral surface area œ " (81) Š2È5‹ œ 81È5 in agreement with the integral value #
11.
dy dx
' œ "# ; S œ 'a 21yÊ1 Š dy dx ‹ dx œ 1 21 b
#
3
1È5 #
(x 1) #
É1 ˆ "# ‰# dx œ
1È5 #
'13 (x 1) dx
œ
1È5 #
#
’ x# x“
$ "
È ˆ 9# 3‰ ˆ "# 1‰‘ œ 1 # 5 (4 2) œ 31È5; Geometry formula: r" œ "# "# œ 1, r# œ 3# "# œ 2, œ slant height œ È(2 1)# (3 1)# œ È5 Ê Frustum surface area œ 1(r" r# ) ‚ slant height œ 1(1 2)È5
œ 31È5 in agreement with the integral value
396
Chapter 6 Applications of Definite Integrals
12. y œ
x #
" #
Ê x œ 2y 1 Ê
dx œ 2; S œ 'c 21x Ê1 Š dy ‹ dy œ '1 21(2y 1)È1 4 dy œ 21È5 '1 (2y 1) dy #
d
dx dy
#
2
2
œ 21È5 cy# yd " œ 21È5 [(4 2) (1 1)] œ 41È5; Geometry formula: r" œ 1, r# œ 3, slant height œ È(2 1)# (3 1)# œ È5 Ê Frustum surface area œ 1(1 3)È5 œ 41È5 in agreement with the integral value 13.
dy dx
#
x# 3
œ
’u œ 1
x% 9
Ê S œ '0
2
x% 9
Ê Š dy dx ‹ œ Ê du œ
4 9
x œ 0 Ê u œ 1, x œ 2 Ê u œ Ä S œ 21 '1
25Î9
14.
œ
1 3
dy dx
œ
" 1 2 4 du œ # 3 1 ˆ 12527 ‰ œ 98811 3 27 #
Ê S œ '3Î4 21Èx É1 œ 21'3Î4 Éx 15Î4
15.
œ
’ˆ 15 4
œ
41 3
(8 1) œ
dy dx
" (2 2x) # È2x x#
œ
ˆ 43
dx;
dx;
#&Î*
u$Î# ‘ "
" 4x
dx
$Î# dx œ 21 ’ 23 ˆx 4" ‰ “
" 4
" ‰$Î# 4
41 3
x$ 9
du œ
x% 9
" 4x
x"Î# Ê Š dy dx ‹ œ 15Î4
É1
25 ‘ 9
u"Î# †
ˆ 125 ‰ 27 1 œ " #
" 4
x$ dx Ê
21 x$ 9
" ‰$Î# “ 4
"&Î% $Î%
41 3
$ ’ˆ 24 ‰
Ê Š dy dx ‹ œ
(1 x)# 2x x#
œ
1“
281 3
œ
#
1x È2x x#
Ê S œ '0 5 21È2x x# É1 1Þ5 Þ
œ 21'0 5 È2x 1Þ5 Þ
(1 x)# 2x x#
È x# 1 2x x# x# 2x È 2x x#
dx
dx
œ 21'0 5 dx œ 21[x]"Þ& !Þ& œ 21 1Þ5 Þ
16.
dy dx
" 2È x 1
œ
#
dy Ê Š dx ‹ œ
" 4(x 1)
Ê S œ '1 21Èx 1 É1 5
œ 21'1 É(x 1) 5
" 4
" 4(x 1)
dx
dx œ 21'1 Éx 5
&
$Î# œ 21 ’ 23 ˆx 54 ‰ “ œ
17.
œ
41 3
œ
1 6
dx dy
‰$Î# ’ˆ 25 4
5 4
41 ˆ 5 ‰$Î# 3 ’ 5 4 " $ $ ˆ 94 ‰$Î# “ œ 431 Š 52$ 32$ ‹
(125 27) œ
981 6
œ
dx ˆ1 45 ‰$Î# “
491 3
% ' œ y# Ê Š dx dy ‹ œ y Ê S œ 0 #
u œ 1 y% Ê du œ 4y$ dy Ê
1
" 4
21 y$ 3
È1 y% dy;
du œ y$ dy; y œ 0
Ê u œ 1, y œ 1 Ê u œ 2d Ä S œ '1 21 ˆ "3 ‰ u"Î# ˆ 4" du‰ 2
œ
1 6
'12 u"Î# du œ 16 32 u$Î# ‘ #" œ 19 ŠÈ8 1‹
Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus 18. x œ ˆ "3 y$Î# y"Î# ‰ Ÿ 0, when 1 Ÿ y Ÿ 3. To get positive area, we take x œ ˆ "3 y$Î# y"Î# ‰ Ê
dx dy
#
œ "# ˆy"Î# y"Î# ‰ Ê Š dx dy ‹ œ
" 4
ay 2 y" b
Ê S œ '1 21 ˆ "3 y$Î# y"Î# ‰ É1 4" ay 2 y" b dy 3
œ 21'1 ˆ 3" y$Î# y"Î# ‰ É 4" ay 2 y" b dy 3
œ 21'1 ˆ "3 y$Î# y"Î# ‰ œ 1'1 ˆ "3 y# 3
2 3
dx dy
œ
" È 4 y
œ 41 '0
15Î4
œ
20.
dx dy
81 3
œ
3
# $
y 1‰ dy œ 1 ’ y9
œ 19 (18 1 3) œ 19.
dy œ 1'1 y"Î# ˆ 3" y 1‰ Šy"Î#
Éay"Î# y"Î# b#
3
#
Ê Š dx dy ‹ œ
Ê S œ '0
15Î4
" 4 y
5È 5 8 ‹
œ
81 3
#
" È2y 1
Ê Š dx dy ‹ œ
Š 40
È 5 5 È 5 ‹ 8
œ
È
È5
1 12
‹œ
#
41 È 2 3
’1$Î# ˆ 58 ‰
" 2y 1
$Î#
“œ
#
1 dy œ ÊŠy'
" #
" 16y' ‹
2
È2
È2
dy dx
œ
" #
aa# x# b
Ê S œ 21'ca Èa# x# É1 a
œ
21 r h
dy dx
#
#
É h h# r
25. y œ cos x Ê
œ
41 È 2 3
5$Î# “
5È 5 ‹ 8È 8
Š1
21 40
" #
" 16y' ‹
dy
dy œ 21'1 ˆy% "4 y# ‰ dy 2
" 4y$ ‹
(8 † 31 5) œ
2531 20
È2
x ax# 1b dx œ 21'0 ax$ xb dx œ 21 ’ x4
"Î#
x# aa # x # b
(2x) œ
x È a# x#
%
#
Ê Š dy dx ‹ œ
È# x# # “!
r h
#
Ê Š dy dx ‹ œ
r# h#
dx œ 21'ca Èaa# x# b x# dx œ 21'ca a dx œ 21a[x]ca a a
a
Ê S œ 21 '0
h
#
#
r h
x É1
r# h#
dx œ 21'0
h
r h
#
#
x É h h# r dx
#
0
#
# È1 sin# x dx ' œ sin x Ê Š dy dx ‹ œ sin x Ê S œ 21 c1Î2 (cos x) #
œ 21 ˆ 44 22 ‰ œ 41
x# aa # x # b
'0h x dx œ 2h1r Èh# r# ’ x# “ h œ 2h1r Èh# r# Š h# ‹ œ 1rÈh# r# dy dx
$Î#
1
œ 21a[a (a)] œ (21a)(2a) œ 41a# x Ê
1‰
È2
23. y œ Èa# x# Ê
r h
" 3
Ê dy œ xÈx# 2 dx Ê ds œ È1 a2x# x% b dx Ê S œ 21'0 x È1 2x# x% dx
$Î#
œ 21'0 xÉax# 1b# dx œ 21'0
24. y œ
dy œ 21'5Î8 È(2y 1) 1 dy
2
#
ax# 2b
" 9
È(4 y) 1 dy
5$Î# “ œ 831 ’ˆ 45 ‰
dy; S œ '1 21y ds œ 21'1 y Šy$
" 4y$ ‹
"
" 3
1‰‘ œ 1 ˆ3
1 dy œ ÊŠy'
"‰ "‰ ˆ " " ‰‘ œ 21 ˆ 31 œ 21 ’ y5 4" y" “ œ 21 ˆ 32 5 8 5 4 5 8 œ
22. y œ
15Î4
15 ‰$Î# 4
1
" 4y$ ‹
dy œ Šy$
&
dy œ 41'0
" 3
Š16È2 5È5‹
21. ds œ Èdx# dy# œ ÊŠy$ " 4y$ ‹
3
351È5 3
"
5 Š 8†2 8†22È 2
dy œ 1 '1 ˆ 3" y 1‰ (y 1) dy
3‰ ˆ "9
9 3
" 4y
21 † 2È4 y É1
Ê S œ '5Î8 21È2y 1 É1
" 2y1
1
œ ÊŠy$
"
È5 y dy œ 41 23 (5 y)$Î# ‘ "&Î% œ 831 ’ˆ5 !
Š 5È 5
41 È 2 3
y“ œ 1 ˆ 27 9
161 9
œ 21'5Î8 È2 y"Î# dy œ 21È2 23 y$Î# ‘ &Î) œ œ
$
y# 3
" ‹ y"Î#
1Î2
397
398
Chapter 6 Applications of Definite Integrals
26. y œ ˆ1 x#Î$ ‰
$Î#
Ê
dy dx
œ
Ê S œ 2'0 21 ˆ1 x#Î$ ‰ 1
3 #
ˆ1 x#Î$ ‰"Î# ˆ 23 x"Î$ ‰ œ
$Î#
#
ˆ1x#Î$ ‰"Î# x"Î$
Ê Š dy dx ‹ œ
1x#Î$ x#Î$
œ
" x#Î$
1
$Î# " É1 ˆ x#Î$ 1‰ dx œ 41'0 ˆ1 x#Î$ ‰ Èx#Î$ dx 1
$Î# œ 41'0 ˆ1 x#Î$ ‰ x"Î$ dx; u œ 1 x#Î$ Ê du œ 23 x"Î$ dx Ê 32 du œ x"Î$ dx; 1
! x œ 0 Ê u œ 1, x œ 1 Ê u œ 0d Ä S œ 41'1 u$Î# ˆ 3# du‰ œ 61 25 u&Î# ‘ " œ 61 ˆ0 25 ‰ œ 0
121 5
# # # È16# y# 27. The area of the surface of one wok is S œ 'c 21x Ê1 Š dx dy ‹ dy. Now, x y œ 16 Ê x œ #
d
Ê
dx dy
œ c7
#
y È16# y#
Ê Š dx dy ‹ œ
c7
; S œ 'c16 21È16# y# É1
y# 16# y#
y# 16# y#
c7
dy œ 21'c16 Èa16# y# b y# dy
œ 21'c16 16 dy œ 321 † 9 œ 2881 ¸ 904.78 cm# . The enamel needed to cover one surface of one wok is
V œ S † 0.5 mm œ S † 0.05 cm œ (904.78)(0.05) cm$ œ 45.24 cm$ . For 5000 woks, we need 5000 † V œ 5000 † 45.24 cm$ œ (5)(45.24)L œ 226.2L Ê 226.2 liters of each color are needed. 28. y œ Èr# x# Ê œ 21'a
abh
dy dx
abh
2x È r# x #
œ
Èar# x# b x# dx œ 21r' a
29. y œ ÈR# x# Ê œ 21'a
œ "#
dy dx
œ "#
2x È R # x#
x Èr# x#
abh
œ
abh
30. (a) x# y# œ 45# Ê x œ È45# y# Ê S œ 'c22 5 21 È45# y# É1 Þ
x# r# x # ;
S œ 21 'a
abh
Èr# x# É1
x# r# x#
dx
dx œ 21rh, which is independent of a. #
x È R # x#
ÈaR# x# b x# dx œ 21R ' a
45
#
Ê Š dx dy ‹ œ
y# 45# y#
dx Ê Š dy ‹ œ
x# R # x# ;
S œ 21'a
abh
ÈR# x# É1
x# R # x#
dx
dx œ 21Rh dx dy
œ
y È45# y#
dy œ 21 '
#
Ê Š dx dy ‹ œ
y# 45# y#
;
Èa45# y# b y# dy œ 21 † 45' 22 5
45
Þ
45 22Þ5
dy
œ (21)(45)(67.5) œ 60751 square feet (b) 19,085 square feet dy ' È1 1 dx œ 21 ' (x)È2 dx 21' xÈ2 dx 31. y œ x Ê Š dy dx ‹ œ 1 Ê Š dx ‹ œ 1 Ê S œ 21 c1 kxk c1 0 #
#
œ 2È21 ’ x# “ 32.
dy dx
œ
x# 3
!
#
"
2
#
!
#
4 9
0
2È21 ’ x# “ œ 2È21 ˆ0 "# ‰ 2È21(2 0) œ 5È21
Ê Š dy dx ‹ œ
Ê du œ
2
x% 9
Ê by symmetry of the graph that S œ 2 'cÈ3 21 Š x9 ‹ É1 0
$
x% 9
dx; ’u œ 1
x$ dx Ê "4 du œ x9 dx; x œ È3 Ê u œ 2, x œ 0 Ê u œ 1“ Ä S œ 41'2 u"Î# ˆ "4 ‰ du 1
$
" œ 1'2 u"Î# du œ 1 23 u$Î# ‘ # œ 1 Š 23 23 È8‹ œ 1
È3
È3
21 3
ŠÈ8 1‹ . If the absolute value bars are dropped the
integral for S œ 'cÈ3 21f(x) ds will equal zero since 'cÈ3 21 Š x9 ‹ É1 $
x% 9
dx is the integral of an odd function
over the symmetric interval È3 Ÿ x Ÿ È3.
33.
dx dt
œ sin t and
dy dt
#
È( sin t)# (cos t)# œ 1 Ê S œ ' 21y ds ‰ Š dy œ cos t Ê Êˆ dx dt dt ‹ œ #
œ '0 21(2 sin t)(1) dt œ 21 c2t cos td #!1 œ 21[(41 1) (0 1)] œ 81# 21
34.
dx dt
x% 9
œ t"Î# and
dy dt
#
#
#
Èt t" œ É t ‰ Š dy œ t"Î# Ê Êˆ dx dt dt ‹ œ
1 t
Ê S œ ' 21x ds
Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus È3
œ '0 21 ˆ 23 t$Î# ‰ É t
#
" t
dt œ
È3
'0
#
21 ˆ 23 t$Î# ‰ É t #
f(t) œ 21 ˆ 23 t$Î# ‰ É t Ê
35.
dx dt
È3
'0
F(t) dt œ
œ 1 and
È2
dy dt
281 9
" t
È3
'0
tÈt# 1 dt; cu œ t# 1 Ê du œ 2t dt; t œ 0 Ê u œ 1,
'14 231 Èu du œ 491 u$Î# ‘ %" œ 2891
’t œ È3 Ê u œ 4“ Ä Note:
41 3
1 t
dt is an improper integral but limb f(t) exists and is equal to 0, where tÄ!
. Thus the discontinuity is removable: define F(t) œ f(t) for t 0 and F(0) œ 0
. #
#
# È2‹ œ Ét# 2È2 t 3 Ê S œ ' 21x ds ‰ Š dy œ t È2 Ê Êˆ dx dt dt ‹ œ Ê1 Št #
œ 'cÈ2 21 Št È2‹ Ét# 2È2 t 3 dt; ’u œ t# 2È2 t 3 Ê du œ Š2t 2È2‹ dt; t œ È2 Ê u œ 1,
* t œ È2 Ê u œ 9“ Ä '1 1Èu du œ 23 1u$Î# ‘ " œ 9
36.
dx dt
œ aa1 cos tb and
dy dt
21 3
(27 1) œ
521 3
#
#
‰ Š dy Éc aa1 cos tb d# aa sin tb# œ a sin t Ê Êˆ dx dt dt ‹ œ
œ Èa2 2 a2 cos t a2 cos2 t a2 sin2 t œ È2a2 2a2 cos t œ aÈ2È1 cos t Ê S œ ' 21y ds œ '0 21 aa1 cos tb † aÈ2È1 cos t dt œ 2È2 1 a2 '0 a1 cos tb3/2 dt 21
37.
dx dt
œ 2 and
21
dy dt
È2# 1# œ È5 Ê S œ ' 21y ds œ ' 21(t 1)È5 dt ‰ Š dy œ 1 Ê Êˆ dx dt dt ‹ œ 0 #
#
1
"
# œ 21È5 ’ t2 t“ œ 31È5. Check: slant height is È5 Ê Area is 1(1 2)È5 œ 31È5 .
!
38.
dx dt
œ h and
dy dt
Èh# r# Ê S œ ' 21y ds œ ' 21rtÈh# r# dt ‰ Š dy œ r Ê Êˆ dx dt dt ‹ œ 0
œ 21rÈh# r#
#
#
1
'01 t dt œ 21rÈh# r# ’ t2 “ " œ 1rÈh# r# . #
!
Check: slant height is Èh# r# Ê Area is
1rÈh# r# . 39. (a) An equation of the tangent line segment is (see figure) y œ f(mk ) f w (mk )(x mk ). When x œ xkc1 we have r" œ f(mk ) f w (mk )(x51 mk ) œ f(mk ) f w (mk ) ˆ ?#xk ‰ œ f(mk ) f w (mk ) when x œ xk we have r# œ f(mk ) f w (mk )(x5 mk ) k œ f(mk ) f w (mk ) ?x # ;
(b) L#k œ (?xk )# (r# r" )#
?x k #
;
#
ˆf w (mk ) ?#xk ‰‘ œ (?xk )# [f w (mk )?xk ]# Ê Lk œ È(?xk )# [f w (mk )?xk ]# , as claimed (c) From geometry it is a fact that the lateral surface area of the frustum obtained by revolving the tangent œ (?xk )# f w (mk )
?x k #
line segment about the x-axis is given by ?Sk œ 1(r" r# )Lk œ 1[2f(mk )] Éa?xk b# [f w (mk )?xk ]# using parts (a) and (b) above. Thus, ?Sk œ 21f(mk ) È1 [f w (mk )]# ?xk . ! ?Sk œ lim ! 21f(mk ) È1 [f w (mk )]# ?xk œ ' 21f(x) È1 [f w (x)]# dx (d) S œ n lim Ä_ nÄ_ a kœ1 kœ1 n
n
b
399
400
Chapter 6 Applications of Definite Integrals È3
40. S œ 'a 21f(x) dx œ '0 b
21 †
dx œ
x È3
1 È3
È3
c x# d ! œ
31 È3
œ È31
41. The centroid of the square is located at (#ß #). The volume is V œ (21) ayb (A) œ (21)(2)(8) œ 321 and the surface area is S œ (21) ayb (L) œ (21)(2) Š4È8‹ œ 32È21 (where È8 is the length of a side). 42. The midpoint of the hypotenuse of the triangle is ˆ 3# ß 3‰ Ê y œ 2x is an equation of the median Ê the line y œ 2x contains the centroid. The point ˆ 3# ß $‰ is 3È 5 #
units from the origin Ê the x-coordinate of the
# centroid solves the equation Ɉx 3# ‰ (2x 3)#
œ
È5 #
Ê ˆx# 3x 94 ‰ a4x# 12x 9b œ
5 4
Ê 5x# 15x 9 œ 1 Ê x# 3x 2 œ (x 2)(x 1) œ 0 Ê x œ 1 since the centroid must lie inside the triangle Ê y œ 2. By the Theorem of Pappus, the volume is V œ (distance traveled by the centroid)(area of the region) œ 21 a5 xb "# (3)(6)‘ œ (21)(4)(9) œ 721
43. The centroid is located at (#ß !) Ê V œ (21) axb (A) œ (21)(2)(1) œ 41# 44. We create the cone by revolving the triangle with vertices (0ß 0), (hß r) and (hß 0) about the y-axis (see the accompanying figure). Thus, the cone has height h and base radius r. By Theorem of Pappus, the lateral surface area swept out by the hypotenuse L is given by S œ 21yL œ 21 ˆ r ‰ Èh# r# #
œ 1rÈr# h# . To calculate the volume we need the position of the centroid of the triangle. From the diagram we see that the centroid lies on the line y œ œ
" 3
Éh#
r# 4
#
r 2h
# x. The x-coordinate of the centroid solves the equation É(x h)# ˆ 2hr x #r ‰
#
#
#
Ê Š 4h4h# r ‹ x# Š 4h 2h r ‹ x
inside the triangle Ê y œ
r 2h
47. V œ 21 yA Ê
4 3
2 ar# 4h# b 9
œ0 Ê xœ
2h 3
or
4h 3
Ê xœ
x œ 3r . By the Theorem of Pappus, V œ 21 ˆ 3r ‰‘ ˆ "# hr‰ œ
45. S œ 21 y L Ê 41a# œ a21yb (1a) Ê y œ 46. S œ 213 L Ê 21 ˆa
r# 4
2a ‰‘ (1a) 1
2a 1,
since the centroid must lie
1r# h.
and by symmetry x œ 0
œ 21a# (1 2)
1ab# œ a21yb ˆ 1#ab ‰ Ê y œ
48. V œ 213A Ê V œ 21 ˆa
" 3
2h 3 ,
4a ‰‘ 1a# Š # ‹ 31
œ
4b 31
and by symmetry x œ 0
1a$ (31 4) 3
49. V œ 213 A œ (21)(area of the region) † (distance from the centroid to the line y œ x a). We must find the 4a ‰ distance from ˆ0ß 31 to y œ x a. The line containing the centroid and perpendicular to y œ x a has slope 1 and contains the point ˆ!ß 34a1 ‰ . This line is y œ x 34a1 . The intersection of y œ x a and y œ x 34a1 is the point ˆ 4a 613a1 ß 4a 613a1 ‰ . Thus, the distance from the centroid to the line y œ x a is Ɉ 4a 613a1 ‰# ˆ 34a1
4a 61
3a1 ‰# 61
œ
È2 (4a 3a1) 61
Ê V œ (21) Š
È2 (4a 3a1) # ‹ Š 1#a ‹ 61
œ
È2 1a$ (4 31) 6
Section 6.6 Work ‰ 50. The line perpendicular to y œ x a and passing through the centroid ˆ!ß 2a 1 has equation y œ x intersection of the two perpendicular lines occurs when x a œ x
Ê xœ
2a 1
2a a1 21
Ê yœ
2a ‰# #
a(21) È 21
#
the distance from the centroid to the line y œ x a is Ɉ 2a 2 1a 0‰ ˆ 2a 2 1a
œ
2a 1 . The 2a a1 21 . Thus
.
1 ) Therefore, by the Theorem of Pappus the surface area is S œ 21 ’ a(2 “ (1a) œ È21a# (2 1). È 21
51. From Example 4 and Pappus's Theorem for Volumes we have the moment about the x-axis is Mx œ y M #
œ ˆ 34a1 ‰ Š 1#a ‹ œ
2a$ 3
.
6.6 WORK 1. The force required to stretch the spring from its natural length of 2 m to a length of 5 m is F(x) œ kx. The work done by F is W œ '0 F(x) dx œ k '0 x dx œ 3
3
k #
$
cx# d ! œ
9k # .
This work is equal to 1800 J Ê
k œ 1800
9 #
Ê k œ 400 N/m 2. (a) We find the force constant from Hooke's Law: F œ kx Ê k œ
Ê kœ
F x
800 4
œ 200 lb/in.
(b) The work done to stretch the spring 2 inches beyond its natural length is W œ '0 kx dx 2
œ 200 '0 x dx œ 200 ’ x# “ œ 200(2 0) œ 400 in † lb œ 33.3 ft † lb 2
#
# !
(c) We substitute F œ 1600 into the equation F œ 200x to find 1600 œ 200x Ê x œ 8 in. 3. We find the force constant from Hooke's law: F œ kx. A force of 2 N stretches the spring to 0.02 m N Ê 2 œ k † (0.02) Ê k œ 100 m . The force of 4 N will stretch the rubber band y m, where F œ ky Ê y œ Ê yœ
4N N 100 m
œ 100 '0
0Þ04
Ê y œ 0.04 m œ 4 cm. The work done to stretch the rubber band 0.04 m is W œ '0
F k
0Þ04
#
x dx œ 100 ’ x# “
!Þ!%
œ
!
(100)(0.04)# #
kx dx
œ 0.08 J
4. We find the force constant from Hooke's law: F œ kx Ê k œ
F x
Ê kœ
90 1
Ê k œ 90
N m. &
The work done to
‰ stretch the spring 5 m beyond its natural length is W œ '0 kx dx œ 90 '0 x dx œ 90 ’ x# “ œ (90) ˆ 25 # œ 1125 J 5
5
#
!
5. (a) We find the spring's constant from Hooke's law: F œ kx Ê k œ
F x
œ
21,714 8 5
œ
21,714 3
Ê k œ 7238
(b) The work done to compress the assembly the first half inch is W œ '0 kx dx œ 7238 '0 0Þ5
#
œ 7238 ’ x# “
!Þ& !
#
œ (7238) (0.5) # œ
(7238)(0.25) #
1Þ0
1Þ0
Þ
Þ
#
¸ 2714 in † lb 6. First, we find the force constant from Hooke's law: F œ kx Ê k œ compresses the scale x œ scale this far is W œ '0
1Î8
in, he/she must weigh F œ kx œ #
kx dx œ 2400 ’ x# “
"Î) !
x dx
¸ 905 in † lb. The work done to compress the assembly the
second half inch is: W œ '0 5 kx dx œ 7238 '0 5 x dx œ 7238 ’ x# “
" 8
lb in
0Þ5
œ
2400 2†64
F x
2,400 ˆ 8" ‰
"Þ! !Þ&
œ
œ
150 " ‰ ˆ 16
7238 #
c1 (0.5)# d œ
(7238)(0.75) #
œ 16 † 150 œ 2,400
lb in .
If someone
œ 300 lb. The work done to compress the
œ 18.75 lb † in. œ
25 16
ft † lb
7. The force required to haul up the rope is equal to the rope's weight, which varies steadily and is proportional to x, the length of the rope still hanging: F(x) œ 0.624x. The work done is: W œ '0 F(x) dx œ '0 0.624x dx 50
50
401
402
Chapter 6 Applications of Definite Integrals #
œ 0.624 ’ x# “
&! !
œ 780 J
8. The weight of sand decreases steadily by 72 lb over the 18 ft, at 4 lb/ft. So the weight of sand when the bag is x ft off the ground is Faxb œ "%% %x. The work done is: W œ 'a F(x) dx œ '0 a"%% %xbdx œ c144x 2x# d ! œ 1944 ft † lb b
18
")
9. The force required to lift the cable is equal to the weight of the cable paid out: F(x) œ (4.5)(180 x) where x is the position of the car off the first floor. The work done is: W œ '0
180
œ 4.5 ’180x
")!
x# # “!
180# # ‹
œ 4.5 Š180#
œ
4.5†180# #
F(x) dx œ 4.5'0
180
(180 x) dx
œ 72,900 ft † lb
10. Since the force is acting toward the origin, it acts opposite to the positive x-direction. Thus F(x) œ xk# . The b work done is W œ 'a xk# dx œ k 'a x"# dx œ k x" ‘ a œ k ˆ b" "a ‰ œ b
b
k(a b) ab
11. The force against the piston is F œ pA. If V œ Ax, where x is the height of the cylinder, then dV œ A dx Ê Work œ ' F dx œ ' pA dx œ 'ap
ap# ßV# b " ßV" b
p dV.
12. pV"Þ% œ c, a constant Ê p œ cV"Þ% . If V" œ 243 in$ and p" œ 50 lb/in$ , then c œ (50)(243)"Þ% œ 109,350 lb. ‘ ˆ 3#"!Þ% Thus W œ '243 109,350V"Þ% dV œ 109,350 œ 109,350 0.4 0.4V!Þ% #%$ $#
32
" ‰ #43!Þ%
ˆ 4" 9" ‰ œ 109,350 0.4
œ (109,350)(5) (0.4)(36) œ 37,968.75 in † lb. Note that when a system is compressed, the work done by the system is negative. 13. Let r œ the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate, the amount of water in the bucket is proportional to a#! xb, the distance the bucket is being raised. The leakage rate of the water is 0.8 lb/ft raised and the weight of the water in the bucket is F œ 0.8a#! xb. So: W œ '0 0.8a#! xb dx œ 0.8 ’20x 20
#! x# # “!
œ 160 ft † lb.
14. Let r œ the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate, the amount of water in the bucket is proportional to a#! xb, the distance the bucket is being raised. The leakage rate of the water is 2 lb/ft raised and the weight of the water in the bucket is F œ 2a#! xb. So: W œ '0 2a#! xb dx œ 2 ’20x 20
#! x# # “!
œ 400 ft † lb.
Note that since the force in Exercise 14 is 2.5 times the force in Exercise 13 at each elevation, the total work is also 2.5 times as great. 15. We will use the coordinate system given. (a) The typical slab between the planes at y and y ?y has a volume of ?V œ (10)(12) ?y œ 120 ?y ft$ . The force F required to lift the slab is equal to its weight: F œ 62.4 ?V œ 62.4 † 120 ?y lb. The distance through which F must act is about y ft, so the work done lifting the slab is about ?W œ force ‚ distance œ 62.4 † 120 † y † ?y ft † lb. The work it takes to lift all 20
the water is approximately W ¸ ! ?W 0 20
œ ! 62.4 † 120y † ?y ft † lb. This is a Riemann sum for 0
the function 62.4 † 120y over the interval 0 Ÿ y Ÿ 20. The work of pumping the tank empty is the limit of these sums:
Section 6.6 Work W œ '0 62.4 † 120y dy œ (62.4)(120) ’ y# “ 20
#
#! !
403
‰ œ (62.4)(120)(200) œ 1,497,600 ft † lb œ (62.4)(120) ˆ 400 #
5 ‰ (b) The time t it takes to empty the full tank with ˆ 11 –hp motor is t œ
W †lb 250 ftsec
œ
1,497,600 ft†lb †lb 250 ftsec
œ 5990.4 sec
œ 1.664 hr Ê t ¸ 1 hr and 40 min (c) Following all the steps of part (a), we find that the work it takes to lower the water level 10 ft is W œ '0 62.4 † 120y dy œ (62.4)(120) ’ y# “ 10
#
œ 1497.6 sec œ 0.416 hr ¸ 25 min (d) In a location where water weighs 62.26
"! !
‰ œ 374,400 ft † lb and the time is t œ œ (62.4)(120) ˆ 100 #
W †lb 250 ftsec
lb ft$ :
a) W œ (62.26)(24,000) œ 1,494,240 ft † lb. b) t œ 1,494,240 œ 5976.96 sec ¸ 1.660 hr Ê t ¸ 1 hr and 40 min 250 In a location where water weighs 62.59
lb ft$
a) W œ (62.59)(24,000) œ 1,502,160 ft † lb b) t œ 1,502,160 œ 6008.64 sec ¸ 1.669 hr Ê t ¸ 1 hr and 40.1 min 250 16. We will use the coordinate system given. (a) The typical slab between the planes at y and y ?y has a volume of ?V œ (20)(12) ?y œ 240 ?y ft$ . The force F required to lift the slab is equal to its weight: F œ 62.4 ?V œ 62.4 † 240 ?y lb. The distance through which F must act is about y ft, so the work done lifting the slab is about ?W œ force ‚ distance 20
œ 62.4 † 240 † y † ?y ft † lb. The work it takes to lift all the water is approximately W ¸ ! ?W 10 20
œ ! 62.4 † 240y † ?y ft † lb. This is a Riemann sum for the function 62.4 † 240y over the interval 10
10 Ÿ y Ÿ 20. The work it takes to empty the cistern is the limit of these sums: W œ '10 62.4 † 240y dy 20
#
œ (62.4)(240) ’ y# “ (b) t œ
W †lb 275 ftsec
œ
#!
œ (62.4)(240)(200 50) œ (62.4)(240)(150) œ 2,246,400 ft † lb
"! 2,246,400 ft†lb 275
¸ 8168.73 sec ¸ 2.27 hours ¸ 2 hr and 16.1 min
(c) Following all the steps of part (a), we find that the work it takes to empty the tank halfway is W œ '10 62.4 † 240y dy œ (62.4)(240) ’ y# “ 15
#
Then the time is t œ
W †lb 275 ftsec
œ
936,000 #75
"& "!
œ (62.4)(240) ˆ 225 #
100 ‰ #
‰ œ 936,000 ft. œ (62.4)(240) ˆ 125 #
¸ 3403.64 sec ¸ 56.7 min
(d) In a location where water weighs 62.26
lb ft$ :
a) W œ (62.26)(240)(150) œ 2,241,360 ft † lb. b) t œ 2,241,360 œ 8150.40 sec œ 2.264 hours ¸ 2 hr and 15.8 min 275 ‰ œ 933,900 ft † lb; t œ 933,900 c) W œ (62.26)(240) ˆ 125 # #75 œ 3396 sec ¸ 0.94 hours ¸ 56.6 min In a location where water weighs 62.59
lb ft$
a) W œ (62.59)(240)(150) œ 2,253,240 ft † lb. b) t œ 2,253,240 œ 8193.60 sec œ 2.276 hours ¸ 2 hr and 16.56 min 275 ‰ œ 938,850 ft † lb; t œ 938,850 c) W œ (62.59)(240) ˆ 125 # 275 ¸ 3414 sec ¸ 0.95 hours ¸ 56.9 min #
17. The slab is a disk of area 1x# œ 1ˆ y# ‰ , thickness ˜y, and height below the top of the tank a"! yb. So the work to pump #
the oil in this slab, ˜W, is 57a"! yb1ˆ y# ‰ . The work to pump all the oil to the top of the tank is W œ '0
10
571 # 4 a"!y
y$ bdy œ
571 4
$
’ "!$y
"! y% % “!
œ 11,8751 ft † lb ¸ 37,306 ft † lb.
404
Chapter 6 Applications of Definite Integrals #
18. Each slab of oil is to be pumped to a height of 14 ft. So the work to pump a slab is a"% yba1bˆ y# ‰ and since the tank is half full and the volume of the original cone is V œ "$ 1r# h œ "$ 1a ba"!b œ with half the volume the cone is filled to a height y, œ
571 "%y$ 4 ’ $
$ È &!! y% “ % !
#&!1 '
#&!1 $
ft3 , half the volume œ $ È &!!
$ œ $" 1 y% y Ê y œ È &!! ft. So W œ '0 #
#&!1 '
ft3 , and
571 # 4 a"%y
y$ b dy
¸ 60,042 ft † lb. #
‰ ?y 19. The typical slab between the planes at y and and y ?y has a volume of ?V œ 1(radius)# (thickness) œ 1 ˆ 20 # œ 1 † 100 ?y ft$ . The force F required to lift the slab is equal to its weight: F œ 51.2 ?V œ 51.2 † 1001 ?y lb Ê F œ 51201 ?y lb. The distance through which F must act is about (30 y) ft. The work it takes to lift all the 30
30
kerosene is approximately W ¸ ! ?W œ ! 51201(30 y) ?y ft † lb which is a Riemann sum. The work to pump the 0
0
tank dry is the limit of these sums: W œ '0 51201(30 y) dy œ 51201 ’30y
$! y# # “!
30
¸ 7,238,229.48 ft † lb
‰ œ (5120)(4501) œ 51201 ˆ 900 #
20. (Alternate Solution) Each method must pump all of the water the 15 ft to the base of the tank. Pumping to the rim requires all the water to be pumped an additional 6 feet. Pumping into the bottom requires that the water be pumped an average of 3 additional feet. Thus pumping through the valve requires È$ fta%1b6 ft3 a'#Þ% lb/ft3 b ¸ 14,115 ft † lb less work and thus less time. lb ft$
21. (a) Follow all the steps of Example 5 but make the substitution of 64.5 W œ '0
8
œ
64.51 4
64.51†8$ 3
(10 y)y# dy œ
64.51 4
$
’ 10y 3
%
y 4
)
“ œ !
64.51 4
$
Š 103†8
%
8 4
for 57
lb ft$ .
Then,
1‰ ‰ a8$ b ˆ 10 ‹ œ ˆ 64.5 4 3 2
œ 21.51 † 8$ ¸ 34,582.65 ft † lb
(b) Exactly as done in Example 5 but change the distance through which F acts to distance ¸ (13 y) ft. Then W œ '0
8
571 4
(13 y)y# dy œ
571 4
$
’ 13y 3
œ (191) a8# b (7)(2) ¸ 53.482.5 ft † lb
) y% 4 “!
œ
571 4
$
Š 133†8
8% 4‹
‰ œ ˆ 5741 ‰ a8$ b ˆ 13 3 2 œ
571†8$ †7 3 †4
22. The typical slab between the planes of y and y?y has a volume of about ?V œ 1(radius)# (thickness) # œ 1 ˆÈy‰ ?y œ xy ?y m$ . The force F(y) is equal to the slab's weight: F(y) œ 10,000 mN$ † ?V œ 110,000y ?y N. The height of the tank is 4# œ 16 m. The distance through which F(y) must act to lift the slab to the level of the top of the tank is about (16 y) m, so the work done lifting the slab is about ?W œ 10,0001y(16 y) ?y N † m. The work done lifting all the slabs from y œ 0 to y œ 16 to the top is 16
approximately W ¸ ! 10,0001y(16 y)?y. Taking the limit of these Riemann sums, we get 0
W œ '0 10,0001y(16 y) dy œ 10,0001'0 a16y y# b dy œ 10,0001 ’ 16y # 16
œ
10,000†1†16$ 6
16
#
"' y$ “ 3 !
$
œ 10,0001 Š 16#
16$ 3 ‹
¸ 21,446,605.9 J
23. The typical slab between the planes at y and y?y has a volume of about ?V œ 1(radius)# (thickness) # œ 1 ˆÈ25 y# ‰ ?y m$ . The force F(y) required to lift this slab is equal to its weight: F(y) œ 9800 † ?V #
œ 98001 ˆÈ25 y# ‰ ?y œ 98001 a25 y# b ?y N. The distance through which F(y) must act to lift the slab to the level of 4 m above the top of the reservoir is about (4 y) m, so the work done is approximately ?W ¸ 98001 a25 y# b (4 y) ?y N † m. The work done lifting all the slabs from y œ 5 m to y œ 0 m is 0
approximately W ¸ ! 98001 a25 y# b (4 y) ?y N † m. Taking the limit of these Riemann sums, we get c5
Section 6.6 Work W œ 'c5 98001 a25 y# b (4 y) dy œ 98001 'c5 a100 25y 4y# y$ b dy œ 98001 ’100C 0
0
œ 98001 ˆ500
25†25 #
4 3
† 125
625 ‰ 4
25 #
y# 34 y$
¸ 15,073,099.75 J
24. The typical slab between the planes at y and y?y has a volume of about ?V œ 1(radius)# (thickness) # œ 1 ˆÈ100 y# ‰ ?y œ 1 a100 y# b ?y ft$ . The force is F(y) œ 56ft$lb † ?V œ 561 a100 y# b ?y lb. The distance through which F(y) must act to lift the slab to the level of 2 ft above the top of the tank is about (12 y) ft, so the work done is ?W ¸ 561 a100 y# b (12 y) ?y lb † ft. The work done lifting all the slabs 10
from y œ 0 ft to y œ 10 ft is approximately W ¸ ! 561 a100 y# b (12 y) ?y lb † ft. Taking the limit of these 0 10 10 Riemann sums, we get W œ '0 561 a100 y# b (12 y) dy œ 561'0 a100 y# b (12 y) dy
œ 561'0 a1200 100y 12y# y$ b dy œ 561 ’1200C 10
œ 561 ˆ12,000
10,000 #
4 † 1000
10,000 ‰ 4
100y# #
12y$ 3
"! y% 4 “!
œ (561) ˆ12 5 4 5# ‰ (1000) ¸ 967,611 ft † lb.
It would cost (0.5)(967,611) œ 483,805¢ œ $4838.05. Yes, you can afford to hire the firm. 25. F œ m œ
" #
dv dt
œ mv
by the chain rule Ê W œ 'x mv x#
dv dx #
"
m cv# (x# ) v (x" )d œ
26. weight œ 2 oz œ
" #
weight 32
œ
" 8
3#
" #56
œ
28. weight œ 1.6 oz œ 0.1 lb Ê m œ
Wœ
" ˆ "# ‰ ˆ 256
" 8
lb Ê m œ
" 8
32
0.1 lb 32 ft/sec#
slugs œ
" 3 #0
œ " #56
14.5 16
6.5 16
lb Ê m œ
lb Ê m œ
14.5 (16)(32)
6.5 (16)(32)
" #56
0.3125 lb 32 ft/sec#
œ
0.3125 32
slugs;
slugs; W œ ˆ "# ‰ ˆ 3"#0 slugs‰ (280 ft/sec)# œ 122.5 ft † lb
slugs; 124 mph œ
"Î6
slugs and v" œ 0 ft/sec. Thus,
1 4
œ
1 4
(124)(5280) (60)(60)
ft † lb. Now W œ
¸ 181.87 ft/sec;
È2 4
The height the bearing reaches is s œ 8È2 t 16t# Ê at t œ
# È (16) Š 42 ‹
œ 2 ft
" #
mv# "# mv"# , where W œ
1 4
ft † lb,
" ft † lb. œ ˆ #" ‰ ˆ #56 slugs‰ v# Ê v œ 8È2 ft/sec. With v œ 0
at the top of the bearing's path and v œ 8È2 32t Ê t œ È Š8È2‹ Š 42 ‹
x#
6.5 slugs; W œ ˆ "# ‰ Š (16)(32) slugs‹ (132 ft/sec)# ¸ 110.6 ft † lb
1Î6
œ
dx œ m "# v# (x)‘ x"
"4.5 slugs; W œ ˆ "# ‰ Š (16)(32) slugs‹ (88 ft/sec)# ¸ 109.7 ft † lb
32. F œ (18 lb/ft)x Ê W œ '0 18x dx œ c9x# d ! 32
dv ‰ dx
slugs‰ (181.87 ft/sec) ¸ 64.6 ft † lb
31. weight œ 6.5 oz œ
mœ
"
#
30. weight œ 14.5 oz œ
" 8
x#
" slugs; W œ ˆ "# ‰ ˆ #56 slugs‰ (160 ft/sec)# ¸ 50 ft † lb
hr 1 min 5280 ft 27. 90 mph œ 901 hrmi † 601 min † 60 sec † 1 mi œ 132 ft/sec; m œ 0.3125 lb ‰ # W œ ˆ "# ‰ ˆ 32 ft/sec# (132 ft/sec) ¸ 85.1 ft † lb
29. weight œ 2 oz œ
dx œ m'x ˆv
mv## "# mv"# , as claimed.
lb; mass œ
2 16
dv dx
sec when the bearing is at the top of its path. È2 4
the bearing reaches a height of
405
! y% 4 “ &
406
Chapter 6 Applications of Definite Integrals
33. (a) From the diagram, rayb œ '! x œ '! É&!# ay 325b# for 325 Ÿ y Ÿ 375 ft. (b) The volume of a horizontal slice of the funnel # is ˜V ¸ 1rayb‘ ˜y #
œ 1”'! É&!# ay 325b# • ˜y (c) The work required to lift the single slice of water is ˜W ¸ 62.4˜Va$(& yb #
œ 62.4a$(& yb1”'! É&!# ay 325b# • ˜y. The total work to pump our the funnel is W #
œ '325 62.4a375 yb1”'! É50# ay 325b# • dy 375
¸ 6.3358 † 10( ft † lb. 34. (a) From the result in Example 6, the work to pump out the throat is 1,353,869,354 ft † lb. Thereform, the total work required to pump out the throat and the funnel is 1,353,869,354 63,358,000 œ 1,417227,354 ft † lb. (b) In horsepower-hours, the work required to pump out the glory hole is 1,417227,354 œ 715.8. Therefore, it would take 1.98†106 715.8 hp†h 1000 hp
œ 0.7158 hours ¸ 43 minutes.
35. We imagine the milkshake divided into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval [!ß (]. The typical slab between the planes at y and y ?y has a volume of about #
17.5 ‰ ?V œ 1(radius)# (thickness) œ 1 ˆ y14 ?y in$ . The force F(y) required to lift this slab is equal to its
weight: F(y) œ
4 9
41 9
?V œ
#
17.5 ‰ ˆ y14 ?y oz. The distance through which F(y) must act to lift this slab to
the level of 1 inch above the top is about (8 y) in. The work done lifting the slab is about 17.5)# ?W œ ˆ 491 ‰ (y14 (8 y) ?y in † oz. The work done lifting all the slabs from y œ 0 to y œ 7 is # 7
approximately W œ ! 0
41 9†14#
(y 17.5)# (8 y) ?y in † oz which is a Riemann sum. The work is the limit of
these sums as the norm of the partition goes to zero: W œ '0
7
œ
41 9†14#
œ
41 9†14#
'07 a2450 26.25y 27y# y$ b dy œ 9†4141 ’
7% 4
$
9†7
26.25 #
41 9†14#
(y 17.5)# (8 y) dy
%
#
’ y4 9y$
26.25 #
y# 2450y“
( !
#
† 7 2450 † 7“ ¸ 91.32 in † oz
36. We fill the pipe and the tank. To find the work required to fill the tank follow Example 6 with radius œ 10 ft. Then ?V œ 1 † 100 ?y ft$ . The force required will be F œ 62.4 † ?V œ 62.4 † 1001 ?y œ 62401 ?y lb. The distance through which F must act is y so the work done lifting the slab is about ?W" œ 62401 † y † ?y lb † ft. The work it takes to 385
385
lift all the water into the tank is: W" ¸ ! ?W" œ ! 62401 † y † ?y lb † ft. Taking the limit we end up with 360
W" œ '360 62401y dy œ 385
# $)& 62401 ’ y# “ $'!
360
œ
62401 #
c385# 360# d ¸ 182,557,949 ft † lb
To find the work required to fill the pipe, do as above, but take the radius to be Then ?V œ 1 †
" 36
$
?y ft and F œ 62.4 † ?V œ
integration: W# ¸ ! ?W# Ê W# œ '0 360 0
360
62.4 36
62.41 36
4 #
in œ
" 6
ft.
?y. Also take different limits of summation and
1y dy œ
62.41 36
#
$'!
’ y# “
!
#
1 ‰ 360 œ ˆ 62.4 Š # ‹ ¸ 352,864 ft † lb. 36
The total work is W œ W" W# ¸ 182,557,949 352,864 ¸ 182,910,813 ft † lb. The time it takes to fill the
Section 6.7 Fluid Pressures and Forces tank and the pipe is Time œ 37. Work œ '6 370 000
35ß780ß000 ß
1000 MG r#
ß
W 1650
¸
¸ 110,855 sec ¸ 31 hr
182,910,813 1650
dr œ 1000 MG '6 370 000
35ß780ß000 ß
ß
" œ (1000) a5.975 † 10#% b a6.672 † 10"" b Š 6,370,000
$&ß()!ß!!!
œ 1000 MG "r ‘ 'ß$(!ß!!!
dr r#
" 35,780,000 ‹
¸ 5.144 ‚ 10"! J
38. (a) Let 3 be the x-coordinate of the second electron. Then r# œ (3 1)# Ê W œ 'c1 F(3) d3 0
œ 'c1 a23(3‚101)# b d3 œ ’ 233‚10" “ 0
#*
#*
!
"
œ a23 ‚ 10#* b ˆ1 #" ‰ œ 11.5 ‚ 10#*
(b) W œ W" W# where W" is the work done against the field of the first electron and W# is the work done against the field of the second electron. Let 3 be the x-coordinate of the third electron. Then r#" œ (3 1)# and r## œ (3 1)# Ê W" œ '3
5
œ a23 ‚ 10
b ˆ "4 #" ‰ œ
#*
23 4
23‚10#* r#"
‚ 10
d3 œ '3
#*
5
23‚10#* (3 ")#
5 , and W# œ '3
&
œ 23 ‚ 10#* ’ 3 " " “ œ a23 ‚ 10#* b ˆ 6" 4" ‰ œ $
#* ‰ #* ‰ W œ W" W# œ ˆ 23 ˆ 23 œ 4 ‚ 10 12 ‚ 10
23 3
d3 œ 23 ‚ 10#* ’ 3 " " “ 23‚10#* r##
23‚10#* 12
5 d3 œ '3 23(3‚10")
#*
(3 2) œ
23 12
#
& $
d3
‚ 10#* . Therefore
‚ 10#* ¸ 7.67 ‚ 10#* J
6.7 FLUID PRESSURES AND FORCES 1. To find the width of the plate at a typical depth y, we first find an equation for the line of the plate's right-hand edge: y œ x 5. If we let x denote the width of the right-hand half of the triangle at depth y, then x œ 5 y and the total width is L(y) œ 2x œ 2(5 y). The depth of the strip is (y). The force exerted by the c2
c2
water against one side of the plate is therefore F œ 'c5 w(y) † L(y) dy œ 'c5 62.4 † (y) † 2(5 y) dy c2
œ 124.8 'c5 a5y y# b dy œ 124.8 5# y# "3 y$ ‘ & œ 124.8 ˆ 5# † 4 #
œ (124.8) ˆ 105 #
" 3
† 8‰ ˆ 5# † 25
" 3
† 125‰‘
œ (124.8) ˆ 315 6 234 ‰ œ 1684.8 lb
117 ‰ 3
2. An equation for the line of the plate's right-hand edge is y œ x 3 Ê x œ y 3. Thus the total width is L(y) œ 2x œ 2(y 3). The depth of the strip is (2 y). The force exerted by the water is F œ 'c3 w(2 y)L(y) dy œ 'c3 62.4 † (2 y) † 2(3 y) dy œ 124.8'c3 a6 y y# b dy œ 124.8 ’6y 0
0
œ (124.8) ˆ18
0
y# #
! y$ 3 “ $
‰ 9‰ œ (124.8) ˆ 27 # œ 1684.8 lb
9 #
3. Using the coordinate system of Exercise 4, we find the equation for the line of the plate's right-hand edge is y œ x 3 Ê x œ y 3. Thus the total width is L(y) œ 2x œ 2(y 3). The depth of the strip changes to (4 y) Ê F œ 'c3 w(4 y)L(y) dy œ 'c3 62.4 † (4 y) † 2(y 3) dy œ 124.8'c3 a12 y y# b dy 0
œ 124.8 ’12y
0
y# #
! y$ 3 “ $
0
œ (124.8) ˆ36
9 #
‰ 9‰ œ (124.8) ˆ 45 # œ 2808 lb
4. Using the coordinate system of Exercise 4, we see that the equation for the line of the plate's right-hand edge remains the same: y œ x 3 Ê x œ 3 y and L(y) œ 2x œ 2(y 3). The depth of the strip changes to (y) Ê F œ 'c3 w(y)L(y) dy œ 'c3 62.4 † (y) † 2(y 3) dy œ 124.8'c3 ay# 3yb dy œ 124.8 ’ y3 3# y# “ 0
œ (124.8) ˆ 27 3
0
27 ‰ #
œ
(124.8)(27)(2 3) 6
0
$
! $
œ 561.6 lb
5. Using the coordinate system of Exercise 4, we find the equation for the line of the plate's right-hand edge to be y œ 2x 4 Ê x œ y # 4 and L(y) œ 2x œ y 4. The depth of the strip is (1 y).
407
408
Chapter 6 Applications of Definite Integrals
(a) F œ 'c4 w(1 y)L(y) dy œ 'c4 62.4 † (1 y)(y 4) dy œ 62.4 'c4 a4 3y y# b dy œ 62.4 ’4y 0
0
œ (62.4) ’(4)(4)
(3)(16) #
(b) F œ (64.0) ’(4)(4)
0
(3)(16) #
64 3 “
œ (62.4) ˆ16 24
64 3 “
œ
(64.0)(120 64) 3
64 ‰ 3
œ
(62.4)(120 64) 3
3y# #
! y$ 3 “ %
œ 1164.8 lb
¸ 1194.7 lb
6. Using the coordinate system given, we find an equation for the line of the plate's right-hand edge to be y œ 2x 4 Ê x œ 4#y and L(y) œ 2x œ 4 y. The depth of the strip is (1 y) Ê F œ '0 w(1 y)(4 y) dy 1
œ 62.4'0 ay# 5y 4b dy œ 62.4 ’ y3 1
$
œ (62.4) ˆ "3
5y# #
4y“
4‰ œ (62.4) ˆ 2 156 24 ‰ œ
5 #
"
! (62.4)(11) 6
œ 114.4 lb
7. Using the coordinate system given in the accompanying figure, we see that the total width is L(y) œ 63 and the depth of the strip is (33.5 y) Ê F œ '0 w(33.5 y)L(y) dy 33
œ '0
33
64 1 #$
64 ‰ † (33.5 y) † 63 dy œ ˆ 12 (63)'0 (33.5 y) dy $ 33
$$ y# # “!
64 ‰ œ ˆ 12 (63) ’33.5y $
œ
(64)(63)(33)(67 33) (#) a12$ b
‰ ’(33.5)(33) œ ˆ 641#†63 $
33# # “
œ 1309 lb
8. (a) Use the coordinate system given in the accompanying ‰ figure. The depth of the strip is ˆ 11 6 y ft Ê F œ '0
11Î6
‰ w ˆ 11 6 y (width) dy
œ (62.4)(width)'0
11Î6
ˆ 11 ‰ 6 y dy
œ (62.4)(width) ’ 11 6 y
""Î' y# # “!
#
‰ † "# “ Ê Fend œ (62.4)(2) ˆ 121 ‰ ˆ "# ‰ ¸ 209.73 lb and Fside œ (62.4)(4) ˆ 121 ‰ ˆ "# ‰ ¸ 419.47 lb œ (62.4)(width) ’ˆ 11 6 36 36 (b) Use the coordinate system given in the accompanying figure. Find Y from the condition that the entire volume of the water is conserved (no spilling): 11 6 †2†4œ 2†2†Y 11 ‰ Ê Y œ 3 ft. The depth of a typical strip is ˆ 11 3 y ft and the total width is L(y) œ 2 ft. Thus, F œ '0
113
‰ w ˆ 11 3 y L(y) dy
11 ‰ œ '0 (62.4) ˆ 11 3 y † 2 dy œ (62.4)(2) ’ 3 y 113
""Î$ y# “ # !
force doubles. 9. Using the coordinate system given in the accompanying figure, we see that the right-hand edge is x œ È1 y# so the total width is L(y) œ 2x œ 2È1 y# and the depth of the strip is (y). The force exerted by the water is therefore F œ 'c1 w † (y) † 2È1 y# dy 0
‰# “ œ œ (62.4)(2) ’ˆ "# ‰ ˆ 11 3
(62.4)(12") 9
¸ 838.93 lb Ê the fluid
Section 6.7 Fluid Pressures and Forces œ 62.4'c1 È1 y# d a1 y# b œ 62.4 ’ 23 a1 y# b 0
$Î# !
“
"
œ (62.4) ˆ 23 ‰ (1 0) œ 416 lb
10. Using the same coordinate system as in Exercise 15, the right-hand edge is x œ È3# y# and the total width is L(y) œ 2x œ 2È9 y# . The depth of the strip is (y). The force exerted by the milk is therefore F œ 'c3 w † (y) † 2È9 y# dy œ 64.5'c3 È9 y# d a9 y# b œ 64.5 ’ 23 a9 y# b 0
0
$Î# !
“
œ (64.5)(18) œ 1161 lb
$
œ (64.5) ˆ 23 ‰ (27 0)
11. The coordinate system is given in the text. The right-hand edge is x œ Èy and the total width is L(y) œ 2x œ 2Èy. (a) The depth of the strip is (2 y) so the force exerted by the liquid on the gate is F œ '0 w(2 y)L(y) dy 1
" œ '0 50(2 y) † 2Èy dy œ 100 '0 (2 y)Èy dy œ 100'0 ˆ2y"Î# y$Î# ‰ dy œ 100 43 y$Î# 25 y&Î# ‘ ! 1
1
1
‰ œ 100 ˆ 43 25 ‰ œ ˆ 100 15 (20 6) œ 93.33 lb
2‰ (b) We need to solve 160 œ '0 w(H y) † 2Èy dy for h. 160 œ 100 ˆ 2H 3 5 Ê H œ 3 ftÞ 1
12. Use the coordinate system given in the accompanying figure. The total width is L(y) œ 1. (a) The depth of the strip is (3 1) y œ (2 y) ft. The force exerted by the fluid in the window is F œ '0 w(2 y)L(y) dy œ 62.4 '0 (2 y) † 1 dy œ (62.4) ’2y 1
1
" y# # “!
œ (62.4) ˆ2 "# ‰ œ
(62.4)(3) #
œ 93.6 lb
(b) Suppose that H is the maximum height to which the tank can be filled without exceeding its design limitation. This means that the depth of a typical strip is (H 1) y and the force is F œ '0 w[(H 1) y]L(y) dy œ Fmax , where 1
Fmax œ 312 lb. Thus, Fmax œ w'0 [(H 1) y] † 1 dy œ (62.4) ’(H 1)y
" y# # “!
1
œ (62.4) ˆH 3# ‰
‰ (2H 3) œ 93.6 62.4H. Then Fmax œ 93.6 62.4H Ê 312 œ 93.6 62.4H Ê H œ œ ˆ 62.4 #
405.6 62.4
œ 6.5 ft
13. Suppose that h is the maximum height. Using the coordinate system given in the text, we find an equation for the line of the end plate's right-hand edge is y œ 5# x Ê x œ 25 y. The total width is L(y) œ 2x œ 45 y and the depth of the typical horizontal strip at level y is (h y). Then the force is F œ '0 w(h y)L(y) dy œ Fmax , h
where Fmax œ 6667 lb. Hence, Fmax œ w'0 (h y) † 45 y dy œ (62.4) ˆ 45 ‰ ' ahy y# b dy h
h
0
œ
# (62.4) ˆ 45 ‰ ’ hy#
h
y$ 3 “0
œ
$ (62.4) ˆ 45 ‰ Š h#
h$ 3‹
œ
(62.4) ˆ 45 ‰ ˆ "6 ‰ h$
$
max ‰ œ (10.4) ˆ 45 ‰ h$ Ê h œ Ɉ 54 ‰ ˆ F10.4
‰ œ $Ɉ 54 ‰ ˆ 6667 10.4 ¸ 9.288 ft. The volume of water which the tank can hold is V œ Height œ h and
" #
(Base) œ
2 5
h Ê Vœ
ˆ 25
#‰
h
#
#
" #
(Base)(Height) † 30, where $
(30) œ 12h ¸ 12(9.288) ¸ 1035 ft .
14. (a) After 9 hours of filling there are V œ 1000 † 9 œ 9000 cubic feet of water in the pool. The level of the water V is h œ Area , where Area œ 50 † 30 œ 1500 Ê h œ 9000 1500 œ 6 ft. The depth of the typical horizontal strip at level y is then (6 y) for the coordinate system given in the text. An equation for the drain plate's right-hand edge is y œ x Ê total width is L(y) œ 2x œ 2y. Thus the force against the drain plate is F œ '0 w(6 y)L(y) dy œ 62.4 '0 (6 y) † 2y dy œ (62.4)(2)'0 a6y y# b œ (62.4)(2) ’ 6y# 1
1
œ (124.8) ˆ3 "3 ‰ œ (124.8) ˆ 83 ‰ œ 332.8 lb
1
#
" y$ 3 “!
409
410
Chapter 6 Applications of Definite Integrals
(b) Suppose that h is the maximum height. Then, the depth of a typical strip is (h y) and the force F œ '0 w(h y)L(y) dy œ Fmax , where Fmax œ 520 lb. Hence, Fmax œ (62.4)'0 (h y) † 2y dy 1
1
œ 124.8'0 ahy y# b dy œ (124.8) ’ hy# 1
Êhœ
27 3
#
" y$ 3 “!
œ (124.8) ˆ h# "3 ‰ œ (20.8)(3h 2) Ê
520 20.8
œ 3h 2
œ 9 ft
15. The pressure at level y is p(y) œ w † y Ê the average pressure is p œ #
œ ˆ wb ‰ Š b# ‹ œ
" b
'0b p(y) dy œ b" '0b w † y dy œ b" w ’ y# “ b #
0
wb #
. This is the pressure at level
b #
, which
is the pressure at the middle of the plate. 16. The force exerted by the fluid is F œ '0 w(depth)(length) dy œ '0 w † y † a dy œ (w † a)'0 y dy œ (w † a) ’ y# “ b
œ
# w Š ab# ‹
b
b
#
b 0
‰ œ ˆ wb # (ab) œ p † Area, where p is the average value of the pressure (see Exercise 21).
17. When the water reaches the top of the tank the force on the movable side is 'c2 (62.4) ˆ2È4 y# ‰ (y) dy 0
œ (62.4)'c2 a4 y# b 0
"Î#
(2y) dy œ (62.4) ’ 23 a4 y# b
$Î# !
“
#
œ (62.4) ˆ 23 ‰ ˆ4$Î# ‰ œ 332.8 ft † lb. The force
compressing the spring is F œ 100x, so when the tank is full we have 332.8 œ 100x Ê x ¸ 3.33 ft. Therefore the movable end does not reach the required 5 ft to allow drainage Ê the tank will overflow. 18. (a) Using the given coordinate system we see that the total width is L(y) œ 3 and the depth of the strip is (3 y).
Thus, F œ '0 w(3 y)L(y) dy œ '0 (62.4)(3 y) † 3 dy 3
3
œ (62.4)(3)'0 (3 y) dy œ (62.4)(3) ’3y 3
$ y# # “!
œ (62.4)(3) ˆ9 9# ‰ œ (62.4)(3) ˆ 9# ‰ œ 842.4 lb
(b) Find a new water level Y such that FY œ (0.75)(842.4 lb) œ 631.8 lb. The new depth of the strip is (Y y) and Y is the new upper limit of integration. Thus, FY œ '0 w(Y y)L(y) dy Y
œ 62.4'0 (Y y) † 3 dy œ (62.4)(3)'0 (Y y) dy œ (62.4)(3) ’Yy Y
Y
Y
y# # “0
œ (62.4)(3) ŠY#
Y# # ‹
#
2FY È6.75 ¸ 2.598 ft. So, ?Y œ 3 Y œ (62.4)(3) Š Y# ‹ . Therefore, Y œ É (62.4)(3) œ É 1263.6 187.2 œ
¸ 3 2.598 ¸ 0.402 ft ¸ 4.8 in 19. Use ac oordinate system with y œ 0 at the bottom of the carton and with L(y) œ 3.75 and the depth of a typical strip being (7.75 y). Then F œ '0
7Þ75
' ‰ w(7.75 y)L(y) dy œ ˆ 64.5 12$ (3.75) 0
7Þ75
‰ (7.75 y) dy œ ˆ 64.5 12$ (3.75) ’7.75y
#
(Þ(& y# # “!
(7.75) ‰ œ ˆ 64.5 ¸ 4.2 lb 12$ (3.75) # 57 ‰ 20. The force against the base is Fbase œ pA œ whA œ w † h † (length)(width) œ ˆ 12 (10)(5.75)(3.5) ¸ 6.64 lb. $
To find the fluid force against each side, use a coordinate system with y œ 0 at the bottom of the can, so that the depth of a of ‰ of ‰ ˆ 57 ‰ ˆ width typical strip is (10 y): F œ '0 w(10 y) ˆ width the side dy œ 12$ the side ’10y 10
"! y# # “!
57 ‰ ˆ width of ‰ ˆ 100 ‰ 57 ‰ 57 ‰ œ ˆ 12 Ê Fend œ ˆ 12 (50)(3.5) ¸ 5.773 lb and Fside œ ˆ 12 (50)(5.75) ¸ 9.484 lb $ $ $ the side #
Chapter 6 Practice Exercises 21. (a) An equation of the right-hand edge is y œ
x Ê xœ
3 #
2 3
y and L(y) œ 2x œ
4y 3
411
. The depth of the strip
is (3 y) Ê F œ '0 w(3 y)L(y) dy œ '0 (62.4)(3 y) ˆ 43 y‰ dy œ (62.4) † ˆ 43 ‰'0 a3y y# b dy 3
3
$ y$ 3 “!
œ (62.4) ˆ 43 ‰ ’ 3# y#
œ (62.4) ˆ 43 ‰ 27 #
3
27 ‘ 3
‰ œ (62.4) ˆ 34 ‰ ˆ 27 6 œ 374.4 lb
(b) We want to find a new water level Y such that FY œ
" #
(374.4) œ 187.2 lb. The new depth of the strip is
(Y y), and Y is the new upper limit of integration. Thus, FY œ '0 w(Y y)L(y) dy Y
œ 62.4'0 (Y y) ˆ 43 y‰ dy œ (62.4) ˆ 43 ‰'0 aYy y# b dy œ (62.4) ˆ 43 ‰ ’Y † Y
Y
œ (62.4) ˆ 29 ‰ Y$ . Therefore Y$ œ
9FY 2†(62.4)
œ
(9)(187.2) 124.8
y# #
Y
y$ 3 “!
$
œ (62.4) ˆ 34 ‰ Š Y2
Y$ 3 ‹
$ $È Ê Y œ É (9)(187.2) 13.5 ¸ 2.3811 ft. So, 124.8 œ
?Y œ 3 Y ¸ 3 2.3811 ¸ 0.6189 ft ¸ 7.5 in. to the nearest half inch. (c) No, it does not matter how long the trough is. The fluid pressure and the resulting force depend only on depthe of the water. ‰ 22. The area of a strip of the face of height ?y and parallel to the base is 100ˆ 26 24 † ?y, where the factor of
26 24
inclination of the face of the dam. With the origin at the bottom of the dam, the force on the face is then: ‰ F œ '0 w(24 y)a100bˆ 26 24 dy œ '('! ’#%y 24
#%
y# # “!
œ '('!Š#%#
#%# # ‹
œ 1,946,880 lb.
CHAPTER 6 PRACTICE EXERCISES # 1. A(x) œ 14 (diameter)# œ 14 ˆÈx x# ‰ œ 14 ˆx 2Èx † x# x% ‰ ; a œ 0, b œ 1
Ê V œ 'a A(x) dx œ b
#
œ
1 4
œ
1 4†70
’ x# 74 x(Î#
È3 4
x& 5 “!
" #
ˆ "#
4 7
5" ‰
91 280 È3 4
ˆ2Èx x‰#
ˆ4x 4xÈx x# ‰ ; a œ 0, b œ 4 È3 4
b
œ
È3 4
œ
32È3 4
’2x# 85 x&Î#
3. A(x) œ 1 4
1 4
œ
(side)# ˆsin 13 ‰ œ
Ê V œ 'a A(x) dx œ
œ
'01 ˆx 2x&Î# x% ‰ dx
(35 40 14) œ
2. A(x) œ œ
1 4 "
ˆ1 1 4
8 5
'04 ˆ4x 4x$Î# x# ‰ dx
% x$ 3 “!
32 ‰ œ
(diameter)# œ
œ
8È 3 15 1 4
È3 4
ˆ32
8†32 5
(15 24 10) œ
64 ‰ 3
8È 3 15
(2 sin x 2 cos x)#
† 4 asin# x 2 sin x cos x cos# xb
œ 1(1 sin 2x); a œ
1 4
,bœ
51 4
Ê V œ 'a A(x) dx œ 1 '1Î4 (1 sin 2x) dx 51Î4
b
œ 1 x
cos 2x ‘ &1Î% # 1Î%
œ 1 ’Š 541
cos 5#1 #
‹ Š 14
cos 1# #
‹“ œ 1 # #
#
%
4. A(x) œ (edge)# œ ŒŠÈ6 Èx‹ 0 œ ŠÈ6 Èx‹ œ 36 24È6 Èx 36x 4È6 x$Î# x# ; a œ 0, b œ 6 Ê V œ 'a A(x) dx œ '0 Š36 24È6 Èx 36x 4È6 x$Î# x# ‹ dx b
6
accounts for the
412
Chapter 6 Applications of Definite Integrals ' x$ 3 “!
œ ’36x 24È6 † 23 x$Î# 18x# 4È6 † 25 x&Î# œ 216 576 648 5. A(x) œ
(diameter)# œ
1 4
72 œ 360
Š2Èx
x# 4‹
#
œ
1728 5
œ
1 4
18001728 5
Š4x x&Î#
œ
6$ 3
72 5
x% 16 ‹ ;
a œ 0, b œ 4 Ê V œ 'a A(x) dx b
'04 Š4x x&Î# 16x ‹ dx œ 14 ’2x# 27 x(Î# 5x†16 “ % œ 14 ˆ32 32 † 87 25 † 32‰ %
œ
1 4
œ
321 4
ˆ1
6. A(x) œ œ
1 4
1728 5
œ 216 16 † È6 È6 † 6 18 † 6# 58 È6 È6 † 6#
È3 4
" #
8 7
25 ‰ œ
81 35
&
(35 40 14) œ
(edge)# sin ˆ 13 ‰ œ
È3 4
!
721 35
2Èx ˆ2Èx‰‘#
ˆ4Èx‰# œ 4È3 x; a œ 0, b œ 1
Ê V œ 'a A(x) dx œ '0 4È3 x dx œ ’2È3 x# “ b
1
" !
œ 2È3
7. (a) .3=5 7/>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
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
4 x$
, r(x) œ
" #
&
" x* 9 “ "
b
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
# x# 4 “"
571 #0
œ 21 ˆ 4# 1‰ ˆ4 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
21†13 5
# # # & 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#
1 x# ‰ dx
œ 21 (1 2 2 1) ˆ4 4 1 4" ‰‘ œ
31 #
51 #
œ
261 5
121 5
Chapter 6 Practice Exercises
413
(d) A+=2/< 7/>29. :
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
1031 20
œ
(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
œ 321 ˆ3
2 5
"3 ‰ œ
321 15
(45 6 5) œ
10881 15
y& 5
23 y$ “
# #
œ 21 ˆ24 † 2
(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$Î# x# ‰ dx œ 21 ’ 45 x&Î# b
œ 21 ˆ 45 † 32
4
64 ‰ 3
œ
4
1281 15
(c) =2/66 7/>29. :
% x$ 3 “!
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
$Î# œ 21 ’ 16 2x# 54 x&Î# 3 x
œ 641 ˆ1 45 ‰ œ
641 5
% x$ 3 “!
4
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
y# 4‹
4 5
† 32
64 ‰ 3
œ 641 ˆ 34 1
dy œ 21'0 Š4y y# y# 4
4 5
y$ 4‹
32 ‰
dy
32 5
2 3
† 8‰
414
Chapter 6 Applications of Definite Integrals œ 21'0 Š4y 2y# 4
y$ 4‹
% y% 16 “ !
dy œ 21 ’2y# 23 y$
œ 21 ˆ32
† 64 16‰ œ 321 ˆ2
2 3
321 3
1‰ œ
8 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 # 1 ˆ 9#1 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% 43 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 †
# &
œ
)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
œ 21'0 a4x 2x# 2x# x$ b dx œ 21'0 ax$ 4x# 4xb dx œ 21 ’ x4 43 x$ 2x# “ œ 21 ˆ4 2
œ
21 3
2
(36 32) œ
#
%
32 3
!
81 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
11Î2
and the x-axis around the x-axis. To find the
volume we use the .3=5 method: V œ 'a 1R# (x) dx œ 'c11Î2 1 ŠÉ12 ˆ1 b
4x# ‰ 121 ‹
#
11Î2
dx œ 1 '11Î2 12 Š1
4x# 121 ‹
dx
Chapter 6 Practice Exercises 11Î2
œ 121'c11Î2 Š1
4x# 121 ‹
œ 1321 ˆ1 "3 ‰ œ 17. y œ x"Î#
x$Î# 3
dx œ 121 ’x
" #
œ
dy dx
$
#
x"Î# "# x"Î# Ê Š dy dx ‹ œ
" 4
ˆ x" 2 x‰ Ê L œ ' É1 4" ˆ x" 2 x‰ dx 1 4
# Ê L œ '1 É 4" ˆ x" 2 x‰ dx œ '1 É 4" ax"Î# x"Î# b dx œ '1 4
œ
" #
ˆ4
4
18. x œ y#Î$ Ê œ '1
8
" #
† 8‰ ˆ2 23 ‰‘ œ
2 3
dx dy
È9x#Î$ 4 3x"Î$
œ
2 3
ˆ2 #
4
14 ‰ 3
x"Î$ Ê Š dx dy ‹ œ " 3
dx œ
œ
4x#Î$ 9
40
" 18
" 18
' u"Î# du œ 13
5 12
x'Î& 58 x%Î& Ê
ˆx"Î# x"Î# ‰ dx œ
dx Ê L œ '1 Ê1 Š dy ‹ dy œ '1 É1 #
8
23 u$Î# ‘ %! œ "$ #
" #
œ
dy dx
" #
" #
2x"Î# 23 x$Î# ‘ % "
10 3
'18 È9x#Î$ 4 ˆx"Î$ ‰ dx; u œ 9x#Î$ 4
x œ 8 Ê u œ 40d Ä L œ
19. y œ
#
4 ‰ 11 ˆ 4 ‰ ˆ 11 ‰ “ œ 1321 ’1 ˆ 363 œ 241 ’ 11 Š 4 ‹“ 2 363 #
œ 881 ¸ 276 in$
2641 3
Ê
""Î# 4x$ 363 “ ""Î#
x"Î& "# x"Î& Ê Š dy dx ‹ œ
" #7
" 4
8
4 9x#Î$
dy
Ê du œ 6y"Î$ dy; x œ 1 Ê u œ 13,
40$Î# 13$Î# ‘ ¸ 7.634
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
32 $# 75 ‰ œ '1 "# ˆx"Î& x"Î& ‰ dx œ "# 56 x'Î& 45 x%Î& ‘ " œ "# ˆ 65 † 2' 54 † 2% ‰ ˆ 56 54 ‰‘ œ "# ˆ 315 6 4
œ
" 48
20. x œ
(1260 450) œ
" 1#
y$
" y
Ê
" % œ '1 É 16 y 2
" #
œ
1710 48
œ
" 4
" y%
dy œ '1 ÊŠ 4" y#
y#
dx dt
" y#
#
dx dy
8 œ ˆ 12 "# ‰ ˆ 1"# 1‰ œ
21.
285 8
dx Ê Š dy ‹ œ
2
7 1#
œ 5 sin t 5 sin 5t and
" #
œ
" y# ‹
" 16 #
y%
" #
" y%
" % Ê L œ '1 Ê1 Š 16 y
dy œ '1 Š 4" y# 2
2
" y# ‹
" y% ‹
dy
# "
13 12
#
‰ Š dy œ 5 cos t 5 cos 5t Ê Êˆ dx dt dt ‹
dy dt
dy œ ’ 1"# y$ y" “
" #
#
œ Éa5 sin t 5 sin 5tb# a5 cos t 5 cos 5tb# œ 5Èsin# 5t #sin t sin 5t sin# t cos# t #cos t cos 5t cos# 5t œ &È# #asin t sin 5t cos t cos 5 tb œ 5È#a" cos %tb œ 5É%ˆ "# ‰a" cos %tb œ "!Èsin# #t œ "!lsin #tl œ "!sin #t (since ! Ÿ t Ÿ 1# ) Ê Length œ '!
1 Î2
22.
dx dt
œ 3t2 12t and
1Î#
"!sin #t dt œ c5 cos #td ! dy dt
œ a&ba"b a&ba"b œ "! #
#
‰ Š dy Éa3t2 12tb# a3t2 12tb# œ È288t# "8t4 œ 3t2 12t Ê Êˆ dx dt dt ‹ œ
œ 3È2 ktkÈ16 t2 Ê Length œ '! 3È2 ktkÈ16 t2 dt œ 3È2'! t È16 t2 dt; ’u œ 16 t2 Ê du œ 2t dt "
"
3È 2 2
Ê "# du œ t dt; t œ 0 Ê u œ 16; t œ 1 Ê u œ 17“; œ
23.
dx d)
3È 2 2
'16"7 Èu du œ 3È2 2 23 u3/2 ‘1617 œ 3È2 2 Š 23 a17b3/2 23 a16b3/2 ‹
† 23 Ša17b3/2 64‹ œ È2Ša17b3/2 64‹ ¸ 8.617.
œ $ sin )and
Ê Length œ '!
dy d)
$1Î2
#
#
‰ Š dy Éa$ sin )b# a$ cos )b# œ È$asin# ) cos# )b œ $ œ $ cos ) Ê Êˆ dx d) d) ‹ œ
$ d) œ $'!
$1Î2
d) œ $ˆ $#1 !‰ œ
*1 #
415
416
Chapter 6 Applications of Definite Integrals t$ 3
24. x œ t# and y œ œ'
t, È3 Ÿ t Ÿ È3 Ê
È3
È 3
Èt% #t# " dt œ
'
dx dt
œ 2t and
dy dt
È3
È 3
Èt% 2t# " dt œ
œ t# " Ê Length œ
'
È3
È 3
Éat# "b# dt œ
'
È3
È 3
Éa2tb# at# "b# dt
È
'È33 at# "b dt œ ’ t3 t“ 3
È3 È 3
œ 4È3 25. 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 #
&
$ ’ x5
2x$ 3
" x$ 3 “ "
œ 3$ ’x
3x“
" "
3 #
$ ax% 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 ax% 2x# 3b dx "
; M œ ' dm œ 3$ 'c1 a1 x# b dx "
. Therefore, the centroid is (xß y) œ ˆ!ß 85 ‰ .
26. 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 x% dx Ê Mx œ ' µ y dm œ
œ
$ #
œ
2$ 10
a2& b œ
32$ 5
$ #
; M œ ' dm œ $
'c22 x% dx œ 10$ cx& d ## 'c22 x# dx œ $ ’ x3 “ # $
#
œ
2$ 3
a2$ b œ
16$ 3
Ê yœ
œ
32†$ †3 5†16†$
œ
6 5
. Therefore, the
dx. Thus, Mx œ ' µ y dm œ
$ #
'04 Š16 16x ‹ dx
Mx M
centroid is(xß y) œ ˆ!ß 65 ‰ . 27. 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‹
µ y dm œ $ †
x# 4 ‹dx,
x# 4,
width: dx,
mass: dm œ $ † dA
dx Ê the moment about the x-axis is #
Š4 x4 ‹ #
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‹
4
4
My M
œ
16†$ †3 32†$
œ
3 2
and y œ
Mx M
œ
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
% x% 16 “ !
32$ 3
‰ . Therefore, the centroid is (xß y) œ ˆ 3# ß 12 5 .
%
Chapter 6 Practice Exercises
417
28. 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 $ #
#
# y% 4 “!
œ $ ’ 23 y$ œ
$ #
ˆ 43†8
yœ
Mx M
2
%
œ
œ
32 ‰ 5
4†$ †3 3†4†$
œ $ ˆ 23 † 8 32$ 15
œ $ ˆ 16 3
16 ‰ 4
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
My M
œ
$ †32†3 15†$ †4
œ
œ 1. Therefore, the centroid is (xß y) œ ˆ 85 ß 1‰ .
29. 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
" #
# y& 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 “!
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 .
30. A typical vertical strip has: center of mass: (µ x ßµ y ) œ ˆxß 2x3$Î# ‰ , length:
3 x$Î#
, width: dx,
area: dA œ x$Î# dx, mass: dm œ $ † dA œ $ † x$Î# dx Ê the moment about the x-axis is µ µ 3 9$ 3 3$ y dm œ #x3$Î# † $ x$Î# dx œ 2x $ dx; the moment about the y-axis is x dm œ x † $ x$Î# dx œ x"Î# dx. 3
3
(a) Mx œ $ '1
9
M œ $ '1
9
(b) Mx œ '1
9
" # 3
ˆ x9$ ‰ dx œ
x$Î# x #
9$ #
#
"
20$ 9
3 ‰ ; My œ $ '1 x ˆ x$Î# dx œ 3$ 2x"Î# ‘ " œ 12$ ;
*
ˆ x9$ ‰ dx œ *
9 #
*
9
dx œ 6$ x"Î# ‘ " œ 4$ Ê x œ
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$Î# dx " x 1 9
œ 6 x"Î# ‘ " œ 12 Ê x œ 31. S œ 'a 21y Ê1 Š dy dx ‹ dx; #
b
*
’ x# “ œ
My M
dy dx
œ
œ
13 3
and y œ
" È2x 1
Mx M
9
œ #
" 3
Ê Š dy dx ‹ œ
" #x 1
Ê S œ '0 21È2x 1 É1 3
" #x 1
dx
2 È ' Èx 1 dx œ 2È21 2 (x 1)$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
32. S œ 'a 21y Ê1 Š dy dx ‹ dx; b
#
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
8 5
and
418
Chapter 6 Applications of Definite Integrals œ
1 6
'01 È1 x% d a1 x% b œ 16 ’ 32 a1 x% b$Î# “ " œ 19 ’2È2 1“ !
33. S œ 'c 21x Ê1 Š dx dy ‹ dy; #
d
dx dy
ˆ "# ‰ (4 2y) È4y y#
œ
2y È4y y#
œ
#
Ê 1 Š dx dy ‹ œ
4y y# 4 4y y# 4y y#
œ
4 4y y#
Ê S œ '1 21 È4y y# É 4y 4 y# dy œ 41'1 dx œ 41 2
2
34. S œ 'c 21x Ê1 Š dx dy ‹ dy; #
d
œ 1'2 È4y 1 dy œ 6
35. x œ
t# #
1 4
dx dy
œ
1 2È y
23 (4y 1)$Î# ‘ ' œ #
and y œ 2t, 0 Ÿ t Ÿ È5 Ê *
œ 21 23 u$Î# ‘ % œ 36. x œ t#
" 2t
761 3
" È2
ŸtŸ1 Ê
1
œ 21 Š2
(125 27) œ
œ t and
dx dt
Ê Surface Area œ '1ÎÈ2 21 ˆt# 1
1 6
dy dt
1 6
" 4y
œ
Ê S œ '2 21Èy † 6
4y 1 4y
(98) œ
È4y 1 È4y
dy
491 3
È5
œ 2 Ê Surface Area œ '0 21(2t)Èt# 4 dt œ '4 21u"Î# du 9
, where u œ t# 4 Ê du œ 2t dt; t œ 0 Ê u œ 4, t œ È5 Ê u œ 9
and y œ 4Èt ,
œ 21 '1ÎÈ2 ˆt#
#
Ê 1 Š dx dy ‹ œ 1
" ‰ˆ 2t 2t
" ‰ 2t#
"‰ #t
dx dt
œ 2t
ʈ2t
" ‰# 2t#
dt œ 21 '1ÎÈ2 ˆ2t$ 1
" 2t#
and
dy dt
œ
2 Èt
Š È2 t ‹ dt œ 21 '1ÎÈ2 ˆt# #
3 #
1
" ‰ Ɉ 2t #t
" ‰# #t#
dt
"
4" t$ ‰ dt œ 21 2" t% 3# t 8" t# ‘ "ÎÈ#
3È 2 4 ‹
37. 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;
38. 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)
39. 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
#
" !
# 2 2 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 #
"
40. 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
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
œ [125x# ]!"Þ# œ 125(1.2)# œ 180 J
1Þ2
1Þ2
Chapter 6 Practice Exercises
419
41. We imagine the water divided into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval [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)
1y# (14 y) dy œ
(62.4)(25)1 16
'08 a14y# y$ b dy œ (62.4) ˆ 25161 ‰ ’ 143 y$ y4 “ ) %
!
¸ 418,208.81 ft † lb
42. The same as in Exercise 41, 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
(62.4)(25)1 16
y# (8 y) dy œ (62.4) ˆ 25161 ‰'0 a8y# y$ b dy œ (62.4) ˆ 25161 ‰ ’ 83 y$ 5
œ (62.4) ˆ 25161 ‰ Š 38 † 5$
5% 4‹
& y% 4 “!
¸ 54,241.56 ft † lb
43. The tank's cross section looks like the figure in Exercise 41 with right edge given by x œ #
horizontal slab has volume ?V œ 1(radius)# (thickness) œ 1 ˆ #y ‰ ?y œ slab is its weight: F(y) œ 60 †
1 4
1 4
10
22,500 ft†lb 275 ft†lb/sec
y# ?y. The force required to lift this
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 1(12 y) Š y4 ‹ dy œ 151 ’ 12y 3 to empty the tank is
y œ y# . A typical
5 10
#
$
"! y% 4 “!
œ 22,5001 ft † lb; the time needed
¸ 257 sec
44. 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 œ 2880 'c4 10È16 y# dy 1140'c4 a16 y# b 0
0
0
œ 22,800 † (area of a quarter circle having radius 4) 23 (1140) ’a16 œ 335,153.25 ft † lb
$Î# ! y# b “ %
"Î#
(2y) dy
œ (22,800)(41) 48,640
strip 45. 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
œ (249.6) ˆ4 83 ‰ œ (249.6) ˆ 43 ‰ œ 332.8 lb strip 46. 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
(75)(3075) 9†#16
¸ 118.63 lb.
# y$ 3 “!
420
Chapter 6 Applications of Definite Integrals
strip 47. F œ 'a W † Š depth ‹ † L(y) dy Ê F œ 62.4'0 (9 y) Š2 † b
4
%
œ 62.4 6y$Î# 25 y&Î# ‘ ! œ (62.4) ˆ6 † 8
2 5
Èy 2 ‹
dy œ 62.4'0 ˆ9y"Î# 3y$Î# ‰ dy 4
‰ (48 † 5 64) œ † 32‰ œ ˆ 62.4 5
(62.4)(176) 5
œ 2196.48 lb
strip 48. 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
49. F œ w" '0 (8 y)(2)(6 y) dy w# 'c6 (8 y)(2)(y 6) dy œ 2w" '0 a48 14y y# b dy 2w# '6 a48 2y y# b dy 6
0
œ 2w" ’48y 7y#
' y$ 3 “!
6
2w# ’48y y#
! y$ 3 “ '
0
œ 216w" 360w#
50. (a) F œ 62.4'0 (10 y) ˆ8 y6 ‰ ˆ y6 ‰‘ dy 6
6
œ
62.4 3
œ
62.4 3
' a240 34y y# b dy 0
’240y 17y#
œ 18,720 lb.
' y$ 3 “!
œ
62.4 3
(1440 612 72)
(b) The centroid ˆ 72 ß 3‰ of the parallelogram is located at the intersection of y œ
6 7
x and y œ 65 x
36 5 .
The centroid of
the triangle is located at (7ß 2). Therefore, F œ (62.4)(7)(36) (62.4)(8)(6) œ (300)(62.4) œ 18,720 lb 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. 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 ‘ #
x
c1
#
0
#
n 1
#n 1 !
Chapter 6 Additional and Advanced Exercises œ1
2 n 1
" #n 1
œ
(n 1)(2n 1) 2(2n ") (n 1) (n 1)(#n 1)
œ
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
1
#
œ
†
2n (n 1)(2n 1)
1
(n 1) 2n
œ
n 2n 1
xn b 1 ‘ " n 1 !
œ
2n# (n 1)(#n 1)
œ 2 ˆ1
" ‰ n 1
. œ
2n n 1.
Therefore,
Ê ˆ!ß #n n 1 ‰ is the location of the centroid. As n Ä _, y Ä
the limiting position of the centroid is ˆ!ß
421
" #
so
"‰ # .
6. 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
# x‰‘ dx
11 80
b 11 ‰# '040 x ˆ9 80 x dx; M œ 'a 1y# dx 40 40 # " ' ˆ ‰ ‘ ‰# dx. œ 1 '0 8"1 ˆ9 11 x dx œ 9 11 80 641 0 80 x
œ
" 641
My M
Thus, x œ
¸
¸ 23.06 (using a calculator to compute
129,700 5623.3
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. 7. (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 . 8. (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&Î# dx œ 4kÈa 27 x(Î# ‘ 0 œ 4ka"Î# † 27 a(Î# œ a
a
8ka 7
a œ 4kÈa'0 x$Î# dx œ 4kÈa 25 x&Î# ‘ 0 œ 4ka"Î# † 25 a&Î# œ a
Ê (xß y) œ
ˆ 5a ‰ 7 ß0
%
8ka$ 5
0
. Also, M œ ' dm œ '0 4kxÈax dx a
My M
. Thus, x œ
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
œ 'c2a Šay# 0
%
œ 8a3
32a& 20a
y# 4a ‹
y% 4a ‹
8a% 3
y# 4a ‹
dy, mass: dm œ $ dA œ kyk Ša
dy œ 'c2a y# Ša 0
dy '0 Šay# 2a
y% 4a ‹
y# 4a ‹
dy '0 y# Ša 2a
dy œ ’ 3a y$
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# 4a ‹
#
#
œ
" 3 #a #
'c02a a16a% y y& b dy 32a" '02a a16a% y y& b dy œ 3#"a
œ
" 32a#
’8a% † 4a#
#
" 32a#
’8a% † 4a#
M œ ' dm œ 'c2a kyk Š 4a 4ay ‹ dy œ #
#
" 4a
dy
’ 3a y$
kyk Ša
#a y& #0a “ !
y# 4a ‹
dy
' kyk a16a% y% b dy
c2a
#
,
2a
" 8a
64a' 6 “
y# 4a ‹
4a# 8a ‹
y# 4a
dy. Thus, Mx œ ' µ y dm
! y& #0a “ #a
œ
2a
a
64a' 6 “
œ
" 16a#
#
’8a% y#
Š32a'
'c2a2a kyk a4a# y# b dy
32a' 3 ‹
œ
! y' 6 “ #a
" 16a#
1 3#a#
’8a% y#
† 32 a32a' b œ
4 3
a% ;
#a y' 6 “!
422
Chapter 6 Applications of Definite Integrals œ
'c02a a4a# y y$ b dy 4a" '02a a4a# y y$ b dy œ 4a" ’2a# y# y4 “ ! %
" 4a
#a
" 4a
œ2† yœ
#
16a% 4 ‹
#
Š2a † 4a
" #a
œ
%
%
$
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
9. (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#
dx $ '0 x aa# x# b a
"Î# a
$ 2 aa • #”
#
$Î#
x# b 3
# $Î#
#
œ ’ab a b
# $Î#
ab b
dx $ 'a x ab# x# b
a
$ 2 ab • #”
b
#
!
0
$ 3
"Î#
“ ’0 aa b
• a
# $Î#
$ 3
$ 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
#
œ
4 31
$
$
a Š bb# a# ‹ œ
dx
b
$Î#
x# b 3
"Î#
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
.
ab b# ‹ a b
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). 10. 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
‰ $'ˆ 24 a "!)ayb "%%
which we solve to get x œ )
a *
and y œ
)a a " b . a
Set
x œ 7 in. (Given). It follows that a œ *, whence y œ
'% *
œ 7 "* in. The distances of the centroid ax, yb from the other sides are easily computed. (Note that if we set y œ 7 in.
above, we will find x œ 7 "* .)
Chapter 6 Additional and Advanced Exercises 11. y œ 2Èx Ê ds œ É "x 1 dx Ê A œ '0 2Èx É "x 1 dx œ 3
4 3
(1 x)$Î# ‘ $ œ !
28 3
12. 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#
13. 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
14. Converting to pounds and feet, 2 lb/in œ "Î#
" 320
t$ 3m
dt œ
4h 3
È3mh
†
12 in 1 ft
2 lb 1 in
" 3m
'
’ t6 “
0
œ 3 ft † lb. Since W œ
t$ 3m
Ê xœ
t% 12m
The work done is
" ‰ œ ˆ 18m (12mh)'Î%
œ 24 lb/ft. Thus, F œ 24x Ê W œ '0
1Î2
24x dx
œ v œ 32t v! . At the top of the ball's path, v œ 0 Ê t œ
ds dt
#
v!#
and the height is s œ 16 ˆ 3v#! ‰ v! ˆ 3v#! ‰ œ
64
œ
3†640 64
œ 30 ft.
15. 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
16. 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 = w
œ
" #
s œ 16t# v! t (since s œ 0 at t œ 0) Ê
œ
dx dt
" " ‰ mv!# "# mv"# , where W œ 3 ft † lb, m œ ˆ 10 lb‰ ˆ 3# ft/sec # "‰ ˆ " #‰ # ˆ slugs, and v" œ 0 ft/sec, we have 3 œ # 3#0 v! Ê v! œ 3 † 640. For the projectile height,
œ c12x# d ! œ
t# †
Ð12mh)"Î%
Cœ0 Ê
#
’ y# “
!
=w #
œ
w
. Therefore the force
= w
'c0w (y)dy
=jw# #
‰ œ ˆ =w # (jw) œ (the average pressure up and down) † (thearea of the plate). 17. (a) We establish a coordinate system as shown. A typical horizontal strip has: center of pressure: (µ x ßµ y ) b œ ˆ # ß y‰ , length: L(y) œ b, width: dy, area: dA
œ b dy, pressure: dp œ = kyk dA œ =b kyk dy 0 0 Ê Fx œ ' µ y dp œ 'ch y † =b kyk dy œ =b 'ch y# dy $
œ =b ’ y3 “ F œ ' dp œ
!
h
$
œ =b ’0 Š 3h ‹“ œ
=bh$ 3
'ch = kyk L(y) dy œ =b 'ch 0
;
0
y dy
v! 3#
C" ;
423
424
Chapter 6 Applications of Definite Integrals #
œ =b ’ y# “
$
! h
œ =b ’0
h# #“
=bh# #
œ
. Thus, y œ
œ
Fx F
Š =3bh ‹ #
Š =bh # ‹
œ
2h 3
Ê the distance below the surface is
(b) A typical horizontal strip has length L(y). By similar triangles from the figure at the right,
L(y) b
œ
y a h
Ê L(y) œ bh (y a). Thus, a typical strip has center of pressure: (µ x ßµ y ) œ (µ x ß y), length: L(y) œ bh (y a), width: dy, area: dA œ bh (y a) dy, pressure: dp œ = kyk dA œ =(y) ˆ bh ‰ (y a) dy œ =b ay# ayb dy Ê F œ ' µ y dp h
x
a
œ 'cÐahÑ y † %
=b h
ay# ayb dy œ
'ÐaahÑ
=b h
a ay$ 3 “ cÐahÑ
ay$ ay# b dy
œ
=b h
’ y4
œ
=b h
’Š a4
œ œ
=b 12h =b 12h
œ
=bh 12
œ
=b h
’Š 3a
œ
=b h
’a
œ
=b 6h
a6a# h 6ah# 2h$ 6a# h 3ah# b œ
œ #
%
a% 3‹
%
Š (a 4 h)
a(a h)$ ‹“ 3
œ
=b h
’a
%
(a h)% 4
a% a(a h)$ “ 3
c3 aa% aa% 4a$ h 6a# h# 4ah$ h% bb 4 aa% a aa$ 3a# h 3ah# h$ bbd a12a$ h 12a# h# 4ah$ 12a$ h 18a# h# 12ah$ 3h% b œ a6a# 8ah 3h# b ; F œ ' dp œ ' = kyk L(y) dy œ $
$
a$ #‹
$
Š (a 3 h)
3a# h 3ah# h$ a$ 3
ˆ 1=#bh ‰ a6a# 8ah 3h# b ˆ =6bh ‰ (3a 2h)
6a 8ah 3h 6a 4h
#
.
a(a h)# ‹“ #
œ
=b h
a$ aa$ 2a# h ah# b “ #
‰ 6a œ ˆ " # Š
#
=b 6h
8ah 3h# ‹ 3a 2h
$
’ (a h)3
œ
=b 6h
a$
=b 12h
a6a# h# 8ah$ 3h% b
=b h
'cÐaahÑ
a$ a(a h)# “ 2
ay# ayb dy œ
=b h
$
’ y3
c2 a3a# h 3ah# h$ b 3 a2a# h ah# bd
a3ah# 2h$ b œ
=bh 6
(3a 2h). Thus, y œ
Ê the distance below the surface is
Fx F
a ay# 2 “ ÐahÑ
2 3
h.
CHAPTER 7 TRANSCENDENTAL FUNCTIONS 7.1 INVERSE FUNCTIONS AND THEIR DERIVATIVES 1. Yes one-to-one, the graph passes the horizontal test. 2. Not one-to-one, the graph fails the horizontal test. 3. Not one-to-one since (for example) the horizontal line y œ # intersects the graph twice. 4. Not one-to-one, the graph fails the horizontal test. 5. Yes one-to-one, the graph passes the horizontal test 6. Yes one-to-one, the graph passes the horizontal test 7. Domain: 0 x Ÿ 1, Range: 0 Ÿ y
9. Domain: 1 Ÿ x Ÿ 1, Range: 1# Ÿ y Ÿ
8. Domain: x 1, Range: y 0
1 #
10. Domain: _ x _, Range: 1# y Ÿ
11. 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)
1 #
426
Chapter 7 Transcendental Functions
12. The graph is symmetric about y œ x.
yœ
" x
Ê xœ
" y
Ê yœ
" x
œ f " (x)
13. 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) 14. Step 1: y œ x# Ê x œ Èy, since x Ÿ !. Step 2: y œ Èx œ f " (x) 15. Step 1: y œ x$ 1 Ê x$ œ y 1 Ê x œ (y 1)"Î$ Step 2: y œ $Èx 1 œ f " (x) 16. Step 1: y œ x# 2x 1 Ê y œ (x 1)# Ê Èy œ x 1, since x 1 Ê x œ 1 Èy Step 2: y œ 1 Èx œ f " (x) 17. Step 1: y œ (x 1)# Ê Èy œ x 1, since x 1 Ê x œ Èy 1 Step 2: y œ Èx 1 œ f " (x) 18. Step 1: y œ x#Î$ Ê x œ y$Î# Step 2: y œ x$Î# œ f " (x) 19. Step 1: y œ x& Ê x œ y"Î& Step 2: y œ &Èx 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
20. 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
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); 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
"Î$
œ ax$ b
"Î$
œx
Section 7.1 Inverse Functions and Their Derivatives 22. Step 1: y œ
" #
x
" #
Ê
7 #
"
xœy
7 #
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# ‰ 23. Step 1: y œ Step 2: y œ
" x#
Ê x# œ
" y
" Èx
œ f " (x)
Ê xœ
7 #
œ x and f " (f(x)) œ 2 ˆ "# x 7# ‰ 7 œ (x 7) 7 œ x
" Èy
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 ‹
24. Step 1: y œ
" x$
Ê x$ œ
" x"Î$ "
Step 2: y œ Domain of f f af " (x)b œ
Šx‹
" y
Ê xœ
(c)
26. (a) y œ
" 5
" $ ax"Î$ b
" x"
œ
œ 2,
df " dx ¹ xœ1
x7 Ê
df ¸ dx xœ1
(c)
œ x since x 0
" y"Î$
œ x and f " (f(x)) œ ˆ x"$ ‰
" 5
œ
œ "5 ,
"
df dx
¹
œ 4,
df " dx ¹ xœ3
œ ˆ x" ‰
"
(b) x #
3 #
xœy7
xœ$%Î&
"Î$
" #
"
(b)
(x) œ 5x 35
œ5
27. (a) y œ 5 4x Ê 4x œ 5 y Ê x œ 54 y4 Ê f " (x) œ df ¸ dx xœ1Î#
" Š "x ‹
: x Á 0, Range of f " : y Á 0;
Ê x œ 5y 35 Ê f (c)
œ
œ $É x" œ f " (x);
25. (a) y œ 2x 3 Ê 2x œ y 3 Ê x œ y# 3# Ê f " (x) œ df ¸ dx xœ1
" É x"#
œ
" 4
(b) 5 4
x 4
œx
427
428
Chapter 7 Transcendental Functions " #
28. (a) y œ 2x# Ê x# œ Ê xœ (c)
df ¸ dx xœ&
" È2
(b)
y
Èy Ê f
"
È x#
(x) œ
œ 4xk xœ5 œ 20,
df " dx ¹ xœ&0
œ
" #È 2
x"Î# ¹
xœ50
" #0
œ
$ $ 29. (a) f(g(x)) œ ˆ $Èx‰ œ x, g(f(x)) œ Èx$ 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 $Èx at (0ß 0)
30. (a) h(k(x)) œ
" 4
ˆ(4x)"Î$ ‰$ 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
31.
df dx
33.
df " dx ¹ x œ 4
df " dx ¹ x œ f(3)
df " dx ¹ x œ f(2)
œ
35. (a) y œ mx Ê x œ
" m
œ
(b) The graph of y œ f 36. 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
32.
df dx
34.
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
37. (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.
" m
and y-intercept mb .
Section 7.1 Inverse Functions and Their Derivatives 38. (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.
39. 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. 40. f(x) is increasing since x# x" Ê
" 3
x#
5 6
" 3
x" 56 ;
df dx
œ
" 3
41. 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 " dx
Ê " 3
œ
df dx
œ 24x# Ê
df dx
œ
" ¸ 24x# 12 Ð1xÑ"Î$
œ
œ3
y"Î$ Ê f " (x) œ
" 3
x"Î$ ;
x#Î$
42. f(x) is decreasing since x# x" Ê 1 8x$# 1 8x"$ ; y œ 1 8x$ Ê x œ "
" ˆ "3 ‰
" 6(" x)#Î$
" #
(1 y)"Î$ Ê f " (x) œ
" #
(1 x)"Î$ ;
œ "6 (1 x)#Î$
43. 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 " dx
œ
" 3(1 x)# ¹ 1cx"Î$ &Î$
44. f(x) is increasing since x# x" Ê x# df dx
œ
5 3
x#Î$ Ê
df " dx
œ
5 3
" ¹ x#Î$ x$Î&
œ
3 5x#Î&
œ
" 3x#Î$
œ "3 x#Î$
&Î$
x" ; y œ x&Î$ Ê x œ y$Î& Ê f " (x) œ x$Î& ; œ
3 5
x#Î&
45. 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. 46. 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# ). 47. 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# )). 48. 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.
429
430
Chapter 7 Transcendental Functions
49. The first integral is the area between f(x) and the x-axis over a Ÿ x Ÿ b. The second integral is the area between f(x) and the y-axis for f(a) Ÿ y Ÿ f(b). The sum of the integrals is the area of the larger rectangle with corners at (0ß 0), (bß 0), (bß f(b)) and (0ß f(b)) minus the area of the smaller rectangle with vertices at (0ß 0), (aß 0), (aß f(a)) and (0ß f(a)). That is, the sum of the integrals is bf(b) af(a).
50. f w axb œ
acx dba aax bbc acx db#
œ
ad bc . acx db#
Thus if ad bc Á !, f w axb is either always positive or always negative. Hence faxb is
either always increasing or always decreasing. If follows that faxb is one-to-one if ad bc Á !. 51. (g ‰ f)(x) œ x Ê g(f(x)) œ x Ê gw (f(x))f w (x) œ 1 52. 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 t
t
t
Ê Sw (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]. 53-60. Example CAS commands: Maple: with( plots );#53 f := x -> sqrt(3*x-2); domain := 2/3 .. 4; x0 := 3; Df := D(f); # (a) plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[1,3], legend=["y=f(x)","y=f '(x)"], title="#53(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="#53(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^p*ln(x); domain := 0..exp(1); fn_list := [seq( f(x,p), p=-2..2 )];
557
558
Chapter 8 Techniques of Integration plot( fn_list, x=domain, y=-50..10, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9], thickness=[3,4,1,2,0], legend=["p= -2","p = -1","p = 0","p = 1","p = 2"], title="#89 (Section 8.8)" ); q1 := Int( f(x,p), x=domain ); q2 := value( q1 ); q3 := simplify( q2 ) assuming p>-1; q4 := simplify( q2 ) assuming p x^p*ln(x); domain := exp(1)..infinity; fn_list := [seq( f(x,p), p=-2..2 )]; plot( fn_list, x=exp(1)..10, y=0..100, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9], thickness=[3,4,1,2,0], legend=["p = -2","p = -1","p = 0","p = 1","p = 2"], title="#90 (Section 8.8)" ); q6 := Int( f(x,p), x=domain ); q7 := value( q6 ); q8 := simplify( q7 ) assuming p>-1; q9 := simplify( q7 ) assuming p x^p*ln(x); domain := 0..infinity; fn_list := [seq( f(x,p), p=-2..2 )]; plot( fn_list, x=0..10, y=-50..50, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9], thickness=[3,4,1,2,0], legend=["p = -2","p = -1","p = 0","p = 1","p = 2"], title="#91 (Section 8.8)" ); q11 := Int( f(x,p), x=domain ): q11 = lhs(i1+i2); `` = rhs(i1+i2); `` = piecewise( p-1, q3+q8 ); `` = piecewise( p-1, infinity );
92.
Example CAS commands: Maple: f := (x,p) -> x^p*ln(abs(x)); domain := -infinity..infinity; fn_list := [seq( f(x,p), p=-2..2 )]; plot( fn_list, x=-4..4, y=-20..10, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9], legend=["p = -2","p = -1","p = 0","p = 1","p = 2"], title="#92 (Section 8.8)" ); q12 := Int( f(x,p), x=domain ); q12p := Int( f(x,p), x=0..infinity ); q12n := Int( f(x,p), x=-infinity..0 ); q12 = q12p + q12n; `` = simplify( q12p+q12n );
Chapter 8 Practice Exercises 89-92. Example CAS commands: Mathematica: (functions and domains may vary) Clear[x, f, p] f[x_]:= xp Log[Abs[x]] int = Integrate[f[x], {x, e, 100)] int /. p Ä 2.5 In order to plot the function, a value for p must be selected. p = 3; Plot[f[x], {x, 2.72, 10}] CHAPTER 8 PRACTICE EXERCISES
1.
# ' xÈ4x# 9 dx; ” u œ 4x 9 • Ä du œ 8x dx
2.
# ' 6xÈ3x# 5 dx; ” u œ 3x 5 • Ä ' Èu du œ 23 u$Î# C œ 23 a3x# 5b$Î# C du œ 6x dx
3.
' x(2x 1)"Î# dx; ” u œ 2x 1 • Ä du œ 2 dx œ
4.
(2x 1)&Î# 10
' Èx
dx; ”
1x
œ
2 3
(2x 1)$Î# 6
" #
" 1#
a4x# 9b
$Î#
C
' ˆ u # 1 ‰ Èu du œ "4 Š' u$Î# du ' u"Î# du‹ œ "4 ˆ 25 u&Î# 23 u$Î# ‰ C
C
uœ1x Ä ' du œ dx •
(1 u) Èu
du œ ' ŠÈu
" Èu ‹
du œ
2 3
u$Î# 2u"Î# C
(1 x)$Î# 2(1 x)"Î# C
5.
' È x dx
6.
' È x dx
7.
' 25ydyy
8.
' 4ydyy ; ”
9.
' Èt
10.
'
11.
' Èu du œ "8 † 23 u$Î# C œ
" 8
8x#
; 1 ”
;”
9 4x#
#
;”
$
%
$ dt 9 4t%
2t dt t% 1
" 16
' Èduu œ 16" † 2u"Î# C œ È8x8 1 C
u œ 9 4x# Ä "8 ' du œ 8x dx •
u œ 25 y# Ä du œ 2y dy •
u œ 4 y% • Ä du œ 4y$ dy
;”
;”
u œ 8x# 1 Ä du œ 16x dx •
" #
" 4
#
du Èu
È9 4x# 4
C
' duu œ #" ln kuk C œ #" ln a25 y# b C
' duu œ 4" ln kuk C œ 4" ln a4 y% b C
u œ 9 4t% " • Ä 16 ' du œ 16t$ dt
u œ t# Ä du œ 2t dt •
œ 8" † 2u"Î# C œ
du Èu
" œ 16 † 2u"Î# C œ
È9 4t% 8
C
' u du1 œ tan" u C œ tan" t# C #
&Î$ " ' z#Î$ ˆz&Î$ 1‰#Î$ dz; ” u œ z5 #Î$ • Ä du œ z dz 3
3 5
' u#Î$ du œ 35 † 35 u&Î$ C œ
9 #5
ˆz&Î$ 1‰&Î$ C
559
560 12.
Chapter 8 Techniques of Integration
z%Î& ' z"Î& ˆ1 z%Î& ‰"Î# dz; ” u œ 41 "Î& • Ä du œ z dz 5
5 4
' u"Î# du œ 54 † 2Èu C œ #5 ˆ1 z%Î& ‰"Î# C
13.
2) d) ' (1sin cos 2) )
14.
) d) ' (1 cos sin ))
15.
u œ 3 4 cos t dt ' 3 sin4tcos • t ;”
16.
' 1cos sin2t dt2t ; ”
17.
' (sin 2x) ecos 2x dx; ” u œ cos 2x • Ä "# ' eu du œ "# eu C œ "# ecos 2x C du œ 2 sin 2x dx
18.
u œ sec x ' (sec x tan x) esec x dx; ” Ä ' eu du œ eu C œ esec x C du œ sec x tan x dx •
19.
' e) sin ae) b cos# ae) b d); ”
20.
) ' e) sec# ae) b d); ” u œ e) • Ä ' sec# u du œ tan u C œ tan ae) b C du œ e d)
21.
' 2xc1 dx œ
23.
' v dvln v ; ”
24.
' v(2 dvln v) ; ” u œ 2 " ln v •
#
;”
u œ 1 cos 2) Ä du œ # sin 2) d) •
"Î#
;”
u œ 1 sin ) Ä du œ cos ) d) •
du œ 4 sin t dt
26.
' sinÈ
#
"
x dx 1 x#
28.
' È dx
œ
49x#
xb ; ”
v
dv
Ä
1x
" 7
'
œ 4" ln kuk C œ 4" ln k3 4 cos tk C
' duu œ 2" ln kuk C œ 2" ln k1 sin 2tk C
' u# du œ "3 u$ C œ "3 cos$ ae) b C
dx É1ˆ x7 ‰#
;”
' 5xÈ2 dx œ
" È2
È2
x
Š 5ln 5 ‹ C
' duu œ ln kuk C œ ln k2 ln vk C
u œ 2 tan" x Ä du œ x#dx1 •
u œ 2x Ä du œ 2 dx •
;”
" 2
du u
' duu œ ln kuk C œ ln kln vk C
u œ sin" x Ä du œ È dx # —
' È 2 dx
œ 2u"Î# C œ 2È1 sin ) C
22.
;–
27.
1 4x#
"
"Î#
C
du œ
' ax 1b adx2tan
' udu
#
u œ cos ˆe) ‰ • Ä du œ sin ˆe) ‰ † e) d)
u œ ln v Ä du œ v" dv •
25.
' duu œ 2u" C œ #(1 "cos 2)) C
Ä 4" '
u œ 1 sin 2t Ä du œ 2 cos 2t dt •
2xc1 ln 2
" #
' duu œ ln kuk C œ ln k2 tan" xk C
' u du œ "# u# C œ "# asin" xb# C
' È du
1 u#
œ sin" u C œ sin" (2x) C
u œ x7 Ä du œ "7 dx •
' È du
1 u#
œ sin" u C œ sin" ˆ x7 ‰ C
Chapter 8 Practice Exercises 29.
30.
31.
'È
' È dt
9 4t#
' 9 dt t
#
" 4
œ
dt 16 9t#
œ
'
" 9
'
" 3
œ
'
dt #
Ê1 Š 3t 4‹
dt
;–
#
Ê1 Š 2t 3‹
dt # 1ˆt‰
;–
3
uœ du œ
;–
uœ du œ
uœ du œ " 3 " 3
3 4 3 4
2 3 2 3
t Ä dt —
t Ä dt —
t Ä dt —
" 2
34.
'
35.
' È dx
36.
'È
37.
2) " " ˆ y 2 ‰ ' y dy4y 8 œ ' (yd(y C 2) 4 œ # tan #
38.
2) " ' t dt4t 5 œ ' (t d(t (t 2) C 2) 1 œ tan
4 dx 5xÈ25x# 16
œ 3'
6 dx xÈ4x# 9
4x x#
œ
œ'
dx 4xx# 3
4 25
'
dx xÉx# 94
œ'
d(x2) È1(x2)#
40.
'
" 5
dv (v 1)Èv# 2v
œ'
d(x1) (x 1)È(x 1)# 1
œ sec" kx 1k C
œ'
d(v 1) (v 1)È(v 1)# 1
œ sec" kv 1k C
6x ' cos# 3x dx œ ' " cos dx œ x# sin126x C #
43.
' sin$ #) d) œ ' ˆ1 cos# #) ‰ ˆsin #) ‰ d); –
tan" (5t) C
cos$
) #
2 cos
) #
u œ cos #) Ä 2 ' a1 u# b du œ du œ "# sin #) d) —
u$ 3
Cœ
2u$ 3
2u C
C
' sin$ ) cos# ) d) œ ' a1 cos# )b (sin )) acos# )b d); ” u& 5
tan" ˆ 3t ‰ C
œ sin" (x 2) C
42.
œ
" 5
" 3
sin" ˆ 2t3 ‰ C
œ sin" ˆ x # 2 ‰ C
2x ' sin# x dx œ ' 1 cos dx œ #x sin42x C #
44.
tan" u C œ
" 2
sin" ˆ 3t4 ‰ C
¸ sec" ¸ 5x 4 C
41.
2 3
" 3
sin" u C œ
" 3
#
dx (x 1)Èx# 2x
œ
œ
" 2
sin" u C œ
#
#
'
œ
œ
#
œ
" 3
¸ œ 2 sec" ¸ 2x 3 C
d(x 2) È4 (x 2)#
#
39.
dx xÉx# 16 25
#
1 u#
tan" u C œ
'
" 5
' È du
" 5
33.
u œ 5t Ä du œ 5 dt •
œ
1 u#
' 1 duu
' 1 dt25t
;”
' È du
" 3
32.
#
' 1 duu
" 3
cos& ) 5
cos$ ) 3
C
u œ cos ) Ä ' a1 u# b u# du œ ' au% u# b du du œ sin ) d) •
561
562 45.
Chapter 8 Techniques of Integration
' tan$ 2t dt œ ' Ä œ
46.
" 4
" #
(tan 2t) asec# 2t 1b dt œ ' tan 2t sec# 2t dt ' tan 2t dt; ”
u œ 2t du œ 2 dt •
' tan u sec# u du "# ' tan u du œ 4" tan# u #" ln kcos uk C œ 4" tan# 2t #" ln kcos 2tk C
tan# 2t
" #
ln ksec 2tk C
' 6 sec% t dt œ 6' atan# t 1b asec# tb dt; ”
u œ tan t Ä 6 ' au# 1b du œ 2u$ 6u C du œ sec# t dt •
œ 2 tan$ t 6 tan t C 47.
' 2 sin dxx cos x œ ' sindx2x œ '
48.
' cos 2xdxsin x œ ' cos2 dx2x ; ” #
#
csc 2x dx œ "# ln kcsc 2x cot 2xk C
u œ 2x Ä du œ 2 dx •
'
du cos u
œ ' sec u du œ ln ksec u tan uk C
œ ln ksec 2x tan 2xk C 49.
'11ÎÎ42 Ècsc# y 1 dy œ '11ÎÎ42 cot y dy œ cln ksin ykd 11ÎÎ24 œ ln 1 ln È"
50.
'13Î14Î4 Ècot# t 1 dt œ '13Î14Î4 csc t dt œ c ln kcsc t cot tkd 311Î4Î4 œ ln ¸csc 341 cot 341 ¸ ln ¸csc 14 cot 14 ¸ È
œ ln ¹È2 1¹ ln ¹È2 1¹ œ ln ¹ È2 " ¹ œ ln »
œ ln È2
ŠÈ2 "‹ ŠÈ2 1‹ #1
21
51.
2
» œ ln Š3 2
È2‹
'01 È1 cos# 2x dx œ '01 ksin 2xk dx œ '01Î2 sin 2x dx '11Î2 sin 2x dx œ cos#2x ‘ 10 Î2 cos#2x ‘ 11Î2 œ ˆ "# "# ‰ "# ˆ "# ‰‘ œ 2
52.
'021 È1 sin# x# dx œ '021 ¸cos x# ¸ dx œ '01 cos x# dx '121 cos x# dx œ 2 sin x# ‘ 10 2 sin x# ‘ 1#1 œ (2 0) (0 2) œ 4
53.
'11ÎÎ22 È1 cos 2t dt œ È2 '11ÎÎ22 ksin tk dt œ 2È2 '01Î2 sin t dt œ ’2È2 cos t“ 1Î2 œ 2È2 [0 (1)] œ 2È2 0
54.
'121 È1 cos 2t dt œ È2 '121 kcos tk dt œ È2 '131Î2 cos t dt È2 '3211Î2 cos t dt œ È2 csin td $11Î2 È2 csin td #$11Î# œ È2 (1 0) È2 [0 (1)] œ 2È2
55.
' xx dx4 œ x ' x4 dx4 œ x 2 tan" ˆ #x ‰ C
56.
' 9xdxx
57.
' 4x2x 13 dx œ ' (2x 1) #x 4 1 ‘ dx œ x x# 2 ln k2x 1k C
58.
' 2xx dx4 œ ' ˆ2 x 8 4 ‰ dx œ 2x 8 ln kx 4k C
59.
' 2yy 41 dy œ '
#
#
#
$
#
œ ' ’ x ax x# 9b9 9x “ dx œ ' ˆx #
9x ‰ x# 9
dx œ
x# #
9 #
ln a9 x# b C
#
#
2y dy y# 4
'
dy y# 4
œ ln ay# 4b
" #
tan" ˆ #y ‰ C
Chapter 8 Practice Exercises 60.
' yy 41 dy œ ' yy dy1 4 ' y dy 1 œ #" ln ay# 1b 4 tan" y C
61.
' Èt 2
62.
' 2t È È1 t
63.
#
#
4 t#
dt œ '
#
#
1 t#
t
#
t dt È4 t#
dt œ '
2t dt È1 t#
'
dt t
œ È4 t# 2 sin" ˆ #t ‰ C
œ 2È1 t# ln ktk C #
#
#
#
cos x) dx cos x dx ' cotcotx x csc ' cos ' (cos1x)(1 dx œ ' cos x sin1 x sin x dx x œ x1 œ cos x x) ' sindx x ' dx œ sin" x cot x x C œ x cot x csc x C œ ' d(sin sin x #
#
#
65.
dt È4 t#
sin x) dx ' tantanx x sec ' sinsin xxdx1 œ ' (sin1x)(1 dx œ ' sin x cos1 xcos x dx x œ sin x x) ' cosdx x ' dx œ cos" x tan x x C œ x tan x sec x C œ ' d(cos cos x #
64.
2'
#
#
' sec (5 3x) dx; ” y œ 5 3x •
Ä
dy œ 3 dx
' sec y † Š dy3 ‹ œ "3 ' sec y dy œ 3" ln ksec y tan yk C
œ "3 ln ksec (5 3x) tan (5 3x)k C 66.
' x csc ax# 3b dx œ "# ' csc ax# 3b d ax# 3b œ "# ln kcsc ax# 3b cot ax# 3bk C
67.
' cot ˆ x4 ‰ dx œ 4 ' cot ˆ x4 ‰ d ˆ x4 ‰ œ 4 ln ¸sin ˆ x4 ‰¸ C
68.
' tan (2x 7) dx œ "# ' tan (2x 7) d(2x 7) œ "# ln kcos (2x 7)k C œ "# ln ksec (2x 7)k C
69.
' xÈ1 x dx; ” u œ 1 x •
Ä ' (1 u)Èu du œ ' ˆu$Î# u"Î# ‰ du œ
du œ dx
$
œ
70.
72.
(1 x)
&Î#
(1 x) 2 3
$Î#
C œ 2 –
' 3xÈ2x 1 dx; ” u œ 2x 1 • du œ 2 dx
œ 71.
2 5
3 10
œ
sec ) tan ) 31
œ
sin ) 2 cos# )
z œ tan ) Ä dz œ sec# ) d) •
" #
32 31
C
&
ŠÈ1 x‹
3
3 ˆÈ2x 1‰ 10
ŠÈ1 x‹
—C
5
&
ˆÈ2x 1‰ #
$
C
' Ètan# ) 1 † sec# ) d) œ ' sec$ ) d) (FORMULA 92)
ln ksec ) tan )k C œ
' a16 z# b$Î# dz; ” z 16 a16 z# b"Î#
' sec ) d)
u&Î# 23 u$Î# C
' 3 ˆ u # 1 ‰ Èu † "# du œ 34 ' ˆu$Î# u"Î# ‰ du œ 34 † 25 u&Î# 34 † 23 u$Î# C
(2x 1)&Î# "# (2x 1)$Î# C œ
' Èz# 1 dz; ”
œ
Ä
2 5
z œ 4 tan ) Ä dz œ 4 sec# ) d) •
zÈ z# 1 #
" #
ln ¹z È1 z# ¹ C
) d) " ' " ' 644 sec cos ) d) œ 16 sin ) C œ sec ) d) œ 16 #
$
z 16È16 z#
C
563
564 73.
Chapter 8 Techniques of Integration
' È25dy y
œ
#
" 5
'
dy É1ˆ 5y ‰#
œ'
du È 1 u#
u œ tan ) , u œ y5 ‘ ; ” Ä du œ sec# ) d) •
' Èsec ) d) #
1 tan# )
œ ' sec ) d)
#
œ ln ksec ) tan )k C" œ ln ¹È1 u# u¹ C" œ ln ºÉ1 ˆ y5 ‰ y5 º C" œ ln ¹
È25 y# y ¹ 5
C"
œ ln ¸y È25 y# ¸ C 74.
' È25dy 9y " 3
Ä 75.
'
76.
' Èx
#
œ
" 5
'
dy Ê1 Š
3y # 5 ‹
œ
" 3
' È du
dx x# È 1 x#
$ dx 1 x#
;”
;”
x œ sin ) Ä dx œ cos ) d) •
x œ sin ) Ä dx œ cos ) d) •
Note: Ans ´
' Èx
# dx 1 x#
œ 78.
" #
" 3
ln ¹È1 u# u¹ C" from Exercise 73
ln ¸È25 9y# 3y¸ C
' sincos))cosd) ) œ ' csc# ) d) œ cot ) C œ È1x x
)
;” " #
x
#È
1 x#
3
' sin )coscos) ) d) œ ' sin$ ) d) œ ' a1 cos# )b (sin )) d); $
u$ 3
C œ cos )
" 3
cos$ ) œ È1 x# "3 a1 x# b
x# 9
sin ) cos ) œ
;”
sin" x #
xÈ 1 x# #
x œ 2 sin ) Ä dx œ 2 cos ) d) •
C
' 2 cos ) † 2 cos ) d) œ 2 ' (1 cos 2)) d) œ 2 ˆ) "# sin 2)‰ C
x œ 3 sec ) Ä' dx œ 3 sec ) tan ) d) •
3 sec ) tan ) d) È9 sec# ) 9 #
œ'
3 sec ) tan ) d) 3 tan )
œ ln ksec ) tan )k C" œ ln º x3 Ɉ x3 ‰ 1º C" œ ln ¹ x
80.
'
12 dx ax# 1b$Î#
;”
C
#
#
' È dx
$Î#
2) ' sin )coscos) ) d) œ ' sin# ) d) œ ' 1 cos d) œ "# ) 4" sin 2) C #
œ 2) 2 sin ) cos ) C œ 2 sin" ˆ x# ‰ xÉ1 ˆ x# ‰ C œ 2 sin" ˆ x# ‰ 79.
C
23 È1 x# C by another method
x œ sin ) Ä dx œ cos ) d) •
' È4 x# dx; ”
#
#
cu œ cos )d Ä ' a1 u# b du œ u
77.
œ
1 u#
x œ sec ) Ä dx œ sec ) tan ) d) •
È x# 9 ¹ 3
#
C
œ ' sec ) d)
cos ) d) ' 12 sectan) tan) ) d) œ ' 12 sin ;” ) $
xÈ 4 x# 2
C" œ ln ¹x Èx# 9¹ C u œ sin ) Ä du œ cos ) d) •
' 12u du #
12 12 x œ 12 C u C œ sin ) C œ È # x 1
81.
' Èww 1 dw; ”
w œ sec ) tan ) ‰ ' tan# ) d) œ ' asec# ) 1b d) Ä ' ˆ sec ) † sec ) tan ) d) œ dw œ sec ) tan ) d) • œ tan ) ) C œ Èw# 1 sec" w C
82.
' Èz z 16 dz; ”
#
#
z œ 4 sec ) Ä dz œ 4 sec ) tan ) d) •
sec ) tan ) d) ' 4 tan )†44sec œ 4' )
tan# ) d) œ 4(tan ) )) C
œ Èz# 16 4 sec" ˆ 4z ‰ C 83. u œ ln (x 1), du œ
dx x1
; dv œ dx, v œ x;
' ln (x 1) dx œ x ln (x 1) ' x x 1 dx œ x ln (x 1) ' dx ' x dx 1 œ x ln (x 1) x ln (x 1) C" œ (x 1) ln (x 1) x C" œ (x 1) ln (x 1) (x 1) C, where C œ C" 1
Chapter 8 Practice Exercises 84. u œ ln x, du œ
dx x
; dv œ x# dx, v œ
" 3
x$ ;
' x# ln x dx œ 3" x$ ln x ' 3" x$ ˆ x" ‰ dx œ x3
$
85. u œ tan" 3x, du œ
3 dx 1 9x#
#
" 6
x$ 9
C
; dv œ dx, v œ x;
' tan" 3x dx œ x tan" 3x ' 13x 9xdx œ x tan" (3x)
ln x
;”
y œ 1 9x# Ä x tan" 3x dy œ 18x dx •
" 6
' dyy
ln a1 9x# b C
86. u œ cos" ˆ x# ‰ , du œ
dx È 4 x#
; dv œ dx, v œ x;
' cos" ˆ x# ‰ dx œ x cos" ˆ x# ‰ ' Èx dx
4 x #
;”
y œ 4 x# Ä x cos" ˆ x# ‰ dy œ 2x dx •
" #
' Èdyy
#
œ x cos" ˆ x# ‰ È4 x# C œ x cos" ˆ x# ‰ 2É1 ˆ x# ‰ C ex
87.
ÐÑ (x 1)# ïïïïî ex ÐÑ 2(x 1) ïïïïî ex ÐÑ 2 ïïïïî ex Ê
0
' (x 1)# ex dx œ c(x 1)# 2(x 1) 2d ex C
sin (1 x)
88. ÐÑ x# ïïïïî ÐÑ 2x ïïïïî ÐÑ 2 ïïïïî
cos (1 x) sin (1 x) cos (1 x) Ê
0
' x# sin (1 x) dx œ x# cos (1 x) 2x sin (1 x) 2 cos (1 x) C
89. u œ cos 2x, du œ 2 sin 2x dx; dv œ ex dx, v œ ex ; I œ ' ex cos 2x dx œ ex cos 2x 2 ' ex sin 2x dx; u œ sin 2x, du œ 2 cos 2x dx; dv œ ex dx, v œ ex ;
I œ ex cos 2x 2 ’ex sin 2x 2 ' ex cos 2x dx“ œ ex cos 2x 2ex sin 2x 4I Ê I œ
ex cos 2x 5
2ex sin 2x 5
90. u œ sin 3x, du œ 3 cos 3x dx; dv œ ec2x dx, v œ "# ec2x ; I œ ' ec2x sin 3x dx œ "# ec2x sin 3x
3 2
' ec2x cos 3x dx;
u œ cos 3x, du œ 3 sin 3x dx; dv œ ec2x dx, v œ "# ec2x ; I œ "# ec2x sin 3x 3# ’ "# ec2x cos 3x Ê Iœ
4 13
3 #
' ec2x sin 3x dx“ œ "# ec2x sin 3x 34 ec2x cos 3x 94 I
2 c2x ˆ "# ec2x sin 3x 34 ec2x cos 3x‰ C œ 13 e sin 3x
91.
' x x 3xdx 2 œ ' x2dx2 ' x dx 1 œ 2 ln kx 2k ln kx 1k C
92.
' x x 4xdx 3 œ #3 ' xdx3 #" ' x dx 1 œ #3 ln kx 3k #" ln kx 1k C
#
#
3 13
ec2x cos 3x C
C
565
566
Chapter 8 Techniques of Integration
93.
' x(xdx 1)
œ ' Š "x
94.
' x x(x11) dx œ ' ˆ x2 1 2x x" ‰ dx œ 2 ln ¸ x" ¸ "x C œ 2 ln kxk "x 2 ln kx 1k C x
95.
' cos )sin )cosd)) 2 ; ccos ) œ yd
#
1 (x 1)# ‹
1 x1
#
Ä '
" 3
œ "3 '
dy y# y 2
C
dy y1
" 3
' y dy # œ 3" ln ¹ yy 12 ¹ C
)2¸ " ¸ cos ) 1 ¸ ln ¸ cos cos ) 1 C œ 3 ln cos ) 2 C
96.
) d) ' sin )cos sin ) 6 ; csin ) œ xd
97.
' 3x x 4xx 4 dx œ '
98.
' x4xdx4x œ ' x4 dx4 œ 2 tan" ˆ #x ‰ C
99.
' (v2v3)8vdv œ #" '
#
4 x
$
$
'
Ä
#
dx '
dx x# x 6
x4 x# 1
œ
" 5
)2¸ ' xdx 2 5" ' xdx 3 œ 5" ln ¸ sin sin ) 3 C
dx œ 4 ln kxk
" #
ln ax# 1b 4 tan" x C
#
$
œ
" x1
#
#
œ
dx œ ln kxk ln kx 1k
" 16
3 Š 4v
5 8(v 2)
" 8(v #) ‹
dv œ 38 ln kvk
5 16
ln kv 2k
&
7) dv ' (v (3v ' (v 2) 1dv ' 1)(v 2)(v 3) œ
101.
' t 4tdt 3 œ "# ' t dt 1 #" ' t dt 3 œ #" tan" t
102.
' t t tdt 2 œ "3 ' t t dt2 3" ' t t dt1 œ 6" ln kt# 2k 6" ln at# 1b C
103.
' x xx x 2 dx œ ' ˆx x 2xx 2 ‰ dx œ ' x dx 32 ' x dx 1 34 ' x dx 2
%
#
#
%
#
#
$
dv v2
'
dv v3
#
3) œ ln ¹ (v (v2)(v 1)# ¹ C
" #È 3
tan" Š Èt 3 ‹ C œ
" #
tan" t
ln kx 2k
2 3
105.
' x x4x4x 3 dx œ ' ˆx x 3x4x 3 ‰ dx œ '
$
107.
$
$
#
#
'
#
#
x #
9 #
ln kx 3k
3 #
2x$ x# 21x 24 dx œ x# 2x 8 2 # x 3x 3 ln kx 4k
dx x ˆ3 È x 1 ‰
œ
" 3
È
(2x 3)
" 3
3 #
' x dx 1 #9 ' xdx3
x ‘ x# 2x 8
dx œ ' (2x 3) dx
" 3
' x dx # 32 ' x dx 4
ln kx 2k C
Ô u œ Èx 1 × Ù Ä ; Ö du œ 2Èdx x1 Õ dx œ 2u du Ø
1" ln ¹ Èxx ¹C 11
x dx
ln kx 1k ln kxk C
ln kx 1k C
'
œ
C
ln kx 1k C
$
'
t È3
#
4 3
' xx 1x dx œ ' ˆ1 xx "x ‰ dx œ ' ’1 x(x " 1) “ dx œ ' dx ' x dx 1 ' dxx œ x
106.
tan"
#
104.
œ
È3 6
#
#
œ
ln kv 2k C
ln ¹ (v 2)v'(v 2) ¹ C
100.
x# #
" 16
2 3
' au udu1b u œ 3" ' u du 1 3" ' u du 1 œ 3" ln ku 1k 3" ln ku 1k C #
Chapter 8 Practice Exercises $ Ô u œ Èx × dx Ö du œ dx Ù Ä ' $ x‰ ; x ˆ1 È 3x#Î$ Õ dx œ 3u# du Ø
108.
'
109.
' e ds 1 ; Ô du œ es ds ×
u œ es 1
110.
s
'
Õ ds œ
È es 1 " ¹ È es 1 1 ¹
' È16y dy y
111. (a)
' È16y dy y
(b)
Ø
du u1
#
; cy œ 4 sin xd Ä 4 '
d a16 y# b È16 y#
œ
(b)
' Èx dx
; cx œ 2 tan yd Ä
(b)
#
#
#
C
' Èx dx
' 4xdxx ' 4xdxx
du ' u a2uu du 1b œ 2 ' (u 1)(u ' u du 1 ' u du 1 œ ln ¸ uu "1 ¸ C 1) œ
u 1
112. (a)
113. (a)
$ x È
œ 3 ln ¸ u u 1 ¸ C œ 3 ln ¹ 1È $ x¹ C
s
œ "# '
4 x#
du u(1 u)
s
#
4 x#
œ 3'
' u(udu 1) œ ' udu1 ' duu œ ln ¸ u u 1 ¸ C œ ln ¸ e e " ¸ C œ ln k1 ecs k C
È s Ôu œ es 1× e ds Ö ; du œ 2Èes 1 Ù Ä Õ ds œ 2u# du Ø
ds È es 1
œ ln
Ä
3u# du u$ (1 u)
" #
œ È16 y# C sin x cos x dx cos x
œ 4 cos x C œ
4È16 y# 4
C œ È16 y# C
' dÈa4 x b œ È4 x# C #
4 x#
œ #" '
d a4 x # b 4 x#
; cx œ 2 sin )d Ä
' 2 tan y2†2secsecy y dy œ 2 ' sec y tan y dy œ 2 sec y C œ È4 x# C #
œ #" ln k4 x# k C
È 2 cos ) d) ' 2 sin4)†cos œ ' tan ) d) œ ln kcos )k C œ ln Š 4 2 x ‹ C ) #
#
œ "# ln k4 x# k C 114. (a)
' È t dt
œ
(b)
' È t dt
; t œ
4t# 1
4t# 1
" 8
' dÈa4t
# 1b 4t# 1
" #
œ
" 4
sec )‘ Ä
È4t# 1 C
'
u œ 9 x# Ä #" ' du œ 2x dx •
" #
sec ) tan )† "# sec ) d) tan )
œ
" 4
' sec# ) d) œ tan4 ) C œ È4t4 1 C #
115.
' 9xdxx
116.
' x a9dx x b œ 9" ' dxx 18" ' 3 dx x 18" ' 3 dx x œ 9" ln kxk 18" ln k3 xk 18" ln k3 xk C
#
;”
du u
œ #" ln kuk C œ ln
" Èu
C œ ln
" È 9 x#
C
#
œ
" 9
ln kxk
" 18
ln k9 x# k C
117.
' 9 dxx
118.
' È dx
119.
' sin3 x cos4 x dx œ ' cos4 xa1 cos2 xbsin x dx œ ' cos4 x sin x dx ' cos6 x sin x dx œ cos5 x cos7 x C
120.
#
œ
9 x#
" 6
;”
' 3 dx x 6" ' 3dxx œ 6" ln k3 xk 6" ln k3 xk C œ 6" ln ¸ xx 33 ¸ C x œ 3 sin ) Ä dx œ 3 cos ) d) •
) ' 33 cos ' d) œ ) C œ sin" x C cos ) d) œ 3
5
' cos5 x sin5 x dx œ ' sin5 x cos4 x cos x dx œ ' sin5 x a1 sin2 xb2 cos x dx œ ' sin5 x cos x dx 2' sin7 x cos x dx ' sin9 x cos x dx œ sin6 x 2sin8 x sin10 x C 6
8
10
7
567
568
Chapter 8 Techniques of Integration
121.
' tan4 x sec2 x dx œ tan5 x C
122.
' tan3 x sec3 x dx œ ' asec2 x 1b sec2 x † sec x † tan x dx œ ' sec4 x † sec x † tan x dx ' sec2 x † sec x † tan x dx
5
œ 123.
sec5 x 5
sec3 x 3
C
' sin 5) cos 6) d) œ "# ' asina)b sina11)bb d) œ "# ' sina)b d) "# ' sina11)b d) œ "# cosa)b ##" cos 11) C œ "# cos )
" ## cos
11) C
124.
' cos 3) cos 3) d) œ "# ' acos 0 cos 6)b d) œ "# '
d)
125.
' É1 cosˆ 2t ‰ dt œ ' È2¸ cos
t¸ 4
126.
' et Ètan2 et 1 dt œ ' k sec et k et dt œ lnk sec et tan et k C 3" 180 Ð%Ñ
127. kEs k Ÿ
(˜x)% M where ˜x œ
dt œ 4È2 ¸ sin
t¸ 4
3" n
œ
2 n
; f(x) œ
" x
" #
' cos 6) d) œ "# ) 121 sin 6) C
C
œ x" Ê f w (x) œ x# Ê f ww (x) œ 2x$ Ê f'''(x) œ 6x%
(x) œ 24x& which is decreasing on [1ß 3] Ê maximum of f Ð%Ñ (x) on [1ß 3] is f Ð%Ñ (1) œ 24 Ê M œ 24. Then " ‰ ˆ 2 ‰% ˆ 768 ‰ ˆ n"% ‰ Ÿ 0.0001 Ê n"% Ÿ (0.0001) ˆ 180 ‰ Ê n% 10,000 ˆ 768 ‰ kEs k Ÿ 0.0001 Ê ˆ 3180 n (24) Ÿ 0.0001 Ê 180 768 180 Ê f
Ê n 14.37 Ê n 16 (n must be even) 10 12
128. kET k Ÿ Ê
2 3n#
129. ˜x œ
(˜x)# M where ˜x œ
Ÿ 10$ Ê
ba n
œ
10 6
3n# #
œ
1 6
10 n #
œ
1000 Ê n Ê
˜x #
œ
1 1#
" n ;0 2000 3
Ÿ f ww (x) Ÿ 8 Ê M œ 8. Then kET k Ÿ 10$ Ê
! mf(xi ) œ 12 Ê T œ ˆ 1 ‰ (12) œ 1 ; 12 iœ0
6
! mf(xi ) œ 18 and iœ0
˜x 3
œ
1 18
Ê
1‰ S œ ˆ 18 (18) œ 1 .
130. ¸f Ð%Ñ (x)¸ Ÿ 3 Ê M œ 3; ˜x œ
2" n
œ
" n
ˆ "n ‰# (8) Ÿ 10$
Ê n 25.82 Ê n 26
x! x" x# x$ x% x& x'
xi 0 1/6 1/3 1/2 21/3 51/6 1
f(xi ) 0 1/2 3/2 2 3/2 1/2 0
m 1 2 2 2 2 2 1
mf(xi ) 0 1 3 4 3 1 0
x! x" x# x$ x% x& x'
xi 0 1/6 1/3 1/2 21/3 51/6 1
f(xi ) 0 1/2 3/2 2 3/2 1/2 0
m 1 4 2 4 2 4 1
mf(xi ) 0 2 3 8 3 2 0
;
6
" 12
" ‰ ˆ " ‰% & . Hence kEs k Ÿ 10& Ê ˆ 2180 Ê n (3) Ÿ 10
" 60n%
Ÿ 10& Ê n%
Ê n 6.38 Ê n 8 (n must be even) 131. yav œ œ œ
" 365 0
21 " '0365 37 sin ˆ 365 ˆ 21 ‰ ‰‘ $'& (x 101)‰ 25‘ dx œ 365 37 ˆ 365 21 cos 365 (x 101) 25x !
" ˆ ‰ 21 ‘ ‰ ˆ ˆ 365 ‰ 21 ‘ ‰‘ 37 ˆ 365 365 21 cos 365 (365 101) 25(365) 37 21 cos 365 (0 101) 25(0) 21 21 21 21 2371 cos ˆ 365 (264)‰ 25 2371 cos ˆ 365 (101)‰ œ 2371 ˆcos ˆ 365 (264)‰ cos ˆ 365 (101)‰‰
25
10& 60
Chapter 8 Practice Exercises
569
37 ¸ #1 (0.16705 0.16705) 25 œ 25° F
132. av(Cv ) œ ¸
" 655
" "3 '20675 c8.27 10& a26T 1.87T# bd dT œ 655 8.27T 10
" 67520
&
T#
0.62333 10&
'(&
T$ ‘ #!
[(5582.25 59.23125 1917.03194) (165.4 0.052 0.04987)] ¸ 5.434;
8.27 10& a26T 1.87T# b œ 5.434 Ê 1.87T# 26T 283,600 œ 0 Ê T ¸
26È676 4(1.87)(283,600) #(1.87)
¸ 396.45° C 133. (a) Each interval is 5 min œ
1 12
hour.
2a2.4b 2a2.3b Þ Þ Þ 2a2.4b 2.3 d œ ‰ (b) a60 mphbˆ 12 29 hours/gal ¸ 24.83 mi/gal 1 2.5 24 c
29 12
¸ 2.42 gal
134. Using the trapezoid rule, ˜x œ 15 Ê ˜x 2 œ 7.5; ! mf(xi ) œ 794.8 Ê Area ¸ a794.8ba7.5b œ 5961 ft2 ;
x! x" x# x$ x% x5 x' x( x)
The cost is Area † a$2.10/ft2 b ¸ a5961 ft2 ba$2.10/ft2 b œ $12,518.10 Ê the job cannot be done for $11,000.
135.
'03 È dx
œ lim c '0 bÄ3
136.
'01 ln x dx œ
b
9 x#
xi 0 15 30 45 60 75 90 105 120
f(xi ) 0 36 54 51 49.5 54 64.4 67.5 42
b œ lim c sin" ˆ x3 ‰‘ 0 œ lim c sin" ˆ 3b ‰ sin" ˆ 30 ‰ œ bÄ3 bÄ3
dx È 9 x#
lim cx ln x xd 1b œ (1 † ln 1 1) lim cb ln b bd œ 1 lim b Ä ! bÄ! bÄ!
m 1 2 2 2 2 2 2 2 1 1 #
mf(xi ) 0 72 108 102 99 108 128.8 135 42 1 #
0œ
ln b Š "b ‹
œ 1 lim bÄ!
œ 1 0 œ 1 137.
'c11 ydy
138.
'c_2 () d1))
#Î$
œ 'c1 0
dy y#Î$
'0
1
œ 'c2
$Î&
)$Î&
lim $Î& ) Ä _ ()1)
1
dy y#Î$
d) () 1)$Î&
1
_
d) )$Î&
diverges Ê
'3_ u 2du2u œ '3_ u du 2 '3_ duu œ
140.
'1_ 4v3v v1
#
_
#
dv œ '1 ˆ "v
1 œ 2 † 3 lim y"Î$ ‘ b œ 6 Š1 lim b"Î$ ‹ œ 6 bÄ! bÄ!
2
139.
$
dy y#Î$
) '2 'c1 () d1) $Î&
_
œ 1 and '2
œ 2 '0
" v#
d) () 1)$Î&
'2 () d1))
$Î&
4 ‰ 4v 1
bÄ_
dv œ lim ln v bÄ_
bÄ_
'0_ x# ex dx œ
" 4
" v
b
ln (4v 1)‘ 1
1 ln 3 œ 1 ln
3 4
lim cx# ex 2xex 2ex d 0 œ lim ab# eb 2beb 2eb b (2) œ 0 2 œ 2 b
bÄ_
bÄ_
142.
0 'c_ xe3x dx œ
143.
_ _ dx dx " ' 'c_ ' _ dx 4x 9 œ 2 0 4x 9 œ # 0 x #
diverges
b lim ln ¸ u u 2 ¸‘ 3 œ lim ln ¸ b b 2 ¸‘ ln ¸ 3 3 2 ¸ œ 0 ln ˆ "3 ‰ œ ln 3
bÄ_
œ lim ln ˆ 4b b 1 ‰ b" ‘ (ln 1 1 ln 3) œ ln 141.
converges if each integral converges, but
_
lim
b Ä _
#
x3 e3x 9" e3x ‘ 0 œ 9" b
#
9 4
œ
lim
b Ä _
" # b lim Ä_
ˆ b3 e3b 9" e3b ‰ œ 9" 0 œ 9"
32 tan" ˆ 2x ‰‘ b œ 3 0
" # b lim Ä_
32 tan" ˆ 2b ‰‘ 3
" 3
tan" (0)
Š b" ‹ Š
" ‹ b#
570
Chapter 8 Techniques of Integration " #
œ
ˆ 23 † 1# ‰ 0 œ
1 6
144.
_ 4 dx 'c_ ' _ 4 dx x 16 œ 2 0 x 16 œ 2
145.
) ) Ä _ È)# 1
#
#
b lim tan" ˆ x4 ‰‘ 0 œ 2 Š lim tan" ˆ b4 ‰‘ tan" (0)‹ œ 2 ˆ 1# ‰ 0 œ 1
bÄ_
_ d)
œ 1 and '6
lim
bÄ_
_
diverges Ê '6
)
d) È)# 1
_
diverges
_
_
146. I œ '0 eu cos u du œ lim ceu cos ud b0 '0 eu sin u du œ 1 lim ceu sin ud b0 '0 aeu b cos u du bÄ_
Ê I œ 1 0 I Ê 2I œ 1 Ê I œ 147.
bÄ_
" #
converges
'1_ lnz z dz œ '1e lnz z dz 'e_ lnz z dz œ ’ (ln2z) “ e #
1
œ _ Ê diverges 148. 0
ect Èt
_
Ÿ ect for t 1 and '1 ect dt converges Ê
149.
_ 2 dx '_ e ec
_
150.
_ dx 'c_ ' 1 dx ' 0 dx '1 x a1 e b œ _ x a1 e b 1 x a1 e b 0
œ 2'0
x
x
#
Š x"# ‹
x Ä 0 ’ x# a1" ex b “ #
' 1 x dxÈx ; – œ
œ lim
xÄ0
converges Ê
4 dx ex
#
x
x # a1 e x b x#
x
#
#
bÄ_
e
'1_ Èect dt converges t
_ 2 dx '_ e ec x
dx x # a1 e x b
x
converges _
'1
œ lim a1 ex b œ 2 and '0
1
xÄ0
dx x#
dx x # a1 e x b
;
diverges Ê
'01 x a1dx e b diverges #
x
_ dx 'c_ x a1 e b diverges
Ê
151.
_
'0
#
x
lim
2 dx ex ecx
b
#
lim ’ (ln#z) “ œ Š 12 0‹ lim ’ (ln2b) "# “
bÄ_
2x$Î# 3
x
u œ Èx — Ä du œ #dx Èx
' u 1†2u udu œ ' ˆ2u# 2u 2 1 2 u ‰ du œ 32 u$ u# 2u 2 ln k1 uk C #
x 2Èx 2 ln ˆ1 Èx‰ C
152.
' x4 x2 dx œ ' ˆx 4xx 42 ‰ dx œ '
153.
'
$
#
#
dx x ax # 1 b #
;”
x œ tan ) Ä dx œ sec# ) d) •
œ ln ksin )k
154.
' cosÈÈx x dx; –
155.
'È
156.
' (tÈ 1) dt ' ”
157.
' È du
dx 2x x#
t# 2t
1 u#
" #
3 #
' x dx 2 #5 ' x dx 2 œ x#
#
3 #
ln kx 2k
) d) ' tansec) sec œ ' cossin))d) œ ' Š 1 sinsin) ) ‹ d(sin )) ) #
$
#
%
#
sin# ) C œ ln ¹ Èxx# 1 ¹ "# Š Èx#x 1 ‹ C
u œ Èx — Ä du œ #dx Èx
œ'
x dx
d(x 1) È1 (x 1)#
' cos uu†2u du œ 2' cos u du œ 2 sin u C œ 2 sin Èx C
œ sin" (x 1) C
u œ t# 2t Ä du œ (2t 2) dt œ 2(t 1) dt •
; cu œ tan )d Ä
" #
' Èduu œ Èu C œ Èt# 2t C
' secsec))d) œ ln ksec ) tan )k C œ ln ¹È1 u# u¹ C #
5 #
ln kx 2k C
Chapter 8 Practice Exercises 158.
' et cos et dt œ sin et C
159.
' 2 cossinx x sin x dx œ ' #
2 csc# x dx '
cos x dx sin# x
' csc x dx œ 2 cot x
" sin x
ln kcsc x cot xk C
œ 2 cot x csc x ln kcsc x cot xk C 160.
sin ) ' cos ' 1 coscos) ) d) œ ' sec# ) d) ' d) œ tan ) ) C ) d) œ
161.
' 819dvv
162.
x) " ' 1cos sinx dxx œ ' 1d(sin (sin x) C sin x œ tan
#
#
#
#
%
œ
" #
' v dv 9 12" ' 3 dv v 1"# ' 3 dv v œ 12" ln ¸ 33 vv ¸ 6" tan" v3 C #
#
#
cos (2) 1)
163. ÐÑ ) ïïïïî ÐÑ 1 ïïïïî
" #
sin (2) 1) "4 cos (2) 1) Ê
0
' ) cos (2) 1) d) œ #) sin (2) 1) "4 cos (2) 1) C
164.
'2_ (x dx1)
165.
' x x 2xdx 1 œ ' ˆx 2 x 3x 2x 2 1 ‰ dx œ '
b œ lim 1 " x ‘ 2 œ lim 1 " b (1)‘ œ 0 1 œ 1
bÄ_
bÄ_
$
#
œ
166.
#
'
#
x# #
2x 3 ln kx 1k
d) É 1 È)
" x1
× Ù Ä 2 ) Õ d) œ 2(x 1) dx Ø
œ
4 3
Š1 È ) ‹
4 Š1 È ) ‹
y œ Èx — Ä dy œ 2dx Èx
"Î#
Cœ4
Ô ŒÉ1 È) 3
Õ
' 2Èsinx secÈxÈdxx ; –
168.
' xx dx16 œ ' ˆx x 16x ‰ dx œ x# 16
169.
'
170.
' ) d2)) 4 œ ' () d1)) 3 œ È33 tan" Š )È " ‹ C
171.
tan x ' cos ' tan x sec# x dx œ ' tan x † d(tan x) œ "# tan# x C x dx œ
172.
&
y†2y dy ' 2 siny sec œ ' 2 sin 2y dy œ cos (2y) C œ cos ˆ2Èx‰ C y
#
%
dy sin y cos y
œ'
' ˆ x#2x 4
2x ‰ x# 4
#
dr (r 1)Èr# 2r
œ'
x# #
#
4 ln ¹ xx# 4¹ C
3
#
'
dx œ
œ ' 2 csc (2y) dy œ ln kcsc (2y) cot (2y)k C
2 dy sin 2y
#
dx (x 1)#
× É1 È) C Ø
167.
%
'
' 2(x È1)x dx œ 2 ' Èx dx 2 ' Èdxx œ 34 x$Î# 4x"Î# C $
$Î#
dx x1
C
x œ 1 È) d) dx œ È
Ô ;Ö
(x 2) dx 3 '
d(r 1) (r 1)È(r 1)# 1
œ sec" kr 1k C
571
572
Chapter 8 Techniques of Integration
173.
' È(r 2) dr
œ'
174.
' 4ydyy
" #
175.
2) d) ' (1sin cos 2) )
r#
%
4r
œ
#
'
(r 2) dr È4 (r 2)#
d ay # b 4 ay # b #
œ
œ #" '
u œ % (r 2)# Ä ' du œ 2(r 2) dr •
du 2È u
œ Èu C œ È4 (r 2)# C
#
tan" Š y# ‹ C
d(1 cos 2)) (1 cos 2))#
œ
" #(1 cos 2))
Cœ
" 4
sec# ) C
176.
'
177.
'11ÎÎ42 È1 cos 4x dx œ È2 '11ÎÎ42 cos 2x dx œ ’ È#2 sin 2x“ 1Î2 œ È#2
178.
' (15)2x1 dx œ "# ' (15)2x1 d(2x 1) œ "# Š 15ln 15b ‹ C
179.
' Èx dx
dx ax # 1 b #
œ'
" 4
;”
œ
dx a 1 x # b#
x 2 a1 x # b
" 4
"¸ ln ¸ xx 1 C (FORMULA 19)
1Î4
2x 1
;”
2x
yœ2x Ä ' dy œ dx •
(2 y) dy Èy
œ
y$Î# 4y"Î# C œ
2 3
2 3
(2 x)$Î# 4(2 x)"Î# C
$
ŠÈ2 x‹
œ 2–
180.
3
' È1v v
#
#
2È2 x— C
' cos )sin†cos)) d) œ ' a1 sinsin ))b d) œ ' csc# ) d) ' d) œ cot ) ) C #
dv; cv œ sin )d Ä
œ sin" v
È 1 v# v
#
#
C
181.
1) " ' y dy2y 2 œ ' (yd(y (y 1) C 1) 1 œ tan
182.
' ln Èx 1 dx; – y œ
#
#
Èx 1
dy œ
Ê œ
" #
dx 2È x 1
— Ä
ln y † 2y dy; u œ ln y, du œ
dy y
; dv œ 2y dy, v œ y#
' 2y ln y dy œ y# ln y ' y dy œ y# ln y "# y# C œ (x 1) ln Èx 1 "# (x 1) C" c(x 1) ln kx 1k xd ˆC" "# ‰ œ
183.
'
184.
'È
185.
' z azz1 4b dz œ "4 ' ˆ "z z"
186.
' x$ ex
187.
' È t dt
188.
'01Î10 È1 cos 5) d) œ È2 '01Î10
)# tan a)$ b d) œ
#
'
x dx 8 2x# x%
œ
" #
" 3
'
d ax # 1 b É 9 ax # 1 b #
#
dx œ
9 4t#
" #
' x# ex
œ "8 '
cx ln kx 1k x ln kx 1kd C
' tan a)$ b d a)$ b œ 3" ln ksec )$ k C
#
#
" #
#
d a x# b œ
d a9 4t# b È9 4t#
œ
" #
z1 ‰ z# 4
" #
sin" Š x
dz œ
" 4
#
1 3 ‹
C
ln kzk
" 4z
ˆx# ex# ex# ‰ C œ
" 8
ln az# 4b #
ax # 1 b e x #
" 8
tan"
z #
C
C
œ 4" È9 4t# C cos ˆ 5#) ‰ d) œ
2È 2 5
sin ˆ 5#) ‰‘ 1Î"! œ !
2È 2 5
ˆsin
1 4
0‰ œ
2 5
Chapter 8 Practice Exercises 189.
) d) ' 1cot sin) d)) œ ' (sin )cos ) a1 sin )b ; ” #
#
' x a1dx x b œ ' dxx ' xx dx1
x œ sin ) Ä dx œ cos ) d) •
#
#
œ ln ksin )k "# ln a1 sin# )b C 190. u œ tan" x, du œ
dx 1 x#
; dv œ
dx x#
, v œ x" ;
' tan x x dx œ "x tan" x ' x a1dx x b œ x" tan" x ' dxx ' 1xdxx "
#
#
œ x" tan" x ln kxk 191.
' tan2ÈÈyy dy ; Èy œ x‘
192.
'e œ
193.
ln a1 x# b C œ tanx
'
et dt dx t (x 1)(x 2) 3et 2 ; ce œ xd Ä t "¸ ˆe "‰ ln ¸ xx # C œ ln et # C
' 4)d))
œ ' ˆ1
#
#
2¸ ¸ )) 2
2x ' 11 cos ' cos 2x dx œ
195.
' cos Èasin
xb dx 1 x#
;–
ln kxk ln È1 x# C
œ'
dx x1
d) œ ' d ) '
'
d) )#
'
dx x#
œ ln kx 1k ln kx 2k C
d) )#
œ ) ln k) 2k ln k) 2k C
C
194.
"
4 ‰ 4 )#
x
' tan x2x†2x dx œ ln ksec xk C œ ln ¸sec Èy¸ C
Ä
2t
œ ) ln
196.
" #
#
"
tan# x dx œ ' asec# x 1b dx œ tan x x C
u œ sin" x Ä du œ È dx # —
' cos u du œ sin u C œ sin asin" xb C œ x C
1x
' sincosx x dxsin x œ ' (sin x)cosa1xdxsin xb œ ' (sincosx) axcosdx xb œ ' sin2 dx2x œ 2' csc 2x dx $
#
#
œ ln kcsc (2x) cot (2x)k C 197.
'
198.
' x x 2 dx œ ' x dx 2 '
sin
x #
cos
x #
dx œ '
" #
sin ˆ x# x# ‰ dx œ
#
ax # 2 b#
œ
" È2
#
tan" Š Èx2 ‹
199.
'
200.
' tan$ t dt œ '
201.
'1_ ln yy dy ; Ö
et dt 1 et
x dx a x # 2 b#
" # ax # 2 b
œ
" È2
" #
' sin x dx œ "# cos x C
tan" Š Èx2 ‹ "# ax# 2b
"
C
C
œ ln a1 et b C (tan t) asec# t 1b dt œ
Ô x œ ln y × Ù Ä dx œ dy y x Õ dy œ e dx Ø
$
b œ lim ˆ 2e 2b
bÄ_
" ‰ 4e2b
'0_ xe†e
x
3x
ˆ0 "4 ‰ œ
tan# t #
' tan t dt œ
ln ksec tk C
_
b dx œ '0 xe2x dx œ lim x# e2x 4" e2x ‘ 0
bÄ_
" 4
202.
' 3 sectanxx sin x dx œ 3 ' cot x dx ' sectanxxdx '
203.
u œ ln (sin v) v dv ' lncot(sinv dvv) œ ' (sincos cos v dv • v) ln (sin v) ; ”
#
#
du œ
tan# t #
sin v
Ä
cos x dx œ 3 ln ksin xk ln ktan xk sin x C
' duu œ ln kuk C œ ln kln (sin v)k C
573
574
Chapter 8 Techniques of Integration
'
204.
œ'
dx (2x 1)Èx# x
œ'
2 dx (2x 1)È4x# 4x
2 dx (2x 1)È(2x 1)# 1
;”
u œ 2x 1 Ä du œ 2 dx •
'
du uÈ u# 1
œ sec" kuk C œ sec" k2x 1k C 205.
' eln Èx dx œ ' Èx dx œ 23 x$Î# C
206.
' e) È3 4e) d); ”
207.
' 1 sin(cos5t dt5t)
208.
' È dv
209.
'
e2v 1
#
;”
;”
5x% 20x$ 60x# 120x 120
'
x œ ev Ä dx œ ev dv • " 3
' È3 u du œ "4 † 23 (3 u)$Î# C œ 6" a3 4e) b$Î# C
dx xÈ x# 1
du 1 u#
œ 5" tan" u C œ 5" tan" (cos 5t) C
œ sec" x C œ sec" aev b C
' (27)3)1 d(3) 1) œ
sin x ÐÑ ïïïïî cos x ÐÑ ïïïïî sin x ÐÑ ïïïïî cos x ÐÑ ïïïïî sin x ÐÑ ïïïïî cos x ÐÑ ïïïïî sin x
x&
" 4
u œ cos 5t Ä "5 ' du œ 5 sin 5t dt •
(27)3)1 d) œ
210.
u œ 4e) Ä du œ 4e) d) •
" 3 ln 27
(27)3)" C œ
" 3
3)b1
Š 27 ln 27 ‹ C
Ê ' x& sin x dx œ x& cos x 5x% sin x 20x$ cos x 60x# sin x 120x cos x
0
120 sin x C u œ Èr dr — Ä du œ 2È r
211.
' 1 drÈr ; –
212.
d ˆx 10x 9‰ 20x ' x 4x 10x dx œ ' x 10x 9 œ ln kx% 10x# 9k C 9
213.
' y (y8 dy 2) œ ' dyy ' 2ydy ' 4ydy ' (y dy 2) œ ln ¹ y y 2 ¹ 2y y2
214.
'
215.
'
216.
' t(1 ln t)Èdt(ln t)(2 ln t) ; ” u œ lndtt •
%
$
%
' 2u1 duu œ ' ˆ2 1 2 u ‰ du œ 2u 2 ln k1 uk C œ 2Èr 2 ln ˆ1 Èr‰ C
#
%
$
(t 1) dt at# 2tb#Î$
#
#
#
;”
8 dm mÈ49m# 4
$
#
u œ t# 2t Ä du œ 2(t 1) dt •
œ
8 7
'
' udu
#Î$
œ
" #
† 3u"Î$ C œ
3 #
at# 2tb
"Î$
¸C œ 4 sec" ¸ 7m #
dm # mÉm# ˆ 27 ‰
du œ
" #
C
t
Ä
' (1 u)Èduu(2 u) œ ' (u 1)Èdu (u 1) 1
œ sec" ku 1k C œ sec" kln t 1k C
#
C
Chapter 8 Practice Exercises
'0x È1 (t 1)% dt and dv œ 3(x 1)# dx, then du œ È1 (x 1)% dx, and v œ (x 1)$ so integration
217. If u œ
by parts Ê '0 3(x 1)# ’'0 È1 (t 1)% dt“ dx œ ’(x 1)$ 1
'0x È1 (t 1)% dt“ "
x
1 È $Î# " '0 (x 1)$ È1 (x 1)% dx œ ’ "6 a1 (x 1)% b “ œ 86 "
!
!
4v$ v 1 v# (v 1) av# 1b
218.
œ
A v #
B v#
C v1
Dv E v# 1 #
Ê 4v$ v 1
œ Av(v 1) av 1b B(v 1) av 1b Cv# av# 1b (Dv E) av# b (v 1) v œ 0: 1 œ B Ê B œ 1; v œ 1: 4 œ 2C Ê C œ 2; coefficient of v% : 0 œ A C D Ê A D œ 2; coefficient of v$ : 4 œ A B E D coefficient of v# : 0 œ A B C E Ê C D œ 4 Ê D œ 2 (summing with previous equation); coefficient of v: 1 œ A B Ê A œ 0; in summary: A œ 0, B œ 1, C œ 2, D œ 2 and E œ 1
'2_ v (v4v 1) vav 1 1b dv œ $
Ê
#
#
œ lim ln (v 1)# bÄ_
#
" v
1) œ lim ’ln Š (b1 b# ‹
œ
bÄ_ 1 #
ln (5)
" #
lim
bÄ_
'2b ˆ v 2 1 v# 1 " v
#
2v ‰ 1 v#
dv
tan" v ln a1 v# b‘ 2 b
" b
tan" b“ ˆln 1
" #
tan" 2 ln 5‰ œ ˆ0 0 1# ‰ ˆ0
" #
tan" 2 ln 5‰
1 #
tan" a;
tan" 2
219. u œ f(x), du œ f w (x) dx; dv œ dx, v œ x;
'13Î12Î2 f(x) dx œ cx f(x)d 311Î2Î2 '13Î12Î2 xf w (x) dx œ 3#1 f ˆ 3#1 ‰ 1# f ˆ 1# ‰‘ '13Î12Î2 cos x dx œ ˆ 31#b
'0a 1dxx
220.
1a ‰ #
csin xd 311Î2Î2 œ 1# a3b ab c(1) 1d œ 1# a3b ab 2
œ ctan" xd 0 œ tan" a; 'a a
#
therefore, tan" a œ
1 #
_
dx 1 x #
œ lim ctan" xd a œ lim atan" b tan" ab œ b
bÄ_
tan" a Ê tan" a œ
1 4
bÄ_
Ê a œ 1 since a 0.
CHAPTER 8 ADDITIONAL AND ADVANCED EXERCISES #
1. u œ asin" xb , du œ
'
#
'
2.
" x
; dv œ dx, v œ x;
asin" xb dx œ x asin" xb ' #
u œ sin" x, du œ
'
2 sin" x dx È 1 x#
2x sin" x dx È 1 x#
dx È 1 x#
2x sin" x dx È 1 x#
;
; dv œ È2x dx # , v œ 2È1 x# ; 1x
œ 2 asin" xb È1 x# ' 2 dx œ 2 asin" xb È1 x# 2x C; therefore
#
#
asin" xb dx œ x asin" xb 2 asin" xb È1 x# 2x C œ
" x
,
" " " x(x 1) œ x x 1 , " " " " x(x 1)(x 2) œ 2x x 1 #(x 2) , " " " " " x(x 1)(x 2)(x 3) œ 6x #(x 1) #(x 2) 6(x 3) , " " " " " x(x 1)(x 2)(x 3)(x 4) œ #4x 6(x 1) 4(x #) 6(x 3) " x(x 1)(x 2) â (x m)
m
œ! kœ0
(") (k!)(m k)!(x k) ; k
therefore '
" 24(x 4)
Ê the following pattern:
dx x(x 1)(x 2) â (x m)
575
576
Chapter 8 Techniques of Integration m
(") œ ! ’ (k!)(m k)! ln kx kk“ C k
kœ0
3. u œ sin" x, du œ
dx È 1 x#
' x sin" x dx œ x#
#
4.
œ
x# #
sin" x
œ
x# #
sin" x
sin" x '
'
z œ Èy
— Ä
dy 2È y
x# #
;
x œ sin ) Ä dx œ cos ) d) •
;”
x "# ˆ #)
sin 2) ‰ 4
'
Cœ
x sin" x dx œ
x# #
sin" x
C
"
Èy
C
) ' 1 dtan) ) œ ' cos cos ' 12coscos#2)) d) œ #" ' (sec 2) 1) d) œ ln ksec 2) 4tan 2)k 2) C ) sin ) d) œ #
#
#
#
x "# œ "# sec ) – dx œ " sec ) tan ) d) — Ä # tan ) ln ksec ) tan )k #
œ
" ‹ 2È 1 x
'
Cœ
" 4
œ
dx 2È x È 1 x
dt t È1 t#
Õ d) œ
' u du 1 #" ' u du 1 #" ' uu du1 œ #" ln ¹ Èu 1
œ
" #
ln Št È1 t# ‹
3e2x
ex b dx 6ex 1
u1œ
Ô
Õ du œ
2 È3
2 È3
sec )
' Š È4
œ
2 3
È3u# 6u 1
œ
" È3
Ä
3u# 6u 1
" È3
3
" È3 " 3
" #
tan" u C œ
œ
" È3
'
sec ) 1‹ (sec )) d) œ
#
Ä
' (u 1)duau 1b #
ln ¸ tansec) ) " ¸ #" ) C
4 3
;
' sec# ) d) È" ' sec ) d) 3
" È3
ln ¹
ln ¹u 1 É(u 1)# 43 ¹ ŠC"
" È3
ln
4 3
dx œ '
" ax# 2x 2b ax# 2x 2b
dx
' ’ x 2x 2x 2 2 (x 1)2 1 x 2x 2x 2# (x 1)2 1 “ dx #
(2u 1) du É(u 1)# 43
ln ¹ex 1 Ée2x 2ex 3" ¹“ C
" ax# 2b# 4x#
" #
† É 34 (u 1)# 1
ln ksec ) tan )k C" œ
’2Ée2x 2ex
' x " 4 dx œ ' " 16
×
sec ) tan ) d) Ø " È3
¹
Ø
du u# 1
C
sin" t C
' È(2u 1) du
tan )
œ
u# 1
u œ ex Ä du œ ex dx •
4 3
%
#
;”
œ
4
u œ tan )
" #
2x
Èx# x ln ¹2x 1 2Èx# x¹
d) ' sincos) ) cos ' tan d)) 1 ; Ô du œ sec# ) d) × ) œ
œ
' Èa2e
;
C
#
t œ sin ) Ä ;” dt œ cos ) d) •
" #
x dx Ɉx "# ‰# "4
" #
2Èx# x ln ¹2x 1 2Èx# x¹
#
; dv œ dx, v œ x;
sec ) tan ) d) ' (sec ) ˆ1)†tan œ #" ' asec# ) sec )b d) )‰
' ln ŠÈx È1 x‹ dx œ x ln ŠÈx È1 x‹ 2
Ê
9.
sin# ) cos ) d) 2 cos )
C
' ln ŠÈx È1 x‹ dx œ x ln ŠÈx È1 x‹ "# ' ÈxxÈdx1 x ; "# '
8.
sin" x '
sin ) cos ) ) 4
' sin" Èy dy œ y sin" Èy Èy È1 y# sin
" 6. u œ ln ŠÈx È1 x‹ , du œ Š Èx dx È1 x ‹ Š #Èx
7.
x# #
' 2z sin" z dz; from Exercise 3, ' z sin" z dz
zÈ1 z# sin" z C Ê 4 " È # È sin y y y sin" Èy # #
z# sin" z #
œy
x# dx 2È 1 x#
# " sin# ) d) œ x# sin" # xÈ1 x# sin" x C 4
' sin" Èy dy; – dz œ œ
5.
; dv œ x dx, v œ
#
#
È3 #
(u 1) É 34 (u 1)# 1¹ C"
È3 # ‹
Chapter 8 Additional and Advanced Exercises
10.
#
" 16
œ
2x 2 " " ln ¹ xx# (x 1) tan" (x 1)d C 2x # ¹ 8 ctan
' x " 1 dx œ 6" '
ˆ x " 1
'
" x1
x2 x# x 1
x2 ‰ x# x 1
dx
œ
" 6
1¸ ln ¸ xx 1
" 1#
' ” x 2x x " 1 ˆ
œ
" 6
"¸ ln ¸ xx 1
" 1#
x" " 2x 1 È ’ln ¹ xx# Š È3 ‹ 2È3 tan" Š 2xÈ3 1 ‹“ C x 1 ¹ 2 3 tan
#
3 # x "# ‰ 34
2x 1 x# x 1
3 ˆx "# ‰# 34 •
dx
#
11. x lim Ä_
'cxx sin t dt œ x lim c cos td xcx œ x lim c cos x cos (x)d œ x lim a cos x cos xb œ x lim 0œ0 Ä_ Ä_ Ä_ Ä_
12.
'x1 cost t dt;
lim
xÄ!
#
lim x 'x
1
xÄ!
lim
xÄ!
cos t t#
lim
tÄ!
1 t x 'x cos dt œ t #
n
nc1
œ '0
1
dy dx
" È n# k#
x
lim
cos t t#
dt
" x
xÄ!
" cos t
lim 'x
1
œ1 Ê
xÄ!
cos t t#
dt diverges since '0
1
œ lim xÄ!
Š cos# x ‹ Š
dt t#
diverges; thus
œ lim cos x œ 1 xÄ!
x " ‹ x#
! ln ˆ1 k ˆ " ‰‰ ˆ " ‰ œ ' ln (1 x) dx; ” œ n lim n n Ä_ 0 n
1
kœ1
nc1
! ŠÈ œ n lim Ä_ kœ0
dx œ csin" xd ! œ
n ‹ ˆ n" ‰ n# k#
nc1
! œ n lim Ä_ kœ0
Î
"
u œ 1 x, du œ dx x œ 0 Ê u œ 1, x œ 1 Ê u œ 2 •
Ê 1 Š dy dx ‹ œ #
1Î2
#
a1 x# b 4x# a1 x # b #
x ' ˆ1 œ '0 Š "1 x# ‹ dx œ 0 1Î2
1Î4
2 ‰ 1 x#
œ ˆ "# ln 3‰ (0 ln 1) œ ln 3
œ
1 2x# x% a 1 x # b#
#
x ' œ Š 11 x# ‹ ; L œ 0
dx œ '0 ˆ1 1Î2
" #
shell ‰ shell 17. V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21xy dx b
1
œ 61 '0 x# È1 x dx; 1
Ê1 ŠÈcos 2t‹ dt œ È2 '0
1Î4
œ1 #
2x 1 x#
ˆ n" ‰
1 #
#
œ
Ñ #
Ï Ê1 ’k Š "n ‹“ Ò
# ' œ Ècos 2x Ê 1 Š dy dx ‹ œ 1 cos 2x œ 2 cos x; L œ 0 1Î%
dy dx
k n
'1
"
" È 1 x #
œ È2 csin td ! 16.
œ lim tÄ!
'12 ln u du œ cu ln u ud #" œ (2 ln 2 2) (ln 1 1) œ 2 ln 2 1 œ ln 4 1
! 14. n lim Ä_ kœ0
15.
t Š cos# t ‹ t
^ dt is an indeterminate 0 † _ form and we apply l'Hopital's rule:
! ln nÉ1 13. n lim Ä_ kœ1 Ä
Š "# ‹
Ô uœ1x × du œ dx Õ x# œ (1 u)# Ø
Ä 61 '1 (1 u)# Èu du 0
œ 61 '1 ˆu"Î# 2u$Î# u&Î# ‰ du 0
!
œ 61 23 u$Î# 45 u&Î# 27 u(Î# ‘ " œ 61 ˆ 23 84 30 ‰ 16 ‰ œ 61 ˆ 70 105 œ 61 ˆ 105 œ 32351
4 5
27 ‰
#
" 1x
#
1Î2
" ‰ 1x
#
Ê1 Š dy dx ‹ dx "Î#
x ¸‘ dx œ x ln ¸ 11 x !
Ècos# t dt
577
578
Chapter 8 Techniques of Integration
18. V œ 'a 1y# dx œ 1 '1 b
4
œ 1 '1 ˆ dx x 4
5 dx x#
25 dx x# (5 x)
dx ‰ 5x
% œ 1 ln ¸ 5 x x ¸ 5x ‘ " œ 1 ˆln 4 54 ‰ 1 ˆln
œ
151 4
" 4
5‰
21 ln 4
shell ‰ shell 19. V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21xex dx b
1
œ 21 cxex ex d "! œ 21
20. V œ '0
ln 2
21(ln 2 x) aex 1b dx
œ 21 '0
ln 2
c(ln 2) ex ln 2 xex xd dx ln 2
œ 21 ’(ln 2) ex (ln 2)x xex ex
x# # “0
œ 21 ’2 ln 2 (ln 2)# 2 ln 2 2
(ln 2)# # “
21(ln 2 1)
#
œ 21 ’ (ln#2) ln 2 1“ 21. (a) V œ '1 1 c1 (ln x)# d dx e
œ 1 cx x(ln x)# d 1 21'1 ln x dx e
e
(FORMULA 110) e œ 1 cx x(ln x)# 2(x ln x x)d 1 e œ 1 cx x(ln x)# 2x ln xd 1 œ 1 ce e 2e (1)d œ 1
(b) V œ '1 1(1 ln x)# dx œ 1'1 c1 2 ln x (ln x)# d dx e
e
œ 1 cx 2(x ln x x) x(ln x)# d 1 21'1 ln x dx e
e
œ 1 cx 2(x ln x x) x(ln x)# 2(x ln x x)d 1 e œ 1 c5x 4x ln x x(ln x)# d 1 œ 1 c(5e 4e e) (5)d œ 1(2e 5) e
22. (a) V œ 1 '0 aey b# 1‘ dy œ 1'0 ae2y 1b dy œ 1 e# y‘ ! œ 1 ’ e# 1 ˆ "# ‰“ œ 1
1
"
2y
#
1 ae # 3 b #
(b) V œ 1'0 aey 1b# dy œ 1'0 ae2y 2ey 1b dy œ 1 e# 2ey y‘ ! œ 1 ’Š e# 2e 1‹ ˆ "# 2‰“ 1
#
1
œ 1 Š e# 2e 5# ‹ œ 23. (a)
lim
x Ä !
x ln x œ 0 Ê
2y
1 ae# 4e 5b #
lim
x Ä !
f(x) œ 0 œ f(0) Ê f is continuous
"
#
Chapter 8 Additional and Advanced Exercises # Ô u œ (ln x) × Ö du œ (2 ln x) dx Ù # 2 2 x # # Ö Ù Ä 1 Œ lim ’ x$ (ln x)# “ ' Š x$ ‹ (2 ln x) (b) V œ '0 1x (ln x) dx; Ö # Ù 3 3 0 dv œ x dx b bÄ! $ x Õ Ø vœ
579
dx x
3
$
œ 1 ”ˆ 83 ‰ (ln 2)# ˆ 23 ‰ lim ’ x3 ln x bÄ!
2
x$ 9 “ b•
#
œ 1 ’ 8(ln3 2)
16(ln 2) 9
16 27 “
24. V œ '0 1( ln x)# dx 1
œ 1 Œ lim cx(ln x)# d b 2'0 ln x dx bÄ! 1
1
œ 21 lim cx ln x xd 1b œ 21 bÄ!
25. M œ '1 ln x dx œ cx ln x xd e1 œ (e e) (0 1) œ 1; e
Mx œ '1 (ln x) ˆ ln#x ‰ dx œ e
œ
" #
e
e
My œ '1 x ln x dx œ ’ x " #
'1e (ln x)# dx
Šcx(ln x)# d 1 2 '1 ln x dx‹ œ e
œ
" #
’x# ln x
My M
therefore, x œ 26. M œ '0
1
e
x# # “1
2 dx È 1 x#
œ
œ
#
e
ln x # “1
" #
’Še#
e# 1 4
" #
" #
'1
e# #‹
and y œ
(e 2);
e
x dx "# “ œ
Mx M
œ
" 4
ae# 1b ;
e2 #
"
œ 2 csin" xd ! œ 1;
My œ '0
" 2x dx È # È1 x# œ 2 ’ 1 x “ œ 2; ! M therefore, x œ My œ 12 and y œ 0 by symmetry 1
27. L œ '1 É1 e
tan" e
œ '1Î4
" x#
dx œ '1
e
(sec )) atan# ) 1b tan )
œ ŠÈ1 e# ln ¹
È x# 1 x
tan" e
d) œ '1Î4
È 1 e# e
dx; ”
tan" e x œ tan ) ' Ä L œ 1Î4 dx œ sec# ) d) •
sec )†sec# ) d) tan ) " e
(tan ) sec ) csc )) d) œ csec ) ln kcsc ) cot )kd 1tanÎ4
# È "e ¹‹ ’È2 ln Š1 È2‹“ œ È1 e# ln Š 1e e "e ‹ È2 ln Š1 È2‹
# # ' È ' yÈ1 e2y dy; ” 28. y œ ln x Ê 1 Š dx dy ‹ œ 1 x Ê S œ 21 c x 1 x dy Ê S œ 21 0 e #
d
1
tan" e
u œ tan ) Ä S œ 21'1 È1 u# du; ” Ä 21 '1Î4 du œ sec# ) d) • e
" e
œ 21 ˆ "# ‰ csec ) tan ) ln ksec ) tan )kd tan 1Î%
u œ ey du œ ey dy •
sec ) † sec# ) d)
œ 1 ’ŠÈ1 e# ‹ e ln ¹È1 e# e¹“ 1 ’È2 † 1 ln ŠÈ2 1‹“
È 1 e# e œ 1 ’eÈ1 e# ln Š È ‹ È 2“ 21
#Î$ 29. L œ 4 '0 Ê1 Š dy y#Î$ œ 1 Ê y œ ˆ1 x#Î$ ‰ dx ‹ dx; x 1
#
$Î#
Ê
dy dx
œ 3# ˆ1 x#Î$ ‰
"Î#
ˆx"Î$ ‰ ˆ 23 ‰
580
Chapter 8 Techniques of Integration #
Ê Š dy dx ‹ œ
x Ê L œ 4 '0 Ê1 Š 1 x#Î$ ‹ dx œ 4'0 1
1 x#Î$ x#Î$
1
#Î$
30. S œ 21 'c1 f(x) É1 cf w (x)d# dx; f(x) œ ˆ1 x#Î$ ‰ 1
$Î# " ‰ dx; – œ 41'0 ˆ1 x#Î$ ‰ ˆ x"Î$ 1
" œ 61 † 25 (1 u)&Î# ‘ ! œ #
31. Š dy dx ‹ œ
" 4x
Ê
dy dx
„" #Èx
œ
$Î#
"
œ 6 x#Î$ ‘ ! œ 6
dx x"Î$
Ê cf w (x)d# 1 œ
Ê S œ 21 'c1 ˆ1 x#Î$ ‰ 1
" x#Î$
$Î#
†
dx Èx#Î$
1 u œ x#Î$ 3 $Î# '1 du œ 61'0 (1 u)$Î# d(1 u) 2 dx — Ä 4 † # 1 0 (1 u) du œ 3 x"Î$
121 5
Ê y œ Èx or y œ Èx, 0 Ÿ x Ÿ 4
32. The integral 'c1 È1 x# dx is the area enclosed by the x-axis and the semicircle y œ È1 x# . This area is half 1
the circle's area, or
1 #
and multiplying by 2 gives 1. The length of the circular arc y œ È1 x# from x œ 1 to
x ' x œ 1 is L œ 'c1 Ê1 Š dy ‹ dx œ 'c1 È dx # dx ‹ dx œ c1 Ê1 Š È #
1
#
1
1
1 x#
1x
circle's circumference. In conclusion, 2 'c1 È1 x# dx œ 'c1 È dx 1
_ 'c_ e xe a
x
b
_
dx œ '_ eae b ex dx
'a0 e ce
œaÄ lim _
a
x
b
" #
(21) œ 1 since L is half the
1
1 x#
33. (b)
œ
.
x
ex dx
lim
b Ä _
'0b e ce a
x
b
ex dx;
u œ ex ” du œ ex dx • Ä lim ' ecu du a Ä _ ea
'1e
b
1
lim
b Ä _
œaÄ lim cecu d 1ea _
ecu du
cecu d e1
b
lim
b Ä _
"e ec aea b ‘ œaÄ lim _
ec ˆeb ‰ "e ‘
lim
b Ä _
œ ˆ "e e! ‰ ˆ0 "e ‰ œ 1 34. u œ
" 1y
lim nÄ_
, du œ (1 dyy)# ; dv œ nync1 dy, v œ yn ;
'01 ny1 y
n1
dy œ n lim ’ y “ '0 Ä _ Œ 1y 1
" #
0œ
1
!
' Ê 0 Ÿ n lim Ä_ 0 œ
"
n
yn 1 y#
dy Ÿ n lim Ä_
36.
1 6
uÐn2ÑÎ2 n#
œ
" È2
" #
" #
n lim Ä_
ˆÈu‰ n2 n#
ŠÈx# a# ‹
Cœ
n#
" œ sin" x# ‘ ! œ '0
1
È2
sin" u# ‘ 0
œ
'01
yn 1 y#
n 1
!
" #
Cœ
œ sin"
dy œ
" '01 yn dy œ n lim ’ y “ œ n lim Ä _ n1 Ä_
35. u œ x# a# Ê du œ 2x dx; ' x ŠÈx# a# ‹ n dx œ "# ' ˆÈu‰ n du œ œ
yn 1 y#
" #
" n 1
dy. Now, 0 Ÿ
' œ 0 Ê n lim Ä_ 0
1
' unÎ2 du œ "# Š u 1 ‹ C, n Á 2 nÎ2 1 n #
n2
C
'1
dx dx È 4 x# 0 È 4 x# x$ 1È2 " " ˆ1‰ " È2 È2 sin # œ È2 4 œ 8
'0
1
dx È4 2x#
œ
" È2
yn 1 y#
È2
'0
du È 4 u#
Ÿ yn nync1 1y
dy
Chapter 8 Additional and Advanced Exercises 37.
'1_ ˆ x ax 1 #"x ‰ dx œ
lim
#
" bÄ_ #
œ lim
ab # 1 b b
’ln
bÄ_ a
integral diverges if a " #
œ Ê
_
'1
ˆln 1
" #
lim ln
bÄ_
ˆ x#ax 1
bÄ_
ab # 1 b b bÄ_
ln 2a “ ; lim " #
ab # 1 b b
a
_ dx xp
a
" #
p " #
bÄ_
b2a bÄ_ b
" b#
" #
'0b ecxt dt œ
bÄ_
" #
" #
" bÄ_ #
lim
ax # 1 b x
b
a
“
1
Ê the improper
”ln
ab # 1 b b
"Î#
ln 2"Î# •
œ lim (b 1)2ac1 œ 0 bÄ_
; in summary, the improper integral
and has the value ln42 cxb
b lim "x ecxt ‘ 0 œ lim Š 1 xe ‹ œ
bÄ_
bÄ_
œ1 Ê
(b1)2a b 1
lim
œ _ Ê the improper integral diverges if a
dx converges only when a œ
b
bÄ_
bÄ_ a
ln x‘ 1 œ lim ’ #" ln
1
œ lim É1
ab # 1 b b bÄ_
" #
œ lim b2 ˆa c 2 ‰ œ _ if a
lim
: 0 Ÿ lim
bÄ_
10 x
œ
" x
if x 0 Ê xG(x) œ x ˆ "x ‰
converges if p 1 and diverges if p Ÿ 1. Thus, p Ÿ 1 for infinite area. The volume of the solid of revolution
_
_ dx
about the x-axis is V œ '1 1 ˆ x"p ‰ dx œ 1 '1 " #
lim #a ln ax# 1b
È b# 1 " # : b lim b Ä_
; for a œ
œ 1 if x 0 39. A œ '1
#
ln 2‰ œ ln42 ; if a
" ‰ #x
38. G(x) œ lim
'1b ˆ x ax 1 #"x ‰ dx œ
#
x2p
which converges if 2p 1 and diverges if 2p Ÿ 1. Thus we want
for finite volume. In conclusion, the curve y œ xcp gives infinite area and finite volume for values of p satisfying
p Ÿ 1.
40. The area is given by the integral A œ '0
1
dx xp
;
p œ 1: A œ lim cln xd œ lim ln b œ _, diverges; bÄ! bÄ! 1 b
p 1: A œ lim cx1p d 1b œ 1 lim b1p œ _, diverges; bÄ! bÄ!
p 1: A œ lim cx1p d 1b œ " lim b1p œ 1 0, converges; thus, p 1 for infinite area. bÄ! bÄ! The volume of the solid of revolution about the x-axis is Vx œ 1'0
1
p
" #
" #
, and diverges if p
dx x2p
which converges if 2p 1 or
. Thus, Vx is infinite whenever the area is infinite (p 1).
_
_
The volume of the solid of revolution about the y-axis is Vy œ 1 '1 cR(y)d# dy œ 1'1 converges if
2 p
dy y2Îp
which
1 Í p 2 (see Exercise 39). In conclusion, the curve y œ xcp gives infinite area and finite
volume for values of p satisfying 1 Ÿ p 2, as described above. ÐÑ
cos 3x
2e2x
ÐÑ
1 3 sin
4e2x
ÐÑ
19 cos 3x
41. e2x
Iœ
e2x 3
581
sin 3x
2e2x 9
cos 3x 49 I Ê
ÐÑ
42. e3x
3x
ÐÑ
4" cos 4x
9e3x
ÐÑ
" 16 sin 4x
3x
Iœ
e2x 9
(3 sin 3x 2 cos 3x) Ê I œ
e2x 13
(3 sin 3x 2 cos 3x) C
sin 4x
3e3x
I œ e4 cos 4x
13 9
3e3x 16
sin 4x
9 16
I Ê
25 16
Iœ
e3x 16
(3 sin 4x 4 cos 4x) Ê I œ
e3x 25
(3 sin 4x 4 cos 4x) C
582
Chapter 8 Techniques of Integration sin 3x
ÐÑ
sin x
3 cos 3x
ÐÑ
cos x
9 sin 3x
ÐÑ
sin x
43.
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 cos 5x
ÐÑ
sin 4x
sin 5x
ÐÑ
"4 cos 4x
25cos 5x
ÐÑ
" 16 sin 4
44.
I œ "4 cos 5x cos 4x Ê Iœ
" 9
ÐÑ
"b cos bx
a# eax
ÐÑ
b"# sin bx
ax
I œ eb cos bx eax a# b#
aeax b#
sin bx
a# b#
I Ê Ša
ÐÑ
" b
a# eax
ÐÑ
b"# cos bx
eax b
sin bx
Ê Iœ
sin 5x sin 4x
eax a# b#
#
b# b# ‹ I
œ
eax b#
(a sin bx b cos bx)
cos bx
aeax
Iœ
5 16
(a sin bx b cos bx) C
ÐÑ
46. eax
9 I Ê 16 I œ "4 cos 5x cos 4x
sin bx
aeax
Ê Iœ
25 16
(4 cos 5x cos 4x 5 sin 5x sin 4x) C ÐÑ
45. eax
sin 5x sin 4x
5 16
aeax b#
sin bx
cos bx
a# b#
I Ê Ša
#
b# b# ‹ I
œ
eax b#
(a cos bx b sin bx)
(a cos bx b sin bx) C
47. ln (ax)
ÐÑ
1
" x
ÐÑ
x
I œ x ln (ax) ' ˆ "x ‰ x dx œ x ln (ax) x C 48. ln (ax)
ÐÑ
x#
" x
ÐÑ
" 3
Iœ
" 3
x$
x$ ln (ax) ' ˆ "x ‰ Š x3 ‹ dx œ $
_
49. (a) >(1) œ '0 et dt œ lim
bÄ_
" 3
x$ ln (ax) 9" x$ C
'0b et dt œ
lim cet d b0 œ lim e"b (1)‘ œ 0 1 œ 1
bÄ_
bÄ_
(b) u œ tx , du œ xtxc1 dt; dv œ ect dt, v œ ect ; x œ fixed positive real _
_
Ê >(x 1) œ '0 tx et dt œ lim ctx et d b0 x '0 tx1 et dt œ lim ˆ beb 0x e! ‰ x>(x) œ x>(x) bÄ_
x
bÄ_
Chapter 8 Additional and Advanced Exercises (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.
for some k 0; from part (b) induction hypothesis definition of factorial
x n n 50. (a) >(x) ¸ ˆ xe ‰ É 2x1 and n>(n) œ n! Ê n! ¸ n ˆ ne ‰ É 2n1 œ ˆ ne ‰ È2n1
(b)
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)"
(c)
n 10
ˆ ne ‰n È2n1 3598695.619
ˆ ne ‰n È2n1 e1Î12n 3628810.051
calculator 3628800
583
584
Chapter 8 Techniques of Integration
NOTES:
CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION 9.1 SLOPE FIELDS AND SEPARABLE DIFFERENTIAL EQUATIONS 1. (a) y œ ex Ê y w œ ex Ê 2y w 3y œ 2 aex b 3ex œ ex (b) y œ ex e3xÎ2 Ê y w œ ex 3# e3xÎ2 Ê 2y w 3y œ 2 ˆex 3# e3xÎ2 ‰ 3 aex e3xÎ2 b œ ex (c) y œ ex Ce3xÎ2 Ê y w œ ex 3# Ce3xÎ2 Ê 2y w 3y œ 2 ˆex 3# Ce3xÎ2 ‰ 3 aex Ce3xÎ2 b œ ex 2. (a) y œ "x Ê y w œ
" x#
(b) y œ x " 3 Ê y w œ (c) y œ 3. y œ
" xC
'1x
" x
et t
Ê yw œ
#
œ ˆ x" ‰ œ y# " (x 3)#
" (x C)#
#
œ ’ (x " 3) “ œ y# #
œ y: 0 STO > x: y (enter) y 0.1*(1 y2 ) STO > y: x 0.1 STO > x: y (enter, 10 times) The last value displayed gives yEuler a1b ¸ 1.3964 The exact solution: dy œ a1 y2 bdx Ê
dy 1 y2
œ dx Ê tan1 y œ x C; tan1 ya0b œ tan1 0 œ 0 œ 0 C Ê C œ 0
596
Chapter 9 Further Applications of Integration Ê tan1 y œ x Ê y œ tan x Ê yexact a"b œ tan 1 ¸ 1.5574
17. (a)
dy dx
œ 2y2 ax 1b Ê
dy y2
œ 2ax 1bdx Ê ' y2 dy œ ' a2x 2bdx Ê y" œ x2 2x C
Initial value: ya2b œ "# Ê 2 œ 22 2a2b C Ê C œ 2 1 Solution: y" œ x2 2x 2 or y œ x2 2x 2
ya3b œ 32 21a3b 2 œ 15 œ 0.2 (b) To find the approximation, set y1 œ 2y2 ax 1b and use EULERT with initial values x œ 2 and y œ "# and step size
0.2 for 5 Points. This gives ya3b ¸ 0.1851; error ¸ 0.0149. (c) Use step size 0.1 for 10 points. This gives ya3b ¸ 0.1929; error ¸ 0.0071. (d) Use step size 0.05 for 20 points. This gives ya3b ¸ 0.1965; error ¸ 0.0035. 18. (a)
dy dx
œy1Ê'
dy y1
œ ' dx Ê ln ky 1k œ x C Ê ky 1k œ exC Ê y 1 œ „ eC ex Ê y œ Aex 1
Initial value: ya0b œ 3 Ê 3 œ Ae0 1 Ê A œ 2 Solution: y œ 2ex 1 ya1b œ 2e 1 ¸ 6.4366 (b) To find the approximation, set y1 œ y 1 and use a graphing calculator or CAS with initial values x œ 0 and y œ 3 and step size 0.2 for 5 Points. This gives ya1b ¸ 5.9766; error ¸ 0.4599 (c) Use step size 0.1 for 10 points. This gives ya1b ¸ 6.1875; error ¸ 0.2491. (d) Use step size 0.05 for 20 points. This gives ya1b ¸ 6.3066; error ¸ 0.1300. 1 2 x2 2x 2 , so ya3b œ 0.2. To find the approximation, let zn œ yn1 2yn1 axn1 1bdx ay2n1 axn1 1b z2n ax2n 1bbdx with initial values x0 œ 2 and y0 œ "# . Use a spreadsheet, graphing
19. The exact solution is y œ yn œ y n 1
and
calculator, or CAS as indicated in parts (a) through (d). (a) Use dx œ 0.2 with 5 steps to obtain ya3b ¸ 0.2024 Ê error ¸ 0.0024. (b) Use dx œ 0.1 with 10 steps to obtain ya3b ¸ 0.2005 Ê error ¸ 0.0005. (c) Use dx œ 0.05 with 20 steps to obtain ya3b ¸ 0.2001 Ê error ¸ 0.0001. (d) Each time the step size is cut in half, the error is reduced to approximately one-fourth of what it was for the larger step size. 20. The exact solution is y œ 2ex 1, so ya1b œ 2e 1 ¸ 6.4366. To find the approximation, let zn œ yn1 ayn1 1bdx and yn œ yn1 ˆ ync1 2zn 2 ‰dx with initial value yn œ 3. Use a spreadsheet, graphing calculator, or CAS as indicated in parts (a) through (d). (a) Use dx œ 0.2 with 5 steps to obtain ya1b ¸ 6.4054 Ê error ¸ 0.0311. (b) Use dx œ 0.1 with 10 steps to obtain ya1b ¸ 6.4282 Ê error ¸ 0.0084 (c) Use dx œ 0.05 with 20 steps to obtain ya1b ¸ 6.4344 Ê error ¸ 0.0022 (d) Each time the step size is cut in half, the error is reduced to approximately one-fourth of what it was for the larger step size. 13-16. Example CAS commands: Maple: ode := diff( y(x), x ) = 2*x*exp(x^2);ic := y(0)=2; xstar := 1; dx := 0.1; approx := dsolve( {ode,ic}, y(x), numeric, method=classical[foreuler], stepsize=dx ): approx(xstar); exact := dsolve( {ode,ic}, y(x) ); eval( exact, x=xstar );
Section 9.3 Euler's Method evalf( % ); 17.
Example CAS commands: Maple: ode := diff( y(x), x ) = 2*y(x)*(x-1);ic := y(2)=-1/2; xstar := 3; exact := dsolve( {ode,ic}, y(x) ); # (a) eval( exact, x=xstar ); evalf( % ); approx1 := dsolve( {ode,ic}, y(x), # (b) numeric, method=classical[foreuler], stepsize=0.2 ): approx1(xstar); approx2 := dsolve( {ode,ic}, y(x), # (c) numeric, method=classical[foreuler], stepsize=0.1 ): approx2(xstar); approx3 := dsolve( {ode,ic}, y(x), # (d) numeric, method=classical[foreuler], stepsize=0.05 ): approx3(xstar);
19.
Example CAS commands: Maple: ode := diff( y(x), x ) = 2*y(x)*(x-1);ic := y(2)=-1/2; xstar := 3; approx1 := dsolve( {ode,ic}, y(x), # (a) numeric, method=classical[heunform], stepsize=0.2 ): approx1(xstar); approx2 := dsolve( {ode,ic}, y(x), # (b) numeric, method=classical[heunform], stepsize=0.1 ): approx2(xstar); approx3 := dsolve( {ode,ic}, y(x), # (c) numeric, method=classical[heunform], stepsize=0.05 ): approx3(xstar);
21.
Example CAS commands: Maple: ode := diff( y(x), x ) = x + y(x);ic := y(0)=-7/10; x0 := -4;x1 := 4;y0 := -4; y1 := 4; b := 1; P1 := DEplot( ode, y(x), x=x0..x1, y=y0..y1, arrows=thin, title="#21(a) (Section 9.3)" ): P1; Ygen := unapply( rhs(dsolve( ode, y(x) )), x,_C1 ); # (b) P2 := seq( plot( Ygen(x,c), x=x0..x1, y=y0..y1, color=blue ), c=-2..2 ): # (c) display( [P1,P2], title="#21(c) (Section 9.3)" ); CC := solve( Ygen(0,C)=rhs(ic), C ); # (d) Ypart := Ygen(x,CC); P3 := plot( Ypart, x=0..b, title="#21(d) (Section 9.3)" ): P3; euler4 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/4 ): # (e) P4 := odeplot( euler4, [x,y(x)], x=0..b, numpoints=4, color=blue ):
597
598
Chapter 9 Further Applications of Integration display( [P3,P4], title="#21(e) (Section 9.3)" ); euler8 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/8 ): # (f) P5 := odeplot( euler8, [x,y(x)], x=0..b, numpoints=8, color=green ): euler16 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/16 ): P6 := odeplot( euler16, [x,y(x)], x=0..b, numpoints=16, color=pink ): euler32 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/32 ): P7 := odeplot( euler32, [x,y(x)], x=0..b, numpoints=32, color=cyan ): display( [P3,P4,P5,P6,P7], title="#21(f) (Section 9.3)" ); , # (g) < 4 | (x1-x0)/ 4 | evalf[5]( abs(1-eval(y(x),euler4(b))/eval(Ypart,x=b))*100 ) >, < 8 | (x1-x0)/ 8 | evalf[5]( abs(1-eval(y(x),euler8(b))/eval(Ypart,x=b))*100 ) >, < 16 | (x1-x0)/16 | evalf[5]( abs(1-eval(y(x),euler16(b))/eval(Ypart,x=b))*100 ) >, < 32 | (x1-x0)/32 | evalf[5]( abs(1-eval(y(x),euler32(b))/eval(Ypart,x=b))*100 ) > >;
13-24. Example CAS commands: Mathematica: (assigned functions, step sizes, and values for initial conditions may vary) For exercises 13 - 20, find the exact solution as follows. Set up two error lists. Clear[x, y, f] f[x_,y_]:= 2 y2 (x 1) a = 2; b = 1/2; xstar = 3; desol=DSolve[{y'[x] == f[x, y[x]], y[a] == b}, y[x], x] //Simplify actual[x_] = desol[[1, 1, 2]]; {xstar, actual[xstar]} errorlisteuler = { }; errorlisteulerimp = { }; pa = Plot[actual[x], {x, a, xstar}] Euler's method with error at x*. The Do command is used with a sequence of commands that are repeated n times. a = 2; b = -1/2; dx = 0.2; xstar = 3; n = (xstar a) /dx; solnslist = {{a,b}}; Do[ {new = b + f[a,b] dx, a = a + dx, b = new, AppendTo[solnslist, {a,b}]},{n}] solnslist error= actual[xstar] solnslist[[n, 2]] relativeerror= error / actual[xstar] AppendTo[errorlisteuler, error] pe = ListPlot[solnslist, PlotStyle Ä {Hue[.4], PointSize[0.02]}] Show[pa, pe] Rerun with different values for dx, starting from largest to smallest. After doing this, observe what happens to the error as the step size decreases by entering the input command: errorlisteuler Improved Euler's method. with error at x* a = 2; b = 1/2; dx = 0.2; xstar = 3; n = (xstar a) /dx; solnslist = {{a,b}}; Do[{new1 = b f[a,b] dx, new2 = b + (f[a, b] f[a+dx, new1])/2 dx, a = a dx, b = new2, AppendTo[solnslist, {a,b}]},{n}] solnslist error= actual[xstar] solnslist[[n, 2]
Section 9.4 Graphical Solutions of Autonomous Differential Equations relativeerror= error / actual[xstar] AppendTo[errorlisteulerimp, error] peimp = ListPlot[solnslist, PlotStyle Ä {Hue[.8], PointSize[0.02]}] Show[pa, peimp] Rerun with different values for dx, starting from largest to smallest. After doing this, observe what happens to the error as the step size decreases by entering the input command: errorlisteulerimp You can also type Show[pa, pe, peimp]. This would be appropriate for a fixed value of dx with each method. You can also make a list of relative errors. Problems 21 - 24 involve use of code from section 9.1 together with the above code for Euler's method. 9.4 GRAPHICAL SOUTIONS OF AUTONOMOUS DIFFERENTIAL EQUATIONS 1. y w œ ay 2bay 3b (a) y œ 2 is a stable equilibrium value and y œ 3 is an unstable equilibrium. (b) yww œ a2y 1by w œ 2ay 2bˆy 12 ‰ay 3b
(c)
2. y w œ ay 2bay 2b (a) y œ 2 is a stable equilibrium value and y œ 2 is an unstable equilibrium. (b) yww œ 2yy w œ 2ay 2byay 2b
(c)
599
600
Chapter 9 Further Applications of Integration
3. y w œ y3 y œ ay 1byay 1b (a) y œ 1 and y œ 1 is an unstable equilibrium and y œ 0 is a stable equilibrium value. (b) yww œ a3y2 1by w œ 3ay 1bŠy
1 È3 ‹yŠy
1 È3 ‹ay
1b
(c)
4. y w œ yay 2b (a) y œ 0 is a stable equilibrium value and y œ 2 is an unstable equilibrium. (b) yww œ a2y 2by w œ 2yay 1bay 2b
(c)
5. y w œ Èy, y 0 (a) There are no equilibrium values. 1 1 w Èy œ "# (b) yww œ 2È y y œ 2È y
Section 9.4 Graphical Solutions of Autonomous Differential Equations (c)
6. y w œ y Èy, y 0 (a) y œ 1 is an unstable equilibrium. (b) yww œ Š1
1 2È y ‹
y w œ Š1
1 ˆ 2È y ‹ y
Èy‰ œ ˆÈy "# ‰ˆÈy 1‰
(c)
7. y w œ ay 1bay 2bay 3b (a) y œ 1 and y œ 3 is an unstable equilibrium and y œ 2 is a stable equilibrium value. (b) yww œ a3y2 12y 11bay 1bay 2bay 3b œ 3ay 1bŠy
(c)
6 È3 ‹ay 3
2bŠy
6 È3 3 ‹ay
3b
601
602
Chapter 9 Further Applications of Integration
8. y w œ y3 y2 œ y2 ay 1b (a) y œ 0 and y œ 1 is an unstable equilibrium. (b) yww œ a3y2 2ybay3 y2 b œ y3 a3y 2bay 1b
(c)
9.
10.
dP dt
œ 1 2P has a stable equilibrium at P œ "# .
dP dt œ Pa1 2Pb has an unstable equilibrium d2 P dP dt2 œ a1 4Pb dt œ Pa1 4Pba1 2Pb
d2 P dt2
œ 2 dP dt œ 2a1 2Pb
at P œ 0 and a stable equilibrium at P œ "# .
Section 9.4 Graphical Solutions of Autonomous Differential Equations
11.
12.
dP dt œ 2PaP 3b has a d2 P dP dt2 œ 2a2P 3b dt œ
dP dt d2 P dt2
stable equilibrium at P œ 0 and an unstable equilibrium at P œ 3. 4Pa2P 3baP 3b
œ 3Pa1 PbˆP "# ‰ has a stable equilibria at P œ 0 and P œ 1 an unstable equilibrium at P œ "# . 3 œ #3 a6P2 6P+1b dP dt œ # PŠP
3 È3 ˆ ‹ P 6
"# ‰ŠP
3 È3 ‹aP 6
1b
603
604
Chapter 9 Further Applications of Integration
13.
Before the catastrophe, the population exhibits logistic growth and Patb Ä M0 , the stable equilibrium. After the catastrophe, the population declines logistically and Patb Ä M1 , the new stable equilibrium. 14.
dP dt
œ rPaM PbaP mb, r, M, m 0
The model has 3 equilibrium oints. The rest oint P œ 0, P œ M are asymptotically stable while P œ m is unstable. For initial populations greater than m, the model predicts P approaches M for large t. For initial populations less than m, the model predicts extinction. Points of inflection occur at P œ a and P œ b where a œ "3 M m ÈM2 mM m2 ‘ and b œ "3 M m ÈM2 mM m2 ‘. (a) The model is reasonable in the sense that if P m, then P Ä 0 as t Ä _; if m P M, then P Ä M as t Ä _; if P M, then P Ä M as t Ä _. (b) It is different if the population falls below m, for then P Ä 0 as t Ä _ (extinction). If is probably a more realistic model for that reason because we know some populations have become extinct after the population level became too low. (c) For P M we see that dP dt œ rPaM PbaP mb is negative. Thus the curve is everywhere decreasing. Moreover, P ´ M is a solution to the differential equation. Since the equation satisfies the existence and uniqueness conditions, solution trajectories cannot cross. Thus, P Ä M as t Ä _. (d) See the initial discussion above. (e) See the initial discussion above. 15.
dv dt
œg
k 2 mv ,
Equilibrium: Concavity:
g, k, m 0 and vatb 0
dv dt
d2 v dt2
œg
k 2 mv
œ 0 Ê v œ É mg k
ˆ k ‰ˆ œ 2ˆ mk v‰ dv dt œ 2 m v g
(a)
(b)
160 (c) vterminal œ É 0.005 œ 178.9
ft s
œ 122 mph
k 2‰ mv
Section 9.4 Graphical Solutions of Autonomous Differential Equations
605
16. F œ Fp Fr ma œ mg kÈv dv kÈ v, va0b œ v0 dt œ g m Thus,
dv dt
‰ , the terminal velocity. If v0 ˆ mg ‰ , the object will fall faster and faster, approaching the œ 0 implies v œ ˆ mg k k 2
2
‰ , the object will slow down to the terminal velocity. terminal velocity; if v0 ˆ mg k 2
17. F œ Fp Fr ma œ 50 5kvk dv 1 dt œ m a50 5kvkb The maximum velocity occurs when
dv dt
œ 0 or v œ 10
ft sec .
18. (a) The model seems reasonable because the rate of spread of a piece of information, an innovation, or a cultural fad is proportional to the product of the number of individuals who have it (X) and those who do not (N X). When X is small, there are only a few individuals to spread the item so the rate of spread is slow. On the other hand, when (N X) is small the rate of spread will be slow because there are only a few indiciduals who can receive it during the interval of time. The rate of spread will be fastest when both X and (N X) are large because then there are a lot of individuals to spread the item and a lot of individuals to receive it. (b) There is a stable equilibrium at X œ N and an unstable equilibrium at X œ 0. d2 X dt2
dX 2 œ k dX dt aN Xb kX dt œ k XaN XbaN 2Xb Ê inflection points at X œ 0, X œ
(c)
œ VL RL i œ RL ˆ VR i‰, V, L, R 0 œ RL ˆ VR i‰ œ 0 Ê i œ VR
19. L di dt Ri œ V Ê
di dt d2 i dt2 œ
Equilibrium: Concavity: Phase Line:
di dt
ˆ R ‰2 ˆ VR i‰ ˆ RL ‰ di dt œ L
N 2,
and X œ N.
606
Chapter 9 Further Applications of Integration
If the switch is closed at t œ 0, then ia0b œ 0, and the graph of the solution looks like this:
As t Ä _, it Ä isteady state œ
V R.
(In the steady state condition, the self-inductance acts like a simple wire connector and, as
a result, the current throught the resistor can be calculated using the familiar version of Ohm's Law.) 20. (a) Free body diagram of he pearl:
(b) Use Newton's Second Law, summing forces in the direction of the acceleration: ˆ m m P ‰g mk v. mg Pg kv œ ma Ê dv dt œ (c) Equilibrium: Ê vterminal œ Concavity:
dv dt
œ
k amPbg mŠ k
v‹ œ 0
am Pbg k
d2 v dt2
ˆ k ‰ am k Pbg v‹ œ mk dv dt œ m Š 2
(d)
(e) The terminal velocity of the pearl is
am Pbg . k
9.5 APPLICATIONS OF FIRST ORDER DIFFERENTIAL EQUATIONS 1. Note that the total mass is 66 7 œ 73 kg, therefore, v œ v0 eakÎmbt Ê v œ 9e3.9tÎ73 3.9tÎ73 (a) satb œ ' 9e3.9tÎ73 dt œ 2190 C 13 e
Since sa0b œ 0 we have C œ
2190 13
3.9tÎ73 ‰ ˆ and lim satb œ lim 2190 œ 13 1 e tÄ_
tÄ_
The cyclist will coast about 168.5 meters. 73 ln 9 (b) 1 œ 9e3.9tÎ73 Ê 3.9t 73 œ ln 9 Ê t œ 3.9 ¸ 41.13 sec It will take about 41.13 seconds.
2190 13
¸ 168.5
Section 9.5 Applications of First-Order Differential Equations 2. v œ v0 eakÎmbt Ê v œ 9ea59,000Î51,000,000bt Ê v œ 9e59tÎ51,000 (a) satb œ ' 9e59tÎ51,000 dt œ 459,0000 e59tÎ51,000 C 59 Since sa0b œ 0 we have C œ
459,0000 59
ˆ1 e59tÎ51,000 ‰ œ and lim satb œ lim 459,0000 59 tÄ_
tÄ_
459,0000 59
¸ 7780 m
The ship will coast about 7780 m, or 7.78 km. ln 9 59t (b) 1 œ 9e59tÎ51,000 Ê 51,000 œ ln 9 Ê t œ 51,000 ¸ 1899.3 sec 59 It will take about 31.65 minutes. 3. The total distance traveled œ v0km Ê a2.75bak39.92b œ 4.91 Ê k œ 22.36. Therefore, the distance traveled is given by the function satb œ 4.91ˆ1 ea22.36/39.92bt ‰. The graph shows satb and the data points.
4.
a0.80ba49.90b œ 1.32 Ê k œ 998 k 33 v0 m k 998 We know that k œ 1.32 and m œ 33a49.9b œ 20 . 33 v0 m ˆ ak/mbt ‰ Using Equation 3, we have: satb œ k 1 e œ 1.32ˆ1 v0 m k
5. (a)
œ coasting distance Ê
dP dt
œ 0.0015Pa150 Pb œ
0.255 150 Pa150
Pb œ
Thus, k œ 0.255 and M œ 150, and P œ Initial condition: Pa0b œ 6 Ê 6 œ Formula: P œ
150 0.255t 1 24ec0.255t Ê 1 24e ln 48 Ê t œ 0.255 ¸ 17.21 weeks 150 125 œ 1 24ec0.255t Ê 1 24e0.255t 120 Ê t œ ln0.255 ¸ 21.28
(b) 100 œ
M 1 Aeckt
150 1 Ae0
150 1 24ec0.255t
k M PaM
œ
e20t/33 ‰ ¸ 1.32a1 e0.606t b
Pb
150 1 Aec0.255t
Ê 1 A œ 25 Ê A œ 24
œ
3 2
Ê 24e0.255t œ
" #
Ê e0.255t œ
" 48
œ
6 5
Ê 24e0.255t œ
" 5
Ê e0.255t œ
" 120
Ê 0.255t œ ln 48 Ê 0.255t œ ln 120
It will take about 17.21 weeks to reach 100 guppies, and about 21.28 weeks to reach 125 guppies. 6. (a)
dP dt
œ 0.0004Pa250 Pb œ
0.1 250 Pa150
Pb œ
Thus, k œ 0.1 and M œ 250, and P œ
k M PaM Pb M œ 1 250 Aec0.1t 1 Aeckt
Initial condition: Pa0b œ 28, where t œ 0 represents the year 1970 250 111 28 œ 1 250 Ae0 Ê 28a1 Ab œ 250 Ê A œ 28 1 œ 14 ¸ 7.9286 Formula: P œ
250 c0.1t 1 111 14 e
or approximately P œ
250 1 7.9286ec0.1t
(b) The population Patb will round to 250 when Patb 249.5 Ê 249.5 œ Ê
a249.5bˆ111ec0.1t ‰ 14
œ 0.5 Ê e0.1t œ
14 55,389
Ê 0.1t œ ln
14 55,389
250 c0.1t 1 111 14 e
Ê 249.5ˆ1
Ê t œ 10 aln 55,389 ln 14b ¸ 82.8.
It will take about 83 years. 7. (a) Using the general solution form Example 2, part (c), dy dt
œ a0.08875 ‚ 107 ba8 ‚ 107 yby Ê yatb œ
M 1 AecrMt
œ
111 0.1t ‰ 14 e
8‚107 1 Aeca!Þ!))(&ba)bt
œ
8‚107 1 Aec0.71t
œ 250
607
608
Chapter 9 Further Applications of Integration Apply the initial condition: ya0b œ 1.6 ‚ 107 œ
(b) yatb œ 4 ‚ 107 œ
8‚107 1A
Ê
8‚107 1 4ec0.71t
8 1.6
8‚107 1 4ec0.71 ¸ 2.69671 ‚ lnˆ 1 ‰ 0.714 ¸ 1.95253 years.
1 œ 4 Ê ya1b œ
Ê 4e0.71t œ 1 Ê t œ
107 kg.
8. (a) If a part of the population leaves or is removed from the environment (e.g., a preserve or a region) each year, then c would represent the rate of reduction of the population due to this removal and/or migration. When grizzly bears become a nuisance (e.g., feeding on livestock) or threaten human safety, they are often relocated to other areas or even eliminated, but only after relocation efforts fail. In addition, bears are killed, sometimes accidently and sometimes maliciously. For an environment that has a capacity of about 100 bears, a realistic value for c would probably be between 0 and 4. (b)
Equilibrium solutions:
dP dt
œ 0 œ 0.001a100 PbP 1 Ê P2 100P 1000 œ 0 Ê Peq ¸ 11.27 (unstable) and
Peq ¸ 88.73 (stable) (c)
For 0 Pa0b Ÿ 11, the bear population will eventually disappear, for 12 Ÿ Pa0b Ÿ 88, the population will grow to about 89, the population will remain at about 89, and for Pa0b 89, the population will decrease to about 89 bears. 9. (a)
dy dt
œ 1 y Ê dy œ a1 ybdt Ê
dy 1y
œ dt Ê ln k1 yk œ t C1 Ê eln k1yk œ etC1 Ê k1 yk œ et eC1
1 y œ „ C2 et Ê y œ Cet 1, where C2 œ eC1 and C œ „ C2 . Apply the initial condition: ya0b œ 1 œ Ce0 1 Ê C œ 2 Ê y œ 2et 1. (b)
dy dt
œ 0.5a400 yby Ê dy œ 0.5a400 yby dt Ê
Example 2, part (c), we obtain Ê ' Š 1y ln¹ y cy400 ¹
Êe Ê
y y400
Êyœ
1 400 y ‹dy
1 1 400 Š y
1 400 y ‹dy
dy a400 yby
œ 0.5 dt. Using the partial fraction decomposition in
œ 0.5 dt Ê Š 1y
1 400 y ‹dy
œ 200 dt
œ ' 200 dt Ê lnkyk lnky 400k œ 200t C1 Ê ln¹ y y400 ¹ œ 200t C1
œ e200tC1 œ e200t eC1 Ê ¹ y y400 ¹ œ C2 e200t (where C2 œ eC1 ) Ê
y y 400
œ „ C2 e200t
œ Ce200t (where C œ „ C2 ) Ê y œ Ce200t y 400 Ce200t Ê a1 Ce200t by œ 400 Ce200t
400 Ce200t Ce200t 1
ya0b œ 2 œ
Êyœ
400 1 Ae0
400 1 C1 ec200t
œ
400 1 Aec200t ,
Ê A œ 199 Ê yatb œ
where A œ C1 . Apply the initial condition:
400 1 199ec200t
Section 9.5 Applications of First-Order Differential Equations 10.
dP dt
œ raM PbP Ê dP œ raM PbP dt Ê
we obtain
1 ˆ1 M P
1 ‰ M P dP
dP aM PbP
œ r dt. Using the partial fraction decomposition in Example 6, part (c),
'
œ r dt Ê ˆ P1
Ê lnkPk lnkP Mk œ arMb Ê ¸ P P M ¸ œ C2 earMbt (where
'
1 ‰ ˆ P1 P 1 M ‰dP œ rM dt M P dP œ rM dt Ê P t C1 Ê ln¸ P P M ¸ œ arMb t C1 Ê eln¸ P c M ¸ œ earMbtC1 œ earMbt eC1 C2 œ eC1 ) Ê PPM œ „ C2 earMbt Ê PPM œ CearMbt (where C œ „ C2 )
Ê P œ CearMbt P M CearMbt Ê ˆ1 CearMbt ‰P œ M CearMbt Ê P œ ÊPœ 11. (a)
dP dt
M , 1 AecarMbt
œ kP2 Ê ' P2 dP œ ' k dt Ê P" œ kt C Ê P œ
Initial condition: Pa0b œ P0 Ê P0 œ C1 Ê C œ
dP dt
œ raM PbaP mb Ê
Ê
ÊPœ
M 1 C1 ecarMbt
aP 100ba1200 Pb dP a1200 PbaP 100b dt
1 P0
" kt C
P0 1 kP0 t
(b) There is a vertical asymptote at t œ 12. (a)
M CearMbt CearMbt 1
where A œ C1 .
Solution: P œ kt a11/P0 b œ
dP dt
1 kpO
œ ra1200 PbaP 100b Ê
œ 1100 r Ê
ˆ 12001 P
1 ‰ dP P 100 dt
1 dP a1200 PbaP 100b dt
œ 1100 r Ê
œrÊ
ˆ 12001 P
1100 dP a1200 PbaP 100b dt
1 ‰ P 100 dP
P 100 1200 P 1100 r t
where C œ „ eC1 Ê P 100 œ 1200Ce1100 r t CPe1100 r t Ê Pa1 Ce1100 r t b œ 1200Ce ÊPœ
1200Ce 100 Ce1100 r t 1
œ
c1100 r t 1200 100 C e 1 C1 ec1100 r t
(b) Apply the initial condition: 300 œ
1200 100Aec1100 r t 1 Aec1100 r t
ÊPœ
1200 100A 1A
œ 1100 r
œ 1100 r dt
Ê ' ˆ 12001 P P1100 ‰dP œ ' 1100 r dt Ê ln a1200 Pb ln aP 100b œ 1100 r t C1 P 100 ¸ P 100 C1 1100 r t ¸ P 100 ¸ Ê ln ¸ 1200 Ê P œ 1100 r t C1 Ê ln 1200 P œ 1100 r t C1 Ê 1200 P œ „ e e 1100 r t
œ Ce1100 r t
100
where A œ C1 .
Ê 300 300A œ 1200 100A Ê A œ
9 2
ÊPœ
2400 900Aec1100 r t Þ 2 9ec1100 r t
(Note that P Ä 1200 as t Ä _.) (c)
dP dt
œ raM PbaP mb Ê
Ê ˆ M 1 P
1 ‰ dP P m dt
1 dP aM PbaP mb dt
œrÊ
œ raM mb Ê ' ˆ M 1 P
Ê ln aM Pb ln aP mb œ aM mb r t Ê
Pm MP
Mm dP aM PbaP mb dt
Ê Pˆ1 Ce
aMmb r t
œ MCe
œ raM mb Ê
'
1 ‰ raM mbdt P m dP œ Pm ¸ ¸ C1 Ê ln M P œ aM mb r t aMmb r t aMmb r t
œ CeaMmb r t where C œ „ eC1 Ê P m œ MCe aMmb r t ‰
mÊPœ
aP mb aM Pb dP aM PbaP mb dt
C1 Ê
Apply the initial condition Pa0b œ P0 MmA 1 A
Ê P0 P0 A œ M mA Ê A œ
M P0 P0 m
(Note that P Ä M as t Ä _ provided P0 m.) 13. y œ mx Ê
y x
orthogonals:
œmÊ dy dx
2
w
œ 0 Ê y w œ yx . So for
œ xy Ê y dy œ x dx Ê
Ê x y œ C1 2
xy y x2
y2 2
x2 2
œC
ÊPœ
Pm MP
œ raM mb
œ „ eC1 eaMmb r t
CPe
MCeaMcmb r t m CeaMcmb r t 1
ÊPœ
caMcmb r t M m Ce 1 C1 ecaMcmb r t
A œ C1 . P0 œ
609
MaP0 mb maM P0 becaMcmb r t aP0 mb aM P0 becaMcmb r t
ÊPœ
M mAecaMcmb r t 1 AecaMcmb r t
610
Chapter 9 Further Applications of Integration
14. y œ cx2 Ê Ê yw œ
y x2
2y x .
œcÊ
x2 y 2xy x4
œ 0 Ê x2 y w œ 2xy
w
So for the orthogonals:
dy dx
x œ 2y
2
Ê 2ydy œ xdx Ê y2 œ x2 C Ê y œ „ É x2 C, 2
C0
1 y2 x2
15. kx2 y2 œ 1 Ê 1 y2 œ kx2 Ê
œk
x a2yby ˆ1 y2 ‰2x Ê œ 0 Ê 2yx2 y w œ a1 y2 ba2xb x ˆ1 y2 ‰a2xb ˆ1 y 2 ‰ Ê y w œ 2xy2 œ xy . So for the orthogonals: 2 ˆ1 y 2 ‰ 2 dy xy dy œ x dx Ê ln y y2 œ x2 C dx œ 1y2 Ê y 2
w
%
2x 16. 2x2 y2 œ c2 Ê 4x 2yy w œ 0 Ê y w œ 4x 2y œ y . For
orthogonals:
dy dx
œ
y 2x
Ê
œ
dy y
dx 2x
Ê ln y œ "# ln x C
Ê ln y œ ln x1/2 ln C1 Ê y œ C1 kxk1/2
17. y œ cex Ê
y ecx
œcÊ
e x y c yae x bac1b ae x b2
w
œ!
Ê ex y w œ yex Ê y w œ y. So for the orthogonals: dy dx
œ
1 y 2
Ê y dy œ dx Ê
y2 2
œxC
Ê y œ 2x C1 Ê y œ „ È2x C1
xŠ 1y ‹y c ln y w
18. y œ ekx Ê ln y œ kx Ê
ln y x
œkÊ
Ê Š xy ‹ y w ln y œ 0 Ê y w œ dy dx
y ln y x .
x y ln y Ê y ln y dy œ x dx 1 2 " 2 ˆ " 2‰ # y ln y 4 ay b œ # x 2 y2 ln y y2 œ x2 C1
œ
Ê Ê
C
x2
œ0
So for the orthogonals:
Section 9.5 Applications of First-Order Differential Equations 2 w 19. 2x2 3y2 œ 5 and y2 œ x3 intersect at a1, 1b. Also, 2x2 3y2 œ 5 Ê 4x 6y y w œ 0 Ê y w œ 4x 6y Ê y a1, 1b œ 3
y21 œ x3 Ê 2y1 y1w œ 3x2 Ê y1w œ x2 2
20. (a) x dx y dy œ 0 Ê
y2 2 w
3x2 2y1
Ê y1w a1, 1b œ 32 . Since y w † y1w œ ˆ 23 ‰ˆ 32 ‰ œ 1, the curves are orthogonal.
œ C is the general equation
of the family with slope y œ xy . For the orthogonals: yw œ
y x
Ê
dy y
œ
dx x
Ê ln y œ ln x C or y œ C1 x
(where C1 œ e Ñ is the general equation of the orthogonals. C
(b) x dy 2y dx œ 0 Ê 2y dx œ x dy Ê Ê "# Š dy y ‹œ
dx x
dy 2y
œ
dx x
Ê "# ln y œ ln x C Ê y œ C1 x2 is
the equation for the solution family. " # ln
y ln x œ C Ê
"y # y
w
Ê slope of orthogonals is
1 x dy dx
œ 0 Ê yw œ
2y x
x œ 2y 2
Ê 2y dy œ x dx Ê y2 œ x2 C is the general equation of the orthogonals.
2". y2 œ 4a2 4ax and y2 œ 4b2 4bx Ê (at intersection) 4a2 4ax œ 4b2 4bx Ê a2 b2 œ xaa bb Ê aa bbaa bb œ aa bbx Ê x œ a b. Now, y2 œ 4a2 4aaa bb œ 4a2 4a2 4ab œ 4ab Ê y œ „ 2Èab. 4a Thus the intersections are at Ša b, „ 2Èab‹. So, y2 œ 4a2 4ax Ê y1w œ 2y which are equal to 4a and È 2Š2
4a 2Š2Èab‹ 4b 2Š2Èab‹
œ
È ba
and
È ba
at the intersections. Also, y œ 4b 4bx Ê 2
2
y2w
œ
4b 2y
which are equal to
œ É ba and É ba at the intersections. ay1w b † ay2w b œ ". Thus the curves are orthogonal.
ab‹
4b 2Š2Èab‹
and
611
612
Chapter 9 Further Applications of Integration
CHAPTER 9 PRACTICE EXERCISES ".
dy dx
œ Èy cos2 Èy Ê
2. y w œ
3yax1b2 y 1
Ê
œ dx Ê 2tanÈy œ x C Ê y œ ˆtan1 ˆ x 2 C ‰‰
dy Èy cos2 Èy
ay 1 b y dy
3. yy w œ secay2 bsec2 x Ê
œ 3ax 1b2 dx Ê y ln y œ ax 1b3 C
y dy secay2 b
sinˆy2 ‰ 2
œ sec2 x dx Ê
œ tan x C Ê sinay2 b œ 2tan x C1
sin x 4. y cos2 axb dy sin x dx œ 0 Ê y dy œ cos 2 axb dx Ê
y2 2
œ cos1axb C Ê y œ „ É cosa2xb C1
2ax 2b3/2 a3x 4b 15 2ax 2b3/2 a3x 4b C“ 15
5. y w œ xey Èx 2 Ê ey dy œ xÈx 2 dx Ê ey œ 2ax 2b3/2 a3x 4b 15
Ê y œ ln’ 6. y w œ xyex Ê 2
dy y
2
C“ Ê y œ ln’
C Ê ey œ
2ax 2b3/2 a3x 4b 15
C
œ ex x dx Ê ln y œ "# ex C 2
2
7. sec x dy x cos2 y dx œ 0 Ê
dy cos2 y
x dx œ sec x Ê tan y œ cos x x sin x C
8. 2x2 dx 3Èy csc x dy œ 0 Ê 3Èy dy œ
2x2 csc x dx
Ê 2y3/2 œ 2a2 x2 bcos x 4x sin x C
Ê y3/2 œ a2 x2 bcos x 2x sin x C1 9. y w œ
ey xy
Ê yey dy œ
Ê ay 1bey œ ln kxk C
dx x
10. y w œ xexy csc y Ê y w œ
x ex ey csc
yÊ
ey csc y dy
œ x ex dx Ê
11. xax 1bdy y dx œ 0 Ê xax 1bdy œ y dx Ê
dy y
œ
ey 2 asin
dx x ax 1 b
y cos yb œ ax 1bex C
Ê ln y œ lnax 1b lnaxb C
Ê ln y œ lnax 1b lnaxb ln C1 Ê ln y œ lnŠ C1 axx 1b ‹ Ê y œ
12. y w œ ay2 1bax1 b Ê
dy
y 2 1
œ
Ê
dx x
1 lnŠ yy c b1‹
2
C1 ax 1b x
1 œ ln x C Ê lnŠ yy 1 ‹ œ 2ln x ln C1 Ê
y1 y1
œ C1 x2
13. 2y w y œ xex/2 Ê y w "# y œ x2 ex/2 .
' ˆ "‰ paxb œ "# , vaxb œ e c # dx œ ex/2 .
ex/2 y w "# ex/2 y œ ˆex/2 ‰ˆ x2 ‰ˆex/2 ‰ œ 14.
w
y 2
x 2
Ê
d ˆ x/2 dx e
y‰ œ
x 2
Ê ex/2 y œ
x2 4
2
C Ê y œ ex/2 Š x4 C‹
y œ ex sin x Ê y w 2y œ 2ex sin x.
paxb œ 2, vaxb œ e' 2dx œ e2x . e2x y w 2e2x y œ 2e2x ex sin x œ 2ex sin x Ê x
d 2x dx ae
yb œ 2ex sin x Ê e2x y œ ex asin x cos xb C
2x
Ê y œ e asin x cos xb Ce
15. xy w 2y œ 1 x1 Ê y w ˆ 2x ‰y œ vaxb œ e2'
dx x
1 x
1 x2 .
2
œ e2ln x œ eln x œ x2 .
x2 y w 2xy œ x 1 Ê
d 2 dx ax yb
œ x 1 Ê x2 y œ
x2 2
xCÊyœ
" #
1 x
C x2
Chapter 9 Practice Exercises
613
16. xy w y œ 2x ln x Ê y w ˆ 1x ‰y œ 2 ln x. vaxb œ e d ˆ1 dx x
' dxx
2 œ eln x œ 1x . ˆ 1x ‰y w ˆ 1x ‰ y œ 2x ln x Ê
† y‰ œ 2x ln x Ê
† y œ c ln x d2 C Ê y œ xc ln x d2 Cx
1 x
17. a1 ex bdy ayex ex bdx œ 0 Ê a1 ex by w ex y œ ex Ê y w œ ' a exdx b
vaxb œ e 1 b ex œ elnae 1b œ ex 1. x cx aex 1by w aex 1bˆ 1 e ex ‰y œ a1e ex b aex 1b Ê Êyœ 18.
dx dy
ecx C ex 1
ex 1 ex y
œ
ecx a1 e x b .
x
œ
e cx C
1by d œ ex Ê aex 1by œ ex C
d aex dx c
1 ex
x 4yey œ 0 Ê x w x œ 4yey . Let vayb œ e
' dy
œ ey . Then ey x w xey œ 4ye2y Ê
d y dy axe b
œ 4ye2y
Ê xey œ a2y 1be2y C Ê x œ a2y 1bey Cey 19. ax 3y2 b dy y dx œ 0 Ê x dy y dx œ 3y2 dy Ê
d dx axyb
œ 3y2 dy Ê xy œ y3 C '
20. y dx a3x y2 cos yb dx œ 0 Ê x w Š 3y ‹x œ y3 cos y. Let vayb œ e
3dy y
3
œ e3ln y œ eln y œ y3 .
Then y3 x w 3y2 x œ cos y and y3 x œ ' cos y dy œ sin y C. So x œ y3 asin y Cb 21.
œ exy2 Ê ey dy œ eax2b dx Ê ey œ eax2b C. We have ya0b œ 2, so e2 œ e2 C Ê C œ 2e2 and e œ eax2b 2e2 Ê y œ lnˆeax2b 2e2 ‰
22.
dy dx
dy dx y
œ
y ln y 1 x2
Ê etan
Ê
c1 a0bC
dy y ln y
œ
dx 1 x2
Ê lnaln yb œ tan1 axb C Ê y œ ee
So y w ax 1b2 Ê yax 1b2 œ
x x1.
' x b2 1 dx
Let vaxb œ e
2 ax 1 b a x
1b2 y œ
x3 3
C Ê y œ ax 1b2 Š x3
y œ ax 1b2 Š x3 3
x2 2 2
x 2
x ax 1 b a x
1b2 Ê
d dx yax
3
x2 2
dy dx
tanc1 axbbln 2
2
œ e2lnax1b œ elnax1b œ ax 1b2 .
2 1b2 ‘ œ xax 1b Ê yax 1b œ ' xax 1bdx
1‹ ' ˆ 2 ‰dx
25.
tanc1 a0bbC
C‹. We have ya0b œ 1 Ê 1 œ C. So
1 2 w ˆ2‰ 24. x dy dx 2y œ x 1 Ê y x y œ x x . Let vaxb œ e
So
. We have ya0b œ e2 Ê e2 œ ee
œ 2 Ê tan1 a0b C œ ln 2 Ê 0 C œ ln 2 Ê C œ ln 2 Ê y œ ee
w ˆ 2 ‰ 23. ax 1b dy dx 2y œ x Ê y x 1 y œ
Ê
tanc1 axbbC
d x4 2 3 2 dx ax yb œ x x Ê x y œ 4 2 4 2x2 1 y œ x4 4x1 2 "# œ x 4x 2
x2 2
CÊyœ
x2 4
x
C x2
œ eln x œ x2 . So x2 y w 2xy œ x3 x 2
"# . We have ya1b œ 1 Ê 1 œ
3x2 y œ x2 . Let vaxb œ e' 3x dx œ ex . So ex y w 3x2 ex y œ x2 ex Ê 2
3
3
3
3
We have ya0b œ 1 Ê e0 a1b œ 13 e0 C Ê 1 œ
3
1 3
d dx axyb
dy ˆy È y ‰
3
" #
3
Ê C œ 14 .
3
œ x2 ex Ê ex y œ 13 ex C.
3
cos x x .
C
3
4 3
Êyœ
1 3
43 ex
3
' 1 dx x œ eln x œ x.
Let vaxb œ e
œ cos x Ê xy œ ' cos x dx Ê xy œ sin x C. We have yˆ 12 ‰ œ 0 Ê ˆ 12 ‰0 œ 1 C
Ê C œ 1. So xy œ 1 sin x Ê y œ 27. x dy ˆy Èy‰dx œ 0 Ê
d x3 dx Še y‹
C Ê C œ 43 and ex y œ 13 ex
26. xdy ay cos xbdx œ 0 Ê xy w y cos x œ 0 Ê y w ˆ 1x ‰y œ So xy w xˆ 1x ‰y œ cos x Ê
3
1 4
œ
dx x
1 sin x x
Ê 2lnˆÈy 1‰ œ ln x C. We have ya1b œ 1 Ê 2 lnŠÈ1 1‹ œ ln 1 C
Ê 2 ln 2 œ C œ ln 22 œ ln 4. So 2 lnˆÈy 1‰ œ ln x ln 4 œ lna4xb Ê lnˆÈy 1‰ œ "# lna4xb œ lna4xb1/2
614
Chapter 9 Further Applications of Integration Ê eln
ˆÈ y 1 ‰
28. y2 dx dy œ So
y3 3
1/2
œ elna4xb Ê Èy 1 œ 2Èx Ê y œ ˆ2Èx 1‰
ex e2x 1
Ê
e2x 1 ex dx
œ ex ex
1 3
œ
dy yc2
Ê
y3 3
2
œ ex ex C. We have ya0b œ 1 Ê
a1 b 3 3
œ e0 e0 C Ê C œ 13 .
Ê y3 œ 3aex ex b 1 Ê y œ c3aex ex b 1 d1/3 xc2
29. xy w ax 2by œ 3x3 ex Ê y w ˆ x x 2 ‰y œ 3x2 ex . Let vaxb œ e' ˆ x ‰dx œ ex2ln x œ xe2 . So x x ex w ex ˆ x 2 ‰ d ˆ y œ 3 Ê dx y † xe2 ‰ œ 3 Ê y † xe2 œ 3x C. We have ya1b œ 0 Ê 0 œ 3a1b C Ê C œ 3 x2 y x2 x Êy†
ex x2
œ 3x 3 Ê y œ x2 ex a3x 3b
30. y dx a3x xy 2bdy œ 0 Ê Payb œ
3 y
x
dx dy
3x xy 2 y
œ0Ê
dx dy
3x y
x œ 2y Ê
dx dy
Š 3y 1‹x œ 2y .
1 Ê ' Paybdy œ 3ln y y Ê vayb œ e3ln yy œ y3 ey
y3 ey x w y3 ey Š 3y 1‹x œ 2y2 ey Ê y3 ey x œ ' 2y2 ey dy œ 2ey ay2 2y 2b C Ê y3 œ Ê y3 œ
2ˆy2 2y 2‰ Cey . x 2 yb1 ˆ ‰ 2 y 2y 2 4e x
We have ya2b œ 1 Ê 1 œ
2a1 2 2b Cec1 2
Ê C œ 4e and
31. To find the approximate values let yn œ yn1 ayn1 cos xn1 ba0.1b with x0 œ 0, y0 œ 0, and 20 steps. Use a spreadsheet, graphing calculator, or CAS to obtain the values in the following table. x y x y 1.1 1.6241 0 0 1.2 1.8319 0.1 0.1000 1.3 2.0513 0.2 0.2095 1.4 2.2832 0.3 0.3285 1.5 2.5285 0.4 0.4568 1.6 2.7884 0.5 0.5946 1.7 3.0643 0.6 0.7418 1.8 3.3579 0.7 0.8986 1.9 3.6709 0.8 1.0649 2.0 4.0057 0.9 1.2411 1.0 1.4273 32. To find the approximate values let zn œ yn1 aa2 yn1 ba2 xn1 3bba0.1b and yn œ yn1 Š a2 ync1 ba2 xnc1 32b a2 zn ba2 xn 3b ‹a0.1b with initial values x0 œ 3, y0 œ 1, and 20 steps. Use a spreadsheet, graphing calculator, or CAS to obtain the values in the following table. x y x y 1.9 5.9686 3 1 1.8 6.5456 2.9 0.6680 1.7 6.9831 2.8 0.2599 1.6 7.2562 2.7 0.2294 1.5 7.3488 2.6 0.8011 1.4 7.2553 2.5 1.4509 1.3 6.9813 2.4 2.1687 1.2 6.5430 2.3 2.9374 1.1 5.9655 2.2 3.7333 1.0 5.2805 2.1 4.5268 2.0 5.2840
Chapter 9 Practice Exercises 2ync1 " xnc1 2ync1 33. To estimate ya3b, let zn œ yn1 Š xncx1nc1 1 ‹a0.05b and yn œ yn1 # Š xnc1 1
xn 2zn xn 1 ‹a0.05b
615
with initial values
x0 œ 0, y0 œ 1, and 60 steps. Use a spreadsheet, graphing calculator, or CAS to obtain ya3b ¸ 0.9063. 34. To estimate ya4b, let zn œ yn1 Š
x2nc1 2ync1 1 ‹a0.05b xnc1
with initial values x0 œ 1, y0 œ 1, and 60 steps. Use a
spreadsheet, graphing calculator, or CAS to obtain ya4b ¸ 4.4974. 35. Let yn œ yn1 ˆ exnc1 b1ync1 b 2 ‰adxb with starting values x0 œ 0 and y0 œ 2, and steps of 0.1 and 0.1. Use a spreadsheet, programmable calculator, or CAS to generate the following graphs. (a)
(b) Note that we choose a small interval of x-values because the y-values decrease very rapidly and our calculator cannot handle the calculations for x Ÿ 1. (This occurs because the analytic solution is y œ 2 lna2 ex b, which has an asymptote at x œ ln 2 ¸ 0.69. Obviously, the Euler approximations are misleading for x Ÿ 0.7.)
y
y
36. Let zn œ yn1 Š eynncc11 xnncc11 ‹adxb and yn œ yn1 #" Š eynncc11 xnncc11 x2
x2
xn2 zn ezn xn ‹adxb
with starting values x0 œ 0 and y0 œ 0,
and steps of 0.1 and 0.1. Use a spreadsheet, programmable calculator, or CAS to generate the following graphs. (a) (b)
37.
x y dy dx
1 1.2 1.4 1.6 1 0.8 0.56 0.28
œ x Ê dy œ x dx Ê y œ
Ê 1 œ
" #
Ê ya2b œ
CÊCœ 2
2 2
3 2
œ
" #
32
x2 2
1.8 0.04
2.0 0.4
C; x œ 1 and y œ 1
Ê yaexactb œ
is the exact value.
x2 2
3 2
616
Chapter 9 Further Applications of Integration
38.
x y œ
dy dx
1 1.2 1.4 1.6 1.8 2.0 1 0.8 0.6333 0.4904 0.3654 0.2544 1 x
Ê dy œ x1 dx Ê y œ lnkxk C; x œ 1 and y œ 1
Ê 1 œ ln 1 C Ê C œ 1 Ê yaexactb œ lnkxk 1 Ê ya2b œ ln 2 1 ¸ 0.3069 is the exact value.
39.
x y
1 1.2 1.4 1.6 1.8 2.0 1 1.2 0.488 1.9046 2.5141 3.4192
œ xy Ê
dy dx
Êyœe
dy y
x2 2 C
œ x dx Ê lnkyk œ x2 2
x2 2
C
x2 2
œ e † eC œ C1 e ; x œ 1 and y œ 1 x2
Ê 1 œ C1 e1/2 Ê C1 œ e1/2 yaexactb œ e1/2 † e 2 œ eˆx 1‰/2 Ê ya2b œ e3/2 ¸ 4.4817 is the exact value. 2
40.
x y
1 1.2 1.4 1.6 1.8 2.0 1 1.2 1.3667 1.5130 1.6452 1.7688
dy y2 1 dx œ y Ê y dy œ dx Ê 2 œ x " " 2 # œ 1 C Ê C œ # Ê y œ
C; x œ 1 and y œ 1
2x 1 Ê yaexactb œ È2x 1 Ê ya2b œ È3 ¸ 1.7321 is the exact value.
41.
dy dx
œ y2 1 Ê y w œ ay 1bay 1b. We have y w œ 0 Ê ay 1b œ 0, ay 1b œ 0 Ê y œ 1, 1.
(a) Equilibrium points are 1 (stable) and 1 (unstable) (b) y w œ y2 1 Ê y ww œ 2yy w Ê y ww œ 2yay2 1b œ 2yay 1bay 1b. So y ww œ 0 Ê y œ 0, y œ 1, y œ 1.
(c)
42.
dy dx
œ y y2 Ê y w œ ya1 yb. We have y w œ 0 Ê ya1 yb œ 0 Ê y œ 0, 1 y œ 0 Ê y œ 0, 1.
(a) The equilibrium points are 0 and 1. So, 0 is unstable and 1 is stable. (b) Let ïî œ increasing, íï œ decreasing. yw ! yw ! yw ! qqíïïïïïqqñqqïïïïïîqqñqqíïïïïïqqpy 0 1 y w œ y y2 Ê y ww œ y w 2yy w Ê y ww œ ay y2 b 2yay y2 b œ y y2 2y2 2y3 Ê y ww œ 2y3 3y2 y œ ya2y2 3y 1b Ê y ww œ ya2y 1bay 1b. So, y ww œ 0 Ê y œ 0, 2y 1 œ 0, y 1 œ 0 Ê y œ 0, y œ "# ,
Chapter 9 Additional and Advanced Exercises
617
y œ 1. Let ïî œ concave up, íï œ concave down. y ww ! y ww ! y ww ! y ww ! qíïïïïïqqñqqïïïïïîqqñqqíïïïïïqqñqqïïïïïîqpy 0 1 1/2 (c)
43. (a) Force œ Mass times Acceleration (Newton's Second Law) or F œ ma. Let a œ
dv dt
œ
dv ds
†
ds dt
œ v dv ds . Then
2 2 ma œ mgR2 s2 Ê a œ gR2 s2 Ê v dv Ê v dv œ gR2 s2 ds Ê ' v dv œ ' gR2 s2 ds ds œ gR s
Ê
v2 2
œ
ÊCœ
gR2 s C1 2 v0 2gR
Ê v2 œ Êv œ 2
(b) If v0 œ È2gR, then v2 œ
2gR2 s 2gR2 s
2gR s
2
2C1 œ
v20
2gR2 s
C. When t œ 0, v œ v0 and s œ R Ê v20 œ
2gR2 R
C
2gR 2
È2gR. Then Ê v œ É 2gR s , since v 0 if v0
ds dt
œ
È2gR2 Ès
Ê Ès ds œ È2gR2 dt
Ê ' s1/2 ds œ ' È2gR2 dt Ê 23 s3/2 œ È2gR2 t C1 Ê s3/2 œ ˆ 32 È2gR2 ‰t C; t œ 0 and s œ R Ê R3/2 œ ˆ 32 È2gR2 ‰a0b C Ê C œ R3/2 Ê s3/2 œ ˆ 32 È2gR2 ‰t R3/2 œ ˆ 32 RÈ2g‰t R3/2 3 œ R3/2 ˆ 32 R1/2 È2g‰t 1 ‘ œ R3/2 ’ Š
44.
v0 m k
a0.86ba30.84b k 0.8866t
œ coasting distance Ê
Ê satb œ 0.97a1 e
È2gR 2R ‹t
2/3 0‰ 0‰ ‘ ‘ Ê s œ R 1 ˆ 3v 1 “ œ R3/2 ˆ 3v 2R t 1 2R t
œ 0.97 Ê k ¸ 27.343. satb œ
v0 m ˆ k 1
eak/mbt ‰ Ê satb œ 0.97ˆ1 ea27.343/30.84bt ‰
b. A graph of the model is shown superimposed on a graph of the data.
CHAPTER 9 ADDITIONAL AND ADVANCED EXERCISES 1. (a)
dy dt
A œ kA V ac yb Ê dy œ k V ay cbdt Ê
dy yc
' œ k A V dt Ê
dy yc
A œ ' k A V dt Ê lnky ck œ k V t C1
Ê y c œ „ eC1 ek V t . Apply the initial condition, ya0b œ y0 Ê y0 œ c C Ê C œ y0 c A
Ê y œ c ay0 cbek V t . A (b) Steady state solution: y_ œ lim yatb œ lim c ay0 cbek V t ‘ œ c ay0 cba0b œ c A
tÄ_
tÄ_
2. Measure the amounts of oxygen involved in mL. Then the inflow of oxygen is 1000 mL/min (Assumed: it will take 5 minutes to deliver the 5L œ 5000mL); the amount of oxygen at t œ 0 is 210 mL; letting A œ the amount of oxygen in the flask, the concentration at time t is A mL/L; the outflow rate of oxygen is A mL/L (lb/sec). The rate of change in A, dA dt , equals the rate of gain (1000 mL/min) minus rate of loss (A mL/min). Thus: dA dA t dt œ 1000 A Ê 1000 A œ dt Ê lnaA 1000b œ t C Ê A 1000 œ Ce . At t œ 0, A œ 210, so C œ 790 and A œ 1000 790et . Thus, Aa5b œ 1000 790e5 ¸ 994.7 mL. The concentration is
994.7 mL 1000 mL
œ 99.47%.
618
Chapter 9 Further Applications of Integration
3. The amount of CO2 in the room at time t is Aatb. The rate of change in the amount of CO2 ,
dA dt
is the rate of internal
production (R1 ) plus the inflow rate (R2 ) minus the outflow rate (R3 ). R1 œ ˆ20
breaths/min ‰ a30 student
R2 œ Š1000
ft3 CO2 ft3 min ‹Š0.0004 min ‹
A R3 œ Š 10,000 ‹1000 œ 0.1A dA dt
3
100 2 studentsbˆ 1728 ft3 ‰Š0.04 ft ftCO ‹ ¸ 1.39 3
œ 0.4
ft3 CO2 min
ft3 CO2 min
ft3 CO2 min
œ 1.39 0.4 0.1A œ 1.79 0.1A Ê Aw 0.1A œ 1.79. Let vatb œ e
' 0.1dt
. We have
' 0.1dt d ‹ dt ŠAe
' 0.1dt
œ 1.79e
Ê Ae0.1t œ ' 1.79e0.1t dt œ 17.9e0.1t C. At t œ 0, A œ a10,000ba0.0004b œ 4 ft3 CO2 Ê C œ 13.9 Ê A œ 17.9 13.9e0.1t . So Aa60b œ 17.9 13.9e0.1a60b ¸ 17.87 ft3 of CO2 in the 10,000 ft3 room. The percent of 17.87 CO2 is 10,000 ‚ 100 œ 0.18% 4.
damvb damvb dm dm dv dm dm dm dv dm dt œ F av ub dt Ê F œ dt av ub dt Ê F œ m dt v dt v dt u dt Ê F œ m dt u dt . dm dt œ b Ê m œ kbkt C. At t œ 0, m œ m0 , so C œ m0 and m œ m0 kbkt. u kb k m0 kbkt dv Thus, F œ am0 kbktb dv dt ukbk œ am0 kbktbkgk Ê dt œ g m0 kbkt Ê v œ gt u lnŠ m0 ‹ C1
v œ 0 at t œ 0 Ê C1 œ 0. So v œ gt u lnŠ m0 m0kbkt ‹ œ t œ 0 Ê y œ "# gt2 c’ t Š
m0 kbkt m0 kbkt kbk ‹ lnŠ m0 ‹
dy dt
Ê y œ ' ’ gt u lnŠ
m0 kbkt m0 ‹
“dt and u œ c, y œ 0 at
“
' 5. (a) Let y be any function such that vaxby œ ' vaxbQaxb dx C, vaxb œ e Paxb dx . Then ' Paxb dx ' d w w Ê v w axb œ œ e Paxb dx Paxb œ vaxbPaxb. dx avaxb † yb œ vaxb † y y † v axb œ vaxbQaxb. We have vaxb œ e
Thus vaxb † y w y † vaxb Paxb œ vaxbQaxb Ê y w y Paxb œ Qaxb Ê the given y is a solution.
(b) If v and Q are continuous on c a, b d and x − aa, bb, then Ê
d dx ’
'xx vatbQatb dt“ œ vaxbQaxb 0
'xx vatbQatb dt œ ' vaxbQaxb dx. So C œ y0 vax0 b ' vaxbQaxb dx. From part (a), vaxby œ ' vaxbQaxb dx C. 0
Substituting for C: vaxby œ ' vaxbQaxb dx y0 vax0 b ' vaxbQaxb dx Ê vaxby œ y0 vax0 b when x œ x0 . 6. (a) y w Paxby œ 0, yax0 b œ 0. Use vaxb œ e' Paxb dx as an integrating factor. Then
d dx avaxbyb
œ 0 Ê vaxby œ C
Ê y œ Ce' Paxb dx and y1 œ C1 e' Paxb dx , y2 œ C# e' Paxb dx , y1 ax0 b œ y2 ax0 b œ 0, y1 y2 œ aC1 C2 be' Paxb dx œ C3 e (b)
' Paxb dx
d y axb dx avaxbc 1
and y1 y2 œ 0 0 œ 0. So y1 y2 is a solution to y w Paxby œ 0 with yax0 b œ 0.
y2 axb db œ
' Paxb dx ' Paxb dx d e aC1 dx Še
C2 b ‘‹ œ
d dx aC1
C2 b œ
d dx aC3 b
œ !.
' dxd avaxbc y1 axb y2 axb dbdx œ avaxbc y1 axb y2 axb db œ ' ! dx œ C ' ' ' ' (c) y1 œ C1 e Paxb dx , y2 œ C# e Paxb dx , y œ y1 y2 . So yax0 b œ 0 Ê C1 e Paxb dx C# e Paxb dx œ ! Ê C1 C2 œ 0 Ê C1 œ C2 Ê y1 axb œ y2 axb for a x b.
CHAPTER 10 CONIC SECTIONS AND POLAR COORDINATES 10.1 CONIC SECTIONS AND QUADRATIC EQUATIONS 1. x œ
y# 8
Ê 4p œ 8 Ê p œ 2; focus is (2ß 0), directrix is x œ 2 #
2. x œ y4 Ê 4p œ 4 Ê p œ 1; focus is (1ß 0), directrix is x œ 1 #
3. y œ x6 Ê 4p œ 6 Ê p œ 4. y œ
x# 2
Ê 4p œ 2 Ê p œ
1 #
3 #
; focus is ˆ!ß 3# ‰ , directrix is y œ
3 #
; focus is ˆ!ß 1# ‰ , directrix is y œ 1#
5.
x# 4
y# 9
œ 1 Ê c œ È4 9 œ È13 Ê foci are Š „ È13ß !‹ ; vertices are a „ 2ß 0b ; asymptotes are y œ „ 3# x
6.
x# 4
y# 9
œ 1 Ê c œ È9 4 œ È5 Ê foci are Š0ß „ È5‹ ; vertices are a0ß „ 3b
7.
x# 2
y# œ 1 Ê c œ È2 1 œ 1 Ê foci are a „ 1ß 0b ; vertices are Š „ È2ß !‹
8.
y# 4
x# œ 1 Ê c œ È4 1 œ È5 Ê foci are Š0ß „ È5‹ ; vertices are a!ß „ 2b ; asymptotes are y œ „ 2x
9. y# œ 12x Ê x œ
y# 1#
Ê 4p œ 12 Ê p œ 3;
focus is ($ß !), directrix is x œ 3
11. x# œ 8y Ê y œ
x# 8
Ê 4p œ 8 Ê p œ 2;
focus is (!ß 2), directrix is y œ 2
#
10. x# œ 6y Ê y œ x6 Ê 4p œ 6 Ê p œ focus is ˆ!ß 3# ‰ , directrix is y œ 3#
#
3 #
;
y 12. y# œ 2x Ê x œ # Ê 4p œ 2 Ê p œ " focus is ˆ # ß !‰ , directrix is x œ "#
" #
;
620
Chapter 10 Conic Sections and Polar Coordinates
13. y œ 4x# Ê y œ
x# ˆ "4 ‰
" 4
Ê 4p œ
Ê pœ
" 16
#
14. y œ 8x# Ê y œ ˆx" ‰ Ê 4p œ
;
8
" ‰ " focus is ˆ!ß 16 , directrix is y œ 16
#
15. x œ 3y# Ê x œ ˆy" ‰ Ê 4p œ 3
focus is ˆ 1"# ß !‰ , directrix is x œ
#
#
" 3
" 1#
y 17. 16x# 25y# œ 400 Ê #x5 16 œ1 Ê c œ Èa# b# œ È25 16 œ 3
#
19. 2x# y# œ 2 Ê x# y# œ 1 Ê c œ Èa# b# œ È2 1 œ 1
" ‰ focus is ˆ!ß 32 , directrix is y œ
Ê pœ
" 1#
;
16. x œ 2y# Ê x œ
y# ˆ "# ‰
Ê 4p œ
" #
" 8
" 3#
Ê pœ
focus is ˆ 8" ß !‰ , directrix is x œ 8"
#
#
x 18. 7x# 16y# œ 112 Ê 16 y7 œ 1 Ê c œ Èa# b# œ È16 7 œ 3
#
#
Ê pœ
20. 2x# y# œ 4 Ê x# y4 œ 1 Ê c œ Èa# b# œ È4 2 œ È2
" 8
;
" 32
;
Section 10.1 Conic Sections and Quadratic Equations #
#
21. 3x# 2y# œ 6 Ê x# y3 œ 1 Ê c œ Èa# b# œ È3 2 œ 1
#
#
23. 6x# 9y# œ 54 Ê x9 y6 œ 1 Ê c œ Èa# b# œ È9 6 œ È3
#
#
x 22. 9x# 10y# œ 90 Ê 10 y9 œ 1 Ê c œ Èa# b# œ È10 9 œ 1
#
#
y x 24. 169x# 25y# œ 4225 Ê 25 169 œ1 Ê c œ Èa# b# œ È169 25 œ 12
#
25. Foci: Š „ È2ß !‹ , Vertices: a „ 2ß 0b Ê a œ 2, c œ È2 Ê b# œ a# c# œ 4 ŠÈ2‹ œ 2 Ê 26. Foci: a!ß „ 4b , Vertices: a0ß „ 5b Ê a œ 5, c œ 4 Ê b# œ 25 16 œ 9 Ê 27. x# y# œ 1 Ê c œ Èa# b# œ È1 1 œ È2 ; asymptotes are y œ „ x
x# 9
#
y# #5
œ1 #
x 28. 9x# 16y# œ 144 Ê 16 y9 œ 1 Ê c œ Èa# b# œ È16 9 œ 5; asymptotes are y œ „ 34 x
x# 4
y# #
œ1
621
622
Chapter 10 Conic Sections and Polar Coordinates
29. y# x# œ 8 Ê y8 x8 œ 1 Ê c œ Èa# b# œ È8 8 œ 4; asymptotes are y œ „ x
# # 30. y# x# œ 4 Ê y4 x4 œ 1 Ê c œ Èa# b# œ È4 4 œ 2È2; asymptotes are y œ „ x
31. 8x# 2y# œ 16 Ê x# y8 œ 1 Ê c œ Èa# b# œ È2 8 œ È10 ; asymptotes are y œ „ 2x
32. y# 3x# œ 3 Ê y3 x# œ 1 Ê c œ Èa# b# œ È3 1 œ 2; asymptotes are y œ „ È3x
# # 33. 8y# 2x# œ 16 Ê y# x8 œ 1 Ê c œ Èa# b# œ È2 8 œ È10 ; asymptotes are y œ „ x
y x 34. 64x# 36y# œ 2304 Ê 36 64 œ 1 Ê c œ Èa# b# œ È36 64 œ 10; asymptotes are y œ „ 4
#
#
#
#
#
#
#
#
3
35. Foci: Š!ß „ È2‹ , Asymptotes: y œ „ x Ê c œ È2 and
a b
œ 1 Ê a œ b Ê c# œ a# b# œ 2a# Ê 2 œ 2a#
Ê a œ 1 Ê b œ 1 Ê y# x# œ 1 36. Foci: a „ 2ß !b , Asymptotes: y œ „ Ê 4œ
4a# 3
" È3
x Ê c œ 2 and
Ê a# œ 3 Ê a œ È3 Ê b œ 1 Ê
x# 3
b a
œ
" È3
Ê bœ
a È3
4 3
Ê c# œ a# b# œ a#
y# œ 1
37. Vertices: a „ 3ß 0b , Asymptotes: y œ „ 43 x Ê a œ 3 and
b a
œ
4 3
Ê bœ
(3) œ 4 Ê
38. Vertices: a!ß „ 2b , Asymptotes: y œ „ 12 x Ê a œ 2 and
a b
œ
1 2
Ê b œ 2(2) œ 4 Ê
x# 9 y# 4
y# 16 x# 16
œ1 œ1
a# 3
œ
4a# 3
Section 10.1 Conic Sections and Quadratic Equations 39. (a) y# œ 8x Ê 4p œ 8 Ê p œ 2 Ê directrix is x œ 2, focus is (#ß !), and vertex is (!ß 0); therefore the new directrix is x œ 1, the new focus is (3ß 2), and the new vertex is (1ß 2)
40. (a) x# œ 4y Ê 4p œ 4 Ê p œ 1 Ê directrix is y œ 1, focus is (!ß 1), and vertex is (!ß 0); therefore the new directrix is y œ 4, the new focus is (1ß 2), and the new vertex is (1ß 3)
41. (a)
x# 16
y# 9
œ 1 Ê center is (!ß 0), vertices are (4ß 0)
and (%ß !); c œ Èa# b# œ È7 Ê foci are ŠÈ7ß 0‹ and ŠÈ7ß !‹ ; therefore the new center is (%ß $), the new vertices are (!ß 3) and (8ß 3), and the new foci are Š4 „ È7ß $‹
42. (a)
x# 9
y# 25
œ 1 Ê center is (!ß 0), vertices are (0ß 5) and (0ß 5); c œ Èa# b# œ È16 œ 4 Ê foci are (!ß 4) and (!ß 4) ; therefore the new center is (3ß 2), the new vertices are (3ß 3) and (3ß 7), and the new foci are (3ß 2) and (3ß 6)
43. (a)
x# 16
y# 9
œ 1 Ê center is (!ß 0), vertices are (4ß 0)
and (4ß 0), and the asymptotes are x4 œ „ y3 or Èa# b# œ È25 œ 5 Ê foci are y œ „ 3x 4 ;cœ (5ß 0) and (5ß 0) ; therefore the new center is (2ß 0), the new vertices are (2ß 0) and (6ß 0), the new foci are (3ß 0) and (7ß 0), and the new asymptotes are yœ „
3(x 2) 4
623
624
Chapter 10 Conic Sections and Polar Coordinates
44. (a)
y# 4
x# 5
œ 1 Ê center is (!ß 0), vertices are (0ß 2)
and (0ß 2), and the asymptotes are yœ „
2x È5
y 2
œ „
x È5
or
; c œ Èa# b# œ È9 œ 3 Ê foci are
(0ß 3) and (0ß 3) ; therefore the new center is (0ß 2), the new vertices are (0ß 4) and (0ß 0), the new foci are (0ß 1) and (0ß 5), and the new asymptotes are 2x y2œ „ È 5 45. y# œ 4x Ê 4p œ 4 Ê p œ 1 Ê focus is ("ß 0), directrix is x œ 1, and vertex is (0ß 0); therefore the new vertex is (2ß 3), the new focus is (1ß 3), and the new directrix is x œ 3; the new equation is (y 3)# œ 4(x 2) 46. y# œ 12x Ê 4p œ 12 Ê p œ 3 Ê focus is (3ß 0), directrix is x œ 3, and vertex is (0ß 0); therefore the new vertex is (4ß 3), the new focus is (1ß 3), and the new directrix is x œ 7; the new equation is (y 3)# œ 12(x 4) 47. x# œ 8y Ê 4p œ 8 Ê p œ 2 Ê focus is (0ß 2), directrix is y œ 2, and vertex is (0ß 0); therefore the new vertex is (1ß 7), the new focus is (1ß 5), and the new directrix is y œ 9; the new equation is (x 1)# œ 8(y 7) Ê focus is ˆ!ß #3 ‰ , directrix is y œ 3# , and vertex is (0ß 0); therefore the new vertex is (3ß 2), the new focus is ˆ3ß "# ‰ , and the new directrix is y œ 7# ; the new equation is
48. x# œ 6y Ê 4p œ 6 Ê p œ
3 #
(x 3)# œ 6(y 2) 49.
x# 6
y# 9
œ 1 Ê center is (!ß 0), vertices are (0ß 3) and (!ß 3); c œ Èa# b# œ È9 6 œ È3 Ê foci are Š!ß È3‹
and Š!ß È3‹ ; therefore the new center is (#ß 1), the new vertices are (2ß 2) and (#ß 4), and the new foci are Š#ß 1 „ È3‹ ; the new equation is 50.
x# 2
(x 2)# 6
(y 1)# 9
œ1
y# œ 1 Ê center is (!ß 0), vertices are ŠÈ2ß !‹ and ŠÈ2ß !‹ ; c œ Èa# b# œ È2 1 œ 1 Ê foci are
(1ß 0) and ("ß !); therefore the new center is (3ß 4), the new vertices are Š3 „ È2ß 4‹ , and the new foci are (2ß 4) and (4ß 4); the new equation is 51.
x# 3
y# #
(x 3)# #
(y 4)# œ 1
œ 1 Ê center is (!ß 0), vertices are ŠÈ3ß !‹ and ŠÈ3ß !‹ ; c œ Èa# b# œ È3 2 œ 1 Ê foci are
(1ß 0) and ("ß !); therefore the new center is (2ß 3), the new vertices are Š2 „ È3ß 3‹ , and the new foci are (1ß 3) and (3ß 3); the new equation is 52.
x# 16
y# #5
(x 2)# 3
(y 3)# #
œ1
œ 1 Ê center is (!ß 0), vertices are (!ß &) and (!ß 5); c œ Èa# b# œ È25 16 œ 3 Ê foci are
(0ß 3) and (0ß 3); therefore the new center is (4ß 5), the new vertices are (4ß 0) and (4ß 10), and the new foci are (4ß 2) and (4ß 8); the new equation is 53.
x# 4
y# 5
(x 4)# 16
(y 5)# #5
œ1
œ 1 Ê center is (!ß 0), vertices are (2ß 0) and (2ß 0); c œ Èa# b# œ È4 5 œ 3 Ê foci are ($ß !) and
(3ß 0); the asymptotes are „
x #
œ
y È5
Ê yœ „
È5x #
; therefore the new center is (2ß 2), the new vertices are
(4ß 2) and (0ß 2), and the new foci are (5ß 2) and (1ß 2); the new asymptotes are y 2 œ „
È5 (x 2) #
; the new
Section 10.1 Conic Sections and Quadratic Equations equation is 54.
x# 16
y# 9
(x 2)# 4
(y 2)# 5
625
œ1
œ 1 Ê center is (!ß 0), vertices are (4ß 0) and (4ß 0); c œ Èa# b# œ È16 9 œ 5 Ê foci are (5ß !)
and (5ß 0); the asymptotes are „
x 4
œ
Ê yœ „
y 3
3x 4
; therefore the new center is (5ß 1), the new vertices are
(1ß 1) and (9ß 1), and the new foci are (10ß 1) and (0ß 1); the new asymptotes are y 1 œ „ the new equation is
(x 5)# 16
(y 1)# 9
3(x 5) 4
;
œ1
55. y# x# œ 1 Ê center is (!ß 0), vertices are (0ß 1) and (0ß 1); c œ Èa# b# œ È1 1 œ È2 Ê foci are Š!ß „ È2‹ ; the asymptotes are y œ „ x; therefore the new center is (1ß 1), the new vertices are (1ß 0) and (1ß 2), and the new foci are Š1ß 1 „ È2‹ ; the new asymptotes are y 1 œ „ (x 1); the new equation is (y 1)# (x 1)# œ 1 56.
y# 3
x# œ 1 Ê center is (!ß 0), vertices are Š0ß È3‹ and Š!ß È3‹ ; c œ Èa# b# œ È3 1 œ 2 Ê foci are (!ß #)
and (!ß 2); the asymptotes are „ x œ
y È3
Ê y œ „ È3x; therefore the new center is (1ß 3), the new vertices
are Š"ß $ „ È3‹ , and the new foci are ("ß &) and (1ß 1); the new asymptotes are y 3 œ „ È3 (x 1); the new equation is
(y 3)# 3
(x 1)# œ 1
57. x# 4x y# œ 12 Ê x# 4x 4 y# œ 12 4 Ê (x 2)# y# œ 16; this is a circle: center at C(2ß 0), a œ 4 58. 2x# 2y# 28x 12y 114 œ 0 Ê x# 14x 49 y# 6y 9 œ 57 49 9 Ê (x 7)# (y 3)# œ 1; this is a circle: center at C(7ß 3), a œ 1 59. x# 2x 4y 3 œ 0 Ê x# 2x 1 œ 4y 3 1 Ê (x 1)# œ 4(y 1); this is a parabola: V(1ß 1), F(1ß 0) 60. y# 4y 8x 12 œ 0 Ê y# 4y 4 œ 8x 12 4 Ê (y 2)# œ 8(x 2); this is a parabola: V(#ß 2), F(!ß #) 61. x# 5y# 4x œ 1 Ê x# 4x 4 5y# œ 5 Ê (x 2)# 5y# œ 5 Ê
(x 2)# 5
y# œ 1; this is an ellipse: the
center is (2ß 0), the vertices are Š2 „ È5ß 0‹ ; c œ Èa# b# œ È5 1 œ 2 Ê the foci are (4ß 0) and (!ß 0) #
62. 9x# 6y# 36y œ 0 Ê 9x# 6 ay# 6y 9b œ 54 Ê 9x# 6(y 3)# œ 54 Ê x6 (y 9 3) œ 1; this is an ellipse: the center is (0ß 3), the vertices are (!ß 0) and (!ß 6); c œ Èa# b# œ È9 6 œ È3 Ê the foci are Š0ß 3 „ È3‹ #
63. x# 2y# 2x 4y œ 1 Ê x# 2x 1 2 ay# 2y 1b œ 2 Ê (x 1)# 2(y 1)# œ 2 # Ê (x1) (y 1)# œ 1; this is an ellipse: the center is (1ß 1), the vertices are Š" „ È2ß "‹ ; 2
c œ Èa# b# œ È2 1 œ 1 Ê the foci are (2ß 1) and (0ß 1) 64. 4x# y# 8x 2y œ 1 Ê 4 ax# 2x 1b y# 2y 1 œ 4 Ê 4(x 1)# (y 1)# œ 4 Ê (x 1)#
(y1)# 4
œ 1; this is an ellipse: the center is (1ß 1), the vertices are (1ß 3) and
626
Chapter 10 Conic Sections and Polar Coordinates
(1ß 1); c œ Èa# b# œ È4 1 œ È3 Ê the foci are Š1ß " „ È3‹ 65. x# y# 2x 4y œ 4 Ê x# 2x 1 ay# 4y 4b œ 1 Ê (x 1)# (y 2)# œ 1; this is a hyperbola: the center is (1ß 2), the vertices are (2ß 2) and (!ß 2); c œ Èa# b# œ È1 1 œ È2 Ê the foci are Š1 „ È2ß #‹ ; the asymptotes are y 2 œ „ (x 1) 66. x# y# 4x 6y œ 6 Ê x# 4x 4 ay# 6y 9b œ 1 Ê (x 2)# (y 3)# œ 1; this is a hyperbola: the center is (2ß 3), the vertices are (1ß 3) and (3ß 3); c œ Èa# b# œ È1 1 œ È2 Ê the foci are Š2 „ È2ß 3‹ ; the asymptotes are y 3 œ „ (x 2) 67. 2x# y# 6y œ 3 Ê 2x# ay# 6y 9b œ 6 Ê
(y 3)# 6
x# 3
œ 1; this is a hyperbola: the center is (!ß $),
the vertices are Š!ß 3 „ È6‹ ; c œ Èa# b# œ È6 3 œ 3 Ê the foci are (0ß 6) and (!ß 0); the asymptotes are y 3 È6
œ „
x È3
Ê y œ „ È2x 3
68. y# 4x# 16x œ 24 Ê y# 4 ax# 4x 4b œ 8 Ê
y# 8
(x 2)# 2
œ 1; this is a hyperbola: the center is (2ß 0),
the vertices are Š2ß „ È8‹ ; c œ Èa# b# œ È8 2 œ È10 Ê the foci are Š2ß „ È10‹ ; the asymptotes are y È8
œ „
x 2 È2
Ê y œ „ 2(x 2)
69.
70.
71.
72.
Section 10.1 Conic Sections and Quadratic Equations
627
74. kx# y# k Ÿ 1 Ê 1 Ÿ x# y# Ÿ 1 Ê 1 Ÿ x# y# and x# y# Ÿ 1 Ê 1 y# x# and x# y# Ÿ 1
73.
75. Volume of the Parabolic Solid: V" œ '0 21x ˆh bÎ2
œ
1hb# 8
76. y œ '
; Volume of the Cone: V# œ w H
x dx œ
w H
#
Š x# ‹ C œ
wx# 2H
" 3
#
1 ˆ b# ‰ h œ
" 3
4h b#
x# ‰ dx œ 21h '0 Šx bÎ2
#
1 Š b4 ‹ h œ
1hb# 12
4x$ b# ‹
; therefore V" œ
C; y œ 0 when x œ 0 Ê 0 œ
w(0)# 2H
#
dx œ 21h ’ x2 3 #
bÎ2
x% b# “ !
V#
C Ê C œ 0; therefore y œ
wx# 2H
is the
equation of the cable's curve 77. A general equation of the circle is x# y# ax by c œ 0, so we will substitute the three given points into a c œ 1 Þ b c œ 1 ß Ê c œ 43 and a œ b œ 73 ; therefore this equation and solve the resulting system: 2a 2b c œ 8 à 3x# 3y# 7x 7y 4 œ 0 represents the circle 78. A general equation of the circle is x# y# ax by c œ 0, so we will substitute each of the three given points 2a 3b c œ 13 Þ into this equation and solve the resulting system:
3a 2b c œ 13 ß Ê a œ 2, b œ 2, and c œ 23; 4a 3b c œ 25 à
therefore x# y# 2x 2y 23 œ 0 represents the circle 79. r# œ (2 1)# (1 3)# œ 13 Ê (x 2)# (y 1)# œ 13 is an equation of the circle; the distance from the center to (1.1ß 2.8) is È(# 1.1)# (1 2.8)# œ È12.85 È13 , the radius Ê the point is inside the circle 80. (x 2)# (y 1)# œ 5 Ê 2(x 2) 2(y 1)
dy dx
œ0 Ê
dy dx
2 # # œ yx 1 ; y œ 0 Ê (x 2) (0 1) œ 5
Ê (x 2)# œ 4 Ê x œ 4 or x œ 0 Ê the circle crosses the x-axis at (4ß 0) and (!ß 0); x œ 0 Ê (0 2)# (y 1)# œ 5 Ê (y 1)# œ 1 Ê y œ 2 or y œ 0 Ê the circle crosses the y-axis at (!ß 2) and (!ß !). At (4ß 0): At (!ß !): At (!ß #):
dy dx dy dx dy dx
2 œ 40 1 œ 2 Ê the tangent line is y œ 2(x 4) or y œ 2x 8 2 œ 00 1 œ 2 Ê the tangent line is y œ 2x
2 œ 02 1 œ 2 Ê the tangent line is y 2 œ 2x or y œ 2x 2
628
Chapter 10 Conic Sections and Polar Coordinates
81. (a) y# œ kx Ê x œ
y# k
; the volume of the solid formed by
Èkx
revolving R" about the y-axis is V" œ '0 œ
1 k#
Èkx
'0
y% dy œ
1x# Èkx 5
#
#
1 Š yk ‹ dy
; the volume of the right
circular cylinder formed by revolving PQ about the y-axis is V# œ 1x# Èkx Ê the volume of the solid formed by revolving R# about the y-axis is V$ œ V# V" œ
41x# Èkx 5
. Therefore we can see the
ratio of V$ to V" is 4:1.
(b) The volume of the solid formed by revolving R# about the x-axis is V" œ '0 1 ŠÈkt‹ dt œ 1k'0 t dt #
x
œ
1kx# #
x
. The volume of the right circular cylinder formed by revolving PS about the x-axis is #
V# œ 1 ŠÈkx‹ x œ 1kx# Ê the volume of the solid formed by revolving R" about the x-axis is 1kx# #
V$ œ V# V" œ 1kx#
œ
1kx# #
. Therefore the ratio of V$ to V" is 1:1.
82. Let P" (pß y" ) be any point on x œ p, and let P(xß y) be a point where a tangent intersects y# œ 4px. Now y# œ 4px Ê 2y
dy dx
œ 4p Ê
dy dx
œ
2p y
Ê y# yy" œ 2px 2p# . Since x œ Ê
" #
tangents from P" are m" œ
y# 4p
2p y" Èy#" 4p#
œ
dy dx
œ
#
y , we have y# yy" œ 2p Š 4p ‹ 2p# Ê y# yy" œ
2y" „ È4y#" 16p# #
y# yy" 2p# œ 0 Ê y œ
y y" x (p)
; then the slope of a tangent line from P" is
and m# œ
" #
2p y
y# 2p#
œ y" „ Èy#" 4p# . Therefore the slopes of the two 2p y" Èy#" 4p#
Ê m" m# œ
4p# y#" ay#" 4p# b
œ 1
Ê the lines are perpendicular 83. Let y œ É1
x# 4
on the interval 0 Ÿ x Ÿ 2. The area of the inscribed rectangle is given by
A(x) œ 2x Š2É1 Ê Aw (x) œ 4É1
x# 4‹ x# 4
œ 4xÉ1
x# É1 x4#
x# 4
(since the length is 2x and the height is 2y)
. Thus Aw (x) œ 0 Ê 4É1
x# 4
x# É1 x4#
œ 0 Ê 4 Š1
x# 4‹
x# œ 0 Ê x# œ 2
Ê x œ È2 (only the positive square root lies in the interval). Since A(0) œ A(2) œ 0 we have that A ŠÈ2‹ œ 4 is the maximum area when the length is 2È2 and the height is È2. 84. (a) Around the x-axis: 9x# 4y# œ 36 Ê y# œ 9 94 x# Ê y œ „ É9 94 x# and we use the positive root #
Ê V œ 2 '0 1 ŠÉ9 94 x# ‹ dx œ 2 '0 1 ˆ9 94 x# ‰ dx œ 21 9x 34 x$ ‘ ! œ 241 2
2
#
(b) Around the y-axis: 9x# 4y# œ 36 Ê x# œ 4 49 y# Ê x œ „ É4 49 y# and we use the positive root #
Ê V œ 2'0 1 ŠÉ4 49 y# ‹ dy œ 2 '0 1 ˆ4 49 y# ‰ dy œ 21 4y 3
85. 9x# 4y# œ 36 Ê y# œ œ
91 4
9x# 36 4
'24 ax# 4b dx œ 941 ’ x3
$
3
4 27
$
y$ ‘ ! œ 161
Ê y œ „ #3 Èx# 4 on the interval 2 Ÿ x Ÿ 4 Ê V œ '2 1 Š #3 Èx# 4‹ dx #
4
%
4x“ œ #
91 4
ˆ 64 ‰ ˆ8 ‰‘ œ 3 16 3 8
91 4
ˆ 56 ‰ 3 8 œ
31 4
(56 24) œ 241
86. x# y# œ 1 Ê x œ „ È1 y# on the interval 3 Ÿ y Ÿ 3 Ê V œ 'c3 1 ˆÈ1 y# ‰ dy œ 2'0 1 ˆÈ1 y# ‰ dy 3
œ 21'0 a1 y# b dy œ 21 ’y 3
$ y$ 3 “!
œ 241
#
3
#
Section 10.1 Conic Sections and Quadratic Equations 87. Let y œ É16
x# on the interval 3 Ÿ x Ÿ 3. Since the plate is symmetric about the y-axis, x œ 0. For a
16 9
É16 aµ x ßµ y b œ xß #
vertical strip:
Ê mass œ dm œ $ dA œ $É16 # É16 16 9 x
µ y dm œ
#
Š$ É16
16 9
16 9
16 9
x#
, length œ É16
16 9
x# , width œ dx Ê area œ dA œ É16
16 9
x# dx
x# ‹ dx œ $ ˆ8 98 x# ‰ dx so the moment of the plate about the x-axis is
3
3
16 9
x# dx. Moment of the strip about the x-axis:
Mx œ ' µ y dm œ 'c3 $ ˆ8 89 x# ‰ dx œ $ 8x M œ 'c3 $ É16
629
8 27
$
x$ ‘ $ œ 32$ ; also the mass of the plate is
# x# dx œ 'c3 4$ É1 ˆ "3 x‰ dx œ 4$ 'c1 3È1 u# du where u œ 3
1
x 3
Ê 3 du œ dx; x œ 3
Ê u œ 1 and x œ 3 Ê u œ 1. Hence, 4$ 'c1 3È1 u# du œ 12$ 'c1 È1 u# du 1
œ 12$ ’ "2 ŠuÈ1 u# sin" u‹“ 88. y œ Èx# 1 Ê
dy dx
" #
œ
È2
1 ' œ É 2x x# 1 Ê S œ 0 #
–
89.
u œ È2x — Ä du œ È2 dx
drA dt
œ
drB dt
Ê
d dt
21 È2
ax# 1b
"
"
1
œ 61$ Ê y œ
"Î#
(2x) œ
x È x# 1
Mx M
œ
32$ 61$
#
œ
Ê Š dy dx ‹ œ
È2
16 31
. Therefore the center of mass is ˆ!ß 3161 ‰ .
x# x # 1
#
dy Ê Ê1 Š dx ‹ œ É1
È2
dy 1 È # ' 21yÊ1 Š dx ‹ dx œ '0 21Èx# 1 É 2x x# 1 dx œ 0 21 2x 1 dx ; #
'02 Èu# 1 du œ È21
#
#
’ " ŠuÈu# 1 ln Šu Èu# 1‹‹“ œ 2 2 !
1 È2
90. (a) tan " œ mL Ê tan " œ f w (x! ) where f(x) œ È4px ; œ
2p y!
" #
(4px)"Î# (4p) œ
(b) tan 9 œ mFP œ
œ
2p È4px
Ê f w (x! ) œ
2p È4px!
Ê tan " œ
(c) tan ! œ
’2È5 ln Š2 È5‹“
(rA rB ) œ 0 Ê rA rB œ C, a constant Ê the points P(t) lie on a hyperbola with foci at A
and B
f w (x) œ
x# x# 1
2p y! . y! 0 y! x! p œ x! p
tan 9 tan " 1 tan 9 tan "
y#! 2p(x! p) y! (x! p 2p)
œ
œ
y! c y2p ‹ ! p ! y! 1 b Š x p ‹ Š y2p ‹ ! !
Šx
4px! 2px! 2p# y! (x! p)
œ
2p(x! p) y! (x! p)
œ
2p y!
91. PF will always equal PB because the string has constant length AB œ FP PA œ AP PB. 92. (a) In the labeling of the accompanying figure we have y 1 œ tan t so the coordinates of A are (1ß tan t). The coordinates of P are therefore (1 rß tan t). Since 1# y# œ (OA)# , we have 1# tan# t œ (1 r)# Ê 1 r œ È1 tan# t œ sec t Ê r œ sec t 1. The coordinates of P are therefore (xß y) œ (sec tß tan t) Ê x# y# œ sec# t tan# t œ 1
630
Chapter 10 Conic Sections and Polar Coordinates
(b) In the labeling of the accompany figure the coordinates of A are (cos tß sin t), the coordinates of C are (1ß tan t), and the coordinates of P are (1 dß tan t). By similar triangles,
d AB
œ
Ê
OC OA
d 1 cos t
œ
È1 tan# t 1
Ê d œ (1 cos t)(sec t) œ sec t 1. The coordinates of P are therefore (sec tß tan t) and P moves on the hyperbola x# y# œ 1 as in part (a).
93. x# œ 4py and y œ p Ê x# œ 4p# Ê x œ „ 2p. Therefore the line y œ p cuts the parabola at points (2pß p) and (2pß p), and these points are È[2p (2p)]# (p p)# œ 4p units apart. 94. x lim Š b x ba Èx# a# ‹ œ Ä_ a œ
b a x lim Ä_
’
x # ax # a # b “ x È x # a#
œ
b a x lim Ä_
b a x lim Ä_
’
Šx Èx# a# ‹ œ
a# “ x È x # a#
b a x lim Ä_
œ0
10.2 CLASSIFYING CONIC SECTIONS BY ECCENTRICITY # y# 1. 16x# 25y# œ 400 Ê #x5 16 œ 1 Ê c œ Èa# b# œ È25 16 œ 3 Ê e œ ca œ 35 ; F a „ 3ß 0b ;
directrices are x œ 0 „
œ „
a e
5 ˆ 35 ‰
œ „
25 3
# x# 2. 7x# 16y# œ 112 Ê 16 y7 œ 1 Ê c œ Èa# b# œ È16 7 œ 3 Ê e œ ca œ 34 ; F a „ 3ß 0b ;
directrices are x œ 0 „
œ „
a e
4 ˆ 34 ‰
œ „
16 3
3. 2x# y# œ 2 Ê x# y2 œ 1 Ê c œ Èa# b# œ È2 1 œ 1 Ê e œ ca œ È12 ; F a0ß „ 1b ; #
directrices are y œ 0 „
a e
œ „
È2 Š È12 ‹
œ „2
–
Šx Èx# a# ‹ Šx Èx# a# ‹ x È x # a#
—
Section 10.2 Classifying Conic Sections by Eccentricity 4. 2x# y# œ 4 Ê
x# #
œ 1 Ê c œ Èa# b#
y# 4
œ È4 2 œ È2 Ê e œ directrices are y œ 0 „
a e
c a
œ
È2 2
; F Š0ß „ È2‹ ;
œ „ È22 œ „ 2È2 Š ‹ 2
# # 5. 3x# 2y# œ 6 Ê x# y3 œ 1 Ê c œ Èa# b# œ È3 2 œ 1 Ê e œ ca œ È13 ; F a0ß „ 1b ;
directrices are y œ 0 „
a e
œ „
È3
œ „3
Š È13 ‹
# x# 6. 9x# 10y# œ 90 Ê 10 y9 œ 1 Ê c œ Èa# b# œ È10 9 œ 1 Ê e œ ca œ È110 ; F a „ 1ß 0b ;
directrices are x œ 0 „
7. 6x# 9y# œ 54 Ê
x# 9
a e
œ „
y# 6
œ È9 6 œ È3 Ê e œ directrices are x œ 0 „
a e
È10 Š È110 ‹
œ „ 10
œ 1 Ê c œ Èa# b# c a
œ
È3 3
; F Š „ È3ß 0‹ ;
œ „ È33 œ „ 3È3 Š ‹ 3
631
632
Chapter 10 Conic Sections and Polar Coordinates
y# x# 8. 169x# 25y# œ 4225 Ê 25 169 œ 1 Ê c œ Èa# b# œ È169 25 œ 12 Ê e œ c œ 12 ; F a0ß „ 12b ; a
directrices are y œ 0 „
a e
œ „
13
13 ˆ 12 ‰ 13
œ „
169 12
x# #7
y# 36
9. Foci: a0ß „ 3b , e œ 0.5 Ê c œ 3 and a œ
c e
œ
3 0.5
œ 6 Ê b# œ 36 9 œ 27 Ê
10. Foci: a „ 8ß 0b , e œ 0.2 Ê c œ 8 and a œ
c e
œ
8 0.#
œ 40 Ê b# œ 1600 64 œ 1536 Ê
œ1 x# 1600
y# 1536
11. Vertices: a0ß „ 70b , e œ 0.1 Ê a œ 70 and c œ ae œ 70(0.1) œ 7 Ê b# œ 4900 49 œ 4851 Ê
œ1
x# 4851
y# 4900
œ1
Šx
9 È5 ‹
12. Vertices: a „ 10ß 0b , e œ 0.24 Ê a œ 10 and c œ ae œ 10(0.24) œ 2.4 Ê b# œ 100 5.76 œ 94.24 x# 100
Ê
y# 94.24
œ1
13. Focus: ŠÈ5ß !‹ , Directrix: x œ Ê eœ
È5 3
. Then PF œ
PF œ
Ê È(x x
256 ‰ 9
Ê
a e
œ
5 9
Šx#
16 3
x
Ê
81 5 ‹
Ê c œ ae œ 4 and 0)#
œ
x# y# œ
16 3
(y
" 4
18 È5
È3 #
¸x
Ê
x# ˆ 64 ‰ 3
4 9
a e
È5 3
16 3
Ê
œ
ae e#
#
œ
ae e#
¹x
Ê
16 3 #
Ê (x 4) y œ y# ˆ 16 ‰ 3
9 È5
9 È5 ¹
x# 9
x# y# œ 4 Ê
œ
16 ¸ 3
Ê
9 È5
PD Ê ÊŠx È5‹ (y 0)# œ
14. Focus: (%ß 0), Directrix: x œ 4)#
Ê c œ ae œ È5 and #
È5 3
Ê x# 2È5 x 5 y# œ
È œ #3 PD 3 ˆ # 32 4 x 3
9 È5
y# 4
4 e# 3 4
Ê
È5 e#
œ
Ê e# œ
9 È5 #
Ê Šx È5‹ y# œ
5 9
5 9
œ1 œ
16 3
ˆx
Ê e# œ
16 ‰# 3
Ê eœ
3 4
È3 #
. Then
#
Ê x 8x 16 y#
œ1
4 " # 15. Focus: (%ß 0), Directrix: x œ 16 Ê c œ ae œ 4 and ae œ 16 Ê ae e# œ 16 Ê e# œ 16 Ê e œ 4 Ê e œ PF œ 1 PD Ê È(x 4)# (y 0)# œ 1 kx 16k Ê (x 4)# y# œ 1 (x 16)# Ê x# 8x 16 y# #
œ
1 4
#
#
ax 32x 256b Ê
3 4
#
#
x y œ 48 Ê
œ
" #
1 È2
. Then PF œ #
1 È2
y# 48
. Then
œ1
#
PD Ê ÊŠx È2‹ (y 0)# œ
Šx 2È2‹ Ê x# 2È2 x 2 y# œ
1 #
4
x# 64
16. Focus: ŠÈ2ß !‹ , Directrix: x œ 2È2 Ê c œ ae œ È2 and Ê eœ
#
" #
a e
œ 2È 2 Ê
1 È2
ae e#
œ 2È 2 Ê
È2 e#
œ 2 È 2 Ê e# œ
#
¹x 2È2¹ Ê Šx È2‹ y#
Šx# 4È2 x 8‹ Ê
" #
x# y# œ 2 Ê
x# 4
y# #
œ1
" #
Section 10.2 Classifying Conic Sections by Eccentricity 17. e œ
Ê take c œ 4 and a œ 5; c# œ a# b#
4 5
Ê 16 œ 25 b# Ê b# œ 9 Ê b œ 3; therefore x# #5
y# 9
œ1
18. The eccentricity e for Pluto is 0.25 Ê e œ
c a
œ 0.25 œ
" 4 #
Ê take c œ 1 and a œ 4; c# œ a# b# Ê 1 œ 16 b # # Ê b# œ 15 Ê b œ È15 ; therefore, x y œ 1 is a 16
15
model of Pluto's orbit.
19. One axis is from A("ß ") to B("ß 7) and is 6 units long; the other axis is from C($ß %) to D(1ß 4) and is 4 units long. Therefore a œ 3, b œ 2 and the major axis is vertical. The center is the point C("ß 4) and the ellipse is given by (x1)# 4
(y4)# 9
œ 1; c# œ a# b# œ 3# 2# œ 5
Ê c œ È5 ; therefore the foci are F Š1ß 4 „ È5‹ , the eccentricity is e œ yœ4„
a e
œ
c a
È5 3
, and the directrices are
œ 4 „ È5 œ 4 „ Š ‹ 3
9È 5 5
.
3
20. Using PF œ e † PD, we have È(x 4)# y# œ œ
4 9
ax# 18x 81b Ê
5 9
2 3
kx 9k Ê (x 4)# y# œ
x# y# œ 20 Ê 5x# 9y# œ 180 or
x# 36
#
y 20
4 9
(x 9)# Ê x# 8x 16 y#
œ 1.
21. The ellipse must pass through (!ß 0) Ê c œ 0; the point (1ß 2) lies on the ellipse Ê a 2b œ 8. The ellipse is tangent to the x-axis Ê its center is on the y-axis, so a œ 0 and b œ 4 Ê the equation is 4x# y# 4y œ 0. Next, 4x# y# 4y 4 œ 4 Ê 4x# (y 24)# œ 4 Ê x#
(y 2)# 4
standard symbols) Ê c# œ a# b# œ 4 1 œ 3 Ê c œ È3 Ê e œ 22. We first prove a result which we will use: let m" , and m# be two nonparallel, nonperpendicular lines. Let ! be the acute angle between the lines. Then tan ! œ 1m" m"mm## . To see this result, let )" be the angle of inclination of the line with slope m" , and )# be the angle of inclination of the line with slope m# . Assume m" m# . Then )" )# and we have ! œ )" )# . Then tan ! œ tan ()" )# ) )" tan )# m" m# œ 1tan tan )" tan )# œ 1 m" m# , since m" œ tan )" and and m# œ tan )# .
œ 1 Ê a œ 2 and b œ 1 (now using the c a
œ
È3 #
.
633
634
Chapter 10 Conic Sections and Polar Coordinates
Now we prove the reflective property of ellipses (see the x# a#
accompanying figure): If # #
# #
# #
b x a y œ a b and y œ
b a
y# b#
œ 1, then Èa# x# Ê yw œ
bx aÈ a# x#
.
Let P(x! ß y! ) be any point on the ellipse Ê yw (x! ) œ
bx! aÉa# x#!
œ
b # x ! a# y!
be the foci. Then mPF" œ
. Let F" (cß 0) and F# (cß 0)
y! x! c
and mPF# œ
y! x! c
. Let ! and
" be the angles between the tangent line and PF" and PF# , respectively. Then tan ! œ
Œc
b# x ! a# y!
cx
y! ! c
b# x! y! ‹ ! (x! c)
Š1 c a# y
Similarly, tan " œ
b# cy!
œ
b# x#! b# x! c a# y#! a # y ! x ! a# y! c b# x! y!
œ
b# x! c ab# x#! a# y#! b a # y ! c aa # b # b x ! y!
È2 1
œ È2 ; asymptotes are y œ „ x; F Š „ È2 ß !‹ ;
directrices are x œ 0 „
a e
œ „
" È2
# x# 24. 9x# 16y# œ 144 Ê 16 y9 œ 1 Ê c œ Èa# b# œ È16 9 œ 5 Ê e œ ca œ 54 ; asymptotes are
y œ „ 34 x; F a „ 5ß !b ; directrices are x œ 0 „ œ „
a e
"6 5
# # 25. y# x# œ 8 Ê y8 x8 œ 1 Ê c œ Èa# b# œ È8 8 œ 4 Ê e œ ca œ È48 œ È2 ; asymptotes are
y œ „ x; F a0ß „ 4b ; directrices are y œ 0 „ œ „
È8 È2
œ „2
b # x! c a# b # a # y ! c c # x ! y !
œ
b# cy!
.
. Since tan ! œ tan " , and ! and " are both less than 90°, we have ! œ " .
23. x# y# œ 1 Ê c œ Èa# b# œ È1 1 œ È2 Ê e œ œ
œ
a e
c a
Section 10.2 Classifying Conic Sections by Eccentricity y# 4
26. y# x# œ 4 Ê
x# 4
œ 1 Ê c œ Èa# b#
œ È4 4 œ 2È2 Ê e œ
c a
œ
2È 2 2
œ È2 ; asymptotes
are y œ „ x; F Š0ß „ 2È2‹ ; directrices are y œ 0 „ œ „
2 È2
a e
œ „ È2
27. 8x# 2y# œ 16 Ê
x# 2
y# 8
œ È2 8 œ È10 Ê e œ
œ 1 Ê c œ Èa# b# c a
œ
È10 È2
œ È5 ; asymptotes
are y œ „ 2x; F Š „ È10ß !‹ ; directrices are x œ 0 „ œ „
È2 È5
635
œ „
a e
2 È10
# 28. y# 3x# œ 3 Ê y3 x# œ 1 Ê c œ Èa# b# œ È3 1 œ 2 Ê e œ ca œ È23 ; asymptotes are
y œ „ È3 x; F a0ß „ 2b ; directrices are y œ 0 „ œ „
È3 Š È23 ‹
œ „
a e
3 #
29. 8y# 2x# œ 16 Ê
y# 2
x# 8
œ È2 8 œ È10 Ê e œ
œ 1 Ê c œ Èa# b# c a
œ
È10 È2
œ È5 ; asymptotes
are y œ „ x# ; F Š0ß „ È10‹ ; directrices are y œ 0 „ œ „
È2 È5
œ „
a e
2 È10
y# x# 30. 64x# 36y# œ 2304 Ê 36 64 œ 1 Ê c œ Èa# b# 5 œ È36 64 œ 10 Ê e œ ca œ 10 6 œ 3 ; asymptotes are
y œ „ 43 x; F a „ 10ß !b ; directrices are x œ 0 „ œ „
6 ˆ 53 ‰
œ „
a e
18 5
31. Vertices a!ß „ 1b and e œ 3 Ê a œ 1 and e œ
c a
œ 3 Ê c œ 3a œ 3 Ê b# œ c# a# œ 9 1 œ 8 Ê y#
x# 8
œ1
636
Chapter 10 Conic Sections and Polar Coordinates
y# 1#
œ1
œ 3 Ê c œ 3a Ê a œ 1 Ê b# œ c# a# œ 9 1 œ 8 Ê x#
y# 8
œ1
32. Vertices a „ 2ß !b and e œ 2 Ê a œ 2 and e œ 33. Foci a „ 3ß !b and e œ 3 Ê c œ 3 and e œ
c a
34. Foci a!ß „ 5b and e œ 1.25 Ê c œ 5 and e œ œ 25 16 œ 9 Ê
#
y 16
#
x 9
œ 2 Ê c œ 2a œ 4 Ê b# œ c# a# œ 16 4 œ 12 Ê
c a
c a
œ 1.25 œ
5 4
Ê cœ
a Ê 5œ
5 4
5 4
x# 4
a Ê a œ 4 Ê b# œ c# a#
œ1
4 # È2 . Then 35. Focus (4ß 0) and Directrix x œ 2 Ê c œ ae œ 4 and ae œ 2 Ê ae e# œ 2 Ê e# œ 2 Ê e œ # Ê e œ PF œ È2 PD Ê È(x 4)# (y 0)# œ È2 kx 2k Ê (x 4)# y# œ 2(x 2)# Ê x# 8x 16 y#
œ 2 ax# 4x 4b Ê x# y# œ 8 Ê
x# 8
y# 8
œ1
36. Focus ŠÈ10ß !‹ and Directrix x œ È2 Ê c œ ae œ È10 and
a e
œ È2 Ê
ae e#
œ È2 Ê
È10 e#
œ È 2 Ê e# œ È 5
#
#
Ê e œ %È5 . Then PF œ %È5 PD Ê ÊŠx È10‹ (y 0)# œ %È5 ¹x È2¹ Ê Šx È10‹ y# #
œ È5 Šx È2‹ Ê x# 2È10 x 10 y# œ È5 Šx# 2È2 x 2‹ Ê Š1 È5‹ x# y# œ 2È5 10 Ê
Š1 È5‹ x# #È5 10
y# 2È5 10
œ1 Ê
x# 2È 5
y# 10 2È5
œ1
37. Focus (2ß 0) and Directrix x œ "# Ê c œ ae œ 2 and
a e
œ
" # #
Ê
ae e#
œ
" #
Ê
PF œ 2PD Ê È(x 2)# (y 0)# œ 2 ¸x "# ¸ Ê (x 2) y# œ 4 ˆx œ 4 ˆx# x "4 ‰ Ê 3x# y# œ 3 Ê x#
#
y 3
2 e# " ‰# #
œ
" #
Ê e# œ 4 Ê e œ 2. Then
Ê x# 4x 4 y#
œ1
6 # È3. Then 38. Focus (6ß 0) and Directrix x œ # Ê c œ ae œ 6 and ae œ # Ê ae e# œ # Ê e# œ # Ê e œ 3 Ê e œ PF œ È3 PD Ê È(x 6)# (y 0)# œ È3 kx 2k Ê (x 6)# y# œ 3(x 2)# Ê x# 12x 36 y# x# 1#
œ 3 ax# 4x 4b Ê 2x# y# œ 24 Ê 39. È(x 1)# (y 3)# œ
3 #
y# 24
œ1
ky 2k Ê x# 2x 1 y# 6y 9 œ
Ê 4 ax# 2x 1b 5 ay# 12y 36b œ 4 4 180 Ê 40. c# œ a# b# Ê b# œ c# a# ; e œ x# a#
#
y b#
œ" Ê
x# a#
#
y a # ae # 1 b
c a
(y6)# 36
9 4
ay# 4y 4b Ê 4x# 5y# 8x 60y 4 œ 0
(x1)# 45
œ1
Ê c œ ea Ê c# œ e# a# Ê b# œ e# a# a# œ a# ae# 1b ; thus,
œ 1; the asymptotes of this hyperbola are y œ „ ae# 1bx Ê as e increases, the
absolute values of the slopes of the asymptotes increase and the hyperbola approaches a straight line. 41. To prove the reflective property for hyperbolas: x# a#
y# b#
œ 1 Ê a# y# œ b# x# a# b# and
dy dx
œ
xb# ya#
.
Let P(x! ß y! ) be a point of tangency (see the accompanying figure). The slope from P to F(cß 0) is x!y! c and from P to F# (cß 0) it is
y! x ! c
. Let the tangent through P meet
the x-axis in point A, and define the angles nF" PA œ ! and nF# PA œ " . We will show that tan ! œ tan " . From the preliminary result in Exercise 22,
Section 10.3 Quadratic Equations and Rotations x b#
tan ! œ
!
x b# y! 1 b Š y! a# ‹ Š x ‹ ! c ! y
tan " œ
y
! Œ y! a# c x! c
! Œ x! c
1 Šx
x! b# y! a#
y! x b# ‹ Š y! a# ‹ !c !
œ
œ
x#! b# x! b# cy#! a# x ! y ! a # y ! a # c x! y! b #
b# y! c
œ
a# b# x! b# c x ! y ! c # y! a# c
œ
b# y! c
. In a similar manner,
. Since tan ! œ tan " , and ! and " are acute angles, we have ! œ " .
42. From the accompanying figure, a ray of light emanating from the focus A that met the parabola at P would be reflected from the hyperbola as if it came directly from B (Exercise 41). The same light ray would be reflected off the ellipse to pass through B. Thus BPC is a straight line. Let " be the angle of incidence of the light ray on the hyperbola. Let ! be the angle of incidence of the light ray on the ellipse. Note that ! " is the angle between the tangent lines to the ellipse and hyperbola at P. Since BPC is a straight line, 2! 2" œ 180°. Thus ! " œ 90°. 10.3 QUADRATIC EQUATIONS AND ROTATIONS 1. x# 3xy y# x œ 0 Ê B# 4AC œ (3)# 4(1)(1) œ 5 0 Ê Hyperbola 2. 3x# 18xy 27y# 5x 7y œ 4 Ê B# 4AC œ (18)# 4(3)(27) œ 0 Ê Parabola 3. 3x# 7xy È17y# œ 1 Ê B# 4AC œ (7)# 4(3) È17 ¸ 0.477 0 Ê Ellipse #
4. 2x# È15 xy 2y# x y œ 0 Ê B# 4AC œ ŠÈ15‹ 4(2)(2) œ 1 0 Ê Ellipse 5. x# 2xy y# 2x y 2 œ 0 Ê B# 4AC œ 2# 4(1)(1) œ 0 Ê Parabola 6. 2x# y# 4xy 2x 3y œ 6 Ê B# 4AC œ 4# 4(2)(1) œ 24 0 Ê Hyperbola 7. x# 4xy 4y# 3x œ 6 Ê B# 4AC œ 4# 4(1)(4) œ 0 Ê Parabola 8. x# y# 3x 2y œ 10 Ê B# 4AC œ 0# 4(1)(1) œ 4 0 Ê Ellipse (circle) 9. xy y# 3x œ 5 Ê B# 4AC œ 1# 4(0)(1) œ 1 0 Ê Hyperbola 10. 3x# 6xy 3y# 4x 5y œ 12 Ê B# 4AC œ 6# 4(3)(3) œ 0 Ê Parabola 11. 3x# 5xy 2y# 7x 14y œ 1 Ê B# 4AC œ (5)# 4(3)(2) œ 1 0 Ê Hyperbola 12. 2x# 4.9xy 3y# 4x œ 7 Ê B# 4AC œ (4.9)# 4(2)(3) œ 0.01 0 Ê Hyperbola 13. x# 3xy 3y# 6y œ 7 Ê B# 4AC œ (3)# 4(1)(3) œ 3 0 Ê Ellipse 14. 25x# 21xy 4y# 350x œ 0 Ê B# 4AC œ 21# 4(25)(4) œ 41 0 Ê Hyperbola 15. 6x# 3xy 2y# 17y 2 œ 0 Ê B# 4AC œ 3# 4(6)(2) œ 39 0 Ê Ellipse
637
638
Chapter 10 Conic Sections and Polar Coordinates
16. 3x# 12xy 12y# 435x 9y 72 œ 0 Ê B# 4AC œ 12# 4(3)(12) œ 0 Ê Parabola 1 1 # Ê !œ 4 ; È È xw sin ! yw cos ! Ê x œ xw #2 yw #2 , y È È È È Š #2 xw #2 yw ‹ Š #2 xw #2 yw ‹ œ 2 Ê "#
17. cot 2! œ yœ Ê
18. cot 2! œ
AC B
AC B
œ
11 1
œ
therefore x œ xw cos ! yw sin !,
œ 0 Ê 2! œ
0 1
œ 0 Ê 2! œ
1 #
Ê !œ
È2 #
œ xw
xw # "# yw # œ 2 Ê xw # yw # œ 4 Ê Hyperbola 1 4
; therefore x œ xw cos ! yw sin !,
È2 w È2 w È2 # y # ,yœ x # È È È È Š #2 xw #2 yw ‹ Š #2 xw #2 yw ‹
y œ xw sin ! yw cos ! Ê x œ xw Ê Š Ê
" #
È2 #
xw
#
È2 #
yw ‹
È2 #
yw
yw
È2 # È Š #2
xw
xw # xw yw "# yw # "# xw # "# yw # "# xw # xw yw "# yw # œ 1 Ê
19. cot 2! œ
AC B
31 2È 3
œ
œ
" È3
1 3
Ê 2! œ
1 6
Ê !œ
È2 #
#
yw ‹ œ 1
xw # "# yw # œ 1 Ê 3xw # yw # œ 2 Ê Ellipse
3 #
; therefore x œ xw cos ! yw sin !,
È3 w È3 w " w 1 w # x # y,yœ # x # y È È Š #3 xw 1# yw ‹ Š 1# xw #3 yw ‹
y œ xw sin ! yw cos ! Ê x œ Ê 3Š
È3 #
#
xw 1# yw ‹ 2È3
8È3 Š "# xw 20. cot 2! œ
AC B
È3 #
œ
È3 #
È3 #
#
yw ‹ 8 Š
È3 #
xw "# yw ‹
yw ‹ œ 0 Ê 4xw # 16yw œ 0 Ê Parabola
12 È 3
œ
" È3
#
1 3
Ê 2! œ
y œ xw sin ! yw cos ! Ê x œ Ê Š
Š 1# xw
xw 1# yw ‹ È3 Š
È3 #
È3 #
1 6
Ê !œ
xw #1 yw , y œ
" #
; therefore x œ xw cos ! yw sin !,
xw È3 #
xw 1# yw ‹ Š 1# xw
È3 #
yw
yw ‹ 2 Š 1# xw
È3 #
#
yw ‹ œ 1 Ê
" #
xw # 5# yw # œ 1
Ê xw # 5yw # œ 2 Ê Ellipse 21. cot 2! œ
AC B
œ
11 2
œ 0 Ê 2! œ
y œ xw sin ! yw cos ! Ê x œ Ê Š
È2 #
xw
#
È2 #
yw ‹ 2 Š
È2 #
È2 #
1 2
Ê !œ È2 #
; therefore x œ xw cos ! yw sin !,
È2 w È2 w # x # y È2 w È2 w È2 w È2 # y ‹Š # x # y ‹ Š #
xw
xw
1 4
yw , y œ
xw
È2 #
#
yw ‹ œ 2 Ê yw # œ 1
Ê Parallel horizontal lines 22. cot 2! œ
AC B
œ
31 2 È 3
œ È"3 Ê 2! œ
y œ xw sin ! yw cos ! Ê x œ Ê 3 Š 1# xw
È3 #
#
1 #
xw
yw ‹ 2È3 Š 1# xw
È3 # È3 #
21 3
Ê !œ
yw , y œ yw ‹ Š
È3 #
È3 #
1 3
; therefore x œ xw cos ! yw sin !,
xw 1# yw
xw 1# yw ‹ Š
È3 #
#
xw 1# yw ‹ œ 1 Ê 4yw # œ 1
Ê Parallel horizontal lines 23. cot 2! œ
AC B
œ
È2 È2 2È 2
y œ xw sin ! yw cos ! Ê x œ È Ê È 2 Š # 2 xw
È2 #
Ê !œ
#
È2 #
xw
È2 #
yw , y œ
È2 #
xw
È2 #
yw ‹ œ 0 Ê 2È2xw # 8È2 yw œ 0 Ê Parabola
yw ‹ 2È2 Š
È2 #
xw
È2 #
yw ‹ 8 Š
24. cot 2! œ
AC B
œ
00 1
8Š
1 2
œ 0 Ê 2! œ
È2 #
xw
œ 0 Ê 2! œ
y œ xw sin ! yw cos ! Ê x œ
È2 #
1 2
xw
; therefore x œ xw cos ! yw sin !,
È2 w È2 w # x # y È2 w È2 w È2 w # y ‹Š # x # y ‹
Ê !œ È2 #
1 4
1 4
È2 Š
È2 #
xw
; therefore x œ xw cos ! yw sin !,
yw , y œ
È2 #
xw
È2 #
yw
È2 #
yw ‹
#
Section 10.3 Quadratic Equations and Rotations Ê Š
È2 #
È2 #
xw
yw ‹ Š
È2 #
xw
È2 #
yw ‹ Š
È2 #
xw
È2 #
yw ‹ Š
È2 #
xw
È2 #
639
yw ‹ 1 œ 0 Ê xw # yw # 2È2 xw 2
œ 0 Ê Hyperbola 25. cot 2! œ
AC B
œ
33 2
œ 0 Ê 2! œ
y œ xw sin ! yw cos ! Ê x œ Ê 3Š
È2 #
xw
È2 #
#
yw ‹ 2 Š
È2 #
È2 #
1 2
xw
xw
1 4
Ê !œ È2 #
È2 #
; therefore x œ xw cos ! yw sin !, È2 #
yw , y œ
yw ‹ Š
È2 #
xw +
xw È2 #
È2 #
yw
yw ‹ 3 Š
È2 #
xw
È2 #
#
yw ‹ œ 19 Ê 4xw # 2yw # œ 19
Ê Ellipse 26. cot 2! œ
AC B
œ
3 (1) 4È 3
œ
" È3
Ê 2! œ
È3 # È3 È 4 3Š #
y œ xw sin ! yw cos ! Ê x œ Ê 3Š
#
È3 #
xw 1# yw ‹
1 3
Ê !œ
xw #1 yw , y œ
1 #
1 6
; therefore x œ xw cos ! yw sin !,
xw
xw 1# yw ‹ Š 1# xw
È3 #
È3 #
yw
yw ‹ Š 1# xw
È3 #
#
yw ‹ œ 7 Ê 5xw # 3yw # œ 7
Ê Hyperbola 27. cot 2! œ
14 2 16
œ
3 4
Ê cos 2! œ
2! and cos ! œ É 1 cos œÉ #
28. cot 2! œ œÉ
AC B
1 ˆ 35 ‰ #
29. tan 2! œ
œ
œ
" 13 w
4" 4
2 È5
1ˆ 35 ‰ #
3 5
2! (if we choose 2! in Quadrant I); thus sin ! œ É 1 cos œÉ 2
œ
2 È5
(or sin ! œ
2 È5
and cos ! œ
1 ˆ 35 ‰ #
œ
" È5
" È5 )
2! œ 34 Ê cos 2! œ 35 (if we choose 2! in Quadrant II); thus sin ! œ É 1 cos 2
2! and cos ! œ É 1 cos œÉ #
1 ˆ 35 ‰ #
œ
1 È5
(or sin ! œ
1 È5
and cos ! œ
2 È5 )
" #
Ê 2! ¸ 26.57° Ê ! ¸ 13.28° Ê sin ! ¸ 0.23, cos ! ¸ 0.97; then Aw ¸ 0.9, Bw ¸ 0.0,
œ
" 5
œ
Cw ¸ 3.1, D ¸ 0.7, Ew ¸ 1.2, and Fw œ 3 Ê 0.9 xw # 3.1 yw # 0.7xw 1.2yw 3 œ 0, an ellipse 30. tan 2! œ
" 2 (3)
Ê 2! ¸ 11.31° Ê ! ¸ 5.65° Ê sin ! ¸ 0.10, cos ! ¸ 1.00; then Aw ¸ 2.1, Bw ¸ 0.0,
Cw ¸ 3.1, Dw ¸ 3.0, Ew ¸ 0.3, and Fw œ 7 Ê 2.1 xw # 3.1 yw # 3.0xw 0.3yw 7 œ 0, a hyperbola 31. tan 2! œ
4 14 w
œ
Ê 2! ¸ 53.13° Ê ! ¸ 26.5(° Ê sin ! ¸ 0.45, cos ! ¸ 0.89; then Aw ¸ 0.0, Bw ¸ 0.0,
4 3
Cw ¸ 5.0, D ¸ 0, Ew ¸ 0, and Fw œ 5 Ê 5.0 yw # 5 œ 0 or yw œ „ 1.0, parallel lines 32. tan 2! œ
12 2 18 w
œ
3 4
Ê 2! ¸ 36.87° Ê ! ¸ 18.43° Ê sin ! ¸ 0.32, cos ! ¸ 0.95; then Aw ¸ 0.0, Bw ¸ 0.0,
Cw ¸ 20.1, D ¸ 0, Ew ¸ 0, and Fw œ 49 Ê 20.1 yw # 49 œ 0, parallel lines 33. tan 2! œ
œ 5 Ê 2! ¸ 78.69° Ê ! ¸ 39.35° Ê sin ! ¸ 0.63, cos ! ¸ 0.77; then Aw ¸ 5.0, Bw ¸ 0.0,
5 3 2
Cw ¸ 0.05, Dw ¸ 5.0, Ew ¸ 6.2, and Fw œ 1 Ê 5.0 xw # 0.05 yw # 5.0xw 6.2yw 1 œ 0, a hyperbola 34. tan 2! œ
7 29 w
œ 1 Ê 2! ¸ 45.00° Ê ! ¸ 22.5° Ê sin ! ¸ 0.38, cos ! ¸ 0.92; then Aw ¸ 0.5, Bw ¸ 0.0,
Cw ¸ 10.4, D ¸ 18.4, Ew ¸ 7.6, and Fw œ 86 Ê 0.5 xw # 10.4ayw b# 18.4xw 7.6yw 86 œ 0, an ellipse 35. ! œ 90° Ê x œ xw cos 90° yw sin 90° œ yw and y œ xw sin 90° yw cos 90° œ xw (a)
w#
x b
#
w#
y a
#
œ1
(b)
w#
y a
#
w#
x b
#
œ1
(c) xw # yw # œ a#
(d) y œ mx Ê y mx œ 0 Ê D œ m and E œ 1; ! œ 90° Ê Dw œ 1 and Ew œ m Ê myw xw œ 0 Ê yw œ m" xw (e) y œ mx b Ê y mx b œ 0 Ê D œ m and E œ 1; ! œ 90° Ê Dw œ 1, Ew œ m and Fw œ b Ê myw xw b œ 0 Ê yw œ m" xw mb
640
Chapter 10 Conic Sections and Polar Coordinates
36. ! œ 180° Ê x œ xw cos 180° yw sin 180° œ xw and y œ xw sin 180° yw cos 180° œ yw (a)
w#
x a
#
w#
y b
#
œ1
(b)
w#
x a
#
w#
y b
#
(c) xw # yw # œ a#
œ1
(d) y œ mx Ê y mx œ 0 Ê D œ m and E œ 1; ! œ 180° Ê Dw œ m and Ew œ 1 Ê yw mxw œ 0 Ê yw œ mxw (e) y œ mx b Ê y mx b œ 0 Ê D œ m and E œ 1; ! œ 180° Ê Dw œ m, Ew œ 1 and Fw œ b Ê yw mxw b œ 0 Ê yw œ mxw b 37. (a) Aw œ cos 45° sin 45° œ Š " w# " # x # Aw œ "# , Cw œ
Ê
(b)
c a
œ
2È 2 #
œ
" #
, Bw œ 0, Cw œ cos 45° sin 45° œ "# , Fw œ 1
yw # œ 1 Ê xw # yw # œ 2 "# (see part (a) above), Dw œ Ew œ Bw œ 0, Fw œ a Ê
38. xy œ 2 Ê xw # yw # œ 4 Ê Ê eœ
È2 È2 # ‹Š # ‹
w#
x 4
w#
y 4
" #
xw # "# yw # œ a Ê xw # yw # œ 2a
œ 1 (see Exercise 37(b)) Ê a œ 2 and b œ 2 Ê c œ È4 4 œ 2È2
œ È2
39. Yes, the graph is a hyperbola: with AC 0 we have 4AC 0 and B# 4AC 0. 40. The one curve that meets all three of the stated criteria is the ellipse x# 4xy 5y# 1 œ 0. The reasoning: The symmetry about the origin means that (xß y) lies on the graph whenever (xß y) does. Adding Ax# Bxy Cy# Dx Ey F œ 0 and A(x)# B(x)(y) C(y)# D(x) E(y) F œ 0 and dividing the result by 2 produces the equivalent equation Ax# Bxy Cy# F œ 0. Substituting x œ 1, y œ 0 (because the point (1ß 0) lies on the curve) shows further that A œ F. Then Fx# Bxy Cy# F œ 0. By implicit differentiation, 2Fx By Bxyw 2Cyyw œ 0, so substituting x œ 2, y œ 1, and yw œ 0 (from Property 3) gives 4F B œ 0 Ê B œ 4F Ê the conic is Fx# 4Fxy Cy# F œ 0. Now substituting x œ 2 and y œ 1 again gives 4F 8F C F œ 0 Ê C œ 5F Ê the equation is now Fx# 4Fxy 5Fy# F œ 0. Finally, dividing through by F gives the equation x# 4xy 5y# 1 œ 0. 41. Let ! be any angle. Then Aw œ cos# ! sin# ! œ 1, Bw œ 0, Cw œ sin# ! cos# ! œ 1, Dw œ Ew œ 0 and Fw œ a# Ê xw # yw # œ a# . 42. If A œ C, then Bw œ B cos 2! (C A) sin 2! œ B cos 2!. Then ! œ
1 4
Ê 2! œ
1 #
Ê Bw œ B cos
1 #
œ 0 so the
xy-term is eliminated. 43. (a) B# 4AC œ 4# 4(1)(4) œ 0, so the discriminant indicates this conic is a parabola (b) The left-hand side of x# 4xy 4y# 6x 12y 9 œ 0 factors as a perfect square: (x 2y 3)# œ 0 Ê x 2y 3 œ 0 Ê 2y œ x 3; thus the curve is a degenerate parabola (i.e., a straight line). 44. (a) B# 4AC œ 6# 4(9)(1) œ 0, so the discriminant indicates this conic is a parabola (b) The left-hand side of 9x# 6xy y# 12x 4y 4 œ 0 factors as a perfect square: (3x y 2)# œ 0 Ê 3x y 2 œ 0 Ê y œ 3x 2; thus the curve is a degenerate parabola (i.e., a straight line).
Section 10.3 Quadratic Equations and Rotations 45. (a) B# 4AC œ 1 4(0)(0) œ 1 Ê hyperbola (b) xy 2x y œ 0 Ê y(x 1) œ 2x Ê y œ (c) y œ
2x x1
Ê
dy dx
œ
2 (x 1)#
and we want
the slope of y œ 2x Ê 2 œ (x#1)
1 dy Š dx ‹
2x x1
œ 2,
#
Ê (x 1)# œ 4 Ê x œ 3 or x œ 1; x œ 3 Ê y œ 3 Ê (3ß 3) is a point on the hyperbola where the line with slope m œ 2 is normal Ê the line is y 3 œ 2(x 3) or y œ 2x 3; x œ 1 Ê y œ 1 Ê (1ß 1) is a point on the hyperbola where the line with slope m œ 2 is normal Ê the line is y 1 œ 2(x 1) or y œ 2x 3 46. (a) False: let A œ C œ 1, B œ 2 Ê B# 4AC œ 0 Ê parabola (b) False: see part (a) above (c) True: AC 0 Ê 4AC 0 Ê B# 4AC 0 Ê hyperbola 47. Assume the ellipse has been rotated to eliminate the xy-term Ê the new equation is Aw xw # Cw yw # œ 1 Ê the semi-axes are É A" and É C" Ê the area is 1 ŠÉ A" ‹ ŠÉ C" ‹ œ w
w
w
w
1 ÈA C
œ Bw # 4Aw Cw œ 4Aw Cw (because Bw œ 0) we find that the area is
w
w
œ
21 È4A C
21 È4AC B#
w
w
. Since B# 4AC
as claimed.
48. (a) Aw Cw œ aA cos# ! B cos ! sin ! C sin# !b aA sin# ! B cos ! sin ! C sin# !b œ A acos# ! sin# !b C asin# ! cos# !b œ A C (b) Dw # Ew # œ (D cos ! E sin !)# (D sin ! E cos !)# œ D# cos# ! 2DE cos ! sin ! E# sin# ! D# sin# ! 2DE sin ! cos ! E# cos# ! œ D# acos# ! sin# !b E# asin# ! cos# !b œ D# E# 49. Bw # 4Aw Cw œ aB cos 2! (C A) sin 2!b# 4 aA cos# ! B cos ! sin ! C sin# !b aA sin# ! B cos ! sin ! C cos# !b œ B# cos# 2! 2B(C A) sin 2! cos 2! (C A)# sin# 2! 4A# cos# ! sin# ! 4AB cos$ ! sin ! 4AC cos% ! 4AB cos ! sin$ ! 4B# cos# ! sin# ! 4BC cos$ ! sin ! 4AC sin% ! 4BC cos ! sin$ ! 4C# cos# ! sin# ! # œ B cos# 2! 2BC sin 2! cos 2! 2AB sin 2! cos 2! C# sin# 2! 2AC sin# 2! A# sin# 2! 4A# cos# ! sin# ! 4AB cos$ ! sin ! 4AC cos% ! 4AB cos ! sin$ ! B# sin# 2! 4BC cos$ ! sin ! 4AC sin% ! 4BC cos ! sin$ ! 4C# cos# ! sin# ! # œ B 2BC(2 sin ! cos !) acos# ! sin# !b 2AB(2 sin ! cos !) acos# ! sin# !b C# a4 sin# ! cos# !b 2AC a4 sin# ! cos# !b A# a4 sin# ! cos# !b 4A# cos# ! sin# ! 4AB cos$ ! sin ! 4AC cos% ! 4AB cos ! sin$ ! 4BC cos$ ! sin ! 4AC sin% ! 4BC cos ! sin$ ! 4C# cos# ! sin# ! # œ B 8AC sin# ! cos# ! 4AC cos% ! 4AC sin% ! œ B# 4AC acos% ! 2 sin# ! cos# ! sin% !b œ B# 4AC acos# ! sin# !b œ B# 4AC
#
641
642
Chapter 10 Conic Sections and Polar Coordinates
10.4 CONICS AND PARAMETRIC EQUATAIONS; THE CYCLOID 1. x œ cos t, y œ sin t, 0 Ÿ t Ÿ 1 Ê cos# t sin# t œ 1 Ê x# y# œ 1
2. x œ sin (21(1 t)), y œ cos (21(1 t)), 0 Ÿ t Ÿ 1 Ê sin# (21(1 t)) cos# (21(1 t)) œ 1 Ê x# y# œ 1
3. x œ 4 cos t, y œ 5 sin t, 0 Ÿ t Ÿ 1
4. x œ 4 sin t, y œ 5 cos t, 0 Ÿ t Ÿ 21
Ê
16 cos# t 16
25 sin# t 25
x# 16
œ1 Ê
y# 25
œ1
5. x œ t, y œ Èt, t 0 Ê y œ Èx
Ê
16 sin# t 16
25 cos# t 25
œ1 Ê
x# 16
6. x œ sec# t 1, y œ tan t, 1# t Ê sec# t 1 œ tan# t Ê x œ y#
7. x œ sec t, y œ tan t, 1# t #
#
#
1 # #
Ê sec t tan t œ 1 Ê x y œ 1
y# #5
œ1
1 #
8. x œ csc t, y œ cot t, 0 t 1 Ê 1 cot# t œ csc# t Ê 1 y# œ x# Ê x# y# œ 1
Section 10.4 Conics and Parametric Equations; The Cycloid 9. x œ t, y œ È4 t# , 0 Ÿ t Ÿ 2 Ê y œ È4 x#
10. x œ t# , y œ Èt% 1, t 0 Ê y œ Èx# 1, x 0
11. x œ cosh t, y œ sinh t, _ 1 _ Ê cosh# t sinh# t œ 1 Ê x# y# œ 1
12. x œ 2 sinh t, y œ 2 cosh t, _ t _ Ê 4 cosh# t 4 sinh# t œ 4 Ê y# x# œ 4
13. Arc PF œ Arc AF since each is the distance rolled and Arc PF œ nFCP Ê Arc PF œ b(nFCP); ArcaAF œ ) b Ê Arc AF œ a) Ê a) œ b(nFCP) Ê nFCP œ nOCG œ
1 #
a b
);
); nOCG œ nOCP nPCE œ nOCP ˆ 1# !‰ . Now nOCP œ 1 nFCP œ 1 ba ). Thus nOCG œ 1 ba ) œ 1 ba )
1 #
1 #
! Ê
! Ê ! œ 1 ba ) ) œ 1
1 # ) ˆ ab b )‰ .
Then x œ OG BG œ OG PE œ (a b) cos ) b cos ! œ (a b) cos ) b cos ˆ1 œ (a b) cos ) b cos ˆ a b b )‰ . Also y œ EG œ CG CE œ (a b) sin ) b sin !
ab b
)‰
œ (a b) sin ) b sin ˆ1 a b b )‰ œ (a b) sin ) b sin ˆ a b b )‰ . Therefore x œ (a b) cos ) b cos ˆ a b b )‰ and y œ (a b) sin ) b sin ˆ a b b )‰ . If b œ 4a , then x œ ˆa 4a ‰ cos ) œ œ œ œ
3a 4 3a 4 3a 4 3a 4
cos )
œ œ œ
3a 4 3a 4 3a 4 3a 4
cos 3) œ
3a 4
cos Š
a ˆ 4a ‰ ˆ 4a ‰
)‹
cos ) 4a (cos ) cos 2) sin ) sin 2))
cos ) a(cos )) acos# ) sin# )b (sin ))(2 sin ) cos ))b a 2a # # 4 cos ) sin ) 4 sin ) cos ) # $ ) cos$ ) 3a 4 (cos )) a1 cos )b œ a cos ); a ˆ 4a ‰ a‰ a 3a a 3a 4 sin ) 4 sin Š ˆ 4a ‰ )‹ œ 4 sin ) 4 sin 3) œ 4
cos ) cos
y œ ˆa œ
a 4 a 4 a 4 a 4
a 4
cos$ )
sin ) 4a a(sin )) acos# ) sin# )b (cos ))(2 sin ) cos ))b sin ) sin ) sin )
a 4 3a 4 3a 4
sin ) cos# ) sin ) cos# )
a 4 a 4 #
sin$ )
2a 4
cos# ) sin )
sin$ )
(sin )) a1 sin )b
a 4
sin$ ) œ a sin$ ).
sin ) 4a (sin ) cos 2) cos ) sin 2))
643
644
Chapter 10 Conic Sections and Polar Coordinates
14. P traces a hypocycloid where the larger radius is 2a and the smaller is a Ê x œ (2a a) cos ) a cos ˆ 2a a a )‰ œ 2a cos ), 0 Ÿ ) Ÿ 21, and y œ (2a a) sin ) a sin ˆ 2a a a )‰ œ a sin ) a sin ) œ 0. Therefore P traces the diameter of the circle back and forth as ) goes from 0 to 21. 15. Draw line AM in the figure and note that nAMO is a right angle since it is an inscribed angle which spans the diameter of a circle. Then AN# œ MN# AM# . Now, OA œ a, AN AM a œ tan t, and a œ sin t. Next MN œ OP Ê OP# œ AN# AM# œ a# tan# t a# sin# t Ê OP œ Èa# tan# t a# sin# t œ (a sin t)Èsec# t 1 œ a sin$ t cos t œ #
x œ OP sin t œ
a sin# t cos t #
. In triangle BPO,
a sin t tan t and
y œ OP cos t œ a sin t Ê x œ a sin# t tan t and y œ a sin# t. 16. Let the x-axis be the line the wheel rolls along with the y-axis through a low point of the trochoid (see the accompanying figure).
Let ) denote the angle through which the wheel turns. Then h œ a) and k œ a. Next introduce xw yw -axes parallel to the xy-axes and having their origin at the center C of the wheel. Then xw œ b cos ! and yw œ b sin !, where ! œ 3#1 ). It follows that xw œ b cos ˆ 3#1 )‰ œ b sin ) and yw œ b sin ˆ 3#1 )‰
œ b cos ) Ê x œ h xw œ a) b sin ) and y œ k yw œ a b cos ) are parametric equations of the trochoid.
# # # 17. D œ É(x 2)# ˆy "# ‰ Ê D# œ (x 2)# ˆy "# ‰ œ (t 2)# ˆt# "# ‰ Ê D# œ t% 4t
Ê
d aD # b dt
17 4
œ 4t$ 4 œ 0 Ê t œ 1. The second derivative is always positive for t Á 0 Ê t œ 1 gives a local
minimum for D# (and hence D) which is an absolute minimum since it is the only extremum Ê the closest point on the parabola is (1ß 1). # # 18. D œ Ɉ2 cos t 34 ‰ (sin t 0)# Ê D# œ ˆ2 cos t 34 ‰ sin# t Ê
d aD # b dt
œ 2 ˆ2 cos t 34 ‰ (2 sin t) 2 sin t cos t œ (2 sin t) ˆ3 cos t 3# ‰ œ 0 Ê 2 sin t œ 0 or 3 cos t Ê t œ 0, 1 or t œ #
#
1 3
,
51 3
. Now
#
#
d aD b dt#
œ 6 cos# t 3 cos t 6 sin# t so that #
#
#
d aD b dt#
3 #
œ0
(0) œ 3 Ê relative
#
maximum, d dtaD# b (1) œ 9 Ê relative maximum, d dtaD# b ˆ 13 ‰ œ 92 Ê relative minimum, and d # aD # b ˆ 5 1 ‰ œ 9# Ê relative minimum. Therefore both t œ 13 and t œ 531 give points on the ellipse dt# 3 È È the point ˆ 34 ß !‰ Ê Š1ß #3 ‹ and Š1ß #3 ‹ are the desired points.
closest to
Section 10.4 Conics and Parametric Equations; The Cycloid 19. (a)
(b)
(c)
20. (a)
(b)
(c)
(b)
(c)
21.
22. (a)
23. (a)
(b)
645
646
Chapter 10 Conic Sections and Polar Coordinates
24. (a)
25. (a)
(b)
(b)
(c)
26. (a)
(b)
(c)
(d)
Section 10.5 Polar Coordinates 10.5 POLAR COORDINATES 1. a, e; b, g; c, h; d, f
2. a, f; b, h; c, g; d, e
3. (a) ˆ2ß 1# 2n1‰ and ˆ2ß 1# (2n 1)1‰ , n an integer
(b) (#ß 2n1) and (#ß (2n 1)1), n an integer (c) ˆ2ß 3#1 2n1‰ and ˆ2ß 3#1 (2n 1)1‰ , n an integer
(d) (#ß (2n 1)1) and (#ß 2n1), n an integer
4. (a) ˆ3ß 14 2n1‰ and ˆ3ß 541 2n1‰ , n an integer (b) ˆ3ß 14 2n1‰ and ˆ3ß 541 2n1‰ , n an integer (c) ˆ3ß 14 2n1‰ and ˆ3ß 341 2n1‰ , n an integer (d) ˆ3ß 14 2n1‰ and ˆ3ß 341 2n1‰ , n an integer
5. (a) x œ r cos ) œ 3 cos 0 œ 3, y œ r sin ) œ 3 sin 0 œ 0 Ê Cartesian coordinates are ($ß 0) (b) x œ r cos ) œ 3 cos 0 œ 3, y œ r sin ) œ 3 sin 0 œ 0 Ê Cartesian coordinates are ($ß 0) (c) x œ r cos ) œ 2 cos 21 œ 1, y œ r sin ) œ 2 sin 21 œ È3 Ê Cartesian coordinates are Š1ß È3‹ 3
(d) x œ r cos ) œ 2 cos
71 3
3
œ 1, y œ r sin ) œ 2 sin
71 3
œ È3 Ê Cartesian coordinates are Š1ß È3‹
(e) x œ r cos ) œ 3 cos 1 œ 3, y œ r sin ) œ 3 sin 1 œ 0 Ê Cartesian coordinates are (3ß 0) (f) x œ r cos ) œ 2 cos 1 œ 1, y œ r sin ) œ 2 sin 1 œ È3 Ê Cartesian coordinates are Š1ß È3‹ 3
3
(g) x œ r cos ) œ 3 cos 21 œ 3, y œ r sin ) œ 3 sin 21 œ 0 Ê Cartesian coordinates are (3ß 0) (h) x œ r cos ) œ 2 cos ˆ 1 ‰ œ 1, y œ r sin ) œ 2 sin ˆ 1 ‰ œ È3 Ê Cartesian coordinates are Š1ß È3‹ 3
6. (a) x œ È2 cos
1 4
œ 1, y œ È2 sin
3
1 4
œ 1 Ê Cartesian coordinates are (1ß 1)
(b) x œ 1 cos 0 œ 1, y œ 1 sin 0 œ 0 Ê Cartesian coordinates are (1ß 0) (c) x œ 0 cos 1# œ 0, y œ 0 sin 1# œ 0 Ê Cartesian coordinates are (!ß 0) (d) x œ È2 cos ˆ 1 ‰ œ 1, y œ È2 sin ˆ 1 ‰ œ 1 Ê Cartesian coordinates are (1ß 1) 4
(e) x œ 3 cos
51 6
œ
4
3È 3 2
, y œ 3 sin
51 6
È
œ 3# Ê Cartesian coordinates are Š 3 # 3 ß 3# ‹
(f) x œ 5 cos ˆtan" 43 ‰ œ 3, y œ 5 sin ˆtan" 43 ‰ œ 4 Ê Cartesian coordinates are ($ß 4) (g) x œ 1 cos 71 œ 1, y œ 1 sin 71 œ 0 Ê Cartesian coordinates are (1ß 0) (h) x œ 2È3 cos 231 œ È3, y œ 2È3 sin 231 œ 3 Ê Cartesian coordinates are ŠÈ3ß 3‹
647
648
Chapter 10 Conic Sections and Polar Coordinates
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
Section 10.5 Polar Coordinates
649
22.
23. r cos ) œ 2 Ê x œ 2, vertical line through (#ß 0)
24. r sin ) œ 1 Ê y œ 1, horizontal line through (0ß 1)
25. r sin ) œ 0 Ê y œ 0, the x-axis
26. r cos ) œ 0 Ê x œ 0, the y-axis
27. r œ 4 csc ) Ê r œ
4 sin )
28. r œ 3 sec ) Ê r œ
Ê r sin ) œ 4 Ê y œ 4, a horizontal line through (0ß 4)
3 cos )
Ê r cos ) œ 3 Ê x œ 3, a vertical line through (3ß 0)
29. r cos ) r sin ) œ 1 Ê x y œ 1, line with slope m œ 1 and intercept b œ 1 30. r sin ) œ r cos ) Ê y œ x, line with slope m œ 1 and intercept b œ 0 31. r# œ 1 Ê x# y# œ 1, circle with center C œ (!ß 0) and radius 1 32. r# œ 4r sin ) Ê x# y# œ 4y Ê x# y# 4y 4 œ 4 Ê x# (y 2)# œ 4, circle with center C œ (0ß 2) and radius 2 33. r œ
5 sin )2 cos )
Ê r sin ) 2r cos ) œ 5 Ê y 2x œ 5, line with slope m œ 2 and intercept b œ 5
34. r# sin 2) œ 2 Ê 2r# sin ) cos ) œ 2 Ê (r sin ))(r cos )) œ 1 Ê xy œ 1, hyperbola with focal axis y œ x )‰ˆ " ‰ 35. r œ cot ) csc ) œ ˆ cos Ê r sin# ) œ cos ) Ê r# sin# ) œ r cos ) Ê y# œ x, parabola with vertex (0ß 0) sin ) sin )
which opens to the right sin ) ‰ 36. r œ 4 tan ) sec ) Ê r œ 4 ˆ cos Ê r cos# ) œ 4 sin ) Ê r# cos# ) œ 4r sin ) Ê x# œ 4y, parabola with #)
vertex œ (!ß 0) which opens upward
37. r œ (csc )) er cos ) Ê r sin ) œ er cos ) Ê y œ ex , graph of the natural exponential function 38. r sin ) œ ln r ln cos ) œ ln (r cos )) Ê y œ ln x, graph of the natural logarithm function 39. r# 2r# cos ) sin ) œ 1 Ê x# y# 2xy œ 1 Ê x# 2xy y# œ 1 Ê (x y)# œ 1 Ê x y œ „ 1, two parallel straight lines of slope 1 and y-intercepts b œ „ 1 40. cos# ) œ sin# ) Ê r# cos# ) œ r# sin# ) Ê x# œ y# Ê kxk œ kyk Ê „ x œ y, two perpendicular lines through the origin with slopes 1 and 1, respectively. 41. r# œ 4r cos ) Ê x# y# œ 4x Ê x# 4x y# œ 0 Ê x# 4x 4 y# œ 4 Ê (x 2)# y# œ 4, a circle with center C(2ß 0) and radius 2
650
Chapter 10 Conic Sections and Polar Coordinates
42. r# œ 6r sin ) Ê x# y# œ 6y Ê x# y# 6y œ 0 Ê x# y# 6y 9 œ 9 Ê x# (y 3)# œ 9, a circle with center C(0ß 3) and radius 3 43. r œ 8 sin ) Ê r# œ 8r sin ) Ê x# y# œ 8y Ê x# y# 8y œ 0 Ê x# y# 8y 16 œ 16 Ê x# (y 4)# œ 16, a circle with center C(0ß 4) and radius 4 44. r œ 3 cos ) Ê r# œ 3r cos ) Ê x# y# œ 3x Ê x# y# 3x œ 0 Ê x# 3x # Ê ˆx 3# ‰ y# œ
9 4
, a circle with center C ˆ 3# ß !‰ and radius
9 4
y# œ
9 4
3 #
45. r œ 2 cos ) 2 sin ) Ê r# œ 2r cos ) 2r sin ) Ê x# y# œ 2x 2y Ê x# 2x y# 2y œ 0 Ê (x 1)# (y 1)# œ 2, a circle with center C(1ß 1) and radius È2 46. r œ 2 cos ) sin ) Ê r# œ 2r cos ) r sin ) Ê x# y# œ 2x y Ê x# 2x y# y œ 0 # Ê (x 1)# ˆy "# ‰ œ 54 , a circle with center C ˆ1ß "# ‰ and radius
È5 #
È
47. r sin ˆ) 16 ‰ œ 2 Ê r ˆsin ) cos 16 cos ) sin 16 ‰ œ 2 Ê #3 r sin ) "# r cos ) œ 2 Ê Ê È3 y x œ 4, line with slope m œ " and intercept b œ 4 È3
È3 #
È3
È
48. r sin ˆ 231 )‰ œ 5 Ê r ˆsin 231 cos ) cos 231 sin )‰ œ 5 Ê #3 r cos ) "# r sin ) œ 5 Ê Ê È3 x y œ 10, line with slope m œ È3 and intercept b œ 10 49. x œ 7 Ê r cos ) œ 7 51. x œ y Ê r cos ) œ r sin ) Ê ) œ
y "# x œ 2
È3 #
x "# y œ 5
50. y œ 1 Ê r sin ) œ 1 1 4
52. x y œ 3 Ê r cos ) r sin ) œ 3
53. x# y# œ 4 Ê r# œ 4 Ê r œ 2 or r œ 2 54. x# y# œ 1 Ê r# cos# ) r# sin# ) œ 1 Ê r# acos# ) sin# )b œ 1 Ê r# cos 2) œ 1 55.
x# 9
y# 4
œ 1 Ê 4x# 9y# œ 36 Ê 4r# cos# ) 9r# sin# ) œ 36
56. xy œ 2 Ê (r cos ))(r sin )) œ 2 Ê r# cos ) sin ) œ 2 Ê 2r# cos ) sin ) œ 4 Ê r# sin 2) œ 4 57. y# œ 4x Ê r# sin# ) œ 4r cos ) Ê r sin# ) œ 4 cos ) 58. x# xy y# œ 1 Ê x# y# xy œ 1 Ê r# r# sin ) cos ) œ 1 Ê r# (1 sin ) cos )) œ 1 59. x# (y 2)# œ 4 Ê x# y# 4y 4 œ 4 Ê x# y# œ 4y Ê r# œ 4r sin ) Ê r œ 4 sin ) 60. (x 5)# y# œ 25 Ê x# 10x 25 y# œ 25 Ê x# y# œ 10x Ê r# œ 10r cos ) Ê r œ 10 cos ) 61. (x 3)# (y 1)# œ 4 Ê x# 6x 9 y# 2y 1 œ 4 Ê x# y# œ 6x 2y 6 Ê r# œ 6r cos ) 2r sin ) 6 62. (x 2)# (y 5)# œ 16 Ê x# 4x 4 y# 10y 25 œ 16 Ê x# y# œ 4x 10y 13 Ê r# œ 4r cos ) 10r sin ) 13 63. (!ß )) where ) is any angle
Section 10.6 Graphing in Polar Coordinates 64. (a) x œ a Ê r cos ) œ a Ê r œ (b) y œ b Ê r sin ) œ b Ê r œ
a cos ) b sin )
Ê r œ a sec ) Ê r œ b csc )
10.6 GRAPHING IN POLAR COORDINATES 1. 1 cos ()) œ 1 cos ) œ r Ê symmetric about the x-axis; 1 cos ()) Á r and 1 cos (1 )) œ 1 cos ) Á r Ê not symmetric about the y-axis; therefore not symmetric about the origin
2. 2 2 cos ()) œ 2 2 cos ) œ r Ê symmetric about the x-axis; 2 # cos ()) Á r and 2 2 cos (1 )) œ 2 2 cos ) Á r Ê not symmetric about the y-axis; therefore not symmetric about the origin
3. 1 sin ()) œ 1 sin ) Á r and 1 sin (1 )) œ 1 sin ) Á r Ê not symmetric about the x-axis; 1 sin (1 )) œ 1 sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin
4. 1 sin ()) œ 1 sin ) Á r and 1 sin (1 )) œ 1 sin ) Á r Ê not symmetric about the x-axis; 1 sin (1 )) œ 1 sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin
5. 2 sin ()) œ 2 sin ) Á r and 2 sin (1 )) œ 2 sin ) Á r Ê not symmetric about the x-axis; 2 sin (1 )) œ 2 sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin
651
652
Chapter 10 Conic Sections and Polar Coordinates
6. 1 2 sin ()) œ 1 2 sin ) Á r and 1 2 sin (1 )) œ 1 2 sin ) Á r Ê not symmetric about the x-axis; 1 2 sin (1 )) œ 1 2 sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin
7. sin ˆ #) ‰ œ sin ˆ #) ‰ œ r Ê symmetric about the y-axis; sin ˆ 21#) ‰ œ sin ˆ 2) ‰ , so the graph is symmetric about the x-axis, and hence the origin.
8. cos ˆ #) ‰ œ cos ˆ #) ‰ œ r Ê symmetric about the x-axis; cos ˆ 21#) ‰ œ cos ˆ 2) ‰ , so the graph is symmetric about the y-axis, and hence the origin.
9. cos ()) œ cos ) œ r# Ê (rß )) and (rß )) are on the graph when (rß )) is on the graph Ê symmetric about the x-axis and the y-axis; therefore symmetric about the origin
10. sin (1 )) œ sin ) œ r# Ê (rß 1 )) and (rß 1 )) are on the graph when (rß )) is on the graph Ê symmetric about the y-axis and the x-axis; therefore symmetric about the origin
Section 10.6 Graphing in Polar Coordinates 11. sin (1 )) œ sin ) œ r# Ê (rß 1 )) and (rß 1 )) are on the graph when (rß )) is on the graph Ê symmetric about the y-axis and the x-axis; therefore symmetric about the origin
12. cos ()) œ cos ) œ r# Ê (rß )) and (rß )) are on the graph when (rß )) is on the graph Ê symmetric about the x-axis and the y-axis; therefore symmetric about the origin
13. Since a „ rß )b are on the graph when (rß )) is on the graph ˆa „ rb# œ 4 cos 2( )) Ê r# œ 4 cos 2)‰ , the graph is symmetric about the x-axis and the y-axis Ê the graph is symmetric about the origin
14. Since (rß )) on the graph Ê (rß )) is on the graph ˆa „ rb# œ 4 sin 2) Ê r# œ 4 sin 2)‰ , the graph is symmetric about the origin. But 4 sin 2()) œ 4 sin 2) Á r# and 4 sin 2(1 )) œ 4 sin (21 2)) œ 4 sin (2)) œ 4 sin 2) Á r# Ê the graph is not symmetric about the x-axis; therefore the graph is not symmetric about the y-axis 15. Since (rß )) on the graph Ê (rß )) is on the graph ˆa „ rb# œ sin 2) Ê r# œ sin 2)‰ , the graph is symmetric about the origin. But sin 2()) œ ( sin 2)) sin 2) Á r# and sin 2(1 )) œ sin (21 2)) œ sin (2)) œ ( sin 2)) œ sin 2) Á r# Ê the graph is not symmetric about the x-axis; therefore the graph is not symmetric about the y-axis 16. Sincea „ rß )b are on the graph when (rß )) is on the graph ˆa „ rb# œ cos 2()) Ê r# œ cos 2)‰, the graph is symmetric about the x-axis and the y-axis Ê the graph is symmetric about the origin.
653
654
Chapter 10 Conic Sections and Polar Coordinates Ê r œ 1 Ê ˆ1ß 1# ‰ , and ) œ 1# Ê r œ 1 )r cos ) Ê ˆ1ß 1# ‰ ; rw œ ddr) œ sin ); Slope œ rr sin cos )r sin )
17. ) œ
1 #
w w
sin )r cos ) œ sin ) cos )r sin ) sin# ˆ 1# ‰(1) cos 1# sin 1# cos 1# (1) sin 1# #
Ê Slope at ˆ1ß
1‰ #
is
œ 1; Slope at ˆ1ß 1# ‰ is
sin# ˆ 1# ‰(1) cos ˆ 1# ‰ sin ˆ 1# ‰ cos ˆ 1# ‰(1) sin ˆ 1# ‰
œ1
18. ) œ 0 Ê r œ 1 Ê ("ß 0), and ) œ 1 Ê r œ 1 dr Ê ("ß 1); rw œ d) œ cos ); r sin )r cos ) cos ) sin )r cos ) r cos )r sin ) œ cos ) cos )r sin ) 0 sin 0(1) cos 0 cos ) sin )r cos ) Ê Slope at ("ß 0) is coscos cos )r sin ) 0(1) sin 0 cos 1 sin 1(1) cos 1 1; Slope at ("ß 1) is cos# 1(1) sin 1 œ 1
Slope œ œ œ
w w
#
#
Ê r œ 1 Ê ˆ"ß 14 ‰ ; ) œ 14 Ê r œ 1 Ê ˆ1ß 14 ‰ ; ) œ 341 Ê r œ 1 Ê ˆ"ß 341 ‰ ; ) œ 341 Ê r œ 1 Ê ˆ1ß 341 ‰ ;
19. ) œ
rw œ
1 4
dr d)
œ 2 cos 2);
Slope œ
r sin )r cos ) r cos )r sin ) w w
Ê Slope at ˆ1ß 14 ‰ is Slope at ˆ1ß 14 ‰ is Slope at ˆ1ß 341 ‰ is Slope at ˆ1ß 341 ‰ is
2 cos 2) sin )r cos ) 2 cos 2) cos )r sin ) 2 cos ˆ 1# ‰ sin ˆ 14 ‰(1) cos ˆ 14 ‰ 2 cos ˆ 1 ‰ cos ˆ 1 ‰(1) sin ˆ 1 ‰
œ
#
4
4
œ 1;
2 cos ˆ 1# ‰ sin ˆ 14 ‰(1) cos ˆ 14 ‰ 2 cos ˆ 1# ‰ cos ˆ 14 ‰(1) sin ˆ 14 ‰
2 cos Š 3#1 ‹ sin Š 341 ‹(1) cos Š 341 ‹ 2 cos Š 3#1 ‹ cos Š 341 ‹(1) sin Š 341 ‹
œ 1;
œ 1;
2 cos Š 3#1 ‹ sin Š 341 ‹(1) cos Š 341 ‹ 2 cos Š 3#1 ‹ cos Š 341 ‹(1) sin Š 341 ‹
œ 1
20. ) œ 0 Ê r œ 1 Ê (1ß 0); ) œ 12 Ê r œ 1 Ê ˆ1ß 12 ‰ ; ) œ 1# Ê r œ 1 Ê ˆ"ß 12 ‰ ; ) œ 1 Ê r œ 1 Ê (1ß 1); rw œ
dr d) œ 2 sin 2); )r cos ) 2 sin 2) sin )r cos ) Slope œ rr sin cos )r sin ) œ 2 sin 2) cos )r sin ) 2 sin 0 sin 0cos 0 Ê Slope at (1ß 0) is 2 sin 0 cos 0sin 0 , which is undefined; 2 sin 2 ˆ 1 ‰ sin ˆ 1 ‰(1) cos ˆ 1 ‰ Slope at ˆ1ß 12 ‰ is 2 sin 2 ˆ 12 ‰ cos ˆ21 ‰(1) sin ˆ 21 ‰ œ 0; w w
2
Slope at ˆ1ß 12 ‰ is Slope at ("ß 1) is
2
2
2 sin 2 ˆ 1# ‰ sin ˆ 1# ‰(1) cos ˆ 1# ‰ 2 sin 2 ˆ 1 ‰ cos ˆ 1 ‰(1) sin ˆ 1 ‰ #
2 sin 21 sin 1cos 1 2 sin 21 cos 1sin 1
#
#
œ 0;
, which is undefined
Section 10.6 Graphing in Polar Coordinates 21. (a)
(b)
22. (a)
(b)
23. (a)
(b)
24. (a)
(b)
25.
655
656
Chapter 10 Conic Sections and Polar Coordinates
26. r œ 2 sec ) Ê r œ
2 cos )
Ê r cos ) œ 2 Ê x œ 2
27.
28.
29. ˆ#ß 341 ‰ is the same point as ˆ2ß 14 ‰ ; r œ 2 sin 2 ˆ 14 ‰ œ 2 sin ˆ 1# ‰ œ 2 Ê ˆ2ß 14 ‰ is on the graph Ê ˆ#ß 341 ‰ is on the graph 30. ˆ "# ß 321 ‰ is the same point as ˆ "# ß 12 ‰ ; r œ sin Š
ˆ 1# ‰ 3 ‹
œ sin
is on the graph 31. 1 cos ) œ 1 cos ) Ê cos ) œ 0 Ê ) œ 1# , 3#1 Ê r œ 1; points of intersection are ˆ"ß 1# ‰ and ˆ"ß 3#1 ‰ . The point of intersection (!ß 0) is found by graphing.
32. 1 sin ) œ 1 sin ) Ê sin ) œ 0 Ê ) œ 0, 1 Ê r œ 1; points of intersection are (1ß 0) and (1ß 1). The point of intersection (!ß 0) is found by graphing.
1 6
œ "# Ê ˆ "# ß 1# ‰ is on the graph Ê ˆ "# ß 3#1 ‰
Section 10.6 Graphing in Polar Coordinates 33. 2 sin ) œ 2 sin 2) Ê sin ) œ sin 2) Ê sin ) œ 2 sin ) cos ) Ê sin ) 2 sin ) cos ) œ 0 Ê (sin ))(1 2 cos )) œ 0 Ê sin ) œ 0 or cos ) œ
, or 13 ; ) œ 0 or 1 Ê r œ 0, Ê r œ È3 , and ) œ 1 Ê r œ È3 ; points of
Ê ) œ 0, 1, )œ
1 3
1 3
" #
3
intersection are (!ß 0), ŠÈ3 ß 13 ‹, and ŠÈ3 ß 13 ‹
34. cos ) œ 1 cos ) Ê 2 cos ) œ 1 Ê cos ) œ
" #
Ê ) œ 13 , 13 Ê r œ "# ; points of intersection are ˆ "# ß 13 ‰ and ˆ "# , 13 ‰ . The point (0ß 0) is found by graphing.
#
35. ŠÈ2‹ œ 4 sin ) Ê
" #
œ sin ) Ê ) œ
1 6
,
51 6
; points
of intersection are ŠÈ2ß 16 ‹ and ŠÈ2ß 561 ‹ . The points ŠÈ2ß 16 ‹ and ŠÈ2ß 561 ‹ are found by graphing.
36. È2 sin ) œ È2 cos ) Ê sin ) œ cos ) Ê ) œ )œ
1 4
#
Ê r œ 1 Ê r œ „ 1 and ) œ
51 4
1 4
,
51 4
;
#
Ê r œ 1
Ê no solution for r; points of intersection are ˆ „ 1ß 14 ‰ . The points (!ß 0) and ˆ „ 1ß 341 ‰ are found by graphing.
37. 1 œ 2 sin 2) Ê sin 2) œ
" #
Ê 2) œ
1 6
,
51 6
,
131 6
,
171 6
1 Ê ) œ 12 , 5121 , 13121 , 17121 ; points of intersection are 1‰ ˆ ˆ1ß 12 , 1ß 5121 ‰ , ˆ1ß 131#1 ‰, and ˆ1ß 171#1 ‰ . No other
points are found by graphing.
657
658
Chapter 10 Conic Sections and Polar Coordinates
38. È2 cos 2) œ È2 sin 2) Ê cos 2) œ sin 2) Ê 2) œ 14 , 541 , 941 , 1341 Ê ) œ 18 , 581 , 981 , )œ
1 91 8 , 8 #
#
Ê r œ 1 Ê r œ „ 1; ) œ
51 8
,
131 8 131 8
;
Ê r œ 1 Ê no solution for r; points of intersection are ˆ1ß 18 ‰ and ˆ1ß 981 ‰ . The point of intersection (!ß 0) is found by graphing.
39. r# œ sin 2) and r# œ cos 2) are generated completely for 0 Ÿ ) Ÿ 1# . Then sin 2) œ cos 2) Ê 2) œ 14 is the only solution on that interval Ê ) œ 18 Ê r# œ sin 2 ˆ 18 ‰ œ È" Ê rœ „
" % È 2
2
" ß 1‹. % È 2 8
; points of intersection are Š „
The point of intersection (!ß 0) is found by graphing.
40. 1 sin
) #
31 #
Ê )œ )œ
71 #
œ 1 cos ,
71 #
) #
;)œ
Ê sin 31 #
Ê r œ 1 cos
intersection are Š"
) #
œ cos
Ê r œ 1 cos 71 4
œ1
È 2 31 # ß # ‹
È2 #
) #
Ê
) #
31 4
œ1
31 71 4 , 4 È2 # ;
œ
; points of
and Š1
È 2 71 # ß # ‹.
three points of intersection (0ß 0) and Š1 „
È2 #
The
ß 1# ‹ are
found by graphing and symmetry.
41. 1 œ 2 sin 2) Ê sin 2) œ
" #
Ê 2) œ
1 6
,
51 6
,
131 6
,
171 6
Ê ) œ 11# , 511# , 131#1 , 171#1 ; points of intersection are ˆ"ß 11# ‰ , ˆ"ß 511# ‰ , ˆ1ß 131#1 ‰ , and ˆ"ß 17121 ‰ . The points of intersection ˆ1ß 711# ‰ , ˆ"ß 111#1 ‰ , ˆ"ß 191#1 ‰ and ˆ"ß 231#1 ‰ are found by graphing and symmetry.
42. r# œ 2 sin 2) is completely generated on 0 Ÿ ) Ÿ " #
1 6
1 #
so
1 that 1 œ 2 sin 2) Ê sin 2) œ Ê 2) œ , 561 Ê ) œ 12 51 1 5 1 ˆ ‰ ˆ ‰ 1# ; points of intersection are 1ß 1# and "ß 1# . The 1 5 1 points of intersection ˆ"ß 1# ‰ and ˆ1ß 1# ‰ are found
,
by graphing.
43. Note that (rß )) and (rß ) 1) describe the same point in the plane. Then r œ 1 cos ) Í 1 cos () 1) œ 1 (cos ) cos 1 sin ) sin 1) œ 1 cos ) œ (1 cos )) œ r; therefore (rß )) is on the graph of r œ 1 cos ) Í (rß ) 1) is on the graph of r œ 1 cos ) Ê the answer is (a).
Section 10.6 Graphing in Polar Coordinates
44. Note that (rß )) and (rß ) 1) describe the same point in the plane. Then r œ cos 2) Í sin ˆ2() 1)) 1# ‰ œ sin ˆ2) 5#1 ‰ œ sin (2)) cos ˆ 5#1 ‰ cos (2)) sin ˆ 5#1 ‰ œ cos 2) œ r; therefore (rß )) is on the graph of r œ sin ˆ2) 1# ‰ Ê the answer is (a).
45.
47. (a)
46.
(b)
(c)
(d)
659
660
Chapter 10 Conic Sections and Polar Coordinates
48. (a)
(b)
(d)
(c)
(e)
#
#
49. (a) r# œ 4 cos ) Ê cos ) œ r4 ; r œ 1 cos ) Ê r œ 1 Š r4 ‹ Ê 0 œ r# 4r 4 Ê (r 2)# œ 0 #
Ê r œ 2; therefore cos ) œ 24 œ 1 Ê ) œ 1 Ê (2ß 1) is a point of intersection (b) r œ 0 Ê 0# œ 4 cos ) Ê cos ) œ 0 Ê ) œ 1# , 3#1 Ê ˆ!ß 1# ‰ or ˆ!ß 3#1 ‰ is on the graph; r œ 0 Ê 0 œ 1 cos ) Ê cos ) œ 1 Ê ) œ 0 Ê (0ß 0) is on the graph. Since (!ß 0) œ ˆ!ß 1# ‰ for polar coordinates, the graphs intersect at the origin. 50. (a) Let r œ f()) be symmetric about the x-axis and the y-axis. Then (rß )) on the graph Ê (rß )) is on the graph because of symmetry about the x-axis. Then (rß ())) œ (rß )) is on the graph because of symmetry about the y-axis. Therefore r œ f()) is symmetric about the origin. (b) Let r œ f()) be symmetric about the x-axis and the origin. Then (rß )) on the graph Ê (rß )) is on the graph because of symmetry about the x-axis. Then (rß )) is on the graph because of symmetry about the origin. Therefore r œ f()) is symmetric about the y-axis. (c) Let r œ f()) be symmetric about the y-axis and the origin. Then (rß )) on the graph Ê (rß )) is on the graph because of symmetry about the y-axis. Then ((r)ß )) œ (rß )) is on the graph because of symmetry about the origin. Therefore r œ f()) is symmetric about the x-axis. 51. The maximum width of the petal of the rose which lies along the x-axis is twice the largest y value of the curve on the interval 0 Ÿ ) Ÿ 14 . So we wish to maximize 2y œ 2r sin ) œ 2 cos 2) sin ) on 0 Ÿ ) Ÿ 14 . Let f()) œ 2 cos 2) sin ) œ 2 a1 2 sin# )b (sin )) œ 2 sin ) 4 sin$ ) Ê f w ()) œ 2 cos ) 12 sin# ) cos ). Then f w ()) œ 0 Ê 2 cos ) 12 sin# ) cos ) œ 0 Ê (cos )) a1 6 sin# )b œ 0 Ê cos ) œ 0 or 1 6 sin# ) œ 0 Ê ) œ sin ) œ
„1 È6 .
Since we want 0 Ÿ ) Ÿ
œ 2 Š È"6 ‹ 4 † interval 0 Ÿ ) Ÿ is
2È 6 9
" œ . We 6È 6 1 4 . Therefore the
1 4
"
, we choose ) œ sin
Š È"6 ‹
$
Ê f()) œ 2 sin ) 4 sin )
can see from the graph of r œ cos 2) that a maximum does occur in the maximum width occurs at ) œ sin" Š È"6 ‹ , and the maximum width
2È 6 9 .
52. We wish to maximize y œ r sin ) œ 2(1 cos ))(sin )) œ 2 sin ) 2 sin ) cos ). Then dy # # # d) œ 2 cos ) 2(sin ))( sin )) 2 cos ) cos ) œ 2 cos ) 2 sin ) 2 cos ) œ 2 cos ) 4 cos ) 2; thus dy # # d) œ 0 Ê 4 cos ) 2 cos ) 2 = 0 Ê 2 cos ) cos ) 1 œ 0 Ê (2 cos ) 1)(cos ) 1) œ 0 Ê cos ) œ or cos ) œ 1 Ê ) œ 13 , 531 , 1. From the graph, we can see that the maximum occurs in the first quadrant so È
we choose ) œ 13 . Then y œ 2 sin 13 2 sin 13 cos 13 œ 3 # 3 . The x-coordinate of this point is x œ r cos È œ 2 ˆ1 cos 13 ‰ ˆcos 13 ‰ œ 3# . Thus the maximum height is h œ 3 # 3 occurring at x œ 3# .
1 3
" #
1 #
or
Section 10.7 Area and Lengths in Polar Coordinates 10.7 AREA AND LENGTHS IN POLAR COORDINATES 1. A œ '0
21
" #
(4 2 cos ))# d) œ '0
21
2) ‰‘ a16 16 cos ) 4 cos# )b d) œ '0 8 8 cos ) 2 ˆ 1 cos d) # 21
" #
œ '0 (9 8 cos ) cos 2)) d) œ 9) 8 sin ) 21
2. A œ '0
21
œ
" #
a#
" #
[a(1 cos ))]# d) œ '0
21
" #
#1
" 2
sin 2)‘ ! œ 181
a# a1 2 cos ) cos# )b d) œ
" #
cos# 2) d) œ '0
4. A œ 2 '1Î4
" #
a2a# cos 2)b d) œ 2a#
1Î4
1Î4
5. A œ '0
1Î2
" #
1Î4
(4 sin 2)) d) œ '0
1Î2
6. A œ (6)(2)'0
1Î6
1 cos 4) #
d) œ
" #
)
sin 4) ‘ 1Î% 4 !
2) ‰ '021 ˆ1 2 cos ) 1 cos d) #
A œ 2 '0
1Î4
1Î#
2 sin 2) d) œ c cos 2)d ! 1Î6
1Î4
2) ‰ 4 ˆ 1 cos d) œ '0 #
1Î4
œ c2) sin
1 6
1Î% 2) d !
or
51 6
œ
1 #
4 sin# ) d)
(2 2 cos 2)) d)
1
" #
; therefore
A œ 1(1)# '1Î6
51Î6
" #
c(2 sin ))# 1# d d)
œ 1 '1Î6 ˆ2 sin# ) "# ‰ d) 51Î6
œ 1 '1Î6 ˆ1 cos 2) "# ‰ d) 51Î6
œ 1 '1Î6 ˆ "# cos 2)‰ d) œ 1 "2 )
sin 2) ‘ &1Î' # 1Î'
œ 1 ˆ 511#
41 3È 3 6
51Î6
" #
œ2 1Î'
8. r œ 1 and r œ 2 sin ) Ê 2 sin ) œ 1 Ê sin ) œ Ê )œ
1 8
(2 sin 3)) d) œ 12 '0 sin 3) d) œ 12 cos3 3) ‘ !
" #
(2 sin ))# d) œ '0
" #
œ
'11ÎÎ44 cos 2) d) œ 2a# sin22) ‘ 1Î%1Î% œ 2a#
7. r œ 2 cos ) and r œ 2 sin ) Ê 2 cos ) œ 2 sin ) Ê cos ) œ sin ) Ê ) œ 14 ; therefore 1Î4
a#
'021 ˆ 3# 2 cos ) "# cos 2)‰ d) œ "# a# 3# ) 2 sin ) 4" sin 2)‘ #!1 œ 3# 1a#
3. A œ 2 '0
œ '0
" #
sin
51 ‰ 3
1 ˆ 12
" #
sin 13 ‰ œ
œ4
661
662
Chapter 10 Conic Sections and Polar Coordinates
9. r œ 2 and r œ 2(1 cos )) Ê 2 œ 2(1 cos )) Ê cos ) œ 0 Ê ) œ „ 1# ; therefore A œ 2 '0
1Î2
œ '0
1Î2
œ '0
1Î2
œ '0
1Î2
" #
[2(1 cos ))]# d) "# area of the circle
4 a1 2 cos ) cos# )b d) ˆ "# 1‰ (2)# 4 ˆ1 2 cos )
1 cos 2) ‰ #
d) 2 1
(4 8 cos ) 2 2 cos 2)) d) 21 1Î#
œ c6) 8 sin ) sin 2)d !
21 œ 51 8
10. r œ 2(1 cos )) and r œ 2(1 cos )) Ê 1 cos ) œ 1 cos ) Ê cos ) œ 0 Ê ) œ 1# or 3#1 ; the graph also gives the point of intersection (0ß 0); therefore A œ 2 '0
1Î2
" #
[2(1 cos ))]# d) 2 '1Î2 "# [2(1 cos ))]# d) 1
œ '0
4 a1 2 cos ) cos# )b d)
œ '0
4 ˆ1 2 cos )
œ '0
(6 8 cos ) 2 cos 2)) d) '1Î2 (6 8 cos ) 2 cos 2)) d)
1Î2
'1Î2 4 a1 2 cos ) cos# )b d) 1
1Î2 1Î2
1 cos 2) ‰ #
d) '1Î2 4 ˆ1 2 cos ) 1
1 cos 2) ‰ #
d)
1
1Î#
œ c6) 8 sin ) sin 2)d !
c6) 8 sin ) sin 2)d 11Î# œ 61 16
11. r œ È3 and r# œ 6 cos 2) Ê 3 œ 6 cos 2) Ê cos 2) œ 1 6
Ê )œ
" #
(in the 1st quadrant); we use symmetry of the
graph to find the area, so A œ 4 '0 ” "# (6 cos 2)) "# ŠÈ3‹ • d) #
1Î6
œ 2 '0 (6 cos 2) 3) d) œ 2 c3 sin 2) 3)d ! 1Î6
1Î'
œ 3È3 1 12. r œ 3a cos ) and r œ a(1 cos )) Ê 3a cos ) œ a(1 cos )) Ê 3 cos ) œ 1 cos ) Ê cos ) œ "# Ê ) œ 13 or 13 ; the graph also gives the point of intersection (0ß 0); therefore A œ 2 '0
1Î3
" #
c(3a cos ))# a# (1 cos ))# d d)
œ '0 a9a# cos# ) a# 2a# cos ) a# cos# )b d) 1Î3
œ '0
1Î3
a8a# cos# ) 2a# cos ) a# b d)
œ '0 c4a# (1 cos 2)) 2a# cos ) a# d d) 1Î3
œ '0 a3a# 4a# cos 2) 2a# cos )b d) 1Î3
1Î$
œ c3a# ) 2a# sin 2) 2a# sin )d !
œ 1a# 2a# ˆ "# ‰ 2a# Š
È3 # ‹
œ a# Š1 1 È3‹
Section 10.7 Area and Lengths in Polar Coordinates
663
13. r œ 1 and r œ 2 cos ) Ê 1 œ 2 cos ) Ê cos ) œ "# Ê )œ
A œ 2'
1
21 3
in quadrant II; therefore c(2 cos ))# 1# d d) œ '21Î3 a4 cos# ) 1b d) 1
" 21Î3 #
œ '21Î3 [2(1 cos 2)) 1] d) œ '21Î3 (1 2 cos 2)) d) 1
1
œ c) sin 2)d 1#1Î$ œ 14. (a) A œ 2 '0
21Î3
œ '0
21Î3
" #
1 3
È3 #
(2 cos ) 1)# d) œ '0
21Î3
a4 cos# ) 4 cos ) 1b d) œ '0
21Î3
#1Î$
(3 2 cos 2) 4 cos )) d) œ c3) sin 2) 4 sin )d ! 3È 3 # ‹
(b) A œ Š21
Š1
3È 3 # ‹
1 6
; therefore A œ '1Î6
51Î6
51 6
or
œ '1Î6 ˆ18 51Î6
9 #
È3 #
4È 3 #
œ 21
3È 3 #
œ 1 3È3 (from 14(a) above and Example 2 in the text) " #
15. r œ 6 and r œ 3 csc ) Ê 6 sin ) œ 3 Ê sin ) œ Ê )œ
œ 21
[2(1 cos 2)) 4 cos ) 1] d)
csc# )‰ d) œ 18)
" #
a6# 9 csc# )b d)
9 #
cot )‘ 1Î'
&1Î'
œ Š151 9# È3‹ Š31 9# È3‹ œ 121 9È3
16. r# œ 6 cos 2) and r œ Ê
sec ) Ê
3 # #
9 4 %
sec# ) œ 6 cos 2) Ê
œ 2 cos% ) cos ) Ê 2 cos ) cos# )
3 8
1 6
(in the first quadrant); thus A œ 2 '0
œ 3 sin 2)
9 4
tan )‘ !
1Î6
1Î'
17. (a) r œ tan ) and r œ Š
œ 3Š
È2 # ‹
È3 # ‹
9 4È 3
œ
3È 3 #
csc ) Ê tan ) œ Š
" ˆ # 6 cos 2) 3È 3 3È 3 4 œ 4
È2 # ‹
È2 # ‹
cos ) Ê 1 cos# ) œ Š
Ê cos# ) Š
È2 # ‹
cos ) 1 œ 0 Ê cos ) œ È2 or 1 4
œ acos# )b a2 cos# ) 1b
9 4
È3 #
(the second equation has no real
sec# )‰ d) œ '0 ˆ6 cos 2) 1Î6
9 4
sec# )‰ d)
csc )
Ê sin# ) œ Š
(use the quadratic formula) Ê ) œ
3 8
or cos# ) œ "4 Ê cos ) œ „
3 4
roots) Ê ) œ
È2 #
œ cos# ) cos 2) Ê
œ 0 Ê 16 cos% ) 8 cos# ) 3 œ 0
3 8
Ê a4 cos# ) 1ba4 cos# ) 3b œ 0 Ê cos# ) œ
9 24
È2 # ‹
cos )
(the solution
in the first quadrant); therefore the area of R" is A" œ '0
1Î4
AO œ Š
" #
È2 # ‹
tan# ) d) œ csc
1 #
œ
È2 #
" #
and OB œ Š
Ê the area of R# is A# œ 2 ˆ "#
1 8
4" ‰ œ
3 #
1 4
'01Î4 asec# ) 1b d)
" #
Š
rœ (b)
lim
lim
œ
" 4
1 4
1Î%
ctan ) )d !
œ
" #
ˆtan
œ 1 Ê AB œ Ê1# Š
1 4
14 ‰ œ
È2 # ‹
#
œ
) Ä 1 Î2 c
r œ sec ) as ) Ä
" ‰ cos ) 1c #
1 8
;
; therefore the area of the region shaded in the text is
1 4
generates the arc OB of r œ tan ) but does not generate the segment AB of the line
tan ) œ _ and the line x œ 1 is r œ sec ) in polar coordinates; then sin ) ˆ cos )
" #
È2 #
csc ). Instead the interval generates the half-line from B to _ on the line r œ
) Ä 1 Î2
=
È2 È2 # ‹Š # ‹
csc
" #
. Note: The area must be found this way since no common interval generates the region. For
example, the interval 0 Ÿ ) Ÿ È2 #
È2 # ‹
œ
œ
lim
) Ä 1 Î2 c
ˆ sincos) ) 1 ‰ œ
lim
) Ä 1 Î2 c
lim
) Ä 1 Î2 c
È2 #
csc ).
(tan ) sec ))
) ‰ ˆ cos sin ) œ 0 Ê r œ tan ) approaches
Ê r œ sec ) (or x œ 1) is a vertical asymptote of r œ tan ). Similarly, r œ sec )
664
Chapter 10 Conic Sections and Polar Coordinates (or x œ 1) is a vertical asymptote of r œ tan ).
18. It is not because the circle is generated twice from ) œ 0 to 21. The area of the cardioid is A œ 2 '0
1
" #
2) (cos ) 1)# d) œ '0 acos# ) 2 cos ) 1b d) œ '0 ˆ 1 cos 2 cos ) 1‰ d) # 1
œ 32) 31 #
sin 2) 4 1 51 œ 4 4
1
2 sin )‘ ! œ
19. r œ )# , 0 Ÿ ) Ÿ È5 Ê
#
È5
È5
œ '0 k)k È)# 4 d) œ (since ) 0) '0
) œ È5 Ê u œ 9“ Ä '4
9
20. r œ
e) È2
,0Ÿ)Ÿ1 Ê
dr d)
" #
œ
1 4
. The area of the circle is A œ 1 ˆ "# ‰ œ
œ 2); therefore Length œ '0
dr d)
È5
31 #
1
Èu du œ
e) È2
" #
Ê the area requested is actually
Éa)# b# (2))# d) œ '
0
) È ) # 4 d ) ; u œ ) # 4 Ê
23 u$Î# ‘ * œ %
È5
" #
È ) % 4) # d)
du œ ) d); ) œ 0 Ê u œ 4,
19 3
; therefore Length œ '0 ÊŠ Èe 2 ‹ Š Èe 2 ‹ d) œ '0 Ê2 Š e# ‹ d) 1
#
)
#
)
1
2)
œ '0 e) d) œ e) ‘ ! œ e1 1 1
1
21. r œ 1 cos ) Ê
dr d)
œ sin ); therefore Length œ '0 È(1 cos ))# ( sin ))# d) 21
1 œ 2 '0 È2 2 cos ) d) œ 2'0 É 4(1 #cos )) d) œ 4 '0 É 1 #cos ) d) œ 4 '0 cos ˆ #) ‰ d) œ 4 2 sin 2) ‘ ! œ 8 1
1
22. r œ a sin#
) #
1
, 0 Ÿ ) Ÿ 1, a 0 Ê
œ '0 Éa# sin% 1
) #
a# sin#
) #
dr d) ) #
cos#
œ a sin
) #
cos
) #
1
# ; therefore Length œ '0 Ɉa sin# #) ‰ ˆa sin 1
d) œ '0 a ¸sin #) ¸ Ésin# 1
) #
) #
cos#
1
6 1 cos )
œ '0
1Î2
,0Ÿ)Ÿ
1 #
É (1 36 cos ))#
œ ˆsince
" 1 cos )
Ê
dr d)
œ
; therefore Length œ '0
1Î2
6 sin ) (1 cos ))#
d) œ 6 '0
1Î2
36 sin# ) a1 cos )b%
" ¸ 1cos ¸ ) É1
0
1Î2
1Î2
1Î2
1Î2
) #
d)
cos# ) sin# ) 0 on 0 Ÿ ) Ÿ 1# ‰ 6 '0 ˆ 1 "cos ) ‰ É 1 2 cos(1)cos d) ) )#
cos ) È ' œ 6 '0 ˆ 1 "cos ) ‰ É (12 2cos ) )# d) œ 6 2 0
œ 3'0 sec$
#
#
6 sin ) ʈ 1 6cos ) ‰ Š (1 cos ))# ‹ d)
sin# ) (1 cos ))#
d) œ 6'0
1Î4
d) (1 cos ))$Î#
œ 6È2 '0
1Î2
1Î% sec$ u du œ (use tables) 6 Œ sec u2tan u ‘ !
d) ˆ2 cos# #) ‰$Î# " #
'01Î4
œ 3'0
1Î2
¸sec$ #) ¸ d)
sec u du
1Î% œ 6 Š È"2 2" ln ksec u tan uk‘ ! ‹ œ 3 ’È2 ln Š1 È2‹“
24. r œ
2 1 cos )
,
1 #
Ÿ)Ÿ1 Ê
4 œ '1Î2 Ê (1 cos ) ) # Š1 1
dr d)
œ
2 sin ) (1 cos ))#
sin# ) ‹ a1 cos )b#
œ ˆsince 1 cos ) 0 on
1 #
sin ) ; therefore Length œ '1Î2 ʈ 1 2cos ) ‰ Š (12cos ))# ‹ d) 1
1
#
œ '1Î2 csc$ ˆ #) ‰ d) œ ˆsince csc 1
) #
cos ) d) È ' È ' œ 2 '1Î2 ˆ 1 "cos ) ‰ É (12 2cos ))# d) œ 2 2 1Î2 (1 cos ))$Î# œ 2 2 1Î2 1
#
d)
cos ) sin Ÿ ) Ÿ 1‰ 2 '1Î2 ˆ 1 "cos ) ‰ É 1 2 cos(1)cos ) )# 1
1
) #
0 on
1 #
1
#
#
) sin d) œ '1Î2 ¸ 1 2cos ) ¸ É (1 (1cos )cos ) )#
#
d) ˆ2 sin# )# ‰$Î#
)
d)
œ '1Î2 ¸csc$ #) ¸ d) 1
Ÿ ) Ÿ 1‰ 2 '1Î4 csc$ u du œ (use tables) 1Î2
# cos #) ‰ d)
d) œ (since 0 Ÿ ) Ÿ 1) a ' sin ˆ #) ‰ d)
1 œ 2a cos 2) ‘ ! œ 2a
23. r œ
) #
Section 10.7 Area and Lengths in Polar Coordinates 1Î# 2Œ csc u2cot u ‘ 1Î%
'11ÎÎ42
" #
1Î#
csc u du œ 2 Š È"2 2" ln kcsc u cot uk‘ 1Î% ‹ œ 2 ’ È"2
" #
ln ŠÈ2 1‹“
œ È2 ln Š1 È2‹ ) 3
25. r œ cos$ œ '0
Ê
dr d)
) 3
œ sin
; therefore Length œ '0
1Î4
) 3
cos#
Écos' ˆ 3) ‰ sin# ˆ 3) ‰ cos% ˆ 3) ‰ d) œ '
1Î4
1Î4 1cos ˆ 2) ‰ 3
#
d) œ
" #
)
3 2
2) ‘ 1Î% 3 !
sin
26. r œ È1 sin 2) , 0 Ÿ ) Ÿ 1È2 Ê 1È 2
Length œ '0 œ '0
È
1 2
É(1 sin 2))
sin 2) ' É 212sin 2) d) œ 0
27. r œ È1 cos 2) Ê 1È 2
œ '0
È
1 2
œ
dr d)
" #
cos# 2) (1 sin 2))
È
d) œ '0
1 2
1È 2
d) œ '0
2)
1È# !
#
sin 2) cos É 1 2 sin 2)1 sin 2)
#
2)
d)
œ 21
È
1È 2
cos 2) ' É 212cos 2 ) d) œ 0
21
É(1 cos 2))
È2 d) œ ’È2 )“
œ 0; Length œ '0 Èa# 0# d) œ '0 kak d) œ ca)d #!1 œ 21a
1È# !
sin# 2) (1 cos 2))
21
œ a sin ); Length œ '0 È(a cos ))# (a sin ))# d) œ '0 Èa# acos# ) sin# )b d)
dr d)
œ a cos ); Length œ '0 È(a cos ))# (a sin ))# d) œ '0 Èa# acos# ) sin# )b d)
1
œ '0 kak d) œ ca)d 1! œ 1a 1
d)
œ 21
dr d)
(b) r œ a cos ) Ê
(c) r œ a sin ) Ê
cos# ˆ 3) ‰ d)
(1 sin 2))"Î# (2 cos 2)) œ (cos 2))(1 sin 2))"Î# ; therefore
È2 d) œ ’È2 )“
#
1Î4
3 8
1 2
#
dr d)
" #
œ
# cos# 3) ‰ d) 0
(1 cos 2))"Î# (2 sin 2)); therefore Length œ '0
cos 2) sin É 1 2 cos 21) cos 2)
28. (a) r œ a Ê
dr d)
1 8
œ
) 3
ˆcos# 3) ‰ Écos# ˆ 3) ‰ sin# ˆ 3) ‰ d) œ '
1Î4
0
œ '0
Ɉcos$ 3) ‰# ˆ sin
1
1
1
œ '0 kak d) œ ca)d 1! œ 1a 1
29. r œ Ècos 2) , 0 Ÿ ) Ÿ œ '0
1Î4
1 4
Ê
dr d)
œ
" #
(cos 2))"Î# ( sin 2))(2) œ
sin 2) Ècos 2)
; therefore Surface Area
sin 2) (21r cos )) ÊŠÈcos 2)‹ Š È ‹ d) œ '0 Š21Ècos 2)‹ (cos ))Écos 2) cos 2) #
#
œ '0 Š21Ècos 2)‹ (cos ))É cos" 2) d) œ '0 1Î4
1Î4
30. r œ È2e)Î2 , 0 Ÿ ) Ÿ
1 #
Ê
dr d)
œ È2 ˆ "# ‰ e)Î2 œ
œ '0 Š21È2 e)Î2 ‹ (sin )) ÊŠÈ2 e)Î2 ‹ Š #
1Î2
1Î4
1Î%
21 cos ) d) œ c21 sin )d ! È2 #
È2 #
sin# 2) cos 2)
d)
œ 1È2
e)Î2 ; therefore Surface Area
e)Î2 ‹ d) œ '0 Š21È2 e)Î2 ‹ (sin )) É2e) "# e) d) #
1Î2
œ '0 Š21È2 e)Î2 ‹ (sin )) É 5# e) d) œ '0 Š21È2 e)Î2 ‹ (sin )) Š È52 e)Î2 ‹ d) œ 21È5 '0 e) sin ) d) 1Î2
1Î2
)
1Î#
œ 21È5 e2 (sin ) cos ))‘ !
œ 1È5 ae1Î2 1b where we integrated by parts
31. r# œ cos 2) Ê r œ „ Ècos 2) ; use r œ Ècos 2) on 0ß 14 ‘ Ê therefore Surface Area œ 2 '0 Š21Ècos 2)‹ (sin )) Écos 2) 1Î4
œ 41 '0 sin ) d) œ 41 c cos )d ! 1Î4
1Î2
È
1Î%
œ 41 ’
È2 #
dr d)
œ
sin# 2) cos 2)
" #
(cos 2))"Î# ( sin 2))(2) œ
d) œ 41 '0
(1)“ œ 21 Š2 È2‹
1Î4
sin 2) Ècos 2)
Ècos 2) (sin )) É
;
" cos 2)
d)
665
666
Chapter 10 Conic Sections and Polar Coordinates
32. r œ 2a cos ) Ê
dr d)
œ 2a sin ); therefore Surface Area œ '0 21(2a cos ))(cos ))È(2a cos ))# (2a sin ))# d) 1
œ 4a1 '0 acos# )b È4a# acos# ) sin# )b d) œ 8a1 '0 acos# )b kak d) œ 8a# 1 '0 cos# ) d) 1
1
1
2) ‰ œ 8a# 1 '0 ˆ 1 cos d) œ 4a# 1 '0 (1 cos 2)) d) œ 4a# 1 ) # 1
1
33. Let r œ f()). Then x œ f()) cos ) Ê
" 2
1
sin 2)‘ ! œ 4a# 1#
‰# œ cf w ()) cos ) f()) sin )d# œ f w ()) cos ) f()) sin ) Ê ˆ dx d)
dx d)
œ cf w ())d# cos# ) 2f w ()) f()) sin ) cos ) [f())]# sin# ); y œ f()) sin ) Ê
dy d)
#
œ f w ()) sin ) f()) cos )
# # w w # w # # Ê Š dy d) ‹ œ cf ()) sin ) f()) cos )d œ cf ())d sin ) 2f ())f()) sin ) cos ) [f())] cos ). Therefore #
# # w # # # # # w # # ˆ dx ‰# Š dy ˆ dr ‰# d) d) ‹ œ cf ())d acos ) sin )b [f())] acos ) sin )b œ cf ())d [f())] œ r d) .
' Ér# ˆ ddr) ‰# d). ‰# Š dy Thus, L œ '! ʈ dx d) d) ‹ d) œ ! "
"
#
'021 a(1 cos )) d) œ 2a1 c) sin )d #!1 œ a 21 rav œ 21"0 '0 a d) œ #"1 ca)d #!1 œ a 1Î2 1Î# rav œ ˆ 1 ‰"ˆ 1 ‰ 'c1Î2 a cos ) d) œ 1" ca sin )d 1Î# œ 2a 1
34. (a) rav œ (b) (c)
" 2 1 0
#
#
35. r œ 2f()), ! Ÿ ) Ÿ " Ê
dr d)
œ 2f w ()) Ê r# ˆ ddr) ‰ œ [2f())]# c2f w ())d# Ê Length œ '! É4[f())]# 4 cf w ())d# d) "
#
œ 2 '! É[f())]# cf w ())d# d) which is twice the length of the curve r œ f()) for ! Ÿ ) Ÿ " . "
36. Again r œ 2f()) Ê r# ˆ ddr) ‰ œ [2f())]2 c2f w ())d# Ê Surface Area œ '! 21[2f()) sin )] É4[f())]# 4 cf w ())d# d) "
#
œ 4 '! 21[f()) sin )] É[f())]# cf w ())d# d) which is four times the area of the surface generated by revolving "
r œ f()) about the x-axis for ! Ÿ ) Ÿ " . '021 r$ cos ) d) 37. x œ œ '021 r# d) 2 3
œ
2 3
2 3
'021 [a(1 cos ))]$ (cos )) d) œ '021 [a(1 cos ))]# d)
2 3
a$
'021 a1 3 cos ) 3 cos# ) cos$ )b (cos )) d) 21 a# '0 a1 2 cos ) cos# )b d)
2) ‰ 2) ‰ 3 a1 sin# )b (cos )) ˆ 1 cos a '0 ’cos ) 3 ˆ 1 cos “ d) # # 21
#
'0
21
2) ‰‘ 1 2 cos ) ˆ 1 cos d) #
œ (After considerable algebra using
" 4 # ' 15 8 1 cos 2A ‰ a 0 ˆ 12 3 cos ) 3 cos 2) 2 cos ) sin ) 12 cos 4)‰ d) 21 # " 3 ' ˆ # 2 cos ) # cos 2)‰ d) 21
the identity cos# A œ
0
œ
#1
" 8 2 2 $ ‘ a 15 12 ) 3 sin ) 3 sin 2) 3 sin ) 48 sin 4) ! #3 ) 2 sin ) "4 sin 2)‘ #1
yœ
!
2 3
2 3
'2a2a "a u$ du 31
œ
0 31
'01 r# d) œ '01 a# d) œ ca# )d !1 œ a# 1; x œ yœ
2 3
œ
5 6
a;
21 2' $ '021 r$ sin ) d) 3 0 [a(1 cos ))] (sin )) d) œ ; u œ a(1 cos )) Ê "a du œ sin ) d); ) œ 0 Ê u œ 2a; 21 31 '0 r # d )
) œ 21 Ê u œ 2ad Ä 38.
œ
‰ a ˆ 15 6 1 31
2' $ '0 r$ sin ) d) 3 0 a sin ) d) œ a# 1 '01 r# d) œ 1
1
2 3
œ 0. Therefore the centroid is aBß yb œ ˆ 56 aß 0‰ 2 3
'01 r$ cos ) d) œ '01 r# d)
a$ c cos )d 1! a# 1
œ
ˆ 43 ‰ a$ a# 1
œ
2 3
'01 a$ cos ) d)
4a 31 .
a# 1
œ
2 3
a$ c sin )d 1! a# 1
œ
0 a# 1
œ 0;
Therefore the centroid is axß yb œ ˆ0ß 34a1 ‰ .
Section 10.8 Conic Sections in Polar Coordinates
667
10.8 CONIC SECTIONS IN POLAR COORDINATES 1. r cos ˆ) 16 ‰ œ 5 Ê r ˆcos ) cos œ 10 Ê y œ È3 x 10 2. r cos ˆ) Ê
È2 #
3. r cos ˆ) Ê 1# x
31 ‰ 4
œ 2 Ê r ˆcos ) cos
x
È2 #
1 6
sin ) sin 16 ‰ œ 5 Ê
31 4
31 ‰ 4
sin ) sin
È3 #
œ2 Ê
È2 #
41 ‰ œ3 Ê 3 È3 # yœ3
r cos )
r ˆcos ) cos
41 3
41 ‰ 3
sin ) sin
È2 #
È3 3
Ê x È 3 y œ 6 Ê y œ
È2 #
r sin ) œ 4 Ê
È2
x
È2 #
È2
È2
Ê
œ 1 Ê r ˆcos ) cos
È2 2
r cos ) Ê y œ x È 2
7. r cos ˆ)
21 ‰ 3
È2 2
" È2
È3 2
31 4
sin ) sin
31 ‰ 4
œ1
r sin ) œ 1 Ê x y œ È2
œ 3 Ê r ˆcos ) cos
Ê r cos ) 1 2
È2 #
x "# y œ 5 Ê È3 x y
r sin ) œ 2
21 3
sin ) sin " #
r sin ) œ 3 Ê x
Ê x È 3 y œ 6 Ê y œ
È3 3
x 2È 3
È3 #
r sin ) œ 3
x 2È3 1 4
sin ) sin 14 ‰ œ 4
y œ 4 Ê È2 x È2 y œ 8 Ê y œ x 4È2
œ È2 Ê x y œ 2 Ê y œ 2 x
31 ‰ 4
r cos )
œ 3 Ê #1 r cos )
5. r cos ˆ) 14 ‰ œ È2 Ê r ˆcos ) cos 14 sin ) sin 14 ‰ œ È2 Ê " r cos ) " r sin ) œ È2 Ê " x
6. r cos ˆ)
È2 #
È3 #
y œ 2 Ê È2 x È2 y œ 4 Ê y œ x 2È2
4. r cos ˆ) ˆ 14 ‰‰ œ 4 Ê r cos ˆ) 14 ‰ œ 4 Ê r ˆcos ) cos Ê
r cos ) "# r sin ) œ 5 Ê
È3 #
21 ‰ 3
œ3
yœ3
y
668
Chapter 10 Conic Sections and Polar Coordinates
8. r cos ˆ) 13 ‰ œ 2 Ê r ˆcos ) cos Ê
1 2
r cos )
È3 2
r sin ) œ 2 Ê
Ê x È3 y œ 4 Ê y œ
È3 3
1 3
sin ) sin 13 ‰ œ 2
" #
x
x
È3 #
yœ2
4È 3 3
È 9. È2 x È2 y œ 6 Ê È2 r cos ) È2 r sin ) œ 6 Ê r Š #2 cos )
È2 #
sin )‹ œ 3 Ê r ˆcos
1 4
cos ) sin
œ 3 Ê r cos ˆ) 14 ‰ œ 3 È 10. È3 x y œ 1 Ê È3 r cos ) r sin ) œ 1 Ê r Š #3 cos )
œ
" #
Ê r cos ˆ) 16 ‰ œ
1 #
sin )‹ œ
" #
Ê r ˆcos
1 6
cos ) sin
1 6
sin )‰
" #
11. y œ 5 Ê r sin ) œ 5 Ê r sin ) œ 5 Ê r sin ()) œ 5 Ê r cos ˆ 1# ())‰ œ 5 Ê r cos ˆ) 1# ‰ œ 5 12. x œ 4 Ê r cos ) œ 4 Ê r cos ) œ 4 Ê r cos () 1) œ 4 13. r œ 2(4) cos ) œ 8 cos )
14. r œ 2(1) sin ) œ 2 sin )
15. r œ 2È2 sin )
16. r œ 2 ˆ "# ‰ cos ) œ cos )
17.
18.
19.
20.
1 4
sin )‰
Section 10.8 Conic Sections in Polar Coordinates 21. (x 6)# y# œ 36 Ê C œ (6ß 0), a œ 6 Ê r œ 12 cos ) is the polar equation
22. (x 2)# y# œ 4 Ê C œ (2ß 0), a œ 2 Ê r œ 4 cos ) is the polar equation
23. x# (y 5)# œ 25 Ê C œ (!ß 5), a œ 5 Ê r œ 10 sin ) is the polar equation
24. x# (y 7)# œ 49 Ê C œ (!ß 7), a œ 7 Ê r œ 14 sin ) is the polar equation
25. x# 2x y# œ 0 Ê (x 1)# y# œ 1 Ê C œ (1ß 0), a œ 1 Ê r œ 2 cos ) is the polar equation
26. x# 16x y# œ 0 Ê (x 8)# y# œ 64 Ê C œ (8ß 0), a œ 8 Ê r œ 16 cos ) is the polar equation
# 27. x# y# y œ 0 Ê x# ˆy "# ‰ œ 4" Ê C œ ˆ!ß "# ‰ , a œ "# Ê r œ sin ) is the
# 28. x# y# 43 y œ 0 Ê x# ˆy 23 ‰ œ 49 Ê C œ ˆ0ß 23 ‰ , a œ 23 Ê r œ 43 sin ) is the
polar equation
polar equation
669
670
Chapter 10 Conic Sections and Polar Coordinates
29. e œ 1, x œ 2 Ê k œ 2 Ê r œ
2(1) 1 (1) cos )
œ
2 1cos )
30. e œ 1, y œ 2 Ê k œ 2 Ê r œ
2(1) 1 (1) sin )
œ
2 1sin )
31. e œ 5, y œ 6 Ê k œ 6 Ê r œ
6(5) 1 5 sin )
32. e œ 2, x œ 4 Ê k œ 4 Ê r œ
4(2) 1 2 cos )
33. e œ "# , x œ 1 Ê k œ 1 Ê r œ
ˆ "# ‰ (1) 1 ˆ "# ‰ cos )
35. e œ "5 , x œ 10 Ê k œ 10 Ê r œ
37. r œ
" 1 cos )
38. r œ
6 2 cos )
œ
30 15 sin )
8 12 cos )
œ
ˆ "4 ‰ (2) 1 ˆ "4 ‰ cos )
34. e œ 4" , x œ 2 Ê k œ 2 Ê r œ
36. e œ "3 , y œ 6 Ê k œ 6 Ê r œ
œ
1 2cos )
œ
ˆ "5 ‰ (10) 1 ˆ "5 ‰ sin )
ˆ "3 ‰ (6) 1 ˆ "3 ‰ sin )
œ
2 4cos )
œ
10 5sin )
6 3sin )
Ê e œ 1, k œ 1 Ê x œ 1
œ
3 1 ˆ "# ‰ cos )
Ê eœ
" #
, k œ 6 Ê x œ 6;
#
a a1 e# b œ ke Ê a ’1 ˆ "# ‰ “ œ 3 Ê
3 4
aœ3
Ê a œ 4 Ê ea œ 2
39. r œ
25 10 5 cos )
Ê eœ
" #
Ê rœ
œ
ˆ #5 ‰
1 ˆ "# ‰ cos )
, k œ 5 Ê x œ 5; a a1 e# b œ ke #
Ê a ’1 ˆ "# ‰ “ œ
40. r œ
ˆ 25 ‰ 10
5 ‰ 1 ˆ 10 cos )
4 22 cos )
Ê rœ
5 #
Ê
2 1cos )
3 4
aœ
5 #
Ê aœ
10 3
Ê ea œ
5 3
Ê e œ 1, k œ 2 Ê x œ 2
Section 10.8 Conic Sections in Polar Coordinates 41. r œ eœ
400 16 8 sin ) " #
Ê rœ
ˆ 400 ‰ 16
8 ‰ 1 ˆ 16 sin )
25 1 ˆ "# ‰ sin )
, k œ 50 Ê y œ 50; a a1 c e# b œ ke #
Ê a ’1 c ˆ "# ‰ “ œ 25 Ê Ê ea œ
42. r œ
Ê rœ
a œ 25 Ê a œ
3 4
100 3
50 3
12 3 3 sin )
Ê rœ
4 1 sin )
Ê e œ 1,
43. r œ
kœ4 Ê yœ4
44. r œ
4 2 sin )
Ê rœ
2 1 ˆ "# ‰ sin )
Ê eœ
" #
,kœ4 #
3 4
Ê rœ
k œ 4 Ê y œ c4
Ê y œ c4; a a1 c e# b œ ke Ê a ’1 c ˆ "# ‰ “ œ 2 Ê
8 2 2 sin )
aœ2 Ê aœ
8 3
Ê ea œ
4 3
45.
46.
47.
48.
4 1 sin )
Ê e œ 1,
671
672
Chapter 10 Conic Sections and Polar Coordinates
49.
50.
51.
52.
53.
54.
55.
56.
57. (a) Perihelion œ a c ae œ a(1 c e), Aphelion œ ea b a œ a(1 b e) (b) Planet Perihelion Aphelion Mercury 0.3075 AU 0.4667 AU Venus 0.7184 AU 0.7282 AU Earth 0.9833 AU 1.0167 AU Mars 1.3817 AU 1.6663 AU Jupiter 4.9512 AU 5.4548 AU Saturn 9.0210 AU 10.0570 AU Uranus 18.2977 AU 20.0623 AU Neptune 29.8135 AU 30.3065 AU Pluto 29.6549 AU 49.2251 AU
Section 10.8 Conic Sections in Polar Coordinates (0.3871) a1 0.2056# b 0.3707 œ 1 0.2056 1 0.2056 cos ) cos ) (0.7233) a1 0.0068# b 0.7233 Venus: r œ 1 0.0068 cos ) œ 1 0.0068 cos ) 0.0167# b 0.9997 Earth: r œ 11a10.0167 cos ) œ 1 0.0617 cos ) a1 0.0934# b 1.511 Mars: r œ (1.524) œ 1 0.0934 1 0.0934 cos ) cos ) (5.203) a1 0.0484# b 5.191 Jupiter: r œ 1 0.0484 cos ) œ 1 0.0484 cos ) a1 0.0543# b 9.511 Saturn: r œ (9.539) œ 1 0.0543 1 0.0543 cos ) cos ) (19.18) a1 0.0460# b 19.14 Uranus: r œ 1 0.0460 cos ) œ 1 0.0460 cos ) a1 0.0082# b 30.06 Neptune: r œ (30.06) œ 1 0.0082 1 0.0082 cos ) cos )
58. Mercury: r œ
59. (a) r œ 4 sin ) Ê r# œ 4r sin ) Ê x# b y# œ 4y; È r œ È3 sec ) Ê r œ cos3) Ê r cos ) œ È3
(b)
#
Ê x œ È3 ; x œ È3 Ê ŠÈ3‹ b y# œ 4y Ê y# c 4y b 3 œ 0 Ê (y c 3)(y c 1) œ 0 Ê y œ 3 or y œ 1. Therefore in Cartesian coordinates, the points of intersection are ŠÈ3ß 3‹ and ŠÈ3ß 1‹. In polar coordinates, 4 sin ) œ È3 sec ) Ê 4 sin ) cos ) œ È3 Ê 2 sin ) cos ) œ 21 3
Ê )œ
1 6
or
1 3
È3 #
;)œ
Ê sin 2) œ 1 6
È3 #
Ê 2) œ
Ê r œ 2, and ) œ
1 3
or
1 3
Ê r œ 2È3 Ê ˆ2ß 16 ‰ and Š2È3ß 13 ‹ are the points of intersection in polar coordinates. 60. (a) r œ 8 cos ) Ê r# œ 8r cos ) Ê x# b y# œ 8x Ê x# c 8x b y# œ 0 Ê (x c 4)# b y# œ 16; r œ 2 sec ) Ê r œ cos2 ) Ê r cos ) œ 2
(b)
Ê x œ 2; x œ 2 Ê 2# c 8(2) b y# œ 0 Ê y# œ 12 Ê y œ „ 2È3. Therefore Š2ß „ 2È3‹
are the points of intersection in Cartesian coordinates. In polar coordinates, 8 cos ) œ 2 sec ) Ê 8 cos# ) œ 2 Ê cos# ) œ "4 Ê cos ) œ „ #" Ê ) œ 13 , 231 , 431 , or 51 3
Ê r œ 4, and ) œ 231 and 431 Ê r œ c4 Ê ˆ4ß 13 ‰ and ˆ4ß 531 ‰ are the points of intersection in polar coordinates. The points ˆc4ß 231 ‰ and ˆc4ß 431 ‰ are the same points. ;)œ
1 3
and
51 3
61. r cos ) œ 4 Ê x œ 4 Ê k œ 4: parabola Ê e œ 1 Ê r œ 62. r cos ˆ) c 1# ‰ œ 2 Ê r ˆcos ) cos Ê rœ
2 1 sin )
1 #
4 1 cos )
b sin ) sin 1# ‰ œ 2 Ê r sin ) œ 2 Ê y œ 2 Ê k œ 2: parabola Ê e œ 1
673
674
Chapter 10 Conic Sections and Polar Coordinates
63. (a) Let the ellipse be the orbit, with the Sun at one focus. rmin Then rmax œ a b c and rmin œ a c c Ê rrmax max rmin œ
(a c) (a c) (a c) (a c)
œ
2c 2a
œ
c a
œe
(b) Let F" , F# be the foci. Then PF" b PF# œ 10 where P is any point on the ellipse. If P is a vertex, then PF" œ a b c and PF# œ a c c Ê (a b c) b (a c c) œ 10 Ê 2a œ 10 Ê a œ 5. Since e œ ca we have 0.2 œ
c 5
Ê c œ 1.0 Ê the pins should be 2 inches apart. 64. e œ 0.97, Major axis œ 36.18 AU Ê a œ 18.09, Minor axis œ 9.12 AU Ê b œ 4.56 (1 AU ¸ 1.49 ‚ 10) km) (a) r œ
ke 1e cos )
œ
(b) ) œ 0 Ê r œ (c) ) œ 1 Ê r œ
(18.09) c1(0.97)# d a a1 e # b 1.07 œ 10.97 1e cos ) œ 10.97 cos ) cos ) AU 1.07 ( 10.97 ¸ 0.5431 AU ¸ 8.09 ‚ 10 km 1.07 * 10.97 ¸ 35.7 AU ¸ 5.32 ‚ 10 km
65. x# b y# c 2ay œ 0 Ê (r cos ))# b (r sin ))# c 2ar sin ) œ 0 Ê r# cos# ) b r# sin# ) c 2ar sin ) œ 0 Ê r# œ 2ar sin ) Ê r œ 2a sin )
66. y# œ 4ax b 4a# Ê (r sin ))# œ 4ar cos ) b 4a# Ê r# sin# ) œ 4ar cos ) b 4a# Ê r# a1 c cos# )b œ 4ar cos ) b 4a# Ê r# c r# cos# ) œ 4ar cos ) b 4a# Ê r# œ r# cos# ) b 4ar cos ) b 4a# Ê r# œ (r cos ) b 2a)# Ê r œ „ (r cos ) b 2a) Ê r c r cos ) œ 2a or 2a r b r cos ) œ c2a Ê r œ 12a cos ) or r œ 1cos ) ; the equations have the same graph, which is a parabola opening to the right 67. x cos ! b y sin ! œ p Ê r cos ) cos ! b r sin ) sin ! œ p Ê r(cos ) cos ! b sin ) sin !) œ p Ê r cos () c !) œ p
#
68. ax# b y# b b 2ax ax# b y# b c a# y# œ 0 Ê Ê Ê Ê Ê Ê Ê
#
ar# b b 2a(r cos )) ar# b c a# (r sin ))# œ 0 r% b 2ar$ cos ) c a# r# sin# ) œ 0 r# cr# b 2ar cos ) c a# a1 c cos# )bd œ 0 (assume r Á 0) r# b 2ar cos ) c a# b a# cos# ) œ 0 ar# b 2ar cos ) b a# cos# )b c a# œ 0 (r b a cos ))# œ a# Ê r b a cos ) œ „ a r œ a(1 c cos )) or r œ ca(1 b cos ));
Chapter 10 Practice Exercises the equations have the same graph, which is a cardioid 69 - 70. Example CAS commands: Maple: with( plots );#69 f := (r,k,e) -> k*e/(1+e*cos(theta)); elist := [3/4,1,5/4]; # (a) P1 := seq( plot( f(r,-2,e), theta=-Pi..Pi, coords=polar ), e=elist ): display( [P1], insequence=true, view=[-20..20,-20..20], title="#69(a) (Section 10.8)\nk=-2" ); P2 := seq( plot( f(r,2,e), theta=-Pi..Pi, coords=polar ), e=elist ): display( [P2], insequence=true, view=[-20..20,-20..20], title="#69(a) (Section 10.8)\nk=2" ); elist2 := [7/6,5/4,4/3,3/2,2,3,5,10,20]; # (b) P3 := seq( plot( f(r,-1,e), theta=-Pi..Pi, coords=polar ), e=elist2 ): display( [P3], insequence=true, view=[-20..20,-20..20], title="#69(b) (Section 10.8)\nk=-1, e>1" ); elist3 := [1/2,1/3,1/4,1/10,1/20]; P4 := seq( plot( f(r,-1,e), theta=-Pi..Pi, coords=polar ), e=elist3 ): display( [P4], insequence=true, title="#69(b) (Section 10.8)\nk=-1, e 1/sqrt(1+x); x0 := -3/4; x1 := 3/4; # Step 1: plot( f(x), x=x0..x1, title="Step 1: #57 (Section 11.9)" ); # Step 2: P1 := unapply( TaylorApproximation(f(x), x = 0, order=1), x ); P2 := unapply( TaylorApproximation(f(x), x = 0, order=2), x ); P3 := unapply( TaylorApproximation(f(x), x = 0, order=3), x ); # Step 3: D2f := D(D(f)); D3f := D(D(D(f))); D4f := D(D(D(D(f)))); plot( [D2f(x),D3f(x),D4f(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green], title="Step 3: #57 (Section 11.9)" ); c1 := x0; M1 := abs( D2f(c1) );
Section 11.9 Convergence of Taylor Series; Error Estimates
749
c2 := x0; M2 := abs( D3f(c2) ); c3 := x0; M3 := abs( D4f(c3) ); # Step 4: R1 := unapply( abs(M1/2!*(x-0)^2), x ); R2 := unapply( abs(M2/3!*(x-0)^3), x ); R3 := unapply( abs(M3/4!*(x-0)^4), x ); plot( [R1(x),R2(x),R3(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green], title="Step 4: #57 (Section 11.9)" ); # Step 5: E1 := unapply( abs(f(x)-P1(x)), x ); E2 := unapply( abs(f(x)-P2(x)), x ); E3 := unapply( abs(f(x)-P3(x)), x ); plot( [E1(x),E2(x),E3(x),R1(x),R2(x),R3(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green], linestyle=[1,1,1,3,3,3], title="Step 5: #57 (Section 11.9)" ); # Step 6: TaylorApproximation( f(x), view=[x0..x1,DEFAULT], x=0, output=animation, order=1..3 ); L1 := fsolve( abs(f(x)-P1(x))=0.01, x=x0/2 ); # (a) R1 := fsolve( abs(f(x)-P1(x))=0.01, x=x1/2 ); L2 := fsolve( abs(f(x)-P2(x))=0.01, x=x0/2 ); R2 := fsolve( abs(f(x)-P2(x))=0.01, x=x1/2 ); L3 := fsolve( abs(f(x)-P3(x))=0.01, x=x0/2 ); R3 := fsolve( abs(f(x)-P3(x))=0.01, x=x1/2 ); plot( [E1(x),E2(x),E3(x),0.01], x=min(L1,L2,L3)..max(R1,R2,R3), thickness=[0,2,4,0], linestyle=[0,0,0,2], color=[red,blue,green,black], view=[DEFAULT,0..0.01], title="#57(a) (Section 11.9)" ); abs(`f(x)`-`P`[1](x) )