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
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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
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Section 1.1 Real Numbers and the Real Line
4 or 2s
29. 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.
1 or ˆ r# 1 ‰
r"
#
Ê r
34.
3r
5
2 or r 1 Ÿ 2
1 or r Ÿ 3; solution intervals (_ß 3] [1ß _)
"
Ê
1 Ê r1
ïïïïïïðqqqqqqðïïïïïïî x
1
7/3
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 #
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3
4
Chapter 1 Preliminaries
41. x# x 0 Ê x# x +
1
4
<
1
4
2
Ê ˆx 12 ‰ <
ʹx 1
4
1
2
¹<
1
2
Ê 12 < x 1
2
<
1
2
Ê 0 < x < 1.
So the solution is the interval (0ß 1)
42. x# x 2
0 Ê x# x +
1
4
9
4
Ê ¹x 1
2
¹
3
2
Ê x
1
2
3
2
or ˆx 12 ‰
3
2
Ê x
2 or x Ÿ 1.
The solution interval is (_ß 1] [2ß _)
43. True if a
0; False if a 0.
44. kx 1k œ 1 x Í k(x 1)k œ 1 x Í 1 x
0 Í xŸ1
45. (1) ka bk œ (a b) or ka bk œ (a b);
both squared equal (a b)#
(2) ab Ÿ kabk œ kak kbk
(3) kak œ a or kak œ a, so kak# œ a# ; likewise, kbk# œ b#
(4) x# Ÿ y# implies Èx# Ÿ Èy# or x Ÿ y for all nonnegative real numbers x and y. Let x œ ka bk and
y œ kak kbk so that ka bk# Ÿ akak kbkb# Ê ka bk Ÿ kak kbk .
46. If a 0 and b 0, then ab 0 and kabk œ ab œ kak kbk .
If a 0 and b 0, then ab 0 and kabk œ ab œ (a)(b) œ kak kbk .
If a 0 and b 0, then ab Ÿ 0 and kabk œ (ab) œ (a)(b) œ kak kbk .
If a 0 and b 0, then ab Ÿ 0 and kabk œ (ab) œ (a)(b) œ kak kbk .
47. 3 Ÿ x Ÿ 3 and x "# Ê "
#
x Ÿ 3.
48. Graph of kxk kyk Ÿ 1 is the interior
of “diamond-shaped" region.
49. Let $ be a real number > 0 and f(x) = 2x + 1. Suppose that | x1 | < $ . Then | x1 | < $ Ê 2| x1 | < 2$ Ê
| 2x # | < 2$ Ê | (2x + 1) 3 | < 2$ Ê | f(x) f(1) | < 2$
50. Let % > 0 be any positive number and f(x) = 2x + 3. Suppose that | x 0 | < % /2. Then 2| x 0 | < % and
| 2x + 3 3 | < %. But f(x) = 2x + 3 and f(0) = 3. Thus | f(x) f(0) | < %.
51. Consider: i) a > 0; ii) a < 0; iii) a = 0.
i) For a > 0, | a | œ a by definition. Now, a > 0 Ê a < 0. Let a = b. By definition, | b | œ b. Since b = a,
| a | œ (a) œ a and | a | œ | a | œ a.
ii) For a < 0, | a | œ a. Now, a < 0 Ê a > 0. Let a œ b. By definition, | b | œ b and thus |a| œ a. So again
| a | œ |a|.
iii) By definition | 0 | œ 0 and since 0 œ 0, | 0 | œ 0. Thus, by i), ii), and iii) | a | œ | a | for any real number.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 1.2 Lines, Circles and Parabolas
Prove | x | > 0 Ê x > a or x < a for any positive number, a.
For x
0, | x | œ x. | x | > a Ê x > a.
For x < 0, | x | œ x. | x | > a Ê x > a Ê x < a.
ii) Prove x > a or x < a Ê | x | > 0 for any positive number, a.
a > 0 and x > a Ê | x | œ x. So x > a Ê | x | > a.
For a > 0, a < 0 and x < a Ê x < 0 Ê | x | œ x. So x < a Ê x > a Ê | x | > a.
52. i)
53. a)
1=1 Ê |1|=1 ʹb
b)
lal
lbl
œ ¹a
†
"
b
¹ œ ¹ a¹
† "b ¹ œ
†¹
"
b
l bl
lbl
¹ œ ¹ a¹
Ê ¹ b¹
† l bl
"
† ¹ b" ¹ œ
œ
lbl
lbl
Ê
†
¹ b ¹ ¹ "b ¹
¹ b¹
œ
¹ b¹
†
¹ b¹ ¹ b¹
Ê ¹ b" ¹ œ "
¹ b¹
lal
lbl
54. Prove Sn œ kan k œ kakn for any real number a and any positive integer n.
ka" k œ kak " œ a, so S" is true. Now, assume that Sk œ ¸ak ¸ œ kak k is true form some positive integer 5 .
Since ka" k œ kak " and ¸ak ¸ œ kak k , we have ¸ak" ¸ œ ¸ak † a" ¸ œ ¸ak ¸ka" k œ kak k kak " œ kak k+" . Thus,
Sk" œ ¸ak" ¸ œ kak k+" is also true. Thus by the Principle of Mathematical Induction, Sn œ l an l œ l a ln
is true for all n positive integers.
1.2 LINES, CIRCLES, AND PARABOLAS
1. ?x œ 1 (3) œ 2, ?y œ 2 2 œ 4; d œ È(?x)# (?y)# œ È4 16 œ 2È5
2. ?x œ $ (1) œ 2, ?y œ 2 (2) œ 4; d œ È(2)# 4# œ 2È5
3. ?x œ 8.1 (3.2) œ 4.9, ?y œ 2 (2) œ 0; d œ È(4.9)# 0# œ 4.9
#
4. ?x œ 0 È2 œ È2, ?y œ 1.5 4 œ 2.5; d œ ÊŠÈ2‹ (2.5)# œ È8.25
5. Circle with center (!ß !) and radius 1.
6. Circle with center (!ß !) and radius È2.
7. Disk (i.e., circle together with its interior points) with center (!ß !) and radius È3.
8. The origin (a single point).
9. m œ
?y
?x
œ
1 2
2 (1)
œ3
perpendicular slope œ "3
10. m œ
?y
?x
œ
# "
2 (2)
œ 34
perpendicular slope œ
4
3
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5
6
Chapter 1 Preliminaries
11. m œ
?y
?x
œ
33
1 2
œ0
12. m œ
14. (a) x œ È2
(b) y œ
# 0
# (#)
; no slope
15. (a) x œ 0
16. (a) x œ 1
(b) y œ È2
(b) y œ 1.3
4
3
œ
perpendicular slope œ 0
perpendicular slope does not exist
13. (a) x œ 1
?y
?x
(b) y œ 0
17. P(1ß 1), m œ 1 Ê y 1 œ 1ax (1)b Ê y œ x
18. P(2ß 3), m œ
"
#
Ê y (3) œ
19. P(3ß 4), Q(2ß 5) Ê m œ
?y
?x
20. P(8ß 0), Q(1ß 3) Ê m œ
œ
?y
?x
"
#
(x 2) Ê y œ
54
2 3
œ
"
#
x4
œ "5 Ê y 4 œ "5 (x 3) Ê y œ "5 x 30
1 (8)
œ
3
7
Ê y0œ
3
7
ax (8)b Ê y œ
3
7
23
5
x
21. m œ 54 , b œ 6 Ê y œ 54 x 6
22. m œ "# , b œ 3 Ê y œ
"
#
23. m œ 0, P(12ß 9) Ê y œ 9
24. No slope, P ˆ "3 ß %‰ Ê x œ
24
7
x3
"
3
25. a œ 1, b œ 4 Ê (0ß 4) and ("ß 0) are on the line Ê m œ
?y
?x
œ
04
1 0
œ 4 Ê y œ 4x 4
26. a œ 2, b œ 6 Ê (2ß 0) and (!ß 6) are on the line Ê m œ
?y
?x
œ
6 0
02
œ 3 Ê y œ 3x 6
27. P(5ß 1), L: 2x 5y œ 15 Ê mL œ 25 Ê parallel line is y (1) œ 25 (x 5) Ê y œ 25 x 1
È
È
È
28. P ŠÈ2ß 2‹ , L: È2x 5y œ È3 Ê mL œ 52 Ê parallel line is y 2 œ 52 Šx ŠÈ2‹‹ Ê y œ 52 x 8
5
29. P(4ß 10), L: 6x 3y œ 5 Ê mL œ 2 Ê m¼ œ "# Ê perpendicular line is y 10 œ "# (x 4) Ê y œ "# x 12
30. P(!ß 1), L: 8x 13y œ 13 Ê mL œ
8
13
13
Ê m¼ œ 13
8 Ê perpendicular line is y œ 8 x 1
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Section 1.2 Lines, Circles and Parabolas
31. x-intercept œ 4, y-intercept œ 3
32. x-intercept œ 4, y-intercept œ 2
33. x-intercept œ È3, y-intercept œ È2
34. x-intercept œ 2, y-intercept œ 3
35. Ax By œ C" Í y œ AB x C"
B
and Bx Ay œ C# Í y œ
B
A
x
C#
A.
Since ˆ AB ‰ ˆ AB ‰ œ 1 is the
product of the slopes, the lines are perpendicular.
36. Ax By œ C" Í y œ AB x slope
AB ,
C"
B
and Ax By œ C# Í y œ AB x C#
B.
Since the lines have the same
they are parallel.
37. New position œ axold ?xß yold ?yb œ (# &ß 3 (6)) œ ($ß 3).
38. New position œ axold ?xß yold ?yb œ (6 (6)ß 0 0) œ (0ß 0).
39. ?x œ 5, ?y œ 6, B(3ß 3). Let A œ (xß y). Then ?x œ x# x" Ê 5 œ 3 x Ê x œ 2 and
?y œ y# y" Ê 6 œ 3 y Ê y œ 9. Therefore, A œ (#ß 9).
40. ?x œ " " œ !, ?y œ ! ! œ !
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7
8
Chapter 1 Preliminaries
41. C(!ß 2), a œ 2 Ê x# (y 2)# œ 4
42. C($ß 0), a œ 3 Ê (x 3)# y# œ 9
43. C(1ß 5), a œ È10 Ê (x 1)# (y 5)# œ 10
44. C("ß "), a œ È2 Ê (x 1)# (y 1)# œ 2
x œ 0 Ê (0 1)# (y 1)# œ 2 Ê (y 1)# œ 1
Ê y 1 œ „ 1 Ê y œ 0 or y œ 2.
Similarly, y œ 0 Ê x œ 0 or x œ 2
#
45. C ŠÈ3ß 2‹ , a œ 2 Ê Šx È3‹ (y 2)# œ 4,
#
x œ 0 Ê Š0 È3‹ (y 2)# œ 4 Ê (y 2)# œ 1
Ê y 2 œ „ 1 Ê y œ 1 or y œ 3. Also, y œ 0
#
#
Ê Šx È3‹ (0 2)# œ 4 Ê Šx È3‹ œ 0
Ê x œ È 3
#
46. C ˆ3ß "# ‰, a œ 5 Ê (x 3)# ˆy "# ‰ œ 25, so
#
x œ 0 Ê (0 3)# ˆy "# ‰ œ 25
#
Ê ˆy "# ‰ œ 16 Ê y "
#
œ „4 Ê yœ
9
#
#
or y œ 7# . Also, y œ 0 Ê (x 3)# ˆ0 "# ‰ œ 25
Ê (x 3)# œ
Ê xœ3„
99
4
3È11
#
Ê x3œ „
3È11
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 1.2 Lines, Circles and Parabolas
47. x# y# 4x 4y % œ 0
Ê x# %B y# 4y œ 4
Ê x# 4x 4 y# 4y 4 œ 4
Ê (x 2)# (y 2)# œ 4 Ê C œ (2ß 2), a œ 2.
48. x# y# 8x 4y 16 œ 0
Ê x# 8x y# 4y œ 16
Ê x# 8x 16 y# 4y 4 œ 4
Ê (x 4)# (y 2)# œ 4
Ê C œ (%ß 2), a œ 2.
49. x# y# 3y 4 œ 0 Ê x# y# 3y œ 4
Ê x# y# 3y 94 œ 25
4
#
Ê x# ˆy 3# ‰ œ
25
4
Ê C œ ˆ0ß 3# ‰ ,
a œ 5# .
50. x# y# 4x #
9
4
#
œ0
Ê x 4x y œ
9
4
#
Ê x# 4x 4 y œ
Ê (x 2)# y# œ
25
4
25
4
Ê C œ (2ß 0), a œ 5# .
51. x# y# 4x 4y œ 0
Ê x# 4x y# 4y œ 0
Ê x# 4x 4 y# 4y 4 œ 8
Ê (x 2)# (y 2)# œ 8
Ê C(2ß 2), a œ È8.
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9
10
Chapter 1 Preliminaries
52. x# y# 2x œ 3
Ê x# 2x 1 y# œ 4
Ê (x 1)# y# œ 4
Ê C œ (1ß 0), a œ 2.
2
53. x œ #ba œ 2(1)
œ1
Ê y œ (1)# 2(1) 3 œ 4
Ê V œ ("ß 4). If x œ 0 then y œ 3.
Also, y œ 0 Ê x# 2x 3 œ 0
Ê (x 3)(x 1) œ 0 Ê x œ 3 or
x œ 1. Axis of parabola is x œ 1.
4
54. x œ #ba œ 2(1)
œ 2
Ê y œ (2)# 4(2) 3 œ 1
Ê V œ (2ß 1). If x œ 0 then y œ 3.
Also, y œ 0 Ê x# 4x 3 œ 0
Ê (x 1)(x 3) œ 0 Ê x œ 1 or
x œ 3. Axis of parabola is x œ 2.
55. x œ #ba œ 2(4 1) œ 2
Ê y œ (2)# 4(2) œ 4
Ê V œ (2ß 4). If x œ 0 then y œ 0.
Also, y œ 0 Ê x# 4x œ 0
Ê x(x 4) œ 0 Ê x œ 4 or x œ 0.
Axis of parabola is x œ 2.
56. x œ #ba œ 2(4 1) œ 2
Ê y œ (2)# 4(2) 5 œ 1
Ê V œ (2ß 1). If x œ 0 then y œ 5.
Also, y œ 0 Ê x# 4x 5 œ 0
Ê x# 4x 5 œ 0 Ê x œ
4 „È 4
#
Ê no x intercepts. Axis of parabola is x œ 2.
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Section 1.2 Lines, Circles and Parabolas
57. x œ #ba œ 2(61) œ 3
Ê y œ (3)# 6(3) 5 œ 4
Ê V œ (3ß %). If x œ 0 then y œ 5.
Also, y œ 0 Ê x# 6x 5 œ 0
Ê (x 5)(x 1) œ 0 Ê x œ 5 or
x œ 1. Axis of parabola is x œ 3.
1
58. x œ #ba œ 2(2)
œ
"
4
#
Ê y œ 2 ˆ "4 ‰ 4" 3 œ 23
8
‰
Ê V œ ˆ "4 ß 23
.
If
x
œ
0
then
y œ 3.
8
Also, y œ 0 Ê 2x# x 3 œ 0
Ê xœ
1„È23
4
Ê no x intercepts.
Axis of parabola is x œ "4 .
1
59. x œ #ba œ 2(1/2)
œ 1
"
#
(1)# (1) 4 œ 72
Ê V œ ˆ"ß 72 ‰ . If x œ 0 then y œ 4.
Ê yœ
Also, y œ 0 Ê
Ê xœ
1 „ È 7
1
"
#
x# x 4 œ 0
Ê no x intercepts.
Axis of parabola is x œ 1.
60. x œ #ba œ 2(21/4) œ 4
Ê y œ "4 (4)# 2(4) 4 œ 8
Ê V œ (4ß 8) . If x œ 0 then y œ 4.
Also, y œ 0 Ê "4 x# 2x 4 œ 0
Ê xœ
2 „ È 8
1/2
œ 4 „ 4È2.
Axis of parabola is x œ 4.
61. The points that lie outside the circle with center (!ß 0) and radius È7.
62. The points that lie inside the circle with center (!ß 0) and radius È5.
63. The points that lie on or inside the circle with center ("ß 0) and radius 2.
64. The points lying on or outside the circle with center (!ß 2) and radius 2.
65. The points lying outside the circle with center (!ß 0) and radius 1, but inside the circle with center (!ß 0),
and radius 2 (i.e., a washer).
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11
12
Chapter 1 Preliminaries
66. The points on or inside the circle centered
at (!ß !) with radius 2 and on or inside the
circle centered at (2ß 0) with radius 2.
67. x# y# 6y 0 Ê x# (y 3)# 9.
The interior points of the circle centered at
(!ß 3) with radius 3, but above the line
y œ 3.
68. x# y# 4x 2y 4 Ê (x 2)# (y 1)# 9.
The points exterior to the circle centered at
(2ß 1) with radius 3 and to the right of the
line x œ 2.
69. (x 2)# (y 1)# 6
70. (x 4)# (y 2)# 16
71. x# y# Ÿ 2, x
72. x# y# 4, (x 1)# (y 3)# 10
1
73. x# y# œ 1 and y œ 2x Ê 1 œ x# 4x# œ 5x#
Ê Šx œ
"
È5
and y œ
2
È5 ‹
or Šx œ È"5 and y œ È25 ‹ .
Thus, A Š È"5 ß È25 ‹ , B Š È"5 ß È25 ‹ are the
points of intersection.
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Section 1.2 Lines, Circles, and Parabolas
74. x y œ 1 and (x 1)# y# œ 1
Ê 1 œ (y)# y# œ 2y#
Ê Šy œ
"
È2
and x œ " Šy œ È"2 and x œ 1 A Š" "
È2
"
È2 ‹
"
È2 ‹ .
ß È"2 ‹ and B Š1 or
Thus,
"
È2
ß È"2 ‹
are intersection points.
75. y x œ 1 and y œ x# Ê x# x œ 1
1 „È 5
.
#
1 È 5
3 È 5
If x œ # , then y œ x 1 œ # .
È
È
If x œ 1# 5 , then y œ x 1 œ 3# 5 .
È
È
È
È
Thus, A Š 1# 5 ß 3# 5 ‹ and B Š 1# 5 ß 3# 5 ‹
Ê x# x 1 œ 0 Ê x œ
are the intersection points.
76. y œ x and C œ (x 1)# Ê (x 1)# œ x
3 „È 5
.
#
È 5 3
3 È 5
x œ # , then y œ x œ # . If
È
È
x œ 3# 5 , then y œ x œ 3# 5 .
È
È
È
Thus, A Š 3# 5 ß 5#3 ‹ and B Š 3# 5
Ê x# 3x " œ 0 Ê x œ
If
È
ß 3# 5 ‹
are the intersection points.
77. y œ 2x# 1 œ x# Ê 3x# œ 1
Ê x œ È"3 and y œ 3" or x œ È"3 and y œ 3" .
Thus, A Š È"3 ß 3" ‹ and B Š È"3 ß 3" ‹ are the
intersection points.
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13
14
Chapter 1 Preliminaries
78. y œ
x#
4
œ (x 1)# Ê 0 œ
#
3x#
4
2x 1
Ê 0 œ 3x 8x 4 œ (3x 2)(x 2)
Ê x œ 2 and y œ
yœ
#
x
4
x#
4
œ 1, or x œ
œ 9" . Thus, A(2ß 1) and
2
3 and
2
B ˆ 3 ß 9" ‰
are the intersection points.
79. x# y# œ 1 œ (x 1)# y#
Ê x# œ (x 1)# œ x# 2x 1
Ê 0 œ 2x 1 Ê x œ "# . Hence
y# œ " x # œ
A Š "# ß
È3
# ‹
and
3
4
or y œ „
È3
#
È
B Š "# ß #3 ‹
. Thus,
are the
intersection points.
80. x# y# œ 1 œ x# y Ê y# œ y
Ê y(y 1) œ 0 Ê y œ 0 or y œ 1.
If y œ 1, then x# œ " y# œ 0 or x œ 0.
If y œ 0, then x# œ 1 y# œ 1 or x œ „ 1.
Thus, A(0ß 1), B("ß 0), and C(1ß 0) are the
intersection points.
81. (a) A ¸ (69°ß 0 in), B ¸ (68°ß .4 in) Ê m œ
(b) A ¸ (68°ß .4 in), B ¸ (10°ß 4 in) Ê m œ
(c) A ¸ (10°ß 4 in), B ¸ (5°ß 4.6 in) Ê m œ
82. The time rate of heat transfer across a material,
to the temperature gradient across the material,
of the material.
?U
?>
œ
X
-kA ?
?B
Ê
?U ÎA
k = ??> X .
?B
68° 69°
.4 0 ¸ 2.5°/in.
10° 68°
4 .4 ¸ 16.1°/in.
5° 10°
4.6 4 ¸ 8.3°/in.
?U
?> , is directly
?X
?B (the slopes
Note that
?U
?>
proportional to the cross-sectional area, A, of the material,
from the previous problem), and to a constant characteristic
and
?X
?B
are of opposite sign because heat flow is toward lower
temperature. So a small value of k corresponds to low heat flow through the material and thus the material is a good
insulator.Since all three materials have the same cross section and the heat flow across each is the same (temperatures are
X
not changing), we may define another constant, K, characteristics of the material: K œ ?"X Þ Using the values of ?
?B from
?B
the prevous problem, fiberglass has the smallest K at 0.06 and thus is the best insulator. Likewise, the wallboard is the
poorest insulator, with K œ 0.4.
83. p œ kd 1 and p œ 10.94 at d œ 100 Ê k œ
10.94"
100
œ 0.0994. Then p œ 0.0994d 1 is the diver's
pressure equation so that d œ 50 Ê p œ (0.0994)(50) 1 œ 5.97 atmospheres.
84. The line of incidence passes through (!ß 1) and ("ß 0) Ê The line of reflection passes through ("ß 0) and (#ß ")
0
Ê m œ 1#
1 œ 1 Ê y 0 œ 1(x 1) Ê y œ x 1 is the line of reflection.
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Section 1.2 Lines, Circles, and Parabolas
85. C œ
5
9
(F 32) and C œ F Ê F œ
86. m œ
37.1
100
œ
14
?x
Ê ?x œ
14
.371 .
5
9
F
160
9
Ê
4
9
15
F œ 160
9 or F œ 40° gives the same numerical reading.
#
14 ‰
Therefore, distance between first and last rows is É(14)# ˆ .371
¸ 40.25 ft.
87. length AB œ È(5 1)# (5 2)# œ È16 9 œ 5
length AC œ È(4 1)# (# #)# œ È9 16 œ 5
length BC œ È(4 5)# (# 5)# œ È1 49 œ È50 œ 5È2 Á 5
#
88. length AB œ Ê(1 0)# ŠÈ3 0‹ œ È1 3 œ 2
length AC œ È(2 0)# (0 0)# œ È4 0 œ 2
#
length BC œ Ê(2 1)# Š0 È3‹ œ È1 3 œ 2
89. Length AB œ È(?x)# (?y)# œ È1# 4# œ È17 and length BC œ È(?x)# (?y)# œ È4# 1# œ È17.
Also, slope AB œ 41 and slope BC œ "4 , so AB ¼ BC. Thus, the points are vertices of a square. The coordinate
increments from the fourth vertex D(xß y) to A must equal the increments from C to B Ê 2 x œ ?x œ 4 and
1 y œ ?y œ " Ê x œ 2 and y œ 2. Thus D(#ß 2) is the fourth vertex.
90. Let A œ (xß 2) and C œ (9ß y) Ê B œ (xß y). Then 9 x œ kADk and 2 y œ kDCk Ê 2(9 x) 2(2 y) œ 56
and 9 x œ 3(2 y) Ê 2(3(2 y)) 2(2 y) œ 56 Ê y œ 5 Ê 9 x œ 3(2 (5)) Ê x œ 12.
Therefore, A œ (12ß 2), C œ (9ß 5), and B œ (12ß 5).
91. Let A("ß "), B(#ß $), and C(2ß !) denote the points.
Since BC is vertical and has length kBCk œ 3, let
D" ("ß 4) be located vertically upward from A and
D# ("ß 2) be located vertically downward from A so
that kBCk œ kAD" k œ kAD# k œ 3. Denote the point
D$ (xß y). Since the slope of AB equals the slope of
3
"
CD$ we have yx
2 œ 3 Ê 3y 9 œ x 2 or
x 3y œ 11. Likewise, the slope of AC equals the slope
0
2
of BD$ so that yx 2 œ 3 Ê 3y œ 2x 4 or 2x 3y œ 4.
Solving the system of equations
x 3y œ ""
we find x œ 5 and y œ 2 yielding the vertex D$ (5ß #).
2x 3y œ 4 92. Let ax, yb, x Á ! and/or y Á ! be a point on the coordinate plane. The slope, m, of the segment a!ß !b to ax, yb is yx . A 90‰
rotation gives a segment with slope mw œ m" œ xy . If this segment has length equal to the original segment, its endpoint
will be ay, xb or ay, xb, the first of these corresponds to a counter-clockwise rotation, the latter to a clockwise
rotation.
(a) ("ß 4);
(b) (3ß 2);
(c) (5ß 2);
(d) (0ß x);
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16
Chapter 1 Preliminaries
(e) (yß 0);
(f) (yß x);
(g) (3ß 10)
93. 2x ky œ 3 has slope 2k and 4x y œ 1 has slope 4. The lines are perpendicular when 2k (4) œ 1 or
k œ 8 and parallel when 2k œ 4 or k œ "# .
94. At the point of intersection, 2x 4y œ 6 and 2x 3y œ 1. Subtracting these equations we find 7y œ 7 or
y œ 1. Substitution into either equation gives x œ 1 Ê (1ß 1) is the intersection point. The line through (1ß 1)
and ("ß #) is vertical with equation x œ 1.
95. Let M(aß b) be the midpoint. Since the two triangles
shown in the figure are congruent, the value a must
lie midway between x" and x# , so a œ x" #x# .
Similarly, b œ
y " y #
# .
96. (a) L has slope 1 so M is the line through P(2ß 1) with slope 1; or the line y œ x 3. At the intersection
point, Q, we have equal y-values, y œ x 2 œ x 3. Thus, 2x œ 1 or x œ "# . Hence Q has coordinates
ˆ "# ß 5# ‰ . The distance from P to L œ the distance from P to Q œ Ɉ #3 ‰# ˆ 3# ‰# œ É 18
4 œ
(b) L has slope 43 so M has slope
3
4
3È 2
# .
and M has the equation 4y 3x œ 12. We can rewrite the equations of
84
the lines as L: x y œ 3 and M: B 43 y œ 4. Adding these we get 25
12 y œ 7 so y œ 25 . Substitution
12
‰
ˆ 12 84 ‰
into either equation gives x œ 43 ˆ 84
25 4 œ 25 so that Q 25 ß 25 is the point of intersection. The distance
3
4
from P to L œ Ɉ4 12 ‰#
25
ˆ6 84 ‰#
25
œ
22
5 .
(c) M is a horizontal line with equation y œ b. The intersection point of L and M is Q("ß b). Thus, the
distance from P to L is È(a 1)# 0# œ ka 1k .
(d) If B œ 0 and A Á 0, then the distance from P to L is ¸ AC x! ¸ as in (c). Similarly, if A œ 0 and B Á 0, the
distance is ¸ CB y! ¸ . If both A and B are Á 0 then L has slope AB so M has slope AB . Thus,
L: Ax By œ C and M: Bx Ay œ Bx! Ay! . Solving these equations simultaneously we find the
point of intersection Q(xß y) with x œ
ACB aAy! Bx! b
A# B#
P to Q equals È(?x)# (?y)# , where (?x)# œ
œ
A# aAx! By! Cb#
aA# B# b#
#
#
BCA aAy! Bx! b
.
A# B#
#
#
#
#
ABy! B x!
Š x! aA B bAAC
‹
# B#
and y œ
#
#
A y! ABx!
, and (?y)# œ Š y! aA B bABC
‹ œ
# B#
#
! Cb
Thus, È(?x)# (?y)# œ É aAx!A#By
œ
B#
kAx! By! Ck
ÈA# B#
The distance from
B# aAx! By! Cb#
.
aA# B# b#
.
1.3 FUNCTIONS AND THEIR GRAPHS
1. domain œ (_ß _); range œ [1ß _)
3. domain œ (!ß _); y in range Ê y œ
2. domain œ [0ß _); range œ (_ß 1]
"
Èt
, t 0 Ê y# œ
"
t
and y ! Ê y can be any positive real number
Ê range œ (!ß _).
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Section 1.3 Functions and Their Graphs
4. domain œ [0ß _); y in range Ê y œ
"
1 È t
17
, t 0. If t œ 0, then y œ 1 and as t increases, y becomes a smaller
and smaller positive real number Ê range œ (0ß 1].
5. 4 z# œ (2 z)(2 z) 0 Í z − [2ß 2] œ domain. Largest value is g(0) œ È4 œ 2 and smallest value is
g(2) œ g(2) œ È0 œ 0 Ê range œ [0ß 2].
6. domain œ (2ß 2) from Exercise 5; smallest value is g(0) œ "# and as 0 z increases to 2, g(z) gets larger and
larger (also true as z 0 decreases to 2) Ê range œ "# ß _‰ .
7. (a) Not the graph of a function of x since it fails the vertical line test.
(b) Is the graph of a function of x since any vertical line intersects the graph at most once.
8. (a) Not the graph of a function of x since it fails the vertical line test.
(b) Not the graph of a function of x since it fails the vertical line test.
9. y œ Ɉ "x ‰ " Ê
(a) No (x !Ñ;
(c) No; if x ",
"
x
"
x
"
! Ê x Ÿ 1 and x !. So,
"Ê
10. y œ É# Èx Ê # Èx
"
x
(b) No; division by ! undefined;
(d) Ð!ß "Ó
" !;
! Ê Èx
! and Èx Ÿ #. Èx
! and Èx Ÿ # Ê x Ÿ %Þ So, ! Ÿ x Ÿ %.
!Êx
(a) No; (b) No; (c) Ò!ß %Ó
#
11. base œ x; (height)# ˆ #x ‰ œ x# Ê height œ
È3
#
x; area is a(x) œ
"
#
(base)(height) œ
"
#
(x) Š
È3
# x‹
œ
È3
4
x# ;
perimeter is p(x) œ x x x œ 3x.
12. s œ side length Ê s# s# œ d# Ê s œ
d
È2
; and area is a œ s# Ê a œ
"
#
d#
13. Let D œ diagonal of a face of the cube and j œ the length of an edge. Then j# D# œ d# and (by Exercise 10)
D# œ 2j# Ê 3j# œ d# Ê j œ
d
È3
. The surface area is 6j# œ
6d#
3
14. The coordinates of P are ˆxß Èx‰ so the slope of the line joining P to the origin is m œ
ˆx, Èx‰ œ ˆ m"# ,
#
œ 2d# and the volume is j$ œ Š d3 ‹
Èx
x
œ
"
Èx
"‰
m .
15. The domain is a_ß _b.
16. The domain is a_ß _b.
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$Î#
œ
d$
3È 3
(x 0). Thus,
.
18
Chapter 1 Preliminaries
17. The domain is a_ß _b.
18. The domain is Ð_ß !Ó.
19. The domain is a_ß !b a!ß _b.
20. The domain is a_ß !b a!ß _b.
21. Neither graph passes the vertical line test
(a)
(b)
22. Neither graph passes the vertical line test
(a)
(b)
Ú xyœ" Þ
Ú yœ1x Þ
or
or
kx yk œ 1 Í Û
Í Û
ß
ß
Ü x y œ " à
Ü y œ " x à
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Section 1.3 Functions and Their Graphs
23.
x
y
0
0
25. y œ œ
1
1
2
0
24.
x
y
0
1
1
0
2
0
"
, x0
26. y œ œ x
x, 0 Ÿ x
3 x, x Ÿ 1
2x, 1 x
27. (a) Line through a!ß !b and a"ß "b: y œ x
Line through a"ß "b and a#ß !b: y œ x 2
x, 0 Ÿ x Ÿ 1
f(x) œ œ
x 2, 1 x Ÿ 2
Ú
Ý
Ý 2, ! Ÿ x "
!ß " Ÿ x #
(b) f(x) œ Û
Ý
Ý 2ß # Ÿ x $
Ü !ß $ Ÿ x Ÿ %
28. (a) Line through a!ß 2b and a#ß !b: y œ x 2
"
Line through a2ß "b and a&ß !b: m œ !& # œ
x #, 0 x Ÿ #
f(x) œ œ "
$ x &$ , # x Ÿ &
"
$
$ !
! Ð"Ñ œ
" $
%
#! œ #
(b) Line through a"ß !b and a!ß $b: m œ
Line through a!ß $b and a#ß "b: m œ
f(x) œ œ
œ "$ , so y œ "$ ax 2b " œ "$ x &
$
$, so y œ $x $
œ #, so y œ #x $
$x $, " x Ÿ !
#x $, ! x Ÿ #
29. (a) Line through a"ß "b and a!ß !b: y œ x
Line through a!ß "b and a"ß "b: y œ "
Line through a"ß "b and a$ß !b: m œ !"
$" œ
Ú
x
" Ÿ x !
"
!xŸ"
f(x) œ Û
Ü "# x $#
"x$
"
#
œ "# , so y œ "# ax "b " œ "# x $
#
(b) Line through a#ß "b and a!ß !b: y œ "# x
Line through a!ß #b and a"ß !b: y œ #x #
Line through a"ß "b and a$ß "b: y œ "
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19
20
Chapter 1 Preliminaries
Ú
"
#x
f(x) œ Û #x #
Ü "
# Ÿ x Ÿ !
!xŸ"
"xŸ$
30. (a) Line through ˆ T# ß !‰ and aTß "b: m œ
Ú
A,
Ý
Ý
Ý
Aß
f(x) œ Û
Aß
Ý
Ý
Ý
Ü Aß
! Ÿ x T#
T
# Ÿ x T
T Ÿ x $#T
$T
# Ÿ x Ÿ #T
x
#
31. (a) From the graph,
(b)
x
#
1
x 0:
x
#
x 0:
x
2
œ T# , so y œ T# ˆx T# ‰ 0 œ T# x "
!, 0 Ÿ x Ÿ T#
#
T
T x ", # x Ÿ T
f(x) œ (b)
"!
TaTÎ#b
4
x
1
x
#
Ê
4
x
1
4
x
Ê x − (2ß 0) (%ß _)
1 4x 0
#
2x8
0 Ê x 2x
0 Ê
(x4)(x2)
#x
0
(x4)(x2)
#x
0
Ê x 4 since x is positive;
1
4
x
0 Ê
x# 2x8
2x
0 Ê
Ê x 2 since x is negative;
sign of (x 4)(x 2)
ïïïïïðïïïïïðïïïïî
2
%
Solution interval: (#ß 0) (%ß _)
3
2
x 1 x 1
3
2
x 1 x 1
32. (a) From the graph,
(b) Case x 1:
Ê x − (_ß 5) (1ß 1)
Ê
3(x1)
x 1
2
Ê 3x 3 2x 2 Ê x 5.
Thus, x − (_ß 5) solves the inequality.
Case 1 x 1:
3
x 1
2
x 1
Ê
3(x1)
x 1
2
Ê 3x 3 2x 2 Ê x 5 which is true
if x 1. Thus, x − (1ß 1) solves the
inequality.
Case 1 x: x3 1 x2 1 Ê 3x 3 2x 2 Ê x 5
which is never true if 1 x, so no solution
here.
In conclusion, x − (_ß 5) (1ß 1).
33. (a) ÚxÛ œ 0 for x − [0ß 1)
(b) ÜxÝ œ 0 for x − (1ß 0]
34. ÚxÛ œ ÜxÝ only when x is an integer.
35. For any real number x, n Ÿ x Ÿ n ", where n is an integer. Now: n Ÿ x Ÿ n " Ê Ðn "Ñ Ÿ x Ÿ n. By
definition: ÜxÝ œ n and ÚxÛ œ n Ê ÚxÛ œ n. So ÜxÝ œ ÚxÛ for all x − d .
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Section 1.3 Functions and Their Graphs
21
36. To find f(x) you delete the decimal or
fractional portion of x, leaving only
the integer part.
37. v œ f(x) œ xÐ"% 2xÑÐ22 2xÑ œ %x$ 72x# $!)x; ! x 7Þ
38. (a) Let h œ height of the triangle. Since the triangle is isosceles, AB # AB # œ 2# Ê AB œ È2Þ So,
#
h# "# œ ŠÈ2‹ Ê h œ " Ê B is at a!ß "b Ê slope of AB œ " Ê The equation of AB is
y œ f(x) œ B "; x − Ò!ß "Ó.
(b) AÐxÑ œ 2x y œ 2xÐx "Ñ œ 2x# #x; x − Ò!ß "Ó.
39. (a) Because the circumference of the original circle was )1 and a piece of length x was removed.
x
x
(b) r œ )1#
1 œ % #1
(c) h œ È"' r# œ É"' ˆ% #
x‰
(d) V œ "$ 1 r# h œ "$ 1ˆ )1#
†
1
x ‰#
#1
œ É"' ˆ16 È"'1x x#
#1
œ
4x
1
x# ‰
%1#
œ É 4x
1 x#
%1#
œ É "'%11#x x#
%1#
œ
È"'1xx#
#1
a)1 xb# È"'1x x#
#%1#
40. (a) Note that 2 mi = 10,560 ft, so there are È)!!# x# feet of river cable at $180 per foot and a"!ß &'! xb feet of land
cable at $100 per foot. The cost is Caxb œ ")!È)!!# x# "!!a"!ß &'! xb.
(b) Ca!b œ $"ß #!!ß !!!
Ca&!!b ¸ $"ß "(&ß )"#
Ca"!!!b ¸ $"ß ")'ß &"#
Ca"&!!b ¸ $"ß #"#ß !!!
Ca#!!!b ¸ $"ß #%$ß ($#
Ca#&!!b ¸ $"ß #()ß %(*
Ca$!!!b ¸ $"ß $"%ß )(!
Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from the
point P.
41. A curve symmetric about the x-axis will not pass the vertical line test because the points ax, yb and ax, yb lie on the same
vertical line. The graph of the function y œ faxb œ ! is the x-axis, a horizontal line for which there is a single y-value, !,
for any x.
42. Pick 11, for example: "" & œ "' Ä # † "' œ $# Ä $# ' œ #' Ä
faxb œ
#ax&b'
#
#'
#
œ "$ Ä "$ # œ "", the original number.
# œ x, the number you started with.
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22
Chapter 1 Preliminaries
1.4 IDENTIFYING FUNCTIONS; MATHEMATICAL MODELS
1. (a) linear, polynomial of degree 1, algebraic.
(c) rational, algebraic.
(b) power, algebraic.
(d) exponential.
2. (a) polynomial of degree 4, algebraic.
(c) algebraic.
(b) exponential.
(d) power, algebraic.
3. (a) rational, algebraic.
(c) trigonometric.
(b) algebraic.
(d) logarithmic.
4. (a) logarithmic.
(c) exponential.
(b) algebraic.
(d) trigonometric.
5. (a) Graph h because it is an even function and rises less rapidly than does Graph g.
(b) Graph f because it is an odd function.
(c) Graph g because it is an even function and rises more rapidly than does Graph h.
6. (a) Graph f because it is linear.
(b) Graph g because it contains a!ß "b.
(c) Graph h because it is a nonlinear odd function.
7. Symmetric about the origin
Dec: _ x _
Inc: nowhere
8. Symmetric about the y-axis
Dec: _ x !
Inc: ! x _
9. Symmetric about the origin
Dec: nowhere
Inc: _ x !
!x_
10. Symmetric about the y-axis
Dec: ! x _
Inc: _ x !
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Section 1.4 Identifying Functions; Mathematical Models
11. Symmetric about the y-axis
Dec: _ x Ÿ !
Inc: ! x _
12. No symmetry
Dec: _ x Ÿ !
Inc: nowhere
13. Symmetric about the origin
Dec: nowhere
Inc: _ x _
14. No symmetry
Dec: ! Ÿ x _
Inc: nowhere
15. No symmetry
Dec: ! Ÿ x _
Inc: nowhere
16. No symmetry
Dec: _ x Ÿ !
Inc: nowhere
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23
24
Chapter 1 Preliminaries
17. Symmetric about the y-axis
Dec: _ x Ÿ !
Inc: ! x _
18. Symmetric about the y-axis
Dec: ! Ÿ x _
Inc: _ x !
19. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, the
function is even.
20. faxb œ x& œ
"
x&
and faxb œ axb& œ
"
a x b&
œ ˆ x"& ‰ œ faxb. Thus the function is odd.
21. Since faxb œ x# " œ axb# " œ faxb. The function is even.
22. Since Òfaxb œ x# xÓ Á Òfaxb œ axb# xÓ and Òfaxb œ x# xÓ Á Òfaxb œ axb# xÓ the function is neither even nor
odd.
23. Since gaxb œ x$ x, gaxb œ x$ x œ ax$ xb œ gaxb. So the function is odd.
24. gaxb œ x% $x# " œ axb% $axb# " œ gaxbß thus the function is even.
25. gaxb œ
"
x# "
26. gaxb œ
x
x# " ;
27. hatb œ
"
t ";
œ
"
axb# "
œ gaxb. Thus the function is even.
gaxb œ x#x" œ gaxb. So the function is odd.
h a t b œ
"
t " ;
h at b œ
"
" t.
Since hatb Á hatb and hatb Á hatb, the function is neither even nor odd.
28. Since l t$ | œ l atb$ |, hatb œ hatb and the function is even.
29. hatb œ 2t ", hatb œ 2t ". So hatb Á hatb. hatb œ 2t ", so hatb Á hatb. The function is neither even nor
odd.
30. hatb œ 2l t l " and hatb œ 2l t l " œ 2l t l ". So hatb œ hatb and the function is even.
31. (a)
The graph supports the assumption that y is proportional to x. The
constant of proportionality is estimated from the slope of the
regression line, which is 0.166.
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Section 1.4 Identifying Functions; Mathematical Models
(b)
25
The graph supports the assumption that y is proportional to x"Î# .
The constant of proportionality is estimated from the slope of the
regression line, which is 2.03.
32. (a) Because of the wide range of values of the data, two graphs are needed to observe all of the points in relation to the
regression line.
The graphs support the assumption that y is proportional to $x . The constant of proportionality is estimated from the
slope of the regression line, which is 5.00.
(b) The graph supports the assumption that y is proportional to ln x. The constant of proportionality is extimated from
the slope of the regression line, which is 2.99.
33. (a) The scatterplot of y œ reaction distance versus x œ speed is
Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which is
approximately 1.1.
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26
Chapter 1 Preliminaries
(b) Calculate x w œ speed squared. The scatterplot of x w versus y œ braking distance is:
Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which
is approximately 0.059.
34. Kepler's 3rd Law is Tadaysb œ !Þ%"R$Î# , R in millions of miles. "Quaoar" is 4 ‚ "!* miles from Earth, or about
4 ‚ "!* *$ ‚ "!' ¸ % ‚ "!* miles from the sun. Let R œ 4000 (millions of miles) and
T œ a!Þ%"ba%!!!b$Î# days ¸ "!$ß (#$ days.
35. (a)
The hypothesis is reasonable.
(b) The constant of proportionality is the slope of the line ¸
(c) y(in.) œ a!Þ)( in./unit massba"$ unit massb œ ""Þ$" in.
36. (a)
)Þ(%" !
"! !
in./unit mass œ !Þ)(% in./unit mass.
(b)
Graph (b) suggests that y œ k x$ is the better model. This graph is more linear than is graph (a).
1.5 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS
1. Df : _ x _, Dg : x
2. Df : x 1
Rf œ Rg : y
0 Ê x
1 Ê Dfbg œ Dfg : x
1, Dg : x 1
0, Rfbg : y È2, Rfg : y
0 Ê x
1. Rf : _ y _, Rg : y
1. Therefore Dfbg œ Dfg : x
0, Rfbg : y
1.
0
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1, Rfg : y
0
Section 1.5 Combining Functions; Shifting and Scaling Graphs
3. Df : _ x _, Dg : _ x _ Ê DfÎg : _ x _ since g(x) Á 0 for any x; DgÎf : _ x _
since f(x) Á 0 for any x. Rf : y œ 2, Rg : y 1, RfÎg : 0 y Ÿ 2, RgÎf : y "#
4. Df : _ x _, Dg : x 0 Ê DfÎg : x 0 since g(x) Á 0 for any x
for any x 0. Rf : y œ 1, Rg : y 1, RfÎg : 0 y Ÿ 1, RgÎf : y "
5. (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
0; DgÎf : x
0 since f(x) Á 0
f(g(0)) œ f(3) œ 2
g(f(0)) œ g(5) œ 22
f(g(x)) œ f(x# 3) œ x# 3 5 œ x# 2
g(f(x)) œ g(x 5) œ (x 5)# 3 œ x# 10x 22
f(f(5)) œ f(0) œ 5
g(g(2)) œ g(1) œ 2
f(f(x)) œ f(x 5) œ (x 5) 5 œ x 10
g(g(x)) œ g(x# 3) œ (x# 3)# 3 œ x% 6x# 6
6. (a) f ˆg ˆ "# ‰‰ œ f ˆ 23 ‰ œ 3"
(b) g ˆf ˆ "# ‰‰ œ g ˆ "# ‰ œ 2
(c) f(g(x)) œ f ˆ x " 1 ‰ œ
"
x 1
1œ
(d) g(f(x)) œ g(x 1) œ
"
(x1) 1
(e) f(f(2)) œ f(1) œ 0
(f) g(g(2)) œ g ˆ "3 ‰ œ
œ
"
4
3
œ
x
x1
"
x
3
4
(g) f(f(x)) œ f(x 1) œ (x 1) 1 œ x 2
"
(h) g(g(x)) œ g ˆ x " 1 ‰ œ " " 1 œ xx # (x Á 1 and x Á 2)
x1
#
7. (a) u(v(f(x))) œ u ˆv ˆ "x ‰‰ œ u ˆ x"# ‰ œ 4 ˆ x" ‰ 5 œ x4# 5
(b) u(f(v(x))) œ u af ax# bb œ u ˆ x"# ‰ œ 4 ˆ x"# ‰ 5 œ x4# 5
#
(c) v(u(f(x))) œ v ˆu ˆ "x ‰‰ œ v ˆ4 ˆ x" ‰ 5‰ œ ˆ 4x 5‰
(d) v(f(u(x))) œ v(f(4x 5)) œ v ˆ 4x " 5 ‰ œ ˆ 4x " 5 ‰
(e) f(u(v(x))) œ f au ax# bb œ f a4 ax# b 5b œ
"
4x# 5
(f) f(v(u(x))) œ f(v(4x 5)) œ f a(4x 5)# b œ
8. (a) h(g(f(x))) œ h ˆg ˆÈx‰‰ œ h Š
Èx
4 ‹
#
œ 4Š
"
(4x 5)#
Èx
4 ‹
8 œ Èx 8
(b) h(f(g(x))) œ h ˆf ˆ x4 ‰‰ œ h ˆÈ x4 ‰ œ 4È x4 8 œ 2Èx 8
4È x 8
œ Èx 2
4
È4x 8
È
œ 4
œ x# 2
(c) g(h(f(x))) œ g ˆh ˆÈx‰‰ œ g ˆ4Èx 8‰ œ
(d) g(f(h(x))) œ g(f(4x 8)) œ g ŠÈ4x 8‹
(e) f(g(h(x))) œ f(g(4x 8)) œ f ˆ 4x 4 8 ‰ œ f(x 2) œ Èx 2
(f) f(h(g(x))) œ f ˆh ˆ x ‰‰ œ f ˆ4 ˆ x ‰ 8‰ œ f(x 8) œ Èx 8
4
4
9. (a) y œ f(g(x))
(c) y œ g(g(x))
(e) y œ g(h(f(x)))
(b) y œ j(g(x))
(d) y œ j(j(x))
(f) y œ h(j(f(x)))
10. (a) y œ f(j(x))
(c) y œ h(h(x))
(e) y œ j(g(f(x)))
(b) y œ h(g(x)) œ g(h(x))
(d) y œ f(f(x))
(f) y œ g(f(h(x)))
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27
28
Chapter 1 Preliminaries
g(x)
f(x)
(f ‰ g)(x)
(a)
x7
Èx
Èx 7
(b)
x2
3x
3(x 2) œ 3x 6
(c)
x#
Èx 5
Èx# 5
(d)
x
x1
x
x1
"
x1
"
x
1
11.
(e)
(f)
x
xc1
x
xc1 1
"
x
(b) af‰gbaxb œ
gaxb"
g ax b
œ
x
x (x1)
œx
x
"
x
12. (a) af‰gbaxb œ lgaxbl œ
œ
x
"
lx "l .
x
x"
Ê"
"
g ax b
œ
x
x"
Ê"
x
x"
œ
"
g ax b
Ê
"
x"
œ
"
gaxb ß so
gaxb œ x ".
(c) Since af‰gbaxb œ Ègaxb œ lxl, gaxb œ x .
(d) Since af‰gbaxb œ fˆÈx‰ œ l x l, faxb œ x# . (Note that the domain of the composite is Ò!ß _Ñ.)
#
The completed table is shown. Note that the absolute value sign in part (d) is optional.
gaxb
faxb
af‰gbaxb
"
"
lxl
x"
lx "l
x"
x
x"
Èx
x#
Èx
x#
x
x"
lxl
lxl
13. (a) fagaxbb œ É 1x 1 œ É 1x x
gafaxbb œ
1
Èx 1
(b) Domain af‰gb: Ð0, _Ñ, domain ag‰f b: Ð1, _Ñ
(c) Range af‰gb: Ð1, _Ñ, range ag‰f b: Ð0, _Ñ
14. (a) fagaxbb œ 1 2Èx x
gafaxbb œ 1 kxk
(b) Domain af‰gb: Ð0, _Ñ, domain ag‰f b: Ð0, _Ñ
(c) Range af‰gb: Ð0, _Ñ, range ag‰f b: Ð_, 1Ñ
15. (a) y œ (x 7)#
(b) y œ (x 4)#
16. (a) y œ x# 3
(b) y œ x# 5
17. (a) Position 4
(b) Position 1
(c) Position 2
(d) Position 3
18. (a) y œ (x 1)# 4
(b) y œ (x 2)# 3
(c) y œ (x 4)# 1
(d) y œ (x 2)#
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Section 1.5 Combining Functions; Shifting and Scaling Graphs
19.
20.
21.
22.
23.
24.
25.
26.
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29
30
Chapter 1 Preliminaries
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 1.5 Combining Functions; Shifting and Scaling Graphs
37.
38.
39.
40.
41.
42.
43.
44.
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31
32
Chapter 1 Preliminaries
45.
46.
47.
48.
49. (a) domain: [0ß 2]; range: [#ß $]
(b) domain: [0ß 2]; range: [1ß 0]
(c) domain: [0ß 2]; range: [0ß 2]
(d) domain: [0ß 2]; range: [1ß 0]
(e) domain: [2ß 0]; range: [!ß 1]
(f) domain: [1ß 3]; range: [!ß "]
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Section 1.5 Combining Functions; Shifting and Scaling Graphs
(g) domain: [2ß 0]; range: [!ß "]
(h) domain: [1ß 1]; range: [!ß "]
50. (a) domain: [0ß 4]; range: [3ß 0]
(b) domain: [4ß 0]; range: [!ß $]
(c) domain: [4ß 0]; range: [!ß $]
(d) domain: [4ß 0]; range: ["ß %]
(e) domain: [#ß 4]; range: [3ß 0]
(f) domain: [2ß 2]; range: [3ß 0]
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33
34
Chapter 1 Preliminaries
(g) domain: ["ß 5]; range: [3ß 0]
(h) domain: [0ß 4]; range: [0ß 3]
51. y œ 3x# 3
52. y œ a2xb# 1 œ %x# 1
53. y œ "# ˆ" 54. y œ 1 "‰
x#
"
axÎ$b#
œ
"
#
œ1
"
#x#
*
x#
55. y œ È%x 1
56. y œ 3Èx 1
#
57. y œ É% ˆ x# ‰ œ "# È16 x#
58. y œ "$ È% x#
59. y œ " a3xb$ œ " 27x$
$
60. y œ " ˆ x# ‰ œ " x$
)
"Î#
"Î#
61. Let y œ È#x " œ faxb and let gaxb œ x"Î# , haxb œ ˆx "# ‰ , iaxb œ È#ˆx "# ‰ , and
"Î#
jaxb œ ’È#ˆx "# ‰ “ œ faBb. The graph of haxb is the graph of gaxb shifted left
"
#
unit; the graph of iaxb is the graph
of haxb stretched vertically by a factor of È#; and the graph of jaxb œ faxb is the graph of iaxb reflected across the x-axis.
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Section 1.5 Combining Functions; Shifting and Scaling Graphs
62. Let y œ È" x
#
œ faxbÞ Let gaxb œ axb"Î# , haxb œ ax #b"Î# , and iaxb œ
"
È # a x
#b"Î# œ È" x
#
35
œ faxbÞ
The graph of gaxb is the graph of y œ Èx reflected across the x-axis. The graph of haxb is the graph of gaxb shifted right
two units. And the graph of iaxb is the graph of haxb compressed vertically by a factor of È#.
63. y œ faxb œ x$ . Shift faxb one unit right followed by a shift two units up to get gaxb œ ax "b3 #.
64. y œ a" Bb$ # œ Òax "b$ a#bÓ œ faxb. Let gaxb œ x$ , haxb œ ax "b$ , iaxb œ ax "b$ a#b, and
jaxb œ Òax "b$ a#bÓ. The graph of haxb is the graph of gaxb shifted right one unit; the graph of iaxb is the graph of
haxb shifted down two units; and the graph of faxb is the graph of iaxb reflected across the x-axis.
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36
Chapter 1 Preliminaries
65. Compress the graph of faxb œ
get haxb œ
"
#x
".
66. Let faxb œ
"
x#
and gaxb œ
#
x#
"
x
horizontally by a factor of 2 to get gaxb œ
"œ
"
#
Š B# ‹
"œ
"
#
ŠxÎÈ#‹
"œ
"
#
’Š"ÎÈ#‹B“
"
#x .
Then shift gaxb vertically down 1 unit to
"Þ Since È# ¸ "Þ%, we see that the graph of
faxb stretched horizontally by a factor of 1.4 and shifted up 1 unit is the graph of gaxb.
$
$
67. Reflect the graph of y œ faxb œ È
x across the x-axis to get gaxb œ È
x.
68. y œ faxb œ a#xb#Î$ œ Òa"ba#bxÓ#Î$ œ a"b#Î$ a#xb#Î$ œ a#xb#Î$ . So the graph of faxb is the graph of gaxb œ x#Î$
compressed horizontally by a factor of 2.
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Section 1.5 Combining Functions; Shifting and Scaling Graphs
69.
70.
71. *x# #&y# œ ##& Ê
x#
&#
73. $x# ay #b# œ $ Ê
x#
"#
75. $ax "b# #ay #b# œ '
Ê
ax " b #
#
ŠÈ#‹
y a#b‘#
#
ŠÈ$‹
y#
$#
a y #b #
#
ŠÈ$‹
œ"
74. ax "b# #y# œ % Ê
y#
%#
x a"b‘#
##
œ"
#
#
76. 'ˆx $# ‰ *ˆy "# ‰ œ &%
#
œ"
x#
#
È
Š (‹
72. "'x# (y# œ ""# Ê
œ"
Ê
’xˆ $# ‰“
$#
ˆy "# ‰#
#
ŠÈ'‹
œ"
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y#
#
ŠÈ#‹
œ"
37
38
77.
Chapter 1 Preliminaries
x#
"'
y#
*
œ " has its center at a!ß !b. Shiftinig 4 units left and 3 units up gives the center at ah, kb œ a%ß $b. So the
equation is
x a4b‘#
4#
ay 3 b #
3#
œ"Ê
ax %b #
4#
a y $b #
3#
œ ". Center, C, is a%ß $b, and major axis, AB, is the segment
from a)ß $b to a!ß $b.
78. The ellipse
x#
%
y#
#&
œ " has center ah, kb œ a!ß !b. Shifting the ellipse 3 units right and 2 units down produces an ellipse
with center at ah, kb œ a$ß #b and an equation
a$ß $b to a$ß (b is the major axis.
ax 3 b#
%
y a#b‘#
#&
œ ". Center, C, is a3ß #b, and AB, the segment from
79. (a) (fg)(x) œ f(x)g(x) œ f(x)(g(x)) œ (fg)(x), odd
(b) Š gf ‹ (x) œ
(c) ˆ gf ‰ (x) œ
(d)
(e)
(f)
(g)
(h)
(i)
f(x)
g(x)
g(x)
f(x)
œ
œ
f(x)
g(x)
g(x)
f(x)
œ Š gf ‹ (x), odd
œ ˆ gf ‰ (x), odd
f # (x) œ f(x)f(x) œ f(x)f(x) œ f # (x), even
g# (x) œ (g(x))# œ (g(x))# œ g# (x), even
(f ‰ g)(x) œ f(g(x)) œ f(g(x)) œ f(g(x)) œ (f ‰ g)(x), even
(g ‰ f)(x) œ g(f(x)) œ g(f(x)) œ (g ‰ f)(x), even
(f ‰ f)(x) œ f(f(x)) œ f(f(x)) œ (f ‰ f)(x), even
(g ‰ g)(x) œ g(g(x)) œ g(g(x)) œ g(g(x)) œ (g ‰ g)(x), odd
80. Yes, f(x) œ 0 is both even and odd since f(x) œ 0 œ f(x) and f(x) œ 0 œ f(x).
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Section 1.6 Trigonometric Functions
81. (a)
(b)
(c)
(d)
82.
1.6 TRIGONOMETRIC FUNCTIONS
1. (a) s œ r) œ (10) ˆ 451 ‰ œ 81 m
radians and
51
4
1 ‰
3. ) œ 80° Ê ) œ 80° ˆ 180°
œ
41
9
2. ) œ
s
r
œ
101
8
œ
51
4
1 ‰
(b) s œ r) œ (10)(110°) ˆ 180°
œ
1101
18
œ
551
9
m
ˆ 180°
‰ œ 225°
1
Ê s œ (6) ˆ 491 ‰ œ 8.4 in. (since the diameter œ 12 in. Ê radius œ 6 in.)
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39
40
Chapter 1 Preliminaries
4. d œ 1 meter Ê r œ 50 cm Ê ) œ
5.
1
)
231
0
1
#
s
r
œ
30
50
31
4
"
È2
È"
2
sin )
0
cos )
1
È
#3
"#
tan )
0
È3
0
und.
"
und.
"
È3
und.
0
1
und.
È 2
cot )
1
#
und.
È23
sec )
csc )
0
"
"
0
"
und.
7. cos x œ 45 , tan x œ 34
9. sin x œ È8
3
, tan x œ È8
"
‰ ¸ 34°
œ 0.6 rad or 0.6 ˆ 180°
1
6.
È2
3#1
)
sin )
"
cos )
!
"
#
tan )
und.
È 3
cot )
!
È"3
sec )
und.
#
csc )
"
È23
2
È5
10. sin x œ
12
13
11. sin x œ È"5 , cos x œ È25
12. cos x œ 13.
14.
15.
1'
È
#3
8. sin x œ
period œ 1
13
, cos x œ
"
È2
&1
'
"
#
È
#3
È"3
"
È"3
È 3
"
È 3
2
È3
È2
È23
#
È2
#
"#
È3
#
"
È5
, tan x œ 12
5
È3
#
, tan x œ
"
È3
period œ 41
16.
period œ 2
17.
period œ 4
18.
period œ 6
period œ 1
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1
%
"
È2
Section 1.6 Trigonometric Functions
19.
20.
period œ 21
period œ 21
21.
22.
period œ 21
period œ 21
23. period œ 1# , symmetric about the origin
24. period œ 1, symmetric about the origin
25. period œ 4, symmetric about the y-axis
26. period œ 41, symmetric about the origin
27. (a) Cos x and sec x are positive in QI and QIV and
negative in QII and QIII. Sec x is undefined when
cos x is 0. The range of sec x is (_ß 1] ["ß _);
the range of cos x is ["ß 1].
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42
Chapter 1 Preliminaries
(b) Sin x and csc x are positive in QI and QII and
negative in QIII and QIV. Csc x is undefined when
sin x is 0. The range of csc x is (_ß 1] [1ß _);
the range of sin x is ["ß "].
28. Since cot x œ
"
tan x
, cot x is undefined when tan x œ 0
and is zero when tan x is undefined. As tan x approaches
zero through positive values, cot x approaches infinity.
Also, cot x approaches negative infinity as tan x
approaches zero through negative values.
29. D: _ x _; R: y œ 1, 0, 1
30. D: _ x _; R: y œ 1, 0, 1
31. cos ˆx 1# ‰ œ cos x cos ˆ 1# ‰ sin x sin ˆ 1# ‰ œ (cos x)(0) (sin x)(1) œ sin x
32. cos ˆx 1# ‰ œ cos x cos ˆ 1# ‰ sin x sin ˆ 1# ‰ œ (cos x)(0) (sin x)(1) œ sin x
33. sin ˆx 1# ‰ œ sin x cos ˆ 1# ‰ cos x sin ˆ 1# ‰ œ (sin x)(0) (cos x)(1) œ cos x
34. sin ˆx 1# ‰ œ sin x cos ˆ 1# ‰ cos x sin ˆ 1# ‰ œ (sin x)(0) (cos x)(1) œ cos x
35. cos (A B) œ cos (A (B)) œ cos A cos (B) sin A sin (B) œ cos A cos B sin A (sin B)
œ cos A cos B sin A sin B
36. sin (A B) œ sin (A (B)) œ sin A cos (B) cos A sin (B) œ sin A cos B cos A (sin B)
œ sin A cos B cos A sin B
37. If B œ A, A B œ 0 Ê cos (A B) œ cos 0 œ 1. Also cos (A B) œ cos (A A) œ cos A cos A sin A sin A
œ cos# A sin# A. Therefore, cos# A sin# A œ 1.
38. If B œ 21, then cos (A 21) œ cos A cos 21 sin A sin 21 œ (cos A)(1) (sin A)(0) œ cos A and
sin (A 21) œ sin A cos 21 cos A sin 21 œ (sin A)(1) (cos A)(0) œ sin A. The result agrees with the
fact that the cosine and sine functions have period 21.
39. cos (1 x) œ cos 1 cos B sin 1 sin x œ (1)(cos x) (0)(sin x) œ cos x
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Section 1.6 Trigonometric Functions
40. sin (21 x) œ sin 21 cos (x) cos (21) sin (x) œ (0)(cos (x)) (1)(sin (x)) œ sin x
41. sin ˆ 3#1 x‰ œ sin ˆ 3#1 ‰ cos (x) cos ˆ 3#1 ‰ sin (x) œ (1)(cos x) (0)(sin (x)) œ cos x
42. cos ˆ 3#1 x‰ œ cos ˆ 3#1 ‰ cos x sin ˆ 3#1 ‰ sin x œ (0)(cos x) (1)(sin x) œ sin x
œ sin ˆ 14 13 ‰ œ sin
44. cos
111
1#
45. cos
1
12
œ cos ˆ 13 14 ‰ œ cos
46. sin
51
1#
œ sin ˆ 231 14 ‰ œ sin ˆ 231 ‰ cos ˆ 14 ‰ cos ˆ 231 ‰ sin ˆ 14 ‰ œ Š
21 ‰
3
cos
œ cos
È
47. cos#
1
8
œ
1cos ˆ 281 ‰
#
œ
1 # 2
#
49. sin#
1
1#
œ
1cos ˆ 211# ‰
#
œ
1 # 3
#
È
1
3
1
4
1
3
cos
cos
21
3
1
4
1
3
È2
È3
# ‹Š # ‹
71
1#
œ cos ˆ 14 1
4
È2
ˆ"‰
# ‹ #
43. sin
sin
sin
1
4
cos ˆ 14 ‰ sin
œŠ
sin
1
3
21
3
œŠ
Š
È2
ˆ "‰
# ‹ #
sin ˆ 14 ‰ œ ˆ "# ‰ Š
œ
2 È 2
4
48. cos#
1
1#
œ
2 È 3
4
50. sin#
1
8
Š
È2
# ‹
œ
È2
È3
# ‹Š # ‹
Š
1cos ˆ 211# ‰
#
1cos ˆ 281 ‰
#
51. tan (A B) œ
sin (AB)
cos (AB)
œ
sin A cos Bcos A cos B
cos A cos Bsin A sin B
œ
sin A cos B
cos A sin B
cos A cos B cos A cos B
sin A sin B
cos A cos B
cos A cos B cos A cos B
œ
tan Atan B
1tan A tan B
52. tan (A B) œ
sin (AB)
cos (AB)
œ
sin A cos Bcos A cos B
cos A cos Bsin A sin B
œ
sin A cos B
cos A sin B
cos A cos B cos A cos B
sin A sin B
cos A cos B
cos A cos B cos A cos B
œ
tan Atan B
1tan A tan B
È 2 È 6
4
œ
È3
È2
# ‹ Š # ‹
È3
È2
# ‹Š # ‹
œ
È 6 È 2
4
œ
ˆ "# ‰ Š
œ
œ
œ
È
1 # 3
#
È
1 # 2
#
œ
œ
1 È 3
2È 2
È2
# ‹
œ
1 È 3
2È 2
2 È 3
4
2 È 2
4
53. According to the figure in the text, we have the following: By the law of cosines, c# œ a# b# 2ab cos )
œ 1# 1# 2 cos (A B) œ 2 2 cos (A B). By distance formula, c# œ (cos A cos B)# (sin A sin B)#
œ cos# A 2 cos A cos B cos# B sin# A 2 sin A sin B sin# B œ 2 2(cos A cos B sin A sin B). Thus
c# œ 2 2 cos (A B) œ 2 2(cos A cos B sin A sin B) Ê cos (A B) œ cos A cos B sin A sin B.
54. (a) cosaA Bb œ cos A cos B sin A sin B
sin ) œ cosˆ 1# )‰ and cos ) œ sinˆ 1# )‰
Let ) œ A B
sinaA Bb œ cos’ 1# aA Bb“ œ cos’ˆ 1# A‰ B“ œ cos ˆ 1# A‰ cos B sin ˆ 1# A‰ sin B
œ sin A cos B cos A sin B
(b) cosaA Bb œ cos A cos B sin A sin B
cosaA aBbb œ cos A cos aBb sin A sin aBb
Ê cosaA Bb œ cos A cos aBb sin A sin aBb œ cos A cos B sin A asin Bb
œ cos A cos B sin A sin B
Because the cosine function is even and the sine functions is odd.
55. c# œ a# b# 2ab cos C œ 2# 3# 2(2)(3) cos (60°) œ 4 9 12 cos (60°) œ 13 12 ˆ "# ‰ œ 7.
Thus, c œ È7 ¸ 2.65.
56. c# œ a# b# 2ab cos C œ 2# 3# 2(2)(3) cos (40°) œ 13 12 cos (40°). Thus, c œ È13 12 cos 40° ¸ 1.951.
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Chapter 1 Preliminaries
57. From the figures in the text, we see that sin B œ hc . If C is an acute angle, then sin C œ hb . On the other hand,
if C is obtuse (as in the figure on the right), then sin C œ sin (1 C) œ hb . Thus, in either case,
h œ b sin C œ c sin B Ê ah œ ab sin C œ ac sin B.
a # b # c #
2ab
By the law of cosines, cos C œ
and cos B œ
a # c # b #
.
2ac
Moreover, since the sum of the
interior angles of a triangle is 1, we have sin A œ sin (1 (B C)) œ sin (B C) œ sin B cos C cos B sin C
#
#
#
#
#
#
b c
c b ˆ h ‰
h ‰
œ ˆ hc ‰ ’ a 2ab
a2a# b# c# c# b# b œ
“ ’ a 2ac
“ b œ ˆ 2abc
ah
bc
Ê ah œ bc sin A.
Combining our results we have ah œ ab sin C, ah œ ac sin B, and ah œ bc sin A. Dividing by abc gives
h
sin A
sin C
sin B
bc œ ðóóóóóóóñóóóóóóóò
a œ c œ b .
law of sines
58. By the law of sines,
Thus sin B œ
3È 3
2È 7
sin A
#
œ
sin B
3
œ
È3/2
c .
By Exercise 55 we know that c œ È7.
¶ 0.982.
59. From the figure at the right and the law of cosines,
b# œ a# 2# 2(2a) cos B
œ a# 4 4a ˆ "# ‰ œ a# 2a 4.
Applying the law of sines to the figure,
Ê
È2/2
a
œ
È3/2
b
sin A
a
œ
sin B
b
Ê b œ É 3# a. Thus, combining results,
a# 2a 4 œ b# œ
3
#
"
#
a# Ê 0 œ
a# 2a 4
Ê 0 œ a# 4a 8. From the quadratic formula and the fact that a 0, we have
aœ
4È4# 4(1)(8)
#
œ
4 È 3 4
#
¶ 1.464.
60. (a) The graphs of y œ sin x and y œ x nearly coincide when x is near the origin (when the calculator
is in radians mode).
(b) In degree mode, when x is near zero degrees the sine of x is much closer to zero than x itself. The
curves look like intersecting straight lines near the origin when the calculator is in degree mode.
61. A œ 2, B œ 21, C œ 1, D œ 1
62. A œ "# , B œ 2, C œ 1, D œ
"
#
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Section 1.6 Trigonometric Functions
63. A œ 12 , B œ 4, C œ 0, D œ
64. A œ
L
21 ,
"
1
B œ L, C œ 0, D œ 0
65. (a) amplitude œ kAk œ 37
(c) right horizontal shift œ C œ 101
(b) period œ kBk œ 365
(d) upward vertical shift œ D œ 25
66. (a) It is highest when the value of the sine is 1 at f(101) œ 37 sin (0) 25 œ 62° F.
The lowest mean daily temp is 37(1) 25 œ 12° F.
(b) The average of the highest and lowest mean daily temperatures œ
The average of the sine function is its horizontal axis, y œ 25.
62°(12)°
#
œ 25° F.
67-70. Example CAS commands:
Maple
f := x -> A*sin((2*Pi/B)*(x-C))+D1;
A:=3; C:=0; D1:=0;
f_list := [seq( f(x), B=[1,3,2*Pi,5*Pi] )];
plot( f_list, x=-4*Pi..4*Pi, scaling=constrained,
color=[red,blue,green,cyan], linestyle=[1,3,4,7],
legend=["B=1","B=3","B=2*Pi","B=3*Pi"],
title="#67 (Section 1.6)" );
Mathematica
Clear[a, b, c, d, f, x]
f[x_]:=a Sin[21/b (x c)] + d
Plot[f[x]/.{a Ä 3, b Ä 1, c Ä 0, d Ä 0}, {x, 41, 41 }]
67. (a) The graph stretches horizontally.
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46
Chapter 1 Preliminaries
(b) The period remains the same: period œ l B l. The graph has a horizontal shift of
"
#
period.
68. (a) The graph is shifted right C units.
(b) The graph is shifted left C units.
(c) A shift of „ one period will produce no apparent shift. l C l œ '
69. The graph shifts upwards l D lunits for D ! and down l D lunits for D !Þ
70. (a) The graph stretches l A l units.
(b) For A !, the graph is inverted.
1.7 GRAPHING WITH CALCULATORS AND COMPUTERS
1-4.
The most appropriate viewing window displays the maxima, minima, intercepts, and end behavior of the graphs and
has little unused space.
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Section 1.7 Graphing with Calculators and Computers
1. d.
2. c.
3. d.
4. b.
5-30.
For any display there are many appropriate display widows. The graphs given as answers in Exercises 530
are not unique in appearance.
5. Ò2ß 5Ó by Ò15ß 40Ó
6. Ò4ß 4Ó by Ò4ß 4Ó
7. Ò2ß 6Ó by Ò250ß 50Ó
8. Ò1ß 5Ó by Ò5ß 30Ó
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48
Chapter 1 Preliminaries
9. Ò4ß 4Ó by Ò5ß 5Ó
10. Ò2ß 2Ó by Ò2ß 8Ó
11. Ò2ß 6Ó by Ò5ß 4Ó
12. Ò4ß 4Ó by Ò8ß 8Ó
13. Ò"ß 'Ó by Ò"ß %Ó
14. Ò"ß 'Ó by Ò"ß &Ó
15. Ò3ß 3Ó by Ò!ß "!Ó
16. Ò"ß #Ó by Ò!ß "Ó
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Section 1.7 Graphing with Calculators and Computers
17. Ò&ß "Ó by Ò&ß &Ó
18. Ò&ß "Ó by Ò#ß %Ó
19. Ò%ß %Ó by Ò!ß $Ó
20. Ò&ß &Ó by Ò#ß #Ó
21. Ò"!ß "!Ó by Ò'ß 'Ó
22. Ò&ß &Ó by Ò#ß #Ó
23. Ò'ß "!Ó by Ò'ß 'Ó
24. Ò$ß &Ó by Ò#ß "!Ó
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49
50
Chapter 1 Preliminaries
25. Ò!Þ!$ß !Þ!$Ó by Ò"Þ#&ß "Þ#&Ó
26. Ò!Þ"ß !Þ"Ó by Ò$ß $Ó
27. Ò$!!ß $!!Ó by Ò"Þ#&ß "Þ#&Ó
28. Ò&!ß &!Ó by Ò!Þ"ß !Þ"Ó
29. Ò!Þ#&ß !Þ#&Ó by Ò!Þ$ß !Þ$Ó
30. Ò!Þ"&ß !Þ"&Ó by Ò!Þ!#ß !Þ!&Ó
31. x# #x œ % %y y# Ê y œ # „ Èx# #x ).
The lower half is produced by graphing
y œ # Èx# #x ).
32. y# "'x# œ " Ê y œ „ È" "'x# . The upper branch
is produced by graphing y œ È" "'x# .
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Section 1.7 Graphing with Calculators and Computers
33.
34.
35.
36.
37.
38Þ
39.
40.
41. (a) y œ "!&*Þ"%x #!(%*(#
(b) m œ "!&*Þ"% dollars/year, which is the yearly increase in compensation.
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51
52
Chapter 1 Preliminaries
(c)
(d) Answers may vary slightly. y œ a"!&*Þ14ba#!"!b #!(%*(# œ $&$ß 899
42. (a) Let C œ cost and x œ year.
C œ a(*'!Þ("bx "Þ' ‚ "!(
(b) Slope represents increase in cost per year
(c) C œ a#'$(Þ"%bx &Þ# ‚ "!'
(d) The median price is rising faster in the northeast (the slope is larger).
43. (a) Let x represent the speed in miles per hour and d the stopping distance in feet. The quadratic regression function is
d œ !Þ!)''x# "Þ*(x &!Þ".
(b)
(c) From the graph in part (b), the stopping distance is about $(! feet when the vehicle is (# mph and it is about &#& feet
when the speed is )& mph.
Algebraically: dquadratic a(#b œ !Þ!)''a(#b# "Þ*(a(#b &!Þ" œ $'(Þ' ft.
dquadratic a)&b œ !Þ!)''a)&b# "Þ*(a)&b &!Þ" œ &##Þ) ft.
(d) The linear regression function is d œ 'Þ)*x "%!Þ% Ê dlinear a(#b œ 'Þ)*a(#b "%!Þ% œ $&&Þ( ft and
dlinear a)&b œ 'Þ)*a)&b "%!Þ% œ %%&Þ# ft. The linear regression line is shown on the graph in part (b). The quadratic
regression curve clearly gives the better fit.
44. (a) The power regression function is y œ %Þ%%'%(x!Þ&""%"% .
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Chapter 1 Practice Exercises
(b)
(c) 15Þ2 km/h
(d) The linear regression function is y œ !Þ*"$'(&x %Þ")**(' and it is shown on the graph in part (b). The linear
regession function gives a speed of "%Þ# km/h when y œ "" m. The power regression curve in part (a) better fits the
data.
CHAPTER 1 PRACTICE EXERCISES
1. ( 2x
$ Ê #x
% Ê x
2.
3x "! Ê x "!
$
3.
"
& ax
#
qqqqqqqqðïïïïïïïî
x
"!
$
"b "% ax #b Ê %ax "b &ax #b
Ê %x % &x "! Ê ' x
4.
x$
#
%$ x Ê $ax $b
Ê $x *
5.
#a% xb
) #x Ê &x
"Êx
qqqqqqqqñïïïïïïïî
x
"
&
"
&
lx " l œ ( Ê x " œ ( or ax "b œ ( Ê x œ ' or x œ )
6. ly $ l % Ê % y $ % Ê " y (
7. ¹" x# ¹ $
#
Ê "
x
#
$# or " x
#
$
#
Ê x# &# or x# "
#
Ê x & or x "
Ê x & or x "
8. ¹ #x$( ¹ Ÿ & Ê & Ÿ
#x(
$
Ÿ & Ê 1& Ÿ #x ( Ÿ 1& Ê 22 Ÿ #x Ÿ 8 Ê "" Ÿ x Ÿ %
9. Since the particle moved to the y-axis, # ?x œ ! Ê ?x œ 2. Since ?y œ 3?x œ 6, the new coordinates
are (x ?xß y ?y) œ (# #ß & ') œ (0ß 11).
10. (a)
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53
54
Chapter 1 Preliminaries
(b)
line
AB
slope
10 1
9
3
2 8 œ 6 œ #
10 6
4
2
2 (4) œ 6 œ 3
6 (3)
9
3
% 2 œ 6 œ #
1 (3)
4
2
82 œ 6 œ 3
66
œ0
% 14
3
BC
CD
DA
CE
BD
is vertical and has no slope
(c) Yes; A, B, C and D form a parallelogram.
3 ˆ 14
‰
(d) Yes. The line AB has equation y 1 œ 3# (x 8). Replacing x by 14
3 gives y œ # 3 8 "
3 ˆ
10 ‰
14
œ # 3 1 œ 5 1 œ 6. Thus, E ˆ 3 ß 6‰ lies on the line AB and the points A, B and E are collinear.
(e) The line CD has equation y 3 œ 3# (x 2) or y œ 3# x. Thus the line passes through the origin.
11. The triangle ABC is neither an isosceles triangle nor is it a right triangle. The lengths of AB, BC and AC are
È53, È72 and È65, respectively. The slopes of AB, BC and AC are 7 , 1 and " , respectively.
#
8
12. P(xß 3x 1) is a point on the line y œ 3x 1. If the distance from P to (!ß 0) equals the distance from P to
($ß %), then x# (3x 1)# œ (x 3)# (3 3x)# Ê x# 9x# 6x 1 œ x# 6x 9 9 18x 9x#
23
ˆ 17 ‰
ˆ 17 23 ‰
Ê 18x œ 17 or x œ 17
18 Ê y œ 3x 1 œ 3 18 1 œ 6 . Thus the point is P 18 ß 6 .
13. y œ $ax "b a'b Ê y œ $x *
14. y œ "# ax "b # Ê y œ "# x $
#
15. x œ !
16. m œ
# '
" a$b
œ
)
%
œ # Ê y œ #ax $b ' Ê y œ #x
17. y œ #
18. m œ
&$
# $
œ
#
&
œ &# Ê y œ &# ax $b $ Ê y œ &# x #"
&
19. y œ $x $
20. Since #x y œ # is equivalent to y œ #x #, the slope of the given line (and hence the slope of the desired line) is 2.
y œ #a x "b " Ê y œ # x &
21. Since %x $y œ "# is equivalent to y œ %$ x %, the slope of the given line (and hence the slope of the desired line) is
%$ . y œ %$ ax 4b "2 Ê y œ %$ x #!
$
22. Since $x &y œ " is equivalent to y œ $& x "& , the slope of the given line is
5$ .
yœ
5$ ax
#b $ Ê y œ
5$ x
"*
$
$
&
and the slope of the perpendicular line is
23. Since "# x "$ y œ " is equivalent to y œ $# x $, the slope of the given line is $# and the slope of the perpendicular line
is #$ . y œ #$ ax "b # Ê y œ #$ x )
$
24. The line passes through a!ß &b and a$ß !b. m œ
! a&b
$!
œ
&
$
Ê y œ $& x &
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Chapter 1 Practice Exercises
25. The area is A œ 1 r# and the circumference is C œ #1 r. Thus, r œ
26. The surface area is S œ %1 r# Ê r œ ˆ %S1 ‰
surface area gives S œ %1 r# œ %1 ˆ $%V1 ‰
"Î#
#Î$
C
#1
#
Ê A œ 1ˆ #C1 ‰ œ
C#
%1 .
$ $V
. The volume is V œ %$ 1 r$ Ê r œ É
%1 . Substitution into the formula for
.
27. The coordinates of a point on the parabola are axß x# b. The angle of inclination ) joining this point to the origin satisfies
the equation tan ) œ
28. tan ) œ
rise
run
œ
h
&!!
x#
x
œ x. Thus the point has coordinates axß x# b œ atan )ß tan# )b.
Ê h œ &!! tan ) ft.
29.
30.
Symmetric about the origin.
Symmetric about the y-axis.
31.
32.
Neither
Symmetric about the y-axis.
33. yaxb œ axb# " œ x# " œ yaxb. Even.
34. yaxb œ axb& axb$ axb œ x& x$ x œ yaxb. Odd.
35. yaxb œ " cosaxb œ " cos x œ yaxb. Even.
36. yaxb œ secaxb tanaxb œ
37. yaxb œ
axb% "
axb$ #axb
œ
x% "
x$ #x
sinaxb
cos# axb
œ
sin x
cos# x
œ sec x tan x œ yaxb. Odd.
%
"
œ xx$ #
x œ yaxb. Odd.
38. yaxb œ " sinaxb œ " sin x. Neither even nor odd.
39. yaxb œ x cosaxb œ x cos x. Neither even nor odd.
40. yaxb œ Éaxb% " œ Èx% " œ yaxb. Even.
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55
56
Chapter 1 Preliminaries
41. (a) The function is defined for all values of x, so the domain is a_ß _b.
(b) Since l x l attains all nonnegative values, the range is Ò#ß _Ñ.
42. (a) Since the square root requires " x !, the domain is Ð_ß "Ó.
(b) Since È" x attains all nonnegative values, the range is Ò#ß _Ñ.
43. (a) Since the square root requires "' x#
!, the domain is Ò%ß %Ó.
(b) For values of x in the domain, ! Ÿ "' x# Ÿ "', so ! Ÿ È"' x# Ÿ %. The range is Ò!ß %Ó.
44. (a) The function is defined for all values of x, so the domain is a_ß _b.
(b) Since $#x attains all positive values, the range is a"ß _b.
45. (a) The function is defined for all values of x, so the domain is a_ß _b.
(b) Since #ex attains all positive values, the range is a$ß _b.
46. (a) The function is equivalent to y œ tan #x, so we require #x Á
k1
#
for odd integers k. The domain is given by x Á
k1
%
for
odd integers k.
(b) Since the tangent function attains all values, the range is a_ß _b.
47. (a) The function is defined for all values of x, so the domain is a_ß _b.
(b) The sine function attains values from " to ", so # Ÿ #sina$x 1b Ÿ # and hence $ Ÿ #sina$x 1b " Ÿ ". The
range is Ò3ß 1Ó.
48. (a) The function is defined for all values of x, so the domain is a_ß _b.
&
(b) The function is equivalent to y œ È
x# , which attains all nonnegative values. The range is Ò!ß _Ñ.
49. (a) The logarithm requires x $ !, so the domain is a$ß _b.
(b) The logarithm attains all real values, so the range is a_ß _b.
50. (a) The function is defined for all values of x, so the domain is a_ß _b.
(b) The cube root attains all real values, so the range is a_ß _b.
51. (a) The function is defined for % Ÿ x Ÿ %, so the domain is Ò%ß %Ó.
(b) The function is equivalent to y œ Èl x l, % Ÿ x Ÿ %, which attains values from ! to # for x in the domain. The
range is Ò!ß #Ó.
52. (a) The function is defined for # Ÿ x Ÿ #, so the domain is Ò#ß #Ó.
(b) The range is Ò"ß "Ó.
53. First piece: Line through a!ß "b and a"ß !b. m œ
Second piece: Line through a"ß "b and a#ß !b. m
faxb œ œ
" x, ! Ÿ x "
# x, " Ÿ x Ÿ #
54. First piece: Line through a!ß !b and a2ß 5b. m œ
Second piece: Line through a2ß 5b and a4ß !b. m
faxb œ 10 5
2 x,
5x
2 ,
!"
"
"! œ " œ
"
"
œ !# " œ "
" Ê y œ x " œ " x
œ " Ê y œ ax "b " œ x # œ # x
5!
5
5
2! œ 2 Ê y œ 2x
5
5
5
œ !4 2 œ 2 œ 2 Ê
y œ 52 ax 2b 5 œ 52 x 10 œ 10 !Ÿx2
(Note: x œ 2 can be included on either piece.)
2ŸxŸ4
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5x
2
Chapter 1 Practice Exercises
55. (a) af‰gba"b œ faga"bb œ fŠ È"" # ‹ œ fa"b œ
(b) ag‰f ba#b œ gafa#bb œ gˆ "2 ‰ œ
(c) af‰f baxb œ fafaxbb œ fˆ "x ‰ œ
"
É "# #
"
"Îx
œ
"
È#Þ&
"
"
œ"
or É &#
œ x, x Á !
(d) ag‰gbaxb œ gagaxbb œ gŠ Èx" # ‹ œ
"
"
É Èx # #
œ
% x#
È
É " #È x #
$
56. (a) af‰gba"b œ faga"bb œ fˆÈ
" "‰ œ fa!b œ # ! œ #
$
(b) ag‰f ba#b œ faga#bb œ ga# #b œ ga!b œ È
!"œ"
(c) af‰f baxb œ fafaxbb œ fa# xb œ # a# xb œ x
$
$
$
È
(d) ag‰gbaxb œ gagaxbb œ gˆÈ
x "‰ œ É
x""
#
57. (a) af‰gbaxb œ fagaxbb œ fˆÈx #‰ œ # ˆÈx #‰ œ x, x #.
ag‰f baxb œ fagaxbb œ ga# x# b œ Èa# x# b # œ È% x#
(b) Domain of f‰g: Ò#ß _ÑÞ
Domain of g‰f: Ò#ß #ÓÞ
(c) Range of f‰g: Ð_ß #ÓÞ
Range of g‰f: Ò!ß #ÓÞ
%
58. (a) af‰gbaxb œ fagaxbb œ fŠÈ" x‹ œ ÉÈ" x œ È
" x.
ag‰f baxb œ fagaxbb œ gˆÈx‰ œ É" Èx
(b) Domain of f‰g: Ð_ß "ÓÞ
Domain of g‰f: Ò!ß "ÓÞ
(c) Range of f‰g: Ò!ß _ÑÞ
Range of g‰f: Ò!ß "ÓÞ
59.
60.
The graph of f# (x) œ f" akxkb is the same as the
graph of f" (x) to the right of the y-axis. The
graph of f# (x) to the left of the y-axis is the
reflection of y œ f" (x), x 0 across the y-axis.
The graph of f# (x) œ f" akxkb is the same as the
graph of f" (x) to the right of the y-axis. The
graph of f# (x) to the left of the y-axis is the
reflection of y œ f" (x), x 0 across the y-axis.
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58
Chapter 1 Preliminaries
61.
62.
The graph of f# (x) œ f" akxkb is the same as the
graph of f" (x) to the right of the y-axis. The
graph of f# (x) to the left of the y-axis is the
reflection of y œ f" (x), x 0 across the y-axis.
It does not change the graph.
63.
64.
The graph of f# (x) œ f" akxkb is the same as the
graph of f" (x) to the right of the y-axis. The
graph of f# (x) to the left of the y-axis is the
reflection of y œ f" (x), x 0 across the y-axis.
65.
66.
Whenever g" (x) is positive, the graph of y œ g# (x)
œ kg" (x)k is the same as the graph of y œ g" (x).
When g" (x) is negative, the graph of y œ g# (x) is
the reflection of the graph of y œ g" (x) across the
x-axis.
67.
The graph of f# (x) œ f" akxkb is the same as the
graph of f" (x) to the right of the y-axis. The
graph of f# (x) to the left of the y-axis is the
reflection of y œ f" (x), x 0 across the y-axis.
It does not change the graph.
Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is
the same as the graph of y œ g" (x). When g" (x) is negative, the
graph of y œ g# (x) is the reflection of the graph of y œ g" (x)
across the x-axis.
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Chapter 1 Practice Exercises
59
68.
Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is
the same as the graph of y œ g" (x). When g" (x) is negative, the
graph of y œ g# (x) is the reflection of the graph of y œ g" (x)
across the x-axis.
69.
70.
period œ 1
period œ 41
71.
72.
period œ 2
period œ 4
73.
74.
period œ 21
period œ 21
75. (a) sin B œ sin
1
3
œ
b
c
œ
b
#
Ê b œ 2 sin
1
3
œ 2Š
È3
# ‹
œ È3. By the theorem of Pythagoras,
a# b# œ c# Ê a œ Èc# b# œ È4 3 œ 1.
(b) sin B œ sin
1
3
œ
b
c
œ
2
c
Ê cœ
2
sin 13
œ È23 œ
Š ‹
#
4
È3
#
. Thus, a œ Èc# b# œ ÊŠ È43 ‹ (2)# œ É 43 œ
76. (a) sin A œ
a
c
Ê a œ c sin A
(b) tan A œ
a
b
Ê a œ b tan A
77. (a) tan B œ
b
a
Ê aœ
(b) sin A œ
a
c
Ê cœ
b
tan B
a
sin A
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2
È3
.
60
Chapter 1 Preliminaries
78. (a) sin A œ
(c) sin A œ
a
c
a
c
œ
È c # b #
c
79. Let h œ height of vertical pole, and let b and c denote the
distances of points B and C from the base of the pole,
measured along the flatground, respectively. Then,
tan 50° œ hc , tan 35° œ hb , and b c œ 10.
Thus, h œ c tan 50° and h œ b tan 35° œ (c 10) tan 35°
Ê c tan 50° œ (c 10) tan 35°
Ê c (tan 50° tan 35°) œ 10 tan 35°
tan 35°
Ê c œ tan10
50°tan 35° Ê h œ c tan 50°
œ
10 tan 35° tan 50°
tan 50°tan 35°
¸ 16.98 m.
80. Let h œ height of balloon above ground. From the figure at
the right, tan 40° œ ha , tan 70° œ hb , and a b œ 2. Thus,
h œ b tan 70° Ê h œ (2 a) tan 70° and h œ a tan 40°
Ê (2 a) tan 70° œ a tan 40° Ê a(tan 40° tan 70°)
70°
œ 2 tan 70° Ê a œ tan 240°tantan
70° Ê h œ a tan 40°
œ
2 tan 70° tan 40°
tan 40°tan 70°
¸ 1.3 km.
81. (a)
(b) The period appears to be 41.
(c) f(x 41) œ sin (x 41) cos ˆ x#41 ‰ œ sin (x 21) cos ˆ x# 21‰ œ sin x cos
since the period of sine and cosine is 21. Thus, f(x) has period 41.
x
#
82. (a)
(b) D œ (_ß 0) (!ß _); R œ [1ß 1]
(c) f is not periodic. For suppose f has period p. Then f ˆ #"1 kp‰ œ f ˆ #"1 ‰ œ sin 21 œ 0 for all
integers k. Choose k so large that
"
#1
kp "
1
Ê 0
"
(1/21)kp
1. But then
f ˆ #"1 kp‰ œ sin Š (1/#1")kp ‹ 0 which is a contradiction. Thus f has no period, as claimed.
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Chapter 1 Additional and Advanced Exercises
CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES
1. (a) The given graph is reflected about the y-axis.
(b) The given graph is reflected about the x-axis.
(c) The given graph is shifted left 1 unit, stretched
vertically by a factor of 2, reflected about the
x-axis, and then shifted upward 1 unit.
2. (a)
(d) The given graph is shifted right 2 units, stretched
vertically by a factor of 3, and then shifted
downward 2 units.
(b)
3. There are (infinitely) many such function pairs. For example, f(x) œ 3x and g(x) œ 4x satisfy
f(g(x)) œ f(4x) œ 3(4x) œ 12x œ 4(3x) œ g(3x) œ g(f(x)).
4. Yes, there are many such function pairs. For example, if g(x) œ (2x 3)$ and f(x) œ x"Î$ , then
(f ‰ g)(x) œ f(g(x)) œ f a(2x 3)$ b œ a(2x 3)$ b
"Î$
œ 2x 3.
5. If f is odd and defined at x, then f(x) œ f(x). Thus g(x) œ f(x) 2 œ f(x) 2 whereas
g(x) œ (f(x) 2) œ f(x) 2. Then g cannot be odd because g(x) œ g(x) Ê f(x) 2 œ f(x) 2
Ê 4 œ 0, which is a contradiction. Also, g(x) is not even unless f(x) œ 0 for all x. On the other hand, if f is
even, then g(x) œ f(x) 2 is also even: g(x) œ f(x) 2 œ f(x) 2 œ g(x).
6. If g is odd and g(0) is defined, then g(0) œ g(0) œ g(0). Therefore, 2g(0) œ 0 Ê g(0) œ 0.
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62
Chapter 1 Preliminaries
7. For (xß y) in the 1st quadrant, kxk kyk œ 1 x
Í x y œ 1 x Í y œ 1. For (xß y) in the 2nd
quadrant, kxk kyk œ x 1 Í x y œ x 1
Í y œ 2x 1. In the 3rd quadrant, kxk kyk œ x 1
Í x y œ x 1 Í y œ 2x 1. In the 4th
quadrant, kxk kyk œ x 1 Í x (y) œ x 1
Í y œ 1. The graph is given at the right.
8. We use reasoning similar to Exercise 7.
(1) 1st quadrant: y kyk œ x kxk
Í 2y œ 2x Í y œ x.
(2) 2nd quadrant: y kyk œ x kxk
Í 2y œ x (x) œ 0 Í y œ 0.
(3) 3rd quadrant: y kyk œ x kxk
Í y (y) œ x (x) Í 0 œ 0
Ê all points in the 3rd quadrant
satisfy the equation.
(4) 4th quadrant: y kyk œ x kxk
Í y (y) œ 2x Í 0 œ x. Combining
these results we have the graph given at the
right:
9. By the law of sines,
10. By the law of sines,
sin 13
È3
œ
sin 14
4
œ
sin A
a
œ
sin A
a
œ
sin B
b
œ
sin B
b
œ
sin 14
b
Ê bœ
sin B
3
Ê sin B œ
È3 sin (1/4)
sin (1/3)
3
4
11. By the law of cosines, a# œ b# c# 2bc cos A Ê cos A œ
sin
œ
È3 Š È2 ‹
È3
#
œ È2.
#
1
4
œ
b # c # a#
2bc
3
4
Š
œ
12. By the law of cosines, c# œ a# b# 2ab cos C œ 2# 3# (2)(2)(3) cos
È2
# ‹
œ
3È 2
8
2# 3# 2#
2(2)(3)
1
4
.
œ 34 .
œ 4 9 12 Š
È2
# ‹
œ 13 6È2 Ê c œ É13 6È2 , since c 0.
#
a # c # b #
4 # 3 #
œ 2(2)(2)(4)
#ac
È135
3È15
121
256 œ 16 œ 16 .
œ
4169
16
#
4 # 5 #
a # b # c #
œ 2(2)(2)(4)
2ab
È231
25
256 œ 16 .
œ
41625
16
13. By the law of cosines, b# œ a# c# 2ac cos B Ê cos B œ
œ
11
16 .
Since 0 B 1, sin B œ È1 cos# B œ É1 14. By the law of cosines, c# œ a# b# 2ab cos C Ê cos C œ
5
œ 16
. Since 0 C 1, sin C œ È1 cos# C œ É1 15. (a) sin# x cos# x œ 1 Ê sin# x œ 1 cos# x œ (1 cos x)(1 cos x) Ê (1 cos x) œ
Ê
1cos x
sin x
œ
sin# x
1cos x
sin x
1cos x
(b) Using the definition of the tangent function and the double angle formulas, we have
tan# ˆ x# ‰ œ
sin# ˆ x# ‰
cos# ˆ #x ‰
œ
"cos Š2 Š #x ‹‹
#
"cos Š2 Š #x ‹‹
#
œ
1cos x
1cos x
.
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Chapter 1 Additional and Advanced Exercises
16. The angles labeled # in the accompanying figure are
equal since both angles subtend arc CD. Similarly, the
two angles labeled ! are equal since they both subtend
arc AB. Thus, triangles AED and BEC are similar which
) b
implies ab c œ 2a cos
a c
Ê (a c)(a c) œ b(2a cos ) b)
Ê a# c# œ 2ab cos ) b#
Ê c# œ a# b# 2ab cos ).
17. As in the proof of the law of sines of Section P.5, Exercise 57, ah œ bc sin A œ ab sin C œ ac sin B
Ê the area of ABC œ "# (base)(height) œ "# ah œ "# bc sin A œ "# ab sin C œ "# ac sin B.
18. As in Section P.5, Exercise 57, (Area of ABC)# œ
œ
"
4
(base)# (height)# œ
"
4
"
4
a# h # œ
a# b# sin# C
a# b# a" cos# Cb . By the law of cosines, c# œ a# b# 2ab cos C Ê cos C œ
Thus, (area of ABC)# œ
œ
"
4
"
16
"
4
a# b# a" cos# Cb œ
#
Š4a# b# aa# b# c# b ‹ œ
"
16
"
4
a# b# Œ" Š a
#
b # c#
‹
#ab
#
œ
a# b#
4
a # b # c#
2ab
Š" #
.
#
aa b c # b
4a# b#
#
‹
ca2ab aa# b# c# bb a2ab aa# b# c# bbd
"
ca(a b)# c# b ac# (a b)# bd œ 16
c((a b) c)((a b) c)(c (a b))(c (a b))d
a
b
c
a
b
c
a
b
c
a
b
c
œ ˆ # ‰ ˆ # ‰ ˆ # ‰ ˆ # ‰‘ œ s(s a)(s b)(s c), where s œ a#bc .
œ
"
16
Therefore, the area of ABC equals Ès(s a)(s b)(s c) .
19. 1.
2.
3.
4.
5.
6.
b c (a c) œ b a, which is positive since a b. Thus, a c b c.
b c (a c) œ b a, which is positive since a b. Thus, a c b c.
c 0 and a b Ê c 0 œ c and b a are positive Ê (b a)c œ bc ac is positive Ê ac bc.
a b and c 0 Ê b a and c are positive Ê (b a)(c) œ ac bc is positive Ê bc ac.
Since a 0, a and "a are positive Ê "a 0.
Since 0 a b, both "a and b" are positive. By (3), a b and "a 0 Ê a ˆ "a ‰ b ˆ "a ‰ or 1 ba
Ê 1 ˆ "b ‰ 7.
b
a
b
a
"
a
"
b
0 Ê
"
b
"
a .
"
a and b" are both negative, i.e.,
0 and b" 0. By (4), a b and
1 Ê 1 ˆ b" ‰ ba ˆ b" ‰ by (4) since b" 0 Ê b" "a .
ab0 Ê
Ê
ˆ b" ‰ by (3) since
"
a
0 Ê b ˆ "a ‰ a ˆ "a ‰
20. (a) If a œ 0, then 0 œ kak kbk Í b Á 0 Í 0 œ kak# kbk# . Since kak# œ kak kak œ ka# k œ a# and
kbk# œ b# we obtain a# b# . If a Á 0 then kak 0 and kak kbk Ê a# b# . On the other hand,
if a# b# then a# œ kak# kbk# œ b# Ê 0 kbk# kak# œ akbk kakb akbk kakb . Since akbk kakb 0
and the product akbk kakb akbk kakb is positive, we must have akbk kakb 0 Ê kbk kak . Thus
kak kbk Í a# b# .
(b) ab Ÿ kabk Ê ab
#
#
2 kabk by Exercise 19(4) above Ê a# 2ab b#
kak œ a# and kbk œ b# . Factoring both sides, (a b)#
kak# 2 kak kbk kbk# , since
#
akak kbkb Ê ka bk
kkak kbkk , by part (a).
21. The fact that ka" a# á an k Ÿ ka" k ka# k á kan k holds for n œ 1 is obvious. It also holds for
n œ 2 by the triangle inequality. We now show it holds for all positive integers n, by induction.
Suppose it holds for n œ k 1: ka" a# á ak k Ÿ ka" k ka# k á kak k (this is the induction
hypothesis). Then ka" a# á ak akb1 k œ kaa" a# á ak b akb1 k Ÿ ka" a# á ak k kakb1 k
(by the triangle inequality) Ÿ ka" k ka# k á kak k kakb1 k (by the induction hypothesis) and the
inequality holds for n œ k 1. Hence it holds for all n by induction.
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63
64
Chapter 1 Preliminaries
22. The fact that ka" a# á an k ka" k ka# k á kan k holds for n œ 1 is obvious. It holds for n œ 2
by Exercise 21(b), since ka" a# k œ ka" (a# )k kka" k ka# kk œ kka" k ka# kk ka" k ka# k .
We now show it holds for all positive integers n by induction.
Suppose the inequality holds for n œ k 1. Then ka" a# á ak k ka" k ka# k á kak k (this is
the induction hypothesis). Thus ka" á ak akb1 k œ kaa" á ak b aakb1 bk
kkaa" á ak bk kakb1 kk (by Exercise 21(b)) œ kka" á ak k kakb1 kk ka" á ak k kakb1 k
ka" k ka# k á kak k kakb1 k (by the induction hypothesis). Hence the inequality holds for all
n by induction.
23. If f is even and odd, then f(x) œ f(x) and f(x) œ f(x) Ê f(x) œ f(x) for all x in the domain of f.
Thus 2f(x) œ 0 Ê f(x) œ 0.
f(x) f((x))
œ f(x) #f(x) œ E(x) Ê E
#
even function. Define O(x) œ f(x) E(x) œ f(x) f(x) #f(x) œ f(x) #f(x) . Then
O(x) œ f(x) #f((x)) œ f(x)# f(x) œ Š f(x) #f(x) ‹ œ O(x) Ê O is an odd function
24. (a) As suggested, let E(x) œ
f(x) f(x)
#
Ê E(x) œ
is an
Ê f(x) œ E(x) O(x) is the sum of an even and an odd function.
(b) Part (a) shows that f(x) œ E(x) O(x) is the sum of an even and an odd function. If also
f(x) œ E" (x) O" (x), where E" is even and O" is odd, then f(x) f(x) œ 0 œ aE" (x) O" (x)b
(E(x) O(x)). Thus, E(x) E" (x) œ O" (x) O(x) for all x in the domain of f (which is the same as the
domain of E E" and O O" ). Now (E E" )(x) œ E(x) E" (x) œ E(x) E" (x) (since E and E" are
even) œ (E E" )(x) Ê E E" is even. Likewise, (O" O)(x) œ O" (x) O(x) œ O" (x) (O(x))
(since O and O" are odd) œ (O" (x) O(x)) œ (O" O)(x) Ê O" O is odd. Therefore, E E" and
O" O are both even and odd so they must be zero at each x in the domain of f by Exercise 23. That is,
E" œ E and O" œ O, so the decomposition of f found in part (a) is unique.
25. y œ ax# bx c œ a Šx# ba x b#
4a# ‹
b#
4a
c œ a ˆx b ‰#
2a
b#
4a
c
(a) If a 0 the graph is a parabola that opens upward. Increasing a causes a vertical stretching and a shift
of the vertex toward the y-axis and upward. If a 0 the graph is a parabola that opens downward.
Decreasing a causes a vertical stretching and a shift of the vertex toward the y-axis and downward.
(b) If a 0 the graph is a parabola that opens upward. If also b 0, then increasing b causes a shift of the
graph downward to the left; if b 0, then decreasing b causes a shift of the graph downward and to the
right.
If a 0 the graph is a parabola that opens downward. If b 0, increasing b shifts the graph upward
to the right. If b 0, decreasing b shifts the graph upward to the left.
(c) Changing c (for fixed a and b) by ?c shifts the graph upward ?c units if ?c 0, and downward ?c
units if ?c 0.
26. (a) If a 0, the graph rises to the right of the vertical line x œ b and falls to the left. If a 0, the graph
falls to the right of the line x œ b and rises to the left. If a œ 0, the graph reduces to the horizontal
line y œ c. As kak increases, the slope at any given point x œ x! increases in magnitude and the graph
becomes steeper. As kak decreases, the slope at x! decreases in magnitude and the graph rises or falls
more gradually.
(b) Increasing b shifts the graph to the left; decreasing b shifts it to the right.
(c) Increasing c shifts the graph upward; decreasing c shifts it downward.
27. If m 0, the x-intercept of y œ mx 2 must be negative. If m 0, then the x-intercept exceeds
Ê 0 œ mx 2 and x "
#
Ê x œ m2 "
#
Ê 0 m 4.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
"
#
Chapter 1 Additional and Advanced Exercises
28. Each of the triangles pictured has the same base
b œ v?t œ v(1 sec). Moreover, the height of each
triangle is the same value h. Thus "# (base)(height) œ
"
#
bh
œ A" œ A# œ A$ œ á . In conclusion, the object sweeps
out equal areas in each one second interval.
29. (a) By Exercise #95 of Section 1.2, the coordinates of P are ˆ a# 0 ß b# 0 ‰ œ ˆ #a ß b# ‰ . Thus the slope
of OP œ
(b)
?y
?x
œ
b/2
a/2
œ
b
a .
b 0
The slope of AB œ 0a œ ba . The line
#
of their slopes is " œ ˆ ba ‰ ˆ ba ‰ œ ba#
segments AB and OP are perpendicular when the product
. Thus, b# œ a# Ê a œ b (since both are positive). Therefore, AB
is perpendicular to OP when a œ b.
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65
66
Chapter 1 Preliminaries
NOTES:
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
CHAPTER 2 LIMITS AND CONTINUITY
2.1 RATES OF CHANGE AND LIMITS
1. (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approaches 1 from the left, g(x)
approaches 1. There is no single number L that all the values g(x) get arbitrarily close to as x Ä 1.
(b) 1
(c) 0
2. (a) 0
(b) 1
(c) Does not exist. As t approaches 0 from the left, f(t) approaches 1. As t approaches 0 from the right, f(t)
approaches 1. There is no single number L that f(t) gets arbitrarily close to as t Ä 0.
3. (a) True
(d) False
(b) True
(e) False
(c) False
(f) True
4. (a) False
(d) True
(b) False
(e) True
(c) True
5.
x
lim
x Ä 0 kx k
x
kx k
does not exist because
x
kx k
œ
x
x
œ 1 if x 0 and
approaches 1. As x approaches 0 from the right,
x
kx k
x
kxk
œ
x
x
œ 1 if x 0. As x approaches 0 from the left,
approaches 1. There is no single number L that all
the function values get arbitrarily close to as x Ä 0.
6. As x approaches 1 from the left, the values of
"
x 1
become increasingly large and negative. As x approaches 1
from the right, the values become increasingly large and positive. There is no one number L that all the
function values get arbitrarily close to as x Ä 1, so lim x" 1 does not exist.
xÄ1
7. Nothing can be said about f(x) because the existence of a limit as x Ä x! does not depend on how the function
is defined at x! . In order for a limit to exist, f(x) must be arbitrarily close to a single real number L when
x is close enough to x! . That is, the existence of a limit depends on the values of f(x) for x near x! , not on the
definition of f(x) at x! itself.
8. Nothing can be said. In order for lim f(x) to exist, f(x) must close to a single value for x near 0 regardless of
xÄ0
the value f(0) itself.
9. No, the definition does not require that f be defined at x œ 1 in order for a limiting value to exist there. If f(1)
is defined, it can be any real number, so we can conclude nothing about f(1) from lim f(x) œ 5.
xÄ1
10. No, because the existence of a limit depends on the values of f(x) when x is near 1, not on f(1) itself. If
lim f(x) exists, its value may be some number other than f(1) œ 5. We can conclude nothing about lim f(x),
xÄ1
xÄ1
whether it exists or what its value is if it does exist, from knowing the value of f(1) alone.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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68
Chapter 2 Limits and Continuity
11. (a) f(x) œ ax# *b/(x 3)
x
3.1
f(x)
6.1
2.9
5.9
x
f(x)
3.01
6.01
3.001
6.001
3.0001
6.0001
3.00001
6.00001
3.000001
6.000001
2.99
5.99
2.999
5.999
2.9999
5.9999
2.99999
5.99999
2.999999
5.999999
The estimate is lim f(x) œ 6.
x Ä $
(b)
(c) f(x) œ
x# 9
x3
œ
(x 3)(x 3)
x3
œ x 3 if x Á 3, and lim (x 3) œ 3 3 œ 6.
x Ä $
12. (a) g(x) œ ax# #b/ Šx È2‹
x
g(x)
1.4
2.81421
1.41
2.82421
1.414
2.82821
1.4142
2.828413
1.41421
2.828423
1.414213
2.828426
(b)
(c) g(x) œ
x# 2
x È2
œ
Šx È2‹ Šx È2‹
Šx È2‹
œ x È2 if x Á È2, and
13. (a) G(x) œ (x 6)/ ax# 4x 12b
x
5.9
5.99
G(x)
.126582 .1251564
x
G(x)
6.1
.123456
6.01
.124843
5.999
.1250156
6.001
.124984
lim
x Ä È#
5.9999
.1250015
6.0001
.124998
Šx È2‹ œ È2 È2 œ 2È2.
5.99999
.1250001
6.00001
.124999
5.999999
.1250000
6.000001
.124999
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 2.1 Rates of Change and Limits
(b)
(c) G(x) œ
x6
ax# 4x 12b
œ
x6
(x 6)(x 2)
œ
"
x#
14. (a) h(x) œ ax# 2x 3b / ax# 4x 3b
x
2.9
2.99
h(x)
2.052631
2.005025
x
h(x)
3.1
1.952380
3.01
1.995024
"
if x Á 6, and lim
x Ä ' x 2
œ
"
' 2
œ "8 œ 0.125.
2.999
2.000500
2.9999
2.000050
2.99999
2.000005
2.999999
2.0000005
3.001
1.999500
3.0001
1.999950
3.00001
1.999995
3.000001
1.999999
(b)
(c) h(x) œ
x# 2x 3
x# 4x 3
œ
(x 3)(x 1)
(x 3)(x 1)
œ
x1
x1
15. (a) f(x) œ ax# 1b / akxk 1b
x
1.1
1.01
f(x)
2.1
2.01
x
f(x)
.9
1.9
.99
1.99
if x Á 3, and lim
x1
x Ä $ x1
œ
31
31
œ
4
#
œ 2.
1.001
2.001
1.0001
2.0001
1.00001
2.00001
1.000001
2.000001
.999
1.999
.9999
1.9999
.99999
1.99999
.999999
1.999999
(b)
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69
70
Chapter 2 Limits and Continuity
(c) f(x) œ
x# "
kx k 1
(x 1)(x 1)
1
œ (x x1)(x
1)
(x 1)
œ x 1, x 0 and x Á 1
, and lim (1 x) œ 1 (1) œ 2.
x Ä 1
œ 1 x, x 0 and x Á 1
16. (a) F(x) œ ax# 3x 2b / a2 kxkb
x
2.1
2.01
F(x)
1.1
1.01
1.9
.9
x
F(x)
1.99
.99
2.001
1.001
2.0001
1.0001
2.00001
1.00001
2.000001
1.000001
1.999
.999
1.9999
.9999
1.99999
.99999
1.999999
.999999
(b)
(c) F(x) œ
x# 3x 2
2 kx k
(x 2)(x 1)
œ (x 2)(x# x")
2x
17. (a) g()) œ (sin ))/)
)
.1
g())
.998334
,
x 0
, and lim (x 1) œ 2 1 œ 1.
x Ä #
œ x 1, x 0 and x Á 2
.01
.999983
.001
.999999
.0001
.999999
.00001
.999999
.000001
.999999
.1
.998334
.01
.999983
.001
.999999
.0001
.999999
.00001
.999999
.000001
.999999
18. (a) G(t) œ (1 cos t)/t#
t
.1
G(t)
.499583
.01
.499995
.001
.499999
.0001
.5
.00001
.5
.000001
.5
.1
.499583
.01
.499995
.001
.499999
.0001
.5
.00001
.5
.000001
.5
)
g())
lim g()) œ 1
)Ä!
(b)
t
G(t)
lim G(t) œ 0.5
tÄ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 2.1 Rates of Change and Limits
(b)
Graph is NOT TO SCALE
19. (a) f(x) œ x"ÎÐ"xÑ
x
.9
f(x)
.348678
x
f(x)
1.1
.385543
.99
.366032
.999
.367695
.9999
.367861
.99999
.367877
.999999
.367879
1.01
.369711
1.001
.368063
1.0001
.367897
1.00001
.367881
1.000001
.367878
lim f(x) ¸ 0.36788
xÄ1
(b)
Graph is NOT TO SCALE. Also the intersection of the axes is not the origin: the axes intersect at the point
(1ß 2.71820).
20. (a) f(x) œ a3x 1b /x
x
.1
f(x)
1.161231
.01
1.104669
.001
1.099215
.0001
1.098672
.00001
1.098618
.000001
1.098612
.1
1.040415
.01
1.092599
.001
1.098009
.0001
1.098551
.00001
1.098606
.000001
1.098611
x
f(x)
lim f(x) ¸ 1.0986
xÄ!
(b)
21. lim 2x œ 2(2) œ 4
22. lim 2x œ 2(0) œ 0
23. lim" (3x 1) œ 3 ˆ "3 ‰ 1 œ 0
24. lim
xÄ#
xÄ
$
xÄ!
1
x Ä 1 3x1
œ
"
3(1) 1
œ #"
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71
72
25.
Chapter 2 Limits and Continuity
lim 3x(2x 1) œ 3(1)(2(1) 1) œ 9
26.
x Ä "
1
#
27. lim1 x sin x œ
xÄ
#
1
#
sin
œ
1
#
28. xlim
Ä1
29. (a)
?f
?x
œ
f(3) f(2)
3#
30. (a)
?g
?x
œ
g(1) g(1)
1 (1)
31. (a)
?h
?t
œ
h ˆ 341 ‰ h ˆ 14 ‰
1
31
4 4
œ
?g
?t
œ
g(1) g(0)
10
(2 1) (2 1)
10
32. (a)
33.
?R
?)
œ
R(2) R(0)
20
34.
?P
?)
œ
P(2) P(1)
21
35. (a)
œ
œ
28 9
1
œ
œ
œ
1 1
2
Q% (18ß 550)
œ
3
3
œ 1
"
1 1
œ
"
1 1
œ
f(1) f(")
1 (1)
œ
20
#
œ1
œ0
(b)
?g
?x
œ
g(0)g(2)
0(2)
œ
04
#
œ 2
(b)
?h
?t
œ
h ˆ 1# ‰ h ˆ 16 ‰
11
#
6
œ
?g
?t
œ
g(1) g(1)
1 (1)
œ
1 1
1
#
œ 14
œ
650 225
20 10
650 375
20 14
650 475
20 16.5
650 550
20 18
Q$ (16.5ß 475)
cos 1
1 1
œ
?f
?x
œ 12
3"
#
(b)
0 È3
1
3
œ
3 È 3
1
(2 1) (2 ")
#1
œ0
œ1
œ22œ0
Slope of PQ œ
Q# (14ß 375)
œ
3(1)#
2(1)1
(b)
(8 16 10)(" % &)
1
Q" (10ß 225)
cos x
1 1
œ
œ 19
È 8 1 È 1
#
Q
3x#
lim
x Ä 1 2x1
?p
?t
œ 42.5 m/sec
œ 45.83 m/sec
œ 50.00 m/sec
œ 50.00 m/sec
(b) At t œ 20, the Cobra was traveling approximately 50 m/sec or 180 km/h.
36. (a)
Slope of PQ œ
Q
Q" (5ß 20)
Q# (7ß 39)
Q$ (8.5ß 58)
Q% (9.5ß 72)
80 20
10 5
80 39
10 7
80 58
10 8.5
80 72
10 9.5
?p
?t
œ 12 m/sec
œ 13.7 m/sec
œ 14.7 m/sec
œ 16 m/sec
(b) Approximately 16 m/sec
37. (a)
(b)
?p
?t
œ
174 62
1994 1992
œ
112
#
œ 56 thousand dollars per year
(c) The average rate of change from 1991 to 1992 is ??pt œ
The average rate of change from 1992 to 1993
is ??pt
œ
62 27
1992 1991
111 62
1993 1992
œ 35 thousand dollars per year.
œ 49 thousand dollars per year.
So, the rate at which profits were changing in 1992 is approximatley "# a35 49b œ 42 thousand dollars per year.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 2.1 Rates of Change and Limits
38. (a) F(x) œ (x 2)/(x 2)
x
1.2
F(x)
4.0
?F
?x
?F
?x
?F
?x
œ
?g
?x
?g
?x
œ
œ
œ
1.1
3.4
1.01
3.04
1.001
3.004
1.0001
3.0004
1
3
4.0 (3)
œ 5.0;
1.2 1
3.04 (3)
œ 4.04;
1.01 1
3.!!!% (3)
œ 4.!!!%;
1.0001 1
?F
?x
?F
?x
œ
œ
3.4 (3)
œ 4.4;
1.1 1
3.004 (3)
œ 4.!!%;
1.001 1
È
g(2) g(1)
œ #21" ¸ 0.414213
21
È1 h"
g(1 h) g(1)
(1 h) 1 œ
h
?g
?x
œ
g(1.5) g(1)
1.5 1
(b) The rate of change of F(x) at x œ 1 is 4.
39. (a)
œ
œ
È1.5 "
0.5
¸ 0.449489
(b) g(x) œ Èx
1h
È1 h
1.1
1.04880
1.01
1.004987
1.001
1.0004998
1.0001
1.0000499
1.00001
1.000005
1.000001
1.0000005
ŠÈ1 h 1‹ /h
0.4880
0.4987
0.4998
0.499
0.5
0.5
(c) The rate of change of g(x) at x œ 1 is 0.5.
(d) The calculator gives lim
hÄ!
40. (a) i)
ii)
(b)
f(3) f(2)
32
f(T) f(2)
T#
œ
œ
""
3
#
1
" "
T #
T#
T
f(T)
af(T) f(2)b/aT 2b
œ
œ
"
6
1
È1 h"
h
œ 6"
#TT
T#
2
#T
œ "# .
œ
2.1
0.476190
0.2381
2T
#T(T 2)
œ
2T
#T(2 T)
2.01
0.497512
0.2488
œ #"T , T Á 2
2.001
0.499750
0.2500
2.0001
0.4999750
0.2500
2.00001
0.499997
0.2500
2.000001
0.499999
0.2500
(c) The table indicates the rate of change is 0.25 at t œ 2.
" ‰
(d) lim ˆ #T
œ 4"
TÄ#
41-46. Example CAS commands:
Maple:
f := x -> (x^4 16)/(x 2);
x0 := 2;
plot( f(x), x œ x0-1..x0+1, color œ black,
title œ "Section 2.1, #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<0 in many CASs. This can be
overcome in Maple by entering the function as f := 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]
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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 #
lim 8(t 5)(t 7) œ 8(6 5)(6 7) œ 8
x3
œ
9.
lim
y Ä & 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 (
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 "
lim (x 3)"*)% œ (4 3)"*)% œ (1)"*)% œ 1
x Ä %
%
lim (5 y)%Î$ œ [5 (3)]%Î$ œ (8)%Î$ œ ˆ(8)"Î$ ‰ œ 2% œ 16
y Ä $
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 "‹
œ lim
3
h Ä 0 È3h1"
h Ä 0 hŠÈ5h 4 2‹
h Ä 0 hŠÈ5h 4 2‹
œ lim
5
4
x5
œ lim
x Ä & (x 5)(x 5)
x3
lim
È3h 1 1
È3h 1 1
œ lim
lim
#
x Ä $ x 4x 3
x Ä &
†
hÄ0
$
#
18. lim
œ
È3h 1 "
h
œ lim
x# 3x "0
x5
œ lim
œ lim
x3
x Ä $ (x 3)(x 1)
œ lim
x Ä &
1
x Ä & x5
(x 5)(x 2)
x5
œ
œ lim
"
55
œ
"
10
1
œ
"
3 1
x Ä $ x 1
œ "2
œ lim (x 2) œ & # œ 7
x Ä &
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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 " t t 2
25.
lim
$
#
x Ä # 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 "
œ lim
xÄ%
œ
x Ä "
35.
x2
x Ä 2 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 2
(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 2
œ
Š2 Èx# 5‹ Š2 Èx# 5‹
x Ä 3
x Ä 3 (x 3) Š2 Èx# 5‹
È x# 5 3
x2
œ
œ
4
3
ax # 8 b *
x Ä 1 (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 2 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 1 È x # ) $
ax 2b ŠÈx# 5 3‹
2 È x# 5
x3
x Ä 3
lim
uÄ1
œ lim
(x 2)(x 2)
lim
x Ä 2
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 1 (x 1) ŠÈx# ) $‹
œ lim
2
4
(x 1) ˆÈx 3 2‰
œ lim
È x# 8 3
x1
œ lim
x(4 x)
x Ä % 2 Èx
x1
31. lim
Èx 3
x Ä * ˆÈ x 3 ‰ ˆ È x 3 ‰
x Ä % 2 Èx
œ
2
œ lim
(v 2) av# 2v 4b
(v
2)(v 2) av# 4b
vÄ#
3
#
1 2
1 2
#
y Ä ! 3y 16
au# "b (u 1)(u 1)
au# u 1b (u 1)
œ
œ
#
x Ä # x
#
#
y Ä ! y a3y 16b
Èx 3
x9
t2
t Ä " t 2
œ lim
uÄ1
12
11
œ
œ lim
2(x 2)
œ lim
30. lim
t2
œ lim
v$ 8
% 16
v
vÄ#
xÄ*
32.
t Ä " (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 # 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 3 (x 3) Š2 Èx# 5‹
x Ä 3 (x 3) Š2 Èx# 5‹
œ lim
3x
x Ä 3 2 È x # 5
œ
6
2 È4
œ
3
2
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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
xÄ%
xÄ%
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 #
x Ä #
x Ä #
x Ä #
lim p(x) † r(x) † s(x) œ ’ lim p(x)“ ’ lim r(x)“ ’ lim s(x)“ œ (4)(0)(3) œ 0
x Ä #
x Ä #
(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 #
x Ä #
44. lim
hÄ!
x Ä #
lim [4p(x) 5r(x)]/s(x) œ ’4 lim p(x) 5 lim r(x)“ ‚ lim s(x) œ [4(4) 5(0)]/3 œ
x Ä #
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 #
h(2 h)
h
hÄ!
x Ä #
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
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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"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‹
È3(0 h) 1 È3(0) 1
h
hÄ!
3h
h Ä ! h ŠÈ3h 1 "‹
œ
h Ä ! È 7 h È 7
48. lim
œ lim
"
œ lim
œ lim
"
#È 7
ŠÈ3h 1 "‹ ŠÈ3h 1 "‹
œ lim
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 1
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
%
xÄ%
œ xÄlim
x lim 2 œ
xÄ%
xÄ%
lim f(x) 5
xÄ%
%#
Ê lim f(x) 5 œ 2(1) Ê lim f(x) œ 2 5 œ 7.
xÄ%
xÄ%
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77
78
Chapter 2 Limits and Continuity
56. (a) 1 œ lim
f(x)
x#
lim f(x)
lim f(x)
#
xÄ#
œ xÄlim
Ê
%
x# œ
(b) 1 œ lim
f(x)
x#
œ ’ lim
x Ä #
x Ä #
xÄ#
f(x)
lim x" “
x “ ’x Ä
#
x Ä #
57. (a) 0 œ 3 † 0 œ ’ lim
xÄ#
lim f(x) œ 4.
x Ä #
œ ’ lim
x Ä #
f(x) ˆ " ‰
x “ #
Ê
lim
x Ä #
f(x)
x
œ 2.
f(x) 5
x # “ ’xlim
Ä#
5
(x 2)“ œ lim ’Š f(x)
x # ‹ (x 2)“ œ lim [f(x) 5] œ lim f(x) 5
f(x) 5
x # “ ’xlim
Ä#
(x 2)“ Ê lim f(x) œ 5 as in part (a).
xÄ#
Ê lim f(x) œ 5.
xÄ#
xÄ#
xÄ#
(b) 0 œ 4 † 0 œ ’ lim
xÄ#
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
x Ê lim x sin
"
x
œ 0 by the sandwich theorem.
x sin
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
xÄ!
theorem since lim x# œ 0.
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.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
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 .
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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
È5 2
# .
È5
#
Ê $œ
È5 2
#
¸ 0.1180, or $ 1 œ 13. Step 1:
Step 2:
kx (1)k $ Ê $ x 1 $ Ê $ 1 x $ 1
7
16
From the graph, $ 1 œ 16
9 Ê $ œ 9 ¸ 0.77, or $ 1 œ 25 Ê
14. Step 1:
¸x "# ¸ $ Ê $ x Step 2:
From the graph, $ thus $ œ 0.00248.
"
#
œ
"
# 1
2.01
$ Ê $ Ê $œ
1
2
"
#
x$
"
#.01
"
#
¸ 0.00248, or $ "
#
œ
È3
#
9
25
Ê $œ
2 È3
#
œ 0.36; thus $ œ
1
1.99
Ê $œ
1
1.99
¸ 0.1340;
9
25
"
#
œ 0.36.
¸ 0.00251;
15. Step 1:
Step 2:
k(x 1) 5k 0.01 Ê kx 4k 0.01 Ê 0.01 x 4 0.01 Ê 3.99 x 4.01
kx 4k $ Ê $ x 4 $ Ê $ 4 x $ 4 Ê $ œ 0.01.
16. Step 1:
k(2x 2) (6)k 0.02 Ê k2x 4k 0.02 Ê 0.02 2x 4 0.02 Ê 4.02 2x 3.98
Ê 2.01 x 1.99
kx (2)k $ Ê $ x 2 $ Ê $ 2 x $ 2 Ê $ œ 0.01.
Step 2:
17. Step 1:
Step 2:
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.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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 1
Step 1:
k(3x 2) 1k 0.03 Ê 0.03 3x 3 0.03 Ê 0.01 x 1 0.01 Ê 1.01 x 0.99.
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82
Chapter 2 Limits and Continuity
kx (1)k $ Ê $ x 1 $ Ê $ " x $ 1.
Then $ " œ 1.01 Ê $ œ 0.01, or $ " œ 0.99 Ê $ œ 0.01; thus $ œ 0.01.
Step 2:
33. lim
x# 4
x Ä # x#
34.
35.
œ lim
xÄ#
#
(x 2)(x 2)
(x 2)
œ lim (x 2) œ # # œ 4, x Á 2
xÄ#
(x 2)(x 2)
(x 2)
Step 1:
¹Š xx 24 ‹
Step 2:
Ê 1.95 x 2.05, x Á 2.
kx 2k $ Ê $ x 2 $ Ê $ # 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 &
x# 6x 5
x5
4¹ 0.05 Ê 0.05 œ lim
x Ä &
(x 5)(x 1)
(x 5)
% 0.05 Ê 3.95 x 2 4.05, x Á 2
œ lim (x 1) œ 4, x Á 5.
x Ä &
(x 5)(x ")
(x 5)
Step 1:
#
¹Š x x 6x5 5 ‹
Step 2:
Ê 5.05 x 4.95, x Á 5.
kx (5)k $ Ê $ x 5 $ Ê $ & x $ &.
Then $ & œ 5.05 Ê $ œ 0.05, or $ & œ 4.95 Ê $ œ 0.05; thus $ œ 0.05.
(4)¹ 0.05 Ê 0.05 4 0.05 Ê 4.05 x 1 3.95, x Á 5
lim È1 5x œ È1 5(3) œ È16 œ 4
x Ä $
Step 1:
¹È1 5x 4¹ 0.5 Ê 0.5 È1 5x 4 0.5 Ê 3.5 È1 5x 4.5 Ê 12.25 1 5x 20.25
Step 2:
Ê 11.25 5x 19.25 Ê 3.85 x 2.25.
kx (3)k $ Ê $ x 3 $ Ê $ $ x $ $.
Then $ $ œ 3.85 Ê $ œ 0.85, or $ $ œ 2.25 Ê 0.75; thus $ œ 0.75.
36. lim
4
xÄ# x
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:
%
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 (# %)# %.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 2.3 Precise Definition of a Limit
Step 2:
41. Step 1:
Step 2:
42. Step 1:
Step 2:
43. Step 1:
Step 2:
kx 0k $ Ê $ x $ .
Then $ œ (# %)# 4 œ %# %% Ê $ œ %% %# , or $ œ (# %)# 4 œ 4% %# . Thus choose
the smaller distance, $ œ 4% %# .
For x Á 1, kx# 1k % Ê % x# " % Ê " % x# " % Ê È1 % kxk È1 %
Ê È" % x È1 % near B œ ".
kx 1k $ Ê $ x 1 $ Ê $ " x $ ".
Then $ " œ È1 % Ê $ œ " È1 %, or $ 1 œ È" % Ê $ œ È" % 1. Choose
$ œ min š" È1 %ß È1 % "›, that is, the smaller of the two distances.
For x Á 2, kx# 4k % Ê % x# 4 % Ê 4 % x# 4 % Ê È4 % kxk È4 %
Ê È4 % x È4 % near B œ 2.
kx (2)k $ Ê $ x 2 $ Ê $ 2 x $ 2.
Then $ 2 œ È% % Ê $ œ È% % #, or $ # œ È% % Ê $ œ # È% %. Choose
$ œ min šÈ% % #ß # È% %› .
¸ "x 1¸ % Ê % "
x
"% Ê "% "
x
"% Ê
"
1%
%
"%,
"
1%.
x
kx 1k $ Ê $ x 1 $ Ê " $ x " $ .
Then " $ œ " " % Ê $ œ " " " % œ " % % , or " $ œ " " % Ê $ œ
Choose $ œ
44. Step 1:
83
"
"%
"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
Step 2:
48. Step 1:
1: l(6x 4) 2l % Ê ! Ÿ 6x 6 % since x
%
#
x !;
1. Thus, " Ÿ x 1 6% .
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 #%.
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84
Chapter 2 Limits and Continuity
Step 2:
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
then by the sandwich theorem, in either case, lim x sin
xÄ!
50. By the figure, x# Ÿ x# sin
"
x
"
x
"
x
x for x 0. Since lim (x) œ lim x œ 0,
xÄ!
œ 0.
xÄ!
Ÿ x# for all x except possibly at x œ 0. Since lim 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
Ê
2É 8.99
1
ŸxŸ
2É 9.01
1
1 x#
4
Ÿ 9.01 Ê
4
1
(8.99) Ÿ x# Ÿ
4
1
(9.01)
or 3.384 Ÿ x Ÿ 3.387. To be safe, the left endpoint was rounded up and the right
endpoint was rounded down.
56. V œ RI Ê
V
R
œ I Ê ¸ VR 5¸ Ÿ 0.1 Ê 0.1 Ÿ
120
R
5 Ÿ 0.1 Ê 4.9 Ÿ
120
R
Ÿ 5.1 Ê
10
49
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
R
1#0
10
51
Ê
Section 2.3 Precise Definition of a Limit
(120)(10)
51
ŸRŸ
(120)(10)
49
85
Ê 23.53 Ÿ R Ÿ 24.48.
To be safe, the left endpoint was rounded up and the right endpoint was rounded down.
57. (a) $ x 1 0 Ê " $ x 1 Ê f(x) œ x. Then kf(x) 2k œ kx 2k œ 2 x 2 1 œ 1. That is,
kf(x) 2k 1 "# no matter how small $ is taken when " $ x 1 Ê lim f(x) Á 2.
xÄ1
(b) 0 x 1 $ Ê " x " $ Ê f(x) œ x 1. Then kf(x) 1k œ k(x 1) 1k œ kxk œ x 1. That is,
kf(x) 1k 1 no matter how small $ is taken when " x " $ Ê lim f(x) Á 1.
xÄ1
(c) $ x 1 ! Ê " $ x 1 Ê f(x) œ x. Then kf(x) 1.5k œ kx 1.5k œ 1.5 x 1.5 1 œ 0.5.
Also, ! x 1 $ Ê 1 x " $ Ê f(x) œ x 1. Then kf(x) 1.5k œ k(x 1) 1.5k œ kx 0.5k
œ x 0.5 " 0.5 œ 0.5. Thus, no matter how small $ is taken, there exists a value of x such that
$ x 1 $ but kf(x) 1.5k "# Ê lim f(x) Á 1.5.
xÄ1
58. (a) For 2 x 2 $ Ê h(x) œ 2 Ê kh(x) 4k œ 2. Thus for % 2, kh(x) 4k
matter how small we choose $ 0 Ê lim h(x) Á 4.
% whenever 2 x 2 $ no
(b) For 2 x 2 $ Ê h(x) œ 2 Ê kh(x) 3k œ 1. Thus for % 1, kh(x) 3k
matter how small we choose $ 0 Ê lim h(x) Á 3.
% whenever 2 x 2 $ no
xÄ#
xÄ#
(c) For 2 $ x 2 Ê h(x) œ x# so kh(x) 2k œ kx# 2k . No matter how small $ 0 is chosen, x# is close to 4
when x is near 2 and to the left on the real line Ê kx# 2k will be close to 2. Thus if % 1, kh(x) 2k %
whenever 2 $ x 2 no mater how small we choose $ 0 Ê lim h(x) Á 2.
xÄ#
59. (a) For 3 $ x 3 Ê f(x) 4.8 Ê kf(x) 4k 0.8. Thus for % 0.8, kf(x) 4k
3 $ x 3 no matter how small we choose $ 0 Ê lim f(x) Á 4.
xÄ$
% whenever
(b) For 3 x 3 $ Ê f(x) 3 Ê kf(x) 4.8k 1.8. Thus for % 1.8, kf(x) 4.8k
no matter how small we choose $ 0 Ê lim f(x) Á 4.8.
% whenever 3 x 3 $
(c) For 3 $ x 3 Ê f(x) 4.8 Ê kf(x) 3k 1.8. Again, for % 1.8, kf(x) 3k
no matter how small we choose $ 0 Ê lim f(x) Á 3.
% whenever $ $ x 3
xÄ$
xÄ$
60. (a) No matter how small we choose $ 0, for x near 1 satisfying " $ x " $ , the values of g(x) are
near 1 Ê kg(x) 2k is near 1. Then, for % œ "# we have kg(x) 2k "# for some x satisfying
" $ x " $ , or ! kx 1k $ Ê
lim g(x) Á 2.
x Ä 1
(b) Yes, lim g(x) œ 1 because from the graph we can find a $ ! such that kg(x) 1k % if ! kx (1)k $ .
x Ä 1
61-66. Example CAS commands (values of del may vary for a specified eps):
Maple:
f := x -> (x^4-81)/(x-3);x0 := 3;
plot( f(x), x=x0-1..x0+1, color=black,
# (a)
title="Section 2.3, #61(a)" );
L := limit( f(x), x=x0 );
# (b)
epsilon := 0.2;
# (c)
plot( [f(x),L-epsilon,L+epsilon], x=x0-0.01..x0+0.01,
color=black, linestyle=[1,3,3], title="Section 2.3, #61(c)" );
q := fsolve( abs( f(x)-L ) = epsilon, x=x0-1..x0+1 ); # (d)
delta := abs(x0-q);
plot( [f(x),L-epsilon,L+epsilon], x=x0-delta..x0+delta, color=black, title="Section 2.3, #61(d)" );
for eps in [0.1, 0.005, 0.001 ] do
# (e)
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86
Chapter 2 Limits and Continuity
q := fsolve( abs( f(x)-L ) = eps, x=x0-1..x0+1 );
delta := abs(x0-q);
head := sprintf("Section 2.3, #61(e)\n epsilon = %5f, delta = %5f\n", eps, delta );
print(plot( [f(x),L-eps,L+eps], x=x0-delta..x0+delta,
color=black, linestyle=[1,3,3], title=head ));
end do:
Mathematica (assigned function and values for x0, eps and del may vary):
Clear[f, x]
y1: œ L eps; y2: œ L eps; x0 œ 1;
f[x_]: œ (3x2 (7x 1)Sqrt[x] 5)/(x 1)
Plot[f[x], {x, x0 0.2, x0 0.2}]
L: œ Limit[f[x], x Ä x0]
eps œ 0.1; del œ 0.2;
Plot[{f[x], y1, y2},{x, x0 del, x0 del}, PlotRange Ä {L 2eps, L 2eps}]
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 #b
2
#
" œ #, lim c f(x) œ $ # œ "
xÄ#
(b) No, lim f(x) does not exist because lim b f(x) Á lim c f(x)
xÄ#
xÄ#
xÄ#
(c) lim c f(x) œ 4# 1 œ 3, lim b f(x) œ 4# " œ $
xÄ%
xÄ%
(d) Yes, lim f(x) œ 3 because 3 œ lim c f(x) œ lim b f(x)
xÄ%
xÄ%
xÄ%
4. (a)
lim f(x) œ
x Ä #b
2
#
œ 1, lim c f(x) œ $ # œ ", f(2) œ 2
xÄ#
(b) Yes, lim f(x) œ 1 because " œ lim b f(x) œ lim c f(x)
xÄ#
xÄ#
xÄ#
(c)
lim c f(x) œ 3 (1) œ 4, lim b f(x) œ 3 (1) œ 4
x Ä "
x Ä "
(d) Yes, lim f(x) œ 4 because 4 œ
x Ä "
lim
x Ä "c
f(x) œ
lim
x Ä "b
f(x)
5. (a) No, lim b f(x) does not exist since sin ˆ "x ‰ does not approach any single value as x approaches 0
xÄ!
(b) lim c f(x) œ lim c 0 œ 0
xÄ!
(c)
xÄ!
lim f(x) does not exist because lim b f(x) does not exist
xÄ!
xÄ!
6. (a) Yes, lim b g(x) œ 0 by the sandwich theorem since Èx Ÿ g(x) Ÿ Èx when x 0
xÄ!
(b) No, lim c g(x) does not exist since Èx is not defined for x 0
xÄ!
(c) No, lim g(x) does not exist since lim c g(x) does not exist
xÄ!
xÄ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 2.4 One-Sided Limits and Limits at Infinity
7. (aÑ
lim f(x) œ " œ lim b f(x)
xÄ1
(c) Yes, lim f(x) œ 1 since the right-hand and left-hand
(b)
x Ä 1c
xÄ1
limits exist and equal 1
8. (a)
(b)
lim f(x) œ 0 œ lim c f(x)
xÄ1
x Ä 1b
(c) Yes, lim f(x) œ 0 since the right-hand and left-hand
xÄ1
limits exist and equal 0
9. (a) domain: 0 Ÿ x Ÿ 2
range: 0 y Ÿ 1 and y œ 2
(b) xlim
f(x) exists for c belonging to
Äc
(0ß 1) ("ß #)
(c) x œ 2
(d) x œ 0
10. (a) domain: _ x _
range: " Ÿ y Ÿ 1
(b) xlim
f(x) exists for c belonging to
Äc
(_ß 1) ("ß ") ("ß _)
(c) none
(d) none
11.
x Ä !Þ&c
lim
13.
x Ä #b
14.
x Ä 1c
15.
lim b
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 1b
"
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 b
hÄ!
œ lim b 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 b
hÄ!
h(h 4)
h ŠÈh# 4h 5 È5‹
œ
04
È5 È5
œ
2
È5
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87
88
16.
Chapter 2 Limits and Continuity
lim
h Ä !c
È6 È5h# 11h 6
h
6 a5h# 11h 6b
œ lim c
hÄ!
17. (a)
19. (a)
) Ä $b
20. (a)
t Ä %b
(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 b È2x œ È2
xÄ1
œ lim c
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 $c
(b)
t Ä %c
Ú) Û
)
lim
œ
2
3
lim at ÚtÛb œ 4 3 œ 1
t cos t
sin t
œ lim ˆ sin xxcos x xÄ!
œk†1œk
3
sin )
4 )lim
Ä! )
œ
"
3
Œ
œ
"
lim
)Ä!c
(where ) œ kt)
(where ) œ 3y)
3
4
œ
sin )
)
"
œ Š lim
‹ Š lim
x Ä ! cos 2x
xÄ!
sin 2x
"
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!
6x# cos x
sin
x sin 2x
xÄ!
x Ä ! sin x cos x
œ
"
"
sin 3h
3 h lim
Ä !c 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 c 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 !c sin 3h
xÄ!
œ lim
œ lim
25. lim
œ 211
È6
œ lim c È2x œ È2
xÄ1
21. lim
24.
lim
(0 11)
È6 È6
œ
(x 3) œ (2) 3 œ 1
lim
x Ä #b
x Ä #c
œ lim b
xÄ1
È2x (x 1)
kx 1 k
lim
x Ä 1c
œ
h(5h 11)
h ŠÈ6 È5h# 11h 6‹
(x 3)
lim
x Ä #b
œ
È2x (x 1)
kx 1 k
lim
(b)
kx 2 k
x2
(x 3)
lim
x Ä 1b
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 c
hÄ!
h ŠÈ6 È5h# 11h 6‹
x Ä #b
(b)
œ lim c Š
hÄ!
2x
œ lim ˆ3 cos x †
xÄ!
x cos x ‰
sin x cos x
œ lim ˆ sinx x †
xÄ!
x
sin x
†
" ‰
cos x
œ ˆ #" † 1‰ (1) œ
2x ‰
sin 2x
"
#
œ3†"†1œ3
lim
x
x Ä ! sin x
œ lim Š sin" x ‹ † lim ˆ cos" x ‰ lim Š sin" x ‹ œ (1)(1) 1 œ 2
xÄ!
30. lim
xÄ!
x
x# x sin x
#x
xÄ!
œ lim ˆ #x xÄ!
xÄ!
"
#
x
"# ˆ sinx x ‰‰ œ 0 "
#
"# (1) œ 0
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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
"
lim
‹
xÄ „_ x
mÎn
(b) 53
3
4
(b)
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) ) †
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 _
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
cos )
3)
œ 0 by the Sandwich Theorem
1 ˆ sint t ‰
1 ˆ cost t ‰
œ
1 ˆ sinr r ‰
2 7r 5 ˆ sinr r ‰
2 3x
5 7x
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
010
10
œ 1
œrÄ
lim_
2
5
2 Š x7$ ‹
1 "x x"# x7$
10
200
œ
(b)
"
#
2
5
(same process as part (a))
œ2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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
_
œ x lim
Ä_
$
& x
È
xÈ
$ xÈ
& x
È
x" x%
x# x$
61. x lim
Ä_
2x&Î$ x"Î$ 7
x)Î& 3x Èx
62. x Ä
lim
_
2
Š "Î#
‹ Š x"# ‹
x
3 7x
œxÄ
lim
_
60. x lim
Ä_
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
_
œ 1
58. x lim
Ä_
œxÄ
lim
_
" ‹
1 Š #Î"&
x
" ‹
1 Š #Î"&
2 Èx
2 Èx
œ x lim
Ä_
2
Š "Î#
‹"
x
2
Š "Î#
‹1
x
œ 1
œ1
x
œ_
" 7
2x"Î"& "*Î"&
x
x)Î&
"
3
1 $Î&
""Î"!
x
"
x#Î$
2
5 3x
"
x"Î$
œ_
x
4x
œ 5#
63. Yes. If lim b f(x) œ L œ lim c f(x), then xlim
f(x) œ L. If lim b f(x) Á lim c f(x), then xlim
f(x) does not exist.
Äa
Äa
xÄa
xÄa
xÄa
xÄa
64. Since xlim
f(x) œ L if and only if lim b f(x) œ L and lim c f(x) œ L, then xlim
f(x) can be found by calculating
Äc
Äc
xÄc
xÄc
lim b f(x).
xÄc
65. If f is an odd function of x, then f(x) œ f(x). Given lim b f(x) œ 3, then lim c f(x) œ $.
xÄ!
xÄ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 2.4 One-Sided Limits and Limits at Infinity
91
66. If f is an even function of x, then f(x) œ f(x). Given lim c f(x) œ 7 then lim b f(x) œ 7. However, nothing
xÄ#
x Ä #
can be said about
67. Yes. If x lim
Ä_
lim
x Ä #c
f(x)
g(x)
f(x) because we don't know lim b f(x).
xÄ#
œ 2 then the ratio of the polynomials' leading coefficients is 2, so x Ä
lim
_
f(x)
g(x)
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 _ 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 &b
74. I œ (% $ ß %) Ê % $ x 4. Also, È% x % Ê % x %# Ê x % %# . Choose $ œ %#
Ê lim È% x œ 0.
x Ä %c
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 ! c 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)
lim
x Ä %!!b
x 2
lim
x Ä #b 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 ! %.
(b)
lim c ÚxÛ œ 399. Just observe that if 399 x 400 then ÚxÛ œ 399. Thus if we choose $ œ ", we have for any
x Ä %!!
number % ! that 400 $ x 400 Ê lÚxÛ 399l œ l399 399l œ ! %.
(c) Since lim b ÚxÛ Á lim c ÚxÛ we conclude that lim ÚxÛ does not exist.
x Ä %!!
78. (a)
x Ä %!!
x Ä %!!
lim f(x) œ lim b Èx œ È0 œ 0; ¸Èx 0¸ % Ê % Èx % Ê ! x %# for x positive. Choose $ œ %#
xÄ!
Ê lim b f(x) œ 0.
x Ä !b
xÄ!
(b)
lim f(x) œ lim c x# sin ˆ x" ‰ œ 0 by the sandwich theorem since x# Ÿ x# sin ˆ x" ‰ Ÿ x# for all x Á 0.
x Ä !c
xÄ!
Since kx# 0k œ kx# 0k œ x# % whenever kxk È%, we choose $ œ È% and obtain ¸x# sin ˆ "x ‰ 0¸ %
if $ x 0.
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92
Chapter 2 Limits and Continuity
(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.
81.
82.
83.
84.
lim
"
x
x sin
xÄ „_
3x 4
x Ä „ _ 2x 5
œ lim
œ
lim
"
)Ä0 )
sin ) œ 1, ˆ) œ x" ‰
3 4x
5
x Ä „ _ 2 x
lim
œ lim
3 4t
t Ä 0 2 5t
œ
80.
3
#
cos
lim
"
x
x Ä _ 1 x"
œ lim c
)Ä!
cos )
1)
œ
"
1
œ 1, ˆ) œ x" ‰
, ˆt œ "x ‰
"Îx
lim ˆ "x ‰ œ lim b zz œ 1, ˆz œ x" ‰
zÄ!
xÄ_
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 b 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 !b 3x
3.
lim
x Ä #c x 2
5.
lim
x Ä )b x8
7.
lim
#
x Ä ( (x7)
3
2x
4
œ _
œ _
œ_
lim
"Î$
x Ä !b 3x
10. (a)
lim
"Î&
x Ä !b x
4
11. lim
#Î&
xÄ! x
13.
œ lim
4
#
x Ä ! ax"Î& b
œ_
Š positive
positive ‹
lim
x Ä !c 2x
positive
Š negative
‹
4.
lim
x Ä $b x 3
Š negative
positive ‹
6.
lim
x Ä &c 2x10
3x
œ_
positive
Š positive
‹
8.
lim #
x Ä ! x (x1)
œ _
(b)
lim
"Î$
x Ä !c 3x
(b)
lim
"Î&
x Ä !c x
œ_
2
positive
Š negative
‹
2.
œ_
2
9. (a)
œ _
Š positive
positive ‹
œ_
5
"
"
2
2
12. lim
"
#Î$
xÄ! x
lim tan x œ _
14.
x Ä ˆ 1# ‰
Š negative
negative ‹
negative
Š positive
†positive ‹
œ _
œ _
œ lim
"
#
x Ä ! ax"Î$ b
œ_
lim sec x œ _
x Ä ˆ #1 ‰
lim (1 csc )) œ _
15.
) Ä !
16.
) Ä !b
lim (2 cot )) œ _ and lim c (2 cot )) œ _, so the limit does not exist
)Ä!
"
œ lim b
xÄ#
"
(x2)(x2)
œ_
Š positive"†positive ‹
"
œ lim c
xÄ#
"
(x2)(x2)
œ _
Š positive†"negative ‹
17. (a)
lim
#
x Ä # b x 4
(b)
lim
#
x Ä # c x 4
(c)
lim
#
x Ä #b x 4
(d)
lim
#
x Ä #c x 4
"
œ
lim
x Ä #b (x2)(x2)
"
œ _
Š positive†"negative ‹
"
œ
lim
x Ä #c (x2)(x2)
"
œ_
Š negative"†negative ‹
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 2.5 Infinite Limits and Vertical Asymptotes
x
œ lim b
xÄ"
x
(x1)(x1)
œ_
positive
Š positive
†positive ‹
x
œ lim c
xÄ"
x
(x1)(x1)
œ _
positive
Š positive
†negative ‹
18. (a)
lim
#
x Ä "b x 1
(b)
lim
#
x Ä "c x 1
(c)
lim
#
x Ä "b x 1
(d)
lim
#
x Ä "c x 1
x
œ
lim
x Ä "b (x1)(x1)
x
œ_
negative
Š positive
†negative ‹
x
œ
lim
x Ä "c (x1)(x1)
x
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)
lim
#
x Ä $È2
(d)
lim
x Ä 1 #
20. (a)
x#
x#
lim
x Ä #b
"
x
"
x
œ
x# 1
2x 4
x# 1
2#Î$
#
œ
"
#
(d)
lim
x Ä !c 2x 4
œ
(b)
(c)
(d)
(e)
22. (a)
x# 3x 2
x$ 2x#
lim b
x# 3x 2
x$ 2x#
#
x 3x 2
x$ 2x#
lim
x Ä #c
lim
xÄ#
#
x 3x 2
x$ 2x#
#
x 3x 2
x$ 2x#
lim
xÄ!
lim
x Ä #b
(c)
x Ä 0c
œ lim b
xÄ#
œ lim c
xÄ#
(d)
x Ä "b
(e)
lim
x Ä !b x(x #)
x# 1
2x 4
lim
x Ä #c
œ _
positive
Š negative
‹
œ0
(x 2)(x 1)
x# (x 2)
œ _
(x 2)(x 1)
x# (x 2)
œ lim b
xÄ#
(x 2)(x 1)
x# (x 2)
œ lim c
xÄ#
œ lim
œ lim
(x 2)(x 1)
x# (x 2)
œ _
xÄ!
x# 3x 2
x$ 4x
lim
x# 3x 2
x$ 4x
x"
2†0
#4
(x 2)(x 1)
x# (x 2)
xÄ#
lim
and
œ lim b
xÄ!
œ lim b
xÄ#
x# 3x 2
x$ 4x
(b)
œ
œ lim
x# 3x 2
x$ 4x
lim
x Ä #b
(x 1)(x 1)
2x 4
(b)
"
4
lim b
xÄ#
3
#
Š positive
positive ‹
œ lim b
xÄ"
xÄ!
œ 2"Î$ 2"Î$ œ 0
œ_
lim
x Ä "b 2x 4
21. (a)
"
#"Î$
ˆ "1 ‰ œ
(c)
x# 1
œ
xÄ#
(x 2)(x ")
x(x #)(x 2)
(x 2)(x ")
œ lim c
xÄ!
œ lim b
xÄ"
(x 2)(x ")
x(x #)(x 2)
(x 2)(x ")
x(x #)(x 2)
œ
x1
x#
œ
"
4
,xÁ2
x1
x#
œ
"
4
,xÁ2
x1
x#
œ
"
4
,xÁ2
†negative
Š negative
positive†negative ‹
œ lim b
xÄ#
lim
x Ä #b x(x #)(x 2)
†negative
Š negative
positive†negative ‹
(x 1)
x(x #)
œ
(x 1)
lim
x Ä #b x(x #)
œ lim c
xÄ!
œ lim b
xÄ"
œ_
(x 1)
x(x #)
œ
negative
Š positive
†positive ‹
x"
negative
Š negative
†positive ‹
œ_
œ
"
8
œ_
(x 1)
x(x #)
œ _
lim
x Ä !c x(x #)
"
#(4)
0
(1)(3)
negative
Š negative
†positive ‹
negative
Š negative
†positive ‹
œ0
so the function has no limit as x Ä 0.
lim 2 23. (a)
t Ä !b
24. (a)
t Ä !b
25. (a)
x Ä !b
(c)
x Ä "b
3 ‘
t"Î$
œ _
"
lim t$Î&
7‘ œ _
lim
2 lim
lim
"
’ x#Î$
2
“
(x 1)#Î$
œ_
lim
"
’ x#Î$
2
“
(x 1)#Î$
œ_
(b)
t Ä !c
(b)
t Ä !c
lim
"
’ x#Î$
2
“
(x 1)#Î$
œ_
(b)
x Ä !c
lim
"
’ x#Î$
2
“
(x 1)#Î$
œ_
(d)
x Ä "c
"
t$Î&
3 ‘
t"Î$
œ_
7‘ œ _
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93
94
Chapter 2 Limits and Continuity
26. (a)
x Ä !b
(c)
x Ä "b
lim
"
’ x"Î$
1
“
(x 1)%Î$
œ_
(b)
x Ä !c
lim
"
’ x"Î$
1
“
(x 1)%Î$
œ _
lim
"
’ x"Î$
1
“
(x 1)%Î$
œ _
(d)
x Ä "c
lim
"
’ x"Î$
1
“
(x 1)%Î$
œ _
27. y œ
"
x1
28. y œ
"
x1
29. y œ
"
#x 4
30. y œ
3
x3
31. y œ
x3
x2
32. y œ
2x
x1
œ1
"
x#
œ#
2
x1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 2.5 Infinite Limits and Vertical Asymptotes
33. y œ
x#
x"
35. y œ
x# %
x"
œx"
37. y œ
x# 1
x
œx
œx1
"
x"
$
x"
"
x
39. Here is one possibility.
34. y œ
x# "
x1
œx"
36. y œ
x2 "
#x %
œ #" x " 38. y œ
x$ 1
x#
œx
#
x1
$
#x %
"
x#
40. Here is one possibility.
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95
96
Chapter 2 Limits and Continuity
41. Here is one possibility.
42. Here is one possibility.
43. Here is one possibility.
44. Here is one possibility.
45. Here is one possibility.
46. Here is one possibility.
"
x#
47. For every real number B 0, we must find a $ 0 such that for all x, 0 kx 0k $ Ê
"
x#
Ê
B ! Í
"
x#
"
x#
#
B0 Í x "
B
"
ÈB
Í kxk . Choose $ œ
"
ÈB
, then 0 kxk $ Ê kxk xÄ!
"
B.
B ! Í lxl "
B.
Choose $ œ
Then ! kx 0k $ Ê lxl "
B
Ê
"
lx l
"
lx l
2
(x 3)#
B ! Í
2
(x 3)#
$ œ É B2 , then 0 kx 3k $ Ê
B0 Í
2
(x 3)#
(x 3)#
2
"
B
Í (x 3)# B 0 so that lim
2
#
x Ä $ (x 3)
2
B
B. Now,
x Ä ! lx l
2
(x 3)#
Now,
#
B ! Í (x 5) Ê kx 5k "
ÈB
Ê
"
(x 5)#
"
B
Í kx 5k B so that lim
"
"
ÈB
#
x Ä & (x 5)
. Choose $ œ
œ _.
B.
Í ! kB $k É B2 . Choose
œ _.
50. For every real number B 0, we must find a $ 0 such that for all x, 0 kx (5)k $ Ê
1
(x 5)#
"
B so that lim
49. For every real number B 0, we must find a $ 0 such that for all x, 0 kx 3k $ Ê
Now,
"
ÈB
B so that lim x"# œ _.
48. For every real number B 0, we must find a $ 0 such that for all x, ! kx 0k $ Ê
"
lx l
B. Now,
"
ÈB
1
(x 5)#
B.
. Then 0 kx (5)k $
œ _.
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Section 2.5 Infinite Limits and Vertical Asymptotes
97
51. (a) We say that f(x) approaches infinity as x approaches x! from the left, and write lim f(x) œ _, if
x Ä x!
for every positive number B, there exists a corresponding number $ 0 such that for all x,
x! $ x x! Ê f(x) B.
(b) We say that f(x) approaches minus infinity as x approaches x! from the right, and write lim f(x) œ _,
x Ä x!
if for every positive number B (or negative number B) there exists a corresponding number $ 0 such
that for all x, x! x x! $ Ê f(x) B.
(c) We say that f(x) approaches minus infinity as x approaches x! from the left, and write lim f(x) œ _,
x Ä x!
if for every positive number B (or negative number B) there exists a corresponding number $ 0 such
that for all x, x! $ x x! Ê f(x) B.
52. For B 0,
"
x
B 0 Í x B" . Choose $ œ B" . Then ! x $ Ê 0 x 53. For B 0,
"
x
B 0 Í x" B 0 Í x Ê B" x Ê
54. For B !,
"
x#
"
x
B so that lim c
xÄ!
"
x
"
B
Ê
"
x#
"
x#
Ê
"
x#
"
1 x#
"
x
œ _.
"
B
Í x 2 B" Í x 2 B" . Choose $ œ B" . Then
"
x#
B 0 so that lim c
xÄ#
"
x#
œ _.
œ _.
B Í 1 x# "
#B . Then " $ x " Ê
"
1 x# B for ! x 1 and
"
x
B so that lim b
xÄ!
B Í ! x 2 B" . Choose $ œ B" . Then # x # $ Ê ! x # $ Ê ! x 2 B ! so that lim b
xÄ#
57. y œ sec x "
x
œ _.
B Í x " # B Í (x 2) 56. For B 0 and ! x 1,
$
Ê
Í B" x. Choose $ œ B" . Then $ x !
2 $ x 2 Ê $ x 2 ! Ê B" x 2 0 Ê
55. For B 0,
"
B
"
B
Í (" x)(" x) B" . Now
$ x 1 0 Ê " x $ x near 1 Ê
"
lim
#
x Ä "c " x
"
#B
1x
#
1 since x 1. Choose
Ê (" x)(" x) B" ˆ 1 # x ‰ B"
œ _.
58. y œ sec x "
x#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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"
B
98
Chapter 2 Limits and Continuity
59. y œ tan x 61. y œ
"
x#
x
È 4 x#
63. y œ x#Î$ 60. y œ
"
x
62. y œ
"
È 4 x#
tan x
64. y œ sin ˆ x# 1 1 ‰
"
x"Î$
2.6 CONTINUITY
1. No, discontinuous at x œ 2, not defined at x œ 2
2. No, discontinuous at x œ 3, " œ lim c g(x) Á g(3) œ 1.5
xÄ$
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 2.6 Continuity
3. Continuous on [1ß 3]
4. No, discontinuous at x œ 1, 1.5 œ lim c k(x) Á lim b k(x) œ !
xÄ"
xÄ"
5. (a) Yes
(b) Yes,
(c) Yes
(d) Yes
6. (a) Yes, f(1) œ 1
lim
x Ä "b
f(x) œ 0
(b) Yes, lim f(x) œ 2
xÄ1
(c) No
(d) No
7. (a) No
(b) No
8. ["ß !) (!ß ") ("ß #) (#ß $)
9. f(2) œ 0, since lim c f(x) œ 2(2) 4 œ 0 œ lim b f(x)
xÄ#
xÄ#
10. f(1) should be changed to 2 œ lim f(x)
xÄ1
11. Nonremovable discontinuity at x œ 1 because lim f(x) fails to exist ( limc f(x) œ 1 and lim b f(x) œ 0).
xÄ"
xÄ1
xÄ"
Removable discontinuity at x œ 0 by assigning the number lim f(x) œ 0 to be the value of f(0) rather than
xÄ!
f(0) œ 1.
12. Nonremovable discontinuity at x œ 1 because lim f(x) fails to exist ( lim c f(x) œ 2 and lim b f(x) œ 1).
xÄ"
xÄ1
xÄ"
Removable discontinuity at x œ 2 by assigning the number lim f(x) œ 1 to be the value of f(2) rather than
xÄ#
f(2) œ 2.
13. Discontinuous only when x 2 œ 0 Ê x œ 2
14. Discontinuous only when (x 2)# œ 0 Ê x œ 2
15. Discontinuous only when x# %x $ œ ! Ê (x 3)(x 1) œ 0 Ê x œ 3 or x œ 1
16. Discontinuous only when x# 3x 10 œ 0 Ê (x 5)(x 2) œ 0 Ê x œ 5 or x œ 2
17. Continuous everywhere. ( kx 1k sin x defined for all x; limits exist and are equal to function values.)
18. Continuous everywhere. ( kxk " Á 0 for all x; limits exist and are equal to function values.)
19. Discontinuous only at x œ 0
20. Discontinuous at odd integer multiples of 1# , i.e., x = (2n ") 1# , n an integer, but continuous at all other x.
21. Discontinuous when 2x is an integer multiple of 1, i.e., 2x œ n1, n an integer Ê x œ
n1
# ,
n an integer, but
continuous at all other x.
22. Discontinuous when
1x
#
is an odd integer multiple of 1# , i.e.,
1x
#
œ (2n 1) 1# , n an integer Ê x œ 2n 1, n an
integer (i.e., x is an odd integer). Continuous everywhere else.
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99
100
Chapter 2 Limits and Continuity
23. Discontinuous at odd integer multiples of 1# , i.e., x = (2n 1) 1# , n an integer, but continuous at all other x.
24. Continuous everywhere since x% 1
and are equal to the function values.
1 and " Ÿ sin x Ÿ 1 Ê 0 Ÿ sin# x Ÿ 1 Ê 1 sin# x
1; limits exist
25. Discontinuous when 2x 3 0 or x 3# Ê continuous on the interval 3# ß _‰ .
26. Discontinuous when 3x 1 0 or x "
3
Ê continuous on the interval 3" ß _‰ .
27. Continuous everywhere: (2x 1)"Î$ is defined for all x; limits exist and are equal to function values.
28. Continuous everywhere: (2 x)"Î& is defined for all x; limits exist and are equal to function values.
29. xlim
sin (x sin x) œ sin (1 sin 1) œ sin (1 0) œ sin 1 œ 0, and function continuous at x œ 1.
Ä1
30. lim sin ˆ 1# cos (tan t)‰ œ sin ˆ 1# cos (tan (0))‰ œ sin ˆ 1# cos (0)‰ œ sin ˆ 1# ‰ œ 1, and function continuous at t œ !.
tÄ!
31. lim sec ay sec# y tan# y 1b œ lim sec ay sec# y sec# yb œ lim sec a(y 1) sec# yb œ sec a(" ") sec# 1b
yÄ1
yÄ1
yÄ1
œ sec 0 œ 1, , and function continuous at y œ ".
32. lim tan 14 cos ˆsin x"Î$ ‰‘ œ tan 14 cos (sin(0))‘ œ tan ˆ 14 cos (0)‰ œ tan ˆ 14 ‰ œ 1, and function continuous at x œ !.
xÄ!
33. lim cos ’ È19 13 sec 2t “ œ cos ’ È19 13 sec 0 “ œ cos
tÄ!
1
È16
œ cos
1
4
œ
È2
# ,
and function continuous at t œ !.
34. lim1 Écsc# x 5È3 tan x œ Écsc# ˆ 16 ‰ 5È3 tan ˆ 16 ‰ œ Ê4 5È3 Š È"3 ‹ œ È9 œ 3, and function continuous at
xÄ
'
x œ 1' .
35. g(x) œ
x# 9
x3
(x 3)(x 3)
(x 3)
œ
36. h(t) œ
t# 3t 10
t#
37. f(s) œ
s$ "
s# 1
38. g(x) œ
œ
œ
œ x 3, x Á 3 Ê g(3) œ lim (x 3) œ 6
(t 5)(t 2)
t#
as# s 1b (s 1)
(s 1)(s 1)
x# 16
x# 3x 4
œ
xÄ$
œ t 5, t Á # Ê h(2) œ lim (t 5) œ 7
tÄ#
œ
(x 4)(x 4)
(x 4)(x 1)
s# s "
s1 ,
œ
x4
x1
s Á 1 Ê f(1) œ lim Š s
sÄ1
#
s1
s1 ‹
4‰
, x Á 4 Ê g(4) œ lim ˆ xx 1 œ
xÄ%
œ
3
#
8
5
39. As defined, lim c f(x) œ (3)# 1 œ 8 and lim b (2a)(3) œ 6a. For f(x) to be continuous we must have
xÄ$
xÄ$
6a œ 8 Ê a œ 43 .
40. As defined,
lim
x Ä #c
g(x) œ 2 and
4b œ 2 Ê b œ "# .
lim
x Ä #b
g(x) œ b(2)# œ 4b. For g(x) to be continuous we must have
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 2.6 Continuity
41. The function can be extended: f(0) ¸ 2.3.
42. The function cannot be extended to be continuous at
x œ 0. If f(0) ¸ 2.3, it will be continuous from the
right. Or if f(0) ¸ 2.3, it will be continuous from the
left.
43. The function cannot be extended to be continuous
at x œ 0. If f(0) œ 1, it will be continuous from
the right. Or if f(0) œ 1, it will be continuous
from the left.
44. The function can be extended: f(0) ¸ 7.39.
101
45. f(x) is continuous on [!ß "] and f(0) 0, f(1) 0
Ê by the Intermediate Value Theorem f(x) takes
on every value between f(0) and f(1) Ê the
equation f(x) œ 0 has at least one solution between
x œ 0 and x œ 1.
46. cos x œ x Ê (cos x) x œ 0. If x œ 1# , cos ˆ 1# ‰ ˆ 1# ‰ 0. If x œ 1# , cos ˆ 1# ‰ for some x between 1
#
and
1
#
1
#
0. Thus cos x x œ 0
according to the Intermediate Value Theorem.
47. Let f(x) œ x$ 15x 1 which is continuous on [4ß 4]. Then f(4) œ 3, f(1) œ 15, f(1) œ 13, and f(4) œ 5.
By the Intermediate Value Theorem, f(x) œ 0 for some x in each of the intervals % x 1, " x 1, and
" x 4. That is, x$ 15x 1 œ 0 has three solutions in [%ß 4]. Since a polynomial of degree 3 can have at most 3
solutions, these are the only solutions.
48. Without loss of generality, assume that a b. Then F(x) œ (x a)# (x b)# x is continuous for all values of
x, so it is continuous on the interval [aß b]. Moreover F(a) œ a and F(b) œ b. By the Intermediate Value
Theorem, since a a # b b, there is a number c between a and b such that F(x) œ a # b .
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102
Chapter 2 Limits and Continuity
49. Answers may vary. Note that f is continuous for every value of x.
(a) f(0) œ 10, f(1) œ 1$ 8(1) 10 œ 3. Since $ 1 10, by the Intermediate Value Theorem, there exists a c
so that ! c 1 and f(c) œ 1.
(b) f(0) œ 10, f(4) œ (4)$ 8(4) 10 œ 22. Since 22 È3 10, by the Intermediate Value
Theorem, there exists a c so that 4 c 0 and f(c) œ È3.
(c) f(0) œ 10, f(1000) œ (1000)$ 8(1000) 10 œ 999,992,010. Since 10 5,000,000 999,992,010, by the
Intermediate Value Theorem, there exists a c so that ! c 1000 and f(c) œ 5,000,000.
50. All five statements ask for the same information because of the intermediate value property of continuous
functions.
(a) A root of f(x) œ x$ 3x 1 is a point c where f(c) œ 0.
(b) The points where y œ x$ crosses y œ 3x 1 have the same y-coordinate, or y œ x$ œ 3x 1
Ê f(x) œ x$ 3x 1 œ 0.
(c) x$ 3x œ 1 Ê x$ 3x 1 œ 0. The solutions to the equation are the roots of f(x) œ x$ 3x 1.
(d) The points where y œ x$ 3x crosses y œ 1 have common y-coordinates, or y œ x$ 3x œ 1
Ê f(x) œ x$ 3x 1 œ !.
(e) The solutions of x$ 3x 1 œ 0 are those points where f(x) œ x$ 3x 1 has value 0.
51. Answers may vary. For example, f(x) œ
sin (x 2)
x2
is discontinuous at x œ 2 because it is not defined there.
However, the discontinuity can be removed because f has a limit (namely 1) as x Ä 2.
52. Answers may vary. For example, g(x) œ
"
x1
has a discontinuity at x œ 1 because lim g(x) does not exist.
x Ä "
Š lim c g(x) œ _ and lim b g(x) œ _.‹
x Ä "
x Ä "
53. (a) Suppose x! is rational Ê f(x! ) œ 1. Choose % œ "# . For any $ 0 there is an irrational number x (actually
infinitely many) in the interval (x! $ ß x! $ ) Ê f(x) œ 0. Then 0 kx x! k $ but kf(x) f(x! )k
œ 1 "# œ %, so x lim
f(x) fails to exist Ê f is discontinuous at x! rational.
Äx
!
On the other hand, x! irrational Ê f(x! ) œ 0 and there is a rational number x in (x! $ ß x! $ ) Ê f(x)
œ 1. Again x lim
f(x) fails to exist Ê f is discontinuous at x! irrational. That is, f is discontinuous at
Äx
!
every point.
(b) f is neither right-continuous nor left-continuous at any point x! because in every interval (x! $ ß x! ) or
(x! ß x! $ ) there exist both rational and irrational real numbers. Thus neither limits lim f(x) and
x Ä x!
lim f(x) exist by the same arguments used in part (a).
x Ä x!
54. Yes. Both f(x) œ x and g(x) œ x g ˆ "# ‰ œ 0 Ê
f(x)
g(x)
"
#
are continuous on [!ß "]. However
f(x)
g(x)
is undefined at x œ
"
#
since
is discontinuous at x œ "# .
55. No. For instance, if f(x) œ 0, g(x) œ ÜxÝ, then h(x) œ 0 aÜxÝb œ 0 is continuous at x œ 0 and g(x) is not.
56. Let f(x) œ
œ
"
x1
"
(x 1) 1
œ
and g(x) œ x 1. Both functions are continuous at x œ 0. The composition f ‰ g œ f(g(x))
"
x
is discontinuous at x œ 0, since it is not defined there. Theorem 10 requires that f(x) be
continuous at g(0), which is not the case here since g(0) œ 1 and f is undefined at 1.
57. Yes, because of the Intermediate Value Theorem. If f(a) and f(b) did have different signs then f would have to
equal zero at some point between a and b since f is continuous on [aß b].
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Section 2.7 Tangents and Derivatives
58. Let f(x) be the new position of point x and let d(x) œ f(x) x. The displacement function d is negative if x is
the left-hand point of the rubber band and positive if x is the right-hand point of the rubber band. By the
Intermediate Value Theorem, d(x) œ 0 for some point in between. That is, f(x) œ x for some point x, which is
then in its original position.
59. If f(0) œ 0 or f(1) œ 1, we are done (i.e., c œ 0 or c œ 1 in those cases). Then let f(0) œ a 0 and f(1) œ b 1
because 0 Ÿ f(x) Ÿ 1. Define g(x) œ f(x) x Ê g is continuous on [0ß 1]. Moreover, g(0) œ f(0) 0 œ a 0 and
g(1) œ f(1) 1 œ b 1 0 Ê by the Intermediate Value Theorem there is a number c in (!ß ") such that
g(c) œ 0 Ê f(c) c œ 0 or f(c) œ c.
60. Let % œ
kf(c)k
#
0. Since f is continuous at x œ c there is a $ 0 such that kx ck $ Ê kf(x) f(c)k %
Ê f(c) % f(x) f(c) %.
If f(c) 0, then % œ "# f(c) Ê
"
#
"
#
If f(c) 0, then % œ f(c) Ê
f(c) f(x) 3
#
3
#
f(c) f(x) f(c) Ê f(x) 0 on the interval (c $ ß c $ ).
"
#
f(c) Ê f(x) 0 on the interval (c $ ß c $ ).
61. By Exercises 52 in Section 2.3, we have xlim
faxb œ L Í lim fac hb œ L.
Äc
hÄ0
Thus, faxb is continuous at x œ c Í xlim
faxb œ facb Í lim fac hb œ facb.
Äc
hÄ0
62. By Exercise 61, it suffices to show that lim sinac hb œ sin c and lim cosac hb œ cos c.
hÄ0
hÄ0
Now lim sinac hb œ lim asin cbacos hb acos cbasin hb‘ œ asin cbŠ lim cos h‹ acos cbŠ lim sin h‹
hÄ0
hÄ0
hÄ0
hÄ0
By Example 6 Section 2.2, lim cos h œ " and lim sin h œ !. So lim sinac hb œ sin c and thus faxb œ sin x is
hÄ0
hÄ0
continuous at x œ c. Similarly,
hÄ0
lim cosac hb œ lim acos cbacos hb asin cbasin hb‘ œ acos cbŠ lim cos h‹ asin cbŠ lim sin h‹ œ cos c.
hÄ0
hÄ0
Thus, gaxb œ cos x is continuous at x œ c.
hÄ0
63. x ¸ 1.8794, 1.5321, 0.3473
64. x ¸ 1.4516, 0.8547, 0.4030
65. x ¸ 1.7549
66. x ¸ 1.5596
67. x ¸ 3.5156
68. x ¸ 3.9058, 3.8392, 0.0667
69. x ¸ 0.7391
70. x ¸ 1.8955, 0, 1.8955
hÄ0
2.7 TANGENTS AND DERIVATIVES
1. P" : m" œ 1, P# : m# œ 5
2. P" : m" œ 2, P# : m# œ 0
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Chapter 2 Limits and Continuity
3. P" : m" œ 5# , P# : m# œ "#
5. m œ lim
hÄ!
4. P" : m" œ 3, P# : m# œ 3
c4 (" h)# d a4 (1)# b
h
a1 2h h# b1
h
hÄ!
œ lim
œ lim
hÄ!
h(# h)
h
œ 2;
at ("ß $): y œ $ #(x (1)) Ê y œ 2x 5,
tangent line
6. m œ lim
hÄ!
c(1 h 1)# 1d c(" ")# 1d
h
h#
œ lim
hÄ! h
œ lim h œ 0; at ("ß "): y œ 1 0(x 1) Ê y œ 1,
hÄ!
tangent line
È
2È 1 h 2È 1
œ lim 2 1 h h 2
h
hÄ!
hÄ!
4(1 h) 4
œ lim
œ lim È1 2h 1
h Ä ! 2h ŠÈ1 h 1‹
hÄ!
7. m œ lim
†
2È 1 h 2
2È 1 h #
œ 1;
at ("ß #): y œ 2 1(x 1) Ê y œ x 1, tangent line
8. m œ lim
hÄ!
"
(1 h)#
("" )#
h
a2h h# b
#
h Ä ! h(1 h)
œ lim
1 (1 h)#
#
h Ä ! h(1h)
2h
lim
# œ 2;
h Ä ! (1 h)
œ lim
œ
at ("ß "): y œ 1 2(x (1)) Ê y œ 2x 3,
tangent line
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 2.7 Tangents and Derivatives
(2 h)$ (2)$
h
9. m œ lim
hÄ!
8 12h 6h# h$ 8
h
œ lim
hÄ!
œ lim a12 6h h# b œ 12;
hÄ!
at (2ß 8): y œ 8 12(x (2)) Ê y œ 12x 16,
tangent line
"
(# h)$
10. m œ lim
h
hÄ!
œ
"2
8(8)
hÄ!
hÄ!
at ˆ#ß "8 ‰ : y œ 8" Ê yœ
11. m œ lim
hÄ!
x
"
#,
12 6h h#
8(2 h)$
œ lim
3
œ 16
;
3
16
8 (# h)$
8h(# h)$
œ lim
a12h 6h# h$ b
8h(# h)$
œ lim
hÄ!
(#" )$
3
16 (x
(2))
tangent line
c(2 h)# 1d 5
h
œ lim
hÄ!
a5 4h h# b 5
h
hÄ!
at (2ß 5): y 5 œ 4(x 2), tangent line
12. m œ lim
hÄ!
c(" h) 2(1 h)# d (1)
h
œ lim
hÄ!
h(4 h)
h
œ lim
a1 h 2 4h 2h# b 1
h
hÄ!
3h
(3 h) 2
3
h
œ lim
hÄ!
(3 h) 3(h 1)
h(h 1)
h Ä ! h(h 1)
at ($ß $): y 3 œ 2(x 3), tangent line
14. m œ lim
hÄ!
8
(2 h)#
2
h
hÄ!
hÄ!
(2 h)$ 8
h
œ lim
hÄ!
œ lim
a8 12h 6h# h$ b 8
h
hÄ!
at (2ß )): y 8 œ 12(t 2), tangent line
16. m œ lim
hÄ!
c(1 h)$ 3(1 h)d 4
h
a1 3h 3h# h$ 3 3hb 4
h
œ lim
hÄ!
at ("ß %): y 4 œ 6(t 1), tangent line
È4 h 2
h
hÄ!
17. m œ lim
È4 h 2
h
hÄ!
œ lim
œ "4 ; at (%ß #): y 2 œ
18. m œ lim
hÄ!
œ
"
È9 3
È(8 h) 1 3
h
"
4
†
È4 h 2
È4 h 2
œ lim
hÄ!
h a12 6h h# b
h
œ lim
œ 3;
œ 2;
8 2 a4 4h h# b
h(2 h)#
hÄ!
8 2(2 h)#
#
h Ä ! h(2 h)
œ lim
at (2ß 2): y 2 œ 2(x 2)
15. m œ lim
2h
œ lim
h(3 2h)
h
œ lim
at ("ß "): y 1 œ 3(x 1), tangent line
13. m œ lim
œ %;
œ lim
2h(4 h)
h(2 h)#
œ
8
4
œ 2;
œ 12;
œ lim
hÄ!
(4 h) 4
h Ä ! h ŠÈ4 h #‹
h a6 3h h# b
h
œ lim
œ 6;
h
h Ä ! h ŠÈ4 h #‹
œ
"
È4 #
(x 4), tangent line
œ lim
hÄ!
È9 h 3
h
œ 6" ; at (8ß 3): y 3 œ
19. At x œ 1, y œ 5 Ê m œ lim
hÄ!
"
6
†
È9 h 3
È9 h 3
œ lim
(9 h) 9
h Ä ! h ŠÈ9 h 3‹
œ lim
h
h Ä ! h ŠÈ9 h 3‹
(x 8), tangent line
5(" h)# 5
h
œ lim
hÄ!
5 a1 2h h# b 5
h
œ lim
hÄ!
5h(2 h)
h
œ 10, slope
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105
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Chapter 2 Limits and Continuity
c1 (2 h)# d (3)
h
20. At x œ 2, y œ 3 Ê m œ lim
hÄ!
21. At x œ 3, y œ
"
#
"
(3 h) 1
Ê m œ lim
#"
h1
h1
22. At x œ 0, y œ 1 Ê m œ lim
hÄ!
hÄ!
hÄ!
(1)
h
a1 4 4h h# b 3
h
2 (2 h)
2h(2 h)
œ lim
h
hÄ!
œ lim
hÄ!
hÄ!
h
œ lim
h Ä ! 2h(2 h)
(h 1) (h ")
h(h 1)
œ lim
œ lim
œ lim
h(4 h)
h
œ 4, slope
œ "4 , slope
2h
h Ä ! h(h 1)
œ 2, slope
c(x h)# 4(x h) 1d ax# 4x 1b
h
hÄ!
a2xh h# 4hb
lim
œ lim (2x h 4) œ 2x
h
hÄ!
hÄ!
23. At a horizontal tangent the slope m œ 0 Ê 0 œ m œ lim
ax# 2xh h# 4x 4h 1b ax# 4x 1b
h
hÄ!
œ lim
œ
4;
2x 4 œ 0 Ê x œ 2. Then f(2) œ 4 8 1 œ 5 Ê (2ß 5) is the point on the graph where there is a
horizontal tangent.
24. 0 œ m œ lim
hÄ!
c(x h)$ 3(x h)d ax$ 3xb
h
3x# h 3xh# h$ 3h
h
œ lim
hÄ!
œ lim
hÄ!
ax$ 3x# h 3xh# h$ 3x 3hb ax$ 3xb
h
œ lim a3x# 3xh h# 3b œ 3x# 3; 3x# 3 œ 0 Ê x œ 1 or x œ 1. Then
hÄ!
f(1) œ 2 and f(1) œ 2 Ê ("ß 2) and ("ß 2) are the points on the graph where a horizontal tangent exists.
"
(x h) 1
25. 1 œ m œ lim
x " 1
h
hÄ!
(x 1) (x h 1)
h(x 1)(x h 1)
œ lim
hÄ!
h
œ lim
h Ä ! h(x 1)(x h 1)
œ (x " 1)#
Ê (x 1)# œ 1 Ê x# 2x œ 0 Ê x(x 2) œ 0 Ê x œ 0 or x œ 2. If x œ 0, then y œ 1 and m œ 1
Ê y œ 1 (x 0) œ (x 1). If x œ 2, then y œ 1 and m œ 1 Ê y œ 1 (x 2) œ (x 3).
26.
"
4
œ m œ lim
Èx h Èx
œ lim
h
y œ 2 "4 (x 4) œ
hÄ!
f(2 h) f(2)
h
x
4
Èx h Èx
h
hÄ!
h Ä ! h ŠÈx h Èx‹
27. lim
œ lim
h
hÄ!
œ
"
#È x
. Thus,
"
4
œ
†
Èx h Èx
Èx h Èx
"
#Èx
(x h) x
œ lim
h Ä ! h ŠÈx h Èx‹
Ê Èx œ 2 Ê x œ 4 Ê y œ 2. The tangent line is
1.
œ lim
hÄ!
a100 4.9(# h)# b a100 4.9(2)# b
h
4.9 a4 4h h# b 4.9(4)
h
œ lim
hÄ!
œ lim (19.6 4.9h) œ 19.6. The minus sign indicates the object is falling downward at a speed of
hÄ!
19.6 m/sec.
f(10 h) f(10)
h
hÄ!
28. lim
3(10 h)# 3(10)#
h
hÄ!
œ lim
29. lim
f(3 h) f(3)
h
œ lim
30. lim
f(2 h) f(2)
h
œ lim
hÄ!
hÄ!
hÄ!
hÄ!
1(3 h)# 1(3)#
h
41
3
œ lim
(2 h)$ 431 (2)$
h
f(0 h) f(0)
h
hÄ!
31. Slope at origin œ lim
3 a20h h# b
h
hÄ!
œ lim
hÄ!
œ lim
1 c9 6h h# 9d
h
41
3
hÄ!
h# sin ˆ "h ‰
h
hÄ!
œ lim
œ 60 ft/sec.
œ lim 1(6 h) œ 61
c12h 6h# h$ d
h
hÄ!
œ lim
hÄ!
41
3
c12 6h h# d œ 161
œ lim h sin ˆ h" ‰ œ 0 Ê yes, f(x) does have a tangent at
hÄ!
the origin with slope 0.
32. lim
hÄ!
g(0 h) g(0)
h
œ lim
hÄ!
h sin ˆ "h ‰
h
œ lim sin h" . Since lim sin
hÄ!
hÄ!
"
h
does not exist, f(x) has no tangent at
the origin.
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Section 2.7 Tangents and Derivatives
33.
lim
h Ä !c
lim
hÄ!
34.
f(0 h) f(0)
h
f(0 h) f(0)
h
œ lim c
hÄ!
1 0
h
œ _, and lim b
hÄ!
f(0 h) f(0)
h
10
h
œ lim b
hÄ!
œ _ Ê yes, the graph of f has a vertical tangent at the origin.
œ _, and lim b U(0 h)h U(0) œ lim b
hÄ!
hÄ!
does not have a vertical tangent at (!ß ") because the limit does not exist.
lim
h Ä !c
œ _. Therefore,
U(0 h) U(0)
h
œ lim c
hÄ!
01
h
11
h
œ 0 Ê no, the graph of f
35. (a) The graph appears to have a cusp at x œ 0.
(b)
lim
h Ä !c
f(0 h) f(0)
h
œ lim c
hÄ!
h#Î& 0
h
œ lim c
hÄ!
"
h$Î&
œ _ and lim b
hÄ!
"
h$Î&
œ _ Ê limit does not exist
Ê the graph of y œ x#Î& does not have a vertical tangent at x œ 0.
36. (a) The graph appears to have a cusp at x œ 0.
(b)
lim
h Ä !c
f(0 h) f(0)
h
œ lim c
hÄ!
h%Î& 0
h
œ lim c
hÄ!
"
h"Î&
œ _ and lim b
hÄ!
"
h"Î&
œ _ Ê limit does not exist
Ê y œ x%Î& does not have a vertical tangent at x œ 0.
37. (a) The graph appears to have a vertical tangent at x œ !.
(b)
f(0 h) f(0)
h
hÄ!
lim
h"Î& 0
h
hÄ!
œ lim
œ lim
"
%Î&
hÄ! h
œ _ Ê y œ x"Î& has a vertical tangent at x œ 0.
38. (a) The graph appears to have a vertical tangent at x œ 0.
(b)
lim
hÄ!
f(0 h) f(0)
h
at x œ 0.
œ lim
hÄ!
h$Î& 0
h
œ lim
"
#Î&
hÄ! h
œ _ Ê the graph of y œ x$Î& has a vertical tangent
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107
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Chapter 2 Limits and Continuity
39. (a) The graph appears to have a cusp at x œ 0.
(b)
lim
h Ä !c
f(0 h) f(0)
h
œ lim c
hÄ!
4h#Î& 2h
h
œ lim c
hÄ!
4
h$Î&
2 œ _ and lim b
hÄ!
4
h$Î&
#œ_
Ê limit does not exist Ê the graph of y œ 4x#Î& 2x does not have a vertical tangent at x œ 0.
40. (a) The graph appears to have a cusp at x œ 0.
(b)
lim
hÄ!
f(0 h) f(0)
h
œ lim
hÄ!
h&Î$ 5h#Î$
h
œ lim h#Î$ hÄ!
5
h"Î$
œ 0 lim
5
"Î$
hÄ! h
y œ x&Î$ 5x#Î$ does not have a vertical tangent at x œ !.
does not exist Ê the graph of
41. (a) The graph appears to have a vertical tangent at x œ 1
and a cusp at x œ 0.
(b) x œ 1:
(1 h)#Î$ (1 h 1)"Î$ "
h
hÄ!
#Î$
"Î$
lim
Ê yœx
x œ 0:
(x 1)
lim f(0 h)h f(0)
hÄ!
(1 h)#Î$ h"Î$ "
h
hÄ!
œ lim
has a vertical tangent at x œ 1;
h#Î$ (h 1)"Î$ (1)"Î$
h
hÄ!
#Î$
"Î$
œ lim
does not exist Ê y œ x
œ _
(x 1)
"
œ lim ’ h"Î$
hÄ!
(h ")"Î$
h
h" “
does not have a vertical tangent at x œ 0.
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Section 2.7 Tangents and Derivatives
42. (a) The graph appears to have vertical tangents at x œ 0 and
x œ 1.
(b) x œ 0:
h"Î$ (h 1)"Î$ (")"Î$
h
hÄ!
f(0 h) f(0)
h
hÄ!
œ lim
f(1 h) f(1)
h
œ lim
lim
œ _ Ê y œ x"Î$ (x 1)"Î$ has a
vertical tangent at x œ 0;
x œ 1:
lim
hÄ!
hÄ!
(1 h)"Î$ (" h 1)"Î$ 1
h
œ _ Ê y œ x"Î$ (x 1)"Î$ has a
vertical tangent at x œ ".
43. (a) The graph appears to have a vertical tangent at x œ 0.
(b)
lim
h Ä !b
f(0 h) f(0)
h
œ lim b
xÄ!
Èh 0
h
œ lim
È kh k 0
f(0 h) f(0)
h
"
h Ä ! Èh
œ lim c
œ lim c
h
hÄ!
hÄ!
Ê y has a vertical tangent at x œ 0.
lim
h Ä !c
œ _;
È kh k
kh k
œ lim c
hÄ!
"
È kh k
œ_
44. (a) The graph appears to have a cusp at x œ 4.
(b)
lim
f(4 h) f(4)
h
œ lim b
hÄ!
Èk4 (4 h)k 0
h
lim
f(4 h) f(4)
h
œ lim c
hÄ!
Èk4 (4 h)k
h
h Ä !b
h Ä !c
œ lim b
hÄ!
œ lim c
hÄ!
È kh k
h
È kh k
lhl
œ lim b
hÄ!
œ lim c
hÄ!
"
Èh
"
È kh k
œ _;
œ _
Ê y œ È% x does not have a vertical tangent at x œ 4.
45-48. Example CAS commands:
Maple:
f := x -> x^3 + 2*x;x0 := 0;
plot( f(x), x=x0-1/2..x0+3, color=black,
# part (a)
title="Section 2.7, #45(a)" );
q := unapply( (f(x0+h)-f(x0))/h, h );
# part (b)
L := limit( q(h), h=0 );
# part (c)
sec_lines := seq( f(x0)+q(h)*(x-x0), h=1..3 );
# part (d)
tan_line := f(x0) + L*(x-x0);
plot( [f(x),tan_line,sec_lines], x=x0-1/2..x0+3, color=black,
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Chapter 2 Limits and Continuity
linestyle=[1,2,5,6,7], title="Section 2.7, #45(d)",
legend=["y=f(x)","Tangent line at x=0","Secant line (h=1)",
"Secant line (h=2)","Secant line (h=3)"] );
Mathematica: (function and value for x0 may change)
Clear[f, m, x, h]
x0 œ p;
f[x_]: œ Cos[x] 4Sin[2x]
Plot[f[x], {x, x0 1, x0 3}]
dq[h_]: œ (f[x0+h] f[x0])/h
m œ Limit[dq[h], h Ä 0]
ytan: œ f[x0] m(x x0)
y1: œ f[x0] dq[1](x x0)
y2: œ f[x0] dq[2](x x0)
y3: œ f[x0] dq[3](x x0)
Plot[{f[x], ytan, y1, y2, y3}, {x, x0 1, x0 3}]
CHAPTER 2 PRACTICE EXERCISES
1. At x œ 1:
Ê
lim
x Ä "c
f(x) œ
lim
x Ä "b
f(x) œ 1
lim f(x) œ 1 œ f(1)
x Ä 1
Ê f is continuous at x œ 1.
At x œ 0: lim c f(x) œ lim b f(x) œ 0 Ê lim f(x) œ 0.
xÄ!
xÄ!
xÄ!
But f(0) œ 1 Á lim f(x)
xÄ!
Ê f is discontinuous at x œ 0.
If we define fa!b œ !, then the discontinuity at x œ ! is
removable.
At x œ 1: lim c f(x) œ 1 and lim b f(x) œ 1
xÄ"
Ê lim f(x) does not exist
xÄ"
xÄ1
Ê f is discontinuous at x œ 1.
2. At x œ 1:
Ê
lim
x Ä "c
f(x) œ 0 and
lim
x Ä "b
f(x) œ 1
lim f(x) does not exist
x Ä "
Ê f is discontinuous at x œ 1.
At x œ 0: lim c f(x) œ _ and lim b f(x) œ _
xÄ!
Ê lim f(x) does not exist
xÄ!
xÄ!
Ê f is discontinuous at x œ 0.
At x œ 1: lim c f(x) œ lim b f(x) œ 1 Ê lim f(x) œ 1.
xÄ"
xÄ1
xÄ"
But f(1) œ 0 Á lim f(x)
xÄ1
Ê f is discontinuous at x œ 1.
If we define fa"b œ ", then the discontinuity at x œ " is
removable.
3. (a)
(b)
lim a3fatbb œ 3 lim fatb œ 3(7) œ 21
t Ä t!
t Ä t!
#
#
lim afatbb œ Š lim fatb‹ œ a(b# œ 49
t Ä t!
t Ä t!
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Chapter 2 Practice Exercises
(c)
(d)
(e)
(f)
111
lim afatb † gatbb œ lim fatb † lim gatb œ (7)(0) œ 0
t Ä t!
t Ä t!
lim fatb
t Ä t! g(t)7
Ät
t Ä t!
lim fatb
œ
Ät
t
t
Ät
t
!
Ät
t
!
7
07
œ
lim gatb lim 7
t
!
Ät
lim fatb
œ
!
lim agatb 7b
!
œ1
lim cos agatbb œ cos Š lim gatb‹ œ cos ! œ 1
t Ä t!
t Ä t!
lim kfatbk œ ¹ lim fatb¹ œ k7k œ 7
t Ä t!
t Ä t!
(g) lim afatb gatbb œ lim fatb lim gatb œ 7 0 œ 7
t Ä t!
(h)
4. (a)
(b)
(c)
(d)
(e)
(f)
t Ä t!
lim Š " ‹
t Ä t! fatb
œ
"
lim fatb
t
Ät
t Ä t!
"
7
œ
!
œ 71
lim g(x) œ lim g(x) œ È2
xÄ!
xÄ!
lim ag(x) † f(x)b œ lim g(x) † lim f(x) œ ŠÈ2‹ ˆ "# ‰ œ
xÄ!
xÄ!
xÄ!
lim af(x) g(x)b œ lim f(x) lim g(x) œ
xÄ!
"
lim
x Ä ! f(x)
œ
xÄ!
"
lim f(x)
œ
xÄ!
"
"
#
xÄ!
œ2
"
#
lim ax f(x)b œ lim x lim f(x) œ 0 xÄ!
xÄ!
f(x)†cos x
x 1
xÄ!
lim
xÄ!
lim f(x)† lim cos x
œ
xÄ!
xÄ!
lim x lim 1
xÄ!
xÄ!
œ
ˆ "# ‰ (1)
01
"
#
È2
#
È2
œ
"
#
œ #"
5. Since lim x œ 0 we must have that lim (4 g(x)) œ 0. Otherwise, if lim (% g(x)) is a finite positive
xÄ!
xÄ!
xÄ!
’ 4xg(x) “
’ 4xg(x) “
œ _ and lim b
œ _ so the limit could not equal 1 as
xÄ!
x Ä 0. Similar reasoning holds if lim (4 g(x)) is a finite negative number. We conclude that lim g(x) œ 4.
number, we would have lim c
xÄ!
xÄ!
6. 2 œ lim
x Ä %
xÄ!
’x lim g(x)“ œ lim x † lim
xÄ!
x Ä %
x Ä %
’ lim g(x)“ œ 4 lim
xÄ!
(since lim g(x) is a constant) Ê lim g(x) œ
xÄ!
xÄ!
2
%
x Ä %
œ #" .
’ lim g(x)“ œ 4 lim g(x)
xÄ!
xÄ!
7. (a) xlim
faxb œ xlim
x"Î$ œ c"Î$ œ facb for every real number c Ê f is continuous on a_ß _b.
Äc
Äc
(b) xlim
gaxb œ xlim
x$Î% œ c$Î% œ gacb for every nonnegative real number c Ê g is continuous on Ò!ß _Ñ.
Äc
Äc
(c) xlim
haxb œ xlim
x#Î$ œ
Äc
Äc
(d) xlim
kaxb œ xlim
x"Î' œ
Äc
Äc
"
c#Î$
"
c"Î'
œ hacb for every nonzero real number c Ê h is continuous on a_ß !b and a_ß _b.
œ kacb for every positive real number c Ê k is continuous on a!ß _b
8. (a) - ˆˆn "# ‰1ß ˆn "# ‰1‰, where I œ the set of all integers.
n−I
(b) - an1ß an 1b1b, where I œ the set of all integers.
n−I
(c) a_ß 1b a1ß _b
(d) a_ß !b a!ß _b
9.
(a)
(b)
10. (a)
(x 2)(x 2)
x# 4x 4
$ 5x# 14x œ lim
x
xÄ!
x Ä ! x(x 7)(x 2)
x2
x2
lim
œ _ and lim b x(x
7)
x Ä !c x(x 7)
œ lim
lim (x 2)(x 2)
x Ä # x(x 7)(x #)
œ lim
lim
#
x 4x 4
xÄ!
lim $
#
x Ä # x 5x 14x
x# x
lim &
%
$
x Ä ! x 2x x
Now lim c
xÄ!
œ
œ lim
x(x 1)
$
#
x Ä ! x ax 2x 1b
1
x# (x 1)
œ _ and lim b
xÄ!
x2
, x Á 2; the limit does not exist because
x2
, x Á 2, and lim
x Ä ! x(x 7)
œ _
x Ä # x(x 7)
œ lim
x1
#
x Ä ! x (x 1)(x 1)
1
x# (x 1)
x2
x Ä # x(x 7)
œ lim
œ _ Ê lim
"
#
x Ä 0 x (x 1)
#
x x
&
%
$
x Ä ! x 2x x
œ
0
2(9)
œ0
, x Á 0 and x Á 1.
œ _.
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112
Chapter 2 Limits and Continuity
(b)
x# x
exist because
$
#
x Ä " x ax 2x 1b
"
1 Èx
1x
œ lim
12. xlim
Äa
x # a#
x % a%
œ xlim
Äa
13. lim
(x h)# x#
h
œ lim
(x h)# x#
h
xÄ!
œ lim
hÄ!
"
15. lim
#x #
16. lim
(# x)$ 8
x
xÄ!
xÄ!
17.
18.
xÄ!
xÄ!
lim
xÄ1
"
x g(x)
3x# 1
g(x)
œ2 Ê
xÄ!
"
x Ä ! 4 #x
xÄ!
xÄ1
5 x#
œ0 Ê
"
#
lim 4 g(x) œ 8, since 2$ œ 8. Then lim b g(x) œ 2.
xÄ!
x Ä !b
Ê È5 lim
x Ä È5
g(x) œ
"
#
Ê
lim
x Ä È5
g(x) œ
%x )
$x $
x Ä #
#!
&!
ˆ" œxÄ
lim
_ $x
%
$x#
"
#
#
#
&
œ
"
x#
24. x lim
œ x lim
œ
Ä _ x # (x "
Ä _ " (x x"#
#x $
##. x Ä
lim
œxÄ
lim
_ &x# (
_ & ) ‰
$x$
!
"!!
$
x#
(
x#
œ
#!
&!
œ!!!œ!
œ!
#
%
x (x
x(
25. x Ä
lim
œxÄ
lim
œ _
_ x 1
_ " "x
$
x x
x"
#'. x lim
œxÄ
lim
œ_
Ä _ "#x$ "#)
_ "# "#)
x$
lsin xl
lsin xl
"
27. x lim
Ÿ x lim
œ ! since int x Ä _ as x Ä _ Êx lim
œ !.
Ä _ gx h
Ä _ gx h
Ä _ gx h
lim
)Ä_
29. x lim
Ä_
lcos ) "l
)
Ÿ lim
"
l#l
)Ä_ )
x sin x #Èx
x sin x
#Î$
È5
lim g(x) œ _ since lim a5 x# b œ 1
# $
28.
"
#
xÄ1
x Ä #
#x $
x
21. x lim
œ x lim
œ
Ä _ &x (
Ä _ & (x
#
œ2 Ê
œ _ Ê lim g(x) œ 0 since lim a3x# 1b œ 4
lim
x Ä # Èg(x)
x
23. x Ä
lim
_
œ "4
œ lim ax# 6x 12b œ 12
(x g(x)) œ
lim
"
#a#
œ lim (2x h) œ h
œ lim
x Ä È&
œ
"
#
hÄ!
"Î$
x Ä È&
œ
œ _.
œ lim (2x h) œ 2x
ax$ 6x# 12x 8b 8
x
œ lim
"
x # a#
œ xlim
Äa
ax# 2hx h# b x#
h
2 (2 x)
2x(# x)
œ lim
"
lim [4 g(x)]"Î$ œ 2 Ê ’ lim b 4 g(x)“
x Ä !b
xÄ!
19. lim
20.
x
ax # a # b
ax # a # b a x # a # b
"
lim
#
x Ä "b x (x 1)
x Ä 1 1 Èx
, x Á 0 and x Á 1. The limit does not
1
#
x Ä " x (x 1)
œ lim
ax# 2hx h# b x#
h
xÄ!
14. lim
"
" Èx
x Ä 1 ˆ1 È x ‰ ˆ 1 È x ‰
hÄ!
œ lim
œ _ and
lim
#
x Ä "c x (x 1)
11. lim
xÄ1
x(x 1)
œ lim
lim
&
%
$
x Ä " x 2x x
œ ! Ê lim
œ x lim
Ä_
)Ä_
" sinx x È#x
" sinx x
&Î$
x x
" x
30. x lim
œ x lim
#x œ
Ä _ x#Î$ cos# x
Ä _Œ " cos#Î$
lcos ) "l
)
œ !.
"!!
"!
œ"
œ
"!
"!
œ"
x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
œ
#
&
Chapter 2 Practice Exercises
31. At x œ 1:
œ
lim
x Ä "c
lim
x Ä "c
lim
x Ä "b
f(x) œ
x ax # 1 b
x# 1
f(x) œ
œ
lim
x ax # 1 b
kx # 1 k
lim
x Ä "c
lim
x Ä "c
x Ä "b
x œ 1, and
x ax # 1 b
kx # 1 k
œ
x ax # 1 b
lim
#
x Ä "b ax "b
œ lim (x) œ (1) œ 1. Since
x Ä 1
lim f(x) Á lim b f(x)
x Ä "c
x Ä "
Ê
lim f(x) does not exist, the function f cannot be
x Ä 1
extended to a continuous function at x œ 1.
At x œ 1:
lim f(x) œ lim c
xÄ"
x Ä "c
#
x ax # 1 b
kx # 1 k
œ lim c
xÄ"
#
x ax # 1 b
ax # 1 b
œ lim c (x) œ 1, and
xÄ"
lim f(x) œ lim b xkaxx# 11k b œ lim b x axx# "1b œ lim b x œ 1. Again lim f(x) does not exist so f
xÄ1
xÄ"
xÄ"
xÄ1
cannot be extended to a continuous function at x œ 1 either.
x Ä "b
32. The discontinuity at x œ 0 of f(x) œ sin ˆ "x ‰ is nonremovable because lim sin
xÄ!
"
x
does not exist.
33. Yes, f does have a continuous extension to a œ 1:
"
define f(1) œ lim xxÈ
œ 43 .
%
x
xÄ1
34. Yes, g does have a continuous extension to a œ 1# :
)
5
g ˆ 1# ‰ œ lim1 45)cos
#1 œ 4 .
)Ä #
35. From the graph we see that lim h(t) Á lim h(t)
tÄ!
tÄ!
so h cannot be extended to a continuous function
at a œ 0.
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113
114
Chapter 2 Limits and Continuity
36. From the graph we see that lim c k(x) Á lim b k(x)
xÄ!
xÄ!
so k cannot be extended to a continuous function
at a œ 0.
37. (a) f(1) œ 1 and f(2) œ 5 Ê f has a root between 1 and 2 by the Intermediate Value Theorem.
(b), (c) root is 1.32471795724
38. (a) f(2) œ 2 and f(0) œ 2 Ê f has a root between 2 and 0 by the Intermediate Value Theorem.
(b), (c) root is 1.76929235424
CHAPTER 2 ADDITIONAL AND ADVANCED EXERCISES
1. (a)
x
xx
0.1
0.7943
0.01
0.9550
0.001
0.9931
10
100
1000
0.3679
0.3679
0.3679
0.0001
0.9991
0.00001
0.9999
Apparently, lim b xx œ 1
xÄ!
(b)
2. (a)
x
ˆ "x ‰"ÎÐln xÑ
Apparently,
"ÎÐln xÑ
lim ˆ " ‰
xÄ_ x
œ 0.3678 œ
"
e
(b)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Chapter 2 Additional and Advanced Exercises 115
3.
lim L œ lim c L! É" vc# œ L! É1 vÄcc# œ L! É1 cc# œ 0
vÄc
The left-hand limit was needed because the function L is undefined if v c (the rocket cannot move faster
than the speed of light).
lim v#
#
v Ä cc
#
4. ¹
Èx
#
1¹ 0.2 Ê 0.2 Èx
#
1 0.2 Ê 0.8 Èx
#
1.2 Ê 1.6 Èx 2.4 Ê 2.56 x 5.76.
¹
Èx
#
1¹ 0.1 Ê 0.1 Èx
#
1 0.1 Ê 0.9 Èx
#
1.1 Ê 1.8 Èx 2.2 Ê 3.24 x 4.84.
5. k10 (t 70) ‚ 10% 10k 0.0005 Ê k(t 70) ‚ 10% k 0.0005 Ê 0.0005 (t 70) ‚ 10% 0.0005
Ê 5 t 70 5 Ê 65° t 75° Ê Within 5° F.
6. We want to know in what interval to hold values of h to make V satisfy the inequality
lV "!!!l œ l$'1h "!!!l Ÿ "!. To find out, we solve the inequality:
**!
l$'1h "!!!l Ÿ "! Ê "! Ÿ $'1h "!!! Ÿ "! Ê **! Ÿ $'1h Ÿ "!"! Ê $'
1 Ÿ hŸ
"!"!
$'1
Ê )Þ) Ÿ h Ÿ )Þ*. where 8.8 was rounded up, to be safe, and 8.9 was rounded down, to be safe.
The interval in which we should hold h is about )Þ* )Þ) œ !Þ" cm wide (1 mm). With stripes 1 mm wide, we can expect
to measure a liter of water with an accuracy of 1%, which is more than enough accuracy for cooking.
7. Show lim f(x) œ lim ax# 7b œ ' œ f(1).
xÄ1
xÄ1
Step 1: kax# 7b 6k % Ê % x# 1 % Ê 1 % x# 1 % Ê È1 % x È1 %.
Step 2: kx 1k $ Ê $ x 1 $ Ê $ " x $ ".
Then $ " œ È1 % or $ " œ È1 %. Choose $ œ min š1 È1 %ß È1 % 1› , then
0 kx 1k $ Ê kax# (b 6k % and lim f(x) œ 6. By the continuity test, f(x) is continuous at x œ 1.
xÄ1
8. Show lim" g(x) œ lim"
xÄ
xÄ
%
"
2x
œ 2 œ g ˆ "4 ‰ .
%
Step 1: ¸ #"x 2¸ % Ê % #"x # % Ê # % #"x # % Ê
Step 2: ¸B "4 ¸ $ Ê $ x 4" $ Ê $ 4" x $ 4" .
Then $ Choose $ œ
"
4
œ
"
4 #%
%
4(#%)
Ê $œ
"
4
"
4 #%
œ
%
4(2 %)
, or $ "
4
œ
, the smaller of the two values. Then 0 ¸x
By the continuity test, g(x) is continuous at x œ
"
4
"
4 #% Ê
4" ¸ $
"
4#%
x
"
4 #%
¸ #"x "
4#%
.
"
4
%
4(2 %)
$œ
œ
Ê
2¸ % and lim"
.
xÄ
%
"
#x
œ 2.
.
9. Show lim h(x) œ lim È2x 3 œ " œ h(2).
xÄ#
xÄ#
Step 1: ¹È2x 3 1¹ % Ê % È2x 3 " % Ê " % È2x 3 " % Ê
(1 %)# $
#
x
(" %)# 3
.
#
Step 2: kx 2k $ Ê $ x 2 $ or $ # x $ #.
(" % )# $
Ê $œ
#
(" % Ñ # $
(" %Ñ# "
#œ
#
#
Then $ # œ
#
Ê $œ
œ%
#
(" %)# $
œ " (1# %)
#
#
%# . Choose $ œ %
œ%
%#
#,
%#
#
, or $ # œ
(" %)# $
#
the smaller of the two values . Then,
! kx 2k $ Ê ¹È2x 3 "¹ %, so lim È2x 3 œ 1. By the continuity test, h(x) is continuous at x œ 2.
xÄ#
10. Show lim F(x) œ lim È9 x œ # œ F(5).
xÄ&
xÄ&
Step 1: ¹È9 x 2¹ % Ê % È9 x # % Ê 9 (2 %)# x * (# %)# .
Step 2: 0 kx 5k $ Ê $ x & $ Ê $ & x $ &.
Then $ & œ * (# %)# Ê $ œ (# %)# % œ %# #%, or $ & œ * (# %)# Ê $ œ % (# %)# œ %# #%.
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116
Chapter 2 Limits and Continuity
Choose $ œ %# #%, the smaller of the two values. Then, ! kx 5k $ Ê ¹È9 x #¹ %, so
lim È9 x œ #. By the continuity test, F(x) is continuous at x œ 5.
xÄ&
11. Suppose L" and L# are two different limits. Without loss of generality assume L# L" . Let % œ
"
3
(L# L" ).
Since x lim
f(x) œ L" there is a $" 0 such that 0 kx x! k $" Ê kf(x) L" k % Ê % f(x) L" %
Äx
!
Ê "3 (L# L" ) L" f(x) "
3
(L# L" ) L" Ê 4L" L# 3f(x) 2L" L# . Likewise, x lim
f(x) œ L#
Ä x!
so there is a $# such that 0 kx x! k $# Ê kf(x) L# k % Ê % f(x) L# %
Ê "3 (L# L" ) L# f(x) 3" (L# L" ) L# Ê 2L# L" 3f(x) 4L# L"
Ê L" 4L# 3f(x) 2L# L" . If $ œ min e$" ß $# f both inequalities must hold for 0 kx x! k $ :
4L" L# 3f(x) 2L" L#
Ê 5(L" L# ) 0 L" L# . That is, L" L# 0 and L" L# 0,
L" %L# 3f(x) 2L# L" a contradiction.
12. Suppose xlim
f(x) œ L. If k œ !, then xlim
kf(x) œ xlim
0 œ ! œ ! † xlim
f(x) and we are done.
Äc
Äc
Äc
Äc
%
If k Á 0, then given any % !, there is a $ ! so that ! lx cl $ Ê lfaxb Ll l5l
Ê lkllfaxb Ll %
Ê lkafaxb Lb| % Ê lakfaxbb akLbl %. Thus, xlim
kf(x) œ kL œ kŠxlim
f(x)‹.
Äc
Äc
13. (a) Since x Ä 0 , 0 x$ x 1 Ê ax$ xb Ä 0 Ê
lim f ax$ xb œ lim c f(y) œ B where y œ x$ x.
yÄ!
x Ä !b
(b) Since x Ä 0 , 1 x x$ 0 Ê ax$ xb Ä 0 Ê
(c) Since x Ä 0 , 0 x% x# 1 Ê ax# x% b Ä 0 Ê
lim f ax$ xb œ lim b f(y) œ A where y œ x$ x.
yÄ!
x Ä !c
lim f ax# x% b œ lim b f(y) œ A where y œ x# x% .
yÄ!
x Ä !b
(d) Since x Ä 0 , 1 x 0 Ê ! x% x# 1 Ê ax# x% b Ä 0 Ê
lim f ax# x% b œ A as in part (c).
x Ä !b
14. (a) True, because if xlim
(f(x) g(x)) exists then xlim
(f(x) g(x)) xlim
f(x) œ xlim
[(f(x) g(x)) f(x)]
Äa
Äa
Äa
Äa
œ xlim
g(x) exists, contrary to assumption.
Äa
(b) False; for example take f(x) œ
"
x
and g(x) œ x" . Then neither lim f(x) nor lim g(x) exists, but
lim (f(x) g(x)) œ lim ˆ "x x" ‰ œ lim 0 œ 0 exists.
xÄ!
xÄ!
xÄ!
xÄ!
xÄ!
(c) True, because g(x) œ kxk is continuous Ê g(f(x)) œ kf(x)k is continuous (it is the composite of continuous
functions).
1, x Ÿ 0
Ê f(x) is discontinuous at x œ 0. However kf(x)k œ 1 is
(d) False; for example let f(x) œ œ
1, x 0
continuous at x œ 0.
15. Show lim f(x) œ lim
x Ä 1
x# "
x Ä 1 x 1
œ lim
x Ä 1
(x 1)(x ")
(x 1)
Define the continuous extension of f(x) as F(x) œ œ
œ #, x Á 1.
x# 1
x1 ,
2
x Á "
. We now prove the limit of f(x) as x Ä 1
, x œ 1
exists and has the correct value.
#
Step 1: ¹ xx 1" (#)¹ % Ê % (x 1)(x ")
(x 1)
# % Ê % (x 1) # %, x Á " Ê % " x % ".
Step 2: kx (1)k $ Ê $ x 1 $ Ê $ " x $ ".
Then $ " œ % " Ê $ œ %, or $ " œ % " Ê $ œ %. Choose $ œ %. Then ! kx (1)k $
#
Ê ¹ xx 1" a#b¹ % Ê
lim F(x) œ 2. Since the conditions of the continuity test are met by F(x), then f(x) has a
x Ä 1
continuous extension to F(x) at x œ 1.
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Chapter 2 Additional and Advanced Exercises 117
16. Show lim g(x) œ lim
xÄ$
xÄ$
x# 2x 3
2x 6
œ lim
xÄ$
(x 3)(x ")
2(x 3)
œ #, x Á 3.
#
x 2x 3
2x 6 ,
Define the continuous extension of g(x) as G(x) œ œ
xÁ3
. We now prove the limit of g(x) as
, xœ3
2
x Ä 3 exists and has the correct value.
Step 1: ¹ x
#
2x 3
#x 6
2¹ % Ê % (x 3)(x ")
2(x 3)
# % Ê % x"
#
# % , x Á $ Ê $ #% x $ #% .
Step 2: kx 3k $ Ê $ x 3 $ Ê $ $ x $ $.
Then, $ $ œ $ #% Ê $ œ #%, or $ $ œ $ #% Ê $ œ #%. Choose $ œ #%. Then ! kx 3k $
Ê ¹x
#
2x 3
2x 6
2¹ % Ê lim
xÄ$
(x 3)(x ")
#(x 3)
œ 2. Since the conditions of the continuity test hold for G(x),
g(x) can be continuously extended to G(x) at B œ 3.
17. (a) Let % ! be given. If x is rational, then f(x) œ x Ê kf(x) 0k œ kx 0k % Í kx 0k %; i.e., choose
$ œ %. Then kx 0k $ Ê kf(x) 0k % for x rational. If x is irrational, then f(x) œ 0 Ê kf(x) 0k %
Í ! % which is true no matter how close irrational x is to 0, so again we can choose $ œ %. In either case,
given % ! there is a $ œ % ! such that ! kx 0k $ Ê kf(x) 0k %. Therefore, f is continuous at
x œ 0.
(b) Choose x œ c !. Then within any interval (c $ ß c $ ) there are both rational and irrational numbers.
If c is rational, pick % œ #c . No matter how small we choose $ ! there is an irrational number x in
(c $ ß c $ ) Ê kf(x) f(c)k œ k0 ck œ c c
#
œ %. That is, f is not continuous at any rational c 0. On
the other hand, suppose c is irrational Ê f(c) œ 0. Again pick % œ #c . No matter how small we choose $ !
there is a rational number x in (c $ ß c $ ) with kx ck œ kxk c
#
œ% Í
œ % Ê f is not continuous at any irrational c 0.
If x œ c 0, repeat the argument picking % œ
nonzero value x œ c.
18. (a) Let c œ
c
#
kc k
#
œ
c
# .
x
c
#
Then kf(x) f(c)k œ kx 0k
3c
#.
Therefore f fails to be continuous at any
m
n
be a rational number in [0ß 1] reduced to lowest terms Ê f(c) œ "n . Pick % œ
"
#n
œ %. Therefore f is discontinuous at x œ c, a rational number.
"
#n .
No matter how
small $ ! is taken, there is an irrational number x in the interval (c $ ß c $ ) Ê kf(x) f(c)k œ ¸0 "n ¸
œ
"
n
(b) Now suppose c is an irrational number Ê f(c) œ 0. Let % 0 be given. Notice that
number reduced to lowest terms with denominator 2 and belonging to [0ß 1];
denominator 3 belonging to [0ß 1];
"
4
and
[0ß 1]; etc. In general, choose N so that
"
N
3
4
with denominator 4 in [0ß 1];
"
3
and
" 2 3
5, 5, 5
2
3
and
"
#
is the only rational
the only rationals with
4
5
with denominator 5 in
% Ê there exist only finitely many rationals in [!ß "] having
denominator Ÿ N, say r" , r# , á , rp . Let $ œ min ekc ri k : i œ 1ß á ß pf . Then the interval (c $ ß c $ )
contains no rational numbers with denominator Ÿ N. Thus, 0 kx ck $ Ê kf(x) f(c)k œ kf(x) 0k
œ kf(x)k Ÿ N" % Ê f is continuous at x œ c irrational.
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118
Chapter 2 Limits and Continuity
(c) The graph looks like the markings on a typical ruler
when the points (xß f(x)) on the graph of f(x) are
connected to the x-axis with vertical lines.
19. Yes. Let R be the radius of the equator (earth) and suppose at a fixed instant of time we label noon as the
zero point, 0, on the equator Ê 0 1R represents the midnight point (at the same exact time). Suppose x"
is a point on the equator “just after" noon Ê x" 1R is simultaneously “just after" midnight. It seems
reasonable that the temperature T at a point just after noon is hotter than it would be at the diametrically
opposite point just after midnight: That is, T(x" ) T(x" 1R) 0. At exactly the same moment in time
pick x# to be a point just before midnight Ê x# 1R is just before noon. Then T(x# ) T(x# 1R) 0.
Assuming the temperature function T is continuous along the equator (which is reasonable), the Intermediate
Value Theorem says there is a point c between 0 (noon) and 1R (simultaneously midnight) such that
T(c) T(c 1R) œ 0; i.e., there is always a pair of antipodal points on the earth's equator where the
temperatures are the same.
#
#
#
#
"
20. xlim
f(x)g(x) œ xlim
af(x) g(x)b‹ Šxlim
af(x) g(x)b‹ “
’af(x) g(x)b af(x) g(x)b “ œ "% ’Šxlim
Äc
Äc %
Äc
Äc
œ "% ˆ$# a"b# ‰ œ #.
21. (a) At x œ 0: lim r (a) œ lim
aÄ!
œ lim
1 (" a)
aÄ!
a Ä ! a ˆ" È1 a‰
At x œ 1:
(b) At x œ 0:
lim
a Ä "b
œ
r (a) œ
" È1 a
a
1
" È1 0
œ lim c
aÄ!
1 (" a)
a ˆ" È1 a‰
œ
"
#
aÄ!
1 (1 a)
lim
a Ä "b a ˆ1 È1 a‰
lim r (a) œ lim c
aÄ!
a Ä !c
È1 a
œ lim Š " a
" È1 a
a
œ lim c
aÄ!
" È1 a
‹ Š " È1 a ‹
a
œ lim
a Ä 1 a ˆ" È1 a‰
È1 a
œ lim c Š " a
aÄ!
a
a ˆ 1 È 1 a ‰
œ lim c
aÄ!
œ
"
" È0
œ1
" È1 a
‹ Š " È1 a ‹
"
œ _ (because the
" È1 a
"
œ _ (because the
" È1 a
denominator is always negative); lim b r (a) œ lim b
aÄ!
aÄ!
is always positive). Therefore, lim r (a) does not exist.
aÄ!
At x œ 1:
lim r (a) œ lim b
a Ä "b
a Ä "
1 È 1 a
a
œ
lim
"
a Ä 1b " È1 a
œ1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
denominator
Chapter 2 Additional and Advanced Exercises 119
(c)
(d)
22. f(x) œ x 2 cos x Ê f(0) œ 0 2 cos 0 œ 2 0 and f(1) œ 1 2 cos (1) œ 1 # 0. Since f(x) is
continuous on [1ß !], by the Intermediate Value Theorem, f(x) must take on every value between [1 #ß #].
Thus there is some number c in [1ß !] such that f(c) œ 0; i.e., c is a solution to x 2 cos x œ 0.
23. (a) The function f is bounded on D if f(x) M and f(x) Ÿ N for all x in D. This means M Ÿ f(x) Ÿ N for all x
in D. Choose B to be max ekMk ß kNkf . Then kf(x)k Ÿ B. On the other hand, if kf(x)k Ÿ B, then
B Ÿ f(x) Ÿ B Ê f(x) B and f(x) Ÿ B Ê f(x) is bounded on D with N œ B an upper bound and
M œ B a lower bound.
(b) Assume f(x) Ÿ N for all x and that L N. Let % œ L # N . Since x lim
f(x) œ L there is a $ ! such that
Äx
!
0 kx x! k $ Ê kf(x) Lk % Í L % f(x) L % Í L Í
LN
#
f(x) 3L N
# .
But L N Ê
LN
#
Ê L
f(x) L b, then a b
24. (a) If a
ML
#
Í
3L M
#
f(x)
ML
# . As in part (b), 0 kx L
M
M, a contradiction.
#
0 Ê ka bk œ a b Ê max (aß b) œ
ab
#
ka b k
#
If a Ÿ b, then a b Ÿ 0 Ê ka bk œ (a b) œ b a Ê max (aß b) œ
œ
2b
#
f(x) L x! k $
ab
ab
2a
# # œ # œ a.
ka b k
ab
œ a # b b # a
# #
œ
œ b.
(b) Let min (aß b) œ
ab
#
ka b k
#
LN
#
N Ê N f(x) contrary to the boundedness assumption
f(x) Ÿ N. This contradiction proves L Ÿ N.
(c) Assume M Ÿ f(x) for all x and that L M. Let % œ
ML
#
LN
#
.
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120
Chapter 2 Limits and Continuity
25. lim œ
xÄ0
sina" cos xb
x
œ lim
œ lim
xÄ0
†
sin x
xÄ0 x
26.
lim
sin x
x Ä 0b sin Èx
œ
sina" cos xb
" cos x
sin x
" cos x
sin x
lim
x Ä 0b B
†
†
" cos x
x
Èx
sin Èx
†
†
œ lim
x
Èx
œ lim
sinasin xb
sin x
28. lim
sinax# xb
x
œ lim
sinax# xb
x# x
† ax "b œ lim
sinax# %b
x Ä 2 x2
œ lim
sinax# %b
#
x Ä 2 x %
† ax 2b œ lim
xÄ0
29. lim
xÄ0
sinˆÈx $‰
x9
xÄ9
30. lim
sin x
x
xÄ0
sinasin xb
sin x
†
sina" cos xb
" cos x
† lim
" cos# x
x Ä 0 xa" cos xb
"
Èx $
† lim
sin x
xÄ0 x
œ " † lim
œ " † " œ ".
sinax# xb
x# x
† lim ax "b œ " † " œ "
sinax# %b
#
x Ä 2 x %
† lim ax 2b œ " † % œ %
xÄ0
sinˆÈx $‰
x Ä 9 Èx $
œ lim
xÄ0
œ " † lim b sin"Èx † lim b Èx œ " † ! † ! œ !.
x Ä 0 Š Èx ‹ x Ä 0
sinasin xb
x
xÄ0
œ lim
œ " † ˆ #! ‰ œ !.
27. lim
xÄ0
" cos x
" cos x
†
xÄ0
xÄ2
sinˆÈx $‰
x Ä 9 Èx $
œ lim
† lim
"
x Ä 9 Èx $
œ"†
"
'
œ
"
'
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
sin# x
x Ä 0 xa" cos xb
CHAPTER 3 DIFFERENTIATION
3.1 THE DERIVATIVE OF A FUNCTION
1. Step 1: f(x) œ 4 x# and f(x h) œ 4 (x h)#
f(x h) f(x)
h
Step 2:
œ
c4 (x h)# d a4 x# b
h
œ
a4 x# 2xh h# b 4 x#
h
œ
2xh h#
h
œ
h(2x h)
h
œ 2x h
Step 3: f w (x) œ lim (2x h) œ 2x; f w ($) œ 6, f w (0) œ 0, f w (1) œ 2
hÄ!
c(x h 1)# 1d c(x 1)# 1d
h
hÄ!
2xh h# 2h
lim
œ lim (2x h 2)
h
hÄ!
hÄ!
2. F(x) œ (x 1)# 1 and F(x h) œ (x h 1)# " Ê Fw (x) œ lim
ax# 2xh h# 2x 2h 1 1b ax# 2x 1 1b
h
œ lim
hÄ!
œ
œ 2(x 1); Fw (1) œ 4, Fw (0) œ 2, Fw (2) œ 2
3. Step 1: g(t) œ
"
t#
and g(t h) œ
"
"
# #
g(t h) g(t)
œ (t h)h t
h
2t h)
2t h
œ h(
(t h)# t# h œ (t h)# t#
Step 2:
2t h
Step 3: gw (t) œ lim
1 z
#z
4. k(z) œ
and k(z h) œ
Œ
œ
œ
# #
h Ä ! (t h) t
"
(t h)#
h
2t
t# †t#
1 (z h)
2(z h)
œ
(" z)(z h)
lim (1 z h)z
#(z h)zh
hÄ!
œ
"
2z#
œ
t# (t h)#
(t h)# †t# œ
2
t$
t# at# 2th h# b
(t h)# †t# †h
œ
œ
2th h#
(t h)# t# h
2
; gw (1) œ 2, gw (2) œ "4 , gw ŠÈ3‹ œ 3È
3
Ê kw (z) œ lim
Š
" (z h) " z ‹
#(z h)
#z
h
hÄ!
#
z h z# zh
lim z z zh
2(z h)zh
hÄ!
h
œ lim
h Ä ! 2(z h)zh
œ lim
"
h Ä ! #(z h)z
; kw (") œ "# , kw (1) œ "# , kw ŠÈ2‹ œ "4
5. Step 1: p()) œ È3) and p() h) œ È3() h)
Step 2:
p() h) p())
h
œ
œ
È3() h) È3)
h
3h
h ŠÈ3) 3h È3)‹
Step 3: pw ()) œ lim
œ
ŠÈ3) 3h È3)‹
œ
3
È3) 3h È3)
3
œ
h Ä ! È3) 3h È3)
†
h
3
È 3) È 3)
œ
3
2È 3 )
ŠÈ3) 3h È3)‹
ŠÈ3) 3h È3)‹
; pw (1) œ
6. r(s) œ È2s 1 and r(s h) œ È2(s h) 1 Ê rw (s) œ lim
hÄ!
œ lim
hÄ!
œ lim
ŠÈ2s h 1 È2s 1‹
h
†
2h
h Ä ! h ŠÈ2s 2h 1 È2s 1‹
œ
"
È2s 1
; rw (0) œ 1, rw (1) œ
ŠÈ2s 2h 1 È2s 1‹
ŠÈ2s 2h 1 È2s 1‹
œ lim
"
È3
6x# h 6xh# 2h$
h
hÄ!
3
#È2
È2s 2h 1 È2s 1
h
œ
2
È2s 1 È2s 1
œ
2
2È2s 1
"
È2
dy
dx
h a6x# 6xh 2h# b
h
hÄ!
œ lim
, pw (3) œ "# , pw ˆ 32 ‰ œ
h Ä ! h ŠÈ2s 2h 1 È2s 1‹
2
7. y œ f(x) œ 2x$ and f(x h) œ 2(x h)$ Ê
œ lim
(3) 3h) 3)
h ŠÈ3) 3h È3)‹
(2s 2h 1) (2s 1)
œ lim
h Ä ! È2s 2h 1 È2s 1
, rw ˆ #" ‰ œ
3
2È 3
œ
2(x h)$ 2x$
h
hÄ!
œ lim
2 ax$ 3x# h 3xh# h$ b 2x$
h
hÄ!
œ lim
œ lim a6x# 6xh 2h# b œ 6x#
hÄ!
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122
Chapter 3 Differentiation
8. r œ
œ
s$
#
1 Ê
”
œ lim
dr
ds
(s h)$
#
t
2t1
Š
œ lim
(t b h)(2t b 1) c t(2t b 2h b 1)
‹
(2t b 2h b 1)(2t b 1)
h
œ
2t# t 2ht h 2t# 2ht t
(2t 2h 1)(2t 1)h
hÄ!
"
"
(2t 1)(2t 1) œ (2t 1)#
dv
dt
œ lim
œ
"
th“ ˆt "t ‰
#
#
th
lim ht h(th h)t
hÄ!
"
Èq 1
11. p œ f(q) œ
œ
œ lim
h
" "
t
h
th
œ
t# 1
t#
"
È(q h) 1
Ê
h Ä ! (2t 2h 1)(2t 1)
Š
h(t h)t t (t h)
‹
(t h)t
h
hÄ!
œ1
"
t#
Š È(q "h) 1 ‹ Š Èq" 1 ‹
œ lim
dp
dq
"
œ lim
œ lim
h
hÄ!
h
hÄ!
Èq 1 Èq h 1
œ lim
h Ä ! hÈ q h 1 È q 1
h
hÄ!
3 #
# s
h
hÄ!
h Ä ! (2t 2h 1)(2t 1)h
and f(q h) œ
3sh h# b œ
t ‰
Š 2(t bt bh)hb 1 ‹ ˆ 2t b
1
œ lim
ds
dt
œ lim
#
1
lim t (thth)t
hÄ!
Èq b 1 c Èq b h b 1
Œ Èq b h b 1 Èq b 1 œ
(t h)(2t 1) t(2t 2h 1)
(2t 2h 1)(2t 1)h
hÄ!
œ lim
h
hÄ!
Ê
œ lim
œ lim
’(t h) œ
th
2(th)1
and r(t h) œ
hÄ!
c(s h)$ 2d cs$ 2d
"
h
# hlim
Ä!
h c3s# 3sh h# d
"
"
œ # lim a3s#
# hlim
h
Ä!
hÄ!
h
hÄ!
"
s$ 3s# h 3sh# h$ 2 s$ 2
# hlim
h
Ä!
9. s œ r(t) œ
10.
$
1• ’ s# 1“
œ
ˆÈ q 1 È q h 1 ‰ ˆ È q 1 È q h 1 ‰
1) (q h 1)
† ˆÈq 1 Èq h 1‰ œ lim hÈq h 1(qÈq 1 ˆÈ q 1 È q h 1 ‰
h Ä ! h Èq h 1 Èq 1
hÄ!
h
"
lim
œ lim Èq h 1 Èq 1 ˆÈq 1 Èq h 1‰
h Ä ! h È q h 1 È q 1 ˆÈ q 1 È q h 1 ‰
hÄ!
"
"
œ
È q 1 È q 1 ˆÈ q 1 È q 1 ‰
2(q 1) Èq 1
dz
dw
œ lim
œ lim
œ
12.
"
Š È3(w h) 2
h
hÄ!
"
È3w 2 ‹
ŠÈ3w 2 È3w 3h 2‹
œ lim
hÈ3w 3h 2 È3w 2
hÄ!
È3w 2 È3w 3h 2
œ lim
h Ä ! hÈ3w 3h 2 È3w 2
†
ŠÈ3w2È3w3h2‹
ŠÈ3w 2 È3w 3h 2‹
œ lim
(3w 2) (3w 3h 2)
œ lim
3
h Ä ! hÈ3w 3h 2 È3w 2 ŠÈ3w 2 È3w 3h 2‹
h Ä ! È3w 3h 2 È3w 2 ŠÈ3w 2 È3w 3h 2‹
œ
9
x
and f(x h) œ (x h) œ
x(x h)# 9x x# (x h) 9(x h)
x(x h)h
œ
h(x# xh 9)
x(x h)h
14. k(x) œ
"
#x
œ lim
hÄ!
œ lim
(# x) (2 x h)
h(2 x)(2 x h)
hÄ!
œ lim
hÄ!
œ
x# xh 9
x(x h)
œ
w
œ
9
(x h)
Ê
f(x h) f(x)
h
œ
’(x h) x$ 2x# h xh# 9x x$ x# h 9x 9h
x(x h)h
; f (x) œ
and k(x h) œ
"
kw (2) œ 16
ds
dt
3
È3w 2 È3w 2 ŠÈ3w 2 È3w 2‹
3
2(3w 2) È3w 2
13. f(x) œ x 15.
œ
#
lim x xh 9
h Ä ! x(x h)
œ
x# 9
x#
9
9
(x b h) “ ’x x “
h
œ
œ1
x# h xh# 9h
x(x h)h
9
x#
; m œ f w (3) œ 0
Š # "x h k(x h) k(x)
œ
lim
h
h
hÄ!
hÄ!
h
"
"
lim
œ lim (2 x)(# x h) œ (2 x)# ;
h Ä ! h(2 x)(2 x h)
hÄ!
"
2 (x h)
c(t h)$ (t h)# d at$ t# b
h
3t# h 3th# h$ 2th h#
h
Ê kw (x) œ lim
œ lim
hÄ!
œ lim
hÄ!
"
#x‹
at$ 3t# h 3th# h$ b at# 2th h# b t$ t#
h
h a3t# 3th h# 2t hb
h
œ lim a3t# 3th h# 2t hb
hÄ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 3.1 The Derivative of a Function
œ 3t# 2t; m œ
16.
dy
dx
ds ¸
dt tœ"
œ5
(x h 1)$ (x 1)$
h
œ lim
hÄ!
(x 1)$ 3(x ")# h 3(x 1)h# h$ (x 1)$
h
œ lim
hÄ!
œ lim c3(x 1)# 3(x 1)h h# d œ 3(x 1)# ; m œ
hÄ!
17. f(x) œ
œ
8
Èx 2
and f(x h) œ
8 ŠÈx 2 Èx h 2‹
hÈ x h 2 È x 2
†
8
È(x h) 2
f(x h) f(x)
h
Ê
ŠÈx 2 Èx h 2‹
œ
ŠÈx 2 Èx h 2‹
œ
8h
hÈx h 2 Èx 2 ŠÈx 2 Èx h 2‹
œ
8
Èx 2 Èx 2 ŠÈx 2 Èx 2‹
œ
œ3
dy
dx ¹ x=#
È(x b h) c 2 Èx c 2
8
œ
8
h
8[(x 2) (x h 2)]
hÈx h 2 Èx 2 ŠÈx 2 Èx h 2‹
8
Ê f w (x) œ lim
h Ä ! Èx h 2 Èx 2 ŠÈx 2 Èx h 2‹
4
(x 2)Èx 2
; m œ f w (6) œ
4
4È 4
œ "# Ê the equation of the tangent
line at (6ß 4) is y 4 œ "# (x 6) Ê y œ "# x $ % Ê y œ "# x (.
18. gw (z) œ lim
ˆ1 È4 (z h)‰ Š1 È4 z‹
h
hÄ!
œ
h
lim
h Ä ! h ŠÈ4 z h È4 z‹
œ
†
h
hÄ!
(4 z h) (4 z)
lim
h Ä ! h ŠÈ4 z h È4 z‹
"
œ "#
2È 4 3
"# z $# # Ê w
ŠÈ4 z h È4 z‹
œ lim
ŠÈ4 z h È4 z‹
ŠÈ4 z h È4 z‹
"
œ lim
h Ä ! ŠÈ4 z h È4 z‹
œ
"
2È 4 z
m œ gw (3) œ
Ê the equation of the tangent line at ($ß #) is w 2 œ "# (z 3)
Êwœ
œ "# z (# .
19. s œ f(t) œ 1 3t# and f(t h) œ 1 3(t h)# œ 1 3t# 6th 3h# Ê
a1 3t# 6th 3h# b a1 3t# b
h
hÄ!
œ lim
20. y œ f(x) œ " œ lim
hÄ!
h
h Ä ! x(x h)h
hÄ!
2È 4 ) 2È 4 ) h
hÈ 4 ) È 4 ) h
œ
œ
"
xh
Ê
"
lim
h Ä ! x(x h)
2
È4 () h)
4(% )) 4(% ) h)
œ lim
h Ä ! 2hÈ4 ) È4 ) h ŠÈ4 ) È4 ) h‹
œ
2
(4 )) Š2È4 )‹
œ
dy
dx
œ
œ lim
"
x#
"
3
Ê
"
(4 ))È4 )
Ê
Šz h Èz h‹ ˆz Èz‰
hÄ!
h
œ 1 lim
(z h) z
h Ä ! h ŠÈz h Èz‹
hÄ!
Š1 dr ¸
d) )œ!
Ê
dr
d)
È œ
È4 c ) c h È4 c )
2
2
h
2
h Ä ! È4 ) È4 ) h ŠÈ4 ) È% ) h‹
œ
"
8
h Èz h Èz
h
hÄ!
œ lim
œ 1 lim
"
h
œ lim
œ lim
"
x h ‹ Š1 x ‹
hÄ!
dy
dx ¹x= 3
f(t h) f(t)
h
œ6
f(x h) f(x)
h
hÄ!
œ lim
22. w œ f(z) œ z Èz and f(z h) œ (z h) Èz h Ê
œ lim
ds ¸
dt t="
œ lim
f() h) f())
œ lim
h
hÄ!
hÄ!
È
È
2È4 ) #È% ) h Š2 % ) 2 4 ) h‹
lim
† È
Š2 4 ) #È4 ) h‹
h Ä ! hÈ 4 ) È 4 ) h
and f() h) œ
2
È4 )
hÄ!
and f(x h) œ 1 " "
x
xh
œ lim
h
21. r œ f()) œ
œ lim
"
x
œ lim (6t 3h) œ 6t Ê
ds
dt
;
"
h Ä ! Èz h Èz
dw
dz
œ lim
hÄ!
œ lim –1 hÄ!
œ"
"
2È z
Ê
f(z h) f(z)
h
Èz h Èz
dw ¸
dz zœ4
h
œ
†
ŠÈz h Èz‹
ŠÈz h Èz‹ —
5
4
" "
fazb faxb
a x #b a z # b
xz
"
z#
x#
23. f w axb œ zlim
œ zlim
œ zlim
œ zlim
œ zlim
Äx zx
Ä x zx
Ä x az xbaz #bax #b
Ä x az xbaz #bax #b
Ä x az #bax #b
œ ax "
#b#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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123
124
Chapter 3 Differentiation
"
"
#
#
#
#
Òax "b az "bÓÒax "b az "bÓ
ax "b az"b
fazb faxb
az"b
ax"b
24. f w axb œ zlim
œ zlim
œ zlim
œ zlim
zx
Ä x zx
Äx
Ä x az xbaz "b# ax "b#
Äx
az xbaz "b# ax "b#
ax zbax z 2b
"ax z 2b
œ zlim
œ zlim
œ
Ä x az xbaz "b# ax "b#
Ä x a z " b # a x "b#
z
"a#x #b
a x "b %
œ
#ax "b
a x "b %
œ
#
a x "b $
x
gazb gaxb
z a x "b x a z " b
z x
"
zc"
x"
25. gw axb œ zlim
œ zlim
œ zlim
œ zlim
œ zlim
Äx zx
Äx zx
Ä x az xbaz "bax "b
Ä x az xbaz "bax "b
Ä x az "bax "b
œ ax "
"b#
gazb gaxb
26. gw axb œ zlim
œ zlim
Äx zx
Äx
"
"
œ zlim
œ
#È x
Ä x Èz Èx
ˆ" Èz‰ˆ" Èx‰
zx
œ zlim
Äx
Èz Èx
zx
†
Èz Èx
Èz Èx
zx
œ zlim
Ä x az x bˆÈ z È x ‰
27. Note that as x increases, the slope of the tangent line to the curve is first negative, then zero (when x œ 0),
then positive Ê the slope is always increasing which matches (b).
28. Note that the slope of the tangent line is never negative. For x negative, f#w (x) is positive but decreasing as x
increases. When x œ 0, the slope of the tangent line to x is 0. For x 0, f#w (x) is positive and increasing. This
graph matches (a).
29. f$ (x) is an oscillating function like the cosine. Everywhere that the graph of f$ has a horizontal tangent we
expect f$w to be zero, and (d) matches this condition.
30. The graph matches with (c).
31. (a) f w is not defined at x œ 0, 1, 4. At these points, the left-hand and right-hand derivatives do not agree.
For example, lim c
xÄ!
f(x) f(0)
x0
œ slope of line joining (%ß 0) and (!ß #) œ
"
#
but lim b
xÄ!
line joining (0ß 2) and ("ß 2) œ 4. Since these values are not equal, f w (0) œ lim
xÄ!
f(x) f(0)
x0
f(x) f(0)
x0
(b)
32. (a)
(b) Shift the graph in (a) down 3 units
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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œ slope of
does not exist.
Section 3.1 The Derivative of a Function
33.
(b) The fastest is between the 20th and 30th days;
slowest is between the 40th and 50th days.
34. (a)
35. Left-hand derivative: For h 0, f(0 h) œ f(h) œ h# (using y œ x# curve) Ê
œ lim c
hÄ!
h# 0
h
œ lim c h œ 0;
hÄ!
Right-hand derivative: For h 0, f(0 h) œ f(h) œ h (using y œ x curve) Ê
œ lim b
hÄ!
Then lim c
hÄ!
h0
h
lim
h Ä !c
œ lim b 1 œ 1;
hÄ!
f(0 h) f(0)
h
Á lim b
hÄ!
f(0 h) f(0)
h
lim
h Ä !b
œ lim c 0 œ 0;
hÄ!
f(1 h) f(1)
h
lim
h Ä !c
Right-hand derivative: When h !, 1 h 1 Ê f(1 h) œ 2(1 h) œ 2 2h Ê
Then lim c
hÄ!
(2 2h)2
h
œ lim b
hÄ!
f(1 h) f(1)
h
2h
h
È1 h "
h
œ lim c
hÄ!
lim
h Ä !b
ŠÈ1 h "‹
h
†
ŠÈ1 h "‹
ŠÈ1 h 1‹
œ lim c
hÄ!
lim
h Ä !c
Then lim c
hÄ!
(2h 1) "
h
f(1 h) f(1)
h
38. Left-hand derivative:
lim
h Ä !c
Right-hand derivative:
œ lim b
hÄ!
Then lim c
hÄ!
h
h(1 h)
œ lim b 2 œ 2;
hÄ!
lim b
hÄ!
œ lim b
hÄ!
f(1 h) f(1)
h
f(1 h) f(1)
h
Á lim b
hÄ!
f(1 h) f(")
h
(1 h) "
h ŠÈ1 h "‹
œ lim c
hÄ!
Á lim b
hÄ!
"
È1 h 1
lim
h Ä !b
Ê the derivative f w (1) does not exist.
œ lim c
f(1 h) f(")
h
"
1h
f(1 h) f(1)
h
f(" h) f(1)
h
Right-hand derivative: When h 0, 1 h 1 Ê f(1 h) œ 2(1 h) 1 œ 2h 1 Ê
œ lim b
hÄ!
22
h
Ê the derivative f w (1) does not exist.
37. Left-hand derivative: When h 0, 1 h 1 Ê f(1 h) œ È1 h Ê
œ lim c
hÄ!
œ lim c
hÄ!
œ lim b 2 œ 2;
hÄ!
f(1 h) f(1)
h
Á lim b
hÄ!
f(0 h) f(0)
h
Ê the derivative f w (0) does not exist.
36. Left-hand derivative: When h !, 1 h 1 Ê f(1 h) œ 2 Ê
œ lim b
hÄ!
f(0 h) f(0)
h
hÄ!
œ lim b
hÄ!
(1 h) "
h
œ lim c 1 œ 1;
" "‹
Š1
h
h
hÄ!
œ lim b
hÄ!
Š
1 (1 h)
1h ‹
h
œ 1;
f(1 h) f(1)
h
Ê the derivative f w (1) does not exist.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
œ #" ;
f("h)f(1)
h
125
126
Chapter 3 Differentiation
39. (a) The function is differentiable on its domain $ Ÿ x Ÿ 2 (it is smooth)
(b) none
(c) none
40. (a) The function is differentiable on its domain # Ÿ x Ÿ 3 (it is smooth)
(b) none
(c) none
41. (a) The function is differentiable on $ Ÿ x 0 and ! x Ÿ 3
(b) none
(c) The function is neither continuous nor differentiable at x œ 0 since lim c f(x) Á lim b f(x)
xÄ!
xÄ!
42. (a) f is differentiable on # Ÿ x 1, " x 0, 0 x 2, and 2 x Ÿ 3
(b) f is continuous but not differentiable at x œ 1: lim f(x) œ 0 exists but there is a corner at x œ 1 since
x Ä 1
œ 3 and lim b f(" h)h f(1) œ 3 Ê f w (1) does not exist
hÄ!
hÄ!
(c) f is neither continuous nor differentiable at x œ 0 and x œ 2:
at x œ 0, lim c f(x) œ 3 but lim b f(x) œ 0 Ê lim f(x) does not exist;
lim c
f(1 h) f(")
h
xÄ!
xÄ0
xÄ!
at x œ 2, lim f(x) exists but lim f(x) Á f(2)
xÄ#
xÄ#
43. (a) f is differentiable on " Ÿ x 0 and 0 x Ÿ 2
(b) f is continuous but not differentiable at x œ 0: lim f(x) œ 0 exists but there is a cusp at x œ 0, so
f(0 h) f(0)
h
hÄ!
f w (0) œ lim
xÄ!
does not exist
(c) none
44. (a) f is differentiable on $ Ÿ x 2, 2 x 2, and 2 x Ÿ 3
(b) f is continuous but not differentiable at x œ 2 and x œ 2: there are corners at those points
(c) none
45. (a) f w (x) œ lim
hÄ!
f(x h) f(x)
h
œ lim
hÄ!
(x h)# ax# b
h
œ lim
hÄ!
x# 2xh h# x#
h
œ lim (2x h) œ 2x
hÄ!
(b)
(c) yw œ 2x is positive for x 0, yw is zero when x œ 0, yw is negative when x 0
(d) y œ x# is increasing for _ x 0 and decreasing for ! x _; the function is increasing on intervals
where yw 0 and decreasing on intervals where yw 0
f(x h) f(x)
h
hÄ!
46. (a) f w (x) œ lim
œ lim
hÄ!
Š xc"
h h
1 ‹
x
œ lim
hÄ!
x (x h)
x(x h)h
œ lim
"
h Ä ! x(x h)
œ
"
x#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 3.1 The Derivative of a Function
(b)
(c) yw is positive for all x Á 0, yw is never 0, yw is never negative
(d) y œ "x is increasing for _ x 0 and ! x _
w
47. (a) Using the alternate formula for calculating derivatives: f (x) œ
œ
$
$
lim z x
z Ä x 3(z x)
œ
az# zx x# b
lim (z x)3(z
x)
zÄx
œ
#
#
lim z zx3 x
zÄx
f(x)
lim f(z)z x
zÄx
#
w
$
Š z3 œ zlim
Äx
x$
3 ‹
zx
œ x Ê f (x) œ x#
(b)
(c) yw is positive for all x Á 0, and yw œ 0 when x œ 0; yw is never negative
(d) y œ
x$
3
is increasing for all x Á 0 (the graph is horizontal at x œ 0) because y is increasing where yw 0; y is
never decreasing
%
48. (a) Using the alternate
œ zlim
Äx
z% x%
4(z x)
œ
%
z
x
Œ4 4
f(z) f(x)
form for calculating derivatives: f (x) œ zlim
œ
lim
zx
zx
Äx
zÄx
(z x) az$ xz# x# z x$ b
z$ xz# x# z x$
$
w
lim
œ zlim
œ x Ê f (x) œ x$
4(z x)
4
zÄx
Äx
w
(b)
(c) yw is positive for x 0, yw is zero for x œ 0, yw is negative for x 0
(d) y œ
x%
4
is increasing on 0 x _ and decreasing on _ x 0
#
#
(xc) ax xc c b
f(x) f(c)
x c
49. yw œ xlim
œ xlim
œ xlim
œ xlim
ax# xc c# b œ 3c# .
xc
xc
Äc
Ä c xc
Äc
Äc
The slope of the curve y œ x$ at x œ c is yw œ 3c# . Notice that 3c# 0 for all c Ê y œ x$ never has a negative
slope.
$
$
50. Horizontal tangents occur where yw œ 0. Thus, yw œ lim
hÄ!
œ lim
hÄ!
2 ŠÈx h Èx‹
h
†
ŠÈx h Èx‹
ŠÈx h Èx‹
œ lim
2È x h 2È x
h
2((x h) x))
h Ä ! h ŠÈx h Èx‹
œ lim
2
h Ä ! Èx h Èx
œ
"
Èx
.
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127
128
Chapter 3 Differentiation
Then yw œ 0 when
51. yw œ lim
hÄ!
œ lim
hÄ!
"
Èx
œ 0 which is never true Ê the curve has no horizontal tangents.
a2(x h)# 13(x h) 5b a2x# 13x 5b
h
4xh 2h# 13h
h
œ lim
hÄ!
2x# 4xh 2h# 13x 13h 5 2x# 13x 5
h
œ lim (4x 2h 13) œ 4x 13, slope at x. The slope is 1 when 4x 13 œ "
hÄ!
Ê 4x œ 12 Ê x œ 3 Ê y œ 2 † 3# 13 † 3 5 œ 16. Thus the tangent line is y 16 œ (1)(x 3)
Ê y œ x "$ and the point of tangency is (3ß 16).
52. For the curve y œ Èx, we have yw œ lim
ŠÈx h Èx‹
hÄ!
"
œ lim
h Ä ! Èx h Èx
œ
"
#Èx
h
†
ŠÈx h Èx‹
ŠÈx h Èx‹
h Ä ! ŠÈx h Èx‹ h
. Suppose ˆ+ß Èa‰ is the point of tangency of such a line and ("ß !) is the point
on the line where it crosses the x-axis. Then the slope of the line is
"
;
2È a
(x h) x
œ lim
using the derivative formula at x œ a Ê
exist: its point of tangency is ("ß "), its slope is
Èa
a1
œ
"
#È a
œ
Èa 0
a (1)
œ
Èa
a1
which must also equal
"
Ê 2a œ a 1 Ê a œ 1.
#Èa
"
# ; and an equation of the line is
Thus such a line does
y1œ
"
#
(x 1)
Ê y œ "# x "# .
53. No. Derivatives of functions have the intermediate value property. The function f(x) œ ÚxÛ satisfies f(0) œ 0
and f(1) œ 1 but does not take on the value "# anywhere in [!ß "] Ê f does not have the intermediate value
property. Thus f cannot be the derivative of any function on [!ß "] Ê f cannot be the derivative of any function
on (_ß _).
54. The graphs are the same. So we know that
for f(x) œ kxk , we have f w (x) œ
kx k
x
.
55. Yes; the derivative of f is f w so that f w (x! ) exists Ê f w (x! ) exists as well.
56. Yes; the derivative of 3g is 3gw so that gw (7) exists Ê 3gw (7) exists as well.
57. Yes, lim
g(t)
t Ä ! h(t)
can exist but it need not equal zero. For example, let g(t) œ mt and h(t) œ t. Then g(0) œ h(0)
œ 0, but lim
g(t)
t Ä ! h(t)
œ lim
tÄ!
mt
t
œ lim m œ m, which need not be zero.
tÄ!
58. (a) Suppose kf(x)k Ÿ x# for " Ÿ x Ÿ 1. Then kf(0)k Ÿ 0# Ê f(0) œ 0. Then f w (0) œ lim
œ lim
hÄ!
f(h) 0
h
œ lim
hÄ!
f(h)
h .
For khk Ÿ 1, h# Ÿ f(h) Ÿ h# Ê h Ÿ
hÄ!
f(h)
h
f(0 h) f(0)
h
Ÿ h Ê f w (0) œ lim
hÄ!
f(h)
h
œ0
by the Sandwich Theorem for limits.
(b) Note that for x Á 0, kf(x)k œ ¸x# sin "x ¸ œ kx# k ksin xk Ÿ kx# k † 1 œ x# (since " Ÿ sin x Ÿ 1). By part (a),
f is differentiable at x œ 0 and f w (0) œ 0.
59. The graphs are shown below for h œ 1, 0.5, 0.1. The function y œ
y œ Èx so that
"
#È x
œ lim
hÄ!
Èx h Èx
h
"
2È x
. The graphs reveal that y œ
is the derivative of the function
Èx h Èx
h
gets closer to y œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
"
#È x
Section 3.1 The Derivative of a Function
as h gets smaller and smaller.
60. The graphs are shown below for h œ 2, 1, 0.5. The function y œ 3x# is the derivative of the function y œ x$ so
that 3x# œ lim
hÄ!
(xh)$ x$
h
. The graphs reveal that y œ
(xh)$ x$
h
gets closer to y œ 3x# as h
gets smaller and smaller.
61. Weierstrass's nowhere differentiable continuous function.
62-67. Example CAS commands:
Maple:
f := x -> x^3 + x^2 - x;
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129
130
Chapter 3 Differentiation
x0 := 1;
plot( f(x), x=x0-5..x0+2, color=black,
title="Section 3_1, #62(a)" );
q := unapply( (f(x+h)-f(x))/h, (x,h) );
# (b)
L := limit( q(x,h), h=0 );
# (c)
m := eval( L, x=x0 );
tan_line := f(x0) + m*(x-x0);
plot( [f(x),tan_line], x=x0-2..x0+3, color=black,
linestyle=[1,7], title="Section 3.1 #62(d)",
legend=["y=f(x)","Tangent line at x=1"] );
Xvals := sort( [ x0+2^(-k) $ k=0..5, x0-2^(-k) $ k=0..5 ] ):
# (e)
Yvals := map( f, Xvals ):
evalf[4](< convert(Xvals,Matrix) , convert(Yvals,Matrix) >);
plot( L, x=x0-5..x0+3, color=black, title="Section 3.1 #62(f)" );
Mathematica: (functions and x0 may vary) (see section 2.5 re. RealOnly ):
<<Miscellaneous`RealOnly`
Clear[f, m, x, y, h]
x0= 1 /4;
f[x_]:=x2 Cos[x]
Plot[f[x], {x, x0 3, x0 3}]
q[x_, h_]:=(f[x h] f[x])/h
m[x_]:=Limit[q[x, h], h Ä 0]
ytan:=f[x0] m[x0] (x x0)
Plot[{f[x], ytan},{x, x0 3, x0 3}]
m[x0 1]//N
m[x0 1]//N
Plot[{f[x], m[x]},{x, x0 3, x0 3}]
3.2 DIFFERENTIATION RULES
1. y œ x# 3 Ê
dy
dx
œ
2. y œ x# x 8 Ê
dy
dx
3. s œ 5t$ 3t& Ê
œ
ds
dt
ax# b d
dx
4
3
6. y œ
x$
3
x$ x Ê
x#
#
dy
dx
(3) œ 2x 0 œ #B Ê
d
dt
a5t$ b dw
dz
d
dt
d# y
dx#
"
4
œ x# x 7. w œ 3z# z" Ê
dw
dz
œ 6z$ z# œ
x
4
8. s œ 2t" 4t# Ê
ds
dt
Ê
œ 2t# 8t$ œ
9. y œ 6x# 10x 5x# Ê
dy
dx
œ 2
d# s
dt#
œ
d
dt
a15t# b d
dt
a15t% b œ 30t 60t$
œ 126z& 42z 42
œ 8x
dy
dx
d# w
dz#
œ 21z' 21z# 42z Ê
œ 4x# 1 Ê
d# y
dx#
œ#
a3t& b œ 15t# 15t% Ê
Ê
d# y
dx#
œ 2x 1 0 œ 2x 1 Ê
4. w œ 3z( 7z$ 21z# Ê
5. y œ
d
dx
d# y
dx#
6
z$
2
t#
œ 2x 1 0 œ 2x 1
8
t$
"
z#
Ê
Ê
d# w
dz#
d# s
dt#
œ 18z% 2z$ œ
œ 4t$ 24t% œ
œ 12x 10 10x$ œ 12x 10 10
x$
Ê
d# y
dx#
18
z%
4
t$
2
z$
24
t%
œ 12 0 30x% œ 12 Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
30
x%
Section 3.2 Differentiation Rules
10. y œ 4 2x x$ Ê
" #
3 s
11. r œ
œ # 3x% œ # dy
dx
5# s" Ê
œ 32 s$ 5# s# œ
dr
ds
12. r œ 12)" 4)$ )% Ê
œ
24
)$
48
)&
dr
d)
3
x%
2
3s$
Ê
d# y
dx#
5
2s#
Ê
12
x&
œ 0 12x& œ
d# r
ds#
œ 12)# 12)% 4)& œ
œ 2s% 5s$ œ
12
)#
12
)%
4
)&
2
s%
d# r
d) #
Ê
5
s$
œ 24)$ 48)& 20)'
20
)'
13. (a) y œ a3 x# b ax$ x 1b Ê yw œ a3 x# b †
ax$ x 1b ax$ x 1b †
d
dx
d
dx
a3 x# b
œ a3 x# b a3x# 1b ax$ x 1b (2x) œ 5x% 12x# 2x 3
(b) y œ x& 4x$ x# 3x 3 Ê yw œ 5x% 12x# 2x 3
14. (a) y œ (x 1) ax# x 1b Ê yw œ (x 1)(2x 1) ax# x 1b (") œ 3x#
(b) y œ (x 1) ax# x 1b œ x$ 1 Ê yw œ 3x#
15. (a) y œ ax# 1b ˆx 5 "x ‰ Ê yw œ ax# 1b †
d
dx
ˆx 5 "x ‰ ˆx 5 x" ‰ †
d
dx
ax# 1b
œ ax# 1b a1 x# b ax 5 x" b (2x) œ ax# 1 1 x# b a2x# 10x 2b œ 3x# 10x 2 (b) y œ x$ 5x# 2x 5 16. y œ ˆx "x ‰ ˆx w
"
x
"
x
Ê yw œ 3x# 10x 2 "
x#
1‰
"
(a) y œ ax x b † a1 x# b ax x" 1b a1 x# b œ 2x 1 (b) y œ x# x "
x
"
x#
Ê yw œ 2x 1 "
x#
2x 5
3x 2 ; use the quotient rule: u œ 2x 5 and
(3x 2)(2) (2x 5)(3)
4 6x 15
œ 6x (3x
œ (3x192)#
(3x 2)#
#)#
17. y œ
œ
18. z œ
2x 1
x# 1
Ê
dz
dx
œ
ax# 1b (2) (2x 1)(2x)
ax # 1 b #
œ
t# "
t# t 2
œ
at "bat "b
at #bat "b
21. v œ (1 t) a1 t# b
22. w œ
x5
2x 7
23. f(s) œ
24. u œ
Ê ww œ
Ès "
Ès 1
NOTE:
d
ds
5x "
#È x
"
du
dx
œ
1 t
1t#
t"
t2,
2x# 2 4x# 2x
a x # 1 b#
Ê
dv
dt
œ
"
#È s
ˆ È s "‰ Š
œ
2x 7 2x 10
(2x 7)#
"
"
ˆÈ s 1 ‰
#
2x# 2x 2
a x # 1 b#
at #ba"b at "ba"b
at 2 b 2
a1 t# b (") (1 t)(2t)
a1 t# b#
Ès ‹ ˆÈs 1‰ Š #Ès ‹
#
œ
œ
vuw uvw
v#
2 a x # x 1b
ax# 1b#
v œ x 0.5 Ê uw œ 2x and vw œ 1 Ê gw (x) œ
t Á " Ê f w (t) œ
(2x 7)(1) (x 5)(2)
(2x 7)#
Ê f w (s) œ
ˆÈ s ‰ œ
Ê
œ
œ
2
x$
v œ 3x 2 Ê uw œ 2 and vw œ 3 Ê yw œ
x# 4
#
x0.5 ; use the quotient rule: u œ x 4 and
#
#
(x 0.5)(2x) ax# 4b (")
x# 4
x4
œ 2x (xx 0.5)
œ x(x #
(x 0.5)#
0.5)#
20. f(t) œ
2
x$
19. g(x) œ
œ
"
x#
œ
œ
œ
œ
t#t"
at 2 b 2
" t# 2t 2t#
a1 t# b#
œ
œ
"
at 2 b 2
t# 2t "
a1 t# b#
17
(2x 7)#
ˆ È s "‰ ˆ È s 1 ‰
2 È s ˆÈ s 1 ‰
#
œ
"
È s ˆÈ s 1 ‰#
from Example 2 in Section 2.1
ˆ2Èx‰ (5) (5x 1) Š È" ‹
x
4x
œ
5x 1
4x$Î#
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vuw uvw
v#
"
x#
131
132
Chapter 3 Differentiation
25. v œ
1 x 4È x
x
Ê vw œ
x Š1 È x ‹ ˆ1 x 4 È x ‰
2
x#
26. r œ 2 Š È" È)‹ Ê rw œ 2 )
27. y œ
"
ax # 1 b a x # x 1 b
#
È)(0) 1 Š
)
"
È) ‹
#
œ
2È x "
x#
"
#È) "
œ )$Î#
"
)"Î#
; use the quotient rule: u œ 1 and v œ ax# 1b ax# x 1b Ê uw œ 0 and
vw œ ax 1b (2x 1) ax# x 1b (2x) œ 2x$ x# 2x 1 2x$ 2x# 2x œ 4x$ 3x# 1
Ê
28. y œ
œ
vuw uvw
v#
œ
dy
dx
(x 1)(x 2)
(x 1)(x #)
œ
œ
0 1 a4x$ 3x# 1b
ax # 1 b # a x # x 1 b#
x# 3x 2
x# 3x 2
4x$ 3x# 1
ax# 1b# ax# x 1b#
œ
ax# 3x 2b (2x 3) ax# 3x 2b (2x 3)
(x 1)# (x 2)#
Ê yw œ
6 ax # 2 b
(x 1)# (x 2)#
29. y œ
"
#
30. y œ
"
1 #0
31. y œ
x $ 7
x
32. s œ
t# 5t 1
œ 1 5t t"# œ 1 5t"
t#
d# s
'
$
6t% œ "!
dt# œ 10t
t$ t%
Ê
x% 3
#
x# x Ê yw œ 2x$ 3x 1 Ê yww œ 6x# 3 Ê ywww œ 12x Ê yÐ%Ñ œ 12 Ê yÐnÑ œ 0 for all n
34. u œ
Ê
"
#4
x& Ê yw œ
x% Ê yww œ
œ x# 7x" Ê
() " ) a ) # ) 1 b
)$
33. r œ
œ
)$ "
)$
dy
dx
"
6
"
#
x$ Ê ywww œ
"
)$
œ"
t# Ê
x# Ê yÐ%Ñ œ x Ê yÐ&Ñ œ 1 Ê yÐnÑ œ 0 for all n
(
x#
œ 2x 7x# œ #x Ê
œ 1 )$ Ê
dr
d)
"
z#
"Ê
d w
dz#
œ 2z$ 0 œ 2z$ œ
"
3
1
#
%
Ê
38. p œ
Ê
d# p
dq#
œ
"
6
#" q% 5q' œ
q# 3
(q 1)$ (q 1)$
dp
dq
q' q# 3q% 3
12q%
œ
"
6
œ
"
#q %
q# "
1#
q# d# p
dq#
"%
x$
&
t#
Ê
d# r
d) #
z Ê
dw
dz
œ z# 0 1 œ z# 1
œ 12)& œ
(c)
d
dx
d
dx
ˆ vu ‰ œ
ˆ uv ‰ œ
vuw uvw
v#
œ q$ œ
w
uv vu
u#
w
Ê
d
dx
ˆ vu ‰¸
x=0
œ
Ê
d
dx
ˆ uv ‰¸
x=0
œ
"#
)&
œ 1 x$
8
3
"
4
dw
dz
œ 4z$ 0 œ 4z$ Ê
4" q% Ê
dp
dq
œ
"
6
d# w
dz#
œ 12z#
q 6" q$ q& œ
œ
q# 3
2q$ 6q
œ
q# 3
2q aq# 3b
œ
"
#q
œ
"
#
q"
"
q$
39. u(0) œ 5, uw (0) œ 3, v(0) œ 1, vw (0) œ 2
d
d
(a) dx
(uv) œ uvw vuw Ê dx
(uv)¸ x=0 œ u(0)vw (0) v(0)uw (0) œ 5 † 2 (1)(3) œ 13
(b)
#
t$
$
)%
&
q6
q# 3
aq$ 3q# 3q 1b aq$ 3q# 3q 1b
œ "# q# œ #"q# Ê
6
#
z$
"
1#
x
x%
3 z œ z" 36. w œ (z 1)(z 1) az# 1b œ az# 1b az# 1b œ z% 1 Ê
3
37. p œ Š q12q
‹ Š q q$ 1 ‹ œ
œ 2 14x$ œ # œ 0 $)% œ $)% œ
#
$
%
ax # x b a x # x 1 b
œ x(x 1) axx% x "b œ x axx% 1b œ x x% x œ
x%
#
du
%
œ 3x% œ $
Ê ddxu# œ 12x& œ "#
dx œ 0 3x
x&
x%
#
d# y
dx#
œ 0 5t# 2t$ œ 5t# 2t$ œ
ds
dt
35. w œ ˆ 13z3z ‰ (3 z) œ ˆ "3 z" 1‰ (3 z) œ z" œ
6x# 12
(x 1)# (x 2)#
œ
v(0)uw (0) u(0)vw (0)
(v(0))#
u(0)vw (0) v(0)uw (0)
(u(0))#
œ
œ
(")(3) (5)(2)
(1)#
(5)(2) (1)(3)
(5)#
œ 7
œ
7
25
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
"
6
q
"
6q$
"
q&
5
Section 3.2 Differentiation Rules
(d)
d
dx
(7v 2u) œ 7vw 2uw Ê
d
dx
133
(7v 2u)¸ x=0 œ 7vw (0) 2uw (0) œ 7 † 2 2(3) œ 20
40. u(1) œ 2, uw (1) œ 0, v(1) œ 5, vw (1) œ 1
d
(a) dx
(uv)¸ x=1 œ u(1)vw (1) v(1)uw (1) œ 2 † (1) 5 † 0 œ 2
(b)
(c)
(d)
v(1)uw (")u(1)vw (1)
(v(1))#
u(1)vw (")v(1)uw (1)
d ˆ v ‰¸
dx u x=1 œ
(u(1))#
d
w
¸
dx (7v 2u) x=1 œ 7v (1) d
dx
ˆ vu ‰¸
x=1
œ
œ
œ
5†02†(1)
(5)#
2†(1)5†0
(2)#
œ
2
25
œ 12
2uw (1) œ 7 † (1) 2 † 0 œ 7
41. y œ x$ 4x 1. Note that (#ß ") is on the curve: 1 œ 2$ 4(2) 1
(a) Slope of the tangent at (xß y) is yw œ 3x# 4 Ê slope of the tangent at (#ß ") is yw (2) œ 3(2)# 4 œ 8. Thus
the slope of the line perpendicular to the tangent at (#ß ") is "8 Ê the equation of the line perpendicular to
to the tangent line at (#ß ") is y 1 œ "8 (x 2) or y œ x8 54 .
(b) The slope of the curve at x is m œ 3x# 4 and the smallest value for m is 4 when x œ 0 and y œ 1.
(c) We want the slope of the curve to be 8 Ê yw œ 8 Ê 3x# 4 œ 8 Ê 3x# œ 12 Ê x# œ 4 Ê x œ „ 2. When
x œ 2, y œ 1 and the tangent line has equation y 1 œ 8(x 2) or y œ 8x 15; when x œ 2,
y œ (2)$ 4(2) 1 œ 1, and the tangent line has equation y 1 œ 8(x 2) or y œ 8x 17.
42. (a) y œ x$ 3x 2 Ê yw œ 3x# 3. For the tangent to be horizontal, we need m œ yw œ 0 Ê 0 œ 3x# 3
Ê 3x# œ 3 Ê x œ „ 1. When x œ 1, y œ 0 Ê the tangent line has equation y œ 0. The line
perpendicular to this line at ("ß !) is x œ 1. When x œ 1, y œ 4 Ê the tangent line has equation
y œ 4. The line perpendicular to this line at ("ß %) is x œ 1.
(b) The smallest value of yw is 3, and this occurs when x œ 0 and y œ 2. The tangent to the curve at (!ß 2)
has slope 3 Ê the line perpendicular to the tangent at (!ß 2) has slope "3 Ê y 2 œ "3 (x 0) or
yœ
43. y œ
4x
x# 1
"
3
x 2 is an equation of the perpendicular line.
Ê
dy
dx
œ
ax# 1b(4) (4x)(2x)
ax # 1 b #
œ
4x# 4 8x#
a x # 1 b#
œ
4 ax# "b
a x # 1 b#
. When x œ 0, y œ 0 and yw œ
4(0 1)
1
œ %, so the tangent to the curve at (!ß !) is the line y œ 4x. When x œ 1, y œ 2 Ê yw œ 0, so the tangent to the
curve at ("ß 2) is the line y œ 2.
44. y œ
8
x# 4
Ê yw œ
ax# 4b(0) 8(2x)
ax # 4 b #
œ
16x
a x # 4 b#
. When x œ 2, y œ 1 and yw œ
16(2)
a 2 # 4 b#
œ "# , so the tangent
line to the curve at (2ß ") has the equation y 1 œ "# (x 2), or y œ x# 2.
45. y œ ax# bx c passes through (!ß !) Ê 0 œ a(0) b(0) c Ê c œ 0; y œ ax# bx passes through ("ß #)
Ê 2 œ a b; yw œ 2ax b and since the curve is tangent to y œ x at the origin, its slope is 1 at x œ 0
Ê yw œ 1 when x œ 0 Ê 1 œ 2a(0) b Ê b œ 1. Then a b œ 2 Ê a œ 1. In summary a œ b œ 1 and c œ 0 so
the curve is y œ x# x.
46. y œ cx x# passes through ("ß !) Ê 0 œ c(1) 1 Ê c œ 1 Ê the curve is y œ x x# . For this curve,
yw œ 1 2x and x œ 1 Ê yw œ 1. Since y œ x x# and y œ x# ax b have common tangents at x œ 0,
y œ x# ax b must also have slope 1 at x œ 1. Thus yw œ 2x a Ê 1 œ 2 † 1 a Ê a œ 3
Ê y œ x# 3x b. Since this last curve passes through ("ß !), we have 0 œ 1 3 b Ê b œ 2. In summary,
a œ 3, b œ 2 and c œ 1 so the curves are y œ x# 3x 2 and y œ x x# .
47. (a) y œ x$ x Ê yw œ 3x# 1. When x œ 1, y œ 0 and yw œ 2 Ê the tangent line to the curve at ("ß !) is
y œ 2(x 1) or y œ 2x 2.
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134
Chapter 3 Differentiation
(b)
(c)
y œ x$ x
Ê x$ x œ 2x 2 Ê x$ 3x 2 œ (x 2)(x 1)# œ 0 Ê x œ 2 or x œ 1. Since
y œ 2x 2 y œ 2a2b 2 œ 6; the other intersection point is (2ß 6)
48. (a) y œ x$ 6x# 5x Ê yw œ 3x# 12x 5. When x œ 0, y œ 0 and yw œ 5 Ê the tangent line to the curve at
(0ß 0) is y œ 5x.
(b)
(c)
y œ x$ 6x# 5x
$
#
$
#
#
Ê x 6x 5x œ 5x Ê x 6x œ 0 Ê x (x 6) œ 0 Ê x œ 0 or x œ 6.
y œ 5x
Since y œ 5a6b œ $!, the other intersection point is (6ß 30).
49. Paxb œ an xn an" xn" â a# x# a" x a! Ê P w axb œ nan xn" an "ban" xn# â #a# x a"
50. R œ M# ˆ C# M‰
3
œ
C
#
51. Let c be a constant Ê
M# "3 M$ , where C is a constant Ê
dc
dx
œ0 Ê
d
dx
(u † c) œ u †
dc
dx
c†
dR
dM
œ CM M#
œu†0c
du
dx
œc
du
dx
du
dx
. Thus when one of the
functions is a constant, the Product Rule is just the Constant Multiple Rule Ê the Constant Multiple Rule is
a special case of the Product Rule.
52. (a) We use the Quotient rule to derive the Reciprocal Rule (with u œ 1):
œ v"# †
dv
dx
d
dx
ˆ "v ‰ œ
v†0 1† dv
dx
v#
d ˆ"‰
" du
dx v v † dx (Product Rule)
dv
v du
dx u dx
, the Quotient Rule.
v#
œu†
53. (a)
d
dx
(uvw) œ
w
"† dv
dx
v#
.
(b) Now, using the Reciprocal Rule and the Product Rule, we'll derive the Quotient Rule:
œ
œ
d
dx
w
œ u † ˆ v#1 ‰ dv
dx ((uv) † w) œ (uv) dw
dx w †
w
d
dx
(uv) œ uv
œ uvw uv w u vw
d
d
%
(b) dx
au" u# u$ u% b œ dx
aau" u# u$ b u% b œ au" u# u$ b du
dx u%
du$
du#
du" ‰
%
ˆ
œ u" u# u$ du
dx u% u" u# dx u$ u" dx u$ u# dx
" du
v dx
dw
dx
d
dx
(Reciprocal Rule) Ê
w ˆu
dv
dx
v
a u" u# u$ b Ê
d
dx
du ‰
dx
d
dx
œ uv
d
dx
ˆ vu ‰ œ
dw
dx
d ˆ
"‰
dx u † v
du
u dv
dx v dx
v#
ˆ uv ‰ œ
wu
dv
dx
wv
a u" u# u$ u% b
(using (a) above)
du$
du#
du"
%
Ê
au" u# u$ u% b œ u" u# u$ du
dx u" u# u% dx u" u$ u% dx u# u$ u% dx
œ u" u# u$ u%w u" u# u$w u% u" u#w u$ u% u"w u# u$ u%
d
Generalizing (a) and (b) above, dx
au" âun b œ u" u# âun" unw u" u# âun# unw " un
d
dx
(c)
á u"w u# âun
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
du
dx
Section 3.3 The Derivative as a Rate of Change
135
54. In this problem we don't know the Power Rule works with fractional powers so we can't use it. Remember
d ˆÈ ‰
"
x œ #È
(from Example 2 in Section 2.1)
dx
x
"
#È x
Èx
#
3È x
#
(a)
d
dx
ˆx$Î# ‰ œ
d
dx
ˆx † x"Î# ‰ œ x †
(b)
d
dx
ˆx&Î# ‰ œ
d
dx
ˆx# † x"Î# ‰ œ x#
d
dx
ˆÈx‰ Èx
d
dx
"
ax# b œ x# † Š #È
‹ Èx † 2x œ
x
(c)
d
dx
ˆx(Î# ‰ œ
d
dx
ˆx$ † x"Î# ‰ œ x$
d
dx
ˆÈx‰ Èx
d
dx
"
ax$ b œ x$ † Š #È
‹ Èx † 3x# œ
x
"
#
x&Î# 3x&Î# œ
d
dx
ˆx(Î# ‰ œ
d
dx
axnÎ# b œ
(d) We have
d
dx
ˆx$Î# ‰ œ
3
#
x"Î# ,
d
dx
d
dx
ˆÈx‰ Èx
ˆx&Î# ‰ œ
whenever n is an odd positive integer
55. p œ
Ê
5
#
d
dx
x$Î# ,
(x) œ x †
km
q
cm hq
#
Èx œ
x&Î# so it appears that
"
#
œ
3
#
x"Î#
x$Î# 2x$Î# œ
n
#
5
#
x$Î#
7
#
x&Î#
xÐnÎ#Ñ"
3.
nRT
an#
Vnb V# . We are holding T constant, and a, b, n, R are
#
#
(V nb)†0 (nRT)(1)
dP
2an#
V (0) aVa#anb# b (2V) œ (VnRT
dV œ
(Vnb)#
nb)# V$
56. Aaqb œ
7
#
Èx † 1 œ
œ akmbq" cm ˆ h# ‰q Ê
dA
dq
also constant so their derivatives are zero
œ akmbq# ˆ #h ‰ œ km
q# h
#
Ê
d# A
dt#
œ #akmbq$ œ
#km
q$
3.3 THE DERIVATIVE AS A RATE OF CHANGE
1. s œ t# $t #, 0 Ÿ t Ÿ #
(a) displacement œ ?s œ s(#) s(0) œ !m #m œ # m, vav œ
(b) v œ
aœ
ds
dt œ #t
d# s
dt# œ #
?s
?t
œ
Ê a(0) œ # m/sec# and a(#) œ # m/sec#
changes direction at t œ
$
#.
2. s œ 't t# , ! Ÿ t Ÿ '
(a) displacement œ ?s œ s(') s(0) œ ! m, vav œ
aœ
œ " m/sec
$ Ê kv(0)k œ l$l œ $ m/sec and kv(#)k œ 1 m/sec;
(c) v œ 0 Ê #t $ œ 0 Ê t œ $# . v is negative in the interval ! t (b) v œ
#
#
ds
dt œ ' d# s
dt# œ #
?s
?t
œ
!
'
$
#
and v is positive when
$
#
t # Ê the body
œ ! m/sec
#> Ê kv(0)k œ l 'l œ ' m/sec and kv(')k œ l'l œ ' m/sec;
Ê a(0) œ # m/sec# and a(') œ # m/sec#
(c) v œ 0 Ê ' #t œ 0 Ê t œ $. v is positive in the interval ! t $ and v is negative when $ t ' Ê the body
changes direction at t œ $.
3. s œ t$ 3t# 3t, 0 Ÿ t Ÿ 3
(a) displacement œ ?s œ s(3) s(0) œ 9 m, vav œ
(b) v œ
ds
dt
?s
?t
œ
9
3
œ 3 m/sec
œ 3t# 6t 3 Ê kv(0)k œ k3k œ 3 m/sec and kv(3)k œ k12k œ 12 m/sec; a œ
#
#
d# s
dt#
œ 6t 6
Ê a(0) œ 6 m/sec and a(3) œ 12 m/sec
(c) v œ 0 Ê 3t# 6t 3 œ 0 Ê t# 2t 1 œ 0 Ê (t 1)# œ 0 Ê t œ 1. For all other values of t in the
interval the velocity v is negative (the graph of v œ 3t# 6t 3 is a parabola with vertex at t œ 1 which
opens downward Ê the body never changes direction).
4. s œ
t%
4
t$ t# , 0 Ÿ t Ÿ $
(a) ?s œ s($) s(0) œ
*
%
m, vav œ
?s
?t
œ
*
%
$
œ
$
%
m/sec
(b) v œ t$ 3t# 2t Ê kv(0)k œ 0 m/sec and kv($)k œ ' m/sec; a œ 3t# 6t 2 Ê a(0) œ 2 m/sec# and
a($) œ "" m/sec#
(c) v œ 0 Ê t$ 3t# 2t œ 0 Ê t(t 2)(t 1) œ 0 Ê t œ 0, 1, 2 Ê v œ t(t 2)(t 1) is positive in the
interval for 0 t 1 and v is negative for 1 t 2 and v is positive for # t $ Ê the body changes direction at
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136
Chapter 3 Differentiation
t œ 1 and at t œ #.
5. s œ
25
t#
5t , 1 Ÿ t Ÿ 5
(a) ?s œ s(5) s(1) œ 20 m, vav œ
(b) v œ
50
t$
5
t#
m/sec#
(c) v œ 0 Ê
505t
t$
6. s œ
25
t5
œ 5 m/sec
Ê kv(1)k œ 45 m/sec and kv(5)k œ
4
25
a(5) œ
20
4
"
5
m/sec; a œ
150
t%
10
t$
Ê a(1) œ 140 m/sec# and
œ 0 Ê 50 5t œ 0 Ê t œ 10 Ê the body does not change direction in the interval
, % Ÿ t Ÿ 0
(a) ?s œ s(0) s(4) œ 20 m, vav œ 20
4 œ 5 m/sec
(b) v œ
a(0)
(c) v œ
25
(t5)# Ê kv(4)k œ
œ 25 m/sec#
0 Ê (t25
5)# œ 0 Ê
25 m/sec and kv(0)k œ " m/sec; a œ
50
(t5)$
Ê a(4) œ 50 m/sec# and
v is never 0 Ê the body never changes direction
7. s œ t$ 6t# 9t and let the positive direction be to the right on the s-axis.
(a) v œ 3t# 12t 9 so that v œ 0 Ê t# 4t 3 œ (t 3)(t 1) œ 0 Ê t œ 1 or 3; a œ 6t 12 Ê a(1)
œ 6 m/sec# and a(3) œ 6 m/sec# . Thus the body is motionless but being accelerated left when t œ 1, and
motionless but being accelerated right when t œ 3.
(b) a œ 0 Ê 6t 12 œ 0 Ê t œ 2 with speed kv(2)k œ k12 24 9k œ 3 m/sec
(c) The body moves to the right or forward on 0 Ÿ t 1, and to the left or backward on 1 t 2. The
positions are s(0) œ 0, s(1) œ 4 and s(2) œ 2 Ê total distance œ ks(1) s(0)k ks(2) s(1)k œ k4k k2k
œ 6 m.
8. v œ t# 4t 3 Ê a œ 2t 4
(a) v œ 0 Ê t# 4t 3 œ 0 Ê t œ 1 or 3 Ê a(1) œ 2 m/sec# and a(3) œ 2 m/sec#
(b) v 0 Ê (t 3)(t 1) 0 Ê 0 Ÿ t 1 or t 3 and the body is moving forward; v 0 Ê (t 3)(t 1) 0
Ê " t 3 and the body is moving backward
(c) velocity increasing Ê a 0 Ê 2t 4 0 Ê t 2; velocity decreasing Ê a 0 Ê 2t 4 0 Ê ! Ÿ t 2
9. sm œ 1.86t# Ê vm œ 3.72t and solving 3.72t œ 27.8 Ê t ¸ 7.5 sec on Mars; sj œ 11.44t# Ê vj œ 22.88t and
solving 22.88t œ 27.8 Ê t ¸ 1.2 sec on Jupiter.
10. (a) v(t) œ sw (t) œ 24 1.6t m/sec, and a(t) œ vw (t) œ sw w (t) œ 1.6 m/sec#
(b) Solve v(t) œ 0 Ê 24 1.6t œ 0 Ê t œ 15 sec
(c) s(15) œ 24(15) .8(15)# œ 180 m
(d) Solve s(t) œ 90 Ê 24t .8t# œ 90 Ê t œ
30„15È2
#
¸ 4.39 sec going up and 25.6 sec going down
(e) Twice the time it took to reach its highest point or 30 sec
11. s œ 15t "# gs t# Ê v œ 15 gs t so that v œ 0 Ê 15 gs t œ 0 Ê gs œ
15
t
. Therefore gs œ
15
20
œ
3
4
œ 0.75 m/sec#
12. Solving sm œ 832t 2.6t# œ 0 Ê t(832 2.6t) œ 0 Ê t œ 0 or 320 Ê 320 sec on the moon; solving
se œ 832t 16t# œ 0 Ê t(832 16t) œ 0 Ê t œ 0 or 52 Ê 52 sec on the earth. Also, vm œ 832 5.2t œ 0
Ê t œ 160 and sm (160) œ 66,560 ft, the height it reaches above the moon's surface; ve œ 832 32t œ 0
Ê t œ 26 and se (26) œ 10,816 ft, the height it reaches above the earth's surface.
13. (a) s œ 179 16t# Ê v œ 32t Ê speed œ kvk œ 32t ft/sec and a œ 32 ft/sec#
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Section 3.3 The Derivative as a Rate of Change
(b) s œ 0 Ê 179 16t# œ 0 Ê t œ É 179
16 ¸ 3.3 sec
È
É 179
(c) When t œ É 179
16 , v œ 32
16 œ 8 179 ¸ 107.0 ft/sec
14. (a)
lim1 v œ lim1 9.8(sin ))t œ 9.8t so we expect v œ 9.8t m/sec in free fall
)Ä
(b) a œ
#
dv
dt
)Ä
#
œ 9.8 m/sec#
(b) between 3 and 6 seconds: $ Ÿ t Ÿ 6
(d)
15. (a) at 2 and 7 seconds
(c)
16. (a) P is moving to the left when 2 t 3 or 5 t 6; P is moving to the right when 0 t 1; P is standing
still when 1 t 2 or 3 t 5
(b)
17. (a)
(c)
(e)
(f)
190 ft/sec
at 8 sec, 0 ft/sec
From t œ 8 until t œ 10.8 sec, a total of 2.8 sec
Greatest acceleration happens 2 sec after launch
(g) From t œ 2 to t œ 10.8 sec; during this period, a œ
(b) 2 sec
(d) 10.8 sec, 90 ft/sec
v(10.8)v(2)
10.82
¸ 32 ft/sec#
18. (a) Forward: 0 Ÿ t 1 and 5 t 7; Backward: 1 t 5; Speeds up: 1 t 2 and 5 t 6;
Slows down: 0 Ÿ t 1, 3 t 5, and 6 t 7
(b) Positive: 3 t 6; negative: 0 Ÿ t 2 and 6 t 7; zero: 2 t 3 and 7 t 9
(c) t œ 0 and 2 Ÿ t Ÿ 3
(d) 7 Ÿ t Ÿ 9
19. s œ 490t# Ê v œ 980t Ê a œ 980
(a) Solving 160 œ 490t# Ê t œ
4
7
sec. The average velocity was
s(4/7)s(0)
4/7
œ 280 cm/sec.
(b) At the 160 cm mark the balls are falling at v(4/7) œ 560 cm/sec. The acceleration at the 160 cm mark
was 980 cm/sec# .
17
(c) The light was flashing at a rate of 4/7
œ 29.75 flashes per second.
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137
138
Chapter 3 Differentiation
20. (a)
(b)
21. C œ position, A œ velocity, and B œ acceleration. Neither A nor C can be the derivative of B because B's
derivative is constant. Graph C cannot be the derivative of A either, because A has some negative slopes while
C has only positive values. So, C (being the derivative of neither A nor B) must be the graph of position.
Curve C has both positive and negative slopes, so its derivative, the velocity, must be A and not B. That
leaves B for acceleration.
22. C œ position, B œ velocity, and A œ acceleration. Curve C cannot be the derivative of either A or B because
C has only negative values while both A and B have some positive slopes. So, C represents position. Curve C
has no positive slopes, so its derivative, the velocity, must be B. That leaves A for acceleration. Indeed, A is
negative where B has negative slopes and positive where B has positive slopes.
23. (a) c(100) œ 11,000 Ê cav œ
#
11,000
100
œ $110
w
(b) c(x) œ 2000 100x .1x Ê c (x) œ 100 .2x. Marginal cost œ cw (x) Ê the marginal cost of producing 100
machines is cw (100) œ $80
(c) The cost of producing the 101st machine is c(101) c(100) œ 100 201
10 œ $79.90
24. (a) r(x) œ 20000 ˆ1 "x ‰ Ê rw (x) œ
(b) rw a"!!b œ
20000
100#
20000
x#
, which is marginal revenue.
œ $#Þ
(c) x lim
rw (x) œ x lim
Ä_
Ä_
will approach zero.
20000
x#
œ 0. The increase in revenue as the number of items increases without bound
25. b(t) œ 10' 10% t 10$ t# Ê bw (t) œ 10% (2) a10$ tb œ 10$ (10 2t)
(a) bw (0) œ 10% bacteria/hr
(b) bw (5) œ 0 bacteria/hr
w
%
(c) b (10) œ 10 bacteria/hr
26. Q(t) œ 200(30 t)# œ 200 a900 60t t# b Ê Qw (t) œ 200(60 2t) Ê Qw (10) œ 8,000 gallons/min is the rate
the water is running at the end of 10 min. Then
Q(10)Q(0)
10
œ 10,000 gallons/min is the average rate the
water flows during the first 10 min. The negative signs indicate water is leaving the tank.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 3.3 The Derivative as a Rate of Change
27. (a) y œ 6 ˆ1 t ‰#
1#
œ 6 Š1 (b) The largest value of
(c)
dy
dt
t
6
t#
144 ‹
Ê
dy
dt
œ
t
12
139
1
is 0 m/h when t œ 12 and the fluid level is falling the slowest at that time. The
smallest value of dy
dt is 1 m/h, when t œ 0, and
dy
In this situation, dt Ÿ 0 Ê the graph of y is
always decreasing. As dy
dt increases in value,
the fluid level is falling the fastest at that time.
the slope of the graph of y increases from 1
to 0 over the interval 0 Ÿ t Ÿ 12.
28. (a) V œ
4
3
1 r$ Ê
(b) When r œ 2,
dV
dr
dV
dr
œ 41 r # Ê
dV ¸
dr r=2
œ 41(2)# œ 161 ft$ /ft
œ 161 so that when r changes by 1 unit, we expect V to change by approximately 161.
Therefore when r changes by 0.2 units V changes by approximately (161)(0.2) œ 3.21 ¸ 10.05 ft$ . Note
that V(2.2) V(2) ¸ 11.09 ft$ .
29. 200 km/hr œ 55 59 m/sec œ
t œ 25, D œ
10
9
#
(25) œ
500
9 m/sec,
6250
9 m
and D œ
10 #
9 t
30. s œ v! t 16t# Ê v œ v! 32t; v œ 0 Ê t œ
v!
32
Ê Vœ
20
9
t. Thus V œ
500
9
Ê
; 1900 œ v! t 16t# so that t œ
Ê v! œ È(64)(1900) œ 80È19 ft/sec and, finally,
80È19 ft
sec
†
60 sec
1 min
†
60 min
1 hr
†
1 mi
5280 ft
v!
32
20
9
tœ
500
9
Ê t œ 25 sec. When
Ê 1900 œ
v!#
3#
¸ 238 mph.
31.
v œ 0 when t œ 6.25 sec
v 0 when 0 Ÿ t 6.25 Ê body moves up; v 0 when 6.25 t Ÿ 12.5 Ê body moves down
body changes direction at t œ 6.25 sec
body speeds up on (6.25ß 12.5] and slows down on [0ß 6.25)
The body is moving fastest at the endpoints t œ 0 and t œ 12.5 when it is traveling 200 ft/sec. It's
moving slowest at t œ 6.25 when the speed is 0.
(f) When t œ 6.25 the body is s œ 625 m from the origin and farthest away.
(a)
(b)
(c)
(d)
(e)
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v!#
64
140
Chapter 3 Differentiation
32.
(a) v œ 0 when t œ
3
#
sec
(b) v 0 when 0 Ÿ t 1.5 Ê body moves down; v 0 when 1.5 t Ÿ 5 Ê body moves up
(c) body changes direction at t œ 3# sec
(d) body speeds up on ˆ 3# ß &‘ and slows down on !ß 3# ‰
(e) body is moving fastest at t œ 5 when the speed œ kv(5)k œ 7 units/sec; it is moving slowest at
t œ 3# when the speed is 0
(f) When t œ 5 the body is s œ 12 units from the origin and farthest away.
33.
6 „ È15
3
6 È15
t
3
(a) v œ 0 when t œ
sec
(b) v 0 when
6 È15
3
Ê body moves left; v 0 when 0 Ÿ t 6 È15
3
or
6 È15
3
tŸ4
Ê body moves right
6 „ È15
sec
3
È
È
Š 6 3 15 ß #‹ Š 6 3 15 ß %“
(c) body changes direction at t œ
(d) body speeds up on
È15
and slows down on ’0ß 6 3
È15
‹ Š#ß 6 3
‹.
(e) The body is moving fastest at t œ 0 and t œ 4 when it is moving 7 units/sec and slowest at t œ
(f) When t œ
6È15
3
the body is at position s ¸ 6.303 units and farthest from the origin.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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6„È15
3
sec
Section 3.4 Derivatives of Trigonometric Functions
141
34.
6 „ È15
3
È
È
v 0 when 0 Ÿ t 6 3 15 or 6 3 15 t Ÿ 4 Ê body is moving left; v 0 when
È
6 È15
t 6 3 15 Ê body is moving right
3
È
body changes direction at t œ 6 „ 3 15 sec
È
È
È
È
body speeds up on Š 6 3 15 ß #‹ Š 6 3 15 ß %“ and slows down on ’!ß 6 3 15 ‹ Š#ß 6 3 15 ‹
(a) v œ 0 when t œ
(b)
(c)
(d)
(e) The body is moving fastest at 7 units/sec when t œ 0 and t œ 4; it is moving slowest and stationary at
tœ
6 „ È15
3
(f) When t œ
6 È15
3
the position is s ¸ 10.303 units and the body is farthest from the origin.
35. (a) It takes 135 seconds.
(b) Average speed œ ??Ft œ
&!
($ !
œ
&
($
¸ !Þ!') furlongs/sec.
(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 fastest 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.‹
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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
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Section 3.4 Derivatives of Trigonometric Functions
23. p œ
œ
sin q cos q
cos q
Ê
œ
dp
dq
(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
#
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
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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.
lim
x Ä 16
È1 cos (1 csc x) œ É1 cos ˆ1 csc ˆ 1 ‰‰ œ È1 cos a1 † a2bb œ È2
6
1 ‰
‘
ˆ 1 ‰
‘
ˆ1‰
‘
41. lim sec cos x 1 tan ˆ 4 sec
x 1 œ sec cos 0 1 tan 4 sec 0 1 œ sec 1 1 tan 4 1 œ sec 1 œ 1
xÄ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 3.4 Derivatives of Trigonometric Functions
x ‰
ˆ 1tan 0 ‰
ˆ 1‰
42. lim sin ˆ tan1xtan
2 sec x œ sin tan 02 sec 0 œ sin # œ 1
xÄ!
43. lim tan ˆ1 tÄ!
sin t ‰
t
œ tan Š1 lim
tÄ!
1) ‰
44. lim cos ˆ sin
) œ cos Š1 lim
)
sin t
t ‹
‹
) Ä ! sin )
)Ä!
œ tan (1 1) œ 0
"
œ cos Œ1 †
lim
"
œ cos ˆ1 † 1 ‰ œ 1
sin )
)Ä!
)
dv
da
ˆ1‰
45. s œ # # sin t Ê v œ ds
dt œ 2 cos t Ê a œ dt œ 2 sin t Ê j œ dt œ 2 cos t. Therefore, velocity œ v 4
œ È2 m/sec; speed œ ¸v ˆ 1 ‰¸ œ È2 m/sec; acceleration œ a ˆ 1 ‰ œ È2 m/sec# ; jerk œ j ˆ 1 ‰ œ È2 m/sec$ .
4
46. s œ sin t cos t Ê v œ
4
œ cos t sin t Ê a œ
ds
dt
v ˆ 14 ‰
velocity œ
œ 0 m/sec; speed œ
1‰
ˆ
jerk œ j 4 œ 0 m/sec$ .
47. lim f(x) œ lim
xÄ!
Ê 9 œ c.
48.
xÄ!
sin# 3x
x#
¸v ˆ 14 ‰¸
dv
dt
4
œ sin t cos t Ê j œ
œ 0 m/sec; acceleration œ
a ˆ 14 ‰
œ cos t sin t. Therefore
È
œ 2 m/sec# ;
da
dt
œ lim 9 ˆ sin3x3x ‰ ˆ sin3x3x ‰ œ 9 so that f is continuous at x œ 0 Ê lim f(x) œ f(0)
xÄ!
xÄ!
lim g(x) œ lim c (x b) œ b and lim b g(x) œ lim b cos x œ 1 so that g is continuous at x œ 0 Ê lim c g(x)
xÄ!
xÄ!
xÄ!
xÄ!
œ lim b g(x) Ê b œ 1. Now g is not differentiable at x œ 0: At x œ 0, the left-hand derivative is
x Ä !c
xÄ!
d
dx
(x b)¸ x=0 œ 1, but the right-hand derivative is
d
dx
(cos x)¸ x=0 œ sin 0 œ 0. The left- and right-hand
derivatives can never agree at x œ 0, so g is not differentiable at x œ 0 for any value of b (including b œ 1).
49.
d***
dx***
d%
dx%
(cos x) œ sin x because
(cos x) œ cos x Ê the derivative of cos x any number of times that is a
multiple of 4 is cos x. Thus, dividing 999 by 4 gives 999 œ 249 † 4 3 Ê
œ
d$
dx$
d#%*†%
’ dx
#%*†%
(cos x)“ œ
50. (a) y œ sec x œ
Ê
d
dx
"
cos x
d$
dx$
Ê
d***
dx***
(cos x)
(cos x) œ sin x.
dy
dx
œ
(cos x)(0) (1)(sin x)
(cos x)#
œ
(sin x)(0) (1)(cos x)
(sin x)#
œ
sin x
cos# x
sin x ‰
œ ˆ cos" x ‰ ˆ cos
x œ sec x tan x
(sec x) œ sec x tan x
(b) y œ csc x œ
Ê
d
dx
d
dx
Ê
dy
dx
œ
cos x
sin# x
" ‰ ˆ cos x ‰
œ ˆ sin
x
sin x œ csc x cot x
(csc x) œ csc x cot x
(c) y œ cot x œ
Ê
"
sin x
cos x
sin x
Ê
dy
dx
#
œ
(sin x)(sin x) (cos x)(cos x)
(sin x)#
œ
sin# xcos# x
sin# x
œ
"
sin# x
œ csc# x
(cot x) œ csc x
51.
As h takes on the values of 1, 0.5, 0.3 and 0.1 the corresponding dashed curves of y œ
closer and closer to the black curve y œ cos x because
d
dx
(sin x) œ
is true as h takes on the values of 1, 0.5, 0.3 and 0.1.
sin x
lim sin (x h)
h
hÄ!
sin (x h) sin x
h
œ cos x. The same
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145
146
Chapter 3 Differentiation
52.
cos (x h) cos x
h
cos x
lim cos (x h)
œ
sin x.
h
hÄ!
As h takes on the values of 1, 0.5, 0.3, and 0.1 the corresponding dashed curves of y œ
get
closer and closer to the black curve y œ sin x because
The
d
dx
(cos x) œ
same is true as h takes on the values of 1, 0.5, 0.3, and 0.1.
53. (a)
The dashed curves of y œ
sinax hb sinax hb
#h
are closer to the black curve y œ cos x than the corresponding dashed
curves in Exercise 51 illustrating that the centered difference quotient is a better approximation of the derivative of
this function.
(b)
The dashed curves of y œ
cosax hb cosax hb
#h
are closer to the black curve y œ sin x than the corresponding dashed
curves in Exercise 52 illustrating that the centered difference quotient is a better approximation of the derivative of
this function.
54. lim
hÄ!
k0 h k k 0 h k
2h
œ lim
xÄ!
k h k k hk
2h
œ lim 0 œ 0 Ê the limits of the centered difference quotient exists even
hÄ!
though the derivative of f(x) œ kxk does not exist at x œ 0.
55. y œ tan x Ê yw œ sec# x, so the smallest value
yw œ sec# x takes on is yw œ 1 when x œ 0;
yw has no maximum value since sec# x has no
largest value on ˆ 1# ß 1# ‰ ; yw is never negative
since sec# x
1.
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Section 3.4 Derivatives of Trigonometric Functions
56. y œ cot x Ê yw œ csc# x so yw has no smallest
value since csc# x has no minimum value on
(!ß 1); the largest value of yw is 1, when x œ 1# ;
the slope is never positive since the largest
value yw œ csc2 x takes on is 1.
57. y œ
sin x
x appears to cross the y-axis at y œ 1, since
lim sin x œ 1; y œ sinx2x appears to cross the y-axis
xÄ! x
at y œ 2, since lim sinx2x œ 2; y œ sinx4x appears to
xÄ!
cross the y-axis at y œ 4, since lim sinx4x œ 4.
xÄ!
However, none of these graphs actually cross the y-axis
since x œ 0 is not in the domain of the functions. Also,
lim
xÄ!
sin 5x
x
sin (3x)
x
œ 5, lim
xÄ!
œ k Ê the graphs of y œ
yœ
sin kx
x
œ 3, and lim
sin kx
x
xÄ!
yœ
sin 5x
x ,
sin (3x)
,
x
and
approach 5, 3, and k, respectively, as
x Ä 0. However, the graphs do not actually cross the
y-axis.
58. (a)
sin h
h
h
1
0.01
0.001
0.0001
ˆ sinh h ‰ ˆ 180
‰
1
.99994923
1
1
1
.017452406
.017453292
.017453292
.017453292
1 ‰
sin ˆh† 180
h
xÄ!
œ lim
sin h°
h
hÄ!
lim
œ lim
1
180
hÄ!
1 ‰
sin ˆh† 180
1 †h
180
œ lim
1 sin )
180
)Ä!
)
œ
1
180
() œ h †
1
180 )
(converting to radians)
(b)
cos h1
h
h
1
0.01
0.001
0.0001
lim
hÄ!
0.0001523
0.0000015
0.0000001
0
cos h1
h
(c) In degrees,
œ 0, whether h is measured in degrees or radians.
d
dx
(sin x) œ lim
hÄ!
œ lim ˆsin x †
hÄ!
cos h 1 ‰
h
sin (x h) sin x
h
lim ˆcos x †
hÄ!
œ lim
hÄ!
sin h ‰
h
(sin x cos h cos x sin h) sin x
h
œ (sin x) † lim ˆ cos hh 1 ‰ (cos x) † lim ˆ sinh h ‰
hÄ!
1 ‰
œ (sin x)(0) (cos x) ˆ 180
œ
(d)
1
180 cos x
cos x
d
In degrees, dx
(cos x) œ lim cos (x h)
œ lim (cos x cos h sinh x sin h) cos x
h
hÄ!
hÄ!
(cos x)(cos h 1) sin x sin h
ˆ
œ lim
œ lim cos x † cos hh 1 ‰ lim ˆsin x † sinh h ‰
h
hÄ!
hÄ!
hÄ!
œ (cos x) lim ˆ
hÄ!
(e)
d#
dx#
d#
dx#
cos h 1 ‰
h
hÄ!
1 ‰
1
(sin x) lim ˆ sinh h ‰ œ (cos x)(0) (sin x) ˆ 180
œ 180
sin x
hÄ!
(sin x) œ
d
dx
1
1 ‰#
ˆ 180
cos x‰ œ ˆ 180
sin x;
(cos x) œ
d
dx
1
1 ‰#
ˆ 180
sin x‰ œ ˆ 180
cos x;
d$
dx$
(sin x) œ
d$
dx$
d
dx
(cos x) œ
#
$
1 ‰
1 ‰
sin x‹ œ ˆ 180
cos x;
Š ˆ 180
d
dx
#
$
1 ‰
1 ‰
cos x‹ œ ˆ 180
sin x
Š ˆ 180
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148
Chapter 3 Differentiation
3.5 THE CHAIN RULE AND PARAMETRIC EQUATIONS
1. f(u) œ 6u 9 Ê f w (u) œ 6 Ê f w (g(x)) œ 6; g(x) œ
œ 6 † 2x$ œ 12x$
"
#
x% Ê gw (x) œ 2x$ ; therefore
dy
dx
œ f w (g(x))gw (x)
2. f(u) œ 2u$ Ê f w (u) œ 6u# Ê f w (g(x)) œ 6(8x 1)# ; g(x) œ 8x 1 Ê gw (x) œ 8; therefore
œ 6(8x 1)# † 8 œ 48(8x 1)#
dy
dx
œ f w (g(x))gw (x)
3. f(u) œ sin u Ê f w (u) œ cos u Ê f w (g(x)) œ cos (3x 1); g(x) œ 3x 1 Ê gw (x) œ 3; therefore
œ f w (g(x))gw (x)
dy
dx
œ (cos (3x 1))(3) œ 3 cos (3x 1)
4. f(u) œ cos u Ê f w (u) œ sin u Ê f w (g(x)) œ sin ˆ 3x ‰ ; g(x) œ
‰ œ "3 sin ˆ 3x ‰
œ sin ˆ 3x ‰ † ˆ "
3
x
3
Ê gw (x) œ "3 ; therefore
dy
dx
œ f w (g(x))gw (x)
5. f(u) œ cos u Ê f w (u) œ sin u Ê f w (g(x)) œ sin (sin x); g(x) œ sin x Ê gw (x) œ cos x; therefore
dy
dx
œ f w (g(x))gw (x) œ (sin (sin x)) cos x
6. f(u) œ sin u Ê f w (u) œ cos u Ê f w (g(x)) œ cos (x cos x); g(x) œ x cos x Ê gw (x) œ 1 sin x; therefore
dy
dx
œ f w (g(x))gw (x) œ (cos (x cos x))(1 sin x)
7. f(u) œ tan u Ê f w (u) œ sec# u Ê f w (g(x)) œ sec# (10x 5); g(x) œ 10x 5 Ê gw (x) œ 10; therefore
dy
dx
œ f w (g(x))gw (x) œ asec# (10x 5)b (10) œ 10 sec# (10x 5)
8. f(u) œ sec u Ê f w (u) œ sec u tan u Ê f w (g(x)) œ sec ax# 7xb tan ax# 7xb ; g(x) œ x# 7x
Ê gw (x) œ 2x 7; therefore
œ f w (g(x))gw (x) œ (2x 7) sec ax# 7xb tan ax# 7xb
dy
dx
9. With u œ (2x 1), y œ u& :
dy
dx
œ
dy du
du dx
œ 5u% † 2 œ 10(2x 1)%
10. With u œ (4 3x), y œ u* :
dy
dx
œ
dy du
du dx
œ 9u) † (3) œ 27(4 3x))
11. With u œ ˆ1 x7 ‰ , y œ ?( :
œ
dy
dx
12. With u œ ˆ x# 1‰ , y œ ?"! :
dy
dx
œ
#
13. With u œ Š x8 x "x ‹ , y œ ?% :
14. With u œ ˆ x5 " ‰
5x ,
y œ ?& :
15. With u œ tan x, y œ sec u:
16. With u œ 1 "x , y œ cot u:
17. With u œ sin x, y œ u$ :
dy
dx
dy
dx
dy
dx
œ
œ
dy du
du dx
dy
dx
œ
œ
dy du
du dx
œ
dy du
du dx
18. With u œ cos x, y œ 5u% :
dy
dx
œ
œ 10u"" † ˆ "# ‰ œ 5 ˆ x# 1‰
dy du
du dx
dy du
du dx
dy du
du dx
dy
dx
)
œ 7u) † ˆ "7 ‰ œ ˆ" x7 ‰
dy du
du dx
œ 4u$ † ˆ x4 1 œ 5u% † ˆ 15 " ‰
5x#
"‰
x#
œ ˆ x5 ""
$
#
œ 4 Š x8 x x" ‹ ˆ x4 1 " ‰%
5x
ˆ1 "‰
x#
œ (sec u tan u) asec# xb œ (sec (tan x) tan (tan x)) sec# x
œ acsc# ub ˆ x"# ‰ œ x"# csc# ˆ1 x" ‰
œ 3u# cos x œ 3 asin# xb (cos x)
dy du
du dx
œ a20u& b (sin x) œ 20 acos& xb (sin x)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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"‰
x#
Section 3.5 The Chain Rule and Parametric Equations
19. p œ È3 t œ (3 t)"Î# Ê
20. q œ È2r r# œ a2r r# b
21. s œ
œ
4
31
4
1
sin 3t 4
51
dp
dt
"Î#
Ê
cos 5t Ê
23. r œ (csc ) cot ))" Ê
dr
d)
dq
dr
(3 t)"Î# †
œ
"
#
d
dt
a2r r# b
(3 t) œ "# (3 t)"Î# œ
"Î#
†
d
dr
ds
dt
œ
cos 3t †
d
dt
(3t) ds
dt
œ cos ˆ 3#1t ‰ †
d
dt
ˆ 3#1t ‰ sin ˆ 3#1t ‰ †
4
31
4
51
"
#
a2r r# b œ
(sin 5t) †
a2r r# b
(5t) œ
d
dt
"
2È 3 t
4
1
"Î#
"r
È2r r#
(2 2r) œ
cos 3t 4
1
sin 5t
d
dt
ˆ 3#1t ‰ œ
31
2
cos ˆ 3#1t ‰ 31
2
sin ˆ 3#1t ‰
œ (csc ) cot ))#
d
d)
(csc ) cot )) œ
csc ) cot ) csc# )
(csc ) cot ))#
œ
csc ) (cot ) csc ))
(csc ) cot ))#
œ (sec ) tan ))#
d
d)
(sec ) tan )) œ
sec ) tan ) sec# )
(sec ) tan ))#
œ
sec ) (tan ) sec ))
(sec ) tan ))#
csc )
csc ) cot )
24. r œ (sec ) tan ))" Ê
œ
"
#
(cos 3t sin 5t)
22. s œ sin ˆ 3#1t ‰ cos ˆ 3#1t ‰ Ê
œ 321 ˆcos 3#1t sin 3#1t ‰
œ
œ
dr
d)
sec )
sec ) tan )
d
d
# d
%
%
#
#
25. y œ x# sin% x x cos# x Ê dy
xb cos# x †
dx œ x dx asin xb sin x † dx ax b x dx acos
d
d
œ x# ˆ4 sin$ x dx
(sin x)‰ 2x sin% x x ˆ2 cos$ x † dx
(cos x)‰ cos# x
d
dx
(x)
œ x# a4 sin$ x cos xb 2x sin% x xa a2 cos$ xb (sin x)b cos# x
œ 4x# sin$ x cos x 2x sin% x 2x sin x cos$ x cos# x
d ˆ"‰
x d
$
$
asin& xb sin& x † dx
x 3 dx acos xb cos x †
œ "x a5 sin' x cos xb asin& xb ˆ x"# ‰ 3x a a3 cos# xb (sin x)b acos$ xb ˆ 3" ‰
26. y œ
"
x
sin& x x
3
cos$ x Ê yw œ
œ 5x sin' x cos x "
x#
" d
x dx
sin& x x cos# x sin x "
3
ˆ x3 ‰
cos$ x
"
" ‰"
7
d
(
'
ˆ
Ê dy
21 (3x 2) 4 #x#
dx œ 21 (3x 2) † dx (3x 2) 7
" ‰# ˆ " ‰
"
'
'
ˆ
#
21 (3x 2) † 3 (1) 4 #x#
x$ œ (3x 2) $
x Š4 "# ‹
27. y œ
œ
d
dx
(1) ˆ4 " ‰#
#x #
†
d
dx
ˆ4 " ‰
#x #
#x
%
28. y œ (5 2x)$ "8 ˆ 2x 1‰ Ê
dy
dx
$
$
œ 3(5 2x)% (2) 84 ˆ x2 1‰ ˆ x2# ‰ œ 6(5 2x)% ˆ x"# ‰ ˆ 2x 1‰
$
œ
6
(5 2x)%
Š 2x 1‹
x#
29. y œ (4x 3)% (x 1)$ Ê
%
dy
dx
œ (4x 3)% (3)(x 1)% †
%
$
d
dx
(x 1) (x 1)$ (4)(4x 3)$ †
$
%
œ (4x 3) (3)(x 1) (1) (x 1) (4)(4x 3) (4) œ 3(4x 3) (x 1)
œ
$
(4x 3)
(x 1)%
c3(4x 3) 16(x 1)d œ
'
30. y œ (2x 5)" ax# 5xb Ê
&
œ 6 ax# 5xb 2 ax# 5xb
(2x 5)#
dy
dx
%
d
dx
$
(4x 3)
16(4x 3) (x 1)$
$
(4x 3) (4x 7)
(x 1)%
&
'
œ (2x 5)" (6) ax# 5xb (2x 5) ax# 5xb (1)(2x 5)# (2)
'
d ˆ
d
31. h(x) œ x tan ˆ2Èx‰ 7 Ê hw (x) œ x dx
tan ˆ2x"Î# ‰‰ tan ˆ2x"Î# ‰ † dx
(x) 0
"
d ˆ "Î# ‰
# ˆ "Î# ‰
"Î#
#
† dx 2x
œ x sec 2x
tan ˆ2x ‰ œ x sec ˆ2Èx‰ † È tan ˆ2Èx‰ œ Èx sec# ˆ2Èx‰ tan ˆ2Èx‰
x
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149
150
Chapter 3 Differentiation
d ˆ
d
32. k(x) œ x# sec ˆ "x ‰ Ê kw (x) œ x# dx
sec x" ‰ sec ˆ x" ‰ † dx
ax# b œ x# sec ˆ x" ‰ tan ˆ x" ‰ †
œ x# sec ˆ "x ‰ tan ˆ x" ‰ † ˆ x"# ‰ 2x sec ˆ x" ‰ œ 2x sec ˆ x" ‰ sec ˆ x" ‰ tan ˆ x" ‰
#
33. f()) œ ˆ 1 sincos) ) ‰ Ê f w ()) œ 2 ˆ 1 sincos) ) ‰ †
œ
(2 sin )) acos ) cos# ) sin# )b
(1 cos ))$
34. g(t) œ ˆ 1 sincost t ‰
œ
"
(2 sin )) (cos ) 1)
(1 cos ))$
œ
Ê gw (t) œ ˆ 1 sincost t ‰
asin# t cos t cos# tb
(1 cos t)#
#
†
(1 cos ))(cos )) (sin ))(sin ))
(1 cos ))#
2 sin )
(1 cos ))#
œ
†
2 sin )
1 cos )
d
dt
#
ˆ 1 sincost t ‰ œ (1 sincost t)# †
(sin t)(sin t) (" cos t)(cos t)
(sin t)#
"
1 cos t
œ
35. r œ sin a)# b cos (2)) Ê
ˆ 1 sincos) ) ‰ œ
d
d)
ˆ x" ‰ 2x sec ˆ x" ‰
d
dx
œ sin a)# b (sin 2))
dr
d)
(2)) cos (2)) acos a)# bb †
d
d)
d
d)
a) # b
œ sin a)# b (sin 2))(2) (cos 2)) acos a)# bb (2)) œ 2 sin a)# b sin (#)) 2) cos (2)) cos a)# b
36. r œ Šsec È)‹ tan ˆ ") ‰ Ê
dr
d)
œ )"# sec È) sec# ˆ ") ‰ 37. q œ sin Š Ètt 1 ‹ Ê
œ cos Š Ètt 1 ‹ †
39. y œ sin# (1t 2) Ê
Èt 1
t
2
t1
dq
dt
"
#È )
tan ˆ ") ‰ sec È) tan È) œ Šsec È)‹ ”
œ cos Š Ètt 1 ‹ †
dq
dt
Èt 1 38. q œ cot ˆ sint t ‰ Ê
œ Šsec È)‹ ˆ sec# ") ‰ ˆ )"# ‰ tan ˆ ") ‰ Šsec È) tan È)‹ Š
d
dt
Èt 1 (1)t †
d
dt
sec# ˆ )" ‰
)# •
ˆÈ t 1 ‰
ˆÈ t 1 ‰
#
1) t
œ cos Š Ètt 1 ‹ Š 2(t
‹ œ Š 2(tt1)2$Î# ‹ cos Š Ètt 1 ‹
2(t 1)$Î#
œ csc# ˆ sint t ‰ †
d
dt
ˆ sint t ‰ œ ˆcsc# ˆ sint t ‰‰ ˆ t cos tt# sin t ‰
œ 2 sin (1t 2) †
dy
dt
Š Ètt 1 ‹ œ cos Š Ètt 1 ‹ †
tan È) tan ˆ ") ‰
#È )
"
‹
#È )
d
dt
sin (1t 2) œ 2 sin (1t 2) † cos (1t 2) †
d
dt
(1t 2)
œ 21 sin (1t 2) cos (1t 2)
40. y œ sec# 1t Ê
dy
dt
œ (2 sec 1t) †
41. y œ (1 cos 2t)% Ê
42. y œ ˆ1 cot ˆ #t ‰‰
œ
#
dy
dt
Ê
(sec 1t) œ (2 sec 1t)(sec 1t tan 1t) †
œ 4(1 cos 2t)& †
dy
dt
ˆ #t ‰
$
ˆ1 cot ˆ t ‰‰
#
csc#
43. y œ sin acos (2t 5)b Ê
d
dt
dy
dt
œ 2 ˆ1 cot ˆ #t ‰‰
(1t) œ 21 sec# 1t tan 1t
(1 cos 2t) œ 4(1 cos 2t)& (sin 2t) †
d
dt
$
œ cos (cos (2t 5)) †
d
dt
cot ˆ #t ‰‰ œ 2 ˆ1 cot ˆ #t ‰‰
$
d
dt
(2t) œ
8 sin 2t
(1 cos 2t)&
† ˆcsc# ˆ #t ‰‰ †
†
dˆ
dt 1
d
dt
cos (2t 5) œ cos (cos (2t 5)) † (sin (2t 5)) †
d
dt
dˆt‰
dt #
(2t 5)
œ 2 cos (cos (2t 5))(sin (2t 5))
ˆ
ˆ t ‰‰ †
44. y œ cos ˆ5 sin ˆ 3t ‰‰ Ê dy
dt œ sin 5 sin 3
5
t
t
œ 3 sin ˆ5 sin ˆ 3 ‰‰ ˆcos ˆ 3 ‰‰
d
dt
$
dy
% ˆ t ‰‘# d † dt 1
dt œ 3 1 tan
1#
#
tan% ˆ 1t# ‰‘ tan$ ˆ 1t# ‰ sec# ˆ 1t# ‰ † 1"# ‘ œ 1 45. y œ 1 tan% ˆ 1t# ‰‘ Ê
œ 12 1 46. y œ
"
6
ˆ5 sin ˆ 3t ‰‰ œ sin ˆ5 sin ˆ 3t ‰‰ ˆ5 cos ˆ 3t ‰‰ †
$
c1 cos# (7t)d Ê
dy
dt
œ
3
6
d
dt
ˆ 3t ‰
#
tan% ˆ 1t# ‰‘ œ 3 1 tan% ˆ 1t# ‰‘ 4 tan$ ˆ 1t# ‰ †
#
tan% ˆ 1t# ‰‘ tan$ ˆ 1t# ‰ sec# ˆ 1t# ‰‘
#
#
d
dt
tan ˆ 1t# ‰‘
c1 cos# (7t)d † 2 cos (7t)(sin (7t))(7) œ 7 c1 cos# (7t)d (cos (7t) sin (7t))
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 3.5 The Chain Rule and Parametric Equations
"Î#
Ê
œ "# a1 cos at# bb
"Î#
47. y œ a1 cos at# bb
dy
dt
2 cos ŒÉ1 Èt
É1 Èt†2Èt
œ
a1 cos at# bb
"Î#
†
d
dt
a1 cos at# bb œ
#
"
#
a1 cos at# bb
œ 4 cos ŒÉ1 Èt †
dy
dt
" ‰#
x
œ
6
x%
œ
"
#x
51. y œ
"
9
œ
"
ˆ1 Èx‰$ Š
ˆ1 x" ‰ "
#
6
x$
ˆ1 " ‰#
x
œ
d
dx
6
x$
ˆ1 x" ‰# ˆ1 x" ‰# †
d
dt
ˆ1 Èt‰
d
dx
ˆ x3# ‰
ˆ1 x" ‰ ˆ x" 1 x" ‰
ˆ1 Èx‰# x"Î#
"
#È x
"
#
1‹ œ
"
#x
"
#
$
x" ˆ1 Èx‰ "# x"Î# ˆ1 Èx‰ 1‘
ˆ1 Èx‰$ Š 3# "
‹
#Èx
cot (3x 1) Ê yw œ 9" csc# (3x 1)(3) œ 3" csc# (3x 1) Ê yww œ ˆ 32 ‰ (csc (3x 1) †
2
3
†
# É 1 È t
#
$
’ˆ1 Èx‰ ˆ "# x$Î# ‰ x"Î# (2) ˆ1 Èx‰ ˆ "# x"Î# ‰“
#
$
$Î# ˆ
1 Èx‰ x" ˆ1 Èx‰ “ œ
’ "
# x
"
#
a t# b ‰
É t Èt
"
#
50. y œ ˆ1 Èx‰ Ê yw œ ˆ1 Èx‰ ˆ "# x"Î# ‰ œ
œ
d
dt
cos ŒÉ1 Èt
œ ˆ x3# ‰ ˆ2 ˆ1 x" ‰ ˆ x"# ‰‰ ˆ x6$ ‰ ˆ1 œ x6$ ˆ1 x" ‰ ˆ1 2x ‰
"
#
ˆsin at# b †
ŒÉ1 Èt œ 4 cos ŒÉ1 Èt †
d
dt
$
#
#
49. y œ ˆ1 "x ‰ Ê yw œ 3 ˆ1 x" ‰ ˆ x"# ‰ œ x3# ˆ1 x" ‰ Ê yww œ ˆ x3# ‰ †
Ê yww œ
"Î#
at b
asin at# bb † 2t œ È1t sin
cos at# b
48. y œ 4 sin ŒÉ1 Èt Ê
œ
"
#
œ
151
csc (3x 1)(csc (3x 1) cot (3x 1) †
d
dx
csc (3x 1))
d
dx
#
(3x 1)) œ 2 csc (3x 1) cot (3x 1)
52. y œ 9 tan ˆ x3 ‰ Ê yw œ 9 ˆsec# ˆ x3 ‰‰ ˆ "3 ‰ œ 3 sec# ˆ x3 ‰ Ê yww œ 3 † 2 sec ˆ x3 ‰ ˆsec ˆ x3 ‰ tan ˆ x3 ‰‰ ˆ "3 ‰ œ 2 sec# ˆ 3x ‰ tan ˆ 3x ‰
53. g(x) œ Èx Ê gw (x) œ
"
#È x
Ê g(1) œ 1 and gw (1) œ
therefore, (f ‰ g)w (1) œ f w (g(1)) † gw (1) œ 5 †
"
#
œ
"
u#
1
10
14
=
"
(1x)#
Ê g(1) œ
"
#
w
and gw (1) œ
"
4
; f(u) œ 1 Ê f w (g(1)) œ f w ˆ #" ‰ œ 4; therefore, (f ‰ g)w (1) œ f (g(1))gw (1) œ 4 †
55. g(x) œ 5Èx Ê gw (x) œ
œ
; f(u) œ u& 1 Ê f w (u) œ 5u% Ê f w (g(1)) œ f w (1) œ 5;
5
#
54. g(x) œ (1 x)" Ê gw (x) œ (1 x)# (1) œ
Ê f w (u) œ
"
#
5
#Èx
Ê g(1) œ 5 and gw (1) œ
5
#
"
4
"
u
œ1
1‰
; f(u) œ cot ˆ 110u ‰ Ê f w (u) œ csc# ˆ 110u ‰ ˆ 10
1
1
1
csc# ˆ 110u ‰ Ê f w (g(1)) œ f w (5) œ 10
csc# ˆ 1# ‰ œ 10
; therefore, (f ‰ g)w (1) œ f w (g(1))gw (1) œ 10
†
5
#
56. g(x) œ 1x Ê gw (x) œ 1 Ê g ˆ "4 ‰ œ 14 and gw ˆ 4" ‰ œ 1; f(u) œ u sec# u Ê f w (u) œ 1 2 sec u † sec u tan u
œ 1 2 sec# u tan u Ê f w ˆg ˆ "4 ‰‰ œ f w ˆ 14 ‰ œ 1 2 sec# 14 tan 14 œ 5; therefore, (f ‰ g)w ˆ 4" ‰ œ f w ˆg ˆ 4" ‰‰ gw ˆ 4" ‰ œ 51
57. g(x) œ 10x# x 1 Ê gw (x) œ 20x 1 Ê g(0) œ 1 and gw (0) œ 1; f(u) œ
œ
2u# 2
au # 1 b #
Ê f w (u) œ
au# 1b(2) (2u)(2u)
au # 1 b #
Ê f w (g(0)) œ f w (1) œ 0; therefore, (f ‰ g)w (0) œ f w (g(0))gw (0) œ 0 † 1 œ 0
"
2
w
x# 1 Ê g (x) œ x$ Ê g(1) œ 0 and
4(u 1)
1 ‰ (u 1)(1) (u 1)(1)
2 ˆ uu œ 2(u(u1)(2)
1 †
(u 1)#
1)$ œ (u 1)$
w
w
w
58. g(x) œ
œ
2u
u # 1
#
1‰
1‰
gw (1) œ 2; f(u) œ ˆ uu Ê f w (u) œ 2 ˆ uu 1
1
Ê f w (g(1)) œ f w (0) œ 4; therefore,
(f ‰ g) (1) œ f (g(1))g (1) œ (4)(2) œ 8
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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d
du
1‰
ˆ uu 1
152
Chapter 3 Differentiation
59. (a) y œ 2f(x) Ê
œ 2f w (x) Ê
dy
dx
(b) y œ f(x) g(x) Ê
(c) y œ f(x) † g(x) Ê
œ 2f w (2) œ 2 ˆ "3 ‰ œ
dy
dx ¹ x=2
œ f w (x) gw (x) Ê
dy
dx
dy
dx
œ
g(x)f w (x) f(x)gw (x)
[g(x)]#
(e) y œ f(g(x)) Ê
dy
dx
œ f w (g(x))gw (x) Ê
(d) y œ
f(x)
g(x)
Ê
"
#
Ê
dy
dx ¹ x=2
œ
(g) y œ (g(x))# Ê
dy
dx
œ 2(g(x))$ † gw (x) Ê
(h) y œ a(f(x))# (g(x))# b
dy
dx ¹ x=2
œ
"
#
(f(x))"Î# † f w (x) œ
"Î#
œ
dy
dx
a(f(2))# (g(2)) b
Ê
dy
dx ¹ x=3
"
#
# a(f(x)) # "Î#
w
Ê
(2) ˆ "3 ‰ (8)(3)
##
œ
"
3
œ f w (g(2))gw (2) œ f w (2)(3) œ
f w (x)
#Èf(x)
dy
dx
œ f(3)gw (3) g(3)f w (3) œ 3 † 5 (4)(21) œ 15 81
dy
dx ¹ x=3
g(2)f w (2) f(2)gw (2)
[g(2)]#
œ
dy
dx ¹ x=2
(f) y œ (f(x))"Î# Ê
Ê
œ f w (3) gw (3) œ 21 5
dy
dx ¹ x=3
œ f(x)gw (x) g(x)f w (x) Ê
dy
dx
2
3
dy
dx ¹ x=2
œ
f w (2)
#Èf(2)
ˆ "3 ‰
œ
œ
(3) œ 1
œ
#È 8
37
6
"
6È 8
œ
"
1 #È 2
œ 2(g(3))$ gw (3) œ 2(4)$ † 5 œ
(g(x))# b
"Î#
œ
È2
24
5
3#
a2f(x) † f w (x) 2g(x) † gw (x)b
a2f(2)f (2) 2g(2)gw (2)b œ
"
#
a8# 2# b
"Î#
ˆ2 † 8 †
"
3
2 † 2 † (3)‰
œ 3È517
60. (a) y œ 5f(x) g(x) Ê
(b) y œ f(x)(g(x))$ Ê
œ 5f w (x) gw (x) Ê
dy
dx
dy
dx ¹ x=1
œ 5f w (1) gw (1) œ 5 ˆ "3 ‰ ˆ 38 ‰ œ 1
œ f(x) a3(g(x))# gw (x)b (g(x))$ f w (x) Ê
dy
dx
dy
dx ¹ x=0
œ $f(0)(g(0))# gw (0) (g(0))$ f w (0)
œ 3(1)(1)# ˆ 3" ‰ (1)$ (5) œ 6
(c) y œ
œ
f(x)
g(x) 1
Ê
(g(x) 1)f w (x) f(x) gw (x)
(g(x) 1)#
œ
dy
dx
(4") ˆ "3 ‰(3) ˆ 83 ‰
(41)#
Ê
dy
dx ¹ x=1
œ
(g(1) 1)f w (1) f(1)gw (1)
(g(1) 1)#
œ1
(d) y œ f(g(x)) Ê
dy
dx
œ f w (g(x))gw (x) Ê
dy
dx ¹ x=0
œ f w (g(0))gw (0) œ f w (1) ˆ "3 ‰ œ ˆ "3 ‰ ˆ 3" ‰ œ 9"
(e) y œ g(f(x)) Ê
dy
dx
œ gw (f(x))f w (x) Ê
dy
dx ¹ x=0
œ gw (f(0))f w (0) œ gw (1)(5) œ ˆ 83 ‰ (5) œ 40
3
(f) y œ ax"" f(x)b
#
Ê
œ 2 ax"" f(x)b
dy
dx
$
a11x"! f w (x)b Ê
dy
dx ¹ x=1
œ 2(1 f(1))$ a11 f w (1)b
"
‰
œ 2(1 3)$ ˆ11 "3 ‰ œ ˆ 42$ ‰ ˆ 32
3 œ 3
(g) y œ f(x g(x)) Ê
œ f w (x g(x)) a1 gw (x)b Ê
dy
dx
dy
dx ¹ x=0
œ f w (0 g(0)) a1 gw (0)b œ f w (1) ˆ1 "3 ‰
œ ˆ "3 ‰ ˆ 43 ‰ œ 49
61.
ds
dt
œ
ds
d)
†
d)
dt :
s œ cos ) Ê
62.
dy
dt
œ
dy
dx
†
dx
dt :
y œ x# 7x 5 Ê
63. With y œ x, we should get
(a) y œ
(b)
u
5
7 Ê
y œ 1 "u Ê
"
œ "
u# † (x 1)#
œ
dy
du
dy
du
œ
dy
dx
"
5
dy
du
dy
dx
œ
"
u#
ds ¸
d) )= 321
œ sin ˆ 3#1 ‰ œ 1 so that
œ 2x 7 Ê
dy
dx ¹ x=1
œ 9 so that
dy
dt
œ
ds
dt
dy
dx
œ
†
ds
d)
dx
dt
†
d)
dt
œ9†
œ 1†5œ 5
"
3
œ3
œ 1 for both (a) and (b):
; u œ 5x 35 Ê
; u œ (x 1)
"
a(x 1)" b#
64. With y œ x$Î# , we should get
(a) y œ u$ Ê
œ sin ) Ê
ds
d)
†
dy
dx
"
(x 1)#
œ
3
#
"
du
dx
œ 5; therefore,
Ê
du
dx
œ (x 1)# †
dy
dy
dx œ du
#
†
œ (x 1) (1) œ
"
(x 1)#
du
"
dx œ 5
"
(x 1)# ;
† 5 œ 1, as expected
therefore
dy
dx
œ
dy
du
†
du
dx
œ 1, again as expected
x"Î# for both (a) and (b):
œ 3u# ; u œ Èx Ê
du
dx
"
#È x
; therefore,
dy
dx
œ
dy
du
†
du
dx
œ 3u# †
œ 3x# ; therefore,
dy
dx
œ
dy
du
†
du
dx
œ
œ
"
#È x
#
œ 3 ˆÈx‰ †
"
#Èx
œ
3
#
Èx,
as expected.
(b) y œ Èu Ê
dy
du
œ
"
#È u
; u œ x$ Ê
du
dx
"
#Èu
† 3x# œ
"
#È x $
again as expected.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
† 3x# œ
3
#
x"Î# ,
Section 3.5 The Chain Rule and Parametric Equations
65. y œ 2 tan ˆ 14x ‰ Ê
(a)
dy
dx ¹ x=1
œ
1
#
dy
dx
œ ˆ2 sec#
1x ‰ ˆ 1 ‰
4
4
œ
1
#
sec#
1x
4
sec# ˆ 14 ‰ œ 1 Ê slope of tangent is 2; thus, y(1) œ 2 tan ˆ 14 ‰ œ 2 and yw (1) œ 1 Ê tangent line is
given by y 2 œ 1(x 1) Ê y œ 1x 2 1
(b) yw œ 1# sec# ˆ 14x ‰ and the smallest value the secant function can have in # x 2 is 1 Ê the minimum
value of yw is 1# and that occurs when 1# œ 1# sec# ˆ 14x ‰ Ê 1 œ sec# ˆ 14x ‰ Ê „ 1 œ sec ˆ 14x ‰ Ê x œ 0.
66. (a) y œ sin 2x Ê yw œ 2 cos 2x Ê yw (0) œ 2 cos (0) œ 2 Ê tangent to y œ sin 2x at the origin is y œ 2x;
y œ sin ˆ x# ‰ Ê yw œ "# cos ˆ x# ‰ Ê yw (0) œ "# cos 0 œ "# Ê tangent to y œ sin ˆ x# ‰ at the origin is
y œ "# x. The tangents are perpendicular to each other at the origin since the product of their slopes is
1.
(b) y œ sin (mx) Ê yw œ m cos (mx) Ê yw (0) œ m cos 0 œ m; y œ sin ˆ mx ‰ Ê yw œ m" cos ˆ mx ‰
Ê yw (0) œ m" cos (0) œ m" . Since m † ˆ m" ‰ œ 1, the tangent lines are perpendicular at the origin.
(c) y œ sin (mx) Ê yw œ m cos (mx). The largest value cos (mx) can attain is 1 at x œ 0 Ê the largest value
yw can attain is kmk because kyw k œ km cos (mx)k œ kmk kcos mxk Ÿ kmk † 1 œ kmk . Also, y œ sin ˆ mx ‰
ˆ x ‰¸ Ÿ ¸ m" ¸ ¸cos ˆ mx ‰¸ Ÿ km" k Ê the largest value yw can attain is ¸ m" ¸ .
Ê yw œ m" cos ˆ mx ‰ Ê kyw k œ ¸ "
m cos m
(d) y œ sin (mx) Ê yw œ m cos (mx) Ê yw (0) œ m Ê slope of curve at the origin is m. Also, sin (mx) completes
m periods on [0ß 21]. Therefore the slope of the curve y œ sin (mx) at the origin is the same as the number
of periods it completes on [0ß 21]. In particular, for large m, we can think of “compressing" the graph of
y œ sin x horizontally which gives more periods completed on [0ß 21], but also increases the slope of the
graph at the origin.
67. x œ cos 2t, y œ sin 2t, 0 Ÿ t Ÿ 1
Ê cos# 2t sin# 2t œ 1 Ê x# y# œ 1
68. x œ cos (1 t), y œ sin (1 t), 0 Ÿ t Ÿ 1
Ê cos# (1 t) sin# (1 t) œ 1
Ê x# y# œ 1, y !
69. x œ 4 cos t, y œ 2 sin t, 0 Ÿ t Ÿ 21
70. x œ 4 sin t, y œ 5 cos t, 0 Ÿ t Ÿ 21
Ê
16 cos# t
16
4 sin# t
4
œ1 Ê
x#
16
y#
4
œ1
Ê
16 sin# t
16
25 cos# t
25
œ1 Ê
x#
16
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
y#
#5
œ1
153
154
Chapter 3 Differentiation
71. x œ 3t, y œ 9t# , _ t _ Ê y œ x#
72. x œ Èt , y œ t, t
0 Ê x œ È y
#
or y œ x , x Ÿ 0
73. x œ 2t 5, y œ 4t 7, _ t _
Ê x 5 œ 2t Ê 2(x 5) œ 4t
Ê y œ 2(x 5) 7 Ê y œ 2x 3
75. x œ t, y œ È1 t# , 1 Ÿ t Ÿ 0
Ê y œ È1 x#
#
Ê y œ 2 23 x, ! Ÿ x Ÿ $
76. x œ Èt 1, y œ Èt, t 0
Ê y# œ t Ê x œ Èy# 1, y
77. x œ sec# t 1, y œ tan t, 1# t #
74. x œ 3 3t, y œ 2t, 0 Ÿ t Ÿ 1 Ê y# œ t
Ê x œ 3 3 ˆ y# ‰ Ê 2x œ 6 3y
#
Ê sec t 1 œ tan t Ê x œ y
1
#
78. x œ sec t, y œ tan t, 1# t #
#
#
0
1
#
#
Ê sec t tan t œ 1 Ê x y œ 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 3.5 The Chain Rule and Parametric Equations
79. (a) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 21
80. (a) x œ a sin t, y œ b cos t,
(b) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 21
(c) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 41
1
#
ŸtŸ
155
51
#
(b) x œ a cos t, y œ b sin t, 0 Ÿ t Ÿ 21
(c) x œ a sin t, y œ b cos t, 1# Ÿ t Ÿ 9#1
(d) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 41
(d) x œ a cos t, y œ b sin t, 0 Ÿ t Ÿ 41
81. Using a"ß $b we create the parametric equations x œ " at and y œ $ bt, representing a line which goes
through a"ß $b at t œ !. We determine a and b so that the line goes through a%ß "b when t œ ".
Since % œ " a Ê a œ &. Since " œ $ b Ê b œ %. Therefore, one possible parameterization is x œ " &t,
y œ $ %t, 0 Ÿ t Ÿ ".
82. Using a"ß $b we create the parametric equations x œ " at and y œ $ bt, representing a line which goes through
a"ß $b at t œ !. We determine a and b so that the line goes through a$ß #b when t œ ". Since $ œ " a Ê a œ %.
Since # œ $ b Ê b œ &. Therefore, one possible parameterization is x œ " %t, y œ $ &t, 0 Ÿ t Ÿ ".
83. The lower half of the parabola is given by x œ y# " for y Ÿ !. Substituting t for y, we obtain one possible
parameterization x œ t# ", y œ t, t Ÿ 0Þ
84. The vertex of the parabola is at a"ß "b, so the left half of the parabola is given by y œ x# #x for x Ÿ ". Substituting
t for x, we obtain one possible parametrization: x œ t, y œ t# #t, t Ÿ ".
85. For simplicity, we assume that x and y are linear functions of t and that the pointax, yb starts at a#ß $b for t œ ! and passes
through a"ß "b at t œ ". Then x œ fatb, where fa!b œ # and fa"b œ ".
Since slope œ ??xt œ "#
"! œ $, x œ fatb œ $t # œ # $t. Also, y œ gatb, where ga!b œ $ and ga"b œ ".
Since slope œ
?y
?t
"3
"!
œ
œ 4. y œ gatb œ %t $ œ $ %t.
One possible parameterization is: x œ # $t, y œ $ %t, t
!.
86. For simplicity, we assume that x and y are linear functions of t and that the pointax, yb starts at a"ß #b for t œ ! and
passes through a!ß !b at t œ ". Then x œ fatb, where fa!b œ " and fa"b œ !.
Since slope œ
Since slope œ
?x
?t
?y
?t
! a"b
"!
!#
"! œ
œ
œ
œ ", x œ fatb œ "t a"b œ " t. Also, y œ gatb, where ga!b œ # and ga"b œ !.
#. y œ gatb œ #t # œ # #t.
One possible parameterization is: x œ " t, y œ # #t, t
87. t œ
Ê
1
4 Ê
dy
dx ¹ tœ 1
x œ 2 cos
d# y
dx#
dyw /dt
dx/dt
œ cot
1
4
1
4
œ È2, y œ 2 sin
1
4
œ È2;
dx
dt
!.
œ 2 sin t,
dy
dt
œ 2 cos t Ê
dy
dx
œ 1; tangent line is y È2 œ 1 Šx È2‹ or y œ x œ
dy/dt
dx/dt
w
È
2 2 ; dy
dt
œ
2 cos t
2 sin t
œ cot t
œ csc# t
4
Ê
88. t œ
Ê
œ
21
3 Ê
dy
dx ¹ tœ 21
œ
csc# t
2 sin t
x œ cos
21
3
"
œ 2 sin
$t Ê
d# y
dx# ¹ tœ 21
œ È 2
4
È3 dx
21
3 œ # ; dt
È3
È 3 x # ‹œ
œ "# , y œ È3 cos
œ È3 ; tangent line is y Š
3
Ê
d# y
dx# ¹ tœ 1
œ sin t,
dy
dt
œ È3 sin t Ê
ˆ "# ‰‘ or y œ È3 x;
dy
dx
œ
È3 sin t
sin t
œ È3
d# y
dx#
œ
œ0
dyw
dt
œ0 Ê
"
œ 1; tangent line is
0
sin t
œ0
3
89. t œ
y
1
4
"
#
Ê xœ
1
4
,yœ
"
#
;
dx
dt
œ 1,
dy
dt
œ 1 † ˆx 4" ‰ or y œ x 4" ;
œ
dyw
dt
"
#Èt
Ê
dy
dx
œ
œ 4" t$Î# Ê
dy/dt
dx/dt
d# y
dx#
œ
œ
1
2È t
dyw /dt
dx/dt
Ê
dy
dx ¹ tœ 1
4
œ
#É "4
œ 4" t$Î# Ê
d# y
dx# ¹ tœ 1
œ 2
4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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156
Chapter 3 Differentiation
90. t œ 3 Ê x œ È3 1 œ 2, y œ È3(3) œ 3;
È
œ 3 Èt3t1 œ
dyw
dt
Ê
œ
dy
dx ¹ tœ3
3 È 3 1
È3(3)
œ
d# y
dx# ¹ tœ3
#
3t
œ 4t,
dx
dt
y 1 œ 1 † (x 5) or y œ x 4;
œ
1
3
1
3
Ê xœ
Ê
sin t
1cos t
93. t œ
Ê
1
3
sin
dy
dx ¹ tœ 1
œ
3
Ê y œ È3x œ
1
(1 cos t)#
1
2 Ê
dy
dx ¹ tœ 1
1È3
3
œ
1
3
sin ˆ 13 ‰
1cos ˆ 13 ‰
2;
d# y
dx# ¹ tœ 1
Ê
œ
Ê
3
2tÈ3t Èt 1
d# y
dx#
dyw
dt
œ 4t$ Ê
dy
dt
dyw
dt
œ 2t Ê
dy
dx
d# y
dx#
œ
œ
(3t)"Î# Ê
dy
dx
ˆ 3# ‰ (3t)"Î#
ˆ "# ‰ (t 1)"Î#
œ
Š 2tÈ3t3Èt b 1 ‹
œ tÈ33t
Š 2Èct1b 1 ‹
œ
dyw /dt
dx/dt
4t$
4t
œ
œ t# Ê
œ
2t
4t
"
#
dy
dx ¹ tœc1
d# y
dx# ¹ tœc1
Ê
œ (1)# œ 1; tangent line is
œ
dy
, y œ 1 cos 13 œ 1 "# œ "# ; dx
dt œ 1 cos t, dt œ sin t Ê
È
Š #3 ‹
È
œ ˆ " ‰ œ È3 ; tangent line is y "# œ È3 Šx 13 #3 ‹
#
œ
"
#
È3
#
(1 cos t)(cos t) (sin t)(sin t)
(1cos t)#
œ
1
1cos t
d# y
dx#
Ê
dyw /dt
dx/dt
œ
œ
dy
dx
œ
dy/dt
dx/dt
1 ‰
ˆ 1 cos
t
1 cos t
œ 4
x œ cos
œ cot
1
2
œ 0, y œ 1 sin
1
#
1
2
œ 2;
œ sin t,
dx
dt
œ 0; tangent line is y œ 2;
sec# t
2 sec# t tan t
œ
"
2 tan t
œ
"
#
cot t Ê
w
dy
dt
dy
dx ¹ tœc 1
4
y (1) œ "# (x 1) or y œ "# x "# ;
Ê
œ
œ
dy/dt
dx/dt
d# y
dx# ¹ tœc 1
œ
dyw
dt
œ
dy
dt
œ cos t Ê
œ csc# t Ê
94. t œ 14 Ê x œ sec# ˆ 14 ‰ 1 œ 1, y œ tan ˆ 14 ‰ œ 1;
dy
dx
3
#
3
2
Ê
œ
œ "3
91. t œ 1 Ê x œ 5, y œ 1;
92. t œ
dy
dt
œ 2; tangent line is y 3 œ 2[x (2)] or y œ 2x 1;
È3t 3 (t 1)"Î# ‘ 3Èt 1 3 (3t)"Î# ‘
#
œ "# (t 1)"Î# ,
dx
dt
"
#
dx
dt
#
d y
dx#
œ
dy
dx
#
csc t
sin t
œ
cos t
sin t
œ cot t
œ csc$ t Ê
d# y
dx# ¹ tœ 1
œ 1
2
œ 2 sec# t tan t,
dy
dt
œ sec# t
cot ˆ 14 ‰ œ #" ; tangent line is
œ "# csc# t Ê
d# y
dx#
œ
"# csc# t
2 sec# t tan t
œ "4 cot$ t
"
4
4
95. s œ A cos (21bt) Ê v œ
ds
dt
œ A sin (21bt)(21b) œ 21bA sin (21bt). If we replace b with 2b to double the
frequency, the velocity formula gives v œ 41bA sin (41bt) Ê doubling the frequency causes the velocity to
# #
double. Also v œ #1bA sin (21bt) Ê a œ dv
dt œ 41 b A cos (21bt). If we replace b with 2b in the
acceleration formula, we get a œ 161# b# A cos (41bt) Ê doubling the frequency causes the acceleration to
$ $
quadruple. Finally, a œ 41# b# A cos (21bt) Ê j œ da
dt œ 81 b A sin (21bt). If we replace b with 2b in the jerk
formula, we get j œ 641$ b$ A sin (41bt) Ê doubling the frequency multiplies the jerk by a factor of 8.
21
21
21 ‰
96. (a) y œ 37 sin 365
(x 101)‘ 25 Ê yw œ 37 cos 365
(x 101)‘ ˆ 365
œ
741
365
21
cos 365
(x 101)‘ .
The temperature is increasing the fastest when yw is as large as possible. The largest value of
21
21
cos 365
(x 101)‘ is 1 and occurs when 365
(x 101) œ 0 Ê x œ 101 Ê on day 101 of the year
( µ April 11), the temperature is increasing the fastest.
1
741
21
‘ 741
(b) yw (101) œ 74
365 cos 365 (101 101) œ 365 cos (0) œ 365 ¸ 0.64 °F/day
97. s œ (" 4t)"Î# Ê v œ
ds
dt
œ
v œ 2(" 4t)"Î# Ê a œ
dv
dt
"
#
(1 4t)"Î# (4) œ 2(1 4t)"Î# Ê v(6) œ 2(" % † 6)"Î# œ
"
#
2
5
m/sec;
œ † 2(1 4t)$Î# (4) œ 4(1 4t)$Î# Ê a(6) œ 4(1 4 † 6)$Î# œ 14#5 m/sec#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 3.5 The Chain Rule and Parametric Equations
98. We need to show a œ
œ
k
2È s
† kÈs œ
99. v proportional to
œ 2sk$Î# †
dx
dt
100. Let
k
Ès
kT
2
k
#
is constant: a œ
dv
dt
œ
dv
ds
†
ds
dt
and
dv
ds
œ
d
ds
ˆkÈs‰ œ
Ê aœ
k
2È s
dv
ds
†
†
ds
dt
ds
dt
œ
dv
ds
†v
which is a constant.
"
Ès
Ê vœ
k
Ès
for some constant k Ê
#
dv
ds
œ 2sk$Î# . Thus, a œ
œ k# ˆ s"# ‰ Ê acceleration is a constant times
œ f(x). Then, a œ
101. T œ 21É Lg Ê
œ
#
dv
dt
dT
dL
dv
dt
œ 21 †
œ
"
#É Lg
dv
dx
†
dx
dt
œ
†
"
g
œ
1
gÉ Lg
dv
dx
† f(x) œ
œ
1
ÈgL
d
dx
"
s#
œ
dv
ds
œ
dv
ds
†v
so a is inversely proportional to s# .
ˆ dx
‰
dt † f(x) œ
. Therefore,
dv
dt
dT
du
œ
d
dx
(f(x)) † f(x) œ f w (x)f(x), as required.
dT
dL
†
dL
du
œ
1
ÈgL
† kL œ
1 kÈ L
Èg
œ
"
#
† 21kÉ Lg
, as required.
102. No. The chain rule says that when g is differentiable at 0 and f is differentiable at g(0), then f ‰ g is
differentiable at 0. But the chain rule says nothing about what happens when g is not differentiable at 0 so
there is no contradiction.
103. The graph of y œ (f ‰ g)(x) has a horizontal tangent at x œ 1 provided that (f ‰ g)w (1) œ 0 Ê f w (g(1))gw (1) œ 0
Ê either f w (g(1)) œ 0 or gw (1) œ 0 (or both) Ê either the graph of f has a horizontal tangent at u œ g(1), or the
graph of g has a horizontal tangent at x œ 1 (or both).
104. (f ‰ g)w (5) 0 Ê f w (g(5)) † gw (5) 0 Ê f w (g(5)) and gw (5) are both nonzero and have opposite signs.
That is, either cf w (g(5)) 0 and gw (5) 0d or cf w (g(5)) 0 and gw (5) 0d .
105. As h Ä 0, the graph of y œ
sin 2(xh)sin 2x
h
approaches the graph of y œ 2 cos 2x because
lim
hÄ!
sin 2(xh)sin 2x
h
œ
d
dx
(sin 2x) œ 2 cos 2x.
106. As h Ä 0, the graph of y œ
cos c(x h)# dcos ax# b
h
#
approaches the graph of y œ 2x sin ax b because
lim
hÄ!
cos c(x h)# dcos ax# b
h
œ
d
dx
ccos ax# bd œ 2x sin ax# b.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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157
158
107.
Chapter 3 Differentiation
dx
dt
œ cos t and
œ 2 cos 2t Ê
dy
dt
dy
dx
Ê 2 cos# t 1 œ 0 Ê cos t œ „
y œ sin 2 ˆ 14 ‰ œ 1 Ê Š
È2
# ß 1‹
œ
dy/dt
dx/dt
"
È2
œ
œ
2 cos 2t
cos t
Ê tœ
1
4
31
4
,
2 a2 cos# t 1b
cos t
51
4
,
108.
dx
dt
œ 2 cos 2t and
dy
dx ¹ tœ0
3 ca2 cos# t 1b (cos t) 2 sin t cos t sin td
2 a2 cos# t1b
œ
dy
dx
œ 3 cos 3t Ê
dy
dt
(3 cos t) a4 cos# t 3b
2 a2 cos# t 1b
œ0 Ê
and y œ sin 3 ˆ 16 ‰ œ 1 Ê Š
œ
œ 2 Ê y œ 2x and
dy
dx
œ
œ
dy/dt
dx/dt
3 cos 3t
2 cos 2t
È3
#
È3
# ß 1‹
1
6
Ê tœ
,
51
6
,
71
6
œ
"
#É È x
†
d
dx
ˆÈx‰ œ
"
#ÉÈx
†
"
#Èx
œ
dy
dx
œ
"
#É x È x
3È x
4È x É È x
œ
3
4
†
d
dx
ˆxÈx‰ Ê
dy
dx
œ
dy
dx
œ
110. From the power rule, with y œ x$Î% , we get
Ê
œ
df
dt
œ
(3 cos t) a4 cos# t 3b
2 a2 cos# t1b
,
111
6
. In the 1st quadrant: t œ
1
6
1
#
,
31
#
and
and
Ê x œ sin 2 ˆ 16 ‰ œ
1
#
, 1,
31
#
dy
dx ¹ tœ0
and t œ 0,
œ
3 cos 0
2 cos 0
1
3
œ
,
21
3
3
#
, 1,
41
3
Ê yœ
,
51
3
3
#
È3
#
x, and
Ê t œ 0 and t œ 1 give
dy
dx ¹ tœ1
"
4
dy
dx
"
#ÉxÈx
œ
"
4
x$Î% . From the chain rule, y œ ÉÈx
x$Î% , in agreement.
œ
3
4
x"Î% . From the chain rule, y œ ÉxÈx
† Šx †
"
#È x
È x‹ œ
"
#ÉxÈx
† ˆ 3# Èx‰ œ
3È x
4É xÈ x
œ 1.27324 sin 2t 0.42444 sin 6t 0.2546 sin 10t 0.18186 sin 14t
df
dt
È2
#
; then
x"Î% , in agreement.
(c) The curve of y œ
œ
3(cos 2t cos t sin 2t sin t)
2 a2 cos# t1b
111. (a)
(b)
1
4
1 give the tangent lines at
œ 3# Ê y œ 3# x
109. From the power rule, with y œ x"Î% , we get
dy
dx
Ê x œ sin
is the point where the graph has a horizontal tangent. At the origin: x œ 0
the tangent lines at the origin. Tangents at the origin:
Ê
1
31
# , 1, # ; thus t œ 0 and t œ
dy
dx ¹ tœ1 œ 2 Ê y œ 2x
(3 cos t) a2 cos# t 1 2 sin# tb
2 a2 cos# t1b
and y œ 0 Ê sin 2t œ 0 and sin 3t œ 0 Ê t œ 0,
3 cos (31)
2 cos (21)
1
4
. In the 1st quadrant: t œ
œ 0 Ê 3 cos t œ 0 or 4 cos# t 3 œ 0: 3 cos t œ 0 Ê t œ
4 cos# t 3 œ 0 Ê cos t œ „
œ
71
4
œ0
is the point where the tangent line is horizontal. At the origin: x œ 0 and y œ 0
Ê sin t œ 0 Ê t œ 0 or t œ 1 and sin 2t œ 0 Ê t œ 0,
the origin. Tangents at origin:
,
2 a2 cos# t 1b
cos t
œ0 Ê
dy
dx
; then
approximates y œ
dg
dt
the best when t is not 1, 1# , 0, 1# , nor 1.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 3.5 The Chain Rule and Parametric Equations
112. (a)
(b)
dh
dt
œ 2.5464 cos (2t) 2.5464 cos (6t) 2.5465 cos (10t) 2.54646 cos (14t) 2.54646 cos (18t)
(c)
111-116. Example CAS commands:
Maple:
f := t -> 0.78540 - 0.63662*cos(2*t) - 0.07074*cos(6*t)
- 0.02546*cos(10*t) - 0.01299*cos(14*t);
g := t -> piecewise( t<-Pi/2, t+Pi, t<0, -t, t<Pi/2, t, Pi-t );
plot( [f(t),g(t)], t=-Pi..Pi );
Df := D(f);
Dg := D(g);
plot( [Df(t),Dg(t)], t=-Pi..Pi );
Mathematica: (functions, domains, and value for t0 may change):
To see the relationship between f[t] and f'[t] in 111 and h[t] in 112
Clear[t, f]
f[t_] = 0.78540 0.63662 Cos[2t] 0.07074 Cos[6t] 0.02546 Cos[10t] 0.01299 Cos[14t]
f'[t]
Plot[{f[t], f'[t]},{t, 1, 1}]
For the parametric equations in 113 - 116, do the following. Do NOT use the colon when defining tanline.
Clear[x, y, t]
t0 = p/4;
x[t_]:=1Cos[t]
y[t_]:=1 Sin[t]
p1=ParametricPlot[{x[t], y[t]},{t, 1, 1}]
yp[t_]:=y'[t]/x'[t]
ypp[t_]:=yp'[t]/x'[t]
yp[t0]//N
ypp[t0]//N
tanline[x_]=y[t0] yp[t0] (x x[t0])
p2=Plot[tanline[x], {x, 0, 1}]
Show[p1, p2]
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159
160
Chapter 3 Differentiation
3.6 IMPLICIT DIFFERENTIATION
1. y œ x*Î% Ê
dy
dx
œ
x&Î%
9
4
3
3. y œ È
2x œ (2x)"Î$ Ê
2. y œ x$Î& Ê
"
3
œ
dy
dx
(2x)#Î$ † 2 œ
5. y œ 7Èx 6 œ 7(x 6)"Î# Ê
œ
dy
dx
6. y œ 2Èx 1 œ 2(x 1)"Î# Ê
7. y œ (2x 5)"Î# Ê
8. y œ (" 6x)#Î$ Ê
9. y œ x ax# 1b
10. y œ x ax# 1b
"Î#
dy
dx
7 #
11. s œ È
t œ t#Î( Ê
7
2È x 6
(1 6x)"Î$ (6) œ 4(1 6x)"Î$
2
3
œ
"Î#
a#xb ax# 1b
$Î#
dy
dt
14. z œ cos ˆ(" 6t)#Î$ ‰ Ê
dz
dt
† " œ ax# 1b
"Î#
"Î#
ax# x# "b œ
† " œ ax# 1b
4
12. r œ È
)$ œ )$Î% Ê
$Î#
dr
d)
2x# 1
È x# 1
ax# x# "b œ
œ 43 )(Î%
œ sin ˆ(" 6t)#Î$ ‰ † 23 (1 6t)"Î$ (') œ 4(1 6t)"Î$ sin ˆ(1 6t)#Î$ ‰
"Î#
Ê f w (x) œ
"
#
ˆ1 x"Î# ‰"Î# ˆ #" x"Î# ‰ œ
Ê gw (x) œ 23 ˆ2x"Î# 1‰
3
17. h()) œ È
1 cos (2)) œ (1 cos 2))"Î$ Ê hw ()) œ
18. k()) œ (sin () 5))&Î% Ê kw ()) œ
5
4
"
3
%Î$
† (1)x$Î# œ
2
3
"
4 ŠÉ1 Èx‹ Èx
œ
"
4 É x ˆ1 È x ‰
ˆ2x"Î# 1‰%Î$ x$Î#
(1 cos 2))#Î$ † (sin 2)) † 2 œ 23 (sin 2))(1 cos 2))#Î$
(sin () 5))"Î% † cos () 5) œ
5
4
cos () 5)(sin () 5))"Î%
19. x# y xy# œ 6:
Step 1:
Šx #
Step 2:
x#
dy
dx
Step 3:
dy
dx
dy
dx
ax# 2xyb œ 2xy y#
Step 4:
dy
dx
œ
y † 2x‹ Šx † 2y
2xy
dy
dx
dy
dx
y# † 1‹ œ 0
œ 2xy y#
2xyy#
x# 2xy
20. x$ y$ œ 18xy Ê 3x# 3y#
dy
dx
œ 18y 18x
dy
dx
Ê a3y# 18xb
dy
dx
œ 18y 3x# Ê
dy
dx
œ
21. 2xy y# œ x y:
Step 1:
Š2x
dy
dx
"
ax# 1b$Î#
œ cos ˆ(2t 5)#Î$ ‰ † ˆ 23 ‰ (2t 5)&Î$ † 2 œ 43 (2t 5)&Î$ cos ˆ(2t 5)#Î$ ‰
15. f(x) œ É1 Èx œ ˆ1 x"Î# ‰
"Î$
"Î#
a#xb ax# 1b
2 &Î(
7 t
13. y œ sin ˆ(2t 5)#Î$ ‰ Ê
16. g(x) œ 2 ˆ2x"Î# 1‰
5"Î%
4x$Î%
(5x)$Î% † 5 œ
x1
Ê yw œ x † ˆ "# ‰ax# 1b
ds
dt
"
4
œ 1(x 1)"Î# œ È "
Ê yw œ x † "# ax# 1b
"Î#
œ
dy
dx
œ "# (2x 5)$Î# † 2 œ (2x 5)$Î#
œ
dy
dx
4
4. y œ È
5x œ (5x)"Î% Ê
(x 6)"Î# œ
7
#
dy
dx
2"Î$
3x#Î$
œ 35 x)Î&
dy
dx
2y‹ 2y
dy
dx
œ1
dy
dx
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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6y x#
y# 6x
Section 3.6 Implicit Differentiation
Step 2:
2x
Step 3:
dy
dx
dy
dx
Step 4:
dy
dx
2y
dy
dx
dy
dx
161
œ 1 2y
(2x 2y 1) œ " 2y
œ
1 2y
2x 2y 1
22. x$ xy y$ œ 1 Ê 3x# y x
dy
dx
3y#
œ 0 Ê a3y# xb
dy
dx
dy
dx
œ y 3x# Ê
dy
dx
y 3x#
3y# x
œ
23. x# (x y)# œ x# y# :
Step 1:
x# ’2(x y) Š1 Step 2:
2x# (x y)
Step 3:
dy
dx
dy
dx
Step 4:
dy
dx
(x y)# (2x) œ 2x 2y
dy
dx ‹“
2y
dy
dx
œ 2x 2x# (x y) 2x(x y)#
c2x# (x y) 2yd œ 2x c1 x(x y) (x y)# d
œ
œ
2x c1 x(x y) (x y)# d
2x# (x y) 2y
œ
dy
dx
x c1 x(x y) (x y)# d
y x# (x y)
œ
x a1 x# xy x# 2xy y# b
x# y x$ y
x 2x$ 3x# y xy#
x# y x$ y
24. (3xy 7)# œ 6y Ê 2(3xy 7) † Š3x
Ê
dy
dx
dy
dx
3y‹ œ 6
[6x(3xy 7) 6] œ 6y(3xy 7) Ê
(x 1) (x 1)
(x 1)#
dy
dx
dy
dx
Ê 2(3xy 7)(3x)
y(3xy 7)
œ x(3xy
7) 1 œ
dy
dx
6
dy
dx
œ 6y(3xy 7)
#
3xy 7y
1 3x# y 7x
25. y# œ
x"
x1
Ê 2y
26. x# œ
xy
xy
Ê x$ x# y œ x y Ê 3x# 2xy x# yw œ 1 yw Ê ax# 1b yw œ 1 3x# 2xy Ê yw œ
dy
dx
œ
27. x œ tan y Ê 1 œ asec# yb
Ê
dy
dx
œ
dy
dx
Ê
2
(x 1)#
œ
"
sec# y
œ
dy
dx
"
y(x 1)#
1 3x# 2xy
x# 1
œ cos# y
dy
dy
dy
#
#
#
28. xy œ cot axyb Ê x dy
dx y œ csc (xy)Šx dx y‹ Ê x dx x csc (xy) dx œ y csc (xy) y
Ê
dy dx x
x csc# (xy)‘ œ y csc# (xy) "‘ Ê
29. x tan (xy) œ ! Ê 1 csec# (xy)d Šy x
œ
1
x sec# (xy)
y
x
œ
cos# (xy)
x
y
x
30. x sin y œ xy Ê 1 (cos y)
œ
dy
dx
dy
dx ‹
œyx
’ "y cos Š "y ‹ sin Š y" ‹ x“ œ y Ê
’sin Š y" ‹ 2y cos Š y" ‹ 2“ œ 2 Ê
33. )"Î# r"Î# œ 1 Ê
"
#
y csc# (xy) "‘
x" csc# (xy)‘
œ yx
œ 0 Ê x sec# (xy)
Ê (cos y x)
dy
dx
"
y#
dy
dx
†
œ
)"Î# "# r"Î# †
dy
dx
dr
d)
œ
dy
dx
sin Š y" ‹ †
dy
dx “
32. y# cos Š "y ‹ œ 2x 2y Ê y# ’sin Š y" ‹ † (1)
dy
dx
œ
dy
dx
œ 1 y sec# (xy) Ê
cos# (xy) y
x
31. y sin Š "y ‹ œ 1 xy Ê y ’cos Š y" ‹ † (1)
dy
dx
dy
dx
dy
dx
œy1 Ê
œ x
y
"
y#
"
y
†
cos Š "y ‹ sin Š "y ‹ x
dy
dx “
œ
cos Š y" ‹ † 2y
dy
dx
dy
dx
y1
cos y x
œ
y Ê
y #
y
sin Š "y ‹ cos Š "y ‹ xy
dy
dx
œ22
dy
dx
Ê
2
sin Š "y ‹ 2y cos Š "y ‹ #
œ0 Ê
dr
d)
"
’ #È
“œ
r
"
#È )
Ê
dr
d)
œ
2È r
2È )
Èr
œÈ
)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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dy
dx
œ
" y sec# (xy)
x sec# (xy)
162
Chapter 3 Differentiation
34. r 2È) œ
"
#
35. sin (r)) œ
)#Î$ 43 )$Î% Ê
3
#
)"Î# œ )"Î$ )"Î% Ê
dr
d)
Ê [cos (r))] ˆr )
dr ‰
d)
œ0 Ê
dr
d)
œ )"Î# )"Î$ )"Î%
dr
d)
[) cos (r))] œ r cos (r)) Ê
r cos (r))
) cos (r))
œ
dr
d)
œ )r ,
cos (r)) Á 0
36. cos r cot ) œ r) Ê (sin r)
dr
d)
csc# ) œ r )
37. x# y# œ 1 Ê 2x 2yyw œ 0 Ê 2yyw œ 2x Ê
y(1)xyw
y#
Ê yww œ
yx Š xy ‹
œ
38. x#Î$ y#Î$ œ 1 Ê
y#
2
3
œ0 Ê
dy
dx
dy
dx
d# y
dx#
since yw œ xy Ê
x"Î$ 23 y"Î$
Ê
dr
d)
dy
dx
Ê
œ
"
3
#Î$ "Î$
x
y
2x 2
2y
39. y# œ x# 2x Ê 2yyw œ 2x 2 Ê yw œ
d# y
dx#
Ê
y# (x 1)#
y$
ww
œy œ
x1
y
œ
œ yw œ xy ; now to find
y # x #
y$
d# y
dx#
y# a"y# b
y$
œ
d
dx
,
œ
#
csc )
œ rsin
r)
dr
d)
ayw b œ
d
dx
Š xy ‹
"
y$
dy
dx
"Î$
"Î$
œ yx"Î$ œ ˆ yx ‰
"Î$
x#Î$
y (x 1)yw
y#
; then yww œ
"
y1
œ
y (x 1) Š xy 1 ‹
y#
œ (y 1)" ; then yww œ (y 1)# † yw
"
(y 1)$
œ yww œ
41. 2Èy œ x y Ê y"Î# yw œ 1 yw Ê yw ˆy"Î# 1‰ œ 1 Ê
dy
dx
œ yw œ
"
y"Î# 1
Èy
Èy 1
œ
; we can
differentiate the equation yw ˆy"Î# 1‰ œ 1 again to find yww : yw ˆ "# y$Î# yw ‰ ˆy"Î# 1‰ yww œ 0
Ê ˆy"Î# 1‰ yww œ
"
#
w # $Î#
Ê
cy d y
d# y
dx#
"
#
œ yww œ
#
"
$Î#
Œ y"Î# 1 y
ay"Î# 1b
œ
"
$
2y$Î# ay"Î# 1b
œ
"
$
# ˆ1 È y ‰
42. xy y# œ 1 Ê xyw y 2yyw œ 0 Ê xyw 2yyw œ y Ê yw (x 2y) œ y Ê yw œ
œ
(x 2y)yw y(1 2yw )
(x 2y)#
œ
2y(x 2y) 2y#
(x 2y)$
œ
œ
y
y
(x 2y) ’ (x 2y) “ y ’1 2 Š (x 2y) ‹“
(x 2y)#
2y# 2xy
(x 2y)$
œ
œ
"
(x 2y)
y
(x2y)
;
d# y
dx#
œ yww
cy(x 2y) y(x 2y) 2y# d
(x 2y)#
2y(x y)
(x 2y)$
#
43. x$ y$ œ 16 Ê 3x# 3y# yw œ 0 Ê 3y# yw œ 3x# Ê yw œ xy# ; we differentiate y# yw œ x# to find yww :
# ww
w
w
w #
# ww
y y y c2y † y d œ 2x Ê y y œ 2x 2y cy d
œ
2xy$ 2x%
y&
Ê
d# y
dx# ¹ (2ß2)
œ
32 32
32
ww
Ê y œ
2x 2y Š
y#
44. xy y# œ 1 Ê xyw y 2yyw œ 0 Ê yw (x 2y) œ y Ê yw œ
since yw k (0 1) œ "# we obtain yww k (0ß1) œ
ß
#
x#
‹
y#
œ
2x y#
2x%
y$
œ 2
(2) ˆ "# ‰ (1)(0)
4
45. y# x# œ y% 2x at (#ß ") and (#ß 1) Ê 2y
dy
dx
;
x"Î$ †ˆ "3 y#Î$ ‰ Œ y"Î$ y"Î$ ˆ "3 x#Î$ ‰
x
40. y# 2x œ 1 2y Ê 2y † yw 2 œ 2yw Ê yw (2y 2) œ 2 Ê yw œ
œ (y 1)# (y 1)" Ê
d# y
dx#
23 y"Î$ ‘ œ 23 x"Î$ Ê yw œ
x"Î$ †ˆ 3" y#Î$ ‰ yw y"Î$ ˆ "3 x#Î$ ‰
œ
x#Î$
"Î$
y
" "Î$ %Î$
"
x
œ 3x
%Î$ 3y"Î$ x#Î$
3 y
[sin r )] œ r csc# ) Ê
œ yww œ
Differentiating again, yww œ
d# y
dx#
dr
d)
y
(x2y)
Ê yww œ
(x 2y) ayw b (y) a1 2yw b
(x 2y)#
œ 4"
2x œ 4y$
dy
dx
2 Ê 2y
dy
dx
4y$
dy
dx
œ 2 2x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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;
Section 3.6 Implicit Differentiation
Ê
dy
dx
a2y 4y$ b œ 2 2x Ê
dy
dx
œ
x"
#y $ y
Ê
dy
dx ¹ (2ß1)
œ 1 and
#
46. ax# y# b œ (x y)# at("ß !) and ("ß 1) Ê 2 ax# y# b Š2x 2y
Ê
and
dy
dx
c2y ax# y# b (x y)d œ 2x ax# y# b (x y) Ê
dy
dx ¹ (1ßc1)
dy
dx
dy
dx ¹ (2ß1)
dy
dx ‹
œ
œ1
œ 2(x y) Š1 2x ax# y# b (x y)
2y ax# y# b (x y)
dy
dx ‹
Ê
dy
dx ¹ (1ß0)
œ 1
œ1
47. x# xy y# œ 1 Ê 2x y xyw 2yyw œ 0 Ê (x 2y)yw œ 2x y Ê yw œ
(a) the slope of the tangent line m œ yw k (2 3) œ
ß
Ê the tangent line is y 3 œ
7
4
(b) the normal line is y 3 œ 47 (x 2) Ê y œ 47 x 7
4
2x y
2y x
;
(x 2) Ê y œ
7
4
x
"
#
29
7
48. x# y# œ 25 Ê 2x 2yyw œ 0 Ê yw œ xy ;
(a) the slope of the tangent line m œ yw k (3 4) œ xy ¹
ß
Ê yœ
3
4
x
(3ß4)
œ
Ê the tangent line is y 4 œ
3
4
3
4
(x 3)
25
4
(b) the normal line is y 4 œ 43 (x 3) Ê y œ 43 x
49. x# y# œ 9 Ê 2xy# 2x# yyw œ 0 Ê x# yyw œ xy# Ê yw œ yx ;
(a) the slope of the tangent line m œ yw k (1ß3) œ yx ¸ (1ß3) œ 3 Ê the tangent line is y 3 œ 3(x 1)
Ê y œ 3x 6
(b) the normal line is y 3 œ "3 (x 1) Ê y œ 3" x 8
3
50. y# 2x 4y " œ ! Ê 2yyw 2 4yw œ 0 Ê 2(y 2)yw œ 2 Ê yw œ
"
y#
;
w
(a) the slope of the tangent line m œ y k (2ß1) œ 1 Ê the tangent line is y 1 œ 1(x 2) Ê y œ x 1
(b) the normal line is y 1 œ 1(x 2) Ê y œ x 3
51. 6x# 3xy 2y# 17y 6 œ 0 Ê 12x 3y 3xyw 4yyw 17yw œ 0 Ê yw (3x 4y 17) œ 12x 3y
12x 3y
Ê yw œ 3x
4y 17 ;
(a) the slope of the tangent line m œ yw k (1ß0) œ
Ê yœ
6
7
x
"2x 3y
3x 4y 17 ¹ (1ß0)
œ
6
7
Ê the tangent line is y 0 œ
6
7
(x 1)
6
7
(b) the normal line is y 0 œ 76 (x 1) Ê y œ 76 x 7
6
52. x# È3xy 2y# œ 5 Ê 2x È3xyw È3y 4yyw œ 0 Ê yw Š4y È3x‹ œ È3y 2x Ê yw œ
(a) the slope of the tangent line m œ yw k ŠÈ3 2‹ œ
ß
È3y 2x
¹
4y È3x ŠÈ3ß2‹
œ 0 Ê the tangent line is y œ 2
(b) the normal line is x œ È3
53. 2xy 1 sin y œ 21 Ê 2xyw 2y 1(cos y)yw œ 0 Ê yw (2x 1 cos y) œ 2y Ê yw œ
(a) the slope of the tangent line m œ yw k ˆ1 12 ‰ œ
ß
2y
2x 1 cos y ¹ ˆ1ß 1 ‰
2
y
1
#
2y
2x 1 cos y
œ 1# Ê the tangent line is
œ 1# (x 1) Ê y œ 1# x 1
(b) the normal line is y 1
#
œ
2
1
(x 1) Ê y œ
2
1
x
2
1
1
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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;
È3y 2x
4y È3x
;
163
164
Chapter 3 Differentiation
54. x sin 2y œ y cos 2x Ê x(cos 2y)2yw sin 2y œ 2y sin 2x yw cos 2x Ê yw (2x cos 2y cos 2x)
œ sin 2y 2y sin 2x Ê yw œ
sin 2y 2y sin 2x
cos 2x 2x cos 2y
(a) the slope of the tangent line m œ yw k ˆ 14
ß
1‰
2
œ
;
sin 2y 2y sin 2x
cos 2x 2x cos 2y ¹ ˆ 1 ß 1 ‰
œ
4 2
1
#
y
œ 2 ˆx 14 ‰ Ê y œ 2x
(b) the normal line is y 1
#
œ "# ˆx 14 ‰ Ê y œ "# x 1
1
#
œ 2 Ê the tangent line is
51
8
55. y œ 2 sin (1x y) Ê yw œ 2 [cos (1x y)] † a1 yw b Ê yw [1 2 cos (1x y)] œ 21 cos (1x y)
Ê yw œ
21 cos (1x y)
1 # cos (1x y)
;
(a) the slope of the tangent line m œ yw k (1 0) œ
ß
21 cos (1x y)
1 2 cos (1x y) ¹(1ß0)
y 0 œ 21(x 1) Ê y œ 21x 21
(b) the normal line is y 0 œ #"1 (x 1) Ê y œ 2x1 œ 21 Ê the tangent line is
"
#1
56. x# cos# y sin y œ 0 Ê x# (2 cos y)(sin y)yw 2x cos# y yw cos y œ 0 Ê yw c2x# cos y sin y cos yd
2x cos# y
2x# cos y sin y cos y
œ 2x cos# y Ê yw œ
;
(a) the slope of the tangent line m œ yw k (0 1) œ
ß
2x cos# y
2x# cos y sin y cos y ¹ (0ß1)
œ 0 Ê the tangent line is y œ 1
(b) the normal line is x œ 0
57. Solving x# xy y# œ 7 and y œ 0 Ê x# œ 7 Ê x œ „ È7 Ê ŠÈ7ß !‹ and ŠÈ7ß !‹ are the points where the
curve crosses the x-axis. Now x# xy y# œ 7 Ê 2x y xyw 2yyw œ 0 Ê (x 2y)yw œ 2x y
2 È 7
y
2x y
È
È
Ê yw œ 2x
x 2y Ê m œ x 2y Ê the slope at Š 7ß !‹ is m œ È7 œ 2 and the slope at Š 7ß !‹ is
È
m œ 2È77 œ 2. Since the slope is 2 in each case, the corresponding tangents must be parallel.
58. x# xy y# œ 7 Ê 2x y x
dy
dx
(a) Solving
dy
dx
2y
dy
dx
œ 0 Ê (x 2y)
dy
dx
œ 2x y Ê
dy
dx
œ
2x y
x 2y
and
dx
dy
œ
x 2y
2x y
;
œ 0 Ê 2x y œ 0 Ê y œ 2x and substitution into the original equation gives
x# x(2x) (2x)# œ 7 Ê 3x# œ 7 Ê x œ „ É 73 and y œ … 2É 73 when the tangents are parallel to the
x-axis.
dx
dy
(b) Solving
#
œ 0 Ê x 2y œ 0 Ê y œ x# and substitution gives x# x ˆ x# ‰ ˆ x# ‰ œ 7 Ê
3x#
4
œ7
Ê x œ „ 2É 73 and y œ … É 73 when the tangents are parallel to the y-axis.
59. y% œ y# x# Ê 4y$ yw œ 2yyw 2x Ê 2 a2y$ yb yw œ 2x Ê yw œ y x2y$ ; the slope of the tangent line at
È3
"
È È
È
Š 43 ß #3 ‹ is y x2y$ ¹ È3 È3 œ È3 4 6È3 œ " 4 3 œ # " 3 œ 1; the slope of the tangent line at Š 43 ß #" ‹
4
#
Œ
is
x
y2y$ ¹
Œ
È3
4
ß
1
2
œ
È3
"
#
4
28
4
ß
2
œ
2È 3
42
#
8
œ È3
60. y# (2 x) œ x$ Ê 2yyw (2 x) y# (1) œ 3x# Ê yw œ
mœ
#
#
y 3x
2y(2 x) ¹ (1ß1)
œ
4
#
y# 3x#
2y(2 x)
; the slope of the tangent line is
œ 2 Ê the tangent line is y 1 œ 2(x 1) Ê y œ 2x 1; the normal line is
y 1 œ "# (x 1) Ê y œ "# x 3
#
61. y% 4y# œ x% 9x# Ê 4y$ yw 8yyw œ 4x$ 18x Ê yw a4y$ 8yb œ 4x$ 18x Ê yw œ
4x$ 18x
4y$ 8y
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
œ
2x$ 9x
2y$ 4y
Section 3.6 Implicit Differentiation
x a2x# 9b
y a2y# 4b
œ
œ m; (3ß 2): m œ
(3)(18 9)
2(8 4)
œ 27
8 ; ($ß #): m œ
; (3ß #): m œ
27
8
27
8
; (3ß #): m œ 27
8
62. x$ y$ 9xy œ 0 Ê 3x# 3y# yw 9xyw 9y œ 0 Ê yw a3y# 9xb œ 9y 3x# Ê yw œ
(a) yw k (4 2) œ
ß
and yw k (2 4) œ
5
4
ß
#
3y x
y# 3x
(b) yw œ 0 Ê
4
5
165
9y 3x#
3y# 9x
œ
3y x#
y# 3x
;
x#
3
œ 0 Ê 3y x# œ 0 Ê y œ
$
#
#
Ê x$ Š x3 ‹ 9x Š x3 ‹ œ 0 Ê x' 54x$ œ 0
Ê x$ ax$ 54b œ 0 Ê x œ 0 or x œ $È54 œ 3 $È2 Ê there is a horizontal tangent at x œ 3 $È2 . To find the
corresponding y-value, we will use part (c).
dx
dy
(c)
œ0 Ê
$
y# 3x
3y x#
œ 0 Ê y# 3x œ 0 Ê y œ „ È3x ; y œ È3x Ê x$ ŠÈ3x‹ 9xÈ3x œ 0
Ê x$ 6È3 x$Î# œ 0 Ê x$Î# Šx$Î# 6È3‹ œ 0 Ê x$Î# œ 0 or x$Î# œ 6È3 Ê x œ 0 or x œ $È108 œ 3 $È4 .
Since the equation x$ y$ 9xy œ 0 is symmetric in x and y, the graph is symmetric about the line y œ x.
That is, if (aß b) is a point on the folium, then so is (bß a). Moreover, if yw k (a b) œ m, then yw k (b a) œ m" .
ß
ß
Thus, if the folium has a horizontal tangent at (aß b), it has a vertical tangent at (bß a) so one might expect
that with a horizontal tangent at x œ $È54 and a vertical tangent at x œ 3 $È4, the points of tangency are
Š $È54ß 3 $È4‹ and Š3 $È4ß $È54‹, respectively. One can check that these points do satisfy the equation
x$ y$ 9xy œ 0.
63. x# 2tx 2t# œ 4 Ê 2x
2y$ 3t# œ 4 Ê 6y#
#
dx
dt
2x 2t
6t œ 0 Ê
dy
dt
#
dx
dt
4t œ 0 Ê (2x 2t)
œ
dy
dt
6t
6y#
œ
t
y#
; thus
œ
dy
dx
#
œ 2x 4t Ê
dx
dt
dy/dt
dx/dt
œ
#
Š yt# ‹
ˆ xx2tt ‰
œ
dx
dt
œ
t(xt)
y# (x2t)
2x4t
2x2t
œ
x2t
x t
;
;tœ2
Ê x 2(2)x 2(2) œ 4 Ê x 4x 4 œ 0 Ê (x 2) œ 0 Ê x œ 2; t œ 2 Ê 2y$ 3(2)# œ 4
Ê 2y$ œ 16 Ê y$ œ 8 Ê y œ 2; therefore
64. x œ É5 Èt Ê
Ê at 1b
dy
dt
œ
"
#È t
therefore,
dy
dx ¹ tœ4
œ
dy
dt
dy
dt
œ
"Î#
"
#Èt y
at 1 b
œ
" #y È t
#tÈt 2Èt
œ
"4
dx
dt
3x"Î#
dx
dt
œ
dy
t
Èy ‹ dt
dy
dx
œ 2t 1 Ê ˆ1 3x"Î# ‰
#
œ
y
2Èt1
Œ 2Èy (t b 1) b 2tÈt b 1 Š
" "Î#
; y(t 1) œ Èt Ê y (t 1) dy
dt œ # t
œ
dy
dt
dx
dt
" #yÈt
" #yÈt 4Èt É5 Èt
#tÈt 2Èt
œ
œ È
†
"
"
# tat" b
4È t É 5 È t
14
3
#
œ
"
t œ 4 Ê x œ É5 È4 œ È3; t œ 4 Ê y(3) œ È4 œ 2
2Èy Ê
dy
dt
œ
cyÈy c 4yÈt b 1
dy/dt
dx/dt
œ0
4È t É 5 È t
; thus
Èt 1 y ˆ " ‰ (t 1)"Î# 2Èy 2t ˆ " y"Î# ‰
Ê ŠÈ t 1 dy
dx
2(2 2)
(2)# (2 2(2))
œ
ˆ "# t"Î# ‰ œ 2Š" 2a2bÈ4‹É& È4
65. x 2x$Î# œ t# t Ê
Ê
ˆ5 Èt‰
y Ê
#ˆ" #yÈt‰É& Èt
;
"t
œ
"
#
œ
dx
dt
dy
dx ¹ tœ2
2t b 1
‹
1 b 3x"Î#
dy
dt
dx
dt
œ 2t 1 Ê
œ0 Ê
Š 2Èct yb 1 2Èy‹
ŠÈt 1 Èt y ‹
œ
dy
dt
œ
dx
dt
Èt 1 2t1
13x"Î#
y
2È t 1
yÈy 4yÈt 1
2Èy (t 1) 2tÈt 1
; yÈt 1 2tÈy œ 4
2Èy Š Èt y ‹
; thus
; t œ 0 Ê x 2x$Î# œ 0 Ê x ˆ1 2x"Î# ‰ œ 0 Ê x œ 0; t œ 0
4È4 4(4)È0 1
Ê yÈ0 1 2(0)Èy œ 4 Ê y œ 4; therefore
dy
dx ¹ tœ0
œ
Œ 2È4(0 1) 2(0)È0 1 2(0) 1
Œ 1 3(0)"Î# œ 6
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dy
dt
œ0
166
Chapter 3 Differentiation
66. x sin t 2x œ t Ê
dx
dt
sin t x cos t 2
t sin t 2t œ y Ê sin t t cos t 2 œ
Ê xœ
1
#
; therefore
œ
dy
dx ¹ tœ1
œ 1 Ê (sin t 2)
dx
dt
dy
dt ;
thus
sin 1 1 cos 1 2
–
1 Š1
# ‹ cos 1
sin 1 2
œ
dy
dx
œ
41 8
21
dx
dt
œ 1 x cos t Ê
sin t t cos t 2
c x cos t ‰
ˆ 1sin
tb2
(c)
(d)
;
œ 4
2x
w
68. 2x# 3y# œ 5 Ê 4x 6yyw œ 0 Ê yw œ 2x
3y Ê y k (1 1) œ 3y ¹
ß
3x#
2y
1 x cos t
sin t2
—
3 #Î$
3, then f w (x) œ x"Î$ and f ww (x) œ "3 x%Î$ so the
# x
9 &Î$
if f(x) œ 10
x 7, then f w (x) œ 3# x#Î$ and f ww (x) œ x"Î$ is true
f ww (x) œ x"Î$ Ê f www (x) œ "3 x%Î$ is true
if f w (x) œ #3 x#Î$ 6, then f ww (x) œ x"Î$ is true
also, y# œ x$ Ê 2yyw œ 3x# Ê yw œ
œ
; t œ 1 Ê x sin 1 2x œ 1
67. (a) if f(x) œ
(b)
dx
dt
Ê yw k (1 1) œ
ß
3x#
2y ¹ (1ß1)
œ
(1ß1)
3
#
claim f ww (x) œ x"Î$ is false
œ 23 and yw k (1 1) œ 2x
3y ¹
ß
3x#
2y ¹ (1ß1)
and yw k (1 1) œ
ß
(1ß1)
œ
2
3
;
œ #3 . Therefore
the tangents to the curves are perpendicular at (1ß 1) and (1ß 1) (i.e., the curves are orthogonal at these two
points of intersection).
69. x# 2xy 3y# œ 0 Ê 2x 2xyw 2y 6yyw œ 0 Ê yw (2x 6y) œ 2x 2y Ê yw œ
tangent line m œ yw k (1ß1) œ
xy
3y x ¹ (1ß1)
xy
3y x
Ê the slope of the
œ 1 Ê the equation of the normal line at (1ß 1) is y 1 œ 1(x 1)
Ê y œ x 2. To find where the normal line intersects the curve we substitute into its equation:
x# 2x(2 x) 3(2 x)# œ 0 Ê x# 4x 2x# 3 a4 4x x# b œ 0 Ê 4x# 16x 12 œ 0
Ê x# 4x 3 œ 0 Ê (x 3)(x 1) œ 0 Ê x œ 3 and y œ x 2 œ 1. Therefore, the normal to the curve
at (1ß 1) intersects the curve at the point (3ß 1). Note that it also intersects the curve at (1ß 1).
70. xy 2x y œ 0 Ê x
dy
dx
y2
dy
dx
œ0 Ê
dy
dx
œ
y2
1x
; the slope of the line 2x y œ 0 is 2. In order to be
parallel, the normal lines must also have slope of 2. Since a normal is perpendicular to a tangent, the slope of
2
"
the tangent is "# . Therefore, y1 x œ # Ê 2y 4 œ 1 x Ê x œ 3 2y. Substituting in the original equation,
y(3 2y) 2(3 2y) y œ 0 Ê y# 4y 3 œ 0 Ê y œ 3 or y œ 1. If y œ 3, then x œ 3 and
y 3 œ 2(x 3) Ê y œ 2x 3. If y œ 1, then x œ 1 and y 1 œ 2(x 1) Ê y œ 2x 3.
71. y# œ x Ê
dy
dx
œ
"
#y
y" 0
x" a
. If a normal is drawn from (aß 0) to (x" ß y" ) on the curve its slope satisfies
œ 2y"
Ê y" œ 2y" (x" a) or a œ x" "# . Since x" 0 on the curve, we must have that a "# . By symmetry, the
two points on the parabola are ˆx" ß Èx" ‰ and ˆx" ß Èx" ‰ . For the normal to be perpendicular,
Èx
Èx
"
"
Š x " a ‹ Š a x " ‹ œ 1 Ê
x"
(a x" )#
Therefore, ˆ "4 ß „ #" ‰ and a œ
72. Ex. 6b.)
Ex. 7a.)
3
4
œ 1 Ê x" œ (a x" )# Ê x" œ ˆx" "
4
and y" œ „ #" .
.
"Î%
y œ a1 x# b has a derivative only on ("ß ") because the function is defined only on ["ß "] and
the slope of the tangent becomes vertical at both x œ 1 and x œ 1.
dy
dx ‹
y$ x#
dy
dx
2xy œ 0 Ê
$
y 2xy
$
#
#
$
œ 3xy
# x# ; also, xy x y œ 6 Ê x a3y b y
dx
dy
#
x" ‰ Ê x" œ
y œ x"Î# has no derivative at x œ 0 because the slope of the graph becomes vertical at x œ 0.
73. xy$ x# y œ 6 Ê x Š3y#
Ê
"
#
#
#
x
œ 3xy
y$ 2xy ; thus
dx
dy
appears to equal
"
dy
dx
dx
dy
dy
dx
a3xy# x# b œ y$ 2xy Ê
x# y Š2x
dx
dy ‹
œ0 Ê
dx
dy
dy
dx
œ
y$ 2xy
3xy# x#
ay$ 2xyb œ 3xy# x#
. The two different treatments view the graphs as functions
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 3.6 Implicit Differentiation
symmetric across the line y œ x, so their slopes are reciprocals of one another at the corresponding points
(aß b) and (bß a).
74. x$ y# œ sin# y Ê 3x# 2y
œ
3x#
2 sin y cos y 2y
appears to equal
dy
dx
œ (2 sin y)(cos y)
; also, x$ y# œ sin# y Ê 3x#
"
dy
dx
dx
dy
dy
dx
Ê
dy
dx
(2y 2 sin y cos y) œ 3x# Ê
2y œ 2 sin y cos y Ê
dx
dy
œ
2 sin y cos y 2y
3x#
75. x% 4y# œ 1:
(a)
%
y œ 14x
Ê dy
dx œ
œ
3x#
2y 2 sin y cos y
; thus
dx
dy
. The two different treatments view the graphs as functions symmetric across the line
y œ x so their slopes are reciprocals of one another at the corresponding points (aß b) and (bß a).
#
dy
dx
(b)
Ê yœ
„ "# È1
% "Î#
x%
„x $
"Î#
a1 x % b
differentiating implicitly, we find, 4x$ 8y dy
dx
dy
4x$
4x$
„x $
Ê dx œ 8y œ
œ
"Î# .
a1 x % b
8 Š„ "# È1 x% ‹
„ "4 a1 x b
a4x$ b œ
76. (x 2)# y# œ 4:
(a) y œ „ È4 (x 2)#
;
œ0
(b)
dy
"
# "Î#
(2(x 2))
dx œ „ # a4 (x 2) b
„(x 2)
œ
; differentiating implicitly,
c4 (x 2)# d"Î#
dy
2(x 2)
2(x 2) 2y dy
dx œ 0 Ê dx œ
2y
(x 2)
(x 2)
„(x 2)
œ y œ
œ
.
„c4 (x 2)# d "Î#
c4 (x #)# d "Î#
Ê
77-84. Example CAS commands:
Maple:
q1 := x^3-x*y+y^3 = 7;
pt := [x=2,y=1];
p1 := implicitplot( q1, x=-3..3, y=-3..3 ):
p1;
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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167
168
Chapter 3 Differentiation
eval( q1, pt );
q2 := implicitdiff( q1, y, x );
m := eval( q2, pt );
tan_line := y = 1 + m*(x-2);
p2 := implicitplot( tan_line, x=-5..5, y=-5..5, color=green ):
p3 := pointplot( eval([x,y],pt), color=blue ):
display( [p1,p2,p3], ="Section 3.6 #77(c)" );
Mathematica: (functions and x0 may vary):
Note use of double equal sign (logic statement) in definition of eqn and tanline.
<<Graphics`ImplicitPlot`
Clear[x, y]
{x0, y0}={1, 1/4};
eqn=x + Tan[y/x]==2;
ImplicitPlot[eqn,{ x, x0 3, x0 3},{y, y0 3, y0 3}]
eqn/.{x Ä x0, y Ä y0}
eqn/.{ y Ä y[x]}
D[%, x]
Solve[%, y'[x]]
slope=y'[x]/.First[%]
m=slope/.{x Ä x0, y[x] Ä y0}
tanline=y==y0 m (x x0)
ImplicitPlot[{eqn, tanline}, {x, x0 3, x0 3},{y, y0 3, y0 + 3}]
3.7 RELATED RATES
1. A œ 1r# Ê
2. S œ 41r# Ê
dA
dt
dS
dt
œ 21r
dr
dt
œ 81r
3. (a) V œ 1r# h Ê
dV
dt
dV
dt
(c) V œ 1r# h Ê
dr
dt
œ 1 r#
œ
4. (a) V œ "3 1r# h Ê
dh
dt
# dh
1r dt
dV
dt œ
" # dh
2
3 1r dt 3 1rh
(b) V œ 1r# h Ê
21rh
dV
dt
" # dh
3 1r dt
dr
dt
(b) V œ "3 1r# h Ê
dV
dt
dV
dt
œ
5. (a)
dV
dt
dV
dt
dR
dt
"
œ 1 volt/sec
(b) dI
dt œ 3 amp/sec
" ˆ dV
dI ‰
" ˆ dV
V dI ‰
‰
ˆ dR ‰ Ê dR
œ R ˆ dI
Ê dR
dt I dt
dt œ I
dt R dt
dt œ I
dt I dt
ˆ " ‰‘ œ ˆ #" ‰ (3) œ 3# ohms/sec, R is increasing
œ "# 1 12
# 3
(d)
6. (a) P œ RI# Ê
dP
dt
œ I#
(b) P œ RI# Ê 0 œ
dP
dt
dR
dt
2RI
œ I#
dR
dt
dr
dt
dr
dt
(c)
(c)
œ 21rh
œ 32 1rh
dr
dt
dI
dt
2RI
dI
dt
Ê
dR
dt
œ 2RI
I#
7. (a) s œ Èx# y# œ ax# y# b
"Î#
Ê
ds
dt
œ
x
dx
Èx# y# dt
(b) s œ Èx# y# œ ax# y# b
"Î#
Ê
ds
dt
œ
x
dx
Èx# y# dt
(c) s œ Èx# y# Ê s# œ x# y# Ê 2s
ds
dt
œ 2x
dx
dt
8. (a) s œ Èx# y# z# Ê s# œ x# y# z# Ê 2s
dI
dt
2 ˆ PI ‰ dI
I#
dt
œ 2P
I$
dI
dt
y
dy
Èx# y# dt
2y
ds
dt
œ
dy
dt
œ 2x
Ê 2s † 0 œ 2x
dx
dt
2y
dy
dt
2z
dx
dt
2y
dy
dt
Ê
dz
dt
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
dx
dt
œ yx
dy
dt
Section 3.7 Related Rates
Ê
ds
dt
œ
x
dx
Èx# y# z# dt
y
dy
Èx# y# z# dt
(b) From part (a) with
dx
dt
œ0 Ê
(c) From part (a) with
ds
dt
œ 0 Ê 0 œ 2x
9. (a) A œ
(c) A œ
"
#
"
#
ab sin ) Ê
ab sin ) Ê
10. Given A œ 1r# ,
dr
dt
dA
dt
dA
dt
œ
œ
"
#
"
#
œ
ds
dt
z
dz
Èx# y# z# dt
y
dy
Èx# y# z# dt
dx
dt
d)
dt
d)
dt
ab cos )
ab cos )
2y
dy
dt
z
dz
Èx# y# z# dt
2z
Ê
dz
dt
(b) A œ
"# b sin )
z dz
x dt
"
#
ab sin ) Ê
dA
dt
"# a sin )
da
dt
œ 0.01 cm/sec, and r œ 50 cm. Since
dx
dt
œ 21r
dA
dt
y dy
x dt
œ0
œ
"
#
ab cos )
d)
dt
"# b sin )
db
dt
dr
dt
, then
dA ¸
dt r=50
" ‰
œ 21(50) ˆ 100
œ 1 cm# /min.
(a)
(b)
(c)
dj
dt
dw
dt œ 2 cm/sec, j œ 12 cm and w œ 5 cm.
dj
dA
#
A œ jw Ê
œ j dw
dt w dt Ê dt œ 12(2) 5(2) œ 14 cm /sec, increasing
dP
dj
dw
P œ 2j 2w Ê dt œ 2 dt 2 dt œ 2(2) 2(2) œ 0 cm/sec, constant
"Î#
"
dj ‰
#
# "Î# ˆ
D œ Èw# j# œ aw# j# b
Ê dD
2w dw
Ê dD
dt œ # aw j b
dt 2j dt
dt
11. Given
œ 2 cm/sec,
dA
dt
œ
(5)(2) (12)(2)
È25 144
12. (a) V œ xyz Ê
dV
dt
œ
dj
w dw
dt j dt
Èw# j#
œ 14
13 cm/sec, decreasing
œ yz
xz
dx
dt
xy
dy
dt
dz
dt
Ê
dV ¸
dt (4ß3ß2)
œ (3)(2)(1) (4)(2)(2) (4)(3)(1) œ 2 m$ /sec
dx
(b) S œ 2xy 2xz 2yz Ê dS
dt œ (2y 2z) dt (2x 2z)
¸
Ê dS
œ (10)(1) (12)(2) (14)(1) œ 0 m# /sec
dt
dy
dt
(2x 2y)
dz
dt
(4ß3ß2)
(c) j œ Èx# y# z# œ ax# y# z# b
Ê
13. Given:
dj ¸
dt (4ß3ß2)
dx
dt
"Î#
dj
dt
Ê
œ
x
dx
Èx# y# z# dt
y
dy
Èx# y# z# dt
z
dz
Èx# y# z# dt
œ Š È429 ‹ (1) Š È329 ‹ (2) Š È229 ‹ (1) œ 0 m/sec
œ 5 ft/sec, the ladder is 13 ft long, and x œ 12, y œ 5 at the instant of time
(a) Since x# y# œ 169 Ê
dy
dt
œ xy
dx
dt
‰
œ ˆ 12
5 (5) œ 12 ft/sec, the ladder is sliding down the wall
(b) The area of the triangle formed by the ladder and walls is A œ
is changing at
(c) cos ) œ
x
13
"
#
xy Ê
dA
dt
œ ˆ "# ‰ Šx
dy
dt
y
dx
dt ‹ .
The area
#
[12(12) 5(5)] œ 119
# œ 59.5 ft /sec.
Ê sin )
14. s# œ y# x# Ê 2s
"
#
ds
dt
d)
dt
œ 2x
†
dx
dt
Ê
d)
dt
"
œ 13 sin
) †
2y
dy
dt
Ê
ds
dt
œ
œ
dx
dt
"
13
"
s
Šx
dx
dt
dx
dt
y
œ ˆ 5" ‰ (5) œ 1 rad/sec
dy
dt ‹
Ê
ds
dt
œ
"
È169
[5(442) 12(481)]
œ 614 knots
15. Let s represent the distance between the girl and the kite and x represents the horizontal distance between the
girl and kite Ê s# œ (300)# x# Ê
ds
dt
œ
x dx
s dt
œ
400(25)
500
œ 20 ft/sec.
"
#
16. When the diameter is 3.8 in., the radius is 1.9 in. and dr
dt œ 3000 in/min. Also V œ 61r Ê
$
ˆ " ‰
Ê dV
dt œ 121(1.9) 3000 œ 0.00761. The volume is changing at about 0.0239 in /min.
17. V œ
(a)
(b)
"
" ˆ 4h ‰#
1 h$
3
3r
4h
#
h œ 1627
3 1r h, h œ 8 (2r) œ 4 Ê r œ 3 Ê V œ 3 1 3
dh ¸
90
ˆ 9 ‰
dt h=4 œ 1614# (10) œ 2561 ¸ 0.1119 m/sec œ 11.19 cm/sec
dr
4 dh
4 ˆ 90 ‰
15
r œ 4h
3 Ê dt œ 3 dt œ 3 2561 œ 321 ¸ 0.1492 m/sec œ 14.92
Ê
dV
dt
œ
dV
dt
œ 121r
161h# dh
9
dt
cm/sec
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
dr
dt
da
dt
169
170
Chapter 3 Differentiation
18. (a) V œ
"
3
1r# h and r œ
Ê Vœ
15h
#
"
3
#
‰ hœ
1 ˆ 15h
#
751h$
4
¸ 0.0113 m/min œ 1.13 cm/min
dr
15 dh
dr ¸
8 ‰
ˆ 15 ‰ ˆ 225
(b) r œ 15h
# Ê dt œ # dt Ê dt h=5 œ #
1 œ
19. (a) V œ
1
3
y# (3R y) Ê
y œ 8 we have
"
1441 (6)
œ
dy
dt
1
3
œ
dV
dt
4
151
c2y(3R y) y# (1)d
œ
œ
dV
dt
2251h# dh
4
dt
Ê
œ
dh ¸
dt h=5
4(50)
2251(5)#
œ
8
2251
¸ 0.0849 m/sec œ 8.49 cm/sec
dy
dt
Ê
dy
dt
" dV
dt
œ 13 a6Ry 3y# b‘
Ê at R œ 13 and
m/min
(b) The hemisphere is on the circle r (13 y)# œ 169 Ê r œ È26y y# m
(c) r œ a26y y# b
5
2881
œ
20. If V œ
4
3
"Î#
Ê
"
#
œ
dr
dt
#
"
241
Ê
a26y y# b
"Î#
(26 2y)
dy
dt
Ê
dr
dt
œ
13 y
dy
È26y y# dt
Ê
dr ¸
dt y=8
œ
13 8
È26†8 64
ˆ #"
‰
41
m/min
1r$ , S œ 41r# , and
dV
dt
œ kS œ 4k1r# , then
dV
dt
œ 41r#
Ê 4k1r# œ 41r#
dr
dt
dr
dt
Ê
dr
dt
œ k, a constant.
Therefore, the radius is increasing at a constant rate.
4
dV
dV
dr
$
$
# dr
3 1r , r œ 5, and dt œ 1001 ft /min, then dt œ 41r dt Ê dt
dr
#
dt œ 81(5)(1) œ 401 ft /min, the rate at which the surface area
21. If V œ
œ 81r
œ 1 ft/min. Then S œ 41r# Ê
dS
dt
is increasing.
22. Let s represent the length of the rope and x the horizontal distance of the boat from the dock.
s ds
s
ds
(a) We have s# œ x# 36 Ê dx
dt œ x dt œ È #
dt . Therefore, the boat is approaching the dock at
s 36
dx ¸
dt s=10
œ
(b) cos ) œ
d)
dt
Ê
10
È10# 36
6
r
œ
(2) œ 2.5 ft/sec.
Ê sin )
6
8 ‰
10# ˆ 10
d)
dt
œ r6#
† (2) œ 3
20
Ê
dr
dt
d)
dt
œ
6
dr
r# sin ) dt
. Thus, r œ 10, x œ 8, and sin ) œ
8
10
rad/sec
23. Let s represent the distance between the bicycle and balloon, h the height of the balloon and x the horizontal
distance between the balloon and the bicycle. The relationship between the variables is s# œ h# x#
" ˆ dh
dx ‰
"
Ê ds
Ê ds
dt œ s h dt x dt
dt œ 85 [68(1) 51(17)] œ 11 ft/sec.
24. (a) Let h be the height of the coffee in the pot. Since the radius of the pot is 3, the volume of the coffee is
dh
dh
" dV
10
V œ 91h Ê dV
dt œ 91 dt Ê the rate the coffee is rising is dt œ 91 dt œ 91 in/min.
(b) Let h be the height of the coffee in the pot. From the figure, the radius of the filter r œ
œ
1 h$
1#
, the volume of the filter. The rate the coffee is falling is
25. y œ QD" Ê
26. (a)
(b)
dc
dt
dc
dt
dp
dt
dy
dt
œ D"
dQ
dt
œ a3x# 12x 15b
#
QD#
dD
dt
œ
"
41
(0) 233
(41)#
(2) œ
dh
dt
466
1681
œ a3(2)# 12(2) 15b (0.1) œ 0.3,
dx
dt
#
œ a3x 12x 45x b
dx
dt
dr
dt
#
#
œ
4 dV
1h# dt
œ
4
#5 1
Ê Vœ
h
#
"
3
1r# h
(10) œ 581 in/min.
L/min Ê increasing about 0.2772 L/min
œ9
dx
dt
œ 9(0.1) œ 0.9,
œ a3(1.5) 12(1.5) 45(1.5) b (0.05) = 1.5625,
dr
dt
dp
dt
œ 70
œ 0.9 0.3 œ 0.6
dx
dt
œ 70(0.05) œ 3.5,
œ 3.5 (1.5625) œ 5.0625
27. Let P(xß y) represent a point on the curve y œ x# and ) the angle of inclination of a line containing P and the
origin. Consequently, tan ) œ
#
and cos )kx=3 œ
x#
y # x #
œ
28. y œ (x)"Î# and tan ) œ
3#
9 # 3 #
y
x
y
x
œ
Ê tan ) œ
"
10
, we have
Ê tan ) œ
(x)"Î#
x
x#
d)
#
x œ x Ê sec ) dt
d) ¸
dt x=3 œ 1 rad/sec.
Ê sec# )
d)
dt
œ
œ
dx
dt
Ê
d)
dt
œ cos# )
dx
dt
. Since
ˆ "# ‰ (x)"Î# (1)x (x)"Î# (1) dx
x#
dt
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dx
dt
œ 10 m/sec
Section 3.7 Related Rates
d)
dt
Ê
d)
dt
cx
È È x
dx
#
œ Œ 2 cx x#
acos )b ˆ dt ‰ . Now, tan ) œ
4
2
œ Š 4 16 ‹ ˆ 45 ‰ (8) œ
2
5
2
4
œ #" Ê cos ) œ È25 Ê cos# ) œ
"
#
a x# y # b
"Î#
Š2x
dx
dt
. Then
rad/sec.
29. The distance from the origin is s œ Èx# y# and we wish to find
œ
4
5
2y
dy
dt ‹¹ (5ß12)
œ
(5)(1) (12)(5)
È25 144
ds ¸
dt (5ß12)
œ 5 m/sec
30. When s represents the length of the shadow and x the distance of the man from the streetlight, then s œ
3
5
x.
(a) If I represents the distance of the tip of the shadow from the streetlight, then I œ s x Ê
œ dx
dt
3 dx
dx ¸
8 ¸ ¸ dx ¸
8
¸
¸
¸
(which is velocity not speed) Ê ¸ dI
œ
œ
œ
k
5
k
œ
8
ft/sec,
the
speed
the
tip
of
the
dt
5 dt
dt
5
dt
5
dI
dt
ds
dt
shadow is moving along the ground.
ds
3 dx
3
dt œ 5 dt œ 5 (5) œ 3 ft/sec, so the length of the shadow is decreasing at a rate of 3 ft/sec.
(b)
31. Let s œ 16t# represent the distance the ball has fallen,
h the distance between the ball and the ground, and I
the distance between the shadow and the point directly
beneath the ball. Accordingly, s h œ 50 and since
the triangle LOQ and triangle PRQ are similar we have
Iœ
œ
30h
50 h
1500
16t# Ê h œ 50 16t# and I œ
30 Ê
dI
dt
œ 1500
8t$ Ê
30 a50 16t# b
50 a50 16t# b
dI ¸
dt t= 12
œ 1500 ft/sec.
32. Let s œ distance of car from foot of perpendicular in the textbook diagram Ê tan ) œ
Ê
d)
dt
œ
#
cos ) ds
132 dt
;
ds
dt
œ 264 and ) œ 0 Ê
d)
dt
4
3
1r$ 43 14$ Ê
thickness of the ice is decreasing at
5
721
Ê sec# )
d)
dt
" ds
13# dt
œ
œ 2 rad/sec. A half second later the car has traveled 132 ft
right of the perpendicular Ê k)k œ 14 , cos# ) œ "# , and
33. The volume of the ice is V œ
s
13#
dV
dt
ds
dt
œ 264 (since s increases) Ê
œ 41r#
dr
dt
Ê
dr ¸
dt r=6
œ
5
721
10
3
œ
in./min when
in/min. The surface area is S œ 41r# Ê
#
œ 10
3 in /min, the outer surface area of the ice is decreasing at
d)
dt
dS
dt
œ 81r
dr
dt
in# /min.
ˆ "# ‰
132
(264) œ 1 rad/sec.
œ 10 in$ /min, the
5 ‰
¸ œ 481 ˆ 72
Ê dS
dt
1
dV
dt
r=6
34. Let s represent the horizontal distance between the car and plane while r is the line-of-sight distance between
r
dr
ds ¸
5
the car and plane Ê 9 s# œ r# Ê ds
dt œ È #
dt Ê dt r=5 œ È16 (160) œ 200 mph
r 9
Ê speed of plane speed of car œ 200 mph Ê the speed of the car is 80 mph.
35. When x represents the length of the shadow, then tan ) œ
We are given that
¸ dx
¸
dt
œ
d)
dt
œ 0.27° œ
#
#
¹ x 80sec ) ddt) ¹¹ d)
Š =
dt
31
2000
31
#000
80
x
Ê sec# )
rad/min. At x œ 60, cos ) œ
and sec ) = 35 ‹
œ
31
16
3
5
d)
dt
œ 80
x#
dx
dt
Ê
dx
dt
œ
x# sec# ) d)
80
dt
Ê
ft/min ¸ 0.589 ft/min ¸ 7.1 in./min.
36. Let A represent the side opposite ) and B represent the side adjacent ). tan ) œ AB Ê sec# ) ddt) œ B" dA
dt 2
d)
" ‰
10
4‰
ˆ
ˆ
‰
‘
ˆ
t Ê at A œ 10 m and B œ 20 m we have cos ) œ 20
œ
and
œ
(
2)
(1)
È
È
dt
#0
400
5
10
œ ˆ "
10 " ‰ ˆ4‰
40
5
5
#
A dB
B# dt
5
"
œ 10
rad/sec œ 18°
1 /sec ¸ 6°/sec
37. Let x represent distance of the player from second base and s the distance to third base. Then
#
.
(a) s œ x 8100 Ê 2s
ds
dt
œ 2x
dx
dt
Ê
ds
dt
œ
x dx
s dt
dx
dt
œ 16 ft/sec
. When the player is 30 ft from first base, x œ 60
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171
172
Chapter 3 Differentiation
Ê s œ 30È13 and
(b) cos )" œ
Ê
(c)
d)"
dt
œ
90
sx
d)"
dt
œ
90
s
œ
œ
ds
dt
Ê sin )"
90
Š30È13‹ (60)
†
ds
dt .
90
s# sin )"
60
30È13
d)"
dt
32
È13
(16) œ
œ 90
s# †
32
† ŠÈ
‹œ
13
8
65
ds
dt
Ê
¸ 8.875 ft/sec
d)"
dt
œ
90
s# sin )"
rad/sec; sin )# œ
90
s
†
ds
dt
œ
90
sx
†
Ê cos )#
. Therefore, x œ 60 and s œ 30È13
ds
dt
d)#
dt
œ 90
s# †
8
Therefore, x œ 60 and s œ 30È13 Ê ddt)# œ 65
rad/sec.
ds
90
x
dx
90
dx
90
† dt œ ˆs# † x ‰ † ˆ s ‰ † ˆ dt ‰ œ ˆ s# ‰ ˆ dt ‰ œ ˆ x# 8100 ‰ dx
dt Ê lim
xÄ!
s
œ lim ˆ x# 908100 ‰ (15) œ 6" rad/sec;
xÄ!
90 ‰
œ ˆ x# 8100
dx
dt
d)#
x Ä ! dt
Ê lim
œ
"
6
d)#
dt
œ
90
s# cos )#
†
ds
dt
ds
dt
Ê
d)#
dt
œ
90
s# cos )#
†
ds
dt
d)"
dt
90 ˆ x ‰ ˆ dx ‰
ˆ s90
‰ ˆ dx
‰
œ Š
#
s# † x ‹ s
dt œ
dt
s
rad/sec
38. Let a represent the distance between point O and ship A, b the distance between point O and ship B, and
D the distance between the ships. By the Law of Cosines, D# œ a# b# 2ab cos 120°
" da
db
db
da ‘
da
db
dD
413
Ê dD
dt œ #D 2a dt 2b dt a dt b dt . When a œ 5, dt œ 14, b œ 3, and dt œ 21, then dt œ 2D
where D œ 7. The ships are moving
dD
dt
œ 29.5 knots apart.
3.8 LINEARIZATION AND DIFFERENTIALS
1. f(x) œ x$ 2x 3 Ê f w (x) œ 3x# 2 Ê L(x) œ f w (2)(x 2) f(2) œ 10(x 2) 7 Ê L(x) œ 10x 13 at x œ 2
2. f(x) œ Èx# 9 œ ax# 9b
"Î#
Ê f w (x) œ ˆ "# ‰ ax# 9b
œ 45 (x 4) 5 Ê L(x) œ 45 x 3. f(x) œ x "
x
9
5
"Î#
(2x) œ
x
È x# 9
Ê L(x) œ f w (4)(x 4) f(4)
at x œ 4
Ê f w (x) œ 1 x# Ê L(x) œ f(1) f w (1)(x 1) œ # !(x 1) œ #
4. f(x) œ x"Î$ Ê f w (x) œ
"
$x#Î$
Ê L(x) œ f w (8)ax a8bb fa8b œ
"
1#
(x 8) 2 Ê L(x) œ
"
1#
x
4
3
5. f(x) œ x# 2x Ê f w (x) œ 2x 2 Ê L(x) œ f w (0)(x 0) f(0) œ 2(x 0) 0 Ê L(x) œ 2x at x œ 0
6. f(x) œ x" Ê f w (x) œ x# Ê L(x) œ f w (1)(x 1) f(1) œ (1)(x 1) 1 Ê L(x) œ x 2 at x œ 1
7. f(x) œ 2x# 4x 3 Ê f w (x) œ 4x 4 Ê L(x) œ f w (1)(x 1) f(1) œ 0(x 1) (5) Ê L(x) œ 5 at x œ 1
8. f(x) œ 1 x Ê f w (x) œ 1 Ê L(x) œ f w (8)(x 8) f(8) œ 1(x 8) 9 Ê L(x) œ x 1 at x œ 8
9. f(x) œ $Èx œ x"Î$ Ê f w (x) œ ˆ "3 ‰ x#Î$ Ê L(x) œ f w (8)(x 8) f(8) œ
10. f(x) œ
x
x1
Ê L(x) œ
Ê f w (x) œ
"
4
x
"
4
(1)(x 1) (")(x)
(x 1)#
œ
"
(x 1)#
"
1#
(x 8) 2 Ê L(x) œ
Ê L(x) œ f w (1)(x 1) f(1) œ
"
4
(x 1) "
#
at x œ 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
"
1#
x
4
3
at x œ 8
Section 3.8 Linearization and Differentials
11. f(x) œ sin x Ê f w (x) œ cos x
(a) L(x) œ f w (0)(x 0) f(0) œ 1(x 0) 0
Ê L(x) œ x at x œ 0
(b) L(x) œ f w (1)(x 1) f(1) œ (1)(x 1) 0
Ê L(x) œ 1 x at x œ 1
12. f(x) œ cos x Ê f w (x) œ sin x
(a) L(x) œ f w (0)(x 0) f(0) œ 0(x 0) 1
Ê L(x) œ 1 at x œ 0
(b) L(x) œ f w ˆ 1# ‰ ˆx 1# ‰ f ˆ 1# ‰
œ (1) ˆx 1# ‰ 0 Ê L(x) œ x at x œ 1#
1
2
13. f(x) œ sec x Ê f w (x) œ sec x tan x
(a) L(x) œ f w (0)(x 0) f(0) œ 0(x 0) 1
Ê L(x) œ 1 at x œ 0
(b) L(x) œ f w ˆ 13 ‰ ˆx 13 ‰ f ˆ 13 ‰
œ 2È3 ˆx 13 ‰ 2 Ê L(x) œ 2 2È3 ˆx 13 ‰
at x œ 13
14. f(x) œ tan x Ê f w (x) œ sec# x
(a) L(x) œ f w (0)(x 0) f(0) œ 1(x 0) 0 œ x
Ê L(x) œ x at x œ 0
(b) L(x) œ f w ˆ 14 ‰ ˆx 14 ‰ f ˆ 14 ‰ œ 2 ˆx 14 ‰ 1
Ê L(x) œ 1 2 ˆx 14 ‰ at x œ 14
15. f w axb œ ka" xbk" . We have fa!b œ " and f w a!b œ k. Laxb œ fa!b f w a!bax !b œ " kax !b œ " kx
'
16. (a) faxb œ a" xb' œ " axb‘ ¸ " 'axb œ " 'x
(b) faxb œ
#
" x
"
œ #" axb‘
(c) faxb œ a" xb
"Î#
(d) faxb œ È" x# œ
¸ #" a"baxb‘ œ # #x
¸ " ˆ "# ‰x œ " x#
"Î#
È#Š" x# ‹ ¸ È#Š"
#
(e) faxb œ a% $xb"Î$ œ %"Î$ ˆ" $x ‰"Î$
%
" x#
# #‹
¸ %"Î$ ˆ" œ È#Š" " $x ‰
$ %
x#
%‹
œ %"Î$ ˆ" x% ‰
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173
174
Chapter 3 Differentiation
(f) faxb œ ˆ" " ‰2Î$
#x
2Î$
œ ’" ˆ # " x ‰“
¸ " $# ˆ # " x ‰ œ " #
'$x
17. (a) (1.0002)&! œ (1 0.0002)&! ¸ 1 50(0.0002) œ 1 .01 œ 1.01
(b) $È1.009 œ (1 0.009)"Î$ ¸ 1 ˆ " ‰ (0.009) œ 1 0.003 œ 1.003
3
18. f(x) œ Èx 1 sin x œ (x 1)"Î# sin x Ê f w (x) œ ˆ "# ‰ (x 1)"Î# cos x Ê Lf (x) œ f w (0)(x 0) f(0)
œ 3 (x 0) 1 Ê Lf (x) œ 3 x 1, the linearization of f(x); g(x) œ Èx 1 œ (x 1)"Î# Ê gw (x)
#
#
œ ˆ "# ‰ (x 1)"Î# Ê Lg (x) œ gw (0)(x 0) g(0) œ
w
w
"
#
(x 0) 1 Ê Lg (x) œ
"
#
x 1, the linearization of g(x);
h(x) œ sin x Ê h (x) œ cos x Ê Lh (x) œ h (0)(x 0) h(0) œ (1)(x 0) 0 Ê Lh (x) œ x, the linearization of
h(x). Lf (x) œ Lg (x) Lh (x) implies that the linearization of a sum is equal to the sum of the linearizations.
19. y œ x$ 3Èx œ x$ 3x"Î# Ê dy œ ˆ3x# #3 x"Î# ‰ dx Ê dy œ Š3x# 3
‹
2È x
dx
"Î#
"Î#
"Î#
20. y œ xÈ1 x# œ x a1 x# b
Ê dy œ ’(1) a1 x# b (x) ˆ "# ‰ a1 x# b
(2x)“ dx
œ a1 x# b
"Î#
#
a1 2x# b
È 1 x#
Ê dy œ Š (2) a1 a1xb x# b(2x)(2x)
‹ dx œ
#
21. y œ
2x
1 x #
22. y œ
2È x
3 ˆ1 È x ‰
Ê dy œ
ca1 x# b x# d dx œ
œ
2x"Î#
3 a1 x"Î# b
"
#
3 È x ˆ1 È x ‰
Ê dy œ Š
dx
2 2x#
a 1 x # b#
dx
x"Î# ˆ3 ˆ1 x"Î# ‰‰ 2x"Î# ˆ #3 x"Î# ‰
9 a1 x"Î# b
#
‹ dx œ
3x"Î# 3 3
#
9 a1 x"Î# b
dx
dx
23. 2y$Î# xy x œ 0 Ê 3y"Î# dy y dx x dy dx œ 0 Ê ˆ3y"Î# x‰ dy œ (1 y) dx Ê dy œ
1 y
3 È y x
24. xy# 4x$Î# y œ 0 Ê y# dx 2xy dy 6x"Î# dx dy œ 0 Ê (2xy 1) dy œ ˆ6x"Î# y# ‰ dx
Ê dy œ
6È x y#
2xy 1
dx
25. y œ sin ˆ5Èx‰ œ sin ˆ5x"Î# ‰ Ê dy œ ˆcos ˆ5x"Î# ‰‰ ˆ 5# x"Î# ‰ dx Ê dy œ
5 cos ˆ5Èx‰
2È x
dx
26. y œ cos ax# b Ê dy œ csin ax# bd (2x) dx œ 2x sin ax# b dx
$
$
$
27. y œ 4 tan Š x3 ‹ Ê dy œ 4 Šsec# Š x3 ‹‹ ax# b dx Ê dy œ 4x# sec# Š x3 ‹ dx
28. y œ sec ax# 1b Ê dy œ csec ax# 1b tan ax# 1bd (2x) dx œ 2x csec ax# 1b tan ax# 1bd dx
29. y œ 3 csc ˆ1 2Èx‰ œ 3 csc ˆ1 2x"Î# ‰ Ê dy œ 3 ˆcsc ˆ1 2x"Î# ‰‰ cot ˆ1 2x"Î# ‰ ˆx"Î# ‰ dx
Ê dy œ È3 csc ˆ1 2Èx‰ cot ˆ1 2Èx‰ dx
x
30. y œ 2 cot Š È"x ‹ œ 2 cot ˆx"Î# ‰ Ê dy œ 2 csc# ˆx"Î# ‰ ˆ #" ‰ ˆx$Î# ‰ dx Ê dy œ
"
È x$
csc# Š È"x ‹ dx
31. f(x) œ x# 2x, x! œ 1, dx œ 0.1 Ê f w (x) œ 2x 2
(a) ?f œ f(x! dx) f(x! ) œ f(1.1) f(1) œ 3.41 3 œ 0.41
(b) df œ f w (x! ) dx œ [2(1) 2](0.1) œ 0.4
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dx
Section 3.8 Linearization and Differentials
175
(c) k?f df k œ k0.41 0.4k œ 0.01
32. f(x) œ 2x# 4x 3, x! œ 1, dx œ 0.1 Ê f w (x) œ 4x 4
(a) ?f œ f(x! dx) f(x! ) œ f(.9) f(1) œ .02
(b) df œ f w (x! ) dx œ [4(1) 4](.1) œ 0
(c) k?f df k œ k.02 0k œ .02
33. f(x) œ x$ x, x! œ 1, dx œ 0.1 Ê f w (x) œ 3x# 1
(a) ?f œ f(x! dx) f(x! ) œ f(1.1) f(1) œ .231
(b) df œ f w (x! ) dx œ [3(1)# 1](.1) œ .2
(c) k?f df k œ k.231 .2k œ .031
34. f(x) œ x% , x! œ 1, dx œ 0.1 Ê f w (x) œ 4x$
(a) ?f œ f(x! dx) f(x! ) œ f(1.1) f(1) œ .4641
(b) df œ f w (x! ) dx œ 4(1)$ (.1) œ .4
(c) k?f df k œ k.4641 .4k œ .0641
35. f(x) œ x" , x! œ 0.5, dx œ 0.1 Ê f w (x) œ x#
(a) ?f œ f(x! dx) f(x! ) œ f(.6) f(.5) œ "3
" ‰
(b) df œ f w (x! ) dx œ (4) ˆ 10
œ 25
(c) k?f df k œ ¸ "3 25 ¸ œ
"
15
36. f(x) œ x$ 2x 3, x! œ 2, dx œ 0.1 Ê f w (x) œ 3x# 2
(a) ?f œ f(x! dx) f(x! ) œ f(2.1) f(2) œ 1.061
(b) df œ f w (x! ) dx œ (10)(0.10) œ 1
(c) k?f df k œ k1.061 1k œ .061
37. V œ
4
3
1r$ Ê dV œ 41r!# dr
38. V œ x$ Ê dV œ 3x!# dx
39. S œ 6x# Ê dS œ 12x! dx
40. S œ 1rÈr# h# œ 1r ar# h# b
Ê
dS
dr
œ
1 ar# h# b 1r#
È r# h #
"Î#
Ê dS œ
, h constant Ê
1 a2r#! h# b
Ér#! h#
dS
dr
œ 1 ar# h# b
"Î#
1r † r ar# h# b
"Î#
dr, h constant
41. V œ 1r# h, height constant Ê dV œ 21r! h dr
42. S œ 21rh Ê dS œ 21r dh
43. Given r œ 2 m, dr œ .02 m
(a) A œ 1r# Ê dA œ 21r dr œ 21(2)(.02) œ .081 m#
1‰
(b) ˆ .08
41 (100%) œ 2%
44. C œ 21r and dC œ 2 in. Ê dC œ 21 dr Ê dr œ
œ 21(5) ˆ 1" ‰ œ 10 in.#
"
1
Ê the diameter grew about
45. The volume of a cylinder is V œ 1r# h. When h is held fixed, we have
dV
dr
2
1
in.; A œ 1r# Ê dA œ 21r dr
œ #1rh, and so dV œ #1rh dr. For h œ $! in.,
r œ ' in., and dr œ !Þ& in., the volume of the material in the shell is approximately dV œ #1rh dr œ #1a'ba$!ba!Þ&b
œ ")!1 ¸ &'&Þ& in$ .
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176
Chapter 3 Differentiation
46. Let ) œ angle of elevation and h œ height of building. Then h œ $!tan ), so dh œ $!sec# ) d). We want ldhl !Þ!%h,
&1
&1
sin )
which gives: l$!sec# ) d)l !Þ!%l$!tan )l Ê cos"# ) ld)l !Þ!%
cos ) Ê ld)l !Þ!%sin ) cos ) Ê ld)l !Þ!%sin "# cos "#
œ !Þ!" radian. The angle should be measured with an error of less than !Þ!" radian (or approximatley !Þ&( degrees),
which is a percentage error of approximately !Þ('%.
47. V œ 1h$ Ê dV œ 31h# dh; recall that ?V ¸ dV. Then k?Vk Ÿ (1%)(V) œ
(1) a1h$ b
100
Ê k31h# dhk Ÿ
of h is
"
3
Ê kdhk Ÿ
"
300
(1) a1h$ b
100
Ê kdVk Ÿ
(1) a1h$ b
100
h œ ˆ "3 %‰ h. Therefore the greatest tolerated error in the measurement
%.
#
48. (a) Let Di represent the inside diameter. Then V œ 1r# h œ 1 ˆ D#i ‰ h œ
1D#i h
4
dV œ 51Di dDi . Recall that ?V ¸ dV. We want k?Vk Ÿ (1%)(V) Ê kdVk Ÿ
Ê 51Di dDi Ÿ
1D#i
40
Ê
dDi
Di
51D#i
#
1 D#
œ 40i
and h œ 10 Ê V œ
" ‰ 51D#i
ˆ 100
Š # ‹
Ê
Ÿ 200. The inside diameter must be measured to within 0.5%.
(b) Let De represent the exterior diameter, h the height and S the area of the painted surface. S œ 1De h Ê dS œ 1hdDe
dDe
Ê dS
S œ De . Thus for small changes in exterior diameter, the approximate percentage change in the exterior diameter
is equal to the approximate percentage change in the area painted, and to estimate the amount of paint required to
within 5%, the tanks's exterior diameter must be measured to within 5%.
49. V œ 1r# h, h is constant Ê dV œ 21rh dr; recall that ?V ¸ dV. We want k?Vk Ÿ
Ê k21rh drk Ÿ
1r# h
1000
Ê kdrk Ÿ
r
#000
"
1000
V Ê kdVk Ÿ
1 r# h
1000
œ (.05%)r Ê a .05% variation in the radius can be tolerated.
50. Volume œ (x ?x)$ œ x$ 3x# (?x) 3x(?x)# (?x)$
51. W œ a b
g
œ a bg" Ê dW œ bg# dg œ bgdg
Ê
#
dWmoon
dWearth
œ
b dg
‹
(5.2)#
b dg
Š # ‹
(32)
Š
#
32 ‰
œ ˆ 5.2
œ 37.87, so a change of
gravity on the moon has about 38 times the effect that a change of the same magnitude has on Earth.
52. (a) T œ 21 Š Lg ‹
"Î#
Ê dT œ 21ÈL ˆ "# g$Î# ‰ dg œ 1ÈL g$Î# dg
(b) If g increases, then dg 0 Ê dT 0. The period T decreases and the clock ticks more frequently. Both
the pendulum speed and clock speed increase.
(c) 0.001 œ 1È100 ˆ980$Î# ‰ dg Ê dg ¸ 0.977 cm/sec# Ê the new g ¸ 979 cm/sec#
53. The error in measurement dx œ (1%)(10) œ 0.1 cm; V œ x$ Ê dV œ 3x# dx œ 3(10)# (0.1) œ 30 cm$ Ê the
30 ‰
percentage error in the volume calculation is ˆ 1000
(100%) œ 3%
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Section 3.8 Linearization and Differentials
54. A œ s# Ê dA œ 2s ds; recall that ?A ¸ dA. Then k?Ak Ÿ (2%)A œ
Ê kdsk Ÿ
s#
(2s)(50)
œ
s
100
œ”
56. V œ
œ
4
3
1 D$
200
#
• a10 %b œ ”
1 r$ œ
4
3
10' 1
#
10' 1
6
$
1 ˆ D# ‰ œ
Ê kdVk Ÿ
1 D$
200
œ
s#
50
Ê kdAk Ÿ
s#
50
Ê k2s dsk Ÿ
s#
50
œ (1%) s Ê the error must be no more than 1% of the true value.
55. Given D œ 100 cm, dD œ 1 cm, V œ
10% 1
#
10' 1
6
2s#
100
4
3
$
1 ˆ D# ‰ œ
1 D$
6
Ê dV œ
1
#
D# dD œ
1
#
(100)# (1) œ
10% 1
#
. Then
dV
V
(100%)
• % œ 3%
1 D$
6
Ê dV œ
#
Ê ¹ 1D# dD¹ Ÿ
1 D#
#
3 ‰ 1D
dD; recall that ?V ¸ dV. Then k?Vk Ÿ (3%)V œ ˆ 100
Š 6 ‹
$
1 D$
#00
Ê kdDk Ÿ
D
100
œ (1%) D Ê the allowable percentage error in
measuring the diameter is 1%.
57. A 5% error in measuring t Ê dt œ (5%)t œ
t
20
32t#
20
. Then s œ 16t# Ê ds œ 32t dt œ 32t ˆ 20t ‰ œ
œ
16t#
10
" ‰
œ ˆ 10
s
œ (10%)s Ê a 10% error in the calculation of s.
58. From Example 8 we have
59. lim
xÄ0
È1 x
1 x#
œ
È1 0
1 #0
dV
V
œ4
dr
r
. An increase of 12.5% in r will give a 50% increase in V.
œ1
60. lim
xÄ0
tan x
x
œ lim ˆ sinx x ‰ ˆ cos" x ‰ œ (1)(1) œ 1
xÄ0
61. E(x) œ f(x) g(x) Ê E(x) œ f(x) m(x a) c. Then E(a) œ 0 Ê f(a) m(a a) c œ 0 Ê c œ f(a). Next
f(x) m(x a) c
f(a)
œ 0 Ê xlim
œ 0 Ê xlim
’ f(x)x xa
a m“ œ 0 (since c œ f(a))
Äa
Äa
Ê f w (a) m œ 0 Ê m œ f w (a). Therefore, g(x) œ m(x a) c œ f w (a)(x a) f(a) is the linear approximation,
as claimed.
we calculate m: xlim
Äa
E(x)
xa
62. (a) i. Qaab œ faab implies that b! œ faab.
ii. Since Qw axb œ b" #b# ax ab, Qw aab œ f w aab implies that b" œ f w aab.
iii. Since Qww axb œ #b# , Qww aab œ f ww aab implies that b2 œ
In summary, b! œ faab, b" œ f w aab, and b2 œ
(b) faxb œ a" xb"
ww
f aa b
# .
ww
f aa b
# .
f w axb œ "a" xb# a"b œ a" xb#
f ww axb œ #a" xb$ a"b œ #a" xb$
Since fa!b œ ", f w a!b œ ", and f ww a!b œ #, the coefficients are b! œ ", b" œ ", b# œ
#
approximation is Qaxb œ " x x .
(c)
#
#
œ ". The quadratic
As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(d) gaxb œ x"
gw axb œ "x#
gww axb œ #x$
Since ga"b œ ", gw a"b œ ", and gww a"b œ # , the coefficients are b! œ ", b" œ ", b# œ
#
#
œ ". The quadratic
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177
178
Chapter 3 Differentiation
approximation is Qaxb œ " ax "b ax "b# .
As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(e) haxb œ a" xb"Î#
hw axb œ "# a" xb"Î#
hww axb œ "% a" xb$Î#
Since ha!b œ ", hw a!b œ "# , and hww a!b œ "% , the coefficients are b! œ ", b" œ "# , b# œ
approximation is Qaxb œ " x
#
#
x
8
"%
2
œ "8 . The quadratic
.
As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(f) The linearization of any differentiable function uaxb at x œ a is Laxb œ uaab uw aabax ab œ b! b" ax ab, where
b! and b" are the coefficients of the constant and linear terms of the quadratic approximation. Thus, the linearization
for faxb at x œ ! is " x; the linearization for gaxb at x œ " is " ax "b or # x; and the linearization for haxb at
x œ ! is " x# .
63. (a) x œ 1
(b) x œ 1; m œ 2.5, e1 ¸ 2.7
x œ 0; m œ 1, e0 œ 1
x œ 1; m œ 0.3, e1 ¸ 0.4
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Section 3.8 Linearization and Differentials
179
64. If f has a horizontal tangent at x œ a, then f w (a) œ 0 and the linearization of f at x œ a is
L(x) œ f(a) f w (a)(x a) œ f(a) 0 † (x a) œ f(a). The linearization is a constant.
65. Find lvl when m œ "Þ!"m! . m œ
Ê lvl œ cÉ" dv œ
"!"$
"!!$
c†m#
Í !
Í
m3! Í" Ì
m#!
m#
m!
#
É" v#
c
Ê dv œ c † "# Š" m#
!
"!"# m#
"!!# !
m! ‰
ˆ "!!
œ
Ê mÉ" m#!
m# ‹
1!!!
"!"$ Ê" "!!#
"!"#
"Î#
v#
c#
œ m! Ê É" v#
c#
œ
m!
m
Ê"
#m#
Š m$! ‹dm, dm œ !Þ!"m! Ê dv œ
v#
c#
c m#!
m$
Ê" œ
m#
!
m#
m!#
m#
Ê v# œ c# Š" m! ‰
ˆ "!!
. mœ
m!#
m# ‹
"!"
"!! m! ,
¸ 0.69c. Body at rest Ê v! œ ! and v œ v! dv
Ê v œ 0.69c.
66. (a) The successive square roots of 2 appear to converge to the number 1. For tenth roots the convergence is more rapid.
(b) Successive square roots of 0.5 also converge to 1. In fact, successive square roots of any positive number converge
to 1.
A graph indicates what is going on:
Starting on the line y œ x, the successive square roots are found by moving to the graph of y œ Èx and then across to
the line y œ x again. From any positive starting value x, the iterates converge to 1.
67-70. Example CAS commands:
Maple:
with(plots):
a:= 1: f:=x -> 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]]
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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 )
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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 Šx1‹ †
$
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‰
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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
"
œ "3 x"Î& y"Î& œ 3(xy)
"Î&
dy
dx
dy
dx
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#
Š10p
dp
dq
2
dp
dq ‹
Ê 23 a5p# 2pb
&Î#
œ
dp
dq
(10p 2)
&Î#
œ a5p3(5p 2p1)b
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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
œ 2t; y œ 3 sin 2x Ê
œ 6 cos a2t# b † 2t Ê
58. t œ au# 2ub
"Î$
œ 2 au# 2ub
"Î$
w
Ê
dt
du
œ
5; thus
"
3
dy
dt ¹ t=0
au# 2ub
ds
du
œ
ds
dt
†
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
w
(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)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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
2
3
dy
dx
œ 2 sin x Ê
a3y# 1b (2 cos x) (2 sin x) Š6y
"
3
œ
"
œ cos ˆÈr 2‰ Š #È
‹
r
† 8 cos ˆs 16 ‰‘
(2)t 1) œ )# Ê
dy
dx
d)
dt
œ
) #
2)t1
; 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 ‹
a3y# 1b#
x#Î$ 3" y#Î$
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
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$
dy
ˆ #Î$ ‰ ˆ 23
dx ‹ y
œ #"
dy
dx
x"Î$ ‰
#
ax#Î$ b
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)
#(th)1 #t1 œ 2t 1 (2t 2h 1)
"
"
and
f(t
h)
œ
Ê
œ
2t 1
#(t h) 1
h
h
(2t 2h 1)(2t 1)h
f(t h) f(t)
2h
2
2
w
œ
Ê
f
(t)
œ
lim
œ
lim
(2t 2h 1)(2t 1)h
(2t 2h 1)(2t 1)
h
hÄ!
h Ä ! (2t 2h 1)(#t 1)
#
(2t 1)#
63. f(t) œ
œ
œ
g(x h) g(x)
h
lim g(x h)h g(x) œ lim
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!
xÄ!
xÄ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Chapter 3 Practice Exercises
185
follows that f is differentiable at x œ 0.
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
Ê 2œ
œ 6x# 6x 12; the tangent is parallel to the x-axis when
dy
dx
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)
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186
Chapter 3 Differentiation
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)
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
(b) The tangent is parallel to the line y œ È2 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
dy
dx
œ
x(1 cos x) (1 sin x)(1)
x#
Ê m" œ
dy
dx ¹ x=1
œ
1 #
1#
œ 1 and m# œ
Since m" œ m"# the tangents intersect at right angles.
75. y œ tan x, 1# x 1
#
Ê
dy
dx
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
œ sin x Ê
dy
dx
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
x1
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
x1
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.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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
1yx
yx
3
2
x"Î# 3y"Î#
y œ 4 "4 (x 1) œ 4" x dy
dx
dy
dx ‹
dy
dx
y œ 2Èxy Ê
dy
dx
œ
y$ a3x# b“ 2y
a3x$ y# 2y 1b œ 1 3x# y$ Ê
dy
dx
2Èxy y
x
œ
x"Î#
2y"Î#
Ê
dy
dx ¹ (1ß4)
dy
dx
dy
dx ¹ (6ß2)
œ
3
4
Ê
œ
dy
dx ¹ (4ß1)
4
5
x
5
4
11
5
.
œ "4 Ê the tangent line is
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
Therefore, the curve has slope "# at ("ß ") but the slope is undefined at ("ß 1).
88. y œ sin (x sin x) Ê
œ2
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
dy
dx ¹ (1ßc1) is
dy
dx
œ " 3x# y$
undefined.
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
"
#
tan
1
3
œ 2 cos$ ˆ 13 ‰ œ
90. x œ " "
t#
"
#
sec t Ê
œ
È3
#
dy
dx
œ
dy/dt
dx/dt
œ
"
#
sec
1
3
and y œ
"
#
sec t tan t
"
#
# sec t
œ
tan t
sec t
œ sin t Ê
œ1 Ê yœ
È3
#
x 4" ;
d# y
dx#
"
4
,yœ"
3
t
Ê
dy
dx
œ
dy/dt
dx/dt
œ
Š t3# ‹
Š t2$ ‹
œ 32 t Ê
dy
dx ¹ tœ2
dy
dx ¹ tœ1Î3
œ
dyw /dt
dx/dt
œ sin
œ
"
#
1
3
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 Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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"
##
œ
5
4
and
187
188
Chapter 3 Differentiation
yœ1
3
#
œ "# Ê y œ 3x "3
4
d# y
dx#
;
œ
dyw /dt
dx/dt
œ
ˆ #3 ‰
Š t2$ ‹
œ
3 $
4 t
d# y
dx# ¹ tœ2
Ê
œ
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!
xÄ!
œ 1. Therefore, to make g continuous at the origin, define g(0) œ 1.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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
2rh 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
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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# r#
b œ 2r and
Ê
œ
da
dt
db
dt
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
"Î#
#h dh œ
1rh
Èr# h# dh.
Height changes from h! to h! dh
1 r h! adhb
Ér# h#!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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.
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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 !.
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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 c 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
is a parabola. At x œ #, the value of the y-coordinate on the parabola is
$
#
#, the graph of f
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.
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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(xh)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)!
ˆm1‰
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
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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 $% ‰
s!
#k
œ
s!
s ! s "
œ
av! b"Î$
"Î$
av! b ˆ $% v! ‰
"Î$
¸ "" hr.
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196
Chapter 3 Differentiation
NOTES:
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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.
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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
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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
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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
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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 ".
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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
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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
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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
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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. Therefore, to minimize the cost of construction, the pipeline should be placed
from the docking facility to point B, 3.58 miles along the shore from point A, and then along the shore from B to the
refinery.
(b) If the per mile cost of underwater construction is p, then 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 be infinite to justify running the pipe directly from the docking facility to point A (i.e., for
xc to be zero). For all values of p !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.
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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.
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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# %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# .
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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.)
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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.
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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 ):
<<Miscellaneous `RealOnly`
Clear[f,x]
a = 1; b = 10/3;
f[x_] =2 2x 3 x2/3
f'[x]
Plot[{f[x], f'[x]}, {x, a, b}]
NSolve[f'[x]==0, x]
{f[a], f[0], f[x]/.%, f[b]//N
In more complicated expressions, NSolve may not yield results. In this case, an approximate solution (say 1.1 here)
is observed from the graph and the following command is used:
FindRoot[f'[x]==0,{x, 1.1}]
4.2 THE MEAN VALUE THEOREM
1. When f(x) œ x# 2x 1 for 0 Ÿ x Ÿ 1, then
2. When f(x) œ x#Î$ for 0 Ÿ x Ÿ 1, then
3. When f(x) œ x "
x
for
"
#
f(1)f(0)
1 0
Ÿ x Ÿ 2, then
f(1)f(0)
1 0
œ f w (c) Ê 1 œ ˆ 32 ‰ c"Î$ Ê c œ
f(2)f(1/2)
21/2
4. When f(x) œ Èx 1 for 1 Ÿ x Ÿ 3, then
œ f w (c) Ê 3 œ 2c 2 Ê c œ #" .
f(3)f(1)
3 1
œ f w (c) Ê 0 œ " œ f w (c) Ê
È2
#
œ
"
c#
8
#7 .
Ê c œ 1.
"
#Èc1
Ê c œ #3 .
5. Does not; f(x) is not differentiable at x œ 0 in ("ß 8).
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Section 4.2 The Mean Value Theorem
6. Does; f(x) is continuous for every point of [0ß 1] and differentiable for every point in (0ß 1).
7. Does; f(x) is continuous for every point of [0ß 1] and differentiable for every point in (0ß 1).
8. Does not; f(x) is not continuous at x œ 0 because lim c f(x) œ 1 Á 0 œ f(0).
xÄ!
9. Since f(x) is not continuous on 0 Ÿ x Ÿ 1, Rolle's Theorem does not apply:
Á 0 œ f(1).
lim f(x) œ lim c x œ 1
x Ä 1c
xÄ1
10. Since f(x) must be continuous at x œ 0 and x œ 1 we have lim b f(x) œ a œ f(0) Ê a œ 3 and
xÄ!
lim c f(x) œ lim b f(x) Ê 1 3 a œ m b Ê 5 œ m b. Since f(x) must also be differentiable at
xÄ1
xÄ1
x œ 1 we have lim c f w (x) œ lim b f w (x) Ê 2x 3k x=1 œ mk x=1 Ê 1 œ m. Therefore, a œ 3, m œ 1 and b œ 4.
xÄ1
xÄ1
11. (a) i
ii
iii
iv
(b) Let r" and r# be zeros of the polynomial P(x) œ xn an-1 xn-1 á a" x a! , then P(r" ) œ P(r# ) œ 0.
Since polynomials are everywhere continuous and differentiable, by Rolle's Theorem Pw (r) œ 0 for some r
between r" and r# , where Pw (x) œ nxn-1 (n 1) an-1 xn-2 á a" .
12. With f both differentiable and continuous on [aß b] and f(r" ) œ f(r# ) œ f(r$ ) œ 0 where r" , r# and r$ are in [aß b],
then by Rolle's Theorem there exists a c" between r" and r# such that f w (c" ) œ 0 and a c# between r# and r$
such that f w (c# ) œ 0. Since f w is both differentiable and continuous on [aß b], Rolle's Theorem again applies and
we have a c$ between c" and c# such that f w w (c$ ) œ 0. To generalize, if f has n1 zeros in [aß b] and f ÐnÑ is
continuous on [aß b], then f ÐnÑ has at least one zero between a and b.
13. Since f ww exists throughout [aß b] the derivative function f w is continuous there. If f w has more than one zero in
[aß b], say f w (r" ) œ f w (r# ) œ 0 for r" Á r# , then by Rolle's Theorem there is a c between r" and r# such that
f ww (c) œ 0, contrary to f ww 0 throughout [aß b]. Therefore f w has at most one zero in [aß b]. The same argument
holds if f ww 0 throughout [aß b].
14. If f(x) is a cubic polynomial with four or more zeros, then by Rolle's Theorem f w (x) has three or more zeros,
f ww (x) has 2 or more zeros and f www (x) has at least one zero. This is a contradiction since f www (x) is a non-zero
constant when f(x) is a cubic polynomial.
15. With f(2) œ 11 0 and f(1) œ 1 0 we conclude from the Intermediate Value Theorem that
f(x) œ x% 3x 1 has at least one zero between 2 and 1. Then 2 x 1 Ê ) x$ 1
Ê 32 4x$ 4 Ê 29 4x$ 3 1 Ê f w (x) 0 for 2 x 1 Ê f(x) is decreasing on [#ß 1]
Ê f(x) œ 0 has exactly one solution in the interval (#ß 1).
16. f(x) œ x$ 4
x#
7 Ê f w (x) œ 3x# 8
x$
0 on (_ß 0) Ê f(x) is increasing on (_ß 0). Also, f(x) 0 if
x 2 and f(x) 0 if 2 x 0 Ê f(x) has exactly one zero in (_ß !).
17. g(t) œ Èt Èt 1 4 Ê gw (t) œ
"
#È t
"
2Èt1
0 Ê g(t) is increasing for t in (!ß _); g(3) œ È3 2 0
and g(15) œ È15 0 Ê g(t) has exactly one zero in (!ß _)Þ
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212
Chapter 4 Applications of Derivatives
18. g(t) œ
"
"t
È1 t 3.1 Ê gw (t) œ
"
("t)#
"
2 È 1 t
0 Ê g(t) is increasing for t in (1ß 1);
g(0.99) œ 2.5 and g(0.99) œ 98.3 Ê g(t) has exactly one zero in (1ß 1).
19. r()) œ ) sin# ˆ 3) ‰ 8 Ê rw ()) œ 1 sin ˆ 3) ‰ cos ˆ 3) ‰ œ 1 "3 sin ˆ 23) ‰ 0 on (_ß _) Ê r()) is
increasing on (_ß _); r(0) œ 8 and r(8) œ sin# ˆ 83 ‰ 0 Ê r()) has exactly one zero in (_ß _).
2
3
20. r()) œ 2) cos# ) È2 Ê rw ()) œ 2 2 sin ) cos ) œ 2 sin 2) 0 on (_ß _) Ê r()) is increasing on
(_ß _); r(#1) œ 41 cos (#1) È2 œ 41 1 È2 0 and r(21) œ 41 1 È2 0 Ê r()) has
exactly one zero in (_ß _).
0 on ˆ!ß 1# ‰ Ê r()) is increasing on ˆ!ß 1# ‰ ;
r(0.1) ¸ 994 and r(1.57) ¸ 1260.5 Ê r()) has exactly one zero in ˆ!ß 1# ‰ .
21. r()) œ sec ) "
)$
5 Ê rw ()) œ (sec ))(tan )) 3
)%
22. r()) œ tan ) cot ) ) Ê rw ()) œ sec# ) csc# ) 1 œ sec# ) cot# ) 0 on ˆ!ß 1# ‰ Ê r()) is increasing
on ˆ0ß 1# ‰ ; r ˆ 14 ‰ œ 14 0 and r(1.57) ¸ 1254.2 Ê r()) has exactly one zero in ˆ!ß 1# ‰ .
23. By Corollary 1, f w (x) œ 0 for all x Ê f(x) œ C, where C is a constant. Since f(1) œ 3 we have C œ 3
Ê f(x) œ 3 for all x.
24. g(x) œ 2x 5 Ê gw (x) œ 2 œ f w (x) for all x. By Corollary 2, f(x) œ g(x) C for some constant C. Then
f(0) œ g(0) C Ê 5 œ 5 C Ê C œ 0 Ê f(x) œ g(x) œ 2x 5 for all x.
25. g(x) œ x# Ê gw (x) œ 2x œ f w (x) for all x. By Corollary 2, f(x) œ g(x) C.
(a) f(0) œ 0 Ê 0 œ g(0) C œ 0 C Ê C œ 0 Ê f(x) œ x# Ê f(2) œ 4
(b) f(1) œ 0 Ê 0 œ g(1) C œ 1 C Ê C œ 1 Ê f(x) œ x# 1 Ê f(2) œ 3
(c) f(2) œ 3 Ê 3 œ g(2) C Ê 3 œ 4 C Ê C œ 1 Ê f(x) œ x# 1 Ê f(2) œ 3
26. g(x) œ mx Ê gw (x) œ m, a constant. If f w (x) œ m, then by Corollary 2, f(x) œ g(x) b œ mx b where
b is a constant. Therefore all functions whose derivatives are constant can be graphed as straight lines
y œ mx b.
27. (a) y œ
x#
#
C
(b) y œ
28. (a) y œ x# C
x$
3
C
(b) y œ x# x C
29. (a) yw œ x# Ê y œ
"
x
C
(b) y œ x "
x
"
#
31. (a) y œ "# cos 2t C
(c) y œ "
#
cos 2t 2 sin
32. (a) y œ tan ) C
C
(c) y œ 5x "
x
C
(b) y œ 2Èx C
(b) y œ 2 sin
t
#
x%
4
(c) y œ x$ x# x C
C
x"Î# Ê y œ x"Î# C Ê y œ Èx C
(c) y œ 2x# 2Èx C
30. (a) yw œ
(c) y œ
t
#
C
C
(b) yw œ )"Î# Ê y œ
2
3
)$Î# C
(c) y œ
33. f(x) œ x# x C; 0 œ f(0) œ 0# 0 C Ê C œ 0 Ê f(x) œ x# x
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3
)$Î# tan ) C
Section 4.2 The Mean Value Theorem
213
34. g(x) œ "x x# C; 1 œ g(1) œ "1 (1)# C Ê C œ 1 Ê g(x) œ x" x# 1
35. r()) œ 8) cot ) C; 0 œ r ˆ 14 ‰ œ 8 ˆ 14 ‰ cot ˆ 14 ‰ C Ê 0 œ 21 1 C Ê C œ 21 1
Ê r()) œ 8) cot ) 21 1
36. r(t) œ sec t t C; 0 œ r(0) œ sec (0) 0 C Ê C œ 1 Ê r(t) œ sec t t 1
37. v œ
ds
dt
œ *Þ)t & Ê s œ %Þ*t# &t C; at s œ "! and t œ ! we have C œ "! Ê s œ %Þ*t# &t "!
38. v œ
ds
dt
œ $#t # Ê s œ "'t# #t C; at s œ % and t œ
39. v œ
ds
dt
œ sina1tb Ê s œ 1" cosa1tb C; at s œ ! and t œ ! we have C œ
40. v œ
ds
dt
œ 12 cosˆ #1t ‰ Ê s œ sinˆ #1t ‰ C; at s œ " and t œ 1# we have C œ " Ê s œ sinˆ #1t ‰ "
"
#
we have C œ " Ê s œ 't# #t "
"
1
Êsœ
" cosa1tb
1
41. a œ $# Ê v œ $#t C" ; at v œ #! and t œ ! we have C" œ #! Ê v œ $#t #! Ê s œ "'t# #!t C# ; at s œ & and
t œ ! we have C# œ & Ê s œ "'t# #!t &
42. a œ 9.8 Ê v œ 9.8t C" ; at v œ $ and t œ ! we have C" œ $ Ê v œ *Þ)t $ Ê s œ %Þ*t# $t C# ; at s œ ! and
t œ ! we have C# œ ! Ê s œ %Þ*t# $t
43. a œ %sina#tb Ê v œ #cosa#tb C" ; at v œ # and t œ ! we have C" œ ! Ê v œ #cosa#tb Ê s œ sina#tb C# ; at s œ $
and t œ ! we have C# œ $ Ê s œ sina#tb $
Ê v œ 1$ sinˆ $1t ‰ C" ; at v œ ! and t œ ! we have C" œ ! Ê v œ 1$ sinˆ $1t ‰ Ê s œ cosˆ $1t ‰ C# ; at
s œ " and t œ ! we have C# œ ! Ê s œ cosˆ $1t ‰
44. a œ
*
ˆ $t ‰
1# cos 1
45. If T(t) is the temperature of the thermometer at time t, then T(0) œ 19° C and T(14) œ 100° C. From the
Mean Value Theorem there exists a 0 t! 14 such that
T(14) T(0)
14 0
œ 8.5° C/sec œ Tw (t! ), the rate at which
the temperature was changing at t œ t! as measured by the rising mercury on the thermometer.
46. Because the trucker's average speed was 79.5 mph, by the Mean Value Theorem, the trucker must have been going that
speed at least once during the trip.
47. Because its average speed was approximately 7.667 knots, and by the Mean Value Theorem, it must have been going that
speed at least once during the trip.
48. The runner's average speed for the marathon was approximately 11.909 mph. Therefore, by the Mean Value Theorem, the
runner must have been going that speed at least once during the marathon. Since the initial speed and final speed are both 0
mph and the runner's speed is continuous, by the Intermediate Value Theorem, the runner's speed must have been 11 mph
at least twice.
49. Let d(t) represent the distance the automobile traveled in time t. The average speed over 0 Ÿ t Ÿ 2 is
d(2) d(0)
#0 .
The Mean Value Theorem says that for some 0 t! 2, dw (t! ) œ
d(2) d(0)
#0 .
The value dw (t! ) is
the speed of the automobile at time t! (which is read on the speedometer).
50. aatb œ vw atb œ "Þ' Ê vatb œ "Þ't C; at a!ß !b we have C œ ! Ê vatb œ "Þ't. When t œ $!, then va$!b œ %) m/sec.
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214
Chapter 4 Applications of Derivatives
51. The conclusion of the Mean Value Theorem yields
52. The conclusion of the Mean Value Theorem yields
"
b
"a
ba
b # a#
ba
b‰
œ c"# Ê c# ˆ a ab
œ a b Ê c œ Èab.
œ 2c Ê c œ
a b
# .
53. f w (x) œ [cos x sin (x 2) sin x cos (x 2)] 2 sin (x 1) cos (x 1) œ sin (x x 2) sin 2(x 1)
œ sin (2x 2) sin (2x 2) œ 0. Therefore, the function has the constant value f(0) œ sin# 1 ¸ 0.7081
which explains why the graph is a horizontal line.
54. (a) faxb œ ax #bax "bxax "bax #b œ x& &x$ %x is one possibility.
(b) Graphing faxb œ x& &x$ %x and f w axb œ &x% "&x# % on Ò$ß $Ó by Ò(ß (Ó we see that each x-intercept of
f w axb lies between a pair of x-intercepts of faxb, as expected by Rolle's Theorem.
(c) Yes, since sin is continuous and differentiable on a _ß _b.
55. faxb must be zero at least once between a and b by the Intermediate Value Theorem. Now suppose that faxb is zero twice
between a and b. Then by the Mean Value Theorem, f w axb would have to be zero at least once between the two zeros of
faxb, but this can't be true since we are given that f w axb Á ! on this interval. Therefore, faxb is zero once and only once
between a and b.
56. Consider the function kaxb œ faxb gaxb. kaxb is continuous
and differentiable on Òa, bÓ, and since kaab œ faab gaab and
kabb œ fabb gabb, by the Mean Value Theorem, there must
be a point c in aa, bb where kw acb œ !. But since
kw acb œ f w acb gw acb, this means that f w acb œ gw acb, and c is a
point where the graphs of f and g have tangent lines with the
same slope, so these lines are either parallel or are the same
line.
57. Yes. By Corollary 2 we have f(x) œ g(x) c since f w (x) œ gw (x). If the graphs start at the same point x œ a,
then f(a) œ g(a) Ê c œ 0 Ê f(x) œ g(x).
58. Let f(x) œ sin x for a Ÿ x Ÿ b. From the Mean Value Theorem there exists a c between a and b such that
sin b sin a
sin a
sin a ¸
œ cos c Ê 1 Ÿ sin bb Ÿ 1 Ê ¸ sin bb Ÿ 1 Ê ksin b sin ak Ÿ kb ak.
ba
a
a
59. By the Mean Value Theorem we have
w
we have f(b) f(a) 0 Ê f (c) 0.
f(b) f(a)
ba
œ f w (c) for some point c between a and b. Since b a 0 and f(b) f(a),
60. The condition is that f w should be continuous over [aß b]. The Mean Value Theorem then guarantees the
existence of a point c in (aß b) such that
f(b) f(a)
ba
w
œ f w (c). If f w is continuous, then it has a minimum and
maximum value on [aß b], and min f w Ÿ f (c) Ÿ max f w , as required.
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Section 4.3 Monotonic Functions and the First Derivative Test
61. f w (x) œ a1 x% cos xb
"
Ê f ww (x) œ a1 x% cos xb
#
a4x$ cos x x% sin xb
#
œ x$ a1 x% cos xb (4 cos x x sin x) 0 for 0 Ÿ x Ÿ 0.1 Ê f w (x) is decreasing when 0 Ÿ x Ÿ 0.1
f(0.1) "
0.1
Ê min f w ¸ 0.9999 and max f w œ 1. Now we have 0.9999 Ÿ
Ÿ 1 Ê 0.09999 Ÿ f(0.1) 1 Ÿ 0.1
Ê 1.09999 Ÿ f(0.1) Ÿ 1.1.
62. f w (x) œ a1 x% b
"
Ê f ww (x) œ a1 x% b
#
a4x$ b œ
4x$
$
a1 x % b
0 for 0 x Ÿ 0.1 Ê f w (x) is increasing when
0 Ÿ x Ÿ 0.1 Ê min f w œ 1 and max f w œ 1.0001. Now we have 1 Ÿ
f(0.1) 2
0.1
Ÿ 1.0001
Ê 0.1 Ÿ f(0.1) 2 Ÿ 0.10001 Ê 2.1 Ÿ f(0.1) Ÿ 2.10001.
63. (a) Suppose x 1, then by the Mean Value Theorem
(b)
f(1)
Mean Value Theorem f(x)x 0
1
f(x) f(1)
Yes. From part (a), lim c x 1
xÄ1
f(x) f(1)
x1
0 Ê f(x) f(1). Suppose x 1, then by the
Ê f(x) f(1). Therefore f(x)
1 for all x since f(1) œ 1.
f(x) f(1)
x1
Ÿ 0 and lim b
0. Since f w (1) exists, these two one-sided
xÄ1
limits are equal and have the value f w (1) Ê f w (1) Ÿ 0 and f w (1) 0 Ê f w (1) œ 0.
64. From the Mean Value Theorem we have
has only one solution c œ q
#p .
f(b) f(a)
ba
œ f w (c) where c is between a and b. But f w (c) œ 2pc q œ 0
(Note: p Á 0 since f is a quadratic function.)
4.3 MONOTONIC FUNCTIONS AND THE FIRST DERIVATIVE TEST
1. (a) f w (x) œ x(x 1) Ê critical points at 0 and 1
(b) f w œ ± ± Ê increasing on (_ß !) and ("ß _), decreasing on (!ß ")
!
"
(c) Local maximum at x œ 0 and a local minimum at x œ 1
2. (a) f w (x) œ (x 1)(x 2) Ê critical points at 2 and 1
(b) f w œ ± ± Ê increasing on (_ß #) and ("ß _), decreasing on (2ß ")
#
"
(c) Local maximum at x œ 2 and a local minimum at x œ 1
3. (a) f w (x) œ (x 1)# (x 2) Ê critical points at 2 and 1
(b) f w œ ± ± Ê increasing on (2ß 1) and ("ß _), decreasing on (_ß 2)
#
"
(c) No local maximum and a local minimum at x œ 2
4. (a) f w (x) œ (x 1)# (x 2)# Ê critical points at 2 and 1
(b) f w œ ± ± Ê increasing on (_ß 2) (#ß ") ("ß _), never decreasing
#
"
(c) No local extrema
5. (a) f w (x) œ (x 1)(x 2)(x 3) Ê critical points at 2, 1 and 3
(b) f w œ ± ± ± Ê increasing on (2ß 1) and ($ß _), decreasing on (_ß 2) and ("ß $)
#
"
$
(c) Local maximum at x œ 1, local minima at x œ 2 and x œ 3
6. (a) f w (x) œ (x 7)(x 1)(x 5) Ê critical points at 5, 1 and 7
(b) f w œ ± ± ± Ê increasing on (5ß 1) and (7ß _), decreasing on (_ß 5) and ("ß 7)
&
"
(
(c) Local maximum at x œ 1, local minima at x œ 5 and x œ 7
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216
Chapter 4 Applications of Derivatives
7. (a) f w (x) œ x"Î$ (x 2) Ê critical points at 2 and 0
(b) f w œ ± )( Ê increasing on (_ß 2) and (0ß _), decreasing on (2ß 0)
!
#
(c) Local maximum at x œ 2, local minimum at x œ 0
8. (a) f w (x) œ x"Î# (x 3) Ê critical points at 0 and 3
(b) f w œ ( ± Ê increasing on ($ß _), decreasing on (0ß 3)
!
$
(c) No local maximum and a local minimum at x œ 3
9. (a) g(t) œ t# 3t 3 Ê gw (t) œ 2t 3 Ê a critical point at t œ 3# ; gw œ ± , increasing on
$Î#
ˆ_ß 3# ‰ , decreasing on ˆ 3# ß _‰
(b) local maximum value of g ˆ 3# ‰ œ
(c) absolute maximum is
21
4
21
4
at t œ 3#
at t œ 3#
(d)
10. (a) g(t) œ 3t# 9t 5 Ê gw (t) œ 6t 9 Ê a critical point at t œ
3
#
; gw œ ± , increasing on
$Î#
ˆ_ß 32 ‰ , decreasing on ˆ 3# ß _‰
(b) local maximum value of g ˆ 3# ‰ œ
(c) absolute maximum is
47
4
at t œ
47
4
at t œ
3
#
3
#
(d)
11. (a) h(x) œ x$ 2x# Ê hw (x) œ 3x# 4x œ x(4 3x) Ê critical points at x œ 0, 43
Ê hw œ ± ± , increasing on ˆ0ß 43 ‰ , decreasing on (_ß !) and ˆ 43 ß _‰
!
%Î$
4
(b) local maximum value of h ˆ 43 ‰ œ 32
27 at x œ 3 ; local minimum value of h(0) œ 0 at x œ 0
(c) no absolute extrema
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Section 4.3 Monotonic Functions and the First Derivative Test
217
(d)
12. (a) h(x) œ 2x$ 18x Ê hw (x) œ 6x# 18 œ 6 Šx È3‹ Šx È3‹ Ê critical points at x œ „ È3
Ê hw œ | | , increasing on Š_ß È3‹ and ŠÈ$ß _‹ , decreasing on ŠÈ$ß È3‹
È$
È $
(b) a local maximum is h ŠÈ3‹ œ 12È3 at x œ È3; local minimum is h ŠÈ3‹ œ 12È3 at x œ È3
(c) no absolute extrema
(d)
13. (a) f()) œ 3)# 4)$ Ê f w ()) œ 6) 12)# œ 6)(1 2)) Ê critical points at ) œ 0,
"
#
Ê f w œ ± ± ,
!
"Î#
increasing on ˆ0ß "# ‰ , decreasing on (_ß !) and ˆ "# ß _‰
(b) a local maximum is f ˆ "# ‰ œ 4" at ) œ #" , a local minimum is f(0) œ 0 at ) œ 0
(c) no absolute extrema
(d)
14. (a) f()) œ 6) )$ Ê f w ()) œ 6 3)# œ 3 ŠÈ2 )‹ ŠÈ2 )‹ Ê critical points at ) œ „ È2 Ê
f w œ ± ± , increasing on ŠÈ2ß È2‹, decreasing on Š_ß È2‹ and ŠÈ2ß _‹
È#
È #
(b) a local maximum is f ŠÈ2‹ œ 4È2 at ) œ È2, a local minimum is f ŠÈ2‹ œ %È2 at ) œ È2
(c) no absolute extrema
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Chapter 4 Applications of Derivatives
(d)
15. (a) f(r) œ 3r$ 16r Ê f w (r) œ 9r# 16 Ê no critical points Ê f w œ , increasing on (_ß _), never
decreasing
(b) no local extrema
(c) no absolute extrema
(d)
16. (a) h(r) œ (r 7)$ Ê hw (r) œ 3(r 7)# Ê a critical point at r œ 7 Ê hw œ ± , increasing on
(
(_ß 7) ((ß _), never decreasing
(b) no local extrema
(c) no absolute extrema
(d)
17. (a) f(x) œ x% 8x# 16 Ê f w (x) œ 4x$ 16x œ 4x(x 2)(x 2) Ê critical points at x œ 0 and x œ „ 2
Ê f w œ ± ± ± , increasing on (#ß !) and (#ß _), decreasing on (_ß 2) and (!ß #)
#
!
#
(b) a local maximum is f(0) œ 16 at x œ 0, local minima are f a „ 2b œ 0 at x œ „ 2
(c) no absolute maximum; absolute minimum is 0 at x œ „ 2
(d)
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Section 4.3 Monotonic Functions and the First Derivative Test
219
18. (a) g(x) œ x% 4x$ 4x# Ê gw (x) œ 4x$ 12x# )x œ 4x(x 2)(x 1) Ê critical points at x œ 0, 1, 2
Ê gw œ ± ± ± , increasing on (0ß 1) and (#ß _), decreasing on (_ß 0) and (1ß #)
!
"
#
(b) a local maximum is g(1) œ 1 at x œ 1, local minima are g(0) œ 0 at x œ 0 and g(2) œ 0 at x œ 2
(c) no absolute maximum; absolute minimum is 0 at x œ 0, 2
(d)
19. (a) H(t) œ
3 %
# t
t' Ê Hw (t) œ 6t$ 6t& œ 6t$ (1 t)(" t) Ê critical points at t œ 0, „ 1
Ê Hw œ ± ± ± , increasing on (_ß 1) and (0ß 1), decreasing on ("ß 0) and ("ß _)
"
!
"
(b) the local maxima are H(1) œ "# at t œ 1 and H(1) œ "# at t œ 1, the local minimum is H(0) œ 0 at t œ 0
(c) absolute maximum is
"
#
at t œ „ 1; no absolute minimum
(d)
20. (a) K(t) œ 15t$ t& Ê Kw (t) œ 45t# 5t% œ 5t# (3 t)(3 t) Ê critical points at t œ 0, „ 3
Ê Kw œ ± ± ± , increasing on (3ß 0) (0ß 3), decreasing on (_ß 3) and (3ß _)
$
!
$
(b) a local maximum is K(3) œ 162 at t œ 3, a local minimum is K(3) œ 162 at t œ 3
(c) no absolute extrema
(d)
21. (a) g(x) œ xÈ8 x# œ x a8 x# b
"Î#
Ê gw (x) œ a8 x# b
"Î#
x ˆ "# ‰ a) x# b
"Î#
(2x) œ
2(2 x)(2 x)
ÊŠ2È2 x‹ Š2È2 x‹
Ê critical points at x œ „ 2, „ 2È2 Ê gw œ ( ± ± Ñ
, increasing on (#ß #), decreasing on
#
#
#È#
#È #
Š#È2ß #‹ and Š#ß #È2‹
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220
Chapter 4 Applications of Derivatives
(b) local maxima are g(2) œ 4 at x œ 2 and g Š2È2‹ œ 0 at x œ 2È2, local minima are g(2) œ 4 at
x œ 2 and g Š2È2‹ œ 0 at x œ 2È2
(c) absolute maximum is 4 at x œ 2; absolute minimum is 4 at x œ 2
(d)
22. (a) g(x) œ x# È5 x œ x# (5 x)"Î# Ê gw (x) œ 2x(5 x)"Î# x# ˆ "# ‰ (5 x)"Î# (1) œ
5x(4x)
2 È 5 x
Ê critical points at x œ 0, 4 and 5 Ê gw œ ± ± Ñ , increasing on (0ß 4), decreasing on (_ß !)
&
!
%
and (%ß &)
(b) a local maximum is g(4) œ 16 at x œ 4, a local minimum is 0 at x œ 0 and x œ 5
(c) no absolute maximum; absolute minimum is 0 at x œ 0, 5
(d)
23. (a) f(x) œ
x# 3
x#
Ê f w (x) œ
2x(x 2) ax# 3b (1)
(x 2)#
œ
(x 3)(x ")
(x #)#
Ê critical points at x œ 1, 3
Ê f w œ ± )( ± , increasing on (_ß 1) and ($ß _), decreasing on ("ß #) and (#ß $),
#
"
$
discontinuous at x œ 2
(b) a local maximum is f(1) œ 2 at x œ 1, a local minimum is f(3) œ 6 at x œ 3
(c) no absolute extrema
(d)
24. (a) f(x) œ
x$
3x# 1
Ê f w (x) œ
3x# a3x# 1b x$ (6x)
a3x# 1b#
œ
3x# ax# 1b
a3x# 1b#
Ê a critical point at x œ 0
w
Ê f œ ± , increasing on (_ß !) (!ß _), and never decreasing
!
(b) no local extrema
(c) no absolute extrema
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Section 4.3 Monotonic Functions and the First Derivative Test
221
(d)
25. (a) f(x) œ x"Î$ (x 8) œ x%Î$ 8x"Î$ Ê f w (x) œ
4
3
x"Î$ 83 x#Î$ œ
w
4(x 2)
3x#Î$
Ê critical points at x œ 0, 2
Ê f œ ± )( , increasing on (#ß !) (!ß _), decreasing on (_ß 2)
!
#
(b) no local maximum, a local minimum is f(2) œ 6 $È2 ¸ 7.56 at x œ 2
(c) no absolute maximum; absolute minimum is 6 $È2 at x œ 2
(d)
26. (a) g(x) œ x#Î$ (x 5) œ x&Î$ 5x#Î$ Ê gw (x) œ
5
3
x#Î$ 10
3
x"Î$ œ
5(x 2)
$
3È
x
Ê critical points at x œ 2 and
x œ 0 Ê gw œ ± )( , increasing on (_ß 2) and (!ß _), decreasing on (2ß !)
!
#
(b) local maximum is g(2) œ 3 $È4 ¸ 4.762 at x œ 2, a local minimum is g(0) œ 0 at x œ 0
(c) no absolute extrema
(d)
27. (a) h(x) œ x"Î$ ax# 4b œ x(Î$ 4x"Î$ Ê hw (x) œ
x œ 0,
„2
È7
7
3
x%Î$ 43 x#Î$ œ
ŠÈ7x 2‹ ŠÈ7x #‹
$ #
3È
x
Ê critical points at
2
Ê hw œ ±
)( ± , increasing on Š_ß È
‹ and Š È27 ß _‹ , decreasing on
7
!
È
È
#Î (
#Î (
2
ß !‹ and Š!ß È27 ‹
ŠÈ
7
2
(b) local maximum is h Š È
‹œ
7
$
24 È
2
7(Î'
¸ 3.12 at x œ
2
È7 ,
the local minimum is h Š È27 ‹ œ $
24 È
2
7(Î'
(c) no absolute extrema
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¸ 3.12
222
Chapter 4 Applications of Derivatives
(d)
28. (a) k(x) œ x#Î$ ax# 4b œ x)Î$ 4x#Î$ Ê kw (x) œ
8
3
x&Î$ 83 x"Î$ œ
8(x 1)(x 1)
$
3È
x
Ê critical points at
x œ 0, „ 1 Ê kw œ ± )( ± , increasing on ("ß !) and ("ß _), decreasing on (_ß 1)
!
"
"
and (!ß 1)
(b) local maximum is k(0) œ 0 at x œ 0, local minima are k a „ 1b œ 3 at x œ „ 1
(c) no absolute maximum; absolute minimum is 3 at x œ „ 1
(d)
29. (a) f(x) œ 2x x# Ê f w (x) œ 2 2x œ 2(1 x) Ê a critical point at x œ 1 Ê f w œ ± ] and f(1) œ 1,
#
"
f(2) œ 0 Ê a local maximum is 1 at x œ 1, a local minimum is 0 at x œ 2
(b) absolute maximum is 1 at x œ 1; no absolute minimum
(c)
30. (a) f(x) œ (x 1)# Ê f w (x) œ 2(x 1) Ê a critical point at x œ 1 Ê f w œ ± ] and f(1) œ 0, f(0) œ 1
!
"
Ê a local maximum is 1 at x œ 0, a local minimum is 0 at x œ 1
(b) no absolute maximum; absolute minimum is 0 at x œ 1
(c)
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Section 4.3 Monotonic Functions and the First Derivative Test
223
31. (a) g(x) œ x# 4x 4 Ê gw (x) œ 2x 4 œ 2(x 2) Ê a critical point at x œ 2 Ê gw œ [ ± and
"
#
g(1) œ 1, g(2) œ 0 Ê a local maximum is 1 at x œ 1, a local minimum is g(2) œ 0 at x œ 2
(b) no absolute maximum; absolute minimum is 0 at x œ 2
(c)
32. (a) g(x) œ x# 6x 9 Ê gw (x) œ 2x 6 œ 2(x 3) Ê a critical point at x œ 3 Ê gw œ [ ± and
%
$
g(4) œ 1, g(3) œ 0 Ê a local maximum is 0 at x œ 3, a local minimum is 1 at x œ 4
(b) absolute maximum is 0 at x œ 3; no absolute minimum
(c)
33. (a) f(t) œ 12t t$ Ê f w (t) œ 12 3t# œ 3(2 t)(2 t) Ê critical points at t œ „ 2 Ê f w œ [ ± ± $
#
#
and f(3) œ 9, f(2) œ 16, f(2) œ 16 Ê local maxima are 9 at t œ 3 and 16 at t œ 2, a local
minimum is 16 at t œ 2
(b) absolute maximum is 16 at t œ 2; no absolute minimum
(c)
34. (a) f(t) œ t$ 3t# Ê f w (t) œ 3t# 6t œ 3t(t 2) Ê critical points at t œ 0 and t œ 2
Ê f w œ ± ± ] and f(0) œ 0, f(2) œ 4, f(3) œ 0 Ê a local maximum is 0 at t œ 0 and t œ 3, a
$
!
#
local minimum is 4 at t œ 2
(b) absolute maximum is 0 at t œ 0, 3; no absolute minimum
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224
Chapter 4 Applications of Derivatives
(c)
x$
3
2x# 4x Ê hw (x) œ x# 4x 4 œ (x 2)# Ê a critical point at x œ 2 Ê hw œ [ ± and
!
#
h(0) œ 0 Ê no local maximum, a local minimum is 0 at x œ 0
(b) no absolute maximum; absolute minimum is 0 at x œ 0
(c)
35. (a) h(x) œ
36. (a) k(x) œ x$ 3x# 3x 1 Ê kw (x) œ 3x# 6x 3 œ 3(x 1)# Ê a critical point at x œ 1
Ê kw œ ± ] and k(1) œ 0, k(0) œ 1 Ê a local maximum is 1 at x œ 0, no local minimum
!
"
(b) absolute maximum is 1 at x œ 0; no absolute minimum
(c)
37. (a) f(x) œ
x
#
2 sin ˆ x# ‰ Ê f w (x) œ
Ê f w œ [ ± ]
!
#1
#1Î$
cos ˆ x# ‰ , f w (x) œ 0 Ê cos ˆ x# ‰ œ "# Ê a critical point at x œ 231
and f(0) œ 0, f ˆ 231 ‰ œ 13 È3, f(21) œ 1 Ê local maxima are 0 at x œ 0 and 1
"
#
at x œ 21, a local minimum is 13 È3 at x œ 231
(b) The graph of f rises when f w 0, falls when f w 0,
and has a local minimum value at the point where f w
changes from negative to positive.
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Section 4.3 Monotonic Functions and the First Derivative Test
225
38. (a) f(x) œ 2 cos x cos# x Ê f w (x) œ 2 sin x 2 cos x sin x œ 2(sin x)(1 cos x) Ê critical points at
x œ 1, 0, 1 Ê f w œ [ ± ] and f(1) œ 1, f(0) œ 3, f(1) œ 1 Ê a local maximum is 1 at
1
1
!
x œ „ 1, a local minimum is 3 at x œ 0
(b) The graph of f rises when f w 0, falls when f w 0,
and has local extreme values where f w œ 0. The
function f has a local minimum value at x œ 0, where
the values of f w change from negative to positive.
39. (a) f(x) œ csc# x 2 cot x Ê f w (x) œ 2(csc x)(csc x)(cot x) 2 acsc# xb œ 2 acsc# xb (cot x 1) Ê a critical
point at x œ 14 Ê f w œ ( ± ) and f ˆ 14 ‰ œ 0 Ê no local maximum, a local minimum is 0 at x œ 14
1
!
1Î%
w
(b) The graph of f rises when f 0, falls when f w 0,
and has a local minimum value at the point where
f w œ 0 and the values of f w change from negative to
positive. The graph of f steepens as f w (x) Ä „ _.
40. (a) f(x) œ sec# x 2 tan x Ê f w (x) œ 2(sec x)(sec x)(tan x) 2 sec# x œ a2 sec# xb (tan x 1) Ê a critical point
at x œ 14 Ê f w œ ( ± ) and f ˆ 14 ‰ œ 0 Ê no local maximum, a local minimum is 0 at x œ 14
1Î#
1Î#
1Î%
w
(b) The graph of f rises when f 0, falls when f w 0,
and has a local minimum value where f w œ 0 and the
values of f w change from negative to positive.
41. h()) œ 3 cos ˆ #) ‰ Ê hw ()) œ 3# sin ˆ #) ‰ Ê hw œ [ ] , (!ß $) and (#1ß 3) Ê a local maximum is 3 at ) œ 0,
!
#1
a local minimum is 3 at ) œ 21
42. h()) œ 5 sin ˆ #) ‰ Ê hw ()) œ
minimum is 0 at ) œ 0
5
#
cos ˆ #) ‰ Ê hw œ [ ] , (!ß 0) and (1ß 5) Ê a local maximum is 5 at ) œ 1, a local
1
!
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226
Chapter 4 Applications of Derivatives
43. (a)
(b)
44. (a)
(b)
(c)
(d)
(c)
45. (a)
(b)
46. (a)
(b)
(d)
47. f(x) œ x$ 3x 2 Ê f w (x) œ 3x# 3 œ 3(x 1)(x 1) Ê f w œ ± ± Ê rising for x œ c œ # since
"
"
f w (x) 0 for x œ c œ 2.
48. f(x) œ ax# bx c œ a ˆx# ba x‰ c œ a Šx# ba x b#
4a# ‹
b#
4a
c œ a ˆx b ‰#
2a
b# 4ac
4a
, a parabola whose
b
vertex is at x œ 2a
. Thus when a 0, f is increasing on ˆ 2ab ß _‰ and decreasing on ˆ_ß #ab ‰ ; when a 0,
f is increasing on ˆ_ß #ab ‰ and decreasing on ˆ #ab ß _‰ . Also note that f w (x) œ 2ax b œ 2a ˆx #ba ‰ Ê for
a 0, f w œ | ; for a 0, f w œ ± .
bÎ2a
bÎ2a
4.4 CONCAVITY AND CURVE SKETCHING
x$
3
x#
#
Ê yw œ x# x 2 œ (x 2)(x 1) Ê yww œ 2x 1 œ 2 ˆx "# ‰ . The graph is rising on
(_ß 1) and (#ß _), falling on ("ß #), concave up on ˆ "# ß _‰ and concave down on ˆ_ß "# ‰ . Consequently,
a local maximum is 3# at x œ 1, a local minimum is 3 at x œ 2, and ˆ "# ß 34 ‰ is a point of inflection.
1. y œ
2x "
3
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Section 4.4 Concavity and Curve Sketching
2. y œ
227
2x# 4 Ê yw œ x$ 4x œ x ax# 4b œ x(x 2)(x 2) Ê yww œ 3x# 4 œ ŠÈ3x 2‹ ŠÈ3x 2‹ . The
x%
4
graph is rising on (2ß 0) and (#ß _), falling on (_ß #) and (!ß #), concave up on Š_ß È23 ‹ and Š È23 ß _‹
and concave down on Š È23 ß È23 ‹ . Consequently, a local maximum is 4 at x œ 0, local minima are 0 at
2
16
x œ „ 2, and Š È23 ß 16
9 ‹ and Š È3 ß 9 ‹ are points of inflection.
3
4
ax# 1b
#Î$
Ê yw œ ˆ 34 ‰ ˆ 23 ‰ ax# 1b
"Î$
minima are 0 at x œ „ 1; yww œ ax# 1b
"Î$
(2x) œ x ax# 1b
"Î$
, yw œ ) ( ± )( "
"
!
Ê the graph is rising on ("ß !) and ("ß _), falling on (_ß ") and (!ß ") Ê a local maximum is 34 at x œ 0, local
3. y œ
(x) ˆ 3" ‰ ax# 1b
%Î$
(2x) œ
x # 3
$
3É
ax # 1 b %
,
yww œ ± ) ( )( ± Ê the graph is concave up on Š_ß È3‹ and ŠÈ3ß _‹, concave
"
"
È$
È $
$
È
down on ŠÈ3ß È3‹ Ê points of inflection at Š „ È3ß $ % ‹
%
x#Î$ ax# 1b, yw œ ± )( ± !
"
"
Ê the graph is rising on (_ß 1) and ("ß _), falling on (1ß ") Ê a local maximum is 27
at
x
œ
1,
a
local
7
4. y œ
9
14
x"Î$ ax# 7b Ê yw œ
3
14
x#Î$ ax# 7b 9
14
x"Î$ (2x) œ
3
#
ww
&Î$
minimum is 27
ax# 1b 3x"Î$ œ 2x"Î$ x&Î$ œ x&Î$ a2x# 1b ,
7 at x œ 1; y œ x
yww œ )( Ê the graph is concave up on (!ß _), concave down on (_ß !) Ê a point of inflection at
!
(!ß !)
5. y œ x sin 2x Ê yw œ 1 2 cos 2x, yw œ [
± ± ] Ê the graph is rising on ˆ 13 ß 13 ‰ , falling
#1Î$
#1Î$
1Î$
1Î$
on ˆ #31 ß 13 ‰ and ˆ 13 ß 231 ‰ Ê local maxima are 231 13 È3
#
at x œ 13 and
21
3
È3
#
È3
#
at x œ 231 and
1
3
È3
#
at x œ
1
3
, local minima are
21
3
; yww œ 4 sin 2x, yww œ [
± ± ± ]
Ê the
!
#1Î$
#1Î$
1Î#
1Î#
graph is concave up on ˆ 1# ß !‰ and ˆ 1# ß 231 ‰ , concave down on ˆ 231 ß 1# ‰ and ˆ!ß 1# ‰ Ê points of inflection at
at x œ
ˆ 1# ß 1# ‰ , (!ß !), and ˆ 1# ß 1# ‰
6. y œ tan x 4x Ê yw œ sec# x 4, yw œ ( ± ± )
Ê the graph is rising on ˆ 12 ß 13 ‰ and
1Î#
1Î#
1Î$
1Î$
ˆ 13 ß 1# ‰ , falling on ˆ 13 ß 13 ‰ Ê a local maximum is È3 431 at x œ 13 , a local minimum is È3 431 at x œ 13 ;
yww œ 2(sec x)(sec x)(tan x) œ 2 asec# xb (tan x), yww œ (
± ) Ê the graph is concave up on ˆ0ß 1# ‰ ,
!
1Î#
1Î#
concave down on ˆ 12 ß 0‰ Ê a point of inflection at (0ß !)
7. If x 0, sin kxk œ sin x and if x 0, sin kxk œ sin (x)
œ sin x. From the sketch the graph is rising on
ˆ 3#1 ß 1# ‰ , ˆ!ß 1# ‰ and ˆ 3#1 ß #1‰ , falling on ˆ21ß 3#1 ‰ ,
ˆ 1# ß !‰ and ˆ 1# ß 3#1 ‰ ; local minima are 1 at x œ „ 3#1
and 0 at x œ !; local maxima are 1 at x œ „
1
#
and
0 at x œ „ #1; concave up on (#1ß 1) and (1ß #1), and
concavedown on (1ß 0) and (!ß 1) Ê points of inflection
are (1ß !) and (1ß !)
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Chapter 4 Applications of Derivatives
8. y œ 2 cos x È2 x Ê yw œ 2 sin x È2, yw œ [ ±
± ± ]
Ê rising on
1
$1Î#
$1Î%
1Î%
&1Î%
ˆ 341 ß 14 ‰and ˆ 541 ß 3#1 ‰ , falling on ˆ1ß 341 ‰ and ˆ 14 ß 541 ‰ Ê local maxima are 2 1È2 at x œ 1, È2 È2
at x œ 14 and 31#
at x œ
31
# ,
È
at x œ 341 and È2 514 2 at x œ 541 ;
Ê concave up on ˆ1ß 1# ‰ and ˆ 1# ß 3#1 ‰ , concave down on
and local minima are È2 yww œ 2 cos x, yww œ [ ± ± ]
1
$1Î#
1Î#
1Î#
ˆ 1# ß 1# ‰ Ê points of inflection at Š 1# ß
È 21
# ‹
and Š 1# ß 31 È 2
4
1È2
4
È 21
# ‹
9. When y œ x# 4x 3, then yw œ 2x 4 œ 2(x 2) and
yww œ 2. The curve rises on (#ß _) and falls on (_ß #).
At x œ 2 there is a minimum. Since yw w 0, the curve is
concave up for all x.
10. When y œ ' 2x x# , then yw œ # 2x œ 2(" x) and
yww œ 2. The curve rises on (_ß 1) and falls on
(1ß _). At x œ 1 there is a maximum. Since yw w 0, the
curve is concave down for all x.
11. When y œ x$ 3x 3, then yw œ 3x# 3 œ 3(x 1)(x 1)
and yww œ 6x. The curve rises on (_ß 1) ("ß _) and
falls on (1ß 1). At x œ 1 there is a local maximum and at
x œ 1 a local minimum. The curve is concave down on
(_ß 0) and concave up on (!ß _). There is a point of
inflection at x œ 0.
12. When y œ x(6 2x)# , then yw œ 4x(6 2x) (' 2x)#
œ 12(3 x)(" x) and yww œ 12(3 x) 12(" x)
œ 24(x 2). The curve rises on (_ß ") ($ß _) and
falls on ("ß $). The curve is concave down on (_ß #) and
concave up on (#ß _). At x œ 2 there is a point of
inflection.
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Section 4.4 Concavity and Curve Sketching
13. When y œ 2x$ 6x# 3, then yw œ 6x# 12x
œ 6x(x 2) and yww œ 12x 12 œ 12(x 1). The
curve rises on (!ß #) and falls on (_ß 0) and (#ß _).
At x œ 0 there is a local minimum and at x œ 2 a local
maximum. The curve is concave up on (_ß ") and
concave down on ("ß _). At x œ 1 there is a point of
inflection.
14. When y œ 1 9x 6x# x$ , then yw œ 9 12x 3x#
œ $(x 3)(B 1) and yww œ 12 6x œ 6(x 2).
The curve rises on ($ß ") and falls on (_ß 3) and
("ß _). At x œ 1 there is a local maximum and at
x œ 3 a local minimum. The curve is concave up on
(_ß 2) and concave down on (#ß _). At x œ 2
there is a point of inflection.
15. When y œ (x 2)$ 1, then yw œ 3(x 2)# and
yww œ 6(x 2). The curve never falls and there are no
local extrema. The curve is concave down on (_ß #)
and concave up on (#ß _). At x œ 2 there is a point
of inflection.
16. When y œ 1 (x 1)$ , then yw œ 3(x 1)# and
yww œ 6(x 1). The curve never rises and there are
no local extrema. The curve is concave up on (_ß 1)
and concave down on ("ß _). At x œ 1 there is a
point of inflection.
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229
230
Chapter 4 Applications of Derivatives
17. When y œ x% 2x# , then yw œ 4x$ 4x œ 4x(x 1)(x 1)
and yww œ 12x# 4 œ 12 Šx "
È 3 ‹ Šx
"
È3 ‹ .
The curve
rises on ("ß !) and ("ß _) and falls on (_ß 1) and (!ß ").
At x œ „ 1 there are local minima and at x œ 0 a local
maximum. The curve is concave up on Š_ß È"3 ‹ and
Š È"3 ß _‹ and concave down on Š È"3 ß È"3 ‹ . At x œ
„"
È3
there are points of inflection.
18. When y œ x% 6x# 4, then yw œ 4x$ 12x
œ 4x Šx È3‹ Šx È3‹ and yww œ 12x# 12
œ 12(x 1)(x 1). The curve rises on Š_ß È3‹
and Š!ß È3‹ , and falls on ŠÈ3ß !‹ and ŠÈ3ß _‹ . At
x œ „ È3there are local maxima and at x œ 0 a local
minimum. The curve is concave up on ("ß ") and concave
down on (_ß 1) and ("ß _). At x œ „ 1 there are points
of inflection.
19. When y œ 4x$ x% , then yw œ 12x# 4x$ œ 4x# ($ x) and
yww œ 24x 12x# œ 12x(2 x). The curve rises on a_ß $b
and falls on a$ß _b. At x œ 3 there is a local maximum, but
there is no local minimum. The graph is concave up on
a!ß #b and concave down on a_ß !b and a#ß _b. There are
inflection points at x œ 0 and x œ 2.
20. When y œ x% 2x$ , then yw œ 4x$ 6x# œ 2x# (2x 3) and
yww œ 12x# 12x œ 12x(x 1). The curve rises on
ˆ 3# ß _‰ and falls on ˆ_ß 32 ‰ . There is a local
minimum at x œ 3# , but no local maximum. The curve is
concave up on (_ß 1) and (!ß _), and concave down on
(1ß 0). At x œ 1 and x œ 0 there are points of inflection.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 4.4 Concavity and Curve Sketching
21. When y œ x& 5x% , then yw œ 5x% 20x$ œ 5x$ (x 4) and
yww œ 20x$ 60x# œ 20x# (x 3). The curve rises on
(_ß !) and (%ß _), and falls on (!ß %). There is a local
maximum at x œ 0, and a local minimum at x œ 4. The
curve is concave down on (_ß 3) and concave up on
(3ß _). At x œ 3 there is a point of inflection.
%
%
$
22. When y œ x ˆ x# 5‰ , then yw œ ˆ x# 5‰ x(4)ˆ x# 5‰ ˆ "# ‰
$
ww
‰
ˆx
‰# ˆ "# ‰ ˆ 5x
‰
œ ˆ x# 5‰ ˆ 5x
# 5 , and y œ 3 # 5
# 5
$
#
ˆ x# 5‰ ˆ 5# ‰ œ 5 ˆ x# 5‰ (x 4). The curve is rising
on (_ß #) and (10ß _), and falling on (#ß 10). There is a
local maximum at x œ 2 and a local minimum at x œ 10.
The curve is concave down on (_ß %) and concave up on
(%ß _). At x œ 4 there is a point of inflection.
23. When y œ x sin x, then yw œ " cos x and yww œ sin x.
The curve rises on (!ß 21). At x œ 0 there is a local and
absolute minimum and at x œ 21 there is a local and absolute
maximum. The curve is concave down on (!ß 1) and concave
up on (1ß #1). At x œ 1 there is a point of inflection.
24. When y œ x sin x, then yw œ " cos x and yww œ sin x.
The curve rises on (!ß 21). At x œ 0 there is a local and
absolute minimum and at x œ 21 there is a local and absolute
maximum. The curve is concave up on (!ß 1) and concave
down on (1ß #1). At x œ 1 there is a point of inflection.
25. When y œ x"Î& , then yw œ
"
5
4 *Î&
x%Î& and yww œ 25
x
.
The curve rises on (_ß _) and there are no extrema.
The curve is concave up on (_ß !) and concave down
on (!ß _). At x œ 0 there is a point of inflection.
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231
232
Chapter 4 Applications of Derivatives
26. When y œ x$Î& , then yw œ
3
5
6 (Î&
x#Î& and yww œ 25
x
.
The curve rises on (_ß _) and there are no extrema.
The curve is concave up on (_ß !) and concave down
on (!ß _). At x œ 0 there is a point of inflection.
27. When y œ x#Î& , then yw œ
2
5
6 )Î&
x$Î& and yww œ 25
x
.
The curve is rising on (0ß _) and falling on (_ß !). At
x œ 0 there is a local and absolute minimum. There is
no local or absolute maximum. The curve is concave
down on (_ß !) and (!ß _). There are no points of
inflection, but a cusp exists at x œ 0.
28. When y œ x%Î& , then yw œ
4
5
4 'Î&
x"Î& and yww œ 25
x
.
The curve is rising on (0ß _) and falling on (_ß !). At
x œ 0 there is a local and absolute minimum. There is
no local or absolute maximum. The curve is concave
down on (_ß !) and (!ß _). There are no points of
inflection, but a cusp exists at x œ 0.
29. When y œ 2x 3x#Î$ , then yw œ 2 2x"Î$ and
yww œ 23 x%Î$ . The curve is rising on (_ß !) and
("ß _), and falling on (!ß "). There is a local maximum
at x œ 0 and a local minimum at x œ 1. The curve is
concave up on (_ß !) and (!ß _). There are no
points of inflection, but a cusp exists at x œ 0.
30. When y œ 5x#Î& 2x, then yw œ 2x$Î& 2 œ 2 ˆx$Î& 1‰
and yww œ 65 x)Î& . The curve is rising on (0ß 1) and
falling on (_ß 0) and ("ß _). There is a local minimum
at x œ 0 and a local maximum at x œ 1. The curve is
concave down on (_ß !) and (!ß _). There are no
points of inflection, but a cusp exists at x œ 0.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 4.4 Concavity and Curve Sketching
31. When y œ x#Î$ ˆ #5 x‰ œ
5 #Î$
x&Î$ , then
# x
yw œ 53 x"Î$ 53 x#Î$ œ 53 x"Î$ (1 x) and
"Î$
yww œ 59 x%Î$ 10
œ 95 x%Î$ (1 2x).
9 x
The curve is rising on (!ß ") and falling on (_ß !) and
("ß _). There is a local minimum at x œ 0 and a local
maximum at x œ 1. The curve is concave up on ˆ_ß "# ‰
and concave down on ˆ "# ß !‰ and (0ß _). There is a point
of inflection at x œ "# and a cusp at x œ 0.
32. When y œ x#Î$ (x 5) œ x&Î$ 5x#Î$ , then
"Î$
yw œ 53 x#Î$ 10
œ 35 x"Î$ (x 2) and
3 x
yww œ
10
9
x"Î$ 10
9
x%Î$ œ
10
9
x%Î$ (x 1). The curve
is rising on (_ß !) and (#ß _), and falling on (!ß #).
There is a local minimum at x œ 2 and a local maximum
at x œ 0. The curve is concave up on ("ß 0) and (!ß _),
and concave down on (_ß 1). There is a point of
inflection at x œ 1 and a cusp at x œ 0.
"Î#
33. When y œ xÈ8 x# œ x a8 x# b , then
yw œ a8 x# b
"Î#
# "Î#
œ a8 x b
(x) ˆ "# ‰ a8 x# b
a8 2x# b œ
yww œ ˆ "# ‰a8 x# b
œ
2x ax# 12b
É a8 x # b $
$#
"Î#
(#x)
2(2 x)(2 x)
ÊŠ2È2 x‹ Š2È2 x‹
and
(2x)a8 2x# b a8 x# b
"#
(4x)
. The curve is rising on (#ß #), and falling
on Š2È2ß 2‹ and Š#ß 2È2‹ . There are local minima
x œ 2 and x œ 2È2, and local maxima at x œ 2È2 and
x œ 2. The curve is concave up on Š2È2ß !‹ and
concave down on Š!ß #È2‹ . There is a point of inflection
at x œ 0.
34. When y œ a2 x# b
$Î#
, then yw œ ˆ 3# ‰ a2 x# b
"Î#
(2x)
œ 3xÈ2 x# œ 3xÊŠÈ2 x‹ ŠÈ2 x‹ and
yww œ (3) a2 x# b
œ
"Î#
6(" x)(1 x)
ÊŠÈ2 x‹ ŠÈ2 x‹
(3x) ˆ "# ‰ a2 x# b
"Î#
(2x)
. The curve is rising on
ŠÈ2ß !‹ and falling on Š!ß È2‹ . There is a local
maximum at x œ 0, and local minima at x œ „ È2. The
curve is concave down on ("ß ") and concave up on
ŠÈ2ß "‹ and Š"ß È2‹ . There are points of inflection at
x œ „ 1.
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233
234
Chapter 4 Applications of Derivatives
x# 3
x # , then
(x 3)(x 1)
and
(x 2)#
35. When y œ
œ
yww œ
yw œ
2x(x 2) ax# 3b (")
(x2)#
(2x 4)(x 2)# ax# 4x 3b# (x 2)
(x 2)%
œ
2
(x 2)$
.
The curve is rising on (_ß ") and ($ß _), and falling on
("ß #) and (#ß $). There is a local maximum at x œ 1 and a
local minimum at x œ 3. The curve is concave down on
(_ß #) and concave up on (#ß _). There are no points
of inflection because x œ 2 is not in the domain.
36. When y œ
œ
3x# ax# 1b
a3x# 1b#
then yw œ
3x# a3x# 1b x$ (6x)
a3x# 1b#
and
#
a12x 6xb a3x# "b 2 a3x# "b(6x)ˆ3x% 3x# ‰
a3x# 1b%
6x(1 x)(" x)
. The curve is rising on (_ß _)
a3x# 1b$
yww œ
œ
x$
3x# 1 ,
$
so
there are no local extrema. The curve is concave up on
(_ß ") and (!ß "), and concave down on ("ß !) and
("ß _). There are points of inflection at x œ ", x œ 0,
and x œ 1.
x# 1, kxk 1
, then
1 x# , kxk 1
2x, kxk 1
2, kxk "
yw œ œ
and yww œ œ
. The
2x, kxk "
#, kxk "
37. When y œ kx# 1k œ œ
curve rises on ("ß !) and ("ß _) and falls on (_ß 1)
and (0ß 1). There is a local maximum at x œ 0 and local
minima at x œ „ 1. The curve is concave up on (_ß 1)
and ("ß _), and concave down on ("ß "). There are no
points of inflection because y is not differentiable at x œ „ 1
(so there is no tangent line at those points).
Ú x# 2x, x 0
38. When y œ kx 2xk œ Û 2x x# , 0 Ÿ x Ÿ 2 , then
Ü x# 2x, x 2
Ú 2x 2, x 0
Ú 2, x 0
w
ww
y œ Û 2 2x, 0 x 2 , and y œ Û 2, 0 x 2 .
Ü 2x 2, x 2
Ü 2, x 2
#
The curve is rising on (!ß 1) and (#ß _), and falling on
(_ß !) and ("ß #). There is a local maximum at x œ 1 and
local minima at x œ 0 and x œ 2. The curve is concave up
on (_ß !) and (#ß _), and concave down on (!ß #).
There are no points of inflection because y is not
differentiable at x œ 0 and x œ 2 (so there is no tangent
at those points).
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 4.4 Concavity and Curve Sketching
39. When y œ Èkxk œ Èx , x 0
, then
È x , x 0
Ú
"
x$Î#
, x0
, x0
#È x
y œ Û "
and yww œ (x)4$Î#
.
, x0
Ü 2 È x , x 0
4
w
Since lim c yw œ _ and lim b yw œ _ there is a
xÄ!
xÄ!
cusp at x œ 0. There is a local minimum at x œ 0, but no
local maximum. The curve is concave down on (_ß !)
and (!ß _). There are no points of inflection.
40. When y œ Èkx 4k œ Ú
Èx 4 , x 4
, then
È4 x , x 4
"
(x 4)$Î#
, x4
Èx 4 , x 4
y œ Û 2 "
and yww œ (4 4x)$Î#
.
, x4
Ü #È 4 x , x 4
4
w
Since lim c yw œ _ and lim b yw œ _ there is a cusp
xÄ4
xÄ4
at x œ 4. There is a local minimum at x œ 4, but no local
maximum. The curve is concave down on (_ß %)
and (%ß _). There are no points of inflection.
41. yw œ 2 x x# œ (1 x)(# x), yw œ ± ± "
#
Ê rising on ("ß #), falling on (_ß 1) and (#ß _)
Ê there is a local maximum at x œ 2 and a local minimum
at x œ 1; yww œ 1 2x, yww œ ± "Î#
"
Ê concave up on ˆ_ß # ‰ , concave down on ˆ "# ß _‰
Ê a point of inflection at x œ
"
#
42. yw œ x# x 6 œ (x 3)(x 2), yw œ ± ± #
$
Ê rising on (_ß #) and (3ß _), falling on (2ß 3)
Ê there is a local maximum at x œ 2 and a local
minimum at x œ 3; yww œ 2x 1, yww œ ± "Î#
"
ˆ
‰
Ê concave up on # ß _ , concave down on ˆ_ß "# ‰
Ê a point of inflection at x œ
"
#
43. yw œ x(x 3)# , yw œ ± ± Ê rising on
!
$
(!ß _), falling on (_ß !) Ê no local maximum, but there
is a local minimum at x œ 0; yww œ (x 3)# x(2)(x 3)
œ 3(x 3)(x 1), yww œ ± ± Ê concave
"
$
up on (_ß ") and ($ß _), concave down on ("ß $) Ê
points of inflection at x œ 1 and x œ 3
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Chapter 4 Applications of Derivatives
44. yw œ x# (2 x), yw œ ± ± Ê rising on
!
#
(_ß #), falling on (2ß _) Ê there is a local maximum at
x œ 2, but no local minimum; yww œ 2x(2 x) x# (1)
œ x(4 3x), yww œ ± ± Ê concave up
!
%Î$
4‰
ˆ
on !ß 3 , concave down on (_ß !) and ˆ 43 ß _‰ Ê points
of inflection at x œ 0 and x œ
4
3
45. yw œ x ax# 12b œ x Šx 2È3‹ Šx 2È3‹ ,
yw œ ±
± ± Ê rising on
!
#È$
#È $
Š2È3ß !‹ and Š#È3ß _‹ , falling on Š_ß #È3‹
and Š!ß #È3‹ Ê a local maximum at x œ 0, local minima
at x œ „ #È3 ; yww œ (1) ax# 12b (x)(2x)
œ 3(x 2)(x 2), yww œ ± ± #
#
Ê concave up on (_ß #) and (#ß _), concave down on
(#ß #) Ê points of inflection at x œ „ 2
46. yw œ (x 1)# (2x 3), yw œ ± ± "
$Î#
3
3‰
ˆ
‰
ˆ
Ê rising on # ß _ , falling on _ß # Ê no local
maximum, a local minimum at x œ 3# ;
yww œ 2(x 1)(2x 3) (x 1)# (2) œ 2(x 1)(3x 2),
yww œ ± ± Ê concave up on
"
#Î$
2
ˆ_ß 3 ‰ and ("ß _), concave down on ˆ 23 ß "‰
Ê points of inflection at x œ 23 and x œ 1
47. yw œ a8x 5x# b (4 x)# œ x(8 5x)(% x)# ,
yw œ ± ± ± Ê rising on ˆ!ß 85 ‰ ,
!
%
)Î&
8
falling on (_ß !) and ˆ 5 ß _‰ Ê a local maximum at
xœ
ww
8
5
, a local minimum at x œ 0;
y œ (8 10x)(4 x)# a8x 5x# b (2)(% x)(1)
œ 4(4 x) a5x# 16x 8b ,
yww œ ±
± ± Ê concave up
%
)#È'
)#È'
&
È6
on Š_ß 8 52
Š 8 52
xœ
&
‹ and Š 8 52
È 6 8 2È 6
ß 5 ‹
8 „ 2È 6
5
È6
ß %‹ , concave down on
and (4ß _) Ê points of inflection at
and x œ 4
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Section 4.4 Concavity and Curve Sketching
48. yw œ ax# 2xb (x 5)# œ x(x 2)(x 5)# ,
yw œ ± ± ± Ê rising on (_ß !)
!
#
&
and (#ß _), falling on (!ß #) Ê a local maximum at x œ 0,
a local minimum at x œ 2;
yww œ (2x 2)(x 5)# ax# 2xb (2)(x 5)
œ 2(x 5) a2x# 8x 5b ,
yww œ ± ± ± Ê concave up
&
%È'
%È'
on
#
4 È6 4 È6
Š # ß 2 ‹
È6
Š_ß % 2
xœ
4 „È 6
2
#
and (5ß _), concave down on
È6
‹ and Š 4 #
ß &‹ Ê points of inflection at
and x œ 5
49. yw œ sec# x, yw œ ( ) Ê rising on ˆ 1# ß 1# ‰ ,
1Î#
1Î#
never falling Ê no local extrema;
yww œ 2(sec x)(sec x)(tan x) œ 2 asec# xb (tan x),
yww œ ( ± ) Ê concave up on ˆ!, 1# ‰,
!
1Î#
1Î#
1
ˆ
concave down on # ß !‰, ! is a opoint of inflection.
50. yw œ tan x, yw œ ( ± ) Ê rising on ˆ0ß 1# ‰ ,
!
1Î#
1Î#
1
ˆ
‰
falling on # ß ! Ê no local maximum, a local minimum
at x œ 0; yww œ sec# x, yww œ ( ) Ê concave up
1Î#
1Î#
1 1‰
ˆ
on # ß # Ê no points of inflection
51. yw œ cot
)
#
, yw œ ( ± ) Ê rising on (!ß 1),
1
!
#1
falling on (1ß #1) Ê a local maximum at ) œ 1, no local
minimum; yww œ "# csc# #) , yww œ ( ) Ê never
!
#1
concave up, concave down on (!ß #1) Ê no points of
inflection
52. yw œ csc#
)
#
, yw œ ( ) Ê rising on (!ß 21), never
!
#1
falling Ê no local extrema;
yww œ 2 ˆcsc #) ‰ ˆcsc #) ‰ ˆcot #) ‰ ˆ "# ‰
œ ˆcsc# #) ‰ ˆcot #) ‰, yww œ ( ± )
1
!
#1
Ê concave up on (1ß #1), concave down on (!ß 1)
Ê a point of inflection at ) œ 1
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Chapter 4 Applications of Derivatives
53. yw œ tan# ) 1 œ (tan ) 1)(tan ) 1),
yw œ ( | ± ) Ê rising on
1Î#
1Î%
1Î#
1Î%
ˆ 1# ß 14 ‰ and ˆ 14 ß 1# ‰ , falling on ˆ 14 ß 14 ‰
Ê a local maximum at ) œ 14 , a local minimum at ) œ 14 ;
yww œ 2 tan ) sec# ), yww œ ( ± )
!
1Î#
1Î#
Ê concave up on ˆ!ß 1# ‰ , concave down on ˆ 1# ß !‰
Ê a point of inflection at ) œ 0
54. yw œ 1 cot# ) œ (" cot ))(1 cot )),
yw œ ( | ± ) Ê rising on ˆ 14 ß 341 ‰ ,
1
!
1Î%
$1Î%
1
3
1
falling on ˆ0ß 4 ‰ and ˆ 4 ß 1‰ Ê a local maximum at
)œ
ww
31
4 ,
a local minimum at ) œ
#
ww
1
4
;
y œ 2(cot )) acsc )b, y œ ( ± )
1
!
1Î#
1
1
Ê concave up on ˆ!ß # ‰ , concave down on ˆ # ß 1‰
Ê a point of inflection at ) œ
1
#
55. yw œ cos t, yw œ [ ± ± ] Ê rising on
!
#1
1Î#
$1Î#
1
3
1
1
3
1
ˆ!ß # ‰ and ˆ # ß 21‰ , falling on ˆ # ß # ‰ Ê local maxima at
tœ
ww
1
#
and t œ 21, local minima at t œ 0 and t œ
ww
y œ sin t, y œ [ ± ]
1
!
#1
Ê concave up on (1ß #1), concave down
on (!ß 1) Ê a point of inflection at t œ 1
31
#
;
56. yw œ sin t, yw œ [ ± ] Ê rising on (!ß 1),
1
!
#1
falling on (1ß 21) Ê a local maximum at t œ 1, local
minima at t œ 0 and t œ 21; yww œ cos t,
yww œ [ ± ± ] Ê concave up on ˆ!ß 1# ‰
!
#1
1Î#
$1Î#
3
1
and ˆ # ß #1‰ , concave down on ˆ 1# ß 3#1 ‰ Ê points
of inflection at t œ
1
#
and t œ
31
#
57. yw œ (x 1)#Î$ , yw œ ) ( Ê rising on
"
(_ß _), never falling Ê no local extrema;
yww œ 23 (x 1)&Î$ , yww œ ) ( "
Ê concave up on (_ß 1), concave down on ("ß _)
Ê a point of inflection and vertical tangent at x œ 1
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Section 4.4 Concavity and Curve Sketching
58. yw œ (x 2)"Î$ , yw œ )( Ê rising on (2ß _),
#
falling on (_ß #) Ê no local maximum, but a local
minimum at x œ 2; yww œ 13 (x 2)%Î$ ,
yww œ )( Ê concave down on (_ß 2) and
#
(#ß _) Ê no points of inflection, but there is a cusp at
xœ2
59. yw œ x#Î$ (x 1), yw œ )( ± Ê rising on
!
"
("ß _), falling on (_ß ") Ê no local maximum, but a
local minimum at x œ 1; yww œ "3 x#Î$ 23 x&Î$
"
3
x&Î$ (x 2), yww œ ± )( !
#
Ê concave up on (_ß 2) and (!ß _), concave down on
(#ß !) Ê points of inflection at x œ 2 and x œ 0, and a
vertical tangent at x œ 0
œ
60. yw œ x%Î& (x 1), yw œ ± )( Ê rising on
!
"
("ß 0) and (!ß _), falling on (_ß ") Ê no local
maximum, but a local minimum at x œ 1;
yww œ "5 x%Î& 45 x*Î& œ "5 x*Î& (x 4),
yww œ )( ± Ê concave up on (_ß 0) and
!
%
(4ß _), concave down on (0ß 4) Ê points of inflection at
x œ 0 and x œ 4, and a vertical tangent at x œ 0
61. yw œ œ
#x, x Ÿ 0 w
, y œ ± Ê rising on
2x, x 0
!
(_ß _) Ê no local extrema; yww œ œ
2, x 0
,
2, x 0
yww œ )( Ê concave up on (!ß _), concave
!
down on (_ß !) Ê a point of inflection at x œ 0
62. yw œ œ
x# , x Ÿ 0 w
, y œ ± Ê rising on
x# , x 0
!
(!ß _), falling on (_ß !) Ê no local maximum, but a
2x, x Ÿ 0
local minimum at x œ 0; yww œ œ
,
2x, x 0
yww œ ± Ê concave up on (_ß _)
!
Ê no point of inflection
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Chapter 4 Applications of Derivatives
63. The graph of y œ f ww (x) Ê the graph of y œ f(x) is concave
up on (!ß _), concave down on (_ß !) Ê a point of
inflection at x œ 0; the graph of y œ f w (x)
Ê yw œ ± ± Ê the graph y œ f(x) has
both a local maximum and a local minimum
64. The graph of y œ f ww (x) Ê yww œ ± Ê the
graph of y œ f(x) has a point of inflection, the graph of
y œ f w (x) Ê yw œ ± ± Ê the graph of
y œ f(x) has both a local maximum and a local minimum
65. The graph of y œ f ww (x) Ê yww œ ± ± Ê the graph of y œ f(x) has two points of inflection, the
graph of y œ f w (x) Ê yw œ ± Ê the graph of
y œ f(x) has a local minimum
66. The graph of y œ f ww (x) Ê yww œ ± Ê the
graph of y œ f(x) has a point of inflection; the graph of
y œ f w (x) Ê yw œ ± ± Ê the graph of
y œ f(x) has both a local maximum and a local minimum
67.
Point
P
Q
R
S
T
yw
!
yww
!
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Section 4.4 Concavity and Curve Sketching
68.
241
69.
70.
71. Graphs printed in color can shift during a press run, so your values may differ somewhat from those given here.
(a) The body is moving away from the origin when kdisplacementk is increasing as t increases, 0 t 2 and
6 t 9.5; the body is moving toward the origin when kdisplacementk is decreasing as t increases, 2 t 6
and 9.5 t 15
(b) The velocity will be zero when the slope of the tangent line for y œ s(t) is horizontal. The velocity is zero
when t is approximately 2, 6, or 9.5 sec.
(c) The acceleration will be zero at those values of t where the curve y œ s(t) has points of inflection. The
acceleration is zero when t is approximately 4, 7.5, or 12.5 sec.
(d) The acceleration is positive when the concavity is up, 4 t 7.5 and 12.5 t 15; the acceleration is
negative when the concavity is down, 0 t 4 and 7.5 t 12.5
72. (a) The body is moving away from the origin when kdisplacementk is increasing as t increases, 1.5 t 4,
10 t 12 and 13.5 t 16; the body is moving toward the origin when kdisplacementk is decreasing as t
increases, 0 t 1.5, 4 t 10 and 12 t 13.5
(b) The velocity will be zero when the slope of the tangent line for y œ s(t) is horizontal. The velocity is zero
when t is approximately 0, 4, 12 or 16 sec.
(c) The acceleration will be zero at those values of t where the curve y œ s(t) has points of inflection. The
acceleration is zero when t is approximately 1.5, 6, 8, 10.5, or 13.5 sec.
(d) The acceleration is positive when the concavity is up, 0 t 1.5, 6 t 8 and 10 t 13.5, the
acceleration is negative when the concavity is down, 1.5 t 6, 8 t 10 and 13.5 t 16.
73. The marginal cost is
dc
dx
which changes from decreasing to increasing when its derivative
d# c
dx#
is zero. This is a
point of inflection of the cost curve and occurs when the production level x is approximately 60 thousand units.
74. The marginal revenue is
dy
dx
d# y
dx# is positive Ê the curve is concave up
#
when ddxy# 0 Ê the curve is concave down
and it is increasing when its derivative
Ê ! t 2 and 5 t 9; marginal revenue is decreasing
Ê 2 t 5 and 9 t 12.
75. When yw œ (x 1)# (x 2), then yww œ 2(x 1)(x 2) (x 1)# . The curve falls on (_ß 2) and rises on
(#ß _). At x œ 2 there is a local minimum. There is no local maximum. The curve is concave upward on (_ß ") and
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242
Chapter 4 Applications of Derivatives
ˆ 53 ß _‰ , and concave downward on ˆ"ß 53 ‰ . At x œ 1 or x œ
5
3
there are inflection points.
76. When yw œ (x 1)# (x 2)(x 4), then yww œ 2(x 1)(x 2)(x 4) (x 1)# (x 4) (x 1)# (x 2)
œ (x 1) c2 ax# 6x 8b ax# 5x 4b ax# 3x 2bd œ 2(x 1) a2x# 10x 11b. The curve rises on
(_ß 2) and (%ß _) and falls on (#ß %). At x œ 2 there is a local maximum and at x œ 4 a local minimum. The
È3 5 È3
ß # ‹
curve is concave downward on (_ß ") and Š 5 2
È
Š 5 # 3 ß _‹ .
At x œ 1,
5 È3
#
and
5 È3
#
È3
and concave upward on Š1ß 5 #
‹ and
there are inflection points.
77. The graph must be concave down for x 0 because
f ww (x) œ x"# 0.
78. The second derivative, being continuous and never zero, cannot change sign. Therefore the graph will always
be concave up or concave down so it will have no inflection points and no cusps or corners.
79. The curve will have a point of inflection at x œ 1 if 1 is a solution of yww œ 0; y œ x$ bx# cx d
Ê yw œ 3x# 2bx c Ê yw w œ 6x 2b and 6(1) 2b œ 0 Ê b œ 3.
80. (a) True. If f(x) is a polynomial of even degree then f w is of odd degree. Every polynomial of odd degree has
at least one real root Ê f w (x) œ 0 for some x œ r Ê f has a horizontal tangent at x œ r.
(b) False. For example, f(x) œ x 1 is a polynomial of odd degree but f w (x) œ 1 is never 0. As another
example, y œ "3 x$ x# x is a polynomial of odd degree, but yw œ x# 2x 1 œ (x 1)# 0 for all x.
81. (a) f(x) œ ax# bx c œ a ˆx# ba x‰ c œ a Šx# ba B b#
4a# ‹
b#
4a
c œ a ˆx b
b
whose vertex is at x œ 2a
Ê the coordinates of the vertex are Š 2a
ß b
#
b ‰#
#a
b# 4ac
4a
a parabola
4ac
4a ‹
(b) The second derivative, f ww (x) œ 2a, describes concavity Ê when a 0 the parabola is concave up and
when a 0 the parabola is concave down.
82. No, f ww (x) could be decreasing to zero at x œ c and then increase again so it would be concave up on every
interval even though f ww (x) œ 0. For example f(x) œ x% is always concave up even though f ww (0) œ 0.
83. A quadratic curve never has an inflection point. If y œ ax# bx c where a Á 0, then yw œ 2ax b and
yww œ 2a. Since 2a is a constant, it is not possible for yww to change signs.
84. A cubic curve always has exactly one inflection point. If y œ ax$ bx# cx d where a Á 0, then
yw œ 3ax# 2bx c and yww œ 6ax 2b. Since 3ab is a solution of yww œ 0, we have that yww changes its sign
b
b
at x œ 3a
and yw exists everywhere (so there is a tangent at x œ 3a
). Thus the curve has an inflection
b
point at x œ 3a
. There are no other inflection points because yww changes sign only at this zero.
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Section 4.4 Concavity and Curve Sketching
85. If y œ x& 5x% 240, then yw œ 5x$ (x 4) and
yww œ 20x# (x 3). The zeros of yw are extrema, and
there is a point of inflection at x œ $Þ
86. If y œ x$ 12x# , then yw œ 3x(x 8) and
yww œ 6(x 4). The zeros of yw and yww are
extrema and points of inflection, respectively.
87. If y œ
ww
4
5
x& 16x# 25, then yw œ 4x ax$ 8b and
y œ 16 ax$ 2b . The zeros of yw and yww are
extrema and points of inflection, respectively.
88. If y œ
w
x%
4
$
x$
3
#
4x# 12x 20, then
y œ x x )x "# œ (x 3)(x 2)# Þ
So y has a local minimum at x œ $ as its only extreme
value. Also yww œ $x# #x ) œ (3x 4)(x 2) and there
are inflection points at both zeros, %$ and 2, of yww .
89. The graph of f falls where f w 0, rises where f w 0,
and has horizontal tangents where f w œ 0. It has local
minima at points where f w changes from negative to
positive and local maxima where f w changes from
positive to negative. The graph of f is concave down
where f ww 0 and concave up where f ww 0. It has an
inflection point each time f ww changes sign, provided a tangent
line exists there.
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243
244
Chapter 4 Applications of Derivatives
90. The graph f is concave down where f ww 0, and concave
up where f ww 0. It has an inflection point each time
f ww changes sign, provided a tangent line exists there.
91. (a) It appears to control the number and magnitude of the
local extrema. If k 0, there is a local maximum to the
left of the origin and a local minimum to the right. The
larger the magnitude of k (k 0), the greater the
magnitude of the extrema. If k 0, the graph has only
positive slopes and lies entirely in the first and third
quadrants with no local extrema. The graph becomes
increasingly steep and straight as k Ä _.
(b) f w (x) œ 3x# k Ê the discriminant 0# 4(3)(k) œ 12k is positive for k 0, zero for k œ 0, and
negative for k 0; f w has two zeros x œ „ É k3 when k 0, one zero x œ 0 when k œ 0 and no real zeros
when k 0; the sign of k controls the number of local extrema.
(c) As k Ä _, f w (x) Ä _ and the graph becomes
increasingly steep and straight. As k Ä _, the
crest of the graph (local maximum) in the second
quadrant becomes increasingly high and the trough
(local minimum) in the fourth quadrant becomes
increasingly deep.
92. (a) It appears to control the concavity and the number of
local extrema.
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Section 4.5 Applied Optimization Problems
245
(b) f(x) œ x% kx$ 6x# Ê f w (x) œ 4x$ 3kx# 12x
Ê f ww (x) œ 12x# 6kx 12 Ê the discriminant is
36k# 4(12)(12) œ 36(k 4)(k 4), so the sign line
of the discriminant is ± ± Ê the
%
%
discriminant is positive when kkk 4, zero when
k œ „ 4, and negative when kkk 4; f ww (x) œ 0 has
two zeros when kkk 4, one zero when k œ „ 4, and
no real zeros for kkk 4; the value of k controls the
number of possible points of inflection.
93. (a) If y œ x#Î$ ax# 2b , then yw œ
yww œ
4
9
4
3
x"Î$ a2x# 1b and
x%Î$ a10x# 1b . The curve rises on
Š È"2 ß 0‹ and Š È"2 ß _‹ and falls on Š_ß È"2 ‹
and Š!ß È"2 ‹ . The curve is concave up on (_ß !)
and (!ß _).
(b) A cusp since lim c yw œ _ and lim b yw œ _.
xÄ!
xÄ!
94. (a) If y œ 9x#Î$ (x 1), then yw œ
15 ˆx 25 ‰
x"Î$
and
10 ˆx "5 ‰
x%Î$
yww œ
. The curve rises on (_ß !) and
ˆ 25 ß _‰ and falls on ˆ!ß 25 ‰ . The curve is concave
down on ˆ_ß "5 ‰ and concave up on ˆ 5" ß !‰ and
(!ß _).
(b) A cusp since lim c yw œ _ and lim b yw œ _.
xÄ!
xÄ!
95. Yes: y œ x# 3 sin 2x Ê yw œ 2x 6 cos 2x. The graph
of yw is zero near 3 and this indicates a horizontal tangent
near x œ 3.
4.5 APPLIED OPTIMIZATION PROBLEMS
1. Let j and w represent the length and width of the rectangle, respectively. With an area of 16 in.# , we have
that (j)(w) œ 16 Ê w œ 16j" Ê the perimeter is P œ 2j 2w œ 2j 32j" and Pw (j) œ 2 w
Solving P (j) œ 0 Ê
2(j 4)(j 4)
j#
32
j#
œ
2 aj# 16b
j#
.
œ 0 Ê j œ 4, 4. Since j 0 for the length of a rectangle, j must be 4 and
w œ 4 Ê the perimeter is 16 in., a minimum since Pww (j) œ
16
j$
0.
2. Let x represent the length of the rectangle in meters (! x %) Then the width is % x and the area is
Aaxb œ xa% xb œ %x x# . Since Aw axb œ % #x, the critical point occurs at x œ #. Since, Aw axb ! for ! x # and
Aw axb ! for # x %, this critical point corresponds to the maximum area. The rectangle with the largest area measures
# m by % # œ # m, so it is a square.
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Chapter 4 Applications of Derivatives
Graphical Support:
3. (a) The line containing point P also contains the points (!ß ") and ("ß !) Ê the line containing P is y œ 1 x
Ê a general point on that line is (xß 1 x).
(b) The area A(x) œ 2x(1 x), where 0 Ÿ x Ÿ 1.
(c) When A(x) œ 2x 2x# , then Aw (x) œ 0 Ê 2 4x œ 0 Ê x œ "# . Since A(0) œ 0 and A(1) œ 0, we conclude
that A ˆ "# ‰ œ "# sq units is the largest area. The dimensions are " unit by "# unit.
4. The area of the rectangle is A œ 2xy œ 2x a12 x# b ,
where 0 Ÿ x Ÿ È12 . Solving Aw (x) œ 0 Ê 24 6x# œ 0
Ê x œ 2 or 2. Now 2 is not in the domain, and since
A(0) œ 0 and A ŠÈ12‹ œ 0, we conclude that A(2) œ 32
square units is the maximum area. The dimensions are 4 units
by 8 units.
5. The volume of the box is V(x) œ x(15 2x)(8 2x)
œ 120x 46x# 4x$ , where 0 Ÿ x Ÿ 4. Solving Vw (x) œ 0
Ê 120 92x 12x# œ 4(6 x)(5 3x) œ 0 Ê x œ 53
or 6, but 6 is not in the domain. Since V(0) œ V(4) œ 0,
$
V ˆ 53 ‰ œ #%&!
#( ¸ *" in must be the maximum volume of
the box with dimensions
14
3
‚
35
3
‚
5
3
inches.
6. The area of the triangle is A œ "# ba œ b# È400 b# , where
b#
0 Ÿ b Ÿ 20. Then dA œ " È400 b# db
œ
200 b#
È400 b#
#
2È400 b#
œ 0 Ê the interior critical point is b œ 10È2.
When b œ 0 or 20, the area is zero Ê A Š10È2‹ is the
maximum area. When a# b# œ 400 and b œ 10È2, the
value of a is also 10È2 Ê the maximum area occurs when
a œ b.
7. The area is A(x) œ x(800 2x), where 0 Ÿ x Ÿ 400.
Solving Aw (x) œ 800 4x œ 0 Ê x œ 200. With
A(0) œ A(400) œ 0, the maximum area is
A(200) œ 80,000 m# . The dimensions are 200 m by 400 m.
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Section 4.5 Applied Optimization Problems
8. The area is 2xy œ 216 Ê y œ
108
x
247
. The amount of fence
needed is P œ 4x 3y œ 4x 324x" , where ! x;
dP
324
#
dx œ 4 x# œ 0 Ê x 81 œ 0 Ê the critical points are
0 and „ 9, but 0 and 9 are not in the domain. Then
Pww (9) 0 Ê at x œ 9 there is a minimum Ê the
dimensions of the outer rectangle are 18 m by 12 m
Ê 72 meters of fence will be needed.
9. (a) We minimize the weight œ tS where S is the surface area, and t is the thickness of the steel walls of the tank. The
surace area is S œ x# %xy where x is the length of a side of the square base of the tank, and y is its depth. The
ˆ # #!!! ‰. Treating the
volume of the tank must be &!!ft$ Ê y œ &!!
x# . Therefore, the weight of the tank is waxb œ t x x
#!!!
thickness as a constant gives ww axb œ tˆ#x x# ‰ for xÞ!. The critical value is at x œ "!. Since
www a"!b œ tˆ# %!!! ‰
"!$
!, there is a minimum at x œ "!. Therefore, the optimum dimensions of the tank are "! ft on
the base edges and & ft deep.
(b) Minimizing the surface area of the tank minimizes its weight for a given wall thickness. The thickness of the steel
walls would likely be determined by other considerations such as structural requirements.
10. (a) The volume of the tank being ""#& ft$ , we have that yx# œ ""#& Ê y œ ""#&
x# . The cost of building the tank is
""#&
$$(&!
#
w
caxb œ &x $!xˆ x# ‰, where ! x. Then c axb œ "!x x# œ ! Ê the critical points are ! and "&, but ! is not
in the domain. Thus, cww a"&b ! Ê at x œ "& we have a minimum. The values of x œ "& ft and y œ & ft will
minimize the cost.
(b) The cost function c œ &ax# %xyb "!xy, can be separated into two items: (1) the cost of the materials and labor to
fabricate the tank, and (2) the cost for the excavation. Since the area of the sides and bottom of the tanks is ax# %xyb,
it can be deduced that the unit cost to fabricate the tanks is $&/ft# . Normally, excavation costs are per unit volume of
‰ #
excavated material. Consequently, the total excavation cost can be taken as "!xy œ ˆ "!
x ax yb. This suggests that the
unit cost of excavation is
$"!Îft#
x
where x is the length of a side of the square base of the tank in feet. For the least
expensive tank, the unit cost for the excavation is
$"!Îft#
"& ft
œ
$!Þ'(
ft$
œ
$")
yd$ .
The total cost of the least expensive tank is
$$$(&, which is the sum of $#'#& for fabrication and $(&! for the excavation.
11. The area of the printing is (y 4)(x 8) œ 50.
‰
Consequently, y œ ˆ x 50
8 4. The area of the paper is
50
A(x) œ x ˆ x 8 4‰ , where 8 x. Then
50
‰
Aw (x) œ ˆ x 50
8 4 x Š (x 8)# ‹ œ
4(x 8)# 400
(x 8)#
œ0
Ê the critical points are 2 and 18, but 2 is not in the
domain. Thus Aww (18) 0 Ê at x œ 18 we have
a minimum. Therefore the dimensions 18 by 9 inches
minimize the amount of paper.
12. The volume of the cone is V œ
Thus, V(y) œ
1
3
"
3
1r# h, where r œ x œ È9 y# and h œ y 3 (from the figure in the text).
a9 y# b (y 3) œ
1
3
a27 9y 3y# y$ b Ê Vw (y) œ
ww
1
3
a9 6y 3y# b œ 1(1 y)(3 y).
The critical points are 3 and 1, but 3 is not in the domain. Thus V (1) œ
we have a maximum volume of V(1) œ
1
3
(8)(4) œ
321
3
1
3
(' 6(1)) 0 Ê at y œ 1
cubic units.
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248
Chapter 4 Applications of Derivatives
13. The area of the triangle is A()) œ
ab cos )
#
ab sin )
#
, where 0 ) 1.
Solving A ()) œ 0 Ê
œ 0 Ê ) œ 1# . Since Aww ())
)
œ ab sin
Ê Aww ˆ 1# ‰ 0, there is a maximum at ) œ 1# .
#
w
14. A volume V œ 1r# h œ 1000 Ê h œ
1000
1 r#
. The amount of
material is the surface area given by the sides and bottom of
#
the can Ê S œ 21rh 1r# œ 2000
r 1r , 0 r. Then
dS
dr
œ 2000
r# 21r œ ! Ê
are 0 and
#
d S
dr#
rœ
œ
4000
r$
10
$
È
1
10
$
È
1
1r$ 1000
r#
œ 0. The critical points
, but 0 is not in the domain. Since
#1 0, we have a minimum surface area when
cm and h œ
1000
1 r#
œ
10
$
È
1
cm. Comparing this result to
the result found in Example 2, if we include both ends of the
can, then we have a minimum surface area when the can is
shorter-specifically, when the height of the can is the same as
its diameter.
15. With a volume of 1000 cm and V œ 1r# h, then h œ
A œ 8r# 21rh œ 8r# 2000
r
. Then Aw (r) œ
but r œ 0 results in no can. Since Aww (r) œ 16
16. (a) The base measures "! #x in. by
"&#x
#
1000
1r# . The amount of aluminum used per can is
8r$ 1000
16r 2000
œ 0 Ê the critical points are 0 and 5,
r# œ 0 Ê
r#
1000
r$ 0 we have a minimum at r œ 5 Ê h œ 40
1 and h:r œ 8:1.
in., so the volume formula is Vaxb œ
xa"!#xba"&#xb
#
œ #x$ #&x# (&x.
(b) We require x !, #x "!, and #x "&. Combining these requirements, the domain is the interval a!ß &b.
(c) The maximum volume is approximately 66.02 in.$ when x ¸ "Þ*' in.
(d) Vw axb œ 'x# &!x (&. The critical point occurs when Vw axb œ !, at x œ
œ
ww
#& „ &È(
, that
'
&! „ Éa&!b# %a'ba(&b
# a 'b
œ
&! „ È(!!
"#
is, x ¸ "Þ*' or x ¸ 'Þ$(. We discard the larger value because it is not in the domain. Since
V axb œ "#x &!, which is negative when x ¸ "Þ*' , the critical point corresponds to the maximum volume. The
maximum volume occurs when x œ
#& &È(
'
¸ "Þ*', which comfimrs the result in (c).
17. (a) The" sides" of the suitcase will measure #% #x in. by ") #x in. and will be #x in. apart, so the volume formula is
Vaxb œ #xa#% #xba") #xb œ )x$ "')x# )'#x.
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Section 4.5 Applied Optimization Problems
249
(b) We require x !, #x "), and #x #%. Combining these requirements, the domain is the interval a!ß *b.
(c) The maximum volume is approximately 1309.95 in.$ when x ¸ 3Þ3* in.
(d) Vw axb œ #%x# $$'x )'% œ #%ax# "%x $'b. The critical point is at x œ
"% „ Éa"%b# %a"ba$'b
# a" b
œ
"% „ È&#
#
œ ( „ È"$, that is, x ¸ $Þ$* or x ¸ "!Þ'". We discard the larger value because it is not in the domain. Since
Vww axb œ #%a#x "%b which is negative when x ¸ $Þ$*, the critical point corresponds to the maximum volume. The
maximum value occurs at x œ ( È"$ ¸ $Þ$*, which confirms the results in (c).
(e) )x$ "')x# )'#x œ ""#! Ê 8ax$ #"x# "!)x "%!b œ ! Ê )ax #bax &bax "%b œ !. Since "% is not in
the fomain, the possible values of x are x œ # in. or x œ & in.
(f) The dimensions of the resulting box are #x in., a#% #xb in., and a") #xb. Each of these measurements must be
positive, so that gives the domain of a!ß *b.
18. If the upper right vertex of the rectangle is located at axß % cos !Þ& xb for ! x 1, then the rectangle has width #x and
height % cos !Þ&x, so the area is Aaxb œ )x cos !Þ&x. Solving Aw axb œ ! graphically for ! x 1, we find that
x ¸ #Þ#"%. Evaluating #x and % cos !Þ&x for x ¸ #Þ#"%, the dimensions of the rectangle are approximately %Þ%$ (width) by
"Þ(* (height), and the maximum area is approximately (Þ*#$.
19. Let the radius of the cylinder be r cm, ! r "!. Then the height is #È"!! r# and the volume is
Varb œ #1r# È"!! r# cm$ . Then, Vw arb œ #1r# Š " ‹a#rb Š#1È"!! r# ‹a#rb
È"!! r#
œ
$
#
#1r %1ra"!! r b
È"!! r#
œ
#
#1ra#!! $r b
È"!! r# .
É $# . Since Vw arb ! for
The critical point for ! r "! occurs at r œ É #!!
$ œ "!
! r "!É #$ and Vw arb ! for "!É #$ r "!, the critical point corresponds to the maximum volume. The
dimensions are r œ "!É #$ ¸ )Þ"' cm and h œ
#!
È$
¸ ""Þ&& cm, and the volume is
%!!!1
$È$
¸ #%")Þ%! cm$ .
20. (a) From the diagram we have 4x j œ 108 and V œ x# j.
The volume of the box is V(x) œ x# (108 4x), where
0 Ÿ x 27. Then
Vw (x) œ 216x 12x# œ 12x(18 x) œ 0
Ê the critical points are 0 and 18, but x œ 0 results in
no box. Since Vww (x) œ 216 24x 0 at x œ 18 we
have a maximum. The dimensions of the box are
18 ‚ 18 ‚ 36 in.
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250
Chapter 4 Applications of Derivatives
#
(b) In terms of length, V(j) œ x# j œ ˆ 1084 j ‰ j. The graph
indicates that the maximum volume occurs near j œ 36,
which is consistent with the result of part (a).
21. (a) From the diagram we have 3h 2w œ 108 and
V œ h# w Ê V(h) œ h# ˆ54 #3 h‰ œ 54h# 3# h$ .
Then Vw (h) œ 108h 9# h# œ
9
#
h(24 h) œ 0
Ê h œ 0 or h œ 24, but h œ 0 results in no box. Since
Vww (h) œ 108 9h 0 at h œ 24, we have a maximum
volume at h œ 24 and w œ 54 3# h œ 18.
(b)
22. From the diagram the perimeter is P œ 2r 2h 1r,
where r is the radius of the semicircle and h is the
height of the rectangle. The amount of light transmitted
proportional to
A œ 2rh "4 1r# œ r(P 2r 1r) 4" 1r#
œ rP 2r# 34 1r# . Then
dA
dr
œ P 4r 3# 1r œ 0
(4 1)P
2P
4P
21 P
8 31 Ê 2h œ P 8 31 8 31 œ 8 31 .
Therefore, 2rh œ 4 8 1 gives the proportions that admit the
#
most light since ddrA# œ 4 3# 1 0.
Ê rœ
23. The fixed volume is V œ 1r# h 23 1r$ Ê h œ
V
1 r#
2r
3
, where h is the height of the cylinder and r is the radius
of the hemisphere. To minimize the cost we must minimize surface area of the cylinder added to twice the
8
#
surface area of the hemisphere. Thus, we minimize C œ 21rh 41r# œ 21r ˆ 1Vr# 2r3 ‰ 41r# œ 2V
r 3 1r .
Then
œ
dC
dr
œ 2V
r# 4V"Î$
1"Î$ †3#Î$
16
3
2†3"Î$ †V"Î$
3†#†1"Î$
1r œ 0 Ê V œ
œ
8
3
‰
1r$ Ê r œ ˆ 3V
81
3"Î$ †2†4†V"Î$ 2†3"Î$ †V"Î$
3†#†1"Î$
‰
œ ˆ 3V
1
"Î$
"Î$
. From the volume equation, h œ
. Since
d# C
dr#
œ
4V
r$
16
3
V
1 r#
2r
3
1 0, these
dimensions do minimize the cost.
24. The volume of the trough is maximized when the area of the cross section is maximized. From the diagram
the area of the cross section is A()) œ cos ) sin ) cos ), 0 ) 1# . Then Aw ()) œ sin ) cos# ) sin# )
œ a2 sin# ) sin ) 1b œ (2 sin ) 1)(sin ) 1) so Aw ()) œ 0 Ê sin ) œ
sin ) Á 1 when 0 ) 1
#
w
. Also, A ()) 0 for 0 ) 1
6
w
and A ()) 0 for
1
6
"
#
or sin ) œ 1 Ê ) œ
)
1
#
because
. Therefore, at ) œ
there is a maximum.
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1
6
1
6
Section 4.5 Applied Optimization Problems
25. (a) From the diagram we have: AP œ x, RA œ ÈL x# ,
PB œ 8.5 x, CH œ DR œ 11 RA œ 11 ÈL x# ,
QB œ Èx# (8.5 x)# , HQ œ 11 CH QB
œ 11 ’11 ÈL x# Èx# (8.5 x)# “
#
#
#
œ ÈL x# Èx# (8.5 x)# , RQ œ RH HQ
#
œ (8.5)# ŠÈL x# Èx# (8.5 x)# ‹ . It
#
#
#
#
follows that RP œ PQ RQ Ê L# œ x# ŠÈL# x# Èx# (x 8.5)# ‹ (8.5)#
Ê L# œ x# L# x# 2ÈL# x# È17x (8.5)# 17x (8.5)# (8.5)#
Ê 17# x# œ 4 aL# x# b a17x (8.5)# b Ê L# œ x# œ
17x$
‰#
17x ˆ 17
#
(b) If f(x) œ
4x$
4x 17
œ
4x$
4x17
œ
(c) When x œ
51
8
. Thus L# is minimized when x œ
cylinder is formed, x œ 21r Ê r œ
Then Vww (x)
17x$
17x (8.5)#
4x# (8x 51)
(4x 17)#
Ê f w (x) 0 when x 51
8
51
8 .
then L ¸ 11.0 in.
26. (a) From the figure in the text we have P œ 2x 2y Ê y œ
Ê V(x) œ
œ
2x$
2x 8.5 .
is minimized, then L# is minimized. Now f w (x) œ
and f w (x) 0 when x 51
8 ,
17# x#
4 c17x (8.5)# d
#
$
18x x
41
œ 13 ˆ3
x
#1
P
#
x. If P œ 36, then y œ 18 x. When the
and h œ y Ê h œ 18 x. The volume of the cylinder is V œ 1r# h
x)
. Solving Vw (x) œ 3x(12
œ 0 Ê x œ 0 or 12; but when x œ 0, there is no cylinder.
41
x‰
ww
# Ê V (12) 0 Ê there is a maximum at x œ 12. The values of x œ 12 cm and
y œ 6 cm give the largest volume.
(b) In this case V(x) œ 1x# (18 x). Solving Vw (x) œ 31x(12 x) œ 0 Ê x œ 0 or 12; but x œ 0 would result in
no cylinder. Then Vww (x) œ 61(6 x) Ê Vww (12) 0 Ê there is a maximum at x œ 12. The values of
x œ 12 cm and y œ 6 cm give the largest volume.
27. Note that h# r# œ $ and so r œ È$ h# . Then the volume is given by V œ 1$ r# h œ 1$ a$ h# bh œ 1h 1$ h$ for
dV
#
#
! h È$, and so dV
dh œ 1 1r œ 1a" r b. The critical point (for h !) occurs at h œ ". Since dh ! for
! h ", and dV ! for " h È$, the critical point corresponds to the maximum volume. The cone of greatest
dh
volume has radius È# m, height "m, and volume
28. (a) f(x) œ x# (b) f(x) œ x# a
x
a
x
#1
$
m$ .
Ê f w (x) œ x# a2x$ ab , so that f w (x) œ 0 when x œ 2 implies a œ 16
Ê f ww (x) œ 2x$ ax$ ab , so that f ww (x) œ 0 when x œ 1 implies a œ 1
29. If f(x) œ x# xa , then f w (x) œ 2x ax# and f ww (x) œ 2 2ax$ . The critical points are 0 and $È #a , but x Á 0.
Now f ww ˆ $È #a ‰ œ 6 0 Ê at x œ $È #a there is a local minimum. However, no local maximum exists for any a.
30. If f(x) œ x$ ax# bx, then f w (x) œ 3x# 2ax b and f ww (x) œ 6x 2a.
(a) A local maximum at x œ 1 and local minimum at x œ 3 Ê f w (1) œ 0 and f w (3) œ 0 Ê 3 2a b œ 0 and
27 6a b œ 0 Ê a œ 3 and b œ 9.
(b) A local minimum at x œ 4 and a point of inflection at x œ 1 Ê f w (4) œ 0 and f ww (1) œ 0 Ê 48 8a b œ 0
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251
252
Chapter 4 Applications of Derivatives
and 6 2a œ 0 Ê a œ 3 and b œ 24.
31. (a) satb œ "'t# *'t ""# Ê vatb œ sw atb œ $#t *'. At t œ !, the velocity is va!b œ *' ft/sec.
(b) The maximum height ocurs when vatb œ !, when t œ $. The maximum height is sa$b œ #&' ft and it occurs at t œ $
sec.
(c) Note that satb œ "'t# *'t ""# œ "'at "bat (b, so s œ ! at t œ " or t œ (. Choosing the positive value
of t, the velocity when s œ ! is va(b œ "#) ft/sec.
32.
Let x be the distance from the point on the shoreline nearest Jane's boat to the point where she lands her boat. Then she
needs to row È% x# mi at 2 mph and walk ' x mi at 5 mph. The total amount of time to reach the village is
faxb œ
È % x#
#
have: #È%x x# œ
'x
&
"
&
hours (! Ÿ x Ÿ '). Then f w axb œ
"
"
# #È% x# a#xb
"
&
œ
x
#È% x#
&" . Solving f w axb œ !, we
Ê &x œ #È% x# Ê #&x# œ %a% x# b Ê #"x# œ "' Ê x œ „
%
È#" .
We discard the negative
value of x because it is not in the domain. Checking the endpoints and critical point, we have fa!b œ #Þ#,
fŠ È%#" ‹ ¸ #Þ"#, and fa'b ¸ $Þ"'. Jane should land her boat
%
È#"
¸ !Þ)( miles donw the shoreline from the point
nearest her boat.
33.
)
x
œ
h
x #(
Êhœ)
œ Ɉ) #"' ‰#
x
#"'
x
and Laxb œ Éh# ax #(b#
ax #(b# when x
minimized when faxb œ ˆ) #"' ‰#
x
!. Note that Laxb is
ax #(b# is
minimized. If f w axb œ !, then
‰ˆ #"'
‰
#ˆ) #"'
x
x# #ax #(b œ !
Ê ax #(bˆ" "(#) ‰
x$
œ ! Ê x œ #( (not acceptable
since distance is never negative or x œ "#. Then La"#b œ È#"*( ¸ %'Þ)( ftÞ
34. (a) From the diagram we have d# œ 4r# w# . The strength of the beam is S œ kwd# œ kw a4r# w# b . When
r œ 6, then S œ 144kw kw$ . Also, Sw (w) œ 144k 3kw# œ 3k a48 w# b so Sw (w) œ 0 Ê w œ „ 4È3 ;
Sww Š4È3‹ 0 and 4È3 is not acceptable. Therefore S Š4È3‹ is the maximum strength. The dimensions
of the strongest beam are 4È3 by 4È6 inches.
(b)
(c)
Both graphs indicate the same maximum value and are consistent with each other. Changing k does not
change the dimensions that give the strongest beam (i.e., do not change the values of w and d that produce
the strongest beam).
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 4.5 Applied Optimization Problems
35. (a) From the situation we have w# œ 144 d# . The stiffness of the beam is S œ kwd$ œ kd$ a144 d# b
#
#
where 0 Ÿ d Ÿ 12. Also, Sw (d) œ 4kd a108 d b Ê critical points at 0, 12, and 6È3. Both d œ 0 and
"Î#
253
,
È144 d#
d œ 12 cause S œ 0. The maximum occurs at d œ 6È3. The dimensions are 6 by 6È3 inches.
(b)
(c)
Both graphs indicate the same maximum value and are consistent with each other. The changing of k has
no effect.
36. (a) s" œ s# Ê sin t œ sin ˆt 13 ‰ Ê sin t œ sin t cos
Ê tœ
1
3
or
1
3
sin
1
3
cos t Ê sin t œ
41
3
(b) The distance between the particles is s(t) œ ks" s# k œ ¸sin t sin ˆt 13 ‰¸ œ
Ê sw (t) œ
are 0, 13 ,
Šsin t È3 cos t‹ Šcos t È3 sin t‹
2 ¹sin t È3 cos t¹
51 41 111
6 , 3 , 6 ,
21; then s(0) œ
since
È3
#
d
dx
kxk œ
x
kx k
"
#
"
#
sin t È3
#
cos t Ê tan t œ È3
¹sin t È3 cos t¹
Ê critical times and endpoints
, s ˆ 13 ‰ œ 0, s ˆ 561 ‰ œ 1, s ˆ 431 ‰ œ 0, s ˆ 1161 ‰ œ 1, s(21) œ
È3
#
Ê the
greatest distance between the particles is 1.
(c) Since sw (t) œ
Šsin t È3 cos t‹ Šcos t È3 sin t‹
2 ¹sin t È3 cos t¹
we can conclude that at t œ
1
3
and
41 w
3 , s (t)
has cusps and
the distance between the particles is changing the fastest near these points.
37. (a) s œ 10 cos (1t) Ê v œ 101 sin (1t) Ê speed œ k101 sin (1t)k œ 101 ksin (1t)k Ê the maximum speed is
101 ¸ 31.42 cm/sec since the maximum value of ksin (1t)k is 1; the cart is moving the fastest at t œ 0.5 sec,
1.5 sec, 2.5 sec and 3.5 sec when ksin (1t)k is 1. At these times the distance is s œ 10 cos ˆ 1# ‰ œ 0 cm and
a œ 101# cos (1t) Ê kak œ 101# kcos (1t)k Ê kak œ 0 cm/sec#
(b) kak œ 101# kcos (1t)k is greatest at t œ 0.0 sec, 1.0 sec, 2.0 sec, 3.0 sec and 4.0 sec, and at these times the
magnitude of the cart's position is ksk œ 10 cm from the rest position and the speed is 0 cm/sec.
38. (a) 2 sin t œ sin 2t Ê 2 sin t 2 sin t cos t œ 0 Ê (2 sin t)(1 cos t) œ 0 Ê t œ k1 where k is a positive
integer
(b) The vertical distance between the masses is s(t) œ ks" s# k œ ˆas" s# b# ‰
Ê sw (t) œ ˆ "# ‰ a(sin 2t 2 sin t)# b
2(cos 2t cos t)(sin 2t 2 sin t)
ksin 2t 2 sin tk
œ
0,
21
3 ,
1,
41
3 ,
œ
"Î#
4(2 cos t 1)(cos t ")(sin t)(cos t 1)
ksin 2t 2 sin tk
3È 3
2 ,
ds
dt
œ
Ê
"
#
a(12 12t)# 64t# b
ds ¸
dt t=0
Ê critical times at
œ 12 knots and
"Î#
, s(1) œ 0, s ˆ 431 ‰
3È 3
#
at t œ
21
3
and
"Î#
[2(12 12t)(12) 128t] œ
ds ¸
dt t=1
3È 3
#
s(21) œ 0 Ê the greatest distance is
39. (a) s œ È(12 12t)# (8t)# œ a(12 12t)# 64t# b
(b)
œ a(sin 2t 2 sin t)# b
(2)(sin 2t 2 sin t)(2 cos 2t 2 cos t)
21; then s(0) œ 0, s ˆ 231 ‰ œ ¸sin ˆ 431 ‰ 2 sin ˆ 231 ‰¸ œ
œ ¸sin ˆ 831 ‰ 2 sin ˆ 431 ‰¸ œ
"Î#
208t 144
È(12 12t)# 64t#
œ 8 knots
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41
3
"Î#
254
Chapter 4 Applications of Derivatives
(c) The graph indicates that the ships did not see
each other because s(t) 5 for all values of t.
(d) The graph supports the conclusions in parts (b)
and (c).
(e)
lim ds
t Ä _ dt
œ
(208t 144)#
É lim 144(
" t)# 64t#
tÄ_
Š208 œ Ë lim
#
144
t ‹
#
t Ä _ 144 Š " 1‹ 64
t
#
œ É 144208 64 œ È208 œ 4È13
which equals the square root of the sums of the squares of the individual speeds.
40. The distance OT TB is minimized when OB is
a straight line. Hence n! œ n" Ê )" œ )# .
41. If v œ kax kx# , then vw œ ka 2kx and vww œ 2k, so vw œ 0 Ê x œ
vww ˆ #a ‰
œ 2k 0. The maximum value of v is
ka#
4
a
#
. At x œ
a
#
there is a maximum since
.
42. (a) According to the graph, yw a!b œ !.
(b) According to the graph, yw aLb œ !.
(c) ya!b œ !, so d œ !. Now yw axb œ $ax# #bx c, so yw a!b œ ! implies that c œ !. There fore, yaxb œ ax$ bx# and
yw axb œ $ax# #bx. Then yaLb œ aL$ bL# œ H and yw aLb œ $aL# #bL œ !, so we have two linear
equations in two unknowns a and b. The second equation gives b œ $aL
# . Substituting into the first equation, we have
aL$ $aL$
#
œ H, or
aL$
#
œ H, so a œ # LH$ . Therefore, b œ $ LH# and the equation for y is
$
#
yaxb œ # LH$ x$ $ LH# x# , or yaxb œ H’#ˆ Lx ‰ $ˆ Lx ‰ “.
43. The profit is p œ nx nc œ n(x c) œ ca(x c)" b(100 x)d (x c) œ a b(100 x)(x c)
œ a (bc 100b)x 100bc bx# . Then pw (x) œ bc 100b 2bx and pww (x) œ 2b. Solving pw (x) œ 0 Ê
x œ #c 50. At x œ #c 50 there is a maximum profit since pww (x) œ 2b 0 for all x.
44. Let x represent the number of people over 50. The profit is p(x) œ (50 x)(200 2x) 32(50 x) 6000
œ 2x# 68x 2400. Then pw (x) œ 4x 68 and pww œ 4. Solving pw (x) œ 0 Ê x œ 17. At x œ 17 there is a
maximum since pww (17) 0. It would take 67 people to maximize the profit.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 4.5 Applied Optimization Problems
45. (a) A(q) œ kmq" cm h# q, where q 0 Ê Aw (q) œ kmq# h
#
œ
hq# 2km
2q#
255
and Aww (q) œ 2kmq$ . The
ww É 2km
É 2km
É 2km
critical points are É 2km
h , 0, and
h , but only
h is in the domain. Then A Š
h ‹ 0 Ê at
q œ É 2km
h there is a minimum average weekly cost.
(b) A(q) œ
(kbq)m
q
cm h# q œ kmq" bm cm h# q, where q 0 Ê Aw (q) œ 0 at q œ É 2km
h as in (a).
Also Aww (q) œ 2kmq$ 0 so the most economical quantity to order is still q œ É 2km
h which minimizes
the average weekly cost.
46. We start with caxb œ the cost of producing x items, x !, and
c ax b
x
œ the average cost of producing x items, assumed
to be differentiable. If the average cost can be minimized, it will be at a production level at which
Ê
w
w
x c ax b c a x b
x#
d c ax b
dx Š x ‹
œ ! (by the quotient rule) Ê x cw axb caxb œ ! (multiply both sides by x# ) Ê cw axb œ
c ax b
x
œ!
where
c axb is the marginal cost. This concludes the proof. (Note: The theorem does not assure a production level that will give a
minimum cost, but rather, it indicates where to look to see if there is one. Find the production levels where the average cost
equals the marginal cost, then check to see if any of them give a mimimum.)
47. The profit p(x) œ r(x) c(x) œ 6x ax$ 6x# 15xb œ x$ 6x# 9x, where x 0. Then
pw (x) œ 3x# 12x 9 œ 3(x 3)(x 1) and pw w (x) œ 6x 12. The critical points are 1 and 3. Thus
pww (1) œ 6 0 Ê at x œ 1 there is a local minimum, and pww (3) œ ' 0 Ê at x œ 3 there is a local maximum.
But p(3) œ 0 Ê the best you can do is break even.
48. The average cost of producing x items is caxb œ
œ x# #!x #!ß !!! Ê c w axb œ #x #! œ ! Ê x œ "!, the
c ax b
x
only critical value. The average cost is ca"!b œ $"*ß *!! per item is a minimum cost because c ww a"!b œ # !.
49. (a) The artisan should order px units of material in order to have enough until the next delivery.
ˆ px ‰
(b) The average number of units in storage until the next delivery is px
# and so the cost of storing then is s # per
‰
day, and the total cost for x days is ˆ px
# sx. When added to the delivery cost, the total cost for delivery and storage
for each cycle is: cost per cycle œ d px
# sx.
ˆd (c) The average cost per day for storage and delivery of materials is: average cost per day œ
To minimize the average cost per day, set the derivative equal to zero.
"
d
dx Šdaxb
ps # ‰
#x
x
ps
# x‹
œ daxb
d
x
#
ps
# x.
ps
# œ
Ê x œ „ É #psd . Only the positive root makes sense in this context so that x‡ œ É #psd . To verify that x‡ gives a
d ˆ
minimum, check the second derivative ’ dx
daxb# ps ‰
# Ҽ
É #psd
œ
#d
x$ º
É #psd
œ
#d
#d
$
ŒÉ ps ! Ê a minimum.
The amount to deliver is px‡ œ É #pd
s .
(d) The line and the hyperbola intersect when
d
x
œ
ps
# x.
Solving for x gives xintersection œ „ É #psd . For x !,
xintersection œ É #psd œ x‡ . From this result, the average cost per day is minimized when the average daily cost of
delivery is equal to the average daily cost of storage.
50. Average Cost:
c ax b
x
c ax b
d#
dx# Š x ‹º
œ
xœ"!!
œ
#!!!
x
%!!!
"!!$
*' %x"Î# Ê
d c ax b
dx Š x ‹
"Î#
œ #!!!
œ ! Ê x œ "!!. Check for a minimum:
x # #x
"!!$Î# œ !Þ!!$ ! Ê a minimum at x œ "!!. At a production level of "!!ß !!! units,
the average cost will be minimized at $"&' per unit.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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!
256
Chapter 4 Applications of Derivatives
51. We have
dR
dM
œ CM M# . Solving
d# R
dM#
œ C #M œ ! Ê M œ C# . Also,
d$ R
dM$
œ # ! Ê at M œ
C
#
there is a
maximum.
52. (a) If v œ cr! r# cr$ , then vw œ 2cr! r 3cr# œ cr a2r! 3rb and vww œ 2cr! 6cr œ 2c ar! 3rb . The solution of
vw œ 0 is r œ 0 or 2r3! , but 0 is not in the domain. Also, vw 0 for r 2r3! and vw 0 for r 2r3! Ê at
rœ
2r!
3
there is a maximum.
(b) The graph confirms the findings in (a).
53. If x 0, then (x 1)#
then Š a
#
#
0 Ê x# 1
#
2x Ê
#
1
b 1
c 1
d "
a ‹Š b ‹Š c ‹Š d ‹
54. (a) f(x) œ
x
È a# x#
Ê f w (x) œ
aa # x # b
"Î#
"Î#
x # aa # x # b
aa # x # b
Ê gw (x) œ
ab# (d x)# b (d x)#
ab# (d x)# b$Î#
ab# (d x)# b
œ
a# x# x#
aa# x# b$Î#
œ
a#
aa# x# b$Î#
0
"Î#
"Î#
(d x)# ab# (d x)# b
b# (d x)#
b #
0 Ê g(x) is a decreasing function of x
ab# (d x)# b$Î#
dt
Since c" , c# 0, the derivative dx
is an increasing function of x (from part (a)) minus a decreasing
dt
d# t
" w
" w
w
function of x (from part (b)): dx œ c"" f(x) c"# g(x) Ê dx
# œ c" f (x) c# g (x) 0 since f (x) dt
gw (x) 0 Ê dx
is an increasing function of x.
œ
(c)
dx
Èb# (d x)#
2. In particular if a, b, c and d are positive integers,
16.
Ê f(x) is an increasing function of x
(b) g(x) œ
x# 1
x
œ
0 and
55. At x œ c, the tangents to the curves are parallel. Justification: The vertical distance between the curves is
D(x) œ f(x) g(x), so Dw (x) œ f w (x) gw (x). The maximum value of D will occur at a point c where Dw œ 0. At
such a point, f w (c) gw (c) œ 0, or f w (c) œ gw (c).
56. (a) f(x) œ 3 4 cos x cos 2x is a periodic function with period 21
(b) No, f(x) œ 3 4 cos x cos 2x œ 3 4 cos x a2 cos# x 1b œ 2 a1 2 cos x cos# xb œ 2(1 cos x)#
Ê f(x) is never negative
0
57. (a) If y œ cot x È2 csc x where 0 x 1, then yw œ (csc x) ŠÈ2 cot x csc x‹. Solving yw œ 0
Ê cos x œ
"
È2
Ê x œ 14 . For 0 x 1
4
we have yw 0, and yw 0 when
1
4
x 1. Therefore, at x œ
there is a maximum value of y œ 1.
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1
4
Section 4.5 Applied Optimization Problems
257
(b)
The graph confirms the findings in (a).
58. (a) If y œ tan x 3 cot x where 0 x Ê xœ „
1
3,
Therefore at x œ
1
#
, then yw œ sec# x 3 csc# x. Solving yw œ 0 Ê tan x œ „ È3
is not in the domain. Also, yww œ 2 sec# x tan x 3 csc# x cot x 0 for all 0 x there is a minimum value of y œ 2È3.
but 1
3
1
3
1
2
.
(b)
The graph confirms the findings in (a).
#
#
59. (a) The square of the distance is Daxb œ ˆx $# ‰ ˆÈx !‰ œ x# #x *% , so Dw axb œ #x # and the critical
point occurs at x œ ". Since Dw axb ! for x " and Dw axb ! for x ", the critical point corresponds to the
minimum distance. The minimum distance is ÈDa"b œ
È&
# .
(b)
The minimum distance is from the point ˆ $# ß !‰ to the point a"ß "b on the graph of y œ Èx, and this occurs at the
value x œ " where Daxb, the distance squared, has its minimum value.
60. (a) Calculus Method:
The square of the distance from the point Š"ß È$‹ to Šxß È"' x# ‹ is given by
#
Daxb œ ax "b# ŠÈ"' x# È$‹ œ x# #x " "' x# #È%) $x# $ œ #x #! #È%) $x# .
Then Dw axb œ # #
È%) $x# a'xb
œ #
'x
È%) $x# .
Solving Dw axb œ ! we have: 'x œ #È%) $x#
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258
Chapter 4 Applications of Derivatives
Ê $'x# œ %a%) $x# b Ê *x# œ %) $x# Ê "#x# œ %) Ê x œ „ #. We discard x œ # as an extraneous solution,
leaving x œ #. Since Dw axb ! for % x # and Dw axb ! for # x %, the critical point corresponds to the
minimum distance. The minimum distance is ÈDa#b œ #.
Geometry Method:
The semicircle is centered at the origin and has radius %. The distance from the origin to Š"ß È$‹ is
#
Ê"# ŠÈ$‹ œ #. The shortest distance from the point to the semicircle is the distance along the radius
containing the point Š"ß È$‹. That distance is % # œ #.
(b)
The minimum distance is from the point Š"ß È$‹ to the point Š#ß #È$‹ on the graph of y œ È"' x# , and this
occurs at the value x œ # where Daxb, the distance squared, has its minimum value.
61. (a) The base radius of the cone is r œ
#1 a x
#1
#
and so the height is h œ Èa# r# œ Éa# ˆ #1a#1 x ‰ . Therefore,
#
#
Vaxb œ 1$ r# h œ 1$ ˆ #1a#1 x ‰ Éa# ˆ #1a#1 x ‰ .
(b) To simplify the calculations, we shall consider the volume as a function of r: volume œ farb œ 1$ r# Èa# r# , where
! r a. f w arb œ
#
$
œ 1$ ” #Èa r# $r# • œ
a r
1 d
#È #
a
$ dr Šr
1ra#a# $r# b
.
$Èa# r#
h œ Èa# r# œ Éa# #a #
$
r# ‹ œ 1$ ”r# †
"
a#rb
# È a # r#
ŠÈa# r# ‹a#rb• œ 1$ ” r
The critical point occurs when r# œ
#
œ É a$ œ
aÈ $
$ .
Using r œ
aÈ '
$
#a #
$ ,
and h œ
$
#raa# r# b
•
Èa# r#
which gives r œ aÉ #$ œ
aÈ $
$ ,
aÈ '
$ .
Then
we may now find the values of r and h
for the given values of a.
hœ
When a œ &: r œ
%È'
$ ,
È
& '
$ ,
When a œ ): r œ
)È'
$ ,
hœ
)È$
$ à
aÈ $
$ ,
the relationship is
When a œ %: r œ
%È$
$ à
È
& $
$ à
hœ
È
When a œ ': r œ # ', h œ #È$à
(c) Since r œ
aÈ '
$
and h œ
r
h
œ È #.
62. (a) Let x! represent the fixed value of x at the point P, so that P has the coordinates ax! ß ab, and let m œ f w ax! b be the
slope of the line RT. Then the equation of the line RT is y œ max x! b a. The y-intercept of this line is
ma! x! b a œ a mx! , and the x-intercept is the solution of max x! b a œ !, or x œ mx!m a . Let O designate
the origin. Then
(Area of triangle RST)
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Section 4.5 Applied Optimization Problems
259
œ #(Area of triangle ORT)
œ # † "# (x-intercept of line RT)(y-intercept of line RT)
œ # † "# ˆ mx!m a ‰aa mx! b
œ mˆ mx!m a ‰ˆ mx!m a ‰
œ mˆ mx!m a ‰
œ mˆx! #
a ‰#
m
Substituting x for x! , f w axb for m, and faxb for a, we have Aaxb œ f w axb”x faxb
f ax b •
w
(b) The domain is the open interval a!ß "!b. To graph, let y" œ faxb œ & &É" y$ œ Aaxb œ y# Šx y"
y# ‹
#
x#
"!! ,
#
.
y# œ f w axb œ NDERay" b, and
. The graph of the area function y$ œ Aaxb is shown below.
The vertical asymptotes at x œ ! and x œ "! correspond to horizontal or vertical tangent lines, which do not form
triangles.
(c) Using our expression for the y-intercept of the tangent line, the height of the triangle is
x
x#
a mx œ faxb f w axb † x œ & " È"!! x# x œ & " È"!! x# #
#
2È"!! x#
2È"!! x#
We may use graphing methods or the analytic method in part (d) to find that the minimum value of Aaxb occurs at
x ¸ )Þ''. Substituting this value into the expression above, the height of the triangle is 15. This is 3 times the
y-coordinate of the center of the ellipse.
(d) Part (a) remains unchanged. Assuming C B, the domain is a!ß Cb. To graph, note that
faxb œ B BÉ" x#
C#
Aaxb œ f w axb”x faxb
f ax b •
œ B BC ÈC# x# and f w axb œ
#
œ
w
Bx
Œx
CÈC# x#
"
#
”Bx
BCxÈC# x#
œ
"
”BCŠC ÈC#
BCxÈC# x#
A axb œ BC †
œ
œ
œ
œ
x# aC# x# b
BCŠC ÈC# x# ‹
x# aC# x# b
BCŠC ÈC# x# ‹
x# aC# x# b
x# aC# x# b$Î#
È C# x #
#
œ
ŠBC BÈC# x# ‹ŠÈC# x# ‹• œ
#
x# ‹• œ
œ
Bx
.
C È C # x#
Bx
x
C È C # x# "
#
”Bx
BCxÈC# x#
Therefore we have
#
ŠBC BÈC# x# ‹ŠÈC# x# ‹
Bx
BCÈC# x# BaC# x# b•
#
#
BCŠC ÈC# x# ‹
xÈC# x#
x ‹
C# x #
#
ŠC ÈC# x# ‹ Šx È
x# aC#
x# b
x ÈC#
C# x #
x# a"b‹
#
x
#
È #
È #
#
#
–#x ŠC C x ‹Š ÈC# x# C x ‹—
#
”#x Š È Cx
#
#
C BC# ŠCÈC# x# ‹
Bx
C
ŠxÈC# x# ‹a#bŠC ÈC# x# ‹Š È
BCŠC ÈC# x# ‹
B BC ÈC# x#
#
œ
w
B
"
C #ÈC# x# a#xb
x#
Cx#
ÈC# x#
CÈC# x# x# aC# x# b•
CÈC# x# C# ‹ œ
BCŠC ÈC# x# ‹
x# aC# x# b$Î#
’Cx# CaC# x# b C# ÈC# x# “
Š#x# C# CÈC# x# ‹
To find the critical points for ! x C, we solve: #x# C# œ CÈC# x# Ê %x% %C# x# C% œ C% C# x#
Ê %x% $C# x# œ ! Ê x# a%x# $C# b œ !. The minimum value of Aaxb for ! x C occurs at the critical point
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260
Chapter 4 Applications of Derivatives
xœ
CÈ$
# ,
or x# œ
$C#
% .
w
The corresponding triangle height is
a mx œ faxb f axb † x
œ B B ÈC# x# C
Bx#
É
C C# $C#
%
#
œB
BÈ #
C
C
œ B BC ˆ C# ‰ œB
B
#
$B
#
x#
BŠ $C% ‹
$C#
%
CÉ C# $BC#
%
C#
#
œ $B
This shows that the traingle has minimum arrea when its height is $B.
^
4.6 INDETERMINATE FORMS AND L'HOPITAL'S
RULE
^
1. l'Hopital:
lim
x2
œ
"
#x ¹xœ#
^
2. l'Hopital:
lim
sin 5x
x
œ
5 cos 5x
¹
1
xœ!
#
x Ä 2 x 4
xÄ0
5x# 3x
7x# 1
^
3. l'Hopital:
lim
xÄ_
x$ 1
$
x Ä 1 4x x 3
#
3
lim ax 2 x "b œ 11
x Ä 1 a4x + 4x + 3b
1 cos x
x#
^
5. l'Hopital:
lim
xÄ!
sin# x
œ lim
2
x Ä ! x a" cos xb
7.
lim
tÄ0
sin t#
t
tÄ0
2x1
cos x
œ
8.
lim
x Ä 1 Î2
9.
lim
) Ä 1 1 )
10.
lim
x Ä 1Î2 1 cos 2x
sin )
œ
11.
sin x cos x
x 1%
x Ä 1Î%
12.
cos x 1
x Ä 1Î$ x $
lim
lim
"
#
œ
sin x
2x
œ x lim
Ä_
œ x lim
Ä_
œ
3
11
œ lim
10
14
œ
cos )
"
œ
lim
x Ä 1 Î2
œ
%x 3
$x # "
"
"
œ
2
"
"
#
or lim
xÄ!
œ x lim
Ä_
%
'x
œ
"
4
œ5†1œ5
5x# 3x
7x# 1
or x lim
Ä_
x $ 1
$
x Ä 1 4x x3
œ
sin 5x
5x
"
x Ä 2 x#
œ x lim
Ä_
5
7
3
x
"
x
œ
5
7
ax "bax# x "b
2
a
x Ä 1 x "ba4x + 4x + 3b
œ lim
1cos x
x#
xb ˆ " cos x ‰
œ lim ” a" xcos
2
" cos x •
xÄ!
"
#
œ ! or x lim
Ä_
#x# 3x
x$ x 1
œ x lim
Ä_
#
x
"
"
x#
3
x#
œ 2
œ"
cos x
2 sin 2x
œ
cos x sin x
"
x Ä 1Î%
lim
lim
5
7
or lim
cos x
2
xÄ!
œ 5 lim
5x Ä 0
sin 5x
x
œ lim
œ0
2
)Ä1
xÄ0
x Ä 2 ax #bax #b
œ lim ”ˆ sinx x ‰ˆ sinx x ‰ˆ " "cos x ‰• œ
xÄ!
lim
) Ä 1Î2 sin x
œ lim
1 sin x
xÄ!
2t cos t#
1
œ lim
œ 5 or lim
x2
œ lim
#
x Ä 2 x 4
3x#
#
x Ä 1 12x 1
œ lim
x2
or lim
œ lim
#x # $ x
x$ x 1
^
6. l'Hopital:
lim
xÄ_
"
4
"0x 3
14x
œ x lim
Ä_
^
4. l'Hopital:
lim
œ
œ
x Ä 1Î$
sin x
"
œ
lim
x Ä 1 Î2
œ
È#
#
sin x
4 cos 2x
È#
#
œ
"
4(1)
œ
"
4
œ È#
È$
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
"
x$
œ
!
"
œ!
^
Section 4.6 Indeterminate Forms and L'Hopital's
Rule
13.
lim
x Ä 1 Î2
#x
14. lim
ˆx 1# ‰ sin x
cos x
x Ä 1 Î2
ˆx 1# ‰ tan x œ
lim
#
œ lim
#x # a$ x " b È x #
x"
xÄ"
x Ä ! 2Èx(
#x# $x$Î# x"Î# #
x"
xÄ"
" ax# 5b"Î# a2xb
#
16. lim
È x# 5 $
x# %
17. lim
È a aa x b a
x
18. lim
10(sin t t)
t$
œ lim
10(cos t ")
$t#
19. lim
x(cos x ")
sin x x
œ lim
xsin x cos x "
cos x "
20. lim
sinaa hb sin a
h
21. lim
rÄ"
a ar n " b
r"
lim
Š "x xÄ0
tÄ0
xÄ0
hÄ0
22.
x Ä !b
œ lim
xÄ#
œ
a
x Ä 0 #Èa# ax
tÄ0
xÄ0
œ lim
hÄ0
aan†rn" b
1
œ lim
rÄ1
"
Èx ‹
"
"!asin tb
't
œ lim
xÄ0
cosaa hb cos a
"
"
œ
"
Èx
#
œ lim
tÄ0
xcos x #sin x
sin x
"
'
"! cos t
'
œ lim
œ
xÄ0
10†"
'
"
xcos x #sin x
sin x
$x &
sin (x
x Ä 0 tan ""x
27. x lim
Ä_
Èx
lim
x Ä !b Èsin x
" Èx
x ‹
tanŠ "x ‹
29.
lim
x Ä 1Î2c tan x
sec x
lim b
xÄ!
cot x
csc x
"
x
rule
È x‰ †
ˆ
œ Š l'Hopital's
does not apply ‹ œ lim b " xÄ!
œ
3
œ!
(cosa(xb
(†"
""†"
œ
"
lim
xÄ!b
lim
œ
9x 1
x1
sin x
x
x Ä 1 Î2 c
œ lim b
xÄ!
œ x lim
Ä_
lim
x Ä „ _ 4x "
œ Éx lim
Ä_
œÊ
xsin x $ cos x
cos x
rÄ1
#
x Ä 0 ""sec a""xb
28.
30.
œ
œ lim
È9x 1
Èx 1
xÄ0
#
1
x
œ_
x # ax # x b
x È x# x
œ x lim
Ä_
xx
x
x
rule
œ "# Š l'Hopital's
is unnecessary ‹
"
x
lim
#
x Ä „ _ #x x #
26. lim
œ lim
œ an lim rn" œ an, where n is a positive integer.
œ lim b Š
xÄ!
24. x lim
x tanˆ "x ‰ œ x lim
Ä_
Ä_
25.
œ &$
œ!
È
" É" œ"
œ "
x x
23. x lim
Šx Èx# x‹ œ x lim
Šx Èx# x‹Š xx ‹ œ x lim
Ä_
Ä_
Ä_
È x# x
œ x lim
Ä_
"
"
œ "# , where a 0.
a
#Èa#
tÄ0
%x *# x"Î# xÄ1
x Ä # 2È x# 5
œ lim
œ
œ!
œ lim
œ lim
2x
œ lim
ˆ 1 x‰ cos x sin xa"b
#
sin x
x Ä 1 Î2
lim
%†!
#†!(
œ
œ lim
15. lim
xÄ#
%È x
œ lim
(
x Ä ! " #Èx
x Ä ! x (È x
œ
x"#
œ x lim
sec# ˆ x" ‰ œ sec# ! œ "
Ä_
(
""
œ Éx lim
Ä_
9
1
œ È9 œ 3
œ É 1" œ 1
x‰
ˆ cos" x ‰ ˆ cos
sin x œ
ˆ cos x ‰
sin x
x"# sec# ˆ x" ‰
Š sin" x ‹
"
lim
x Ä 1Î2c sin x
œ1
œ lim b cos x œ 1
xÄ!
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
È x# x
È x#
œ
$
"
œ$
261
262
Chapter 4 Applications of Derivatives
31. Part (b) is correct because part (a) is neither in the
0
0
nor
_
_
^
form and so l'Hopital's
rule may not be used.
32. Answers may vary.
(a) faxb œ $x ";gaxb œ x
lim
x Ä _ g(x)
œ x lim
Ä_
$x "
x
#
lim
x Ä _ g(x)
f(x)
œ x lim
Ä_
x"
x#
f(x)
œ x lim
Ä_
x#
x"
f(x)
(b) faxb œ x ";gaxb œ x
#
(c) faxb œ x ;gaxb œ x "
lim
x Ä _ g(x)
$
"
œ$
œ x lim
Ä_
"
#x
œ!
œ x lim
Ä_
#x
"
œ_
œ x lim
Ä_
33. If f(x) is to be continuous at x œ 0, then lim f(x) œ f(0) Ê c œ f(0) œ lim
œ lim
xÄ0
27 sin 3x
30x
œ lim
34. (a) For x Á 0, f w (x) œ
œ
x2
x1
œ
81 cos 3x
30
xÄ0
02
01
d
dx
xÄ0
œ
27
10
xÄ0
9x 3 sin 3x
5x$
œ lim
xÄ0
9 9 cos 3x
15x#
.
(x 2) œ 1 and gw (x) œ
(x 1) œ 1. Therefore, lim
d
dx
f w (x)
w
x Ä 0 g (x)
œ 2.
œ
1
1
œ 1, while lim
f(x)
x Ä 0 g(x)
^
(b) This does not contradict l'Hopital's
rule because neither f nor g is differentiable at x œ 0
^
(as evidenced by the fact that neither is continuous at x œ 0), so l'Hopital's
rule does not apply.
35. The graph indicates a limit near 1. The limit leads to the
indeterminate form 00 : lim
xÄ1
2x# 3x$Î# x"Î# 2
x1
xÄ1
4 9# #"
œ 4 1 5 œ 1
1
œ lim
œ
2x# (3x 1) Èx 2
x 1
4x œ lim
xÄ1
9
#
x"Î# 1
"
#
x"Î#
36. (a)
(b) The limit leads to the indeterminate form _ _:
È #
lim Šx Èx# x‹ œ x lim
Šx Èx# x‹Š x Èx# x ‹ œ x lim
Š
xÄ_
Ä_
Ä_
x
œ x lim
Ä_
"
" É" 37. Graphing faxb œ
'
"
x
" cos x'
x"#
œ
"
" È" !
x # ax # x b
‹
x È x# x
œ x lim
Ä_
x
x È x# x
œ "#
on th window Ò"ß "Ó by Ò!Þ&ß "Ó it appears that lim faxb œ !. However, we see that if we let
" cos u
#
uÄ0 u
u œ x , then lim faxb œ lim
xÄ0
x x
38. (a) We seek c in a#ß !b so that
f ac b
g ac b
w
w
œ lim
sin u
u Ä 0 #u
œ
œ lim
fa!b fa#b
ga!b ga#b
cos u
uÄ0 #
œ
!#
!%
œ "# .
xÄ0
œ #" . Since f w acb œ " and gw acb œ #c we have that
Ê c œ ".
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
"
#c
œ #"
(b) We seek c in any open interval aaß bb so that
(c) We seek c in a!ß $b so that
(Note that c œ
" È$(
$
œ
w
fa$b fa!b
ga$b ga!b
w
œ
œ
$ !
*!
fabb faab
g ab b g a a b
œ
œ "$ Ê
ba
b # a#
c# %
#c
œ
ba
ab abab ab
ba
#
œ
"
ba
Ê
"
#c
œ
"
ba
œ "$ Ê $c# #c "# œ ! Ê c œ
"x
)
" cos )
) sin ) ,
œ
œ
)Ä_
since the coordinates of C are acos )ß sin )b. Hence, " x œ
)Ä_
”
)a" cos )b
) sin )
a" cos )b• œ lim
)Ä_
As ) Ä _, a" cos )b oscillates between ! and #, and so it is bounded. Since lim
)Ä_
a" cos )b” ) )sin ) "•
ˆ ) )sin ) "‰ œ " " œ !,
a" cos )b” ) )sin ) "• œ !. Geometrically, this means that as ) Ä _, the distance between points P and D
lim
)Ä_
approaches 0.
40. Throughout this problem note that r# œ y# ", r y and that both r Ä _ and y Ä _ as ) Ä 1# Þ
(b)
" È$(
.
$
←→
←→ ←→
where E is the point on AB such that CE ¼ AB :
CE
EB
a" xb a" cos )b‘ œ lim
(c) We have that lim
(a)
Êcœ
)a" cos )b
) sin ) .
)a" cos )b
) sin ) " cos )
) cos ) sin ) sin )
) #sin )
lim a" xb œ lim ) sin ) œ lim
œ lim
œ lim ) cos sin
" cos )
sin )
)
)Ä0
)Ä0
)Ä0
)Ä0
)Ä0
)a sin )b cos ) #cos )
) sin ) $cos )
!$
œ lim
œ lim
œ " œ$
cos )
cos )
)Ä0
)Ä0
Thus
(b)
w
w
263
is not in the given interval a!ß $b.)
PA
AB
39. (a) By similar triangles,
f ac b
g ac b
f ac b
g ac b
Section 4.7 Newton's Method
"
lim
ryœ
lim
r # y# œ
) Ä 1 Î2
) Ä 1 Î2
lim
) Ä 1 Î 2 r y
lim
) Ä 1 Î2
œ!
"œ"
(c) We have that r$ y$ œ ar ybar# ry y# b œ
Since
lim
) Ä 1 Î2
$y †
y
r
œ
lim
) Ä 1 Î2
r# ry y#
ry
$sin ) † y œ _ we have that
y # y †y y #
r
$
lim
) Ä 1 Î2
œ
$y #
r
$
œ $y † yr .
r y œ _.
4Þ7 NEWTON'S METHOD
1. y œ x# x 1 Ê yw œ 2x 1 Ê xnb1 œ xn Ê x# œ
2
3
4 2
9 3 1
4
3 1
Ê x# œ 2 Ê x# œ
42"
4 1
œ
5
3
2
3
4 6 9
129
œ
2
3
x#n xn 1
# xn 1
; x! œ 1 Ê x " œ 1 "
#1
¸ .61905; x! œ 1 Ê x" œ 1 #"7 11
"
3 3
13
21
œ
2
3
1 1 1
#1
¸ 1.66667
2. y œ x$ 3x 1 Ê yw œ 3x# 3 Ê xnb1 œ xn Ê x# œ "3 œ
1 1 1
#1
œ 3" "
90
x$n 3xn 1
3xn# 3
; x! œ 0 Ê x" œ 0 "
3
œ "3
œ 29
90 ¸ 0.32222
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
œ 2
264
Chapter 4 Applications of Derivatives
x%n xn 3
4xn$ 1
3. y œ x% x 3 Ê yw œ 4x$ 1 Ê xnb1 œ xn Ê x# œ
6
5
1296 6
625 5 3
864
125 1
œ 2 Ê x# œ # œ
6
5
12967501875
4320625
1623
3#1
œ 2 œ
6
5
Ê x# œ "# œ
¸ 2.41667
5
#
"
1#
5
4
113
2000
œ
2500113
2000
œ
2387
#000
6. From Exercise 5, xnb1 œ xn 625512
2000
œ 54 œ 54 œ
6
5
¸ 1.16542; x! œ 1 Ê x" œ 1 "13
4 1
2xn xn# 1
22xn
x%n 2
4xn$ ; x!
00"
#0
; x! œ 0 Ê x" œ 0 œ 1 Ê x" œ 1 "2
4
œ
5
4
44"
#4
œ "#
œ
5
#
Ê x# œ
5
4
Ê x# œ
625
256 2
125
16
œ
5
#
5
4
; x! œ 1 Ê x" œ 1 "2
4
œ 1 "
4
œ 54 Ê x# œ 54 625
256 2
125
16
f axn b
f axn b
w
gives x" œ x! Ê x# œ x! Ê xn œ x! for all n
0. That is, all of
the approximations in Newton's method will be the root of f(x) œ 0.
8. It does matter. If you start too far away from x œ
1
#
, the calculated values may approach some other root.
Starting with x! œ 0.5, for instance, leads to x œ 1# as the root, not x œ
œh
Èh
Š
f(x! )
f w (x! )
if x! œ h 0 Ê x" œ x! œ h Èh
Š "
È‹
2
œh
1
#
.
f(h)
f w (h)
œ h ŠÈh‹ Š2Èh‹ œ h;
"
Èh ‹
#
625512
2000
¸ 1.1935
7. f(x! ) œ 0 and f w (x! ) Á ! Ê xnb1 œ xn 9. If x! œ h 0 Ê x" œ x! 5 25
4 1
#5
¸ 1.1935
x%n 2
4xn$
113
2000
5763
4945
œ 15# ¸ .41667; x! œ 2 Ê x" œ 2 5. y œ x% 2 Ê yw œ 4x$ Ê xnb1 œ xn œ
œ
1 1 3
4 1
51
œ 31
¸ 1.64516
11
31
4. y œ 2x x# 1 Ê yw œ 2 2x Ê xnb1 œ xn 1 "4 "
œ #"
#1
20254
29
œ #5 1"# œ 12
1 #
171
4945
; x! œ 1 Ê x " œ 1 f(x! )
f w (x! )
œ h f(h)
f w (h)
œ h ŠÈh‹ Š2Èh‹ œ h.
h
"Î$
10. f(x) œ x"Î$ Ê f w (x) œ ˆ "3 ‰ x#Î$ Ê xnb1 œ xn ˆ " ‰xn #Î$
xn
3
œ 2xn ; x! œ 1 Ê x" œ 2, x# œ 4, x$ œ 8, and
x% œ 16 and so forth. Since kxn k œ 2lxnc1 l we may conclude
that n Ä _ Ê kxn k Ä _.
11. i) is equivalent to solving x$ $x " œ !.
ii) is equivalent to solving x$ $x " œ !.
iii) is equivalent to solving x$ $x " œ !.
iv) is equivalent to solving x$ $x " œ !.
All four equations are equivalent.
12. f(x) œ x 1 0.5 sin x Ê f w (x) œ 1 0.5 cos x Ê xnb1 œ xn xn 1 0.5 sin xn
1 0.5 cos xn
; if x! œ 1.5, then
x" œ 1.49870
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 4.7 Newton's Method
13. For x! œ !Þ$, the procedure converges to the root !Þ$##")&$&ÞÞÞÞ
(a)
(b)
(c)
(d) Values for x will vary. One possible choice is x! œ !Þ1.
(e) Values for x will vary.
14. (a) f(x) œ x$ 3x 1 Ê f w (x) œ 3x# 3 Ê xnb1 œ xn x$n 3xn 1
3xn# 3
Ê the two negative zeros are 1.53209
and 0.34730
(b) The estimated solutions of x$ 3x 1 œ 0 are
1.53209, 0.34730, 1.87939.
(c) The estimated x-values where
g(x) œ 0.25x% 1.5x# x 5 has horizontal tangents
are the roots of gw (x) œ x$ 3x 1, and these are
1.53209, 0.34730, 1.87939.
15. f(x) œ tan x 2x Ê f w (x) œ sec# x 2 Ê xnb1 œ xn tan axn b 2xn
sec# axn b
; x! œ 1 Ê x" œ 12920445
Ê x# œ 1.155327774 Ê x16 œ x17 œ 1.165561185
16. f(x) œ x% 2x$ x# 2x 2 Ê f w (x) œ 4x$ 6x# 2x 2 Ê xnb1 œ xn x%n 2xn$ xn# 2xn 2
4xn$ 6xn# 2xn 2
if x! œ 0.5, then x% œ 0.630115396; if x! œ 2.5, then x% œ 2.57327196
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;
265
266
Chapter 4 Applications of Derivatives
17. (a) The graph of f(x) œ sin 3x 0.99 x# in the window
2 Ÿ x Ÿ 2, 2 Ÿ y Ÿ 3 suggests three roots.
However, when you zoom in on the x-axis near x œ 1.2,
you can see that the graph lies above the axis there.
There are only two roots, one near x œ 1, the other
near x œ 0.4.
(b) f(x) œ sin 3x 0.99 x# Ê f w (x) œ 3 cos 3x 2x
Ê xnb1 œ xn sin (3xn ) 0.99xn#
3 cos (3xn ) 2xn
and the solutions
are approximately 0.35003501505249 and
1.0261731615301
18. (a) Yes, three times as indicted by the
graphs
(b) f(x) œ cos 3x x Ê f w (x)
œ 3 sin 3x 1 Ê xnb1
œ xn cos a3xn b xn
3 sin a3xn b 1
; at
approximately 0.979367,
0.887726, and 0.39004 we have
cos 3x œ x
19. f(x) œ 2x% 4x# 1 Ê f w (x) œ 8x$ 8x Ê xnb1 œ xn 2x%n 4xn# 1
8xn$ 8xn
; if x! œ 2, then x' œ 1.30656296; if
x! œ 0.5, then x$ œ 0.5411961; the roots are approximately „ 0.5411961 and „ 1.30656296 because f(x) is
an even function.
20. f(x) œ tan x Ê f w (x) œ sec# x Ê xnb1 œ xn tan axn b
sec# axn b
; x! œ 3 Ê x" œ 3.13971 Ê x# œ 3.14159 and we
approximate 1 to be 3.14159.
21. From the graph we let x! œ 0.5 and f(x) œ cos x 2x
Ê xnb1 œ xn cos axn b 2xn
sin axn b 2
Ê x" œ .45063
Ê x# œ .45018 Ê at x ¸ 0.45 we have cos x œ 2x.
22. From the graph we let x! œ 0.7 and f(x) œ cos x x
Ê xnb1 œ xn xn cos axn b
1 sin axn b
Ê x" œ .73944
Ê x# œ .73908 Ê at x ¸ 0.74 we have cos x œ x.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 4.7 Newton's Method
23. If f(x) œ x$ 2x 4, then f(1) œ 1 0 and f(2) œ 8 0 Ê by the Intermediate Value Theorem the equation
x$ 2x 4 œ 0 has a solution between 1 and 2. Consequently, f w (x) œ 3x# 2 and xnb1 œ xn x$n 2xn 4
3x#n 2
.
Then x! œ 1 Ê x" œ 1.2 Ê x# œ 1.17975 Ê x$ œ 1.179509 Ê x% œ 1.1795090 Ê the root is approximately
1.17951.
24. We wish to solve 8x% 14x$ 9x# 11x 1 œ 0. Let f(x) œ 8x% 14x$ 9x# 11x 1, then
f w (x) œ 32x$ 42x# 18x 11 Ê xnb1 œ xn x!
1.0
0.1
0.6
2.0
8x%n 14xn$ 9xn# 11xn 1
3#xn$ 42xn# 18xn 11
.
approximation of corresponding root
0.976823589
0.100363332
0.642746671
1.983713587
25. f(x) œ 4x% 4x# Ê f w (x) œ 16x$ 8x Ê xib1 œ xi faxi b
f w axi b
œ xi xi$ xi
.
%x#i #
Iterations are performed using the
procedure in problem 13 in this section.
(a) For x! œ # or x! œ !Þ), xi Ä " as i gets large.
(b) For x! œ !Þ& or x! œ !Þ#&, xi Ä ! as i gets large.
(c) For x! œ !Þ) or x! œ #, xi Ä " as i gets large.
(d) (If your calculator has a CAS, put it in exact mode, otherwise approximate the radicals with a decimal value.)
For x! œ x! œ
È
721
È21
7
or x! œ or x! œ
È
721
È21
7 ,
Newton's method does not converge. The values of xi alternate between
as i increases.
26. (a) The distance can be represented by
#
D(x) œ É(x 2)# ˆx# "# ‰ , where x
0. The
distance D(x) is minimized when
#
f(x) œ (x 2)# ˆx# "# ‰ is minimized. If
#
f(x) œ (x 2)# ˆx# "# ‰ , then
f w (x) œ 4 ax$ x 1b and f w w (x) œ 4 a3x# 1b 0.
Now f w (x) œ 0 Ê x$ x 1 œ 0 Ê x ax# 1b œ 1
Ê x œ x#"1 .
(b) Let g(x) œ
"
x # 1
Ê xnb1 œ xn x œ ax# 1b
"
Πx# 1 xn n
Î
2xn Ñ
#
Ï Šx#n 1‹ 1 Ò
"
#
x Ê gw (x) œ ax# 1b (2x) 1 œ
2x
ax # 1 b #
1
; x! œ 1 Ê x% œ 0.68233 to five decimal places.
27. f(x) œ (x 1)%! Ê f w (x) œ 40(x 1)$* Ê xnb1 œ xn axn 1b%!
40 axn 1b$*
œ
39xn "
40
. With x! œ 2, our computer
gave x)( œ x)) œ x)* œ â œ x#!! œ 1.11051, coming within 0.11051 of the root x œ 1.
28. f(x) œ 4x% 4x# Ê f w (x) œ 16x$ 8x œ 8x a2x# 1b Ê xnb1 œ xn xn ax#n "b
2 a2x#n 1b
; if x! œ .65, then
x"# ¸ .000004, if x! œ .7, then x"# œ 1.000004; if x! œ .8, then x' œ 1.000000. NOTE:
29. f(x) œ x$ 3.6x# 36.4 Ê f w (x) œ 3x# 7.2x Ê xnb1 œ xn x$n 3.6xn# 36.4
3xn# 7.2xn
È21
7
¸ .654654
; x! œ 2 Ê x" œ 2.5303
Ê x# œ 2.45418225 Ê x$ œ 2.45238021 Ê x% œ 2.45237921 which is 2.45 to two decimal places. Recall that
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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267
268
Chapter 4 Applications of Derivatives
x œ 10% cH$ O d Ê cH$ O d œ (x) a10% b œ (2.45) a10% b œ 0.000245
30. Newton's method yields the following:
the initial value
2
i
the approached value
1
5.55931i
È3 i
29.5815 17.0789i
4.8 ANTIDERIVATIVES
1. (a) x#
(b)
x$
3
(c)
x$
3
x# x
2. (a) 3x#
(b)
x)
8
(c)
x)
8
3x# 8x
3. (a) x$
(b) x3
4. (a) x#
(b) x4 $
$
(c) x3 x# 3x
#
x$
3
(c)
x#
#
x#
2
x
5. (a)
"
x
(b)
5
x
(c) 2x 6. (a)
"
x#
(b)
"
4x#
(c)
x%
4
(c)
2
3
Èx$ 2Èx
(c)
3
4
x%Î$ 3# x#Î$
5
x
"
#x #
7. (a) Èx$
(b) Èx
8. (a) x%Î$
(b)
9. (a) x#Î$
(b) x"Î$
(c) x"Î$
10. (a) x"Î#
(b) x"Î#
(c) x$Î#
11. (a) cos (1x)
(b) 3 cos x
(c)
12. (a) sin (1x)
(b) sin ˆ 1#x ‰
(c) ˆ 12 ‰ sin ˆ 1#x ‰ 1 sin x
13. (a) tan x
(b) 2 tan ˆ x3 ‰
‰
(c) 23 tan ˆ 3x
#
14. (a) cot x
‰
(b) cot ˆ 3x
#
(c) x 4 cot (2x)
15. (a) csc x
(b)
"
5
csc (5x)
(c) 2 csc ˆ 1#x ‰
16. (a) sec x
(b)
4
3
sec (3x)
(c)
17.
' (x 1) dx œ
19.
'
x#
#
xC
ˆ3t# #t ‰ dt œ t$ t#
4
C
"
#
x#Î$
18.
' (5 6x) dx œ 5x 3x# C
20.
'
#
Š t# 4t$ ‹ dt œ
t$
6
cos (1x)
1
2
1
t% C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
cos (3x)
sec ˆ 1#x ‰
Section 4.8 Antiderivatives
21.
'
a2x$ 5x 7b dx œ
23.
'
ˆ x"# x# 3" ‰ dx œ ' ˆx# x# 3" ‰ dx œ
24.
'
ˆ "5 25.
' x"Î$ dx œ
27.
'
ˆÈx $Èx‰ dx œ ' ˆx"Î# x"Î$ ‰ dx œ
28.
'
Š
29.
'
Š8y 30.
'
Èx
#
2
x$
Š "7 "
#
x% 5# x# 7x C
x"
1
2x‰ dx œ ' ˆ 5" 2x$ 2x‰ dx œ
x#Î$
2
3
2
Èx ‹
2
‹
y"Î%
1
‹
y&Î%
Cœ
3
#
'
22.
x$
3
3" x C œ x" #
"
5
x Š 2x# ‹ x#Î$ C
dx œ ' ˆ "# x"Î# 2x"Î# ‰ dx œ
dy œ ' ˆ8y 2y"Î% ‰ dy œ
dy œ ' ˆ "7 y&Î% ‰ dy œ
"
7
3
#
8y#
#
"
#
x%Î$
Cœ
4
3
2x#
#
Cœ
' x&Î% dx œ
26.
x$Î#
a1 x# 3x& b dx œ x "3 x$ "# x' C
2
3
5
x
x$
3
x"Î%
"4
"
x#
x
3
C
x# C
Cœ
4
% x
È
C
x$Î# 34 x%Î$ C
$Î#
"Î#
#
#
Šx3 ‹ 2 Šx" ‹ C œ
"
3
x$Î# 4x"Î# C
$Î%
2 Š y 3 ‹ C œ 4y# 83 y$Î% C
4
"Î%
y Š y 1 ‹ C œ
4
y
7
C
4
y"Î%
31.
'
2x a1 x$ b dx œ ' a2x 2x# b dx œ
32.
'
x$ (x 1) dx œ ' ax# x$ b dx œ
33.
'
tÈtÈt
t#
34.
'
4 È t
t$
35.
' 2 cos t dt œ 2 sin t C
36.
' 5 sin t dt œ 5 cos t C
37.
' 7 sin 3) d) œ 21 cos 3) C
38.
' 3 cos 5) d) œ 35 sin 5) C
39.
' 3 csc# x dx œ 3 cot x C
40.
' sec3# x dx œ tan3 x C
41.
'
42.
'
43.
' a4 sec x tan x 2 sec# xb dx œ 4 sec x 2 tan x C
44.
'
45.
' asin 2x csc# xb dx œ "# cos 2x cot x C
46.
' (2 cos 2x 3 sin 3x) dx œ sin 2x cos 3x C
47.
'
1 cos 4t
#
dt œ ' ˆ "# "
#
cos 4t‰ dt œ
"
#
t "# ˆ sin4 4t ‰ C œ
t
2
sin 4t
8
C
48.
'
1 cos 6t
#
dt œ ' ˆ "# "
#
cos 6t‰ dt œ
"
#
t "# ˆ sin6 6t ‰ C œ
t
2
sin 6t
12
C
$Î#
dt œ ' Š t4$ csc ) cot )
#
"
2
dt œ ' Š t t# t"Î#
t# ‹
t"Î#
t$ ‹
2x#
#
x"
1
"
2 Š x1 ‹ C œ x# #
Š x# ‹ C œ "x dt œ ' ˆt"Î# t$Î# ‰ dt œ
t"Î#
"
#
2
x
C
"
#x #
C
"Î#
Š t " ‹ C œ 2Èt #
t
dt œ ' ˆ4t$ t&Î# ‰ dt œ 4 Š #
‹ Š t 3 ‹ C œ t2# #
$Î#
#
d) œ "# csc ) C
acsc# x csc x cot xb dx œ #" cot x "
#
2
5
sec ) tan ) d) œ
2
5
2
Èt
C
2
3t$Î#
C
sec ) C
csc x C
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269
270
Chapter 4 Applications of Derivatives
49.
' a1 tan# )b d) œ ' sec# ) d) œ tan ) C
50.
' a2 tan# )b d) œ ' a1 1 tan# )b d) œ ' a1 sec# )b d) œ ) tan ) C
51.
' cot# x dx œ ' acsc# x 1b dx œ cot x x C
52.
' a1 cot# xb dx œ ' a1 acsc# x 1bb dx œ ' a2 csc# xb dx œ 2x cot x C
53.
' cos ) (tan ) sec )) d) œ ' (sin ) 1) d) œ cos ) ) C
54.
'
csc )
csc ) sin )
55.
d
dx
2)
C‹ œ
Š (7x28
56.
d
dx
Š (3 x3 5)
57.
d
dx
ˆ "5 tan (5x 1) C‰ œ
58.
d
dx
ˆ3 cot ˆ x 3 " ‰ C‰ œ 3 ˆcsc# ˆ x 3 " ‰‰ ˆ "3 ‰ œ csc# ˆ x 3 " ‰
59.
d
dx
#
ˆ x"
‰
œ
1 C œ (1)(1)(x 1)
)
)‰
'
‰ ˆ sin
d) œ ' ˆ csc csc
) sin )
sin ) d) œ
%
"
4(7x 2)$ (7)
28
d) œ '
# (3)
asec# (5x 1)b (5) œ sec# (5x 1)
#
"
(x 1)#
x#
#
Š x# sin x C‹ œ
(b) Wrong:
d
dx
(x cos x C) œ cos x x sin x Á x sin x
62. (a) Wrong:
(b) Right:
(c) Right:
2x
#
d
dx
ˆ xx 1 C‰ œ
cos x œ x sin x x#
#
$
Š sec3 ) C‹ œ
3 sec# )
3
ˆ "# tan# ) C‰ œ "# (2 tan )) sec# ) œ tan ) sec# )
ˆ "# sec# ) C‰ œ "# (2 sec )) sec ) tan ) œ tan ) sec# )
$
3(2x 1)# (2)
3
Š (2x 3 1) C‹ œ
(b) Wrong:
d
dx
a(2x 1)$ Cb œ 3(2x 1)# (2) œ 6(2x 1)# Á 3(2x 1)#
d
dx
ax# x Cb
(b) Wrong:
d
dx
Šax# xb
d
dx
œ 2(2x 1)# Á (2x 1)#
a(2x 1)$ Cb œ 6(2x 1)#
64. (a) Wrong:
(c) Right:
"
(x 1)#
(sec ) tan )) œ sec$ ) tan ) Á tan ) sec# )
d
dx
d
dx
œ
cos x Á x sin x
63. (a) Wrong:
(c) Right:
(x 1)(") x(1)
(x 1)#
(x cos x sin x C) œ cos x x sin x cos x œ x sin x
d
d)
d
d)
d
d)
sin x 60.
d
dx
(c) Right:
d) œ ' sec# ) d) œ tan ) C
‹ œ (3x 5)#
61. (a) Wrong:
d
dx
"
cos# )
œ (7x 2)$
C‹ œ Š (3x 35)
"
5
"
1sin# )
"Î#
"Î#
œ
"
#
ax# x Cb
C‹ œ
$
"
#
ax# xb
"
Œ 3 ŠÈ2x 1‹ C œ
d
dx
"Î#
"Î#
(2x 1) œ
(2x 1) œ
2x 1
2 È x# x C
2x 1
2 È x# x
ˆ 3" (2x 1)$Î# C‰ œ
3
6
Á È2x 1
Á È2x 1
(2x 1)"Î# (2) œ È2x 1
65. Graph (b), because
dy
dx
œ 2B Ê y œ x# C. Then y(1) œ 4 Ê C œ 3.
66. Graph (b), because
dy
dx
œ B Ê y œ "# x# C. Then y(1) œ 1 Ê C œ
3
#
.
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Section 4.8 Antiderivatives
271
67.
dy
dx
œ 2x 7 Ê y œ x# 7x C; at x œ 2 and y œ 0 we have 0 œ 2# 7(2) C Ê C œ 10 Ê y œ x# 7x 10
68.
dy
dx
œ 10 x Ê y œ 10x Ê y œ 10x 69.
dy
dx
œ
"
x#
x#
#
C; at x œ 0 and y œ 1 we have 1 œ 10(0) 0#
#
C Ê C œ 1
1
x œ x# x Ê y œ x" Ê y œ x" 70.
x#
#
#
x
#
"
#
or y œ x" #
x
#
x#
#
C; at x œ 2 and y œ 1 we have 1 œ 2" "
#
2#
#
C Ê C œ "#
dy
dx
œ 9x# 4x 5 Ê y œ 3x$ 2x# 5x C; at x œ 1 and y œ 0 we have 0 œ 3(1)$ 2(1)# 5(1) C
dy
dx
œ $x#Î$ Ê y œ
Ê C œ 10 Ê y œ 3x$ 2x# 5x 10
71.
$x"Î$
"
$
C œ *; at x œ 9x"Î$ C; at x œ " and y œ & we have & œ *(")"Î$ C Ê C œ %
Ê y œ 9x"Î$ %
"
#È x
œ
"
#
x"Î# Ê y œ x"Î# C; at x œ 4 and y œ 0 we have 0 œ 4"Î# C Ê C œ 2 Ê y œ x"Î# 2
72.
dy
dx
œ
73.
ds
dt
œ 1 cos t Ê s œ t sin t C; at t œ 0 and s œ 4 we have 4 œ 0 sin 0 C Ê C œ 4 Ê s œ t sin t 4
74.
ds
dt
œ cos t sin t Ê s œ sin t cos t C; at t œ 1 and s œ 1 we have 1 œ sin 1 cos 1 C Ê C œ 0
Ê s œ sin t cos t
75.
dr
d)
œ 1 sin 1) Ê r œ cos (1)) C; at r œ 0 and ) œ 0 we have 0 œ cos (10) C Ê C œ " Ê r œ cos (1)) 1
76.
dr
d)
œ cos 1) Ê r œ
77.
dv
dt
œ
78.
dv
dt
œ 8t csc# t Ê v œ 4t# cot t C; at v œ 7 and t œ
"
#
"
1
sin(1)) C; at r œ 1 and ) œ 0 we have 1 œ
sec t tan t Ê v œ
"
#
sec t C; at v œ 1 and t œ 0 we have 1 œ
1
#
Ê v œ 4t# cot t 7 1#
79.
d# y
dx#
Ê
œ 2 6x Ê
dy
dx
œ 2x 3x# C" ; at
dy
dx
#
"
1
dy
dx
sin (10) C Ê C œ " Ê r œ
"
#
"
#
sec (0) C Ê C œ
"
1
Ê vœ
sin (1)) 1
"
#
sec t "
#
#
we have 7 œ 4 ˆ 1# ‰ cot ˆ 1# ‰ C Ê C œ 7 1#
œ 4 and x œ 0 we have 4 œ 2(0) 3(0)# C" Ê C" œ 4
œ 2x 3x 4 Ê y œ x# x$ 4x C# ; at y œ 1 and x œ 0 we have 1 œ 0# 0$ 4(0) C# Ê C# œ 1
Ê y œ x# x$ 4x 1
80.
d# y
dx#
œ0 Ê
dy
dx
œ C" ; at
dy
dx
œ 2 and x œ 0 we have C" œ 2 Ê
dy
dx
œ 2 Ê y œ 2x C# ; at y œ 0 and x œ 0 we
have 0 œ 2(0) C# Ê C# œ 0 Ê y œ 2x
81.
d# r
dt#
œ
d# s
dt#
œ
2
t$
œ 2t$ Ê
dr
dt
œ t# C" ; at
dr
dt
œ 1 and t œ 1 we have 1 œ (1)# C" Ê C" œ 2 Ê
dr
dt
œ t# 2
Ê r œ t" 2t C# ; at r œ 1 and t œ 1 we have 1 œ 1" 2(1) C# Ê C# œ 2 Ê r œ t" 2t 2 or
r œ "t 2t 2
82.
3t
8
Ê
ds
dt
œ
3t#
16
C" ; at
s œ 4 and t œ 4 we have 4 œ
ds
dt
$
4
16
œ 3 and t œ 4 we have 3 œ
C# Ê C# œ 0 Ê s œ
3(4)#
16
C" Ê C" œ 0 Ê
ds
dt
œ
3t#
16
$
t
16
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Ê sœ
t$
16
C# ; at
272
83.
Chapter 4 Applications of Derivatives
d$ y
dx$
œ6 Ê
Ê
dy
dx
d# y
dx#
#
œ 6x C" ; at
œ 3x 8x C# ; at
d$ )
dt$
œ0 Ê
#
dy
dx
œ 8 and x œ 0 we have 8 œ 6(0) C" Ê C" œ 8 Ê
œ 0 and x œ 0 we have 0 œ 3(0)# 8(0) C# Ê C# œ 0 Ê
d# y
dx# œ 6x 8
dy
#
dx œ 3x 8x
$
#
Ê y œ x 4x C$ ; at y œ 5 and x œ 0 we have 5 œ 0$ 4(0)# C$ Ê C$ œ 5 Ê y œ x 4x 5
84.
$
d# y
dx#
d# )
dt#
d# )
dt#
œ C" ; at
d# )
d)
dt# œ 2 Ê dt
"
d)
#
dt œ 2t # Ê ) œ t
) œ t# "# t È2
œ 2 and t œ 0 we have
have "# œ 2(0) C# Ê C# œ "# Ê
È2 œ 0# " (0) C$ Ê C$ œ È2 Ê
#
œ "# and t œ 0 we
"# t C$ ; at ) œ È2 and t œ 0 we have
œ 2t C# ; at
d)
dt
85. yÐ%Ñ œ sin t cos t Ê ywww œ cos t sin t C" ; at ywww œ 7 and t œ 0 we have 7 œ cos (0) sin (0) C"
Ê C" œ 6 Ê ywww œ cos t sin t 6 Ê yww œ sin t cos t 6t C# ; at yww œ 1 and t œ 0 we have
1 œ sin (0) cos (0) 6(0) C# Ê C# œ 0 Ê yww œ sin t cos t 6t Ê yw œ cos t sin t 3t# C$ ;
at yw œ 1 and t œ 0 we have 1 œ cos (0) sin (0) 3(0)# C$ Ê C$ œ 0 Ê yw œ cos t sin t 3t#
Ê y œ sin t cos t t$ C% ; at y œ 0 and t œ 0 we have 0 œ sin (0) cos (0) 0$ C% Ê C% œ 1
Ê y œ sin t cos t t$ 1
86. yÐ%Ñ œ cos x 8 sin (2x) Ê ywww œ sin x 4 cos (2x) C" ; at ywww œ 0 and x œ 0 we have
0 œ sin (0) % cos (2(0)) C" Ê C" œ 4 Ê ywww œ sin x 4 cos (2x) 4 Ê yww œ cos x 2 sin (2x) 4x C# ;
at yww œ 1 and x œ 0 we have 1 œ cos (0) 2 sin (2(0)) 4(0) C# Ê C# œ 0 Ê yww œ cos x 2 sin (2x) 4x
Ê yw œ sin x cos (2x) 2x# C$ ; at yw œ 1 and x œ 0 we have 1 œ sin (0) cos (2(0)) 2(0)# C$ Ê C$ œ 0
Ê yw œ sin x cos (2x) 2x# Ê y œ cos x "# sin (2x) 23 x$ C% ; at y œ 3 and x œ 0 we have
3 œ cos (0) "
#
sin (2(0)) 23 (0)$ C% Ê C% œ 4 Ê y œ cos x "
#
sin (2x) 23 x$ 4
87. m œ yw œ 3Èx œ 3x"Î# Ê y œ 2x$Î# C; at (*ß 4) we have 4 œ 2(9)$Î# C Ê C œ 50 Ê y œ 2x$Î# 50
88. (a)
d# y
dx#
œ 6x Ê
dy
dx
œ 3x# C" ; at yw œ 0 and x œ 0 we have 0 œ 3(0)# C" Ê C" œ 0 Ê
dy
dx
œ 3x#
Ê y œ x$ C# ; at y œ 1 and x œ 0 we have C# œ 1 Ê y œ x$ 1
(b) One, because any other possible function would differ from x$ 1 by a constant that must be zero because
of the initial conditions
89.
dy
dx
œ 1 34 x"Î$ Ê y œ ' ˆ1 34 x"Î$ ‰ dx œ x x%Î$ C; at (1ß 0.5) on the curve we have 0.5 œ 1 1%Î$ C
Ê C œ 0.5 Ê y œ x x%Î$ 90.
dy
dx
œ x 1 Ê y œ ' (x 1) dx œ
Ê C œ "# Ê y œ
91.
dy
dx
"
#
#
x
#
x
x#
#
x C; at (1ß 1) on the curve we have 1 œ
(")#
#
(1) C
"
#
œ sin x cos x Ê y œ ' (sin x cos x) dx œ cos x sin x C; at (1ß 1) on the curve we have
" œ cos (1) sin (1) C Ê C œ 2 Ê y œ cos x sin x 2
92.
dy
dx
œ
"
#Èx
1 sin 1x œ
"
#
x"Î# 1 sin 1x Ê y œ ' ˆ #" x"Î# sin 1x‰ dx œ x"Î# cos 1x C; at (1ß #) on the
curve we have 2 œ 1"Î# cos 1(1) C Ê C œ 0 Ê y œ Èx cos 1x
93. (a)
ds
dt
œ 9.8t 3 Ê s œ 4.9t# 3t C; (i) at s œ 5 and t œ 0 we have C œ 5 Ê s œ 4.9t# 3t 5;
displacement œ s(3) s(1) œ ((4.9)(9) 9 5) (4.9 3 5) œ 33.2 units; (ii) at s œ 2 and t œ 0 we have
C œ 2 Ê s œ 4.9t# 3t 2; displacement œ s(3) s(1) œ ((4.9)(9) 9 2) (4.9 3 2) œ 33.2 units;
(iii) at s œ s! and t œ 0 we have C œ s! Ê s œ 4.9t# 3t s! ; displacement œ s(3) s(1)
œ ((4.9)(9) 9 s! ) (4.9 3 s! ) œ 33.2 units
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Section 4.8 Antiderivatives
273
(b) True. Given an antiderivative f(t) of the velocity function, we know that the body's position function is
s œ f(t) C for some constant C. Therefore, the displacement from t œ a to t œ b is (f(b) C) (f(a) C)
œ f(b) f(a). Thus we can find the displacement from any antiderivative f as the numerical difference
f(b) f(a) without knowing the exact values of C and s.
94. a(t) œ vw (t) œ 20 Ê v(t) œ 20t C; at (0ß !) we have C œ 0 Ê v(t) œ 20t. When t œ 60, then v(60) œ 20(60)
œ 1200 m/sec.
95. Step 1:
d# s
dt#
œ k Ê
ds
dt
œ kt C" ; at
ds
dt
œ 88 and t œ 0 we have C" œ 88 Ê
#
ds
dt
œ kt 88 Ê
#
s œ k Š t# ‹ 88t C# ; at s œ 0 and t œ 0 we have C# œ 0 Ê s œ kt# 88t
Step 2:
ds
dt
œ 0 Ê 0 œ kt 88 Ê t œ
Step 3: 242 œ
96.
d# s
dt#
œ k Ê
Ê
ds
dt
‰
k ˆ 88
k
#
#
#
‰ Ê 242 œ (88)
88 ˆ 88
k
2k œ ' k dt œ kt C; at
ds
dt
œ kt 44 Ê s œ #
Ê s œ kt# 44t. Then
Ê 968
k 1936
k
88
k
œ 45 Ê
kt#
#
ds
dt
(88)#
k
Ê 242 œ
(88)#
2k
Ê k œ 16
œ 44 when t œ 0 we have 44 œ k(0) C Ê C œ 44
#
44t C" ; at s œ 0 when t œ 0 we have 0 œ k(0)
# 44(0) C" Ê C" œ 0
ds
44
dt œ 0 Ê kt 44 œ 0 Ê t œ k
968
968
ft
k œ 45 Ê k œ 45 ¸ 21.5 sec2 .
97. (a) v œ ' a dt œ ' ˆ15t"Î# 3t"Î# ‰ dt œ 10t$Î# 6t"Î# C;
‰
and s ˆ 44
k œ
ds
dt
‰
k ˆ 44
k
#
#
‰
44 ˆ 44
k œ 45
(1) œ 4 Ê 4 œ 10(1)$Î# 6(1)"Î# C Ê C œ 0
Ê v œ 10t$Î# 6t"Î#
(b) s œ ' v dt œ ' ˆ10t$Î# 6t"Î# ‰ dt œ 4t&Î# 4t$Î# C; s(1) œ 0 Ê 0 œ 4(1)&Î# 4(1)$Î# C Ê C œ 0
Ê s œ 4t&Î# 4t$Î#
98.
d# s
dt#
œ 5.2 Ê
ds
dt
œ 5.2t C" ; at
ds
dt
œ 0 and t œ 0 we have C" œ 0 Ê
ds
dt
œ 5.2t Ê s œ 2.6t# C# ; at s œ 4
4
and t œ 0 we have C# œ 4 Ê s œ 2.6t# 4. Then s œ 0 Ê 0 œ 2.6t# 4 Ê t œ É 2.6
¸ 1.24 sec, since t 0
99.
d# s
dt#
œa Ê
ds
dt
œ ' a dt œ at C;
when t œ 0 Ê s! œ
a(0)#
#
ds
dt
œ v! when t œ 0 Ê C œ v! Ê
v! (0) C" Ê C" œ s! Ê s œ
#
at
#
100. The appropriate initial value problem is: Differential Equation:
s œ s! when t œ 0. Thus,
Ê
ds
dt
ds
dt
œ ' g dt œ gt C" ;
œ gt v! . Thus s œ ' agt v! b dt œ Thus s œ "# gt# v! t s!.
ds
dt
œ at v! Ê s œ
at#
#
v! t C" ; s œ s!
v! t s!
d# s
dt#
œ g with Initial Conditions:
ds
dt (0) œ v! Ê v! œ (g)(0) "
"
#
# gt v! t C# ; s(0) œ s! œ #
ds
dt
œ v! and
C" Ê C" œ v!
(g)(0)# v! (0) C# Ê C# œ s!
' f(x) dx œ 1 Èx C" œ Èx C
(b) ' g(x) dx œ x 2 C" œ x C
(c) ' f(x) dx œ ˆ1 Èx‰ C" œ Èx C
(d) ' g(x) dx œ (x 2) C" œ x C
(e) ' [f(x) g(x)] dx œ ˆ1 Èx‰ (x 2) C" œ x Èx C
(f) ' [f(x) g(x)] dx œ ˆ1 Èx‰ (x 2) C" œ x Èx C
101. (a)
102. Yes. If F(x) and G(x) both solve the initial value problem on an interval I then they both have the same first
derivative. Therefore, by Corollary 2 of the Mean Value Theorem there is a constant C such that
F(x) œ G(x) C for all x. In particular, F(x! ) œ G(x! ) C, so C œ F(x! ) G(x! ) œ 0. Hence F(x) œ G(x)
for all x.
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274
Chapter 4 Applications of Derivatives
103 106 Example CAS commands:
Maple:
with(student):
f := x -> cos(x)^2 + sin(x);
ic := [x=Pi,y=1];
F := unapply( int( f(x), x ) + C, x );
eq := eval( y=F(x), ic );
solnC := solve( eq, {C} );
Y := unapply( eval( F(x), solnC ), x );
DEplot( diff(y(x),x) = f(x), y(x), x=0..2*Pi, [[y(Pi)=1]],
color=black, linecolor=black, stepsize=0.05, title="Section 4.8 #103" );
Mathematica: (functions and values may vary)
The following commands use the definite integral and the Fundamental Theorem of calculus to construct the solution
of the initial value problems for exercises 103 - 105.
Clear[x, y, yprime]
yprime[x_] = Cos[x]2 Sin[x];
initxvalue = 1; inityvalue = 1;
y[x_] = Integrate[yprime[t], {t, initxvalue, x}] inityvalue
If the solution satisfies the differential equation and initial condition, the following yield True
yprime[x]==D[y[x], x] //Simplify
y[initxvalue]==inityvalue
Since exercise 106 is a second order differential equation, two integrations will be required.
Clear[x, y, yprime]
y2prime[x_] = 3 Exp[x/2] 1;
initxval = 0; inityval = 4; inityprimeval = 1;
yprime[x_] = Integrate[y2prime[t],{t, initxval, x}] inityprimeval
y[x_] = Integrate[yprime[t], {t, initxval, x}] inityval
Verify that y[x] solves the differential equation and initial condition and plot the solution (red) and its derivative (blue).
y2prime[x]==D[y[x], {x, 2}]//Simplify
y[initxval]==inityval
yprime[initxval]==inityprimeval
Plot[{y[x], yprime[x]}, {x, initxval 3, initxval 3}, PlotStyle Ä {RGBColor[1,0,0], RGBColor[0,0,1]}]
CHAPTER 4 PRACTICE EXERCISES
1. No, since f(x) œ x$ 2x tan x Ê f w (x) œ 3x# 2 sec# x 0 Ê f(x) is always increasing on its domain
cos x
2. No, since g(x) œ csc x 2 cot x Ê gw (x) œ csc x cot x 2 csc# x œ sin
#x 2
sin# x
œ sin"# x (cos x 2) 0
Ê g(x) is always decreasing on its domain
3. No absolute minimum because x lim
(7 x)(11 3x)"Î$ œ _. Next f w (x) œ
Ä_
(11 3x)"Î$ (7 x)(11 3x)#Î$ œ
w
(11 3x) (7 x)
(11 3x)#Î$
œ
4(1 x)
(11 3x)#Î$
Ê x œ 1 and x œ
11
3
are critical points.
w
Since f 0 if x 1 and f 0 if x 1, f(1) œ 16 is the absolute maximum.
4. f(x) œ
ax b
x# 1
Ê f w (x) œ
We require also that f(3)
w
#a$x "bax $b
ax # 1 b #
#
a ax# 1b 2x(ax b)
ab
œ aaxax#2bx
1 b#
ax # 1 b#
œ 1. Thus " œ 3a8b Ê 3a b œ
w
"
; f w (3) œ 0 Ê '%
(*a 'b a) œ ! Ê &a $b œ !.
). Solving both equations yields a œ 6 and b œ 10. Now,
so that f œ ± ± ± ± . Thus f w changes sign at x œ $ from
1
1
3
1/3
positive to negative so there is a local maximum at x œ $ which has a value f(3) œ 1.
f (x) œ
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Chapter 4 Practice Exercises
275
5. Yes, because at each point of [!ß "Ñ except x œ 0, the function's value is a local minimum value as well as a
local maximum value. At x œ 0 the function's value, 0, is not a local minimum value because each open
interval around x œ 0 on the x-axis contains points to the left of 0 where f equals 1.
6. (a) The first derivative of the function f(x) œ x$ is zero at x œ 0 even though f has no local extreme value at
x œ 0.
(b) Theorem 2 says only that if f is differentiable and f has a local extreme at x œ c then f w (c) œ 0. It does not
assert the (false) reverse implication f w (c) œ 0 Ê f has a local extreme at x œ c.
7. No, because the interval 0 x 1 fails to be closed. The Extreme Value Theorem says that if the function is
continuous throughout a finite closed interval a Ÿ x Ÿ b then the existence of absolute extrema is guaranteed on
that interval.
8. The absolute maximum is k1k œ 1 and the absolute minimum is k0k œ 0. This is not inconsistent with the Extreme Value
Theorem for continuous functions, which says a continuous function on a closed interval attains its extreme values on that
interval. The theorem says nothing about the behavior of a continuous function on an interval which is half open and half
closed, such as Ò"ß "Ñ, so there is nothing to contradict.
9. (a) There appear to be local minima at x œ 1.75
and 1.8. Points of inflection are indicated at
approximately x œ 0 and x œ „ 1.
(b) f w (x) œ x( 3x& 5x% 15x# œ x# ax# 3b ax$ 5b. The pattern yw œ ± ± ± ± $È
!
È$
È $
&
$È
indicates a local maximum at x œ
5 and local minima at x œ „ È3 .
(c)
10. (a) The graph does not indicate any local
extremum. Points of inflection are indicated at
approximately x œ $% and x œ ".
(b) f w (x) œ x( 2x% 5 10
x$
œ x$ ax$ 2b ax( 5b . The pattern f w œ )( ± ± indicates
(È
$È
!
&
#
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276
Chapter 4 Applications of Derivatives
a local maximum at x œ (È5 and a local minimum at x œ $È2 .
(c)
11. (a) g(t) œ sin# t 3t Ê gw (t) œ 2 sin t cos t 3 œ sin (2t) 3 Ê gw 0 Ê g(t) is always falling and hence must
decrease on every interval in its domain.
(b) One, since sin# t 3t 5 œ 0 and sin# t 3t œ 5 have the same solutions: f(t) œ sin# t 3t 5 has the same
derivative as g(t) in part (a) and is always decreasing with f(3) 0 and f(0) 0. The Intermediate Value
Theorem guarantees the continuous function f has a root in [$ß 0].
12. (a) y œ tan ) Ê
dy
d)
œ sec# ) 0 Ê y œ tan ) is always rising on its domain Ê y œ tan ) increases on every
interval in its domain
(b) The interval 14 ß 1‘ is not in the tangent's domain because tan ) is undefined at ) œ
1
#
. Thus the tangent
need not increase on this interval.
13. (a) f(x) œ x% 2x# 2 Ê f w (x) œ 4x$ 4x. Since f(0) œ 2 0, f(1) œ 1 0 and f w (x) 0 for 0 Ÿ x Ÿ 1, we
may conclude from the Intermediate Value Theorem that f(x) has exactly one solution when 0 Ÿ x Ÿ 1.
È
(b) x# œ 2 „ 4 8 0 Ê x# œ È3 1 and x 0 Ê x ¸ È.7320508076 ¸ .8555996772
#
14. (a) y œ
x
x1
Ê yw œ
"
(x 1)#
0, for all x in the domain of
x
x1
Ê yœ
x
x1
is increasing in every interval in
its domain
(b) y œ x$ 2x Ê yw œ 3x# 2 0 for all x Ê the graph of y œ x$ 2x is always increasing and can never
have a local maximum or minimum
15. Let V(t) represent the volume of the water in the reservoir at time t, in minutes, let V(0) œ a! be the initial
amount and V(1440) œ a! (1400)(43,560)(7.48) gallons be the amount of water contained in the reservoir
after the rain, where 24 hr œ 1440 min. Assume that V(t) is continuous on [!ß 1440] and differentiable on
(!ß 1440). The Mean Value Theorem says that for some t! in (!ß 1440) we have Vw (t! ) œ
œ
a! (1400)(43,560)(7.48) a!
1440
œ
456,160,320 gal
1440 min
V(1440) V(0)
1440 0
œ 316,778 gal/min. Therefore at t! the reservoir's volume
was increasing at a rate in excess of 225,000 gal/min.
16. Yes, all differentiable functions g(x) having 3 as a derivative differ by only a constant. Consequently, the
d
difference 3x g(x) is a constant K because gw (x) œ 3 œ dx
(3x). Thus g(x) œ 3x K, the same form as F(x).
x
1
x
1
x 1 œ 1 x 1 Ê x 1 differs from x 1
(x 1) x(1)
d ˆ x ‰
d ˆ " ‰
œ (x " 1)# œ dx
dx x 1 œ
(x 1)#
x1 .
17. No,
18. f w (x) œ gw (x) œ
2x
ax # 1 b #
by the constant 1. Both functions have the same derivative
Ê f(x) g(x) œ C for some constant C Ê the graphs differ by a vertical shift.
19. The global minimum value of
"
#
occurs at x œ #.
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Chapter 4 Practice Exercises
277
20. (a) The function is increasing on the intervals Ò$ß #Ó and Ò"ß #Ó.
(b) The function is decreasing on the intervals Ò#ß !Ñ and Ð!ß "Ó.
(c) The local maximum values occur only at x œ #, and at x œ #; local minimum values occur at x œ $ and at x œ "
provided f is continuous at x œ !.
21. (a) t œ 0, 6, 12
(b) t œ 3, 9
(c) 6 t 12
(d) 0 t 6, 12 t 14
22. (a) t œ 4
(b) at no time
(c) 0 t 4
(d) 4 t 8
23.
24.
25.
26.
27.
28.
29.
30.
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278
Chapter 4 Applications of Derivatives
31.
32.
33. (a) yw œ 16 x# Ê yw œ ± ± Ê the curve is rising on (%ß %), falling on (_ß 4) and (%ß _)
%
%
Ê a local maximum at x œ 4 and a local minimum at x œ 4; yww œ 2x Ê yww œ ± Ê the curve
!
is concave up on (_ß !), concave down on (!ß _) Ê a point of inflection at x œ 0
(b)
34. (a) yw œ x# x 6 œ (x $)(x 2) Ê yw œ ± ± Ê the curve is rising on (_ß 2) and ($ß _),
#
$
falling on (#ß $) Ê local maximum at x œ 2 and a local minimum at x œ 3; yww œ 2x 1
Ê yww œ ± Ê concave up on ˆ "# ß _‰ , concave down on ˆ_ß "# ‰ Ê a point of inflection at x œ "#
"Î#
(b)
35. (a) yw œ 6x(x 1)(x 2) œ 6x$ 6x# 12x Ê yw œ ± ± ± Ê the graph is rising on ("ß !)
"
!
#
and (#ß _), falling on (_ß 1) and (!ß #) Ê a local maximum at x œ 0, local minima at x œ 1 and
x œ 2; yww œ 18x# 12x 12 œ 6 a3x# 2x 2b œ 6 Šx yww œ ±
on
±
"È(
$
È
È
Š 1 3 7 ß 1 3 7 ‹
"È(
$
1 È7
3 ‹ Šx
Ê the curve is concave up on Š_ß
Ê points of inflection at x œ
1 È7
3 ‹
1 È7
3 ‹
Ê
È7
and Š 1 3
1 „ È7
3
(b)
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ß _‹ , concave down
Chapter 4 Practice Exercises
279
36. (a) yw œ x# (6 4x) œ 6x# 4x$ Ê yw œ ± ± Ê the curve is rising on ˆ_ß #3 ‰, falling on ˆ #3 ß _‰
!
$Î#
3
ww
Ê a local maximum at x œ # ; y œ 12x 12x# œ 12x(" x) Ê yww œ ± ± Ê concave up on
!
"
(!ß "), concave down on (_ß !) and ("ß _) Ê points of inflection at x œ 0 and x œ 1
(b)
37. (a) yw œ x% 2x# œ x# ax# 2b Ê yw œ ± ± ± Ê the curve is rising on Š_ß È2‹ and
!
È#
È #
ŠÈ2ß _‹ , falling on ŠÈ2ß È2‹ Ê a local maximum at x œ È2 and a local minimum at x œ È2 ;
yww œ 4x$ 4x œ 4x(x 1)(x 1) Ê yww œ ± ± ± Ê concave up on ("ß 0) and ("ß _),
"
!
"
concave down on (_ß 1) and (0ß 1) Ê points of inflection at x œ 0 and x œ „ 1
(b)
38. (a) yw œ 4x# x% œ x# a4 x# b Ê yw œ ± ± ± Ê the curve is rising on (2ß 0) and (0ß 2),
#
!
#
falling on (_ß 2) and (#ß _) Ê a local maximum at x œ 2, a local minimum at x œ 2; yww œ 8x 4x$
œ 4x a2 x# b Ê yww œ ± ± ± Ê concave up on Š_ß È2‹ and Š0ß È2‹ , concave
!
È#
È #
down on ŠÈ2ß 0‹ and ŠÈ2ß _‹ Ê points of inflection at x œ 0 and x œ „ È2
(b)
39. The values of the first derivative indicate that the curve is rising on (!ß _) and falling on (_ß 0). The slope
of the curve approaches _ as x Ä ! , and approaches _ as x Ä 0 and x Ä 1. The curve should therefore
have a cusp and local minimum at x œ 0, and a vertical tangent at x œ 1.
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280
Chapter 4 Applications of Derivatives
40. The values of the first derivative indicate that the curve is rising on ˆ!ß "# ‰ and ("ß _), and falling on (_ß !)
and ˆ "# ß "‰ . The derivative changes from positive to negative at x œ "# , indicating a local maximum there. The
slope of the curve approaches _ as x Ä 0 and x Ä 1 , and approaches _ as x Ä 0 and as x Ä 1 ,
indicating cusps and local minima at both x œ 0 and x œ 1.
41. The values of the first derivative indicate that the curve is always rising. The slope of the curve approaches _
as x Ä 0 and as x Ä 1, indicating vertical tangents at both x œ 0 and x œ 1.
È33
42. The graph of the first derivative indicates that the curve is rising on Š!ß 17 16
on (_ß !) and
xœ
17 È33
16
È
È
Š 17 16 33 ß 17 16 33 ‹
Ê a local maximum at x œ
17 È33
16
È33
‹ and Š 17 16
ß _‹ , falling
, a local minimum at
. The derivative approaches _ as x Ä 0 and x Ä 1, and approaches _ as x Ä 0 ,
indicating a cusp and local minimum at x œ 0 and a vertical tangent at x œ 1.
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Chapter 4 Practice Exercises
43. y œ
x1
x3
45. y œ
x# 1
x
œ1
œx
4
x3
"
x
44. y œ
2x
x5
œ2
46. y œ
x# x 1
x
10
x5
œx1
"
x
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281
282
Chapter 4 Applications of Derivatives
47. y œ
x$ 2
#x
œ
49. y œ
x# 4
x# 3
œ1
51. lim
xÄ"
52. lim
54. lim
tan x
x
œ
tan x
sin# x
sinamxb
x Ä 1Î#c
58.
xÄ"
axa"
b"
x Ä " bx
x Ä ! sinanxb
57. lim
œ lim
œ lim
#
x Ä ! tanax b
56. lim
"
x
"
x# 3
xa "
x Ä ! x sin x
55. lim
x # $x %
x"
b
x Ä " x "
53. xlim
Ä1
x#
#
tan 1
1
#x $
"
œ
x% 1
x#
œ x# 50. y œ
x#
x# 4
œ1
"
x#
4
x# 4
œ&
a
b
œ!
œ lim
sec# x
x Ä ! " cos x
œ lim
#sin x†cos x
# #
x Ä ! #x sec ax b
œ lim
xÄ!
m cosamxb
n cosanxb
œ
"
""
œ
"
#
sina#xb
œ lim
# #
x Ä ! #x sec ax b
œ
cosa$xb
x Ä 1Î#c cosa(xb
Èx
cos x
59. lim acsc x cot xb œ lim
xÄ!
#cosa#xb
œ lim
# #
#
# #
x Ä ! #x a#sec ax btanax b†#xb #sec ax b
œ
#
! #†"
œ"
m
n
seca(xbcosa$xb œ lim
lim Èx sec x œ lim b
x Ä !b
xÄ!
xÄ!
48. y œ
œ
!
"
$sina$xb
œ lim
x Ä 1Î#c (sina(xb
œ
$
(
œ!
" cos x
sin x
œ lim
sin x
x Ä ! cos x
œ
!
"
œ!
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Chapter 4 Practice Exercises
60. lim ˆ x"% xÄ!
61.
#
œ lim Š " x%x ‹ œ lim a" x# b †
xÄ!
xÄ!
"
x%
œ lim a" x# b œ lim
ŠÈx# x " Èx# x‹ œ lim ŠÈx# x " Èx# x‹ †
xÄ_
œ lim È # #x " È #
xÄ_
x x" x x
Notice that x œ Èx# for x ! so this is equivalent to
lim
#x "
x
#
x
x
" É x # x
É
#
x
x#
$
œ lim
xÄ_
$
lim Š x#x " x#x " ‹ œ lim
xÄ_
xÄ_
"#
"
œ lim #%
œ
lim
œ!
x
xÄ_
x Ä _ #x
"
%
xÄ! x
xÄ!
xÄ_
œ lim
xÄ_
62.
"‰
x#
œ"†_œ_
Èx# x "Èx# x
È x# x " È x# x
# x"
œ È"# È" œ "
"
É" x x"# É" x"
x $ a x # "b x $ a x # " b
ax# "bax# "b
œ lim
xÄ_
#x $
x% "
œ lim
xÄ_
'x#
%x $
œ lim
xÄ_
"#x
"#x#
63. (a) Maximize f(x) œ Èx È36 x œ x"Î# (36 x)"Î# where 0 Ÿ x Ÿ 36
Ê f w (x) œ
"
#
x"Î# "# (36 x)"Î# (1) œ
È36 x Èx
#Èx È36 x
Ê derivative fails to exist at 0 and 36; f(0) œ 6,
and f(36) œ 6 Ê the numbers are 0 and 36
(b) Maximize g(x) œ Èx È36 x œ x"Î# (36 x)"Î# where 0 Ÿ x Ÿ 36
Ê gw (x) œ
"
#
x"Î# "# (36 x)"Î# (1) œ
È36 x Èx
#Èx È36 x
Ê critical points at 0, 18 and 36; g(0) œ 6,
g(18) œ 2È18 œ 6È2 and g(36) œ 6 Ê the numbers are 18 and 18
64. (a) Maximize f(x) œ Èx (20 x) œ 20x"Î# x$Î# where 0 Ÿ x Ÿ 20 Ê f w (x) œ 10x"Î# 3# x"Î#
œ
20 3x
#È x
œ 0 Ê x œ 0 and x œ
œ
40È20
3È 3
Ê the numbers are
20
3
20
3
‰ É 20
ˆ
are critical points; f(0) œ f(20) œ 0 and f ˆ 20
3 œ
3 20 and
40
3
.
(b) Maximize g(x) œ x È20 x œ x (20 x)"Î# where 0 Ÿ x Ÿ 20 Ê gw (x) œ
Ê È20 x œ
"
#
Ê xœ
the numbers must be
65. A(x) œ
"
#
79
4
and
79
4 .
"
4 .
The critical points are x œ
79
4
2È20 x 1
#È20 x
‰
and x œ 20. Since g ˆ 79
4 œ
(2x) a27 x# b for 0 Ÿ x Ÿ È27
Ê Aw (x) œ 3(3 x)(3 x) and Aw w (x) œ 6x.
The critical points are 3 and 3, but 3 is not in the
domain. Since Aw w (3) œ 18 0 and A ŠÈ27‹ œ 0,
the maximum occurs at x œ 3 Ê the largest area is
A(3) œ 54 sq units.
66. The volume is V œ x# h œ 32 Ê h œ 32
x# . The
32 ‰
#
ˆ
surface area is S(x) œ x 4x x# œ x# 128
x ,
where x 0 Ê Sw (x) œ
20 ‰
3
2(x 4) ax# 4x 16b
x#
Ê the critical points are 0 and 4, but 0 is not in the
domain. Now Sw w (4) œ 2 256
4$ 0 Ê at x œ 4 there
is a minimum. The dimensions 4 ft by 4 ft by 2 ft
minimize the surface area.
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81
4
œ0
and g(20) œ 20,
283
284
Chapter 4 Applications of Derivatives
#
67. From the diagram we have ˆ h# ‰ r# œ ŠÈ3‹
Ê r# œ
12h#
4
#
. The volume of the cylinder is
#
V œ 1r# h œ 1 Š 12 4 h ‹ h œ
1
4
0 Ÿ h Ÿ 2È3 . Then Vw (h) œ
a12h h$ b , where
31
4
(2 h)(2 h)
Ê the critical points are 2 and 2, but 2 is not in
the domain. At h œ 2 there is a maximum since
Vw w (2) œ 31 0. The dimensions of the largest
cylinder are radius œ È2 and height œ 2.
68. From the diagram we have x œ radius and
y œ height œ 12 2x and V(x) œ "3 1x# (12 2x), where
0 Ÿ x Ÿ 6 Ê Vw (x) œ 21x(4 x) and Vw w (4) œ 81. The
critical points are 0 and 4; V(0) œ V(6) œ 0 Ê x œ 4
gives the maximum. Thus the values of r œ 4 and
h œ 4 yield the largest volume for the smaller cone.
‰ , where p is the profit on grade B tires and 0 Ÿ x Ÿ 4. Thus
69. The profit P œ 2px py œ 2px p ˆ 40510x
x
Pw (x) œ
2p
(5 x)#
ax# 10x 20b Ê the critical points are Š5 È5‹, 5, and Š5 È5‹ , but only Š5 È5‹ is in
the domain. Now Pw (x) 0 for 0 x Š5 È5‹ and Pw (x) 0 for Š5 È5‹ x 4 Ê at x œ Š5 È5‹ there
is a local maximum. Also P(0) œ 8p, P Š5 È5‹ œ 4p Š5 È5‹ ¸ 11p, and P(4) œ 8p Ê at x œ Š5 È5‹ there
is an absolute maximum. The maximum occurs when x œ Š5 È5‹ and y œ 2 Š5 È5‹ , the units are
hundreds of tires, i.e., x ¸ 276 tires and y ¸ 553 tires.
70. (a) The distance between the particles is lfatbl where fatb œ cos t cosˆt 1% ‰. Then, f w atb œ sin t sinˆt 1% ‰.
Solving f w atb œ ! graphically, we obtain t ¸ "Þ"(), t ¸ %Þ$#!, and so on.
Alternatively, f w atb œ ! may be solved analytically as follows. f w atb œ sin’ˆt 1) ‰ 1) “ sin’ˆt 1) ‰ 1) “
œ ’sinˆt 1) ‰cos 1) cosˆt 1) ‰sin 1) “ ’sinˆt 1) ‰cos 1) cosˆt 1) ‰sin 1) “ œ #sin 1) cosˆt 1) ‰
so the critical points occur when cosˆt 1) ‰ œ !, or t œ
$1
)
k1. At each of these values, fatb œ „ cos $)1
¸ „ !Þ('& units, so the maximum distance between the particles is !Þ('& units.
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Chapter 4 Practice Exercises
285
(b) Solving cos t œ cos ˆt 1% ‰ graphically, we obtain t ¸ #Þ(%*, t ¸ &Þ)*!, and so on.
Alternatively, this problem can be solved analytically as follows.
cos t œ cos ˆt 1% ‰
cos’ˆt 1) ‰ 1) “ œ cos’ˆt 1) ‰ 1) “
cosˆt 1) ‰cos 1) sinˆt 1) ‰sin 1) œ cosˆt 1) ‰cos 1) sinˆt 1) ‰sin 1)
#sin ˆt 1) ‰sin 1) œ !
sin ˆt 1) ‰ œ !
tœ
The particles collide when t œ
(1
)
(1
)
k1
¸ #Þ(%*. (plus multiples of 1 if they keep going.)
71. The dimensions will be x in. by "! #x in. by "' #x in., so Vaxb œ xa"! #xba"' #xb œ %x$ &#x# "'!x for
! x &. Then Vw axb œ "#x# "!%x "'! œ %ax #ba$x #!b , so the critical point in the correct domain is x œ #.
This critical point corresponds to the maximum possible volume because Vw axb ! for ! x # and Vw axb ! for
2 x &. The box of largest volume has a height of 2 in. and a base measuring 6 in. by 12 in., and its volume is 144 in.$
Graphical support:
72. The length of the ladder is d" d# œ 8 sec ) 6 csc ). We
wish to maximize I()) œ 8 sec ) 6 csc ) Ê Iw ())
œ 8 sec ) tan ) 6 csc ) cot ). Then Iw ()) œ 0
Ê 8 sin$ ) 6 cos$ ) œ 0 Ê tan ) œ
$
È
6
#
Ê
d" œ 4 É4 $È36 and d# œ $È36 É4 $È36
Ê the length of the ladder is about
Š4 $È36‹ É4 $È36 œ Š4 $È36‹
$Î#
¸ "*Þ( ft.
73. g(x) œ 3x x$ 4 Ê g(2) œ 2 0 and g(3) œ 14 0 Ê g(x) œ 0 in the interval [#ß 3] by the Intermediate
Value Theorem. Then gw (x) œ 3 3x# Ê xnb1 œ xn 3xn x$n 4
33xn#
; x! œ 2 Ê x" œ 2.22 Ê x# œ 2.196215, and
so forth to x& œ 2.195823345.
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Chapter 4 Applications of Derivatives
74. g(x) œ x% x$ 75 Ê g(3) œ 21 0 and g(4) œ 117 0 Ê g(x) œ 0 in the interval [$ß %] by the Intermediate
Value Theorem. Then gw (x) œ 4x$ 3x# Ê xnb1 œ xn x%n x$n 75
4xn$ 3xn#
; x! œ 3 Ê x" œ 3.259259
Ê x# œ 3.229050, and so forth to x& œ 3.22857729.
75.
' ax$ 5x 7b dx œ
76.
' Š8t$ t# t‹ dt œ 8t4% t6$ t## C œ 2t% t6$ t## C
77.
' ˆ3Èt t4# ‰ dt œ ' ˆ3t"Î# 4t# ‰ dt œ 3t$Î# 4t"1 C œ 2t$Î# 4t C
78.
' Š #È" t t3% ‹ dt œ ' ˆ #" t"Î# 3t% ‰ dt œ #" Œ t"Î# (3t$3) C œ Èt t"$ C
x%
4
5x#
#
7x C
#
Š 3# ‹
"
#
79. Let u œ r 5 Ê du œ dr
' ar dr5b
œ'
#
du
u#
u"
1
œ ' u# du œ
C œ u" C œ ar " 5b C
80. Let u œ r È2 Ê du œ dr
'
6 dr
$
Šr È2‹
œ 6'
dr
$
Šr È2‹
œ 6'
du
u$
81. Let u œ )# 1 Ê du œ 2) d) Ê
'
"
#
3)È)# 1 d) œ ' Èu ˆ #3 du‰ œ
82. Let u œ 7 )2 Ê du œ 2) d) Ê
'È)
d) œ '
7 ) 2
"
Èu
ˆ #" du‰ œ
"
#
x$ a 1 x % b
"Î%
#
C
du œ ) d)
3
#
"
#
3
ŠrÈ2‹
$Î#
$Î#
' u"Î# du œ 3# Œ u$Î#
C œ a ) # 1b C
3 C œ u
#
du œ ) d)
"Î#
' u"Î# du œ #" Œ u"Î#
C œ È7 )2 C
" C œ u
#
83. Let u œ 1 x% Ê du œ 4x$ dx Ê
'
#
œ 6' u$ du œ 6 Š u# ‹ C œ 3u# C œ "
4
du œ x$ dx
dx œ ' u"Î% ˆ "4 du‰ œ
"
4
$Î%
" $Î%
' u"Î% du œ 4" Œ u$Î%
C œ 3" a1 x% b C
3 C œ 3 u
4
84. Let u œ 2 x Ê du œ dx Ê du œ dx
' (2 x)$Î& dx œ ' u$Î& ( du) œ ' u$Î& du œ u
)Î&
Š 85 ‹
85. Let u œ
'
s
10
sec# 10s
Ê du œ
"
10
C œ 58 u)Î& C œ 58 (2 x))Î& C
ds Ê 10 du œ ds
ds œ ' asec# ub (10 du) œ 10 ' sec# u du œ 10 tan u C œ 10 tan
86. Let u œ 1s Ê du œ 1 ds Ê
"
1
s
10
C
du œ ds
' csc# 1s ds œ ' acsc# ub ˆ 1" du‰ œ 1" ' csc# u du œ 1" cot u C œ 1" cot 1s C
87. Let u œ È2 ) Ê du œ È2 d) Ê
' csc È2) cot È2) d) œ '
"
È2
du œ d)
(csc u cot u) Š È"2 du‹ œ
"
È2
(csc u) C œ È"2 csc È2) C
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Chapter 4 Additional and Advanced Exercises
)
3
88. Let u œ
'
sec
)
3
tan
89. Let u œ
'
Ê du œ
x
4
)
3
"
3
d) Ê 3 du œ d)
d) œ ' (sec u tan u)(3 du) œ 3 sec u C œ 3 sec
Ê du œ
"
4
)
3
C
dx Ê 4 du œ dx
2u ‰
dx œ ' asin# ub (4 du) œ ' 4 ˆ 1 cos
du œ 2' (1 cos 2u) du œ 2 ˆu #
œ 2u sin 2u C œ 2 ˆ x4 ‰ sin 2 ˆ x4 ‰ C œ x# sin x# C
sin#
x
4
90. Let u œ
'
cos#
œ
x
#
91. y œ '
x
#
x
#
"
#
Ê du œ
"
#
2u ‰
dx œ ' acos# ub (2 du) œ ' 2 ˆ 1 cos
du œ ' (1 cos 2u) du œ u #
x# "
x#
dx œ ' a1 x# b dx œ x x" C œ x y œ 1 when x œ 1 Ê
"
x
œ ' Š15Èt 3
Èt ‹
"
3
2
1
1
"‰
x#
Ê
"
x
C; y œ 1 when x œ 1 Ê 1 dx œ ' ax# 2 x# b dx œ
C œ 1 Ê C œ 3" Ê y œ
$
x
3
x$
3
2x x" C œ
2x dt œ ' ˆ15t"Î# 3t"Î# ‰ dt œ 10t$Î# 6t"Î# C;
dr
dt œ
&Î#
œ 4t&Î# 4t$Î# 8t C; r œ 0 when t œ 1 Ê 4(1)
r œ 4t&Î# 4t$Î# 8t
d# r
dt#
C
dr
dt
"
x
C œ 1
x$
3
2x "
x
C;
"
3
œ 8 when t œ 1
10t$Î# 6t"Î# 8 Ê r œ ' ˆ10t$Î# 6t"Î# 8‰ dt
4(1)$Î# 8(1) C" œ 0 Ê C" œ 0. Therefore,
œ ' cos t dt œ sin t C; rw w œ 0 when t œ 0 Ê sin 0 C œ 0 Ê C œ 0. Thus,
dr
dt
1
1
1
Ê 10(1)$Î# 6(1)"Î# C œ 8 Ê C œ 8. Thus
94.
sin 2u
#
sin x C
#
92. y œ ' ˆx x" ‰ dx œ ' ˆx# 2 dr
dt
C
dx Ê 2 du œ dx
Ê C œ 1 Ê y œ x 93.
sin 2u ‰
#
œ ' sin t dt œ cos t C" ; rw œ 0 when t œ 0 Ê 1 C" œ 0 Ê C" œ 1. Then
d# r
dt# œ sin t
dr
dt œ cos t 1
Ê r œ ' (cos t 1) dt œ sin t t C# ; r œ 1 when t œ 0 Ê 0 0 C# œ 1 Ê C# œ 1. Therefore,
r œ sin t t 1
CHAPTER 4 ADDITIONAL AND ADVANCED EXERCISES
1. If M and m are the maximum and minimum values, respectively, then m Ÿ f(x) Ÿ M for all x − I. If m œ M
then f is constant on I.
3x 6, 2 Ÿ x 0
has an absolute minimum value of 0 at x œ 2 and an absolute
9 x# , 0 Ÿ x Ÿ 2
maximum value of 9 at x œ 0, but it is discontinuous at x œ 0.
2. No, the function f(x) œ œ
3. On an open interval the extreme values of a continuous function (if any) must occur at an interior critical
point. On a half-open interval the extreme values of a continuous function may be at a critical point or at the
closed endpoint. Extreme values occur only where f w œ 0, f w does not exist, or at the endpoints of the interval.
Thus the extreme points will not be at the ends of an open interval.
4. The pattern f w œ ± ± ± ± indicates a local maximum at x œ 1 and a local
"
#
$
%
minimum at x œ 3.
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287
288
Chapter 4 Applications of Derivatives
5. (a) If yw œ 6(x 1)(x 2)# , then yw 0 for x 1 and yw 0 for x 1. The sign pattern is
f w œ ± ± Ê f has a local minimum at x œ 1. Also yww œ 6(x 2)# 12(x 1)(x 2)
"
#
œ 6(x 2)(3x) Ê yw w 0 for x 0 or x 2, while yww 0 for 0 x 2. Therefore f has points of inflection
at x œ 0 and x œ 2. There is no local maximum.
(b) If yw œ 6x(x 1)(x 2), then yw 0 for x 1 and 0 x 2; yw 0 for " x 0 and x 2. The sign
sign pattern is yw œ ± ± ± . Therefore f has a local maximum at x œ 0 and
"
!
#
È7
local minima at x œ 1 and x œ 2. Also, yww œ ") ’x Š 1 $
1 È 7
$
x
1 È 7
$
È7
‹“ ’x Š 1 $
‹“ , so yww 0 for
and yww 0 for all other x Ê f has points of inflection at x œ
6. The Mean Value Theorem indicates that
f(6) f(0)
60
1 „È 7
$
.
œ f w (c) Ÿ 2 for some c in (0ß 6). Then f(6) f(0) Ÿ 12
indicates the most that f can increase is 12.
7. If f is continuous on [aß c) and f w (x) Ÿ 0 on [aß c), then by the Mean Value Theorem for all x − [aß c) we have
f(c) f(x)
cx
f(c). Also if f is continuous on (cß b] and f w (x)
Ÿ 0 Ê f(c) f(x) Ÿ 0 Ê f(x)
f(x) f(c)
xc
all x − (cß b] we have
0 Ê f(x) f(c)
0 Ê f(x)
f(c). Therefore f(x)
8. (a) For all x, (x 1)# Ÿ 0 Ÿ (x 1)# Ê a1 x# b Ÿ 2x Ÿ a1 x# b Ê "# Ÿ
(b) There exists c − (aß b) such that
Ê kf(b) f(a)k Ÿ
"
#
c
1 c#
œ
f(b) f(a)
ba
f(a)
¸ c ¸
Ê ¹ f(b)b a ¹ œ 1 c# Ÿ
"
#
0 on (cß b], then for
f(c) for all x − [aß b].
x
1 x#
Ÿ
"
#
.
, from part (a)
kb ak .
9. No. Corollary 1 requires that f w (x) œ 0 for all x in some interval I, not f w (x) œ 0 at a single point in I.
10. (a) h(x) œ f(x)g(x) Ê hw (x) œ f w (x)g(x) f(x)gw (x) which changes signs at x œ a since f w (x), gw (x) 0 when
x a, f w (x), gw (x) 0 when x a and f(x), g(x) 0 for all x. Therefore h(x) does have a local maximum
at x œ a.
(b) No, let f(x) œ g(x) œ x$ which have points of inflection at x œ 0, but h(x) œ x' has no point of inflection
(it has a local minimum at x œ 0).
11. From (ii), f(1) œ
lim
xÄ „_
f(x) œ
12.
dy
dx
œ 0 Ê a œ 1; from (iii), either 1 œ x lim
f(x) or 1 œ x Ä
lim
f(x). In either case,
Ä_
_
x"
#
x Ä „ _ bx cx #
1 "x
x
c 2x
xÄ „_
lim
" a
bc#
œ ! and if c œ 0,
œ 3x# 2kx 3 œ 0 Ê x œ
1 "x
bx
c 2x
xÄ „_
1 x"
then lim
bx
2x
xÄ „_
œ
lim
lim
2k „ È4k# 36
6
œ " Ê b œ 0 and c œ ". For if b œ ", then
œ
lim
xÄ „_
1 x"
2
x
œ „ _. Thus a œ 1, b œ 0, and c œ 1.
Ê x has only one value when 4k# 36 œ 0 Ê k# œ 9 or
k œ „ 3.
13. The area of the ?ABC is A(x) œ
w
where 0 Ÿ x Ÿ 1. Thus A (x) œ
"
#
(2) È1 x# œ a1 x# b
x
È 1 x#
"Î#
,
Ê 0 and „ 1 are
critical points. Also A a „ 1b œ 0 so A(0) œ 1 is the
maximum. When x œ 0 the ?ABC is isosceles since
AC œ BC œ È2 .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Chapter 4 Additional and Advanced Exercises
f (c h) f (c)
œ f ww (c) Ê for % œ "# kf ww (c)k 0
h
hÄ0
Ê ¹ f (ch)h f (c) f ww (c)¹ "# kf ww (c)k . Then f w (c) œ
w
14. lim
w
w
w
3
#
f ww (c) 0 Ê "# kf ww (c)k f (c h)
f ww (c) "# kf ww (c)k . If f ww (c) 0, then
h
f (c h)
"# f ww (c) 0; likewise if f ww (c) 0, then 0 "#
h
w
Ê f ww (c) "# kf ww (c)k Ê
there exists a $ 0 such that 0 khk $
w
w
f (c h)
h
"
#
f ww (c) w
kf ww (c)k
kf ww (c)k œ f ww (c)
f ww (c) f (c h)
h
w
3
#
f ww (c).
(a) If f ww (c) 0, then $ h 0 Ê f (c h) 0 and 0 h $ Ê f w (c h) 0. Therefore, f(c) is a local
maximum.
(b) If f ww (c) 0, then $ h 0 Ê f w (c h) 0 and 0 h $ Ê f w (c h) 0. Therefore, f(c) is a local
minimum.
15. The time it would take the water to hit the ground from height y is É 2y
g , where g is the acceleration of
gravity. The product of time and exit velocity (rate) yields the distance the water travels:
È64(h y) œ 8 É 2 ahy y# b
D(y) œ É 2y
g
g
are critical points. Now D(0) œ 0,
D ˆ h# ‰
16. From the figure in the text, tan (" )) œ
give
ba
h
œ
tan " 1 ha tan "
a
h
h tan " a
h a tan "
œ
œ
"Î#
, 0 Ÿ y Ÿ h Ê Dw (y) œ 4 É g2 ahy y# b
8h
Èg
ba
h ;
"Î#
(h 2y) Ê 0,
and D(h) œ 0 Ê the best place to drill the hole is at y œ
tan (" )) œ
tan " tan )
1 tan " tan )
. Solving for tan " gives tan " œ
; and tan ) œ
bh
h# a(b a)
a
h
h
#
h
#
and h
.
. These equations
or
#
ah a(b a)b tan " œ bh. Differentiating both sides with respect to h gives
2h tan " ah# a(b a)b sec# "
d"
dh
œ b. Then
d"
dh
bh
œ 0 Ê 2h tan " œ b Ê 2h Š h# a(b
a) ‹ œ b
Ê 2bh# œ bh# ab(b a) Ê h# œ a(b a) Ê h œ Èa(a b) .
17. The surface area of the cylinder is S œ 21r# 21rh. From
the diagram we have Rr œ H H h Ê h œ RH R rH and
S(r) œ 21r(r h) œ 21r ˆr H r HR ‰
œ 21 ˆ1 HR ‰ r# 21Hr, where 0 Ÿ r Ÿ R.
Case 1: H R Ê S(r) is a quadratic equation containing
the origin and concave upward Ê S(r) is maximum at
r œ R.
Case 2: H œ R Ê S(r) is a linear equation containing the
origin with a positive slope Ê S(r) is maximum at
r œ R.
Case 3: H R Ê S(r) is a quadratic equation containing the origin and concave downward. Then
dS
H‰
dS
H‰
RH
ˆ
ˆ
dr œ 41 1 R r 21H and dr œ 0 Ê 41 1 R r 21H œ 0 Ê r œ 2(H R) . For simplification
we let r‡ œ
RH
2(H R)
.
(a) If R H 2R, then 0 H 2R Ê H 2(H R) Ê
(b) If H œ 2R, then r‡ œ
#
2R
2R
RH
2(H R)
R which is impossible.
œ R Ê S(r) is maximum at r œ R.
(c) If H 2R, then 2R H 2H Ê H 2(H R) Ê
S(r) is a maximum at r œ r‡ œ
RH
2(H R)
H
2(H R)
1 Ê
RH
2(H R)
R Ê r‡ R. Therefore,
.
Conclusion: If H − (0ß R] or H œ 2R, then the maximum surface area is at r œ R. If H − (Rß 2R), then r R
which is not possible. If H − (2Rß _), then the maximum is at r œ r‡ œ 2(HRH
R) .
18. f(x) œ mx 1 "
x
Ê f w (x) œ m minimum. If f Š È"m ‹
"
x#
and f w w (x) œ
2
x$
0 when x 0. Then f w (x) œ 0 Ê x œ
0, then Èm 1 Èm œ 2Èm 1
0 Ê m
"
4
"
Èm
yields a
. Thus the smallest acceptable value
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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289
290
Chapter 4 Applications of Derivatives
for m is
19. (a)
(b)
(c)
"
4
.
lim
xÄ!
#sina&xb
$x
œ lim
xÄ!
x
x Ä ! sin# È#x
lim
x Ä 1/2
"
xÄ!
(g)
(h)
cos$ x
#
sinax# b
lim
x Ä ! xsin x
lim
xÄ!
x$ )
20. (a) x lim
Ä_
(b) x lim
Ä_
"!
$
$sina&xbsina$xb &cosa&xbcosa$xb
$cosa$xb
xÄ!
È #x
"
œ lim
#sinÈ#x cosÈ#x
xÄ!
È#x
xÄ!
sinŠ#È#x‹
œ
&
$
"
È#x
œ lim
x Ä ! cosŠ#È#x‹ È#
#x
"sin x
cos x
lim " cos#x
x Ä ! " sec x
œ
cos x
œ lim
œ !Þ
x Ä 1/2 sin x
lim " cos# x
x Ä ! tan x
œ
lim cos x# "
x Ä ! tan x
œ lim
sin x
#
x Ä ! # tan x sec x
œ lim
sin x
xÄ!
# sin x
cos$ x
œ
œ #"
#x cosax# b
œ lim
x Ä ! xcos xsin x
sec x "
x#
lim #
x Ä # x %
†"œ
"
#
œ
x Ä 1/2
œ
"!
$
œ
œ lim
œ lim
asec x tan xb œ lim
lim x sin x
x Ä ! x tan x
œ lim
(f)
sina&xbcosa$xb
sina$xb
lim x csc# È#x œ lim
x Ä ! cosŠ#È#x‹†#
(e)
xÄ! $
xÄ!
xÄ!
"! sina&xb
a&xb
œ lim
lim sina&xbcota$xb œ lim
xÄ!
œ lim
(d)
#sina&xb
$
& a &x b
œ lim
xÄ!
œ lim
xÄ#
sec x tan x
#x
xÄ!
ax #bax# #x %b
ax #bax #b
œ x lim
Ä_
#
sec x tan x sec x
#
xÄ!
x # #x %
x#
œ lim
xÄ#
œ x lim
œ x lim
Ä _ Èx &
Ä_
#x
x (È x
$
œ lim
Èx &
Èx
Èx
Èx &
Èx &
a#x# bsinax# b #cosax# b
xsin x#cos x
œ lim
É" x&
" È&x
#x
x
x( x
x
È œ x lim
Ä_
œ
#
" (É x"
œ
œ
"
"
œ"
œ
#
"!
œ
"!
#
#
#
œ"
œ
%%%
%
"
#
œ$
œ#
21. (a) The profit function is Paxb œ ac exbx aa bxb œ ex# ac bbx a. Pw axb œ #ex c b œ !
Ê x œ c#eb . Pww axb œ #e ! if e ! so that the profit function is maximized at x œ c #e b .
(b) The price therefore that corresponds to a production level yeilding a maximum profit is
p¹
xœ c#eb
œ c eˆ c #e b ‰ œ
c b
#
dollars.
#
(c) The weekly profit at this production level is Paxb œ eˆ c #e b ‰ ac bbˆ c #e b ‰ a œ
ac b b #
%e
#
a.
(d) The tax increases cost to the new profit function is Faxb œ ac exbx aa bx txb œ ex ac b tbx a.
bc
cbt
ww
Now Fw axb œ #ex c b t œ ! when x œ t #
e œ
#e . Since F axb œ #e ! if e !, F is maximized
when x œ c #be t units per week. Thus the price per unit is p œ c eˆ c #be t ‰ œ c #b t dollars. Thus, such a tax
increases the cost per unit by
cbt
#
The x-intercept occurs when
"
x
cb
#
œ
t
#
dollars if units are priced to maximize profit.
22. (a)
$œ!Ê
(b) By Newton's method, xn" œ xn œ xn xn $x#n
œ #xn $xn#
faxn b
f ax n b .
w
"
x
œ $ Ê x œ $" .
Here f w axn b œ x#
n œ
"
x#n .
" $
So xn" œ xn xn" œ xn Š x"n $‹x#n
x#
n
œ xn a# $xn b.
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Chapter 4 Additional and Advanced Exercises
23. x" œ x! and
a
q"
x!
fax! b
f w ax ! b
q
qx!q xq! a
qxq!"
x a
œ x! ! q" œ
qx
!
with weights m! œ
In the case where x! œ
a
xq!"
q"
q
q
x aq "b a
œ ! q"
œ x! Š q q " ‹ a
Š"‹
xq!" q
qx!
and m" œ "q .
we have xq! œ a and x" œ
q"
a
a
"
q" Š q ‹ q" Š q ‹
x!
x!
œ
291
so that x" is a weighted average of x!
q"
a
q" Š q
x!
q" ‹ œ
a
q" .
x!
#
dy
d y
24. We have that ax hb# ay hb# œ r# and so #ax hb #ay hb dy
dx œ ! and # # dx #ay hb dx# œ ! hold.
dy
x y dx
dy .
" dx
dy
Thus #x #y dy
dx œ #h #h dx , by the former. Solving for h, we obtain h œ
#
d y
equation yields # # dy
dx #y dx# #Œ
dy
x y dx
dy " dx
œ !. Dividing by 2 results in " Substituting this into the second
dy
dx
#
y ddxy# Œ
dy
x y dx
dy " dx
œ !.
25. (a) aatb œ sww atb œ k ak !b Ê sw atb œ kt C" , where sw a!b œ )) Ê C" œ )) Ê sw atb œ kt )). So
satb œ
kt#
#
kt#
#
))t C# where sa!b œ ! Ê C# œ ! so satb œ
))t œ "!!. Solving for t we obtain t œ
kŠ )) È))# #!!k
‹
k
))#
#!!
so that k œ
)) œ ! or kŠ )) )) „ È))# #!!k
.
k
È))# #!!k
‹
k
kt#
#
))t. Now satb œ "!! when
At such t we want sw atb œ !, thus
)) œ !. In either case we obtain ))# #!!k œ !
¸ $)Þ(# ft/sec# .
(b) The initial condition that sw a!b œ %% ft/sec implies that sw atb œ kt %% and satb œ
w
The car is stopped at a time t such that s atb œ kt %% œ ! Ê t œ
‰
sˆ %%
k
œ
k ˆ %% ‰#
#
k
‰
%%ˆ %%
k
œ
%%#
#k
*')
k
œ
œ
‰
*')ˆ #!!
))#
%%
k .
kt#
#
%%t where k is as above.
At this time the car has traveled a distance
œ #& feet. Thus halving the initial velocity quarters
stopping distance.
26. haxb œ f # axb g# axb Ê hw axb œ #faxbf w axb #gaxbgw axb œ #faxbf w axb gaxbgw axb‘ œ #faxbgaxb gaxbafaxbb‘
œ # † ! œ !. Thus haxb œ c, a constant. Since ha!b œ &, haxb œ & for all x in the domain of h. Thus ha"!b œ &.
27. Yes. The curve y œ x satisfies all three conditions since
dy
dx
œ " everywhere, when x œ !, y œ !, and
d# y
dx#
œ ! everywhere.
28. yw œ $x# # for all x Ê y œ x$ #x C where " œ "$ # † " C Ê C œ % Ê y œ x$ #x %.
29. sww atb œ a œ t# Ê v œ sw atb œ
maximum for this t‡ . Since satb
t‡ œ a$Cb"Î$ . So
ÊCœ
a%bb$Î%
$ .
a$Cb"Î$ ‘%
12
t$
w
‡
‡
$ C. We seek v! œ s a!b œ C. We know that sat b œ b for some t and s is at a
%
%
œ 12t Ct k and sa!b œ ! we have that satb œ 12t Ct and also sw at‡ b œ ! so that
Ca$Cb"Î$ œ b Ê a$Cb"Î$ ˆC Thus v! œ sw a!b œ
a%bb$Î%
$
œ
#È# $Î%
.
$ b
$C ‰
"#
œ b Ê a$Cb"Î$ ˆ $%C ‰ œ b Ê $"Î$ C%Î$ œ
30. (a) sww atb œ t"Î# t"Î# Ê vatb œ sw atb œ #$ t$Î# #t"Î# k where va!b œ k œ
(b) satb œ
% &Î#
"& t
%
%$ t$Î# %$ t k# where sa!b œ k# œ "&
. Thus satb œ
%
# $Î#
#t"Î#
$ Ê vatb œ $ t
% &Î#
%
%$ t$Î# %$ t "&
.
"& t
31. The graph of faxb œ ax# bx c with a ! is a parabola opening upwards. Thus faxb
one real value of x. The solutions to faxb œ ! are, by the quadratic equation
#
Thus we require
#
a#bb %ac Ÿ ! Ê b ac Ÿ !.
#
Thus aa" b" a# b# Þ Þ Þ an bn b aa#"
a##
ÞÞÞ an# bab"#
b##
ÞÞÞ bn# b
!.
Ÿ ! by Exercise 31.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
%$ Þ
! for all x if faxb œ ! for at most
#b „ Éa#bb# %ac
.
#a
32. (a) Clearly faxb œ aa" x b" b# Þ Þ Þ aan x bn b# ! for all x. Expanding we see
faxb œ aa#" x# #a" b" x b"# b Þ Þ Þ aan# x# #an bn x bn# b
œ aa#" a## Þ Þ Þ an# bx# #aa" b" a# b# Þ Þ Þ an bn bx ab"# b## Þ Þ Þ bn# b
%b
$
292
Chapter 4 Applications of Derivatives
Thus aa" b" a# b# Þ Þ Þ an bn b# Ÿ aa#" a## Þ Þ Þ an# bab"# b## Þ Þ Þ bn# b.
(b) Referring to Exercise 31: It is clear that faxb œ ! for some real x Í b# %ac œ !, by quadratic formula.
Now notice that this implies that
faxb œ aa" x b" b# Þ Þ Þ aan x bn b#
œ aa#" a## Þ Þ Þ an# bx# #aa" b" a# b# Þ Þ Þ an bn bx ab"# b## Þ Þ Þ bn# b œ !
Í aa" b" a# b# Þ Þ Þ an bn b# aa#" a## Þ Þ Þ an# bab"# b## Þ Þ Þ bn# b œ !
Í aa" b" a# b# Þ Þ Þ an bn b# œ aa#" a## Þ Þ Þ an# bab"# b## Þ Þ Þ bn# b
But now faxb œ ! Í ai x bi œ ! for all i œ "ß #ß Þ Þ Þ ß n Í ai x œ bi œ ! for all i œ "ß #ß Þ Þ Þ ß n.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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CHAPTER 5 INTEGRATION
5.1 ESTIMATING WITH FINITE SUMS
1. faxb œ x#
Since f is increasing on Ò!ß "Ó, we use left endpoints to obtain
lower sums and right endpoints to obtain upper sums.
"
#
iœ!
$
"
#
œ "# Š!# ˆ "# ‰ ‹ œ
#
"
4
œ 4" Š!# ˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ ‹ œ
(a) ˜x œ
"!
#
œ
"
#
and xi œ i˜x œ
i
#
Ê a lower sum is !ˆ #i ‰ †
(b) ˜x œ
"!
%
œ
"
%
and xi œ i˜x œ
i
%
Ê a lower sum is !ˆ 4i ‰ †
(c) ˜x œ
"!
#
œ
"
#
and xi œ i˜x œ
i
#
Ê an upper sum is !ˆ #i ‰ †
(d) ˜x œ
"!
%
œ
"
%
and xi œ i˜x œ
i
%
Ê an upper sum is !ˆ 4i ‰ †
2. faxb œ x$
iœ!
2
iœ1
%
#
"
)
#
#
#
#
"
#
#
œ "# Šˆ "# ‰ +1# ‹ œ
#
"
4
#
#
#
œ 4" Šˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ +1# ‹ œ
iœ"
"
%
†
(
)
"
%
$! ‰
† ˆ "'
œ
œ
(
$#
&
)
"&
$#
Since f is increasing on Ò!ß "Ó, we use left endpoints to obtain
lower sums and right endpoints to obtain upper sums.
"
$
iœ!
$
"
#
œ "# Š!$ ˆ "# ‰ ‹ œ
$
"
4
œ 4" Š!$ ˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ ‹ œ
(a) ˜x œ
"!
#
œ
"
#
and xi œ i˜x œ
i
#
Ê a lower sum is !ˆ #i ‰ †
(b) ˜x œ
"!
%
œ
"
%
and xi œ i˜x œ
i
%
Ê a lower sum is !ˆ 4i ‰ †
(c) ˜x œ
"!
#
œ
"
#
and xi œ i˜x œ
i
#
Ê an upper sum is !ˆ #i ‰ †
(d) ˜x œ
"!
%
œ
"
%
and xi œ i˜x œ
i
%
Ê an upper sum is !ˆ 4i ‰ †
iœ!
2
iœ1
%
iœ"
$
"
"'
$
$
$
$'
#&'
$
"
#
$
œ "# Šˆ "# ‰ +1$ ‹ œ
$
"
4
$
$
$
œ 4" Šˆ 4" ‰ ˆ #" ‰ ˆ 4$ ‰ +1$ ‹ œ œ
"
#
†
*
)
œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
œ
*
'%
"!!
#&'
œ
*
"'
#&
'%
294
Chapter 5 Integration
3. faxb œ
"
x
Since f is decreasing on Ò!ß "Ó, we use left endpoints to obtain
upper sums and right endpoints to obtain lower sums.
#
(a) ˜x œ
&"
#
œ # and xi œ " i˜x œ " #i Ê a lower sum is ! x"i † # œ #ˆ $" &" ‰ œ
(b) ˜x œ
&"
%
œ 1 and xi œ " i˜x œ " i Ê a lower sum is
(c) ˜x œ
&"
#
œ # and xi œ " i˜x œ " #i Ê an upper sum is ! x"i † # œ #ˆ" $" ‰ œ
(d) ˜x œ
&"
%
œ 1 and xi œ " i˜x œ " i Ê an upper sum is
4. faxb œ % x#
iœ"
%
!"
xi
iœ"
"
† " œ "ˆ #" iœ!
$
!"
xi
iœ!
"
$
† " œ "ˆ" "
#
"
%
"'
"&
&" ‰ œ
"
$
((
'!
)
$
"% ‰ œ
#&
"#
Since f is increasing on Ò#ß !Ó and decreasing on Ò!ß #Ó, we use
left endpoints on Ò#ß !Ó and right endpoints on Ò!ß #Ó to obtain
lower sums and use right endpoints on Ò#ß !Ó and left endpoints
on Ò!ß #Ó to obtain upper sums.
(a) ˜x œ
# a#b
#
œ # and xi œ # i˜x œ # #i Ê a lower sum is # † ˆ% a#b# ‰ # † a% ## b œ !
(b) ˜x œ
# a#b
%
œ " and xi œ # i˜x œ # i Ê a lower sum is !ˆ% axi b# ‰ † " !ˆ% axi b# ‰ † "
"
%
iœ!
iœ$
œ "ˆˆ% a#b# ‰ ˆ% a"b# ‰ a% "# b a% ## b‰ œ '
(c) ˜x œ
# a#b
#
œ # and xi œ # i˜x œ # #i Ê a upper sum is # † ˆ% a!b# ‰ # † a% !# b œ "'
(d) ˜x œ
# a#b
%
œ " and xi œ # i˜x œ # i Ê a upper sum is !ˆ% axi b# ‰ † " !ˆ% axi b# ‰ † "
#
$
iœ"
iœ#
œ "ˆˆ% a"b# ‰ a% !# b a% !# b a% "# b‰ œ "%
5. faxb œ x#
œ
"
#
Using 4 rectangles Ê ˜x œ " % ! œ
Ê "% ˆfˆ ") ‰ fˆ $) ‰ fˆ &) ‰ fˆ () ‰‰
"
%
Using 2 rectangles Ê ˜x œ
#
#
œ "# Šˆ "% ‰ ˆ $% ‰ ‹ œ
#
#
"!
$#
œ
#
"!
#
&
"'
#
œ "% Šˆ ") ‰ ˆ $) ‰ ˆ &) ‰ ˆ () ‰ ‹ œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Ê "# ˆfˆ "% ‰ fˆ $% ‰‰
#"
'%
Section 5.1 Estimating with Finite Sums
6. faxb œ x$
œ
"
#
Using 4 rectangles Ê ˜x œ " % ! œ
Ê "% ˆfˆ ") ‰ fˆ $) ‰ fˆ &) ‰ fˆ () ‰‰
"
%
$
$
œ "# Šˆ "% ‰ ˆ $% ‰ ‹ œ
$
$
$
$
(
œ "% Š " $ )&
‹œ
$
7. faxb œ
"
x
"!
#
Using 2 rectangles Ê ˜x œ
#)
# † '%
%*'
% † )$
œ
œ
Using 2 rectangles Ê ˜x œ
œ #ˆ "# "% ‰ œ $#
Ê "# ˆfˆ "% ‰ fˆ $% ‰‰
(
$#
"#%
)$
&"
#
œ
$"
"#)
œ # Ê #afa#b fa%bb
Using 4 rectangles Ê ˜x œ & % " œ "
Ê "ˆfˆ $# ‰ fˆ &# ‰ fˆ (# ‰ fˆ *# ‰‰
œ "ˆ #$ 8. faxb œ % x#
#
&
#
(
#* ‰ œ
"%))
$†&†(†*
Using 2 rectangles Ê ˜x œ
œ #a$ $b œ "#
295
œ
# a#b
#
%*'
&†(†*
œ
%*'
$"&
œ # Ê #afa"b fa"bb
Using 4 rectangles Ê ˜x œ # %a#b œ "
Ê "ˆfˆ $# ‰ fˆ "# ‰ fˆ "# ‰ fˆ $# ‰‰
#
#
#
#
œ "ŠŠ% ˆ $# ‰ ‹ Š% ˆ "# ‰ ‹ Š% ˆ "# ‰ ‹ Š% ˆ $# ‰ ‹‹
œ "' ˆ *% † # "% † #‰ œ "' "!
# œ ""
9. (a) D ¸ (0)(1) (12)(1) (22)(1) (10)(1) (5)(1) (13)(1) (11)(1) (6)(1) (2)(1) (6)(1) œ 87 inches
(b) D ¸ (12)(1) (22)(1) (10)(1) (5)(1) (13)(1) (11)(1) (6)(1) (2)(1) (6)(1) (0)(1) œ 87 inches
10. (a) D ¸ (1)(300) (1.2)(300) (1.7)(300) (2.0)(300) (1.8)(300) (1.6)(300) (1.4)(300) (1.2)(300)
(1.0)(300) (1.8)(300) (1.5)(300) (1.2)(300) œ 5220 meters (NOTE: 5 minutes œ 300 seconds)
(b) D ¸ (1.2)(300) (1.7)(300) (2.0)(300) (1.8)(300) (1.6)(300) (1.4)(300) (1.2)(300) (1.0)(300)
(1.8)(300) (1.5)(300) (1.2)(300) (0)(300) œ 4920 meters (NOTE: 5 minutes œ 300 seconds)
11. (a) D ¸ (0)(10) (44)(10) (15)(10) (35)(10) (30)(10) (44)(10) (35)(10) (15)(10) (22)(10)
(35)(10) (44)(10) (30)(10) œ 3490 feet ¸ 0.66 miles
(b) D ¸ (44)(10) (15)(10) (35)(10) (30)(10) (44)(10) (35)(10) (15)(10) (22)(10) (35)(10)
(44)(10) (30)(10) (35)(10) œ 3840 feet ¸ 0.73 miles
12. (a) The distance traveled will be the area under the curve. We will use the approximate velocities at the
midpoints of each time interval to approximate this area using rectangles. Thus,
D ¸ (20)(0.001) (50)(0.001) (72)(0.001) (90)(0.001) (102)(0.001) (112)(0.001) (120)(0.001)
(128)(0.001) (134)(0.001) (139)(0.001) ¸ 0.967 miles
(b) Roughly, after 0.0063 hours, the car would have gone 0.484 miles, where 0.0060 hours œ 22.7 sec. At 22.7
sec, the velocity was approximately 120 mi/hr.
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296
Chapter 5 Integration
13. (a) Because the acceleration is decreasing, an upper estimate is obtained using left end-points in summing
acceleration † ?t. Thus, ?t œ 1 and speed ¸ [32.00 19.41 11.77 7.14 4.33](1) œ 74.65 ft/sec
(b) Using right end-points we obtain a lower estimate: speed ¸ [19.41 11.77 7.14 4.33 2.63](1)
œ 45.28 ft/sec
(c) Upper estimates for the speed at each second are:
t
0
1
2
3
4
5
v
0
32.00 51.41 63.18 70.32
74.65
Thus, the distance fallen when t œ 3 seconds is s ¸ [32.00 51.41 63.18](1) œ 146.59 ft.
14. (a) The speed is a decreasing function of time Ê right end-points give an lower estimate for the height (distance)
attained. Also
t
0
1
2
3
4
5
v
400
368
336
304
272
240
gives the time-velocity table by subtracting the constant g œ 32 from the speed at each time increment
?t œ 1 sec. Thus, the speed ¸ 240 ft/sec after 5 seconds.
(b) A lower estimate for height attained is h ¸ [368 336 304 272 240](1) œ 1520 ft.
15. Partition [!ß #] into the four subintervals [0ß 0.5], [0.5ß 1], [1ß 1.5], and [1.5ß 2]. The midpoints of these
subintervals are m" œ 0.25, m# œ 0.75, m$ œ 1.25, and m% œ 1.75. The heights of the four approximating
1
125
343
$
$
rectangles are f(m" ) œ (0.25)$ œ 64
, f(m# ) œ (0.75)$ œ 27
64 , f(m$ ) œ (1.25) œ 64 , and f(m% ) œ (1.75) œ 64
Notice that the average value is approximated by
œ
"
length of [!ß#]
†”
"
#
$
$
$
$
’ˆ 4" ‰ ˆ #" ‰ ˆ 34 ‰ ˆ #" ‰ ˆ 54 ‰ ˆ #" ‰ ˆ 74 ‰ ˆ #" ‰“ œ
$"
"'
approximate area under
• . We use this observation in solving the next several exercises.
curve f(x) œ x$
16. Partition [1ß 9] into the four subintervals ["ß $], [3ß &], [&ß (], and [(ß *]. The midpoints of these subintervals are
m" œ 2, m# œ 4, m$ œ 6, and m% œ 8. The heights of the four approximating rectangles are f(m" ) œ "# ,
f(m# ) œ "4 , f(m$ ) œ 6" , and f(m% ) œ 8" . The width of each rectangle is ?x œ 2. Thus,
Area ¸ 2 ˆ "# ‰ 2 ˆ 4" ‰ 2 ˆ 6" ‰ 2 ˆ 8" ‰ œ
Ê average value ¸
25
1#
area
length of ["ß*]
œ
ˆ 25
‰
12
8
œ
25
96 .
17. Partition [0ß 2] into the four subintervals [0ß 0.5], [0.5ß 1], [1ß 1.5], and [1.5ß 2]. The midpoints of the subintervals
are m" œ 0.25, m# œ 0.75, m$ œ 1.25, and m% œ 1.75. The heights of the four approximating rectangles are
"
#
f(m" ) œ
œ
"
#
"
#
sin#
1
4
œ
"
#
"
#
œ 1, and f(m% ) œ
œ 1, f(m# ) œ
"
2
sin#
71
4
œ
sin#
"
#
Š È"2 ‹ œ 1. The width of each rectangle is ?x œ #" . Thus,
31
4
œ
"
#
"
#
œ 1, f(m$ ) œ
"
2
sin#
51
4
œ
"
#
Š È"2 ‹
#
"
2
#
Area ¸ (1 1 1 1) ˆ "# ‰ œ 2 Ê average value ¸
area
length of [0ß2]
œ
2
#
œ 1.
18. Partition [0ß 4] into the four subintervals [0ß 1], [1ß 2ß ], [2ß 3], and [3ß 4]. The midpoints of the subintervals
are m" œ "# , m# œ #3 , m$ œ 5# , and m% œ 7# . The heights of the four approximating rectangles are
f(m" ) œ 1 Šcos Š
%
1 ˆ "# ‰
4 ‹‹
œ 1 ˆcos ˆ 18 ‰‰ œ 0.27145 (to 5 decimal places),
f(m# ) œ 1 Šcos Š
%
1 ˆ 3# ‰
4 ‹‹
œ 1 ˆcos ˆ 381 ‰‰ œ 0.97855, f(m3 ) œ 1 Šcos Š
%
%
œ 0.97855, and f(m% ) œ 1 Šcos Š
%
1 ˆ 7# ‰
4 ‹‹
%
1 ˆ #5 ‰
4 ‹‹
œ 1 ˆcos ˆ 581 ‰‰
%
%
œ 1 ˆcos ˆ 781 ‰‰ œ 0.27145. The width of each rectangle is
?x œ ". Thus, Area ¸ (0.27145)(1) (0.97855)(1) (0.97855)(1) (0.27145)(1) œ 2.5 Ê average
2.5
5
value ¸ lengtharea
of [0ß4] œ 4 œ 8 .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 5.1 Estimating with Finite Sums
297
19. Since the leakage is increasing, an upper estimate uses right endpoints and a lower estimate uses left
endpoints:
(a) upper estimate œ (70)(1) (97)(1) (136)(1) (190)(1) (265)(1) œ 758 gal,
lower estimate œ (50)(1) (70)(1) (97)(1) (136)(1) (190)(1) œ 543 gal.
(b) upper estimate œ (70 97 136 190 265 369 516 720) œ 2363 gal,
lower estimate œ (50 70 97 136 190 265 369 516) œ 1693 gal.
(c) worst case: 2363 720t œ 25,000 Ê t ¸ 31.4 hrs;
best case: 1693 720t œ 25,000 Ê t ¸ 32.4 hrs
20. Since the pollutant release increases over time, an upper estimate uses right endpoints and a lower estimate
uses left endpoints:
(a) upper estimate œ (0.2)(30) (0.25)(30) (0.27)(30) (0.34)(30) (0.45)(30) (0.52)(30) œ 60.9 tons
lower estimate œ (0.05)(30) (0.2)(30) (0.25)(30) (0.27)(30) (0.34)(30) (0.45)(30) œ 46.8 tons
(b) Using the lower (best case) estimate: 46.8 (0.52)(30) (0.63)(30) (0.70)(30) (0.81)(30) œ 126.6 tons,
so near the end of September 125 tons of pollutants will have been released.
#
21. (a) The diagonal of the square has length 2, so the side length is È#. Area œ ŠÈ#‹ œ #
(b) Think of the octagon as a collection of 16 right triangles with a hypotenuse of length 1 and an acute angle measuring
#1
1
"' œ ) .
Area œ "'ˆ " ‰ˆsin 1 ‰ˆcos 1 ‰ œ % sin 1 œ #È# ¸ #Þ)#)
#
)
)
%
(c) Think of the 16-gon as a collection of 32 right triangles with a hypotenuse of length 1 and an acute angle measuring
#1
1
$# œ "' .
Area œ $#ˆ " ‰ˆsin 1 ‰ˆcos 1 ‰ œ ) sin 1 œ #È# ¸ $Þ!'"
#
"'
"'
)
(d) Each area is less than the area of the circle, 1. As n increases, the area approaches 1.
22. (a) Each of the isosceles triangles is made up of two right triangles having hypotenuse 1 and an acute angle measuring
#1
1
1 ‰ˆ
ˆ " ‰ˆ
cos 1n ‰ œ "# sin #n1 .
#n œ n . The area of each isosceles triangle is AT œ # # sin n
(b) The area of the polygon is AP œ nAT œ
n
#
sin
#1
n ,
n
nÄ_ #
so lim
sin
#1
n
œ lim 1 †
nÄ_
sin #n1
ˆ #n1 ‰
œ1
(c) Multiply each area by r# .
AT œ "# r# sin #n1
AP œ n# r# sin
lim AP œ 1r
#
#1
n
nÄ_
23-26. 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));
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298
Chapter 5 Integration
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 );
Mathematica: (assigned function and values for a and b may vary):
Symbols for 1, Ä , powers, roots, fractions, etc. are available in Palettes (under File).
Never insert a space between the name of a function and its argument.
Clear[x]
f[x_]:=x Sin[1/x]
{a,b}={1/4, 1}
Plot[f[x],{x, a, b}]
The following code computes the value of the function for each interval midpoint and then finds the average. Each
sequence of commands for a different value of n (number of subdivisions) should be placed in a separate cell.
n =100; dx = (b a) /n;
values = Table[N[f[x]], {x, a dx/2, b, dx}]
average=Sum[values[[i]],{i, 1, Length[values]}] / n
n =200; dx = (b a) /n;
values = Table[N[f[x]],{x, a + dx/2, b, dx}]
average=Sum[values[[i]],{i, 1, Length[values]}] / n
n =1000; dx = (b a) /n;
values = Table[N[f[x]],{x, a dx/2, b, dx}]
average=Sum[values[[i]],{i, 1, Length[values]}] / n
FindRoot[f[x] == average,{x, a}]
5.2 SIGMA NOTATION AND LIMITS OF FINITE SUMS
2
1. !
kœ1
3
2. !
kœ1
6k
k1
œ
6(1)
11
6(2)
21
œ
6
2
k1
k
œ
11
1
21
2
31
3
12
3
œ7
œ0
1
2
2
3
œ
7
6
4
3. ! cos k1 œ cos (11) cos (21) cos (31) cos (41) œ 1 1 1 1 œ 0
kœ1
5
4. ! sin k1 œ sin (11) sin (21) sin (31) sin (41) sin (51) œ 0 0 0 0 0 œ 0
kœ1
3
5. ! (1)kb1 sin
kœ1
1
k
œ (1)"" sin
1
1
(1)#" sin
1
#
(")$" sin
1
3
œ 01
È3
#
œ
È3 2
#
4
6. ! (1)k cos k1 œ (1)" cos (11) (1)# cos (21) (1)$ cos (31) (1)% cos (41)
kœ1
œ (1) 1 (1) 1 œ 4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 5.2 Sigma Notation and Limits of Finite Sums
6
7. (a) ! 2kc1 œ 2"" 2#" 2$" 2%" 2&" 2'" œ 1 2 4 8 16 32
kœ1
5
(b) ! 2k œ 2! 2" 2# 2$ 2% 2& œ 1 2 4 8 16 32
kœ0
4
(c) ! 2k1 œ 2"" 2!" 2"" 2#" 2$" 2%" œ 1 2 4 8 16 32
kœ"
All of them represent 1 2 4 8 16 32
6
8. (a) ! (2)k1 œ (2)"" (2)#" (2)$" (2)%" (2)&" (2)'" œ 1 2 4 8 16 32
kœ1
5
(b) ! (1)k 2k œ (1)! 2! Ð")" 2" (1)# 2# (1)$ 2$ (1)% 2% (1)& 2& œ 1 2 4 8 16 32
kœ0
3
(c) ! (1)k1 2k2 œ Ð")#" 2## (")"" 2"# (")!" 2!# (1)"" 2"# (")#" 2##
kœ2
(1)$" 2$# œ 1 2 4 8 16 32;
(a) and (b) represent 1 2 4 8 16 32; (c) is not equivalent to the other two
4
9. (a) !
kœ2
2
(b) !
kœ0
1
(c) !
kœ"
(")k"
k1
(1)#"
21
œ
(")k
k1
œ
(1)!
01
(")k
k2
œ
(1)"
1 2
(")$"
31
(")"
11
(")!
02
(")#
21
(")%"
41
œ 1 œ1
(")"
12
"
#
œ 1 "
#
"
#
"
3
"
3
"
3
(a) and (c) are equivalent; (b) is not equivalent to the other two.
4
10. (a) ! (k 1)# œ (1 1)# (2 1)# (3 1)# (4 1)# œ 0 1 4 9
kœ1
3
(b) ! (k 1)# œ (1 1)# (0 1)# (1 1)# (2 1)# (3 1)# œ 0 1 4 9 16
kœ1
"
(c) ! k# œ (3)# (2)# (1)# œ 9 4 1
kœ3
(a) and (c) are equivalent to each other; (b) is not equivalent to the other two.
6
4
4
12. ! k#
11. ! k
kœ1
13. !
kœ1
5
5
15. ! (1)k1
14. ! 2k
kœ1
kœ1
n
kœ1
"
k
5
16. ! (1)k
kœ1
n
17. (a) ! 3ak œ 3 ! ak œ 3(5) œ 15
kœ1
n
(b) !
kœ1
n
bk
6
œ
"
6
kœ1
n
! bk œ
kœ1
"
6
(6) œ 1
n
n
kœ1
n
kœ1
n
kœ1
n
kœ1
n
kœ1
n
kœ1
(c) ! (ak bk ) œ ! ak ! bk œ 5 6 œ 1
(d) ! (ak bk ) œ ! ak ! bk œ 5 6 œ 11
n
(e) ! (bk 2ak ) œ ! bk 2 ! ak œ 6 2(5) œ 16
kœ1
kœ1
kœ1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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"
#k
k
5
299
300
Chapter 5 Integration
n
n
kœ1
n
kœ1
n
18. (a) ! 8ak œ 8 ! ak œ 8(0) œ 0
n
n
kœ1
kœ1
(c) ! (ak 1) œ ! ak ! 1 œ 0 n œ n
kœ1
10
19. (a) ! k œ
kœ1
10(10 1)
#
n
(b) ! 250bk œ 250 ! bk œ 250(1) œ 250
kœ1
n
n
kœ1
n
kœ1
kœ1
kœ1
(d) ! (bk 1) œ ! bk ! 1 œ " n
10
(b) ! k# œ
œ 55
kœ1
10(10 1)(2(10) 1)
6
œ 385
13(13 1)(2(13) 1)
6
œ 819
#
10
(c) ! k$ œ ’ 10(10# 1) “ œ 55# œ 3025
kœ1
13
20. (a) ! k œ
kœ1
13(13 1)
#
13
(b) ! k# œ
œ 91
kœ1
#
13
(c) ! k$ œ ’ 13(13# 1) “ œ 91# œ 8281
kœ1
7
7
kœ1
kœ1
6
6
6
kœ1
kœ1
kœ1
6
6
6
kœ1
kœ1
kœ1
5
21. ! 2k œ 2 ! k œ 2 Š 7(7 # ") ‹ œ 56
23. ! a3 k# b œ ! 3 ! k# œ 3(6) 24. ! ak# 5b œ ! k# ! 5 œ
22. !
kœ1
6(6 ")(2(6) 1)
6
6(6 ")(2(6) 1)
6
1k
15
œ
1
15
5
!kœ
kœ1
1
15
Š 5(5 # 1) ‹ œ 1
œ 73
5(6) œ 61
5
5
5
5
kœ1
kœ1
kœ1
kœ1
7
7
7
7
kœ1
kœ1
kœ1
kœ1
1)
25. ! k(3k 5) œ ! a3k# 5kb œ 3 ! k# 5 ! k œ 3 Š 5(5 1)(2(5)
‹ 5 Š 5(5 # 1) ‹ œ 240
6
1)
26. ! k(2k 1) œ ! a2k# kb œ 2 ! k# ! k œ 2 Š 7(7 1)(2(7)
‹
6
5
k$
225
27. !
kœ1
7
kœ1
#
7
28. Œ! k !
kœ1
29. (a)
$
5
Œ! k œ
kœ1
k$
4
"
2 #5
7
5
5
kœ1
kœ1
#
œ Œ! k kœ1
$
! k $ Œ! k œ
"
4
"
#25
#
! k$ œ Š 7(7 1) ‹ #
kœ1
(b)
œ 308
$
Š 5(5 # 1) ‹ Š 5(5 # 1) ‹ œ 3376
#
7
7(7 1)
#
"
4
#
Š 7(7 # 1) ‹ œ 588
(c)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 5.2 Sigma Notation and Limits of Finite Sums
30. (a)
(b)
(c)
31. (a)
(b)
(c)
32. (a)
(b)
(c)
301
33. kx" x! k œ k1.2 0k œ 1.2, kx# x" k œ k1.5 1.2k œ 0.3, kx$ x# k œ k2.3 1.5k œ 0.8, kx% x$ k œ k2.6 2.3k œ 0.3,
and kx& x% k œ k3 2.6k œ 0.4; the largest is lPl œ 1.2.
34. kx" x! k œ k1.6 (2)k œ 0.4, kx# x" k œ k0.5 (1.6)k œ 1.1, kx$ x# k œ k0 (0.5)k œ 0.5,
kx% x$ k œ k0.8 0k œ 0.8, and kx& x% k œ k1 0.8k œ 0.2; the largest is lPl œ 1.1.
35. faxb œ " x#
Since f is decreasing on Ò !, 1Ó we use left endpoints to obtain
upper sums. ˜x œ " n ! œ "n and xi œ i˜x œ ni . So an upper sum
n"
is ! a" x#i b "n œ
iœ!
œ
n$
n$
n
"! #
i
n$
iœ!
"
n
n"
! Š" ˆ i ‰# ‹ œ
n
iœ!
œ"
an " b n a# an " b " b
'n $
# $n n"#
. Thus,
'
n"
lim ! a" x#i b "n œ lim Œ"
nÄ_ iœ!
nÄ_
"
n$
n"
! an# i# b
iœ!
œ"
#n$ $n# n
'n $
œ"
# $n n"#
'
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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œ"
"
$
œ
#
$
302
Chapter 5 Integration
Since f is increasing on Ò !, $Ó we use right endpoints to obtain
upper sums. ˜x œ $ n ! œ $n and xi œ i˜x œ $ni . So an upper
36. faxb œ #x
n
n
sum is !#xi ˆ n$ ‰ œ ! 'ni †
iœ"
iœ"
n
Thus,
37. faxb œ x# "
lim ! 'i
nÄ_ iœ" n
†
$
n
œ
#
œ lim *n n# *n
nÄ_
$
n
n
")
n#
")
n#
!i œ
iœ"
n an " b
#
†
œ
*n# *n
n#
œ lim ˆ* n* ‰ œ *.
nÄ_
Since f is increasing on Ò !, $Ó we use right endpoints to obtain
upper sums. ˜x œ $ n ! œ $n and xi œ i˜x œ $ni . So an upper
n
n
#
sum is !ax#i "b $n œ !Šˆ $ni ‰ "‹ n$ œ
œ
iœ"
n
#( ! #
i n$
n
iœ"
iœ"
†nœ
#( nan "ba#n "b
‹
n$ Š
'
n
!Š *i## "‹
n
$
n
iœ"
$
*
") #(
*a#n$ $n# nb
n n#
$œ
$Þ Thus,
#n $
#
n
#(
") n n*#
lim !ax#i "b $n œ lim Œ
$ œ
#
nÄ_ iœ"
nÄ_
œ
38. faxb œ $x#
Since f is increasing on Ò !, "Ó we use right endpoints to obtain
upper sums. ˜x œ " n ! œ "n and xi œ i˜x œ ni . So an upper sum
n
n
iœ"
iœ"
#
is !$x#i ˆ "n ‰ œ !$ˆ ni ‰ ˆ n" ‰ œ
œ
#n$ $n# n
#n $
œ lim Œ
nÄ_
39. faxb œ x x# œ xa" xb
œ
# $n n"#
#
# $n n"#
#
#
#
œ
n
$
n$
! i# œ
iœ"
$
n$
† Š nan "ba' #n "b ‹
n
. Thus, lim !$x#i ˆ "n ‰
nÄ_ iœ"
œ ".
Since f is increasing on Ò !, "Ó we use right endpoints to obtain
upper sums. ˜x œ " n ! œ "n and xi œ i˜x œ ni . So an upper sum
n
n
#
is !axi xi# b n" œ !Š ni ˆ ni ‰ ‹ n" œ
iœ"
iœ"
œ
" n a n "b
‹
n# Š
#
œ
" "n
#
n"$ Š nan "ba' #n "b ‹
n
# $ "
œ lim ”Š
nÄ_
40. faxb œ $x #x#
* $ œ "#.
n
'
"
#
"
n
n#
n
"!
i
n#
iœ"
n
"! #
i
n$
iœ"
n# n
#n#
#n $ $ n # n
'n$
œ
. Thus, lim !axi x#i b "n
nÄ_ iœ"
‹Œ
$
n
# n"#
'
• œ
"
#
#
'
œ &' .
Since f is increasing on Ò !, "Ó we use right endpoints to obtain
upper sums. ˜x œ " n ! œ "n and xi œ i˜x œ ni . So an upper sum
n
n
#
is !a$xi #x#i b "n œ !Š $ni #ˆ ni ‰ ‹ n" œ
iœ"
iœ"
œ
$ n a n "b
‹
n# Š
#
œ
$ $n
#
œ lim ”Š
nÄ_
n#$ Š nan "ba' #n "b ‹
n
# $ "
$
#
n
$
$
n
n#
œ
n
$!
i
n#
iœ"
$n# $n
#n#
. Thus, lim !a$xi #x#i b "n
‹Œ
nÄ_ iœ"
$
n
# n"#
$
• œ
$
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
#
$
œ
"$
' .
n
#! #
i
n$
iœ"
#n# $n "
$n#
Section 5.3 The Definite Integral
303
5.3 THE DEFINITE INTEGRAL
1.
'02 x# dx
2.
'"! 2x$ dx
3.
'(& ax# 3xb dx
4.
'"% "x dx
5.
'#$ 1 " x dx
6.
'0" È4 x# dx
7.
'! Î% (sec x) dx
8.
'0 Î% (tan x) dx
1
1
9. (a)
(c)
(e)
(f)
10. (a)
(b)
(c)
(d)
(e)
(f)
11. (a)
(c)
12. (a)
(c)
13. (a)
(b)
14. (a)
(b)
"
&
'#2 g(x) dx œ 0
(b) ' g(x) dx œ ' g(x) dx œ 8
&
"
2
2
&
&
2
'" 3f(x) dx œ 3'" f(x) dx œ 3(4) œ 12
(d) ' f(x) dx œ ' f(x) dx ' f(x) dx œ 6 (4) œ 10
#
"
"
&
&
&
'" [f(x) g(x)] dx œ '" f(x) dx '" g(x) dx œ 6 8 œ 2
'"& [4f(x) g(x)] dx œ 4 '"& f(x) dx '"& g(x) dx œ 4(6) 8 œ 16
'"* 2f(x) dx œ 2 '"* f(x) dx œ 2(1) œ 2
'(* [f(x) h(x)] dx œ '(*f(x) dx '(* h(x) dx œ 5 4 œ 9
'(* [2f(x) 3h(x)] dx œ 2 '(* f(x) dx 3 '(* h(x) dx œ 2(5) 3(4) œ 2
'*"f(x) dx œ '"* f(x) dx œ (1) œ 1
'"( f(x) dx œ '"* f(x) dx '(* f(x) dx œ 1 5 œ 6
'*( [h(x) f(x)] dx œ '(* [f(x) h(x)] dx œ '(* f(x) dx '(* h(x) dx œ 5 4 œ 1
'"2 f(u) du œ '"2 f(x) dx œ 5
'#" f(t) dt œ '"2 f(t) dt œ 5
'!$ g(t) dt œ '$! g(t) dt œ È2
'$! [g(x)] dx œ '$! g(x) dx œ È2
(b)
(d)
(b)
(d)
'"2 È3 f(z) dz œ È3 '"2 f(z) dz œ 5È3
'"2 [f(x)] dx œ '"2 f(x) dx œ 5
'$! g(u) du œ '$! g(t) dt œ È2
'$! Èg(r)2 dr œ È"2 '$! g(t) dt œ Š È"2 ‹ ŠÈ2‹ œ 1
'$% f(z) dz œ '!% f(z) dz '!$ f(z) dz œ 7 3 œ 4
'%$ f(t) dt œ '$% f(t) dt œ 4
'"$ h(r) dr œ '"$ h(r) dr '"" h(r) dr œ 6 0 œ 6
"
$
$
' h(u) du œ Œ ' h(u) du œ ' h(u) du œ 6
$
"
"
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
304
Chapter 5 Integration
15. The area of the trapezoid is A œ
œ
"
#
(5 2)(6) œ 21 Ê
œ 21 square units
"
#
(3 1)(1) œ 2 Ê
(B b)h
"
#
(B b)h
'# ˆ #x 3‰ dx
%
16. The area of the trapezoid is A œ
œ
"
#
'"Î#
$Î#
(2x 4) dx
œ 2 square units
17. The area of the semicircle is A œ
œ
9
#
1 Ê
"
#
1r# œ
1(3)#
'$$ È9 x# dx œ 9# 1 square units
18. The graph of the quarter circle is A œ
œ 41 Ê
"
#
"
4
1 r# œ
"
4
1(4)#
'%! È16 x# dx œ 41 square units
19. The area of the triangle on the left is A œ
"
#
bh œ
œ 2. The area of the triangle on the right is A œ
œ
Ê
"
#
(1)(1) œ
"
#.
"
#
"
#
(2)(2)
bh
Then, the total area is 2.5
'# kxk dx œ 2.5 square units
"
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 5.3 The Definite Integral
20. The area of the triangle is A œ
Ê
"
#
bh œ
'" a1 kxkb dx œ 1 square unit
"
21. The area of the triangular peak is A œ
"
#
(2)(1) œ 1
"
#
bh œ
"
#
(2)(1) œ 1.
The area of the rectangular base is S œ jw œ (2)(1) œ 2.
Then the total area is 3 Ê
'"" a2 kxkb dx œ 3 square units
22. y œ 1 È1 x# Ê y 1 œ È1 x#
Ê (y 1)# œ 1 x# Ê x# (y 1)# œ 1, a circle with
center (!ß ") and radius of 1 Ê y œ 1 È1 x# is the
upper semicircle. The area of this semicircle is
A œ "# 1r# œ "# 1(1)# œ 1# . The area of the rectangular base
is A œ jw œ (2)(1) œ 2. Then the total area is 2 Ê
23.
'"" Š1 È1 x# ‹ dx œ 2 1# square units
'!b x2 dx œ "# (b)( b2 ) œ b4
#
1
#
24.
'!b 4x dx œ "# b(4b) œ 2b#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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305
306
Chapter 5 Integration
25.
'ab 2s ds œ "# b(2b) "# a(2a) œ b# a#
27.
'"
29.
'1#1 ) d) œ (2#1)
31.
'0
33.
'!"Î# t# dt œ ˆ 3‰
œ
35.
'a#a x dx œ (2a)#
37.
'!
39.
'$" 7 dx œ 7(1 3) œ 14
41.
'!2 5x dx œ 5 '!2 x dx œ 5 ’ 2#
43.
'!2 (2t 3) dt œ 2 '"" t dt '!2 3 dt œ 2 ’ 2#
44.
'!
45.
'#" ˆ1 #z ‰ dz œ '#" 1 dz '#" #z dz œ '#" 1 dz "# '"# z dz œ 1[1 2] "# ’ 2# 1# “ œ " "# ˆ 3# ‰ œ 74
46.
'$! (2z 3) dz œ '$! 2z dz '$! 3 dz œ 2 '!$ z dz '$! 3 dz œ 2 ’ 3#
47.
'"# 3u# du œ 3 '"# u# du œ 3 ”'!# u# du '!" u# du• œ 3 Š’ 23
È#
3
È
7
ŠÈ2‹
#
#
(1)#
#
œ
1#
#
œ
31 #
#
"
#
28.
'!Þ&#Þ& x dx œ (2.5)#
30.
'È& # # r dr œ Š5È#2‹
x dx œ
3
7‹
ŠÈ
œ
32.
'!!Þ$ s# ds œ (0.3)3
3
34.
'!1Î# )# d) œ ˆ 3‰
7
3
"
24
a#
#
œ
36.
'a
$
b‹
ŠÈ
3
œ
b
3
38.
'!$b x# dx œ (3b)3
40.
'!2 È2 dx œ È2 (# !) œ 2È2
#
0#
#“
œ 10
42.
'$& 8x dx œ "8 '$& x dx œ 8" ’ 5#
#
Št È2‹ dt œ
È2
'!
È2
t dt '
!
0#
#“
(0.5)#
#
#
$
1 $
#
È$a
3a#
#
$
x# dx œ
#
È
$
#
#
È2
'ab 3t dt œ "# b(3b) "# a(3a) œ 3# ab# a# b
#
x dx œ
" $
#
$
È
b
26.
œ3
#
ŠÈ2‹
#
œ 24
œ 0.009
œ
1$
#4
#
ŠÈ3a‹
x dx œ
#
$
a#
#
œ a#
œ 9b$
#
3#
#“
œ
16
16
œ1
3(2 0) œ 4 6 œ 2
#
È2 dt œ
ŠÈ2‹
–
#
0#
#—
È2 ’È2 0“ œ 1 2 œ 1
#
#
$
0$
3“
$
’ "3 0#
#“
#
3[0 3] œ 9 9 œ 0
0$
3 “‹
$
œ 3 ’ 23 Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
1$
3“
œ 3 ˆ 73 ‰ œ 7
Section 5.3 The Definite Integral
48.
'"Î#" 24u# du œ 24 '"Î#"
49.
'!# a3x# x 5b dx œ 3 '!# x# dx '!# x dx '!# 5 dx œ 3 ’ 23
50.
'"! a3x# x 5b dx œ '!" a3x# x 5b dx œ ”3 '!" x# dx '!" x dx '!" 5 dx•
u# du œ 24 –
'!"
u# du '!"Î#
$
u# du— œ 24 ” 13 $
$
0$
3‹
b0
n
œ
œ ’3 Š 13 51. Let ?x œ
#
Š 1# 0#
#‹
5(1 0)“ œ ˆ 3# 5‰ œ
0$
3“
ˆ "# ‰$
3 •
#
’ 2# œ 24 ’
0#
#“
ˆ 78 ‰
3
“œ7
5[2 0] œ (8 2) 10 œ 0
7
#
and let x! œ 0, x" œ ?x,
b
n
x# œ 2?xß á ß xn" œ (n 1)?x, xn œ n?x œ b.
Let the ck 's be the right end-points of the subintervals
Ê c" œ x" , c# œ x# , and so on. The rectangles
defined have areas:
f(c" ) ?x œ f(?x) ?x œ 3(?x)# ?x œ 3(?x)$
f(c# ) ?x œ f(2?x) ?x œ 3(2?x)# ?x œ 3(2)# (?x)$
f(c$ ) ?x œ f(3?x) ?x œ 3(3?x)# ?x œ 3(3)# (?x)$
ã
f(cn ) ?x œ f(n?x) ?x œ 3(n?x)# ?x œ 3(n)# (?x)$
n
n
kœ1
n
kœ1
Then Sn œ ! f(ck ) ?x œ ! 3k# (?x)$
$
1)
œ 3(?x) ! k# œ 3 Š bn$ ‹ Š n(n 1)(2n
‹
6
$
kœ1
œ
$
b
#
ˆ2 52. Let ?x œ
3
n
b0
n
"‰
n#
œ
Ê
b
n
'!b 3x# dx œ n lim
Ä_
b$
#
ˆ2 3
n
"‰
n#
œ b$ .
and let x! œ 0, x" œ ?x,
x# œ 2?xß á ß xn" œ (n 1)?x, xn œ n?x œ b.
Let the ck 's be the right end-points of the subintervals
Ê c" œ x" , c# œ x# , and so on. The rectangles
defined have areas:
f(c" ) ?x œ f(?x) ?x œ 1(?x)# ?x œ 1(?x)$
f(c# ) ?x œ f(2?x) ?x œ 1(2?x)# ?x œ 1(2)# (?x)$
f(c$ ) ?x œ f(3?x) ?x œ 1(3?x)# ?x œ 1(3)# (?x)$
ã
f(cn ) ?x œ f(n?x) ?x œ 1(n?x)# ?x œ 1(n)# (?x)$
n
n
Then Sn œ ! f(ck ) ?x œ ! 1k# (?x)$
kœ1
n
kœ1
1)
œ 1(?x)$ ! k# œ 1 Š bn$ ‹ Š n(n 1)(2n
‹
6
$
kœ1
œ
1b
6
$
ˆ2 3
n
"‰
n#
Ê
'!b 1x# dx œ n lim
Ä_
1 b$
6
ˆ2 3
n
"‰
n#
œ
1 b$
3 .
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307
308
Chapter 5 Integration
b0
n
53. Let ?x œ
œ
b
n
and let x! œ 0, x" œ ?x,
x# œ 2?xß á ß xn" œ (n 1)?x, xn œ n?x œ b.
Let the ck 's be the right end-points of the subintervals
Ê c" œ x" , c# œ x# , and so on. The rectangles
defined have areas:
f(c" ) ?x œ f(?x) ?x œ 2(?x)(?x) œ 2(?x)#
f(c# ) ?x œ f(2?x) ?x œ 2(2?x)(?x) œ 2(2)(?x)#
f(c$ ) ?x œ f(3?x) ?x œ 2(3?x)(?x) œ 2(3)(?x)#
ã
f(cn ) ?x œ f(n?x) ?x œ 2(n?x)(?x) œ 2(n)(?x)#
n
n
Then Sn œ ! f(ck ) ?x œ ! 2k(?x)#
kœ1
n
kœ1
œ 2(?x)# ! k œ 2 Š bn# ‹ Š n(n 2 1) ‹
#
kœ1
œ b# ˆ1 "n ‰ Ê
b0
n
54. Let ?x œ
œ
'!b 2x dx œ n lim
Ä_
b
n
b# ˆ1 n" ‰ œ b# .
and let x! œ 0, x" œ ?x,
x# œ 2?xß á ß xn" œ (n 1)?x, xn œ n?x œ b.
Let the ck 's be the right end-points of the subintervals
Ê c" œ x" , c# œ x# , and so on. The rectangles
defined have areas:
"
#
‰
f(c" ) ?x œ f(?x) ?x œ ˆ ?x
# 1 (?x) œ # (?x) ?x
2
?
x
"
f(c# ) ?x œ f(2?x) ?x œ ˆ # 1‰ (?x) œ # (2)(?x)# ?x
f(c$ ) ?x œ f(3?x) ?x œ ˆ 3?# x 1‰ (?x) œ
"
#
(3)(?x)# ?x
f(cn ) ?x œ f(n?x) ?x œ ˆ n?# x 1‰ (?x) œ
"
#
(n)(?x)# ?x
ã
n
n
kœ1
kœ1
Then Sn œ ! f(ck ) ?x œ ! ˆ "# k(?x)# ?x‰ œ
œ
"
4
b# ˆ1 n1 ‰ b Ê
55. av(f) œ Š È3" 0 ‹
œ
'! ˆ x# 1‰ dx œ n lim
Ä_
b
"
È3
È$
'!
È$
'!
x# dx "
#
n
n
kœ1
kœ1
(?x)# ! k ?x ! 1 œ
ˆ 4" b# ˆ1 n" ‰ b‰ œ
"
4
"
#
Š bn# ‹ Š n(n 2 1) ‹ ˆ bn ‰ (n)
#
b# b.
ax# 1b dx
È$
'!
"
È3
1 dx
$
œ
"
È3
ŠÈ3‹
3
56. av(f) œ ˆ 3 " 0 ‰
$
"
È3
ŠÈ3 0‹ œ 1 1 œ 0.
'!$ Š x# ‹ dx œ 3" ˆ #" ‰ '!$ x# dx
#
#
œ "6 Š 33 ‹ œ 3# ; x# œ 3# .
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Section 5.3 The Definite Integral
'!" a3x# 1b dx œ
"
"
œ 3 ' x# dx ' 1 dx œ 3 Š 13 ‹ (1 0)
!
!
57. av(f) œ ˆ 1 " 0 ‰
$
œ #.
'!" a3x# 3b dx œ
"
"
œ 3 ' x# dx ' 3 dx œ 3 Š 13 ‹ 3(1 0)
!
!
58. av(f) œ ˆ 1 " 0 ‰
$
œ #.
'!$ (t 1)# dt
$
$
$
œ 3" ' t# dt 23 ' t dt 3" ' 1 dt
!
!
!
59. av(f) œ ˆ 3 " 0 ‰
œ
"
3
$
#
Š 33 ‹ 32 Š 3# 60. av(f) œ Š 1 1(2) ‹
0#
#‹
3" (3 0) œ 1.
'#" at# tb dt
'#" t# dt 3" '#" t dt
"
#
œ "3 ' t# dt 3" '
t# dt 3" Š 1#
!
!
œ
"
3
#
œ
"
3
$
Š 13 ‹ 3" Š (32) ‹ $
61. (a) av(g) œ Š 1 "(1) ‹
"
#
œ
3
#
(2)#
# ‹
.
'"" akxk 1b dx
'"! (x 1) dx "# '!" (x 1) dx
!
!
"
"
œ "# ' x dx "# ' 1 dx "# ' x dx "# ' 1 dx
"
"
!
!
œ
"
#
#
œ "# Š 0# (1)#
# ‹
#
"# (0 (1)) "# Š 1# 0#
#‹
"# (1 0)
œ "# .
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309
310
Chapter 5 Integration
'"$ akxk 1b dx œ #" '"$ (x 1) dx
$
$
œ "# ' x dx "# ' 1 dx œ "# Š 3# 1# ‹ "# (3 1)
"
"
(b) av(g) œ ˆ 3 " 1 ‰
#
#
œ 1.
(c) av(g) œ Š 3 "(1) ‹
œ
"
4
"
4
'"$ akxk 1b dx
'"" akxk 1b dx 4" '"$ akxk 1b dx
"
4
(see parts (a) and (b) above).
62. (a) av(h) œ Š 0 "(1) ‹
'"0 kxk dx œ '"0 (x) dx
œ
œ
(1 2) œ
'"0 x dx œ 0#
#
(b) av(h) œ ˆ 1 " 0 ‰
#
œ Š "# (1)#
#
œ "# .
'0" kxk dx œ '0" x dx
0#
#‹
œ "# .
(c) av(h) œ Š 1 "(1) ‹
'"" kxk dx
'"0 kxk dx '0" kxk dx
œ
"
#
Œ
œ
"
#
ˆ "# ˆ "# ‰‰ œ "# (see parts (a) and (b)
above).
63. To find where x x# 0, let x x# œ 0 Ê x(1 x) œ 0 Ê x œ 0 or x œ 1. If 0 x 1, then 0 x x# Ê a œ 0
and b œ 1 maximize the integral.
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Section 5.3 The Definite Integral
311
64. To find where x% 2x# Ÿ 0, let x% 2x# œ 0 Ê x# ax# 2b œ 0 Ê x œ 0 or x œ „ È2. By the sign graph,
0 0 0 , we can see that x% 2x# Ÿ 0 on ’È2ß È2“ Ê a œ È2 and b œ È2
!
È#
È #
minimize the integral.
"
1 x #
65. f(x) œ
is decreasing on [0ß 1] Ê maximum value of f occurs at 0 Ê max f œ f(0) œ 1; minimum value of f
occurs at 1 Ê min f œ f(1) œ
Ê
"
#
Ÿ
'0" 1 " x
"
1 1#
œ
"
#
. Therefore, (1 0) min f Ÿ
dx Ÿ 1. That is, an upper bound œ 1 and a lower bound œ
#
66. See Exercise 65 above. On [0ß 0.5], max f œ
'0
0.5
(0.5 0) min f Ÿ
"
1 1#
min f œ
Then
"
4
2
5
"
1 0#
'0
0.5
"
1 x#
dx '0.5 1 " x
"
#
dx Ÿ
67. 1 Ÿ sin ax# b Ÿ 1 for all x Ê (1 0)(1) Ÿ
"
#
"
1 (0.5)#
œ 1, min f œ
f(x) dx Ÿ (0.5 0) max f Ê
Ÿ
2
5
'0
'0.5 1 " x
"
œ 0.5. Therefore (1 0.5) min f Ÿ
Ÿ
'0" 1 " x
2
5
Ê
0.5
#
"
1 x#
#
"
#
dx Ÿ (1 0) max f
.
œ 0.8. Therefore
dx Ÿ
"
#
. On [0.5ß 1], max f œ
dx Ÿ (1 0.5) max f Ê
13
20
Ÿ
'0 1 " x
"
#
dx Ÿ
9
10
"
4
Ÿ
"
1 (0.5)#
'0.5 1 1 x
"
#
dx Ÿ
œ 0.8 and
2
5
.
.
'0" sin ax# b dx Ÿ (1 0)(1) or '0"sin x# dx Ÿ 1
Ê
'0"sin x# dx cannot
equal 2.
68. f(x) œ Èx 8 is increasing on [!ß "] Ê max f œ f(1) œ È1 8 œ 3 and min f œ f(0) œ È0 8 œ 2È2 .
Therefore, (1 0) min f Ÿ
'0" Èx 8 dx Ÿ (1 0) max f
0 on [aß b], then min f
69. If f(x)
a Ê ba
Then b
Ê 2È 2 Ÿ
'0" Èx 8 dx Ÿ 3.
0 on [aß b]. Now, (b a) min f Ÿ
0 and max f
0 Ê (b a) min f
0 Ê
'ab f(x) dx
0.
70. If f(x) Ÿ 0 on [aß b], then min f Ÿ 0 and max f Ÿ 0. Now, (b a) min f Ÿ
b
a Ê ba
71. sin x Ÿ x for x
Ÿ0 Ê
72. sec x
0 Ê (b a) max f Ÿ 0 Ê
1
x#
#
Ê
'ab f(x) dx Ÿ 0.
x#
#‹
Exercise 69) since [0ß 1] is contained in ˆ 1# ß 1# ‰ Ê
#
Ê
#
0 on ˆ 1# ß 1# ‰ Ê
'0" ’sec x Š1 x# ‹“ dx
#
'0"sec x dx '0" Š1 x# ‹ dx
'0" sec x dx '0" 1 dx "# '0" x# dx
#
Ê
'0" sec x dx
0 Ê
$
(1 0) "# Š 13 ‹ Ê
'ab f(x) dx is a constant K. Thus'ab av(f) dx œ 'ab K dx
'ab av(f) dx œ (b a)K œ (b a) † b " a 'ab f(x) dx œ 'ab f(x) dx.
73. Yes, for the following reasons: av(f) œ
0 (see
'0" sec x dx
Thus a lower bound is 76 .
œ K(b a) Ê
Then
0 Ê
#
on ˆ 1# ß 1# ‰ Ê sec x Š1 '0" Š1 x# ‹ dx
'ab f(x) dx Ÿ (b a) max f.
'0" (sin x x) dx Ÿ 0 (see Exercise 70) Ê '0" sin x dx '0" x dx
'0" sin x dx Ÿ Š 1# 0# ‹ Ê '0" sin x dx Ÿ "# . Thus an upper bound is "# .
0 Ê sin x x Ÿ 0 for x
'0" sin x dx Ÿ '0" x dx
'ab f(x) dx Ÿ (b a) max f.
"
ba
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'0" sec x dx
7
6.
312
Chapter 5 Integration
74. All three rules hold. The reasons: On any interval [aß b] on which f and g are integrable, we have:
(a) av(f g) œ
"
ba
'ab [f(x) g(x)] dx œ b " a ”'ab f(x) dx 'ab g(x) dx• œ b " a 'ab f(x) dx b " a 'ab g(x) dx
œ av(f) av(g)
(b) av(kf) œ
(c) av(f) œ
"
ba
"
ba
'ab kf(x) dx œ b " a ”k 'ab f(x) dx• œ k ” b " a 'ab f(x) dx• œ k av(f)
'ab f(x) dx Ÿ b " a 'ab g(x) dx since f(x) Ÿ g(x) on [aß b], and b " a 'ab g(x) dx œ av(g).
Therefore, av(f) Ÿ av(g).
ba
n and let ck be the right
n ab a b
× and ck œ a kabn ab .
n
75. Consider the partition P that subdivides the interval Òa, bÓ into n subintervals of width ˜x œ
endpoint of each subinterval. So the partition is P œ Öa, a n
n
kœ"
kœ"
b
We get the Riemann sum ! fack b˜x œ ! c †
this expression remains cab ab. Thus,
ba
n
œ
ba
n ,
n
c ab a b !
"
n
kœ"
a
œ
#a b a b
,
n
c ab a b
n
...,a
† n œ cab ab. As n Ä _ and mPm Ä !
'a c dx œ cab ab.
76. Consider the partition P that subdivides the interval Òa, bÓ into n subintervals of width ˜x œ
endpoint of each subinterval. So the partition is P œ
n
n
We get the Riemann sum ! fack b˜x œ ! ck# ˆ b n a ‰
œ
n
ba ! #
a
n Œ
kœ"
kœ"
n
#a a b a b !
k
n
kœ"
œ ab aba# aab ab# †
n"
n
kœ"
n
ab a b # ! #
k n#
kœ"
ab a b $
'
†
œ
b a
n
ba
n
and let ck be the right
Öa, a b n a , a #abn ab , . . ., a nabn ab × and ck œ a kabn ab .
n
n
#
#
#
œ b n a ! Ša kabn ab ‹ œ bn a ! Ša# #akabn ab k abn# ab ‹
kœ"
kœ"
† na# an "ba#n "b
n#
#a a b a b#
n#
†
n a n "b
#
ab a b $
n$
†
nan "ba#n "b
'
$
"
" n"
ab ab$ # n n#
†
" '
"
ab a b $
†#
'
b
$
$
x# dx œ b$ a$ .
a
œ ab aba# aab ab# †
As n Ä _ and mPm Ä ! this expression has value ab aba# aab ab# † " œ ba# a$ ab# #a# b a$ "$ ab$ $b# a $ba# a$ b œ
b$
$
a$
$.
Thus,
'
77. (a) U œ max" ?x max# ?x á maxn ?x where max" œ f(x" ), max# œ f(x# ), á , maxn œ f(xn ) since f is
increasing on [aß b]; L œ min" ?x min# ?x á minn ?x where min" œ f(x! ), min# œ f(x" ), á ,
minn œ f(xnc1 ) since f is increasing on [aß b]. Therefore
U L œ (max" min" ) ?x (max# min# ) ?x á (maxn minn ) ?x
œ (f(x" ) f(x! )) ?x (f(x# ) f(x" ))?x á (f(xn ) f(xnc1 )) ?x œ (f(xn ) f(x! )) ?x œ (f(b) f(a)) ?x.
(b) U œ max" ?x" max# ?x# á maxn ?xn where max" œ f(x" ), max# œ f(x# ), á , maxn œ f(xn ) since f
is increasing on[aß b]; L œ min" ?x" min# ?x# á minn ?xn where
min" œ f(x! ), min# œ f(x" ), á , minn œ f(xnc1 ) since f is increasing on [aß b]. Therefore
U L œ (max" min" ) ?x" (max# min# ) ?x# á (maxn minn ) ?xn
œ (f(x" ) f(x! )) ?x" (f(x# ) f(x" ))?x# á (f(xn ) f(xnc1 )) ?xn
Ÿ (f(x" ) f(x! )) ?xmax (f(x# ) f(x" )) ?xmax á (f(xn ) f(xnc1 )) ?xmax . Then
U L Ÿ (f(xn ) f(x! )) ?xmax œ (f(b) f(a)) ?xmax œ kf(b) f(a)k ?xmax since f(b) f(a). Thus
lim (U L) œ lim (f(b) f(a)) ?xmax œ 0, since ?xmax œ lPl .
lPl Ä 0
lPl Ä 0
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Section 5.3 The Definite Integral
313
78. (a) U œ max" ?x max# ?x á maxn ?x where
max" œ f(x! ), max# œ f(x" ), á , maxn œ f(xnc" )
since f is decreasing on [aß b];
L œ min" ?x min# ?x á minn ?x where
min" œ f(x" ), min# œ f(x# )ß á , minn œ f(xn )
since f is decreasing on [aß b]. Therefore
U L œ (max" min" ) ?x (max# min# ) ?x
á (maxn minn ) ?x
œ (f(x! ) f(x" )) ?x (f(x" ) f(x# ))?x
á (f(xn" ) f(xn )) ?x œ (f(x! ) f(xn )) ?x
œ (f(a) f(b)) ?x.
(b) U œ max" ?x" max# ?x# á maxn ?xn where max" œ f(x! ), max# œ f(x" ), á , maxn œ f(xn" ) since
f is decreasing on[aß b]; L œ min" ?x" min# ?x# á minn ?xn where
min" œ f(x" ), min# œ f(x# ), á , minn œ f(xn ) since f is decreasing on [aß b]. Therefore
U L œ (max" min" ) ?x" (max# min# ) ?x# á (maxn minn ) ?xn
œ (f(x! ) f(x" )) ?x" (f(x" ) f(x# ))?x# á (f(xn" ) f(xn )) ?xn
Ÿ (f(x! ) f(xn )) ?xmax œ (f(a) f(b) ?xmax œ kf(b) f(a)k ?xmax since f(b) Ÿ f(a). Thus
lim (U L) œ lim kf(b) f(a)k ?xmax œ 0, since ?xmax œ lPl .
lPl Ä 0
lPl Ä 0
79. (a) Partition 0ß 1# ‘ into n subintervals, each of length ?x œ
x# œ 2?x, á , xn œ n?x œ
1
#.
1
#n
with points x! œ 0, x" œ ?x,
Since sin x is increasing on 0ß 1# ‘ , the upper sum U is the sum of the areas
of the circumscribed rectangles of areas f(x" ) ?x œ (sin ?x)?x, f(x# ) ?x œ (sin 2?x) ?x, á , f(xn ) ?x
œ (sin n?x) ?x. Then U œ (sin ?x sin 2?x á sin n?x) ?x œ ”
œ”
1 cos ˆˆn " ‰ 1 ‰
cos 4n
1
# 2n
1
• ˆ #n ‰
# sin 4n
'!
1 cos ˆ 1 1 ‰‰
1 ˆcos 4n
#
4n
1
4n sin 4n
œ
1 cos ˆ 1 1 ‰
cos 4n
#
4n
sin 1
Š 14n ‹
4n
1Î#
(b) The area is
œ
cos ?#x cosˆ ˆn #" ‰ ?x‰
• ?x
# sin ?#x
sin x dx œ n lim
Ä_
1 cos ˆ 1 1 ‰
cos 4n
#
4n
sin 1
Š 14n ‹
œ
1 cos 1#
1
œ 1.
4n
n
80. (a) The area of the shaded region is !˜xi † mi which is equal to L.
iœ"
n
(b) The area of the shaded region is !˜xi † Mi which is equal to U.
iœ"
(c) The area of the shaded region is the difference in the areas of the shaded regions shown in the second part of the figure
and the first part of the figure. Thus this area is U L.
n
n
iœ"
iœ"
81. By Exercise 80, U L œ !˜xi † Mi !˜xi † mi where Mi œ maxÖfaxb on the ith subinterval× and
n
n
iœ"
iœ"
mi œ minÖfaxb on the ith subinterval×. Thus U L œ !aMi mi b˜xi !% † ˜xi provided ˜xi $ for each
n
n
iœ"
iœ"
i œ "ß Þ Þ Þ , n. Since !% † ˜xi œ % !˜xi œ %ab ab the result, U L %ab ab follows.
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314
Chapter 5 Integration
82. The car drove the first 150 miles in 5 hours and the
second 150 miles in 3 hours, which means it drove 300
miles in 8 hours, for an average of 300
8 mi/hr
œ 37.5 mi/hr. In terms of average values of functions,
the function whose average value we seek is
30, 0 Ÿ t Ÿ 5
v(t) œ œ
, and the average value is
50, 5 1 Ÿ 8
(30)(5) (50)(3)
8
œ 37.5.
83-88. Example CAS commands:
Maple:
with( plots );
with( Student[Calculus1] );
f := x -> 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]
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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 ‹
œ
133
4
œ8
%
œ Š 24 2# 3(2)‹ Š (42) (2)# 3(2)‹ œ 12
%
"
$
ˆx# Èx‰ dx œ ’ x3 23 x$Î# “ œ ˆ "3 23 ‰ 0 œ 1
!
0
0
#
4#
4‹
œ Š 3(4)
# x# 3x“
2
5.
5
&
x$Î# dx œ 25 x&Î# ‘ ! œ
32
1
(5)&Î# 0 œ 2(5)$Î# œ 10È5
$#
'cc x2
2
2
5
x'Î& dx œ 5x"Î& ‘ " œ ˆ #5 ‰ (5) œ
1
8.
%
x%
16 “ !
#
dx œ ’ 3x# 2
4.
œ Š5(4) #
dx œ
5
#
'cc 2x# dx œ c2x" d "
ˆ 2 ‰ ˆ 2 ‰
# œ 1 # œ 1
1
2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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
18.
'cÎ Î% ˆ4 sec# t t1 ‰ dt œ 'Î Î% a4 sec# t 1t# b dt œ 4 tan t 1t ‘ 11Î%Î$
1Î#
1Î#
$
a8y# sin yb dy œ ’ 8y3 cos y“
1Î#
1Î#
1
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
'
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 È2 %È8 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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
sin 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%
1Î#
d
dx
ˆsin Èx‰ œ cos Èx ˆ "# x"Î# ‰
sin x
Èu du œ
t%
!
tan )
!
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
Œ'
œ sin$ x 1 Ê
cos Èx
2È x
'
t%
!
t%
u"Î# du œ 23 u$Î# ‘ ! œ
2
3
at% b
$Î#
0œ
2 '
3 t
Ê
d
dt
Œ'
Œ
'
t%
!
Èu du œ
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)
'
Œ
tan )
!
sec# y dy œ asec# (tan ))b ˆ dd) (tan ))‰ œ asec# (tan ))b sec# )
31. y œ
'
33. y œ
'È sin t# dt œ '
34. y œ
'
35. y œ
'
36. y œ
'
x
!
È1 t# dt Ê
Èx
!
x
x#
cos Èt dt Ê
sin x
dt
È1 t#
!
!
tan x
dt
1 t#
, kxk Ê
dy
dx
œ È1 x#
dy
dx
!
!
0#
#‹
cos x dx œ [sin x]!
cos Èx
2È x
œ
(b)
1Î#
%
#
%
sin 0 œ 1
Èx
'
'
!
dy
dx
1
#
sin t# dt Ê
32. y œ
dy
dx
'
1
x
"
t
dt Ê
dy
dx
œ
"
x
,x0
#
d ˆÈ ‰‰
œ Šsin ˆÈx‰ ‹ ˆ dx
x œ (sin x) ˆ "# x"Î# ‰ œ 2sinÈxx
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
cos x
kcos xk
œ
cos x
cos x
œ 1 since kxk "
d
‰ ˆ dx
œ ˆ 1 tan
(tan x)‰ œ ˆ sec"# x ‰ asec# xb œ 1
#x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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 œ
'c! ˆx"Î$ x‰dx '!" ˆx"Î$ x‰dx '") ˆx"Î$ x‰dx
"
%Î$
’ 34
x
!
x#
# “ "
œ ’Š 34 (0)%Î$ ’ 34 x%Î$ 0#
#‹
"
x#
# “!
’ 43 x%Î$ (1)#
# ‹“
Š 34 (1)%Î$ ’Š 34 (1)%Î$ 1#
#‹
Š 34 (0)%Î$ 0#
# ‹“
’Š 34 (8)%Î$ 8#
#‹
Š 34 (1)%Î$ 1#
# ‹“
œ
"
4
"
4
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
!
1Î4
sec ) tan ) d) œ [sec )]!1Î%
œ (sec 0) ˆsec ˆ 14 ‰‰ œ È2 1. Therefore the area of the shaded region on 14 ß !‘ is
1È2
4
On 0ß 14 ‘ : The area of the rectangle bounded by ) œ 14 , ) œ 0, y œ È2, and y œ 0 is È2 ˆ 14 ‰ œ
under the curve y œ sec ) tan ) is
of the shaded region on !ß 14 ‘ is
È
'
1Î4
!
1È2
4
1Î%
sec ) tan ) d) œ [sec )]!
œ sec
1
4
ŠÈ2 1‹ .
1È2
4
. The area
sec 0 œ È2 1. Therefore the area
ŠÈ2 1‹ . Thus, the area of the total shaded region is
È
1È2
#
Š 1 4 2 È2 1‹ Š 1 4 2 È2 1‹ œ
.
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
!
1Î4
"
$
œ
dt 3 œ 0 3 œ 3 Ê (d) is a solution to this problem.
dy
dx
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
dt 3 Ê
"
x
1
1
Thus, the total
. Therefore the area of the shaded region is ˆ2 1# ‰ '
"
1 t
$
5
3
47. y œ
x
$
!
2
3
'
!
. The
!
ˆ 1‰
sec# t dt œ [tan t]
1Î% œ tan 0 tan 4 œ 1. The area
'! a1 t# b dt œ ’t t3 “ " œ Š1 13 ‹ Š0 03 ‹ œ 32 .
area under the curves on 14 ß "‘ is 1 1
#
'cc sec t dt 4 œ 0 4 œ 4
1
1
!
5
3
œ
"
3
1
#
.
Ê (c) is a solution to this problem.
sec t dt 4 œ 0 4 œ 4 Ê (b) is a solution to this problem.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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319
320
Chapter 5 Integration
50. y œ
'
51. y œ
'
53. s œ
'
x
"
" t
"
"
t
dt 3 œ 0 3 œ 3 Ê (a) is a solution to this problem.
'
'c ÎÎ
b 2
b 2
$
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
4hx$
3b# “ bÎ2
ˆh ˆ 4h
‰ # ‰ dx œ ’hx b# x
œ ˆ bh
# !
"
54. v œ
4h ˆ b# ‰
3b#
$
'
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 *.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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
' x %t# dt
! t "
x$
xÄ!
x
66.
1
x#
"
k
x#
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
Ê f w (x) œ 1 (x9 1) œ
9
x 2
Ê f w (1) œ 3; f(1) œ 2 '#"" 1 9 t dt œ 2 0 œ 2;
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.
1 Îk
cos kx‘ !
%
L(x) œ 3(x 1) f(1) œ 3(x 1) 2 œ 3x 5
g(x) œ 3 '
sin kx dx œ 2
k
xÄ!
f(t) dt œ x cos 1x Ê f(x) œ
'# " 1 9 t dt
!
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
70. The limit is 3x#
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Chapter 5 Integration
71-74. Example CAS commands:
Maple:
with( plots );
f := x -> x^3-4*x^2+3*x;
a := 0;
b := 4;
F := unapply( int(f(t),t=a..x), x );
# (a)
p1 := plot( [f(x),F(x)], x=a..b, legend=["y = f(x)","y = F(x)"], title="#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)<0, x );
df := D(f);
# (d)
p3 := plot( [df(x),F(x)], x=a..b, legend=["y = f '(x)","y = F(x)"], title="#71(d) (Section 5.4)" ):
p3;
q2 := solve( df(x)=0, x );
pts2 := [ seq( [x,F(x)], x=remove(has,evalf([q2]),I) ) ];
p4 := plot( pts2, style=point, color=blue, symbolsize=18, symbol=diamond, legend="(x,f(x)) where f '(x)=0" ):
display( [p3,p4], title="71(d) (Section 5.4)" );
75-78. Example CAS commands:
Maple:
a := 1;
u := 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)<0, x );
d2F := D(dF);
# (c)
solve( d2F(x)=0, x );
plot( F(x), x=-1..1, title="#75(d) (Section 5.4)" );
79.
Example CAS commands:
Maple:
f := `f`;
q1 := Diff( Int( f(t), t=a..u(x) ), x );
d1 := value( q1 );
80.
Example CAS commands:
Maple:
f := `f`;
q2 := Diff( Int( f(t), t=a..u(x) ), x,x );
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 5.5 Indefinite Integrals and the Substitution Rule
value( q2 );
71-80. Example CAS commands:
Mathematica: (assigned function and values for a, and b may vary)
For transcendental functions the FindRoot is needed instead of the Solve command.
The Map command executes FindRoot over a set of initial guesses
Initial guesses will vary as the functions vary.
Clear[x, f, F]
{a, b}= {0, 21}; f[x_] = Sin[2x] Cos[x/3]
F[x_] = Integrate[f[t], {t, a, x}]
Plot[{f[x], F[x]},{x, a, b}]
x/.Map[FindRoot[F'[x]==0, {x, #}] &,{2, 3, 5, 6}]
x/.Map[FindRoot[f'[x]==0, {x, #}] &,{1, 2, 4, 5, 6}]
Slightly alter above commands for 75 - 80.
Clear[x, f, F, u]
a=0; f[x_] = x2 2x 3
u[x_] = 1 x2
F[x_] = Integrate[f[t], {t, a, u(x)}]
x/.Map[FindRoot[F'[x]==0,{x, #}] &,{1, 2, 3, 4}]
x/.Map[FindRoot[F''[x]==0,{x,#}] &,{1, 2, 3, 4}]
After determining an appropriate value for b, the following can be entered
b = 4;
Plot[{F[x], {x, a, b}]
5.5 INDEFINTE INTEGRALS AND THE SUBSTITUTION RULE
"
3
1. Let u œ 3x Ê du œ 3 dx Ê
'
sin 3x dx œ '
"
3
du œ dx
sin u du œ 3" cos u C œ 3" cos 3x C
"
4
2. Let u œ 2x# Ê du œ 4x dx Ê
'
x sin a2x b dx œ '
"
4
#
sin u du œ 4" cos u C œ 4" cos 2x# C
"
#
3. Let u œ 2t Ê du œ 2 dt Ê
'
sec 2t tan 2t dt œ '
"
#
'
ˆ1 cos
du œ dt
sec u tan u du œ
4. Let u œ 1 cos 2t Ê du œ
t ‰#
#
"
#
ˆsin #t ‰ dt œ ' 2u# du œ
28(7x 2)& dx œ '
"
7
"
7
x$ ax% 1b dx œ '
#
"
4
2
3
"
#
sec u C œ
u$ C œ
t
2
2
3
sec 2t C
dt
ˆ1 cos #t ‰$ C
du œ dx
(28)u& du œ ' 4u& du œ u% C œ (7x 2)% C
6. Let u œ x% " Ê du œ 4x$ dx Ê
'
"
#
dt Ê 2 du œ sin
t
#
sin
5. Let u œ 7x 2 Ê du œ 7 dx Ê
'
du œ x dx
u# du œ
u$
1#
"
4
du œ x$ dx
Cœ
"
1#
$
ax% 1b C
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324
Chapter 5 Integration
7. Let u œ 1 r$ Ê du œ 3r# dr Ê 3 du œ 9r# dr
' È9r
#
dr
1 r$
œ ' 3u"Î# du œ 3(2)u"Î# C œ 6 a1 r$ b
"Î#
C
8. Let u œ y% 4y# 1 Ê du œ a4y$ 8yb dy Ê 3 du œ 12 ay$ 2yb dy
'
12 ay% 4y# 1b ay$ 2yb dy œ ' 3u# du œ u$ C œ ay% 4y# 1b C
#
$
9. Let u œ x$Î# 1 Ê du œ
'
x"Î# dx Ê
Èx sin# ˆx$Î# 1‰ dx œ '
10. Let u œ "x Ê du œ
'
3
#
"
x#
2
3
2
3
du œ Èx dx
sin# u du œ
2
3
"
4
ˆ #u sin 2u‰ C œ
"
3
ˆx$Î# 1‰ "
6
sin ˆ2x$Î# 2‰ C
dx
cos# ˆ x" ‰ dx œ ' cos# aub du œ
"
œ 2x
"4 sin ˆ 2x ‰ C
"
x#
'
cos# aub du œ ˆ u# "
4
"
sin 2u‰ C œ 2x
"
4
sin ˆ x2 ‰ C
11. (a) Let u œ cot 2) Ê du œ 2 csc# 2) d) Ê "# du œ csc# 2) d)
'
csc# 2) cot 2) d) œ '
"
#
#
#
u du œ "# Š u# ‹ C œ u4 C œ 4" cot# 2) C
(b) Let u œ csc 2) Ê du œ 2 csc 2) cot 2) d) Ê "# du œ csc 2) cot 2) d)
'
csc# 2) cot 2) d) œ ' "# u du œ "# Š u# ‹ C œ u4 C œ 4" csc# 2) C
#
"
5
12. (a) Let u œ 5x 8 Ê du œ 5 dx Ê
'
dx
È5x8
œ'
"
5
Š È"u ‹ du œ
"
#
(b) Let u œ È5x 8 Ê du œ
'
dx
È5x8
œ'
du œ
2
5
2
5
'
"
5
#
du œ dx
u"Î# du œ
"
5
ˆ2u"Î# ‰ C œ
(5x 8)"Î# (5) dx Ê
uCœ
2
5
2
5
du œ
2
5
u"Î# C œ
2
5
È5x 8 C
dx
È5x8
È5x 8 C
13. Let u œ 3 2s Ê du œ 2 ds Ê "# du œ ds
'
È3 2s ds œ ' Èu ˆ " du‰ œ " ' u"Î# du œ ˆ " ‰ ˆ 2 u$Î# ‰ C œ " (3 2s)$Î# C
#
#
#
3
3
"
#
14. Let u œ 2x 1 Ê du œ 2 dx Ê
'
(2x 1) dx œ ' u
$
$
ˆ "#
du‰ œ
15. Let u œ 5s 4 Ê du œ 5 ds Ê
'
"
È5s 4
ds œ '
"
Èu
ˆ 5" du‰ œ
"
5
"
#
'
"
5
'
du œ dx
%
u$ du œ ˆ "# ‰ Š u4 ‹ C œ
"
8
(2x 1)% C
du œ ds
u"Î# du œ ˆ 5" ‰ ˆ2u"Î# ‰ C œ
2
5
È5s 4 C
16. Let u œ 2 x Ê du œ dx Ê du œ dx
'
3
(2 x)#
dx œ '
3(du)
u#
œ 3 ' u# du œ 3 Š u1 ‹ C œ
"
3
2 x
C
17. Let u œ 1 )# Ê du œ 2) d) Ê "# du œ ) d)
'
%
&Î%
) È1 )# d) œ ' %Èu ˆ "# du‰ œ "# ' u"Î% du œ ˆ "# ‰ ˆ 45 u&Î% ‰ C œ 25 a1 )# b C
18. Let u œ )# 1 Ê du œ 2) d) Ê 4 du œ 8) d)
' 8) $È)# 1 d) œ ' $Èu (4 du) œ 4 ' u"Î$ du œ 4 ˆ 3 u%Î$ ‰ C œ 3 a)# 1b%Î$ C
4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 5.5 Indefinite Integrals and the Substitution Rule
19. Let u œ 7 3y# Ê du œ 6y dy Ê "# du œ 3y dy
'
$Î#
3yÈ7 3y# dy œ ' Èu ˆ "# du‰ œ "# ' u"Î# du œ ˆ "# ‰ ˆ 23 u$Î# ‰ C œ 3" a7 3y# b C
20. Let u œ 2y# 1 Ê du œ 4y dy
'
4y dy
È2y# 1
œ'
"
Èu
du œ ' u"Î# du œ 2u"Î# C œ 2È2y# 1 C
"
2È x
21. Let u œ 1 Èx Ê du œ
'
dx œ '
"
È x ˆ" È x ‰ #
œ 2u C œ
2 du
u#
"
2È x
22. Let u œ 1 Èx Ê du œ
'
ˆ1 È x ‰ $
Èx
dx Ê 2 du œ
"
3
cos (3z 4) dz œ ' (cos
u) ˆ "3
sin (8z 5) dz œ ' (sin
u) ˆ "8
'
"
3
ˆ1 Èx‰% C
"
3
' cos u du œ 3" sin u C œ 3" sin (3z 4) C
du œ dz
du‰ œ
25. Let u œ 3x 2 Ê du œ 3 dx Ê
"
#
dx
du œ dz
du‰ œ
"
8
dx
C
"
Èx
dx œ ' u$ (2 du) œ 2 ˆ 4" u% ‰ C œ
24. Let u œ 8z 5 Ê du œ 8 dz Ê
'
2
1 È x
dx Ê 2 du œ
23. Let u œ 3z 4 Ê du œ 3 dz Ê
'
"
Èx
"
8
'
"
8
sin u du œ
(cos u) C œ 8" cos (8z 5) C
du œ dx
sec# (3x 2) dx œ ' asec# ub ˆ "3 du‰ œ
"
3
'
"
3
sec# u du œ
tan u C œ
"
3
tan (3x 2) C
26. Let u œ tan x Ê du œ sec# x dx
'
tan# x sec# x dx œ ' u# du œ
27. Let u œ sin ˆ x3 ‰ Ê du œ
'
r$
18
1 Ê du œ
"
#
r#
6
$
cos ˆ x3 ‰ dx Ê 3 du œ cos ˆ x3 ‰ dx
"
#
sin' ˆ x3 ‰ C
sec# ˆ x# ‰ dx Ê 2 du œ sec# ˆ x# ‰ dx
"
4
tan) ˆ x# ‰ C
dr Ê 6 du œ r# dr
r % Š7 r&
10
$
r&
10 ‹
'
$
'
Ê du œ "# r% dr Ê 2 du œ r% dr
dr œ ' u$ (2 du) œ 2 ' u$ du œ 2 Š u4 ‹ C œ "# Š7 %
31. Let u œ x$Î# 1 Ê du œ
'
tan$ x C
r
r
r# Š 18
1‹ dr œ ' u& (6 du) œ 6 ' u& du œ 6 Š u6 ‹ C œ Š 18
1‹ C
&
30. Let u œ 7 '
"
3
tan( ˆ x# ‰ sec# ˆ x# ‰ dx œ ' u( (2 du) œ 2 ˆ 8" u) ‰ C œ
29. Let u œ
'
u$ C œ
sin& ˆ x3 ‰ cos ˆ x3 ‰ dx œ ' u& (3 du) œ 3 ˆ 6" u' ‰ C œ
28. Let u œ tan ˆ x# ‰ Ê du œ
'
"
3
"
3
3
#
x"Î# dx Ê
2
3
r&
10 ‹
%
C
du œ x"Î# dx
x"Î# sin ˆx$Î# 1‰ dx œ ' (sin u) ˆ 23 du‰ œ
2
3
'
sin u du œ
2
3
(cos u) C œ 23 cos ˆx$Î# 1‰ C
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325
326
Chapter 5 Integration
32. Let u œ x%Î$ 8 Ê du œ
'
4
3
x"Î$ dx Ê
3
4
du œ x"Î$ dx
x"Î$ sin ˆx%Î$ 8‰ dx œ ' (sin u) ˆ 34 du‰ œ
3
4
'
sin u du œ
3
4
(cos u) C œ 34 cos ˆx%Î$ 8‰ C
33. Let u œ sec ˆv 1# ‰ Ê du œ sec ˆv 1# ‰ tan ˆv 1# ‰ dv
'
sec ˆv 1# ‰ tan ˆv 1# ‰ dv œ ' du œ u C œ sec ˆv 1# ‰ C
34. Let u œ csc ˆ v # 1 ‰ Ê du œ "# csc ˆ v # 1 ‰ cot ˆ v # 1 ‰ dv Ê 2 du œ csc ˆ v # 1 ‰ cot ˆ v # 1 ‰ dv
'
csc ˆ v # 1 ‰ cot ˆ v # 1 ‰ dv œ ' 2 du œ 2u C œ 2 csc ˆ v # 1 ‰ C
35. Let u œ cos (2t 1) Ê du œ 2 sin (2t 1) dt Ê "# du œ sin (2t 1) dt
'
sin (2t 1)
cos# (2t 1)
dt œ ' #"
du
u#
œ
"
#u
Cœ
"
# cos (2t 1)
C
36. Let u œ 2 sin t Ê du œ cos t dt
'
dt œ '
6 cos t
(2 sin t)$
6
u$
du œ 6 ' u$ du œ 6 Š u# ‹ C œ 3(2 sin t)# C
#
37. Let u œ cot y Ê du œ csc# y dy Ê du œ csc# y dy
'
Ècot y csc# y dy œ ' Èu (du) œ ' u"Î# du œ 23 u$Î# C œ 23 (cot y)$Î# C œ 23 acot$ yb"Î# C
38. Let u œ sec z Ê du œ sec z tan z dz
'
sec z tan z
Èsec z
39. Let u œ
'
"
t#
"
t
dz œ '
"
Èu
du œ ' u"Î# du œ 2u"Î# C œ 2Èsec z C
"
Èt
"
)#
" "Î#
# t
dt Ê 2 du œ
sin
"
)
"
)
cos
Ê du œ ˆcos ") ‰ ˆ )"# ‰ d) Ê du œ
"
)
cos È)
È) sin# È)
"
)#
cos
d) œ ' u du œ #" u# C œ #" sin#
42. Let u œ csc È) Ê du œ Šcsc È) cot È)‹ Š
'
"
Èt
dt
cos ˆÈt 3‰ dt œ ' (cos u)(2 du) œ 2 ' cos u du œ 2 sin u C œ 2 sin ˆÈt 3‰ C
41. Let u œ sin
'
dt
cos ˆ "t 1‰ dt œ ' (cos u)(du) œ ' cos u du œ sin u C œ sin ˆ "t 1‰ C
40. Let u œ Èt 3 œ t"Î# 3 Ê du œ
'
"
t#
1 œ t" 1 Ê du œ t# dt Ê du œ
d) œ '
"
È)
"
‹
#È )
"
)
"
)
d)
C
d) Ê 2 du œ
"
È)
cot È) csc È) d)
cot È) csc È) d) œ ' 2 du œ 2u C œ 2 csc È) C œ 2
sin È)
C
43. Let u œ s$ 2s# 5s 5 Ê du œ a3s# 4s 5b ds
' as$ 2s# 5s 5b a3s# 4s 5b ds œ '
44.
u du œ
u#
#
Let u œ )% 2)# 8) 2 Ê du œ a4)$ 4) 8b d) Ê
'
as$ 2s# 5s 5b
#
Cœ
a)% 2)# 8) 2b a)$ ) 2b d) œ ' u ˆ "4 du‰ œ
"
4
'
"
4
#
C
du œ a)$ ) 2b d)
u du œ
"
4
#
Š u# ‹ C œ
ˆ) % 2 ) # 8 ) 2 ‰ #
8
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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C
Section 5.5 Indefinite Integrals and the Substitution Rule
"
4
45. Let u œ 1 t% Ê du œ 4t$ dt Ê
'
t$ a1 t% b dt œ ' u$ ˆ "4 du‰ œ
$
46. Let u œ 1 '
"
x
Ê du œ
É x x& 1 dx œ '
"
x#
"
x#
"
4
du œ t$ dt
ˆ 4" u% ‰ C œ
"
16
%
a 1 t% b C
dx
É x x 1 dx œ '
"
x#
"
x
É1 dx œ ' Èu du œ ' u"Î# du œ
2
3
u$Î# C œ
2
3
ˆ1 "x ‰$Î# C
47. Let u œ x# ". Then du œ #xdx and "# du œ xdx and x# œ u ". Thus ' x$ Èx# " dx œ ' au "b "# Èu du
œ
"
#
' au$Î# u"Î# bdu œ "# ’ #& u&Î# #$ u$Î# “ C œ "& u&Î# "$ u$Î# C œ "& ax# "b&Î# "$ ax# "b$Î# C
48. Let u œ x$ " Ê du œ $x# dx and x$ œ u ". So ' $B& Èx$ " dx œ ' au "bÈu du œ ' au$Î# u"Î# bdu
œ #& u&Î# #$ u$Î# C œ #& ax$ "b
&Î#
#$ ax$ "b
$Î#
C
49. (a) Let u œ tan x Ê du œ sec# x dx; v œ u$ Ê dv œ 3u# du Ê 6 dv œ 18u# du; w œ 2 v Ê dw œ dv
'
18 tan# x sec# x
dx œ
a2 tan$ xb#
6
œ 2 u$ C
$
'
'
18u#
du œ
a 2 u $ b#
6
2 tan$ x C
#
#
œ
6 dv
(2 v)#
œ'
6 dw
w#
œ 6 ' w# dw œ 6w" C œ # 6 v C
(b) Let u œ tan x Ê du œ 3 tan x sec x dx Ê 6 du œ 18 tan# x sec# x dx; v œ 2 u Ê dv œ du
'
18 tan# x sec# x
a2 tan$ xb#
dx œ '
œ'
6 du
(2 u)#
6 dv
v#
6
œ v6 C œ 2 6 u C œ # tan
$x C
(c) Let u œ 2 tan$ x Ê du œ 3 tan# x sec# x dx Ê 6 du œ 18 tan# x sec# x dx
'
18 tan# x sec# x
a2 tan$ xb#
dx œ '
6 du
u#
6
œ u6 C œ 2 tan
$x C
50. (a) Let u œ x 1 Ê du œ dx; v œ sin u Ê dv œ cos u du; w œ 1 v# Ê dw œ 2v dv Ê
'
"
#
Èw dw œ
"
3
w$Î# C œ
"
3
a 1 v# b
$Î#
Cœ
"
3
a1 sin# ub
$Î#
#
Cœ
(b) Let u œ sin (x 1) Ê du œ cos (x 1) dx; v œ 1 u Ê dv œ 2u du Ê
'
È1 sin# (x 1) sin (x 1) cos (x 1) dx œ ' u È1 u# du œ '
œ ˆ "# ˆ 23 ‰ v$Î# ‰ C œ
"
3
v$Î# C œ
"
3
a1 u # b
$Î#
Cœ
"
3
(c) Let u œ 1 sin (x 1) Ê du œ 2 sin (x 1) cos (x 1) dx Ê
'
È1 sin# (x 1) sin (x 1) cos (x 1) dx œ '
œ
"
3
a1 sin# (x 1)b
$Î#
"
6
dr œ ' Š
cos Èu
ˆ"
È u ‹ 1#
sin È)
É) cos$ È)
d) œ
'
$Î#
dv œ u du
Èv dv œ '
$Î#
"
#
v"Î# dv
C
"
#
du œ sin (x 1) cos (x 1) dx
"
#
u"Î# du œ
"
#
ˆ 23 u$Î# ‰ C
"
1#
du œ (2r 1) dr; v œ Èu Ê dv œ
"
#È u
du Ê
du‰ œ ' (cos v) ˆ 6" dv‰ œ
"
6
sin v C œ
"
6
sin Èu C
sin È3(2r 1)# 6 C
52. Let u œ cos È) Ê du œ Šsin È)‹ Š
'
a1 sin# (x 1)b
du
(2r 1) cos È3(2r 1)# 6
È3(2r 1)# 6
œ
Èu du œ '
"
3
C
51. Let u œ 3(2r 1)# 6 Ê du œ 6(2r 1)(2) dr Ê
"
1#Èu
"
#
"
#
"
#
a1 sin# (x 1)b
#
'
dw œ v dv
È1 sin# (x 1) sin (x 1) cos (x 1) dx œ ' È1 sin# u sin u cos u du œ ' vÈ1 v# dv
œ'
œ
"
#
sin È)
È) Écos$ È)
"
‹
#È )
d) œ '
d) Ê 2 du œ
2 du
u$Î#
sin È)
È)
d)
œ 2 ' u$Î# du œ 2 ˆ2u"Î# ‰ C œ
4
Èu
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
C
"
6
dv
C
327
328
Chapter 5 Integration
œ
4
Écos È)
C
53. Let u œ 3t# 1 Ê du œ 6t dt Ê 2 du œ 12t dt
s œ ' 12t a3t# 1b dt œ ' u$ (2 du) œ 2 ˆ "4 u% ‰ C œ
$
s œ 3 when t œ 1 Ê 3 œ
"
#
"
#
u% C œ
%
"
#
a3t# 1b C;
(3 1)% C Ê 3 œ 8 C Ê C œ 5 Ê s œ
"
#
%
a3t# 1b 5
54. Let u œ x# 8 Ê du œ 2x dx Ê 2 du œ 4x dx
y œ ' 4x ax# 8b
"Î$
dx œ ' u"Î$ (2 du) œ 2 ˆ 3# u#Î$ ‰ C œ 3u#Î$ C œ 3 ax# 8b
y œ 0 when x œ 0 Ê 0 œ 3(8)#Î$ C Ê C œ 12 Ê y œ 3 ax# 8b
55. Let u œ t 1
1#
#Î$
#Î$
C;
12
Ê du œ dt
s œ ' 8 sin# ˆt dt œ ' 8 sin# u du œ 8 ˆ u# "4 sin 2u‰ C œ 4 ˆt 11# ‰ 2 sin ˆ2t 16 ‰ C;
s œ 8 when t œ 0 Ê 8 œ 4 ˆ 11# ‰ 2 sin ˆ 16 ‰ C Ê C œ 8 13 1 œ 9 13
Ê s œ 4ˆt 11# ‰ 2 sin ˆ2t 16 ‰ 9 13 œ 4t 2 sin ˆ2t 16 ‰ 9
56. Let u œ
1
4
1‰
1#
) Ê du œ d)
r œ ' 3 cos# ˆ 14 )‰ d) œ ' 3 cos# u du œ 3 ˆ u# sin 2u‰ C œ 3# ˆ 14 )‰ 43 sin ˆ 1# 2)‰ C;
C Ê C œ 1# 43 Ê r œ 3# ˆ 14 )‰ 43 sin ˆ 1# 2)‰ 1# when ) œ 0 Ê 18 œ 381 43 sin 1#
Ê r œ 3# ) 34 sin ˆ 1# 2)‰ 18 43 Ê r œ
rœ
1
8
57. Let u œ 2t ds
dt
1
#
3
2
)
3
4
"
4
cos 2) 1
8
3
4
3
4
Ê du œ 2 dt Ê 2 du œ 4 dt
œ ' 4 sin ˆ2t 1# ‰ dt œ ' (sin u)(2 du) œ 2 cos u C" œ 2 cos ˆ2t 1# ‰ C" ;
at t œ 0 and
ds
dt
œ 100 we have 100 œ 2 cos ˆ 1# ‰ C" Ê C" œ 100 Ê
œ 2 cos ˆ2t 1# ‰ 100
ds
dt
Ê s œ ' ˆ2 cos ˆ2t 1# ‰ 100‰ dt œ ' (cos u 50) du œ sin u 50u C# œ sin ˆ2t 1# ‰ 50 ˆ2t 1# ‰ C# ;
at t œ 0 and s œ 0 we have 0 œ sin ˆ 1# ‰ 50 ˆ 1# ‰ C# Ê C# œ 1 251
Ê s œ sin ˆ2t 1# ‰ 100t 251 (1 251) Ê s œ sin ˆ2t 1# ‰ 100t 1
58. Let u œ tan 2x Ê du œ 2 sec# 2x dx Ê 2 du œ 4 sec# 2x dx; v œ 2x Ê dv œ 2 dx Ê
dy
dx
œ ' 4 sec# 2x tan 2x dx œ ' u(2 du) œ u# C" œ tan# 2x C" ;
at x œ 0 and
dy
dx
œ 4 we have 4 œ 0 C" Ê C" œ 4 Ê
Ê y œ ' asec# 2x 3b dx œ ' asec# v 3b ˆ "# dv‰ œ
at x œ 0 and y œ 1 we have 1 œ
"
#
dy
dx
"
#
"
#
dv œ dx
œ tan# 2x 4 œ asec# 2x 1b 4 œ sec# 2x 3
tan v 3# v C# œ
(0) 0 C# Ê C# œ 1 Ê y œ
"
#
"
#
tan 2x 3x C# ;
tan 2x 3x 1
59. Let u œ 2t Ê du œ 2 dt Ê 3 du œ 6 dt
s œ ' 6 sin 2t dt œ ' (sin u)(3 du) œ 3 cos u C œ 3 cos 2t C;
at t œ 0 and s œ 0 we have 0 œ 3 cos 0 C Ê C œ 3 Ê s œ 3 3 cos 2t Ê s ˆ 1# ‰ œ 3 3 cos (1) œ 6 m
60. Let u œ 1t Ê du œ 1 dt Ê 1 du œ 1# dt
v œ ' 1# cos 1t dt œ ' (cos u)(1 du) œ 1 sin u C" œ 1 sin (1t) C" ;
at t œ 0 and v œ 8 we have 8 œ 1(0) C" Ê C" œ 8 Ê v œ
ds
dt
œ 1 sin (1t) 8 Ê s œ ' (1 sin (1t) 8) dt
œ ' sin u du 8t C# œ cos (1t) 8t C# ; at t œ 0 and s œ 0 we have 0 œ 1 C# Ê C# œ 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 5.6 Substitution and Area Between Curves
329
Ê s œ 8t cos (1t) 1 Ê s(1) œ 8 cos 1 1 œ 10 m
61. All three integrations are correct. In each case, the derivative of the function on the right is the integrand on
the left, and each formula has an arbitrary constant for generating the remaining antiderivatives. Moreover,
sin# x C" œ 1 cos# x C" Ê C# œ 1 C" ; also cos# x C# œ cos#2x "# C# Ê C$ œ C# "# œ C" "# .
62. Both integrations are correct. In each case, the derivative of the function on the right is the integrand on the
left, and each formula has an arbitrary constant for generating the remaining antiderivatives. Moreover,
#
tan# x
sec# x1
ˆC "# ‰
C œ sec# x ðñò
# Cœ
#
a constant
63. (a) Š
"
60
"
‹
0
'
0
1Î60
"Î'!
Vmax sin 1201t dt œ 60 Vmax ˆ 120" 1 ‰ cos (1201t)‘ !
œ V#1max [1 1] œ 0
(b) Vmax œ È2 Vrms œ È2 (240) ¸ 339 volts
(c)
'
1Î60
0
aVmax b# sin# 1201t dt œ aVmax b#
aVmax b
#
œ
#
"Î'!
2401t‘ !
t ˆ 240" 1 ‰ sin
'
1Î60
0
œ
ˆ 1 cos# 2401t ‰ dt œ
aVmax b
#
#
aVmax b#
#
œ V#max
1 [cos 21 cos 0]
'
1Î60
0
(1 cos 2401t) dt
"
ˆ 60
ˆ 240" 1 ‰ sin (41)‰ ˆ0 ˆ #40" 1 ‰ sin (0)‰‘ œ
5.6 SUBSTITUTION AND AREA BETWEEN CURVES
1. (a) Let u œ y 1 Ê du œ dy; y œ 0 Ê u œ 1, y œ 3 Ê u œ 4
'
3
0
Èy 1 dy œ
'
4
1
%
u"Î# du œ 23 u$Î# ‘ " œ ˆ 23 ‰ (4)$Î# ˆ 23 ‰ (1)$Î# œ ˆ 23 ‰ (8) ˆ 23 ‰ (1) œ
14
3
(b) Use the same substitution for u as in part (a); y œ 1 Ê u œ 0, y œ 0 Ê u œ 1
'c Èy 1 dy œ '
0
1
1
0
"
u"Î# du œ 23 u$Î# ‘ ! œ ˆ 23 ‰ (1)$Î# 0 œ
2
3
2. (a) Let u œ 1 r# Ê du œ 2r dr Ê "# du œ r dr; r œ 0 Ê u œ 1, r œ 1 Ê u œ 0
'
1
0
r È1 r# dr œ
'
0
!
"# Èu du œ "3 u$Î# ‘ " œ 0 ˆ 3" ‰ (1)$Î# œ
1
"
3
(b) Use the same substitution for u as in part (a); r œ 1 Ê u œ 0, r œ 1 Ê u œ 0
'c r È1 r# dr œ '
1
1
0
0
"# Èu du œ 0
3. (a) Let u œ tan x Ê du œ sec# x dx; x œ 0 Ê u œ 0, x œ
'
1Î4
0
tan x sec# x dx œ
'
1
0
"
#
u du œ ’ u# “ œ
!
1#
#
0œ
1
4
Ê uœ1
"
#
(b) Use the same substitution as in part (a); x œ 14 Ê u œ 1, x œ 0 Ê u œ 0
'c Î
0
1 4
tan x sec# x dx œ
' u du œ ’ u# “ !
0
#
"
1
œ0
"
#
œ "#
4. (a) Let u œ cos x Ê du œ sin x dx Ê du œ sin x dx; x œ 0 Ê u œ 1, x œ 1 Ê u œ 1
'
1
0
3 cos# x sin x dx œ
$
$
' 3u# du œ cu$ d "
" œ (1) a(1) b œ 2
1
1
(b) Use the same substitution as in part (a); x œ 21 Ê u œ 1, x œ 31 Ê u œ 1
'
31
21
3 cos# x sin x dx œ
'
1
1
3u# du œ 2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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aVmax b#
1#0
330
Chapter 5 Integration
"
4
5. (a) u œ 1 t% Ê du œ 4t$ dt Ê
'
1
0
'
$
t$ a1 t% b dt œ
2
#
%
"
4
1
du œ t$ dt; t œ 0 Ê u œ 1, t œ 1 Ê u œ 2
2%
16
u
u$ du œ ’ 16
“ œ
"
1%
16
œ
15
16
(b) Use the same substitution as in part (a); t œ 1 Ê u œ 2, t œ 1 Ê u œ 2
'c
1
'
$
t$ a1 t% b dt œ
1
2
"
4
2
u$ du œ 0
"
#
6. (a) Let u œ t# 1 Ê du œ 2t dt Ê
'
È7
t at# 1b
0
"Î$
'
dt œ
8
)
"
#
1
du œ t dt; t œ 0 Ê u œ 1, t œ È7 Ê u œ 8
u"Î$ du œ ˆ "# ‰ ˆ 34 ‰ u%Î$ ‘ " œ ˆ 38 ‰ (8)%Î$ ˆ 38 ‰ (1)%Î$ œ
45
8
(b) Use the same substitution as in part (a); t œ È7 Ê u œ 8, t œ 0 Ê u œ 1
'cÈ
0
t at# 1b
7
"Î$
dt œ
'
1
"
8 #
"
#
7. (a) Let u œ 4 r# Ê du œ 2r dr Ê
' a4 5rr b
1
dr œ 5
# #
1
'
5
"
#
5
'
u"Î$ du œ 8
1
"
#
u"Î$ du œ 45
8
du œ r dr; r œ 1 Ê u œ 5, r œ 1 Ê u œ 5
u# du œ 0
(b) Use the same substitution as in part (a); r œ 0 Ê u œ 4, r œ 1 Ê u œ 5
'
1
5r
#
0 a4 r# b
dr œ 5
'
5
"
#
4
&
u# du œ 5 "# u" ‘ % œ 5 ˆ "# (5)" ‰ 5 ˆ "# (4)" ‰ œ
8. (a) Let u œ 1 v$Î# Ê du œ
'
1
10Èv
a1 v$Î# b
0
#
dv œ
'
2
"
u#
1
3
#
v"Î# dv Ê
ˆ 20
‰
3 du œ
'
20
3
20
3
2
1
"
8
du œ 10Èv dv; v œ 0 Ê u œ 1, v œ 1 Ê u œ 2
20 "
1‘
"‘#
u# du œ 20
3 u " œ 3 # 1 œ
10
3
(b) Use the same substitution as in part (a); v œ 1 Ê u œ 2, v œ 4 Ê u œ 1 4$Î# œ 9
'
4
10Èv
#
1 a1 v$Î# b
dv œ
'
9
"
u#
2
20 " ‘ *
20 ˆ "
1‰
20 ˆ
7 ‰
ˆ 20
‰
3 du œ 3 u # œ 3 9 2 œ 3 18 œ
70
#7
9. (a) Let u œ x# 1 Ê du œ 2x dx Ê 2 du œ 4x dx; x œ 0 Ê u œ 1, x œ È3 Ê u œ 4
'
È3
0
4x
È x# 1
dx œ
'
4
2
1 Èu
'
du œ
4
1
%
2u"Î# du œ 4u"Î# ‘ " œ 4(4)"Î# 4(1)"Î# œ 4
(b) Use the same substitution as in part (a); x œ È3 Ê u œ 4, x œ È3 Ê u œ 4
È3
'cÈ
4x
3 È x# 1
dx œ
'
4
4
2
Èu
du œ 0
10. (a) Let u œ x% 9 Ê du œ 4x$ dx Ê
'
1
0
x$
È x% 9
dx œ
'
10
9
"
4
"
4
du œ x$ dx; x œ 0 Ê u œ 9, x œ 1 Ê u œ 10
"!
u"Î# du œ 4" (2)u"Î# ‘ * œ
"
#
(10)"Î# #" (9)"Î# œ
È10 3
#
(b) Use the same substitution as in part (a); x œ 1 Ê u œ 10, x œ 0 Ê u œ 9
'c
0
x$
1 È x% 9
dx œ
'
9
"
10 4
u"Î# du œ '
9
10
"
4
11. (a) Let u œ 1 cos 3t Ê du œ 3 sin 3t dt Ê
'
1Î6
0
(1 cos 3t) sin 3t dt œ
'
1
0
"
3
'
1Î6
(1 cos 3t) sin 3t dt œ
'
1
2
"
3
"
3
3 È10
#
du œ sin 3t dt; t œ 0 Ê u œ 0, t œ
#
"
u du œ ’ 3" Š u# ‹ “ œ
(b) Use the same substitution as in part (a); t œ
1Î3
u"Î# du œ
1
6
!
"
6
(1)# 6" (0)# œ
Ê u œ 1, t œ
#
#
u du œ ’ 3" Š u# ‹ “ œ
"
"
6
1
3
1
6
Ê u œ 1 cos
"
6
Ê u œ 1 cos 1 œ 2
(2)# 6" (1)# œ
"
2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
1
#
œ1
Section 5.6 Substitution and Area Between Curves
12. (a) Let u œ 2 tan
'c
0
1Î2
t
#
"
#
Ê du œ
ˆ2 tan #t ‰ sec#
'
dt œ
t
#
sec#
2
1
t
#
dt Ê 2 du œ sec#
'c
1Î2
ˆ2 tan #t ‰ sec#
'
dt œ 2
t
#
1
'
0
cos z
È4 3 sin z
dz œ
'
4
"
Èu
4
1
#
Ê u œ 2 tan ˆ 41 ‰ œ 1, t œ 0 Ê u œ 2
#
3
1
#
Ê u œ 1, t œ
1
#
Ê uœ3
$
u du œ cu# d " œ 3# 1# œ 8
"
3
13. (a) Let u œ 4 3 sin z Ê du œ 3 cos z dz Ê
21
dt; t œ
u (2 du) œ cu# d " œ 2# 1# œ 3
(b) Use the same substitution as in part (a); t œ
1Î2
t
#
331
du œ cos z dz; z œ 0 Ê u œ 4, z œ 21 Ê u œ 4
ˆ 3" du‰ œ 0
(b) Use the same substitution as in part (a); z œ 1 Ê u œ 4 3 sin (1) œ 4, z œ 1 Ê u œ 4
'c
1
cos z
1 È4 3 sin z
dz œ
'
4
"
Èu
4
ˆ 3" du‰ œ 0
14. (a) Let u œ 3 2 cos w Ê du œ 2 sin w dw Ê "# du œ sin w dw; w œ 1# Ê u œ 3, w œ 0 Ê u œ 5
'c
0
'
dw œ
sin w
#
1Î2 (3 2 cos w)
5
u# ˆ #" du‰ œ
3
"
#
&
cu" d $ œ
"
#
"
ˆ "5 "3 ‰ œ 15
(b) Use the same substitution as in part (a); w œ 0 Ê u œ 5, w œ
'
1Î2
!
sin w
(3 2 cos w)#
'
dw œ
3
5
u# ˆ #" du‰ œ
"
#
5
' u# du œ
1
#
Ê uœ3
"
15
3
15. Let u œ t& 2t Ê du œ a5t% 2b dt; t œ 0 Ê u œ 0, t œ 1 Ê u œ 3
'
1
0
16. Let u œ 1 Èy Ê du œ
'
'
Èt& 2t a5t% 2b dt œ
4
dy
#
1 2 È y ˆ1 È y ‰
œ
'
3
"
#
2 u
3
0
$
u"Î# du œ 23 u$Î# ‘ ! œ
; y œ 1 Ê u œ 2, y œ 4 Ê u œ 3
dy
2È y
du œ
(3)$Î# 23 (0)$Î# œ 2È3
2
3
'
3
2
$
u# du œ cu" d # œ ˆ 13 ‰ ˆ 12 ‰ œ
"
6
17. Let u œ cos 2) Ê du œ 2 sin 2) d) Ê "# du œ sin 2) d); ) œ 0 Ê u œ 1, ) œ
'
1Î6
!
cos$ 2) sin 2) d) œ
18. Let u œ tan ˆ 6) ‰ Ê du œ
u œ tan
'
31Î2
1
1
4
'
1Î2
1
"
6
u$ ˆ "# du‰ œ "#
!
#
u$ du œ ’ 2" Š u# ‹“
"Î#
"
œ
cot& ˆ 6) ‰ sec# ˆ 6) ‰ d) œ
'
"
#
4 ˆ 1# ‰
1
!
"
4(1)#
"
È3
œ
,)œ
%
"
"
u
3
3
3
&
È3 u (6 du) œ ’6 Š 4 ‹“ "ÎÈ$ œ 2u% ‘ "ÎÈ$ œ 2(1)% # Š "
"
#
3
4
31
#
Ê
(1 sin 2t)$Î# cos 2t dt œ
%
È3 ‹
'
9
1
"
4
5u"Î% ˆ "4 du‰ œ
'
1
0
œ 12
du œ sin t dt; t œ 0 Ê u œ 5 4 cos 0 œ 1, t œ 1 Ê
5
4
'
1
9
*
u"Î% du œ 54 ˆ 45 u&Î% ‰‘ " œ 9&Î% 1 œ $&Î# "
20. Let u œ 1 sin 2t Ê du œ 2 cos 2t dt Ê "# du œ cos 2t dt; t œ 0 Ê u œ 1, t œ
'
1Î
5 (5 4 cos t)"Î% sin t dt œ
1Î4
Ê u œ cos 2 ˆ 16 ‰ œ
œ1
u œ 5 4 cos 1 œ 9
1
1
1
6
sec# ˆ 6) ‰ d) Ê 6 du œ sec# ˆ 6) ‰ d); ) œ 1 Ê u œ tan ˆ 16 ‰ œ
19. Let u œ 5 4 cos t Ê du œ 4 sin t dt Ê
'
'
1Î2
1
4
Ê uœ0
!
"# u$Î# du œ "2 ˆ 25 u&Î# ‰‘ " œ ˆ 15 (0)&Î# ‰ ˆ 15 (1)&Î# ‰ œ
"
5
21. Let u œ 4y y# 4y$ 1 Ê du œ a4 2y 12y# b dy; y œ 0 Ê u œ 1, y œ 1 Ê u œ 4(1) (1)# 4(1)$ 1 œ 8
'
!
1
a4y y# 4y$ 1b
#Î$
a12y# 2y 4b dy œ
'
1
8
)
u#Î$ du œ 3u"Î$ ‘ " œ 3(8)"Î$ 3(1)"Î$ œ 3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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332
Chapter 5 Integration
"
3
22. Let u œ y$ 6y# 12y 9 Ê du œ a3y# 12y 12b dy Ê
Ê uœ4
'
1
!
ay$ 6y# 12y 9b
œ
'
$
)"Î# d) Ê
È) cos# ˆ)$Î# ‰ d) œ
24. Let u œ 1 'c Î
1 2
1
"
#
3
#
#
!
œ
ay# 4y 4b dy œ
'
4
9
%
"
3
u"Î# du œ 3" ˆ2u"Î# ‰‘ * œ
2
3
(4)"Î# 32 (9)"Î# œ
2
3
(2 3)
2
3
23. Let u œ )$Î# Ê du œ
È1
"Î#
du œ ay# 4y 4b dy; y œ 0 Ê u œ 9, y œ 1
"
t
'
"
4
!
du œ È) d); ) œ 0 Ê u œ 0, ) œ $È1# Ê u œ 1
cos# u ˆ 23 du‰ œ 23 ˆ #u "
4
1
sin 2u‰‘ ! œ
ˆ 1# 2
3
"
4
sin 21‰ 32 (0) œ
1
3
Ê du œ t# dt; t œ 1 Ê u œ 0, t œ #" Ê u œ 1
t# sin# ˆ1 "t ‰ dt œ
1
2
3
'! sin# u du œ ˆ u2 4" sin 2u‰‘ "
œ ’ˆ #" 4" sin (2)‰ ˆ #0 4" sin 0‰“
!
1
sin 2
25. Let u œ 4 x# Ê du œ 2x dx Ê "# du œ x dx; x œ 2 Ê u œ 0, x œ 0 Ê u œ 4, x œ 2 Ê u œ 0
Aœ
'c
0
2
xÈ4 x# dx %
œ 23 u$Î# ‘ ! œ
2
3
'
2
!
xÈ4 x# dx œ (4)$Î# 23 (0)$Î# œ
'
4
!
"# u"Î# du '
0
4
'
"# u"Î# du œ 2
4
"
! #
u"Î# du œ
'
!
4
u"Î# du
16
3
26. Let u œ 1 cos x Ê du œ sin x dx; x œ 0 Ê u œ 0, x œ 1 Ê u œ 2
'
!
1
(1 cos x) sin x dx œ
'
2
!
#
#
u du œ ’ u2 “ œ
!
2#
#
0#
#
œ2
27. Let u œ 1 cos x Ê du œ sin x dx Ê du œ sin x dx; x œ 1 Ê u œ 1 cos (1) œ 0, x œ 0
Ê u œ 1 cos 0 œ 2
Aœ
'c
0
1
3 (sin x) È1 cos x dx œ '
2
!
28. Let u œ 1 1 sin x Ê du œ 1 cos x dx Ê
Because of symmetry about x œ 1# , A œ 2
œ
'
!
1
3u"Î# (du) œ 3
"
1
'c
'
!
2
#
u"Î# du œ 2u$Î# ‘ ! œ 2(2)$Î# 2(0)$Î# œ 2&Î#
du œ cos x dx; x œ 1# Ê u œ 1 1 sin ˆ 1# ‰ œ 0, x œ 0 Ê u œ 1
0
1
1Î2 #
(cos x) (sin (1 1 sin x)) dx œ 2
'
1
(" cos 2x)
#
!
dx œ
"
#
'
1
!
(1 cos 2x) dx œ
"
#
30. For the sketch given, a œ 13 , b œ 13 ; f(t) g(t) œ
Aœ
œ
"
#
1
!
1
#
(sin u) ˆ 1" du‰
sin u du œ [cos u]1! œ (cos 1) (cos 0) œ 2
29. For the sketch given, a œ 0, b œ 1; f(x) g(x) œ 1 cos# x œ sin# x œ
Aœ
'
x "
#
sin 2x ‘ 1
#
!
œ
"
#
1 cos 2x
;
#
[(1 0) (0 0)] œ
sec# t a4 sin# tb œ
"
#
1
#
sec# t 4 sin# t;
2t)
'c ÎÎ ˆ "# sec# t 4 sin# t‰ dt œ "# ' ÎÎ sec# t dt 4' ÎÎ sin# t dt œ "# ' ÎÎ sec# t dt 4' ÎÎ (" cos
dt
#
1 3
1 3
1 3
1Î3
1Î3
1 3
1 3
1 3
1 3
'c Î sec# t dt 2' Î (1 cos 2t) dt œ
1 3
"
#
[tan
1Î$
t]1Î$
2[t 1 3
1 3
1 3
1Î$
sin 2t
# ]1Î$
1 3
œ È3 4 †
1
3
È3 œ
31. For the sketch given, a œ 2, b œ 2; f(x) g(x) œ 2x# ax% 2x# b œ 4x# x% ;
Aœ
'c
2
2
$
a4x# x% b dx œ ’ 4x3 #
x&
5 “ #
œ ˆ 32
3 32 ‰
5
ˆ 32 ‰‘ œ
32
3 5
64
3
64
5
œ
320192
15
œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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128
15
41
3
Section 5.6 Substitution and Area Between Curves
333
32. For the sketch given, c œ 0, d œ 1; f(y) g(y) œ y# y$ ;
Aœ
'
1
!
ay# y$ b dy œ
'
1
!
y# dy '
!
1
$
"
%
"
(" 0)
3
y$ dy œ ’ y3 “ ’ y4 “ œ
!
!
(" 0)
4
"
3
œ
"
4
œ
"
1#
33. For the sketch given, c œ 0, d œ 1; f(y) g(y) œ a12y# 12y$ b a2y# 2yb œ 10y# 12y$ 2y;
Aœ
'
1
!
a10y# 12y$ 2yb dy œ
'
‰
œ ˆ 10
3 0 (3 0) (1 0) œ
!
1
10y# dy '
1
!
12y$ dy '
1
!
"
"
"
$‘
12 % ‘
2 #‘
2y dy œ 10
3 y ! 4 y ! # y !
4
3
34. For the sketch given, a œ 1, b œ 1; f(x) g(x) œ x# a2x% b œ x# 2x% ;
Aœ
'c ax# 2x% b dx œ ’ x3
1
$
1
"
2x&
5 “ "
œ ˆ "3 25 ‰ 3" ˆ 52 ‰‘ œ
35. We want the area between the line y œ 1, 0 Ÿ x Ÿ 2, and the curve y œ
'
(formed by y œ x and y œ 1) with base 1 and height 1. Thus, A œ
œ ˆ2 8 ‰
1#
"
#
œ2
2
3
"
#
œ
!
2
2
3
x#
4,
4
5
œ
1012
15
œ
22
15
738?= the area of a triangle
Š1 x#
4‹
dx "# (1)(1) œ ’x #
x$
1# “ !
"
#
5
6
36. We want the area between the x-axis and the curve y œ x# , 0 Ÿ x Ÿ 1 :6?= the area of a triangle (formed by x œ 1,
x y œ 2, and the x-axis) with base 1 and height 1. Thus, A œ
'
1
!
$
"
x# dx "# (1)(1) œ ’ x3 “ !
"
#
œ
"
3
"
#
œ
5
6
37. AREA œ A1 A2
A1: For the sketch given, a œ 3 and we find b by solving the equations y œ x# 4 and y œ x# 2x
simultaneously for x: x# 4 œ x# 2x Ê 2x# 2x 4 œ 0 Ê 2(x 2)(x 1) Ê x œ 2 or x œ 1 so
b œ 2: f(x) g(x) œ ax# 4b ax# 2xb œ 2x# 2x 4 Ê A1 œ
$
œ ’ 2x3 2x
#
#
4x“
#
$
‰
œ ˆ 16
3 4 8 (18 9 12) œ 9 16
3
œ
'cc a2x# 2x 4b dx
2
3
11
3 ;
A2: For the sketch given, a œ 2 and b œ 1: f(x) g(x) œ ax# 2xb ax# 4b œ 2x# 2x 4
'c a2x# 2x 4b dx œ ’ 2x3
1
Ê A2 œ $
2
œ 23 1 4 16
3
x# 4x“
"
#
‰
œ ˆ 23 1 4‰ ˆ 16
3 48
4 8 œ 9;
Therefore, AREA œ A1 A2 œ
11
3
9œ
38
3
38. AREA œ A1 A2
A1: For the sketch given, a œ 2 and b œ 0: f(x) g(x) œ a2x$ x# 5xb ax# 3xb œ 2x$ 8x
Ê A1 œ
'c a2x$ 8xb dx œ ’ 2x4
0
%
2
!
8x#
# “ #
œ 0 (8 16) œ 8;
A2: For the sketch given, a œ 0 and b œ 2: f(x) g(x) œ ax# 3xb a2x$ x# 5xb œ 8x 2x$
Ê A2 œ
'
2
0
#
a8x 2x$ b dx œ ’ 8x2 Therefore, AREA œ A1 A2 œ 16
#
2x%
4 “!
œ (16 8) œ 8;
39. AREA œ A1 A2 A3
A1: For the sketch given, a œ 2 and b œ 1: f(x) g(x) œ (x 2) a4 x# b œ x# x 2
Ê A1 œ
'cc ax# x 2b dx œ ’ x3
1
$
2
x#
#
2x“
"
#
œ ˆ "3 "
#
2‰ ˆ 83 4
2
4‰ œ
7
3
"
#
œ
143
6
œ
1"
6 ;
"
#
œ 9# ;
A2: For the sketch given, a œ 1 and b œ 2: f(x) g(x) œ a4 x# b (x 2) œ ax# x 2b
Ê A2 œ 'c
2
1
$
ax# x 2b dx œ ’ x3 x#
#
2x“
#
"
œ ˆ 83 4
#
4‰ ˆ 13 1
2
2‰ œ 3 8 Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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334
Chapter 5 Integration
A3: For the sketch given, a œ 2 and b œ 3: f(x) g(x) œ (x 2) a4 x# b œ x# x 2
Ê A3 œ
'
3
$
x#
#
ax# x 2b dx œ ’ x3 2
Therefore, AREA œ A1 A2 A3 œ
11
6
9
#
$
2x“ œ ˆ 27
3 #
9
#
ˆ9 9
#
6‰ ˆ 38 83 ‰ œ 9 5
6
œ
4
2
4‰ œ 9 9
#
38 ;
49
6
40. AREA œ A1 A2 A3
$
A1: For the sketch given, a œ 2 and b œ 0: f(x) g(x) œ Š x3 x‹ Ê A1 œ
"
3
'c
0
2
ax$ 4xb dx œ
"
3
%
’ x4 2x# “
!
x
3
for x:
xœ
f(x) g(x) œ
"
3
œ
Ê
x
3
x$
3
xœ0 Ê
4
3
x
3
"
3
ax$ 4xb
x$
3
x and y œ
x
3
$
'
0
2
ax$ 4xb dx œ
"
3
'
0
2
a4x x$ b œ
$
Ê A3 œ
"
3
'
3
2
ax$ 4xb dx œ
"
3
Therefore, AREA œ A1 A2 A3 œ
$
%
’ x4 2x# “ œ
#
4
3
4
3
25
12
œ
"
3
x
3
œ
"
3
ax$ 4xb
ˆ 81
‰ ˆ 16
‰‘ œ
4 2†9 4 8
3225
1#
œ
"
3
ˆ 81
‰
4 14 œ
19
4
41. a œ 2, b œ 2;
f(x) g(x) œ 2 ax# 2b œ 4 x#
'c a4 x# bdx œ ’4x x3 “ #
2
$
#
2
8‰
œ 2 † ˆ 24
3 3 œ
œ ˆ8 83 ‰ ˆ8 83 ‰
32
3
42. a œ 1, b œ 3;
f(x) g(x) œ a2x x# b (3) œ 2x x# 3
'c a2x x# 3b dx œ ’x# x3
3
Ê Aœ
œ ˆ9 $
1
27
3
9‰ ˆ1 1
3
3‰ œ 11 43. a œ 0, b œ 2;
f(x) g(x) œ 8x x% Ê A œ
#
œ ’ 8x2 #
x&
“
5 !
œ 16 32
5
œ
'
2
0
"
3
3x“
œ
$
"
32
3
a8x x% b dx
80 32
5
œ
48
5
44. Limits of integration: x# 2x œ x Ê x# œ 3x
Ê x(x 3) œ 0 Ê a œ 0 and b œ 3;
f(x) g(x) œ x ax# 2xb œ 3x x#
Ê Aœ
œ
27
#
"
3
(8 4) œ 43 ;
A3: For the sketch given, a œ 2 and b œ 3: f(x) g(x) œ Š x3 x‹ Ê Aœ
simultaneously
(x 2)(x 2) œ 0 Ê x œ 2, x œ 0, or x œ 2 so b œ 2:
Š x3 x‹ œ 3" ax$ 4xb Ê A2 œ 3"
x
3
43 x œ
œ 0 3" (4 8) œ 43 ;
#
A2: For the sketch given, a œ 0 and we find b by solving the equations y œ
x$
3
x$
3
œ
'
0
9œ
3
#
a3x x# b dx œ ’ 3x2 27 18
#
œ
$
x$
3 “!
9
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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25
12 ;
’2x# #
x%
4 “!
Section 5.6 Substitution and Area Between Curves
45. Limits of integration: x# œ x# 4x Ê 2x# 4x œ 0
Ê 2x(x 2) œ 0 Ê a œ 0 and b œ 2;
f(x) g(x) œ ax# 4xb x# œ 2x# 4x
Ê Aœ
'
2
0
œ 16
3 #
4x#
2 “!
$
a2x# 4xb dx œ ’ 2x
3 œ
16
#
32 48
6
œ
8
3
46. Limits of integration: 7 2x# œ x# 4 Ê 3x# 3 œ 0
Ê 3(x 1)(x 1) œ 0 Ê a œ 1 and b œ 1;
f(x) g(x) œ a7 2x# b ax# 4b œ 3 3x#
Ê Aœ
'c a3 3x# b dx œ 3 ’x x3 “ "
1
$
"
1
œ 3 ˆ1 "3 ‰ ˆ1 3" ‰‘ œ 6 ˆ 23 ‰ œ 4
47. Limits of integration: x% 4x# 4 œ x#
Ê x% 5x# 4 œ 0 Ê ax# 4b ax# 1b œ 0
Ê (x 2)(x 2)(x 1)(x 1) œ ! Ê x œ 2, 1, 1, 2;
f(x) g(x) œ ax% 4x# 4b x# œ x% 5x# 4 and
g(x) f(x) œ x# ax% 4x# 4b œ x% 5x# 4
Ê Aœ
'
2
1
'cc ax% 5x# 4bdx 'c ax% 5x# 4bdx
1
2
1
%
#
ax 5x 4bdx
&
œ ’ x5 œ ˆ "5 œ
1
5
3
60
5
5x$
3
4x“
"
&
’ x5 #
4‰ ˆ 32
5 60
3
œ
300180
15
40
3
5x$
3
4x“
8‰ ˆ 5" 5
3
"
&
"
’ 5x 5x$
3
4‰ ˆ 5" 4x“
5
3
#
"
4‰ ˆ 32
5 40
3
8‰ ˆ 5" œ8
48. Limits of integration: xÈa# x# œ 0 Ê x œ 0 or
Èa# x# œ 0 Ê x œ 0 or a# x# œ 0 Ê x œ a, 0, a;
Aœ
œ
œ
"
#
"
3
'c xÈa# x# dx '
0
a
0
’ 23 aa# x# b
# $Î#
aa b
a
$Î# !
“
"
3
a
"# ’ 23 aa# x# b
# $Î#
’ aa b
xÈa# x# dx
“œ
2a
3
$Î# a
“
!
$
49. Limits of integration: y œ Èkxk œ Èx, x Ÿ 0
and
Èx, x 0
5y œ x 6 or y œ x5 65 ; for x Ÿ 0: Èx œ x5 65
Ê 5Èx œ x 6 Ê 25(x) œ x# 12x 36
Ê x# 37x 36 œ 0 Ê (x 1)(x 36) œ 0
Ê x œ 1, 36 (but x œ 36 is not a solution);
for x 0: 5Èx œ x 6 Ê 25x œ x# 12x 36
Ê x# 13x 36 œ 0 Ê (x 4)(x 9) œ 0
Ê x œ 4, 9; there are three intersection points and
Aœ
'c ˆ x 5 6 Èx‰dx '
0
1
0
4
ˆ x 5 6 Èx‰dx '
4
9
ˆÈ x x6‰
5
dx
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5
3
4‰
335
336
Chapter 5 Integration
#
œ ’ (x 106) 23 (x)$Î# “
œ ˆ 36
10 25
10
!
%
#
’ (x 106) 23 x$Î# “ ’ 23 x$Î# "
23 ‰ ˆ 100
10 2
3
† 4$Î# !
36
10
0‰ ˆ 32 † 9$Î# 50. Limits of integration:
x# 4, x Ÿ 2 or x
y œ kx# 4k œ œ
4 x# , 2 Ÿ x Ÿ 2
for x Ÿ 2 and x
#
2: x# 4 œ
#
#
x#
2
*
(x 6)#
10 “ %
225
10
2
3
† 4$Î# 100 ‰
10
œ 50
10 20
3
2
4
Ê 2x 8 œ x 8 Ê x œ 16 Ê x œ „ 4;
x#
#
for 2 Ÿ x Ÿ 2: 4 x# œ
#
4 Ê 8 2x# œ x# 8
Ê x œ 0 Ê x œ 0; by symmetry of the graph,
Aœ2
'
2
0
#
’Š x2 4‹ a4 x# b“dx 2
œ 2 ˆ 8# 0‰ 2 ˆ32 '
4
2
#
#
!
16 68 ‰ œ 40 64
6
$
’Š x2 4‹ ax# 4b“dx œ 2 ’ x2 “ 2 ’8x 56
3
œ
%
x$
6 “#
64
3
51. Limits of integration: c œ 0 and d œ 3;
f(y) g(y) œ 2y# 0 œ 2y#
'
Ê Aœ
0
3
$
$
2y# dy œ ’ 2y3 “ œ 2 † 9 œ 18
!
52. Limits of integration: y# œ y 2 Ê (y 1)(y 2) œ 0
Ê c œ 1 and d œ 2; f(y) g(y) œ (y 2) y#
Ê Aœ
'c ay 2 y# b dy œ ’ y#
2
#
1
2y œ ˆ 4# 4 83 ‰ ˆ "# 2 3" ‰ œ 6 8
3
"
#
#
y$
3 “ "
2
"
3
œ
9
#
53. Limits of integration: 4x œ y# 4 and 4x œ 16 y
Ê y# 4 œ 16 y Ê y# y 20 œ 0 Ê
(y 5)(y 4) œ 0 Ê c œ 4 and d œ 5;
#
f(y) g(y) œ ˆ 164y ‰ Š y 44 ‹ œ
Ê Aœ
"
4
y$
3
'c ay# y 20b dy
5
4
y#
#
œ
"
4
’
œ
"
4
"
4
ˆ 125
3 189
ˆ 3 œ
y# y20
4
20y“
&
%
25
‰
100
4" ˆ 64
2
3
9
243
‰
180
œ
2
8
16
#
80‰
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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œ
5
3
Section 5.6 Substitution and Area Between Curves
54. Limits of integration: x œ y# and x œ 3 2y#
Ê y# œ 3 2y# Ê 3y# œ 3 Ê 3(y 1)(y 1) œ 0
Ê c œ 1 and d œ 1; f(y) g(y) œ a3 2y# b y#
œ 3 3y# œ 3 a1 y# b Ê A œ 3
œ 3 ’y "
y$
3 “ "
œ 3 ˆ1 "‰
3
'c a1 y# b dy
1
1
3 ˆ1 3" ‰
œ 3 † 2 ˆ1 "3 ‰ œ 4
55. Limits of integration: x œ y# and x œ 2 3y#
Ê y# œ 2 3y# Ê 2y# 2 œ 0
Ê 2(y 1)(y 1) œ 0 Ê c œ 1 and d œ 1;
f(y) g(y) œ a2 3y# b ay# b œ 2 2y# œ 2 a1 y# b
Ê Aœ2
'c a1 y# b dy œ 2 ’y y3 “ "
1
$
"
1
œ 2 ˆ1 "3 ‰ 2 ˆ1 3" ‰ œ 4 ˆ 23 ‰ œ
8
3
56. Limits of integration: x œ y#Î$ and x œ 2 y%
Ê y#Î$ œ 2 y% Ê c œ 1 and d œ 1;
f(y) g(y) œ a2 y% b y#Î$
'c ˆ2 y% y#Î$ ‰ dy
1
Ê Aœ
1
œ ’2y y&
5
35 y&Î$ “
"
"
œ ˆ2 "5 35 ‰ ˆ2 œ 2 ˆ2 "5 35 ‰ œ 12
5
"
5
35 ‰
57. Limits of integration: x œ y# 1 and x œ kyk È1 y#
Ê y# 1 œ kyk È1 y# Ê y% 2y# 1 œ y# a1 y# b
Ê y% 2y# 1 œ y# y% Ê 2y% 3y# 1 œ 0
Ê a2y# 1b ay# 1b œ 0 Ê 2y# 1 œ 0 or y# 1 œ 0
Ê y# œ
"
#
or y# œ 1 Ê y œ „
„È 2
#
È2
#
or y œ „ 1.
are not solutions Ê y œ „ 1;
for 1 Ÿ y Ÿ 0, f(x) g(x) œ yÈ1 y# ay# 1b
Substitution shows that
œ 1 y# y a1 y# b
"Î#
, and by symmetry of the graph,
'c ’1 y# y a1 y# b"Î# “ dy
"Î#
œ 2' a1 y# b dy 2 ' y a1 y# b dy
c
c
0
Aœ2
1
0
0
1
œ 2 ’y 1
$
y
3
“
!
"
# $Î#
2 ˆ "# ‰ ” 2 a1 3y b •
!
"
œ 2 (! 0) ˆ1 3" ‰‘ ˆ 23 0‰ œ 2
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337
338
Chapter 5 Integration
58. AREA œ A1 A2
Limits of integration: x œ 2y and x œ y$ y# Ê
y$ y# œ 2y Ê y ay# y 2b œ y(y 1)(y 2) œ 0
Ê y œ 1, 0, 2:
for 1 Ÿ y Ÿ 0, f(y) g(y) œ y$ y# 2y
'c
0
Ê A1 œ
1
œ 0 ˆ "4 "
3
%
y$
3
ay$ y# 2yb dy œ ’ y4 1‰ œ
y# “
!
"
5
12 ;
for 0 Ÿ y Ÿ 2, f(y) g(y) œ 2y y$ y#
'! a2y y$ y# b dy œ ’y# y4
2
Ê A2 œ
Ê ˆ4 %
#
y$
3 “!
38 ‰ 0 œ 38 ;
16
4
Therefore, A1 A2 œ
5
12
8
3
œ
37
12
59. Limits of integration: y œ 4x# 4 and y œ x% 1
Ê x% 1 œ 4x# 4 Ê x% 4x# 5 œ 0
Ê ax# 5b (x 1)(x 1) œ 0 Ê a œ 1 and b œ 1;
f(x) g(x) œ 4x# 4 x% 1 œ 4x# x% 5
Ê Aœ
'c a4x# x% 5b dx œ ’ 4x3
1
$
1
œ ˆ 43 "
5
"
5
5‰ ˆ 43 5‰ œ 2 ˆ 43 "
x&
5
5x“
"
5
5‰ œ
"
104
15
60. Limits of integration: y œ x$ and y œ 3x# 4
Ê x$ 3x# 4 œ 0 Ê ax# x 2b (x 2) œ 0
Ê (x 1)(x 2)# œ 0 Ê a œ 1 and b œ 2;
f(x) g(x) œ x$ a3x# 4b œ x$ 3x# 4
Ê Aœ
'c ax$ 3x# 4b dx œ ’ x4
œ ˆ 16
4 24
3
2
%
1
8‰ ˆ 41 " 4‰ œ
3x$
3
4x“
#
"
27
4
61. Limits of integration: x œ 4 4y# and x œ 1 y%
Ê 4 4y# œ 1 y% Ê y% 4y# 3 œ 0
Ê Šy È3‹ Šy È3‹ (y 1)(y 1) œ 0 Ê c œ 1
0; f(y) g(y) œ a4 4y# b a1 y% b
and d œ 1 since x
'c a3 4y# y% b dy
1
œ 3 4y# y% Ê A œ
œ ’3y 4y$
3
"
y&
5 “ "
1
œ 2ˆ3 4
3
5" ‰ œ
56
15
#
62. Limits of integration: x œ 3 y# and x œ y4
#
Ê 3 y# œ y4 Ê
3y#
4
3œ0 Ê
3
4
(y 2)(y 2) œ 0
#
Ê c œ 2 and d œ 2; f(y) g(y) œ a3 y# b Š y4 ‹
œ 3 Š1 œ 3 ˆ2 y#
4‹
8 ‰
12
'c Š1 y4 ‹ dy œ 3 ’y 1y# “ #
Ê Aœ3
ˆ 2 2
#
$
#
2
8 ‰‘
12
œ 3 ˆ4 16 ‰
12
œ 12 4 œ 8
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 5.6 Substitution and Area Between Curves
63. a œ 0, b œ 1; f(x) g(x) œ 2 sin x sin 2x
Ê Aœ
'
1
0
(2 sin x sin 2x) dx œ 2 cos x cos 2x ‘ 1
2
!
œ 2(1) "# ‘ ˆ2 † 1 "# ‰ œ 4
64. a œ 13 , b œ 13 ; f(x) g(x) œ 8 cos x sec# x
Ê Aœ
œ Š8 †
'c
1Î3
1Î3
È3
#
1Î$
a8 cos x sec# xb dx œ [8 sin x tan x] 1Î$
È3
#
È3‹ Š8 †
È3‹ œ 6È3
‰
65. a œ 1, b œ 1; f(x) g(x) œ a1 x# b cos ˆ 1x
#
Ê Aœ
œ ˆ1 "
3
'c 1 x# cos ˆ 1#x ‰‘ dx œ ’x x3
1
$
1
12 ‰ ˆ1 "
3
12 ‰ œ 2 ˆ 23 12 ‰ œ
2
1
4
3
sin ˆ 1#x ‰“
"
"
4
1
66. A œ A1 A2
a" œ 1, b" œ 0 and a# œ 0, b# œ 1;
f" (x) g" (x) œ x sin ˆ 1#x ‰ and f# (x) g# (x) œ sin ˆ 1#x ‰ x
Ê by symmetry about the origin,
A" A# œ 2A" Ê A œ 2
œ 2 ’ 12 cos ˆ 1#x ‰ "
x#
# “!
œ 2 ˆ 12 "# ‰ œ 2 ˆ 4211 ‰ œ
'
1
0
sin ˆ 1x
‰
‘
# x dx
œ 2 ˆ 12 † 0 "# ‰ ˆ 12 † 1 0‰‘
4 1
1
67. a œ 14 , b œ 14 ; f(x) g(x) œ sec# x tan# x
Ê Aœ
œ
'c
1Î4
1Î4
'c
1Î4
1Î4
asec# x tan# xb dx
csec# x asec# x 1bd dx
1Î4
1Î%
œ ' 1 † dx œ [x]1Î% œ
1Î4
1
4
ˆ 14 ‰ œ
1
#
68. c œ 14 , d œ 14 ; f(y) g(y) œ tan# y a tan# yb œ 2 tan# y
œ 2 asec# y 1b Ê A œ
'c
1Î4
1Î4
2 asec# y 1b dy
1Î%
œ 2[tan y y]1Î% œ 2 ˆ1 14 ‰ ˆ1 14 ‰‘
œ 4 ˆ1 14 ‰ œ 4 1
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339
340
Chapter 5 Integration
69. c œ 0, d œ 1# ; f(y) g(y) œ 3 sin yÈcos y 0 œ 3 sin yÈcos y
Ê Aœ3
'
1Î2
0
1Î#
sin yÈcos y dy œ 3 23 (cos y)$Î# ‘ !
œ 2(0 1) œ 2
"Î$
‰
70. a œ 1, b œ 1; f(x) g(x) œ sec# ˆ 1x
3 x
Ê Aœ
œ
Š 13
'c sec# ˆ 13x ‰ x"Î$ ‘ dx œ 13 tan ˆ 13x ‰ 43 x%Î$ ‘ ""
1
1
È3 3 ‹ ’ 3 ŠÈ3‹ 3 “ œ
4
1
4
6È 3
1
71. A œ A" A#
Limits of integration: x œ y$ and x œ y Ê y œ y$
Ê y$ y œ 0 Ê y(y 1)(y 1) œ 0 Ê c" œ 1, d" œ 0
and c# œ 0, d# œ 1; f" (y) g" (y) œ y$ y and
f# (y) g# (y) œ y y$ Ê by symmetry about the origin,
A" A# œ 2A# Ê A œ 2
œ 2 ˆ "# 4" ‰ œ
'
1
0
#
ay y$ b dy œ 2 ’ y# "
#
"
y%
4 “!
72. A œ A" A#
Limits of integration: y œ x$ and y œ x& Ê x$ œ x&
Ê x& x$ œ 0 Ê x$ (x 1)(x 1) œ 0 Ê a" œ 1, b" œ 0
and a# œ 0, b# œ 1; f" (x) g" (x) œ x$ x& and
f# (x) g# (x) œ x& x$ Ê by symmetry about the origin,
A" A# œ 2A# Ê A œ 2
œ 2 ˆ "4 6" ‰ œ
'
0
1
%
ax$ x& b dx œ 2 ’ x4 "
6
73. A œ A" A#
Limits of integration: y œ x and y œ
$
"
x#
Ê xœ
"
x# ,
"
x'
6 “!
xÁ0
Ê x œ 1 Ê x œ 1 , f" (x) g" (x) œ x 0 œ x
Ê A" œ
'
0
1
#
"
x dx œ ’ x2 “ œ "# ; f# (x) g# (x) œ
œ x# Ê A# œ
A œ A" A# œ
'
"
#
!
2
1
"
x#
0
#
"
"
‘
x# dx œ "
x " œ # 1 œ #;
"
#
œ1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 5.6 Substitution and Area Between Curves
74. Limits of integration: sin x œ cos x Ê x œ
1
4;
and b œ
œŠ
È2
#
1Î4
0
1Î%
(cos x sin x) dx œ [sin x cos x]!
È2
# ‹
Ê aœ0
f(x) g(x) œ cos x sin x
'
Ê Aœ
1
4
341
(0 1) œ È2 1
75. (a) The coordinates of the points of intersection of the
line and parabola are c œ x# Ê x œ „ Èc and y œ c
(b) f(y) g(y) œ Èy ˆÈy‰ œ 2Èy Ê the area of the
lower section is, AL œ
œ2
'
c
0
'
0
c
[f(y) g(y)] dy
Èy dy œ 2 23 y$Î# ‘ ! œ
c
c$Î# . The area of the
4
3
entire shaded region can be found by setting c œ 4: A œ ˆ 43 ‰ 4$Î# œ 43†8 œ 32
3 . Since we want c to divide the region
32
4
$Î#
into subsections of equal area we have A œ 2AL Ê 3 œ 2 ˆ 3 c ‰ Ê c œ 4#Î$
(c) f(x) g(x) œ c x# Ê AL œ
œ
4
3
Èc
Èc
c
c
'cÈ [f(x) g(x)] dx œ 'cÈ ac x# b dx œ ’cx x3 “ È
$
c
cÈc
œ 2 ’c$Î# c$Î# . Again, the area of the whole shaded region can be found by setting c œ 4 Ê A œ
condition A œ 2AL , we get
4
3
$Î#
c
œ
#Î$
Ê cœ4
32
3
32
3 .
c$Î#
3 “
From the
as in part (b).
76. (a) Limits of integration: y œ 3 x# and y œ 1
Ê 3 x# œ 1 Ê x# œ 4 Ê a œ 2 and b œ 2;
f(x) g(x) œ a3 x# b (1) œ 4 x#
Ê Aœ
'c a4 x# b dx œ ’4x x3 “ #
2
$
#
32
3
1
œ ˆ8 83 ‰ ˆ8 83 ‰ œ 16 16
3
œ
(b) Limits of integration: let x œ 0 in y œ 3 x#
Ê y œ 3; f(y) g(y) œ È3 y ˆÈ3 y‰
œ 2(3 y)"Î#
Ê Aœ2
'c (3 y)"Î# dy œ 2 'c (3 y)"Î# (1) dy œ (2) ’ 2(3 3y)
œ ˆ 43 ‰ (8) œ
3
3
1
1
“
$
"
œ ˆ 43 ‰ 0 (3 1)$Î# ‘
32
3
77. Limits of integration: y œ 1 Èx and y œ
Ê 1 Èx œ
$Î#
2
Èx
2
Èx
, x Á 0 Ê Èx x œ 2 Ê x œ (2 x)#
Ê x œ 4 4x x# Ê x# 5x 4 œ 0
Ê (x 4)(x 1) œ 0 Ê x œ 1, 4 (but x œ 4 does not
satisfy the equation); y œ È2x and y œ x4 Ê È2x œ x4
Ê 8 œ xÈx Ê 64 œ x$ Ê x œ 4.
Therefore, AREA œ A" A# : f" (x) g" (x) œ ˆ1 x"Î# ‰ Ê A" œ
'
œ ˆ1 "8 ‰ 0 œ
2
3
0
œ ˆ4 † 2 1
ˆ1 x"Î# x4 ‰ dx œ ’x 23 x$Î# 16 ‰
8
37
24 ; f# (x)
"
x#
8 “!
g# (x) œ 2x"Î# ˆ4 "8 ‰ œ 4 15
8
œ
17
8 ;
x
4
x
4
Ê A# œ
'
1
4
ˆ2x"Î# 4x ‰ dx œ ’4x"Î# Therefore, AREA œ A" A# œ
37
24
17
8
œ
3751
24
œ
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88
24
%
x#
8 “"
œ
11
3
342
Chapter 5 Integration
78. Limits of integration: (y 1)# œ 3 y Ê y# 2y 1
œ 3 y Ê y# y 2 œ 0 Ê (y 2)(y 1) œ 0
Ê y œ 2 since y 0; also, 2Èy œ 3 y
Ê 4y œ 9 6y y# Ê y# 10y 9 œ 0
Ê (y 9)(y 1) œ 0 Ê y œ 1 since y œ 9 does not
satisfy the equation;
AREA œ A" A#
f" (y) g" (y) œ 2Èy 0 œ 2y"Î#
Ê A" œ 2
Ê A# œ
'
1
0
'
2
1
"
$Î#
y"Î# dy œ 2 ’ 2y3 “ œ 43 ; f# (y) g# (y) œ (3 y) (y 1)#
!
#
c3 y (y 1) d dy œ 3y "# y# "3 (y 1)$ ‘ " œ ˆ6 2 3" ‰ ˆ3 #
Therefore, A" A# œ
4
3
7
6
œ
œ
15
6
"
#
80. A œ
'
b
a
2f(x) dx '
a
b
'
0
a
'
b
a
a$
3‹
aa# x# b dx œ 2 a# x "3 x$ ‘ ! œ 2 Ša$ a
a$
$
Š 4a3 ‹
(2a) aa# b œ a$ ; limit of ratio œ lim b
aÄ!
f(x) dx œ 2
0‰ œ 1 "
3
"
#
œ 67 ;
5
2
79. Area between parabola and y œ a# : A œ 2
Area of triangle AOC:
"
#
f(x) dx '
b
a
f(x) dx œ
'
b
a
œ
3
4
0œ
4a$
3 ;
which is independent of a.
f(x) dx œ 4
81. Neither one; they are both zero. Neither integral takes into account the changes in the formulas for the
region's upper and lower bounding curves at x œ 0. The area of the shaded region is actually
Aœ
'c [x (x)] dx '
0
1
0
1
[x (x)] dx œ
'c 2x dx '
0
1
0
1
2x dx œ #.
82. It is sometimes true. It is true if f(x) g(x) for all x between a and b. Otherwise it is false. If the graph of f
lies below the graph of g for a portion of the interval of integration, the integral over that portion will be
negative and the integral over [aß b] will be less than the area between the curves (see Exercise 53).
83. Let u œ 2x Ê du œ 2 dx Ê
'
3
sin 2x
x
1
dx œ
'
6
2
sin u
ˆ u# ‰
"
#
ˆ "# du‰ œ
du œ dx; x œ 1 Ê u œ 2, x œ 3 Ê u œ 6
'
6
du œ cF(u)d '# œ F(6) F(2)
sin u
u
2
84. Let u œ 1 x Ê du œ dx Ê du œ dx; x œ 0 Ê u œ 1, x œ 1 Ê u œ 0
'
0
1
f(1 x) dx œ
'
1
0
f(u) ( du) œ '
0
1
f(u) du œ
'
1
f(u) du œ
0
'
0
1
f(x) dx
85. (a) Let u œ x Ê du œ dx; x œ 1 Ê u œ 1, x œ 0 Ê u œ 0
f odd Ê f(x) œ f(x). Then
'c f(x) dx œ '
0
1
0
1
f(u) ( du) œ
'
1
œ 3
(b) Let u œ x Ê du œ dx; x œ 1 Ê u œ 1, x œ 0 Ê u œ 0
f even Ê f(x) œ f(x). Then
'c f(x) dx œ '
0
1
1
0
0
f(u) ( du) œ '
f(u) ( du) œ
1
0
f(u) du œ
'
0
1
'
1
0
f(u) du œ '
0
1
f(u) du
f(u) du œ 3
'c f(x) dx when f is odd. Let u œ x Ê du œ dx Ê du œ dx and x œ a Ê u œ a and x œ !
Ê u œ !. Thus ' f(x) dx œ ' f(u) du œ ' f(u) du œ ' f(u) du œ ' f(x) dx.
c
Thus ' f(x) dx œ ' f(x) dx ' f(x) dx œ ' f(x) dx ' f(x) dx œ !.
c
c
0
86. (a) Consider
a
0
0
a
0
a
a
0
a
a
a
a
0
a
0
a
0
a
0
a
0
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 5.6 Substitution and Area Between Curves
'c
1/2
(b)
1/2
1Î#
sin x dx œ [cos x]1Î# œ cos ˆ 1# ‰ cos ˆ 1# ‰ œ ! ! œ !.
87. Let u œ a x Ê du œ dx; x œ 0 Ê u œ a, x œ a Ê u œ 0
Iœ
'
' f(u)f(af(au) duu) œ ' f(x)f(af(ax) dxx)
dx
f(x)f(ax)
' f(x)f(af(ax) dxx) œ ' f(x)
' dx œ [x]! œ a 0 œ a.
I I œ ' f(x)f(x)
f(ax) f(ax) dx œ
a
f(x) dx
0 f(x)f(ax)
œ
'
0
f(au)
a f(au)f(u)
a
Ê
a
0
0
a
a
a
a
0
0
Therefore, 2I œ a Ê I œ
88. Let u œ
'
a
( du) œ
xy
x
"
t
xy
t
dt œ
a
#
0
.
t
Ê du œ xy
t# dt Ê xy du œ
'
y
1
u" du œ 0
'
y
1
"
u
du œ
'
y
1
"
t
"
u
dt Ê u" du œ
du œ
'
1
y
"
t
"
t
dt; t œ x Ê u œ y, t œ xy Ê u œ 1. Therefore,
dt
89. Let u œ x c Ê du œ dx; x œ a c Ê u œ a, x œ b c Ê u œ b
' cc
b c
a c
90. (a)
f(x c) dx œ
'
a
b
f(u) du œ
'
a
b
f(x) dx
(b)
(c)
91-94. Example CAS commands:
Maple:
f := 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
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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
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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 u"Î# ‘ " œ 3 1 œ 2
6. Let u œ x# 1 Ê du œ 2x dx Ê
'
(10) 25 œ 0
! 3ak œ 3 ! ak œ 3(0) œ 0
4. (a)
'
5
#
"
#
du œ x dx; x œ 1 Ê u œ 0, x œ 3 Ê u œ 8
)
u"Î$ ˆ "# du‰ œ 38 u%Î$ ‘ ! œ
3
8
(16 0) œ 6
Ê 2 du œ dx; x œ 1 Ê u œ 1# , x œ 0 Ê u œ 0
x
2
' cos ˆ x# ‰ dx œ ' Î (cos u)(2 du) œ [2 sin u]!1Î# œ 2 sin 0 2 sin ˆ 1# ‰ œ 2(0 (1)) œ 2
0
0
1
1 2
8. Let u œ sin x Ê du œ cos x dx; x œ 0 Ê u œ 0, x œ
'
1Î2
0
(sin x)(cos x) dx œ
(e)
10. (a)
(c)
(e)
#
u du œ ’ u2 “ œ
!
Ê uœ1
"
#
2
2
5
2
2
(c)
0
"
'c f(x) dx œ "3 'c 3 f(x) dx œ 3" (12) œ 4
(b) ' f(x) dx œ ' f(x) dx ' f(x) dx œ 6 4 œ 2
c
c
c
' g(x) dx œ 'c g(x) dx œ 2
(d) ' (1 g(x)) dx œ 1 ' g(x) dx œ 1(2) œ 21
c
c
'c Š f(x) 5 g(x) ‹ dx œ 5" 'c f(x) dx 5" 'c g(x) dx œ 5" (6) 5" (2) œ 85
2
9. (a)
'
1
1
#
'
'
'
2
5
5
2
5
5
5
2
2
2
2
0
0
2
' 7 g(x) dx œ "7 (7) œ 1
f(x) dx œ ' f(x) dx œ 1
[g(x) 3 f(x)] dx œ ' g(x) dx 3'
g(x) dx œ
2
"
7
(b)
0
2
(d)
0
2
0
5
2
0
0
2
'
'
2
2
2
5
5
2
2
2
1
2
0
g(x) dx œ
'
0
2
g(x) dx È2 f(x) dx œ È2
'
0
2
'
0
1
g(x) dx œ 1 2 œ 1
f(x) dx œ È2 (1) œ 1È2
f(x) dx œ 1 31
11. x# 4x 3 œ 0 Ê (x 3)(x 1) œ 0 Ê x œ 3 or x œ 1;
Area œ
'
0
1
ax# 4x 3b dx "
$
'
3
1
$
ax# 4x 3b dx
œ ’ x3 2x# 3x“ ’ x3 2x# 3x“
!
$
"
$
œ ’Š "3 2(1)# 3(1)‹ 0“
$
$
’Š 33 2(3)# 3(3)‹ Š 13 2(1)# 3(1)‹“
œ ˆ "3 1‰ 0 ˆ 3" 1‰‘ œ
8
3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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345
346
Chapter 5 Integration
12. 1 x#
4
œ 0 Ê 4 x# 0 Ê x œ „ 2;
Area œ
'c Š1 x4 ‹ dx '
2
2
œ ’x 2
#
x$
12 “ #
2$
12 ‹
œ ’Š2 3
#
’x Š1 x#
4‹
dx
$
x$
12 “ #
Š2 (2)$
12 ‹“
œ 43 ˆ 43 ‰‘ ˆ 34 43 ‰ œ
’Š3 3$
12 ‹
2$
12 ‹“
Š2 13
4
13. 5 5x#Î$ œ 0 Ê 1 x#Î$ œ 0 Ê x œ „ 1;
Area œ
'c ˆ5 5x#Î$ ‰ dx '
1
1
1
8
ˆ5 5x#Î$ ‰ dx
"
)
œ 5x 3x&Î$ ‘ " 5x 3x&Î$ ‘ "
œ ˆ5(1) 3(1)&Î$ ‰ ˆ5(1) 3(1)&Î$ ‰‘
ˆ5(8) 3(8)&Î$ ‰ ˆ5(1) 3(1)&Î$ ‰‘
œ [2 (2)] [(40 96) 2] œ 62
14. 1 Èx œ 0 Ê x œ 1;
Area œ
œ
œ
œ
'
0
1
ˆ1 Èx‰ dx 4
1
ˆ1 Èx‰ dx
x 23 x$Î# ‘ " x 23 x$Î# ‘ %
!
"
ˆ1 23 (1)$Î# ‰ 0‘ ˆ4 23
"
ˆ4 16
‰ "‘
3 3 3 œ 2
15. f(x) œ x, g(x) œ
œ
'
'
2
1
ˆx "‰
x#
œ
'
œ
Š 42
1
Šx a œ 1, b œ 2 Ê A œ
'
b
[f(x) g(x)] dx
a
#
#
dx œ ’ x# x" “ œ ˆ 4# "# ‰ ˆ "# 1‰ œ 1
16. f(x) œ x, g(x) œ
2
"
x# ,
(4)$Î# ‰ ˆ1 23 (1)$Î# ‰‘
"
"
Èx
"
Èx ‹dx
, a œ 1, b œ 2 Ê A œ
#
œ ’ x# 2Èx“
2È2‹ ˆ "# 2‰ œ
'
a
b
[f(x) g(x)] dx
#
"
7 4 È 2
#
'
#
17. f(x) œ ˆ1 Èx‰ , g(x) œ 0, a œ 0, b œ 1 Ê A œ
œ
'
0
1
ˆ1 2x"Î# x‰ dx œ ’x 43 x$Î# #
"
x#
# “!
x%
#
"
x(
7 “!
œ1
"
#
"
7
œ
a
œ1
18. f(x) œ a1 x$ b , g(x) œ 0, a œ 0, b œ 1 Ê A œ
œ ’x b
'
a
b
'
[f(x) g(x)] dx œ
4
3
"
#
œ
"
6
1
0
ˆ1 Èx‰# dx œ
(6 8 3) œ
[f(x) g(x)] dx œ
'
0
1
'
0
1
ˆ1 2Èx x‰ dx
"
6
#
a1 x$ b dx œ
'
0
9
14
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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1
a1 2x$ x' b dx
Chapter 5 Practice Exercises
19. f(y) œ 2y# , g(y) œ 0, c œ 0, d œ 3
Ê Aœ
œ2
'
3
'
d
c
y# dy œ
0
'
[f(y) g(y)] dy œ
2
3
3
0
a2y# 0b dy
$
cy$ d ! œ 18
20. f(y) œ 4 y# , g(y) œ 0, c œ 2, d œ 2
Ê Aœ
'
d
c
#
y$
3 “ #
œ ’4y 'c a4 y# b dy
2
[f(y) g(y)] dy œ
œ 2 ˆ8 8‰
3
2
œ
32
3
y#
4
21. Let us find the intersection points:
y 2
4
œ
Ê y# y 2 œ 0 Ê (y 2)(y 1) œ 0 Ê y œ 1
or y œ 2 Ê c œ 1, d œ 2; f(y) œ
Ê Aœ
'
d
c
y 2
4
2
#
1
œ
"
4
'c ay 2 y# b dy œ 4" ’ y#
œ
"
4
ˆ 4# 4 83 ‰ ˆ "# 2 3" ‰‘ œ
#
1
y#
4
'c Š y4 2 y4 ‹ dy
[f(y) g(y)] dy œ
2
, g(y) œ
2y 9
8
y# 4
4
22. Let us find the intersection points:
#
y$
3 “ "
œ
y 16
4
Ê y# y 20 œ 0 Ê (y 5)(y 4) œ 0 Ê y œ 4
or y œ 5 Ê c œ 4, d œ 5; f(y) œ
Ê Aœ
'
d
c
[f(y) g(y)] dy œ
y 16
4
, g(y) œ
y# 4
4
'c Š y 416 y 4 4 ‹ dy
5
#
4
œ
"
4
'c ay 20 y# b dy œ "4 ’ y#
œ
"
4
"
4
125 ‰
ˆ 25
‰‘
ˆ "#6 80 64
# 100 3
3
9
"
9
"
ˆ # 180 63‰ œ 4 ˆ # 117‰ œ 8 (9 234) œ
œ
5
#
20y 4
23. f(x) œ x, g(x) œ sin x, a œ 0, b œ
Ê Aœ
'
b
a
#
[f(x) g(x)] dx œ
œ ’ x# cos x“
1Î%
!
#
œ Š 31# '
1Î4
(x sin x) dx
1
24. f(x) œ 1, g(x) œ ksin xk , a œ 1# , b œ
Ê Aœ
œ
'c
œ2
0
'
a
b
[f(x) g(x)] dx œ
(1 sin x) dx 1Î2
1Î2
'
0
'
0
1Î2
243
8
1
4
0
È2
# ‹
&
y$
3 “ %
'c
1Î2
1Î2
1
2
a1 ksin xkb dx
(1 sin x) dx
1Î#
(1 sin x) dx œ 2[x cos x]!
œ 2 ˆ 1# 1‰ œ 1 2
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347
348
Chapter 5 Integration
25. a œ 0, b œ 1, f(x) g(x) œ 2 sin x sin 2x
'
Ê Aœ
1
0
(2 sin x sin 2x) dx œ 2 cos x cos 2x ‘ 1
#
!
œ 2 † (1) "# ‘ ˆ2 † 1 "# ‰ œ 4
26. a œ 13 , b œ 13 , f(x) g(x) œ 8 cos x sec# x
'c
1Î3
Ê Aœ
œ Š8 †
1Î3
È3
#
1Î$
a8 cos x sec# xb dx œ [8 sin x tan x]1Î$
È3‹ Š8 †
È3
#
È3‹ œ 6È3
27. f(y) œ Èy, g(y) œ 2 y, c œ 1, d œ 2
'
Ê Aœ
œ
'
1
2
d
c
[f(y) g(y)] dy œ
'
2
1
Èy (2 y)‘ dy
ˆÈy 2 y‰ dy œ ’ 23 y$Î# 2y œ Š 43 È2 4 2‹ ˆ 23 2 "# ‰ œ
4
3
#
y#
# “"
È2 7
6
œ
8 È 2 7
6
28. f(y) œ 6 y, g(y) œ y# , c œ 1, d œ 2
Ê Aœ
'
œ ’6y y#
#
œ4
c
7
3
d
[f(y) g(y)] dy œ
"
#
#
y$
3 “"
œ
'
2
1
a6 y y# b dy
œ ˆ12 2 83 ‰ ˆ6 24143
6
œ
"
#
3" ‰
13
6
29. f(x) œ x$ 3x# œ x# (x 3) Ê f w (x) œ 3x# 6x œ 3x(x 2) Ê f w œ ± ± !
#
Ê f(0) œ 0 is a maximum and f(2) œ 4 is a minimum. A œ ‰
œ ˆ 81
4 27 œ
30. A œ
'
a
0
4
3
a#
6
"# ‰ œ
'
a
&Î$
A# œ
"
y#
# “!
œ
0
(6 8 3) œ
"
10
'
0
1
a
x#
# “0
$
!
œ a# 34 Èa † aÈa a#
6
ˆy#Î$ y‰ dy
; the area below the x-axis is
'c ˆy#Î$ y‰ dy œ ’ 3y5
0
%
ax$ 3x# b dx œ ’ x4 x$ “
ˆa 2Èa x"Î# x‰ dx œ ’ax 43 Èa x$Î# 31. The area above the x-axis is A" œ
œ ’ 3y5 0
3
27
4
ˆa"Î# x"Î# ‰# dx œ
œ a# ˆ1 '
&Î$
1
Ê the total area is A" A# œ
!
y#
# “ "
œ
11
10
6
5
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
a#
#
Chapter 5 Practice Exercises
'
32. A œ
1Î4
0
'
31Î2
51Î4
(cos x sin x) dx '
51Î4
1Î4
(sin x cos x) dx
1Î%
(cos x sin x) dx œ [sin x cos x]!
&1Î%
$1Î#
[ cos x sin x]1Î% [sin x cos x]&1Î%
œ ’Š
È2
#
È2
# ‹
’(1 0) Š
33. y œ x# 34. y œ
'
x
0
'
x
"
1 t
È2
#
dt Ê
dy
dx
È2
# ‹“
'
0
0
È2
# ‹
È2
#
Š
œ
8È 2
#
2 œ 4È2 2
"
x
Ê
d# y
dx#
œ 2x ˆ1 2Èsec t‰ dt Ê
xœ0 Ê yœ
È2
#
(0 1)“ ’Š
œ2
"
x#
; y(1) œ 1 œ 1 2Èsec x Ê
dy
dx
d# y
dx#
ˆ1 2Èsec t‰ dt œ 0 and x œ 0 Ê
dy
dx
36. y œ
'c È2 sin# t dt 2 so that dydx œ È2 sin# x; x œ 1
sin t
t
dt 3 Ê
dy
dx
œ
;xœ5 Ê yœ
sin x
x
5
5
sin t
t
1
1
"
t
dt œ 1 and yw (1) œ 2 1 œ 3
œ 1 2Èsec 0 œ 3
'
x
'
œ 2 ˆ "# ‰ (sec x)"Î# (sec x tan x) œ Èsec x (tan x);
35. y œ
5
'
È2
# ‹“
dt 3 œ 3
x
Ê yœ
1
'cc È2 sin# t dt 2 œ 2
1
1
37. Let u œ cos x Ê du œ sin x dx Ê du œ sin x dx
' 2(cos x)"Î# sin x dx œ ' 2u"Î# ( du) œ 2 ' u"Î# du œ 2 Š u"Î#" ‹ C œ 4u"Î# C
#
œ 4(cos x)"Î# C
38. Let u œ tan x Ê du œ sec# x dx
' (tan x)$Î# sec# x dx œ ' u$Î# du œ ˆu"Î#"‰ C œ 2u"Î# C œ (tanx)2 "Î# C
#
39. Let u œ 2) 1 Ê du œ 2 d) Ê
"
#
du œ d)
' [2) 1 2 cos (2) 1)] d) œ ' (u 2 cos u) ˆ "# du‰ œ u4
#
œ )# ) sin (2) 1) C, where C œ C" 40. Let u œ 2) 1 Ê du œ 2 d) Ê
'Š
œ
41.
42.
"
#
"
È 2 ) 1
"
#
2 sec# (#) 1)‹ d) œ
"
4
sin u C" œ
(2)1)#
4
sin (2) 1) C"
is still an arbitrary constant
du œ d)
' Š È"u 2 sec# u‹ ˆ #" du‰ œ #" ' ˆu"Î# 2 sec# u‰ du
"Î#
Š u " ‹ "# (2 tan u) C œ u"Î# tan u C œ (2) 1)"Î# tan (2) 1) C
#
' ˆt 2t ‰ ˆt 2t ‰ dt œ ' ˆt# t4 ‰ dt œ ' at# 4t# b dt œ t3$ 4 Š t"1 ‹ C œ t3$ 4t C
#
t#
' (t1)t%#1 dt œ ' t#t%2t dt œ ' ˆ t"# t2$ ‰ dt œ ' at# 2t$ b dt œ (t"1) 2 Š #
‹ C œ "t t"# C
43. Let u œ #t$Î# Ê du œ $Èt dt Ê "$ du œ Èt dt
' Èt sin ˆ#t$Î# ‰dt œ "$ ' sin u du œ "$ cos u C œ "$ cosˆ#t$Î# ‰ C
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349
350
Chapter 5 Integration
44. Let u œ " sec ) Ê du œ sec ) tan) d) Ê
' sec ) tan) È" sec ) d) œ ' u"Î# du œ #$ u$Î# C
œ #$ a" sec )b$Î# C
'c a3x# 4x 7b dx œ cx$ 2x# 7xd "" œ c1$ 2(1)# 7(1)d c(1)$ 2(1)# 7(1)d œ 6 (10) œ 16
1
45.
1
46.
'
47.
'
48.
'
49.
'
1
0
"
a8s$ 12s# 5b ds œ c2s% 4s$ 5sd ! œ c2(1)% 4(1)$ 5(1)d 0 œ 3
2
4
#
1 v
27
1
'
dv œ
2
#
4v# dv œ c4v" d " œ ˆ #4 ‰ ˆ 14 ‰ œ 2
1
#(
x%Î$ dx œ 3x"Î$ ‘ " œ 3(27)"Î$ ˆ3(1)"Î$ ‰ œ 3 ˆ "3 ‰ 3(1) œ 2
4
dt
1 tÈt
œ
'
4
œ
dt
t$Î#
1
'
4
1
%
50. Let x œ 1 Èu Ê dx œ
'
4
ˆ1 Èu‰"Î#
Èu
1
du œ
'
3
2
2
È4
t$Î# dt œ 2t"Î# ‘ " œ
"
#
u"Î# du Ê 2 dx œ
du
Èu
(2)
È1
œ1
; u œ 1 Ê x œ 2, u œ 4 Ê x œ 3
$
x"Î# (2 dx) œ 2 ˆ 23 ‰ x$Î# ‘ # œ
4
3
ˆ3$Î# ‰ 43 ˆ2$Î# ‰ œ 4È3 83 È2 œ
4
3
Š3È3 2È2‹
51. Let u œ 2x 1 Ê du œ 2 dx Ê 18 du œ 36 dx; x œ 0 Ê u œ 1, x œ 1 Ê u œ 3
'
1
36 dx
$
0 (2x1)
œ
'
3
# $
$
9 ‘
ˆ 9 ‰ ˆ 9 ‰
18u$ du œ ’ "8u
2 “ œ u # " œ 3 # 1 # œ 8
"
1
52. Let u œ 7 5r Ê du œ 5 dr Ê "5 du œ dr; r œ 0 Ê u œ 7, r œ 1 Ê u œ 2
'
1
dr
$
È
(7 5r)#
0
œ
'
'
1
(7 5r)#Î$ dr œ
0
2
7
#
u#Î$ ˆ 5" du‰ œ 5" 3u"Î$ ‘ ( œ
53. Let u œ 1 x#Î$ Ê du œ 23 x"Î$ dx Ê 3# du œ x"Î$ dx; x œ
x œ 1 Ê u œ 1 1#Î$ œ 0
'
1
1Î8
œ
x"Î$ ˆ1 x#Î$ ‰
$Î#
'
dx œ
0
1Î2
0
x$ a1 9x% b
" ˆ 25 ‰
œ 18
16
"Î#
$Î#
dx œ
'
0
1
sin# 5r dr œ
56. Let u œ 4t '
1Î%
0
œ
1
8
1
4
$Î%
"
36
#Î$
œ
3
4
,
!
&Î#
œ 35 u&Î# ‘ $Î% œ 35 (0)&Î# ˆ 35 ‰ ˆ 34 ‰
'
51
0
"
16
"
16
"
5
"Î#
#
1
8
' Î acos
1 4
"
%
Ê u œ 1 9 ˆ #" ‰ œ
25
16
#&Î"'
" "Î# ‘
œ 18
u
"
du œ dr; r œ 0 Ê u œ 0, r œ 1 Ê u œ 51
Ê du œ 4 dt Ê
œ
#&Î"'
"
#
"
90
asin# ub ˆ "5 du‰ œ
31Î4
du œ x$ dx; x œ 0 Ê u œ 1, x œ
"
"
u$Î# ˆ 36
du‰ œ ’ 36
Š u " ‹“
"
ˆ 18
(1)"Î# ‰ œ
cos# ˆ4t 14 ‰ dt œ
25Î16
1
55. Let u œ 5r Ê du œ 5 dr Ê
'
!
#
Ê u œ 1 ˆ 8" ‰
27È3
160
54. Let u œ 1 9x% Ê du œ 36x$ dx Ê
'
&Î#
u$Î# ˆ #3 du‰ œ ’ˆ 32 ‰ Š u 5 ‹“
3Î4
"
8
Š $È7 $È2‹
3
5
"
4
#
"
5
u2 sin 2u ‘ &1
4
!
œ ˆ 1# sin 101 ‰
#0
du œ dt; t œ 0 Ê u œ 14 , t œ
ub ˆ "4 du‰ œ
"
4
u2 sin 2u ‘ $1Î%
4
1Î%
œ
ˆ0 1
4
"
4
sin 0 ‰
20
Ê uœ
Š 381 œ
1
#
31
4
sin ˆ 3#1 ‰
‹
4
4" Š 18 Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
sin ˆ 1# ‰
‹
4
Chapter 5 Practice Exercises
'
1Î$
57.
'
31Î4
58.
0
1Î$
sec# ) d) œ [tan )]!
1Î4
31
1
1
3
tan 0 œ È3
$1Î%
csc# x dx œ [cot x]1Î% œ ˆ cot
59. Let u œ
'
œ tan
cot#
"
6
Ê du œ
x
6
dx œ
x
6
'
1Î6
31 ‰
4
ˆ cot 14 ‰ œ 2
dx Ê 6 du œ dx; x œ 1 Ê u œ 16 , x œ 31 Ê u œ
1Î2
351
'
6 cot# u du œ 6
1Î2
1
#
1Î#
acsc# u 1b du œ [6(cot u u)]1Î' œ 6 ˆ cot
1Î6
1
#
1# ‰ 6 ˆcot
1
6
16 ‰
œ 6È3 21
60. Let u œ
'
1
0
tan#
)
3
)
3
œ 3 tan
"
3
Ê du œ
d) œ
1
3
'
0
1
d) Ê 3 du œ d); ) œ 0 Ê u œ 0, ) œ 1 Ê u œ
)
3
ˆsec#
1‰ d) œ
'
1Î3
0
1
3
1Î$
3 asec# u 1b du œ [3 tan u 3u]!
3 ˆ 13 ‰‘ (3 tan 0 0) œ 3È3 1
'c
sec x tan x dx œ [sec x]!1Î$ œ sec 0 sec ˆ 13 ‰ œ 1 2 œ 1
'
csc z cot z dz œ [csc z]1Î% œ ˆ csc
0
61.
62.
1Î3
31Î4
1Î4
$1Î%
31 ‰
4
ˆ csc 14 ‰ œ È2 È2 œ 0
63. Let u œ sin x Ê du œ cos x dx; x œ 0 Ê u œ 0, x œ
'
1Î2
0
'
5(sin x)$Î# cos x dx œ
1
0
1
#
Ê uœ1
"
"
5u$Î# du œ 5 ˆ 25 ‰ u&Î# ‘ ! œ 2u&Î# ‘ ! œ 2(1)&Î# 2(0)&Î# œ 2
64. Let u œ 1 x# Ê du œ 2x dx Ê du œ 2x dx; x œ 1 Ê u œ 0, x œ 1 Ê u œ 0
'c
1
1
'
2x sin a1 x# b dx œ
0
sin u du œ 0
0
"
3
65. Let u œ sin 3x Ê du œ 3 cos 3x dx Ê
œ 1
'c ÎÎ
1 2
15 sin% 3x cos 3x dx œ
1 2
du œ cos 3x dx; x œ 1# Ê u œ sin ˆ 3#1 ‰ œ 1, x œ
1
#
&
&
' 15u% ˆ "3 du‰ œ ' 5u% du œ cu& d "
" œ (1) (1) œ 2
1
1
1
1
66. Let u œ cos ˆ x# ‰ Ê du œ "# sin ˆ x# ‰ dx Ê 2 du œ sin ˆ x# ‰ dx; x œ 0 Ê u œ cos ˆ 0# ‰ œ 1, x œ
œ
'
"
#
21Î3
cos% ˆ x# ‰ sin ˆ x# ‰ dx œ
0
'
1
1Î2
$
u% (2 du) œ ’2 Š u3 ‹“
67. Let u œ 1 3 sin# x Ê du œ 6 sin x cos x dx Ê
Ê u œ 1 3 sin#
'
1Î2
0
3 sin x cos x
È1 3 sin# x
1
#
œ4
dx œ
'
4
"
Èu
1
ˆ #" du‰ œ
'
4
1
"
#
'
0
1Î4
1
4
Ê u œ 1 7 tan
sec# x
(1 7 tan x)#Î$
dx œ
'
1
1
4
8
"
#
"Î#
"
œ
2
3
ˆ "# ‰$ 32 (1)$ œ
2
3
(8 1) œ
du œ 3 sin x cos x dx; x œ 0 Ê u œ 1, x œ
"Î#
%
#
"
21
3
21
Ê u œ cos Š #3 ‹
14
3
1
#
%
u"Î# du œ ’ 2" Š u " ‹“ œ u"Î# ‘ " œ 4"Î# 1"Î# œ 1
"
7
68. Let u œ 1 7 tan x Ê du œ 7 sec# x dx Ê
xœ
Ê u œ sin ˆ 3#1 ‰
du œ sec# x dx; x œ 0 Ê u œ 1 7 tan 0 œ 1,
œ8
"
u#Î$
ˆ 7" du‰ œ
'
1
8
"
7
"Î$
)
3
"
)
u#Î$ du œ ’ 7" Š u " ‹“ œ 37 u"Î$ ‘ " œ
3
7
(8)"Î$ 37 (1)"Î$ œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
3
7
352
Chapter 5 Integration
69. Let u œ sec ) Ê du œ sec ) tan ) d); ) œ 0 Ê u œ sec 0 œ 1, ) œ
'
1Î3
0
tan )
È2 sec )
"
È2
œ
'
d) œ
"Î#
#
#
"
'
1Î3
0
’ ˆu " ‰ “ œ ’
1Î3
sec ) tan )
sec ) tan )
d) œ
È2 (sec ))$Î#
sec ) È2 sec )
0
#
2
2
2
È2u “ œ È2(2) Š È2(1) ‹ œ
"
cos Èt
2È t
70. Let u œ sin Èt Ê du œ ˆcos Èt‰ ˆ "# t"Î# ‰ dt œ
#
1
4
tœ
'
1 Î4
#
Ê u œ sin
71. (a) av(f) œ
'
"
1 (1)
'c
"
k (k)
(b) av(f) œ
"
#k
œ
1
"
1Î2 Èu
1
1
'c
k
k
(2 du) œ 2
73. favw œ
'
a
b
"
È2 u$Î#
du œ
"
È2
'
1
2
œ2
u$Î# du
È2 1
dt Ê 2 du œ
cos Èt
Èt
dt; t œ
1#
36
Ê u œ sin
1
6
œ
"
#
,
"
"
"
#k
’ mx2 bx“
"
#
’ mx2 bx“
(2bk) œ b
"
30
1
1
3
u"Î# du œ 4Èu‘ "Î# œ 4È1 4É #" œ 2 Š2 È2‹
#
(mx b) dx œ
3
0
3
0
0
"
ba
1
1Î2
"
#
a
(b) yav
'
(mx b) dx œ
' È3x dx œ "3 '
œ a " 0 ' Èax dx œ "a '
72. (a) yav œ
'
Ê u œ sec
œ1
dt œ
cos Èt
Ét sin Èt
1# Î36
1
#
d) œ
2
1
3
È3 x"Î# dx œ
a
Èa x"Î# dx œ
0
"
b a
Èaxf w (x) dx œ
[f(x)]ab œ
"
b a
k
ck
#
#
m(1)
b(1)‹“ œ
’Š m(1)
2 b(1)‹ Š #
œ
"
#
œ
"
#k
#
"
#
(2b) œ b
#
m(k)
b(k)‹“
’Š m(k)
2 b(k)‹ Š #
È3
3
23 x$Î# ‘ $ œ
!
È3
3
23 (3)$Î# 23 (0)$Î# ‘ œ
È3
3
Š2È3‹ œ 2
Èa
a
23 x$Î# ‘ a œ
!
Èa
a
ˆ 23 (a)$Î# 23 (0)$Î# ‰ œ
Èa
a
ˆ 32 aÈa‰ œ
[f(b) f(a)] œ
f(b) f(a)
ba
2
3
a
so the average value of f w over [aß b] is the
slope of the secant line joining the points (aß f(a)) and (bß f(b)), which is the average rate of change of f over [aß b].
74. Yes, because the average value of f on [aß b] is
and the average value of the function is
"
#
'
a
"
ba
'
a
b
f(x) dx. If the length of the interval is 2, then b a œ 2
b
f(x) dx.
75. We want to evaluate
"
$'& !
'
$'&
!
f(x) dx œ
"
$'&
'
$'&
!
#1
Œ$(sin” $'& ax "!"b• #&dx œ
#1
Notice that the period of y œ sin” $'&
ax "!"b• is
length 365. Thus the value of
76.
"
'(&#!
œ
'#
'(&
!
$(
$'&
'
$'&
!
"
'&& Œ”)Þ#(a'(&b
#'a'(&b
#†"!&
#1
$'&
"Þ)(a'(&b
$†"!&
$
'
!
$'&
#1
sin” $'&
ax "!"b•dx #&
$'&
'
$'&
!
dx
œ $'& and that we are integrating this function over an iterval of
#1
ax "!"b•dx sin” $'&
a)Þ#( "!& a#'T "Þ)(T# bbdT œ
#
#1
$(
$'&
"
'&& ”)Þ#(T
• ”)Þ#(a#!b #&
$'&
#'T#
#†"!&
#'a#!b
#†"!&
#
'
!
$'&
dx is
"Þ)(T$
$†"!& •
"Þ)(a#!b
$†"!&
$
$(
$'&
†!
#&
$'&
† $'& œ #&.
'(&
#!
• ¸
"
'&& a$(#%Þ%%
"'&Þ%!b
œ &Þ%$ œ the average value of Cv on [20, 675]. To find the temperature T at which Cv œ &Þ%$, solve
&Þ%$ œ )Þ#( "!& a#'T "Þ)(T# b for T. We obtain "Þ)(T# #'T #)%!!! œ !
ÊTœ
#' „ Éa#'b# %a"Þ)(ba#)%!!!b
È#"#%**'
œ #' „ $Þ(%
.
#a"Þ)(b
‰
So T œ $)#Þ)# or T œ $*'Þ(#. Only T œ $*'Þ(# lies in the
interval [20, 675], so T œ $*'Þ(# C.
77.
dy
dx
œ È# cos$ x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Chapter 5 Practice Exercises
78.
dy
dx
œ È# cos$ a(x# b †
79.
dy
dx
œ
d
dx Œ
80.
dy
dx
œ
d
dx Œ
d
#
dx a(x b
œ "%xÈ# cos$ a(x# b
' x $ ' t dt œ $'x
%
1
353
%
'sec# x t " " dt œ dxd Œ'#sec x t " " dt œ sec "x " dxd asec xb œ sec" xsectan xx
#
#
#
#
81. Yes. The function f, being differentiable on [aß b], is then continuous on [aß b]. The Fundamental Theorem of
Calculus says that every continuous function on [aß b] is the derivative of a function on [aß b].
82. The second part of the Fundamental Theorem of Calculus states that if F(x) is an antiderivative of f(x) on
[aß b], then
'
0
1
'
b
a
f(x) dx œ F(b) F(a). In particular, if F(x) is an antiderivaitve of È1 x% on [0ß 1], then
È1 x% dx œ F(1) F(0).
83. y œ
' x È1 t# dt œ ' x È1 t# dt
84. y œ
'
1
1
0
"
#
cos x 1 t
dt œ '
0
cos x
"
1 t#
dt Ê
Ê
dy
dx
œ
d
dx
dy
dx
œ
d
dx
”
' x È1 t# dt• œ dxd ”' x È1 t# dt• œ È1 x#
”
'
1
cos x
0
"
d
‰ ˆ dx
œ ˆ 1 cos
(cos x)‰ œ ˆ sin"# x ‰ ( sin x) œ
#x
1
"
1 t#
"
sin x
d
dt• œ dx
”
'
0
cos x
"
1 t#
dt•
œ csc x
85. We estimate the area A using midpoints of the vertical intervals, and we will estimate the width of the parking lot on each
interval by averaging the widths at top and bottom. This gives the estimate
A ¸ "& † ˆ ! # $' $' # &% &% # &" &" #%*Þ& %*Þ&# &% &% #'%Þ% '%Þ% # '(Þ& '(Þ&# %# ‰
A ¸ &*'" ft# . The cost is Area † ($2.10/ft# ) ¸ a5961 ft# b a$2.10/ft# b œ $12,518.10 Ê the job cannot be done for $11,000.
86. (a) Before the chute opens for A, a œ 32 ft/sec# . Since the helicopter is hovering, v! œ 0 ft/sec
Ê v œ ' 32 dt œ 32t v! œ 32t. Then s! œ 6400 ft Ê s œ ' 32t dt œ 16t# s! œ 16t# 6400.
At t œ 4 sec, s œ 16(4)# 6400 œ 6144 ft when A's chute opens;
(b) For B, s! œ 7000 ft, v! œ 0, a œ 32 ft/sec# Ê v œ ' 32 dt œ 32t v! œ 32t Ê s œ ' 32t dt
œ 16t# s! œ 16t# 7000. At t œ 13 sec, s œ 16(13)# 7000 œ 4296 ft when B's chute opens;
(c) After the chutes open, v œ 16 ft/sec Ê s œ ' 16 dt œ 16t s! . For A, s! œ 6144 ft and for B,
s! œ 4296 ft. Therefore, for A, s œ 16t 6144 and for B, s œ 16t 4296. When they hit the ground,
4296
s œ 0 Ê for A, 0 œ 16t 6144 Ê t œ 6144
16 œ 384 seconds, and for B, 0 œ 16t 4296 Ê t œ 16
œ 268.5 seconds to hit the ground after the chutes open. Since B's chute opens 58 seconds after A's opens
Ê B hits the ground first.
87. av(I) œ
œ
"
30
"
30
'
30
0
(1200 40t) dt œ
"
30
$!
c1200t 20t# d! œ
"
30
ca(1200(30) 20(30)# b a1200(0) 20(0)# bd
(18,000) œ 600; Average Daily Holding Cost œ (600)($0.03) œ $18
88. av(I) œ
"
14
'
0
14
(600 600t) dt œ
"
14
"%
c600t 300t# d! œ
"
14
c600(14) 300(14)# 0d œ 4800; Average Daily
Holding Cost œ (4800)($0.04) œ $192
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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354
Chapter 5 Integration
'
"
30
89. av(I) œ
30
#
0
$
"
30
Š450 t# ‹ dt œ
’450t t6 “
œ (300)($0.02) œ $6
œ
"
60
'
"
60
90. av(I) œ
60
0
40È15
3
’600(60) (60)
$Î#
'
"
60
Š600 20È15t‹ dt œ
0
"
60
0“ œ
60
$!
!
"
30
œ
30$
6
’450(30) 0“ œ 300; Average Daily Holding Cost
Š600 20È15 t"Î# ‹ dt œ
"
60
’600t 20È15 ˆ 23 ‰ t$Î# “
'!
!
ˆ36,000 ˆ 320
‰ 15# ‰ œ 200; Average Daily Holding Cost
3
œ (200)($0.005) œ $1.00
CHAPTER 5 ADDITIONAL AND ADVANCED EXERCISES
'
1. (a) Yes, because
1
0
(b) No. For example,
4È 2
3
œ
'
1
0
'
"
7
f(x) dx œ
1
0
"
7
7f(x) dx œ
(7) œ 1
'
"
8x dx œ c4x# d ! œ 4, but
1
0
"
È8x dx œ ’2È2 Š x$Î#
œ
3 ‹“
2
2
5
5
5
2
2
2
2
2
œ432œ9
'c f(x) dx œ 4 3 œ 7 2 œ 'c g(x) dx
(c) False:
5
5
2
2
œ
'
x
0
sin ax
a
sin ax
a
'
0
f(t) cos at dt '
d
Πdx
œ cos ax
x
0
'
x
0
f(x) dx 5
2
5
0
2
x
0
0
x
x
0
0
'
f(t) cos at dt x
0
f(t) sin at dt (f(x) cos ax) sin ax
sin ax
a
x
cos ax
a
'
x
0
d
Πdx
'
0
x
f(t) sin at dt
f(t) sin at dt cos ax
a
(f(x) sin ax)
x
0
x
#
#
0
x
0
'
d
(sin ax) Πdx
a cos ax
'
0
x
0
x
0
f(t) sin at dt œ a sin ax
4. x œ
'
"
#
x
0
y
"
0 È1 4t#
Ê 1œ
œ
'
'
f(t) cos at dt dt Ê
"
È14y#
a1 4y# b
'
0
x
d
dx
Š dy
dx ‹ Ê
"Î#
(x) œ
dy
dx
x
0
'
d
dx
0
'
0
y
x
'
'
"
È1 4t#
dy
4y Š dx
‹
È1 4y#
0
x
0
x
f(t) cos at dt a cos ax
'
0
x
f(t) sin at dt f(x).
f(t) cos at dt f(x)
f(t) sin at dt œ f(x). Note also that yw (0) œ y(0) œ 0.
dt œ
œ È1 4y# . Then
(8y) Š dy
dx ‹ œ
f(t) sin at dt
f(t) cos at dt (cos ax)f(x) cos ax
f(t) sin at dt a sin ax
cos ax
a
x
0
f(t) sin at dt (sin ax)f(x) sin ax œ a sin ax
Therefore, yww a# y œ a cos ax
a# Πsinaax
2
2
x
"
a
f(t) cos at dt sin ax
œ cos ax
dy
dx
5
5
' f(t) cos at dt sin ax ' f(t) sin at dt. Next,
d y
' f(t) cos at dt (cos ax) Πdxd ' f(t) cos at dt a cos ax '
dx œ a sin ax
Ê
'c g(x) dx
' f(t) sin ax cos at dt "a ' f(t) cos ax sin at dt
cos ax '
' f(t) cos at dt
f(t) sin at dt Ê dy
a
dx œ cos ax Œ
f(t) sin a(x t) dt œ
x
2
5
'c [f(x) g(x)] dx 0 Ê 'c [g(x) f(x)] dx 0.
Ê ' [g(x) f(x)] dx 0 which is a contradiction.
c
Ê
On the other hand, f(x) Ÿ g(x) Ê [g(x) f(x)]
"
a
ˆ1$Î# 0$Î# ‰
5
5
(b) True:
4È 2
3
Á È4
' f(x) dx œ ' f(x) dx œ 3
'c [f(x) g(x)] dx œ 'c f(x) dx 'c g(x) dx œ 'c f(x) dx '
2. (a) True:
3. y œ
!
#
œ
'
d
dy
”
d# y
dx#
œ
4y ˆÈ1 4y# ‰
È1 4y#
y
0
"
È1 4t#
d
dx
ˆÈ1 4y# ‰ œ
œ 4y. Thus
dt• Š dy
dx ‹ from the chain rule
d# y
dx#
d
dy
ˆÈ1 4y# ‰ Š dy
dx ‹
œ 4y, and the constant of
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Chapter 5 Additional and Advanced Exercises
355
proportionality is 4.
'
5. (a)
x#
f(t) dt œ x cos 1x Ê
0
cos 1x 1x sin 1x
.
2x
Ê f ax# b œ
'
(b)
f(x)
0
d
dx
$
t# dt œ ’ t3 “
f(x)
!
œ
"
3
'
x#
f(t) dt œ cos 1x 1x sin 1x Ê f ax# b (2x) œ cos 1x 1x sin 1x
0
cos 21 21 sin 21
4
Thus, x œ 2 Ê f(4) œ
"
3
(f(x))$ Ê
"
4
œ
(f(x))$ œ x cos 1x Ê (f(x))$ œ 3x cos 1x Ê f(x) œ $È3x cos 1x
Ê f(4) œ $È3(4) cos 41 œ $È12
6.
'
a
f(x) dx œ
0
a#
#
a
#
Ê f(a) œ Fw (a) œ a 7.
'
b
1
1
#
sin a "
#
cos a. Let F(a) œ
sin a a
#
1
#
cos a f(x) dx œ Èb# 1 È2 Ê f(b) œ
d
db
'
a
0
f(t) dt Ê f(a) œ Fw (a). Now F(a) œ
sin a Ê f ˆ 1# ‰ œ
'
b
1
f(x) dx œ
"
#
side of the equation is:
œ
d
dx
œ
'
0
'
”x
0
x
f(u) du• '
d
dx
”
d
dx
'
x
0
x
0
f(u)(x u) du• œ
u f(u) du œ
'
x
0
' ”'
dy
dx
d
dx
0
'
x
0
0
u
"
#
1
#
sin
"Î#
'
0
x
d
dx
(2b) œ
f(t) dt• du• œ
f(u) x du d
f(u) du x ” dx
'
'
0
cos
1
#
1
#
a
#
sin a sin
1
#
Ê f(x) œ
b
È b# 1
œ
1
#
1
#
cos a
"
#
1
#
œ
"
#
x
È x# 1
x
f(t) dt; the derivative of the right
x
0
u f(u) du
f(u) du• xf(x) œ
'
0
x
f(u) du xf(x) xf(x)
x
f(u) du. Since each side has the same derivative, they differ by a constant, and since both sides equal 0
when x œ 0, the constant must be 0. Therefore,
9.
”
ab# 1b
x
d
dx
8. The derivative of the left side of the equation is:
1
#
ˆ 1# ‰
#
a#
#
' ”'
x
0
0
u
f(t) dt• du œ
'
0
x
f(u)(x u) du.
œ 3x# 2 Ê y œ ' a3x# 2b dx œ x$ 2x C. Then (1ß 1) on the curve Ê 1$ 2(1) C œ 1 Ê C œ 4
Ê y œ x$ 2x 4
10. The acceleration due to gravity downward is 32 ft/sec# Ê v œ ' 32 dt œ 32t v! , where v! is the initial
velocity Ê v œ 32t 32 Ê s œ ' (32t 32) dt œ 16t# 32t C. If the release point, at t œ !, is s œ 0, then
C œ 0 Ê s œ 16t# 32t. Then s œ 17 Ê 17 œ 16t# 32t Ê 16t# 32t 17 œ 0. The discriminant of this
quadratic equation is 64 which says there is no real time when s œ 17 ft. You had better duck.
11.
'c f(x) dx œ 'c x#Î$ dx '
œ
œ
œ
12.
3
0
8
8
3
4 dx
0
35 x&Î$ ‘ ! [4x]!$
)
ˆ0 35 (8)&Î$ ‰ (4(3)
36
5
0) œ
'c f(x) dx œ 'c Èx dx '
3
0
4
4
!
3
0
$
œ 23 (x)$Î# ‘ % ’ x3 4x“
96
5
12
ax# 4b dx
$
!
$
œ 0 ˆ 23 (4)$Î# ‰‘ ’ Š 33 4(3)‹ 0 “
œ
16
3
3œ
7
3
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356
13.
Chapter 5 Integration
'
2
0
'
g(t) dt œ
1
t dt 0
"
#
'
2
1
sin 1t dt
#
œ ’ t2 “ 1" cos 1t‘ "
!
œ ˆ "# 0‰ 1" cos 21 ˆ 1" cos 1‰‘
"
#
œ
14.
'
2
0
2
1
h(z) dz œ
'
1
0
'
È1 z dz 1
2
(7z 6)"Î$ dz
"
#
3
œ 23 (1 z)$Î# ‘ ! 14
(7z 6)#Î$ ‘ "
œ 23 (1 1)$Î# ˆ 23 (1 0)$Î# ‰‘
3
14
(7(2) 6)#Î$ 6
3 ‰
55
œ ˆ 7 14
œ 42
3
14
(7(1) 6)#Î$ ‘
2
3
'c f(x) dx œ 'cc dx 'c a1 x# b dx '
2
15.
1
2
1
2
1
"
x$
3 “ "
œ [x]"
# ’x 1$
3‹
16.
ˆ 23 ‰ 4 2 œ
2
3
'c h(r) dr œ 'c r dr '
2
0
1
1
#
œ ’ r2 “
!
"
2
3
Š Š1 1œ
a1 r# b dr "
ba
1$
3‹
'
2
1
7
6
'
b
a
f(x) dx œ
"
#0
'
2
0
f(x) dx œ
#
’Š 1# 0‹ Š 2# 2‹ Š 1# 1‹“ œ
'
20. f(x) œ
'
x
"
t
1/x
"
ba
sin x
" "
t 1 t#
"
sin x
'
a
'ÈÈ sin t# dt
22. f(x) œ
'
y
x
xb3
f(x) dx œ
y
"
x
"
30
'
3
0
"
#
”
'
1
0
x dx '
2
1
(x 1) dx• œ
"
#
#
"
#
’ x2 “ #" ’ x2 x“
!
#
"
"
#
f(x) dx œ
"
3
”
'
dx ‰
d ˆ " ‰‰
ˆ dx
Š "" ‹ ˆ dx
œ
x
x
0
"
x
1
dx '
1
2
0 dx x ˆ x"# ‰ œ
"
x
'
3
2
dx• œ
"
x
œ
#
"
3
[1 0 0 3 2] œ
2
3
2
x
"
d
"
d
‰ ˆ dx
‰ ˆ dx
dt Ê f w (x) œ ˆ 1 sin
(sin x)‰ ˆ 1 cos
(cos x)‰ œ
#x
#x
21. g(y) œ
2
b
dt Ê f w (x) œ
cos x
"
cos x
dr
0‹ a2 1b
#
18. Ave. value œ
œ
13
3
"
#
19. f(x) œ
’2(2) 2(1)“
!
(1)#
# ‹
17. Ave. value œ
"
#
(1)$
3 ‹•
Š1 ’r r3 “ [r]#"
œ "# œ
1
0
$
œ Š0 2 dx
[2x]#"
œ a1 (2)b ”Š1 œ1
2
1
#
cos x
cos# x
sin x
sin# x
d ˆ
d ˆ
Èy‰‹ œ
Ê gw (y) œ Šsin ˆ2Èy‰ ‹ Š dy
2Èy‰‹ Šsin ˆÈy‰ ‹ Š dy
sin 4y
Èy
sin y
2È y
d
‰
t(5 t) dt Ê f w (x) œ (x 3)(& (x 3)) ˆ dx
(x 3)‰ x(5 x) ˆ dx
dx œ (x 3)(2 x) x(5 x)
œ 6 x x# 5x x# œ 6 6x. Thus f w (x) œ 0 Ê 6 6x œ 0 Ê x œ 1. Also, f ww (x) œ 6 ! Ê x œ 1 gives a
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Chapter 5 Additional and Advanced Exercises
maximum.
23. Let f(x) œ x& on [0ß 1]. Partition [0ß 1] into n subintervals with ?x œ
10
n
œ "n . Then "n , n2 , á ,
_
n
n
&
are the
right-hand endpoints of the subintervals. Since f is increasing on [0ß 1], U œ ! Š nj ‹ ˆ "n ‰ is the upper sum for
_
œ
j 1
&
! Š j ‹ ˆ"‰ œ
f(x) œ x& on [0ß 1] Ê n lim
n
n
Ä_
jœ1
œ
'
1
'
"
x& dx œ ’ x6 “ œ
!
0
&
lim " ’ˆ n" ‰
nÄ_ n
ˆ n2 ‰&
&
á ˆ nn ‰ “ œ n lim
’1
Ä_
&
2& á n&
“
n'
"
6
24. Let f(x) œ x$ on [0ß 1]. Partition [0ß 1] into n subintervals with ?x œ
10
n
œ "n . Then "n , n2 , á ,
_
n
n
$
are the
right-hand endpoints of the subintervals. Since f is increasing on [0ß 1], U œ ! Š nj ‹ ˆ "n ‰ is the upper sum for
_
œ
j 1
$
! Š j ‹ ˆ " ‰ œ lim
f(x) œ x$ on [0ß 1] Ê n lim
n
n
Ä_
nÄ_
jœ1
œ
'
0
1
%
"
x$ dx œ ’ x4 “ œ
!
"
n
$
$
$
’ˆ n" ‰ ˆ n2 ‰ á ˆ nn ‰ “ œ n lim
’1
Ä_
$
2$ á n$
“
n%
"
4
25. Let y œ f(x) on [0ß 1]. Partition [0ß 1] into n subintervals with ?x œ
10
n
œ "n . Then "n , 2n , á ,
_
n
n
are the
right-hand endpoints of the subintervals. Since f is continuous on [!ß 1], ! f Š nj ‹ ˆ "n ‰ is a Riemann sum of
œ
j 1
_
! f Š j ‹ ˆ " ‰ œ lim
y œ f(x) on [0ß 1] Ê n lim
n
n
Ä_
nÄ_
jœ1
"
n
'
1
"
26. (a) n lim
[2 4 6 á 2n] œ n lim
Ä _ n#
Ä_
on [0ß 1] (see Exercise 25)
"
n
n2 "
(b) n lim
c1"& 2"& á n"& d œ n lim
Ä _ n"'
Ä_
"&
f(x) œ x on [0ß 1] (see Exercise 25)
"
n
"&
"&
"&
’ˆ 1n ‰ ˆ 2n ‰ á ˆ nn ‰ “ œ
'
'
f ˆ n" ‰ f ˆ n2 ‰ á f ˆ nn ‰‘ œ
4
n
6
n
á œ
2n ‘
n
0
1
0
f(x) dx
"
2x dx œ cx# d! œ 1, where f(x) œ 2x
'
0
1
"'
"
"
16 ,
x"& dx œ ’ x16 “ œ
!
where
1
"
" (c) n lim
sin 1n sin 2n1 á sin nn1 ‘ œ
sin n1 dx œ 1" cos 1x‘ ! œ 1" cos 1 ˆ 1" cos 0‰
Ä_ n
0
œ 12 , where f(x) œ sin 1x on [0ß 1] (see Exercise 25)
"
(d) n lim
c1"& 2"& á n"& d œ Šn lim
Ä _ n"(
Ä_
" ‰
ˆ
œ 0 16 œ 0 (see part (b) above)
"
n"&
(e) n lim
Ä_
c1"& 2"& á n"& d œ n lim
Ä_
œ Šn lim
n‹ Šn lim
Ä_
Ä_
"
n"'
"
"
n ‹ Šn lim
Ä _ n"'
n
n"'
c1"& 2"& á n"& d‹ œ Šn lim
Ä_
"
n‹
'
1
0
x"& dx
c1"& 2"& á n"& d
c1"& 2"& á n"& d‹ œ Šn lim
n‹
Ä_
'
0
1
x"& dx œ _ (see part (b) above)
27. (a) Let the polygon be inscribed in a circle of radius r. If we draw a radius from the center of the circle (and
the polygon) to each vertex of the polygon, we have n isosceles triangles formed (the equal sides are equal
to r, the radius of the circle) and a vertex angle of )n where )n œ 2n1 . The area of each triangle is
An œ
" #
# r
sin )n Ê the area of the polygon is A œ nAn œ
(b) n lim
A œ n lim
Ä_
Ä_
nr#
#
sin
21
n
œ n lim
Ä_
n1r#
21
sin
21
n
nr#
#
nr#
21
# sin n .
sin ˆ 2n1 ‰
#
ˆ 2n1 ‰ œ a1r b
sin )n œ
œ n lim
a1 r # b
Ä_
lim
2 1 În Ä 0
sin ˆ 2n1 ‰
ˆ 2n1 ‰
œ 1 r#
'x cos 2t dt " œ sin x ' x cos 2t dt " Ê yw œ cos x cosa2xb; when x œ 1 we have
1
yw œ cos 1 cosa21b œ " " œ #. And yww œ sin x 2sina2xb; when x œ 1, y œ sin 1 ' cos 2t dt "
x
28. y œ sin x 1
1
œ ! ! " œ ".
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358
Chapter 5 Integration
'
ga$b œ '
29. (a) ga"b œ
(b)
1
1
$
1
(c) ga"b œ
fatb dt œ !
fatb dt œ "# a#ba"b œ "
' fatb dt œ ' fatb dt œ "% a1 ## b œ 1
1
1
1
1
(d) gw axb œ faxb œ ! Ê x œ $, ", $ and the sign chart for gw axb œ faxb is
relative maximum at x œ ".
(e) gw a"b œ fa"b œ # is the slope and ga"b œ
± ± ± . So g has a
3
1
3
' fatb dt œ 1, by (c). Thus the equation is y 1 œ #ax "b
1
1
y œ #x # 1 .
(f) gww axb œ f w axb œ ! at x œ " and gww axb œ f w axb is negative on a$ß "b and positive on a"ß "b so there is an
inflection point for g at x œ ". We notice that gww axb œ f w axb ! for x on a"ß #b and gww axb œ f w axb ! for x on
a#ß %b, even though gww a#b does not exist, g has a tangent line at x œ #, so there is an inflection point at x œ #.
(g) g is continuous on Ò$ß %Ó and so it attains its absolute maximum and minimum values on this interval. We saw in (d)
that gw axb œ ! Ê x œ $, ", $. We have that
ga$b œ
' $ fatb dt œ '$" fatb dt œ 1##
1
#
œ #1
' fatb dt œ !
$
ga$b œ ' fatb dt œ "
%
ga%b œ ' fatb dt œ " "# † " † " œ "#
ga"b œ
1
1
1
1
Thus, the absolute minimum is #1 and the absolute maximum is !. Thus, the range is Ò#1ß !Ó.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Chapter 5 Additional and Advanced Exercises
NOTES:
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359
360
Chapter 5 Integration
NOTES:
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CHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS
6.1 VOLUMES BY SLICING AND ROTATION ABOUT AN AXIS
1. (a) A œ 1(radius)# and radius œ È1 x# Ê A(x) œ 1 a1 x# b
(b) A œ width † height, width œ height œ 2È1 x# Ê A(x) œ 4 a1 x# b
(diagonal)#
;
#
(c) A œ (side)# and diagonal œ È2(side) Ê A œ
(d) A œ
È3
4
diagonal œ 2È1 x# Ê A(x) œ 2 a1 x# b
(side)# and side œ 2È1 x# Ê A(x) œ È3 a1 x# b
2. (a) A œ 1(radius)# and radius œ Èx Ê A(x) œ 1x
(b) A œ width † height, width œ height œ 2Èx Ê A(x) œ 4x
(diagonal)#
;
#
(c) A œ (side)# and diagonal œ È2(side) Ê A œ
(d) A œ
È3
4
(side)# and side œ 2Èx Ê A(x) œ È3x
(diagonal)#
#
3. A(x) œ
diagonal œ 2Èx Ê A(x) œ 2x
œ
ˆ È x ˆ È x ‰ ‰ #
#
œ 2x (see Exercise 1c); a œ 0, b œ 4;
V œ 'a A(x) dx œ '0 2x dx œ cx# d ! œ 16
b
4
1(diameter)#
4
4. A(x) œ
œ
%
1 c a2 x # b x # d
4
#
œ
1c2 a1 x# bd
4
#
œ 1 a1 2x# x% b ; a œ 1, b œ 1;
V œ 'a A(x) dx œ 'c1 1 a1 2x# x% b dx œ 1 ’x 23 x$ b
1
"
x&
5 “ "
#
œ 21 ˆ1 2
3
5" ‰ œ
161
15
#
5. A(x) œ (edge)# œ ’È1 x# ŠÈ1 x# ‹“ œ Š2È1 x# ‹ œ 4 a1 x# b ; a œ 1, b œ 1;
V œ 'a A(x) dx œ 'c1 4a1 x# b dx œ 4 ’x b
1
#
(diagonal)#
#
6. A(x) œ
œ
œ
#
Š2È1 x# ‹
V œ 'a A(x) dx œ 2'c1 a1 x# b dx œ 2 ’x 1
"
#
7. (a) STEP 1) A(x) œ
#
"
x$
3 “ "
(side) † (side) † ˆsin 13 ‰ œ
STEP 2) a œ 0, b œ 1
œ 8 ˆ1 "3 ‰ œ
16
3
#
’È1 x# ŠÈ1 x# ‹“
b
"
x$
3 “ "
"
#
œ 2 a1 x# b (see Exercise 1c); a œ 1, b œ 1;
œ 4 ˆ1 "3 ‰ œ
8
3
† Š2Èsin x‹ † Š2Èsin x‹ ˆsin 13 ‰ œ È3 sin x
STEP 3) V œ 'a A(x) dx œ È3 '0 sin x dx œ ’È3 cos x“ œ È3(1 1) œ 2È3
1
1
b
!
#
(b) STEP 1) A(x) œ (side) œ Š2Èsin x‹ Š2Èsin x‹ œ 4 sin x
STEP 2) a œ 0, b œ 1
STEP 3) V œ 'a A(x) dx œ '0 4 sin x dx œ c4 cos xd 1! œ 8
1
b
#
8. (a) STEP 1) A(x) œ 1(diameter)
œ 14 (sec x tan x)# œ
4
sin x ‘
œ 14 sec# x asec# x 1b 2 cos
#x
STEP 2) a œ 13 , b œ
asec# x tan# x 2 sec x tan xb
1
3
STEP 3) V œ 'a A(x) dx œ 'c1Î3
b
1
4
1Î3
1
4
ˆ2 sec# x 1 2 sin x ‰
cos# x
dx œ
1
4
2 tan x x 2 ˆ cos" x ‰‘1Î$
1Î$
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Chapter 6 Applications of Definite Integrals
œ
1
4
’2È3 1
3
2 Š ˆ "" ‰ ‹ Š2È3 #
1
3
2 Š ˆ "" ‰ ‹‹“ œ
#
(b) STEP 1) A(x) œ (edge)# œ (sec x tan x)# œ ˆ2 sec# x 1 2
STEP 2) a œ 13 , b œ
1
3
STEP 3) V œ 'a A(x) dx œ 'c1Î3 ˆ2 sec# x 1 1Î3
b
1
4
9. A(y) œ
(diameter)# œ
1
4
#
ŠÈ5y# 0‹ œ
c œ 0, d œ 2; V œ 'c A(y) dy œ '0
d
#
&
œ ’ˆ 541 ‰ Š y5 ‹“ œ
!
"
#
10. A(y) œ
2
1
4
51
4
51
4
Š4È3 21
3 ‹
sin x ‰
cos# x
dx œ 2 Š2È3 13 ‹ œ 4È3 21
3
y% ;
y% dy
a2& 0b œ 81
"
#
(leg)(leg) œ
#
È1 y# ˆÈ1 y# ‰‘ œ
V œ 'c A(y) dy œ 'c1 2a1 y# b dy œ 2 ’y d
2 sin x ‰
cos# x
1
4
1
"
y$
3 “ "
"
#
#
ˆ2È1 y# ‰ œ 2 a1 y# b ; c œ 1, d œ 1;
œ 4 ˆ1 "3 ‰ œ
8
3
11. (a) It follows from Cavalieri's Principle that the volume of a column is the same as the volume of a right
prism with a square base of side length s and altitude h. Thus, STEP 1) A(x) œ (side length)# œ s# ;
STEP 2) a œ 0, b œ h; STEP 3) V œ 'a A(x) dx œ '0 s# dx œ s# h
b
h
(b) From Cavalieri's Principle we conclude that the volume of the column is the same as the volume of the
prism described above, regardless of the number of turns Ê V œ s# h
12. 1) The solid and the cone have the same altitude of 12.
2) The cross sections of the solid are disks of diameter
x ˆ x# ‰ œ x# . If we place the vertex of the cone at the
origin of the coordinate system and make its axis of
symmetry coincide with the x-axis then the cone's cross
sections will be circular disks of diameter
x
ˆ x‰ x
4 4 œ # (see accompanying figure).
3) The solid and the cone have equal altitudes and identical
parallel cross sections. From Cavalieri's Principle we
conclude that the solid and the cone have the same
volume.
13. R(x) œ y œ 1 œ 1 ˆ2 4
2
14. R(y) œ x œ
3y
#
x
#
8 ‰
12
#
Ê V œ '0 1[R(x)]# dx œ 1'0 ˆ1 x# ‰ dx œ 1'0 Š1 x 2
œ
2
2
x#
4‹
dx œ 1 ’x x#
#
21
3
‰ dy œ 1'
Ê V œ '0 1[R(y)]# dy œ 1'0 ˆ 3y
#
0
2
15. R(x) œ tan ˆ 14 y‰ ; u œ
1
4
2
y Ê du œ
1
4
#
2
9
4
#
y# dy œ 1 34 y$ ‘ ! œ 1 †
3
4
dy Ê 4 du œ 1 dy; y œ 0 Ê u œ 0, y œ 1 Ê u œ
#
x$
12 “ !
† 8 œ 61
1
4
;
1Î%
V œ '0 1[R(y)]# dy œ 1'0 tan ˆ 14 y‰‘ dy œ 4 '0 tan# u du œ 4 '0 a1 sec# ub du œ 4[u tan u]!
1
1
#
1Î4
1Î4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 6.1 Volumes by Slicing and Rotation About an Axis
œ 4 ˆ 14 1 0‰ œ 4 1
1
#
16. R(x) œ sin x cos x; R(x) œ 0 Ê a œ 0 and b œ
œ 1'0
1Î2
xœ
1
#
(sin x cos x)# dx œ 1 '0
1Î2
Ê u œ 1‘ Ä V œ 1'0
1
"
8
(sin 2x)#
4
are the limits of integration; V œ '0
1Î2
dx; u œ 2x Ê du œ 2 dx Ê
sin# u du œ
1
8
#u
"
4
sin
1
2u‘ !
œ
1
8
ˆ 1#
du
8
œ
dx
4
1[R(x)]# dx
; x œ 0 Ê u œ 0,
0‰ 0‘ œ
1#
16
17. R(x) œ x# Ê V œ '0 1[R(x)]# dx œ 1 '0 ax# b dx
2
2
œ 1 '0 x% dx œ 1 ’ x5 “ œ
2
#
&
#
321
5
!
18. R(x) œ x$ Ê V œ '0 1[R(x)]# dx œ 1'0 ax$ b dx
2
2
œ 1 '0 x' dx œ 1 ’ x7 “ œ
2
(
#
!
#
1281
7
19. R(x) œ È9 x# Ê V œ 'c3 1[R(x)]# dx œ 1 'c3 a9 x# b dx
3
$
x$
3 “ $
œ 1 ’9x 3
œ 21 9(3) 27 ‘
3
œ 2 † 1 † 18 œ 361
20. R(x) œ x x# Ê V œ '0 1[R(x)]# dx œ 1'0 ax x# b dx
1
1
œ 1'0 ax# 2x$ x% b dx œ 1 ’ x3 1
œ 1 ˆ 13 $
"
#
5" ‰ œ
1
30
(10 15 6) œ
21. R(x) œ Ècos x Ê V œ '0
1Î2
1Î#
œ 1 csin xd !
2x%
4
1
30
#
"
x&
5 “!
1[R(x)]# dx œ 1'0 cos x dx
1Î2
œ 1(1 0) œ 1
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364
Chapter 6 Applications of Definite Integrals
1Î4
1Î4
22. R(x) œ sec x Ê V œ 'c1Î4 1[R(x)]# dx œ 1 '1Î4 sec# x dx
1Î%
œ 1 ctan xd 1Î% œ 1[1 (1)] œ 21
23. R(x) œ È2 sec x tan x Ê V œ
œ1
'01Î4 1[R(x)]# dx
'01Î4 ŠÈ2 sec x tan x‹# dx
œ 1 '0 Š2 2È2 sec x tan x sec# x tan# x‹ dx
1Î4
œ 1 Œ'0 2 dx 2È2 '0 sec x tan x dx 1Î4
1Î%
œ 1 Œ[2x]!
'01Î4 (tan x)# sec# x dx
1Î4
1Î%
2È2 [sec x]!
$
’ tan3 x “
1Î%
!
œ 1 ’ˆ 1# 0‰ 2È2 ŠÈ2 1‹ "3 a1$ 0b“ œ 1 Š 1# 2È2 11
3 ‹
24. R(x) œ 2 2 sin x œ 2(1 sin x) Ê V œ '0 1[R(x)]# dx
1Î2
œ 1 '0 4(1 sin x)# dx œ 41 '0 a1 sin# x 2 sin xb dx
1Î2
1Î2
œ 41'0 1 "# (1 cos 2x) 2 sin x‘ dx
1Î2
œ 41'0 ˆ 3# 1Î2
2 sin x‰
cos 2x
2
1Î#
œ 41 3# x sin42x 2 cos x‘ !
œ 41 ˆ 341 0 0‰ (0 0 2)‘ œ 1(31 8)
25. R(y) œ È5 † y# Ê V œ 'c1 1[R(y)]# dy œ 1 'c1 5y% dy
1
1
"
œ 1 cy& d " œ 1[1 (1)] œ 21
26. R(y) œ y$Î# Ê V œ '0 1[R(y)]# dy œ 1'0 y$ dy
2
%
2
#
œ 1 ’ y4 “ œ 41
!
27. R(y) œ È2 sin 2y Ê V œ '0 1[R(y)]# dy
1Î2
œ 1'0 2 sin 2y dy œ 1 c cos 2yd !
1Î2
1Î#
œ 1[1 (1)] œ 21
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Section 6.1 Volumes by Slicing and Rotation About an Axis
28. R(y) œ Écos
1y
4
Ê V œ 'c2 1[R(y)]# dy
0
œ 1 'c2 cos ˆ 14y ‰ dy œ 4 sin
0
29. R(y) œ
2
y1
1y ‘ !
4 #
œ 4[0 (1)] œ 4
Ê V œ '0 1[R(y)]# dy œ 41 '0
3
3
"
(y 1)#
dy
$
"
‘
œ 41 ’ y"
1 “ œ 41 4 (1) œ 31
!
30. R(y) œ
È2y
y # 1
Ê V œ '0 1[R(y)]# dy œ 1'0 2y ay# 1b
1
1
#
dy;
#
cu œ y 1 Ê du œ 2y dy; y œ 0 Ê u œ 1, y œ 1 Ê u œ 2d
Ä V œ 1'1 u# du œ 1 "u ‘ " œ 1 #" (1)‘ œ
2
#
1
#
31. For the sketch given, a œ 1# , b œ 1# ; R(x) œ 1, r(x) œ Ècos x; V œ 'a 1 a[R(x)]# [r(x)]# b dx
b
œ 'c1Î2 1(1 cos x) dx œ 21'0 (1 cos x) dx œ 21[x sin x]!
1Î2
1Î2
1Î#
œ 21 ˆ 1# 1‰ œ 1# 21
32. For the sketch given, c œ 0, d œ 14 ; R(y) œ 1, r(y) œ tan y; V œ 'c 1 a[R(y)]# [r(y)]# b dy
d
œ 1'0 a1 tan# yb dy œ 1 '0 a2 sec# yb dy œ 1[2y tan y]!
1Î4
1Î4
33. r(x) œ x and R(x) œ 1 Ê V œ
œ '0 1 a1 x# b dx œ 1 ’x 1
1Î%
œ 1 ˆ 1# 1‰ œ
1#
#
1
'01 1 a[R(x)]# [r(x)]# b dx
"
x$
3 “!
œ 1 ˆ1 "3 ‰ 0‘ œ
21
3
34. r(x) œ 2Èx and R(x) œ 2 Ê V œ '0 1 a[R(x)]# [r(x)]# b dx
1
œ 1'0 (4 4x) dx œ 41’x 1
"
x#
# “!
œ 41 ˆ1 "# ‰ œ 21
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365
366
Chapter 6 Applications of Definite Integrals
35. r(x) œ x# 1 and R(x) œ x 3
Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx
2
œ 1'c1 ’(x 3)# ax# 1b “ dx
2
#
œ 1 'c1 cax# 6x 9b ax% 2x# 1bd dx
2
œ 1 'c1 ax% x# 6x 8b dx
2
&
œ 1 ’ x5 x$
3
œ 1 ˆ 32
5 8
3
#
6x#
#
8x“
24
#
16‰ ˆ 5" "
"
3
6
#
‰
ˆ 5†30533 ‰ œ
8‰‘ œ 1 ˆ 33
5 3 28 3 8 œ 1
36. r(x) œ 2 x and R(x) œ 4 x#
Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx
2
œ 1'c1 ’a4 x# b (2 x)# “ dx
2
#
œ 1 'c1 ca16 8x# x% b a4 4x x# bd dx
2
œ 1'c1 a12 4x 9x# x% b dx
2
œ 1 ’12x 2x# 3x$ œ 1 ˆ24 8 24 #
x&
5 “ "
32 ‰
5
ˆ12 2 3 "5 ‰‘ œ 1 ˆ15 33 ‰
5
œ
1081
5
37. r(x) œ sec x and R(x) œ È2
Ê V œ 'c1Î4 1 a[R(x)]# [r(x)]# b dx
1Î4
œ 1 'c1Î4 a2 sec# xb dx œ 1[2x tan x]1Î%
1Î4
1Î%
œ 1 ˆ 1# 1‰ ˆ 1# 1‰‘ œ 1(1 2)
38. R(x) œ sec x and r(x) œ tan x
Ê V œ '0 1 a[R(x)]# [r(x)]# b dx
1
œ 1 '0 asec# x tan# xb dx œ 1 '0 1 dx œ 1[x]!" œ 1
1
1
39. r(y) œ 1 and R(y) œ 1 y
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
1
œ 1'0 c(1 y)# 1d dy œ 1 '0 a1 2y y# 1b dy
1
1
œ 1 '0 a2y y# b dy œ 1 ’y# 1
"
y$
3 “!
œ 1 ˆ1 3" ‰ œ
41
3
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5
Section 6.1 Volumes by Slicing and Rotation About an Axis
40. R(y) œ 1 and r(y) œ 1 y Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
1
œ 1'0 c1 (1 y)# d dy œ 1'0 c1 a1 2y y# bd dy
1
1
œ 1'0 a2y y# b dy œ 1 ’y# 1
"
y$
3 “!
œ 1 ˆ1 "3 ‰ œ
21
3
41. R(y) œ 2 and r(y) œ Èy
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
4
œ 1'0 (4 y) dy œ 1 ’4y 4
%
y#
2 “!
œ 1(16 8) œ 81
42. R(y) œ È3 and r(y) œ È3 y#
È3
Ê V œ '0
È3
œ 1 '0
$
œ 1 ’ y3 “
1 a[R(y)]# [r(y)]# b dy
È3
c3 a3 y# bd dy œ 1'0
È$
!
y# dy
œ 1È3
43. R(y) œ 2 and r(y) œ 1 Èy
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
1
œ 1'0 ’4 ˆ1 Èy‰ “ dy
1
#
œ 1 '0 ˆ4 1 2Èy y‰ dy
1
œ 1 '0 ˆ3 2Èy y‰ dy
1
œ 1 ’3y 43 y$Î# œ 1 ˆ3 "
y#
# “!
"# ‰ œ 1 ˆ 18683 ‰ œ
4
3
71
6
44. R(y) œ 2 y"Î$ and r(y) œ 1
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
1
#
œ 1'0 ’ˆ2 y"Î$ ‰ 1“ dy
1
œ 1'0 ˆ4 4y"Î$ y#Î$ 1‰ dy
1
œ 1 '0 ˆ3 4y"Î$ y#Î$ ‰ dy
1
œ 1 ’3y 3y%Î$ "
3y&Î$
5 “!
œ 1 ˆ3 3 53 ‰ œ
31
5
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367
368
Chapter 6 Applications of Definite Integrals
45. (a) r(x) œ Èx and R(x) œ 2
Ê V œ '0 1 a[R(x)]# [r(x)]# b dx
4
œ 1'0 (4 x) dx œ 1 ’4x 4
(b) r(y) œ 0 and R(y) œ y#
%
x#
# “!
œ 1(16 8) œ 81
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
2
œ 1'0 y% dy œ 1 ’ y5 “ œ
2
&
#
!
321
5
#
(c) r(x) œ 0 and R(x) œ 2 Èx Ê V œ '0 1 a[R(x)]# [r(x)]# b dx œ 1'0 ˆ2 Èx‰ dx
4
œ 1'0 ˆ4 4Èx x‰ dx œ 1 ’4x 4
4
8x$Î#
3
%
x#
# “!
œ 1 ˆ16 64
3
16 ‰
#
œ
81
3
(d) r(y) œ 4 y# and R(y) œ 4 Ê V œ '0 1 a[R(y)]# [r(y)]# b dy œ 1 '0 ’16 a4 y# b “ dy
2
2
œ 1 '0 a16 16 8y# y% b dy œ 1 '0 a8y# y% b dy œ 1 ’ 83 y$ 2
2
46. (a) r(y) œ 0 and R(y) œ 1 #
y&
5 “!
#
œ 1 ˆ 64
3 32 ‰
5
œ
2241
15
y
#
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy
2
#
œ 1'0 ˆ1 y# ‰ dy œ 1'0 Š1 y 2
œ 1 ’y 2
y#
#
#
y$
12 “ !
œ 1 ˆ# (b) r(y) œ 1 and R(y) œ 2 4
2
8 ‰
12
y#
4‹
œ
dy
21
3
y
#
#
Ê V œ '0 1 a[R(y)]# [r(y)]# b dy œ 1 '0 ’ˆ2 y# ‰ 1“ dy œ 1 '0 Š4 2y 2
2
œ 1'0 Š3 2y 2
y#
4‹
dy œ 1 ’3y y# #
y$
12 “ !
2
œ 1 ˆ6 4 8 ‰
12
œ 1 ˆ2 23 ‰ œ
y#
4
1‹ dy
81
3
47. (a) r(x) œ 0 and R(x) œ 1 x#
Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx
1
œ 1 'c1 a1 x# b dx œ 1 'c1 a1 2x# x% b dx
1
œ 1 ’x 1
#
2x$
3
"
x&
5 “ "
103 ‰
œ 21 ˆ 1515
œ
œ 21 ˆ1 2
3
15 ‰
161
15
(b) r(x) œ 1 and R(x) œ 2 x# Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx œ 1 'c1 ’a2 x# b 1“ dx
1
1
œ 1 'c1 a4 4x# x% 1b dx œ 1'c1 a3 4x# x% b dx œ 1 ’3x 43 x$ 1
œ
21
15
1
(45 20 3) œ
561
15
#
"
x&
5 “ "
œ 21 ˆ3 4
3
15 ‰
2
3
15 ‰
(c) r(x) œ 1 x# and R(x) œ 2 Ê V œ 'c1 1 a[R(x)]# [r(x)]# b dx œ 1 'c1 ’4 a1 x# b “ dx
1
1
œ 1 'c1 a4 1 2x# x% b dx œ 1'c1 a3 2x# x% b dx œ 1 ’3x 23 x$ 1
œ
21
15
1
(45 10 3) œ
641
15
#
"
x&
5 “ "
œ 21 ˆ3 Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 6.1 Volumes by Slicing and Rotation About an Axis
369
48. (a) r(x) œ 0 and R(x) œ hb x h
Ê V œ '0 1 a[R(x)]# [r(x)]# b dx
b
#
œ 1 '0 ˆ hb x h‰ dx
b
œ 1'0 Š hb# x# b
#
$
x
œ 1h# ’ 3b
# x#
b
2h#
b
x h# ‹ dx
b
x“ œ 1h# ˆ b3 b b‰ œ
!
1 h# b
3
#
(b) r(y) œ 0 and R(y) œ b ˆ1 yh ‰ Ê V œ '0 1 a[R(y)]# [r(y)]# b dy œ 1b# '0 ˆ1 yh ‰ dy
h
œ 1b# '0 Š1 h
2y
h
y#
h# ‹
dy œ 1b# ’y y#
h
h
h
y$
3h# “ !
1 b# h
3
œ 1b# ˆh h 3h ‰ œ
49. R(y) œ b Èa# y# and r(y) œ b Èa# y#
Ê V œ 'ca 1 a[R(y)]# [r(y)]# b dy
a
œ 1 'ca ’ˆb Èa# y# ‰ ˆb Èa# y# ‰ “ dy
#
a
#
œ 1 'ca 4bÈa# y# dy œ 4b1'ca Èa# y# dy
a
a
1a#
#
œ 4b1 † area of semicircle of radius a œ 4b1 †
œ 2a# b1#
50. (a) A cross section has radius r œ È#y and area 1r# œ #1y. The volume is '0 #1ydy œ 1 cy# d ! œ #&1.
&
(b) Vahb œ ' Aahbdh, so
dV
dh
œ Aahb. Therefore
For h œ %, the area is #1a%b œ )1, so
dh
dt
œ
dV
dt
"
)1
œ
dV
dh
†
œ Aahb †
dh
dt
$
$
)1
† $ units
sec œ
hca
51. (a) R(y) œ Èa# y# Ê V œ 1'ca aa# y# b dy œ 1 ’a# y œ 1 ’a# h "3 ah$ 3h# a 3ha# a$ b dV
$
dt œ 0.2 m /sec
dV
#
dh œ 101h 1h
(b) Given
Ê
and a œ 5 m, find
Ê
dV
dt
œ
dV
dh
†
dh
dt
a$
3“
œ 1 Ša# h †
so
dh
dt
œ
"
A ah b
†
dV
dt .
units$
sec .
hca
y$
3 “ ca
h$
3
dh
dt ,
&
œ 1 ’a# h a$ h# a ha# ‹ œ
(h a)$
3
Ša$ a$
3 ‹“
1h# (3a h)
3
#
From part (a), V(h) œ 1h (153 h) œ 51h# 13h
dh ¸
0.2
"
"
œ 1h(10 h) dh
dt Ê dt hœ4 œ 41(10 4) œ (201)(6) œ 1#01 m/sec.
dh ¸
dt hœ4 .
$
52. Suppose the solid is produced by revolving y œ 2 x about
the y-axis. Cast a shadow of the solid on a plane parallel to
the xy-plane.
Use an approximation such as the Trapezoid Rule, to
#
estimate 'a 1cRaybd# dy ¸ ! 1Œ #k ˜y.
b
n
d^
kœ"
53. The cross section of a solid right circular cylinder with a cone removed is a disk with radius R from which a disk of radius
h has been removed. Thus its area is A" œ 1R# 1h# œ 1 aR# h# b . The cross section of the hemisphere is a disk of
#
radius ÈR# h# . Therefore its area is A# œ 1 ŠÈR# h# ‹ œ 1 aR# h# b . We can see that A" œ A# . The altitudes of
both solids are R. Applying Cavalieri's Principle we find
Volume of Hemisphere œ (Volume of Cylinder) (Volume of Cone) œ a1R# b R "3 1 aR# b R œ
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3
1 R$ .
370
Chapter 6 Applications of Definite Integrals
54. R(x) œ
rx
h
Ê V œ '0 1[R(x)]# dx œ 1'0
h
h # #
r x
dx œ
h#
1r#
h#
h
$
#
$
’ x3 “ œ Š 1hr# ‹ Š h3 ‹ œ
!
"
3
1r# h, the volume of
a cone of radius r and height h.
c7
c7
55. R(y) œ È256 y# Ê V œ 'c16 1[R(y)]# dy œ 1'c16 a256 y# b dy œ 1 ’256y œ 1 ’(256)(7) 56. R(x) œ
œ
1
144
x
1#
7$
3
Š(256)(16) 16$
3 ‹“
$
È36 x# Ê V œ ' 1[R(x)]# dx œ 1'
0
0
6
'
x&
5 “!
’12x$ œ
1
144
6
Š12 † 6$ 6&
5‹
œ
16$
3 ‹
œ 1 Š 73 256(16 7) 1 †6 $
144
x#
144
ˆ12 a36 x# b dx œ
36 ‰
5
1
144
(
y$
3 “ "'
œ 10531 cm$ ¸ 3308 cm$
'06 a36x# x% b dx
1 ‰ ˆ 6036 ‰
œ ˆ 196
œ
144
5
361
5
cm$ . The plumb bob will
weigh about W œ (8.5) ˆ 3651 ‰ ¸ 192 gm, to the nearest gram.
57. (a) R(x) œ kc sin xk , so V œ 1'0 [R(x)]# dx œ 1'0 (c sin x)# dx œ 1'0 ac# 2c sin x sin# xb dx
1
1
œ 1'0 ˆc# 2c sin x '1
1
(b)
1
1cos 2x ‰
dx œ 1 0 ˆc# "# 2c sin x cos#2x ‰ dx
#
1
œ 1 ˆc# "# ‰ x 2c cos x sin42x ‘ ! œ 1 ˆc# 1 1# 2c 0‰ (0 2c 0)‘ œ 1 ˆc# 1 1# 4c‰ . Let
2
V(c) œ 1 ˆc# 1 1# 4c‰ . We find the extreme values of V(c): dV
dc œ 1(2c1 4) œ 0 Ê c œ 1 is a critical
#
#
point, and V ˆ 12 ‰ œ 1 ˆ 14 1# 18 ‰ œ 1 ˆ 1# 14 ‰ œ 1# 4; Evaluate V at the endpoints: V(0) œ 1# and
#
#
V(1) œ 1 ˆ 3# 1 4‰ œ 1# (4 1)1. Now we see that the function's absolute minimum value is 1# 4,
taken on at the critical point c œ 12 . (See also the accompanying graph.)
#
From the discussion in part (a) we conclude that the function's absolute maximum value is 1# , taken on at
the endpoint c œ 0.
(c) The graph of the solid's volume as a function of c for
0 Ÿ c Ÿ 1 is given at the right. As c moves away from
[!ß "] the volume of the solid increases without bound.
If we approximate the solid as a set of solid disks, we
can see that the radius of a typical disk increases without
bounds as c moves away from [0ß 1].
58. (a) R(x) œ 1 œ 1 'c4 Š1 4
œ 1 ’x x$
24
œ 21 ˆ4 Ê V œ 'c4 1[R(x)]# dx
4
x#
16
8
3
x#
16 ‹
#
dx œ 1'c4 Š1 4
%
x&
5†16# “ %
45 ‰ œ
21
15
œ 21 Š4 x#
8
x%
16# ‹
4$
24
4&
5†16# ‹
(60 40 12) œ
641
15
dx
ft$
(b) The helicopter will be able to fly ˆ 64151 ‰ (7.481)(2) ¸ 201 additional miles.
6.2 VOLUME BY CYLINDRICAL SHELLS
1. For the sketch given, a œ 0, b œ 2;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x Š1 b
2
x#
4‹
dx œ 21'0 Šx x#
4‹
dx œ 21'0 Š2x 2
x$
4‹
#
dx œ 21 ’ x# œ 21 † 3 œ 61
2. For the sketch given, a œ 0, b œ 2;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x Š2 b
2
2
x$
4‹
#
x%
16 “ !
dx œ 21 ’x# œ 21 ˆ 4# #
x%
16 “ !
œ 21(4 1) œ 61
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16 ‰
16
Section 6.2 Volume by Cylindrical Shells
3. For the sketch given, c œ 0, d œ È2;
È2
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0
d
È2
21y † ay# b dy œ 21'0
%
y$ dy œ 21 ’ y4 “
4. For the sketch given, c œ 0, d œ È3;
È3
È3
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y † c3 a3 y# bd dy œ 21 '0
d
È#
!
œ 21
%
y$ dy œ 21 ’ y4 “
È3
!
œ
5. For the sketch given, a œ 0, b œ È3;
È3
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x † ŠÈx# 1‹ dx;
b
’u œ x# 1 Ê du œ 2x dx; x œ 0 Ê u œ 1, x œ È3 Ê u œ 4“
Ä V œ 1'1 u"Î# du œ 1 23 u$Î# ‘ " œ
%
4
21
3
ˆ4$Î# 1‰ œ ˆ 231 ‰ (8 1) œ
141
3
6. For the sketch given, a œ 0, b œ 3;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x Š Èx9x
‹ dx;
$9
b
3
cu œ x$ 9 Ê du œ 3x# dx Ê 3 du œ 9x# dx; x œ 0 Ê u œ 9, x œ 3 Ê u œ 36d
Ä V œ 21 '9 3u"Î# du œ 61 2u"Î# ‘ * œ 121 ŠÈ36 È9‹ œ 361
$'
36
7. a œ 0, b œ 2;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x x ˆ x2 ‰‘ dx
b
2
œ '0 21x# †
2
3
#
dx œ 1 '0 3x# dx œ 1 cx$ d ! œ 81
2
#
8. a œ 0, b œ 1;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x ˆ2x x2 ‰ dx
b
1
œ 1 '0 2 Š 3x# ‹ dx œ 1 ' 3x# dx œ 1 cx$ d ! œ 1
1
1
#
"
0
9. a œ 0, b œ 1;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x c(2 x) x# d dx
b
1
œ 21'0 a2x x# x$ b dx œ 21 ’x# 1
œ 21 ˆ1 "
3
4" ‰ œ 21 ˆ 12 124 3 ‰ œ
x$
3
101
12
œ
"
x%
4 “!
51
6
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91
#
371
372
Chapter 6 Applications of Definite Integrals
10. a œ 0, b œ 1;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x ca2 x# b x# d dx
b
1
œ 21'0 x a2 2x# b dx œ 41'0 ax x$ b dx
1
1
"
x%
4 “!
#
œ 41 ’ x# œ 41 ˆ "2 4" ‰ œ 1
11. a œ 0, b œ 1;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x Èx (2x 1)‘ dx
b
1
"
œ 21'0 ˆx$Î# 2x# x‰ dx œ 21 25 x&Î# 23 x$ "# x# ‘ !
1
œ 21 ˆ 25 2
3
15 ‰
"# ‰ œ 21 ˆ 12 20
œ
30
71
15
12. a œ ", b œ 4;
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '1 21x ˆ 32 x"Î# ‰ dx
b
4
œ 31'1 x"Î# dx œ 31 23 x$Î# ‘ " œ 21 ˆ4$Î# "‰
%
4
œ 21(8 1) œ 141
13. (a) xf(x) œ œ
xf(x) œ œ
x†
sin x, 0 x Ÿ 1
0xŸ1
Ê xf(x) œ œ
; since sin 0 œ 0 we have
0, x œ 0
x, x œ 0
sin x
x ,
sin x, 0 x Ÿ 1
Ê xf(x) œ sin x, 0 Ÿ x Ÿ 1
sin x, x œ 0
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x † f(x) dx and x † f(x) œ sin x, 0 Ÿ x Ÿ 1 by part (a)
1
b
Ê V œ 21'0 sin x dx œ 21[ cos x]1! œ 21( cos 1 cos 0) œ 41
1
tan# x
x ,
tan# x, 0 x Ÿ 1/4
0 x Ÿ 14
Ê xg(x) œ œ
; since tan 0 œ 0 we have
0, x œ 0
x † 0, x œ 0
tan# x, 0 x Ÿ 1/4
Ê xg(x) œ tan# x, 0 Ÿ x Ÿ 1/4
xg(x) œ œ
tan# x, x œ 0
14. (a) xg(x) œ œ
x†
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x † g(x) dx and x † g(x) œ tan# x, 0 Ÿ x Ÿ 1/4 by part (a)
1Î4
b
Ê V œ 21'0 tan# x dx œ 21'0 asec# x 1b dx œ 21[tan x x]!
1Î4
1Î4
1Î%
œ 21 ˆ1 14 ‰ œ
41 1 #
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 6.2 Volume by Cylindrical Shells
15. c œ 0, d œ 2;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y Èy (y)‘ dy
d
2
œ 21'0 ˆy$Î# y# ‰ dy œ 21 ’ 2y5
2
&Î#
&
œ 21 ” 25 ŠÈ2‹ œ
161
15
2$
3•
#
y$
3 “!
È
œ 21 Š 8 5 2 83 ‹ œ 161 Š
È2
5
"3 ‹
Š3È2 5‹
16. c œ 0, d œ 2;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y cy# (y)ddy
d
2
œ 21'0 ay$ y# b dy œ 21 ’ y4 2
%
œ 161 ˆ 56 ‰ œ
401
3
#
y$
3 “!
œ 161 ˆ 24 "3 ‰
17. c œ 0, d œ 2;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y a2y y# bdy
d
2
œ 21'0 a2y# y$ b dy œ 21 ’ 2y3 2
$
œ 321 ˆ "3 4" ‰ œ
321
12
œ
81
3
#
y%
4 “!
œ 21 ˆ 16
3 "6 ‰
4
18. c œ 0, d œ 1;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y a2y y# ybdy
d
1
œ 21'0 y ay y# b dy œ 21'0 ay# y$ b dy
1
1
$
œ 21 ’ y3 "
y%
“
4 !
œ 21 ˆ 13 "4 ‰ œ
1
6
19. c œ 0, d œ 1;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ 21'0 y[y (y)]dy
d
1
œ 21'0 2y# dy œ
1
41
3
"
cy$ d ! œ
41
3
20. c œ 0, d œ 2;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21 yˆy y2 ‰dy
d
2
œ 21 '0
2
y2
2 dy
1
œ 13 c y3 d ! œ
81
3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
373
374
Chapter 6 Applications of Definite Integrals
21. c œ 0, d œ 2;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y c(2 y) y# d dy
d
2
œ 21 '0 a2y y# y$ b dy œ 21 ’y# 2
œ 21 ˆ4 8
3
16 ‰
4
1
6
œ
y$
3
(48 32 48) œ
#
y%
4 “!
161
3
22. c œ 0, d œ 1;
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y c(2 y) y# d dy
d
1
œ 21'0 a2y y# y$ b dy œ 21 ’y# 1
œ 21 ˆ1 14 ‰ œ
1
3
1
6
(12 4 3) œ
y$
3
51
6
"
y%
4 “!
shell ‰
shell
23. (a) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y † 12 ay# y$ b dy œ 241 '0 ay$ y% b dy œ 241 ’ y4 d
1
œ 241 ˆ 14 15 ‰ œ
241
20
œ
1
"
y&
5 “!
%
61
5
shell ‰
shell
(b) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21(1 y) c12 ay# y$ bd dy œ 241'0 (1 y) ay# y$ b dy
d
1
1
œ 241'0 ay# 2y$ y% b dy œ 241 ’ y3 1
$
y%
2
"
y&
5 “!
œ 241 ˆ "3 1
2
" ‰
51 ‰ œ 241 ˆ 30
œ
41
5
shell ‰
shell
(c) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21 ˆ 85 y‰ c12 ay# y$ bd dy œ 241 '0 ˆ 85 y‰ ay# y$ b dy
d
1
œ 241'0 ˆ 85 y# 1
œ
241
12
13
5
1
8 $
y$ y% ‰ dy œ 241 ’ 15
y 13
20
y% "
y&
5 “!
8
œ 241 ˆ 15
13
20
241
60
15 ‰ œ
(32 39 12)
œ 21
shell ‰
shell
(d) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21 ˆy 25 ‰ c12 ay# y$ bd dy œ 241'0 ˆy 25 ‰ ay# y$ b dy
d
1
1
2 $
œ 241'0 ˆy$ y% 25 y# 25 y$ ‰ dy œ 241'0 ˆ 25 y# 35 y$ y% ‰ dy œ 241 ’ 15
y 1
1
2
œ 241 ˆ 15
3
20
15 ‰ œ
241
60
(8 9 12) œ
241
12
2
%
œ 21 ’ y4 #
y'
24 “ !
%
2'
24 ‹
œ 21 Š 24 #
%
œ 321 ˆ 4" 4 ‰
24
dy œ '0 21y Šy# 2
y#
# ‹“
2
œ 21 '0 Š2y# 2
y%
2
y$ y&
4‹
#
$
dy œ 21 ’ 2y3 y%
4
2
œ 21'0 Š5y# 54 y% y$ 2
y&
4‹
#
$
dy œ 21 ’ 5y3 y%
4
#
y'
#4 “ !
œ 21 ˆ 16
3 œ
2
œ 21'0 Šy$ 2
y&
4
58 y# 5
32
#
%
y% ‹ dy œ 21 ’ y4 %
y'
#4
5y$
#4
2
y#
# ‹“
16
4
64 ‰
24
2
y'
#4 “ !
y#
# ‹“
32
10
dy œ '0 21(5 y) Šy# #
shell ‰
shell
(d) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21 ˆy 58 ‰ ’ y# Š y4 d
81
3
y%
4‹
%
5y&
20
2
dy œ '0 21(2 y) Šy# shell ‰
shell
(c) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21(5 y) ’ y# Š y4 d
dy œ 21'0 Šy$ y#
# ‹“
%
y&
10
y%
4‹
2 ‰
œ 321 ˆ 4" 6" ‰ œ 321 ˆ 24
œ
shell ‰
shell
(b) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21(2 y) ’ y# Š y4 d
"
y&
5 “!
y% œ 21
shell ‰
shell
24. (a) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21y ’ y# Š y4 d
3
20
œ 21 ˆ 40
3 160
20
16
4
dy œ '0 21 ˆy 58 ‰ Šy# #
5y&
160 “ !
2
œ 21 ˆ 16
4 64
24
40
24
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
81
5
y%
4‹
64 ‰
24
dy
dy
œ 81
y%
4‹
160 ‰
160
dy
œ 41
y&
4‹
dy
Section 6.2 Volume by Cylindrical Shells
375
shell ‰
shell
25. (a) About x-axis: V œ 'c 21 ˆ radius
Š height
‹dy
d
œ '0 21yˆÈy y‰dy œ 21'0 ˆy$Î# y# ‰dy
1
1
"
œ 21 #& y&Î# "$ y$ ‘ ! œ 21ˆ #& "$ ‰ œ
#1
"&
shell ‰
shell
About y-axis: V œ 'a 21 ˆ radius
Š height
‹dx
b
œ '0 21xax x# bdx œ 21'0 ax2 x3 bdx
1
1
$
œ 21’ x$ "
x%
% “!
œ 21ˆ "$ "% ‰ œ
1
'
(b) About x-axis: Raxb œ x and raxb œ x# Ê V œ 'a 1Raxb# raxb# ‘dx œ '0 1cx# x% ddx
b
$
œ 1’ x$ "
x&
& “!
œ 1ˆ "$ "& ‰ œ
1
#1
"&
About y-axis: Rayb œ Èy and rayb œ y Ê V œ 'c 1Rayb# rayb# ‘dy œ '0 1cy y2 ddy
d
#
œ 1’ y# "
y$
$ “!
œ 1ˆ "# "$ ‰ œ
1
1
'
#
26. (a) V œ 'a 1Raxb# raxb# ‘dx œ 1'0 ’ˆ #x #‰ x# “dx
%
b
œ 1'0 ˆ $% x# #x %‰dx œ 1’ x% x# %x“
%
$
œ 1a"' "' "'b œ "'1
%
!
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹dx œ '0 #1xˆ x# # x‰dx
%
b
œ '0 #1xˆ# x# ‰dx œ #1'0 Š#x %
%
œ #1’x# %
x$
' “!
'% ‰
'
œ #1ˆ"' œ
x#
# ‹dx
$#1
$
shell ‰
shell
(c) V œ 'a 21 ˆ radius
Š height
‹dx œ '0 #1a% xbˆ x# # x‰dx œ '0 #1a% xbˆ# x# ‰dx œ #1'0 Š) %x %
b
œ #1’)x #x# %
x$
“
' !
œ #1ˆ$# $# %
'% ‰
'
%
x#
# ‹dx
'%1
$
œ
#
(d) V œ 'a 1Raxb# raxb# ‘dx œ 1'0 ’a) xb# ˆ' #x ‰ “dx œ 1'0 ’a'% "'x x# b Š$' 'x x% ‹“dx
%
b
%
#
1'0 ˆ $% x# "!x #)‰dx œ 1’ x% &x# #)x“ œ 1"' a&ba"'b a(ba"'b‘ œ 1a$ba"'b œ %)1
%
%
$
!
shell ‰
shell
27. (a) V œ 'c 21 ˆ radius
Š height
‹ dy œ '1 21y(y 1) dy
d
2
œ 21'1 ay# yb dy œ 21 ’ y3 2
$
#
y#
# “"
œ 21 ˆ 83 42 ‰ ˆ "3 #" ‰‘
œ 21 ˆ 73 2 "# ‰ œ 13 (14 12 3) œ
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx
51
3
b
œ '1 21x(2 x) dx œ 21'1 a2x x# b dx œ 21 ’x# 2
2
œ 21 ˆ 12 3 8 ‰ ˆ 3 3 " ‰‘ œ 21 ˆ 34 32 ‰ œ
41
3
#
x$
3 “"
œ 21 ˆ4 83 ‰ ˆ1 "3 ‰‘
shell ‰
shell
' ˆ 203 ‰
(c) V œ 'a 21 ˆ radius
Š height
‹ dx œ '1 21 ˆ 10
3 x (2 x) dx œ 21 1
b
2
#
" $‘
8 #
ˆ 40
œ 21 20
3 x 3 x 3 x " œ 21
3 2
32
3
38 ‰ ˆ 20
3 8
3
16
3
x x# ‰ dx
3" ‰‘ œ 21 ˆ 33 ‰ œ 21
shell ‰
shell
(d) V œ 'c 21 ˆ radius
Š height
‹ dy œ '1 21(y 1)(y 1) dy œ 21'1 (y 1)# œ 21 ’ (y31) “ œ
d
2
2
$
#
"
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
21
3
376
Chapter 6 Applications of Definite Integrals
shell ‰
shell
28. (a) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21yay# 0b dy
d
2
œ 21'0 y$ dy œ 21 ’ y4 “ œ 21 Š 24 ‹ œ 81
2
%
#
%
!
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx
b
œ '0 21x ˆ2 Èx‰ dx œ 21'0 ˆ2x x$Î# ‰ dx
4
4
%
2 †2 &
5 ‹
œ 21 x# 25 x&Î# ‘ ! œ 21 Š16 œ 21 ˆ16 64 ‰
5
21
5
œ
321
5
(80 64) œ
shell ‰
shell
(c) V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21(4 x) ˆ2 Èx‰ dx œ 21'0 ˆ8 4x"Î# 2x x$Î# ‰ dx
b
4
4
%
œ 21 8x 83 x$Î# x# 25 x&Î# ‘ ! œ 21 ˆ32 64
3
16 64 ‰
5
œ
21
15
(240 320 192) œ
21
15
shell ‰
shell
(d) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21(2 y) ay# b dy œ 21 '0 a2y# y$ b dy œ 21 ’ 23 y$ d
2
œ 21 ˆ 16
3 16 ‰
4
œ
321
12
2
81
3
(4 3) œ
(112) œ
2241
15
#
y%
4 “!
shell ‰
shell
29. (a) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21yay y$ b dy
d
1
œ '0 21 ay# y% b dy œ 21 ’ y3 1
œ
$
41
15
"
y&
“
5 !
œ 21 ˆ "3 "5 ‰
shell ‰
shell
(b) V œ 'c 21 ˆ radius
Š height
‹ dy
d
œ '0 21(1 y) ay y$ b dy
1
œ 21 '0 ay y# y$ y% b dy œ 21 ’ y# 1
#
y$
3
y%
4
"
y&
5 “!
œ 21 ˆ "# "
3
"
4
5" ‰ œ
21
60
(30 20 15 12) œ
71
30
shell ‰
shell
30. (a) V œ 'c 21 ˆ radius
Š height
‹dy
d
œ '0 21y c1 ay y$ bddy
1
œ 21 '0 ay y# y% b dy œ 21 ’ y# 1
#
œ 21 ˆ "# œ
"
3
5" ‰ œ
21
30
y$
3
"
y&
5 “!
(15 10 6)
111
15
(b) Use the washer method:
V œ 'c 1 cR# (y) r# (y)d dy œ '0 1 ’1# ay y$ b “ dy œ 1 '0 a1 y# y' 2y% b dy œ 1 ’y d
1
œ 1 ˆ1 "
3
"
7
25 ‰ œ
1
105
1
#
(105 35 15 42) œ
y$
3
y(
7
971
105
"
2y&
5 “!
(c) Use the washer method:
V œ 'c 1 cR# (y) r# (y)d dy œ '0 1 ’c1 ay y$ bd 0“ dy œ 1'0 ’1 2 ay y$ b ay y$ b “ dy
d
1
œ 1'0 a1 y# y' 2y 2y$ 2y% b dy œ 1 ’y 1
œ
1
210
(70 30 105 2 † 42) œ
1
#
y$
3
y(
7
y# #
y%
#
1211
210
"
2y&
5 “!
œ 1 ˆ1 "
3
"
7
1
"
#
25 ‰
shell ‰
shell
(d) V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21(1 y) c1 ay y$ bd dy œ 21 '0 (1 y) a1 y y$ b dy
d
1
1
œ 21'0 a1 y y$ y y# y% b dy œ 21'0 a1 2y y# y$ y% b dy œ 21 ’y y# 1
œ 21 ˆ1 1 1
"
3
"
4
5" ‰ œ
21
60
(20 15 12) œ
231
30
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
y$
3
y%
4
"
y&
5 “!
Section 6.2 Volume by Cylindrical Shells
shell ‰
shell
31. (a) V œ 'c 21 ˆ radius
Š height
‹dy œ '0 21y ˆÈ8y y# ‰ dy
d
2
œ 21'0 Š2È2 y$Î# y$ ‹ dy œ 21 ’ 4 5 2 y&Î# È
2
#
y%
4 “!
&
œ 21 4È2†ŠÈ2‹
2%
4
5
œ 21 † 4 ˆ 85 1‰ œ
81
5
$
œ 21 Š 4†52 (8 5) œ
4 †4
4 ‹
241
5
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x ŠÈx b
4
&
œ 21 Š 2†52 4%
3# ‹
'
œ 21 Š 25 2)
32 ‹
œ
1†2(
160
x#
8‹
dx œ 21'0 Šx$Î# (32 20) œ
4
1†2* †3
160
œ
1†2% †3
5
œ
x$
8‹
dx œ 21 ’ 25 x&Î# 481
5
shell ‰
shell
32. (a) V œ 'a 21 ˆ radius
Š height
‹ dx
b
œ '0 21x ca2x x# b xd dx
1
œ 21 '0 x ax x# b dx œ 21'0 ax# x$ b dx
1
$
1
œ 21 ’ x3 "
x%
4 “!
œ 21 ˆ "3 4" ‰ œ
1
6
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21(1 x) ca2x x# b xd dx œ 21'0 (1 x) ax x# b dx
b
1
1
œ 21 '0 ax 2x# x$ b dx œ 21 ’ x2 23 x$ 1
"
x%
4 “!
#
œ 21 ˆ 12 2
3
"4 ‰ œ
21
1#
(6 8 3) œ
1
6
33. (a) V œ 'a 1 cR# (x) r# (x)d dx œ 1 '1Î16 ˆx"Î# 1‰ dx
b
1
"
œ 1 2x"Î# x‘"Î"' œ 1 (2 1) ˆ2 †
œ 1 ˆ1 7 ‰
16
œ
"
4
" ‰‘
16
91
16
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dy œ '1 21y Š y"% b
2
œ 21'1 ˆy$ 2
y ‰
16
dy œ 21 ’ 12 y# œ 21 ˆ "8 8" ‰ ˆ #" œ
21
32
(8 1) œ
91
16
" ‰‘
3#
d
2
œ 1 "3 y$ œ
1
48
y ‘#
16 "
y#
32 “ "
" ‰
32
"
16 ‹
dy
"
œ 1 ˆ 24
8" ‰ ˆ 3" (2 6 16 3) œ
dy
#
œ 21 ˆ 4" 34. (a) V œ 'c 1 cR# (y) r# (y)d dy œ '1 1 Š y"% "
16 ‹
" ‰‘
16
111
48
shell ‰
shell
(b) V œ 'a 21 ˆ radius
Š height
‹ dx œ '1Î4 21x Š È"x "‹ dx
b
1
œ 21 '1Î4 ˆx"Î# x‰ dx œ 21 ’ 23 x$Î# 1
œ 21 ˆ 23 "# ‰ ˆ 23 †
"
8
" ‰‘
3#
"
x#
2 “ "Î%
œ 1 ˆ 43 1 "
6
" ‰
16
œ
1
48
(4 † 16 48 8 3) œ
111
48
35. (a) H3=k: V œ V" V#
V" œ 'a 1[R" (x)]# dx and V# œ 'a 1[R# (x)]# with R" (x) œ É x 3 2 and R# (x) œ Èx,
b"
b#
"
#
a" œ 2, b" œ 1; a# œ 0, b# œ 1 Ê two integrals are required
(b) [ +=2/<: V œ V" V#
V" œ 'a 1 a[R" (x)]# [r" (x)]# b dx with R" (x) œ É x 3 2 and r" (x) œ 0; a" œ 2 and b" œ 0;
b"
"
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
%
x%
32 “ !
377
378
Chapter 6 Applications of Definite Integrals
V# œ 'a 1 a[R# (x)]# [r# (x)]# b dx with R# (x) œ É x 3 2 and r# (x) œ Èx; a# œ 0 and b# œ 1
b#
#
Ê two integrals are required
shell ‰
shell
shell
(c) W2/66: V œ 'c 21 ˆ radius
Š height
‹ dy œ 'c 21y Š height
‹ dy where shell height œ y# a3y# 2b œ 2 2y# ;
d
d
c œ 0 and d œ 1. Only 98/ integral is required. It is, therefore preferable to use the =2/66 method.
However, whichever method you use, you will get V œ 1.
36. (a) H3=k: V œ V" V# V$
Vi œ 'c 1[Ri (y)]# dy, i œ 1, 2, 3 with R" (y) œ 1 and c" œ 1, d" œ 1; R# (y) œ Èy and c# œ 0 and d# œ 1;
di
i
R$ (y) œ (y)"Î% and c$ œ 1, d$ œ 0 Ê three integrals are required
(b) [ +=2/<: V œ V" V#
Vi œ 'c 1a[Ri (y)]# [ri (y)]# b dy, i œ 1, 2 with R" (y) œ 1, r" (y) œ Èy, c" œ 0 and d" œ 1;
di
i
R# (y) œ 1, r# (y) œ (y)"Î% , c# œ 1 and d# œ 0 Ê two integrals are required
shell ‰
shell
shell
(c) W2/66: V œ 'a 21 ˆ radius
Š height
‹dx œ 'a 21xŠ height
‹dx, where shell height œ x# ax% b œ x# x% ,
b
b
a œ 0 and b œ 1 Ê only one integral is required. It is, therefore preferable to use the =2/66 method.
However, whichever method you use, you will get V œ 561 .
6.3 LENGTHS OF PLANE CURVES
1.
dx
dt
œ 1 and
dy
dt
#
#
È(1)# (3)# œ È10
‰ Š dy
œ 3 Ê Êˆ dx
dt
dt ‹ œ
Ê Length œ '2/3 È10 dt œ È10 ctd 12/3 œ È10 Š 23 È10‹ œ
1
2.
dx
dt
œ sin t and
dy
dt
5È10
3
#
#
È( sin t)# (1 cos t)# œ È2 2 cos t
‰ Š dy
œ 1 cos t Ê Êˆ dx
dt
dt ‹ œ
cos t ‰
È2 ' É sin# t dt
Ê Length œ '! È2 2 cos t dt œ È2 '! Ɉ 11 cos t (1 cos t) dt œ
1 cos t
!
1
œ È2'!
1
sin t
È1 cos t
1
1
0 on [0ß 1]); [u œ 1 cos t Ê du œ sin t dt; t œ 0 Ê u œ 0,
dt (since sin t
#
t œ 1 Ê u œ 2] Ä È2 '! u"Î# du œ È2 2u"Î# ‘ ! œ 4
2
3.
dx
dt
œ 3t# and
dy
dt
#
#
‰ Š dy
Éa3t# b# (3t)# œ È9t% 9t# œ 3tÈt# 1 Šsince t
œ 3t Ê Êˆ dx
dt
dt ‹ œ
È3
Ê Length œ '! 3tÈt# 1 dt; ’u œ t# 1 Ê
Ä
4.
dx
dt
3
#
0 on ’0ß È3“‹
du œ 3t dt; t œ 0 Ê u œ 1, t œ È3 Ê u œ 4“
'14 3# u"Î# du œ u$Î# ‘ %" œ (8 1) œ 7
œ t and
dy
dt
#
#
Èt# (2t 1) œ È(t 1)# œ kt 1k œ t 1 since 0 Ÿ t Ÿ 4
‰ Š dy
œ (2t 1)"Î# Ê Êˆ dx
dt
dt ‹ œ
Ê Length œ '0 (t 1) dt œ ’ t2 t“ œ (8 4) œ 12
4
%
#
!
5.
dx
dt
œ (2t 3)"Î# and
dy
dt
#
#
È(2t 3) (1 t)# œ Èt# 4t 4 œ kt 2k œ t 2
‰ Š dy
œ 1 t Ê Êˆ dx
dt
dt ‹ œ
since 0 Ÿ t Ÿ 3 Ê Length œ '0 (t 2) dt œ ’ t2 2t“ œ
$
#
$
!
21
#
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Section 6.3 Lengths of Plane Curves
6.
dx
dt
œ 8t cos t and
dy
dt
œ k8tk œ 8t since 0 Ÿ t Ÿ
7.
dy
dx
œ
"
3
† 3# ax# 2b
#
#
È(8t cos t)# (8t sin t)# œ È64t# cos# t 64t# sin# t
‰ Š dy
œ 8t sin t Ê Êˆ dx
dt
dt ‹ œ
"Î#
Ê Length œ '0 8t dt œ c4t# d !
1Î2
1
#
1Î#
œ 1#
† 2x œ Èax# 2b † x
Ê L œ '0 È1 ax# 2b x# dx œ '0 È1 2x# x% dx
$
3
œ '0 Éa1 x# b# dx œ '0 a1 x# b dx œ ’x $
œ3
8.
dy
dx
œ
3
#
œ 12
27
3
Èx Ê L œ ' É1 94 x dx; u œ 1 94 x
0
4
Ê du œ
dx Ê
9
4
du œ dx; x œ 0 Ê u œ 1; x œ 4
4
9
Ê u œ 10d Ä L œ '1 u"Î# ˆ 49 du‰ œ
10
œ
9.
dx
dy
8
27
œ y# #
"
4y#
%
Ê Š dx
dy ‹ œ y 3
œ '1 Éy% 3
"
#
œ '1 ÊŠy# 3
$
œ ’ y3 dx
dy
œ
4
9
23 u$Î# ‘ "!
"
Š10È10 1‹
Ê L œ '1 É1 y% 10.
$
x$
3 “!
$
"
#
"
16y%
"
4y# ‹
$
y"
4 “"
#
"
#
"
16y%
"
#
"
16y%
dy
dy
dy œ '1 Šy# 3
" ‰
1#
œ ˆ 27
3 "
4y# ‹
dy
ˆ 3" 4" ‰ œ 9 #
y"Î# "# y"Î# Ê Š dx
dy ‹ œ
"
4
"
1#
"
3
"
4
œ9
(1 4 3)
1#
œ9
(2)
1#
œ
53
6
Šy 2 y" ‹
Ê L œ '1 Ê1 "4 Šy 2 y" ‹ dy
9
œ '1 Ê "4 Šy 2 y" ‹ dy œ '1
9
œ
"
#
9
"
#
"
Èy ‹
Ê ŠÈ y *
$Î#
$
"
dx
dy
dy
'19 ˆy"Î# y"Î# ‰ dy œ "# 23 y$Î# 2y"Î# ‘ *"
œ ’ y 3 y"Î# “ œ Š 33 3‹ ˆ "3 1‰ œ 11 11.
#
œ y$ "
4y$
#
'
Ê Š dx
dy ‹ œ y Ê L œ '1 É1 y' 2
œ '1 Éy' 2
œ '1 Šy$ 2
œ Š 16
4 "
2
y$
4 ‹
"
(16)(2) ‹
"
16y'
"
2
"
16y'
"
#
32
3
dy
2
%
œ
"
16y'
dy œ '1 ÊŠy$ dy œ ’ y4 "
3
y$
4 ‹
#
dy
#
y#
8 “"
ˆ "4 "8 ‰ œ 4 "
32
"
4
"
8
œ
128184
32
œ
123
32
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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379
380
12.
Chapter 6 Applications of Definite Integrals
dx
dy
œ
y#
#
"
#y #
#
Ê Š dx
dy ‹ œ
"
4
ay% 2 y% b
Ê L œ '2 É1 "4 ay% 2 y% b dy
3
œ '2 É "4 ay% 2 y% b dy
3
13.
œ
"
#
'23 Éay# y# b# dy œ "# '23 ay# y# b dy
œ
"
#
’ y3 y" “ œ
dy
dx
$
$
"
#
#
"‰
ˆ 27
ˆ 8 " ‰‘ œ
3 3 3 #
#
#Î$
œ x"Î$ "4 x"Î$ Ê Š dy
dx ‹ œ x
Ê L œ '1 É1 x#Î$ 8
œ '1 Éx#Î$ 8
"
#
x#Î$
16
"
#
x#Î$
16
"
#
"
#
ˆ 26
3 8
3
#" ‰ œ
"
#
ˆ6 #" ‰ œ
13
4
x#Î$
16
dx
dx
#
œ '1 Ɉx"Î$ "4 x"Î$ ‰ dx œ '1 ˆx"Î$ "4 x"Î$ ‰ dx
8
8
)
œ 34 x%Î$ 38 x#Î$ ‘ " œ
œ
14.
dy
dx
3
8
2x%Î$ x#Î$ ‘ )
"
3
8
ca2 † 2% 2# b (2 1)d œ
œ x# 2x 1 œ (1 x)# #
2
œ '0 É(1 x)% "
#
œ '0 Ê’(1 x)# 2
œ '0 ’(1 x)# 2
"
#
(1x)%
16
#
(1x)#
“
4
(1x)#
“
4
(1x)%
16
99
8
"
"
4 (1x)#
%
Ê Š dy
dx ‹ œ (1 x) Ê L œ '0 É1 (1 x)% 2
(32 4 3) œ
œ x# 2x 1 4
(4x4)#
"
"
4 (1x)#
3
8
"
#
"
16(1x)%
dx
dx
dx
dx; cu œ 1 x Ê du œ dx; x œ 0 Ê u œ 1, x œ 2 Ê u œ 3d
Ä L œ '1 ˆu# "4 u# ‰ du œ ’ u3 "4 u" “ œ ˆ9 3
$
$
"
15.
dx
dy
" ‰
1#
ˆ 3" 4" ‰ œ
108143
12
œ
106
12
œ
53
6
#
%
œ Èsec% y 1 Ê Š dx
dy ‹ œ sec y 1
1Î4
1Î4
Ê L œ '1Î4 È1 asec% y 1b dy œ '1Î4 sec# y dy
1Î%
œ ctan yd 1Î% œ 1 (1) œ 2
16.
dy
dx
#
%
œ È3x% 1 Ê Š dy
dx ‹ œ 3x 1
c1
c1
Ê L œ 'c2 È1 a3x% 1b dx œ 'c2 È3 x# dx
$
œ È3 ’ x3 “
"
#
œ
È3
3
c1 (2)$ d œ
È3
3
(" 8) œ
7È 3
3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 6.3 Lengths of Plane Curves
17. (a)
dy
dx
#
(b)
#
œ 2x Ê Š dy
dx ‹ œ 4x
Ê L œ 'c1 Ê1 Š dy
dx ‹ dx
#
2
œ 'c1 È1 4x# dx
2
(c) L ¸ 6.13
18. (a)
dy
dx
#
(b)
%
œ sec# x Ê Š dy
dx ‹ œ sec x
Ê L œ 'c1Î3 È1 sec% x dx
0
(c) L ¸ 2.06
19. (a)
dx
dy
#
(b)
#
œ cos y Ê Š dx
dy ‹ œ cos y
Ê L œ '0 È1 cos# y dy
1
(c) L ¸ 3.82
20. (a)
dx
dy
#
œ È1y y# Ê Š dx
dy ‹ œ
1Î2
Ê L œ 'c1Î2 É1 œ 'c1Î2 a1 y# b
1Î2
"Î#
y#
a1 y # b
y#
1 y#
(b)
1Î2
dy œ '1Î2 É 1 " y# dy
dy
(c) L ¸ 1.05
21. (a) 2y 2 œ 2
dx
dy
#
#
Ê Š dx
dy ‹ œ (y 1)
(b)
Ê L œ 'c1 È1 (y 1)# dy
3
(c) L ¸ 9.29
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381
382
Chapter 6 Applications of Definite Integrals
22. (a)
dy
dx
#
(b)
#
#
œ cos x - cos x + x sin x Ê Š dy
dx ‹ œ x sin x
Ê L œ '0 È1 x# sin# x dx
1
(c) L ¸ 4.70
23. (a)
dy
dx
#
(b)
#
œ tan x Ê Š dy
dx ‹ œ tan x
# x cos# x
Ê L œ '0 È1 tan# x dx œ '0 É sin cos
dx
#x
1Î6
œ '0
1Î6
1Î6
œ '0 sec x dx
1Î6
dx
cos x
(c) L ¸ 0.55
24. (a)
dx
dy
#
(b)
#
œ Èsec# y 1 Ê Š dx
dy ‹ œ sec y 1
Ê L œ 'c1Î3 È1 asec# y 1b dy
1Î4
1Î4
1Î4
œ 'c1Î3 ksec yk dy œ '1Î3 sec y dy
(c) L ¸ 2.20
25. È2 x œ '0 Ê1 Š dy
dt ‹ dt, x
#
x
#
0 Ê È2 œ Ê1 Š dy
dx ‹ Ê
dy
dx
œ „ 1 Ê y œ f(x) œ „ x C where C is any
real number.
26. (a) From the accompanying figure and definition of the
differential (change along the tangent line) we see that
dy œ f w (xkc1 ) ˜ xk Ê length of kth tangent fin is
È( ˜ xk )# (dy)# œ È( ˜ xk )# [f w (xkc1 ) ˜ xk ]# .
n
n
! (length of kth tangent fin) œ lim ! È( ˜ xk )# [f w (xk1 ) ˜ xk ]#
(b) Length of curve œ n lim
Ä_
nÄ_
kœ1
! È1 [f w (xk1 )]# ˜ xk œ ' È1 [f w (x)]# dx
œ n lim
Ä_
a
n
kœ1
b
kœ1
#
"
27. (a) Š dy
dx ‹ correspondes to 4x here, so take
So y œ Èx from ("ß ") to (4ß 2).
dy
dx
as
"
.
#È x
Then y œ Èx C and since ("ß ") lies on the curve, C œ 0.
(b) Only one. We know the derivative of the function and the value of the function at one value of x.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 6.3 Lengths of Plane Curves
#
28. (a) Š dx
dy ‹ correspondes to
So y œ
"
y%
here, so take
dy
dx
as
"
y# .
Then x œ y" C and, since (!ß ") lies on the curve, C œ 1
"
"x.
(b) Only one. We know the derivative of the function and the value of the function at one value of x.
29. (a)
dx
dt
œ 2 sin 2t and
dy
dt
Ê Length œ '0 2 dt œ c2td !
1Î2
(b)
dx
dt
œ 1 cos 1t and
1Î#
dy
dt
#
#
È(2 sin 2t)# (2 cos 2t)# œ 2
‰ Š dy
œ 2 cos 2t Ê Êˆ dx
dt
dt ‹ œ
œ1
#
#
È(1 cos 1t)# (1 sin 1t)# œ 1
‰ Š dy
œ 1 sin 1t œ ʈ dx
dt
dt ‹ œ
Ê Length œ 'c1Î2 1 dt œ c1td "Î# œ 1
1Î2
30. x œ a() sin )) Ê
Ê
dy
d)
dx
d)
"Î#
‰# œ a# a1 2 cos ) cos# )b and y œ a(1 cos ))
œ a(1 cos )) Ê ˆ dx
d)
#
#
' ˆ dx ‰# Š dyd) ‹ d) œ '0 È2a# (1 cos )) d)
œ a sin ) Ê Š dy
d) ‹ œ a sin ) Ê Length œ 0 Ê d)
#
#
21
œ aÈ2'0 È2 É 1 #cos ) d) œ 2a '0 ¸sin #) ¸ d) œ 2a '0 sin
21
21
21
)
#
21
#1
d) œ 4a cos 2) ‘ ! œ 8a
31-36. Example CAS commands:
Maple:
with( plots );
with( Student[Calculus1] );
with( student );
f := x -> 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 );
# (b)
# (a)
# (c)
37-40. Example CAS commands:
Maple:
with( plots );
with( student );
x := t -> t^3/3;
y := t -> t^2/2;
a := 0;
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)];
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383
384
Chapter 6 Applications of Definite Integrals
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
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
#
!†mL†2m
m2m
L
6
and y œ
œ
4
#
œ 8; M œ '0 4 dx œ [4x]#! œ 4 † 2 œ 8 Ê x œ
2L
3
‰
Ê ˆ L6 ß 2L
3 is the center of mass location.
2
M!
M
œ1
(9 1) œ 16; M œ '1 4 dx œ [4x]$" œ 12 4 œ 8 Ê x œ
3
M!
M
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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œ
16
8
œ2
Section 6.4 Moments and Centers of Mass
7. M! œ '0 x ˆ1 x3 ‰ dx œ '0 Šx 3
œ3
3
œ
9
6
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 ‹
%
x#
8 “!
15
9
$
x$
9 “!
#
dx œ ’ x# œ
x#
4‹
œ
27 ‰
9
M œ '0 ˆ1 3x ‰ dx œ ’x 3
15
# ;
16
8
%
x$
12 “ !
œ ˆ16 64 ‰
12
œ
œ
œ6 Ê xœ
dx œ '1 ˆx x"Î# ‰ dx œ ’ x# 4
#
M!
M
%
2x$Î#
3 “"
32
3 †6
œ ˆ8 œ 16 16 ‰
3
M!
M
10. M! œ '1Î4 x † 3 ˆx$Î# x&Î# ‰ dx œ 3'1Î4 ˆx"Î# x$Î# ‰ dx œ 3 2x"Î# 1
1
œ 6 14 œ 20 Ê x œ
M!
M
œ
2 ‘"
3x$Î# "Î%
œ
9
3
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
œ
32
3 ;
ˆ 73
‰
6
5
œ
2 ‘"
x"Î# "Î%
œ
15
#
14
3
2
1
1
2
2
œ
23
6 ;
Ê xœ
M!
M
‰ ˆ 72 ‰ œ
œ ˆ 23
6
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
4528
6
œ
9
20
2
1
2
3
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# 1
œ 16 †
ˆ "# 32 ‰ œ
4
œ 3(4 1) œ 9; M œ 3'1Î4 ˆx$Î# x&Î# ‰ dx œ 3 x"Î#2 16
3
16
9
%
M œ '1 ˆ1 x"Î# ‰ dx œ x 2x"Î# ‘ " œ (4 4) (1 2) œ 5 Ê x œ
1
$
x#
6 “!
5
3
dx œ ’x# œ8
œ ˆ 92 385
!
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 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$ ‹
‰
mass is the point (xß y) œ ˆ!ß 12
5 .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
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
$ † 5$ . Therefore y œ
4
3
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
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
Ê (xß y) œ ˆ0ß 85 ‰ is the center of mass.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
4$
3
;
Section 6.4 Moments and Centers of Mass
387
17. The typical 29<3D98>+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<3D98>+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
ˆ 1# 0‰ ˆ 1# ‰‘ œ
$1
4
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,
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
$
#
1Î4
#
sec% x dx. Mx œ 'c1Î4 µ
y dm œ
$
#
'11ÎÎ44 sec% x dx
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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1Î2
; M œ ' dm œ $ '1Î2 cos x dx
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
$
1Î%
1 Î4
Therefore, y œ
Mx
M
œ ˆ 43$ ‰ ˆ 2"$ ‰ œ
2
3
$
3
$ œ
4$
3
#$ [tan x]1Î%
; M œ ' dm œ $ 'c1Î4 sec# x dx œ $ [tan x]1Î4 œ $ [1 (1)] œ 2$ .
1Î4
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
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
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
œ 2$ x" ‘ " œ 2$ ˆ "a 1‰ œ
a
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
%
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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"
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
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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#
!
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
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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)
#
#
1
sin 2) ‘ 1Î#
#
!
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
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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
(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
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
(b)
(tan x) È1 sec% x dx
(c) S ¸ 3.84
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393
394
2. (a)
Chapter 6 Applications of Definite Integrals
dy
dx
#
(b)
2
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
(b)
#
#
#
Ê S œ 21'1 ˆ3 x"Î# ‰ É1 a1 3x"Î# b dx
4
(c) S ¸ 63.37
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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
(x 1)
#
É1 ˆ "# ‰# dx œ
1È5
#
'13 (x 1) dx
œ
1È5
#
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
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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 ‰$Î# “
491
3
%
'
œ y# Ê Š dx
dy ‹ œ y Ê S œ 0
#
1
u œ 1 y% Ê du œ 4y$ dy Ê
"
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$Î# ‘ #" œ 19 ŠÈ8 1‹
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Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus
18. x œ ˆ "3 y$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$Î# y"Î# ‰ É 4" ay 2 y" b dy
3
Éay"Î# y"Î# b#
œ 21'1 ˆ "3 y$Î# y"Î# ‰
3
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# #
#
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
"
2y 1
$Î#
’1$Î# ˆ 58 ‰
“œ
#
1 dy œ ÊŠy' "
#
"
16y' ‹
2
È2
È2
dy
dx
œ
"
#
aa# x# b
Ê S œ 21'ca Èa# x# É1 a
$Î#
œ
21 r
h
dy
dx
#
#
É h h# r
25. y œ cos x Ê
œ
41 È 2
3
5È 5
‹
8È 8
Š1 21
40
"
#
"
16y' ‹
dy
dy œ 21'1 ˆy% "4 y# ‰ dy
2
"
4y$ ‹
(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#
# “!
œ 21 ˆ 44 22 ‰ œ 41
x#
aa # x # b
dx œ 21'ca Èaa# x# b x# dx œ 21'ca a dx œ 21a[x]ca a
a
a
r
h
#
Ê Š dy
dx ‹ œ
r#
h#
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
'0h x dx œ 2h1r Èh# r# ’ x# “ h œ 2h1r Èh# r# Š h# ‹ œ 1rÈh# r#
#
#
dy
dx
5$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
15Î4
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
dy œ 41'0
15 ‰$Î#
4
1
"
4y$ ‹
dy œ Šy$ &
"
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#
#
0
#
#
È1 sin# x dx
'
œ sin x Ê Š dy
dx ‹ œ sin x Ê S œ 21 c1Î2 (cos x)
#
1Î2
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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
œ 21'a
2x
È r# x #
œ
Èar# x# b x# dx œ 21r'
a
29. y œ ÈR# x# Ê
abh
œ "#
dy
dx
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
45
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
S œ 'c22Þ5 21 È45# y# É1 #
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#
#
Ê Š dx
dy ‹ œ
y#
45# y#
;
dy œ 21 '22Þ5 Èa45# y# b y# dy œ 21 † 45'22Þ5 dy
45
45
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
x%
9
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
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Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus
34.
dx
dt
œ t"Î# and
È3
dy
dt
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$Î# ‰ É t
Ê
35.
dx
dt
È3
'0
F(t) dt œ
œ 1 and
È2
dy
dt
#
281
9
"
t
41
3
È3
'0
1
t
Ê S œ ' 21x ds
tÈt# 1 dt; cu œ t# 1 Ê du œ 2t dt; t œ 0 Ê u œ 1,
'14 231 Èu du œ 491 u$Î# ‘ %" œ 2891
’t œ È3 Ê u œ 4“ Ä
Note:
#
#
Èt t" œ É t
‰ Š dy
œ t"Î# Ê Êˆ dx
dt
dt ‹ œ
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" )#
;
#
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
#
?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 .
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399
400
Chapter 6 Applications of Definite Integrals
! ?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
È3
40. S œ 'a 21f(x) dx œ '0
b
21 †
b
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 $‰ is
3È 5
#
units from the origin Ê the x-coordinate of the
#
centroid solves the equation Ɉx 3# ‰ (2x 3)#
œ
È5
#
Ê ˆx# 3x 94 ‰ a4x# 12x 9b œ
5
4
Ê 5x# 15x 9 œ 1
Ê x# 3x 2 œ (x 2)(x 1) œ 0 Ê x œ 1 since the centroid must lie inside the triangle Ê y œ 2. By the
Theorem of Pappus, the volume is V œ (distance traveled by the centroid)(area of the region) œ 21 a5 xb "# (3)(6)‘
œ (21)(4)(9) œ 721
43. The centroid is located at (#ß !) Ê V œ (21) axb (A) œ (21)(2)(1) œ 41#
44. We create the cone by revolving the triangle with vertices
(0ß 0), (hß r) and (hß 0) about the x-axis (see the accompanying
figure). Thus, the cone has height h and base radius r. By
Theorem of Pappus, the lateral surface area swept out by the
hypotenuse L is given by S œ 21yL œ 21 ˆ r ‰ Èh# r#
#
œ 1rÈr# h# . To calculate the volume we need the position
of the centroid of the triangle. From the diagram we see that
the centroid lies on the line y œ
œ
"
3
Éh# r#
4
#
r
2h
#
#
x. The x-coordinate of the centroid solves the equation É(x h)# ˆ 2hr x #r ‰
#
#
Ê Š 4h4h# r ‹ x# Š 4h 2h r ‹ x inside the triangle Ê y œ
r
2h
47. V œ 21 yA Ê
4
3
2 ar# 4h# b
9
œ0 Ê xœ
2h
3
or
4h
3
Ê xœ
x œ 3r . By the Theorem of Pappus, V œ 21 ˆ 3r ‰‘ ˆ "# hr‰ œ
45. S œ 21 y L Ê 41a# œ a21yb (1a) Ê y œ
46. S œ 213 L Ê 21 ˆa r#
4
2a ‰‘
(1a)
1
2a
1,
since the centroid must lie
1r# h.
and by symmetry x œ 0
œ 21a# (1 2)
1ab# œ a21yb ˆ 1#ab ‰ Ê y œ
48. V œ 213A Ê V œ 21 ˆa "
3
2h
3 ,
4a ‰‘ 1a#
Š # ‹
31
œ
4b
31
and by symmetry x œ 0
1a$ (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
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Section 6.6 Work
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
‰
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
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401
402
Chapter 6 Applications of Definite Integrals
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
#
œ 0.624 ’ x# “
&!
!
50
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% #%$
$#
32
" ‰
#43!Þ%
ˆ 4" 9" ‰
œ 109,350
0.4
œ (109,350)(5)
(0.4)(36) œ 37,968.75 in † lb. Note that when a system is compressed, the work done by the system is negative.
13. Let r œ the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate,
the amount of water in the bucket is proportional to a#! xb, the distance the bucket is being raised. The leakage rate of
the water is 0.8 lb/ft raised and the weight of the water in the bucket is F œ 0.8a#! xb. So:
W œ '0 0.8a#! xb dx œ 0.8 ’20x 20
#!
x#
# “!
œ 160 ft † lb.
14. Let r œ the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate,
the amount of water in the bucket is proportional to a#! xb, the distance the bucket is being raised. The leakage rate of
the water is 2 lb/ft raised and the weight of the water in the bucket is F œ 2a#! xb. So:
W œ '0 2a#! xb dx œ 2 ’20x 20
#!
x#
# “!
œ 400 ft † lb.
Note that since the force in Exercise 14 is 2.5 times the force in Exercise 13 at each elevation, the total work is also 2.5
times as great.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 6.6 Work
403
15. We will use the coordinate system given.
(a) The typical slab between the planes at y and y ?y has
a volume of ?V œ (10)(12) ?y œ 120 ?y ft$ . 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:
W œ '0 62.4 † 120y dy œ (62.4)(120) ’ y# “
20
#
#!
!
‰ 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
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
404
Chapter 6 Applications of Definite Integrals
lb
ft$ :
(d) In a location where water weighs 62.26
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
lb
ft$
In a location where water weighs 62.59
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
$
’ "!$y 571
4
"!
y%
% “!
œ 11,8751 ft † lb ¸ 37,306 ft † lb.
#
18. Each slab of oil is to be pumped to a height of 14 ft. So the work to pump a slab is a"% yba1bˆ y# ‰ and since the tank is
half full and the volume of the original cone is V œ "$ 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 30
¸ 7,238,229.48 ft † lb
$!
y#
# “!
‰ œ (5120)(4501)
œ 51201 ˆ 900
#
20. (Alternate Solution) Each method must pump all of the water the 15 ft to the base of the tank. Pumping to the rim requires
all the water to be pumped an additional 6 feet. Pumping into the bottom requires that the water be pumped an average of 3
additional feet. Thus pumping through the valve requires È$ fta%1b6 ft3 a'#Þ% lb/ft3 b ¸ 14,115 ft † lb less work and thus
less time.
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 lb
ft$
8%
4‹
for 57
lb
ft$ .
Then,
1‰
‰
œ ˆ 64.5
a8$ b ˆ 10
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
œ (191) a8# b (7)(2) ¸ 53.482.5 ft † lb
$
’ 13y
3 )
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
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 6.6 Work
405
?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
16
10,000†1†16$
6
#
"'
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
W œ 'c5 98001 a25 y# b (4 y) dy œ 98001 'c5 a100 25y 4y# y$ b dy œ 98001 ’100C 0
0
25†25
#
œ 98001 ˆ500 † 125 4
3
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$lb † ?V œ 561 a100 y# b ?y lb. The
distance through which F(y) must act to lift the slab to the level of 2 ft above the top of the tank is about
(12 y) ft, so the work done is ?W ¸ 561 a100 y# b (12 y) ?y lb † ft. The work done lifting all the slabs
10
from y œ 0 ft to y œ 10 ft is approximately W ¸ ! 561 a100 y# b (12 y) ?y lb † ft. Taking the limit of these
0
Riemann sums, we get W œ '0 561 a100 y b (12 y) dy œ 561'0 a100 y# b (12 y) dy
10
10
#
œ 561'0 a1200 100y 12y# y$ 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 $4838.05. Yes, you can afford to hire the firm.
25. F œ m
œ
"
#
dv
dt
œ mv
by the chain rule Ê W œ 'x mv
x#
dv
dx
#
"
m cv# (x# ) v (x" )d œ
26. weight œ 2 oz œ
"
#
weight
32
œ
"
8
3#
œ
"
#56
28. weight œ 1.6 oz œ 0.1 lb Ê m œ
"
8
x#
"
dv ‰
dx
dx œ m "# v# (x)‘ x"
x#
"
slugs; W œ ˆ "# ‰ ˆ #56
slugs‰ (160 ft/sec)# ¸ 50 ft † lb
hr
1 min
5280 ft
27. 90 mph œ 901 hrmi † 601 min
† 60
sec † 1 mi œ 132 ft/sec; m œ
0.3125 lb ‰
#
W œ ˆ "# ‰ ˆ 32
ft/sec# (132 ft/sec) ¸ 85.1 ft † lb
29. weight œ 2 oz œ
dx œ m'x ˆv
mv## "# mv"# , as claimed.
lb; mass œ
2
16
dv
dx
lb Ê m œ
"
8
0.1 lb
32 ft/sec#
slugs œ
32
œ
"
#56
"
3 #0
0.3125 lb
32 ft/sec#
œ
0.3125
32
slugs;
slugs; W œ ˆ "# ‰ ˆ 3"#0 slugs‰ (280 ft/sec)# œ 122.5 ft † lb
slugs; 124 mph œ
(124)(5280)
(60)(60)
¸ 181.87 ft/sec;
"
W œ ˆ "# ‰ ˆ 256
slugs‰ (181.87 ft/sec)# ¸ 64.6 ft † lb
30. weight œ 14.5 oz œ
31. weight œ 6.5 oz œ
14.5
16
6.5
16
lb Ê m œ
lb Ê m œ
14.5
(16)(32)
6.5
(16)(32)
"4.5
slugs; W œ ˆ "# ‰ Š (16)(32)
slugs‹ (88 ft/sec)# ¸ 109.7 ft † lb
6.5
slugs; W œ ˆ "# ‰ Š (16)(32)
slugs‹ (132 ft/sec)# ¸ 110.6 ft † lb
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
!
y%
4 “ &
406
Chapter 6 Applications of Definite Integrals
32. F œ (18 lb/ft)x Ê W œ '0 18x dx œ c9x# d !
1Î6
mœ
"
8
32
œ
"
#56
"Î6
slugs and v" œ 0 ft/sec. Thus,
1
4
œ
1
4
ft † lb. Now W œ
È2
4
È2
4 ‹
(16) Š
È2
4 ‹
1
4
ft † lb,
sec when the bearing is at the top of its path.
È2
4
The height the bearing reaches is s œ 8È2 t 16t# Ê at t œ
Š8È2‹ Š
mv# "# mv"# , where W œ
"
ft † lb. œ ˆ #" ‰ ˆ #56
slugs‰ v# Ê v œ 8È2 ft/sec. With v œ 0
at the top of the bearing's path and v œ 8È2 32t Ê t œ
#
"
#
the bearing reaches a height of
œ 2 ft
33. (a) From the diagram,
rayb œ '! x œ '! É&!# ay 325b#
for 325 Ÿ y Ÿ 375 ft.
(b) The volume of a horizontal slice of the funnel
#
is ˜V ¸ 1rayb‘ ˜y
#
œ 1”'! É&!# ay 325b# • ˜y
(c) The work required to lift the single slice of
water is ˜W ¸ 62.4˜Va$(& yb
#
œ 62.4a$(& yb1”'! É&!# ay 325b# • ˜y.
The total work to pump our the funnel is W
#
œ '325 62.4a375 yb1”'! É50# ay 325b# • dy
375
¸ 6.3358 † 10( ft † lb.
34. (a) From the result in Example 6, the work to pump out the throat is 1,353,869,354 ft † lb. Therefor, the total work
required to pump out the throat and the funnel is 1,353,869,354 63,358,000 œ 1,417227,354 ft † lb.
(b) In horsepower-hours, the work required to pump out the glory hole is 1,417227,354
œ 715.8. Therefore, it would take
1.98†106
715.8 hp†h
1000 hp
œ 0.7158 hours ¸ 43 minutes.
35. We imagine the milkshake divided into thin slabs by planes perpendicular to the y-axis at the points of a
partition of the interval [!ß (]. The typical slab between the planes at y and y ?y has a volume of about
#
17.5 ‰
?V œ 1(radius)# (thickness) œ 1 ˆ y14
?y in$ . The force F(y) required to lift this slab is equal to its
weight: F(y) œ
4
9
?V œ
41
9
#
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
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 6.7 Fluid Pressures and Forces
385
385
360
360
lift all the water into the tank is: W" ¸ ! ?W" œ ! 62401 † y † ?y lb † ft. Taking the limit we end up with
W" œ '360 62401y dy œ 62401 ’ y# “
385
#
$)&
$'!
62401
#
œ
c385# 360# d ¸ 182,557,949 ft † lb
4
#
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
360
0
62.4
36
62.41
36
"
6
in œ
ft.
?y. Also take different limits of summation and
1y dy œ
62.41
36
#
$'!
#
1 ‰ 360
œ ˆ 62.4
Š # ‹ ¸ 352,864 ft † lb.
36
’ y# “
!
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
W
tank and the pipe is Time œ 1650
¸ 182,910,813
¸ 110,855 sec ¸ 31 hr
1650
37. Work œ '6 370 000
35ß780ß000
ß
1000 MG
r#
ß
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 ")#
, and W# œ '
d3 œ 23 ‚ 10#* ’ 3 " " “
23‚10#*
r##
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
5
23‚10#*
12
d3 œ '
23‚10#*
#
3 (3 " )
5
(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
# 117 ‰
3
"
3
† 8‰ ˆ 5# † 25 "
3
† 125‰‘
œ (124.8) ˆ 315 6 234 ‰ œ 1684.8 lb
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 9
#
0
y#
#
!
y$
3 “ $
‰
9‰ œ (124.8) ˆ 27
# œ 1684.8 lb
3. Using the coordinate system of Exercise 4, we find the equation for the line of the plate's right-hand edge is
y 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 $
œ (124.8) ˆ36 0
9
#
‰
9‰ œ (124.8) ˆ 45
# œ 2808 lb
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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407
408
Chapter 6 Applications of Definite Integrals
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
0
œ (124.8) ˆ 27
3 œ
27 ‰
#
0
(124.8)(27)(2 3)
6
$
!
$
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).
(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 5
#
5y#
#
4y“
4‰ œ (62.4) ˆ 2 156 24 ‰ œ
"
!
(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#
# “!
‰# “ œ
œ (62.4)(2) ’ˆ "# ‰ ˆ 11
3
(62.4)(12")
9
force doubles.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
¸ 838.93 lb Ê the fluid
Section 6.7 Fluid Pressures and Forces
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 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 $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
$
max ‰
‹ œ (62.4) ˆ 45 ‰ ˆ "6 ‰ h$ œ (10.4) ˆ 45 ‰ h$ Ê h œ Ɉ 54 ‰ ˆ F10.4
"
#
(Base)(Height) † 30, where
h Ê V œ ˆ 25 h# ‰ (30) œ 12h# ¸ 12(9.288)# ¸ 1035 ft$ .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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409
410
Chapter 6 Applications of Definite Integrals
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
1
#
œ (124.8) ˆ3 "3 ‰ œ (124.8) ˆ 83 ‰ œ 332.8 lb
"
y$
3 “!
(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# ‹
œ
ˆ wb
‰
# (ab)
b
b
#
b
0
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 a coordinate 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 #
(7.75)
‰
œ ˆ 64.5
¸ 4.2 lb
12$ (3.75)
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
(Þ(&
y#
# “!
Chapter 6 Practice Exercises
411
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
#
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
. 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 ‹
$
$È
Ê Y œ É (9)(187.2)
13.5 ¸ 2.3811 ft. So,
124.8 œ
?Y œ 3 Y ¸ 3 2.3811 ¸ 0.6189 ft ¸ 7.5 in. to the nearest half inch.
(c) No, it does not matter how long the trough is. The fluid pressure and the resulting force depend only on depth of the
water.
‰
22. The area of a strip of the face of height ?y and parallel to the base is 100ˆ 26
24 † ?y, where the factor of
26
24
inclination of the face of the dam. With the origin at the bottom of the dam, the force on the face is then:
‰
F œ '0 w(24 y)a100bˆ 26
24 dy œ '('! ’#%y 24
#%
y#
# “!
œ '('!Š#%# #%#
# ‹
œ 1,946,880 lb.
CHAPTER 6 PRACTICE EXERCISES
#
1. A(x) œ 14 (diameter)# œ 14 ˆÈx x# ‰
œ 14 ˆx 2Èx † x# x% ‰ ; a œ 0, b œ 1
Ê V œ 'a A(x) dx œ
b
œ
1
4
œ
1
4†70
#
’ x# 74 x(Î# È3
4
x&
5 “!
(35 40 14) œ
2. A(x) œ
œ
1
4
"
"
#
'01 ˆx 2x&Î# x% ‰ dx
1
4
œ
ˆ "# 4
7
5" ‰
91
280
(side)# ˆsin 13 ‰ œ
È3
4
ˆ2Èx x‰#
ˆ4x 4xÈx x# ‰ ; a œ 0, b œ 4
Ê V œ 'a A(x) dx œ
È3
4
b
œ
È3
4
œ
32È3
4
’2x# 85 x&Î# ˆ1 8
5
'04 ˆ4x 4x$Î# x# ‰ dx
%
x$
3 “!
32 ‰ œ
8È 3
15
œ
È3
4
ˆ32 8†32
5
(15 24 10) œ
64 ‰
3
8È 3
15
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
accounts for the
412
Chapter 6 Applications of Definite Integrals
3. A(x) œ
œ
1
4
1
4
(diameter)# œ
1
4
(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
#
cos 1#
#
‹ Š 14 ‹“ 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
œ ’36x 24È6 † 23 x$Î# 18x# 4È6 † 25 x&Î# œ 216 576 648 5. A(x) œ
(diameter)# œ
1
4
72 œ 360 Š2Èx x#
4‹
#
œ
1728
5
œ
1
4
œ 216 16 † È6 È6 † 6 18 † 6# 58 È6 È6 † 6# 18001728
5
Š4x x&Î# œ
6$
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
'
x$
3 “!
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
&
"
x*
9 “ "
#
œ 181 25 9" ‘ œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
21†13
5
œ
261
5
121
5
Chapter 6 Practice Exercises
8. (a) A+=2/< 7/>29. :
R(x) œ
, r(x) œ
4
x$
"
#
#
#
#
&
Ê V œ 'a 1cR# (x) r# (x)d dx œ '1 1 ’ˆ x4$ ‰ ˆ "# ‰ “ dx œ 1 16
x4 ‘ "
5 x
b
2
"‰
"
ˆ 16 " ‰‘ œ 1 ˆ 10
œ 1 ˆ 5†16
32 # 5 4
(b) =2/66 7/>29. :
V œ 21'1 x ˆ x4$ "# ‰ dx œ 21 ’4x" 2
(c) =2/66 7/>29. :
"
#
#
x#
4 “"
4" ‰ œ
16
5
1
20
(2 10 64 5) œ
b
2
4
x
x
(d) A+=2/< 7/>29. :
#
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$ œ 21 ’ x4# 413
2
4
x#
51
#
1 x# ‰ dx
œ 21 (1 2 2 1) ˆ4 4 1 4" ‰‘ œ
31
#
V œ 'a 1cR# (x) r# (x)d dx
b
#
œ 1 '1 ’ˆ 7# ‰ ˆ4 2
dx
œ
491
4
161'1 a1 2x$ x' b dx
œ
491
4
161 ’x x# œ
491
4
491
4
491
4
161 ˆ2 4" 5†"3# ‰ ˆ1 1 5" ‰‘
"
161 ˆ 4" 160
5" ‰
œ
œ
9.
4 ‰#
x$ “
2
161
160
#
x&
5 “"
(40 1 32) œ
491
4
711
10
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
#
y&
5
23 y$ “
10881
15
#
#
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‹
321
3
$
dy œ 21 ’ y3 %
y%
16 “ !
y#
4‹
dy
œ 21 ˆ 64
3 64 ‰
4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
32
5
2
3
† 8‰
414
Chapter 6 Applications of Definite Integrals
(b) =2/66 7/>29. :
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21x ˆ2Èx x‰ dx œ 21'0 ˆ2x$Î# x# ‰ dx œ 21 ’ 45 x&Î# b
4
œ 21 ˆ 45 † 32 64 ‰
3
œ
4
1281
15
%
x$
3 “!
(c) =2/66 7/>29. :
shell ‰
shell
V œ 'a 21 ˆ radius
Š height
‹ dx œ '0 21(4 x) ˆ2Èx x‰ dx œ 21'0 ˆ8x"Î# 4x 2x$Î# x# ‰ dx
b
4
$Î#
œ 21 ’ 16
2x# 54 x&Î# 3 x
œ 641 ˆ1 45 ‰ œ
641
5
4
%
x$
3 “!
œ 21 ˆ 16
3 † 8 32 (d) =2/66 7/>29. :
shell ‰
shell
V œ 'c 21 ˆ radius
Š height
‹ dy œ '0 21(4 y) Šy d
4
œ 21'0 Š4y 2y# 4
y$
4‹
y#
4‹
%
y%
16 “ !
dy œ 21 ’2y# 23 y$ 4
5
† 32 64 ‰
3
œ 641 ˆ 34 1 dy œ 21'0 Š4y y# y# 4
œ 21 ˆ32 2
3
4
5
y$
4‹
32 ‰
dy
† 64 16‰ œ 321 ˆ2 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
œ 1 4x 4 cos x 1
x
#
sin 2x ‘ 1
4
!
1
œ 1 ˆ41 4 1
#
0‰ (0 4 0 0)‘ œ
œ
1cos 2x ‰
dx
#
9
1
1 ˆ # 8‰ œ 1#
1 Š3È31‹
3
(91 16)
13. (a) .3=5 7/>29. :
V œ 1'0 ax# 2xb dx œ 1'0 ax% 4x$ 4x# b dx œ 1 ’ x5 x% 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
#
%
!
81
3
32
3
8‰
(d) A+=2/< 7/>29. :
V œ 1 '0 c2 ax# 2xbd dx 1'0 2# dx œ 1'0 ’4 4 ax# 2xb ax# 2xb “ dx 81
2
2
#
2
#
œ 1'0 a4 4x# 8x x% 4x$ 4x# b dx 81 œ 1'0 ax% 4x$ 8x 4b dx 81
2
&
2
#
‰
œ 1 ’ x5 x% 4x# 4x“ 81 œ 1 ˆ 32
5 16 16 8 81 œ
!
1
5
(32 40) 81 œ
721
5
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)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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401
5
œ
321
5
Chapter 6 Practice Exercises
15. The material removed from the sphere consists of a cylinder
and two "caps." From the diagram, the height of the cylinder
#
is 2h, where h# 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
ft$ . Therefore,
Vremoved œ Vcyl #Vcap œ '1 "!1
3
#)1
3
œ
ft$ .
16. We rotate the region enclosed by the curve y œ É12 ˆ1 4x# ‰
121
and the x-axis around the x-axis. To find the
11Î2
volume we use the .3=5 method: V œ 'a 1R# (x) dx œ 'c11Î2 1 ŠÉ12 ˆ1 b
œ 121'c11Î2 Š1 11Î2
4x#
121 ‹
œ 1321 ˆ1 "3 ‰ œ
17. y œ x"Î# x$Î#
3
Ê
ˆ4 "
#
œ
#
x"Î# "# x"Î# Ê Š dy
dx ‹ œ
"
4
œ '1
8
dx
dy
È9x#Î$ 4
3x"Î$
œ
5
12
2
3
#
4
4
ˆ2 14 ‰
3
#
x"Î$ Ê Š dx
dy ‹ œ
œ
4x#Î$
9
"
#
ˆx"Î# x"Î# ‰ dx œ
"
#
2x"Î# 23 x$Î# ‘ %
"
10
3
dx
Ê L œ '1 Ê1 Š dy
‹ dy œ '1 É1 #
8
8
4
9x#Î$
dy
'1 È9x#Î$ 4 ˆx"Î$ ‰ dx; u œ 9x#Î$ 4 Ê du œ 6y"Î$ dy; x œ 1
40
" '
" 2 $Î# ‘ %!
Ä L œ 18
u"Î# du œ 18
œ #"7 40$Î# 13$Î# ‘ ¸ 7.634
3 u
"$
13
dx œ
x œ 8 Ê u œ 40d
19. y œ
"
#
† 8‰ ˆ2 23 ‰‘ œ
18. x œ y#Î$ Ê
8
"
3
x'Î& 58 x%Î& Ê
dx
ˆ x" 2 x‰ Ê L œ ' É1 4" ˆ x" 2 x‰ dx
1
4
2
3
4x#
121 ‹
œ 881 ¸ 276 in$
4
"
#
11Î2
dx œ 1 '11Î2 12 Š1 4 ‰ 11
ˆ 4 ‰ ˆ 11
‰ “ œ 1321 ’1 ˆ 363
œ 241 ’ 11
Š 4 ‹“
2 363
#
#
Ê L œ '1 É 4" ˆ x" 2 x‰ dx œ '1 É 4" ax"Î# x"Î# b dx œ '1
œ
#
$
4x$
363 “ ""Î#
dx œ 121 ’x 2641
3
dy
dx
""Î#
4x# ‰
121 ‹
#
"
#
œ
dy
dx
x"Î& "# x"Î& Ê Š dy
dx ‹ œ
"
4
Ê u œ 13,
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
œ '1
32
œ
"
48
20. x œ
"
#
ˆx
"Î&
x
"Î& ‰
(1260 450) œ
"
1#
y$ "
y
Ê
" %
œ '1 É 16
y 2
"
#
dx œ
œ
1710
48
dx
dt
x
'Î&
5
4
$#
x%Î& ‘ "
#
dx
dy
œ
"
4
"
y%
dy œ '1 ÊŠ 4" y# y# 8
œ ˆ 12
"# ‰ ˆ 1"# 1‰ œ
21.
" 5
# 6
285
8
"
y#
dx
Ê Š dy
‹ œ
2
7
1#
œ 5 sin t 5 sin 5t and
dy
dt
"
#
œ
"
y# ‹
"
16
#
œ
"
#
ˆ 65
y% "
#
'
†2 "
y%
5
4
ˆ 56
5 ‰‘
4
œ
"
#
ˆ 315
6 " %
Ê L œ '1 Ê1 Š 16
y dy œ '1 Š 4" y# 2
†2
%‰
2
"
y# ‹
dy œ ’ 1"# y$ y" “
"
#
75 ‰
4
"
y% ‹
dy
#
"
13
12
#
‰ Š dy
œ 5 cos t 5 cos 5t Ê Êˆ dx
dt
dt ‹
#
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# )
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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415
416
Chapter 6 Applications of Definite Integrals
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
† 23 Ša17b3/2 64‹ œ È2Ša17b3/2 64‹ ¸ 8.617.
œ $ sin ) and
Ê Length œ '!
dy
d)
$1Î2
#
24. x œ t and y œ
œ'
'16"7 Èu du œ 3È2 2 23 u3/2 ‘1617 œ 3È2 2 Š 23 a17b3/2 23 a16b3/2 ‹
t$
3
$ d) œ $'!
$1Î2
d) œ $ˆ $#1 !‰ œ
t, È3 Ÿ t Ÿ È3 Ê
È3
È 3
#
#
‰ Š dy
Éa$ sin )b# a$ cos )b# œ È$asin# ) cos# )b œ $
œ $ cos ) Ê Êˆ dx
d)
d) ‹ œ
Èt% #t# " dt œ
'
dx
dt
*1
#
œ 2t and
dy
dt
È3
È 3
Èt% 2t# " dt œ
œt
'
#
" Ê Length œ '
È3
È 3
È3
È 3
Éat# "b# dt œ
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
#
3
#
$ ax# 3b a1 x# b dx œ
&
$ ’ x5 œ 3$ ’x 2x$
3
"
x$
3 “ "
3x“
"
"
3
#
$ ax% 2x# 3b dx Ê Mx œ ' µ
y dm œ
œ 3$ ˆ 5" 2
3
3‰ œ
œ 6$ ˆ1 "3 ‰ œ 4$ Ê y œ
Mx
M
œ
3$
15
(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
Mx
M
œ
32†$ †3
5†16†$
œ
centroid is(xß y) œ ˆ!ß 65 ‰ .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
6
5
. Therefore, the
Chapter 6 Practice Exercises
417
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‹
x#
4 ‹dx,
width: dx,
mass: dm œ $ † dA
dx Ê the moment about the x-axis is
#
Š4 x4 ‹
µ
y dm œ $ †
x#
4,
x#
4‹
Š4 #
$
#
dx œ
moment about the y-axis is µ
x dm œ $ Š4 œ
$
2
’16x %
x&
5†16 “ !
$
#
œ
64 #
œ
16†$ †3
32†$
œ
Mx
M
and y œ
3
2
œ
dx. Thus, Mx œ ' µ
y dm œ
4
4
My
M
x$
4‹
‹ † x dx œ $ Š4x x
4
œ $ (32 16) œ 16$ ; M œ ' dm œ $ '0 Š4 Ê xœ
dx; the
; My œ ' µ
x dm œ $ '0 Š4x 128$
5
œ
64 ‘
5
x%
16 ‹
Š16 128†$ †3
5†32†$
x#
4‹
œ
dx œ $ ’4x 12
5
%
x$
12 “ !
x$
4‹
dx œ $ ’2x# œ $ ˆ16 64 ‰
1#
$
#
'04 Š16 16x ‹ dx
%
%
x%
16 “ !
32$
3
œ
‰
. Therefore, the centroid is (xß y) œ ˆ 3# ß 12
5 .
28. A typical 29<3D98>+6 strip has:
#
center of mass: (µ
x ßµ
y ) œ Š y # 2y ß y‹ , length: 2y y# ,
width: dy, area: dA œ a2y y# b dy, mass: dm œ $ † dA
œ $ a2y y# b dy; the moment about the x-axis is
µ
y dm œ $ † y † a2y y# b dy œ $ a2y# y$ b ; the moment
#
about the y-axis is µ
x dm œ $ † ay 2yb † a2y y# b dy
#
œ a4y y b dy Ê Mx œ ' µ
y dm œ $ '0 a2y# y$ b dy
$
#
#
œ $ ’ 23 y$ œ
$
#
ˆ 43†8
yœ
Mx
M
2
%
œ
#
y%
4 “!
32 ‰
5
4†$ †3
3†4†$
œ
œ $ ˆ 23 † 8 32$
15
16 ‰
4
œ $ ˆ 16
3 16 ‰
4
œ
$ †16
12
œ
4$
3
; My œ ' µ
x dm œ
$
#
'02 a4y# y% b dy œ #$ ’ 34 y$ y5 “ #
&
$
; M œ ' dm œ $ '0 a2y y# b dy œ $ ’y# y3 “ œ $ ˆ4 83 ‰ œ
#
2
!
!
4$
3
Ê xœ
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 Šy
#
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
4
15
(11) œ
2'
6‹
‰ ˆ 38 ‰ œ
œ ˆ 44
15
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# 2
44
15
y%
4
4
5
"
#
#
y&
5 “!
œ ˆ4 8
3
16 ‰
4
œ ˆ 16
3 œ
8
3
24
5
area: dA œ
x$Î#
32 ‰
5
œ 16 ˆ "3 "
#
dx, mass: dm œ $ † dA œ $ †
x$Î#
25 ‰
’ 43 y$ y% y&
5
2
Ê xœ
My
M
‰ ˆ 83 ‰ œ
œ ˆ 24
5
9
5
‰
. Therefore, the center of mass is (xß y) œ ˆ 95 ß 11
10 .
3
"
4
; M œ ' dm œ '0 (1 y) a2y y# b dy
30. A typical vertical strip has: center of mass: (µ
x ßµ
y ) œ ˆxß 2x3$Î# ‰ , length:
3
16
4
a4y# 4y$ y% y& b dy œ
86 ‰ œ 4 ˆ2 45 ‰ œ
#
y%
4 “!
a4y# 4y$ y% y& b dy
3
x$Î#
, width: dx,
dx Ê the moment about the x-axis is
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
and y œ
Mx
M
#
y'
6 “!
8
5
and
418
Chapter 6 Applications of Definite Integrals
µ
y dm œ
9$
2x$
3
† $ x$Î#
dx œ
3
#x$Î#
(a) Mx œ $ '1
9
M œ $ '1
9
(b) Mx œ '1
9
"
#
9$
#
ˆ x9$ ‰ dx œ
#
*
20$
9
’ x# “ œ
"
*
ˆ x9$ ‰ dx œ
9
#
*
dx.
*
My
M
œ
12$
4$
œ 3 and y œ
Mx
M
œ
ˆ 209$ ‰
4$
œ
5
9
*
3 ‰
3 ‰
"x ‘ * œ 4; My œ ' x# ˆ $Î#
dx œ 2x$Î# ‘ " œ 52; M œ '1 x ˆ x$Î#
dx
"
x
1
9
œ 6 x"Î# ‘ " œ 12 Ê x œ
31. S œ 'a 21y Ê1 Š dy
dx ‹ dx;
#
b
3$
x"Î#
3 ‰
; My œ $ '1 x ˆ x$Î#
dx œ 3$ 2x"Î# ‘ " œ 12$ ;
9
dx œ 6$ x"Î# ‘ " œ 4$ Ê x œ
3
x$Î#
x
#
3
dx; the moment about the y-axis is µ
x dm œ x † $ x$Î#
dx œ
My
M
dy
dx
œ
œ
13
3
and y œ
Mx
M
9
œ
"
3
#
"
È2x 1
Ê Š dy
dx ‹ œ
Ê S œ '0 21È2x 1 É1 3
"
#x 1
"
#x 1
dx
2
È ' Èx 1 dx œ 2È21 2 (x 1)$Î# ‘ $ œ 2È21 † 2 (8 1) œ
œ 21'0 È2x 1 É 2x
2x1 dx œ 2 21 0
3
3
!
3
3
32. S œ 'a 21y Ê1 Š dy
dx ‹ dx;
#
b
œ
1
6
dy
dx
%
'
œ x# Ê Š dy
dx ‹ œ x Ê S œ 0 21 †
#
1
x$
3
È1 x% dx œ
1
6
281È2
3
'01 È1 x% a4x$ b dx
'01 È1 x% d a1 x% b œ 16 ’ 32 a1 x% b$Î# “ " œ 19 ’2È2 1“
!
33. S œ 'c 21x Ê1 Š dx
dy ‹ dy;
#
d
dx
dy
œ
ˆ "# ‰ (4 2y)
È4y y#
2y
È4y y#
œ
#
Ê 1 Š dx
dy ‹ œ
4y y# 4 4y y#
4y y#
œ
4
4y y#
Ê S œ '1 21 È4y y# É 4y 4 y# dy œ 41'1 dx œ 41
2
2
34. S œ 'c 21x Ê1 Š dx
dy ‹ dy;
#
d
œ 1'2 È4y 1 dy œ
6
35. x œ
t#
#
1
4
dx
dy
œ
*
36. x œ t# "
2t
761
3
"
È2
1
œ 21 Š2 (125 27) œ
œ t and
ŸtŸ1 Ê
Ê Surface Area œ '1ÎÈ2 21 ˆt# 1
dx
dt
1
6
dy
dt
1
6
"
4y
œ
Ê S œ '2 21Èy †
6
4y 1
4y
(98) œ
È4y 1
È4y
dy
491
3
È5
œ 2 Ê Surface Area œ '0 21(2t)Èt# 4 dt œ '4 21u"Î# du
9
, where u œ t# 4 Ê du œ 2t dt; t œ 0 Ê u œ 4, t œ È5 Ê u œ 9
and y œ 4Èt ,
œ 21 '1ÎÈ2 ˆt# #
Ê 1 Š dx
dy ‹ œ 1 23 (4y 1)$Î# ‘ ' œ
#
and y œ 2t, 0 Ÿ t Ÿ È5 Ê
œ 21 23 u$Î# ‘ % œ
1
2È y
" ‰ˆ
2t
2t
" ‰
2t#
"‰
#t
dx
dt
œ 2t ʈ2t " ‰#
2t#
dt œ 21 '1ÎÈ2 ˆ2t$ 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
x ‰
9500
lb. The work done is W œ 'a F(x) dx
b
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Chapter 6 Practice Exercises
œ '0
4750
6400 ˆ1 x ‰
9500
dx œ 6400 ’x œ 22,800,000 ft † lb
%(&!
x#
2†9500 “ !
œ 6400 Š4750 4750#
4†4750 ‹
419
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
#
!
#
x#
20 ’ # “
"
W œ '1 kx dx œ k '1 x dx œ
2
"
2
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
1Þ2
1Þ2
œ [125x# ]!"Þ# œ 125(1.2)# œ 180 J
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 œ y# . A typical
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
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
"Î#
(2y) dy
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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420
Chapter 6 Applications of Definite Integrals
œ 22,800 † (area of a quarter circle having radius 4) 23 (1140) ’a16 y# b
$Î# !
“
œ 335,153.25 ft † lb
%
œ (22,800)(41) 48,640
strip
45. F œ 'a W † Š depth
‹ † L(y) dy Ê F œ 2 '0 (62.4)(2 y)(2y) dy œ 249.6'0 a2y y# b dy œ 249.6 ’y# b
2
2
#
y$
3 “!
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
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$Î# 25 y&Î# ‘ ! œ (62.4) ˆ6 † 8 2
5
Èy
2 ‹
(75)(3075)
9†#16
¸ 118.63 lb.
dy œ 62.4'0 ˆ9y"Î# 3y$Î# ‰ dy
4
‰ (48 † 5 64) œ
† 32‰ œ ˆ 62.4
5
(62.4)(176)
5
œ 2196.48 lb
strip
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
œ
6
62.4
3
' a240 34y y# b dy
0
œ
62.4
3
’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
x
1
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
where C
x
1.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Chapter 6 Additional and Advanced Exercises
421
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 ‘
#
œ1
2
n1
"
#n 1
œ
x
c1 #
(n 1)(2n 1) 2(2n ") (n 1)
(n 1)(#n 1)
œ
0 #
2n# 3n 1 4n 2 n 1
(n 1)(#n 1)
Also, M œ 'c1 dA œ 'c1 a1 xn b dx œ 2 '0 a1 xn b dx œ 2 x 1
yœ
Mx
M
œ
1
#
2n
(n 1)(2n 1)
†
1
(n 1)
2n
œ
n
2n 1
xn b 1 ‘ "
n1 !
n1
œ
2n#
(n 1)(#n 1)
œ 2 ˆ1 " ‰
n1
#n 1 !
.
œ
2n
n1.
Therefore,
Ê ˆ!ß #n n 1 ‰ is the location of the centroid. As n Ä _, y Ä
"
#
so
the limiting position of the centroid is ˆ!ß "# ‰ .
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
11
80
#
x‰‘ dx
b
11 ‰#
'040 x ˆ9 80
x dx; M œ 'a 1y# dx
40
40
‰‘# dx œ 64"1 ' ˆ9 11
‰# dx.
œ 1 '0 8"1 ˆ9 11
80 x
80 x
0
œ
"
641
My
M
Thus, x œ
¸
129,700
5623.3
¸ 23.06 (using a calculator to compute
the integrals). By symmetry about the x-axis, y œ 0 so the center of mass is about 23 ft from the top of the pole.
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
œ 4kÈa'0 x$Î# dx œ 4kÈa 25 x&Î# ‘ 0 œ 4ka"Î# † 25 a&Î# œ
a
a
8ka
7
%
8ka$
5
0
. Also, M œ ' dm œ '0 4kxÈax dx
a
. Thus, x œ
My
M
œ
8ka%
7
†
5
8ka$
œ
5
7
a
‰
Ê (xß y) œ ˆ 5a
7 ß 0 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 y#
4a ‹
dy, mass: dm œ $ dA œ kyk Ša y#
4a ‹
dy. Thus, Mx œ ' µ
y dm
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
y#
4a
,
422
Chapter 6 Applications of Definite Integrals
œ 'c2a y kyk Ša 2a
œ 'c2a Šay# 0
%
32a&
20a
œ 8a3 y#
4a ‹
0
dy '0 Šay# 2a
y%
4a ‹
8a%
3
dy œ 'c2a y# Ša #
œ
%
&
'c2a a16a y y b dy œ
"
32a#
’8a% † 4a# 64a'
6 “
"
32a#
'0
2a
"
32a#
2a
#
c2a
"
3 #a #
a16a% y y& b dy œ
64a'
6 “
"
16a#
œ
’8a% y# 32a'
3 ‹
Š32a' œ
œ2†
16a%
4 ‹
#
Š2a † 4a "
#a
œ
!
y'
6 “ #a
1
3#a#
’8a% y# † 32 a32a' b œ
4
3
#a
y'
6 “!
a% ;
'c2a2a kyk a4a# y# b dy
"
4a
#a
#
dy
"
16a#
%
"
4a
#a
y&
“
#0a !
y#
4a ‹
kyk Ša 'c02a a4a# y y$ b dy 4a" '02a a4a# y y$ b dy œ 4a" ’2a# y# y4 “ !
"
4a
yœ
#
4a#
8a ‹
’ 3a y$ ' kyk a16a% y% b dy
#
’8a% † 4a# M œ ' dm œ 'c2a kyk Š 4a 4ay ‹ dy œ
y&
#0a “ #a
dy
2a
#
"
3 #a #
#
y#
4a ‹
!
dy œ ’ 3a y$ 'c2a2a kyk ay# 4a# b Š 4a 4a y ‹ dy œ 32a"
2a
2a
"
8a
0
dy '0 y# Ša œ 0; My œ ' µ
x dm œ 'c2a Š y
32a&
#0a
œ
œ
y%
4a ‹
y#
4a ‹
%
%
$
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
a
dx $ 'a x ab# x# b
$ 2 ab
• #”
b
#
!
0
$
3
"Î#
# $Î#
$
3
“ ’0 aa b
•
a
$
3
“ ’0 ab# a# b
#
#
$Î#
We calculate the mass geometrically: M œ $ A œ $ Š 14b ‹ $ Š 14a ‹ œ
œ
$ ab $ a $ b
3
yœ
(b) lim
œ
Mx
M
4
b Ä a 31
†
4
$1 ab# a# b
4 aa# abb# b
31(ab)
Ša
#
ab b#
‹
ab
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
.
œ ˆ 341 ‰ Š a
#
a# a#
‹
aa
#
œ ˆ 341 ‰ Š 3a
2a ‹ œ
position of the centroid as b Ä a. This is the centroid of a circle of radius a (and we note the two circles
coincide when b œ a).
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Chapter 6 Additional and Advanced Exercises
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
‰
$'ˆ 24
a "!)ayb
"%%
which we solve to get x œ ) a
*
and y œ
)a a " b
.
a
Set
x œ 7 in. (Given). It follows that a œ *, whence y œ
œ
7 "*
'%
*
in. The distances of the centroid ax, yb from the other sides are easily computed. (Note that if we set y œ 7 in.
above, we will find x œ 7 "* .)
11. y œ 2Èx Ê ds œ É "x 1 dx Ê A œ '0 2Èx É "x 1 dx œ
3
4
3
(1 x)$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
t# †
t$
3m
dt œ
4h
3
È3mh
†
12 in
1 ft
2 lb
1 in
"
3m
'
’ t6 “
Ð12mh)"Î%
0
Cœ0 Ê
dx
dt
œ
t$
3m
Ê xœ
The work done is
" ‰
œ ˆ 18m
(12mh)'Î%
œ 24 lb/ft. Thus, F œ 24x Ê W œ '0
1Î2
24x dx
"Î#
"
"
‰
œ c12x# d ! œ 3 ft † lb. Since W œ "# mv!# "# mv"# , where W œ 3 ft † lb, m œ ˆ 10
lb‰ ˆ 3# ft/sec
#
"
œ 320
slugs, and v" œ 0 ft/sec, we have 3 œ ˆ #" ‰ ˆ 3#"0 v#! ‰ Ê v!# œ 3 † 640. For the projectile height,
s œ 16t# v! t (since s œ 0 at t œ 0) Ê ds
dt œ v œ 32t v! . At the top of the ball's path, v œ 0 Ê
#
and the height is s œ 16 ˆ 3v#! ‰ v! ˆ 3v#! ‰ œ
v!#
64
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
t%
12m
¸ 2329.6 lb
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
tœ
v!
3#
C" ;
423
424
Chapter 6 Applications of Definite Integrals
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
#
!
œ
w
=jw#
#
. The
=
w
average force on one side of the plate is Fav œ
œ
=
w
#
’ y# “
!
=w
#
œ
w
. Therefore the force
'c0w (y)dy
=jw#
#
‰
œ ˆ =w
# (jw) œ (the average pressure up and down) † (the area 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
ˆ
‰
œ # 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 “
!
œ =b ’0 Š 3h ‹“ œ
=bh$
3
'c0h = kyk L(y) dy œ =b 'c0h
F œ ' dp œ
œ
$
h
# !
=b ’ y# “
h
œ =b ’0 h#
#“
=bh#
#
œ
;
y dy
$
. 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$
a
=b
12h
a6a# h# 8ah$ 3h% b
=b
h
'cÐahÑ
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
.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
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)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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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 $Èx 1 œ f " (x)
16. Step 1: y œ x# 2x 1 Ê y œ (x 1)# Ê Èy œ x 1, since x
1 Ê x œ 1 Èy
Step 2: y œ 1 Èx œ f " (x)
17. Step 1: y œ (x 1)# Ê Èy œ x 1, since x
Step 2: y œ Èx 1 œ f
"
1 Ê x œ È y 1
(x)
18. Step 1: y œ x#Î$ 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
f af
"
(x)b œ ˆx
"Î% ‰%
0, Range of f " : y
œ x and f
"
0;
(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
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
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
" df "
œ 5,
dx
¹
œ 4,
df "
dx ¹ xœ3
œ ˆ x" ‰
"
œx
(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
(b)
5
4
x
4
"
4
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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
"
(x) œ
È x#
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
"
m
and y-intercept mb .
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.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
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$
"
9x#Î$
œ
œ
"
9
df "
dx
Ê
"
3
œ
df
dx
œ 24x# Ê
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
df "
"
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.
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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.
<<Miscellaneous `RealOnly`
Clear[x, y]
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 7.1 Inverse Functions and Their Derivatives
431
{a,b} = {2, 1}; x0 = 1/2 ;
f[x_] = (3x 2) / (2x 11)
Plot[{f[x], f'[x]}, {x, a, b}]
solx = Solve[y == f[x], x]
g[y_] = x /. solx[[1]]
y0 = f[x0]
ftan[x_] = y0 f'[x0] (x-x0)
gtan[y_] = x0 1/ f'[x0] (y y0)
Plot[{f[x], ftan[x], g[x], gtan[x], Identity[x]},{x, a, b},
Epilog Ä Line[{{x0, y0},{y0, x0}}], PlotRange Ä {{a,b},{a,b}}, AspectRatio Ä Automatic]
61-62. Example CAS commands:
Maple:
with( plots );
eq := cos(y) = x^(1/5);
domain := 0 .. 1;
x0 := 1/2;
f := unapply( solve( eq, y ), x ); # (a)
Df := D(f);
plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[1,3], legend=["y=f(x)","y=f '(x)"],
title="#62(a) (Section 7.1)" );
q1 := solve( eq, 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="#62(e) (Section 7.1)" );
Mathematica: (assigned function and values for a, b, and x0 may vary)
For problems 61 and 62, the code is just slightly altered. At times, different "parts" of solutions need to be used, as in the
definitions of f[x] and g[y]
Clear[x, y]
{a,b} = {0, 1}; x0 = 1/2 ;
eqn = Cos[y] == x1/5
soly = Solve[eqn, y]
f[x_] = y /. soly[[2]]
Plot[{f[x], f'[x]}, {x, a, b}]
solx = Solve[eqn, x]
g[y_] = x /. solx[[1]]
y0 = f[x0]
ftan[x_] = y0 f'[x0] (x x0)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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432
Chapter 7 Transcendental Functions
gtan[y_] = x0 1/ f'[x0] (y y0)
Plot[{f[x], ftan[x], g[x], gtan[x], Identity[x]},{x, a, b},
Epilog Ä Line[{{x0, y0},{y0, x0}}], PlotRange Ä {{a, b}, {a, b}}, AspectRatio Ä Automatic]
7.2 NATURAL LOGARITHMS
1. (a) ln 0.75 œ ln
(b) ln
4
9
"
#
3
4
œ ln 3 ln 4 œ ln 3 ln 2# œ ln 3 2 ln 2
œ ln 4 ln 9 œ ln 2# ln 3# œ 2 ln 2 2 ln 3
œ ln 1 ln 2 œ ln 2
(d) ln $È9 œ "3 ln 9 œ
(e) ln 3È2 œ ln 3 ln 2"Î# œ ln 3 "# ln 2
(f) ln È13.5 œ " ln 13.5 œ " ln 27 œ " aln 3$ ln 2b œ " (3 ln 3 ln 2)
(c) ln
#
2. (a) ln
"
125
#
(e) ln 0.056 œ ln
ln 35 ln
ln 25
"
7
7
125
3
#
"
#
2
3
ln 3
(b) ln 9.8 œ ln
49
5
œ ln 7# ln 5 œ 2 ln 7 ln 5
(d) ln 1225 œ ln 35# œ 2 ln 35 œ 2 ln 5 2 ln 7
œ ln 7 ln 5$ œ ln 7 3 ln 5
ln 5 ln 7 ln 7
# ln 5
œ
ln 3# œ
#
ln 7
"
#
œ
3. (a) ln sin ) ln ˆ sin5 ) ‰ œ ln (c)
#
œ ln 1 3 ln 5 œ 3 ln 5
(c) ln 7È7 œ ln 7$Î# œ
(f)
#
"
3
sin )
Š sin5 ) ‹ #
" ‰
(b) ln a3x# 9xb ln ˆ 3x
œ ln Š 3x 3x 9x ‹ œ ln (x 3)
œ ln 5
#
ln a4t% b ln 2 œ ln È4t% ln 2 œ ln 2t# ln 2 œ ln Š 2t# ‹ œ ln at# b
4. (a) ln sec ) ln cos ) œ ln [(sec ))(cos ))] œ ln 1 œ 0
(b) ln (8x 4) ln 2# œ ln (8x 4) ln 4 œ ln ˆ 8x 4 4 ‰ œ ln (2x 1)
$
"Î$
")
(c) 3 ln Èt# 1 ln (t 1) œ 3 ln at# 1b ln (t 1) œ 3 ˆ "3 ‰ ln at# 1b ln (t 1) œ ln Š (t (t1)(t
‹
1)
œ ln (t 1)
1 ‰
5. y œ ln 3x Ê yw œ ˆ 3x
(3) œ
7. y œ ln at# b Ê
9. y œ ln
3
x
10. y œ ln
10
x
œ ˆ t"# ‰ (2t) œ
dy
dt
œ ln 3x" Ê
dy
dx
œ ln 10x" Ê
11. y œ ln () 1) Ê
13. y œ ln x$ Ê
dy
dx
15. y œ t(ln t)# Ê
dy
d)
17. y œ
x%
4
ln x dy
dt
8. y œ ln ˆt$Î# ‰ Ê
2
t
dy
dx
dy
dx
" ‰ ˆ 3 "Î# ‰
œ ˆ t$Î#
œ
# t
3
2t
œ ˆ 10x"" ‰ a10x# b œ x"
"
)1
dy
dt
12. y œ ln (2) 2) Ê
14. y œ (ln x)$ Ê
3
x
œ (ln t)# 2t(ln t) †
Ê
dy
dt
œ ˆ 3x"" ‰ a3x# b œ x"
œ ˆ x"$ ‰ a3x# b œ
"
#(ln t)"Î#
x%
16
" ‰
6. y œ ln kx Ê yw œ ˆ kx
(k) œ x
œ ˆ ) " 1 ‰ (1) œ
16. y œ tÈln t œ t(ln t)"Î# Ê
œ (ln t)"Î# "
x
d
dt
(ln t) œ (ln t)# œ (ln t)"Î# "# t(ln t)"Î# †
œ x$ ln x x%
4
†
"
x
4x$
16
d
dt
2t ln t
t
dy
dx
dy
d)
œ ˆ #) " 2 ‰ (2) œ
œ 3(ln x)# †
d
dx
œ (ln t)# 2 ln t
(ln t) œ (ln t)"Î# t(ln t)"Î#
#t
œ x$ ln x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
"
)1
(ln x) œ
3(ln x)#
x
Section 7.2 Natural Logarithms
18. y œ
x$
3
ln x x$
9
19. y œ
ln t
t
Ê
20. y œ
" ln t
t
Ê
21. y œ
ln x
1 ln x
Ê yw œ
(1 ln x) ˆ "x ‰ (ln x) ˆ x" ‰
(1 ln x)#
22. y œ
x ln x
1 ln x
Ê yw œ
(1ln x) ˆln x x† "x ‰ (x ln x) ˆ x" ‰
(1ln x)#
dy
dt
Ê
œ x# ln x dy
dx
œ
t ˆ "t ‰ (ln t)(1)
t#
dy
dt
œ
23. y œ ln (ln x) Ê yw œ ˆ ln"x ‰ ˆ "x ‰ œ
"
ln (ln x)
†
25. y œ )[sin (ln )) cos (ln ))] Ê
†
"
x
3x#
9
œ x# ln x
1 ln t
t#
œ
t ˆ "t ‰ (" ln t)(1)
t#
24. y œ ln (ln (ln x)) Ê yw œ
x$
3
" 1 ln t
t#
œ
"
x
œ
œ lnt# t
lnxx lnxx
(1 ln x)#
œ
œ
"
x(1 ln x)#
(" ln x)# ln x
(1 ln x)#
œ1
ln x
(1 ln x)#
"
x ln x
d
dx
(ln (ln x)) œ
"
ln (ln x)
†
"
ln x
†
(ln x) œ
d
dx
"
x (ln x) ln (ln x)
œ [sin (ln )) cos (ln ))] ) cos (ln )) †
dy
d)
œ sin (ln )) cos (ln )) cos (ln )) sin (ln )) œ 2 cos (ln ))
26. y œ ln (sec ) tan )) Ê
dy
d)
"
xÈ x 1
"
#
27. y œ ln
28. y œ
"
#
29. y œ
1 ln t
1 ln t
ln
1x
1x
œ ln x œ
Ê
"
#
dy
dt
"
#
(1 ln t)#
"
t"Î#
31. y œ ln (sec (ln ))) Ê
œ
"
#
Èsin ) cos )
1 2 ln )
"
#
œ
"Î#
Ê
œ
œ
dy
dt
† #" t"Î# œ
dy
d)
"
#
"
#
œ
"
#
1 " x ˆ 1 " x ‰ (1)‘ œ
" ln t " ln t
t
t
t
ˆln t"Î# ‰"Î# †
œ
t
(1 ln t)#
"
#
1x1x
’ (1
x)(" x) “ œ
2
t(1 ln t)#
ˆln t"Î# ‰ œ
"
#
ˆln t"Î# ‰"Î# †
sec (ln )) tan (ln ))
sec (ln ))
†
d
d)
d
dt
"
t"Î#
†
d
dt
ˆt"Î# ‰
"
sec (ln ))
†
d
d)
(sec (ln ))) œ
dy
d)
œ
"
#
(ln )) œ
)
ˆ cos
sin ) tan (ln ))
)
sin ) ‰
cos )
2
)
1 # ln )
"
#
ln (1 x) Ê yw œ
5†2x
x# 1
[5 ln (x 1) 20 ln (x 2)] Ê yw œ
"
#
#" ˆ 1 " x ‰ (1) œ
ˆ x 5 1 20 ‰
x#
œ
5
#
10x
x# 1
"
#(1 x)
4(x 1)
’ (x(x2)1)(x
2) “
2
œ 5# ’ (x 3x1)(x
#) “
35. y œ 'x#Î2 ln Èt dt Ê
x#
"
1 x#
4
)(1 2 ln )) “
&
34. y œ ln É (x(x2)1)#! œ
œ sec )
"
4tÉln Èt
x 1b
33. y œ ln Š aÈ
‹ œ 5 ln ax# 1b 1x
&
sec )(tan ) sec ))
tan ) sec )
(ln sin ) ln cos )) ln (1 2 ln )) Ê
’cot ) tan ) #
œ
sin (ln )) † ") ‘
1) x
3x 2
ln (x 1) Ê yw œ x" #" ˆ x " 1 ‰ œ 2(x
2x(x 1) œ 2x(x 1)
‰
(1ln t) ˆ "t ‰ (1 ln t) ˆ "
t
œ
ˆln t"Î# ‰"Î# †
32. y œ ln
sec ) tan ) sec# )
sec ) tan )
cln (1 x) ln (1 x)d Ê yw œ
30. y œ Éln Èt œ ˆln t"Î# ‰
œ
œ
"
)
dy
dx
œ Šln Èx# ‹ †
d
dx
#
ax# b Šln É x# ‹ †
d
dx
#
Š x# ‹ œ 2x ln kxk x ln
kx k
È2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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433
434
Chapter 7 Transcendental Functions
Èx
36. y œ 'Èx ln t dt Ê
$
œ
$
ln È
x
ˆ $Èx‰ ˆln Èx‰ †
d
dx
d
dx
ˆÈx‰ œ ˆln $Èx‰ ˆ "3 x#Î$ ‰ ˆln Èx‰ ˆ #" x"Î# ‰
ln Èx
2È x
$
3 È x#
œ ˆln $Èx‰ †
dy
dx
37.
2
'cc32 x" dx œ cln kxkd #
$ œ ln 2 ln 3 œ ln 3
38.
'c01 3x3# dx œ cln k3x 2kd !" œ ln 2 ln 5 œ ln 52
39.
' y 2y25 dy œ ln ky# 25k C
40.
' 4r8r5 dr œ ln k4r# 5k C
41.
42.
#
#
'01 2sincost t dt œ cln k2 cos tkd !1 œ ln 3 ln 1 œ ln 3; or let u œ 2 cos t Ê du œ sin t dt with t œ 0
1
3
Ê u œ 1 and t œ 1 Ê u œ 3 Ê '0 2sincost t dt œ '1 "u du œ cln kukd $" œ ln 3 ln 1 œ ln 3
'01Î3 14 4sincos) ) d) œ cln k1 4 cos )kd !1Î$ œ ln k1 2k œ ln 3 œ ln "3 ; or let u œ 1 4 cos ) Ê du œ 4 sin ) d)
1Î3
c1
"
with ) œ 0 Ê u œ 3 and ) œ 13 Ê u œ 1 Ê '0 14 4sincos) ) d) œ 'c3 u" du œ cln kukd "
$ œ ln 3 œ ln 3
43. Let u œ ln x Ê du œ
'1
2
2 ln x
x
dx œ '0
ln 2
'2
dx
x ln x
œ 'ln 2
ln 4
"
u
ln 2
'2
dx
x(ln x)#
'2
dx; x œ 2 Ê u œ ln 2 and x œ 4 Ê u œ ln 4;
#
"
x
dx; x œ 2 Ê u œ ln 2 and x œ 4 Ê u œ ln 4;
œ 'ln 2 u# du œ "u ‘ ln 2 œ ln"4 ln 4
ln 4
46. Let u œ ln x Ê du œ
16
"
x
4‰
ln 2
ˆ 2 ln 2 ‰ œ ln 2
du œ cln ud lnln 42 œ ln (ln 4) ln (ln 2) œ ln ˆ ln
ln 2 œ ln Š ln 2 ‹ œ ln ln 2
45. Let u œ ln x Ê du œ
4
dx; x œ 1 Ê u œ 0 and x œ 2 Ê u œ ln 2;
2u du œ cu# d 0 œ (ln 2)#
44. Let u œ ln x Ê du œ
4
"
x
dx
2xÈln x
œ
"
#
'ln 2
ln 16
"
x
"
ln #
œ ln"## "
ln 2
œ 2 ln" # "
ln #
œ
"
# ln 2
œ
"
ln 4
dx; x œ 2 Ê u œ ln 2 and x œ 16 Ê u œ ln 16;
u"Î# du œ u"Î# ‘ ln 2 œ Èln 16 Èln 2 œ È4 ln 2 Èln 2 œ 2Èln 2 Èln 2 œ Èln 2
ln 16
47. Let u œ 6 3 tan t Ê du œ 3 sec# t dt;
t
' 6 3sec
' duu œ ln kuk C œ ln k6 3 tan tk C
3 tan t dt œ
#
48. Let u œ 2 sec y Ê du œ sec y tan y dy;
' sec# ysectanyy dy œ ' duu œ ln kuk C œ ln k2 sec yk C
49. Let u œ cos
x
#
Ê du œ "# sin
sin
'01Î2 tan x# dx œ '01Î2 cos
x
#
dx Ê 2 du œ sin
1Î
50. Let u œ sin t Ê du œ cos t dt; t œ
'11ÎÎ42 cot t dt œ '11ÎÎ42
51. Let u œ sin
)
3
cos t
sin t
Ê du œ
'1Î2 2 cot 3) d) œ '1Î2
1
1
4
du
u
dx; x œ 0 Ê u œ 1 and x œ
È2
œ c2 ln kukd 11Î
Ê uœ
"
È2
and t œ
1
#
"
dt œ '1ÎÈ2 du
u œ cln kukd "ÎÈ# œ ln
1
"
3
1 2 cos
sin
È2
dx œ 2 '1
x
#
x
#
x
#
)
3
cos
)
3
)
3
d) Ê 6 du œ 2 cos
È3Î2
d) œ 6 '1Î2
du
u
)
3
œ 2 ln
Ê uœ
"
È2
œ 2 ln È2 œ ln 2
Ê u œ 1;
"
È2
d) ; ) œ
È3Î2
"
È2
1
#
œ ln È2
1
#
Ê uœ
œ 6 cln kukd 1Î2 œ 6 Šln
È3
#
"
#
and ) œ 1 Ê u œ
È3
#
ln "# ‹ œ 6 ln È3 œ ln 27
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
;
;
Section 7.2 Natural Logarithms
1
1#
52. Let u œ cos 3x Ê du œ 3 sin 3x dx Ê 2 du œ 6 sin 3x dx; x œ 0 Ê u œ 1 and x œ
1ÎÈ2
'01Î12 6 tan 3x dx œ '01Î12 6cossin3x3x dx œ 2 '1
53.
'
œ'
dx
2Èx 2x
dx
;
2 È x ˆ1 È x ‰
du
u
È2
œ 2 cln kukd 11Î
let u œ 1 Èx Ê du œ
"
#È x
œ 2 ln
"
È2
"
È2
;
ln 1 œ 2 ln È2 œ ln 2
' 2Èx ˆdx1 Èx‰ œ '
dx;
Ê uœ
œ ln kuk C
du
u
œ ln ¸1 Èx¸ C œ ln ˆ1 Èx‰ C
54. Let u œ sec x tan x Ê du œ asec x tan x sec# xb dx œ (sec x)(tan x sec x) dx Ê sec x dx œ
'
sec x dx
Èln (sec x tan x)
œ'
du
uÈln u
œ ' (ln u)"Î# †
"
#
55. y œ Èx(x 1) œ (x(x 1))"Î# Ê ln y œ
Ê yw œ ˆ "# ‰ Èx(x 1) ˆ x" " ‰
x 1
œ
"
#
56. y œ Èax# 1b (x 1)# Ê ln y œ
Ê yw œ Èax# +1b (x 1)# ˆ x# x 1 57. y œ É t t 1 œ ˆ t t 1 ‰
Ê
dy
dt
œ
"
#
"Î#
É t t 1 ˆ "t "
#
Ê ln y œ
" ‰
t1
œ
"
#
dy
dt
du œ 2(ln u)"Î# C œ 2Èln (sec x tan x) C
ln (x(x 1)) Ê 2 ln y œ ln (x) ln (x 1) Ê
Èx(x 1) (2x 1)
2x(x 1)
" ‰
x1
dy
d)
" dy
y dt
dy
d)
62. y œ
Ê
dy
dt
"
#
œ
ˆ x#2x 1 #
"
x 1
ˆ "t a2x# x 1b kx 1k
Èx# 1 (x 1)
" ‰
t1
" dy
y dt
œ #" ˆ "t " ‰
t1
"
#
ln () 3) ln (sin )) Ê
" dy
y d)
œ
"
#() 3)
#
)
œ (tan )) È2) 1 Š sec
tan ) "
#) 1 ‹
œ asec# )b È2) 1 œ t(t 1)(t+2) ˆ "t œ
"
t1
" ‰
t#
cos )
sin )
"t "
t1
" ‘
t#
ln (2) 1) Ê
" dy
y d)
œ
sec# )
tan )
)5
) cos ) Ê ln y œ ln () 5) ln )
dy
ˆ )5 ‰ ˆ ) " 5 ") tan )‰
d) œ ) cos )
ˆ #" ‰ ˆ #) 2 1 ‰
tan )
È 2) 1
" dy
y dt
œ
"
t
"
t1
"
t#
t(t 2) t(t 1)
œ t(t 1)(t 2) ’ (t 1)(t t(t2)1)(t
“ œ 3t# 6t 2
2)
Ê ln y œ ln 1 ln t ln (t 1) ln (t 2) Ê
"
t(t 1)(t 2)
"
#
œ
"
t(t 1)(t #)
" dy
y dt
œ "t "
t1
"
t#
t(t 2) t(t 1)
’ (t 1)(t t(t2)1)(t
“
2)
#
Ê
2 ‰
x1
œ at$3t3t#6t2t2b#
63. y œ
"
x
œ È) 3 (sin )) ’ 2() " 3) cot )“
"
t(t 1)(t 2)
dy
dt
œ
2t 1
2 at# tb$Î#
61. y œ t(t 1)(t 2) Ê ln y œ ln t ln (t 1) ln (t 2) Ê
Ê
w
"
2Èt (t 1)$Î#
60. y œ (tan )) È2) 1 œ (tan ))(2) 1)"Î# Ê ln y œ ln (tan )) Ê
"
#
œ
[ln t ln (t 1)] Ê
59. y œ È) 3 (sin )) œ () 3)"Î# sin ) Ê ln y œ
Ê
w
y
y
#
[ln t ln (t 1)] Ê
"
#
2y
y
2x "
2Èx(x 1)
œ Èax# 1b (x 1)# ’ axx#x1b (xx 1)1 “ œ
É t t 1 ’ t(t " 1) “ œ
"
œ "# É t(t 1 1) ’ t(t2t 1)
“œ
œ
cln ax# 1b 2 ln (x 1)d Ê
58. y œ É t(t 1 1) œ [t(t 1)]"Î# Ê ln y œ
Ê
"
u
du
u ;
ln (cos )) Ê
" dy
y d)
œ
"
)5
"
)
sin )
cos )
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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435
436
Chapter 7 Transcendental Functions
64. y œ
) sin )
Èsec ) Ê ln y
dy
) sin ) ˆ "
d) œ Èsec ) )
Ê
65. y œ
Ê
"
#
œ ln ) ln (sin )) cot ) "
#
"!
"
#
"!
$
$
2) ˆ "
É x(x
x# 1
x "
3
Ê yw œ
"
3
1b 2
3
ln (x 1) Ê
cln x ln (x 2) ln ax# 1bd Ê
"
x#
$
"
3
$
1)(x 2) ˆ "
É ax(x
x# 1b (2x 3) x "
3
cos )
sin )
(sec ))(tan ))
2 sec ) “
yw
y
yw
y
"
x
œ
œ
5
x1
x
x# 1
2
3(x 1)
5
2x 1
5 ‰
2x 1
1)(x 2)
68. y œ É ax(x
x# 1b (2x 3) Ê ln y œ
Ê yw œ
œ ’ ") [10 ln (x 1) 5 ln (2x 1)] Ê
(x 1) ˆ 5
Ê yw œ É (2x
x1 1)&
2)
67. y œ É x(x
x# 1 Ê ln y œ
" dy
y d)
tan )‰
xÈ x# 1
Ê ln y œ ln x "# ln ax# (x 1)#Î$
È #
yw œ x(x x1)#Î$1 ’ "x x# x 1 3(x 2 1) “
(x 1)
66. y œ É (2x
1)& Ê ln y œ
ln (sec )) Ê
yw
y
œ
"
3
ˆ "x "
x#
2x ‰
x# 1
2x ‰
x# 1
cln x ln (x 1) ln (x 2) ln ax# 1b ln (2x 3)d
"
x1
"
x#
2x
x# 1
2 ‰
2x 3
sin x
1
w
w
69. (a) f(x) œ ln (cos x) Ê f w (x) œ cos
x œ tan x œ 0 Ê x œ 0; f (x) 0 for 4 Ÿ x 0 and f (x) 0 for
0 x Ÿ 13 Ê there is a relative maximum at x œ 0 with f(0) œ ln (cos 0) œ ln 1 œ 0; f ˆ 14 ‰ œ ln ˆcos ˆ 14 ‰‰
œ ln Š È"2 ‹ œ #" ln 2 and f ˆ 13 ‰ œ ln ˆcos ˆ 13 ‰‰ œ ln
xœ
1
3
"
#
œ ln 2. Therefore, the absolute minimum occurs at
with f ˆ 13 ‰ œ ln 2 and the absolute maximum occurs at x œ 0 with f(0) œ 0.
(b) f(x) œ cos (ln x) Ê f w (x) œ
sin (ln x)
x
œ 0 Ê x œ 1; f w (x) 0 for
"
#
Ÿ x 1 and f w (x) 0 for 1 x Ÿ 2
Ê there is a relative maximum at x œ 1 with f(1) œ cos (ln 1) œ cos 0 œ 1; f ˆ "# ‰ œ cos ˆln ˆ "# ‰‰
œ cos ( ln 2) œ cos (ln 2) and f(2) œ cos (ln 2). Therefore, the absolute minimum occurs at x œ
x œ 2 with f ˆ "# ‰ œ f(2) œ cos (ln 2), and the absolute maximum occurs at x œ 1 with f(1) œ 1.
"
#
and
70. (a) f(x) œ x ln x Ê f w (x) œ 1 "x ; if x 1, then f w (x) 0 which means that f(x) is increasing
(b) f(1) œ 1 ln 1 œ 1 Ê f(x) œ x ln x 0, if x 1 by part (a) Ê x ln x if x 1
71.
'15 (ln 2x ln x) dx œ '15 ( ln x ln 2 ln x) dx œ (ln 2)'15 dx œ (ln 2)(5 1) œ ln 2% œ ln 16
72. A œ
'c01Î4 tan x dx '01Î3 tan x dx œ '01Î4 cossinxx dx '01Î3 cossinxx dx œ cln kcos xkd !1Î% cln kcos xkd !1Î$
œ Šln 1 ln
"
È2 ‹
ˆln
"
#
ln 1‰ œ ln È2 ln 2 œ
73. V œ 1'0 Š Èy2 1 ‹ dy œ 41 '0
#
3
3
74. V œ 1 '1Î6 cot x dx œ 1 '1Î6
1Î2
1Î2
"
y 1
cos x
sin x
3
#
ln 2
dy œ 41 cln ky 1kd $! œ 41(ln 4 ln 1) œ 41 ln 4
1Î#
dx œ 1 cln (sin x)d 1Î' œ 1 ˆln 1 ln "# ‰ œ 1 ln 2
75. V œ 21'1Î2 x ˆ x"# ‰ dx œ 21 '1Î2 x" dx œ 21 cln kxkd #"Î# œ 21 ˆln 2 ln #" ‰ œ 21(2 ln 2) œ 1 ln 2% œ 1 ln 16
2
2
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Section 7.2 Natural Logarithms
437
76. V œ 1 '0 Š Èx9x
‹ dx œ 271'0 dx œ 271 cln ax$ 9bd ! œ 271(ln 36 ln 9)
$9
#
3
3
$
œ 271(ln 4 ln 9 ln 9) œ 271 ln 4 œ 541 ln 2
77. (a) y œ
x#
8
œ '4
#
ln x Ê 1 ayw b# œ 1 ˆ x4 x" ‰ œ 1 Š x 4x 4 ‹ œ Š x 4x 4 ‹ Ê L œ '4 É1 ayw b# dx
8 #
x 4
(b) x œ
#
#
4x
ˆ y4 ‰#
#
#
8
dx œ '4 ˆ x4 "x ‰ dx œ ’ x8 ln kxk“ œ (8 ln 8) (2 ln 4) œ 6 ln 2
8
)
#
%
2
ln ˆ y4 ‰
Ê
dx
dy
œ
y
8
2
y
'
Ê L œ '4 Ê1 Š dx
dy ‹ dy œ 4
#
12
#
#
y
y
2
Ê 1 Š dx
dy ‹ œ 1 Š 8 y ‹ œ 1 Š
12 #
y 16
8y
#
y
dy œ '4 Š y8 2y ‹ dy œ ’ 16
2 ln y“
12
#
16
8y ‹
"#
%
#
œ Šy
#
16
8y ‹
#
œ (9 2 ln 12) (1 2 ln 4)
œ 8 2 ln 3 œ 8 ln 9
78. L œ '1 É1 2
"
x#
dx Ê
dy
dx
œ
"
x
Ê y œ ln kxk C œ ln x C since x 0 Ê 0 œ ln 1 C Ê C œ 0 Ê y œ ln x
" ‰ ˆ"‰
79. (a) My œ '1 x ˆ x" ‰ dx œ 1, Mx œ '1 ˆ 2x
x dx œ
2
Ê xœ
2
My
M
œ
"
ln 2
¸ 1.44 and y œ
Mx
M
œ
ˆ "4 ‰
ln 2
"
#
'12 x"
" ‘
dx œ 2x
œ 4" , M œ '1
"
2
#
#
dx œ cln kxkd #" œ ln 2
"
x
¸ 0.36
(b)
80. (a) My œ '1 x Š È"x ‹ dx œ '1 x"Î# dx œ
16
œ
16
'
cln kxkd "'
" œ ln 4, M œ 1
16
"
#
"
Èx
2
3
"
"
x$Î# ‘ "' œ 42; Mx œ ' Š #È
x ‹ Š Èx ‹ dx œ
"
1
16
"'
dx œ 2x"Î# ‘ " œ 6 Ê x œ
My
M
œ 7 and y œ
Mx
M
"
#
œ
'116 x" dx
ln 4
6
"
"
4
' $Î# dx
(b) My œ '1 x Š È"x ‹ Š È4x ‹ dx œ 4'1 dx œ 60, Mx œ '1 Š #È
x ‹ Š Èx ‹ Š Èx ‹ dx œ # 1 x
16
16
16
œ 4 x"Î# ‘" œ 3, M œ '1 Š È"x ‹ Š È4x ‹ dx œ 4'1
"'
yœ
81.
dy
dx
82.
d# y
dx#
Mx
M
œ1
"
x
œ
16
16
"
x
16
dx œ c4 ln kxkd "'
" œ 4 ln 16 Ê x œ
My
M
œ
15
ln 16
and
3
4 ln 16
at ("ß 3) Ê y œ x ln kxk C; y œ 3 at x œ 1 Ê C œ 2 Ê y œ x ln kxk 2
œ sec# x Ê
dy
dx
œ tan x C and 1 œ tan 0 C Ê
dy
dx
œ tan x 1 Ê y œ ' (tan x 1) dx
œ ln ksec xk x C" and 0 œ ln ksec 0k 0 C" Ê C" œ 0 Ê y œ ln ksec xk x
83. (a) L(x) œ f(0) f w (0) † x, and f(x) œ ln (1 x) Ê f w (x)k xœ0 œ
ww
(b) Let faxb œ lnax "b. Since f axb œ
"
ax"
b#
" ¸
1x xœ0
œ 1 Ê L(x) œ ln 1 1 † x Ê L(x) œ x
! on Ò!ß !Þ"Ó, the graph of f is concave down on this interval and the
largest error in the linear approximation will occur when x œ !Þ". This error is !Þ" lna"Þ"b ¸ !Þ!!%'* to five
decimal places.
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438
Chapter 7 Transcendental Functions
(c) The approximation y œ x for ln (1 x) is best for smaller
positive values of x; in particular for 0 Ÿ x Ÿ 0.1 in the
graph. As x increases, so does the error x ln (1 x).
From the graph an upper bound for the error is
0.5 ln (1 0.5) ¸ 0.095; i.e., kE(x)k Ÿ 0.095 for
0 Ÿ x Ÿ 0.5. Note from the graph that 0.1 ln (1 0.1)
¸ 0.00469 estimates the error in replacing ln (1 x) by
x over 0 Ÿ x Ÿ 0.1. This is consistent with the estimate
given in part (b) above.
84. For all positive values of x,
d ln a
x
dx c
dœ
1
a
x
† xa2 œ 1x and
d
ln a
dx c
ln x d œ 0 1
x
œ 1x . Since ln xa and ln a ln x have
the same derivative, then ln xa œ ln a ln x C for some constant C. Since this equation holds for all positve values of x,
it must be true for x œ 1 Ê ln 1x œ ln 1 ln x C œ 0 ln x C Ê ln 1x œ ln x C. By part 3 we know that
ln 1x œ ln x Ê C œ 0 Ê ln xa œ ln a ln x.
85. y œ ln kx Ê y œ ln x ln k; thus the graph of
y œ ln kx is the graph of y œ ln x shifted vertically
by ln k, k 0.
86. To turn the arches upside down we would use the
formula y œ ln ksin xk œ ln ksin" xk .
87. (a)
(b) yw œ
cos x
asin x .
Since lsin xl and lcos xl are less than
or equal to 1, we have for a "
"
"
w
a" Ÿ y Ÿ a" for all x.
Thus, lim yw œ ! for all x Ê the graph of y looks
aÄ_
more and more horizontal as a Ä _.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 7.3 The Exponential Function
88. (a) The graph of y œ Èx ln x appears to be concave
upward for all x 0.
(b) y œ Èx ln x Ê yw œ
"
#È x
"
x
Ê yww œ 4x"$Î# "
x#
œ
"
x#
Š
Èx
4
1‹ œ 0 Ê Èx œ 4 Ê x œ 16.
Thus, yww 0 if 0 x 16 and yww 0 if x 16 so a point of inflection exists at x œ 16. The graph of
y œ Èx ln x closely resembles a straight line for x 10 and it is impossible to discuss the point of
inflection visually from the graph.
7.3 THE EXPONENTIAL FUNCTION
#
(b) e ln x œ
1. (a) eln 7.2 œ 7.2
#
#
2. (a) eln ax y b œ x# y#
(b) e ln 0Þ3 œ
3. (a) 2 ln Èe œ 2 ln e"Î# œ (2) ˆ "# ‰ ln e œ 1
#
"
eln x#
œ
"
eln 0 3
Þ
"
x#
œ
(c) eln xln y œ elnÐxÎyÑ œ
"
0.3
x
y
(c) eln 1xln 2 œ elnÐ1xÎ2Ñ œ
1x
#
(b) ln aln ee b œ ln (e ln e) œ ln e œ 1
#
(c) ln eax y b œ ax# y# b ln e œ x# y#
4. (a) ln ˆesec ) ‰ œ (sec ))(ln e) œ sec )
x
(b) ln eae b œ aex b (ln e) œ ex
#
(c) ln ˆe2 ln x ‰ œ ln Šeln x ‹ œ ln x# œ 2 ln x
5. ln y œ 2t 4 Ê eln y œ e2t4 Ê y œ e2t4
6. ln y œ t 5 Ê eln y œ et5 Ê y œ et5
7. ln (y 40) œ 5t Ê eln Ðy40) œ e5t Ê y 40 œ e5t Ê y œ e5t 40
"
8. ln (1 2y) œ t Ê eln Ð12y) œ et Ê 1 2y œ et Ê 2y œ et 1 Ê y œ Š e # ‹
t
1‰
9. ln (y 1) ln 2 œ x ln x Ê ln (y 1) ln 2 ln x œ x Ê ln ˆ y 2x
œ x Ê eln ˆ
y1‰
2x
œ ex Ê
Ê y 1 œ 2xex Ê y œ 2xex 1
#
y1
#x
œ ex
10. ln ay# 1b ln (y 1) œ ln (sin x) Ê ln Š yy 1" ‹ œ ln (sin x) Ê ln (y 1) œ ln (sin x) Ê eln Ðy1Ñ œ eln Ðsin xÑ
Ê y 1 œ sin x Ê y œ sin x 1
11. (a) e2k œ 4 Ê ln e2k œ ln 4 Ê 2k ln e œ ln 2# Ê 2k œ 2 ln 2 Ê k œ ln 2
(b) 100e10k œ 200 Ê e10k œ 2 Ê ln e10k œ ln 2 Ê 10k ln e œ ln 2 Ê 10k œ ln 2 Ê k œ
(c) ekÎ1000 œ a Ê ln ekÎ1000 œ ln a Ê
12. (a) e5k œ
"
4
k
1000
ln e œ ln a Ê
k
1000
œ ln a Ê k œ 1000 ln a
Ê ln e5k œ ln 4" Ê 5k ln e œ ln 4 Ê 5k œ ln 4 Ê k œ ln54
(b) 80ek œ 1 Ê ek œ 80" Ê ln ek œ ln 80" Ê k ln e œ ln 80 Ê k œ ln 80
k
(c) eÐln 0Þ8Ñk œ 0.8 Ê ˆeln 0Þ8 ‰ œ 0.8 Ê (0.8)k œ 0.8 Ê k œ 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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ln 2
10
439
440
Chapter 7 Transcendental Functions
13. (a) e0Þ3t œ 27 Ê ln e0Þ3t œ ln 3$ Ê (0.3t) ln e œ 3 ln 3 Ê 0.3t œ 3 ln 3 Ê t œ 10 ln 3
(b) ekt œ "# Ê ln ekt œ ln 2" œ kt ln e œ ln 2 Ê t œ lnk2
t
(c) eÐln 0Þ2Ñt œ 0.4 Ê ˆeln 0Þ2 ‰ œ 0.4 Ê 0.2t œ 0.4 Ê ln 0.2t œ ln 0.4 Ê t ln 0.2 œ ln 0.4 Ê t œ
ln 0.4
ln 0.2
14. (a) e0Þ01t œ 1000 Ê ln e0Þ01t œ ln 1000 Ê (0.01t) ln e œ ln 1000 Ê 0.01t œ ln 1000 Ê t œ 100 ln 1000
"
(b) ekt œ 10
Ê ln ekt œ ln 10" œ kt ln e œ ln 10 Ê kt œ ln 10 Ê t œ lnk10
"
#
(c) eÐln 2Ñt œ
Èt
15. e
t
Ê ˆeln 2 ‰ œ 2" Ê 2t œ 2" Ê t œ 1
Èt
œ x# Ê ln e
#
œ ln x# Ê Èt œ 2 ln x Ê t œ 4(ln x)#
#
#
16. ex e2x1 œ et Ê ex 2x1 œ et Ê ln ex 2x1 œ ln et Ê t œ x# 2x 1
17. y œ e5x Ê yw œ e5x
d
dx
18. y œ e2xÎ3 Ê yw œ e2xÎ3
(5x) Ê yw œ 5e5x
d
dx
19. y œ e57x Ê yw œ e57x
#
d
dx
ˆ 2x
‰ Ê yw œ
3
2
3
e2xÎ3
(5 7x) Ê yw œ 7e57x
#
20. y œ eˆ4Èxx ‰ Ê yw œ eˆ4Èxx ‰
d
dx
ˆ4Èx x# ‰ Ê yw œ Š È2 2x‹ eˆ4Èxx# ‰
x
21. y œ xex ex Ê yw œ aex xex b ex œ xex
22. y œ (1 2x) e2x Ê yw œ 2e2x (1 2x)e2x
d
dx
(2x) Ê yw œ 2e2x 2(1 2x) e2x œ 4xe2x
23. y œ ax# 2x 2b ex Ê yw œ (2x 2)ex ax# 2x 2b ex œ x# ex
24. y œ a9x# 6x 2b e3x Ê yw œ (18x 6)e3x a9x# 6x 2b e3x
d
dx
(3x) Ê yw œ (18x 6)e3x 3 a9x# 6x 2b e3x
# 3x
œ 27x e
25. y œ e) (sin ) cos )) Ê yw œ e) (sin ) cos )) e) (cos ) sin )) œ 2e) cos )
26. y œ ln ˆ3)e) ‰ œ ln 3 ln ) ln e) œ ln 3 ln ) ) Ê
#
27. y œ cos Še) ‹ Ê
dy
d)
28. y œ )$ e#) cos 5) Ê
#
œ sin Še) ‹
dy
d)
d
d)
dy
d)
#
œ
"
)
#
#
Še) ‹ œ Š sin Še) ‹‹ Še) ‹
œ a3)# b ˆe#) cos 5)‰ a)$ cos 5)b e#)
œ )# e#) (3 cos 5) 2) cos 5) 5) sin 5))
29. y œ ln a3tet b œ ln 3 ln t ln et œ ln 3 ln t t Ê
dy
dt
œ
"
t
d
d)
d
d)
#
#
a)# b œ 2)e) sin Še) ‹
(2)) 5(sin 5)) ˆ)$ e#) ‰
1t
t
1œ
30. y œ ln a2et sin tb œ ln 2 ln et ln sin t œ ln 2 t ln sin t Ê
œ
1
dy
dt
œ 1 ˆ sin" t ‰
d
dt
(sin t) œ 1 cos t sin t
sin t
31. y œ ln
e)
1 e)
œ ln e) ln ˆ1 e) ‰ œ ) ln ˆ1+e) ‰ Ê
dy
d)
œ 1 ˆ 1 " e) ‰
d
d)
ˆ 1 e) ‰ œ 1 e)
1 e)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
œ
"
1 e)
cos t
sin t
Section 7.3 The Exponential Function 441
32. y œ ln
È)
1 È)
œ ln È) ln Š1 È)‹ Ê
"
"
œ Š È" ‹ Š È
‹ Š 1 "È) ‹ Š #È
‹œ
)
# )
)
33. y œ eÐcos tln tÑ œ ecos t eln t œ tecos t Ê
34. y œ esin t aln t# 1b Ê
35.
'0ln x sin et dt
dy
dt
œ Š È" ‹
)
Š1 È)‹ È)
dy
dt
2) Š1 È)‹
œ
d
d)
ŠÈ)‹ Š 1 "È) ‹
"
#) Š1 È)‹
œ ecos t tecos t
d
dt
œ
d
d)
Š1 È ) ‹
"
#) a1)"Î# b
(cos t) œ (1 t sin t) ecos t
œ esin t (cos t) aln t# 1b 2t esin t œ esin t aln t# 1b (cos t) 2t ‘
Ê yw œ ˆsin eln x ‰ †
d
dx
36. y œ 'e4Èx ln t dt Ê yw œ aln e2x b †
e2x
dy
d)
(ln x) œ
d
dx
sin x
x
ae2x b Šln e4Èx ‹ †
d
dx
Še4Èx ‹ œ (2x) a2e2x b ˆ4Èx‰ Še4Èx ‹ †
d
dx
ˆ4Èx‰
œ 4xe2x 4Èx e4Èx Š È2x ‹ œ 4xe2x 8e4Èx
37. ln y œ ey sin x Ê Š y" ‹ yw œ ayw ey b (sin x) ey cos x Ê yw Š y" ey sin x‹ œ ey cos x
Ê yw Š 1 yey sin x ‹ œ ey cos x Ê yw œ
y
38. ln xy œ exy Ê ln x ln y œ exy Ê
Ê yw Š 1 ye
y
xby
‹œ
xex b y "
x
Ê yw œ
yey cos x
1 yey sin x
"
x
Š y" ‹ yw œ a1 yw b exy Ê yw Š y" exy ‹ œ exy y axex b y "b
x a1 yex b y b
39. e2x œ sin (x 3y) Ê 2e2x œ a1 3yw b cos (x 3y) Ê 1 3yw œ
w
Ê y œ
"
x
2e2x
cos (x 3y)
Ê 3yw œ
2e2x
cos (x 3y)
1
2e2x cos (x 3y)
3 cos (x 3y)
40. tan y œ ex ln x Ê asec# yb yw œ ex 41.
' ae3x 5ex b dx œ e3
43.
"
x
Ê yw œ
axex "b cos# y
x
42.
' a2ex 3e2x b dx œ 2ex #3 e2x C
'lnln23 ex dx œ cex d lnln 32 œ eln 3 eln 2 œ 3 2 œ 1
44.
'lnln32 ex dx œ cex d 0 ln 2 œ e! eln 2 œ 1 2 œ 1
45.
' 8eÐx1Ñ dx œ 8eÐx1Ñ C
46.
' 2eÐ2x1Ñ dx œ eÐ2x1Ñ C
47.
'lnln49 exÎ2 dx œ 2exÎ2 ‘ lnln 94 œ 2 eÐln 9ÑÎ2 eÐln 4)Î2 ‘ œ 2 ˆeln 3 eln 2 ‰ œ 2(3 2) œ 2
48.
'0ln 16 exÎ4 dx œ 4exÎ4 ‘ ln0 16 œ 4 ˆeÐln 16ÑÎ4 e0 ‰ œ 4 ˆeln 2 1‰ œ 4(2 1) œ 4
3x
5ex C
49. Let u œ r"Î# Ê du œ "# r"Î# dr Ê 2 du œ r"Î# dr;
' eÈÈrr dr œ ' er"Î# † r"Î# dr œ 2 ' eu du œ 2eu C œ 2er"Î# C œ 2eÈr C
50. Let u œ r"Î# Ê du œ "# r"Î# dr Ê 2 du œ r"Î# dr;
' eÈÈrr dr œ ' er"Î# † r"Î# dr œ 2 ' eu du œ 2er"Î# C œ 2eÈr C
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442
Chapter 7 Transcendental Functions
51. Let u œ t# Ê du œ 2t dt Ê du œ 2t dt;
' 2tet
dt œ ' eu du œ eu C œ et C
#
#
"
4
52. Let u œ t% Ê du œ 4t$ dt Ê
'
%
t$ et dt œ
"
x
53. Let u œ
' ex#
"
4
du œ t$ dt;
' eu du œ 4" et% C
Ê du œ x"# dx Ê du œ
"
x#
dx;
dx œ ' eu du œ eu C œ e1Îx C
1Îx
54. Let u œ x# Ê du œ 2x$ dx Ê
' ex$Î #
dx œ ' ex † x$ dx œ
#
1 x
"
#
"
#
du œ x$ dx;
' eu du œ "# eu C œ "# ex# C œ "# e1Îx# C
55. Let u œ tan ) Ê du œ sec# ) d); ) œ 0 Ê u œ 0, ) œ
'0
1 Î4
ˆ1 etan ) ‰ sec# ) d) œ '
1Î4
0
1
4
Ê u œ 1;
sec# ) d) '0 eu du œ ctan )d 0
1
1 Î4
ceu d "! œ tan ˆ 14 ‰ tan (0)‘ ae" e! b
œ (1 0) (e 1) œ e
56. Let u œ cot ) Ê du œ csc# ) d); ) œ
'1Î4 ˆ1 ecot ) ‰ csc# ) d) œ '1Î4
1 Î2
1Î2
1
4
Ê u œ 1, ) œ
1
2
Ê u œ 0;
csc# ) d) '1 eu du œ c cot )d 1Î4 ceu d !" œ cot ˆ 12 ‰ cot ˆ 14 ‰‘ ae! e" b
0
1Î2
œ (0 1) (1 e) œ e
57. Let u œ sec 1t Ê du œ 1 sec 1t tan 1t dt Ê
' esec Ð1tÑ sec (1t) tan (1t) dt œ 1" ' eu du œ e1
u
du
1
œ sec 1t tan 1t dt;
Cœ
esec a1tb
1
C
58. Let u œ csc (1 t) Ê du œ csc (1 t) cot (1 t) dt;
' ecsc Ð1tÑ csc (1 t) cot (1 t) dt œ ' eu du œ eu C œ ecsc Ð1tÑ C
59. Let u œ ev Ê du œ ev dv Ê 2 du œ 2ev dv; v œ ln
1
6
Ê u œ 16 , v œ ln
1
#
Ê u œ 1# ;
'lnlnÐÐ11ÎÎ62ÑÑ 2ev cos ev dv œ 2 '11ÎÎ62 cos u du œ c2 sin ud 11ÎÎ26 œ 2 sin ˆ 1# ‰ sin ˆ 16 ‰‘ œ 2 ˆ1 "# ‰ œ 1
#
#
60. Let u œ ex Ê du œ 2xex dx; x œ 0 Ê u œ 1, x œ Èln 1 Ê u œ eln 1 œ 1;
Èln 1
'0
2xex cos Šex ‹ dx œ '1 cos u du œ csin ud 1" œ sin (1) sin (1) œ sin (1) ¸ 0.84147
#
1
#
61. Let u œ 1 er Ê du œ er dr;
' 1 e e
r
dr œ '
x
dx œ '
r
62.
' 1 " e
x
let u œ e
63.
"
u
du œ ln kuk C œ ln a1 er b C
ecx
ecx 1
dx;
1 Ê du œ ex dx Ê du œ ex dx;
'
ecx
ecx 1
dx œ '
dy
dt
œ et sin aet 2b Ê y œ ' et sin aet 2b dt;
"
u
du œ ln kuk C œ ln aex 1b C
let u œ et 2 Ê du œ et dt Ê y œ ' sin u du œ cos u C œ cos aet 2b C; y(ln 2) œ 0
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Section 7.3 The Exponential Function 443
Ê cos ˆeln 2 2‰ C œ 0 Ê cos (2 2) C œ 0 Ê C œ cos 0 œ 1; thus, y œ 1 cos aet 2b
64.
dy
dt
œ et sec# a1et b Ê y œ ' et sec# a1et b dt;
let u œ 1et Ê du œ 1et dt Ê 1" du œ et dt Ê y œ 1" ' sec# u du œ 1" tan u C
œ 1" tan a1et b C; y(ln 4) œ 12 Ê 1" tan ˆ1eln 4 ‰ C œ 12 Ê 1" tan ˆ1 † 4" ‰ C œ 12
Ê 1" (1) C œ
65.
d# y
dx#
œ 2ex Ê
x
Ê y œ 2e
66.
dy
dx
2
1
Ê C œ 13 ; thus, y œ
œ 2ex C; x œ 0 and
3
1
dy
dx
"
1
tan a1et b
œ 0 Ê 0 œ 2e! C Ê C œ 2; thus
dy
dx
x
!
2x 1 œ 2 aex xb 1
2x C" ; x œ 0 and y œ 1 Ê 1 œ 2e C" Ê C" œ 1 Ê y œ 2e
d# y
dy
" 2t
" #
2t
Ê dy
dt# œ 1 e
dt œ t # e C; t œ 1 and dt œ 0 Ê 0 œ 1 # e C Ê
dy
" 2t
" #
" #
" 2t
ˆ" #
‰
dt œ t # e # e 1 Ê y œ # t 4 e # e 1 t C" ; t œ 1 and y œ
"
" #
" #
" 2t
"
#
Ê C" œ # 4 e Ê y œ # t 4 e ˆ # e 1‰ t ˆ #" 4" e# ‰
Cœ
œ 2ex 2
"
#
e# 1; thus
"
#
1 Ê " œ
4" e# #" e# 1 C"
67. f(x) œ ex 2x Ê f w (x) œ ex 2; f w (x) œ 0 Ê ex œ 2 Ê x œ ln 2; f(0) œ 1, the absolute maximum;
f(ln 2) œ 2 2 ln 2 ¸ 0.613706, the absolute minimum; f(1) œ e 2 ¸ 0.71828, a relative or local maximum
since f ww (x) œ ex is always positive.
68. The function f(x) œ 2esin ÐxÎ2Ñ has a maximum whenever sin
x
#
œ 1 and a minimum whenever sin
x
#
œ 1.
Therefore the maximums occur at x œ 1 2k(21) and the minimums occur at x œ 31 2k(21), where k is any
integer. The maximum is 2e ¸ 5.43656 and the minimum is 2e ¸ 0.73576.
69. f(x) œ x# ln
"
x
Ê f w (x) œ 2x ln
"
x
x# Š "" ‹ ax# b œ 2x ln
x
"
x
ln x œ "# . Since x œ 0 is not in the domain of f, x œ e"Î# œ
f w (x) 0 for x "
Èe
of f assumed at x œ
. Therefore, f Š È"e ‹ œ
"
Èe
"
e
ln Èe œ
"
e
x œ x(2 ln x 1); f w (x) œ 0 Ê x œ 0 or
"
Èe
. Also, f w (x) 0 for 0 x ln e"Î# œ
"
#e
ln e œ
"
#e
"
Èe
and
is the absolute maximum value
.
70. f(x) œ (x 3)# ex Ê f w (x) œ 2(x 3) ex (x 3)# ex
œ (x 3) ex (2 x 3) œ (x 1)(x 3) ex ; thus
f w (x) 0 for x 1 or x 3, and f w (x) 0 for
1 x 3 Ê f(1) œ 4e ¸ 10.87 is a local maximum and
f(3) œ 0 is a local minimum. Since f(x) 0 for all x,
f(3) œ 0 is also an absolute minimum.
71.
'0ln 3 ae2x ex b dx œ ’ e#
72.
'02 ln 2 ˆexÎ2 exÎ2 ‰ dx œ 2exÎ2 2exÎ2 ‘ 20 ln 2 œ ˆ2eln 2 2e ln 2 ‰ a2e! 2e! b œ (4 1) (2 2) œ 5 4 œ 1
2x
73. L œ '0 É1 1
ex
4
dx Ê
ex “
dy
dx
œ
ln 3
0
exÎ2
#
!
œ Š e # eln 3 ‹ Š e# e! ‹ œ ˆ 9# 3‰ ˆ "# 1‰ œ
2 ln 3
8
#
2œ2
Ê y œ exÎ2 C; y(0) œ 0 Ê 0 œ e0 C Ê C œ 1 Ê y œ exÎ2 1
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444
Chapter 7 Transcendental Functions
74. S œ 21'0 ˆ e
ln 2
œ 21'0
ln 2
œ
œ
75. (a)
1
#
1
#
ˆe
y
y
É1 ˆ ey #ecy ‰# dy œ 21 '
0
ecy ‰
#
ecy ‰
#
ln 2
Ɉ ey #ecy ‰# dy œ 21 '0ln 2 ˆ e
y
ˆe
y
e cy ‰
#
e cy ‰ #
#
É1 "4 ae2y 2 e2y b dy
dy œ
1
#
'0ln 2 ae2y 2 e2y bdy
"# e2y 2y "# e2y ‘ ln 2 œ 1# ˆ "# e2 ln 2 2 ln 2 "# e2 ln 2 ‰ ˆ "# 0 "# ‰‘
0
ˆ "# † 4 2 ln 2 "# † 4" ‰ œ 1# ˆ2 8" 2 ln 2‰ œ 1 ˆ 15
‰
16 ln 2
"
x
(x ln x x C) œ x †
d
dx
(b) average value œ
œ
"
e 1
'1
e
"
e1
(e e 1) œ
76. average value œ
"
2 1
ln x 1 0 œ ln x
ln x dx œ
"
e1
cx ln x xd e1 œ
"
e1
[(e ln e e) (1 ln 1 1)]
"
e1
'12 "x dx œ cln kxkd #" œ ln 2 ln 1 œ ln 2
77. (a) f(x) œ ex Ê f w (x) œ ex ; L(x) œ f(0) f w (0)(x 0) Ê L(x) œ 1 x
(b) f(0) œ 1 and L(0) œ 1 Ê error œ 0; f(0.2) œ e0 2 ¸ 1.22140 and L(0.2) œ 1.2 Ê error ¸ 0.02140
(c) Since yww œ ex 0, the tangent line
approximation always lies below the curve y œ ex .
Thus L(x) œ x 1 never overestimates ex .
Þ
78. (a) ex ex œ eÐxxÑ œ e! œ 1 Ê ex œ
x" x#
(b) y œ ae b
"
ex
ex"
ex#
for all x;
œ ex" ˆ e"x# ‰ œ ex" ex# œ ex" x#
Ê ln y œ x# ln e œ x# x" œ x" x# Ê eln y œ ex" x# Ê y œ ex" x# Ê aex" bx# œ ex" x#
x"
79. f(x) œ ln(x) 1 Ê f w (x) œ
"
x
Ê x n 1 œ xn ln (xn ) 1
ˆ x1 ‰
n
Ê xn1 œ xn c2 ln (xn )d . Then x" œ 2
Ê x# œ 2.61370564, x$ œ 2.71624393 and x& œ 2.71828183, where we have used Newton's method.
80. eln x œ x and ln aex b œ x for all x 0
81. Note that y œ ln x and ey œ x are the same curve; '1 ln x dx œ area under the curve between 1 and a;
a
'0ln a ey dy œ area to the left of the curve between 0 and ln a.
a
ln a
Ê '1 ln x dx '0 ey dy œ a ln a.
The sum of these areas is equal to the area of the rectangle
82. (a) y œ ex Ê yww œ ex 0 for all x Ê the graph of y œ ex is always concave upward
(b) area of the trapezoid ABCD 'ln a ex dx area of the trapezoid AEFD Ê
ln b
'ln a ex dx Š e
ln b
ln a
e
#
ln b
‹ (ln b ln a). Now
"
#
"
#
(AB CD)(ln b ln a)
(AB CD) is the height of the midpoint
M œ eÐln aln bÑÎ2 since the curve containing the points B and C is linear Ê eÐln aln bÑÎ2 (ln b ln a)
'ln a ex dx Š e
ln b
ln a
eln b
‹ (ln
#
b ln a)
'ln a ex dx œ cex d lnln ba œ eln b eln a œ b a, so part (b) implies that
ln b
(c)
eÐln aln bÑÎ2 (ln b ln a) b a Š e
Ê eln aÎ2 † eln bÎ2 ba
ln b ln a
ab
#
ln a
eln b
‹ (ln
#
b ln a) Ê eÐln aln bÑÎ2 Ê Èeln a Èeln b ba
ln b ln a
ab
#
ba
ln b ln a
Ê Èab ab
#
ba
ln b ln a
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ab
#
Section 7.4 ax and loga x
7.4 ax and loga x
1. (a) 5log5 7 œ 7
(b) 8log8
È2 œ È 2
#
Þ
(e) log3 È3 œ log3 3"Î# œ
(d) log4 16 œ log4 4 œ 2 log4 4 œ 2 † 1 œ 2
(f) log4 ˆ "4 ‰ œ log4 4" œ 1 log4 4 œ 1 † 1 œ 1
2. (a) 2log2 3 œ 3
(c) 1.3log1 3 75 œ 75
(b) 10log10 Ð1Î2Ñ œ
(d) log11 121 œ log11 11# œ 2 log11 11 œ 2 † 1 œ 2
(e) log121 11 œ log121 121"Î# œ ˆ "# ‰ log121 121 œ ˆ "# ‰ † 1 œ
(f) log3 ˆ "9 ‰ œ log3 3# œ 2 log3 3 œ 2 † 1 œ 2
"
#
log3 3 œ
"
#
"
#
"
#
†1œ
œ 0.5
(c) 1log1 7 œ 7
"
#
3. (a) Let z œ log4 x Ê 4z œ x Ê 22z œ x Ê a2z b# œ x Ê 2z œ Èx
(b) Let z œ log3 x Ê 3z œ x Ê a3z b# œ x# Ê 32z œ x# Ê 9z œ x#
(c) log2 aeÐln 2Ñ sin x b œ log2 2sin x œ sin x
4. (a) Let z œ log5 a3x# b Ê 5z œ 3x# Ê 25z œ 9x%
(b) loge aex b œ x
x
x
x
x
(c) log4 ˆ2e sin x ‰ œ log4 4ˆe sin x‰Î# œ e sin
#
5. (a)
(c)
6. (a)
(b)
(c)
log2 x
ln x
ln x
ln x ln 3
ln 3
log3 x œ ln # ƒ ln 3 œ ln # † ln x œ ln 2
logx a
ln a
ln a
ln a ln x#
2 ln x
logx# a œ ln x ƒ ln x# œ ln x † ln a œ ln x
(b)
œ
ln b
ln a
ƒ
ln a
ln b
œ
ln b
ln a
†
b‰
œ ˆ ln
ln a
ln b
ln a
œ
ln x
ln #
ƒ
ln x
ln 8
œ
ln x
ln #
†
ln 8
ln x
œ
3 ln 2
ln 2
œ3
œ2
log9 x
ln x
ln x
ln x
ln 3
1
log3 x œ ln 9 ƒ ln 3 œ 2 ln 3 † ln x œ 2
logÈ10 x
ˆ "# ‰ ln 2
ln x
ln x
ln x
logÈ2 x œ ln È10 ƒ ln È2 œ ˆ "# ‰ ln 10 † ln x
loga b
logb a
log2 x
log8 x
œ
ln 2
ln 10
#
7. 3log3 Ð7Ñ 2log2 Ð5Ñ œ 5log5 ÐxÑ Ê 7 5 œ x Ê x œ 12
8. 8log8 Ð3Ñ eln 5 œ x# 7log7 Ð3xÑ Ê 3 5 œ x# 3x Ê 0 œ x# 3x 2 œ (x 1)(x 2) Ê x œ 1 or x œ 2
#
9. 3log3 ax b œ 5eln x 3 † 10log10 Ð2Ñ Ê x# œ 5x 6 Ê x# 5x 6 œ 0 Ê (x 2)(x 3) œ 0 Ê x œ 2 or x œ 3
10. ln e 42 log4 ÐxÑ œ
#
"
x
log10 100 Ê 1 4log4 ax
c# b
œ
#
"
x
log10 10# Ê 1 x# œ ˆ x" ‰ (2) Ê 1 "
x#
2
x
œ0
Ê x 2x 1 œ 0 Ê (x 1) œ 0 Ê x œ 1
12. y œ 3cx Ê yw œ 3cx (ln 3)(1) œ 3cx ln 3
11. y œ 2x Ê yw œ 2x ln 2
13. y œ 5Ès Ê
#
14. y œ 2s Ê
dy
ds
dy
ds
œ 5Ès (ln 5) ˆ "# s"Î# ‰ œ Š 2lnÈ5s ‹ 5Ès
#
#
œ 2s (ln 2)2s œ aln 2# b Šs2s ‹ œ (ln 4)s2s
15. y œ x1 Ê yw œ 1xÐ11Ñ
17. y œ (cos ))
È2 Ê
dy
d)
#
16. y œ t1e Ê
dy
dt
œ (1 e) te
È
œ È2 (cos ))Š 2c1‹ (sin ))
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445
446
Chapter 7 Transcendental Functions
18. y œ (ln ))1 Ê
œ 1(ln ))Ð11Ñ ˆ ") ‰ œ
dy
d)
1(ln ))Ð11Ñ
)
19. y œ 7sec ) ln 7 Ê
dy
d)
œ a7sec ) ln 7b(ln 7)(sec ) tan )) œ 7sec ) (ln 7)# (sec ) tan ))
20. y œ 3tan ) ln 3 Ê
dy
d)
œ a3tan ) ln 3b(ln 3) sec# ) œ 3tan ) (ln 3)# sec# )
21. y œ 2sin 3t Ê
dy
dt
œ a2sin 3t ln 2b(cos 3t)(3) œ (3 cos 3t) a2sin 3t b (ln 2)
22. y œ 5c cos 2t Ê
dy
dt
œ a5c cos 2t ln 5b(sin 2t)(2) œ (2 sin 2t) a5c cos 2t b (ln 5)
23. y œ log2 5) œ
ln 5)
ln #
Ê
24. y œ log3 (1 ) ln 3) œ
25. y œ
ln x
ln 4
26. y œ
x ln e
ln #5
ln x#
ln 4
œ
ln x
2 ln 5
œ
ln (1 ) ln 3)
ln 3
2
ln x
ln 4
œ ˆ ln"# ‰ ˆ 5") ‰ (5) œ
dy
d)
x
# ln 5
ln x
ln 4
Ê
œ3
ln x
2 ln 5
"
) ln #
œ ˆ ln"3 ‰ ˆ 1 )" ln 3 ‰ (ln 3) œ
dy
d)
Ê yw œ
ln x
ln 4
"
1 ) ln 3
3
x ln 4
œ ˆ # ln" 5 ‰ (x ln x) Ê yw œ ˆ # ln" 5 ‰ ˆ1 "x ‰ œ
x1
2x ln 5
27. y œ log2 r † log4 r œ ˆ lnln #r ‰ ˆ lnln 4r ‰ œ
ln# r
(ln 2)(ln 4)
Ê
dy
dr
"
ˆ"‰
œ ’ (ln 2)(ln
4) “ (2 ln r) r œ
2 ln r
r(ln 2)(ln 4)
28. y œ log3 r † log9 r œ ˆ lnln 3r ‰ ˆ lnln 9r ‰ œ
ln# r
(ln 3)(ln 9)
Ê
dy
dr
"
ˆ"‰
œ ’ (ln 3)(ln
9) “ (2 ln r) r œ
2 ln r
r(ln 3)(ln 9)
1‰
29. y œ log3 Šˆ xx ‹œ
1
ln 3
Ê
dy
dx
œ
"
x1
"
x1
30. y œ log5 Ɉ 3x7x 2 ‰
œ
"
#
ln 7x "
#
œ
1
ln ˆ xx b
c1‰
ln 3
œ
ln 3
Ðln 5ÑÎ2
œ log5 ˆ 3x7x 2 ‰
œ
ln 5
ln (3x 2) Ê
)‰
32. y œ log7 ˆ sin e) #cos
œ
)
)
dy
d)
œ
cos )
(sin ))(ln 7)
33. y œ log5 ex œ
ln ex
ln 5
œ
# #
2
x ln 2
œ
dy
dx
7
2†7x
dy
d)
ln ˆ 3x7x 2 ‰
Ðln 5ÑÎ2
ln 5
3
2†(3x 2)
œ
œ ˆ ln#5 ‰ ”
(3x 2) 3x
2x(3x 2)
œ
ln ˆ 3x7x 2 ‰
ln 5
dy
dt
œ
4(x 1) x
2x(x 1)(ln 2)
œ
"
#
ln ˆ 3x7x # ‰
"
x(3x 2)
"
ln 7
cos (log7 ))
2)
"
ln 5
œ
œ
2 ln x 2 ln 2 "# ln (x 1)
ln 2
3x 4
2x(x 1) ln #
œ c3Ðln tÑÎÐln 2Ñ (ln 3)d ˆ t ln" 2 ‰ œ
3 ln (log2 t)
ln 8
•œ
ln ) ‰
ˆ ln ) ‰‘ ˆ ) ln" 7 ‰ œ sin (log7 )) œ sin ˆ ln
7 ) cos ln 7
ln x# ln e# ln 2 ln Èx 1
ln 2
35. y œ 3log2 t œ 3Ðln tÑÎÐln 2Ñ Ê
œ
Ê yw œ
"
# (ln 2)(x 1)
36. y œ 3 log8 (log2 t) œ
1‰
œ ln ˆ xx 1 œ ln (x 1) ln (x 1)
ln (sin )) ln (cos )) ln e) ln 2)
)) ) ) ln 2
œ ln (sin )) ln (cos
ln 7
ln 7
sin )
"
ln 2
ˆ " ‰
(cos ))(ln 7) ln 7 ln 7 œ ln 7 (cot ) tan ) 1 ln
x
ln 5
34. y œ log2 Š 2Èx xe 1 ‹ œ
Ê yw œ
ln 3
2
(x 1)(x 1)
ln ) ‰
31. y œ ) sin (log7 )) œ ) sin ˆ ln
Ê
7
Ê
1
(ln 3) ln Š xx b
c1‹
3 ln ˆ lnln 2t ‰
ln 8
Ê
dy
dt
"
t
alog2 3b 3log2 t
"
ˆ " ‰
œ ˆ ln38 ‰ ’ (ln t)/(ln
2) “ t ln # œ
3
t(ln t)(ln 8)
"
t(ln t)(ln #)
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Section 7.4 ax and loga x
ln 8 ln ˆtln 2 ‰
ln #
37. y œ log2 a8tln 2 b œ
38. y œ
t ln Šˆeln 3 ‰
sin t
‹
ln 3
t ln ˆ3sin t ‰
ln 3
œ
3 ln 2 (ln 2)(ln t)
ln #
œ
œ
œ 3 ln t Ê
œ t sin t Ê
t(sin t)(ln 3)
ln 3
40. y œ xÐx"Ñ Ê ln y œ ln xÐx"Ñ œ (x 1) ln x Ê
Ê yw œ xÐx1Ñ ˆ1 t
"
x
w
dy
dt
dy
dt
Ê yw œ (x 1)x x x 1 ln (x 1)‘
œ ln x (x 1) ˆ x" ‰ œ ln x 1 w
y
y
"
x
ln x‰
" dy
y dt
t
œ ˆ #" ‰ (ln t) ˆ #t ‰ ˆ "t ‰ œ
ln t
#
"
#
t
œ ˆÈt‰ ˆ ln# t "# ‰
42. y œ tÈt œ tˆt
Ê
"
(x 1)
œ ln (x 1) x †
y
y
41. y œ ˆÈt‰ œ ˆt"Î# ‰ œ ttÎ# Ê ln y œ ln ttÎ# œ ˆ #t ‰ ln t Ê
Ê
"
t
œ sin t t cos t
dy
dt
39. y œ (x 1)x Ê ln y œ ln (x 1)x œ x ln (x 1) Ê
œ
dy
dt
"Î# ‰
Ê ln y œ ln tˆt
œ ˆt"Î# ‰ (ln t) Ê
"Î# ‰
" dy
y dt
t2
œ Š ln2È
‹ tÈ t
t
43. y œ (sin x)x Ê ln y œ ln (sin x)x œ x ln (sin x) Ê
w
44. y œ xsin x Ê ln y œ ln xsin x œ (sin x)(ln x) Ê
y
y
œ ˆ "# t"Î# ‰ (ln t) t"Î# ˆ "t ‰ œ
ln t2
2È t
x‰
œ ln (sin x) x ˆ cos
Ê yw œ (sin x)x cln (sin x) x cot xd
sin x
w
y
y
œ (cos x)(ln x) (sin x) ˆ x" ‰ œ
sin x x (ln x)(cos x)
x
Ê yw œ xsin x ’ sin x x(lnx x)(cos x) “
45. y œ xln x , x 0 Ê ln y œ (ln x)# Ê
w
y
y
#
œ 2(ln x) ˆ "x ‰ Ê yw œ axln x b Š lnxx ‹
w
46. y œ (ln x)ln x Ê ln y œ (ln x) ln (ln x) Ê
y
y
œ ˆ "x ‰ ln (ln x) (ln x) ˆ ln"x ‰
d
dx
(ln x) œ
ln (ln x)
x
Ê yw œ Š ln (ln xx) " ‹ (ln x)ln x
47.
' 5x dx œ ln5 5 C
49.
'01 2c
50.
x
)
'c02 5c
)
48.
d) œ '2 ˆ "5 ‰ d) œ –
0
)
'1
52. Let u œ x"Î# Ê du œ
'14 È2Èx dx œ '14 2x
x
"Î#
"
#
!
"
#
ln Š "# ‹
"
ln Š "# ‹
!
)
ln Š 5" ‹ —
#
x2ax b dx œ '1 ˆ "# ‰ 2u du œ
2
— œ
Š "5 ‹
51. Let u œ x# Ê du œ 2x dx Ê
È2
"
)
"
#‹
ln Š "# ‹
1
Š
)
d) œ '0 ˆ "# ‰ d) œ –
œ
' (1.3)x dx œ ln(1.3)
(1.3) C
x
"
#
œ
ln Š "# ‹
"
2(ln 1 ln 2)
œ
"
# ln 2
c#
œ
#
"
ln Š 5" ‹
Š 5" ‹
ln Š 5" ‹
œ
"
ln Š 5" ‹
(1 25) œ
24
ln 1 ln 5
"
#
du œ x dx; x œ 1 Ê u œ 1, x œ È2 Ê u œ 2;
"
#
ln2 # ‘ # œ ˆ 2 ln" 2 ‰ a2# 2" b œ
"
u
x"Î# dx Ê 2 du œ
dx
Èx
œ
24
ln 5
"
ln #
; x œ 1 Ê u œ 1, x œ 4 Ê u œ 2;
† x"Î# dx œ 2'1 2u du œ ’ 2ln # “ œ ˆ ln"# ‰ a2$ 2# b œ
2
Ðu1Ñ #
"
447
4
ln #
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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"
x
448
Chapter 7 Transcendental Functions
53. Let u œ cos t Ê du œ sin t dt Ê du œ sin t dt; t œ 0 Ê u œ 1, t œ
'0
1Î2
7cos t sin t dt œ '1 7u du œ 0
7 ‘!
ln 7 "
u
œ ˆ ln"7 ‰ a7! 7b œ
54. Let u œ tan t Ê du œ sec# t dt; t œ 0 Ê u œ 0, t œ
"
3‹
ln Š 3" ‹
'01Î4 ˆ 3" ‰tan t sec# t dt œ '01 ˆ 3" ‰u du œ – Š
" du
u dx
55. Let u œ x2x Ê ln u œ 2x ln x Ê
"
u
1
4
1
#
Ê u œ 0;
6
ln 7
Ê u œ 1;
!
"
"
"
"
— œ ˆ ln 3 ‰ ’ˆ 3 ‰ ˆ 3 ‰ “ œ
2
3 ln 3
!
œ 2 ln x (2x) ˆ x" ‰ Ê
du
dx
œ 2u(ln x 1) Ê
x œ 2 Ê u œ 2% œ 16, x œ 4 Ê u œ 4) œ 65,536;
"
#
du œ x2x (1 ln x) dx;
'24 x2x (1 ln x) dx œ "# '1665 536 du œ "# cud 6516 536 œ "# (65,536 16) œ 65,520
œ 32,760
#
ß
"
x
56. Let u œ ln x Ê du œ
'1
2 ln x
2
x
dx œ '0
ln 2
È3 dx œ
ß
dx; x œ 1 Ê u œ 0, x œ 2 Ê u œ ln 2;
2u du œ ln2 # ‘ 0 œ ˆ ln"# ‰ a2ln 2 2! b œ
u
ln 2
2ln 2 "
ln #
57.
'
C
58.
'
59.
'03 ŠÈ2 1‹ xÈ2 dx œ ’xŠÈ2"‹ “ $ œ 3ŠÈ2"‹
60.
'1e xÐln 2Ñ1 dx œ lnx # ‘ e1 œ e
61.
'
3x
È
!
x ‰ ˆ"‰
dx œ ' ˆ lnln10
x dx; u œ ln x Ê du œ
log10 x
x
"
x
È
xŠ 2c1‹ dx œ
È
x 2
È2
C
ln 2
1ln 2
ln 2
ln 2
#
'14 logx x dx œ '14 ˆ lnln #x ‰ ˆ x" ‰ dx; u œ ln x Ê du œ x" dx; x œ 1 Ê u œ 0, x œ 4 Ê u œ ln 4‘
4
x‰ ˆ"‰
' ln 4 ˆ ln"# ‰ u du œ ˆ ln"# ‰ #" u# ‘ ln0 4 œ ˆ ln"# ‰ #" (ln 4)# ‘ œ (ln2 ln4)# œ (lnln4)4 œ ln 4
Ä '1 ˆ ln
ln #
x dx œ 0
2
2
"
#
(2 ln 2)# œ 2(ln 2)#
64.
'1e 2 ln 10 x(log
65.
'02 logx (x # 2) dx œ ln"# '02 cln (x 2)d ˆ x " # ‰ dx œ ˆ ln"# ‰ ’ (ln (x # 2)) “ # œ ˆ ln"# ‰ ’ (ln#4)
10
x)
dx œ '1
e
(ln 10)(2 ln x)
(ln 10)
ˆ x" ‰ dx œ c(ln x)# d e1 œ (ln e)# (ln 1)# œ 1
#
2
#
'110Î10 log
œ
10
(10x)
x
dx œ
#
ˆ ln"010 ‰ ’ 4(ln#010) “
(ln 2)#
# “
"0
ln 10
œ
3
#
"!
'110Î10 cln (10x)d ˆ 10x" ‰ dx œ ˆ ln"010 ‰ ’ (ln (10x))
“
#0
#
"Î"!
#
œ ˆ ln"010 ‰ ’ (ln #100)
0
(ln 1)#
# “
œ # ln 10
'09 2 logx (x1 1) dx œ ln210 '09 ln (x 1) ˆ x " 1 ‰ dx œ ˆ ln210 ‰ ’ (ln (x# 1)) “ * œ ˆ ln210 ‰ ’ (ln 210)
#
10
!
'23 2 logx (x1 1) dx œ ln22 '23 ln (x 1) ˆ x" 1 ‰ dx œ ˆ ln22 ‰ ’ (ln (x#1)) “ $ œ ˆ ln22 ‰ ’ (ln22)
2
(ln 2)#
# “
ln 2
œ ln 10
68.
#
!
œ ˆ ln"# ‰ ’ 4(ln# 2) 67.
#
4
x
"
"
x‰
#‘ %
#
#
#
'14 ln 2 log
' 4 ln x
"
dx œ '1 ˆ lnx2 ‰ ˆ ln
x
ln # dx œ 1 x dx œ # (ln x) " œ # c(ln 4) (ln 1) d œ # (ln 4)
œ
66.
2 1
ln 2
dx‘
#
63.
œ
' ˆ lnln10x ‰ ˆ x" ‰ dx œ ln"10 ' u du œ ˆ ln"10 ‰ ˆ #" u# ‰ C œ 2(lnlnx)10 C
Ä
62.
3xŠ 3b1‹
È 3 1
#
#
#
#
(ln ")#
# “
(ln ")#
# “
œ ln 2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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œ
"
ln #
Section 7.4 ax and loga x
69.
'
‰ ˆ x" ‰ dx œ (ln 10) ' ˆ ln"x ‰ ˆ x" ‰ dx; u œ ln x Ê du œ
œ ' ˆ lnln10
x
dx
x log10 x
Ä (ln 10) ' ˆ ln"x ‰ ˆ "x ‰ dx œ (ln 10) '
'
71.
'1ln x "t dt œ cln ktkd ln1 x œ ln kln xk ln 1 œ ln (ln x), x 1
72.
'1e "t dt œ cln ktkd e1
73.
'11/x "t dt œ cln ktkd 1"Îx œ ln ¸ x" ¸ ln 1 œ aln 1 ln kxkb ln 1 œ ln x, x 0
74.
"
ln a
œ (ln 8)#
dx
x ‰#
x ˆ ln
ln 8
'
(ln x)#
x
dx‘
du œ (ln 10) ln kuk C œ (ln 10) ln kln xk C
70.
dx
x (log8 x)#
œ'
"
u
"
x
dx œ (ln 8)#
(ln x)"
1
#
C œ (lnln 8)x C
x
x
œ ln ex ln 1 œ x ln e œ x
'1x "t dt œ ln"a ln ktk‘ x1 œ lnln xa lnln 1a œ loga x, x 0
75. A œ 'c2 1 2xx# dx œ 2'0
2
2
Ä A œ 2'1
5
76. A œ '1
1
"
u
2x
1 x#
dx; cu œ 1 x# Ê du œ 2x dx; x œ 0 Ê u œ 1, x œ 2 Ê u œ 5d
du œ 2 cln kukd &" œ 2(ln 5 ln 1) œ 2 ln 5
2Ð1xÑ dx œ 2
'
x
1
ˆ " ‰x
1 #
dx œ 2 –
Š "# ‹
ln Š "# ‹ —
"
œ ln2# ˆ "# 2‰ œ ˆ ln2# ‰ ˆ 3# ‰ œ
3
ln #
"
77. Let cH$ O d œ x and solve the equations 7.37 œ log10 x and 7.44 œ log10 x. The solutions of these equations
are 10(Þ$( and 10(Þ%% . Consequently, the bounds for cH$ O d are c10(Þ%% ß 10(Þ$( d .
78. pH œ log10 a4.8 ‚ 10) b œ (log10 4.8) 8 œ 7.32
79. Let O œ original sound level œ 10 log10 aI ‚ 10"# b db from Equation (6) in the text. Solving
O 10 œ 10 log10 akI ‚ 10"# b for k Ê 10 log10 aI ‚ 10"# b 10 œ 10 log10 akI ‚ 10"# b Ê log10 aI ‚ 10"# b 1
k
œ log10 akI ‚ 10"# b Ê log10 aI ‚ 10"# b 1 œ log10 k log10 aI ‚ 10"# b Ê 1 œ log10 k Ê 1 œ lnln10
Ê ln k œ ln 10 Ê k œ 10
80. Sound level with 10I œ 10 log10 a10I ‚ 10"# b œ 10 clog10 10 log10 aI ‚ 10"# bd œ 10 10 log10 aI ‚ 10"# b
œ original sound level 10 Ê an increase of 10 db
81. (a) If x œ cH$ O d and S x œ cOH d , then x(S x) œ 10"% Ê S œ x and
d# S
dx#
œ
2†10"%
x$
10c"%
x
Ê
dS
dx
œ1
œ
ln b
ln a
10"%
x#
0 Ê a minimum exists at x œ 10(
(b) pH œ log10 a10( b œ 7
(c)
cOHc d
cH $ O d
œ
Sx
x
œ
Šx 10"%
x ‹ x
x
œ
10"%
x#
Ê the ratio
82. Yes, it's true for all positive values of a and b: loga b œ
cOHc d
c H $ O d
ln b
ln a
equals 1 at x œ 10(
and logb a œ
ln a
ln b
Ê
"
logb a
œ loga b
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449
450
Chapter 7 Transcendental Functions
83. From zooming in on the graph at the right, we estimate
the third root to be x ¸ 0.76666
84. The functions f(x) œ xln 2 and g(x) œ 2ln x appear to
have identical graphs for x 0. This is no accident,
because xln 2 œ eln 2 ln x œ aeln 2 bln x œ 2ln x .
†
85. (a) f(x) œ 2x Ê f w (x) œ 2x ln 2; L(x) œ a2! ln 2b x 2! œ x ln 2 1 ¸ 0.69x 1
(b)
86. (a) f(x) œ log3 x Ê f w (x) œ
"
x ln 3
, and f(3) œ
ln 3
ln 3
Ê L(x) œ
"
3 ln 3
(x 3) ln 3
ln 3
œ
x
3 ln 3
"
ln 3
1
¸ 0.30x 0.09
(b)
87. (a) log3 8 œ
(c) log20 17
ln 8
ln 3 ¸
17
œ ln
ln #0
1.89279
(b) log7 0.5 œ
¸ 0.94575
(d) log0 5 7 œ
Þ
(e) ln x œ (log10 x)(ln 10) œ 2.3 ln 10 ¸ 5.29595
(g) ln x œ (log2 x)(ln 2) œ 1.5 ln 2 ¸ 1.03972
88. (a)
89.
ln 10
ln #
d ˆ " #
dx # x
† log10 x œ
ln 10
ln #
†
k‰ œ x and
Since x †
"
x
ln x
ln 10
d
dx aln
œ
ln x
ln #
œ log2 x
ln 0.5
ln 7 ¸ 0.35621
ln 7
ln 0.5 ¸ 2.80735
(f) ln x œ (log2 x)(ln 2) œ 1.4 ln 2 ¸ 0.97041
(h) ln x œ (log10 x)(ln 10) œ 0.7 ln 10 ¸ 1.61181
(b)
ln a
ln b
† loga x œ
ln a
ln b
†
ln x
ln a
œ
ln x
ln b
œ logb x
x cb œ x" .
œ " for any x Á !, these two curves will have perpendicular tangent lines.
90. eln x œ x for x ! and lnaex b œ x for all x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 7.5 Exponential Growth and Decay
"
x
91. Using Newton's Method: faxb œ lnaxb " Ê f w axb œ
Ê xn" œ xn lnaxn b"
"
x8
451
Ê xn" œ xn ’# lnaxn b“.
Then, x1 œ 2, x2 œ 2.61370564, x3 œ 2.71624393, and x& œ 2.71828183. Many other methods may be used. For example,
graph y œ ln x " and determine the zero of y.
92. (a) The point of tangency is apß ln pb and mtangent œ
"
p
since
œ x" . The tangent line passes through a!ß !b Ê the
dy
dx
equation of the tangent line is y œ "p x. The tangent line also passes throughapß ln pb Ê ln p œ "p p œ " Ê p œ e, and
the tangent line equation is y œ "e x.
(b)
d# y
dx#
œ x"# for x Á ! Ê y œ ln x is concave downward over its domain. Therefore, y œ ln x lies below the graph of
y œ "e x for all x !, x Á e, and ln x for x !, x Á e.
x
e
(c) Multiplying by e, e ln x x or ln x x.
e
(d) Exponentiating both sides of ln xe x, we have eln x ex , or xe ex for all positive x Á e.
(e) Let x œ 1 to see that 1e e1 . Therefore, e1 is bigger.
e
7.5 EXPONENTIAL GROWTH AND DECAY
1. (a) y œ y! ekt Ê 0.99y! œ y! e1000k Ê k œ
ln 0.99
1000
¸ 0.00001
(b) 0.9 œ eÐ0Þ00001)t Ê (0.00001)t œ ln (0.9) Ê t œ
(c) y œ y! eÐ20ß000Ñk ¸ y! e0Þ2 œ y! (0.82) Ê 82%
2. (a)
dp
dh
ln (0.9)
0.00001
œ kp Ê p œ p! ekh where p! œ 1013; 90 œ 1013e20k Ê k œ
(b) p œ 1013e6Þ05 ¸ 2.389 millibars
900 ‰
(c) 900 œ 1013eÐ0Þ121Ñh Ê 0.121h œ ln ˆ 1013
Ê hœ
3.
dy
dt
¸ 10,536 years
ln (90) ln (1013)
20
ln (1013) ln (900)
0.121
¸ 0.121
¸ 0.977 km
œ 0.6y Ê y œ y! e0Þ6t ; y! œ 100 Ê y œ 100e0Þ6t Ê y œ 100e0Þ6 ¸ 54.88 grams when t œ 1 hr
4. A œ A! ekt Ê 800 œ 1000e10k Ê k œ
ln (0.8)
10
Ê A œ 1000eÐln (0Þ8ÑÎ10Ñt , where A represents the amount of
sugar that remains after time t. Thus after another 14 hrs, A œ 1000eÐln Ð0Þ8ÑÎ10Ñ24 ¸ 585.35 kg
5. L(x) œ L! ekx Ê
L!
#
œ L! e18k Ê ln
is one-tenth of the surface value,
L!
10
"
#
œ 18k Ê k œ
ln 2
18
¸ 0.0385 Ê L(x) œ L! e0Þ0385x ; when the intensity
œ L! ec0Þ0385x Ê ln 10 œ 0.0385x Ê x ¸ 59.8 ft
6. V(t) œ V! etÎ40 Ê 0.1V! œ V! etÎ40 when the voltage is 10% of its original value Ê t œ 40 ln (0.1)
¸ 92.1 sec
7. y œ y! ekt and y! œ 1 Ê y œ ekt Ê at y œ 2 and t œ 0.5 we have 2 œ e0Þ5k Ê ln 2 œ 0.5k Ê k œ
Therefore, y œ eÐln 4Ñt Ê y œ e24 ln 4 œ 424 œ 2.81474978 ‚ 1014 at the end of 24 hrs
ln 2
0.5
œ ln 4.
8. y œ y! ekt and y(3) œ 10,000 Ê 10,000 œ y! e3k ; also y(5) œ 40,000 œ y! e5k . Therefore y! e5k œ 4y! e3k
Ê e5k œ 4e3k Ê e2k œ 4 Ê k œ ln 2. Thus, y œ y! eÐln 2Ñt Ê 10,000 œ y! e3 ln 2 œ y! eln 8 Ê 10,000 œ 8y!
Ê y! œ 10,000
œ 1250
8
9. (a) 10,000ekÐ1Ñ œ 7500 Ê ek œ 0.75 Ê k œ ln 0.75 and y œ 10,000eÐln 0Þ75Ñt . Now 1000 œ 10,000eÐln 0Þ75Ñt
0.1
Ê ln 0.1 œ (ln 0.75)t Ê t œ lnln0.75
¸ 8.00 years (to the nearest hundredth of a year)
(b) 1 œ 10,000eÐln 0Þ75Ñt Ê ln 0.0001 œ (ln 0.75)t Ê t œ
ln 0.0001
ln 0.75
¸ 32.02 years (to the nearest hundredth of a
year)
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452
Chapter 7 Transcendental Functions
10. (a) There are (60)(60)(24)(365) œ 31,536,000 seconds in a year. Thus, assuming exponential growth,
P œ 257,313,431ekt and 257,313,432 œ 257,313,431eÐ14kÎ31ß536ß000Ñ Ê ln Š 257,313,432
257,313,431 ‹ œ
14k
31,536,000
Ê k ¸ 0.0087542
(b) P œ 257,313,431eÐ0.0087542Ña"&b ¸ 293,420,847 (to the nearest integer). Answers will vary considerably
with the number of decimal places retained.
11. 0.9P! œ P! ek Ê k œ ln 0.9; when the well's output falls to one-fifth of its present value P œ 0.2P!
0.2
Ê 0.2P! œ P! eÐln 0Þ9Ñt Ê 0.2 œ eÐln 0Þ9Ñt Ê ln (0.2) œ (ln 0.9)t Ê t œ ln
ln 0.9 ¸ 15.28 yr
12. (a)
dp
dx
"
œ 100
p Ê
dp
p
"
"
œ 100
dx Ê ln p œ 100
x C Ê p œ eÐ0Þ01xCÑ œ eC e0Þ01x œ C" e0Þ01x ;
p(100) œ 20.09 Ê 20.09 œ C" eÐ0Þ01ÑÐ100Ñ Ê C" œ 20.09e ¸ 54.61 Ê p(x) œ 54.61e0Þ01x (in dollars)
(b) p(10) œ 54.61eÐ0Þ01ÑÐ10Ñ œ $49.41, and p(90) œ 54.61eÐ0Þ01ÑÐ90Ñ œ $22.20
(c) r(x) œ xp(x) Ê rw (x) œ p(x) xpw (x);
pw (x) œ .5461e0Þ01x Ê rw (x)
œ (54.61 .5461x)e0Þ01x . Thus, rw (x) œ 0
Ê 54.61 œ .5461x Ê x œ 100. Since rw 0
for any x 100 and rw 0 for x 100, then
r(x) must be a maximum at x œ 100.
13. (a) A! eÐ0Þ04Ñ5 œ A! e0Þ2
(b) 2A! œ A! eÐ0Þ04Ñt Ê ln 2 œ (0.04)t Ê t œ
Ê tœ
ln 3
0.04
ln 2
0.04
¸ 17.33 years; 3A! œ A! eÐ0Þ04Ñt Ê ln 3 œ (0.04)t
¸ 27.47 years
14. (a) The amount of money invested A! after t years is A(t) œ A! et
(b) If A(t) œ 3A! , then 3A! œ A! et Ê ln 3 œ t or t ¸ 1.099 years
(c) At the beginning of a year the account balance is A! et , while at the end of the year the balance is A! eÐt1Ñ .
The amount earned is A! eÐt1Ñ A! et œ A! et (e 1) ¸ 1.7 times the beginning amount.
15. A(100) œ 90,000 Ê 90,000 œ 1000erÐ100Ñ Ê 90 œ e100r Ê ln 90 œ 100r Ê r œ
16. A(100) œ 131,000 Ê 131,000 œ 1000e100r Ê ln 131 œ 100r Ê r œ
ln 131
100
17. y œ y! e0Þ18t represents the decay equation; solving (0.9)y! œ y! e0Þ18t Ê t œ
18. A œ A! ekt and
Ê tœ
"
#
ln 0.05
0.00499
"
#
A! œ A! e139k Ê
20. (a) A œ A! ekt Ê
(b)
ln (0.5)
139
¸ 0.0450 or 4.50%
¸ 0.04875 or 4.875%
ln (0.9)
0.18
¸ 0.585 days
¸ 0.00499; then 0.05A! œ A! e0Þ00499t
¸ 600 days
19. y œ y! ekt œ y! eÐkÑÐ3ÎkÑ œ y! e3 œ
"
k
œ e139k Ê k œ
ln 90
100
"
#
y!
e$
y!
20
œ e2Þ645k Ê k œ
œ (0.05)(y! ) Ê after three mean lifetimes less than 5% remains
ln 2
#.645
¸ 0.262
¸ 3.816 years
ln 2 ‰
ln 2 ‰
(c) (0.05)A œ A exp ˆ 2.645
t Ê ln 20 œ ˆ 2.645
t Ê tœ
2.645 ln 20
ln #
¸ 11.431 years
21. T Ts œ (T! Ts ) ekt , T! œ 90°C, Ts œ 20°C, T œ 60°C Ê 60 20 œ 70e10k Ê
Ê kœ
ln ˆ 74 ‰
10
4
7
œ e10k
¸ 0.05596
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 7.6 Relative Rates of Growth
(a) 35 20 œ 70e0Þ05596t Ê t ¸ 27.5 min is the total time Ê it will take 27.5 10 œ 17.5 minutes longer to reach
35°C
(b) T Ts œ (T! Ts ) ekt , T! œ 90°C, Ts œ 15°C Ê 35 15 œ 105e0Þ05596t Ê t ¸ 13.26 min
22. T 65° œ (T! 65°) ekt Ê 35° 65° œ (T! 65°) e10k and 50° 65° œ (T! 65°) e20k . Solving
30° œ (T! 65°) e10k and 15° œ (T! 65°) e20k simultaneously Ê (T! 65°) e10k œ 2(T! 65°) e20k
ln 2
Ê e10k œ 2 Ê k œ ln102 and 30° œ T!e10k65° Ê 30° e10 ˆ 10 ‰ ‘ œ T! 65° Ê T! œ 65° 30° ˆeln 2 ‰ œ 65° 60° œ 5°
23. T Ts œ (T! Ts ) eckt Ê 39 Ts œ (46 Ts ) ec10k and 33 Ts œ (46 Ts ) ec20k Ê
33Ts
46Ts
œ ec20k œ aec10k b# Ê
33Ts
46Ts
39Ts
46Ts
œ ec10k and
#
Ts
#
#
œ Š 39
46Ts ‹ Ê (33 Ts )(46 Ts ) œ (39 Ts ) Ê 1518 79Ts Ts
œ 1521 78Ts T#s Ê Ts œ 3 Ê Ts œ 3°C
24. Let x represent how far above room temperature the silver will be 15 min from now, y how far above room
temperature the silver will be 120 min from now, and t! the time the silver will be 10°C above room
temperature. We then have the following time-temperature table:
time in min.
0
20 (Now) 35
140
t!
temperature
Ts 70° Ts 60° Ts x Ts y Ts 10°
" ‰
T Ts œ (T! Ts ) eckt Ê (60 Ts ) Ts œ c(70 Ts ) Ts d ec20k Ê 60 œ 70ec20k Ê k œ ˆ 20
ln ˆ 67 ‰
¸ 0.00771
(a) T Ts œ (T! Ts ) ec0Þ00771t Ê (Ts x) Ts œ c(70 Ts ) Ts d eÐ0Þ00771ÑÐ35Ñ Ê x œ 70e0Þ26985 ¸ 53.44°C
(b) T Ts œ (T! Ts ) e0Þ00771t Ê (Ts y) Ts œ c(70 Ts ) Ts d eÐ0Þ00771ÑÐ140Ñ Ê y œ 70e1Þ0794 ¸ 23.79°C
(c) T Ts œ (T! Ts ) e0Þ00771t Ê (Ts 10) Ts œ c(70 Ts ) Ts d eÐ0Þ00771Ñ t! Ê 10 œ 70e0Þ00771t!
"
‰ ln ˆ "7 ‰ œ 252.39 Ê 252.39 20 ¸ 232 minutes from now the
Ê ln ˆ "7 ‰ œ 0.00771t! Ê t! œ ˆ 0.00771
silver will be 10°C above room temperature
25. From Example 5, the half-life of carbon-14 is 5700 yr Ê
Ê c œ c! e0Þ0001216t Ê (0.445)c! œ c! e0Þ0001216t Ê t œ
"
ckÐ5700Ñ
# c! œ c! e
ln (0.445)
0.0001216 ¸ 6659
Ê kœ
ln 2
5700
¸ 0.0001216
years
26. From Exercise 25, k ¸ 0.0001216 for carbon-14.
(a) c œ c! e0Þ0001216t Ê (0.17)c! œ c! e0Þ0001216t Ê t ¸ 14,571.44 years Ê 12,571 BC
(b) (0.18)c! œ c! e0Þ0001216t Ê t ¸ 14,101.41 years Ê 12,101 BC
(c) (0.16)c! œ c! e0Þ0001216t Ê t ¸ 15,069.98 years Ê 13,070 BC
27. From Exercise 25, k ¸ 0.0001216 for carbon-14. Thus, c œ c! e0Þ0001216t Ê (0.995)c! œ c! e0Þ0001216t
Ê tœ
ln (0.995)
0.0001216
¸ 41 years old
7.6 RELATIVE RATES OF GROWTH
lim x 3 œ lim e"x œ 0
x Ä _ ex
xÄ_
$
#x
#
x cos x
2x
(b) slower, lim x esin
œ lim 3x 2 sin
œ lim 6x 2excos 2x œ lim 6 4esin
œ 0 by the
x
x
ex
xÄ_
xÄ_
xÄ_
xÄ_
Sandwich Theorem because e2x Ÿ 6 4exsin 2x Ÿ 10
lim e2x œ 0 œ lim "e0x
ex for all reals and x Ä
_
xÄ_
1. (a) slower,
"Î#
Èx
lim
œ lim xex œ lim
x Ä _ ex
xÄ_
xÄ_
x
ˆ 4e ‰x œ _ since
(d) faster, lim 4ex œ lim
xÄ_
xÄ_
(c) slower,
Š "# ‹ x"Î#
ex
4
e
œ
lim
xÄ_
"
#Èx ex
œ0
1
x
(e) slower,
lim
xÄ_
Š 3# ‹
ex
œ
x
lim ˆ 3 ‰ œ 0 since
x Ä _ 2e
3
2e
1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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453
454
Chapter 7 Transcendental Functions
(f) slower,
(g) same,
lim
xÄ_
lim
xÄ_
x
Š e# ‹
ex
(h) slower,
lim
xÄ_
2. (a) slower,
lim
xÄ_
(b) slower,
œ
lim
xÄ_
lim
xÄ_
Š "x ‹
ex
(c) slower,
exÎ2
ex
lim
xÄ_
œ È0 œ 0
œ
lim
xÄ_
"
exÎ2
œ
lim
xÄ_
"
#
œ
log10 x
ex
lim
xÄ_
10x% 30x 1
ex
x ln x x
ex
œ
œ
lim
xÄ_
È 1 x%
ex
œ
œ0
œ
ln x
(ln 10) ex
lim
xÄ_
œ
"
x
lim
xÄ_
40x$ 30
ex
x (ln x 1)
ex
lim
xÄ_
"
xex
"
#
œ
œ
œ
(ln 10) ex
lim
xÄ_
lim
xÄ_
ln
lim
xÄ_
"20x#
ex
œ
"
(ln 10)xex
240x
ex
lim
xÄ_
x 1 x Š "x ‹
ex
œ
œ0
lim
xÄ_
œ
240
ex
lim
xÄ_
ln x 1 1
ex
œ
œ0
ln x
ex
lim
xÄ_
œ0
œ É lim
xÄ_
1 x%
e2x
œ É lim
xÄ_
4x$
2e2x
œ É lim
xÄ_
12x#
4e2x
œ É lim
xÄ_
24x
8e2x
œ É lim
xÄ_
x
Š 5# ‹
x
5
lim
œ lim ˆ #5e ‰ œ 0 since 2e
1
x Ä _ ex
xÄ_
x
e
"
(e) slower, lim
œ lim e2x œ 0
x Ä _ ex
xÄ_
xex
(f) faster, lim
œ
lim x œ _
x Ä _ ex
xÄ_
(g) slower, since for all reals we have 1 Ÿ cos x Ÿ 1 Ê e" Ÿ ecos x Ÿ e" Ê
(d) slower,
e"
ex
e"
ex ,
e"
ex
Ÿ
ecos x
ex
lim
œ 0 œ lim
so by the Sandwich Theorem we conclude that lim
xÄ_
xÄ_
xÄ_
x1
e
"
"
"
(h) same, lim
œ lim eÐx x 1Ñ œ lim e œ e
x Ä _ ex
xÄ_
xÄ_
3. (a) same,
lim
xÄ_
(b) faster, lim
xÄ_
(c) same,
lim
xÄ_
(d) same,
lim
xÄ_
x# 4x
x#
lim 2x 4 œ lim
x Ä _ 2x
xÄ_
$
œ lim ax 1b œ _
xÄ_
œ
x& x#
x#
È x% x$
x#
(x 3)#
x#
œ É lim
xÄ_
œ
lim
xÄ_
x% x$
x%
2(x 3)
#x
œ
Ÿ
ecos x
ex
e"
ex
and also
œ0
œ1
2
#
œ É lim ˆ1 "x ‰ œ È1 œ 1
xÄ_
lim
xÄ_
2
#
œ1
Š "x ‹
lim x ln x œ lim lnxx œ lim
œ0
x Ä _ x#
xÄ_
xÄ_ 1
x
# x
x
2) 2
(f) faster, lim 2x# œ lim (ln2x
œ lim (ln 2)# 2 œ _
xÄ_
xÄ_
xÄ_
$ x
(g) slower, lim x xe# œ lim exx œ lim e"x œ 0
xÄ_
xÄ_
xÄ_
#
(h) same, lim 8x
œ
lim
8
œ
8
#
xÄ_ x
xÄ_
(e) slower,
# È
x x
" ‰
lim
œ lim ˆ1 x$Î#
œ1
x Ä _ x#
xÄ_
"0x#
(b) same, lim
œ lim 10 œ 10
x Ä _ x#
xÄ_
# x
x e
(c) slower, lim
œ lim e"x œ 0
x Ä _ x#
xÄ_
4. (a) same,
#
#
Š ln x ‹
ln 10
lim log10 x œ lim
œ ln"10 lim
x Ä _ x#
x Ä _ x#
xÄ_
x $ x #
(e) faster, lim
œ lim (x 1) œ _
x Ä _ x#
xÄ_
(d) slower,
2 ln x
x#
œ
2
ln 10
lim
xÄ_
Š x" ‹
2x
œ
"
ln 10
x
"
Š 10
‹
lim
œ lim 10"x x# œ 0
x Ä _ x#
xÄ_
x
(1.1)x
(g) faster, lim
œ lim (ln 1.1)(1.1)
œ lim
#x
x Ä _ x#
xÄ_
xÄ_
x# 100x
100
ˆ
‰
(h) same, lim
œ
lim
1
œ
1
x#
x
xÄ_
xÄ_
(f) slower,
(ln 1.1)# (1.1)x
#
œ_
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
lim
xÄ_
"
x#
œ0
24
16e2x
Section 7.6 Relative Rates of Growth
5. (a) same,
lim
xÄ_
log3 x
ln x
(b) same,
lim
xÄ_
ln 2x
ln x
(c) same,
lim
xÄ_
(d) faster,
lim
xÄ_
(e) faster,
(f) same,
œ
œ
ln Èx
ln x
x
Š ln
ln 3 ‹
lim
xÄ_
ln x
ˆ #2x ‰
ˆ x" ‰
lim
xÄ_
œ
œ
lim
xÄ_
lim
xÄ_
x
ln x
œ
lim
xÄ_
lim
xÄ_
5 ln x
ln x
œ
Š"‹
lim
xÄ_
"
ln 3
œ
"
ln 3
"
#
œ
"
#
œ1
Š "# ‹ ln x
lim
xÄ_
Èx
ln x
œ
455
œ
ln x
lim
xÄ_
Š "# ‹ x"Î#
x"Î#
ln x
œ
lim
xÄ_
"
Š "x ‹
œ
lim x œ _
xÄ_
Š x" ‹
œ
lim
xÄ_
x
#Èx
œ
Èx
#
lim
xÄ_
œ_
lim 5 œ 5
xÄ_
x
lim
œ lim x ln" x œ 0
x Ä _ ln x
xÄ_
x
x
(h) faster, lim lne x œ lim ˆe" ‰ œ lim xex œ _
xÄ_
xÄ_ x
xÄ_
(g) slower,
6. (a) same,
lim
xÄ_
(b) same,
lim
xÄ_
lim
xÄ_
(d) slower,
lim
xÄ_
lim
xÄ_
(g) slower,
lim
xÄ_
lim
xÄ_
lim
xÄ_
ln x
œ
lim
xÄ_
ecx
ln x
œ
œ
"
ln #
œ
ln x
œ
ln x
"
‹
x#
lim
xÄ_
lim
xÄ_
Š È"x ‹
Š
Š lnlnx2 ‹
œ
x2 ln x
ln x
lim
xÄ_
(f) slower,
(h) same,
œ
log10 10x
ln x
(c) slower,
(e) faster,
#
log2 x#
ln x
Š lnln10x
10 ‹
ln x
œ
"
ˆÈx‰ (ln x)
"
x# ln x
lim
xÄ_
"
ln 10
ln x#
ln x
lim
xÄ_
"
ex ln x
ln (ln x)
ln x
œ
lim
xÄ_
ln (2x5)
ln x
œ
lim
xÄ_
"
ln #
ln 10x
ln x
œ
lim
xÄ_
"
ln 10
2 ln x
ln x
lim
xÄ_
œ
"
ln #
"0
Š 10x
‹
Š x" ‹
lim 2 œ
xÄ_
œ
"
ln 10
2
ln #
lim 1 œ
xÄ_
"
ln 10
œ0
œ0
lim ˆ x 2‰ œ Š lim
x Ä _ ln x
xÄ_
lim
xÄ_
œ
x
ln x ‹
2 œ lim
xÄ_
"
Š "x ‹ 2 œ Š lim x‹ 2 œ _
xÄ_
œ0
"/x
Š ln
x‹
Š x" ‹
Š 2x25 ‹
Š x" ‹
œ
lim
xÄ_
œ
"
ln x
œ0
2x
2x5
lim
xÄ_
œ
lim
xÄ_
2
#
œ
lim 1 œ 1
xÄ_
x
7.
e
lim
œ lim exÎ2 œ _ Ê ex grows faster than exÎ2 ; since for x ee we have ln x e and lim (lnexx)
x Ä _ exÎ2 x Ä _
xÄ_
x
x
x
œ lim ˆ lne x ‰ œ _ Ê (ln x)x grows faster than ex ; since x ln x for all x 0 and lim (lnxx)x œ lim ˆ lnxx ‰
xÄ_
xÄ_
xÄ_
œ _ Ê xx grows faster than (ln x)x . Therefore, slowest to fastest are: exÎ2 , ex , (ln x)x , xx so the order is d, a, c, b
8.
2)
lim (ln 2) œ lim (ln (ln 2))(ln
œ lim (ln (ln 2))# (ln 2) œ (ln (ln# 2))
lim (ln 2)x œ 0
#x
x Ä _ x#
xÄ_
xÄ_
xÄ_
#
Ê (ln 2)x grows slower than x# ; lim x2x œ lim (ln2x#)2x œ lim (ln 2)2 # #x œ 0 Ê x# grows slower than 2x ;
xÄ_
xÄ_
xÄ_
x
x
lim 2ex œ lim ˆ 2e ‰ œ 0 Ê 2x grows slower than ex . Therefore, the slowest to the fastest is: (ln 2)x , x# , 2x
xÄ_
xÄ_
and ex so the order is c, b, a, d
x
x
#
x
x
#
lim x œ 1
xÄ_ x
(b) false; lim x x 5 œ 1" œ 1
xÄ_
(c) true; x x 5 Ê xx 5 1 if x 1 (or sufficiently large)
9. (a) false;
(d) true; x 2x Ê
(e) true;
lim
xÄ_
(f) true;
x ln x
x
ex
e2x
x
2x
1 if x 1 (or sufficiently large)
œ lim
xÄ0
œ1
ln x
x
"
ex
1
œ0
Èx
x
œ1
"
Èx
2 if x 1 (or sufficiently large)
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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456
Chapter 7 Transcendental Functions
(g) false;
È x# 5
x
(h) true;
"
Šx
3‹
10. (a) true;
ln x
ln 2x
lim
xÄ_
Š x" ‹
œ
œ
Š "x ‹
lim
xÄ_
È(x 5)#
x
œ
Š #2x ‹
x5
x
œ1
lim 1 œ 1
xÄ_
5
x
6 if x 1 (or sufficiently large)
1 if x 1 (or sufficiently large)
x
x3
Š "x "‹
x#
œ 1 "x 2 if x 1 (or sufficiently large)
"
Š ‹
(b) true;
x
(c) false;
lim
xÄ_
Š "x "
‹
x#
Š "x ‹
(d) true; 2 cos x Ÿ 3 Ê
e x
ex
x
(e) true;
œ1
(f) true;
lim
xÄ_
(g) true; lnln(lnxx) (h) false;
lim
xÄ_
x
ex
and
2 cos x
#
x
ex Ä
Ÿ
3
#
if x is sufficiently large
0 as x Ä _ Ê 1 Š "x ‹
lim ln x œ lim
xÄ_ x
xÄ_
œ 1 if x is sufficiently large
x ln x
x#
ln x
ln x
lim ˆ1 "x ‰ œ 1
xÄ_
œ
œ
ln x
ln ax# 1b
œ
lim
xÄ_
Š "x ‹
Š
2x
‹
x# 1
œ
1
x
ex
2 if x is sufficiently large
œ0
lim
xÄ_
x# "
#x #
œ
lim ˆ " xÄ_ #
lim
xÄ_
f(x)
g(x)
œLÁ0 Ê
lim
xÄ_
f(x)
L¹ 1 if x is sufficiently large Ê L 1 ¹ g(x)
f(x)
g(x)
L1 Ê
f(x)
g(x)
11. If f(x) and g(x) grow at the same rate, then
Ê f œ O(g). Similarly,
g(x)
f(x)
" ‰
#x#
g(x)
f(x)
œ
œ
"
L
"
#
Á 0. Then
Ÿ kLk 1 if x is sufficiently large
Ÿ ¸ L" ¸ 1 Ê g œ O(f).
12. When the degree of f is less than the degree of g since in that case
lim
xÄ_
f(x)
g(x)
œ 0.
f(x)
lim
œ 0 when the degree of f is smaller
x Ä _ g(x)
(the ratio of the leading coefficients) when the degrees are the same.
13. When the degree of f is less than or equal to the degree of g since
than the degree of g, and
lim
xÄ_
f(x)
g(x)
œ
a
b
14. Polynomials of a greater degree grow at a greater rate than polynomials of a lesser degree. Polynomials of the
same degree grow at the same rate.
15.
lim
xÄ_
œ
16.
17.
ln (x ")
ln x
lim
xÄ_
lim
xÄ_
œ
x
x 999
ln (x a)
ln x
œ
lim
xÄ_
"
Šx
1‹
Š x" ‹
œ
lim
xÄ_
x
x 1
œ
lim
xÄ_
œ
lim
xÄ_
x
x a
œ
lim
xÄ_
ln (x 999)
ln x
Š x "999 ‹
"
1
œ 1 and
"
1
œ 1. Therefore, the relative rates are the same.
lim
xÄ_
œ
lim
xÄ_
Š x" ‹
œ1
lim
xÄ_
" ‹
Šx
a
Š x" ‹
È10x 1
È x 1
lim
œ É lim "0xx 1 œ È10 and lim
œ É lim x x 1 œ È1 œ 1. Since the growth rate
Èx
xÄ_
xÄ_
x Ä _ Èx
xÄ_
is transitive, we conclude that È10x 1 and Èx 1 have the same growth rate ˆthat of Èx‰ .
È x% x
x#
È
%
$
x x
œ É lim x x% x œ 1 and lim
œ É lim x x% x œ 1. Since the growth rate is
x#
xÄ_
xÄ_
xÄ_
È
%
%
$
transitive, we conclude that Èx x and x x have the same growth rate athat of x# b .
18.
lim
xÄ_
19.
lim
xÄ_
xn
ex
œ
%
lim
xÄ_
nxnc1
ex
œá œ
lim
xÄ_
n!
ex
%
$
œ 0 Ê xn œ o aex b for any non-negative integer n
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 7.6 Relative Rates of Growth
20. If p(x) œ an xn an1 xn1 á a" x a! , then
457
nc1
n
lim p(x) œ an lim xex an1 lim xex á
x Ä _ ex
xÄ_
xÄ_
a" lim exx a! lim e"x where each limit is zero (from Exercise 19). Therefore, lim p(x)
œ0
xÄ_
xÄ_
x Ä _ ex
x
Ê e grows faster than any polynomial.
21. (a)
x1În
ln x
lim
xÄ_
œ
lim
xÄ_
xÐ1nÑÎn
n ˆ "x ‰
œ ˆ "n ‰
lim x1În œ _ Ê ln x œ o ˆx1În ‰ for any positive integer n
xÄ_
'
(b) ln ae17ß000ß000 b œ 17,000,000 Še"(‚"! ‹
1Î10'
œ e"( ¸ 24,154,952.75
(c) x ¸ 3.430631121 ‚ 10"&
(d) In the interval c3.41 ‚ 10"& ß 3.45 ‚ 10"& d we have
ln x œ 10 ln (ln x). The graphs cross at about
3.4306311 ‚ 10"& .
Š "x ‹
22.
lim
xÄ_
œ
23. (a)
œ
ln x
an xn anc1 xn1 á a" x a!
lim
xÄ_
"
aan b anxn b
lim
nÄ_
lim
xÄ_
lim Š ln x ‹
xÄ_ xn
an1
Ša n x á a"
xn1
a!
xn
‹
œ
lim
xÄ_ – nxn1 —
an
œ 0 Ê ln x grows slower than any non-constant polynomial (n
œ
n log2 n
n alog2 nb#
slower than n (log2 n)# ;
Š "n ‹
(b)
œ 0 Ê n log2 n grows
n log2 n
n$Î#
lim
nÄ_
œ
lim
nÄ_
n
Š ln
ln 2 ‹
n"Î#
"
lim
œ ln2# lim n"Î#
œ0
n Ä _ ˆ "# ‰ n"Î#
nÄ_
$Î#
Ê n log2 n grows slower than n . Therefore, n log2 n
grows at the slowest rate Ê the algorithm that takes
O(n log2 n) steps is the most efficient in the long run.
œ
"
ln #
"
log2 n
lim
nÄ_
1)
#
24. (a)
lim
nÄ_
(log2 n)#
n
œ
lim
nÄ_
œ
2
(ln 2)#
œ
2(ln n) Š "n ‹
(ln 2)#
lim
nÄ_
than n;
lim
nÄ_
œ
lim
nÄ_
n
Š ln
ln 2 ‹
n"Î#
œ
Š "n ‹
1
œ
lim
nÄ_
lim
nÄ_
(ln n)#
n(ln 2)#
(b)
ln n
n
œ 0 Ê (log2 n)# grows slower
"
ln #
Š"‹
n
2
(ln 2)#
(log2 n)#
Èn log2 n
œ
n
Š ln
ln 2 ‹
œ
lim
nÄ_
lim
nÄ_
log2 n
Èn
ln n
n"Î#
n
"
lim
œ ln2# lim n"Î#
œ 0 Ê (log2 n)# grows slower than Èn log2 n. Therefore (log2 n)# grows
x Ä _ ˆ "# ‰ n"Î#
nÄ_
at the slowest rate Ê the algorithm that takes O a(log2 n)# b steps is the most efficient in the long run.
œ
"
ln #
lim
nÄ_
25. It could take one million steps for a sequential search, but at most 20 steps for a binary search because
2"* œ 524,288 1,000,000 1,048,576 œ 2#! .
26. It could take 450,000 steps for a sequential search, but at most 19 steps for a binary search because
2") œ 262,144 450,000 524,288 œ 2"* .
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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458
Chapter 7 Transcendental Functions
7.7 INVERSE TRIGONOMETRIC FUNCTIONS
1. (a)
1
4
(b) 13
3. (a) 16
1
6
(c)
2. (a) 14
(b)
1
4
(c) 13
1
3
(b)
(c) 16
4. (a)
1
6
(b) 14
(c)
1
3
5. (a)
1
3
(b)
31
4
(c)
1
6
6. (a)
21
3
(b)
1
4
(c)
51
6
7. (a)
31
4
(b)
1
6
(c)
21
3
8. (a)
1
4
(b)
51
6
(c)
1
3
9. (a)
1
4
(b) 13
(c)
1
6
10. (a) 14
(b)
1
3
(c) 16
11. (a)
31
4
(b)
(c)
21
3
12. (a)
1
4
(b)
51
6
(c)
csc ! œ
13
5 ,
1
6
5 ‰
13. ! œ sin" ˆ 13
Ê cos ! œ
12
13 ,
tan ! œ
5
12 ,
sec ! œ
13
12 ,
and cot ! œ
14. ! œ tan" ˆ 43 ‰ Ê sin ! œ 45 , cos ! œ 35 , sec ! œ 53 , csc ! œ 54 , and cot ! œ
15. ! œ sec" Š È5‹ Ê sin ! œ
16. ! œ sec" Š
17. sin Šcos"
È13
# ‹
È2
# ‹
Ê sin ! œ
œ sin ˆ 14 ‰ œ
2
È5 ,
cos ! œ È15 , tan ! œ 2, csc ! œ
3
È13 ,
12
5
3
4
È5
2 ,
cos ! œ È213 , tan ! œ 3# , csc ! œ
1
3
and cot ! œ 12
È13
3 ,
and cot ! œ 23
18. sec ˆcos" #" ‰ œ sec ˆ 13 ‰ œ 2
"
È2
19. tan ˆsin" ˆ "# ‰‰ œ tan ˆ 16 ‰ œ È"3
20. cot Šsin" Š
È3
# ‹‹
œ cot ˆ 13 ‰ œ È"3
21. csc asec" 2b cos Štan" ŠÈ3‹‹ œ csc ˆcos" ˆ "# ‰‰ cos ˆ 13 ‰ œ csc ˆ 13 ‰ cos ˆ 13 ‰ œ
2
È3
"
#
œ
4 È3
2È 3
22. tan asec" 1b sin acsc" a2bb œ tan ˆcos" "1 ‰ sin ˆsin" ˆ 12 ‰‰ œ tan (0) sin ˆ 16 ‰ œ 0 ˆ #" ‰ œ #"
23. sin ˆsin" ˆ "# ‰ cos" ˆ "# ‰‰ œ sin ˆ 16 21 ‰
3
œ sin ˆ 1# ‰ œ 1
24. cot ˆsin" ˆ "# ‰ sec" 2‰ œ cot ˆ 16 cos" ˆ "# ‰‰ œ cot ˆ 16 13 ‰ œ cot ˆ 1# ‰ œ 0
25. sec atan" 1 csc" 1b œ sec ˆ 14 sin" 11 ‰ œ sec ˆ 14 1# ‰ œ sec ˆ 341 ‰ œ È2
26. sec Šcot" È3 csc" (1)‹ œ sec ˆ 16 sin" ( "1 )‰ œ sec ˆ 1# 27. sec" ˆsec ˆ 16 ‰‰ œ sec" Š È23 ‹ œ cos" Š
28. cot" ˆcot ˆ 14 ‰‰ œ cot" (1) œ
È3
# ‹
œ
1
3
12 ‰ œ sec ˆ 13 ‰ œ 2
1
6
31
4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 7.7 Inverse Trigonometric Functions
29. ! œ tan"
x
#
indicates the diagram
30. ! œ tan" 2x indicates the diagram
Ê sec ˆtan" x# ‰ œ sec ! œ
È x# 4
#
Ê sec atan" 2xb œ sec ! œ È4x# 1
31. ! œ sec" 3y indicates the diagram
Ê tan asec" 3yb œ tan ! œ È9y# 1
32. ! œ sec"
Ê tan ˆsec" y5 ‰ œ tan ! œ
y
5
indicates the diagram
Èy# 25
5
Ê cos asin" xb œ cos ! œ È1 x#
33. ! œ sin" x indicates the diagram
34. ! œ cos" x indicates the diagram
Ê tan acos" xb œ tan ! œ
35. ! œ tan" Èx# 2x indicates the diagram
È 1 x#
x
Ê sin Štan" Èx# 2x‹
Èx# 2x
x1
œ sin ! œ
36. ! œ tan"
x
È x# 1
37. ! œ sin"
2y
3
38. ! œ sin"
y
5
indicates the diagram
39. ! œ sec"
x
4
indicates the diagram
indicates the diagram
indicates the diagram
Ê sin Štan"
Ê cos ˆsin"
2y ‰
3
x
È x# 1 ‹
œ cos ! œ
Ê cos ˆsin" y5 ‰ œ cos ! œ
œ sin ! œ
È9 4y#
3
È25 y#
5
Ê sin ˆsec" x4 ‰ œ sin ! œ
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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x
È2x# 1
Èx# 16
x
459
460
Chapter 7 Transcendental Functions
40. ! œ sec"
È x# 4
x
È x# 4
‹
x
Ê sin Šsec"
indicates the diagram
œ sin ! œ
sin" x œ
1
#
42.
43. x lim
tan" x œ
Ä_
1
#
44. x Ä
lim
tan" x œ 1#
_
45. x lim
sec" x œ
Ä_
1
#
46. x Ä
lim
sec" x œ x Ä
lim
cos" ˆ "x ‰ œ
_
_
41.
lim
x Ä 1c
47. x lim
csc" x œ x lim
sin" ˆ "x ‰ œ 0
Ä_
Ä_
49. y œ cos" ax# b Ê
dy
dx
œ
51. y œ sin" È2t Ê
dy
dt
œ
2x
É 1 ax # b#
È2
#
Ê1 ŠÈ2t‹
53. y œ sec" (2s 1) Ê
dy
ds
54. y œ sec" 5s Ê
5
k5sk È(5s)# 1
dy
ds
œ
55. y œ csc" ax# 1b Ê
56. y œ csc" ˆ x# ‰ Ê
dy
dx
œ
œ
œ
2x
È 1 x%
50. y œ cos" ˆ "x ‰ œ sec" x Ê
œ
È2
È1 2t#
52. y œ sin" (1 t) Ê
#
57. y œ sec" ˆ "t ‰ œ cos" t Ê
œ
#
59. y œ cot" Èt œ cot" t"Î# Ê
œ
62. y œ tan" (ln x) Ê
dy
dx
œ
63. y œ csc" aet b Ê
dy
dt
œ
Š
"
‹
1 x#
tanc" x
œ
œ
ˆ "x ‰
œ
1 (ln x)#
œ
œ
œ
"
k2s 1k Ès# s
2x
ax# 1b Èx% 2x#
"
kx k É x
2
# 4 œ kx k È x # 4
4
ˆ 2t3 ‰
#
#
#
¹ t3 ¹ ÊŠ t3 ‹ 1
Š "# ‹ tc"Î#
1 at"Î# b
dy
dt
#
œ
œ
œ
2t
%
t# É t 9 9
œ
6
t Èt% 9
"
#Èt(1 t)
Š "# ‹ (t 1)c"Î#
1 c(t 1)"Î# d
#
œ
"
2Èt 1 (1 t 1)
œ
"
2tÈt 1
"
atan" xb a1x# b
œ
et
ke t k É a e t b # 1
1
#
"
x c1 (ln x)# d
œ
"
Èe2t 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
"
kx k È x # 1
"
È1 (1 t)#
"
È1 t#
dy
dt
dy
dt
œ
60. y œ cot" Èt 1 œ cot" (t 1)"Î# Ê
dy
dx
œ
2x
Ɉ x# ‰# 1
dy
dt
2
k2s 1k È4s# 4s
kx # 1 k É a x # 1 b # 1
58. y œ sin" ˆ t3# ‰ œ csc" Š t3 ‹ Ê
61. y œ ln atan" xb Ê
œ
dy
dt
dy
dx
"
ksk È25s# 1
œ
Š "# ‹
¸x¸
cos" x œ 1
48. x Ä
lim
csc" x œ x Ä
lim
sin" ˆ "x ‰ œ 0
_
_
2
k2s 1k È(2s 1)# 1
œ
dy
dx
lim
x Ä 1 b
2
È x# 4
œ
"
È2t t#
Section 7.7 Inverse Trigonometric Functions
64. y œ cos" aet b Ê
dy
dt
et
É1 aet b#
œ
œ
e t
È1 e2t
"Î#
65. y œ sÈ1 s# cos" s œ s a1 s# b cos" s Ê
s#
È 1 s#
œ È1 s# "
È 1 s#
œ È1 s# 66. y œ Ès# 1 sec" s œ as# 1b
œ
"Î#
s# 1
È 1 s#
sec" s Ê
dy
ds
œ
dy
dx
œ a1 s# b
1 s # s# 1
È 1 s#
s ˆ "# ‰ a1 s# b
"Î#
(2s) "Î#
(2s) "
ks k È s # 1
œ
s
È s# 1
"
x È x# 1
"
kx k È x # 1
œ sin" x 1
#
tan" ax" b tan" x Ê
x
È 1 x#
x
È 1 x#
dy
dx
œ
Š "# ‹ ax# 1b
"Î#
(2x)
#
1 ’ax# 1b"Î# “
"
kx k È x # 1
"Î#
Ê
dy
dx
œ0
xc#
1 ax" b#
œ sin" x x Š È
dy
dx
"
1 x#
"
‹
1 x#
œ
"
x# 1
ˆ #" ‰ a1 x# b
œ
dy
dx
2x
4
x#
tan" ˆ #x ‰ x –
Š "# ‹
1 ˆ #x ‰
#
—œ
2x
x# 4
tan" ˆ #x ‰ œ tan" ˆ x# ‰
71.
'È"
72.
'È
dx œ sin" ˆ x3 ‰ C
9 x#
"
1 4x#
"
#
œ
73.
' 17 " x
74.
' 9 "3x
75.
'
sin" u C œ
#
dx œ '
#
dx œ
"
È2
dx
xÈ5x# 4
œ
'01 È4 ds
4 s#
"
#
"
3
"
#
"
#
ŠÈ17‹ x#
'
œ'
dx
xÈ25x# 2
œ
' È1 2(2x)
"
#
dx œ
dx œ
dx œ
du
uÈ u# 4
' È du
1 u #
, where u œ 2x and du œ 2 dx
"
È17
tan"
"
3È 3
x
È17
C
tan" Š Èx3 ‹ C œ
È3
9
tan" Š Èx3 ‹ C
, where u œ 5x and du œ 5 dx
sec" ¹ Èu ¹ C œ
2
œ'
"
#
sin" (2x) C
"
#
ŠÈ3‹ x#
du
uÈ u# 2
dx œ
#
"
È2
5x
sec" ¹ È
¹C
2
, where u œ È5x and du œ È5 dx
sec" ¸ u# ¸ C œ
"
#
"
1 x#
sec" ¹
"
œ 4 sin" #s ‘ ! œ 4 ˆsin"
È5x
# ¹
"
#
C
sin" 0‰ œ 4 ˆ 16 0‰ œ
21
3
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
œ0
"Î#
œ sin" x
70. y œ ln ax# 4b x tan" ˆ x# ‰ Ê
77.
"
k sk È s # 1
œ 0, for x 1
69. y œ x sin" x È1 x# œ x sin" x a1 x# b
'
s ks k 1
ks k È s # 1
68. y œ cot" ˆ x" ‰ tan" x œ
76.
"
È 1 s#
2s#
È1 s#
œ
œ ˆ "# ‰ as# 1b
"Î#
67. y œ tan" Èx# 1 csc" x œ tan" ax# 1b csc" x Ê
œ
"Î#
2x
4 x#
(2x)
461
462
78.
'03
Chapter 7 Transcendental Functions
È2Î4
œ
ds
È9 4s#
"
#
'03
3
œ "# sin" u3 ‘ 0
79.
80.
'02 8 dt2t
"
È2
'02
œ ’ È"2 †
"
È8
#
'c22
œ
œ ’ È"3 †
81.
È2Î2
'cc1
È2Î2
È
"
#
È #
œ csec" kukd #
82.
È
È2
'c2Î32Î3 yÈ9ydy 1 œ '2
#
È #
' È1 34(rdr 1)
œ
84.
3
#
85.
' 2 (xdx 1)
œ
86.
œ
87.
#
"
3
"
œ
tan
"
#
"
#
†
#È$
#È$
3
#
du
È 4 u#
u
È2
1
8
"
4
Štan"
2È 2
È8
tan" 0‹ œ
"
4
atan" 1 tan" 0b œ
"
4
ˆ 14 0‰ œ
œ
"
#È3
’tan" È3 tan" ŠÈ3‹“ œ
"
#È3
13 ˆ 13 ‰‘ œ
' 1 duu
#
, where u œ 2y and du œ 2 dy; y œ 1 Ê u œ 2, y œ 1
4
1
3
È2
#
œ 11#
, where u œ 3y and du œ 3 dy; y œ 23 Ê u œ 2, y œ 1
4
1
3
Ê u œ È 2
È2
3
Ê u œ È 2
œ 11#
sin" 2(r 1) C
, where u œ r 1 and du œ dr
"
sec
¸ u# ¸
œ 2 '1
1
"
È2
tan" Š xÈ1 ‹ C
2
, where u œ 3x 1 and du œ 3 dx
uCœ
"
3
tan" (3x 1) C
du
, where u œ 2x uÈ u# 4
"
C œ 4 sec" ¸ 2x # 1 ¸ C
du
, where u œ x
uÈu# 25
C œ 5" sec" ¸ x5 3 ¸ C
du
1 u#
1
16
1
3È 3
, where u œ 2(r 1) and du œ 2 dr
Cœ
sec" ¸ u5 ¸ #
ˆ 14 0‰ œ
, where u œ x 1 and du œ dx
#
'c11ÎÎ22 12cos(sin) d)))
"
#
3È 2
#
C œ 6 sin" ˆ r# 1 ‰ C
' (x 3)È(xdx 3) 25 œ '
"
5
sin" 0‹ œ
Ê uœ
, where u œ È3t and du œ È3 dt; t œ 2 Ê u œ 2È3, t œ 2 Ê u œ 2È3
du
È 1 u#
du
2 u#
tan"
È2
#
œ sec" ¹È2¹ sec" k2k œ
#
œ
89.
œ'
œ
!
du
uÈ u# 1
' (2x 1)Èdx(2x 1) 4 œ "# '
œ
88.
sin" u#
"
È2
"
3
'
œ6'
#
#
' 1 (3xdx 1)
3
#
#È #
du
uÈ u# 1
sin" u C œ
' È4 6 (rdr 1)
œ6
œ
#
Šsin"
œ sec" ¹È2¹ sec" k2k œ
œ csec" kukd #
83.
u
È8 “
tan" u# “
È2
"
#
3È 2
4
, where u œ È2t and du œ È2 dt; t œ 0 Ê u œ 0, t œ 2 Ê u œ 2È2
du
4 u#
œ 'c2
dy
yÈ4y# 1
œ
tan"
, where u œ 2s and du œ 2 ds; s œ 0 Ê u œ 0, s œ
du
È 9 u#
du
8 u#
'c22È33
"
È3
œ
dt
4 3t#
È2
È2Î4
1 and du œ 2 dx
3 and du œ dx
, where u œ sin ) and du œ cos ) d); ) œ 1# Ê u œ ", ) œ
1
#
Êuœ"
"
œ c2 tan" ud " œ 2 atan" 1 tan" (1)b œ 2 14 ˆ 14 ‰‘ œ 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 7.7 Inverse Trigonometric Functions
90.
'11ÎÎ64
œ
91.
'0ln
œ 'È3
1
csc# x dx
1 (cot x)#
È3
"
c tan" ud È$
È3
œ '1
ex dx
1 e2x
È$
'1e
1Î%
' Èy1 dy y
"
#
œ
94.
œ tan" È3 tan" 1 œ
1
3
1Î4
du
1 u# , where u œ ln t and
1Î%
tan" ud ! œ 4 ˆtan" 14 tan" 0‰
œ
%
"
#
' È du
1 u#
#
#
1 u#
œ
Ê u œ È3 , x œ
1
4
du œ
œ
"
t
1
1#
dt; t œ 1 Ê u œ 0, t œ e1Î4 Ê u œ
œ 4 tan"
1
4
1
4
"
#
sin" y# C
, where u œ tan y and du œ sec# y dy
œ sin" u C œ sin" (tan y) C
95.
'È
96.
' È dx
97.
'01
dx
x# 4x 3
2x x#
œ'
œ'
6 dt
È3 2t t#
dx
È1 ax# 4x 4b
dx
È1 ax# 2x 1b
œ'
œ'
dx
È1 (x 2)#
dx
È1 (x 1)#
œ sin" (x 2) C
œ sin" (x 1) C
œ 6 '1 È4 at#dt 2t 1b œ 6 '1 È2# dt(t 1)# œ 6 sin" ˆ t # 1 ‰‘ "
0
0
!
œ 6 sin" ˆ "# ‰ sin" 0‘ œ 6 ˆ 16 0‰ œ 1
98.
'11Î2
6 dt
È3 4t 4t#
dt
1 ‰‘
œ 3'1Î2 È4 a4t2#dt 4t 1b œ 3 '1Î2 È2# 2 (2t
œ 3 sin" ˆ 2t #
"Î#
1)#
1
1
œ 3 sin" ˆ "# ‰ sin" 0‘ œ 3 ˆ 16 0‰ œ
99.
#
#
"
1
#
' y dy2y 5 œ ' 4 y dy 2y 1 œ ' # (ydy 1)
#
#
œ
"
#
tan" ˆ y # 1 ‰ C
100.
' y 6ydy 10 œ ' 1 ay dy 6y 9b œ ' 1 (ydy 3)
101.
'12 x 8 2xdx 2 œ 8'12 1 ax dx 2x 1b œ 8'12 1 (xdx 1)
#
#
#
#
#
œ tan" (y 3) C
#
#
œ 8 ctan" (x 1)d "
œ 8 atan" 1 tan" 0b œ 8 ˆ 14 0‰ œ 21
102.
'24
2 dx
x# 6x 10
"
œ 2 ctan
103.
'
œ 2'2
4
1
dx
1 ax# 6x 9b
tan" (1)d œ 2 14
dx
œ
(x 1)Èx# 2x
œ ' Èdu# ,
u u 1
"
œ sec
'
1
4
Ê uœ1
1
1#
, where u œ y# and du œ 2y dy
sin" u C œ
' Èsec1 ytandy y œ ' È du
1
3
1
6
, where u œ ex and du œ ex dx; x œ 0 Ê u œ 1, x œ ln È3 Ê u œ È3
œ 4'0
4 dt
t a1 ln# tb
œ c4
93.
where u œ cot x and du œ csc# x dx; x œ
œ tan" 1 tan" È3 œ 14 du
1 u#
œ ctan" ud "
92.
du
1 u# ,
œ 2'2
4
dx
1 (x 3)#
%
œ 2 ctan" (x 3)d #
ˆ 14 ‰‘ œ 1
dx
(x 1)Èx# 2x 1 1
œ'
dx
(x 1)È(x 1)# 1
where u œ x 1 and du œ dx
kuk C œ sec" kx 1k C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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463
464
104.
Chapter 7 Transcendental Functions
'
dx
(x 2)Èx# 4x 3
œ'
"
uÈ u# 1
"
œ sec
105.
sinc" x
' Èe
œ'
kuk C œ sec" kx 2k C
c" x
dx œ ' eu du, where u œ cos" x and du œ
1 x#
' aÈsinc xb
"
108.
u$
3
c" x
C
dx œ ' u# du, where u œ sin" x and du œ
Cœ
asin" xb
3
$
dx
È 1 x#
C
2
3
u$Î# C œ
2
3
atan" xb
' atanc" yb"a1 y#b dy œ '
Š
"
‹
1 y#
'
$Î#
Cœ
dy œ '
tan" y
"
œ ln kuk C œ ln ktan
110.
dx
È 1 x#
' È1tan "x# x dx œ ' u"Î# du, where u œ tan" x and du œ 1 dxx#
œ
109.
#
1 x#
œ
dx
È 1 x#
C
œ eu C œ ecos
107.
dx
(x 2)È(x 2)# 1
dx œ ' eu du, where u œ sin" x and du œ
1 x#
cos" x
' Èe
œ'
du, where u œ x 2 and du œ dx
œ eu C œ esin
106.
dx
(x 2)Èx# 4x 4 1
"
u
2
3
Éatan" xb$ C
du, where u œ tan" y and du œ
dy
1 y#
yk C
"
"
asinc" yb È1 y#
dy œ
' Ésin" y# dy œ ' u" du, where u œ sin" y and du œ È1dy y#
1
y
œ ln kuk C œ ln ksin" yk C
111.
'È22
sec# asec" xb
xÈ x# 1
dx œ '1Î4 sec# u du, where u œ sec" x and du œ
1Î3
1Î$
œ ctan ud 1Î4 œ tan
112.
'22ÎÈ3
cos asecc" xb
xÈ x# 1
114.
sinc" 5x
x
xÄ0
lim b
xÄ1
œ lim
È x# 1
sec" x
1
3
sin
ŠÈ
5
‹
1 25x#
xÄ0
œ lim b
xÄ1
1
2 tanc" 3x#
7x#
xÄ0
lim
œ lim
xÄ0
1
6
Š
14x
1
4
,xœ2 Ê uœ
1
3
dx
xÈ x# 1
;xœ
1
6
,xœ2 Ê uœ
1
3
œ È3 1
œ
2
È3
Ê uœ
È3 "
#
œ lim b
xÄ1
tanc" a2x" b
x"
12x
‹
1 9x%
; x œ È2 Ê u œ
œ5
"Î#
ax # 1 b
sec" x
115. x lim
x tan" ˆ 2x ‰ œ x lim
Ä_
Ä_
116.
1
4
1Î3
1Î$
lim
tan
dx œ '1Î6 cos u du, where u œ sec" x and du œ
œ csin ud 1Î' œ sin
113.
1
3
dx
xÈ x# 1
Š "# ‹ ax# 1b
Œ
œ x lim
Ä_
œ lim
6
"Î#
"
x
k k
È
Š
%
x Ä 0 7 a1 9x b
x# 1
(2x)
2x# ‹
1 4x#
x#
œ
œ lim b x kxk œ 1
xÄ1
œ x lim
Ä_
2
14x#
œ2
6
7
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 7.7 Inverse Trigonometric Functions
"
#
117. If y œ ln x œ Š x" x
1 x#
ln a1 x# b "
x a1 x # b
tanc" x
x
tan" x
x# ‹
C, then dy œ – x" dx œ
x
1 x#
Š
x
‹ tan" x
1 x#
x#
x a1 x# b x$ x atan" xba1 x# b
x # a1 x # b
dx œ
465
— dx
tan" x
x#
dx,
which verifies the formula
118. If y œ
x%
4
cos" 5x 5
4
'È
x%
1 25x#
%
dx, then dy œ ’x$ cos" 5x Š x4 ‹ Š È
5
‹
1 25x#
45 Š È
x%
‹“
1 25x#
dx
œ ax$ cos" 5xb dx, which verifies the formula
#
119. If y œ x asin" xb 2x 2È1 x# sin" x C, then
c"
#
dy œ ’asin" xb 2x asin xb 2 2x sin" x 2È1 x# Š
È 1 x#
È 1 x#
"
È 1 x # ‹“
#
dx œ asin" xb dx, which verifies
the formula
120. If y œ x ln aa# x# b 2x 2a tan" ˆ xa ‰ C, then dy œ –ln aa# x# b #
2x#
a# x#
2
2
#
1 Š x# ‹ —
dx
a
#
x
#
#
œ ’ln aa# x# b 2 Š aa# x# ‹ 2“ dx œ ln aa x b dx, which verifies the formula
121.
dy
dx
œ
"
È 1 x#
122.
dy
dx
œ
"
x# 1
Ê dy œ
dx
È 1 x#
Ê y œ sin" x C; x œ 0 and y œ 0 Ê 0 œ sin" 0 C Ê C œ 0 Ê y œ sin" x
1 Ê dy œ ˆ 1 " x# 1‰ dx Ê y œ tan" (x) x C; x œ 0 and y œ 1 Ê 1 œ tan" 0 0 C
Ê C œ 1 Ê y œ tan" (x) x 1
123.
dy
dx
œ
œ1
124.
dy
dx
œ
"
Ê dy œ Èdx#
Ê y œ sec"
xÈ x# 1
x x 1
13 œ 231 Ê y œ sec" (x) 231 , x 1
"
1 x#
2
È 1 x#
Ê dy œ Š 1 " x# kxk C; x œ 2 and y œ 1 Ê 1 œ sec" 2 C Ê C œ 1 sec" 2
2
È 1 x# ‹
dx Ê y œ tan" x 2 sin" x C; x œ 0 and y œ 2
Ê 2 œ tan" 0 2 sin" 0 C Ê C œ 2 Ê y œ tan" x 2 sin" x 2
125. The angle ! is the large angle between the wall and the right end of the blackboard minus the small angle
x ‰
between the left end of the blackboard and the wall Ê ! œ cot" ˆ 15
cot" ˆ x3 ‰ .
126. V œ 1'0 c2# (sec y)# d dy œ 1 c4y tan yd !
1Î3
1Î$
œ 1 Š 431 È3‹
127. V œ ˆ "3 ‰ 1r# h œ ˆ 3" ‰ 1(3 sin ))# (3 cos )) œ 91 acos ) cos$ )b, where 0 Ÿ ) Ÿ
œ ! Ê sin ) œ 0 or cos ) œ „
"
È3
1
#
Ê
dV
d)
œ 91(sin )) a1 3 cos# )b
Ê the critical points are: 0, cos" Š È"3 ‹ , and cos" Š È"3 ‹ ; but
cos" Š È"3 ‹ is not in the domain. When ) œ 0, we have a minimum and when ) œ cos" Š È"3 ‹ ¸ 54.7°, we
have a maximum volume.
‰
128. 65° (90° " ) (90° !) œ 180° Ê ! œ 65° " œ 65° tan" ˆ 21
50 ¸ 65° 22.78° ¸ 42.22°
129. Take each square as a unit square. From the diagram we have the following: the smallest angle ! has a
tangent of 1 Ê ! œ tan" 1; the middle angle " has a tangent of 2 Ê " œ tan" 2; and the largest angle #
has a tangent of 3 Ê # œ tan" 3. The sum of these three angles is 1 Ê ! " # œ 1
Ê tan" 1 tan" 2 tan" 3 œ 1.
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466
Chapter 7 Transcendental Functions
130. (a) From the symmetry of the diagram, we see that 1 sec" x is the vertical distance from the graph of
y œ sec" x to the line y œ 1 and this distance is the same as the height of y œ sec" x above the x-axis at
x; i.e., 1 sec" x œ sec" (x).
(b) cos" (x) œ 1 cos" x, where 1 Ÿ x Ÿ 1 Ê cos" ˆ "x ‰ œ 1 cos" ˆ "x ‰, where x 1 or x Ÿ 1
Ê sec" (x) œ 1 sec" x
131. sin" (1) cos" (1) œ
1
#
0œ
1
#
; sin" (0) cos" (0) œ 0 1
#
œ
1
#
; and sin" (1) cos" (1) œ 1# 1 œ 1# .
If x − ("ß 0) and x œ a, then sin" (x) cos" (x) œ sin" (a) cos" (a) œ sin" a a1 cos" ab
œ 1 asin" a cos" ab œ 1 1# œ 1# from Equations (3) and (4) in the text.
"
x
Ê tan ! œ x and tan " œ
132.
Ê
1
#
œ ! " œ tan" x tan"
"
x
.
133. (a) Defined; there is an angle whose tangent is 2.
(b) Not defined; there is no angle whose cosine is 2.
134. (a) Not defined; there is no angle whose cosecant is "# .
(b) Defined; there is an angle whose cosecant is 2.
135. (a) Not defined; there is no angle whose secant is 0.
(b) Not defined; there is no angle whose sine is È2.
136. (a) Defined; there is an angle whose cotangent is "# .
(b) Not defined; there is no angle whose cosine is 5.
x ‰
137. !(x) œ cot" ˆ 15
cot" ˆ x3 ‰ , x 0 Ê !w (x) œ
15
225 x#
3
9 x#
œ
15 a9 x# b 3 a225 x# b
a225 x# b a9 x# b
; solving
!w (x) œ 0 Ê 135 15x# 675 3x# œ 0 Ê x œ 3È5 ; !w (x) 0 when 0 x 3È5 and !w (x) 0 for
x 3È5 Ê there is a maximum at 3È5 ft from the front of the room
138. From the accompanying figure, ! " ) œ 1, cot ! œ
2x
Ê ) œ 1 cot" x cot" (2
1
1 (2 x)# a1 x# b
"
"
1 x# 1 (2 x)# œ a1 x# b c1 (2 x)# d
and cot " œ
Ê
œ
d)
dx
œ
4 4x
a1 x# b c1 (2 x)# d
; solving
d)
dx
œ 0 Ê x œ 1;
Ê at x œ 1 there is a maximum ) œ 1 cot
"
139. Yes, sin" x and cos" x differ by the constant
x
1
x)
d)
dx
0 for ! x 1 and
"
1 cot
(2 1) œ 1 d)
dx 0 for
1
1
1
4 4 œ #
1
#
140. Yes, the derivatives of y œ cos" x C and y œ cos" (x) C are both
141. csc" u œ
1
#
sec" u Ê
d
dx
acsc" ub œ
d
dx
x1
ˆ 1# sec" u‰ œ 0 du
dx
ku k È u # 1
"
È 1 x#
œ
du
dx
ku k È u # 1
, ku k 1
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Section 7.7 Inverse Trigonometric Functions
467
142. y œ tan" x Ê tan y œ x Ê
Ê asec# yb
œ
"
1 x#
dy
dx
œ1 Ê
d
d
dx (tan y) œ dx (x)
"
"
#
sec# y œ È
Š 1 x# ‹
œ
dy
dx
, as indicated by the triangle
143. f(x) œ sec x Ê f w (x) œ sec x tan x Ê
df "
dx ¹xœb
œ
"
df
dx ¹x œ f " abb
œ
"
secasec" bbtanasec" bb
Since the slope of sec" x is always positive, we the right sign by writing
144. cot" u œ
1
#
tan" u Ê
d
dx
acot" ub œ
d
dx
du
dx
ˆ 1# tan" u‰ œ 0 145. The functions f and g have the same derivative (for x
xœ
"
.
b Š„ È b # " ‹
"
.
lx l È x # "
du
1 u#
0), namely
d
"
dx sec
œ
œ 1 dxu#
"
Èx (x 1)
. The functions therefore differ
by a constant. To identify the constant we can set x equal to 0 in the equation f(x) œ g(x) C, obtaining
1‰
" È
sin" (1) œ 2 tan" (0) C Ê 1# œ 0 C Ê C œ 1# . For x 0, we have sin" ˆ xx x
1 œ 2 tan
146. The functions f and g have the same derivative for x 0, namely
"
1 x#
. The functions therefore differ by a
constant for x 0. To identify the constant we can set x equal to 1 in the equation f(x) œ g(x) C, obtaining
sin" Š È" ‹ œ tan" 1 C Ê
2
È3
1
4
œ
1
4
C Ê C œ 0. For x 0, we have sin"
È3
147. V œ 1 'cÈ3Î3 Š È1" x# ‹ dx œ 1 'È3Î3
#
œ 1 13 ˆ 16 ‰‘ œ
1Î2
"
È 1 x#
149. (a) A(x) œ
1
4
œ tan"
È3
dx œ 1 ctan" xd È3Î3 œ 1 ’tan" È3 tan" Š
"
x
.
È3
3 ‹“
1#
#
148. y œ È1 x# œ a1 x# b
œ 2 '0
"
1 x#
"
È x# 1
"Î#
Ê yw œ ˆ "# ‰ a1 x# b
"Î#
1Î2
dx œ 2 csin" xd 0 œ 2 ˆ 16 0‰ œ
(diameter)# œ
1
4
#
"
"
1 x#
1#
#
Š È
#
"
‹
“
#
1x
œ
"
1 x#
; L œ 'c1Î2 É1 ayw b# dx
1Î2
1
3
’ È1" x# Š È1" x# ‹“ œ
œ 1 ctan" xd " œ (1)(2) ˆ 14 ‰ œ
(b) A(x) œ (edge)# œ ’ È
(2x) Ê 1 ayw b# œ
4
1 x#
1
1 x#
Ê V œ 'a A(x) dx œ 'c1
b
1
1 dx
1 x#
Ê V œ 'a A(x) dx œ 'c1 14dxx#
b
1
"
œ 4 ctan" xd " œ 4 ctan" (1) tan" (1)d œ 4 14 ˆ 14 ‰‘ œ 21
150. (a) A(x) œ
1
4
È2Î2
œ 'cÈ2Î2
(b) A(x) œ
(diameter)# œ
1
È 1 x#
(diagonal)#
2
1
4
Š
2
%
È
1 x#
#
0‹ œ
1
4
È2Î2
ŠÈ
4
‹
1 x#
dx œ 1 csin" xd È2Î2 œ 1 ’sin" Š
œ
È2Î2
1
2
Š
2
%
È
1 x#
#
0‹ œ
2
È 1 x#
È2
# ‹
œ
1
È 1 x#
Ê V œ 'a A(x) dx
sin" Š
b
È2
# ‹“
œ 1 14 ˆ 14 ‰‘ œ
È2Î2
Ê V œ 'a A(x) dx œ 'cÈ2Î2
b
2
È 1 x#
1#
#
dx
œ 2 csin" xd È2Î2 œ 2 ˆ 14 † 2‰ œ 1
151. (a) sec" 1.5 œ cos"
(c) cot" 2 œ
1
#
tan
"
1.5
"
¸ 0.84107
2 ¸ 0.46365
152. (a) sec" (3) œ cos" ˆ "3 ‰ ¸ 1.91063
(c) cot
"
(2) œ
1
#
" ‰
(b) csc" (1.5) œ sin" ˆ 1.5
¸ 0.72973
tan
"
" ‰
(b) csc" 1.7 œ sin" ˆ 1.7
¸ 0.62887
(2) ¸ 2.67795
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1
#
.
468
Chapter 7 Transcendental Functions
153. (a) Domain: all real numbers except those having
the form 1# k1 where k is an integer.
Range: 1# y 1
#
(b) Domain: _ x _; Range: _ y _
The graph of y œ tan" (tan x) is periodic, the
graph of y œ tan atan" xb œ x for _ Ÿ x _.
154. (a) Domain: _ x _; Range: 1# Ÿ y Ÿ
1
#
(b) Domain: " Ÿ x Ÿ 1; Range: " Ÿ y Ÿ 1
The graph of y œ sin" (sin x) is periodic; the
graph of y œ sin asin" xb œ x for " Ÿ x Ÿ 1.
155. (a) Domain: _ x _; Range: 0 Ÿ y Ÿ 1
(b) Domain: 1 Ÿ x Ÿ 1; Range: " Ÿ y Ÿ 1
The graph of y œ cos" (cos x) is periodic; the
graph of y œ cos acos" xb œ x for " Ÿ x Ÿ 1.
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Section 7.7 Inverse Trigonometric Functions
156. Since the domain of sec" x is (_ß 1] ["ß _), we
have sec asec" xb œ x for kxk 1. The graph of
y œ sec asec" xb is the line y œ x with the open
line segment from ("ß ") to ("ß ") removed.
157. The graphs are identical for y œ 2 sin a2 tan" xb
œ 4 csin atan" xbd ccos atan" xbd œ 4 Š È
œ
4x
x# 1
x
‹ Š Èx#" 1 ‹
x# 1
from the triangle
158. The graphs are identical for y œ cos a2 sec" xb
œ cos# asec" xb sin# asec" xb œ
œ
2 x #
x#
"
x#
x# 1
x#
from the triangle
159. The values of f increase over the interval ["ß 1] because
f w 0, and the graph of f steepens as the values of f w
increase towards the ends of the interval. The graph of f
is concave down to the left of the origin where f ww 0,
and concave up to the right of the origin where f ww 0.
There is an inflection point at x œ 0 where f ww œ 0 and
f w has a local minimum value.
160. The values of f increase throughout the interval (_ß _)
because f w 0, and they increase most rapidly near the
origin where the values of f w are relatively large. The
graph of f is concave up to the left of the origin where
f ww 0, and concave down to the right of the origin
where f ww 0. There is an inflection point at x œ 0
where f ww œ 0 and f w has a local maximum value.
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469
470
Chapter 7 Transcendental Functions
7.8 HYPERBOLIC FUNCTIONS
#
1. sinh x œ 34 Ê cosh x œ È1 sinh# x œ É1 ˆ 34 ‰ œ É1 coth x œ
2. sinh x œ
sech x œ
3. cosh x œ
œ
8
17
"
tanh x
coth x œ
œ
, and csch x œ
4
5
Ê cosh x œ È1 sinh# x œ É1 4
3
"
cosh x
œ
3
5
, and csch x œ
"
sinh x
œ
16
9
"
sin x
œ É 25
9 œ
25
œ É 16
œ
5
4
œ
17
8
, sech x œ
5
3
, tanh x œ
"
tanh x
œ
5. 2 cosh (ln x) œ 2 Š e
6. sinh (2 ln x) œ
"
cosh x
, sech x œ
13
12
"
cosh x
ln x
ecln x
‹
#
e2 ln x ec2 ln x
#
œ
œ
5
13
œ eln x "
eln x
#
#
eln x eln x
#
œ
15
17
, and csch x œ
, and csch x œ
œ
œx
Šx# x"# ‹
#
7. cosh 5x sinh 5x œ
e5x e5x
#
e5x e5x
#
œ e5x
8. cosh 3x sinh 3x œ
e3x e3x
#
e3x e3x
#
œ e3x
ecx
#
e x e cx ‰ %
#
9. (sinh x cosh x)% œ ˆ e
x
œ 35 ,
sinh x
cosh x
œ
ˆ 43 ‰
ˆ 53 ‰
4
5
, coth x œ
"
tanh x
œ
5
4
8
15
, tanh x œ
sinh x
cosh x
œ
8 ‰
ˆ 15
ˆ 17
‰
15
œ
"
sinh x
œ
"
sinh x
œ
15
8
12
5
, tanh x œ
sinh x
cosh x
œ
ˆ 12
‰
5
ˆ 13
‰
5
œ
5
12
"
x
œ
x% "
#x#
œ aex b% œ e4x
10. ln (cosh x sinh x) ln (cosh x sinh x) œ ln acosh# x sinh# xb œ ln 1 œ 0
11. (a) sinh 2x œ sinh (x x) œ sinh x cosh x cosh x sinh x œ 2 sinh x cosh x
(b) cosh 2x œ cosh (x x) œ cosh x cosh x sinh x sin x œ cosh# x sinh# x
ecx ‰#
#
"
"
!
4 a4e b œ 4
12. cosh# x sinh# x œ ˆ e
œ
"
4
a2ex b a2ex b œ
13. y œ 6 sinh
14. y œ
"
#
x
3
Ê
dy
dx
x
ˆe
x
e cx ‰ #
#
16. y œ t# tanh
"
t
"
4
dy
dx
œ
"
#
caex ex b aex ex bd caex ex b aex ex bd
(4) œ 1
x
3
[cosh (2x 1)](2) œ cosh (2x 1)
15. y œ 2Èt tanh Èt œ 2t"Î# tanh t"Î# Ê
œ sech# Èt œ
œ 6 ˆcosh x3 ‰ ˆ "3 ‰ œ 2 cosh
sinh (2x 1) Ê
dy
dt
œ sech# ˆt"Î# ‰‘ ˆ "# t"Î# ‰ ˆ2t"Î# ‰ ˆtanh t"Î# ‰ ˆt"Î# ‰
tanh Èt
Èt
œ t# tanh t" Ê
17. y œ ln (sinh z) Ê
dy
dz
œ
cosh z
sinh z
dy
dt
œ csech# at" bd at# b at# b (2t) atanh t" b œ sech#
œ coth z
,
3
4
É 144
, x 0 Ê sinh x œ Ècosh# x 1 œ É 169
25 1 œ
25 œ
13
5
ˆ 34 ‰
ˆ 54 ‰
sinh x
cosh x
œ 34
#
"
tanh x
œ
, tanh x œ
17 ‰
289
64
, x 0 Ê sinh x œ Ècosh# x 1 œ Ɉ 15
1 œ É 225
1 œ É 225
œ
17
15
, coth x œ
4. cosh x œ
"
cosh x
œ 35 , sech x œ
9
16
18. y œ ln (cosh z) Ê
dy
dz
œ
"
t
sinh z
cosh z
2t tanh
œ tanh z
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"
t
12
13
,
Section 7.8 Hyperbolic Functions 471
19. y œ (sech ))(1 ln sech )) Ê
dy
d)
) tanh ) ‰
œ ˆ sech
(sech )) ( sech ) tanh ))(1 ln sech ))
sech )
dy
d)
) coth ) ‰
œ (csch )) ˆ csch
(1 ln csch ))( csch ) coth ))
csch )
œ sech ) tanh ) (sech ) tanh ))(1 ln sech )) œ (sech ) tanh ))[1 (1 ln sech ))]
œ (sech ) tanh ))(ln sech ))
20. y œ (csch ))(1 ln csch )) Ê
œ csch ) coth ) (1 ln csch ))(csch ) coth )) œ (csch ) coth ))(1 1 ln csch )) œ (csch ) coth ))(ln csch ))
21. y œ ln cosh v "
#
tanh# v Ê
"
#
coth# v Ê
dy
dv
œ
dy
dv
œ
sinh v
cosh v
ˆ "# ‰ (2 tanh v) asech# vb œ tanh v (tanh v) asech# vb
cosh v
sinh v
ˆ "# ‰ (2 coth v) a csch# vb œ coth v (coth v) acsch# vb
œ (tanh v) a1 sech# vb œ (tanh v) atanh# vb œ tanh$ v
22. y œ ln sinh v œ (coth v) a1 csch# vb œ (coth v) acoth# vb œ coth$ v
23. y œ ax# 1b sech (ln x) œ ax# 1b ˆ eln x 2ec ln x ‰ œ ax# 1b ˆ x 2xc" ‰ œ ax# 1b ˆ x#2x1 ‰ œ 2x Ê
dy
dx
œ2
2
4x
#
24. y œ a4x# 1b csch (ln 2x) œ a4x# 1b ˆ eln 2x 2e ln 2x ‰ œ a4x# 1b Š 2x (2x)
" ‹ œ a4x 1b ˆ 4x# 1 ‰
œ 4x Ê
dy
dx
œ4
25. y œ sinh" Èx œ sinh" ˆx"Î# ‰ Ê
dy
dx
Š "# ‹ xc"Î#
œ
É1 ax"Î# b#
26. y œ cosh" 2Èx 1 œ cosh" ˆ2(x 1)"Î# ‰ Ê
27. y œ (1 )) tanh" ) Ê
dy
d)
) # 2)
) # 2 )
œ
"
#È x È 1 x
dy
d)
œ
"
#Èx(1 x)
(2) Š "# ‹ (x 1)c"Î#
œ
Éc2(x 1)"Î# d# 1
œ (1 )) ˆ 1 " )# ‰ (1) tanh" ) œ
28. y œ a)# 2)b tanh" () 1) Ê
œ
dy
dx
œ
"
1)
"
Èx 1 È4x 3
œ
"
È4x# 7x 3
tanh" )
œ a)# 2)b ’ 1 ()" 1)# “ (2) 2) tanh" () 1)
(2) 2) tanh" () 1) œ (2) 2) tanh" () 1) 1
29. y œ (1 t) coth" Èt œ (1 t) coth" ˆt"Î# ‰ Ê
30. y œ a1 t# b coth" t Ê
dy
dt
dy
dt
œ (1 t) –
Š "# ‹ tc"Î#
1 at"Î# b
#
"
ˆt"Î# ‰ œ
"
È 1 x#
— (1) coth
"
#È t
coth" Èt
œ a1 t# b ˆ 1 " t# ‰ (2t) coth" t œ 1 2t coth" t
31. y œ cos" x x sech" x Ê
œ
dy
dx
"
È 1 x#
’x Š
"
‹
xÈ 1 x#
(1) sech" x“ œ
"
È 1 x#
sech" x
œ sech" x
32. y œ ln x È1 x# sech" x œ ln x a1 x# b
œ
"
x
a1 x# b
"
33. y œ csch
"Î#
Š
ˆ "# ‰) Ê
"
‹
xÈ 1 x#
ˆ "# ‰ a1 x# b
"Î#
"Î#
sech" x Ê
dy
dx
(2x) sech" x œ
"
x
"
x
x
È 1 x#
sech" x œ
)
dy
d)
œ
’ln Š "# ‹“ Š "# ‹
)
Š "# ‹
) #
"
Ë 1 ”Š # ‹
•
œ ln (1) ln (2)
œ
#)
Ê1 Š "# ‹
ln 2
#)
Ê1 Š "# ‹
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x
È 1 x#
sech" x
472
Chapter 7 Transcendental Functions
34. y œ csch" 2) Ê
œ
dy
d)
35. y œ sinh" (tan x) Ê
œ
dy
dx
36. y œ cosh" (sec x) Ê
(ln 2) 2)
2 ) É 1 a2 ) b
œ
dy
dx
ln 2
È1 2#)
œ
sec# x
Èsec# x
(sec x)(tan x)
Èsec# x 1
œ
(sec x)(tan x)
Ètan# x
(b) If y œ sin" (tanh x) C, then
x#
#
œ
sec# x
È1 (tan x)#
37. (a) If y œ tan" (sinh x) C, then
38. If y œ
#
dy
dx
dy
dx
œ
œ
œ
sec# x
ksec xk
œ
œ
ksec xk ksec xk
ksec xk
(sec x)(tan x)
ktan xk
œ ksec xk
œ sec x, 0 x 1
#
cosh x
cosh x
1 sinh# x œ cosh# x œ sech x, which verifies the formula
#
sech x
sech# x
È1 tanh# x œ sech x œ sech x, which verifies the formula
sech" x "# È1 x# C, then
œ x sech" x dy
dx
x#
#
Š
"
‹
xÈ 1 x#
2x
4È 1 x#
œ x sech" x,
which verifies the formula
x# "
#
39. If y œ
coth" x x
#
C, then
dy
dx
œ x coth" x Š x
#
" ˆ " ‰
# ‹ 1 x#
"
#
œ x coth" x, which verifies
the formula
40. If y œ x tanh" x "
#
ln a1 x# b C, then
dy
dx
œ tanh" x x ˆ 1 " x# ‰ #" ˆ 12xx# ‰ œ tanh" x, which verifies
the formula
41.
'
sinh 2x dx œ
œ
42.
'
sinh
x
5
cosh u
#
"
#
'
sinh u du, where u œ 2x and du œ 2 dx
Cœ
cosh 2x
#
C
dx œ 5 ' sinh u du, where u œ
œ 5 cosh u C œ 5 cosh
43.
'
44.
'
'
"
5
and du œ
4 cosh (3x ln 2) dx œ
tanh
x
7
4
3
sinh u C œ
dx œ 7 '
sinh u
cosh u
4
3
4
3
'
dx
C
6 cosh ˆ x# ln 3‰ dx œ 12 ' cosh u du, where u œ
œ 12 sinh u C œ 12 sinh ˆ x# ln 3‰ C
œ
45.
x
5
x
5
x
#
ln 3 and du œ
"
#
dx
cosh u du, where u œ 3x ln 2 and du œ 3 dx
sinh (3x ln 2) C
du, where u œ
x
7
and du œ
"
7
dx
œ 7 ln kcosh uk C" œ 7 ln ¸cosh x7 ¸ C" œ 7 ln ¹ e
xÎ7
exÎ7
¹
#
C" œ 7 ln ¸exÎ7 exÎ7 ¸ 7 ln 2 C"
œ 7 ln kexÎ7 exÎ7 k C
46.
'
coth
)
È3
d) œ È 3 '
cosh u
sinh u
du, where u œ
œ È3 ln ksinh uk C" œ È3 ln ¹sinh
œ È3 ln ¹e)Î
47.
'
È$
e)Î
È$
)
È3
and du œ
)
È3 ¹
d)
È3
C" œ È3 ln ¹ e
¹ È3 ln 2 C" œ È3 ln ¹e)Î
È3
È$ e)ÎÈ$
)Î
#
e)Î
È3
¹ C"
¹C
sech# ˆx "# ‰ dx œ ' sech# u du, where u œ ˆx "# ‰ and du œ dx
œ tanh u C œ tanh ˆx "# ‰ C
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Section 7.8 Hyperbolic Functions 473
48.
'
csch# (5 x) dx œ ' csch# u du, where u œ (5 x) and du œ dx
œ ( coth u) C œ coth u C œ coth (5 x) C
49.
'
dt œ 2 ' sech u tanh u du, where u œ Èt œ t"Î# and du œ
sech Èt tanh Èt
Èt
dt
2È t
œ 2( sech u) C œ 2 sech Èt C
50.
'
csch (ln t) coth (ln t)
t
dt œ ' csch u coth u du, where u œ ln t and du œ
dt
t
œ csch u C œ csch (ln t) C
51.
x
'lnln24 coth x dx œ 'lnln24 cosh
' 15Î8
sinh x dx œ 3Î4
"
u
"&Î)
¸
¸3¸
¸ 15 4 ¸
du œ cln kukd $Î% œ ln ¸ 15
8 ln 4 œ ln 8 † 3 œ ln
where u œ sinh x, du œ cosh x dx, the lower limit is sinh (ln 2) œ
limit is sinh (ln 4) œ
52.
eln 4 ec ln 4
#
œ
4 Š "4 ‹
#
œ
17Î8
sinh 2x
" '
'0ln 2 tanh 2x dx œ '0ln 2 cosh
2x dx œ # 1
eln 2 ec ln 2
#
œ
2 Š "# ‹
5
#
,
œ
3
4
and the upper
ln ˆ 17
‰
‘
8 ln 1 œ
"
#
ln
#
15
8
"
u
du œ
"
#
"(Î)
cln kukd "
"
#
œ
17
8
, where
u œ cosh 2x, du œ 2 sinh (2x) dx, the lower limit is cosh 0 œ 1 and the upper limit is cosh (2 ln 2) œ cosh (ln 4)
œ
53.
eln 4 ec ln 4
#
c2 ln 2
#
'0ln 2 4ec
)
ln 2‹ Š e
#
e2 ln 2
# ‹
ec)
‹
#
c ln 2
d) œ 'c ln 4 ae2) 1b d) œ e# )‘ c ln 4
"
ln 4‹ œ ˆ 8" ln 2‰ ˆ 32
ln 4‰ œ
Š0 )
e0
# ‹“
ec)
‹
#
ln 2
2)
3
32
ln 2 2 ln 2 œ
d) œ 2 '0 a1 ec2) b d) œ 2 ’) ln 2
œ 2 ˆln 2 "
8
"# ‰ œ 2 ln 2 "
4
3
3#
ln 2
ec#)
# “0
ln 2
1 œ ln 4 3
4
'c11ÎÎ44 cosh (tan )) sec# ) d) œ '11 cosh u du œ csinh ud "" œ sinh (1) sinh (1) œ Š e" #e" ‹ Š e" # e" ‹
e" , where u œ tan ), du œ sec# ) d), the lower limit is tan ˆ 14 ‰ œ 1 and the upper
'01Î2 2 sinh (sin )) cos ) d) œ 2'01 sinh u du
"
œee
œ 2 ccosh ud "! œ 2(cosh 1 cosh 0) œ 2 Š e #e
c"
1‹
2, where u œ sin ), du œ cos ) d), the lower limit is sin 0 œ 0 and the upper limit is sin ˆ 1# ‰ œ 1
'12 cosht(ln t) dt œ '0ln 2 cosh u du œ csinh ud ln0 2 œ sinh (ln 2) sinh (0) œ e
u œ ln t, du œ
58.
c2 ln 4
ln 2
e e" e" e
œe
#
1
limit is tan ˆ 4 ‰ œ 1
57.
17
8
sinh ) d) œ '0 4ec) Š e
œ
56.
œ
)
œ 2 ’Šln 2 55.
#
'cclnln42 2e) cosh ) d) œ 'cclnln42 2e) Š e
œ Še
54.
4 Š "4 ‹
œ
"
t
ln 2
ec ln 2
#
0œ
2
#
"
#
œ
, where
dt, the lower limit is ln 1 œ 0 and the upper limit is ln 2
2
Èx
#
"
'14 8 cosh
dx œ 16'1 cosh u du œ 16 csinh ud #" œ 16(sinh 2 sinh 1) œ 16 ’Š e #e ‹ Š e #e ‹“
Èx
#
œ 8 ae# e# e e" b , where u œ Èx œ x"Î# , du œ
"
#
x"Î# dx œ
dx
2È x
, the lower limit is È1 œ 1 and the upper
limit is È4 œ 2
59.
3
4
'c0ln 2 cosh# ˆ x# ‰ dx œ 'c0ln 2 cosh#x " dx œ "# 'c0ln 2 (cosh x 1) dx œ "# csinh x xd c0 ln 2
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474
60.
Chapter 7 Transcendental Functions
œ
"
#
[(sinh 0 0) (sinh ( ln 2) ln 2)] œ
œ
"
#
ˆ1 "
4
ln 2‰ œ
3
8
"
#
ln 2 œ
"
#
’(0 0) Š e
c ln 2 eln 2
#
ln 2‹“ œ
"
#
–
Š "# ‹ 2
#
ln 2—
ln È2
3
8
'0ln 10 4 sinh# ˆ x# ‰ dx œ '0ln 10 4 ˆ cosh#x 1 ‰ dx œ 2'0ln 10 (cosh x 1) dx œ 2 csinh x xd ln0 10
œ 2[(sinh (ln 10) ln 10) (sinh 0 0)] œ eln 10 ec ln 10 2 ln 10 œ 10 5
25
61. sinh" ˆ 1#5 ‰ œ ln Š 12
É 144
1‹ œ ln ˆ 32 ‰
63. tanh" ˆ "# ‰ œ
"
#
65. sech" ˆ 35 ‰ œ ln Š
67. (a)
'02
È3
dx
È 4 x#
"
10
2 ln 10 œ 9.9 2 ln 10
62. cosh" ˆ 53 ‰ œ ln Š 53 É 25
9 1‹ œ ln 3
(1/2)
ln 3
ln Š 11 (1/2) ‹ œ #
64. coth" ˆ 45 ‰ œ
1È1 (9/25)
‹
(3/5)
66. csch" Š È"3 ‹ œ ln È3 2
œ sinh" x# ‘ 0
œ ln 3
È3
"
#
ln Š (9/4)
(1/4) ‹ œ
"
#
ln 9 œ ln 3
È4/3
Š1/È3‹ œ ln ŠÈ3 2‹
œ sinh" È3 sinh 0 œ sinh" È3
(b) sinh" È3 œ ln ŠÈ3 È3 1‹ œ ln ŠÈ3 2‹
68. (a)
'01Î3 È 6 dx
œ 2'0
1
1 9x#
dx
È a# u# ,
"
sinh" ud !
œ c2
where u œ 3x, du œ 3 dx, a œ 1
œ 2 asinh" 1 sinh" 0b œ 2 sinh" 1
(b) 2 sinh" 1 œ 2 ln Š1 È1# 1‹ œ 2 ln Š1 È2‹
69. (a)
'52Î4
#
"
1 x#
dx œ ccoth" xd &Î% œ coth" 2 coth"
(b) coth" 2 coth"
70. (a)
'01Î2
71. (a)
'13ÎÎ513
"
#
"
#
œ
ln 3 ln ˆ 9/4
‰‘ œ
1/4
"Î#
"
1 x #
(b) tanh"
5
4
dx œ ctanh" xd !
"
#
œ
(1/2)
ln Š 11 (1/2) ‹ œ
dx
xÈ1 16x#
œ '4Î5
12Î13
œ tanh"
"
#
œ c sech" ud 4Î5 œ sech"
(b) sech"
12
13
sech"
œ ln Š 13 È169 144
‹
1#
œ ln ˆ2 † 23 ‰ œ ln
72. (a)
(b)
73. (a)
'12
"
#
dx
xÈ 4 x#
ˆcsch"
'01
"
3
tanh" 0 œ tanh"
"
#
where u œ 4x, du œ 4 dx, a œ 1
12
13
œ ln Š
4
5
ln
ln 3
du
,
u È a# u#
12Î13
"
#
"
#
5
4
sech"
4
5
1È1 (12/13)#
‹
(12/13)
ln Š 5 È25 16
‹
4
ln Š
1È1 (4/5)#
‹
(4/5)
œ ln ˆ 5 4 3 ‰ ln ˆ 1312 5 ‰ œ ln 2 ln
4
3
#
œ "# csch" ¸ x# ¸‘ " œ "# ˆcsch" 1 csch" "# ‰ œ
"
#
csch" 1‰ œ
cos x
È1 sin# x
3
#
dx œ '0
0
"
#
’ln Š2 "
È 1 u#
È5/4
(1/2) ‹
ln Š1 È2‹“ œ
"
#
ˆcsch"
"
#
"
#
5
ln Š 21 ‹
È2
csch" 1‰
È
!
du œ csinh" ud ! œ sinh" 0 sinh" 0 œ 0, where u œ sin x, du œ cos x dx
(b) sinh" 0 sinh" 0 œ ln Š0 È0 1‹ ln Š0 È0 1‹ œ 0
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Section 7.8 Hyperbolic Functions 475
74. (a)
'1e xÈ1 dx(ln x)
œ '0
1
#
du
È a# u#
, where u œ ln x, du œ
"
x
dx, a œ 1
"
œ csinh" ud ! œ sinh" 1 sinh" 0 œ sinh" 1
(b) sinh" 1 sinh" 0 œ ln Š1 È1# 1‹ ln Š0 È0# 1‹ œ ln Š1 È2‹
f(x) f(x)
. Then E(x) O(x) œ f(x) #f(x) f(x) #f(x)
#
f(x) f((x))
œ 2f(x)
œ f(x) #f(x) œ E(x) Ê E(x) is even, and
# œ f(x). Also, E(x) œ
#
O(x) œ f(x) #f((x)) œ f(x) #f(x) œ O(x) Ê O(x) is odd. Consequently, f(x) can
75. (a) Let E(x) œ
f(x) f(x)
#
and O(x) œ
be written as
a sum of an even and an odd function.
(b) f(x) œ
f(x) f(x)
#
because
Thus, if f is even f(x) œ
f(x) f(x)
œ
#
2f(x)
# 0 and
76. y œ sinh" x Ê x œ sinh y Ê x œ
Ê ey œ
2x „ È4x# 4
#
f(x) f(x)
#
2f(x)
#
0 if f is even and f(x) œ
if f is odd, f(x) œ 0 ey ecy
#
Ê 2x œ ey "
ey
because
f(x) f(x)
#
œ 0 if f is odd.
Ê 2xey œ e2y 1 Ê e2y 2xey 1 œ 0
Ê ey œ x Èx# 1 Ê sinh" x œ y œ ln Šx Èx# 1‹ . Since ey 0, we cannot
choose ey œ x Èx# 1 because x Èx# 1 0.
É gk
77. (a) v œ É mg
k tanhŒ
m t Ê
dv
dt
# É gk
# É gk
É gk
œ É mg
k ”sech Œ
m t•Œ
m œ g sech Œ
m t.
# É gk
# É gk
#
Thus m dv
dt œ mg sech Œ
m t œ mgŒ" tanh Œ
m t œ mg kv . Also, since tanh x œ ! when x œ !, v œ !
when t œ !.
(b)
mg
mg
mg
É kg
lim v œ lim É mg
lim tanh ΃ kg
k tanh Œ
m t œ É k t Ä
m t œ É k (1) œ É k
tÄ_
_
tÄ_
160
(c) É 0.005
œ É 160,000
œ
5
400
È5
œ 80È5 ¸ 178.89 ft/sec
78. (a) s(t) œ a cos kt b sin kt Ê
ds
dt
œ ak sin kt bk cos kt Ê
d# s
dt#
œ ak# cos kt bk# sin kt
œ k# (a cos kt b sin kt) œ k# s(t) Ê acceleration is proportional to s. The negative constant k#
implies that the acceleration is directed toward the origin.
(b) s(t) œ a cosh kt b sinh kt Ê
ds
dt
œ ak sinh kt bk cosh kt Ê
d# s
dt#
œ ak# cosh kt bk# sinh kt
œ k# (a cosh kt b sinh kt) œ k# s(t) Ê acceleration is proportional to s. The positive constant k# implies
that the acceleration is directed away from the origin.
79.
dy
dx
œ
"
xÈ 1 x#
x
È 1 x#
Ê yœ'
dx '
"
xÈ 1 x#
x
È 1 x#
dx Ê y œ sech" (x) È1 x# C; x œ 1 and
y œ 0 Ê C œ 0 Ê y œ sech" (x) È1 x#
80. To find the length of the curve: y œ
"
a
cosh ax Ê yw œ sinh ax Ê L œ '0 È1 (sinh ax)# dx
b
Ê L œ '0 cosh ax dx œ "a sinh ax‘ 0 œ
b
b
œ a"# sinh ax‘ 0 œ
b
"
a#
"
a
sinh ab. Then the area under the curve is A œ '0
b
sinh ab œ ˆ "a ‰ ˆ "a sinh ab‰ which is the area of the rectangle of height
"
a
"
a
and length L
as claimed.
81. V œ 1'0 acosh# x sinh# xb dx œ 1'0 1 dx œ 21
2
2
82. V œ 21 '0
ln
È3
sech# x dx œ 21 ctanh xd ln0
È3
È3 Š1/È3‹
œ 21 – È
3 Š1/È3‹ —
œ1
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cosh ax dx
476
Chapter 7 Transcendental Functions
83. y œ
"
#
cosh 2x Ê yw œ sinh 2x Ê L œ '0
ln
œ ’ "# Š e
2x
ec2x
‹“
#
0
ln
È5
"
4
œ
ˆ5 "5 ‰ œ
È5
È1 (sinh 2x)# dx œ '
0
ln
È5
cosh 2x dx œ "# sinh 2x‘ 0
ln
È5
6
5
84. (a) Let the point located at (cosh uß 0) be called T. Then A(u) œ area of the triangle ?OTP minus the area
under the curve y œ Èx# 1 from A to T Ê A(u) œ
"
#
(b) A(u) œ
œ
"
#
cosh u sinh u '1
cosh u
cosh# u (c) Aw (u) œ
85. y œ 4 cosh
x
4
"
#
"
#
Èx# 1 dx Ê Aw (u) œ
sinh# u sinh# u œ
Ê A(u) œ
u
#
"
#
"
#
'1cosh u Èx# 1 dx.
cosh u sinh u "
#
acosh# u sinh# ub ŠÈcosh# u 1‹ (sinh u)
acosh# u sinh# ub œ ˆ "# ‰ (1) œ
"
#
C, and from part (a) we have A(0) œ 0 Ê C œ 0 Ê A(u) œ
u
#
Ê u œ 2A
dy
# ˆx‰
# ˆx‰
'
Ê 1 Š dy
dx ‹ œ 1 sinh
4 œ cosh
4 ; the surface area is S œ c ln 16 21yÊ1 Š dx ‹ dx
#
#
ln 81
ln 81
œ 81 'c ln 16 cosh# ˆ x4 ‰ dx œ 41 'c ln 16 ˆ1 cosh #x ‰ dx œ 41 x 2 sinh x# ‘ c ln 16
œ 41 ˆln 81 2 sinh ˆ ln#81 ‰‰ ˆ ln 16 2 sinh ˆ ln# 16 ‰‰‘ œ 41 cln (81 † 16) 2 sinh (ln 9) 2 sinh (ln 4)d
15 ‰
œ 41 cln (9 † 4)# aeln 9 ec ln 9 b aeln 4 ec ln 4 bd œ 41 2 ln 36 ˆ9 "9 ‰ ˆ4 "4 ‰‘ œ 41 ˆ4 ln 6 80
9 4
ln 81
œ 41 ˆ4 ln 6 ln 81
320 135 ‰
36
œ 161 ln 6 4551
9
86. (a) y œ cosh x Ê ds œ È(dx)# (dy)# œ È(dx)# asinh# xb (dx)# œ cosh x dx; Mx œ 'c ln 2 y ds
ln 2
ln 2
ln 2
ln 2
œ ' cosh x ds œ '
cosh# x dx œ ' (cosh 2x 1) dx œ sinh# 2x x‘ œ "4 aeln 4 ec ln 4 b ln 2
ln 2
c ln 2
œ
15
16
c ln 2
0
0
ln 2
ln 2
ln 2; M œ 2 '0 È1 sinh# x dx œ 2 '0 cosh x dx œ 2 csinh xd ln0 2 œ eln 2 ec ln 2 œ 2 "# œ 3# .
Therefore, y œ
Mx
M
œ
ˆ 15
‰
16 ln 2
ˆ #3 ‰
œ
5
8
ln 4
3
, and by symmetry x œ 0.
(b) x œ 0, y ¸ 1.09
87. (a) y œ
H
w
‰
cosh ˆ w
H x Ê tan 9 œ
dy
dx
‰ w
ˆ w ‰‘ œ sinh ˆ w
‰
œ ˆH
w
H sinh H x
H x
(b) The tension at P is given by T cos 9 œ H Ê T œ H sec 9 œ HÈ1 tan# 9 œ HÉ1 ˆsinh
w
H
#
x‰
‰
ˆH‰
ˆw ‰
œ H cosh ˆ w
H x œ w w cosh H x œ wy
"
a
sinh" as; y œ
"
a
cosh ax œ
89. (a) Since the cable is 32 ft long, s œ 16 and x œ 15. From Exercise 88, x œ
"
a
sinh" as Ê 15a œ sinh" 16a
88. s œ
œ
"
a
sinh ax Ê sinh ax œ as Ê ax œ sinh" as Ê x œ
Èsinh# ax 1 œ
"
a
Èa# s# 1 œ És# "
a
"
a
Ècosh# ax
"
a#
Ê sinh 15a œ 16a.
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Chapter 7 Practice Exercises
(b) The intersection is near (0.042ß 0.672).
(c) Newton's method indicates that at a ¸ 0.0417525 the curves y œ 16a and y œ sinh 15a intersect.
"
‰ ¸ 47.90 lb
(d) T œ wy ¸ (2 lb) ˆ 0.0417525
(e) The sag is "a cosha"&ab "
a
¸ 4.85 ft.
CHAPTER 7 PRACTICE EXERCISES
1. y œ 10ecxÎ5 Ê
dy
dx
È2x Ê
2. y œ È2 e
3. y œ
"
4
"
16
xe4x œ (10) ˆ 5" ‰ exÎ5 œ 2exÎ5
È2x œ 2eÈ2x
œ ŠÈ2‹ ŠÈ2‹ e
dy
dx
dy
dx
œ
"
4
Ê
dy
dx
œ x# ca2x# b e2x d e2x (2x) œ (2 2x)e2x œ 2e2Îx (1 x)
e4x Ê
4. y œ x# ec2Îx œ x# e2x
"
cx a4e4x b e4x (1)d dy
d)
œ
2(sin ))(cos ))
sin# )
6. y œ ln asec# )b Ê
dy
d)
œ
2(sec ))(sec ) tan ))
sec# )
#
#
ln Š x# ‹
ln #
8. y œ log5 (3x 7) œ
9. y œ 8ct Ê
11. y œ 5x3Þ6 Ê
dy
dt
ln (3x7)
ln 5
dy
dx
œ
Ê
œ
"
ln #
dy
dx
2 cos )
sin )
"
"
œ 2 cot )
œ 2 tan )
Š x# ‹ œ
x
#
2
(ln 2)x
œ ˆ ln"5 ‰ ˆ 3x37 ‰ œ
œ 8ct (ln 8)(1) œ 8ct (ln 8)
3
(ln 5)(3x7)
10. y œ 92t Ê
dy
dt
œ 92t (ln 9)(2) œ 92t (2 ln 9)
œ 5(3.6)x2Þ6 œ 18x2Þ6
dy
dx
È2 Ê
12. y œ È2 x
Ê
a4e4x b œ xe4x 4" e4x 4" e4x œ xe4x
"
5. y œ ln asin# )b Ê
7. y œ log2 Š x# ‹ œ
"
16
dy
dx
È
È
œ ŠÈ2‹ ŠÈ2‹ xŠ 21‹ œ 2xŠ 21‹
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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477
478
Chapter 7 Transcendental Functions
13. y œ (x 2)xb2 Ê ln y œ ln (x 2)xb2 œ (x 2) ln (x 2) Ê
Ê
dy
dx
w
y
y
œ (x 2)xb2 cln (x 2) 1d
" ‰
œ (x 2) ˆ x#
(1) ln (x 2)
w
14. y œ 2(ln x)xÎ2 Ê ln y œ ln c2(ln x)xÎ2 d œ ln (2) ˆ x# ‰ ln (ln x) Ê
Ê yw œ # ln" x ˆ "# ‰ ln (ln x)‘ 2 (ln x)xÎ2 œ (ln x)xÎ2 ln (ln x) 15. y œ sin" È1 u# œ sin" a1 u# b
œ
u
uÈ 1 u#
"
È 1 u#
œ
Ê
dy
du
œ
"
#
a1 u # b
1 Î2
"Î#
#
u
È 1 u # È 1 a1 u # b
œ
“
u
ku k È 1 u #
œ
œ
dy
dv
" v$Î#
#
É1 av"Î# b#
"
2v$Î# È1 v"
œ
œ
"
1
2v$Î# É v v
È v
2v$Î# Èv 1
œ
"
2vÈv 1
17. y œ ln acos" xb Ê yw œ
Š È "
1 x#
cos" x
‹
œ
"
È1 x# cos" x
18. y œ z cos" z È1 z# œ z cos" z a1 z# b
"
œ cos
z
z
È 1 z#
"
œ cos
z
È 1 z#
"Î#
Ê
œ cos" z dy
dz
dy
dt
20. y œ a1 t# b cot" 2t Ê
œ 2t cot" 2t a1 t# b ˆ 1 24t# ‰
dy
dt
œ tan" t t ˆ 1 " t# ‰ ˆ #" ‰ ˆ "t ‰ œ tan" t 21. y œ z sec" z Èz# 1 œ z sec" z az# 1b
z
kz k È z # 1
"
sec
z
È z# 1
z
È 1 z#
zœ
ˆ "# ‰ a1 z# b
"Î#
(2z)
z
19. y œ t tan" t ˆ "# ‰ ln t Ê
œ
" ‘
ln x
(2u)
Ê1 ’a1 u# b
ˆ"‰
œ 0 ˆ x# ‰ ’ lnx x “ (ln (ln x)) ˆ "# ‰
,0u1
16. y œ sin" Š È"v ‹ œ sin" v"Î# Ê
œ
"Î#
y
y
1z
È z# 1
"Î#
"
sec
Ê
dy
dz
œ zŠ
t
1 t#
"
‹
kz k È z # 1
"
2t
asec" zb (1) #" az# 1b
"Î#
(2z)
z, z 1
22. y œ 2Èx 1 sec" Èx œ 2(x 1)"Î# sec" ˆx"Î# ‰
Ê
dy
dx
Š " ‹ x"Î#
œ 2 –ˆ "# ‰ (x 1)"Î# sec" ˆx"Î# ‰ (x 1)"Î# È# È
x
23. y œ csc" (sec )) Ê
24. y œ a1 x# b etan
25. y œ
2 ax# 1b
Ècos 2x
" x
dy
d)
œ
sec ) tan )
ksec )k Èsec# ) 1
Ê yw œ 2xetan
" x
tan" x
a1 x# b Š e1 x# ‹ œ 2xetan
#
2 ax # 1 b
Ècos 2x
"! 3x 4
"! 3x 4
É
26. y œ É
2x 4 Ê ln y œ ln
2x 4 œ
"
10
)
œ ktan
tan )k œ 1, 0 ) Ê ln y œ ln Š 2Èaxcos2x1b ‹ œ ln (2) ln ax# 1b Ê yw œ ˆ x#2x 1 tan 2x‰ y œ
Ê yw œ
x 1 —
ˆ 3x 3 4 " ‰
x2 y
"
#
œ 2Š
sec" Èx
2È x 1
"
#x ‹
œ
sec" Èx
Èx 1
" x
etan
" x
ln (cos 2x) Ê
cln (3x 4) ln (2x 4)d Ê
"! 3x 4
ˆ " ‰ ˆ 3x 3 4 œÉ
2x 4 10
"
x
1
#
yw
y
œ0
2x
x # 1
2 sin 2x)
ˆ #" ‰ (cos
2x
ˆ x#2x 1 tan 2x‰
"
10
yw
y
œ
"
10
ˆ 3x 3 4 2 ‰
2x 4
" ‰
x2
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Chapter 7 Practice Exercises
&
1)(t 1)
dy
"
27. y œ ’ (t(t 2)(t 3) “ Ê ln y œ 5 cln (t 1) ln (t 1) ln (t 2) ln (t 3)d Ê Š y ‹ Š dt ‹
œ 5 ˆ t " 1 28. y œ
Ê
"
t1
2u2u
Èu# 1 Ê ln y
dy
2u2u ˆ "
du œ Èu# 1 u
È)
29. y œ (sin ))
Ê
"
t#
dy
d)
" ‰
t3
Ê
dy
dt
&
1)(t 1)
ˆ t " 1 œ 5 ’ (t(t 2)(t 3) “
œ ln 2 ln u u ln 2 ln 2 "
#
"
t1
"
t#
ln au# 1b Ê Š "y ‹ Š dy
du ‹ œ
"
u
" ‰
t3
ln 2 #" ˆ u#2u 1 ‰
u ‰
u# 1
È) ˆ cos ) ‰ " )"Î# ln (sin ))
Ê ln y œ È) ln (sin )) Ê Š "y ‹ Š dy
d) ‹ œ
sin )
#
œ (sin ))
È)
ŠÈ) cot ) ln (sin ))
‹
2È )
30. y œ (ln x)1Îln x Ê ln y œ ˆ ln"x ‰ ln (ln x) Ê
w
y
y
œ ˆ ln"x ‰ ˆ ln"x ‰ ˆ x" ‰ ln (ln x) ’ (ln"x)# “ ˆ x" ‰
ln (ln x)
Ê yw œ (ln x)1Îln x ’ 1 x(ln
x)# “
31.
' ex sin aex b dx œ ' sin u du, where u œ ex and du œ ex dx
œ cos u C œ cos aex b C
32.
' et cos a3et 2b dt œ 3" ' cos u du, where u œ 3et 2 and du œ 3et dt
œ
33.
"
3
sin u C œ
"
3
sin a3et 2b C
' ex sec# aex 7b dx œ ' sec# u du, where u œ ex 7 and du œ ex dx
œ tan u C œ tan aex 7b C
34.
' ey csc aey 1b cot aey 1b dy œ ' csc u cot u du, where u œ ey 1 and du œ ey dy
œ csc u C œ csc aey 1b C
35.
' asec# xb etan x dx œ ' eu du, where u œ tan x and du œ sec# x dx
œ eu C œ etan x C
36.
' acsc# xb ecot x dx œ ' eu du, where u œ cot x and du œ csc# x dx
œ eu C œ ecot x C
' c1
37.
'c11
38.
'1e Èlnx x dx œ '01 u"Î# du, where u œ ln x, du œ "x dx; x œ 1
"
"
"
3x 4 dx œ 3 c7 u du, where u œ 3x 4, du œ 3 dx; x œ 1 Ê
"
"
ln 7
œ "3 cln kukd "
( œ 3 cln k1k ln k7kd œ 3 [0 ln 7] œ 3
œ
39.
23 u$Î# ‘ "
!
œ 23 1$Î# 23 0$Î# ‘ œ
u œ 7, x œ 1 Ê u œ 1
Ê u œ 0, x œ e Ê u œ 1
2
3
sin ˆ ‰
"
'01 tan ˆ x3 ‰ dx œ '01 cos
' 1Î2 "
ˆx‰
ˆx‰
ˆ ‰ dx œ 3 1 u du, where u œ cos 3 , du œ 3 sin 3 dx; x œ 0
x
3
x
3
Ê uœ
œ 3 cln
"Î#
k u kd "
œ 3 ln ¸ "# ¸ ln k1k‘ œ 3 ln
"
#
"
#
$
œ ln 2 œ ln 8
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Ê u œ 1, x œ 1
479
480
40.
Chapter 7 Transcendental Functions
'11ÎÎ64
2 cot 1x dx œ 2'1Î6
1Î4
cos 1x
sin 1x
dx œ
2
1
È2
'11ÎÎ2
"
u
du, where u œ sin 1x, du œ 1 cos 1x dx; x œ
Ê uœ
œ
41.
'04
È
cln kukd 11ÎÎ2 2 œ
2
1
c9
dt œ 'c25
2t
t# 25
"
u
2
1
’ln ¹ È"2 ¹ ln ¸ #" ¸“ œ
2
1
ln 1 "
#
"Î#
43.
'
"
u
#" ln 2‘ œ
1
6
œ ln ¸ "# ¸ ln k2k‘ œ ln 1 ln 2 ln 2 œ 2 ln 2 œ ln 4
sin u
cos u
du, where u œ ln v and du œ
"
v
dv
œ ln kcos uk C œ ln kcos (ln v)k C
44.
'
"
v ln v
dv œ '
"
u
du, where u œ ln v and du œ
"
v
dv
œ ln kuk C œ ln kln vk C
45.
'
(ln x)$
x
œ
46.
'
ln (x 5)
x5
œ
47.
dx œ ' u$ du, where u œ ln x and du œ
u#
#
'
"
r
"
#
C œ (ln x)
#
"
x
dx
C
dx œ ' u du, where u œ ln (x 5) and du œ
#
u
#
Cœ
cln (x5)d
2
#
"
x 5
dx
C
csc# (1 ln r) dr œ ' csc# u du, where u œ 1 ln r and du œ
"
r
dr
œ cot u C œ cot (1 ln r) C
48.
'
cos (1 ln v)
v
dv œ ' cos u du, where u œ 1 ln v and du œ "v dv
œ sin u C œ sin (1 ln v) C
49.
'
#
œ
50.
'
"
#
x3x dx œ
"
# ln 3
'
3u du, where u œ x# and du œ 2x dx
a3u b C œ
"
# ln 3
#
Š3x ‹ C
2tan x sec# x dx œ ' 2u du, where u œ tan x and du œ sec# x dx
œ
"
ln #
a2u b C œ
dx œ 3'1
2tan x
ln #
C
51.
'17
52.
"
"
È
'132 5x" dx œ "5 '132 x" dx œ 5" cln kxkd $#
" œ 5 aln 32 ln 1b œ 5 ln 32 œ ln Š 32‹ œ ln 2
3
x
7
"
x
ln 2
1
9
25
du, where u œ 1 sin t, du œ cos t dt; t œ 1# Ê u œ 2, t œ
dv œ ' tan u du œ '
tan (ln v)
v
2
1
du, where u œ t# 25, du œ 2t dt; t œ 0 Ê u œ 25, t œ 4 Ê u œ 9
'c11ÎÎ62 1 cossint t dt œ '21Î2
œ cln kukd #
Ê uœ
"
È2
ln 2 ln 1 ln 2‘ œ
œ cln kukd *
#& œ ln k9k ln k25k œ ln 9 ln 25 œ ln
42.
"
6
dx œ 3 cln kxkd (" œ 3 aln 7 ln 1b œ 3 ln 7
&
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Ê uœ
"
#
"
#
,xœ
"
4
Chapter 7 Practice Exercises
53.
15
'14 ˆ x8 #"x ‰ dx œ #" '14 ˆ 4" x x" ‰ dx œ #" 8" x# ln kxk‘ %" œ #" ˆ 168 ln 4‰ ˆ 8" ln 1‰‘ œ 16
#" ln 4
œ
54.
ln È4 œ
15
16
ˆln 8 'c0ln 2
e2w dw œ
'0ln 9 e
)
ˆln 8 21 ‰
#
œ
2
3
(ln 8) 7 œ ln ˆ8#Î$ ‰ 7 œ ln 4 7
"
#
Ê u œ 1, x œ 1 Ê u œ 0
œ ae! e" b œ e 1
"
#
'ln0Ð1Î4Ñ eu du, where u œ 2w, du œ 2 dw; w œ ln 2
"
#
ceu d 0ln Ð1Î4Ñ œ
ce! eln Ð1Î4Ñ d œ
"
#
ˆ1 4" ‰ œ
Ê u œ ln "4 , w œ 0 Ê u œ 0
3
8
2
3
u"Î# ‘ "'
%
œ ˆ16"Î# 4"Î# ‰ œ ˆ
2
3
2‰ ˆ"
3
4
"‰
#
œ ˆ
Ê u œ 4, r œ ln 5 Ê u œ 16
2‰ ˆ
4" ‰
3
œ
8
2
3
u$Î# ‘ ) œ
!
ˆ8$Î# 0$Î# ‰ œ
2
3
2
3
ˆ2*Î# 0‰ œ
2""Î#
3
œ
32È2
3
'1e "x (1 7 ln x)"Î$ dx œ 7" '18 u"Î$ du, where u œ 1 7 ln x, du œ 7x dx, x œ 1
'ee
#
"
6
ae) 1b"Î# d) œ '0 u"Î# du, where u œ e) 1, du œ e) d); ) œ 0 Ê u œ 0, ) œ ln 9 Ê u œ 8
œ
60.
2
3
'1ln 5 er a3er 1b$Î# dr œ "3 '416 u$Î# du, where u œ 3er 1, du œ 3er dr; r œ 0
œ
59.
12‰ œ
ceu d !"
œ
58.
3
#
'cc21 eÐx1Ñ dx œ '10 eu du, where u œ (x 1), du œ dx; x œ 2
œ
57.
ln 2
#
2
3
œ
56.
15
16
'18 ˆ 3x2 x8 ‰ dx œ 23 '18 ˆ "x 12x# ‰ dx œ 23 cln kxk 12x" d )" œ 23 ˆln 8 128 ‰ (ln 1 12)‘
œ
55.
481
"
xÈln x
3 #Î$ ‘ )
14 u
"
œ
3
14
3 ‰
ˆ8#Î$ 1#Î$ ‰ œ ˆ 14
(4 1) œ
dx œ 'e (ln x)"Î#
e#
"
x
Ê u œ 1, x œ e Ê u œ 8
9
14
dx œ '1 u"Î# du, where u œ ln x, du œ
2
"
x
dx; x œ e Ê u œ 1, x œ e# Ê u œ 2
#
œ 2 u"Î# ‘ " œ 2 ŠÈ2 1‹ œ 2È2 2
61.
'13 [ln (vv 11)]
#
dv œ '1 [ln (v 1)]#
3
"
v1
dv œ 'ln 2 u# du, where u œ ln (v 1), du œ
ln 4
"
v 1
dv;
v œ 1 Ê u œ ln 2, v œ 3 Ê u œ ln 4;
œ
62.
"
3
$ ln 4
cu d ln 2 œ
"
3
$
$
c(ln 4) (ln 2) d œ
"
3
$
$
c(2 ln 2) (ln 2) d œ
(ln 2)$
3
(8 1) œ
7
3
(ln 2)$
'24 (1 ln t)(t ln t) dt œ '24 (t ln t)(1 ln t) dt œ '24lnln24 u du, where u œ t ln t, du œ ˆ(t) ˆ "t ‰ (ln t)(1)‰ dt
œ (1 ln t) dt; t œ 2 Ê u œ 2 ln 2, t œ 4
Ê u œ 4 ln 4
œ
63.
cu# d 2 ln 2 œ
4 ln 4
"
#
c(4 ln 4)# (2 ln 2)# d œ
"
#
c(8 ln 2)# (2 ln 2)# d œ
(2 ln 2)#
#
(16 1) œ 30 (ln 2)#
'18 log) ) d) œ ln"4 '18 (ln )) ˆ ") ‰ d) œ ln"4 '0ln 8 u du, where u œ ln ), du œ ") d), ) œ 1
4
œ
64.
"
#
'1e
"
# ln 4
cu # d
8(ln 3)(log3 ))
)
ln 8
!
"
ln 16
œ
d) œ '1
c(ln 8)# 0# d œ
e
8(ln 3)(ln ))
)(ln 3)
#
#
#
(3 ln 2)
4 ln 2
œ
Ê u œ 0, ) œ 8 Ê u œ ln 8
9 ln 2
4
d) œ 8 '1 (ln )) ˆ ") ‰ d) œ 8'0 u du, where u œ ln ), du œ
e
1
"
)
d) ;
) œ 1 Ê u œ 0, ) œ e Ê u œ 1
œ
"
4 cu # d !
œ 4 a1 0 b œ 4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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482
65.
Chapter 7 Transcendental Functions
'c33ÎÎ44 È
6
9 4x#
3Î4
3Î2
dx œ 3 '3Î4 È3# 2 (2x)# dx œ 3'3Î2 È
"
3# u#
du, where u œ 2x, du œ 2 dx;
x œ 34 Ê u œ 3# , x œ
$Î#
Ê uœ
3
4
3
#
œ 3 sin" ˆ u3 ‰‘ $Î# œ 3 sin" ˆ "# ‰ sin" ˆ "# ‰‘ œ 3 16 ˆ 16 ‰‘ œ 3 ˆ 13 ‰ œ 1
66.
'c11ÎÎ55
6
È4 25x#
œ
67.
'c22
6
5
3
4 3t#
dx œ
"
sin
'11ÎÎ55
6
5
ˆ u2 ‰‘ "
"
dt œ È3 'c2
œ
2
6
5
"
sin
69.
'
"
3 t#
dt œ 'È3
3
"
yÈ4y# 1
œ sec
'
71.
'È2Î23Î3
24
yÈy# 16
#
È3
"
sin
" ‰‘
#
ˆ
È3
"
2# u#
È3
#
du, where u œ 5x, du œ 5 dx;
œ
6
5
16
ˆ
1 ‰‘
6
œ
ˆ 13 ‰
6
5
È3
#
dy œ '
"
uÈ u# 1
3
È3 œ
"
È3
Štan" È3 tan" 1‹ œ
"
yÈ y# 4#
3
'È2
dy œ 'È2Î3 k3yk È(3y)
# 1 dy œ
"
kyk È5y# 3
ˆ 13 14 ‰ œ
È 31
36
du, where u œ 2y and du œ 2 dy
dy œ 24 ˆ 4" sec" ¸ y4 ¸‰ C œ 6 sec" ¸ y4 ¸ C
2Î3
2
œ csec" ud È2 œ ’sec" 2 sec" È2“ œ
È È5
"
È3
1
È3
k2yk C
2
'cc2/È65Î
Ê uœ1
21
5
13 ˆ 13 ‰‘ œ
"
ku k È u # 1
du, where u œ 3y, du œ 3 dy;
yœ
72.
œ
1
5
du, where u œ È3t, du œ È3 dt;
’tan" ŠÈ3‹ tan" ŠÈ3‹“ œ
dt œ ’ È"3 tan" Š Èt 3 ‹“
kuk C œ sec
"
kyk È9y# 1
"
2# u#
t œ 2 Ê u œ 2È3, t œ 2 Ê u œ 2È3
2
(2y)È(2y)# 1
"
dy œ 24 '
ˆ "# ‰
dt œ È3 'c2È3
#
2# ŠÈ3t‹
ŠÈ3‹ t#
dy œ '
"
70.
"
'11 È
2
È3
2
'È33
6
5
x œ 15 Ê u œ 1, x œ
œ È3 "# tan" ˆ u2 ‰‘ c2È3 œ
68.
dx œ
5
È2# (5x)#
È6ÎÈ5
dy œ 'c2/È5
1
3
1
4
œ
#
#
È5y ÊŠÈ5y‹ ŠÈ3‹
Ê u œ È2, y œ
2
3
Ê uœ2
1
12
È6
È5
È2
3
dy œ 'c2
"
#
du,
uÊu# ŠÈ3‹
È
œ ’ È"3 sec" ¹ Èu3 ¹“
73.
'
dx œ '
"
È2x x#
cÈ6
c2
where u œ È5y, du œ È5 dy; y œ È25 Ê u œ 2, y œ È65 Ê u œ È6
œ
"
È1 ax# 2x 1b
1
È3
’sec" È2 sec"
dx œ '
"
È1 (x 1)#
2
È3 “
dx œ '
œ
"
È3
"
È 1 u#
ˆ 14 16 ‰ œ
"
È3
3121 21 ‘
1#
œ
1
12È3
œ
È 3 1
36
du, where u œ x 1 and
du œ dx
œ sin" u C œ sin" (x 1) C
74.
'È
"
x# 4x 1
dx œ '
"
È3 ax# 4x 4b
dx œ '
"
#
ÊŠÈ3‹ (x 2)#
dx œ '
"
#
du
ÊŠÈ3‹ u#
where u œ x 2 and du œ dx
"
œ sin
Š Èu3 ‹
"
C œ sin
Š xÈ32 ‹
C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Chapter 7 Practice Exercises
75.
'cc21 v 4v2 5 dv œ 2 'cc21 1 av " 4v 4b dv œ 2 'cc21 1 (v" 2)
#
#
"
œ 2 ctan
76.
77.
'c11
3
4v# 4v 4
dv œ
"
œ 2 atan
3
4
'c11
2u
’ È23 tan" Š È
‹“
3
œ
3
4
œ
È 31
4
'
"
ud !
"
(t 1)Èt# 2t 8
#
3
4
$Î#
"Î#
dt œ '
Šv#
1 tan
"
"
v 4" ‹
dv œ
0b œ
3
4
dv œ 2'0
1
"
1 u#
483
du,
2 ˆ 14
where u œ v 2, du œ dv; v œ 2 Ê u œ 0, v œ 1 Ê u œ 1
0‰ œ 1#
'c11
È3
Œ
#
"
#
Šv dv œ
#
"
#‹
3
4
'c31ÎÎ22
Œ
È3
#
"
du
#
u#
"
#
where u œ v , du œ dv; v œ 1 Ê u œ "# , v œ 1 Ê u œ
œ
È3
#
È3
#
’tan" È3 tan" Š È"3 ‹“ œ
"
(t 1)Èat# 2t 1b 9
dt œ '
13 ˆ 16 ‰‘ œ
dt œ '
"
(t 1)È(t 1)# 3#
"
uÈ u# 3#
È3
#
ˆ 261 16 ‰ œ
È3
#
†
3
#
1
#
du
where u œ t 1 and du œ dt
œ
78.
'
"
3
"
sec
"
(3t 1)È9t# 6t
¸ u3 ¸
Cœ
dt œ '
"
3
"
sec
¸ t31 ¸
C
"
(3t 1)Èa9t# 6t 1b 1
dt œ '
"
(3t 1)È(3t 1)# 1#
dt œ
"
3
'
"
uÈ u# 1
du
where u œ 3t 1 and du œ 3 dt
œ
"
3
sec" kuk C œ
"
3
sec" k3t 1k C
79. 3y œ 2y1 Ê ln 3y œ ln 2y1 Ê y(ln 3) œ (y 1) ln 2 Ê (ln 3 ln 2)y œ ln 2 Ê ˆln 3# ‰ y œ ln 2 Ê y œ
ln 2
ln Š 3# ‹
80. 4y œ 3y2 Ê ln 4y œ ln 3y2 Ê y ln 4 œ (y 2) ln 3 Ê 2 ln 3 œ (ln 3 ln 4)y Ê (ln 12)y œ 2 ln 3
9
Ê y œ lnln12
81. 9e2y œ x# Ê e2y œ
x#
9
#
#
Ê ln e2y œ ln Š x9 ‹ Ê 2y(ln e) œ ln Š x9 ‹ Ê y œ
82. 3y œ 3 ln x Ê ln 3y œ ln (3 ln x) Ê y ln 3 œ ln (3 ln x) Ê y œ
ln (3 ln x)
ln 3
œ
"
#
#
#
ln Š x9 ‹ œ ln É x9 œ ln ¸ x3 ¸ œ ln kxk ln 3
ln 3 ln (ln x)
ln 3
83. ln (y 1) œ x ln y Ê eln Ðy1Ñ œ eÐxln yÑ œ ex eln y Ê y 1 œ yex Ê y yex œ 1 Ê y a1 ex b œ 1
Ê y œ 1 " ex
84. ln (10 ln y) œ ln 5x Ê eln Ð10 ln yÑ œ eln 5x Ê 10 ln y œ 5x Ê ln y œ
85. The limit leads to the indeterminate form 00 : lim
10x 1
x
86. The limit leads to the indeterminate form 00 : lim
3) 1
)
87. The limit leads to the indeterminate form 00 : lim
2sin x 1
ex 1
xÄ0
)Ä0
xÄ0
)Ä0
2c sin x "
ex 1
xÄ0
89. The limit leads to the indeterminate form 00 : lim
(ln 3)3)
1
œ lim
xÄ0
88. The limit leads to the indeterminate form 00 : lim
5 5 cos x
x
x Ä 0 e x1
Ê eln y œ exÎ2 Ê y œ exÎ2
(ln 10)10x
1
œ lim
xÄ0
œ lim
x
#
œ lim
xÄ0
œ lim
xÄ0
œ ln 10
œ ln 3
2sin x (ln 2)(cos x)
ex
œ ln 2
2c sin x (ln 2)( cos x)
ex
5 sin x
ex 1
œ lim
xÄ0
œ ln 2
5 cos x
ex
œ5
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484
Chapter 7 Transcendental Functions
4 4ex
xex
90. The limit leads to the indeterminate form 00 : lim
xÄ0
91. The limit leads to the indeterminate form 00 :
t ln (1 2t)
t#
lim tÄ!
sin# (1x)
92. The limit leads to the indeterminate form 00 : lim
xc4
x Ä 4 e 3x
œ lim
xÄ4
1 sin (21x)
exc4 1
21# cos (21x)
exc4
œ lim
xÄ4
94. The limit leads to the indeterminate form
yÄ!
eyc"
_
_:
indeterminate form
_
_:
tÄ!
lim e1Îy ln y œ
yÄ!
ln a1 3xc" b
x"
lim b
Š
xÄ!
œ x lim
Ä_
x
Š ln
ln 2 ‹
x
Š ln
ln 3 ‹
3x# ‹
1 3x"
xc#
œ x lim
Ä_
x#
x # 1
œ x lim
Ä_
ln 3
ln 2
œ x lim
Ä_
ˆ x ‰
xex
ex
100
œ x lim
xex œ x lim
Ä _ 100x
Ä _ 100
x
tan" x œ _ Ê faster
(e) x lim
Ä_
csc" x
Š "x ‹
œ x lim
Ä_
(f) x lim
Ä_
sinh x
ex
œxÄ
lim_
$
#
lim 10x ex 2x
xÄ_
(f) x lim
Ä_
tanc" Š "x ‹
Š "x ‹
sinc" Š "x ‹
"
‹
x#
sech x
ecx
yÄ!
œ
lim yÄ!
sin" ax" b
x"
aex ecx b
#e x
2x
#x
œ1
y"
" ˆy# ‰
e y
œ
#
lim 30x ex 4x
xÄ_
œ x lim
Ä_
œ
tan" ax" b
x"
" "
lim sin x#ax b
xÄ_
œ x lim
Ä_
Š ex b2ex ‹
e cx
œ
the limit leads to the
œ0 Ê
ln a1 3xc" b
;
x"
the limit leads to the
ˆ1 x3 ‰x œ lim eln fÐxÑ œ e! œ 1
x Ä !b
lim
x Ä !b
Ê same rate
ln 3
ln #
œ x lim
" œ 1 Ê same rate
Ä_
œ _ Ê faster
œ x lim
Ä_
œxÄ
lim_
œ
ln a1 3x" b
;
x"
lim ln f(x) œ lim x Ä !
xÄ!
3x
x3
Šx# ‹
#
Ê1 Šx" ‹
x#
"ec2x
#
3cx
ˆ 23 ‰x œ 0 Ê slower
2cx œ x lim
Ä_
ln 2x
ln 2 ln x
ˆ ln 2
œ x lim
ln x# œ x lim
Ä _ 2 (ln x)
Ä _ # ln x
Š
ln y
eyc"
lim et
1
ˆ1 x3 ‰x œ lim eln fÐxÑ œ e$
œ 3 Ê x lim
Ä_
xÄ_
3
œ lim b
xÄ!
(c) x lim
Ä_
(d) x lim
Ä_
(e) x lim
Ä_
tÄ!
Ê x lim
ln f(x) œ x lim
Ä_
Ä_
x Ä _ 1 3x
x
x Š "x ‹
(d) x lim
Ä_
t
tÄ!
(b) x lim
Ä_
(c)
œ _
21(sin 1x)(cos 1x)
exc4 1
œ lim
xÄ4
t
x
96. Let f(x) œ ˆ1 3x ‰ Ê ln f(x) œ x ln ˆ1 3x ‰ Ê
98. (a) x lim
Ä_
(b) x lim
Ä_
2t
lim Š et "t ‹ œ lim Š e "
t ‹ œ lim #
Š 3x " ‹
1 3x
œ lim
c
x #
indeterminate form 00 : x lim
Ä_
log2 x
log3 x
Š1 1 b2 2t ‹
lim
t Ä !
œ0
y
x
95. Let f(x) œ ˆ1 3x ‰ Ê ln f(x) œ
97. (a) x lim
Ä_
œ
œ 4
œ 21 #
93. The limit leads to the indeterminate form 00 :
œ lim 4ex
ex xex
œ lim
xÄ0
œ
"
#
x#
"
œ 1 Ê same rate
Ê same rate
œ x lim
Ä_
#
Š x # ‹
1x
"
Ê 1 Š x# ‹
Ê same rate
"# ‰ œ
lim 60xex4
xÄ_
œ x lim
Ä_
"
#
œ x lim
Ä_
60
ex
œ x lim
Ä_
œ 0 Ê slower
"
1 "#
œ 1 Ê same rate
x
x#
œ x lim
Ä_
œ x lim
Ä_
É1 ˆx" ‰# 2x$
2
e c x ae x e c x b
œ x lim
Ä_
x
2É1 x"#
œ _ Ê faster
ˆ 2 ‰ œ 2 Ê same rate
œ x lim
Ä _ 1 ec2x
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Chapter 7 Practice Exercises
99. (a)
(b)
" "‹
x%
œ 1 x"# Ÿ 2 for x sufficiently large Ê true
Š "# ‹
x
Š
x#
Š
"
x#
Š
"
‹
x%
"
‹
x%
œ x# 1 M for any positive integer M whenever x ÈM Ê false
(c) x lim
Ä_
x
x ln x
œ x lim
Ä_
(d) x lim
Ä_
ln (ln x)
ln x
œ x lim
Ä_
(e)
(f)
tan" x
1
cosh x
ex
Š
100. (a)
Š "#
x
Ÿ
œ
Š
x
"
%‹
df
dx
"
‹
x%
œ 0 Ê grows slower Ê true
1) œ 1 if x 0 Ê true
ˆ " ‰ œ 0 Ê true
œ x lim
Ä _ x# 1
Š "x ‹
ln x
œ0 Ê
x 1 œ x lim
Ä_ 1
ln 2
ln x 1 Ÿ 1 1 œ 2 if x
secc" x
œ
1
sinh x
"
ex œ #
"
ln x
œ x lim
Ä_
ˆ "x ‰
Ÿ 1 if x 0 Ê true
cos" Š " ‹
(f)
101.
"
x # 1
œ
x
(b) x lim
Ä _ Š "# (e)
œ 1 Ê the same growth rate Ê false
"
x
"
Šx‹
– ln x —
1
# for all x Ê true
"
c2x b Ÿ " (1 # a1 e
#
"
‹
x%
"% ‹
x
(c) x lim
Ä_
(d) lnln2x
x œ
"
1
x
1
ˆ 1# ‰
1
Ÿ
a1 ec2x b Ÿ
"
#
102. y œ f(x) Ê y œ 1 x œ fÐln 2Ñ
"
x
Ê
f af " (x)b œ 1 "
"
Šx
1‹
f w (x) œ x"# Ê
df "
dx ¹ fÐxÑ
2 Ê true
if x 1 Ê true
if x 0 Ê true
c"
œ ex 1 Ê Š dfdx ‹
1
#
œ
true
"
x
œ
"
"
Ê Š dfdx ‹
df
Š dx
‹
x œ ln 2
œy1 Ê xœ
œ 1 (x 1) œ x;
œ
"
y 1
df c"
dx ¹ fÐxÑ
œ
x œ fÐln 2Ñ
œ
"
aex 1bx œ ln 2
Ê f " (x) œ
"
(x 1)# ¹ fÐxÑ
œ
"
x 1
œ
"
#1
œ
; f " (f(x)) œ
"
’Š1 x" ‹1“
#
"
3
"
Š1 "x ‹1
œ x# ;
"
f w (x)
2 ‰
103. y œ x ln 2x x Ê yw œ x ˆ 2x
ln (2x) 1 œ ln 2x;
solving yw œ 0 Ê x œ
x
and
"
#
"
#
; yw 0 for x Ê relative minimum of "#
f ˆ #e ‰
"
#
and yw 0 for
at x œ "# ; f ˆ #"e ‰ œ "e
œ 0 Ê absolute minimum is "# at x œ
the absolute maximum is 0 at x œ
"
#
and
e
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
œ
"
Š x" ‹
œ x and
485
486
Chapter 7 Transcendental Functions
104. y œ 10x(2 ln x) Ê yw œ 10(2 ln x) 10x ˆ "x ‰
œ 20 10 ln x 10 œ 10(1 ln x); solving yw œ 0
Ê x œ e; yw 0 for x e and yw 0 for x e
Ê relative maximum at x œ e of 10e; y ! on Ð!ß e# Ó and
y ae# b œ 10e# (2 2 ln e) œ 0 Ê absolute minimum is 0
at x œ e# and the absolute maximum is 10e at x œ e
105. A œ '1
e
dx œ '0 2u du œ cu# d ! œ 1, where
1
2 ln x
x
"
x
u œ ln x and du œ
106. (a) A" œ '10
20
"
x
(b) A" œ 'ka
kb
"
dx; x œ 1 Ê u œ 0, x œ e Ê u œ 1
dx œ cln kxkd #!
"! œ ln 20 ln 10 œ ln
20
10
œ ln 2, and A# œ '1
kb
dx œ cln kxkd ka
œ ln kb ln ka œ ln
kb
ka
œ ln
"
x
2
b
a
"
x
dx œ cln kxkd #" œ ln 2 ln 1 œ ln 2
œ ln b ln a, and A# œ 'a
b
"
x
dx œ cln kxkd ab
œ ln b ln a
107. y œ ln x Ê
dy
dx
108. y œ 9ecxÎ3 Ê
Ê
dx ¸
dt xœ9
œ
dy
dx
"
x
Š "4 ‹ É9 œ
Š
xœ
"
È2
;
dA
dx
dA
dx
œ
Ê
dy dx
dx dt
œ
dx
dt
"
4
œ
(dy/dt)
(dy/dx)
œ ˆ "x ‰ Èx œ
Ê
dx
dt
œ
"
Èx
Ê
Š "4 ‹ È9 y
3exÎ3
œ
dy
dt ¹ e#
"
e
ln x
x
and dA
dx
#
Ê
m/sec
; x œ 9 Ê y œ 9e$
Èe$ Èe$ 1 ¸ 5 ft/sec
#
"
È2
and
units long by y œ e"Î# œ
0 for x e
dy
dt
œ ecx (x)(2x) ecx œ ecx a1 2x# b . Solving
0 for x 110. A œ xy œ x ˆ lnx#x ‰ œ
dA
dx
9
e$
3
‹
e$
#
"
È2
dy
dt
œ 3exÎ3 ;
109. A œ xy œ xecx Ê
Ê xœ
;
dA
dx
œ
"
Èe
"
x#
dA
dx
#
0 for 0 x "
È2
dA
dx
œ 0 Ê 1 2x# œ 0
Ê absolute maximum of
"
È2
e"Î# œ
"
È2e
at
units high.
ln x
x#
œ
1ln x
x#
. Solving
dA
dx
0 for x e Ê absolute maximum of
œ 0 Ê 1 ln x œ 0 Ê x œ e;
ln e
e
œ
"
e
at x œ e units long and y œ
high.
111. K œ ln (5x) ln (3x) œ ln 5 ln x ln 3 ln x œ ln 5 ln 3 œ ln
5
3
112. (a) No, there are two intersections: one at x œ 2
and the other at x œ 4
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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"
e#
units
Chapter 7 Practice Exercises
(b) Yes, because there is only one intersection
113.
log4 x
log2 x
œ
x
Š ln
ln 4 ‹
x
Š ln
ln # ‹
114. (a) f(x) œ
œ
ln 2
ln x
ln x
ln 4
†
ln 2
ln x
, g(x) œ
œ
ln 2
ln 4
œ
ln 2
2 ln 2
œ
"
#
ln x
ln #
(b) f is negative when g is negative, positive when g is
positive, and undefined when g œ 0; the values of f
decrease as those of g increase
115. (a) y œ
Ê yw œ
ln x
Èx
"
xÈ x
ln x
2x$Î#
œ
2 ln x
2xÈx
Ê yww œ 34 x&Î# (2 ln x) "# x&Î# œ x&Î# ˆ 34 ln x 2‰ ;
solving yw œ 0 Ê ln x œ 2 Ê x œ e# ; yw 0 for x e# and
and yw 0 for x e# Ê a maximum of 2e ; yww œ 0
Ê ln x œ 83 Ê x œ e)Î$ ; the curve is concave down on
ˆ0ß e)Î$ ‰ and concave up on ˆe)Î$ ß _‰; so there is an
inflection point at ˆe)Î$ ß
#
) ‰
.
$e%Î$
x #
(b) y œ ex Ê yw œ 2xe
#
Ê yww œ 2ex 4x# ex
#
#
œ a4x# 2bex ; solving yw œ 0 Ê x œ 0; yw 0 for
x 0 and yw 0 for x 0 Ê a maximum at x œ 0 of
e! œ 1; there are points of inflection at x œ „ È"2 ; the
curve is concave down for È"2 x "
È2
and concave
up otherwise.
(c) y œ (1 x) ecx Ê yw œ ecx (1 x) ecx œ xecx
Ê yww œ ecx xecx œ (x 1) ecx ; solving yw œ 0
Ê xecx œ 0 Ê x œ 0; yw 0 for x 0 and yw 0
for x 0 Ê a maximum at x œ 0 of (1 0) e! œ 1;
there is a point of inflection at x œ 1 and the curve is
concave up for x 1 and concave down for x 1.
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487
488
Chapter 7 Transcendental Functions
116. y œ x ln x Ê yw œ ln x x ˆ "x ‰ œ ln x 1; solving yw œ 0
Ê ln x 1 œ 0 Ê ln x œ 1 Ê x œ e" ; yw 0 for
x e" and yw 0 for x e" Ê a minimum of e" ln e"
œ "e at x œ e" . This minimum is an absolute minimum
since yww œ
"
x
is positive for all x !.
117. Since the half life is 5700 years and A(t) œ A! ekt we have
Ê kœ
ln (0.5)
5700
Ê ln (0.1) œ
A!
#
"
#
œ A! e5700k Ê
œ e5700k Ê ln (0.5) œ 5700k
ln Ð0Þ5Ñ
. With 10% of the original carbon-14 remaining we have 0.1A! œ A! e 5700
ln (0.5)
5700
t Ê tœ
(5700) ln (0.1)
ln (0.5)
t
ln Ð0Þ5Ñ
Ê 0.1 œ e 5700
t
¸ 18,935 years (rounded to the nearest year).
118. T Ts œ (To Ts ) eckt Ê 180 40 œ (220 40) eckÎ4 , time in hours, Ê k œ 4 ln ˆ 79 ‰ œ 4 ln ˆ 97 ‰ Ê 70 40
œ (220 40) ec4 ln Ð9Î7Ñ t Ê t œ
ln 6
4 ln ˆ 97 ‰
¸ 1.78 hr ¸ 107 min, the total time Ê the time it took to cool from
180° F to 70° F was 107 15 œ 92 min
x ‰
119. ) œ 1 cot" ˆ 60
cot" ˆ 53 œ 30 ’ 60# 2 x# "
30# (50 x)# “ ;
x ‰
30 ,
solving
100 20È17 is not in the domain;
d)
dx
0 x 50 Ê
d)
dx
d)
dx
œ
1
Š 60
‹
x ‰#
1 ˆ 60
Š
"
30 ‹
cx‹
1 Š 5030
#
œ 0 Ê x# 200x 3200 œ 0 Ê x œ 100 „ 20È17, but
d)
dx
0 for x 20 Š5 È17‹ and
0 for 20 Š5 È17‹ x 50
Ê x œ 20 Š5 È17‹ ¸ 17.54 m maximizes )
120. v œ x# ln ˆ "x ‰ œ x# (ln 1 ln x) œ x# ln x Ê
Ê 2 ln x 1 œ 0 Ê ln x œ maximum at x œ e1Î2 ;
r
h
dv
dx
Ê x œ ec1Î2 ;
"
#
œ 2x ln x x# ˆ "x ‰ œ x(2 ln x 1); solving
dv
dx
0 for x e1Î2 and
œ x and r œ 1 Ê h œ e1Î2 œ Èe ¸ 1.65 cm
dv
dx
dv
dx
œ0
0 for x e1Î2 Ê a relative
CHAPTER 7 ADDITIONAL AND ADVANCED EXERCISES
lim '0
b
1.
bÄ1
2. x lim
Ä_
"
x
"
È 1 x#
dx œ lim csin" xd 0 œ lim asin" b sin" 0b œ lim asin" b 0b œ lim sin" b œ
bÄ1
bÄ1
bÄ1
bÄ1
b
'0x tan" t dt œ x lim
Ä_
œ x lim
Ä_
tan" x
1
œ
" x"Î# sec# Èx
#
" x"Î#
#
4. y œ ax ex b2Îx Ê ln y œ
x 2Îx
Ê x lim
ax e b
Ä_
ˆ " 5. x lim
Ä _ n1
"
n#
x
_
ˆ_
form‰
1
#
1Îx
3. y œ ˆcos Èx‰ Ê ln y œ
œ "# lim b
xÄ!
'0x tanc" t dt
"
x
ln ˆcos Èx‰
x
ln ˆcos Èx‰ and lim b
xÄ!
œ "# Ê
lim
x Ä !b
ˆcos Èx‰1Îx œ e1Î2 œ
2 ln ax ex b
x
Ê x lim
ln y œ x lim
Ä_
Ä_
y
#
œ x lim
e
œ
e
Ä_
á " ‰
#n
œ lim b
xÄ!
2 a1 e x b
x ex
sin Èx
2Èx cos Èx
œ
"
#
lim
x Ä !b
tan Èx
Èx
"
Èe
œ x lim
Ä_
2ex
1 ex
œ x lim
Ä_
2ex
ex
œ2
1
ˆ"‰
œ x lim
ˆ n" ‰ – 1 " — á ˆ n" ‰ – 1 " —
Ä _ n – 1 Š"‹ —
12Š ‹
1nŠ ‹
n
n
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n
1
#
Chapter 7 Additional and Advanced Exercises
which can be interpreted as a Riemann sum with partitioning ?x œ
œ '0
1
6. x lim
Ä_
ˆ " Ê x lim
Ä _ n1
"
n
Ê x lim
Ä_
"
n
t
(c)
8. (a)
1
t
lim A(t) œ lim a1 ect b œ 1
tÄ_
tÄ_
V(t)
t Ä _ A(t)
limb V(t)
A(t)
lim
œ lim
tÄ_
œ limb
tÄ!
tÄ!
1
#
ˆ1 ec2t ‰
1 ect
1
#
œ
ˆ1 ec2t ‰
1 ect
1
#
ln 2
ln a
œ 0;
lim loga 2 œ lim c
aÄ1
ln 2
ln a
œ _;
lim loga 2 œ lim b
aÄ1
ln 2
ln 1
œ _;
lim loga 2 œ a lim
Ä_
ln 2
ln a
œ0
a Ä 1b
aÄ_
9. A" œ '1
e
2 log2 x
x
#
e
x)
œ ’ (ln
2 ln # “ œ
1
dx œ
"
# ln 2
2
ln 2
"
1 x#
#
a 1 e ct b a 1 e c t b
a1 e ct b
œ limb
tÄ!
1
#
1
#
a1 ec2t b
a1 ect b œ 1
(b)
'1e lnxx dx œ ’ (lnlnx)2 “ e œ ln"# ; A# œ '1e 2 log4
#
1
4
x
dx œ
2
ln 4
'1e lnxx dx
Ê A" : A# œ 2 : 1
10. y œ tan" x tan" ˆ "x ‰ Ê yw œ
"
1 x#
1
œ limb
tÄ!
lim loga 2 œ lim b
aÄ!
a Ä !b
a Ä 1c
œ
" ‰
#n
ce1În e2În á ed œ '0 ex dx œ cex d "! œ e 1
t
7. A(t) œ '0 ecx dx œ cecx d t0 œ 1 ect , V(t) œ 1'0 ec2x dx œ 1# ec2x ‘ 0 œ
(b)
á ˆ n" ‰ eÐ1ÎnÑ ˆ n" ‰ e2Ð1ÎnÑ á ˆ n" ‰ enÐ1ÎnÑ ‘ which can be interpreted as a
ce1În e2În á ed œ x lim
Ä_
Riemann sum with partitioning ?x œ
(a)
"
n#
dx œ cln (1 x)d "! œ ln 2
"
1x
"
n
"
n
489
"
1 x#
Š
"
‹
x#
Š1 x"# ‹
œ 0 Ê tan" x tan" ˆ x" ‰ is a constant
and the constant is 1# for x 0; it is 1# for x 0 since
tan" x tan" ˆ "x ‰ is odd. Next the
lim
x Ä !b
tan" x tan" ˆ "x ‰‘ œ ! 1
#
œ
1
#
and
lim
x Ä !c
ˆtan" x tan" ˆ x" ‰‰ œ 0 ˆ 1# ‰ œ 1#
11. ln xax b œ xx ln x and ln axx bx œ x ln xx œ x# ln x; then, xx ln x œ x# ln x Ê axx x# bln x œ ! Ê xx œ x# or ln x œ !Þ
x
ln x œ ! Ê x œ "; xx œ x# Ê x ln x œ 2 ln x Ê x œ 2. Therefore, xax b œ axx bx when x œ 2 or x œ ".
x
12. In the interval 1 x 21 the function sin x 0
Ê (sin x)sin x is not defined for all values in that
interval or its translation by 21.
13. f(x) œ egÐxÑ Ê f w (x) œ egÐxÑ gw (x), where gw (x) œ
x
1 x%
Ê f w (2) œ e! ˆ 1 2 16 ‰ œ
2
17
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490
Chapter 7 Transcendental Functions
14. (a)
df
dx
œ
2 ln ex
ex
(b) f(0) œ '1
1
(c)
df
dx
† ex œ 2x
2 ln t
t
dt œ 0
œ 2x Ê f(x) œ x# C; f(0) œ 0 Ê C œ 0 Ê f(x) œ x# Ê the graph of f(x) is a parabola
15. Triangle ABD is an isosceles right triangle with its right angle at B and an angle of measure
1
4
therefore have
tan" 3"
œ nDAB œ nDAE nCAB œ
tan" #"
1
4
at A. We
.
16. (a) The figure shows that lne e ln11 Ê 1 ln e e ln 1 Ê ln e1 ln 1e Ê e1 1e
(b) y œ lnxx Ê yw œ ˆ "x ‰ ˆ x" ‰ lnx#x Ê 1 x#ln x ; solving yw œ 0 Ê ln x œ 1 Ê x œ e; yw 0 for x e and
yw 0 for 0 x e Ê an absolute maximum occurs at x œ e
17. The area of the shaded region is '0 sin" x dx œ '0 sin" y dy, which is the same as the area of the region to
1
1
the left of the curve y œ sin x (and part of the rectangle formed by the coordinate axes and dashed lines y œ 1,
x œ 1# ) . The area of the rectangle is
1
#
1
#
œ '0 sin" x dx '0 sin x dx Ê
1Î2
1
œ '0 sin" y dy '0
1Î2
1
'0
1Î2
sin x dx œ
18. (a) slope of L$ slope of L# slope of L" Ê
"
b
1
#
sin x dx, so we have
'0 sin" x dx.
1
ln b ln a
ba
"
a
(b) area of small (shaded) rectangle area under curve area of large rectangle
"
b
Ê
(b a) 'a
b
"
x
dx "
a
"
b
(b a) Ê
ln b ln a
ba
"
a
19. (a) g(x) h(x) œ 0 Ê g(x) œ h(x); also g(x) h(x) œ 0 Ê g(x) h(x) œ 0 Ê g(x) h(x) œ 0
Ê g(x) œ h(x); therefore h(x) œ h(x) Ê h(x) œ 0 Ê g(x) œ 0
(b)
f(x) f(x)
#
f(x) f(x)
#
œ
œ
cfE (x) fO (x)d b fE (x) fO (x)‘
œ fE (x) fO (x) # fE (x) fO (x) œ fE (x);
#
cfE (x) fO (x)d cfE (x) fO (x)d
œ fE (x) fO (x) # fE (x) fO (x) œ fO (x)
#
(c) Part b Ê such a decomposition is unique.
20. (a) g(0 0) œ
g(0) g(0)
1 g(0) g(0)
#
Ê c1 g# (0)d g(0) œ 2g(0) Ê g(0) g$ (0) œ 2g(0) Ê g$ (0) g(0) œ 0
Ê g(0) cg (0) 1d œ 0 Ê g(0) œ 0
g(x) b g(h)
’
“ g(x)
g(x h) g(x)
g(x) g# (x) g(h)
œ lim 1 c g(x) g(h)
œ lim g(x) g(h)
h
h
h
c
1
g(x) g(h)d
hÄ0
hÄ0
hÄ0
g(h)
1 g# (x)
#
#
lim ’ h “ ’ 1 g(x) g(h) “ œ 1 † c1 g (x)d œ 1 g (x) œ 1 [g(x)]#
hÄ0
(b) gw (x) œ lim
œ
(c)
dy
dx
œ 1 y# Ê
dy
1 y#
"
Ê C œ 0 Ê tan
œ dx Ê tan" y œ x C Ê tan" (g(x)) œ x C; g(0) œ 0 Ê tan" 0 œ 0 C
(g(x)) œ x Ê g(x) œ tan x
21. M œ '0
1
œ
"
2
"
xd ! œ 1#
1 x# dx œ 2 ctan
ln 2
ln 4
ˆ 1 ‰ œ 1 ; y œ 0 by symmetry
#
and My œ '0
1
2x
1 x#
"
dx œ cln a1 x# bd ! œ ln 2 Ê x œ
My
M
"
22. (a) V œ 1 '1Î4 Š #È
‹ dx œ
x
1
4
'14Î4 x" dx œ 14 cln kxkd %"Î% œ 14 ˆln 4 ln 4" ‰ œ 14 ln 16 œ 14 ln a2% b œ 1 ln 2
"
(b) My œ '1Î4 x Š #È
‹ dx œ
x
1
2
63
'14Î4 x"Î# dx œ "3 x$Î# ‘ %"Î% œ ˆ 83 24" ‰ œ 64#4 1 œ 24
;
4
4
#
"
"
Mx œ '1Î4 "# Š #È
‹ Š 2È
‹ dx œ
x
x
4
1
8
'14Î4
"
x
%
dx œ 8" ln kxk‘ "Î% œ
"
8
ln 16 œ
"
#
ln 2;
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Chapter 7 Additional and Advanced Exercises
M œ '1Î4
4
yœ
Mx
M
"
#È x
dx œ '1Î4 "2 x"Î# dx œ x"Î# ‘ "Î% œ 2 %
4
œ ˆ "# ln 2‰ ˆ 32 ‰ œ
"
#
œ
3
#
; therefore, x œ
ds
dt
œ ks Ê
ds
s
63 ‰ ˆ 2 ‰
œ ˆ 24
3 œ
œ
21
1#
ln 2
3
23. A(t) œ A! ert ; A(t) œ 2A! Ê 2A! œ A! ert Ê ert œ 2 Ê rt œ ln 2 Ê t œ
24.
My
M
ln 2
r
Ê t¸
.7
r
œ
70
100r
œ
70
(r%)
œ k dt Ê ln s œ kt C Ê s œ s! ekt
Ê the 14th century model of free fall was exponential;
note that the motion starts too slowly at first and then
becomes too fast after about 7 seconds
25. (a) L œ k ˆ a bR%cot ) b csc ) ‰
r%
Ê
dL
d)
#
œ k Š b csc
R%
)
b csc ) cot )
‹;
r%
solving
dL
d)
œ0
Ê r% b csc# ) bR% csc ) cot ) œ 0 Ê (b csc )) ar% csc ) R% cot )b œ 0; but b csc ) Á 0 since
)Á
1
#
Ê r% csc ) R% cot ) œ 0 Ê cos ) œ
r%
R%
%
Ê ) œ cos" Š Rr % ‹ , the critical value of )
%
(b) ) œ cos" ˆ 56 ‰ ¸ cos" (0.48225) ¸ 61°
26. Two views of the graph of y œ 1000 1 (.99)x "x ‘ are shown below.
(a) At about x œ 11 there is a minimum
(b) There is no maximum; however, the curve is asymptotic to y œ 1000. The curve is near 1000 when
x 643.
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7
4
and
491
492
Chapter 7 Transcendental Functions
NOTES:
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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CHAPTER 8 TECHNIQUES OF INTEGRATION
8.1 BASIC INTEGRATION FORMULAS
u œ 8x# 1
Ä
du œ 16x dx •
1.
' È16x dx
2.
' È3 cos x dx
3.
'
4.
' cot$ y csc# y dy; ”
5.
'01 8x16xdx2 ; Ô
6.
'11ÎÎ43
7.
8x# 1
;”
1 3 sin x
' Èduu œ 2Èu C œ 2È1 3 sin x C
u œ 1 3 sin x
Ä
du œ 3 cos x dx •
u œ sin v
Ä
du œ cos v dv •
'
3Èu du œ 3 † 23 u$Î# C œ 2(sin v)$Î# C
u œ cot y
Ä
du œ csc# y dy •
'
u$ ( du) œ u4 C œ
3Èsin v cos v dv; ”
#
'
;”
' Èduu œ 2Èu C œ 2È8x# 1 C
%
cot% y
4
C
u œ 8x# 2
×
10
"!
Ä '2 du
du œ 16x dx
u œ cln kukd # œ ln 10 ln 2 œ ln 5
Õ x œ 0 Ê u œ 2, x œ 1 Ê u œ 10 Ø
sec# z dz
tan z
Ô
;Ö
dx
È x ˆÈ x 1 ‰
Õz œ
1
4
u œ tan z
×
Ù Ä
du œ sec# z dz
1
È
Ê u œ 1, z œ 3 Ê u œ 3 Ø
Ô u œ Èx " ×
"
Ö
Ù
; Ö du œ #Èx dx Ù Ä
dx
Õ 2 du œ Èx Ø
8.
' x dxÈx œ '
9.
' cot (3 7x) dx; ” u œ 3 7x •
Ä "7
10.
' csc (1x 1) dx; ” u œ 1x 1 •
Ä
du œ 1 dx
"
u
È
du œ cln kukd 1 3 œ ln È3 ln 1 œ ln È3
' 2 udu œ 2 ln kuk C œ 2 ln ˆÈx 1‰ C
Ô u œ Èx 1 ×
Ö du œ " dx Ù
dx
Ù Ä
#È x
È x ˆÈ x 1 ‰ ; Ö
dx
Õ 2 du œ Èx Ø
du œ 7 dx
È3
'1
' 2 udu œ 2 ln kuk C œ 2 ln ¸Èx 1¸ C
' cot u du œ 7" ln ksin uk C œ 7" ln ksin (3 7x)k C
' csc u † du1 œ "1 ln kcsc u cot uk C
œ 1" ln kcsc (1x 1) cot (1x 1)k C
11.
' e) csc ˆe) 1‰ d); ” u œ e
12.
' cot (3 x ln x) dx; ” u œ 3 dxln x •
1
Ä
du œ e) d) •
)
du œ
Ä
' csc u du œ ln kcsc u cot uk C œ ln ¸csc ˆe) 1‰ cot ˆe) 1‰¸ C
' cot u du œ ln ksin uk C œ ln ksin (3 ln x)k C
x
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494
Chapter 8 Techniques of Integration
u œ 3t
Ä
du œ dt3 —
13.
' sec 3t dt; –
' 3 sec u du œ 3 ln ksec u tan uk C œ 3 ln ¸sec 3t tan 3t ¸ C
14.
' x sec ax# 5b dx; ” u œ x
#
' "# sec u du œ "# ln ksec u tan uk C
5
Ä
du œ 2x dx •
œ
"
#
ln ksec ax# 5b tan ax# 5bk C
15.
' csc (s 1) ds; ” u œ s 1 •
16.
' )"
du œ ds
csc
#
Èln 2
"
)
d); –
u œ ")
Ä
du œ )d# ) —
' csc u du œ ln kcsc u cot uk C œ ln ¸csc ") cot ") ¸ C
u œ x#
×
ln 2
du œ 2x dx
Ä '0 eu du œ ceu d ln0 2 œ eln 2 e! œ 2 1 œ 1
2xe dx;
Õ x œ 0 Ê u œ 0, x œ Èln 2 Ê u œ ln 2 Ø
Ô
17.
'0
18.
'11Î2 sin (y) ecos y dy; Ô
19.
' etan v sec# v dv; ”
20.
' eÈÈtdt ; –
21.
' 3x1 dx; ” u œ x 1 •
22.
' 2x
23.
' 2È#Èwdw ; –
24.
' 102) d); ”
25.
' 1 9du9u
26.
4 dx
' 1 (2x
1)
27.
'01Î6 È dx
x#
Õy œ
t
u œ Èt
— Ä
du œ 2dt
Èt
u œ ln x
• Ä
du œ dx
x
dx; ”
w
#
;”
1
#
Ä
u œ 2)
Ä
du œ 2 d) •
1 9x#
;
' 3u du œ ˆ ln"3 ‰ 3u C œ 3lnÐ 3 Ñ C
x 1
' 2u du œ ln2 # C œ 2ln # C
u œ Èw
— Ä
du œ #dw
Èw
;”
' eu du œ eu C œ etan v C
' 2eu du œ 2eu C œ 2eÈt C
x œ 3u
Ä
dx œ 3 du •
#
u œ cos y
×
c1
0
du œ sin y dy
Ä '0 eu du œ 'c1 eu du œ ceu d !" œ 1 e" œ
Ê u œ 0, y œ 1 Ê u œ 1 Ø
u œ tan v
Ä
du œ sec# v dv •
du œ dx
ln x
' csc u du œ ln kcsc u cot uk C œ ln kcsc (s 1) cot (s 1)k C
Ä
u
ln x
' 2u du œ ln2 # C œ 2lnÈ#
u
w
C
' "# 10u du œ # 10ln 10 C œ "# Š ln1010 ‹ C
2)
u
' 13dxx
u œ 2x 1
Ä
du œ 2 dx •
#
œ 3 tan" x C œ 3 tan" 3u C
' 12duu
#
œ 2 tan" u C œ 2 tan" (2x 1) C
u œ 3x
du œ 3 dx
Õ x œ 0 Ê u œ 0, x œ " Ê u œ
6
Ô
×
"
#
Ø
Ä '0
1Î2
"
du
3 È 1 u#
"Î#
œ 3" sin" u‘ !
œ
"
3
ˆ 16 0‰ œ
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1
18
e"
e
Section 8.1 Basic Integration Formulas
28.
'01 È dt
29.
' È2s ds
30.
'
31.
'
32.
'
33.
' e dxe
34.
' È dy
35.
'1e
"
œ sin" #t ‘ ! œ sin" ˆ "# ‰ 0 œ
4 t#
;”
1 s%
u œ s#
Ä
du œ 2s ds •
2 dx
xÈ1 4 ln# x
6 dx
xÈ25x# 1
"
3
e2y 1
dx
x cos (ln x)
œ ln ¸sec
ex dx
e2x 1
œ'
1 Î3
1
3
œ
6 dx
5xÉx# #"5
6
5
œ sin" u C œ sin" s# C
' È du
1 u#
œ sin" u C œ sin" (2 ln x) C
† 5 sec" k5xk C œ 6 sec" k5xk C
sec" ¸ 3r ¸ C
œ'
x
1 u#
u œ 2 ln x
Ä
du œ 2 xdx •
œ'
œ
dr
r È r# 9
x
;”
' È du
1
6
;”
u œ ex
Ä'
du œ ex dx •
ey dy
e y É ae y b #
1
;”
du
u# 1
u œ ey
Ä
du œ ey dy •
'
œ tan" u C œ tan" ex C
du
uÈ u# 1
u œ ln x
du œ dx
;
x
Õ x œ 1 Ê u œ 0, x œ e1Î3 Ê u œ
Ô
œ sec" kuk C œ sec" ey C
×
1
3
Ä '0
1Î3
Ø
du
cos u
œ '0
1Î3
1Î$
sec u du œ cln ksec u tan ukd !
tan 13 ¸ ln ksec 0 tan 0k œ ln Š2 È3‹ ln (1) œ ln Š2 È3‹
u œ ln# x
Ä
du œ 2x ln x dx •
36.
' x ln4xx dxln x œ ' x a1lnx4dxln xb ; ”
37.
'12
38.
'24 x 26xdx 10 œ 2 '24 (x dx3) 1 ; Ô
39.
'È
40.
' È d)
41.
'
#
#
' #"
du
1 4u
œ
"
8
ln k1 4uk C œ
"
8
ln a1 4 ln# xb C
uœx1
×
1
"
du œ dx
Ä 8'0 1 duu# œ 8 ctan" ud !
Õ x œ 1 Ê u œ 0, x œ 2 Ê u œ " Ø
œ 8 atan" 1 tan" 0b œ 8 ˆ 14 0‰ œ 21
x#
8 dx
2x 2
œ 8'1
2
dx
1 (x 1)#
;
Ô
uœx3
×
1
du œ dx
Ä 2'c1
Õ x œ 2 Ê u œ 1, x œ 4 Ê u œ 1 Ø
œ 2 ctan" 1 tan" (1)d œ 2 14 ˆ 14 ‰‘ œ 1
#
#
dt
t# 4t 3
2) ) #
œ'
œ'
dx
(x 1)Èx# 2x
dt
È1 (t 2)#
d)
È1 () 1)#
œ'
;”
;”
uœt2
Ä'
du œ dt •
uœ)1
Ä'
du œ d) •
dx
(x 1)È(x 1)# 1
;”
du
È 1 u#
du
È 1 u#
du
u# 1
"
œ 2 ctan" ud "
œ sin" u C œ sin" (t 2) C
œ sin" u C œ sin" () 1) C
uœx1
Ä'
du œ dx •
du
uÈ u# 1
œ sec" kuk C œ sec" kx 1k C,
kuk œ kx 1k 1
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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495
496
42.
Chapter 8 Techniques of Integration
'
dx
(x 2)Èx# 4x 3
œ'
dx
(x 2)È(x 2)# 1
;”
uœx2
Ä
du œ dx •
'
du
uÈ u# 1
œ sec" kuk C
œ sec" kx 2k C, kuk œ kx 2k 1
43.
' (sec x cot x)# dx œ ' asec# x 2 sec x cot x cot# xb dx œ ' sec# x dx ' 2 csc x dx ' acsc# x 1b dx
œ tan x 2 ln kcsc x cot xk cot x x C
44.
' (csc x tan x)# dx œ ' acsc# x 2 csc x tan x tan# xb dx œ ' csc# x dx ' 2 sec x dx ' asec# x 1b dx
œ cot x 2 ln ksec x tan xk tan x x C
45.
' csc x sin 3x dx œ ' (csc x)(sin 2x cos x sin x cos 2x) dx œ ' (csc x) a2 sin x cos# x sin x cos 2xb dx
œ ' a2 cos# x cos 2xb dx œ ' [(1 cos 2x) cos 2x] dx œ ' (1 2 cos 2x) dx œ x sin 2x C
46.
' (sin 3x cos 2x cos 3x sin 2x) dx œ ' sin (3x 2x) dx œ ' sin x dx œ cos x C
47.
' x x 1 dx œ ' ˆ1 x "1 ‰ dx œ x ln kx 1k C
48.
' x x 1 dx œ ' ˆ1 x " 1 ‰ dx œ x tan" x C
49.
'È32
50.
'c31 4x2x 37 dx œ 'c31 (2x 3) 2x 2 3 ‘ dx œ cx# 3x ln k2x 3kd $" œ (9 9 ln 9) (1 3 ln 1) œ ln 9 4
51.
' 4t t t4 16t dt œ ' (4t 1) t 4 4 ‘ dt œ 2t# t 2 tan" ˆ #t ‰ C
52.
' 2) 2)7) 5 7) d) œ ' a)# ) 1b 2) 5 5 ‘ d) œ )3
53.
' È1 x
54.
' x È2Èx 1 dx œ '
55.
'01Î4
56.
'01Î2 124x8x
#
#
#
dx œ 'È2 ˆ2x 3
2x$
x# 1
2x ‰
x# 1
dx œ cx# ln kx# 1kd È2 œ (9 ln 8) (2 ln 1) œ 7 ln 8
3
#
$
#
#
#
$
#
1 x#
2x
$
dx œ '
dx
È 1 x#
x1
1 sin x
cos# x
#
'
dx
2È x 1
dx
x
5
#
dx œ '0 asec# x sec x tan xb dx œ ctan x sec xd !
1Î4
dx œ '0 ˆ 1 24x# 1Î2
ln k2) 5k C
1Î%
8x ‰
1 4x#
œ Š1 È2‹ (! 1) œ È2
"Î#
dx œ ctan" (2x) ln k1 4x# kd !
1
4
ln 2
sin x)
' 1 dxsin x œ ' a(11 sinsin x)xb dx œ ' (1cos
' asec# x sec x tan xb dx œ tan x sec x C
x dx œ
#
#
58. 1 cos x œ 1 cos ˆ2 † #x ‰ œ 2 cos#
59.
)
œ (x 1)"Î# ln kxk C
œ atan" 1 ln 2b atan" 0 ln 1b œ
57.
)#
#
œ sin" x È1 x# C
x dx
È 1 x#
'
' sec ) " tan ) d) œ '
d); ”
x
#
Ê
' 1 dxcos x œ ' 2 cosdx ˆ ‰ œ #" ' sec# ˆ #x ‰ dx œ tan x# C
u œ 1 sin )
Ä
du œ cos ) d) •
#
x
#
' duu œ ln kuk C œ ln k1 sin )k C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 8.1 Basic Integration Formulas
60.
61.
62.
' csc ) " cot ) d) œ ' 1 sincos) ) d); ”
#
#
1
' 1 "csc x dx œ ' sinsinx x 1 dx œ ' ˆ1 sin x" 1 ‰ dx œ ' Š1 (sin x sin1)x(sin
x 1) ‹ dx
1 sin x ‰
cos# x
dx œ ' ˆ1 sec# x '021 É 1 #cos x dx œ '021 ¸sin x# ¸ dx; ”
œ (2)(2) œ 4
64.
' udu œ ln kuk C œ ln k1 cos )k C
cos x ‰
' 1 "sec x dx œ ' coscosx x 1 dx œ ' ˆ1 cos x" 1 ‰ dx œ ' ˆ1 1 sincosx x ‰ dx œ ' ˆ1 csc# x sin
x dx
œ ' a1 csc# x csc x cot xb dx œ x cot x csc x C
œ ' ˆ1 63.
u œ 1 cos )
Ä
du œ sin ) d) •
sin x ‰
cos# x
dx œ ' a1 sec# x sec x tan xb dx œ x tan x sec x C
21
sin #x 0
#1
Ä '0 sin ˆ x# ‰ dx œ 2 cos x# ‘ ! œ 2(cos 1 cos 0)
•
x
for 0 Ÿ # Ÿ 21
'01 È1 cos 2x dx œ '01 È2 ksin xk dx; ”
1
1
sin x 0
Ä È2 '0 sin x dx œ ’È2 cos x“
•
for 0 Ÿ x Ÿ 1
!
œ È2 (cos 1 cos 0) œ 2È2
65.
'11Î2 È1 cos 2t dt œ '11Î2 È2 kcos tk dt; ”
œ È2 ˆsin 1 sin
66.
1‰
#
cos t Ÿ 0
Ä
for 1# Ÿ t Ÿ 1 •
'11Î2 È2 cos t dt œ ’È2 sin t“ 1
1Î#
œ È2
'c01 È1 cos t dt œ 'c01 È2 ¸cos #t ¸ dt; ”
0
cos #t
0
• Ä 'c1 È2 cos
for 1 Ÿ t Ÿ 0
t
#
dt œ ’2È2 sin #t “
!
1
œ 2È2 sin 0 sin ˆ 1# ‰‘ œ 2È2
67.
'c01 È1 cos# ) d) œ 'c01 ksin )k d); ”
0
sin ) Ÿ 0
Ä 'c1 sin ) d) œ ccos )d !1 œ cos 0 cos (1)
•
for 1 Ÿ ) Ÿ 0
œ 1 (1) œ 2
cos ) Ÿ 0
Ä
for 1# Ÿ ) Ÿ 1 •
68.
'11Î2 È1 sin# ) d) œ '11Î2 kcos )k d); ”
69.
'c11ÎÎ44 Ètan# y 1 dy œ '11ÎÎ44 ksec yk dy; ”
sec y 0
for 14 Ÿ y Ÿ
'11Î2 cos ) d) œ c sin )d 11Î# œ sin 1 sin 1# œ 1
1Î4
1
4
• Ä '1Î4 sec y dy œ cln ksec y tan ykd 1Î%
1Î%
œ ln ¹È2 1¹ ln ¹È2 1¹
70.
'c01Î4 Èsec# y 1 dy œ '01Î4 ktan yk dy; ”
tan y Ÿ 0
Ä
for 14 Ÿ y Ÿ 0 •
'01Î4 tan y dy œ cln kcos ykd !1Î% œ ln Š È"
œ ln È2
71.
'13Î14Î4 (csc x cot x)# dx œ '13Î14Î4 acsc# x 2 csc x cot x cot# xb dx œ '13Î14Î4 a2 csc# x 1 2 csc x cot xb dx
$1Î%
œ c2 cot x x 2 csc xd 1Î% œ ˆ2 cot
œ ’2(1) 31
4
2 ŠÈ2‹“ ’2(1) 31
4
31
4
1
4
2 ŠÈ 2‹ “ œ 4 2 csc
31 ‰
4
ˆ2 cot
1
4
1
4
2 csc 14 ‰
1
#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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2
‹
497
498
72.
Chapter 8 Techniques of Integration
1Î%
2x ‰‘
'01Î4 (sec x 4 cos x)# dx œ '01Î4 sec# x 8 16 ˆ 1 cos
dx œ ctan x 16x 4 sin 2xd !
#
œ ˆtan
73.
1
4
41 4 sin 1# ‰ (tan 0 0 4 sin 0) œ 5 41
' cos ) csc (sin )) d); ” u œ sin ) • Ä ' csc u du œ ln kcsc u cot uk C
du œ cos ) d)
œ ln kcsc (sin )) cot (sin ))k C
74.
' ˆ1 "x ‰ cot (x ln x) dx; ” u œ x ln" x • Ä ' cot u du œ ln ksin uk C œ ln ksin (x ln x)k C
du œ ˆ1 ‰ dx
x
75.
' (csc x sec x)(sin x cos x) dx œ ' (1 cot x tan x 1) dx œ ' cot x dx ' tan x dx
œ ln ksin xk ln kcos xk C
76.
'
77.
' Èy6(1dy y) ; –
78.
'
79.
'
3 sinhˆ x2 ln 5‰dx œ ”
dx
xÈ4x# 1
u œ x2 ln 5
œ ' ' sinh u du œ 6 cosh u C œ 6 coshˆ x# ln 5‰ C
2 du œ dx •
u œ Èy
"
— Ä
du œ 2È
y dy
œ'
2 dx
2xÈ(2x)# 1
7 dx
(x 1)Èx# 2x 48
œ'
;”
' 121duu œ 12 tan" u C œ 12 tan" Èy C
#
u œ 2x
Ä'
du œ 2 dx •
7 dx
(x 1)È(x 1)# 49
;”
du
uÈ u# 1
œ sec" kuk C œ sec" k2xk C
uœx1
Ä
du œ dx •
'
7 du
uÈu# 49
œ7†
"
7
sec" ¸ 7u ¸ C
œ sec" ¸ x 7 " ¸ C
80.
'
dx
(2x 1)È4x# 4x
œ
"
#
œ'
dx
(2x 1)È(2x 1)# 1
;”
u œ 2x 1
Ä'
du œ 2 dx •
du
2uÈu# 1
œ
"
#
sec" kuk C
sec" k2x 1k C
81.
' sec# t tan (tan t) dt; ” u œ tan# t • Ä ' tan u du œ ln kcos uk C œ ln ksec uk C œ ln ksec (tan t)k C
du œ sec t dt
82.
'
dx
xÈ $ x#
83. (a)
(b)
œ "$ csch" ¹ Èx$ ¹ C
' cos$ ) d) œ ' (cos )) a1 sin# )b d); ” u œ sin ) • Ä ' a1 u# b du œ u u3$ C œ sin ) "3 sin$ ) C
du œ cos ) d)
' cos& ) d) œ ' (cos )) a1 sin# )b# d) œ ' a1 u# b# du œ ' a1 2u# u% b du œ u 23 u$ u5& C
œ sin ) (c)
84. (a)
2
3
sin$ ) sin& ) C
' cos* ) d) œ ' acos) )b (cos )) d) œ ' a1 sin# )b% (cos )) d)
' sin$ ) d) œ ' a1 cos# )b (sin )) d); ”
œ cos ) (b)
"
5
"
3
cos$ ) C
2
3
cos$ ) u œ cos )
Ä ' a1 u# b ( du) œ
du œ sin ) d) •
u$
3
uC
' sin& ) d) œ ' a1 cos# )b# (sin )) d) œ ' a1 u# b# ( du) œ ' a1 2u# u% b du
œ cos ) "
5
cos& ) C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 8.1 Basic Integration Formulas
' sin( ) d) œ ' a1 u# b$ ( du) œ ' a1 3u# 3u% u' b du œ cos ) cos$ ) 35 cos& ) cos7( ) C
'
(d) ' sin"$ ) d) œ ' asin"# )b (sin )) d) œ ' a1 cos# )b (sin )) d)
(c)
85. (a)
' tan$ ) d) œ ' asec# ) 1b (tan )) d) œ ' sec# ) tan ) d) ' tan ) d) œ "# tan# ) ' tan ) d)
œ
"
#
tan# ) ln kcos )k C
' tan& ) d) œ ' asec# ) 1b atan$ )b d) œ ' tan$ ) sec# ) d) ' tan$ ) d) œ "4 tan% ) ' tan$ ) d)
(c) ' tan( ) d) œ ' asec# ) 1b atan& )b d) œ ' tan& ) sec# ) d) ' tan& ) d) œ "6 tan' ) ' tan& ) d)
(d) ' tan2kb1 ) d) œ ' asec# ) 1b atan2kc1 )b d) œ ' tan2kc1 ) sec# ) d) ' tan2kc1 ) d);
(b)
u œ tan )
” du œ sec# ) d) • Ä
86. (a)
' u2kc1 du ' tan2kc1 ) d) œ
"
2k
u2k ' tan2kc1 ) d) œ
tan2k ) ' tan2kc1 ) d)
"
#k
' cot$ ) d) œ ' acsc# ) 1b (cot )) d) œ ' cot ) csc# ) d) ' cot ) d) œ "# cot# ) ' cot ) d)
œ "# cot# ) ln ksin )k C
' cot& ) d) œ ' acsc# ) 1b acot$ )b d) œ ' cot$ ) csc# ) d) ' cot$ ) d) œ "4 cot% ) ' cot$ ) d)
(c) ' cot( ) d) œ ' acsc# ) 1b acot& )b d) œ ' cot& ) csc# ) d) ' cot& ) d) œ "6 cot' ) ' cot& ) d)
(d) ' cot2kb1 ) d) œ ' acsc# ) 1b acot2kc1 )b d) œ ' cot2kc1 ) csc# ) d) ' cot2kc1 ) d);
(b)
u œ cot )
2kc1
du ' cot2kc1 ) d) œ #"k u2k ' cot2kc1 ) d)
” du œ csc# ) d) • Ä ' u
"
œ 2k
cot2k ) ' cot2kc1 ) d)
87. A œ 'c1Î4 (2 cos x sec x) dx œ c2 sin x ln ksec x tan xkd 1Î%
1Î4
1Î%
œ ’È2 ln ŠÈ2 1‹“ ’È2 ln ŠÈ2 1‹“
#
È 2 1‹
Š
È
œ 2È2 ln Š È2 " ‹ œ 2È2 ln 21
21
œ 2È2 ln Š3 2È2‹
88. A œ '1Î6 (csc x sin x) dx œ c ln kcsc x cot xk cos xd 1Î'
1Î2
1Î#
œ ln k1 0k ln ¹2 È3¹ 1Î4
È3
#
œ ln Š2 È3‹ 1Î4
È3
#
1Î4
1Î4
89. V œ 'c1Î4 1(2 cos x)# dx '1Î4 1 sec# x dx œ 41 '1Î4 cos# x dx 1'1Î4 sec# x dx
œ 21 'c1Î4 (1 cos 2x) dx 1 ctan xd 1Î% œ 21 x 1Î4
1Î%
"
#
1Î%
sin 2x‘ 1Î% 1[1 (1)]
œ 21 ˆ 14 "# ‰ ˆ 14 "# ‰‘ 21 œ 21 ˆ 1# 1‰ 21 œ 1#
90. V œ '1Î6 1 csc# x dx '1Î6 1 sin# x dx œ 1 '1Î6 csc# x dx 1Î2
1Î2
1Î#
œ 1 c cot xd 1Î' œ 1È3 1
#
Š 261 1
#
x È3
4 ‹
"
#
1Î2
1Î#
1
#
sin 2x‘ 1Î' œ 1 ’0 ŠÈ3‹“ '11ÎÎ62 (1 cos 2x) dx
1
#
’ˆ 1# 0‰ Š 16 "
#
†
È3
# ‹“
È
œ 1 Š 7 8 3 16 ‹
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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499
500
Chapter 8 Techniques of Integration
91. y œ ln (cos x) Ê
dy
dy
sin x
#
#
'
œ cos
x Ê Š dx ‹ œ tan x œ sec x 1; L œ a Ê1 Š dx ‹ dx
#
dy
dx
#
b
1Î$
œ '0 È1 asec# x 1b dx œ '0 sec x dx œ cln ksec x tan xkd ! œ ln ¹2 È3¹ ln k1 0k œ ln Š2 È3‹
1Î3
1Î3
92. y œ ln (sec x) Ê
œ
dy
dx
sec x tan x
sec x
dy
#
#
'
Ê Š dy
dx ‹ œ tan x œ sec x 1; L œ a Ê1 Š dx ‹ dx
#
œ '0 sec x dx œ cln ksec x tan xkd !
1Î4
1Î%
1Î4
93. Mx œ 'c1Î4 ˆ "# sec x‰ (sec x) dx œ
œ
"
#
1Î%
ctan xd 1Î% œ
"
#
b
#
œ ln ¹È2 1¹ ln k1 0k œ ln ŠÈ2 1‹
'11ÎÎ44 sec# x dx
"
#
c1 (1)d œ 1;
M œ 'c1Î4 sec x dx œ cln ksec x tan xkd 1Î%
1Î4
1Î%
È
œ ln ¹È2 1¹ ln ¹È2 1¹ œ ln Š È2" ‹
2 1
#
œ ln ŠÈ2 "‹
#1
œ ln Š3 2
È2‹ ; x œ 0 by
symmetry of the region, and y œ
94. Mx œ '1Î6 ˆ "# csc x‰ (csc x) dx œ
51Î6
œ
"
#
&1Î'
c cot xd 1Î' œ
"
#
œ
Mx
M
"
#
"
ln Š3 2È2‹
'15Î16Î6 csc# x dx
’ ŠÈ3‹ ŠÈ3‹“ œ È3;
M œ '1Î6 csc x dx œ c ln kcsc x cot xkd 1Î'
51Î6
&1Î'
È
3
œ ln ¹2 È3¹ Š ln ¹2 È3¹‹ œ ln ¹ 22 ¹
È3
#
œ ln Š2 È3‹
43
œ 2 ln Š2 of the region, and y œ
95.
Mx
M
œ
È 3‹ ; x œ
1
#
by symmetry
È3
2 ln Š2 È3‹
x cot x ‰
x cot x
' csc x dx œ ' (csc x)(1) dx œ ' (csc x) ˆ csc
' csccscx x csc
dx;
csc x cot x dx œ
cot x
#
u œ csc x cot x
” du œ a csc x cot x csc# xb dx • Ä
96. cax# 1b (x 1)d
#Î$
1‰
œ (x 1)# ˆ xx 1
' udu œ ln kuk C œ ln kcsc x cot xk C
œ c(x 1)(x 1)# d
#Î$
#Î$
œ (x 1)# ˆ1 œ (x 1)#Î$ (x 1)%Î$ œ (x 1)# (x 1)#Î$ (x 1)#Î$ ‘
2 ‰#Î$
x1
u œ x " 1
du œ (x " 1)# dx —
(a)
' cax# 1b (x 1)d#Î$ dx œ '
(b)
' (1 2u)#Î$ du œ 3# (1 2u)"Î$ C œ 3# ˆ1 x2 1 ‰"Î$ C œ #3 ˆ xx 11 ‰"Î$ C
' cax# 1b (x 1)d#Î$ dx œ ' (x 1)# ˆ xx 11 ‰#Î$ dx; u œ ˆ xx 11 ‰k
(x 1)# ˆ1 2 ‰#Î$
x 1
dx; –
Ä
Ê du œ k ˆ xx 11 ‰
œ
(x 1)#
2k
œ
"
#k
kc1
c(x 1) (x 1)d
(x 1)#
k c1
(x 1)
dx œ 2k (x
1)kb1 dx; dx œ
1 ‰1ck
1 ‰#Î$
ˆ xx du; then, ' ˆ xx 1
1
"
#k
1 ‰1ck
ˆ xx du œ
1
(x 1)#
#k
"
#k
1 ‰kc1
ˆ xx du
1
' ˆ xx 11 ‰Ð1Î3kÑ du
' ˆ xx 11 ‰kÐ1Î3k1Ñ du œ #"k ' uÐ1Î3k1Ñ du œ #"k (3k) u1Î3k C œ #3 u1Î3k C œ 3# ˆ xx 11 ‰"Î$ C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 8.2 Integration by Parts
(c)
' cax# 1b (x 1)d#Î$ dx œ ' (x 1)# ˆ xx 11 ‰#Î$ dx;
"
Ô u œ tan x ×
x œ tan u
Ä
Õ dx œ du# Ø
cos u
' (tan u" 1)
#
tan u 1 ‰#Î$ ˆ du ‰
'
ˆ tan
u1
cos# u œ
"
(sin u cos u)#
sin u cos u œ sin u sin ˆ 1# u‰ œ 2 sin 14 cos ˆu 14 ‰
– sin u cos u œ sin u sin ˆ 1 u‰ œ 2 cos 1 sin ˆu 1 ‰ — Ä
4
4
#
u cos u ‰#Î$
ˆ sin
du;
sin u cos u
sin ˆu 1 ‰ #Î$
' 2 cos ˆ"u 1 ‰ ’ cos
du
ˆu 1 ‰ “
4
#
4
4
1
"Î$
œ
"
#
u tan 4
' tan#Î$ ˆu 14 ‰ sec# ˆu 14 ‰ du œ 3# tan"Î$ ˆu 14 ‰ C œ 3# ’ 1tan
C
tan u tan 14 “
œ
3
#
1 ‰"Î$
ˆ xx C
1
(d) u œ tan" Èx Ê tan u œ Èx Ê tan# u œ x Ê dx œ 2 tan u ˆ cos"# u ‰ du œ
sin# u cos# u
cos# u
x 1 œ tan# u 1 œ
1 2 cos# u
cos# u
œ
; x 1 œ tan# u 1 œ
#Î$
ub
' (x 1)#Î$ (x 1)%Î$ dx œ ' a12 cos
# #Î$ †
#
œ ' a1 2 cos# ub
œ
a1 2 cos# ub
3
#
acos ub
#Î$
"Î$
(e) u œ tan" ˆ x # 1 ‰ Ê
'
(x 1)
#Î$
Cœ
%Î$
3
#
–
Š 1 2 cos
#
#u
cos u
Š
Cœ
" ‹ —
cos# u
dx œ ' (tan u)
#Î$
3
#
(tan u 1)
2d(cos u)
cos$ u
1 ‰"Î$
ˆ xx C
1
%Î$
†2
#
œ 2d(tan u);
2 du
cos# u
† 2 † d(tan u)
œ
"
#
' ˆ1 tan u" 1 ‰#Î$ d ˆ1 tan u" 1 ‰ œ #3 ˆ1 tan u" 1 ‰"Î$ C œ #3 ˆ1 x 2 1 ‰"Î$ C
œ
3
#
1 ‰"Î$
ˆ xx C
1
œ'
du
%Î$
(sin u)"Î$ ˆ2#Î$ cos u# ‰
œ
'
3
#
' cax
œ
u
#
'
œ "#
du
"Î$
&Î$
2 ˆsin #u ‰ ˆcos #u ‰
œ ' tan"Î$ ˆ u# ‰ d ˆtan u# ‰ œ 3# tan#Î$
œ
œ
sin u du
$
Éacos# u 1b# (cos u 1)#
Cœ
u)
2d(cos
cos$ u ;
' a1 2 cos# ub#Î$ † d a1 2 cos# ub
"Î$
‹
†
œ tan u Ê x 1 œ 2(tan u 1) Ê dx œ
"
Ô u œ cos x ×
Ä '
(f)
x œ cos u
Õ dx œ sin u du Ø
(g)
"
#
† (2) † cos u † d(cos u) œ
x1
#
(x 1)
"
acos# ub%Î$
2 sin u
cos$ u du œ
cos# u sin# u
œ cos"# u ;
cos# u
3
#
sin u du
%Î$
asin%Î$ ub ˆ2#Î$ cos #u ‰
' Š cos
sin
u
#
u
#
‹
"Î$
du
ˆcos# #u ‰
ˆ tan# u# ‰"Î$ C œ
3
#
u 1 ‰"Î$
ˆ cos
C
cos u 1
3
#
1 ‰"Î$
ˆ xx C
1
" ‰"Î$
ˆ xx C
1
#
'
1b (x 1)d
#Î$
sinh u du
$
È
asinh% ub (cosh u1)#
"
Ô u œ cosh x ×
dx;
Ä
x œ cosh u
Õ dx œ sinh u Ø
'
œ
"
#
'
du
$
É
(sinh u) ˆ4 cosh% #u ‰
"Î$
œ ' ˆtanh u# ‰
d ˆtanh u# ‰ œ
3
#
œ
ˆtanh u# ‰#Î$ C œ
sinh u du
$
Éacosh# u1b# (cosh u1)#
'
3
#
du
$
É
sinh ˆ #u ‰ cosh& ˆ #u ‰
u 1 ‰"Î$
ˆ cosh
Cœ
cosh u 1
8.2 INTEGRATION BY PARTS
1. u œ x, du œ dx; dv œ sin
'
x sin
x
#
dx œ 2x cos
x
#
x
#
dx, v œ 2 cos
) cos 1) d) œ
)
1
;
' ˆ2 cos x# ‰ dx œ 2x cos ˆ x# ‰ 4 sin ˆ x# ‰ C
2. u œ ), du œ d); dv œ cos 1) d), v œ
'
x
#
sin 1) '
"
1
"
1
sin 1);
sin 1) d) œ
)
1
sin 1) "
1#
cos 1) C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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501
502
Chapter 8 Techniques of Integration
3.
cos t
ÐÑ
t# ïïïïî
ÐÑ
2t ïïïïî
ÐÑ
2 ïïïïî
sin t
cos t
sin t
0
4.
'
t# cos t dt œ t# sin t 2t cos t 2 sin t C
'
x# sin x dx œ x# cos x 2x sin x 2 cos x C
sin x
ÐÑ
x# ïïïïî
ÐÑ
2x ïïïïî
ÐÑ
2 ïïïïî
cos x
sin x
cos x
0
5. u œ ln x, du œ
dv œ x dx, v œ
dx
x ;
'1 x ln x dx œ ’ x#
2
6. u œ ln x, du œ
ln x“ '1
#
#
dx
x ;
dx
x
ln x“ '1
e
e %
x
1
dy
1 y #
dx
x
4
;
#
#
œ 2 ln 2 ’ x4 “ œ 2 ln 2 "
x%
4
dv œ x$ dx, v œ
%
7. u œ tan" y, du œ
#
"
'1 x$ ln x dx œ ’ x4
e
2 #
x
x#
#
3
4
œ ln 4 3
4
;
œ
e%
4
%
e
x
’ 16
“ œ
1
3e% 1
16
; dv œ dy, v œ y;
' tan" y dy œ y tan" y ' a1ydyy b œ y tan" y #" ln a1 y# b C œ y tan" y ln È1 y# C
#
8. u œ sin" y, du œ
dy
È 1 y#
; dv œ dy, v œ y;
' sin" y dy œ y sin" y ' Èy1 dy y
#
œ y sin" y È1 y# C
9. u œ x, du œ dx; dv œ sec# x dx, v œ tan x;
' x sec# x dx œ x tan x ' tan x dx œ x tan x ln kcos xk C
10.
' 4x sec# 2x dx; [y œ 2x]
' y sec# y dy œ y tan y ' tan y dy œ y tan y ln ksec yk C
Ä
œ 2x tan 2x ln ksec 2xk C
ex
11.
ÐÑ
x$ ïïïïî
ÐÑ
3x# ïïïïî
ÐÑ
6x ïïïïî
ÐÑ
6 ïïïïî
0
ex
ex
ex
ex
'
x$ ex dx œ x$ ex 3x# ex 6xex 6ex C œ ax$ 3x# 6x 6b ex C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 8.2 Integration by Parts
ecp
12.
ÐÑ
ïïïïî
ÐÑ
4p$ ïïïïî
ÐÑ
12p# ïïïïî
ÐÑ
24p ïïïïî
ÐÑ
24 ïïïïî
p%
ecp
ecp
ecp
ecp
ecp
'
0
p% ecp dp œ p% ecp 4p$ ecp 12p# ecp 24pecp 24ecp C
œ ap% 4p$ 12p# 24p 24b ecp C
ex
13.
ÐÑ
x# 5x ïïïïî ex
ÐÑ
2x 5 ïïïïî ex
ÐÑ
2
ïïïïî ex
' ax# 5xb ex dx œ ax# 5xb ex (2x 5)ex 2ex C œ x# ex 7xex 7ex C
0
œ ax# 7x 7b ex C
er
14.
ÐÑ
r# r 1 ïïïïî er
ÐÑ
2r 1
ïïïïî er
ÐÑ
2
ïïïïî er
0
' ar# r 1b er dr œ ar# r 1b er (2r 1) er 2er C
œ car# r 1b (2r 1) 2d er C œ ar# r 2b er C
ex
15.
x&
5x%
20x$
60x#
120x
120
0
ÐÑ
ïïïïî ex
ÐÑ
ïïïïî ex
ÐÑ
ïïïïî ex
ÐÑ
ïïïïî ex
ÐÑ
ïïïïî ex
ÐÑ
ïïïïî ex
' x& ex dx œ x& ex 5x% ex 20x$ ex 60x# ex 120xex 120ex C
œ ax& 5x% 20x$ 60x# 120x 120b ex C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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503
504
Chapter 8 Techniques of Integration
e4t
16.
ÐÑ
t# ïïïïî
ÐÑ
2t ïïïïî
ÐÑ
2 ïïïïî
"
4
e4t
"
16
e4t
"
64
e4t
' t# e4t dt œ t4 e4t 162t e4t 642 e4t C œ t4 e4t 8t e4t 3"# e4t C
#
0
#
œ Š t4 17.
"
4t
3# ‹ e
C
sin 2)
ÐÑ
)# ïïïïî "2 cos 2)
ÐÑ
2) ïïïïî "4 sin 2)
ÐÑ
2 ïïïïî "8 cos 2)
'01Î2 )# sin 2) d) œ ’ )#
#
0
œ ’
18.
t
8
#
1#
8
† (1) 1
4
cos 2) †0
"
4
)
#
sin 2) "
4
cos 2)“
† (1)“ 0 0 "
4
1Î#
!
† 1‘ œ
1#
8
"
#
œ
1# 4
8
cos 2x
ÐÑ
x$ ïïïïî "2 sin 2x
ÐÑ
3x# ïïïïî "4 cos 2x
ÐÑ
6x ïïïïî "8 sin 2x
ÐÑ
"
6 ïïïïî 16
cos 2x
'01Î2 x$ cos 2x dx œ ’ x#
$
0
œ
19. u œ sec" t, du œ
dt
tÈt# 1
'22ÎÈ3 t sec" t dt œ ’ t#
#
œ
51
9
’ "# Èt# 1“
20. u œ sin" ax# b , du œ
È2
'01Î
1$
’ 16
†0
; dv œ t dt, v œ
sec" t“
#
#ÎÈ$
œ
2x dx
È 1 x %
#
’È1 x% “
!
#
dt
tÈt# 1
œ
†0
1
1#
3
8
œ ˆ2 †
"# ŠÈ3 É 43 1‹ œ
; dv œ 2x dx, v œ x# ;
"ÎÈ#
31
8
3x#
4
cos 2x 3x
4
sin 2x 3
8
cos 2x“
† (1)“ 0 0 0 3
8
1Î#
!
#
1
† 1‘ œ 316
3
4
œ
3 a4 1 # b
16
;
"ÎÈ#
1
1#
† (1) '2ÎÈ3 Š t# ‹
2x sin" ax# b dx œ cx# sin" ax# bd !
œ
t#
#
2
#ÎÈ$
51
9
31 #
16
sin 2x '0
É 34 1 œ
È2
1Î
x# †
51
9
1
3
2
3
† 16 ‰ '2ÎÈ3
"# ŠÈ3 2x dx
È 1 x%
2
È3
3 ‹
œ
œ ˆ "# ‰ ˆ 16 ‰ '0
t dt
2Èt# 1
51
9
È3
3
œ
51 3È 3
9
È 2 d ˆ1 x % ‰
1Î
2È 1 x%
16È312
1#
21. I œ ' e) sin ) d); cu œ sin ), du œ cos ) d); dv œ e) d), v œ e) d Ê I œ e) sin ) ' e) cos ) d);
cu œ cos ), du œ sin ) d); dv œ e) d), v œ e) d Ê I œ e) sin ) Še) cos ) ' e) sin ) d)‹
œ e) sin ) e) cos ) I Cw Ê 2I œ ae) sin ) e) cos )b Cw Ê I œ
"
#
ae) sin ) e) cos )b C, where C œ
another arbitrary constant
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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w
C
#
is
Section 8.2 Integration by Parts
22. I œ ' ecy cos y dy; cu œ cos y, du œ sin y dy; dv œ ecy dy, v œ ecy d
Ê I œ ecy cos y ' aecy b (sin y) dy œ ecy cos y ' ecy sin y dy; cu œ sin y, du œ cos y dy;
dv œ ecy dy, v œ ecy d Ê I œ ecy cos y Šecy sin y ' aey b cos y dy‹ œ ecy cos y ecy sin y I Cw
Ê 2I œ ecy (sin y cos y) Cw Ê I œ
"
#
aecy sin y ecy cos yb C, where C œ
23. I œ ' e2x cos 3x dx; u œ cos 3x; du œ 3 sin 3x dx, dv œ e2x dx; v œ
is another arbitrary constant
e2x ‘
Ê
' e2x sin 3x dx; u œ sin 3x, du œ 3 cos 3x, dv œ e2x dx; v œ "# e2x ‘
I œ "# e2x cos 3x 3# Š "# e2x sin 3x 3# ' e2x cos 3x dx‹ œ "# e2x cos 3x 34 e2x sin 3x 94 I Cw
Ê
13
4
Ê Iœ
24.
"
#
w
C
#
"
#
e2x cos 3x Iœ
"
#
3
#
e2x cos 3x 34 e2x sin 3x Cw Ê
e2x
13
(3 sin 3x 2 cos 3x) C, where C œ
4
13
Cw
' ec2x sin 2x dx; [y œ 2x] Ä "# ' ecy sin y dy œ I; cu œ sin y, du œ cos y dy; dv œ ecy dy, v œ ecyd
Ê I œ "# Šecy sin y ' ecy cos y dy‹ cu œ cos y, du œ sin y; dv œ ecy dy, v œ ecy d
Ê I œ "# ecy sin y "# Šecy cos y ' aecy b ( sin y) dy‹ œ "# ecy (sin y cos y) I Cw
c2x
Ê 2I œ "# ecy (sin y cos y) Cw Ê I œ "4 ecy (sin y cos y) C œ e 4 (sin 2x cos 2x) C, where
Cœ
w
C
#
#
25.
' eÈ3sb9 ds; ” 3s 92 œ x
ds œ
2
3
' xex dx œ
2
3
3
x dx •
Ä
' ex † 23 x dx œ 23 ' xex dx; cu œ x, du œ dx; dv œ ex dx, v œ ex d ;
Šxex ' ex dx‹ œ
2
3
axex ex b C œ
2
3
ŠÈ3s 9 eÈ3sb9 eÈ3sb9 ‹ C
26. u œ x, du œ dx; dv œ È1 x dx, v œ 23 È(1 x)$ ;
'01 xÈ1 x dx œ 23 È(1 x)$ x‘ "! 23 '01 È(1 x)$ dx œ 23 25 (1 x)&Î# ‘ "! œ 154
27. u œ x, du œ dx; dv œ tan# x dx, v œ ' tan# x dx œ '
œ tan x x;'0
1Î3
œ
1
3
1Î$
x tan# x dx œ cx(tan x x)d !
ŠÈ3 13 ‹ ln
28. u œ ln ax x# b, du œ
œ x ln ax x# b '
"
#
1#
18
œ
(2x 1) dx
x x #
(2x 1) dx
x 1
1È3
3
ln 2 sin# x
cos# x
dx œ '
" cos# x
cos# x
'0 (tan x x) dx œ
1Î3
1
3
dx œ '
dx
cos# x
' dx
ŠÈ3 13 ‹ ’ln kcos xk 1#
18
; dv œ dx, v œ x; ' ln ax x# b dx œ x ln ax x# b '
œ x ln ax x# b '
2(x 1) "
x 1
2x "
x(x 1)
† x dx
dx œ x ln ax x# b 2x ln kx 1k C
u œ ln x
29.
1Î$
x#
# “!
' sin (ln x) dx; Ô du œ "x dx ×
Ä ' (sin u) eu du. From Exercise 21, ' (sin u) eu du œ eu ˆ sin u # cos u ‰ C
Õ dx œ eu du Ø
œ "# cx cos (ln x) x sin (ln x)d C
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505
506
Chapter 8 Techniques of Integration
u œ ln z
30.
' z(ln z)# dz; Ô du œ "z dz ×
Õ dz œ eu du Ø
e2u
ÐÑ
u# ïïïïî
ÐÑ
2u ïïïïî
ÐÑ
2 ïïïïî
"
2
e2u
"
4
e2u
"
8
e2u
'
0
Ä
' eu † u# † eu du œ ' e2u † u# du;
u# e2u du œ
œ
z#
4
u#
#
#
e2u #u e2u "4 e2u C œ
e2u
4
c2u# 2u 1d C
c2(ln z) 2 ln z 1d C
31. (a) u œ x, du œ dx; dv œ sin x dx, v œ cos x;
S" œ '0 x sin x dx œ [x cos x]!1 '0 cos x dx œ 1 [sin x]1! œ 1
1
1
(b) S# œ '1 x sin x dx œ ”[x cos x]#11 '1 cos x dx• œ c31 [sin x]1#1 d œ 31
21
21
(c) S$ œ '21 x sin x dx œ [x cos x]$#11 '21 cos x dx œ 51 [sin x]$#11 œ 51
31
31
Ðn1Ñ1
(d) S8" œ (1)nb1 'n1
x sin x dx œ (1)nb1 c[x cos x]Ðnn11Ñ1 [sin x]Ðnn11Ñ1 d
œ (1)nb1 c(n 1)1(1)n n1(1)nb1 d 0 œ (2n 1)1
32. (a) u œ x, du œ dx; dv œ cos x dx, v œ sin x;
31Î2
1‰
S" œ '1Î2 x cos x dx œ ”[x sin x]311Î2Î2 '1Î2 sin x dx• œ ˆ 31
# # [cos x]1Î2 œ 21
31Î2
31Î2
(b) S# œ '31Î2 x cos x dx œ [x sin x]&$11ÎÎ22 '31Î2 sin x dx œ 5#1 ˆ 3#1 ‰‘ [cos x]&$11ÎÎ22 œ 41
51Î2
51Î2
(c) S$ œ '51Î2 x cos x dx œ ”[x sin x](&11ÎÎ22 '51Î2 sin x dx• œ ˆ 7#1 71Î2
71Î2
Ð2n1Ñ1Î2
51 ‰
#
[cos x](&11ÎÎ22 œ 61
Ð2n1Ñ1Î2
n1Ñ1Î2
n'
(d) Sn œ (1)n 'Ð2n1Ñ1Î2 x cos x dx œ (1)n ”[x sin x]Ð#
sin x dx•
Ð2n1Ñ1Î2 Ð2n1Ñ1Î2
œ (1)n ’ (2n# 1)1 (1)n 33. V œ '0
ln 2
(2n1)1
#
n1Ñ1Î2
(1)nc1 “ [cos x]Ð#
Ð2n1Ñ1Î2 œ
"
#
(2n1 1 2n1 1) œ 2n1
21(ln 2 x) ex dx œ 21 ln 2 '0 ex dx 21'0 xex dx
ln 2
ln 2
œ (21 ln 2) cex d ln0 2 21 Œcxex d ln0 2 '0 ex dx
ln 2
œ 21 ln 2 21 ˆ2 ln 2 cex d ln0 2 ‰ œ 21 ln 2 21 œ 21(1 ln 2)
34. (a) V œ '0 21xecx dx œ 21 Œcxecx d "! '0 ecx dx
1
1
œ 21 Š "e cecx d "! ‹ œ 21 ˆ "e œ 21 "
e
1‰
41
e
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
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Section 8.2 Integration by Parts
(b) V œ '0 21(1 x)ecx dx; u œ 1 x, du œ dx; dv œ ecx dx,
1
v œ ecx ; V œ 21 ”c(1 x) aecx bd "! '0 ecx dx•
1
œ 21 ’[0 1(1)] cecx d "! “ œ 21 ˆ1 "
e
35. (a) V œ '0 21x cos x dx œ 21 Œ[x sin x] !
1Î2
1Î#
1‰ œ
21
e
'0 sin x dx
1Î2
1Î#
œ 21 Š 1# [cos x] ! ‹ œ 21 ˆ 1# 0 1‰ œ 1(1 2)
(b) V œ '0 21 ˆ 1# x‰ cos x dx; u œ
1Î2
V œ 21 ˆ 1# x‰ sin
1Î#
x‘ !
1
#
x, du œ dx; dv œ cos x dx, v œ sin x;
21'0 sin x dx œ 0 21[ cos x] !
1Î2
1Î#
œ 21(0 1) œ 21
36. (a) V œ '0 21x(x sin x) dx;
1
sin x
ÐÑ
x# ïïïïî cos x
ÐÑ
2x ïïïïî sin x
ÐÑ
2 ïïïïî cos x
0
Ê V œ 21'0 x# sin x dx œ 21 cx# cos x 2x sin x 2 cos xd ! œ 21 a1# 4b
1
1
(b) V œ '0 21(1 x)x sin x dx œ 21# '0 x sin x dx 21 '0 x# sin x dx œ 21# [x cos x sin x]1! a21$ 81b
1
1
1
œ 81
37. av(y) œ
œ
"
1
"
#1
'021 2ect cos t dt
ect ˆ sin t # cos t ‰‘ #1
!
(see Exercise 22) Ê av(y) œ
"
#1
a1 ec21 b
'021 4ect (sin t cos t) dt
21
21
œ 12 '0 ect sin t dt 12 '0 ect cos t dt
38. av(y) œ
œ
œ
2
1
2
1
"
#1
ect ˆ sin t# cos t ‰ ect ˆ sin t # cos t ‰‘ #1
!
cect sin td #!1 œ 0
39. I œ ' xn cos x dx; cu œ xn , du œ nxn" dx; dv œ cos x dx, v œ sin xd
Ê I œ xn sin x ' nxn" sin x dx
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507
508
Chapter 8 Techniques of Integration
40. I œ ' xn sin x dx; cu œ xn , du œ nxn" dx; dv œ sin x dx, v œ cos xd
Ê I œ xn cos x ' nxn" cos x dx
41. I œ ' xn eax dx; u œ xn , du œ nxn" dx; dv œ eax dx, v œ "a eax ‘
ÊIœ
xn eax ax
a e
' xn" eax dx, a Á !
n
a
42. I œ ' aln xbn dx; ’u œ aln xbn , du œ
naln xbn"
x
dx; dv œ " dx, v œ x“
Ê I œ xaln xbn ' naln xbn" dx
43.
' sin" x dx œ x sin" x ' sin y dy œ x sin" x cos y C œ x sin" x cos asin" xb C
44.
' tan" x dx œ x tan" x ' tan y dy œ x tan" x ln kcos yk C œ x tan" x ln kcos atan" xbk C
45.
' sec" x dx œ x sec" x ' sec y dy œ x sec" x ln ksec y tan yk C
œ x sec" x ln ksec asec" xb tan asec" xbk C œ x sec" x ln ¹x Èx# 1¹ C
46.
' log2 x dx œ x log2 x ' 2y dy œ x log2 x ln2 # C œ x log2 x lnx# C
y
47. Yes, cos" x is the angle whose cosine is x which implies sin acos" xb œ È1 x# .
48. Yes, tan" x is the angle whose tangent is x which implies sec atan" xb œ È1 x# .
49. (a)
' sinh" x dx œ x sinh" x ' sinh y dy œ x sinh" x cosh y C œ x sinh" x cosh asinh" xb C;
check: d cx sinh" x cosh asinh" xb Cd œ ’sinh" x x
È 1 x#
œ sinh" x dx
(b)
' sinh" x dx œ x sinh" x '
œ x sinh" x a1 x# b
"
check: d ’x sinh
50. (a)
"Î#
x ŠÈ
"
‹
1 x#
dx œ x sinh" x "
#
sinh asinh" xb
dx
' a1 x# b"Î# 2x dx
C
x a1 x# b
"Î#
C“ œ ’sinh" x x
È 1 x#
x
È 1 x# “
dx œ sinh" x dx
' tanh" x dx œ x tanh" x ' tanh y dy œ x tanh" x ln kcosh yk C
œ x tanh" x ln kcosh atanh" xbk C;
check: d cx tanh" x ln kcosh atanh" xbk Cd œ ’tanh" x "
œ tanh
(b)
"
È 1 x# “
x
x
1 x#
x ‘
1 x#
dx œ tanh
' tanh" x dx œ x tanh" x ' 1 x x
check: d x tanh" x "
#
"
x
1 x#
sinh atanh" xb
"
cosh atanh" xb 1 x# “
dx
x dx
dx œ x tanh" x #" ' 12xx# dx œ x tanh" x #" ln k1 x# k C
ln k1 x# k C‘ œ tanh" x 1 x x# 1 x x# ‘ dx œ tanh" x dx
#
8.3 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS
1.
5x 13
(x 3)(x 2)
Ê
œ
A
x3
B
x2
Ê 5x 13 œ A(x 2) B(x 3) œ (A B)x (2A 3B)
ABœ5
Ê B œ (10 13) Ê B œ 3 Ê A œ 2; thus,
2A 3B œ 13 5x 13
(x 3)(x 2)
œ
2
x3
3
x#
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
Section 8.3 Integration of Rational Functions by Partial Fractions
2.
5x 7
x# 3x 2
Ê
3.
œ
A
x2
x4
(x 1)#
œ
A
x1
B
x1
Ê 5x 7 œ A(x 1) B(x 2) œ (A B)x (A 2B)
x4
(x 1)#
2x 2
x# 2x 1
œ
1
x1
2x 2
(x 1)#
œ
œ
z1
z# (z 1)
5x 7
x# 3x 2
œ
3
x2
A
x1
B
(x 1)#
Ê 2x 2 œ A(x 1) B œ Ax (A B) Ê
2x 2
x# 2x 1
œ
2
x1
Aœ2
A B œ 2 4
(x 1)#
œ
A
z
B
z#
"
z# z 6
œ
z
z$ z# 6z
œ
"
(z 3)(z 2)
œ
A
z3
B
z#
z"
z# (z 1)
ABœ0
Ê 5B œ 1 Ê B œ "5 Ê A œ 5" ; thus,
2A 3B œ 1 t# 8
t# 5t 6
œ1
5t 2
t# 5t 6
(after long division);
Ê B œ 12 Ê A œ 17; thus,
t% 9
t% 9t#
9t# 9
t% 9t#
œ1
œ1
t# 8
t# 5t 6
5t 2
t# 5t 6
9t# 9
t# at# 9b
#
œ1
œ
2
z
"
z#
2
z1
Ê 1 œ A(z 2) B(z 3) œ (A B)z (2A 3B)
œ
17
t3
œ
z
z$ z# 6z
5t 2
(t 3)(t 2)
Ê 5t 2 œ A(t 2) B(t 3) œ (A B)t (2A 3B) Ê
8.
Aœ1
Ê A œ 1 and B œ 3;
A B œ 4
z C 1 Ê z 1 œ Az(z 1) B(z 1) Cz# Ê z 1 œ (A C)z# (A B)z B
ACœ0 Þ
Ê
7.
2
x1
3
(x 1)#
Ê A B œ 1 ß
Ê B œ 1 Ê A œ 2 Ê C œ 2; thus,
B œ 1 à
6.
Ê x 4 œ A(x 1) B œ Ax (A B) Ê
B
(x 1)#
Ê A œ 2 and B œ 4; thus,
5.
ABœ5
Ê B œ 2 Ê A œ 3; thus,
A 2B œ 7 thus,
4.
5x 7
(x 2)(x 1)
œ
œ
A
t3
"
5
z3
"5
z2
B
t2
ABœ5
Ê B œ (10 2) œ 12
2A 3B œ 2 12
t2
(after long division);
9t# 9
t# at# 9b
œ
A
t
B
t#
CtD
t# 9
#
Ê 9t# 9 œ At at# 9b B at 9b (Ct D)t# œ (A C)t$ (B D)t 9At 9B
ACœ0 Þ
á
á
%
B D œ 9
Ê
Ê A œ 0 Ê C œ 0; B œ 1 Ê D œ 10; thus, t%t 9t9# œ 1 t"# t#109
ß
9A œ 0 á
á
9B œ 9 à
9.
"
1 x#
'
10.
œ
dx
1 x#
"
x# 2x
A
1x
œ
A
x
Ê 1 œ A(1 x) B(1 x); x œ 1 Ê A œ
B
1x
'
dx
1x
B
x2
Ê 1 œ A(x 2) Bx; x œ 0 Ê A œ
"
#
œ
"
#
'
dx
1x
œ
"
#
"
#
; x œ 1 Ê B œ
"
#
;
cln k1 xk ln k1 xkd C
' x dx 2x œ #" ' dxx #" ' x dx 2 œ #" cln kxk ln kx 2kd C
"
#
; x œ 2 Ê B œ #" ;
#
11.
x4
x# 5x 6
'x
12.
#
x4
5x 6
2x 1
x# 7x 12
'x
#
œ
A
x6
dx œ
2
7
œ
2x 1
7x 12
A
x4
B
x1
Ê x 4 œ A(x 1) B(x 6); x œ 1 Ê B œ
; x œ 6 Ê A œ
5
7
2
7
œ
2
7
;
œ 7 ; x œ 4 Ê A œ
9
1
' x dx 6 75 ' x dx 1 œ 72 ln kx 6k 75 ln kx 1k C œ 7" ln k(x 6)# (x 1)& k C
B
x3
dx œ 9 '
Ê 2x 1 œ A(x 3) B(x 4); x œ 3 Ê B œ
dx
x4
7'
dx
x3
œ 9 ln kx 4k 7 ln kx 3k C œ ln
7
1
4)*
¹ (x
(x 3)( ¹
C
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley
www.MathSchoolinternational.com
œ 9;
509
510
1