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Calculus 9e Purcell-Varberg-Rigdon (Solution)

CHAPTER
0
0.1 Concepts Review
1. rational numbers
Preliminaries
1 ⎡ 2 1 ⎛ 1 1 ⎞⎤
1
8. − ⎢ − ⎜ − ⎟ ⎥ = −
3 ⎣ 5 2 ⎝ 3 5 ⎠⎦
⎡ 2 1 ⎛ 5 3 ⎞⎤
3 ⎢ − ⎜ − ⎟⎥
⎣ 5 2 ⎝ 15 15 ⎠ ⎦
2. dense
1 ⎡ 2 1 ⎛ 2 ⎞⎤
1 ⎡2 1 ⎤
= − ⎢ − ⎜ ⎟⎥ = − ⎢ − ⎥
3 ⎣ 5 2 ⎝ 15 ⎠ ⎦
3 ⎣ 5 15 ⎦
1⎛ 6 1 ⎞
1⎛ 5 ⎞
1
=− ⎜ − ⎟=− ⎜ ⎟=−
3 ⎝ 15 15 ⎠
3 ⎝ 15 ⎠
9
3. If not Q then not P.
4. theorems
2
Problem Set 0.1
1. 4 − 2(8 − 11) + 6 = 4 − 2(−3) + 6
= 4 + 6 + 6 = 16
2. 3 ⎡⎣ 2 − 4 ( 7 − 12 ) ⎤⎦ = 3[ 2 − 4(−5) ]
= 3[ 2 + 20] = 3(22) = 66
3.
–4[5(–3 + 12 – 4) + 2(13 – 7)]
= –4[5(5) + 2(6)] = –4[25 + 12]
= –4(37) = –148
4.
5 [ −1(7 + 12 − 16) + 4] + 2
= 5 [ −1(3) + 4] + 2 = 5 ( −3 + 4 ) + 2
= 5 (1) + 2 = 5 + 2 = 7
5.
6.
7.
5 1 65 7 58
– =
– =
7 13 91 91 91
3
3 1 3
3 1
+ − =
+ −
4 − 7 21 6 −3 21 6
42 6
7
43
=− +
−
=−
42 42 42
42
1 ⎡1 ⎛ 1 1 ⎞ 1⎤ 1 ⎡1 ⎛ 3 – 4 ⎞ 1⎤
=
⎜ – ⎟+
⎜
⎟+
3 ⎢⎣ 2 ⎝ 4 3 ⎠ 6 ⎥⎦ 3 ⎢⎣ 2 ⎝ 12 ⎠ 6 ⎥⎦
1 ⎡1 ⎛ 1 ⎞ 1⎤
= ⎢ ⎜– ⎟+ ⎥
3 ⎣ 2 ⎝ 12 ⎠ 6 ⎦
1⎡ 1
4⎤
= ⎢– + ⎥
3 ⎣ 24 24 ⎦
1⎛ 3 ⎞ 1
= ⎜ ⎟=
3 ⎝ 24 ⎠ 24
Instructor’s Resource Manual
2
2
14 ⎛ 2 ⎞
14 ⎛ 2 ⎞
14 6
⎜
⎟ = ⎜ ⎟ = ⎛⎜ ⎞⎟
9.
21 ⎜ 5 − 1 ⎟
21 ⎜ 14 ⎟
21 ⎝ 14 ⎠
3⎠
⎝
⎝ 3 ⎠
2
=
14 ⎛ 3 ⎞
2⎛ 9 ⎞ 6
⎜ ⎟ = ⎜ ⎟=
21 ⎝ 7 ⎠
3 ⎝ 49 ⎠ 49
⎛2
⎞ ⎛ 2 35 ⎞ ⎛ 33 ⎞
⎜ − 5⎟ ⎜ − ⎟ ⎜ − ⎟
7
⎠ = ⎝ 7 7 ⎠ = ⎝ 7 ⎠ = − 33 = − 11
10. ⎝
6
2
⎛ 1⎞ ⎛7 1⎞
⎛6⎞
⎜1 − ⎟ ⎜ − ⎟
⎜ ⎟
⎝ 7⎠ ⎝7 7⎠
⎝7⎠
7
11 – 12 11 – 4
7
7
21
7
7
=
= 7 =
11.
11 + 12 11 + 4 15 15
7 21 7 7
7
1 3 7 4 6 7 5
− +
− +
5
12. 2 4 8 = 8 8 8 = 8 =
1 3 7 4 6 7 3 3
+ −
+ −
2 4 8 8 8 8 8
13. 1 –
1
1
2 3 2 1
=1– =1– = – =
1
3
3 3 3 3
1+ 2
2
14. 2 +
15.
(
3
5
1+
2
5+ 3
3
3
= 2+
2 5
7
−
2 2
2
6 14 6 20
= 2+ = + =
7 7 7 7
= 2+
)(
) ( 5) – ( 3)
5– 3 =
2
2
=5–3= 2
Section 0.1
1
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16.
(
5− 3
) = ( 5)
2
2
−2
( 5 )( 3 ) + ( 3 )
2
27.
= 5 − 2 15 + 3 = 8 − 2 15
17. (3x − 4)( x + 1) = 3 x 2 + 3 x − 4 x − 4
= 3x2 − x − 4
18. (2 x − 3)2 = (2 x − 3)(2 x − 3)
12
4
2
+
x + 2x x x + 2
12
4( x + 2)
2x
=
+
+
x( x + 2) x( x + 2) x( x + 2)
12 + 4 x + 8 + 2 x 6 x + 20
=
=
x( x + 2)
x( x + 2)
2(3 x + 10)
=
x( x + 2)
2
+
= 4 x2 − 6 x − 6 x + 9
= 4 x 2 − 12 x + 9
19.
28.
(3x – 9)(2 x + 1) = 6 x 2 + 3 x –18 x – 9
2
y
+
2(3 y − 1) (3 y + 1)(3 y − 1)
2(3 y + 1)
2y
=
+
2(3 y + 1)(3 y − 1) 2(3 y + 1)(3 y − 1)
=
2
= 6 x –15 x – 9
20. (4 x − 11)(3x − 7) = 12 x 2 − 28 x − 33 x + 77
= 12 x 2 − 61x + 77
21. (3t 2 − t + 1) 2 = (3t 2 − t + 1)(3t 2 − t + 1)
4
3
2
3
2
2
y
+
6 y − 2 9 y2 −1
2
= 9t − 3t + 3t − 3t + t − t + 3t − t + 1
=
6y + 2 + 2y
8y + 2
=
2(3 y + 1)(3 y − 1) 2(3 y + 1)(3 y − 1)
=
2(4 y + 1)
4y +1
=
2(3 y + 1)(3 y − 1) (3 y + 1)(3 y − 1)
= 9t 4 − 6t 3 + 7t 2 − 2t + 1
0⋅0 = 0
b.
0
is undefined.
0
c.
0
=0
17
d.
3
is undefined.
0
e.
05 = 0
f. 170 = 1
29. a.
22. (2t + 3)3 = (2t + 3)(2t + 3)(2t + 3)
= (4t 2 + 12t + 9)(2t + 3)
= 8t 3 + 12t 2 + 24t 2 + 36t + 18t + 27
= 8t 3 + 36t 2 + 54t + 27
23.
x 2 – 4 ( x – 2)( x + 2)
=
= x+2, x ≠ 2
x–2
x–2
24.
x 2 − x − 6 ( x − 3)( x + 2)
=
= x+2, x ≠3
x−3
( x − 3)
25.
t 2 – 4t – 21 (t + 3)(t – 7)
=
= t – 7 , t ≠ −3
t +3
t +3
26.
2
2x − 2x
3
2
2
x − 2x + x
=
2 x(1 − x)
2
x( x − 2 x + 1)
−2 x( x − 1)
=
x( x − 1)( x − 1)
2
=−
x −1
Section 0.1
0
= a , then 0 = 0 ⋅ a , but this is meaningless
0
because a could be any real number. No
0
single value satisfies = a .
0
30. If
31.
.083
12 1.000
96
40
36
4
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
32.
.285714
7 2.000000
14
60
56
40
35
50
49
10
7
30
28
2
33.
.142857
21 3.000000
21
90
84
60
42
180
168
120
105
150
147
3
34.
.294117...
17 5.000000... → 0.2941176470588235
34
160
153
70
68
20
17
30
17
130
119
11
Instructor’s Resource Manual
35.
3.6
3 11.0
9
20
18
2
36.
.846153
13 11.000000
10 4
60
52
80
78
20
13
70
65
50
39
11
37. x = 0.123123123...
1000 x = 123.123123...
x = 0.123123...
999 x = 123
123 41
x=
=
999 333
38. x = 0.217171717 …
1000 x = 217.171717...
10 x = 2.171717...
990 x = 215
215 43
x=
=
990 198
39. x = 2.56565656...
100 x = 256.565656...
x = 2.565656...
99 x = 254
254
x=
99
40. x = 3.929292…
100 x = 392.929292...
x = 3.929292...
99 x = 389
389
x=
99
Section 0.1
3
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
41. x = 0.199999...
100 x = 19.99999...
52.
10 x = 1.99999...
90 x = 18
18 1
x=
=
90 5
54.
55.
10 x = 3.99999...
90 x = 36
36 2
x=
=
90 5
56.
43. Those rational numbers that can be expressed
by a terminating decimal followed by zeros.
⎛1⎞
p
1
= p ⎜ ⎟ , so we only need to look at . If
q
q
⎝q⎠
q = 2n ⋅ 5m , then
n
m
1 ⎛1⎞ ⎛1⎞
= ⎜ ⎟ ⋅ ⎜ ⎟ = (0.5)n (0.2)m . The product
q ⎝ 2⎠ ⎝5⎠
of any number of terminating decimals is also a
n
m
terminating decimal, so (0.5) and (0.2) ,
and hence their product,
decimal. Thus
1
, is a terminating
q
p
has a terminating decimal
q
expansion.
45. Answers will vary. Possible answer: 0.000001,
1
≈ 0.0000010819...
π 12
46. Smallest positive integer: 1; There is no
smallest positive rational or irrational number.
47. Answers will vary. Possible answer:
3.14159101001...
48. There is no real number between 0.9999…
(repeating 9's) and 1. 0.9999… and 1 represent
the same real number.
49. Irrational
50. Answers will vary. Possible answers:
−π and π , − 2 and 2
51. ( 3 + 1)3 ≈ 20.39230485
4
Section 0.1
2− 3
)
4
≈ 0.0102051443
53. 4 1.123 – 3 1.09 ≈ 0.00028307388
42. x = 0.399999…
100 x = 39.99999...
44.
(
( 3.1415 )−1/ 2 ≈ 0.5641979034
8.9π2 + 1 – 3π ≈ 0.000691744752
4 (6π 2
− 2)π ≈ 3.661591807
57. Let a and b be real numbers with a b} . Since
a nonempty set of integers that is bounded
below contains a least element, there is a
k 0 ∈ S such that k 0 / n > b but
(k 0 − 1) / n ≤ b . Then
k0 − 1 k0 1
1
=
− >b− > a
n
n n
n
k 0 −1
k 0 −1
Thus, a < n ≤ b . If n < b , then choose
r=
k 0 −1
n
. Otherwise, choose r =
k0 − 2
n
.
1
<r.
n
Given a < b , choose r so that a < r1 < b . Then
choose r2 , r3 so that a < r2 < r1 < r3 < b , and so
on.
Note that a < b −
58. Answers will vary. Possible answer: ≈ 120 in 3
ft
= 21,120, 000 ft
mi
equator = 2π r = 2π (21,120, 000)
≈ 132, 700,874 ft
59. r = 4000 mi × 5280
60. Answers will vary. Possible answer:
beats
min
hr
day
× 60
× 24
× 365
× 20 yr
70
min
hr
day
year
= 735,840, 000 beats
2
⎛ 16
⎞
61. V = πr 2 h = π ⎜ ⋅12 ⎟ (270 ⋅12)
⎝ 2
⎠
≈ 93,807, 453.98 in.3
volume of one board foot (in inches):
1× 12 × 12 = 144 in.3
number of board feet:
93,807, 453.98
≈ 651, 441 board ft
144
Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
b. Every circle has area less than or equal to
9π. The original statement is true.
62. V = π (8.004) 2 (270) − π (8)2 (270) ≈ 54.3 ft.3
63. a.
If I stay home from work today then it
rains. If I do not stay home from work,
then it does not rain.
b. If the candidate will be hired then she
meets all the qualifications. If the
candidate will not be hired then she does
not meet all the qualifications.
64. a.
c.
Some real number is less than or equal to
its square. The negation is true.
71. a.
True; If x is positive, then x 2 is positive.
b. False; Take x = −2 . Then x 2 > 0 but
x<0.
If I pass the course, then I got an A on the
final exam. If I did not pass the course,
thn I did not get an A on the final exam.
c.
2
e.
2
a + b = c . If a triangle is not a right
2
2
72. a.
2
triangle, then a + b ≠ c .
c.
If angle ABC is an acute angle, then its
measure is 45o. If angle ABC is not an
acute angle, then its measure is not 45o.
2
2
2
The statement, converse, and
contrapositive are all true.
True; 1/ 2n can be made arbitrarily close
to 0.
73. a.
If n is odd, then there is an integer k such
that n = 2k + 1. Then
n 2 = (2k + 1) 2 = 4k 2 + 4k + 1
= 2(2k 2 + 2k ) + 1
The statement and contrapositive are true.
The converse is false.
Some isosceles triangles are not
equilateral. The negation is true.
b. All real numbers are integers. The original
statement is true.
c.
70. a.
True; Let x be any number. Take
1
1
y = + 1 . Then y > .
x
x
e.
b.
b. The statement, converse, and
contrapositive are all false.
69. a.
True; x + ( − x ) < x + 1 + ( − x ) : 0 < 1
2
b. The statement, converse, and
contrapositive are all true.
68. a.
True; Let y be any positive number. Take
y
x = . Then 0 < x < y .
2
d. True; 1/ n can be made arbitrarily close
to 0.
b. If a < b then a < b. If a ≥ b then
a ≥ b.
67. a.
<x
b. False; There are infinitely many prime
numbers.
b. If the measure of angle ABC is greater than
0o and less than 90o, it is acute. If the
measure of angle ABC is less than 0o or
greater than 90o, then it is not acute.
66. a.
1
4
y = x 2 + 1 . Then y > x 2 .
If a triangle is a right triangle, then
2
1
2
. Then x =
2
d. True; Let x be any number. Take
b. If I take off next week, then I finished my
research paper. If I do not take off next
week, then I did not finish my research
paper.
65. a.
False; Take x =
Some natural number is larger than its
square. The original statement is true.
Prove the contrapositive. Suppose n is
even. Then there is an integer k such that
n = 2k . Then n 2 = (2k )2 = 4k 2 = 2(2k 2 ) .
Thus n 2 is even.
Parts (a) and (b) prove that n is odd if and
74.
only if n 2 is odd.
75. a.
b.
243 = 3 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3
124 = 4 ⋅ 31 = 2 ⋅ 2 ⋅ 31 or 22 ⋅ 31
Some natural number is not rational. The
original statement is true.
Instructor’s Resource Manual
Section 0.1
5
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5100 = 2 ⋅ 2550 = 2 ⋅ 2 ⋅1275
c.
82. a.
= 2 ⋅ 2 ⋅ 3 ⋅ 425 = 2 ⋅ 2 ⋅ 3 ⋅ 5 ⋅ 85
= 2 ⋅ 2 ⋅ 3 ⋅ 5 ⋅ 5 ⋅17 or 22 ⋅ 3 ⋅ 52 ⋅17
c.
76. For example, let A = b ⋅ c 2 ⋅ d 3 ; then
A2 = b 2 ⋅ c 4 ⋅ d 6 , so the square of the number
is the product of primes which occur an even
number of times.
77.
p
p2
;2 =
; 2q 2 = p 2 ; Since the prime
2
q
q
2
factors of p must occur an even number of
p
times, 2q2 would not be valid and = 2
q
must be irrational.
3=
p
p2
; 3=
; 3q 2 = p 2 ; Since the prime
q
q2
factors of p 2 must occur an even number of
times, 3q 2 would not be valid and
e.
f.
83. a.
p
= 3
q
x = 2.4444...;
10 x = 24.4444...
x = 2.4444...
9 x = 22
22
x=
9
2
3
n = 1: x = 0, n = 2: x = , n = 3: x = – ,
3
2
5
n = 4: x =
4
3
The upper bound is .
2
2
Answers will vary. Possible answer: An
example
is S = {x : x 2 < 5, x a rational number}.
Here the least upper bound is 5, which is
real but irrational.
must be irrational.
79. Let a, b, p, and q be natural numbers, so
b. –2
d. 1
2=
78.
–2
a
b
p
a p aq + bp
are rational. + =
This
q
b q
bq
sum is the quotient of natural numbers, so it is
also rational.
and
b. True
0.2 Concepts Review
1. [−1,5); (−∞, −2]
2. b > 0; b < 0
p
80. Assume a is irrational, ≠ 0 is rational, and
q
p r
q⋅r
is
= is rational. Then a =
q s
p⋅s
rational, which is a contradiction.
a⋅
81. a.
– 9 = –3; rational
b.
3
0.375 = ; rational
8
c.
(3 2)(5 2) = 15 4 = 30; rational
d.
(1 + 3)2 = 1 + 2 3 + 3 = 4 + 2 3;
irrational
3. (b) and (c)
4. −1 ≤ x ≤ 5
Problem Set 0.2
1. a.
b.
c.
d.
6
Section 0.2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
–3 < 1 – 6 x ≤ 4
9. –4 < –6 x ≤ 3
e.
2
1 ⎡ 1 2⎞
> x ≥ – ; ⎢– , ⎟
3
2 ⎣ 2 3⎠
f.
2. a.
c.
(2, 7)
(−∞, −2]
b.
d.
[−3, 4)
[−1, 3]
10.
3. x − 7 < 2 x − 5
−2 < x;( − 2, ∞)
4 < 5 − 3x < 7
−1 < −3x < 2
1
2 ⎛ 2 1⎞
> x > − ; ⎜− , ⎟
3
3 ⎝ 3 3⎠
4. 3x − 5 < 4 x − 6
1 < x; (1, ∞ )
11. x2 + 2x – 12 < 0;
x=
5.
7 x – 2 ≤ 9x + 3
–5 ≤ 2 x
= –1 ± 13
(
7. −4 < 3 x + 2 < 5
−6 < 3 x < 3
−2 < x < 1; (−2, −1)
)
(
)
⎡ x – –1 + 13 ⎤ ⎡ x – –1 – 13 ⎤ < 0;
⎣
⎦⎣
⎦
5 ⎡ 5 ⎞
x ≥ – ; ⎢– , ∞ ⎟
2 ⎣ 2 ⎠
6. 5 x − 3 > 6 x − 4
1 > x;(−∞,1)
–2 ± (2)2 – 4(1)(–12) –2 ± 52
=
2(1)
2
( –1 –
13, – 1 + 13
)
12. x 2 − 5 x − 6 > 0
( x + 1)( x − 6) > 0;
(−∞, −1) ∪ (6, ∞)
13. 2x2 + 5x – 3 > 0; (2x – 1)(x + 3) > 0;
⎛1 ⎞
(−∞, −3) ∪ ⎜ , ∞ ⎟
⎝2 ⎠
8. −3 < 4 x − 9 < 11
6 < 4 x < 20
3
⎛3 ⎞
< x < 5; ⎜ ,5 ⎟
2
⎝2 ⎠
14.
⎛ 3 ⎞
(4 x + 3)( x − 2) < 0; ⎜ − , 2 ⎟
⎝ 4 ⎠
15.
Instructor’s Resource Manual
4 x2 − 5x − 6 < 0
x+4
≤ 0; [–4, 3)
x–3
Section 0.2
7
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16.
3x − 2
2⎤
⎛
≥ 0; ⎜ −∞, ⎥ ∪ (1, ∞)
x −1
3⎦
⎝
3
>2
x+5
20.
3
−2 > 0
x+5
17.
2
−5 < 0
x
2 − 5x
< 0;
x
⎛2 ⎞
(– ∞, 0) ∪ ⎜ , ∞ ⎟
⎝5 ⎠
18.
3 − 2( x + 5)
>0
x+5
2
<5
x
7
≤7
4x
7
−7 ≤ 0
4x
7 − 28 x
≤ 0;
4x
1
( −∞, 0 ) ∪ ⎡⎢ , ∞ ⎞⎟
⎣4 ⎠
−2 x − 7
7⎞
⎛
> 0; ⎜ −5, − ⎟
2⎠
x+5
⎝
21. ( x + 2)( x − 1)( x − 3) > 0; (−2,1) ∪ (3,8)
3⎞ ⎛1 ⎞
⎛
22. (2 x + 3)(3x − 1)( x − 2) < 0; ⎜ −∞, − ⎟ ∪ ⎜ , 2 ⎟
2⎠ ⎝3 ⎠
⎝
3⎤
⎛
23. (2 x - 3)( x -1)2 ( x - 3) ≥ 0; ⎜ – ∞, ⎥ ∪ [3, ∞ )
2⎦
⎝
24. (2 x − 3)( x − 1) 2 ( x − 3) > 0;
19.
( −∞,1) ∪ ⎛⎜1,
3⎞
⎟ ∪ ( 3, ∞ )
⎝ 2⎠
1
≤4
3x − 2
1
−4≤ 0
3x − 2
1 − 4(3 x − 2)
≤0
3x − 2
9 − 12 x
2 ⎞ ⎡3 ⎞
⎛
≤ 0; ⎜ −∞, ⎟ ∪ ⎢ , ∞ ⎟
3x − 2
3 ⎠ ⎣4 ⎠
⎝
25.
x3 – 5 x 2 – 6 x < 0
x( x 2 – 5 x – 6) < 0
x( x + 1)( x – 6) < 0;
(−∞, −1) ∪ (0, 6)
26. x3 − x 2 − x + 1 > 0
( x 2 − 1)( x − 1) > 0
( x + 1)( x − 1) 2 > 0;
(−1,1) ∪ (1, ∞)
8
Section 0.2
27. a.
False.
c.
False.
b.
True.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
28. a.
True.
c.
False.
29. a.
b.
True.
33. a.
( x + 1)( x 2 + 2 x – 7) ≥ x 2 – 1
x3 + 3 x 2 – 5 x – 7 ≥ x 2 – 1
x3 + 2 x 2 – 5 x – 6 ≥ 0
( x + 3)( x + 1)( x – 2) ≥ 0
[−3, −1] ∪ [2, ∞)
⇒ Let a < b , so ab 0 , we can divide by 2b to get
b−a > 0.
( x 2 − 4)( x 2 + 2) ≥ 0
( x 2 + 2)( x + 2)( x − 2) ≥ 0
b. We can divide or multiply an inequality by
any positive number.
a
1 1
a 1 and 2x + 1 < 3
3x > –6 and 2x < 2
x > –2 and x < 1; (–2, 1)
( x 2 − 4)( x 2 − 1) < 0
( x + 2)( x + 1)( x − 1)( x − 2) < 0
(−2, −1) ∪ (1, 2)
34. a.
32. a.
3x + 7 > 1 and 2x + 1 < –4
5
x > –2 and x < – ; ∅
2
1 ⎞
⎛ 1
,
⎟
⎜
2
.
01
1
.
99 ⎠
⎝
2 x − 7 > 1 or 2 x + 1 < 3
b.
x > 4 or x < 1
(−∞,1) ∪ (4, ∞)
2.99 <
1
< 3.01
x+2
2.99( x + 2) < 1 < 3.01( x + 2)
2.99 x + 5.98 < 1 and 1 < 3.01x + 6.02
− 4.98 and
− 5.02
x<
x>
2 x − 7 ≤ 1 or 2 x + 1 < 3
2 x ≤ 8 or 2 x < 2
2.99
5.02
4.98
−
<x<−
3.01
2.99
⎛ 5.02 4.98 ⎞
,−
⎜−
⎟
⎝ 3.01 2.99 ⎠
x ≤ 4 or x < 1
(−∞, 4]
c.
1
< 2.01
x
1
1
<x<
2.01
1.99
2 x > 8 or 2 x < 2
b.
1.99 <
1.99 x < 1 < 2.01x
1.99 x < 1 and 1 < 2.01x
1
and x > 1
x<
1.99
2.01
b. 3x + 7 > 1 and 2x + 1 > –4
3x > –6 and 2x > –5
5
x > –2 and x > – ; ( −2, ∞ )
2
c.
( x 2 + 1)2 − 7( x 2 + 1) + 10 < 0
2 x − 7 ≤ 1 or 2 x + 1 > 3
3.01
2 x ≤ 8 or 2 x > 2
x ≤ 4 or x > 1
(−∞, ∞)
35.
x − 2 ≥ 5;
x − 2 ≤ −5 or x − 2 ≥ 5
x ≤ −3 or x ≥ 7
(−∞, −3] ∪ [7, ∞)
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Section 0.2
9
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
36.
x + 2 < 1;
–1 < x + 2 < 1
43.
–3 < x < –1
(–3, –1)
37.
4 x + 5 ≤ 10;
−10 ≤ 4 x + 5 ≤ 10
−15 ≤ 4 x ≤ 5
−
38.
15
5 ⎡ 15 5 ⎤
≤ x ≤ ; ⎢− , ⎥
4
4 ⎣ 4 4⎦
2 x – 1 > 2;
2x – 1 < –2 or 2x – 1 > 2
2x < –1 or 2x > 3;
1
3 ⎛
1⎞ ⎛3 ⎞
x < – or x > , ⎜ – ∞, – ⎟ ∪ ⎜ , ∞ ⎟
2
2 ⎝
2⎠ ⎝2 ⎠
39.
40.
2x
−5 ≥ 7
7
2x
2x
− 5 ≤ −7 or
−5 ≥ 7
7
7
2x
2x
≤ −2 or
≥ 12
7
7
x ≤ −7 or x ≥ 42;
(−∞, −7] ∪ [42, ∞)
x
+1 < 1
4
x
−1 < + 1 < 1
4
x
−2 < < 0;
4
–8 < x < 0; (–8, 0)
41. 5 x − 6 > 1;
5 x − 6 < −1 or 5 x − 6 > 1
5 x < 5 or 5 x > 7
7
⎛7 ⎞
x < 1 or x > ;(−∞,1) ∪ ⎜ , ∞ ⎟
5
⎝5 ⎠
42.
2 x – 7 > 3;
2x – 7 < –3 or 2x – 7 > 3
2x < 4 or 2x > 10
x < 2 or x > 5; (−∞, 2) ∪ (5, ∞)
44.
1
− 3 > 6;
x
1
1
− 3 < −6 or − 3 > 6
x
x
1
1
+ 3 < 0 or − 9 > 0
x
x
1 + 3x
1− 9x
< 0 or
> 0;
x
x
⎛ 1 ⎞ ⎛ 1⎞
⎜ − , 0 ⎟ ∪ ⎜ 0, ⎟
⎝ 3 ⎠ ⎝ 9⎠
5
> 1;
x
5
5
2 + < –1 or 2 + > 1
x
x
5
5
3 + < 0 or 1 + > 0
x
x
3x + 5
x+5
< 0 or
> 0;
x
x
⎛ 5 ⎞
(– ∞, – 5) ∪ ⎜ – , 0 ⎟ ∪ (0, ∞)
⎝ 3 ⎠
2+
45. x 2 − 3x − 4 ≥ 0;
x=
3 ± (–3)2 – 4(1)(–4) 3 ± 5
=
= –1, 4
2(1)
2
( x + 1)( x − 4) = 0; (−∞, −1] ∪ [4, ∞)
4 ± (−4)2 − 4(1)(4)
=2
2(1)
( x − 2)( x − 2) ≤ 0; x = 2
46. x 2 − 4 x + 4 ≤ 0; x =
47. 3x2 + 17x – 6 > 0;
x=
–17 ± (17) 2 – 4(3)(–6) –17 ± 19
1
=
= –6,
2(3)
6
3
⎛1 ⎞
(3x – 1)(x + 6) > 0; (– ∞, – 6) ∪ ⎜ , ∞ ⎟
⎝3 ⎠
48. 14 x 2 + 11x − 15 ≤ 0;
−11 ± (11) 2 − 4(14)(−15) −11 ± 31
=
2(14)
28
3 5
x=− ,
2 7
3 ⎞⎛
5⎞
⎛
⎡ 3 5⎤
⎜ x + ⎟ ⎜ x − ⎟ ≤ 0; ⎢ − , ⎥
2 ⎠⎝
7⎠
⎝
⎣ 2 7⎦
x=
49. x − 3 < 0.5 ⇒ 5 x − 3 < 5(0.5) ⇒ 5 x − 15 < 2.5
50. x + 2 < 0.3 ⇒ 4 x + 2 < 4(0.3) ⇒ 4 x + 18 < 1.2
10
Section 0.2
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51.
x−2 <
52.
x+4 <
ε
6
ε
2
⇒ 6 x − 2 < ε ⇒ 6 x − 12 < ε
59.
x –1 < 2 x – 6
( x –1) 2 < (2 x – 6)2
⇒ 2 x + 4 < ε ⇒ 2x + 8 < ε
x 2 – 2 x + 1 < 4 x 2 – 24 x + 36
3x 2 – 22 x + 35 > 0
53. 3x − 15 < ε ⇒ 3( x − 5) < ε
(3x – 7)( x – 5) > 0;
⇒ 3 x−5 < ε
⇒ x−5 <
54.
ε
3
;δ =
7⎞
⎛
⎜ – ∞, ⎟ ∪ (5, ∞)
3⎠
⎝
ε
3
4 x − 8 < ε ⇒ 4( x − 2) < ε
60.
55.
ε
4
;δ =
4 x2 − 4 x + 1 ≥ x2 + 2 x + 1
4
3x2 − 6 x ≥ 0
3 x( x − 2) ≥ 0
(−∞, 0] ∪ [2, ∞)
⇒ 6 x+6 <ε
ε
6
;δ =
2
ε
6 x + 36 < ε ⇒ 6( x + 6) < ε
⇒ x+6 <
2x −1 ≥ x + 1
(2 x − 1)2 ≥ ( x + 1)
⇒ 4 x−2 <ε
⇒ x−2 <
x –1 < 2 x – 3
ε
6
61.
2 2 x − 3 < x + 10
4 x − 6 < x + 10
56. 5 x + 25 < ε ⇒ 5( x + 5) < ε
(4 x − 6) 2 < ( x + 10)2
⇒ 5 x+5 <ε
16 x 2 − 48 x + 36 < x 2 + 20 x + 100
⇒ x+5 <
ε
5
;δ =
ε
15 x 2 − 68 x − 64 < 0
5
(5 x + 4)(3 x − 16) < 0;
57. C = π d
C – 10 ≤ 0.02
πd – 10 ≤ 0.02
10 ⎞
⎛
π ⎜ d – ⎟ ≤ 0.02
π⎠
⎝
10 0.02
d–
≤
≈ 0.0064
π
π
We must measure the diameter to an accuracy
of 0.0064 in.
58. C − 50 ≤ 1.5,
5
( F − 32 ) − 50 ≤ 1.5;
9
5
( F − 32 ) − 90 ≤ 1.5
9
F − 122 ≤ 2.7
⎛ 4 16 ⎞
⎜– , ⎟
⎝ 5 3⎠
3x − 1 < 2 x + 6
62.
3x − 1 < 2 x + 12
(3x − 1) 2 < (2 x + 12)2
9 x 2 − 6 x + 1 < 4 x 2 + 48 x + 144
5 x 2 − 54 x − 143 < 0
( 5 x + 11)( x − 13) < 0
⎛ 11 ⎞
⎜ − ,13 ⎟
⎝ 5
⎠
We are allowed an error of 2.7 F.
Instructor’s Resource Manual
Section 0.2
11
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
63.
x < y ⇒ x x ≤ x y and x y < y y
2
⇒ x < y
Order property: x < y ⇔ xz < yz when z is positive.
2
Transitivity
(x
⇒ x2 < y 2
Conversely,
2
2
(x
2
x2 < y 2 ⇒ x < y
2
= x
2
2
= x2
)
)
2
2
⇒ x – y <0
Subtract y from each side.
⇒ ( x – y )( x + y ) < 0 Factor the difference of two squares.
⇒ x – y <0
⇒ x < y
64. 0 < a < b ⇒ a =
( a) < ( b)
2
a <
2
This is the only factor that can be negative.
Add y to each side.
( a)
2
and b =
( b)
2
, so
67.
, and, by Problem 63,
x +9
x–2
a + b + c = ( a + b) + c ≤ a + b + c
≤ a+b+c
66.
1
x2 + 3
−
⎛
1
1
1 ⎞
=
+⎜−
⎟
⎜
2
x +2
x
+ 2 ⎟⎠
x +3 ⎝
≤
1
2
x +3
+−
68.
1
x +2
1
=
+
2
x +2
x +3
1
2
+
x 2 + 3 ≥ 3 and x + 2 ≥ 2, so
1
2
x +3
1
2
x +3
12
≤
1
1
1
≤ , thus,
and
x +2 2
3
+
1
1 1
≤ +
x +2 3 2
Section 0.2
x2 + 9
x
2
x +9
x
+
–2
2
x ≤ 2 ⇒ x2 + 2 x + 7 ≤ x2 + 2 x + 7
Thus,
x2 + 2 x + 7
2
x +1
= x2 + 2 x + 7
1
2
x +1
≤ 15 ⋅1 = 15
1
x +2
x +3
by the Triangular Inequality, and since
1
1
x 2 + 3 > 0, x + 2 > 0 ⇒
> 0,
> 0.
2
x
+2
x +3
≤
x + (–2)
≤ 4 + 4 + 7 = 15
1
and x 2 + 1 ≥ 1 so
≤ 1.
2
x +1
1
=
=
x +9
x +2
2
≤
+
=
2
2
2
x + 9 x + 9 x + 9 x2 + 9
1
1
Since x 2 + 9 ≥ 9,
≤
2
x +9 9
x +2 x +2
≤
9
x2 + 9
x
+2
x–2
≤
2
9
x +9
b ⇒ a< b.
of absolute values.
c.
x +9
2
a – b ≥ a – b ≥ a – b Use Property 4
b.
2
x–2
a – b = a + (–b) ≤ a + –b = a + b
65. a.
x–2
69.
1 3 1 2 1
1
x + x + x+
2
4
8
16
1
1
1
1
≤ x 4 + x3 + x 2 + x +
2
4
8
16
1 1 1 1
≤ 1+ + + +
since x ≤ 1.
2 4 8 16
1
1
1
1
≤ 1.9375 < 2.
So x 4 + x3 + x 2 + x +
2
4
8
16
x4 +
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x < x2
70. a.
77.
x − x2 < 0
x(1 − x) < 0
x < 0 or x > 1
1 11
≤
R 60
2
x <x
b.
2
x −x<0
x( x − 1) < 0
0 < x <1
R≥
60
11
1
1
1
1
≥
+
+
R 20 30 40
1 6+4+3
≥
R
120
120
R≤
13
71. a ≠ 0 ⇒
2
1⎞
1
⎛
0 ≤ ⎜ a – ⎟ = a2 – 2 +
a
⎝
⎠
a2
1
1
or a 2 +
≥2.
so, 2 ≤ a 2 +
2
a
a2
Thus,
72. a < b
a + a < a + b and a + b < b + b
2a < a + b < 2b
a+b
a<
<b
2
60
120
≤R≤
11
13
78. A = 4π r 2 ; A = 4π (10)2 = 400π
4π r 2 − 400π < 0.01
4π r 2 − 100 < 0.01
73. 0 < a < b
r 2 − 100 <
a 2 < ab and ab < b 2
74.
1 1 1
1
≤ +
+
R 10 20 30
1 6+3+ 2
≤
R
60
0.01
4π
0.01 2
0.01
< r − 100 <
4π
4π
a 2 < ab < b 2
−
a < ab < b
0.01
0.01
< r < 100 +
4π
4π
δ ≈ 0.00004 in
100 −
(
)
1
1
( a + b ) ⇔ ab ≤ a 2 + 2ab + b2
2
4
1 2 1
1 2 1 2
⇔ 0 ≤ a − ab + b = a − 2ab + b 2
4
2
4
4
1
2
⇔ 0 ≤ (a − b) which is always true.
4
ab ≤
(
)
0.3 Concepts Review
1.
75. For a rectangle the area is ab, while for a
2
⎛ a+b⎞
square the area is a 2 = ⎜
⎟ . From
⎝ 2 ⎠
1
⎛ a+b⎞
(a + b) ⇔ ab ≤ ⎜
⎟
2
⎝ 2 ⎠
so the square has the largest area.
Problem 74,
ab ≤
76. 1 + x + x 2 + x3 + … + x99 ≤ 0;
(−∞, −1]
Instructor’s Resource Manual
( x + 2)2 + ( y − 3)2
2. (x + 4)2 + (y – 2)2 = 25
2
⎛ −2 + 5 3 + 7 ⎞
3. ⎜
,
⎟ = (1.5,5)
2 ⎠
⎝ 2
4.
d −b
c−a
Section 0.3
13
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem Set 0.3
5. d1 = (5 + 2) 2 + (3 – 4)2 = 49 + 1 = 50
d 2 = (5 − 10)2 + (3 − 8)2 = 25 + 25 = 50
1.
d3 = (−2 − 10)2 + (4 − 8)2
= 144 + 16 = 160
d1 = d 2 so the triangle is isosceles.
6. a = (2 − 4)2 + (−4 − 0) 2 = 4 + 16 = 20
b = (4 − 8)2 + (0 + 2)2 = 16 + 4 = 20
c = (2 − 8)2 + (−4 + 2) 2 = 36 + 4 = 40
d = (3 – 1)2 + (1 – 1)2 = 4 = 2
a 2 + b 2 = c 2 , so the triangle is a right triangle.
2.
7. (–1, –1), (–1, 3); (7, –1), (7, 3); (1, 1), (5, 1)
8.
( x − 3) 2 + (0 − 1) 2 = ( x − 6)2 + (0 − 4) 2 ;
x 2 − 6 x + 10 = x 2 − 12 x + 52
6 x = 42
x = 7 ⇒ ( 7, 0 )
d = (−3 − 2)2 + (5 + 2)2 = 74 ≈ 8.60
3.
⎛ –2 + 4 –2 + 3 ⎞ ⎛ 1 ⎞
9. ⎜
,
⎟ = ⎜1, ⎟ ;
2 ⎠ ⎝ 2⎠
⎝ 2
2
25
⎛1
⎞
d = (1 + 2)2 + ⎜ – 3 ⎟ = 9 +
≈ 3.91
2
4
⎝
⎠
⎛1+ 2 3 + 6 ⎞ ⎛ 3 9 ⎞
10. midpoint of AB = ⎜
,
⎟=⎜ , ⎟
2 ⎠ ⎝2 2⎠
⎝ 2
⎛ 4 + 3 7 + 4 ⎞ ⎛ 7 11 ⎞
midpoint of CD = ⎜
,
⎟=⎜ , ⎟
2 ⎠ ⎝2 2 ⎠
⎝ 2
2
⎛ 3 7 ⎞ ⎛ 9 11 ⎞
d = ⎜ − ⎟ +⎜ − ⎟
⎝2 2⎠ ⎝2 2 ⎠
2
= 4 + 1 = 5 ≈ 2.24
d = (4 – 5)2 + (5 + 8) 2 = 170 ≈ 13.04
4.
11 (x – 1)2 + (y – 1)2 = 1
12. ( x + 2)2 + ( y − 3)2 = 42
( x + 2)2 + ( y − 3)2 = 16
13. ( x − 2) 2 + ( y + 1) 2 = r 2
(5 − 2)2 + (3 + 1) 2 = r 2
r 2 = 9 + 16 = 25
( x − 2) 2 + ( y + 1) 2 = 25
d = (−1 − 6)2 + (5 − 3) 2 = 49 + 4 = 53
≈ 7.28
14
Section 0.3
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
14. ( x − 4) 2 + ( y − 3) 2 = r 2
21. 4 x 2 + 16 x + 15 + 4 y 2 + 6 y = 0
(6 − 4) 2 + (2 − 3) 2 = r 2
3
9⎞
9
⎛
4( x 2 + 4 x + 4) + 4 ⎜ y 2 + y + ⎟ = −15 + 16 +
2
16 ⎠
4
⎝
2
r = 4 +1 = 5
2
3⎞
13
⎛
4( x + 2)2 + 4 ⎜ y + ⎟ =
4⎠
4
⎝
( x − 4)2 + ( y − 3)2 = 5
⎛ 1+ 3 3 + 7 ⎞
15. center = ⎜
,
⎟ = (2, 5)
2 ⎠
⎝ 2
1
1
radius =
(1 – 3)2 + (3 – 7)2 =
4 + 16
2
2
1
=
20 = 5
2
2
2
( x – 2) + ( y – 5) = 5
2
3⎞
13
⎛
( x + 2)2 + ⎜ y + ⎟ =
4⎠
16
⎝
3⎞
⎛
center = ⎜ −2, − ⎟ ; radius =
4⎠
⎝
105
+ 4 y2 + 3 y = 0
16
3
9 ⎞
⎛
2
4( x + 4 x + 4) + 4 ⎜ y 2 + y + ⎟
4
64 ⎠
⎝
105
9
=−
+ 16 +
16
16
22. 4 x 2 + 16 x +
16. Since the circle is tangent to the x-axis, r = 4.
( x − 3)2 + ( y − 4) 2 = 16
17. x 2 + 2 x + 10 + y 2 – 6 y –10 = 0
2
3⎞
⎛
4( x + 2)2 + 4 ⎜ y + ⎟ = 10
8
⎝
⎠
x2 + 2 x + y 2 – 6 y = 0
2
( x 2 + 2 x + 1) + ( y 2 – 6 y + 9) = 1 + 9
3⎞
5
⎛
( x + 2)2 + ⎜ y + ⎟ =
8⎠
2
⎝
( x + 1) 2 + ( y – 3) 2 = 10
3⎞
5
10
⎛
center = ⎜ −2, − ⎟ ; radius =
=
8
2
2
⎝
⎠
center = (–1, 3); radius = 10
x 2 + y 2 − 6 y = 16
18.
x 2 + ( y 2 − 6 y + 9) = 16 + 9
23.
2 –1
=1
2 –1
24.
7−5
=2
4−3
25.
–6 – 3 9
=
–5 – 2 7
26.
−6 + 4
=1
0−2
27.
5–0
5
=–
0–3
3
28.
6−0
=1
0+6
x 2 + ( y − 3) 2 = 25
center = (0, 3); radius = 5
19. x 2 + y 2 –12 x + 35 = 0
x 2 –12 x + y 2 = –35
( x 2 –12 x + 36) + y 2 = –35 + 36
( x – 6) 2 + y 2 = 1
center = (6, 0); radius = 1
x 2 + y 2 − 10 x + 10 y = 0
20.
2
29.
y − 2 = −1( x − 2)
y − 2 = −x + 2
x+ y−4 = 0
30.
y − 4 = −1( x − 3)
y − 4 = −x + 3
2
( x − 10 x + 25) + ( y + 10 y + 25) = 25 + 25
x+ y−7 = 0
( x − 5) 2 + ( y + 5)2 = 50
center = ( 5, −5 ) ; radius = 50 = 5 2
13
4
31.
y = 2x + 3
2x – y + 3 = 0
32.
Instructor’s Resource Manual
y = 0x + 5
0x + y − 5 = 0
Section 0.3
15
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
33. m =
8–3 5
= ;
4–2 2
5
y – 3 = ( x – 2)
2
2 y – 6 = 5 x – 10
c.
3 y = –2 x + 6
y=–
x − 4y + 0 = 0
2
y + 3 = – ( x – 3)
3
2
y = – x –1
3
d.
3
m= ;
2
3
( x – 3)
2
3
15
y= x–
2
2
y+3=
2
1
2
35. 3y = –2x + 1; y = – x + ; slope = – ;
3
3
3
1
y -intercept =
3
–1 – 2
3
=– ;
3 +1
4
3
y + 3 = – ( x – 3)
4
3
3
y=– x–
4
4
36. −4 y = 5 x − 6
5
3
y = − x+
4
2
5
3
slope = − ; y -intercept =
4
2
e.
m=
37. 6 – 2 y = 10 x – 2
–2 y = 10 x – 8
y = –5 x + 4;
slope = –5; y-intercept = 4
f.
x=3
38. 4 x + 5 y = −20
5 y = −4 x − 20
4
y = − x−4
5
4
slope = − ; y -intercept = − 4
5
39. a.
b.
m = 2;
y + 3 = 2( x – 3)
y = 2x – 9
1
m=– ;
2
1
y + 3 = – ( x – 3)
2
1
3
y=– x–
2
2
16
Section 0.3
2
x + 2;
3
2
m=– ;
3
5x – 2 y – 4 = 0
2 −1 1
34. m =
= ;
8−4 4
1
y − 1 = ( x − 4)
4
4y − 4 = x − 4
2x + 3 y = 6
40. a.
g. y = –3
3 x + cy = 5
3(3) + c(1) = 5
c = −4
b.
c=0
c.
2 x + y = −1
y = −2 x − 1
m = −2;
3x + cy = 5
cy = −3x + 5
3
5
y = − x+
c
c
3
−2 = −
c
3
c=
2
d. c must be the same as the coefficient of x,
so c = 3.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
e.
y − 2 = 3( x + 3);
1
perpendicular slope = − ;
3
1
3
− =−
3
c
c=9
3
( x + 2)
2
3
y = x+2
2
y +1 =
b.
c.
2x + 3 y = 4
9 x – 3 y = –15
= –11
11x
x = –1
–3(–1) + y = 5
3
41. m = ;
2
42. a.
45. 2 x + 3 y = 4
–3x + y = 5
m = 2;
kx − 3 y = 10
−3 y = − kx + 10
10
k
y = x−
3
3
k
= 2; k = 6
3
1
m=− ;
2
k
1
=−
3
2
3
k=−
2
2x + 3 y = 6
3 y = −2 x + 6
2
y = − x + 2;
3
3 k 3
9
m= ; = ; k=
2 3 2
2
y=2
Point of intersection: (–1, 2)
3 y = –2 x + 4
2
4
y = – x+
3
3
3
m=
2
3
y − 2 = ( x + 1)
2
3
7
y = x+
2
2
46. 4 x − 5 y = 8
2 x + y = −10
4x − 5 y = 8
−4 x − 2 y = 20
− 7 y = 28
y = −4
4 x − 5(−4) = 8
4 x = −12
x = −3
Point of intersection: ( −3, −4 ) ;
4x − 5 y = 8
−5 y = −4 x + 8
y=
43. y = 3(3) – 1 = 8; (3, 9) is above the line.
b−0
b
=−
0−a
a
b
bx
x y
y = − x + b;
+ y = b; + = 1
a
a
a b
44. (a, 0), (0, b); m =
Instructor’s Resource Manual
m=−
4
8
x−
5
5
5
4
5
y + 4 = − ( x + 3)
4
5
31
y = − x−
4
4
Section 0.3
17
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
47. 3x – 4 y = 5
2x + 3y = 9
9 x – 12 y = 15
8 x + 12 y = 36
17 x
= 51
x=3
3(3) – 4 y = 5
–4 y = –4
y =1
Point of intersection: (3, 1); 3x – 4y = 5;
–4 y = –3x + 5
3
5
y= x–
4
4
4
m=–
3
4
y – 1 = – ( x – 3)
3
4
y = – x+5
3
48. 5 x – 2 y = 5
2x + 3y = 6
15 x – 6 y = 15
4 x + 6 y = 12
19 x
= 27
27
x=
19
⎛ 27 ⎞
2⎜ ⎟ + 3y = 6
⎝ 19 ⎠
60
3y =
19
20
y=
19
⎛ 27 20 ⎞
Point of intersection: ⎜ , ⎟ ;
⎝ 19 19 ⎠
5x − 2 y = 5
–2 y = –5 x + 5
5
5
y= x–
2
2
2
m=–
5
20
2⎛
27 ⎞
y–
= – ⎜x− ⎟
19
5⎝
19 ⎠
2
54 20
y = – x+ +
5
95 19
2
154
y = − x+
5
95
18
Section 0.3
⎛ 2 + 6 –1 + 3 ⎞
,
49. center: ⎜
⎟ = (4, 1)
2 ⎠
⎝ 2
⎛ 2+ 6 3+3⎞
midpoint = ⎜
,
⎟ = (4, 3)
2 ⎠
⎝ 2
inscribed circle: radius = (4 – 4)2 + (1 – 3)2
= 4=2
2
( x – 4) + ( y – 1)2 = 4
circumscribed circle:
radius = (4 – 2)2 + (1 – 3)2 = 8
( x – 4)2 + ( y –1)2 = 8
50. The radius of each circle is 16 = 4. The centers
are (1, −2 ) and ( −9,10 ) . The length of the belt is
the sum of half the circumference of the first
circle, half the circumference of the second circle,
and twice the distance between their centers.
1
1
L = ⋅ 2π (4) + ⋅ 2π (4) + 2 (1 + 9)2 + (−2 − 10)2
2
2
= 8π + 2 100 + 144
≈ 56.37
51. Put the vertex of the right angle at the origin
with the other vertices at (a, 0) and (0, b). The
⎛a b⎞
midpoint of the hypotenuse is ⎜ , ⎟ . The
⎝ 2 2⎠
distances from the vertices are
2
2
a⎞ ⎛
b⎞
⎛
⎜a – ⎟ +⎜0 – ⎟ =
2
2⎠
⎝
⎠ ⎝
=
2
2
a⎞ ⎛
b⎞
⎛
⎜0 – ⎟ + ⎜b – ⎟ =
2⎠ ⎝
2⎠
⎝
=
2
2
a⎞ ⎛
b⎞
⎛
⎜0 – ⎟ + ⎜0 – ⎟ =
2⎠ ⎝
2⎠
⎝
=
a 2 b2
+
4
4
1 2
a + b2 ,
2
a 2 b2
+
4
4
1 2
a + b 2 , and
2
a 2 b2
+
4
4
1 2
a + b2 ,
2
which are all the same.
52. From Problem 51, the midpoint of the
hypotenuse, ( 4,3, ) , is equidistant from the
vertices. This is the center of the circle. The
radius is 16 + 9 = 5. The equation of the
circle is
( x − 4) 2 + ( y − 3) 2 = 25.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
53. x 2 + y 2 – 4 x – 2 y – 11 = 0
( x 2 – 4 x + 4) + ( y 2 – 2 y + 1) = 11 + 4 + 1
( x – 2)2 + ( y – 1)2 = 16
x 2 + y 2 + 20 x – 12 y + 72 = 0
( x 2 + 20 x + 100) + ( y 2 – 12 y + 36)
= –72 + 100 + 36
2
2
( x + 10) + ( y – 6) = 64
center of first circle: (2, 1)
center of second circle: (–10, 6)
d = (2 + 10)2 + (1 – 6) 2 = 144 + 25
= 169 = 13
However, the radii only sum to 4 + 8 = 12, so
the circles must not intersect if the distance
between their centers is 13.
54. x 2 + ax + y 2 + by + c = 0
⎛ 2
a2 ⎞ ⎛ 2
b2 ⎞
⎜ x + ax +
⎟ + ⎜ y + by + ⎟
⎜
4 ⎟⎠ ⎜⎝
4 ⎟⎠
⎝
= −c +
a 2 b2
+
4
4
2
2
a⎞ ⎛
b⎞
a 2 + b 2 − 4c
⎛
⎜x+ ⎟ +⎜ y+ ⎟ =
2⎠ ⎝
2⎠
4
⎝
2
2
a + b − 4c
> 0 ⇒ a 2 + b 2 > 4c
4
55. Label the points C, P, Q, and R as shown in the
figure below. Let d = OP , h = OR , and
a = PR . Triangles ΔOPR and ΔCQR are
similar because each contains a right angle and
they share angle ∠QRC . For an angle of
30 ,
a 1
d
3
and = ⇒ h = 2a . Using a
=
h 2
h
2
56. The equations of the two circles are
( x − R)2 + ( y − R)2 = R 2
( x − r )2 + ( y − r )2 = r 2
Let ( a, a ) denote the point where the two
circles touch. This point must satisfy
(a − R)2 + (a − R)2 = R 2
R2
2
⎛
2⎞
a = ⎜⎜ 1 ±
⎟R
2 ⎟⎠
⎝
(a − R)2 =
⎛
2⎞
Since a < R , a = ⎜⎜1 −
⎟ R.
2 ⎟⎠
⎝
At the same time, the point where the two
circles touch must satisfy
(a − r )2 + (a − r )2 = r 2
⎛
2⎞
a = ⎜⎜ 1 ±
⎟r
2 ⎟⎠
⎝
⎛
2⎞
Since a > r , a = ⎜⎜ 1 +
⎟ r.
2 ⎟⎠
⎝
Equating the two expressions for a yields
⎛
⎛
2⎞
2⎞
⎜⎜1 − 2 ⎟⎟ R = ⎜⎜ 1 + 2 ⎟⎟ r
⎝
⎠
⎝
⎠
2
2
1−
2
r=
R=
2
1+
2
r=
1− 2 +
⎛
2⎞
⎜⎜ 1 −
⎟⎟
2
⎝
⎠
R
⎛
⎞⎛
2
2⎞
⎜⎜ 1 +
⎟⎜ 1 −
⎟
2 ⎟⎜
2 ⎟⎠
⎝
⎠⎝
1
2R
1
2
r = (3 − 2 2) R ≈ 0.1716 R
1−
property of similar triangles, QC / RC = 3 / 2 ,
2
3
4
=
→ a = 2+
a−2
2
3
By the Pythagorean Theorem, we have
d = h 2 − a 2 = 3a = 2 3 + 4 ≈ 7.464
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Section 0.3
19
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
57. Refer to figure 15 in the text. Given ine l1 with
slope m, draw ABC with vertical and
horizontal sides m, 1.
Line l2 is obtained from l1 by rotating it
around the point A by 90° counter-clockwise.
Triangle ABC is rotated into triangle AED .
We read off
1
1
slope of l2 =
=− .
m
−m
60. See the figure below. The angle at T is a right
angle, so the Pythagorean Theorem gives
( PM + r )2 = ( PT )2 + r 2
⇔ ( PM )2 + 2rPM + r 2 = ( PT )2 + r 2
⇔ PM ( PM + 2r ) = ( PT )2
PM + 2r = PN so this gives ( PM )( PN ) = ( PT ) 2
58. 2 ( x − 1)2 + ( y − 1)2 = ( x − 3) 2 + ( y − 4)2
4( x 2 − 2 x + 1 + y 2 − 2 y + 1)
= x 2 − 6 x + 9 + y 2 − 8 y + 16
3x 2 − 2 x + 3 y 2 = 9 + 16 − 4 − 4;
2
17
x + y2 = ;
3
3
1⎞
17
1
⎛ 2 2
2
⎜x − x+ ⎟+ y = +
3
9⎠
3 9
⎝
3x 2 − 2 x + 3 y 2 = 17; x 2 −
2
1⎞
52
⎛
2
⎜x− ⎟ + y =
3⎠
9
⎝
B = (6)2 + (8)2 = 100 = 10
⎛ 52 ⎞
⎛1 ⎞
center: ⎜ , 0 ⎟ ; radius: ⎜⎜
⎟⎟
⎝3 ⎠
⎝ 3 ⎠
59. Let a, b, and c be the lengths of the sides of the
right triangle, with c the length of the
hypotenuse. Then the Pythagorean Theorem
says that a 2 + b 2 = c 2
Thus,
πa 2 πb 2 πc 2
+
=
or
8
8
8
2
61. The lengths A, B, and C are the same as the
corresponding distances between the centers of
the circles:
A = (–2)2 + (8)2 = 68 ≈ 8.2
2
1 ⎛a⎞
1 ⎛b⎞
1 ⎛c⎞
π⎜ ⎟ + π⎜ ⎟ = π⎜ ⎟
2 ⎝2⎠
2 ⎝2⎠
2 ⎝2⎠
C = (8)2 + (0)2 = 64 = 8
Each circle has radius 2, so the part of the belt
around the wheels is
2(2π − a − π ) + 2(2π − b − π ) + 2(2π − c − π )
= 2[3π - (a + b + c)] = 2(2π ) = 4π
Since a + b + c = π , the sum of the angles of a
triangle.
The length of the belt is ≈ 8.2 + 10 + 8 + 4π
≈ 38.8 units.
2
2
1 ⎛ x⎞
π ⎜ ⎟ is the area of a semicircle with
2 ⎝2⎠
diameter x, so the circles on the legs of the
triangle have total area equal to the area of the
semicircle on the hypotenuse.
From a 2 + b 2 = c 2 ,
3 2
3 2
3 2
a +
b =
c
4
4
4
3 2
x is the area of an equilateral triangle
4
with sides of length x, so the equilateral
triangles on the legs of the right triangle have
total area equal to the area of the equilateral
triangle on the hypotenuse of the right triangle.
20
Section 0.3
62 As in Problems 50 and 61, the curved portions
of the belt have total length 2π r. The lengths
of the straight portions will be the same as the
lengths of the sides. The belt will have length
2π r + d1 + d 2 + … + d n .
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
63. A = 3, B = 4, C = –6
3(–3) + 4(2) + (–6) 7
d=
=
5
(3) 2 + (4)2
64. A = 2, B = −2, C = 4
d=
2(4) − 2(−1) + 4)
2
(2) + (2)
2
=
14
8
=
7 2
2
65. A = 12, B = –5, C = 1
12(–2) – 5(–1) + 1 18
d=
=
13
(12) 2 + (–5) 2
66. A = 2, B = −1, C = −5
d=
2(3) − 1(−1) − 5
2
(2) + (−1)
2
=
2
5
=
2 5
5
67. 2 x + 4(0) = 5
5
x=
2
d=
2
( 52 ) + 4(0) – 7 =
(2)2 + (4) 2
2
20
=
5
5
68. 7(0) − 5 y = −1
1
y=
5
⎛1⎞
7(0) − 5 ⎜ ⎟ − 6
7
7 74
⎝5⎠
d=
=
=
2
2
74
74
(7) + (−5)
−2 − 3
5
3
= − ; m = ; passes through
1+ 2
3
5
⎛ −2 + 1 3 − 2 ⎞ ⎛ 1 1 ⎞
,
⎜
⎟ = ⎜− , ⎟
2 ⎠ ⎝ 2 2⎠
⎝ 2
1 3⎛
1⎞
y− = ⎜x+ ⎟
2 5⎝
2⎠
3
4
y = x+
5
5
69. m =
Instructor’s Resource Manual
0–4
1
= –2; m = ; passes through
2–0
2
⎛0+2 4+0⎞
,
⎜
⎟ = (1, 2)
2 ⎠
⎝ 2
1
y – 2 = ( x – 1)
2
1
3
y = x+
2
2
6–0
1
m=
= 3; m = – ; passes through
4–2
3
⎛ 2+4 0+6⎞
,
⎜
⎟ = (3, 3)
2 ⎠
⎝ 2
1
y – 3 = – ( x – 3)
3
1
y = – x+4
3
1
3
1
x+ = – x+4
2
2
3
5
5
x=
6
2
x=3
1
3
y = (3) + = 3
2
2
center = (3, 3)
70. m =
71. Let the origin be at the vertex as shown in the
figure below. The center of the circle is then
( 4 − r , r ) , so it has equation
( x − (4 − r ))2 + ( y − r )2 = r 2 . Along the side of
length 5, the y-coordinate is always
3
4
times
the x-coordinate. Thus, we need to find the
value of r for which there is exactly one x2
⎛3
⎞
solution to ( x − 4 + r ) 2 + ⎜ x − r ⎟ = r 2 .
⎝4
⎠
Solving for x in this equation gives
16 ⎛
⎞
x = ⎜ 16 − r ± 24 − r 2 + 7r − 6 ⎟ . There is
25 ⎝
⎠
(
)
exactly one solution when −r 2 + 7 r − 6 = 0,
that is, when r = 1 or r = 6 . The root r = 6 is
extraneous. Thus, the largest circle that can be
inscribed in this triangle has radius r = 1.
Section 0.3
21
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
72. The line tangent to the circle at ( a, b ) will be
The slope of PS is
1
[ y1 + y4 − ( y1 + y2 )] y − y
2
2
= 4
. The slope of
1
x
−
x
4
2
x
+
x
−
x
+
x
(
)
[1 4 1 2]
2
1
[ y3 + y4 − ( y2 + y3 )] y − y
2
. Thus
QR is 2
= 4
1
x
−
x
[ x3 + x4 − ( x2 + x3 )] 4 2
2
PS and QR are parallel. The slopes of SR and
y −y
PQ are both 3 1 , so PQRS is a
x3 − x1
parallelogram.
perpendicular to the line through ( a, b ) and the
center of the circle, which is ( 0, 0 ) . The line
through ( a, b ) and ( 0, 0 ) has slope
0−b b
a
r2
= ; ax + by = r 2 ⇒ y = − x +
0−a a
b
b
a
so ax + by = r 2 has slope − and is
b
perpendicular to the line through ( a, b ) and
m=
( 0, 0 ) ,
so it is tangent to the circle at ( a, b ) .
73. 12a + 0b = 36
a=3
32 + b 2 = 36
b = ±3 3
3x – 3 3 y = 36
x – 3 y = 12
3x + 3 3 y = 36
x + 3 y = 12
74. Use the formula given for problems 63-66, for
( x, y ) = ( 0, 0 ) .
77. x 2 + ( y – 6) 2 = 25; passes through (3, 2)
tangent line: 3x – 4y = 1
The dirt hits the wall at y = 8.
A = m, B = −1, C = B − b;(0, 0)
d=
m(0) − 1(0) + B − b
m2 + (−1) 2
=
B−b
m2 + 1
75. The midpoint of the side from (0, 0) to (a, 0) is
⎛0+a 0+0⎞ ⎛ a ⎞
,
⎜
⎟ = ⎜ , 0⎟
2 ⎠ ⎝2 ⎠
⎝ 2
The midpoint of the side from (0, 0) to (b, c) is
⎛0+b 0+c⎞ ⎛b c ⎞
,
⎜
⎟=⎜ , ⎟
2 ⎠ ⎝2 2⎠
⎝ 2
c–0
c
m1 =
=
b–a b–a
c –0
c
m2 = 2
=
; m1 = m2
b–a
b–a
2
0.4 Concepts Review
1. y-axis
2.
( 4, −2 )
3. 8; –2, 1, 4
4. line; parabola
Problem Set 0.4
1. y = –x2 + 1; y-intercept = 1; y = (1 + x)(1 – x);
x-intercepts = –1, 1
Symmetric with respect to the y-axis
2
76. See the figure below. The midpoints of the
sides are
⎛ x + x y + y3 ⎞
⎛ x + x y + y2 ⎞
P⎜ 1 2 , 1
, Q⎜ 2 3 , 2
,
⎟
2 ⎠
2 ⎟⎠
⎝ 2
⎝ 2
⎛ x + x y + y4 ⎞
R⎜ 3 4 , 3
, and
2 ⎟⎠
⎝ 2
⎛ x + x y + y4 ⎞
S⎜ 1 4 , 1
.
2 ⎟⎠
⎝ 2
22
Section 0.4
Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2. x = − y 2 + 1; y -intercepts = −1,1;
x-intercept = 1 .
Symmetric with respect to the x-axis.
3. x = –4y2 – 1; x-intercept = –1
Symmetric with respect to the x-axis
4.
y = 4 x 2 − 1; y -intercept = −1
1 1
y = (2 x + 1)(2 x − 1); x-intercepts = − ,
2 2
Symmetric with respect to the y-axis.
5. x2 + y = 0; y = –x2
x-intercept = 0, y-intercept = 0
Symmetric with respect to the y-axis
6. y = x 2 − 2 x; y -intercept = 0
y = x(2 − x); x-intercepts = 0, 2
7
7. 7x2 + 3y = 0; 3y = –7x2; y = – x 2
3
x-intercept = 0, y-intercept = 0
Symmetric with respect to the y-axis
8. y = 3x 2 − 2 x + 2; y -intercept = 2
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9. x2 + y2 = 4
x-intercepts = -2, 2; y-intercepts = -2, 2
Symmetric with respect to the x-axis, y-axis,
and origin
10. 3x 2 + 4 y 2 = 12; y-intercepts = − 3, 3
x-intercepts = −2, 2
Symmetric with respect to the x-axis, y-axis,
and origin
11. y = –x2 – 2x + 2: y-intercept = 2
2± 4+8 2±2 3
x-intercepts =
=
= –1 ± 3
–2
–2
24
Section 0.4
12. 4 x 2 + 3 y 2 = 12; y -intercepts = −2, 2
x-intercepts = − 3, 3
Symmetric with respect to the x-axis, y-axis,
and origin
13. x2 – y2 = 4
x-intercept = -2, 2
Symmetric with respect to the x-axis, y-axis,
and origin
14. x 2 + ( y − 1)2 = 9; y -intercepts = −2, 4
x-intercepts = −2 2, 2 2
Symmetric with respect to the y-axis
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15. 4(x – 1)2 + y2 = 36;
y-intercepts = ± 32 = ±4 2
x-intercepts = –2, 4
Symmetric with respect to the x-axis
18. x 4 + y 4 = 1; y -intercepts = −1,1
x-intercepts = −1,1
Symmetric with respect to the x-axis, y-axis,
and origin
16. x 2 − 4 x + 3 y 2 = −2
19. x4 + y4 = 16; y-intercepts = −2, 2
x-intercepts = −2, 2
Symmetric with respect to the y-axis, x-axis
and origin
x-intercepts = 2 ± 2
Symmetric with respect to the x-axis
17. x2 + 9(y + 2)2 = 36; y-intercepts = –4, 0
x-intercept = 0
Symmetric with respect to the y-axis
Instructor’s Resource Manual
20. y = x3 – x; y-intercepts = 0;
y = x(x2 – 1) = x(x + 1)(x – 1);
x-intercepts = –1, 0, 1
Symmetric with respect to the origin
Section 0.4
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21. y =
1
2
; y-intercept = 1
x +1
Symmetric with respect to the y-axis
2
24. 4 ( x − 5 ) + 9( y + 2) 2 = 36; x-intercept = 5
25. y = (x – 1)(x – 2)(x – 3); y-intercept = –6
x-intercepts = 1, 2, 3
22. y =
x
; y -intercept = 0
x +1
x-intercept = 0
Symmetric with respect to the origin
2
26. y = x2(x – 1)(x – 2); y-intercept = 0
x-intercepts = 0, 1, 2
23. 2 x 2 – 4 x + 3 y 2 + 12 y = –2
2( x 2 – 2 x + 1) + 3( y 2 + 4 y + 4) = –2 + 2 + 12
2( x – 1)2 + 3( y + 2)2 = 12
y-intercepts = –2 ±
30
3
x-intercept = 1
27. y = x 2 ( x − 1)2 ; y-intercept = 0
x-intercepts = 0, 1
26
Section 0.4
Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
28. y = x 4 ( x − 1)4 ( x + 1)4 ; y -intercept = 0
x-intercepts = −1, 0,1
Symmetric with respect to the y-axis
Intersection points: (0, 1) and (–3, 4)
32. 2 x + 3 = −( x − 1) 2
29.
x + y = 1; y-intercepts = –1, 1;
x-intercepts = –1, 1
Symmetric with respect to the x-axis, y-axis
and origin
2 x + 3 = − x2 + 2 x − 1
x2 + 4 = 0
No points of intersection
33. −2 x + 3 = −2( x − 4)2
30.
x + y = 4; y-intercepts = –4, 4;
x-intercepts = –4, 4
Symmetric with respect to the x-axis, y-axis
and origin
−2 x + 3 = −2 x 2 + 16 x − 32
2 x 2 − 18 x + 35 = 0
x=
18 ± 324 – 280 18 ± 2 11 9 ± 11
=
=
;
4
4
2
⎛ 9 – 11
⎞
, – 6 + 11 ⎟⎟ ,
Intersection points: ⎜⎜
⎝ 2
⎠
⎛ 9 + 11
⎞
, – 6 – 11 ⎟⎟
⎜⎜
⎝ 2
⎠
31.
− x + 1 = ( x + 1)2
− x + 1 = x2 + 2 x + 1
x 2 + 3x = 0
x( x + 3) = 0
x = 0, −3
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37. y = 3x + 1
34. −2 x + 3 = 3x 2 − 3 x + 12
x 2 + 2 x + (3 x + 1) 2 = 15
3x 2 − x + 9 = 0
No points of intersection
x 2 + 2 x + 9 x 2 + 6 x + 1 = 15
10 x 2 + 8 x − 14 = 0
2(5 x 2 + 4 x − 7) = 0
−2 ± 39
≈ −1.65, 0.85
5
Intersection points:
⎛ −2 − 39 −1 − 3 39 ⎞
,
⎜
⎟ and
⎜
⎟
5
5
⎝
⎠
⎛ −2 + 39 −1 + 3 39 ⎞
,
⎜
⎟
⎜
⎟
5
5
⎝
⎠
[ or roughly (–1.65, –3.95) and (0.85, 3.55) ]
x=
35. x 2 + x 2 = 4
x2 = 2
x=± 2
(
)(
Intersection points: – 2, – 2 ,
2, 2
)
38. x 2 + (4 x + 3) 2 = 81
x 2 + 16 x 2 + 24 x + 9 = 81
17 x 2 + 24 x − 72 = 0
−12 ± 38
≈ −2.88, 1.47
17
Intersection points:
⎛ −12 − 38 3 − 24 38 ⎞
,
⎜
⎟ and
⎜
⎟
17
17
⎝
⎠
⎛ −12 + 38 3 + 24 38 ⎞
,
⎜
⎟
⎜
⎟
17
17
⎝
⎠
[ or roughly ( −2.88, −8.52 ) , (1.47,8.88 ) ]
x=
36. 2 x 2 + 3( x − 1)2 = 12
2 x 2 + 3 x 2 − 6 x + 3 = 12
5x2 − 6 x − 9 = 0
6 ± 36 + 180 6 ± 6 6 3 ± 3 6
=
=
10
10
5
Intersection points:
⎛ 3 − 3 6 −2 − 3 6 ⎞ ⎛ 3 + 3 6 −2 + 3 6 ⎞
,
,
⎜⎜
⎟⎟ , ⎜⎜
⎟⎟
5
5
⎝ 5
⎠ ⎝ 5
⎠
x=
28
Section 0.4
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
39. a.
y = x 2 ; (2)
(
b.
ax3 + bx 2 + cx + d , with a > 0 : (1)
c.
ax3 + bx 2 + cx + d , with a < 0 : (3)
d.
y = ax3 , with a > 0 : (4)
40. x 2 + y 2 = 13;(−2, −3), (−2,3), (2, −3), (2,3)
2
2
2
2
d1 = (2 + 2) + (−3 + 3) = 4
2
2
d3 = (2 − 2) + (3 + 3) = 6
Three such distances.
(
)
)(
)
(
)
d1 = (–2 – 2) 2 + ⎡1 + 21 – 1 + 13 ⎤
⎣
⎦
(
21 – 13
)
)
)
2
2
(
)
d3 = (−2 + 2)2 + ⎡1 + 21 − 1 − 21 ⎤
⎣
⎦
(
21 + 21
)
2.
2
=
( 2 21)
2
)
2
2
f (1) = 1 – 12 = 0
f (–2) = 1 – (–2)2 = –3
c.
f (0) = 1 – 02 = 1
d.
f (k ) = 1 – k 2
e.
f (–5) = 1 – (–5) 2 = –24
f.
1 15
⎛1⎞
⎛1⎞
f ⎜ ⎟ =1– ⎜ ⎟ =1– =
16 16
⎝4⎠
⎝4⎠
g.
f (1 + h ) = 1 − (1 + h ) = −2h − h 2
h.
f (1 + h ) − f (1) = −2h − h 2 − 0 = −2h − h 2
i.
f ( 2 + h ) − f ( 2) = 1 − ( 2 + h ) + 3
2
2
d 4 = (−2 − 2)2 + ⎡⎣1 − 21 − (1 + 13) ⎤⎦
(
2
2
= −4h − h 2
= 50 + 2 273 ≈ 9.11
(
)
d5 = (−2 − 2)2 + ⎡1 − 21 − 1 − 13 ⎤
⎣
⎦
= 16 +
(
13 − 21
)
2
2
2. a.
b.
F (1) = 13 + 3 ⋅1 = 4
F ( 2) = ( 2)3 + 3( 2) = 2 2 + 3 2
=5 2
= 50 − 2 273 ≈ 4.12
3
c.
Instructor’s Resource Manual
2
b.
= 2 21 ≈ 9.17
= 16 + − 21 − 13
( 2 13 )
f (2u ) = 3(2u ) 2 = 12u 2 ; f ( x + h) = 3( x + h)2
1. a.
= 50 + 2 273 ≈ 9.11
= 0+
=
0.5 Concepts Review
2
(
21 + 13
2
Problem Set 0.5
d 2 = (–2 – 2)2 + ⎡1 + 21 – 1 – 13 ⎤
⎣
⎦
(
)
= 2 13 ≈ 7.21
Four such distances ( d 2 = d 4 and d1 = d5 ).
2
= 50 – 2 273 ≈ 4.12
= 16 +
13 + 13
4. even; odd; y-axis; origin
21 , 2, 1 + 13 , 2, 1 – 13
= 16 +
(
3. asymptote
41. x2 + 2x + y2 – 2y = 20; –2, 1 + 21 ,
)(
= 0+
2
1. domain; range
d 2 = (2 + 2) + (−3 − 3) = 52 = 2 13
( –2, 1 –
)
d6 = (2 − 2)2 + ⎡1 + 13 − 1 − 13 ⎤
⎣
⎦
⎛1⎞ ⎛1⎞
⎛ 1 ⎞ 1 3 49
F ⎜ ⎟ = ⎜ ⎟ + 3⎜ ⎟ =
+ =
⎝4⎠ ⎝4⎠
⎝ 4 ⎠ 64 4 64
Section 0.5
29
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
d.
F (1 + h ) = (1 + h ) + 3 (1 + h )
3
f.
Φ ( x 2 + x) =
= 1 + 3h + 3h 2 + h3 + 3 + 3h
= 4 + 6h + 3h 2 + h3
e.
F (1 + h ) − 1 = 3 + 6h + 3h + h
f.
F ( 2 + h) − F ( 2)
2
=
3
2
= 15h + 6h + h
3. a.
b.
G (0) =
d.
e.
f.
4. a.
b.
30
0.25 − 3
1
f ( x) =
c.
f (3 + 2) =
3
1
= –1
0 –1
1
f (0.25) =
b.
π −3
G( y ) =
G (– x) =
c.
1
1
=–
– x –1
x +1
7. a.
2
1
x
=
– 1 1 – x2
1
2
≈ 0.841
≈ −3.293
(12.26) 2 + 9
12.26 – 3
≈ 1.199
1
–t
1
2
c.
⎛1⎞
Φ⎜ ⎟ =
⎝2⎠
d.
Φ (u + 1) =
e.
Φ( x2 ) =
+
( 12 )
=
2
1
2
=
x2
=
x2 + y2 = 1
c.
x = 2 y +1
x2 = 2 y + 1
≈ 1.06
(u + 1) + (u + 1) 2
( x2 ) + ( x2 )2
Section 0.5
–t
u +1
; undefined
b. xy + y + x = 1
y(x + 1) = 1 – x
1– x
1– x
y=
; f ( x) =
x +1
x +1
t2 – t
3
4
1
2
3– 3
y = ± 1 – x 2 ; not a function
=2
–t + (– t ) 2
( 3)2 + 9
f ( 3) =
y 2 = 1– x 2
1
x2
1 + 12
Φ (–t ) =
is not
y 2 –1
⎛ 1 ⎞
G⎜ ⎟ =
⎝ x2 ⎠
Φ (1) =
=
3+ 2 −3
−0.25
0.79 – 3
b. f(12.26) =
1
− 2.75
≈ 2.658
(0.79) 2 + 9
f(0.79) =
1
G (1.01) =
= 100
1.01 – 1
2
1
=
1
=2
1
G (0.999) =
= –1000
0.999 –1
6. a.
c.
x2 + x
defined
= ( 2 + h ) + 3 ( 2 + h ) − ⎡ 23 − 3 ( 2 ) ⎤
⎣
⎦
= 8 + 12h + 6h 2 + h3 + 6 + 3h − 14
x2 + x
x 4 + 2 x3 + 2 x 2 + x
3
5. a.
( x 2 + x) + ( x 2 + x) 2
=
y=
u 2 + 3u + 2
x2 + x4
x
u +1
d.
x2 – 1
x2 – 1
; f ( x) =
2
2
y
y+1
xy + x = y
x = y – xy
x = y(1 – x)
x
x
; f ( x) =
y=
1– x
1– x
x=
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8. The graphs on the left are not graphs of
functions, the graphs on the right are graphs of
functions.
9.
f (a + h) – f (a) [2(a + h) 2 – 1] – (2a 2 – 1)
=
h
h
4ah + 2h 2
=
= 4a + 2h
h
x 2 – 9 ≥ 0; x 2 ≥ 9; x ≥ 3
Domain: {x ∈
d.
F (a + h) – F (a ) 4(a + h) – 4a
=
h
h
=
4a3 + 12a 2 h + 12ah 2 + 4h3 – 4a3
h
=
12a 2 h + 12ah 2 + 4h3
h
= x − 4 x + hx − 2h + 4
h
−3h
=
2
h( x − 4 x + hx − 2h + 4)
3
=–
2
x – 4 x + hx – 2h + 4
a+h
a + h+ 4
: y ≤ 5}
14. a.
b.
f ( x) =
4 – x2
=
4 – x2
( x – 3)( x + 2)
x2 – x – 6
Domain: {x ∈ : x ≠ −2, 3}
G ( y ) = ( y + 1) –1
1
≥ 0; y > –1
y +1
3
3
g ( x + h) – g ( x) x + h –2 – x –2
=
h
h
3x − 6 − 3x − 3h + 6
G ( a + h) – G ( a )
=
h
H ( y ) = – 625 – y 4
Domain: { y ∈
Domain: { y ∈
: y > −1}
c.
φ (u ) = 2u + 3
(all real numbers)
Domain:
d.
F (t ) = t 2 / 3 – 4
(all real numbers)
Domain:
2
12.
: x ≥ 3}
3
= 12a 2 + 12ah + 4h 2
11.
ψ ( x) = x 2 – 9
625 – y 4 ≥ 0; 625 ≥ y 4 ; y ≤ 5
3
10.
c.
15. f(x) = –4; f(–x) = –4; even function
– a +a 4
h
2
a + 4a + ah + 4h − a 2 − ah − 4a
a 2 + 8a + ah + 4h + 16
h
4h
=
=
=
13. a.
h(a 2 + 8a + ah + 4h + 16)
4
a 2 + 8a + ah + 4h + 16
F ( z) = 2 z + 3
2z + 3 ≥ 0; z ≥ –
⎧
Domain: ⎨ z ∈
⎩
b.
16. f(x) = 3x; f(–x) = –3x; odd function
g (v ) =
3
2
3⎫
:z≥− ⎬
2⎭
1
4v – 1
4v – 1 = 0; v =
⎧
Domain: ⎨v ∈
⎩
1
4
1⎫
:v≠ ⎬
4⎭
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17. F(x) = 2x + 1; F(–x) = –2x + 1; neither
20. g (u ) =
18. F ( x) = 3x – 2; F (– x) = –3x – 2; neither
21. g ( x) =
19. g ( x) = 3x 2 + 2 x – 1; g (– x) = 3 x 2 – 2 x – 1 ;
neither
32
Section 0.5
22. φ ( z ) =
u3
u3
; g (– u ) = – ; odd function
8
8
x
2
x –1
; g (– x) =
–x
2
x –1
; odd
2z +1
–2 z + 1
; φ (– z ) =
; neither
z –1
–z –1
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23.
f ( w) = w – 1; f (– w) = – w – 1; neither
2
2
24. h( x ) = x + 4; h(– x) = x + 4; even
function
26. F (t ) = – t + 3 ; F (– t ) = – –t + 3 ; neither
27. g ( x) =
x
x
; g (− x ) = − ; neither
2
2
28. G ( x) = 2 x − 1 ; G (− x) = −2 x + 1 ; neither
25.
f ( x) = 2 x ; f (– x) = –2 x = 2 x ; even
function
⎧1
if t ≤ 0
⎪
29. g (t ) = ⎨t + 1 if 0 < t < 2
⎪2
⎩t – 1 if t ≥ 2
Instructor’s Resource Manual
neither
Section 0.5
33
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
⎧⎪ – x 2 + 4 if x ≤ 1
30. h( x ) = ⎨
if x > 1
⎪⎩3x
neither
35. Let y denote the length of the other leg. Then
x2 + y 2 = h2
y 2 = h2 − x 2
y = h2 − x 2
L ( x ) = h2 − x 2
36. The area is
1
1
A ( x ) = base × height = x h 2 − x 2
2
2
37. a.
31. T(x) = 5000 + 805x
Domain: {x ∈ integers: 0 ≤ x ≤ 100}
T ( x) 5000
u ( x) =
=
+ 805
x
x
Domain: {x ∈ integers: 0 < x ≤ 100}
P ( x) = 6 x – (400 + 5 x( x – 4))
32. a.
= 6 x – 400 – 5 x( x – 4)
E(x) = 24 + 0.40x
b. 120 = 24 + 0.40x
0.40x = 96; x = 240 mi
38. The volume of the cylinder is πr 2 h, where h is
the height of the cylinder. From the figure,
2
2
2 ⎛ h⎞
2 h
2
= 3r ;
r + ⎜ ⎟ = (2r ) ;
⎝ 2⎠
4
h = 12r 2 = 2r 3.
V (r ) = πr 2 (2r 3) = 2πr 3 3
P(200) ≈ −190 ; P (1000 ) ≈ 610
b.
c. ABC breaks even when P(x) = 0;
6 x – 400 – 5 x( x – 4) = 0; x ≈ 390
33. E ( x) = x – x 2
y
0.5
39. The area of the two semicircular ends is
0.5
1
x
−0.5
1
exceeds its square by the maximum amount.
2
34. Each side has length
p
. The height of the
3
πd 2
.
4
1 – πd
.
2
2
πd 2
d – πd 2
⎛ 1 – πd ⎞ πd
A(d ) =
+d⎜
+
⎟=
4
4
2
⎝ 2 ⎠
The length of each parallel side is
2d – πd 2
4
Since the track is one mile long, π d < 1, so
1
1⎫
⎧
d < . Domain: ⎨d ∈ : 0 < d < ⎬
π
π⎭
⎩
=
3p
.
6
1 ⎛ p ⎞⎛ 3p ⎞
3 p2
A( p ) = ⎜ ⎟ ⎜⎜
⎟⎟ =
2 ⎝ 3 ⎠⎝ 6 ⎠
36
triangle is
34
Section 0.5
Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
40. a.
1
3
A(1) = 1(1) + (1)(2 − 1) =
2
2
42. a.
f(x + y) = 2(x + y) = 2x + 2y = f(x) + f(y)
b.
f ( x + y ) = ( x + y )2 = x 2 + 2 xy + y 2
≠ f ( x) + f ( y )
c.
f(x + y) = 2(x + y) + 1 = 2x + 2y + 1
≠ f(x) + f(y)
d. f(x + y) = –3(x + y) = –3x – 3y = f(x) + f(y)
b.
1
A(2) = 2(1) + (2)(3 − 1) = 4
2
c.
A(0) = 0
d.
1
1
A(c) = c(1) + (c)(c + 1 − 1) = c 2 + c
2
2
e.
43. For any x, x + 0 = x, so
f(x) = f(x + 0) = f(x) + f(0), hence f(0) = 0.
Let m be the value of f(1). For p in N,
p = p ⋅1 = 1 + 1 + ... + 1, so
f(p) = f(1 + 1 + ... + 1) = f(1) + f(1) + ... + f(1)
= pf(1) = pm.
⎛1⎞ 1 1
1
1 = p ⎜ ⎟ = + + ... + , so
p
⎝ p⎠ p p
⎛1 1
1⎞
m = f (1) = f ⎜ + + ... + ⎟
p
p
p⎠
⎝
⎛1⎞
⎛1⎞
⎛1⎞
⎛1⎞
= f ⎜ ⎟ + f ⎜ ⎟ + ... + f ⎜ ⎟ = pf ⎜ ⎟ ,
⎝ p⎠
⎝ p⎠
⎝ p⎠
⎝ p⎠
⎛1⎞ m
hence f ⎜ ⎟ = . Any rational number can
⎝ p⎠ p
be written as
f.
Domain: {c ∈
Range: { y ∈
41. a.
b.
: c ≥ 0}
: y ≥ 0}
B (0) = 0
1 1 1
⎛1⎞ 1
B ⎜ ⎟ = B (1) = ⋅ =
2 6 12
⎝2⎠ 2
p
with p, q in N.
q
⎛1⎞ 1 1
p
1
= p ⎜ ⎟ = + + ... + ,
q
q
⎝q⎠ q q
⎛ p⎞
⎛1 1
1⎞
so f ⎜ ⎟ = f ⎜ + + ... + ⎟
q
q
q
q⎠
⎝ ⎠
⎝
⎛1⎞
⎛1⎞
⎛1⎞
= f ⎜ ⎟ + f ⎜ ⎟ + ... + f ⎜ ⎟
⎝q⎠
⎝q⎠
⎝q⎠
⎛1⎞
⎛m⎞
⎛ p⎞
= pf ⎜ ⎟ = p ⎜ ⎟ = m ⎜ ⎟
⎝q⎠
⎝q⎠
⎝q⎠
c.
Instructor’s Resource Manual
Section 0.5
35
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
44. The player has run 10t feet after t seconds. He
reaches first base when t = 9, second base when
t = 18, third base when t = 27, and home plate
when t = 36. The player is 10t – 90 feet from
first base when 9 ≤ t ≤ 18, hence
46. a.
x
f(x)
–4
–6.1902
–3
0.4118
–2
13.7651
–1
9.9579
0
0
1
–7.3369
2
–17.7388
if 18 < t ≤ 27
3
–0.4521
if 27 < t ≤ 36
4
4.4378
b.
902 + (10t − 90)2 feet from home plate. The
player is 10t – 180 feet from second base when
18 ≤ t ≤ 27, thus he is
90 – (10t – 180) = 270 – 10t feet from third base
and 902 + (270 − 10t ) 2 feet from home plate.
The player is 10t – 270 feet from third base
when 27 ≤ t ≤ 36, thus he is
90 – (10t – 270) = 360 – 10t feet from home
plate.
a.
b.
45. a.
b.
⎧10t
⎪
2
2
⎪ 90 + (10t − 90)
s=⎨
⎪ 902 + (270 − 10t ) 2
⎪
⎪⎩360 – 10t
if 0 ≤ t ≤ 9
⎧180 − 180 − 10t
⎪
⎪
⎪
s = ⎨ 902 + (10t − 90) 2
⎪
2
2
⎪ 90 + (270 − 10t )
⎪
⎪⎩
if 0 ≤ t ≤ 9
if 9 < t ≤ 18
or 27 < t ≤ 36
47.
if 9 < t ≤ 18
if 18 < t ≤ 27
f(1.38) ≈ 0.2994
f(4.12) ≈ 3.6852
x
f(x)
–4
–4.05
–3
–3.1538
a.
–2
–2.375
b. f(x) = 0 when x ≈ –1.1, 1.7, 4.3
f(x) ≥ 0 on [–1.1, 1.7] ∪ [4.3, 5]
–1
–1.8
0
–1.25
1
–0.2
2
1.125
3
2.3846
4
3.55
Section 0.5
Range: {y ∈ R: –22 ≤ y ≤ 13}
48.
a.
36
f(1.38) ≈ –76.8204
f(4.12) ≈ 6.7508
f(x) = g(x) at x ≈ –0.6, 3.0, 4.6
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
b. f(x) ≥ g(x) on [-0.6, 3.0] ∪ [4.6, 5]
c.
f ( x) – g ( x)
= x3 – 5 x 2 + x + 8 – 2 x 2 + 8 x + 1
= x3 – 7 x 2 + 9 x + 9
b. On ⎡⎣ −6, −3) , g increases from
13
g ( −6 ) = ≈ 4.3333 to ∞ . On ( 2, 6⎤⎦ , g
3
26
≈ 2.8889 . On
decreased from ∞ to
9
( −3, 2 ) the maximum occurs around
Largest value f (–2) – g (–2) = 45
x = 0.1451 with value 0.6748 . Thus, the
range is ( −∞, 0.6748⎦⎤ ∪ ⎣⎡ 2.8889, ∞ ) .
49.
c.
x 2 + x – 6 = 0; (x + 3)(x – 2) = 0
Vertical asymptotes at x = –3, x = 2
d. Horizontal asymptote at y = 3
0.6 Concepts Review
1. ( x 2 + 1)3
a.
x-intercept: 3x – 4 = 0; x =
4
3
3⋅ 0 – 4
2
=
y-intercept:
2
0 +0–6 3
2. f(g(x))
3. 2; left
4. a quotient of two polynomial functions
b.
c.
x 2 + x – 6 = 0; (x + 3)(x – 2) = 0
Vertical asymptotes at x = –3, x = 2
Problem Set 0.6
1. a.
( f + g )(2) = (2 + 3) + 22 = 9
d. Horizontal asymptote at y = 0
50.
a.
( f ⋅ g )(0) = (0 + 3)(02 ) = 0
c.
( g f )(3) =
d.
( f g )(1) = f (12 ) = 1 + 3 = 4
e.
( g f )(1) = g (1 + 3) = 4 2 = 16
f.
( g f )(–8) = g (–8 + 3) = (–5) 2 = 25
2. a.
x-intercepts:
3x 2 – 4 = 0; x = ±
b.
4
2 3
=±
3
3
2
y-intercept:
3
32
9 3
= =
3+3 6 2
( f – g )(2) = (22 + 2) –
12 + 1
c.
1
⎡ 2 ⎤
⎛1⎞
g 2 (3) = ⎢
⎥ = ⎜ 3⎟ = 9
+
3
3
⎣
⎦
⎝ ⎠
2
Instructor’s Resource Manual
2
b.
( f g )(1) =
2
1+ 3
=
2
2 28
=6– =
2+3
5 5
2
4
=4
2
Section 0.6
37
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2
d.
e.
f.
3. a.
(f
⎛ 2 ⎞ ⎛1⎞ 1 3
g )(1) = f ⎜
⎟=⎜ ⎟ + =
⎝1+ 3 ⎠ ⎝ 2 ⎠ 2 4
2
2
=
2+3 5
⎛ 2 ⎞
( g g )(3) = g ⎜
⎟=
⎝ 3+3⎠
2
2 3
= 10 =
1
+3 3 5
3
= x2 + 2 x – 3
6
c.
(Ψ Φ )(r ) = Ψ (r + 1) =
d.
Φ 3 ( z ) = ( z 3 + 1) 3
e.
(Φ – Ψ )(5t) = [(5t) 3 +1] –
1
3
2
= g[(x 2 + 1) 2 + 1] = g( x 4 + 2x 2 + 2)
= ( x 4 + 2x 2 + 2) 2 + 1
= x 8 + 4x 6 + 8x 4 + 8x 2 + 5
7. g(3.141) ≈ 1.188
8. g(2.03) ≈ 0.000205
r 3 +1
1/ 3
9. ⎡ g 2 (π ) − g (π ) ⎤
⎣
⎦
≈ 4.789
1
5t
1/ 3
2
= ⎡⎢(11 − 7π ) − 11 − 7π ⎤⎥
⎣
⎦
10. [ g 3 (π) – g (π)]1/ 3 = [(6π – 11)3 – (6π – 11)]1/ 3
≈ 7.807
1
5t
⎛1⎞
((Φ – Ψ ) Ψ )(t ) = (Φ – Ψ )⎜ ⎟
⎝t⎠
3
1 1
⎛ 1⎞
= ⎜ ⎟ + 1– 1 = 3 + 1 – t
⎝ t⎠
t
t
11. a.
b.
12. a.
b.
4
= x + 3x + 3x + 1
( g g g )( x) = ( g g )( x 2 + 1)
1
t
b.
4. a.
f ) ( x) = g ⎜⎛ x 2 − 4 ⎟⎞ = 1 + x 2 − 4
⎝
⎠
6. g 3 (x) = (x 2 +1) 3 = (x 4 + 2x 2 + 1)(x 2 + 1)
3
f.
2
= 1 + x2 – 4
1
⎛1⎞ ⎛1⎞
(Φ Ψ )(r ) = Φ⎜ ⎟ = ⎜ ⎟ + 1 = 3 + 1
r
r
r
⎝ ⎠ ⎝ ⎠
= 125t3 + 1 –
g ) ( x) = f ( 1 + x ) = 1 + x − 4
(f
(g
( g f )(1) = g (12 + 1) =
(Φ + Ψ )(t ) = t 3 + 1 +
5.
g ( x) = x , f ( x) = x + 7
g (x) = x15 , f (x) = x 2 + x
2
f ( x) =
x
2 x2 – 1
x
Domain: (– ∞, – 1] ∪ [1, ∞)
3
, g ( x) = x 2 + x + 1
( f ⋅ g )( x) =
b.
4
⎛2⎞
f 4 ( x) + g 4 ( x) = ⎛⎜ x 2 – 1 ⎞⎟ + ⎜ ⎟
⎝
⎠ ⎝x⎠
16
= ( x 2 – 1)2 +
x4
Domain: (– ∞, 0 ) ∪ (0, ∞ )
2
c.
⎛2⎞
⎛2⎞
g )( x) = f ⎜ ⎟ = ⎜ ⎟ – 1 =
⎝ x⎠
⎝ x⎠
Domain: [–2, 0) ∪ (0, 2]
d.
( g f )( x) = g ⎛⎜ x 2 – 1 ⎞⎟ =
⎝
⎠
(f
4
f ( x) =
13. p = f
1
, g (x) = x 3 + 3 x
x
g h if f(x) =1/ x , g ( x) = x ,
h( x ) = x 2 + 1
p= f
g h if f ( x) = 1/ x , g(x) = x + 1,
h( x) = x 2
4
–1
x2
14. p = f g h l if f ( x) = 1/ x , g ( x) = x ,
2
h(x) = x + 1, l( x) = x
2
2
x –1
Domain: (– ∞ , –1) ∪ (1, ∞ )
38
Section 0.6
Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15. Translate the graph of g ( x) = x to the right 2
units and down 3 units.
17. Translate the graph of y = x 2 to the right 2
units and down 4 units.
18. Translate the graph of y = x 3 to the left 1 unit
and down 3 units.
16. Translate the graph of h( x) = x to the left 3
units and down 4 units.
19. ( f + g )( x) =
x–3
+ x
2
20. ( f + g )( x) = x + x
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Section 0.6
39
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21. F (t ) =
t –t
24. a.
t
F(x) – F(–x) is odd because
F(–x) – F(x) = –[F(x) – F(–x)]
b. F(x) + F(–x) is even because
F(–x) + F(–(–x)) = F(–x) + F(x)
= F(x) + F(–x)
c.
22. G (t ) = t − t
25. Not every polynomial of even degree is an
even function. For example f ( x) = x 2 + x is
neither even nor odd. Not every polynomial of
odd degree is an odd function. For example
g ( x) = x 3 + x 2 is neither even nor odd.
26. a.
23. a.
Even;
(f + g)(–x) = f(–x) + g(–x) = f(x) + g(x)
= (f + g)(x) if f and g are both even
functions.
b. Odd;
(f + g)(–x) = f(–x) + g(–x) = –f(x) – g(x)
= –(f + g)(x) if f and g are both odd
functions.
c.
Even;
( f ⋅ g )(− x) = [ f (− x)][ g (− x)]
= [ f ( x)][ g ( x)] = ( f ⋅ g )( x)
if f and g are both even functions.
d. Even;
( f ⋅ g )(− x) = [ f (− x)][ g (− x)]
= [− f ( x)][− g ( x)] = [ f ( x)][ g ( x)]
= ( f ⋅ g )( x)
if f and g are both odd functions.
e.
40
F ( x ) – F (– x)
F ( x ) + F (– x)
is odd and
is
2
2
even.
F ( x ) − F (− x) F ( x) + F (− x) 2 F ( x)
+
=
= F ( x)
2
2
2
Neither
b.
PF
c.
RF
d.
PF
e.
RF
f.
Neither
27. a.
P = 29 – 3(2 + t ) + (2 + t )2
= t + t + 27
b. When t = 15, P = 15 + 15 + 27 ≈ 6.773
28. R(t) = (120 + 2t + 3t2 )(6000 + 700t )
= 2100 t3 + 19, 400t 2 + 96, 000t + 720, 000
⎧⎪400t
29. D(t ) = ⎨
2
2
⎪⎩ (400t ) + [300(t − 1)]
if 0 < t < 1
if t ≥ 1
if 0 < t < 1
⎧⎪ 400t
D(t ) = ⎨
2
⎪⎩ 250, 000t − 180, 000t + 90, 000 if t ≥ 1
30. D(2.5) ≈ 1097 mi
Odd;
( f ⋅ g )(− x) = [ f (− x)][ g (− x)]
= [ f ( x)][− g ( x)] = −[ f ( x)][ g ( x)]
= −( f ⋅ g )( x)
if f is an even function and g is an odd
function.
Section 0.6
Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
31.
(axcx –+ab ) + b
(axcx –+ab ) – a
36.
⎛ ax + b ⎞ a
f ( f ( x)) = f ⎜
⎟=
⎝ cx – a ⎠ c
=
a 2 x + ab + bcx – ab
=
x(a 2 + bc)
=x
acx + bc – acx + a 2
a 2 + bc
2
If a + bc = 0 , f(f(x)) is undefined, while if
x = a , f(x) is undefined.
c
⎛ x –3 – 3 ⎞
⎛ ⎛ x – 3⎞⎞
x +1
⎟
(
(
(
)))
32. f f f x = f ⎜ f ⎜
⎟ ⎟ = f ⎜⎜ x –3
⎟
+
1
x
1
+
⎠⎠
⎝ ⎝
⎝ x +1
⎠
⎛ x – 3 – 3x – 3 ⎞
⎛ –2 x – 6 ⎞
⎛ –x – 3 ⎞
= f⎜
⎟= f ⎜
⎟= f ⎜
⎟
⎝ x – 3 + x +1 ⎠
⎝ 2x – 2 ⎠
⎝ x –1 ⎠
– x–3
– 3 – x – 3 – 3x + 3 – 4 x
= –xx––13
=
=
=x
– x – 3+ x –1
–4
+1
x –1
If x = –1, f(x) is undefined, while if x = 1,
f(f(x)) is undefined.
33. a.
b.
⎛1⎞
f⎜ ⎟=
⎝ x⎠
34. a.
b.
1
x
–1
=
1
1– x
1
f1 ( f 2 ( x)) = ;
x
f1 ( f3 ( x)) = 1 − x;
1
;
1− x
x −1
f1 ( f5 ( x)) =
;
x
x
;
f1 ( f6 ( x)) =
x −1
f1 ( f 4 ( x)) =
f 2 ( f1 ( x)) =
f 2 ( f 2 ( x)) =
x
x –1
x
–1
x –1
= x;
1
;
1− x
1
f 2 ( f 4 ( x)) =
= 1 − x;
f 2 ( f 6 ( x)) =
x
=x
x – x +1
1
x −1
x
1
x
x −1
=
x
;
x –1
=
x –1
;
x
f3 ( f1 ( x)) = 1 − x;
⎛ 1 ⎞
⎛ x – 1⎞
⎟⎟ = f ⎜
f ⎜⎜
⎟=
f
(
x
)
⎝ x ⎠
⎝
⎠
=1–x
1/ x
1 / x −1
x –1
x
x –1
–1
x
=
x –1
x –1– x
1 x −1
;
=
x
x
f3 ( f3 ( x)) = 1 – (1 – x) = x;
f3 ( f 2 ( x)) = 1 −
1
x
=
;
1 – x x –1
x –1 1
= ;
f3 ( f5 ( x)) = 1 –
x
x
x
1
f3 ( f 6 ( x)) = 1 –
=
;
x –1 1– x
f3 ( f 4 ( x)) = 1 –
1
=
x−x
f ( f ( x)) = f ( x /( x − 1)) =
=
1
x
f 2 ( f3 ( x)) =
f 2 ( f 5 ( x )) =
f (1 / x) =
1
;
x
1
1
1− x
⎛ x ⎞
f ( f ( x)) = f ⎜
⎟=
⎝ x – 1⎠
=
c.
1
x
f1 ( f1 ( x)) = x;
x /( x − 1)
x
x( x − 1) + 1 − x
35. ( f1 ( f 2 f3 ))( x) = f1 (( f 2 f3 )( x))
= f1 ( f 2 ( f3 ( x)))
(( f1 f 2 ) f3 )( x) = ( f1 f 2 )( f3 ( x))
= f1 ( f 2 ( f3 ( x)))
= ( f1 ( f 2 f3 ))( x)
x
−1
x −1
1
;
1− x
1
x
;
f 4 ( f 2 ( x)) =
=
1
1− x x −1
f 4 ( f1 ( x)) =
f 4 ( f3 ( x)) =
f 4 ( f 4 ( x)) =
f 4 ( f5 ( x)) =
f 4 ( f 6 ( x)) =
Instructor’s Resource Manual
1
1
= ;
1 – (1 – x) x
1
1 – 1–1x
1
1–
x –1
x
=
1− x
x –1
=
;
x
1− x −1
=
x
= x;
x − ( x − 1)
1
x −1
=
= 1 – x;
x
1 – x –1 x − 1 − x
Section 0.6
41
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
a.
x −1
f5 ( f1 ( x)) =
;
x
1 −1
f5 ( f 2 ( x)) = x
= 1 − x;
42
b.
=
f 5 ( f 5 ( x)) =
x –1
–1
x
x –1
x
x −1− x
1
=
=
;
x −1
1– x
f 5 ( f 6 ( x)) =
x
–1
x –1
x
x –1
=
f3
f4
f5
f6
f3 )
f4 )
f5 )
f6 )
= ( f 4 f 4 ) ( f5 f 6 )
= f5 f 2 = f3
c.
x − ( x − 1) 1
= ;
x
x
If F
f 6 = f 1 , then F = f 6 .
d. If G f 3
G = f5.
1
;
1– x
f 6 = f 1 , then G f 4 = f 1 so
If f 2 f 5 H = f 5 , then f 6 H = f 5 so
H = f3.
37.
f 6 ( f3 ( x)) =
x –1
1– x
;
=
1– x –1
x
f 6 ( f 4 ( x)) =
1
1– x
1
–1
1– x
=
1
1
= ;
1 − (1 − x) x
f 6 ( f 5 ( x)) =
x –1
x
x –1
–1
x
=
x −1
= 1 – x;
x −1− x
f 6 ( f 6 ( x )) =
x
x –1
x
–1
x –1
=
x
=x
x − ( x − 1)
38.
f1
f2
f3
f4
f5
f6
f1
f1
f2
f3
f4
f5
f6
f2
f2
f1
f4
f3
f6
f5
f3
f3
f5
f1
f6
f2
f4
f4
f4
f6
f2
f5
f1
f3
f5
f5
f3
f6
f1
f4
f2
f6
f6
f4
f5
f2
f3
f1
Section 0.6
f1 f 2
= (((( f 2
e.
=
f3 ) f3 ) f3 ) f3 )
= ((((( f 1 f 2 ) f 3 ) f 4 ) f 5 ) f 6 )
1 − (1 − x)
= x;
1
x
;
x –1
–1
f3
= f1 f 3 = f 3
1
–1
1– x
1
1– x
1
x
f3
= (( f3 f3 ) f3 )
f 5 ( f 4 ( x)) =
f 6 ( f 2 ( x)) =
f3
= ((( f1 f3 ) f3 ) f3 )
1 – x –1
x
f5 ( f3 ( x)) =
=
;
1– x
x –1
1
x
f3
= (((( f 3
1
x
f 6 ( f1 ( x)) =
f3
39.
Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Problem Set 0.7
40.
41. a.
b.
1. a.
⎛ π ⎞ π
30 ⎜
⎟=
⎝ 180 ⎠ 6
b.
⎛ π ⎞ π
45 ⎜
⎟=
⎝ 180 ⎠ 4
c.
π
⎛ π ⎞
–60 ⎜
⎟=–
3
⎝ 180 ⎠
d.
⎛ π ⎞ 4π
240 ⎜
⎟=
⎝ 180 ⎠ 3
e.
37 π
⎛ π ⎞
–370 ⎜
⎟=–
18
⎝ 180 ⎠
f.
⎛ π ⎞ π
10 ⎜
⎟=
⎝ 180 ⎠ 18
2. a.
c.
3
4
c.
1 ⎛ 180 ⎞
⎟ = –60°
– π⎜
3 ⎝ π ⎠
d.
4 ⎛ 180 ⎞
⎟ = 240°
π⎜
3 ⎝ π ⎠
e.
–
f.
3 ⎛ 180 ⎞
π⎜
⎟ = 30°
18 ⎝ π ⎠
3. a.
4
x
4. r = (–4) + 3 = 5; cos θ = = –
5
r
2
Instructor’s Resource Manual
⎛ π ⎞
33.3 ⎜
⎟ ≈ 0.5812
⎝ 180 ⎠
⎛ π ⎞
46 ⎜
⎟ ≈ 0.8029
⎝ 180 ⎠
c.
⎛ π ⎞
–66.6 ⎜
⎟ ≈ –1.1624
⎝ 180 ⎠
d.
⎛ π ⎞
240.11⎜
⎟ ≈ 4.1907
⎝ 180 ⎠
e.
⎛ π ⎞
–369 ⎜
⎟ ≈ –6.4403
⎝ 180 ⎠
f.
⎛ π ⎞
11⎜
⎟ ≈ 0.1920
⎝ 180 ⎠
3. odd; even
2
35 ⎛ 180 ⎞
⎟ = –350°
π⎜
18 ⎝ π ⎠
b.
1. (– ∞ , ∞ ); [–1, 1]
2. 2 π ; 2 π ; π
⎛ 180 ⎟⎞
π⎜
= 135°
⎝ π ⎠
b.
42.
0.7 Concepts Review
7 ⎛ 180 ⎞
⎟ = 210°
π⎜
6 ⎝ π ⎠
Section 0.7
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4. a.
⎛ 180 ⎟⎞
3.141⎜
≈ 180°
⎝ π ⎠
Thus
b.
⎛ 180 ⎟⎞
6. 28⎜
≈ 359. 8°
⎝ π ⎠
c.
⎛ 180 ⎟⎞
≈ 286.5°
5. 00⎜
⎝ π ⎠
d.
⎛ 180 ⎟⎞
0. 001⎜
≈ 0 .057°
⎝ π ⎠
e.
⎛ 180 ⎟⎞
–0.1⎜
≈ –5.73°
⎝ π ⎠
f.
⎛ 180 ⎟⎞
36. 0⎜
≈ 2062.6 °
⎝ π ⎠
5. a.
56. 4 tan34. 2°
≈ 68.37
sin 34.1°
b.
cos
tan (0.452) ≈ 0.4855
d.
sin (–0.361) ≈ –0.3532
6. a.
b.
7. a.
π
=
π
3
=
3
.
2
234.1sin(1.56)
≈ 248.3
cos(0.34 )
sin 2 (2.51) + cos(0.51) ≈ 1.2828
56. 3 tan34. 2°
≈ 46.097
sin 56.1°
Referring to Figure 2, it is clear that sin
sin 35°
⎛⎜
⎞⎟
≈ 0. 0789
⎝ sin 26° + cos 26° ⎠
Identity, cos 2
π
6
Section 0.7
= 1 − sin 2
π
π
π
2
=1
= 0 . The rest of the values are
2
obtained using the same kind of reasoning in
the second quadrant.
and cos
8. Referring to Figure 2, it is clear that
sin 0 = 0 and cos 0 = 1 . If the angle is π / 6 ,
then the triangle in the figure below is
1
1
equilateral. Thus, PQ = OP = . This
2
2
π 1
implies that sin = . By the Pythagorean
6 2
44
= cos
and by the Pythagorean Identity, sin
3
b.
6
π
3
. The results
2
=
2
were derived in the text.
4
4
2
If the angle is π / 3 then the triangle in the
π 1
figure below is equilateral. Thus cos =
3 2
sin
5.34 tan 21.3°
≈ 0.8845
sin 3.1°+ cot 23.5°
c.
π
9. a.
⎛π ⎞
sin ⎜ ⎟
⎛π⎞
⎝6⎠ = 3
tan ⎜ ⎟ =
3
⎝ 6 ⎠ cos ⎛ π ⎞
⎜ ⎟
6
⎝ ⎠
2
3
⎛1⎞
= 1− ⎜ ⎟ = .
6
4
⎝2⎠
1
= –1
cos(π)
b.
sec(π) =
c.
1
⎛ 3π ⎞
sec ⎜ ⎟ =
=– 2
4
⎝ ⎠ cos 3π
4
d.
1
⎛π⎞
csc ⎜ ⎟ =
=1
2
⎝ ⎠ sin π
( )
(2)
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e.
f.
10. a.
b.
( )
( )
b. cos 3t = cos(2t + t ) = cos 2t cos t – sin 2t sin t
π
⎛ π ⎞ cos 4
cot ⎜ ⎟ =
=1
⎝ 4 ⎠ sin π
4
= (2 cos 2 t – 1) cos t – 2sin 2 t cos t
= 2 cos3 t – cos t – 2(1 – cos 2 t ) cos t
( )
( )
π
⎛ π ⎞ sin – 4
tan ⎜ – ⎟ =
= –1
⎝ 4 ⎠ cos – π
4
( )
( )
= 2 cos3 t – cos t – 2 cos t + 2 cos3 t
= 4 cos3 t – 3cos t
c.
π
⎛ π ⎞ sin 3
tan ⎜ ⎟ =
= 3
⎝ 3 ⎠ cos π
3
= 2(2sin x cos x)(2 cos 2 x –1)
= 2(4sin x cos3 x – 2sin x cos x)
= 8sin x cos3 x – 4sin x cos x
1
⎛π⎞
sec ⎜ ⎟ =
=2
⎝ 3 ⎠ cos π
3
( )
d.
( )
( )
c.
π
3
⎛ π ⎞ cos 3
cot ⎜ ⎟ =
=
3
⎝ 3 ⎠ sin π
3
d.
1
⎛π⎞
csc ⎜ ⎟ =
= 2
⎝ 4 ⎠ sin π
e.
π
3
⎛ π ⎞ sin – 6
tan ⎜ – ⎟ =
=–
π
6
3
⎝
⎠ cos –
6
f.
⎛ π⎞ 1
cos ⎜ – ⎟ =
⎝ 3⎠ 2
13. a.
b.
(4)
( )
( )
c.
d.
11. a.
(1 + sin z )(1 – sin z ) = 1 – sin 2 z
1
= cos 2 z =
sin 4 x = sin[2(2 x)] = 2sin 2 x cos 2 x
(1 + cos θ )(1 − cos θ ) = 1 − cos 2 θ = sin 2 θ
sin u cos u
+
= sin 2 u + cos 2 u = 1
csc u sec u
(1 − cos 2 x)(1 + cot 2 x) = (sin 2 x)(csc2 x)
2 ⎛ 1 ⎞
= sin x ⎜ 2 ⎟ = 1
⎝ sin x ⎠
⎛ 1
⎞
sin t (csc t – sin t ) = sin t ⎜
– sin t ⎟
⎝ sin t
⎠
2
2
= 1– sin t = cos t
1 – csc 2 t
csc 2 t
= – cos 2 t = –
2
sec z
b.
(sec t –1)(sec t + 1) = sec 2 t –1 = tan 2 t
c.
sec t – sin t tan t =
=
d.
12. a.
=–
14. a.
cot 2 t
csc 2 t
=–
cos 2 t
sin 2 t
1
sin 2 t
1
sec 2 t
y = sin 2x
1
sin 2 t
–
cos t cos t
1 – sin 2 t cos 2 t
=
= cos t
cos t
cos t
sec2 t – 1
sec 2 t
sin 2 v +
=
tan 2 t
sec 2 t
1
2
=
sin 2 t
cos 2 t
1
cos 2 t
= sin 2 t
= sin 2 v + cos 2 v = 1
sec v
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b.
y = 2 sin t
b. y = 2 cos t
c.
π⎞
⎛
y = cos ⎜ x − ⎟
4⎠
⎝
c.
y = cos 3t
d.
y = sec t
d.
⎛ π⎞
y = cos ⎜ t + ⎟
⎝ 3⎠
15. a.
y = csc t
46
Section 0.7
x
2
Period = 4π , amplitude = 3
16. y = 3 cos
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17. y = 2 sin 2x
Period = π , amplitude = 2
21. y = 21 + 7 sin( 2 x + 3)
Period = π , amplitude = 7, shift: 21 units up,
3
units left
2
π⎞
⎛
22. y = 3cos ⎜ x – ⎟ – 1
2⎠
⎝
18. y = tan x
Period = π
Period = 2 π , amplitude = 3, shifts:
π
units
2
right and 1 unit down.
19. y = 2 +
1
cot(2 x)
6
Period =
π
2
, shift: 2 units up
π⎞
⎛
23. y = tan ⎜ 2 x – ⎟
⎝
3⎠
π
π
units right
Period = , shift:
6
2
20. y = 3 + sec( x − π )
Period = 2π , shift: 3 units up, π units right
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Section 0.7
47
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
π⎞
⎛
24. a. and g.: y = sin ⎜ x + ⎟ = cos x = – cos(π – x)
2⎠
⎝
π⎞
⎛
b. and e.: y = cos ⎜ x + ⎟ = sin( x + π)
2⎠
⎝
= − sin(π − x )
π⎞
⎛
c. and f.: y = cos ⎜ x − ⎟ = sin x
2⎠
⎝
= − sin( x + π)
π⎞
⎛
d. and h.: y = sin ⎜ x − ⎟ = cos( x + π)
2⎠
⎝
= cos( x − π)
–t sin (–t) = t sin t; even
25. a.
b.
sin (– t ) = sin t ; even
c.
1
csc(– t ) =
= – csc t; odd
sin(– t )
2
(π)
( π)
=
32. a.
sin(cos(–t)) = sin(cos t); even
f.
–x + sin(–x) = –x – sin x = –(x + sin x); odd
cot(–t) + sin(–t) = –cot t – sin t
= –(cot t + sin t); odd
26. a.
b.
sin 3 (–t ) = – sin 3 t ; odd
c.
sec(– t) =
1
= sec t; even
cos(–t )
sin 4 (– t ) = sin 4 t ; even
d.
e.
cos(sin(–t)) = cos(–sin t) = cos(sin t); even
f.
(– x )2 + sin(– x ) = x 2 – sin x; neither
2
27. cos 2
π ⎛
π ⎞ ⎛1⎞
1
= ⎜ cos ⎟ = ⎜ ⎟ =
3 ⎝
3⎠
4
⎝2⎠
2
28. sin 2
2
2
π ⎛ π⎞
1
⎛1⎞
= ⎜ sin ⎟ = ⎜ ⎟ =
6 ⎝
6⎠
4
⎝2⎠
3
2
2– 2
4
sin(x – y) = sin x cos(–y) + cos x sin(–y)
= sin x cos y – cos x sin y
b.
cos(x – y) = cos x cos(–y) – sin x sin (–y)
= cos x cos y + sin x sin y
c.
tan( x – y ) =
2
e.
π
3
1 – cos 4 1 – 2
π 1 – cos 2 8
31. sin
=
=
=
8
2
2
2
2
=
sin(−t ) = – sin t = sin t ; even
d.
π
1 + cos 6 1 + 2
π 1 + cos 2 12
=
=
=
30. cos
12
2
2
2
2+ 3
=
4
2
tan x + tan(– y )
1 – tan x tan(– y )
tan x – tan y
1 + tan x tan y
tan t + tan π
tan t + 0
=
1 – tan t tan π 1 – (tan t )(0)
= tan t
33. tan(t + π) =
34. cos( x − π ) = cos x cos(−π ) − sin x sin(−π )
= –cos x – 0 · sin x = –cos x
35. s = rt = (2.5 ft)( 2π rad) = 5π ft, so the tire
goes 5π feet per revolution, or 1 revolutions
5π
per foot.
ft ⎞
⎛ 1 rev ⎞ ⎛ mi ⎞ ⎛ 1 hr ⎞ ⎛
⎜
⎟ ⎜ 60 ⎟ ⎜
⎟ ⎜ 5280 ⎟
5
ft
hr
60
min
mi
π
⎝
⎠⎝
⎠⎝
⎠⎝
⎠
≈ 336 rev/min
36. s = rt = (2 ft)(150 rev)( 2π rad/rev) ≈ 1885 ft
37. r1t1 = r2 t2 ; 6(2π)t1 = 8(2π)(21)
t1 = 28 rev/sec
38. Δy = sin α and Δx = cos α
Δy sin α
m=
=
= tan α
Δx cos α
3
1
π⎞ ⎛1⎞
3 π ⎛
29. sin 6 = ⎜ sin 6 ⎟ = ⎜ 2 ⎟ = 8
⎝
⎠ ⎝ ⎠
48
Section 0.7
Instructor’s Resource Manual
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39. a.
b.
tan α = 3
π
α=
3
3x + 3 y = 6
3 y = – 3x + 6
y=–
3
3
x + 2; m = –
3
3
3
3
tan α = –
α=
44. Divide the polygon into n isosceles triangles by
drawing lines from the center of the circle to
the corners of the polygon. If the base of each
triangle is on the perimeter of the polygon, then
2π
.
the angle opposite each base has measure
n
Bisect this angle to divide the triangle into two
right triangles (See figure).
5π
6
40. m1 = tan θ1 and m2 = tan θ 2
tan θ 2 + tan(−θ1 )
tan θ = tan(θ 2 − θ1 ) =
1 − tan θ 2 tan(−θ1 )
=
tan θ 2 − tan θ1
m − m1
= 2
1 + tan θ 2 tan θ1 1 + m1m2
π b
π
π h
=
so b = 2r sin and cos = so
n 2r
n
n r
π
h = r cos .
n
π
P = nb = 2rn sin
n
π
π
⎛1 ⎞
A = n ⎜ bh ⎟ = nr 2 cos sin
n
n
⎝2 ⎠
sin
3–2
1
=
1 + 3(2 ) 7
θ ≈ 0.1419
41. a.
tan θ =
b.
tan θ =
–1 – 12
1+
( 12 ) (–1)
= –3
θ ≈ 1.8925
c.
2x – 6y = 12
2x + y = 0
–6y = –2x + 12y = –2x
1
y= x–2
3
1
m1 = , m2 = –2
3
–2 – 13
= –7; θ ≈ 1.7127
tan θ =
1 + 13 (–2)
()
42. Recall that the area of the circle is π r 2 . The
measure of the vertex angle of the circle is 2π .
Observe that the ratios of the vertex angles
must equal the ratios of the areas. Thus,
t
A
=
, so
2π π r 2
1
A = r 2t .
2
43. A =
45. The base of the triangle is the side opposite the
t
angle t. Then the base has length 2r sin
2
(similar to Problem 44). The radius of the
t
semicircle is r sin and the height of the
2
t
triangle is r cos .
2
A=
1⎛
t ⎞⎛
t ⎞ π⎛
t⎞
⎜ 2r sin ⎟⎜ r cos ⎟ + ⎜ r sin ⎟
2⎝
2 ⎠⎝
2⎠ 2⎝
2⎠
2
t
t πr 2
t
= r 2 sin cos +
sin 2
2
2
2
2
1
(2)(5) 2 = 25cm 2
2
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Section 0.7
49
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
x
x
x
x
46. cos cos cos cos
2
4
8
16
1⎡
3
1 ⎤1 ⎡
3
1 ⎤
= ⎢cos x + cos x ⎥ ⎢cos x + cos x ⎥
⎣
⎦
⎣
2
4
4 2
16
16 ⎦
1⎡
3
1 ⎤⎡
3
1 ⎤
= ⎢ cos x + cos x ⎥ ⎢cos x + cos x ⎥
4⎣
4
4 ⎦ ⎣ 16
16 ⎦
1⎡
3
3
3
1
= ⎢ cos x cos x + cos x cos x
4⎣
4
16
4
16
3
1
1 ⎤
1
+ cos x cos x + cos x cos x ⎥
16
4
16 ⎦
4
1 ⎡1 ⎛
15
9 ⎞ 1⎛
13
11 ⎞
= ⎢ ⎜ cos + cos x ⎟ + ⎜ cos x + cos x ⎟
4 ⎣2 ⎝
16
16 ⎠ 2 ⎝
16
16 ⎠
1⎛
7
1 ⎞ 1⎛
5
3 ⎞⎤
+ ⎜ cos x + cos x ⎟ + ⎜ cos x + cos x ⎟⎥
2⎝
16
16 ⎠ 2 ⎝
16
16 ⎠ ⎦
1⎡
15
13
11
9
= ⎢cos x + cos x + cos x + cos x
8⎣
16
16
16
16
7
5
3
1 ⎤
+ cos x + cos x + cos x + cos x ⎥
16
16
16
16 ⎦
49. As t increases, the point on the rim of the
wheel will move around the circle of radius 2.
a.
x(2) ≈ 1.902
y (2) ≈ 0.618
x(6) ≈ −1.176
y (6) ≈ −1.618
x(10) = 0
y (10) = 2
x(0) = 0
y (0) = 2
b.
⎛π ⎞
⎛π ⎞
x(t ) = −2 sin ⎜ t ⎟, y (t ) = 2 cos⎜ t ⎟
⎝5 ⎠
⎝5 ⎠
c.
The point is at (2, 0) when
is , when t =
π
5
t=
π
2
; that
5
.
2
2π
. When
10
you add functions that have the same
frequency, the sum has the same frequency.
50. Both functions have frequency
47. The temperature function is
⎛ 2π ⎛ 7 ⎞ ⎞
T (t ) = 80 + 25 sin ⎜⎜
⎜ t − ⎟ ⎟⎟ .
⎝ 12 ⎝ 2 ⎠ ⎠
The normal high temperature for November
15th is then T (10.5) = 67.5 °F.
a.
y (t ) = 3sin(π t / 5) − 5cos(π t / 5)
+2sin((π t / 5) − 3)
48. The water level function is
⎛ 2π
⎞
F (t ) = 8.5 + 3.5 sin ⎜
(t − 9) ⎟ .
⎝ 12
⎠
The water level at 5:30 P.M. is then
F (17.5) ≈ 5.12 ft .
b.
50
Section 0.7
y (t ) = 3cos(π t / 5 − 2) + cos(π t / 5)
+ cos((π t / 5) − 3)
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51.
a.
C sin(ωt + φ ) = (C cos φ )sin ωt + (C sin φ ) cos ω t. Thus A = C ⋅ cos φ and B = C ⋅ sin φ .
b.
A2 + B 2 = (C cos φ )2 + (C sin φ ) 2 = C 2 (cos 2 φ ) + C 2 (sin 2 φ ) = C 2
Also,
c.
B C ⋅ sin φ
=
= tan φ
A C ⋅ cos φ
A1 sin(ωt + φ1 ) + A2 sin(ωt + φ 2 ) + A3 (sin ωt + φ 3 )
= A1 (sin ωt cos φ1 + cos ωt sin φ1 )
+ A2 (sin ωt cos φ 2 + cos ωt sin φ 2 )
+ A3 (sin ωt cos φ 3 + cos ωt sin φ 3 )
= ( A1 cos φ1 + A2 cos φ 2 + A3 cos φ 3 ) sin ωt
+ ( A1 sin φ1 + A2 sin φ 2 + A3 sin φ 3 ) cos ωt
= C sin (ωt + φ )
where C and φ can be computed from
A = A1 cos φ1 + A2 cos φ2 + A3 cos φ3
B = A1 sin φ1 + A2 sin φ2 + A3 sin φ3
as in part (b).
d.
Written response. Answers will vary.
52. ( a.), (b.), and (c.) all look similar to this:
d.
53. a.
b.
c.
e.
The windows in (a)-(c) are not helpful because
the function oscillates too much over the
domain plotted. Plots in (d) or (e) show the
behavior of the function.
Instructor’s Resource Manual
The plot in (a) shows the long term behavior of
the function, but not the short term behavior,
whereas the plot in (c) shows the short term
behavior, but not the long term behavior. The
plot in (b) shows a little of each.
Section 0.7
51
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54. a.
h( x ) = ( f g ) ( x )
3
cos(100 x) + 2
100
=
2
⎛ 1 ⎞
2
⎜
⎟ cos (100 x) + 1
100
⎝
⎠
j ( x) = ( g f )( x) =
56.
⎧
2
f ( x ) = ⎪( x − 2n ) ,
⎨
⎪0.0625,
⎩
1
1⎤
⎡
x ∈ ⎢ 2n − , 2n + ⎥
4
4⎦
⎣
otherwise
where n is an integer.
y
1
3x + 2 ⎞
⎛
cos ⎜100
⎟
100
x2 + 1 ⎠
⎝
0.5
b.
0.25
−2
c.
−1
1
x
2
0.8 Chapter Review
Concepts Test
1. False:
2. True:
⎧
1⎞
⎡
⎪ 4 x − x + 1 : x ∈ ⎢ n, n + ⎟
4⎠
⎪
⎣
55. f ( x ) = ⎨
⎪ − 4 x − x + 7 : x ∈ ⎡ n + 1 , n + 1⎞
⎟
⎢
⎪ 3
3
4
⎣
⎠
⎩
where n is an integer.
(
)
(
p and q must be integers.
p1 p2 p1q2 − p2 q1
−
=
; since
q1 q2
q1q2
p1 , q1 , p2 , and q2 are integers, so
are p1q2 − p2 q1 and q1q2 .
3. False:
If the numbers are opposites
(– π and π ) then the sum is 0,
which is rational.
4. True:
Between any two distinct real
numbers there are both a rational
and an irrational number.
5. False:
0.999... is equal to 1.
6. True:
( am ) = ( an )
7. False:
(a * b) * c = abc ; a *(b * c) = ab
8. True:
Since x ≤ y ≤ z and x ≥ z , x = y = z
)
y
2
1
−1
1
x
9. True:
52
Section 0.8
n
m
= a mn
c
x
would
2
be a positive number less than x .
If x was not 0, then ε =
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10. True:
y − x = −( x − y ) so
( x − y )( y − x) = ( x − y )(−1)( x − y )
20. True:
since 1 + r ≥ 1 − r ,
= (−1)( x − y ) .
2
11. True:
12. True:
14. True:
15. False:
16. False:
17. True:
If r > 1, r = r , and 1 − r = 1 − r , so
−( x − y ) 2 ≤ 0.
1
1
1
.
=
≤
1− r 1− r 1+ r
[ a, b] and [b, c ]
If r < −1, r = − r and 1 − r = 1 + r ,
a
1 1
> 1; <
b
b a
a < b < 0; a < b;
so
share point b in
21. True:
If (a, b) and (c, d) share a point then
c < b so they share the infinitely
many points between b and c.
If x and y are the same sign, then
x – y = x– y . x– y ≤ x+ y
opposite signs then either
x – y = x – (– y ) = x + y
For example, if x = −3 , then
− x = − ( −3) = 3 = 3 which does
(x > 0, y < 0) or
x – y = –x – y = x + y
not equal x.
(x < 0, y > 0). In either case
x – y = x+ y .
For example, take x = 1 and
y = −2 .
4
x < y ⇔ x < y
If either x = 0 or y = 0, the
inequality is easily seen to be true.
4
4
22. True:
x = x and y = y , so x < y
4
4
4
x + y = −( x + y )
4
23. True:
If r = 0, then
1
1
1
=
=
= 1.
1+ r 1 – r 1 – r
For any r, 1 + r ≥ 1 – r . Since
r < 1, 1 – r > 0 so
1
1
;
≤
1+ r 1 – r
If y is positive, then x =
x2 =
= − x + (− y ) = x + y
19. True:
1
1
1
≤
=
.
1− r 1− r 1+ r
when x and y are the same sign, so
x – y ≤ x + y . If x and y have
x 2 = x = − x if x < 0.
4
18. True:
1
1
.
≤
1− r 1+ r
( x − y ) 2 ≥ 0 for all x and y, so
common.
13. True:
If r > 1, then 1 − r < 0. Thus,
( y)
x3 =
= y.
(3 y )
3
=y
24. True:
For example x 2 ≤ 0 has solution
[0].
25. True:
x 2 + ax + y 2 + y = 0
x 2 + ax +
If –1 < r < 0, then r = – r and
2
a2
1 a2 1
+ y2 + y + =
+
4
4 4 4
2
a⎞ ⎛
a2 + 1
1⎞
⎛
+
+
+
=
x
y
⎜
⎟ ⎜
⎟
2⎠ ⎝
2⎠
4
⎝
is a circle for all values of a.
1 – r = 1 + r , so
1
1
1
=
≤
.
1+ r 1 – r 1 – r
1 – r = 1 – r , so
y satisfies
For every real number y, whether it
is positive, zero, or negative, the
cube root x = 3 y satisfies
also, –1 < r < 1.
If 0 < r < 1, then r = r and
2
26. False:
If a = b = 0 and c < 0 , the equation
does not represent a circle.
1
1
1
.
≤
=
1+ r 1 – r 1 – r
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Section 0.8
53
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27. True;
28. True:
29. True:
30. True:
31. True:
32. True:
3
( x − a)
4
3
3a
y = x − + b;
4
4
If x = a + 4:
3
3a
y = (a + 4) –
+b
4
4
3a
3a
=
+3–
+b = b+3
4
4
y −b =
= –( x + 3)( x + 1)
If ab > 0, a and b have the same
sign, so (a, b) is in either the first or
third quadrant.
The domain does not include
π
nπ + where n is an integer.
2
43. True:
The domain is ( − ∞, ∞) and the
range is [−6, ∞) .
44. False:
The range is ( − ∞, ∞) .
45. False:
The range ( − ∞, ∞) .
46. True:
If f(x) and g(x) are even functions,
f(x) + g(x) is even.
f(–x) + g(–x) = f(x) + g(x)
47. True:
If f(x) and g(x) are odd functions,
f(–x) + g(–x) = –f(x) – g(x)
= –[f(x) + g(x)], so f(x) + g(x) is odd
48. False:
If f(x) and g(x) are odd functions,
f(–x)g(–x) = –f(x)[–g(x)] = f(x)g(x),
so
f(x)g(x) is even.
49. True:
If f(x) is even and g(x) is odd,
f(–x)g(–x) = f(x)[–g(x)]
= –f(x)g(x), so f(x)g(x) is odd.
50. False:
If f(x) is even and g(x) is odd,
f(g(–x)) = f(–g(x)) = f(g(x)); while if
f(x) is odd and g(x) is even,
f(g(–x)) = f(g(x)); so f(g(x)) is even.
51. False:
If f(x) and g(x) are odd functions,
f ( g (− x)) = f(–g(x)) = –f(g(x)), so
f(g(x)) is odd.
Let x = ε / 2. If ε > 0 , then x > 0
and x < ε .
If ab = 0, a or b is 0, so (a, b) lies
on
the x-axis or the y-axis. If a
= b = 0,
(a, b) is the origin.
−( x 2 + 4 x + 3) ≥ 0 on −3 ≤ x ≤ −1 .
y1 = y2 , so ( x1 , y1 ) and ( x2 , y2 )
d = [(a + b) – (a – b)]2 + (a – a) 2
34. False:
The equation of a vertical line
cannot be written in point-slope
form.
35. True:
This is the general linear equation.
36. True:
Two non-vertical lines are parallel
if and only if they have the same
slope.
37. False:
The slopes of perpendicular lines
are
negative reciprocals.
38. True:
If a and b are rational and
( a, 0 ) , ( 0, b ) are the intercepts, the
slope is −
b
which is rational.
a
52. True:
f (– x) =
2(– x)3 + (– x)
ax + y = c ⇒ y = − ax + c
ax − y = c ⇒ y = ax − c
(a )(− a) ≠ −1.
(unless a = ±1 )
54
f ( x) = –( x 2 + 4 x + 3)
42. False:
= (2b) 2 = 2b
39. False:
41. True:
The equation is
(3 + 2m) x + (6m − 2) y + 4 − 2m = 0
which is the equation of a straight
line unless 3 + 2m and 6m − 2 are
both 0, and there is no real number
m such that
3 + 2m = 0 and 6m − 2 = 0.
If the points are on the same line,
they have equal slope. Then the
reciprocals of the slopes are also
equal.
are on the same horizontal line.
33. True:
40. True:
Section 0.8
=−
(– x)2 + 1
=
–2 x3 – x
x2 + 1
2 x3 + x
x2 + 1
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53. True:
f (–t ) =
=
(sin(–t )) 2 + cos(– t )
tan(– t ) csc(– t )
(− sin t )2 + cos t (sin t )2 + cos t
=
– tan t (– csc t )
tan t csc t
54. False:
f(x) = c has domain ( − ∞, ∞) and
the only value of the range is c.
55. False:
f(x) = c has domain ( − ∞, ∞) , yet
the range has only one value, c.
g (−1.8) =
57. True:
(f
g )( x) = ( x 3 ) 2 = x 6
(g
f )( x) = ( x 2 )3 = x 6
(f
g )( x) = ( x 3 ) 2 = x 6
f ( x) ⋅ g ( x) = x x = x
2 3
59. False:
60. True:
61. True:
5
b.
cos x
sin x
cos(− x)
cot(− x) =
sin(− x)
cos x
=
= − cot x
− sin x
2
1 ⎞ ⎛ 1⎞
1⎞
25
⎛
⎛
⎜ n + ⎟ ; ⎜ 1 + ⎟ = 2; ⎜ 2 + ⎟ = ;
2⎠
4
n ⎠ ⎝ 1⎠
⎝
⎝
1 ⎞
⎛
⎜ –2 + ⎟
–2 ⎠
⎝
–2
=
4
25
2
(n 2 – n + 1)2 ; ⎡ (1)2 – (1) + 1⎤ = 1;
⎣
⎦
2
⎡ (2) 2 – (2) + 1⎤ = 9;
⎣
⎦
2
⎡ (–2)2 – (–2) + 1⎤ = 49
⎣
⎦
c.
43 / n ; 43 /1 = 64; 43 / 2 = 8; 4 –3 / 2 =
d.
n
1
1
; 1 = 1;
n
1
−2
1
= 2
−2
f
The domain of
excludes any
g
values where g = 0.
f(a) = 0
Let F(x) = f(x + h), then
F(a – h) = f(a – h + h) = f(a) = 0
1
n
1. a.
−1.8
= −0.9 = −1
2
56. True:
58. False:
Sample Test Problems
2. a.
1
1
2
=
=
;
2
2
2
1 1 ⎞⎛
1 1⎞
⎛
⎜1 + + ⎟⎜ 1 − + ⎟
⎝ m n ⎠⎝ m n ⎠
63. False:
The domain of the tangent function
π
excludes all nπ + where n is an
2
integer.
The cosine function is periodic, so
cos s = cos t does not necessarily
imply s = t; e.g.,
cos 0 = cos 2π = 1 , but 0 ≠ 2π .
Instructor’s Resource Manual
=
c.
1 1
+
m
n
=
1 1
1− +
m n
mn + n + m
=
mn − n + m
1+
2
x
2
x
−
− 2
x + 1 x − x − 2 x + 1 ( x − 2)( x + 1)
=
3
2
3
2
−
−
x +1 x − 2
x +1 x − 2
=
62. False:
−1
cot x =
b.
1
8
2( x − 2) − x
3 ( x − 2) − 2( x + 1)
x−4
x −8
(t 3 − 1) (t − 1)(t 2 + t + 1) 2
=
= t + t +1
t −1
t −1
3. Let a, b, c, and d be integers.
a+ c
a
c
ad + bc
b d
which is rational.
=
+
=
2
2b 2d
2bd
Section 0.8
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4. x = 4.1282828…
1000 x = 4128.282828…
10 x = 41.282828…
13. 21t 2 – 44t + 12 ≤ –3; 21t 2 – 44t + 15 ≤ 0;
t=
990 x = 4087
4087
x=
990
44 ± 442 – 4(21)(15) 44 ± 26 3 5
=
= ,
2(21)
42
7 3
⎛ 3 ⎞⎛ 5 ⎞
⎡3 5⎤
⎜ t – ⎟ ⎜ t – ⎟ ≤ 0; ⎢ , ⎥
⎝ 7 ⎠⎝ 3 ⎠
⎣7 3⎦
5. Answers will vary. Possible answer:
13
≈ 0.50990...
50
2x −1
1⎞
⎛
> 0; ⎜ −∞, ⎟ ∪ ( 2, ∞ )
x−2
2⎠
⎝
14.
2
⎛ 3 8.15 × 104 − 1.32 ⎞
⎜
⎟
⎠ ≈ 545.39
6. ⎝
3.24
7.
(π –
2.0
)
2.5
15. ( x + 4)(2 x − 1) 2 ( x − 3) ≤ 0;[−4,3]
– 3 2.0 ≈ 2.66
16. 3x − 4 < 6; −6 < 3 x − 4 < 6; −2 < 3x < 10;
8. sin
2
( 2.45 ) + cos ( 2.40 ) − 1.00 ≈ −0.0495
2
9. 1 – 3 x > 0
3x < 1
1
x<
3
1⎞
⎛
⎜ – ∞, ⎟
3⎠
⎝
10. 6 x + 3 > 2 x − 5
4 x > −8
x > −2; ( −2, ∞ )
11. 3 − 2 x ≤ 4 x + 1 ≤ 2 x + 7
3 − 2 x ≤ 4 x + 1 and 4 x + 1 ≤ 2 x + 7
6 x ≥ 2 and 2 x ≥ 6
1
⎡1 ⎤
x ≥ and x ≤ 3; ⎢ , 3⎥
3
⎣3 ⎦
12. 2 x 2 + 5 x − 3 < 0;(2 x − 1)( x + 3) < 0;
1 ⎛
1⎞
−3 < x < ; ⎜ −3, ⎟
2 ⎝
2⎠
−
17.
2
10 ⎛ 2 10 ⎞
< x < ;⎜ − , ⎟
3
3 ⎝ 3 3⎠
3
≤2
1– x
3
–2≤0
1– x
3 – 2(1 – x)
≤0
1– x
2x +1
≤ 0;
1– x
1⎤
⎛
⎜ – ∞, – ⎥ ∪ (1, ∞ )
2⎦
⎝
18. 12 − 3 x ≥ x
(12 − 3 x)2 ≥ x 2
144 − 72 x + 9 x 2 ≥ x 2
8 x 2 − 72 x + 144 ≥ 0
8( x − 3)( x − 6) ≥ 0
(−∞,3] ∪ [6, ∞)
19. For example, if x = –2, −(−2) = 2 ≠ −2
− x ≠ x for any x < 0
20. If − x = x, then x = x.
x≥0
56
Section 0.8
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⎛ 2 + 10 0 + 4 ⎞
,
27. center = ⎜
⎟ = (6, 2)
2 ⎠
⎝ 2
1
1
radius =
(10 – 2) 2 + (4 – 0) 2 =
64 + 16
2
2
=2 5
21. |t – 5| = |–(5 – t)| = |5 – t|
If |5 – t| = 5 – t, then 5 − t ≥ 0.
t ≤5
22. t − a = −(a − t ) = a − t
If a − t = a − t , then a − t ≥ 0.
t≤a
circle: ( x – 6)2 + ( y – 2) 2 = 20
23. If x ≤ 2, then
28. x 2 + y 2 − 8 x + 6 y = 0
x 2 − 8 x + 16 + y 2 + 6 y + 9 = 16 + 9
0 ≤ 2 x 2 + 3 x + 2 ≤ 2 x 2 + 3 x + 2 ≤ 8 + 6 + 2 = 16
( x − 4) 2 + ( y + 3) 2 = 25;
1
1
≤ . Thus
also x + 2 ≥ 2 so
2
x +2 2
2
2 x2 + 3x + 2
x2 + 2
= 2 x2 + 3x + 2
⎛1⎞
≤ 16 ⎜ ⎟
2
⎝2⎠
x +2
1
=8
24. a.
The distance between x and 5 is 3.
b. The distance between x and –1 is less than
or equal to 2.
c.
The distance between x and a is greater
than b.
center = ( 4, −3) , radius = 5
x2 − 2 x + y 2 + 2 y = 2
29.
x2 − 2 x + 1 + y2 + 2 y + 1 = 2 + 1 + 1
( x − 1) 2 + ( y + 1) 2 = 4
center = (1, –1)
x 2 + 6 x + y 2 – 4 y = –7
x 2 + 6 x + 9 + y 2 – 4 y + 4 = –7 + 9 + 4
( x + 3)2 + ( y – 2)2 = 6
center = (–3, 2)
d = (–3 – 1) 2 + (2 + 1)2 = 16 + 9 = 5
25.
30. a.
d ( A, B ) = (1 + 2) 2 + (2 − 6)2
3x + 2 y = 6
2 y = −3 x + 6
3
y = − x+3
2
3
m=−
2
3
y − 2 = − ( x − 3)
2
3
13
y = − x+
2
2
= 9 + 16 = 5
d ( B, C ) = (5 − 1)2 + (5 − 2)2
= 16 + 9 = 5
d ( A, C ) = (5 + 2)2 + (5 − 6) 2
= 49 + 1 = 50 = 5 2
( AB) + ( BC )2 = ( AC ) 2 , so ΔABC is a right
triangle.
2
⎛1+ 7 2 + 8 ⎞
,
26. midpoint: ⎜
⎟ = ( 4,5 )
2 ⎠
⎝ 2
d = (4 − 3)2 + (5 + 6)2 = 1 + 121 = 122
Instructor’s Resource Manual
Section 0.8
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b.
2
m= ;
3
b.
3x – 2 y = 5
–2 y = –3 x + 5
3
5
y = x– ;
2
2
3
m=
2
3
y –1 = ( x + 2)
2
3
y = x+4
2
c.
3 x + 4y = 9
4y = –3x + 9;
4
3
9
y = – x+ ; m =
3
4
4
4
y –1 = ( x + 2)
3
4
11
y = x+
3
3
2
( x − 1)
3
2
5
y = x−
3
3
y +1 =
c.
y=9
d. x = –2
e.
contains (–2, 1) and (0, 3); m =
3 –1
;
0+2
y=x+3
3 +1 4
11 − 3 8 4
= ; m2 =
= = ;
5−2 3
11 − 5 6 3
11 + 1 12 4
m3 =
=
=
11 − 2 9 3
m1 = m2 = m3 , so the points lie on the same
line.
32. m1 =
d. x = –3
33. The figure is a cubic with respect to y.
The equation is (b) x = y 3 .
34. The figure is a quadratic, opening downward,
with a negative y-intercept. The equation is (c)
y = ax 2 + bx + c. with a < 0, b > 0, and c < 0.
35.
31. a.
58
3 –1 2
m=
= ;
7+2 9
2
y –1 = ( x + 2)
9
2
13
y = x+
9
9
Section 0.8
Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
36.
x2 − 2 x + y 2 = 3
x2 − 2 x + 1 + y 2 = 4
( x − 1)2 + y 2 = 4
40. 4 x − y = 2
y = 4 x − 2;
1
4
contains ( a, 0 ) , ( 0, b ) ;
m=−
ab
=8
2
ab = 16
16
b=
a
1
b−0
b
=− =− ;
0−a
4
a
a = 4b
⎛ 16 ⎞
a = 4⎜ ⎟
⎝a⎠
37.
a 2 = 64
a =8
b=
41. a.
b.
38.
c.
39. y = x2 – 2x + 4 and y – x = 4;
x + 4 = x2 − 2 x + 4
x 2 − 3x = 0
x( x – 3) = 0
points of intersection: (0, 4) and (3, 7)
f (1) =
1 1
1
– =–
1+1 1
2
1
1
⎛ 1⎞
f ⎜– ⎟ =
–
=4
1
1
2
–
+
1
–
⎝
⎠
2
2
f(–1) does not exist.
1
1
1 1
= –
–
t –1+1 t –1 t t –1
d.
f (t – 1) =
e.
1
1
t
⎛1⎞
f ⎜ ⎟=
– =
–t
1
1
⎝ t ⎠ t +1 t 1+ t
42. a.
b.
c.
Instructor’s Resource Manual
16
1
= 2; y = − x + 2
8
4
g (2) =
2 +1 3
=
2
2
⎛1⎞
g⎜ ⎟ =
⎝ 2⎠
1
2
+1
1
2
=3
2 + h +1
– 22+1
g ( 2 + h ) – g ( 2)
= 2+ h
h
h
h
2 h + 6 – 3h – 6
– 2( h + 2)
–1
2( h + 2)
=
=
=
h
h
2(h + 2)
43. a.
{x ∈
: x ≠ –1, 1}
b.
{x ∈
: x ≤ 2}
Section 0.8
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44. a.
b.
f (– x) =
3(– x)
(– x) + 1
2
=–
3x
x +1
2
; odd
46.
g (– x) = sin(– x) + cos(– x)
= − sin x + cos x = sin x + cos x; even
c.
h(– x) = (– x)3 + sin(– x) = – x3 – sin x ; odd
d.
k (– x) =
45. a.
(– x )2 + 1
– x + (– x) 4
=
x2 + 1
x + x4
; even
47. V(x) = x(32 – 2x)(24 – 2x)
Domain [0, 12]
2
f (x) = x – 1
48. a.
1⎞
13
⎛
( f + g )(2) = ⎜ 2 – ⎟ + (22 + 1) =
2⎠
2
⎝
b.
15
⎛3⎞
( f ⋅ g )(2) = ⎜ ⎟ (5) =
2
⎝2⎠
c.
(f
g )(2) = f (5) = 5 –
d.
(g
13
⎛3⎞ ⎛3⎞
f )(2) = g ⎜ ⎟ = ⎜ ⎟ + 1 =
2
2
4
⎝ ⎠ ⎝ ⎠
e.
1⎞
⎛
f 3 (–1) = ⎜ –1 + ⎟ = 0
1⎠
⎝
1 24
=
5 5
2
b.
x
g(x) = 2
x +1
3
2
f.
49. a.
c.
60
⎧ x2
h(x) = ⎨
⎩6 – x
Section 0.8
⎛3⎞
f 2 (2) + g 2 (2) = ⎜ ⎟ + (5) 2
⎝2⎠
9
109
= + 25 =
4
4
y=
1 2
x
4
if 0 ≤ x ≤ 2
if x > 2
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
b.
y=
1
( x + 2)2
4
53. a.
b.
sin (–t) = –sin t = –0.8
sin 2 t + cos2 t = 1
cos 2 t = 1 – (0.8)2 = 0.36
cos t = –0. 6
c.
1
y = –1 + ( x + 2) 2
4
c.
sin 2t = 2 sin t cos t = 2(0.8)(–0.6) = –0.96
d.
tan t =
e.
⎛π ⎞
cos ⎜ – t ⎟ = sin t = 0.8
⎝2 ⎠
f.
sin(π + t ) = − sin t = −0.8
sin t
0.8
4
=
= – ≈ –1.333
cos t –0.6
3
54. sin 3t = sin(2t + t ) = sin 2t cos t + cos 2t sin t
= 2sin t cos 2 t + (1 – 2sin 2 t ) sin t
= 2sin t (1 – sin 2 t ) + sin t – 2sin 3 t
= 2sin t – 2sin 3 t + sin t – 2sin 3 t
= 3sin t – 4sin 3 t
55. s = rt
⎛ rev ⎞⎛ rad ⎞ ⎛ 1 min ⎞
= 9 ⎜ 20
⎟⎜ 2π
⎟⎜
⎟ (1 sec) = 6π
⎝ min ⎠⎝ rev ⎠ ⎝ 60 sec ⎠
≈ 18.85 in.
50. a.
b.
c.
(−∞,16]
f
Review and Preview Problems
g = 16 – x 4 ; domain [–2, 2]
g f = ( 16 – x ) 4 = (16 – x) 2 ;
domain (−∞,16]
(note: the simplification
( 16 – x ) 4 = (16 – x) 2 is only true given
the restricted domain)
51.
f ( x) = x , g ( x) = 1 + x, h( x) = x 2 , k(x) = sin x,
F ( x) = 1 + sin 2 x = f g h k
52. a.
sin(570°) = sin(210°) = –
1
2
b.
⎛ 9π ⎞
⎛π⎞
cos ⎜ ⎟ = cos ⎜ ⎟ = 0
⎝ 2 ⎠
⎝2⎠
c.
3
⎛ 13π ⎞
⎛ π⎞
cos ⎜ –
⎟ = cos ⎜ − ⎟ =
⎝ 6 ⎠
⎝ 6⎠ 2
Instructor’s Resource Manual
1. a)
b)
2. a)
b)
0 < 2 x < 4; 0 < x < 2
−6 < x < 16
13 < 2 x < 14; 6.5 < x < 7
−4 < − x / 2 < 7; − 14 < x < 8
3. x − 7 = 3 or x − 7 = −3
x = 10 or
x=4
4. x + 3 = 2 or
x = −1 or
x + 3 = −2
x = −5
5. x − 7 = 3 or x − 7 = −3
x = 10 or
x=4
6. x − 7 = d or x − 7 = − d
x = 7 + d or x = 7 − d
7. a)
x − 7 < 3 and x − 7 > −3
x < 10 and
x>4
4 < x < 10
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b)
c)
d)
8. a)
x − 7 ≤ 3 and x − 7 ≥ −3
x ≤ 10 and
x≥4
4 ≤ x ≤ 10
x − 7 ≤ 1 and
x ≤ 8 and
6≤ x≤8
x − 2 < 1 and x − 2 > −1
x < 3 and
x >1
1< x < 3
x − 2 < 0.1 and x − 2 > −0.1
x < 2.1 and
x > 1.9
1.9 < x < 2.1
d)
x − 2 < 0.01 and x − 2 > −0.01
x < 2.01 and
x > 1.99
1.99 < x < 2.01
11. a)
g ( 0.999 ) = −0.000333556
g (1.001) = 0.000333111
g (1.1) = 0.03125
x − 7 < 0.1 and x − 7 > −0.1
x < 7.1 and
x > 6.9
6.9 < x < 7.1
c)
10. a)
g ( 0.99 ) = −0.0033557
g (1.01) = 0.00331126
x − 2 ≥ 1 or x − 2 ≤ −1
x ≥ 3 or
x ≤1
b)
g ( 2) =
12. a)
x − 1 ≠ 0; x ≠ 1
2 x 2 − x − 1 ≠ 0; x ≠ 1, − 0.5
x≠0
b)
g ( 0 ) = −1
g ( 0.9 ) = −0.0357143
x − 7 ≥ −1
x≥6
b)
9. a)
b)
b)
1
= −1
−1
0.1
= −1
F ( −0.1) =
−0.1
0.01
F ( −0.01) =
= −1
−0.01
0.001
F ( −0.001) =
= −1
−0.001
0.001
F ( 0.001) =
=1
0.001
0.01
F ( 0.01) =
=1
0.01
0.01
F ( 0.1) =
=1
0.01
1
F (1) = = 1
1
F ( −1) =
G ( −1) = 0.841471
G ( −0.1) = 0.998334
G ( −0.01) = 0.999983
x≠0
0 −1
f ( 0) =
=1
0 −1
0.81 − 1
f ( 0.9 ) =
= 1.9
0.9 − 1
0.9801 − 1
= 1.99
f ( 0.99 ) =
0.99 − 1
0.998001 − 1
= 1.999
f ( 0.999 ) =
.999 − 1
1.002001 − 1
= 2.001
f (1.001) =
1.001 − 1
1.0201 − 1
= 2.01
f (1.01) =
1.01 − 1
1.21 − 1
= 2.1
f (1.1) =
1.1 − 1
4 −1
=3
f ( 2) =
2 −1
1
5
G ( −0.001) = 0.99999983
G ( 0.001) = 0.99999983
G ( 0.01) = 0.999983
G ( 0.1) = 0.998334
G (1) = 0.841471
13. x − 5 < 0.1 and x − 5 > −0.1
x < 5.1 and
x > 4.9
4.9 < x < 5.1
14. x − 5 < ε
and x − 5 > −ε
x < 5 + ε and
x > 5−ε
5−ε < x < 5+ε
15. a.
True.
b. False: Choose a = 0.
c.
True.
d. True
16. sin ( c + h ) = sin c cos h + cos c sin h
62
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
CHAPTER
Limits
9.
1.1 Concepts Review
x3 – 4 x 2 + x + 6
x → –1
x +1
lim
( x + 1)( x 2 – 5 x + 6)
x → –1
x +1
1. L; c
= lim
2. 6
= lim ( x 2 – 5 x + 6)
x → –1
2
3. L; right
= (–1) – 5(–1) + 6
4. lim f ( x) = M
= 12
x →c
Problem Set 1.1
x2
= lim( x 2 + 2 x –1) = –1
x →0
1. lim( x – 5) = –2
x →0
x →3
2. lim (1 – 2t ) = 3
11.
t → –1
3.
4.
lim ( x 2 + 2 x − 1) = (−2) 2 + 2(−2) − 1 = −1
= –t – t = –2t
lim ( x 2 + 2t − 1) = (−2) 2 + 2t − 1 = 3 + 2t
x →−2
(
2
) ( ( −1)
6. lim t 2 − x 2 =
t →−1
12.
) ( ( −1) − 1) = 0
5. lim t 2 − 1 =
t →−1
2
)
x2 – 4
( x – 2)( x + 2)
= lim
x→2 x – 2
x→2
x–2
= lim( x + 2)
x2 – 9
x →3 x – 3
( x – 3)( x + 3)
= lim
x →3
x–3
= lim( x + 3)
lim
x →3
− x2 = 1 − x2
7. lim
=3+3=6
13.
x→2
lim
(t + 4)(t − 2) 4
(3t − 6) 2
t →2
= lim
=2+2=4
8.
x2 – t 2
( x + t )( x – t )
= lim
x→–t x + t
x→ – t
x+t
= lim ( x – t )
lim
x→ –t
x →−2
(
x 4 + 2 x3 – x 2
10. lim
(t − 2) 2 t + 4
9(t − 2) 2
t →2
t 2 + 4t – 21
t → –7
t+7
(t + 7)(t – 3)
= lim
t → –7
t+7
= lim (t – 3)
t+4
9
= lim
lim
t →2
=
2+4
6
=
9
9
t → –7
= –7 – 3 = –10
14.
(t − 7)3
t −7
lim
t →7+
= lim
t →7
+
= lim
t →7+
(t − 7) t − 7
t −7
t −7
= 7−7 = 0
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63
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or by any means, without permission in writing from the publisher.
15. lim
x 4 –18 x 2 + 81
x →3
( x – 3)
2
= lim
x →3
( x – 3) 2 ( x + 3) 2
= lim
x →3
( x – 3)
( x 2 – 9) 2
( x – 3)
= lim( x + 3)2 = (3 + 3) 2
2
lim
t →0
2
(3u + 4)(2u – 2)3
(u –1) 2
u →1
= lim
( x − sin x ) 2 / x 2
x
21.
x →3
1.
0.0251314
0.1
2.775 × 10−6
8(3u + 4)(u –1)3
0.01
2.77775 × 10−10
(u –1) 2
0.001
2.77778 × 10−14
–1.
–0.1
0.0251314
2.775 × 10−6
–0.01
2.77775 × 10−10
–0.001
2.77778 × 10−14
= 36
16. lim
1 − cos t
=0
2t
u →1
= lim 8(3u + 4)(u – 1) = 8[3(1) + 4](1 – 1) = 0
u →1
17.
(2 + h) 2 − 4
4 + 4h + h 2 − 4
= lim
h→0
h→0
h
h
lim
h 2 + 4h
= lim(h + 4) = 4
h →0
h →0
h
= lim
lim
( x – sin x) 2
x2
x →0
18.
( x + h) 2 − x 2
x 2 + 2 xh + h 2 − x 2
= lim
h→0
h →0
h
h
lim
h 2 + 2 xh
= lim(h + 2 x) = 2 x
h →0
h →0
h
= lim
sin x
2x
x
19.
2
(1 − cos x ) / x
x
22.
0.211322
0.1
0.00249584
0.01
0.001
0.0000249996
2.5 × 10−7
–1.
0.211322
–0.1
0.00249584
0.0000249996
2.5 × 10−7
0.420735
0.1
0.499167
0.01
0.499992
–0.01
0.001
0.49999992
–0.001
–1.
0.420735
–0.1
0.499167
–0.01
0.499992
–0.001
0.49999992
0.01
0.001
x2
x →0
2
(t − 1) /(sin(t − 1))
t
23.
=0
1.1
2.1035
1.01
2.01003
1.001
2.001
0.229849
0
1.1884
0.0249792
0.9
1.90317
0.00249998
0.99
1.99003
0.999
1.999
1− cos t
2t
t
0.1
(1 – cos x) 2
3.56519
sin x
= 0.5
x →0 2 x
1.
lim
2.
lim
0.00024999998
2
1.
1.
20.
=0
t −1
=2
− 1)
2
lim
64
–1.
–0.229849
–0.1
–0.0249792
–0.01
–0.00249998
–0.001
–0.00024999998
Section 1.1
t →1 sin(t
Instructor’s Resource Manual
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or by any means, without permission in writing from the publisher.
x −sin( x − 3) − 3
x −3
x
24.
4.
1. + π4
0.1 +
0.158529
3.1
2.
lim
x→ π
4
(1 + sin( x − 3π / 2)) /( x − π )
x
1. + π
0.4597
0.1 + π
0.0500
0.01 + π
0.0050
0.001 + π
0.0005
–0.0050
–0.001 + π
–0.0005
1 + sin ( x − 32π )
x−π
x →π
=0
−0.1 +
0.0000210862
2.12072 × 10−7
π
2
0.536908
π
2
0.00226446
π
2
π
2
0.0000213564
2.12342 × 10−7
2 − 2sin u
lim
=0
π
u→
3u
2
29. a.
–0.896664
0.01
–0.989967
0.001
–0.999
d.
–1.
–1.64209
e.
–0.1
–1.09666
f.
–0.01
–1.00997
–0.001
–1.001
lim f ( x) = 2
x → –3
b. f(–3) = 1
c.
g.
= –1
h.
i.
Instructor’s Resource Manual
0.00199339
−0.001 +
0.1
1
t
0.11921
−0.01 +
0.357907
1 – cot t
= 0.25
(2 − 2sin u ) / 3u
0.001 +
1.
t →0
(tan x − 1)2
0.01 + π2
(1 − cot t ) /(1 / t )
lim
(x − )
0.2505
0.1 + π2
t
26.
0.255008
1. + π2
–0.0500
–0.01 + π
lim
0.300668
u
28.
–0.4597
–0.1 + π
0.674117
π 2
4
−1. + π2
–1. + π
0.2495
4
−0.001 + π4
x – sin( x – 3) – 3
=0
lim
x →3
x–3
25.
0.245009
π
−0.01 + π4
0.0000166666
1.66667 × 10−7
2.999
4
−0.1 + π4
0.00166583
2.99
0.201002
π
−1. + π4
0.158529
2.9
4
0.001 +
0.0000166666
1.66667 × 10−7
3.001
0.0320244
π
0.01 +
0.00166583
3.01
( x − π / 4) 2 /(tan x − 1) 2
x
27.
f(–1) does not exist.
lim f ( x) =
x → –1
5
2
f(1) = 2
lim f(x) does not exist.
x→1
lim f ( x) = 2
x →1–
lim f ( x) = 1
x →1+
lim f ( x ) =
+
x →−1
5
2
Section 1.1
65
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or by any means, without permission in writing from the publisher.
lim f ( x) does not exist.
b.
lim f ( x) does not exist.
b.
f(–3) = 1
c.
f(1) = 2
c.
f(–1) = 1
d.
30. a.
d.
x → –3
lim f ( x) = 2
x → –1
e.
f(1) = 1
f.
lim f ( x) does not exist.
g.
h.
i.
31. a.
b.
c.
d.
e.
f.
32. a.
b.
c.
x →1
lim f ( x) = 2
x →1+
34.
x →1
lim f ( x) = 1
x →1–
lim f ( x) does not exist.
x →1+
lim f ( x ) = 2
x →−1+
a.
f(–3) = 2
f(3) is undefined.
x →1
b. g(1) does not exist.
lim f ( x) = 2
c.
x → –3−
lim f ( x) = 4
x → –3+
d.
lim f ( x) does not exist.
x → –3
lim g ( x) = 0
35.
lim g ( x ) = 1
x→2
lim g ( x ) = 1
x → 2+
f ( x) = x – ⎣⎡[ x ]⎦⎤
lim f ( x) does not exist.
x →3+
lim f ( x) = −2
x → –1−
lim f ( x) = −2
x → –1+
lim f ( x) = −2
x → –1
d. f (–1) = –2
e.
lim f ( x) = 0
f.
f (1) = 0
x →1
a.
b.
33.
c.
d.
a.
66
f(0) = 0
lim f ( x) does not exist.
x →0
lim f ( x ) = 1
x →0 –
lim f ( x) =
x→ 1
2
1
2
lim f ( x) = 0
x →0
Section 1.1
Instructor’s Resource Manual
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or by any means, without permission in writing from the publisher.
f ( x) =
36.
41. lim f ( x) exists for a = –1, 0, 1.
x
x
x→a
42. The changed values will not change lim f ( x) at
x→a
any a. As x approaches a, the limit is still a 2 .
43. a.
x −1
lim
x →1
lim
x −1
x −1
−
x →1
b.
f (0) does not exist.
a.
lim f ( x) does not exist.
b.
x →0
lim
−
x →1
lim
c.
x −1
x −1
x −1
does not exist.
= −1 and lim
+
x →1
x −1
=1
= −1
x2 − x − 1 − 1
x −1
x →1−
x −1
= −3
lim f ( x ) = –1
c.
x →0 –
d.
⎡ 1
1 ⎤
lim ⎢
−
⎥ does not exist.
− x −1
x − 1 ⎥⎦
x →1 ⎢
⎣
d.
lim f ( x) = 1
x→ 1
2
44. a.
x2 − 1
37. lim
does not exist.
x →1 x − 1
lim
x →1−
x →0
= lim
c1f
lim dd gg does not exist.
+ x
x →0 e h
x+2− 2
x
c.
lim x(−1)ed
( x + 2 − 2)( x + 2 + 2)
d.
x →0
x+2−2
x( x + 2 + 2)
= lim
x →0
39. a.
b.
c1/ x f
hg
x →0
= lim
x →0
1
x
2
=
=
=
4
0+2 + 2 2 2
x+2+ 2
=0
c1/ x f
hg
x →0
+
45. a) 1
x( x + 2 + 2)
1
+
lim a x b (−1)ed
x( x + 2 + 2)
x →0
= lim
x − a xb = 0
b.
x2 − 1
x2 − 1
=2
= −2 and lim
x −1
x →1+ x − 1
38. lim
lim
x →1+
b) 0
−1
c)
=0
d)
−1
1
46. a) Does not exist
c)
lim f ( x) does not exist.
1
b) 0
d) 0.556
x →1
lim f ( x) = 0
47. lim x does not exist since
x →0
x →0
40.
x is not defined
for x < 0.
48.
lim x x = 1
x → 0+
49. lim
x →0
x =0
x
50. lim x = 1
x →0
sin 2 x 1
=
x →0 4 x
2
51. lim
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or by any means, without permission in writing from the publisher.
52. lim
x →0
7. If x is within 0.001 of 2, then 2x is within 0.002
of 4.
sin 5 x 5
=
3x
3
⎛1⎞
53. lim cos ⎜ ⎟ does not exist.
x →0
⎝ x⎠
⎛1⎞
54. lim x cos ⎜ ⎟ = 0
x →0
⎝ x⎠
x3 − 1
55. lim
56. lim
x →0
57.
=6
2x + 2 − 2
x →1
x sin 2 x
sin( x 2 )
lim
x →2–
58. lim
+
x →1
8. If x is within 0.0005 of 2, then x2 is within 0.002
of 4.
=2
x2 – x – 2
= –3
x–2
2
1/( x −1)
1+ 2
=0
59. lim x ; The computer gives a value of 0, but
x →0
lim
x →0−
9. If x is within 0.0019 of 2, then
0.002 of 4.
8 x is within
x does not exist.
1.2 Concepts Review
1. L – ε ; L + ε
2. 0 < x – a < δ ; f ( x) – L < ε
10. If x is within 0.001 of 2, then
3.
ε
8
is within 0.002
x
of 4.
3
4. ma + b
Problem Set 1.2
1. 0 < t – a < δ ⇒ f (t ) – M < ε
2. 0 < u – b < δ ⇒ g (u ) – L < ε
2x – 1+ 1 < ε ⇔ 2x < ε
3. 0 < z – d < δ ⇒ h( z ) – P < ε
⇔ 2 x <ε
4. 0 < y – e < δ ⇒ φ ( y ) – B < ε
⇔ x <
5. 0 < c – x < δ ⇒ f ( x) – L < ε
6. 0 < t – a < δ ⇒ g (t ) – D < ε
68
11. 0 < x – 0 < δ ⇒ (2 x – 1) – (–1) < ε
Section 1.2
ε
2
ε
δ = ;0 < x –0 <δ
2
(2 x – 1) – (–1) = 2 x = 2 x < 2δ = ε
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or by any means, without permission in writing from the publisher.
12. 0 < x + 21 < δ ⇒ (3x – 1) – (–64) < ε
3 x – 1 + 64 < ε ⇔ 3 x + 63 < ε
⇔ 3( x + 21) < ε
2 x 2 – 11x + 5
(2 x – 1)( x – 5)
–9 <ε ⇔
–9 <ε
x–5
x–5
⇔ 3 x + 21 < ε
⇔ x + 21 <
ε
⇔ 2x – 1 – 9 < ε
3
⇔ 2( x – 5) < ε
ε
δ = ; 0 < x + 21 < δ
3
(3 x – 1) – (–64) = 3 x + 63 = 3 x + 21 < 3δ = ε
x 2 – 25
13. 0 < x – 5 < δ ⇒
– 10 < ε
x–5
x 2 – 25
( x – 5)( x + 5)
– 10 < ε ⇔
– 10 < ε
x–5
x–5
⇔ x + 5 – 10 < ε
⇔ x–5 <
ε
2
ε
δ = ;0 < x –5 <δ
2
2 x – 11x + 5
(2 x – 1)( x – 5)
–9 =
–9
x–5
x–5
2
= 2 x – 1 – 9 = 2( x – 5) = 2 x – 5 < 2δ = ε
16. 0 < x – 1 < δ ⇒
⇔ x–5 <ε
2x – 2 < ε
2x – 2 < ε
δ = ε; 0 < x – 5 < δ
( 2 x – 2 )( 2 x + 2 )
⇔
x – 25
( x – 5)( x + 5)
– 10 =
– 10 = x + 5 – 10
x–5
x–5
2x + 2
2
2x – 2
⇔
2x + 2
= x–5 <δ =ε
2
⇔2
2x – x
14. 0 < x – 0 < δ ⇒
− (−1) < ε
x
2 x2 – x
x(2 x – 1)
+1 < ε ⇔
+1 < ε
x
x
⇔ 2x < ε
⇔ 2 x <ε
ε
2
ε
δ = ;0 < x –0 <δ
2
2 x2 – x
x(2 x – 1)
− (−1) =
+ 1 = 2x – 1+ 1
x
x
<ε
x –1
2x + 2
<ε
2ε
; 0 < x –1 < δ
2
( 2 x – 2)( 2 x + 2)
2x − 2 =
2x + 2
=
2x – 2
2x + 2
2 x –1
2 x – 1 2δ
≤
<
=ε
2x + 2
2
2
17. 0 < x – 4 < δ ⇒
2x – 1
x–3
= 2 x = 2 x < 2δ = ε
⇔
⇔
⇔
Instructor’s Resource Manual
<ε
δ=
⇔ 2x – 1 +1 < ε
⇔ x <
2 x 2 – 11x + 5
–9 <ε
x–5
15. 0 < x – 5 < δ ⇒
2x – 1
x–3
– 7 <ε ⇔
– 7 <ε
2 x – 1 – 7( x – 3)
x–3
<ε
( 2 x – 1 – 7( x – 3))( 2 x – 1 + 7( x – 3))
x – 3( 2 x – 1 + 7( x – 3))
2 x – 1 – (7 x – 21)
x – 3( 2 x – 1 + 7( x – 3))
–5( x – 4)
x – 3( 2 x – 1 + 7( x – 3))
<ε
<ε
<ε
Section 1.2
69
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or by any means, without permission in writing from the publisher.
⇔ x−4 ⋅
To bound
5
x − 3( 2 x − 1 + 7( x − 3))
x – 3( 2 x – 1 + 7( x – 3))
1
2
1
2
, agree that
7
9
< x < , so
2
2
5
x – 3( 2 x – 1 + 7( x – 3))
hence x − 4 ⋅
19. 0 < x – 1 < δ ⇒
< 1.65 and
x − 3( 2 x − 1 + 7( x − 3))
<ε
2x −1
5
− 7 = x−4 ⋅
x −3
x − 3( 2 x − 1 + 7( x − 3))
< x – 4 (1.65) < 1. 65δ ≤ ε
1
1
ε
since δ = only when ≤
so 1.65δ ≤ ε .
2
2 1. 65
14 x 2 – 20 x + 6
–8 < ε
x –1
14 x 2 – 20 x + 6
2(7 x – 3)( x – 1)
–8 <ε ⇔
–8 <ε
x –1
x –1
⇔ 2(7 x – 3) – 8 < ε
⇔ 14( x – 1) < ε
⇔ 14 x – 1 < ε
δ=
ε
14
( x – 1)2
–4 <ε
–4 <ε
⇔ 10 x – 6 – 4 < ε
⇔ 10 x – 1 < ε
⇔ x –1 <
1.65
For whatever ε is chosen, let δ be the smaller of
1
ε
and
.
1.65
2
⎧1
ε ⎫
δ = min ⎨ ,
⎬, 0 < x – 4 < δ
⎩ 2 1. 65 ⎭
⇔ x –1 <
(10 x – 6)( x – 1)2
–4 <ε
⇔ 10( x – 1) < ε
5
18. 0 < x – 1 < δ ⇒
( x – 1) 2
( x –1)2
⇔
ε
⇔ x–4 <
10 x3 – 26 x 2 + 22 x – 6
10 x3 – 26 x 2 + 22 x – 6
5
δ ≤ . If δ ≤ , then
0.65 <
<ε
ε
14
δ=
ε
10
ε
10
; 0 < x –1 < δ
10 x3 – 26 x 2 + 22 x – 6
( x – 1) 2
–4 =
(10 x – 6)( x – 1) 2
( x – 1) 2
–4
= 10 x − 6 − 4 = 10( x − 1)
= 10 x − 1 < 10δ = ε
20. 0 < x – 1 < δ ⇒ (2 x 2 + 1) – 3 < ε
2 x2 + 1 – 3 = 2 x2 – 2 = 2 x + 1 x – 1
To bound 2 x + 2 , agree that δ ≤ 1 .
x – 1 < δ implies
2x + 2 = 2x – 2 + 4
≤ 2x – 2 + 4
<2+4=6
ε
⎧ ε⎫
δ ≤ ; δ = min ⎨1, ⎬; 0 < x – 1 < δ
6
⎩ 6⎭
(2 x + 1) – 3 = 2 x 2 – 2
2
⎛ε ⎞
= 2x + 2 x −1 < 6 ⋅ ⎜ ⎟ = ε
⎝6⎠
; 0 < x –1 < δ
14 x 2 – 20 x + 6
2(7 x – 3)( x – 1)
–8 =
–8
x –1
x –1
= 2(7 x – 3) – 8
= 14( x – 1) = 14 x – 1 < 14δ = ε
70
Section 1.2
Instructor’s Resource Manual
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or by any means, without permission in writing from the publisher.
21. 0 < x + 1 < δ ⇒ ( x 2 – 2 x – 1) – 2 < ε
x2 – 2 x – 1 – 2 = x2 – 2 x – 3 = x + 1 x – 3
To bound x – 3 , agree that δ ≤ 1 .
x + 1 < δ implies
⎛1⎞
25. For all x ≠ 0 , 0 ≤ sin 2 ⎜ ⎟ ≤ 1 so
⎝x⎠
1
⎛ ⎞
x 4 sin 2 ⎜ ⎟ ≤ x 4 for all x ≠ 0 . By Problem 18,
⎝ x⎠
4
lim x = 0, so, by Problem 20,
x→0
4
2 ⎛ 1⎞
lim x sin ⎜ ⎟ = 0.
⎝ x⎠
x→0
x – 3 = x + 1 – 4 ≤ x + 1 + –4 < 1 + 4 = 5
ε
⎧ ε⎫
δ ≤ ; δ = min ⎨1, ⎬ ; 0 < x + 1 < δ
⎩ 5⎭
5
26. 0 < x < δ ⇒
( x – 2 x – 1) – 2 = x 2 – 2 x – 3
2
= x +1 x – 3 < 5⋅
ε
5
x –0 =
x = x <ε
2
For x > 0, ( x ) = x.
=ε
x < ε ⇔ ( x )2 = x < ε 2
δ = ε 2; 0 < x < δ ⇒ x < δ = ε 2 = ε
22. 0 < x < δ ⇒ x 4 – 0 = x 4 < ε
x 4 = x x3 . To bound x3 , agree that
3
δ ≤ 1. x < δ ≤ 1 implies x3 = x ≤ 1 so
δ ≤ ε.
δ = min{1, ε }; 0 < x < δ ⇒ x 4 = x x3 < ε ⋅1
27.
lim x : 0 < x < δ ⇒ x – 0 < ε
x →0 +
For x ≥ 0 , x = x .
δ = ε; 0 < x < δ ⇒ x – 0 = x = x < δ = ε
Thus, lim+ x = 0.
x→0
lim x : 0 < 0 – x < δ ⇒ x – 0 < ε
=ε
23. Choose ε > 0. Then since lim f ( x) = L, there is
x →c
some δ1 > 0 such that
0 < x – c < δ1 ⇒ f ( x ) – L < ε .
x →0 –
For x < 0, x = – x; note also that
since x ≥ 0.
x = x
δ = ε ;0 < − x < δ ⇒ x = x = − x < δ = ε
Since lim f (x) = M, there is some δ 2 > 0 such
Thus, lim– x = 0,
that 0 < x − c < δ 2 ⇒ f ( x) − M < ε .
since lim x = lim x = 0, lim x = 0.
x→c
Let δ = min{δ1 , δ2 } and choose x 0 such that
0 < x0 – c < δ .
Thus, f ( x0 ) – L < ε ⇒ −ε < f ( x0 ) − L < ε
⇒ − f ( x0 ) − ε < − L < − f ( x0 ) + ε
⇒ f ( x0 ) − ε < L < f ( x0 ) + ε .
Similarly,
f ( x0 ) − ε < M < f ( x0 ) + ε .
Thus,
−2ε < L − M < 2ε . As ε ⇒ 0, L − M → 0, so
L = M.
24. Since lim G(x) = 0, then given any ε > 0, we
x→0
x →0 +
x →0
x →0 –
28. Choose ε > 0. Since lim g( x) = 0 there is some
x→ a
δ1 > 0 such that
0 < x – a < δ1 ⇒ g(x ) − 0 <
ε.
B
Let δ = min{1, δ1} , then f ( x) 0 such that whenever
x – c < δ , G ( x) < ε .
Take any ε > 0 and the corresponding δ that
works for G(x), then x – c < δ implies
F ( x) – 0 = F ( x) ≤ G ( x ) < ε since
lim G(x) = 0.
x→c
Thus, lim F( x) = 0.
x→c
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or by any means, without permission in writing from the publisher.
29. Choose ε > 0. Since lim f ( x) = L, there is a
1.3 Concepts Review
x→ a
δ > 0 such that for 0 < x – a < δ , f ( x) – L < ε .
That is, for
a − δ < x < a or a < x < a + δ ,
L − ε < f ( x) < L + ε .
Let f(a) = A,
M = max { L − ε , L + ε , A } , c = a – δ,
d = a + δ. Then for x in (c, d), f ( x) ≤ M , since
either x = a, in which case
f ( x) = f (a ) = A ≤ M or 0 < x – a < δ so
1. 48
2. 4
3. – 8; – 4 + 5c
4. 0
Problem Set 1.3
1. lim (2 x + 1)
= lim 2 x + lim 1
x→1
x →1
α
and δ = min{δ1 , δ 2} where
2
0 < x – a < δ1 ⇒ f ( x) – L < ε and
2.
x→ –1
L – ε < f(x) < L + ε and M – ε < g(x) < M + ε.
Combine the inequalities and use the fact
that f ( x) ≤ g ( x) to get
L – ε < f(x) ≤ g(x) < M + ε which leads to
L – ε < M + ε or L – M < 2ε.
However,
L – M = α > 2ε
which is a contradiction.
Thus L ≤ M .
x+6
x – 4x + x 2 + x + 6
an asymptote at x ≈ 3.49.
c.
2
= 3(–1) – 1 = 2
3. lim [(2 x +1)( x – 3)]
+ 1 has
1
, then 2.75 < x < 3
4
or 3 < x < 3.25 and by graphing
6
x→0
= lim (2 x +1) ⋅ lim (x – 3)
x→ 0
x→ 0
⎞ ⎛
⎞
⎛
= ⎜ lim 2 x + lim 1⎟ ⋅ ⎜ lim x – lim 3⎟
⎝ x→ 0
x→ 0 ⎠ ⎝ x→0
x→ 0 ⎠
⎞ ⎛
⎞
⎛
= ⎜ 2 lim x + lim 1⎟ ⋅ ⎜ lim x – lim 3⎟
⎠
⎝ x →0
⎝
x→ 0
x→0
x→ 0 ⎠
= [2(0) +1](0 – 3) = –3
lim [(2 x 2 + 1)(7 x 2 + 13)]
x→ 2
x→ 2
If δ ≤
y = g ( x) =
2, 1
= lim (2 x 2 + 1) ⋅ lim (7 x 2 + 13)
x 4 – 4x 3 + x 2 + x + 6
3
2
8
x→–1
⎛
⎞
= 3⎜ lim x ⎟ – lim 1
⎝ x→ –1 ⎠
x →–1
4.
x 3 – x 2 – 2x – 4
3
x→–1
x→ –1
32. For every ε > 0 and δ > 0 there is some x with
0 < x – c < δ such that f ( x ) – L > ε .
4
5
= 3 lim x 2 – lim 1
31. (b) and (c) are equivalent to the definition of
limit.
b. No, because
lim (3x 2 – 1)
x→ –1
= lim 3x 2 – lim 1
Thus, for 0 < x – a < δ ,
g(x) =
2,1
x→1
= 2(1) + 1 = 3
0 < x – a < δ 2 ⇒ g ( x) – M < ε .
33. a.
3
x→1
= 2 lim x + lim 1
30. Suppose that L > M. Then L – M = α > 0. Now
take ε <
4
x→1
L − ε < f ( x) < L + ε and f ( x) < M .
x→ 2
4, 5
3
2, 1
6
4, 3
⎛
⎞ ⎛
⎞
= ⎜ 2 lim x 2 + lim 1⎟ ⋅ ⎜ 7 lim x 2 + lim 13 ⎟ 8,1
x→ 2 ⎠ ⎝ x→ 2
x→ 2 ⎠
⎝ x→ 2
2
2
⎡ ⎛
⎤
⎡
⎤
⎞
⎞
⎛
= ⎢2⎜ lim x ⎟ + 1⎥ ⎢7⎜ lim x ⎟ + 13⎥
2
⎢⎣ ⎝ x → 2 ⎠
⎥⎦ ⎢⎣ ⎝ x → 2 ⎠
⎥⎦
= [2( 2 ) 2 + 1][7( 2 ) 2 + 13] = 135
x3 − x 2 − 2 x − 4
x 4 − 4 x3 + x 2 + x + 6
on the interval [2.75, 3.25], we see that
0<
x3 – x 2 – 2 x – 4
<3
x 4 – 4 x3 + x 2 + x + 6
so m must be at least three.
72
Section 1.3
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or by any means, without permission in writing from the publisher.
2x + 1
x→2 5 – 3x
lim (2 x + 1)
= x→2
lim (5 – 3 x)
9.
7
5. lim
4, 5
x→2
=
3, 1
lim 5 – lim 3 x
x→2
13
4, 3
⎡
⎤
= ⎢2 lim t 3 + lim 15⎥
⎣ t→ –2
t→ –2 ⎦
lim 2 x + lim 1
x→2
8
t→ –2
⎡
⎤
= ⎢ lim (2t3 + 15) ⎥
⎣t→–2
⎦
x→2
=
lim (2t 3 +15)13
13
2 lim x + 1
x→2
8
3
⎡
⎤
= ⎢ 2 ⎛⎜ lim t ⎞⎟ + lim 15⎥
⎣⎢ ⎝ t → –2 ⎠ t → –2 ⎦⎥
x→2
2, 1
= [2(–2) 3 + 15]13 = –1
2
5 – 3 lim x
13
x→2
2(2) + 1
=
= –5
5 – 3(2 )
10.
=
3
6.
4x +1
lim
7
x → –3 7 – 2 x 2
lim (4 x + 1)
lim (7 – 2 x )
lim 4 x 3 + lim 1
x → –3
lim 7 – lim 2 x 2
x → –3
=
x → –3
2
x →3
13
5, 3
= 3 lim x – lim 5
2, 1
x →3
lim
x → –3
=
5x2 + 2 x
9
2
lim (5 x + 2 x )
x → –3
= 5 lim x 2 + 2 lim x
x → –3
x → –3
⎛ 4 lim y 3 + 8 lim y ⎞
⎜ y →2
y →2 ⎟
=⎜
⎟
y + lim 4 ⎟
⎜ ylim
→
y
→
2
2
⎝
⎠
8, 1
1/ 3
= 3(3) – 5 = 2
8.
4, 3
13
lim (3 x – 5)
x →3
x →3
7
⎡ lim (4 y 3 + 8 y ) ⎤
⎢ y →2
⎥
=⎢
( y + 4) ⎥⎥
⎢ ylim
→
2
⎣
⎦
9
7. lim 3 x – 5
2
9
⎛
4 y3 + 8 y ⎞
= ⎜ lim
⎟
⎜ y →2 y + 4 ⎟
⎝
⎠
3
=
⎛ 4 y3 + 8 y ⎞
lim ⎜
⎟
y →2 ⎜ y + 4 ⎟
⎝
⎠
1/ 3
x → –3
4⎛⎜ lim x ⎞⎟ + 1
x → –3 ⎠
= ⎝
2
7 – 2⎛⎜ lim x ⎞⎟
⎝ x → –3 ⎠
4(–3)3 + 1 107
=
=
11
7 – 2(–3) 2
2
1/ 3
11.
8
7 – 2 lim x 2
8
w→ –2
= –3(–2)3 + 7(–2) 2 = 2 13
3, 1
x → –3
4 lim x 3 + 1
4, 3
= –3 ⎛⎜ lim w ⎞⎟ + 7 ⎛⎜ lim w ⎞⎟
⎝ w→ –2 ⎠
⎝ w→ –2 ⎠
x → –3
=
lim (–3w3 + 7 w2 )
3
4, 5
2
9
w→ –2
w→ –2
x → –3
x → –3
–3w3 + 7 w2
= –3 lim w3 + 7 lim w2
3
=
lim
w→ –2
4, 3
3
⎡ ⎛
⎤
⎞
⎢ 4 ⎜ lim y ⎟ + 8 lim y ⎥
y →2 ⎥
⎢ y →2 ⎠
=⎢ ⎝
⎥
lim y + 4
⎢
⎥
y →2
⎢
⎥
⎣
⎦
2
1/ 3
⎡ 4(2)3 + 8(2) ⎤
=⎢
⎥
2+4
⎣⎢
⎦⎥
=2
8
2
= 5 ⎛⎜ lim x ⎞⎟ + 2 lim x
x → –3
⎝ x→ –3 ⎠
2
= 5(–3)2 + 2(–3) = 39
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Section 1.3
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or by any means, without permission in writing from the publisher.
12.
lim (2 w 4 – 9 w 3 +19)–1 /2
1
= lim
w→ 5
7
4
2w − 9 w3 + 19
lim 1
w→5
4
=
1, 9
3
2w – 9 w + 19
lim
w→ 5
x→2
19. lim
lim (2w – 9 w3 + 19)
w→ 5
1
lim 2 w4 − lim 9 w3 + lim 19
w→5
w→5
4
2 lim w − 9 lim w3 + 19
w→5
1
4
=
3
1
144
x→2
=
x2 − 4
2
x +4
=
(
lim ( x
2
) 4−4
=
=0
+ 4) 4 + 4
lim x − 4
x→2
2
x→2
= lim ( x − 3) = −1
x→2
( x − 3)( x + 1)
x2 − 2 x − 3
= lim
x →−1
x →−1
x +1
( x + 1)
lim
= lim ( x − 3) = −4
16.
17.
lim
x →−1
x2 + x
x2 + 1
(
lim ( x
2
=
) 0
= =0
+ 1) 2
lim x + x
x →−1
x →−1
2
( x − 1)( x − 2)( x − 3)
x−3
= lim
x →−1 ( x − 1)( x − 2)( x + 7)
x →−1 x + 7
lim
=
−1 − 3
2
=−
−1 + 7
3
2
2
u –u– 6
u– x x+2
= lim
=
5
u → –2 u – 3
u →–2
( u + 2 )( u – x )
u→ –2 ( u + 2)(u – 3)
= lim
x 2 + ux – x – u
( x – 1)( x + u)
= lim
2
x→1 x + 2 x – 3
x →1 ( x – 1)( x + 3)
x + u 1+ u u + 1
= lim
=
=
4
x→1 x + 3 1+ 3
2 x2 – 6 xπ + 4 π2
2( x – π)( x – 2 π)
x→ π
x –π
x→ π ( x – π)( x + π)
2( x – 2π) 2(π – 2 π)
= lim
=
= –1
π+π
x→ π x + π
23. lim
24.
2
lim
2
= lim
(w + 2)(w 2 – w – 6)
w 2 + 4w + 4
( w + 2) 2 ( w – 3)
= lim
= lim ( w – 3)
( w + 2 )2
w→ –2
w→ –2
= –2 – 3 = –5
w→ –2
25. lim
x→a
f 2 ( x) + g 2 ( x)
lim f 2 ( x) + lim g 2 ( x)
x→a
x→a
2
= ⎛⎜ lim f ( x) ⎞⎟ + ⎛⎜ lim g ( x) ⎞⎟
⎝ x →a
⎠ ⎝ x→a
⎠
2
= (3) 2 + (–1)2 = 10
[2 f ( x) – 3 g ( x)]
2 f ( x) – 3g ( x ) xlim
= →a
x → a f ( x) + g ( x)
lim [ f ( x) + g ( x)]
26. lim
x→a
2 lim f ( x) – 3 lim g ( x)
x→a
x→a
lim f ( x) + lim g ( x)
x→a
Section 1.3
= lim
u2 – ux + 2u – 2 x
lim
=
74
( x + 2)( x − 1)
( x + 1)( x − 1)
( x + 3)( x – 17)
x→ –3 x – 4 x – 21
x→ –3 ( x + 3)( x – 7)
x – 17 –3 – 17
= lim
=
=2
–3 – 7
x→ –3 x – 7
=
x →−1
x →1
22. lim
1
12
( x − 3)( x − 2 )
x2 − 5x + 6
14. lim
= lim
x→2
x→2
x−2
( x − 2)
15.
2
2(5)4 − 9(5)3 + 19
lim
13.
21.
2 ⎛⎜ lim w ⎞⎟ − 9 ⎛⎜ lim w ⎞⎟ + 19
⎝ w→ 5 ⎠
⎝ w→5 ⎠
1
= lim
x 2 – 14 x – 51
lim
8
w→5
=
=
1,3
20.
w→5
1
=
2
x −1
x + 2 1+ 2 3
= lim
=
=
x →1 x + 1
1+1 2
4,5
4
=
x2 + x − 2
x →1
1
=
x 2 + 7 x + 10
( x + 2)( x + 5)
= lim
x→2
x
→
2
x+2
x+2
= lim( x + 5) = 7
18. lim
w→ 5
x→a
=
2(3) – 3(–1) 9
=
3 + (–1)
2
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or by any means, without permission in writing from the publisher.
27. lim 3 g ( x) [ f ( x) + 3] = lim 3 g ( x) ⋅ lim [ f ( x) + 3]
x→a
x→a
x→a
= 3 lim g ( x) ⋅ ⎡ lim f ( x) + lim 3⎤ = 3 – 1 ⋅ (3 + 3)
⎢⎣ x → a
x→a
x→a ⎥
⎦
= –6
28. lim [ f ( x) – 3]4 = ⎡⎢ lim ( f ( x) – 3) ⎤⎥
x→a
⎣ x→a
⎦
4
= lim f (t ) + 3 lim g (t )
t →a
⎛
⎞
30. lim [ f (u) + 3g(u)] = ⎜ lim [ f (u) + 3g(u)]⎟
⎝
⎠
u →a
u →a
3
3
3
⎡
⎤
= ⎢ lim f (u ) + 3 lim g(u) ⎥ = [3 + 3( –1)]3 = 0
⎣u→ a
⎦
u →a
3x 2
– 12
3( x – 2 )(x + 2)
31. lim
= lim
x
–
2
x –2
x→2
x→2
= 3 lim (x + 2) = 3(2 + 2) = 12
x→2
(3x 2 + 2 x + 1) – 17
3x 2 + 2 x – 16
= lim
x–2
x–2
x→2
x →2
(3 x + 8)( x – 2)
= lim
= lim (3 x + 8)
x–2
x→2
x →2
= 3 lim x + 8 = 3(2) + 8 = 14
x→2
1
2
= lim
2– x
2x
= lim
–
34.
–
3
4
x–2
3( 4 – x 2 )
= lim
4x2
–3( x + 2 )( x – 2 )
= lim
4x2
x→2
x–2
x–2
–3 ⎛⎜ lim x + 2 ⎞⎟
–3( x + 2)
⎠ = –3(2 + 2)
= lim
= ⎝ x →2
2
2
x→2
4x
4(2)2
4 ⎛⎜ lim x ⎞⎟
⎝ x→2 ⎠
3
=–
4
x→2
x→c
x→c
exist δ 2 and δ 3 such that 0 < x – c < δ 2 ⇒
g ( x) – M <
ε
and 0 < x – c < δ 3 ⇒
L + M +1
ε
L + M +1
. Let
δ = min{δ1 , δ 2 , δ 3 }, then 0 < x – c < δ ⇒
f ( x) g ( x) – LM ≤ g ( x) f ( x) – L + L g ( x ) – M
< ( M + 1)
ε
L + M +1
+L
ε
L + M +1
=ε
Hence,
lim f ( x) g ( x) = LM = ⎛⎜ lim f ( x) ⎞⎟ ⎛⎜ lim g ( x) ⎞⎟
⎝ x →c
⎠ ⎝ x →c
⎠
x→2
36. Say lim g ( x ) = M , M ≠ 0 , and choose
x →c
1
M
.
2
There is some δ1 > 0 such that
ε1 =
1
M or
2
1
1
M < g ( x) < M + M .
2
2
1
1
1
1
M − M ≥ M and M + M ≥ M
2
2
2
2
1
2
1
so g ( x) > M and
<
g ( x)
M
2
M−
x →2
lim
0 < x – c < δ1 , g ( x) < M + 1. Choose ε > 0.
Since lim f (x) = L and lim g(x) = M, there
0 < x − c < δ1 ⇒ g ( x) − M < ε1 =
x–2
2x
x – 2 x→2 x – 2 x →2 x – 2
1
–1
–1
1
= lim –
=
=
=–
2 lim x 2(2)
4
x→2 2 x
x→2
3
x2
x →c
x →c
32. lim
–
f ( x) g ( x) – LM ≤ g ( x) f ( x) – L + L g ( x ) – M
as shown in the text. Choose ε 1 = 1. Since
lim g ( x) = M , there is some δ1 > 0 such that if
f ( x) – L <
= 3 + 3 –1 = 6
33. lim
x→c
M – 1 ≤ M + 1 and M + 1 ≤ M + 1 so for
29. lim ⎡⎣ f (t ) + 3g (t ) ⎤⎦ = lim f (t ) + 3 lim g (t )
t →a
t →a
t →a
1
x
x→c
0 < x – c < δ1 , g ( x) – M < ε1 = 1 or
M – 1 < g(x) < M + 1
4
= ⎡⎢ lim f ( x) – lim 3⎤⎥ = (3 – 3) 4 = 0
x →a ⎦
⎣ x→a
t →a
35. Suppose lim f (x) = L and lim g(x) = M.
Choose ε > 0.
Since lim g(x) = M there is δ 2 > 0 such that
x→c
0 < x − c < δ 2 ⇒ g ( x) − M <
Let δ = min{δ1 , δ 2}, then
0< x–c <δ ⇒
=
1 2
M .
2
1
1
M – g ( x)
–
=
g ( x) M
g ( x) M
1
2
2 1 2
⋅ M ε
g ( x) − M <
g ( x) − M =
2
M g ( x)
M
M2 2
=ε
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Section 1.3
75
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or by any means, without permission in writing from the publisher.
Thus, lim
x→c
1
1
1
=
=
.
g(x) M lim g (x)
43.
x→c
Using statement 6 and the above result,
f ( x)
1
lim
= lim f ( x) ⋅ lim
x →c g ( x )
x →c
x →c g ( x )
lim f ( x )
1
.
= lim f ( x) ⋅
= x →c
lim g ( x ) lim g ( x)
x →c
x →c
x→c
x →3+
x→ c
⇔ lim f (x) – lim L = 0
⇔ lim [ f (x) – L] = 0
45.
x→c
2
⎡
⎤
38. lim f (x) = 0 ⇔ ⎢ lim f (x) ⎥ = 0
⎣
⎦
x→c
x→c
x2 – 9
x →3+
46.
2
⇔ lim f ( x) = 0
x2 – 9
x+3
32 – 9
=0
3+3
=
lim
x →1–
x →c
x→c
( x – 3) x 2 – 9
( x – 3) x 2 – 9
= lim
x →3+ ( x – 3)( x + 3)
x →3+
44.
x→ c
x2 – 9
= lim
= lim
x →c
37. lim f (x) = L ⇔ lim f ( x) = lim L
x–3
lim
lim
x → 2+
1+ x
1+1
2
=
=
4 + 4 x 4 + 4(1)
8
( x 2 + 1) x
(3 x − 1)
=
2
(22 + 1) 2
(3 ⋅ 2 − 1)
2
lim ( x − x ) = lim x − lim
x →3−
x →3−
x →3−
=
5⋅ 2
5
2
=
2
5
x = 3− 2 =1
x→c
lim f 2 ( x) = 0
⇔
47.
x →c
⇔ lim
x →c
f 2 ( x) = 0
48.
⇔ lim f ( x) = 0
lim
x
= –1
x
lim
x 2 + 2 x = 32 + 2 ⋅ 3 = 15
x →0 –
x →3+
x→c
2
39. lim x = ⎛⎜ lim x ⎞⎟ =
x →c
⎝ x →c ⎠
lim x
x →c
2
=
lim x 2
x →c
2
x +1
x–5
, g ( x) =
and c = 2, then
x–2
x–2
lim [ f (x) + g (x)] exists, but neither
x→c
lim f (x) nor lim g(x) exists.
x→c
2
, g ( x) = x, and c = 0, then
x
lim [ f (x) ⋅ g( x)] exists, but lim f (x) does
b. If f ( x) =
x→c
x→c
not exist.
41.
42.
76
lim
x → –3+
3+ x
3–3
=
=0
x
–3
π3 + x3
=
x
x → – π+
lim
Section 1.3
1
f ( x)
lim g ( x) = 0 ⇔ lim
1
=0
f ( x)
⇔
If f ( x) =
x→c
f ( x) g ( x) = 1; g ( x) =
x →a
= ⎛⎜ lim x ⎞⎟ = c 2 = c
⎝ x →c ⎠
40. a.
49.
π3 + (– π)3
=0
–π
x →a
1
=0
lim f ( x)
x→a
No value satisfies this equation, so lim f ( x)
x→ a
must not exist.
1⎞
⎛ x
50. R has the vertices ⎜ ± , ± ⎟
⎝ 2
2⎠
Each side of Q has length
x 2 + 1 so the
perimeter of Q is 4 x 2 + 1. R has two sides of
length 1 and two sides of length
x 2 so the
perimeter of R is 2 + 2 x 2 .
lim
x →0 +
=
perimeter of R
2 x2 + 2
= lim
perimeter of Q x →0+ 4 x 2 + 1
2 02 + 2
2
4 0 +1
=
2 1
=
4 2
Instructor’s Resource Manual
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or by any means, without permission in writing from the publisher.
NO = (0 – 0)2 + (1 – 0)2 = 1
51. a.
4. lim
OP = ( x – 0)2 + ( y – 0) 2 = x 2 + y 2
= x2 + x
NP = ( x – 0)2 + ( y – 1)2 = x 2 + y 2 – 2 y + 1
= x + x − 2 x +1
MO = (1 – 0) 2 + (0 – 0) 2 = 1
y2 + x2 – 2 x + 1
= x2 − x + 1
perimeter of ΔNOP
lim
x →0+ perimeter of ΔMOP
= lim
=
1+ 1
1+ 1
1 + x2 + x + x2 – x + 1
=1
x
1
(1)( x) =
2
2
1
x
Area of ΔMOP = (1)( y ) =
2
2
b. Area of ΔNOP =
x
area of ΔNOP
x
= lim 2 = lim
x
+
+
+
MOP
area
of
Δ
x
x →0
x →0
x →0
lim
2
= lim
x →0+
x =0
1.4 Concepts Review
1. 0
0
=0
1
5. lim
sin x 1
sin x 1
1
= lim
= ⋅1 =
2x
2 x →0 x
2
2
sin 3θ
3 sin 3θ 3
sin 3θ
= lim ⋅
= lim
3θ
2 θ →0 3θ
θ →0 2θ
θ →0 2
3
3
= ⋅1 =
2
2
6. lim
sin 3θ
sin 3θ
cos θ sin 3θ
= lim sin θ = lim
θ → 0 tan θ
θ →0
θ →0
sin θ
cos θ
⎡
sin 3θ 1 ⎤
= lim ⎢cos θ ⋅ 3 ⋅
⋅ sin θ ⎥
θ →0⎢
3θ
θ ⎥
⎣
⎦
⎡
sin 3θ 1
= 3 lim ⎢cos θ ⋅
⋅ sin θ
θ →0 ⎢
3θ
θ
⎣
⎤
⎥ = 3 ⋅1 ⋅1 ⋅1 = 3
⎥⎦
sin 5θ
sin 5θ
tan 5θ
= lim cos 5θ = lim
θ → 0 sin 2θ
θ → 0 sin 2θ
θ → 0 cos 5θ sin 2θ
sin 5θ 1 2θ ⎤
⎡ 1
= lim ⎢
⋅5⋅
⋅ ⋅
5θ 2 sin 2θ ⎥⎦
θ →0 ⎣ cos 5θ
5
sin 5θ 2θ ⎤
⎡ 1
= lim ⎢
⋅
⋅
2 θ →0 ⎣ cos 5θ 5θ sin 2θ ⎥⎦
5
5
= ⋅1⋅1⋅1 =
2
2
8. lim
cot πθ sin θ
= lim
θ →0
θ →0
2 sec θ
9. lim
cos πθ
sin πθ
sin θ
2
cos θ
cos πθ sin θ cos θ
2sin πθ
⎡ cos πθ cos θ sin θ 1 πθ ⎤
= lim ⎢
⋅
⋅ ⋅
θ π sin πθ ⎥⎦
2
θ →0 ⎣
1
sin θ πθ ⎤
⎡
=
⋅
lim ⎢cos πθ cos θ ⋅
θ sin πθ ⎥⎦
2 π θ →0 ⎣
1
1
⋅1⋅1⋅1⋅1 =
=
2π
2π
= lim
2. 1
θ →0
3. the denominator is 0 when t = 0 .
4. 1
Problem Set 1.4
cos x 1
= =1
x →0 x + 1 1
sin 2 3t
9t sin 3t sin 3t
= lim ⋅
⋅
= 0 ⋅1 ⋅1 = 0
t →0
t
→
0
2t
2 3t
3t
1. lim
2.
3 x tan x
3x (sin x / cos x)
3x
= lim
= lim
x →0
x → 0 cos x
sin x
sin x
7. lim
1 + x2 + x + x2 + x – 2 x + 1
x → 0+
=
x →0
2
MP = ( x – 1)2 + ( y – 0) 2 =
x →0
lim θ cosθ =
θ →π / 2
π
2
10. lim
⋅0 = 0
cos 2 t
cos 2 0
1
=
=
=1
t →0 1 + sin t 1 + sin 0 1 + 0
3. lim
tan 2 3t
sin 2 3t
= lim
t →0
t →0 (2t )(cos 2 3t )
2t
11. lim
= lim
t →0
3(sin 3t ) sin 3t
⋅
= 0 ⋅1 = 0
3t
2 cos 2 3t
tan 2t
0
=
=0
t → 0 sin 2t − 1
−1
12. lim
Instructor’s Resource Manual
Section 1.4
77
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or by any means, without permission in writing from the publisher.
sin(3t ) + 4t
4t ⎞
⎛ sin 3t
= lim ⎜
+
⎟
t →0 ⎝ t sec t
t sec t
t sec t ⎠
sin 3t
4t
= lim
+ lim
t →0 t sec t t →0 t sec t
sin 3t
= lim 3cos t ⋅
+ lim 4 cos t
t →0
t →0
3t
= 3 ⋅1 + 4 = 7
13. lim
t →0
14.
sin 2 θ
sin θ sin θ
lim
= lim
2
θ
θ →0 θ
θ →0 θ
sin θ
sin θ
= lim
× lim
= 1× 1 = 1
θ
θ →0
θ →0 θ
19. lim 1 +
x →0
sin x
=2
x
20. The result that lim cos t = 1 was established in
t →0
the proof of the theorem. Then
lim cos t = lim cos(c + h)
t →c
h →0
= lim (cos c cos h − sin c sin h)
h →0
= lim cos c lim cos h − sin c lim sin h
15. lim x sin (1/ x ) = 0
h →0
x →0
h →0
h→0
= cos c
lim sin t
sin t t →c
sin c
=
=
= tan c
t → c cos t
lim cos t cos c
21. lim tan t = lim
t →c
t →c
lim cot t = lim
t →c
(
)
16. lim x sin 1/ x 2 = 0
x →0
t →c
lim cos t
cos t t →c
cos c
=
=
= cot c
sin t lim sin t sin c
t →c
1
1
=
= sec c
cos t cos c
1
1
lim csc t = lim
=
= csc c
t →c
t →c sin t
sin c
22. lim sec t = lim
t →c
t →c
23. BP = sin t , OB = cos t
area( ΔOBP) ≤ area (sector OAP)
≤ area (ΔOBP) + area( ABPQ)
(
)
17. lim 1 − cos 2 x / x = 0
x →0
1
1
1
OB ⋅ BP ≤ t (1) 2 ≤ OB ⋅ BP + (1 – OB ) BP
2
2
2
1
1
1
sin t cos t ≤ t ≤ sin t cos t + (1 – cos t ) sin t
2
2
2
t
≤ 2 – cos t
sin t
1
sin t
1
π
π
≤
≤
for − < t < .
2 – cos t
t
cos t
2
2
1
sin t
1
lim
≤ lim
≤ lim
t →0 2 – cos t t →0 t
t →0 cos t
sin t
≤1
1 ≤ lim
t →0 t
sin t
= 1.
Thus, lim
t →0 t
cos t ≤
18. lim cos 2 x = 1
x →0
78
Section 1.4
Instructor’s Resource Manual
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or by any means, without permission in writing from the publisher.
24. a.
Written response
6.
b.
c.
1
1
AB ⋅ BP = (1 − cos t ) sin t
2
2
sin t (1 − cos t )
=
2
1
1
t sin t cos t
E = t (1)2 – OB ⋅ BP = –
2
2
2
2
D sin t (1 – cos t )
=
E
t – sin t cos t
D=
⎛D⎞
lim ⎜ ⎟ = 0.75
+
t →0 ⎝ E ⎠
3
2 x – 100 x
x →∞
8.
10.
1. x increases without bound; f(x) gets close to L as
x increases without bound
4. x = 6; vertical
sin 2 θ
lim
θ →∞ θ 2 – 5
1.
2.
3.
4.
5.
lim
x →∞
x
1
= lim
=1
x – 5 x →∞ 1 – 5
lim
x
x2
x →∞ 5 – x3
lim
t2
t →–∞ 7 − t
lim
t →–∞ t
2
= lim
x →∞ 5
x3
x →∞
= lim
θ →∞ 1 – 5
θ2
x
2
= lim
= 0 so lim
sin 2 θ
θ →∞ θ 2 – 5
=0
3 x3 / 2 + 3 x
2 x3 / 2
x →∞
3
=
2
πx3 + 3x
πx3 + 3 x
lim 3
= 3 lim
x →∞
x →∞ 2 x3 + 7 x
2 x3 + 7 x
3
x2
+ 72
x
π+
2
=3
π
2
=0
2
13.
1
t →–∞ 7
t2
θ2
= lim
2 x3
3+ 3
x →∞
–1
=π
; 0 ≤ sin 2 θ ≤ 1 for all θ and
3 x3 + 3 x
lim
= 3 lim
1
x
π
θ →–∞ 1 – 5
θ
1
1
x →∞
12.
x →∞
3– 1 3
3 x3 – x 2
x =
= lim
lim
3
2
x → ∞ πx – 5 x
x→∞π– 5 π
x
= lim
Problem Set 1.5
1
1
=
2 – 100
2
x
= lim
= lim
θ → – ∞ θ 5 – 5θ 4
θ →∞ θ 2 – 5
11.
2
πθ 5
lim
lim
2. f(x) increases without bound as x approaches c
from the right; f(x) decreases without bound as x
approaches c from the left
3. y = 6; horizontal
x3
7. lim
9.
1.5 Concepts Review
1
x2
lim
= lim
=1
2
8
x → ∞ x – 8 x + 15 x → ∞ 1 – + 15
x x2
−1
= −1
t
1
= lim
=1
– 5 t →–∞ 1 – 5
t
= 3 lim
14.
x2
x2
= lim
x →∞ ( x – 5)(3 – x) x →∞ − x 2 + 8 x − 15
Instructor’s Resource Manual
1
x2
+8
x →∞ 1 + 4
x2
lim
1
= lim
= –1
8
x → ∞ −1 + − 15
x x2
2
1 + 8x
1 + 8x
lim 3
= 3 lim
2
x →∞ x + 4
x →∞ x 2 + 4
lim
x →∞
=
= 38 =2
x2 + x + 3
=
( x –1)( x + 1)
lim
x →∞
1 + 1x + 32
x
1 – 12
lim
x2 + x + 3
x →∞
x 2 –1
= 1 =1
x
15.
lim
n →∞
n
1
1
= lim
=
2n + 1 n→∞ 2 + 1 2
n
Section 1.5
79
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or by any means, without permission in writing from the publisher.
16. lim
n →∞
n2
n2 + 1
1
= lim
n →∞
1+
=
1
n2
1
=1
1+ 0
23.
lim n
n2
n
∞
n →∞
= lim
=
=
=∞
1
n →∞ n + 1 n →∞
1 ⎞ 1+ 0
⎛
1+
lim 1 +
n n →∞ ⎜⎝ n ⎟⎠
17. lim
18. lim
n →∞
n
n2 + 1
n →∞
1+
lim
=
1
n2
x →∞
x →∞
0
=0
1+ 0
x →∞
x2 + 3
= lim
2 + 1x
x →∞
x 2 +3
x
=
20.
21.
2
1
x →∞
1+
2
3
x2
lim
x →∞
x →∞
= lim
x →∞
2 x 2 + 3 – (2 x 2 – 5)
2 x2 + 3 + 2 x2 – 5
8
2
2 x + 3 + 2 x2 − 5
8
x
2+
3
x2
x →∞
8
x
2 x 2 +3 + 2 x 2 –5
x2
+ 2–
22.
lim ⎛⎜ x 2 + 2 x − x ⎞⎟
⎠
x →∞ ⎝
⎛ x 2 + 2 x – x ⎞⎛ x 2 + 2 x + x ⎞
⎜
⎟⎜
⎟
⎠⎝
⎠
= lim ⎝
2
x →∞
x + 2x + x
= lim
x →∞
= lim
x →∞
2
x + 2x – x
2
= lim
2x
x 2 + 2 x + x x→∞ x 2 + 2 x + x
2
2
= =1
1+ 2 +1 2
n →∞
+
bn
xn
1
1
1+ 2
n
=
=
a0
b0
1
1+ 0
=1
t → –3+
29. As t → 3– , t 2 → 9 while 9 – t 2 → 0+.
t →3–
t2
9 – t2
=∞
+
30. As x → 3 5 , x 2 → 52 / 3 while 5 – x3 → 0 – .
x→3 5
5
x2
= lim
an
xn
t2 – 9
(t + 3)(t – 3)
= lim
+
+
t +3
t → –3 t + 3
t → –3
= lim (t – 3) = –6
lim
=0
+ …+
bn –1
x n –1
+
lim
lim
= lim
an –1
x n –1
27. As x → 4+ , x → 4 while x – 4 → 0 .
x
lim
=∞
+
x
–
4
x →4
28.
⎛ 2 x 2 + 3 – 2 x 2 – 5 ⎞⎛ 2 x 2 + 3 + 2 x 2 – 5 ⎞
⎜
⎟⎜
⎟
⎠⎝
⎠
= lim ⎝
2
2
x →∞
2x + 3 + 2x – 5
= lim
+ …+
+
+ 12
lim ⎛⎜ 2 x 2 + 3 – 2 x 2 – 5 ⎞⎟
⎠
x →∞ ⎝
x →∞
= –∞
n2
∞
n
n3/ 2
= lim
= =∞
3
n →∞
1
2
1
n + 2n + 1
1+ 2 + 3
n
n
26. lim
2x +1
x
= lim
= lim
=0
x+4
x →∞ 1 + 4
x →∞ 1 + 4
x
x
= lim
b0 +
b1
x
n +1
n →∞
2
x
y→–∞ 1 –
2
=2
2 x +1
x2
a1
x
2
2 + 1x
= lim
a0 +
n
25. lim
19. For x > 0, x = x 2 .
2x + 1
= lim
b0 x n + b1 x n –1 +…+ bn –1 x + bn
n →∞
lim
y2 – 2 y + 2
a0 x n + a1 x n –1 +…+ an –1 x + an
= lim
1
n
= lim
24.
lim
y→– ∞
1
y2
2+ 2
y y2
9y +
9 y3 + 1
x2
+
5 – x3
= –∞
31. As x → 5– , x 2 → 25, x – 5 → 0 – , and
3 – x → –2.
x2
lim
=∞
x →5 – ( x – 5)(3 – x)
32. As θ → π+ , θ 2 → π2 while sin θ → 0− .
θ2
= −∞
θ →π+ sin θ
lim
x
80
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or by any means, without permission in writing from the publisher.
−
33. As x → 3− , x3 → 27, while x − 3 → 0 .
43.
3
lim
x →3−
x
= −∞
x−3
π+
π2
, πθ →
while cos θ → 0 – .
2
2
πθ
= –∞
lim
π + cos θ
34. As θ →
θ→
35.
3
3
= 0, lim
= 0;
x +1
x→ – ∞ x + 1
Horizontal asymptote y = 0.
3
3
lim
= ∞, lim
= – ∞;
x → –1+ x + 1
x → –1– x + 1
Vertical asymptote x = –1
lim
x →∞
2
lim
x →3–
x2 – x – 6
( x + 2)( x – 3)
= lim
x–3
x–3
x →3–
= lim ( x + 2) = 5
x →3 –
36.
x2 + 2 x – 8
lim
2
x → 2+
= lim
x → 2+
x –4
x+4 6 3
= lim
= =
x → 2+ x + 2 4 2
( x + 4)( x – 2)
( x + 2)( x – 2)
37. For 0 ≤ x < 1 , x = 0 , so for 0 < x < 1,
thus lim
x →0 +
x
x
44.
x
x
lim
3
x →∞ ( x + 1)
2
3
= 0, lim
x → – ∞ ( x + 1) 2
= 0;
Horizontal asymptote y = 0.
3
3
lim
= ∞, lim
= ∞;
2
2
–
+
x → –1 ( x + 1)
x → –1 ( x + 1)
Vertical asymptote x = –1
=0
=0
38. For −1 ≤ x < 0 , x = −1 , so for –1 < x < 0,
x
1
thus lim
= ∞.
−
x
x
x
x →0
1
(Since x < 0, – > 0. )
x
x
=−
39. For x < 0, x = – x, thus
lim
x →0 –
x
x
= lim
x →0 –
–x
= –1
x
40. For x > 0, x = x, thus lim
x →0 +
45.
x
x
= lim
x →0 +
x
=1
x
41. As x → 0 – , 1 + cos x → 2 while sin x → 0 – .
1 + cos x
lim
= –∞
–
sin x
x →0
lim
x →∞
2x
2
= lim
= 2,
x – 3 x→∞ 1 – 3
x
2x
2
lim
= lim
= 2,
x →−∞ x – 3 x →−∞ 1 – 3
x
Horizontal asymptote y = 2
2x
2x
lim
= ∞, lim
= – ∞;
x →3+ x – 3
x →3– x – 3
Vertical asymptote x = 3
42. –1 ≤ sin x ≤ 1 for all x, and
1
sin x
lim = 0, so lim
= 0.
x →∞ x
x →∞ x
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or by any means, without permission in writing from the publisher.
46.
lim
3
2
x →∞ 9 –
3
= 0, lim
2
49. f ( x ) = 2 x + 3 –
= 0;
x→ – ∞ 9 – x
x
Horizontal asymptote y = 0
3
3
lim
= – ∞, lim
= ∞,
2
2
–
+
x →3 9 – x
x →3 9 – x
3
3
lim
= ∞, lim
= – ∞;
2
2
x → –3+ 9 – x
x → –3– 9 – x
Vertical asymptotes x = –3, x = 3
50.
14
x →∞ 2 x
2
f ( x) = 3x + 4 –
4x + 3
x2 + 1
, thus
We say that lim f ( x) = – ∞ if to each
x →c +
negative number M there corresponds a δ > 0
such that 0 < x – c < δ ⇒ f(x) < M.
14
= 0, lim
, thus
⎡ 4x + 3⎤
lim [ f ( x) – (3 x + 4)] = lim ⎢ –
⎥
x →∞
x →∞ ⎣ x 2 + 1 ⎦
⎡ 4+ 3 ⎤
x x2 ⎥
= lim ⎢ –
=0.
⎢
x →∞
1 + 12 ⎥
⎢⎣
⎥
x ⎦
The oblique asymptote is y = 3x + 4.
= 0;
x→ – ∞ 2 x2 + 7
+7
Horizontal asymptote y = 0
2
Since 2x + 7 > 0 for all x, g(x) has no vertical
asymptotes.
lim
x –1
1 ⎤
⎡
lim [ f ( x) – (2 x + 3)] = lim ⎢ –
⎥=0
3
x →∞
x →∞ ⎣ x –1 ⎦
The oblique asymptote is y = 2x + 3.
51. a.
47.
1
3
b. We say that lim f ( x) = ∞ if to each
x →c –
positive number M there corresponds a δ > 0
such that 0 < c – x < δ ⇒ f(x) > M.
We say that lim f ( x) = ∞ if to each
52. a.
x →∞
positive number M there corresponds an
N > 0 such that N < x ⇒ f(x) > M.
b. We say that lim f ( x ) = ∞ if to each
x → –∞
positive number M there corresponds an
N < 0 such that x < N ⇒ f(x) > M.
53. Let ε > 0 be given. Since lim f ( x ) = A, there is
x →∞
48.
lim
x →∞
lim
x→ – ∞
2x
2
x +5
= lim
2x
x2 + 5
x →∞
= lim
2
1+
x→ – ∞
5
x2
=
2
– 1+
5
x2
2
1
=
a corresponding number M1 such that
= 2,
2
– 1
ε
x > M1 ⇒ f ( x) – A < . Similarly, there is a
2
= –2
Since x 2 + 5 > 0 for all x, g(x) has no vertical
asymptotes.
ε
number M2 such that x > M 2 ⇒ g ( x) – B < .
2
Let M = max{M1 , M 2 } , then
x > M ⇒ f ( x) + g ( x) – ( A + B)
= f ( x) – A + g ( x) – B ≤ f ( x) – A + g ( x) – B
ε
ε
=ε
2 2
Thus, lim [ f ( x) + g ( x)] = A + B
<
+
x →∞
54. Written response
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or by any means, without permission in writing from the publisher.
55. a.
lim sin x does not exist as sin x oscillates
x →∞
56.
between –1 and 1 as x increases.
1
, then as x → ∞, u → 0+.
x
1
lim sin = lim sin u = 0
x u →0 +
x →∞
b. Let u =
c.
1
Let u = , then as x → ∞, u → 0+.
x
1
1
sin u
lim x sin = lim sin u = lim
=1
+
x u → 0+ u
u
x →∞
u →0
d. Let u =
lim x
3/ 2
x →∞
e.
h.
59.
60.
3/ 2
1 − v2 / c2
v →c
3x 2 + x +1 3
=
2
2
x →∞ 2x –1
2 x 2 – 3x
lim
=
2
5x + 1
x→ – ∞
2
5
3
lim ⎛⎜ 2 x 2 + 3x – 2 x 2 – 5 ⎞⎟ = –
⎠
2 2
x→ – ∞ ⎝
2x +1
lim
x →∞
3x 2 + 1
sin u
2
=
3
10
⎡⎛ 1 ⎞⎛ sin u ⎞⎤
⎟⎟⎜
= lim+ ⎢⎜⎜
⎟⎥ = ∞
u →0 ⎣
⎢⎝ u ⎠⎝ u ⎠⎦⎥
As x → ∞, sin x oscillates between –1 and 1,
1
while x –1/ 2 =
→ 0.
x
62.
⎛ 1⎞
lim ⎜1 + ⎟ = e ≈ 2.718
x⎠
x →∞ ⎝
–1/ 2
sin x = 0
1
, then
x
⎛π 1⎞
⎛π
⎞
lim sin ⎜ + ⎟ = lim+ sin ⎜ + u ⎟
x→∞
x
6
6
u
→
0
⎝
⎠
⎝
⎠
π 1
= sin =
6 2
1
1⎞
⎛
→ ∞, so lim sin ⎜ x + ⎟
x
x⎠
x →∞
⎝
does not exist. (See part a.)
1⎞
1
1
⎛
sin ⎜ x + ⎟ = sin x cos + cos x sin
x⎠
x
x
⎝
⎡ ⎛
1⎞
⎤
lim ⎢sin ⎜ x + ⎟ – sin x ⎥
x
x →∞ ⎣
⎝
⎠
⎦
⎡
1 ⎞
1⎤
⎛
= lim ⎢sin x ⎜ cos –1⎟ + cos x sin ⎥
x ⎠
x⎦
x →∞ ⎣
⎝
1
1
As x → ∞, cos → 1 so cos –1 → 0.
x
x
1
From part b., lim sin = 0.
x
x →∞
As x → ∞ both sin x and cos x oscillate
between –1 and 1.
⎡ ⎛
1⎞
⎤
lim ⎢sin ⎜ x + ⎟ – sin x ⎥ = 0.
x⎠
x →∞ ⎣
⎝
⎦
Instructor’s Resource Manual
=1
x
⎛ 1⎞
63. lim ⎜ 1 + ⎟
x⎠
x →∞ ⎝
Let u =
As x → ∞, x +
=∞
lim
⎛ 1⎞
lim ⎜ 1 + ⎟
x⎠
x →∞ ⎝
lim x
g.
1
⎛1⎞
sin = lim+ ⎜ ⎟
x u →0 ⎝ u ⎠
58.
v →c
61.
x →∞
f.
1
, then
x
57.
m0
lim− m(v) = lim−
64.
65.
66.
67.
68.
⎛ 1⎞
lim ⎜1 + ⎟
x⎠
x →∞ ⎝
70.
71.
=∞
sin x
=1
sin x – 3
lim
x →3–
= –1
x–3
sin x – 3
lim
x →3–
tan( x – 3)
= –1
lim
cos( x – 3)
= –∞
x–3
lim
cos x
= –1
x – π2
x →3–
x→ π
2
69.
x2
+
lim (1 + x )
x →0 +
1
x
= e ≈ 2.718
lim (1 + x )1/ x = ∞
x → 0+
lim (1 + x ) x = 1
x →0+
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or by any means, without permission in writing from the publisher.
13.
1.6 Concepts Review
t →3+
lim f (t ) = lim (t – 3) = 0
t →3 –
1. lim f ( x)
x →c
t →3–
lim f (t ) = f (3); continuous
t →3
2. every integer
3.
lim f (t ) = lim (3 – t ) = 0
t →3+
lim f ( x) = f (a); lim f ( x) = f (b)
x→a+
14.
x →b –
lim f (t ) = lim (3 – t )2 = 0
t →3+
t →3+
lim f (t ) = lim (t 2 – 9) = 0
t →3–
4. a; b; f(c) = W
t →3 –
lim f (t ) = f (3); continuous
t →3
15. lim f ( x) = −2 = f (3); continuous
Problem Set 1.6
t →3
1. lim[( x – 3)( x – 4)] = 0 = f (3); continuous
x →3
2. lim ( x 2 – 9) = 0 = g (3); continuous
x →3
3
3. lim
x →3 x – 3
and h(3) do not exist, so h(x) is not
continuous at 3.
16. g is discontinuous at x = –3, 4, 6, 8; g is left
continuous at x = 4, 8; g is right continuous at
x = –3, 6
17. h is continuous on the intervals
(−∞, −5), [ −5, 4] , (4, 6), [ 6,8] , (8, ∞)
x 2 – 49
( x – 7)( x + 7)
= lim
= lim ( x + 7)
x–7
x →7 x – 7
x →7
x →7
= 7 + 7 = 14
Define f(7) = 14.
18. lim
4. lim t – 4 and g(3) do not exist, so g(t) is not
t →3
continuous at 3.
t –3
and h(3) do not exist, so h(t) is not
t –3
continuous at 3.
5. lim
2 x 2 –18
2( x + 3)( x – 3)
= lim
3– x
x →3 3 – x
x →3
= lim[–2( x + 3)] = –2(3 + 3) = –12
19. lim
t →3
x →3
Define f(3) = –12.
6. h(3) does not exist, so h(t) is not continuous at 3.
7. lim t = 3 = f (3); continuous
t →3
20. lim
t →3
21. lim
t →1
t 3 – 27
(t – 3)(t 2 + 3t + 9)
= lim
t –3
t →3 t – 3
t →3
= lim(t 2 + 3t + 9) = (3)2 + 3(3) + 9 = 27 = r (3)
22.
12. From Problem 11, lim r (t ) = 27, so r(t) is not
t →3
continuous at 3 because lim r (t ) ≠ r (3).
t →3
t –1
–1)( t + 1)
1
Define H(1) = .
2
t →1 (t
11. lim
continuous
t –1
( t –1)( t + 1)
= lim
t –1 t →1 (t –1)( t + 1)
= lim
10. f(3) does not exist, so f(x) is not continuous at 3.
t →3
=1
θ
Define g(0) = 1
θ →0
8. lim t – 2 = 1 = g (3); continuous
9. h(3) does not exist, so h(t) is not continuous at 3.
sin(θ )
= lim
t →1
1
t +1
=
1
2
x4 + 2 x2 – 3
( x 2 –1)( x 2 + 3)
= lim
x +1
x +1
x → –1
x → –1
lim
( x + 1)( x – 1)( x 2 + 3)
x +1
x → –1
= lim
= lim [( x – 1)( x 2 + 3)]
x → –1
= (–1 – 1)[(–1)2 + 3] = –8
Define φ(–1) = –8.
84
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or by any means, without permission in writing from the publisher.
23.
⎛ x2 – 1 ⎞
⎛ ( x – 1)( x + 1) ⎞
lim sin ⎜
⎟ = lim sin ⎜
⎟
⎜
⎟
x +1
x → –1
⎠
⎝ x + 1 ⎠ x→ –1 ⎝
= lim sin( x –1) = sin(–1 – 1) = sin(−2) = – sin 2
37.
x → –1
Define F(–1) = –sin 2.
24. Discontinuous at x = π ,30
25.
33 – x 2
(π – x)( x – 3)
Discontinuous at x = 3, π
f ( x) =
38.
26. Continuous at all points
27. Discontinuous at all θ = nπ + π where n is any
2
integer.
28. Discontinuous at all u ≤ −5
39.
29. Discontinuous at u = –1
30. Continuous at all points
31. G ( x) =
1
(2 – x)(2 + x)
Discontinuous on (−∞, −2] ∪ [2, ∞)
32. Continuous at all points since
lim f ( x) = 0 = f (0) and lim f ( x) = 1 = f (1).
x →0
40.
x →1
33. lim g ( x ) = 0 = g (0)
x →0
lim g ( x) = 1, lim g ( x) = –1
x →1+
x →1–
lim g(x ) does not exist, so g(x) is discontinuous
x→1
at x = 1.
34. Discontinuous at every integer
35. Discontinuous at t = n +
1
where n is any integer
2
Discontinuous at all points except x = 0, because
lim f ( x ) ≠ f (c) for c ≠ 0 . lim f ( x ) exists only
x →c
x →c
at c = 0 and lim f ( x) = 0 = f (0) .
x →0
36.
41. Continuous.
42. Discontinuous: removable, define f (10) = 20
43. Discontinuous: removable, define f (0) = 1
44. Discontinuous: nonremovable.
45. Discontinuous, removable, redefine g (0) = 1
46. Discontinuous: removable, define F (0) = 0
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or by any means, without permission in writing from the publisher.
47. Discontinuous: nonremovable.
48. Discontinuous: removable, define f (4) = 4
49. The function is continuous on the intervals
( 0,1] , (1, 2], (2,3], …
52. Let f ( x) = x3 + 3 x − 2. f is continuous on [0, 1].
f(0) = –2 < 0 and f(1) = 2 > 0. Thus, there is at
least one number c between 0 and 1 such that
x 3 + 3x − 2 = 0.
53. Because the function is continuous on [ 0,2π ] and
(cos 0)03 + 6sin 5 0 – 3 = –3 < 0,
Cost $
0.60
(cos 2π)(2π)3 + 6sin 5 (2π) – 3 = 8π3 – 3 > 0, there
is at least one number c between 0 and 2π such
0.48
that (cos t )t 3 + 6sin 5 t – 3 = 0.
0.72
0.36
54. Let f ( x ) = x − 7 x + 14 x − 8 . f(x) is
continuous at all values of x.
f(0) = –8, f(5) = 12
Because 0 is between –8 and 12, there is at least
one number c between 0 and 5 such that
3
0.24
0.12
1
3
5
2
4
6
Length of call in minutes
50. The function is continuous on the intervals
[0, 200], (200,300], (300, 400], …
2
f ( x ) = x 3 − 7 x 2 + 14 x − 8 = 0 .
This equation has three solutions (x = 1,2,4)
Cost $
80
60
40
55. Let f ( x ) = x − cos x. . f(x) is continuous at all
20
100 200 300 400 500
Miles Driven
51. The function is continuous on the intervals
(0, 0.25], (0.25, 0.375], (0.375, 0.5], …
values of x ≥ 0.
f(0) = –1, f(π/2) = π / 2
Because 0 is between –1 and π / 2 , there is at
least one number c between 0 and π/2 such that
f ( x ) = x − cos x = 0.
The interval [0.6,0.7] contains the solution.
Cost $
4
3
2
1
0.25
0.5
0.75
Miles Driven
1
56. Let f ( x) = x5 + 4 x3 – 7 x + 14
f(x) is continuous at all values of x.
f(–2) = –36, f(0) = 14
Because 0 is between –36 and 14, there is at least
one number c between –2 and 0 such that
f ( x) = x5 + 4 x3 – 7 x + 14 = 0.
86
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or by any means, without permission in writing from the publisher.
57. Suppose that f is continuous at c, so
lim f ( x) = f (c). Let x = c + t, so t = x – c, then
x →c
as x → c , t → 0 and the statement
lim f ( x) = f (c) becomes lim f (t + c ) = f (c).
x →c
t →0
Suppose that lim f (t + c) = f (c) and let x = t +
t→ 0
c, so t = x – c. Since c is fixed, t → 0 means that
x → c and the statement lim f (t + c) = f (c)
t →0
becomes lim f ( x) = f (c) , so f is continuous at
x →c
c.
58. Since f(x) is continuous at c,
lim f ( x) = f (c) > 0. Choose ε = f ( c ) , then
x →c
there exists a δ > 0 such that
0 < x − c < δ ⇒ f ( x) − f (c) < ε .
Thus, f ( x ) − f ( c ) > −ε = − f ( c ) , or f ( x ) > 0 .
Since also f ( c ) > 0 , f ( x ) > 0 for all x in
(c − δ , c + δ ).
59. Let g(x) = x – f(x). Then,
g(0) = 0 – f(0) = –f(0) ≤ 0 and g(1) = 1 – f(1) ≥ 0
since 0 ≤ f(x) ≤ 1 on [0, 1] . If g(0) = 0, then
f(0) = 0 and c = 0 is a fixed point of f. If g(1) = 0,
then f(1) = 1 and c = 1 is a fixed point of f. If
neither g(0) = 0 nor g(1) = 0, then g(0) < 0 and
g(1) > 0 so there is some c in [0, 1] such that
g(c) = 0. If g(c) = 0 then c – f(c) = 0 or
f(c) = c and c is a fixed point of f.
60. For f(x) to be continuous everywhere,
f(1) = a(1) + b = 2 and f(2) = 6 = a(2) + b
a+b=2
2a + b = 6
– a = –4
a = 4, b = –2
63. Let f(x) be the difference in times on the hiker’s
watch where x is a point on the path, and suppose
x = 0 at the bottom and x = 1 at the top of the
mountain.
So f(x) = (time on watch on the way up) – (time
on watch on the way down).
f(0) = 4 – 11 = –7, f(1) = 12 – 5 = 7. Since time is
continuous, f(x) is continuous, hence there is
some c between 0 and 1 where f(c) = 0. This c is
the point where the hiker’s watch showed the
same time on both days.
⎡ π⎤
64. Let f be the function on ⎢0, 2 ⎥ such that f(θ) is
⎣
⎦
the length of the side of the rectangle which
makes angle θ with the x-axis minus the length of
the sides perpendicular to it. f is continuous on
⎡ π⎤
⎢0, 2 ⎥ . If f(0) = 0 then the region is
⎣
⎦
circumscribed by a square. If f(0) ≠ 0, then
⎛π ⎞
observe that f (0) = − f ⎜ ⎟ . Thus, by the
⎝2⎠
Intermediate Value Theorem, there is an angle
θ 0 between 0 and
π
such that f (θ 0 ) = 0.
2
Hence, D can be circumscribed by a square.
65. Yes, g is continuous at R .
lim g ( r ) =
r →R−
= lim g ( r )
GMm
r →R+
R2
66. No. By the Intermediate Value Theorem, if f
were to change signs on [a,b], then f must be
0 at some c in [a,b]. Therefore, f cannot
change sign.
67. a.
f(x) = f(x + 0) = f(x) + f(0), so f(0) = 0. We
want to prove that lim f (x) = f (c), or,
x→c
equivalently, lim [ f (x) – f (c)] = 0. But
x→c
61. For x in [0, 1], let f(x) indicate where the string
originally at x ends up. Thus f(0) = a, f(1) = b.
f(x) is continuous since the string is unbroken.
Since 0 ≤ a, b ≤ 1 , f(x) satisfies the conditions of
Problem 59, so there is some c in [0, 1] with
f(c) = c, i.e., the point of string originally at c
ends up at c.
62. The Intermediate Value Theorem does not imply
the existence of a number c between –2 and 2
such that f (c ) = 0. The reason is that the
function f ( x ) is not continuous on [ −2, 2] .
Instructor’s Resource Manual
f(x) – f(c) = f(x – c), so
lim[ f ( x) – f (c)] = lim f ( x – c). Let
x →c
x →c
h = x – c then as x → c, h → 0 and
lim f ( x – c) = lim f (h) = f (0) = 0. Hence
x →c
h →0
lim f (x) = f (c) and f is continuous at c.
x→c
Thus, f is continuous everywhere, since c
was arbitrary.
b. By Problem 43 of Section 0.5, f(t) = mt for
all t in Q. Since g(t) = mt is a polynomial
function, it is continuous for all real
numbers. f(t) = g(t) for all t in Q, thus
f(t) = g(t) for all t in R, i.e. f (t ) = mt.
Section 1.6
87
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or by any means, without permission in writing from the publisher.
68. If f(x) is continuous on an interval then
lim f ( x) = f (c) for all points in the interval:
x →c
lim f ( x) = f (c) ⇒ lim f ( x)
x →c
x →c
= lim
x →c
f 2 ( x) = ⎛⎜ lim f ( x) ⎞⎟
⎝ x →c
⎠
⎡ 3 3⎤
Domain: ⎢ – , ⎥ ;
⎣ 4 4⎦
3⎫
⎧ 3
Range: ⎨ – , 0, ⎬
4
4⎭
⎩
2
= ( f (c))2 = f (c )
⎧ 1 if x ≥ 0
69. Suppose f ( x) = ⎨
. f(x) is
⎩−1 if x < 0
discontinuous at x = 0, but g(x) = f ( x) = 1 is
b. At x = 0
3
3
c. If x = 0, f ( x) = 0 , if x = – , f ( x) = – and
4
4
3
3
3
3
if x = , f ( x) = , so x = − , 0, are
4
4
4
4
fixed points of f.
continuous everywhere.
70. a.
1.7 Chapter Review
Concepts Test
1. False.
Consider f ( x ) = x at x = 2.
2. False:
c may not be in the domain of f(x), or
it may be defined separately.
3. False:
c may not be in the domain of f(x), or
it may be defined separately.
b. If r is any rational number, then any deleted
interval about r contains an irrational
1
number. Thus, if f (r ) = , any deleted
q
interval about r contains at least one point c
1
1
such that f (r ) – f (c) = – 0 = . Hence,
q
q
lim f (x) does not exist.
x→r
If c is any irrational number in (0, 1), then as
p
p
x = → c (where
is the reduced form
q
q
of the rational number) q → ∞, so
f ( x) → 0 as x → c. Thus,
lim f ( x) = 0 = f (c) for any irrational
x →c
4. True.
By definition, where c = 0, L = 0.
5. False:
If f(c) is not defined, lim f ( x ) might
x→c
exist; e.g., f ( x) =
88
Suppose the block rotates to the left. Using
3
geometry, f ( x) = – . Suppose the block
4
rotates to the right. Using geometry,
3
f ( x) = . If x = 0, the block does not rotate,
4
so f(x) = 0.
Section 1.7
x –4
.
x+2
x2 – 4
= −4.
x →−2 x + 2
f(–2) does not exist, but lim
6. True:
x 2 − 25
( x − 5)( x + 5)
= lim
x−5
x →5 x − 5
x →5
= lim ( x + 5) = 5 + 5 = 10
lim
x →5
7. True:
8. False:
number c.
71. a.
2
9. False:
10. True:
Substitution Theorem
lim
x →0
sin x
=1
x
The tangent function is not defined for
all values of c.
sin x
,
cos x
then cos x ≠ 0 , and Theorem A.7
applies..
If x is in the domain of tan x =
Instructor’s Resource Manual
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or by any means, without permission in writing from the publisher.
11. True:
Since both sin x and cos x are
continuous for all real numbers, by
Theorem C we can conclude that
f ( x) = 2 sin 2 x − cos x is also
continuous for all real numbers.
12. True.
By definition, lim f ( x ) = f ( c ) .
13. True.
2 ∈ [1,3]
14. False:
25. True:
x→2
such that 0 < x − 2 < δ ⇒
f ( x ) − f (2) < 0. 001 f (2), or
−0. 001 f (2 ) < f ( x ) − f (2 )
< 0.001f(2)
Thus, 0.999f(2) < f(x) < 1.001f(2) and
f(x) < 1.001f(2) for 0 < x − 2 < δ .
Since f(2) < 1.001f(2), as f(2) > 0,
f(x) < 1.001f(2) on (2 − δ , 2 + δ ).
x →c
lim may not exist
x →0 −
Choose ε = 0. 001 f (2) then since
lim f ( x ) = f (2), there is some δ
26. False:
That lim [ f ( x ) + g ( x )] exists does
x→c
15. False:
Consider f ( x) = sin x.
16. True.
By the definition of continuity on an
interval.
17. False:
Since −1 ≤ sin x ≤ 1 for all x and
1
sin x
lim = 0 , we get lim
=0.
x →∞ x
x →∞ x
18. False.
It could be the case where
lim f ( x ) = 2
not imply that lim f ( x ) and
x→c
lim g( x ) exist; e.g., f ( x) =
x→c
g ( x) =
The graph has many vertical
asymptotes; e.g., x = ± π/2, ± 3π/2,
± 5π/2, …
20. True:
x = 2 ; x = –2
21. True:
As x → 1+ both the numerator and
denominator are positive. Since the
numerator approaches a constant and
the denominator approaches zero, the
limit goes to + ∞ .
22. False:
lim f ( x) must equal f(c) for f to be
Squeeze Theorem
28. True:
A function has only one limit at a
point, so if lim f ( x ) = L and
x→ a
lim f ( x ) = M , L = M
x→ a
29. False:
24. True:
x +x–6
and
x–2
5
x, then f(x) ≠ g(x) for all x,
2
but lim f ( x ) = lim g ( x ) = 5.
x→ 2
x→2
30. False:
If f(x) < 10, lim f ( x ) could equal 10
x→2
if there is a discontinuity point (2, 10).
For example,
– x3 + 6 x 2 − 2 x − 12
f ( x) =
< 10 for
x–2
all x, but lim f ( x) = 10.
lim f ( x) = f ⎛⎜ lim x ⎞⎟ = f (c), so f is
⎝ x →c ⎠
continuous at x = c.
x →2.3
2
g ( x) =
x →c
x
= 1 = f ( 2.3)
2
x →c
example, if f ( x) =
x →c
lim
That f(x) ≠ g(x) for all x does not
imply that lim f ( x) ≠ lim g ( x). For
x →c
continuous at x = c.
23. True:
x+7
for c = −2 .
x+2
27. True:
x →−∞
19. False:
x–3
and
x+2
x →2
31. True:
lim f ( x) = lim
x →a
x →a
f 2 ( x)
2
= ⎡⎢ lim f ( x) ⎤⎥ = (b)2 = b
⎣ x→a
⎦
32. True:
Instructor’s Resource Manual
If f is continuous and positive on
[a, b], the reciprocal is also
continuous, so it will assume all
1
1
and
.
values between
f ( a)
f (b )
Section 1.7
89
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or by any means, without permission in writing from the publisher.
Sample Test Problems
14.
x−2 2−2 0
=
= =0
x →2 x + 2
2+2 4
1. lim
u 2 – 1 12 − 1
=
=0
1+1
u →1 u + 1
u2 – 1
(u – 1)(u + 1)
= lim
= lim (u + 1)
u –1
u →1 u – 1
u →1
u →1
=1+1=2
u +1
u +1
= lim
u →1 (u + 1)(u – 1)
–1
does not exist
1 – 2x
5. lim
= lim
x –2
x
= lim
x →2 ( x – 2)( x + 2)
x→2 x 2
–4
1
1
=
=
2 (2 + 2 ) 8
z2 – 4
1
u →1 u – 1
= lim
x→2
1 − cos 2 x
2 1 − cos 2 x
= lim
3x
2x
x →0
x →0 3
2
1 − cos 2 x 2
= lim
= ×0 = 0
3 x →0
2x
3
1
x ( x + 2)
1
1−
x −1
x = 1+ 0 = 1
17. lim
= lim
2 1+ 0
x →∞ x + 2
x →∞
1+
x
18. Since −1 ≤ sin t ≤ 1 for all t and lim
= lim
get lim
t →∞
19. lim
sin x
tan x
1
cos x
= lim
= lim
x → 0 sin 2 x
x → 0 2 sin x cos x
x → 0 2 cos 2 x
1
1
=
=
2
2 cos 0 2
y →1 y 2
–1
( y – 1)( y 2 + y + 1)
y →1 ( y – 1)( y + 1)
t+2
( t − 2 )2
20.
= lim
21.
y 2 + y + 1 12 + 1 + 1 3
=
=
1+1
2
y +1
y →1
x–4
x –2
x→4
= lim
( x – 2)( x + 2)
x →4
x –2
= lim ( x + 2) = 4 + 2 = 4
x→4
12.
13.
90
x
lim
x →0 –
x
lim
x →(1/ 2)+
lim
t →2 –
(
= lim
x →0 –
–x
= lim (–1) = –1
x x →0 –
4x = 2
cos x
= ∞ , because as x → 0+ , cos x → 1
x →0 + x
while the denominator goes to 0 from the right.
lim
Section 1.7
−
x →π / 4−
tan 2 x = ∞ because as x → (π / 4 ) ,
−
2 x → (π / 2 ) , so tan 2 x → ∞.
22.
1 + sin x
= ∞ , because as x → 0+ ,
+
x
x →0
lim
1 + sin x → 1 while the denominator goes to
0 from the right.
| 2 x − 6 |< ε ⇔ 2 | x − 3 |< ε
⇔| x − 3 |<
ε
2
. Choose δ =
ε
2
.
Let ε > 0. Choose δ = ε / 2. Thus,
t − t ) = lim t − lim t = 1 − 2 = −1
t →2 –
= ∞ because as t → 0, t + 2 → 4
23. Preliminary analysis: Let ε > 0. We need to find
a δ > 0 such that
0 <| x − 3 |< δ ⇒| ( 2 x + 1) − 7 |< ε .
cos x
does not exist.
x →0 x
10. lim
11.
sin t
=0.
t
lim
= lim
9. lim
= 0 , we
while the denominator goes to 0 from the right.
7. lim
y3 – 1
1
t →∞ t
t →2
8. lim
1− x
= −1 since x − 1 < 0 as
x −1
16. lim
z → 2 ( z + 3)( z
z →2 z 2
x →1−
sin 5 x
5 sin 5 x
= lim
x →0 3 x
x →0 3 5 x
5
sin 5 x 5
5
= lim
= ×1 =
3 x →0 5 x
3
3
;
( z + 2)( z – 2)
– 2)
+z–6
z +2 2 + 2 4
=
=
= lim
5
z→ 2 z + 3 2 + 3
6. lim
x −1
= lim
15. lim
3. lim
u →1 u 2
x →1−
x → 1−
2. lim
4. lim
x −1
lim
t →2–
( 2 x + 1) − 7
= 2 x − 6 = 2 x − 3 < 2 (ε / 2 ) = ε .
Instructor’s Resource Manual
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or by any means, without permission in writing from the publisher.
24. a. f(1) = 0
b.
c.
d.
28.
lim f ( x) = lim (1 – x) = 0
x →1+
x →1+
lim f ( x) = lim x = 1
x →1–
x →1–
lim f ( x) = –1 because
x → –1
lim f ( x) = lim x3 = –1 and
x → –1–
x → –1–
lim f ( x) = lim x = –1
x → –1+
25. a.
x → –1+
f is discontinuous at x = 1 because f(1) = 0,
but lim f (x ) does not exist. f is
x→1
discontinuous at x = –1 because f(–1) does
not exist.
b. Define f(–1) = –1
26. a.
b.
27. a.
0 < u – a < δ ⇒ g (u ) – M < ε
0 < a – x < δ ⇒ f ( x) – L < ε
lim[2 f ( x) – 4 g ( x)]
Horizontal: lim
c.
d.
x →3
x2 – 9
lim g ( x)
= lim g ( x )( x + 3)
x – 3 x →3
x →3
= lim g ( x ) ⋅ lim ( x + 3) = –2 ⋅ (3 + 3) = –12
x →3
+1
Horizontal: lim
x2
x →∞ x 2
= lim
x2
+ 1 x →−∞ x 2 + 1
y = 1 is a horizontal asymptote.
g(3) = –2
lim g ( f ( x)) = g ⎛⎜ lim f ( x) ⎞⎟ = g (3) = –2
x →3
⎝ x →3
⎠
lim
x →3
2
f ( x) – 8 g ( x)
2
= ⎡⎢ lim f ( x) ⎤⎥ – 8 lim g ( x)
x →3
⎣ x →3
⎦
33. Vertical: x = 1, x = −1 because lim
x →1+
lim
x →3
g ( x) – g (3)
f ( x)
=
–2 – g (3)
3
lim
x →−1−
x2
x2 − 1
Horizontal: lim
= 1 , so
x2
x2 − 1
x2
= lim
x2
− 1 x→−∞ x 2 − 1
y = 1 is a horizontal asymptote.
=
= 0 , so
=∞
=∞
x →∞ x 2
= (3) 2 – 8(–2) = 5
f.
x
x →−∞ x 2
32. Vertical: None, denominator is never 0.
and
e.
= lim
+1
y = 0 is a horizontal asymptote.
= 2(3) – 4(–2) = 14
x →3
x
x →∞ x 2
= 2 lim f ( x) – 4 lim g ( x )
b.
30. Let f ( x) = x5 – 4 x3 – 3 x + 1
f(2) = –5, f(3) = 127
Because f(x) is continuous on [2, 3] and
f(2) < 0 < f(3), there exists some number c
between 2 and 3 such that f(c) = 0.
31. Vertical: None, denominator is never 0.
x →3
x →3
29. a(0) + b = –1 and a(1) + b = 1
b = –1; a + b = 1
a–1=1
a=2
= 1 , so
−2 − (−2)
3
=0
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Section 1.7
91
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or by any means, without permission in writing from the publisher.
34. Vertical: x = 2, x = −2 because
x
lim
3
2
x → 2+
x −4
= ∞ and
x →∞ x 2
x
lim
3
x →−∞ x 2
−4
asymptotes.
x
lim
x →−2−
x3
Horizontal: lim
−4
2. a.
3
2
x −4
=∞
= ∞ and
= −∞ , so there are no horizontal
35. Vertical: x = ±π / 4, ± 3π / 4, ± 5π / 4,… because
lim tan 2 x = ∞ and similarly for other odd
x →π / 4−
g ( 2 ) = 1/ 2
b.
g ( 2.1) = 1/ 2.1 ≈ 0.476
c.
g ( 2.1) − g ( 2 ) = 0.476 − 0.5 = −0.024
d.
g ( 2.1) − g ( 2 )
2.1 − 2
−0.024
= −0.24
0.1
=
e.
g ( a + h ) = 1/ ( a + h )
f.
g ( a + h ) − g ( a ) = 1/ ( a + h ) − 1/ a =
multiples of π / 4.
Horizontal: None, because lim tan 2 x and
x →∞
g.
36. Vertical: x = 0, because
sin x
1 sin x
lim
= lim
=∞.
2
+
+
x
x →0
x →0 x x
h.
lim tan 2 x do not exist.
g (a + h) − g (a)
(a + h) − a
x →−∞
3. a.
Horizontal: y = 0, because
lim
x →∞
sin x
x2
= lim
sin x
x →−∞
x2
1. a.
f ( 2.1) = 2.12 = 4.41
c.
f ( 2.1) − f ( 2 ) = 4.41 − 4 = 0.41
d.
e.
f.
g.
h.
f ( 2.1) − f ( 2 )
2.1 − 2
0.41
=
= 4.1
0.1
f ( a + h ) = ( a + h ) = a 2 + 2ah + h 2
2
f ( a + h ) − f ( a ) = a 2 + 2ah + h 2 − a 2
= 2ah + h 2
f (a + h) − f (a)
(a + h) − a
lim
h→0
2ah + h 2
=
= 2a + h
h
f (a + h) − f ( a)
(a + h) − a
g (a + h) − g (a)
(a + h) − a
h→0
h →0
c.
F ( 2.1) − F ( 2 ) = 1.449 − 1.414 = 0.035
d.
F ( 2.1) − F ( 2 )
2.1 − 2
Review and Preview
=
0.035
= 0.35
0.1
e.
F (a + h) = a + h
f.
F (a + h) − F (a) = a + h − a
g.
h.
F (a + h) − F (a)
lim
a+h − a
h
=
( a + h) − a
F (a + h) − F (a)
h→0
= lim
(
(a + h) − a
a+h − a
h →0
h
= lim
h →0
h →0
= lim
h →0
92
−1
a2
F ( 2 ) = 2 ≈ 1.414
h
= lim
= lim ( 2a + h ) = 2a
=
−1
a (a + h)
F ( 2.1) = 2.1 ≈ 1.449
f ( 2 ) = 22 = 4
b.
lim
h
=
b.
= 0.
Review and Preview Problems
=
−h
a (a + h)
−h
a (a + h)
h
(
(
(
h→0
)(
a+h + a
a+h + a
a+h−a
a+h + a
h
a+h + a
1
a+h + a
a+h − a
h
= lim
=
)
)
)
)
1
2 a
=
a
2a
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or by any means, without permission in writing from the publisher.
4. a.
G ( 2) = ( 2) + 1 = 8 + 1 = 9
3
10.
b.
G ( 2.1) = ( 2.1) + 1 = 9.261 + 1 = 10.261
c.
G ( 2.1) − G ( 2 ) = 10.261 − 9 = 1.261
d.
e.
3
G ( 2.1) − G ( 2 )
2.1 − 2
=
1.261
= 12.61
0.1
4
32π
3
cm3
V0 = π ( 2 ) =
3
3
4
62.5π 125π
3
=
cm3
V1 = π ( 2.5 ) =
3
3
6
125π
32π
ΔV = V1 − V0 =
cm3 −
cm3
6
3
61
= π cm3 ≈ 31.940 cm3
6
11. a.
G ( a + h) = ( a + h) + 1
3
= a 3 + 3a 2 h + 3ah 2 + h3 + 1
f.
G ( a + h ) − G ( a ) = ⎡( a + h ) + 1⎤ − ⎡⎣ a + 1⎤⎦
⎣
⎦
3
3
) (
)
(
b.
d = 6002 + 4002
= 721 miles
c.
d = 6752 + 5002
= 840 miles
= a3 + 3a 2 h + 3ah 2 + h3 + 1 − a 3 + 1
2
2
= 3a h + 3ah + h
g.
G ( a + h) − G ( a)
(a + h) − a
3
=
North plane has traveled 600miles. East
plane has traveled 400 miles.
3a 2 h + 3ah 2 + h3
h
= 3a 2 + 3ah + h 2
h.
lim
h→0
G ( a + h) − G ( a)
(a + h) − a
= lim 3a 2 + 3ah + h 2
h →0
= 3a 2
5. a.
( a + b )3 = a3 + 3a 2b +
b.
( a + b ) 4 = a 4 + 4 a 3b +
c.
( a + b )5 = a 5 + 5 a 4 b +
6.
( a + b )n = a n + na n −1b +
7. sin ( x + h ) = sin x cos h + cos x sin h
8. cos ( x + h ) = cos x cos h − sin x sin h
9. a.
The point will be at position (10, 0 ) in all
three cases ( t = 1, 2,3 ) because it will have
made 4, 8, and 12 revolutions respectively.
b. Since the point is rotating at a rate of 4
revolutions per second, it will complete 1
1
revolution after second. Therefore, the
4
point will first return to its starting position
1
at time t = .
4
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or by any means, without permission in writing from the publisher.
2
CHAPTER
2.1 Concepts Review
The Derivative
4.
1. tangent line
2. secant line
3.
f (c + h ) − f ( c )
h
4. average velocity
Problem Set 2.1
1. Slope =
2. Slope =
5–3
2 – 32
=4
6–4
= –2
4–6
Slope ≈ 1.5
5.
3.
Slope ≈
Slope ≈ −2
5
2
6.
Slope ≈ –
94
Section 2.1
3
2
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7. y = x 2 + 1
[(2.01)3 − 1.0] − 7
2.01 − 2
0.120601
=
0.01
= 12.0601
d.
msec =
e.
mtan = lim
a., b.
f (2 + h) – f (2)
h
h →0
[(2 + h)3 – 1] – (23 − 1)
h
h →0
= lim
12h + 6h 2 + h3
h
h→0
= lim
c.
m tan = 2
d.
msec =
e.
(1.01)2 + 1.0 − 2
1.01 − 1
0.0201
=
.01
= 2.01
f (1 + h) – f (1)
h
h →0
mtan = lim
[(1 + h)2 + 1] – (12 + 1)
h
h →0
= lim
2 + 2h + h 2 − 2
h
h →0
h(2 + h)
= lim
h
h →0
= lim (2 + h) = 2
= lim
h(12 + 6h + h 2 )
h
h→0
= 12
= lim
9. f (x) = x 2 – 1
f (c + h ) – f (c )
mtan = lim
h
h→0
[(c + h)2 – 1] – (c 2 – 1)
h
h→0
= lim
c 2 + 2ch + h 2 – 1 – c 2 + 1
h
h→0
h(2c + h)
= lim
= 2c
h
h→0
At x = –2, m tan = –4
x = –1, m tan = –2
x = 1, m tan = 2
x = 2, m tan = 4
= lim
h →0
3
8. y = x – 1
a., b.
10. f (x) = x 3 – 3x
f (c + h ) – f (c )
mtan = lim
h
h→0
[(c + h)3 – 3(c + h)] – (c3 – 3c)
h
h→0
= lim
c3 + 3c 2 h + 3ch 2 + h3 – 3c – 3h – c3 + 3c
h
h→0
= lim
h(3c 2 + 3ch + h 2 − 3)
= 3c 2 – 3
h
h→0
At x = –2, m tan = 9
x = –1, m tan = 0
x = 0, m tan = –3
x = 1, m tan = 0
x = 2, m tan = 9
= lim
c.
m tan = 12
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11.
13. a.
16(12 ) –16(02 ) = 16 ft
b.
16(22 ) –16(12 ) = 48 ft
c.
Vave =
d.
f ( x) =
mtan
1
x +1
f (1 + h) – f (1)
= lim
h
h→0
−
= lim 2+ h 2
h
h →0
− 2(2h+ h)
= lim
h
h →0
1
= lim −
h→0 2(2 + h)
1
e.
b.
1
4
1
1
y – = – ( x –1)
2
4
=–
1
x –1
f (0 + h) − f (0)
= lim
h
h →0
1 +1
= lim h −1
h
h →0
12. f (x) =
mtan
= lim
16(3.01) 2 − 16(3)2
3.01 − 3
0.9616
=
0.01
= 96.16 ft/s
Vave =
f (t ) = 16t 2 ; v = 32c
v = 32(3) = 96 ft/s
1
14. a.
h
h −1
h →0
h
1
= lim
h →0 h − 1
= −1
y + 1 = –1(x – 0); y = –x – 1
Vave =
d.
(32 + 1) – (22 + 1)
= 5 m/sec
3– 2
[(2.003)2 + 1] − (22 + 1)
2.003 − 2
0.012009
=
0.003
= 4.003 m/sec
Vave =
Vave =
c.
144 – 64
= 80 ft/sec
3–2
[(2 + h) 2 + 1] – (22 + 1)
2+h–2
4h + h 2
h
= 4 +h
=
f (t ) = t2 + 1
f (2 + h) – f (2)
v = lim
h
h →0
[(2 + h)2 + 1] – (22 + 1)
h
h →0
= lim
4h + h 2
h
h →0
= lim (4 + h)
= lim
h →0
=4
96
Section 2.1
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15. a.
f (α + h) – f (α )
h
v = lim
h →0
2(α + h) + 1 – 2α + 1
h
h →0
= lim
2α + 2h + 1 – 2α + 1
h
= lim
h →0
= lim
( 2α + 2h + 1 – 2α + 1)( 2α + 2h + 1 + 2α + 1)
h( 2α + 2h + 1 + 2α + 1)
h →0
2h
= lim
2α + 2h + 1 + 2α + 1)
h →0 h(
2
=
2α + 1 + 2α + 1
1
b.
2α + 1
=
=
1
2α + 1
ft/s
1
2
2α + 1 = 2
3
2
The object reaches a velocity of 1 ft/s when t = 3 .
2
2
2 α + 1= 4; α =
16. f (t ) = – t2 + 4 t
18. a.
[–(c + h)2 + 4(c + h)] – (– c 2 + 4c)
h
h →0
v = lim
– c 2 – 2ch – h 2 + 4c + 4h + c 2 – 4c
h
h →0
h(–2c – h + 4)
= lim
= –2c + 4
h
h →0
–2c + 4 = 0 when c = 2
The particle comes to a momentary stop at
t = 2.
b.
1000(2.5)2 – 1000(2)2 2250
=
= 4500
2.5 – 2
0.5
c.
f (t ) = 1000t 2
= lim
17. a.
b.
c.
1000(2 + h)2 − 1000(2) 2
h
h→0
r = lim
4000 + 4000h + 1000h 2 – 4000
h
h→0
h(4000 + 1000h)
= lim
= 4000
h
h→0
= lim
⎡1
⎤ ⎡1 2 ⎤
2
⎢ 2 (2.01) + 1⎥ – ⎢ 2 (2) + 1⎥ = 0.02005 g
⎣
⎦ ⎣
⎦
rave
0.02005
=
= 2.005 g/hr
2.01 – 2
1 2
t +1
2
⎡ 1 (2 + h)2 + 1⎤ – ⎡ 1 22 + 1⎤
2
⎦ ⎣2
⎦
r = lim ⎣
h
h →0
f (t ) =
= lim
h→0
= lim
2 + 2h + 12 h 2 + 1 − 2 − 1
(
h 2 + 12 h
h
h→0
At t = 2, r = 2
h
)=2
Instructor’s Resource Manual
1000(3)2 – 1000(2)2 = 5000
19. a.
b.
dave =
53 – 33 98
=
= 49 g/cm
5–3
2
f (x) = x 3
(3 + h)3 – 33
h
h →0
d = lim
27 + 27h + 9h 2 + h3 – 27
h
h→0
= lim
h(27 + 9h + h 2 )
= 27 g/cm
h
h→0
= lim
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20. MR = lim
h→0
R (c + h ) – R (c )
h
[0.4(c + h) – 0.001(c + h)2 ] – (0.4c – 0.001c 2 )
h
h→0
= lim
0.4c + 0.4h – 0.001c 2 – 0.002ch – 0.001h 2 – 0.4c + 0.001c 2
h
h→0
h(0.4 – 0.002c – 0.001h)
= lim
= 0.4 – 0.002c
h
h→0
When n = 10, MR = 0.38; when n = 100, MR = 0.2
= lim
2(1 + h)2 – 2(1) 2
h
h →0
21. a = lim
2 + 4h + 2h 2 – 2
h
h→0
h(4 + 2h)
= lim
=4
h
h→0
= lim
22. r = lim
h→0
p (c + h ) – p (c )
h
[120(c + h)2 – 2(c + h)3 ] – (120c 2 – 2c3 )
h
h →0
= lim
h(240c – 6c 2 + 120h – 6ch – 2h 2 )
h
h →0
= lim
= 240c – 6c 2
When
t = 10, r = 240(10) – 6(10) 2 = 1800
t = 20, r = 240(20) – 6(20)2 = 2400
t = 40, r = 240(40) – 6(40)2 = 0
100 – 800
175
=–
≈ –29.167
24 – 0
6
29,167 gal/hr
700 – 400
≈ −75
At 8 o’clock, r ≈
6 − 10
75,000 gal/hr
23. rave =
24. a. The elevator reached the seventh floor at time
t = 80 . The average velocity is
v avg = (84 − 0) / 80 = 1.05 feet per second
b. The slope of the line is approximately
60 − 12
= 1.2 . The velocity is
55 − 15
approximately 1.2 feet per second.
98
Section 2.1
c. The building averages 84/7=12 feet from
floor to floor. Since the velocity is zero for
two intervals between time 0 and time 85, the
elevator stopped twice. The heights are
approximately 12 and 60. Thus, the elevator
stopped at floors 1 and 5.
25. a.
A tangent line at t = 91 has slope
approximately (63 − 48) /(91 − 61) = 0.5 . The
normal high temperature increases at the rate
of 0.5 degree F per day.
b. A tangent line at t = 191 has approximate
slope (90 − 88) / 30 ≈ 0.067 . The normal
high temperature increases at the rate of
0.067 degree per day.
c. There is a time in January, about January 15,
when the rate of change is zero. There is also
a time in July, about July 15, when the rate of
change is zero.
d. The greatest rate of increase occurs around
day 61, that is, some time in March. The
greatest rate of decrease occurs between day
301 and 331, that is, sometime in November.
26. The slope of the tangent line at t = 1930 is
approximately (8 − 6) /(1945 − 1930) ≈ 0.13 . The
rate of growth in 1930 is approximately 0.13
million, or 130,000, persons per year. In 1990,
the tangent line has approximate slope
(24 − 16) /(20000 − 1980) ≈ 0.4 . Thus, the rate of
growth in 1990 is 0.4 million, or 400,000,
persons per year. The approximate percentage
growth in 1930 is 0.107 / 6 ≈ 0.018 and in 1990 it
is approximately 0.4 / 20 ≈ 0.02 .
27. In both (a) and (b), the tangent line is always
positive. In (a) the tangent line becomes steeper
and steeper as t increases; thus, the velocity is
increasing. In (b) the tangent line becomes flatter
and flatter as t increases; thus, the velocity is
decreasing.
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28.
2.2 Concepts Review
1
f (t ) = t 3 + t
3
f ( c + h ) – f (c )
h
h →0
3
1
⎡ ( c + h ) + (c + h ) ⎤ – 1 c 3 + c
3
⎦ 3
= lim ⎣
h
h→0
current = lim
(
= lim
(
)
) = c2 + 1
h c 2 + ch + 13 h 2 + 1
h
h→0
When t = 3, the current =10
c 2 + 1 = 20
2
c = 19
c = 19 ≈ 4.4
A 20-amp fuse will blow at t = 4.4 s.
1.
f (c + h) – f (c) f (t ) – f (c)
;
h
t –c
2.
f ′(c )
3. continuous; f ( x) = x
4.
=
f ′(1) = lim
h →0
f (1 + h) – f (1)
h
(1 + h)2 – 12
2h + h 2
= lim
= lim
h
h
h→0
h →0
= lim (2 + h ) = 2
h→0
2.
f ′(2) = lim
h →0
f (2 + h) – f (2)
h
[2(2 + h)]2 – [2(2)]2
h
h→0
4
1
30. V = π r 3 , r = t
3
4
1
V = π t3
48
rate =
dy
dx
Problem Set 2.2
1.
29. A = πr 2 , r = 2t
A = 4πt2
4π(3 + h)2 – 4π(3)2
rate = lim
h
h →0
h(24π + 4πh)
= lim
= 24π km2/day
h
h→0
f '( x);
= lim
16h + 4h 2
= lim (16 + 4h) = 16
h
h→0
h →0
= lim
1
(3 + h)3 − 33 27
π lim
= π
h
48 h→0
48
3.
9
π inch 3 / sec
16
m tan = 7
b. m tan = 0
c.
m tan = –1
d. m tan = 17. 92
32. y = f ( x) = sin x sin 2 x
h →0
5h + h 2
= lim (5 + h) = 5
h
h→0
h →0
= lim
4.
f ′(4) = lim
h →0
2
a.
m tan = –1.125 b. m tan ≈ –1.0315
c.
m tan = 0
d. m tan ≈ 1.1891
33. s = f (t ) = t + t cos 2 t
At t = 3, v ≈ 2.818
(t + 1)3
t+2
At t = 1.6, v ≈ 4.277
34. s = f (t ) =
Instructor’s Resource Manual
f (3 + h) – f (3)
h
[(3 + h)2 – (3 + h)] – (32 – 3)
= lim
h
h→0
31. y = f ( x) = x 3 – 2 x 2 + 1
a.
f ′(3) = lim
= lim
h →0
1
3+ h
f (4 + h) – f (4)
h
1
– 4–1
h
= lim
h →0
3–(3+ h )
3(3+ h )
h
–1
h →0 3(3 + h)
= lim
1
=–
9
s ( x + h) – s ( x )
h
h →0
[2( x + h) + 1] – (2 x + 1)
= lim
h
h →0
2h
= lim
=2
h →0 h
5. s ′( x) = lim
Section 2.2
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6.
f ( x + h) – f ( x )
h
[α ( x + h) + β ] – (α x + β )
= lim
h
h →0
αh
= lim
=α
h →0 h
f ′( x) = lim
12. g ′( x) = lim
h →0
h →0
[( x + h)4 + ( x + h) 2 ] – ( x 4 + x 2 )
h
h →0
= lim
4hx3 + 6h 2 x 2 + 4h3 x + h 4 + 2hx + h 2
h
h →0
= lim
= lim (4 x3 + 6hx 2 + 4h 2 x + h3 + 2 x + h)
r ( x + h) – r ( x )
7. r ′( x) = lim
h
h →0
h →0
3
= 4x + 2x
[3( x + h)2 + 4] – (3 x 2 + 4)
= lim
h
h →0
6 xh + 3h 2
= lim (6 x + 3h) = 6 x
h
h →0
h →0
f ′( x) = lim
h →0
f ( x + h) – f ( x )
h
[( x + h)2 + ( x + h) + 1] – ( x 2 + x + 1)
h
h →0
= lim
= lim
h →0
9.
2 xh + h + h
= lim (2 x + h + 1) = 2 x + 1
h
h →0
2
f ′( x) = lim
h →0
h( x + h) – h( x )
h
h →0
⎡⎛ 2
2 ⎞ 1⎤
= lim ⎢⎜
– ⎟⋅ ⎥
h → 0 ⎣⎝ x + h x ⎠ h ⎦
13. h′( x) = lim
= lim
8.
f ( x + h) – f ( x )
h
[a( x + h) 2 + b( x + h) + c] – (ax 2 + bx + c)
h
h →0
= lim
2axh + ah 2 + bh
= lim (2ax + ah + b)
h
h →0
h →0
= 2ax + b
⎡ –2h 1 ⎤
–2
2
= lim ⎢
⋅ ⎥ = lim
=–
h → 0 ⎣ x ( x + h ) h ⎦ h →0 x ( x + h )
x2
S ( x + h) – S ( x )
h
h →0
⎡⎛
1
1 ⎞ 1⎤
= lim ⎢⎜
–
⎟⋅ ⎥
h →0 ⎣⎝ x + h + 1 x + 1 ⎠ h ⎦
14. S ′( x) = lim
⎡
–h
1⎤
= lim ⎢
⋅ ⎥
h→0 ⎣ ( x + 1)( x + h + 1) h ⎦
–1
1
= lim
=−
h→0 ( x + 1)( x + h + 1)
( x + 1) 2
= lim
10.
f ′( x) = lim
h →0
f ( x + h) – f ( x )
h
( x + h) 4 – x 4
h
h →0
= lim
4hx3 + 6h 2 x 2 + 4h3 x + h 4
h
h →0
= lim
= lim (4 x3 + 6hx 2 + 4h 2 x + h3 ) = 4 x3
h →0
11.
F ( x + h) – F ( x )
h
h →0
15. F ′( x) = lim
⎡⎛
6
6 ⎞ 1⎤
–
= lim ⎢⎜
⎟⋅ ⎥
h→0 ⎢⎜⎝ ( x + h) 2 + 1 x 2 + 1 ⎟⎠ h ⎥
⎣
⎦
2
2
⎡ 6( x + 1) – 6( x + 2hx + h 2 + 1) 1 ⎤
= lim ⎢
⋅ ⎥
h ⎥⎦
h→0 ⎢⎣ ( x 2 + 1)( x 2 + 2hx + h 2 + 1)
⎡
–12hx – 6h 2
1⎤
= lim ⎢
⋅ ⎥
h→0 ⎢ ( x 2 + 1)( x 2 + 2hx + h 2 + 1) h ⎥
⎣
⎦
= lim
f ′( x) = lim
h →0
h →0 ( x 2
f ( x + h) – f ( x )
h
[( x + h)3 + 2( x + h)2 + 1] – ( x3 + 2 x 2 + 1)
= lim
h
h →0
3hx 2 + 3h 2 x + h3 + 4hx + 2h 2
h
h →0
= lim
= lim (3x + 3hx + h + 4 x + 2h) = 3 x + 4 x
2
h →0
2
g ( x + h) – g ( x )
h
2
–12 x – 6h
+ 1)( x + 2hx + h + 1)
2
2
=−
12 x
( x + 1)2
2
F ( x + h) – F ( x )
h
h →0
⎡⎛ x + h –1 x –1 ⎞ 1 ⎤
= lim ⎢⎜
–
⎟⋅ ⎥
h →0 ⎣ ⎝ x + h + 1 x + 1 ⎠ h ⎦
16. F ′( x) = lim
⎡ x 2 + hx + h –1 – ( x 2 + hx – h –1) 1 ⎤
= lim ⎢
⋅ ⎥
h ⎦⎥
( x + h + 1)( x + 1)
h →0 ⎢
⎣
⎡
2h
1⎤
2
= lim ⎢
⋅ ⎥=
h→0 ⎣ ( x + h + 1)( x + 1) h ⎦ ( x + 1) 2
100
Section 2.2
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G ( x + h) – G ( x )
h
⎡⎛ 2( x + h) –1 2 x –1 ⎞ 1 ⎤
= lim ⎢⎜
–
⎟⋅
x – 4 ⎠ h ⎥⎦
h→0 ⎣⎝ x + h – 4
17. G ′( x ) = lim
h →0
⎡ 2 x 2 + 2hx − 9 x − 8h + 4 − (2 x 2 + 2hx − 9 x − h + 4) 1 ⎤
⎡
–7h
1⎤
⋅ ⎥ = lim ⎢
⋅ ⎥
h
( x + h − 4)( x − 4)
⎥⎦ h→0 ⎣ ( x + h – 4)( x – 4) h ⎦
–7
7
= lim
=–
h→0 ( x + h – 4)( x – 4)
( x – 4)2
= lim ⎢
h →0 ⎢⎣
G ( x + h) – G ( x )
h
h →0
18. G ′( x ) = lim
⎡⎛
⎡ (2 x + 2h)( x 2 – x ) – 2 x( x 2 + 2 xh + h 2 – x – h) 1 ⎤
2( x + h)
2x ⎞ 1 ⎤
= lim ⎢⎜
⋅ ⎥ = lim ⎢
–
⋅ ⎥
⎟
h ⎦⎥
h→0 ⎢⎝⎜ ( x + h) 2 – ( x + h) x 2 – x ⎠⎟ h ⎥
( x 2 + 2hx + h 2 – x – h)( x 2 – x)
⎣
⎦ h→0 ⎣⎢
⎡
–2h 2 x – 2hx 2
1⎤
= lim ⎢
⋅ ⎥
2
2
2
h→0 ⎣⎢ ( x + 2hx + h – x – h)( x – x ) h ⎦⎥
–2hx – 2 x 2
= lim
h→0 ( x 2
=
+ 2hx + h 2 – x – h)( x 2 – x)
–2 x 2
2
( x – x)
2
19. g ′( x) = lim
h →0
=–
2
( x – 1) 2
g ( x + h) – g ( x )
h
3( x + h) – 3x
h
h→0
= lim
= lim
( 3x + 3h – 3x )( 3x + 3h + 3 x )
h( 3 x + 3h + 3x )
h→0
3h
= lim
h→0 h(
3 x + 3h + 3 x )
20. g ′( x) = lim
h →0
= lim
h →0
3
3x + 3h + 3x
=
3
2 3x
g ( x + h) – g ( x )
h
⎡⎛
1
1 ⎞ 1⎤
–
= lim ⎢⎜
⎟⋅ ⎥
h→0 ⎢⎜⎝ 3( x + h)
3 x ⎟⎠ h ⎦⎥
⎣
⎡ 3x – 3x + 3h 1 ⎤
= lim ⎢
⋅ ⎥
h ⎦⎥
h→0 ⎣⎢
9 x ( x + h)
⎡ ( 3 x – 3 x + 3h )( 3 x + 3 x + 3h ) 1 ⎤
= lim ⎢
⋅ ⎥
h ⎦⎥
h→0 ⎣⎢
9 x( x + h)( 3x + 3x + 3h )
= lim
h→0 h
–3h
9 x( x + h)( 3x + 3x + 3h )
Instructor’s Resource Manual
=
–3
3x ⋅ 2 3x
=–
1
2 x 3x
Section 2.2
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21. H ′( x ) = lim
h →0
H ( x + h) – H ( x )
h
⎡⎛
3
–
= lim ⎢⎜
h→0 ⎣⎝ x + h – 2
⎞ 1⎤
⎟⋅ ⎥
x – 2 ⎠ h⎦
3
⎡3 x – 2 – 3 x + h – 2 1 ⎤
= lim ⎢
⋅ ⎥
h→0 ⎣⎢
( x + h – 2)( x – 2) h ⎦⎥
= lim
h→0
3( x – 2 – x + h – 2)( x – 2 + x + h – 2)
h ( x + h – 2)( x – 2)( x – 2 + x + h – 2)
−3h
= lim
h→0 h[( x – 2)
–3
= lim
h→0 ( x – 2)
=–
x + h – 2 + ( x + h – 2) x – 2]
x + h – 2 + ( x + h – 2) x – 2
3
2( x – 2) x – 2
22. H ′( x) = lim
h →0
=−
3
2( x − 2)3 2
H ( x + h) – H ( x )
h
( x + h) 2 + 4 – x 2 + 4
h
h→0
= lim
⎛ x 2 + 2hx + h 2 + 4 – x 2 + 4 ⎞ ⎛ x 2 + 2hx + h 2 + 4 + x 2 + 4 ⎞
⎜
⎟⎜
⎟
⎠⎝
⎠
= lim ⎝
h →0
2
2
2
⎛
⎞
h ⎜ x + 2hx + h + 4 + x + 4 ⎟
⎝
⎠
= lim
h→0
= lim
2hx + h 2
h ⎛⎜ x 2 + 2hx + h 2 + 4 + x 2 + 4 ⎞⎟
⎝
⎠
2x + h
h →0
x 2 + 2hx + h 2 + 4 + x 2 + 4
2x
x
=
=
2
2
2 x +4
x +4
23. f ′( x) = lim
t→x
f (t ) – f ( x)
t–x
(t − 3t ) – ( x – 3 x)
t–x
t→x
= lim
2
2
t 2 – x 2 – (3t – 3x)
t–x
t→x
(t – x)(t + x) – 3(t – x)
= lim
t–x
t→x
(t – x)(t + x – 3)
= lim
= lim (t + x – 3)
t–x
t→x
t→x
=2x–3
= lim
24.
f ′( x) = lim
t→x
3
f (t ) – f ( x)
t–x
(t + 5t ) – ( x3 + 5 x)
t–x
t→x
= lim
t 3 – x3 + 5t – 5 x
t–x
t→x
= lim
(t – x)(t 2 + tx + x 2 ) + 5(t – x)
t–x
t→x
= lim
(t – x)(t 2 + tx + x 2 + 5)
t–x
t→x
= lim
= lim (t 2 + tx + x 2 + 5) = 3x 2 + 5
t→x
102
Section 2.2
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25.
f (t ) – f ( x)
t–x
t→x
⎡⎛ t
x ⎞ ⎛ 1 ⎞⎤
= lim ⎢⎜
–
⎟⎜
⎟⎥
t → x ⎣⎝ t – 5 x – 5 ⎠ ⎝ t – x ⎠ ⎦
f ′( x) = lim
38. The slope of the tangent line is always −1 .
tx – 5t – tx + 5 x
t → x (t – 5)( x – 5)(t – x)
= lim
= lim
t → x (t
=−
26.
–5(t – x)
–5
= lim
– 5)( x – 5)(t – x) t → x (t – 5)( x – 5)
5
( x − 5)
2
39. The derivative is positive until x = 0 , then
becomes negative.
f (t ) – f ( x)
t–x
t→x
⎡⎛ t + 3 x + 3 ⎞ ⎛ 1 ⎞ ⎤
= lim ⎢⎜
–
⎟⎜
⎟
x ⎠ ⎝ t – x ⎠ ⎥⎦
t → x ⎣⎝ t
f ′( x) = lim
3x – 3t
–3
3
= lim
=–
t → x xt (t – x ) t → x xt
x2
= lim
27. f (x) = 2 x 3 at x = 5
40. The derivative is negative until x = 1 , then
becomes positive.
28. f (x) = x 2 + 2 x at x = 3
29. f (x) = x 2 at x = 2
30. f (x) = x 3 + x at x = 3
31. f (x) = x 2 at x
32. f (x) = x 3 at x
33. f (t ) =
2
at t
t
41. The derivative is −1 until x = 1 . To the right of
x = 1 , the derivative is 1. The derivative is
undefined at x = 1 .
34. f(y) = sin y at y
35. f(x) = cos x at x
36. f(t) = tan t at t
37. The slope of the tangent line is always 2.
42. The derivative is −2 to the left of x = −1 ; from
−1 to 1, the derivative is 2, etc. The derivative is
not defined at x = −1, 1, 3 .
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Section 2.2
103
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43. The derivative is 0 on ( −3, −2 ) , 2 on ( −2, −1) , 0
on ( −1, 0 ) , −2 on ( 0,1) , 0 on (1, 2 ) , 2 on ( 2,3)
53.
and 0 on ( 3, 4 ) . The derivative is undefined at
x = −2, − 1, 0, 1, 2, 3 .
1
1
Δy x +Δx +1 – x +1
=
Δx
Δx
⎛ x + 1 – ( x + Δx + 1) ⎞ ⎛ 1 ⎞
=⎜
⎟⎜ ⎟
⎝ ( x + Δx + 1)( x + 1) ⎠ ⎝ Δx ⎠
=
– Δx
( x + Δx + 1)( x + 1)Δx
=–
1
( x + Δx + 1)( x + 1)
⎡
⎤
dy
1
1
= lim −
=−
dx Δx →0 ⎢⎣ ( x + Δx + 1)( x + 1) ⎥⎦
( x + 1) 2
44. The derivative is 1 except at x = −2, 0, 2 where
it is undefined.
1
⎛ 1⎞
− ⎜1 + ⎟
x + Δx ⎝ x ⎠
Δx
−Δx
1
1
−
x ( x + Δx )
1
= x + Δx x =
=−
Δx
Δx
x ( x + Δx )
Δy
54.
=
Δx
1+
dy
1
1
= lim −
=− 2
dx Δx →0 x ( x + Δx )
x
55.
45. Δy = [3(1.5) + 2] – [3(1) + 2] = 1.5
46. Δy = [3(0.1) 2 + 2(0.1) + 1] – [3(0.0) 2 + 2(0.0) + 1]
= 0.23
x 2 + xΔx − x + x + Δx − 1 − ⎡ x 2 + xΔx − x + x − Δx − 1⎤ 1
⎣
⎦×
2
Δx
x + x Δx + x + x + Δ x + 1
2Δx
1
2
= 2
×
= 2
Δ
x
x + xΔx + x + x + Δx + 1
x + x Δx + x + x + Δ x + 1
2
2
2
dy
= lim
=
=
dx Δx →0 x 2 + xΔx + x + x + Δx + 1 x 2 + 2 x + 1 ( x + 1)2
47. Δy = 1/1.2 – 1/1 = – 0.1667
=
48. Δy = 2/(0.1+1) – 2/(0+1) = – 0.1818
49. Δy =
3
3
–
≈ 0.0081
2.31 + 1 2.34 + 1
50. Δy = cos[2(0.573)] – cos[2(0.571)] ≈ –0.0036
51.
52.
Δy ( x + Δx) 2 – x 2 2 xΔx + (Δx) 2
=
=
= 2 x + Δx
Δx
Δx
Δx
dy
= lim (2 x + Δx) = 2 x
dx Δx →0
56.
Δy
=
Δx
( x + Δx ) 2 − 1 − x 2 − 1
x + Δx
Δx
x
(
)
3 x 2 Δx + 3x(Δx)2 – 6 xΔx – 3(Δx) 2 + Δx3
=
Δx
⎡ x ( x + Δx )2 − x − ( x + Δx ) x 2 − 1 ⎤
⎥× 1
=⎢
⎢
⎥ Δx
x ( x + Δx )
⎣
⎦
2
2
3
⎡ x x + 2 xΔx + Δx − x − x + x 2 Δx − x − Δx
=⎢
⎢
x 2 + x Δx
⎣⎢
= 3x 2 + 3xΔx – 6 x – 3Δx + (Δx)2
=
Δy [( x + Δx)3 – 3( x + Δx) 2 ] – ( x3 – 3 x 2 )
=
Δx
Δx
dy
= lim (3 x 2 + 3 xΔx – 6 x – 3Δx + (Δx)2 )
dx Δx→0
= 3x2 – 6 x
104
x + Δx − 1 x − 1
−
Δy x + Δx + 1 x + 1
=
Δx
Δx
( x + 1)( x + Δx − 1) − ( x − 1)( x + Δx + 1) 1
=
×
Δx
( x + Δx + 1)( x + 1)
Section 2.2
(
( ))
x 2 Δx + x ( Δx ) + Δx
2
x + x Δx
2
×
(
) ⎤⎥ × 1
⎥ Δx
⎦⎥
1
x 2 + x Δx + 1
= 2
Δx
x + x Δx
dy
x 2 + xΔx + 1 x 2 + 1
= lim
= 2
Δ
x
→
0
dx
x 2 + x Δx
x
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57.
1
f ′(0) ≈ – ; f ′(2) ≈ 1
2
2
f ′(5) ≈ ; f ′(7) ≈ –3
3
58. g ′(–1) ≈ 2; g ′(1) ≈ 0
g ′(4) ≈ –2; g ′(6) ≈ –
1
3
63. The derivative is 0 at approximately t = 15 and
t = 201 . The greatest rate of increase occurs at
about t = 61 and it is about 0.5 degree F per day.
The greatest rate of decrease occurs at about
t = 320 and it is about 0.5 degree F per day. The
derivative is positive on (15,201) and negative on
(0,15) and (201,365).
59.
64. The slope of a tangent line for the dashed
function is zero when x is approximately 0.3 or
1.9. The solid function is zero at both of these
points. The graph indicates that the solid
function is negative when the dashed function
has a tangent line with a negative slope and
positive when the dashed function has a tangent
line with a positive slope. Thus, the solid
function is the derivative of the dashed function.
60.
61. a.
5
3
f (2) ≈ ; f ′(2) ≈
2
2
f (0.5) ≈ 1.8; f ′(0.5) ≈ –0.6
b.
2.9 − 1.9
= 0.5
2.5 − 0.5
c.
x=5
d. x = 3, 5
e.
x = 1, 3, 5
f.
x=0
g.
x ≈ −0.7,
3
and 5 < x < 7
2
62. The derivative fails to exist at the corners of the
graph; that is, at t = 10, 15, 55, 60, 80 . The
derivative exists at all other points on the interval
(0,85) .
Instructor’s Resource Manual
65. The short-dash function has a tangent line with
zero slope at about x = 2.1 , where the solid
function is zero. The solid function has a tangent
line with zero slope at about x = 0.4, 1.2 and 3.5.
The long-dash function is zero at these points.
The graph shows that the solid function is
positive (negative) when the slope of the tangent
line of the short-dash function is positive
(negative). Also, the long-dash function is
positive (negative) when the slope of the tangent
line of the solid function is positive (negative).
Thus, the short-dash function is f, the solid
function is f ' = g , and the dash function is g ' .
66. Note that since x = 0 + x, f(x) = f(0 + x) = f(0)f(x),
hence f(0) = 1.
f ( a + h) – f ( a )
f ′(a ) = lim
h
h →0
f ( a ) f ( h) – f ( a )
= lim
h
h→0
f ( h) – 1
f (h) – f (0)
= f (a ) lim
= f (a) lim
h
h
h →0
h →0
= f (a ) f ′(0)
f ′ ( a) exists since f ′ (0 ) exists.
Section 2.2
105
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
67. If f is differentiable everywhere, then it is
continuous everywhere, so
lim − f ( x ) = lim− ( mx + b ) = 2 m + b = f (2) = 4
x→2
b. If f is an even function,
f (t ) − f ( x0 )
f ′(– x0 ) = lim
. Let u = –t, as
t + x0
t →− x0
x→ 2
f (−u ) − f ( x0 )
−u + x0
u → x0
and b = 4 – 2m.
For f to be differentiable everywhere,
f ( x) − f (2)
f ′(2) = lim
must exist.
x−2
x→2
f ( x) − f (2)
x2 − 4
= lim
= lim ( x + 2) = 4
x−2
x → 2+
x → 2+ x − 2
x → 2+
f ( x) − f (2)
mx + b − 4
lim
= lim
−
−
x
x−2
−
2
x→2
x→2
above, then f ′(− x0 ) = lim
f (u ) − f ( x0 )
f (u ) − f ( x0 )
= − lim
u − x0
u → x0 −(u − x0 )
u → x0
= − f ′ (x 0 ) = −m.
= lim
lim
mx + 4 − 2m − 4
m( x − 2)
= lim
=m
−
x−2
x−2
x →2
Thus m = 4 and b = 4 – 2(4) = –4
= lim
x → 2−
68.
69.
f ( x + h) – f ( x ) + f ( x ) – f ( x – h )
2h
h →0
⎡ f ( x + h ) – f ( x ) f ( x – h) – f ( x ) ⎤
= lim ⎢
+
⎥
2h
–2h
h→0 ⎣
⎦
1
f ( x + h) – f ( x ) 1
f [ x + (– h)] – f ( x)
= lim
+ lim
h
2 h →0
2 – h →0
–h
1
1
= f ′( x) + f ′( x ) = f ′( x).
2
2
For the converse, let f (x) = x . Then
h – –h
h–h
f s (0) = lim
= lim
=0
2h
h→0
h →0 2h
but f ′ (0) does not exist.
f s ( x) = lim
70. Say f(–x) = –f(x). Then
f (– x + h) – f (– x)
f ′(– x) = lim
h
h →0
– f ( x – h) + f ( x )
f ( x – h) – f ( x )
= lim
= – lim
h
h
h→0
h →0
f [ x + (– h)] − f ( x)
= lim
= f ′( x) so f ′ ( x ) is
–h
– h →0
an even function if f(x) is an odd function.
Say f(–x) = f(x). Then
f (– x + h) – f (– x)
f ′(– x) = lim
h
h →0
f ( x – h) – f ( x )
= lim
h
h→0
f [ x + (– h)] – f ( x)
= – lim
= – f ′( x) so f ′ (x)
–h
– h→0
is an odd function if f(x) is an even function.
71.
f (t ) − f ( x0 )
, so
t − x0
0
f (t ) − f (− x0 )
f ′(− x0 ) = lim
t − (− x0 )
t →− x
0
f ′( x0 ) = lim
t→x
f (t ) − f (− x0 )
= lim
t + x0
t →− x 0
a. If f is an odd function,
f (t ) − [− f (− x0 )]
f ′(− x0 ) = lim
t + x0
t →− x0
f (t ) + f (− x0 )
.
t + x0
Let u = –t. As t → − x0 , u → x 0 and so
f (−u ) + f ( x0 )
f ′(− x0 ) = lim
−u + x0
u → x0
a.
0< x<
8 ⎛ 8⎞
; ⎜ 0, ⎟
3 ⎝ 3⎠
b.
0≤ x≤
8 ⎡ 8⎤
; 0,
3 ⎢⎣ 3 ⎥⎦
c.
A function f(x) decreases as x increases when
f ′ ( x ) < 0.
a.
π < x < 6.8
c.
A function f(x) increases as x increases when
f ′ ( x ) > 0.
= lim
t →− x0
72.
− f (u ) + f ( x0 )
−[ f (u ) − f ( x0 )]
= lim
u
x
−
(
−
)
−(u − x0 )
u → x0
u → x0
0
= lim
f (u ) − f ( x0 )
= f ′( x0 ) = m.
u − x0
u → x0
= lim
106
Section 2.2
b. π < x < 6.8
Instructor’s Resource Manual
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2.3 Concepts Review
13. Dx ( x 4 + x3 + x 2 + x + 1)
1. the derivative of the second; second;
f (x) g′ (x ) + g(x) f ′( x)
2. denominator; denominator; square of the
g ( x) f ′( x) – f ( x) g ′( x)
denominator;
g 2 ( x)
3. nx
n– 1
h ; nx
= 3Dx ( x 4 ) – 2 Dx ( x3 ) – 5Dx ( x 2 )
= 12 x3 – 6 x 2 –10 x + π
15. Dx (πx 7 – 2 x5 – 5 x –2 )
= πDx ( x7 ) – 2 Dx ( x5 ) – 5 Dx ( x –2 )
1. Dx (2 x ) = 2 Dx ( x ) = 2 ⋅ 2 x = 4 x
2
= π(7 x6 ) – 2(5 x 4 ) – 5(–2 x –3 )
= 7 πx 6 –10 x 4 + 10 x –3
2. Dx (3x3 ) = 3Dx ( x3 ) = 3 ⋅ 3x 2 = 9 x 2
16. Dx ( x12 + 5 x −2 − πx −10 )
3. Dx (πx ) = πDx ( x) = π ⋅1 = π
4. Dx (πx ) = πDx ( x ) = π ⋅ 3 x = 3πx
5.
14. Dx (3x 4 – 2 x3 – 5 x 2 + πx + π2 )
= 3(4 x3 ) – 2(3 x 2 ) – 5(2 x) + π(1) + 0
Problem Set 2.3
3
= 4 x3 + 3 x 2 + 2 x + 1
+ πDx ( x) + Dx (π2 )
n –1
4. kL(f); L(f) + L(g); Dx
2
= Dx ( x 4 ) + Dx ( x3 ) + Dx ( x 2 ) + Dx ( x) + Dx (1)
3
2
2
Dx (2 x –2 ) = 2 Dx ( x –2 ) = 2(–2 x –3 ) = –4 x –3
6. Dx (–3 x –4 ) = –3Dx ( x –4 ) = –3(–4 x –5 ) = 12 x –5
⎛π⎞
7. Dx ⎜ ⎟ = πDx ( x –1 ) = π(–1x –2 ) = – πx –2
⎝x⎠
π
=– 2
x
⎛α ⎞
8. Dx ⎜ ⎟ = α Dx ( x –3 ) = α (–3x –4 ) = –3α x –4
⎝ x3 ⎠
3α
=–
x4
⎛ 100 ⎞
9. Dx ⎜
= 100 Dx ( x –5 ) = 100(–5 x –6 )
5 ⎟
⎝ x ⎠
500
= –500 x –6 = –
x6
⎛ 3α ⎞ 3α
3α
10. Dx ⎜
Dx ( x –5 ) =
=
(–5 x –6 )
5⎟
4
⎝ 4x ⎠ 4
15α –6
15α
=–
x =–
4
4 x6
11. Dx ( x 2 + 2 x) = Dx ( x 2 ) + 2 Dx ( x ) = 2 x + 2
= Dx ( x12 ) + 5Dx ( x −2 ) − πDx ( x −10 )
= 12 x11 + 5(−2 x −3 ) − π(−10 x −11 )
= 12 x11 − 10 x −3 + 10πx −11
⎛ 3
⎞
17. Dx ⎜ + x –4 ⎟ = 3Dx ( x –3 ) + Dx ( x –4 )
3
⎝x
⎠
9
= 3(–3 x –4 ) + (–4 x –5 ) = –
– 4 x –5
x4
18. Dx (2 x –6 + x –1 ) = 2 Dx ( x –6 ) + Dx ( x –1 )
= 2(–6 x –7 ) + (–1x –2 ) = –12 x –7 – x –2
⎛2 1 ⎞
19. Dx ⎜ –
= 2 Dx ( x –1 ) – Dx ( x –2 )
2⎟
x
x ⎠
⎝
2
2
= 2(–1x –2 ) – (–2 x –3 ) = –
+
2
x
x3
⎛ 3
1 ⎞
–3
–4
20. Dx ⎜ –
⎟ = 3 Dx ( x ) – Dx ( x )
3
x4 ⎠
⎝x
9
4
= 3(–3 x –4 ) – (–4 x –5 ) = –
+
4
x
x5
⎛ 1
⎞ 1
21. Dx ⎜ + 2 x ⎟ = Dx ( x –1 ) + 2 Dx ( x)
⎝ 2x
⎠ 2
1
1
= (–1x –2 ) + 2(1) = –
+2
2
2 x2
12. Dx (3x 4 + x3 ) = 3Dx ( x 4 ) + Dx ( x3 )
= 3(4 x3 ) + 3x 2 = 12 x3 + 3 x 2
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⎛ 2 2⎞ 2
⎛2⎞
22. Dx ⎜ – ⎟ = Dx ( x –1 ) – Dx ⎜ ⎟
⎝ 3x 3 ⎠ 3
⎝3⎠
2
2
= (–1x –2 ) – 0 = –
3
3x 2
23. Dx [ x( x 2 + 1)] = x Dx ( x 2 + 1) + ( x 2 + 1) Dx ( x)
= x(2 x) + ( x + 1)(1) = 3 x + 1
2
2
26. Dx [(–3 x + 2)2 ]
= (–3 x + 2) Dx (–3 x + 2) + (–3 x + 2) Dx (–3x + 2)
= (–3x + 2)(–3) + (–3x + 2)(–3) = 18x – 12
27. Dx [( x 2 + 2)( x3 + 1)]
= ( x 2 + 2) Dx ( x3 + 1) + ( x3 + 1) Dx ( x 2 + 2)
= ( x 2 + 2)(3x 2 ) + ( x3 + 1)(2 x)
24. Dx [3 x( x3 –1)] = 3 x Dx ( x3 –1) + ( x3 –1) Dx (3 x)
= 3x(3 x 2 ) + ( x3 –1)(3) = 12 x3 – 3
25. Dx [(2 x + 1) ]
= (2 x + 1) Dx (2 x + 1) + (2 x + 1) Dx (2 x + 1)
= (2 x + 1)(2) + (2 x + 1)(2) = 8 x + 4
2
= 3x 4 + 6 x 2 + 2 x 4 + 2 x
= 5x4 + 6 x2 + 2 x
28. Dx [( x 4 –1)( x 2 + 1)]
= ( x 4 –1) Dx ( x 2 + 1) + ( x 2 + 1) Dx ( x 4 –1)
= ( x 4 –1)(2 x) + ( x 2 + 1)(4 x3 )
= 2 x5 – 2 x + 4 x5 + 4 x3 = 6 x 5 + 4 x3 – 2 x
29. Dx [( x 2 + 17)( x3 – 3 x + 1)]
= ( x 2 + 17) Dx ( x3 – 3 x + 1) + ( x3 – 3x + 1) Dx ( x 2 + 17)
= ( x 2 + 17)(3 x 2 – 3) + ( x3 – 3x + 1)(2 x)
= 3x 4 + 48 x 2 – 51 + 2 x 4 – 6 x 2 + 2 x
= 5 x 4 + 42 x 2 + 2 x – 51
30. Dx [( x 4 + 2 x)( x3 + 2 x 2 + 1)] = ( x 4 + 2 x) Dx ( x3 + 2 x 2 + 1) + ( x3 + 2 x 2 + 1) Dx ( x 4 + 2 x)
= ( x 4 + 2 x)(3 x 2 + 4 x) + ( x3 + 2 x 2 + 1)(4 x3 + 2)
= 7 x6 + 12 x5 + 12 x3 + 12 x 2 + 2
31. Dx [(5 x 2 – 7)(3x 2 – 2 x + 1)] = (5 x 2 – 7) Dx (3x 2 – 2 x + 1) + (3x 2 – 2 x + 1) Dx (5 x 2 – 7)
= (5 x 2 – 7)(6 x – 2) + (3 x 2 – 2 x + 1)(10 x)
= 60 x3 – 30 x 2 – 32 x + 14
32. Dx [(3 x 2 + 2 x)( x 4 – 3 x + 1)] = (3 x 2 + 2 x) Dx ( x 4 – 3 x + 1) + ( x 4 – 3 x + 1) Dx (3x 2 + 2 x)
= (3 x 2 + 2 x)(4 x3 – 3) + ( x 4 – 3 x + 1)(6 x + 2)
= 18 x5 + 10 x 4 – 27 x 2 – 6 x + 2
⎛ 1 ⎞ (3x 2 + 1) Dx (1) – (1) Dx (3 x 2 + 1)
33. Dx ⎜
⎟=
(3 x 2 + 1)2
⎝ 3x2 + 1 ⎠
=
(3 x 2 + 1)(0) – (6 x)
(3 x + 1)
2
2
=–
6x
(3x + 1) 2
2
⎛ 2 ⎞ (5 x 2 –1) Dx (2) – (2) Dx (5 x 2 –1)
34. Dx ⎜
⎟=
(5 x 2 –1) 2
⎝ 5 x 2 –1 ⎠
=
(5 x 2 –1)(0) – 2(10 x)
2
(5 x –1)
108
Section 2.3
2
=–
20 x
(5 x 2 –1)2
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⎛
⎞ (4 x 2 – 3x + 9) Dx (1) – (1) Dx (4 x 2 – 3 x + 9)
1
35. Dx ⎜
⎟=
(4 x 2 – 3x + 9)2
⎝ 4 x 2 – 3x + 9 ⎠
=
=
(4 x 2 – 3x + 9)(0) – (8 x – 3)
(4 x – 3 x + 9)
−8 x + 3
2
=–
2
8x − 3
(4 x – 3x + 9)2
2
(4 x 2 – 3x + 9)2
⎛
⎞ (2 x3 – 3 x) Dx (4) – (4) Dx (2 x3 – 3 x)
4
36. Dx ⎜
⎟ =
(2 x3 – 3 x)2
⎝ 2 x3 – 3x ⎠
=
(2 x3 – 3 x)(0) – 4(6 x 2 – 3)
(2 x3 – 3 x)2
=
–24 x 2 + 12
(2 x3 – 3x) 2
⎛ x –1 ⎞ ( x + 1) Dx ( x –1) – ( x –1) Dx ( x + 1)
37. Dx ⎜
⎟=
⎝ x +1⎠
( x + 1)2
=
( x + 1)(1) – ( x –1)(1)
( x + 1)
2
=
2
( x + 1)2
⎛ 2 x –1 ⎞ ( x –1) Dx (2 x –1) – (2 x –1) Dx ( x –1)
38. Dx ⎜
⎟=
⎝ x –1 ⎠
( x –1) 2
=
( x –1)(2) – (2 x –1)(1)
( x –1)
2
=–
1
( x –1) 2
⎛ 2 x 2 – 1 ⎞ (3 x + 5) Dx (2 x 2 –1) – (2 x 2 –1) Dx (3 x + 5)
39. Dx ⎜
⎟ =
⎜ 3x + 5 ⎟
(3 x + 5)2
⎝
⎠
=
=
(3 x + 5)(4 x) – (2 x 2 – 1)(3)
(3x + 5) 2
6 x 2 + 20 x + 3
(3x + 5)2
⎛ 5 x – 4 ⎞ (3 x 2 + 1) Dx (5 x – 4) – (5 x – 4) Dx (3x 2 + 1)
40. Dx ⎜
⎟=
(3 x 2 + 1) 2
⎝ 3x2 + 1 ⎠
=
=
(3 x 2 + 1)(5) – (5 x – 4)(6 x)
(3x 2 + 1)2
−15 x 2 + 24 x + 5
(3x 2 + 1)2
⎛ 2 x 2 – 3 x + 1 ⎞ (2 x + 1) Dx (2 x 2 – 3x + 1) – (2 x 2 – 3x + 1) Dx (2 x + 1)
41. Dx ⎜
⎟ =
⎜ 2x +1 ⎟
(2 x + 1)2
⎝
⎠
=
=
(2 x + 1)(4 x – 3) – (2 x 2 – 3 x + 1)(2)
(2 x + 1)2
4 x2 + 4 x – 5
(2 x + 1) 2
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⎛ 5 x 2 + 2 x – 6 ⎞ (3 x – 1) Dx (5 x 2 + 2 x – 6) – (5 x 2 + 2 x – 6) Dx (3 x – 1)
42. Dx ⎜
⎟=
⎜
3 x – 1 ⎟⎠
(3 x – 1)2
⎝
=
=
(3 x – 1)(10 x + 2) – (5 x 2 + 2 x – 6)(3)
(3x – 1)2
15 x 2 – 10 x + 16
(3x – 1)2
⎛ x 2 – x + 1 ⎞ ( x 2 + 1) Dx ( x 2 – x + 1) – ( x 2 – x + 1) Dx ( x 2 + 1)
43. Dx ⎜
⎟ =
⎜ x2 + 1 ⎟
( x 2 + 1)2
⎝
⎠
=
=
( x 2 + 1)(2 x – 1) – ( x 2 – x + 1)(2 x)
( x 2 + 1)2
x2 – 1
( x 2 + 1)2
⎛ x 2 – 2 x + 5 ⎞ ( x 2 + 2 x – 3) Dx ( x 2 – 2 x + 5) – ( x 2 – 2 x + 5) Dx ( x 2 + 2 x – 3)
44. Dx ⎜
⎟=
⎜ x2 + 2 x – 3 ⎟
( x 2 + 2 x – 3) 2
⎝
⎠
=
=
45. a.
( x 2 + 2 x – 3)(2 x – 2) – ( x 2 – 2 x + 5)(2 x + 2)
( x 2 + 2 x – 3) 2
4 x 2 – 16 x – 4
( x 2 + 2 x – 3) 2
( f ⋅ g )′(0) = f (0) g ′(0) + g (0) f ′(0)
= 4(5) + (–3)(–1) = 23
b.
( f + g )′(0) = f ′(0) + g ′(0) = –1 + 5 = 4
c.
( f g )′(0) =
=
g 2 (0)
–3(–1) – 4(5)
(–3)
46. a.
g (0) f ′(0) – f (0) g ′(0)
2
=–
17
9
( f – g )′(3) = f ′(3) – g ′(3) = 2 – (–10) = 12
b.
( f ⋅ g )′(3) = f (3) g ′(3) + g (3) f ′(3) = 7(–10) + 6(2) = –58
c.
( g f )′(3) =
f (3) g ′(3) – g (3) f ′(3)
2
f (3)
=
7(–10) – 6(2)
(7)
2
=–
82
49
47. Dx [ f ( x )]2 = Dx [ f ( x ) f ( x)]
= f ( x) Dx [ f ( x)] + f ( x) Dx [ f ( x)]
= 2 ⋅ f ( x ) ⋅ Dx f ( x )
48. Dx [ f ( x) g ( x)h( x)] = f ( x) Dx [ g ( x)h( x)] + g ( x)h( x) Dx f ( x)
= f ( x)[ g ( x) Dx h( x) + h( x) Dx g ( x)] + g ( x)h( x) Dx f ( x)
= f ( x) g ( x) Dx h( x ) + f ( x)h( x) Dx g ( x) + g ( x)h( x) Dx f ( x)
110
Section 2.3
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54. Proof #1:
49. Dx ( x 2 – 2 x + 2) = 2 x – 2
At x = 1: m tan = 2 (1) – 2 = 0
Tangent line: y = 1
Dx [ f ( x) − g ( x) ] = Dx [ f ( x) + (−1) g ( x) ]
= Dx [ f ( x) ] + Dx [ (−1) g ( x) ]
⎛ 1 ⎞ ( x + 4) Dx (1) – (1) Dx ( x + 4)
50. Dx ⎜
⎟=
( x 2 + 4)2
⎝ x2 + 4 ⎠
2
=
( x 2 + 4)(0) – (2 x)
( x 2 + 4)2
At x = 1: mtan = −
=–
2x
( x 2 + 4) 2
2(1)
=–
2
25
(1 + 4)
1
2
Tangent line: y – = – ( x –1)
5
25
2
7
y = – x+
25
25
2
2
51. Dx ( x3 – x 2 ) = 3 x 2 – 2 x
The tangent line is horizontal when m tan = 0:
mtan = 3x 2 – 2 x = 0
x(3 x − 2) = 0
2
x = 0 and x =
3
4 ⎞
⎛2
(0, 0) and ⎜ , – ⎟
⎝ 3 27 ⎠
⎛1
⎞
52. Dx ⎜ x3 + x 2 – x ⎟ = x 2 + 2 x –1
3
⎝
⎠
mtan = x + 2 x –1 = 1
2
x2 + 2 x – 2 = 0
–2 ± 4 – 4(1)(–2) –2 ± 12
x=
=
2
2
= –1 – 3, –1 + 3
x = –1 ± 3
5
5
⎛
⎞ ⎛
⎞
⎜ −1 + 3, − 3 ⎟ , ⎜ −1 − 3, + 3 ⎟
3
3
⎝
⎠ ⎝
⎠
53.
= Dx f ( x) − Dx g ( x)
2
y = 100 / x5 = 100 x −5
y ' = −500 x −6
Set y ' equal to −1 , the negative reciprocal of
the slope of the line y = x . Solving for x gives
x = ±5001/ 6 ≈ ±2.817
y = ±100(500)−5 / 6 ≈ ±0.563
Proof #2:
Let F ( x) = f ( x) − g ( x) . Then
F '( x) = lim
[ f ( x + h) − g ( x + h) ] − [ f ( x ) − g ( x ) ]
h →0
h
⎡ f ( x + h) − f ( x ) g ( x + h ) − g ( x ) ⎤
= lim ⎢
−
⎥
h →0 ⎣
h
h
⎦
= f '( x) − g '( x)
55. a.
Dt (–16t 2 + 40t + 100) = –32t + 40
v = –32(2) + 40 = –24 ft/s
b. v = –32t + 40 = 0
t=5 s
4
56. Dt (4.5t 2 + 2t ) = 9t + 2
9t + 2 = 30
28
t=
s
9
57. mtan = Dx (4 x – x 2 ) = 4 – 2 x
The line through (2,5) and (x 0 , y0 ) has slope
y0 − 5
.
x0 − 2
4 – 2 x0 =
4 x0 – x0 2 – 5
x0 – 2
–2 x02 + 8 x0 – 8 = – x0 2 + 4 x0 – 5
x0 2 – 4 x0 + 3 = 0
( x0 – 3)( x0 –1) = 0
x 0 = 1, x0 = 3
At x 0 = 1: y0 = 4(1) – (1)2 = 3
mtan = 4 – 2(1) = 2
Tangent line: y – 3 = 2(x – 1); y = 2x + 1
At x0 = 3 : y0 = 4(3) – (3)2 = 3
mtan = 4 – 2(3) = –2
Tangent line: y – 3 = –2(x – 3); y = –2x + 9
The points are (2.817,0.563) and
(−2.817,−0.563) .
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58. Dx ( x 2 ) = 2 x
The line through (4, 15) and ( x0 , y0 ) has slope
61. The watermelon has volume
of the rind is
y0 − 15
. If (x 0 , y0 ) is on the curve y = x 2 , then
x0 − 4
mtan = 2 x0 =
4 3
πr ; the volume
3
3
V=
x02 –15
.
x0 – 4
r ⎞
4 3 4 ⎛
271 3
πr – π ⎜ r – ⎟ =
πr .
3
3 ⎝ 10 ⎠
750
At the end of the fifth week r = 10, so
271 2 271
542π
DrV =
πr =
π(10)2 =
≈ 340 cm3
250
250
5
per cm of radius growth. Since the radius is
growing 2 cm per week, the volume of the rind is
542π
(2) ≈ 681 cm3 per
growing at the rate of
5
week.
2 x0 2 – 8 x0 = x02 –15
x0 2 – 8 x0 + 15 = 0
( x0 – 3)( x0 – 5) = 0
At x0 = 3 : y0 = (3)2 = 9
She should shut off the engines at (3, 9). (At
x 0 = 5 she would not go to (4, 15) since she is
moving left to right.)
2.4 Concepts Review
59. Dx (7 – x ) = –2 x
The line through (4, 0) and ( x0 , y0 ) has
2
slope
1.
y0 − 0
. If the fly is at ( x0 , y0 ) when the
x0 − 4
spider sees it, then mtan = –2 x0 =
2
7 – x0 – 0
.
x0 – 4
–2 x02 + 8 x0 = 7 – x02
x 02 – 8x 0 + 7 = 0
( x0 – 7)( x0 –1) = 0
At x0 = 1: y0 = 6
d = (4 – 1)2 + (0 – 6) 2 = 9 + 36 = 45 = 3 5
≈ 6. 7
They are 6.7 units apart when they see each
other.
1
⎛ 1⎞
60. P(a, b) is ⎜ a, ⎟ . Dx y = –
so the slope of
a
⎝
⎠
x2
1
the tangent line at P is – . The tangent line is
a2
1
1
1
y– =–
( x – a ) or y = –
( x – 2a ) which
2
a
a
a2
has x-intercept (2a, 0).
1
1
d (O, P ) = a 2 + , d ( P, A) = (a – 2a )2 +
2
a
a2
= a2 +
1
= d (O, P ) so AOP is an isosceles
a2
triangle. The height of AOP is a while the base,
1
OA has length 2a, so the area is (2 a)(a) = a2 .
2
112
Section 2.4
sin( x + h) – sin( x)
h
2. 0; 1
3. cos x; –sin x
4. cos
π 1
3 1⎛
π⎞
= ;y–
= ⎜x– ⎟
3 2
2
2⎝
3⎠
Problem Set 2.4
1. Dx(2 sin x + 3 cos x) = 2 Dx(sin x) + 3 Dx(cos x)
= 2 cos x – 3 sin x
2. Dx (sin 2 x) = sin x Dx (sin x ) + sin x Dx (sin x)
= sin x cos x + sin x cos x = 2 sin x cos x = sin 2x
3. Dx (sin 2 x + cos 2 x) = Dx (1) = 0
4. Dx (1 – cos 2 x) = Dx (sin 2 x)
= sin x Dx (sin x) + sin x Dx (sin x)
= sin x cos x + sin x cos x
= 2 sin x cos x = sin 2x
⎛ 1 ⎞
5. Dx (sec x) = Dx ⎜
⎟
⎝ cos x ⎠
cos x Dx (1) – (1) Dx (cos x )
=
cos 2 x
sin x
1 sin x
=
=
⋅
= sec x tan x
2
x cos x
cos
cos x
Instructor’s Resource Manual
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⎛ 1 ⎞
6. Dx (csc x) = Dx ⎜
⎟
⎝ sin x ⎠
sin x Dx (1) − (1) Dx (sin x )
=
sin 2 x
– cos x
–1 cos x
=
=
⋅
= – csc x cot x
2
sin
x sin x
sin x
⎛ sin x ⎞
7. Dx (tan x) = Dx ⎜
⎟
⎝ cos x ⎠
cos x Dx (sin x) − sin x Dx (cos x)
=
cos 2 x
=
cos 2 x + sin 2 x
cos2 x
1
=
cos 2 x
=
− sin x – cos x
2
2
=
sin x
=
cos x(cos x – sin x) – (– sin 2 x – sin x cos x)
cos 2 x + sin 2 x
cos2 x
=
cos 2 x
cos 2 x
1
cos x
− cos x −
−
sin x
sin x sin 2 x
12. Dx ( sin x tan x ) = sin xDx [ tan x ] + tan xDx [sin x ]
2
2
cos 2 x
1
tan 2 x
⎛
sin 2 x sin x
1 ⎞ ⎛ sin 2 x ⎞
–
–
= ⎜ sin x –
⎟ ÷⎜
⎟
2
cos x cos x cos x ⎠ ⎝ cos 2 x ⎠
⎝
= sin x ( − sin x ) + cos x ( cos x ) = cos 2 x − sin 2 x
⎛ sin x + cos x ⎞
9. Dx ⎜
⎟
cos x
⎝
⎠
cos x Dx (sin x + cos x) − (sin x + cos x) Dx (cos x)
=
cos 2 x
=
tan x(cos x – sin x) – sec2 x(sin x + cos x)
11. Dx ( sin x cos x ) = sin xDx [ cos x ] + cos xDx [sin x ]
–(sin x + cos x)
sin x
1
=–
= – csc2 x
2
sin x
=
=
= sec2 x
2
⎛ sin x + cos x ⎞
Dx ⎜
⎟
tan x
⎝
⎠
tan x Dx (sin x + cos x) − (sin x + cos x) Dx (tan x )
=
tan 2 x
⎛
sin 2 x sin x
1 ⎞⎛ cos 2 x ⎞
= ⎜ sin x −
−
−
⎟⎜
⎟
2
cos x cos x cos x ⎠⎝ sin 2 x ⎠
⎝
⎛ cos x ⎞
8. Dx (cot x) = Dx ⎜
⎟
⎝ sin x ⎠
sin x Dx (cos x) − cos x Dx (sin x)
=
sin 2 x
2
10.
= sec2 x
(
)
= sin x sec2 x + tan x ( cos x )
⎛ 1 ⎞ sin x
= sin x ⎜
( cos x )
⎟+
⎝ cos 2 x ⎠ cos x
= tan x sec x + sin x
⎛ sin x ⎞ xDx ( sin x ) − sin xDx ( x )
13. Dx ⎜
⎟=
x2
⎝ x ⎠
x cos x − sin x
=
x2
⎛ 1 − cos x ⎞ xDx (1 − cos x ) − (1 − cos x ) Dx ( x )
14. Dx ⎜
⎟=
x
x2
⎝
⎠
x sin x + cos x − 1
=
x2
15. Dx ( x 2 cos x) = x 2 Dx (cos x) + cos x Dx ( x 2 )
= − x 2 sin x + 2 x cos x
⎛ x cos x + sin x ⎞
16. Dx ⎜
⎟
x2 + 1
⎝
⎠
=
=
=
( x 2 + 1) Dx ( x cos x + sin x) − ( x cos x + sin x) Dx ( x 2 + 1)
( x 2 + 1) 2
( x 2 + 1)(– x sin x + cos x + cos x) – 2 x( x cos x + sin x)
( x 2 + 1)2
– x3 sin x – 3 x sin x + 2 cos x
( x 2 + 1) 2
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y = tan 2 x = (tan x)(tan x)
17.
b. Dt(20 sin t) = 20 cos t
π
π
At t = : rate = 20 cos = 10 3 ≈ 17. 32 ft/s
6
6
Dx y = (tan x)(sec 2 x ) + (tan x)(sec 2 x)
= 2 tan x sec 2 x
18.
2
Dx y = (sec 2 x) sec x tan x + (sec x) Dx (sec2 x)
25.
When y = 0 , y = tan 0 = 0 and y ' = sec 2 0 = 1 .
The tangent line at x = 0 is y = x .
= sec3 x tan x + 2sec3 x tan x
= 3sec2 x tan x
26.
= 2 tan x sec 2 x
Now, sec2 x is never 0, but tan x = 0 at
x = kπ where k is an integer.
27.
π
At x = : mtan = –2;
4
y=1
= 9 ⎡sin 2 x − cos 2 x ⎤
⎣
⎦
= 9 [ − cos 2 x ]
The tangent line is horizontal when y ' = 0 or, in
this case, where cos 2 x = 0 . This occurs when
21. Dx sin 2 x = Dx (2sin x cos x)
= 2 ⎣⎡sin x Dx cos x + cos x Dx sin x ⎦⎤
x=
= −2sin x + 2 cos x
2
2
28.
22. Dx cos 2 x = Dx (2 cos x − 1) = 2 Dx cos x − Dx 1
2
2
= −2sin x cos x
23. Dt (30sin 2t ) = 30 Dt (2sin t cos t )
(
= 30 −2sin 2 t + 2 cos 2 t
)
= 60 cos 2t
30sin 2t = 15
1
sin 2t =
2
π
→ t=
π
6
12
π
⎛ π ⎞
At t = ; 60 cos ⎜ 2 ⋅ ⎟ = 30 3 ft/sec
12
⎝ 12 ⎠
The seat is moving to the left at the rate of 30 3
ft/s.
24. The coordinates of the seat at time t are
(20 cos t, 20 sin t).
a.
π
π⎞
⎛
⎜ 20 cos , 20sin ⎟ = (10 3, 10)
6
6⎠
⎝
≈ (17.32, 10)
Section 2.4
y = 9sin x cos x
y ' = 9 [sin x(− sin x) + cos x(cos x) ]
π⎞
⎛
Tangent line: y –1 = –2 ⎜ x – ⎟
4⎠
⎝
2t =
y = tan 2 x = (tan x)(tan x)
y ' = (tan x)(sec 2 x) + (tan x)(sec2 x)
19. Dx(cos x) = –sin x
At x = 1: mtan = – sin1 ≈ –0.8415
y = cos 1 ≈ 0.5403
Tangent line: y – 0.5403 = –0.8415(x – 1)
20. Dx (cot x) = – csc2 x
y = tan x
y ' = sec2 x
= sec3 x tan x + sec x(sec x ⋅ sec x tan x
+ sec x ⋅ sec x tan x)
114
The fastest rate 20 cos t can obtain is
20 ft/s.
c.
y = sec x = (sec x)(sec x)
3
π
4
+k
π
2
where k is an integer.
f ( x) = x − sin x
f '( x) = 1 − cos x
f '( x) = 0 when cos x = 1 ; i.e. when x = 2kπ
where k is an integer.
f '( x) = 2 when x = (2k + 1)π where k is an
integer.
29. The curves intersect when 2 sin x = 2 cos x,
sin x = cos x at x = π for 0 < x < π .
4
2
π
Dx ( 2 sin x) = 2 cos x ; 2 cos = 1
4
π
Dx ( 2 cos x) = – 2 sin x ; − 2 sin = −1
4
1(–1) = –1 so the curves intersect at right angles.
30. v = Dt (3sin 2t ) = 6 cos 2t
At t = 0: v = 6 cm/s
π
t = : v = −6 cm/s
2
t = π : v = 6 cm/s
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sin( x + h) 2 – sin x 2
h
h→0
31. Dx (sin x 2 ) = lim
sin( x 2 + 2 xh + h 2 ) – sin x 2
h
h→0
sin x 2 cos(2 xh + h 2 ) + cos x 2 sin(2 xh + h 2 ) – sin x 2
sin x 2 [cos(2 xh + h 2 ) – 1] + cos x 2 sin(2 xh + h 2 )
= lim
= lim
h →0
h
h
h→0
= lim
2
⎡
cos(2 xh + h 2 ) – 1
2 sin(2 xh + h ) ⎤
cos
x
= lim(2 x + h) ⎢sin x 2
+
⎥ = 2 x(sin x 2 ⋅ 0 + cos x 2 ⋅1) = 2 x cos x 2
h →0
2 xh + h 2
2 xh + h 2 ⎥⎦
⎢⎣
sin(5( x + h)) – sin 5 x
h
h →0
sin(5 x + 5h) – sin 5 x
= lim
h
h →0
sin 5 x cos 5h + cos 5 x sin 5h – sin 5 x
= lim
h
h →0
cos 5h – 1
sin 5h ⎤
⎡
= lim ⎢sin 5 x
+ cos 5 x
h
h ⎥⎦
h→0 ⎣
cos 5h – 1
sin 5h ⎤
⎡
= lim ⎢5sin 5 x
+ 5cos 5 x
5h
5h ⎥⎦
h→0 ⎣
= 0 + 5cos 5 x ⋅1 = 5cos 5 x
32. Dx (sin 5 x) = lim
33. f(x) = x sin x
a.
34.
f ( x ) = cos3 x − 1.25cos 2 x + 0.225
x0 ≈ 1.95
f ′ (x 0 ) ≈ –1. 24
2.5 Concepts Review
1. Dt u; f ′( g (t )) g ′(t )
2. Dv w; G ′( H ( s )) H ′( s )
b. f(x) = 0 has 6 solutions on [π , 6π ]
f ′ (x) = 0 has 5 solutions on [π , 6π ]
c.
f(x) = x sin x is a counterexample.
Consider the interval [ 0, π ] .
f ( −π ) = f (π ) = 0 and f ( x ) = 0 has
exactly two solutions in the interval (at 0 and
π ). However, f ' ( x ) = 0 has two solutions
in the interval, not 1 as the conjecture
indicates it should have.
d. The maximum value of f ( x) – f ′( x) on
[π , 6π ] is about 24.93.
3. ( f ( x)) 2 ;( f ( x)) 2
2
2
4. 2 x cos( x );6(2 x + 1)
Problem Set 2.5
1. y = u15 and u = 1 + x
Dx y = Du y ⋅ Dx u
= (15u14 )(1)
= 15(1 + x )14
2. y = u5 and u = 7 + x
Dx y = Du y ⋅ Dx u
= (5u 4 )(1)
= 5(7 + x)4
3. y = u5 and u = 3 – 2x
Dx y = Du y ⋅ Dx u
= (5u 4 )(–2) = –10(3 – 2 x) 4
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4. y = u7 and u = 4 + 2 x 2
Dx y = Du y ⋅ Dx u
10. y = cos u and u = 3 x 2 – 2 x
Dx y = Du y ⋅ Dx u
= (–sin u)(6x – 2)
= (7u 6 )(4 x) = 28 x(4 + 2 x 2 )6
= –(6 x – 2) sin(3x 2 – 2 x)
5. y = u11 and u = x3 – 2 x 2 + 3 x + 1
Dx y = Du y ⋅ Dx u
11. y = u 3 and u = cos x
Dx y = Du y ⋅ Dx u
= (11u10 )(3x 2 – 4 x + 3)
= (3u 2 )(– sin x)
= 11(3 x 2 – 4 x + 3)( x3 – 2 x 2 + 3x + 1)10
= –3sin x cos 2 x
6. y = u –7 and u = x 2 – x + 1
Dx y = Du y ⋅ Dx u
12. y = u 4 , u = sin v, and v = 3 x 2
Dx y = Du y ⋅ Dv u ⋅ Dx v
= (–7u –8 )(2 x – 1)
= (4u 3 )(cos v )(6 x)
= –7(2 x – 1)( x 2 – x + 1) –8
= 24 x sin 3 (3 x 2 ) cos(3 x 2 )
7. y = u –5 and u = x + 3
Dx y = Du y ⋅ Dx u
x +1
x –1
Dx y = Du y ⋅ Dx u
13. y = u 3 and u =
= (–5u –6 )(1) = –5( x + 3) –6 = –
5
( x + 3)6
= (3u 2 )
8. y = u and u = 3x + x – 3
Dx y = Du y ⋅ Dx u
–9
( x –1) 2
2
⎛ x +1⎞
= 3⎜
⎟
⎝ x –1⎠
= (–9u –10 )(6 x + 1)
= –9(6 x + 1)(3 x 2 + x – 3) –10
=–
( x –1) Dx ( x + 1) – ( x + 1) Dx ( x –1)
–2
⎜⎜
2
⎝ ( x – 1)
14. y = u −3 and u =
9(6 x + 1)
⎞
6( x + 1) 2
⎟⎟ = −
( x – 1)4
⎠
x−2
x−π
Dx y = Du y ⋅ Dx u
(3 x 2 + x – 3)10
= (−3u −4 ) ⋅
9. y = sin u and u = x + x
Dx y = Du y ⋅ Dx u
= (cos u)(2x + 1)
2
( x − π) Dx ( x − 2) − ( x − 2) Dx ( x − π)
⎛ x−2⎞
= −3 ⎜
⎟
⎝ x−π⎠
= (2 x + 1) cos( x + x)
2
15. y = cos u and u =
2⎛
( x − π) 2
−4
(2 − π)
( x − π) 2
= −3
( x − π)2
( x − 2)4
(2 − π)
3x 2
x+2
Dx y = Du y ⋅ Dx u = (– sin u )
( x + 2) Dx (3 x 2 ) – (3x 2 ) Dx ( x + 2)
( x + 2) 2
⎛ 3x 2 ⎞ ( x + 2)(6 x) – (3x 2 )(1)
⎛ 3x 2 ⎞
3 x 2 + 12 x
= – sin ⎜
=–
sin ⎜
⎟
⎟
⎜ x+2⎟
⎜
⎟
( x + 2)2
( x + 2)2
⎝
⎠
⎝ x+2⎠
16. y = u 3 , u = cos v, and v =
x2
1– x
Dx y = Du y ⋅ Dv u ⋅ Dx v = (3u 2 )(− sin v)
(1 – x) Dx ( x 2 ) – ( x 2 ) Dx (1 − x)
(1 – x) 2
⎛ x2 ⎞ ⎛ x2 ⎞
⎛ x 2 ⎞ ⎛ x 2 ⎞ (1 – x)(2 x) – ( x 2 )(–1)
–3(2 x – x 2 )
=
= –3cos 2 ⎜
cos 2 ⎜
⎟ sin ⎜
⎟
⎟ sin ⎜
⎟
⎜1– x ⎟ ⎜1– x ⎟
⎜1– x ⎟ ⎜1– x ⎟
(1 – x)2
(1 – x)2
⎝
⎠ ⎝
⎠
⎝
⎠ ⎝
⎠
116
Section 2.5
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17. Dx [(3 x – 2)2 (3 – x 2 ) 2 ] = (3 x – 2)2 Dx (3 – x 2 )2 + (3 – x 2 ) 2 Dx (3 x – 2)2
= (3 x – 2)2 (2)(3 – x 2 )(–2 x) + (3 – x 2 ) 2 (2)(3 x – 2)(3)
= 2(3 x − 2)(3 − x 2 )[(3 x − 2)(−2 x) + (3 − x 2 )(3)] = 2(3 x − 2)(3 − x 2 )(9 + 4 x − 9 x 2 )
18. Dx [(2 – 3x 2 )4 ( x 7 + 3)3 ] = (2 – 3 x 2 )4 Dx ( x 7 + 3)3 + ( x 7 + 3)3 Dx (2 – 3x 2 ) 4
= (2 – 3 x 2 )4 (3)( x 7 + 3) 2 (7 x 6 ) + ( x 7 + 3)3 (4)(2 – 3 x 2 )3 (–6 x) = 3x (3 x 2 – 2)3 ( x 7 + 3) 2 (29 x 7 – 14 x5 + 24)
⎡ ( x + 1)2 ⎤ (3x – 4) Dx ( x + 1)2 – ( x + 1)2 Dx (3x – 4)
(3 x – 4)(2)( x + 1)(1) – ( x + 1)2 (3) 3 x 2 – 8 x – 11
19. Dx ⎢
=
=
⎥=
(3x – 4) 2
(3x – 4) 2
(3 x – 4)2
⎣⎢ 3x – 4 ⎦⎥
=
( x + 1)(3 x − 11)
(3x − 4)2
⎡ 2 x – 3 ⎤ ( x 2 + 4) 2 Dx (2 x – 3) – (2 x – 3) Dx ( x 2 + 4)2
20. Dx ⎢
⎥=
2
2
( x 2 + 4) 4
⎣⎢ ( x + 4) ⎦⎥
=
( x 2 + 4) 2 (2) – (2 x – 3)(2)( x 2 + 4)(2 x)
( x 2 + 4) 4
(
)(
) (
=
)
−6 x 2 + 12 x + 8
( x 2 + 4)3
(
′
21. y ′ = 2 x 2 + 4 x 2 + 4 = 2 x 2 + 4 (2 x ) = 4 x x 2 + 4
)
22. y ′ = 2(x + sin x )(x + sin x )′ = 2(x + sin x )(1 + cos x )
3
2
⎛ 3t – 2 ⎞
⎛ 3t – 2 ⎞ (t + 5) Dt (3t – 2) – (3t – 2) Dt (t + 5)
23. Dt ⎜
`⎟ = 3 ⎜
⎟
t
5
+
⎝
⎠
⎝ t +5 ⎠
(t + 5)2
2
51(3t – 2)2
⎛ 3t – 2 ⎞ (t + 5)(3) – (3t – 2)(1)
=
= 3⎜
⎟
⎝ t +5 ⎠
(t + 5) 4
(t + 5)2
⎛ s 2 – 9 ⎞ ( s + 4) Ds ( s 2 – 9) – ( s 2 – 9) Ds ( s + 4)
( s + 4)(2s ) – ( s 2 – 9)(1) s 2 + 8s + 9
24. Ds ⎜
=
=
⎟=
⎜ s+4 ⎟
( s + 4)2
( s + 4) 2
( s + 4)2
⎝
⎠
d ⎛ (3t − 2)3 ⎞
25.
⎜
⎟=
dt ⎜⎝ t + 5 ⎟⎠
=
26.
(t + 5)
d
d
(3t − 2)3 − (3t − 2)3 (t + 5)
(t + 5)(3)(3t – 2)2 (3) – (3t – 2)3 (1)
dt
dt
=
(t + 5)2
(t + 5) 2
(6t + 47)(3t – 2)2
(t + 5) 2
d
(sin 3 θ ) = 3sin 2 θ cos θ
dθ
3
2
2
dy d ⎛ sin x ⎞
⎛ sin x ⎞ d sin x
⎛ sin x ⎞
= ⎜
= 3⎜
27.
⎟ = 3⎜
⎟ ⋅
⎟ ⋅
dx dx ⎝ cos 2 x ⎠
⎝ cos 2 x ⎠ dx cos 2 x
⎝ cos 2 x ⎠
(cos 2 x)
d
d
(sin x) − (sin x) (cos 2 x)
dx
dx
cos 2 2 x
2
2
3
⎛ sin x ⎞ cos x cos 2 x + 2sin x sin 2 x 3sin x cos x cos 2 x + 6sin x sin 2 x
= 3⎜
=
⎟
⎝ cos 2 x ⎠
cos 2 2 x
cos 4 2 x
=
3(sin 2 x)(cos x cos 2 x + 2sin x sin 2 x)
cos 4 2 x
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28.
dy d
d
d
= [sin t tan(t 2 + 1)] = sin t ⋅ [tan(t 2 + 1)] + tan(t 2 + 1) ⋅ (sin t )
dt dt
dt
dt
= (sin t )[sec 2 (t 2 + 1)](2t ) + tan(t 2 + 1) cos t = 2t sin t sec2 (t 2 + 1) + cos t tan(t 2 + 1)
2
29.
2
⎛ x 2 + 1 ⎞ ( x + 2) Dx ( x 2 + 1) – ( x 2 + 1) Dx ( x + 2)
⎛ x 2 + 1 ⎞ 2 x 2 + 4 x – x 2 – 1 3( x 2 + 1)2 ( x 2 + 4 x – 1)
f ′( x) = 3 ⎜
=
= 3⎜
⎟
⎟
⎜ x+2 ⎟
⎜ x+2 ⎟
( x + 2) 2
( x + 2) 2
( x + 2)4
⎝
⎠
⎝
⎠
f ′(3) = 9.6
30. G ′(t ) = (t 2 + 9)3 Dt (t 2 – 2)4 + (t 2 – 2) 4 Dt (t 2 + 9)3 = (t 2 + 9)3 (4)(t 2 – 2)3 (2t ) + (t 2 – 2) 4 (3)(t 2 + 9)2 (2t )
= 2t (7t 2 + 30)(t 2 + 9)2 (t 2 – 2)3
G ′(1) = –7400
31. F ′(t ) = [cos(t 2 + 3t + 1)](2t + 3) = (2t + 3) cos(t 2 + 3t + 1) ;
F ′(1) = 5cos 5 ≈ 1.4183
32. g ′( s ) = (cos πs ) Ds (sin 2 πs ) + (sin 2 πs ) Ds (cos πs ) = (cos πs )(2sin πs )(cos πs )(π) + (sin 2 πs )(– sin πs )(π)
= π sin πs[2 cos 2 πs – sin 2 πs ]
⎛1⎞
g ′ ⎜ ⎟ = –π
⎝2⎠
33. Dx [sin 4 ( x 2 + 3x)] = 4sin 3 ( x 2 + 3x) Dx sin( x 2 + 3x) = 4sin 3 ( x 2 + 3 x) cos( x 2 + 3 x) Dx ( x 2 + 3 x)
= 4sin 3 ( x 2 + 3 x) cos( x 2 + 3x)(2 x + 3) = 4(2 x + 3) sin 3 ( x 2 + 3x) cos( x 2 + 3 x)
34. Dt [cos5 (4t – 19)] = 5cos 4 (4t – 19) Dt cos(4t – 19) = 5cos 4 (4t – 19)[– sin(4t – 19)]Dt (4t – 19)
4
= –5cos 4 (4t – 19) sin(4t – 19)(4) = –20 cos (4t – 19) sin(4t – 19)
35. Dt [sin 3 (cos t )] = 3sin 2 (cos t ) Dt sin(cos t ) = 3sin 2 (cos t ) cos(cos t ) Dt (cos t )
2
= 3sin 2 (cos t ) cos(cos t )(– sin t ) = –3sin t sin (cos t ) cos(cos t )
⎡
⎛ u + 1 ⎞⎤
⎛ u +1⎞
⎛ u + 1 ⎞⎤ ⎛ u + 1 ⎞
3 ⎛ u +1⎞
3 ⎛ u +1⎞ ⎡
36 . Du ⎢cos4 ⎜
⎟ ⎥ = 4 cos ⎜
⎟ Du cos ⎜
⎟ = 4 cos ⎜
⎟ ⎢ – sin ⎜
⎟ ⎥ Du ⎜
⎟
⎝ u –1 ⎠ ⎦
⎝ u –1 ⎠
⎝ u –1 ⎠
⎝ u –1 ⎠ ⎣
⎝ u –1 ⎠ ⎦ ⎝ u –1 ⎠
⎣
8
⎛ u +1⎞ ⎛ u +1⎞
⎛ u + 1 ⎞ ⎛ u + 1 ⎞ (u –1) Du (u + 1) – (u + 1) Du (u –1)
cos3 ⎜
=
= –4 cos3 ⎜
⎟ sin ⎜
⎟
⎟ sin ⎜
⎟
2
2
u
–1
u
–1
⎝ u –1 ⎠ ⎝ u –1 ⎠
⎝
⎠ ⎝
⎠
(u –1)
(u –1)
37. Dθ [cos 4 (sin θ 2 )] = 4 cos3 (sin θ 2 ) Dθ cos(sin θ 2 ) = 4 cos3 (sin θ 2 )[– sin(sin θ 2 )]Dθ (sin θ 2 )
= –4 cos3 (sin θ 2 ) sin(sin θ 2 )(cosθ 2 ) Dθ (θ 2 ) = –8θ cos3 (sin θ 2 ) sin(sin θ 2 )(cos θ 2 )
38. Dx [ x sin 2 (2 x)] = x Dx sin 2 (2 x) + sin 2 (2 x) Dx x = x[2sin(2 x) Dx sin(2 x)] + sin 2 (2 x)(1)
= x[2sin(2 x ) cos(2 x) Dx (2 x)] + sin 2 (2 x) = x[4sin(2 x) cos(2 x)] + sin 2 (2 x) = 2 x sin(4 x) + sin 2 (2 x)
39. Dx {sin[cos(sin 2 x)]} = cos[cos(sin 2 x)]Dx cos(sin 2 x) = cos[cos(sin 2 x)][– sin(sin 2 x)]Dx (sin 2 x)
= – cos[cos(sin 2 x)]sin(sin 2 x)(cos 2 x) Dx (2 x) = –2 cos[cos(sin 2 x)]sin(sin 2 x)(cos 2 x)
40. Dt {cos 2 [cos(cos t )]} = 2 cos[cos(cos t )]Dt cos[cos(cos t )] = 2 cos[cos(cos t )]{– sin[cos(cos t )]}Dt cos(cos t )
= –2 cos[cos(cos t )]sin[cos(cos t )][– sin(cos t )]Dt (cos t ) = 2 cos[cos(cos t )]sin[cos(cos t )]sin(cos t )(– sin t )
= –2sin t cos[cos(cos t )]sin[cos(cos t )]sin(cos t )
118
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41. ( f + g )′(4) = f ′(4) + g ′(4)
≈
42.
(f
53.
1 3
+ ≈2
2 2
′
′
− 2g ) ( 2) = f ′ ( 2) − ( 2g ) ( 2)
54.
= f ′ ( 2) − 2g′ ( 2)
= 1 − 2 ( 0) = 1
d
d
cos ( F ( x ) ) = − sin ( F ( x ) ) F ( x )
dx
dx
= − F ′ ( x ) sin ( F ( x ) )
55. Dx ⎣⎡ tan ( F ( 2 x ) ) ⎦⎤ = sec 2 ( F ( 2 x ) ) Dx ⎡⎣ F ( 2 x ) ⎤⎦
43. ( fg )′ (2 ) = ( fg ′ + gf ′)(2) = 2(0) + 1(1) = 1
44. ( f g )′(2) =
d
d
F ( cos x ) = F ′ ( cos x ) ( cos x )
dx
dx
= − sin xF ′ ( cos x )
= sec2 ( F ( 2 x ) ) × F ′ ( 2 x ) × Dx [ 2 x ]
= 2 F ′ ( 2 x ) sec2 ( F ( 2 x ) )
g (2) f ′(2) – f (2) g ′(2)
2
g (2)
≈
(1)(1) – (3)(0)
(1) 2
=1
56.
d
d
⎡⎣ g ( tan 2 x ) ⎤⎦ = g ' ( tan 2 x ) ⋅ tan 2 x
dx
dx
(
45. ( f D g )′(6) = f ′( g (6)) g ′(6)
= f ′(2) g ′(6) ≈ (1)(−1) = –1
= 2 g ' ( tan 2 x ) sec2 2 x
57. Dx ⎡⎣ F ( x ) sin 2 F ( x ) ⎤⎦
= F ( x ) × Dx ⎡⎣sin 2 F ( x ) ⎤⎦ + sin 2 F ( x ) × Dx F ( x )
= F ( x ) × 2sin F ( x ) × Dx ⎡⎣sin F ( x ) ⎤⎦
46. ( g D f )′(3) = g ′( f (3)) f ′(3)
3
⎛3⎞
= g ′(4) f ′(3) ≈ ⎜ ⎟ (1) =
2
2
⎝ ⎠
47. D x F (2 x ) = F ′(2 x )D x (2 x ) = 2 F ′(2 x )
48.
(
)
(
) (
+ F ′ ( x ) sin 2 F ( x )
= F ( x ) × 2sin F ( x ) × cos ( F ( x ) ) × Dx ⎣⎡ F ( x ) ⎦⎤
)
Dx F x 2 +1 = F ′ x 2 +1 Dx x 2 +1
(
)
+ F ′ ( x ) sin 2 F ( x )
= 2 xF ′ x 2 + 1
[
= 2 F ( x ) F ′ ( x ) sin F ( x ) cos F ( x )
]
49. Dt (F (t ))−2 = −2(F (t ))−3 F ′(t )
50.
51.
52.
+ F ′ ( x ) sin 2 F ( x )
d ⎡ 1 ⎤
−3
⎢
⎥ = −2(F (z )) F ′(z )
dz ⎣⎢ (F (z ))2 ⎦⎥
58. Dx ⎣⎡sec3 F ( x ) ⎦⎤ = 3sec2 ⎡⎣ F ( x ) ⎤⎦ Dx ⎡⎣sec F ( x ) ⎤⎦
= 3sec2 ⎡⎣ F ( x ) ⎤⎦ sec F ( x ) tan F ( x ) Dx [ x ]
d ⎡
d
2
1 + F ( 2 z ) ) ⎤ = 2 (1 + F ( 2 z ) ) (1 + F ( 2 z ) )
(
⎢
⎥
⎦
dz ⎣
dz
′
= 2 (1 + F ( 2 z ) ) ( 2 F ( 2 z ) ) = 4 (1 + F ( 2 z ) ) F ′ ( 2 z )
⎡
d ⎢ 2
1
y +
⎢
dy
F y2
⎣
( )
⎤
⎥ = 2 y + d ⎡ F y2
⎥
dy ⎢⎣
⎦
( ( ))
( ) dyd y
= 2 y − F ′ y2
⎛
F ′ y2
⎜
= 2 y ⎜1 −
⎜
F y2
⎝
( )
( ( ))
2
2 yF ′ y
2
= 3F ′ ( x ) sec3 F ( x ) tan F ( x )
59. g ' ( x ) = − sin f ( x ) Dx f ( x ) = − f ′ ( x ) sin f ( x )
g ′ ( 0 ) = − f ′ ( 0 ) sin f ( 0 ) = −2sin1 ≈ −1.683
−1 ⎤
( )
= 2y −
( F ( y ))
⎞
⎟
2 ⎟
⎟
⎠
⎥
⎦
60. G ′ ( x ) =
2
2
=
d
d
x − x (1 + sec F ( 2 x ) )
(1 + sec F ( 2 x ) ) dx
dx
(1 + sec F ( 2 x ) )
2
(1 + sec F ( 2 x ) ) − 2 xF ′ ( 2 x ) sec F ( 2 x ) tan F ( 2 x )
2
(1 + sec F ( 2 x ) )
G′ ( 0) =
=
Instructor’s Resource Manual
)
= g ' ( tan 2 x ) sec2 2 x ⋅ 2
1 + sec F ( 0 ) − 0
(1 + sec F ( 0 ) )
2
=
1 + sec F ( 0 )
(1 + sec F ( 0 ) )
2
1
1
=
≈ −0.713
1 + sec F ( 0 ) 1 + sec 2
Section 2.5
119
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
61. F ′ ( x ) = − f ( x ) g ′ ( x ) sin g ( x ) + f ′ ( x ) cos g ( x )
c. Dt L =
F ′ (1) = − f (1) g ′ (1) sin g (1) + f ′ (1) cos g (1)
= −2 (1) sin 0 + −1cos 0 = −1
=
62. y = 1 + x sin 3 x; y ′ = 3x cos 3 x + sin 3 x
y ′ (π / 3) = 3
π
cos 3
π
+ sin
3
3
y − 1 = −π x − π / 3
π
3
=
y = −π x − π / 3 + 1
The line crosses the x-axis at x =
=
3−π
.
3
(
) ( ) ( ) ( )
( x + 1)( x + 1) + 3x ( x + 1) ( x + 1)
3
2
64. y ′ = x 2 + 1 2 x 4 + 1 x3 + 3 x 2 + 1 x x 4 + 1
= 2 x3
4
3
2
y ′ (1) = 2 ( 2 )( 2 ) + 3 (1)( 2 )
3
2
4
2
2
2
2
69. a.
b.
(
)
−3
( 2 x ) = −4 x ( x 2 + 1)
y ′ (1) = −4 (1)(1 + 1)
1
1
1
y− = − x+ ,
4
2
2
66. y ′ = 3 ( 2 x + 1)
2
−3
70. a.
−3
b.
= −1/ 2
( 2 ) = 6 ( 2 x + 1)
c.
2
2
y − 1 = 6 x − 0, y = 6 x + 1
The line crosses the x-axis at x = −1/ 6 .
(
)
y ′ (1) = −4 ( 2 )
−3
−3
( 2 x ) = −4 x ( x 2 + 1)
16 cos 2 2t + 49sin 2 2t
π
33
: rate =
≈ 5.8 ft/sec.
8
16 ⋅ 12 + 49 ⋅ 12
(10 cos8π t ,10sin 8π t )
Dt (10sin 8πt ) = 10 cos(8πt ) Dt (8πt )
(cos 2t, sin 2t)
(0 – cos 2t ) 2 + ( y – sin 2t )2 = 52 , so
Dt ⎛⎜ sin 2t + 25 − cos 2 2t ⎞⎟
⎝
⎠
1
= 2 cos 2t +
⋅ 4 cos 2t sin 2t
2 25 − cos 2 2t
⎛
sin 2t
= 2 cos 2t ⎜ 1 +
⎜
25 – cos 2 2t
⎝
−3
⎞
⎟
⎟
⎠
71. 60 revolutions per minute is 120π radians per
minute or 2π radians per second.
= −1/ 2
1
1
1
1
3
= − x+ , y = − x+
4
2
2
2
4
Set y = 0 and solve for x. The line crosses the
y−
a.
(cos 2π t ,sin 2π t )
x-axis at x = 3 / 2 .
b.
(0 – cos 2πt ) 2 + ( y – sin 2πt )2 = 52 , so
2
68. a.
y = sin 2πt + 25 – cos 2 2πt
2
2
⎛x⎞ ⎛ y⎞
⎛ 4 cos 2t ⎞ ⎛ 7 sin 2t ⎞
⎜ ⎟ +⎜ ⎟ = ⎜
⎟ +⎜
⎟
⎝4⎠ ⎝7⎠
⎝ 4 ⎠ ⎝ 7 ⎠
= cos 2 2t + sin 2 2t = 1
b.
L = ( x – 0)2 + ( y – 0) 2 = x 2 + y 2
= (4 cos 2t )2 + (7 sin 2t )2
= 16 cos 2 2t + 49sin 2 2t
120
16 cos 2 2t + 49sin 2 2t
33sin 4t
y = sin 2t + 25 – cos 2 2t
1
3
y = − x+
2
4
y ′ ( 0 ) = 6 (1) = 6
67. y ′ = −2 x 2 + 1
2 16 cos 2 2t + 49sin 2 2t
−16sin 4t + 49sin 4t
= 80π cos(8πt )
At t = 1: rate = 80π ≈ 251 cm/s
P is rising at the rate of 251 cm/s.
( 2 )2 = 32 + 48 = 80
y − 32 = 80 x − 1, y = 80 x + 31
65. y ′ = −2 x 2 + 1
2 16 cos 2 2t + 49sin 2 2t
–64 cos 2t sin 2t + 196sin 2t cos 2t
At t =
63. y = sin 2 x; y ′ = 2sin x cos x = sin 2 x = 1
x = π / 4 + kπ , k = 0, ± 1, ± 2,...
Dt (16cos2 2t + 49sin 2 2t )
32 cos 2t Dt (cos 2t ) + 98sin 2t Dt (sin 2t )
= −π + 0 = −π
=
1
2 16cos2 2t + 49sin 2 2t
Section 2.5
2
c.
Dt ⎛⎜ sin 2πt + 25 − cos 2 2πt ⎞⎟
⎝
⎠
= 2π cos 2πt
+
1
2 25 − cos 2 2πt
⋅ 4π cos 2πt sin 2πt
⎛
sin 2πt
= 2π cos 2πt ⎜ 1 +
⎜
25 – cos 2 2πt
⎝
⎞
⎟
⎟
⎠
Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
72. The minute hand makes 1 revolution every hour,
so at t minutes after the hour, it makes an angle
πt
of
radians with the vertical. By the Law of
30
Cosines, the length of the elastic string is
πt
s = 102 + 102 – 2(10)(10) cos
30
= 10 2 – 2 cos
ds
= 10 ⋅
dt
74. From Problem 73,
Using a computer algebra system or graphing
ds
ds
for 0 ≤ t ≤ 60 ,
is largest
utility to view
dt
dt
when t ≈ 7.5. Thus, the distance between the tips
of the hands is increasing most rapidly at about
12:08.
πt
30
π
πt
sin
15
30
πt
2 2 – 2 cos
30
1
75.
⋅
3
π
3
Dx(sin x) = cos x, Dx(sin 2x) = 2cos 2x, so at x 0 ,
the tangent lines to y = sin x and y = sin 2x have
1
⎛ 1⎞
slopes of m1 = and m2 = 2 ⎜ – ⎟ = –1,
2
⎝ 2⎠
respectively. From Problem 40 of Section 0.7,
m – m1
where θ is the angle between
tan θ = 2
1 + m1m2
At 12:15, the string is stretching at the rate of
π sin π2
π
=
≈ 0.74 cm/min
3 2 – 2 cos π2 3 2
73. The minute hand makes 1 revolution every hour,
so at t minutes after noon it makes an angle of
πt
radians with the vertical. Similarly, at t
30
minutes after noon the hour hand makes an angle
πt
of
with the vertical. Thus, by the Law of
360
Cosines, the distance between the tips of the
hands is
⎛ πt πt ⎞
s = 62 + 82 – 2 ⋅ 6 ⋅ 8cos ⎜ –
⎟
⎝ 30 360 ⎠
11πt
360
ds
1
44π 11πt
=
⋅
sin
360
dt 2 100 – 96 cos 11πt 15
360
=
1
[if sin x0 ≠ 0]
2
x0 =
πt
2 – 2 cos 30
= 100 – 96 cos
sin x0 = sin 2 x0
sin x0 = 2sin x0 cos x0
cos x0 =
πt
π sin 30
=
πt
22π sin 11
ds
360
=
.
dt 15 100 – 96 cos 11πt
360
πt
22π sin 11
360
πt
15 100 – 96 cos 11
360
At 12:20,
π
22π sin 11
ds
18
=
≈ 0.38 in./min
dt 15 100 – 96 cos 11π
18
the tangent lines. tan θ =
( )
1 + 12 (–1)
=
– 32
1
2
= –3,
so θ ≈ –1.25. The curves intersect at an angle of
1.25 radians.
76.
1
t
AB = OA sin
2
2
2
1
t
t
t
D = OA cos ⋅ AB = OA cos sin
2
2
2
2
E = D + area (semi-circle)
2
2
t
t 1 ⎛1
⎞
= OA cos sin + π ⎜ AB ⎟
2
2 2 ⎝2
⎠
2
2
t
t 1
t
= OA cos sin + πOA sin 2
2
2 2
2
2
t⎛
t 1
t⎞
= OA sin ⎜ cos + π sin ⎟
2⎝
2 2
2⎠
t
cos 2
D
=
E cos t + 1 π sin t
2 2
2
D
1
=
=1
lim
+
E
+
1
0
t →0
lim
t →π −
Instructor’s Resource Manual
–1 – 12
D
cos(t / 2)
= lim
−
E t →π cos(t / 2) + π sin(t / 2)
2
0
=
=0
π
0+
2
Section 2.5
121
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
81. [ f ( f ( f ( f (0))))]′
= f ′( f ( f ( f (0)))) ⋅ f ′( f ( f (0))) ⋅ f ′( f (0)) ⋅ f ′(0)
= 2 ⋅ 2 ⋅2 ⋅ 2 = 16
77. y = u and u = x 2
Dx y = Du y ⋅ Dx u
=
1
2 u
2 x
x2 – 1
78. Dx x – 1 =
2
=
x2 – 1
2
x2 – 1
(2 x) =
80. a.
b.
sin x
82. a.
Dx ( x 2 – 1)
b.
2
x –1
79. Dx sin x =
sin x
x
x
=
x
x
=
2
2 x x2 – 1
x –1
=
2x
⋅ 2x =
sin x
sin x
Dx (sin x)
cos x = cot x sin x
c.
( )
( ) ( )
Dx L x 2 = L ' x 2 Dx x 2 =
1
x
2
⋅ 2x =
2
x
d [2]
f = f '( f ( x)) ⋅ f '( x)
dx
d [1]
= f '( f [1] ) ⋅
f ( x)
dx
d [3]
f = f '( f ( f ( x))) ⋅ f '( f ( x)) ⋅ f '( x)
dx
d [1]
= f '( f [2] ( x)) ⋅ f '( f [1] ( x)) ⋅
f ( x)
dx
d [2]
= f '( f [2] ( x)) ⋅
f ( x)
dx
Conjecture:
d [n]
d [ n −1]
f ( x) = f '( f [ n −1] ( x)) ⋅
f
( x)
dx
dx
Dx L(cos 4 x) = sec4 x Dx (cos 4 x )
= sec4 x(4 cos3 x) Dx (cos x )
= 4sec4 x cos3 x(− sin x)
1
⋅ cos3 x ⋅ ( − sin x )
cos 4 x
= –4sec x sin x = −4 tan x
= 4⋅
⎛ f ( x) ⎞
⎛
1 ⎞
−1
−1
−1
83. Dx ⎜
⎟ = Dx ⎜ f ( x ) ⋅
⎟ = Dx f ( x) ⋅ ( g ( x)) = f ( x) Dx ( g ( x)) + ( g ( x)) Dx f ( x)
g ( x) ⎠
⎝ g ( x) ⎠
⎝
(
)
(
)
= f ( x) ⋅ (−1)( g ( x)) −2 Dx g ( x) + ( g ( x)) −1 Dx f ( x) = − f ( x)( g ( x)) −2 Dx g ( x) + ( g ( x))−1 Dx f ( x)
=
=
Dx f ( x ) − f ( x ) Dx g ( x ) g ( x ) Dx f ( x ) − f ( x ) Dx g ( x ) g ( x ) Dx f ( x )
+
⋅
=
+
=
g ( x) g ( x)
g ( x)
g 2 ( x)
g 2 ( x)
g ( x)
g 2 ( x)
g ( x) Dx f ( x) − f ( x) Dx g ( x)
− f ( x) Dx g ( x)
2
+
g 2 ( x)
)) f ′ ( f ( f ( x ))) f ′ ( f ( x )) f ′ ( x )
( (
g ′ ( x ) = f ′ ( f ( f ( f ( x )))) f ′ ( f ( f ( x ))) f ′ ( f ( x )) f ′ ( x )
84. g ′ ( x ) = f ′ f f ( f ( x ) )
1
1
(
1
1
1
)
= f ′ f ( f ( x2 ) ) f ′ ( f ( x2 ) ) f ′ ( x2 ) f ′ ( x1 ) = f ′ ( f ( x1 ) ) f ′ ( x1 ) f ′ ( x2 ) f ′ ( x1 )
= ⎡⎣ f ′ ( x1 ) ⎤⎦ ⎡⎣ f ′ ( x2 ) ⎤⎦
2
( (
g ′ ( x2 ) = f ′ f f ( f ( x2 ) )
(
)
2
) ) f ′ ( f ( f ( x2 ) ) ) f ′ ( f ( x2 ) ) f ′ ( x2 )
= f ′ f ( f ( x1 ) ) f ′ ( f ( x1 ) ) f ′ ( x1 ) f ′ ( x2 ) = f ′ ( f ( x2 ) ) f ′ ( x2 ) f ′ ( x1 ) f ′ ( x2 )
= ⎡⎣ f ′ ( x1 ) ⎤⎦ ⎡⎣ f ′ ( x2 ) ⎤⎦ = g ′ ( x1 )
2
122
Section 2.5
2
Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2.6 Concepts Review
1.
f ′′′( x), Dx3 y,
2.
ds ds d 2 s
;
;
dt dt dt 2
3.
f ′ (t ) > 0
d3y
dx
3
3.
d2y
, y '''
dx 2
d3y
dx3
4.
dx 2
d3y
Problem Set 2.6
dx
dy
1.
= 3x2 + 6 x + 6
dx
d y
dx 2
d3y
dx3
5.
= 6x + 6
= 162
= –100(3 – 5 x)3 (–5) = 500(3 – 5 x)3
= 1500(3 – 5 x) 2 (–5) = –7500(3 – 5 x)2
dy
= 7 cos(7 x)
dx
dx 2
d3y
dx
dy
2.
= 5 x 4 + 4 x3
dx
d2y
= 20x 3 +12 x 2
dx 2
d3y
= 60 x 2 + 24 x
3
dx
6.
3
d2y
=6
= 18(3 x + 5)(3) = 162 x + 270
dy
= 5(3 – 5 x)4 (–5) = –25(3 – 5 x)4
dx
d2y
4. 0; < 0
2
dy
= 3(3 x + 5) 2 (3) = 9(3x + 5) 2
dx
3
= –7 2 sin(7 x)
= –73 cos(7 x) = –343cos(7 x)
dy
= 3x 2 cos( x3 )
dx
d2y
dx 2
d3y
dx
3
= 3 x 2 [–3x 2 sin( x3 )] + 6 x cos( x3 ) = –9 x 4 sin( x3 ) + 6 x cos( x3 )
= –9 x 4 cos( x3 )(3 x 2 ) + sin( x3 )(–36 x3 ) + 6 x[– sin( x3 )(3 x 2 )] + 6 cos( x3 )
= –27 x 6 cos( x3 ) – 36 x3 sin( x3 ) –18 x3 sin( x3 ) + 6 cos( x3 ) = (6 – 27 x 6 ) cos( x3 ) – 54 x3 sin( x3 )
7.
dy ( x –1)(0) – (1)(1)
1
=
=–
2
dx
( x –1)
( x –1)2
d2y
dx
2
d3y
dx3
=−
( x –1)2 (0) – 2( x –1)
( x –1)
=
4
=
8.
d2y
2
( x –1)
dy (1 – x )(3) – (3x )(–1)
3
=
=
2
dx
(1 – x)
( x – 1)2
3
dx
2
( x − 1)3 (0) − 2[3( x − 1) 2 ]
d3y
( x − 1)6
dx3
=−
6
( x − 1)
4
Instructor’s Resource Manual
=
( x – 1) 2 (0) – 3[2( x – 1)]
( x – 1)
=−
=
4
=–
6
( x – 1)3
( x − 1)3 (0) − 6(3)( x − 1) 2
( x − 1)6
18
( x − 1) 4
Section 2.6
123
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9.
f ′( x) = 2 x; f ′′( x) = 2; f ′′(2) = 2
12.
10.
f ′( x) = 15 x 2 + 4 x + 1
f ′′( x) = 30 x + 4
f ′′(2) = 64
f ′(u ) =
f ′′(u ) =
=
11.
f ′(t ) = –
f ′′(t ) =
2
t2
4
(5 – u )(4u ) – (2u 2 )(–1)
(5 – u ) 2
=
20u – 2u 2
(5 – u ) 2
(5 – u )2 (20 – 4u ) – (20u – 2u 2 )2(5 – u )(–1)
(5 – u )4
100
(5 – u )3
f ′′(2) =
100
3
3
=
100
27
3
t
4 1
f ′′(2) = =
8 2
13.
f ′(θ ) = –2(cos θπ) –3 (– sin θπ)π = 2 π(cos θ π) –3 (sin θ π)
f ′′(θ ) = 2π[(cos θπ) –3 (π)(cos θπ) + (sin θπ)(–3)(cosθπ) –4 (– sin θπ)(π)] = 2π2 [(cos θπ)−2 + 3sin 2 θπ(cosθπ) −4 ]
f ′′(2) = 2π2 [1 + 3(0)(1)] = 2π2
14.
⎛ π ⎞⎛ π ⎞
⎛π⎞ ⎛ π⎞
⎛π⎞
⎛π⎞
f ′(t ) = t cos ⎜ ⎟ ⎜ – ⎟ + sin ⎜ ⎟ = ⎜ – ⎟ cos ⎜ ⎟ + sin ⎜ ⎟
2
⎝ t ⎠⎝ t ⎠
⎝t⎠ ⎝ t⎠
⎝t⎠
⎝t⎠
π2
⎛ π⎞⎡
⎛ π ⎞ ⎛ π ⎞⎤ ⎛ π ⎞
⎛π⎞ ⎛ π ⎞
⎛π⎞
⎛π⎞
f ′′(t ) = ⎜ – ⎟ ⎢ – sin ⎜ ⎟ ⎜ – ⎟ ⎥ + ⎜ ⎟ cos ⎜ ⎟ + ⎜ – ⎟ cos ⎜ ⎟ = –
sin ⎜ ⎟
2
2
2
3
⎝ t ⎠⎣
⎝ t ⎠ ⎝ t ⎠⎦ ⎝ t ⎠
⎝t⎠ ⎝ t ⎠
⎝t⎠
⎝t⎠
t
f ′′(2) = –
15.
π2
π2
⎛π⎞
≈ –1.23
sin ⎜ ⎟ = –
8
8
⎝2⎠
f ′( s ) = s (3)(1 – s 2 )2 (–2 s ) + (1 – s 2 )3 = –6s 2 (1 – s 2 ) 2 + (1 – s 2 )3 = –7 s 6 + 15s 4 – 9 s 2 + 1
f ′′( s ) = –42 s5 + 60 s3 –18s
f ′′(2) = –900
16.
f ′( x) =
f ′′( x) =
( x –1)2( x + 1) – ( x + 1)2
( x –1)2
x2 – 2 x – 3
( x –1)2
( x –1) 2 (2 x – 2) – ( x 2 – 2 x – 3)2( x –1)
( x –1)
f ′′(2) =
=
8
13
4
=
( x –1)(2 x – 2) – ( x 2 – 2 x – 3)(2)
( x –1)
=
8
( x –1)3
=8
17. Dx ( x n ) = nx n –1
Dx2 ( x n ) = n(n –1) x n –2
Dx3 ( x n ) = n(n –1)(n – 2) x n –3
Dx4 ( x n ) = n(n – 1)(n – 2)(n – 3) x n –4
#
n −1 n
Dx ( x ) = n(n –1)(n – 2)(n – 3)...(2) x
18. Let k < n.
Dxn ( x k ) = Dxn − k [ Dxk ( x k )] = Dx (k !) = 0
so Dxn [an x n –1 +…+ a1 x + a0 ] = 0
19. a.
Section 2.6
Dx4 (3x3 + 2 x –19) = 0
b.
11
10
D12
x (100 x − 79 x ) = 0
c.
2
5
D11
x ( x – 3) = 0
Dxn ( x n ) = n(n –1)(n – 2)(n – 3)...2(1) x 0 = n!
124
3
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1
⎛1⎞
20. Dx ⎜ ⎟ = –
⎝ x⎠
x2
2
⎛1⎞
Dx2 ⎜ ⎟ = Dx (– x –2 ) = 2 x –3 =
x
⎝ ⎠
x3
3(2)
⎛1⎞
Dx3 ⎜ ⎟ = Dx (2 x –3 ) = –
⎝ x⎠
x4
⎛ 1 ⎞ 4(3)(2)
Dx4 ⎜ ⎟ =
⎝ x⎠
x5
n
⎛ 1 ⎞ (−1) n !
Dxn ⎜ ⎟ =
⎝x⎠
x n +1
21.
f ′( x) = 3 x 2 + 6 x – 45 = 3( x + 5)( x − 3)
3(x + 5)(x – 3) = 0
x = –5, x = 3
f ′′( x) = 6 x + 6
f ′′(–5) = –24
f ′′(3) = 24
22. g ′(t ) = 2at + b
g ′′(t ) = 2a
g ′′(1) = 2a = −4
a = −2
g ′(1) = 2a + b = 3
2(–2) + b = 3
b=7
g (1) = a + b + c = 5
( −2 ) + ( 7 ) + c = 5
c=0
23. a.
v(t ) =
a(t ) =
b.
3t 2 – 12t > 0
3t(t – 4) > 0; (−∞, 0) ∪ (4, ∞)
c.
3t 2 – 12t < 0
(0, 4)
d. 6t – 12 < 0
6t < 12
t < 2; (−∞, 2)
e.
25. a.
a(t ) =
= 6t – 18
c.
3t 2 –18t + 24 < 0
(2, 4)
d. 6t – 18 < 0
6t < 18
t < 3; (−∞,3)
ds
= 12 – 4t
dt
d 2s
dt 2
= –4
26. a.
e.
a(t ) =
dt 2
e.
12 – 4t < 0
t > 3; (3, ∞)
v(t ) =
d 2s
3t 2 –18t + 24 > 0
3(t – 2)(t – 4) > 0
(−∞, 2) ∪ (4, ∞)
ds
= 3t 2 –12t
dt
d 2s
dt 2
v(t ) =
a(t ) =
ds
= 6t 2 – 6
dt
d 2s
dt 2
= 12t
b.
6t 2 – 6 > 0
6(t + 1)(t – 1) > 0
(−∞, −1) ∪ (1, ∞)
c.
6t 2 – 6 < 0
(–1, 1)
d. a(t) = –4 < 0 for all t
24. a.
ds
= 3t 2 –18t + 24
dt
b.
b. 12 – 4t > 0
4t < 12
t < 3; ( −∞,3)
c.
v(t ) =
d. 12t < 0
t<0
The acceleration is negative for negative t.
= 6t –12
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e.
29. v(t ) =
ds
= 2t 3 –15t 2 + 24t
dt
d 2s
a(t ) =
27. a.
ds
16
= 2t –
dt
t2
v(t ) =
d 2s
a(t ) =
dt
b.
2
= 2+
32
t3
16
>0
t2
2t 3 – 16
> 0; (2, ∞)
t2
2t –
16
< 0; (0, 2)
c.
2t –
d.
<0
t3
2t3 + 32 < 0; The acceleration is not
t3
negative for any positive t.
t2
32
2+
e.
v(t ) =
a(t ) =
30. v(t ) =
dt
b.
1–
4
t2
t2 – 4
t2
c.
1–
4
t
d.
8
3
2
2
=
d 2s
2
=
1
(12t 2 – 84t + 120)
10
dt
1
(12t 2 – 84t + 120) = 0
10
12
(t − 2)(t − 5) = 0
10
t = 2, t = 5
v(2) = 10.4, v(5) = 5
ds1
= 4 – 6t
dt
ds
v2 (t ) = 2 = 2t – 2
dt
31. v1 (t ) =
4 – 6t = 2t – 2
8t = 6
3
t = sec
4
8
t3
b.
> 0; (2, ∞)
< 0; (0, 2)
< 0; The acceleration is not negative for
t
any positive t.
4 – 6t = 2t – 2 ; 4 – 6t = –2t + 2
t=
>0
e.
ds 1
= (4t 3 – 42t 2 + 120t )
dt 10
a(t ) =
ds
4
=1–
dt
t2
d 2s
= 6t 2 – 30t + 24
6t 2 – 30t + 24 = 0
6(t – 4)(t – 1) = 0
t = 4, 1
v(4) = –16, v(1) = 11
a.
28. a.
dt 2
c.
1
3
sec and t = sec
2
4
4t – 3t 2 = t 2 – 2t
4t 2 – 6t = 0
2t(2t – 3) = 0
t = 0 sec and t =
3
sec
2
ds1
= 9t 2 – 24t + 18
dt
ds
v2 (t ) = 2 = –3t 2 + 18t –12
dt
32. v1 (t ) =
9t 2 – 24t + 18 = –3t 2 + 18t –12
12t 2 – 42t + 30 = 0
2t 2 – 7t + 5 = 0
(2t – 5)(t – 1) = 0
5
t = 1,
2
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33. a.
v(t) = –32t + 48
initial velocity = v0 = 48 ft/sec
b. –32t + 48 = 0
3
t = sec
2
c.
s = –16(1.5) 2 + 48(1.5) + 256 = 292 ft
d.
–16t + 48t + 256 = 0
2
–48 ± 48 – 4(–16)(256)
≈ –2.77, 5.77
–32
The object hits the ground at t = 5.77 sec.
t=
e.
v(5.77) ≈ –137 ft/sec;
speed = −137 = 137 ft/sec.
(t – 4)(t + 2)
<0
t < –2, 1 < t < 4; (−∞, −2) ∪ (1, 4)
38. Point slowing down when
d
v(t ) < 0
dt
v(t ) a (t )
d
v(t ) =
dt
v(t )
v (t )
signs.
48 – 32t = 0
t = 1.5
b. v(1) = 16 ft/sec upward
48t –16t 2 = 0
–16t(–3 + t) = 0
t = 3 sec
35. v(t ) = v0 – 32t
v0 – 32t = 0
t=
(t – 4)(t + 2) (6t – 6)
v (t ) a(t )
s = 48(1.5) –16(1.5)2 = 36 ft
c.
3t 2 – 6t – 24
d 2
3t – 6t – 24 =
(6t – 6)
dt
3t 2 – 6t – 24
(t – 4)(t + 2)
=
(6t – 6)
(t – 4)(t + 2)
2
34. v(t) = 48 –32t
a.
37. v(t ) = 3t 2 – 6t – 24
v0
32
2
⎛v ⎞
⎛v ⎞
v0 ⎜ 0 ⎟ –16 ⎜ 0 ⎟ = 5280
⎝ 32 ⎠
⎝ 32 ⎠
v02 v0 2
–
= 5280
32 64
v02
= 5280
64
v0 = 337,920 ≈ 581 ft/sec
36. v(t ) = v0 + 32t
v0 + 32t = 140
v0 + 32(3) = 140
v0 = 44
< 0 when a(t) and v(t) have opposite
39. Dx (uv) = uv′ + u ′v
Dx2 (uv) = uv ′′ + u ′v ′ + u ′v ′ + u ′′v
= uv ′′ + 2u ′v ′ + u ′′v
Dx3 (uv) = uv ′′′ + u ′v′′ + 2(u ′v′′ + u ′′v′) + u ′′v′ + u ′′′v
= uv′′′ + 3u ′v′′ + 3u ′′v′ + u ′′′v
Dxn (uv) =
n
⎛n⎞
∑ ⎜ k ⎟ Dxn−k (u ) Dxk (v)
k =0 ⎝
⎠
⎛ n⎞
where ⎜ ⎟ is the binomial coefficient
⎝ k⎠
n!
.
(n – k )!k !
⎛ 4⎞
40. Dx4 ( x 4 sin x ) = ⎜ ⎟ Dx4 ( x 4 ) Dx0 (sin x)
⎝0⎠
⎛ 4⎞
⎛ 4⎞
+ ⎜ ⎟ Dx3 ( x 4 ) D1x (sin x) + ⎜ ⎟ Dx2 ( x 4 ) Dx2 (sin x)
⎝1⎠
⎝ 2⎠
⎛ 4⎞
⎛ 4⎞
+ ⎜ ⎟ D1x ( x 4 ) Dx3 (sin x) + ⎜ ⎟ Dx0 ( x 4 ) Dx4 (sin x)
⎝ 3⎠
⎝ 4⎠
= 24sin x + 96 x cos x − 72 x 2 sin x
−16 x3 cos x + x 4 sin x
s = 44(3) + 16(3) 2 = 276 ft
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4. 2 x + 2α 2 yDx y = 0
41. a.
Dx y = –
2x
2α y
2
=–
x
α2y
5. x(2 y ) Dx y + y 2 = 1
Dx y =
f ′′′ (2.13) ≈ –1. 2826
b.
1 – y2
2 xy
6. 2 x + 2 x 2 Dx y + 4 xy + 3 x Dx y + 3 y = 0
Dx y (2 x 2 + 3 x) = –2 x – 4 xy – 3 y
42. a.
Dx y =
–2 x – 4 xy – 3 y
2 x2 + 3x
7. 12 x 2 + 7 x(2 y ) Dx y + 7 y 2 = 6 y 2 Dx y
12 x 2 + 7 y 2 = 6 y 2 Dx y – 14 xyDx y
f ′′′(2.13) ≈ 0.0271
b.
2.7 Concepts Review
1.
Dx y =
x 2 Dx y – 2 xyDx y = y 2 – 2 xy
9
Dx y =
x –3
dy
dx
3. x(2 y )
4.
9.
dy
dy dy
+ y2 + 3y2
–
= 3x2
dx
dx dx
p p q –1 5 2
x
; ( x – 5 x)2 / 3 (2 x – 5)
q
3
6 y 2 – 14 xy
8. x 2 Dx y + 2 xy = y 2 + x(2 y ) Dx y
3
2. 3 y 2
12 x 2 + 7 y 2
1
2 5 xy
y 2 – 2 xy
x 2 – 2 xy
⋅ (5 x Dx y + 5 y ) + 2 Dx y
= 2 y Dx y + x(3 y 2 ) Dx y + y 3
5x
2 5 xy
= y3 –
Dx y + 2 Dx y – 2 y Dx y – 3 xy 2 Dx y
5y
2 5 xy
Problem Set 2.7
1. 2 y Dx y – 2 x = 0
Dx y =
–18 x
9x
=–
8y
4y
3. x Dx y + y = 0
y
Dx y = –
x
128
Dx y =
2x x
=
2y y
2. 18 x + 8 y Dx y = 0
Dx y =
y3 –
Section 2.7
10. x
5x
2 5 xy
1
2 y +1
x
2 y +1
Dx y =
5y
2 5 xy
+ 2 – 2 y – 3 xy 2
Dx y + y + 1 = x Dx y + y
Dx y – x Dx y = y – y + 1
y – y +1
x
2 y +1
–x
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11. x Dx y + y + cos( xy )( x Dx y + y ) = 0
x Dx y + x cos( xy ) Dx y = – y – y cos( xy )
Dx y =
17.
– y – y cos( xy )
y
=–
x + x cos( xy )
x
2 –1/ 3 2 –1/ 3
x
– y
y′ – 2 y′ = 0
3
3
2 –1/ 3
⎛2
⎞
= y ′ ⎜ y –1/ 3 + 2 ⎟
x
3
⎝3
⎠
y′ =
12. – sin( xy 2 )(2 xy Dx y + y 2 ) = 2 yDx y + 1
–2 xy sin( xy 2 ) Dx y – 2 y Dx y = 1 + y 2 sin( xy 2 )
Dx y =
At (1, –1), y ′ =
1 + y sin( xy )
2
2
–2 xy sin( xy 2 ) – 2 y
2
3
2
y ′( x3 + 3xy 2 ) = –3 x 2 y – y 3
y′ =
2
–3 x y – y
3
18.
36
9
=–
28
7
9
Tangent line: y – 3 = – ( x – 1)
7
At (1, 3), y ′ = –
y′ =
– y cos( xy )
y cos( xy )
=
x cos( xy ) – 1 1 – x cos( xy )
⎛π ⎞
At ⎜ , 1⎟ , y ′ = 0
⎝2 ⎠
π⎞
⎛
Tangent line: y – 1 = 0 ⎜ x – ⎟
2⎠
⎝
y=1
– y2
1 + 2 xy
–1
17
2
=–
2
17
2
( x – 4)
17
dy
1
= 5x2 / 3 +
dx
2 x
20.
dy 1 –2 / 3
1
= x
– 7 x5 / 2 =
– 7 x5 / 2
3 2
dx 3
3 x
21.
1
1
dy 1 –2 / 3 1 –4 / 3
= x
=
– x
–
3
3
3
dx 3
3 x2 3 x4
22.
dy 1
1
= (2 x + 1) –3 / 4 (2) =
4
dx 4
2 (2 x + 1)3
y ′[1 – 2 xy sin( xy 2 )] = y 2 sin( xy 2 ) – 6 x
y 2 sin( xy 2 ) – 6 x
6
= –6
1
Tangent line: y – 0 = –6(x – 1)
y ′ + 2 xyy ′ + y 2 = 0
19.
16. y ′ + [– sin( xy 2 )][2 xyy ′ + y 2 ] + 6 x = 0
1 – 2 xy sin( xy 2 )
1
( x –1)
2
Tangent line: y –1 = –
23.
y′ =
2 y
At (4, 1), y ′ =
– xy 2 – 2 y
15. cos( xy )( xy ′ + y ) = y ′
y ′[ x cos( xy ) – 1] = – y cos( xy )
1
2
2 y
y ′(2 x 2 y + 4 x – 12) = –2 xy 2 – 4 y
=
2 x 2 y + 4 x – 12 x 2 y + 2 x – 6
At (2, 1), y ′ = –2
Tangent line: y – 1 = –2( x – 2)
1
y′ =
14. x 2 (2 y ) y ′ + 2 xy 2 + 4 xy ′ + 4 y = 12 y ′
y′ =
=
⎛ 1
⎞
+ 2 xy ⎟ = – y 2
y′ ⎜
⎜2 y
⎟
⎝
⎠
x3 + 3 xy 2
–2 xy 2 – 4 y
2
3
4
3
Tangent line: y + 1 =
13. x y ′ + 3 x y + y + 3xy y ′ = 0
3
2 x –1/ 3
3
2 y –1/ 3 + 2
3
dy 1
= (3 x 2 – 4 x) –3 / 4 (6 x – 4)
dx 4
6x – 4
3x – 2
=
=
2
3
4
4
4 (3 x – 4 x)
2 (3 x 2 – 4 x)3
24.
dy 1 3
= ( x – 2 x) –2 / 3 (3 x 2 – 2)
dx 3
25.
dy d
= [( x3 + 2 x)−2 / 3 ]
dx dx
At (1, 0), y ′ = –
2
6 x2 + 4
= – ( x3 + 2 x) –5 / 3 (3 x 2 + 2) = −
3
3 3 ( x 3 + 2 x )5
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26.
5
dy
= – (3 x – 9) –8 / 3 (3) = –5(3 x – 9) –8 / 3
3
dx
dy
1
27.
(2 x + cos x)
=
dx 2 x 2 + sin x
2 x + cos x
=
2 x 2 + sin x
28.
x 2 cos x + 2 x sin x
2
−5
5 x
dy
=0
dx
dy
2x + 4
x+2
=−
=−
dx
2y
y
2x + 4 + 2 y
The tangent line at ( x0 , y0 ) has equation
x +2
y – y0 = − 0
( x – x0 ) which simplifies to
y0
4
dy 1
= (1 + sin 5 x) –3 / 4 (cos 5 x)(5)
dx 4
5cos 5 x
=
4 4 (1 + sin 5 x)3
( x0 , y0 ) is on the circle, x0 2 + y02 = –3 – 4 x0 ,
so the equation of the tangent line is
– yy0 – 2 x0 – 2 x – xx0 = 3.
dy [1 + cos( x 2 + 2 x)]–3 / 4 [– sin( x 2 + 2 x)(2 x + 2)]
=
dx
4
3
If (0, 0) is on the tangent line, then x0 = – .
2
Solve for y0 in the equation of the circle to get
( x + 1) sin( x 2 + 2 x)
2 [1 + cos( x + 2 x )]
2
4
3
dy (tan 2 x + sin 2 x) –1/ 2 (2 tan x sec 2 x + 2 sin x cos x)
=
dx
2
=
tan x sec2 x + sin x cos x
33. s 2 + 2 st
ds – s – 3t
s + 3t
=
=−
dt
2 st
2 st
dt
dt
s 2 + 2st + 3t 2
=0
ds
ds
dt 2
( s + 3t 2 ) = –2 st
ds
dt
2 st
=−
2
ds
s + 3t 2
Section 2.7
3
. Put these values into the equation of
2
the tangent line to get that the tangent lines are
3 y + x = 0 and 3 y – x = 0.
y0 = ±
36. 16( x 2 + y 2 )(2 x + 2 yy ′) = 100(2 x – 2 yy ′)
y ′(4 x 2 y + 4 y 3 + 25 y ) = 25 x – 4 x3 – 4 xy 2
ds
+ 3t 2 = 0
dt
2
2 x0 – yy0 – 2 x – xx0 + y02 + x02 = 0. Since
32 x3 + 32 x 2 yy ′ + 32 xy 2 + 32 y 3 y ′ = 200 x – 200 yy ′
tan 2 x + sin 2 x
2
130
(x + 2)2 + y2 = 1
−5
dy d
= [( x 2 sin x) –1/ 3 ]
dx dx
1
= – ( x 2 sin x) –4 / 3 ( x 2 cos x + 2 x sin x)
3
=−
32.
5
2 x 2 cos x
33 ( x sin x)
31.
y
35.
2 x cos x – x 2 sin x
=–
30.
dx
dx
+ 6x2
dy
dy
dx
1
=
dy 2 x cos( x 2 ) + 6 x 2
dy
1
[ x 2 (– sin x) + 2 x cos x]
=
dx 2 x 2 cos x
=
29.
34. 1 = cos( x 2 )(2 x )
2
2
y′ =
25 x – 4 x3 – 4 xy 2
4 x 2 y + 4 y 3 + 25 y
The slope of the normal line = –
=
1
y′
4 x 2 y + 4 y 3 + 25 y
4 x3 + 4 xy 2 – 25 x
65 13
=
45 9
13
Normal line: y – 1 = ( x – 3)
9
At (3, 1), slope =
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xy ′ + y + 3 y 2 y ′ = 0
37. a.
y ′′(6 y 2 – x 2 ) =
y ′( x + 3 y 2 ) = – y
y′ = –
y
y ′′ =
x + 3y2
⎛ –y
xy ′′ + ⎜
⎜ x + 3y2
⎝
b.
2
⎛ –y
+6 y ⎜
⎜ x + 3 y2
⎝
⎞
⎟⎟ = 0
⎠
2y
6 y3
xy ′′ + 3 y 2 y ′′ –
+
x + 3 y2
y ′′( x + 3 y 2 ) =
y ′′( x + 3 y 2 ) =
y ′′ =
⎞ ⎛ –y ⎞
2
⎟⎟ + ⎜⎜
⎟ + 3 y y ′′
2⎟
⎠ ⎝ x + 3y ⎠
2y
x + 3y2
2 xy
–
( x + 3 y 2 )2
6 y3
( x + 3 y 2 )2
(x + 3y )
2 2
2 xy
( x + 3 y 2 )3
2x
x
=–
y
2y
2 + 2[ yy ′′ + ( y ′)2 ] = 0
2
⎛ x⎞
2 + 2 yy ′′ + 2 ⎜ – ⎟ = 0
⎝ y⎠
2 yy ′′ = −2 −
2 x2
y2
41. 3x 2 + 3 y 2 y ′ = 3( xy ′ + y )
y ′(3 y 2 – 3x) = 3 y – 3 x 2
6 x – 8( yy ′′ + ( y ′)2 ) = 0
⎛ 3x2
6 x – 8 yy ′′ – 8 ⎜
⎜ 8y
⎝
6 x – 8 yy ′′ –
9 x4
8 y2
48 xy 2 − 9 x 4
8 y2
y′ =
2
⎞
⎟ =0
⎟
⎠
y – x2
y2 – x
⎛3 3⎞
At ⎜ , ⎟ , y ′ = –1
⎝2 2⎠
Slope of the normal line is 1.
3
3⎞
⎛
Normal line: y – = 1⎜ x – ⎟ ; y = x
2
2⎠
⎝
This line includes the point (0, 0).
=0
= 8 yy ′′
48 xy 2 – 9 x 4
42. xy ′ + y = 0
64 y 3
y
x
2 x − 2 yy ′ = 0
y′ = –
39. 2( x 2 y ′ + 2 xy ) – 12 y 2 y ′ = 0
2 x 2 y ′ – 12 y 2 y ′ = –4 xy
y′ =
40. 2 x + 2 yy ′ = 0
1 x2
y 2 + x2
−
=−
y y3
y3
25
At (3, 4), y ′′ = −
64
3x2
8y
y ′′ =
(6 y 2 – x 2 )3
−120
At (2, 1), y ′′ =
= −15
8
y ′′ = −
38. 3x 2 – 8 yy ′ = 0
y′ =
(6 y 2 – x 2 ) 2
72 y 5 − 6 x 4 y − 24 x 2 y 3
y′ = –
=0
72 y 5 − 6 x 4 y − 24 x 2 y 3
x
y
The slopes of the tangents are negative
reciprocals, so the hyperbolas intersect at right
angles.
y′ =
2 xy
6 y2 – x2
2( x 2 y ′′ + 2 xy ′ + 2 xy ′ + 2 y ) – 12[ y 2 y ′′ + 2 y ( y ′) 2 ] = 0
2 x 2 y ′′ − 12 y 2 y ′′ = −8 xy ′ − 4 y + 24 y ( y ′)2
y ′′(2 x 2 – 12 y 2 ) = −
y ′′(2 x 2 – 12 y 2 ) =
16 x 2 y
6 y2 – x2
– 4y +
96 x 2 y3
(6 y 2 − x 2 )2
12 x 4 y + 48 x 2 y 3 − 144 y5
(6 y 2 – x 2 ) 2
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131
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43. Implicitly differentiate the first equation.
4 x + 2 yy ′ = 0
2x
y′ = –
y
Implicitly differentiate the second equation.
2 yy ′ = 4
2
y
Solve for the points of intersection.
y′ =
45. x 2 – x(2 x) + 2(2 x) 2 = 28
7 x 2 = 28
x2 = 4
x = –2, 2
Intersection point in first quadrant: (2, 4)
y1′ = 2
2 x – xy2′ – y + 4 yy2′ = 0
y2′ (4 y – x) = y – 2 x
y – 2x
4y – x
2 x2 + 4 x = 6
y2′ =
2( x 2 + 2 x – 3) = 0
(x + 3)(x – 1) = 0
x = –3, x = 1
x = –3 is extraneous, and y = –2, 2 when x = 1.
The graphs intersect at (1, –2) and (1, 2).
At (1, –2): m1 = 1, m2 = –1
At (1, 2): m1 = –1, m2 = 1
At (2, 4): m1 = 2, m2 = 0
44. Find the intersection points:
x2 + y 2 = 1 → y 2 = 1 − x2
( x − 1)2 + y 2 = 1
( x − 1)2 + (1 − x 2 ) = 1
x2 − 2 x + 1 + 1 − x2 = 1
0–2
= –2; θ = π + tan –1 (–2) ≈ 2.034
1 + (0)(2)
46. The equation is mv 2 – mv02 = kx02 – kx 2 .
Differentiate implicitly with respect to t to get
dx
dv
dx
2mv
= –2kx . Since v =
this simplifies
dt
dt
dt
dv
dv
to 2mv
= –2kxv or m = – kx.
dt
dt
47. x 2 – xy + y 2 = 16 , when y = 0,
⇒
x=
1
2
⎛1 3⎞
⎛1
3⎞
Points of intersection: ⎜⎜ ,
⎟⎟ and ⎜⎜ , –
⎟
2 ⎟⎠
⎝2 2 ⎠
⎝2
Implicitly differentiate the first equation.
2 x + 2 yy ′ = 0
x
y
Implicitly differentiate the second equation.
2( x –1) + 2 yy ′ = 0
y′ = –
y′ =
tan θ =
1– x
y
x 2 = 16
x = –4, 4
The ellipse intersects the x-axis at (–4, 0) and
(4, 0).
2 x – xy ′ – y + 2 yy ′ = 0
y ′(2 y – x) = y – 2 x
y′ =
y – 2x
2y – x
At (–4, 0), y ′ = 2
At (4, 0), y ′ = 2
Tangent lines: y = 2(x + 4) and y = 2(x – 4)
⎛1 3⎞
1
1
At ⎜⎜ ,
⎟⎟ : m1 = – 3 , m2 = 3
2
2
⎝
⎠
1 + 1
2
π
3
3
tan θ =
= 3 = 3 → θ=
2
3
1+ 1 − 1
3
( )( )
3
3
⎛1
3⎞
1
1
, m2 = –
At ⎜⎜ , –
⎟⎟ : m1 =
2 ⎠
3
3
⎝2
1
1
2
− −
−
3
3
3
tan θ =
=
=− 3
2
1
1
1+
–
3
( )( )
3
θ=
132
3
2π
3
Section 2.7
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
48. x 2 + 2 xy
Problem Set 2.8
dx
dx
– 2 xy – y 2
=0
dy
dy
dx 2 xy – x 2
=
dy 2 xy – y 2
2 xy – x 2
2 xy – y 2
= 0 if x(2y – x) = 0, which occurs
x
when x = 0 or y = . There are no points on
2
x
2
2
x y – xy = 2 where x = 0. If y = , then
2
2
x⎞
x3 x3 x3
⎛ x⎞
=
2 = x ⎜ ⎟ – x⎜ ⎟ =
–
so x = 2,
2
4
4
⎝2⎠
⎝2⎠
2
y = = 1.
2
The tangent line is vertical at (2, 1).
2⎛
49. 2 x + 2 y
dy
dy
x
= 0;
=–
dx
dx
y
The tangent line at ( x0 , y0 ) has slope –
dx
=3
dt
dV
dx
= 3x2
dt
dt
dV
When x = 12,
= 3(12)2 (3) = 1296 in.3/s.
dt
1. V = x3 ;
dx
(2 xy – y 2 ) = 2 xy – x 2 ;
dy
x0
,
y0
hence the equation of the tangent line is
x
y – y0 = – 0 ( x – x0 ) which simplifies to
y0
yy0 + xx0 – ( x02 + y02 ) = 0 or yy0 + xx0 = 1
since ( x0 , y0 ) is on x 2 + y 2 = 1 . If (1.25, 0) is
on the tangent line through ( x0 , y0 ) , x0 = 0.8.
Put this into x 2 + y 2 = 1 to get y0 = 0.6, since
y0 > 0. The line is 6y + 8x = 10. When x = –2,
13
13
y = , so the light bulb must be
units high.
3
3
4 3 dV
πr ;
=3
3
dt
dV
dr
= 4πr 2
dt
dt
2. V =
When r = 3, 3 = 4π(3)2
dr
1
=
≈ 0.027 in./s
dt 12π
dx
= 400
dt
dy
dx
2y
= 2x
dt
dt
dy x dx
=
mi/hr
dt y dt
3. y 2 = x 2 + 12 ;
When x = 5, y = 26,
1.
du
;t = 2
dt
2. 400 mi/hr
3. negative
4. negative; positive
Instructor’s Resource Manual
dy
5
=
(400)
dt
26
≈ 392 mi/h.
1
r 3
3h
4. V = πr 2 h; = ; r =
3
h 10
10
2
1 ⎛ 3h ⎞
3πh3 dV
V = π⎜ ⎟ h =
;
= 3, h = 5
3 ⎝ 10 ⎠
100 dt
dV 9πh 2 dh
=
dt
100 dt
When h = 5, 3 =
2.8 Concepts Review
dr
.
dt
9π(5)2 dh
100 dt
dh 4
=
≈ 0.42 cm/s
dt 3π
dx
dy
= 300,
= 400,
dt
dt
ds
dx
dy
2s = 2( x + 300) + 2 y
dt
dt
dt
ds
dx
dy
s = ( x + 300) + y
dt
dt
dt
5. s 2 = ( x + 300)2 + y 2 ;
When x = 300, y = 400, s = 200 13 , so
ds
200 13 = (300 + 300)(300) + 400(400)
dt
ds
≈ 471 mi/h
dt
Section 2.8
133
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6. y 2 = x 2 + (10)2 ;
dy
=2
dt
dy
dx
= 2x
dt
dt
When y = 25, x ≈ 22.9, so
dx y dy
25
=
≈
(2) ≈ 2.18 ft/s
dt x dt 22.9
2y
dx
=1
dt
dx
dy
0 = 2x + 2 y
dt
dt
7. 202 = x 2 + y 2 ;
When x = 5, y = 375 = 5 15 , so
dy
x dx
5
=–
=–
(1) ≈ –0.258 ft/s
dt
y dt
5 15
The top of the ladder is moving down at
0.258 ft/s.
8.
dV
dh
= –4 ft3/h; V = πhr 2 ;
= –0.0005 ft/h
dt
dt
V
dA
dV V dh
A = πr 2 = = Vh –1 , so
.
–
= h –1
h
dt
dt h 2 dt
When h = 0.001 ft, V = π(0.001)(250) 2 = 62.5π
dA
= 1000(–4) –1, 000, 000(62.5π)(–0.0005)
dt
= –4000 + 31,250 π ≈ 94,175 ft2/h.
(The height is decreasing due to the spreading of
the oil rather than the bacteria.)
and
1
d r
9. V = πr 2 h; h = = , r = 2h
3
4 2
1
4
dV
V = π(2h) 2 h = πh3 ;
= 16
3
3
dt
dV
dh
= 4πh 2
dt
dt
dh
When h = 4, 16 = 4π(4) 2
dt
dh 1
=
≈ 0.0796 ft/s
dt 4π
10. y 2 = x 2 + (90)2 ;
hx
40 x
(20);
= , x = 8h
2
5 h
dV
V = 10h(8h) = 80h 2 ;
= 40
dt
dV
dh
= 160h
dt
dt
dh
When h = 3, 40 = 160(3)
dt
dh 1
=
ft/min
dt 12
11. V =
12. y = x 2 – 4;
dx
=5
dt
dy
1
dx
x
dx
=
(2 x) =
dt 2 x 2 – 4
dt
x 2 – 4 dt
dy
3
15
When x = 3,
=
≈ 6.7 units/s
(5) =
2
dt
5
3 –4
dr
= 0.02
dt
dA
dr
= 2πr
dt
dt
dA
When r = 8.1,
= 2π(0.02)(8.1) = 0.324π
dt
≈ 1.018 in.2/s
13. A = πr 2 ;
dx
dy
= 30,
= 24
dt
dt
ds
dx
dy
2s = 2 x + 2( y + 48)
dt
dt
dt
ds
dx
dy
s = x + ( y + 48)
dt
dt
dt
At 2:00 p.m., x = 3(30) = 90, y = 3(24) = 72,
so s = 150.
ds
(150) = 90(30) + (72 + 48)(24)
dt
ds 5580
=
= 37.2 knots/h
dt 150
14. s 2 = x 2 + ( y + 48) 2 ;
dx
=5
dt
dy
dx
= 2x
dt
dt
When y = 150, x = 120, so
dy x dx 120
=
=
(5) = 4 ft/s
dt y dt 150
2y
134
Section 2.8
Instructor’s Resource Manual
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15. Let x be the distance from the beam to the point
opposite the lighthouse and θ be the angle
between the beam and the line from the
lighthouse to the point opposite.
x dθ
tan θ = ;
= 2(2π) = 4π rad/min,
1 dt
dθ dx
sec2 θ
=
dt dt
1
1
5
At x = , θ = tan –1 and sec2 θ = .
2
2
4
dx 5
= (4π) ≈ 15.71 km/min
dt 4
4000
x
dθ
4000 dx
sec2 θ
=−
dt
x 2 dt
1 dθ
1
4000
and x =
=
≈ 7322.
When θ = ,
2 dt 10
tan 12
16. tan θ =
dx
≈ sec2
dt
1 ⎛ 1 ⎞ ⎡ (7322) 2 ⎤
⎥
⎜ ⎟ ⎢−
2 ⎝ 10 ⎠ ⎣⎢ 4000 ⎦⎥
≈ –1740 ft/s or –1186 mi/h
The plane’s ground speed is 1186 mi/h.
17. a.
Let x be the distance along the ground from
the light pole to Chris, and let s be the
distance from Chris to the tip of his shadow.
6
30
x
By similar triangles, =
, so s =
s x+s
4
ds 1 dx dx
and
=
.
= 2 ft/s, hence
dt 4 dt dt
ds 1
= ft/s no matter how far from the light
dt 2
pole Chris is.
b. Let l = x + s, then
dl dx ds
1 5
=
+
= 2 + = ft/s.
dt dt dt
2 2
c.
The angular rate at which Chris must lift his
head to follow his shadow is the same as the
rate at which the angle that the light makes
with the ground is decreasing. Let θ be the
angle that the light makes with the ground at
the tip of Chris' shadow.
6
dθ
6 ds
tan θ = so sec2 θ
=–
and
s
dt
s 2 dt
6
( )
1
2
2
2
1
dθ
⎛1⎞
=–
⎜ ⎟=– .
2
24
dt
⎝ ⎠
6
Chris must lift his head at the rate of
1
rad/s.
24
18. Let θ be the measure of the vertex angle, a be the
measure of the equal sides, and b be the measure
of the base. Observe that b = 2a sin
θ
θ
2
and the
height of the triangle is a cos .
2
1⎛
θ ⎞⎛
θ⎞ 1
A = ⎜ 2a sin ⎟ ⎜ a cos ⎟ = a 2 sin θ
2⎝
2 ⎠⎝
2⎠ 2
dθ
1
1
A = (100)2 sin θ = 5000sin θ ;
=
dt 10
2
dA
dθ
= 5000 cos θ
dt
dt
π dA
π ⎞⎛ 1 ⎞
⎛
When θ = ,
= 5000 ⎜ cos ⎟ ⎜ ⎟ = 250 3
6 dt
6 ⎠ ⎝ 10 ⎠
⎝
≈ 433 cm 2 min .
19. Let p be the point on the bridge directly above
the railroad tracks. If a is the distance between p
da
and the automobile, then
= 66 ft/s. If l is the
dt
distance between the train and the point directly
dl
= 88 ft/s. The distance from the
below p, then
dt
train to p is 1002 + l 2 , while the distance from
p to the automobile is a. The distance between
the train and automobile is
2
D = a 2 + ⎛⎜ 1002 + l 2 ⎞⎟ = a 2 + l 2 + 1002 .
⎝
⎠
dD
1
dl ⎞
⎛ da
=
⋅ ⎜ 2a
+ 2l ⎟
dt 2 a 2 + l 2 + 1002 ⎝ dt
dt ⎠
=
a da
+ l dl
dt
dt
. After 10 seconds, a = 660
a 2 + l 2 + 1002
and l = 880, so
dD
660(66) + 880(88)
=
≈ 110 ft/s.
dt
6602 + 8802 + 1002
6 cos 2 θ ds ds 1
dθ
= ft/s
=–
.
dt
dt dt 2
s2
π
When s = 6, θ = , so
4
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135
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
h
20. V = πh ⋅ (a 2 + ab + b 2 ); a = 20, b = + 20,
3
4
2
⎞
1 ⎛
h
V = πh ⎜ 400 + 5h + 400 +
+ 10h + 400 ⎟
⎜
⎟
3 ⎝
16
⎠
3
1 ⎛
h ⎞
= π ⎜ 1200h + 15h 2 + ⎟
⎜
3 ⎝
16 ⎟⎠
dV 1 ⎛
3h 2 ⎞ dh
= π ⎜ 1200 + 30h +
⎟
dt 3 ⎜⎝
16 ⎟⎠ dt
dV
= 2000,
When h = 30 and
dt
1 ⎛
675 ⎞ dh 3025π dh
2000 = π ⎜1200 + 900 +
=
⎟
3 ⎝
4 ⎠ dt
4 dt
dh 320
=
≈ 0.84 cm/min.
dt 121π
⎡ h ⎤ dV
21. V = πh 2 ⎢ r – ⎥ ;
= –2, r = 8
⎣ 3 ⎦ dt
πh3
πh3
= 8πh 2 –
3
3
dV
dh
dh
= 16πh
– πh 2
dt
dt
dt
dh
When h = 3, –2 = [16π(3) – π(3)2 ]
dt
dh –2
=
≈ –0.016 ft/hr
dt 39π
V = πrh 2 –
22. s 2 = a 2 + b 2 − 2ab cos θ ;
dθ
π 11π
rad/h
= 2π – =
a = 5, b = 4,
dt
6
6
s 2 = 41 – 40 cos θ
ds
dθ
2s = 40sin θ
dt
dt
π
At 3:00, θ = and s = 41 , so
2
ds
⎛ π ⎞ ⎛ 11π ⎞ 220π
2 41 = 40sin ⎜ ⎟ ⎜
⎟=
dt
3
⎝ 2 ⎠⎝ 6 ⎠
ds
≈ 18 in./hr
dt
23. Let P be the point on the ground where the ball
hits. Then the distance from P to the bottom of
the light pole is 10 ft. Let s be the distance
between P and the shadow of the ball. The height
of the ball t seconds after it is dropped is
64 –16t 2 .
136
Section 2.8
48
By similar triangles,
64 – 16t
(for t > 1), so s =
10t 2 – 40
1 – t2
2
=
10 + s
s
.
ds 20t (1 – t 2 ) – (10t 2 – 40)(–2t )
60t
=
=–
2
2
dt
(1 – t )
(1 – t 2 )2
The ball hits the ground when t = 2,
The shadow is moving
ds
120
=–
.
dt
9
120
≈ 13.33 ft/s.
9
h⎞
⎛
24. V = πh 2 ⎜ r – ⎟ ; r = 20
3⎠
⎝
h⎞
π
⎛
V = πh 2 ⎜ 20 – ⎟ = 20πh 2 − h3
3⎠
3
⎝
dV
dh
= (40πh − πh 2 )
dt
dt
dh
At 7:00 a.m., h = 15,
≈ −3, so
dt
dV
= (40π(15) − π(15) 2 )(−3) ≈ −1125π ≈ −3534.
dt
Webster City residents used water at the rate of
2400 + 3534 = 5934 ft3/h.
25. Assuming that the tank is now in the shape of an
upper hemisphere with radius r, we again let t be
the number of hours past midnight and h be the
height of the water at time t. The volume, V, of
water in the tank at that time is given by
2
π
V = π r 3 − ( r − h) 2 ( 2r + h )
3
3
16000
π
and so V =
π − (20 − h)2 ( 40 + h )
3
3
from which
dV
π
dh 2π
dh
= − (20 − h)2
+
(20 − h) ( 40 + h )
dt
3
dt
3
dt
dV
At t = 7 ,
≈ −525π ≈ −1649
dt
Thus Webster City residents were using water at
the rate of 2400 + 1649 = 4049 cubic feet per
hour at 7:00 A.M.
26. The amount of water used by Webster City can
be found by:
usage = beginning amount + added amount
− remaining amount
Thus the usage is
≈ π (20)2 (9) + 2400(12) − π (20)2 (10.5) ≈ 26,915 ft 3
over the 12 hour period.
Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
27. a.
dx
= 2 ft/s. Let y
dt
y
18
216
=
, so y =
.
be the height of the opposite end of the ladder. By similar triangles,
2
12
144 + x
144 + x 2
dy
216
dx
216 x
dx
=–
2x
=–
2 3/ 2
2 3 / 2 dt
dt
dt
2(144 + x )
(144 + x )
Let x be the distance from the bottom of the wall to the end of the ladder on the ground, so
When the ladder makes an angle of 60° with the ground, x = 4 3 and
b.
d2y
dt 2
Since
d2y
dt 2
=
=
d ⎛
216 x
dx ⎞ d ⎛
216 x
⎜⎜ –
⎟ = ⎜–
dt ⎝ (144 + x 2 )3 / 2 dt ⎟⎠ dt ⎜⎝ (144 + x 2 )3 / 2
dy
216(4 3)
=–
⋅ 2 = –1.125 ft/s.
dt
(144 + 48)3 / 2
⎞ dx
216 x
d2x
⋅
⎟⎟ –
2 3/ 2
dt 2
⎠ dt (144 + x )
dx
d2x
= 2,
= 0, thus
dt
dt 2
( )
⎡ –216(144 + x 2 )3 / 2 dx + 216 x 3
2
dt
=⎢
2 3
⎢
(144 + x )
⎢⎣
–216(144 + x 2 ) + 648 x 2 ⎛ dx ⎞
⎜ ⎟
⎝ dt ⎠
(144 + x 2 )5 / 2
2
=
⎤
144 + x 2 (2 x) dx
dt ⎥ dx
⎥ dt
⎥⎦
432 x 2 – 31,104 ⎛ dx ⎞
⎜ ⎟
(144 + x 2 )5 / 2 ⎝ dt ⎠
2
When the ladder makes an angle of 60° with the ground,
d 2 y 432 ⋅ 48 – 31,104 2
=
(2) ≈ –0.08 ft/s2
dt 2
(144 + 48)5 / 2
28. a.
If the ball has radius 6 in., the volume of the
water in the tank is
V = 8πh 2 –
πh3 4 ⎛ 1 ⎞
– π⎜ ⎟
3
3 ⎝2⎠
3
dV
= k (4πr 2 )
dt
a.
πh3 π
–
3
6
dV
dh
dh
= 16πh
– πh 2
dt
dt
dt
V=
4 3
πr
3
dV
dr
= 4πr 2
dt
dt
= 8πh 2 –
This is the same as in Problem 21, so
29.
k (4πr 2 ) = 4πr 2
dh
is
dt
dr
dt
dr
=k
dt
again –0.016 ft/hr.
b. If the ball has radius 2 ft, and the height of
the water in the tank is h feet with 2 ≤ h ≤ 3 ,
the part of the ball in the water has volume
4
4 – h ⎤ (6 – h)h 2 π
⎡
π(2)3 – π(4 – h) 2 ⎢ 2 –
=
.
3
3 ⎥⎦
3
⎣
The volume of water in the tank is
πh3 (6 – h)h 2 π
V = 8πh 2 –
–
= 6h 2 π
3
3
dV
dh
= 12hπ
dt
dt
dh
1 dV
=
dt 12hπ dt
dh
1
When h = 3,
=
(–2) ≈ –0.018 ft/hr.
dt 36π
Instructor's Resource Manual
b. If the original volume was V0 , the volume
after 1 hour is
was r0 = 3
8
V0 . The original radius
27
3
V0 while the radius after 1
4π
8
3
2
dr
is
V0 ⋅
= r0 . Since
dt
27
4π 3
dr
1
constant,
= – r0 unit/hr. The snowball
dt
3
will take 3 hours to melt completely.
hour is r1 = 3
Section 2.8
137
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30. PV = k
dV
dP
P
+V
=0
dt
dt
dP
≈ –30, V = 300
dt
dV
V dP
300
=–
=–
(–30) ≈ 134 in.3/min
dt
P dt
67
At t = 6.5, P ≈ 67,
31. Let l be the distance along the ground from the
brother to the tip of the shadow. The shadow is
3
5
or
controlled by both siblings when =
l l+4
l = 6. Again using similar triangles, this occurs
y 6
when
= , so y = 40. Thus, the girl controls
20 3
the tip of the shadow when y ≥ 40 and the boy
controls it when y < 40.
Let x be the distance along the ground from the
dx
= –4
light pole to the girl.
dt
4
20
5
When y ≥ 40,
=
or y = x.
3
y
y–x
20
20
3
( x + 4).
=
or y =
17
y
y – ( x + 4)
x = 30 when y = 40. Thus,
⎧ 4
if x ≥ 30
⎪⎪ 3 x
y=⎨
⎪ 20 ( x + 4) if x < 30
⎪⎩ 17
and
⎧ 4 dx
if x ≥ 30
dy ⎪⎪ 3 dt
=⎨
dt ⎪ 20 dx
if x < 30
⎪⎩ 17 dt
Hence, the tip of the shadow is moving at the rate
4
16
ft/s when the girl is at least 30 feet
of (4) =
3
3
from the light pole, and it is moving
20
80
ft/s when the girl is less than 30 ft
(4) =
17
17
from the light pole.
When y < 40,
Problem Set 2.9
1. dy = (2x + 1)dx
2. dy = (21x 2 + 6 x)dx
3. dy = –4(2 x + 3) –5 (2)dx = –8(2 x + 3) –5 dx
4. dy = –2(3 x 2 + x + 1) –3 (6 x + 1)dx
= –2(6 x + 1)(3x 2 + x + 1) –3 dx
5. dy = 3(sin x + cos x)2 (cos x – sin x) dx
6. dy = 3(tan x + 1) 2 (sec2 x)dx
= 3sec2 x(tan x + 1)2 dx
3
7. dy = – (7 x 2 + 3x –1) –5 / 2 (14 x + 3)dx
2
3
= − (14 x + 3)(7 x 2 + 3 x − 1) −5 2 dx
2
1
8. dy = 2( x10 + sin 2 x )[10 x9 +
⋅ (cos 2 x )(2)]dx
2 sin 2 x
⎛
cos 2 x ⎞ 10
= 2 ⎜ 10 x9 +
⎟ ( x + sin 2 x )dx
sin 2 x ⎠
⎝
9. ds =
=
10. a.
b.
3 2
(t – cot t + 2)1/ 2 (2t + csc2 t )dt
2
3
(2t + csc2 t ) t 2 – cot t + 2dt
2
dy = 3 x 2 dx = 3(0.5)2 (1) = 0.75
dy = 3x 2 dx = 3(–1)2 (0.75) = 2.25
11.
2.9 Concepts Review
1.
f ′( x)dx
2. Δy; dy
12. a.
dy = –
x
3. Δx is small.
4. larger ; smaller
dx
b.
dy = –
dx
x
138
Section 2.9
2
2
=–
=–
0.5
(1)2
= –0.5
0.75
(–2)2
= –0.1875
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13.
20. y = 3 x ; dy =
1 –2 / 3
1
x
dx =
dx;
3 2
3
3 x
x = 27, dx = –0.09
1
dy =
(–0.09) ≈ –0.0033
33 (27)2
3
26.91 ≈ 3 27 + dy = 3 – 0.0033 = 2.9967
21. V =
4 3
πr ; r = 5, dr = 0.125
3
dV = 4πr 2 dr = 4π(5)2 (0.125) ≈ 39.27 cm3
Δy = (1.5)3 – (0.5)3 = 3.25
14. a.
Δy = (–0.25)3 – (–1)3 = 0.984375
b.
Δy =
15. a.
b.
V≈
Δy = [(2.88) – 3] – [(3) – 3] = –0.7056
dy = 2xdx = 2(3)(–0.12) = –0.72
2
Δy = [(3) 4 + 2(3)] – [(2)4 + 2(2)] = 67
17. a.
dy = (4 x3 + 2)dx = [4(2)3 + 2](1) = 34
b.
Δy = [(2.005)4 + 2(2.005)] – [(2)4 + 2(2)]
≈ 0.1706
dy = (4 x3 + 2)dx = [4(2)3 + 2](0.005) = 0.17
18. y = x ; dy =
dy =
1
2 400
1
2 x
dx; x = 400, dx = 2
(2) = 0.05
402 ≈ 400 + dy = 20 + 0.05 = 20.05
19. y = x ; dy =
dy =
1
2 36
1
2 x
dx; x = 36, dx = –0.1
(–0.1) ≈ –0.0083
35.9 ≈ 36 + dy = 6 – 0.0083 = 5.9917
Instructor’s Resource Manual
4 3
πr ; r = 6 ft = 72in., dr = –0.3
3
dV = 4πr 2 dr = 4π(72)2 (–0.3) ≈ –19,543
Δy = [(2.5) 2 – 3] – [(2) 2 – 3] = 2.25
dy = 2xdx = 2(2)(0.5) = 2
2
dV = 3 x 2 dx = 3( 3 40)2 (0.5) ≈ 17.54 in.3
23. V =
1
1
Δy =
+ = –0.3
–1.25 2
b.
16. a.
1 1
1
– =–
1.5 1
3
22. V = x3 ; x = 3 40, dx = 0.5
4
π(72)3 –19,543
3
≈ 1,543,915 in 3 ≈ 893 ft 3
24. V = πr 2 h; r = 6 ft = 72in., dr = −0.05,
h = 8ft = 96in.
dV = 2πrhdr = 2π(72)(96)(−0.05) ≈ −2171in.3
About 9.4 gal of paint are needed.
25. C = 2π r ; r = 4000 mi = 21,120,000 ft, dr = 2
dC = 2π dr = 2π (2) = 4π ≈ 12.6 ft
L
; L = 4, dL = –0.03
32
π
2π 1
dT =
⋅ ⋅ dL =
dL
32 L
2 L 32
26. T = 2π
32
dT =
π
(–0.03) ≈ –0.0083
32(4)
The time change in 24 hours is
(0.0083)(60)(60)(24) ≈ 717 sec
27. V =
4 3 4
πr = π(10)3 ≈ 4189
3
3
dV = 4πr 2 dr = 4π(10) 2 (0.05) ≈ 62.8 The
volume is 4189 ± 62.8 cm3.
The absolute error is ≈ 62.8 while the relative
error is 62.8 / 4189 ≈ 0.015 or 1.5% .
Section 2.9
139
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28. V = πr 2 h = π(3) 2 (12) ≈ 339
dV = 24πrdr = 24π(3)(0.0025) ≈ 0.565
The volume is 339 ± 0.565 in.3
The absolute error is ≈ 0.565 while the relative
error is 0.565 / 339 ≈ 0.0017 or 0.17% .
29. s = a 2 + b 2 – 2ab cos θ
= 1512 + 1512 – 2(151)(151) cos 0.53 ≈ 79.097
s = 45, 602 – 45, 602 cos θ
ds =
1
2 45, 602 – 45, 602 cos θ
22,801sin θ
=
45, 602 – 45, 602 cosθ
dθ
22,801sin 0.53
=
⋅ 45, 602sin θ dθ
45, 602 – 45, 602 cos 0.53
(0.005) ≈ 0.729
s ≈ 79.097 ± 0.729 cm
The absolute error is ≈ 0.729 while the relative
error is 0.729 / 79.097 ≈ 0.0092 or 0.92% .
1
1
ab sin θ = (151)(151) sin 0.53 ≈ 5763.33
2
2
22,801
A=
sin θ ;θ = 0.53, dθ = 0.005
2
22,801
dA =
(cos θ )dθ
2
22,801
=
(cos 0.53)(0.005) ≈ 49.18
2
A ≈ 5763.33 ± 49.18 cm2
The absolute error is ≈ 49.18 while the relative
error is 49.18 / 5763.33 ≈ 0.0085 or 0.85% .
30. A =
31. y = 3 x 2 – 2 x + 11; x = 2, dx = 0.001
dy = (6x – 2)dx = [6(2) – 2](0.001) = 0.01
d2y
= 6, so with Δx = 0.001,
dx 2
1
Δy – dy ≤ (6)(0.001) 2 = 0.000003
2
32. Using the approximation
f ( x + Δx) ≈ f ( x) + f '( x)Δx
we let x = 1.02 and Δx = −0.02 . We can rewrite
the above form as
f ( x) ≈ f ( x + Δx) − f '( x)Δx
which gives
f (1.02) ≈ f (1) − f '(1.02)( −0.02)
= 10 + 12(0.02) = 10.24
140
Section 2.9
33. Using the approximation
f ( x + Δx) ≈ f ( x) + f '( x)Δx
we let x = 3.05 and Δx = −0.05 . We can rewrite
the above form as
f ( x) ≈ f ( x + Δx) − f '( x)Δx
which gives
f (3.05) ≈ f (3) − f '(3.05)(−0.05)
1
= 8 + (0.05) = 8.0125
4
34. From similar triangles, the radius at height h is
2
1
4
h. Thus, V = πr 2 h = πh3 , so
5
3
75
4
dV =
πh 2 dh. h = 10, dh = –1:
25
4
dV =
π(100)(−1) ≈ −50 cm3
25
The ice cube has volume 33 = 27 cm3 , so there is
room for the ice cube without the cup
overflowing.
4
35. V = πr 2 h + πr 3
3
4
V = 100πr 2 + πr 3 ; r = 10, dr = 0.1
3
dV = (200πr + 4πr 2 )dr
= (2000π + 400π)(0.1) = 240π ≈ 754 cm3
36. The percent increase in mass is
m ⎛ v2 ⎞
dm = – 0 ⎜ 1 – ⎟
2 ⎜⎝ c 2 ⎟⎠
m v ⎛ v2
= 0 ⎜1 –
c 2 ⎜⎝ c 2
⎞
⎟
⎟
⎠
–3 / 2
dm
.
m
⎛ 2v ⎞
⎜ – 2 ⎟ dv
⎝ c ⎠
–3 / 2
dv
–1
dm v ⎛ v 2 ⎞
v ⎛ c2 ⎞
=
1 – ⎟ dv = ⎜
⎜
⎟ dv
m c 2 ⎜⎝ c 2 ⎟⎠
c 2 ⎜⎝ c 2 − v 2 ⎟⎠
v
=
dv
2
c − v2
v = 0.9c, dv = 0.02c
0.9c
0.018
dm
(0.02c) =
=
≈ 0.095
0.19
m c 2 − 0.81c 2
The percent increase in mass is about 9.5.
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37.
f ( x) = x 2 ; f '( x) = 2 x; a = 2
The linear approximation is then
L( x) = f (2) + f '(2)( x − 2)
39. h( x) = sin x; h '( x) = cos x; a = 0
The linear approximation is then
L( x) = 0 + 1( x − 0) = x
= 4 + 4( x − 2) = 4 x − 4
38. g ( x) = x cos x; g '( x) = − x sin x + 2 x cos x
a =π /2
The linear approximation is then
2
2
40. F ( x ) = 3x + 4; F '( x) = 3; a = 3
The linear approximation is then
L( x) = 13 + 3( x − 3) = 13 + 3 x − 9
= 3x + 4
2
π⎞
⎛π ⎞ ⎛
L( x) = 0 + − ⎜ ⎟ ⎜ x − ⎟
2⎠
⎝2⎠ ⎝
=−
π2
4
L( x) = 0 + −
=−
π2
4
x+
π3
8
π ⎛
π⎞
⎜x− ⎟
4 ⎝
2⎠
2
x+
π3
8
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Section 2.9
141
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
41.
f ( x ) = 1 − x2 ;
(
45.
)
−1/ 2
1
1 − x2
( −2 x )
2
−x
=
, a=0
1 − x2
The linear approximation is then
L ( x ) = 1 + 0 ( x − 0) = 1
f ′( x) =
f (x ) = mx + b; f ′(x ) = m
The linear approximation is then
L(x ) = ma + b + m(x − a ) = am + b + mx − ma
f ( x ) = L(x )
= mx + b
46. L ( x ) − f ( x ) = a +
=
x
2 a
(
=
)
2 a
2 a
( x − a) −
x
a x−2 a x +a
=
2
2 a
− x+
x− a
1
2
≥0
47. The linear approximation to f ( x ) at a is
L( x) = f (a) + f '(a)( x − a)
42. g ( x ) =
x
1 − x2
= a 2 + 2a ( x − a )
;
(1 − x ) − x ( −2 x ) = 1 + x
g '( x) =
(1 − x )
(1 − x )
2
2
2 2
2 2
,a =
= 2ax − a 2
Thus,
1
2
(
f ( x) − L( x) = x 2 − 2ax − a 2
)
= x 2 − 2ax + a 2
The linear approximation is then
2 20 ⎛
1 ⎞ 20
4
L(x ) = +
x−
⎜x− ⎟ =
3 9 ⎝
2⎠ 9
9
= ( x − a)2
≥0
48.
f (x ) = (1 + x )α , f ′(x ) = α (1 + x )α −1 , a = 0
The linear approximation is then
L(x ) = 1 + α (x ) = αx + 1
y
5
43. h(x ) = x sec x; h ′(x ) = sec x + x sec x tan x, a = 0
The linear approximation is then
L(x ) = 0 + 1(x − 0) = x
−5
5
x
−5
α = −2
y
44. G (x ) = x + sin 2 x; G ′(x ) = 1 + 2 cos 2 x , a = π / 2
The linear approximation is then
π
π⎞
⎛
L(x ) = + (− 1)⎜ x − ⎟ = − x + π
2
2⎠
⎝
5
−5
5
−5
142
Section 2.9
x
α = −1
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49. a. lim ε ( h ) = lim ( f ( x + h ) − f ( x ) − f ′ ( x ) h )
y
h→0
⎡ f ( x + h) − f ( x)
⎤
= lim ⎢
− f ′ ( x )⎥
h→0 h
h
⎣
⎦
= f ′( x) − f ′( x) = 0
b. lim
−5
5
x
−5
ε (h)
α = −0.5
y
2.10 Chapter Review
Concepts Test
5
−5
5
1. False:
If f ( x) = x3 , f '( x) = 3 x 2 and the
tangent line y = 0 at x = 0 crosses the
curve at the point of tangency.
2. False:
The tangent line can touch the curve
at infinitely many points.
3. True:
mtan = 4 x3 , which is unique for each
value of x.
4. False:
mtan = – sin x, which is periodic.
5. True:
If the velocity is negative and
increasing, the speed is decreasing.
6. True:
If the velocity is negative and
decreasing, the speed is increasing.
7. True:
If the tangent line is horizontal, the
slope must be 0.
8. False:
f ( x) = ax 2 + b, g ( x) = ax 2 + c,
b ≠ c . Then f ′( x) = 2ax = g ′( x), but
f(x) ≠ g(x).
9. True:
Dx f ( g ( x)) = f ′( g ( x)) g ′( x); since
g(x) = x, g ′( x) = 1, so
Dx f ( g ( x)) = f ′( g ( x)).
x
−5
α =0
y
5
−5
5
x
−5
α = 0.5
y
5
−5
5
x
−5
α =1
10. False:
Dx y = 0 because π is a constant, not
a variable.
11. True:
Theorem 3.2.A
12. True:
The derivative does not exist when the
tangent line is vertical.
13. False:
( f ⋅ g )′( x) = f ( x) g ′( x) + g ( x) f ′( x)
14. True:
Negative acceleration indicates
decreasing velocity.
y
5
−5
h →0
= f ( x) − f ( x) − f ′( x) 0 = 0
5
5
−5
Instructor’s Resource Manual
x
α =2
Section 2.10
143
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15. True:
16. False:
If f ( x) = x3 g ( x), then
29. True:
Dx f ( x) = x3 g ′( x) + 3x 2 g ( x)
Dx2 (sin x ) = – sin x;
= x 2 [ xg ′( x) + 3 g ( x)].
Dx3 (sin x) = – cos x;
Dx4 (sin x) = sin x;
Dx y = 3 x 2 ; At (1, 1):
mtan = 3(1) = 3
Tangent line: y – 1 = 3(x – 1)
Dx5 (sin x) = cos x
2
17. False:
30. False:
Dx3 (cos x) = sin x;
Since D1x+3 (cos x) = D1x (sin x),
Dxn +3 (cos x) = Dxn (sin x).
8
Dx25 y = 0.
31. True:
19. True:
f ( x) = ax n ; f ′( x) = anx n –1
20. True:
Dx
21. True:
h′( x) = f ( x) g ′( x) + g ( x) f ′( x)
h′(c) = f (c) g ′(c) + g (c) f ′(c)
= f(c)(0) + g(c)(0) = 0
22. True:
Dx4 (cos x ) = Dx [ Dx3 (cos x)] = Dx (sin x)
The degree of y = ( x + x) is 24, so
3
f ( x) g ( x) f ′( x) – f ( x) g ′( x)
=
g ( x)
g 2 ( x)
sin x – sin
⎛π⎞
f ′ ⎜ ⎟ = lim
x – π2
⎝ 2 ⎠ x→ π
32. True:
33. True:
( π2 )
x→ π
2
23. True:
tan x 1
sin x
= lim
3 x →0 x cos x
x →0 3 x
1
1
= ⋅1 =
3
3
lim
ds
= 15t 2 + 6 which is greater
dt
than 0 for all t.
v=
V=
4 3
πr
3
dV
dr
= 4πr 2
dt
dt
dV
dr
3
=
= 3, then
so
If
dt 4πr 2
dt
dr
> 0.
dt
2
= lim
Dx (cos x ) = – sin x;
Dx2 (cos x) = – cos x;
Dx y = f ( x) g ′( x) + g ( x) f ′( x)
Dx2 y = f ( x) g ′′( x) + g ′( x) f ′( x)
+ g ( x) f ′′( x) + f ′( x) g ′( x)
= f ( x) g ′′( x) + 2 f ′( x) g ′( x) + f ′′( x) g ( x)
18. True:
Dx (sin x ) = cos x;
sin x –1
x – π2
d 2r
D 2 (kf ) = kD 2 f and
dt 2
D2 ( f + g ) = D2 f + D2 g
=–
d 2r
dr
so
<0
dt 2
2πr 3 dt
3
24. True:
h′( x) = f ′( g ( x)) ⋅ g ′( x)
h′(c) = f ′( g (c)) ⋅ g ′(c) = 0
34. True:
When h > r, then
25. True:
( f D g )′(2) = f ′( g (2)) ⋅ g ′(2)
= f ′(2) ⋅ g ′(2) = 2 ⋅ 2 = 4
35. True:
V=
26. False:
27. False:
28. True:
144
Consider f ( x) = x . The curve
always lies below the tangent.
The rate of volume change depends
on the radius of the sphere.
dr
=4
dt
dc
dr
= 2π = 2π(4) = 8π
dt
dt
c = 2π r ;
Section 2.10
d 2h
dt 2
>0
4 3
πr , S = 4πr 2
3
dV = 4πr 2 dr = S ⋅ dr
If Δr = dr, then dV = S ⋅ Δr
36. False:
dy = 5 x 4 dx, so dy > 0 when dx > 0,
but dy < 0 when dx < 0.
37. False:
The slope of the linear approximation
is equal to
f '(a ) = f '(0) = − sin(0) = 0 .
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Sample Test Problems
3( x + h)3 – 3x3
9 x 2 h + 9 xh 2 + 3h3
= lim (9 x 2 + 9 xh + 3h 2 ) = 9 x 2
= lim
h
h
h →0
h →0
h →0
1. a.
f ′( x) = lim
b.
f ′( x) = lim
[2( x + h)5 + 3( x + h)] – (2 x5 + 3 x)
10 x 4 h + 20 x3 h 2 + 20 x 2 h3 + 10 xh 4 + 2h5 + 3h
= lim
h
h
h →0
h →0
= lim (10 x 4 + 20 x3 h + 20 x 2 h 2 + 10 xh3 + 2h 4 + 3) = 10 x 4 + 3
h →0
1
3( x + h )
– 31x
⎛
⎞
⎡
⎤1
1
1
h
= lim ⎢ –
⎥ = lim – ⎜ 3x( x + h) ⎟ = – 2
h →0 ⎝
h →0 ⎣ 3( x + h) x ⎦ h
3x
⎠
c.
f ′( x) = lim
d.
⎡⎛
⎡ 3x 2 + 2 – 3( x + h) 2 – 2 1 ⎤
1
1 ⎞ 1⎤
f ′( x) = lim ⎢⎜
= lim ⎢
⋅ ⎥
–
⎥
⎟
2
2
h →0 ⎢⎜⎝ 3( x + h) 2 + 2 3 x 2 + 2 ⎟⎠ h ⎥
⎣
⎦ h→0 ⎣⎢ (3( x + h) + 2)(3x + 2) h ⎦⎥
h →0
h
⎡
–6 xh – 3h 2
1⎤
–6 x – 3h
6x
=–
= lim ⎢
⋅ ⎥ = lim
2
2
2
2
2
h→0 ⎢ (3( x + h) + 2)(3 x + 2) h ⎥
0
h
→
(3( x + h) + 2)(3x + 2)
(3 x + 2)2
⎣
⎦
e.
f ′( x) = lim
3( x + h) – 3 x
( 3x + 3h – 3x )( 3x + 3h + 3 x )
= lim
h
h →0
h( 3 x + 3h + 3x )
= lim
3h
h →0
h →0 h(
f.
g.
3x + 3h + 3x
=
3
2 3x
sin[3( x + h)] – sin 3x
sin(3x + 3h) – sin 3 x
= lim
h
h
h →0
sin 3 x cos 3h + sin 3h cos 3 x – sin 3x
sin 3 x(cos 3h –1)
sin 3h cos 3 x
= lim
+ lim
= lim
h
h
h
h →0
h →0
h →0
cos 3h –1
sin 3h
sin 3h
= 3sin 3 x lim
+ cos 3 x lim
= (3sin 3x)(0) + (cos 3 x)3 lim
= (cos 3x)(3)(1) = 3cos 3 x
3h
h →0
h →0 h
h →0 3h
h →0
⎛ ( x + h) 2 + 5 – x 2 + 5 ⎞ ⎛ ( x + h) 2 + 5 + x 2 + 5 ⎞
⎜
⎟⎜
⎟
( x + h) 2 + 5 – x 2 + 5
⎠⎝
⎠
f ′( x) = lim
= lim ⎝
h
h →0
h →0
h ⎛⎜ ( x + h)2 + 5 + x 2 + 5 ⎞⎟
⎝
⎠
h→0
2. a.
h →0
3
f ′( x) = lim
= lim
h.
3 x + 3h + 3 x )
= lim
2 xh + h 2
h ⎛⎜ ( x + h) 2 + 5 + x 2 + 5 ⎞⎟
⎝
⎠
= lim
h→0
2x + h
( x + h) + 5 + x + 5
2
2
=
2x
2 x +5
2
=
x
x +5
2
cos[π( x + h)] – cos πx
cos(πx + πh) – cos πx
cos πx cos πh – sin πx sin πh – cos πx
= lim
= lim
h
h
h
h→0
h→0
1 – cos πh ⎞
sin πh ⎞
⎛
⎛
= lim ⎜ – π cos πx
⎟ − lim ⎜ π sin πx
⎟ = (–π cos πx)(0) – (π sin πx) = – π sin πx
πh
πh ⎠
h→0 ⎝
⎠ h→0 ⎝
f ′( x) = lim
h →0
2t 2 – 2 x 2
2(t – x)(t + x)
= lim
t–x
t–x
t→x
t→x
= 2 lim (t + x) = 2(2 x) = 4 x
g ′( x) = lim
t→x
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b.
(t 3 + t ) – ( x3 + x)
t–x
t→x
g ′( x) = lim
= lim
t→x
g.
(t – x)(t + tx + x ) + (t – x)
t–x
2
2
t→x
c.
g ′( x) = lim
t→x
t–x
= lim
x–t
t → x tx(t – x)
t→x
= lim
–1
1
=–
t → x tx
x2
= lim
d.
h.
+ 1)( x 2 + 1)(t – x)
–( x + t )(t – x)
= lim
+ 1)( x 2 + 1)(t – x)
–( x + t )
2x
= lim
=–
2
t → x (t 2 + 1)( x 2 + 1)
( x + 1)2
t → x (t 2
e.
t– x
t→x t – x
g ′( x ) = lim
= lim
( t – x )( t + x )
t→x
= lim
t → x (t
=
f.
(t – x)( t + x )
t–x
– x)( t + x )
= lim
t→x
1
3. a.
t+ x
1
sin πt – sin πx
t–x
t→x
Let v = t – x, then t = v + x and as
t → x, v → 0.
sin πt – sin πx
sin π(v + x) – sin πx
lim
= lim
t–x
v
t→x
v →0
sin πv cos πx + sin πx cos πv – sin πx
= lim
v
v →0
sin πv
cos πv –1 ⎤
⎡
= lim ⎢ π cos πx
+ π sin πx
πv
πv ⎥⎦
v →0 ⎣
= π cos πx ⋅1 + π sin πx ⋅ 0 = π cos πx
Other method:
Use the subtraction formula
π(t + x)
π(t − x)
sin πt – sin πx = 2 cos
sin
2
2
Section 2.10
t 3 + C + x3 + C
=
3x 2
2 x3 + C
cos 2t – cos 2 x
t–x
Let v = t – x, then t = v + x and as
t → x, v → 0.
cos 2t – cos 2 x
cos 2(v + x) – cos 2 x
lim
= lim
t–x
v
t→x
v →0
cos 2v cos 2 x – sin 2v sin 2 x – cos 2 x
= lim
v
v →0
cos 2v –1
sin 2v ⎤
⎡
– 2sin 2 x
= lim ⎢ 2 cos 2 x
2v
2v ⎥⎦
v →0 ⎣
= 2 cos 2 x ⋅ 0 – 2sin 2 x ⋅1 = –2sin 2 x
Other method:
Use the subtraction formula
cos 2t − cos 2 x = −2sin(t + x) sin(t − x).
g ′( x) = lim
t→x
f(x) = 3x at x = 1
f ( x) = 4 x3 at x = 2
c.
f ( x) = x3 at x = 1
d. f(x) = sin x at x = π
e.
f ( x) =
4
at x
x
f.
f(x) = –sin 3x at x
g.
f(x) = tan x at x =
h.
4. a.
b.
146
t 2 + tx + x 2
b.
2 x
g ′( x) = lim
(t – x) ⎛⎜ t 3 + C + x3 + C ⎞⎟
⎝
⎠
t→x
x2 – t 2
t → x (t 2
t 3 – x3
= lim
⎡⎛ 1
1 ⎞ ⎛ 1 ⎞⎤
g ′( x) = lim ⎢⎜
–
⎟⎜
⎟⎥
t → x ⎣⎝ t 2 + 1 x 2 + 1 ⎠ ⎝ t – x ⎠ ⎦
= lim
t→x
t 3 + C – x3 + C
t–x
⎛ t 3 + C – x3 + C ⎞ ⎛ t 3 + C + x3 + C ⎞
⎜
⎟⎜
⎟
⎠⎝
⎠
= lim ⎝
t→x
3
3
⎛
⎞
(t – x) ⎜ t + C + x + C ⎟
⎝
⎠
= lim (t 2 + tx + x 2 + 1) = 3x 2 + 1
1– 1
t x
g ′( x) = lim
f ( x) =
1
x
f ′(2) ≈ –
f ′(6) ≈
π
4
at x = 5
3
4
3
2
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c.
Vavg =
6 – 32
7–3
=
9
8
11.
d
d.
f (t 2 ) = f ′(t 2 )(2t )
dt
⎛2⎞ 8
At t = 2, 4 f ′(4) ≈ 4 ⎜ ⎟ =
⎝3⎠ 3
e.
f.
d 2
[ f (t )] = 2 f (t ) f ′(t )
dt
At t = 2,
⎛ 3⎞
2 f (2) f ′(2) ≈ 2(2) ⎜ – ⎟ = –3
⎝ 4⎠
=
=
13.
⎛ 3 ⎞⎛ 3 ⎞ 9
≈ ⎜ – ⎟⎜ – ⎟ =
⎝ 4 ⎠ ⎝ 4 ⎠ 16
5. Dx (3x ) = 15 x
14.
4
6. Dx ( x3 – 3 x 2 + x –2 ) = 3 x 2 – 6 x + (–2) x –3
−3 x 2 + 10 x + 3
(2) + 2t + 6
+ 2t + 6
⎞ d 2
d ⎛
1
⎜
⎟ = ( x + 4) –1/ 2
⎜
2
dx ⎝ x + 4 ⎟⎠ dx
1
= – ( x 2 + 4) –3 / 2 (2 x)
2
x
=–
2
( x + 4)3
d
dx
x2 – 1
3
x –x
=
d 1
d −1 2
1
=
=−
x
dx x dx
2 x3 2
= – sin θ – 3[sin θ (2)(cos θ )(– sin θ ) + cos3 θ ]
= – sin θ + 6sin 2 θ cos θ – 3cos3 θ
16.
( x 2 + 1) 2
d
[sin(t 2 ) – sin 2 (t )] = cos(t 2 )(2t ) – (2sin t )(cos t )
dt
= 2t cos(t 2 ) – sin(2t )
⎛ 4t − 5 ⎞ (6t 2 + 2t )(4) – (4t – 5)(12t + 2)
9. Dt ⎜
⎟=
(6t 2 + 2t )2
⎝ 6t 2 + 2t ⎠
=
2t + 6
2 2t + 6
Dθ2 (sin θ + cos3 θ )
2
⎛ 3 x – 5 ⎞ ( x 2 + 1)(3) – (3 x – 5)(2 x)
8. Dx ⎜
⎟=
( x 2 + 1)2
⎝ x2 + 1 ⎠
=
t
1
= cosθ – 3sin θ cos 2 θ
7. Dz ( z + 4 z + 2 z ) = 3z + 8 z + 2
2
( x3 + x ) 2
15. Dθ (sin θ + cos3 θ ) = cos θ + 3cos 2 θ (– sin θ )
= 3x 2 – 6 x – 2 x –3
3
−4 x 4 + 10 x 2 + 2
12. Dt (t 2t + 6) = t
d
( f ( f (t ))) = f ′( f (t )) f ′(t )
dt
At t = 2, f ′( f (2)) f ′(2) = f ′(2) f ′(2)
5
d ⎛ 4 x 2 – 2 ⎞ ( x3 + x)(8 x) – (4 x 2 – 2)(3 x 2 + 1)
⎜
⎟=
dx ⎜⎝ x3 + x ⎟⎠
( x3 + x ) 2
−24t 2 + 60t + 10
(6t 2 + 2t ) 2
17. Dθ [sin(θ 2 )] = cos(θ 2 )(2θ ) = 2θ cos(θ 2 )
18.
d
(cos3 5 x) = (3cos 2 5 x)(– sin 5 x )(5)
dx
= –15cos 2 5 x sin 5 x
10. Dx (3x + 2) 2 / 3 =
2
(3 x + 2) –1/ 3 (3)
3
= 2(3 x + 2) –1/ 3
2
Dx2 (3x + 2) 2 / 3 = – (3x + 2) –4 / 3 (3)
3
= –2(3x + 2) –4 / 3
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19.
d
[sin 2 (sin(πθ ))] = 2sin(sin(πθ )) cos(sin(πθ ))(cos(πθ ))(π) = 2π sin(sin(πθ )) cos(sin(πθ )) cos(πθ )
dθ
20.
d
[sin 2 (cos 4t )] = 2sin(cos 4t ) ( cos(cos 4t ) ) (– sin 4t )(4) = –8sin(cos 4t ) cos(cos 4t ) sin 4t
dt
21. Dθ tan 3θ = (sec 2 3θ )(3) = 3sec 2 3θ
22.
23.
d ⎛ sin 3 x ⎞ (cos 5 x 2 )(cos 3 x)(3) – (sin 3 x)(– sin 5 x 2 )(10 x ) 3cos 5 x 2 cos 3 x + 10 x sin 3x sin 5 x 2
=
⎜
⎟=
dx ⎝ cos 5 x 2 ⎠
cos 2 5 x 2
cos 2 5 x 2
f ′( x) = ( x 2 –1)2 (9 x 2 – 4) + (3 x3 – 4 x)(2)( x 2 –1)(2 x) = ( x 2 –1)2 (9 x 2 – 4) + 4 x( x 2 –1)(3 x3 – 4 x)
f ′(2) = 672
24. g ′( x) = 3cos 3 x + 2(sin 3 x)(cos 3 x)(3) = 3cos 3x + 3sin 6 x
g ′′( x) = –9sin 3 x + 18cos 6 x
g ′′(0) = 18
25.
d ⎛ cot x ⎞ (sec x 2 )(– csc 2 x) – (cot x)(sec x 2 )(tan x 2 )(2 x) – csc2 x – 2 x cot x tan x 2
=
⎜
⎟=
dx ⎝ sec x 2 ⎠
sec x 2
sec 2 x 2
⎛ 4t sin t ⎞ (cos t – sin t )(4t cos t + 4sin t ) – (4t sin t )(– sin t – cos t )
26. Dt ⎜
⎟=
⎝ cos t – sin t ⎠
(cos t – sin t )2
=
27.
4t cos 2 t + 2sin 2t – 4sin 2 t + 4t sin 2 t
(cos t – sin t ) 2
=
4t + 2sin 2t – 4sin 2 t
(cos t – sin t )2
f ′( x) = ( x – 1)3 2(sin πx – x)(π cos πx – 1) + (sin πx – x) 2 3( x – 1)2
= 2( x – 1)3 (sin πx – x)(π cos πx – 1) + 3(sin πx – x) 2 ( x – 1) 2
f ′(2) = 16 − 4π ≈ 3.43
28. h′(t ) = 5(sin(2t ) + cos(3t )) 4 (2 cos(2t ) – 3sin(3t ))
h′′(t ) = 5(sin(2t ) + cos(3t )) 4 (−4sin(2t ) – 9 cos(3t )) + 20(sin(2t ) + cos(3t ))3 (2 cos(2t ) – 3sin(3t )) 2
h′′(0) = 5 ⋅14 ⋅ (−9) + 20 ⋅13 ⋅ 22 = 35
29. g ′(r ) = 3(cos 2 5r )(– sin 5r )(5) = –15cos 2 5r sin 5r
g ′′(r ) = –15[(cos 2 5r )(cos 5r )(5) + (sin 5r )2(cos 5r )(– sin 5r )(5)] = –15[5cos3 5r – 10(sin 2 5r )(cos 5r )]
g ′′′(r ) = –15[5(3)(cos 2 5r )(– sin 5r )(5) − (10sin 2 5r )(− sin 5r )(5) − (cos 5r )(20sin 5r )(cos 5r )(5)]
= –15[−175(cos 2 5r )(sin 5r ) + 50sin 3 5r ]
g ′′′(1) ≈ 458.8
30.
f ′(t ) = h′( g (t )) g ′(t ) + 2 g (t ) g ′(t )
31. G ′( x ) = F ′(r ( x ) + s ( x))(r ′( x) + s ′( x)) + s ′( x)
G ′′( x) = F ′(r ( x) + s ( x))(r ′′( x) + s ′′( x)) + (r ′( x) + s ′( x)) F ′′(r ( x) + s ( x))(r ′( x) + s ′( x)) + s ′′( x)
= F ′(r ( x) + s ( x))(r ′′( x ) + s ′′( x)) + (r ′( x) + s ′( x))2 F ′′(r ( x) + s ( x)) + s ′′( x)
148
Section 2.10
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32. F ′( x) = Q ′( R ( x)) R ′( x) = 3[ R ( x)]2 (– sin x)
b.
= –3cos 2 x sin x
33. F ′( z ) = r ′( s ( z )) s ′( z ) = [3cos(3s ( z ))](9 z 2 )
= 27 z 2 cos(9 z 3 )
34.
dy
= 2( x – 2)
dx
2x – y + 2 = 0; y = 2x + 2; m = 2
1
2( x – 2) = –
2
7
x=
4
128t – 16t 2 = 0
–16t(t – 8) = 0
The object hits the ground when t = 8s
v = 128 – 32(8) = –128 ft/s
39. s = t 3 – 6t 2 + 9t
ds
v(t ) =
= 3t 2 – 12t + 9
dt
a(t ) =
d 2s
dt 2
= 6t –12
a.
3t 2 – 12t + 9 < 0
3(t – 3)(t – 1) < 0
1 < t < 3; (1,3)
b.
3t 2 – 12t + 9 = 0
3(t – 3)(t – 1) = 0
t = 1, 3
a(1) = –6, a(3) = 6
c.
6t – 12 > 0
t > 2; (2, ∞)
2
1 ⎛7 1 ⎞
⎛7
⎞
y = ⎜ – 2⎟ = ; ⎜ , ⎟
16 ⎝ 4 16 ⎠
⎝4
⎠
35. V =
4 3
πr
3
dV
= 4πr 2
dr
dV
= 4π(5) 2 = 100π ≈ 314 m3 per
dr
meter of increase in the radius.
When r = 5,
4
dV
36. V = πr 3 ;
= 10
3
dt
dV
dr
= 4πr 2
dt
dt
dr
When r = 5, 10 = 4π(5)
dt
dr
1
=
≈ 0.0318 m/h
dt 10π
2
1
6 b
3h
bh(12); = ; b =
2
4 h
2
dV
⎛ 3h ⎞
V = 6 ⎜ ⎟ h = 9h 2 ;
=9
2
dt
⎝ ⎠
dV
dh
= 18h
dt
dt
dh
When h = 3, 9 = 18(3)
dt
dh 1
= ≈ 0.167 ft/min
dt 6
37. V =
38. a.
v = 128 – 32t
v = 0, when t = 4s
s = 128(4) – 16(4) 2 = 256 ft
Instructor’s Resource Manual
40. a.
Dx20 ( x19 + x12 + x5 + 100) = 0
b.
Dx20 ( x 20 + x19 + x18 ) = 20!
c.
Dx20 (7 x 21 + 3 x 20 ) = (7 ⋅ 21!) x + (3 ⋅ 20!)
d.
Dx20 (sin x + cos x) = Dx4 (sin x + cos x)
= sin x + cos x
e.
Dx20 (sin 2 x) = 220 sin 2 x
= 1,048,576 sin 2x
f.
41. a.
b.
20
⎛ 1 ⎞ (–1) (20!) 20!
Dx20 ⎜ ⎟ =
=
⎝ x⎠
x 21
x 21
dy
=0
dx
dy –( x – 1) 1 – x
=
=
dx
y
y
2( x –1) + 2 y
x(2 y )
dy
dy
+ y 2 + y (2 x) + x 2
=0
dx
dx
dy
(2 xy + x 2 ) = –( y 2 + 2 xy )
dx
dy
y 2 + 2 xy
=−
dx
x 2 + 2 xy
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c.
3x 2 + 3 y 2
dy
dy
= x3 (3 y 2 ) + 3 x 2 y 3
dx
dx
a.
dy = –
b.
dy = –
dy
(3 y 2 – 3 x3 y 2 ) = 3x 2 y3 – 3x 2
dx
dy 3x 2 y 3 – 3x 2 x 2 y 3 – x 2
=
=
dx 3 y 2 – 3 x3 y 2 y 2 – x3 y 2
d.
e.
⎡ dy
⎤
x cos( xy ) ⎢ x + y ⎥ + sin( xy ) = 2 x
⎣ dx
⎦
dy
x 2 cos( xy )
= 2 x – sin( xy ) – xy cos( xy )
dx
dy 2 x – sin( xy ) – xy cos( xy )
=
dx
x 2 cos( xy )
⎛ dy
⎞
x sec 2 ( xy ) ⎜ x + y ⎟ + tan( xy ) = 0
dx
⎝
⎠
dy
x 2 sec2 ( xy )
= –[tan( xy ) + xy sec2 ( xy )]
dx
dy
tan( xy ) + xy sec ( xy )
=–
dx
x 2 sec2 ( xy )
2
45. a.
43. dy = [π cos(π x) + 2 x]dx ; x = 2, dx = 0.01
dy = [π cos(2π ) + 2(2)](0.01) = (4 + π )(0.01)
≈ 0.0714
dy
dy
+ y 2 + 2 y[2( x + 2)] + ( x + 2)2 (2)
=0
dx
dx
dy
[2 xy + 2( x + 2) 2 ] = –[ y 2 + 2 y (2 x + 4)]
dx
dy –( y 2 + 4 xy + 8 y )
=
dx
2 xy + 2( x + 2) 2
= 2(3)(4) + 3(2) 2 (5) = 84
b.
c.
Section 2.10
d
[ f ( x) g ( x)] = f ( x) g ′( x) + g ( x) f ′( x)
dx
f (2) g ′(2) + g (2) f ′(2) = (3)(5) + (2)(4) = 23
d
[ f ( g ( x))] = f ′( g ( x)) g ′( x)
dx
f ′( g (2)) g ′(2) = f ′(2) g ′(2) = (4)(5) = 20
Dx [ f 2 ( x)] = 2 f ( x) f ′( x)
dx
=2
dt
dx
dy
0 = 2x + 2 y
dt
dt
dy
x dx
=–
dt
y dt
When y= 5, x = 12, so
12
24
dy
= – (2) = –
= –4.8 ft/s
dt
5
5
dx
y dx
,
= 400
x dt
y = x sin15°
47. sin15° =
dy
dx
= sin15°
dt
dt
dy
= 400sin15° ≈ 104 mi/hr
dt
48. a.
b.
150
d 2
[ f ( x) + g 3 ( x)]
dx
46. (13) 2 = x 2 + y 2 ;
1
3
Since ( y1′ )( y2′ ) = –1 at (1, 2), the tangents are
perpendicular.
2 xy + 2( x + 2)2
When x = –2, y = ±1
(–0.01)
= 2(3)(–1) + 2(4)2 = 26
At (1, 2): y2′ = –
dy = –
2(–2)(–1) + 2(–2 + 2) 2
= 0.0025
= 2 f (2) f ′′(2) + 2[ f ′(2)]2
2x
3y
y 2 + 4 xy + 8 y
(–1)2 + 4(–2)(–1) + 8(–1)
Dx2 [ f 2 ( x)] = 2[ f ( x) f ′′( x) + f ′( x) f ′( x)]
At (1, 2): y1′ = 3
4 x + 6 yy2′ = 0
44. x(2 y )
(–0.01)
2 f (2) f ′(2) + 3g 2 (2) g ′(2)
6x 2
y1′ =
y
y2′ = –
2(–2)(1) + 2(–2 + 2)2
= –0.0025
= 2 f ( x) f ′( x) + 3g 2 ( x) g ′( x)
d.
42. 2 yy1′ = 12 x 2
(1)2 + 4(–2)(1) + 8(1)
2
Dx ( x ) = 2 x ⋅
x
x
2
=
2( x )
x
=
2x2
= 2x
x
x
x ⎛⎜ x ⎞⎟ − x
⎛
x
⎞
x−x
=
=0
Dx2 x = Dx ⎜ ⎟ = ⎝ ⎠
2
x
x2
⎝ x⎠
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c.
Dx3 x = Dx ( Dx2 x ) = Dx (0) = 0
d.
Dx2 (
b.
x ( x − 1)( x − 2 ) = 0
2
cos θ = cot θ sin θ
x = 0, x = 1 or x = 2
The split points are 0, 1, and 2. The expression
on the left can only change signs at the split
points. Check a point in the intervals ( −∞, 0 ) ,
(− sin θ ) = − tan θ cosθ
( 0,1) , (1, 2 ) , and ( 2, ∞ ) . The solution set is
{ x | x ≤ 0 or 1 ≤ x ≤ 2} , or ( −∞, 0] ∪ [1, 2] .
x ) = Dx (2 x) = 2
Dθ sin θ =
49. a.
3. x ( x − 1)( x − 2 ) ≤ 0
Dθ cos θ =
sin θ
sin θ
cos θ
cos θ
1
( x + 1)−1/ 2 ; a = 3
2
L( x) = f (3) + f '(3)( x − 3)
−5 −4 −3 −2 −1 0
f ( x) = x + 1; f '( x) = −
50. a.
4.
5
)
x ( x + 1)( x + 2 ) = 0
x = 0, x = −1, x = −2
The split points are 0, −1 , and −2 . The
expression on the left can only change signs at
the split points. Check a point in the intervals
( −∞, −2 ) , ( −2, −1) , ( −1, 0 ) , and ( 0, ∞ ) . The
solution set is { x | −2 ≤ x ≤ −1 or x ≥ 0} , or
[ −2, −1] ∪ [0, ∞ ) .
−5 −4 −3 −2 −1 0
5.
( x − 2 )( x − 3) < 0
( x − 2 )( x − 3) = 0
x = 2 or x = 3
The split points are 2 and 3. The expression on
the left can only change signs at the split points.
Check a point in the intervals ( −∞, 2 ) , ( 2,3) ,
and ( 3, ∞ ) . The solution set is { x | 2 < x < 3} or
( 2,3) .
2.
4
x ( x + 1)( x + 2 ) ≥ 0
f ( x) = x cos x; f '( x) = − x sin x + cos x; a = 1
L( x) = f (1) + f '(1)( x − 1)
= cos1 + (− sin1 + cos1)( x − 1)
= cos1 − (sin1) x + sin1 + (cos1) x − cos1
= (cos1 − sin1) x + sin1
≈ −0.3012 x + 0.8415
−2 −1 0
3
x3 + 3x 2 + 2 x ≥ 0
(
Review and Preview Problems
1.
2
x x 2 + 3x + 2 ≥ 0
1
= 4 + − (4) −1/ 2 ( x − 3)
2
1
3
1
11
= 2− x+ = − x+
4
4
4
4
b.
1
x ( x − 2)
x2 − 4
x ( x − 2)
( x − 2 )( x + 2 )
1
2
3
4
5
≥0
≥0
The expression on the left is equal to 0 or
undefined at x = 0 , x = 2 , and x = −2 . These
are the split points. The expression on the left can
only change signs at the split points. Check a
point in the intervals: ( −∞, −2 ) , ( −2, 0 ) , ( 0, 2 ) ,
and ( 2, ∞ ) . The solution set is
1
2
3
4
5 6
7
8
x2 − x − 6 > 0
( x − 3)( x + 2 ) > 0
( x − 3)( x + 2 ) = 0
{ x | x < −2 or 0 ≤ x < 2 or x > 2} , or
( −∞, −2 ) ∪ [0, 2 ) ∪ ( 2, ∞ ) .
−5 −4 −3 −2 −1 0
1
2
3
4
5
x = 3 or x = −2
The split points are 3 and −2 . The expression on
the left can only change signs at the split points.
Check a point in the intervals ( −∞, −2 ) , ( −2,3) ,
and ( 3, ∞ ) . The solution set is
{ x | x < −2 or x > 3} , or ( −∞, −2 ) ∪ ( 3, ∞ ) .
−5 −4 −3 −2 −1 0
1
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3
4
5
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6.
x2 − 9
x2 + 2
( x − 3)( x + 3)
>0
>0
x2 + 2
The expression on the left is equal to 0 at x = 3 ,
and x = −3 . These are the split points. The
expression on the left can only change signs at
the split points. Check a point in the intervals:
( −∞, −3) , ( −3, 3) , and ( 3, ∞ ) . The solution set
is { x | x < −3 or x > 3} , or ( −∞, −3) ∪ ( 3, ∞ ) .
−5 −4 −3 −2 −1 0
1
2
3
4
5
7.
f ' ( x ) = 4 ( 2 x + 1) ( 2 ) = 8 ( 2 x + 1)
8.
f ' ( x ) = cos (π x ) ⋅ π = π cos (π x )
9.
f ' ( x ) = x 2 − 1 ⋅ − sin ( 2 x ) ⋅ 2 + cos ( 2 x ) ⋅ ( 2 x )
3
(
(
)
3
)
= −2 x 2 − 1 sin ( 2 x ) + 2 x cos ( 2 x )
10.
f '( x) =
=
11.
x ⋅ sec x tan x − sec x ⋅1
x2
sec x ( x tan x − 1)
f '( x) =
)
=
13.
(
1
1 + sin 2 x
2
sin x cos x
)
−1/ 2
an integer.
17. The line y = 2 + x has slope 1, so any line parallel
to this line will also have a slope of 1.
For the tangent line to y = x + sin x to be parallel
to the given line, we need its derivative to equal 1.
y ' = 1 + cos x = 1
cos x = 0
The tangent line will be parallel to y = 2 + x
( x ) ⋅ 12 x
2
.
= x ( 9 − 2 x )( 24 − 2 x )
( 2sin x )( cos x )
19. Consider the diagram:
1 + sin 2 x
f ' ( x ) = cos
π
18. Length: 24 − 2x
Width: 9 − 2x
Height: x
Volume: l ⋅ w ⋅ h = ( 24 − 2 x )( 9 − 2 x ) x
f ' ( x ) = 2 ( tan 3 x ) ⋅ sec 2 3 x ⋅ 3
= 6 sec 2 3 x ( tan 3 x )
12.
16. The tangent line is horizontal when the derivative
is 0.
y ' = 1 + cos x
The tangent line is horizontal whenever
cos x = −1 . That is, for x = ( 2k + 1) π where k is
whenever x = ( 2k + 1)
x2
(
15. The tangent line is horizontal when the derivative
is 0.
y ' = 2 tan x ⋅ sec 2 x
2 tan x sec x = 0
2sin x
=0
cos 2 x
The tangent line is horizontal whenever
sin x = 0 . That is, for x = kπ where k is an
integer.
1
−1/ 2
=
x
cos x
2 x
(note: you cannot cancel the x here because it
is not a factor of both the numerator and
denominator. It is the argument for the cosine in
the numerator.)
4− x
His distance swimming will be
14.
1
cos 2 x
−1/ 2
f ' ( x ) = ( sin 2 x )
⋅ cos 2 x ⋅ 2 =
2
sin 2 x
12 + x 2 = x 2 + 1 kilometers. His distance
running will be 4 − x kilometers.
Using the distance traveled formula, d = r ⋅ t , we
d
solve for t to get t = . Andy can swim at 4
r
kilometers per hour and run 10 kilometers per
hour. Therefore, the time to get from A to D will
be
152
Review and Preview
x2 + 1 4 − x
+
hours.
4
10
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20. a.
f ( 0 ) = 0 − cos ( 0 ) = 0 − 1 = −1
f (π ) = π − cos (π ) = π − ( −1) = π + 1
Since x − cos x is continuous, f ( 0 ) < 0 ,
and f (π ) > 0 , there is at least one point c
.in the interval ( 0, π ) where f ( c ) = 0 .
(Intermediate Value Theorem)
b.
⎛π ⎞ π
⎛π ⎞ π
f ⎜ ⎟ = − cos ⎜ ⎟ =
⎝2⎠ 2
⎝2⎠ 2
f ' ( x ) = 1 + sin x
⎛π ⎞
⎛π ⎞
f ' ⎜ ⎟ = 1 + sin ⎜ ⎟ = 1 + 1 = 2
⎝2⎠
⎝2⎠
The slope of the tangent line is m = 2 at the
⎛π π ⎞
point ⎜ , ⎟ . Therefore,
⎝2 2⎠
y−
c.
π⎞
π
⎛
= 2 ⎜ x − ⎟ or y = 2 x − .
2
2⎠
2
⎝
π
2x −
2x =
x=
π
2
= 0.
π
2
π
4
The tangent line will intersect the x-axis at
x=
21. a.
π
4
.
The derivative of x 2 is 2x and the
derivative of a constant is 0. Therefore, one
possible function is f ( x ) = x 2 + 3 .
b. The derivative of − cos x is sin x and the
derivative of a constant is 0. Therefore, one
possible function is f ( x ) = − ( cos x ) + 8 .
c.
The derivative of x3 is 3x 2 , so the
1
derivative of x3 is x 2 . The derivative of
3
1
x 2 is 2x , so the derivative of x 2 is x .
2
The derivative of x is 1, and the derivative of
a constant is 0. Therefore, one possible
1
1
function is x3 + x 2 + x + 2 .
3
2
22. Yes. Adding 1 only changes the constant term in
the function and the derivative of a constant is 0.
Therefore, we would get the same derivative
regardless of the value of the constant.
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CHAPTER
Applications of the
Derivative
3
3.1 Concepts Review
1. continuous; closed and bounded
2. extreme
3. endpoints; stationary points; singular points
4.
3
7. Ψ ′( x) = 2 x + 3; 2x + 3 = 0 when x = – .
2
3
Critical points: –2, – , 1
2
9
⎛ 3⎞
Ψ (–2) = –2, Ψ ⎜ – ⎟ = – , Ψ (1) = 4
2
4
⎝
⎠
Maximum value = 4, minimum value = –
f ′(c) = 0; f ′(c) does not exist
9
4
1
6
8. G ′( x) = (6 x 2 + 6 x –12) = ( x 2 + x – 2);
5
5
Problem Set 3.1
x 2 + x – 2 = 0 when x = –2, 1
Critical points: –3, –2, 1, 3
9
7
G (–3) = , G (–2) = 4, G (1) = – , G (3) = 9
5
5
Maximum value = 9,
7
minimum value = –
5
1. Endpoints: −2 , 4
Singular points: none
Stationary points: 0, 2
Critical points: −2, 0, 2, 4
2. Endpoints: −2 , 4
Singular points: 2
Stationary points: 0
Critical points: −2, 0, 2, 4
9.
3. Endpoints: −2 , 4
Singular points: none
Stationary points: −1, 0,1, 2,3
Critical points: −2, −1, 0,1, 2,3, 4
f ′( x) = 3 x 2 – 3; 3x 2 – 3 = 0 when x = –1, 1.
Critical points: –1, 1
f(–1) = 3, f(1) = –1
No maximum value, minimum value = –1
(See graph.)
4. Endpoints: −2 , 4
Singular points: none
Stationary points: none
Critical points: −2, 4
5.
f ′( x) = 2 x + 4; 2 x + 4 = 0 when x = –2.
Critical points: –4, –2, 0
f(–4) = 4, f(–2) = 0, f(0) = 4
Maximum value = 4, minimum value = 0
10.
1
6. h′( x) = 2 x + 1; 2 x + 1 = 0 when x = – .
2
1
Critical points: –2, – , 2
2
1
⎛ 1⎞
h(–2) = 2, h ⎜ – ⎟ = – , h(2) = 6
2
4
⎝
⎠
Maximum value = 6, minimum value = –
154
Section 3.1
f ′( x) = 3 x 2 – 3; 3x 2 – 3 = 0 when x = –1, 1.
3
Critical points: – , –1, 1, 3
2
⎛ 3 ⎞ 17
f ⎜ – ⎟ = , f (–1) = 3, f (1) = –1, f (3) = 19
⎝ 2⎠ 8
Maximum value = 19, minimum value = –1
1
4
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11. h′(r ) = −
1
2
; h′(r ) is never 0; h′(r ) is not defined
15. g ′( x) = −
x → 0+
2x
; −
2 2
2x
(1 + x )
(1 + x 2 ) 2
Critical points: –3, 0, 1
1
1
g(–3) =
, g(0) = 1, g(1) =
10
2
f '( x) = 4x − 4x
1
10
16.
3
(
)
= 4 x x2 − 1
= 4 x ( x − 1)( x + 1)
4 x ( x − 1)( x + 1) = 0 when x = 0,1, −1 .
Critical points: −2, −1, 0,1, 2
f ( −2 ) = 10 ; f ( −1) = 1 ; f ( 0 ) = 2 ; f (1) = 1 ;
f ( 2 ) = 10
Maximum value: 10
Minimum value: 1
14.
f ' ( x ) = 5 x 4 − 25 x 2 + 20
(
)
= 5 ( x 2 − 4 )( x 2 − 1)
4
2
= 5 x − 5x + 4
= 5 ( x − 2 )( x + 2 )( x − 1)( x + 1)
5 ( x − 2 )( x + 2 )( x − 1)( x + 1) = 0 when
x = −2, −1,1, 2
Critical points: −3, −2, −1,1, 2
19
41
f ( −3) = −79 ; f ( −2 ) = − ; f ( −1) = − ;
3
3
35
13
f (1) =
; f ( 2) =
3
3
35
Maximum value:
3
Minimum value: −79
Instructor’s Resource Manual
2x
(1 + x 2 ) 2
= 0 when x = 0.
= 0 when x = 0
Maximum value = 1, minimum value =
13.
; −
As x → ∞, g ( x) → 0+ ; as x → −∞, g ( x) → 0+.
Maximum value = 1, no minimum value
(See graph.)
No maximum value, no minimum value.
12. g ′( x) = −
2 2
(1 + x )
Critical point: 0
g(0) = 1
r
when r = 0, but r = 0 is not in the domain on
[–1, 3] since h(0) is not defined.
Critical points: –1, 3
Note that lim h(r ) = −∞ and lim h( x) = ∞.
x → 0−
2x
f ′( x) =
1 − x2
(1 + x 2 )2
;
1 − x2
= 0 when x = –1, 1
(1 + x 2 )2
Critical points: –1, 1, 4
1
1
4
f (−1) = − , f (1) = , f (4) =
2
2
17
1
Maximum value = ,
2
1
minimum value = –
2
17. r ′(θ ) = cos θ ; cos θ = 0 when θ =
π
+ kπ
2
π π
Critical points: – ,
4 6
1
⎛ π⎞
⎛ π⎞ 1
r⎜− ⎟ = −
, r⎜ ⎟ =
2
⎝ 4⎠
⎝6⎠ 2
1
1
Maximum value = , minimum value = –
2
2
18. s ′(t ) = cos t + sin t ; cos t + sin t = 0 when
π
+ k π.
4
3π
Critical points: 0,
,π
4
⎛ 3π ⎞
s(0) = –1, s ⎜ ⎟ = 2, s (π ) = 1 .
⎝ 4 ⎠
Maximum value = 2,
minimum value = –1
tan t = –1 or t = –
Section 3.1
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19. a ′( x) =
x –1
; a ′( x) does not exist when x = 1.
x –1
Critical points: 0, 1, 3
a(0) = 1, a(1) = 0, a(3) = 2
Maximum value = 2, minimum value = 0
20.
f ′( s ) =
25. g ' (θ ) = θ 2 ( sec θ tan θ ) + 2θ sec θ
= θ sec θ (θ tan θ + 2 )
θ sec θ (θ tan θ + 2 ) = 0 when θ = 0 .
Consider the graph:
y
3(3s – 2)
; f ′( s ) does not exist when s = 2 .
3
3s – 2
Critical points: −1, 2 , 4
3
2
= 0, f(4) = 10
f(–1) = 5, f
3
Maximum value = 10, minimum value = 0
1
( )
21. g ′( x) =
2
; s ′(t ) does not exist when t = 0.
5t
Critical points: –1, 0, 32
s(–1) = 1, s(0) = 0, s(32) = 4
Maximum value = 4, minimum value = 0
Critical points: −
− sin t = 0 when
t = 0, π , 2π ,3π , 4π ,5π , 6π , 7π ,8π
Critical points: 0, π , 2π ,3π , 4π ,5π , 6π , 7π ,8π
H ( 0 ) = 1 ; H (π ) = −1 ; H ( 2π ) = 1 ;
24. g ' ( x ) = 1 − 2 cos x
1
when
2
5π π π 5π
, − , , , 2π
3
3 3 3
⎛ 5π ⎞ −5π
g ( −2π ) = −2π ; g ⎜ −
− 3;
⎟=
3
⎝ 3 ⎠
π
⎛ π⎞
⎛π ⎞ π
g⎜− ⎟ = − + 3 ; g⎜ ⎟ = − 3 ;
3
⎝ 3⎠
⎝3⎠ 3
5
π
5
π
⎛ ⎞
g⎜
+ 3 ; g ( 2π ) = 2π
⎟=
⎝ 3 ⎠ 3
5π
+ 3
Maximum value:
3
5π
− 3
Minimum value: −
3
156
Section 3.1
, 0,
π
4
26. h ' ( t ) =
π2 2
16
; Minimum value: 0
5
( 2 + t ) ⎛⎜ t 2 / 3 ⎞⎟ − t 5/ 3 (1)
⎝3
⎠
( 2 + t )2
⎛5
⎞
⎛ 10 2 ⎞
t2/3 ⎜ (2 + t ) − t ⎟ t2/3 ⎜ + t ⎟
⎝3
⎠=
⎝ 3 3 ⎠
=
( 2 + t )2
( 2 + t )2
=
2t 2 / 3 ( t + 5 )
3( 2 + t )
2
h ' ( t ) is undefined when t = −2 and h ' ( t ) = 0
Maximum value: 1
Minimum value: −1
Critical points: −2π , −
4
2
Maximum value:
H ( 6π ) = 1 ; H ( 7π ) = −1 ; H ( 8π ) = 1
5π π π 5π
,− , ,
3
3 3 3
π
⎛ π⎞ π 2
⎛π ⎞ π 2
g⎜− ⎟ =
; g ( 0) = 0 ; g ⎜ ⎟ =
4
16
16
⎝
⎠
⎝4⎠
H ( 3π ) = −1 ; H ( 4π ) = 1 ; H ( 5π ) = −1 ;
1 − 2 cos x = 0 → cos x =
x
2
3/ 5
23. H ' ( t ) = − sin t
x=−
π
4
−1
1
; f ′( x) does not exist when x = 0.
3x2 / 3
Critical points: –1, 0, 27
g(–1) = –1, g(0) = 0, g(27) = 3
Maximum value = 3, minimum value = –1
22. s ′(t ) =
− π4
when t = 0 or t = −5 . Since −5 is not in the
interval of interest, it is not a critical point.
Critical points: −1, 0,8
h ( −1) = −1 ; h ( 0 ) = 0 ; h ( 8 ) = 16
5
; Minimum value: −1
Maximum value: 16
5
27. a.
f ′( x) = 3 x 2 –12 x + 1;3x 2 –12 x + 1 = 0
when x = 2 –
33
33
and x = 2 +
.
3
3
Critical points: –1, 2 –
33
33
,2+
,5
3
3
⎛
33 ⎞
f(–1) = –6, f ⎜⎜ 2 –
⎟ ≈ 2.04,
3 ⎟⎠
⎝
⎛
33 ⎞
f ⎜⎜ 2 +
⎟ ≈ –26.04, f(5) = –18
3 ⎟⎠
⎝
Maximum value ≈ 2.04;
minimum value ≈ −26.04
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b.
g ′( x) =
( x3 – 6 x 2 + x + 2)(3x 2 – 12 x + 1)
3
2
x – 6x + x + 2
;
29. Answers will vary. One possibility:
y
5
33
g '( x) = 0 when x = 2 –
and
3
33
. g ′( x) does not exist when
3
f(x) = 0; on [–1, 5], f(x) = 0 when
x ≈ –0.4836 and x ≈ 0.7172
33
,
Critical points: –1, –0.4836, 2 –
3
x = 2+
33
, 5
3
g(–1) = 6, g(–0.4836) = 0,
⎛
33 ⎞
g ⎜⎜ 2 –
⎟ ≈ 2.04, g(0.7172) = 0,
3 ⎟⎠
⎝
⎛
33 ⎞
g ⎜⎜ 2 +
⎟ ≈ 26.04, g(5) = 18
3 ⎟⎠
⎝
Maximum value ≈ 26.04,
minimum value = 0
0.7172, 2 +
28. a.
f ′( x) = x cos x; on [–1, 5], x cos x = 0 when
π
3π
x = 0, x = , x =
2
2
π 3π
Critical points: –1, 0, , , 5
2 2
⎛π⎞
f(–1) ≈3.38, f(0) = 3, f ⎜ ⎟ ≈ 3.57,
⎝2⎠
⎛ 3π ⎞
f ⎜ ⎟ ≈ –2.71, f(5) ≈ −2.51
⎝ 2 ⎠
Maximum value ≈ 3.57,
minimum value ≈–2.71
b.
g ′( x) =
(cos x + x sin x + 2)( x cos x)
;
cos x + x sin x + 2
π
3π
, x=
2
2
g ′( x) does not exist when f(x) = 0;
on [–1, 5], f(x) = 0 when x ≈ 3.45
π
3π
Critical points: –1, 0, , 3.45, , 5
2
2
⎛π⎞
g(–1) ≈ 3.38, g(0) = 3, g ⎜ ⎟ ≈ 3.57,
⎝2⎠
⎛ 3π ⎞
g(3.45) = 0, g ⎜ ⎟ ≈ 2.71, g(5) ≈ 2.51
⎝ 2 ⎠
Maximum value ≈ 3.57;
minimum value = 0
g ′( x ) = 0 when x = 0, x =
Instructor’s Resource Manual
−5
5
x
−5
30. Answers will vary. One possibility:
y
5
5
x
−5
31. Answers will vary. One possibility:
y
5
5
x
−5
32. Answers will vary. One possibility:
y
5
5
x
−5
Section 3.1
157
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
33. Answers will vary. One possibility:
y
3.2 Concepts Review
1. Increasing; concave up
5
2.
f ′( x) > 0; f ′′( x) < 0
3. An inflection point
5
x
−5
4.
f ′′(c) = 0; f ′′(c) does not exist.
Problem Set 3.2
34. Answers will vary. One possibility:
y
5
5
x
−5
35. Answers will vary. One possibility:
y
1.
1
2. g ′( x) = 2 x – 1; 2x – 1 > 0 when x > . g(x) is
2
⎡1 ⎞
increasing on ⎢ , ∞ ⎟ and decreasing on
⎣2 ⎠
1⎤
⎛
⎜ – ∞, ⎥ .
2⎦
⎝
3. h′(t ) = 2t + 2; 2t + 2 > 0 when t > –1. h(t) is
increasing on [–1, ∞ ) and decreasing on
( −∞ , –1].
4.
5
5
x
−5
36. Answers will vary. One possibility:
y
5
−5
x
f ′( x) = 3x 2 ; 3 x 2 > 0 for x ≠ 0 .
f(x) is increasing for all x.
5. G ′( x ) = 6 x 2 – 18 x + 12 = 6( x – 2)( x – 1)
Split the x-axis into the intervals (– ∞ , 1), (1, 2),
(2, ∞ ).
3
3
⎛3⎞
Test points: x = 0, , 3; G ′(0) = 12, G ′ ⎜ ⎟ = – ,
2
2
⎝2⎠
G ′(3) = 12
G(x) is increasing on (– ∞ , 1] ∪ [2, ∞ ) and
decreasing on [1, 2].
6.
5
f ′( x) = 3; 3 > 0 for all x. f(x) is increasing
for all x.
f ′(t ) = 3t 2 + 6t = 3t (t + 2)
Split the x-axis into the intervals (– ∞ , –2),
(–2, 0), (0, ∞ ).
Test points: t = –3, –1, 1; f ′(–3) = 9,
f ′(–1) = –3, f ′(1) = 9
f(t) is increasing on (– ∞ , –2] ∪ [0, ∞ ) and
decreasing on [–2, 0].
7. h′( z ) = z 3 – 2 z 2 = z 2 ( z – 2)
Split the x-axis into the intervals (– ∞ , 0), (0, 2),
(2, ∞ ).
Test points: z = –1, 1, 3; h′(–1) = –3, h′(1) = –1,
h′(3) = 9
h(z) is increasing on [2, ∞ ) and decreasing on
(– ∞ , 2].
158
Section 3.2
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8.
f ′( x) =
2– x
16.
3
x
Split the x-axis into the intervals (– ∞ , 0), (0, 2),
(2, ∞ ).
Test points: –1, 1, 3; f ′(–1) = –3, f ′(1) = 1,
1
27
f(x) is increasing on (0, 2] and decreasing on
(– ∞ , 0) ∪ [2, ∞ ).
f ′(3) = –
9. H ′(t ) = cos t ; H ′(t ) > 0 when 0 ≤ t <
π
and
2
17. F ′′( x) = 2sin 2 x – 2 cos 2 x + 4 = 6 – 4 cos 2 x;
6 – 4 cos 2 x > 0 for all x since 0 ≤ cos 2 x ≤ 1.
F(x) is concave up for all x; no inflection points.
18. G ′′( x) = 48 + 24 cos 2 x – 24sin 2 x
= 24 + 48cos 2 x; 24 + 48cos 2 x > 0 for all x.
G(x) is concave up for all x; no inflection points.
3π
< t ≤ 2π.
2
⎡ π ⎤ ⎡ 3π
⎤
H(t) is increasing on ⎢ 0, ⎥ ∪ ⎢ , 2π ⎥ and
⎣ 2⎦ ⎣ 2
⎦
⎡ π 3π ⎤
decreasing on ⎢ , ⎥ .
⎣2 2 ⎦
10. R ′(θ ) = –2 cos θ sin θ ; R ′(θ ) > 0 when
and
π
<θ < π
2
f ′′( x) = 12 x 2 + 48 x = 12 x( x + 4); f ′′( x) > 0 when
x < –4 and x > 0.
f(x) is concave up on (– ∞ , –4) ∪ (0, ∞ ) and
concave down on (–4, 0); inflection points are
(–4, –258) and (0, –2).
19.
f ′( x) = 3 x 2 – 12; 3 x 2 – 12 > 0 when
x < –2 or x > 2.
f(x) is increasing on (– ∞ , –2] ∪ [2, ∞ ) and
decreasing on [–2, 2].
f ′′( x) = 6 x; 6x > 0 when x > 0. f(x) is concave up
on (0, ∞ ) and concave down on (– ∞ , 0).
3π
< θ < 2π.
2
⎡ π ⎤ ⎡ 3π
⎤
R( θ ) is increasing on ⎢ , π ⎥ ∪ ⎢ , 2π ⎥ and
⎣2 ⎦ ⎣ 2
⎦
⎡ π ⎤ ⎡ 3π ⎤
decreasing on ⎢ 0, ⎥ ∪ ⎢ π, ⎥ .
⎣ 2⎦ ⎣ 2 ⎦
11.
f ′′( x) = 2; 2 > 0 for all x. f(x) is concave up for all
x; no inflection points.
12. G ′′( w) = 2; 2 > 0 for all w. G(w) is concave up for
all w; no inflection points.
13. T ′′(t ) = 18t ; 18t > 0 when t > 0. T(t) is concave up
on (0, ∞ ) and concave down on (– ∞ , 0);
(0, 0) is the only inflection point.
14.
f ′′( z ) = 2 –
6
z
4
=
2
z4
( z 4 – 3); z 4 – 3 > 0 for
z < – 4 3 and z > 4 3.
f(z) is concave up on (– ∞, – 4 3) ∪ ( 4 3, ∞) and
concave down on (– 4 3, 0) ∪ (0, 4 3); inflection
20. g ′( x) = 12 x 2 – 6 x – 6 = 6(2 x + 1)( x – 1); g ′( x) > 0
1
or x > 1. g(x) is increasing on
2
1⎤
⎛
⎡ 1 ⎤
⎜ – ∞, – ⎥ ∪ [1, ∞) and decreasing on ⎢ – , 1⎥ .
2⎦
⎣ 2 ⎦
⎝
g ′′( x) = 24 x – 6 = 6(4 x – 1); g ′′( x) > 0 when
when x < –
1
x> .
4
⎛1 ⎞
g(x) is concave up on ⎜ , ∞ ⎟ and concave down
⎝4 ⎠
1⎞
⎛
on ⎜ – ∞, ⎟ .
4⎠
⎝
1 ⎞
1 ⎞
⎛
⎛4
points are ⎜ – 4 3, 3 –
⎟ and ⎜ 3, 3 –
⎟.
3⎠
3⎠
⎝
⎝
15. q ′′( x ) = 12 x 2 – 36 x – 48; q ′′( x) > 0 when x < –1
and x > 4.
q(x) is concave up on (– ∞ , –1) ∪ (4, ∞ ) and
concave down on (–1, 4); inflection points are
(–1, –19) and (4, –499).
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21. g ′( x) = 12 x3 – 12 x 2 = 12 x 2 ( x – 1); g ′( x) > 0
when x > 1. g(x) is increasing on [1, ∞ ) and
decreasing on (−∞,1].
g ′′( x) = 36 x 2 – 24 x = 12 x(3 x – 2); g ′′( x) > 0
2
when x < 0 or x > . g(x) is concave up on
3
⎛2 ⎞
⎛ 2⎞
(– ∞, 0) ∪ ⎜ , ∞ ⎟ and concave down on ⎜ 0, ⎟ .
3
⎝
⎠
⎝ 3⎠
23. G ′( x ) = 15 x 4 – 15 x 2 = 15 x 2 ( x 2 – 1); G ′( x) > 0
when x < –1 or x > 1. G(x) is increasing on
(– ∞ , –1] ∪ [1, ∞ ) and decreasing on [–1, 1].
G ′′( x) = 60 x3 – 30 x = 30 x(2 x 2 – 1);
1 ⎞
⎛
Split the x-axis into the intervals ⎜ −∞, −
⎟,
2⎠
⎝
⎛ 1
⎞ ⎛ 1 ⎞ ⎛ 1
⎞
, 0 ⎟ , ⎜ 0,
, ∞ ⎟.
⎜−
⎟, ⎜
2 ⎠ ⎝
2⎠ ⎝ 2
⎝
⎠
1 1
Test points: x = –1, – , , 1; G ′′(–1) = –30,
2 2
1 ⎞ 15
1⎞
15
⎛
⎛
G ′′ ⎜ – ⎟ = , G ′′ ⎜ ⎟ = – , G ′′(1) = 30.
2
⎝ 2⎠ 2
⎝2⎠
⎛ 1
⎞ ⎛
, 0⎟ ∪ ⎜
G(x) is concave up on ⎜ –
2 ⎠ ⎝
⎝
1
⎛
⎞ ⎛
concave down on ⎜ – ∞, –
⎟ ∪ ⎜ 0,
2⎠ ⎝
⎝
⎞
, ∞ ⎟ and
2
⎠
1 ⎞
⎟.
2⎠
1
22. F ′( x) = 6 x5 – 12 x3 = 6 x3 ( x 2 – 2)
Split the x-axis into the intervals (– ∞ , − 2) ,
(− 2, 0), (0, 2), ( 2, ∞) .
Test points: x = –2, –1, 1, 2; F ′(–2) = –96,
F ′(–1) = 6, F ′(1) = –6, F ′(2) = 96
F(x) is increasing on [– 2, 0] ∪ [ 2, ∞) and
decreasing on (– ∞, – 2] ∪ [0, 2]
4
2
2
2
2
F ′′( x) = 30 x – 36 x = 6 x (5 x – 6); 5 x – 6 > 0
6
6
.
or x >
5
5
⎛
6⎞ ⎛ 6 ⎞
, ∞ ⎟ and
F(x) is concave up on ⎜⎜ – ∞, –
⎟∪⎜
5 ⎟⎠ ⎜⎝ 5 ⎟⎠
⎝
⎛ 6 6⎞
concave down on ⎜⎜ – ,
⎟⎟ .
⎝ 5 5⎠
when x < –
160
Section 3.2
24. H ′( x ) =
2x
; H ′( x) > 0 when x > 0.
( x + 1)2
H(x) is increasing on [0, ∞ ) and decreasing on
(– ∞ , 0].
2(1 – 3 x 2 )
H ′′( x) =
; H ′′( x) > 0 when
( x 2 + 1)3
–
1
3
2
<x<
1
3
.
⎛
H(x) is concave up on ⎜ –
⎝
1 ⎞ ⎛
⎛
down on ⎜ – ∞, –
⎟∪⎜
3⎠ ⎝
⎝
1 ⎞
⎟ and concave
3⎠
1
⎞
, ∞ ⎟.
3 ⎠
1
3
,
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
25.
f ′( x) =
cos x
2 sin x
; f ′( x) > 0 when 0 < x <
π
. f(x)
2
⎡ π⎤
is increasing on ⎢ 0, ⎥ and decreasing on
⎣ 2⎦
⎡π ⎤
⎢ 2 , π⎥ .
⎣
⎦
f ′′( x) =
– cos 2 x – 2sin 2 x
; f ′′( x) < 0 for all x in
4sin 3 / 2 x
(0, ∞ ). f(x) is concave down on (0, π ).
4
; 3x – 4 > 0 when x > .
3
2 x–2
g(x) is increasing on [2, ∞ ).
3x – 8
8
g ′′( x) =
; 3x – 8 > 0 when x > .
3/ 2
3
4( x – 2)
26. g ′( x) =
–2(5 x + 1)
f ′′( x) =
9 x4 / 3
; –2(5x + 1) > 0 when
1
x < – , f ′′( x) does not exist at x = 0.
5
1
8
Test points: –1, – , 1; f ′′(–1) = ,
10
9
104 / 3
4
⎛ 1⎞
f ′′ ⎜ – ⎟ = –
, f (1) = – .
10
9
3
⎝
⎠
1⎞
⎛
f(x) is concave up on ⎜ – ∞, – ⎟ and concave
5⎠
⎝
⎛ 1 ⎞
down on ⎜ – , 0 ⎟ ∪ (0, ∞).
⎝ 5 ⎠
3x – 4
⎛8 ⎞
g(x) is concave up on ⎜ , ∞ ⎟ and concave down
⎝3 ⎠
⎛ 8⎞
on ⎜ 2, ⎟ .
⎝ 3⎠
28. g ′( x) =
4( x + 2)
; x + 2 > 0 when x > –2, g ′( x)
3x 2 / 3
does not exist at x = 0.
Split the x-axis into the intervals ( −∞, −2 ) ,
(–2, 0), (0, ∞ ).
Test points: –3, –1, 1; g ′(–3) = –
4
5/3
3
,
2
; 2 – 5x > 0 when x < , f ′( x )
5
3x
does not exist at x = 0.
Split the x-axis into the intervals ( − ∞, 0),
4
, g ′(1) = 4.
3
g(x) is increasing on [–2, ∞ ) and decreasing on
(– ∞ , –2].
4( x – 4)
g ′′( x ) =
; x – 4 > 0 when x > 4, g ′′( x)
9 x5 / 3
does not exist at x = 0.
20
Test points: –1, 1, 5; g ′′(–1) = ,
9
4
4
g ′′(1) = – , g ′′(5) =
.
3
9(5)5 / 3
⎛ 2⎞ ⎛2 ⎞
⎜ 0, ⎟ , ⎜ , ∞ ⎟ .
⎝ 5⎠ ⎝5 ⎠
g(x) is concave up on (– ∞ , 0) ∪ (4, ∞ ) and
concave down on (0, 4).
g ′(–1) =
27.
f ′( x) =
2 – 5x
1/ 3
1
7
Test points: –1, , 1; f ′(−1) = – ,
5
3
⎛1⎞ 35 ′
f ′⎜ ⎟ =
, f (1) = –1.
⎝5⎠ 3
⎡ 2⎤
f(x) is increasing on ⎢ 0, ⎥ and decreasing on
⎣ 5⎦
⎡2 ⎞
(– ∞, 0] ∪ ⎢ , ∞ ⎟ .
⎣5 ⎠
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
29.
35.
f ( x) = ax 2 + bx + c; f ′( x) = 2ax + b;
f ′′( x) = 2a
An inflection point would occur where f ′′( x) = 0 ,
or 2a = 0. This would only occur when a = 0, but
if a = 0, the equation is not quadratic. Thus,
quadratic functions have no points of inflection.
36.
f ( x) = ax3 + bx 2 + cx + d ;
f ′( x) = 3ax 2 + 2bx + c; f ′′( x) = 6ax + 2b
An inflection point occurs where f ′′( x) = 0 , or
6ax + 2b = 0.
The function will have an inflection point at
b
x = – , a ≠ 0.
3a
30.
31.
37. Suppose that there are points x1 and x2 in I
where f ′( x1 ) > 0 and f ′( x2 ) < 0. Since f ′ is
continuous on I, the Intermediate Value Theorem
says that there is some number c between x1 and
x2 such that f ′(c) = 0, which is a contradiction.
Thus, either f ′( x) > 0 for all x in I and f is
increasing throughout I or f ′( x) < 0 for all x in I
and f is decreasing throughout I.
32.
38. Since x 2 + 1 = 0 has no real solutions, f ′( x )
exists and is continuous everywhere.
x 2 – x + 1 = 0 has no real solutions. x 2 – x + 1 > 0
and x 2 + 1 > 0 for all x, so f ′( x) > 0 for all x.
Thus f is increasing everywhere.
39. a.
Let f ( x) = x 2 and let I = [ 0, a ] , a > y .
f ′( x) = 2 x > 0 on I. Therefore, f(x) is
increasing on I, so f(x) < f(y) for x < y.
33.
b. Let f ( x) = x and let I = [ 0, a ] , a > y .
1
> 0 on I. Therefore, f(x) is
2 x
increasing on I, so f(x) < f(y) for x < y.
f ′( x) =
c.
34.
40.
1
and let I = [0, a], a > y.
x
1
f ′( x) = −
< 0 on I. Therefore f(x) is
x2
decreasing on I, so f(x) > f(y) for x < y.
Let f ( x) =
f ′( x) = 3ax 2 + 2bx + c
In order for f(x) to always be increasing, a, b, and
c must meet the condition 3ax 2 + 2bx + c > 0 for
all x. More specifically, a > 0 and b 2 − 3ac < 0.
162
Section 3.2
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41.
f ′′( x) =
3b – ax
4x
5/ 2
. If (4, 13) is an inflection point
45. a.
3b – 4a
b
= 0. Solving these
and
4 ⋅ 32
2
39
13
equations simultaneously, a =
and b = .
2
8
then 13 = 2a +
42.
f ( x) = a ( x − r1 )( x − r2 )( x − r3 )
f ′( x) = a[( x − r1 )(2 x − r2 − r3 ) + ( x − r2 )( x − r3 )]
b.
f ′( x) < 0 : (1.3, 5.0)
c.
f ′′( x) < 0 : (−0.25, 3.1) ∪ (6.5, 7]
d.
1
x
f ′( x) = cos x – sin
2
2
e.
1
x
f ′′( x) = − sin x − cos
4
2
2
f ′( x) = a[3 x − 2 x(r1 + r2 + r3 ) + r1r2 + r2 r3 + r1r3 ]
f ′′( x) = a[6 x − 2(r1 + r2 + r3 )]
a[6 x − 2(r1 + r2 + r3 )] = 0
r +r +r
6 x = 2(r1 + r2 + r3 ); x = 1 2 3
3
43. a.
b.
[ f ( x ) + g ( x)]′ = f ′( x) + g ′( x).
Since f ′( x) > 0 and g ′( x) > 0 for all x,
f ′( x) + g ′( x) > 0 for all x. No additional
conditions are needed.
[ f ( x) ⋅ g ( x)]′ = f ( x) g ′( x) + f ′( x) g ( x).
f ( x) g ′( x) + f ′( x) g ( x) > 0 if
f ( x) > −
c.
44. a.
b.
c.
f ′( x)
g ( x) for all x.
g ′( x)
[ f ( g ( x))]′ = f ′( g ( x)) g ′( x).
Since f ′( x) > 0 and g ′( x) > 0 for all x,
f ′( g ( x)) g ′( x) > 0 for all x. No additional
conditions are needed.
[ f ( x) + g ( x)]′′ = f ′′( x) + g ′′( x).
Since f ′′( x) > 0 and g ′′ > 0 for all x,
f ′′( x) + g ′′( x) > 0 for all x. No additional
conditions are needed.
[ f ( x ) ⋅ g ( x)]′′ = [ f ( x) g ′( x) + f ′( x) g ( x)]′
= f ( x) g ′′( x) + f ′′( x) g ( x) + 2 f ′( x) g ′( x).
The additional condition is that
f ( x) g ′′( x) + f ′′( x) g ( x) + 2 f ′( x) g ′( x) > 0
for all x is needed.
[ f ( g ( x))]′′ = [ f ′( g ( x)) g ′( x)]′
46. a.
b.
f ′( x) < 0 : (2.0, 4.7) ∪ (9.9, 10]
c.
f ′′( x) < 0 : [0, 3.4) ∪ (7.6, 10]
⎡ 2
⎛ x ⎞ ⎛ x ⎞⎤
⎛ x⎞
d. f ′( x) = x ⎢ − cos ⎜ ⎟ sin ⎜ ⎟ ⎥ + cos2 ⎜ ⎟
⎝ 3 ⎠ ⎝ 3 ⎠⎦
⎝3⎠
⎣ 3
⎛ x⎞ x
⎛ 2x ⎞
= cos 2 ⎜ ⎟ − sin ⎜ ⎟
3
3
⎝ ⎠
⎝ 3 ⎠
= f ′( g ( x)) g ′′( x) + f ′′( g ( x))[ g ′( x)]2 .
The additional condition is that
f ′′( g ( x))[ g ′( x)]2
for all x.
f ′( g ( x)) > −
g ′′( x)
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e.
f ′′( x) = −
2x
⎛ 2x ⎞ 2 ⎛ 2x ⎞
cos ⎜ ⎟ − sin ⎜ ⎟
9
⎝ 3 ⎠ 3 ⎝ 3 ⎠
c.
d 3s
dt 3
< 0,
d 2s
dt 2
>0
s
47.
f ′( x) > 0 on (–0.598, 0.680)
f is increasing on [–0.598, 0.680].
48.
f ′′( x) < 0 when x > 1.63 in [–2, 3]
f is concave down on (1.63, 3).
49. Let s be the distance traveled. Then
speed of the car.
a.
t
Concave up.
ds
is the
dt
d.
d 2s
dt 2
= 10 mph/min
s
ds
= ks, k a constant
dt
s
t
Concave up.
Concave up.
b.
d 2s
dt 2
t
e.
ds
d 2s
are approaching zero.
and
dt
dt 2
s
>0
s
Concave down.
Concave up.
164
Section 3.2
t
t
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f.
ds
is constant.
dt
c.
s
Neither concave up nor down.
50. a.
dV
dh
d 2h
= k,
> 0,
<0
dt
dt
dt 2
Concave down.
h ( t)
t
dV
= k <0, V is the volume of water in the
dt
tank, k is a constant.
Neither concave up nor down.
t
d.
dI d 2 I
,
> 0 in the future
dt dt 2
where I is inflation.
I ( t) = k now, but
I ( t)
v(t)
t
t
b.
e.
dV
1
1
= 3 – = 2 gal/min
dt
2
2
Neither concave up nor down.
P ( t)
v(t)
t
Instructor’s Resource Manual
dp
d2 p
dp
< 0, but
> 0 and at t = 2:
>0.
2
dt
dt
dt
where p is the price of oil.
Concave up.
2
t
Section 3.2
165
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f.
dT
d 2T
> 0,
< 0 , where T is David’s
dt
dt 2
temperature.
Concave down.
c.
T ( t)
dP
d 2P
> 0,
< 0 , where P is world
dt
dt 2
population.
Concave down.
P ( t)
t
51. a.
dC
d 2C
> 0,
> 0 , where C is the car’s cost.
dt
dt 2
Concave up.
t
d.
C ( t)
dθ
d 2θ
> 0,
> 0 , where θ is the angle that
dt
dt 2
the tower makes with the vertical.
Concave up.
θ( t)
t
b. f(t) is oil consumption at time t.
df
d2 f
< 0,
>0
dt
dt 2
Concave up.
f( t)
t
e.
P = f(t) is profit at time t.
dP
d 2P
> 0,
<0
dt
dt 2
Concave down.
P ( t)
t
t
166
Section 3.2
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f.
R is revenue at time t.
dP
>0
P < 0,
dt
Could be either concave up or down.
P
t
54. The height is always increasing so h '(t ) > 0 . The
rate of change of the height decreases for the first
50 minutes and then increases over the next 50
minutes. Thus h ''(t ) < 0 for 0 ≤ t ≤ 50 and
h ''(t ) > 0 for 50 < t ≤ 100 .
P
t
52. a.
R(t) ≈ 0.28, t < 1981
b. On [1981, 1983],
R(1983) ≈ 0.36
53.
dR
d 2R
> 0,
>0,
dt
dt 2
dV
= 2 in 3 / sec
dt
The cup is a portion of a cone with the bottom cut
off. If we let x represent the height of the missing
cone, we can use similar triangles to show that
x x+5
=
3
3.5
3.5 x = 3 x + 15
0.5 x = 15
x = 30
Similar triangles can be used again to show that, at
any given time, the radius of the cone at water
level is
h + 30
r=
20
Therefore, the volume of water can be expressed
as
55. V = 3t , 0 ≤ t ≤ 8 . The height is always increasing,
so h '(t ) > 0. The rate of change of the height
decreases from time t = 0 until time t1 when the
water reaches the middle of the rounded bottom
part. The rate of change then increases until time
t2 when the water reaches the middle of the neck.
Then the rate of change decreases until t = 8 and
the vase is full. Thus, h ''(t ) > 0 for t1 < t < t2 and
h ''(t ) < 0 for t2 < t < 8 .
h ( t)
24
t1
t2
8
t
π (h + 30) 3
45π
−
.
1200
2
We also know that V = 2t from above. Setting the
two volume equations equal to each other and
2400
t + 27000 − 30 .
solving for h gives h = 3
V=
π
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167
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56. V = 20 − .1t , 0 ≤ t ≤ 200 . The height of the water
is always decreasing so h '(t ) < 0 . The rate of
change in the height increases (the rate is negative,
and its absolute value decreases) for the first 100
days and then decreases for the remaining time.
Therefore we have h ''(t ) > 0 for 0 < t < 100 , and
h ''(t ) < 0 for 100 < t < 200 .
3.3 Concepts Review
1. maximum
2. maximum; minimum
3. maximum
4. local maximum, local minimum, 0
Problem Set 3.3
1.
f ′( x) = 3 x 2 –12 x = 3 x( x – 4)
Critical points: 0, 4
f ′( x) > 0 on (– ∞ , 0), f ′( x) < 0 on (0, 4),
f ′( x) > 0 on (4, ∞ )
f ′′( x) = 6 x –12; f ′′(0) = –12, f ′′(4) = 12.
Local minimum at x = 4;
local maximum at x = 0
2.
f ′( x) = 3 x 2 –12 = 3( x 2 – 4)
Critical points: –2, 2
f ′( x) > 0 on (– ∞ , –2), f ′( x) < 0 on (–2, 2),
f ′( x) > 0 on (2, ∞ )
f ′′( x) = 6 x; f ′′(–2) = –12, f ′′(2) = 12
Local minimum at x = 2;
local maximum at x = –2
57. a. The cross-sectional area of the vase is
approximately equal to ΔV and the
corresponding radius is r = ΔV / π . The
table below gives the approximate values for r.
The vase becomes slightly narrower as you
move above the base, and then gets wider as
you near the top.
Depth
V
A ≈ ΔV
r = ΔV / π
1
4
4
1.13
2
8
4
1.13
3
11
3
0.98
4
14
3
0.98
5
20
6
1.38
6
28
8
1.60
b. Near the base, this vase is like the one in part
(a), but just above the base it becomes larger.
Near the middle of the vase it becomes very
narrow. The top of the vase is similar to the
one in part (a).
168
Depth
V
A ≈ ΔV
r = ΔV / π
1
4
4
1.13
2
9
5
1.26
3
12
3
0.98
4
14
2
0.80
5
20
6
1.38
6
28
8
1.60
Section 3.3
3.
4.
⎛ π⎞
f ′(θ ) = 2 cos 2θ ; 2 cos 2θ ≠ 0 on ⎜ 0, ⎟
⎝ 4⎠
No critical points; no local maxima or minima on
⎛ π⎞
⎜ 0, ⎟ .
⎝ 4⎠
1
1
1
+ cos x; + cos x = 0 when cos x = – .
2
2
2
2π 4π
,
Critical points:
3 3
⎛ 2π ⎞
⎛ 2π 4π ⎞
f ′( x) > 0 on ⎜ 0,
⎟ , f ′( x) < 0 on ⎜ ,
⎟,
3
⎝
⎠
⎝ 3 3 ⎠
⎛ 4π
⎞
f ′( x) > 0 on ⎜ , 2π ⎟
⎝ 3
⎠
f ′( x) =
3
3
⎛ 2π ⎞
⎛ 4π ⎞
f ′′( x) = – sin x; f ′′ ⎜ ⎟ = –
, f ′′ ⎜ ⎟ =
3
2
3
2
⎝ ⎠
⎝ ⎠
4π
; local maximum at
Local minimum at x =
3
2π
x=
.
3
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5. Ψ ′(θ ) = 2sin θ cosθ
−
π
<θ <
9. h ' ( y ) = 2 y +
π
2
2
Critical point: 0
⎛ π ⎞
Ψ ′(θ ) < 0 on ⎜ − , 0 ⎟ , Ψ ′(θ ) > 0 on
⎝ 2 ⎠
⎛ π⎞
⎜ 0, ⎟ ,
⎝ 2⎠
( x 2 + 4 ) ⋅1 − x ( 2 x ) = 4 − x 2
2
2
( x2 + 4)
( x2 + 4)
Critical points: −2, 2
f ' ( x ) < 0 on ( −∞, −2 ) and ( 2, ∞ ) ;
f ' ( x ) > 0 on ( −2, 2 )
(
2 x x 2 − 12
( x2 + 4)
( )
)
1+ z2 ) ( 2z ) − z2 ( 2z )
(
2z
g '( z ) =
=
2
2 2
(1 + z )
(1 + z 2 )
Critical point: z = 0
g ' ( z ) < 0 on ( −∞, 0 )
g ' ( z ) > 0 on ( 0, ∞ )
g '' ( z ) =
(
)
−2 3 z 2 − 1
(
2
)
z +1
Local minima at −
10.
f '( x) =
3
g '' ( 0 ) = 2
Local minima at z = 0 .
Instructor’s Resource Manual
(x
2
)
3
16
=6
4
= 2+
3
4
2
+ 1 ( 3) − ( 3 x + 1)( 2 x )
(x
2
)
+1
2
=
3 − 2 x − 3x 2
(x
2
)
+1
2
The only critical points are stationary points. Find
these by setting the numerator equal to 0 and
solving.
3 − 2 x − 3x2 = 0
a = −3, b = −2, c = 3
3
1
1
f '' ( −2 ) = ; f '' ( 2 ) = −
16
16
Local minima at x = −2 ; Local maxima at x = 2
8.
4
2
⎛ 34⎞
2
h⎜ −
⎟ = 2− 3
⎝ 2 ⎠
− 24
r ′′( x ) = 12 x 2 ; r ′′(0) = 0; the Second Derivative
Test fails.
Local minimum at z = 0; no local maxima
f '' ( x ) =
3
3
⎛
4⎞
h ' ( y ) < 0 on ⎜ −∞, −
⎟
2 ⎠
⎝
⎛ 34 ⎞
, 0 ⎟ and ( 0, ∞ )
h ' ( y ) > 0 on ⎜ −
⎝ 2
⎠
2
h '' ( y ) = 2 − 3
y
6. r ′( z ) = 4 z 3
Critical point: 0
r ′( z ) < 0 on (−∞, 0);
r ′( z ) > 0 on (0, ∞)
f '( x) =
y2
Critical point: −
Ψ ′′(θ ) = 2 cos 2 θ – 2sin 2 θ ; Ψ ′′(0) = 2
Local minimum at x = 0
7.
1
x=
2±
( −2 )2 − 4 ( −3)( 3) 2 ± 40
=
2 ( −3)
−6
=
−1 ± 10
3
−1 − 10
−1 + 10
and
3
3
⎛
−1 − 10 ⎞
f ' ( x ) < 0 on ⎜⎜ −∞,
⎟⎟ and
3
⎝
⎠
⎛ −1 + 10 ⎞
, ∞ ⎟⎟ .
⎜⎜
3
⎝
⎠
⎛ −1 − 10 −1 + 10 ⎞
f ' ( 0 ) > 0 on ⎜⎜
,
⎟⎟
3
3
⎝
⎠
Critical points:
f '' ( x ) =
(
)
2 3x3 + 3x 2 − 9 x − 1
(x
2
)
+1
3
⎛ −1 − 10 ⎞
f '' ⎜⎜
⎟⎟ ≈ 0.739
3
⎝
⎠
⎛ −1 + 10 ⎞
f '' ⎜⎜
⎟⎟ ≈ −2.739
3
⎝
⎠
Local minima at x =
−1 − 10
;
3
Local maxima at x =
−1 + 10
3
Section 3.3
169
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11.
f ′( x) = 3 x 2 – 3 = 3( x 2 – 1)
Critical points: –1, 1
f ′′( x) = 6 x; f ′′(–1) = –6, f ′′(1) = 6
Local minimum value f(1) = –2;
local maximum value f(–1) = 2
8/3
⎛ ⎛ 2 ⎞5 / 3 ⎞
6 ⎛ 15 ⎞
; r ′′⎜ – ⎜ ⎟ ⎟ = – ⎜ ⎟
⎜ ⎝ 15 ⎠ ⎟
25 ⎝ 2 ⎠
25s8 / 5
⎝
⎠
⎛ ⎛ 2 ⎞5 3 ⎞
r ′( s ) < 0 on ⎜ − ⎜ ⎟ , 0 ⎟ , r ′( s ) > 0 on (0, ∞)
⎜ ⎝ 15 ⎠
⎟
⎝
⎠
12. g ′( x) = 4 x3 + 2 x = 2 x(2 x 2 + 1)
Critical point: 0
g ′′( x) = 12 x 2 + 2; g ′′(0) = 2
Local minimum value g(0) = 3; no local maximum
values
13. H ′( x ) = 4 x3 – 6 x 2 = 2 x 2 (2 x – 3)
Local minimum value r(0) = 0; local maximum
value
5/3
2/3
2/3
⎛ ⎛ 2 ⎞5 / 3 ⎞
3⎛ 2 ⎞
⎛ 2⎞
⎛ 2⎞
r ⎜ – ⎜ ⎟ ⎟ = –3 ⎜ ⎟ + ⎜ ⎟
= ⎜ ⎟
⎜ ⎝ 15 ⎠ ⎟
5 ⎝ 15 ⎠
⎝ 15 ⎠
⎝ 15 ⎠
⎝
⎠
17.
3
Critical points: 0,
2
H ′′( x) = 12 x 2 – 12 x = 12 x( x – 1); H ′′(0) = 0,
⎛3⎞
H ′′ ⎜ ⎟ = 9
⎝2⎠
18.
⎛ 3⎞
H ′( x) < 0 on (−∞, 0), H ′( x) < 0 on ⎜ 0, ⎟
⎝ 2⎠
27
⎛3⎞
Local minimum value H ⎜ ⎟ = – ; no local
2
16
⎝ ⎠
maximum values (x = 0 is neither a local
minimum nor maximum)
14.
; g ′(t ) does not exist at t = 2.
3(t – 2)1/ 3
Critical point: 2
2
2
g ′(1) = , g ′(3) = –
3
3
No local minimum values; local maximum value
g(2) = π .
16. r ′( s ) = 3 +
⎛ 2⎞
s = –⎜ ⎟
⎝ 15 ⎠
2
5s 3 / 5
5/3
=
15s3 / 5 + 2
5s 3 / 5
, r ′( s ) does not exist at s = 0.
⎛ 2⎞
Critical points: – ⎜ ⎟
⎝ 15 ⎠
170
; r ′( s ) = 0 when
Section 3.3
5/3
,0
1
t2
No critical points
No local minimum or maximum values
f ′( x) =
x( x 2 + 8)
( x 2 + 4)3 / 2
Critical point: 0
f ′( x) < 0 on (−∞, 0), f ′( x) > 0 on (0, ∞)
Local minimum value f(0) = 0, no local maximum
values
1
; Λ ′(θ ) does not exist at
1 + sin θ
3π
, but Λ (θ ) does not exist at that point
2
either.
No critical points
No local minimum or maximum values
θ=
20.
2
f ′(t ) = 1 +
19. Λ ′(θ ) = –
f ′( x) = 5( x – 2) 4
Critical point: 2
f ′′( x) = 20( x – 2)3 ; f ′′(2) = 0
f ′( x) > 0 on (−∞, 2), f ′( x) > 0 on (2, ∞)
No local minimum or maximum values
15. g ′(t ) = –
6
r ′′( s ) = –
g ′(θ ) =
sin θ cos θ
π 3π
; g ′(θ ) = 0 when θ = ,
;
sin θ
2 2
g ′(θ ) does not exist at x = π .
⎛ π⎞
Split the x -axis into the intervals ⎜ 0, ⎟ ,
⎝ 2⎠
⎛ π ⎞ ⎛ 3π ⎞ ⎛ 3π
⎞
⎜ , π ⎟ , ⎜ π, ⎟ , ⎜ , 2π ⎟ .
2
2
2
⎝
⎠ ⎝
⎠ ⎝
⎠
π 3π 5π 7 π
⎛π⎞ 1
Test points: , , , ; g ′ ⎜ ⎟ =
,
4 4 4 4
2
⎝4⎠
1
1
⎛ 3π ⎞
⎛ 5π ⎞ 1
⎛ 7π ⎞
g′⎜ ⎟ = –
, g′⎜ ⎟ =
, g′⎜ ⎟ = –
2
2
2
⎝ 4 ⎠
⎝ 4 ⎠
⎝ 4 ⎠
Local minimum value g( π ) = 0; local maximum
⎛π⎞
⎛ 3π ⎞
values g ⎜ ⎟ = 1 and g ⎜ ⎟ = 1
⎝2⎠
⎝ 2 ⎠
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21.
f ' ( x ) = 4 ( sin 2 x )( cos 2 x )
4 ( sin 2 x )( cos 2 x ) = 0 when x =
( 2k − 1) π
kπ
where k is an integer.
x=
2
Critical points: 0, π4 , π2 , 2
4
⎛π ⎞
Minimum value: f ( 0 ) = f ⎜ ⎟ = 0
⎝2⎠
π
⎛ ⎞
Maximum value: f ⎜ ⎟ = 1
⎝4⎠
22.
f '( x) =
(
(x
2
+4
)
f ' ( x ) = 0 when x = 2 or x = −2 . (there are no
23. g ' ( x ) =
(
− x x3 − 64
( x3 + 32 )
1
2
)
2
g ' ( x ) = 0 when x = 0 or x = 4 .
Critical points: 0, 4
1
g ( 0) = 0 ; g ( 4) =
6
As x approaches ∞ , the value of g approaches 0
but never actually gets there.
1
Maximum value: g ( 4 ) =
6
Minimum value: g ( 0 ) = 0
24. h ' ( x ) =
– 4 = 0 when x =
9
16
9
.
16
⎛ 9⎞
⎛9
⎞
F ′( x) > 0 on ⎜ 0, ⎟ , F ′( x) < 0 on ⎜ , ∞ ⎟
⎝ 16 ⎠
⎝ 16 ⎠
9
⎛
⎞
F decreases without bound on ⎜ , ∞ ⎟ . No
⎝ 16 ⎠
⎛9⎞ 9
minimum values; maximum value F ⎜ ⎟ =
⎝ 16 ⎠ 4
)
Maximum value: f ( 2 ) =
3
26. From Problem 25, the critical points are 0 and
2
singular points)
Critical points: 0, 2 (note: −2 is not in the given
domain)
1
f ( 0 ) = 0 ; f ( 2 ) = ; f ( x ) → 0 as x → ∞ .
2
Minimum value: f ( 0 ) = 0
x
– 4;
x
9
Critical points: 0, , 4
16
⎛9⎞ 9
F(0) = 0, F ⎜ ⎟ = , F(4) = –4
⎝ 16 ⎠ 4
Minimum value F(4) = –4; maximum value
⎛9⎞ 9
F⎜ ⎟=
⎝ 16 ⎠ 4
or
⎛π ⎞
⎛π ⎞
f (0) = 0 ; f ⎜ ⎟ = 1 ; f ⎜ ⎟ = 0 ;
⎝4⎠
⎝2⎠
f ( 2 ) ≈ 0.5728
−2 x 2 − 4
3
25. F ′( x) =
27.
f ′( x) = 64(−1)(sin x)−2 cos x
+27(−1)(cos x)−2 (− sin x)
=−
=
64 cos x
2
sin x
+
27 sin x
cos 2 x
2
2
(3sin x − 4 cos x)(9 sin x + 12 cos x sin x + 16 cos x )
2
2
sin x cos x
⎛ π⎞
On ⎜ 0, ⎟ , f ′( x) = 0 only where 3sin x = 4cos x;
⎝ 2⎠
4
tan x = ;
3
4
x = tan −1 ≈ 0.9273
3
Critical point: 0.9273
For 0 < x < 0.9273, f ′( x) < 0, while for
0.9273 < x <
π
2
, f '( x) > 0
4 ⎞ 64 27
⎛
Minimum value f ⎜ tan −1 ⎟ =
+
= 125;
3
3⎠ 4
⎝
5
5
no maximum value
−2 x
( x2 + 4)
2
h ' ( x ) = 0 when x = 0 . (there are no singular
points)
Critical points: 0
Since h ' ( x ) < 0 for x > 0 , the function is always
decreasing. Thus, there is no minimum value.
1
Maximum value: h ( 0 ) =
4
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171
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28. g ′( x) = 2 x +
= 2x +
(8 − x)2 (32 x) − (16 x 2 )2(8 − x)(−1)
(8 − x)
256 x
=
(8 − x )3
32.
(1, 2) ∪ (3, 4). Thus, the function has a local
minimum at x = 1,3 and a local maximum at
x = 2, 4 .
2 x[(8 − x)3 + 128]
(8 − x)3
For x > 8, g ′( x) = 0 when (8 − x)3 + 128 = 0;
(8 − x)3 = −128; 8 − x = − 3 128 ;
33.
(−∞,1) ∪ (1, 2) ∪ (2,3) ∪ (4, ∞) Thus, the function
has a local minimum at x = 4 and a local
maximum at x = 3 .
g ′( x) < 0 on (8, 8 + 4 3 2),
29. H ' ( x ) =
(
)
f '( x) = 0 at x = 1, 2,3, 4 ; f '( x) is negative on
(3, 4) and positive on
x = 8 + 4 3 2 ≈ 13.04
g ′( x) > 0 on (8 + 4 3 2, ∞)
g(13.04) ≈ 277 is the minimum value
f '( x) = 0 at x = 1, 2,3, 4 ; f '( x) is negative on
(−∞,1) ∪ (2,3) ∪ (4, ∞) and positive on
4
34. Since f ' ( x ) ≥ 0 for all x, the function is always
2 x x2 − 1
increasing. Therefore, there are no local extrema.
x2 − 1
35. Since f ' ( x ) ≥ 0 for all x, the function is always
H ' ( x ) = 0 when x = 0 .
increasing. Therefore, there are no local extrema.
H ' ( x ) is undefined when x = −1 or x = 1
Critical points: −2 , −1 , 0, 1, 2
H ( −2 ) = 3 ; H ( −1) = 0 ; H ( 0 ) = 1 ; H (1) = 0 ;
H ( 2) = 3
Minimum value: H ( −1) = H (1) = 0
Maximum value: H ( −2 ) = H ( 2 ) = 3
36.
f ' ( x ) = 0 at x = 0, A, and B .
f ' ( x ) is negative on ( −∞, 0 ) and ( A, B )
f ' ( x ) is positive on ( 0, A ) and ( B, ∞ )
Therefore, the function has a local minimum at
x = 0 and x = B , and a local maximum at x = A .
37. Answers will vary. One possibility:
y
30. h ' ( t ) = 2t cos t 2
h ' ( t ) = 0 when t = 0 , t =
t=
10π
2
3π
5π
(Consider t = , t =
, and t 2 =
)
2
2
2
2π
6π
10π
Critical points: 0,
,
,
,π
2
2
2
⎛ 2π ⎞
⎛ 6π ⎞
h ( 0) = 0 ; h ⎜
⎟ = 1; h⎜
⎟ = −1 ;
⎝ 2 ⎠
⎝ 2 ⎠
2
π
2π
6π
, t=
, and
2
2
3
6 x
2
⎛ 10π ⎞
h⎜
⎟ = 1 ; h (π ) ≈ −0.4303
⎝ 2 ⎠
⎛ 6π ⎞
Minimum value: h ⎜
⎟ = −1
⎝ 2 ⎠
⎛ 2π ⎞
⎛ 10π ⎞
Maximum value: h ⎜
⎟ = h⎜
⎟ =1
⎝ 2 ⎠
⎝ 2 ⎠
31.
5
−5
38. Answers will vary. One possibility:
y
5
3
6 x
−5
f '( x) = 0 when x = 0 and x = 1 . On the interval
(−∞, 0) we get f '( x) < 0 . On (0, ∞) , we get
f '( x) > 0 . Thus there is a local min at x = 0 but
no local max.
172
Section 3.3
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39. Answers will vary. One possibility:
y
5
3
6 x
43. The graph of f is a parabola which opens up.
B
f ' ( x ) = 2 Ax + B = 0 → x = −
2A
f '' ( x ) = 2 A
Since A > 0 , the graph of f is always concave up.
There is exactly one critical point which yields the
minimum of the graph.
2
⎛ B ⎞
⎛ B ⎞
⎛ B ⎞
f ⎜−
⎟ = A⎜ −
⎟ + B⎜−
⎟+C
⎝ 2A ⎠
⎝ 2A ⎠
⎝ 2A ⎠
B2 B2
−
+C
4A 2A
B 2 − 2 B 2 + 4 AC
=
4A
B 2 − 4 AC
4 AC − B 2
=
=−
4A
4A
−5
=
40. Answers will vary. One possibility:
y
5
(
)
If f ( x ) ≥ 0 with A > 0 , then − B 2 − 4 AC ≥ 0 ,
3
6 x
−5
41. Answers will vary. One possibility:
y
or B 2 − 4 AC ≤ 0 .
⎛ B ⎞
If B 2 − 4 AC ≤ 0 , then we get f ⎜ −
⎟≥0
⎝ 2A ⎠
⎛ B ⎞
Since 0 ≤ f ⎜ −
⎟ ≤ f ( x ) for all x, we get
⎝ 2A ⎠
f ( x ) ≥ 0 for all x.
44. A third degree polynomial will have at most two
extrema.
f ' ( x ) = 3 Ax 2 + 2 Bx + C
5
f '' ( x ) = 6 Ax + 2 B
3
6 x
Critical points are obtained by solving f ' ( x ) = 0 .
3 Ax 2 + 2 Bx + C = 0
−5
x=
42. Answers will vary. One possibility:
y
=
5
−2 B ± 4 B 2 − 12 AC
6A
−2 B ± 2 B 2 − 3 AC
6A
− B ± B 2 − 3 AC
3A
To have a relative maximum and a relative
minimum, we must have two solutions to the
above quadratic equation. That is, we must have
B 2 − 3 AC > 0 .
=
3
−5
6 x
The two solutions would be
− B − B 2 − 3 AC
3A
− B + B 2 − 3 AC
. Evaluating the second
3A
derivative at each of these values gives:
and
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173
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
⎛ − B − B 2 − 3 AC
f '' ⎜
⎜
3A
⎝
⎞
⎟
⎟
⎠
⎛ − B − B 2 − 3 AC
= 6A⎜
⎜
3A
⎝
Problem Set 3.4
⎞
⎟ + 2B
⎟
⎠
= −2 B − 2 B 2 − 3 AC + 2 B
= −2 B 2 − 3 AC
and
⎛ − B + B 2 − 3 AC
f '' ⎜
⎜
3A
⎝
⎞
⎟
⎟
⎠
⎛ − B + B 2 − 3 AC
= 6A⎜
⎜
3A
⎝
⎞
⎟ + 2B
⎟
⎠
= −2 B + 2 B 2 − 3 AC + 2 B
= 2 B 2 − 3 AC
If B 2 − 3 AC > 0 , then −2 B 2 − 3 AC exists and
is negative, and 2 B 2 − 3 AC exists and is
positive.
Thus, from the Second Derivative Test,
− B − B 2 − 3 AC
would yield a local maximum
3A
− B + B 2 − 3 AC
would yield a local
3A
minimum.
and
45.
f ′′′(c) > 0 implies that f ′′ is increasing at c, so f
is concave up to the right of c (since f ′′( x) > 0 to
the right of c) and concave down to the left of c
(since f ′′( x) < 0 to the left of c). Therefore f has a
point of inflection at c.
3.4 Concepts Review
1. 0 < x < ∞
n
2
i =1
4. marginal revenue; marginal cost
174
Section 3.4
256
Q = x2 + y 2 = x2 +
x2
dQ
512
= 2x –
dx
x3
512
=0
2x –
x3
x 4 = 256
x = ±4
The critical points are –4, 4.
dQ
dQ
< 0 on (– ∞ , –4) and (0, 4).
> 0 on
dx
dx
(–4, 0) and (4, ∞ ).
When x = –4, y = 4 and when x = 4, y = –4. The
two numbers are –4 and 4.
2. Let x be the number.
Q = x – 8x
x will be in the interval (0, ∞ ).
dQ 1 –1/ 2
= x
–8
dx 2
1 –1/ 2
x
–8 = 0
2
x –1/ 2 = 16
1
x=
256
dQ
> 0 on
dx
1 ⎞
dQ
⎛
⎛ 1
⎞
< 0 on ⎜
, ∞ ⎟.
⎜ 0,
⎟ and
dx
⎝ 256 ⎠
⎝ 256 ⎠
1
.
Q attains its maximum value at x =
256
3. Let x be the number.
200
2. 2x +
x
3. S = ∑ ( yi − bxi )
1. Let x be one number, y be the other, and Q be the
sum of the squares.
xy = –16
16
y=–
x
The possible values for x are in (– ∞ , 0) or (0, ∞) .
Q = 4 x – 2x
x will be in the interval (0, ∞ ).
dQ 1 –3 / 4
= x
–2
dx 4
1 –3 / 4
x
–2=0
4
x –3 / 4 = 8
1
x=
16
dQ
dQ
⎛ 1⎞
⎛1
⎞
> 0 on ⎜ 0, ⎟ and
< 0 on ⎜ , ∞ ⎟
16
16
dx
dx
⎝
⎠
⎝
⎠
1
Q attains its maximum value at x = .
16
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4. Let x be one number, y be the other, and Q be the
sum of the squares.
xy = –12
12
y=–
x
The possible values for x are in (– ∞ , 0) or (0, ∞) .
Q = x2 + y2 = x2 +
144
x2
dQ
288
= 2x –
dx
x3
288
2x –
=0
x3
x 4 = 144
x = ±2 3
The critical points are –2 3, 2 3
dQ
< 0 on (– ∞, – 2 3) and (0, 2 3).
dx
dQ
> 0 on (–2 3, 0) and (2 3, ∞).
dx
When x = –2 3, y = 2 3 and when
x = 2 3, y = –2 3.
The two numbers are –2 3 and 2 3.
5. Let Q be the square of the distance between (x, y)
and (0, 5).
Q = ( x – 0)2 + ( y – 5)2 = x 2 + ( x 2 – 5)2
= x 4 – 9 x 2 + 25
dQ
= 4 x3 – 18 x
dx
4 x3 – 18 x = 0
2 x(2 x 2 – 9) = 0
x = 0, ±
3
2
3 ⎞
dQ
⎛
< 0 on ⎜ – ∞, –
⎟ and
dx
2⎠
⎝
dQ
⎛ 3
⎞
⎛
> 0 on ⎜ –
, 0 ⎟ and ⎜
dx
2
⎝
⎠
⎝
3 ⎞
⎛
⎜ 0,
⎟.
2⎠
⎝
3
⎞
, ∞ ⎟.
2 ⎠
3
9
3
When x = –
, y = and when x =
,
2
2
2
9
y= .
2
⎛ 3 9⎞
⎛ 3 9⎞
The points are ⎜ –
, ⎟ and ⎜
, ⎟.
2 2⎠
⎝
⎝ 2 2⎠
Instructor’s Resource Manual
6. Let Q be the square of the distance between (x, y)
and (10, 0).
Q = ( x – 10)2 + ( y – 0) 2 = (2 y 2 – 10) 2 + y 2
= 4 y 4 – 39 y 2 + 100
dQ
= 16 y 3 – 78 y
dy
16 y 3 – 78 y = 0
2 y (8 y 2 – 39) = 0
y = 0, ±
dQ
dy
dQ
dy
39
2 2
⎛
⎛
39 ⎞
39 ⎞
< 0 on ⎜⎜ – ∞, –
⎟⎟ and ⎜⎜ 0,
⎟⎟ .
2 2⎠
⎝
⎝ 2 2⎠
⎛
⎛ 39
⎞
39 ⎞
, 0 ⎟⎟ and ⎜⎜
, ∞ ⎟⎟ .
> 0 on ⎜⎜ –
⎝ 2 2 ⎠
⎝2 2
⎠
When y = –
y=
39
2 2
39
2 2
,x=
,x=
39
and when
4
39
.
4
⎛ 39
⎛ 39 39 ⎞
39 ⎞
The points are ⎜⎜ , –
⎟⎟ and ⎜⎜ ,
⎟⎟ .
2 2⎠
⎝ 4
⎝ 4 2 2⎠
7. x ≥ x 2 if 0 ≤ x ≤ 1
f ( x) = x − x 2 ; f ′( x ) = 1 − 2 x;
f ′( x) = 0 when x =
1
2
1
Critical points: 0, , 1
2
1
⎛1⎞ 1
f(0) = 0, f(1) = 0, f ⎜ ⎟ = ; therefore,
2
4
2
⎝ ⎠
exceeds its square by the maximum amount.
8. For a rectangle with perimeter K and width x, the
K
− x . Then the area is
length is
2
⎛K
⎞ Kx
A = x⎜ − x⎟ =
− x2 .
⎝2
⎠ 2
dA K
dA
K
= − 2 x;
= 0 when x =
dx 2
dx
4
K K
Critical points: 0, ,
4 2
K2
K
K
, A = 0; at x = , A =
.
At x = 0 or
16
2
4
The area is maximized when the width is one
fourth of the perimeter, so the rectangle is a
square.
Section 3.4
175
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9. Let x be the width of the square to be cut out and V
the volume of the resulting open box.
V = x (24 − 2 x)2 = 4 x3 − 96 x 2 + 576 x
dV
= 12 x 2 − 192 x + 576 = 12( x − 12)( x − 4);
dx
12(x – 12)(x – 4) = 0; x = 12 or x = 4.
Critical points: 0, 4, 12
At x = 0 or 12, V = 0; at x = 4, V = 1024.
3
The volume of the largest box is 1024 in.
10. Let A be the area of the pen.
dA
A = x(80 − 2 x) = 80 x − 2 x 2 ;
= 80 − 4 x;
dx
80 − 4 x = 0; x = 20
Critical points: 0, 20, 40.
At x = 0 or 40, A = 0; at x = 20, A = 800.
The dimensions are 20 ft by 80 – 2(20) = 40 ft,
with the length along the barn being 40 ft.
11. Let x be the width of each pen, then the length
along the barn is 80 – 4x.
dA
A = x(80 − 4 x) = 80 x − 4 x 2 ;
= 80 − 8 x;
dx
dA
= 0 when x = 10.
dx
Critical points: 0, 10, 20
At x = 0 or 20, A = 0; at x = 10, A = 400.
The area is largest with width 10 ft
and length 40 ft.
12. Let A be the area of the pen. The perimeter is
100 + 180 = 280 ft.
y + y – 100 + 2x = 180; y = 140 – x
dA
A = x(140 − x ) = 140 x − x 2 ;
= 140 − 2 x;
dx
140 − 2 x = 0; x = 70
Since 0 ≤ x ≤ 40 , the critical points are 0 and 40.
When x = 0, A = 0. When x = 40, A = 4000. The
dimensions are 40 ft by 100 ft.
900
x
The possible values for x are in (0, ∞ ).
2700
⎛ 900 ⎞
Q = 4x + 3 y = 4x + 3⎜
⎟ = 4x +
x
⎝ x ⎠
dQ
2700
= 4−
dx
x2
2700
4–
=0
x2
13. xy = 900; y =
x 2 = 675
x = ±15 3
x = 15 3 is the only critical point in (0, ∞ ).
176
Section 3.4
dQ
< 0 on (0, 15 3) and
dx
dQ
> 0 on (15 3, ∞).
dx
900
When x = 15 3, y =
= 20 3.
15 3
Q has a minimum when x = 15 3 ≈ 25.98 ft and
y = 20 3 ≈ 34.64 ft.
300
x
The possible values for x are in (0, ∞ ).
1200
Q = 6x + 4 y = 6x +
x
dQ
1200
=6–
dx
x2
1200
=0
6–
x2
14. xy = 300; y =
x 2 = 200
x = ±10 2
x = 10 2 is the only critical point in (0, ∞ ).
dQ
dQ
< 0 on (0, 10 2) and
> 0 on (10 2, ∞)
dx
dx
300
When x = 10 2, y =
= 15 2.
10 2
Q has a minimum when x = 10 2 ≈ 14.14 ft and
y = 15 2 ≈ 21.21 ft.
300
x
The possible values for x are in (0, ∞).
15. xy = 300; y =
Q = 3(6x + 2y) + 2(2y) = 18x + 10y = 18x +
3000
x
dQ
3000
= 18 –
dx
x2
3000
=0
18 –
x2
500
x2 =
3
x=±
x=
10 5
3
10 5
3
is the only critical point in (0, ∞).
⎛ 10 5 ⎞
⎜⎜ 0,
⎟ and
3 ⎟⎠
⎝
⎛ 10 5 ⎞
dQ
, ∞ ⎟⎟ .
> 0 on ⎜⎜
dx
⎝ 3
⎠
dQ
< 0 on
dx
Instructor’s Resource Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
When x =
10 5
3
,y=
300
10 5
3
= 6 15
Q has a minimum when x =
10 5
3
≈ 12.91 ft and
x
=
y
y = 6 15 ≈ 23.24 ft.
900
16. xy = 900; y =
x
The possible values for x are in (0, ∞ ).
3600
Q = 6x + 4 y = 6x +
x
dQ
3600
=6–
dx
x2
3600
6–
=0
x2
x 2 = 600
x = ±10 6
x = 10 6 is the only critical point in (0, ∞ ).
dQ
dQ
< 0 on (0, 10 6) and
> 0 on (10 6, ∞).
dx
dx
900
When x = 10 6, y =
= 15 6
10 6
Q has a minimum when x = 10 6 ≈ 24.49 ft and
y = 15 6 ≈ 36.74.
x 2
= .
y 3
Suppose that each pen has area A.
A
xy = A; y =
x
The possible values for x are in (0, ∞ ).
4A
Q = 6x + 4 y = 6x +
x
dQ
4A
=6–
dx
x2
4A
6–
=0
x2
2A
x2 =
3
It appears that
x=±
x=
dQ
dx
dQ
dx
2A
,y=
3
When x =
2A
3
3A
2
=
A
2A
3
=
3A
2
2
3
17. Let D be the square of the distance.
⎛ x2
⎞
D = ( x − 0) + ( y − 4) = x + ⎜
− 4⎟
⎜ 4
⎟
⎝
⎠
2
=
2
2
2
x4
− x 2 + 16
16
dD x3
x3
=
− 2 x;
− 2 x = 0; x( x 2 − 8) = 0
dx
4
4
x = 0, x = ± 2 2
Critical points: 0, 2 2, 2 3
Since D is continuous and we are considering a
closed interval for x, there is a maximum and
minimum value of D on the interval. These
extrema must occur at one of the critical points.
At x = 0, y = 0, and D = 16. At x = 2 2, y = 2,
and D = 12. At x = 2 3 , y = 3, and D = 13.
Therefore, the point on y =
(
)
x2
closest to ( 0, 4 ) is
4
P 2 2, 2 and the point farthest from ( 0, 4 ) is
Q ( 0, 0 ) .
18. Let r1 and h1 be the radius and altitude of the
outer cone; r2 and h2 the radius and altitude of
the inner cone.
3V1
1
V1 = πr12 h1 is fixed. r1 =
πh1
3
By similar triangles
h1 – h2 r2
=
(see figure).
h1
r1
2A
3
2A
is the only critical point on (0, ∞ ).
3
⎛
2A ⎞
< 0 on ⎜⎜ 0,
⎟ and
3 ⎟⎠
⎝
⎛ 2A ⎞
, ∞ ⎟⎟ .
> 0 on ⎜⎜
⎝ 3
⎠
Instructor’s Resource Manual
⎛ h
r2 = r1 ⎜ 1 – 2
h1
⎝
⎞
⎟=
⎠
3V1
πh1
⎛ h2 ⎞
⎜1 – ⎟
h1 ⎠
⎝
Section 3.4
177
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2
1
1 ⎡ 3V1 ⎛ h2 ⎞ ⎤
V2 = πr22 h2 = π ⎢
⎜ 1 – ⎟ ⎥ h2
3
3 ⎢⎣ πh1 ⎝
h1 ⎠ ⎦⎥
2
2
h2 ⎛ h2 ⎞
π 3V1h2 ⎛ h2 ⎞
⋅
⎜ 1 – ⎟ = V1 ⎜ 1 – ⎟
3 πh1 ⎝
h1 ⎠
h1 ⎝
h1 ⎠
h
Let k = 2 , the ratio of the altitudes of the cones,
h1
=
then V2 = V1k (1 – k ) 2 .
dV2
= V1 (1 – k ) 2 – 2kV1 (1 – k ) = V1 (1 – k )(1 – 3k )
dk
dV2
1
0 < k < 1 so
= 0 when k = .
3
dk
d 2V2
d 2V2
1
= V1 (6k − 4);
< 0 when k =
2
2
3
dk
dk
1
The altitude of the inner cone must be the
3
altitude of the outer cone.
19. Let x be the distance from P to where the woman
lands the boat. She must row a distance of
x 2 + 4 miles and walk 10 – x miles. This will
x 2 + 4 10 – x
+
hours;
3
4
1
x
0 ≤ x ≤ 10. T ′( x) =
– ; T ′( x) = 0
3 x2 + 4 4
take her T ( x) =
when x =
T (0) =
6
7
.
19
hr = 3 hr 10 min ≈ 3.17 hr ,
6
⎛ 6 ⎞ 15 + 7
T⎜
≈ 2.94 hr,
⎟=
6
⎝ 7⎠
shore from P.
6
x=
2491
13
⎛ 6 ⎞
T (0) =
≈ 0.867 hr; T ⎜
⎟ ≈ 0.865 hr;
15
⎝ 2491 ⎠
T (10) ≈ 3.399 hr
 ≈ 0.12 mi down
x 2 + 4 10 – x
+
, 0 ≤ x ≤ 10.
20
4
x
1
– ; T ′( x) = 0 has no solution.
T ′( x ) =
2
20 x + 4 4
21. T ( x) =
T (0) =
2 10 13
+ =
hr = 2 hr, 36 min
20 4
5
104
≈ 0.5 hr
20
She should take the boat all the way to town.
T (10) =
22. Let x be the length of cable on land, 0 ≤ x ≤ L.
Let C be the cost.
C = a ( L − x) 2 + w2 + bx
dC
a( L − x)
=−
+b
dx
( L − x ) 2 + w2
−
a( L − x)
( L − x ) 2 + w2
+ b = 0 when
(a 2 − b 2 )( L − x) 2 = b 2 w2
6
7
≈ 2.27 mi down the
x = L−
aw
d 2C
=
bw
2
a – b2
ft on land;
ft under water
aw2
dx 2 [( L − x)2 + w2 ]3 2
minimizes the cost.
Section 3.4
2491
the shore from P.
a 2 – b2
178
6
She should land the boat
b 2 [( L − x) 2 + w2 ] = a 2 ( L − x)2
104
T (10) =
≈ 3.40 hr
3
She should land the boat
x 2 + 4 10 – x
+
, 0 ≤ x ≤ 10.
3
50
x
1
T ′( x) =
– ; T ′( x) = 0 when
3 x 2 – 4 50
20. T ( x) =
> 0 for all x, so this
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
23. Let the coordinates of the first ship at 7:00 a.m. be
(0, 0). Thus, the coordinates of the second ship at
7:00 a.m. are (–60, 0). Let t be the time in hours
since 7:00 a.m. The coordinates of the first and
second ships at t are (–20t, 0) and
( −60 + 15
=
x=
) + ( 0 + 15 2t )
= (1300 + 600 2 ) t 2 − ( 2400 + 1800 2 ) t + 3600
dD
= 2 (1300 + 600 2 ) t − ( 2400 + 1800 2 )
dt
2 (1300 + 600 2 ) t − ( 2400 + 1800 2 ) = 0 when
t=
12 + 9 2
13 + 6 2
2
D is the minimum at t =
12 + 9 2
13 + 6 2
since
d D
dt 2
=
2
a a −x
ab
2
bx 2
2
a a −x
2
+
>0
⎛ a ⎞
b⎜
⎟
⎝ 2⎠
dA
< 0 on
dx
2
=−
b
a
⎛ a ⎞
⎜ 0,
⎟ and
2⎠
⎝
2
2
h
⎛h⎞
V = πx 2 h; r 2 = x 2 + ⎜ ⎟ ; x 2 = r 2 −
4
⎝2⎠
2
3
⎛
h ⎞
πh
V = π ⎜ r 2 − ⎟ h = πhr 2 −
⎜
⎟
4
4
⎝
⎠
dV
3πh 2
2 3r
= πr 2 −
; V ′ = 0 when h = ±
4
3
dh
Since
d 2V
dh
2
when h =
a2 − x2
Find the x-intercept, x0 , of the tangent line
through the point (x, y):
y–0
bx
=–
x – x0
a a2 – x2
=−
3πh
, the volume is maximized
2
2 3r
.
3
⎛2 3 ⎞ 2 π
V = π ⎜⎜
r ⎟⎟ r −
⎝ 3 ⎠
=
ay a 2 – x 2
a2 – x2
a2
+x=
+x=
bx
x
x
Compute the Area A of the resulting triangle and
maximize:
−1
1
a 3b
a 3b ⎛
2
2⎞
=
A = x0 y0 =
⎜x a −x ⎟
⎠
2
2 ⎝
2 x a2 − x2
x0 =
Instructor’s Resource Manual
b 2 ⎛ a ⎞
b
a −⎜
⎟ =
a
2
⎝ 2⎠
25. Let x be the radius of the base of the cylinder and
h the height.
b 2
a − x2
a
−2 ⎛
dA
a 3b ⎛
x2
2
2⎞
2
2
=−
⎜ x a − x ⎟ ⎜⎜ a − x −
⎠ ⎝
dx
2 ⎝
a2 − x2
2
;y=
dA
⎛ a
⎞
, a ⎟ , so A is a minimum at
> 0 on ⎜
dx
⎝ 2 ⎠
a
x=
. Then the equation of the tangent line is
2
b⎛
a ⎞ b
y = − ⎜x−
or bx + ay − ab 2 = 0 .
⎟+
a⎝
2⎠
2
b 2
a − x 2 (positive
a
square root since the point is in the first quadrant).
Compute the slope of the tangent line:
bx
y′ = −
.
a a2 − x2
Find the y-intercept, y0 , of the tangent line
through the point (x, y):
y0 − y
bx
=−
0− x
a a2 − x2
+y=
(2 x 2 − a 2 ) = 0 when
2
a
Note that
24. Write y in terms of x: y =
bx 2
2 3/ 2
(2 x 2 – a 2 )
⎛ a ⎞
a a2 − ⎜
⎟
⎝ 2⎠
for all t.
The ships are closest at 8:09 A.M.
y0 =
2
y′ = −
≈ 1.15 hrs or 1 hr, 9 min
2
2 3/ 2
2 x (a − x )
square of the distances at t.
2
2
2 x (a − x )
2
)
(
2
a 3b
2t , − 15 2t respectively. Let D be the
D = −20t + 60 − 15 2t
a 3b
( r)
2 3
3
3
4
2 π 3 3 2 π 3 3 4π 3 3
r −
r =
r
3
9
9
26. Let r be the radius of the circle, x the length of the
rectangle, and y the width of the rectangle.
2
2
x2 y2
⎛ x⎞ ⎛ y⎞
;
P = 2x + 2y; r 2 = ⎜ ⎟ + ⎜ ⎟ ; r 2 =
+
4
4
⎝2⎠ ⎝ 2⎠
⎞
⎟
⎟
⎠
y = 4r 2 − x 2 ; P = 2 x + 2 4r 2 − x 2
dP
2x
= 2−
;
dx
4r 2 − x 2
Section 3.4
179
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2x
2−
2
4r − x
= 0; 2 4r 2 − x 2 = 2 x;
2
2
16r − 4 x 2 = 4 x 2 ; x = ± 2r
d 2P
dx 2
8r 2
=−
(4r 2 − x 2 )3 2
2
300 3 400
−
≈ 10.874 .
11
11
Critical points: x = 0, 10.874, 25
At x = 0, A ≈ 481; at x = 10.874, A ≈ 272; at
x = 25, A = 625.
A ' ( x ) = 0 when x =
< 0 when x = 2r ;
a.
2
y = 4 r − 2r = 2 r
The rectangle with maximum perimeter is a square
with side length 2r .
27. Let x be the radius of the cylinder, r the radius of
the sphere, and h the height of the cylinder.
A = 2π xh ; r 2 = x 2 +
A = 2π r 2 −
(
h2
h2
; x = r2 −
4
4
h2
h4
h = 2π h 2 r 2 −
4
4
2
3
)
so A is a maximum when h = 2r.
r
2
.
28. Let x be the distance from I1.
kI
kI 2
Q= 1+
2
x
( s − x)2
2kI 2
dQ −2kI1
=
+
3
dx
( s − x )3
x
2kI1
x
x=
3
+
2kI 2
( s − x)
3
= 0;
x3
I
= 1;
I2
( s − x)
3
s 3 I1
3
d 2Q
6kI1
+
⎛V ⎞
4kV
= 2.7 kx 2 + 4kx ⎜ ⎟ = 2.7 kx 2 +
2
x
⎝x ⎠
dC
4kV dC
= 5.4kx −
;
= 0 when x ≈ 0.9053 V
dx
x 2 dx
V
y≈
≈ 1.22 3 V
(0.9053 V )2
31. Let r be the radius of the cylinder and h the height
of the cylinder.
V − 23 πr 3
V
2
2
V = πr 2 h + πr 3 ; h =
=
− r
2
2 3
3
πr
πr
Let k be the cost per square foot of the cylindrical
wall. The cost is
C = k (2πrh) + 2k (2πr 2 )
⎛ 2V 8πr 2
⎛
⎞
2 ⎞
⎛ V
= k ⎜ 2πr ⎜
− r ⎟ + 4πr 2 ⎟ = k ⎜
+
⎜
3
⎝ πr 2 3 ⎠
⎝
⎠
⎝ r
⎞
⎟
⎟
⎠
dC
⎛ 2V 16πr ⎞ ⎛ 2V 16πr ⎞
= k ⎜−
+
+
⎟; k ⎜ −
⎟=0
3 ⎠ ⎝ r2
3 ⎠
dr
⎝ r2
I1 + 3 I 2
=
30. Let x be the length of the sides of the base, y be
the height of the box, and k be the cost per square
inch of the material in the sides of the box.
The cost is C = 1.2kx 2 + 1.5kx 2 + 4kxy
dA
dA
> 0 on (0, 2r ) and
< 0 on ( 2r , 2r ),
dh
dh
−
b. For maximum area, the wire should not be
cut; it should be bent to form a square.
V = x 2 y;
dA π 2r h − h
; A′ = 0 when h = 0, ± 2r
=
dh
2 2 h4
h r − 4
The dimensions are h = 2r , x =
For minimum area, the cut should be
approximately 4(10.874) ≈ 43.50 cm from
one end and the shorter length should be bent
to form the square.
6kI 2
dx 2
x 4 ( s − x) 4
minimizes the sum.
> 0, so this point
29. Let x be the length of a side of the square, so
100 − 4 x
is the side of the triangle, 0 ≤ x ≤ 25
3
1 ⎛ 100 − 4 x ⎞ 3 ⎛ 100 − 4 x ⎞
A = x2 + ⎜
⎟
⎜
⎟
2⎝
3
3
⎠ 2 ⎝
⎠
2⎞
⎛
3 10, 000 − 800 x + 16 x
= x2 +
⎜
⎟
⎟
4 ⎜⎝
9
⎠
when r 3 =
h=
1/ 3
3V
1 ⎛ 3V ⎞
,r = ⎜
⎟
8π
2⎝ π ⎠
1/ 3
1 ⎛ 3V ⎞
− ⎜
⎟
2
/
3
3⎝ π ⎠
π 3πV
4V
( )
13
⎛ 3V ⎞
=⎜
⎟
⎝ π ⎠
For a given volume V, the height of the cylinder is
1/ 3
⎛ 3V ⎞
⎜
⎟
⎝ π ⎠
1/ 3
and the radius is
1 ⎛ 3V ⎞
⎜
⎟
2⎝ π ⎠
.
dA
200 3 8 3
= 2x −
+
x
dx
9
9
180
Section 3.4
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
32.
dx
= 2 cos 2t − 2 3 sin 2t ;
dt
dx
1
= 0 when tan2t=
;
dt
3
π
2t = + πn for any integer n
6
π π
t= + n
12 2
π π
When t = + n ,
12 2
⎛π
⎞
⎛π
⎞
x = sin ⎜ + πn ⎟ + 3 cos ⎜ + πn ⎟
⎝6
⎠
⎝6
⎠
π
π
= sin cos πn + cos sin πn
6
6
π
π
+ 3(cos cos πn − sin sin πn)
6
6
1
3
= (−1)n + (−1)n = 2.
2
2
The farthest the weight gets from the origin is 2
units.
2 Ar
2A
= 2r +
Q = 2r + rθ = 2r +
2
r
r
dQ
2A
= 2−
; Q′ = 0 when r = A
dr
r2
2A
θ=
=2
( A )2
d 2Q
dr
2
=
4A
r3
cos θ =
3
3
h
w
h
3
2/3
h
+ w2 / 3
,
w
+ w2 / 3
⎛ h 2 / 3 + w2 / 3
h 2 / 3 + w2 / 3 ⎞⎟
+ w⎜
3
3
⎟
⎜
h
w
⎠
⎝
h
⎛
x = h⎜
⎜
⎝
3
; sin θ =
2/3
⎞
⎟
⎟
⎠
= ( h 2 / 3 + w 2 / 3 )3 / 2
35. x is limited by 0 ≤ x ≤ 12 .
A = 2 x(12 − x 2 ) = 24 x − 2 x3 ;
dA
= 24 − 6 x 2 ;
dx
24 − 6 x 2 = 0; x = −2, 2
Critical points: 0, 2, 12.
When x = 0 or 12, A = 0.
When x = 2, y = 12 − (2)2 = 8.
The dimensions are 2x = 2(2) = 4 by 8.
36. Let the x-axis lie on the diameter of the semicircle
and the y-axis pass through the middle.
Then the equation y = r 2 − x 2 describes the
semicircle. Let (x, y) be the upper-right corner of
the rectangle. x is limited by 0 ≤ x ≤ r .
2
r θ
2A
; θ=
2
r2
The perimeter is
33. A =
tan θ =
> 0, so this minimizes the perimeter.
34. The distance from the fence to the base of the
h
.
ladder is
tan θ
The length of the ladder is x.
h +w
h
cos θ = tan θ
; x cos θ =
+ w;
x
tan θ
h
w
+
x=
sin θ cos θ
A = 2 xy = 2 x r 2 − x 2
dA
2x2
2
= 2 r 2 − x2 −
=
(r 2 − 2 x 2 )
2
2
2
2
dx
r −x
r −x
2
r
(r 2 − 2 x 2 ) = 0; x =
2
2
2
r −x
r
Critical points: 0,
,r
2
r
When x = 0 or r, A = 0. When x =
, A = r2.
2
2
r
⎛ r ⎞
y = r2 − ⎜
⎟ =
2
2
⎝
⎠
The dimensions are
r
2
by
2r
2
.
dx
h cosθ w sin θ w sin 3 θ − h cos3 θ
=−
+
;
=0
dθ
sin 2 θ cos 2 θ
sin 2 θ cos 2 θ
h
when tan 3 θ =
w
θ = tan −1 3
h
w
Instructor’s Resource Manual
Section 3.4
181
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
37. If the end of the cylinder has radius r and h is the
height of the cylinder, the surface area is
A
A = 2πr 2 + 2πrh so h =
– r.
2πr
The volume is
⎛ A
⎞ Ar
–r⎟ =
– πr 3 .
V = πr 2 h = πr 2 ⎜
2
2
π
r
⎝
⎠
A
A
V ′(r ) = – 3πr 2 ; V ′(r ) = 0 when r =
,
6π
2
V ′′(r ) = −6πr , so the volume is maximum when
r=
h=
A
.
6π
sin t =
h
, so h = r cos t,
r
1 2
r – h 2 , and
r
r 2 – h 2 = r sin t
Area of submerged region = tr 2 – h r 2 – h 2
= tr 2 – (r cos t )(r sin t ) = r 2 (t – cos t sin t )
A = area of exposed wetted region
= r 2 (π – π cos 2 t – t + cos t sin t )
dA
= r 2 (2π cos t sin t –1 + cos 2 t – sin 2 t )
dt
38. The ellipse has equation
y = ± b2 –
b2 x2
a2
=±
= r 2 (2π cos t sin t – 2sin 2 t )
b 2
a – x2
a
= 2r 2 sin t (π cos t – sin t )
⎛ b 2
⎞
Let ( x, y ) = ⎜ x,
a – x 2 ⎟ be the upper right⎝ a
⎠
hand corner of the rectangle (use a and b positive).
Then the dimensions of the rectangle are 2x by
2b 2
a – x 2 and the area is
a
4bx 2
A( x) =
a – x2 .
a
4b 2
4bx 2
4b(a 2 – 2 x 2 )
=
A′( x) =
a – x2 –
;
a
a a2 – x2
a a2 – x2
a
A′( x) = 0 when x =
, so the corner is at
2
⎛ a b ⎞
,
⎜
⎟ . The corners of the rectangle are at
⎝ 2 2⎠
b ⎞
⎛ a b ⎞ ⎛ a b ⎞ ⎛ a
,
,
,–
⎜
⎟, ⎜ –
⎟, ⎜ –
⎟,
2 2⎠ ⎝
2
2⎠
⎝ 2 2⎠ ⎝
b ⎞
⎛ a
,−
⎜
⎟.
2⎠
⎝ 2
diagonal is d = l 2 + w2 , so l = d 2 – w2 . The
area is A = lw = w d 2 – w2 .
w2
d 2 – w2
d
A′( w) = 0 when w =
and so
2
=
d 2 – 2 w2
d 2 – w2
;
d2
d
⎛ d ⎞
=
. A′( w) > 0 on ⎜ 0,
⎟ and
2
2
2⎠
⎝
Section 3.4
this is
1
r
r 2 – h2
h
r
= π or h =
r
1 + π2
.
41. The carrying capacity of the gutter is maximized
when the area of the vertical end of the gutter is
maximized. The height of the gutter is 3sin θ . The
area is
⎛1⎞
A = 3(3sin θ ) + 2 ⎜ ⎟ (3cos θ )(3sin θ )
⎝2⎠
= 9sin θ + 9 cos θ sin θ .
dA
= 9 cos θ + 9(− sin θ ) sin θ + 9 cosθ cosθ
dθ
= 9(2 cos 2 θ + cosθ − 1)
39. If the rectangle has length l and width w, the
A′( w) = d 2 – w2 –
dA
= 0 only when
dt
π cos t = sin t or tan t = π . In terms of r and h,
Since 0 < t < π ,
= 9(cos θ − sin 2 θ + cos 2 θ )
The dimensions are a 2 and b 2 .
182
40. Note that cos t =
= πr 2 – πh 2 – r 2 (t – cos t sin t )
A
A
–r =2
= 2r
2πr
6π
l = d2 –
⎛ d
⎞
A′( w) < 0 on ⎜
, d ⎟ . Maximum area is for a
⎝ 2 ⎠
square.
1
π
2 cos 2 θ + cos θ − 1 = 0; cos θ = −1, ; θ = π,
2
3
π
Since 0 ≤ θ ≤ , the critical points are
2
π
π
0, , and .
3
2
When θ = 0 , A = 0.
π
27 3
,A=
≈ 11.7.
3
4
π
When θ = , A = 9.
2
When θ =
The carrying capacity is maximized when θ =
π
.
3
Instructor’s Resource Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
42. The circumference of the top of the tank is the
circumference of the circular sheet minus the arc
length of the sector,
20π − 10θ meters. The radius of the top of the
20π − 10θ 5
= (2π − θ ) meters. The
tank is r =
2π
π
slant height of the tank is 10 meters, so the height
of the tank is
2
5θ ⎞
5
⎛
h = 102 − ⎜ 10 − ⎟ =
4πθ − θ 2 meters.
π ⎠
π
⎝
2
1
1 ⎡5
⎤ ⎡5
⎤
V = πr 2 h = π ⎢ (2π − θ ) ⎥ ⎢
4πθ − θ 2 ⎥
3
3 ⎣π
⎦ ⎣π
⎦
125
(2π − θ )2 4πθ − θ 2
=
3π2
dV 125 ⎛
2
=
⎜ 2(2π − θ )(−1) 4πθ − θ
dθ 3π2 ⎝
(2π − θ )2 12 (4π − 2θ ) ⎞
⎟
+
⎟
4πθ − θ 2
⎠
( )
=
125(2π − θ )
3π
2
4πθ − θ
125(2π − θ )
3π
2
4πθ − θ
2
2
(3θ 2 − 12πθ + 4π2 ) ;
(3θ 2 − 12πθ + 4π2 ) = 0
2π − θ = 0 or 3θ 2 − 12πθ + 4π2 = 0
2 6
2 6
π, θ = 2π +
π
3
3
Since 0 < θ < 2π, the only critical point is
θ = 2π, θ = 2π −
2 6
π . A graph shows that this maximizes
3
the volume.
2π −
44. Let x be the length of the edges of the cube. The
1
surface area of the cube is 6x 2 so 0 ≤ x ≤
.
6
The surface area of the sphere is 4πr 2 , so
1 – 6x2
4π
4
1
V = x3 + πr 3 = x3 +
(1 – 6 x 2 )3 / 2
3
6 π
⎛
dV
3
1 – 6x2
= 3x2 –
x 1 – 6 x 2 = 3x ⎜ x –
⎜
dx
π
π
⎝
dV
1
= 0 when x = 0,
dx
6+π
1
V (0) =
≈ 0.094 m3 .
6 π
6 x 2 + 4πr 2 = 1, r =
⎞
⎟
⎟
⎠
3/ 2
1 ⎛
6 ⎞
⎛ 1 ⎞
–3 / 2
V⎜
+
⎟ = (6 + π)
⎜1 –
⎟
6
+
π⎠
6 π⎝
⎝ 6+π ⎠
1
⎛ π⎞
= ⎜ 1 + ⎟ (6 + π) –3 / 2 =
≈ 0.055 m3
6 6+π
⎝ 6⎠
For maximum volume: no cube, a sphere of radius
1
≈ 0.282 meters.
2 π
For minimum volume: cube with sides of length
1
≈ 0.331 meters,
6+π
1
sphere of radius
≈ 0.165 meters
2 6+π
45. Consider the figure below.
43. Let V be the volume. y = 4 – x and z = 5 – 2x.
x is limited by 0 ≤ x ≤ 2.5 .
V = x(4 − x)(5 − 2 x) = 20 x − 13 x 2 + 2 x3
dV
= 20 − 26 x + 6 x 2 ; 2(3x 2 − 13 x + 10) = 0;
dx
2(3 x − 10)( x − 1) = 0;
10
x = 1,
3
Critical points: 0, 1, 2.5
At x = 0 or 2.5, V = 0. At x = 1, V = 9.
Maximum volume when x = 1, y = 4 – 1 = 3, and
z = 5 – 2(1) = 3.
a.
y = x 2 − (a − x )2 = 2ax − a 2
Area of A = A =
=
1
(a − x) 2ax − a 2
2
( )
1 (a − x) 1 (2a )
dA
1
2
2
=−
2ax − a + 2
dx
2
2ax − a 2
=
Instructor’s Resource Manual
1
(a − x) y
2
a 2 − 32 ax
2ax − a 2
Section 3.4
183
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
a 2 − 32 ax
= 0 when x =
2a
.
3
2ax − a 2
dA
dA
⎛ a 2a ⎞
⎛ 2a ⎞
> 0 on ⎜ , ⎟ and
< 0 on ⎜ , a ⎟ ,
dx
dx
⎝2 3 ⎠
⎝ 3
⎠
2a
so x =
maximizes the area of triangle A.
3
c.
b. Triangle A is similar to triangle C, so
ax
ax
w=
=
y
2ax − a 2
Area of B = B =
1
ax 2
xw =
2
2 2ax − a 2
⎛ 2 x 2ax − a 2 − x 2
dB a ⎜
= ⎜
dx 2 ⎜
2ax − a 2
⎜
⎝
a
2 ax − a 2
⎞
⎟
⎟
⎟
⎟
⎠
z = x 2 + w2 = x 2 +
=
2ax – a 2
2ax3
2ax – a 2
dz 1 2ax − a 2
=
dx 2
2ax3
=
a2 x2
⎛ 6ax 2 (2ax − a 2 ) − 2ax3 (2a) ⎞
⎜
⎟
⎜
⎟
(2ax − a 2 ) 2
⎝
⎠
4a 2 x3 − 3a3 x 2
2ax3 (2ax − a 2 )3
dz
3a
3a
= 0 when x = 0,
→ x=
dx
4
4
dz
dz
⎛ a 3a ⎞
⎛ 3a ⎞
< 0 on ⎜ , ⎟ and
> 0 on ⎜ , a ⎟ ,
dx
2
4
dx
⎝
⎠
⎝ 4 ⎠
3a
so x =
minimizes length z.
4
a ⎛ 2 x(2ax − a 2 ) − ax 2 ⎞ a ⎛ 3ax 2 − 2 xa 2 ⎞
⎜
⎟= ⎜
⎟
2 ⎜⎝ (2ax − a 2 )3 / 2 ⎟⎠ 2 ⎜⎝ (2ax − a 2 )3 / 2 ⎟⎠
a 2 ⎛ 3x 2 − 2 xa ⎞
2a
⎜
⎟ = 0 when x = 0,
2
3
/
2
⎜
⎟
2 ⎝ (2ax − a )
3
⎠
2a
.
Since x = 0 is not possible, x =
3
dB
dB
⎛ a 2a ⎞
⎛ 2a ⎞
< 0 on ⎜ , ⎟ and
> 0 on ⎜ , a ⎟ ,
dx
dx
⎝2 3 ⎠
⎝ 3
⎠
2a
minimizes the area of triangle B.
so x =
3
=
46. Let 2x be the length of a bar and 2y be the width of a bar.
θ 1
θ⎞ a ⎛ θ
θ⎞
⎛ 1
⎛π θ ⎞
x = a cos ⎜ − ⎟ = a ⎜
cos +
sin ⎟ =
⎜ cos + sin ⎟
2
2⎠
2
2⎠
2
2⎝
⎝4 2⎠
⎝ 2
θ 1
θ⎞ a ⎛ θ
θ⎞
⎛ 1
⎛π θ ⎞
y = a sin ⎜ − ⎟ = a ⎜
cos −
sin ⎟ =
⎜ cos − sin ⎟
4
2
2
2
2
2⎠
2
2⎝
⎝
⎠
⎝ 2
⎠
Compute the area A of the cross and maximize.
⎡ a ⎛
θ
θ ⎞⎤ ⎡ a ⎛ θ
θ ⎞⎤ ⎡ a ⎛ θ
θ ⎞⎤
A = 2(2 x)(2 y ) − (2 y ) 2 = 8 ⎢
⎜ cos + sin ⎟ ⎥ ⎢
⎜ cos − sin ⎟ ⎥ − 4 ⎢
⎜ cos − sin ⎟ ⎥
2
2 ⎠⎦ ⎣ 2 ⎝
2
2 ⎠⎦
2
2 ⎠⎦
⎣ 2⎝
⎣ 2⎝
θ
θ⎞
θ
θ⎞
⎛
⎛
= 4a 2 ⎜ cos 2 − sin 2 ⎟ − 2a 2 ⎜1 − 2 cos sin ⎟ = 4a 2 cosθ − 2a 2 (1 − sin θ )
2
2⎠
2
2⎠
⎝
⎝
dA
1
= −4a 2 sin θ + 2a 2 cosθ ; −4a 2 sin θ + 2a 2 cos θ = 0 when tan θ = ;
dθ
2
1
2
sin θ =
, cos θ =
5
5
d2A
dθ
2
< 0 when tan θ =
2
1
, so this maximizes the area.
2
⎛ 2 ⎞
1 ⎞ 10a 2
2⎛
A = 4a 2 ⎜
–
2
a
1
–
– 2a 2 = 2a 2 ( 5 – 1)
⎟
⎜
⎟=
5⎠
5
⎝ 5⎠
⎝
184
Section 3.4
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47. a.
L′(θ ) = 15(9 + 25 − 30 cosθ )−1/ 2 sin θ = 15(34 − 30 cosθ )−1/ 2 sin θ
L′′(θ ) = −
15
(34 − 30 cosθ )−3 / 2 (30sin θ ) sin θ + 15(34 − 30 cosθ )−1/ 2 cosθ
2
= −225(34 − 30 cos θ ) −3 / 2 sin 2 θ + 15(34 − 30 cos θ ) −1/ 2 cosθ
= 15(34 − 30 cos θ )−3 / 2 [−15sin 2 θ + (34 − 30 cos θ ) cosθ ]
= 15(34 − 30 cosθ ) −3 / 2 [−15sin 2 θ + 34 cos θ − 30 cos 2 θ ]
= 15(34 − 30 cos θ ) −3 / 2 [−15 + 34 cos θ − 15cos 2 θ ]
= −15(34 − 30 cos θ )−3 / 2 [15cos 2 θ − 34 cosθ + 15]
L′′ = 0 when cos θ =
34 ± (34)2 − 4(15)(15) 5 3
= ,
2(15)
3 5
⎛3⎞
⎝ ⎠
θ = cos −1 ⎜ ⎟
5
⎛
⎛
⎛ 3 ⎞⎞
⎛ 3 ⎞⎞
L′ ⎜ cos −1 ⎜ ⎟ ⎟ = 15 ⎜ 9 + 25 − 30 ⎜ ⎟ ⎟
5
⎝ ⎠⎠
⎝ 5 ⎠⎠
⎝
⎝
−1/ 2
⎛4⎞
⎜ ⎟=3
⎝5⎠
1/ 2
⎛
⎛ 3 ⎞⎞ ⎛
⎛ 3 ⎞⎞
L ⎜ cos−1 ⎜ ⎟ ⎟ = ⎜ 9 + 25 − 30 ⎜ ⎟ ⎟
=4
⎝ 5 ⎠⎠ ⎝
⎝ 5 ⎠⎠
⎝
φ = 90° since the resulting triangle is a 3-4-5 right triangle.
b.
L′(θ ) = 65(25 + 169 − 130 cos θ )−1/ 2 sin θ = 65(194 − 130 cosθ ) −1/ 2 sin θ
L′′(θ ) = −
65
(194 − 130 cos θ )−3 / 2 (130sin θ ) sin θ + 65(194 − 130 cos θ ) −1/ 2 cosθ
2
= −4225(194 − 130 cos θ )−3 / 2 sin 2 θ + 65(194 − 130 cosθ ) −1/ 2 cosθ
= 65(194 − 130 cosθ )−3 / 2 [−65sin 2 θ + (194 − 130 cos θ ) cosθ ]
= 65(194 − 130 cos θ )−3 / 2 [−65sin 2 θ + 194 cos θ − 130 cos 2 θ ]
= 65(194 − 130 cos θ )−3 / 2 [−65cos 2 θ + 194 cosθ − 65]
= −65(194 − 130 cosθ ) −3 / 2 [65cos 2 θ − 194 cos θ + 65]
L′′ = 0 when cos θ =
194 ± (194)2 − 4(65)(65) 13 5
= ,
2(65)
5 13
⎛5⎞
⎟
⎝ 13 ⎠
θ = cos −1 ⎜
1/ 2
⎛
⎛
⎛ 5 ⎞⎞
⎛ 5 ⎞⎞
L′ ⎜ cos −1 ⎜ ⎟ ⎟ = 65 ⎜ 25 + 169 − 130 ⎜ ⎟ ⎟
13
⎝
⎠
⎝ 13 ⎠ ⎠
⎝
⎠
⎝
⎛ 12 ⎞
⎜ ⎟=5
⎝ 13 ⎠
1/ 2
⎛
⎛ 5 ⎞⎞ ⎛
⎛ 5 ⎞⎞
L ⎜ cos−1 ⎜ ⎟ ⎟ = ⎜ 25 + 169 − 130 ⎜ ⎟ ⎟
= 12
⎝ 13 ⎠ ⎠ ⎝
⎝ 13 ⎠ ⎠
⎝
φ = 90° since the resulting triangle is a 5-12-13 right triangle.
c.
When the tips are separating most rapidly, φ = 90°, L = m 2 − h 2 , L′ = h
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Section 3.4
185
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d.
L′(θ ) = hm(h 2 + m2 − 2hm cos θ )−1/ 2 sin θ
L′′(θ ) = −h 2 m2 (h 2 + m 2 − 2hm cos θ ) −3 / 2 sin 2 θ + hm(h 2 + m 2 − 2hm cosθ )−1/ 2 cos θ
= hm(h 2 + m 2 − 2hm cos θ )−3 / 2 [− hm sin 2 θ + (h 2 + m 2 ) cos θ − 2hm cos 2 θ ]
= hm(h 2 + m 2 − 2hm cosθ ) −3 / 2 [− hm cos 2 θ + (h 2 + m 2 ) cos θ − hm]
= −hm(h 2 + m2 − 2hm cos θ )−3 / 2 [hm cos 2 θ − (h 2 + m2 ) cosθ + hm]
L′′ = 0 when hm cos 2 θ − (h 2 + m2 ) cos θ + hm = 0
(h cos θ − m)(m cos θ − h) = 0
cos θ =
m h
,
h m
Since h < m, cos θ =
h
⎛h⎞
so θ = cos −1 ⎜ ⎟ .
m
⎝m⎠
⎛
⎛
⎛ h ⎞⎞
⎛ h ⎞⎞
L′ ⎜ cos −1 ⎜ ⎟ ⎟ = hm ⎜ h 2 + m2 − 2hm ⎜ ⎟ ⎟
⎝ m ⎠⎠
⎝ m ⎠⎠
⎝
⎝
1/ 2
⎛
⎛ h ⎞⎞ ⎛
⎛ h ⎞⎞
L ⎜ cos −1 ⎜ ⎟ ⎟ = ⎜ h 2 + m2 − 2hm ⎜ ⎟ ⎟
m
⎝ ⎠⎠ ⎝
⎝ m ⎠⎠
⎝
−1/ 2
m2 − h2
= hm(m2 − h 2 ) −1/ 2
m
m2 − h2
=h
m
= m2 − h2
Since h 2 + L2 = m 2 , φ = 90°.
48. We are interested in finding the global extrema for
the distance of the object from the observer. We
will obtain this result by considering the squared
distance instead. The squared distance can be
expressed as
2
1
⎛
⎞
D( x) = ( x − 2)2 + ⎜ 100 + x − x 2 ⎟
10
⎝
⎠
The first and second derivatives are given by
1 3 3 2
D '( x) =
x − x − 36 x + 196 and
25
5
3 2
D ''( x) =
x − 10 x − 300
25
Using a computer package, we can solve the
equation D '( x) = 0 to find the critical points. The
critical points are x ≈ 5.1538,36.148 . Using the
second derivative we see that
D ''(5.1538) ≈ −38.9972 (max) and
(
)
D ''(36.148) ≈ 77.4237 (min)
Therefore, the position of the object closest to the
observer is ≈ ( 36.148,5.48 ) while the position of
the object farthest from the person is
≈ (5.1538,102.5) .
(Remember to go back to the original equation for
the path of the object once you find the critical
points.)
186
Section 3.4
49. Here we are interested in minimizing the distance
between the earth and the asteroid. Using the
coordinates P and Q for the two bodies, we can
use the distance formula to obtain a suitable
equation. However, for simplicity, we will
minimize the squared distance to find the critical
points. The squared distance between the objects
is given by
D(t ) = (93cos(2π t ) − 60 cos[2π (1.51t − 1)]) 2
+ (93sin(2π t ) − 120sin[2π (1.51t − 1)])2
The first derivative is
D '(t ) ≈ −34359 [ cos(2π t )][ sin(9.48761t ) ]
+ [ cos(9.48761t ) ][(204932sin(9.48761t )
−141643sin(2π t ))]
Plotting the function and its derivative reveal a
periodic relationship due to the orbiting of the
objects. Careful examination of the graphs reveals
that there is indeed a minimum squared distance
(and hence a minimum distance) that occurs only
once. The critical value for this occurrence is
t ≈ 13.82790355 . This value gives a squared
distance between the objects of ≈ 0.0022743
million miles. The actual distance is ≈ 0.047851
million miles ≈ 47,851 miles.
Instructor’s Resource Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
50. Let x be the width and y the height of the flyer.
51. Consider the following sketch.
1 inch
1 inch
2 inches
2 inches
We wish to minimize the area of the flyer,
A = xy .
As it stands, A is expressed in terms of two
variables so we need to write one in terms of the
other.
The printed area of the flyer has an area of 50
square inches. The equation for this area is
( x − 2 )( y − 4 ) = 50
We can solve this equation for y to obtain
50
y=
+4
x−2
Substituting this expression for y in our equation
for A, we get A in terms of a single variable, x.
A = xy
⎛ 50
⎞ 50 x
= x⎜
+ 4⎟ =
+ 4x
⎝ x−2
⎠ x−2
The allowable values for x are 2 < x < ∞ ; we want
to minimize A on the open interval ( 2, ∞ ) .
dA ( x − 2 ) 50 − 50 x
−100
=
+4=
+4
2
dx
( x − 2 )2
( x − 2)
=
4 x 2 − 16 x − 84
( x − 2)
2
=
4 ( x − 7 )( x + 3)
( x − 2 )2
The only critical points are obtained by solving
dA
= 0 ; this yields x = 7 and x = −3 . We reject
dx
x = −3 because it is not in the feasible domain
dA
dA
( 2, ∞ ) . Since < 0 for x in ( 2, 7 ) and > 0
dx
dx
for x in ( 7, ∞ ) , we conclude that A attains its
minimum value at x = 7 . This value of x makes
y = 14 . So, the dimensions for the flyer that will
use the least amount of paper are 7 inches by 14
inches.
Instructor’s Resource Manual
By similar triangles,
27t
x=
t + 64
2
t + 64
1728
(t + 64)3 / 2
dt 2
27 − t + 64
27t 2
t 2 + 64
27 t 2 + 64 −
2
d2x
2
=
−1 =
−5184t
(
27 4 5
(4 5)
t
2
t + 64
.
1728
2
(t + 64)3 / 2
−1
− 1 = 0 when t = 4 5
;
d2x
(t 2 + 64)5 / 2 dt 2 t = 4
Therefore
x=
=
−t
2
dx
=
dt
x
2
)
<0
5
− 4 5 = 5 5 ≈ 11.18 ft is the
+ 64
maximum horizontal overhang.
52. a.
b. There are only a few data points, but they do
seem fairly linear.
c.
The data values can be entered into most
scientific calculators to utilize the Least
Squares Regression feature. Alternately one
could use the formulas for the slope and
intercept provided in the text. The resulting
line should be y = 0.56852 + 2.6074 x
Section 3.4
187
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
d. Using the result from c., the predicted number
of surface imperfections on a sheet with area
2.0 square feet is
y = 0.56852 + 2.6074(2.0) = 5.7833 ≈ 6
since we can't have partial imperfections
53. a.
dS d n
2
=
∑ [ yi − (5 + bxi )]
db db i =1
54. C(x) = 7000 + 100x
55. n = 100 + 10
250 – p (n)
n
so p(n) = 300 –
5
2
R (n) = np (n) = 300n –
n2
2
56. P (n) = R (n) – C (n)
n2
– (7000 + 100n)
2
n
= 300n –
n
= −7000 + 200n –
d
2
= ∑ [ yi − (5 + bxi ) ]
db
i =1
= ∑ 2( yi − 5 − bxi )(− xi )
n2
2
i =1
⎡n
= 2 ⎢ ∑ − xi yi + 5 xi + bxi 2
⎢⎣ i =1
(
57.
⎤
)⎥⎥⎦
n
n
n
i =1
i =1
i =1
= −2∑ xi yi + 10∑ xi + 2b∑ xi 2
Setting
dS
= 0 gives
db
n
n
n
0 = −2∑ xi yi + 10∑ xi + 2b∑ xi
i =1
n
i =1
Estimate n ≈ 200
P ′(n) = 200 – n; 200 – n = 0 when n = 200.
P ′′(n) = –1, so profit is maximum at n = 200.
2
i =1
n
n
i =1
i =1
0 = −∑ xi yi + 5∑ xi + b∑ xi 2
i =1
n
n
58.
n
b∑ xi 2 =∑ xi yi − 5∑ xi
i =1
i =1
n
i =1
n
∑ xi yi − 5∑ xi
b = i =1
i =1
n
∑ xi 2
i =1
You should check that this is indeed the value
of b that minimizes the sum. Taking the
second derivative yields
n
d 2S
= 2∑ xi 2
db 2
i =1
which is always positive (unless all the x
values are zero). Therefore, the value for b
above does minimize the sum as required.
b. Using the formula from a., we get that
(2037) − 5(52)
b=
≈ 3.0119
590
c.
188
The Least Squares Regression line is
y = 5 + 3.0119 x
Using this line, the predicted total number of
labor hours to produce a lot of 15 brass
bookcases is
y = 5 + 3.0119(15) ≈ 50.179 hours
Section 3.4
59.
C ( x) 100
=
+ 3.002 – 0.0001x
x
x
C ( x)
= 2.9045 or $2.90 per unit.
When x = 1600,
x
dC
= 3.002 − 0.0002 x
dx
C ′(1600) = 2.682 or $2.68
C (n) 1000
n
=
+
n
n
1200
C ( n)
≈ 1.9167 or $1.92 per unit.
When n = 800,
n
dC
n
=
dn 600
C ′(800) ≈ 1.333 or $1.33
60. a.
dC
= 33 − 18 x + 3 x 2
dx
d 2C
d 2C
= 0 when x = 3
dx 2
d 2C
< 0 on (0, 3),
> 0 on (3, ∞)
dx 2
dx 2
Thus, the marginal cost is a minimum when
x = 3 or 300 units.
dx 2
d 2C
b.
= −18 + 6 x;
33 − 18(3) + 3(3) 2 = 6
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
61. a.
R ( x) = xp( x) = 20 x + 4 x 2 −
65. The revenue function would be
R ( x ) = x ⋅ p ( x ) = 200 x − 0.15 x 2 . This, together
x3
3
with the cost function yields the following profit
function:
⎧⎪−5000 + 194 x − 0.148 x 2 if 0 ≤ x ≤ 500
P ( x) = ⎨
2
⎪⎩−9000 + 194 x − 0.148 x if 500 < x ≤ 750
dR
= 20 + 8 x − x 2 = (10 − x )( x + 2 )
dx
b. Increasing when
dR
>0
dx
20 + 8 x − x 2 > 0 on [0, 10)
Total revenue is increasing if 0 ≤ x ≤ 10.
c.
d 2R
dx 2
d 3R
dx
3
= 8 – 2 x;
= −2;
d 2R
dx 2
a.
= 0 when x = 4
dR
is maximum at x = 4.
dx
1/ 2
x ⎞
⎛
62. R ( x) = x ⎜ 182 − ⎟
36 ⎠
⎝
1⎛
dR
x ⎞
= x ⎜182 − ⎟
2⎝
36 ⎠
dx
−1/ 2
1/ 2
x ⎞
⎛ 1 ⎞ ⎛
⎜ − ⎟ + ⎜182 − ⎟
36 ⎠
⎝ 36 ⎠ ⎝
−1 2
x ⎞
x ⎞
⎛
⎛
= ⎜ 182 − ⎟
⎜ 182 − ⎟
36 ⎠
24 ⎠
⎝
⎝
dR
= 0 when x = 4368
dx
x1 = 4368; R(4368) ≈ 34, 021.83
At x1 ,
At x1 ,
P ( 655 ) = 54,574.30 ; P ( 750 ) = 53, 250
The profit is maximized if the company
produces 500 chairs. The current machine can
handle this work, so they should not buy the
new machine.
dR
=0.
dx
800 x
− 3x
63. R( x) =
x+3
dR ( x + 3)(800) − 800 x
2400
=
−3 =
− 3;
2
dx
( x + 3)
( x + 3)2
dR
= 0 when x = 20 2 − 3 ≈ 25
dx
x1 = 25; R (25) ≈ 639.29
dR
=0.
dx
64. p( x) = 12 − (0.20)
( x − 400)
= 20 − 0.02 x
10
R ( x) = 20 x − 0.02 x 2
dR
dR
= 20 − 0.04 x;
= 0 when x = 500
dx
dx
Total revenue is maximized at x1 = 500 .
Instructor’s Resource Manual
The only difference in the two pieces of the
profit function is the constant. Since the
derivative of a constant is 0, we can say that
on the interval 0 < x < 750 ,
dP
= 194 − 0.296 x
dx
There are no singular points in the given
interval. To find stationary points, we solve
dP
=0
dx
194 − 0.296 x = 0
−0.296 x = −194
x ≈ 655
Thus, the critical points are 0, 500, 655, and
750.
P ( 0 ) = −5000 ; P ( 500 ) = 55, 000 ;
b. Without the new machine, a production level
of 500 chairs would yield a maximum profit
of $55,000.
66. The revenue function would be
R ( x ) = x ⋅ p ( x ) = 200 x − 0.15 x 2 . This, together
with the cost function yields the following profit
function:
⎧⎪−5000 + 194 x − 0.148 x 2 if 0 ≤ x ≤ 500
P ( x) = ⎨
2
⎪⎩−8000 + 194 x − 0.148 x if 500 < x ≤ 750
a.
The only difference in the two pieces of the
profit function is the constant. Since the
derivative of a constant is 0, we can say that
on the interval 0 < x < 750 ,
dP
= 194 − 0.296 x
dx
There are no singular points in the given
interval. To find stationary points, we solve
dP
=0
dx
194 − 0.296 x = 0
−0.296 x = −194
x ≈ 655
Section 3.4
189
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Thus, the critical points are 0, 500, 655, and 750.
P ( 0 ) = −5000 ; P ( 500 ) = 55, 000 ;
b.
F (b) =
a 2 + 2ab + b 2
4b
P ( 655 ) = 55,574.30 ; P ( 750 ) = 54, 250
As b → 0+ , a 2 + 2ab + b 2 → a 2 while
The profit is maximized if the company produces
655 chairs. The current machine cannot handle
this work, so they should buy the new machine.
4b → 0+ , thus lim F (b) = ∞ which is not
b → 0+
close to a.
2
a + 2a + b
a 2 + 2ab + b 2
b
lim
=
=∞ ,
4b
4
b →∞
b →∞
so when b is very large, F(b) is not close to a.
2(a + b)(4b) – 4(a + b)2
F ′(b) =
16b 2
4b 2 – 4a 2 b 2 – a 2
=
=
;
16b 2
4b 2
F ′(b) = 0 when b 2 = a 2 or b = a since a and
b are both positive.
( a + a ) 2 4a 2
F (a) =
=
=a
4a
4a
b. With the new machine, a production level of
655 chairs would yield a maximum profit of
$55,574.30.
lim
67. R ( x) = 10 x − 0.001x 2 ; 0 ≤ x ≤ 300
P ( x) = (10 x – 0.001x 2 ) – (200 + 4 x – 0.01x 2 )
= –200 + 6 x + 0.009 x 2
dP
dP
= 6 + 0.018 x;
= 0 when x ≈ −333
dx
dx
Critical numbers: x = 0, 300; P(0) = –200;
P(300) = 2410; Maximum profit is $2410
at x = 300.
⎧⎪200 + 4 x − 0.01x 2 if 0 ≤ x ≤ 300
68. C ( x) = ⎨
2
if 300 < x ≤ 450
⎪⎩800 + 3 x − 0.01x
⎧⎪−200 + 6 x + 0.009 x 2 if 0 ≤ x ≤ 300
P( x) = ⎨
2
if 300 < x ≤ 450
⎪⎩−800 + 7 x + 0.009 x
There are no stationary points on the interval
[0, 300]. On [300, 450]:
dP
dP
= 7 + 0.018 x;
= 0 when x ≈ −389
dx
dx
The critical numbers are 0, 300, 450.
P(0) = –200, P(300) = 2410, P(450) = 4172.5
Monthly profit is maximized at x = 450,
P(450) = 4172.50
Thus a ≤
ab ≤
( a + b) 2
for all b > 0 or
4b
( a + b) 2
which leads to
4
ab ≤
3
c.
1⎛ a+b+c⎞
( a + b + c )3
Let F (b) = ⎜
⎟ =
b⎝
3
27b
⎠
F ′(b) =
=
=
3(a + b + c)2 (27b) – 27(a + b + c)3
27 2 b 2
(a + b + c) 2 [3b – (a + b + c)]
27b 2
(a + b + c)2 (2b – a – c)
27b 2
F ′(b) = 0 when b =
;
a+c
.
2
2 ⎛a+c a+c⎞
⎛a+c⎞
F⎜
⋅⎜
+
⎟=
⎟
6 ⎠
⎝ 2 ⎠ a+c ⎝ 3
3
=
Section 3.4
3
2
3
2
a2 1
b2 ⎛ a b ⎞
⎛a–b⎞
– ab +
=⎜ – ⎟ =⎜
⎟
4 2
4 ⎝ 2 2⎠
⎝ 2 ⎠
Since a square can never be negative, this is
always true.
190
2
⎛a+c⎞
From (b), ac ≤ ⎜
⎟ , thus
⎝ 2 ⎠
a2 1
b2
=
+ ab +
4 2
4
This is true if
2
3
2 ⎛ 3(a + c) ⎞
2 ⎛a+c⎞
⎛a+c⎞
⎜
⎟ =
⎜
⎟ =⎜
⎟
a+c⎝ 6 ⎠
a+c⎝ 2 ⎠
⎝ 2 ⎠
2
a 2 + 2ab + b 2
⎛ a+b⎞
ab ≤ ⎜
⎟ =
4
⎝ 2 ⎠
0≤
3
1⎛ a+b+c⎞
⎛a+c⎞
Thus ⎜
⎟ ≤ ⎜
⎟ for all b > 0.
b⎝
3
⎝ 2 ⎠
⎠
2
69. a.
a+b
.
2
1⎛ a+b+c⎞
⎛ a+b+c⎞
ac ≤ ⎜
⎟ or abc ≤ ⎜
⎟
b⎝
3
3
⎠
⎝
⎠
which gives the desired result
a+b+c
(abc)1/ 3 ≤
.
3
3
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70. Let a = lw, b = lh, and c = hw, then
S = 2(a + b + c) while V 2 = abc. By problem
a+b+c
69(c), (abc)1/ 3 ≤
so
3
2(a + b + c) S
(V 2 )1/ 3 ≤
= .
2⋅3
6
In problem 1c, the minimum occurs, hence
a+c
equality holds, when b =
. In the result used
2
from Problem 69(b), equality holds when c = a,
a+a
= a, so a = b = c. For the boxes,
thus b =
2
this means l = w = h, so the box is a cube.
3.5 Concepts Review
1. f(x); –f(x)
Critical points: –
1
2
,
f ′( x) > 0 when x < –
1
2
1
2
or x >
1
2
1 ⎤ ⎡ 1
⎛
⎞
, ∞ ⎟ and
f(x) is increasing on ⎜ – ∞, –
⎥∪⎢
2⎦ ⎣ 2
⎝
⎠
⎡ 1 1 ⎤
,
decreasing on ⎢ –
⎥.
⎣ 2 2⎦
⎛ 1 ⎞
Local minimum f ⎜
⎟ = – 2 –10 ≈ –11.4
⎝ 2⎠
⎛ 1 ⎞
Local maximum f ⎜ –
⎟ = 2 –10 ≈ –8.6
2⎠
⎝
f ′′( x) = 12 x; f ′′( x) > 0 when x > 0. f(x) is concave
up on (0, ∞ ) and concave down on
(– ∞ , 0); inflection point (0, –10).
2. decreasing; concave up
3. x = –1, x = 2, x = 3; y = 1
4. polynomial; rational.
Problem Set 3.5
1. Domain: (−∞, ∞) ; range: (−∞, ∞)
Neither an even nor an odd function.
y-intercept: 5; x-intercept: ≈ –2.3
f ′( x) = 3x 2 – 3; 3 x 2 – 3 = 0 when x = –1, 1
Critical points: –1, 1
f ′( x) > 0 when x < –1 or x > 1
f(x) is increasing on (– ∞ , –1] ∪ [1, ∞ ) and
decreasing on [–1, 1].
Local minimum f(1) = 3;
local maximum f(–1) = 7
f ′′( x) = 6 x; f ′′( x) > 0 when x > 0.
f(x) is concave up on (0, ∞ ) and concave down
on (– ∞ , 0); inflection point (0, 5).
3. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ )
Neither an even nor an odd function.
y-intercept: 3; x-intercepts: ≈ –2.0, 0.2, 3.2
f ′( x) = 6 x 2 – 6 x –12 = 6( x – 2)( x + 1);
f ′( x) = 0 when x = –1, 2
Critical points: –1, 2
f ′( x) > 0 when x < –1 or x > 2
f(x) is increasing on (– ∞ , –1] ∪ [2, ∞ ) and
decreasing on [–1, 2].
Local minimum f(2) = –17;
local maximum f(–1) = 10
1
f ′′( x) = 12 x − 6 = 6(2 x − 1); f ′′( x) > 0 when x > .
2
⎛1 ⎞
f(x) is concave up on ⎜ , ∞ ⎟ and concave down
⎝2 ⎠
1
⎛
⎞
⎛1 7⎞
on ⎜ – ∞, ⎟ ; inflection point: ⎜ , – ⎟
2⎠
⎝
⎝2 2⎠
2. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ )
Neither an even nor an odd function.
y-intercept: –10; x-intercept: 2
f ′( x) = 6 x 2 – 3 = 3(2 x 2 –1); 2 x 2 –1 = 0 when
x=–
1
2
,
1
2
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4. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ )
Neither an even nor an odd function
y-intercept: –1; x-intercept: 1
f ′( x) = 3( x –1)2 ; f ′( x) = 0 when x = 1
Critical point: 1
f ′( x) > 0 for all x ≠ 1
f(x) is increasing on (– ∞ , ∞ )
No local minima or maxima
f ′′( x) = 6( x –1); f ′′( x) > 0 when x > 1.
f(x) is concave up on (1, ∞ ) and concave down
on (– ∞ , 1); inflection point (1, 0)
H ′(t ) > 0 for –
1
2
< t < 0 or
1
2
< t.
⎡ 1
⎤ ⎡ 1
⎞
H(t) is increasing on ⎢ –
, 0⎥ ∪ ⎢
, ∞ ⎟ and
2
2
⎣
⎦ ⎣
⎠
1 ⎤ ⎡ 1 ⎤
⎛
decreasing on ⎜ – ∞, –
⎥ ∪ ⎢ 0,
⎥
2⎦ ⎣
2⎦
⎝
1 ⎛ 1 ⎞
1
⎛ 1 ⎞
Global minima f ⎜ –
⎟=– , f ⎜
⎟=– ;
4 ⎝ 2⎠
4
2⎠
⎝
Local maximum f(0) = 0
H ′′(t ) = 12t 2 – 2 = 2(6t 2 –1); H ′′ > 0 when
t<–
1
6
or t >
1
6
⎛
H(t) is concave up on ⎜ – ∞, –
⎝
⎛ 1
and concave down on ⎜ –
,
6
⎝
5. Domain: (– ∞ , ∞ ); range: [0, ∞ )
Neither an even nor an odd function.
y-intercept: 1; x-intercept: 1
G ′( x ) = 4( x – 1)3 ; G ′( x) = 0 when x = 1
Critical point: 1
G ′( x) > 0 for x > 1
G(x) is increasing on [1, ∞ ) and decreasing on
(– ∞ , 1].
Global minimum f(1) = 0; no local maxima
G ′′( x) = 12( x –1) 2 ; G ′′( x) > 0 for all x ≠ 1
G(x) is concave up on (– ∞ , 1) ∪ (1, ∞ ); no
inflection points
⎡ 1 ⎞
6. Domain: (– ∞ , ∞ ); range: ⎢ – , ∞ ⎟
⎣ 4 ⎠
⎞
1 ⎞ ⎛ 1
, ∞⎟
⎟∪⎜
6⎠ ⎝ 6
⎠
1 ⎞
⎟ ; inflection
6⎠
⎛ 1
5 ⎞
⎛ 1 5 ⎞
points H ⎜ –
, − ⎟ and H ⎜
, ⎟
36
6
⎝
⎠
⎝ 6 36 ⎠
7.
Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ )
Neither an even nor an odd function.
y-intercept: 10; x-intercept: 1 –111/ 3 ≈ –1.2
f ′( x) = 3 x 2 – 6 x + 3 = 3( x –1) 2 ; f ′( x) = 0 when
x = 1.
Critical point: 1
f ′( x) > 0 for all x ≠ 1.
f(x) is increasing on (– ∞ , ∞ ) and decreasing
nowhere.
No local maxima or minima
f ′′( x) = 6 x – 6 = 6( x –1); f ′′( x) > 0 when x > 1.
f(x) is concave up on (1, ∞ ) and concave down
on (– ∞ , 1); inflection point (1, 11)
H (–t ) = (– t )2 [(– t )2 – 1] = t 2 (t 2 – 1) = H (t ); even
function; symmetric with respect to the y-axis.
y-intercept: 0; t-intercepts: –1, 0, 1
H ′(t ) = 4t 3 – 2t = 2t (2t 2 – 1); H ′(t ) = 0 when
t=–
1
2
, 0,
1
2
Critical points: –
192
Section 3.5
1
2
, 0,
1
2
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8.
⎡ 16 ⎞
Domain: (– ∞ , ∞ ); range: ⎢ – , ∞ ⎟
⎣ 3
⎠
4(– s )4 – 8(– s ) 2 –12 4s 4 – 8s 2 –12
=
3
3
= F ( s ); even function; symmetric with respect
to the y-axis
y-intercept: –4; s-intercepts: – 3, 3
16
16
16
F ′( s ) = s3 – s = s ( s 2 –1); F ′( s ) = 0
3
3
3
when s = –1, 0, 1.
Critical points: –1, 0, 1
F ′( s ) > 0 when –1 < x < 0 or x > 1.
F(s) is increasing on [–1, 0] ∪ [1, ∞ ) and
decreasing on (– ∞ , –1] ∪ [0, 1]
16
16
Global minima F (−1) = − , F (1) = − ; local
3
3
maximum F(0) = –4
16
1⎞
⎛
F ′′( s ) = 16 s 2 − = 16 ⎜ s 2 − ⎟ ; F ′′( s ) > 0
3
3⎠
⎝
1
1
when s < –
or s >
3
3
1 ⎞ ⎛ 1
⎛
⎞
F(s) is concave up on ⎜ −∞, −
, ∞⎟
⎟∪⎜
3⎠ ⎝ 3 ⎠
⎝
⎛ 1 1 ⎞
,
and concave down on ⎜ –
⎟;
3 3⎠
⎝
inflection points
128 ⎞ ⎛ 1
128 ⎞
⎛ 1
F⎜–
,−
,−
⎟, F ⎜
⎟
27
27 ⎠
3
3
⎝
⎠ ⎝
F (– s ) =
down on (–1, ∞ ); no inflection points (–1 is not
in the domain of g).
x
1
lim
= lim
= 1;
x →∞ x + 1 x →∞ 1 + 1
x
x
1
= lim
= 1;
x→ – ∞ x + 1
x→ – ∞ 1 + 1
lim
x
horizontal asymptote: y = 1
As x → –1– , x + 1 → 0 – so
as x → –1+ , x + 1 → 0+ so
lim
x
= ∞;
x +1
lim
x
= – ∞;
x +1
x → –1–
x → –1+
vertical asymptote: x = –1
10. Domain: (– ∞ , 0) ∪ (0, ∞ );
range: (– ∞ , –4 π ] ∪ [0, ∞ )
Neither an even nor an odd function
No y-intercept; s-intercept: π
g ′( s ) =
s 2 – π2
s2
; g ′( s ) = 0 when s = – π , π
Critical points: −π , π
g ′( s ) > 0 when s < – π or s > π
g(s) is increasing on (−∞, −π ] ∪ [π , ∞) and
decreasing on [– π , 0) ∪ (0, π ].
Local minimum g( π ) = 0;
local maximum g(– π ) = –4 π
9. Domain: (– ∞ , –1) ∪ (–1, ∞ );
range: (– ∞ , 1) ∪ (1, ∞ )
Neither an even nor an odd function
y-intercept: 0; x-intercept: 0
1
g ′( x) =
; g ′( x) is never 0.
( x + 1) 2
No critical points
g ′( x) > 0 for all x ≠ –1.
g(x) is increasing on (– ∞ , –1) ∪ (–1, ∞ ).
No local minima or maxima
2
g ′′( x) = –
; g ′′( x) > 0 when x < –1.
( x + 1)3
g(x) is concave up on (– ∞ , –1) and concave
Instructor’s Resource Manual
g ′′( s) =
2π2
s3
; g ′′( s ) > 0 when s > 0
g(s) is concave up on (0, ∞ ) and concave down
on (– ∞ , 0); no inflection points (0 is not in the
domain of g(s)).
g ( s ) = s – 2π +
π2
; y = s – 2π is an oblique
s
asymptote.
As s → 0 – , ( s – π) 2 → π2 , so lim g ( s ) = – ∞;
s →0 –
Section 3.5
193
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as s → 0+ , ( s – π)2 → π2 , so lim g ( s ) = ∞;
No vertical asymptotes
s →0 +
s = 0 is a vertical asymptote.
⎡ 1 1⎤
11. Domain: (– ∞ , ∞ ); range: ⎢ – , ⎥
⎣ 4 4⎦
–x
x
f (– x) =
=–
= – f ( x); odd
x2 + 4
(– x) 2 + 4
function; symmetric with respect to the origin.
y-intercept: 0; x-intercept: 0
4 – x2
f ′( x) =
; f ′( x) = 0 when x = –2, 2
( x 2 + 4)2
Critical points: –2, 2
f ′( x) > 0 for –2 < x < 2
f(x) is increasing on [–2, 2] and
decreasing on (– ∞ , –2] ∪ [2, ∞ ).
1
Global minimum f (–2) = – ; global maximum
4
1
f (2) =
4
2 x( x 2 – 12)
f ′′( x) =
; f ′′( x) > 0 when
( x 2 + 4)3
–2 3 < x < 0 or x > 2 3
f(x) is concave up on (–2 3, 0) ∪ (2 3, ∞) and
concave down on (– ∞, – 2 3) ∪ (0, 2 3);
⎛
3⎞
inflection points ⎜⎜ −2 3, −
⎟ , (0, 0) ,
8 ⎟⎠
⎝
⎛
3⎞
⎜⎜ 2 3,
⎟
8 ⎟⎠
⎝
lim
x →∞ x 2
lim
x
+4
x→ – ∞ x 2
= lim
x
+4
1
x
x →∞ 1 + 4
x2
1
x
= lim
= 0;
x→ – ∞ 1 + 4
x2
12. Domain: (– ∞ , ∞ ); range: [0, 1)
(–θ ) 2
θ2
Λ (–θ ) =
=
= Λ (θ ); even
(–θ )2 + 1 θ 2 + 1
function; symmetric with respect to the y-axis.
y-intercept: 0; θ -intercept: 0
2θ
Λ ′(θ ) =
; Λ ′(θ ) = 0 when θ = 0
2
(θ + 1)2
Critical point: 0
Λ ′(θ ) > 0 when θ > 0
Λ(θ) is increasing on [0, ∞ ) and
decreasing on (– ∞ , 0].
Global minimum Λ(0) = 0; no local maxima
2(1 – 3θ 2 )
Λ ′′(θ ) =
; Λ ′′(θ ) > 0 when
(θ 2 + 1)3
1
1
<θ <
–
3
3
⎛ 1 1 ⎞
Λ(θ) is concave up on ⎜ –
,
⎟ and
3 3⎠
⎝
1 ⎞ ⎛ 1
⎛
⎞
concave down on ⎜ – ∞, –
, ∞ ⎟;
⎟∪⎜
3⎠ ⎝ 3 ⎠
⎝
⎛ 1 1⎞ ⎛ 1 1⎞
inflection points ⎜ –
, ⎟, ⎜
, ⎟
3 4⎠ ⎝ 3 4⎠
⎝
lim
θ2
θ →∞ θ 2 + 1
θ2
= lim
1
= 1;
θ →∞ 1 + 1
θ2
1
= 1;
θ →–∞ θ + 1 θ →–∞ 1 + 1
2
lim
2
= lim
θ
y = 1 is a horizontal asymptote. No vertical
asymptotes
= 0;
y = 0 is a horizontal asymptote.
194
Section 3.5
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13. Domain: (– ∞ , 1) ∪ (1, ∞ );
range (– ∞ , 1) ∪ (1, ∞ )
Neither an even nor an odd function
y-intercept: 0; x-intercept: 0
1
; h′( x) is never 0.
h( x ) = −
( x − 1)2
No critical points
h′( x) < 0 for all x ≠ 1.
h(x) is increasing nowhere and
decreasing on (– ∞ , 1) ∪ (1, ∞ ).
No local maxima or minima
2
; h′′( x) > 0 when x > 1
h′′( x) =
( x – 1)3
(−∞, 0] and decreasing on [0, ∞) . Global
maximum P (0) = 1 ; no local minima.
P ''( x) =
6x2 − 2
( x 2 + 1)3
P ''( x) > 0 on (−∞, −1/ 3) ∪ (1/ 3, ∞) (concave
up) and P ''( x) < 0 on (−1/ 3,1/ 3) (concave
down).
⎛ 1 3⎞
, ⎟
Inflection points: ⎜ ±
3 4⎠
⎝
No vertical asymptotes.
lim P ( x) = 0; lim P ( x) = 0
x →∞
x →−∞
y = 0 is a horizontal asymptote.
h( x) is concave up on (1, ∞ ) and concave down
on (– ∞ , 1); no inflection points (1 is not in the
domain of h( x) )
lim
x →∞
x
1
= lim
= 1;
x – 1 x→∞ 1 – 1
x
x
1
lim
= lim
= 1;
x →−∞ x − 1 x →−∞ 1 − 1
x
y = 1 is a horizontal asymptote.
As x → 1– , x – 1 → 0 – so lim
x
= – ∞;
x –1
as x → 1+ , x – 1 → 0+ so lim
x
= ∞;
x –1
x →1–
x →1+
x = 1 is a vertical asymptote.
14. Domain: ( −∞, ∞ )
Range: ( 0,1]
Even function since
1
1
P(− x) =
=
= P( x)
2
2
(− x) + 1 x + 1
so the function is symmetric with respect to the
y-axis.
y-intercept: y = 1
x-intercept: none
−2 x
P '( x) =
; P '( x) is 0 when x = 0 .
2
( x + 1) 2
critical point: x = 0
P '( x) > 0 when x < 0 so P ( x) is increasing on
Instructor’s Resource Manual
15. Domain: (– ∞ , –1) ∪ (–1, 2) ∪ (2, ∞ );
range: (– ∞ , ∞ )
Neither an even nor an odd function
3
y-intercept: – ; x-intercepts: 1, 3
2
2
3x – 10 x + 11
; f ′( x) is never 0.
f ′( x) =
( x + 1) 2 ( x – 2) 2
No critical points
f ′( x) > 0 for all x ≠ –1, 2
f(x) is increasing on
(– ∞ , –1) ∪ (–1, 2) ∪ ( 2, ∞ ) .
No local minima or maxima
–6 x3 + 30 x 2 – 66 x + 42
; f ′′( x) > 0 when
f ′′( x) =
( x + 1)3 ( x – 2)3
x < –1 or 1 < x < 2
f(x) is concave up on (– ∞ , –1) ∪ (1, 2) and
concave down on (–1, 1) ∪ (2, ∞ );
inflection point f(1) = 0
( x – 1)( x – 3)
x2 – 4 x + 3
= lim
lim
x →∞ ( x + 1)( x – 2) x →∞ x 2 – x – 2
= lim
1 – 4x +
x →∞ 1 – 1
x
–
3
x2
2
x2
= 1;
1 – 4x +
( x – 1)( x – 3)
lim
= lim
x → – ∞ ( x + 1)( x – 2) x → – ∞ 1 – 1 –
x
3
x2
2
x2
= 1;
y =1 is a horizontal asymptote.
As x → –1– , x – 1 → –2, x – 3 → –4,
x – 2 → –3, and x + 1 → 0 – so
lim f ( x) = ∞;
x → –1–
Section 3.5
195
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as x → –1+ , x – 1 → –2, x – 3 → –4,
x – 2 → –3, and x + 1 → 0+ , so
lim f ( x) = – ∞
x → –1+
As x → 2 – , x – 1 → 1, x – 3 → –1, x + 1 → 3, and
x – 2 → 0 – , so lim f ( x) = ∞; as
x→2–
x → 2+ , x – 1 → 1, x – 3 → –1, x + 1 → 3, and
x – 2 → 0+ , so lim f ( x) = – ∞
x → 2+
x = –1 and x = 2 are vertical asymptotes.
17. Domain: ( −∞,1) ∪ (1, ∞ )
Range:
( −∞, ∞ )
Neither even nor odd function.
y-intercept: y = 6 ; x-intercept: x = −3, 2
g '( x) =
x2 − 2 x + 5
; g '( x) is never zero. No
( x − 1) 2
critical points.
g '( x) > 0 over the entire domain so the function
is always increasing. No local extrema.
−8
f ''( x) =
; f ''( x) > 0 when
( x − 1)3
x < 1 (concave up) and f ''( x) < 0 when
x > 1 (concave down); no inflection points.
No horizontal asymptote; x = 1 is a vertical
asymptote; the line y = x + 2 is an oblique (or
slant) asymptote.
16. Domain: ( −∞, 0 ) ∪ ( 0, ∞ )
Range: (−∞, −2] ∪ [2, ∞)
Odd function since
(− z )2 + 1
z2 +1
w(− z ) =
=−
= − w( z ) ; symmetric
−z
z
with respect to the origin.
y-intercept: none
x-intercept: none
1
w '( z ) = 1 −
; w '( z ) = 0 when z = ±1 .
z2
critical points: z = ±1 . w '( z ) > 0 on
(−∞, −1) ∪ (1, ∞) so the function is increasing on
(−∞, −1] ∪ [1, ∞) . The function is decreasing on
[−1, 0) ∪ (0,1) .
local minimum w(1) = 2 and local maximum
w(−1) = −2 . No global extrema.
w ''( z ) =
2
> 0 when z > 0 . Concave up on
z3
(0, ∞) and concave down on ( −∞, 0 ) .
18. Domain: (– ∞ , ∞ ); range [0, ∞ )
3
3
f (– x) = – x = x = f ( x); even function;
symmetric with respect to the y-axis.
y-intercept: 0; x-intercept: 0
2⎛ x ⎞
f ′( x) = 3 x ⎜⎜ ⎟⎟ = 3x x ; f ′( x) = 0 when x = 0
⎝ x⎠
Critical point: 0
f ′( x) > 0 when x > 0
f(x) is increasing on [0, ∞ ) and decreasing on
(– ∞ , 0].
Global minimum f(0) = 0; no local maxima
3x 2
2
f ′′( x) = 3 x +
= 6 x as x 2 = x ;
x
f ′′( x) > 0 when x ≠ 0
f(x) is concave up on (– ∞ , 0) ∪ (0, ∞ ); no
inflection points
No horizontal asymptote; x = 0 is a vertical
asymptote; the line y = z is an oblique (or slant)
asymptote.
No inflection points.
196
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19. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ )
R (– z ) = – z – z = – z z = – R ( z ); odd function;
symmetric with respect to the origin.
y-intercept: 0; z-intercept: 0
z2
2
R ′( z ) = z +
= 2 z since z 2 = z for all z;
z
R ′( z ) = 0 when z = 0
Critical point: 0
R ′( z ) > 0 when z ≠ 0
R(z) is increasing on (– ∞ , ∞ ) and decreasing
nowhere.
No local minima or maxima
2z
R′′( z ) = ; R′′( z ) > 0 when z > 0.
z
R(z) is concave up on (0, ∞ ) and concave down
on (– ∞ , 0); inflection point (0, 0).
21. Domain: (– ∞ , ∞ ); range: [0, ∞ )
Neither an even nor an odd function.
Note that for x ≤ 0, x = – x so x + x = 0, while
for x > 0, x = x so
x +x
2
if x ≤ 0
= x.
⎧⎪0
g ( x) = ⎨ 2
⎪⎩3x + 2 x if x > 0
y-intercept: 0; x-intercepts: ( − ∞, 0]
if x ≤ 0
⎧0
g ′( x) = ⎨
x
x>0
6
2
if
+
⎩
No critical points for x > 0.
g(x) is increasing on [0, ∞ ) and decreasing
nowhere.
⎧0 if x ≤ 0
g ′′( x) = ⎨
⎩6 if x > 0
g(x) is concave up on (0, ∞ ); no inflection points
20. Domain: (– ∞ , ∞ ); range: [0, ∞ )
H (– q) = (– q )2 – q = q 2 q = H (q); even
function; symmetric with respect to the y-axis.
y-intercept: 0; q-intercept: 0
q3 3q3
2
H ′(q) = 2q q +
=
= 3q q as q = q 2
q
q
for all q; H ′(q ) = 0 when q = 0
Critical point: 0
H ′(q ) > 0 when q > 0
H(q) is increasing on [0, ∞ ) and
decreasing on (– ∞ , 0].
Global minimum H(0) = 0; no local maxima
3q 2
H ′′(q) = 3 q +
= 6 q ; H ′′(q ) > 0 when
q
q ≠ 0.
H(q) is concave up on (– ∞ , 0) ∪ (0, ∞ ); no
inflection points.
Instructor’s Resource Manual
22. Domain: (– ∞ , ∞ ); range: [0, ∞ )
Neither an even nor an odd function. Note that
x –x
= – x , while for
for x < 0, x = – x so
2
x −x
x ≥ 0, x = x so
= 0.
2
⎧⎪− x3 + x 2 − 6 x if x < 0
h( x ) = ⎨
if x ≥ 0
⎪⎩0
y-intercept: 0; x-intercepts: [0, ∞ )
⎧⎪−3 x 2 + 2 x − 6 if x < 0
h′( x) = ⎨
if x ≥ 0
⎪⎩0
No critical points for x < 0
h(x) is increasing nowhere and decreasing on
(– ∞ , 0].
⎧−6 x + 2 if x < 0
h′′( x) = ⎨
if x ≥ 0
⎩0
h(x) is concave up on (– ∞ , 0); no inflection
Section 3.5
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points
23. Domain: (– ∞ , ∞ ); range: [0, 1]
f (− x) = sin(− x) = − sin x = sin x = f ( x); even
function; symmetric with respect to the y-axis.
y-intercept: 0; x-intercepts: k π where k is any
integer.
sin x
π
f ′( x) =
cos x; f ′( x) = 0 when x = + k π
sin x
2
and f ′( x) does not exist when x = k π , where k
is any integer.
π
kπ
and kπ + , where k is any
Critical points:
2
2
integer; f ′( x) > 0 when sin x and cos x are either
both positive or both negative.
π⎤
⎡
f(x) is increasing on ⎢ k π, k π + ⎥ and decreasing
2⎦
⎣
π
⎡
⎤
on ⎢ k π + , (k + 1)π ⎥ where k is any integer.
2
⎣
⎦
Global minima f(k π ) = 0; global maxima
π⎞
⎛
f ⎜ k π + ⎟ = 1, where k is any integer.
2⎠
⎝
f ′′( x) =
24. Domain: [2k π , (2k + 1) π ] where k is any
integer; range: [0, 1]
Neither an even nor an odd function
y-intercept: 0; x-intercepts: k π , where k is any
integer.
π
cos x
f ′( x) =
; f ′( x) = 0 when x = 2k π +
2
2 sin x
while f ′( x) does not exist when x = k π , k any
integer.
π
Critical points: k π, 2k π + where k is any
2
integer
π
f ′( x) > 0 when 2k π < x < 2k π +
2
π⎤
⎡
f(x) is increasing on ⎢ 2k π, 2k π + ⎥ and
2⎦
⎣
π
⎡
⎤
decreasing on ⎢ 2k π + , (2k + 1)π ⎥ , k any
2
⎣
⎦
integer.
Global minima f(k π ) = 0; global maxima
π⎞
⎛
f ⎜ 2k π + ⎟ = 1, k any integer
2⎠
⎝
f ′′( x) =
– cos 2 x – 2sin 2 x
=
–1 – sin 2 x
4sin 3 / 2 x
4sin 3 / 2 x
1 + sin 2 x
=–
;
4sin 3 / 2 x
f ′′( x) < 0 for all x.
f(x) is concave down on (2k π , (2k + 1) π );
no inflection points
cos 2 x sin 2 x
−
sin x
sin x
⎛
1 ⎞⎟ ⎛ sin x ⎞
+ sin x cos x ⎜ −
⎜
⎟ (cos x)
⎜ sin x 2 ⎟ ⎜⎝ sin x ⎟⎠
⎝
⎠
=
cos 2 x sin 2 x cos 2 x
sin 2 x
−
−
=−
= − sin x
sin x
sin x
sin x
sin x
f ′′( x) < 0 when x ≠ k π , k any integer
f(x) is never concave up and concave down on
(k π , (k + 1) π ) where k is any integer.
No inflection points
25. Domain: (−∞, ∞)
Range: [0,1]
Even function since
h(−t ) = cos 2 (−t ) = cos 2 t = h(t )
so the function is symmetric with respect to the
y-axis.
y-intercept: y = 1 ; t-intercepts: x =
π
2
+ kπ
where k is any integer.
h '(t ) = −2 cos t sin t ; h '(t ) = 0 at t =
Critical points: t =
198
Section 3.5
kπ
2
kπ
.
2
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h '(t ) > 0 when kπ +
π
< t < (k + 1)π . The
2
function is increasing on the intervals
[ kπ + (π / 2), (k + 1)π ] and decreasing on the
g '(t ) = 2
=2
intervals [ kπ , kπ + (π / 2) ] .
Global maxima h ( kπ ) = 1
=2
⎛π
⎞
Global minima h ⎜ + kπ ⎟ = 0
⎝2
⎠
h ''(t ) = 2sin 2 t − 2 cos 2 t = −2(cos 2t )
π
π⎞
⎛
h ''(t ) < 0 on ⎜ kπ − , kπ + ⎟ so h is concave
4
4⎠
⎝
π
3π ⎞
⎛
down, and h ''(t ) > 0 on ⎜ kπ + , kπ +
⎟ so h
4
4 ⎠
⎝
is concave up.
⎛ kπ π 1 ⎞
+ , ⎟
Inflection points: ⎜
⎝ 2 4 2⎠
No vertical asymptotes; no horizontal
asymptotes.
cos4 t + sin t (3) cos 2 t sin t
cos6 t
cos 2 t + 3sin 2 t
cos 4 t
1 + 2sin 2 t
>0
cos 4 t
over the entire domain. Thus the function is
π
π⎞
⎛
concave up on ⎜ kπ − , kπ + ⎟ ; no inflection
2
2⎠
⎝
points.
No horizontal asymptotes; t =
π
2
+ kπ are
vertical asymptotes.
27. Domain: ≈ (– ∞ , 0.44) ∪ (0.44, ∞ );
range: (– ∞ , ∞ )
Neither an even nor an odd function
26. Domain: all reals except t =
π
2
+ kπ
Range: [0, ∞)
y-intercepts: y = 0 ; t-intercepts: t = kπ where k
is any integer.
Even function since
g (−t ) = tan 2 (−t ) = (− tan t )2 = tan 2 t
so the function is symmetric with respect to the
y-axis.
2sin t
g '(t ) = 2sec 2 t tan t =
; g '(t ) = 0 when
cos3 t
t = kπ .
Critical points: kπ
π⎞
⎡
g ( t ) is increasing on ⎢ kπ , kπ + ⎟ and
2⎠
⎣
π
⎛
⎤
decreasing on ⎜ kπ − , kπ ⎥ .
2
⎝
⎦
Global minima g (kπ ) = 0 ; no local maxima
y-intercept: 0; x-intercepts: 0, ≈ 0.24
f ′( x) =
74.6092 x3 – 58.2013 x 2 + 7.82109 x
(7.126 x – 3.141) 2
;
f ′( x) = 0 when x = 0, ≈ 0.17, ≈ 0.61
Critical points: 0, ≈ 0.17, ≈ 0.61
f ′( x) > 0 when 0 < x < 0.17 or 0.61 < x
f(x) is increasing on ≈ [0, 0.17] ∪ [0.61, ∞ )
and decreasing on
(– ∞ , 0] ∪ [0.17, 0.44) ∪ (0.44, 0.61]
Local minima f(0) = 0, f(0.61) ≈ 0.60; local
maximum f(0.17) ≈ 0.01
f ′′( x) =
531.665 x3 – 703.043 x 2 + 309.887 x – 24.566
(7.126 x – 3.141)
3
;
f ′′( x) > 0 when x < 0.10 or x > 0.44
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f(x) is concave up on (– ∞ , 0.10) ∪ (0.44, ∞ )
30.
and concave down on (0.10, 0.44);
inflection point ≈ (0.10, 0.003)
5.235 x3 − 1.245 x 2
5.235 x 2 − 1.245 x
= lim
=∞
x →∞ 7.126 x − 3.141
x →∞ 7.126 − 3.141
lim
x
so f(x) does not have a horizontal asymptote.
As x → 0.44 – , 5.235 x3 – 1.245 x 2 → 0.20 while
7.126 x – 3.141 → 0 – , so
lim
x →0.44 –
31.
f ( x) = – ∞;
as x → 0.44+ , 5.235 x3 – 1.245 x 2 → 0.20 while
7.126 x – 3.141 → 0+ , so
lim
x →0.44+
f ( x) = ∞;
32.
x ≈ 0.44 is a vertical asymptote of f(x).
33.
28.
34.
29.
200
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35.
40. Let f ( x) = ax 2 + bx + c, then f ′( x) = 2ax + b
and f ′′( x) = 2a. An inflection point occurs
where f ′′( x) changes from positive to negative,
but 2a is either always positive or always
negative, so f(x) does not have any inflection
points.
( f ′′( x) = 0 only when a = 0, but then f(x) is not a
quadratic curve.)
36. y ′ = 5( x – 1) 4 ; y ′′ = 20( x – 1)3 ; y ′′( x) > 0
when x > 1; inflection point (1, 3)
At x = 1, y ′ = 0, so the linear approximation is a
horizontal line.
41. Let f ( x ) = ax3 + bx 2 + cx + d , then
f ′( x) = 3ax 2 + 2bx + c and f ′′( x) = 6ax + 2b. As
long as a ≠ 0 , f ′′( x) will be positive on one
side of x =
x=
3
b
and negative on the other side.
3a
b
is the only inflection point.
3a
42. Let f ( x) = ax 4 + bx3 + cx 2 + dx + c, then
f ′( x) = 4ax3 + 3bx 2 + 2cx + d and
37.
f ′′( x) = 12ax 2 + 6bx + 2c = 2(6ax 2 + 3bx + c)
Inflection points can only occur when f ′′( x)
changes sign from positive to negative and
f ′′( x) = 0. f ′′( x) has at most 2 zeros, thus f(x)
has at most 2 inflection points.
43. Since the c term is squared, the only difference
occurs when c = 0. When c = 0,
y = x2 x2 = x
38.
3
which has domain (– ∞ , ∞ )
and range [0, ∞ ). When c ≠ 0, y = x 2 x 2 – c 2
has domain (– ∞ , –|c|] ∪ [|c|, ∞ ) and
range [0, ∞ ).
39.
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The only extremum points are ± c . For c = 0 ,
there is one minimum, for c ≠ 0 there are two.
No maxima, independent of c. No inflection
points, independent of c.
44.
f ( x) =
cx
4 + (cx)
f ′( x) =
2
=
cx
; f ′( x) = 0 when x = ±
(4 + c 2 x 2 )2
unless c = 0, in which case f(x) = 0 and
f ′( x) = 0.
2
c
⎡ 2 2⎤
If c > 0, f(x) is increasing on ⎢ – , ⎥ and
⎣ c c⎦
2⎤ ⎡2 ⎞
⎛
decreasing on ⎜ – ∞, – ⎥ ∪ ⎢ , ∞ ⎟ , thus, f(x) has
c⎦ ⎣c ⎠
⎝
1
⎛ 2⎞
a global minimum at f ⎜ – ⎟ = – and a global
4
⎝ c⎠
2
1
⎛ ⎞
maximum of f ⎜ ⎟ = .
⎝c⎠ 4
2⎤ ⎡ 2 ⎞
⎛
If c < 0, f(x) is increasing on ⎜ – ∞, ⎥ ∪ ⎢ – , ∞ ⎟
c⎦ ⎣ c ⎠
⎝
⎡2 2⎤
and decreasing on ⎢ , – ⎥ . Thus, f(x) has a
⎣c c⎦
1
⎛ 2⎞
global minimum at f ⎜ – ⎟ = – and a global
4
⎝ c⎠
2
1
⎛ ⎞
maximum at f ⎜ ⎟ = .
⎝c⎠ 4
f ′′( x) =
2c3 x(c 2 x 2 – 12)
(4 + c 2 x 2 )3
points at x = 0, ±
202
Section 3.5
f ( x) =
f ′( x) =
1
2
(cx – 4)2 + cx 2
, then
2cx(7 − 2cx 2 )
[(cx 2 – 4) 2 + cx 2 ]2
;
If c > 0, f ′( x) = 0 when x = 0, ±
4 + c2 x2
c(4 – c 2 x 2 )
45.
7
.
2c
If c < 0, f ′( x) = 0 when x = 0.
Note that f(x) =
1
(a horizontal line) if c = 0.
16
If c > 0, f ′( x) > 0 when x < −
7
and
2c
7
, so f(x) is increasing on
2c
⎛
7 ⎤ ⎡
7 ⎤
⎜⎜ −∞, −
⎥ ∪ ⎢ 0,
⎥ and decreasing on
2c ⎦ ⎣
2c ⎦
⎝
⎡
⎞
7 ⎤ ⎡ 7
, 0⎥ ∪ ⎢
, ∞ ⎟⎟ . Thus, f(x) has local
⎢−
⎣ 2c ⎦ ⎣ 2 c ⎠
⎛
⎛ 7 ⎞ 4
7 ⎞ 4
maxima f ⎜⎜ −
and
⎟⎟ = , f ⎜⎜
⎟⎟ =
⎝ 2c ⎠ 15
⎝ 2c ⎠ 15
1
. If c < 0, f ′( x) > 0
local minimum f (0) =
16
when x < 0, so f(x) is increasing on (– ∞ , 0] and
decreasing on [0, ∞ ). Thus, f(x) has a local
1
. Note that f(x) > 0 and has
maximum f (0) =
16
horizontal asymptote y = 0.
0< x<
, so f(x) has inflection
2 3
, c ≠ 0
c
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46.
f ( x) =
1
2
x + 4x + c
. By the quadratic formula,
x 2 + 4 x + c = 0 when x = –2 ± 4 – c . Thus f(x)
has vertical asymptote(s) at x = –2 ± 4 – c
–2 x – 4
; f ′( x) = 0
when c ≤ 4. f ′( x) =
2
( x + 4 x + c) 2
when x = –2, unless c = 4 since then x = –2 is a
vertical asymptote.
For c ≠ 4, f ′( x) > 0 when x < –2, so f(x) is
increasing on (– ∞ , –2] and decreasing on
[–2, ∞ ) (with the asymptotes excluded). Thus
1
f(x) has a local maximum at f (–2) =
. For
c–4
2
c = 4, f ′( x) = –
so f(x) is increasing on
( x + 2)3
(– ∞ , –2) and decreasing on (–2, ∞ ).
If c < 0 :
⎡ (4k + 1)π ( 4k − 1) π ⎤
f ( x ) is decreasing on ⎢
,
⎥
2c
⎣ 2c
⎦
⎡ ( 4k − 1) π ( 4k − 3) π ⎤
f ( x ) is increasing on ⎢
,
⎥
2c
2c
⎣
⎦
f ( x ) has local minima at x =
maxima at x =
( 4k − 3 ) π
2c
( 4k − 1)
2c
π and local
where k is an integer.
If c = 0 , f ( x ) = 0 and there are no extrema.
If c > 0 :
⎡ ( 4k − 3) π ( 4k − 1) π ⎤
f ( x ) is decreasing on ⎢
,
⎥
2c
2c
⎣
⎦
⎡ ( 4k − 1) π (4k + 1)π ⎤
f ( x ) is increasing on ⎢
,
2c
2c ⎥⎦
⎣
( 4k − 1)
f ( x ) has local minima at x =
π and
2c
( 4k − 3 ) π
where k is an
local maxima at x =
2c
integer.
y
−3π −2π
π
−π
2π
3π x
c = −2
c
y
47.
f ( x ) = c + sin cx .
Since c is constant for all x and sin cx is continuous
everywhere, the function f ( x ) is continuous
everywhere.
f ' ( x ) = c ⋅ cos cx
−3π −2π
(
f ' ( x ) = 0 when cx = k + 12
where k is an integer.
f '' ( x ) = −c 2 ⋅ sin cx
((
) )
)π
( (
(
)
or x = k + 12 πc
π
−π
c
2π
3π x
c = −1
) )
k
f '' k + 12 πc = −c 2 ⋅ sin c ⋅ k + 12 πc = −c 2 ⋅ ( −1)
In general, the graph of f will resemble the graph of
y = sin x . The period will decrease as c increases
and the graph will shift up or down depending on
whether c is positive or negative.
If c = 0 , then f ( x ) = 0 .
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y
Justification:
f (1) = g (1) = 1
c=0
−3π −2π
π
−π
2π
3π x
f (− x) = g ((− x)4 ) = g ( x 4 ) = f ( x)
f is an even function; symmetric with respect to
the y-axis.
f
f
f
f
y
c
−π
'( x) = 0 for x = −1, 0,1 since f ' is continuous.
f ''( x) < 0 for x on ( x0 ,1)
2π
3π x
y
−3π −2π
'( x) < 0 for x on (−∞, −1) ∪ (−1, 0)
f ''( x) = 0 for x = −1, 0,1
f ''( x) > 0 for x on (0, x0 ) ∪ (1, ∞)
c=1
π
−π
'( x) > 0 for x on (0,1) ∪ (1, ∞)
f ''( x) = g ''( x 4 )16 x6 + g '( x)12 x 2
c
−3π −2π
'( x) = g '( x 4 )4 x3
c=2
π
2π
3π x
48. Since we have f ''' ( c ) > 0 , we know that f ' ( x )
is concave up in a neighborhood around x = c .
Since f ' ( c ) = 0 , we then know that the graph of
Where x0 is a root of f ''( x) = 0 (assume that
there is only one root on (0, 1)).
50. Suppose H ′′′(1) < 0, then H ′′( x) is decreasing in
a neighborhood around x = 1. Thus, H ′′( x) > 0
to the left of 1 and H ′′( x) < 0 to the right of 1, so
H(x) is concave up to the left of 1 and concave
down to the right of 1. Suppose H ′′′(1) > 0, then
H ′′( x) is increasing in a neighborhood around
x = 1. Thus, H ′′( x) < 0 to the left of 1 and
H ′′( x) > 0 to the right of 1, so H(x) is concave
up to the right of 1 and concave down to the left
of 1. In either case, H(x) has a point of inflection
at x = 1 and not a local max or min.
51. a.
f ' ( x ) must be positive in that neighborhood.
This means that the graph of f must be increasing
to the left of c and increasing to the right of c.
Therefore, there is a point of inflection at c.
Not possible; F ′( x) > 0 means that F(x) is
increasing. F ′′( x) > 0 means that the rate at
which F(x) is increasing never slows down.
Thus the values of F must eventually
become positive.
b. Not possible; If F(x) is concave down for all
x, then F(x) cannot always be positive.
49.
c.
204
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52. a.
53. a.
No global extrema; inflection point at (0, 0)
f ′( x) = 2 cos x – 2 cos x sin x
= 2 cos x(1 – sin x);
b.
π π
f ′( x) = 0 when x = – ,
2 2
′′
f ( x) = –2sin x – 2 cos 2 x + 2sin 2 x
= 4sin 2 x – 2sin x – 2; f ′′( x) = 0 when
1
or sin x = 1 which occur when
2
π 5π π
x= – ,– ,
6
6 2
⎛ π⎞
Global minimum f ⎜ – ⎟ = –2; global
⎝ 2⎠
⎛π⎞
maximum f ⎜ ⎟ = 2; inflection points
⎝2⎠
π
1
1
⎛
⎞
⎛ 5π ⎞
f ⎜– ⎟ = – , f ⎜– ⎟ = –
4 ⎝ 6 ⎠
4
⎝ 6⎠
sin x = –
No global maximum; global minimum at
(0, 0); no inflection points
c.
Global minimum
f(– π ) = –2 π + sin (– π ) = –2 π ≈ –6.3;
b.
global maximum
f (π ) = 2π + sin π = 2π ≈ 6.3 ;
inflection point at (0, 0)
f ′( x) = 2 cos x + 2sin x cos x
= 2 cos x(1 + sin x); f ′( x) = 0 when
d.
π π
x=– ,
2 2
f ′′( x) = –2sin x + 2 cos 2 x – 2sin 2 x
Global minimum
sin(– π)
= – π ≈ 3.1; global
f (– π) = – π –
2
sin π
= π ≈ 3.1;
maximum f (π) = π +
2
inflection point at (0, 0).
Instructor’s Resource Manual
= –4sin 2 x – 2sin x + 2; f ′′( x) = 0 when
sin x = –1 or sin x =
1
which occur when
2
π π 5π
x=– , ,
2 6 6
⎛ π⎞
Global minimum f ⎜ – ⎟ = –1; global
⎝ 2⎠
⎛π⎞
maximum f ⎜ ⎟ = 3; inflection points
⎝2⎠
π
5
⎛ ⎞
⎛ 5π ⎞ 5
f ⎜ ⎟= , f ⎜ ⎟= .
⎝6⎠ 4 ⎝ 6 ⎠ 4
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205
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f ′( x) = 2 cos 2 x + 3sin 3 x
Using the graphs, f(x) has a global minimum
at f(2.17) ≈ –1.9 and a global maximum at
f(0.97) ≈ 1.9
f ′′( x) = –4sin 2 x + 9 cos 3 x; f ′′( x) = 0 when
c.
π π
x = – , and when
2 2
x ≈ –2.469, –0.673, 0.413, 2.729.
⎛ π ⎞ ⎛π ⎞
Inflection points: ⎜ − , 0 ⎟ , ⎜ , 0 ⎟ ,
⎝ 2 ⎠ ⎝2 ⎠
≈ ( −2.469, 0.542 ) , ( −0.673, −0.542 ) ,
f ′( x) = –2sin 2 x + 2sin x
= –4sin x cos x + 2sin x = 2sin x(1 – 2 cos x);
π
π
, 0, , π
3
3
f ′′( x) = –4 cos 2 x + 2 cos x; f ′′( x) = 0 when
x ≈ –2.206, –0.568, 0.568, 2.206
⎛ π⎞
⎛π⎞
Global minimum f ⎜ – ⎟ = f ⎜ ⎟ = –1.5;
3
⎝
⎠
⎝3⎠
Global maximum f(– π ) = f( π ) = 3;
Inflection points: ≈ ( −2.206, 0.890 ) ,
f ′( x) = 0 when x = – π, –
( 0.413, 0.408 ) , ( 2.729, −0.408)
54.
( −0.568, −1.265 ) , ( 0.568, −1.265) ,
( 2.206, 0.890 )
d.
y
55.
5
f ′( x) = 3cos 3 x – cos x; f ′( x) = 0 when
3cos 3x = cos x which occurs when
π π
x = – , and when
2 2
x ≈ –2.7, –0.4, 0.4, 2.7
f ′′( x) = –9sin 3x + sin x which occurs when
x = – π , 0, π and when
x ≈ –2.126, –1.016, 1.016, 2.126
⎛π⎞
Global minimum f ⎜ ⎟ = –2;
⎝2⎠
⎛ π⎞
global maximum f ⎜ – ⎟ = 2;
⎝ 2⎠
Inflection points: ≈ ( −2.126, 0.755 ) ,
( −1.016, 0.755 ) , ( 0, 0 ) , (1.016, −0.755 ) ,
( 2.126, −0.755)
−5
5
x
−5
a.
f is increasing on the intervals ( −∞, −3]
and [ −1, 0] .
f is decreasing on the intervals [ −3, −1]
and [ 0, ∞ ) .
b.
f is concave down on the intervals
( −∞, −2 ) and ( 2, ∞ ) .
f is concave up on the intervals ( −2, 0 )
and ( 0, 2 ) .
e.
206
Section 3.5
c.
f attains a local maximum at x = −3 and
x = 0.
f attains a local minimum at x = −1 .
d.
f has a point of inflection at x = −2 and
x = 2.
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y
56.
59. a.
5
−5
5
x
f ′( x) =
; f ′( x) is never 0,
x 2 – 6 x + 40
and always positive, so f(x) is increasing for
all x. Thus, on [–1, 7], the global minimum is
f(–1) ≈ –6.9 and the global maximum if
f(7) ≈ 48.0.
2 x3 − 18 x 2 + 147 x – 240
f ′′( x) =
; f ′′( x) = 0
( x 2 – 6 x + 40)3 / 2
when x ≈ 2.02; inflection point
f(2.02) ≈ 11.4
−5
a.
f is increasing on the interval [ −1, ∞ ) .
f is decreasing on the interval ( −∞, −1]
b.
f is concave up on the intervals ( −2, 0 )
and ( 2, ∞ ) .
f is concave down on the interval ( 0, 2 ) .
c.
f does not have any local maxima.
f attains a local minimum at x = −1 .
d.
f has inflection points at x = 0 and
x = 2.
2 x 2 – 9 x + 40
b.
57.
Global minimum f(0) = 0; global maximum
f(7) ≈ 124.4; inflection point at x ≈ 2.34,
f(2.34) ≈ 48.09
c.
58.
No global minimum or maximum; no
inflection points
d.
Global minimum f(3) ≈ –0.9;
global maximum f(–1) ≈ 1.0 or f(7) ≈ 1.0;
Inflection points at x ≈ 0.05 and x ≈ 5.9,
f(0.05) ≈ 0.3, f(5.9) ≈ 0.3.
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3.6 Concepts Review
60. a.
1. continuous; (a, b); f (b) – f (a) = f ′(c)(b – a)
2.
f ′( x) = 3 x 2 –16 x + 5; f ′( x) = 0 when
1
x = , 5.
3
Global minimum f(5) = –46;
⎛1⎞
global maximum f ⎜ ⎟ ≈ 4.8
⎝3⎠
8
f ′′( x) = 6 x –16; f ′′( x) = 0 when x = ;
3
inflection point:
(
8 , −20.6
3
f ′(0) does not exist.
3. F(x) = G(x) + C
4. x 4 + C
Problem Set 3.6
1.
f ′( x) =
x
x
f (2) – f (1) 2 –1
=
=1
2 –1
1
c
= 1 for all c > 0, hence for all c in (1, 2)
c
)
b.
Global minimum when x ≈ –0.5 and
x ≈ 1.2, f(–0.5) ≈ 0, f(1.2) ≈ 0;
global maximum f(5) = 46
Inflection point: ( −0.5, 0 ) , (1.2, 0 ) ,
( 83 , 20.6)
2. The Mean Value Theorem does not apply
because g ′(0) does not exist.
c.
No global minimum or maximum;
inflection point at
x ≈ −0.26, f (−0.26) ≈ −1.7
3.
d.
f ′( x) = 2 x + 1
f (2) – f (–2) 6 – 2
=
=1
2 – (–2)
4
2c + 1 = 1 when c = 0
No global minimum, global maximum when
x ≈ 0.26, f(0.26) ≈ 4.7
Inflection points when x ≈ 0.75 and
x ≈ 3.15, f(0.75) ≈ 2.6, f(3.15) ≈ –0.88
208
Section 3.6
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4. g ′( x) = 3( x + 1)2
7.
g (1) – g (–1) 8 – 0
=
=4
1 – (–1)
2
3(c + 1)2 = 4 when c = –1 +
2
3
≈ 0.15
1
1
f ′( z ) = (3 z 2 + 1) = z 2 +
3
3
f (2) – f (–1) 2 – (–2) 4
=
=
2 – (–1)
3
3
1 4
= when c = –1, 1, but −1 is not in
3 3
(−1, 2) so c = 1 is the only solution.
c2 +
5. H ′( s ) = 2s + 3
H (1) – H (–3) 3 – (–1)
=
=1
1 – (–3)
1 – (–3)
2c + 3 = 1 when c = –1
8. The Mean Value Theorem does not apply
because F(t) is not continuous at t = 1.
9. h′( x) = –
6. F ′( x) = x 2
( )
8
8
F (2) – F (–2) 3 – – 3
4
=
=
2 – (–2)
4
3
c2 =
4
2
≈ ±1.15
when c = ±
3
3
Instructor’s Resource Manual
3
( x – 3) 2
h(2) – h(0) –2 – 0
=
= –1
2–0
2
3
–
= –1 when c = 3 ± 3,
(c – 3) 2
c = 3 – 3 ≈ 1.27 (3 + 3 is not in (0, 2).)
Section 3.6
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10. The Mean Value Theorem does not apply
because f(x) is not continuous at x = 3.
5 2/3
x
3
g (1) – g (–1) 1 – (–1)
=
=1
1 – (–1)
2
14. g ′( x) =
5 2/3
⎛3⎞
c
= 1 when c = ± ⎜ ⎟
3
⎝5⎠
11. h′(t ) =
3/ 2
≈ ±0.46
2
1/ 3
3t
h(2) – h(0) 22 / 3 − 0
=
= 2 –1/ 3
2–0
2
16
2
= 2 –1/ 3 when c =
≈ 0.59
1/ 3
27
3c
15. S ′(θ ) = cos θ
S (π) – S (– π) 0 – 0
=
=0
π – (– π)
2π
π
cos c = 0 when c = ± .
2
12. The Mean Value Theorem does not apply
because h′(0) does not exist.
16. The Mean Value Theorem does not apply
because C (θ ) is not continuous at θ = −π , 0, π .
5 2/3
x
3
g (1) – g (0) 1 – 0
=
=1
1– 0
1
13. g ′( x) =
5 2/3
⎛3⎞
c
= 1 when c = ± ⎜ ⎟
3
⎝5⎠
⎛3⎞
c=⎜ ⎟
⎝5⎠
210
3/ 2
3/ 2
,
⎛ ⎛ 3 ⎞3 / 2
⎞
is not in (0, 1). ⎟
≈ 0.46, ⎜ – ⎜ ⎟
⎜ ⎝5⎠
⎟
⎝
⎠
Section 3.6
17. The Mean Value Theorem does not apply
π
because T (θ ) is not continuous at θ = .
2
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18. The Mean Value Theorem does not apply
because f(x) is not continuous at x = 0.
22. By the Mean Value Theorem
f (b) − f (a)
= f ′(c) for some c in (a, b).
b−a
0
Since f(b) = f(a),
= f ′(c); f ′(c) = 0 .
b−a
23.
19.
f ′( x) = 1 –
1
x2
5
f (2) – f (1) 2 – 2 1
=
=
2 –1
1
2
1 1
1–
= when c = ± 2, c = 2 ≈ 1.41
c2 2
(c = – 2 is not in (1, 2).)
24.
f (8) − f (0)
1
=−
8−0
4
There are three values for c such that
1
f ′(c) = − .
4
They are approximately 1.5, 3.75, and 7.
f ′( x) = 2α x + β
f (b) – f (a )
1
[α (b 2 – a 2 ) + β (b – a )]
=
b–a
b–a
= α ( a + b) + β
2α c + β = α (a + b) + β when c =
a+b
which is
2
the midpoint of [a, b].
20. The Mean Value Theorem does not apply
because f(x) is not continuous at x = 2.
25. By the Monotonicity Theorem, f is increasing on
the intervals (a, x0 ) and ( x0 , b) .
To show that f ( x0 ) > f ( x) for x in (a, x0 ) ,
consider f on the interval (a, x0 ] .
f satisfies the conditions of the Mean Value
Theorem on the interval [ x, x0 ] for x in (a, x0 ) .
So for some c in ( x, x0 ),
f ( x0 ) − f ( x) = f ′(c)( x0 − x) .
Because
f ′(c) > 0 and x0 − x > 0, f ( x0 ) − f ( x ) > 0,
so f ( x0 ) > f ( x ) .
Similar reasoning shows that
f ( x) > f ( x0 ) for x in ( x0 , b) .
Therefore, f is increasing on (a, b).
26. a.
21. The Mean Value Theorem does not apply
because f is not differentiable at x = 0 .
f ′( x) = 3 x 2 > 0 except at x = 0 in (– ∞ , ∞ ).
f ( x) = x3 is increasing on (– ∞ , ∞ ) by
Problem 25.
b.
f ′( x) = 5 x 4 > 0 except at x = 0 in (– ∞ , ∞ ).
f ( x) = x5 is increasing on (– ∞ , ∞ ) by
Problem 25.
c.
⎪⎧3 x 2 x ≤ 0
> 0 except at x = 0 in
f ′( x) = ⎨
x>0
⎪⎩1
(– ∞ , ∞ ).
⎧⎪ x3 x ≤ 0
is increasing on
f ( x) = ⎨
x>0
⎪⎩ x
(– ∞ , ∞ ) by Problem 25.
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27. s(t) is defined in any interval not containing t = 0.
1
s ′(c) = –
< 0 for all c ≠ 0. For any a, b with
c2
a 0 while s ′(c) < 0, hence
s(b) – s(a) < 0, or s(b) < s(a).
Thus, s(t) is decreasing on any interval not
containing t = 0.
28. s ′(c) = –
2
< 0 for all c > 0. If 0 < a < b, the
c3
Mean Value Theorem says
s (b) – s (a) = s ′(c)(b – a ) for some c in (a, b).
Since a < b, b – a > 0 while s ′(c) < 0, hence
s(b) – s(a) < 0, or s(b) < s(a). Thus, s(t) is
decreasing on any interval to the right of the
origin.
29. F ′( x) = 0 and G ( x ) = 0; G ′( x) = 0 .
By Theorem B,
F(x) = G(x) + C, so F(x) = 0 + C = C.
30. F ( x ) = cos 2 x + sin 2 x; F (0) = 12 + 02 = 1
F ′( x) = 2 cos x (− sin x) + 2sin x(cos x) = 0
By Problem 29, F(x) = C for all x.
Since F(0) = 1, C = 1, so sin 2 x + cos 2 x = 1 for
all x.
31. Let G ( x) = Dx; F ′( x) = D and G ′( x) = D .
By Theorem B, F(x) = G(x) + C; F(x) = Dx + C.
32. F ′( x) = 5; F (0) = 4
F(x) = 5x + C by Problem 31.
F(0) = 4 so C = 4.
F(x) = 5x + 4
33. Since f(a) and f(b) have opposite signs, 0 is
between f(a) and f(b). f(x) is continuous on [a, b],
since it has a derivative. Thus, by the
Intermediate Value Theorem, there is at least one
point c,
a < c < b with f(c) = 0.
Suppose there are two points, c and c ′, c < c ′ in
(a, b) with f (c) = f (c′) = 0. Then by Rolle’s
Theorem, there is at least one number d in (c, c ′)
with f ′(d ) = 0. This contradicts the given
information that f ′( x) ≠ 0 for all x in [a, b], thus
there cannot be more than one x in [a, b] where
f(x) = 0.
212
Section 3.6
34.
f ′( x) = 6 x 2 – 18 x = 6 x( x – 3); f ′( x) = 0 when
x = 0 or x = 3.
f(–1) = –10, f(0) = 1 so, by Problem 33, f(x) = 0
has exactly one solution on (–1, 0).
f(0) = 1, f(1) = –6 so, by Problem 33, f(x) = 0 has
exactly one solution on (0, 1).
f(4) = –15, f(5) = 26 so, by Problem 33, f(x) = 0
has exactly one solution on (4, 5).
35. Suppose there is more than one zero between
successive distinct zeros of f ′ . That is, there are
a and b such that f(a) = f(b) = 0 with a and b
between successive distinct zeros of f ′ . Then by
Rolle’s Theorem, there is a c between a and b
such that f ′(c) = 0 . This contradicts the
supposition that a and b lie between successive
distinct zeros.
36. Let x1 , x2 , and x3 be the three values such that
g ( x1 ) = g ( x2 ) = g ( x3 ) = 0 and
a ≤ x1 < x2 < x3 ≤ b . By applying Rolle’s
Theorem (see Problem 22) there is at least one
number x4 in ( x1 , x2 ) and one number x5 in
( x2 , x3 ) such that g ′( x4 ) = g ′( x5 ) = 0 . Then by
applying Rolle’s Theorem to g ′( x) , there is at
least one number x6 in ( x4 , x5 ) such that
g ′′( x6 ) = 0 .
37. f(x) is a polynomial function so it is continuous
on [0, 4] and f ′′( x) exists for all x on (0, 4).
f(1) = f(2) = f(3) = 0, so by Problem 36, there are
at least two values of x in [0, 4] where f ′( x) = 0
and at least one value of x in [0, 4] where
f ′′( x) = 0.
38. By applying the Mean Value Theorem and taking
the absolute value of both sides,
f ( x2 ) − f ( x1 )
= f ′(c) , for some c in ( x1 , x2 ) .
x2 − x1
Since f ′( x) ≤ M for all x in (a, b),
f ( x2 ) − f ( x1 )
≤ M ; f ( x2 ) − f ( x1 ) ≤ M x2 − x1 .
x2 − x1
39.
f ′( x) = 2 cos 2 x; f ′( x) ≤ 2
f ( x2 ) − f ( x1 )
x2 − x1
= f ′( x) ;
f ( x2 ) − f ( x1 )
x2 − x1
≤2
f ( x2 ) − f ( x1 ) ≤ 2 x2 − x1 ;
sin 2 x2 − sin 2 x1 ≤ 2 x2 − x1
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40. a.
b.
44. Let f ( x) = x so f ′( x) =
1
. Apply the Mean
2 x
Value Theorem to f on the interval [x, x + 2] for
x > 0.
1
1
for some c in
Thus x + 2 − x =
(2) =
2 c
c
1
1
1
<
<
(x, x + 2). Observe
.
x+2
c
x
1
Thus as x → ∞,
→ 0.
c
1
Therefore lim x + 2 − x = lim
=0.
x →∞
x →∞ c
(
)
45. Let f(x) = sin x. f ′( x) = cos x, so
f ′( x) = cos x ≤ 1 for all x.
41. Suppose f ′( x) ≥ 0 . Let a and b lie in the interior
of I such that b > a. By the Mean Value Theorem,
there is a point c between a and b such that
f (b) − f (a ) f (b) − f (a)
f ′(c) =
;
≥0.
b−a
b−a
Since a < b, f(b) ≥ f(a), so f is nondecreasing.
Suppose f ′( x) ≤ 0. Let a and b lie in the interior
of I such that b > a. By the Mean Value Theorem,
there is a point c between a and b such that
f (b) − f (a ) f (b) − f (a)
;
f ′(c) =
≤ 0 . Since
b−a
b−a
a < b, f(a) ≥ f(b), so f is nonincreasing.
42. [ f 2 ( x)]′ = 2 f ( x) f ′( x)
Because f(x) ≥ 0 and f ′( x) ≥ 0 on I , [ f 2 ( x)]′ ≥ 0
on I.
As a consequence of the Mean Value Theorem,
f 2 ( x2 ) − f 2 ( x1 ) ≥ 0 for all x2 > x1 on I.
Therefore f 2 is nondecreasing.
43. Let f(x) = h(x) – g(x).
f ′( x) = h′( x) − g ′( x); f ′( x) ≥ 0 for all x in
(a, b) since g ′( x) ≤ h′( x ) for all x in (a, b), so f is
nondecreasing on (a, b) by Problem 41. Thus
x1 < x2 ⇒ f ( x1 ) ≤ f ( x2 );
h( x1 ) − g ( x1 ) ≤ h( x2 ) − g ( x2 );
g ( x2 ) − g ( x1 ) ≤ h( x2 ) − h( x1 ) for all x1 and x2
in (a, b).
Instructor’s Resource Manual
By the Mean Value Theorem,
f ( x) − f ( y )
= f ′(c) for some c in (x, y).
x− y
Thus,
f ( x) − f ( y )
x− y
= f ′(c) ≤ 1;
sin x − sin y ≤ x − y .
46. Let d be the difference in distance between horse
A and horse B as a function of time t.
Then d ′ is the difference in speeds.
Let t0 and t1 and be the start and finish times of
the race.
d (t0 ) = d (t1 ) = 0
By the Mean Value Theorem,
d (t1 ) − d (t0 )
= d ′(c) for some c in (t0 , t1 ) .
t1 − t0
Therefore d ′(c) = 0 for some c in (t0 , t1 ) .
47. Let s be the difference in speeds between horse A
and horse B as function of time t.
Then s ′ is the difference in accelerations.
Let t2 be the time in Problem 46 at which the
horses had the same speeds and let t1 be the
finish time of the race.
s (t2 ) = s (t1 ) = 0
By the Mean Value Theorem,
s (t1 ) − s (t2 )
= s ′(c) for some c in (t2 , t1 ) .
t1 − t2
Therefore s ′(c) = 0 for some c in (t2 , t1 ) .
Section 3.6
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48. Suppose x > c. Then by the Mean Value
Theorem,
f ( x) − f (c ) = f ′(a )( x − c) for some a in (c, x) .
Since f is concave up, f ′′ > 0 and by the
Monotonicity Theorem f ′ is increasing.
Therefore f ′(a ) > f ′(c) and
f ( x) − f (c ) = f ′(a )( x − c) > f ′(c)( x − c)
f ( x) > f (c ) + f ′(c )( x − c), x > c
Suppose x < c. Then by the Mean Value
Theorem,
f (c) − f ( x) = f ′(a )(c − x) for some a in ( x, c) .
Since f is concave up, f ′′ > 0 , and by the
Monotonicity Theorem f ′ is increasing.
Therefore, f ′(c) > f ′(a) and
f (c) − f ( x ) = f ′(a )(c − x) < f ′(c)(c − x) .
− f ( x) < − f (c ) + f ′(c)(c − x)
f ( x) > f (c) − f ′(c)(c − x)
f ( x) > f (c ) + f ′(c )( x − c), x < c
Therefore f ( x) > f (c ) + f ′(c )( x − c), x ≠ c .
49. Fix an arbitrary x.
f ( y) − f ( x)
f ' ( x ) = lim
= 0 , since
y→x
y−x
f ( y) − f ( x)
y−x
≤M y−x .
So, f ' ≡ 0 → f = constant .
50.
1/ 3
f ( x) = x
on [0, a] or [–a, 0] where a is any
positive number. f ′(0) does not exist, but f(x)
has a vertical tangent line at x = 0.
51. Let f(t) be the distance traveled at time t.
f (2) − f (0) 112 − 0
=
= 56
2−0
2
By the Mean Value Theorem, there is a time c
such that f ′(c) = 56.
At some time during the trip, Johnny must have
gone 56 miles per hour.
214
Section 3.6
52. s is differentiable with s (0) = 0 and s (18) = 20 so
we can apply the Mean Value Theorem. There
exists a c in the interval ( 0,18 ) such that
v(c) = s '(c) =
(20 − 0)
≈ 1.11 miles per minute
(18 − 0 )
≈ 66.67 miles per hour
53. Since the car is stationary at t = 0 , and since v is
1
continuous, there exists a δ such that v(t ) <
2
for all t in the interval [0, δ ] . v(t ) is therefore
1
1 δ
and s (δ ) < δ ⋅ = . By the Mean
2
2 2
Value Theorem, there exists a c in the interval
(δ , 20) such that
less than
δ⎞
⎛
⎜ 20 − ⎟
2⎠
⎝
v(c) = s '(c) =
(20 − δ )
20 − δ
>
20 − δ
= 1 mile per minute
= 60 miles per hour
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54. Given the position function s ( t ) = at 2 + bt + c ,
the car’s instantaneous velocity is given by the
function s ' ( t ) = 2at + b .
A+ B
.
2
Thus, the car’s instantaneous velocity at the
midpoint of the interval is given by
⎛ A+ B ⎞
⎛ A+ B ⎞
s '⎜
⎟ = 2a ⎜
⎟+b
⎝ 2 ⎠
⎝ 2 ⎠
= a ( A + B) + b
The midpoint of the interval [ A, B ] is
The car’s average velocity will be its change in
position divided by the length of the interval.
That is,
2
2
s ( B ) − s ( A ) ( aB + bB + c ) − ( aA + bA + c )
=
B− A
B− A
2
2
aB − aA + bB − bA
=
B− A
=
=
(
)
a B 2 − A2 + b ( B − A )
B− A
a ( B − A )( B + A ) + b ( B − A )
B− A
= a ( B + A) + b
= a ( A + B) + b
This is the same result as the instantaneous
velocity at the midpoint.
3.7 Concepts Review
1. slowness of convergence
2. root; Intermediate Value
3. algorithms
4. fixed point
Problem Set 3.7
1. Let f ( x) = x3 + 2 x – 6.
f(1) = –3, f(2) = 6
hn
mn
f (mn )
1
0.5
1.5
0.375
2
0.25
1.25
–1.546875
n
3
0.125
1.375
–0.650391
4
0.0625
1.4375
–0.154541
5
0.03125
1.46875
0.105927
6
0.015625
1.45312
–0.0253716
7
0.0078125
1.46094
0.04001
8
0.00390625
1.45703
0.00725670
9 0.00195312
r ≈ 1.46
1.45508
–0.00907617
2. Let f ( x) = x 4 + 5 x3 + 1.
f(–1) = –3, f(0) = 1
n
hn
mn
f (mn )
1
0.5
–0.5
0.4375
2
0.25
–0.75
–0.792969
3
0.125
–0.625
–0.0681152
4
0.0625
–0.5625
0.21022
5
0.03125
–0.59375
0.0776834
6
0.015625
–0.609375
0.00647169
7
0.0078125
–0.617187
–0.0303962
8
0.00390625
–0.613281
–0.011854
9 0.00195312
r ≈ –0.61
–0.611328
–0.00266589
3. Let f ( x ) = 2 cos x − sin x .
f (1) ≈ 0.23913 ; f ( 2 ) ≈ −1.74159
n
hn
mn
f ( mn )
1
0.5
1.5
−0.856021
2
0.25
1.25
−0.318340
3
0.125
1.125
−0.039915
4
0.0625
1.0625
0.998044
5 0.03125 1.09375
0.029960
6 0.01563 1.109375 −0.004978
r ≈ 1.11
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Section 3.7
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4. Let f ( x ) = x − 2 + 2 cos x
f (1) = 1 − 2 + 2 cos (1) ≈ 0.080605
f ( 2 ) = 2 − 2 + 2 cos ( 2 ) ≈ −0.832294
n
xn
1
1
2
0.8636364
n
hn
mn
f ( mn )
3
0.8412670
1
0.5
1.5
−0.358526
4
0.8406998
5
0.8406994
2
0.25
1.25
−0.119355
3
0.125
1.125
−0.012647
4
0.0625
1.0625
0.035879
5 0.03125 1.09375
0.012065
6 0.01563 1.109375 −0.000183
r ≈ 1.11
6
r ≈ 0.84070
0.8406994
7. Let f ( x ) = x − 2 + 2 cos x .
y
5
5. Let f ( x) = x3 + 6 x 2 + 9 x + 1 = 0 .
−5
5
x
−5
f ' ( x ) = 1 − 2sin x
f ′( x) = 3x 2 + 12 x + 9
n
xn
1
4
2 3.724415
n
xn
1
0
4 3.698154
2
–0.1111111
3
–0.1205484
5 3.698154
r ≈ 3.69815
4
–0.1206148
5
r ≈ –0.12061
–0.1206148
3 3.698429
8. Let f ( x ) = 2 cos x − sin x .
y
5
6. Let f ( x) = 7 x3 + x – 5
−5
5
x
−5
f ' ( x ) = −2sin x − cos x
f ′( x) = 21x 2 + 1
n
xn
1
0.5
2 1.1946833
3 1.1069244
4 1.1071487
5 1.1071487
r ≈ 1.10715
216
Section 3.7
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9. Let f(x) = cos x – 2x.
f ′( x) = 4 x3 – 24 x 2 + 44 x – 24
Note that f(2) = 0.
f ′( x) = – sin x – 2
n
xn
1
0.5
2
0.4506267
3
0.4501836
4
r ≈ 0.45018
0.4501836
n
xn
1
0.5
2
0.575
3
0.585586
4
0.585786
n
xn
1
3.5
2
3.425
3
3.414414
4
3.414214
5
3.414214
r = 2, r ≈ 0.58579, r ≈ 3.41421
12. Let f ( x) = x 4 + 6 x3 + 2 x 2 + 24 x – 8.
10. Let f ( x ) = 2 x − sin x − 1 .
y
5
−5
5
x
−5
f ' ( x ) = 2 − cos x
f ′( x) = 4 x3 + 18 x 2 + 4 x + 24
n
xn
n
xn
1
–6.5
1
1
2
–6.3299632
3
–6.3167022
4
–6.3166248
5
–6.3166248
n
xn
1
0.5
2
0.3286290
3
0.3166694
4
0.3166248
2 0.891396
3 0.887866
4 0.887862
5 0.887862
r ≈ 0.88786
11. Let f ( x) = x 4 – 8 x3 + 22 x 2 – 24 x + 8.
5
0.3166248
r ≈ –6.31662, r ≈ 0.31662
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Section 3.7
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13. Let f ( x ) = 2 x 2 − sin x .
16. Let f ( x) = x 4 – 47 .
y
f ′( x) = 4 x3
2
−2
2
x
−2
4
f ' ( x ) = 4 x − cos x
n
1
xn
0.5
17.
n
xn
1
2.5
2
2.627
3
2.618373
4
2.618330
5
2.618330
47 ≈ 2.61833
f ( x ) = x 4 + x3 + x 2 + x is continuous on the
given interval.
2 0.481670
3 0.480947
4 0.480946
r ≈ 0.48095
From the graph of f, we see that the maximum
value of the function on the interval occurs at the
right endpoint. The minimum occurs at a
stationary point within the interval. To find
where the minimum occurs, we solve f ' ( x ) = 0
14. Let f ( x ) = 2 cot x − x .
y
2
on the interval [ −1,1] .
−2
2
x
f ' ( x ) = −2 csc 2 x − 1
xn
1
1
Using Newton’s Method to solve g ( x ) = 0 , we
get:
−2
n
f ' ( x ) = 4 x3 + 3x 2 + 2 x + 1 = g ( x )
n
xn
1
0
2
−0.5
3
−0.625
4 −0.60638
2 1.074305
5 −0.60583
3 1.076871
6 −0.60583
4 1.076874
r ≈ 1.07687
Minimum: f ( −0.60583) ≈ −0.32645
Maximum: f (1) = 4
15. Let f ( x) = x3 – 6.
f ′( x) = 3 x 2
3
218
n
xn
1
1.5
2
1.888889
3
1.819813
4
1.817125
5
1.817121
6
1.817121
6 ≈ 1.81712
Section 3.7
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18.
f ( x) =
x3 + 1
x4 + 1
n
is continuous on the given
interval.
xn
n
xn
1 4.712389
1 7.853982
2 4.479179
2 7.722391
3 4.793365
3 7.725251
4 4.493409
4 7.725252
5 4.493409
5 7.725252
Minimum: f ( 4.493409 ) ≈ −0.21723
From the graph of f, we see that the maximum
and minimum will both occur at stationary points
within the interval. The minimum appears to
occur at about x = −1.5 while the maximum
appears to occur at about x = 0.8 . To find the
stationary points, we solve f ' ( x ) = 0 .
f '( x) =
(
(x
− x2 x4 + 4 x − 3
4
)
+1
2
20.
f ( x ) = x 2 sin
x
is continuous on the given
2
interval.
) = g ( x)
Using Newton’s method to solve g ( x ) = 0 on
the interval, we use the starting values of −1.5
and 0.8 .
n
xn
n
xn
1
−1.5
1
0.8
From the graph of f, we see that the minimum
value and maximum value on the interval will
occur at stationary points within the interval. To
find these points, we need to solve f ' ( x ) = 0 on
the interval.
2 −1.680734
2 0.694908
3 −1.766642
3 0.692512
4 −1.783766
4 0.692505
x
x
x 2 cos + 4 x sin
2
2 = g ( x)
f '( x) =
2
Using Newton’s method to solve g ( x ) = 0 on
5 −1.784357
5 0.692505
the interval, we use the starting values of
6 −1.784358
13π
.
4
7 −1.784358
Maximum: f ( 0.692505 ) ≈ 1.08302
n
Minimum: f ( −1.78436 ) ≈ −0.42032
19.
Maximum: f ( 7.725252 ) ≈ 0.128375
f ( x) =
sin x
is continuous on the given interval.
x
xn
n
3π
and
2
xn
1 4.712389
1 10.210176
2 4.583037
2 10.174197
3 4.577868
3 10.173970
4 4.577859
4 10.173970
5 4.577859
Minimum: f (10.173970 ) ≈ −96.331841
From the graph of f, we see that the minimum
value and maximum value on the interval will
occur at stationary points within the interval. To
find these points, we need to solve f ' ( x ) = 0 on
Maximum: f ( 4.577859 ) ≈ 15.78121
the interval.
x cos x − sin x
= g ( x)
f '( x) =
x2
Using Newton’s method to solve g ( x ) = 0 on
the interval, we use the starting values of
5π
.
2
Instructor’s Resource Manual
3π
and
2
Section 3.7
219
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21. Graph y = x and y = 0.8 + 0.2 sin x.
b. Let
f (i ) = 20i (1 + i ) 24 − (1 + i )24 + 1
= (1 + i )24 (20i − 1) + 1 .
Then
f ′(i ) = 20(1 + i ) 24 + 480i (1 + i )23 − 24(1 + i )23
= (1 + i )23 (500i − 4), so
in +1 = in −
xn +1 = 0.8 + 0.2sin xn
Let x1 = 1.
n
f (in )
(1 + in ) 24 (20in − 1) + 1
= in −
f ′(in )
(1 + in )23 (500in − 4)
⎡ 20i 2 + 19in − 1 + (1 + in ) −23 ⎤
= in − ⎢ n
⎥.
500in − 4
⎣⎢
⎦⎥
xn
1
2
3
4
5
6
7
x ≈ 0.9643
1
0.96829
0.96478
0.96439
0.96434
0.96433
0.96433
c.
22.
n
in
1
0.012
2
0.0165297
3
0.0152651
4
0.0151323
5
0.0151308
6
0.0151308
i = 0.0151308
r = 18.157%
24. From Newton’s algorithm, xn +1 – xn = −
f ( xn )
.
f ′( xn )
lim ( xn +1 – xn ) = lim xn +1 – lim xn
xn → x
xn +1 = xn –
f ( xn )
x 1/ 3
= xn – n
1 x –2 / 3
f ′( xn )
3 n
= xn – 3 xn = –2 xn
Thus, every iteration of Newton’s Method gets
further from zero. Note that xn +1 = (–2) n +1 x0 .
Newton’s Method is based on approximating f by
its tangent line near the root. This function has a
vertical tangent at the root.
23. a.
For Tom’s car, P = 2000, R = 100, and
k = 24, thus
100 ⎡
1 ⎤
2000 =
⎢1 −
⎥ or
i ⎣⎢ (1 + i ) 24 ⎦⎥
20i = 1 −
1
(1 + i )24
, which is equivalent to
xn → x
xn → x
=x–x =0
f ( xn )
exists if f and f ′ are continuous at
lim
xn → x f ′( xn )
x and f ′( x ) ≠ 0.
f ( xn )
f (x )
=
= 0, so f ( x ) = 0.
f ′( x )
xn → x f ′( xn )
x is a solution of f(x) = 0.
Thus, lim
25. xn +1 =
n
1
xn + 1.5cos xn
2
xn
n
1
xn
5 0.914864
2 0.905227
6 0.914856
3 0.915744
4 0.914773
x ≈ 0.91486
7 0.914857
20i (1 + i ) 24 − (1 + i )24 + 1 = 0 .
220
Section 3.7
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26. xn +1 = 2 − sin x
29. a.
n
xn
n
1
2
xn
n
xn
5 1.10746
9
1.10603
2 1.09070
6 1.10543
10 1.10607
3 1.11305
4 1.10295
x ≈ 1.10606
7 1.10634
8 1.10612
11 1.10606
12 1.10606
27. xn +1 = 2.7 + xn
x ≈ 0.5
n
xn
1
1
2
1.923538
3
2.150241
4
2.202326
5
2.214120
6
2.216781
7
2.217382
8
2.217517
9
2.217548
10
2.217554
11
2.217556
12
2.217556
b.
c.
xn +1 = 2( xn – xn2 )
n
xn
1
0.7
2
0.42
3
0.4872
4
0.4996723
5
0.4999998
6
0.5
7
0.5
x = 2( x – x 2 )
2 x2 – x = 0
x(2x – 1) = 0
1
x = 0, x =
2
x ≈ 2.21756
30. a.
28. xn +1 = 3.2 + xn
n
xn
1
47
2
7.085196
3
3.207054
4
2.531216
5
2.393996
6
2.365163
7
2.359060
8
2.357766
9
2.357491
10
2.357433
11
2.357421
12
2.357418
13
2.357418
x ≈ 2.35742
Instructor’s Resource Manual
x ≈ 0.8
b.
xn +1 = 5( xn – xn2 )
n
xn
1
0.7
2
1.05
3
–0.2625
4
–1.657031
5
–22.01392
6
–2533.133
Section 3.7
221
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c.
x = 5( x – x 2 )
33. a.
1
x2 = 1 + = 2
1
1
3
x3 = 1 +
= = 1.5
1 + 11 2
5x2 – 4 x = 0
x(5x – 4) = 0
4
x = 0, x =
5
31. a.
x1 = 1
x1 = 0
x4 = 1 +
x2 = 1 = 1
x3 = 1 + 1 = 2 ≈ 1.4142136
x5 = 1 +
x4 = 1 + 1 + 1 ≈ 1.553774
1
1
1+ 1
1+ 1
x = 1+ x
x2 = 1 + x
1+
1 ± 1 + 4 ⋅1⋅1 1 ± 5
=
2
2
Taking the minus sign gives a negative
solution for x, violating the requirement that
1+ 5
x ≥ 0 . Hence, x =
≈ 1.618034 .
2
c.
(1 + 5 ) / 2 ≈ 1.618034 .
x1 = 0
x2 = 5 ≈ 2.236068
x5 = 5 + 5 + 5 + 5 ≈ 2.7880251
b.
x = 5 + x , and x must satisfy x ≥ 0
x2 = 5 + x
1
x
1 ± 1 + 4 ⋅1⋅1 1 ± 5
=
2
2
Taking the minus sign gives a negative
solution for x, violating the requirement that
1+ 5
≈ 1.618034 .
x ≥ 0 . Hence, x =
2
Let
1
x = 1+
.
1
1+
1+"
1
Then x satisfies the equation x = 1 + .
x
From part (b) we know that x must equal
(1 + 5 ) / 2 ≈ 1.618034 .
x3 = 5 + 5 ≈ 2.689994
x4 = 5 + 5 + 5 ≈ 2.7730839
1
1 + 11
x=
the equation x = 1 + x . From part (b) we
know that x must equal
32. a.
34. a.
Suppose r is a root. Then r = r –
(
f (r )
.
f ′(r )
f (r )
= 0, so f(r) = 0.
f ′(r )
Suppose f(r) = 0. Then r –
x2 − x − 5 = 0
1 ± 1 + 4 ⋅1 ⋅ 5 1 ± 21
x=
=
2
2
Taking the minus sign gives a negative
solution for x, violating the requirement that
x ≥ 0 . Hence,
8
= 1.6
5
x2 − x − 1 = 0
x=
Let x = 1 + 1 + 1 +… . Then x satisfies
x = 1+
=
x2 = x + 1
x2 − x − 1 = 0
c.
1
1+
b.
5
≈ 1.6666667
3
1
x5 = 1 + 1 + 1 + 1 ≈ 1.5980532
b.
=
so r is a root of x = x –
f (r )
= r – 0 = r,
f ′(r )
f ( x)
.
f ′( x)
)
x = 1 + 21 / 2 ≈ 2.7912878
c.
Let x = 5 + 5 + 5 + … . Then x satisfies
the equation x = 5 + x .
From part (b) we know that x must equal
(1 +
222
)
21 / 2 ≈ 2.7912878
Section 3.7
Instructor’s Resource Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
b. If we want to solve f(x) = 0 and f ′( x) ≠ 0 in
f ( x)
= 0 or
f ′( x)
[a, b], then
x= x–
f ( x)
= g ( x) .
f ′( x)
=
f ( x) f ′′( x)
[ f ′( x)]2
f (r ) f ′′(r )
and g '(r ) =
35. a.
starting with x =
f ′( x)
f ( x)
+
f ′′( x)
f ′( x) [ f ′( x)]2
g ′( x) = 1 –
[ f ′(r )]2
= 0.
The algorithm computes the root of
1
1
– a = 0 for x1 close to .
a
x
b. Let f ( x) =
⎛ π⎞
On the interval ⎜ 0, ⎟ , there is only one
⎝ 2⎠
stationary point (check graphically). We will use
Newton’s Method to find the stationary point,
π
4
≈ 0.785398 .
n
xn
π
1 4 ≈ 0.785398
2
3
0.862443
0.860335
4
0.860334
5
0.860334
x ≈ 0.860334 will maximize the area of the
rectangle in quadrant I, and subsequently the
larger rectangle as well.
y = cos x = cos ( 0.860334 ) ≈ 0.652184
The maximum area of the larger rectangle is
AL = ( 2 x ) y ≈ 2 ( 0.860334 )( 0.652184 )
1
– a.
x
≈ 1.122192 square units
x2
f ( x)
= – x + ax 2
f ′( x)
The recursion formula is
f ( xn )
xn +1 = xn –
= 2 xn – axn 2 .
′
f ( xn )
37. The rod that barely fits around the corner will
touch the outside walls as well as the inside
corner.
F
E
36. We can start by drawing a diagram:
C
θ
y
b
2
D
θ
(x, y)
− π2
0
x
cos x − x sin x = 0
Instructor’s Resource Manual
B
a
π
2
x
From symmetry, maximizing the area of the
entire rectangle is equivalent to maximizing the
area of the rectangle in quadrant I. The area of
the rectangle in quadrant I is given by
A = xy
= x cos x
To find the maximum area, we first need the
⎛ π⎞
stationary points on the interval ⎜ 0, ⎟ .
⎝ 2⎠
A ' ( x ) = cos x − x sin x
Therefore, we need to solve
A'( x) = 0
6.2 feet
f ′( x) = –
1
A
8.6 feet
As suggested in the diagram, let a and b represent
the lengths of the segments AB and BC, and let θ
denote the angles ∠DBA and ∠FCB . Consider
the two similar triangles ΔADB and ΔBFC ;
these have hypotenuses a and b respectively. A
little trigonometry applied to these angles gives
8.6
6.2
= 8.6sec θ and b =
= 6.2 csc θ
a=
cos θ
sin θ
Note that the angle θ determines the position of
the rod. The total length of the rod is then
L = a + b = 8.6sec θ + 6.2 csc θ
⎛ π⎞
The domain for θ is the open interval ⎜ 0, ⎟ .
⎝ 2⎠
Section 3.7
223
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
The derivative of L is
8.6sin 3 θ − 6.2 cos3 θ
L ' (θ ) =
sin 2 θ ⋅ cos 2 θ
Thus, L ' (θ ) = 0 provided
8.6sin 3 θ − 6.2 cos3 θ = 0
8.6sin 3 θ = 6.2 cos3 θ
sin 3 θ
6.2
8.6
6.2
tan 3 θ =
8.6
cos3 θ
=
tan θ =
3
Note that the angle θ determines the position of
the rod. The total length of the rod is then
L = a + b = 8csc θ + 8csc ( 75 − θ )
6.2
8.6
⎛ π⎞
On the interval ⎜ 0, ⎟ , there will only be one
⎝ 2⎠
solution to this equation. We will use Newton’s
6.2
= 0 starting with
method to solve tan θ − 3
8.6
θ1 =
π
4
As suggested in the diagram, let a and b represent
the lengths of the segments AB and BC, and let θ
denote the angle ∠ABD . Consider the two right
triangles ΔADB and ΔCEB ; these have
hypotenuses a and b respectively. A little
trigonometry applied to these angles gives
8
= 8csc θ and
a=
sin θ
8
b=
= 8csc ( 75 − θ )
sin ( 75 − θ )
The domain for θ is the open interval ( 0, 75 ) .
A graph of of L indicates there is only one
extremum (a minimum) on the interval.
.
The derivative of L is
n
θn
π
1 4 ≈ 0.78540
2
3
0.73373
0.73098
4
0.73097
L ' (θ ) =
(
8 sin 2 θ ⋅ cos (θ − 75 ) − cos θ ⋅ sin 2 (θ − 75 )
)
sin θ ⋅ sin (θ − 75 )
We will use Newton’s method to solve
L ' (θ ) = 0 starting with θ1 = 40 .
5
0.73097
Note that θ ≈ 0.73097 minimizes the length of
the rod that does not fit around the corner, which
in turn maximizes the length of the rod that will
fit around the corner (verify by using the Second
Derivative Test).
L ( 0.73097 ) = 8.6sec ( 0.73097 ) + 6.2 csc ( 0.73097 )
≈ 20.84
Thus, the length of the longest rod that will fit
around the corner is about 20.84 feet.
38. The rod that barely fits around the corner will
touch the outside walls as well as the inside
corner.
C
8 feet
2
n
θn
1
40
2
2 37.54338
3 37.50000
4
37.5
Note that θ = 37.5° minimizes the length of the
rod that does not fit around the corner, which in
turn maximizes the length of the rod that will fit
around the corner (verify by using the Second
Derivative Test).
L ( 37.5 ) = 8csc ( 37.5 ) + 8csc ( 75 − 37.5 )
= 16 csc ( 37.5 )
≈ 26.28
Thus, the length of the longest rod that will fit
around the corner is about 26.28 feet.
b
E
75 − θ
B
105
o
θ
a
A
224
8 feet
Section 3.7
D
Instructor’s Resource Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2x2
+ x + 42 = 0 to
25
find the value for x when the object hits the
ground. We want the value to be positive, so we
use the quadratic formula, keeping only the
positive solution.
39. We can solve the equation −
x=
−1 − 12 − 4 ( −0.08 )( 42 )
2 ( −0.08 )
= 30
3.8 Concepts Review
1. rx r −1 ;
r −1
2. r [ f ( x) ]
f ′( x); [ f ( x) ] f ′( x)
r
3. u = x 4 + 3 x 2 + 1, du = (4 x3 + 6 x)dx
∫ (x
We are interested in the global extrema for the
distance of the object from the observer. We
obtain the same extrema by considering the
squared distance
D( x) = ( x − 3)2 + (42 + x − .08 x 2 ) 2
A graph of D will help us identify a starting point
for our numeric approach.
x r +1
+ C , r ≠ −1
r +1
=
4
+ 3x 2 + 1)8 (4 x3 + 6 x)dx = ∫ u8 du
u9
( x 4 + 3 x 2 + 1)9
+C =
+C
9
9
4. c1 ∫ f ( x)dx + c2 ∫ g ( x)dx
Problem Set 3.8
From the graph, it appears that D (and thus the
distance from the observer) is maximized at
about x = 7 feet and minimized just before the
object hits the ground at about x = 28 feet.
The first derivative is given by
16 3 12 2 236
D '( x) =
x − x −
x + 78 .
625
25
25
a.
We will use Newton’s method to find the
stationary point that yields the minimum
distance, starting with x1 = 28 .
n
xn
1
28
2 28.0280
3 28.0279
4 28.0279
x ≈ 28.0279; y ≈ 7.1828
The object is closest to the observer when it
is at the point ( 28.0279, 7.1828 ) .
b. We will use Newton’s method to find the
stationary point that yields the maximum
distance, starting with x1 = 7 .
n
xn
1
7
2 6.7726
3 6.7728
4 6.7728
x ≈ 6.7728; y ≈ 45.1031
The object is closest to the observer when it
is at the point ( 6.7728, 45.1031) .
Instructor’s Resource Manual
1.
∫ 5dx = 5 x + C
2.
∫ ( x − 4)dx = ∫ xdx − 4∫ 1dx
=
x2
− 4x + C
2
3.
2
2
∫ ( x + π)dx = ∫ x dx + π∫ 1dx =
4.
∫ ( 3x
6.
7.
8.
)
+ 3 dx = 3∫ x 2 dx + 3 ∫ 1dx
3
x
+ 3 x + C = x3 + 3 x + C
3
=3
5.
2
x3
+ πx + C
3
5/ 4
∫ x dx =
x9 / 4
9
4
+C =
4 9/4
x
+C
9
⎛ x5 / 3
⎞
2/3
2/3
⎜
3
x
dx
=
3
x
dx
=
3
+
C
1⎟
∫
∫
⎜ 5
⎟
⎝ 3
⎠
9
= x5 / 3 + C
5
∫3
1
x
∫ 7x
2
dx = ∫ x −2 / 3 dx = 3 x1/ 3 + C = 33 x + C
−3 / 4
dx = 7 ∫ x −3 / 4 dx = 7(4 x1/ 4 + C1 )
= 28 x1/ 4 + C
9.
∫ (x
2
− x)dx = ∫ x 2 dx − ∫ x dx =
x3 x 2
−
+C
3
2
Section 3.8
225
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10.
∫ (3x
2
− πx)dx = 3∫ x 2 dx − π∫ x dx
17.
πx 2
=x −
+C
2
= x4 +
∫ (4 x
=
5
− x3 )dx = 4 ∫ x5 dx − ∫ x3 dx
18.
100
x101 x100
+
+C
101 100
5
8
4
( x3 + 5x2 − 3x + 3 )⎤⎦⎥ dx
= ∫ ( x5 + 5 x 4 − 3 x3 + 3 x 2 ) dx
=
dx − 2 ∫ x
−1
2 x1/ 2
1
2
= 2 2x −
226
+
4x
4
+C
Section 3.8
∫(z +
∫
=
24.
(
2z
)
2
)∫
( z 2 + 1)2
z
2
( )
2
1 + 2 ) z3
(
2
+C
z dz =
2
dz = ∫ ⎡ 1 + 2 z ⎤ dz
⎣
⎦
dz = ∫
3
z4 + 2z2 + 1
z
dz
= ∫ z 7 / 2 dz + 2∫ z 3 / 2 + ∫ z −1/ 2 dz
dx
3 x −4
+C
−4
3
x 4 x3 / 2
x 4 2 x3
+
+C =
+
+C
3
4
4
3
(
−2
⎛ 2x 3 ⎞
16. ∫ ⎜⎜
+ ⎟⎟ dx = ∫
x5 ⎠
⎝ x
=
)
+ x dx = ∫ x3 dx + ∫ x1/ 2 dx
= 1+ 2
23.
3x
2x
−
+C
−1
−2
3 1
=− +
+C
x x2
=
3
x3 x 2
+
+C
3
2
2
22.
⎛ 3
2 ⎞
−2
−3
∫ ⎜⎝ x2 − x3 ⎟⎠ dx = ∫ (3x − 2 x ) dx
= 3∫ x
x4 1
+ +C
4 x
∫(x
x6
3x4
3 x3
=
+ x5 −
+
+C
6
4
3
−3
x 4 x −1
−
+C
4
−1
21. Let u = x + 1; then du = dx.
u3
( x + 1)3
2
2
∫ ( x + 1) dx = ∫ u du = 3 + C = 3 + C
2
−2
dx = ∫ ( x3 − x −2 ) dx
20.
= ∫ x5 dx + 5∫ x 4 dx −3∫ x3 dx + 3 ∫ x 2 dx
15.
x
3
2
27 x
x
45 x
2x
+
−
+
+C
8
2
4
2
∫ ⎡⎣⎢ x
x6 − x
2
2
∫ ( x + x) dx = ∫ x dx + ∫ x dx =
3
6
3x 2
+C
2
19.
∫ (27 x + 3x − 45 x + 2 x)dx
= 27 ∫ x7 dx + 3∫ x5 dx − 45∫ x3 dx + 2 ∫ x dx
=
14.
=
+ x99 )dx = ∫ x100 dx + ∫ x99 dx
7
∫
x
= ∫ x3 dx − ∫ x −2 dx =
2 x6 x 4
−
+C
3
4
∫ (x
=
13.
dx = ∫ (4 x3 + 3 x) dx
= 4 ∫ x3 dx + 3∫ x dx
⎛ x6
⎞ ⎛ x4
⎞
= 4 ⎜ + C1 ⎟ − ⎜
+ C2 ⎟
⎜ 6
⎟ ⎜ 4
⎟
⎝
⎠ ⎝
⎠
12.
3
⎛ x3
⎞
⎛ x2
⎞
= 3 ⎜ + C1 ⎟ − π ⎜
+ C2 ⎟
⎜ 3
⎟
⎜ 2
⎟
⎝
⎠
⎝
⎠
3
11.
∫
4 x6 + 3x 4
∫
2 9/ 2 4 5/ 2
z
+ z
+ 2 z1/ 2 + C
9
5
s( s + 1)2
s
ds = ∫
s3 + 2 s 2 + s
s
ds
= ∫ s5 / 2 ds + 2∫ s3 / 2 ds + ∫ s1/ 2 ds
)
2 x −1/ 2 + 3 x −5 dx
=
25.
2s 7 / 2 4s5 / 2 2s3 / 2
+
+
+C
7
5
3
∫ (sin θ − cosθ )dθ = ∫ sin θ dθ − ∫ cosθ dθ
= − cos θ − sin θ + C
Instructor’s Resource Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
26.
∫ (t
=
2
− 2 cos t )dt = ∫ t 2 dt − 2∫ cos t dt
t3
− 2sin t + C
3
∫
27. Let g ( x) = 2 x + 1 ; then g ′( x) = 2 .
∫(
)
2 x +1
3
2 dx = ∫ [ g ( x) ] g ′( x)dx
[ g ( x)]4 + C = (
=
3
)
2 x +1
4
4
4
+C
28. Let g ( x) = πx3 + 1 ; then g ′( x) = 3πx 2 .
∫ (πx
=
3
+ 1)4 3πx 2 dx = ∫ [ g ( x) ] g ′( x) dx
4
[ g ( x)]5 + C = (πx3 + 1)5 + C
5
5
29. Let u = 5 x3 + 3x − 8 ; then du = (15 x 2 + 3) dx .
∫ (5 x
2
+ 1)(5 x3 + 3x − 8)6 dx
1
= ∫ (15 x 2 + 3)(5 x3 + 3 x − 8)6 dx
3
⎞
1
1 ⎛ u7
= ∫ u 6 du = ⎜
+ C1 ⎟
⎟
3
3 ⎜⎝ 7
⎠
=
(5 x3 + 3x − 8)7
+C
21
30. Let u = 5 x3 + 3x − 2; then du = (15 x 2 + 3)dx .
∫ (5x
2
+ 1) 5 x3 + 3 x − 2 dx
1
= ∫ (15 x 2 + 3) 5 x3 + 3 x − 2 dx
3
1 1/ 2
1⎛ 2
⎞
= ∫ u du = ⎜ u 3 / 2 + C1 ⎟
3
3⎝ 3
⎠
2
= (5 x3 + 3x − 2)3 / 2 + C
9
2
(5 x3 + 3x − 2)3 + C
=
9
31. Let u = 2t 2 − 11; then du = 4t dt .
3
2t 2 − 11 dt = ∫ (4t )(2t 2 − 11)1/ 3 dt
4
3 1/ 3
3 ⎛ 3 4/3
⎞
= ∫ u du = ⎜ u
+ C1 ⎟
4
4⎝4
⎠
9
= (2t 2 − 11) 4 / 3 + C
16
9
= 3 (2t 2 − 11) 4 + C
16
∫ 3t
32. Let u = 2 y 2 + 5; then du = 4 y dy
3
Instructor’s Resource Manual
3y
3
dy = ∫ (4 y )(2 y 2 + 5) −1/ 2 dy
4
2y + 5y
2
3 −1/ 2
3
u
du = (2u1/ 2 + C1 )
∫
4
4
3
2
2y + 5 + C
=
2
=
33. Let u = x3 + 4 ; then du = 3 x 2 dx .
1 2 3
2
3
∫ x x + 4 dx = ∫ 3 3x x + 4 dx
1
1
= ∫ u du = ∫ u1/ 2 du
3
3
1 ⎛ 2 3/ 2
⎞
= ⎜ u + C1 ⎟
3⎝ 3
⎠
3/
2
2
= x3 + 4
+C
9
(
)
34. Let u = x 4 + 2 x 2 ; then
(
)
(
)
du = 4 x3 + 4 x dx = 4 x 3 + x dx .
( x3 + x ) x4 + 2 x2 dx
1
= ∫ ⋅ 4 ( x3 + x ) x 4 + 2 x 2 dx
4
∫
1
1
u du = ∫ u1/ 2 du
∫
4
4
1 ⎛ 2 3/ 2
⎞
= ⎜ u + C1 ⎟
4⎝3
⎠
3/
2
1 4
= x + 2 x2
+C
6
=
(
)
35. Let u = 1 + cos x ; then du = − sin x dx .
4
4
∫ sin x (1 + cos x ) dx = − ∫ − sin x (1 + cos x ) dx
⎛1
⎞
= − ∫ u 4 du = − ⎜ u 5 + C1 ⎟
⎝5
⎠
1
5
= − (1 + cos x ) + C
5
36. Let u = 1 + sin 2 x ; then du = 2sin x cos x dx .
2
∫ sin x cos x 1 + sin x dx
1
⋅ 2sin x cos x 1 + sin 2 x dx
2
1
1
= ∫ u du = ∫ u1/ 2 du
2
2
1 ⎛ 2 3/ 2
⎞
= ⎜ u + C1 ⎟
2⎝3
⎠
3/
2
1
= 1 + sin 2 x
+C
3
=∫
(
)
Section 3.8
227
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
37.
38.
3 2
x + x + C1
2
⎛3
⎞
f ( x) = ∫ ⎜ x 2 + x + C1 ⎟ dx
2
⎝
⎠
1 3 1 2
= x + x + C1 x + C2
2
2
f ′( x) = ∫ (3 x + 1)dx =
f ′( x) = ∫ (−2 x + 3) dx = − x 2 + 3x + C1
44. The Quotient Rule for derivatives says
⎤ g ( x) f ′( x) − f ( x) g ′( x)
d ⎡ f ( x)
+ C⎥ =
.
⎢
dx ⎣ g ( x)
g 2 ( x)
⎦
Thus,
40.
41.
f ′( x) = 3 x 2 , g ′( x) = −(2 x + 5)−3 / 2
=−
3 7/3
x
+ C1
7
9 10 / 3
⎛3
⎞
f ( x) = ∫ ⎜ x7 / 3 + C1 ⎟ dx =
x
+ C1 x + C2
70
⎝7
⎠
228
Section 3.8
1
(2 x + 5)3 / 2
⎡
− x3
3x2 ⎤
+
⎢
∫ ⎢ (2 x + 5)3 / 2 2 x + 5 ⎥⎥ dx
⎣
⎦
= ∫ [ f ( x) g ′( x) + g ( x) f ′( x) ] dx
= f ( x) g ( x) + C = x3 (2 x + 5)−1/ 2 + C
f ′′( x) = x + x −3
43. The Product Rule for derivatives says
d
[ f ( x) g ( x) + C ] = f ( x) g ′( x) + f ′( x) g ( x) .
dx
Thus,
∫ [ f ( x) g ′( x) + f ′( x) g ( x)]dx = f ( x) g ( x) + C .
1
46. Let f ( x) = x3 , g ( x) = (2 x + 5)−1/ 2 .
f ′( x) = ∫ x 4 / 3 dx =
3
f ′( x) = 2∫ ( x + 1)1/ 3 dx = ( x + 1)4 / 3 + C1
2
⎡3
⎤
f ( x) = ∫ ⎢ ( x + 1)4 / 3 + C1 ⎥ dx
⎣2
⎦
9
= ( x + 1)7 / 3 + C1 x + C2
14
f ( x)
+C .
g ( x)
= x2 x − 1 + C
f ′( x) = ∫ ( x + x −3 )dx =
42.
g ( x)
dx =
2 x −1
⎡ x2
⎤
∫ ⎢⎢ 2 x − 1 + 2 x x − 1⎥⎥ dx
⎣
⎦
= ∫ [ f ( x ) g ′( x) + f ′( x) g ( x) ] dx = f ( x) g ( x) + C
f ′( x) = ∫ x1/ 2 dx =
x 2 x −2
−
+ C1
2
2
1
⎛1
⎞
f ( x) = ∫ ⎜ x 2 − x −2 + C1 ⎟ dx
2
⎝2
⎠
1 3 1 −1
= x + x + C1 x + C2
6
2
1 3 1
= x +
+ C1 x + C2
6
2x
2
f ′( x) = 2 x, g ′( x) =
1
3
= − x3 + x 2 + C1 x + C2
3
2
2 3/ 2
x
+ C1
3
⎛2
⎞
f ( x) = ∫ ⎜ x3 / 2 + C1 ⎟ dx
3
⎝
⎠
4 5/ 2
= x
+ C1 x + C2
15
g ( x) f ′( x) − f ( x) g ′( x)
45. Let f ( x) = x 2 , g ( x) = x − 1 .
f ( x) = ∫ (− x 2 + 3 x + C1 )dx
39.
∫
=
47.
x3
2x + 5
+C
d
∫ f ′′( x)dx = ∫ dx f ′( x)dx = f ′( x) + C
f ′( x) = x3 + 1 +
∫ f ′′( x)dx =
48.
3 x3
2 x3 + 1
5 x3 + 2
2 x3 + 1
=
5 x3 + 2
2 x3 + 1
so
+C .
⎞
d ⎛ f ( x)
+C⎟
⎜⎜
⎟
dx ⎝ g ( x)
⎠
=
=
g ( x) f ′( x) − f ( x) 12 [ g ( x)]−1/ 2 g ′( x)
g ( x)
2 g ( x) f ′( x) − f ( x) g ′( x)
2[ g ( x)]3 / 2
Thus,
2 g ( x) f '( x) − f ( x) g '( x)
∫
2 [ g ( x)]
3/ 2
=
f ( x)
g ( x)
+C
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49. The Product Rule for derivatives says that
d m
[ f ( x) g n ( x) + C ]
dx
54. a. F1 ( x) = ∫ ( x sin x)dx = sin x − x cos x + C1
F2 ( x) = ∫ (sin x − x cos x + C1 )dx
= −2 cos x − x sin x + C1 x + C2
= f m ( x )[ g n ( x)]′ + [ f m ( x)]′ g n ( x)
= f m ( x)[ng n −1 ( x) g ′( x)] + [mf m −1 ( x) f ′( x)]g n ( x)
F3 ( x) = ∫ (−2 cos x − x sin x + C1 x + C2 )dx
= f m −1 ( x) g n −1 ( x)[nf ( x) g ′( x) + mg ( x) f ′( x)] .
Thus,
m −1
n −1
∫ f ( x) g ( x)[nf ( x) g ′( x) + mg ( x) f ′( x)]dx
1
= x cos x − 3sin x + C1 x 2 + C2 x + C3
2
1
F4 ( x) = ∫ ( x cos x − 3sin x + C1 x 2 + C2 x + C3 )dx
2
1
1
= x sin x + 4 cos x + C1 x3 + C2 x 2 + C3 x + C4
6
2
= f m ( x) g n ( x) + C .
50. Let u = sin[( x 2 + 1)4 ];
Cn x16− n
n =1 (16 − n)!
16
b.
then du = cos ⎡( x 2 + 1)4 ⎤ 4( x 2 + 1)3 (2 x)dx .
⎣
⎦
F16 ( x) = x sin x + 16 cos x + ∑
du = 8 x cos ⎡ ( x 2 + 1)4 ⎤ ( x 2 + 1)3 dx
⎣
⎦
( x 2 + 1)4 ⎤ cos ⎡ ( x 2 + 1) 4 ⎤ ( x 2 + 1)3 x dx
⎦
⎣
⎦
4
⎛
⎞
1
1
1 u
= ∫ u 3 ⋅ du = ∫ u 3 du = ⎜
+ C1 ⎟
⎟
8
8
8 ⎜⎝ 4
⎠
∫ sin
3⎡
⎣
If x < 0, then x = − x and
∫
1. differential equation
2. function
3. separate variables
sin 4 ⎡ ( x 2 + 1)4 ⎤
⎣
⎦ +C
=
32
51. If x ≥ 0, then x = x and
3.9 Concepts Review
4. −32t + v0 ; − 16t 2 + v0t + s0
1 2
+C .
∫ x dx = 2 x
1 2
+C .
∫ x dx = − 2 x
Problem Set 3.9
1.
⎧1 2
if x ≥ 0
⎪⎪ 2 x + C
x dx = ⎨
⎪− 1 x 2 + C if x < 0
⎪⎩ 2
u 1 − cos u
,
=
2
2
1 − cos 2 x
1
1
2
∫ sin x dx =∫ 2 dx = 2 x − 4 sin 2 x + C .
2.
52. Using sin 2
53. Different software may produce different, but
equivalent answers. These answers were
produced by Mathematica.
a.
∫ 6sin ( 3( x − 2) ) dx = −2 cos ( 3( x − 2) ) + C
b.
∫ sin
c.
∫ (x
3⎛
2
x⎞
1
⎛x⎞ 9
⎛ x⎞
⎜ ⎟ dx = cos ⎜ ⎟ − cos ⎜ ⎟ + C
6
2
2
2
⎝ ⎠
⎝ ⎠
⎝6⎠
cos 2 x + x sin 2 x)dx =
Instructor’s Resource Manual
3.
−2 x
−x
dy
=
=
dx 2 1 − x 2
1 − x2
dy x
−x
x
+ =
+
=0
dx y
1 − x2
1 − x2
dy
=C
dx
dy
− x + y = −Cx + Cx = 0
dx
dy
= C1 cos x − C2 sin x;
dx
d2y
dx 2
d2y
= −C1 sin x − C2 cos x
+y
dx 2
= (−C1 sin x − C2 cos x) + (C1 sin x + C2 cos x) = 0
x 2 sin 2 x
+C
2
Section 3.9
229
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
4. For y = sin(x + C),
dy
= cos( x + C )
dx
7.
2
∫ y dy = ∫ x dx
⎛ dy ⎞
2
2
2
⎜ ⎟ + y = cos ( x + C ) + sin ( x + C ) = 1
dx
⎝ ⎠
dy
= 0.
For y = ±1,
dx
y2
x2
+ C1 =
+ C2
2
2
y 2 = x2 + C
2
⎛ dy ⎞
2
2
2
⎜ ⎟ + y = 0 + (±1) = 1
dx
⎝ ⎠
5.
dy
= x2 + 1
dx
y = ± x2 + C
At x = 1, y = 1:
1 = ± 1 + C ; C = 0 and the square root is
positive.
dy = ( x 2 + 1) dx
y = x 2 or y = x
∫ dy = ∫ ( x
2
+ 1) dx
3
x
y + C1 =
+ x + C2
3
8.
y=
4 = (1 + C )2 / 3 ; C = 7
y = ( x3 / 2 + 7)2 / 3
dy = ( x −3 + 2) dx
y=−
1
2
+ 2) dx
x −2
+ 2 x + C2
2
+ 2x + C
2x
At x = 1, y = 3:
1
3
3 = − + 2 + C; C =
2
2
1
3
y=−
+ 2x +
2
2 x2
230
y = ( x3 / 2 + C ) 2 / 3
At x = 1, y = 4:
dy
= x −3 + 2
dx
y + C1 = −
y dy = ∫ x dx
y 3 / 2 = x3 / 2 + C
x
1
+x−
3
3
−3
x
y
2 3/ 2
2
y
+ C1 = x3 / 2 + C2
3
3
3
∫ dy = ∫ ( x
dy
=
dx
∫
x3
y=
+ x+C
3
At x = 1, y = 1:
1
1
1 = + 1 + C; C = −
3
3
6.
dy x
=
dx y
Section 3.9
9.
dz 2 2
=t z
dt
∫z
−2
dz = ∫ t 2 dt
− z −1 + C1 =
t3
+ C2
3
1
t3
C − t3
= − + C3 =
z
3
3
3
z=
C − t3
1
At t = 1, z = :
3
1
3
=
; C − 1 = 9; C = 10
3 C −1
3
z=
10 − t 3
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
10.
dy
= y4
dt
∫y
−
−4
13.
dy = ∫ dt
1
3 y3
1
(2 x + 1)4 2 dx
∫
2
1 (2 x + 1)5
(2 x + 1)5
=
+C =
+C
2
5
10
At x = 0, y = 6:
1
59
6 = + C; C =
10
10
y = ∫ (2 x + 1)4 dx =
+ C1 = t + C2
1
y=−
3
3t + C
At t = 0, y = 1:
C = –1
1
y=−
3
3t − 1
y=
14.
ds
11.
= 16t 2 + 4t − 1
dt
∫ ds = ∫ (16t
s + C1 =
2
+ 4t − 1) dt
16 3
t + 2t 2 − t + C2
3
16 3
t + 2t 2 − t + C
3
At t = 0, s = 100:C = 100
16
s = t 3 + 2t 2 − t + 100
3
−
u
−3
2u 2
−2
+ C1 =
t4 t2
− + C2
4 2
15.
t4
= t − +C
2
2
−1/ 2
⎛
⎞
t4
u = ⎜t2 − + C ⎟
⎜
⎟
2
⎝
⎠
At t = 0, u = 4:
1
4 = C −1/ 2 ; C =
16
⎛
t4 1 ⎞
u = ⎜t2 − + ⎟
⎜
2 16 ⎟⎠
⎝
−1/ 2
Instructor’s Resource Manual
10
( x + 2)5 + C
At x = 0, y = 1:
10
1=
; C = 10 − 32 = −22
32 + C
10
y=
2
( x + 2)5 − 22
du = ∫ (t 3 − t ) dt
1
dy
= − y 2 x( x 2 + 2)4
dx
1
− ∫ y −2 dy = ∫ 2 x( x 2 + 2)4 dx
2
1
1 ( x 2 + 2)5
+ C1 =
+ C2
2
5
y
y=
du
= u 3 (t 3 − t )
dt
∫u
(2 x + 1)5 59 (2 x + 1)5 + 59
+
=
10
10
10
1 ( x 2 + 2)5 + C
=
y
10
s=
12.
dy
= (2 x + 1) 4
dx
2
dy
= 3x
dx
y = ∫ 3 x dx =
3 2
x +C
2
At (1, 2):
3
2 = +C
2
1
C=
2
3 2 1 3x 2 + 1
y= x + =
2
2
2
Section 3.9
231
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16.
dy
= 3y2
dx
∫y
−
−2
19. v = ∫ (2t + 1)1/ 3 dt =
3
= (2t + 1)4 / 3 + C1
8
3
3
v0 = 0 : 0 = + C1 ; C1 = −
8
8
3
3
v = (2t + 1) 4 / 3 −
8
8
3
3
s = ∫ (2t + 1) 4 / 3 dt − ∫ 1dt
8
8
3
3
= ∫ (2t + 1)4 3 2dt − ∫ 1dt
16
8
9
3
(2t + 1)7 3 − t + C2
=
112
8
9
1111
s0 = 10 :10 =
+ C2 ; C2 =
112
112
9
3 1111
(2t + 1)7 3 − t +
s=
112
8
112
3
3
At t = 2: v = (5) 4 3 − ≈ 2.83
8
8
9
6 1111
s=
(5)7 3 − +
≈ 12.6
112
8 112
dy = 3∫ dx
1
+ C1 = 3 x + C2
y
1
= −3x + C
y
1
C − 3x
At (1, 2):
1
2=
C −3
7
C=
2
1
2
=
y=
7 − 3x
7 − 6x
2
y=
17. v = ∫ t dt =
t2
+ v0
2
t2
+3
2
⎛ t2
⎞
t3
s = ∫ ⎜ + 3 ⎟ dt = + 3t + s0
⎜2
⎟
6
⎝
⎠
v=
20. v = ∫ (3t + 1) −3 dt =
t3
t3
+ 3t + 0 = + 3t
6
6
At t = 2:
v = 5 cm/s
22
s=
cm
3
s=
18. v = ∫ (1 + t )−4 dt = −
v0 = 0 : 0 = −
v=−
3(1 + t )3
1
3(1 + 0)
1
3(1 + t )
1
3
+
3
+C
+ C; C =
1
3
1
3
⎛
1
1⎞
1
1
s = ∫⎜−
+ ⎟ dt =
+ t+C
2 3
⎜ 3(1 + t )3 3 ⎟
6(1 + t )
⎝
⎠
1
1
59
s0 = 10 :10 =
+ (0) + C ; C =
2 3
6
6(1 + 0)
1 59
+ t+
3
6
6(1 + t )
At t = 2:
1 1 26
cm/s
v=− + =
81 3 81
1 2 59 284
cm
s=
+ +
=
54 3 6
27
s=
232
1
2
Section 3.9
1
(2t + 1)1/ 3 2dt
2∫
1
(3t + 1) −3 3dt
3∫
1
= − (3t + 1)−2 + C1
6
1
25
v0 = 4 : 4 = − + C1; C1 =
6
6
1
25
v = − (3t + 1)−2 +
6
6
1
25
s = − ∫ (3t + 1)−2 dt + ∫ dt
6
6
1
25
−2
= − ∫ (3t + 1) 3dt + ∫ dt
18
6
1
25
= (3t + 1) −1 + t + C2
18
6
1
1
s0 = 0 : 0 = + C2 ; C2 = −
18
18
1
25
1
s = (3t + 1)−1 + t −
18
6
18
1 −2 25
≈ 4.16
At t = 2: v = − (7) +
6
6
1
25 1
s = (7) −1 + − ≈ 8.29
18
3 18
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21. v = –32t + 96,
2
27. vesc = 2 gR
2
s = −16t + 96t + s0 = −16t + 96t
v = 0 at t = 3
At t = 3, s = −16(32 ) + 96(3) = 144 ft
22. a =
dv
=k
dt
v = ∫ k dt = kt + v0 =
ds
;
dt
k
k
s = ∫ (kt + v0 )dt = t 2 + v0t + s0 = t 2 + v0t
2
2
v0
v = 0 when t = − . Then
k
s=
23.
2
v2
k ⎛ v0 ⎞ ⎛ v02 ⎞
− ⎟ +⎜− ⎟ = − 0 .
⎜
2 ⎝ k ⎠ ⎜⎝ k ⎟⎠
2k
dv
= −5.28
dt
28. v0 = 60 mi/h = 88 ft/s
v = 0 = –11t + 88; t = 8 sec
11
s ( t ) = − t 2 + 88t
2
11 2
s ( 8 ) = − ( 8 ) + 88 ( 8 ) = 352 feet
2
The shortest distance in which the car can be
braked to a halt is 352 feet.
29. a =
∫ dv = −∫ 5.28dt
v=
For the Moon, vesc ≈ 2(0.165)(32)(1080 ⋅ 5280)
≈ 7760 ft/s ≈ 1.470 mi/s.
For Venus, vesc ≈ 2(0.85)(32)(3800 ⋅ 5280)
≈ 33,038 ft/s ≈ 6.257 mi/s.
For Jupiter, vesc ≈ 194,369 ft/s ≈ 36.812 mi/s.
For the Sun, vesc ≈ 2,021,752 ft/s
≈ 382.908 mi/s.
ds
= −5.28t + v0 = –5.28t + 56
dt
30. 75 =
∫ ds = ∫ (−5.28t + 56)dt
s = −2.64t 2 + 56t + s0 = −2.64t 2 + 56t + 1000
When t = 4.5, v = 32.24 ft/s and s = 1198.54 ft
24. v = 0 when t =
−56
≈ 10.6061 . Then
−5.28
s ≈ −2.64(10.6061) 2 + 56(10.6061) + 1000
≈ 1296.97 ft
25.
4 3
πr and S = 4πr 2 ,
3
dr
dr
4πr 2
= −k 4πr 2 so
= −k .
dt
dt
∫ dr = − ∫ k dt
Since V =
r = –kt + C
2 = –k(0) + C and 0.5 = –k(10) + C, so
3
3
C = 2 and k =
. Then, r = − t + 2 .
20
20
8
(3.75) 2 + v0 (3.75) + 0; v0 = 5 ft/s
2
31. For the first 10 s, a =
dv
= 6t , v = 3t 2 , and
dt
s = t 3 . So v(10) = 300 and s(10) = 1000. After
dv
10 s, a =
= −10 , v = –10(t – 10) + 300, and
dt
s = −5(t − 10)2 + 300(t − 10) + 1000. v = 0 at
t = 40, at which time s = 5500 m.
32. a.
dV
= −kS
dt
26. Solving v = –136 = –32t yields t =
dv Δv 60 − 45
=
=
= 1.5 mi/h/s = 2.2 ft/s2
dt Δt
10
After accelerating for 8 seconds, the velocity
is 8 · 3 = 24 m/s.
b. Since acceleration and deceleration are
constant, the average velocity during those
times is
24
= 12 m/s . Solve 0 = –4t + 24 to get the
2
24
time spent decelerating. t =
= 6 s;
4
d = (12)(8) + (24)(100) + (12)(6) = 2568 m.
17
.
4
2
⎛ 17 ⎞
⎛ 17 ⎞
Then s = 0 = −16 ⎜ ⎟ + (0) ⎜ ⎟ + s0 , so
⎝ 4⎠
⎝ 4⎠
s0 = 289 ft.
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Section 3.9
233
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3.10 Chapter Review
15. True:
Concepts Test
1. True:
Max-Min Existence Theorem
2. True:
Since c is an interior point and f is
differentiable ( f ′(c) exists), by the
Critical Point Theorem, c is a
stationary point ( f ′(c) = 0).
lim
x →∞ 1 – x 2
=
4. False:
increasing for all x, but f ' ( x ) does
16. True:
not exist at x = 0.
5. True:
f ′( x) = 18 x5 + 16 x3 + 4 x;
17. True:
f ′′( x) = 90 x 4 + 48 x 2 + 4 , which is
greater than zero for all x.
18. False:
6. False:
8. False:
If f ′′(c) = 0 , c is a candidate, but not
necessarily an inflection point. For
example, if f ( x) = x 4 , P ′′(0) = 0 but
x = 0 is not an inflection point.
10. True:
f ( x) = ax 2 + bx + c;
f ′( x) = 2ax + b; f ′′( x) = 2a
12. True:
The function is differentiable on
(0, 2).
f ′( x) =
23. False:
lim (2 x3 + x + tan x) = ∞ while
The rectangle will have minimum
perimeter if it is a square.
K
A = xy = K; y =
x
2 K dP
2K d 2 P 4K
= 2−
=
;
;
x dx
x 2 dx 2
x3
dP
d 2P
= 0 and
>0
dx
dx 2
when x = K , y = K .
−
lim (2 x3 + x + tan x) = −∞.
+
Instructor's Resource Manual
3 3
,
.
3 3
dy
d2y
= cos x;
= − sin x; –sin x = 0
dx
dx 2
has infinitely many solutions.
P = 2x +
At x = 3 there is a removable
discontinuity.
x
so f ′(0) does not exist.
x
For example if f ( x) = x 4 ,
f ′(0) = f ′′(0) = 0 but f has a
minimum at x = 0.
x →−∞
14. False:
3x 2 + 2 x + sin x
sin x
– (3 x + 2) =
;
x
x
sin x
sin x
lim
= 0 and lim
= 0.
x →∞ x
x→ – ∞ x
21. False:
x →∞
x →− π
2
x
–1
Let g(x) = D where D is any number.
Then g ′( x) = 0 and so, by Theorem B
of Section 3.6,
f(x) = g(x) + C = D + C, which is a
constant, for all x in (a, b).
lim (2 x3 + x) = ∞ while
x→ π
2
x→ – ∞ 1
x2
20. True:
lim (2 x3 + x) = −∞
13. True:
= lim
1 + 12
There are two points: x = −
If f(x) is increasing for all x in [a, b],
the maximum occurs at b.
tan 2 x has a minimum value of 0.
This occurs whenever x = kπ where k
is an integer.
–1
19. False:
22. True:
11. False:
x →∞ 1
x2
1
= –1.
–1
For example, f ( x) = x is increasing
on [–1, 1] but f ′(0) = 0.
When f ′( x) > 0, f ( x) is increasing.
9. True:
x2 + 1
3
7. True:
= lim
1
x2
1
= –1 and
–1
lim
For example, let f(x) = sin x.
f ( x) = x1/ 3 is continuous and
1+
x→ – ∞ 1 – x 2
=
3. True:
x2 + 1
24. True:
By the Mean Value Theorem, the
derivative must be zero between each
pair of distinct x-intercepts.
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25. True:
26. False:
If f ( x1 ) < f ( x2 ) and g ( x1 ) < g ( x2 )
for x1 < x2 ,
f ( x1 ) + g ( x1 ) < f ( x2 ) + g ( x2 ), so
f + g is increasing.
28. False:
29. True:
b 2 – 3ac < 0 there are no critical
points.)
On an open interval, no local maxima
can come from endpoints, so there can
be at most one local maximum in an
open interval.
Let f(x) = g(x) = 2x, f ′( x) > 0 and
g ′( x) > 0 for all x, but
f ( x) g ( x) = 4 x 2 is decreasing on
(– ∞ , 0).
27. True:
is an inflection point while if
Since f ′′( x) > 0, f ′( x) is increasing
for x ≥ 0. Therefore, f ′( x) > 0 for x
in [0, ∞ ), so f(x) is increasing.
If f(3) = 4, the Mean Value Theorem
requires that at some point c in [0, 3],
f (3) – f (0) 4 –1
=
= 1 which
f ′(c) =
3–0
3–0
does not contradict that f ′( x) ≤ 2 for
all x in [0, 3].
If the function is nondecreasing,
f ′( x) must be greater than or equal to
zero, and if f ′( x) ≥ 0, f is
nondecreasing. This can be seen using
the Mean Value Theorem.
34. True:
f ′( x) = a ≠ 0 so f(x) has no local
minima or maxima. On an open
interval, no local minima or maxima
can come from endpoints, so f(x) has
no local minima.
35. True:
Intermediate Value Theorem
36. False:
The Bisection Method can be very
slow to converge.
37. False:
xn +1 = xn –
38. False:
Newton’s method can fail to exist for
several reasons (e.g. if f’(x) is 0 at or
near r). It may be possible to achieve
convergence by selecting a different
starting value.
39. True:
From the Fixed-point Theorem, if g is
continuous on [ a, b ] and
f ( xn )
= –2 xn .
f ′( xn )
30. True:
However, if the constant is 0, the
functions are the same.
a ≤ g ( x ) ≤ b whenever a ≤ x ≤ b ,
31. False:
For example, let f ( x) = e .
then there is at least one fixed point
on [ a, b ] . The given conditions satisfy
x
these criteria.
lim e x = 0, so y = 0 is a horizontal
x →−∞
32. True:
33. True:
asymptote.
40. True:
If f(c) is a global maximum then f(c)
is the maximum value of f on
(a, b) ↔ S where (a, b) is any interval
containing c and S is the domain of f.
Hence, f(c) is a local maximum value.
The Bisection Method always
converges as long as the function is
continuous and the values of the
function at the endpoints are of
opposite sign.
41. True:
Theorem 3.8.C
42. True:
Obtained by integrating both sides of
the Product Rule
43. True:
(− sin x) 2 = sin 2 x = 1 − cos 2 x
44. True:
If F ( x) = ∫ f ( x) dx, f ( x) is a
f ′( x) = 3ax 2 + 2bx + c; f ′( x) = 0
–b ± b 2 – 3ac
by the
3a
Quadratic Formula. f ′′( x) = 6ax + 2b
so
⎛ – b ± b 2 – 3ac ⎞
⎟ = ±2 b 2 – 3ac .
f ′′ ⎜
⎜
⎟
3a
⎝
⎠
when x =
Thus, if b 2 – 3ac > 0, one critical
point is a local maximum and the
other is a local minimum.
derivative of F(x).
45. False:
f ( x) = x 2 + 2 x + 1 and
g ( x) = x 2 + 7 x − 5 are a counterexample.
(If b2 – 3ac = 0 the only critical point
236
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46. False:
The two sides will in general differ by
a constant term.
47. True:
At any given height, speed on the
downward trip is the negative of
speed on the upward.
5.
2.
3.
f ′( x) = 2 x – 2; 2x – 2 = 0 when x = 1.
Critical points: 0, 1, 4
f(0) = 0, f(1) = –1, f(4) = 8
Global minimum f(1) = –1;
global maximum f(4) = 8
f ′(t ) = –
1
f ′( z ) = –
;–
2
z
3
;–
6.
2
z3
f ′( s ) = 1 – 1 = 0.
Critical points: 1 and all s in [–1, 0]
f(1) = 2, f(s) = 0 for s in [–1, 0]
Global minimum f(s) = 0, –1 ≤ s ≤ 0;
global maximum f(1) = 2.
7.
8.
2
;–
1
Global minimum f (–2) = ; no global
4
maximum.
Instructor's Resource Manual
f ′(u ) =
u (7u – 12)
2/3
; f ′(u ) = 0 when u = 0,
Critical points: –1, 0,
12
7
12
, 2, 3
7
f (–1) = 3 –3 ≈ –1.44, f (0) = 0,
⎛ 12 ⎞ 144 3 2
f ⎜ ⎟=
– ≈ –1.94, f(2) = 0, f(3) = 9
7
⎝ 7 ⎠ 49
⎛ 12 ⎞
Global minimum f ⎜ ⎟ ≈ –1.94;
⎝7⎠
global maximum f(3) = 9
2
is never 0.
x
x3
Critical point: –2
1
f (–2) =
4
f ′( x) > 0 for x < 0, so f is increasing.
3
f ′( x) = 12 x3 – 12 x 2 = 12 x 2 ( x – 1); f ′( x) = 0
when x = 0, 1
Critical points: –2, 0, 1, 3
f(–2) = 80, f(0) = 0, f(1) = –1, f(3) = 135
Global minimum f(1) = –1;
global maximum f(3) = 135
3(u – 2)
f ′(2) does not exist.
1
Global minimum f (–2) = ;
4
⎛ 1⎞
global maximum f ⎜ – ⎟ = 4.
⎝ 2⎠
f ′( x) = –
s
; f ′( s ) does not exist when s = 0.
s
is never 0.
1
Critical points: –2, –
2
1 ⎛ 1⎞
f (–2) = , f ⎜ – ⎟ = 4
4 ⎝ 2⎠
4.
f ′( s) = 1 +
For s < 0, s = – s so f(s) = s – s = 0 and
1
is never 0.
t
t2
Critical points: 1, 4
1
f(1) = 1, f (4) =
4
1
Global minimum f (4) = ;
4
global maximum f(1) = 1.
2
x
; f ′( x) does not exist at x = 0.
x
1
Critical points: – , 0, 1
2
⎛ 1⎞ 1
f ⎜ – ⎟ = , f (0) = 0, f (1) = 1
⎝ 2⎠ 2
Global minimum f(0) = 0;
global maximum f(1) = 1
Sample Test Problems
1.
f ′( x) =
9.
f ′( x) = 10 x 4 – 20 x3 = 10 x3 ( x – 2);
f ′( x) = 0 when x = 0, 2
Critical points: –1, 0, 2, 3
f(–1) = 0, f(0) = 7, f(2) = –9, f(3) = 88
Global minimum f(2) = –9;
global maximum f(3) = 88
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237
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10. f ′( x) = 3( x – 1)2 ( x + 2)2 + 2( x – 1)3 ( x + 2)
14.
= ( x – 1)2 ( x + 2)(5 x + 4); f ′( x) = 0 when
f ′′( x) = 72 x 7 ; f ′′( x) < 0 when x < 0.
f(x) is increasing on (– ∞ , ∞ ) and concave down
on (– ∞ , 0).
4
x = –2, – , 1
5
4
Critical points: –2, – , 1 , 2
5
26, 244
⎛ 4⎞
≈ –8.40,
f(–2) = 0, f ⎜ – ⎟ = –
3125
⎝ 5⎠
f(1) = 0, f(2) = 16
⎛ 4⎞
Global minimum f ⎜ – ⎟ ≈ –8.40;
⎝ 5⎠
global maximum f(2) = 16
π
11. f ′(θ ) = cos θ ; f ′(θ ) = 0 when θ = in
2
⎡ π 4π ⎤
⎢4 , 3 ⎥
⎣
⎦
π π 4π
Critical points: , ,
4 2 3
⎛π⎞ 1
⎛π⎞
f ⎜ ⎟=
≈ 0.71, f ⎜ ⎟ = 1,
4
2
⎝ ⎠
⎝2⎠
15.
16.
π π 5π
in [0, π ]
, ,
6 2 6
π π 5π
Critical points: 0, , , , π
6 2 6
1 ⎛π⎞
⎛π⎞
f(0) = 0, f ⎜ ⎟ = – , f ⎜ ⎟ = 0,
4 ⎝2⎠
⎝6⎠
1
⎛ 5π ⎞
f ⎜ ⎟ = – , f( π ) = 0
6
4
⎝ ⎠
1
1
⎛π⎞
⎛ 5π ⎞
Global minimum f ⎜ ⎟ = – or f ⎜ ⎟ = – ;
4
4
⎝6⎠
⎝ 6 ⎠
⎛π⎞
global maximum f(0) = 0, f ⎜ ⎟ = 0, or
⎝2⎠
f( π ) = 0
13.
f ′( x) = −6 x 2 − 6 x + 12 = –6(x + 2)(x – 1);
f ′( x) > 0 when –2 < x < 1.
1
x>− .
2
f(x) is increasing on [–2, 1] and concave down on
⎛ 1 ⎞
⎜− , ∞⎟ .
⎝ 2 ⎠
17.
f ′( x) = 4 x3 – 20 x 4 = 4 x3 (1 – 5 x); f ′( x) > 0
when 0 < x <
1
.
5
f ′′( x) = 12 x 2 – 80 x3 = 4 x 2 (3 – 20 x); f ′′( x) < 0
when x >
3
.
20
f(x) is increasing on ⎡ 0, 1 ⎤ and concave down on
⎣ 5⎦
⎛ 3
⎞
⎜ , ∞ ⎟.
20
⎝
⎠
f ′(θ ) = 2sin θ cosθ – cosθ = cosθ (2sin θ – 1);
f ′(θ ) = 0 when θ =
f ′( x) = 3 x 2 – 3 = 3( x 2 – 1); f ′( x) > 0 when
x < –1 or x > 1.
f ′′( x) = 6 x; f ′′( x) < 0 when x < 0.
f(x) is increasing on (– ∞ , –1] ∪ [1, ∞ ) and
concave down on (– ∞ , 0).
f ′′( x) = −12 x − 6 = –6(2x + 1); f ′′( x) < 0 when
3
⎛ 4π ⎞
f ⎜ ⎟=–
≈ –0.87
2
⎝ 3 ⎠
⎛ 4π ⎞
Global minimum f ⎜ ⎟ ≈ –0.87;
⎝ 3 ⎠
⎛π⎞
global maximum f ⎜ ⎟ = 1
⎝2⎠
12.
f ′( x) = 9 x8 ; f ′( x) > 0 for all x ≠ 0.
18.
f ′( x) = 3 x 2 – 6 x 4 = 3 x 2 (1 – 2 x 2 ); f ′( x) > 0
when –
1
2
< x < 0 and 0 < x <
1
2
.
f ′′( x) = 6 x – 24 x3 = 6 x(1 – 4 x 2 ); f ′′( x) < 0 when
1
1
< x < 0 or x > .
2
2
⎡ 1 1 ⎤
f(x) is increasing on ⎢ –
,
⎥ and concave
2 2⎦
⎣
⎛ 1 ⎞ ⎛1 ⎞
down on ⎜ – , 0 ⎟ ∪ ⎜ , ∞ ⎟ .
⎝ 2 ⎠ ⎝2 ⎠
–
3
f ′( x) = 3 – 2 x; f ′( x ) > 0 when x < .
2
f ′′( x) = –2; f ′′( x) is always negative.
3⎤
⎛
f(x) is increasing on ⎜ – ∞, ⎥ and concave down
2⎦
⎝
on (– ∞ , ∞ ).
238
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19.
f ′( x) = 3 x 2 – 4 x3 = x 2 (3 – 4 x); f ′( x) > 0 when
4
f ′′( x) = 6 x – 8; f ′′( x) > 0 when x > .
3
⎛4 ⎞
f(x) is concave up on ⎜ , ∞ ⎟ and concave down
⎝3 ⎠
4⎞
⎛ 4 128 ⎞
⎛
on ⎜ – ∞, ⎟ ; inflection point ⎜ , −
⎟
27 ⎠
3⎠
⎝3
⎝
3
x< .
4
f ′′( x) = 6 x – 12 x 2 = 6 x(1 – 2 x); f ′′( x) < 0 when
x < 0 or x >
1
.
2
3⎤
⎛
f(x) is increasing on ⎜ – ∞, ⎥ and concave down
4⎦
⎝
⎛1 ⎞
on (– ∞, 0) ∪ ⎜ , ∞ ⎟ .
⎝2 ⎠
20. g ′(t ) = 3t 2 –
1
t
2
; g ′(t ) > 0 when 3t 2 >
1
t2
or
1
1
1
t 4 > , so t < –
or t >
.
1/ 4
1/ 4
3
3
3
1 ⎤ ⎡ 1
⎛
⎞
g ′(t ) is increasing on ⎜ – ∞, –
∪
, ∞⎟
1/ 4 ⎥ ⎢ 1/ 4
3 ⎦ ⎣3
⎝
⎠
⎡ 1
⎞ ⎛
1 ⎤
, 0 ⎟ ∪ ⎜ 0,
and decreasing on ⎢ –
.
1/ 4
1/ 4 ⎥
⎣ 3
⎠ ⎝ 3 ⎦
⎛ 1 ⎞
1
=
+ 31/ 4 ≈ 1.75;
Local minimum g ⎜
1/ 4 ⎟
3/ 4
⎝3 ⎠ 3
local maximum
⎛
1 ⎞
1
=–
g⎜–
– 31/ 4 ≈ –1.75
1/ 4 ⎟
3/ 4
3
⎝ 3 ⎠
2
g ′′(t ) = 6t + ; g ′′(t ) > 0 when t > 0. g(t) has no
t3
inflection point since g(0) does not exist.
21.
22.
f ′( x) = –
f ′′( x) =
8x
( x + 1)2
2
; f ′( x) = 0 when x = 0.
8(3x 2 – 1)
; f ′′(0) = –8, so f(0) = 6 is a
( x 2 + 1)3
local maximum. f ′( x) > 0 for x < 0 and
f ′( x) < 0 for x > 0 so
f(0) = 6 is a global maximum value. f(x) has no
minimum value.
23.
f ′( x) = 4 x3 – 2; f ′( x) = 0 when x =
1
3
.
2
f ′′( x) = 12 x 2 ; f ′′( x) = 0 when x = 0.
⎛ 1 ⎞ 12
f ′′ ⎜
=
> 0, so
3 ⎟
2/3
⎝ 2⎠ 2
⎛ 1 ⎞
1
2
3
–
f⎜
is a global
=
=–
3 ⎟
4/3
1/ 3
4/3
2
2
⎝ 2⎠ 2
minimum.
f ′′( x) > 0 for all x ≠ 0; no inflection points
No horizontal or vertical asymptotes
f ′( x) = 2 x ( x – 4) + x 2 = 3 x 2 – 8 x = x (3 x – 8);
8
3
⎡8 ⎞
f(x) is increasing on (– ∞, 0] ∪ ⎢ , ∞ ⎟ and
⎣3 ⎠
⎡ 8⎤
decreasing on ⎢ 0, ⎥
⎣ 3⎦
256
⎛8⎞
Local minimum f ⎜ ⎟ = –
≈ –9.48;
27
⎝3⎠
local maximum f(0) = 0
f ′( x) > 0 when x < 0 or x >
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24.
Vertical asymptote x = 3
f ′( x) = 2( x 2 – 1)(2 x) = 4 x( x 2 – 1) = 4 x3 – 4 x;
f ′( x) = 0 when x = –1, 0, 1.
f ′′( x) = 12 x 2 – 4 = 4(3 x 2 – 1); f ′′( x ) = 0 when
x=±
1
.
3
f ′′(–1) = 8, f ′′(0) = –4, f ′′(1) = 8
Global minima f(–1) = 0, f(1) = 0;
local maximum f(0) = 1
⎛ 1 4⎞
Inflection points ⎜ ±
, ⎟
3 9⎠
⎝
No horizontal or vertical asymptotes
27.
f ′( x) = 12 x3 – 12 x 2 = 12 x 2 ( x – 1); f ′( x) = 0
when x = 0, 1.
f ′′( x) = 36 x 2 – 24 x = 12 x(3 x – 2); f ′′( x) = 0
2
.
3
f ′′(1) = 12, so f(1) = –1 is a minimum.
Global minimum f(1) = –1; no local maxima
⎛ 2 16 ⎞
Inflection points (0, 0), ⎜ , − ⎟
⎝ 3 27 ⎠
No horizontal or vertical asymptotes.
when x = 0,
25.
26.
f ′( x) =
3x – 6
; f ′( x) = 0 when x = 2, but x = 2
2 x–3
is not in the domain of f(x). f ′( x ) does not exist
when x = 3.
3( x – 4)
f ′′( x) =
; f ′′( x) = 0 when x = 4.
4( x – 3)3 / 2
Global minimum f(3) = 0; no local maxima
Inflection point (4, 4)
No horizontal or vertical asymptotes.
f ′( x) = –
f ′′( x) =
1
( x – 3) 2
; f ′( x) < 0 for all x ≠ 3.
28.
f ′( x) = 1 +
1
; f ′( x) > 0 for all x ≠ 0.
x2
2
f ′′( x) = – ; f ′′( x) > 0 when x < 0 and
x3
f ′′( x) < 0 when x > 0.
No local minima or maxima
No inflection points
1
f ( x) = x – , so
x
⎛ 1⎞
lim [ f ( x) − x] = lim ⎜ − ⎟ = 0 and y = x is an
x →∞
x →∞ ⎝ x ⎠
oblique asymptote.
Vertical asymptote x = 0
2
; f ′′( x) > 0 when x > 3.
( x – 3)3
No local minima or maxima
No inflection points
1 – 2x
x–2
lim
= lim
=1
x →∞ x – 3 x →∞ 1 – 3
x
Horizontal asymptote y = 1
240
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29.
f ′( x) = 3 +
f ′′( x) = –
1
x
2
2
; f ′( x ) > 0 for all x ≠ 0.
31.
π 3π
x=– ,
.
4 4
f ′′( x) = – cos x + sin x; f ′′( x) = 0 when
; f ′′( x) > 0 when x < 0 and
x3
f ′′( x) < 0 when x > 0
No local minima or maxima
No inflection points
1
f ( x) = 3x – , so
x
⎛ 1⎞
lim [ f ( x) − 3 x] = lim ⎜ − ⎟ = 0 and y = 3x is an
x →∞
x →∞ ⎝ x ⎠
oblique asymptote.
Vertical asymptote x = 0
30.
f ′( x) = –
12
; f ′′( x) > 0 for all x ≠ −1 .
( x + 1)4
No local minima or maxima
No inflection points
lim f ( x) = 0, lim f ( x) = 0, so y = 0 is a
x →∞
3π π
, .
4 4
⎛ π⎞
⎛ 3π ⎞
f ′′ ⎜ – ⎟ = – 2, f ′′ ⎜ ⎟ = 2
⎝ 4⎠
⎝ 4 ⎠
⎛ 3π ⎞
Global minimum f ⎜ ⎟ = – 2;
⎝ 4 ⎠
⎛ π⎞
global maximum f ⎜ – ⎟ = 2
⎝ 4⎠
⎛ 3π ⎞ ⎛ π ⎞
Inflection points ⎜ − , 0 ⎟ , ⎜ , 0 ⎟
⎝ 4 ⎠ ⎝4 ⎠
x=–
4
; f ′( x ) > 0 when x < −1 and
( x + 1)3
f ′( x) < 0 when x > –1.
f ′′( x) =
f ′( x) = – sin x – cos x; f ′( x) = 0 when
x→ – ∞
horizontal asymptote.
Vertical asymptote x = –1
Instructor's Resource Manual
32.
f ′( x) = cos x – sec 2 x; f ′( x ) = 0 when x = 0
f ′′( x) = – sin x – 2sec2 x tan x
= – sin x(1 + 2sec3 x )
f ′′( x) = 0 when x = 0
No local minima or maxima
Inflection point f(0) = 0
π π
Vertical asymptotes x = – ,
2 2
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33.
f ′( x) = x sec 2 x + tan x; f ′( x ) = 0 when x = 0
f ′′( x) = 2sec 2 x(1 + x tan x); f ′′( x) is never 0 on
⎛ π π⎞
⎜– , ⎟.
⎝ 2 2⎠
f ′′(0) = 2
Global minimum f(0) = 0
36.
f ′( x) = –2sin x – 2 cos x; f ′( x) = 0 when
π 3π
x=– ,
.
4 4
f ′′( x) = –2 cos x + 2sin x; f ′′( x) = 0 when
3π π
, .
4 4
⎛ π⎞
⎛ 3π ⎞
f ′′ ⎜ – ⎟ = –2 2, f ′′ ⎜ ⎟ = 2 2
⎝ 4⎠
⎝ 4 ⎠
⎛ 3π ⎞
Global minimum f ⎜ ⎟ = –2 2;
⎝ 4 ⎠
⎛ π⎞
global maximum f ⎜ – ⎟ = 2 2
⎝ 4⎠
⎛ 3π ⎞ ⎛ π ⎞
Inflection points ⎜ − , 0 ⎟ , ⎜ , 0 ⎟
⎝ 4 ⎠ ⎝4 ⎠
x=–
34.
f ′( x) = 2 + csc2 x; f ′( x ) > 0 on (0, π )
f ′′( x) = –2 cot x csc2 x; f ′′( x) = 0 when
π
⎛π ⎞
; f ′′( x) > 0 on ⎜ , π ⎟
2
⎝2 ⎠
⎛π ⎞
Inflection point ⎜ , π ⎟
⎝2 ⎠
x=
35.
f ′( x) = cos x – 2 cos x sin x = cos x(1 – 2sin x);
π π π 5π
f ′( x) = 0 when x = – , , ,
2 6 2 6
f ′′( x) = – sin x + 2sin 2 x – 2 cos 2 x; f ′′( x ) = 0
when x ≈ –2.51, –0.63, 1.00, 2.14
3
⎛ π⎞
⎛π⎞
⎛π⎞
f ′′ ⎜ – ⎟ = 3, f ′′ ⎜ ⎟ = – , f ′′ ⎜ ⎟ = 1,
2
6
2
⎝
⎠
⎝ ⎠
⎝2⎠
3
⎛ 5π ⎞
f ′′ ⎜ ⎟ = –
6
2
⎝ ⎠
⎛ π⎞
Global minimum f ⎜ – ⎟ = –2,
⎝ 2⎠
π
⎛ ⎞
local minimum f ⎜ ⎟ = 0;
⎝2⎠
⎛ π ⎞ 1 ⎛ 5π ⎞ 1
global maxima f ⎜ ⎟ = , f ⎜ ⎟ =
⎝6⎠ 4 ⎝ 6 ⎠ 4
Inflection points (–2.51, –0.94),
(–0.63, –0.94), (1.00, 0.13), (2.14, 0.13)
242
Section 3.10
37.
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
38.
=
=
⎛
⎛ 1⎞ ⎞
–( x 2 + 64) + ⎜1 + ⎟ x3 ⎟
⎜
⎝ x⎠ ⎠
x 2 + 64 ⎝
1
x2
x3 − 64
x 2 x 2 + 64
x3 − 64
= 0; x = 4
x 2 x 2 + 64
dp
dp
< 0 if x < 4,
> 0 if x > 4
dx
dx
⎛ 1⎞
When x = 4, p = ⎜ 1 + ⎟ 16 + 64 ≈ 11.18 ft.
⎝ 4⎠
39.
40. Let x be the length of a turned up side and let l be
the (fixed) length of the sheet of metal.
V = x (16 − 2 x )l = 16 xl − 2 x 2 l
42. Let x be the width and y the height of a page.
A = xy. Because of the margins,
27
(y – 4)(x – 3) = 27 or y =
+4
x−3
27 x
A=
+ 4 x;
x −3
dA ( x − 3)(27) − 27 x
81
=
+4= −
+4
2
dx
( x − 3)
( x − 3) 2
dA
3 15
= 0 when x = − ,
dx
2 2
dV
= 16l − 4 xl ; V ′ = 0 when x = 4
dx
d 2V
d2A
= −4l ; 4 inches should be turned up for
dx
dx 2
each side.
41. Let p be the length of the plank and let x be the
distance from the fence to where the plank
touches the ground.
See the figure below.
2
x=
43.
p
x 2 + 64
=
x +1
x
⎛ 1⎞
p = ⎜ 1 + ⎟ x 2 + 64
⎝ x⎠
Minimize p:
dp
1
x
⎛ 1⎞
=−
x 2 + 64 + ⎜ 1 + ⎟
2
2
dx
⎝ x ⎠ x + 64
x
Instructor's Resource Manual
162
( x − 3)
3
;
d2A
dx
2
> 0 when x =
15
2
15
; y = 10
2
1 2
πr h = 128π
2
256
h=
r2
Let S be the surface area of the trough.
256π
S = πr 2 + πrh = πr 2 +
r
dS
256π
= 2πr −
dr
r2
256π
2πr −
= 0; r 3 = 128, r = 4 3 2
2
r
Since
By properties of similar triangles,
=
d 2S
2
> 0 when r = 4 3 2 , r = 4 3 2
dr
minimizes S.
256
h=
= 83 2
2
43 2
(
)
Section 3.10
243
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
⎧x 3
if − 2 < x < 0
⎪⎪ 2 + 2
44. f ′( x) = ⎨
⎪− x + 2 if 0 < x < 2
⎪⎩
3
x 3
+ = 0; x = −3 , which is not in the domain.
2 2
x+2
−
= 0; x = −2, which is not in the domain.
3
g ′( x) =
c.
( x − 1) − ( x + 1)
=
−2
( x − 1)
( x − 1) 2
g (3) − g (2) 2 − 3
=
= −1
3− 2
1
−2
= –1; c = 1 ± 2
(c − 1)2
2
Only c = 1 + 2 is in the interval (2, 3).
Critical points: x = –2, 0, 2
f(–2) = 0, f(0) = 2, f(2) = 0
Minima f(–2) = 0, f(2) = 0, maximum f(0) = 2.
⎧1
if − 2 < x < 0
⎪⎪ 2
′′
f ( x) = ⎨
⎪− 1 if 0 < x < 2
⎪⎩ 3
Concave up on (–2, 0), concave down on (0, 2)
46.
dy
= 4 x3 − 18 x 2 + 24 x − 3
dx
d2y
2
45. a.
= 12 x 2 − 36 x + 24; 12( x 2 − 3x + 2) = 0 when
dx
x = 1, 2
Inflection points: x = 1, y = 5
and x = 2, y = 11
dy
=7
Slope at x = 1:
dx x =1
Tangent line: y – 5 = 7(x – 1); y = 7x – 2
dy
=5
Slope at x = 2:
dx x = 2
Tangent line: y – 11 = 5(x – 2); y = 5x + 1
f ′( x) = x 2
f (3) − f (−3) 9 + 9
=
=3
3 − (−3)
6
c 2 = 3; c = − 3, 3
47.
b. The Mean Value Theorem does not apply
because F ′(0) does not exist.
244
Section 3.10
48.
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
49. Let f ( x ) = 3 x − cos 2 x ; a1 = 0 , b1 = 1 .
f ( 0 ) = −1 ; f (1) ≈ 3.4161468
n
hn
1
0.5
2
0.25
3
0.125
4
0.0625
5
0.03125
6 0.015625
7 0.0078125
8 0.0039063
9 0.0019532
10 0.0009766
11 0.0004883
12 0.0002442
13 0.0001221
14 0.0000611
15 0.0000306
16 0.0000153
17 0.0000077
18 0.0000039
19 0.0000020
20 0.0000010
21 0.0000005
22 0.0000003
23 0.0000002
x ≈ 0.281785
mn
0.5
0.25
0.375
0.3125
0.28125
0.296875
0.2890625
0.2851563
0.2832031
0.2822266
0.2817383
0.2819824
0.2818604
0.2817994
0.2817689
0.2817842
0.2817918
0.2817880
0.2817861
0.2817852
0.2817847
0.2817845
0.2817846
f ( mn )
0.9596977
−0.1275826
0.3933111
0.1265369
−0.0021745
0.0617765
0.0296988
0.0137364
0.0057745
0.0017984
−0.0001884
0.0008049
0.0003082
0.0000600
−0.0000641
−0.0000018
0.0000293
0.0000138
0.0000061
0.0000022
0.0000004
−0.0000006
−0.0000000
50. f(x) = 3x – cos 2x, f ′( x) = 3 + 2sin 2 x
Let x1 = 0.5 .
n
xn
1
2
3
4
5
0.5
0.2950652
0.2818563
0.2817846
0.2817846
x ≈ 0.281785
51. xn +1 =
cos 2 xn
3
n
xn
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
0.5
0.18010
0.311942
0.270539
0.285718
0.280375
0.282285
0.281606
0.281848
0.281762
0.281793
0.281782
0.281786
0.281784
0.281785
0.281785
x ≈ 0.2818
52. y = x and y = tan x
Let x1 =
11π
.
8
f(x) = x – tan x, f ′( x) = 1 – sec2 x .
n
1
2
3
4
5
6
7
8
xn
11π
8
4.64661795
4.60091050
4.54662258
4.50658016
4.49422443
4.49341259
4.49340946
x ≈ 4.4934
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Section 3.10
245
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
53.
3
2
∫ ( x − 3x + 3 x ) dx
(
58. Let u = cos x ; then du = − sin x dx or
−du = sin x dx .
)
= ∫ x3 − 3x 2 + 3x1/ 2 dx
∫
1 4
2
x − x3 + 3 ⋅ x3/ 2 + C
4
3
1 4
= x − x3 + 2 x3/ 2 + C
4
cos 4 x sin x dx = ∫ ( cos x ) sin x dx
4
=
54.
∫
2 x 4 − 3x 2 + 1
x2
= ∫ u 4 ⋅ −du
= − ∫ u 4 du
1
= − u5 + C
5
1
= − cos5 x + C
5
dx
(
)
= ∫ 2 x 2 − 3 + x −2 dx
2 3
x − 3 x − x −1 + C
3
2 x3
1
2 x4 − 9 x2 − 3
=
− 3x − + C or
+C
x
3
3x
=
55.
∫
y 3 − 9 y sin y + 26 y −1
dy
y
=∫
=
2
2
2
2
∫ ( x + 1) tan ( 3x + 6 x ) sec ( 3x + 6 x ) dx
1 2
1
u du = u 3 + C
6∫
18
1
= tan 3 3 x 2 + 6 x + C
18
=
(
( y 2 − 9sin y + 26 ) dy
1 3
y + 9 cos y + 26 y + C
3
56. Let u = y 2 − 4 ; then du = 2 ydy or
∫
59. u = tan(3 x 2 + 6 x ), du = (6 x + 6) sec2 (3x 2 + 6 x)
1
du = ydy .
2
60. u = t 4 + 9, du = 4t 3 dt
1
du
t3
4
dt
=
∫ t4 + 9 ∫ u
1
u −1/ 2 du
4∫
1
= ⋅ 2u1/ 2 + C
4
1 4
t +9 +C
=
2
=
1
u ⋅ du
2
y y 2 − 4 dy = ∫
1
u1/ 2 du
∫
2
1 2 3/ 2
= ⋅ u +C
2 3
3/ 2
1 2
= y −4
+C
3
=
(
2
∫ z ( 2 z − 3)
1/ 3
1
du = zdz .
4
1
dz = ∫ u1/ 3 ⋅ du
4
1
= ∫ u1/ 3 du
4
1 3 4/3
= ⋅ u +C
4 4
4/3
3
=
+C
2z2 − 3
16
(
246
61. Let u = t 5 + 5 ; then du = 5t 4 dt or
)
57. Let u = 2 z 2 − 3 ; then du = 4 zdz or
Section 3.10
)
4 5
∫ t ( t + 5)
2/3
dt = ∫
1
du = t 4 dt .
5
1 2/3
u du
5
1
u 2 / 3 du
∫
5
1 3 5/3
= ⋅ u +C
5 5
5/ 3
3 5
=
+C
t +5
25
=
(
)
)
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
62. Let u = x 2 + 4 ; then du = 2 x dx or
∫
1
du = xdx .
2
66. Let u = y 3 − 3 y ; then
(
)
(
1 du
dx = ∫
2
2
u
x +4
∫
1
u −1/ 2 du
2∫
1
= ⋅ 2u1/ 2 + C
2
=
y2 −1
(y
3
− 3y
)
2
dy =
∫
1
u −2 du
3∫
1
= ⋅ − u −1 + C
3
1
1
=− ⋅ 3
+C
3 y − 3y
1
du = x 2 dx .
3
=−
x2
1 du
dx = ∫
3
3
u
x +9
1
u −1/ 2 du
3∫
1
= ⋅ 2u1/ 2 + C
3
2 3
=
x +9 +C
3
=
1
1
u2
du
68.
2
du
u3
69.
∫ dy = ∫ sin x dx
x +1
dx
y = 2 x + 1 + 14
70.
∫ sin y dy = ∫ dx
− cos y = x + C
x = –1 – cos y
71.
∫ dy = ∫
2t − 1 dt
1
y = (2t − 1)3 2 + C
3
1
y = (2t − 1)3 2 − 1
3
72.
∫y
−4
−
1
−
dy = ∫ t 2 dt
3 y3
1
3 y3
y=3
Instructor's Resource Manual
1
∫ dy = ∫
y = 2 x +1 + C
= ∫ u −3 du
1
= − u −2 + C
2
1
=−
+C
2
2 ( 2 y − 1)
+C
y = − cos x + C
y = –cos x + 3
65. Let u = 2 y − 1 ; then du = 2dy .
∫ ( 2 y − 1)3 dy = ∫
3y − 9 y
1
5
u −1/ 5 du =
(2 y 3 + 3 y 2 + 6 y )4 / 5 + C
6∫
24
= ∫ u −2 du
= −u −1 + C
1
=−
+C
y +1
1
3
67. u = 2 y 3 + 3 y 2 + 6 y, du = (6 y 2 + 6 y + 6) dy
64. Let u = y + 1 ; then du = dy .
∫ ( y + 1)2 dy = ∫
1 du
3 ∫ u2
=
= x2 + 4 + C
63. Let u = x3 + 9 ; then du = 3 x 2 dx or
)
du = 3 y 2 − 3 dy = 3 y 2 − 1 dy .
x
=
t3
+C
3
=
t3 2
−
3 3
1
2 − t3
Section 3.10
247
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
73.
∫ 2 y dy = ∫ (6 x − x
3
7. Aregion = 0.5 (1 + 1.5 + 2 + 2.5 ) = 3.5
)dx
1 4
x +C
4
1
y 2 = 3x 2 − x 4 + 9
4
1
y = 3x 2 − x 4 + 9
4
8. Aregion = 0.5 (1.5 + 2 + 2.5 + 3) = 4.5
y 2 = 3x 2 −
74.
9. Aregion = Arect + Atri = 1x +
1
1
x ⋅ x = x2 + x
2
2
1
1
1
10. Aregion = bh = x ⋅ xt = x 2t
2
2
2
∫ cos y dy = ∫ x dx
11. y = 5 − x; Aregion = Arect + Atri
x2
+C
2
⎛ x2 ⎞
y = sin −1 ⎜ ⎟
⎜ 2 ⎟
⎝ ⎠
sin y =
= 2 ( 2) +
1
( 2 )( 2 ) = 6
2
12. Aregion = Arect + Atri
75. s ( t ) = −16t 2 + 48t + 448; s = 0 at t = 7;
v ( t ) = s ' ( t ) = −32t + 48
= 1(1) +
1
(1)( 7 ) = 4.5
2
when t = 7, v = –32(7) + 48 = –176 ft/s
Review and Preview Problems
1
1
3 2
1. Aregion = bh = aa sin 60o =
a
2
2
4
⎛1
⎞
⎛ 1 ⎞⎛ 3 ⎞
2. Aregion = 6 ⎜ base × height ⎟ = 6 ⎜ a ⎟ ⎜⎜
a⎟
⎝2
⎠
⎝ 2 ⎠ ⎝ 2 ⎟⎠
=
3 3 2
a
2
2
a
⎛1
⎞
3. Aregion = 10 ⎜ base × height ⎟ = 5 cot 36D
2
4
⎝
⎠
5
= a 2 cot 36D
4
1 ⎛ 8.5 ⎞
4. Aregion = Arect + Atri = 17 ( 8.5 ) + 17 ⎜
2 ⎝ tan 45o ⎟⎠
= 216.75
1
2
5. Aregion = Arect + Asemic. = 3.6 ⋅ 5.8 + π (1.8 )
2
≈ 25.97
⎛1
⎞
6. Aregion = A#5 + 2 Atri = 25.97 + 2 ⎜ ⋅1.2 ⎟ 5.8
2
⎝
⎠
= 32.93
248
Review and Preview
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
33. a.
b.
∫ 10V
−1/ 2
dV = ∫ C1dt ; 20 V = C1t + C2 ;
V(0) = 1600: C2 = 20 ⋅ 40 = 800;
V(40) = 0: C1 = −
V (t ) =
b. Since the trip that involves 1 min more travel
time at speed vm is 0.6 mi longer,
vm = 0.6 mi/min
= 36 mi/h.
c.
From part b, vm = 0.6 mi/min. Note that the
average speed during acceleration and
v
deceleration is m = 0.3 mi/min. Let t be the
2
time spent between stop C and stop D at the
constant speed vm , so
0.6t + 0.3(4 – t)= 2 miles. Therefore,
2
t = 2 min and the time spent accelerating
3
is
4 − 2 23
a=
=
2
0.6 − 0
2
3
2
min.
3
c.
36. a.
b.
1
2
(−20t + 800)2 = ( 40 − t )
400
2
V (10) = ( 40 − 10 ) = 900 cm3
dP
= C1 3 P , P(0) = 1000, P(10) = 1700
dt
where t is the number of years since 1980.
3
dP = ∫ C1dt ; P 2 / 3 = C1t + C2
2
3
P(0) = 1000: C2 = ⋅10002 / 3 = 150
2
∫P
−1/ 3
P(10) = 1700: C1 =
3 ⋅17002 / 3
2
− 150
10
≈ 6.3660
P = (4.2440t + 100)3 / 2
c.
4000 = (4.2440t + 100)3 / 2
40002 / 3 − 100
≈ 35.812
4.2440
t ≈ 36 years, so the population will reach
4000 by 2016.
= 0.9 mi/min 2 .
t=
dh
= 4 , so h(t ) = 4t + C1 . Set
dt
t = 0 at the time when Victoria threw the ball, and
height 0 at the ground, then h(t) = 4t + 64. The
34. For the balloon,
height of the ball is given by s (t ) = −16t 2 + v0t ,
since s0 = 0 . The maximum height of the ball is
v
when t = 0 , since then s ′(t ) = 0 . At this time
32
2
⎛v ⎞
⎛v ⎞
⎛v ⎞
h(t) = s(t) or 4 ⎜ 0 ⎟ + 64 = −16 ⎜ 0 ⎟ + v0 ⎜ 0 ⎟ .
32
32
⎝ ⎠
⎝ ⎠
⎝ 32 ⎠
Solve this for v0 to get v0 ≈ 68.125 feet per
second.
35. a.
800
= −20
40
dV
= C1 h where h is the depth of the
dt
V
.
water. Here, V = πr 2 h = 100h , so h =
100
37. Initially, v = –32t and s = −16t 2 + 16 . s = 0 when
t = 1. Later, the ball falls 9 ft in a time given by
3
0 = −16t 2 + 9 , or s, and on impact has a
4
⎛3⎞
velocity of −32 ⎜ ⎟ = −24 ft/s. By symmetry,
⎝4⎠
24 ft/s must be the velocity right after the first
bounce. So
a.
for 0 ≤ t < 1
⎧−32t
v(t ) = ⎨
32(
1)
24
for 1 < t ≤ 2.5
−
t
−
+
⎩
b.
9 = −16t 2 + 16 ⇒ t ≈ 0.66 sec; s also equals 9
at the apex of the first rebound at t = 1.75 sec.
dV
V
= C1
, V(0) = 1600,
dt
10
V(40) = 0.
Hence
234
Section 3.9
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4
CHAPTER
4.1 Concepts Review
1. 2 ⋅
The Definite Integral
8
5.
m =1
5(6)
= 30; 2(5) = 10
2
= (−1)1 2−1 + (−1) 2 20 + (−1)3 21
+(−1) 4 22 + (−1)5 23 + (−1)6 24
2. 3(9) – 2(7) = 13; 9 + 4(10) = 49
+(−1)7 25 + (−1)8 26
3. inscribed; circumscribed
1
= − + 1 − 2 + 4 − 8 + 16 − 32 + 64
2
85
=
2
4. 0 + 1 + 2 + 3 = 6
Problem Set 4.1
1.
6
6
6
k =1
k =1
k =1
∑ (k − 1) = ∑ k − ∑ 1
∑ i2 =
i =1
1
1
1
1
3. ∑
=
+
+
k =1 k + 1 1 + 1 2 + 1 3 + 1
1
1
1
1
+
+
+
4 +1 5 +1 6 +1 7 +1
1 1 1 1 1 1 1
= + + + + + +
2 3 4 5 6 7 8
1443
=
840
481
=
280
(−1)5 25 (−1)6 26 (−1)7 27
+
+
6
7
8
1154
=−
105
7.
8
∑ (l + 1)2 = 42 + 52 + 62 + 72 + 82 + 92 = 271
l =3
6
6
n =1
n =1
∑ n cos(nπ) = ∑ ( −1)
n
⋅n
= –1 + 2 – 3 + 4 – 5 + 6
=3
+
4.
(−1)3 23 (−1) 4 24
+
4
5
+
6(7)(13)
= 91
6
7
∑
=
6(7)
− 6(1)
2
= 15
6
(−1)k 2k
k =3 ( k + 1)
7
6.
=
2.
∑ (−1)m 2m−2
⎛ kπ ⎞
k sin ⎜ ⎟
⎝ 2 ⎠
k =−1
6
8.
∑
⎛ π⎞
⎛π⎞
= − sin ⎜ − ⎟ + sin ⎜ ⎟ + 2sin(π)
⎝ 2⎠
⎝2⎠
π
3
⎛ ⎞
⎛ 5π ⎞
+3sin ⎜ ⎟ + 4sin(2π) +5sin ⎜ ⎟ + 6sin(3π)
⎝ 2 ⎠
⎝ 2 ⎠
=1+1+0–3+0+5+0
=4
41
9. 1 + 2 + 3 + " + 41 = ∑ i
i =1
25
10. 2 + 4 + 6 + 8 + " + 50 = ∑ 2i
i =1
11. 1 +
1 1
1 100 1
+ +" +
=∑
2 3
100 i =1 i
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1 1 1
1 100 (−1)i +1
12. 1 − + − + " −
=∑
2 3 4
100 i =1 i
10
20.
i =1
10
= ∑ (4i 2 − i − 3)
50
13. a1 + a3 + a5 + a7 + " + a99 = ∑ a2i −1
i =1
14.
∑ [(i − 1)(4i + 3)]
i =1
10
10
10
i =1
i =1
i =1
= 4∑ i 2 − ∑ i − ∑ 3
f ( w1 )Δx + f ( w2 )Δx + " + f ( wn )Δx
= 4(385) – 55 – 3(10)
= 1455
n
= ∑ f ( wi )Δx
i =1
10
21.
10
15.
∑ (ai + bi )
i =1
10
10
i =1
i =1
k =1
22.
∑ (3an + 2bn )
10
10
n =1
n =1
= 3∑ an + 2 ∑ bn
23.
= 3(40) + 2(50)
= 220
∑ (a p +1 − b p +1 )
p =0
10
p =1
= 40 − 50
= –10
24.
∑ (aq − bq − q)
n
n
n
i =1
i =1
i =1
i =1
=
2n(n + 1)(2n + 1) 3n(n + 1)
−
+n
6
2
=
2n3 + 3n 2 + n 3n 2 + 3n
−
+n
3
2
=
4n3 − 3n 2 − n
6
n
n
10
10
q =1
q =1
q =1
i =1
∑ aq − ∑ bq − ∑ q
10(11)
2
= −65
25.
100
∑ (3i − 2)
i =1
i =1
i =1
∑ (2i − 3)2 = ∑ (4i 2 − 12i + 9)
i =1
n
n
= 4∑ i 2 − 12∑ i + ∑ 9
10
100
k =1
n
q =1
100
k =1
i =1
10
= 40 − 50 −
10
n
10
∑ a p − ∑ bp
p =1
k =1
10
∑ (2i 2 − 3i + 1) = 2∑ i 2 − 3∑ i + ∑1
9
19.
10
∑ 5k 2 (k + 4) = ∑ (5k 3 + 20k 2 )
= 5(3025) + 20(385)
= 22,825
n =1
=
k =1
= 5 ∑ k 3 + 20 ∑ k 2
10
18.
k =1
k =1
= 90
=
10
∑ k3 −∑ k2
10
= 40 + 50
17.
10
= 3025 − 385
= 2640
= ∑ ai + ∑ bi
16.
∑ (k 3 − k 2 ) =
= 3∑ i − ∑ 2
= 3(5050) − 2(100)
= 14,950
i =1
i =1
=
4n(n + 1)(2n + 1) 12n(n + 1)
−
+ 9n
6
2
=
4n3 − 12n 2 + 11n
3
S = 1 + 2 + 3 + " + (n − 2) + (n − 1) + n
+ S = n + (n − 1) + (n − 2) + " + 3 + 2 + 1
2S = (n + 1) + (n + 1) + (n + 1) + " + (n + 1) + (n + 1) + (n + 1)
2S = n(n + 1)
n(n + 1)
S=
2
250
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26. S − rS = a + ar + ar 2 + " + ar n
− (ar + ar 2 + " + ar n + ar n +1 )
27. a.
= a − ar n +1
(1)
1−
⎛1⎞
∑ ⎜⎝ 2 ⎟⎠ = 12
k =0
2
10
k
10
k
11
10
⎛1⎞
⎛1⎞
∑ ⎜⎝ 2 ⎟⎠ = 1 − ⎜⎝ 2 ⎟⎠
k =1
n +1
a − ar
= S (1 − r ); S =
1− r
10
b.
∑ 2k =
k =0
10
10
⎛1⎞
= 2 − ⎜ ⎟ , so
⎝2⎠
=
1023
.
1024
1 − 211
= 211 − 1, so
−1
∑ 2k = 211 − 2 = 2046 .
k =1
28.
S = a + (a + d ) + (a + 2d ) + "" + [ a + (n − 2)d ] + [ a + (n − 1)d ] + (a + nd )
+ S = (a + nd ) + [ a + (n − 1)d ] + [ a + (n − 2)d ] + " + (a + 2d ) + (a + d ) + a
2S = (2a + nd ) + (2a + nd ) + (2a + nd ) + " + (2a + nd ) + (2a + nd ) + (2a + nd )
2S = (n + 1)(2a + nd)
(n + 1)(2a + nd )
S=
2
( i + 1)3 − i3 = 3i 2 + 3i + 1
29.
n
∑ ⎡⎣( i + 1)
i =1
3
n
(
)
− i 3 ⎤ = ∑ 3i 2 + 3i + 1
⎦ i =1
n
n
n
i =1
i =1
i =1
( n + 1)3 − 13 = 3∑ i 2 + 3∑ i + ∑1
n
n ( n + 1)
i =1
2
n3 + 3n 2 + 3n = 3∑ i 2 + 3
+n
n
2n3 + 6n 2 + 6n = 6∑ i 2 + 3n 2 + 3n + 2n
i =1
2n + 3n + n
= ∑ i2
6
i =1
3
2
n ( n + 1)( 2n + 1)
6
n
n
= ∑ i2
i =1
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( i + 1)4 − i 4 = 4i3 + 6i 2 + 4i + 1
30.
∑ ⎡⎣(i + 1)4 − i 4 ⎤⎦ =∑ ( 4i3 + 6i 2 + 4i + 1)
n
n
i =1
i =1
n
n
n
n
i =1
n
i =1
i =1
i =1
(n + 1)4 − 14 = 4∑ i3 + 6∑ i 2 + 4∑ i + ∑1
n 4 + 4 n3 + 6 n 2 + 4 n = 4∑ i 3 + 6
i =1
n(n + 1)(2n + 1)
n(n + 1)
+4
+n
6
2
n
Solving for
∑ i3 gives
i =1
(
n
) (
)
4∑ i3 = n 4 + 4n3 + 6n 2 + 4n − 2n3 + 3n 2 + n − 2n 2 + 2n − n
i =1
n
4∑ i 3 = n 4 + 2 n3 + n 2
i =1
n 4 + 2n3 + n 2 ⎡ n ( n + 1) ⎤
=⎢
⎥
∑i =
4
⎣ 2 ⎦
i =1
n
2
3
( i + 1)5 − i5 = 5i 4 + 10i3 + 10i 2 + 5i + 1
31.
n
∑ ⎡⎣⎢( i + 1)
5
i =1
n
n
n
n
n
i =1
2
i =1
i =1
− i5 ⎤⎥ =5∑ i 4 + 10∑ i3 + 10∑ i 2 + 5∑ i + ∑ 1
⎦
i =1
n
i =1
( n + 1)5 − 15 = 5∑ i 4 + 10
n
2
( n + 1)
4
i =1
n
n5 + 5n 4 + 10n3 + 10n 2 + 5n = 5∑ i 4 + 52 n 2 ( n + 1)
i =1
2
+ 10
n(n + 1)(2n + 1)
n(n + 1)
+5
+n
6
2
5
+ 10
n ( n + 1) (2n + 1) + n(n + 1) + n
6
2
n
Solving for
∑ i4
yields
i =1
n
∑ i 4 = 15 ⎡⎣n5 + 52 n4 + 53 n3 − 16 n ⎤⎦ =
i =1
n(n + 1)(2n + 1)(3n 2 + 3n − 1)
30
32. Suppose we have a ( n + 1) × n grid. Shade in
n + 1 – k boxes in the kth column. There are n columns, and the shaded area is 1 + 2 + " + n . The shaded area is
n(n + 1)
n(n + 1)
. Thus, 1 + 2 + " + n =
.
2
2
n(n + 1)
. From the diagram the area is
Suppose we have a square grid with sides of length 1 + 2 + " + n =
2
also half the area of the grid or
2
2
⎡ n(n + 1) ⎤
⎡ n(n + 1) ⎤
13 + 23 + " + n3 or ⎢
. Thus, 13 + 23 + " + n3 = ⎢
⎥ .
⎥
⎣ 2 ⎦
⎣ 2 ⎦
33. x =
1
55
(2 + 5 + 7 + 8 + 9 + 10 + 14) =
≈ 7.86
7
7
2
1 ⎡⎛
55 ⎞ ⎛
55 ⎞ ⎛
55 ⎞ ⎛
55 ⎞
55 ⎞ ⎛
55 ⎞ ⎛
55 ⎞
⎛
s = ⎢⎜ 2 − ⎟ + ⎜ 5 − ⎟ + ⎜ 7 − ⎟ + ⎜ 8 − ⎟ + ⎜ 9 − ⎟ + ⎜ 10 − ⎟ + ⎜ 14 − ⎟
7 ⎢⎝
7 ⎠ ⎝
7 ⎠ ⎝
7 ⎠ ⎝
7 ⎠
7
7
7 ⎠
⎝
⎠ ⎝
⎠ ⎝
⎣
2
2
2
2
2
2
2⎤
608
⎥ =
≈ 12.4
49
⎥⎦
252
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34. a.
x = 1, s 2 = 0
b.
x = 1001, s 2 = 0
c.
x=2
2
s 2 = 13 ⎡(1 − 2)2 + (2 − 2) 2 + (3 − 2)2 ⎤ = 13 ⎡⎢( −1) + 02 + 12 ⎤⎥ = 13 ( 2 ) = 32
⎣
⎦
⎣
⎦
d.
x = 1, 000, 002
s 2 = 13 ⎡(−1)2 + 02 + 12 ⎤ = 32
⎣
⎦
35. a.
b.
n
n
n
i =1
i =1
i =1
∑ ( xi − x ) = ∑ xi − ∑ x = nx − nx = 0
1 n
1 n
( xi − x )2 = ∑ ( xi2 − 2 x xi + x 2 )
∑
n i =1
n i =1
1 n
2x n
1 n
= ∑ xi2 −
xi + ∑ x 2
∑
n i =1
n i =1
n i =1
s2 =
=
1 n 2 2x
1
xi −
(nx ) + (nx 2 )
∑
n i =1
n
n
⎛1 n
⎞
⎛1 n
⎞
= ⎜ ∑ xi2 ⎟ − 2 x 2 + x 2 = ⎜ ∑ xi2 ⎟ − x 2
⎜n
⎟
⎜n
⎟
⎝ i =1 ⎠
⎝ i =1 ⎠
36. The variance of n identical numbers is 0. Let c be the constant. Then
2
2
s 2 = 1n ⎡⎢( c − c ) + ( c − c ) + " + (c − c)2 ⎤⎥ = 0
⎣
⎦
n
37. Let S (c) = ∑ ( xi − c)2 . Then
i =1
d n
( xi − c )2
∑
dc i =1
S '(c) =
n
d
( xi − c )2
dc
i =1
=∑
n
= ∑ 2( xi − c)(−1)
i =1
n
= −2∑ xi + 2nc
i =1
S ''(c) = 2n
Set S '(c) = 0 and solve for c :
n
−2∑ xi + 2nc = 0
c
i =1
n
= 1n xi
i =1
∑
=x
Since S ''( x) = 2n > 0 we know that x minimizes S (c) .
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i
The number of gifts given on the nth day is
38. a.
∑m=
m =1
i (i + 1)
.
2
i (i + 1)
= 364 .
i =1 2
12
The total number of gifts is
∑
i (i + 1)
.
i =1 2
n
b. For n days, the total number of gifts is
∑
i (i + 1) n i 2 n i 1 n 2 1 n
1 ⎡ n(n + 1)(2n + 1) ⎤ 1 ⎡ n(n + 1) ⎤
∑ 2 = ∑ 2 +∑ 2 = 2 ∑ i + 2 ∑ i = 2 ⎢⎣
⎥+ 2⎢ 2 ⎥
6
⎦
⎣
⎦
i =1
i =1
i =1
i =1
i =1
1
1
⎛ 2n + 1 ⎞ 1
= n(n + 1) ⎜
+ 1⎟ = n(n + 1)(2n + 4) = n(n + 1)(n + 2)
4
6
⎝ 3
⎠ 12
n
39. The bottom layer contains 10 · 16 = 160 oranges, the next layer contains 9 · 15 = 135 oranges, the third layer
contains 8 · 14 = 112 oranges, and so on, up to the top layer, which contains 1 · 7 = 7 oranges. The stack contains
1 · 7 + 2 · 8+ ... + 9· 15 + 10 · 16
10
=
∑ i(6 + i) = 715 oranges.
i =1
50
40. If the bottom layer is 50 oranges by 60 oranges, the stack contains
∑ i(10 + i) = 55, 675.
i =1
41. For a general stack whose base is m rows of n oranges with m ≤ n, the stack contains
m
m
m
i =1
i =1
i =1
∑ i ( n − m + i ) = ( n − m)∑ i + ∑ i 2
m(m + 1) m(m + 1)(2m + 1)
= ( n − m)
+
2
6
m(m + 1)(3n − m + 1)
=
6
42.
1
1
1
1
+
+
+" +
1⋅ 2 2 ⋅ 3 3 ⋅ 4
n(n + 1)
1 ⎞
⎛ 1⎞ ⎛ 1 1⎞ ⎛1 1⎞
⎛1
= ⎜1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + " + ⎜ −
⎟
2
2
3
3
4
n
n
+1⎠
⎝
⎠ ⎝
⎠ ⎝
⎠
⎝
1
= 1−
n +1
43. A =
1⎡ 3
5⎤ 7
1+ + 2 + ⎥ =
2 ⎢⎣ 2
2⎦ 2
44. A =
1⎡ 5 3 7
9 5 11 ⎤ 15
1+ + + + 2 + + + ⎥ =
⎢
4⎣ 4 2 4
4 2 4⎦ 4
45. A =
1 ⎡3
5 ⎤ 9
+ 2 + + 3⎥ =
⎢
2 ⎣2
2 ⎦ 2
46. A =
1 ⎡5 3 7
9 5 11 ⎤ 17
+ + + 2 + + + + 3⎥ =
4 ⎢⎣ 4 2 4
4 2 4
⎦ 4
254
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47. A =
2
2
⎞⎤
⎞ ⎛1
1 ⎡⎛ 1 2 ⎞ ⎛ 1 ⎛ 1 ⎞
1 ⎛ 9 3 17 ⎞ 23
⎞ ⎛1 ⎛3⎞
2
⎜
⎜
⎟
+
⋅
+ 1⎟ ⎥ = ⎜ 1 + + + ⎟ =
⋅
+
+
⋅
+
+
⋅
+
0
1
1
1
1
⎜
⎟
⎜
⎟ ⎜ ⎜ ⎟
⎜
⎟
⎢
⎜
⎟
⎟
2⎝ 8 2 8 ⎠ 8
2
2
2 ⎣⎝ 2
2
2
2
⎠ ⎝ ⎝ ⎠
⎠ ⎝ ⎝ ⎠
⎠ ⎦⎥
⎠ ⎝
48. A =
2
⎤
⎛1 3 2 ⎞ 1
⎞ 1
1 ⎡⎛ 1 ⎛ 1 ⎞
1 9 3 17
31
⎢⎜ ⋅ ⎜ ⎟ + 1⎟ + ⎛⎜ ⋅12 + 1⎞⎟ + ⎜ ⋅ ⎜⎛ ⎟⎞ + 1⎟ + ⎜⎛ ⋅ 22 + 1⎟⎞ ⎥ = ⎛⎜ + + + 3 ⎞⎟ =
⎜
⎟
⎜
⎟
2⎝8 2 8
2⎢ 2 ⎝2⎠
2
2
⎠ ⎥⎦
⎠ ⎝2 ⎝2⎠
⎠ 8
⎠ ⎝
⎠ ⎝
⎣⎝
49.
A = 1(1 + 2 + 3) = 6
50.
A=
1 ⎡⎛ 3 ⎞
⎛ 5 ⎞
⎜ 3 ⋅ − 1⎟ + ( 3 ⋅ 2 − 1) + ⎜ 3 ⋅ − 1⎟ + (3 · 3 – 1)] =
2 ⎢⎣⎝ 2 ⎠
⎝ 2 ⎠
1⎛7
13 ⎞ 23
⎜ + 5 + + 8⎟ =
2⎝2
2
⎠ 2
51.
2
⎤
⎞ ⎛ 7 2 ⎞ ⎛ 5 2 ⎞ ⎛ 8 2 ⎞ ⎛ 17 2 ⎞
1 ⎡⎛ ⎛ 13 ⎞
⎢⎜ ⎜ ⎟ − 1⎟ + ⎜ ⎛⎜ ⎞⎟ − 1⎟ + ⎜ ⎛⎜ ⎞⎟ − 1⎟ + ⎜ ⎛⎜ ⎞⎟ − 1⎟ + ⎜ ⎛⎜ ⎞⎟ − 1⎟ + (32 − 1) ⎥
⎟ ⎜⎝ 3 ⎠
⎟ ⎜⎝ 2 ⎠
⎟ ⎜⎝ 3 ⎠
⎟ ⎜⎝ 6 ⎠
⎟
6 ⎢⎜ ⎝ 6 ⎠
⎥⎦
⎠ ⎝
⎠ ⎝
⎠ ⎝
⎠ ⎝
⎠
⎣⎝
1 ⎛ 133 40 21 55 253 ⎞ 1243
= ⎜
+
+ + +
+ 8⎟ =
6 ⎝ 36 9
4 9
36
⎠ 216
A=
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52.
2
2
2
⎛
⎞ ⎛
⎞ ⎛
⎞
1⎡
4
4
3
3
2
2
⎢(3(−1)2 + (−1) + 1) + ⎜ 3 ⎛⎜ − ⎞⎟ + ⎛⎜ − ⎞⎟ + 1⎟ + ⎜ 3 ⎛⎜ − ⎞⎟ + ⎛⎜ − ⎞⎟ + 1⎟ + ⎜ 3 ⎛⎜ − ⎞⎟ + ⎛⎜ − ⎞⎟ + 1⎟ + (3(0) 2 + 0 + 1)
⎜ ⎝ 5⎠ ⎝ 5⎠ ⎟ ⎜ ⎝ 5⎠ ⎝ 5⎠ ⎟ ⎜ ⎝ 5⎠ ⎝ 5⎠ ⎟
5⎢
⎝
⎠ ⎝
⎠ ⎝
⎠
⎣
2
2
2
2
⎤
⎛ ⎛1⎞ 1 ⎞ ⎛ ⎛2⎞
2 ⎞ ⎛ ⎛ 3⎞
3 ⎞⎛ ⎛ 4 ⎞
4 ⎞
+ ⎜ 3 ⎜ ⎟ + + 1⎟ + ⎜ 3 ⎜ ⎟ + + 1⎟ + ⎜ 3 ⎜ ⎟ + + 1⎟ ⎜ 3 ⎜ ⎟ + + 1⎟ + (3(1) 2 + 1 + 1) ⎥
⎜ 5
5 ⎟ ⎜ ⎝5⎠
5 ⎟ ⎜ ⎝5⎠
5 ⎟⎜ ⎝ 5 ⎠
5 ⎟
⎥⎦
⎝ ⎝ ⎠
⎠ ⎝
⎠ ⎝
⎠⎝
⎠
1
= [3 + 2.12 + 1.48 + 1.08 + 1 + 1.32 + 1.88 + 2.68 + 3.72 + 5] = 4.656
5
A=
1
i
, xi =
n
n
i
⎛
⎞⎛ 1 ⎞ i 2
f ( xi )Δx = ⎜ + 2 ⎟ ⎜ ⎟ =
+
⎝n
⎠ ⎝ n ⎠ n2 n
⎡⎛ 1 2 ⎞ ⎛ 2 2 ⎞
1 5
⎛ n 2 ⎞⎤
1
n(n + 1)
+
+
+
+" + ⎜
+
(1 + 2 + 3 + " + n) + 2 =
=
+2 =
+
A( Sn ) = ⎢⎜
2 n⎟ ⎜ 2 n⎟
2 n ⎟⎥
2
2
2n 2
2n
⎠ ⎝n
⎠
⎝n
⎠⎦ n
⎣⎝ n
⎛ 1 5⎞ 5
lim A( Sn ) = lim ⎜ + ⎟ =
n →∞
n→∞ ⎝ 2n 2 ⎠ 2
53. Δx =
1
i
, xi =
n
n
⎡ 1 ⎛ i ⎞2 ⎤ ⎛ 1 ⎞ i 2
1
f ( xi )Δx = ⎢ ⋅ ⎜ ⎟ + 1⎥ ⎜ ⎟ =
+
3
2
n
n
n
⎢⎣ ⎝ ⎠
⎥⎦ ⎝ ⎠ 2n
⎡⎛ 12
⎛ n2 1 ⎞⎤
1 ⎞ ⎛ 22 1 ⎞
1 2
+ ⎟+⎜
+ ⎟ +" + ⎜
+ ⎟⎥ =
(1 + 22 + 32 + " + n 2 ) + 1
A( Sn ) = ⎢⎜
3
3 n⎟ ⎜
3 n⎟
3 n⎟
⎜
⎜
⎢⎣⎝ 2n
⎠ ⎝ 2n
⎠
⎝ 2n
⎠ ⎥⎦ 2n
1 ⎡ n(n + 1)(2n + 1) ⎤
1 ⎡ 2n3 + 3n 2 + n ⎤
1 ⎡
3 1 ⎤
=
=
+
1
⎢
⎥ + 1 = ⎢2 + + 2 ⎥ + 1
⎥
3 ⎢⎣
3
6
12 ⎢⎣
n n ⎦
12 ⎣
⎦
n
2n
⎥⎦
⎡1⎛
3 1 ⎞ ⎤ 7
lim A( Sn ) = lim ⎢ ⎜ 2 + +
⎟ + 1⎥ =
n n2 ⎠ ⎦ 6
n →∞
n →∞ ⎣12 ⎝
54. Δx =
256
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2
2i
, xi = −1 +
n
n
⎡ ⎛
2i ⎞ ⎤ ⎛ 2 ⎞ 8i
f ( xi )Δx = ⎢ 2 ⎜ −1 + ⎟ + 2 ⎥ ⎜ ⎟ =
n ⎠ ⎦ ⎝ n ⎠ n2
⎣ ⎝
1
i
, xi =
n
n
⎡ ⎛ i ⎞3 i ⎤ ⎛ 1 ⎞ i 3
i
f ( xi )Δx = ⎢⎜ ⎟ + ⎥ ⎜ ⎟ =
+
4
n
n
n
⎝
⎠
⎝
⎠
n
n2
⎣⎢
⎦⎥
55. Δx =
58. Δx =
⎡⎛ 8 ⎞ ⎛ 16 ⎞
⎛ 8n ⎞ ⎤
A( Sn ) = ⎢⎜ ⎟ + ⎜ ⎟ + " + ⎜ ⎟ ⎥
2
2
⎝ n2 ⎠⎦
⎣⎝ n ⎠ ⎝ n ⎠
8
8 ⎡ n(n + 1) ⎤
=
(1 + 2 + 3 + " + n) =
⎢
⎥
2
n
n2 ⎣ 2 ⎦
⎡ n2 + n ⎤
4
= 4⎢
⎥ = 4+
2
n
⎣⎢ n ⎦⎥
A( Sn ) =
59.
60.
1
i
, xi =
n
n
3
3
⎛ i ⎞ ⎛1⎞ i
f ( xi )Δx = ⎜ ⎟ ⎜ ⎟ =
⎝ n ⎠ ⎝ n ⎠ n4
⎡1
1 3
1 3 ⎤
A( Sn ) = ⎢ (13 ) +
(2 ) + " +
(n ) ⎥
4
4
n
n4
⎣n
⎦
n
=
1 ⎡ n(n + 1) ⎤
⎢
⎥
n4 ⎣ 2 ⎦
1 ⎡ n(n + 1) ⎤
1 ⎡ n(n + 1) ⎤
⎥ + 2⎢ 2 ⎥
4 ⎢⎣
2
⎦
⎦
n
n ⎣
2
i 2
⎡i
⎤1
+
f (ti )Δt = ⎢ + 2 ⎥ =
⎣n
⎦ n n2 n
⎛1 1 ⎞
=⎜ + ⎟+2
⎝ 2 2n ⎠
1
5
lim A( Sn ) = + 2 =
2
2
n →∞
1
The object traveled 2 ft.
2
⎛8 4
4 ⎞ 8
lim A( Sn ) = lim ⎜ + +
⎟= .
n →∞
n →∞ ⎝ 3 n 3n 2 ⎠ 3
⎛ 8 ⎞ 16
By symmetry, A = 2 ⎜ ⎟ =
.
⎝3⎠ 3
(13 + 23 + " + n3 ) =
(1 + 2 + " + n)
⎡ n2 + n ⎤
=⎢
⎥+2
2
⎣⎢ 2n ⎦⎥
4 ⎡ 2n3 + 3n 2 + n ⎤ 8 4
4
= ⎢
⎥= + + 2
3
n
3 ⎣⎢
3
n
3n
⎦⎥
4
n2
n
n
2
⎛ i 2⎞ 1 n
A( Sn ) = ∑ ⎜
+ ⎟ = ∑i + ∑
2 n
2
⎠ n i =1 i =1 n
i =1 ⎝ n
1 ⎡ n(n + 1) ⎤
=
⎢
⎥+2
n2 ⎣ 2 ⎦
2
⎛ 2i ⎞ ⎛ 2 ⎞ 8i
f ( xi )Δx = ⎜ ⎟ ⎜ ⎟ =
⎝ n ⎠ ⎝ n ⎠ n3
⎡⎛ 8 ⎞ ⎛ 8(22 ) ⎞
⎛ 8n 2 ⎞ ⎤
A( Sn ) = ⎢⎜ ⎟ + ⎜
+ ⋅⋅⋅+ ⎜
⎟
⎟⎥
3
⎜ 3 ⎟
⎜ n3 ⎟ ⎥
⎝
⎠⎦
⎣⎢⎝ n ⎠ ⎝ n ⎠
8 2
8 ⎡ n(n + 1)(2n + 1) ⎤
=
(1 + 22 + ⋅⋅⋅ + n 2 ) =
⎢
⎥
3
6
⎦
n
n3 ⎣
1
1
n 2 + 2n + 1 n 2 + n 3n 2 + 4n + 1 3 1
1
+
=
= + +
2
2
2
4
n
4n
2n
4n
4n 2
3
lim A( Sn ) =
4
n →∞
56. First, consider a = 0 and b = 2.
2
2i
Δx = , xi =
n
n
=
(13 + 23 + " + n3 ) +
=
4⎞
⎛
lim A( Sn ) = lim ⎜ 4 + ⎟ = 4
n⎠
n→∞ ⎝
57. Δx =
n
4
2
=
n →∞
2
1
⎡ 1 ⎛ i ⎞2 ⎤ 1
i2
1
+
f (ti )Δt = ⎢ ⎜ ⎟ + 1⎥ =
3
n
⎢⎣ 2 ⎝ n ⎠
⎥⎦ n 2n
n ⎛ 2
1i
1⎞
1 n 2 n 1
A( Sn ) = ∑ ⎜
i +∑
+ ⎟=
⎜ 3 n ⎟ 2 n3 ∑
i =1 ⎝ 2n
i =1
i =1 n
⎠
1 ⎡ n(n + 1)(2n + 1) ⎤
1 ⎡
3 1 ⎤
=
⎥ + 1 = 12 ⎢ 2 + n + 2 ⎥ + 1
3 ⎢⎣
6
⎦
2n
n ⎦
⎣
1
7
(2) + 1 = ≈ 1.17
12
6
n →∞
The object traveled about 1.17 feet.
lim A( Sn ) =
1⎡ 2 1 ⎤
1 ⎡ n 4 + 2n3 + n 2 ⎤
⎢
⎥ = ⎢1 + + 2 ⎥
4
4
4⎣ n n ⎦
n ⎣⎢
⎦⎥
1⎡ 2 1 ⎤ 1
⎢1 + n + 2 ⎥ = 4
n →∞ 4 ⎣
n ⎦
lim A( Sn ) = lim
n →∞
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
65. a.
A02 ( x3 ) =
23+1
=4
3 +1
b.
A12 ( x3 ) =
23+1 13+1
1 15
−
= 4− =
3 +1 3 +1
4 4
c.
A12 ( x5 ) =
25+1 15+1 32 1 63
−
=
− =
5 +1 5 +1 3 6 6
2
61. a.
3 2
⎛ ib ⎞ ⎛ b ⎞ b i
f ( xi )Δx = ⎜ ⎟ ⎜ ⎟ =
⎝ n ⎠ ⎝ n ⎠ n3
A0b =
b3
=
6
b3 n
∑ i2
n3
=
i =1
b3 ⎡ n(n + 1)(2n + 1) ⎤
⎢
⎥
6
⎦
n3 ⎣
3 1 ⎤
⎡
⎢2 + n + 2 ⎥
n ⎦
⎣
lim A0b =
n →∞
2b3 b3
=
6
3
=
21
= 10.5
2
b. Since a ≥ 0, A0b = A0a + Aab , or
Aab
=
A0b
−
A0a
b3 a 3
=
− .
3
3
d.
A05 =
53 125
=
3
3
b.
A14 =
43 13 63
− =
= 21
3 3
3
c.
A25 =
53 23 117
−
=
= 39
3
3
3
64. a.
Δx =
b
bi
, xi =
n
n
m
m +1 m
⎛ bi ⎞ ⎛ b ⎞ b i
f ( xi )Δx = ⎜ ⎟ ⎜ ⎟ =
⎝ n ⎠ ⎝n⎠
n m +1
b m +1 n m
A( Sn ) =
∑i
n m +1 i =1
=
⎤
b m +1 ⎡ n m +1
+ Cn ⎥
⎢
m +1 m + 1
n
⎢⎣
⎥⎦
b m +1 b m +1Cn
=
+
m +1
n m +1
A0b ( x m ) = lim A( Sn ) =
n →∞
lim
Cn
n →∞ n m +1
29+1 1024
=
= 102.4
9 +1
10
66. Inscribed:
Consider an isosceles triangle formed by one side
of the polygon and the center of the circle. The
2π
. The length of the base
angle at the center is
n
π
π
is 2r sin . The height is r cos . Thus the area
n
n
π
π 1 2
2π
2
of the triangle is r sin cos = r sin
.
n
n 2
n
2π ⎞ 1
2π
⎛1
An = n ⎜ r 2 sin ⎟ = nr 2 sin
n ⎠ 2
n
⎝2
53 33 98
62.
=
−
=
≈ 32.7
3 3
3
The object traveled about 32.7 m.
A35
63. a.
A02 ( x9 ) =
Circumscribed:
Consider an isosceles triangle formed by one side
of the polygon and the center of the circle. The
2π
. The length of the base
angle at the center is
n
π
is 2r tan . The height is r. Thus the area of the
n
π
triangle is r 2 tan .
n
π
π
⎛
⎞
Bn = n ⎜ r 2 tan ⎟ = nr 2 tan
n⎠
n
⎝
⎛ sin 2π ⎞
1 2
2π
n ⎟
nr sin
= lim πr 2 ⎜
⎜ 2π ⎟
n n→∞
n→∞ 2
⎝ n ⎠
lim An = lim
n →∞
b m +1
m +1
= πr 2
= 0 since Cn is a polynomial in n
of degree m.
lim Bn = lim nr 2 tan
n →∞
n→∞
π
π
πr 2 ⎛ sin n ⎞
⎜
⎟
= lim
n n→∞ cos π ⎜ π ⎟
n⎝ n ⎠
= πr 2
b. Notice that Aab ( x m ) = A0b ( x m ) − A0a ( x m ) .
Thus, using part a, Aab ( x m ) =
b m +1 a m +1
−
.
m +1 m +1
258
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4.2 Concepts Review
1. Riemann sum
2. definite integral;
b
∫a
f ( x )dx
3. Aup − Adown
4. 8 −
1 15
=
2 2
Problem Set 4.2
1. RP = f (2)(2.5 − 1) + f (3)(3.5 − 2.5) + f (4.5)(5 − 3.5) = 4(1.5) + 3(1) + (–2.25)(1.5) = 5.625
2. RP = f(0.5)(0.7 – 0) + f(1.5)(1.7 – 0.7) + f(2)(2.7 – 1.7) + f(3.5)(4 – 2.7)
= 1.25(0.7) + (–0.75)(1) + (–1)(1) + 1.25(1.3) = 0.75
5
3. RP = ∑ f ( xi )Δxi = f (3)(3.75 − 3) + f (4)(4.25 − 3.75) + f (4.75)(5.5 − 4.25) + f (6)(6 − 5.5) + f (6.5)(7 − 6)
i =1
= 2(0.75) + 3(0.5) + 3.75(1.25) + 5(0.5) + 5.5(1) = 15.6875
4
4. RP = ∑ f ( xi )Δxi = f (−2)(−1.3 + 3) + f (−0.5)(0 + 1.3) + f (0)(0.9 − 0) + f (2)(2 − 0.9)
i =1
= 4(1.7) + 3.25(1.3) + 3(0.9) + 2(1.1) = 15.925
8
5. RP = ∑ f ( xi )Δxi = [ f (−1.75) + f (−1.25) + f (−0.75) + f (−0.25) + f (0.25) + f (0.75) + f (1.25) + f (1.75) ] (0.5)
i =1
= [–0.21875 – 0.46875 – 0.46875 – 0.21875 + 0.28125 + 1.03125 + 2.03125 + 3.28125](0.5) = 2.625
6
6. RP = ∑ f ( xi )Δxi = [ f (0.5) + f (1) + f (1.5) + f (2) + f (2.5) + f (3) ] (0.5)
i =1
= [1.5 + 5 + 14.5 + 33 + 63.5 + 109](0.5) = 113.25
7.
8.
3
x3 dx
2
( x + 1)3 dx
∫1
∫0
11. Δx =
2
2i
, xi =
n
n
f ( xi ) = xi + 1 =
n
9.
10.
1
x
2
∫−1 1 + x dx
π
∫0 (sin x)
dx
n
⎡
⎛ 2 ⎞⎤ 2
∑ f ( xi )Δx = ∑ ⎢⎣1 + i ⎜⎝ n ⎟⎠⎥⎦ n
i =1
i =1
n
2
4
2
4 ⎡ n(n + 1) ⎤
∑1 + ∑ i = n (n) + n2 ⎢⎣ 2 ⎥⎦
n i =1 n 2 i =1
⎛ 1⎞
= 2 + 2 ⎜1 + ⎟
⎝ n⎠
2
⎡
⎛ 1 ⎞⎤
⎢ 2 + 2 ⎜1 + n ⎟ ⎥ = 4
∫0 ( x + 1)dx = nlim
→∞ ⎣
⎝
⎠⎦
=
2
2i
+1
n
n
Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12. Δx =
2
2i
, xi =
n
n
14. Δx =
2
2
2
3i ⎞
36i 27i 2
⎛
f ( xi ) = 3 ⎜ −2 + ⎟ + 2 = 14 −
+
n⎠
n
⎝
n2
n
n ⎡
⎛ 36 ⎞ ⎛ 27 ⎞ ⎤ 3
∑ f ( xi )Δx = ∑ ⎢14 − ⎜⎝ n ⎟⎠ i + ⎜⎝ n2 ⎟⎠ i 2 ⎥ n
⎦
i =1
i =1 ⎣
4i
⎛ 2i ⎞
f ( xi ) = ⎜ ⎟ + 1 =
+1
⎝n⎠
n2
n
n ⎡
⎛ 4 ⎞⎤ 2
∑ f ( xi )Δx = ∑ ⎢1 + i 2 ⎜⎝ n2 ⎟⎠⎥ n
⎦
i =1
i =1 ⎣
=
2 n
8 n 2 2
8
+
i = ( n) +
1
∑
∑
3
n i =1 n i =1
n
n3
⎡ n(n + 1)(2n + 1) ⎤
⎢
⎥
6
⎣
⎦
4⎛
3 1 ⎞
= 2+ ⎜2+ +
⎟
3⎝
n n2 ⎠
2
∫0
3
3i
, xi = −2 +
n
n
=
3 n
108 n
81 n 2
−
i
+
14
∑ n 2 ∑ n3 ∑ i
n i =1
i =1
i =1
= 42 −
⎡
4⎛
3 1 ⎞ ⎤ 14
( x 2 + 1) dx = lim ⎢ 2 + ⎜ 2 + +
⎟⎥ =
n n2 ⎠ ⎦ 3
3⎝
n →∞ ⎣
3 1 ⎞
⎛ 1 ⎞ 27 ⎛
= 42 − 54 ⎜1 + ⎟ + ⎜ 2 + +
⎟
n n2 ⎠
⎝ n⎠ 2 ⎝
1
∫−2 (3x
3
3i
13. Δx = , xi = −2 +
n
n
3i ⎞
6i
⎛
f ( xi ) = 2 ⎜ −2 + ⎟ + π = π − 4 +
n⎠
n
⎝
n
6i ⎤ 3
⎡
Δ
=
f
(
x
)
x
∑ i
∑ ⎢⎣ π − 4 + n ⎥⎦ n
i =1
i =1
n
3
18 n
18 ⎡ n(n + 1) ⎤
= ∑ (π − 4) + ∑ i = 3(π − 4) +
⎢
⎥
2
n i =1
n i =1
n2 ⎣ 2 ⎦
⎛ 1⎞
= 3π − 12 + 9 ⎜ 1 + ⎟
⎝ n⎠
⎡
⎛
1 ⎞⎤
⎢3π − 12 + 9 ⎜ 1 + n ⎟ ⎥
∫−2 (2 x + π) dx = nlim
→∞ ⎣
⎝
⎠⎦
= 3π − 3
2
+ 2) dx
⎡
3 1 ⎞⎤
⎛ 1 ⎞ 27 ⎛
= lim ⎢ 42 − 54 ⎜1 + ⎟ + ⎜ 2 + +
⎟ ⎥ = 15
n n2 ⎠ ⎦
n→∞ ⎣
⎝ n⎠ 2 ⎝
n
1
108 ⎡ n(n + 1) ⎤ 81 ⎡ n(n + 1)(2n + 1) ⎤
⎢
⎥+ ⎢
⎥
6
⎦
n 2 ⎣ 2 ⎦ n3 ⎣
5
5i
, xi =
n
n
5i
f ( xi ) = 1 +
n
15. Δx =
n
⎡
⎛ 5 ⎞⎤ 5
f
x
x
(
)
Δ
=
∑ i
∑ ⎢⎣1 + i ⎜⎝ n ⎟⎠⎥⎦ n
i =1
i =1
n
5
25 n
25 ⎡ n(n + 1) ⎤
= ∑1 +
∑ i = 5 + n2 ⎢⎣ 2 ⎥⎦
n i =1 n 2 i =1
n
= 5+
25 ⎛ 1 ⎞
⎜1 + ⎟
2 ⎝ n⎠
5
⎡
⎢5 +
∫0 ( x + 1) dx = nlim
→∞ ⎣
16. Δx =
25 ⎛ 1 ⎞ ⎤ 35
⎜1 + ⎟ =
2 ⎝ n ⎠ ⎥⎦ 2
20
20i
, xi = −10 +
n
n
2
20i ⎞ ⎛
20i ⎞
380i 400i 2
⎛
f ( xi ) = ⎜ −10 +
+
⎟ + ⎜ −10 +
⎟ = 90 −
n ⎠ ⎝
n ⎠
n
⎝
n2
n
n ⎡
20 n
7600 n
8000 n 2
⎛ 380 ⎞ 2 ⎛ 400 ⎞ ⎤ 20
f
x
x
i
i
=
−
i
+
(
)
90
90
Δ
=
−
+
∑ i
∑⎢
∑
∑ n3 ∑ i
⎜ 2 ⎟⎥
⎜
⎟
n i =1
⎝ n ⎠
n 2 i =1
⎝ n ⎠⎦ n
i =1
i =1
i =1 ⎣
= 1800 −
10
∫−10 ( x
2
7600 ⎡ n(n + 1) ⎤ 8000 ⎡ n(n + 1)(2n + 1) ⎤
3 1 ⎞
⎛ 1 ⎞ 4000 ⎛
⎥+ 3 ⎢
⎥ = 1800 − 3800 ⎜ 1 + n ⎟ + 3 ⎜ 2 + n + 2 ⎟
2 ⎢⎣
2
6
⎝
⎠
⎦
⎣
⎦
n
n
n ⎠
⎝
⎡
3 1 ⎞ ⎤ 2000
⎛ 1 ⎞ 4000 ⎛
+ x) dx = lim ⎢1800 − 3800 ⎜1 + ⎟ +
⎜ 2 + + 2 ⎟⎥ =
n n ⎠⎦
3 ⎝
3
n→∞ ⎣
⎝ n⎠
260
Section 4.2
Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21. The area under the curve is equal to the area of a
17.
semi-circle:
A
∫− A
A2 − x 2 dx = 12 π A2 .
5
∫0 f ( x) dx
1
1
27
(1)(2) + 1(2) + 3(2) + (3)(3) =
2
2
2
=
18.
22. The area under the curve is equal to the area of a
triangle:
y
4
2
⎛1⎞
∫−4 f ( x ) dx = 2 ⎜⎝ 2 ⎟⎠ 4 ⋅ 4 = 16
4
2
2
∫0
f ( x) dx =
4 x
1
1
9
(1)( 3) + (1)( 2 ) + (1)( 2 ) =
2
2
2
23. s ( 4 ) = ∫ v ( t ) dt =
4
0
1 ⎛ 4 ⎞ 2
4⎜ ⎟ =
2 ⎝ 60 ⎠ 15
4
1
24. s ( 4 ) = ∫ v ( t ) dt = 4 + 4 ( 9 − 1) = 20
0
2
19.
25. s ( 4 ) = ∫ v ( t ) dt =
4
0
1
2 (1) + 2 (1) = 3
2
4
1
2
26. s ( 4 ) = ∫ v ( t ) dt = π ( 2 ) + 0 = π
0
4
27.
2
∫0
f ( x) dx =
1
1
1 π
(π ⋅12 ) + (1)(1) = +
4
2
2 4
20.
t
s(t)
20
40
40
80
60
120
80
160
100
200
120
240
1
1
f ( x) dx = − (π ⋅ 22 ) − (2)(2) − (2)(4)
4
2
= −π − 8
2
∫−2
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28.
t
s(t)
20
10
40
40
60
90
80
160
100
250
120
360
e.
f.
g.
1
1
(3)(3) + (3)(3) = 9
2
2
∫−3
3
x dx =
3
x x dx =
∫−3
(−3)3 (3)3
+
=0
3
3
2
0
1
2
∫−1 x a x b dx = − ∫−1 x dx + 0∫0 x dx + ∫1
x dx
1
1
= − (1)(1) + 1(1) + (1)(1) = 1
2
2
h.
2
0
1
2
2
2
∫−1 x a xb dx = − ∫−1 x dx + 0∫0 x dx
2
29.
30.
t
s(t)
20
20
40
80
60
160
80
240
100
320
120
400
+ ∫ x 2 dx
1
=−
t
s(t)
20
20
32. a.
1
∫−1
13 ⎛ 23 13 ⎞
+⎜ − ⎟ = 2
3 ⎜⎝ 3 3 ⎟⎠
f ( x) dx = 0 because this is an odd
function.
40
60
60
80
80
60
100
0
120
-100
b.
∫−1
1
g ( x ) dx = 3 + 3 = 6
c.
∫−1
1
f ( x) dx = 3 + 3 = 6
d.
∫−1 [ − g ( x)] dx = −3 + (−3) = −6
e.
∫−1
1
1
xg ( x) dx = 0 because xg(x) is an odd
function.
f.
1
∫−1
f 3 ( x) g ( x) dx = 0 because f 3 ( x) g ( x)
is an odd function.
31. a.
b.
33. RP =
3
∫−3a x b dx = (−3 − 2 − 1 + 0 + 1 + 2)(1) = −3
3
∫−3a xb
2
+(−1) 2 + 0 + 1 + 4](1) = 19
⎡1
⎤
c.
∫−3 ( x − a x b) dx = 6 ⎢⎣ 2 (1)(1) ⎥⎦ = 3
d.
∫−3 ( x − a x b)
3
2
1
dx = 6 ∫ x 2 dx = 6 ⋅
0
(
1 n 2
∑ xi − xi2−1
2 i =1
)
1⎡ 2
( x1 − x02 ) + ( x22 − x12 ) + ( x32 − x22 )
⎣
2
+ " + ( xn2 − xn2−1 ) ⎤
⎦
1 2
= ( xn − x02 )
2
1 2
= (b − a 2 )
2
1
1
lim (b 2 − a 2 ) = (b 2 − a 2 )
2
n →∞ 2
=
dx = [(−3)2 + (−2) 2
3
=
1 n
∑ ( xi + xi −1 )( xi − xi −1 )
2 i =1
13
=2
3
262
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(
)
12
⎡1
⎤
34. Note that xi = ⎢ xi2−1 + xi −1 xi + xi2 ⎥
⎣3
⎦
1/ 2
⎡1
⎤
≥ ⎢ ( xi2−1 + xi2−1 + xi2−1 ⎥
⎣3
⎦
= xi −1 and
1/ 2
⎡1
⎤
xi = ⎢ ( xi2−1 + xi −1 xi + xi2 ⎥
⎣3
⎦
1/ 2
⎡1
⎤
≤ ⎢ ( xi2 + xi2 + xi2 ) ⎥
⎣3
⎦
= xi .
n
R p = ∑ xi2 Δxi
i =1
n
1
= ∑ ( xi2 + xi −1 xi + xi2−1 )( xi − xi −1 )
i =1 3
=
n
1
∑ ( xi3 − xi3−1 )
3 i =1
1
= ⎡ ( x13 − x03 ) + ( x23 − x13 ) + ( x33 − x23 )
3⎣
+ " + ( xn3 − xn3−1 ) ⎤
⎦
1 3
1
= ( xn − x03 ) = (b3 − a3 )
3
3
35. Left:
2
∫0
Right:
( x3 + 1) dx = 5.24
2
∫0
Midpoint:
36. Left:
( x + 1) dx = 6.84
3
2
∫0
37. Left:
1
∫0 cos x dx ≈ 0.8638
Right:
1
∫0 cos x dx ≈ 0.8178
Midpoint:
1
∫0 cos x dx ≈ 0.8418
⎛1⎞
⎜ ⎟ dx ≈ 1.1682
⎝ x⎠
3 ⎛1⎞
Right: ∫ ⎜ ⎟ dx ≈ 1.0349
1 ⎝ x⎠
3 ⎛1⎞
Midpoint: ∫ ⎜ ⎟ dx ≈ 1.0971
1 ⎝ x⎠
38. Left:
3
∫1
39. Partition [0, 1] into n regular intervals, so
1
P = .
n
i 1
If xi = + , f ( xi ) = 1 .
n 2n
n
∑
P →0
lim
i =1
n
1
∑ n =1
n →∞
f ( xi )Δxi = lim
i =1
i 1
If xi = + , f ( xi ) = 0 .
n πn
n
n
i =1
i =1
∑ f ( xi )Δxi = nlim
∑0 = 0
→∞
P →0
lim
Thus f is not integrable on [0, 1].
( x3 + 1) dx = 5.98
1
∫0 tan x dx ≈ 0.5398
Right:
1
∫0 tan x dx ≈ 0.6955
Midpoint:
1
∫0 tan x dx ≈ 0.6146
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4.3 Concepts Review
5. A( x ) =
1
ax 2
x ( ax ) =
2
2
6. A( x ) =
1
1
2
( x − 2)(−1 + x / 2) = ( x − 2 ) , x ≥ 2
2
4
1. 4(4 – 2) = 8; 16(4 – 2) = 32
2. sin 3 x
3.
4
∫1
f ( x) dx ;
5
∫2
x dx
4. 5
Problem Set 4.3
1. A( x) = 2 x
7.
2. A( x) = ax
3. A( x) =
1
2
( x − 1)2 ,
⎧2 x
⎪
⎪⎪2 + ( x − 1)
A( x ) = ⎨3 + 2( x − 2)
⎪5 + ( x − 3)
⎪
⎪⎩etc.
0 ≤ x ≤1
1< x ≤ 2
2< x≤3
3< x≤ 4
x ≥1
8.
4. If 1 ≤ x ≤ 2 , then A( x) =
If 2 ≤ x , then A( x) = x −
1
2
3
2
( x − 1)2 .
⎧ 1 x2
⎪2
⎪ 1 + 1 (3 − x)( x − 1)
⎪2 2
⎪1 + 1 ( x − 2) 2
⎪
A( x ) = ⎨ 2
⎪ 3 + 1 (5 − x)( x − 3)
⎪2 2
⎪2+ 1 ( x − 4)2
⎪ 2
⎪⎩etc.
0 ≤ x ≤1
1< x ≤ 2
2< x≤3
3< x≤ 4
4< x≤5
264
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9.
10.
2
∫1
2
∫0
2 f ( x) dx = 2∫
2
f ( x) dx = 2(3) = 6
1
2 f ( x) dx = 2∫
2
0
f ( x) dx
1
2
= 2 ⎡⎢ ∫ f ( x) dx + ∫ f ( x)dx ⎤⎥ = 2(2 + 3) = 10
1
⎣ 0
⎦
11.
∫0 [ 2 f ( x) + g ( x)] dx = 2∫0
2
2
2
f ( x) dx + ∫ g ( x) dx
0
1
2
2
= 2 ⎡⎢ ∫ f ( x) dx + ∫ f ( x) dx ⎤⎥ + ∫ g ( x) dx
0
1
0
⎣
⎦
= 2(2 + 3) + 4 = 14
12.
1
1
x
x
22. G ′( x) = Dx ⎡⎢ ∫ xt dt ⎤⎥ = Dx ⎡⎢ x ∫ t dt ⎤⎥
⎣1
⎦
⎣ 1
⎦
x
⎡ ⎡ 2⎤ ⎤
⎡ ⎛ x2 − 1 ⎞⎤
t
= Dx ⎢ x ⎢ ⎥ ⎥ = Dx ⎢ x ⎜
⎟⎥
⎜ 2 ⎟⎥
⎢ ⎢2⎥ ⎥
⎢
⎣
⎦
⎝
⎠
⎣
⎦
1
⎣⎢
⎦⎥
⎛ x3 x ⎞ 3
1
= Dx ⎜ − ⎟ = x 2 −
⎜ 2 2⎟ 2
2
⎝
⎠
1
∫0 [2 f ( s) + g ( s)] ds = 2∫0 f (s) ds + ∫0 g (s) ds
= 2(2) + (–1) = 3
13.
π/4
21. G ′( x) = Dx ⎡⎢ ∫
( s − 2) cot(2 s )ds ⎤⎥
x
⎣
⎦
x
⎡
⎤
= Dx ⎢ − ∫
( s − 2) cot(2 s )ds ⎥
⎣ π/4
⎦
= −( x − 2) cot(2 x)
1
2
∫2 [2 f ( s) + 5 g ( s)] ds = −2∫1
2
f ( s ) ds − 5∫ g ( s ) ds
⎡ x2
⎤
23. G ′( x ) = Dx ⎢ ∫ sin t dt ⎥ = 2 x sin( x 2 )
1
⎢⎣
⎥⎦
1
2
1
= −2(3) − 5 ⎡⎢ ∫ g ( s ) ds − ∫ g ( s ) ds ⎤⎥
0
⎣ 0
⎦
= –6 – 5[4 + 1] = –31
⎡ x2 + x
24. G ′( x ) = Dx ⎢ ∫
1
⎣⎢
⎤
2 z + sin z dz ⎥
⎦⎥
= (2 x + 1) 2( x 2 + x) + sin( x 2 + x)
14.
15.
∫1 [3 f ( x) + 2 g ( x)] dx = 0
1
25.
∫0 [3 f (t ) + 2 g (t )] dt
2
t2
x
G ( x) = ∫ 2
−x
1+ t2
t2
0
=∫ 2
−x
1
2
2
= 3 ⎡⎢ ∫ f (t ) dt + ∫ f (t ) dt ⎤⎥ + 2 ∫ g (t ) dt
1
0
⎣ 0
⎦
= 3(2 + 3) + 2(4) = 23
= −∫
1+ t2
2
∫0
2
+π∫ dt
=
0
= 3 (2 + 3) + 2(4) + 2π = 5 3 + 4 2 + 2π
(
)
x
19. G ′( x) = Dx ⎡⎢ ∫ 2t 2 + t dt ⎤⎥ = 2 x 2 + x
0
⎣
⎦
x
20. G ′( x) = Dx ⎡⎢ ∫ cos3 (2t ) tan(t ) dt ⎤⎥
⎣1
⎦
= cos3 (2 x) tan( x)
0
1+ t2
1+ t2
dt
x
t2
0
1+ t2
dt + ∫
dt
2
1
2
2
= 3 ⎡⎢ ∫ f (t ) dt + ∫ f (t ) dt ⎤⎥ + 2 ∫ g (t ) dt
0
1
0
⎣
⎦
1
x
18. G ′( x) = Dx ⎡⎢ ∫ 2t dt ⎤⎥ = Dx ⎡⎢ − ∫ 2t dt ⎤⎥ = −2 x
x
1
⎣
⎦
⎣
⎦
t2
− x2 )
(
x2
G '( x) = −
−2 x ) +
(
2
1 + x2
1 + ( − x2 )
⎡ 3 f (t ) + 2 g (t ) + π ⎤ dt
⎣
⎦
x
17. G ′( x) = Dx ⎡⎢ ∫ 2t dt ⎤⎥ = 2 x
⎣1
⎦
x
dt + ∫
t2
− x2
0
16.
dt
2 x5
1 + x4
+
x2
1 + x2
sin x 5 ⎤
26. G ( x) = Dx ⎡⎢ ∫
t dt ⎥
cos
x
⎣
⎦
sin x 5
0
⎡
= Dx ⎢ ∫
t dt + ∫
t 5 dt ⎤⎥
0
cos
x
⎣
⎦
sin x 5
cos x 5 ⎤
⎡
= Dx ⎢ ∫
t dt − ∫
t dt ⎥
0
⎣ 0
⎦
= sin 5 x cos x + cos5 x sin x
27.
f ′( x) =
x
1 + x2
; f ′′ ( x ) =
1
( x + 1)
2
3/ 2
So, f(x) is increasing on [0, ∞) and concave up
on (0, ∞ ).
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28.
f ′( x) =
1+ x
1+ x
35.
2
(1 + x ) − (1 + x ) 2 x = − x + 2 x − 1
f ′′ ( x ) =
( x + 1)
( x + 1)
2
2
2
2
2
2
So, f(x) is increasing on [0, ∞ ) and concave up on
( 0, −1 + 2 ) .
29.
f ′ ( x ) = cos x; f ′′ ( x ) = − sin x
4
⎡ π ⎤ ⎡ 3π 5π ⎤
So, f(x) is increasing on ⎢0, ⎥ , ⎢ , ⎥ ,... and
⎣ 2⎦ ⎣ 2 2 ⎦
concave up on (π , 2π ) , ( 3π , 4π ) ,... .
30.
∫0
2
4
0
2
f ( x) dx = ∫ (2 − x)dx + ∫ ( x − 2) dx
= 2+2 = 4
36.
f ′ ( x ) = x + sin x; f ′′ ( x ) = 1 + cos x
So, f(x) is increasing on ( 0, ∞ ) and concave up
on ( 0, ∞ ) .
31.
1
1
; f ′′ ( x ) = − 2
x
x
So, f(x) is increasing on (0, ∞) and never
concave up.
f ′( x) =
32. f(x) is increasing on x ≥ 0 and concave up on
( 0,1) , ( 2,3) ,...
∫0 ( 3 + x − 3 ) dx
3
4
= ∫ ( 3 + x − 3 ) dx + ∫ ( 3 + x − 3 ) dx
0
3
4
3
4
0
3
= ∫ ( 6 − x ) dx + ∫ x dx =
37. a.
33.
27 7
+ = 17
2 2
Local minima at 0, ≈ 3.8, ≈ 5.8, ≈ 7.9,
≈ 9.9;
local maxima at ≈ 3.1, ≈ 5, ≈ 7.1, ≈ 9, 10
b. Absolute minimum at 0, absolute maximum
at ≈ 9
c.
≈ (0.7, 1.5), (2.5, 3.5), (4.5, 5.5), (6.5, 7.5),
(8.5, 9.5)
d.
4
∫0
2
4
0
2
f ( x) dx = ∫ 2 dx + ∫ x dx = 4 + 6 = 10
34.
38. a.
Local minima at 0, ≈ 1.8, ≈ 3.8, ≈ 5.8;
local maxima at ≈ 1, ≈ 2.9, ≈ 5.2, ≈ 10
b. Absolute minimum at 0, absolute maximum
at 10
c.
4
∫0
1
2
4
0
1
2
f ( x) dx = ∫ dx + ∫ x dx + ∫ (4 − x) dx
(0.5, 1.5), (2.2, 3.2), (4.2,5.2), (6.2,7.2),
(8.2, 9.2)
= 1 + 1.5 + 2.0 = 4.5
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d.
39.
a.
f.
0
0
b.
(
)
F (0) = ∫ t 4 + 1 dt = 0
41.
1 ≤ 1 + x4 ≤ 1 + x4 .
y = F ( x)
1
)
y = 15 x5 + x + C
c.
0=
42.
1
+0+C
40.
a.
1
4
∫0 ( 4 + x
1
)
1
G (0) = ∫ sin t dt = 0
G (2π ) = ∫
0
1
(
)
4 + x 2 dx ≤ ∫ 4 + x 2 dx
0
4 + x 2 dx ≤
= 3 + 65 =
43.
21
5
5 ≤ f ( x) ≤ 69 so
( 5 + x3 ) dx ≤ 4 ⋅ 69
20 ≤ ∫ ( 5 + x3 ) dx ≤ 276
0
4⋅5 ≤ ∫
0
2π
6
5
) dx = ∫01( 3 + 1 + x4 ) dx
1
1
= ∫ 3 dx + ∫ (1 + x 4 ) dx
0
0
4
x
G ( x) = ∫ sin t dt
1 + x 4 dx ≤
21
0
5
Here, we have used the result from problem 39:
1
6
+ 1 dx = F (1) = 15 + 1 = .
5
5
0
0
b.
1
2≤∫
Thus y = F ( x) = 15 x5 + x
∫0 ( x
1
On the interval [0,1], 2 ≤ 4 + x 4 ≤ 4 + x 4 .
Thus
∫0 2 dx ≤ ∫0
C=0
d.
0
0
Now apply the initial condition y (0) = 0 :
1 05
5
1
1 + x 4 dx ≤ ∫ (1 + x 4 ) dx
By problem 39d, 1 ≤ ∫
dy = x + 1 dx
4
1
∫0 dx ≤ ∫0
dy
= F '( x) = x 4 + 1
dx
(
t ≤ t . Since 1 + x 4 ≥ 1 for all x,
For t ≥ 1 ,
4
0
4
sin t dt = 0
Let y = G ( x) . Then
dy
= G '( x) = sin x .
dx
dy = sin x dx
y = − cos x + C
c.
d.
e.
Apply the initial condition
0 = y (0) = − cos 0 + C . Thus, C = 1 ,
and hence y = G ( x) = 1 − cos x .
π
∫0 sin x dx = G (π ) = 1 − cos π = 2
44.
On [2,4], 85 ≤ ( x + 6 ) ≤ 105 . Thus,
5
2 ⋅ 85 ≤ ∫
4
2
( x + 6 )5 dx ≤ 2 ⋅105
65,536 ≤ ∫
4
2
( x + 6 )5 dx ≤ 200, 000
G attains the maximum of 2 when
x = π ,3π .
G attains the minimum of 0 when
x = 0, 2π , 4π
Inflection points of G occur at
π 3π 5π 7π
x= , , ,
2 2 2 2
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45.
On [1,5],
2
2
2
3+ ≤ 3+ ≤ 3+
5
1
x
5⎛
2⎞
⎛ 17 ⎞
4 ⎜ ⎟ ≤ ∫ ⎜ 3 + ⎟ dx ≤ 4 ⋅ 5
1
5
x⎠
⎝ ⎠
⎝
5⎛
68
2⎞
≤ ∫ ⎜ 3 + ⎟ dx ≤ 20
1
5
x⎠
⎝
48.
On [0.2,0.4],
0.002 + 0.0001cos 2 0.4 ≤ 0.002 + 0.0001cos 2 x
≤ 0.002 + 0.0001cos 2 0.2
(
0.2 0.002 + 0.0001cos 2 0.4
)
( 0.002 + 0.0001cos2 x ) dx
≤ 0.2 ( 0.002 + 0.0001cos 2 0.2 )
≤∫
0.4
0.2
Thus,
0.000417 ≤ ∫
0.4
0.2
( 0.002 + 0.0001cos2 x ) dx
≤ 0.000419
46.
On [10, 20],
5
5
1 ⎞
1⎞
⎛
⎛ 1⎞
⎛
⎜1 + ⎟ ≤ ⎜ 1 + ⎟ ≤ ⎜1 + ⎟
20
x
10
⎝
⎠
⎝
⎠
⎝
⎠
5
5
5
20 ⎛
1⎞
⎛ 21 ⎞
⎛ 11 ⎞
10 ⎜ ⎟ ≤ ∫ ⎜ 1 + ⎟ dx ≤ 10 ⎜ ⎟
10
⎝ 20 ⎠
⎝ x⎠
⎝ 10 ⎠
5
5
20 ⎛
4, 084,101
1⎞
161, 051
≤ ∫ ⎜ 1 + ⎟ dx ≤
10
320, 000
10, 000
⎝ x⎠
20 ⎛
49.
47.
1
∫
x 1+ t
51.
∫1
8π
π
(5 + 201 sin 2 x ) dx ≤ 101
5
dt . Then
dt
2+t
1 1+ t
1 ⎡ x 1+ t
⎤
= lim
dt − ∫
dt
⎢∫
0 2 + t ⎥⎦
x →1 x − 1 ⎣ 0 2 + t
F ( x) − F (1)
= lim
x −1
1
→
x
1+1 2
= F '(1) =
=
2 +1 3
( 4π ) (5) ≤ ∫4π ( 5 + 201 sin 2 x ) dx ≤ ( 4π ) ( 5 + 201 )
4π
2+t
50.
1 sin 2 x ≤ 5 + 1
5 ≤ 5 + 20
20
8π
0
lim
On [ 4π ,8π ]
20π ≤ ∫
x 1+ t
1 x 1+ t
F ( x) − F (0)
dt = lim
∫
x−0
x →0 x 0 2 + t
x →0
1+ 0 1
= F '(0) =
=
2+0 2
5
1⎞
1 + ⎟ dx ≤ 16.1051
⎜
10 ⎝
x⎠
12.7628 ≤ ∫
Let F ( x) = ∫
lim
x →1 x − 1 1
x
f (t ) dt = 2 x − 2
Differentiate both sides with respect to x:
d x
d
f (t ) dt = ( 2 x − 2 )
dx ∫1
dx
f ( x) = 2
If such a function exists, it must satisfy
f ( x) = 2 , but both sides of the first equality
may differ by a constant yet still have equal
derivatives. When x = 1 the left side is
1
∫1 f (t ) dt = 0 and the right side is 2 ⋅1 − 2 = 0 .
Thus the function f ( x) = 2 satisfies
x
∫1
f (t ) dt = 2 x − 2 .
268
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52.
x
∫0
f (t ) dt = x 2
59.
Differentiate both sides with respect to x:
d x
d 2
f (t ) dt =
x
∫
0
dx
dx
f ( x) = 2 x
53.
x2
∫0
f (t ) dt =
60.
Differentiate both sides with respect to x:
( )
d x2
d 1 3
f (t ) dt =
x
∫
0
dx
dx 3
( ) ( 2x) = x
x
f ( x2 ) =
2
f x
2
2
f ( x) =
No such function exists. When x = 0 the left
side is 0, whereas the right side is 1
55.
True; by Theorem B (Comparison Property)
56.
False. a = –1, b = 2, f(x) = x is a
counterexample.
57.
False. a = –1, b = 1, f(x) = x is a
counterexample.
62.
b
False. a = 0, b = 1, f(x) = 0, g(x) = –1 is a
counterexample.
⎧⎪2 + ( t − 2 ) , t ≤ 2
v (t ) = ⎨
⎪⎩ 2 − ( t − 2 ) , t > 2
t≤2
⎧ t,
=⎨
−
>2
4
,
t
t
⎩
s ( t ) = ∫ v ( u ) du
t
⎧ t
0≤t ≤2
⎪ ∫0 u du ,
=⎨
t
2
⎪ u du + ( 4 − u ) du, t > 2
∫2
⎩ ∫0
⎧t2
0≤t≤2
⎪ ,
⎪2
=⎨
2
⎪2 + ⎡⎢ 4t − t ⎤⎥ , t > 2
⎪ ⎢
2 ⎥⎦
⎩ ⎣
x
2
False; A counterexample is f ( x ) = 0 for all x,
except f (1) = 1 . Thus,
b
0
54.
58.
∫ a f ( x)dx − ∫ a g ( x)dx
= ∫ ba [ f ( x) − g ( x )]dx
61.
1 x3
3
True.
⎧
t2
,
⎪
⎪
2
=⎨
t2
⎪
⎪⎩−4 + 4t − 2
0≤t≤2
t>2
t2
− 4t + 4 = 0; t = 4 + 2 2 ≈ 6.83
2
∫0 f ( x ) dx = 0 , but f is
2
not identically zero.
⎧ t
⎪ ∫ 5 du,
0 ≤ t ≤ 100
⎪ 0
⎪⎪ 100
t ⎛
u ⎞
du
100 < t ≤ 700
a. s ( t ) = ⎨ ∫ 5 du + ∫ ⎜ 6 −
0
100 ⎝
100 ⎟⎠
⎪
⎪ 100
700 ⎛
t
u ⎞
du + ∫ ( −1) du, t > 700
⎪ ∫ 5 du + ∫ ⎜ 6 −
⎟
0
100
700
100 ⎠
⎝
⎩⎪
⎧
⎪
⎪5t ,
0 ≤ t ≤ 100
⎪
t
⎪
⎡
u2 ⎤
⎪
= ⎨500 + ⎢6u −
100 < t ≤ 700
⎥
200 ⎦⎥
⎪
⎣⎢
100
⎪
700
2
⎪
⎡
u ⎤
⎪500 + ⎢6u −
⎥ − ( t − 700 ) t > 700
200 ⎦⎥
⎪⎩
⎣⎢
100
0 ≤ t ≤ 100
⎧5t ,
⎪
2
t
⎪
= ⎨−50 + 6t −
, 100 < t ≤ 700
200
⎪
⎪2400 − t ,
t > 700
⎩
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b. v ( t ) > 0 for 0 ≤ t < 600 and v ( t ) < 0 for
t > 600 . So, t = 600 is the point at which
the object is farthest to the right of the origin.
At t = 600 , s ( t ) = 1750 .
c. s ( t ) = 0 = 2400 − t ; t = 2400
− f ( x ) ≤ f ( x) ≤ f ( x) , so
63.
b
∫a −
b
∫a
b
f ( x) dx ≤ ∫ f ( x) dx ⇒
∫−1 (3x
4.
∫1
5.
3
⎡ 1⎤
⎛ 1⎞
∫1 w2 dw = ⎢⎣− w ⎥⎦1 = ⎜⎝ − 4 ⎟⎠ − (−1) = 4
6.
⎡ 1⎤
8
⎛ 1⎞
∫1 t 3 dt = ⎢⎣ − t 2 ⎥⎦ = ⎜⎝ − 9 ⎟⎠ − (−1) = 9
1
b
f ( x ) dx ≥ ∫ f ( x) dx,
a
we can conclude that
b
∫a
f ( x) dx ≤ ∫
b
f ( x) dx
a
If x > a ,
64.
3.
2
2
2
(4 x3 + 7) dx = ⎡ x 4 + 7 x ⎤
⎣
⎦1
= (16 + 14) – (1 + 7) = 22
x
∫a
f ′( x ) dx ≤ M ( x − a) by the
7.
∫x
f ( x) dx = − ∫
x
a
f ′( x) dx ≥ − M ( x − a ) by
8.
4
1
4
3
2
3
4
4
16
⎡2
⎤
⎛2 ⎞
t dt = ⎢ t 3 / 2 ⎥ = ⎜ ⋅ 8 ⎟ − 0 =
3
⎣3
⎦0 ⎝ 3 ⎠
8 3
1
⎡3
⎤
⎛3
⎞ ⎛ 3 ⎞ 45
w dw = ⎢ w4 / 3 ⎥ = ⎜ ⋅16 ⎟ − ⎜ ⋅1⎟ =
⎝4
⎠ ⎝4 ⎠ 4
⎣4
⎦1
∫0
Boundedness Property. If x < a ,
a
2
− 2 x + 3) dx = ⎡ x3 − x 2 + 3 x ⎤
⎣
⎦ −1
= (8 – 4 + 6) – (–1 –1 – 3) = 15
2
b
f ( x ) dx ≥ − ∫ f ( x) dx
and combining this with
b
⎡ x5 ⎤
32 1 33
4
∫−1 x dx = ⎢⎢ 5 ⎥⎥ = 5 + 5 = 5
⎣ ⎦ −1
2
a
a
∫a
2
2.
∫
8
the Boundedness Property. Thus
x
∫a
f ′( x) dx ≤ M x − a .
From Problem 63,
x
∫a
x
∫a
f ′( x) dx ≥
9.
x
∫a
f ′( x) dx .
f ′( x ) dx = f ( x) − f (a) ≥ f ( x) − f (a)
Therefore, f ( x) − f (a) ≤ M x − a or
f ( x) ≤ f (a ) + M x − a .
10.
−2
⎡ y3
⎛ 2 1 ⎞
1 ⎤
∫−4 ⎜⎜ y + y3 ⎟⎟ dy = ⎢⎢ 3 − 2 y 2 ⎥⎥
⎝
⎠
⎣
⎦ −4
⎛ 8 1 ⎞ ⎛ 64 1 ⎞ 1783
= ⎜− − ⎟−⎜− − ⎟ =
96
⎝ 3 8 ⎠ ⎝ 3 32 ⎠
−2
4
∫1
s4 − 8
s2
4
4
ds = ∫ ( s − 8s
2
−2
1
⎡ s3 8 ⎤
) ds = ⎢ + ⎥
⎢⎣ 3 s ⎥⎦1
⎛ 64
⎞ ⎛1
⎞
= ⎜ + 2 ⎟ − ⎜ + 8 ⎟ = 15
⎝ 3
⎠ ⎝3 ⎠
4.4 Concepts Review
π/2
cos x dx = [sin x ]0
π/2
2sin t dt = [ −2 cos t ]π / 6 = 0 + 3 = 3
11.
∫0
12.
∫π / 6
π/2
=1–0=1
1. antiderivative; F(b) – F(a)
2. F(b) – F(a)
3. F (d ) − F (c )
13.
4.
∫
2
1
1 4
u du
3
Problem Set 4.4
14.
2
⎡ x4 ⎤
1. ∫ x dx = ⎢ ⎥ = 4 − 0 = 4
0
⎣⎢ 4 ⎦⎥ 0
2
3
π/2
1
⎡2 5
⎤
4
2
3
∫0 (2 x − 3x + 5) dx = ⎢⎣ 5 x − x + 5 x ⎥⎦0
22
⎛2
⎞
= ⎜ −1+ 5⎟ − 0 =
5
5
⎝
⎠
1
1
⎡3 7/3 3 4/3⎤
4/3
1/ 3
∫0 ( x − 2 x ) dx = ⎢⎣ 7 x − 2 x ⎥⎦ 0
15
⎛3 3⎞
= ⎜ − ⎟−0 = −
7
2
14
⎝
⎠
1
270
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15. u = 3x + 2, du = 3 dx
1
2 3/ 2
2
3/ 2
∫ u ⋅ 3 du = 9 u + C = 9 (3x + 2) + C
16. u = 2x – 4, du = 2 dx
3 4/3
3
1/ 3 1
4/3
∫ u ⋅ 2 du = 8 u + C = 8 (2 x − 4) + C
17. u = 3x + 2, du = 3 dx
1
1
1
∫ cos(u) ⋅ 3 du = 3 sin u + C = 3 sin(3x + 2) + C
18. u = 2x – 4, du = 2 dx
1
1
∫ sin u ⋅ 2 du = − 2 cos u + C
1
= − cos(2 x − 4) + C
2
19. u = 6x – 7, du = 6dx
1
1
∫ sin u ⋅ 6 du = − 6 cos u + C
1
= − cos(6 x − 7) + C
6
20. u = πv − 7, du = π dv
1
1
1
∫ cos u ⋅ π du = π sin u + C = π sin(πv − 7) + C
25. u = x 2 + 4, du = 2 x dx
1
1
= − cos( x 2 + 4) + C
2
26. u = x3 + 5, du = 3 x 2 dx
1
27. u = x 2 + 4, du =
22. u = x3 + 5, du = 3 x 2 dx
1
1 10
1 3
10
∫ u ⋅ 3 du = 30 u + C = 30 ( x + 5) + C
9
1
x
x +4
2
3
+ 5) + C
dx
∫ sin u du = − cos u + C = − cos
x2 + 4 + C
2z
3
28. u = z 2 + 3, du =
dz
2
3 2
⎛
⎞
3⎜ z + 3 ⎟
⎝
⎠
3
3
3
3 2
∫ cos u ⋅ 2 du = 2 sin u + C = 2 sin z + 3 + C
29. u = ( x3 + 5)9 ,
du = 9( x3 + 5)8 (3x 2 )dx = 27 x 2 ( x3 + 5)8 dx
1
1
∫ cos u ⋅ 27 du = 27 sin u + C
21. u = x 2 + 4, du = 2 x dx
1
1
1
u ⋅ du = u 3 / 2 + C = ( x 2 + 4)3 / 2 + C
2
3
3
1
∫ cos u ⋅ 3 du = 3 sin u + C = 3 sin( x
=
∫
1
∫ sin(u ) ⋅ 2 du = − 2 cos u + C
1
sin ⎡( x3 + 5)9 ⎤ + C
⎣
⎦
27
30. u = (7 x 7 + π)9 , du = 441x 6 (7 x7 + π)8 dx
1
1
∫ sin u ⋅ 441 du = − 441 cos u + C
=−
1
cos(7 x 7 + π)9 + C
441
31. u = sin( x 2 + 4), du = 2 x cos( x 2 + 4) dx
23. u = x + 3, du = 2 x dx
2
7
−12 / 7 1
⋅ du = − u −5 / 7 + C
∫u
2
10
7 2
= − ( x + 3)−5 / 7 + C
10
1
1
u ⋅ du = u 3 / 2 + C
2
3
3/ 2
1⎡
= sin( x 2 + 4) ⎤
+C
⎣
⎦
3
∫
32. u = cos(3 x7 + 9)
24. u = 3 v + π, du = 2 3v dv
2
∫u
=
7/8
4
15
⋅
1
2 3
(
3
du =
4
15 3
3 v2 + π
)
15 / 8
u15 / 8 + C
+C
du = −21x 6 sin(3 x7 + 9) dx
1
⎛ 1 ⎞
u ⋅ ⎜ − ⎟ du = − u 4 / 3 + C
21
28
⎝
⎠
4/3
1 ⎡
=−
+C
cos(3x 7 + 9) ⎤
⎦
28 ⎣
∫
3
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271
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
33.
u = cos( x3 + 5), du = −3 x 2 sin( x3 + 5) dx
= 2∫
u = tan( x −3 + 1) , du = −3 x −4 sec2 ( x −3 + 1) dx
⎡4
⎤ ⎡4
⎤
= ⎢ (125) ⎥ − ⎢ (125) ⎥ = 0
⎣3
⎦ ⎣3
⎦
3
∫−3
9
7 + 2t 2 (8t ) dt = 2 ∫
25
25
5
⎛ 1⎞
u ⋅ ⎜ − ⎟ du = − u 6 / 5 + C
3
18
⎝
⎠
6/5
5 ⎡
=−
+C
tan( x −3 + 1) ⎤
⎣
⎦
18
∫
35.
u = 7 + 2t 2 , du = 4t dt
1
⎛ 1⎞
⋅ ⎜ − ⎟ du = − u10 + C
3
30
⎝
⎠
1
= − cos10 ( x3 + 5) + C
30
∫u
34.
41.
42.
x2 + 1
∫1
=
x3 + 3 x
43.
1 16 −1/ 2
⎡2
⎤
u
du = ⎢ u1/ 2 ⎥
∫
4
3
⎣3
⎦4
u = cos x, du = − sin x dx
π/2
∫0
u = x3 + 1, du = 3 x 2 dx
∫−1
x3 + 1 (3x 2 ) dx = ∫
1
0
u = t + 2, du = dt
1
5 −2
u du
1
∫−1 (t + 2)2 dt = ∫
⎡ 1⎤
= ⎢− ⎥
⎣ u ⎦1
π/2
=
39.
9
1 π /2 2
1 −1
sin 3 x ( 3cos 3 x ) dx = ∫ u 2 du
∫
3 0
3 0
⎡ u3 ⎤
1
⎛ 1⎞
= ⎢ ⎥ = ⎜− ⎟−0 = −
9
⎝ 9⎠
⎢⎣ 9 ⎥⎦ 0
9
45.
1
=∫
u = 3x + 1, du = 3 dx
8
1 8
1 25
∫5 3x + 1 dx = 3 ∫5 3x + 1 ⋅ 3dx = 3 ∫16 u du
11
02
2
+ 2 x)2 dx
( x 2 + 2 x)2 2( x + 1) dx
3
=
25
u = 2x + 2, du = 2 dx
7
1
1 7
2
∫1 2 x + 2 dx = 2 ∫1 2 x + 2 dx
16
1 16
= ∫ u −1/ 2 du = ⎡⎣ u ⎤⎦ = 4 − 2 = 2
4
2 4
u = x 2 + 2 x, du = (2 x + 2) dx = 2( x + 1) dx
∫0 ( x + 1)( x
⎡2
⎤
⎡2
⎤ ⎡2
⎤ 122
= ⎢ u 3 / 2 ⎥ = ⎢ (125) ⎥ − ⎢ (64) ⎥ =
9
9
9
9
⎣
⎦16 ⎣
⎦ ⎣
⎦
40.
sin 2 3 x cos 3x dx
−1
⎡2
⎤
udu = ⎢ u 3 / 2 ⎥
1
3
⎣
⎦1
2
2
52
⎡
⎤ ⎡
⎤
= ⎢ (27) ⎥ − ⎢ (1) ⎥ =
⎣3
⎦ ⎣3 ⎦ 3
y − 1 dy = ∫
u = sin 3 x, du = 3cos 3 x dx
∫0
u = y – 1, du = dy
10
cos 2 x ( − sin x ) dx
0
4
⎡ 1⎤
= ⎢ − ⎥ − [ −1] =
5
⎣ 5⎦
∫2
π /2
0
0 2
u du
1
44.
5
cos 2 x sin x dx = − ∫
⎡ u3 ⎤
= −∫
= ⎢− ⎥
⎢⎣ 3 ⎥⎦1
⎛ 1⎞ 1
= 0−⎜− ⎟ =
⎝ 3⎠ 3
1
⎡2
⎤
udu = ⎢ u 3 / 2 ⎥
3
⎣
⎦0
⎛2
⎞ ⎛2 ⎞ 2
= ⎜ ⋅13 / 2 ⎟ − ⎜ ⋅ 0 ⎟ =
⎝3
⎠ ⎝3 ⎠ 3
3
1 3 3x2 + 3
dx
3 ∫1 x3 + 3 x
⎛2 ⎞ ⎛2 ⎞ 8
= ⎜ ⋅6⎟ − ⎜ ⋅ 2⎟ =
⎝3 ⎠ ⎝3 ⎠ 3
2047
⎡1
⎤ ⎡1
⎤
= ⎢ (2)11 ⎥ − ⎢ (1)11 ⎥ =
11
11
11
⎣
⎦ ⎣
⎦
0
dx =
36
u = x 2 + 1, du = 2 x dx
2
38.
⎡4
⎤
u du = ⎢ u 3 / 2 ⎥
3
⎣
⎦ 25
u = x3 + 3x, du = (3 x 2 + 3) dx
3
⎡ u11 ⎤
2 10
2
10
∫0 ( x + 1) (2 x)dx = ∫1 u du = ⎢⎢ 11 ⎥⎥
⎣
⎦1
37.
7 + 2t 2 ⋅ ( 4t ) dt
−3
25
5
1
36.
3
⎡ u3 ⎤
1 3 2
9
u du = ⎢ ⎥ =
∫
2 0
6
2
⎣⎢ ⎦⎥
0
46.
u = x − 1, du =
4
∫1
( x − 1)3
x
1
2 x
dx = 2 ∫
4
1
dx
( x − 1)3
2 x
dx
1
1 3
u du
0
= 2∫
⎡u4 ⎤
1
= 2⎢ ⎥ =
⎢⎣ 4 ⎥⎦ 0 2
272
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Instructor’s Resource Manual
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47.
u = sin θ , du = cos θ dθ
1/ 2 3
u du
0
∫
48.
3/2
1
50.
⎡u
=⎢ ⎥
⎣⎢ 4 ⎦⎥ 0
=
56.
u = π sin θ , du = π cos θ dθ
1 π
1
π
cos u du = [ sin u ]−π = 0
π ∫−π
π
1
1
−0 =
64
64
u = cos θ , du = − sin θ dθ
−∫
49.
4 ⎤1/ 2
57.
1 cos1 3
1 ⎡u4 ⎤
− ∫
u du = − ⎢ ⎥
2 1
2 ⎣⎢ 4 ⎦⎥
3/2
1
1⎛4 ⎞ 1
u −3 du = ⎡u −2 ⎤
= ⎜ − 1⎟ =
⎣
⎦
1
2
2⎝3 ⎠ 6
u = 3 x − 3, du = 3dx
1 0
1
1
0
cos u du = [sin u ]−3 = (0 − sin(−3))
3 ∫−3
3
3
sin 3
=
3
u = 2πx, du = 2πdx
1 π
1
1
π
sin u du = − [ cos u ]0 = − (−1 − 1)
2π ∫0
2π
2π
1
=
π
u = cos( x 2 ), du = −2 x sin( x 2 )dx
cos1
=−
1
=
58.
cos 4 1 1
+
8
8
1 − cos 1
8
4
u = sin( x3 ), du = 3x 2 cos( x3 )dx
3
1 sin( π3 / 8) 2
1 ⎡ 3 ⎤ sin( π / 8)
3 / 8) u du = ⎣u ⎦
∫
sin(
−
π
− sin( π3 / 8)
3
9
3
2sin 3 ⎛⎜ π8 ⎞⎟
⎝ ⎠
=
9
59.
a. Between 0 and 3, f ( x) > 0 . Thus,
3
51.
∫0 f ( x) dx > 0 .
u = πx 2 , du = 2πx dx
1 π
1
1
π
sin u du = − [ cos u ]0 = − (−1 − 1)
2π ∫0
2π
2π
1
=
π
52.
b. Since f is an antiderivative of f ' ,
3
∫0 f '( x) dx = f (3) − f (0)
= 0 − 2 = −2 < 0
u = 2 x5 , du = 10 x 4 dx
c.
1 2π5
1
2 π5
cos u du = [sin u ]0
∫
10 0
10
1
1
= (sin(2π5 ) − 0) = sin(2π5 )
10
10
53.
54.
u = 2 x, du = 2dx
1 π/ 2
1 π/2
cos u du + ∫
sin u du
2 ∫0
2 0
1
π/2 1
π/2
= [sin u ]0 − [ cos u ]0
2
2
1
1
= (1 − 0) − (0 − 1) = 1
2
2
= −1 − 0 = −1 < 0
d. Since f is concave down at 0, f ''(0) < 0 .
3
∫0 f '''( x) dx = f ''(3) − f ''(0)
= 0 − (negative number) > 0
60.
a. On [ 0, 4] , f ( x) > 0 . Thus,
u = cos x, du = − sin x dx
4
∫0
f ( x) dx > 0 .
b. Since f is an antiderivative of f ' ,
4
∫0
u = 3 x, du = 3dx; v = 5 x, dv = 5dx
1 3π / 2
1 5π / 2
cos u du + ∫
sin v dv
∫
3
/
2
−
π
3
5 −5π / 2
1
1
3π / 2
5π / 2
= [sin u ]−3π / 2 − [ cos v ]−5π / 2
3
5
1
1
2
= [(−1) − 1] − [0 − 0] = −
3
5
3
55.
3
∫0 f ''( x) dx = f '(3) − f '(0)
f '( x) dx = f (4) − f (0)
= 1 − 2 = −1 < 0
c.
4
∫0
f ''( x) dx = f '(4) − f '(0)
=
d.
4
∫0
1
9
− (−2) = > 0
4
4
f '''( x) dx = f ''(4) − f ''(0)
= ( negative ) − ( positive ) < 0
− ∫ sin u du = [ cos u ] = 1 − cos1
0
1
0
1
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61.
∫ V ′ (t ) = ∫
1 2
t +C
2
V ( 0 ) = C = 0 since no water has leaked out at
V (t ) =
( 20 − t ) dt = 20t −
67.
a.
(a )(a n ) + An + Bn = (b)(b n ) or
Bn + An = b n +1 − a n +1 . Thus
b n
bn
x dx + n n
a
a
∫
1
Time to drain: 20t − t 2 = 200; t = 20 hours.
2
b.
1
⎡
1
t2 ⎤
219
V (1) − V ( 0 ) = ∫ V ′ ( t ) dt = ⎢t −
⎥ =
0
220
220
⎣⎢
⎦⎥ 0
y dy = b n +1 − a n +1
n
b
b
n
∫a x dx + ∫an n y dy
n
b
⎛ b n +1 a n +1 ⎞ ⎛ n n +1
n n +1 ⎞
b −
a ⎟
=⎜
−
⎟+
⎜ n + 1 n + 1 ⎟ ⎜⎝ n + 1
n +1
⎠
⎝
⎠
2
T⎛
t ⎞
T
dt = T −
55 = V (T ) − V ( 0 ) = ∫ ⎜ 1 −
⎟
0 ⎝
110 ⎠
220
T ≈ 110 hrs
=
Use a midpoint Riemann sum with n = 12
partitions.
(n + 1)b n +1 − (n + 1)a n +1
= b n +1 − a n +1
n +1
b
c. Bn = ∫ x n dx =
a
12
V = ∑ f ( xi ) Δxi
=
i =1
≈ 1(5.4 + 6.3 + 6.4 + 6.5 + 6.9 + 7.5 + 8.4
64.
∫
b
⎡ x n +1 ⎤
⎡ n ( n +1) / n ⎤
y
=⎢
⎥ +⎢
⎥ n
⎦a
⎢⎣ n + 1 ⎥⎦ a ⎣ n + 1
10 ⎛
t ⎞
201
V (10 ) − V ( 9 ) = ∫ ⎜ 1 −
dt =
⎟
9 ⎝
110 ⎠
220
63.
n
b
x n dx = Bn ; ∫ n n y dy = An
a
Using Figure 3 of the text,
1
time t = 0 . Thus, V ( t ) = 20t − t 2 , so
2
V ( 20 ) − V (10 ) = 200 − 150 = 50 gallons.
62.
b
∫a
1 ⎡ n +1 ⎤ b
x
⎦a
n +1 ⎣
1
(b n +1 − a n +1 )
n +1
bn
n
an
bn
+ 8.4 + 8.0 + 7.5 + 7.0 + 6.5)
= 84.8
An = ∫
Use a midpoint Riemann sum with n = 10
partitions.
n
b n +1 − a n +1
n +1
n
(b n +1 − a n +1 ) = An
nBn =
n +1
=
10
V = ∑ f ( xi ) Δxi
(
⎡ n ( n +1) / n ⎤
y dy = ⎢
y
⎥ n
⎣ n +1
⎦a
)
i =1
⎛ 6200 + 6300 + 6500 + 6500 + 6600 ⎞
≈ 1⎜
⎟
⎝ + 6700 + 6800 + 7000 + 7200 + 7200 ⎠
68.
Use a midpoint Riemann sum with n = 12
partitions.
12
E = ∑ P ( ti ) Δti
G ( a ) = 0 . Thus, C = − F (a) and
i =0
≈ 2(3.0 + 3.0 + 3.8 + 5.8 + 7.8 + 6.9
+ 6.5 + 6.3 + 7.2 + 8.2 + 8.7 + 5.4)
= 145.2
66.
δ ( x ) = m′ ( x ) = 1 +
2
G ( x ) = F ( x) − F (a) . Now choose x = b to
obtain
b
∫a
x
4
mass = ∫ δ ( x ) dx = m ( 2 ) =
0
a
dy
= G '( x) = f ( x)
dx
dy = f ( x) dx
Let F be any antiderivative of f . Then
G ( x ) = F ( x ) + C . When x = a , we must have
= 67, 000
65.
x
Let y = G ( x) = ∫ f (t ) dt . Then
f (t ) dt = G ( b ) = F (b) − F (a)
3
5
2
69.
3 2
x dx
0
⎡ x3 ⎤
= ⎢ ⎥ = 9−0 = 9
⎣⎢ 3 ⎦⎥ 0
2 3
x dx
0
⎡ x4 ⎤
= ⎢ ⎥ = 4−0 = 4
⎢⎣ 4 ⎥⎦ 0
∫
2
70.
∫
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71.
72.
73.
π
∫0
π
sin x dx = [ − cos x ]0 = 1 + 1 = 2
77.
First Fundamental Theorem of Calculus.
Since G is differentiable at c , G is
continuous there. Now suppose c = a .
2
1 2 1 3⎤
⎡
2
∫0 (1 + x + x ) dx = ⎢⎣ x + 2 x + 3 x ⎥⎦0
8⎞
20
⎛
= ⎜2+ 2+ ⎟−0 =
3⎠
3
⎝
2
x →c
2
1 2
74
1
1
⎡1 ⎤
dx = ⎢ x3 ⎥ = = 0.333
⎣ 3 ⎦0 3
4
x dx
1
⎡1
⎤
−3 ⎢ (2)(2) + (4)(4) ⎥
2
⎣2
⎦
= –24
d ⎛1
⎞ 1 ⎛ x⎞ x
⎜ x x ⎟ = x⎜ ⎟ + = x
dx ⎝ 2
⎠ 2 ⎝ x⎠ 2
b
∫a
76.
Then
x
∫a
b
1
⎡1
⎤
x dx = ⎢ x x ⎥ = ( b b − a a )
2
⎣
⎦a 2
For b > 0, if b is an integer,
i =1
x →a +
x →a +
Thus,
a
lim G ( x) = 0 = ∫ f (t ) dt = G (a)
x→a+
(b − 1)b
.
2
If b is not an integer, let n = ab b . Then
a
Therefore G is right-continuous at x = a .
Now, suppose c = b . Then
∫
lim G ( x) = lim
x →b −
b
x
x →b −
f (t ) dt
As before,
(b − x) f (m) ≤ G ( x) ≤ (b − x) f ( M ) so we
can apply the Squeeze Theorem again to
obtain
lim (b − x) f (m) ≤ lim G ( x)
x →b −
x →b −
≤ lim (b − x) f ( M )
∫0 a x b dx = 0 + 1 + 2 + ⋅⋅⋅ + (b − 1)
b −1
x
a
( x − a ) f ( m) ≤ G ( x ) ≤ ( x − a ) f ( M )
By the Squeeze Theorem
lim ( x − a ) f (m) ≤ lim G ( x )
b
= ∑i =
x
a
f (m) dt ≤ ∫ f (t ) dt ≤ ∫ f ( M ) dt
≤ lim ( x − a ) f ( M )
4
= 2 [ (−2 − 1 + 0 + 1 + 2 + 3)(1) ]
75.
Min-Max Existence Theorem) m and M such
that f (m) ≤ f ( x) ≤ f ( M ) for all x in [ a, b ] .
x→a+
∫−2 ( 2 a x b − 3 x ) dx = 2∫−2 a xb dx − 3∫−2
4
f (t ) dt . Since f is
continuous on [ a, b ] , there exist (by the
n
⎛ 1− 0 ⎞ ⎛ 1 ⎞ 1
0
i
+
=
∑ ⎜⎝ n ⎟⎠ ⎜⎝ n ⎟⎠ n3 ∑ i 2 , which for
i =1
i =1
77
= 0.385 .
n = 10 equals
200
∫0 x
x
∫
x→a a
Then lim G ( x) = lim
The right-endpoint Riemann sum is
n
a. Let c be in ( a, b ) . Then G '(c) = f (c) by the
x →b −
Thus
b
lim G ( x) = 0 = ∫ f (t ) dt = G (b)
x →b −
b
Therefore, G is left-continuous at x = b .
b
∫0 a x b dx = 0 + 1 + 2 + ⋅⋅⋅ + (n − 1) + n(b − n)
(n − 1)n
+ n(b − n)
2
(ab b − 1) ab b
=
+ ab b (b − ab b) .
2
=
b. Let F be any antiderivative of f. Note that G
is also an antiderivative of f. Thus,
F ( x) = G ( x) + C . We know from part (a)
that G ( x) is continuous on [ a, b ] . Thus
F ( x ) , being equal to G ( x) plus a constant,
is also continuous on [ a, b ] .
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78.
x
⎧1, x > 0
Let f ( x) = ⎨
and F ( x) = ∫ f (t ) dt .
−1
⎩0, x ≤ 0
If x < 0 , then F ( x) = 0 . If x ≥ 0 , then
F ( x) = ∫
x
−1
0
6.
f (t ) dt
x
= ∫ 0 dt + ∫ 1 dt
−1
0
= 0+ x = x
Thus,
⎧ x, x ≥ 0
F ( x) = ⎨
⎩0, x < 0
which is continuous everywhere even though
f ( x) is not continuous everywhere.
7.
8.
1 b
f ( x ) dx
b − a ∫a
1
[sin π − sin 0] = 0
π
1 π
1
π
sin x dx = ( − cos x )0
π − 0 ∫0
π
1
2
= − ( −1 − 1) =
π
1
π − 0 ∫0
9.
=
0; 2∫ f ( x ) dx
π
1
cos x dx = [sin x ]0
π ∫0
π
π
2. f ( c )
3.
1 π
=
4.5 Concepts Review
1.
1 2
( x + x ) dx
2 + 3 ∫−3
2
1 0
= ⎛⎜ ∫ ( − x + x ) dx + ∫ 2 x dx ⎞⎟
0
5 ⎝ −3
⎠
1 ⎡ 2 ⎤2 4
= x
=
5 ⎣ ⎦0 5
1
π
π
x cos x 2 dx =
1 ⎛1
2⎞
⎜ sin x ⎟
π ⎝2
⎠0
π
(0 − 0) = 0
2
0
10.
4. f ( x + p ) = f ( x ) ; period
π/2
1
sin 2 x cos x dx
∫
0
π /2−0
π/ 2
=
2 ⎡1 3 ⎤
sin x ⎥
π ⎢⎣ 3
⎦0
2
3π
=
Problem Set 4.5
1.
1 3 3
4 x dx
3 − 1 ∫1
3
1
= ⎡ x4 ⎤
2 ⎣ ⎦1
11.
= 40
(
1 2
y 1 + y2
2 − 1 ∫1
=
4
2.
1 4 2
1 ⎡5 ⎤
5 x dx = ⎢ x3 ⎥ = 35
∫
1
4 −1
3 ⎣ 3 ⎦1
(
⎡1
dy = ⎢ 1 + y 2
⎣8
3
1 3
x
1
1
dx = ⎡ x 2 + 16 ⎤ =
∫
⎦⎥ 0 3
3 − 0 0 x 2 + 16
3 ⎣⎢
x2
1 2
1 ⎡2 3
⎤
dx = ⎢
x + 16 ⎥
∫
0
3
2−0
2 ⎣3
⎦0
x + 16
1
2
=
24 − 4 =
6 −2
3
3
(
5.
)
(
π /4
1
1 ⎡1
⎤
tan x sec2 x =
tan 2 x ⎥
∫
0
π / 4 −1
π / 4 − 1 ⎢⎣ 2
⎦0
=
2
π −4
(1 − 0 ) =
2
π −4
=
(
π /2
13.
)
1 1
( 2 + x ) dx
1 + 2 ∫−2
1
1 0
= ⎡⎢ ∫ ( 2 − x ) dx + ∫ ( 2 + x ) dx ⎤⎥
−
2
0
3⎣
⎦
0
1⎫
1 ⎧⎡
= ⎨ 2 x − 12 x 2 ⎤ + ⎡ 2 x + 12 x 2 ⎤ ⎬
⎣
⎦
⎣
⎦
−2
0⎭
3⎩
)
1
17
−2(−2) + 12 (−2)2 + 2 + 12 =
3
6
) ⎥⎦1
625
609
−2 =
= 76.125
8
8
2
4.
4 ⎤2
π /4
12.
3.
)
3
1 π / 2 sin z
4
dz = ⎡ −2 cos z ⎤
∫
⎦
π
/
4
π /4
π⎣
z
π /4
8
cos π / 4 − cos π / 2 ≈ 0.815
=
π
14.
(
)
1 π / 2 sin v cos v
dv
π / 2 ∫0
1 + cos 2 v
2⎡
− 1 + cos 2 v
π ⎢⎣
2
=
−1 + 2
=
π
(
π /2
⎤
⎥⎦ 0
)
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15.
x + 1 dx = c + 1 ( 3 − 0 )
3
∫0
24.
3
⎡2
3/ 2 ⎤
⎢ 3 ( x + 1) ⎥ = 3 c + 1
⎣
⎦0
115
≈ 1.42
14 / 3 = 3 c + 1; c =
81
2 3
x dx
0
∫
2
⎡1 ⎤
= c3 ( 2 − 0 ) ; ⎢ x 4 ⎥ = 2c3
⎣ 4 ⎦0
c = 3 2 ≈ 1.26
25.
∫1 ( ax + b ) dx = ( ac + b )( 4 − 1)
4
4
16.
1
∫−1 x
5
⎡a 2
⎤
⎢ 2 x + bx ⎥ = 3ac + 3b; c = 2
⎣
⎦1
dx = c 2 (1 − ( −1) )
2
1
3
⎡1 3 ⎤
2
⎢ 3 x ⎥ = 2c ; c = ± 3 ≈ ±0.58
⎣
⎦ −1
17.
∫−4 (1 − x
3
2
26.
) dx = (1 − c ) (3 + 4)
b 2
y dy
0
∫
c=
2
b
⎡1 ⎤
= c 2 ( b − 0 ) ; ⎢ y 3 ⎥ = bc 2
⎣ 3 ⎦0
b
3
3
⎡ 1 3⎤
2
⎢ x − 3 x ⎥ = 7 − 7c
⎣
⎦ −4
B
27.
39
c=±
≈ ±2.08
3
18.
∫A ( ax + b ) dx = f ( c )
B−A
B
⎡a 2
⎤
⎢ 2 x + bx ⎥
⎣
⎦A
= ac + b
B−A
a
( B − A)( B + A) + b ( B − A)
2
= ac + b
B− A
a
a
B + A + b = ac + b;
2
2
1
1
c = B + A = ( A + B) / 2
2
2
∫0 x (1 − x ) dx = c (1 − c )(1 − 0 )
1
1
⎡ − x 2 ( 2 x − 3) ⎤
⎢
⎥ = c − c2
6
⎢⎣
⎥⎦ 0
c=
19.
3± 3
≈ 0.21 or 0.79
6
2
⎡x x ⎤
x dx = c ( 2 − 0 ) ; ⎢
⎥ = 2 c ; c =1
⎣ 2 ⎦0
2
∫0
2
2
⎡x x ⎤
x dx = c ( 2 + 2 ) ; ⎢
⎥ = 4 c ; c = −1,1
⎣ 2 ⎦ −2
20.
∫−2
21.
∫−π sin z dz = sin c (π + π )
π
[ − cos z ]π−π
22.
28.
b
⎡1 3 ⎤
2
2
2
∫0 ay dy = ac ( b − 0 ) ; ⎢⎣ 3 ay ⎥⎦0 = abc
b
c=
b 3
3
29.. Using c = π yields 2π (5) 4 = 1250π ≈ 3927
= 2π sin c; c = 0
π
∫0 cos 2 y dy = ( cos 2c )(π − 0 )
π
π 3π
⎡ sin 2 y ⎤
⎢ 2 ⎥ = π cos 2c; c = 4 , 4
⎣
⎦0
23.
2
∫0
(v
2
)
(
)
− v dv = c − c ( 2 − 0 )
2
30.
(
)
Using c = 0.8 yields 2 3 + sin 0.82 ≈ 7.19
2
⎡1 3 1 2 ⎤
2
⎢ 3 v − 2 v ⎥ = 2c − 2c
⎣
⎦0
c=
21 + 3
≈ 1.26
6
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31.
Using c = 0.5 yields 2
2
1 + 0.5
2
= 3.2
38.
3π
∫−
=
39.
3π 2
x cos( x3 ) dx
0
(
3π
2⎡
2
= sin 3 3 π3
sin( x3 ) ⎤
⎣
⎦
0
3
3
π
∫−π (sin x + cos x)
=∫
π
2
)
dx
(sin 2 x + 2sin x cos x + cos 2 x) dx
−π
π
π
π
−π
−π
= ∫ (1 + 2sin x cos x) dx = ∫ dx + ∫ sin 2 x dx
5
32.
x 2 cos( x3 ) dx = 2∫
3π
⎛ 16 ⎞
Using c = 15 yields ⎜ ⎟ (20 − 10) ≈ 13.8 .
⎝ 15 ⎠
−π
π
= 2 ∫ dx + 0 = 2[ x]0π = 2π
0
40.
π/2
∫−π / 2 z sin
2
( z 3 ) cos( z 3 )dz = 0 , since
(− z ) sin 2 [(− z )3 ]cos[(− z )3 ]
= − z sin 2 (− z 3 ) cos(− z 3 )
= − z[− sin( z 3 )]2 cos( z 3 )
33.
34.
A rectangle with height 25 and width 7 has
approximately the same area as that under the
curve. Thus
1 7
H (t ) dt ≈ 25
7 ∫0
a. A rectangle with height 28 and width 24 has
approximately the same area as that under
the curve. Thus,
24
1
T (t ) dt ≈ 28
24 − 0 ∫0
b. Yes. The Mean Value Theorem for Integrals
guarantees the existence of a c such that
24
1
T (t ) dt = T (c)
∫
0
24 − 0
The figure indicates that there are actually
two such values of c, roughly, c = 11 and
c = 16 .
35.
π
∫−π
π
π
−π
0
= − z sin 2 ( z 3 ) cos( z 3 )
41.
=∫
= 2 [ x]
42.
43.
1
−1
x3 dx
⎡ x3 ⎤
8
+0+ 2⎢ ⎥ +0 =
3
3
⎣⎢ ⎦⎥ 0
100
∫−100 (v + sin v + v cos v + sin
3
v)5 dv = 0
∫−1 ( x
1
3
)
1
1
0
−1
+ x3 dx = 2∫ x3 dx + ∫
x3 dx
1
⎡ x4 ⎤
1
= 2⎢ ⎥ +0 =
4
2
⎣⎢ ⎦⎥ 0
∫−π / 4 ( x sin
π/4
5
)
2
x + x tan x dx = 0
2
since − x sin 5 (− x) + − x tan(− x)
2
= − x sin 5 x − x tan x
odd.
sin x
x 2 dx + ∫
= −(v + sin v + v cos v + sin 3 v)5
∫−1 (1 + x2 )4 dx = 0 , since the integrand is
π/2
1
−1
= (−v − sin v − v cos v − sin 3 v)5
3
∫−π / 2 1 + cos x dx = 0 , since the integrand is odd.
x dx + ∫
since (−v + sin(−v) − v cos(−v) + sin 3 (−v))5
44.
37.
1
−1
1
π
x
+ x3 ) dx
2
dx + ∫
1
0
= 0 + 2 [sin x ]0 = 0
36.
1
−1
(sin x + cos x) dx = ∫ sin x dx + 2 ∫ cos x dx
1
1
∫−1 (1 + x + x
45.
−a
∫−b
f ( x) dx = ∫
b
a
−a
∫−b
f ( x) dx when f is even.
f ( x) dx = − ∫
b
a
f ( x) dx when f is odd.
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
46. u = − x, du = − dx
b. ku =
−b
b
∫a f ( − x ) dx = − ∫−a
=∫
−a
−b
f (u ) du = ∫
f (u ) du
−a
−b
c. Note that
1 b
1 a
u=
u ( x ) dx =
u ( x ) dx , so
∫
b−a a
a − b ∫b
we can assume a < b .
1 b
1 b
u=
u dx ≤
v dx = v
b − a ∫a
b − a ∫a
f ( x) dx since the variable
used in the integration is not important.
47.
4π
∫0
cos x dx = 8∫
π/2
0
cos x dx
π/ 2
= 8 [sin x ]0
=8
48. Since sin x is periodic with period 2π , sin 2x is
periodic with period π .
4π
∫0
sin 2 x dx = 8∫
π2
0
π2
⎡ 1
⎤
= 8 ⎢ − cos 2 x ⎥
⎣ 2
⎦0
49.
1+π
∫1
sin x dx = ∫
π
0
k b
1 b
u dx =
ku dx = ku
∫
a
b−a
b − a ∫a
54.
a. V = 0 by periodicity.
b. V = 0 by periodicity.
sin 2 x dx
2
c. Vrms
=∫
φ +1
φ
Vˆ 2 sin 2 (120π t + φ ) dt
1
= ∫ Vˆ 2 sin 2 (120π t ) dt
= –4(–1 – 1) = 8
0
by periodicity.
u = 120π t , du = 120π dt
π
sin x dx = ∫ sin x dx
0
1 120π ˆ 2 2
V sin u du
120π ∫0
120π
1 ⎤
Vˆ 2 ⎡ 1
cos
sin
=
−
u
u
+
u
120π ⎢⎣ 2
2 ⎥⎦ 0
1
= Vˆ 2
2
2
Vrms
=
π
= [ − cos x ]0 = 2
Vˆ 2
2
ˆ
V = 120 2 ≈ 169.71 Volts
d. 120 =
50.
2 +π / 2
∫2
1+π
∫1
Since f is continuous on a closed interval [ a, b ]
there exist (by the Min-Max Existence Theorem)
an m and M in [ a, b ] such that
π
π /2
0
0
cos x dx = ∫ cos x dx = 2∫
π /2
53.
55.
sin 2 x dx
1
[ − cos 2 x ]0π / 2 = 1
2
= 2 [sin x ]0
52.
π/2
0
=
51.
sin 2 x dx = ∫
f (m) ≤ f ( x) ≤ f ( M ) for all x in [ a, b ] . Thus
b
∫a
cos x dx
= 2 (1 − 0 ) = 2
The statement is true. Recall that
1 b
f =
f ( x) dx .
b − a ∫a
b
b
b
1 b
∫a fdx = f ∫a dx = b − a ∫a f ( x)dx ⋅ ∫a dx
b
1 b
=
f ( x)dx ⋅ (b − a ) = ∫ f ( x) dx
∫
a
a
b−a
b
a
(b − a) f (m) ≤ ∫ f ( x) dx ≤ (b − a ) f ( M )
a
1 b
f ( x) dx ≤ f ( M )
b − a ∫a
Since f is continuous, we can apply the
Intermediate Value Theorem and say that f takes
on every value between f (m) and f ( M ) . Since
f ( m) ≤
1 b
f ( x) dx is between f (m) and f ( M ) ,
b − a ∫a
there exists a c in [ a, b ] such that
All the statements are true.
1 b
1 b
a. u + v =
u dx +
v dx
∫
b−a a
b − a ∫a
1 b
(u + v) dx = u + v
=
b − a ∫a
b
a
b
f (m) dx ≤ ∫ f ( x) dx ≤ ∫ f ( M ) dx
f (c ) =
56. a.
2π
∫0
1 b
f ( x) dx .
b − a ∫a
(sin 2 x + cos 2 x) dx = ∫
2π
0
2π
dx = [ x ]0 = 2π
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58.
b.
a. Odd
b. 2π
c. This function cannot be integrated in closed
form. We can only simplify the integrals
using symmetry and periodicity, and
approximate them numerically.
Note that
∫−a f ( x ) dx = 0
a
π +a
∫π −a f ( x ) dx = 0
since f is odd, and
since
f (π + x ) = − f (π − x ) .
π /2
∫0
f ( x ) dx =
π
J1 (1) ≈ 0.69 (Bessel
2
function)
π /2
∫−π / 2 f ( x ) dx = 0
3π / 2
c. 2π = ∫
= 2∫
=∫
0
2π
0
2π
0
57.
2π
cos x dx + ∫
2
2π
0
sin x dx
2π
∫0
cos 2 x dx, thus
3π / 2
cos 2 x dx
13π / 6
f ( x ) dx = 0
0
4π / 3
10π / 3
4π / 3
∫13π / 6 f ( x ) dx = ∫π / 6 f ( x ) dx ≈ 1.055
b. 2π
c. On [ 0, π ] , sin x = sin x .
59.
u = cos x , du = − sin x dx
∫ f ( x ) dx = ∫
a. Written response.
sin x ⋅ sin ( cos x ) dx
f ( x ) dx = 1 − cos1 ≈ 0.46
∫0
f ( x ) dx
π
3π / 2
0
π
f ( x ) dx = ∫ f ( x ) dx + ∫
3π / 2
∫−3π / 2 f ( x ) dx = 2∫0
f ( x ) dx
f ( x ) dx
= 2 ( cos1 − 1) ≈ −0.92
f ( x ) dx = 0
⎛c ⎞
f ⎜ x ⎟ dx
⎝b ⎠
a
b
∫
c2 0
f ( x) dx +
a 2 + b2 c
=
c
2
b2 c
∫
c2 0
f ( x) dx
c
∫0 f ( x) dx = ∫0 f ( x) dx
60.
If f is odd, then f (− x ) = − f ( x) and we can
write
0
⎛ 3⎞
⎛1⎞
⎟⎟ + cos ⎜ ⎟
⎝2⎠
⎝ 2 ⎠
4π / 3
≈ −0.44
4π / 3
∫13π / 6 f ( x ) dx = ∫π / 6 f ( x ) dx ≈ −0.44
since
a 2 + b 2 = c 2 from the triangle.
∫−a f ( x) dx = ∫−a [ − f (− x)] dx = ∫a
∫π / 6 f ( x ) dx = 2 cos1 − cos ⎜⎜
10π / 3
f ( x) dx
∫0 g ( x) dx + ∫0 h( x) dx
a2 c
=
= cos1 − 1 ≈ −0.46
3π / 2
∫0
c
Thus,
= 2 (1 − cos1) ≈ 0.92
3π / 2
a2 c
b
b
b2 c
f ( x) dx = ∫ f ( x) dx
0 c
c
c2 0
=∫
∫ f ( x ) dx = − cos ( cos x ) + C
π /2
c
a ⎛c ⎞
f ⎜ x ⎟ dx
c ⎝a ⎠
c2
b
b b
B = ∫ h( x) dx = ∫
0
0 c
Likewise, on [π , 2π ] ,
π /2
a
0
a
a
f ( x) dx =
0 c
c
=∫
= cos ( cos x ) + C
∫−π / 2 f ( x ) dx = 2∫0
a
0
b. A = ∫ g ( x) dx = ∫
= − ∫ sin u du = cos u + C
2π
2π
f ( x ) dx = ∫
f ( x ) dx = 0
∫π / 6 f ( x ) dx ≈ 1.055 (numeric integration)
a. Even
∫0
f ( x ) dx ≈ 0.69
2π
∫π / 6
sin x dx = π
π /2
π /2
0
∫−3π / 2 f ( x ) dx = 0 ; ∫0
2
∫0
f ( x ) dx = ∫
∫0
2
0
0
a
a
0
0
f (u ) du
= − ∫ f (u ) du = − ∫ f ( x) dx
On the second line, we have made the
substitution u = − x .
280
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4.6 Concepts Review
1. 1, 2, 2, 2, …, 2, 1
2. 1, 4, 2, 4, 2, …, 4, 1
3. n 4
4. large
Problem Set 4.6
1.
f ( x) =
1
2
;h =
x
x0 = 1.00
3 –1
= 0.25
8
f ( x0 ) = 1
x5 = 2.25
f ( x5 ) ≈ 0.1975
x1 = 1.25
f ( x1 ) = 0.64
x6 = 2.50
f ( x6 ) = 0.16
x2 = 1.50
f ( x2 ) ≈ 0.4444
x7 = 2.75
f ( x7 ) ≈ 0.1322
x3 = 1.75
f ( x3 ) ≈ 0.3265
x8 = 3.00
f ( x8 ) ≈ 0.1111
x4 = 2.00
f ( x4 ) = 0.25
Left Riemann Sum:
3
1
∫1 x2 dx ≈ 0.25[ f ( x0 ) + f ( x1 ) + …+ f ( x7 )] ≈ 0.7877
Right Riemann Sum:
1
∫1 x2 dx ≈ 0.25[ f ( x1 ) + f ( x2 ) + ... + f ( x8 )] ≈ 0.5655
3
0.25
[ f ( x0 ) + 2 f ( x1 ) + …+ 2 f ( x7 ) + f ( x8 )] ≈ 0.6766
2
3 1
0.25
Parabolic Rule: ∫
dx ≈
[ f ( x0 ) + 4 f ( x1 ) + 2 f ( x2 ) +…+ 4 f ( x7 ) + f ( x8 )] ≈ 0.6671
1 x2
3
Trapezoidal Rule:
3
1
∫1 x2 dx ≈
Fundamental Theorem of Calculus:
2.
f ( x) =
1
3
;h =
x
x0 = 1.00
3
1
2
⎡ 1⎤
∫1 x2 dx = ⎢⎣ – x ⎥⎦1 = – 3 + 1 = 3 ≈ 0.6667
3
1
3 –1
= 0.25
8
f ( x0 ) = 1
x5 = 2.25
f ( x5 ) ≈ 0.0878
x1 = 1.25
f ( x1 ) = 0.5120
x6 = 2.50
f ( x6 ) = 0.0640
x2 = 1.50
f ( x2 ) ≈ 0.2963
x7 = 2.75
f ( x7 ) ≈ 0.0481
x3 = 1.75
f ( x3 ) ≈ 0.1866
x8 = 3.00
f ( x8 ) ≈ 0.0370
x4 = 2.00
f ( x4 ) = 0.1250
Left Riemann Sum:
3
Right Riemann Sum:
Trapezoidal Rule:
Parabolic Rule:
3
1
∫1 x3 dx ≈ 0.25[ f ( x0 ) + f ( x1 ) + …+ f ( x7 )] ≈ 0.5799
3
1
∫1 x3 dx ≈ 0.25[ f ( x1 ) + f ( x2 ) + ... + f ( x8 )] ≈ 0.3392
3
1
∫1 x3 dx ≈
1
∫1 x3 dx ≈
0.25
[ f ( x0 ) + 2 f ( x1 ) + …+ 2 f ( x7 ) + f ( x8 )] ≈ 0.4596
2
0.25
[ f ( x0 ) + 4 f ( x1 ) + 2 f ( x2 ) + …+ 4 f ( x7 ) + f ( x8 )] ≈ 0.4455
3
Fundamental Theorem of Calculus:
3
∫1
3
4
⎡ 1 ⎤
dx = ⎢ −
= ≈ 0.4444
3
2⎥
9
x
⎣ 2 x ⎦1
1
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f ( x) = x ; h =
3.
x0 = 0.00
x1 = 0.25
x2 = 0.50
x3 = 0.75
x4 = 1.00
2–0
= 0.25
8
f ( x0 ) = 0
f ( x1 ) = 0.5
f ( x2 ) ≈ 0.7071
f ( x3 ) ≈ 0.8660
f ( x4 ) = 1
Left Riemann Sum:
2
∫0
= 1.25
= 1.50
= 1.75
= 2.00
f ( x5 ) ≈ 1.1180
f ( x6 ) ≈ 1.2247
f ( x7 ) ≈ 1.3229
f ( x8 ) ≈ 1.4142
x dx ≈ 0.25[ f ( x0 ) + f ( x1 ) + …+ f ( x7 )] ≈ 1.6847
x dx ≈ 0.25[ f ( x1 ) + f ( x2 ) + ... + f ( x8 )] ≈ 2.0383
2
∫0
Right Riemann Sum:
x5
x6
x7
x8
0.25
[ f ( x0 ) + 2 f ( x1 ) + …+ 2 f ( x7 ) + f ( x8 )] ≈ 1.8615
2
2
0.25
[ f ( x1 ) + 4 f ( x2 ) + 2 f ( x3 ) + …+ 4 f ( x7 ) + f ( x8 )] ≈ 1.8755
Parabolic Rule: ∫ xdx ≈
0
3
Trapezoidal Rule:
2
∫0
xdx ≈
Fundamental Theorem of Calculus:
4.
2
∫0
2
4 2
⎡2
⎤
xdx = ⎢ x3 / 2 ⎥ =
≈ 1.8856
3
3
⎣
⎦0
x0 = 1.00
3 –1
= 0.25
8
f ( x0 ) ≈ 1.4142
x5 = 2.25
f ( x5 ) ≈ 5.5400
x1 = 1.25
f ( x1 ) ≈ 2.0010
x6 = 2.50
f ( x6 ) ≈ 6.7315
x2 = 1.50
f ( x2 ) ≈ 2.7042
x3 = 1.75
f ( x3 ) ≈ 3.5272
x7 = 2.75
x8 = 3.00
f ( x7 ) ≈ 8.0470
f ( x8 ) ≈ 9.4868
x4 = 2.00
f ( x4 ) ≈ 4.4721
f ( x) = x x 2 + 1; h =
Left Riemann Sum:
3
∫1 x
3
x 2 + 1 dx ≈ 0.25[ f ( x0 ) + f ( x1 ) + " + f ( x7 )] ≈ 8.6093
∫1 x
Right Riemann Sum:
x 2 + 1 dx ≈ 0.25[ f ( x1 ) + f ( x2 ) + ... + f ( x8 )] ≈ 10.6274
0.25
[ f ( x0 ) + 2 f ( x1 ) + " + 2 f ( x7 ) + f ( x8 )] ≈ 9.6184
2
3
0.25
[ f ( x0 ) + 4 f ( x1 ) + 2 f ( x2 ) + " + 4 f ( x7 ) + f ( x8 )] ≈ 9.5981
Parabolic Rule: ∫ x x 2 + 1dx ≈
1
3
Trapezoidal Rule:
3
∫1 x
x 2 + 1 dx ≈
Fundamental Theorem of Calculus:
5.
(
)
5
f ( x) = x x 2 + 1 ; h =
x0 = 0.00
x1 = 0.125
x2 = 0.250
x3 = 0.375
x4 = 0.500
3
(
)
1
⎡1 2
2
3/ 2 ⎤
∫1 x x + 1dx = ⎢⎣ 3 ( x + 1) ⎥⎦1 = 3 10 10 – 2 2 ≈ 9.5981
3
1− 0
= 0.125
8
f ( x0 ) = 0
f ( x1 ) ≈ 0.1351
f ( x2 ) ≈ 0.3385
f ( x3 ) ≈ 0.7240
f ( x4 ) ≈ 1.5259
x5 = 0.625
f ( x5 ) ≈ 3.2504
x6 = 0.750
f ( x6 ) ≈ 6.9849
x7 = 0.875
x8 = 1.000
f ( x7 ) ≈ 15.0414
f ( x8 ) = 32
282
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Left Riemann Sum:
∫0 x ( x
1
2
)
5
+ 1 dx ≈ 0.125[ f ( x0 ) + f ( x1 ) + " + f ( x7 )] ≈ 3.4966
∫0 x ( x
1
Right Riemann Sum:
∫0 x ( x
2
)
5
+ 1 dx ≈ 0.125[ f ( x1 ) + f ( x2 ) + ... + f ( x8 )] ≈ 7.4966
)
5
0.125
[ f ( x0 ) + 2 f ( x1 ) + " + 2 f ( x7 ) + f ( x8 )] ≈ 5.4966
2
5
1
0.125
Parabolic Rule: ∫ x x 2 + 1 dx ≈
[ f ( x0 ) + 4 f ( x1 ) + 2 f ( x2 ) + " + 4 f ( x7 ) + f ( x8 )] ≈ 5.2580
0
3
Trapezoidal Rule:
1
2
(
+ 1 dx ≈
)
6.
(
)
(
)
1
5
6⎤
⎡1 2
2
∫0 x x + 1 dx = ⎢⎣12 x + 1 ⎥⎦ = 5.25
0
1
Fundamental Theorem of Calculus:
x0 = 1.000
4 −1
= 0.375
8
f ( x0 ) ≈ 2.8284
x5 = 2.875
f ( x5 ) ≈ 7.6279
x1 = 1.375
f ( x1 ) ≈ 3.6601
x6 = 3.250
f ( x6 ) ≈ 8.7616
x2 = 1.750
f ( x2 ) ≈ 4.5604
x3 = 2.125
f ( x3 ) ≈ 5.5243
x7 = 3.625
x8 = 4.000
f ( x7 ) ≈ 9.9464
f ( x8 ) ≈ 11.1803
x4 = 2.500
f ( x4 ) ≈ 6.5479
f ( x) = ( x + 1)
3/ 2
;h =
Left Riemann Sum:
4
∫1 ( x + 1)
3/ 2
4
dx ≈ 0.375[ f ( x0 ) + f ( x1 ) + " + f ( x7 )] ≈ 18.5464
∫1 ( x + 1)
Right Riemann Sum:
3/ 2
dx ≈ 0.375[ f ( x1 ) + f ( x2 ) + ... + f ( x8 )] ≈ 21.6784
0.375
[ f ( x0 ) + 2 f ( x1 ) + " + 2 f ( x7 ) + f ( x8 )] ≈ 20.1124
2
4
0.375
3/ 2
Parabolic Rule: ∫ ( x + 1) dx ≈
[ f ( x0 ) + 4 f ( x1 ) + 2 f ( x2 ) + " + 4 f ( x7 ) + f ( x8 )] ≈ 20.0979
1
3
Trapezoidal Rule:
∫1 ( x + 1)
4
3/ 2
dx ≈
Fundamental Theorem of Calculus:
4
∫1
( x + 1)
3/ 2
4
⎡2
5/ 2 ⎤
dx = ⎢ ( x + 1) ⎥ ≈ 20.0979
5
⎣
⎦1
7.
LRS
RRS
MRS
Trap
Parabolic
n=4
0.5728
0.3728
0.4590
0.4728
0.4637
n=8
0.5159
0.4159
0.4625
0.4659
0.4636
n = 16
0.4892
0.4392
0.4634
0.4642
0.4636
LRS
RRS
MRS
Trap
Parabolic
n=4
1.2833
0.9500
1.0898
1.1167
1.1000
n=8
1.1865
1.0199
1.0963
1.1032
1.0987
n = 16
1.1414
1.0581
1.0980
1.0998
1.0986
8.
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9.
LRS
RRS
MRS
Trap
Parabolic
n=4
2.6675
3.2855
2.9486
2.9765
2.9580
n=8
2.8080
3.1171
2.9556
2.9625
2.9579
n = 16
2.8818
3.0363
2.9573
2.9591
2.9579
10.
11.
LRS
RRS
MRS
Trap
Parabolic
n=4
10.3726
17.6027
13.6601
13.9876
13.7687
n=8
12.0163
15.6314
13.7421
13.8239
13.7693
n = 16
12.8792
14.6867
13.7625
13.7830
13.7693
f ′( x) = −
1
x
2
; f ′′ ( x ) =
2
x3
The largest that f ′′ ( c ) can be on [1,3] occurs when c = 1 , and f ′′ (1) = 2
( 3 − 1)3
400
Round up: n = 12
( 2 ) ≤ 0.01; n ≥
3
12n 2
31
0.167
∫1 x dx ≈ 2 [ f ( x0 ) + 2 f ( x1 ) + " + 2 f ( x11 ) + f ( x12 )]
≈ 1.1007
12.
f ′( x) = −
1
(1 + x )
2
; f ′′ ( x ) =
2
(1 + x )3
The largest that f ′′ ( c ) can be on [1,3] occurs when c = 1 , and f ′′ (1) =
1
.
4
( 3 − 1)3 ⎛ 1 ⎞
100
⎜ 4 ⎟ ≤ 0.01; n ≥ 6 Round up: n = 5
12n ⎝ ⎠
3 1
0.4
∫1 1 + x dx ≈ 2 [ f ( x0 ) + 2 f ( x1 ) + " + 2 f ( x4 ) + f ( x5 )]
≈ 0.6956
2
13.
f ′( x) =
1
2 x
; f ′′ ( x ) = −
1
4 x3 / 2
The largest that f ′′ ( c ) can be on [1, 4] occurs when c = 1 , and f ′′ (1) =
1
.
4
( 4 − 1)3 ⎛ 1 ⎞
900
⎜ ⎟ ≤ 0.01; n ≥ 16 Round up: n = 8
12n 2 ⎝ 4 ⎠
4
0.375
∫1 x dx ≈ 2 [ f ( x0 ) + 2 f ( x1 ) + " + 2 f ( x7 ) + f ( x8 )]
≈ 4.6637
284
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14.
1
f ′( x) =
2 x +1
; f ′′ ( x ) = −
1
4 ( x + 1)
3/ 2
The largest that f ′′ ( c ) can be on [1,3] occurs when c = 1 , and f ′′ (1) =
( 3 − 1)3 ⎛
12n
15.
2
4
f ( ) ( x) =
1
2
x
24
4 × 23 / 2
.
100
⎞
Round up: n = 3
⎟ ≤ 0.01; n ≥
12
2
⎠
3
0.667
∫1 x + 1 dx ≈ 2 [ f ( x0 ) + 2 f ( x1 ) + 2 f ( x2 ) + f ( x3 )]
≈ 3.4439
1
⎜
⎝ 4 × 23 / 2
f ′( x) = −
1
; f ′′ ( x ) =
2
x
3
6
; f ′′′ ( x ) = −
x4
;
x5
4
4
The largest that f ( ) ( c ) can be on [1,3] occurs when c = 1 , and f ( ) (1) = 24 .
( 4 − 1)5
( 24 ) ≤ 0.01; n ≈ 4.545 Round up to even: n = 6
180n 4
31
0.333
∫1 x dx ≈ 3 [ f ( x0 ) + 4 f ( x1 ) + ... + 4 f ( x5 ) + f ( x6 )]
≈ 1.0989
16.
f ′( x) =
f ′′′ ( x ) =
1
2 x +1
; f ′′ ( x ) = −
3
8 ( x + 1)
5/ 2
1
4 ( x + 1)
4
; f ( ) ( x) = −
3/ 2
;
15
16 ( x + 1)
7/2
4
4
The largest that f ( ) ( c ) can be on [ 4,8] occurs when c = 4 , and f ( ) ( 4 ) =
3
400 5
.
(8 − 4 )5 ⎛
3 ⎞
⎜
⎟ ≤ 0.01; n ≈ 1.1753 Round up to even: n = 2
180n ⎝ 400 5 ⎠
8
2
∫4 x + 1 dx ≈ 3 ⎡⎣ f ( x0 ) + 4 f ( x1 ) + f ( x2 )⎤⎦ ≈ 10.5464
4
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17.
m+ h
⎡a 3 b 2
⎤
2
∫m – h (ax + bx + c)dx = ⎢⎣ 3 x + 2 x + cx ⎥⎦ m – h
a
b
a
b
= ( m + h )3 + ( m + h ) 2 + c ( m + h ) – ( m – h ) 3 – ( m – h ) 2 – c ( m – h )
3
2
3
2
a
b
h
= (6m 2 h + 2h3 ) + (4mh) + c(2h) = [a(6m2 + 2h 2 ) + b(6m) + 6c]
3
2
3
m+ h
h
[ f (m − h) + 4 f (m) + f (m + h)]
3
h
= [a(m – h)2 + b(m – h) + c + 4am 2 + 4bm + 4c + a (m + h)2 + b(m + h) + c]
3
h
= [a(6m2 + 2h 2 ) + b(6m) + 6c]
3
18. a.
To show that the Parabolic Rule is exact, examine it on the interval [m – h, m + h].
Let f ( x) = ax3 + bx 2 + cx + d , then
m+h
∫m−h
f ( x) dx
a⎡
b
c
(m + h)4 − (m − h) 4 ⎤ + ⎡ (m + h)3 − (m − h)3 ⎤ + ⎡ (m + h) 2 − (m − h) 2 ⎤ + d [(m + h) − (m − h)]
⎣
⎦
⎣
⎦
⎦
4
3
2⎣
a
b
c
= (8m3 h + 8h3 m) + (6m2 h + 2h3 ) + (4mh) + d (2h).
4
3
2
The Parabolic Rule with n = 2 gives
m+ h
h
2 3
3
3
2
∫m−h f ( x) dx = 3 [ f (m − h) + 4 f (m) + f (m + h)] = 2am h + 2amh + 2bm h + 3 bh + 2chm + 2dh
a
b
c
= (8m3 h + 8mh3 ) + (6m 2 h + 2h3 ) + (4mh) + d (2h)
4
3
2
which agrees with the direct computation. Thus, the Parabolic Rule is exact for any cubic polynomial.
=
b. The error in using the Parabolic Rule is given by En = −
(l − k )5
4
f (4) (m) for some m between l and k.
180n
2
(3)
′
′′
However, f ( x) = 3ax + 2bx + c, f ( x) = 6ax + 2b, f ( x) = 6a, and f (4) ( x) = 0 , so En = 0.
19. The left Riemann sum will be smaller than
∫a f ( x ) dx .
b
If the function is increasing, then f ( xi ) < f ( xi +1 ) on the interval [ xi , xi +1 ] . Therefore, the left Riemann sum will
underestimate the value of the definite integral. The following example illustrates this behavior:
If f is increasing, then f ′ ( c ) > 0 for any c ∈ ( a, b ) . Thus, the error En =
( b − a )2
2n
f ′ ( c ) > 0 . Since the error is
positive, then the Riemann sum must be less than the integral.
286
Section 4.6
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20. The right Riemann sum will be larger than
∫a f ( x ) dx .
b
If the function is increasing, then f ( xi ) < f ( xi +1 ) on the interval [ xi , xi +1 ] . Therefore, the right Riemann sum
will overestimate the value of the definite integral. The following example illustrates this behavior:
If f is increasing, then f ′ ( c ) > 0 for any c ∈ ( a, b ) . Thus, the error En = −
( b − a )2
2n
f ′ ( c ) < 0 . Since the error is
negative, then the Riemann sum must be greater than the integral.
21. The midpoint Riemann sum will be larger than
∫a f ( x ) dx .
b
If f is concave down, then f ′′ ( c ) < 0 for any c ∈ ( a, b ) . Thus, the error En =
( b − a )3
24n 2
f ′′ ( c ) < 0 . Since the error
is negative, then the Riemann sum must be greater than the integral.
22. The Trapezoidal Rule approximation will be smaller than
∫a f ( x ) dx .
b
If f is concave down, then f ′′ ( c ) < 0 for any c ∈ ( a, b ) . Thus, the error En = −
( b − a )3
12n 2
error is positive, then the Trapezoidal Rule approximation must be less than the integral.
f ′′ ( c ) > 0 . Since the
23. Let n = 2.
f ( x) = x k ; h = a
x0 = – a
f ( x0 ) = – a k
x1 = 0
f ( x1 ) = 0
x2 = a
f ( x2 ) = a k
a
∫– a x
k
a
dx ≈ [– a k + 2 ⋅ 0 + a k ] = 0
2
a
1
1
⎡ 1 k +1 ⎤
k +1
k +1
k
k +1
k +1
∫– a x dx = ⎢⎣ k + 1 x ⎥⎦ – a = k + 1 [a – (– a) ] = k + 1[a – a ] = 0
A corresponding argument works for all n.
a
24. a.
T ≈ 48.9414; f ′( x) = 4 x3
[4(3)3 – 4(1)3 ](0.25)2
≈ 48.9414 – 0.5417 = 48.3997
12
The correct value is 48.4 .
T–
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b. T ≈ 1.9886; f ′( x) = cos x
T–
( )
π
[cos π – cos 0] 12
2
12
The correct value is 2.
≈ 1.999987
25. The integrand is increasing and concave down. By problems 19-22,
LRS < TRAP < MRS < RRS.
26. The integrand is increasing and concave up. By problems 19-22,
LRS < MRS < TRAP < RRS
27. A ≈
10
[75 + 2 ⋅ 71 + 2 ⋅ 60 + 2 ⋅ 45 + 2 ⋅ 45 + 2 ⋅ 52 + 2 ⋅ 57 + 2 ⋅ 60 + 59] = 4570 ft2
2
3
28. A ≈ [23 + 4 ⋅ 24 + 2 ⋅ 23 + 4 ⋅ 21 + 2 ⋅ 18 + 4 ⋅15 + 2 ⋅ 12 + 4 ⋅11 + 2 ⋅10 + 4 ⋅ 8 + 0] = 465 ft2
3
V = A ⋅ 6 ≈ 2790 ft3
20
[0 + 4 ⋅ 7 + 2 ⋅12 + 4 ⋅18 + 2 ⋅ 20 + 4 ⋅ 20 + 2 ⋅17 + 4 ⋅10 + 0] = 2120 ft2
3
4 mi/h = 21,120 ft/h
(2120)(21,120)(24) = 1,074,585,600 ft3
29. A ≈
30. Using a right-Riemann sum,
Distance = ∫
24
0
v(t ) dt ≈ ∑ v(ti ) Δt
3
60
852
= 14.2 miles
60
5. False:
The two sides will in general differ by
a constant term.
6. True:
At any given height, speed on the
downward trip is the negative of
speed on the upward.
7. True:
a1 + a0 + a2 + a1 + a3 + a2
31. Using a right-Riemann sum,
Water Usage = ∫
120
0
f ( x) = x 2 + 2 x + 1 and
g ( x) = x 2 + 7 x − 5 are a
counterexample.
i =1
= ( 31 + 54 + 53 + 52 + 35 + 31 + 28 )
=
4. False:
8
F (t ) dt
10
+ " + an −1 + an − 2 + an + an −1
= a0 + 2a1 + 2a2 + " + 2an −1 + an
≈ ∑ F (ti ) Δt = 12(71 + 68 + " + 148)
i =1
= 13, 740 gallons
8. True:
100
100
100
i =1
i =1
i =1
∑ (2i − 1) = 2∑ i − ∑ 1
2(100)(100 + 1)
=
− 100 = 10, 000
2
4.7 Chapter Review
Concepts Test
1. True:
Theorem 4.3.D
2. True:
Obtained by integrating both sides of
the Product Rule
3. True:
If F ( x) = ∫ f ( x) dx, f ( x) is a
derivative of F(x).
288
Section 4.7
9. True:
10
10
10
100
i =1
i =1
i =1
i =1
∑ (ai + 1)2 = ∑ ai2 + 2∑ ai + ∑ 1
= 100 + 2(20) + 10 = 150
10. False:
f must also be continuous except at a
finite number of points on [a, b].
11. True:
The area of a vertical line segment is 0.
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12. False:
∫ −1 x dx
13. False:
A counterexample is
⎧0, x ≠ 0
f ( x) = ⎨
⎩1, x = 0
1
with
is a counterexample.
∫−1 ⎡⎣ f ( x )⎤⎦
1
2
28. False:
f ( x) = x3 is a counterexample.
29. False:
f ( x) = x is a counterexample.
30. True:
All rectangles have height 4,
regardless of xi .
31. True:
F (b) − F (a ) = ∫ F ′( x ) dx
dx = 0 .
If f ( x ) is continuous, then
[ f ( x)]2 ≥ 0 , and if [ f ( x)]2 is greater
than 0 on [a, b], the integral will be
also.
b
a
b
= ∫ G ′( x ) dx = G (b) − G (a )
a
a
∫−a
f ( x) dx = 2∫
a
14. False:
Dx ⎡⎢ ∫ f ( z ) dz ⎤⎥ = f ( x)
⎣ a
⎦
32. False:
15. True:
sin x + cos x has period 2π , so
33. False:
z (t ) = t 2 is a counterexample.
34. False:
∫0
35. True:
Odd-exponent terms cancel
themselves out over the interval, since
they are odd.
36. False:
a = 0, b = 1, f(x) = –1, g(x) = 0 is a
counterexample.
37. False:
a = 0, b = 1, f(x) = –1, g(x) = 0 is a
counterexample.
x
x + 2π
∫x
16. True:
lim kf ( x) = k lim f ( x) and
x→a
x→a
lim [ f ( x) + g ( x) ]
x→a
= lim f ( x) + lim g ( x ) when all the
x→a
x →a
limits exist.
17. True:
sin13 x is an odd function.
18. True:
Theorem 4.2.B
19. False:
The statement is not true if c > d.
20. False:
⎡ x2 1
⎤
2x
Dx ⎢ ∫
dt ⎥ =
0 1+ t2
⎥⎦ 1 + x 2
⎣⎢
f ( x) dx because f
is even.
(sin x + cos x) dx
is independent of x.
0
38. True:
b
f ( x ) dx = F (b) − F (0)
a1 + a2 + a3 + " + an
≤ a1 + a2 + a3 + " + an because
any negative values of ai make the
left side smaller than the right side.
39. True:
Note that − f ( x ) ≤ f ( x ) ≤ f ( x )
21. True:
Both sides equal 4.
22. True:
Both sides equal 4.
40. True:
Definition of Definite Integral
23. True:
If f is odd, then the accumulation
41. True:
Definition of Definite Integral
42. False:
Consider
43. True.
Right Riemann sum always bigger.
44. True.
Midpoint of x coordinate is midpoint
of y coordinate.
45. False.
Trapeziod rule overestimates integral.
46. True.
Parabolic Rule gives exact value for
quadratic and cubic functions.
function F ( x ) = ∫ f ( t ) dt is even,
and use Theorem 4.3.B.
x
0
and so is F ( x ) + C for any C.
24. False:
f ( x) = x 2 is a counterexample.
25. False:
f ( x) = x 2 is a counterexample.
26. False:
f ( x) = x 2 is a counterexample.
27. False:
f ( x) = x 2 , v(x) = 2x + 1 is a
counterexample.
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∫
( )
cos x 2 dx
Section 4.7
289
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Sample Test Problems
12. u = 2 y 3 + 3 y 2 + 6 y, du = (6 y 2 + 6 y + 6) dy
1
5
⎡1
⎤
1. ⎢ x 4 − x3 + 2 x3 / 2 ⎥ =
⎣4
⎦0 4
( y2 + y +1)
5
1 355 −1/ 5
u
du
∫1 5 2 y3 +3 y 2 +6 y dy = 6 ∫11
(
355
=
2
1⎤
13
⎡2
2. ⎢ x3 − 3 x − ⎥ =
3
x
⎣
⎦1 6
1 ⎡5 4/5 ⎤
5
3554 / 5 − 114 / 5
u ⎥
=
⎢
6 ⎣4
24
⎦11
13.
⎡⎛ i ⎞ 2 ⎤ ⎛ 1 ⎞ 7
∑ ⎢⎢⎜⎝ 2 ⎟⎠ − 1⎥⎥ ⎜⎝ 2 ⎟⎠ = 4
i =1 ⎣
⎦
14.
f ′( x) =
15.
∫0 (2 −
)
4
π
⎡1
26 ⎤
50 26 π
3. ⎢ y 3 + 9 cos y − ⎥ =
− +
− 9 cos1
3
y
3 π
3
⎣
⎦1
3
9
77 77
⎡1
⎤
4. ⎢ ( y 2 − 4)3 / 2 ⎥ = −8 3 +
3
3
⎣
⎦4
(
8
−15 −125 + 3 5
⎡3
⎤
5. ⎢ (2 z 2 − 3)4 / 3 ⎥ =
16
⎣16
⎦2
π/2
⎡ 1
⎤
6. ⎢ − cos5 x ⎥
⎣ 5
⎦0
=
)
1
5
7. u = tan(3 x 2 + 6 x ), du = (6 x + 6) sec 2 (3x 2 + 6 x)
1 2
1
u du = u 3 + C
6∫
18
π
1 ⎡ 3 2
1
tan (3x + 6 x) ⎤ = tan 3 (3π2 + 6π )
⎦ 0 18
18 ⎣
8. u = t 4 + 9, du = 4t 3 dt
25
1 25 −1/ 2
1
u
du = ⎡u1/ 2 ⎤ = 1
∫
⎦9
4 9
2⎣
=∫
3
11.
∫ ( x + 1) sin ( x
2
(
290
Section 4.7
(x+5−4
)
x + 1 dx
3
5
16.
1 5 2 3
1 ⎡2
⎤
3x x − 4 dx = ⎢ ( x3 − 4)3 / 2 ⎥
∫
5−2 2
3 ⎣3
⎦2
= 294
17.
4⎛
1 ⎞
∫2 ⎜⎝ 5 − x 2 ⎟⎠ dx =
18.
∑ (3i − 3i −1 )
4
1⎤
39
⎡
⎢5 x + x ⎥ = 4
⎣
⎦2
n
)
i =1
+ 2 x + 3 dx
= (3 − 1) + (32 − 3) + (33 − 32 ) + " + (3n − 3n−1 )
)
1
= ∫ sin x 2 + 2 x + 3 ( 2 x + 2 ) dx
2
1
= ∫ sin u du
2
1
= − cos x 2 + 2 x + 3 + C
2
(
x + 1) 2 dx
8
5
⎡1
⎤
= ⎢ x 2 + 5 x − ( x + 1)3 / 2 ⎥ =
2
3
⎣
⎦0 6
1 ⎡3 5
3 ⎡ 5/3 5/3 ⎤
⎤
(t + 5)5 / 3 ⎥ =
37
−6
≈ 46.9
⎢
⎦
5 ⎣5
⎦1 25 ⎣
⎡
⎤
1
4
10. ⎢
⎥ =
3
⎢⎣ 9 y − 3 y ⎥⎦ 2 27
3
0
2
9.
3
1
1
, f ′(7) =
x+3
10
)
= 3n − 1
19.
10
10
10
i =1
i =1
i =1
∑ (6i 2 − 8i) = 6∑ i 2 − 8∑ i
⎡10(11)(21) ⎤
⎡10(11) ⎤
= 6⎢
⎥ − 8 ⎢ 2 ⎥ = 1870
6
⎣
⎦
⎣
⎦
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20. a.
1 1 1 13
+ + =
2 3 4 12
e.
b. 1 + 0 + (–1) + (–2) + (–3) + (–4) = –9
c.
2
f (− x) dx = − ∫
f ( x) dx = −2
0
24. a.
2
2
+0−
−1 = 0
2
2
1+
78
21. a.
−2
∫0
1
∑n
n=2
50
b.
∑ nx2n
4
∫0
n =1
22.
x − 1 dx =
1
1
(1)(1) + (3)(3) = 5
2
2
b.
2
⎡
⎛ 3i ⎞ ⎤ ⎛ 3 ⎞
⎢
−
16
∑
⎜ ⎟ ⎥⎜ ⎟
n →∞ i =1 ⎢
⎝ n ⎠ ⎥⎦ ⎝ n ⎠
⎣
⎧⎪ n ⎡ 48 27 ⎤ ⎫⎪
= lim ⎨∑ ⎢ − i 2 ⎥ ⎬
3
n→∞ ⎩
⎦ ⎭⎪
⎪ i =1 ⎣ n n
n
A = lim
4
∫0 a x b dx = 1 + 2 + 3 = 6
c.
⎧
9⎡
3 1 ⎤⎫
= lim ⎨48 − ⎢ 2 + + ⎥ ⎬
2⎣
n n2 ⎦ ⎭
n→∞ ⎩
= 48 – 9 = 39
∫1
f ( x) dx = ∫
0
25. a.
b.
c.
d.
∫1
2
∫0
2
= −2∫ f ( x ) dx = 8
2
0
∫−2
d.
∫−2 [ f ( x) + f (− x)] dx
0
f (u ) du = 3(2) = 6
∫0 [ 2 g ( x) − 3 f ( x)] dx
2
2
2
0
0
= 2 ∫ g ( x) − 3∫
2
c.
f ( x) dx = − ∫ f ( x) dx = −4
2
f ( x) dx = 2(−4) = −8
2
f ( x) dx
1
0
2
0
2
0
3 f (u ) du = 3∫
f ( x) dx = 2∫
∫−2 f ( x ) dx = −∫−2 f ( x ) dx
= –4 + 2 = –2
0
2
∫−2
b. Since f ( x ) ≤ 0 , f ( x ) = − f ( x ) and
f ( x) dx + ∫
1
4
4
⎧
27 ⎡ n(n + 1)(2n + 1) ⎤ ⎫
= lim ⎨48 − ⎢
⎥⎬
6
n→∞ ⎩
⎦⎭
n3 ⎣
2
4
⎡1 2 ⎤
⎢2 x ⎥ − 6 = 8 – 6 = 2
⎣
⎦0
⎧⎪ 48 n
27 n ⎫⎪
= lim ⎨ ∑1 − ∑ i 2 ⎬
3
n→∞ ⎩
⎪ n i =1 n i =1 ⎭⎪
23. a.
4
∫0 ( x − a x b) dx = ∫0 x dx − ∫0 a x b dx
g ( x) dx = 0
2
= 2∫
2
0
f ( x) dx + 2 ∫
2
0
f ( x) dx
= 4(–4) = –16
f ( x) dx
= 2(–3) – 3(2) = –12
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∫0 [ 2 g ( x) + 3 f ( x)] dx
2
e.
2
2
0
0
= 2 ∫ g ( x) dx + 3∫
0
26.
27.
2
∫−2
f.
100
∫−100 ( x
−1
∫−4
g ( x) dx = − ∫ g ( x) dx = −5
0
3
G ′( x) = ∫ f (t ) dt
e.
G ( x) = ∫
g ( x ) dg (u )
f.
G ( x) = ∫
−x
G ′( x ) =
G ′( x) =
x
0
0
du = [ g (u )]0g ( x )
du
= g ( g ( x)) − g (0)
G ′( x) = g ′( g ( x)) g ′( x)
0
x
f (−t ) dt = ∫ f (u )(− du )
0
G ′( x) = − f ( x )
30. a.
x dx =
∫1 x
3
1
26
dx = ⎡ x3 ⎤ =
⎣
⎦
1
3
3
4
1
x +1
2
b.
2 ⎡ 3 / 2 ⎤ 4 16
=
x
⎦0 3
3⎣
∫0
3 2
2x
x +1
4
3x2
x +1
6
31.
−
2x 1
dt − ∫
dt
1 t
t
1
1
⋅5 − ⋅2 = 0
f ′(x) =
5x
2x
f ( x) = ∫
1
x +1
2
5x 1
2x
t
dt = ∫
5x 1
1
G ′( x ) = sin 2 x
29. a.
G ′( x ) = f ( x + 1) − f ( x )
b.
32. Left Riemann Sum:
1
2
∫1 1 + x4 dx ≈ 0.125[ f ( x0 ) + f ( x1 ) + …+ f ( x7 )] ≈ 0.2319
Right Riemann Sum:
2
2
1
∫1 1 + x4 dx ≈
f ''(c) =
En = −
1
∫1 1 + x4 dx ≈ 0.125[ f ( x1 ) + f ( x2 ) +…+ f ( x8 )] ≈ 0.1767
Midpoint Riemann Sum:
33.
1
f ( x)
x
x
G ′( x ) =
c.
f ( z ) dz +
0
c2 = 7
c = − 7 ≈ −2.65
b.
x2 0
= − ∫ f (u ) du
−1
⎡ x3 ⎤ = 9c 2
⎣ ⎦ −4
28. a.
x
d.
+ sin 5 x) dx = 0
3 x 2 dx = 3c 2 (−1 + 4)
∫
G ′( x) = −
f ( x) dx
= 2(5) + 3(–4) = –2
1
c.
2
1
∫1 1 + x4 dx ≈ 0.125[ f ( x0.5 ) + f ( x1.5 ) + …+ f ( x7.5 )] ≈ 0.2026
0.125
[ f ( x0 ) + 2 f ( x1 ) + …+ 2 f ( x7 ) + f ( x8 )] ≈ 0.2043
2
4c 2 (5c 4 − 3)
(1 + c 4 )3
(2 − 1)3
(12)82
≤
f ''(c) =
(
(4)(22 ) (5)(24 ) − 3
(1 + 1 )
4
3
) = 154
1
154
f ''(c) ≤
≈ 0.2005
(12)(64)
768
Remark: A plot of f '' shows that in fact f '' ( c ) < 1.5 , so En < 0.002 .
292
Section 4.7
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34.
4
1
∫0 1 + 2 x dx ≈
4
f ( ) (c) =
En = −
35.
n2 >
384
(1 + 2c )5
≤ 384
( 4 − 0 )5
5
( 4 ) c ≤ 4 ⋅ 384 = 8
f
⋅
(
)
180 ⋅ 84
180 ⋅ 84 15
4c 2 (5c 4 − 3)
f ''(c) =
En = −
0.5
[ f ( x0 ) + 4 f ( x1 ) + 2 f ( x2 ) +…+ 4 f ( x7 ) + f ( x8 )] ≈ 1.1050
3
(1 + c )
4 3
(2 − 1)3
12n
2
(
(4)(22 ) (5)(24 ) + 3
≤
(1 + 1 )
4
1
f ''(c) =
12n
2
3
f ''(c) ≤
166
12n 2
) = 166
< 0.0001
166
≈ 138,333 so n > 138,333 ≈ 371.9 Round up to n = 372 .
(12)(0.0001)
Remark: A plot of f '' shows that in fact f '' ( c ) < 1.5 which leads to n = 36 .
36.
4
f ( ) (c) =
En = −
n4 >
384
(1 + 2c )5
≤ 384
( 4 − 0 )5
5
( 4 ) c ≤ 4 ⋅ 384 < 0.0001
f
⋅
(
)
180 ⋅ n 4
180 ⋅ n 4
45 ⋅ 384
≈ 21,845,333 , so n ≈ 68.4 . Round up to n = 69 .
180 ( 0.0001)
37. The integrand is decreasing and concave up. Therefore, we get:
Midpoint Rule, Trapezoidal rule, Left Riemann Sum
Review and Preview Problems
6.
2
1.
1 ⎛1⎞
1 1 1
−⎜ ⎟ = − =
2 ⎝2⎠
2 4 4
2. x − x 2
(
( x + h − x )2 + ( x + h )2 − x 2
(
= h 2 + 2 xh + h 2
)
)
2
2
7. V = (π ⋅ 22 )0.4 = 1.6π
3. the distance between (1, 4 ) and ( 3 4, 4) is 3 4 -1
4. the distance between
⎛y ⎞
3 y , y is 3 y − y
⎜ , y ⎟ and
4
⎝4 ⎠
(
)
8. V = [π (42 − 12 )]1 = 15π
9. V = [π (r22 − r12 )]Δx
10. V = [π (52 − 4.52 )]6 = 28.5π
5. the distance between (2,4) and (1,1) is
(2 − 1)2 + (4 − 1) 2 = 10
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2
⎡ x5 x 4
⎤
11. ∫ x − 2 x + 2 dx = ⎢ −
+ 2x⎥
−1
2
⎢⎣ 5
⎥⎦ −1
2
=
12.
(
4
)
3
12 ⎛ 27 ⎞ 51
−⎜− ⎟ =
5 ⎝ 10 ⎠ 10
∫0 y
3
3
3
dy = ⋅ y 5 3 = ⋅ 35 3 ≈ 3.74
0
5
5
2⎛
2
3 23
⎡
x2 x4 ⎞
x3 x5 ⎤
16
13. ∫ ⎜ 1 −
+
⎟ dx = ⎢ x − + ⎥ =
⎟
0⎜
2
16
6
80
⎝
⎠
⎣⎢
⎦⎥ 0 15
14. Let u = 1 + 94 x; then du = 94 dx and
∫
1+
9
4
4 2 32
x dx = ∫ u du =
u +C
4
9
93
8 ⎛ 9 ⎞
=
⎜1 + x ⎟
27 ⎝ 4 ⎠
3
2
+C
4
Thus,
=
294
4
∫1
3 ⎤
⎡
9
8 ⎛ 9 ⎞ 2⎥
⎢
1 + x dx =
1+ x
⎢ 27 ⎜⎝ 4 ⎟⎠ ⎥
4
⎣
⎦1
8 ⎛ 3 2 133 2 ⎞
⎜ 10 −
⎟ ≈ 7.63
27 ⎜⎝
8 ⎟⎠
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5
CHAPTER
Applications of the Integral
6. To find the intersection points, solve
5.1 Concepts Review
1.
b
∫a
f ( x)dx; − ∫
b
a
x + 4 = x2 − 2 .
f ( x)dx
x2 − x − 6 = 0
(x + 2)(x – 3) = 0
x = –2, 3
Slice vertically.
ΔA ≈ ⎡ ( x + 4) − ( x 2 − 2) ⎤ Δx = (− x 2 + x + 6)Δx
⎣
⎦
2. slice, approximate, integrate
3. g ( x) − f ( x); f ( x) = g ( x)
4.
∫c [ q( y) − p( y)] dy
d
3
3
1
⎡ 1
⎤
A = ∫ (− x 2 + x + 6) dx = ⎢ − x3 + x 2 + 6 x ⎥
−2
2
⎣ 3
⎦ −2
Problem Set 5.1
1. Slice vertically.
9
⎛
⎞ ⎛8
⎞ 125
⎜ −9 + + 18 ⎟ − ⎜ + 2 − 12 ⎟ =
2
6
⎝
⎠ ⎝3
⎠
ΔA ≈ ( x 2 + 1)Δx
2
2
⎡1
⎤
A = ∫ ( x 2 + 1)dx = ⎢ x3 + x ⎥ = 6
–1
3
⎣
⎦ −1
7. Solve x3 − x 2 − 6 x = 0 .
2. Slice vertically.
ΔA ≈ ( x3 − x + 2)Δx
2
33
2
1
⎡1
⎤
A = ∫ ( x3 − x + 2)dx = ⎢ x 4 − x 2 + 2 x ⎥ =
−1
2
⎣4
⎦ −1 4
A=∫
−2
A = A1 + A2
=∫
2
1
⎡1
⎤
( x + x + 2)dx = ⎢ x3 + x 2 + 2 x ⎥
2
⎣3
⎦ −2
0
0
ΔA ≈ −( x 2 + 2 x − 3)Δx = (− x 2 − 2 x + 3)Δx
1
32
⎡ 1
⎤
(− x 2 − 2 x + 3)dx = ⎢ − x3 − x 2 + 3x ⎥ =
−3
⎣ 3
⎦ −3 3
5. To find the intersection points, solve 2 – x 2 = x .
x + x−2 = 0
(x + 2)(x – 1) = 0
x = –2, 1
Slice vertically.
ΔA ≈ ⎡ (2 − x 2 ) − x ⎤ Δx = (− x 2 − x − 2)Δx
⎣
⎦
2
3
⎡ ⎛
8
⎞ ⎤ ⎡ 81
⎤
= ⎢ 0 − ⎜ 4 + − 12 ⎟ ⎥ + ⎢ − + 9 + 27 − 0 ⎥
3
⎠⎦ ⎣ 4
⎦
⎣ ⎝
16 63 253
= +
=
3 4
12
8. To find the intersection points, solve
− x + 2 = x2 .
x2 + x − 2 = 0
(x + 2)(x – 1) = 0
x = –2, 1
Slice vertically.
ΔA ≈ ⎡ (− x + 2) − x 2 ⎤ Δx = (− x 2 − x + 2)Δx
⎣
⎦
1
1
1
⎡ 1
⎤
A = ∫ (– x 2 – x + 2)dx = ⎢ – x3 – x 2 + 2 x ⎥
–2
3
2
⎣
⎦ −2
1
1
8
9
⎛
⎞ ⎛
⎞
= ⎜ − − + 2⎟ − ⎜ − 2 − 4⎟ =
⎝ 3 2
⎠ ⎝3
⎠ 2
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3
( x3 − x 2 − 6 x)dx + ∫ (− x3 + x 2 + 6 x)dx
1
1
⎡ 1
⎤
⎡1
⎤
+ ⎢ − x 4 + x3 + 3 x 2 ⎥
= ⎢ x 4 − x3 − 3 x 2 ⎥
4
3
4
3
⎣
⎦ −2 ⎣
⎦0
4. Slice vertically.
1
0
−2
2
⎛8
⎞ ⎛ 8
⎞ 40
= ⎜ + 2 + 4⎟ − ⎜ − + 2 − 4⎟ =
3
3
⎝
⎠ ⎝
⎠ 3
A=∫
ΔA1 ≈ ( x3 − x 2 − 6 x)Δx
ΔA2 ≈ −( x3 − x 2 − 6 x)Δx = (− x3 + x 2 + 6 x)Δx
3. Slice vertically.
ΔA ≈ ⎡ ( x 2 + 2) − (− x) ⎤ Δx = ( x 2 + x + 2)Δx
⎣
⎦
2
x( x 2 − x − 6) = 0
x(x + 2)(x – 3) = 0
x = –2, 0, 3
Slice vertically.
1
1
⎡ 1
⎤
(− x 2 − x + 2)dx = ⎢ − x3 − x 2 + 2 x ⎥
−2
3
2
⎣
⎦ −2
1
1
8
9
⎛
⎞ ⎛
⎞
= ⎜ − − + 2⎟ − ⎜ − 2 − 4⎟ =
⎝ 3 2
⎠ ⎝3
⎠ 2
A=∫
1
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9. To find the intersection points, solve
13.
y +1 = 3 – y .
2
y2 + y − 2 = 0
(y + 2)(y – 1) = 0
y = –2, 1
Slice horizontally.
ΔA ≈ ⎡ (3 − y 2 ) − ( y + 1) ⎤ Δy = (− y 2 − y + 2)Δy
⎣
⎦
1
1
1
1
A = ∫ (– y 2 – y + 2)dy = ⎡⎢ – y 3 – y 2 + 2 y ⎤⎥
–2
2
⎣ 3
⎦ −2
⎛ 1 1
⎞ ⎛8
⎞ 9
= ⎜ − − + 2⎟ − ⎜ − 2 − 4⎟ =
3
2
3
⎝
⎠ ⎝
⎠ 2
ΔA ≈ −( x − 4)( x + 2)Δx = (− x 2 + 2 x + 8)Δx
3
3
⎡ 1
⎤
A = ∫ (− x 2 + 2 x + 8)dx = ⎢ − x3 + x 2 + 8 x ⎥
0
⎣ 3
⎦0
= –9 + 9 + 24 = 24
Estimate the area to be (3)(8) = 24.
10. To find the intersection point, solve y 2 = 6 − y .
y2 + y − 6 = 0
(y + 3)(y – 2) = 0
y = –3, 2
Slice horizontally.
ΔA ≈ ⎡ (6 − y ) − y 2 ⎤ Δy = (− y 2 − y + 6)Δy
⎣
⎦
14.
2
22
2
1
⎡ 1
⎤
A = ∫ ( − y 2 − y + 6)dy = ⎢ − y 3 − y 2 + 6 y ⎥ =
0
3
2
⎣ 3
⎦0
ΔA ≈ −( x 2 − 4 x − 5)Δx = (− x 2 + 4 x + 5)Δx
11.
4
⎡ 1
⎤
(− x 2 + 4 x + 5)dx = ⎢ − x3 + 2 x 2 + 5 x ⎥
−1
⎣ 3
⎦ −1
⎛ 64
⎞ ⎛1
⎞ 100
= ⎜ − + 32 + 20 ⎟ − ⎜ + 2 − 5 ⎟ =
≈ 33.33
3
3
3
⎝
⎠ ⎝
⎠
1
⎛ 1⎞
Estimate the area to be (5) ⎜ 6 ⎟ = 32 .
2
⎝ 2⎠
A=∫
1 ⎞
⎛
ΔA ≈ ⎜ 3 − x 2 ⎟ Δx
3 ⎠
⎝
4
15.
3
3⎛
1 ⎞
1 ⎤
⎡
A = ∫ ⎜ 3 − x 2 ⎟ dx = ⎢3 x − x3 ⎥ = 9 − 3 = 6
0 ⎝
3 ⎠
9 ⎦0
⎣
Estimate the area to be (3)(2) = 6.
12.
1
ΔA ≈ − ( x 2 − 7)Δx
4
2
2 1
1 ⎡1
⎤
A = ∫ − ( x 2 − 7)dx = − ⎢ x3 − 7 x ⎥
0
4
4 ⎣3
⎦0
1⎛8
⎞ 17
= − ⎜ − 14 ⎟ =
≈ 2.83
4⎝3
⎠ 6
ΔA ≈ (5 x − x 2 )Δx
3
3
1 ⎤
⎡5
A = ∫ (5 x − x 2 )dx = ⎢ x 2 − x3 ⎥ ≈ 11.33
1
3 ⎦1
⎣2
⎛ 1⎞
Estimate the area to be (2) ⎜ 5 ⎟ = 11 .
⎝ 2⎠
296
Section 5.1
⎛ 1⎞
Estimate the area to be (2) ⎜ 1 ⎟ = 3 .
⎝ 2⎠
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16.
19.
ΔA ≈ [ x − ( x − 3)( x − 1) ] Δx
ΔA1 ≈ − x3 Δx
= ⎡ x − ( x 2 − 4 x + 3) ⎤ Δx = (− x 2 + 5 x − 3)Δx
⎣
⎦
To find the intersection points, solve
x = (x – 3)(x – 1).
ΔA2 ≈ x Δx
3
A = A1 + A2 = ∫
0
−3
0
3
− x3 dx + ∫ x3 dx
0
x2 − 5x + 3 = 0
3
⎡ 1 ⎤
⎡1 ⎤
⎛ 81 ⎞ ⎛ 81 ⎞ 81
= ⎢− x4 ⎥ + ⎢ x4 ⎥ = ⎜ ⎟ + ⎜ ⎟ =
⎣ 4 ⎦ −3 ⎣ 4 ⎦ 0 ⎝ 4 ⎠ ⎝ 4 ⎠ 2
= 40.5
Estimate the area to be (3)(7) + (3)(7) = 42.
5 ± 25 − 12
2
5 ± 13
x=
2
x=
17.
A=∫
5+ 13
2
5− 13
2
(− x 2 + 5 x − 3)dx
5+ 13
5
13 13
⎡ 1
⎤ 2
= ⎢ − x3 + x 2 − 3 x ⎥
=
≈ 7.81
−
5
13
2
6
⎣ 3
⎦
2
ΔA1 ≈ − x Δx
3
Estimate the area to be
ΔA2 ≈ 3 x Δx
A = A1 + A2 = ∫
0
−2
2
− 3 x dx + ∫ 3 x dx
0
1
(4)(4) = 8 .
2
20.
⎛ 3 3 2 ⎞ ⎛ 33 2 ⎞
⎡ 3
⎤
⎡3
⎤
= ⎢− x4 / 3 ⎥ + ⎢ x4 / 3 ⎥ = ⎜
⎟+⎜
⎟
⎣ 4
⎦ −2 ⎣ 4
⎦ 0 ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
0
2
= 33 2 ≈ 3.78
Estimate the area to be (2)(1) + (2)(1) = 4.
18.
ΔA ≈ ⎣⎡ x − ( x − 4) ⎦⎤ Δx =
(
)
x − x + 4 Δx
To find the intersection point, solve
x = ( x − 4) .
x = ( x − 4)2
x 2 − 9 x + 16 = 0
ΔA ≈ −( x − 10)Δx = (10 − x )Δx
9
9
2
⎡
⎤
A = ∫ (10 − x ) dx = ⎢10 x − x3 2 ⎥
0
3
⎣
⎦0
= 90 – 18 = 72
Estimate the area to be 9 · 8 =72.
Instructor’s Resource Manual
9 ± 81 − 64
2
9 ± 17
x=
2
⎛
9 − 17
9 + 17 ⎞
is extraneous so x =
. ⎟⎟
⎜⎜ x =
2
2
⎝
⎠
x=
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A=∫
9+ 17
2
0
(
x2 + 5x + 6 = 0
(x + 3)(x + 2) = 0
x = –3, –2
)
x − x + 4 dx
9+ 17
1
⎡2
⎤ 2
= ⎢ x3 / 2 − x 2 + 4 x ⎥
2
⎣3
⎦0
2 ⎛ 9 + 17 ⎞
= ⎜⎜
⎟
3 ⎝ 2 ⎟⎠
3/ 2
2 ⎛ 9 + 17 ⎞
= ⎜⎜
⎟
3 ⎝ 2 ⎟⎠
3/ 2
A=∫
−2
23
17
−
≈ 15.92
4
4
Estimate the area to be
(− x 2 − 5 x − 6)dx
5
⎡ 1
⎤
= ⎢ − x3 − x 2 − 6 x ⎥
2
⎣ 3
⎦ −3
45
⎛8
⎞ ⎛
⎞ 1
= ⎜ − 10 + 12 ⎟ − ⎜ 9 − + 18 ⎟ = ≈ 0.17
3
2
⎝
⎠ ⎝
⎠ 6
1 ⎛
2⎞ 1
Estimate the area to be (1) ⎜ 5 − 4 ⎟ = .
2 ⎝
3⎠ 6
2
⎛ 9 + 17 ⎞
1 ⎛ 9 + 17 ⎞
− ⎜⎜
⎟⎟ + 4 ⎜⎜
⎟⎟
2⎝ 2 ⎠
⎝ 2 ⎠
+
−2
−3
1 ⎛ 1 ⎞⎛ 1 ⎞
1
⎜ 5 ⎟ ⎜ 5 ⎟ = 15 .
2 ⎝ 2 ⎠⎝ 2 ⎠
8
23.
21.
ΔA ≈ (8 y − y 2 )Δy
To find the intersection points, solve
ΔA ≈ ⎡ − x 2 − ( x 2 − 2 x) ⎤ Δx = (−2 x 2 + 2 x)Δx
⎣
⎦
To find the intersection points, solve
8 y − y2 = 0 .
y(8 – y) = 0
y = 0, 8
− x2 = x2 − 2 x .
2 x2 − 2 x = 0
2x(x – 1) = 0
x = 0, x = 1
8
8
1 ⎤
⎡
A = ∫ (8 y − y 2 ) dy = ⎢ 4 y 2 − y 3 ⎥
0
3 ⎦0
⎣
512 256
= 256 −
=
≈ 85.33
3
3
Estimate the area to be (16)(5) = 80.
1
1
⎡ 2
⎤
A = ∫ (−2 x 2 + 2 x )dx = ⎢ − x3 + x 2 ⎥
0
⎣ 3
⎦0
2
1
= − + 1 = ≈ 0.33
3
3
⎛ 1 ⎞⎛ 1 ⎞ 1
Estimate the area to be ⎜ ⎟⎜ ⎟ = .
⎝ 2 ⎠⎝ 2 ⎠ 4
24.
22.
ΔA ≈ (3 − y )( y + 1)Δy = (− y 2 + 2 y + 3)Δy
ΔA ≈ ⎡ ( x 2 − 9) − (2 x − 1)( x + 3) ⎤ Δx
⎣
⎦
= ⎡( x 2 − 9) − (2 x 2 + 5 x − 3) ⎤ Δx
⎣
⎦
= (− x 2 − 5 x − 6)Δx
To find the intersection points, solve
3
⎡ 1
⎤
(− y 2 + 2 y + 3)dy = ⎢ − y 3 + y 2 + 3 y ⎥
−1
⎣ 3
⎦ −1
⎛1
⎞ 32
= (−9 + 9 + 9) − ⎜ + 1 − 3 ⎟ =
≈ 10.67
⎝3
⎠ 3
⎛ 1⎞
Estimate the area to be (4) ⎜ 2 ⎟ = 10 .
⎝ 2⎠
A=∫
3
(2 x − 1)( x + 3) = x 2 − 9 .
298
Section 5.1
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25.
27.
ΔA ≈ ⎡(3 − y 2 ) − 2 y 2 ⎤ Δy = (−3 y 2 + 3)Δy
⎣
⎦
To find the intersection points, solve
ΔA ≈ ⎡ (−6 y 2 + 4 y ) − (2 − 3 y ) ⎤ Δy
⎣
⎦
2 y2 = 3 − y2 .
= (−6 y 2 + 7 y − 2)Δy
To find the intersection points, solve
3y2 − 3 = 0
3(y + 1)(y – 1) = 0
y = –1, 1
−6 y 2 + 4 y = 2 − 3 y.
6 y2 − 7 y + 2 = 0
(2 y − 1)(3 y − 2) = 0
−1
1 2
y= ,
2 3
2/3
7
⎡
⎤
(−6 y 2 + 7 y − 2)dy = ⎢ −2 y 3 + y 2 − 2 y ⎥
1/ 2
2
⎣
⎦1/ 2
1
16
14
4
1
7
⎛
⎞ ⎛
⎞
≈ 0.0046
= ⎜ − + − ⎟ − ⎜ − + − 1⎟ =
216
⎝ 27 9 3 ⎠ ⎝ 4 8 ⎠
Estimate the area to be
1 ⎛ 1 ⎞⎛ 1 ⎞ 1 ⎛ 1 ⎞⎛ 1 ⎞
1
.
⎜ ⎟⎜ ⎟ − ⎜ ⎟⎜ ⎟ =
2 ⎝ 2 ⎠⎝ 5 ⎠ 2 ⎝ 2 ⎠⎝ 6 ⎠ 120
A=∫
1
1
(−3 y 2 + 3)dy = ⎡ − y 3 + 3 y ⎤
⎣
⎦ −1
= (–1 + 3) – (1 – 3) = 4
Estimate the value to be (2)(2) = 4.
A=∫
2/3
26.
28.
ΔA ≈ ⎡(8 − 4 y 4 ) − (4 y 4 ) ⎤ Δy = (8 − 8 y 4 )Δy
⎣
⎦
To find the intersection points, solve
4 y4 = 8 − 4 y4 .
8 y4 = 8
y4 = 1
y = ±1
1
8 ⎤
⎡
(8 − 8 y 4 )dy = ⎢8 y − y 5 ⎥
−1
5 ⎦ −1
⎣
8⎞ ⎛
8 ⎞ 64
⎛
= ⎜ 8 − ⎟ − ⎜ −8 + ⎟ =
= 12.8
5⎠ ⎝
5⎠ 5
⎝
⎛ 1⎞
Estimate the area to be (8) ⎜ 1 ⎟ = 12 .
⎝ 2⎠
ΔA ≈ ⎡ ( y + 4) − ( y 2 − 2 y ) ⎤ Δy = (− y 2 + 3 y + 4)Δy
⎣
⎦
To find the intersection points, solve
A=∫
y2 − 2 y = y + 4 .
y2 − 3y − 4 = 0
(y + 1)(y – 4) = 0
y = –1, 4
1
4
3
⎡ 1
⎤
(− y 2 + 3 y + 4)dy = ⎢ − y 3 + y 2 + 4 y ⎥
−1
2
⎣ 3
⎦ −1
⎛ 64
⎞ ⎛1 3
⎞ 125
= ⎜ − + 24 + 16 ⎟ − ⎜ + − 4 ⎟ =
≈ 20.83
3
3
2
6
⎝
⎠ ⎝
⎠
Estimate the area to be (7)(3) = 21.
A=∫
4
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29.
30.
y = x3
y = x+6
2y + x = 0
An equation of the line through (–1, 4) and (5, 1)
1
7
is y = − x + . An equation of the line through
2
2
(–1, 4) and (2, –2) is y = –2x + 2. An equation of
the line through (2, –2) and (5, 1) is y = x – 4.
Two integrals must be used. The left-hand part of
the triangle has area
2 ⎡ 1
2 ⎛3
7
3⎞
⎤
∫−1 ⎢⎣− 2 x + 2 − (−2 x + 2) ⎥⎦ dx = ∫−1 ⎜⎝ 2 x + 2 ⎟⎠ dx .
The right-hand part of the triangle has area
5⎡ 1
5⎛ 3
7
15 ⎞
⎤
∫2 ⎣⎢− 2 x + 2 − ( x − 4)⎥⎦ dx = ∫2 ⎝⎜ − 2 x + 2 ⎠⎟ dx .
The triangle has area
2 ⎛3
5⎛ 3
3⎞
15 ⎞
∫−1⎜⎝ 2 x + 2 ⎟⎠ dx + ∫2 ⎜⎝ − 2 x + 2 ⎟⎠ dx
Let R1 be the region bounded by 2y + x = 0,
y = x + 6, and x = 0.
0 ⎡
⎛ 1 ⎞⎤
A( R1 ) = ∫ ⎢ ( x + 6) − ⎜ − x ⎟ ⎥ dx
−4 ⎣
⎝ 2 ⎠⎦
⎛3
⎞
⎜ x + 6 ⎟ dx
⎝2
⎠
Let R2 be the region bounded by y = x + 6,
=∫
0
−4
y = x3 , and x = 0.
2
2
A( R2 ) = ∫ ⎡ ( x + 6) − x3 ⎤ dx = ∫ (− x3 + x + 6)dx
⎦
0 ⎣
0
A( R ) = A( R1 ) + A( R2 )
=∫
0
−4
2
⎛3
⎞
3
⎜ x + 6 ⎟ dx + ∫0 (− x + x + 6)dx
⎝2
⎠
0
2
31.
9
∫−1 (3t
2
5
3 ⎤
15 ⎤
⎡3
⎡ 3
= ⎢ x2 + x ⎥ + ⎢− x2 + x ⎥
2 ⎦ −1 ⎣ 4
2 ⎦2
⎣4
27 27 27
=
+
=
= 13.5
4
4
2
2
1
⎡3
⎤
⎡ 1
⎤
= ⎢ x2 + 6 x ⎥ + ⎢− x4 + x2 + 6 x ⎥
2
⎣4
⎦ −4 ⎣ 4
⎦0
= 12 + 10 = 22
9
− 24t + 36)dt = ⎡t 3 − 12t 2 + 36t ⎤ = (729 – 972 + 324) – (–1 – 12 – 36) = 130
⎣
⎦ −1
The displacement is 130 ft. Solve 3t 2 − 24t + 36 = 0 .
3(t – 2)(t – 6) = 0
t = 2, 6
⎧⎪3t 2 − 24t + 36 t ≤ 2, t ≥ 6
V (t ) = ⎨
2
⎪⎩−3t + 24t − 36 2 < t < 6
9
∫−1 3t
2
− 24t + 36 dt = ∫
2
−1
2
6
9
2
6
(3t 2 − 24t + 36) dt + ∫ (−3t 2 + 24t − 36) dt + ∫ (3t 2 − 24t + 36) dt
6
9
= ⎡t 3 − 12t 2 + 36t ⎤ + ⎡ −t 3 + 12t 2 − 36t ⎤ + ⎡t 3 − 12t 2 + 36t ⎤ = 81 + 32 + 81 = 194
⎣
⎦ −1 ⎣
⎦2 ⎣
⎦6
The total distance traveled is 194 feet.
32.
300
3π / 2
1 ⎞ 3π
⎛1
⎞
⎡1 1
⎤
⎛ 3π 1 ⎞ ⎛
∫0 ⎜⎝ 2 + sin 2t ⎟⎠ dt = ⎣⎢ 2 t − 2 cos 2t ⎦⎥ 0 = ⎜⎝ 4 + 2 ⎟⎠ − ⎜⎝ 0 − 2 ⎟⎠ = 4 + 1
3π
1
3π
+ 1 ≈ 3.36 feet . Solve + sin 2t = 0 for 0 ≤ t ≤
.
The displacement is
4
2
2
1
7π 11π
7π 11π
,
sin 2t = − ⇒ 2t =
,
⇒t=
6 6
2
12 12
3π / 2
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7 π 11π
3π
⎧1
0≤t≤
,
≤t≤
⎪⎪ 2 + sin 2t
1
12
12
2
+ sin 2t = ⎨
1
7
11
π
π
2
⎪− − sin 2t
<t <
⎪⎩ 2
12
12
3π / 2 1
7 π /12 ⎛ 1
11π /12 ⎛ 1
3π / 2 ⎛ 1
⎞
⎞
⎞
∫0 2 + sin 2t dt = ∫0 ⎜⎝ 2 + sin 2t ⎟⎠ dt + ∫7π /12 ⎜⎝ − 2 − sin 2t ⎟⎠ dt + ∫11π /12 ⎜⎝ 2 + sin 2t ⎟⎠ dt
7 π /12
⎡1 1
⎤
= ⎢ t − cos 2t ⎥
⎣2 2
⎦0
11π /12
3π / 2
⎡ 1 1
⎤
⎡1 1
⎤
+ ⎢ − t + cos 2t ⎥
+ ⎢ t − cos 2t ⎥
⎣ 2 2
⎦ 7 π /12 ⎣ 2 2
⎦11π /12
⎛ 7π
3 1⎞ ⎛ π
3 ⎞ ⎛ 7π
3 1 ⎞ 5π
= ⎜⎜
+
+ ⎟⎟ + ⎜⎜ − +
+
+ ⎟⎟ =
+ 3 +1
⎟⎟ + ⎜⎜
⎝ 24 4 2 ⎠ ⎝ 6 2 ⎠ ⎝ 24 4 2 ⎠ 12
5π
The total distance traveled is
+ 3 + 1 ≈ 4.04 feet.
12
33. s (t ) = ∫ v(t )dt = ∫ (2t − 4)dt = t 2 − 4t + C
Slicing the region horizontally, the area is
1
1
5
5
⎛ 1 ⎞
∫1/ 36 y dy + ⎜⎝ 36 ⎟⎠ (5) . Since 36 < 12 the
c.
Since s(0) = 0, C = 0 and s (t ) = t 2 − 4t. s = 12
when t = 6, so it takes the object 6 seconds to get
s = 12.
⎧4 − 2t 0 ≤ t < 2
2t − 4 = ⎨
⎩2t − 4 2 ≤ t
line that bisects the area is between y =
and y = 1, so we find d such that
1
1 1
5 1 1
∫d y dy = 12 ; ∫d y dy = ⎡⎣ 2 y ⎤⎦ d
2
2
2t − 4 dt = ⎡ −t 2 + 4t ⎤ = 4, so the object
⎣
⎦0
travels a distance of 4 cm in the first two seconds.
∫0
x
∫2
= 2−2 d ; 2−2 d =
x
2t − 4 dt = ⎡t 2 − 4t ⎤ = x 2 − 4 x + 4
⎣
⎦2
361
≈ 0.627 .
576
The line y = 0.627 approximately bisects the
area.
takes 2 + 2 2 ≈ 4.83 seconds to travel a total
distance of 12 centimeters.
6
6
1
5
⎡ 1⎤
A = ∫ x −2 dx = ⎢ − ⎥ = − + 1 =
1
6
6
⎣ x ⎦1
b. Find c so that
c −2
∫1 x
dx =
c
5
.
12
1
⎡ 1⎤
= ⎢− ⎥ = 1 −
∫
c
⎣ x ⎦1
1 5
12
1− = , c =
c 12
7
12
The line x =
bisects the area.
7
c −2
x dx
1
5
;
12
d=
x 2 − 4 x + 4 = 8 when x = 2 + 2 2, so the object
34. a.
1
36
35. Equation of line through (–2, 4) and (3, 9):
y=x+6
Equation of line through (2, 4) and (–3, 9):
y = –x + 6
0
3
–3
0
A( A) = ∫ [9 – (– x + 6)]dx + ∫ [9 – ( x + 6)]dx
0
3
= ∫ (3 + x)dx + ∫ (3 – x)dx
–3
0
0
3
1 ⎤
1 ⎤
9 9
⎡
⎡
= ⎢3 x + x 2 ⎥ + ⎢3 x – x 2 ⎥ = + = 9
2
2
⎣
⎦ −3 ⎣
⎦0 2 2
A( B ) = ∫
–2
–3
[(– x + 6) – x 2 ]dx
0
+ ∫ [(– x + 6) – ( x + 6)]dx
–2
=∫
–2
–3
0
(– x 2 – x + 6)dx + ∫ (–2 x)dx
–2
−2
0
1
37
⎡ 1
⎤
= ⎢ – x3 – x 2 + 6 x ⎥ + ⎡ – x 2 ⎤ =
⎣
⎦
−2
2
6
⎣ 3
⎦ −3
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A(C ) = A( B) =
π/6
37
(by symmetry)
6
0
2
–2
0
A( D) = ∫ [( x + 6) – x 2 ]dx + ∫ [(– x + 6) – x 2 ]dx
0
2
1
1
⎡ 1
⎤
⎡ 1
⎤
= ⎢ – x3 + x 2 + 6 x ⎥ + ⎢ – x3 – x 2 + 6 x ⎥
2
2
⎣ 3
⎦ −2 ⎣ 3
⎦0
44
=
3
A(A) + A(B) + A(C) + A(D) = 36
3
3
1 ⎤
⎡
A( A + B + C + D) = ∫ (9 – x 2 )dx = ⎢9 x – x3 ⎥
–3
3 ⎦ −3
⎣
= 36
⎡1
⎤
= ⎢ x + cos x ⎥
⎣2
⎦0
5π / 6
1 ⎤
⎡
+ ⎢ − cos x − x ⎥
2 ⎦π / 6
⎣
13π / 6
17 π / 6
1 ⎤
⎡1
⎤
⎡
+ ⎢ x + cos x ⎥
+ ⎢ − cos x − x ⎥
2 ⎦13π / 6
⎣2
⎦ 5π / 6 ⎣
⎛π
3 ⎞ ⎛
π⎞ ⎛
2π ⎞
= ⎜⎜ +
− 1⎟⎟ + ⎜ 3 − ⎟ + ⎜ 3 +
⎟
12
2
3
3 ⎠
⎝
⎠
⎝
⎝
⎠
π⎞ π 7 3
⎛
+⎜ 3 − ⎟ = +
− 1 ≈ 5.32
3 ⎠ 12
2
⎝
5.2 Concepts Review
36. Let f(x) be the width of region 1 at every x.
b
ΔA1 ≈ f ( x)Δx, so A1 = ∫ f ( x)dx .
a
Let g(x) be the width of region 2 at every x.
b
ΔA2 ≈ g ( x)Δx, so A2 = ∫ g ( x)dx .
a
Since f(x) = g(x) at every x in [a, b],
b
b
a
a
A1 = ∫ f ( x)dx = ∫ g ( x)dx = A2 .
1. πr 2 h
2. π( R 2 − r 2 )h
3. πx 4 Δx
4. π[( x 2 + 2)2 − 4]Δx
37. The height of the triangular region is given by
for 0 ≤ x ≤ 1 . We need only show that the
height of the second region is the same in order
to apply Cavalieri'’s Principle. The height of the
second region is
h2 = ( x 2 − 2 x + 1) − ( x 2 − 3 x + 1)
= x2 − 2 x + 1 − x2 + 3x − 1
= x for 0 ≤ x ≤ 1.
Since h1 = h2 over the same closed interval, we
can conclude that their areas are equal.
Problem Set 5.2
1. Slice vertically.
ΔV ≈ π( x 2 + 1) 2 Δx = π( x 4 + 2 x 2 + 1)Δx
2
V = π∫ ( x 4 + 2 x 2 + 1)dx
0
2
2
⎡1
⎤
⎛ 32 16
⎞ 206π
= π ⎢ x5 + x3 + x ⎥ = π ⎜ + + 2 ⎟ =
3
15
⎣5
⎦0
⎝ 5 3
⎠
≈ 43.14
38. Sketch the graph.
2. Slice vertically.
ΔV ≈ π(− x 2 + 4 x)2 Δx = π( x 4 − 8 x3 + 16 x 2 )Δx
3
V = π∫ ( x 4 − 8 x3 + 16 x 2 )dx
0
3
16 ⎤
⎡1
= π ⎢ x5 − 2 x 4 + x3 ⎥
3 ⎦0
⎣5
1
17π
.
for 0 ≤ x ≤
2
6
π 5π 13π 17π
x= , ,
,
6 6 6
6
The area of the trapped region is
π/6 ⎛ 1
5π / 6 ⎛
1⎞
⎞
∫0 ⎜⎝ 2 − sin x ⎟⎠ dx + ∫π / 6 ⎜⎝ sin x − 2 ⎟⎠ dx
13π / 6 ⎛ 1
17 π / 6 ⎛
1⎞
⎞
+∫
− sin x ⎟ dx + ∫
⎜ sin x − ⎟ dx
π
5π / 6 ⎜⎝ 2
13
/
6
2⎠
⎠
⎝
Solve sin x =
302
Section 5.2
⎛ 243
⎞
= π⎜
− 162 + 144 ⎟
5
⎝
⎠
153π
=
≈ 96.13
5
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3. a.
Slice vertically.
6.
ΔV ≈ π(4 − x ) Δx = π(16 − 8 x + x )Δx
2 2
2
4
2
V = π∫ (16 – 8 x 2 + x 4 )dx
0
=
256π
≈ 53.62
15
b. Slice horizontally.
x = 4– y
Note that when x = 0, y = 4.
ΔV ≈ π
(
4− y
)
2
ΔV ≈ π( x3 )2 Δx = πx6 Δx
Δy = π(4 − y )Δy
3
3
2187π
⎡1 ⎤
V = π∫ x 6 dx = π ⎢ x7 ⎥ =
≈ 981.52
0
7
7
⎣
⎦0
4
4
1 ⎤
⎡
V = π∫ (4 – y )dy = π ⎢ 4 y – y 2 ⎥
0
2 ⎦0
⎣
= π(16 − 8) = 8π ≈ 25.13
4. a.
7.
Slice vertically.
ΔV ≈ π(4 − 2 x)2 Δx
0≤ x≤2
2
2
⎡ 1
⎤
V = π∫ (4 − 2 x)2 dx = π ⎢ − (4 − 2 x)3 ⎥
0
⎣ 6
⎦0
32π
=
≈ 33.51
3
2
⎛ 1 ⎞
⎛1⎞
ΔV ≈ π ⎜ ⎟ Δx = π ⎜ ⎟ Δx
⎝x⎠
⎝ x2 ⎠
b. Slice vertically.
y
x = 2−
2
V = π∫
4
2
≈ 0.79
2
y⎞
⎛
ΔV ≈ π ⎜ 2 − ⎟ Δy
2
⎝
⎠
0 ≤ y ≤ 4
4
⎡ 1⎤
⎛ 1 1⎞ π
dx = π ⎢ − ⎥ = π ⎜ − + ⎟ =
2
⎣ x ⎦2
⎝ 4 2⎠ 4
x
1
8.
4
2
3
⎡ 2⎛
y⎞
y⎞ ⎤
V = π∫ ⎜ 2 − ⎟ dy = π ⎢− ⎜ 2 − ⎟ ⎥
0⎝
2⎠
3
2⎠ ⎥
⎣⎢ ⎝
⎦0
16π
=
≈ 16.76
3
4⎛
5.
ΔV ≈ π( x3 / 2 ) 2 Δx = πx3Δx
3
3
⎡1 ⎤
⎛ 81 16 ⎞
V = π∫ x3 dx = π ⎢ x 4 ⎥ = π ⎜ − ⎟
2
⎣ 4 ⎦2
⎝ 4 4⎠
65π
=
≈ 51.05
4
⎛ x2
ΔV ≈ π ⎜
⎜ π
⎝
V =∫
4
0
2
⎞
x4
Δx
⎟ Δx =
⎟
π
⎠
4
x4
1 ⎡1 ⎤
1024
dx = ⎢ x5 ⎥ =
≈ 65.19
π
π ⎣ 5 ⎦0
5π
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12.
9.
2
2
⎛ 1 ⎞
⎛2⎞
ΔV ≈ π ⎜ ⎟ Δy = 4π ⎜
Δy
⎜ y 2 ⎟⎟
⎝ y⎠
⎝
⎠
ΔV ≈ π ⎛⎜ 9 − x 2 ⎞⎟ Δx = π(9 − x 2 )Δx
⎝
⎠
V = π∫
3
−2
3
1 ⎤
⎡
(9 − x 2 )dx = π ⎢9 x − x3 ⎥
3 ⎦ −2
⎣
V = 4π ∫
2
⎡
8 ⎞ ⎤ 100π
⎛
= π ⎢(27 − 9) − ⎜ −18 + ⎟ ⎥ =
≈ 104.72
3 ⎠⎦
3
⎝
⎣
10.
6
=
6
⎡ 1⎤
⎛ 1 1⎞
dy = 4π ⎢ − ⎥ = 4π ⎜ − + ⎟
2
⎝ 6 2⎠
y
⎣ y ⎦2
1
4π
≈ 4.19
3
13.
ΔV ≈ π( x 2 / 3 ) 2 Δx = πx 4 / 3Δx
V = π∫
27
1
=
(
ΔV ≈ π 2 y
27
x
4/3
⎡3
⎤
⎛ 6561 3 ⎞
dx = π ⎢ x 7 / 3 ⎥ = π ⎜
− ⎟
7
7⎠
⎣
⎦1
⎝ 7
6558π
≈ 2943.22
7
)
2
Δy = 4πy Δy
4
4
⎡1 ⎤
V = 4π ∫ y dy = 4π ⎢ y 2 ⎥ = 32π ≈ 100.53
0
⎣ 2 ⎦0
14.
11.
ΔV ≈ π( y 2 ) 2 Δy = πy 4 Δy
3 4
y dy
0
V = π∫
304
3
243π
⎡1 ⎤
= π ⎢ y5 ⎥ =
≈ 152.68
5
⎣ 5 ⎦0
Section 5.2
ΔV ≈ π( y 2 / 3 )2 Δy = πy 4 / 3 Δy
27
6561π
⎡3
⎤
y 4 / 3 dy = π ⎢ y 7 / 3 ⎥ =
0
7
⎣7
⎦0
≈ 2944.57
V = π∫
27
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18. Sketch the region.
15.
y
y = 6 x2
8
y = 6x
−1
1
ΔV ≈ π( y 3 / 2 )2 Δy = πy 3Δy
V = π∫
9
0
9
6561π
⎡1 ⎤
y dy = π ⎢ y 4 ⎥ =
≈ 5153.00
4
4
⎣
⎦0
3
16.
x
−4
To find the intersection points, solve 6 x = 6 x 2 .
6( x 2 − x) = 0
6x(x – 1) = 0
x = 0, 1
ΔV ≈ π ⎡ (6 x) 2 − (6 x 2 ) 2 ⎤ Δx = 36π( x 2 − x 4 )Δx
⎣
⎦
1
1
1 ⎤
⎡1
V = 36π ∫ ( x 2 − x 4 )dx = 36π ⎢ x3 − x5 ⎥
0
3
5 ⎦0
⎣
⎛ 1 1 ⎞ 24π
= 36π ⎜ − ⎟ =
≈ 15.08
5
⎝3 5⎠
19. Sketch the region.
2
ΔV ≈ π ⎛⎜ 4 − y 2 ⎞⎟ Δy = π(4 − y 2 )Δy
⎝
⎠
2
1 ⎤
⎡
(4 − y 2 )dy = π ⎢ 4 y − y 3 ⎥
−2
3 ⎦ −2
⎣
V = π∫
2
8 ⎞ ⎤ 32π
⎡⎛ 8 ⎞ ⎛
= π ⎢⎜ 8 − ⎟ − ⎜ −8 + ⎟ ⎥ =
≈ 33.51
3
3 ⎠⎦
3
⎠ ⎝
⎣⎝
17. The equation of the upper half of the ellipse is
y = b 1−
V = π∫
a
x2
a2
b2
−a a2
or y =
b 2
a − x2 .
a
(a 2 − x 2 )dx
3 ⎤a
b2 π ⎡ 2
x
⎢a x − ⎥
2
3 ⎦⎥
a ⎣⎢
−a
2 ⎡⎛
3⎞ ⎛
b π
a
a3 ⎞ ⎤ 4
a3 − ⎟ − ⎜ − a3 + ⎟ ⎥ = ab 2 π
=
⎢
⎜
3 ⎟⎠ ⎜⎝
3 ⎟⎠ ⎥⎦ 3
a 2 ⎢⎣⎜⎝
=
To find the intersection points, solve
x
=2 x.
2
x2
= 4x
4
x 2 − 16 x = 0
x(x – 16) = 0
x = 0, 16
⎡
ΔV ≈ π ⎢ 2 x
⎢⎣
(
)
2
⎛ x⎞
−⎜ ⎟
⎝2⎠
2⎤
⎛
x2 ⎞
⎥ Δx = π ⎜ 4 x − ⎟ Δx
⎜
4 ⎟⎠
⎥⎦
⎝
16
⎛
⎡
x2 ⎞
x3 ⎤
V = π ∫ ⎜ 4 x − ⎟ dx = π ⎢ 2 x 2 − ⎥
0 ⎜
4 ⎟⎠
12 ⎦⎥
⎝
⎣⎢
0
1024 ⎞ 512π
⎛
= π ⎜ 512 −
≈ 536.17
⎟=
3 ⎠
3
⎝
16
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20. Sketch the region.
3 2
1
x + 3, y = x 2 + 5
16
16
Sketch the region.
22. y =
r
1 ⎤
⎡
(r 2 − x 2 )dx = π ⎢ r 2 x − x3 ⎥
r −h
3 ⎦ r −h
⎣
V = π∫
r
To find the intersection point, solve
3 2
1
x + 3 = x2 + 5 .
16
16
1 2
x −2 = 0
8
1
= πh 2 (3r − h)
3
21. Sketch the region.
y
x 2 − 16 = 0
(x + 4)(x – 4) = 0
x = –4, 4
2
2
4 ⎡⎛ 1
⎞ ⎛ 3
⎞ ⎤
V = π∫ ⎢⎜ x 2 + 5 − 2 ⎟ − ⎜ x 2 + 3 − 2 ⎟ ⎥ dx
0 ⎢⎝ 16
⎠ ⎝ 16
⎠ ⎦⎥
⎣
4
2
−1
1
2
4 ⎡⎛ 1
3
⎞
x4 − x2 + 9 ⎟
= π ∫ ⎢⎜
0 ⎣⎝ 256
8
⎠
x
⎛ 9 4 3 2 ⎞⎤
x − x + 1⎟⎥ dx
−⎜
8
⎝ 256
⎠⎦
y
y
.
To find the intersection points, solve =
4
2
y2 y
=
16 4
y2 − 4 y = 0
y(y – 4) = 0
y = 0, 4
2
⎡⎛
⎤
⎛ y y2 ⎞
y ⎞ ⎛ y ⎞2 ⎥
⎢
ΔV ≈ π ⎜
− ⎜ ⎟ Δy = π ⎜ −
⎟ Δy
⎟
⎜ 4 16 ⎟
⎢⎜ 2 ⎟ ⎝ 4 ⎠ ⎥
⎝
⎠
⎝
⎠
⎣
⎦
4
4⎛
1 5⎤
1
⎡
⎞
x
= π ∫ ⎜ 8 − x 4 ⎟ dx = π ⎢8 x −
0⎝
32 ⎠
160 ⎥⎦ 0
⎣
32 ⎞ 128π
⎛
= π ⎜ 32 − ⎟ =
≈ 80.42
5 ⎠
5
⎝
23.
4
⎡ y 2 y3 ⎤
4 ⎛ y y2 ⎞
V = π∫ ⎜ −
⎟ dy = π ⎢ − ⎥
0 ⎜ 4 16 ⎟
⎝
⎠
⎣⎢ 8 48 ⎥⎦ 0
2π
=
≈ 2.0944
3
The square at x has sides of length 2 4 − x 2 , as
shown.
2
⎛ 2 4 − x 2 ⎞ dx = 2 4(4 − x 2 )dx
⎜
⎟
∫−2
−2 ⎝
⎠
V =∫
2
2
⎡
x3 ⎤
⎡⎛ 8 ⎞ ⎛
8 ⎞ ⎤ 128
= 4 ⎢ 4 x − ⎥ = 4 ⎢ ⎜ 8 − ⎟ − ⎜ −8 + ⎟ ⎥ =
3
3
3 ⎠⎦
3
⎠ ⎝
⎣⎝
⎣⎢
⎦⎥ −2
≈ 42.67
306
Section 5.2
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24. The area of each cross section perpendicular to
1
the x-axis is (4) ⎛⎜ 2 4 − x 2 ⎞⎟ = 4 4 − x 2 .
⎠
2 ⎝
The area of a semicircle with radius 2 is
2
∫−2
Thus, the volume inside the “+” for two cylinders
of radius r and length L is
V = vol. of cylinders - vol. of common region
⎛2 ⎞
= 2(π r 2 L) − 8 ⎜ r 3 ⎟
⎝3 ⎠
16
= 2π r 2 L − r 3
3
4 − x 2 dx = 2π .
2
V = ∫ 4 4 − x 2 dx = 4(2π) = 8π ≈ 25.13
−2
25. The square at x has sides of length
V =∫
π/2
−π / 2
cos x .
cos xdx = [sin x]π−π/ 2/ 2 = 2
26. The area of each cross section perpendicular to
the x-axis is [(1 − x 2 ) − (1 − x 4 )]2 = x8 − 2 x 6 + x 4 .
1
V = ∫ ( x8 − 2 x6 + x 4 )dx
−1
1
16
2
1 ⎤
⎡1
= ⎢ x9 − x 7 + x5 ⎥ =
≈ 0.051
7
5 ⎦ −1 315
⎣9
30. From Problem 28, the volume of one octant of
2
the common region is r 3 . We can find the
3
volume of the “T” similarly. Since the “T” has
one-half the common region of the “+” in
Problem 28, the volume of the “T” is given by
V = vol. of cylinders - vol. of common region
⎛2 ⎞
=(π r 2 )( L1 + L2 ) − 4 ⎜ r 3 ⎟
⎝3 ⎠
With r = 2, L1 = 12, and L2 = 8 (inches), the
volume of the “T” is
V = vol. of cylinders - vol. of common region
27. The square at x has sides of length 1 − x 2 .
⎛2 ⎞
=(π r 2 )( L1 + L2 ) − 4 ⎜ r 3 ⎟
⎝3 ⎠
⎛2 ⎞
= (π 22 )(12 + 8) − 4 ⎜ 23 ⎟
⎝3 ⎠
64
= 80π −
in 3
3
3 ⎤1
⎡
1
x
2
V = ∫ (1 − x 2 )dx = ⎢ x − ⎥ =
≈ 0.67
0
3 ⎦⎥
3
⎣⎢
0
28. From Problem 27 we see that horizontal cross
sections of one octant of the common region are
squares. The length of a side at height y is
r − y where r is the common radius of the
cylinders. The volume of the “+” can be found
by adding the volumes of each cylinder and
subtracting off the volume of the common region
(which is counted twice). The volume of one
octant of the common region is
r 2
1 2 r
2
2
∫0 (r − y )dy = r y − 3 y |0
1
2
= r3 − r3 = r3
3
3
Thus, the volume of the “+” is
V = vol. of cylinders - vol. of common region
2
≈ 229.99 in 3
2
⎛2 ⎞
=2(π r 2 l ) − 8 ⎜ r 3 ⎟
⎝3 ⎠
128
⎛2
⎞
= 2π (22 )(12) − 8 ⎜ (2)3 ⎟ = 96π −
3
3
⎝
⎠
≈ 258.93 in 2
31. From Problem 30, the general form for the
volume of a “T” formed by two cylinders with
the same radius is
V = vol. of cylinders - vol. of common region
⎛2 ⎞
=(π r 2 )( L1 + L2 ) − 4 ⎜ r 3 ⎟
⎝3 ⎠
8
= π r 2 ( L1 + L2 ) − r 3
3
32. The area of each cross section perpendicular to
the x-axis is
1 ⎡1
π
2 ⎢⎣ 2
(
)
⎤
x − x2 ⎥
⎦
2
π 4
( x − 2 x5 / 2 + x).
8
π 1
V = ∫ ( x 4 − 2 x5 / 2 + x)dx
8 0
=
1
29. Using the result from Problem 28, the volume of
one octant of the common region in the “+” is
r 2
1 2 r
2
2
∫0 (r − y )dy = r y − 3 y |0
1
2
= r3 − r3 = r3
3
3
Instructor’s Resource Manual
=
π ⎡1 5 4 7 / 2 1 2 ⎤
9π
x − x
+ x ⎥ =
≈ 0.050
⎢
8 ⎣5
7
2 ⎦ 0 560
Section 5.2
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33. Sketch the region.
8
3
⎡ 24
⎤
= π ⎢ y5 / 3 − y 7 / 3 ⎥
7
⎣5
⎦0
⎛ 768 384 ⎞ 3456π
= π⎜
−
≈ 310.21
⎟=
7 ⎠
35
⎝ 5
b. Revolving about the line y = 8, the radius of
the disk at x is 8 − x3 = 8 − x3 / 2 .
4
a.
V = π∫ (8 − x3 / 2 )2 dx
0
Revolving about the line x = 4, the radius of
4
= π∫ (64 − 16 x3 / 2 + x3 )dx
the disk at y is 4 − 3 y 2 = 4 − y 2 / 3 .
0
4
32
1 ⎤
⎡
= π ⎢64 x − x5 / 2 + x 4 ⎥
5
4 ⎦0
⎣
1024
⎡
⎤ 576π
= π ⎢ 256 −
+ 64 ⎥ =
≈ 361.91
5
5
⎣
⎦
8
V = π∫ (4 − y 2 / 3 ) 2 dy
0
8
= π∫ (16 − 8 y 2 / 3 + y 4 / 3 )dy
0
8
24 5 / 3 3 7 / 3 ⎤
⎡
= π ⎢16 y −
y
+ y ⎥
5
7
⎣
⎦0
768 384 ⎞
⎛
= π ⎜ 128 −
+
⎟
5
7 ⎠
⎝
1024π
=
≈ 91.91
35
b. Revolving about the line y = 8, the inner
radius of the disk at x is 8 − x3 = 8 − x3 / 2 .
4
V = π∫ ⎡82 − (8 − x3 / 2 )2 ⎤ dx
⎦
0⎣
4
= π∫ (16 x3 / 2 − x3 )dx
0
4
1 ⎤
⎡ 32
⎛ 1024
⎞
= π ⎢ x5 / 2 − x 4 ⎥ = π ⎜
− 64 ⎟
5
4 ⎦0
⎣5
⎝
⎠
704π
=
≈ 442.34
5
34. Sketch the region.
35. The area of a quarter circle with radius 2 is
2
∫0
2
∫0
4 − y 2 dy = π .
⎡ 2 4 − y 2 + 4 − y 2 ⎤ dy
⎢⎣
⎥⎦
= 2∫
2
0
2
4 − y 2 dy + ∫ (4 − y 2 )dy
0
2
1 ⎤
⎡
⎛ 8⎞
= 2π + ⎢ 4 y − y 3 ⎥ = 2π + ⎜ 8 − ⎟
3 ⎦0
3⎠
⎣
⎝
16
= 2π + ≈ 11.62
3
36. Let the x-axis lie along the diameter at the base
perpendicular to the water level and slice
perpendicular to the x-axis. Let x = 0 be at the
center. The slice has base length 2 r 2 − x 2 and
hx
.
height
r
2h r
V=
x r 2 − x 2 dx
r ∫0
r
2h ⎡ 1 2
2h ⎛ 1 3 ⎞ 2 2
2 3/ 2 ⎤
=
− r −x
⎥ = r ⎜3r ⎟ = 3r h
r ⎢⎣ 3
⎝
⎠
⎦0
(
)
37. Let the x-axis lie on the base perpendicular to the
diameter through the center of the base. The slice
a.
Revolving about the line x = 4, the inner
radius of the disk at y is 4 − 3 y 2 = 4 − y 2 / 3 .
8⎡ 2
4
0 ⎢⎣
V = π∫
8
(
− 4− y
) ⎥⎦ dy
2/3 2 ⎤
= π∫ (8 y 2 / 3 − y 4 / 3 )dy
0
308
Section 5.2
at x is a rectangle with base of length 2 r 2 − x 2
and height x tan θ .
r
V = ∫ 2 x tan θ r 2 − x 2 dx
0
r
⎡ 2
⎤
= ⎢ − tan θ (r 2 − x 2 )3 / 2 ⎥
⎣ 3
⎦0
2
= r 3 tan θ
3
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38. a.
y
k
Slice horizontally.
1
3
3 2
r⋅
r=
r .
2
2
4
The center of an equilateral triangle is
is A =
x=4
2
2 3
1
⋅
r=
r from a vertex. Then the
3 2
3
height of a regular tetrahedron is
⎛ y⎞
⎛ y⎞
ΔV ≈ π ⎜⎜ 4 ⎟⎟ Δy = π ⎜⎜
⎟⎟ Δy
⎝ k⎠
⎝ k⎠
If the depth of the tank is h, then
V = π∫
h
0
=
2π
2
⎛ 1 ⎞
h = r2 − ⎜
r⎟ =
⎝ 3 ⎠
h
y
π ⎡ 2 3/ 2 ⎤
dy =
y ⎥
k
k ⎢⎣ 3
⎦0
V=
h3 / 2 .
3 k
The volume as a function of the depth of the
2π 3 / 2
y
tank is V ( y ) =
3 k
dy −m k
dy
=
= − m y and
dt
π
dt
k
which is constant.
π
y
39. Let A lie on the xy-plane. Suppose ΔA = f ( x)Δx
where f(x) is the length at x, so A = ∫ f ( x)dx .
Slice the general cone at height z parallel to A.
The slice of the resulting region is Az and ΔAz
is a region related to f(x) and Δ x by similar
triangles:
⎛ z⎞
⎛ z⎞
ΔAz = ⎜ 1 − ⎟ f ( x) ⋅ ⎜1 − ⎟ Δx
⎝ h⎠
⎝ h⎠
2
⎛ z⎞
= ⎜ 1 − ⎟ f ( x ) Δx
⎝ h⎠
⎛ z⎞
Therefore, Az = ⎜ 1 − ⎟
⎝ h⎠
2
∫
2
⎛ z⎞
f ( x)dx = ⎜ 1 − ⎟ A.
⎝ h⎠
2
2
3
r.
1
2 3
Ah =
r
3
12
40. If two solids have the same cross sectional area at
every x in [a, b], then they have the same volume.
41. First we examine the cross-sectional areas of
each shape.
Hemisphere: cross-sectional shape is a circle.
dV
= −m y .
b. It is given that
dt
dV
π 1/ 2 dy
From part a,
y
.
=
dt
dt
k
Thus,
2 2
r =
3
2
h⎛
z⎞
⎛ z⎞
ΔV ≈ Az Δz = A ⎜1 − ⎟ Δz V = A∫ ⎜ 1 − ⎟ dz
0⎝
h⎠
⎝ h⎠
The radius of the circle at height y is r 2 − y 2 .
Therefore, the cross-sectional area for the
hemisphere is
Ah = π ( r 2 − y 2 )2 = π (r 2 − y 2 )
Cylinder w/o cone: cross-sectional shape is a
washer. The outer radius is a constant , r. The
inner radius at height y is equal to y. Therefore,
the cross-sectional area is
A2 = π r 2 − π y 2 = π (r 2 − y 2 ) .
Since both cross-sectional areas are the same, we
can apply Cavaleri’s Principle. The volume of
the hemisphere of radius r is
V = vol. of cylinder - vol. of cone
1
= π r 2h − π r 2h
3
2 2
= πr h
3
With the height of the cylinder and cone equal to
r, the volume of the hemisphere is
2
2
V = π r 2 (r ) = π r 3
3
3
h
⎡ h ⎛ z ⎞3 ⎤
1
= A ⎢− ⎜ 1 − ⎟ ⎥ = Ah.
3
⎢⎣ 3 ⎝ h ⎠ ⎥⎦
0
a.
A = πr 2
1
1
V = Ah = πr 2 h
3
3
b. A face of a regular tetrahedron is an
equilateral triangle. If the side of an
equilateral triangle has length r, then the area
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
5.3 Concepts Review
3. a, b.
1. 2πx f ( x)Δx
2
2
0
0
2. 2π∫ x 2 dx; π∫ (4 − y 2 )dy
2
3. 2π∫ (1 + x) x dx
0
2
4. 2π∫ (1 + y )(2 − y )dy
c. ΔV ≈ 2πx x Δx = 2πx3 / 2 Δx
0
3
3
⎡2
⎤
d, e. V = 2π ∫ x3 / 2 dx = 2π ⎢ x5 / 2 ⎥
0
⎣5
⎦0
Problem Set 5.3
=
1. a, b.
36 3
π ≈ 39.18
5
4. a,b.
⎛1⎞
c. ΔV ≈ 2πx ⎜ ⎟ Δx = 2πΔx
⎝ x⎠
c. ΔV ≈ 2πx(9 − x 2 )Δx = 2π(9 x − x3 )Δx
d,e. V = 2π ∫ dx = 2π [ x ]1 = 6π ≈ 18.85
4
4
1
2. a, b.
3
3
1 ⎤
⎡9
d, e. V = 2π ∫ (9 x − x3 )dx = 2π ⎢ x 2 − x 4 ⎥
0
4 ⎦0
⎣2
81
81
81
π
⎛
⎞
= 2π ⎜ − ⎟ =
≈ 127.23
2
⎝ 2 4⎠
5. a, b.
c. ΔV ≈ 2πx ( x 2 )Δx = 2πx3Δx
1
1
π
⎡1 ⎤
d, e. V = 2π∫ x3 dx = 2π ⎢ x 4 ⎥ = ≈ 1.57
0
⎣ 4 ⎦0 2
c.
ΔV ≈ 2π(5 − x) x Δx
= 2π(5 x1/ 2 − x3 / 2 )Δx
5
d, e. V = 2π ∫ (5 x1/ 2 − x3 / 2 )dx
0
5
2
⎡10
⎤
= 2π ⎢ x3 / 2 − x5 / 2 ⎥
5
⎣3
⎦0
⎛ 50 5
⎞ 40 5
= 2π ⎜⎜
− 10 5 ⎟⎟ =
π ≈ 93.66
3
⎝ 3
⎠
310
Section 5.3
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6. a, b.
c. ΔV ≈ 2πx (3 x − x 2 )Δx = 2π(3 x 2 − x3 )Δx
3
3
1 ⎤
⎡
d, e. V = 2π ∫ (3 x 2 − x3 )dx = 2π ⎢ x3 − x 4 ⎥
0
4 ⎦0
⎣
81 ⎞ 27π
⎛
= 2π ⎜ 27 − ⎟ =
≈ 42.41
4⎠
2
⎝
9. a, b.
c. ΔV ≈ 2π(3 − x)(9 − x 2 )Δx
= 2π(27 − 9 x − 3 x 2 + x3 )Δx
3
d, e. V = 2π ∫ (27 − 9 x − 3 x 2 + x3 )dx
0
3
9
1 ⎤
⎡
= 2π ⎢ 27 x − x 2 − x3 + x 4 ⎥
2
4 ⎦0
⎣
81
81 ⎞ 135π
⎛
= 2π ⎜ 81 − − 27 + ⎟ =
≈ 212.06
2
4⎠
2
⎝
c. ΔV ≈ 2πy ( y 2 )Δy = 2πy 3Δy
7. a, b.
1
1
π
⎡1 ⎤
d, e. V = 2π ∫ y 3 dy = 2π ⎢ y 4 ⎥ = ≈ 1.57
0
⎣ 4 ⎦0 2
10. a, b.
⎡⎛ 1
⎤
⎞
c. ΔV ≈ 2πx ⎢⎜ x3 + 1⎟ − (1 − x) ⎥ Δx
4
⎠
⎣⎝
⎦
1
⎛
⎞
= 2 π ⎜ x 4 + x 2 ⎟ Δx
⎝4
⎠
1⎛1 4
x
0 ⎜⎝ 4
d, e. V = 2π ∫
c. ΔV ≈ 2πy
2⎞
+ x ⎟ dx
⎠
1
1 ⎤
⎡1
⎛ 1 1⎞
= 2π ⎢ x 5 + x 3 ⎥ = 2π ⎜ + ⎟
3 ⎦0
⎣ 20
⎝ 20 3 ⎠
23π
=
≈ 2.41
30
(
)
y + 1 Δ y = 2 π( y 3 / 2 + y ) Δ y
4
d, e. V = 2π ∫ ( y 3 / 2 + y ) dy
0
4
1 ⎤
⎡2
⎛ 64
⎞
= 2π ⎢ y 5 / 2 + y 2 ⎥ = 2π ⎜ + 8 ⎟
2 ⎦0
⎣5
⎝ 5
⎠
208π
=
≈ 130.69
5
8. a, b.
11. a, b.
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15.
c. ΔV ≈ 2π(2 − y ) y 2 Δy = 2π(2 y 2 − y 3 )Δy
2
2
1 ⎤
⎡2
d, e. V = 2π ∫ (2 y 2 − y 3 )dy = 2π ⎢ y 3 − y 4 ⎥
0
4 ⎦0
⎣3
⎛ 16
⎞ 8π
= 2π ⎜ − 4 ⎟ =
≈ 8.38
3
⎝
⎠ 3
12. a, b.
c. ΔV ≈ 2π(3 − y )
(
(
)
2 y + 1 Δy
V = 2π ∫
2
0
(3 + 3
3 ⎛ 1 ⎞
3 1
V = 2π∫ x ⎜ ⎟ dx = 2π ∫
dx
1 ⎝ x3 ⎠
1 x2
)
)
2 y1/ 2 − y − 2 y 3 / 2 dy
2
x3
b.
c.
d.
⎡
1
2 2 5/ 2 ⎤
= 2π ⎢3 y + 2 2 y 3 / 2 − y 2 −
y ⎥
2
5
⎣
⎦0
1
A=∫
= 2π 3 + 3 2 y1/ 2 − y − 2 y 3 / 2 Δy
d, e.
3
1
a.
16 ⎞ 88π
⎛
= 2π ⎜ 6 + 8 − 2 − ⎟ =
≈ 55.29
5⎠
5
⎝
dx
2
⎤
3 ⎡⎛ 1
⎞
V = π ∫ ⎢⎜ + 1⎟ − (−1)2 ⎥ dx
1 ⎢⎝ x 3
⎥⎦
⎠
⎣
3⎛ 1
2 ⎞
= π∫ ⎜
+ ⎟ dx
1 ⎝ x6
x3 ⎠
3
⎛ 1 ⎞
V = 2π ∫ (4 − x) ⎜ ⎟ dx
1
⎝ x3 ⎠
3⎛ 4
1 ⎞
= 2π ∫ ⎜ − ⎟ dx
1 ⎝ x3 x 2 ⎠
16.
13. a.
b
π ∫ ⎡ f ( x) − g ( x)
2
⎣
a
2⎤
⎦
dx
b.
2π∫ x [ f ( x) − g ( x) ] dx
c.
2π∫ ( x − a ) [ f ( x) − g ( x) ] dx
d.
2π∫ (b − x) [ f ( x) − g ( x) ] dx
a.
A = ∫ ( x3 + 1) dx
d
π∫ ⎡ f ( y ) 2 − g ( y ) 2 ⎤ dy
⎦
c ⎣
b.
V = 2π ∫ x( x3 + 1)dx = 2π ∫ ( x 4 + x)dx
y [ f ( y ) − g ( y ) ] dy
c.
2
V = π ∫ ⎡ ( x3 + 2)2 − (−1)2 ⎤
⎦
0 ⎣
14. a.
b
a
b
a
b
a
d
b.
2π ∫
c.
2π∫ (3 − y ) [ f ( y ) − g ( y ) ] dy
c
2
0
2
2
0
0
2
= π ∫ ( x 6 + 4 x3 + 3)dx
d
0
c
d.
2
V = 2π ∫ (4 − x)( x3 + 1)dx
0
2
= 2π ∫ (− x 4 + 4 x3 − x + 4)dx
0
312
Section 5.3
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
17. To find the intersection point, solve
y=
y=
y3
.
32
21. To find the intersection point, solve
sin( x 2 ) = cos( x 2 ) .
tan( x 2 ) = 1
y6
1024
π
4
π
x=
2
x2 =
y − 1024 y = 0
6
y ( y 5 − 1024) = 0
y = 0, 4
4 ⎛
y3 ⎞
V = 2π∫ y ⎜ y − ⎟ dy
0 ⎜
32 ⎟⎠
⎝
0
4⎛
y ⎞
= 2π ∫ ⎜ y 3 / 2 −
⎟ dy
0 ⎜
32 ⎟⎠
⎝
0
4
22. V = 2π ∫
⎛
y3 ⎞
18. V = 2π∫ (4 − y ) ⎜ y −
⎟ dy
⎜
0
32 ⎟⎠
⎝
4⎛
y3 y 4 ⎞
= 2π∫ ⎜ 4 y1/ 2 − y 3 / 2 −
+
⎟ dy
0 ⎜
8 32 ⎟⎠
⎝
= 2π ∫
= 2π ∫
0
23. a.
(2 x + x sin x)dx
2 x dx + 2π ∫
2π
0
x sin x dx
2π
⎞ dx
⎟
⎠
The curves intersect when x = 0 and x = 1.
1
1
0
0
V = π ∫ [ x 2 − ( x 2 )2 ] dx = π ∫ ( x 2 − x 4 )dx
2
20. y = ± a 2 − x 2 , − a ≤ x ≤ a
1
1 ⎤
⎡1
⎛ 1 1 ⎞ 2π
= π ⎢ x3 − x5 ⎥ = π ⎜ − ⎟ =
≈ 0.42
5 ⎦0
⎣3
⎝ 3 5 ⎠ 15
b.
1
1
0
0
V = 2π ∫ x( x − x 2 )dx = 2π ∫ ( x 2 − x3 )dx
1
1 ⎤
⎡1
⎛1 1⎞ π
= 2π ⎢ x3 − x 4 ⎥ = 2π ⎜ − ⎟ =
4 ⎦0
⎣3
⎝3 4⎠ 6
≈ 0.52
c.
Slice perpendicular to the line y = x. At
(a, a), the perpendicular line has equation
y = −( x − a ) + a = − x + 2a . Substitute
y = –x + 2a into y = x 2 and solve for x ≥ 0 .
x a 2 − x 2 dx
a
⎛1
⎞
⎡ 1
⎤
= 4πb ⎜ πa 2 ⎟ − 4π ⎢ − (a 2 − x 2 )3 2 ⎥ = 2π2 a 2b
2
3
⎝
⎠
⎣
⎦ −a
(Note that the area of a semicircle with radius a is
a
1 2
2
2
∫−a a − x dx = 2 πa .)
Instructor’s Resource Manual
)
2 − 1 ≈ 1.30
= 2π(4π2 ) + 2π(−2π) = 4π 2 (2π − 1) ≈ 208.57
b
−a
(
x(2 + sin x)dx
2π
b
⎡ 1
⎤
= 4π∫ x b 2 − x 2 dx = 4π ⎢ − (b 2 − x 2 )3 / 2 ⎥
a
3
⎣
⎦a
4
π
⎡1 2
⎤
= 4π ⎢ (b − a 2 )3 / 2 ⎥ =
(b 2 − a 2 )3 / 2
⎣3
⎦ 3
−a
π/2
= 2π ⎡ x 2 ⎤ + 2π [sin x − x cos x ]0
⎣ ⎦0
y = − b 2 − x 2 , and x = a. When R is revolved
about the y-axis, it produces the desired solid.
b
V = 2π∫ x ⎛⎜ b 2 − x 2 + b 2 − x 2 ⎞⎟ dx
a ⎝
⎠
a
2π
0
2π
19. Let R be the region bounded by y = b − x ,
a 2 − x 2 dx − 4π ∫
2π
0
4
⎡ 8 3 / 2 2 5 / 2 y 4 y5 ⎤
= 2π ⎢ y
− y
−
+
⎥
5
32 160 ⎥⎦
⎣⎢ 3
0
32 ⎞ 208π
⎛ 64 64
= 2π ⎜ − − 8 + ⎟ =
≈ 43.56
5
5 ⎠
15
⎝ 3
a
V = 2π∫ (b − x) ⎛⎜ 2 a 2 − x 2
−a
⎝
⎡ x cos( x 2 ) − x sin( x 2 ) ⎤ dx
⎣
⎦
⎡⎛ 1
1 ⎞ 1⎤
= 2π ⎢⎜
+
⎟− ⎥ = π
⎣⎝ 2 2 2 2 ⎠ 2 ⎦
4
2
x ⎡ cos( x 2 ) − sin( x 2 ) ⎤ dx
⎣
⎦
1
⎡1
⎤
= 2π ⎢ sin( x 2 ) + cos( x 2 ) ⎥
2
2
⎣
⎦0
⎡2
y5 ⎤
⎛ 64 32 ⎞ 64π
= 2π ⎢ y 5 / 2 −
⎥ = 2π ⎜ − ⎟ =
5
160
5 ⎠
5
⎝ 5
⎣⎢
⎦⎥ 0
≈ 40.21
a
π /2
= 2π ∫
4
= 4πb ∫
π/2
V = 2π ∫
x 2 + x − 2a = 0
−1 ± 1 + 8a
2
−1 + 1 + 8a
x=
2
x=
Section 5.3
313
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
Substitute into y = –x + 2a, so
1 + 4a − 1 + 8a
. Find an expression for
y=
2
r 2 , the square of the distance from (a, a) to
⎛ −1 + 1 + 8a 1 + 4a − 1 + 8a ⎞
,
⎜⎜
⎟⎟ .
2
2
⎝
⎠
⎡
−1 + 1 + 8a ⎤
r = ⎢a −
⎥
2
⎣
⎦
24. ΔV ≈ 4πx 2 Δx
r
r
4
⎡1 ⎤
V = 4π∫ x 2 dx = 4π ⎢ x3 ⎥ = πr 3
0
⎣ 3 ⎦0 3
2
V=
2
⎡ 1 + 4a − 1 + 8a ⎤
+ ⎢a −
⎥
2
⎣
⎦
⎡ 2a + 1 − 1 + 8a ⎤
=⎢
⎥
2
⎣
⎦
r2
S
r2
S Δx
r
∫0
r
x 2 dx =
S ⎡1 3 ⎤
1
x ⎥ = rS
2 ⎢⎣ 3
⎦0 3
r
2
5.4 Concepts Review
2
1. Circle
⎡ 2a + 1 − 1 + 8a ⎤
+ ⎢−
⎥
2
⎣
⎦
⎡ 2 a + 1 − 1 + 8a ⎤
= 2⎢
⎥
2
⎣
⎦
x2
25. ΔV ≈
x 2 + y 2 = 16 cos 2 t + 16sin 2 t = 16
2
2. x ; x 2 + 1
2
3.
= 2a 2 + 6a + 1 − 2a 1 + 8a − 1 + 8a
2
2
∫a [ f ′(t )] + [ g ′(t )] dt
b
4. Mean Value Theorem (for derivatives)
ΔV ≈ πr Δa
2
1
V = π ∫ (2a 2 + 6a + 1
0
Problem Set 5.4
− 2a 1 + 8a − 1 + 8a ) da
1
1
⎡2
⎤
= π ⎢ a3 + 3a 2 + a − (1 + 8a)3 / 2 ⎥
12
⎣3
⎦0
1.
f ( x) = 4 x3 / 2 , f ′( x ) = 6 x1/ 2
L=∫
5
1/ 3
1
−π ∫ 2a 1 + 8a da
1 + (6 x1/ 2 ) 2 dx = ∫
5
1/ 3
1 + 36 x dx
5
⎡1 2
⎤
= ⎢ ⋅ (1 + 36 x)3 / 2 ⎥
⎣ 36 3
⎦1/ 3
1
=
181 181 − 13 13 ≈ 44.23
54
0
⎡⎛ 2
9 ⎞ ⎛ 1 ⎞⎤
= π ⎢⎜ + 3 + 1 − ⎟ − ⎜ − ⎟ ⎥
4 ⎠ ⎝ 12 ⎠ ⎦
⎣⎝ 3
(
1
−π ∫ 2a 1 + 8a da
)
0
=
1
5π
− π ∫ 2a 1 + 8a da
0
2
To integrate
1
∫0 2a
1 + 8a da , use the
2.
f ( x) =
L=∫
2
1
substitution u = 1 + 8a.
1
9 1
1
∫0 2a 1 + 8a da = ∫1 4 (u − 1) u 8 du
1 9 3 / 2 1/ 2
=
(u
− u )du
32 ∫1
=∫
2
1
2 2
( x + 1)3 / 2 , f ′( x) = 2 x( x 2 + 1)1/ 2
3
2
1 + ⎡ 2 x ( x 2 + 1)1/ 2 ⎤ dx
⎣
⎦
2
4 x 4 + 4 x 2 + 1 dx = ∫ (2 x 2 + 1)dx
1
2
⎡2
⎤
⎛ 16
⎞ ⎛ 2 ⎞ 17
= ⎢ x3 + x ⎥ = ⎜ + 2 ⎟ − ⎜ + 1⎟ =
≈ 5.67
⎣3
⎦1 ⎝ 3
⎠ ⎝3 ⎠ 3
9
=
1 ⎡ 2 5/ 2 2 3/ 2 ⎤
u
− u ⎥
32 ⎢⎣ 5
3
⎦1
=
1 ⎡⎛ 486
⎞ ⎛ 2 2 ⎞ ⎤ 149
− 18 ⎟ − ⎜ − ⎟ ⎥ =
⎜
⎢
32 ⎣⎝ 5
⎠ ⎝ 5 3 ⎠ ⎦ 60
V=
314
5π 149π π
−
=
≈ 0.052
2
60
60
Section 5.4
Instructor’s Resource Manual
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
3.
f ( x) = (4 − x 2 / 3 )3 / 2 ,
f ′( x) =
3
⎛ 2
⎞
(4 − x 2 / 3 )1/ 2 ⎜ − x −1/ 3 ⎟
2
⎝ 3
⎠
g ( y) =
= − x −1/ 3 (4 − x 2 / 3 )1/ 2
L=∫
2
1 + ⎡ − x −1/ 3 (4 − x 2 / 3 )1/ 2 ⎤ dx
⎣
⎦
8
1
=∫
8
4x
1
y5
1
+
30 2 y 3
6. x =
−2 / 3
8
dx = ∫ 2 x
−1/ 3
1
L=∫
3
1
dx
8
⎡3
⎤
= 2 ⎢ x 2 / 3 ⎥ = 3(4 − 1) = 9
2
⎣
⎦1
=∫
y5
1
y4
3
+
, g ′( y ) =
−
3
30 2 y
6 2 y4
⎛ y4
3
1+ ⎜
−
⎜ 6 2 y4
⎝
2
⎞
⎟ dy
⎟
⎠
2
⎛ y4
3 ⎞
+
⎜
⎟ dy
⎜ 6 2 y4 ⎟
⎝
⎠
3
y8 1
9
+ +
dy = ∫
1
36 2 4 y8
3
1
3
4.
⎡ y5
3 ⎛ y4
3 ⎞
1 ⎤
=∫ ⎜
+
dy = ⎢ −
⎟
⎥
4
3
1 ⎜ 6
2 y ⎟⎠
⎝
⎣⎢ 30 2 y ⎦⎥1
⎛ 81 1 ⎞ ⎛ 1 1 ⎞ 1154
= ⎜ − ⎟−⎜ − ⎟ =
≈ 8.55
⎝ 10 54 ⎠ ⎝ 30 2 ⎠ 135
x 4 + 3 x3 1
f ( x) =
=
+
6x
6 2x
f ′( x) =
L=∫
⎛ x2
1
1+ ⎜
−
⎜ 2 2 x2
⎝
3
1
=∫
x2
1
−
2 2 x2
2
⎞
⎟ dx
⎟
⎠
3
x4 1
1
+ +
dx = ∫
1
4 2 4 x4
3
1
7.
⎛ x2
1
+
⎜
⎜ 2 2 x2
⎝
2
⎞
⎟ dx
⎟
⎠
3
⎡ x3 1 ⎤
3 ⎛ x2
1 ⎞
dx = ⎢ − ⎥
=∫ ⎜
+
⎟
1 ⎜ 2
2 x 2 ⎟⎠
⎝
⎣⎢ 6 2 x ⎦⎥1
⎛ 9 1 ⎞ ⎛ 1 1 ⎞ 14
= ⎜ − ⎟−⎜ − ⎟ =
≈ 4.67
⎝2 6⎠ ⎝6 2⎠ 3
dx 2 dy
=t ,
=t
dt
dt
L=∫
y4
1
y3 1
5. g ( y ) =
+
, g ′( y ) =
−
16 2 y 2
4 y3
L=∫
−2
−3
=∫
−2
−3
(t 2 )2 + (t )2 dt = ∫
1
t 4 + t 2 dt
0
1
(
)
1
1
⎡1
⎤
= ∫ t t 2 + 1 dt = ⎢ (t 2 + 1)3 / 2 ⎥ = 2 2 − 1
0
3
⎣
⎦0 3
≈ 0.61
2
⎛ y3 1 ⎞
1+ ⎜
−
⎟ dy
⎜ 4 y3 ⎟
⎝
⎠
−2
y6 1 1
+ +
dy = ∫
6
−3
16 2 y
1
0
2
⎛ y3 1 ⎞
+
⎜
⎟ dy
⎜ 4 y3 ⎟
⎝
⎠
8.
−2
⎛ y3 1 ⎞
⎡ y4
1 ⎤
dy = − ⎢ −
= ∫ −⎜
+
⎟
⎥
3⎟
2
−3 ⎜ 4
y ⎠
⎝
⎣⎢ 16 2 y ⎦⎥ −3
−2
⎡⎛ 1 ⎞ ⎛ 81 1 ⎞ ⎤ 595
= − ⎢⎜ 1 − ⎟ − ⎜ − ⎟ ⎥ =
≈ 4.13
⎣⎝ 8 ⎠ ⎝ 16 18 ⎠ ⎦ 144
dx
dy
= 6t ,
= 6t 2
dt
dt
L=∫
4
1
(6t )2 + (6t 2 )2 dt = ∫
4
1
36t 2 + 36t 4 dt
4
4
= ∫ 6t 1 + t 2 dt = ⎡ 2(1 + t 2 )3 / 2 ⎤
⎣
⎦1
1
(
)
= 2 17 17 − 2 2 ≈ 134.53
Instructor’s Resource Manual
Section 5.4
315
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
9.
12. x = y +
3
2
3
g ( y ) = y + , g ′( y ) = 1
2
L=∫
3
1
3
1 + (1)2 = 2 ∫ dy = 2 2
1
3 5
= .
2 2
3 9
At y = 3, x = 3 + = .
2 2
At y = 1, x = 1 +
dx
dy
= 4 cos t ,
= −4sin t
dt
dt
L=∫
π
0
=∫
π
2
⎛9 5⎞
d = ⎜ − ⎟ + (3 − 1) 2 = 8 = 2 2
⎝2 2⎠
(4 cos t ) 2 + (−4sin t )2 dt
π
16 cos 2 t + 16sin 2 t dt = ∫ 4dt
0
0
= 4π ≈ 12.57
13.
10.
dx
dy
= 1,
= 2t
dt
dt
L=∫
2
0
−3
−2
with n = 8,
−3
2−0 ⎡
⎛1⎞
⎛1⎞
⎛3⎞
f (0) + 4 f ⎜ ⎟ + 2 f ⎜ ⎟ + 4 f ⎜ ⎟
3 × 8 ⎢⎣
⎝4⎠
⎝2⎠
⎝4⎠
⎛5⎞
⎛3⎞
⎛7⎞
+2 f (1) + 4 f ⎜ ⎟ + 2 f ⎜ ⎟ + 4 f ⎜ ⎟ + f (2)]
⎝4⎠
⎝2⎠
⎝4⎠
1
≈ [1 + 4 × 1.118 + 2 × 1.4142
12
+ 4 × 1.8028 + 2 × 2.2361
L≈
dx
dy
= 2 5 cos 2t ,
= −2 5 sin 2t
dt
dt
=∫
π/4
0
(2
5 cos 2t
) + ( −2
2
5 sin 2t
20 cos 2 2t + 20sin 2 2t dt = ∫
3
1
3
1 + (2)2 dx = 5 ∫ dx = 2 5
1
At x = 1, y = 2(1) + 3 = 5.
At x = 3, y = 2(3) + 3 = 9.
d = (3 − 1) + (9 − 5) = 20 = 2 5
2
2
dt
+ 4 × 2.6926 + 2 × 3.1623
2 5 dt
+ 4 × 3.6401 + 4.1231] ≈ 4.6468
14.
f ( x) = 2 x + 3, f ′( x) = 2
L=∫
)
π/ 4
0
5π
=
≈ 3.51
2
11.
1 + 4t 2 dt
Let f (t ) = 1 + 4t 2 . Using the Parabolic Rule
−2
0
π/ 4
2
0
−1
−1
L=∫
12 + (2t )2 dt = ∫
dx
dy
1
= 2t ,
=
dt
dt 2 t
L≈∫
4
1
2
4
1
⎛ 1 ⎞
2
(2t ) 2 + ⎜
⎟ dt = ∫1 4t + dt
4t
⎝2 t ⎠
Let f (t ) = 4t 2 +
2
1
. Using the Parabolic Rule
4t
with n = 8,
4 −1 ⎡
⎛ 11 ⎞
⎛ 14 ⎞
f (1) + 4 f ⎜ ⎟ + 2 f ⎜ ⎟
⎢
3× 8 ⎣
⎝8⎠
⎝8⎠
⎛ 17 ⎞
⎛ 20 ⎞
⎛ 23 ⎞
⎛ 26 ⎞
+4 f ⎜ ⎟ + 2 f ⎜ ⎟ + 4 f ⎜ ⎟ + 2 f ⎜ ⎟
8
8
8
⎝ ⎠
⎝ ⎠
⎝ ⎠
⎝ 8 ⎠
⎤ 1
⎛ 29 ⎞
+4 f ⎜ ⎟ + f (4) ⎥ ≈ ( 2.0616 + 4 × 2.8118
⎝ 8 ⎠
⎦ 8
+2 × 3.562 + 4 × 4.312 + 2 × 5.0621 + 4 × 5.8122
2 × 6.5622 + 4 × 7.3122 + 8.0623) ≈ 15.0467
L≈
316
Section 5.4
Instructor’s Resource Manual
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material
may be reproduced, in any form or by any means, without permission in writing from the publisher.
15.
dx
dy
= cos t ,
= −2sin 2t
dt
dt
L=∫
π 2
0
=∫
π 2
0
( cos t )2 + ( −2sin 2t )2 dt
dx
dy
= 1,
= sec 2 t
dt
dt
π 4
0
π 4
0
9a 2 cos 2 t sin 2 t (sin 2 t + cos 2 t ) dt
π/ 2
⎡ 1
⎤
3a cos t sin tdt = 3a ⎢ − cos 2 t ⎥
0
2
⎣
⎦0
(The integral can also be evaluated as
=
3a
2
π/ 2
⎡1
⎤
3a ⎢ sin 2 t ⎥
with the same result.)
⎣2
⎦0
The total length is 6a.
18. a.
( )
p = aθ
OT = length PT
b. From Figure 18 of the text,
PQ PQ
QC QC
sin θ =
=
and cos θ =
=
.
a
a
PC
PC
Therefore PQ = a sin θ and QC = a cos θ .
x = OT − PQ = aθ − a sin θ = a (θ − sin θ )
y = CT − CQ = a − a cos θ = a (1 − cos θ )
1 + sec 4 t dt
Let f (t ) = 1 + sec4 t . Using the Parabolic
⎛π ⎞
f (0) + 4 f ⎜ ⎟
⎢
3× 8 ⎣
⎝ 32 ⎠
⎛ 2π ⎞
⎛ 3π ⎞
⎛ 4π ⎞
⎛ 5π ⎞
+2 f ⎜
⎟+4f ⎜ ⎟+2f ⎜
⎟+4f ⎜
⎟
⎝ 32 ⎠
⎝ 32 ⎠
⎝ 32 ⎠
⎝ 32 ⎠
⎛ 6π ⎞
⎛ 7π ⎞
⎛ π ⎞⎤
+2 f ⎜
⎟+4f ⎜
⎟ + f ⎜ ⎟⎥
⎝ 32 ⎠
⎝ 32 ⎠
⎝ 4 ⎠⎦
≈
9a 2 cos 2 t sin 4 t + 9a 2 sin 2 t cos 4 t dt
π/ 2
c.
12 + (sec 2 t )2 dt = ∫
Rule with n = 8, L ≈
π/ 2
=∫
cos 2 t + 4sin 2 2t dt
π 2−0 ⎡
L=∫
=∫
0
⎛π ⎞
⎛ 2π ⎞
f (0) + 4 f ⎜ ⎟ + 2 f ⎜
⎟
3 × 8 ⎢⎣
16
⎝ ⎠
⎝ 16 ⎠
⎛ 3π ⎞
⎛ 4π ⎞
⎛ 5π ⎞
⎛ 6π ⎞
+4 f ⎜ ⎟ + 2 f ⎜
⎟+4f ⎜
⎟+2f ⎜
⎟
⎝ 16 ⎠
⎝ 16 ⎠
⎝ 16 ⎠
⎝ 16 ⎠
⎛ 7π ⎞
⎛ π ⎞⎤ π
+4 f ⎜
⎟ + f ⎜ ⎟⎥ ≈ [1 + 4 × 1.2441
16
⎝
⎠
⎝ 2 ⎠ ⎦ 48
+ 2 × 1.6892 + 4 × 2.0262 + 2 × 2.1213 + 4 × 1.9295
+2 × 1.4651 + 4 × 0.7898 + 0) ≈ 2.3241
16.
π/ 2
0
Let f (t ) = cos 2 t + 4sin 2 2t . Using the
Parabolic Rule with n = 8,
L≈
=∫
π 4−0 ⎡
π
[1.4142 + 4 × 1.4211 + 2 × 1.4425 + 4 × 1.4807
96
+2 × 1.5403 + 4 × 1.6288 + 2 × 1.7585
+4 × 1.9495 + 2.2361] ≈ 1.278
17.
19. From Problem 18,
x = a(θ − sin θ ), y = a (1 − cos θ )
dx
dy
= a(1 − cos θ ),
= a sin θ so
dθ
dθ
2
2
⎛ dx ⎞ ⎛ dy ⎞
2
2
⎜
⎟ +⎜
⎟ = [ a (1 − cos θ ) ] + [ a sin θ ]
⎝ dθ ⎠ ⎝ dθ ⎠
= a 2 − 2a 2 cos θ + a 2 cos 2 θ + a 2 sin 2 θ
= 2a 2 − 2a 2 cos θ = 2a 2 (1 − cos θ )
1 − cos θ
⎛θ ⎞
= 4a 2 sin 2 ⎜ ⎟ .
2
⎝2⎠
The length of one arch of the cycloid is
= 4a 2
2π
∫0
2π
⎛θ ⎞
⎛θ ⎞
4a 2 sin 2 ⎜ ⎟ dθ = ∫ 2a sin ⎜ ⎟ dθ
0
2
⎝ ⎠
⎝2⎠
2π
θ⎤
⎡
= 2a ⎢ −2 cos ⎥ = 2a(2 + 2) = 8a
2 ⎦0
⎣
dx
dy
= 3a cos t sin 2 t ,
= −3a sin t cos 2 t
dt
dt
The first quadrant length is L
=∫
π/ 2
0
(3a cos t sin 2 t ) 2 + (−3a sin t cos 2 t )2 dt
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20. a.
Using θ = ω t , the point P is at x = aω t − a sin(ω t ), y = a − a cos(ω t ) at time t.
dx
= aω − aω cos(ω t ) = aω (1 − cos(ω t ))
dt
dy
= aω sin(ω t )
dt
2
ds
⎡ dy ⎤ ⎡ dx ⎤
= ⎢ ⎥ +⎢ ⎥
dt
⎣ dt ⎦ ⎣ dt ⎦
2
= a 2ω 2 sin 2 (ω t ) + a 2ω 2 − 2a 2ω 2 cos(ω t ) + a 2ω 2 cos 2 (ω t ) = 2a 2ω 2 − 2a 2ω 2 cos(ω t )
1
ωt
ωt
(1 − cos(ω t )) = 2aω sin 2
= 2aω sin
2
2
2
= 2aω
ωt
π
= 1, which occurs when t = (2k + 1). The speed is a minimum when
2
ω
2k π
ωt
.
sin
= 0, which occurs when t =
2
ω
b. The speed is a maximum when sin
c.
21. a.
From Problem 18a, the distance traveled by the wheel is aθ, so at time t, the wheel has gone aθ = aω t miles.
Since the car is going 60 miles per hour, the wheel has gone 60t miles at time t. Thus, aω = 60 and the
maximum speed of the bug on the wheel is 2aω = 2(60) = 120 miles per hour.
dy
= x3 − 1
dx
L=∫
2
b.
2
1 + x3 − 1 dx = ∫ x3 / 2 dx
1
1
2
(
)
2
⎡2
⎤
= ⎢ x5 / 2 ⎥ = 4 2 − 1 ≈ 1.86
⎣5
⎦1 5
b.
L=∫
4π
0
2 − 2 cos t dt = ∫
4π
0
=∫
23.
π/6
318
1
1
0
0
1
4
Section 5.4
3
−2
π/3
x
f ( x) = 25 − x 2 , f ′( x) = −
A = 2π ∫
1 + 64sin cos x − 1 dx
π/3
0
⎡1 ⎤
= 12 37 π ⎢ x 2 ⎥ = 6 37π ≈ 114.66
⎣ 2 ⎦0
⎡ 8
⎤
8sin x cos 2 xdx = ⎢ − cos3 x ⎥
π/6
⎣ 3
⎦π / 6
1
= − + 3 ≈ 1.40
3
=∫
−1
0
f ( x) = 6 x, f ′( x) = 6
dy
= 64sin 2 x cos 4 x − 1
dx
L=∫
0
A = 2π∫ 6 x 1 + 36 dx = 12 37 π ∫ x dx
2π
2
1
at dt = ∫ at dt − ∫ at dt
a a
⎡a ⎤ ⎡a ⎤
= ⎢ t2 ⎥ − ⎢ t2 ⎥ = + = a
⎣ 2 ⎦ 0 ⎣ 2 ⎦ −1 2 2
⎛t⎞
2 sin ⎜ ⎟ dt
⎝2⎠
24.
π/3
a 2 t 2 cos 2 t + a 2t 2 sin 2 tdt
1
t⎤
⎛t⎞
⎡
L = 4 ∫ sin ⎜ ⎟ dt = ⎢ −8cos ⎥
0
2 ⎦0
⎝2⎠
⎣
= 8 + 8 = 16
22. a.
1
−1
⎛t⎞
sin ⎜ ⎟ is positive for 0 < t < 2 π , and
⎝2⎠
by symmetry, we can double the integral
from 0 to 2 π .
2π
1
−1
f ′(t ) = 1 − cos t , g ′(t ) = sin t
L=∫
dx
= − a sin t + a sin t + at cos t = at cos t
dt
dy
= a cos t − a cos t + at sin t = at sin t
dt
= 2π ∫
3
−2
3
= 2π ∫
−2
25 − x 2 1 +
25 − x 2
x2
25 − x 2
dx
25 − x 2 + x 2 dx
5dx = 10π[ x]3−2 = 50π ≈ 157.08
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
25.
f ( x) =
x3
, f ′( x) = x 2
3
7
A = 2π ∫
1
3
x
3
26.
f '( x ) = − x(r 2 − x 2 )−1/ 2
1 + x 4 dx
7
π
= 250 2 − 2 2
9
(
= 2π ∫
r
)
−r
r
= 2π ∫
f ( x) =
x6 + 2
2
=
−r
= 2π ∫
2
3 ⎛ x4
1 ⎞ x6 1
1
= 2π ∫ ⎜
+
+ +
dx
⎟
2⎟ 4
1 ⎜ 8
2
4x ⎠
4 x6
⎝
3 ⎛ x4
1 ⎞ ⎛ x3
1 ⎞
= 2π ∫ ⎜
+
⎟ ⎜ + 3 ⎟ dx
2
1 ⎜ 8
4 x ⎟⎠ ⎜⎝ 2 2 x ⎟⎠
⎝
3 ⎛ x7 3x
1 ⎞
= 2π ∫ ⎜
+ +
⎟ dx
1 ⎜ 16 16 8 x5 ⎟
⎝
⎠
r
−r
r
x4
1
x3
1
+
, f ′( x ) =
−
8 4 x2
2 2 x3
⎛ x3
3 ⎛ x4
1 ⎞
1 ⎞
+
1+ ⎜ −
A = 2π ∫ ⎜
⎟
⎟
2⎟
3⎟
⎜
1 ⎜ 8
4x ⎠
⎝
⎝ 2 2x ⎠
= 2π ∫
−r
2
r 2 − x 2 1 + ⎡ − x(r 2 − x 2 )−1/ 2 ⎤ dx
⎣
⎦
r 2 − x 2 1 + x 2 (r 2 − x 2 )−1 dx
( r 2 − x2 )(1 + x2 (r 2 − x2 )−1 )dx
r 2 − x 2 + x 2 dx
r
r 2 dx = 2π ∫ rdx = 2π rx |−r r = 4π r 2
−r
30. x = f (t ) = r cos t
y = g (t ) = r sin t
f '(t ) = −r sin t
g '(t ) = r cos t
π
A = 2π ∫ r sin t (−r sin t ) 2 + (r cos t )2 dt
0
π
= 2π ∫ r sin t r 2 sin 2 t + r 2 cos 2 tdt
0
π
= 2π ∫ r sin t r 2 dt
0
π
= 2π ∫ r 2 sin tdt = −2π r 2 cos t |π0
3
⎡ x8 3 x 2
1 ⎤
= 2π ⎢
+
−
⎥
4
⎣⎢128 32 32 x ⎦⎥1
0
2
= −2r (−1 − 1) = 4π r 2
⎡⎛ 6561 27
1 ⎞ ⎛ 1
3
1 ⎞⎤
= 2π ⎢⎜
+
−
+ − ⎟⎥
⎟−⎜
⎣⎝ 128 32 2592 ⎠ ⎝ 128 32 32 ⎠ ⎦
8429π
=
≈ 326.92
81
31. a.
The base circumference is equal to the arc
length of the sector, so 2πr = θ l. Therefore,
2πr
θ=
.
l
b. The area of the sector is equal to the lateral
surface area. Therefore, the lateral surface
1
1 ⎛ 2πr ⎞
area is l 2θ = l 2 ⎜
⎟ = πrl .
2
2 ⎝ l ⎠
dx
dy
= 1,
= 3t 2
dt
dt
1
A = 2π∫ t 3 1 + 9t 4 dt
0
1
π
⎡1
⎤
= 2π ⎢ (1 + 9t 4 )3 / 2 ⎥ =
10 10 − 1
⎣ 54
⎦ 0 27
≈ 3.56
28.
r
248π 2
≈ 122.43
9
8x
27.
A = 2π ∫
−r
⎡1
⎤
= 2π ⎢ (1 + x 4 )3 / 2 ⎥
18
⎣
⎦1
=
29. y = f ( x) = r 2 − x 2
(
)
dx
dy
= −2t ,
=2
dt
dt
c.
Assume r2 > r1 . Let l1 and l2 be the slant
heights for r1 and r2 , respectively. Then
A = πr2l2 − πr1l1 = πr2 (l1 + l ) − πr1l1 .
From part a, θ =
1
1
0
0
A = 2π∫ 2t 4t 2 + 4 dt = 8π∫ t t 2 + 1 dt
1
8π
⎡1
⎤
= 8π ⎢ (t 2 + 1)3 2 ⎥ =
(2 2 − 1) ≈ 15.32
⎣3
⎦0 3
2πr2 2πr2 2πr1
=
=
.
l2
l1 + l
l1
Solve for l1 : l1r2 = l1r1 + lr1
l1 (r2 − r1 ) = lr1
l1 =
lr1
r2 − r1
⎛ lr
⎞
⎛ lr ⎞
A = πr2 ⎜ 1 + l ⎟ − πr1 ⎜ 1 ⎟
r
−
r
⎝ 2 1 ⎠
⎝ r2 − r1 ⎠
⎡r + r ⎤
= π(lr1 + lr2 ) = 2π ⎢ 1 2 ⎥ l
⎣ 2 ⎦
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32. Put the center of a circle of radius a at (a, 0).
Revolving the portion of the circle from x = b to
x = b + h about the x-axis results in the surface in
question. (See figure.)
34.
dx
dy
= −a sin t ,
= a cos t
dt
dt
Since the circle is being revolved about the line
x = b, the surface area is
A = 2π ∫
= 2πa ∫
2π
0
2π
0
(b − a cos t ) a 2 sin 2 t + a 2 cos 2 tdt
(b − a cos t )dt
= 2πa[bt − a sin t ]02π = 4π2 ab
35. a.
The equation of the top half of the circle is
y = a 2 − ( x − a)2 .
−( x − a )
dy
=
dx
b.
a − ( x − a)2
2
A = 2π ∫
b+h
a 2 − ( x − a)2 1 +
b
= 2π ∫
b+h
( x − a)2
a 2 − ( x − a)2
dx
a 2 − ( x − a )2 + ( x − a) 2 dx
b
b+ h
a dx = 2πa[ x]bb + h = 2 π ah
= 2π ∫
b
c.
A right circular cylinder of radius a and height h
has surface area 2 π ah.
33. a.
dx
dy
= a (1 − cos t ),
= a sin t
dt
dt
A = 2π ∫
2π
0
d.
a(1 − cos t ) ⋅
a 2 (1 − cos t )2 + a 2 sin 2 t dt
= 2πa ∫
2π
0
(1 − cos t ) 2a 2 − 2a 2 cos t dt
= 2 2πa 2 ∫
2π
0
b.
(1 − cos t )3 / 2 dt
e.
⎛t⎞
1 − cos t = 2sin 2 ⎜ ⎟ , so
⎝2⎠
⎛t⎞
sin 3 ⎜ ⎟ dt
⎝2⎠
2
π
t
t
⎛ ⎞
⎛ ⎞
= 8πa 2 ∫ sin ⎜ ⎟ sin 2 ⎜ ⎟ dt
0
⎝2⎠
⎝2⎠
2π
⎛ t ⎞⎡
⎛ t ⎞⎤
= 8πa 2 ∫ sin ⎜ ⎟ ⎢1 − cos 2 ⎜ ⎟ ⎥ dt
0
⎝ 2⎠⎣
⎝ 2 ⎠⎦
A = 2 2πa 2 ∫
2π 3 / 2
0
2
f.
2π
⎡
⎛t⎞ 2
⎛ t ⎞⎤
= 8πa 2 ⎢ −2 cos ⎜ ⎟ + cos3 ⎜ ⎟ ⎥
⎝2⎠ 3
⎝ 2 ⎠⎦0
⎣
⎡⎛
2⎞ ⎛
2 ⎞⎤
64 2
= 8πa 2 ⎢⎜ 2 − ⎟ − ⎜ −2 + ⎟ ⎥ =
πa
3
3⎠ ⎝
3 ⎠⎦
⎣⎝
320
Section 5.4
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f ′(t ) = −3sin t , g ′(t ) = 3cos t
36. a.
L=∫
=∫
2π
0
2π
9sin 2 t + 9 cos 2 tdt
3dt = 3[t ]02 π = 6π ≈ 18.850
0
f ′(t ) = −3sin t , g ′(t ) = cos t
b.
L=∫
2π
0
9sin 2 t + cos 2 tdt ≈ 13.365
f ′(t ) = cos t − t sin t , g ′(t ) = t cos t + sin t
c.
L=∫
=∫
6π
0
6π
0
(cos t − t sin t )2 + (t cos t + sin t )2 dt
1 + t 2 dt ≈ 179.718
f ′(t ) = − sin t , g ′(t ) = 2 cos 2t
d.
L=∫
2π
0
sin 2 t + 4 cos 2 2t dt ≈ 9.429
f ′(t ) = −3sin 3t , g ′(t ) = 2 cos 2t
e.
L=∫
2π
0
9sin 2 3t + 4 cos 2 2t dt ≈ 15.289
f ′(t ) = − sin t , g ′(t ) = π cos πt
f.
L=∫
40
0
sin t + π cos πt dt ≈ 86.58
2
2
2
5.5 Concepts Review
b
1. F ⋅ (b − a); ∫ F ( x) dx
a
2. 30 · 10 = 300
3. the depth of that part of the surface
4. δ hA
Problem Set 5.5
1
⎛1⎞
1. F ⎜ ⎟ = 6; k ⋅ = 6, k = 12
2
2
⎝ ⎠
F(x) = 12x
W =∫
1/ 2
0
1/ 2
12 x dx = ⎡ 6 x 2 ⎤
⎣
⎦0
=
3
= 1.5 ft-lb
2
2. From Problem 1, F(x) = 12x.
2
2
W = ∫ 12 x dx = ⎡ 6 x 2 ⎤ = 24 ft-lb
⎣
⎦0
0
3. F(0.01) = 0.6; k = 60
F(x) = 60x
W =∫
0.02
0
0.02
60 x dx = ⎡30 x 2 ⎤
⎣
⎦0
= 0.012 Joules
4. F(x) = kx and let l be the natural length of the
spring.
37.
W =∫
9 −l
8− l
9 −l
⎡1
⎤
kx dx = ⎢ kx 2 ⎥
2
⎣
⎦ 8−l
1 ⎡
k (81 − 18l + l 2 ) − (64 − 16l + l 2 ) ⎤
⎦
2 ⎣
1
= k (17 − 2l ) = 0.05
2
0.1
.
Thus, k =
17 − 2l
=
y = x, y ′ = 1 ,
L=∫
1
0
1
2dx = ⎡⎣ 2 x ⎤⎦ = 2 ≈ 1.41421
0
y = x2 , y′ = 2 x , L = ∫
1
1 + 4 x 2 dx ≈ 1.47894
0
y = x 4 , y ′ = 4 x3 , L = ∫
1
0
1 + 16 x6 dx ≈ 1.60023
y = x , y ′ = 10 x9 ,
10
L=∫
1
1 + 10018 dx ≈ 1.75441
0
100
y=x
L=∫
1
0
, y ′ = 100 x99 ,
1 + 10, 000 x198 dx ≈ 1.95167
When n = 10,000 the length will be close to 2.
Instructor’s Resource Manual
W =∫
10 −l
9 −l
10−l
⎡1
⎤
kx dx = ⎢ kx 2 ⎥
2
⎣
⎦ 9 −l
1 ⎡
k (100 − 20l + l 2 ) − (81 − 18l + l 2 ) ⎤
⎦
2 ⎣
1
= k (19 − 2l ) = 0.1
2
0.2
.
Thus, k =
19 − 2l
0.1
0.2
15
Solving
=
,l =
.
17 − 2l 19 − 2l
2
Thus k = 0.05, and the natural length is 7.5 cm.
=
Section 5.5
321
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
11. A slab of thickness Δy at height y has width
d
d
⎡1
⎤
5. W = ∫ kxdx = ⎢ kx 2 ⎥
0
2
⎣
⎦0
1
1
= k (d 2 − 0) = kd 2
2
2
6. F (8) = 2; k16 = 2, k =
1
8
27
1 4/3
1 ⎡3
6561
⎤
s
ds = ⎢ s 7 / 3 ⎥ =
0 8
8 ⎣7
56
⎦0
≈ 117.16 inch-pounds
W =∫
27
2
⎡1 ⎤
9 s ds = 9 ⎢ s 2 ⎥ = 18 ft-lb
0
⎣ 2 ⎦0
7. W = ∫
2
1
3
4
=
15
5
1 ⎤
⎡
(62.4) ⎢36 y + y 2 − y 3 ⎥
2
2
3 ⎦0
⎣
15
64 ⎞
⎛
(62.4) ⎜ 144 + 40 − ⎟
2
3 ⎠
⎝
= 76,128 ft-lb
=
8. One spring will move from 2 feet beyond its
natural length to 3 feet beyond its natural length.
The other will move from 2 feet beyond its
natural length to 1 foot beyond its natural length.
3
3
y + 3 and length 10. The slab will be lifted a
4
⎛3
⎞
distance 9 – y. ΔW ≈ δ ⋅10 ⋅ ⎜ y + 3 ⎟ Δy (9 − y )
⎝4
⎠
15
= δ (36 + 5 y − y 2 )Δy
2
4 15
δ (36 + 5 y − y 2 )dy
W =∫
0 2
1
W = ∫ 6 s ds + ∫ 6 s ds = ⎡3s 2 ⎤ + ⎡3s 2 ⎤
⎣
⎦2 ⎣
⎦2
2
2
= 3(9 – 4) + 3(1 – 4) = 6 ft-lb
9. A slab of thickness Δy at height y has width
4
4 − y and length 10. The slab will be lifted a
5
distance 10 – y.
4 ⎞
⎛
ΔW ≈ δ ⋅10 ⋅ ⎜ 4 − y ⎟ Δy (10 − y )
5 ⎠
⎝
12. A slab of thickness Δy at height y has width
2 6 y − y 2 and length 10. The slab will be lifted
a distance 8 – y.
ΔW ≈ δ ⋅10 ⋅ 2 6 y − y 2 Δy (8 − y )
= 20δ 6 y − y 2 (8 − y )Δy
3
W = ∫ 20δ 6 y − y 2 (8 − y )dy
0
= 20δ ∫
3
0
6 y − y 2 (3 − y ) dy
+20δ ∫
3
0
6 y − y 2 (5)dy
3
3
⎡1
⎤
= 20δ ⎢ (6 y − y 2 )3 / 2 ⎥ +100δ ∫ 6 y − y 2 dy
0
⎣3
⎦0
= 8δ ( y 2 − 15 y + 50)Δy
5
W = ∫ 8δ ( y 2 − 15 y + 50) dy
0
Notice that
5
15
⎡1
⎤
= 8(62.4) ⎢ y 3 − y 2 + 50 y ⎥
3
2
⎣
⎦0
125
375
⎛
⎞
= 8(62.4) ⎜
−
+ 250 ⎟ = 52,000 ft-lb
2
⎝ 3
⎠
10. A slab of thickness Δy at height y has width
4
y and length 10. The slab will be lifted a
3
distance 8 – y.
4 ⎞
⎛
ΔW ≈ δ ⋅10 ⋅ ⎜ 4 − y ⎟ Δy (8 − y )
3 ⎠
⎝
40
= δ (24 − 11y + y 2 )Δy
3
3 40
W =∫
δ (24 − 11 y + y 2 )dy
0 3
4−
3
∫0
6 y − y 2 dy is the area of a
quarter of a circle with radius 3.
⎛1 ⎞
W = 20δ (9) + 100δ ⎜ π9 ⎟
⎝4 ⎠
= (62.4)(180 + 225 π ) ≈ 55,340 ft-lb
13. The volume of a disk with thickness Δy is
16πΔy . If it is at height y, it will be lifted a
distance 10 – y.
ΔW ≈ δ 16πΔy (10 − y ) = 16πδ (10 − y )Δy
10
1 ⎤
⎡
16πδ (10 − y )dy = 16π(50) ⎢10 y − y 2 ⎥
0
2 ⎦0
⎣
= 16 π (50)(100 – 50) ≈ 125,664 ft-lb
W =∫
10
3
322
=
40
11
1 ⎤
⎡
(62.4) ⎢ 24 y − y 2 + y 3 ⎥
3
2
3 ⎦0
⎣
=
40
99
⎛
⎞
(62.4) ⎜ 72 − + 9 ⎟ = 26,208 ft-lb
3
2
⎝
⎠
Section 5.5
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14. The volume of a disk with thickness Δx at height
2
x is π(4 + x) Δx . It will be lifted a distance of
10 – x.
ΔW ≈ δπ(4 + x)2 Δx(10 − x)
= πδ (160 + 64 x + 2 x 2 − x3 )Δx
W =∫
10
19. The total work is equal to the work W1 to haul the
load by itself and the work W2 to haul the rope by
itself.
W1 = 200 ⋅ 500 = 100, 000 ft-lb
Let y = 0 be the bottom of the shaft. When the
rope is at y, ΔW2 ≈ 2Δy (500 − y ) .
πδ (160 + 64 x + 2 x 2 − x3 )dx
0
10
2
1 ⎤
⎡
= π(50) ⎢160 x + 32 x 2 + x3 − x 4 ⎥
3
4 ⎦0
⎣
2000
⎛
⎞
= π(50) ⎜ 1600 + 3200 +
− 2500 ⎟
3
⎝
⎠
≈ 466,003 ft-lb
15. The total force on the face of the piston is A · f(x)
if the piston is x inches from the cylinder head.
The work done by moving the piston from
x1 to x2 is W = ∫
x2
x1
x
A ⋅ f ( x)dx = A∫ 2 f ( x)dx .
x1
This is the work done by the gas in moving the
piston. The work done by the piston to compress
the gas is the opposite of this or A∫
x1
x2
10
10
1 ⎤
⎡
W2 = ∫ (10 − y )dy = ⎢10 y − y 2 ⎥
0
2 ⎦0
⎣
= 100 – 50 = 50 ft-lb
W = W1 + W2 = 250 ft-lb
f ( x)dx .
A = 1; p(v) = cv −1.4
f ( x) = cx −1.4
21.
16
2
= 16, x2 = = 2
1
1
W =∫
16
cx −1.4 dx
2
1.4
= 40(16) (−2.5)(16
≈ 2075.83 in.-lb
40002
−2
W =∫
)
1.4
dx
8
= 2c ⎡ −1.25(2 x)−0.4 ⎤
⎣
⎦1
−0.4
−0.4
= 80(16) (−1.25)(16
≈ 2075.83 in.-lb
−2
= 5000 , k = 80,000,000,000
4200 80, 000, 000, 000
22. F ( x) =
16
2
x1 =
= 8, x2 = = 1
2
2
1
; f (4000) = 5000
x2
dx
4200
f ( x) = c(2 x) −1.4
W = 2 ∫ c(2 x)
x2
⎡ 1⎤
= 80, 000, 000, 000 ⎢ − ⎥
⎣ x ⎦ 4000
20, 000, 000
=
≈ 952,381 mi-lb
21
A = 2; p (v) = cv −1.4
−1.4
k
4000
17. c = 40(16)1.4
8
f ( x) =
k
16
= c ⎡ −2.5 x −0.4 ⎤
⎣
⎦2
−0.4
−0.4
500
20. The total work is equal to the work W1 to lift the
monkey plus the work W2 to lift the chain.
W1 = 10 ⋅ 20 = 200 ft-lb
Let y = 20 represent the top. As the monkey
climbs the chain, the piece of chain at height y
(0 ≤ y ≤ 10) will be lifted 20 – 2y ft.
1
ΔW2 ≈ Δy (20 − 2 y ) = (10 − y )Δy
2
16. c = 40(16)1.4
x1 =
500
1 ⎤
⎡
2(500 − y )dy = 2 ⎢500 y − y 2 ⎥
0
2 ⎦0
⎣
= 2(250,000 – 125,000) = 250,000 ft-lb
W = W1 + W2 = 100, 000 + 250, 000
= 350,000 ft-lb
W2 = ∫
)
k
x2
where x is the distance between the
charges. F (2) = 10;
k
= 10, k = 40
4
5
⎡ 40 ⎤
dx = ⎢ − ⎥ = 32 ergs
1 x2
⎣ x ⎦1
W =∫
5 40
18. 80 lb/in.2 = 11,520 lb/ft2
c =11,520(1)1.4 = 11,520
ΔW ≈ p (v)Δv = 11,520v −1.4 Δv
4
4
W = ∫ 11,520v −1.4 dv = ⎡ −28,800v −0.4 ⎤
⎣
⎦1
1
= −28,800(4−0.4 − 1−0.4 ) ≈ 12,259 ft-lb
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Section 5.5
323
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
23. The relationship between the height of the bucket
1
and time is y = 2t, so t = y . When the bucket is
2
a height y, the sand has been leaking out of the
1
bucket for y seconds. The weight of the bucket
2
3
⎛1 ⎞
and sand is 100 + 500 − 3 ⎜ y ⎟ = 600 − y.
2
⎝2 ⎠
3 ⎞
⎛
ΔW ≈ ⎜ 600 − y ⎟ Δy
2 ⎠
⎝
80
80 ⎛
3 ⎞
3 ⎤
⎡
W = ∫ ⎜ 600 − y ⎟ dy = ⎢600 y − y 2 ⎥
0 ⎝
2 ⎠
4 ⎦0
⎣
= 48,000 – 4800 = 43,200 ft-lb
24. The total work is equal to the work W1 needed to
fill the pipe plus the work W2 needed to fill the
tank.
2
δπy
⎛1⎞
ΔW1 = δπ ⎜ ⎟ Δy ( y ) =
Δy
2
4
⎝ ⎠
W1 = ∫
30 δπy
0
4
dy =
( 62.4 ) π ⎡ 1
27. Place the equilateral triangle in the coordinate
system such that the vertices are
(
)
(−3, 0), (3, 0) and 0, −3 3 .
The equation of the line in Quadrant I is
y= 3 ⋅ x − 3 3 or x =
= −2δ ∫
0
−3
⎞⎞
+ 3 ⎟ ⎟ dy
⎠⎠
⎝ ⎝ 3
δ (− y ) ⎜ 2 ⎜
⎛ y2
⎞
+ 3 y ⎟ dy
⎜
⎜
⎟
3
⎝ 3
⎠
0
(30 ≤ y ≤ 50) is π r 2 where
⎡ y3 3 y 2 ⎤
= −2δ ⎢
+
⎥
2 ⎥⎦
⎢⎣ 3 3
−3
r = 102 − (40 − y )2 = − y 2 + 80 y − 1500 .
= 1684.8 pounds
W2 = ∫
50
30
3
2
δπ(− y + 80 y − 1500 y )dy
50
80 3
⎡ 1
⎤
= (62.4)π ⎢ − y 4 +
y − 750 y 2 ⎥
3
⎣ 4
⎦ 30
10, 000, 000
⎡⎛
⎞
= (62.4)π ⎢⎜ −1,562,500 +
− 1,875, 000 ⎟
3
⎠
⎣⎝
− ( −202,500 + 720, 000 − 675, 000 ) ⎤⎦
≈ 10,455,220 ft-lb
W = W1 + W2 ≈ 10, 477, 274 ft-lb
with 0 ≤ y ≤ 3. The force against this rectangle
at depth 3 – y is ΔF ≈ δ (3 − y )(6)Δy . Thus,
3
⎡
y2 ⎤
F = ∫ δ (3 − y )(6) dy = 6δ ⎢3 y − ⎥
0
2 ⎥⎦
⎢⎣
0
= 6 ⋅ 62.4 ( 4.5 ) = 1684.8 pounds
Section 5.5
= −2 ⋅ 62.4(0 − 13.5)
3
28. Place the right triangle in the coordinate system
such that the vertices are (0,0), (3,0) and (0,-4).
The equation of the line in Quadrant IV is
4
3
y = x − 4 or x = y + 3.
3
4
⎛3
⎞
ΔF ≈ δ (3 − y ) ⎜ y + 3 ⎟ Δy and
⎝4
⎠
0 ⎛
3
3 ⎞
F = ∫ δ ⎜ 9 − y − y 2 ⎟ dy
−4 ⎝
4
4 ⎠
0
25. Let y measure the height of a narrow rectangle
3
+ 3.
3
⎛ ⎛ y
0
30
ΔW2 = δπr 2 Δy y = δπ(− y3 + 80 y 2 − 1500 y )Δy
y
⎛ ⎛ y
⎞⎞
ΔF ≈ δ (− y ) ⎜ 2 ⎜
+ 3 ⎟ ⎟ Δy and
⎠⎠
⎝ ⎝ 3
−3 3
≈ 22, 054 ft-lb
The cross sectional area at height y feet
324
3
⎡
3
y2 ⎤
F = ∫ δ (5 − y )(6) dy = 6δ ⎢5 y − ⎥
0
2 ⎥⎦
⎣⎢
0
= 6 ⋅ 62.4 ⋅10.5 = 3931.2 pounds
F=∫
2⎤
⎢2 y ⎥
⎣
⎦0
4
26. Let y measure the height of a narrow rectangle
with 0 ≤ y ≤ 3. The force against this rectangle
at depth 5 – y is ΔF ≈ δ (5 − y )(6)Δy . Thus,
⎡
3 y 2 y3 ⎤
= δ ⎢9 y −
− ⎥ = 62.4 ⋅ 26
8
4 ⎥⎦
⎣⎢
−4
= 1622.4 pounds
29. ΔF ≈ δ (1 − y )
1
(
( y ) Δy; F = ∫01δ (1 − y) ( y ) dy
)
= δ ∫ y1 2 − y 3 2 dy
0
1
2
⎡2
⎤
⎛ 4⎞
= δ ⎢ y3 2 − y5 2 ⎥ = 62.4 ⎜ ⎟
5
⎣3
⎦0
⎝ 15 ⎠
= 16.64 pounds
Instructor’s Resource Manual
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
30. Place the circle in the coordinate system so that
the center is (0.0). The equation of the circle is
33. We can position the x-axis along the bottom of
the pool as shown:
20
x 2 + y 2 = 16 and in Quadrants I and IV,
x = 16 − y 2 . ΔF ≈ δ (6 − y ) ⎛⎜ 2 16 − y 2 ⎞⎟ Δy
⎝
⎠
4
2⎞
⎛
F = ∫ δ (6 − y ) ⎜ 2 16 − y ⎟ dy
−4
⎝
⎠
Using a CAS, F ≈ 18,819 pounds.
⎛a ⎞
ΔF ≈ δ (b − y ) ⎜ y ⎟ Δy and
⎝b ⎠
b
⎛a ⎞
F = ∫ δ (b − y ) ⎜ y ⎟ dy
0
⎝b ⎠
a ⎞
= δ ∫ ⎜ y − y 2 ⎟ dy = δ
0⎝
b ⎠
10
x
From the diagram, we let h = the depth of an
arbitrary slice along the width of the bottom of
the pool.
20
4
Using the Pythagorean Theorem, we can find that
the length of the bottom of the pool is
202 + 42 = 416 = 4 26
Next, we need to get h in terms of x. This can be
done by using similar triangles to set up a
proportion.
b
⎡ ay 2 ay 3 ⎤
−
⎢
⎥
3b ⎥⎦
⎢⎣ 2
0
⎛ ab 2 ab 2 ⎞
ab 2
=δ ⎜
−
⎟ =δ
⎜ 2
3 ⎟⎠
6
⎝
For the lower right triangle II,
a ⎞
⎛
ΔF ≈ δ (b − y ) ⎜ a − y ⎟ dy and
b ⎠
⎝
b
a ⎞
⎛
F = ∫ δ (b − y ) ⎜ a − y ⎟ dy
0
b ⎠
⎝
h−4
b
2
⎡
⎛ 2
ay 3 ⎤
2 ab
= δ ⎢ aby − ay 2 +
⎥ = δ ⎜⎜ ab − ab +
3b ⎦⎥
3
⎣⎢
⎝
0
4
x
4 26
b ⎛
a ⎞
= ∫ δ ⎜ ab − 2ay + y 2 ⎟ dy
0 ⎝
b ⎠
⎞
⎟
⎟
⎠
2
ab
3
The total force on one half of the dam is twice the
=δ
h
Δx
31. Place a rectangle in the coordinate system such
that the vertices are (0,0), (0,b), (a,0) and (a,b).
The equation of the diagonal from (0,0) to (a,b)
b
a
is y = x or x = y. For the upper left triangle I,
a
b
b⎛
8
4
ab 2
δ
3 = 2.
total force on the other half since
ab 2
δ
6
h−4
x
=
4
4 26
→
h = 4+
x
26
ΔF = δ ⋅ h ⋅ ΔA
F=
∫
4 26
⎛
δ ⎜4+
⎝
0
x ⎞
⎟ (10 ) dx
26 ⎠
x ⎞
⎛
62.4 ⎜ 4 +
⎟ (10 ) dx
0
26 ⎠
⎝
4 26 ⎛
x ⎞
= 624
⎜4+
⎟ dx
0
26 ⎠
⎝
=
∫
4 26
∫
4 26
32. Consider one side of the cube and place the
vertices of this square on (0,0), (0,2), (2,0) and
(2,2).
2
ΔF ≈ δ (102 − y )(2)Δy; F = ∫ 2δ (102 − y ) dy
0
2 ⎤2
⎡
x2 ⎤
= 624 ⎢ 4 x +
⎥
2 26 ⎦ 0
⎣
(
)
(
= 624 16 26 + 8 26 = 624 24 26
= 14,976 26 lb
)
( ≈ 76,362.92 lb )
⎡
y
= 2δ ⎢102 y − ⎥ = 2 ⋅ 62.4 ⋅ 202 = 25, 209.6
2 ⎦⎥
⎣⎢
0
The force on all six sides would be 6(25,209.6) =
151,257.6 pounds.
Instructor’s Resource Manual
Section 5.5
325
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
34. If we imagine unrolling the cylinder so we have a
flat sheet, then we need to find the total force
against one side of a rectangular plate as if it had
been submerged in the oil. The rectangle would
be 2π ( 5 ) = 10π feet wide and 6 feet high.
Thus, the total lateral force is given by
F=
∫
6
0
∫
6
0
y dy = ⎡⎣ 250π y 2 ⎤⎦
6
0
= 250π ( 36 ) = 9000π lbs ( ≈ 28, 274.33 lb)
35. Let W1 be the work to lift V to the surface and
W2 be the work to lift V from the surface to 15
feet above the surface. The volume displaced by
the buoy y feet above its original position is
1 ⎛
a
π⎜ a −
3 ⎝
h
2
3
1
y⎞
⎞
⎛
y ⎟ (h − y ) = πa 2 h ⎜ 1 − ⎟ .
3
⎠
⎝ h⎠
δ
The weight displaced is
3
a2 h =
3
y⎞
⎛
πa 2 h ⎜ 1 − ⎟ .
⎝ h⎠
Note by Archimede’s Principle m =
δ
3
πa 2 h or
3m
, so the displaced weight is
δπ
3
y⎞
⎛
m ⎜1 − ⎟ .
⎝ h⎠
3
3
⎛
⎛ ⎛
y⎞ ⎞
y⎞ ⎞
⎛
ΔW1 ≈ ⎜ m − m ⎜1 − ⎟ ⎟ Δy = m ⎜ 1 − ⎜ 1 − ⎟ ⎟ Δy
⎜
⎜ ⎝ h⎠ ⎟
⎝ h ⎠ ⎟⎠
⎝
⎝
⎠
3⎞
⎛
h
y⎞
⎛
W1 = m ∫ ⎜ 1 − ⎜1 − ⎟ ⎟ dy
0⎜
h⎠ ⎟
⎝ ⎝
⎠
h
4
⎡
h⎛
y⎞ ⎤
3mh
= m ⎢ y + ⎜1 − ⎟ ⎥ =
4⎝ h⎠ ⎥
4
⎢⎣
⎦0
W2 = m ⋅15 = 15m
W = W1 + W2 =
326
Section 5.5
3mh
+ 15m
4
4
W1 = ∫ δ 40(10 − y )dy
0
4
50 ⋅ y ⋅10π dy
= 500π
36. First calculate the work W1 needed to lift the
contents of the bottom tank to 10 feet.
ΔW1 ≈ δ 40Δy (10 − y )
⎡ 1
⎤
= (62.4)(40) ⎢ − (10 − y )2 ⎥
⎣ 2
⎦0
= (62.4)(40)(–18 + 50) = 79,872 ft-lb
Next calculate the work W2 needed to fill the top
tank. Let y be the distance from the bottom of the
top tank.
ΔW2 ≈ δ (36π)Δy y
Solve for the height of the top tank:
160 40
36πh = 160; h =
=
36π 9π
W2 = ∫
40 / 9 π
0
δ 36πy dy
40 / 9 π
⎡1 ⎤
= (62.4)(36π) ⎢ y 2 ⎥
⎣ 2 ⎦0
⎛ 800 ⎞
= (62.4)(36π) ⎜
⎟ ≈ 7062 ft-lbs
⎝ 81π2 ⎠
W = W1 + W2 ≈ 86,934 ft-lbs
225
⎛1
⎞
37. Since δ ⎜ πa 2 ⎟ (8) = 300, a =
.
2πδ
⎝3
⎠
When the buoy is at z feet (0 ≤ z ≤ 2) below
floating position, the radius r at the water level is
225 ⎛ 8 + z ⎞
⎛8+ z ⎞
r =⎜
⎟a =
⎜
⎟.
2πδ ⎝ 8 ⎠
⎝ 8 ⎠
⎛1
⎞
F = δ ⎜ πr 2 ⎟ (8 + z ) − 300
3
⎝
⎠
75
3
=
(8 + z ) − 300
128
2 ⎡ 75
⎤
W =∫ ⎢
(8 + z )3 − 300 ⎥ dz
0 ⎣128
⎦
2
⎡ 75
⎤
=⎢
(8 + z ) 4 − 300 z ⎥
⎣ 512
⎦0
⎛ 46,875
⎞
=⎜
− 600 ⎟ − (600 − 0)
⎝ 32
⎠
8475
=
≈ 264.84 ft-lb
32
Instructor’s Resource Manual
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
6. M y = (−3) ⋅ 5 + (−2) ⋅ 6 + 3 ⋅ 2 + 4 ⋅ 7 + 7 ⋅1 = 14
5.6 Concepts Review
1. right;
M x = 2 ⋅ 5 + (−2) ⋅ 6 + 5 ⋅ 2 + 3 ⋅ 7 + (−1) ⋅1 = 28
m = 5 + 6 + 2 + 7 + 1 = 21
My 2
M
4
= ,y= x =
x=
3
3
m
m
4 ⋅1 + 6 ⋅ 3
= 2.2
4+6
2. 2.5; right; x(1+x); 1 + x
3. 1; 3
4.
24 40
;
16 16
The second lamina balances at x = 3, y = 1 .
The first lamina has area 12 and the second
lamina has area 4.
12 ⋅1 + 4 ⋅ 3 24
12 ⋅ 3 + 4 ⋅1 40
x=
=
,y=
=
12 + 4
16
12 + 4
16
Problem Set 5.6
1. x =
7. Consider two regions R1 and R2 such that R1 is
bounded by f(x) and the x-axis, and R2 is
bounded by g(x) and the x-axis. Let R3 be the
region formed by R1 − R2 . Make a regular
partition of the homogeneous region R3 such
that each sub-region is of width , Δx and let x be
the distance from the y-axis to the center of mass
of a sub-region. The heights of R1 and R2 at x
are approximately f(x) and g(x) respectively. The
mass of R3 is approximately
Δm = Δm1 − Δm2
≈ δ f ( x ) Δx − δ g ( x ) Δ x
2 ⋅ 5 + (−2) ⋅ 7 + 1 ⋅ 9 5
=
5+7+9
21
= δ [ f ( x) − g ( x)]Δx
where δ is the density. The moments for R3 are
approximately
M x = M x ( R1 ) − M x ( R2 )
2. Let x measure the distance from the end where
John sits.
180 ⋅ 0 + 80 ⋅ x + 110 ⋅12
=6
180 + 80 + 110
80x + 1320 = 6 · 370
80x = 900
x = 11.25
Tom should be 11.25 feet from John, or,
equivalently, 0.75 feet from Mary.
7
3.
∫ x
x= 0
7
∫0
7
4. x =
(
=
7
x dx
=
x dx
⎡ 2 x5 / 2 ⎤
⎣5
⎦0
7
⎡ 2 x3 / 2 ⎤
⎣3
⎦0
x(1 + x3 )dx
∫0
7
3
∫0 (1 + x )dx
49
2
+ 16,807
5
(7 +
2401
4
)
)=
=
=
δ
[ f ( x)]2 Δx − [ g ( x)]2 Δx
2
2
δ⎡
( f ( x)) 2 − ( g ( x))2 ⎤ Δx
≈ xδ f ( x)Δx − xδ g ( x)Δx
= xδ [ f ( x) − g ( x)]Δx
Taking the limit of the regular partition as
Δx → 0 yields the resulting integrals in
Figure 10.
( 49 7 ) = 21
(7 7 ) 5
⎡ 1 x 2 + 1 x5 ⎤
5
⎣2
⎦0
δ
⎦
2⎣
M y = M y ( R1 ) − M y ( R2 )
2
5
2
3
7
=
≈
8.
7
⎡ x + 1 x4 ⎤
4
⎣
⎦0
33,859
10
2429
4
=
9674
≈ 5.58
1735
5. M y = 1 ⋅ 2 + 7 ⋅ 3 + (−2) ⋅ 4 + (−1) ⋅ 6 + 4 ⋅ 2 = 17
f ( x) = 2 − x; g ( x) = 0
M x = 1 ⋅ 2 + 1 ⋅ 3 + (−5) ⋅ 4 + 0 ⋅ 6 + 6 ⋅ 2 = −3
m = 2 + 3 + 4 + 6 + 2 = 17
My
M
3
= 1, y = x = −
x=
17
m
m
∫ x[(2 − x) − 0]dx
x= 0
2
∫0 [(2 − x) − 0]dx
Instructor’s Resource Manual
2
2
2
∫ [2 x − x ]dx
= 0
2
∫0 [2 − x]dx
Section 5.6
327
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
2
⎛ 2 1 3⎞
⎜x − x ⎟
3 ⎠0
⎝
10.
8
3
=
=
2
4
−
2
1 2⎞
⎛
⎜ 2x − x ⎟
2 ⎠0
⎝
2
=
3
4−
1 2
2
2
∫0 [(2 − x) − 0 ]dx
2
y=
2
∫ [(2 − x) − 0]dx
4
∫
x= 0
0
2
∫ [4 − 4 x + x
= 0
2
]dx
4
2
⎛
2 1 3⎞
8
⎜ 4x − 2x + x ⎟
3 ⎠0 8 − 8 + 3
⎝
=
=
4
4
2
=
3
=
( 13 x2 ) dx = 13 ∫04 x3dx
4 1 2
x dx
3
∫0
4
1 ⎡ 1 x4 ⎤
3 ⎣4
⎦0
4
1 ⎡ 1 x3 ⎤
3 ⎣3
⎦0
y=
9.
=
x
=
64
3
64
9
1 4
3 0
∫
=3
( )
2
1 4 1 x 2 dx
2 0 3
4 1 2
x dx
0 3
∫
∫
512
45
64
9
=
x 2 dx
=
1 4 x 4 dx
18 0
64
9
∫
=
4
1 ⎡ 1 x5 ⎤
18 ⎣ 5
⎦0
64
9
8
5
11.
x = 0 (by symmetry)
y=
2
1
2 − 2
2
∫
∫−
=
∫
2
1
2 − 2
2
(2 − x 2 )2 dx
(2 − x 2 )dx
1
(4 − 4 x 2 + x 4 )dx
1 3⎤
⎡
⎢⎣ 2 x − 3 x ⎥⎦
−
=
328
1 ⎡ 4 x − 4 x3
2⎣
3
2
2
+ 15 x5 ⎤
⎦−
8 2
3
Section 5.6
x=
2
2
=
32 2
15
8 2
3
=
4
5
x( x3 )dx
∫0
1 3
∫0 x dx
=
1 4
x
0
∫
1
dx
1
⎡ 1 x4 ⎤
⎣ 4 ⎦0
=
⎡ 1 x5 ⎤
⎣ 5 ⎦0
1
4
=
1
5
1
4
=
4
5
1
1 1 3 2
1 1 6
1 x7 ⎤
( x ) dx
x dx ⎡ 14
∫
∫
⎣
⎦
0
0
0
y= 2
= 2
=
1 3
1
1
4
4
∫ x dx
0
=
1
14
1
4
=
2
7
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12.
y=
⎡
∫ ⎢⎣( 2 x )
1 4
2 1
4
⎤
− (2 x − 4)2 ⎥ dx
⎦
2
⎡ 2 x − ( 2 x − 4 ) ⎤ dx
⎣
⎦
∫1
4
=
2 ∫ (− x 2 + 5 x − 4)dx
1
19
3
4
=
x = 0 (by symmetry)
2
1 2 ⎡ − 1 ( x 2 − 10) ⎤ dx
⎥
2 ∫−2 ⎢ 2
⎣
⎦
y=
2
⎡ − 1 ( x 2 − 10) ⎤ dx
∫−2 ⎣ 2
⎦
(
=
− 18 ∫
2
−2
)
19
3
=
9
19
3
=
27
19
14.
( x 4 − 20 x 2 + 100)dx
− 12 ∫
2
−2
( x 2 − 10)dx
2
=
2 ⎡ − 13 x3 + 52 x 2 − 4 x ⎤
⎣
⎦1
x3 + 100 x ⎤
− 18 ⎡ 15 x5 − 20
3
⎣
⎦ −2
2
− 12 ⎡ 13 x3 − 10 x ⎤
⎣
⎦ −2
=
− 574
15
52
3
=−
287
130
To find the intersection points, x 2 = x + 3 .
x2 − x − 3 = 0
13.
x=
1 ± 13
2
(1+ 13 )
x( x + 3 − x 2 )dx
∫(1− 213 )
x=
2
(1+ 13 )
∫(1− 213 )
To find the intersection point, solve
2x − 4 = 2 x .
x−2 = x
x2 − 4 x + 4 = x
4
x=
∫1
4
∫1 ⎡⎣ 2
4
=
2
(1+ 13 )
⎡ 1 x 2 + 3 x − 1 x3 ⎤ 2
3
⎣2
⎦ (1− 13 )
2
3/ 2
(1+ 13 )
⎡ 1 x3 + 3 x 2 − 1 x 4 ⎤ 2
2
4
⎣3
⎦ (1− 13 )
x − (2 x − 4) ⎤⎦ dx
− x + 2 x)dx
4
2 ∫ ( x1/ 2 − x + 2)dx
4
2 ⎡ 52 x5 / 2 − 13 x3 + x 2 ⎤
⎣
⎦1
4
2 ⎡ 23 x3 / 2 − 12 x 2 + 2 x ⎤
⎣
⎦1
2
=
2
13 13
6
=
(1+ 13 )
1
=
( x 2 + 3x − x3 )dx
∫(1− 213 )
x ⎡⎣ 2 x − (2 x − 4) ⎤⎦ dx
2∫ ( x
1
2
(1+ 13 )
=
x2 − 5x + 4 = 0
(x – 4)(x – 1) = 0
x = 4 (x = 1 is extraneous.)
( x + 3 − x 2 )dx
=
64
5
19
3
=
192
95
13 3
12
13 13
6
=
1
2
1
2
⎡ ( x + 3)2 − ( x 2 ) 2 ⎤ dx
⎦
2 ∫(1− 13 ) ⎣
y=
2
1+ 13
(
)
2
1− 13
)
∫(
( x + 3 − x 2 )dx
2
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
(1+ 13 )
1
2
∫(
2
1− 13
)
y = 1± 5
( x2 + 6 x + 9 − x4 )
x=
2
=
13 13
6
1 ⎡ 1 x3
2 ⎣3
+ 3x
2
(
∫
∫1−
)
=
)
∫
∫1−
13 13
6
143 13
30
13 13
6
=
11
=
5
1 ⎡ − 1 y5
2⎣ 5
5
1+ 5
∫1−
1+ 5
∫
= 1−
=
To find the intersection points, solve y 2 = 2 .
y=± 2
2
1
⎡ 22
2 − 2 ⎣
2
∫−
2
− ( y 2 )2 ⎤ dy
⎦
(2 − y 2 )dy
2
1 ⎡ 4 y − 1 y5 ⎤
2⎣
5
⎦− 2
8 2
3
=
16 2
5
8 2
3
=
∫
2
1
2 − 2
=
6
5
16.
To find the intersection points, solve
y2 − 3y − 4 = − y .
y2 − 2 y − 4 = 0
330
2 ± 20
2
Section 5.6
(4 − y 4 )dy
⎡2 y − 1 y3 ⎤
3
⎣
⎦−
y = 0 (by symmetry)
y=
1+ 5
+ 32 y 4 − 12 y 2 − 16 y ⎤
⎦1−
5
=
−20 5
5
20 5
3
= −3
1+ 5
=
(− y 2 + 2 y + 4)dy
1+ 5
∫
y = 1−
x=
5
+ 6 y 3 − 24 y − 16)dy
⎡ − 1 y3 + y 2 + 4 y ⎤
⎣ 3
⎦1−
15.
∫
− ( y 2 − 3 y − 4) 2 ⎤ dy
⎦
⎡ (− y ) − ( y 2 − 3 y − 4) ⎤ dy
⎣
⎦
5
1 1+ 5 (− y 4
2 1− 5
1+ 5
2
=
=
(
1+ 13
5⎤
1
2
+ 9x − 5 x
⎦ 1− 13
1 1+ 5 ⎡ ( − y ) 2
2 1− 5 ⎣
1+ 5
2
2
5
5
y ⎡ (− y ) − ( y 2 − 3 y − 4) ⎤ dy
⎣
⎦
⎡ (− y ) − ( y 2 − 3 y − 4) ⎤ dy
⎣
⎦
(− y 3 + 2 y 2 + 4 y )dy
20 5
3
1+ 5
⎡ − 1 y 4 + 2 y3 + 2 y 2 ⎤
3
⎣ 4
⎦1− 5
20 5
3
=
20 5
3
20 5
3
=1
17. We let δ be the density of the regions and Ai be
the area of region i.
Region R1 :
1
m( R1 ) = δ A1 = δ (1/ 2)(1)(1) = δ
2
1
1 3
1
1
x( x)dx 3 x |
∫
2
0
=
= 3=
x1 = 0
1
1
1
3
∫0 xdx 1 x 2 | 2
2
0
Since R1 is symmetric about the line y = 1 − x ,
the centroid must lie on this line. Therefore,
2 1
y1 = 1 − x1 = 1 − = ; and we have
3 3
1
M y ( R1 ) = x2 ⋅ m( R1 ) = δ
3
1
M x ( R1 ) = y2 ⋅ m( R1 ) = δ
6
Region R2 :
m( R2 ) = δ A2 = δ (2)(1) = 2δ
By symmetry we get
1
x 2 = 2 and y2 = .
2
Thus,
M y ( R2 ) = x2 ⋅ m( R2 ) = 4δ
M x ( R2 ) = y2 ⋅ m( R2 ) = δ
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
18. We can obtain the mass and moments for the
whole region by adding the individual regions.
Using the results from Problem 17 we get that
1
5
m = m( R1 ) + m( R2 ) = δ + 2δ = δ
2
2
1
13
M y = M y ( R1 ) + M y ( R2 ) = δ + 4δ = δ
3
3
1
7
M x = M x ( R1 ) + M x ( R2 ) = δ + δ = δ
6
6
Therefore, the centroid is given by
13
δ
My
26
= 3 =
x=
5
15
m
δ
2
7
δ
Mx 6
7
=
=
y=
5
15
m
δ
2
b
20. m( R1 ) = δ ∫ (h( x) − g ( x))dx
a
b
m( R2 ) = δ ∫ ( g ( x) − f ( x))dx
a
M x ( R1 ) =
δ
c
M x ( R2 ) =
a
Now,
b
m( R3 ) = δ ∫ (h( x) − f ( x))dx
a
b
= δ ∫ (h( x) − g ( x) + g ( x) − f ( x))dx
2
2
2
2
=
=
(( g ( x)) − ( f ( x)) )dx
+
a
b
c
a
b
δ
(( g ( x)) 2 − ( f ( x))2 )dx
c
(( g ( x))
2 ∫b
2
− ( f ( x))2 )dx
− ( f ( x)) 2 )dx
= M x ( R1 ) + M x ( R2 )
c
M y ( R3 ) = δ ∫ x( g ( x) − f ( x))dx
a
b
= δ ∫ x( g ( x) − f ( x))dx
a
c
+δ ∫ x( g ( x) − f ( x))dx
b
= M y ( R1 ) + M y ( R2 )
δ
b
δ
b
((h( x))2 − ( g ( x)) 2 + ( g ( x))2 − ( f ( x)) 2 )dx
((h( x))
2 ∫a
2
(( g ( x))
2 ∫a
2
− ( g ( x)) 2 )dx
− ( f ( x)) 2 )dx
a
b
b
a
a
= M y ( R1 ) + M y ( R2 )
= m( R1 ) + m( R2 )
+
b
2 ∫a
− ( f ( x))2 )dx
= δ ∫ x(h( x) − g ( x))dx +δ ∫ x( g ( x) − f ( x))dx
= δ ∫ ( g ( x) − f ( x))dx + δ ∫ ( g ( x) − f ( x))dx
2
δ
2
= δ ∫ x(h( x) − g ( x) + g ( x) − f ( x))dx
m( R3 ) = δ ∫ ( g ( x) − f ( x))dx
b
b
((h( x))
2 ∫a
a
b
c
(( g ( x))
2 ∫a
δ
b
Now,
δ
a
M y ( R3 ) = δ ∫ x(h( x) − f ( x))dx
b
=
b
a
= M x ( R1 ) + M x ( R2 )
M y ( R2 ) = δ ∫ x( g ( x) − f ( x))dx
c
a
b
= δ ∫ (h( x) − g ( x))dx +δ ∫ ( g ( x) − f ( x))dx
a
c
2 ∫a
− ( f ( x)) 2 )dx
M y ( R2 ) = δ ∫ x( g ( x) − f ( x))dx
M y ( R1 ) = δ ∫ x( g ( x) − f ( x))dx
δ
2
a
b
b
M x ( R3 ) =
(( g ( x))
2 ∫a
− ( g ( x)) 2 )dx
b
(( g ( x)) − ( f ( x)) )dx
2 ∫b
2
M y ( R1 ) = δ ∫ x(h( x) − g ( x))dx
b
2 ∫a
b
M x ( R3 ) =
m( R2 ) = δ ∫ ( g ( x) − f ( x))dx
M x ( R1 ) =
δ
((h( x))
2 ∫a
= m( R1 ) + m( R2 )
a
c
b
b
M x ( R2 ) =
b
19. m( R1 ) = δ ∫ ( g ( x) − f ( x))dx
δ
δ
21. Let region 1 be the region bounded by x = –2,
x = 2, y = 0, and y = 1, so m1 = 4 ⋅1 = 4 .
1
. Therefore
2
= y1m1 = 2 .
By symmetry, x1 = 0 and y1 =
M1 y = x1m1 = 0 and M1x
Let region 2 be the region bounded by x = –2,
x = 1, y = –1, and y = 0, so m2 = 3 ⋅1 = 3 .
1
1
and y2 = − . Therefore
2
2
3
3
M 2 y = x2 m2 = − and M 2 x = y2 m2 = − .
2
2
M1 y + M 2 y − 32
3
x=
=
=−
m1 + m2
7
14
By symmetry, x2 = −
1
y=
Instructor’s Resource Manual
M 1x + M 2 x 2 1
= =
m1 + m2
7 14
Section 5.6
331
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
22. Let region 1 be the region bounded by x = –3,
x = 1, y = –1, and y = 4, so m1 = 20 . By
3
. Therefore,
2
M1 y = x1 m1 = −20 and M1x = y1 m1 = 30 . Let
region 2 be the region bounded by x = –3, x = –2,
y = –3, and y = –1, so m2 = 2 . By symmetry,
symmetry, x = −1 and y1 =
5
and y2 = −2 . Therefore,
2
M 2 y = x2 m2 = −5 and M 2 x = y2 m2 = −4 . Let
region 3 be the region bounded by x = 0, x = 1,
y = –2, and y = –1, so m3 = 1 . By symmetry,
x2 = −
1
3
and y3 = − . Therefore,
2
2
1
3
M 3 y = x3 m3 = and M 3 x = y3 m3 = − .
2
2
49
M1 y + M 2 y + M 3 y − 2
49
x=
=
=−
23
46
m1 + m2 + m3
x3 =
49
M + M 2 x + M 3x
49
y = 1x
= 2 =
23 46
m1 + m2 + m3
23. Let region 1 be the region bounded by x = –2,
x = 2, y = 2, and y = 4, so m1 = 4 ⋅ 2 = 8 . By
symmetry, x1 = 0 and y1 = 3 . Therefore,
M1 y = x1m1 = 0 and M1x = y1m1 = 24 . Let
region 2 be the region bounded by x = –1,
x = 2, y = 0, and y = 2, so m2 = 3 ⋅ 2 = 6 . By
1
symmetry, x2 = and y2 = 1 . Therefore,
2
M 2 y = x2 m2 = 3 and M 2 x = y2 m2 = 6 . Let
region 3 be the region bounded by x = 2, x = 4,
y = 0, and y = 1, so m3 = 2 ⋅1 = 2 . By symmetry,
1
. Therefore, M 3 y = x3 m3 = 6
2
and M 3 x = y3 m3 = 1 .
M1 y + M 2 y + M 3 y
9
=
x=
m1 + m2 + m3
16
x3 = 3 and y2 =
y=
M1x + M 2 x + M 3 x 31
=
m1 + m2 + m3
16
24. Let region 1 be the region bounded by x = –3,
x = –1, y = –2, and y = 1, so m1 = 6 . By
1
. Therefore,
2
M1 y = x1 m1 = −12 and M1x = y1 m1 = −3 . Let
region 2 be the region bounded by x = –1, x = 0,
y = –2, and y = 0, so m2 = 2 . By symmetry,
symmetry, x1 = −2 and y1 = −
x2 = −
332
1
and y2 = −1 . Therefore,
2
Section 5.6
M 2 y = x2 m2 = −1 and M 2 x = y2 m2 = −2 . Let
region 3 be the remaining region, so m3 = 22 .
1
. Therefore,
2
M 3 y = x3 m3 = 44 and M 3 x = y3 m3 = −11 .
M1 y + M 2 y + M 3 y 31
=
x=
m1 + m2 + m3
30
By symmetry, x3 = 2 and y3 = −
y=
M 1x + M 2 x + M 3 x
16
8
=− =−
m1 + m2 + m3
30
15
1
1
1
⎡1 ⎤
25. A = ∫ x3 dx = ⎢ x 4 ⎥ =
0
4
⎣
⎦0 4
4
From Problem 11, x = .
5
1⎛
4 ⎞ 2π
V = A(2πx ) = ⎜ 2π ⋅ ⎟ =
4⎝
5⎠ 5
Using cylindrical shells:
1
1
1
2π
⎡1 ⎤
V = 2π∫ x ⋅ x3 dx = 2π ∫ x 4 dx = 2π ⎢ x5 ⎥ =
0
0
5
⎣ 5 ⎦0
26. The area of the region is πa 2 . The centroid is the
center (0, 0) of the circle. It travels a distance of
2 π (2a) = 4 π a. V = 4π2 a3
27. The volume of a sphere of radius a is
4 3
πa . If
3
the semicircle y = a 2 − x 2 is revolved about
the x-axis the result is a sphere of radius a. The
centroid of the region travels a distance of 2πy .
The area of the region is
1 2
πa . Pappus's
2
Theorem says that
4
⎛1
⎞
(2πy ) ⎜ πa 2 ⎟ = π2 a 2 y = πa3 .
2
3
⎝
⎠
4a
y=
, x = 0 (by symmetry)
3π
28. Consider a slice at x rotated about the y-axis.
b
ΔV = 2π xh( x)Δx , so V = 2π∫ xh( x)dx .
a
b
Δm ≈ h( x)Δx , so m = ∫ h( x)dx = A .
a
b
ΔM y ≈ xh( x)Δx so M y = ∫ xh( x)dx .
a
,
b
x=
My
∫ xh( x)dx
= a
m
A
The distance traveled by the centroid is 2πx .
b
(2πx ) A = 2π∫ xh( x)dx
a
Therefore, V = 2πxA .
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
29. a.
ΔV ≈ 2π( K − y ) w( y )Δy
π
, so the centroid travels a distance
2n
π
of 2πr cos
.
2n
Thus, by Pappus's Theorem, the volume of
the resulting solid is
π ⎞⎛ 2
π
π ⎞
⎛
⎜ 2πr cos ⎟ ⎜ 2r n sin cos ⎟
2
n
2
n
2
n⎠
⎝
⎠⎝
π
π
= 4πr 3 n sin cos 2
.
2n
2n
r cos
d
V = 2π∫ ( K − y ) w( y )dy
c
b.
d
Δm ≈ w( y )Δy , so m = ∫ w( y )dy = A .
c
d
ΔM x ≈ yw( y )Δy , so M x = ∫ yw( y )dy .
c
d
∫
y= c
yw( y )dy
A
The distance traveled by the centroid is
2π( K − y ) .
2π( K − y ) A = 2π( KA − M x )
b.
d
d
= 2π ⎛⎜ ∫ Kw( y )dy − ∫ yw( y )dy ⎞⎟
c
c
⎝
⎠
Therefore, V = 2π( K − y ) A .
32. a.
π
since
2
π
f (sin x) = f ( sin(π − x ) ) . Thus x = .
2
h
π
∫ x f (sin x) dx = π
x= 0
π
2
∫0 f (sin x)dx
Therefore
π
∫0
A=
Instructor’s Resource Manual
The graph of f (sin x) on [0, π ] is
symmetric about the line x =
b 3⎤
1
⎡1
= ⎢ by 2 −
y ⎥ = bh 2
3h ⎦ 0 6
⎣2
M
h
y= x =
m
3
The area of a regular polygon P of 2n sides
π
π
is 2r 2 n sin cos . (To find this consider
2n
2n
the isosceles triangles with one vertex at the
center of the polygon and the other vertices
on adjacent corners of the polygon. Each
π
such triangle has base of length 2r sin
2n
π ⎞
and height r cos . ⎟ Since P is a regular
2n ⎠
polygon the centroid is at its center. The
distance from the centroid to any side is
π
= 2π2 r 3
2n
solid is (πr 2 )(2πr ) = 2π2 r 3 which agrees
with the results from the polygon.
The length of a segment at y is b −
31. a.
2π2 r 3 cos 2
circle of area πr 2 whose centroid (= center)
travels a distance of 2πr, the volume of the
1
m = bh
2
1
bh ; the distance traveled by the
2
h⎞
⎛
centroid is 2π ⎜ k − ⎟ .
3⎠
⎝
h ⎞ ⎛ 1 ⎞ πbh
⎛
V = 2π ⎜ k − ⎟ ⎜ bh ⎟ =
( 3k − h )
3 ⎠⎝ 2 ⎠
3
⎝
π
2n
π
π
cos 2
2n
2n
As n → ∞ , the regular polygon approaches
a circle. Using Pappus's Theorem on the
c
b.
sin 2πn
n →∞
= 2π∫ ( K − y ) w( y )dy
b
y.
h
b ⎞
b ⎞
⎛
⎛
ΔM x ≈ y ⎜ b − y ⎟ Δy = ⎜ by − y 2 ⎟ Δy
h ⎠
h ⎠
⎝
⎝
h⎛
b 2⎞
M x = ∫ ⎜ by − y ⎟ dy
0⎝
h ⎠
n →∞
lim
d
30. a.
lim 4πr 3 n sin
b.
x f (sin x)dx =
π π
f (sin x)dx
2 ∫0
sin x cos 4 x = sin x(1 − sin 2 x) 2 , so
f ( x) = x(1 − x 2 )2 .
π
π π
sin x cos 4 x dx
2 ∫0
∫0
x sin x cos 4 x dx =
=
π⎡ 1
π
⎤
− cos5 x ⎥ =
2 ⎢⎣ 5
⎦0 5
π
Section 5.6
333
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
33. Consider the region S – R.
1 1 ⎡ g 2 ( x) −
2 0⎣
f ( x) ⎤ dx
⎦ ≥y
R
S−R
∫
yS − R =
1 1
2 0
∫
=
2
f 2 ( x)dx
∫
8
2
g ( x) dx
≥
1 1
2 0
∫
x ≈ i =1
=
8
1695
≈ 23.38
72.5
8
∑ hi
(1/ 2)(6.52 + 82 + " + 102 + 82 )
(6.5 + 8 + " + 10 + 8)
i =1
=
6.5 + 8 + " + 10 + 8
1 8
∑ ((hi − 4)2 − (−4)2 )
2 i =1
8
∑ hi
i =1
2
(1/ 2)((2.5 − (−4) 2 ) + " + (42 − (−4) 2 ))
6.5 + 8 + " + 10 + 8
45.875
=
≈ 0.633
72.5
A quick computation will show that these values
agree with those in Problem 34 (using a different
reference point).
Now consider the whole lamina as R3 , the
circular hole as R2 , and the remaining lamina as
R1 . We can find the centroid of R1 by noting
that
M x ( R1 ) = M x ( R3 ) − M x ( R2 )
and similarly for M y ( R1 ) .
From symmetry, we know that the centroid of a
circle is at the center. Therefore, both
M x ( R2 ) and M y ( R2 ) must be zero in our case.
i =1
=
y≈
=
6.5 + 8 + " + 10 + 8
∑ hi
( −25)(6.5) + ( −15)(8) + " + (5)(10) + (10)(8)
−480
=
≈ −6.62
72.5
R
(5)(6.5) + (10)(8) + " + (35)(10) + (40)(8)
1 8
∑ (hi )2
2 i =1
=
f ( x)dx
8
∑ xi hi
i =1
2
34. To approximate the centroid, we can lay the
figure on the x-axis (flat side down) and put the
shortest side against the y-axis. Next we can use
the eight regions between measurements to
approximate the centroid. We will let hi ,the
height of the ith region, be approximated by the
height at the right end of the interval. Each
interval is of width Δx = 5 cm. The centroid can
be approximated as
y≈
∑ hi
R
S
yS ≥ y R
=
8
∑ xi hi
x ≈ i =1
1
1 1⎡ 2
1
R ∫ g ( x) − f 2 ( x) ⎤ dx ≥ ( S − R) ∫ f 2 ( x)dx
⎣
⎦
0
0
2
2
1
1 1⎡ 2
1
2
R ∫ g ( x) − f ( x) ⎤ dx + R ∫ f 2 ( x)dx
⎣
⎦
0
0
2
2
1 2
1
1 1 2
≥ ( S − R ) ∫ f ( x)dx + R ∫ f ( x)dx
0
2
2 0
1
1
1
1
R g 2 ( x)dx ≥ S ∫ f 2 ( x)dx
2 ∫0
2 0
1 1
2 0
35. First we place the lamina so that the origin is
centered inside the hole. We then recompute the
centroid of Problem 34 (in this position) as
335.875
≈ 4.63
72.5
This leads to the following equations
M y ( R3 ) − M y ( R2 )
x=
m( R3 ) − m( R2 )
=
=
δΔx(−480)
δΔx(72.5) − δπ (2.5)2
−2400
≈ −7
342.87
y=
=
M x ( R3 ) − M x ( R2 )
m( R3 ) − m( R2 )
δΔx(45.875)
δΔx(72.5) − δπ (2.5) 2
229.375
≈ 0.669
342.87
Thus, the centroid is 7 cm above the center of the
hole and 0.669 cm to the right of the center of the
hole.
=
334
Section 5.6
Instructor’s Resource Manual
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
36. This problem is much like Problem 34 except we
don’t have one side that is completely flat. In
this problem, it will be necessary, in some
regions, to find the value of g(x) instead of just
f(x) – g(x). We will use the 19 regions in the
figure to approximate the centroid. Again we
choose the height of a region to be approximately
the value at the right end of that region. Each
region has a width of 20 miles. We will place the
north-east corner of the state at the origin.
The centroid is approximately
19
∑ xi ( f ( xi ) − g ( xi ))
x≈
2.
a.
5
b.
i =1
+ 3(0.05) + 4(0.5)
= 0.6
3.
a.
P ( X ≥ 2) = P (2) = 0.2
b.
E ( X ) = −2(0.2) + (−1)(0.2) + 0(0.2)
+ 1(0.2) + 2(0.2)
=0
a.
P ( X ≥ 2) = P (2) = 0.1
b.
E ( X ) = −2(0.1) + (−1)(0.2) + 0(0.4)
+1(0.2) + 2(0.1)
∑ ( f ( xi ) − g ( xi ))
i =1
y≈
4.
1 19
∑ [( f ( xi ))2 − ( g ( xi ))2 ]
2 i =1
19
∑ ( f ( xi ) − g ( xi ))
=0
5.
a.
i =1
1⎡
(1452 − 132 ) + (1492 − 102 ) + " + (852 − 852 ) ⎤
⎣
⎦
2
=
(145 − 13) + (149 − 19) + " + (85 − 85)
230,805
=
≈ 83.02
2780
This would put the geographic center of Illinois
just south-east of Lincoln, IL.
b.
2.
sum, integral
3.
∫0 f ( x) dx
4.
cumulative distribution function
E ( X ) = 1(0.4) + 2(0.2) + 3(0.2) + 4(0.2)
= 2.2
6.
a.
P ( X ≥ 2) = P (100) + P(1000)
= 0.018 + 0.002 = 0.02
E ( X ) = −0.1(0.98) + 100(0.018)
+ 1000(0.002)
= 3.702
5.7 Concepts Review
discrete, continuous
P ( X ≥ 2) = P (2) + P(3) + P (4)
= 0.2 + 0.2 + 0.2
= 0.6
b.
1.
E ( X ) = ∑ xi pi
= 0(0.7) + 1(0.15) + 2(0.05)
i =1
19
(20)(145 − 13) + (40)(149 − 10) + " (380)(85 − 85)
=
(145 − 13) + (149 − 19) + " (85 − 85)
482,860
=
≈ 173.69
2780
P ( X ≥ 2) = P (2) + P(3) + P(4)
= 0.05 + 0.05 + 0.05 = 0.15
7.
a.
P ( X ≥ 2) = P (2) + P (3) + P(4)
=
3 2 1
6
+ + =
= 0.6
10 10 10 10
5
b.
E ( X ) = 1(0.4) + 2(0.3) + 3(0.2) + 4(0.1) = 2
Problem Set 5.7
1.
a.
P ( X ≥ 2) = P (2) + P(3) = 0.05 + 0.05 = 0.1
b.
E ( X ) = ∑ xi pi
4
i =1
= 0(0.8) + 1(0.1) + 2(0.05) + 3(0.05)
= 0.35
Instructor’s Resource Manual
Section 5.7
335
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
8.
a.
P ( X ≥ 2) = P (2) + P(3) + P (4)
=
b.
0
(−1)
(−2)
5
+
+
=
= 0.5
10
10
10
10
2
2
2
b. E ( X ) = 0(0.4) + 1(0.1) + 2(0) + 3(0.1) + 4(0.4)
=2
9.
20
P( X ≥ 2) = ∫
b.
⎡ x2 ⎤
1
E ( X ) = ∫ x ⋅ dx = ⎢ ⎥
0
20
⎢⎣ 40 ⎥⎦
c.
2
c.
20
20
20
0
x⋅
3
x(20 − x) dx
4000
(
)
=
3 20
20 x 2 − x3 dx
4000 ∫0
=
3 ⎡ 20 x3 x 4 ⎤
− ⎥ = 10
⎢
4000 ⎣⎢ 3
4 ⎥⎦
0
20
1
1
dx =
⋅18 = 0.9
20
20
a.
E( X ) = ∫
= 10
For 0 ≤ x ≤ 20
x 3
F ( x) = ∫
t (20 − t ) dt
0 4000
x
3 ⎡ 2 t3 ⎤
3 2
1 3
=
x −
x
⎢10t − ⎥ =
4000 ⎢⎣
3 ⎥⎦
400
4000
0
0
For x between 0 and 20,
x 1
1
x
F ( x) = ∫
dt =
⋅x =
0 20
20
20
13.
a.
4
P ( X ≥ 2) = ∫
2
3 2
x (4 − x) dx
64
4
10.
a.
1
1
P ( X ≥ 2) = ∫
dx =
⋅18 = 0.45
2 40
40
b.
E( X ) = ∫
c.
3 ⎡ 4 x3 x 4 ⎤
=
− ⎥ = 0.6875
⎢
64 ⎢⎣ 3
4 ⎥⎦
2
20
20
−20
x⋅
⎡ x2 ⎤
1
dx = ⎢ ⎥
40
⎣⎢ 80 ⎦⎥
20
b.
= 5−5 = 0
−20
4
E( X ) = ∫ x ⋅
0
3 2
x (4 − x) dx
64
4
3 4
3 ⎡ 4 x5 ⎤
3
4
=
4
x
−
x
dx
=
⎢ x − ⎥ = 2.4
64 ∫0
64 ⎣⎢
5 ⎥⎦
0
(
For −20 ≤ x ≤ 20 ,
x 1
1
1
1
( x + 20) =
F ( x) = ∫
dt =
x+
−20 40
40
40
2
c.
)
For 0 ≤ x ≤ 4
x
11.
a.
P ( X ≥ 2) = ∫
8
2
3 2
3 ⎡ 4t 3 t 4 ⎤
t (4 − t ) dt =
− ⎥
⎢
0 64
64 ⎢⎣ 3
4 ⎥⎦
0
1 3
3 4
= x −
x
16
256
3
x(8 − x) dx
256
F ( x) = ∫
8
=
b.
3 ⎡ 2 x3 ⎤
3
27
⋅ 72 =
⎢4 x − ⎥ =
256 ⎣⎢
3 ⎦⎥
256
32
2
8
E( X ) = ∫ x ⋅
0
3
x(8 − x) dx
256
(
14.
a.
)
=
3 8 2
8 x − x3 dx
256 ∫0
=
3 ⎡ 8 x3 x 4 ⎤
− ⎥ =4
⎢
256 ⎣⎢ 3
4 ⎥⎦
0
b.
a.
336
x
20
2
3
x(20 − x) dx
4000
3 ⎤ 20
3 ⎡ 2 x
⎢10 x − ⎥ = 0.972
4000 ⎢⎣
3 ⎥⎦
2
Section 5.7
1
(8 − x) dx
32
8
E( X ) = ∫ x ⋅
0
1
(8 − x) dx
32
8
P ( X ≥ 2) = ∫
=
2
1 ⎡ 2 x3 ⎤
8
=
⎢4 x − ⎥ =
32 ⎢⎣
3 ⎥⎦
3
3 ⎤x
3
3 ⎡ 2 t
t (8 − t ) dt =
⎢ 4t − ⎥
0 256
256 ⎢⎣
3 ⎦⎥
0
3 2
1 3
=
x −
x
64
256
12.
8
8
For 0 ≤ x ≤ 8
F ( x) = ∫
P ( X ≥ 2) = ∫
1 ⎡
x2 ⎤
9
=
⎢8 x − ⎥ =
32 ⎢⎣
2 ⎥⎦
16
2
8
c.
x
0
c.
For 0 ≤ x ≤ 8
x
F ( x) = ∫
x
0
=
1
1 ⎡
t2 ⎤
(8 − t ) dt =
⎢8t − ⎥
32
32 ⎣⎢
2 ⎦⎥
0
1
1
x − x2
4
64
Instructor’s Resource Manual
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
15.
a.
P ( X ≥ 2) = ∫
4π
⎛πx ⎞
sin ⎜
⎟ dx
8
⎝ 4 ⎠
2
E( X ) = ∫ x ⋅
c.
For 1 ≤ x ≤ 9
1
4
=
π⎡ 4
π x⎤
1
1
− cos ⎥ = − (−1 − 0) =
⎢
8⎣ π
4 ⎦2
2
2
c.
π
⎛πx ⎞
sin ⎜
⎟ dx
8
⎝ 4 ⎠
Using integration by parts or a CAS,
E(X ) = 2 .
4
E( X ) = ∫ x ⋅
0
=−
19.
For 0 ≤ x ≤ 4
16.
a.
xπ
c.
Proof of F ′( x) = f ( x) :
Fundamental Theorem of Calculus,
F ′( x) = f ( x).
Proof of F ( A) = 0 and F ( B ) = 1:
A
4π
F ( A) = ∫ f ( x) dx = 0;
4
F ( B ) = ∫ f ( x) dx = 1
A
B
A
Proof of P (a ≤ X ≤ b) = F (b) − F (a ) :
b
P (a ≤ X ≤ b) = ∫ f ( x) dx = F (b) − F (a) due to
a
π
⎛π x ⎞
cos ⎜
⎟ dx
8
⎝ 8 ⎠
Using a CAS, E ( X ) ≈ 1.4535
4
E( X ) = ∫ x ⋅
the Second Fundamental Theorem of Calculus.
0
20.
a.
For 0 ≤ x ≤ 4
⎡ ⎛ π t ⎞⎤
⎛ πt ⎞
cos ⎜ ⎟ dt = ⎢sin ⎜ ⎟ ⎥
8
⎝ 8 ⎠
⎣ ⎝ 8 ⎠⎦0
b.
2
E( X ) = ∫
4
1
=
c.
4
a +b
1
1 ⎛ a+b
⎞
dx =
− a⎟
⎜
a
b−a
b−a⎝ 2
⎠
1 b−a 1
=
⋅
=
b−a 2
2
4
1
⎡ 4⎤
dx = ⎢ − ⎥ =
2
⎣ 3x ⎦ 2 3
3x
4
c.
a.
4
2
P ( X ≥ 2) = ∫
9
2
=
1
1 ⎡ x2 ⎤
dx =
⎢ ⎥
b−a
b − a ⎣⎢ 2 ⎦⎥
b
a
b −a
a+b
=
2(b − a )
2
2
F ( x) = ∫
x
a
2
1
1
x−a
dt =
( x − a) =
b−a
b−a
b−a
1
−4 4
⎡ 4⎤
+
dt = ⎢ − ⎥ =
t
3
⎣
⎦1 3x 3
3t
4x − 4
=
3x
x
a
=
For 1 ≤ x ≤ 4
1
b
E( X ) = ∫ x ⋅
4
4
ln 4 ≈ 1.85
3
F ( x) = ∫
18.
b.
⎡4
⎤
x⋅
dx = ⎢ ln x ⎥
2
⎣3
⎦1
3x
4
a+b
.
2
=∫ 2
⎛πx ⎞
= sin ⎜
⎟
⎝ 8 ⎠
P ( X ≥ 2) = ∫
The midpoint of the interval [a,b] is
a+b⎞
a+b⎞
⎛
⎛
P⎜ X <
⎟ = P⎜ X ≤
⎟
2 ⎠
2 ⎠
⎝
⎝
x
xπ
0
a.
81 81x 2 − 81
=
80
80 x 2
x
⎛πx⎞
P ( X ≥ 2) = ∫
cos ⎜
⎟ dx
2 8
⎝ 8 ⎠
F ( x) = ∫
17.
80 x 2
+
A
⎡ ⎛ π x ⎞⎤
π
π
1
= ⎢sin ⎜
⎟ ⎥ = sin − sin = 1 −
2
4
2
⎣ ⎝ 8 ⎠⎦ 2
b.
81
By definition, F ( x) = ∫ f (t ) dt. By the First
x
π ⎡ −4
πt ⎤
⎛πt ⎞
sin ⎜ ⎟ dt = ⎢ cos ⎥
0 8
π
4
8
4 ⎦0
⎝ ⎠
⎣
1⎛
πx ⎞
1
πx 1
= − ⎜ cos
− 1⎟ = − cos
+
2⎝
4
2
4 2
⎠
F ( x) = ∫
x
⎡ 81 ⎤
F ( x) = ∫
dt = ⎢ −
⎥
1 40t 3
⎣ 80t 2 ⎦1
81
x
b.
9
⎡ 81 ⎤
dx = ⎢ −
⎥ = 1.8
3
⎣ 40 x ⎦1
40 x
81
9
b.
9
⎡ 81 ⎤
dx = ⎢ −
⎥
3
40 x
⎣ 80 x 2 ⎦ 2
81
77
≈ 0.24
320
Instructor’s Resource Manual
21.
The median will be the solution to the
x
1
dx = 0.5 .
equation ∫ 0
a b−a
1
( x0 − a ) = 0.5
b−a
b−a
x0 − a =
2
a+b
x0 =
2
Section 5.7
337
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
22.
23.
15 2
x (4 − x) 2 is
512
symmetric about the line x = 2. Consequently,
P ( X ≤ 2) = 0.5 and 2 must be the median of X.
The graph of f ( x) =
c.
Since the PDF must integrate to one, solve
5
∫0 kx(5 − x) dx = 1.
4
=
5
⎡ 5kx 2 kx3 ⎤
−
⎢
⎥ =1
3 ⎥⎦
⎢⎣ 2
0
125k 125k
−
=1
2
3
375k − 250k = 6
k=
24.
Solve
k∫
5
0
5
∫0
d.
x2
=
8
kx 2 (5 − x) 2 dx = 1
e.
∫0 k ( 2 − x − 2 ) dx = 1
4
Due to the symmetry about the line x = 2, the
solution can be found by solving
2
2∫ kx dx = 1
2
0
=1
P (3 ≤ X ≤ 4) = ∫
3
4
4
( 2 − x − 2 ) dx
(2 − ( x − 2)) dx =
4
=
338
41
3
41
1⎡
1
x2 ⎤
⎢4 x − ⎥ =
4 ⎣⎢
2 ⎦⎥
8
3
Section 5.7
x dx + ∫
x1
2
4
(4 − t ) dt
x
t2 ⎤
x2 3 ⎞
1⎡
1 ⎛
+ ⎢ 4t − ⎥ = + ⎜ x −
− ⎟
4 ⎣⎢
2 ⎥⎦
2 ⎝⎜
8 2 ⎠⎟
2
if x < 0
if 0 ≤ x ≤ 2
if 2 < x ≤ 4
if x > 4
Using a similar procedure as shown in part
(a), the PDF for Y is
1
f ( y) =
(120 − y − 120 )
14, 400
y
1
t dt
If 0 ≤ y < 120, F ( y ) = ∫
0 14, 400
If 120 < y ≤ 240,
y
1
1
F ( y) = + ∫
(240 − t ) dt
2 120 14, 400
1
k=
4
=∫
0
2
4
y
4k = 1
b.
21
⎡ t2 ⎤
y2
=⎢
⎥ =
⎣⎢ 28,800 ⎦⎥ 0 28,800
0
k ⋅ x2
2
x
+ x −1
8
0
⎧
⎪
x2
⎪
⎪⎪
8
F ( x) = ⎨
2
⎪ x
⎪− 8 + x − 1
⎪
⎩⎪1
=−
( 25x2 − 10x3 + x4 ) dx = 1
Solve
0
0
6
125
x
⎡t2 ⎤
x2
t dt = ⎢ ⎥ =
4
8
⎣⎢ 8 ⎦⎥ 0
x1
If 0 ≤ x ≤ 2, F ( x) = ∫
If 2 < x ≤ 4, F ( x) = ∫
5
a.
1 3 2 1 ⎡ 2 x3 ⎤
2 4
x + ⎢2 x − ⎥ = + = 2
12 0 4 ⎣⎢
3 ⎦⎥
3 3
2
⎡ 25 x3 5 x 4 x5 ⎤
k⎢
−
+ ⎥ =1
2
5 ⎥⎦
⎢⎣ 3
0
625
k =1
6
6
k=
625
25.
4
1
E ( X ) = ∫ x ⋅ (2 − x − 2 ) dx
0
4
2
4
1
1
= ∫ x ⋅ (2 + ( x − 2)) dx + ∫ x ⋅ (2 − ( x − 2)) dx
0
2
4
4
1 2 2
1 4
= ∫ x dx + ∫ (4 x − x 2 ) dx
4 0
4 2
1 4
(4 − x) dx
4 ∫3
1
1 ⎡
t2 ⎤
= +
⎢ 240t − ⎥
2 14, 400 ⎣⎢
2 ⎦⎥
y
120
=
2
1 y
y
3
y2
y
+ −
− =−
+ −1
2 60 28,800 2
28,800 60
0
⎧
if y < 0
⎪
y2
⎪
if 0 ≤ y ≤ 120
⎪⎪
28,800
F ( x) = ⎨
y2
y
⎪
−
⎪ 28,800 + 60 − 1 if 120 < y ≤ 240
⎪
if y > 240
⎪⎩1
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
26.
a.
Solve
180
∫0
Alternatively, we can proceed as follows:
kx 2 (180 − x) dx = 1.
Solve
180
b.
⎡
x4 ⎤
=1
k ⎢60 x3 − ⎥
4 ⎥⎦
⎢⎣
0
1
k=
87, 480, 000
P (100 ≤ X ≤ 150)
=∫
150
100
k ≈ 1.132096857 × 1029
8
y
⎛ 3
⎞
− t ⎟ dt
FY ( y ) = ∫ k ⋅ t 6 ⎜
0
⎝ 127 ⎠
Using a CAS,
1
x 2 (180 − x) dx
87, 480, 000
FY ( y ) ≈ (7.54731× 1027 ) y 7 ( y8 − 0.202475 y 7
150
E( X ) = ∫
180
0
x⋅
∫0
8
⎛ 3
⎞
k ⋅ y6 ⎜
− y ⎟ dy = 1 using a
127
⎝
⎠
CAS.
+0.01802 y 6 − 0.000923 y 5
4⎤
⎡
1
3 x
=
≈ 0.468
⎢ 60 x − ⎥
87, 480, 000 ⎢⎣
4 ⎥⎦
100
c.
3 127
+0.00003 y 4 − (6.17827 × 10−7 ) y 3
+(8.108 × 10−9 ) y 2 − (6.156 × 10−11 ) y
1
x 2 (180 − x ) dx
87, 480, 000
+2.07746 × 10−13 )
180
=
27.
a.
5⎤
⎡
1
4 x
⎢ 45 x − ⎥
87, 480, 000 ⎣⎢
5 ⎦⎥
Solve
0.6
∫0
= 108
0
28.
a.
0.6 6
x (0.6 − x)8 dx
0
k∫
8
=1
= 1− k ∫
P ( X ≥ 100) = k ∫
c.
0.6
0
E( X ) = k ∫
200
0
x ⋅ x 2 (200 − x)8 dx
= 50 using a CAS
x (0.6 − x)8 dx
d.
E( X ) = ∫
x (200 − x)8 dx
≈ 0.0327 using a CAS.
≈ 0.884 using a CAS
c.
200 2
100
0.45 6
0.35
kx 2 (200 − x)8 dx = 1.
b. The probability that a batch is not accepted is
Using a CAS, k ≈ 95,802,719
b. The probability that a unit is scrapped is
1 − P (0.35 ≤ X ≤ 0.45)
200
∫0
Using a CAS, k ≈ 2.417 × 10−23
kx (0.6 − x) dx = 1.
6
Solve
x ⋅ kx6 (0.6 − x)8 dx
F ( x ) = ∫ (2.417 × 10−23 )t 2 (200 − t )8 dx
x
0
Using a CAS, F(x) ≈ (2.19727 × 10−24 ) x3 ⋅
( x8 − 1760 x 7 + 136889 x 6 − (6.16 × 108 ) x5
0.6 7
x (0.6 − x)8 dx
0
= k∫
+(1.76 × 1011 ) x 4 − (3.2853 × 1013 ) x3
≈ 0.2625
+(3.942 × 1015 ) x 2 − (2.816 × 1017 ) x
d.
x
F ( x ) = ∫ 95,802, 719t 6 (0.6 − t )8 dt
+9.39 × 1018 )
0
Using a CAS,
F ( x ) ≈ 6,386,850 x 7 ( x8 − 5.14286 x 7
Solve
100
∫0
kx 2 (100 − x)8 dx. Using a CAS,
+ 11.6308 x 6 − 15.12 x5 + 12.3709 x 4
k = 4.95 × 10−20
− 6.53184 x3 + 2.17728 x 2
F ( x) = ∫
− 0.419904 x + 0.36)
e.
e.
If X = measurement in mm, and Y =
measurement in inches, then Y = X / 25.4 .
Thus,
FY ( y ) = P (Y ≤ y ) = P ( X / 25.4 ≤ y )
= P ( X ≤ 25.4 y ) = F ( 25.4 y )
where F ( x ) is given in part (d).
Instructor’s Resource Manual
y
0
( 4.95 ×10−20 ) t 2 (100 − t )8 dt
Using a CAS,
F ( x ) ≈ (4.5 × 10−21 ) x3
( x8 − 880 x 7 + 342, 222 x 6 − (7.7 × 107 ) x5
+ (1.1× 1010 ) x 4 − (1.027 × 1012 ) x3
+ (6.16 × 1013 ) x 2 − (2.2 × 1015 ) x
+ 3.667 × 1016 )
Section 5.7
339
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
29.
The PDF for the random variable X is
⎧1 if 0 ≤ x ≤ 1
f ( x) = ⎨
⎩0 otherwise
From Problem 20, the CDF for X is F ( x ) = x
33.
Y is the distance from (1, X ) to the origin, so
Y=
(1 − 0 )2 + ( X − 0 )2 = 1 + X 2
if x < 0
⎧ 0
⎪ 0.8 if 0 ≤ x < 1
⎪⎪
F ( x) = ⎨ 0.9 if 1 ≤ x < 2
⎪0.95 if 2 ≤ x < 3
⎪
if x > 3
⎪⎩ 1
F(x)
1.0
Here we have a one-to-one transformation from
{
}
0.8
the set { x : 0 ≤ x ≤ 1} to y :1 ≤ y ≤ 2 . For
0.6
every 1 < a < b < 2 , the event a < Y < b will
occur when, and only when,
0.4
a2 − 1 < X < b2 − 1 .
If we let a = 1 and b = y , we can obtain the
CDF for Y.
( 1 − 1 ≤ X ≤ y − 1)
= P (0 ≤ X ≤ y − 1)
= F ( y − 1) = y − 1
P (1 ≤ Y ≤ y ) = P
2
0.2
0
2
2
2
34.
2
To find the PDF, we differentiate the CDF with
respect to y.
d
1
1
y
PDF =
y2 −1 = ⋅
⋅2y =
dy
2 y2 −1
y2 −1
if 2 ≤ x < 3
if 3 ≤ x < 4
if x ≥ 4
0.6
0.4
0.2
0
35.
1
4
P (Y < 2) = P(Y ≤ 2) = F (2) = 1
b.
P (0.5 < Y < 0.6) = F (0.6) − F (0.5)
=
5
x
1.2 1
1
−
=
1.6 1.5 12
By the defintion of a complement of a set,
A ∪ Ac = S , where S denotes the sample space.
c.
f ( y ) = F ′( y ) =
d.
E (Y ) = ∫ y ⋅
Since P ( S ) = 1, P ( A ∪ A ) = 1.
c
Since P ( A ∪ Ac ) = P( A) + P( Ac ),
P ( A) + P( A ) = 1 and P( A ) = 1 − P ( A).
c
32.
3
2
a.
equivalent.
31.
if 1 ≤ x < 2
0.8
x
P (a < X ≤ b) and P (a ≤ X < b), are
if 0 ≤ x < 1
1.0
x
expressions, P( a < X < b), P( a ≤ X ≤ b),
if x < 0
⎧ 0
⎪ 0.7
⎪
⎪⎪0.85
F ( x) = ⎨
⎪ 0.9
⎪0.95
⎪
⎪⎩1
P ( X = x) = ∫ f (t ) dt = 0. Consequently,
P ( X < c) = P ( X ≤ c). As a result, all four
x
3
2
F (x )
Therefore, for 0 ≤ y ≤ 2 the PDF and CDF are
respectively
y
and G ( y ) = y 2 − 1 .
g ( y) =
2
y −1
30.
1
c
2
( y + 1)2
1
2
0
( y + 1) 2
, 0 ≤ y ≤1
dy ≈ 0.38629
P ( X ≥ 1) = 1 − P ( X < 1)
For Problem 1, 1 − P ( X < 1) = 1 − P ( X = 0)
= 1 − 0.8 = 0.2
For Problem 2, 1 − P ( X < 1) = 1 − P ( X = 0)
= 1 − 0.7 = 0.3
For Problem 5, 1 − P ( X < 1) = 1 − 0 = 1
340
Section 5.7
Instructor’s Resource Manual
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
36.
a.
P ( Z > 1) = 1 − P ( Z ≤ 1) = 1 − F (1)
= 1−
b.
1 8
=
9 9
Concepts Test
P (1 < Z < 2) = P (1 ≤ Z ≤ 2) = F (2) − F (1)
=
4 1 1
− =
9 9 3
f ( z ) = F ′( z ) =
d.
E (Z ) = ∫ z ⋅
2z
,0≤ z≤3
9
3
0
⎡ 2 z3 ⎤
2z
dz = ⎢
⎥ =2
9
⎣⎢ 27 ⎦⎥
0
37.
38.
39.
1. False:
π
∫0 cos x dx = 0
because half of the area
lies above the x-axis and half below the xaxis.
c.
3
5.8 Chapter Review
15 2
x (4 − x)2 dx = 2
512
4
15 2
x (4 − x)2 dx
and E(X 2 ) = ∫ x 2 ⋅
0
512
32
≈ 4.57 using a CAS
=
7
2. True: The integral represents the area of the
region in the first quadrant if the center of
the circle is at the origin.
3. False: The statement would be true if either
f(x) ≥ g(x) or g(x) ≥ f(x) for
a ≤ x ≤ b. Consider Problem 1 with f(x)
= cos x and g(x) = 0.
4
E( X ) = ∫ x ⋅
0
3
x(8 − x) dx = 19.2 and
256
8
3
E ( X 3 ) = ∫ x3 ⋅
x (8 − x) dx = 102.4
0
256
using a CAS
8
E ( X 2 ) = ∫ x2 ⋅
0
V ( X ) = E ⎡( X − μ ) 2 ⎤ , where μ = E ( X ) = 2
⎣
⎦
4
15 2
4
V ( X ) = ∫ ( x − 2) 2 ⋅
x (4 − x) 2 dx =
0
512
7
4. True: The area of a cross section of a cylinder
will be the same in any plane parallel to
the base.
5. True: Since the cross sections in all planes
parallel to the bases have the same area,
the integrals used to compute the volumes
will be equal.
6. False: The volume of a right circular cone of
1
radius r and height h is πr 2 h . If the
3
radius is doubled and the height halved
2
the volume is πr 2 h.
3
7. False: Using the method of shells,
1
V = 2π∫ x(− x 2 + x)dx . To use the
0
40.
41.
3
x ( 8 − x ) dx = 4
256
8
3
16
V ( X ) = ∫ ( x − 4) 2 ⋅
x(8 − x) dx =
0
256
5
μ = E(X ) = ∫ x⋅
8
0
2
E ⎡⎢( X − μ ) ⎤⎥ = E ( X 2 − 2 X μ + μ 2 )
⎣
⎦
= E ( X 2 ) − E ( 2 X μ ) + E (μ 2 )
= E ( X 2 ) − 2μ ⋅ E ( X ) + μ 2
= E ( X ) − 2μ + μ since E ( X ) = μ
2
2
2
= E( X 2 ) − μ 2
For Problem 37, V ( X ) = E ( X 2 ) − μ 2 and
32 2 4
using previous results, V ( X ) =
−2 =
7
7
method of washers we need to solve
y = − x 2 + x for x in terms of y.
8. True: The bounded region is symmetric about
1
the line x = . Thus the solids obtained
2
by revolving about the lines
x = 0 and x = 1 have the same volume.
9. False: Consider the curve given by x =
y=
cos t
,
t
sin t
,2≤t <∞.
t
10. False: The work required to stretch a spring 2
inches beyond its natural length is
2
∫0 kx dx = 2k ,
while the work required to
stretch it 1 inch beyond its natural length
1
1
is ∫ kx dx = k .
0
2
Instructor’s Resource Manual
Section 5.8
341
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
11. False: If the cone-shaped tank is placed with the
point downward, then the amount of
water that needs to be pumped from near
the bottom of the tank is much less than
the amount that needs to be pumped from
near the bottom of the cylindrical tank.
12. False: The force depends on the depth, but the
force is the same at all points on a surface
as long as they are at the same depth.
13. True:
This is the definition of the center of
mass.
14. True: The region is symmetric about the point
( π , 0).
15. True: By symmetry, the centroid is on the line
π
x = , so the centroid travels a distance
2
⎛π⎞
of 2π ⎜ ⎟ = π2 .
⎝2⎠
1
2. V = π∫ ( x − x 2 )2 dx
0
1
= π∫ ( x 2 − 2 x3 + x 4 )dx
0
1
1
1 ⎤
π
⎡1
= π ⎢ x3 − x 4 + x 5 ⎥ =
2
5 ⎦ 0 30
⎣3
1
1
0
0
3. V = 2π∫ x( x − x 2 )dx = 2π ∫ ( x 2 − x3 )dx
1
1 ⎤
π
⎡1
= 2π ⎢ x3 − x 4 ⎥ =
4 ⎦0 6
⎣3
1
4. V = π∫ ⎡( x − x 2 + 2)2 − (2)2 ⎤ dx
⎦
0⎣
1
= π∫ ( x 4 − 2 x3 − 3x 2 + 4 x) dx
0
1
1
7π
⎡1
⎤
= π ⎢ x5 − x 4 − x3 + 2 x 2 ⎥ =
2
⎣5
⎦ 0 10
1
16. True: At slice y, ΔA ≈ (9 − y 2 )Δy .
17. True: Since the density is proportional to the
square of the distance from the midpoint,
equal masses are on either side of the
midpoint.
5. V = 2π∫ (3 − x)( x − x 2 )dx
0
1
= 2π∫ ( x3 − 4 x 2 + 3 x) dx
0
1
4
3 ⎤
5π
⎡1
= 2π ⎢ x 4 − x3 + x 2 ⎥ =
4
3
2
⎣
⎦0 6
1
18. True: See Problem 30 in Section 5.6.
19. True: A discrete random variable takes on a
finite number of possible values, or an
infinite set of possible outcomes provided
that these outcomes can be put in a list
such as {x1, x2, …}.
20. True: The computation of E(X) would be the
same as the computation for the center of
mass of the wire.
21. True:
E ( X ) = 5 ⋅1 = 5
22. True:
If F ( x) = ∫ f (t ) dt , then F ′( x) = f ( x)
x
A
1
P ( X = 1) = P (1 ≤ X ≤ 1) = ∫ f ( x) dx = 0
1
Sample Test Problems
1
1
1 ⎤
1
⎡1
1. A = ∫ ( x − x 2 )dx = ⎢ x 2 − x3 ⎥ =
0
3 ⎦0 6
⎣2
342
Section 5.8
y=
=
∫0
1
2
∫0 ( x − x )dx
1
=
1 1 ( x − x 2 ) 2 dx
2 0
1
2
∫
∫0 ( x − x
⎡ 1 x3 − 1 x 4 ⎤
4
⎣3
⎦0
1
⎡ 1 x 2 − 1 x3 ⎤
3
⎣2
⎦0
=
)dx
1 ⎡ 1 x3
2 ⎣3
=
1
2
1
− 12 x 4 + 15 x5 ⎤
⎦0
1
⎡ 1 x 2 − 1 x3 ⎤
3
⎣2
⎦0
1
10
1
.
6
1
1
From Problem 6, x = and y = .
2
10
1
1
π
⎛ ⎞⎛ ⎞
V ( S1 ) = 2π ⎜ ⎟ ⎜ ⎟ =
⎝ 10 ⎠ ⎝ 6 ⎠ 30
⎛ 1 ⎞⎛ 1 ⎞ π
V ( S2 ) = 2π ⎜ ⎟⎜ ⎟ =
⎝ 2 ⎠⎝ 6 ⎠ 6
⎛1
⎞ ⎛ 1 ⎞ 7π
V ( S3 ) = 2 π ⎜ + 2 ⎟ ⎜ ⎟ =
10
⎝
⎠ ⎝ 6 ⎠ 10
1 ⎞⎛ 1 ⎞ 5π
⎛
V ( S4 ) = 2π ⎜ 3 − ⎟⎜ ⎟ =
2 ⎠⎝ 6 ⎠ 6
⎝
7. From Problem 1, A =
by the First Fundamental Theorem of
Calculus.
23. True:
6. x =
x( x − x 2 )dx
Instructor’s Resource Manual
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
8. 8 = F(8) = 8k, k = 1
4
8
a.
b.
12.
8
1
⎡1 ⎤
W = ∫ x dx = ⎢ x 2 ⎥ = (64 − 4)
2
2
⎣
⎦2 2
= 30 in.-lb
−4
=
4
9. W = ∫ (62.4)(52 )π(10 − y )dy
0
1 ⎤
⎡
(10 − y )dy = 1560π ⎢10 y − y 2 ⎥
2 ⎦0
⎣
= 65,520π ≈ 205,837 ft-lb
6
10. The total work is equal to the work W1 to pull up
the object to the top without the cable and the
work W2 to pull up the cable.
W1 = 200 ⋅100 = 20,000 ft-lb
120 6
= lb/ft.
The cable weighs
100 5
6
6
ΔW2 = Δy ⋅ y = y Δy
5
5
11. a.
=
=
∫
=2
− ( x 2 )2 ⎤ dx
⎦
∫0 (4 x − x
1 4
2 0
64
3
32
3
2
)dx
(16 x 2 − x 4 ) dx
32
3
1 ⎡ 16 x3
2⎣3
4
− 15 x5 ⎤
⎦0
32
3
=
1024
15
32
3
=
32
5
4
13. V = π∫ ⎡ (4 x) 2 − ( x 2 ) 2 ⎤ dx
⎦
0 ⎣
4
1 ⎤
2048π
⎡16
= π ⎢ x3 − x5 ⎥ =
5 ⎦0
15
⎣3
Using Pappus’s Theorem:
32
.
From Problem 11, A =
3
32
From Problem 12, y =
.
5
⎛ 32 ⎞ ⎛ 32 ⎞ 2048π
V = 2πy ⋅ A = 2π ⎜ ⎟ ⎜ ⎟ =
15
⎝ 5 ⎠⎝ 3 ⎠
4x = x 2 .
x2 − 4 x = 0
x(x – 4) = 0
x = 0, 4
14. a.
4
4
1 ⎤
⎡
A = ∫ (4 x − x 2 )dx = ⎢ 2 x 2 − x3 ⎥
0
3 ⎦0
⎣
64 ⎞ 32
⎛
= ⎜ 32 − ⎟ =
3 ⎠ 3
⎝
b. To find the intersection points, solve
y
= y.
4
y2
=y
16
(See example 4, section 5.5). Think of
cutting the barrel vertically and opening the
lateral surface into a rectangle as shown in
the sketch below.
At depth 3 – y, a narrow rectangle has width
16π , so the total force on the lateral surface
is (δ = density of water = 62.4 lbs 3 )
ft
y 2 − 16 y = 0
y(y – 16) = 0
y = 0, 16
16
y⎞
1 ⎤
⎛
⎡2
y − ⎟ dy = ⎢ y 3 / 2 − y 2 ⎥
0 ⎜⎝
4⎠
8 ⎦0
⎣3
⎛ 128
⎞ 32
=⎜
− 32 ⎟ =
⎝ 3
⎠ 3
Instructor’s Resource Manual
∫
=
4
100
16
32
3
0
To find the intersection points, solve
A=∫
(4 x 2 − x3 ) dx
= π∫ (16 x 2 − x 4 ) dx
6
6 ⎡1 ⎤
y dy = ⎢ y 2 ⎥
0
5
5 ⎣ 2 ⎦0
= 6000 ft-lb
W = W1 + W2 = 26,000 ft-lb
100
32
3
4
1
⎡ (4 x) 2
2 0 ⎣
4
6
0
W2 = ∫
⎡ 4 x3 − 1 x 4 ⎤
4
⎣3
⎦0
y=
6
= 1560π∫
2
4
⎡1 ⎤
x dx = ⎢ x 2 ⎥ = 8 in.-lb
0
⎣ 2 ⎦0
W =∫
4
∫ x(4 x − x )dx = ∫0
x= 0
4
2
∫0 (4 x − x ) dx
3
3
∫0 δ (3 − y)(16π ) dy =16πδ ∫0 (3 − y) dy
3
⎡
y2 ⎤
= 16πδ ⎢3 y − ⎥ = 16πδ (4.5) ≈ 14,114.55 lbs.
2 ⎦⎥
⎣⎢
0
Section 5.8
343
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
b. All points on the bottom of the barrel are at
the same depth; thus the total force on the
bottom is simply the weight of the column of
water in the barrel, namely
F = π (82 )(3)δ ≈ 37, 638.8 lbs.
2
1
=∫
4
1
1 + [ g ′( x) ] dx
L2 = ∫
2
a
23. A1 = 2π ∫ f ( x) 1 + [ f ′( x) ] dx
3⎛
1
1
1
x + +
dx = ∫ ⎜ x 2 +
1 ⎝
2 16 x 4
4 x2
3
2
L4 = f (b) − g (b)
Total length = L1 + L2 + L3 + L4
⎛
1 ⎞
1 + ⎜ x2 −
⎟ dx
4 x2 ⎠
⎝
3
1 + [ f ′( x) ] dx
b
a
b
L3 = f (a) − g (a)
dy
1
15.
= x2 −
dx
4 x2
L=∫
22. L1 = ∫
⎞
⎟ dx
⎠
3
1⎤
1 ⎞ ⎛ 1 1 ⎞ 53
⎡1
⎛
= ⎢ x3 − ⎥ = ⎜ 9 − ⎟ − ⎜ − ⎟ =
3
4
x
12
⎣
⎦1 ⎝
⎠ ⎝3 4⎠ 6
A3 = π ⎡ f 2 (a) − g 2 (a) ⎤
⎣
⎦
A4 = π ⎡ f 2 (b) − g 2 (b) ⎤
⎣
⎦
Total surface area = A1 + A2 + A3 + A4 .
P ( X ≥ 1) = P(1 ≤ X ≤ 2)
=∫
2
1
=
3
t 4 + 2t 2 + 1 dt = 2∫
0
3
0
b.
=
3
18. A = ∫
b
a
20. V = 2π ∫ ( x − a ) [ f ( x) − g ( x) ] dx
b
a
(
= 0.8
b.
(
)
x
(
21. M y = δ ∫ x [ f ( x) − g ( x) ] dx
a
Mx =
344
δ
b
⎡ f 2 ( x) − g 2 ( x ) ⎤ dx
⎦
2 ∫a ⎣
Section 5.8
x
)
1⎛
x4 ⎞ 2
x4
⎜ 8x − ⎟ = x −
12 ⎜⎝
4 ⎟⎠ 3
48
P ( X ≤ 3) = F (3) = 1 −
f ( x) = F ′( x) = 2 ⋅
0≤ x≤6
b
2
1
1 ⎡
x5 ⎤
8 − x3 dx = ⎢ 4 x 2 − ⎥
12
12 ⎣⎢
5 ⎦⎥
0
1
1 ⎡
t4 ⎤
8 − t 3 dt = ⎢8t − ⎥
0 12
12 ⎢⎣
4 ⎦⎥
0
F ( x) = ∫
=
25. a.
)
85
≈ 0.332
256
0
d.
2
1
1 ⎡
x4 ⎤
8 − x3 dx = ⎢8 x − ⎥
12
12 ⎢⎣
4 ⎥⎦
1
2
[ f ( x) − g ( x)] dx
b
19. V = π ∫ ⎡ f 2 ( x) − g 2 ( x) ⎤ dx
⎦
a⎣
)
c. E ( X ) = ∫ x ⋅
2
1 ⎤
⎡
= ⎢9 x − x3 ⎥ = (27 − 9) − (−27 + 9) = 36
3 ⎦ −3
⎣
0.5
0
⎛ 9 − x 2 ⎞ dx = 3 (9 − x 2 ) dx
⎜
⎟
∫−3
−3 ⎝
⎠
3
(
P (0 ≤ X < 0.5) = P(0 ≤ X ≤ 0.5)
3
17. V = ∫
2
1
1 ⎡
x4 ⎤
8 − x3 dx = ⎢8 x − ⎥
12
12 ⎢⎣
4 ⎥⎦
1
17
≈ 0.354
48
=∫
(t 2 + 1)dt
⎡1
⎤
= 2 ⎢ t3 + t ⎥ = 4 3
⎣3
⎦0
2
a
24. a.
L = 2∫
2
A2 = 2π ∫ g ( x) 1 + [ g ′( x) ] dx
16.
The loop is − 3 ≤ t ≤ 3 . By symmetry, we can
double the length of the loop from t = 0 to
dx
dy 2
= 2t ;
= t −1
t = 3,
dt
dt
b
a
b
(6 − 3) 2 3
=
36
4
1
6− x
,
(6 − x) =
36
18
6 ⎛6− x⎞
c. E ( X ) = ∫ x ⋅ ⎜
⎟ dx
0
⎝ 18 ⎠
6
1 ⎡
x3 ⎤
= ⎢3 x 2 − ⎥ = 2
18 ⎢⎣
3 ⎥⎦
0
Instructor’s Resource Manual
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
Review and Preview Problems
1. By the Power Rule
1
x −2 +1 x −1
1
−2
∫ x 2 dx = ∫ x dx = −2 + 1 = −1 = − x + C
2. By the Power Rule
1
2
x −1.5+1
x −0.5
−1.5
=
=
=
=−
+C
dx
x
dx
∫ x1.5
∫
−1.5 + 1 −0.5
x
3. By the Power Rule
1
x −1.01+1
x −0.01
100
−1.01
∫ x1.01 dx = ∫ x dx = −1.01 + 1 = −0.01 = − x0.01 + C
4. By the Power Rule
1
x −0.99 +1
x 0.01
−0.99
=
=
=
= 100 x 0.01 + C
dx
x
dx
∫ x0.99
∫
−0.99 + 1 0.01
1
10. a. (1 + )1 = 21 = 2
1
10
b. (1 +
1 10 ⎛ 11 ⎞
) = ⎜ ⎟ ≈ 2.593742
10
⎝ 10 ⎠
c. (1 +
1 100 ⎛ 101 ⎞
) =⎜
⎟
100
⎝ 100 ⎠
d. (1 +
1 1000 ⎛ 1001 ⎞
)
=⎜
⎟
1000
⎝ 1000 ⎠
100
1000
2
b. (1 +
c. (1 +
5. F (1) = ∫1 dt = 0
t
7. Let g ( x ) = x 2 ; then by the Chain Rule and
problem 6,
Dx F ( x 2 ) = Dx F ( g ( x )) = F ′( g ( x )) g ′( x )
⎛ 1
=⎜ 2
⎝x
2
⎞
⎟ (2 x) = x
⎠
1
1
1
1
5
1
1
c. (1 + )
10
5
⎛6⎞
= ⎜ ⎟ = 2.48832
⎝5⎠
1
10
1
1
d. (1 + )
50
1
50
10
⎛ 11 ⎞
= ⎜ ⎟ ≈ 2.593742
⎝ 10 ⎠
⎛ 51 ⎞
=⎜ ⎟
⎝ 50 ⎠
50
1
1
e. (1 +
)
100
1
100
2
1
10
2
2
)
10
1
10
1 ⎞
⎛
= ⎜1 + ⎟
⎝ 20 ⎠
d. (1 +
2
1
1
50 )
50
e. (1 +
2
1
1
100 )
100
2
≈ 2.70481
2
≈ 2.71152
10
1⎞
⎛
⎛ 11 ⎞
= ⎜ 1 + ⎟ = ⎜ ⎟ ≈ 2.593742
10
⎝
⎠
⎝ 10 ⎠
20
⎛ 21 ⎞
=⎜ ⎟
⎝ 20 ⎠
100
1 ⎞
⎛
= ⎜1 +
⎟
⎝ 100 ⎠
1 ⎞
⎛
= ⎜1 +
⎟
⎝ 200 ⎠
20
≈ 2.6533
100
⎛ 101 ⎞
=⎜
⎟
⎝ 100 ⎠
200
⎛ 201 ⎞
=⎜
⎟
⎝ 200 ⎠
200
b. (1 +
2 10 2
5
) = (1.2 ) ≈ 2.48832
10
c. (1 +
2 100 2
50
)
= (1.02 ) ≈ 2.691588
100
d. (1 +
2 1000 2
500
)
= (1.002 ) ≈ 2.715569
1000
13. We know from trigonometry that, for any x and
any integer k , sin( x + 2kπ ) = sin( x) . Since
= 21 = 2
1
b. (1 + )
5
2
1
1
5)
5
2 1
12. a. (1 + ) 2 = 3 ≈ 1.732051
1
8. Let h( x ) = x3 ; then by the Chain Rule and
problem 6,
x3 1
Dx ∫1 dt = Dx F ( h( x )) = F ′( h( x ))h′( x )
t
3
⎛ 1 ⎞
= ⎜ 3 ⎟ (3 x 2 ) =
x
⎝x ⎠
9. a. (1 + 1)
≈ 2.7169239
1 2
⎛3⎞
11. a. (1 + ) 1 = ⎜ ⎟ = 2.25
2
⎝2⎠
11
6. By the First Fundamental Theorem of Calculus
d x1
1
F ′( x ) =
dt =
∫
1
dx t
x
≈ 2.704814
≈ 2.691588
100
⎛ 101 ⎞
=⎜
⎟
⎝ 100 ⎠
Instructor’s Resource Manual
⎛π ⎞ 1
⎛ 5π ⎞ 1
and sin ⎜ ⎟ = ,
sin ⎜ ⎟ =
⎝6⎠ 2
⎝ 6 ⎠ 2
1
π
12k + 1
π
sin( x) =
if x = + 2kπ =
2
6
6
5π
12k + 5
π
or x =
+ 2kπ =
6
6
where k is any integer.
≈ 2.704814
Review and Preview
345
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
14. We know from trigonometry that, for any x and
any integer k , cos( x + 2kπ ) = cos( x) . Since
cos(π ) = -1, cos( x) = −1
if x = π + 2kπ = (2k + 1)π where k is any integer.
15. We know from trigonometry that, for any x and
any integer k , tan( x + kπ ) = tan( x) . Since
π
4k + 1
⎛π ⎞
π
tan ⎜ ⎟ = 1, tan( x ) = 1 if x = + kπ =
4
4
4
⎝ ⎠
where k is any integer.
16. Since sec( x) =
y ' = xy 2 → dy = xy 2 dx
21.
1
y2
∫
−
opp = x 2 − 1 , adj = 1 , hypot = x so that
18. In the triangle, relative to θ ,
opp = x , adj = 1 − x 2 , hypot = 1 so that
x
sin θ = x cos θ = 1 − x 2 tan θ =
1 − x2
cot θ =
1 − x2
x
1
sec θ =
1− x
2
csc θ =
y2
= ∫ xdx
1 1 2
= x +C
y 2
When x = 0 and y = 1 we get C = −1 . Thus,
x2 − 2
1 1 2
= x −1 =
y 2
2
2
y=− 2
x −2
22. y ' =
1
x2 − 1
sin θ =
cos θ =
tan θ = x 2 − 1
x
x
x
1
cot θ =
sec θ = x csc θ =
x2 − 1
x2 − 1
dy
−
1
, sec( x) is never 0 .
cos( x)
17. In the triangle, relative to θ ,
dy = xdx
cos x
y
→ dy =
cos x
dx
y
y dy = cos x dx
∫
ydy = ∫ cos x dx
1 2
y = sin x + C
2
When x = 0 and y = 4 we get C = 8 . Thus,
1 2
y = sin x + 8
2
y 2 = 2sin x + 16
1
x
19. In the triangle, relative to θ ,
opp = 1 , adj = x , hypot = 1 + x 2 so that
sin θ =
1
1+ x
2
x
cos θ =
cot θ = x sec θ =
1+ x
1 + x2
x
2
tan θ =
1
x
csc θ = 1 + x 2
20. In the triangle, relative to θ ,
opp = 1 − x 2 , adj = x , hypot = 1 so that
sin θ = 1 − x 2
cot θ =
346
x
1 − x2
cos θ = x tan θ =
sec θ =
1 − x2
x
1
1
csc θ =
x
1 − x2
Review and Preview
Instructor’s Resource Manual
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may be reproduced, in any form or by any means, without permission in writing from the publisher.
Transcendental
Functions
6
CHAPTER
6.1 Concepts Review
x1
1.
∫1 t dt; (0, ∞); (–∞, ∞)
2.
1
x
3.
1
ln(3x – 2)
2
1
1
3
= ⋅
Dx (3 x – 2) =
2 3x – 2
2(3 x – 2)
6. Dx ln 3 x – 2 = Dx
1
; ln x + C
x
7.
dy
1 3
= 3⋅ =
dx
x x
8.
dy
1
= x 2 ⋅ + 2 x ⋅ ln x = x(1 + 2 ln x)
dx
x
4. ln x + ln y; ln x – ln y; r ln x
9. z = x 2 ln x 2 + (ln x)3 = x 2 ⋅ 2 ln x + (ln x)3
Problem Set 6.1
1. a.
b.
⎛3⎞
ln1.5 = ln ⎜ ⎟ = ln 3 − ln 2 = 0.406
⎝2⎠
c.
ln 81 = ln 3 = 4 ln 3 = 4(1.099) = 4.396
d.
ln 2 = ln 21/ 2 =
e.
f.
dz
2
1
= x 2 ⋅ + 2 x ⋅ 2 ln x + 3(ln x) 2 ⋅
dx
x
x
3
= 2 x + 4 x ln x + (ln x)2
x
ln 6 = ln (2 · 3) = ln 2 + ln 3
= 0.693 + 1.099 = 1.792
4
10. r =
1 –2
x – (ln x)3
2
dr –2 –3
1
1 3(ln x) 2
=
x – 3(ln x)2 ⋅ = − −
x
dx 2
x
x3
1
1
ln 2 = (0.693) = 0.3465
2
2
⎛ 1 ⎞
ln ⎜ ⎟ = – ln 36 = – ln(22 ⋅ 32 )
⎝ 36 ⎠
= −2 ln 2 − 2 ln 3 = −3.584
=
11. g ′( x) =
ln 48 = ln(24 ⋅ 3) = 4 ln 2 + ln 3 = 3.871
2. a.
1.792
b. 0.405
c.
4.394
d. 0.3466
e.
–3.584
f. 3.871
=
1
x + 3x + π
2
⋅ Dx ( x 2 + 3 x + π ) =
=
2x + 3
x + 3x + π
=
1
3x + 2 x
3
Dx (3x3 + 2 x)
9 x2 + 2
3 x3 + 2 x
5. Dx ln( x – 4)3 = Dx 3ln( x – 4)
1
3
= 3⋅
Dx ( x – 4) =
x–4
x–4
Instructor’s Resource Manual
x2 + 1
⎡ 1 2
⎤
–1/ 2
⋅ 2x⎥
⎢1 + 2 ( x – 1)
⎣
⎦
x + x –1
1
2
1
x −1
2
2
13.
4. Dx ln(3x3 + 2 x) =
⎡ 1 2
⎤
1 + ( x + 1) –1/ 2 ⋅ 2 x ⎥
⎢
⎦
x + x2 + 1 ⎣ 2
1
1
12. h′( x) =
3. Dx ln( x 2 + 3 x + π)
=
3
ln x
⎛ 1⎞
+ ln
=
+ (– ln x)3
2
2 ⎜⎝ x ⎟⎠
2
x ln x
x ⋅ 2 ln x
ln x
14.
1
f ( x) = ln 3 x = ln x
3
1 1 1
f ′( x) = ⋅ =
3 x 3x
1
1
=
f ′(81) =
3 ⋅ 81 243
1
(– sin x ) = – tan x
cos x
⎛π⎞
⎛π⎞
f ′ ⎜ ⎟ = − tan ⎜ ⎟ = −1 .
⎝4⎠
⎝4⎠
f ′( x) =
Section 6.1
347
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
15. Let u = 2x + 1 so du = 2 dx.
1
1 1
∫ 2 x + 1 dx = 2 ∫ u du
1
1
= ln u + C = ln 2 x + 1 + C
2
2
22. Let u = 2t 2 + 4t + 3 so du = (4t + 4)dt .
t +1
1
1
ln u + C = ln 2t 2 + 4t + 3 + C
4
4
=
16. Let u = 1 – 2x so du = –2dx.
1
1 1
∫ 1 – 2 x dx = – 2 ∫ u du
1
1
= – ln u + C = – ln 1 – 2 x + C
2
2
17. Let u = 3v 2 + 9v so du = 6v + 9.
6v + 9
1
∫ 3v2 + 9v dv = ∫ u du = ln u + C
18. Let u = 2 z 2 + 8 so du = 4z dz.
1 1
z
∫ 2 z 2 + 8 dz = 4 ∫ u du
1
1
= ln u + C = ln 2 z 2 + 8 + C
4
4
19. Let u = ln x so du =
)
⎡1
⎤
2
∫0 2t 2 + 4t + 3dt = ⎢⎣ 4 ln 2t + 4t + 3 ⎥⎦0
1
1
9
1
ln 9 – ln 3 = ln 4 = ln 4 3 = ln 3
4
4
3
4
=
23. By long division,
x2
1
∫ x − 1 dx = ∫ x dx + ∫ 1 dx + ∫ x − 1 dx
x2
=
+ x + ln x − 1 + C
2
so
–1
∫ x(ln x)2 dx = – ∫ u
1
dx
x
1
dx .
x
–2
21. Let u = 2 x5 + π so du = 10 x 4 dx .
x4
1 1
∫ 2 x5 + π dx = 10 ∫ u du
1
1
= ln u + C = ln 2 x5 + π + C
10
10
3
∫0
x4
3
⎡1
⎤
dx = ⎢ ln 2 x5 + π ⎥
5
10
⎣
⎦0
2x + π
1
486 + π
= [ln(486 + π) – ln π] = ln 10
≈ 0.5048
10
π
348
Section 6.1
x2 + x x 3
3
= + +
2 x − 1 2 4 4(2 x − 1)
so
3
3
x2 + x
x
∫ 2 x − 1 dx = ∫ 2 dx + ∫ 4 dx + ∫ 4(2 x − 1) dx
25. By long division,
x4
256
= x3 − 4 x 2 + 16 x − 64 +
x+4
x+4
du
1
1
+C =
+C
u
ln x
=
24. By long division,
3
1
x2 3
dx
+ x+ ∫
4 4
4 2x −1
Let u = 2 x − 1 ; then du = 2dx . Hence
1
1 1
1
∫ 2 x − 1 dx = 2 ∫ u du = 2 ln u + C
1
= ln 2 x − 1 + C
2
2
x +x
x2 3
3
+ x + ln 2 x − 1 + C
and ∫
dx =
2x −1
4 4
8
= u 2 + C = (ln x) 2 + C
20. Let u = ln x, so du =
x2
1
= x +1+
x −1
x −1
=
2 ln x
dx = 2∫ udu
x
∫
1
t +1
1
= ln 3v 2 + 9v + C
(
1 1
∫ 2t 2 + 4t + 3 dt = 4 ∫ u du
so
x4
∫ x + 4 dx =
∫x
=
3
dx − ∫ 4 x 2 dx + ∫ 16 xdx − ∫ 64dx + 256∫
x 4 4 x3
−
+ 8 x 2 − 64 x + 256 ln x + 4 + C
4
3
26. By long division,
∫
1
dx
x+4
x3 + x 2
4
= x2 − x + 2 −
so
x+2
x+2
x3 + x 2
1
dx = ∫ x 2 dx − ∫ xdx + ∫ 2dx − 4∫
dx
x+2
x+2
x3 x 2
=
−
+ 2 x − 4 ln x + 2 + C
3
2
Instructor’s Resource Manual
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
27. 2 ln( x + 1) – ln x = ln( x + 1) 2 – ln x = ln
28.
( x + 1)2
x
1
1
ln( x – 9) + ln x = ln x – 9 – ln x
2
2
x–9
x–9
= ln
= ln
x
x
1
31. ln y = ln( x + 11) – ln( x3 – 4)
2
1 dy
1
1
1
=
⋅1 – ⋅
⋅ 3x2
3
y dx x + 11
2 x –4
=
⎡ 1
3x2 ⎤
dy
= y⋅⎢
–
⎥
3
dx
⎣⎢ x + 11 2( x – 4) ⎦⎥
29. ln(x – 2) – ln(x + 2) + 2 ln x
= ln( x – 2) – ln( x + 2) + ln x 2 = ln
1
3x2
–
x + 11 2( x3 – 4)
x 2 ( x – 2)
x+2
=
30. ln( x 2 – 9) – 2 ln( x – 3) – ln( x + 3)
= ln( x 2 − 9) − ln( x − 3)2 − ln( x + 3)
=–
x2 – 9
1
= ln
= ln
2
x–3
( x – 3) ( x + 3)
x + 11 ⎡ 1
3x2 ⎤
–
⎢
⎥
3
x3 – 4 ⎣⎢ x + 11 2( x – 4) ⎦⎥
x3 + 33x 2 + 8
2( x3 – 4)3 / 2
32. ln y = ln( x 2 + 3x ) + ln( x – 2) + ln( x 2 + 1)
1 dy
2x + 3
1
2x
=
+
+
y dx x 2 + 3 x x – 2 x 2 + 1
dy
⎛ 2x + 3
1
2x ⎞
4
3
2
= ( x 2 + 3 x)( x – 2)( x 2 + 1) ⎜
+
+
⎟ = 5 x + 4 x –15 x + 2 x – 6
2
2
dx
x
–
2
+
+
x
3
x
x
1
⎝
⎠
1
1
ln( x + 13) – ln( x – 4) – ln(2 x + 1)
2
3
1 dy
1
1
2
=
–
–
y dx 2( x + 13) x – 4 3(2 x + 1)
33. ln y =
⎡
⎤
dy
x + 13
1
1
2
10 x 2 + 219 x – 118
=
–
–
=
–
dx ( x – 4) 3 2 x + 1 ⎢⎣ 2( x + 13) x – 4 3(2 x + 1) ⎥⎦
6( x – 4)2 ( x + 13)1/ 2 (2 x + 1)4 / 3
2
1
ln( x 2 + 3) + 2 ln(3 x + 2) – ln( x + 1)
3
2
1 dy 2 2 x
2⋅3
1
= ⋅
+
–
y dx 3 x 2 + 3 3x + 2 2( x + 1)
34. ln y =
dy ( x 2 + 3)2 / 3 (3 x + 2)2
=
dx
x +1
⎡ 4x
6
1 ⎤ (3 x + 2)(51x3 + 70 x 2 + 97 x + 90)
–
+
⎢ 2
⎥=
6( x 2 + 3)1/ 3 ( x + 1)3 / 2
⎢⎣ 3( x + 3) 3 x + 2 2( x + 1) ⎥⎦
35.
36.
y = ln x is reflected across the y-axis.
The y-values of y = ln x are multiplied by
since ln x =
Instructor’s Resource Manual
1
,
2
1
ln x.
2
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37.
y = ln x is reflected across the x-axis since
⎛1⎞
ln ⎜ ⎟ = – ln x.
⎝ x⎠
38.
42. Let r(x) = rate of transmission
1
= kx 2 ln = −kx 2 ln x.
x
⎛1⎞
r ′( x) = −2kx ln x − kx 2 ⎜ ⎟ = − kx(2 ln x + 1)
⎝ x⎠
1
1
r ′( x) = 0 if ln x = − , or − ln x = , so
2
2
1 1
ln = .
x 2
1
1
ln1.65 ≈ , so x ≈
≈ 0.606.
1.65
2
⎛ 1⎞
r ′′( x) = −k (2 ln x + 1) − kx ⎜ 2 ⋅ ⎟ = –k(2 ln x + 3)
⎝ x⎠
′′
r (0.606) ≈ −2k < 0 since k > 0, so
x ≈ 0.606 gives the maximum rate of
transmission.
43. ln 4 > 1
so ln 4m = m ln 4 > m ⋅1 = m
Thus x > 4m ⇒ ln x > m
so lim ln x = ∞
x →∞
y = ln x is shifted two units to the right.
1
so z → ∞ as x → 0+
x
⎛1⎞
Then lim ln x = lim ln ⎜ ⎟ = lim (– ln z )
+
→∞
z
⎝ z ⎠ z →∞
x →0
= – lim ln z = – ∞
44. Let z =
39.
z →∞
45.
y = ln cos x + ln sec x
= ln cos x + ln
1
cos x
⎛ π π⎞
= ln cos x − ln cos x = 0 on ⎜ − , ⎟
⎝ 2 2⎠
40. Since ln is continuous,
sin x
sin x
lim ln
= ln lim
= ln1 = 0
x
x →0
x →0 x
41. The domain is ( 0, ∞ ) .
⎛1⎞
f ′( x) = 4 x ln x + 2 x 2 ⎜ ⎟ − 2 x = 4 x ln x
⎝x⎠
f ' ( x ) = 0 if ln x = 0 , or x = 1 .
f ' ( x ) < 0 for x < 1 and f ' ( x ) > 0 for x > 1
so f(1) = –1 is a minimum.
350
Section 6.1
x
1
x1
1
1
x1
1
1
x1
∫1/ 3 t dt = 2∫1 t dt
x1
∫1/ 3 t dt + ∫1 t dt = 2∫1 t dt
∫1/ 3 t dt = ∫1 t dt
–∫
1/ 3 1
1
t
dt = ∫
x1
1
t
dt
1
– ln = ln x
3
ln 3 = ln x
x=3
46. a.
1 1
<
for t > 1,
t
t
x1
x 1
x
so ln x = ∫ dt < ∫
dt = ∫ t –1/ 2 dt
1 t
1 t
1
x
= ⎡⎣ 2 t ⎤⎦ = 2( x –1)
1
so ln x < 2( x – 1)
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
b. If x > 1, 0 < ln x < 2( x – 1) ,
x →∞
and lim
x →∞
∫π 4
ln x
2( x + 1)
≤ lim
=0
x
x
x →∞
1, 000, 000
≈ 72,382
ln1, 000, 000
c
⎛ ax – b ⎞
⎛ ax − b ⎞
f ( x) = ln ⎜
⎟ = c ln ⎜
⎟
+
ax
b
⎝
⎠
⎝ ax + b ⎠
=
a 2 – b2
[ln(ax – b) – ln(ax + b)]
2ab
f ′( x) =
=
f ′(1) =
b.
a 2 – b2 ⎡ a
a ⎤
–
⎢
2ab ⎣ ax – b ax + b ⎦⎥
a 2 − b2
2ab
⎡
⎤
2ab
a 2 – b2
⎢ (ax − b)(ax + b) ⎥ = 2 2
⎣
⎦ a x – b2
a 2 − b2
a 2 − b2
f ′( x) = cos 2 u ⋅
3
4
π
= ⎡⎣ln tan x ⎤⎦π 3 = ln tan π 3 − ln tan π 4
4
cos x
= ln 1 + sin x + C = ln(1 + sin x) + C
(since 1 + sin x ≥ 0 for all x ).
4
1
= 3cos 2 (0) = 3
dx
2πx
4
dx = ⎡⎢ π ln x 2 + 4 ⎤⎥
⎣
⎦
1
x +4
= π ln 20 − π ln 5 = π ln 4 ≈ 4.355
4
∫1
2
x2
x2 1
– ln x =
– ln x
4
4 2
dy 2 x 1 1 x 1
=
– ⋅ = –
dx
4 2 x 2 2x
54. y =
2
∫1
2
1
x + x –1
2 ⋅1 + 1
f ′(1) = cos [ln(1 + 1 –1)] ⋅
2
1 + 1 –1
x2 + 4
= π ln x 2 + 4 + C
=∫
2
2πx
Let u = x 2 + 4 so du = 2x dx.
2πx
1
∫ x2 + 4 dx = π∫ u du = π ln u + C
du
dx
2x + 1
4
1
53. V = 2π∫ xf ( x)dx = ∫
L=
2
1
∫ 1 + sin x dx = ∫ u du = ln u + C
=1
= cos 2 [ln( x 2 + x –1)] ⋅
2
= ⎡−
⎣ ln cos x + ln sin x ⎤⎦π
52. Let u = 1 + sin x ; then du = cos x dx so that
⎡ 1
1
1 ⎤ 1
⎥⋅
= lim ⎢
+
+ ⋅⋅⋅ +
n⎥ n
n→∞ ⎢1 + 1 1 + 2
+
1
n
n⎦
⎣ n
n ⎛
⎞
21
1
1
⎟ ⋅ = ∫ dx = ln 2 ≈ 0.693
= lim ∑ ⎜
i
1
x
n→∞ i =1 ⎜ 1 + ⎟ n
n⎠
⎝
49. a.
3 sec x csc x dx
= ln( 3) − ln1 = 0.5493 − 0 = 0.5493
ln x
= 0.
x
1
1⎤
⎡ 1
+
+ ⋅⋅⋅ + ⎥
47. lim ⎢
2n ⎦
n →∞ ⎣ n + 1 n + 2
48.
π
π
ln x 2( x – 1)
<
.
so 0 <
x
x
Hence 0 ≤ lim
51. From Ex 10,
2
2
2
⎛ dy ⎞
⎛x 1 ⎞
1 + ⎜ ⎟ dx = ∫ 1 + ⎜ – ⎟ dx
1
⎝ dx ⎠
⎝ 2 2x ⎠
2
2⎛ x
1 ⎞
⎛x 1 ⎞
⎜ + ⎟ dx = ∫1 ⎜ + ⎟ dx
⎝ 2 2x ⎠
⎝ 2 2x ⎠
2
⎤
1 ⎡ x2
1⎡
⎛1
⎞⎤
= ⎢ + ln x ⎥ = ⎢ 2 + ln 2 − ⎜ + ln1⎟ ⎥
2 ⎢⎣ 2
⎝2
⎠⎦
⎥⎦1 2 ⎣
3 1
= + ln 2 ≈ 1.097
4 2
50. From Ex 9,
π
∫0 3 tan x dx = ⎡−
⎣ ln cos x
π
⎤⎦ 0 3
= ln cos 0 − ln cos π 3
⎛ 1 ⎞
= ln(1) − ln(0.5) = ln ⎜
⎟
⎝ 0.5 ⎠
= ln 2 ≈ 0.69315
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55.
b.
c.
1 1
1
+ + ⋅⋅⋅ + = the lower approximate area
2 3
n
1
1
1 + + ⋅⋅⋅ +
= the upper approximate area
2
n –1
ln n = the exact area under the curve
58. a.
Thus,
1 1
1
1 1
1
+ + ⋅⋅⋅ + < ln n < 1 + + + ⋅⋅⋅ +
.
2 3
n
2 3
n −1
y1
56.
ln y – ln x
=
y–x
∫1
t
dt – ∫
x1
1
y–x
= the average value of
t
dt
∫x t dt
y–x
1
on [x, y].
t
1
Since is decreasing on the interval [x, y], the
t
average value is between the minimum value of
1
1
and the maximum value of .
y
x
57. a.
1 + 1.5sin x
(1.5 + sin x)2
On [0,3π ], f ′′( x) = 0 when x ≈ 3.871,
5.553.
Inflection points are (3.871, –0.182),
(5.553, –0.182).
3π
∫0
ln(1.5 + sin x)dx ≈ 4.042
sin(ln x)
x
On [0.1, 20], f ′( x) = 0 when x = 1.
Critical points: 0.1, 1, 20
f(0.1) ≈ –0.668, f(1) = 1, f(20) ≈ –0.989
On [0.1, 20], the maximum value point is
(1, 1) and minimum value point is
(20, –0.989).
f ′( x) = −
b. On [0.01, 0.1], f ′( x) = 0 when x ≈ 0.043.
f(0.01) ≈ –0.107, f(0.043) ≈ –1
On [0.01, 20], the maximum value point is
(1, 1) and the minimum value point is
(0.043, –1).
y1
=
f ′′( x) = −
20
c.
∫0.1 cos(ln x)dx ≈ −8.37
a.
∫0 ⎢⎣ x ln ⎜⎝ x ⎟⎠ − x
59.
1
cos x
⋅ cos x =
1.5 + sin x
1.5 + sin x
f ′( x) = 0 when cos x = 0.
f ′( x) =
π 3π 5π
Critical points: 0, , , , 3π
2 2 2
f(0) ≈ 0.405,
⎛π⎞
⎛ 3π ⎞
f ⎜ ⎟ ≈ 0.916, f ⎜ ⎟ ≈ −0.693,
⎝2⎠
⎝ 2 ⎠
⎛ 5π ⎞
f ⎜ ⎟ ≈ 0.916, f (3π) ≈ 0.405.
⎝ 2 ⎠
On [0,3π ], the maximum value points are
⎛π
⎞ ⎛ 5π
⎞
⎜ , 0.916 ⎟ , ⎜ , 0.916 ⎟ and the minimum
2
2
⎝
⎠ ⎝
⎠
⎛ 3π
⎞
value point is ⎜ , −0.693 ⎟ .
2
⎝
⎠
1⎡
⎛1⎞
2
5
⎛ 1 ⎞⎤
ln ⎜ ⎟ ⎥ dx =
≈ 0.139
x
36
⎝ ⎠⎦
b. Maximum of ≈ 0.260 at x ≈ 0.236
60.
a.
1
∫0 [ x ln x −
x ln x]dx =
7
≈ 0.194
36
b. Maximum of ≈ 0.521 at x ≈ 0.0555
352
Section 6.1
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6.2 Concepts Review
1.
11.
f ′( z ) = 2( z – 1) > 0 for z > 1
f(z) is increasing at z = 1 because f(1) = 0 and
f(z) > 0 for z > 1. Therefore, f(z) is strictly
increasing on z ≥ 1 and so it has an inverse.
12.
f ′( x) = 2 x + 1 > 0 for x ≥ 2 . f(x) is strictly
increasing on x ≥ 2 and so it has an inverse.
13.
f ′( x) = x 4 + x 2 + 10 > 0 for all real x. f(x) is
strictly increasing and so it has an inverse.
f ( x1 ) ≠ f ( x2 )
2. x; f –1 ( y )
3. monotonic; strictly increasing; strictly decreasing
4. ( f –1 )′( y ) =
1
f ′( x)
Problem Set 6.2
14.
1. f(x) is one-to-one, so it has an inverse.
Since f (4) = 2, f −1 (2) = 4 .
2. f(x) is one-to-one, so it has an inverse.
Since f(1) = 2, f −1 (2) = 1 .
3. f(x) is not one-to-one, so it does not have an
inverse.
4. f(x) is not one-to-one, so it does not have an
inverse.
5. f(x) is one-to-one, so it has an inverse.
Since f(–1.3) ≈ 2, f −1 (2) ≈ −1.3 .
6. f(x) is one-to-one, so it has an inverse. Since
1
⎛1⎞
f ⎜ ⎟ = 2, f −1 (2) = .
2
2
⎝ ⎠
7.
8.
9.
f ′( x) = –5 x 4 – 3x 2 = –(5 x 4 + 3 x 2 ) < 0 for all
x ≠ 0. f(x) is strictly decreasing at x = 0 because
f(x) > 0 for x < 0 and f(x) < 0 for x > 0. Therefore
f(x) is strictly decreasing for x and so it has an
inverse.
f ′( x) = 7 x 6 + 5 x 4 > 0 for all x ≠ 0.
f(x) is strictly increasing at x = 0 because f(x) > 0
for x > 0 and f(x) < 0 for x < 0. Therefore f(x) is
strictly increasing for all x and so it has an
inverse.
f ′(θ ) = – sin θ < 0 for 0 < θ < π
f (θ) is decreasing at θ = 0 because f(0) = 1 and
f(θ) < 1 for 0 < θ < π . f(θ) is decreasing at
θ = π because f( π ) = –1 and f(θ) > –1 for
0 < θ < π . Therefore f(θ) is strictly decreasing
on 0 ≤ θ ≤ π and so it has an inverse.
10.
f ′( x) = – csc 2 x < 0 for 0 < x <
f(x) is decreasing on 0 < x <
1
r
r
1
f (r ) = ∫ cos 4 tdt = – ∫ cos 4 tdt
π
f ′(r ) = – cos 4 r < 0 for all r ≠ k π + , k any
2
integer.
π
f(r) is decreasing at r = k π + since f ′(r ) < 0
2
on the deleted neighborhood
π
π
⎛
⎞
⎜ k π + − ε , k π + + ε ⎟ . Therefore, f(r) is
2
2
⎝
⎠
strictly decreasing for all r and so it has an
inverse.
15. Step 1:
y=x+1
x=y–1
Step 2: f –1 ( y ) = y – 1
Step 3: f –1 ( x) = x – 1
Check:
f –1 ( f ( x)) = ( x + 1) – 1 = x
f ( f –1 ( x)) = ( x – 1) + 1 = x
16. Step 1:
x
y = – +1
3
x
– = y –1
3
x = –3(y – 1) = 3 – 3y
Step 2: f –1 ( y ) = 3 – 3 y
Step 3: f –1 ( x) = 3 – 3 x
Check:
⎛ x ⎞
f –1 ( f ( x)) = 3 – 3 ⎜ – + 1⎟ = 3 + ( x – 3) = x
⎝ 3 ⎠
–(3 – 3 x)
+ 1 = (–1 + x) + 1 = x
f ( f –1 ( x)) =
3
π
2
π
and so it has an
2
inverse.
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353
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
17. Step 1:
y = x + 1 (note that y ≥ 0 )
x + 1 = y2
x = 2+
x = y 2 – 1, y ≥ 0
Step 2: f –1 ( y ) = y 2 – 1, y ≥ 0
Step 3: f
Check:
f
–1
–1
x–2=
1
y2
1
y2
,y>0
Step 2: f –1 ( y ) = 2 +
( x) = x – 1, x ≥ 0
2
Step 3: f –1 ( x) = 2 +
( f ( x)) = ( x + 1) – 1 = ( x + 1) – 1 = x
2
f ( f –1 ( x)) = ( x 2 – 1) + 1 = x 2 = x = x
18. Step 1:
y = – 1 – x (note that y ≤ 0 )
1– x = – y
f –1 ( f ( x)) = 2 +
1
(
1
x –2
)
2
= 2+
1
( x 1–2 )
1
⎛2+
⎜
⎝
1 ⎞–2
⎟
x2 ⎠
=
1
= x2
⎛ 1 ⎞
⎜ 2⎟
⎝x ⎠
Step 2: f
( y) = 1 – y , y ≤ 0
Step 3: f
Check:
–1
( x) = 1 – x 2 , x ≤ 0
2
f –1 ( f ( x)) = 1 – (– 1 – x ) 2 = 1 – (1 – x) = x
f ( f –1 ( x)) = – 1 – (1 – x 2 ) = – x 2 = – x
= –(–x) = x
21. Step 1:
y = 4 x 2 , x ≤ 0 (note that y ≥ 0 )
x2 =
y
4
x=–
y
y
=−
, negative since x ≤ 0
4
2
y
2
x
Step 3: f –1 ( x) = −
2
Check:
Step 2: f –1 ( y ) = −
19. Step 1:
1
x–3
1
x–3= –
y
y=–
x = 3–
1
y
Step 2: f
f –1 ( f ( x)) = –
–1
f(f
1
x
( x)) = –
1
– x1–3
1
(3 – )
1
x
⎛
x⎞
x
( x)) = 4 ⎜⎜ –
⎟⎟ = 4 ⋅ = x
4
⎝ 2 ⎠
y = ( x – 3)2 , x ≥ 3 (note that y ≥ 0 )
= 3 + ( x – 3) = x
x–3=
y
x = 3+ y
1
=–
=x
1
–
–3
x
20. Step 1:
1
(note that y > 0)
y=
x–2
1
y2 =
x–2
Section 6.2
–1
22. Step 1:
Check:
f –1 ( f ( x)) = 3 –
4 x2
= – x 2 = – x = –(– x) = x
2
2
1
( y) = 3 –
y
Step 3: f –1 ( x) = 3 –
354
,x>0
x2
= x =x
–1
f(f
1
= 2 + (x – 2) = x
1 – x = (– y ) 2 = y 2
–1
,y>0
y2
Check:
f ( f –1 ( x)) =
x = 1 – y2 , y ≤ 0
1
Step 2: f –1 ( y ) = 3 + y
Step 3: f –1 ( x) = 3 + x
Check:
f –1 ( f ( x)) = 3 + ( x – 3)2 = 3 + x – 3
= 3 + ( x – 3) = x
f ( f –1 ( x)) = [(3 + x ) – 3]2 = ( x )2 = x
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23. Step 1:
y = ( x –1)
x –1 = 3 y
Step 3: f –1 ( x) =
x = 1+ 3 y
Step 3: f –1 ( x) = 1 + 3 x
f(f
–1
–1
( f ( x)) = 1 + 3 ( x –1) = 1 + ( x –1) = x
3
( x)) = [(1 + x ) –1] = ( x ) = x
3
3
3
3
24. Step 1:
y = x5 / 2 , x ≥ 0
x= y
Step 3: f –1 ( x) = x 2 / 5
Check:
f ( f –1 ( x)) = ( x 2 / 5 )5 / 2 = x
25. Step 1:
x –1
y=
x +1
xy + y = x –1
x – xy = 1 + y
1+ y
x=
1– y
3 ⎤1/ 3
( )
1/ 3
⎥⎦
=
3⎤
⎡
1 – ⎢ xx +–11 ⎥
⎣
⎦
x +1 + x – 1 2x
=
=
=x
x +1 – x +1 2
1+
1–
x –1
x +1
x –1
x +1
⎛ 2 x1/ 3 ⎞
=⎜
⎟ = ( x1/ 3 )3 = x
⎜ 2 ⎟
⎝
⎠
27. Step 1:
x3 + 2
y=
x3 + 1
x3 y – x3 = 2 – y
x3 =
2– y
y –1
1/ 3
⎛2– y⎞
x=⎜
⎟
⎝ y –1 ⎠
1+ y
( y) =
1− y
1/ 3
1+ x
1– x
⎛2– y⎞
Step 2: f –1 ( y ) = ⎜
⎟
⎝ y –1 ⎠
x –1
x +1
x –1
x +1
⎛2– x⎞
Step 3: f –1 ( x) = ⎜
⎟
⎝ x –1 ⎠
Check:
Check:
1/ 3
f –1 ( f ( x)) =
( x)) =
1+
1–
1+ x
1– x
1+ x
1– x
=
x + 1 + x –1 2 x
=
=x
x +1 – x +1 2
1 + x –1 + x 2 x
=
=
=x
+1 1+ x +1 – x 2
–1
26. Step 1:
3
⎛ x –1 ⎞
y=⎜
⎟
⎝ x +1⎠
x –1
y1/ 3 =
x +1
xy1/ 3 + y1/ 3 = x –1
x – xy1/ 3 = 1 + y1/ 3
x=
( xx+–11 )
x3 y + y = x3 + 2
Step 3: f –1 ( x) =
f(f
⎡
1+ ⎢
⎣
f –1 ( f ( x)) =
3
f –1 ( f ( x)) = ( x5 / 2 ) 2 / 5 = x
–1
1 – x1/ 3
3
Step 2: f –1 ( y ) = y 2 / 5
Step 2: f
1 + x1/ 3
⎛ 1+ x1/ 3 – 1 ⎞
3
⎛ 1 + x1/ 3 – 1 + x1/ 3 ⎞
⎜ 1– x1/ 3
⎟
–1
f ( f ( x)) = ⎜
⎟
⎟ = ⎜⎜
1/ 3
1 + x1/ 3 + 1 – x1/ 3 ⎟⎠
⎜⎜ 1+ x + 1 ⎟⎟
⎝
⎝ 1– x1/ 3
⎠
2/5
–1
1 – y1/ 3
Check:
Step 2: f –1 ( y ) = 1 + 3 y
Check: f
1 + y1/ 3
Step 2: f –1 ( y ) =
3
1 + y1/ 3
1 – y1/ 3
Instructor’s Resource Manual
1/ 3
⎛ 2 – x3 + 2 ⎞
⎜
x3 +1 ⎟
f –1 ( f ( x)) = ⎜
⎟
3
⎜⎜ x + 2 –1 ⎟⎟
⎝ x3 +1
⎠
1/ 3
⎛ 2 x3 + 2 – x3 – 2 ⎞
=⎜
⎟
⎜ x3 + 2 – x3 –1 ⎟
⎝
⎠
1/ 3
⎛ x3 ⎞
=⎜ ⎟
⎜ 1 ⎟
⎝ ⎠
=x
( )
3
⎡ 2– x 1/ 3 ⎤
x +2
⎢⎣ x –1
⎥⎦ + 2 2–
–1
= x –1
f ( f ( x)) =
3
2– x + 1
⎡ 2– x 1/ 3 ⎤
x –1
1
+
⎢⎣ x –1
⎥⎦
2 – x + 2x – 2 x
=
= =x
2 – x + x –1 1
( )
Section 6.2
355
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
28. Step 1:
⎛ x3 + 2 ⎞
y=⎜
⎟
⎜ x3 + 1 ⎟
⎝
⎠
1/ 5
y
29. By similar triangles,
5
This gives
x3 + 2
=
V=
3
V
27
h3 =
4π
x3 + 1
x3 y1/ 5 + y1/ 5 = x3 + 2
x3 y1/ 5 – x3 = 2 – y1/ 5
x3 =
h = 33
2 – y1/ 5
y1/ 5 – 1
⎛2– y
Step 2: f –1 ( y ) = ⎜
⎟
⎜ y1/ 5 – 1 ⎟
⎝
⎠
1/ 3
3
=
4π h3
27
V
4π
v
v2
v2
H = s (v0 / 32) = v0 0 − 16 0 = 0
32
322 64
Check:
v02 = 64 H
1/ 3
5 ⎤1/ 5 ⎫
⎧
⎡
⎪ 2 – ⎢⎛ x3 + 2 ⎞ ⎥ ⎪
⎜ 3 ⎟
⎪⎪
⎢⎣⎝ x +1 ⎠ ⎥⎦ ⎪⎪
–1
f ( f ( x)) = ⎨
⎬
1/ 5
⎪ ⎡ ⎛ 3 ⎞5 ⎤
⎪
⎪ ⎢ ⎜ x 3+ 2 ⎟ ⎥ – 1 ⎪
⎪⎩ ⎢⎣⎝ x +1 ⎠ ⎥⎦
⎪⎭
v0 = 8 H
31.
f ′( x) = 4 x + 1; f ′( x) > 0 when x > −
1
and
4
1
f ′( x) < 0 when x < − .
4
1/ 3
1/ 3
⎛ 2 x3 + 2 – x3 – 2 ⎞
=⎜
⎟
3
⎜ 3
⎟
⎝ x + 2 – x –1 ⎠
1/ 3
⎛ x3 ⎞
=⎜ ⎟
⎜ 1 ⎟
⎝ ⎠
)
s (t ) = v0t − 16t 2 . The ball then reaches a height
of
⎛ 2 – x1/ 5 ⎞
( x) = ⎜
⎟
⎜ x1/ 5 – 1 ⎟
⎝
⎠
⎛ 2 – x3 + 2 ⎞
⎜
x3 +1 ⎟
=⎜
⎟
3
⎜⎜ x + 2 – 1 ⎟⎟
⎝ x3 +1
⎠
(
v
t = 0 . The position function is
32
1/ 5 ⎞1/ 3
Step 3: f
=
π 4h 2 / 9 h
30. v = v0 − 32t
v = 0 when v0 = 32t , that is, when
1/ 3
⎛ 2 – y1/ 5 ⎞
x=⎜
⎟
⎜ y1/ 5 – 1 ⎟
⎝
⎠
–1
π r 2h
r 4
2h
= . Thus, r =
h 6
3
=x
5
⎧⎡
⎫
1/ 3 ⎤ 3
⎪ ⎢⎛ 2– x1/ 5 ⎞ ⎥ + 2 ⎪
⎜
⎟
⎪⎪ ⎢⎝ x1/ 5 –1 ⎠ ⎥
⎪⎪
⎦
f ( f –1 ( x)) = ⎨ ⎣
⎬
3
⎪ ⎡⎛ 1/ 5 ⎞1/ 3 ⎤
⎪
2– x
⎪ ⎢⎜ 1/ 5 ⎟ ⎥ + 1 ⎪
⎩⎪ ⎣⎢⎝ x –1 ⎠ ⎦⎥
⎭⎪
1⎤
⎛
The function is decreasing on ⎜ −∞, − ⎥ and
4⎦
⎝
⎡ 1 ⎞
increasing on ⎢ − , ∞ ⎟ . Restrict the domain to
⎣ 4 ⎠
1⎤
⎛
⎡ 1 ⎞
⎜ −∞, − ⎥ or restrict it to ⎢ − , ∞ ⎟ .
4⎦
⎣ 4 ⎠
⎝
Then f −1 ( x) =
f −1 ( x ) =
1
(−1 − 8 x + 33) or
4
1
(−1 + 8 x + 33).
4
5
⎛ 2– x1/ 5
⎞
5
⎛ 2 – x1/ 5 + 2 x1/ 5 – 2 ⎞
⎜ x1/ 5 –1 + 2 ⎟
=⎜
=
⎜
⎟
⎟
1/ 5
⎜ 2 – x1/ 5 + x1/ 5 – 1 ⎟
⎜⎜ 2– x + 1 ⎟⎟
⎝
⎠
1/
5
⎝ x –1
⎠
5
⎛ x1/ 5 ⎞
=⎜
⎟ =x
⎜ 1 ⎟
⎝
⎠
356
Section 6.2
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
32.
f ′( x) = 2 x − 3; f ′( x) > 0 when x >
3
2
36.
( f −1 )′(3) ≈
1
2
3
and f ′( x) < 0 when x < .
2
⎛
3⎤
The function is decreasing on ⎜ −∞, ⎥ and
2⎦
⎝
⎡3 ⎞
increasing on ⎢ , ∞ ⎟ . Restrict the domain to
⎣2 ⎠
⎛
3⎤
⎡3 ⎞
⎜ −∞, ⎥ or restrict it to ⎢ , ∞ ⎟ . Then
2⎦
⎣2 ⎠
⎝
1
(3 − 4 x + 5) or
2
1
f −1 ( x) = (3 + 4 x + 5).
2
f −1 ( x ) =
37.
f ′( x) = 15 x 4 + 1 and y = 2 corresponds to x = 1,
so ( f −1 )′(2) =
33.
38.
f ′( x) = 5 x 4 + 5 and y = 2 corresponds to x = 1,
so ( f −1 )′(2) =
39.
1
1
1
=
= .
f ′(1) 15 + 1 16
1
1
1
=
=
f ′(1) 5 + 5 10
π
,
4
1
1
1
⎛π⎞
so ( f −1 )′(2) =
=
= cos 2 ⎜ ⎟
2
π
π
2
⎝4⎠
f′ 4
2sec 4
f ′( x) = 2sec2 x and y = 2 corresponds to x =
( )
(f
34.
−1
)′(3) ≈
=
1
3
( f −1 )′(3) ≈ −
1
2
40.
( )
1
.
4
f ′( x) =
1
2 x +1
so ( f −1 )′(2) =
and y = 2 corresponds to x = 3,
1
= 2 3 +1 = 4 .
f ′(3)
41. ( g –1 D f –1 )(h( x)) = ( g –1 D f –1 )( f ( g ( x)))
= g –1 D [ f –1 ( f ( g ( x)))] = g –1 D [ g ( x)] = x
Similarly,
h(( g –1 D f –1 )( x)) = f ( g (( g –1 D f –1 )( x)))
= f ( g ( g –1 ( f –1 ( x )))) = f ( f –1 ( x)) = x
Thus h –1 = g –1 D f –1
35.
( f −1 )′(3) ≈ −
1
3
Instructor’s Resource Manual
Section 6.2
357
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
42. Find f −1 ( x) :
y=
44. a.
1
1
, x=
x
y
f −1 ( y ) =
1
y
f −1 ( x ) =
1
x
Find g −1 ( x) :
y = 3x + 2
y−2
x=
3
y−2
g −1 ( y ) =
3
x
−
2
g −1 ( x) =
3
c.
1
3x + 2
(1)
−2
⎛1⎞
h −1 ( x) = g −1 ( f −1 ( x)) = g −1 ⎜ ⎟ = x
3
⎝ x⎠
⎛ 1 ⎞ (3x + 2) − 2 3x
h −1 (h( x)) = h −1 ⎜
=
=x
⎟=
3
3
⎝ 3x + 2 ⎠
dy − b
cy − a
f −1 ( x) = −
dx − b
cx − a
( )
+(d 2 − bc) x − bd = 0
(ac + dc) x 2 + (d 2 − a 2 ) x + (− ab − bd ) = 0
Setting the coefficients equal to 0 gives three
requirements:
(1) a = –d or c = 0
(2) a = ±d
(3) a = –d or b = 0
( )
( )
If a = d, then f = f −1 requires b = 0 and
43. f has an inverse because it is monotonic
(increasing):
ax
= x . If a = –d, there are
d
no requirements on b and c (other than
c = 0, so f ( x) =
f ′( x) = 1 + cos 2 x > 0
a.
( f −1 )′( A) =
b.
( f −1 )′( B ) =
1
( )
f ′ π2
1
=
( )
f ′ 56π
If f = f −1 , then for all x in the domain we
have:
ax + b dx − b
+
=0
cx + d cx − a
(ax + b)(cx – a) + (dx – b)(cx + d) = 0
acx 2 + (bc − a 2 ) x − ab + dcx 2
⎛ 1 −2⎞
1
1
⎟=
h(h −1 ( x)) = h ⎜ x
=
=x
1
⎜ 3 ⎟ ⎡ 1 − 2⎤ + 2
x
⎝
⎠ ⎣ x
⎦
c.
f −1 ( y ) = −
b. If bc – ad = 0, then f(x) is either a constant
function or undefined.
h( x ) = f ( g ( x)) = f (3x + 2) =
=
ax + b
cx + d
cxy + dy = ax + b
(cy – a)x = b – dy
b − dy
dy − b
x=
=−
cy − a
cy − a
y=
=
1
( )
1 + cos 2 π2
1
( )
1 + cos 2 56π
bc − ad ≠ 0 ). Therefore, f = f −1 if a = –d
or if f is the identity function.
=1
=
1
45.
7
4
2
7
( f −1 )′(0) =
1
1
1
=
=
2
f ′(0)
2
1 + cos (0)
1 −1
∫0 f
( y ) dy = (Area of region B)
= 1 – (Area of region A)
1
2 3
= 1 − ∫ f ( x) dx = 1 − =
0
5 5
358
Section 6.2
Instructor’s Resource Manual
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
46.
a
∫0
f ( x)dx = the area bounded by y = f(x), y = 0,
47. Given p > 1, q > 1,
and x = a [the area under the curve].
b
f –1 ( y )dy = the area bounded by x = f –1 ( y )
solving
x = 0, and y = b.
ab = the area of the rectangle bounded by x = 0,
x = a, y = 0, and y = b.
Case 1: b > f(a)
1
=
p –1
∫0
1 1
+ = 1, and f ( x) = x p –1 ,
p q
1 1
q
+ = 1 for p gives p =
, so
p q
q –1
1
q
–1
q –1
=
1
=
⎡ q –( q –1) ⎤
⎣⎢ q –1 ⎦⎥
q –1
= q – 1.
1
1
Thus, if y = x p –1 then x = y p –1 = y q –1 , so
f –1 ( y ) = y q –1.
By Problem 44, since f ( x) = x p −1 is strictly
a
b
0
0
increasing for p > 1, ab ≤ ∫ x p –1dx + ∫ y q –1dy
a
The area above the curve is greater than the area
of the part of the rectangle above the curve, so
the total area represented by the sum of the two
integrals is greater than the area ab of the
rectangle.
Case 2: b = f(a)
b
⎡xp ⎤
⎡ yq ⎤
ab ≤ ⎢ ⎥ + ⎢ ⎥
⎣⎢ p ⎦⎥ 0 ⎣⎢ q ⎥⎦ 0
a p bq
ab ≤
+
p
q
6.3 Concepts Review
1. increasing; exp
2. ln e = 1; 2.72
3. x; x
The area represented by the sum of the two
integrals = the area ab of the rectangle.
Case 3: b < f(a)
4. e x ; e x + C
Problem Set 6.3
1. a.
20.086
b. 8.1662
c.
e 2 ≈ e1.41 ≈ 4.1
d.
ecos(ln 4) ≈ e0.18 ≈ 1.20
2. a.
The area below the curve is greater than the area
of the part of the rectangle which is below the
curve, so the total area represented by the sum of
the two integrals is greater than the area ab of the
rectangle.
ab ≤ ∫ f ( x) dx + ∫ f −1 ( y ) dy with equality
a
b
0
0
holding when b = f(a).
Instructor’s Resource Manual
b.
3
e3ln 2 = eln(2 ) = eln 8 = 8
e
ln 64
2
1/ 2 )
= eln(64
= eln 8 = 8
3
3. e3ln x = eln x = x3
4. e –2 ln x = eln x
−2
= x −2 =
1
x2
Section 6.3
359
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5. ln ecos x = cos x
19.
6. ln e –2 x –3 = –2 x – 3
ex
ex
x
=
eln x
2 – y ln x
=
eln x
e
2
x2
=
y ln x
e
ln x y
x2
=
x
y
= x 2– y
11. Dx e x + 2 = e x + 2 Dx ( x + 2) = e x + 2
12. Dx e
=e
2 x2 – x
2 x2 – x
14. Dx
– 1
2
e x
2 x2 – x
=e
x+ 2
Dx (2 x – x)
=−
Dx x + 2 =
e
x+2
2 x+2
⋅ 2x
ln x 2
x2
–3
=
x2
xe
x
+
−2
2
1 ⎤
⎥ = Dx e x + Dx e − x
2
e x ⎥⎦
−2
Dx x −2 + e− x Dx [− x 2 ]
−2
⋅ (−2 x −3 ) + e− x ⋅ (−2 x)
2
2
x2
2e1
3
−
2x
ex
2
21. Dx [e xy + xy ] = Dx [2]
e xy ( xDx y + y ) + ( xDx y + y ) = 0
xe xy Dx y + ye xy + xDx y + y = 0
xe xy Dx y + xDx y = – ye xy – y
– 1
2
2e x
x3
= Dx x 2 = 2 x
1
x
x
x (ln x ) ⋅1 – x ⋅
x
x
= e ln x ⋅
16. Dx e ln x = e ln x Dx
2
ln x
(ln x)
=
x2
x
⎞
⎟
⎝ x ⎠
15. Dx e2 ln x = Dx e
⎡
20. Dx ⎢e1
⎢⎣
= ex
2
– 1
⎛ 1
2
= e x Dx ⎜ –
2
– 1
2
=e x
2
= x ex +
= ex
(4 x –1)
x+2
13. Dx e
=e
2
] = Dx (e x )1 2 + Dx e
2
2
1 x2 −1 2
Dx e x + e x Dx x 2
(e )
2
2
2
1 2
x
= (e x )−1 2 e x Dx x 2 + e x ⋅
2
x2
2 x
1 2
= (e x )1 2 2 x + e x ⋅
x
2
2
9. eln 3+ 2 ln x = eln 3 ⋅ e2 ln x = 3 ⋅ eln x = 3 x 2
10. eln x
x2
=
7. ln( x3e –3 x ) = ln x3 + ln e –3 x = 3ln x – 3 x
8. e x –ln x =
2
Dx [ e x + e
x
e ln x (ln x –1)
Dx y =
− ye xy – y
xe
xy
+x
=–
y (e xy + 1)
x (e
xy
+ 1)
=–
y
x
22. Dx [e x + y ] = Dx [4 + x + y ]
e x + y (1 + Dx y ) = 1 + Dx y
e x + y + e x + y Dx y = 1 + Dx y
e x + y Dx y – Dx y = 1 – e x + y
Dx y =
1 – e x+ y
e x + y –1
= –1
23. a.
(ln x) 2
17. Dx ( x3e x ) = x3 Dx e x + e x Dx ( x3 )
= x3e x + e x ⋅ 3x 2 = x 2 e x ( x + 3)
18. Dx e x
= ex
= ex
3 ln x
3 ln x
Dx ( x3 ln x)
3 ln x ⎛ 3
1
2⎞
⎜ x ⋅ + ln x ⋅ 3x ⎟
x
⎝
⎠
3 ln x
= x2e x
360
= ex
( x 2 + 3x 2 ln x)
3 ln x
The graph of y = e x is reflected across the
x-axis.
(1 + 3ln x)
Section 6.3
Instructor’s Resource Manual
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
b.
26.
f ( x) = e
−x
2
Domain = (−∞, ∞)
1
1 −x
f ′( x) = − e 2 , f ′′( x) = e 2
2
4
Since f ′( x) < 0 for all x, f is decreasing on
(−∞, ∞) .
Since f ′′( x) > 0 for all x, f is concave upward on
(−∞, ∞) .
Since f and f ′ are both monotonic, there are no
extreme values or points of inflection.
−x
The graph of y = e x is reflected across the
y-axis.
y
24. a – b ⇒ e
increasing function.
–a
>e
–b
x
, since e is an
8
25.
f ( x) = e
Domain = (−∞, ∞)
2x
f ′( x) = 2e , f ′′( x) = 4e2 x
Since f ′( x) > 0 for all x, f is increasing on
(−∞, ∞) .
Since f ′′( x) > 0 for all x, f is concave upward on
(−∞, ∞) .
Since f and f ′ are both monotonic, there are no
extreme values or points of inflection.
2x
4
−5
27.
f ( x) = xe − x
Domain = (−∞, ∞)
f ′( x) = (1 − x)e− x ,
y
(−∞,1)
+
−
x
f′
f ′′
8
x
5
f ′′( x) = ( x − 2)e − x
1
0
−
(1, 2)
−
−
(2, ∞)
−
+
2
−
0
f is increasing on (−∞,1] and decreasing on
[1, ∞) . f has a maximum at (1, 1 )
e
f is concave up on (2, ∞) and concave down on
4
−2
2
x
(−∞, 2) . f has a point of inflection at (2, 2
y
e2
)
5
−3
8
x
−5
Instructor’s Resource Manual
Section 6.3
361
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
28.
f ( x) = e x + x
Domain = (−∞, ∞)
30.
f ( x) = ln(2 x − 1) . Since 2 x − 1 > 0 if and only if
f ′( x) = e x + 1 , f ′′( x) = e x
Since f ′( x) > 0 for all x, f is increasing on
x>
(−∞, ∞) .
Since f ′′( x) > 0 for all x, f is concave upward on
(−∞, ∞) .
Since f and f ′ are both monotonic, there are no
extreme values or points of inflection.
(2 x − 1) 2
Since f ′( x) > 0 for all domain values, f is
1
2
1
2
, domain = ( , ∞)
f ′( x) =
2
,
2x −1
f ′′( x) =
−4
1
2
increasing on ( , ∞) .
Since f ′′( x) < 0 for all domain values, f is
1
2
concave downward on ( , ∞) .
y
Since f and f ′ are both monotonic, there are no
extreme values or points of inflection.
5
y
−5
5
5
x
−5
x
8
29.
f ( x) = ln( x 2 + 1) Since x 2 + 1 > 0 for all x,
domain = (−∞, ∞)
f ′( x) =
2x
,
x +1
2
f ′′( x) =
−2( x − 1)
( x 2 + 1) 2
−5
2
x ( −∞ , −1) −1 ( −1,0) 0 (0,1) 1 (1,∞ )
f′
0
−
−
−
+
+
+
f ′′
0
0
−
+
+
+
−
f is increasing on (0, ∞) and decreasing on
(−∞, 0) . f has a minimum at (0, 0)
f is concave up on (−1,1) and concave down on
(−∞, −1) ∪ (1, ∞) . f has points of inflection at
(−1, ln 2) and (1, ln 2)
y
31.
f ( x) = ln(1 + e x ) Since 1 + e x > 0 for all x,
domain = (−∞, ∞)
f ′( x) =
ex
,
f ′′( x) =
ex
1 + ex
(1 + e x ) 2
Since f ′( x) > 0 for all x, f is increasing on
(−∞, ∞) .
Since f ′′( x) > 0 for all x, f is concave upward on
(−∞, ∞) .
Since f and f ′ are both monotonic, there are no
extreme values or points of inflection.
y
5
5
−5
5
x
−5
5
x
−5
−5
362
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32.
f ( x) = e1− x
2
f ′( x) = −2 xe1− x ,
f ′′( x) = (4 x 2 − 2)e1− x
2
x
( −∞ , −
f′
+
y
Domain = (−∞, ∞)
2
2
) −
2
2
(−
+
2
2
,0) 0 (0, )
2
2
2
2
+
−
−
0
3
2
(
2
,∞ )
2
−
−1
f ′′
+
−
0
−
−
+
0
f is increasing on (−∞, 0] and decreasing on
[0, ∞) . f has a maximum at (0, e)
f is concave up on (−∞, −
concave down on (−
inflection at (−
2
2
2
2
,
2
2
) ∪ ( , ∞ ) and
2
2
2
).
2
, e ) and (
−3
34.
f ( x) = e x − e− x
f ′( x) = e + e
x
f has points of
2
2
x
4
2
, e)
y
−x
Domain = (−∞, ∞)
,
f ′′( x) = e x − e− x
x (−∞, 0) 0 (0, ∞)
f′
+
+
+
f ′′
−
+
0
f is increasing on (−∞, ∞) and so has no extreme
values. f is concave up on (0, ∞) and concave
down on (−∞, 0) . f has a point of inflection at
(0, 0)
3
y
−3
3
x
3
−3
33.
f ( x) = e − ( x − 2)
2
Domain = (−∞, ∞)
−3
3
x
f ′( x) = (4 − 2 x)e − ( x − 2) ,
2
f ′′( x) = (4 x 2 − 16 x + 14)e− ( x − 2)
2
−3
Note that 4 x 2 − 16 x + 14 = 0 when
x=
4± 2
≈ 2 ± 0.707
2
x ( −∞ ,1.293) ≈1.293 (1.293,2) 2 (2,2.707) ≈ 2.707 (2.707,∞ )
f′
+
+
+
−
−
−
0
f ′′
+
−
−
−
+
0
0
f is increasing on (−∞, 2] and decreasing on
[2, ∞) . f has a maximum at (2,1)
f is concave up on
(−∞, 4−2 2 ) ∪ ( 4+2 2 , ∞) and
concave down on
(
4− 2 4+ 2
,
) . f has points
2
2
4− 2
of inflection at ( 2
,
1
) and
e
Instructor’s Resource Manual
(
4+ 2
2
,
1
).
e
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
35.
f ( x) = ∫0 e − t dt
x
2
f ′( x) = e− x ,
2
Domain = (−∞, ∞)
f ′′( x) = −2 xe− x
2
x (−∞, 0) 0 (0, ∞)
f′
+
+
+
f ′′
+
0
−
f is increasing on (−∞, ∞) and so has no extreme
values. f is concave up on (−∞, 0) and concave
down on (0, ∞) . f has a point of inflection at
(0, 0)
37. Let u = 3x + 1, so du = 3dx.
1 3 x +1
1 u
1 u
3 x +1
∫ e dx = 3 ∫ e 3dx = 3 ∫ e du = 3 e + C
1
= e3 x +1 + C
3
38. Let u = x 2 − 3, so du = 2x dx.
1 x 2 −3
1 u
x 2 −3
∫ xe dx = 2 ∫ e 2 x dx = 2 ∫ e du
1
1 2
= eu + C = e x −3 + C
2
2
y
39. Let u = x 2 + 6 x , so du = (2x + 6)dx.
1
1
x2 +6 x
dx = ∫ eu du = eu + C
∫ ( x + 3)e
2
2
1 x2 +6 x
= e
+C
2
3
−3
3
x
40. Let u = e x − 1, so du = e x dx .
ex
−3
36.
f ( x) = ∫0 te− t dt
x
f ′( x) = xe
−x
,
( −∞ ,0)
−
+
x
f′
f ′′
1
∫ e x − 1dx = ∫ u du = ln u + C = ln e
Domain = (−∞, ∞)
f ′′( x) = (1 − x)e
0
0
+
(0,1)
+
+
−x
1
+
0
(1,∞ )
+
−
f is increasing on [0, ∞) and decreasing on
(−∞, 0] . f has a minimum at (0, 0)
f is concave up on (−∞,1) and concave down on
(1, ∞) . f has a point of inflection at
1
(1, ∫ te−t dt ) .
0
Note: It can be shown with techniques in
2
1
Chapter 7 that ∫0 te− t dt = 1 − ≈ 0.264
e
y
42.
∫e
x +e x
∫e
x
x
43. Let u = 2x + 3, so du = 2dx
1 u
1 u
1 2 x +3
2 x +3
∫ e dx = 2 ∫ e du = 2 e + C = 2 e + C
∫
9
x
⋅ ee dx = ∫ eu du = eu + C = ee + C
x
∫
1
1
1
⎡1
⎤
= ⎢ e2 x +3 ⎥ = e5 – e3
2
⎣2
⎦0 2
1 3 2
e (e − 1) ≈ 64.2
2
44. Let u =
−2
x
dx = ∫ e x ⋅ ee dx
Let u = e x , so du = e x dx.
=
−3
−1 + C
1
1
41. Let u = − , so du =
dx .
x
x2
e−1/ x
u
u
−1/ x
∫ x 2 dx = ∫ e du = e + C = e + C
1 2 x +3
e
dx
0
4
x
e3 / x
2
3
3
, so du = − dx.
x
x2
dx = –
1 u
1
e du = – eu + C
3∫
3
x
1
= – e3 / x + C
3
2 e3 / x
2
1 3/ 2 1 3
⎡ 1 3/ x ⎤
∫1 x 2 dx = ⎢⎣ – 3 e ⎥⎦1 = – 3 e + 3 e ≈ 5.2
364
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
45. V = π∫
ln 3
0
(e x )2 dx = π∫
ln 3 2 x
e dx
0
ln 3
⎡1
⎤
= π ⎢ e2 x ⎥
⎣2
⎦0
1 ⎞
⎛1
= π ⎜ e2 ln 3 − e0 ⎟ = 4π ≈ 12.57
2 ⎠
⎝2
2
46. V = ∫ 2πxe− x dx .
1
y = et cos t , so dy = (et cos t − et sin t )dt
Let u = − x 2 , so du = –2x dx.
2
ds = dx 2 + dy 2
2
−x
−x
u
∫ 2πxe dx = −π∫ e (−2 x)dx = −π∫ e du
= et (sin t + cos t )2 + (cos t − sin t )2 dt
2
= −πeu + C = −πe− x + C
∫0 2πxe
− x2
= et 2sin 2 t + 2 cos 2 tdt = 2et dt
The length of the curve is
1
⎡ 2⎤
dx = −π ⎢ e− x ⎥ = – π(e−1 − e0 )
⎣
⎦0
π
∫0
−1
= π(1 − e ) ≈ 1.99
1
1− e
1− e
( x − 0);
−1 =
⇒ y −1 =
1− 0 e
e
e
1− e
y=
x +1
e
=
53. a.
⎤
⎞
⎡1 − e 2
⎤
x + 1⎟ − e− x ⎥ dx = ⎢
x + x + e− x ⎥
2
e
⎠
⎣
⎦0
⎦
1− e
1
3−e
=
+1+ −1 =
≈ 0.052
2e
e
2e
48.
f ′( x) =
=
=
(e x –1)2
e x − 1 − xe x
(e x − 1) 2
e x − 1 − xe x
(e x − 1) 2
=−
−
−
1
1 − e− x
1
ex −1
–
1
1 – e– x
(– e
= lim
x →0+
b.
e x − 1 − xe x − (e x − 1)
(e x − 1) 2
Dx [1 + (ln x) 2 ]
ln x
f ′( x) =
=
= lim
x →0+
∞
.
∞
1
x
2 ln x ⋅ 1x
1
=0
2 ln x
x →∞ 1 + (ln x) 2
)(–1)
⎛ 1 ⎞
⎜ x⎟
⎝e ⎠
=
Dx ln x
lim
–x
is of the form
2
x →0+ 1 + (ln x)
x →0+
e
(e x –1)(1) – x(e x )
ln x
lim
= lim
1
1 ⎡⎛ 1 − e
∫0 ⎢⎣⎜⎝
π
2et dt = 2 ⎡ et ⎤ = 2(eπ − 1) ≈ 31.312
⎣ ⎦0
52. Use x = 30, n = 8, and k = 0.25.
(kx) n e− kx (0.25 ⋅ 30)8 e−0.25⋅30
≈ 0.14
Pn ( x) =
=
n!
8!
⎛ 1⎞
47. The line through (0, 1) and ⎜1, ⎟ has slope
⎝ e⎠
1 −1
e
e0.3 ≈ 1.3498588 by direct calculation
51. x = et sin t , so dx = (et sin t + et cos t )dt
0
1
⎧ ⎡⎛ 0.3 ⎞ 0.3 ⎤ 0.3 ⎫
50. e0.3 ≈ ⎨ ⎢⎜
+ 1⎟
+ 1⎥
+ 1⎬ ( 0.3) + 1
⎠ 3
⎦ 2
⎩ ⎣⎝ 4
⎭
= 1.3498375
1
= lim
=0
x →∞ 2 ln x
[1 + (ln x) 2 ] ⋅ 1x – ln x ⋅ 2 ln x ⋅ 1x
[1 + (ln x) 2 ]2
1 – (ln x )2
x[1 + (ln x) 2 ]2
f ′( x) = 0 when ln x = ±1 so x = e1 = e
xe x
(e x − 1) 2
When x > 0, f ′( x) < 0, so f(x) is decreasing for
x > 0.
49. a. Exact:
10! = 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1
= 3, 628,800
Approximate:
10
⎛ 10 ⎞
10! ≈ 20π ⎜ ⎟
⎝ e ⎠
⎛ 60 ⎞
b. 60! ≈ 120π ⎜ ⎟
⎝ e ⎠
≈ 3,598, 696
1
e
ln e
or x = e –1 =
f (e) =
1 + (ln e)
=
1
1+1
2
=
1
2
ln 1e
–1
1
⎛1⎞
f ⎜ ⎟=
=
=–
2
2
2
⎝ e ⎠ 1 + ln 1
1 + (–1)
e
( )
Maximum value of
value of −
60
2
1
at x = e; minimum
2
1
at x = e−1.
2
≈ 8.31× 1081
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Section 6.3
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
F ( x) = ∫
c.
x2
ln t
1 + (ln t )2
1
ln x 2
F ′( x) =
1 + (ln x 2 )2
F ′( e ) =
ln( e )
y-axis so the area is
⎧⎪ 3
2
2
2 ⎨ ∫ 2 [e− x − 2e x (2 x 2 − 1)] dx
0
⎪⎩
dt
⋅ 2x
2
1 + [ln( e ) ]
2 2
⋅2 e =
1
1 + 12
+∫
⋅2 e
= e ≈ 1.65
e x0 – 0
= f ′( x0 ) = e x0 ⇒ e x0 = x0 e x0 ⇒ x0 = 1
x0 – 0
1
a.
b.
2
⎫⎪
(2 x 2 − 1) − e− x ] dx ⎬
⎪⎭
x →∞
f ′( x) = x p e – x (–1) + e – x ⋅ px p –1
b.
= x p –1e – x ( p – x)
f ′( x) = 0 when x = p
so the line is y = e x0 x or y = ex.
59.
⎡
ex 2 ⎤
A = ∫ (e – ex)dx = ⎢ e x –
⎥
0
2 ⎥⎦
⎣⎢
0
e
e
= e − − (e0 − 0) = –1 ≈ 0.36
2
2
− x2
lim x p e – x = 0
58. a.
x
3 [2e
2
≈ 4.2614
54. Let ( x0 , e x0 ) be the point of tangency. Then
1
3
lim ln( x 2 + e – x ) = ∞ (behaves like − x )
x→ – ∞
lim ln( x 2 + e – x ) = ∞ (behaves like 2 ln x )
x →∞
60.
1
V = π∫ [(e x )2 – (ex) 2 ]dx
0
2 3 ⎤1
⎡1
1
e x
= π∫ (e2 x – e2 x 2 )dx = π ⎢ e2 x –
⎥
0
3 ⎥⎦
⎢⎣ 2
0
–1
–1
f ′( x) = –(1 + e x ) –2 ⋅ e x (– x –2 )
=
e1/ x
x 2 (1 + e1/ x )2
⎡1
π
e2 ⎛ 1 ⎞ ⎤
= π ⎢ e2 − − ⎜ e 0 ⎟ ⎥ = (e2 – 3) ≈ 2.30
6
3 ⎝ 2 ⎠ ⎥⎦
⎢⎣ 2
⎛
3
1 ⎞
3
⎛
1 ⎞
∫−3 exp ⎜⎝ − x2 ⎟⎠ dx = 2∫0 exp ⎜⎝ − x2 ⎟⎠ dx ≈ 3.11
55. a.
8π −0.1x
∫0
b.
e
sin x dx ≈ 0.910
lim (1 + x)1 x = e ≈ 2.72
56. a.
x →0
lim (1 + x)−1 x =
b.
x →0
1
≈ 0.368
e
a.
57.
f ( x) = e
− x2
f ′( x) = −2 xe− x
b.
2
2
2
366
c.
2
e− x = 2e− x (2 x 2 − 1); 1 = 4 x 2 − 2;
d.
3
2
Both graphs are symmetric with respect to the
e.
4 x 2 − 3 = 0, x = ±
Section 6.3
lim f ( x) = 1
x →0 –
2
f ′′( x) = −2e− x + 4 x 2 e− x = 2e− x (2 x 2 − 1)
y = f(x) and y = f ′′( x) intersect when
2
lim f ( x) = 0
x →0 +
lim f ( x) =
x →±∞
1
2
lim f ′( x) = 0
x →0
f has no minimum or maximum values.
Instructor’s Resource Manual
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
x + 3 = 5x
3
x=
4
6.4 Concepts Review
3 ln π
1. e
; e x ln a
2. e
3.
9. log5 12 =
ln x
ln a
ln12
≈ 1.544
ln 5
10. log 7 0.11 =
4. ax a −1 ; a x ln a
Problem Set 6.4
ln 0.11
≈ –1.1343
ln 7
11. log11 (8.12)1/ 5 =
1 ln 8.12
≈ 0.1747
5 ln11
12. log10 (8.57)7 = 7
ln 8.57
≈ 6.5309
ln10
1. 2 x = 8 = 23 ; x = 3
2. x = 52 = 25
3. x = 43 / 2 = 8
14. x ln 5 = ln 13
ln13
x=
≈ 1.5937
ln 5
4. x = 64
4
x = 4 64 = 2 2
⎛ x⎞ 1
5. log9 ⎜ ⎟ =
⎝3⎠ 2
x
= 91/ 2 = 3
3
x=
1
2x
1
2⋅4
3
=
1
128
7. log 2 ( x + 3) – log 2 x = 2
log 2
x+3
=2
x
x+3
= 22 = 4
x
x + 3 = 4x
x=1
8. log5 ( x + 3) – log5 x = 1
log5
15. (2s – 3) ln 5 = ln 4
ln 4
2s – 3 =
ln 5
1⎛
ln 4 ⎞
s = ⎜3 +
⎟ ≈ 1.9307
2⎝
ln 5 ⎠
16.
x=9
6. 43 =
13. x ln 2 = ln 17
ln17
x=
≈ 4.08746
ln 2
x+3
=1
x
x+3 1
=5 =5
x
Instructor’s Resource Manual
1
θ –1
ln12 = ln 4
ln12
= θ –1
ln 4
ln12
θ = 1+
≈ 2.7925
ln 4
17. Dx (62 x ) = 62 x ln 6 ⋅ Dx (2 x) = 2 ⋅ 62 x ln 6
18. Dx (32 x
2 –3 x
) = 32 x
= (4 x – 3) ⋅ 32 x
19. Dx log3 e x =
2 –3 x
2 –3 x
ln 3 ⋅ Dx (2 x 2 – 3 x)
ln 3
1
⋅ Dx e x
e ln 3
ex
1
=
=
≈ 0.9102
x
e ln 3 ln 3
Alternate method:
x
Dx log3 e x = Dx ( x log3 e) = log3 e
=
ln e
1
=
≈ 0.9102
ln 3 ln 3
Section 6.4
367
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
20. Dx log10 ( x3 + 9) =
=
21.
3x
( x + 9) ln10
3
⋅ Dx ( x3 + 9)
26.
( x3 + 9) ln10
1
(1) + ln( z + 5) ⋅ 3z ln 3
z +5
⎡ 1
⎤
= 3z ⎢
+ ln( z + 5) ln 3⎥
⎣z +5
⎦
=
=–
1
1
(θ 2 – θ ) ln 3
ln 3
=
⋅ Dθ θ 2 – θ
ln10
ln10
2θ –1
2 θ 2 –θ
ln 3
ln10
27.
23. Let u = x 2 so du = 2xdx.
2
x
∫ x ⋅ 2 dx =
1 u
1 2u
2
du
=
⋅
+C
2∫
2 ln 2
2
24. Let u = 5x – 1, so du = 5 dx.
1
1 10u
u
5 x –1
10
10
dx
=
du
=
⋅
+C
∫
5∫
5 ln10
105 x –1
+C
5ln10
5
∫
x
x
=
1
2 x
dx.
5u
+C
ln 5
2⋅5 x
+C
ln 5
4
⎡5 x ⎤
5 ⎞
⎛ 25
∫1 x dx = 2 ⎢⎢ ln 5 ⎥⎥ = 2 ⎜⎝ ln 5 − ln 5 ⎟⎠
⎣
⎦1
40
=
≈ 24.85
ln 5
368
Section 6.4
2
2
d ( x2 )
d
= 10( x ) ln10 x 2 = 10( x ) 2 x ln10
10
dx
dx
d 2 10 d 20
(x ) =
x = 20 x19
dx
dx
2
dy d
= [10( x ) + ( x 2 )10 ]
dx dx
d
d
sin 2 x = 2sin x sin x = 2sin x cos x
dx
dx
d sin x
d
2
= 2sin x ln 2 sin x = 2sin x ln 2 cos x
dx
dx
dy d
= (sin 2 x + 2sin x )
dx dx
= 2sin x cos x + 2sin x cos x ln 2
dx = 2∫ 5u du = 2 ⋅
45 x
–3 x
2
28.
25. Let u = x , so du =
3x
= 10( x ) 2 x ln10 + 20 x19
2
2x
2 x –1
=
+C =
+C
2 ln 2
ln 2
=
10 –3 x
+C
3ln10
⎡103 x –10 –3 x ⎤
Thus, ∫ (10 + 10 )dx = ⎢
⎥
0
⎢⎣ 3ln10 ⎥⎦ 0
1 ⎛
1 ⎞ 999,999
=
⎜1000 –
⎟=
3ln10 ⎝
1000 ⎠ 3000 ln10
≈ 144.76
) = Dθ (θ 2 – θ ) log10 3
ln 3 1 2
⋅ (θ – θ ) –1/ 2 (2θ –1)
ln10 2
=
1
0
103 x
+C
3ln10
Now let u = –3x, so du = –3dx.
1
1 10u
–3 x
u
10
–
10
–
dx
=
du
=
⋅
+C
∫
3∫
3 ln10
= 3z ⋅
= Dθ
1
0
+ 10 –3 x )dx = ∫ 103 x dx + ∫ 10 –3 x dx
=
Dz [3z ln( z + 5)]
2 –θ
3x
Let u = 3x, so du = 3dx.
1
1 10u
3x
u
10
10
dx
=
du
=
⋅
+C
∫
3∫
3 ln10
2
22. Dθ log10 (3θ
1
∫0 (10
29.
d π+1
x
= (π + 1) x π
dx
d
(π + 1) x = (π + 1) x ln(π + 1)
dx
dy d π+1
= [x
+ (π + 1) x ]
dx dx
= (π + 1) x π + (π + 1) x ln(π + 1)
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30.
x
x
d (e x )
d
= 2(e ) ln 2 e x = 2(e ) e x ln 2
2
dx
dx
d e x
(2 ) = (2e ) x ln 2e = (2e ) x e ln 2
dx
dy d (e x )
= [2
+ (2e ) x ]
dx dx
35.
f ′( x) = (− ln 2)2 , f ′′( x) = (ln 2) 2 2− x
Since f ′( x) < 0 for all x, f is decreasing on
(−∞, ∞) .
Since f ′′( x) > 0 for all x, f is concave upward on
(−∞, ∞) .
Since f and f ′ are both monotonic, there are no
extreme values or points of inflection.
x
2 +1)
y
2
dy
d
= e(ln x ) ln( x +1) [(ln x) ln( x 2 + 1)]
dx
dx
2
1
2x ⎤
⎡
= e(ln x ) ln( x +1) ⎢ ln( x 2 + 1) + ln x
⎥
2
x + 1⎦
⎣x
4
⎛ ln( x 2 + 1) 2 x ln x ⎞
= ( x 2 + 1)ln x ⎜
+
⎟
⎜
x
x 2 + 1 ⎟⎠
⎝
32. y = (ln x 2 ) 2 x +3 = e(2 x +3) ln(ln x
dy
=e
dx
(2 x +3) ln(ln x 2 )
f ( x) = xsin x = esin x ln x
d
(sin x ln x )
dx
⎡
⎤
⎛1⎞
= esin x ln x ⎢ (sin x) ⎜ ⎟ + (cos x)(ln x ) ⎥
⎝ x⎠
⎣
⎦
f ′( x) = esin x ln x
⎛ sin x
⎞
= xsin x ⎜
+ cos x ln x ⎟
x
⎝
⎠
sin1
⎛
⎞
+ cos1ln1⎟ = sin1 ≈ 0.8415
f ′(1) = 1sin1 ⎜
⎝ 1
⎠
34.
f (e) = πe ≈ 22.46
g (e) = e π ≈ 23.14
g(e) is larger than f(e).
d
f ′( x) = π x = π x ln π
dx
−3
9
2)
d
[(2 x + 3) ln(ln x 2 )]
dx
2 ⎡
1 1
⎤
= e(2 x +3) ln(ln x ) ⎢ 2 ln(ln x 2 ) + (2 x + 3)
(2 x) ⎥
2 2
ln x x
⎣
⎦
⎡
⎤
2 x +3 ⎢
2x + 3 ⎥
2
ln
(2
ln
x
)
= (2 ln x)
+
⎢
x ln x ⎥
⎢
⎥⎦
2
2
ln x
ln x
⎣
33.
Domain = (−∞, ∞)
−x
= 2(e ) e x ln 2 + (2e ) x e ln 2
31. y = ( x 2 + 1)ln x = e(ln x ) ln( x
f ( x) = 2− x = e(ln 2)( − x )
x
−2
36.
f ( x) = x 2− x
Domain = (−∞, ∞)
f ′( x) = [1 − (ln 2) x]2− x ,
f ′′( x) = (ln 2)[(ln 2) x − 2]2− x
1
)
ln 2
x
( −∞ ,
f′
+
f ′′
−
1
ln 2
1 2
,
)
ln 2 ln 2
2
ln 2
0
−
−
−
−
−
0
+
(
(
2
,∞ )
ln 2
⎛
1 ⎤
f is increasing on ⎜ −∞,
⎥ and decreasing on
ln
2⎦
⎝
⎡ 1
⎞
1 1
, ∞ ⎟ . f has a maximum at (
,
)
⎢
ln
2
ln
2 (e ln 2)
⎣
⎠
f is concave up on (
(−∞,
(
2
, ∞) and concave down on
ln 2
2
) . f has a point of inflection at
ln 2
2 2
,
)
(e 2 ln 2)
ln 2
y
3
f ′(e) = πe ln π ≈ 25.71
g ′( x) =
d π
x = πx π−1
dx
−2
8
x
g ′(e) = πe π−1 ≈ 26.74
g ′(e) is larger than f ′(e) .
−3
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37.
y
ln( x 2 + 1)
. Since
f ( x) = log 2 ( x + 1) =
ln 2
2
5
x 2 + 1 > 0 for all x, domain = (−∞, ∞)
⎛ 2 ⎞⎛ x ⎞
⎛ 2 ⎞ ⎛ 1 − x2
f ′( x) = ⎜
⎟⎜ 2
⎟ ⎜⎜ 2
⎟ , f ′′( x) = ⎜
2
⎝ ln 2 ⎠ ⎝ x + 1 ⎠
⎝ ln 2 ⎠ ⎝ ( x + 1)
x
⎞
⎟⎟
⎠
−5
5
x
(−∞, −1) −1 (−1, 0) 0 (0,1) 1 (1, ∞)
f′
f ′′
−
−
−
0
+
+
+
−
0
+
+
+
0
−
f is increasing on [0, ∞) and decreasing on
(−∞, 0] . f has a minimum at (0, 0)
f is concave up on (−1,1) and concave down on
(−∞, −1) ∪ (1, ∞) . f has points of inflection at
(−1,1) and (1,1)
−5
39.
f ( x) = ∫1 2− t dt
x
2
f ′( x) = 2− x ,
Domain = (−∞, ∞)
f ′′( x) = −2(ln 2) x 2− x
2
2
(−∞, 0) 0 (0, ∞)
x
f′
f ′′
+
+
+
+
0
−
f is increasing on (−∞, ∞) and so has no extreme
values.
f is concave up on (−∞, 0) and concave down on
(0, ∞) . f has a point of inflection at
y
5
(0, ∫1 2− t dt ) ≈ (0, −0.81)
0
−5
5
2
x
y
5
−5
38.
f ( x) = x log3 ( x 2 + 1) =
x ln( x 2 + 1)
. Since
ln 3
−5
5
x
x 2 + 1 > 0 for all x, domain = (−∞, ∞)
f ′( x ) =
x
⎤
1 ⎡ 2 x2
2 ⎡ x3 + 3 x ⎤
⎢
⎢
⎥
+ ln( x 2 +1) ⎥ , f ′′( x) =
ln 3 ⎢ x 2 +1
ln 3 ⎢ x 2 +1 ⎥
⎥⎦
⎣
⎣
⎦
−5
(−∞, 0) 0 (0, ∞)
f′
+
0
+
f ′′
−
0
+
f is increasing on (−∞, ∞) and so has no extreme
values. f is concave up on (0, ∞) and concave
down on (−∞, 0) . f has a point of inflection at
(0, 0)
370
Section 6.4
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
40.
x
f ( x) = ∫0 log10 (t 2 + 1)dt . Since log10 (t 2 + 1) has
domain = (−∞, ∞) , f also has domain = (−∞, ∞)
ln( x 2 + 1)
f ′( x) = log10 ( x + 1) =
,
ln10
2
⎛ 1 ⎞ ⎛ 2x ⎞
f ′′( x) = ⎜
⎟⎜ 2
⎟
⎝ ln10 ⎠ ⎝ x + 1 ⎠
x
P=
105.75
≈ 4636 lb/in.2
121.3
45. If r is the ratio between the frequencies of
successive notes, then the frequency of C = r12
(the frequency of C). Since C has twice the
(−∞, 0) 0 (0, ∞)
f′
f ′′
44. 115 = 20 log10 (121.3P )
log10 (121.3P ) = 5.75
+
0
+
frequency of C, r = 21/12 ≈ 1.0595
−
0
+
Frequency of C = 440(21/12 )3 = 440 4 2 ≈ 523.25
f is increasing on (−∞, ∞) and so has no extreme
values.
f is concave up on (0, ∞) and concave down on
(−∞, 0) . f has a point of inflection at (0, 0)
46. Assume log 2 3 =
p
where p and q are integers,
q
q ≠ 0 . Then 2 p q = 3 or 2 p = 3q. But
2 p = 2 ⋅ 2 … 2 (p times) and has only powers of 2
y
as factors and 3q = 3 ⋅ 3…3 (q times) and has
only powers of 3 as factors.
5
2 p = 3q only for p = q = 0 which contradicts our
assumption, so log 2 3 cannot be rational.
−5
5
x
If y = C ⋅ x d , then ln y = ln C + d ln x, so the
ln y vs. ln x plot will be linear.
−5
41. log1/ 2 x =
47. If y = A ⋅ b x , then ln y = ln A + x ln b, so the
ln y vs. x plot will be linear.
ln x
ln x
=
= − log 2 x
1
ln 2 − ln 2
42.
48. WRONG 1:
y = f ( x) g ( x )
y ′ = g ( x) f ( x) g ( x ) −1 f ′( x)
WRONG 2:
y = f ( x) g ( x )
y ′ = f ( x) g ( x ) (ln f ( x)) ⋅ g ′( x) = f ( x ) g ( x ) g ′( x) ln f ( x)
RIGHT:
y = f ( x) g ( x ) = e g ( x ) ln f ( x )
43. M = 0.67 log10 (0.37 E ) + 1.46
log10 (0.37 E ) =
E=
10
M − 1.46
0.67
M −1.46
0.67
0.37
Evaluating this expression for M = 7 and M = 8
y ′ = e g ( x ) ln f ( x )
d
[ g ( x) ln f ( x)]
dx
⎡
⎤
1
= f ( x) g ( x ) ⎢ g ′( x) ln f ( x) + g ( x)
f ′( x) ⎥
f ( x)
⎣
⎦
= f ( x) g ( x ) g ′( x ) ln f ( x) + f ( x) g ( x ) −1 g ( x) f ′( x)
Note that RIGHT = WRONG 2 + WRONG 1.
gives E ≈ 5.017 × 108 kW-h and
E ≈ 1.560 × 1010 kW-h, respectively.
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
49.
f ( x) = ( x x ) x = x ( x
f ( x) = x( x
2)
2)
≠ x( x
x)
= g ( x)
51. a.
2 ln x
= ex
2 ln x
d 2
( x ln x)
dx
2
1⎞
⎛
= e x ln x ⎜ 2 x ln x + x 2 ⋅ ⎟
x⎠
⎝
f ′( x) = e x
2
= x ( x ) (2 x ln x + x)
x
x
g ( x) = x ( x ) = e x ln x
Using the result from Example 5
⎛d x
⎞
x
⎜ x = x (1 + ln x) ⎟ :
dx
⎝
⎠
x ln x d
x
x
g ′( x) = e
( x ln x)
dx
x
1⎤
⎡
= e x ln x ⎢ x x (1 + ln x) ln x + x x ⋅ ⎥
x⎦
⎣
x
1
⎡
⎤
= x ( x ) x x ⎢(1 + ln x ) ln x + ⎥
x⎦
⎣
x
1
⎡
⎤
= x x + x ⎢ ln x + (ln x)2 + ⎥
x⎦
⎣
50.
f ( x) =
f ′( x) =
lim f ( x) = lim e
x →∞
c.
ax +1
(a x + 1) 2
=
2a x ln a
(a x + 1)2
Since a is positive, a x is always positive.
(a x + 1) 2 is also always positive, thus f ′( x) > 0
if ln a > 0 and f ′( x) < 0 if ln a < 0. f(x) is either
always increasing or always decreasing,
depending on a, so f(x) has an inverse.
ax −1
y=
ax +1
y (a x + 1) = a x − 1
a x ( y − 1) = −1 − y
ax =
1+ y
1− y
x ln a = ln
1+ y
1− y
1+ y
x=
372
ln 1− y
ln a
= log a
1+ y
1− y
f −1 ( y ) = log a
1+ y
1− y
f −1 ( x) = log a
1+ x
1− x
Section 6.4
x →∞
g ( x)
x →∞
= 0.
b. Again let g(x) = ln f(x) = a ln x – x ln a.
Since y = ln x is an increasing function, f(x)
is maximized when g(x) is maximized.
a ⎞
⎛a⎞
⎛
g ′( x) = ⎜ ⎟ − ln a, so g ′( x) > 0 on ⎜ 0,
⎟
⎝ x⎠
⎝ ln a ⎠
⎛ a
⎞
and g ′( x) < 0 on ⎜
, ∞ ⎟.
⎝ ln a ⎠
Therefore, g(x) (and hence f(x)) is
a
.
maximized at x0 =
ln a
a x −1
(a x + 1)a x ln a − (a x − 1)a x ln a
⎛ xa ⎞
Let g(x) = ln f(x) = ln ⎜ ⎟ = a ln x − x ln a .
⎜ ax ⎟
⎝ ⎠
a
⎛ ⎞
g ′( x) = ⎜ ⎟ − ln a
⎝ x⎠
a
g ′( x) < 0 when x >
, so as x → ∞ g(x)
ln a
a
is decreasing. g ′′( x) = −
, so g(x) is
x2
concave down. Thus, lim g ( x) = −∞, so
Note that x a = a x is equivalent to g(x) = 0.
a
By part b., g(x) is maximized at x0 =
.
ln a
If a = e, then
⎛ e ⎞
g ( x0 ) = g ⎜
⎟ = g (e) = e ln e − e ln e = 0.
⎝ ln e ⎠
Since g ( x) < g ( x0 ) = 0 for all x ≠ x0 , the
equation g(x) = 0 (and hence x a = a x ) has
just one positive solution. If a ≠ e , then
⎛ a ⎞
⎛ a ⎞ a
g ( x0 ) = g ⎜
(ln a )
⎟ = a ln ⎜
⎟−
⎝ ln a ⎠
⎝ ln a ⎠ ln a
⎡ ⎛ a ⎞ ⎤
= a ⎢ ln ⎜
⎟ − 1⎥ .
⎣ ⎝ ln a ⎠ ⎦
a
Now
> e (justified below), so
ln a
a
⎡
⎤
g ( x0 ) = a ⎢ln
− 1⎥ > a (ln e − 1) = 0. Since
⎣ ln a ⎦
g ′( x) > 0 on (0, x0 ), g ( x0 ) > 0, and
lim g ( x) = −∞, g(x) = 0 has exactly one
x →0
solution on (0, x0 ).
Since g ′( x) < 0 on ( x0 , ∞) ,
g ( x0 ) > 0, and lim g ( x) = −∞, g(x) = 0 has
x →∞
exactly one solution on ( x0 , ∞). Therefore,
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the equation g(x) = 0 (and hence x a = a x )
has exactly two positive solutions.
a
To show that
> e when a ≠ e :
ln a
x
Consider the function h( x ) =
, for x > 1.
ln x
h′( x) =
ln( x)(1) − x
( ) = ln x − 1
u →∞ u + 1
u
53.
= lim (− x) = 0
x →0+
x →0 +
g ′( x) = 1 + ln x
Since g ′( x) < 0 on ( 0,1/ e ) and g ′( x) > 0 on
π
implies x < e . In particular, π < e .
(1/ e, ∞ ) , g(x) has its minimum at
u −x
fu ( x ) = x e
()
(u, ∞ ), fu ( x) attains its maximum at x0 = u.
fu (u ) > fu (u + 1) means
u u e −u > (u + 1)u e −(u +1) .
x = 1e .
Therefore, f(x) has its minimum at (e −1 , e −1/ e ) .
Note: this point could also be written as
1⎞
⎛1
e
⎜ , 1e ⎟ .
⎜e
⎟
⎝
⎠
fu′ ( x ) = uxu −1e− x − xu e− x = (u − x) xu −1e− x
Since fu′ ( x ) > 0 on (0, u) and fu′ ( x ) < 0 on
b.
1
x
Therefore, lim x x = e0 = 1 .
Therefore, when x ≠ e , ln x < ln e , which
52. a.
x →0+
1
x
1
+
−
x →0
x2
= lim
x
e
f ( x) = x x = e x ln x
x →0 +
d. For the case a = e, part c. shows that
g ( x) = e ln x − x ln e < 0 for x ≠ e .
x
u
Let g ( x) = x ln x.
Using L’Hôpital’s Rule,
ln x
lim g ( x) = lim
(ln x) 2
(ln x)2
Note that h′( x) < 0 on (1, e) and h′( x) > 0
on (e, ∞ ), so h(x) has its minimum at (e, e).
x
> e for all x ≠ e , x > 1.
Therefore
ln x
e
e = e , this implies that
⎛ u +1⎞
⎛ 1⎞
lim ⎜
⎟ = e, i.e., lim ⎜ 1 + ⎟ = e .
u⎠
u →∞ ⎝ u ⎠
u →∞ ⎝
1
x
e
u
Since lim
54.
eu +1
u
⎛ u +1⎞
gives e > ⎜
⎟ .
u
⎝ u ⎠
u
fu +1 (u + 1) > fu +1 (u ) means
Multiplying by
(u + 1)u +1 e −(u +1) > u u +1e−u .
eu +1
⎛ u +1⎞
gives ⎜
Multiplying by
⎟
u +1
⎝ u ⎠
u
Combining the two inequalities,
u
⎛ u +1⎞
⎛ u +1⎞
⎜
⎟ <e<⎜
⎟
⎝ u ⎠
⎝ u ⎠
(2.4781, 15.2171), (3, 27)
u +1
u +1
.
> e.
55.
4 π sin x
∫0
x
dx ≈ 20.2259
56.
u +1
c.
⎛ u +1⎞
From part b., e < ⎜
⎟ .
⎝ u ⎠
u
Multiplying by
gives
u +1
u
u
⎛ u +1⎞
e<⎜
⎟ .
u +1
⎝ u ⎠
u
⎛ u +1⎞
We showed ⎜
⎟ < e in part b., so
⎝ u ⎠
u
u
⎛ u +1⎞
e<⎜
⎟ < e.
u +1
⎝ u ⎠
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57. a.
In order of increasing slope, the graphs
represent the curves y = 2 x , y = 3x , and
y = 4 x.
b. ln y is linear with respect to x, and at x = 0,
y = 1 since C = 1.
c.
6.5 Concepts Review
1. ky; ky ( L − y )
2. 23 = 8
3. half-life
4.
(1 + h )1/ h
The graph passes through the points (0.2, 4)
and (0.6, 8). Thus, 4 = Cb0.2 and 8 = Cb0.6 .
Dividing the second equation by the first,
gets 2 = b0.4 so b = 25 2.
Therefore C = 23 2.
Problem Set 6.5
1. k = −6 , y0 = 4, so y = 4e −6t
2. k = 6, y0 = 1, so y = e6t
58. The graph of the equation whose log-log plot has
negative slope contains the points (2, 7) and (7,
2).
r
7 ⎛2⎞
Thus, 7 = C 2 and 2 = C 7 , so = ⎜ ⎟ .
2 ⎝7⎠
7
2
ln 7 − ln 2
ln = r ln ⇒ r =
= −1 and C = 14.
2
7
ln 2 − ln 7
r
r
Hence, one equation is y = 14 x −1.
The graph of one equation contains the points
(7, 30) and (10, 70). Thus, 30 = C 7 and
r
r
3 ⎛7⎞
70 = C10r , so = ⎜ ⎟
7 ⎝ 10 ⎠
3
7
ln 3 − ln 7
ln = r ln ⇒ r =
≈ 2.38 and
7
10
ln 7 − ln10
C ≈ 30 ⋅ 7 −2.38 ≈ 0.29 . Hence, another equation is
y = 0.29 x 2.38 .
The graph of another equation contains the points
(1, 2) and (7, 5). Thus, 2 = C1 and 5 = C 7 , so
C = 2 and
ln 5 − ln 2
ln 5 − ln 2 = r ln 7 ⇒ r =
≈ 0.47.
ln 7
r
r
.
Hence, the last equation is y = 2 x
The given answers are only approximate.
Student answers may also vary.
0.47
3. k = 0.005, so y = y0 e0.005t
y (10) = y0 e0.005(10) = y0 e0.05
2
y (10) = 2 ⇒ y0 =
0.05
e
2 0.005t
y=
e
= 2e0.005t −0.05 = 2e0.005(t −10)
e0.05
4. k = –0.003, so y = y0 e –0.003t
y (–2) = y0 e(–0.003)(–2) = y0 e0.006
3
y (−2) = 3 ⇒ y0 =
0.006
e
3
y=
e –0.003t = 3e –0.003t –0.006 = 3e –0.003(t + 2)
e0.006
5. y0 = 10, 000, y(10) = 20,000
20, 000 = 10, 000e k (10)
2 = e10k
ln 2 = 10k;
k=
ln 2
10
y = 10, 000e((ln 2) /10)t = 10, 000 ⋅ 2t /10
After 25 days, y = 10, 000 ⋅ 22.5 ≈ 56,568.
6. Since the growth is exponential and it doubles in
10 days (from t = 0 to t = 10), it will always
double in 10 days.
7. 3 y0 = y0 e((ln 2) /10)t
3 = e((ln 2) /10)t
ln 2
ln 3 =
t
10
10 ln 3
t=
≈ 15.8 days
ln 2
374
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8. Let P(t) = population (in millions) in
year 1790 + t.
In 1960, t = 170.
P (t ) = P0 ekt
178 = 3.9e170k
45.64 = e170k
ln 45.64
k=
≈ 0.02248
170
In 2000, t = 210
P (210) ≈ 3.9e0.02248⋅210 ≈ 438
The model predicts that the population will be about
438 million. The actual number, 275 million, is
quite a bit smaller because the rate of growth has
declined in recent decades.
9. 1 year: (4.5 million) (1.032) ≈ 4.64 million
2 years: (4.5 million) (1.032) 2 ≈ 4.79 million
10 years: (4.5 million) (1.032)10 ≈ 6.17 million
100 years: (4.5 million) (1.032)100 ≈ 105 million
10. y = y0 ekt
1.032 A = Aek (1)
k = ln1.032 ≈ 0.03150
At t = 100, y = 4.5e(0.03150)(100) ≈ 105 .
After 100 years, the population will be about
105 million.
11. The formula to use is y = y0 e kt , where y =
population after t years, y0 =population at time t =
0, and k is the rate of growth. We are given
235, 000 = y0 e k (12) and
164, 000 = y0 e k (5)
Dividing one equation by the other yields
1.43293 = e12 k −5k = e7 k or
ln(1.43293)
k=
≈ 0.0513888
7
235, 000
Thus y0 = 12(0.0513888) = 126,839.
e
12. The formula to use is y = y0 e kt , where y = mass t
months after initial measurement, y0 = mass at time
of initial measurement, and k is the rate of growth.
We are given
6.76 = 4e k (4) so that
1 ⎛ 6.76 ⎞ 0.5247
k = ln ⎜
≈ 0.1312
⎟=
4 ⎝ 4 ⎠
4
13.
1
= e k (700) and y0 = 10
2
–ln 2 = 700k
ln 2
k=−
≈ −0.00099
700
y = 10e−0.00099t
At t = 300, y = 10e−0.00099⋅300 ≈ 7.43.
After 300 years there will be about 7.43 g.
14. 0.85 = e k (2)
ln 0.85 = 2k
ln 0.85
k=
≈ −0.0813
2
1
= e −0.0813t
2
– ln 2 = −0.0813t
ln 2
t=
≈ 8.53
0.0813
The half-life is about 8.53 days.
15. The basic formula is y = y0 e kt . If t* denotes the
half-life of the material, then (see Example 3)
1
ln(0.5)
= e kt* or k =
.
Thus
2
t*
−0.693
−0.693
= −0.0229 and k S =
= −0.0241
30.22
28.8
To find when 1% of each material will remain, we
ln(0.01)
. Thus
use 0.01y0 = y0 ekt or t =
k
−4.6052
tC =
≈ 201 years (2187) and
−0.0229
−4.6052
tS =
≈ 191 years (2177)
−0.0241
kC =
16. The basic formula is y = y0 e kt . We are given
15.231 = y0 ek (2) and 9.086 = y0 ek (8)
Dividing one equation by the other gives
15.231 k (2) − k (8)
=e
= ek ( −6) so k = −0.0861
9.086
15.231
Thus y0 = ( −.0861)(2) ≈ 18.093 grams.
e
To find the half-life:
t* =
ln(0.5) −0.693
=
≈ 8 days
−0.0861
k
Thus, 6 months before the initial measurement, the
mass was y = 4e(0.1312)( −6) ≈ 1.82 grams. The
tumor would have been detectable at that time.
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17.
22. From Example 4, T (t ) = T1 + (T0 − T1 )e kt . In this
1
= e5730 k
2
ln 12
k=
≈ −1.210 × 10−4
5730
problem, 250 = T (15) = 40 + (350 − 40)e k (15) so
( )
0.7 y0 = y0 e( −1.210×10
t=
⎛ 210 ⎞
ln ⎜
⎟
310 ⎠
k= ⎝
= −0.026 ; the brownies will be
15
−4 )t
110D F when 110 = 40 + (310)e−0.026 t or
ln 0.7
≈ 2950
−1.210 × 10−4
The fort burned down about 2950 years ago.
18.
⎛ 70 ⎞
ln ⎜
⎟
310 ⎠
t= ⎝
= 57.2 min.
−0.026
1
= e5730 k
2
ln 12
≈ −1.210 × 10−4
k=
5730
23. From Example 4, T (t ) = T1 + (T0 − T1 )e kt .
Let w = the time of death; then
82 = T (10 − w) = 70 + (98.6 − 70)ek (10 − w)
( )
0.51 y0 = y0 e( −1.210×10
t=
76 = T (11 − w) = 70 + (98.6 − 70)ek (11− w)
−4 )t
≈ 5565
−1.210 × 10−4
The body was buried about 5565 years ago.
6 = 28.6ek (11− w)
Dividing: 2 = e k ( −1) or k = ln (0.5) = −0.693
19. From Example 4, T (t ) = T1 + (T0 − T1 )e kt . In this
problem, 200 = T (0.5) = 75 + (300 − 75)e
k (0.5)
12 = 28.6ek (10 − w)
or
ln 0.51
so
⎛ 125 ⎞
ln ⎜
⎟
225 ⎠
k= ⎝
= −1.1756 and
0.5
T (3) = 75 + 225e( −1.1756)(3) = 81.6D F
20. From Example 4, T (t ) = T1 + (T0 − T1 )e kt . In this
problem, 0 = T (5) = 24 + (−20 − 24)ek (5) so
⎛ −24 ⎞
ln ⎜
⎟
−44 ⎠
k= ⎝
= −0.1212 ; the thermometer will
5
register 20D C when 20 = 24 + (−44)e −0.1212 t or
⎛ −4 ⎞
ln ⎜
⎟
−44 ⎠
t= ⎝
= 19.78 min.
−0.1212
21. From Example 4, T (t ) = T1 + (T0 − T1 )e kt . In this
problem, 70 = T (5) = 90 + (26 − 90)e k (5) so
⎛ −20 ⎞
ln ⎜
⎟
−64 ⎠
k= ⎝
= −0.2326 and
5
T (10) = 90 − 64e( −0.2326)(10) = 90 − 64(0.0977) = 83.7D C
To find w :
⎛ 12 ⎞
ln ⎜
⎟
28.6 ⎠
= 1.25
12 = 28.6e−0.693(10 − w) so 10 − w = ⎝
−0.693
Therefore w = 10 − 1.25 = 8.75 = 8 : 45 pm .
24. a.
From example 4 of this section,
dT
= k (T − T1 ) or
dt
dT
∫ T − T = k dt or ln T(t)-T1 = kt + C
1
This gives
T (t ) − T1 = e kt eC . Now, if T0 is
the temperature at t = 0, T0 − T1 = eC and the
Law of Cooling becomes
T (t ) − T1 = T0 − T1 e kt . Note that T (t ) is
always between T0 and T1 so that
T (t ) − T1 and T0 − T1 always have the same
sign; this simplifies the Law of Cooling to
T (t ) − T1 = (T0 − T1 )e kt
or
T (t ) = T1 + (T0 − T1 )e kt
b. Since T (t ) is always between T0 and T1 , it
follows that e kt =
T (t ) − T1
< 1 so that k < 0 .
T0 − T1
Hence
lim T (t ) = T1 + (T0 − T1 ) lim e kt = T1 + 0 = T1
t →∞
376
Section 6.5
t →∞
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25. a.
($375)(1.035) 2 ≈ $401.71
⎛ 0.035 ⎞
($375) ⎜ 1 +
⎟
12 ⎠
⎝
24
b.
⎛ 0.035 ⎞
($375) ⎜ 1 +
⎟
365 ⎠
⎝
730
c.
d.
($375)e0.035⋅2 ≈ $402.19
26. a.
≈ $402.15
≈ $402.19
($375)(1.046)2 = $410.29
⎛ 0.046 ⎞
($375) ⎜ 1 +
⎟
12 ⎠
⎝
24
b.
⎛ 0.046 ⎞
($375) ⎜ 1 +
⎟
365 ⎠
⎝
730
c.
d.
($375)e0.046⋅2 ≈ $411.14
≈ $411.06
≈ $411.13
12t
27. a.
⎛ 0.06 ⎞
⎜1 +
⎟
12 ⎠
⎝
=2
=2
ln 2
ln 2
12t =
≈ 11.58
so t =
ln1.005
12 ln1.005
It will take about 11.58 years or
11 years, 6 months, 29 days.
12t
1.005
ln 2
b. e
≈ 11.55
=2 ⇒ t=
0.06
It will take about 11.55 years
or 11 years, 6 months, and 18 days.
0.06t
28. $20, 000(1.025)5 ≈ $22, 628.16
29. 1626 to 2000 is 374 years.
y = 24e
0.06 ⋅ 374
34.
dy
= ky ( L – y )
dt
1
dy = kdt
y( L – y)
⎡1
⎤
1
⎢ Ly + L( L – y ) ⎥ dy = kdt
⎣
⎦
1 ⎛1
1 ⎞
⎜ +
⎟ dy = ∫ kdt
∫
L ⎝ y L– y⎠
1
[ln y – ln L − y ] = kt + C1
L
y
ln
= Lkt + LC1
L– y
y
y
= e Lkt + LC1 = e LC1 ⋅ e Lkt , so
= Ce Lkt
L– y
L− y
⎛ Note that: C = Ce0 = Ce Lk ⋅0 ⎞
⎜
⎟
y0 ⎟
y (0)
⎜
=
=
.
⎜
L – y (0) L – y0 ⎟⎠
⎝
y = LCe Lkt – yCe Lkt
y + yCe Lkt = LCe Lkt
y=
LCe Lkt
1 + Ce
Lkt
=
LC
LC
=
+ C C + e – Lkt
1
e Lkt
y
=
L ⋅ L –0y
y0
L – y0
35. y =
=
≈ $133.6 billion
+e
0
– Lkt
=
Ly0
y0 + ( L – y0 )e – Lkt
16 ( 6.4 )
6.4 + (16 − 6.4)e−16(0.00186)t
102.4
6.4 + 9.6e −0.02976t
y
30. $100(1.04)969 ≈ $3.201× 1018
20
31. 1000e(0.05)(1) = $1051.27
10
32. A0 e(0.05)(1) = 1000
A0 = 1000e −0.05 ≈ $951.23
33. If t is the doubling time, then
−50
150 t
t
p ⎞
⎛
⎜1 +
⎟ =2
⎝ 100 ⎠
p ⎞
⎛
t ln ⎜ 1 +
⎟ = ln 2
100
⎝
⎠
ln 2
ln 2 100 ln 2 70
t=
≈
=
≈
p
p
p
p
ln 1 + 100
100
(
)
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
39. Let y = population in millions, t = 0 in 1985,
a = 0.012, b = 0.06 , y0 = 10
lim (1 + x)1000 = 11000 = 1
36. a.
x →0
dy
= 0.012 y + 0.06
dt
0.06 ⎞ 0.012t 0.06
⎛
y = ⎜ 10 +
= 15e0.012t – 5
–
⎟e
0.012
0.012
⎝
⎠
From 1985 to 2010 is 25 years. At t = 25,
lim 11/ x = lim 1 = 1
b.
x →0
x →0
lim (1 + ε )1/ x = lim (1 + ε )n = ∞
c.
x →0 +
n →∞
1
lim (1 + ε )1/ x = lim
d.
x →0 −
y = 15e0.012⋅25 − 5 ≈ 15.25. The population in 2010
will be about 15.25 million.
=0
n→∞ (1 + ε ) n
lim (1 + x)1/ x = e
e.
x →0
1
lim (1 − x)1/ x = lim
37. a.
1 (− x)
x →0 [1 + ( − x )]
x →0
3
lim (1 + 3 x)
x →0
n
⎛n+2⎞
⎛ 2⎞
lim ⎜
⎟ = lim ⎜ 1 + ⎟
n⎠
n →∞ ⎝ n ⎠
n→∞ ⎝
c.
1
e
1 ⎤
⎡
= lim ⎢(1 + 3x) 3 x ⎥ = e3
x →0 ⎣⎢
⎦⎥
1/ x
b.
=
40. Let N(t) be the number of people who have heard
dN
= k (L − N ) .
the news after t days. Then
dt
1
∫ L − N dN = ∫ k dt
–ln(L – N) = kt + C
L − N = e− kt −C
N = L − Ae− kt
N(0) = 0, ⇒ A = L
n
N (t ) = L(1 − e− kt ) .
= lim (1 + 2 x)
1/ x
x →0 +
N (5) =
2
1 ⎤
⎡
= lim ⎢(1 + 2 x) 2 x ⎥ = e 2
x →0+ ⎣⎢
⎦⎥
⎛ n −1 ⎞
lim ⎜
⎟
n →∞ ⎝ n ⎠
d.
2n
⎛ 1⎞
= lim ⎜ 1 − ⎟
n⎠
n →∞ ⎝
1
= e −5k
2
ln 1
k = 2 ≈ 0.1386
−5
2n
N (t ) = L(1 − e−0.1386t )
= lim (1 − x)2 / x
0.99 L = L(1 − e−0.1386t )
x →0+
1 ⎤
⎡
= lim ⎢ (1 − x) − x ⎥
x →0+ ⎣⎢
⎦⎥
38.
−2
=
dy
= ay + b
dt
dy
∫ y + b = ∫ a dt
a
ln y +
b
= at + C
a
b
b
= e at +C ; y + = Aeat
a
a
b
y = Aeat −
a
b
b
y0 = A − ⇒ A = y0 +
a
a
b ⎞ at b
⎛
y = ⎜ y0 + ⎟ e −
a⎠
a
⎝
y+
378
Section 6.5
L
L
⇒ = L(1 − e −5k )
2
2
0.01 = e −0.1386t
ln 0.01
≈ 33
t=
−0.1386
99% of the people will have heard about the scandal
after 33 days.
1
e2
41. If f(t) = e kt , then
42.
f ′(t ) kekt
=
=k.
f (t )
e kt
f ( x) = an x n + an –1 x n –1 + ⋅⋅⋅ + a1 x + a0
lim
x →∞
f ′( x)
f ( x)
= lim
nan x n –1 + (n – 1)an –1 x n –2 + ⋅⋅⋅ + a1
an x n + an –1 x + ⋅⋅⋅ + a1 x + a0
x →∞
= lim
nan
x
x →∞ a
n
+
+
( n –1) an –1
x2
an –1
x
+ ⋅⋅⋅ +
+ ⋅⋅⋅ +
a1
n
x –1
+
a1
xn
a0
xn
=0
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43.
f ′( x)
= k > 0 can be written as
f ( x)
e.
1 dy
= k where y = f(x).
y dx
t = 66 , which is year 2070.
The population will equal the 2004 value of
6.4 billion when 0.0132t − 0.0001t 2 = 0
f ′( x)
1 dy
= k < 0 can be written as
= k where
f ( x)
y dx
dy
y = f(x).
= k dx has the solution y = Cekx .
y
Thus, f ( x) = Cekx which represents exponential
decay since k < 0.
45. Maximum population:
640 acres 1 person
⋅
13,500, 000 mi 2 ⋅
1 acre
1 mi 2
2
t = 0 or t = 132 .
The model predicts that the population will
return to the 2004 level in year 2136.
47. a.
b.
c.
= 1.728 × 1010 people
Let t = 0 be in 2004.
y ' = ( 0.0132 − 0.0001t ) y
dy
= ( 0.0132 − 0.0001t ) y
dt
dy
= ( 0.0132 − 0.0001t ) dt
y
⎛ 1.728 ⋅10 ⎞
ln ⎜
⎜ 6.4 ⋅109 ⎟⎟
⎠ ≈ 75.2 years from 2004, or
t= ⎝
0.0132
sometime in the year 2079.
y = C1e0.0132t −0.00005t
10
c.
k = 0.0132 − 0.0001t
ln y = 0.0132t − 0.00005t 2 + C0
(6.4 × 109 )e0.0132t = 1.728 × 1010
b.
)
t = 0.0132 / 0.0002 = 66
Thus, the equation f ( x) = Ce kx represents
exponential growth since k > 0.
46. a.
(
0.0132 = 0.0002t
dy
= k dx has the solution y = Cekx .
y
44.
The maximum population will occur when
d
0.0132t − 0.0001t 2 = 0
dt
2
The initial condition y (0) = 6.4 implies that
C1 = 6.4 . Thus y = 6.4e0.0132t −0.00005t
2
y
d.
k = 0.0132 − 0.0002t
20
y ' = ( 0.0132 − 0.0002t ) y
dy
= ( 0.0132 − 0.0002t ) y
dt
dy
= ( 0.0132 − 0.0002t ) dt
y
10
100
ln y = 0.0132t − 0.0001t 2 + C0
y = C1e0.0132t − 0.0001t
e.
2
The initial condition y (0) = 6.4 implies that
C1 = 6.4 . Thus y = 6.4e0.0132t −0.0001t
d.
y
10
2
200
t
300
The maximum population will occur when
d
0.0132t − 0.00005t 2 = 0
dt
0.0132 = 0.0001t
(
)
t = 0.0132 / 0.0001 = 132
t = 132 , which is year 2136
The population will equal the 2004 value of
6.4 billion when 0.0132t − 0.00005t 2 = 0
t = 0 or t = 264 .
The model predicts that the population will
return to the 2004 level in year 2268.
5
50
100
150
Instructor’s Resource Manual
t
Section 6.5
379
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
E ( x + h) – E ( x )
h
h →0
E ( x ) E ( h) – E ( x )
= lim
h
h →0
E ( h) – 1
E ( h) – 1
= lim E ( x) ⋅
= E ( x) lim
h
h
h →0
h →0
E ( x) = E ( x + 0) = E ( x) ⋅ E (0)
6.6 Concepts Review
48. E ′( x) = lim
1. exp
2. y exp
3.
so E (0) = 1.
E (h) – E (0)
h
E (0 + h) – E (0)
= E ( x) lim
= E ( x) ⋅ E ′(0)
h
h →0
= kE(x) where k = E ′(0) .
Thus, E ′( x) = E ( x) lim
Hence, E ( x) = E 0 e
Check: E (u + v) = e
= E (0)e
k (u + v )
= e ku ⋅ ekv = E (u ) ⋅ E (v)
49.
=e
kx
= 1⋅ e
ku + kv
kx
( ∫ P( x)dx )
1 d ⎛
; ⎜
x dx ⎝
y⎞
2
⎟ = 1; x + Cx
x⎠
4. particular
h →0
kx
( ∫ P( x)dx )
Problem Set 6.6
=e .
kx
1. Integrating factor is e x .
D ( ye x ) = 1
y = e– x ( x + C)
2. The left-hand side is already an exact derivative.
D[ y ( x + 1)] = x 2 – 1
y=
3. y ′ +
Exponential growth:
In 2010 (t = 6): 6.93 billion
In 2040 (t = 36): 10.29 billion
In 2090 (t = 86): 19.92 billion
Logistic growth:
In 2010 (t = 6): 7.13 billion
In 2040 (t = 36): 10.90 billion
In 2090 (t = 86): 15.15 billion
50. a.
b.
lim (1 + x)1/ x = e
x →0
lim (1 – x)1/ x =
x →0
x3 – 3 x + C
3( x + 1)
x
2
y=
ax
1– x
1 – x2
Integrating factor:
x
exp ∫
dx = exp ⎡ ln(1 – x 2 ) –1/ 2 ⎤
⎣
⎦
1 – x2
= (1 – x 2 ) –1/ 2
D[ y (1 – x 2 ) –1/ 2 ] = ax(1 – x 2 ) –3 / 2
Then y (1 – x 2 ) –1/ 2 = a (1 – x 2 ) –1/ 2 + C , so
y = a + C (1 – x 2 )1/ 2 .
4. Integrating factor is sec x.
1
e
D[ y sec x] = sec 2 x
y = sin x + C cos x
5. Integrating factor is
1
.
x
⎡ y⎤
D ⎢ ⎥ = ex
⎣x⎦
y = xe x + Cx
380
Section 6.6
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6. y ′ – ay = f ( x)
14. Integrating factor is sin 2 x.
Integrating factor: e ∫
– adx
D[ ye – ax ] = e – ax f ( x)
Then ye – ax = ∫ e – ax f ( x)dx, so
y = e ax ∫ e – ax f ( x)dx .
7. Integrating factor is x. D[yx] = 1; y = 1 + Cx –1
8. Integrating factor is ( x + 1) 2 .
D[ y ( x + 1) 2 ] = ( x + 1)5
⎛1⎞
y = ⎜ ⎟ ( x + 1)4 + C ( x + 1) –2
⎝6⎠
9. y ′ + f ( x) y = f ( x)
f ( x ) dx
Integrating factor: e ∫
f ( x ) dx ⎤
f ( x ) dx
D ⎡ ye ∫
= f ( x )e ∫
⎣⎢
⎦⎥
Then ye ∫
f ( x ) dx
y = 1 + Ce
= e∫
– ∫ f ( x ) dx
D[ y sin 2 x] = 2sin 2 x cos x
= e – ax
f ( x ) dx
+ C , so
2
y sin 2 x = sin 3 x + C
3
2
C
y = sin x +
3
sin 2 x
2
5
y = sin x + csc2 x
3
12
⎛π ⎞
goes through ⎜ , 2 ⎟ .
⎝6 ⎠
15. Let y denote the number of pounds of chemical A
after t minutes.
dy ⎛ lbs ⎞ ⎛ gal ⎞ ⎛ y lbs ⎞ ⎛ 3 gal ⎞
= ⎜2
⎟⎜3
⎟⎜
⎟–⎜
⎟
dt ⎝ gal ⎠ ⎝ min ⎠ ⎝ 20 gal ⎠ ⎝ min ⎠
3y
=6–
lb/min
20
3
y′ +
y=6
20
( 3 / 20 ) dt = e3t / 20
Integrating factor: e ∫
D[ ye3t / 20 ] = 6e3t / 20
.
Then ye3t / 20 = 40e3t / 20 + C. t = 0, y = 10
⇒ C = –30.
2x
10. Integrating factor is e .
D[ ye 2 x ] = xe 2 x
Therefore, y (t ) = 40 – 30e –3t / 20 , so
⎛1⎞
⎛1⎞
y = ⎜ ⎟ x – ⎜ ⎟ + Ce –2 x
⎝2⎠
⎝4⎠
y (20) = 40 – 30e –3 ≈ 38.506 lb.
1
⎡ y⎤
11. Integrating factor is . D ⎢ ⎥ = 3 x 2 ; y = x 4 + Cx
x
⎣x⎦
16.
dy
y
⎛ y ⎞
= (2)(4) – ⎜
=8
⎟ (4) or y ′ +
dt
50
⎝ 200 ⎠
Integrating factor is et / 50 .
y = x 4 + 2 x goes through (1, 3).
D[ yet / 50 ] = 8et / 50
y (t ) = 400 + Ce – t / 50
12. y ′ + 3 y = e2 x
3dx
= e3 x
Integrating factor: e ∫
y (t ) = 400 – 350e – t / 50 goes through (0, 50).
D[ ye3 x ] = e5 x
y (40) = 400 – 350e –0.8 ≈ 242.735 lb of salt
Then ye3 x =
ye3 x =
e5 x
4
+ C. x = 0, y = 1 ⇒ C = , so
5
5
e5 x 4
+ .
5 5
⎡
⎤
⎡ 3 ⎤
dy
y
= 4–⎢
⎥ (6) or y ′ + ⎢
⎥y=4
dt
⎣ (120 – 2t ) ⎦
⎣ (60 – t ) ⎦
Integrating factor is (60 – t ) –3 .
e 2 x + 4e –3 x
is the particular
5
solution through (0, 1).
Therefore, y =
13. Integrating factor: xe
17.
x
d [ yxe x ] = 1 ; y = e – x (1 + Cx –1 ); y = e – x (1 – x –1 )
goes through (1, 0).
Instructor’s Resource Manual
D[ y (60 – t ) –3 ] = 4(60 – t ) –3
y (t ) = 2(60 – t ) + C (60 – t )3
⎛ 1 ⎞
3
y (t ) = 2(60 – t ) – ⎜
⎟ (60 − t ) goes through
⎝ 1800 ⎠
(0, 0).
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18.
dy
–2 y
2
=
y = 0.
or y ′ +
dt 50 + t
50 + t
Integrating factor:
2
⎛
⎞
exp ⎜ ∫
dt ⎟ = e 2 ln(50+t ) = (50 + t )2
⎝ 50 + t ⎠
23. Let y be the number of gallons of pure alcohol in the
tank at time t.
a.
Integrating factor is e0.05t .
D[ y (50 + t )2 ] = 0
y (t ) = 25 + Ce –0.05t ; y = 100, t = 0, C = 75
Then y (50 + t )2 = C. t = 0, y = 30 ⇒ C = 75000
y (t ) = 25 + 75e –0.05t ; y = 50, t = T,
T = 20(ln 3) ≈ 21.97 min
Thus, y (50 + t )2 = 75, 000.
If y = 25, 25(50 + t ) 2 = 75, 000, so
t = 3000 – 50 ≈ 4.772 min.
19. I ′ + 106 I = 1
Integrating factor = exp(106 t )
dy
⎛ 5 ⎞
= 5(0.25) – ⎜
⎟ y = 1.25 – 0.05 y
dt
⎝ 100 ⎠
y′ =
b. Let A be the number of gallons of pure alcohol
drained away.
200
(100 – A) + 0.25A = 50 ⇒ A =
3
It took
200
3
minutes for the draining and the
5
same amount of time to refill, so
2 200
80
3
T=
=
≈ 26.67 min.
5
3
D[ I exp(106 t )] = exp(106 t )
I (t ) = 10 –6 + C exp(–106 t )
( )
I (t ) = 10 –6 [1 – exp(–106 t )] goes through (0, 0).
20. 3.5I ′ = 120sin 377t
c.
⎛ 240 ⎞
I′ = ⎜
⎟ sin 377t
⎝ 7 ⎠
⎛ 240 ⎞
I = ⎜–
⎟ cos 377t + C
⎝ 2639 ⎠
⎛ 240 ⎞
I (t ) = ⎜
⎟ (1 – cos 377t ) through (0, 0).
⎝ 2639 ⎠
21.
1000 I = 120 sin 377t
I(t) = 0.12 sin 377t
200
3
5
c>
d.
D[ xe
t / 50
]=0
x = Ce – t / 50
x(t ) = 50e – t / 50 satisfies t = 0, x = 50.
⎛ 50e – t / 50 ⎞
dy
⎛ y ⎞
= 2⎜
⎟ – 2⎜
⎟
⎜
⎟
dt
⎝ 200 ⎠
⎝ 100 ⎠
⎛ 1 ⎞
– t / 50
y′ + ⎜
⎟y=e
⎝ 100 ⎠
+
200
3
c
< 20(ln 3).
10
≈ 7.7170
(3ln 3 – 2)
y ′ = 4(0.25) – 0.05 y = 1 – 0.05 y
Solving for y, as in part a, yields
y = 20 + 80e –0.05t . The drain is closed when
t = 0.8T . We require that
dx
2x
=–
22.
dt
100
1 ⎞
⎛
x′ + ⎜ ⎟ x = 0
⎝ 50 ⎠
Integrating factor is et / 50 .
c would need to satisfy
(20 + 80e −0.05⋅0.8T ) + 4 ⋅ 0.25 ⋅ 0.2T = 50,
or 400e –0.04T + T = 150.
24. a.
v ′ + av = – g
Integrating factor: eat
d
(veat ) = – geat
dt
– g at
–g
veat = ∫ – geat dt =
e + C; v =
+ Ce – at
a
a
v = v0 , t = 0
e at (v ′ + av ) = – ge at ;
D[ yet /100 ] = e – t /100
–g
g
+ C ⇒ C = v0 +
a
a
–g ⎛
g⎞
Therefore, v =
+ ⎜ v0 + ⎟ e – at , so
a ⎝
a⎠
y (t ) = e – t /100 (C – 100e – t /100 )
v(t ) = v∞ + (v0 – v∞ )e – at .
Integrating factor is et /100 .
v0 =
y (t ) = e – t /100 (250 – 100e – t /100 ) satisfies t = 0,
y = 150.
382
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b.
dy
= v∞ + (v0 – v∞ )e – at , so
dt
(v0 − v∞ )e− at
+ C.
a
–(v0 – v∞ )
+C
y = y0 , t = 0 ⇒ y0 =
a
v –v
⇒ C = y0 + 0 ∞
a
y = v∞ ⋅ t −
(v0 – v∞ )e – at ⎛
v –v ⎞
+ ⎜ y0 + 0 ∞ ⎟
a
a ⎠
⎝
v –v
= y0 + v∞ t + 0 ∞ (1 – e – at )
a
y = v∞ t –
25. a.
32
v∞ = –
= –640
0.05
26. For t in [0, 15],
–32
v∞ =
= –320.
0.10
v(t ) = (0 + 320)e –0.1t – 320 = 320(e –0.1t –1);
v(15) = 320(e –1.5 –1) ≈ –248.6
y (t ) = 8000 – 320t + 10(320)(1 – e –0.1t );
y (15) = 3200(2 – e –1.5 ) ≈ 5686
Let t be the number of seconds after the parachute
opens that it takes Megan to reach the ground.
32
= –20.
For t in [15, 15+T], v∞ = –
1.6
0 = y (T + 15)
= [3200(2 – e –1.5 )]
–20T + (0.625)[320(e –1.5 – 1) + 20](1 – e –1.6T )
v(t ) = [120 − (−640)]e−0.05t + (−640) = 0 if
≈ 5543 – 20T –142.9e−1.6T ≈ 5543 – 20T [since
⎛ 19 ⎞
t = 20 ln ⎜ ⎟ .
⎝ 16 ⎠
y (t ) = 0 + (–640)t
T > 50, so e –1.6T < 10 –35 (very small)]
Therefore, T ≈ 277, so it takes Megan about
292 s (4 min, 52 s) to reach the ground.
⎛ 1 ⎞
–0.05t
+⎜
)
⎟ [120 – (–640)](1 – e
⎝ 0.05 ⎠
27. a.
= –640t + 15, 200(1 – e –0.05t )
Therefore, the maximum altitude is
⎛
⎛ 19 ⎞ ⎞
⎛ 19 ⎞ 45, 600
y ⎜ 20 ln ⎜ ⎟ ⎟ = −12,800 ln ⎜ ⎟ +
19
⎝ 16 ⎠ ⎠
⎝ 16 ⎠
⎝
≈ 200.32 ft
b. –640T + 15, 200(1 – e –0.05T ) = 0;
95 – 4T – 95e –0.05T = 0
b.
⎛ dy y ⎞
e − ln x +C ⎜ − ⎟ = x 2 e − ln x +C
⎝ dx x ⎠
⎛ dy y ⎞
e − ln x eC ⎜ − ⎟ = x 2 eC e− ln x
⎝ dx x ⎠
1 C dy
1
1
− yeC
= x 2 eC
e
2
x
dx
x
x
d ⎛ C1 ⎞
y ⎟ = xeC
⎜e
dx ⎝
x ⎠
y
eC = eC ∫ x dx
x
y x2
=
+ C1
x
2
y=
28. e ∫
x3
+ C1 x
2
P ( x ) dx +C
P ( x ) dx + C
dy
+ P ( x )e ∫
y
dx
= Q ( x )e ∫
P ( x ) dx +C
P ( x ) dx + C
d ⎛ ∫ P ( x ) dx +C ⎞
y ⎟ = Q ( x )e ∫
⎜e
⎟
dx ⎜⎝
⎠
ye ∫
P ( x ) dx +C
= ∫ Q ( x)e ∫
P ( x ) dx C
e dx + C1
− P ( x ) dx
∫ P ( x ) dx dx
y=e ∫
∫ Q ( x )e
− P ( x ) dx
+ C2 e ∫
Instructor’s Resource Manual
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
5.
6.7 Concepts Review
1. slope field
2. tangent line
3. yn −1 + hf ( xn −1 , yn −1 )
The oblique asymptote is y = x .
4. underestimate
6.
Problem Set 6.7
1.
The oblique asymptote is y = 3 + x / 2 .
lim y ( x ) = 12 and y (2) ≈ 10.5
7.
x →∞
2.
lim y ( x) = ∞ and y (2) ≈ 16
x →∞
dy 1
1
= y; y (0) =
dx 2
2
dy 1
= dx
y 2
ln y =
3.
x
+C
2
y = C1e x / 2
To find C1 , apply the initial condition:
lim y ( x) = 0 and y (2) ≈ 6
x →∞
1
= y (0) = C1e0 = C1
2
1
y = ex / 2
2
4.
lim y ( x) = ∞ and y (2) ≈ 13
x →∞
384
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3
2
3⎞
⎛
e x y '+ ye x = ⎜ 2 x + ⎟ e x
2⎠
⎝
d x
3⎞
⎛
e y = ⎜ 2x + ⎟ ex
dx
2⎠
⎝
3⎞
⎛
e x y = ∫ ⎜ 2 x + ⎟ e x dx
2⎠
⎝
y '+ y = 2 x +
8.
( )
dy
= − y;
dx
dy
= −dx
y
y (0) = 4
3
Integrate by parts: let u = 2 x + ,
2
dv = e x dx . Then du = 2dx and v = e x .
Thus,
3⎞
⎛
e x y = ⎜ 2 x + ⎟ e x − ∫ 2e x dx
2⎠
⎝
3⎞
⎛
e x y = ⎜ 2 x + ⎟ e x − 2e x + C
2⎠
⎝
1
y = 2 x − + Ce− x
2
To find C, apply the initial condition:
1
1
3 = y (0) = 0 − + Ce −0 = C −
2
2
7
Thus C = , so the solution is
2
1 7 −x
y = 2x − + e
2 2
ln y = − x + C
y = C1e − x
To find C1 , apply the initial condition:
4 = y (0) = C1e −0 = C1
y = 4e − x
9.
y '+ y = x + 2
1dx
= ex .
The integrating factor is e ∫
Note: Solutions to Problems 22-28 are given along with
the corresponding solutions to 11-16.
e x y '+ ye x = e x ( x + 2)
( )
d x
e y = ( x + 2) e x
dx
11., 22.
e y = ∫ ( x + 2) e dx
x
0.0
Integrate by parts: let u = x + 2, dv = e x dx .
Then du = dx and v = e x . Thus
e x y = ( x + 2)e x − ∫ e x dx
e y = ( x + 2)e − e + C
x
xn
x
x
x
−x
y = x + 2 − 1 + Ce
To find C , apply the initial condition:
4 = y (0) = 0 + 1 + Ce−0 = 1 + C → C = 3
Thus, y = x + 1 + 3e
−x
12., 23.
4.2
4.44
0.4
5.88
6.5712
0.6
8.232
9.72538
0.8
11.5248
14.39356
1.0
16.1347
21.30246
xn
Euler's
Method yn
2.0
Improved Euler
Method yn
2.0
0.0
Instructor’s Resource Manual
Improved Euler
Method yn
3.0
0.2
.
10.
Euler's
Method yn
3.0
0.2
1.6
1.64
0.4
1.28
1.3448
0.6
1.024
1.10274
0.8
0.8195
0.90424
1.0
0.65536
0.74148
Section 6.7
385
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13., 24.
xn
0.0
14., 25.
Improved Euler
Method yn
0.0
0.2
0.0
0.02
0.4
0.04
0.08
0.6
0.12
0.18
0.8
0.24
0.32
1.0
0.40
0.50
xn
Euler's
Method yn
0.0
Improved Euler
Method yn
0.0
17. a.
y0 = 1
y1 = y0 + hf ( x0 , y0 )
= y0 + hy0 = (1 + h) y0
y2 = y1 + hf ( x1 , y1 ) = y1 + hy1
= (1 + h) y1 = (1 + h)2 y0
y3 = y2 + hf ( x2 , y2 ) = y2 + hy2
= (1 + h) y2 = (1 + h)3 y0
#
yn = yn −1 + hf ( xn −1 , yn −1 ) = yn −1 + hyn −1
= (1 + h) yn −1 = (1 + h) n y0 = (1 + h )
n
0.2
0.0
0.004
0.4
0.008
0.024
Let N = 1/ h . Then y N is an approximation
to the solution at x = Nh = (1/ h)h = 1 . The
exact solution is y (1) = e . Thus,
0.6
0.040
0.076
(1 + 1/ N ) N
0.8
0.112
0.176
we know that lim (1 + 1/ N )
0.0
1.0
15., 26.
Euler's
Method yn
0.0
xn
1.0
0.240
Euler's
Method yn
1.0
b.
1.2
1.2
1.244
1.4
1.488
1.60924
1.6
1.90464
2.16410
1.8
2.51412
3.02455
2.0
3.41921
4.391765
N
N →∞
0.340
Improved Euler
Method yn
1.0
≈ e for large N. From Chapter 7,
=e.
18. y0 = y ( x0 ) = 0
y1 = y0 + hf ( x0 ) = 0 + hf ( x0 ) = hf ( x0 )
y2 = y1 + hf ( x1 ) = hf ( x0 ) + hf ( x1 )
= h ( f ( x0 ) + f ( x1 ) )
y3 = y2 + hf ( x2 )
= h [ f ( x0 ) + f ( x1 ) ] + hf ( x2 )
3−1
= h [ f ( x0 ) + f ( x1 ) + f ( x2 ) ] = h ∑ f ( xi )
i =0
At the nth step of Euler's method,
n −1
yn = yn −1 + hf ( xn −1 ) = h ∑ f ( xi )
16., 27.
xn
1.0
Euler's
Method yn
2.0
Improved Euler
Method yn
2.0
19. a.
∫
x1
x0
y '( x)dx = ∫
i =0
x1
2
sin x dx
x0
1.2
1.2
1.312
y ( x1 ) − y ( x0 ) ≈ ( x1 − x0 ) sin x02
1.4
0.624
0.80609
y ( x1 ) − y (0) = h sin x02
1.6
0.27456
0.46689
1.8
0.09884
0.25698
y ( x1 ) − 0 ≈ 0.1sin 02
y ( x1 ) ≈ 0
2.0
0.02768
0.13568
b.
x2
∫x0
y '( x)dx = ∫
x2
sin x 2 dx
x0
y ( x2 ) − y ( x0 ) ≈ ( x1 − x0 ) sin x02
+ ( x2 − x1 ) sin x12
y ( x2 ) − y (0) = h sin x02 + h sin x12
y ( x2 ) − 0 ≈ 0.1sin 02 + 0.1sin 0.12
y ( x2 ) ≈ 0.00099998
386
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c.
x3
x3
∫x0 y '( x)dx = ∫x0 sin x
2
x3
+ ( x2 − x1 ) x1 + 1 + ( x3 − x2 ) x2 + 1
+ ( x2 − x1 ) sin x12 + ( x3 − x2 ) sin x12
y ( x3 ) − y (0) = h sin x02 + h sin x12 + h sin x22
y ( x3 ) − y (0) = 0.1 0 + 1 + 0.1 0.1 + 1
y ( x3 ) − 0 ≈ 0.1sin 02 + 0.1sin 0.12
+ 0.1 0.2 + 1
y ( x3 ) ≈ 0.314425
Continuing in this fashion, we have
+ 0.1sin 0.22
y ( x3 ) ≈ 0.004999
Continuing in this fashion, we have
xn
∫x0
y '( x)dx = ∫
xn
∫x0
xn
sin x 2 dx
x0
n −1
i =0
y ( xn ) ≈ h ∑ xi −1 + 1
i =0
i =0
y ( xn ) ≈ h ∑ f ( xi −1 )
When n = 10 , this becomes
y ( x10 ) = y (1) ≈ 1.198119
i =0
When n = 10 , this becomes
y ( x10 ) = y (1) ≈ 0.269097
Δy 1
= [ f ( x0 , y0 ) + f ( x1 + yˆ1 )]
Δx 2
y1 − y0 Δy 1
=
= [ f ( x0 , y0 ) + f ( x1 + yˆ1 )] ⇒
h
Δx 2
2( y1 − y0 ) = h[ f ( x0 , y0 ) + f ( x1 + yˆ1 )] ⇒
21. a.
n −1
The result y ( xn ) ≈ h ∑ f ( xi −1 ) is the same as
b.
i =0
that given in Problem 18. Thus, when f ( x, y )
depends only on x , then the two methods (1)
Euler's method for approximating the solution
to y ' = f ( x) at xn , and (2) the left-endpoint
Riemann sum for approximating
xn
∫0
f ( x) dx ,
c.
are equivalent.
x1
x1
h
y1 − y0 = [ f ( x0 , y0 ) + f ( x1 + yˆ1 )] ⇒
2
h
y1 = y0 + [ f ( x0 , y0 ) + f ( x1 + yˆ1 )]
2
1. xn −1 + h
2. yn −1 + hf ( xn −1 , yn −1 )
x + 1 dx
h
3. yn −1 + [ f ( xn −1 , yn −1 ) + f ( xn , yˆ n )]
2
y ( x1 ) − y ( x0 ) ≈ ( x1 − x0 ) x0 + 1
y ( x1 ) − y (0) = h x0 + 1
y ( x1 ) − 0 ≈ 0.1 0 + 1
22-27.
See problems 11-16
h
0.2
Error from
Euler's
Method
0.229962
Error from
Improved
Euler Method
0.015574
0.1
0.124539
0.004201
0.05
0.064984
0.001091
y ( x2 ) − y (0) = h x0 + 1 + h x1 + 1
0.01
0.013468
0.000045
y ( x2 ) − 0 ≈ 0.1 0 + 1 + 0.1 0.1 + 1
0.005
0.006765
0.000011
y ( x1 ) ≈ 0.1
b.
xn
x + 1 dx
x0
n −1
n −1
n −1
∫x0 y '( x)dx = ∫x0
y '( x)dx = ∫
y ( xn ) − y ( x0 ) ≈ ∑ ( xi +1 − xi ) xi −1 + 1
y ( xn ) − y ( x0 ) ≈ ∑ ( xi +1 − xi ) sin xi2
20. a.
x + 1 dx
y ( x3 ) − y ( x0 ) ≈ ( x1 − x0 ) x0 + 1
y ( x3 ) − y ( x0 ) ≈ ( x1 − x0 ) sin x02
d.
x3
∫x0 y '( x)dx = ∫x0
c.
dx
x2
∫x0
y '( x)dx = ∫
x2
x0
x + 1 dx
y ( x2 ) − y ( x0 ) ≈ ( x1 − x0 ) x0 + 1
+ ( x2 − x1 ) x1 + 1
y ( x2 ) ≈ 0.204881
28.
For Euler's method, the error is halved as the step
size h is halved. Thus, the error is proportional to h.
For the improved Euler method, when h is halved,
the error decreases to approximately one-fourth of
what is was. Hence, for the improved Euler
method, the error is proportional to h 2
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6.8 Concepts Review
⎡ π π⎤
1. ⎢ – , ⎥ ; arcsin
⎣ 2 2⎦
⎛ π π⎞
2. ⎜ – , ⎟ ; arctan
⎝ 2 2⎠
3. 1
14. sec(arccos 0.5111) =
=
1
cos(arccos 0.5111)
1
≈ 1.957
0.5111
⎛ 1 ⎞
15. sec –1 (–2.222) = cos –1 ⎜
⎟ ≈ 2.038
⎝ –2.222 ⎠
4. π
Problem Set 6.8
16. tan −1 (−60.11) ≈ –1.554
⎛ 2⎞ π
π
2
1. arccos ⎜⎜
⎟⎟ = since cos =
4
2
⎝ 2 ⎠ 4
⎛
3⎞
π
3
⎛ π⎞
2. arcsin ⎜⎜ –
⎟⎟ = – since sin ⎜ – ⎟ = –
3
2
⎝ 3⎠
⎝ 2 ⎠
⎛
3⎞
π
3
⎛ π⎞
3. sin –1 ⎜⎜ –
⎟⎟ = – 3 since sin ⎜⎝ – 3 ⎟⎠ = – 2
2
⎝
⎠
⎛
2⎞
π
2
⎛ π⎞
4. sin –1 ⎜⎜ –
⎟⎟ = – since sin ⎜ – ⎟ = –
2
4
4
2
⎝
⎠
⎝
⎠
5. arctan( 3) =
π
⎛π⎞
since tan ⎜ ⎟ = 3
3
⎝3⎠
⎛1⎞ π
⎛π⎞ 1
6. arcsec(2) = arccos ⎜ ⎟ = since cos ⎜ ⎟ = , so
⎝2⎠ 3
⎝3⎠ 2
⎛π⎞
sec ⎜ ⎟ = 2
⎝3⎠
π
1
⎛ 1⎞
⎛ π⎞
7. arcsin ⎜ – ⎟ = – since sin ⎜ – ⎟ = –
6
2
⎝ 2⎠
⎝ 6⎠
⎛
3⎞
π
3
⎛ π⎞
8. tan –1 ⎜⎜ –
⎟⎟ = – since tan ⎜ – ⎟ = –
3
6
6
3
⎝
⎠
⎝
⎠
9. sin(sin –1 0.4567) = 0.4567 by definition
10. cos(sin –1 0.56) = 1 − sin 2 (sin −1 0.56)
= 1 – (0.56) 2 ≈ 0.828
11. sin −1 (0.1113) ≈ 0.1115
12. arccos(0.6341) ≈ 0.8840
388
1 ⎞
⎛
13. cos(arccot 3.212) = cos ⎜ arctan
⎟
3.212 ⎠
⎝
≈ cos 0.3018 ≈ 0.9548
Section 6.8
17. cos(sin(tan −1 2.001)) ≈ 0.6259
18. sin 2 (ln(cos 0.5555)) ≈ 0.02632
19. θ = sin −1
x
8
20. θ = tan −1
x
6
21. θ = sin −1
5
x
22. θ = cos −1
9
x
or θ = sec−1
9
x
23. Let θ1 be the angle opposite the side of length 3,
and θ 2 = θ1 – θ , so θ = θ1 – θ 2 . Then tan θ1 =
3
x
1
3
1
and tan θ 2 = . θ = tan –1 – tan –1 .
x
x
x
24. Let θ1 be the angle opposite the side of length 5,
and θ 2 = θ1 − θ , and y the length of the unlabeled
side. Then θ = θ1 − θ 2 and y = x 2 − 25.
tan θ1 =
5
=
y
θ = tan −1
5
, tan θ 2 =
2
=
y
2
x − 25
x − 25
5
2
− tan −1
2
2
x − 25
x − 25
2
2
,
⎡
⎡
⎛ 2 ⎞⎤
⎛ 2 ⎞⎤
25. cos ⎢ 2sin –1 ⎜ – ⎟ ⎥ = 1 – 2sin 2 ⎢sin –1 ⎜ – ⎟ ⎥
⎝ 3 ⎠⎦
⎝ 3 ⎠⎦
⎣
⎣
2
1
⎛ 2⎞
= 1– 2⎜ – ⎟ =
9
⎝ 3⎠
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()
()
2 tan ⎡ tan –1 13 ⎤
⎡
⎣
⎦
–1 ⎛ 1 ⎞ ⎤
26. tan ⎢ 2 tan ⎜ ⎟ ⎥ =
⎝ 3 ⎠ ⎦ 1 – tan 2 ⎡ tan –1 1 ⎤
⎣
3 ⎦
⎣
=
2 ⋅ 13
()
1 – 13
2
=
3
4
⎡
⎡
⎡
⎡
⎛3⎞
⎛ 5 ⎞⎤
⎛ 3 ⎞⎤
⎛ 5 ⎞⎤
⎛ 3 ⎞⎤ ⎡
⎛ 5 ⎞⎤
27. sin ⎢ cos –1 ⎜ ⎟ + cos –1 ⎜ ⎟ ⎥ = sin ⎢ cos –1 ⎜ ⎟ ⎥ cos ⎢cos –1 ⎜ ⎟ ⎥ + cos ⎢ cos –1 ⎜ ⎟ ⎥ sin ⎢ cos –1 ⎜ ⎟ ⎥
5
13
5
13
5
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝ 13 ⎠ ⎦
⎣
⎦
⎣
⎦
⎣
⎦
⎣
⎦ ⎣
2
2
56
⎛3⎞ 5 3
⎛5⎞
= 1– ⎜ ⎟ ⋅ + 1– ⎜ ⎟ =
65
⎝ 5 ⎠ 13 5
⎝ 13 ⎠
⎡
⎡
⎡
⎡
⎛4⎞
⎛ 12 ⎞ ⎤
⎛ 4 ⎞⎤
⎛ 12 ⎞ ⎤
⎛ 4 ⎞⎤ ⎡
⎛ 12 ⎞ ⎤
28. cos ⎢cos –1 ⎜ ⎟ + sin –1 ⎜ ⎟ ⎥ = cos ⎢cos –1 ⎜ ⎟ ⎥ cos ⎢sin –1 ⎜ ⎟ ⎥ – sin ⎢ cos –1 ⎜ ⎟ ⎥ sin ⎢sin –1 ⎜ ⎟ ⎥
5
13
5
13
5
⎝ ⎠
⎝ ⎠⎦
⎝ ⎠⎦
⎝ ⎠⎦
⎝ ⎠⎦ ⎣
⎝ 13 ⎠ ⎦
⎣
⎣
⎣
⎣
2
=
2
4
16
⎛ 12 ⎞
⎛ 4 ⎞ 12
⋅ 1– ⎜ ⎟ – 1– ⎜ ⎟ ⋅ = –
5
13
5
13
65
⎝ ⎠
⎝ ⎠
29. tan(sin –1 x) =
sin(sin –1 x)
cos(sin –1 x)
1
30. sin(tan –1 x) =
csc(tan
=
1
1+
1
tan 2 (tan –1 x )
=
–1
x
=
1 – x2
1
=
1 + cot 2 (tan –1 x)
x)
1
1+
1
x2
34. a.
=
b.
x
x2 + 1
31. cos(2sin –1 x) = 1 – 2sin 2 (sin –1 x) = 1 – 2 x 2
32. tan(2 tan
33. a.
b.
–1
x) =
2 tan(tan –1 x)
1 – tan 2 (tan –1 x )
lim tan –1 x =
x →∞
=
π
lim tan x = – since
2
x→ – ∞
lim tan θ = −∞
θ →−π / 2+
⎛1⎞
lim sec –1 x = lim cos –1 ⎜ ⎟
x→ – ∞
x→ – ∞
⎝x⎠
π
= lim cos –1 z =
–
2
z →0
Let L = lim sin −1 x . Since
2x
1 – x2
π
since lim tan θ = ∞
2
θ →π / 2−
–1
35. a.
⎛1⎞
lim sec –1 x = lim cos –1 ⎜ ⎟
x →∞
x →∞
⎝x⎠
π
= lim cos –1 z =
2
z → 0+
x →1−
sin(sin −1 x) = x,
lim sin(sin −1 x) = lim x = 1 .
x →1−
x →1−
Thus, since sin is continuous, the Composite
Limit Theorem gives us
lim sin(sin −1 x) = lim sin( L) ; hence
x →1−
x →1−
sin L = 1 and since the range of sin −1 is
π
⎡ π π⎤
⎢− 2 , 2 ⎥ , L = 2 .
⎣
⎦
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b. Let L = lim sin −1 x . Since
sin(sin
x →−1+
−1
39. y = ln(2 + sin x) . Let u = 2 + sin x ; then
y = ln u so by the Chain Rule
x ) = x,
lim sin(sin
−1
x →−1+
dy dy du ⎛ 1 ⎞ du ⎛
1
⎞
=
=⎜ ⎟
=⎜
⎟ ⋅ cos x
dx du dx ⎝ u ⎠ dx ⎝ 2 + sin x ⎠
cos x
=
2 + sin x
x) = lim x = −1 .
x →−1+
Thus, since sin is continuous, the
Composite Limit Theorem gives us
lim sin(sin −1 x) = lim sin( L) ;
x →−1+
x →−1+
40.
hence
sin L = −1 and since the range of sin −1 is
π
⎡ π π⎤
⎢− 2 , 2 ⎥ , L = − 2 .
⎣
⎦
41.
36. No. Since sin −1 x is not defined on (1, ∞) ,
42.
lim sin −1 x does not exist so neither can the
x →1+
two-sided limit lim sin −1 x .
x →1
f ′(c) =
1
1 − c2
d
– csc x cot x – csc2 x
[– ln(csc x + cot x)] = –
dx
csc x + cot x
csc x(cot x + csc x)
=
= csc x
cot x + csc x
d
1
4x
sin –1 (2 x 2 ) =
⋅ 4x =
dx
1 – (2 x 2 ) 2
1 – 4 x4
44.
d
1
ex
arccos(e x ) = –
⋅ ex = –
dx
1 – (e x ) 2
1 – e2 x
45.
d 3
ex
[ x tan –1 (e x )] = x3 ⋅
+ 3x 2 tan –1 (e x )
dx
1 + (e x ) 2
. Hence, lim f ′(c) = ∞ so
c →1−
d
sec x tan x + sec2 x
ln(sec x + tan x) =
dx
sec x + tan x
(sec x)(tan x + sec x)
=
= sec x
sec x + tan x
43.
37. Let f ( x) = y = sin −1 x ; then the slope of the
tangent line to the graph of y at c is
d tan x
d
e
= e tan x
tan x = e tan x sec 2 x
dx
dx
that the tangent lines approach the vertical.
⎡ xe x
⎤
= x2 ⎢
+ 3 tan –1 (e x ) ⎥
2x
⎣⎢1 + e
⎦⎥
38.
46.
d x
2x
(e arcsin x 2 ) = e x ⋅
+ e x arcsin x 2
2 2
dx
1 – (x )
⎛ 2x
⎞
= ex ⎜
+ arcsin x 2 ⎟
⎜
⎟
4
⎝ 1– x
⎠
47.
3(tan –1 x) 2
d
1
=
(tan –1 x)3 = 3(tan –1 x)2 ⋅
dx
1 + x2
1 + x2
48.
d
d sin(cos −1 x) d 1 – x 2
tan(cos –1 x) =
=
dx
dx cos(cos −1 x) dx
x
=
x ⋅ 12 ⋅
1
1– x 2
(–2 x) – 1 – x 2 ⋅1
x2
– x – (1 – x )
1
=
=–
2
2
2
x 1– x
x 1 – x2
2
390
Section 6.8
2
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49.
50.
d
sec –1 ( x3 ) =
dx
x3
( x3 ) 2 – 1
3
⋅ 3x2 =
d
(sec –1 x)3 = 3(sec –1 x) 2 ⋅
dx
x
=
x
54. y = x arc sec( x 2 + 1)
x6 – 1
dy
⎡d
⎤ ⎛d ⎞
= x ⎢ arcsec(x 2 + 1) ⎥ + ⎜ x ⎟ ⋅ arcsec(x 2 + 1)
dx
⎣ dx
⎦ ⎝ dx ⎠
⎡
⎢
= x⎢
2
⎢⎣ x + 1
1
2
(
x –1
3(sec –1 x)2
x
51.
1
(
1
(
3(1 + sin –1 x)2
1 – x2
; then y = sin −1 ( u ( x) ) so by the
⎛ −2 x
⋅⎜
2
2 ⎜ 2
⎛ 1 ⎞ ⎝ ( x + 4)
1− ⎜ 2
⎟
⎝ x +4⎠
1
⎞
⎟⎟ =
⎠
⎛
⎞ ⎛ −2 x ⎞
( x 2 + 4)
⎜
⎟⋅⎜
⎟=
2⎟
⎜ 4
⎟ ⎜ 2
2
⎝ x + 8 x + 15 ⎠ ⎝ ( x + 4) ⎠
−2 x
( x 2 + 4) x 4 + 8 x 2 + 15
(
53. y = tan −1 ln x 2
)
Let u = x , v = ln u ; then y = tan −1 ( v(u ( x)) ) so
2
by the Chain Rule:
dy dy dv du
1
1
=
=
⋅ ⋅ 2x =
dx dv du dx 1 + v 2 u
1
1
⋅ ⋅ 2x =
2 2
1 + (ln x ) x 2
2
x[1 + (ln x 2 )2 ]
)
⎡
⎤
2x
⎢
⎥
2
=⎢
⎥ + arcsec(x + 1)
2
2
⎢⎣ x + 1 x + 2 ⎥⎦
(
x +4
Chain Rule:
dy dy du
1
du
=
=
⋅
=
2 dx
dx du dx
1− u
2
)
⎤
⎥
2
⎥ + arcsec(x + 1)
⎥⎦
⎡
⎤
2 x2
⎢
⎥
2
=⎢
⎥ + arcsec(x + 1)
2
2
⎢⎣ x + 1 ⋅ x x + 2 ⎥⎦
⎛ 1 ⎞
52. y = sin −1 ⎜
⎟
⎝ x2 + 4 ⎠
Let u =
)
⎡
2x2
⎢
=⎢
2
4
2
⎢⎣ x + 1 x + 2 x
x 2 –1
d
1
(1 + sin –1 x)3 = 3(1 + sin –1 x)2 ⋅
dx
1 – x2
=
⎤
⎥
2
⎥ + 1 ⋅ arcsec(x + 1)
2
2
( x + 1) − 1 ⎥
⎦
2x
55.
)
∫ cos 3x dx
Let u = 3 x, du = 3dx ; then
1
∫ cos 3x dx = 3 ∫ cos 3x (3dx) =
1
1
1
cos u du = sin u + C = sin 3 x + C
∫
3
3
3
56. Let u = x 2 , so du = 2 x dx .
1
sin( x 2 ) ⋅ 2 x dx
2∫
1
1
= ∫ sin u du = − cos u + C
2
2
1
2
= − cos( x ) + C
2
∫ x sin( x
2
)dx =
57. Let u = sin 2x, so du = 2 cos 2x dx.
1
∫ sin 2 x cos 2 x dx = 2 ∫ sin 2 x(2 cos 2 x)dx
1
= ∫ u du
2
u2
1
=
+ C = sin 2 2 x + C
4
4
58. Let u = cos x, so du = − sin x dx .
sin x
1
∫ tan x dx = ∫ cos x dx = −∫ cos x (− sin x)dx
1
= − ∫ du = − ln u + C = − ln cos x + C
u
= ln sec x + C
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
65. Let u = 2x, so du = 2 dx.
1
1
1
∫ 1 + 4 x2 dx = 2 ∫ 1 + (2 x)2 2dx
59. Let u = e2 x , so du = 2e2 x dx .
1
2x
2x
2x
2x
∫ e cos(e )dx = 2 ∫ cos(e )(2e )dx
1
= ∫ cos u du
2
1
1
= sin u + C = sin(e2 x ) + C
2
2
1
1
1
du = arctan u + C
∫
2
2 1+ u
2
1
= arctan 2 x + C
2
=
1
⎡1
2x
2x ⎤
∫ cos(e ) dx = ⎢⎣ 2 sin(e ) ⎥⎦0
1
⎡1
⎤
= ⎢ sin(e2 ) − sin(e0 ) ⎥
2
⎣2
⎦
1 2x
e
0
66. Let u = e x , so du = e x dx .
ex
∫ 1 + e2 x
60. Let u = sin x, so du = cos x dx.
u3
sin 3 x
2
2
sin
cos
x
x
dx
=
u
du
=
+
C
=
+C
∫
∫
3
3
π/2
∫0
61.
π/2
⎡ sin 3 x ⎤
sin 2 x cos x dx = ⎢
⎥
⎢⎣ 3 ⎥⎦ 0
1
2/2
∫0
1– x
∫
dx
2
2
x x2 − 1
= sec
−1
= cos
63.
2 − sec
=∫
1
1
−0 =
3
3
−1
−1 ⎛ 1 ⎞
⎜ ⎟ − cos
⎝2⎠
2
x2 − 1
x
1
1
2
∫
3
dx
1
⎛ 3 ⎞
x ⎟⎟
1−⎜⎜
⎝ 2 ⎠
2
dx
3
3
x, du =
dx ; then
2
2
1
1 ⎛ 2 ⎞
1
dx =
du
⎜
⎟∫
2
2 3 ⎝ 3 ⎠ 1− u2
⎛ 3 ⎞
1−⎜
x⎟
∫
= ⎡sec−1 x ⎤
⎣
⎦
2
2
−1 ⎛
2⎞ π π π
⎜⎜
⎟⎟ = − =
⎝ 2 ⎠ 3 4 12
1
x ⎤ = tan −1 1 − tan −1 (−1)
⎦ −1
π ⎛ π⎞ π
−⎜− ⎟ =
4 ⎝ 4⎠ 2
∫
x
12 − 9 x
2
dx . Let u = 12 − 9 x 2 , du = −18 x dx;
then
x
1
1
(−18 dx)
∫
18 12 − 9 x 2
12 − 9 x 2
1 1
⎛ 1⎞
=− ∫
du = ⎜ − ⎟ (2 u ) + C
18 u
⎝ 18 ⎠
∫
=−
64. Let u = cos θ , so du = − sin θ dθ .
⎟
⎠
⎛ 3 ⎞
1
1
x ⎟⎟ + C
= sin −1 u + C = sin −1 ⎜⎜
3
3
⎝ 2 ⎠
2
sin θ
1
⎛ 3 ⎞
12⎜ 1− x 2 ⎟
⎝ 4 ⎠
du
Let u =
68.
−1
x 2
dx = ∫
⎜ 2
⎝
dx
2
2
=
2 3
∫−11 + x 2 dx = ⎡⎣ tan
=
12 − 9 x
dx = [arcsin x]0 2 / 2
1
1
=
1
67. ∫
π
2
– arcsin 0 =
2
4
= arcsin
62.
2
1
dx = ∫
1 + (e )
1 + u2
= arctan u + C = arctan ex + C
sin e2 − sin1
≈ 0.0262
2
=
ex
dx = ∫
dx = −
12 − 9 x 2
+C
9
1
∫ 1 + cos2 θ dθ = − ∫ 1 + cos2 θ (− sin θ )dθ
= −∫
1
1+ u
2
du = − tan −1 u + C
= − tan −1 (cos θ ) + C
π/2
∫0
sin θ
= − tan −1 0 + tan −1 1 = −0 +
392
π/2
dθ = ⎡ − tan −1 (cosθ ) ⎤
2
⎣
⎦0
1 + cos θ
Section 6.8
π π
=
4 4
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
69. ∫
1
x − 6 x + 13
1
2
1
dx = ∫
( x − 6 x + 9) + 4
2
73. The top of the picture is 7.6 ft above eye level,
and the bottom of the picture is 2.6 ft above eye
level. Let θ1 be the angle between the viewer’s
line of sight to the top of the picture and the
horizontal. Then call θ 2 = θ1 − θ , so θ = θ1 − θ 2 .
dx
=∫
dx
( x − 3)2 + 4
Let u = x − 3, du = dx, a = 2; then
1
−1 ⎛ u ⎞
∫ ( x − 3)2 + 4 dx = ∫ u 2 + a 2 du = a tan ⎜⎝ a ⎟⎠ + C
1
=
70.
1
1
⎛ x−3⎞
tan −1 ⎜
⎟+C
2
⎝ 2 ⎠
1
1
∫ 2 x2 + 8 x + 25 dx = ∫ 2( x 2 + 4 x + 4 + 17 ) dx =
2
1
2∫
1
⎛ 17 ⎞
⎟
⎝ 2 ⎠
( x + 2) 2 + ⎜
2
dx
17
; then
2
1
1
dx = ∫
du =
2 u2 + a2
Let u = x + 2, du = dx, a =
1
1
2 ∫ ( x + 2) 2 + 17
2
⎛ x+2⎞
1 1
1
2
⎛u⎞
⋅ tan −1 ⎜ ⎟ + C = ⋅
tan −1 ⎜
⎟+C
⎜ 17 ⎟
2 a
2 17
⎝a⎠
2 ⎠
⎝
=
71.
∫
⎡ 34 ⋅ ( x + 2) ⎤
34
tan −1 ⎢
⎥+C
34
17
⎣
⎦
1
x 4 x2 − 9
then
∫
∫
dx . Let u = 2 x, du = 2 dx, a = 3 ;
1
dx = ∫
x 4x − 9
2
1
du =
u u 2 − a2
1 −1 ⎛ 2 x
sec ⎜
3
⎝ 3
72.
∫
x +1
dx = ∫
1
2 x 4 x2 − 9
(2 dx) =
1 −1 ⎛ u ⎞
sec ⎜ ⎟ + C =
a
⎝a⎠
⎞
⎟+C
⎠
x
x +1
4 − 9x
2
dx = −
⎡ π⎤
74. a. Restrict 2x to [0, π ] , i.e., restrict x to ⎢ 0, ⎥ .
⎣ 2⎦
Then y = 3 cos 2x
y
= cos 2 x
3
y
2 x = arccos
3
1
y
x = f –1 ( y ) = arccos
2
3
1
x
f –1 ( x) = arccos
2
3
π
π⎤
⎡
b. Restrict 3x to ⎢ – , ⎥ , i.e., restrict x to
⎣ 2 2⎦
⎡ π π⎤
⎢– 6 , 6 ⎥
⎣
⎦
Then y = 2 sin 3x
y
= sin 3 x
2
y
3x = arcsin
2
1
y
x = f –1 ( y ) = arcsin
3
2
1
x
f –1 ( x) = arcsin
3
2
c.
dx + ∫
1
dx
4 − 9 x2
4 − 9 x2
4 − 9 x2
These integrals are evaluated the same as those in
problems 67 and 68 (with a constant of 4 rather
than 12). Thus
∫
7.6
2.6
; tan θ 2 =
;
b
b
7.6
2.6
θ = tan −1
− tan −1
b
b
If b = 12.9, θ ≈ 0.3335 or 19.1° .
tan θ1 =
⎛ π π⎞
Restrict x to ⎜ – , ⎟
⎝ 2 2⎠
1
y = tan x
2
2y = tan x
x = f –1 ( y ) = arctan 2 y
f –1 ( x) = arctan 2 x
1
1
⎛ 3x ⎞
4 − 9 x 2 + sin −1 ⎜ ⎟ + C
9
3
⎝ 2 ⎠
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1
2⎞ ⎛2 ⎞
⎛
d. Restrict x to ⎜ – ∞, – ⎟ ∪ ⎜ , ∞ ⎟ so
is
x
π⎠ ⎝π ⎠
⎝
⎛ π ⎞ ⎛ π⎞
restricted to ⎜ – , 0 ⎟ ∪ ⎜ 0, ⎟
⎝ 2 ⎠ ⎝ 2⎠
1
then y = sin
x
1
= arcsin y
x
1
x = f −1 ( y ) =
arcsin y
f −1 ( x) =
2 ⋅ 14
1–
( )
1 2
4
=
( )
( )
47
=
52
⎡
⎛1⎞
⎛ 5 ⎞⎤
tan ⎢3 tan –1 ⎜ ⎟ + tan –1 ⎜ ⎟ ⎥
4
⎝ ⎠
⎝ 99 ⎠ ⎦
⎣
( )⎤⎦ + tan ⎡⎣ tan –1 ( 995 )⎤⎦
( 14 )⎤⎦ tan ⎡⎣ tan –1 ( 995 )⎤⎦
tan ⎡3 tan –1 14
⎣
=
1 – tan ⎡3 tan –1
⎣
5
+ 99
4913
π
=
=
= 1 = tan
47
5
4
1 – 52 ⋅ 99 4913
π
⎛1⎞
⎛ 5 ⎞
Thus, 3 tan –1 ⎜ ⎟ + tan –1 ⎜ ⎟ = tan –1 (1) = .
4
⎝4⎠
⎝ 99 ⎠
⎡
⎛ 1 ⎞⎤
2 tan ⎢ tan –1 ⎜ ⎟ ⎥
⎡
⎛ 1 ⎞⎤
⎝ 5 ⎠⎦
⎣
76. tan ⎢ 2 tan –1 ⎜ ⎟ ⎥ =
5
⎡
⎛ 1 ⎞⎤
⎝ ⎠⎦
⎣
1 – tan 2 ⎢ tan –1 ⎜ ⎟ ⎥
⎝ 5 ⎠⎦
⎣
=
2 ⋅ 15
()
1 – 15
2
=
5
12
⎡
⎡
⎛ 1 ⎞⎤
⎛ 1 ⎞⎤
tan ⎢ 4 tan –1 ⎜ ⎟ ⎥ = tan ⎢ 2 ⋅ 2 tan –1 ⎜ ⎟ ⎥
5
⎝
⎠
⎝ 5 ⎠⎦
⎣
⎦
⎣
394
⎡
⎡
⎛ 1 ⎞⎤
⎛ 1 ⎞⎤
tan ⎢ 4 tan –1 ⎜ ⎟ ⎥ – tan ⎢ tan –1 ⎜
⎟⎥
⎝ 5 ⎠⎦
⎝ 239 ⎠ ⎦
⎣
⎣
=
⎡
⎡
⎛ 1 ⎞⎤
⎛ 1 ⎞⎤
1 + tan ⎢ 4 tan –1 ⎜ ⎟ ⎥ tan ⎢ tan –1 ⎜
⎟⎥
5
⎝
⎠
⎝ 239 ⎠ ⎦
⎣
⎦
⎣
1
–
28,561
π
= 119 239 =
= 1 = tan
120
1
4
1 + 119 ⋅ 239 28,561
π
⎛1⎞
⎛ 1 ⎞
−1
Thus, 4 tan –1 ⎜ ⎟ – tan –1 ⎜
⎟ = tan (1) =
4
⎝5⎠
⎝ 239 ⎠
77.
8
15
( )
( )
47
52
⎡
⎛1⎞
⎛ 1 ⎞⎤
tan ⎢ 4 tan –1 ⎜ ⎟ – tan –1 ⎜
⎟⎥
5
⎝ ⎠
⎝ 239 ⎠ ⎦
⎣
120
⎡
⎡
⎛ 1 ⎞⎤
⎛1⎞
⎛ 1 ⎞⎤
tan ⎢3 tan –1 ⎜ ⎟ ⎥ = tan ⎢ 2 tan –1 ⎜ ⎟ + tan –1 ⎜ ⎟ ⎥
4
4
⎝ ⎠⎦
⎝ ⎠
⎝ 4 ⎠⎦
⎣
⎣
tan ⎡ 2 tan –1 14 ⎤ + tan ⎡ tan –1 14 ⎤
⎣
⎦
⎣
⎦
=
−
1
–1
1
1
⎤ tan ⎡ tan
⎤
1 – tan ⎡ 2 tan
4 ⎦
4 ⎦
⎣
⎣
8 +1
= 15 4
8 ⋅1
1 – 15
4
( )
1
arcsin x
⎡
⎛ 1 ⎞⎤
2 tan ⎢ tan –1 ⎜ ⎟ ⎥
⎡
⎛ 1 ⎞⎤
⎝ 4 ⎠⎦
⎣
75. tan ⎢ 2 tan –1 ⎜ ⎟ ⎥ =
4
⎡
⎛ 1 ⎞⎤
⎝ ⎠⎦
⎣
1 – tan 2 ⎢ tan –1 ⎜ ⎟ ⎥
⎝ 4 ⎠⎦
⎣
=
⎡
⎛ 1 ⎞⎤
2 tan ⎢ 2 tan –1 ⎜ ⎟ ⎥
5
2 ⋅ 12
120
⎝ 5 ⎠⎦
⎣
=
=
=
2
119
⎡
⎛ 1 ⎞⎤
5
1 – tan 2 ⎢ 2 tan –1 ⎜ ⎟ ⎥ 1 – 12
5
⎝ ⎠⎦
⎣
Section 6.8
Let θ represent ∠DAB, then ∠CAB is
θ
2
. Since
b
ΔABC is isosceles, AE =
b
θ
b
, cos = 2 =
and
2
2 a 2a
b
. Thus sector ADB has area
2a
1⎛
–1 b ⎞ 2
2
–1 b
. Let φ represent
⎜ 2 cos
⎟ b = b cos
2⎝
2a ⎠
2a
θ = 2 cos –1
∠DCB, then ∠ACB is
φ
φ
2
and ∠ECA is
φ
4
, so
b
b
b
= 2 =
and φ = 4sin –1 . Thus sector
4 a 2a
2a
1⎛
b
b
⎞
DCB has area ⎜ 4sin –1 ⎟ a 2 = 2a 2 sin –1 .
2⎝
2a ⎠
2a
These sectors overlap on the triangles ΔDAC and
ΔCAB, each of which has area
sin
2
1
1
1
4a 2 – b 2
⎛b⎞
AB h = b a 2 – ⎜ ⎟ = b
.
2
2
2
2
⎝2⎠
The large circle has area πb 2 , hence the shaded
region has area
b
b 1
– 2a 2 sin –1
πb 2 – b 2 cos –1
+ b 4a 2 – b 2
2a
2a 2
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y = sin(arcsin x) is the line y = x, but only
defined for −1 ≤ x ≤ 1 .
78.
y = arcsin(sin x) is defined for all x, but only the
π
π
portion for – ≤ x ≤ is the line y = x.
2
2
They have the same graph.
Conjecture: arcsin x = arctan
x
1 – x2
for
–1 < x < 1
Proof: Let θ = arcsin x, so x = sin θ.
x
sin θ
sin θ
Then
=
=
= tan θ
1 – x2
1 – sin 2 θ cos θ
so θ = arctan
x
1 – x2
.
81.
∫
=∫
dx
dx
=∫
( )
2⎤
⎡
a 2 ⎢1 – ax ⎥
⎣
⎦
1
dx
1
dx
⋅
=∫ ⋅
2
a
a
1 – ax
1 – ax
a2 – x2
( )
( )
2
since a > 0
x
1
, so du = dx.
a
a
1
dx
du
=∫
= sin −1 u + C
∫a⋅
2
2
1− u
1 – ax
Let u =
( )
79.
= sin −1
x
+C
a
82. Dx sin −1
It is the same graph as y = arccos x.
π
Conjecture: – arcsin x = arccos x
2
π
Proof: Let θ = – arcsin x
2
⎛π
⎞
Then x = sin ⎜ − θ ⎟ = cos θ
2
⎝
⎠
so θ = arccos x.
80.
=
=
=
x
=
a
1
a2 – x2
a2
⋅
1
1–
1
=
a
( )
x 2
a
⋅
1
a
a
a2 – x2
⋅
1
a
1
⋅ , since a > 0
a
a –x
1
a
2
2
a2 – x2
x
1
, so du = dx
a
a
dx
1
1
1
∫ a 2 + x 2 = a ∫ x 2 a dx
1+
83. Let u =
(a)
1
1
1
du = tan −1 u + C
a ∫ 1+ u2
a
1
⎛ x⎞
= tan −1 ⎜ ⎟ + C
a
⎝a⎠
=
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395
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
x
, so du = (1/ a ) dx. Since a > 0,
a
dx
1
1
1
dx
∫ 2 2 = a∫
2
a
x
x
x x −a
−1
a
a
84. Let u =
86.
1
a − x2
πa 2
2
This result is expected because the integral
should be half the area of a circle with radius a.
= a 2 sin −1 (1) =
(See
87. Let θ be the angle subtended by viewer’s eye.
⎛ 12 ⎞
⎛2⎞
θ = tan −1 ⎜ ⎟ − tan −1 ⎜ ⎟
b
⎝ ⎠
⎝b⎠
⎛ 12 ⎞
dθ
1
1 ⎛ 2 ⎞
=
⎜−
⎟−
⎜− ⎟
db 1 + 12 2 ⎜⎝ b 2 ⎟⎠ 1 + 2 2 ⎝ b 2 ⎠
b
b
Problem 67).
⎤
d ⎡x 2
a2
x
a − x2 +
sin −1 + C ⎥
⎢
dx ⎢⎣ 2
2
a
⎥⎦
1 2
x
1
( −2 x )
=
a − x2 +
2
2 2 a2 − x2
a2
+
2
=
1
a2 − x2
a 2 − x 2 dx
⎡a
⎤
a2
0 2 a2
sin −1 (1) −
a −
sin −1 (0) ⎥
= 2 ⎢ (0) +
2
2
2
⎣⎢ 2
⎦⎥
1
1
du
a ∫ u u2 −1
x
1
1
= sec −1 u + C = sec−1 + C
a
a
a
2
a
0
a
=
d
⎛ x⎞
sin −1 ⎜ ⎟ =
dx
⎝a⎠
a 2 − x 2 dx = 2 ∫
⎡x 2
a2
x⎤
a − x2 +
sin −1 ⎥
= 2⎢
a ⎦⎥
2
⎣⎢ 2
0
( ) ( )
85. Note that
a
∫−a
( )
=
+0
2
−
( )
12
=
10(24 − b 2 )
b 2 + 4 b 2 + 144 (b 2 + 4)(b 2 + 144)
dθ
Since
> 0 for b in ⎡⎣ 0, 2 6
db
dθ
and
< 0 for b in 2 6, ∞ , the angle is
db
)
1 2
1 − x2 + a2
= a2 − x2
a − x2 +
2
2 a2 − x2
(
)
maximized for b = 2 6 ≈ 4.899 .
The ideal distance is about 4.9 ft from the wall.
88. a.
⎛ x⎞
⎝ ⎠
⎛
−1
dθ ⎜
=⎜
dt ⎜
⎜ 1 − bx
⎝
( )
⎛
b.
⎛ x⎞
⎝ ⎠
θ = cos −1 ⎜ ⎟ − cos−1 ⎜ ⎟
b
a
2
⎞
⎛
⎟ ⎛ 1 ⎞⎛ dx ⎞ ⎜
−1
⎟ ⎜ ⎟⎜ ⎟ − ⎜
⎟⎟ ⎝ b ⎠⎝ dt ⎠ ⎜⎜ 1 − x
a
⎠
⎝
( )
⎞ dx
⎟
⎟ dt
⎠
⎞
⎛ x⎞
⎟ − sin −1 ⎜ ⎟
⎜ 2
⎟
2
⎝b⎠
⎝ b −x ⎠
θ = tan −1 ⎜
a+x
⎛
⎜
dθ ⎜
1
=⎜
dt ⎜ ⎛
a+ x
1+
⎜⎜ ⎜ b2 − x 2
⎝ ⎝
⎞
⎟
⎠
2
⎡⎛
b2 − x2
= ⎢⎜
2
2
⎜ 2
⎣⎢⎝ b − x + (a + x )
⎞
⎟ ⎛ b2 − x2 + (a + x) x
⎟⎜
b2 − x2
⎟⎜
2
2
b −x
⎟⎜
⎟⎟ ⎜⎝
⎠
Section 6.8
⎞
⎛
⎞
⎟ ⎛ dx ⎞ ⎜
⎟ ⎛ 1 ⎞ ⎛ dx ⎞
1
⎟⎜ ⎟− ⎜
⎟⎜ ⎟⎜ ⎟
⎟ ⎝ dt ⎠ ⎜
x 2 ⎟⎟ ⎝ b ⎠ ⎝ dt ⎠
⎜
−
1
⎟
b
⎝
⎠
⎠
⎞⎛ b 2 + ax ⎞
1
⎟⎜ 2
⎟−
⎟⎜ (b − x 2 )3 / 2 ⎟
2
b − x2
⎠⎝
⎠
⎡
b 2 + ax
1
=⎢
−
⎢ (b 2 + a 2 + 2ax) b 2 − x 2
b2 − x2
⎣
396
2
⎞
⎟ ⎛ 1 ⎞ ⎛ dx ⎞ ⎛
1
1
−
⎟⎜ ⎟⎜ ⎟ = ⎜
⎟⎟ ⎝ a ⎠ ⎝ dt ⎠ ⎜⎝ a 2 − x 2
b2 − x2
⎠
( )
⎤ dx
⎥
⎦⎥ dt
⎤ dx ⎡
a 2 + ax
⎥
= ⎢−
⎥ dt ⎢ (b 2 + a 2 + 2ax) b 2 − x 2
⎦
⎣
⎤ dx
⎥
⎥ dt
⎦
Instructor’s Resource Manual
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
89. Let h(t) represent the height of the elevator (the
number of feet above the spectator’s line of sight) t
seconds after the line of sight passes horizontal, and
let θ (t ) denote the angle of elevation.
⎛ 15t ⎞
−1 ⎛ t ⎞
Then h(t) = 15t, so θ (t ) = tan −1 ⎜
⎟ = tan ⎜ ⎟ .
⎝ 60 ⎠
⎝4⎠
dθ
1 ⎛1⎞
4
=
⎜ ⎟=
dt 1 + t 2 ⎝ 4 ⎠ 16 + t 2
(4)
dθ
4
1
=
=
radians per second or
dt 16 + 62 13
about 4.41° per second.
At t = 6,
90. Let x(t) be the horizontal distance from the observer
to the plane, in miles, at time t., in minutes.
Let t = 0 when the distance to the plane is 3 miles.
Then
x(0) = 32 − 22 = 5 . The speed of the plane is 10
miles per minute, so x(t ) = 5 − 10t. The angle of
⎛ 2 ⎞
−1 ⎛
elevation is θ (t ) = tan −1 ⎜
⎟ = tan ⎜
x
(
t
)
⎝
⎝
⎠
⎛
−2
1
dθ
so
=
⎜⎜
2
2
dt
⎝ ( 5 − 10t )
1 + 2 / 5 − 10t
( (
=
20
( 5 − 10t ) 2 + 4
When t = 0,
91.
))
⎞
⎟,
5 − 10t ⎠
⎞
⎟⎟ (−10)
⎠
2
.
dθ 20
=
≈ 2.22 radians per minute.
dt
9
Let x represent the position on the shoreline and let
θ represent the angle of the beam (x = 0 and θ = 0
92 Let x represent the length of the rope and let θ
represent the angle of depression of the rope.
⎛8⎞
Then θ = sin −1 ⎜ ⎟ , so
⎝x⎠
dθ
1
8 dx
8
dx
=
−
=−
.
2 dt
2
2
dt
dt
x
8
−
x
x
64
1− x
( )
dx
= −5 , we obtain
dt
dθ
8
8
=−
(−5) = .
2
dt
51
17 17 − 64
The angle of depression is increasing at a rate of
8 / 51 ≈ 0.16 radians per second.
When x = 17 and
93. Let x represent the distance to the center of the earth
and let θ represent the angle subtended by the
⎛ 6376 ⎞
earth. Then θ = 2sin −1 ⎜
⎟ , so
⎝ x ⎠
dθ
1
⎛ 6376 ⎞ dx
=2
⎜− 2 ⎟
2
dt
⎝ x ⎠ dt
1 − 6376
x
(
)
dx
x x 2 − 63762 dt
When she is 3000 km from the surface
dx
x = 3000 + 6376 = 9376 and
= −2 . Substituting
dt
dθ
≈ 3.96 × 10−4 radians
these values, we obtain
dt
per second.
=−
12, 752
when the light is pointed at P). Then
dθ
1
1 dx
2 dx
⎛ x⎞
θ = tan −1 ⎜ ⎟ , so
=
=
2
dt 1 + x 2 dt 4 + x 2 dt
⎝2⎠
(2)
When x = 1,
dx
dθ
2
= 5π, so
=
(5π) = 2π The beacon
dt
dt 4 + 12
revolves at a rate of 2π radians per minute or 1
revolution per minute.
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397
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6.9 Concepts Review
1.
e x – e– x e x + e– x
;
2
2
2. cosh 2 x − sinh 2 x = 1
3. the graph of x 2 − y 2 = 1 , a hyperbola
4. catenary; a hanging cable or chain
Problem Set 6.9
1. cosh x + sinh x =
=
e x + e– x e x – e– x
+
2
2
2e x
= ex
2
2. cosh 2 x + sinh 2 x =
=
2e 2 x
= e2 x
2
3. cosh x – sinh x =
=
e 2 x + e –2 x e2 x – e –2 x
+
2
2
e x + e– x e x – e– x
–
2
2
2e – x
= e– x
2
4. cosh 2 x – sinh 2 x =
e 2 x + e –2 x e2 x – e –2 x 2e –2 x
–
=
= e –2 x
2
2
2
ex – e– x e y + e– y ex + e– x e y – e– y
⋅
+
⋅
2
2
2
2
x+ y
x– y
– x+ y
– x– y
x+ y
x– y
– x+ y
– x– y
e
+e
–e
–e
e
–e
+e
–e
=
+
4
4
2e x + y – 2e –( x + y ) e x + y – e –( x + y )
=
=
= sinh( x + y )
4
2
5. sinh x cosh y + cosh x sinh y =
ex – e– x e y + e– y ex + e– x e y – e– y
⋅
–
⋅
2
2
2
2
e x+ y + e x – y – e – x+ y – e– x – y e x+ y – e x – y + e – x+ y – e – x – y
=
–
4
4
x– y
– x+ y
x– y
–( x – y )
2e
– 2e
e
–e
=
=
= sinh( x – y )
4
2
6. sinh x cosh y – cosh x sinh y =
398
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ex + e– x e y + e– y ex – e– x e y – e– y
⋅
+
⋅
2
2
2
2
e x+ y + e x – y + e – x+ y + e– x – y e x+ y – e x – y – e – x+ y + e – x – y
=
+
4
4
x+ y
– x– y
x+ y
–( x + y )
2e
+ 2e
e
+e
=
=
= cosh( x + y )
4
2
7. cosh x cosh y + sinh x sinh y =
ex + e– x e y + e– y ex – e– x e y – e– y
⋅
–
⋅
2
2
2
2
x+ y
x– y
– x+ y
– x– y
x+ y
x– y
– x+ y
– x– y
e
+e
+e
+e
e
–e
–e
+e
=
–
4
4
2e x – y + 2e – x + y e x – y + e –( x – y )
=
=
= cosh( x – y )
4
2
8. cosh x cosh y – sinh x sinh y =
sinh y
9.
sinh x +
tanh x + tanh y
cosh x cosh y
=
1 + tanh x tanh y 1 + sinh x ⋅ sinh y
cosh x cosh y
sinh x cosh y + cosh x sinh y sinh( x + y )
=
cosh x cosh y + sinh x sinh y cosh( x + y )
= tanh (x + y)
=
sinh x
16. Dx cosh 3 x = 3cosh 2 x sinh x
17. Dx cosh(3x + 1) = sinh(3 x + 1) ⋅ 3 = 3sinh(3 x + 1)
18. Dx sinh( x 2 + x) = cosh( x 2 + x) ⋅ (2 x + 1)
= (2 x + 1) cosh( x 2 + x)
sinh y
–
tanh x – tanh y
cosh x cosh y
10.
=
1 – tanh x tanh y 1 – sinh x ⋅ sinh y
cosh x cosh y
sinh x cosh y – cosh x sinh y sinh( x – y )
=
=
cosh x cosh y – sinh x sinh y cosh( x – y )
= tanh(x – y)
11. 2 sinh x cosh x = sinh x cosh x + cosh x sinh x
= sinh (x + x) = sinh 2x
12. cosh 2 x + sinh 2 x = cosh x cosh x + sinh x sinh x
= cosh( x + x) = cosh 2 x
19. Dx ln(sinh x) =
1
cosh x
⋅ cosh x =
sinh x
sinh x
= coth x
1
(–csch 2 x)
coth x
sinh x
1
1
=–
⋅
=–
2
cosh x sinh x
sinh x cosh x
= − csch x sech x
20. Dx ln(coth x) =
21. Dx ( x 2 cosh x) = x 2 ⋅ sinh x + cosh x ⋅ 2 x
= x 2 sinh x + 2 x cosh x
2
13. Dx sinh x = 2sinh x cosh x = sinh 2 x
14. Dx cosh 2 x = 2 cosh x sinh x = sinh 2 x
22. Dx ( x –2 sinh x) = x –2 ⋅ cosh x + sinh x ⋅ (–2 x –3 )
= x −2 cosh x − 2 x −3 sinh x
15. Dx (5sinh 2 x) = 10sinh x ⋅ cosh x = 5sinh 2 x
23. Dx (cosh 3x sinh x) = cosh 3 x ⋅ cosh x + sinh x ⋅ sinh 3x ⋅ 3 = cosh 3x cosh x + 3sinh 3x sinh x
24. Dx (sinh x cosh 4 x) = sinh x ⋅ sinh 4 x ⋅ 4 + cosh 4 x ⋅ cosh x = 4 sinh x sinh 4x + cosh x cosh 4x
25. Dx (tanh x sinh 2 x) = tanh x ⋅ cosh 2 x ⋅ 2 + sinh 2 x ⋅ sech 2 x = 2 tanh x cosh 2 x + sinh 2 x sech 2 x
26. Dx (coth 4 x sinh x) = coth 4 x ⋅ cosh x + sinh x(–csch 2 4 x) ⋅ 4 = cosh x coth 4 x – 4sinh x csch 2 4 x
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1
27. Dx sinh –1 ( x 2 ) =
2 2
(x ) +1
1
28. Dx cosh –1 ( x3 ) =
( x3 ) 2 –1
29. Dx tanh –1 (2 x – 3) =
2x
⋅ 2x =
x4 + 1
1
1 – (2 x – 3)
2
1
=
1 – (4 x –12 x + 9)
⎛ 5
⋅ −
2 ⎜
⎝ x6
1 – ⎛⎜ 15 ⎞⎟
⎝x ⎠
1
(3 x) –1
1
5 2
(x ) +1
1
cosh
2
2
–1
⋅
x
=
2
2
–4 x + 12 x – 8
1
=–
2
2( x – 3 x + 2)
5x4
x10 ⎛ 5 ⎞
⎞
⋅⎜ − ⎟ = –
⎟ = 10
x10 –1
⎠ x – 1 ⎝ x6 ⎠
⋅ 3 + cosh –1 (3 x) ⋅1 =
2
32. Dx ( x 2 sinh –1 x5 ) = x 2 ⋅
33. Dx ln(cosh –1 x) =
x6 –1
⋅2 =
⎛ 1 ⎞
30. Dx coth –1 ( x5 ) = Dx tanh −1 ⎜ ⎟ =
⎝ x5 ⎠
31. Dx [ x cosh –1 (3 x)] = x ⋅
3x2
⋅ 3x2 =
3x
+ cosh –1 3 x
2
9x – 1
⋅ 5 x 4 + sinh –1 x5 ⋅ 2 x =
5 x6
10
x
+1
+ 2 x sinh –1 x5
38. Let u = 3x + 2, so du = 3 dx.
1
1
∫ sinh(3x + 2)dx = 3 ∫ sinh u du = 3 cosh u + C
1
= cosh(3x + 2) + C
3
1
x2 – 1
1
x 2 – 1 cosh –1 x
34. cosh –1 (cos x) does not have a derivative, since
Du cosh −1 u is only defined for u > 1 while
cos x ≤ 1 for all x.
35. Dx tanh(cot x ) = sech 2 (cot x ) ⋅ (– csc2 x)
39. Let u = πx 2 + 5, so du = 2πxdx .
∫ x cosh(πx
=
2
+ 5)dx =
1
1
sinh u + C =
sinh(πx 2 + 5) + C
2π
2π
= – csc2 x sech 2 (cot x)
⎛ 1 ⎞
36. Dx coth –1 (tanh x) = Dx tanh –1 ⎜
⎟
⎝ tanh x ⎠
–1
= Dx tanh (coth x)
=
1
1 – (coth x)2
(–csch 2 x) =
–csch 2 x
–csch 2 x
=1
⎡1
⎤ ln 3
cosh 2 xdx = ⎢ sinh 2 x ⎥
⎣2
⎦ 0
2
ln
3
–2
ln
3
0
–0
1⎛e
–e
e –e ⎞
–
= ⎜
⎟
⎟
2 ⎜⎝
2
2
⎠
37. Area =
=
400
40. Let u = z , so du =
∫
cosh z
z
1
2 z
dz .
dz = 2∫ cosh u du = 2sinh u + C
= 2sinh z + C
41. Let u = 2 z1/ 4 , so du =
ln 3
∫0
1
cosh u du
2π ∫
∫
sinh(2 z1/ 4 )
4 3
z
1
1
⋅ 2 z –3 / 4 dz =
dz.
4
4
2 z3
dz = 2 ∫ sinh u du = 2 cosh u + C
= 2 cosh(2 z1/ 4 ) + C
1 ln 9 ln 19
1⎛
1 ⎞ 20
(e − e ) = ⎜ 9 − ⎟ =
4
4⎝
9⎠
9
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48. tanh x = 0 when sinh x = 0, which is when x = 0.
42. Let u = e x , so du = e x dx .
∫e
x
sinh e x dx = ∫ sinh u du = cosh u + C
= cosh e x + C
= cosh(sin x) + C
44. Let u = ln(cosh x), so
1
⋅ sinh x = tanh x dx .
cosh x
∫ tanh x ln(cosh x)dx = ∫ u du =
u2
+C
2
sinh x 2
∫ x coth x
2
⋅ cosh x 2 ⋅ 2 xdx = 2 x coth x 2 dx .
ln(sinh x 2 )dx =
1
1 u2
u du = ⋅
+C
∫
2
2 2
ln 5
ln 5
∫– ln 5 cosh 2 x dx = 2∫0
=
1
∫0 π cosh
2
x dx =
π 1
(1 + cosh 2 x)dx
2 ∫0
sinh 2 x ⎤
π⎡
x+
⎢
2⎣
2 ⎥⎦ 0
π ⎛ sinh 2
⎞
− 0⎟
⎜1 +
2⎝
2
⎠
π π sinh 2
= +
≈ 4.42
2
4
50. Volume =
cosh 2 x dx
47. Note that the graphs of y = sinh x and y = 0
intersect at the origin.
ln 2
49. Volume =
2
sinh x dx = [cosh x]ln
0
1⎞
1
eln 2 + e− ln 2 e0 + e0 1 ⎛
−
= ⎜ 2 + ⎟ −1 =
2⎝
2⎠
4
2
2
= π∫
ln10
∫0
ln10 ⎛ e x
0
⎡1
⎤
= 2 ⎢ sinh 2 x ⎥
2
⎣
⎦0
1
= sinh(2 ln 5) = (e2 ln 5 − e−2 ln 5 )
2
1
ln
1
1⎛
1 ⎞
= (eln 25 − e 25 ) = ⎜ 25 − ⎟
2
2⎝
25 ⎠
312
= 12.48
=
25
∫0
8
0
= π∫
ln 5
Area =
8
0
=
1
= [ln(sinh x 2 )]2 + C
4
46. Area =
sinh x
dx
cosh x
Let u = cosh x, so du = sinh xdx.
sinh x
1
2∫
dx = 2∫ du = 2 ln u + C
cosh x
u
8 sinh x
8
2∫
dx = ⎡⎣ 2 ln cosh x ⎤⎦
0
0 cosh x
= 2(ln cosh 8 − ln1) = 2 ln(cosh 8) ≈ 14.61
=
45. Let u = ln(sinh x 2 ) , so
1
8
1
1
= [ln(cosh x)]2 + C
2
du =
0
∫−8 (− tanh x) dx + ∫0 tanh x dx
= 2 ∫ tanh x dx = 2 ∫
43. Let u = sin x, so du = cos x dx
∫ cos x sinh(sin x)dx = ∫ sinh u du = cosh u + C
du =
Area =
⎜
⎜
⎝
ln10 e 2 x
0
π sinh 2 xdx
− e− x
2
2
⎞
⎟ dx
⎟
⎠
– 2 + e –2 x
π ln10 2 x
dx = ∫
(e – 2 + e –2 x )dx
4
4 0
ln10
=
1
π ⎡ 1 2x
⎤
e – 2 x – e –2 x ⎥
⎢
4 ⎣2
2
⎦0
π
= [e2 x – 4 x – e –2 x ]ln10
0
8
π⎛
1 ⎞
= ⎜ 100 – 4 ln10 –
⎟ ≈ 35.65
8⎝
100 ⎠
51. Note that 1 + sinh 2 x = cosh 2 x and
1 + cosh 2 x
cosh 2 x =
2
Surface area =
1
∫0 2πy
( ) dx
dy 2
1 + dx
1
= ∫ 2π cosh x 1 + sinh 2 x dx
0
1
= ∫ 2π cosh x cosh x dx
0
1
= ∫ π(1 + cosh 2 x )dx
0
1
π
π
⎡
⎤
= ⎢ πx + sinh 2 x ⎥ = π + sinh 2 ≈ 8.84
2
2
⎣
⎦0
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2
1
⎛ dy ⎞
2
∫0 2πy 1 + ⎜⎝ dx ⎟⎠ dx = ∫0 2π sinh x 1 + cosh xdx
Let u = cosh x, so du = sinh x dx
⎡u
⎤
2
2
2 1
2
∫ 2π sinh x 1 + cosh xdx = 2π∫ 1 + u du = 2π ⎢⎣ 2 1 + u + 2 ln u + 1 + u + C ⎥⎦
52. Surface area =
1
= π cosh x 1 + cosh 2 x + π ln cosh x + 1 + cosh 2 x + C (The integration of
∫
1 + u 2 du is shown in Formula 44 of
the Tables in the back of the text, which is covered in Chapter 8.)
1
⎡
2
2 ⎤
2
∫0 2π sinh x 1 + cosh xdx = π ⎢⎣cosh x 1 + cosh x + ln cosh x + 1 + cosh x ⎥⎦0
⎡
⎤
= π ⎢cosh1 1 + cosh 2 1 + ln cosh1 + 1 + cos 2 1 − 2 + ln 1 + 2 ⎥ ≈ 5.53
⎣
⎦
1
(
)
⎛x⎞
53. y = a cosh ⎜ ⎟ + C
⎝a⎠
dy
⎛x⎞
= sinh ⎜ ⎟
dx
⎝a⎠
d2y
dx
2
=
1
⎛x⎞
cosh ⎜ ⎟
a
⎝a⎠
d2y
2
1
⎛ dy ⎞
1+ ⎜ ⎟ .
2
a
⎝ dx ⎠
dx
⎛x⎞
⎛ x⎞
⎛ x⎞
Note that 1 + sinh 2 ⎜ ⎟ = cosh 2 ⎜ ⎟ and cosh ⎜ ⎟ > 0. Therefore,
⎝a⎠
⎝a⎠
⎝a⎠
We need to show that
=
2
2
1
1
⎛ dy ⎞
⎛ x⎞ 1
⎛ x⎞ 1
⎛x⎞ d y
1+ ⎜ ⎟ =
1 + sinh 2 ⎜ ⎟ =
cosh 2 ⎜ ⎟ = cosh ⎜ ⎟ =
a
a
⎝ dx ⎠
⎝a⎠ a
⎝a⎠ a
⎝ a ⎠ dx 2
54. a.
⎛ x⎞
The graph of y = b − a cosh ⎜ ⎟ is symmetric about the y-axis, so if its width along the
⎝a⎠
⎛a⎞
x-axis is 2a, its x-intercepts are (±a, 0). Therefore, y (a ) = b − a cosh ⎜ ⎟ = 0, so b = a cosh1 ≈ 1.54308a.
⎝a⎠
b. The height is y (0) ≈ 1.54308a − a cosh 0 = 0.54308a .
c.
If 2a = 48, the height is about 0.54308a = (0.54308)(24) ≈ 13 .
55. a.
b. Area under the curve is
24
∫−24
24
⎡
⎡
⎛ x ⎞⎤
⎛ x ⎞⎤
≈ 422
⎢37 − 24 cosh ⎜ 24 ⎟ ⎥dx = ⎢37 x − 576sinh ⎜ 24 ⎟ ⎥
⎝ ⎠⎦
⎝ ⎠ ⎦ −24
⎣
⎣
Volume is about (422)(100) = 42,200 ft3.
402
Section 6.9
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c.
Length of the curve is
24
∫−24
24
2
24
24
⎡
⎛ dy ⎞
⎛ x ⎞
⎛ x ⎞
⎛ x ⎞⎤
1 + ⎜ ⎟ dx = ∫
1 + sinh 2 ⎜ ⎟dx = ∫ cosh ⎜ ⎟ dx = ⎢ 24sinh ⎜ ⎟ ⎥
= 48sinh1 ≈ 56.4
24
24
−
−
⎝ 24 ⎠ ⎦ −24
⎝ dx ⎠
⎝ 24 ⎠
⎝ 24 ⎠
⎣
Surface area ≈ (56.4)(100) = 5640 ft 2
cosh t
56. Area =
cosh t
1
1
1
⎡1
⎤
x 2 − 1 dx = cosh t sinh t − ⎢ x x 2 − 1 − ln x + x 2 − 1 ⎥
cosh t sinh t − ∫
1
2
2
2
⎣2
⎦1
1
1
⎡1
⎤
cosh t sinh t − ⎢ cosh t cosh 2 t − 1 − ln cosh t + cosh 2 t − 1 − 0 ⎥
2
2
⎣2
⎦
1
1
1
1
t
= cosh t sinh t − cosh t sinh t + ln cosh t + sinh t = ln et =
2
2
2
2
2
=
57. a.
⎛ ex – e– x ex + e– x
(sinh x + cosh x) = ⎜
+
⎜
2
2
⎝
r
sinh rx + cosh rx =
b.
c.
( cos x + i sin x )
cos rx + i sin rx =
d.
( cos x − i sin x )
r
r
⎞
⎟ = e – rx
⎟
⎠
erx + e – rx e rx – e – rx 2e – rx
–
=
= e – rx
2
2
2
r
⎞
⎛ 2eix
⎟ =⎜
⎜ 2
⎟
⎝
⎠
r
⎞
⎟ = eirx
⎟
⎠
eirx + e−irx
eirx − e−irx 2eirx
+i
=
= eirx
2
2i
2
⎛ eix + e −ix
eix − e−ix
=⎜
−i
⎜
2
2i
⎝
cos rx − i sin rx =
58. a.
r
⎞
⎛ 2e – x
⎟ =⎜
⎜ 2
⎟
⎝
⎠
⎛ eix + e−ix
eix − e−ix
=⎜
+i
⎜
2
2i
⎝
r
r
erx – e – rx e rx + e – rx 2e rx
+
=
= e rx
2
2
2
⎛ e x + e– x e x – e– x
(cosh x – sinh x)r = ⎜
–
⎜
2
2
⎝
cosh rx – sinh rx =
r
⎞
⎛ 2e x ⎞
⎟ =⎜
⎟ = erx
⎟
⎜ 2 ⎟
⎠
⎝
⎠
r
⎞
⎛ 2e −ix
⎟ =⎜
⎜ 2
⎟
⎝
⎠
r
⎞
⎟ = e−irx
⎟
⎠
eirx + e −irx
eirx − e−irx 2e−irx
=
= e −irx
−i
2
2i
2
gd (– t ) = tan –1[sinh(–t )]
= tan –1 (– sinh t ) = – tan –1 (sinh t ) = − gd (t )
so gd is odd.
1
cosh t
Dt [ gd (t )] =
⋅ cosh t =
2
1 + sinh t
cosh 2 t
= sech t > 0 for all t, so gd is increasing.
Dt2 [ gd (t )] = Dt (sech t ) = −sech t tanh t
Dt2 [ gd (t )] = 0 when tanh t = 0, since
sech t > 0 for all t. tanh t = 0 at t = 0 and
tanh t < 0 for t < 0, thus Dt2 [ gd (t )] > 0 for
t < 0 and Dt2 [ gd (t )] < 0 for t > 0. Hence
gd(t) has an inflection point at
b. If y = tan –1 (sinh t ) then tan y = sinh t so
tan y
sin y =
2
tan y + 1
=
sinh t
sinh 2 t + 1
sinh t
= tanh t so y = sin –1 (tanh t )
cosh t
1
Also, Dt y =
⋅ cosh t
1 + sinh 2 t
cosh t
1
=
=
= sech t ,
2
cosh
t
cosh t
=
t
so y = ∫ sech u du by the Fundamental
0
Theorem of Calculus.
(0, gd(0)) = (0, tan −1 0) = (0, 0).
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59. Area =
x
∫0 cosh t dt = [sinh t ]0 = sinh x
x
61.
Arc length =
x
1 + [ Dt cosh t ]2 dt = ∫
∫0
x
0
1 + sinh 2 tdt
= ∫ cosh t dt = [ sinh t ]0 = sinh x
x
x
0
60. From Problem 54, the equation of an inverted
x
catenary is y = b − a cosh . Given the
a
information about the Gateway Arch, the curve
passes through the points (±315, 0) and (0, 630).
315
Thus, b = a cosh
and 630 = b – a, so
a
b = a + 630.
315
a + 630 = a cosh
⇒ a ≈ 128, so b ≈ 758 .
a
x
The equation is y = 758 − 128cosh
.
128
6.10 Chapter Review
The functions y = sinh x and y = ln( x + x 2 + 1)
are inverse functions.
62. y = gd ( x) = tan –1 (sinh x)
tan y = sinh x
x = gd –1 ( y ) = sinh –1 (tan y )
Thus, y = gd –1 ( x) = sinh –1 (tan x)
9. True:
= 4 + ( x − 4) = x
and
Concepts Test
1. False:
2. True:
ln 0 is undefined.
d2y
dx
3. True:
2
e3
∫1
=−
1
x2
g ( f ( x)) = ln(4 + e x − 4) = ln e x = x
< 0 for all x > 0.
e3
1
dt = ⎡⎣ ln t ⎤⎦ = ln e3 − ln1 = 3
t
1
4. False:
The graph is intersected at most once
by every horizontal line.
5. True:
The range of y = ln x is the set of all
real numbers.
6. False:
⎛x⎞
ln x − ln y = ln ⎜ ⎟
⎝ y⎠
7. False:
4 ln x = ln( x 4 )
8. True:
x +1
ln(2e
) – ln(2e ) = ln
= ln e = 1
404
f ( g ( x)) = 4 + eln( x − 4)
Section 6.10
x
10. False:
exp( x + y ) = exp x exp y
11. True:
ln x is an increasing function.
12. False:
Only true for x > 1, or ln x > 0.
13. True:
e z > 0 for all z.
14. True:
e x is an increasing function.
15. True:
lim (ln sin x − ln x)
x →0 +
⎛ sin x ⎞
= lim ln ⎜
⎟ = ln1 = 0
+
⎝ x ⎠
x →0
2e x +1
16. True:
π 2 = e 2 ln π
17. False:
ln π is a constant so
2e x
18. True:
d
ln π = 0.
dx
d
(ln 3 x + C )
dx
1
d
= (ln x + ln 3 + C ) =
dx
x
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
19. True:
e is a number.
e x > 1 and e− x < 1 < e x , thus
20. True:
exp[ g ( x)] ≠ 0 because 0 is not in the
e x − e− x = e x − e− x < e x = e .
x
range of the function y = e x .
If x < 0, e − x > 1 and e x < 1 < e − x ,
thus
21. False:
Dx ( x x ) = x x (1 + ln x)
22. True:
2 ( tan x + sec x ) '− ( tan x + sec x )
(
= 2 sec2 x + sec x tan x
e x − e − x = −(e x − e − x )
2
)
2
2
− tan x − 2 tan x sec x − sec x
23. True:
x
32. False:
= sec 2 x − tan 2 x = 1
The integrating factor is
e∫
24. True:
= e− x − e x < e− x = e .
( )
4 / x dx
= e4 ln x = eln x
but
4
= x4
The solution is y ( x ) = e−4 ⋅ e2x . Thus,
The solution is y ( x ) = e2x , so
26. False:
sin ( arcsin(2) ) is undefined
27. False:
arcsin(sin 2π) = arcsin 0 = 0
28. True:
sinh x is increasing.
29. False:
cosh x is not increasing.
30. True:
cosh(0) = 1 = e0
If x > 0, e x > 1 while e− x < 1 < e x so
1
1
cosh x = (e x + e− x ) < (2e x )
2
2
⎛ sin x ⎞
lim ln ⎜
⎟ = ln1 = 0
⎝ x ⎠
x →0
35. True:
36. False:
37. True:
sinh x ≤
1 x
e is equivalent to
2
e x − e− x ≤ e . When x = 0,
x
sinh x = 0 <
1 0 1
e = . If x > 0,
2
2
Instructor’s Resource Manual
+
cosh x > 1 for x ≠ 0 , while sin −1 u is
only defined for −1 ≤ u ≤ 1.
sinh x
; sinh x is an odd
cosh x
function and cosh x is an even
function.
tanh x =
38. False:
Both functions satisfy y ′′ − y = 0 .
39. True:
ln 3100 = 100 ln 3 > 100 ⋅1 since
ln 3 > 1.
40. False:
ln(x – 3) is not defined for x < 3.
41. True:
y triples every time t increases by t1.
42. False:
x(0) = C;
x
31. True:
π
, since
2
lim tan x = −∞ .
lim tan −1 x = −
x →−∞
x →− π
2
x
= e− x = e .
eln 3 + e − ln 3
2
1⎛
1⎞ 5
= ⎜3+ ⎟ =
2⎝
3⎠ 3
34. False:
= e x = e . If x < 0, –x > 0 and
e− x > 1 while e x < 1 < e− x so
1
1
cosh x = (e x + e− x ) < (2e− x )
2
2
cos −1 12
cosh(ln 3) =
y ' ( x ) = 2e2 x . In general, Euler’s
method will underestimate the
solution if the slope of the solution is
increasing as it is in this case.
( )=1
( ) 2
sin −1 12
33. False:
slope = 2e −4 ⋅ e 2 x and at x = 2 the
slope is 2.
25. False:
⎛1⎞
tan −1 ⎜ ⎟ ≈ 0.4636
⎝ 2⎠
1
C = Ce− kt when
2
1
1
= e− kt , so ln = −kt or
2
2
1
ln
− ln 2 ln 2
t= 2 =
=
−k
−k
k
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405
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
43. True:
( y (t ) + z (t ))′ = y ′(t ) + z ′(t )
= ky (t ) + kz (t ) = k ( y (t ) + z (t ))
44. False:
Only true if C = 0;
( y1 (t ) + y2 (t ))′ = y1′ (t ) + y2′ (t )
= ky1 (t ) + C + ky2 (t ) + C
= k ( y1 (t ) + y2 (t )) + 2C .
45. False:
8.
9.
Use the substitution u = –h.
10.
12
⎛ 0.06 ⎞
e0.05 ≈ 1.051 < ⎜ 1 +
⎟
12 ⎠
⎝
46. False:
Sample Test Problems
x4
1. ln
= 4 ln x − ln 2
2
d
x4 d
4
ln
= (4 ln x − ln 2) =
dx
2 dx
x
12.
4.
d
d 5
1
log10 ( x5 − 1) =
( x − 1)
5
dx
( x − 1) ln10 dx
=
6.
d ln cot x d
e
= cot x = − csc2 x
dx
dx
7.
406
d
d
sech 2 x
2 tanh x = 2sech 2 x
x=
dx
dx
x
Section 6.10
3
1 − 3x 2 3x
=
ex
e x e2 x − 1
=
(
3x
1
( )
sin 2 2x
1
( )
sin 2 2x
)
2
d
3x
dx
3
3x − 9 x 2
1
d x
e
(e x ) 2 − 1 dx
1
e2 x − 1
x
2
d
⎛x⎞
sin 2 ⎜ ⎟
dx
⎝2⎠
⎛ x⎞ d
⎛ x⎞
2sin ⎜ ⎟ sin ⎜ ⎟
⎝ 2 ⎠ dx
⎝2⎠
⎡
⎛ x ⎞⎤ 1
⎛ x⎞
⎛ x⎞
⎢ 2sin ⎜ 2 ⎟ ⎥ 2 cos ⎜ 2 ⎟ = cot ⎜ 2 ⎟
⎝
⎠
⎝
⎠
⎝ ⎠
⎣
⎦
13.
d
3
15e5 x
3ln(e5 x + 1) =
(5e5 x ) =
dx
e5 x + 1
e5 x + 1
14.
d
ln(2 x3 − 4 x + 5)
dx
=
( x5 − 1) ln10
d
d
tan(ln e x ) =
tan x = sec2 x
dx
dx
1−
( )
=
5x4
5.
2
d
1
⎛ x⎞
ln sin 2 ⎜ ⎟ =
2
2
dx
⎝ ⎠ sin
=
2
d x2 −4 x
d 2
e
= e x −4 x
( x − 4 x)
dx
dx
x2 −4 x
2
= sec x
sec 2 x
d
sec−1 e x =
dx
ex
=
d
d
2.
sin 2 ( x3 ) = 2sin( x3 ) sin( x3 )
dx
dx
d
= 2sin( x3 ) cos( x3 ) x3 = 6 x 2 sin( x3 ) cos( x3 )
dx
= (2 x − 4)e
sec 2 x
=
d
2sin −1 3 x =
dx
=
11.
3.
tan 2 x + 1
d
tan x
dx
tan x + 1
2
≈ 1.062
If Dx (a x ) = a x ln a = a x , then
ln a = 1, so a = e.
47. True:
sec2 x
u →0
by Theorem 6.5.A.
1
d
sinh −1 (tan x) =
dx
=
lim (1 – h) –1/ h = lim (1 + u )1 u = e
h →0
d
d
1
cos x
tanh −1 (sin x) =
sin x =
2 dx
dx
1 − sin x
1 − sin 2 x
cos x
=
= sec x
cos 2 x
15.
d
6 x2 − 4
(2 x3 − 4 x + 5) =
2 x3 − 4 x + 5 dx
2 x3 − 4 x + 5
1
d
d
cos e x = − sin e x
e x
dx
dx
d
= (− sin e x )e x
x
dx
=−
e x sin e x
2 x
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16.
17.
d
1 d
ln(tanh x) =
tanh x
dx
tanh x dx
1
sech 2 x = csch x sech x
=
tanh x
23.
d
−2
d
x
2 cos −1 x =
dx
1 − ( x ) 2 dx
=
−2
1
1− x 2 x
1⎞
⎛
= x1+ x ⎜ ln x + 1 + ⎟
x⎠
⎝
1
=−
x − x2
24.
18.
d ⎡ 3x
d
4 + (3 x)4 ⎤ = (64 x + 81x 4 )
⎣
⎦
dx
dx
20.
21.
25. Let u = 3x – 1, so du = 3 dx.
1 3 x −1
1 u
3 x −1
∫ e dx = 3 ∫ e 3dx = 3 ∫ e du
1
1
= eu + C = e3 x −1 + C
3
3
Check:
d ⎛ 1 3 x −1
d
⎞ 1
+ C ⎟ = e3 x −1 (3 x − 1) = e3 x −1
⎜ e
dx ⎝ 3
dx
⎠ 3
d
d
2 csc eln x =
2 csc x
dx
dx
d
= −2 csc x cot x
x
dx
=−
csc x cot x
x
d
(log10 2 x) 2 / 3
dx
2
d
= (log10 2 x) −1/ 3 (log10 2 + log10 x )
dx
3
2
1
= (log10 2 x) −1/ 3
3
x ln10
2
=
3
3x ln10 log10 2 x
d
4 tan 5 x sec5 x
dx
= 20sec 5 x(sec2 5 x + tan 2 5 x)
= 20sec 5 x(2sec 2 5 x − 1)
⎛ x2
d
tan −1 ⎜
⎜ 2
dx
⎝
26. Let u = sin 3x, so du = 3 cos 3x dx.
1
1
∫ 6 cot 3x dx = 2∫ sin 3x 3cos 3x dx = 2∫ u du
= 2 ln u + C = 2 ln sin 3 x + C
Check:
d
2 d
(2 ln sin 3x + C ) =
sin 3 x
sin 3 x dx
dx
2(3cos 3x)
=
= 6 cot 3 x
sin 3 x
27. Let u = e x , so du = e x dx .
= 20sec2 5 x sec 5 x + 20 tan 5 x sec 5 x tan 5 x
22
d
d
(1 + x 2 )e = e(1 + x 2 )e −1 (1 + x 2 )
dx
dx
= 2 xe(1 + x 2 )e −1
= 64 x ln 64 + 324 x3
19.
d 1+ x d (1+ x ) ln x
x
= e
dx
dx
d
= e(1+ x ) ln x [(1 + x) ln x]
dx
⎡
⎛ 1 ⎞⎤
= x1+ x ⎢(1)(ln x) + (1 + x) ⎜ ⎟ ⎥
⎝ x ⎠⎦
⎣
⎞
1
d ⎛ x2 ⎞
⎟=
⎜ ⎟
2
⎟
⎜ ⎟
⎠ ⎛ x 2 ⎞ + 1 dx ⎝ 2 ⎠
⎜ 2 ⎟
⎝ ⎠
4x
x
=
=
⎛ x4 ⎞ + 1 x4 + 4
⎜ 4 ⎟
⎝ ⎠
⎛x
= tan −1 ⎜
⎜ 2
⎝
Instructor’s Resource Manual
x
sin e x dx = ∫ sin u du = − cos u + C
= − cos e x + C
Check:
d
d
(− cos e x + C ) = (sin e x ) e x = e x sin e x
dx
dx
28. Let u = x 2 + x − 5, so du = (2 x + 1)dx .
6x + 3
1
∫ x2 + x − 5 dx = 3∫ x2 + x − 5 (2 x + 1)dx
⎛ x2 ⎞⎤
⎛ x2 ⎞
⎛ 4x ⎞
d ⎡
⎢ x tan −1 ⎜ ⎟ ⎥ = (1) tan −1 ⎜ ⎟ + ( x) ⎜ 4
⎟
⎜
⎟
⎜
⎟
dx ⎢⎣
⎝ x +4⎠
⎝ 2 ⎠ ⎥⎦
⎝ 2 ⎠
2
∫e
⎞ 4x
⎟+ 4
⎟ x +4
⎠
2
1
= 3∫ du = 3ln u + C = 3ln x 2 + x − 5 + C
u
Check:
3
d
d 2
3ln x 2 + x − 5 + C =
( x + x − 5)
2
dx
x + x − 5 dx
6x + 3
=
2
x + x −5
(
)
Section 6.10
407
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
29. Let u = e x +3 + 1, so du = e x +3 dx .
e
x+2
1
1
∫ e x +3 + 1 dx = e ∫ e x +3 + 1 e
x +3
33. Let u = ln x, so du =
dx =
1 1
du
e∫u
−1
e x + 3 e −1
e x +3 + 1
=
∫ 4 x cos x
2
du = − tan −1 u + C = − tan −1 (ln x ) + C
1+ u2
Check:
d
1
d
[− tan −1 (ln x ) + C ] = −
ln x
2
dx
1 + (ln x) dx
=
e x +3 + 1
π
4
f ′( x) > 0 when cos x > sin x which occurs when
π
π
≤x< .
2
4
f ′′( x) = – sin x – cos x; f ′′( x) = 0 when
–
⎞d
⎟ 2x
⎟ dx
⎠
= tan −1 (sin x) + C
Check:
d ⎡ −1
1
d
tan (sin x ) + C ⎤ =
sin x
2
⎣
⎦
dx
1 + sin x dx
cos x
=
1 + sin 2 x
Section 6.10
f ′( x) = cos x – sin x; f ′( x) = 0 when tan x = 1,
x=
32. Let u = sin x, so du = cos x dx.
cos x
1
−1
∫ 1 + sin 2 x dx = ∫ 1 + u 2 du = tan u + C
408
( x − 3)dx = ∫ sech 2 u du = tanh u + C
= tanh( x − 3) + C
Check:
d
d
[tanh( x − 3)] = sech 2 ( x − 3) ( x − 3)
dx
dx
35.
du
= 2sin −1 u + C = 2sin −1 2 x + C
Check:
⎛
d
1
(2sin −1 2 x + C ) = 2 ⎜
⎜ 1 − (2 x) 2
dx
⎝
4
=
1 − 4x 2
2
= sech 2 ( x − 3)
31. Let u = 2x, so du = 2 dx.
4
1
dx = 2∫
2dx
∫
1 − 4 x2
1 − (2 x) 2
1− u2
x + x(ln x) 2
∫ sech
= 2sin u + C = 2sin x 2 + C
Check:
d
d 2
(2sin x 2 + C ) = 2 cos x 2
x = 4 x cos x 2
dx
dx
= 2∫
−1
34. Let u = x – 3, so du = dx.
2
dx = 2 ∫ (cos x )2 x dx = 2 ∫ cos u du
1
1
= −∫
e x+ 2
30. Let u = x 2 , so du = 2x dx.
1
∫ x + x(ln x)2 dx = −∫ 1 + (ln x)2 ⋅ x dx
1
ln(e x +3 + 1)
= ln u + C =
+C
e
e
Check:
⎞ 1
d ⎛ ln(e x +3 + 1)
1
d x +3
+C⎟ =
+ 1)
(e
⎜
⎟ e e x +3 + 1 dx
dx ⎜⎝
e
⎠
=
1
dx .
x
1
π
4
f ′′( x) > 0 when cos x < –sin x which occurs
tan x = –1, x = –
π
π
≤x<– .
2
4
⎡ π π⎤
Increasing on ⎢ – , ⎥
⎣ 2 4⎦
⎡π π⎤
Decreasing on ⎢ , ⎥
⎣4 2⎦
⎛ π π⎞
Concave up on ⎜ – , – ⎟
⎝ 2 4⎠
⎛ π π⎞
Concave down on ⎜ – , ⎟
⎝ 4 2⎠
⎛ π ⎞
Inflection point at ⎜ – , 0 ⎟
⎝ 4 ⎠
⎛π
⎞
Global maximum at ⎜ , 2 ⎟
⎝4
⎠
when –
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
⎛ π
⎞
Global minimum at ⎜ – , –1⎟
⎝ 2
⎠
37. a.
f ′( x) = 5 x 4 + 6 x 2 + 4 ≥ 4 > 0 for all x, so
f(x) is increasing.
b. f(1) = 7, so g(7) = f −1 (7) = 1.
c.
38.
36.
f ( x) =
f ′( x) =
1
1
=
f ′(1) 15
1
= e10 k
2
ln 12
k=
≈ −0.06931
10
( )
x2
ex
e x (2 x) − x 2 (e x )
x 2
=
2x − x
y = 100e−0.06931t
2
1 = 100e−0.06931t
x
(e )
e
f is increasing on [0, 2] because f ′( x) > 0 on
(0, 2).
f is decreasing on (−∞, 0] ∪ [2, ∞) because
f ′( x) < 0 on (−∞, 0) ∪ (2, ∞).
f ′′( x) =
g ′(7) =
e x (2 − 2 x) − (2 x − x 2 )e x
x2 − 4x + 2
=
(e x ) 2
ex
Inflection points are at
4 ± 16 − 4 ⋅ 2
x=
= 2± 2 .
2
The graph of f is concave up on
(−∞, 2 − 2) ∪ (2 + 2, ∞) because f ′′( x) > 0
on these intervals.
The graph of f is concave down on
(2 − 2, 2 + 2) because f ′′( x) < 0 on this
interval.
The absolute minimum value is f(0) = 0.
4
The relative maximum value is f (2) = .
e2
The inflection points are
⎛
⎛
6−4 2 ⎞
6+4 2 ⎞
⎜⎜ 2 − 2,
⎟⎟ and ⎜⎜ 2 + 2,
⎟⎟ .
e 2− 2 ⎠
e 2+ 2 ⎠
⎝
⎝
t=
ln
( 1001 )
≈ 66.44
−0.06931
It will take about 66.44 years.
39.
xn
yn
1.0
2.0
1.2
2.4
1.4 2.976
1.6 3.80928
1.8 5.02825
2.0 6.83842
40. Let x be the horizontal distance from the airplane
dx
= 300.
to the searchlight,
dt
500
500
tan θ =
, so θ = tan −1
.
x
x
dθ
1
⎛ 500 ⎞ dx
=
⎜−
⎟
2
dt 1 + 500 ⎝ x 2 ⎠ dt
(x)
=−
500
dx
x + 250, 000 dt
2
When θ = 30°, x =
500
= 500 3 and
tan 30°
dθ
500
(300)
=−
dt
(500 3)2 + (500)2
300
3
=−
= − . The angle is decreasing at the
2000
20
rate of 0.15 rad/s ≈ 8.59°/s.
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Section 6.10
409
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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
45. (Linear first-order) y ′ + 2 xy = 2 x
41. y = (cos x)sin x = esin x ln(cos x )
dy
d
= esin x ln(cos x ) [sin x ln(cos x)]
dx
dx
⎡
⎤
⎛ 1 ⎞
= esin x ln(cos x ) ⎢ cos x ln(cos x) + (sin x ) ⎜
⎟ ( − sin x) ⎥
⎝ cos x ⎠
⎣
⎦
2
⎡
sin x ⎤
= (cos x)sin x ⎢ cos x ln(cos x) −
⎥
cos x ⎥⎦
⎢⎣
dy 0
At x = 0,
= 1 (1ln1 − 0) = 0 .
dx
The tangent line has slope 0, so it is horizontal:
y = 1.
42. Let t represent the number of years since 1990.
2
2 xdx
= ex
Integrating factor: e ∫
2
2
2
Therefore, y = 1 + 2e – x .
46. Integrating factor is e – ax .
D[ ye – ax ] = 1; y = eax ( x + C )
47. Integrating factor is e –2 x .
48. a.
Q′(t ) = 3 – 0.02Q
b.
Q′(t ) + 0.02Q = 3
(0.03365)(20)
y (20) = 10, 000e
≈ 19, 601
The population will be about 19,600.
43. Integrating factor is x . D [ yx ] = 0; y = Cx
Integrating factor is e0.02t
D[Qe0.02t ] = 3e0.02t
−1
Q(t ) = 150 + Ce –0.02t
Q(t ) = 150 – 30e –0.02t goes through (0, 120).
44. Integrating factor is x 2 .
⎛1⎞
D[ yx 2 ] = x3 ; y = ⎜ ⎟ x 2 + Cx –2
⎝4⎠
Review and Preview Problems
5.
c.
Q → 150 g, as t → ∞ .
∫
sin t
cos t
1
u = cos t
du = −sin t dt
2 x dx = ∫ sin u du = − cos u + C =
∫ usin
2
2
= 2x
du = 2 dx
ln
1
− cos 2 x + C
2
2.
3t
∫ u =e3t
dt =
du = 3 dt
1 u
1
1
e du = eu + C = e3t + C
3∫
3
3
1
1
3. ∫ x sin x dx = ∫ sin u du = − cos u + C =
2
2
2
u=x
2
2
∫ u = 3 x2
dx =
7.
∫ xu =xx2 ++2 2 dx = 2 ∫
u = sin x
du = cos x dx
du = 6 x dx
1
2
u3
sin 3 x
+C =
+C
3
3
1 3
u du = u 2 + C
3
du = 2 x dx
=
1 u
1
1 2
e du = eu + C = e3 x + C
∫
6
6
6
1
1
+ C = ln
+ C = ln sec t + C
u
cos t
2
2
∫ sin x cos x dx = ∫ u du =
1
− cos x 2 + C
2
xe3 x
1
dt = − ∫ du = − ln u + C =
u
6.
du = 2 x dx
4.
2
y = 1 + Ce – x
If x = 0, y = 3, then 3 = 1 + C, so C = 2.
y = 10, 000e0.03365t
1
2
D[ ye –2 x ] = e – x ; y = – e x + Ce2 x
14, 000 = 10, 000e10 k
ln(1.4)
k=
≈ 0.03365
10
1.
2
D[ ye x ] = 2 xe x ; ye x = e x + C ;
8.
∫
(
1 2
x +2
3
x
2
x +1
u = x 2 +1
du = 2 x dx
3
) 2 +C
dx =
1 1
1
du = ln u + C
∫
2 u
2
= ln u + C = ln x 2 + 1 + C
410
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⎡ ⎛1⎞
⎤
f ′( x) = ⎢ x ⎜ ⎟ + (ln x)(1) ⎥ − 1 = ln x
x
⎣ ⎝ ⎠
⎦
21.
⎡ x
⎤
−2 x
+ (1) arcsin x ⎥ +
10. f '( x) = ⎢
2
⎢⎣ 1 − x
⎥⎦ 2 1 − x 2
= arcsin x
22.
9.
11.
a
f ′( x) = ⎡ (−2 x)(cos x) + (− x 2 )(− sin x) ⎤ +
⎣
⎦
[(2)(sin x) + (2 x)(cos x)] + [ 2(− sin x)]
f ′( x) = e x ( cos x + sin x ) + e x (sin x − cos x)
= 2e sin x
x
13. cos 2 x = 1 − 2sin 2 x ; thus sin 2 x =
⎛ 1 + cos 2 x ⎞
15. cos 4 x = (cos 2 x)2 = ⎜
⎟
2
⎝
⎠
24.
1 − cos 2 x
2
14. cos 2 x = 2 cos 2 x − 1 ; thus cos 2 x =
1 + cos 2 x
2
2
a
1
1
⇒ ⎡ −e − x ⎤ = ⇒
⎣
⎦0 2
2
⎡ −e − a + 1⎤ = 1 ⇒ 1 = 1 ⇒
⎣
⎦ 2
ea 2
a −x
∫0 e
dx =
1
1 x − (1 − x)
2x −1
− =
=
1− x x
(1 − x) x
x(1 − x)
7
8
7( x − 3) + 8( x + 2)
+
=
=
5( x + 2) 5( x − 3)
5( x + 2)( x − 3)
15 x − 5
5(3 x − 1)
=
=
5( x + 2)( x − 3) 5( x + 2)( x − 3)
(3 x − 1)
( x + 2)( x − 3)
1
1
3
25. − −
+
x 2( x + 1) 2( x − 3)
=
sin(u + v) + sin(u − v)
⇒
2
sin(7 x) + sin(− x) sin 7 x − sin x
=
sin 3 x cos 4 x =
2
2
−2( x + 1)( x − 3) − x( x − 3) + 3 x( x + 1)
2 x( x + 1)( x − 3)
−2( x 2 − 2 x − 3) − ( x 2 − 3 x) + (3 x 2 + 3 x)
2 x( x + 1)( x − 3)
10 x + 6
2(5 x + 3)
=
=
=
2 x( x + 1)( x − 3) 2 x( x + 1)( x − 3)
(5 x + 3)
x( x + 1)( x − 3)
16. sin u cos v =
cos(u + v) + cos(u − v)
⇒
2
cos(8 x) + cos(−2 x)
cos 3x cos 5 x =
2
cos8 x + cos 2 x
=
2
tan 2 t = a ⋅ tan t
e a = 2 ⇒ a = ln 2
23.
= x 2 sin x
12.
(a sec t )2 − a 2 − = a 2 (sec2 t − 1) =
=
17. cos u cos v =
26.
1
1
(2000 − y ) + y
2000
+
=
=
y 2000 − y
y (2000 − y )
y (2000 − y )
cos(u − v) − cos(u + v)
⇒
2
cos(− x) − cos(5 x)
sin 2 x sin 3x =
2
cos x − cos 5 x
=
2
18. sin u sin v =
19.
a 2 − (a sin t )2 = a 2 (1 − sin 2 t ) =
a cos 2 t = a cos t
20.
a 2 + (a tan t )2 = a 2 (1 + tan 2 t ) =
a sec 2 t = a sec t
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CHAPTER
7
Techniques of Integration
7.1 Concepts Review
5.
1. elementary function
2.
∫u
5
1
–1 ⎛
x⎞
⎜ ⎟+C
2
⎝ ⎠
6. u = 2 + e x , du = e x dx
du
ex
∫ 2 + e x dx = ∫
3. e x
4.
dx
∫ x2 + 4 = 2 tan
du
u
= ln u + C
2 3
u du
1
∫
= ln 2 + e x + C
= ln(2 + e x ) + C
Problem Set 7.1
1.
∫ ( x – 2)
2.
∫
5
1
dx = ( x – 2)6 + C
6
3 x dx =
1
2
3x ⋅ 3dx = (3x)3 / 2 + C
∫
3
9
3. u = x 2 + 1, du = 2 x dx
When x = 0, u = 1 and when x = 2, u = 5 .
2
∫0 x( x
2
+ 1)5 dx =
1 2 2
( x + 1)5 (2 x dx)
2 ∫0
8.
1 5
= ∫ u 5 du
2 1
15624
= 1302
12
4. u = 1 – x 2 , du = –2 x dx
When x = 0, u = 1 and when x = 1, u = 0 .
1
∫0
x 1 – x 2 dx = –
1 0
= − ∫ u1/ 2 du
2 1
1 1
= ∫ u1/ 2 du
2 0
2t 2
∫ 2t 2 + 1
dt = ∫
= ∫ dt – ∫
5
⎡ u6 ⎤
56 − 16
=⎢ ⎥ =
12
⎣⎢ 12 ⎦⎥1
=
7. u = x 2 + 4, du = 2x dx
x
1 du
∫ x2 + 4 dx = 2 ∫ u
1
= ln u + C
2
1
= ln x 2 + 4 + C
2
1
= ln( x 2 + 4) + C
2
1 1
1 − x 2 (−2 x dx)
2 ∫0
2t 2 + 1 − 1
2t 2 + 1
dt
1
dt
2t + 1
u = 2t , du = 2dt
1
1
du
t–∫
dt = t –
∫
2
2 1 + u2
2t + 1
1
=t–
tan –1 ( 2t ) + C
2
2
9. u = 4 + z 2 , du = 2z dz
∫ 6z
4 + z 2 dz = 3∫ u du
= 2u 3 / 2 + C
= 2(4 + z 2 )3 / 2 + C
1
1
⎡1
⎤
= ⎢ u3 / 2 ⎥ =
⎣3
⎦0 3
412
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10. u = 2t + 1, du = 2dt
5
5 du
∫ 2t + 1 dt = 2 ∫ u
16. u = 1 – x , du = –
3 / 4 sin
∫0
= 5 u +C
= 5 2t + 1 + C
= 2∫
tan z
∫ cos2 z dz = ∫ tan z sec
2
1
2
1 2
u +C
2
1
= tan 2 z + C
2
17.
12. u = cos z, du = –sin z dz
∫e
cos z
sin z dz = – ∫ ecos z (– sin z dz )
18.
= − ∫ eu du = −eu + C
= – ecos z + C
13. u = t , du =
∫
1
2 t
dt
sin t
dt = 2 ∫ sin u du
t
= –2 cos u + C
= –2 cos t + C
14. u = x 2 , du = 2x dx
2 x dx
du
=∫
∫
4
1– x
1 – u2
= sin –1 u + C
= sin ( x ) + C
–1
2
15. u = sin x, du = cos x dx
π / 4 cos x
2 / 2 du
∫0 1 + sin 2 x dx = ∫0 1 + u 2
= [tan −1 u ]0 2 / 2
2
2
≈ 0.6155
1/ 2
1
sin u du
sin u du
1⎞
⎛
= −2 ⎜ cos1 − cos ⎟
2⎠
⎝
≈ 0.6746
z dz = ∫ u du
=
dx
= [−2 cos u ]11/ 2
z dz
u = tan z, du = sec2 z dz
∫ tan z sec
dx = –2 ∫
1– x
1/ 2
11.
1– x
1
2 1– x
19.
3x2 + 2 x
1
∫ x + 1 dx = ∫ (3x – 1)dx + ∫ x + 1 dx
3
= x 2 – x + ln x + 1 + C
2
1
x3 + 7 x
2
∫ x – 1 dx = ∫ ( x + x + 8)dx + 8∫ x – 1 dx
1
1
= x3 + x 2 + 8 x + 8ln x – 1 + C
3
2
u = ln 4 x 2 , du =
2
dx
x
sin(ln 4 x 2 )
1
∫ x dx = 2 ∫ sin u du
1
= – cos u + C
2
1
= – cos(ln 4 x 2 ) + C
2
20. u = ln x, du =
1
dx
x
sec 2 (ln x )
1
dx = ∫ sec2 u du
2x
2
1
= tan u + C
2
1
= tan(ln x) + C
2
∫
21. u = e x , du = e x dx
= tan −1
6e x
∫
1 − e2 x
dx = 6 ∫
du
1− u2
du
= 6sin −1 u + C
= 6sin −1 (e x ) + C
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22. u = x 2 , du = 2x dx
x
1
du
∫ x4 + 4 dx = 2 ∫ 4 + u 2
1
u
= tan −1 + C
4
2
⎛
1
x2 ⎞
= tan –1 ⎜ ⎟ + C
⎜ 2 ⎟
4
⎝ ⎠
27.
= x − ln sin x + C
23. u = 1 – e2 x , du = –2e 2 x dx
3e2 x
∫
1– e
2x
dx = –
3 du
2∫ u
= –3 u + C
= –3 1 – e2 x + C
24.
x3
∫ x4 + 4
dx =
1 4 x3
dx
4 ∫ x4 + 4
28. u = cos(4t – 1), du = –4 sin(4t – 1)dt
sin(4t − 1)
sin(4t − 1)
∫ 1 − sin 2 (4t − 1) dt =∫ cos2 (4t − 1) dt
1 1
=− ∫
du
4 u2
1
1
= u −1 + C = sec(4t − 1) + C
4
4
29. u = e x , du = e x dx
∫e
1
= ln x 4 + 4 + C
4
1
= ln( x 4 + 4) + C
4
25.
1
t2
∫0 t 3
dt =
26.
2
1 1 t2
2t 3 dt
2 ∫0
π / 6 cos x
2
(– sin x dx)
0
sin x dx = – ∫
cos x ⎤ π / 6
⎡ 2
= ⎢–
⎥
⎣⎢ ln 2 ⎦⎥ 0
1
=–
(2 3 / 2 – 2)
ln 2
2−2 3/2
ln 2
≈ 0.2559
=
sec e x dx = ∫ sec u du
= ln sec e x + tan e x + C
30. u = e x , du = e x dx
∫e
1
π / 6 cos x
∫0
x
= ln sec u + tan u + C
⎡ 3t ⎤
⎥ = 3 – 1
=⎢
⎢ 2 ln 3 ⎥
2 ln 3 2 ln 3
⎣
⎦0
1
=
≈ 0.9102
ln 3
2
sin x − cos x
⎛ cos x ⎞
dx = ∫ ⎜1 −
⎟ dx
sin x
⎝ sin x ⎠
u = sin x, du = cos x dx
sin x − cos x
du
∫ sin x dx = x − ∫ u
= x − ln u + C
∫
x
sec 2 (e x )dx = ∫ sec 2 u du = tan u + C
= tan(e x ) + C
31.
∫
sec3 x + esin x
dx = ∫ (sec2 x + esin x cos x) dx
sec x
= tan x + ∫ esin x cos x dx
u = sin x, du = cos x dx
tan x + ∫ esin x cos x dx = tan x + ∫ eu du
= tan x + eu + C = tan x + esin x + C
32. u = 3t 2 − t − 1 ,
1
du = (3t 2 − t − 1)−1/ 2 (6t − 1)dt
2
∫
(6t − 1) sin 3t 2 − t − 1
3t 2 − t − 1
= –2 cos u + C
dt = 2∫ sin u du
= −2 cos 3t 2 − t − 1 + C
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33. u = t 3 − 2 , du = 3t 2 dt
∫
t 2 cos(t 3 − 2)
dt =
39. u = 3 y 2 , du = 6 y dy
1 cos u
du
3 ∫ sin 2 u
∫
sin (t − 2)
v = sin u, dv = cos u du
1 cos u
1
1
du = ∫ v −2 dv = − v −1 + C
3 ∫ sin 2 u
3
3
1
=−
+C
3sin u
1
=−
+C .
3sin(t 3 − 2)
34.
∫
2 3
1 + cos 2 x
2
sin 2 x
dx = ∫
1
2
sin 2 x
dx + ∫
cos 2 x
sin 2 2 x
= ∫ csc2 2 x dx + ∫ cot 2 x csc 2 x dx
1
1
= − cot 2 x − csc 2 x + C
2
2
35. u = t 3 − 2 , du = 3t 2 dt
∫
t 2 cos 2 (t 3 − 2)
dt =
1 cos 2 u
du
3 ∫ sin 2 u
sin 2 (t 3 − 2)
1
1
= ∫ cot 2 u du = ∫ (csc2 u –1)du
3
3
1
= [− cot u − u ] + C1
3
1
= [− cot(t 3 − 2) − (t 3 − 2)] + C1
3
1
= − [cot(t 3 − 2) + t 3 ] + C
3
36. u = 1 + cot 2t, du = −2 csc2 2t
∫
csc2 2t
1 + cot 2t
dt = −
1 1
du
2∫ u
=− u +C
= − 1 + cot 2t + C
37. u = tan −1 2t , du =
e tan
−1
1 + 4t 2
dt
2t
1 u
∫ 1 + 4t 2 dt = 2 ∫ e du
−1
1
1
= eu + C = e tan 2t + C
2
2
38. u = −t 2 − 2t − 5 ,
du = (–2t – 2)dt = –2(t + 1)dt
1
− t 2 − 2t − 5
= − ∫ eu du
∫ (t + 1)e
2
1 u
1 − t 2 − 2t − 5
= − e +C = − e
+C
2
2
Instructor’s Resource Manual
16 − 9 y 4
dy =
1
1
du
∫
6
42 − u 2
1
⎛u⎞
= sin −1 ⎜ ⎟ + C
6
⎝4⎠
⎛ 3y2
1
= sin −1 ⎜
⎜ 4
6
⎝
⎞
⎟+C
⎟
⎠
40. u = 3x, du = 3 dx
∫ cosh 3x dx
dx
1
1
(cosh u )du = sinh u + C
3∫
3
1
= sinh 3 x + C
3
=
41. u = x3 , du = 3 x 2 dx
1
2
3
∫ x sinh x dx = 3 ∫ sinh u du
1
= cosh u + C
3
1
= cosh x3 + C
3
42. u = 2x, du = 2 dx
5
5
1
dx = ∫
du
∫
2
9 − 4 x2
32 − u 2
5
⎛u⎞
= sin −1 ⎜ ⎟ + C
2
⎝3⎠
5 −1 ⎛ 2 x ⎞
= sin ⎜ ⎟ + C
2
⎝ 3 ⎠
43. u = e3t , du = 3e3t dt
∫
2
y
e3t
4−e
6t
dt =
1
1
du
3 ∫ 22 − u 2
1
⎛u⎞
= sin −1 ⎜ ⎟ + C
3
⎝2⎠
⎛ e3t
1
= sin −1 ⎜
⎜ 2
3
⎝
⎞
⎟+C
⎟
⎠
44. u = 2t, du = 2dt
dt
1
1
= ∫
du
∫
2
2t 4t − 1 2 u u 2 − 1
1
= ⎡sec −1 u ⎤ + C
⎦
2⎣
1
= sec −1 2t + C
2
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415
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45. u = cos x, du = –sin x dx
π/2
0
sin x
1
∫0 16 + cos2 x dx = − ∫1 16 + u 2 du
1
1
=∫
du
0 16 + u 2
50.
51.
1
1
1
= ln(e4 + 1) − ln(e 2 ) − ln 2
2
2
2
4
⎛
⎞
⎛
⎞
e +1
1
= ⎜ ln ⎜
⎟ − 2 ⎟ ≈ 0.6625
2 ⎜⎝ ⎜⎝ 2 ⎟⎠ ⎟⎠
47.
1
=
48.
( x + 1) 2 + 22
d ( x + 1)
1
1
1
( x – 2)2 + ( 5)2
d ( x – 2)
⎛ x–2⎞
tan –1 ⎜
⎟+C
5
⎝ 5 ⎠
–( x – 3)2 + 52
x +1
52 – ( x – 3)2
18 x + 18
1
(
52.
)
3– x
∫
16 + 6 x – x
2
dx =
1
6 – 2x
dx
2 ∫ 16 + 6 x – x 2
= 16 + 6 x – x 2 + C
53. u = 2t , du = 2dt
dt
du
∫ 2 =∫ 2 2
t 2t – 9
u u –3
⎛ 2t ⎞
1
⎟+C
= sec –1 ⎜
⎜ 3 ⎟
3
⎝
⎠
dx
dx
Section 7.1
tan x
∫
2
sec x – 4
sin x
dx = ∫
cos x
tan x
dx
cos x sec2 x – 4
dx
1 – 4 cos 2 x
u = 2 cos x, du = –2 sin x dx
sin x
1
1
dx = − ∫
du
∫
2
2
1 − 4 cos x
1− u2
1
1
= − sin −1 u + C = – sin –1 (2 cos x) + C
2
2
2
⎛ dy ⎞
1 + ⎜ ⎟ dx
a
⎝ dx ⎠
dx
∫ 9 x2 + 18 x + 10 = ∫ 9 x2 + 18 x + 9 + 1
1
= tan –1 (3x + 3) + C
3
54.
L=∫
(3x + 3)2 + 12
u = 3x + 3, du = 3 dx
dx
1
du
∫ (3x + 3)2 + 12 = 3 ∫ u 2 + 12
416
dx
=∫
55. The length is given by
1
=∫
dx
∫ 9 x2 + 18 x + 10 dx = 18 ∫ 9 x2 + 18 x + 10 dx
=∫
∫ x2 – 4 x + 9 dx = ∫ x2 – 4 x + 4 + 5 dx
=
49.
1
1
⎛ x +1⎞
tan –1 ⎜
⎟+C
2
⎝ 2 ⎠
=∫
–( x – 6 x + 9 – 25)
1
∫ x2 + 2 x + 5 dx = ∫ x2 + 2 x + 1 + 4 dx
=∫
2
1
ln 9 x 2 + 18 x + 10 + C
18
1
= ln 9 x 2 + 18 x + 10 + C
18
− e−2 x
1 e4 + 1 1
= ln
− ln 2
2
2
e2
dx
=∫
=
= 2(e2 x − e −2 x )dx
1 e2 + e−2 1
du
∫0 e2 x + e−2 x dx = 2 ∫2
u
1
1
1
e2 + e −2
= ⎡⎣ ln u ⎤⎦
= ln e2 + e −2 − ln 2
2
2
2
2
2
⎛ x –3⎞
= sin –1 ⎜
⎟+C
⎝ 5 ⎠
46. u = e2 x + e−2 x , du = (2e2 x − 2e −2 x )dx
1 e2 x
16 + 6 x – x
=∫
1
⎡1
⎡1
⎤
⎛ u ⎞⎤
⎛1⎞ 1
= ⎢ tan −1 ⎜ ⎟ ⎥ = ⎢ tan −1 ⎜ ⎟ − tan −1 0 ⎥
⎝ 4 ⎠⎦0 ⎣ 4
⎝4⎠ 4
⎣4
⎦
1
⎛1⎞
= tan −1 ⎜ ⎟ ≈ 0.0612
4
⎝4⎠
dx
∫
b
=∫
π/ 4
=∫
π/ 4
0
0
π/ 4
2
⎡ 1
⎤
1+ ⎢
(− sin x) ⎥ dx
⎣ cos x
⎦
1 + tan 2 x dx = ∫
π/ 4
0
sec2 x dx
=∫
sec x dx = ⎡⎣ln sec x + tan x ⎤⎦ 0π / 4
= ln
2 + 1 − ln 1 = ln
0
2 + 1 ≈ 0.881
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56. sec x =
=
1
1 + sin x
=
cos x cos x(1 + sin x)
sin x + sin 2 x + cos 2 x sin x(1 + sin x) + cos 2 x
=
cos x(1 + sin x)
cos x(1 + sin x)
sin x
cos x
+
cos x 1 + sin x
cos x ⎞
⎛ sin x
∫ sec x = ∫ ⎜⎝ cos x + 1 + sin x ⎟⎠dx
sin x
cos x
dx + ∫
dx
=∫
cos x
1 + sin x
For the first integral use u = cos x, du = –sin x dx,
and for the second integral use v = 1 + sin x,
dv = cos x dx.
sin x
cos x
du
dv
∫ cos x dx + ∫ 1 + sin x dx = – ∫ u + ∫ v
= – ln u + ln v + C
=
= – ln cos x + ln 1 + sin x + C
= ln
57. u = x – π , du = dx
2 π x sin x
π (u + π) sin(u + π)
∫0 1 + cos2 x dx = ∫– π 1 + cos2 (u + π) du
π (u + π) sin u
=∫
du
– π 1 + cos 2 u
π u sin u
π π sin u
=∫
du + ∫
du
2
– π 1 + cos u
– π 1 + cos 2 u
π u sin u
∫– π 1 + cos2 u du = 0 by symmetry.
π π sin u
π π sin u
∫– π 1 + cos2 u du = 2∫0 1 + cos2 u du
v = cos u, dv = –sin u du
–1 π
1
1
−2∫
dv = 2π ∫
dv
2
1 1+ v
–1 1 + v 2
⎡ π ⎛ π ⎞⎤
= 2π[tan −1 v]1−1 = 2π ⎢ − ⎜ − ⎟ ⎥
⎣ 4 ⎝ 4 ⎠⎦
⎛π⎞
= 2π ⎜ ⎟ = π2
⎝2⎠
1 + sin x
+C
cos x
= ln sec x + tan x + C
58.
3π
4 ⎛x+
– π ⎜⎝
4
V = 2π ∫
π⎞
⎟ sin x – cos x dx
4⎠
π
, du = dx
4
π
π⎞
π⎞
π⎞
⎛
⎛
⎛
V = 2π∫ 2π ⎜ u + ⎟ sin ⎜ u + ⎟ – cos ⎜ u + ⎟ du
– ⎝
2⎠
4⎠
4⎠
⎝
⎝
2
u= x–
π
π⎞ 2
2
2
2
⎛
= 2π∫ 2π ⎜ u + ⎟
sin u +
cos u –
cos u +
sin u du
– ⎝
2⎠ 2
2
2
2
2
π
π
π
π⎞
⎛
= 2π∫ 2π ⎜ u + ⎟ 2 sin u du = 2 2π ∫ 2π u sin u du + 2π 2 ∫ 2π sin u du
− ⎝
−
−
2⎠
2
2
2
π
2 2π∫ 2π u sin u du = 0 by symmetry. Therefore,
−
2
π
π
V = 2π2 2 ∫ 2 sin u du = 2 2π2 [− cos u ]02 = 2 2π2
0
Instructor’s Resource Manual
Section 7.1
417
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
6. u = x
dv = sin 2x dx
1
du = dx
v = – cos 2 x
2
1
1
∫ x sin 2 x dx = – 2 x cos 2 x – ∫ – 2 cos 2 x dx
1
1
= – x cos 2 x + sin 2 x + C
2
4
7.2 Concepts Review
1. uv – ∫ v du
2. x; sin x dx
3. 1
4. reduction
7. u = t – 3
dv = cos (t – 3)dt
du = dt
v = sin (t – 3)
∫ (t – 3) cos(t – 3)dt = (t – 3) sin(t – 3) – ∫ sin(t – 3)dt
Problem Set 7.2
v = ex
du = dx
∫ xe
x
= (t – 3) sin (t – 3) + cos (t – 3) + C
dv = e x dx
1. u = x
dx = xe − ∫ e dx = xe − e + C
x
x
x
x
dv = e3 x dx
1
du = dx
v = e3 x
3
1
1 3x
3x
3x
∫ xe dx = 3 xe − ∫ 3 e dx
1
1
= xe3 x − e3 x + C
3
9
2. u = x
dv = e5t +π dt
1
du = dt
v = e5t +π
5
1 5t +π
1 5t +π
5t +π
∫ te dt = 5 te – ∫ 5 e dt
1
1
= te5t +π – e5t +π + C
5
25
3. u = t
4. u = t + 7 dv = e2t +3 dt
1
du = dt
v = e 2t + 3
2
1
1 2t + 3
2t + 3
2t + 3
∫ (t + 7)e dt = 2 (t + 7)e – ∫ 2 e dt
1
1
= (t + 7)e2t +3 – e2t +3 + C
2
4
t
13
= e 2 t + 3 + e 2t + 3 + C
2
4
5. u = x
dv = cos x dx
du = dx
v = sin x
∫ x cos x dx = x sin x – ∫ sin x dx
= x sin x + cos x + C
418
Section 7.2
8. u = x – π
dv = sin(x)dx
du = dx
v = –cos x
(
x
–
π
)
sin(
x
)
dx = –( x – π) cos x + ∫ cos x dx
∫
= ( π – x) cos x + sin x + C
dv = t + 1 dt
9. u = t
v=
du = dt
2
(t + 1)3 / 2
3
2
2
t + 1 dt = t (t + 1)3 / 2 – ∫ (t + 1)3 / 2 dt
3
3
2
4
= t (t + 1)3 / 2 – (t + 1)5 / 2 + C
3
15
∫t
dv = 3 2t + 7dt
3
v = (2t + 7)4 / 3
du = dt
8
3
3
4/3
4/3
3
∫ t 2t + 7dt = 8 t (2t + 7) – ∫ 8 (2t + 7) dt
3
9
(2t + 7)7 / 3 + C
= t (2t + 7)4 / 3 –
8
112
10. u = t
11. u = ln 3x
1
du = dx
x
dv = dx
v=x
1
∫ ln 3x dx = x ln 3x −∫ x x dx
12. u = ln(7 x5 )
du =
5
dx
x
∫ ln(7 x
5
= x ln 3 x − x + C
dv = dx
v=x
5
)dx = x ln(7 x5 ) – ∫ x dx
x
= x ln(7 x5 ) – 5 x + C
Instructor Solutions Manual
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13. u = arctan x
1
du =
dx
1 + x2
dv = dx
1
du = dt
t
v=x
x
∫ arctan x = x arctanx − ∫ 1 + x2 dx
1 2x
= x arctan x − ∫
dx
2 1 + x2
1
= x arctan x − ln(1 + x 2 ) + C
2
14. u = arctan 5x
5
du =
dx
1 + 25 x 2
5x
∫ arctan 5x dx = x arctan 5x – ∫ 1 + 25 x2 dx
1 50 x dx
= x arctan 5 x – ∫
10 1 + 25 x 2
1
= x arctan 5 x – ln(1 + 25 x 2 ) + C
10
dx
x2
1
1
du = dx
v=–
x
x
ln x
ln x
1⎛1⎞
∫ x2 dx = – x – ∫ – x ⎜⎝ x ⎟⎠ dx
ln x 1
=–
– +C
x
x
16. u = ln 2 x5
du =
x2
1
x2
1
v=−
x
5
dx
x
3 ln 2 x5
∫2
dv =
dx
3
3 1
⎡ 1
⎤
dx = ⎢ − ln 2 x5 ⎥ + 5∫
dx
2 x2
⎣ x
⎦2
3
5⎤
⎡ 1
= ⎢ − ln 2 x5 − ⎥
x ⎦2
⎣ x
5⎞ ⎛ 1
5⎞
⎛ 1
= ⎜ − ln(2 ⋅ 35 ) − ⎟ − ⎜ − ln(2 ⋅ 25 ) − ⎟
3⎠ ⎝ 2
2⎠
⎝ 3
1
5
5
5
= − ln 2 − ln 3 − + 3ln 2 +
3
3
3
2
8
5
5
= ln 2 − ln 3 + ≈ 0.8507
3
3
6
Instructor's Resource Manual
e
∫1
e
v=x
dv =
2
v = t3 / 2
3
e2
⎡2
⎤
t ln t dt = ⎢ t 3 / 2 ln t ⎥ – ∫ t1/ 2 dt
⎣3
⎦1 1 3
e
=
2 3/ 2
2
⎡4
⎤
e ln e – ⋅1ln1 − ⎢ t 3 / 2 ⎥
3
3
9
⎣
⎦1
=
2 3/ 2
4
4 2
4
e
− 0 − e3 / 2 + = e3 / 2 + ≈ 1.4404
3
9
9 9
9
dv = dx
15. u = ln x
dv = t dt
17. u = ln t
dv = 2 xdx
1
v = (2 x)3 / 2
3
18. u = ln x3
3
du = dx
x
5
∫1
5
5
⎡1
⎤
2 x ln x3 dx = ⎢ (2 x)3 2 ln x3 ⎥ − ∫ 23 2 x dx
1
⎣3
⎦1
5
1
25 / 2 3 / 2 ⎤
= ⎡ (2 x)3 / 2 ln x3 −
x
3
⎢⎣ 3
⎥⎦1
1
25 2 3 / 2 ⎛ 1 3 2 3 25 2 ⎞
= (10)3 2 ln 53 −
5
− ⎜ (2) ln1 −
⎟
⎜3
3
3
3 ⎟⎠
⎝
=−
4 2 32 4 2
5 +
+ 103 2 ln 5 ≈ 31.699
3
3
dv = z 3dz
1
v = z4
4
1
1 4 1
3
4
∫ z ln z dz = 4 z ln z − ∫ 4 z ⋅ z dz
1
1
= z 4 ln z − ∫ z 3 dz
4
4
1 4
1
= z ln z − z 4 + C
4
16
19. u = ln z
1
du = dz
z
20. u = arctan t
1
du =
dt
1+ t2
dv = t dt
1
v = t2
2
1
∫ t arctan t dt = 2 t
2
arctan t –
1 t2
dt
2 ∫ 1+ t2
1
1 1+ t2 −1
= t 2 arctan t − ∫
dt
2
2 1+ t2
1
1
1
1
= t 2 arctan t − ∫ dt + ∫
dt
2
2
2 1+ t2
1
1 1
= t 2 arctan t − t + arctan t + C
2
2 2
Section 7.2
419
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
⎛1⎞
21. u = arctan ⎜ ⎟
dv = dt
⎝t ⎠
1
v=t
du = –
dt
1+ t2
t
⎛1⎞
⎛1⎞
∫ arctan ⎜⎝ t ⎟⎠ dt = t arctan ⎜⎝ t ⎟⎠ + ∫ 1 + t 2 dt
⎛1⎞ 1
= t arctan ⎜ ⎟ + ln(1 + t 2 ) + C
⎝t ⎠ 2
dv = t 5 dt
7
dt
t
1
v = t6
6
du =
1
7
ln(t 7 )dt = t 6 ln(t 7 ) – ∫ t 5 dt
6
6
1
7
= t 6 ln(t 7 ) – t 6 + C
6
36
∫t
5
dv = csc2 x dx
v = − cot x
23. u = x
du = dx
π/2
22. u = ln(t 7 )
∫π / 6 x csc
π/ 2
x dx = [ − x cot x ]π / 6 + ∫
2
π/ 2
π/6
cot x dx = ⎣⎡− x cot x + ln sin x ⎤⎦
π2
π6
π
π
1
π
= − ⋅ 0 + ln1 +
3 − ln =
+ ln 2 ≈ 1.60
2
6
2 2 3
dv = sec2 x dx
v = tan x
24. u = x
du = dx
π4
∫π 6 x sec
=
2
π4
x dx = [ x tan x ]π 6 − ∫
π4
π6
π
2 ⎛ π
3⎞
π4
= + ln
− ⎜⎜
+ ln
tan x dx = ⎡⎣ x tan x + ln cos x ⎤⎦
⎟
π6
4
2 ⎝6 3
2 ⎟⎠
π
π
1 2
−
+ ln ≈ 0.28
4 6 3 2 3
25. u = x3
dv = x 2 x3 + 4dx
2
v = ( x3 + 4)3 / 2
du = 3 x 2 dx
9
2
2 2 3
2 3 3
4 3
5
3
3 3
3/ 2
3/ 2
3/ 2
5/ 2
∫ x x + 4dx = 9 x ( x + 4) – ∫ 3 x ( x + 4) dx = 9 x ( x + 4) – 45 ( x + 4) + C
dv = x 6 x 7 + 1 dx
26. u = x7
v=
du = 7 x 6 dx
∫x
13
x7 + 1dx =
2 7 7
2
2
4 7
( x + 1)5 / 2 + C
x ( x + 1)3 / 2 – ∫ x6 ( x7 + 1)3 / 2 dx = x 7 ( x 7 + 1)3 / 2 –
21
3
21
105
dv =
4
27. u = t
v=
du = 4t 3 dt
t7
∫ (7 – 3t 4 )3 / 2
420
2 7
( x + 1)3 / 2
21
dt =
Section 7.2
t3
(7 – 3t 4 )3 / 2
1
dt
6(7 – 3t 4 )1/ 2
t4
6(7 – 3t 4 )1/ 2
–
t3
t4
2
1
dt
=
+ (7 – 3t 4 )1/ 2 + C
4 1/ 2 9
3 ∫ (7 – 3t 4 )1/ 2
6(7 – 3t )
Instructor Solutions Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
28. u = x 2
dv = x 4 – x 2 dx
1
du = 2x dx
v = – (4 – x 2 )3 / 2
3
1
1 2
2
3
2
2
2 3/ 2 2
2 3/ 2
2 3/ 2
2 5/ 2
∫ x 4 – x dx = – 3 x (4 – x ) + 3 ∫ x(4 – x ) dx = – 3 x (4 – x ) – 15 (4 – x ) + C
29. u = z 4
dv =
v=
du = 4 z 3 dz
z7
∫ (4 – z 4 )2
dz =
z3
(4 – z 4 ) 2
dz
1
4(4 – z 4 )
z4
4
4(4 – z )
−∫
z3
4– z
4
dz =
z4
1
+ ln 4 – z 4 + C
4(4 – z ) 4
4
30. u = x
dv = cosh x dx
du = dx
v = sinh x
x
cosh
x
dx
=
x
sinh
x – ∫ sinh x dx = x sinh x – cosh x + C
∫
31. u = x
dv = sinh x dx
du = dx
v = cosh x
∫ x sinh x dx = x cosh x – ∫ cosh x dx = x cosh x – sinh x + C
32. u = ln x
dv = x –1/ 2 dx
1
du = dx
v = 2 x1/ 2
x
ln x
1
∫ x dx = 2 x ln x – 2∫ x1/ 2 dx = 2 x ln x – 4 x + C
33. u = x
dv = (3 x + 10)49 dx
1
(3x + 10)50
150
x
1
x
1
49
50
50
50
51
∫ x(3x + 10) dx = 150 (3x + 10) – 150 ∫ (3x + 10) dx = 150 (3x + 10) – 22,950 (3x + 10) + C
du = dx
34. u = t
du = dt
v=
dv = ( t − 1) dt
12
v=
1
( t − 1)13
13
1
1 1
⎡t
13 ⎤
13
12
∫0 t (t − 1) dt = ⎢⎣13 ( t − 1) ⎥⎦0 − 13 ∫0 ( t − 1) dt
1
1
1
1
⎡t
13
= ⎢ ( t − 1) −
( t − 1)14 ⎤⎥ =
13
182
182
⎣
⎦0
dv = 2 dx
1 x
du = dx
v=
2
ln 2
x x
1
x
x
∫ x2 dx = ln 2 2 – ln 2 ∫ 2 dx
x x
1
=
2 –
2x + C
2
ln 2
(ln 2)
35. u = x
x
dv = a z dz
1 z
du = dz
v=
a
ln a
z z
1
z
z
∫ za dz = ln a a – ln a ∫ a dz
z z
1
a –
az + C
=
2
ln a
(ln a)
36. u = z
37. u = x 2
du = 2x dx
∫x
dv = e x dx
v = ex
e dx = x 2 e x − ∫ 2 xe x dx
2 x
u=x
dv = e x dx
du = dx
v = ex
∫x
(
e dx = x 2 e x − 2 xe x − ∫ e x dx
2 x
)
= x 2 e x − 2 xe x + 2e x + C
Instructor's Resource Manual
Section 7.2
421
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
41. u = et
2
dv = xe x dx
1 2
v = ex
2
38. u = x 4
du = 4 x3 dx
du = et dt
∫e
5 x2
1 4 x2
3 x2
∫ x e dx = 2 x e – ∫ 2 x e dx
∫ ln
2
v = –cos t
t
t
∫ e cos t dt = e sin t − ⎡⎣−e cos t + ∫ e cos t dt ⎤⎦
t
2
∫ ln
2
t
∫e
∫e
at
dv = sin t dt
v = –cos t
sin t dt = – e cos t + a ∫ e at cos t dt
at
dv = dz
u = e at
v=z
du = ae dt
dv = cos t dt
at
(
z dz = z ln z – 2 z ln z – ∫ dz
t
1 t
e (sin t + cos t ) + C
2
du = ae dt
v = sin t
∫ e sin t dt = – e cos t + a ( e sin t – a ∫ e sin t dt )
at
at
at
2 at
∫ e sin t dt = – e cos t + ae sin t – a ∫ e sin t dt
at
)
= z ln 2 z – 2 z ln z + 2 z + C
at
at
at
dv = dx
(1 + a 2 ) ∫ e at sin t dt = – e at cos t + ae at sin t + C
v=x
∫e
du =
40 ln x 20
dx
x
∫ ln
x dx = x ln 2 x 20 – 40 ∫ ln x 20 dx
at
sin t dt =
– e at cos t
a2 + 1
+
aeat sin t
a2 + 1
+C
2 20
u = ln x 20
20
du =
dx
x
∫ ln
cos t dt =
t
t
at
z dz = z ln 2 z – 2 ∫ ln z dz
40. u = ln 2 x 20
t
42. u = e at
v=z
2
t
∫ e cos t dt = e sin t + e cos t − ∫ e cos t dt
2∫ et cos t dt = et sin t + et cos t + C
dv = dz
u = ln z
1
du = dz
z
dv = sin t dt
du = e dt
v = ex
1 4 x2 ⎛ 2 x2
⎞
x2
5 x2
∫ x e dx = 2 x e – ⎜⎝ x e − ∫ 2 xe dx ⎟⎠
2
2
2
1
= x4e x – x2e x + e x + C
2
39. u = ln 2 z
2 ln z
du =
dz
z
t
t
dv = 2 xe x dx
du = 2x dx
v = sin t
cos t dt = e sin t − ∫ et sin t dt
t
u = et
2
u = x2
dv = cos t dt
43. u = x 2
du = 2 x dx
dv = dx
v=x
(
x dx = x ln 2 x 20 – 40 x ln x 20 – 20∫ dx
2 20
)
= x ln 2 x 20 – 40 x ln x 20 + 800 x + C
∫x
2
2
v = sin x
cos x dx = x 2 sin x − ∫ 2 x sin x dx
u = 2x
du = 2dx
∫x
dv = cos x dx
dv = sin x dx
v = − cos x
(
cos x dx = x 2 sin x − −2 x cos x + ∫ 2 cos x dx
)
= x sin x + 2 x cos x − 2sin x + C
2
44. u = r 2
du = 2r dr
∫r
2
sin r dr = – r 2 cos r + 2∫ r cos r dr
u=r
du = dr
∫r
422
2
dv = sin r dr
v = –cos r
dv = cos r dr
v = sin r
(
)
sin r dr = – r 2 cos r + 2 r sin r – ∫ sin r dr = – r 2 cos r + 2r sin r + 2 cos r + C
Section 7.2
Instructor Solutions Manual
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
45. u = sin(ln x)
dv = dx
1
du = cos(ln x) ⋅ dx
x
v=x
∫ sin(ln x)dx = x sin(ln x) − ∫ cos(ln x) dx
u = cos (ln x)
dv = dx
1
du = − sin(ln x) ⋅ dx
x
v=x
∫ sin(ln x)dx = x sin(ln x) − ⎡⎣ x cos(ln x) − ∫ − sin(ln x)dx ⎤⎦
∫ sin(ln x)dx = x sin(ln x) − x cos(ln x) − ∫ sin(ln x)dx
2∫ sin(ln x)dx = x sin(ln x) − x cos(ln x) + C
x
∫ sin(ln x)dx = 2 [sin(ln x) − cos(ln x)] + C
46. u = cos(ln x)
dv = dx
1
du = – sin(ln x) dx
x
v=x
∫ cos(ln x)dx = x cos(ln x) + ∫ sin(ln x)dx
u = sin(ln x)
dv = dx
1
du = cos(ln x) dx
x
v=x
∫ cos(ln x)dx = x cos(ln x) + ⎡⎣ x sin(ln x) – ∫ cos(ln x)dx ⎤⎦
2∫ cos(ln x)dx = x[cos(ln x ) + sin(ln x )] + C
x
∫ cos(ln x)dx = 2 [cos(ln x) + sin(ln x)] + C
47. u = (ln x)3
du =
3ln 2 x
dx
x
∫ (ln x)
3
dv = dx
v=x
dx = x(ln x)3 – 3∫ ln 2 x dx
= x ln 3 x – 3( x ln 2 x – 2 x ln x + 2 x + C )
= x ln 3 x – 3x ln 2 x + 6 x ln x − 6 x + C
48. u = (ln x)4
du =
4 ln 3 x
dx
x
∫ (ln x)
4
dv = dx
v=x
dx = x (ln x )4 – 4∫ ln 3 x dx = x ln 4 x – 4( x ln 3 x – 3 x ln 2 x + 6 x ln x − 6 x + C )
= x ln 4 x – 4 x ln 3 x + 12 x ln 2 x − 24 x ln x + 24 x + C
49. u = sin x
dv = sin(3x)dx
1
du = cos x dx
v = – cos(3 x)
3
1
1
∫ sin x sin(3 x)dx = – 3 sin x cos(3 x) + 3 ∫ cos x cos(3x)dx
u = cos x
dv = cos(3x)dx
1
v = sin(3 x)
du = –sin x dx
3
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1 ⎡1
1
⎤
1
∫ sin x sin(3x)dx = – 3 sin x cos(3x) + 3 ⎢⎣ 3 cos x sin(3x) + 3 ∫ sin x sin(3x)dx ⎥⎦
1
1
1
= – sin x cos(3 x ) + cos x sin(3 x) + ∫ sin x sin(3 x)dx
3
9
9
8
1
1
sin x sin(3 x) dx = – sin x cos(3 x) + cos x sin(3 x) + C
9∫
3
9
3
1
∫ sin x sin(3x)dx = – 8 sin x cos(3x) + 8 cos x sin(3x) + C
50. u = cos (5x)
dv = sin(7x)dx
1
du = –5 sin(5x)dx v = – cos(7 x)
7
1
5
∫ cos(5 x) sin(7 x)dx = – 7 cos(5 x) cos(7 x) – 7 ∫ sin(5 x) cos(7 x)dx
u = sin(5x)
dv = cos(7x)dx
1
v = sin(7 x)
du = 5 cos(5x)dx
7
1
5 ⎡1
5
⎤
∫ cos(5 x) sin(7 x)dx = – 7 cos(5 x) cos(7 x) – 7 ⎢⎣ 7 sin(5 x) sin(7 x) – 7 ∫ cos(5 x) sin(7 x)dx ⎥⎦
1
5
25
= – cos(5 x) cos(7 x) – sin(5 x) sin(7 x) + ∫ cos(5 x) sin(7 x)dx
7
49
49
24
1
5
cos(5 x ) sin(7 x)dx = – cos(5 x ) cos(7 x ) – sin(5 x ) sin(7 x ) + C
∫
49
7
49
7
5
∫ cos(5 x) sin(7 x)dx = – 24 cos(5 x) cos(7 x) – 24 sin(5 x) sin(7 x) + C
51. u = eα z
dv = sin βz dz
1
v = – cos β z
du = α eα z dz
αz
∫e
β
sin β z dz = –
αz
du = α eα z dz
αz
=–
⎤
1 αz
α ⎡1
α
e cos β z + ⎢ eα z sin β z − ∫ eα z sin β z dz ⎥
β
β ⎣β
β
⎦
1 αz
α αz
α2 αz
e cos β z +
e sin β z –
∫ e sin β z dz
β
β
2
αz
424
β
β
sin β z dz = −
β 2 +α2
∫e
β
dv = cos βz dz
1
v = sin β z
u=e
∫e
1 αz
α
e cos β z + ∫ eα z cos β z dz
β2
αz
∫e
β2
sin β z dz = –
sin β z dz =
Section 7.2
–β
α2 + β2
1 αz
α αz
e cos β z +
e sin β z + C
β
eα z cos β z +
β2
α
α2 + β 2
eα z sin β z + C =
eα z (α sin β z – β cos β z )
α2 + β2
+C
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
52. u = eα z
dv = cos β z dz
du = α eα z dz
αz
∫e
v=
cos β z dz =
=
β
dv = sin βz dz
1
v = – cos β z
β
cos β z dz =
⎤
1 αz
α⎡ 1
α
e sin β z − ⎢ − eα z cos β z + ∫ eα z cos β z dz ⎥
β⎣ β
β
⎦
β
1 αz
α αz
α2 αz
e sin β z +
e cos β z –
∫ e cos β z dz
β
β2
α2 + β2
β
2
αz
∫e
α2 + β2
+C
dv = xα dx
1
dx
x
v=
xα +1
, α ≠ –1
α +1
xα +1
xα +1
xα +1
1
α
x
ln x –
x
dx
=
ln
–
+ C , α ≠ –1
α +1
α +1 ∫
α +1
(α + 1)2
α
∫ x ln x dx =
dv = xα dx
54. u = (ln x)2
2 ln x
dx
x
du =
α αz
1
e cos β z + eα z sin β z + C
2
β
β
eα z (α cos β z + β sin β z )
53. u = ln x
du =
β2
cos β z dz =
αz
∫ e cos β z dz =
α
2
∫ x (ln x) dx =
=
sin β z
1 αz
α
e sin β z – ∫ eα z sin β z dz
du = α eα z dz
αz
β
β
u = eα z
∫e
1
v=
xα +1
, α ≠ –1
α +1
xα +1
2
xα +1
xα +1 ⎤
2 ⎡ xα +1
α
2
(ln x)2 –
ln
x
x
dx
(ln
)
ln
=
x
−
x
−
⎢
⎥+C
α +1
α +1 ∫
α +1
α + 1 ⎣⎢ α + 1
(α + 1) 2 ⎦⎥
xα +1
xα +1
xα +1
(ln x) 2 – 2
ln x + 2
+ C , α ≠ –1
α +1
(α + 1)2
(α + 1)3
Problem 53 was used for
55. u = xα
α βx
e
dx =
v=
xα e β x
β
56. u = xα
du = α xα –1dx
α
∫x
ln x dx.
dv = e β x dx
du = α xα –1dx
∫x
α
∫x
sin β x dx = –
–
1 βx
e
β
α α –1 β x
x e dx
β∫
dv = sin βx dx
1
v = – cos β x
β
xα cos β x
β
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+
α α –1
cos β x dx
x
β∫
Section 7.2
425
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
57. u = xα
dv = cos βx dx
1
v = sin β x
du = α xα –1dx
α
∫x
β
α
x sin β x
cos β x dx =
β
–
α α –1
sin β x dx
x
β∫
58. u = (ln x)α
dv = dx
α –1
α (ln x)
du =
α
∫ (ln x)
v=x
dx
x
dx = x(ln x)α – α ∫ (ln x)α –1 dx
59. u = (a 2 – x 2 )α
dv = dx
2 α –1
du = –2α x(a – x )
2
∫ (a
2
v=x
dx
– x 2 )α dx = x (a 2 – x 2 )α + 2α ∫ x 2 (a 2 – x 2 )α –1 dx
60. u = cosα –1 x
dv = cos x dx
α –2
du = –(α – 1) cos
α
∫ cos
v = sin x
x sin x dx
x dx = cosα –1 x sin x + (α – 1) ∫ cosα –2 x sin 2 x dx
= cosα −1 x sin x + (α − 1) ∫ cosα − 2 x(1 − cos 2 x) dx = cosα –1 x sin x + (α – 1) ∫ cosα –2 x dx – (α – 1) ∫ cosα x dx
α ∫ cosα x dx = cosα −1 x sin x + (α − 1) ∫ cosα − 2 x dx
α
∫ cos x dx =
cosα –1 x sin x α – 1
α –2
+
∫ cos x dx
α
α
61. u = cosα –1 β x
dv = cos βx dx
du = – β (α – 1) cosα –2 β x sin β x dx
α
∫ cos β x dx =
=
=
β
cosα –1 β x sin β x
β
∫x
sin β x
+ (α – 1) ∫ cosα –2 β x sin 2 β x dx
+ (α – 1) ∫ cosα –2 β x dx – (α – 1) ∫ cosα β x dx
β
+ (α − 1) ∫ cosα − 2 β x dx
cosα –1 β x sin β x α – 1
α –2
+
∫ cos β x dx
αβ
α
1 4 3x 4 3 3x
1
4 ⎡1
⎤
x e – ∫ x e dx = x 4 e3 x – ⎢ x3e3 x – ∫ x 2 e3 x dx ⎥
3
3 ⎣3
3
3
⎦
4
4 ⎡1
2
1
4
4
8 ⎡1
1
⎤
⎤
– x3e3 x + ⎢ x 2 e3 x – ∫ xe3 x dx ⎥ = x 4 e3 x – x3e3 x + x 2 e3 x – ⎢ xe3 x – ∫ e3 x dx ⎥
9
3 ⎣3
3
3
9
9
9
3
3
⎣
⎦
⎦
4 3 3x 4 2 3x 8 3x 8 3x
– x e + x e –
xe + e + C
9
9
27
81
e dx =
4 3x
1 4 3x
x e
3
1
= x 4 e3 x
3
426
1
β
+ (α − 1) ∫ cosα − 2 β x(1 − cos 2 β x) dx
cosα −1 β x sin β x
α
∫ cos β x dx =
=
β
cosα −1 β x sin β x
α ∫ cosα β x =
62.
cosα –1 β x sin β x
v=
Section 7.2
Instructor Solutions Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
63.
64.
1 4
4
1
4⎡ 1
⎤
x sin 3 x – ∫ x3 sin 3 x dx = x 4 sin 3 x – ⎢ – x3 cos 3x + ∫ x 2 cos 3 x dx ⎥
3
3
3
3⎣ 3
⎦
1
4
4 ⎡1
2
⎤
= x 4 sin 3 x + x3 cos 3 x – ⎢ x 2 sin 3x − ∫ x sin 3 x dx ⎥
3
9
3 ⎣3
3
⎦
1 4
4 3
4 2
8⎡ 1
1
⎤
= x sin 3 x + x cos 3 x – x sin 3 x + ⎢ – x cos 3 x + ∫ cos 3 x dx ⎥
3
9
9
9⎣ 3
3
⎦
1 4
4 3
4 2
8
8
= x sin 3 x + x cos 3 x – x sin 3 x –
x cos 3 x + sin 3 x + C
3
9
9
27
81
∫x
4
cos 3 x dx =
1
5
1
5⎡ 1
3
⎤
cos5 3 x sin 3 x + ∫ cos 4 3x dx = cos5 3x sin 3x + ⎢ cos3 3x sin 3 x + ∫ cos 2 3 x dx ⎥
18
6 ⎣12
4
18
6
⎦
1
5
5
1
1
⎡
⎤
= cos5 3 x sin 3 x + cos3 3 x sin 3 x + ⎢ cos 3 x sin 3 x + ∫ dx ⎥
18
72
8 ⎣6
2
⎦
x
1
5
5
5
= cos5 3x sin 3x + cos3 3 x sin 3 x + cos 3 x sin 3 x + + C
18
72
48
16
∫ cos
6
3 x dx =
65. First make a sketch.
From the sketch, the area is given by
e
∫1 ln x dx
u = ln x
1
du = dx
x
dv = dx
v=x
∫1 ln x dx = [ x ln x ]1 − ∫1 dx = [ x ln x − x]1 = (e – e) – (1 · 0 – 1) = 1
e
e
e
e
e
66. V = ∫ π(ln x) 2 dx
1
u = (ln x)2
du =
2 ln x
dx
x
dv = dx
v=x
e
e
e
e
⎛
⎞
π∫ (ln x) 2 dx = π ⎜ ⎡ x(ln x) 2 ⎤ − 2∫ ln x dx ⎟ = π ⎡ x(ln x)2 − 2( x ln x − x) ⎤ = π[ x (ln x) 2 − 2 x ln x + 2 x]1e
⎦1
1
1
⎣
⎦1
⎝⎣
⎠
= π[(e − 2e + 2e) − (0 − 0 + 2)] = π(e − 2) ≈ 2.26
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
67.
9
∫0 3e
–x/3
9
9
⎛ 1 ⎞
dx = –9∫ e – x / 3 ⎜ – dx ⎟ = –9[e – x / 3 ]90 = – + 9 ≈ 8.55
0
3
⎝
⎠
e3
9
9
0
0
68. V = ∫ π(3e – x / 3 ) 2 dx = 9π ∫ e –2 x / 3 dx
27π –2 x / 3 9
27π 27 π
⎛ 3⎞ 9
⎛ 2 ⎞
= 9π ⎜ – ⎟ ∫ e –2 x / 3 ⎜ – dx ⎟ = –
[e
]0 = –
+
≈ 42.31
0
2
3
2
2
⎝
⎠
⎝
⎠
2e 6
69.
π/4
∫0
( x cos x – x sin x)dx = ∫
π/ 4
0
x cos x dx – ∫
π/4
0
x sin x dx
π4
π4
π4
⎛
⎞
π4
= ⎜ ⎣⎡ x sin x ⎦⎤ 0 − ∫ sin x dx ⎟ − ⎛⎜ [ − x cos x ]0 + ∫ cos x dx ⎞⎟
0
0
⎠
⎝
⎠ ⎝
2π
–1 ≈ 0.11
4
Use Problems 60 and 61 for ∫ x sin x dx and ∫ x cos x dx.
= [ x sin x + cos x + x cos x – sin x]0π / 4 =
⎛ x⎞
x sin ⎜ ⎟ dx
⎝2⎠
x
u=x
dv = sin dx
2
x
du = dx v = –2 cos
2
2π
2π
⎛⎡
⎛
2π
x⎤
x ⎞
x⎤ ⎞
⎡
V = 2π ⎜ ⎢ –2 x cos ⎥ + ∫ 2 cos dx ⎟ = 2π ⎜ 4π + ⎢ 4sin ⎥ ⎟ = 8π2
0
⎜⎣
⎜
2 ⎦0
2 ⎟
2 ⎦0 ⎟
⎣
⎝
⎠
⎝
⎠
70. V = 2π∫
2π
0
71.
e
∫1 ln x
2
e
dx = 2 ∫ ln x dx
1
u = ln x
1
du = dx
x
dv = dx
v=x
(
)
e
e ⎞
⎛
2∫ ln x dx = 2 ⎜ [ x ln x]1e − ∫ dx ⎟ = 2 e − [ x]1e = 2
1
1
⎝
⎠
e
∫1 x ln x
2
e
dx = 2 ∫ x ln x dx
1
u = ln x
1
du = dx
x
dv = x dx
1
v = x2
2
e
e
⎛
⎞
⎛1
e
e1
⎡1
⎤
⎡1 ⎤ ⎞ 1
2∫ x ln x dx = 2 ⎜ ⎢ x 2 ln x ⎥ – ∫ x dx ⎟ = 2 ⎜ e2 – ⎢ x 2 ⎥ ⎟ = (e2 + 1)
1
⎜⎣2
⎟
⎜2
⎦1 1 2
⎣ 4 ⎦1 ⎟⎠ 2
⎝
⎠
⎝
428
Section 7.2
Instructor Solutions Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1 e
(ln x)2 dx
2 ∫1
u = (ln x)2
du =
dv = dx
2 ln x
dx
x
v=x
e
1 e
1
1
(ln x)2 dx = ⎛⎜ [ x(ln x) 2 ]1e – 2∫ ln x dx ⎞⎟ = (e – 2)
1
2 ∫1
2⎝
⎠ 2
x=
y=
72. a.
1 (e 2
2
+ 1)
=
2
1 (e – 2)
2
2
=
e2 + 1
4
e–2
4
u = cot x
dv = csc2 x dx
du = – csc2 x dx
v = –cot x
∫ cot x csc
2
x dx = − cot 2 x − ∫ cot x csc2 x dx
2∫ cot x csc2 x dx = − cot 2 x + C
∫
1
cot x csc2 x dx = − cot 2 x + C
2
b. u = csc x
du = –cot x csc x dx
∫ cot x csc
2
dv = cot x csc x dx
v = –csc x
x dx = − csc2 x − ∫ cot x csc2 x dx
2∫ cot x csc2 x dx = − csc2 x + C
∫
c.
73. a.
1
cot x csc2 x dx = − csc 2 x + C
2
1
1
1
1
– cot 2 x = – (csc 2 x – 1) = – csc2 x +
2
2
2
2
p ( x ) = x3 − 2 x
g ( x) = e x
All antiderivatives of g ( x) = e x
∫ (x
b.
3
− 2 x)e x dx = ( x3 − 2 x)e x − (3x 2 − 2)e x + 6 xe x − 6e x + C
p( x) = x 2 − 3x + 1
g(x) = sin x
G1 ( x) = − cos x
G2 ( x) = − sin x
G3 ( x) = cos x
∫ (x
2
− 3 x + 1) sin x dx = ( x 2 − 3x + 1)(− cos x) − (2 x − 3)(− sin x) + 2 cos x + C
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Section 7.2
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
74. a.
We note that the nth arch extends from x = 2π (n − 1) to x = π (2n − 1) , so the area of the nth arch is
A(n) =
π (2 n −1)
∫2π (n−1) x sin x dx .
u=x
Using integration by parts:
dv = sin x dx
du = dx
v = − cos x
π (2 n −1)
π (2 n −1)
π (2 n −1)
1)
1)
A(n) = ∫2ππ(2( nn−−1)
x sin x dx = − x cos x 2π ( n −1) − ∫2ππ(2( nn−−1)
− cos x dx = − x cos x 2π ( n −1) + sin x 2π ( n −1)
[
]
[
]
[
]
= [ −π (2n − 1) cos(π (2n − 1)) + 2π (n − 1) cos(2π (n − 1)) ] + [sin(π (2n − 1)) − sin(2π (n − 1)) ]
= −π (2n − 1)(−1) + 2π (n − 1)(1) + 0 − 0 = π [ (2n − 1) + (2n − 2) ] .
b.
V = 2π ∫
So
A(n) = (4n − 3)π
3π 2
2π
x sin x dx
u = x2
dv = sin x dx
du = 2x dx
v = –cos x
3π
3π
3π
⎛
⎞
⎛
⎞
V = 2π ⎜ ⎡ – x 2 cos x ⎤ + ∫ 2 x cos x dx ⎟ = 2π ⎜ 9π2 + 4π2 + ∫ 2 x cos x dx ⎟
⎣
⎦
π
2
π
2
2π
⎝
⎠
⎝
⎠
u = 2x
dv = cos x dx
du = 2 dx
v = sin x
3π
⎛
⎞
V = 2π ⎜13π2 + [2 x sin x]32ππ – ∫ 2sin x ⎟
2π
⎝
⎠
(
)
= 2π 13π2 + [2 cos x]32ππ = 2π(13π2 – 4) ≈ 781
dv = sin nx dx
75. u = f(x)
1
v = − cos nx
n
du = f ′( x)dx
π
an =
⎤
1⎡ ⎡ 1
1 π
⎤
cos(nx) f ′( x)dx ⎥
− cos(nx) f ( x) ⎥ + ∫
⎢
⎢
n −π
π⎣ ⎣ n
⎦ −π ⎦
Term 1
Term 1 =
Term 2
1
1
cos(nπ)( f (−π) − f (π)) = ± ( f (−π) − f (π))
n
n
Since f ′( x) is continuous on [– ∞ , ∞ ], it is bounded. Thus,
π
∫– π cos(nx) f ′( x)dx
is bounded so
π
1 ⎡
± ( f (−π) − f (π)) + ∫ cos(nx) f ′( x) dx ⎥⎤ = 0.
⎢
−π
n→∞ πn ⎣
⎦
lim an = lim
n →∞
1/ n
Gn [(n + 1)(n + 2) ⋅⋅⋅ (n + n)]1 n ⎡⎛ 1 ⎞⎛ 2 ⎞ ⎛ n ⎞ ⎤
=
76.
= ⎢⎜1 + ⎟⎜ 1 + ⎟ …⎜1 + ⎟ ⎥
n
[n n ]1 n
⎣⎝ n ⎠⎝ n ⎠ ⎝ n ⎠ ⎦
⎛ G ⎞ 1 ⎡⎛ 1 ⎞⎛ 2 ⎞ ⎛ n ⎞ ⎤
ln ⎜ n ⎟ = ln ⎢⎜1 + ⎟⎜ 1 + ⎟ …⎜ 1 + ⎟ ⎥
⎝ n ⎠ n ⎣⎝ n ⎠⎝ n ⎠ ⎝ n ⎠ ⎦
=
1⎡ ⎛ 1⎞
⎛ 2⎞
⎛ n ⎞⎤
ln ⎜1 + ⎟ + ln ⎜ 1 + ⎟ + ⋅⋅⋅ + ln ⎜1 + ⎟ ⎥
n ⎢⎣ ⎝ n ⎠
n
⎝
⎠
⎝ n ⎠⎦
⎛G
lim ln ⎜ n
n →∞ ⎝ n
⎛G
lim ⎜ n
n →∞ ⎝ n
430
2
⎞
⎟ = ∫1 ln x dx = 2 ln 2 –1
⎠
4
⎞ 2 ln 2–1
= 4e –1 =
⎟=e
e
⎠
Section 7.2
Instructor Solutions Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
77. The proof fails to consider the constants when integrating 1t .
The symbol
∫ (1 t ) dt is a
family of functions, all of who whom have derivative 1t . We know that any two of these
functions will differ by a constant, so it is perfectly correct (notationally) to write
78.
∫ (1 t ) dt = ∫ (1 t ) dt + 1
d 5x
[e (C1 cos 7 x + C2 sin 7 x) + C3 ] = 5e5 x (C1 cos 7 x + C2 sin 7 x) + e5 x (–7C1 sin 7 x + 7C2 cos 7 x)
dx
= e5 x [(5C1 + 7C2 ) cos 7 x + (5C2 – 7C1 ) sin 7 x]
Thus, 5C1 + 7C2 = 4 and 5C2 – 7C1 = 6.
Solving, C1 = –
11
29
; C2 =
37
37
79. u = f(x)
du = f ′( x)dx
b
∫a
dv = dx
v=x
f ( x)dx = [ xf ( x)]a – ∫ xf ′( x)dx
b
b
a
Starting with the same integral,
u = f(x)
dv = dx
du = f ′( x)dx
v=x–a
b
∫a
f ( x) dx = [ ( x – a) f ( x) ]a – ∫ ( x – a) f ′( x)dx
b
b
a
80. u = f ′( x)
du = f ′′( x)dx
dv = dx
v=x–a
b
f (b) – f (a ) = ∫ f ′( x)dx = [ ( x – a) f ′( x) ]a – ∫ ( x – a ) f ′′( x)dx = f ′(b)(b – a ) – ∫ ( x – a ) f ′′( x)dx
b
b
b
a
a
a
Starting with the same integral,
u = f ′( x)
dv = dx
du = f ′′( x)dx
v=x–b
f (b) − f (a) = ∫ f ′( x)dx = [ ( x – b) f ′( x) ]a – ∫ ( x – b) f ′′( x)dx = f ′(a)(b − a) – ∫ ( x – b) f ′′( x)dx
b
b
a
b
b
a
a
81. Use proof by induction.
t
t
a
a
n = 1: f (a) + f ′(a )(t – a) + ∫ (t – x) f ′′( x)dx = f (a ) + f ′(a )(t – a ) + [ f ′( x)(t – x)]ta + ∫ f ′( x)dx
= f (a) + f ′(a)(t – a) – f ′(a)(t – a) + [ f ( x)]ta = f (t )
Thus, the statement is true for n = 1. Note that integration by parts was used with u = (t – x), dv = f ′′( x)dx.
Suppose the statement is true for n.
n
t (t – x ) n ( n +1)
f (i ) ( a )
(t – a)i + ∫
( x)dx
f (t ) = f (a ) + ∑
f
a
i!
n!
i =1
Integrate
(t – x)n ( n +1)
∫a n ! f ( x)dx by parts.
t
u = f ( n +1) ( x)
dv =
(t – x)n
dx
n!
du = f ( n + 2) ( x)
v=–
(t – x)n +1
(n + 1)!
t
⎡ (t – x)n +1 ( n +1) ⎤
t (t – x) n +1 ( n + 2)
(t – x) n ( n +1)
=
+
(
)
–
(
)
( x)dx
f
x
dx
f
x
⎢
⎥
∫a n !
∫a (n + 1)! f
⎣⎢ (n + 1)!
⎦⎥
t
a
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Section 7.2
431
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
=
t (t – x ) n +1 ( n + 2)
(t – a )n +1 ( n +1)
(a) + ∫
( x)dx
f
f
a (n + 1)!
(n + 1)!
t (t – x) n +1 ( n + 2)
f (i ) ( a )
(t – a) n +1 ( n +1)
(t – a)i +
(a) + ∫
( x)dx
f
f
a ( n + 1)!
(n + 1)!
i!
i =1
n +1 (i )
t (t – x ) n +1 ( n + 2)
f (a)
(t – a)i + ∫
( x)dx
= f (a) + ∑
f
a ( n + 1)!
i!
i =1
n
Thus f (t ) = f (a ) + ∑
Thus, the statement is true for n + 1.
82. a.
1
B (α , β ) = ∫ xα −1 (1 − x) β −1 dx where α ≥ 1, β ≥ 1
0
x = 1 – u, dx = –du
1 α −1
∫0 x
(1 − x) β −1 dx = ∫ (1 − u )α −1 (u ) β −1 (− du ) = ∫ (1 − u )α −1 u β −1du = B ( β , α )
0
1
1
0
Thus, B(α, β) = B(β, α).
b.
1
B (α , β ) = ∫ xα −1 (1 − x) β −1 dx
0
α −1
dv = (1 − x) β −1 dx
u=x
du = (α − 1) xα − 2 dx
v=−
1
β
(1 − x) β
1
⎡ 1
⎤ α − 1 1 α −2
α − 1 1 α −2
B (α , β ) = ⎢ − xα −1 (1 − x) β ⎥ +
x
(1 − x) β dx =
x
(1 − x) β dx
∫
∫
0
0
β
β
β
⎣
⎦0
α −1
=
B (α − 1, β + 1)
(*)
β
Similarly,
1
B (α , β ) = ∫ xα −1 (1 − x) β −1 dx
0
u = (1 − x)
β −1
dv = xα −1dx
du = − ( β − 1) (1 − x) β − 2 dx
v=
1 α
x
α
1
β −1 1 α
β −1
β −1 1 α
⎡1
⎤
x (1 − x) β − 2 dx =
B (α + 1, β − 1)
B (α , β ) = ⎢ xα (1 − x) β −1 ⎥ +
x (1 − x) β − 2 dx =
∫
∫
0
0
α
α
α
⎣α
⎦0
c.
Assume that n ≤ m. Using part (b) n times,
( n − 1) (n − 2)
n −1
B (n, m) =
B (n − 1, m + 1) =
B (n − 2, m + 2)
m
m(m + 1)
=…=
( n − 1) (n − 2) ( n − 3)…⋅ 2 ⋅1
m(m + 1) ( m + 2 )… (m + n − 2)
1
B(1, m + n − 1) = ∫ (1 − x)m+ n − 2 dx = −
0
Thus, B (n, m) =
Section 7.2
1
1
[(1 − x)m+ n −1 ]10 =
m + n −1
m + n −1
( n − 1) (n − 2) ( n − 3)…⋅ 2 ⋅1
m(m + 1) ( m + 2 ) … (m + n − 2) ( m + n − 1)
If n > m, then B (n, m) = B (m, n) =
432
B(1, m + n − 1).
( n − 1)!( m − 1)!
(n + m − 1)!
=
( n − 1)!( m − 1)! ( n − 1)!( m − 1)!
(m + n − 1)!
=
(n + m − 1)!
by the above reasoning.
Instructor Solutions Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
dv = f ′′(t )dt
v = f ′(t )
83. u = f(t)
du = f ′(t )dt
b
∫a
f ′′(t ) f (t )dt = [ f (t ) f ′(t ) ]a – ∫ [ f ′(t )]2 dt
b
b
a
b
b
a
a
= f (b) f ′(b) − f (a) f ′(a) − ∫ [ f ′(t )]2 dt = − ∫ [ f ′(t )]2 dt
b
[ f ′(t )]2 ≥ 0, so − ∫ [ f ′(t )]2 ≤ 0 .
a
84.
x⎛ t
⎞
∫0 ⎜⎝ ∫0 f ( z )dz ⎟⎠ dt
t
u = ∫ f ( z )dz dv = dt
0
du = f(t)dt
v=t
x⎛ t
⎞
⎡
t
⎤
x
x
x
x
∫0 ⎜⎝ ∫0 f ( z )dz ⎟⎠ dt = ⎣⎢t ∫0 f ( z )dz ⎦⎥0 – ∫0 t f (t )dt = ∫0 x f ( z)dz – ∫0 t f (t )dt
By letting z = t,
x⎛ t
x
x
∫0 x f ( z )dz = ∫0 x f (t )dt ,
⎞
x
so
x
x
∫0 ⎜⎝ ∫0 f ( z )dz ⎟⎠ dt = ∫0 x f (t )dt – ∫0 t f (t )dt = ∫0 ( x – t ) f (t )dt
x t1
t
⋅⋅⋅ n −1
0 0
0
85. Let I = ∫
∫
∫
f (tn ) dtn ...dt2 dt1 be the iterated integral. Note that for i ≥ 2, the limits of integration of the
integral with respect to ti are 0 to ti −1 so that any change of variables in an outer integral affects the limits, and
hence the variables in all interior integrals. We use induction on n, noting that the case n = 2 is solved in the
previous problem.
Assume we know the formula for n − 1 , and we want to show it for n.
x t1 t2
t
⋅⋅⋅ n −1
0 0 0
0
I =∫
∫ ∫
∫
where F ( tn −1 ) = ∫
tn −1
0
t
t
0
0
tn − 2
0
f (tn ) dtn ...dt3 dt2 dt1 = ∫ 1 ∫ 2 ⋅⋅⋅∫
F (tn −1 ) dtn −1...dt3 dt2 dt1
f ( tn ) dn .
By induction,
x
1
n−2
I=
F ( t1 )( x − t1 ) dt1
∫
0
−
n
2
!
(
)
u = F ( t1 ) = ∫ f ( tn ) dtn ,
t1
0
du = f ( t1 ) dt1 ,
I=
dv = ( x − t1 )
v=−
n−2
1
( x − t1 )n −1
n −1
t1 = x
t
1 ⎧⎪ ⎡ 1
1 x
⎪⎫
n −1
f ( t1 )( x − t1 ) dt1 ⎬ .
( x − t1 )n −1 ∫01 f ( tn ) dtn ⎤⎥ +
⎨⎢−
∫
0
( n − 2 )! ⎩⎪ ⎣ n − 1
⎦ t1 = 0 n − 1
⎭⎪
x
1
f (t1 )( x − t1 ) n −1 dt1
∫
0
(n − 1)!
(note: that the quantity in square brackets equals 0 when evaluated at the given limits)
=
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Section 7.3
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
86. Proof by induction.
n = 1:
u = P1 ( x)
du =
∫e
x
dv = e x dx
dP1 ( x)
dx
dx
v = ex
P1 ( x)dx = e x P1 ( x) – ∫ e x
Note that
dP1 ( x)
dP ( x )
dP ( x)
dx = e x P1 ( x) – 1 ∫ e x dx = e x P1 ( x) – e x 1
dx
dx
dx
dP1 ( x)
is a constant.
dx
Suppose the formula is true for n. By using integration by parts with u = Pn +1 ( x) and dv = e x dx,
∫e
x
Pn +1 ( x)dx = e x Pn +1 ( x) – ∫ e x
Note that
dPn +1 ( x)
dx
dx
dPn +1 ( x)
is a polynomial of degree n, so
dx
j +1
j
n
n
⎡
⎤
Pn +1 ( x )
j d
x
x
x
j d ⎛ dPn +1 ( x ) ⎞ ⎥
x
x
⎢
(
)
(
)
(
1)
1
=
−
−
=
−
−
e
P
x
dx
e
P
x
e
e
P
x
e
(
)
(
)
∑
∑
n +1
n +1
⎜ dx ⎟
∫ n+1
j
⎢⎣ j = 0
⎠ ⎥⎦
dx ⎝
dx j +1
j =0
n +1
d j Pn +1 ( x )
j =1
dx j
= e x Pn +1 ( x) + e x ∑ (−1) j
87.
n +1
d j Pn +1 ( x)
j =0
dx j
= e x ∑ (−1) j
4
d j (3 x 4 + 2 x 2 )
j =0
dx j
x
j
4
2 x
∫ (3x + 2 x )e dx = e ∑ (–1)
= e x [3x 4 + 2 x 2 – 12 x3 – 4 x + 36 x 2 + 4 – 72 x + 72]
= e x (3x 4 – 12 x3 + 38 x 2 – 76 x + 76)
7.3 Concepts Review
1.
∫
1 + cos 2 x
dx
2
2.
∫ (1 – sin
3.
∫ sin
2
2
2
x) cos x dx
x(1 – sin 2 x) cos x dx
1
[cos(m + n) x + cos(m − n) x ]
2
Problem Set 7.3
∫ sin
2
x dx = ∫
1 ⎛ 1 – cos 2u ⎞
⎜
⎟ du
6 ∫⎝
2
⎠
1
=
(1 – 2 cos 2u + cos 2 2u )du
24 ∫
1
1
1
=
du –
2 cos 2u du + ∫ (1 + cos 4u )du
∫
∫
24
24
48
3
1
1
=
du –
2 cos 2u du +
4 cos 4u du
∫
∫
48
24
192 ∫
3
1
1
sin 24 x + C
= (6 x) – sin12 x +
48
24
192
3
1
1
= x – sin12 x +
sin 24 x + C
8
24
192
=
4. cos mx cos nx =
1.
2. u = 6x, du = 6 dx
1
4
4
∫ sin 6 x dx = 6 ∫ sin u du
1 – cos 2 x
dx
2
1
1
dx – ∫ cos 2 x dx
∫
2
2
1
1
= x – sin 2 x + C
2
4
=
434
Section 7.2
Instructor Solutions Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
3.
∫ sin x dx = ∫ sin x(1 − cos x)dx
= ∫ sin x dx − ∫ sin x cos 2 x dx
3
2
5.
cos5 θ dθ = ∫
π/2
0
=∫
π/ 2
0
1
= − cos x + cos3 x + C
3
4.
π/2
∫0
(1 – sin 2 θ )2 cosθ dθ
(1 – 2sin 2 θ + sin 4 θ ) cos θ dθ
π/2
2
1
⎡
⎤
= ⎢sin θ – sin 3 θ + sin 5 θ ⎥
3
5
⎣
⎦0
2
1
8
⎛
⎞
= ⎜1 – + ⎟ – 0 =
15
⎝ 3 5⎠
∫ cos x dx =
= ∫ cos x (1 − sin 2 x)dx
= ∫ cos x dx − ∫ cos x sin 2 x dx
3
1
= sin x − sin 3 x + C
3
6.
π/2
∫0
sin 6 θ dθ = ∫
π / 2 ⎛ 1 – cos 2θ
0
⎜
⎝
2
3
⎞
⎟ dθ
⎠
1 π/ 2
(1 – 3cos 2θ + 3cos 2 2θ – cos3 2θ )dθ
∫
0
8
1 π/ 2
3 π/2
3 π/ 2 2
1 π/2 3
= ∫
dθ – ∫
2 cos 2θ dθ + ∫
cos 2θ – ∫
cos 2θ dθ
0
0
0
8
16
8
8 0
1
3
3 π / 2 ⎛ 1 + cos 4θ ⎞
1 π/2
2
= [θ ]0π / 2 – [sin 2θ ]0π / 2 + ∫
⎜
⎟ dθ – ∫0 (1 – sin 2θ ) cos 2θ dθ
0
8
16
8
2
8
⎝
⎠
1 π 3 π/ 2
3 π/2
1 π/2
1 π/2 2
dθ + ∫
4 cos 4θ dθ – ∫
2 cos 2θ dθ + ∫
sin 2θ ⋅ 2 cos 2θ dθ
= ⋅ + ∫
0
0
0
8 2 16
64
16
16 0
π 3π 3
1
1
5π
= + + [sin 4θ ]0π / 2 – [sin 2θ ]0π / 2 + [sin 3 2θ ]0π / 2 =
16 32 64
16
48
32
=
7.
∫ sin
=–
8.
5
4 x cos 2 4 x dx = ∫ (1 – cos 2 4 x) 2 cos 2 4 x sin 4 x dx = ∫ (1 – 2 cos 2 4 x + cos 4 4 x) cos 2 4 x sin 4 x dx
1
1
1
1
(cos 2 4 x – 2 cos 4 4 x + cos6 4 x)(–4sin 4 x)dx = – cos3 4 x + cos5 4 x – cos7 4 x + C
12
10
28
4∫
∫ (sin
3
2t ) cos 2tdt = ∫ (1 – cos 2 2t )(cos 2t )1/ 2 sin 2t dt = –
1
[(cos 2t )1/ 2 – (cos 2t )5 / 2 ](–2sin 2t )dt
2∫
1
1
= – (cos 2t )3 / 2 + (cos 2t )7 / 2 + C
3
7
9.
∫ cos
3
3θ sin –2 3θ dθ = ∫ (1 – sin 2 3θ ) sin –2 3θ cos 3θ dθ =
1
(sin −2 3θ − 1)3cos 3θ dθ
3∫
1
1
= – csc3θ – sin 3θ + C
3
3
10.
∫ sin
1/ 2
=
2 z cos3 2 z dz = ∫ (1 – sin 2 2 z ) sin1/ 2 2 z cos 2 z dz
1
1
1
(sin1/ 2 2 z – sin 5 / 2 2 z )2 cos 2 z dz = sin 3 / 2 2 z – sin 7 / 2 2 z + C
∫
3
7
2
Instructor’s Resource Manual
Section 7.3
435
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2
11.
2
1
⎛ 1 – cos 6t ⎞ ⎛ 1 + cos 6t ⎞
4
4
2
4
∫ sin 3t cos 3t dt = ∫ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ dt = 16 ∫ (1 – 2 cos 6t + cos 6t )dt
1 ⎡
1
1
1
⎤
= ∫ ⎢1 – (1 + cos12t ) + (1 + cos12t )2 ⎥ dt = – ∫ cos12t dt + ∫ (1 + 2 cos12t + cos 2 12t )dt
16 ⎣
4
16
64
⎦
1
1
1
1
=–
12 cos12t dt + ∫ dt +
12 cos12t dt +
(1 + cos 24t )dt
192 ∫
64
384 ∫
128 ∫
1
1
1
1
1
3
1
1
sin12t + t +
sin12t +
sin 24t + C =
sin12t +
sin 24t + C
=–
t+
t–
192
64 384
128 3072
128 384
3072
3
12.
13.
1
⎛ 1 + cos 2θ ⎞ ⎛ 1 – cos 2θ ⎞
6
2
3
4
∫ cos θ sin θ dθ = ∫ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ dθ = 16 ∫ (1 + 2 cos 2θ – 2 cos 2θ – cos 2θ )dθ
1
1
1
1
= ∫ dθ + ∫ 2 cos 2θ dθ – ∫ (1 – sin 2 2θ ) cos 2θ dθ – ∫ (1 + cos 4θ )2 dθ
16
16
8
64
1
1
1
1
1
= ∫ dθ + ∫ 2 cos 2θ dθ – ∫ 2 cos 2θ dθ + ∫ 2sin 2 2θ cos 2θ dθ – ∫ (1 + 2 cos 4θ + cos 2 4θ )dθ
16
16
16
16
64
1
1
1
1
1
4 cos 4θ dθ –
(1 + cos8θ )dθ
= ∫ dθ + ∫ sin 2 2θ ⋅ 2 cos 2θ dθ – ∫ dθ –
16
16
64
128 ∫
128 ∫
1
1
1
1
1
1
θ−
sin 4θ −
sin 8θ + C
= θ + sin 3 2θ − θ −
16
48
64
128
128
1024
5
1
1
1
=
θ + sin 3 2θ –
sin 4θ –
sin 8θ + C
128
48
128
1024
1
=
14.
1
∫ sin 4 y cos 5 y dy = 2 ∫ [sin 9 y + sin(− y)] dy = 2 ∫ (sin 9 y − sin y)dy
1⎛ 1
1
1
⎞
⎜ − cos 9 y + cos y ⎟ + C = cos y − cos 9 y + C
2⎝ 9
2
18
⎠
1
∫ cos y cos 4 y dy = 2 ∫ [cos 5 y + cos(−3 y)]dy
=
1
1
1
1
sin 5 y − sin(−3 y) + C = sin 5 y + sin 3 y + C
10
6
10
6
2
15.
16.
1
⎛ 1 – cos w ⎞ ⎛ 1 + cos w ⎞
4 ⎛ w⎞
2⎛ w⎞
2
3
∫ sin ⎜⎝ 2 ⎟⎠ cos ⎜⎝ 2 ⎟⎠ dw = ∫ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ dw = 8 ∫ (1 – cos w – cos w + cos w)dw
1 ⎡
1
1 ⎡1 1
⎤
⎤
= ∫ ⎢1 – cos w – (1 + cos 2w) + (1 – sin 2 w) cos w⎥ dw = ∫ ⎢ – cos 2 w – sin 2 w cos w⎥dw
8 ⎣
2
8
2
2
⎦
⎣
⎦
1
1
1
= w – sin 2 w – sin 3 w + C
16
32
24
1
∫ sin 3t sin t dt = ∫ − 2 [cos 4t − cos 2t ] dt
(
)
1
cos 4tdt − ∫ cos 2tdt
2 ∫
1⎛1
1
⎞
= − ⎜ sin 4t − sin 2t ⎟ + C
2⎝4
2
⎠
1
1
= − sin 4t + sin 2t + C
8
4
=−
436
Section 7.3
Instructor's Resource Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
17.
∫ x cos
2
u=x
x sin x dx
19.
du = 1 dx
∫ tan
4
( )( )
= ∫ ( tan 2 x ) (sec2 x − 1) dx
= ∫ ( tan 2 x sec2 x − tan 2 x ) dx
x dx = ∫ tan 2 x tan 2 x dx
dv = cos x sin x dx
2
1
v = − ∫ (cos x) 2 (− sin x) dx = − cos3 x
3
t =cos x
Thus
∫ x cos
2
⎤
1⎡
3
2
⎢ − x cos x + ∫ (cos x − cos x sin x) dx ⎥ =
3⎣
t =sin x
⎦
1⎡
1
⎤
− x cos3 x + sin x − sin 3 x ⎥ + C
3 ⎢⎣
3
⎦
∫ x sin x cos x dx
3
u=x
1
= tan 3 x − tan x + x + C
3
x sin x dx =
1
1
x(− cos3 x) − ∫ (1)(− cos3 x) dx =
3
3
1⎡
3
3
− x cos x + ∫ cos x dx ⎤ =
⎦
3⎣
1⎡
− x cos3 x + ∫ cos x (1 − sin 2 x) dx ⎤ =
⎦
3⎣
18.
= ∫ tan 2 x sec2 x dx − ∫ (sec2 x − 1)dx
20.
∫ cot
4
t =sin x
= ∫ cot 2 x csc2 x dx − ∫ (csc2 x − 1)dx
1
= − cot 3 x + cot x + x + C
3
( )
= ∫ ( tan x ) ( sec2 x − 1) dx
21. tan 3 x = ∫ ( tan x ) tan 2 x dx
=
Thus
∫ x sin
3
x cos x dx =
1
1
x( sin 4 x ) − ∫ (1)( sin 4 x) dx =
4
4
1⎡
4
2 2
x sin x − ∫ (sin x) dx ⎤ =
⎦
4⎣
1⎡
1
⎤
x sin 4 x − ∫ (1 − cos 2 x) 2 dx ⎥ =
4 ⎢⎣
4
⎦
)
( )
= ∫ ( cot 2 x csc2 x − cot 2 x ) dx
dv = sin 3 x cos x dx
1 4
sin x
4
)(
= ∫ cot 2 x (csc2 x − 1) dx
du = 1 dx
v = ∫ (sin x)3 (cos x) dx =
(
x dx = ∫ cot 2 x cot 2 x dx
22.
1
tan 2 x + ln cos x + C
2
(
= ∫ ( cot 2t ) ( csc2 2t − 1)dt
∫ cot
3
)
2t dt = ∫ ( cot 2t ) cot 2 2t dt
= ∫ cot 2t csc2 2t dt − ∫ cot 2t dt
1
1
= − cot 2 2t − ln sin 2t + C
4
2
1⎡
1
⎤
x sin 4 x − ∫ (1 − 2 cos 2 x + cos 2 2 x) dx ⎥ =
⎢
4⎣
4
⎦
1⎡
1
1
1
⎤
x sin 4 x − x + sin 2 x − ∫ (1 + cos 4 x) dx ⎥ =
4 ⎢⎣
4
4
8
⎦
1⎡
3
1
1
⎤
x sin 4 x − x + sin 2 x − sin 4 x ⎥ + C
⎢
4⎣
8
4
32
⎦
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23.
5 ⎛θ
⎞
⎜ ⎟ dθ
⎝2⎠
dθ
⎛θ ⎞
u = ⎜ ⎟ ; du =
2
2
⎝ ⎠
∫ tan
∫ tan
5 ⎛θ
⎞
5
⎜ ⎟ dθ = 2∫ tan u du
⎝2⎠
(
)(
)
= 2∫ tan 3 u sec2 u − 1 du
= 2∫ tan 3 u sec2 u du − 2∫ tan 3 u du
(
)
= 2∫ tan 3 u sec2 u du − 2∫ tan u sec2 u − 1 du
= 2∫ tan 3 u sec2 u du − 2∫ tan u sec2 u du + 2∫ tan u du
=
24.
∫ cot
5
θ
1
⎛θ ⎞
⎛θ ⎞
tan 4 ⎜ ⎟ − tan 2 ⎜ ⎟ − 2 ln cos + C
2
2
⎝2⎠
⎝2⎠
2t dt
u = 2t ; du = 2dt
1
5
5
∫ cot 2t dt = 2 ∫ cot u du
1
1
= ∫ cot 3 u cot 2 u du = ∫ cot 3 u csc2 − 1 du
2
2
1
1
= ∫ cot 3 u csc 2 u du − ∫ cot 3 u du
2
2
1
1
3
2
= ∫ cot u csc u du − ∫ ( cot u ) csc2 u − 1 du
2
2
1
1
1
3
2
= ∫ cot u csc u du − ∫ ( cot u ) csc2 u du + ∫ cot u
2
2
2
1 4
1 2
1
= − cot u + cot u + ln sin u + C
8
4
2
1 4
1 2
1
= − cot 2t + cot 2t + ln sin 2t + C
8
4
2
(
(
(
(
25.
∫ tan
−3
)(
)(
)(
)(
)
)
)
)
(
)(
)
(
(
)
)
(
)( )( )
= ∫ ( tan −3 x )(1 + tan 2 x )( sec2 x ) dx
x sec4 xdx = ∫ tan −3 x sec2 x sec2 x dx
= ∫ tan −3 x sec2 x dx + ∫ ( tan x )
−1
sec2 x dx
1
= − tan −2 x + ln tan x + C
2
26.
∫ tan
−3 / 2
(
)( )( )
= ∫ ( tan −3 / 2 x )(1 + tan 2 x )( sec2 x ) dx
x sec4 x dx = ∫ tan −3 / 2 x sec2 x sec2 x
= ∫ tan −3 / 2 x sec2 x dx + ∫ tan1/ 2 x sec 2 x dx
2
= −2 tan −1/ 2 x + tan 3 / 2 x + C
3
438
Section 7.3
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27.
∫ tan
3
x sec2 x dx
Let u = tan x . Then du = sec 2 x dx .
28.
1 4
1
u + C = tan 4 x + C
4
4
∫ tan
3
x sec2 x dx = ∫ u 3 du =
∫ tan
3
x sec−1/ 2 x dx = ∫ tan 2 x sec−3 / 2 x(sec x tan x)dx
(
)(
)
= ∫ sec 2 x − 1 sec−3 / 2 x ( sec x tan x ) dx
= ∫ sec
1/ 2
=
29.
x ( sec x tan x ) dx − ∫ sec−3 / 2 x ( sec x tan x ) dx
2 3/ 2
sec
x + 2sec−1/ 2 x + C
3
π
1 π
1⎡ 1
1
⎤
(cos[(m + n) x] + cos[(m − n) x])dx = ⎢
sin[(m + n) x] +
sin[(m − n) x]⎥
∫– π
∫
π
–
2
2 ⎣m + n
m−n
⎦ −π
= 0 for m ≠ n, since sin k π = 0 for all integers k.
π
cos mx cos nx dx =
If we let u =
πx
then du =
π
30.
dx . Making the substitution and changing the limits as necessary, we get
L
L
L
mπ x
nπ x
L π
∫− L cos L cos L dx = π ∫−π cos mu cos nu du = 0 (See Problem 29
31.
∫0 π( x + sin x)
π
2
π
π
π
0
0
dx = π∫ ( x 2 + 2 x sin x + sin 2 x) dx = π∫ x 2 dx + 2π∫ x sin x dx +
0
π
π π
(1 − cos 2 x)dx
2 ∫0
π
1
1
π
1
5
⎡1 ⎤
⎤
π π⎡
= π ⎢ x3 ⎥ + 2π [sin x − x cos x ]0 + ⎢ x − sin 2 x ⎥ = π4 + 2π(0 + π − 0) + (π − 0 − 0) = π4 + π2 ≈ 57.1437
2⎣
2
2
3
2
⎣ 3 ⎦0
⎦0 3
Use Formula 40 with u = x for
π/2
32. V = 2π∫
0
∫ x sin x dx
x sin 2 ( x 2 )dx
u = x 2 , du = 2x dx
V = π∫
π/2
0
33. a.
sin 2 u du = π∫
π / 2 1 – cos 2u
0
2
π/ 2
1
⎡1
⎤
du = π ⎢ u – sin 2u ⎥
4
⎣2
⎦0
=
π2
≈ 2.4674
4
⎞
π
1 π
1 N
1 π⎛ N
f ( x) sin(mx)dx = ∫ ⎜ ∑ an sin(nx) ⎟ sin(mx)dx = ∑ an ∫ sin(nx) sin(mx) dx
∫
⎟
−π
π
π −π
π −π ⎜⎝ n =1
n =1
⎠
From Example 6,
⎧0 if n ≠ m
π
∫−π sin(nx) sin(mx)dx = ⎨⎩π if n = m
so every term in the sum is 0 except for when n = m.
If m > N, there is no term where n = m, while if m ≤ N, then n = m occurs. When n = m
π
an ∫ sin(nx) sin(mx) dx = am π so when m ≤ N,
−π
1 π
1
f ( x) sin(mx) dx = ⋅ am ⋅ π = am .
π ∫−π
π
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Section 7.3
439
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N
⎞⎛ N
⎞
π
1 π 2
1 N
1 π⎛ N
sin(
)
sin(
)
(
)
f
x
dx
a
nx
a
mx
dx
=
a
=
⎜
⎟
⎜
⎟
∑
∑
∑
n ∑ am ∫−π sin(nx) sin( mx ) dx
n
m
∫
∫
⎜
⎟
⎜
⎟
−π
−π
π n =1 m =1
π
π
⎝ n =1
⎠ ⎝ m =1
⎠
b.
From Example 6, the integral is 0 except when m = n. When m = n, we obtain
N
1 N
an (an π) = ∑ an2 .
∑
π n =1
n =1
Proof by induction
x
x
n = 1: cos = cos
2
2
Assume true for k ≤ n.
34. a.
⎡
x
x
x
x
1
3
2n –1 ⎤ 1
x
= ⎢ cos
cos cos " cos ⋅ cos
x + cos
x + " + cos
x⎥
cos
+
1
–1
n
n
n
n
n
n
n
2
4
2
2
2
2
2
2 +1
⎢⎣
⎥⎦ 2
Note that
k ⎞⎛
1 ⎞ 1⎡
2k + 1
2k –1 ⎤
⎛
⎜ cos n x ⎟⎜ cos n +1 x ⎟ = ⎢cos n +1 x + cos n +1 x ⎥ , so
2 ⎠⎝
2
2
2
⎝
⎠ 2⎣
⎦
⎡
1
3
2n –1 ⎤ ⎛
1
⎢ cos n x + cos n x + " + cos n x ⎥ ⎜ cos n +1
2
2
2
2
⎣⎢
⎦⎥ ⎝
⎡
1
3
2n +1 –1 ⎤ 1
⎞ 1
x⎟
x + cos
x + " + cos
x⎥
= ⎢cos
⎠ 2n –1 ⎢⎣
2n +1
2n +1
2n +1 ⎦⎥ 2n
⎡
⎡
1
3
2n –1 ⎤ 1
1
1
3
2n –1 ⎤ x
lim ⎢ cos
x + cos
x + " + cos
x⎥
x + cos
x + " + cos
x⎥
= lim ⎢cos
n →∞ ⎢
2n
2n
2n ⎦⎥ 2n –1 x n→∞ ⎣⎢
2n
2n
2n ⎦⎥ 2n –1
⎣
b.
=
1 x
cos t dt
x ∫0
1 x
1
sin x
cos t dt = [sin t ]0x =
∫
0
x
x
x
c.
35. Using the half-angle identity cos
x
1 + cos x
=
, we see that since
2
2
π
cos
π
2
= cos 2 =
4
2
2
cos
π
1 + 22
π
= cos 2 =
=
8
4
2
cos
π
1 + 2+2 2
π
= cos 2 =
=
16
8
2
Thus,
2+ 2
,
2
⎛π⎞
⎛π⎞
⎛π⎞
2 2+ 2 2+ 2+ 2
" = cos ⎜ 2 ⎟ cos ⎜ 2 ⎟ cos ⎜ 2 ⎟"
⋅
⋅
⎜2⎟
⎜4⎟
⎜8⎟
2
2
2
⎝ ⎠
⎝ ⎠
⎝ ⎠
⎛π⎞
⎛π⎞
⎛ π
= lim cos ⎜ 2 ⎟ cos ⎜ 2 ⎟" cos ⎜ 2
⎜2⎟
⎜4⎟
⎜ 2n
n→∞
⎝ ⎠
⎝ ⎠
⎝
440
2+ 2+ 2
, etc.
2
Section 7.3
( )
⎞ sin π2
2
⎟=
=
π
⎟
π
2
⎠
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
36. Since (k − sin x)2 = (sin x − k ) 2 , the volume of S is
π
π
0
0
= πk 2 ∫ dx − 2k π ∫ sin x dx +
π
∫0 π(k − sin x)
π
= π ∫ (k 2 − 2k sin x + sin 2 x) dx
2
0
π
π π
1
⎤
π
π π⎡
(1 − cos 2 x) dx = πk 2 [ x ]0 + 2k π [ cos x ]0 + ⎢ x − sin 2 x ⎥
2⎣
2
2 ∫0
⎦0
π
π2
(π − 0) = π2 k 2 − 4k π +
2
2
2
2
π
, then f ′(k ) = 2π2 k − 4π and f ′(k ) = 0 when k = .
Let f (k ) = π2 k 2 − 4k π +
π
2
2
The critical points of f(k) on 0 ≤ k ≤ 1 are 0, , 1.
π
= π2 k 2 + 2k π(−1 − 1) +
f (0) =
π2
≈ 4.93,
2
π
π
⎛2⎞
f ⎜ ⎟ = 4−8+
≈ 0.93, f (1) = π2 − 4π +
≈ 2.24
2
2
⎝π⎠
2
S has minimum volume when k =
a.
2
2
.
π
b. S has maximum volume when k = 0.
7.4 Concepts Review
1.
4. u = x + 4, u 2 = x + 4, 2u du = dx
x 2 + 3x
(u 2 – 4)2 + 3(u 2 – 4)
2u du
∫ x+4
u
2
10
= 2 ∫ (u 4 – 5u 2 + 4)du = u 5 – u 3 + 8u + C
5
3
2
10
= ( x + 4)5 / 2 – ( x + 4)3 / 2 + 8( x + 4)1/ 2 + C
5
3
x–3
2. 2 sin t
3. 2 tan t
4. 2 sec t
Problem Set 7.4
1. u = x + 1, u 2 = x + 1, 2u du = dx
∫x
x + 1dx = ∫ (u 2 – 1)u (2u du )
2
2
= ∫ (2u 4 – 2u 2 )du = u 5 – u 3 + C
5
3
2
5/ 2 2
3/ 2
= ( x + 1)
+C
– ( x + 1)
5
3
5. u = t , u 2 = t , 2u du = dt
2 dt
2 2u du
∫1 t + e = ∫1 u + e = 2∫1
2
2 e
= 2 ∫ du – 2∫
du
1
1 u+e
3
=
3 7 3π 4
u – u +C
7
4
3
3π
( x + π) 7 / 3 – ( x + π ) 4 / 3 + C
7
4
3. u = 3t + 4, u 2 = 3t + 4, 2u du = 3 dt
1 (u 2
3
− 4) 23 u du
t dt
2
=∫
= ∫ (u 2 – 4)du
u
9
3t + 4
2 3 8
=
u – u+C
27
9
2
8
=
(3t + 4)3 / 2 – (3t + 4)1/ 2 + C
27
9
∫
Instructor’s Resource Manual
u+e−e
du
u+e
1
= 2( 2 – 1) – 2e[ln( 2 + e) – ln(1 + e)]
⎛ 2 +e⎞
= 2 2 – 2 – 2e ln ⎜⎜
⎟⎟
⎝ 1+ e ⎠
x + πdx = ∫ (u 3 – π)u (3u 2 du )
= ∫ (3u 6 – 3πu 3 )du =
2
2
= 2[u ]1 2 – 2e ⎡⎣ln u + e ⎤⎦
2. u = 3 x + π , u 3 = x + π, 3u 2 du = dx
∫x
dx = ∫
u = t , u 2 = t , 2u du = dt
6.
1
t
= 2∫
1
1
u
∫0 t + 1 dt = ∫0 u 2 + 1 (2u du )
u2
0 u2
+1
du = 2∫
1
1 u2
0
= 2 ∫ du – 2∫
1
1
+1−1
u +1
2
du
du = 2[u ]10 – 2[tan –1 u ]10
+1
π
= 2 – 2 tan 1 = 2 – ≈ 0.4292
2
0 u2
0
–1
Section 7.4
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7. u = (3t + 2)1/ 2 , u 2 = 3t + 2, 2u du = 3dt
1
⎛2
⎞
dt = ∫ (u 2 – 2)u 3 ⎜ u du ⎟
3
⎝3
⎠
2
2
4
= ∫ (u 6 – 2u 4 )du = u 7 − u 5 + C
9
63
45
2
4
= (3t + 2)7 / 2 – (3t + 2)5 / 2 + C
63
45
∫ t (3t + 2)
3/ 2
8. u = (1 – x)1/ 3 , u 3 = 1 – x, 3u 2 du = – dx
∫ x(1 – x)
2/3
dx = ∫ (1 – u )u (–3u )du
3
2
2
3
3
= 3∫ (u 7 – u 4 )du = u8 − u 5 + C
8
5
3
3
= (1 – x)8 / 3 – (1 – x)5 / 3 + C
8
5
9. x = 2 sin t, dx = 2 cos t dt
∫
4 – x2
2 cos t
dx = ∫
(2 cos t dt )
x
2sin t
1 – sin 2 t
dt = 2 ∫ csc t dt – 2 ∫ sin t dt
= 2∫
sin t
= 2 ln csc t − cot t + 2 cos t + C
= 2 ln
2 − 4 – x2
+ 4 – x2 + C
x
10. x = 4sin t , dx = 4 cos t dt
∫
x 2 dx
16 – x 2
= 16 ∫
sin 2 t cos t
dt
cos t
= 16 ∫ sin 2 t dt = 8∫ (1 – cos 2t )dt
= 8t – 4sin 2t + C = 8t − 8sin t cos t + C
2
⎛ x ⎞ x 16 – x
= 8sin –1 ⎜ ⎟ –
+C
2
⎝4⎠
12. t = sec x, dt = sec x tan x dx
π
Note that 0 ≤ x < .
2
t 2 – 1 = tan x = tan x
dt
3
∫2
t
=∫
2
2
t –1
sec –1 (3)
π/3
=∫
sec –1 (3)
π/3
sec x tan x
sec 2 x tan x
dx
cos x dx
–1
= [sin x]sec
π/3
(3)
= sin[sec−1 (3)] − sin
π
3
⎡
3 2 2
3
⎛ 1 ⎞⎤
= sin ⎢ cos −1 ⎜ ⎟ ⎥ −
=
–
≈ 0.0768
3
2
⎝ 3 ⎠⎦ 2
⎣
13. t = sec x, dt = sec x tan x dx
π
Note that < x ≤ π.
2
t 2 – 1 = tan x = – tan x
–3
∫–2
=∫
t2 –1
t
3
sec –1 (–3)
2π / 3
sec –1 (–3)
– tan x
2π / 3
sec3 x
dt = ∫
– sin 2 x dx = ∫
sec x tan x dx
sec –1 (–3) ⎛ 1
2π / 3
1⎞
⎜ cos 2 x – ⎟ dx
2⎠
⎝2
sec –1 (–3)
1 ⎤
⎡1
= ⎢ sin 2 x – x ⎥
2 ⎦ 2π / 3
⎣4
sec –1 (–3)
1 ⎤
⎡1
= ⎢ sin x cos x – x ⎥
2 ⎦ 2π / 3
⎣2
=–
2 1
3 π
– sec –1 (–3) +
+ ≈ 0.151252
9 2
8 3
14. t = sin x, dt = cos x dx
t
dt = ∫ sin x dx = –cos x + C
∫
1– t2
= – 1– t2 + C
11. x = 2 tan t , dx = 2sec t dt
2
dx
2sec 2 t dt
1
∫ ( x 2 + 4)3 / 2 = ∫ (4sec2 t )3 / 2 = 4 ∫ cos t dt
=
442
1
x
sin t + C =
+C
4
4 x2 + 4
Section 7.4
15. z = sin t, dz = cos t dt
2z – 3
dz = ∫ (2sin t – 3)dt
∫
1 – z2
= –2 cos t – 3t + C
= –2 1 – z 2 – 3sin –1 z + C
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16. x = π tan t, dx = π sec 2 t dt
πx – 1
∫
x +π
2
=π
2
2
dx = ∫ (π2 tan t – 1) sec t dt
∫ tan t sec t dt – ∫ sec t dt
= π2 sec t – ln sec t + tan t + C
= π x 2 + π2 − ln
πx − 1
π
∫0
x 2 + π2
1 2
x
x + π2 + + C
π
π
= 3 x 2 + 2 x + 5 – 3ln
⎤
x 2 + π2 x ⎥
+
π
π⎥
⎦0
= ( 2 – 1)π2 – ln( 2 + 1) ≈ 3.207
17. x 2 + 2 x + 5 = x 2 + 2 x + 1 + 4 = ( x + 1)2 + 4
u = x + 1, du = dx
dx
du
=∫
∫ 2
x + 2x + 5
u2 + 4
u = 2 tan t, du = 2sec 2 t dt
u2 + 4
= ∫ sec t dt = ln sec t + tan t + C1
= ln
u2 + 4 u
+ + C1
2
2
= ln
x2 + 2 x + 5 + x + 1
+ C1
2
= ln
x2 + 2 x + 5 + x + 1 + C
18. x 2 + 4 x + 5 = x 2 + 4 x + 4 + 1 = ( x + 2)2 + 1
u = x + 2, du = dx
dx
du
=∫
∫ 2
x + 4x + 5
u2 + 1
u = tan t , du = sec2 t dt
∫
∫
du
u2 + 1
dx
= ∫ sec t dt = ln sec t + tan t + C
x + 4x + 5
= ln
2
= ln u 2 + 1 + u + C
x2 + 4 x + 5 + x + 2 + C
x2 + 2 x + 5 + x + 1 + C
π
= [ 2π2 – ln( 2 + 1)] – [π2 − ln1]
du
= 3 u 2 + 4 – 3ln u 2 + 4 + u + C
dx
⎡
= ⎢ π x 2 + π2 – ln
⎢
⎣
∫
19. x 2 + 2 x + 5 = x 2 + 2 x + 1 + 4 = ( x + 1)2 + 4
u = x + 1, du = dx
3x
3u – 3
dx = ∫
du
∫ 2
x + 2x + 5
u2 + 4
u
du
du – 3∫
= 3∫
2
u +4
u2 + 4
(Use the result of Problem 17.)
20. x 2 + 4 x + 5 = x 2 + 4 x + 4 + 1 = ( x + 2)2 + 1
u = x + 2, du = dx
2x – 1
2u − 5
dx = ∫
du
∫ 2
x + 4x + 5
u2 + 1
2u du
du
=∫
– 5∫
u2 + 1
u2 + 1
(Use the result of Problem 18.)
= 2 u 2 + 1 – 5ln u 2 + 1 + u + C
= 2 x 2 + 4 x + 5 – 5ln
x2 + 4 x + 5 + x + 2 + C
21. 5 − 4 x − x 2 = 9 − (4 + 4 x + x 2 ) = 9 − ( x + 2)2
u = x + 2, du = dx
∫
5 – 4 x – x 2 dx = ∫ 9 – u 2 du
u = 3 sin t, du = 3 cos t dt
∫
9 − u 2 du =9 ∫ cos 2 t dt =
9
(1 + cos 2t )dt
2∫
9⎛ 1
9
⎞
⎜ t + sin 2t ⎟ + C = (t + sin t cos t ) + C
2⎝ 2
2
⎠
9
⎛u⎞ 1
= sin –1 ⎜ ⎟ + u 9 – u 2 + C
2
⎝3⎠ 2
9
⎛ x+2⎞ x+2
= sin –1 ⎜
5 – 4 x – x2 + C
⎟+
2
2
⎝ 3 ⎠
=
22. 16 + 6 x – x 2 = 25 − (9 − 6 x + x 2 ) = 25 – ( x – 3) 2
u = x – 3, du = dx
dx
du
=∫
∫
16 + 6 x – x 2
25 – u 2
u = 5 sin t, du = 5 cos t
du
⎛u⎞
= dt = t + C = sin –1 ⎜ ⎟ + C
∫
2 ∫
⎝5⎠
25 − u
⎛ x –3⎞
= sin –1 ⎜
⎟+C
⎝ 5 ⎠
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23. 4 x – x 2 = 4 − (4 − 4 x + x 2 ) = 4 – ( x – 2)2
u = x – 2, du = dx
dx
du
=∫
∫
2
4x – x
4 – u2
u = 2 sin t, du = 2 cos t dt
du
⎛u⎞
= dt = t + C = sin –1 ⎜ ⎟ + C
∫
2 ∫
⎝2⎠
4−u
⎛ x–2⎞
= sin –1 ⎜
⎟+C
⎝ 2 ⎠
25. x 2 + 2 x + 2 = x 2 + 2 x + 1 + 1 = ( x + 1) 2 + 1
u = x + 1, du = dx
2x + 1
2u – 1
∫ x2 + 2 x + 2 dx = ∫ u 2 + 1 du
2u
du
du – ∫
=∫
2
2
u +1
u +1
= ln u 2 + 1 – tan –1 u + C
)
= ln x 2 + 2 x + 2 − tan −1 ( x + 1) + C
26. x 2 – 6 x + 18 = x 2 – 6 x + 9 + 9 = ( x – 3)2 + 9
u = x – 3, du = dx
2x – 1
2u + 5
∫ x2 – 6 x + 18 dx = ∫ u 2 + 9 du
2u du
du
=∫
+ 5∫
2
2
u +9
u +9
5
⎛u⎞
= ln u 2 + 9 + tan −1 ⎜ ⎟ + C
3
⎝3⎠
5
⎛ x−3⎞
= ln x 2 − 6 x + 18 + tan −1 ⎜
⎟+C
3
⎝ 3 ⎠
)
)
2
1⎛
1
⎞
27. V = π∫ ⎜
dx
0 ⎝ x 2 + 2 x + 5 ⎟⎠
1⎡
2
⎤
= π∫ ⎢
⎥ dx
2
0 ( x + 1) + 4
⎣⎢
⎦⎥
444
1
Section 7.4
π/4
–1
=
π ⎡1 1
⎤
t + sin 2t ⎥
8 ⎢⎣ 2 4
⎦ tan –1 (1/ 2)
=
π ⎡1 1
⎤
t + sin t cos t ⎥
8 ⎢⎣ 2 2
⎦ tan –1 (1/ 2)
=
π ⎡⎛ π 1 ⎞ ⎛ 1
–1 1 1 ⎞ ⎤
+ ⎟
⎜ + ⎟ – ⎜ tan
8 ⎢⎣⎝ 8 4 ⎠ ⎝ 2
2 5 ⎠ ⎥⎦
=
π⎛1 π
–1 1 ⎞
⎜ + – tan
⎟ ≈ 0.082811
16 ⎝ 10 4
2⎠
π/4
28. V = 2π∫
1
1
+ 2x + 5
1
x
= 2π ∫
dx
0 ( x + 1) 2 + 4
0 x2
= 2π ∫
x +1
1
0 ( x + 1) 2
+4
x dx
dx – 2π ∫
1
1
0 ( x + 1) 2
+4
dx
1
1
⎡1
⎡1
⎤
⎛ x + 1 ⎞⎤
= 2π ⎢ ln[( x + 1) 2 + 4]⎥ – 2π ⎢ tan –1 ⎜
⎟⎥
2
2
⎣
⎦0
⎝ 2 ⎠⎦0
⎣
1⎤
⎡
= π[ln 8 – ln 5] – π ⎢ tan –1 1 – tan –1 ⎥
2⎦
⎣
1⎞
⎛ 8 π
= π ⎜ ln – + tan –1 ⎟ ≈ 0.465751
5
4
2⎠
⎝
29. a.
(
(
2
⎛ 1 ⎞
2sec2 t dt
tan (1/ 2) ⎜⎝ 4sec 2 t ⎟⎠
π π/4
1
π π/ 4
= ∫ –1
cos 2 t dt
dt = ∫ –1
2
8 tan (1/ 2) sec t
8 tan (1/ 2)
π π/ 4
⎛1 1
⎞
= ∫ –1
⎜ + cos 2t ⎟ dt
8 tan (1/ 2) ⎝ 2 2
⎠
V = π∫
π/ 4
24. 4 x – x 2 = 4 − (4 − 4 x + x 2 ) = 4 – ( x – 2)2
u = x – 2, du = dx
x
u+2
dx = ∫
du
∫
2
4x – x
4 – u2
– u du
du
= –∫
+ 2∫
2
4–u
4 – u2
(Use the result of Problem 23.)
⎛u⎞
= – 4 – u 2 + 2sin –1 ⎜ ⎟ + C
⎝2⎠
⎛ x–2⎞
= – 4 x – x 2 + 2sin –1 ⎜
⎟+C
⎝ 2 ⎠
(
x + 1 = 2 tan t, dx = 2sec 2 t dt
u = x 2 + 9, du = 2 x dx
x dx
1 du 1
= ln u + C
u
2
1
1
= ln x 2 + 9 + C = ln x 2 + 9 + C
2
2
∫ x2 + 9 = 2 ∫
(
)
b. x = 3 tan t, dx = 3sec 2 t dt
x dx
∫ x2 + 9 = ∫ tan t dt
= – ln cos t + C
⎛
⎞
3
⎟+C
+ C1 = − ln ⎜
⎜ 2
⎟ 1
x2 + 9
⎝ x +9 ⎠
= ln ⎛⎜ x 2 + 9 ⎞⎟ − ln 3 + C1
⎝
⎠
1
= ln ( x 2 + 9)1/ 2 + C = ln x 2 + 9 + C
2
= − ln
(
3
)
(
)
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32. The equation of the circle with center (–a, 0) is
30. u = 9 + x 2 , u 2 = 9 + x 2 , 2u du = 2x dx
3
∫0
x3 dx
9 + x2
=∫
x2
3
0
9 + x2
x dx = ∫
3 2
3
u2 − 9
udu
u
3 2
⎡ u3
⎤
(u 2 − 9) du = ⎢ – 9u ⎥
3
⎣⎢ 3
⎦⎥ 3
≈ 5.272
=∫
31. a.
3 2
= 18 – 9 2
( x + a )2 + y 2 = b 2 , so y = ± b 2 – ( x + a )2 . By
symmetry, the area of the overlap is four times
the area of the region bounded by x = 0, y = 0,
and y = b 2 – ( x + a )2 dx .
A = 4∫
b–a
x + a = b sin t, dx = b cos t dt
π/ 2
u = 4 – x 2 , u 2 = 4 – x 2 , 2u du = –2x dx
A = 4∫
4 – x2
4 − x2
u 2 du
dx
x
dx
–
=
=
∫ x
∫ x2
∫ 4 – u2
−4 + 4 − u 2
1
=∫
du = −4 ∫
du + ∫ du
2
4−u
4 − u2
1 u+2
= −4 ⋅ ln
+u+C
4 u−2
= 2b 2 ∫
= − ln
4 − x2 + 2
+ 4− x +C
2
4 − x2 − 2
b. x = 2 sin t, dx = 2 cos t dt
∫
4 – x2
cos 2 t
dx = 2 ∫
dt
sin t
x
= 2∫
b 2 – ( x + a) 2 dx
0
sin –1 ( a / b )
π/2
sin –1 ( a / b )
(1 + cos 2t )dt
π/2
⎡ 1
⎤
= 2b 2 ⎢t + sin 2t ⎥
⎣ 2
⎦ sin –1 ( a / b )
= 2b 2 [t + sin t cos t ]π / 2–1
sin
(a / b)
⎡
2
2 ⎞⎤
π ⎛
⎛ a ⎞ a b – a ⎟⎥
= 2b 2 ⎢ – ⎜ sin –1 ⎜ ⎟ +
⎢2 ⎜
⎟⎥
b
⎝b⎠ b
⎝
⎠⎦
⎣
a
⎛ ⎞
= πb 2 – 2b 2 sin –1 ⎜ ⎟ – 2a b 2 – a 2
⎝b⎠
33. a.
(1 – sin 2 t )
dt
sin t
b 2 cos 2 t dt
The coordinate of C is (0, –a). The lower arc
of the lune lies on the circle given by the
equation x 2 + ( y + a)2 = 2a 2 or
= 2 ∫ csc t dt – 2∫ sin t dt
y = ± 2a 2 – x 2 – a. The upper arc of the
lune lies on the circle given by the equation
= 2 ln csc t − cot t + 2 cos t + C
x 2 + y 2 = a 2 or y = ± a 2 – x 2 .
= 2 ln
2
4 − x2
−
+ 4 − x2 + C
x
x
= 2 ln
2− 4− x
x
A=∫
=∫
+ 4 − x2 + C
− ln
= ln
= ln
4− x +2
4 − x2 − 2
= ln
4− x −2
4 − x2 + 2
For
2
(2 − 4 − x )
4− x −4
2
= ln
2a 2 – x 2 dx + 2a 2
a 2 – x 2 dx is the area of a
a
∫– a
2a 2 – x 2 dx, let
x = 2a sin t , dx = 2a cos t dt
( 4 − x 2 + 2)( 4 − x 2 − 2)
2 2
a
∫– a
a
–a
semicircle with radius a, so
a
πa 2
2
2
a
x
dx
=
–
.
∫– a
2
2
( 4 − x − 2)
2
⎛ 2 − 4 − x2
= ln ⎜
⎜
x
⎝
a 2 – x 2 dx – ∫
Note that
To reconcile the answers, note that
2
a
–a
2
a
a 2 – x 2 dx – ∫ ⎛⎜ 2a 2 – x 2 – a ⎞⎟ dx
–a ⎝
⎠
a
–a
a
(2 − 4 − x )
2 2
− x2
∫– a
π/4
–π / 4
2a 2 cos 2 t dt
π/4
⎡ 1
⎤
(1 + cos 2t )dt = a 2 ⎢t + sin 2t ⎥
–π / 4
⎣ 2
⎦–π / 4
= a2 ∫
2
2
⎞
⎟ = 2 ln 2 − 4 − x
⎟
x
⎠
2a 2 – x 2 dx = ∫
=
π/4
πa 2
+ a2
2
A=
⎞
πa 2 ⎛ πa 2
–⎜
+ a 2 ⎟ + 2a 2 = a 2
⎟
2 ⎜⎝ 2
⎠
Thus, the area of the lune is equal to the area
of the square.
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
b. Without using calculus, consider the
following labels on the figure.
7.5 Concepts Review
1. proper
2. x – 1 +
5
x +1
3. a = 2; b = 3; c = –1
4.
Area of the lune = Area of the semicircle of
radius a at O + Area (ΔABC) – Area of the
sector ABC.
1
1⎛π⎞
A = πa 2 + a 2 – ⎜ ⎟ ( 2a )2
2
2⎝ 2⎠
1 2
1
= πa + a 2 – πa 2 = a 2
2
2
Note that since BC has length 2a, the
π
measure of angle OCB is , so the measure
4
π
of angle ACB is .
2
Problem Set 7.5
1.
34. Using reasoning similar to Problem 33 b, the area is
1 2 1
1⎛
a⎞
πa + (2a ) b 2 – a 2 – ⎜ 2sin –1 ⎟ b 2
2
2
2⎝
b⎠
1 2
a
= πa + a b 2 – a 2 – b 2 sin –1 .
2
b
dy
a2 – x2
a2 – x2
; y = ∫–
=–
dx
dx
x
x
x = a sin t, dx = a cos t dt
a cos t
cos 2 t
y = ∫–
a cos t dt = – a ∫
dt
a sin t
sin t
= –a∫
1 – sin 2 t
dt = a ∫ (sin t – csc t )dt
sin t
= a ( – cos t − ln csc t − cot t ) + C
a2 – x2
x
⎛
⎞
a2 – x2
a
a2 − x2 ⎟
y = a⎜ –
− ln −
+C
⎜
⎟
a
x
x
⎝
⎠
cos t =
a2 – x2
a
, csc t = , cot t =
a
x
= − a 2 − x 2 − a ln
1
A
B
= +
(
1)
x x+
x x +1
1 = A(x + 1) + Bx
A = 1, B = –1
1
1
1
∫ x( x + 1) dx = ∫ x dx – ∫ x + 1 dx
= ln x – ln x + 1 + C
2.
35.
A
B
Cx + D
+
+
x –1 ( x –1)2 x 2 + 1
3.
2
2
A
B
= +
x + 3 x x ( x + 3) x x + 3
2 = A(x + 3) + Bx
2
2
A= ,B= –
3
3
2
2 1
2 B
∫ x2 + 3x dx = 3 ∫ x dx – 3 ∫ x + 3 dx
2
2
= ln x – ln x + 3 + C
3
3
=
2
3
3
A
B
=
+
(
x
+
1)(
x
–
1)
x
+
1
x
–1
x –1
3 = A(x – 1) + B(x + 1)
3
3
A= – ,B=
2
2
3
3 1
3 1
∫ x2 – 1 dx = – 2 ∫ x + 1 dx + 2 ∫ x – 1 dx
3
3
= – ln x + 1 + ln x – 1 + C
2
2
2
=
a − a2 − x2
+C
x
Since y = 0 when x = a,
0 = 0 – a ln 1 + C, so C = 0.
y = – a 2 − x 2 − a ln
446
Section 7.5
a − a2 – x2
x
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4.
5.
5x
=
5x
=
5
2 x( x + 3)
2x + 6x
2 x ( x + 3)
A
B
= +
x x+3
5
= A( x + 3) + Bx
2
5
5
A= ,B=–
6
6
5x
5 1
5 1
∫ 2 x3 + 6 x2 = 6 ∫ x dx – 6 ∫ x + 3 dx
5
5
= ln x – ln x + 3 + C
6
6
3
2
x − 11
2
x − 11
A
B
=
+
x + 3 x − 4 ( x + 4)( x − 1) x + 4 x − 1
x – 11 = A(x – 1) + B(x + 4)
A =3, B = –2
x − 11
1
1
∫ x2 + 3x − 4 dx = 3∫ x + 4 dx − 2∫ x − 1 dx
2
7.
x–7
8.
=
x–7
A
B
=
+
x – x – 12 ( x – 4)( x + 3) x – 4 x + 3
x – 7 = A(x + 3) + B(x – 4)
3
10
A= – ,B=
7
7
x–7
3
1
10
1
∫ x2 – x – 12 dx = – 7 ∫ x – 4 dx + 7 ∫ x + 3 dx
3
10
= – ln x – 4 + ln x + 3 + C
7
7
2
=
3x − 13
A
B
=
+
( x + 5)( x − 2) x + 5 x − 2
x + 3 x − 10
3x − 13 = A( x − 2) + B ( x + 5)
A = 4, B = –1
1
1
3 x − 13
∫ x2 + 3x − 10 dx = 4∫ x + 5 dx − ∫ x − 2 dx
= 4 ln x + 5 − ln x − 2 + C
x+π
=
A
B
x+π
=
+
( x – 2π)( x – π) x – 2π x – π
x – 3πx + 2π
x + π = A( x − π ) + B ( x − 2π )
A = 3, B = –2
3
2
x+π
∫ x2 – 3πx + 2π2 dx = ∫ x – 2π dx – ∫ x – π dx
2
2
= 3ln x – 2π – 2 ln x – π + C
9.
= 3ln x + 4 − 2 ln x − 1 + C
6.
3x − 13
2
=
2 x + 21
A
B
2 x + 21
=
+
2 x + 9 x – 5 (2 x – 1)( x + 5) 2 x – 1 x + 5
2x + 21 = A(x + 5) + B(2x – 1)
A = 4, B = –1
2 x + 21
4
1
∫ 2 x2 + 9 x – 5 dx = ∫ 2 x – 1 dx – ∫ x + 5 dx
2
=
= 2 ln 2 x – 1 – ln x + 5 + C
10.
2 x 2 − x − 20
=
2( x 2 + x − 6) − 3x − 8
x2 + x − 6
x2 + x − 6
3x + 8
= 2−
2
x + x−6
A
B
3x + 8
3x + 8
=
+
=
2
x + x − 6 ( x + 3)( x − 2) x + 3 x − 2
3x + 8 = A(x – 2) + B(x + 3)
1
14
A= ,B=
5
5
∫
2 x 2 − x − 20
dx
x2 + x − 6
1
1
14
1
= ∫ 2 dx − ∫
dx − ∫
dx
5 x+3
5 x−2
1
14
= 2 x − ln x + 3 − ln x − 2 + C
5
5
11.
A
B
17 x – 3
=
+
3
x
–
2
x
+1
(3
x
–
2)(
x
+
1)
3x + x – 2
17x – 3 = A(x + 1) + B(3x – 2)
A = 5, B = 4
5
17 x – 3
5
4
∫ 3x2 + x – 2 dx = ∫ 3x – 2 dx + ∫ x + 1 dx = 3 ln 3x – 2 + 4 ln x + 1 + C
17 x – 3
2
=
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
12.
13.
5– x
=
5– x
A
B
=
+
( x – π)( x – 4) x – π x – 4
x – x(π + 4) + 4π
5 – x = A(x – 4) + B(x – π )
5–π
1
A=
,B=
π–4
4–π
5– x
5–π
1
1
1
5–π
1
∫ x2 − x(π + 4) + 4π dx = π – 4 ∫ x – π dx + 4 – π ∫ x – 4 dx = π – 4 ln x – π + 4 – π ln x – 4 + C
2
2 x2 + x − 4
x3 − x 2 − 2 x
=
A
B
C
2 x2 + x − 4
= +
+
x( x + 1)( x − 2) x x + 1 x − 2
2 x 2 + x − 4 = A( x + 1)( x − 2) + Bx( x − 2) + Cx( x + 1)
A = 2, B = –1, C = 1
2 x2 + x − 4
2
1
1
∫ x3 − x2 − 2 x dx = ∫ x dx − ∫ x + 1 dx + ∫ x − 2 dx = 2 ln x − ln x + 1 + ln x − 2 + C
14.
7 x2 + 2 x – 3
A
B
C
=
+
+
(2 x – 1)(3x + 2)( x – 3) 2 x – 1 3 x + 2 x – 3
7 x 2 + 2 x – 3 = A(3 x + 2)( x – 3) + B(2 x – 1)( x – 3) + C (2 x – 1)(3 x + 2)
A=
1
1
6
, B = – ,C =
35
7
5
7 x2 + 2 x – 3
1
1
1
1
6
1
∫ (2 x – 1)(3x + 2)( x – 3) dx = 35 ∫ 2 x – 1 dx – 7 ∫ 3x + 2 dx + 5 ∫ x – 3 dx
=
15.
1
1
6
ln 2 x –1 – ln 3 x + 2 + ln x – 3 + C
70
21
5
6 x 2 + 22 x − 23
(2 x − 1)( x 2 + x − 6)
=
6 x 2 + 22 x − 23
A
B
C
=
+
+
(2 x − 1)( x + 3)( x − 2) 2 x − 1 x + 3 x − 2
6 x 2 + 22 x − 23 = A( x + 3)( x − 2) + B (2 x − 1)( x − 2) + C (2 x − 1)( x + 3)
A = 2, B = –1, C = 3
6 x 2 + 22 x − 23
2
1
3
∫ (2 x − 1)( x2 + x − 6) dx = ∫ 2 x − 1 dx − ∫ x + 3 dx + ∫ x − 2 dx = ln 2 x − 1 − ln x + 3 + 3ln x − 2 + C
16.
⎞
1 ⎛ x3 − 6 x 2 + 11x − 6 ⎞ 1 ⎛
x 2 − 3x + 2
= ⎜1 +
⎜
⎟
⎟
3
2
3
2
3
2
⎜
⎟
⎜
4 x − 28 x + 56 x − 32 4 ⎝ x − 7 x + 14 x − 8 ⎠ 4 ⎝ x − 7 x + 14 x − 8 ⎠⎟
1⎛
( x − 1)( x − 2) ⎞ 1 ⎛
1 ⎞
= ⎜1 +
⎟ = ⎜1 +
⎟
4 ⎝ ( x − 1)( x − 2)( x − 4) ⎠ 4 ⎝ x − 4 ⎠
x3 − 6 x 2 + 11x − 6
x3 – 6 x 2 + 11x – 6
=
1
1
1
∫ 4 x3 – 28 x2 + 56 x – 32 dx = ∫ 4 dx + 4 ∫ x – 4 dx
448
Section 7.5
=
1
1
x + ln x – 4 + C
4
4
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17.
x3
x +x–2
3x – 2
3x – 2
= x –1 +
x +x–2
3x – 2
A
B
=
=
+
2
(
2)(
–
1)
2
–1
x
+
x
x
+
x
x +x–2
3x – 2 = A(x – 1) + B(x + 2)
8
1
A= ,B=
3
3
2
2
x3
8
1
1
1
1 2
8
1
– x + ln x + 2 + ln x – 1 + C
3
3
∫ x2 + x – 2 dx = ∫ ( x − 1) dx + 3 ∫ x + 2 dx + 3 ∫ x − 1 dx = 2 x
18.
19.
x3 + x 2
= x – 4+
14 x + 24
( x + 3)( x + 2)
x + 5x + 6
14 x + 24
A
B
=
+
( x + 3)( x + 2) x + 3 x + 2
14x + 24 = A(x + 2) + B(x + 3)
A = 18, B = –4
18
4
1 2
x3 + x 2
∫ x2 + 5 x + 6 dx = ∫ ( x − 4) dx + ∫ x + 3 dx – ∫ x + 2 dx = 2 x − 4 x + 18ln x + 3 – 4 ln x + 2 + C
2
x4 + 8x2 + 8
x3 − 4 x
= x+
12 x 2 + 8
x( x + 2)( x – 2)
12 x 2 + 8
A
B
C
= +
+
x( x + 2)( x – 2) x x + 2 x – 2
12 x 2 + 8 = A( x + 2)( x – 2) + Bx( x – 2) + Cx( x + 2)
A = –2, B = 7, C = 7
1
1
1
1 2
x4 + 8x2 + 8
∫ x3 – 4 x dx = ∫ x dx – 2∫ x dx + 7∫ x + 2 dx + 7∫ x – 2 dx = 2 x – 2 ln x + 7 ln x + 2 + 7 ln x – 2 + C
20.
x 6 + 4 x3 + 4
x3 – 4 x 2
272 x 2 + 4
2
x ( x – 4)
=
272 x 2 + 4
= x3 + 4 x 2 + 16 x + 68 +
x3 – 4 x 2
A B
C
+
+
x x2 x – 4
272 x 2 + 4 = Ax( x – 4) + B ( x – 4) + Cx 2
1
1089
A = – , B = –1, C =
4
4
x 6 + 4 x3 + 4
1 1
1
1089
1
3
2
∫ x3 – 4 x2 dx = ∫ ( x + 4 x + 16 x + 68) dx – 4 ∫ x dx – ∫ x2 dx + 4 ∫ x − 4 dx
1
4
1
1 1089
= x 4 + x3 + 8 x 2 + 68 x – ln x + +
ln x – 4 + C
x
4
3
4
4
21.
x +1
=
A
B
+
x − 3 ( x − 3)2
( x − 3)
x + 1 = A(x – 3) + B
A = 1, B = 4
x +1
1
4
4
∫ ( x − 3)2 dx = ∫ x − 3 dx + ∫ ( x − 3)2 dx = ln x − 3 − x − 3 + C
2
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22.
23.
5x + 7
5x + 7
=
=
A
B
+
x + 2 ( x + 2) 2
x + 4 x + 4 ( x + 2)
5x + 7 = A(x + 2) + B
A = 5, B = –3
5x + 7
5
3
3
∫ x2 + 4 x + 4 dx = ∫ x + 2 dx − ∫ ( x + 2)2 dx = 5ln x + 2 + x + 2 + C
2
2
3x + 2
x + 3x + 3x + 1
3
2
3x + 2
=
( x + 1)
3
=
A
B
C
+
+
x + 1 ( x + 1)2 ( x + 1)3
3x + 2 = A( x + 1) + B ( x + 1) + C
A = 0, B = 3, C = –1
3x + 2
3
1
3
1
∫ x3 + 3x2 + 3x + 1 dx = ∫ ( x + 1)2 dx − ∫ ( x + 1)3 dx = − x + 1 + 2( x + 1)2 + C
2
24.
x6
A
B
C
D
E
F
G
+
+
+
+
+
+
2
2
3
4
x – 2 ( x – 2)
1 – x (1 – x)
( x – 2) (1 – x)
(1 – x)
(1 – x)
(1 – x)5
A = 128, B = –64, C = 129, D = –72, E = 30, F = –8, G = 1
⎡ 128
x6
64
129
72
30
8
1 ⎤
∫ ( x – 2)2 (1 – x)5 dx = ∫ ⎢⎢ x – 2 – ( x – 2)2 + 1 – x – (1 – x)2 + (1 – x)3 − (1 – x)4 + (1 – x)5 ⎥⎥ dx
⎣
⎦
2
=
5
= 128ln x – 2 +
25.
3 x 2 − 21x + 32
x − 8 x + 16 x
3
2
64
72
15
8
1
–129 ln 1 – x +
−
+
−
+C
2
3
1 – x (1 – x)
x–2
3(1 – x)
4(1 – x) 4
=
3 x 2 − 21x + 32
x( x − 4)
2
=
A
B
C
+
+
x x − 4 ( x − 4)2
3x − 21x + 32 = A( x − 4) 2 + Bx( x − 4) + Cx
A = 2, B = 1, C = –1
3x 2 − 21x + 32
2
1
1
1
∫ x3 − 8 x2 + 16 dx = ∫ x dx + ∫ x − 4 dx − ∫ ( x − 4)2 dx = 2 ln x + ln x − 4 + x − 4 + C
2
26.
27.
450
x 2 + 19 x + 10
=
x 2 + 19 x + 10
=
A B C
D
+
+
+
x x 2 x3 2 x + 5
2 x + 5x
x (2 x + 5)
A = –1, B = 3, C = 2, D = 2
x 2 + 19 x + 10
⎛ 1 3
2
2 ⎞
3 1
∫ 2 x4 + 5 x3 dx = ∫ ⎝⎜ – x + x2 + x3 + 2 x + 5 ⎠⎟ dx = – ln x – x – x2 + ln 2 x + 5 + C
4
3
2 x2 + x – 8
=
3
2x2 + x – 8
=
A Bx + C
+
x x2 + 4
x + 4x
x( x + 4)
A = –2, B = 4, C = 1
2 x2 + x – 8
1
4x +1
1
2x
1
∫ x3 + 4 x dx = –2∫ x dx + ∫ x2 + 4 dx = −2∫ x dx + 2∫ x 2 + 4 dx + ∫ x 2 + 4 dx
1
⎛ x⎞
= –2 ln x + 2 ln x 2 + 4 + tan –1 ⎜ ⎟ + C
2
⎝2⎠
3
Section 7.5
2
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28.
3x + 2
3x + 2
=
x( x + 2) + 16 x x( x + 4 x + 20)
1
1
13
A = , B = – ,C =
10
10
5
2
2
=
A
Bx + C
+
2
x x + 4 x + 20
1 x + 13
– 10
1 1
1 1
14
1
1
2x + 4
5
dx
+
∫ x( x + 2)2 + 16 x
∫ x2 + 4 x + 20 dx = 10 ∫ x dx + 5 ∫ ( x + 2)2 + 16 dx − 20 ∫ x2 + 4 x + 20 dx
10 ∫ x
1
7
⎛ x+2⎞ 1
2
= ln x + tan –1 ⎜
⎟ – ln x + 4 x + 20 + C
10
10
⎝ 4 ⎠ 20
3x + 2
29.
30.
31.
dx =
A
Bx + C
+
2
x
–
1
(2 x − 1)( x + 9)
x2 + 9
A = –4, B = 3, C = 0
2 x 2 – 3 x – 36
1
3x
3
2
∫ (2 x – 1)( x 2 + 9) dx = –4∫ 2 x – 1 dx + ∫ x 2 + 9 dx = –2 ln 2 x – 1 + 2 ln x + 9 + C
2 x 2 – 3x – 36
2
1
=
1
=
x –16 ( x − 2)( x + 2)( x 2 + 4)
A
B
Cx + D
=
+
+
x – 2 x + 2 x2 + 4
1
1
1
A = , B = – , C = 0, D = –
32
32
8
1
1
1
1
1
1
1
1
1
1
–1 ⎛ x ⎞
∫ x4 – 16 dx = 32 ∫ x – 2 dx – 32 ∫ x + 2 dx − 8 ∫ x2 + 4 dx = 32 ln x – 2 – 32 ln x + 2 – 16 tan ⎜⎝ 2 ⎟⎠ + C
4
1
( x – 1) ( x + 4)
2
2
=
A
B
C
D
+
+
+
x – 1 ( x – 1) 2 x + 4 ( x + 4)2
2
1
2
1
, B = ,C =
,D=
125
25
125
25
1
2
1
1
1
2
1
1
1
∫ ( x –1)2 ( x + 4)2 dx = – 125 ∫ x –1 dx + 25 ∫ ( x –1)2 dx + 125 ∫ x + 4 dx + 25 ∫ ( x + 4)2 dx
2
1
2
1
=–
+
+C
ln x – 1 –
ln x + 4 –
125
25( x – 1) 125
25( x + 4)
A= –
32.
x3 – 8 x 2 – 1
( x + 3)( x 2 – 4 x + 5)
–7 x 2 + 7 x – 16
( x + 3)( x – 4 x + 5)
2
A= –
= 1+
=
–7 x 2 + 7 x – 16
( x + 3)( x 2 − 4 x + 5)
A
Bx + C
+
2
x + 3 x – 4x + 5
50
41
14
, B = – ,C =
13
13
13
⎡ 50 ⎛ 1 ⎞ – 41 x + 14 ⎤
13
13
dx
=
∫ ( x + 3)( x2 – 4 x + 5) ∫ ⎢⎢1 – 13 ⎜⎝ x + 3 ⎟⎠ + x2 – 4 x + 5 ⎥⎥ dx
⎣
⎦
50
1
68
1
41
2x − 4
= ∫ dx − ∫
dx − ∫
dx − ∫
dx
13 x + 3
13 ( x − 2) 2 + 1
26 x 2 − 4 x + 5
x3 – 8 x 2 – 1
= x–
50
68
41
ln x + 3 – tan –1 ( x – 2) – ln x 2 – 4 x + 5 + C
13
13
26
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33. x = sin t, dx = cos t dt
(sin 3 t − 8sin 2 t − 1) cos t
∫ (sin t + 3)(sin 2 t − 4sin t + 5)
dt = ∫
x3 − 8 x 2 − 1
( x + 3)( x 2 − 4 x + 5)
dx
50
68
41
ln x + 3 − tan −1 ( x − 2) − ln x 2 − 4 x + 5 + C
13
13
26
which is the result of Problem 32.
(sin 3 t – 8sin 2 t – 1) cos t
50
68
41
–1
2
∫ (sin t + 3)(sin 2 t – 4sin t + 5) dt = sin t – 13 ln sin t + 3 – 13 tan (sin t – 2) – 26 ln sin t – 4sin t + 5 + C
= x−
34. x = sin t, dx = cos t dt
cos t
1
1
1
1
−1 ⎛ x ⎞
∫ sin 4 t − 16 dt = ∫ x4 − 16 dx = 32 ln x − 2 − 32 ln x + 2 − 16 tan ⎜⎝ 2 ⎟⎠ + C
which is the result of Problem 30.
cos t
1
1
1
–1 ⎛ sin t ⎞
∫ sin 4 t – 16 dt = 32 ln sin t – 2 – 32 ln sin t + 2 – 16 tan ⎜⎝ 2 ⎟⎠ + C
35.
x3 – 4 x
=
Ax + B
+
Cx + D
( x + 1)
x + 1 ( x 2 + 1) 2
A = 1, B = 0, C = –5, D = 0
2
2
2
x3 – 4 x
x
x
∫ ( x2 + 1)2 dx = ∫ x2 + 1 dx − 5∫ ( x 2 + 1)2 dx
36. x = cos t, dx = –sin t dt
(sin t )(4 cos 2 t –1)
∫ (cos t )(1 + 2 cos2 t + cos4 t )
4 x2 − 1
1
5
ln x 2 + 1 +
+C
2
2
2( x + 1)
4 x 2 –1
x(1 + 2 x 2 + x 4 )
dx
A Bx + C Dx + E
+
+
x x 2 + 1 ( x 2 + 1) 2
x(1 + 2 x + x ) x( x + 1)
A = –1, B = 1, C = 0, D = 5, E = 0
⎡ 1
x
5x ⎤
1
5
1
5
2
−∫ ⎢− +
+
+ C = ln cos t − ln cos 2 t + 1 +
+C
⎥ dx = ln x − ln x + 1 +
2
2
2
2
x
2
2
x + 1 ( x + 1) ⎦⎥
2( x + 1)
2(cos 2 t + 1)
⎣⎢
2
37.
4
4 x2 − 1
dt = – ∫
=
=
2 x3 + 5 x 2 + 16 x
2
=
2
=
x(2 x 2 + 5 x + 16)
=
2 x 2 + 5 x + 16
=
Ax + B
+
Cx + D
x + 8 x + 16 x
x( x + 8 x + 16)
( x + 4)
x + 4 ( x 2 + 4)2
A = 0, B = 2, C = 5, D = 8
2 x3 + 5 x 2 + 16 x
2
5x + 8
2
5x
8
∫ x5 + 8 x3 + 16 x dx = ∫ x 2 + 4 dx + ∫ ( x2 + 4)2 dx = ∫ x 2 + 4 dx + ∫ ( x2 + 4)2 dx + ∫ ( x2 + 4)2 dx
8
dx, let x = 2 tan θ, dx = 2sec2 θ dθ .
To integrate ∫
2
2
( x + 4)
5
3
4
16sec2 θ
2
2
2
2
⎛1 1
⎞
θ dθ = ∫ ⎜ + cos 2θ ⎟ dθ
2
2
⎝
⎠
x
x
1
1
1
1
1
+C
= θ + sin 2θ + C = θ + sin θ cos θ + C = tan –1 +
2
2 x2 + 4
2
4
2
2
8
∫ ( x2 + 4)2 dx = ∫ 16sec4 θ dθ = ∫ cos
∫
452
2 x3 + 5 x 2 + 16 x
x + 8 x + 16 x
5
3
Section 7.5
dx = tan –1
2
x
5
1
x
x
3
x
2x – 5
–
+ tan –1 +
+ C = tan –1 +
+C
2
2
2 2( x + 4) 2
2 x +4
2
2 2( x 2 + 4)
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38.
x –17
x –17
A
B
=
+
( x + 4)( x – 3) x + 4 x – 3
=
x + x –12
A = 3, B = –2
6
x –17
2
6⎛
2 ⎞
3
∫4 x2 + x –12 dx = ∫4 ⎜⎝ x + 4 – x – 3 ⎟⎠ dx
6
4
= ⎡⎣3ln x + 4 – 2 ln x – 3 ⎤⎦ = (3ln10 – 2 ln 3) – (3ln 8 – 2 ln1)
= 3ln10 – 2 ln 3 – 3ln 8 ≈ –1.53
39. u = sin θ, du = cos θ dθ
π/4
cos θ
∫0
(1 – sin θ )(sin θ + 1)
2
2
1
=
2
dθ = ∫
1
1/ 2
0
(1 – u )(u + 1)
2
2
2
2
du = ∫
2
1/ 2
40.
41.
du
1
2
u +1
2
du +
1 1/
2 ∫0
1
2
(u + 1) 2
2
du
1/ 2
⎡1 1+ u 1
⎤
u
= ⎢ ln
+ tan –1 u +
⎥
2
4(u + 1) ⎥⎦ 0
⎢⎣ 8 1 − u 2
2 +1
1
1
1
+ tan –1
+
≈ 0.65
2 −1 2
2 6 2
1
(To integrate
∫ (u 2 + 1)2 du, let u = tan t.)
3x + 13
A
B
3 x + 13
=
+
( x + 3)( x + 1) x + 3 x + 1
=
x + 4x + 3
A = –2, B = 5
5 3 x + 13
2
∫1
(1 – u )(1 + u )(u 2 + 1)2
1
1 1/
du + ∫
1+ u
4 0
⎡ 1
1
1
1⎛
u ⎞⎤
= ⎢ – ln 1 – u + ln 1 + u + tan –1 u + ⎜ tan –1 u +
⎟⎥
2
8
4
4⎝
u + 1 ⎠⎦0
⎣ 8
1
= ln
8
1
1/ 2
0
A
B
Cu + D Eu + F
+
+
+
1 – u 1 + u u 2 + 1 (u 2 + 1) 2
(1 – u )(u + 1)
1
1
1
1
A = , B = , C = 0, D = , E = 0, F =
8
8
4
2
1/ 2
1
1 1/ 2 1
1 1/
∫0 (1 − u 2 )(u 2 + 1)2 du = 8 ∫0 1 − u du + 8 ∫0
2
2
x + 4x + 3
2
5
1
dx = ⎡⎣ –2 ln x + 3 + 5ln x + 1 ⎤⎦ = –2 ln 8 + 5 ln 6 + 2 ln 4 – 5 ln 2 = 5ln 3 − 2 ln 2 ≈ 4.11
dy
= y (1 − y ) so that
dt
1
∫ y(1 − y) dy = ∫ 1 dt = t + C1
a.
Using partial fractions:
1
A
B
A(1 − y ) + By
= +
=
⇒
y (1 − y ) y 1 − y
y (1 − y )
A + ( B − A) y = 1 + 0 y ⇒ A = 1, B − A = 0 ⇒ A = 1, B = 1 ⇒
1
1
1
= +
y (1 − y ) y 1 − y
⎛1
⎛ y ⎞
1 ⎞
Thus: t + C1 = ∫ ⎜ +
⎟ dy = ln y − ln(1 − y ) = ln ⎜
⎟ so that
⎝ y 1− y ⎠
⎝1− y ⎠
y
et
= et +C1 = Cet or y (t ) = 1
1− y
+et
(C =eC1 )
C
Since y (0) = 0.5, 0.5 =
b.
y (3) =
e3
1 + e3
1
1
+1
C
or C = 1 ; thus y (t ) =
et
1+et
≈ 0.953
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453
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
42.
so that
y
= e 2.4t +8000C1 = Ce 2.4t or
8000 − y
(C =e8000C1 )
dy 1
y (12 − y ) so that
=
dt 10
1
1
1
∫ y(12 − y) dy = ∫ 10 dt = 10 t + C1
y (t ) =
a. Using partial fractions:
1
A
B
A(12 − y ) + By
= +
=
⇒
y (12 − y ) y 12 − y
y (12 − y )
thus y (t ) =
1
1
⇒ A= , B=
12
12
1
1
1
⇒
=
+
y (12 − y ) 12 y 12(12 − y )
⎛ 1
⎞
1
1
t + C1 = ∫ ⎜
+
⎟ dy =
10
⎝ 12 y 12(12 − y ) ⎠
1
1 ⎛ y ⎞
[ln y − ln(12 − y)] = ln ⎜
⎟ so that
12
12 ⎝ 12 − y ⎠
y
= e1.2t +12C1 = Ce1.2t or
12 − y
(C =e12C1 )
y (t ) =
b. y (3) =
44.
a.
12e
1 1.2t
+e
C
12
1
+1
C
or C = 0.2 ;
43.
12e3.6
5 + e3.6
8000e7.2
7 + e7.2
≈ 7958.4
Using partial fractions:
1
A
B
= +
y (4000 − y ) y 4000 − y
A(4000 − y ) + By
=
y (4000 − y )
1
1
, B=
4000
4000
⎤
1
1 ⎡1
1
⇒
=
+
y (4000 − y ) 4000 ⎢⎣ y (4000 − y ) ⎥⎦
Thus:
dy
= 0.0003 y (8000 − y ) so that
dt
1
∫ y(8000 − y) dy = ∫ 0.0003 dt = 0.0003t + C1
Section 7.5
8000e2.4t
7+e2.4t
⇒ A=
⎛1
⎞
1
1
⎜ +
⎟ dy =
4000 ∫ ⎝ y (4000 − y ) ⎠
0.001t + C1 =
⎛
⎞
1
1
y
ln ⎜
[ln y − ln(4000 − y)] =
⎟
4000
4000 ⎝ 4000 − y ⎠
so that
y
= e4t + 4000C1 = Ce4t
or
4000 − y
(C =e4000C1 )
a. Using partial fractions:
1
A
B
A(8000 − y ) + By
= +
=
y (8000 − y ) y 8000 − y
y (8000 − y )
⇒ 8000 A + ( B − A) y = 1 + 0 y
⇒ 8000 A = 1, B − A = 0
1
1
⇒ A=
, B=
8000
8000
⎤
1
1 ⎡1
1
⇒
=
+
y (8000 − y ) 8000 ⎢⎣ y (8000 − y ) ⎥⎦
Thus:
⎛1
⎞
1
1
0.0003t + C1 =
⎜ +
⎟ dy =
∫
8000 ⎝ y (8000 − y ) ⎠
454
1
;
7
⇒ 4000 A = 1, B − A = 0
≈ 10.56
⎛
⎞
1
1
y
ln ⎜
[ln y − ln(8000 − y)] =
⎟
8000
8000 ⎝ 8000 − y ⎠
or C =
1
+1
C
⇒ 4000 A + ( B − A) y = 1 + 0 y
12e1.2t
thus y (t ) =
5+e1.2t
b. y (3) =
8000
dy
= 0.001 y (4000 − y ) so that
dt
1
∫ y(4000 − y) dy = ∫ 0.001 dt = 0.001t + C1
1.2t
Since y (0) = 2.0, 2.0 =
1
+e2.4t
C
Since y (0) = 1000, 1000 =
12 A + ( B − A) y = 1 + 0 y ⇒ 12 A = 1, B − A = 0
Thus:
8000e 2.4t
y (t ) =
4000e 4t
1
+ e 4t
C
Since y (0) = 100, 100 =
thus y (t ) =
b.
y (3) =
4000
1
+1
C
or C =
1
;
39
4000e 4t
39+e4t
4000e12
39 + e12
≈ 3999.04
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45.
dy
= ky ( L − y ) so that
dt
1
∫ y( L − y) dy = ∫ k dt = kt + C1
Using partial fractions:
1
A
B
A( L − y ) + By
= +
=
⇒
y( L − y) y L − y
y( L − y)
Thus if y0 < L , then y (t ) < L for all positive t
(see note at the end of problem 45 solution) and
so the graph will be concave up as long as
L − 2 y > 0 ; that is, as long as the population is
less than half the capacity.
49. a.
LA + ( B − A) y = 1 + 0 y ⇒ LA = 1, B − A = 0 ⇒
A=
1
1
1
1 ⎡1
1 ⎤
= ⎢ +
, B= ⇒
L
L
y ( L − y ) L ⎣ y L − y ⎥⎦
Thus: kt + C1 =
dy
∫ y(16 – y) = ∫ kdt
1 ⎛1
1 ⎞
⎜ +
⎟ dy =
L ∫⎝ y L − y ⎠
1 ⎛1
1 ⎞
⎜ +
⎟ dy = kt + C
16 ∫ ⎝ y 16 – y ⎠
1
1 ⎛ y ⎞
[ln y − ln( L − y )] = ln ⎜
⎟ so that
L
L ⎝ L− y⎠
1
( ln y – ln 16 – y ) = kt + C
16
y
ln
= 16kt + C
16 – y
y
Le kLt
= e kLt + LC1 = Ce kLt or y (t ) = 1
L− y
+ekLt
(C =e LC1 )
C
If y0 = y (0) =
L
1
+1
C
then
final formula is y (t ) =
1 L − y0
=
; so our
y0
C
Le kLt
⎛ L − y0 ⎞ kLt
⎜
⎟+e
⎝ y0 ⎠
(Note: if y0 < L , then u =
ekLt
u + ekLt
y
= Ce16kt
16 – y
.
y(50) = 4:
( 1 ln 7 )t ( 1 ln 7 )t
( 1 ln 7 )t
16e 50 3
16
y=
=
1 ln 7 t
–
( ) 1 + 7e ( 501 ln 73 )t
7 + e 50 3
7 y = 16e 50 3 – ye 50 3
b.
y (90) =
c.
9=
L − y0
47. If y0 < L , then y ′(0) = ky0 ( L − y0 ) > 0 and the
population is increasing initially.
48. The graph will be concave up for values of t that
make y ′′(t ) > 0 . Now
dy ′ d
y ′′(t ) =
= [ ky ( L − y ) ] =
dt dt
k [ − yy ′ + ( L − y ) y ′] = k [ ky ( L − y ) ][ L − 2 y ]
Instructor’s Resource Manual
)
(
46. Since y ′(0) = ky0 ( L − y0 ) is negative if y0 > L ,
the population would be decreasing at time
t = 0. Further, since
L
L
lim y (t ) = lim
=
=L
⎛
⎞
0 +1
t →∞
t →∞ ⎜ L − y0 ⎟
+1
is monotonic as t → ∞ ,we conclude
y0 ekLt
that the population would decrease toward a
limiting value of L.
1 1 800k
1
7
, so k =
ln
= e
3 7
800 3
7
y
1 1 ln t
= e 50 3
16 – y 7
< 1 ; thus y (t ) < L for all t)
(no matter how y0 and L compare), and since
1
y
1
= C;
= e16 kt
7
16 – y 7
y(0) = 2:
L − y0
> 0 and
y0
⎜ y ekLt ⎟
⎝ 0
⎠
dy
= ky (16 – y )
dt
dy
= kdt
y (16 – y )
16
(
)
– 1 ln 7 90
1 + 7e 50 3
≈ 6.34 billion
16
( )
– ( 1 ln 7 )t 16
7e 50 3 = –1
1 + 7e
(
)
– 1 ln 7 t
50 3
– 1 ln 7 t
e 50 3
–
9
=
1
9
( 501 ln 73 ) t = ln 91
⎛ ln 1 ⎞
t = –50 ⎜ 9 ⎟ ≈ 129.66
⎜ ln 7 ⎟
⎝ 3⎠
The population will be 9 billion in 2055.
Section 7.5
455
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50. a.
51. a. Separating variables, we obtain
dx
= k dt
(a − x)(b − x)
dy
= ky (10 – y )
dt
dy
= kdt
y (10 – y )
1 ⎛1
1
⎜ +
10 ∫ ⎝ y 10 –
ln
1
A
B
=
+
(a − x )(b − x) a − x b − x
⎞
⎟ dy = ∫ kdt
y⎠
1
1
,B=
a −b
a −b
dx
∫ (a − x)(b − x)
A=−
y
= 10kt + C
10 – y
y
= Ce10kt
10 – y
1 ⎛
1
1 ⎞
+
⎜−
⎟ dx = ∫ k dt
∫
a −b ⎝ a − x b− x ⎠
ln a − x − ln b − x
= kt + C
a−b
1
a−x
= kt + C
ln
a−b b− x
a−x
= Ce( a −b) kt
b−x
a
Since x = 0 when t = 0, C = , so
b
a ( a −b) kt
a − x = (b − x) e
b
⎛ a
⎞
a 1 − e( a −b ) kt = x ⎜1 − e( a −b) kt ⎟
⎝ b
⎠
=
1
y
1
= C;
= e10 kt
4
10 – y 4
y(0) = 2:
2 1 500k
1
8
,k =
ln
= e
3 4
500 3
y(50) = 4:
( 501 ln 83 )t
y
1
= e
10 – y 4
( 501 ln 83 )t – ye( 501 ln 83 )t
( 1 ln 8 )t
10e 50 3
10
y=
=
1 ln 8 t
–
( ) 1 + 4e ( 501 ln 83 )t
4 + e 50 3
4 y = 10e
b.
c.
y (90) =
9=
10
(
)
– 1 ln 8 90
1 + 4e 50 3
10
(
≈ 5.94 billion
)
– 1 ln 8 t
1 + 4e 50 3
)
(
– 1 ln 8 t
10
4e 50 3 = – 1
9
(
8
)
– 1 ln t
1
e 50 3 =
36
1
⎛ 1 8⎞
– ⎜ ln ⎟ t = ln
36
⎝ 50 3 ⎠
⎞
⎟ ≈ 182.68
⎟
⎠
The population will be 9 billion in 2108.
t=
⎛ ln 1
–50 ⎜ 36
⎜ ln 8
⎝ 3
(
x(t ) =
)
a(1 − e( a −b) kt )
1 − ba e( a −b) kt
=
ab(1 − e( a −b) kt )
b − ae( a −b) kt
b. Since b > a and k > 0, e( a −b ) kt → 0 as
t → ∞ . Thus,
ab(1)
x→
=a.
b−0
c.
x(t ) =
8(1 − e−2kt )
4 − 2e−2kt
x(20) = 1, so 4 − 2e−40k = 8 − 8e−40k
6e−40k = 4
1 2
k = − ln
40 3
t / 20
e−2kt = et / 20 ln 2 / 3 = eln(2 / 3)
⎛2⎞
=⎜ ⎟
⎝3⎠
t / 20
( 23 ) ⎞⎟⎠
t / 20
( 23 )
3⎞
⎛
4 ⎜ 1 − ( 23 ) ⎟
⎠ = 38 ≈ 1.65 grams
x(60) = ⎝
3
23
2 − ( 23 )
⎛
4 ⎜1 −
x(t ) = ⎝
2−
456
Section 7.5
t / 20
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
d. If a = b, the differential equation is, after
separating variables
dx
= k dt
(a − x) 2
dx
∫ (a − x)2 = ∫ k dt
1
= kt + C
a−x
1
=a−x
kt + C
1
x(t ) = a −
kt + C
1
Since x = 0 when t = 0, C = , so
a
a
1
x(t ) = a −
= a−
akt + 1
kt + 1
a
1 ⎞
⎛
⎛ akt ⎞
= a ⎜1 −
⎟ = a⎜
⎟.
⎝ akt + 1 ⎠
⎝ akt + 1 ⎠
52. Separating variables, we obtain
dy
= k dt .
( y − m)( M − y )
1
A
B
=
+
( y − m)( M − y ) y − m M − y
1
1
,B=
M −m
M −m
⎛ 1
1
1 ⎞
dy
∫ ( y − m)(M − y) = M − m ∫ ⎜⎝ y − m + M − y ⎟⎠ dy
A=
= ∫ k dt
ln y − m − ln M − y
= kt + C
M −m
1
y−m
= kt + C
ln
M −m M − y
y−m
= Ce( M − m) kt
M−y
y − m = ( M − y )Ce
m + MCe( M − m) kt
1 + Ce( M − m) kt
1
C
D
=
+
( A − y )( B + y ) A − y B + y
1
1
,D=
A+ B
A+ B
⎛ 1
dy
1
1 ⎞
∫ ( A − y)( B + y) = A + B ∫ ⎜⎝ A − y + B + y ⎟⎠ dy
C=
= ∫ k dt
− ln( A − y ) + ln( B + y )
= kt + C
A+ B
1
B+ y
ln
= kt + C
A+ B A− y
B+ y
= Ce( A+ B ) kt
A− y
B + y = ( A − y )Ce( A+ B ) kt
y (1 + Ce( A+ B ) kt ) = ACe( A+ B ) kt − B
y (t ) =
ACe( A+ B ) kt − B
1 + Ce( A+ B ) kt
54. u = sin x, du = cos x dx
π/2
1
cos x
1
∫π / 6 sin x(sin 2 x + 1)2 dx = ∫12 u (u 2 + 1)2 du
1
A Bu + C Du + E
= +
+
2
2
u u 2 + 1 (u 2 + 1)2
u (u + 1)
A = 1, B = –1, C = 0, D = –1, E = 0
1
1
∫12 u (u 2 + 1)2 du
11
1 u
1
u
= ∫1 du − ∫1
du − ∫1
du
2
2
2
2u
2 u +1
2 (u + 1)
1
⎡
⎤
1
1
= ⎢ln u − ln(u 2 + 1) +
⎥
2
2(u 2 + 1) ⎦⎥ 1
⎣⎢
2
( M − m) kt
y (1 + Ce( M − m) kt = m + MCe( M − m) kt
y=
53. Separating variables, we obtain
dy
= k dt
( A − y )( B + y )
=
1
1 ⎛ 1 1 5 2⎞
= 0 − ln 2 + − ⎜ ln − ln + ⎟ ≈ 0.308
2
4 ⎝ 2 2 4 5⎠
me−( M − m) kt + MC
e−( M − m) kt + C
− ( M − m ) kt
As t → ∞, e
→ 0 since M > m.
MC
= M as t → ∞ .
Thus y →
C
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Section 7.5
457
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
7.6 Concepts Review
5. Trig identity cos 2 u =
1 + cos 2u
and
2
1. substitution
substitution.
2. 53
⎛ 1 + cos 4 x ⎞
4
∫ cos 2 x dx = ∫ ⎜⎝ 2 ⎟⎠ dx =
2
3. approximation
⎡
⎤
1 ⎢
⎥
2
1 + 2 cos 4 x + cos 4 x ⎥ dx =
∫
⎢
4
u = 4x
⎢⎣ du = 4 dx
⎥⎦
4. 0
Problem Set 7.6
Note: Throughout this section, the notation Fxxx
refers to integration formula number xxx in the
back of the book.
1. Integration by parts.
dv = e −5 x
u=x
1
v = − e−5 x
5
1
1 −5 x
−5 x
−5 x
∫ xe dx = − 5 xe − ∫ − 5 e dx
1
1
= − xe−5 x − e−5 x + C
5
25
1 −5 x ⎛
1⎞
=− e
⎜x+ ⎟+C
5
5⎠
⎝
du = 1 dx
6. Substitution
3
3
∫ sin x cos x dx = ∫ u du =
u = sin x
du = cos x dx
2. Substitution
x
1 1
2
∫ x2 + 9 dx = 2 ∫ u du = ln u + C = ln x + 9 + C
(
)
u = x 2 +9
du = 2 x dx
3. Substitution
2
∫1
ln x
dx =
x
u = ln x
du = 1 dx
ln 2
∫
0
ln 2
⎡ u2 ⎤
u du = ⎢ ⎥
⎢⎣ 2 ⎥⎦ 0
=
( ln 2 )2
2
≈ 0.2402
x
4. Partial fractions
x
x
∫ x2 − 5 x + 6 dx =∫ ( x − 3)( x − 2) dx
x
A
B
=
+
=
( x − 3)( x − 2) ( x − 3) ( x − 2)
A( x − 2) + B ( x − 3) ( A + B ) x + (−2 A − 3B)
=
⇒
( x − 3)( x − 2)
( x − 3)( x − 2)
A + B = 1, − 2 A − 3B = 0 ⇒ A = 3, B = −2
x
3
⎡
⎤
⎢
⎥
1⎢
1
⎛ 1 + cos8 x ⎞ ⎥
x + sin 4 x + ∫ ⎜
⎟ dx ⎥ =
4⎢
2
2
⎝
⎠ ⎥
⎢
v = 8x
⎢⎣
⎥⎦
dv = 8dx
1⎡
1
1
1
⎤
x + sin 4 x + x + sin 8 x ⎥ + C =
⎢
4⎣
2
2
16
⎦
1
[ 24 x + 8sin 4 x + sin 8 x ] + C
64
u4
sin 4 x
+C =
+C
4
4
7. Partial fractions
1
1
∫ x2 + 6 x + 8 dx =∫ ( x + 4)( x + 2) dx
A
B
1
=
+
=
( x + 4)( x + 2) ( x + 4) ( x + 2)
A( x + 2) + B ( x + 4) ( A + B ) x + (2 A + 4 B)
=
⇒
( x + 4)( x + 2)
( x + 4)( x + 2)
1
1
A + B = 0, 2 A + 4 B = 1 ⇒ A = − , B =
2
2
2
1
1 2⎛ 1
1 ⎞
∫1 x 2 + 6 x + 8 = 2 ∫1 ⎜⎝ x + 2 − x + 4 ⎟⎠ dx
=
1
1 ⎡ ( x + 2)
2
⎡ ln x + 2 − ln x + 4 ⎤⎦ = ⎢ ln
1
2⎣
2 ⎣ ( x + 4)
=
1⎛ 4
3 ⎞ 1 10
⎜ ln − ln ⎟ = ln ≈ 0.0527
2⎝ 6
5⎠ 2 9
2
⎤
⎥
⎦1
2
∫ x2 − 5 x + 6 dx =∫ ( x − 3) − ( x − 2) dx =
3ln x − 3 − 2 ln x − 2 = ln
458
Section 7.6
( x − 3)3
( x − 2) 2
+C
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8. Partial fractions
1
1
∫ 1 − t 2 dt =∫ (1 − t )(1 + t ) dt
1
A
B
=
+
=
(1 − t )(1 + t ) (1 − t ) (1 + t )
A(1 + t ) + B (1 − t ) ( A − B)t + ( A + B)
=
⇒
(1 − t )(1 + t )
(1 − t )(1 + t )
1
1
A + B = 1, A − B = 0 ⇒ A = , B =
2
2
1
1
1
1
1 ⎞
⎛ 1
∫0 2 1 − t 2 dt = 2 ∫0 2 ⎜⎝ 1 − t + 1 + t ⎟⎠ dx
1
1
⎡−
ln 1 − t + ln 1 + t ⎤⎦ 2 =
⎣
0
2
13. a. Formula 96
∫x
3x + 1 dx
=
2
135
F 96
a =3, b =1
( 9 x − 2 )( 3x + 1)
9. Substitution
5
7
∫0 x u =x +x +22dx = ∫
2
(u 2 − 2)(u )2u du =
⎡ u 5 2u 3 ⎤
7
2u 4 − 4u 2 du =2 ⎢ −
⎥
2
3 ⎥⎦
⎢⎣ 5
2 ⎡ 5
3u − 10u 3 ⎤
⎣
⎦
15
7
=
2
10. Substitution
4
1
∫3 t − 2t dt = ∫
∫ u =e e x , 3due = e+x1dx dx = ∫ u
x
3u + 1 du
x
(
)(
2 ⎣⎡ln u − 2 ⎤⎦
=
2
) 2 +C
2
9e x − 2 3e x + 1
135
3
14. a. Formula 96
∫ 2t (3 − 4t ) dt = 2∫ t (3 − 4t ) dt =
F 96
a =−4,
b =3
2
∫
−π
6 u2
2
= 2 ln
1
du = 2∫
du =
6 u−2
8
−u
≈ 1.223
6 −2
cos 2 x sin x dx = 0
12. Use of symmetry; substitution
∫0
sin 2 x dx = 8∫
π
0
4∫
π
0
4
sin 2 x dx =
1
(2 cos t + 1)(3 − 4 cos t ) 2 + C
20
15. a. Substitution, Formula 18
dx
1
du
∫ 9 − 16 x 2 = 4 ∫ 9 − u 2 F=18
= 4 [ − cos u ]
0
2
∫
ex
9 − 16e
=4
dx =
2x
u = 4e x , du = 4e x dx
1
du
4 ∫ 9 − u2
=
part a.
1
4e x + 3
+C
ln
24 4e x − 3
16. a. Substitution, Formula 18
dx
dx
5
du
∫ 5 x 2 − 11 = − ∫ 11 − 5 x 2 = − 5 ∫ 11 − u 2
u = 5x,
du = 5 dx
− 5 11
ln
5 22
=
Instructor’s Resource Manual
a =3
1 ⎡1 u + 3 ⎤
1
4x + 3
+C =
+C
ln
ln
⎢
⎥
4 ⎣6 u − 3 ⎦
24 4 x − 3
u =2 x
du = 2 dx
π
2 sin u du
part a.
3
b. Substitution, Formula 18
8−2
2
2π
3 ⎤
⎡ 2
2⎢
(−12t − 6)(3 − 4t ) 2 ⎥ + C =
⎣ 240
⎦
3
1
− (2t + 1)(3 − 4t ) 2 + C
10
u = 4 x , du = 4 dx
u
8
11. Use of symmetry; this is an odd function, so
π
=
F 96
a =3, b =1
u = cos t , du = −sin t dt
7
2 ⎡
77 7 + 8 2 ⎤⎦ ≈ 28.67
15 ⎣
u = 2t , u 2 = 2t
u du = dt
8
6
+C
b. Substitution; Formula 96
∫ cos t 3 − 4 cos t sin t dt = − ∫ u 3 − 4u du =
u2 = x+2
2u du = dx
∫
2
b. Substitution; Formula 96
1
1 ⎡ (1 + t ) ⎤ 2
⎢ln
⎥ ≈ 0.5493
2 ⎣ (1 − t ) ⎦ 0
3
55
ln
110
5 x + 11
5 x − 11
5 x − 11
5 x + 11
=
F 18
a = 11
+C
+C
Section 7.6
459
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b. Substitution, Formula 18
∫
x dx
5
du
=−
∫
4
10 11 − u 2
5 x − 11
u = 5x 2 ,
du = 2 5 x dx
=
F 18
a = 11
5 x 2 − 11
55
ln
220
19. a. Substitution, Formula 45
dx
3
du
=
∫
∫
3
5 + 3x 2
5 + u2
5 x + 11
2
u = 3x
du = 3 dx
a =3
du = 2 dx
⎛ 2x ⎞
1⎛
2
2 ⎞ 81 2
sin −1 ⎜⎜
⎟⎟ + C
⎜ x(4 x − 9) 9 − 2 x ⎟ +
⎠
16 ⎝
32
⎝ 3 ⎠
b. Substitution, Formula 57
2 2
2
x 9 − 2sin 2 x dx =
u 9 − u 2 du
∫ sin xucos
4 ∫
= 2 sin x
du = 2 cos x dx
a =3
81 2 −1 ⎛ 2 sin x ⎞
sin ⎜⎜
⎟⎟ + C
32
3
⎝
⎠
16 − 3t
16 − u
dt = ∫
du =
t
u
F 55
a=4
4 + 16 − 3t 2
+C
3t
16 − 3t 6
t 2 16 − 3t 6
dt = ∫
dt =
t
t3
u = 3t 3
du = 3 3 t 2 dt
1 16 − u 2
du =
3∫
u
F 55
a=4
⎧
1⎪
4 + 16 − 3t 6
6
⎨ 16 − 3t − 4 ln
3⎪
3 t3
⎩
460
Section 7.6
3
ln 3 x 2 + 5 + 3 x 4 + C
6
20. a. Substitution; Formula 48
5 2
2
2
2
∫ t u =3 5+t5t dt = 25 ∫ u 3 + u du
=
F 48
a= 3
⎧⎛ 5 ⎞
⎫
t ⎟⎟ 10t 2 + 3 ⎛⎜ 3 + 5t 2 ⎞⎟ − ⎪
⎪⎜⎜
5 ⎪⎝ 8 ⎠
⎝
⎠ ⎪
⎨
⎬+C =
25 ⎪ 9
⎪
2
⎪ 8 ln 5 t + 3 + 5t
⎪
⎩
⎭
1
5t (10t 2 + 3) 3 + 5t 2 − 9 5 ln 5t + 3 + 5t 2 + C
200
(
)
}
b. Substitution; Formula 48
∫t
8
3 + 5t 6 dt = ∫ t 6 3 + 5t 6 t 2 dt =
u = 5 t3
du = 3 5 t 2 dt
5 2
u 3 + u 2 du
75 ∫
b. Substitution, Formula 55
∫
a= 5
2
u = 3t
du = 3 dt
16 − 3t 2 − 4ln
u = 3x 2
du = 2 3 x dx
=
F 45
{
18. a. Substitution, Formula 55
∫
b. Substitution, Formula 45
x
3
du
dx =
∫
∫
4
6
5 + 3x
5 + u2
du = 5 dt
1⎛
2
2 ⎞
=
⎜ sin x(4sin x − 9) 9 − 2sin x ⎟
⎠
F 57 16 ⎝
2
a= 5
3
ln 3 x + 5 + 3 x 2 + C
3
+C
17. a. Substitution, Formula 57
2 2
2
2
2
∫ x u =9 2−x2 x dx = 4 ∫ u 9 − u du F=57
+
=
F 45
⎫
⎪
⎬+C
⎪⎭
=
F 48
a= 3
⎧⎛ 5 3 ⎞
⎫
t ⎟⎟ 10t 6 + 3 ⎛⎜ 3 + 5t 6 ⎞⎟ − ⎪
⎪⎜⎜
5 ⎪⎝ 8 ⎠
⎝
⎠ ⎪
⎨
⎬+C =
75 ⎪ 9
⎪
3
6
⎪ 8 ln 5 t + 3 + 5t
⎪
⎩
⎭
⎧5t 3 (10t 6 + 3) 3 + 5t 6 − ⎫
1 ⎪
⎪
⎨
⎬+C
600 ⎪9 5 ln 5t 3 + 3 + 5t 6 ⎪
⎩
⎭
(
)
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21. a. Complete the square; substitution;
Formula 45.
dt
dt
du
=∫
=∫
=
∫ 2
2
t + 2t − 3
u 2 − 4 aF=452
(t + 1) − 4
b. Substitution, Formula 98
sin t cos t
u
∫ 3sin t + 5 = ∫ 3u + 5 du
u = sin t
du = cos t dt
2
(3sin t − 10) 3sin t + 5 + C
27
u = t +1
du = dt
ln (t + 1) + t 2 + 2t − 3 + C
b. Complete the square; substitution;
Formula 45.
dt
dt
=∫
=
∫ 2
3 2 29
t + 3t − 5
(t + ) −
2
u = t+ 32
du = dt
∫
du
u2 −
29
4
=
4
24. a. Formula 100a
dz
5
=
ln
∫ z 5 − 4 z F100
a 5
u = cos x
du = −sin x dx
3
2
−
( x + 1) − 4
x2 + 2 x − 3
dx = ∫
dx =
x +1
x +1
∫
u2 − 4
du =
F 47
u
⎛ x +1⎞
x 2 + 2 x − 3 − 2sec−1 ⎜
⎟+C
⎝ 2 ⎠
b. Complete the square; substitution;
Formula 47.
( x − 2) 2 − 4
x2 − 4 x
dx = ∫
dx =
x−2
x−2
u = x−2
du = dx
∫
u2 − 4
du =
F 47
u
x − 4 x − 2sec
23. a. Formula 98
y
∫ 3 y + 5 dy
25. Substitution; Formula 84
1
2
3t dt = ∫ sinh 2 u du =
∫ sinh
F 84
3
u = 3t
1⎛ 1
3 ⎞
1
⎜ sinh 6t − t ⎟ + C = ( sinh 6t − 6t ) + C
3⎝ 4
2 ⎠
12
26. Substitution; Formula 82
sech x
∫ x dx = 2∫ sech u du F=82
u= x
du =
1
dx
2 x
2 tan −1 sinh
x−2⎞
⎜
⎟+C
⎝ 2 ⎠
x +C
2
F 98 27
a =3, b =5
(3 y − 10) 3 y + 5 + C
a=2
b =1
u = cos t
du = −sin t dt
−1 ⎛
=
+C
27. Substitution; Formula 98
cos t sin t
u
∫ 2 cos t + 1 dt = − ∫ 2u + 1 du F=98
a=2
2
5 − 4 cos x − 5
+C =
du = 3 dt
a=2
∫
5 − 4 cos x + 5
5 − 4 cos x + 5
=
F 100 a
a = −4
b=5
5 − 4 cos x − 5
5
ln
5
5
ln
5
2
u = x +1
du = dx
+C
5 − 4z + 5
b. Substitution, Formula 100a
sin x
du
∫ cos x 5 − 4 cos x dx = − ∫ u 5 − 4u
22. a. Complete the square; substitution;
Formula 47.
∫
5 − 4z − 5
a = −4
b =5
ln (t + ) + t 2 + 3t − 5 + C
F 45
a = 29 2
=
F 98
a =3, b =5
1
− (2 cos t − 2) 2 cos t + 1 + C =
6
1
(1 − cos t ) 2 cos t + 1 + C
3
28. Substitution; Formula 96
∫ cos t sin t 4 cos t − 1 dt = − ∫ u 4u − 1 du =
u = cos t
du = −sin t dt
−
Instructor’s Resource Manual
F 96
a=4
b = −1
3
1
(6 cos t + 1)(4 cos t − 1) 2 + C
60
Section 7.6
461
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
29. Substitution; Formula 99, Formula 98
cos 2 t sin t
u2
dt
=
−
∫ cos t + 1
∫ u + 1 du F=99
n=2
a =1
b =1
u = cos t
du = −sin t dt
2⎡
u
⎤
du ⎥ =
− ⎢u 2 u + 1 − 2 ∫
5⎣
u + 1 ⎦ F 98
2⎡
⎛2
⎞⎤
− ⎢u 2 u + 1 − 2 ⎜ (u − 2) u + 1 ⎟ ⎥ + C =
5⎣
3
⎝
⎠⎦
2
4
⎡
⎤
cos t + 1 ⎢ cos 2 t − (cos t − 2) ⎥ + C
−
5
3
⎣
⎦
n =3
a =3
n=2
a =3
⎡1⎛
1 ⎡
x
x
dx ⎞ ⎤ ⎤
⎢
+ 3⎢ ⎜
+∫
⎟⎥ ⎥
2
2
2
36 ⎢ (9 + x )
9 + x 2 ⎟⎠ ⎦⎥ ⎦⎥
⎢⎣18 ⎜⎝ (9 + x )
⎣
1 ⎧⎪
x
x
⎛ x ⎞ ⎫⎪
+
+ tan −1 ⎜ ⎟ ⎬ + C
⎨
F 17 36 ⎪ (9 + x 2 ) 2
⎝ 3 ⎠ ⎭⎪
6 ⋅ (9 + x 2 )
⎩
=
a =3
31. Using a CAS, we obtain:
2
π cos x
∫0 1 + sin x dx = π − 2 ≈ 1.14159
32. Using a CAS, we obtain:
3
x dx ≈ 0.76803
33. Using a CAS, we obtain:
π /2
231π
12
∫0 sin x dx = 2048 ≈ 0.35435
34. Using a CAS, we obtain:
π
3π
4 x
∫0 cos 2 dx = 8 ≈ 1.17810
35. Using a CAS, we obtain:
4
t
∫1 1 + t 8 dt ≈ 0.11083
36. Using a CAS, we obtain:
3 4 −x / 2
∫0 x e
39. Using a CAS, we obtain:
2
3 x + 2x −1
∫2 x 2 − 2 x + 1 dx = 4 ln ( 2 ) + 2 ≈ 4.77259
40. Using a CAS, we obtain:
3
π
du
−1
∫1 u 2u − 1 = 2 tan 5 − 2 ≈ 0.72973
41.
1 ⎡
x
dx ⎤
+ 3∫
⎢
⎥ =
2
2
36 ⎢⎣ (9 + x )
(9 + x 2 )2 ⎥⎦ F 95
1
38. Using a CAS, we obtain:
π /4
x3
∫−π / 4 4 + tan x dx ≈ −0.00921
( )
30. Formula 95, Formula 17
1
∫ (9 + x2 )3 dx F=95
∫0 sech
37. Using a CAS, we obtain:
π /2
1
∫0 1 + 2 cos5 x dx ≈ 1.10577
dx = 768 − 3378e −3/ 2 ≈ 14.26632
c
1
c
∫0 x + 1 dx F=3 ⎡⎣ln x + 1 ⎤⎦ 0 = ln(c + 1)
ln(c + 1) = 1 ⇒ c + 1 = e ⇒
c = e − 1 ≈ 1.71828
42. Formula 17
c 2
−1 c
−1
∫0 x2 + 1 dx F=17 ⎡⎣ 2 tan x ⎤⎦0 = 2 tan c
1
2 tan −1 c = 1 ⇒ tan −1 c = ⇒
2
1
c = tan ≈ 0.5463
2
43. Substitution; Formula 65
∫ ln( x + 1) dx = ∫ ln u du =
F 65
u = x +1
du = dx
( x + 1) [ ln( x + 1) − 1] . Thus
∫0 ln( x + 1) dx =( x + 1) [ln( x + 1) − 1]0 =
c
c
(c + 1) ln(c + 1) − c and
(c + 1) ln(c + 1) − c = 1 ⇒ ln(c + 1) = 1 ⇒
c + 1 = e ⇒ c = e − 1 ≈ 1.71828
44. Substitution ; Formula 3
c
x
1 c 2 +1 1
dx
du =
=
∫0 x2 + 1
2 ∫1
u
u = x 2 +1
du = 2 x dx
2
1
1
[ln u ]1c +1 = ln(c 2 + 1)
2
2
1
2
ln(c + 1) = 1 ⇒ c 2 + 1 = e2 ⇒
2
c = e2 − 1 ≈ 2.528
462
Section 7.6
Instructor’s Resource Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
45. There is no antiderivative that can be expressed
in terms of elementary functions; an
approximation for the integral, as well as a
process such as Newton’s Method, must be used.
Several approaches are possible. c ≈ 0.59601
46. Integration by parts; partial fractions; Formula 17
a.
∫ ln( x
3
+ 1) dx = x ln( x3 + 1) − 3∫
u = ln( x3 +1)
du =
3x
x3
x3 + 1
3
+ 1) dx =
⎡ x ln( x3 + 1) − 3x + ln( x + 1) −
⎢
⎢ 1 ln( x 2 − x + 1) + 3 tan −1 2 ( x − 1 )
⎢⎣ 2
2
3
(
1 ⎞
⎛
x ln( x3 + 1) − 3∫ ⎜1 −
⎟ dx =
3
⎝ x +1⎠
⎛
⎞
1
x ln( x3 + 1) − 3x + 3∫ ⎜
dx
⎜ ( x + 1)( x 2 − x + 1) ⎟⎟
⎝
⎠
1
( x + 1)( x − x + 1)
2
=
( x + 1)( x 2 − x + 1)
A + C = 1 B + C = A A = −B ⇒
1
1
2
A=
B=−
C= .
3
3
3
Therefore
1
3∫
dx =
( x + 1)( x 2 − x + 1)
1
x−2
∫ x + 1 dx − ∫ x2 − x + 1 dx =
x−2
ln x + 1 − ∫
dx =
1
3
( x − )2 +
2
1
2
du = dx
and G ′(c) = ln(c3 + 1) we get
n
an
⇒
3
2
2 3
u +
4
c
3
4
5
3
+ 1) dx = 1 ⇒ c ≈ 1.6615
47. There is no antiderivative that can be expressed
in terms of elementary functions; an
approximation for the integral, as well as a
process such as Newton’s Method, must be used.
Several approaches are possible. c ≈ 0.16668
48. There is no antiderivative that can be expressed
in terms of elementary functions; an
approximation for the integral, as well as a
process such as Newton’s Method, must be used.
Several approaches are possible. c ≈ 0.2509
du =
F 17
(
2
2.0000 1.6976 1.6621 1.6615 1.6615
∫0 ln( x
4
1
ln x + 1 − ln x 2 − x + 1 + 3 tan −1
2
1
Therefore
u = x−
u−
)
)
(
A
Bx + C
+
=
x + 1 x2 − x + 1
( A + B) x 2 + ( B + C − A) x + ( A + C )
c
⎤
⎥ =
⎥
⎥⎦ 0
)
(
2
ln x + 1 − ∫
c
∫0 ln( x
⎧
⎛
c +1 ⎞ ⎫
⎪c(ln(c3 + 1) − 3) + ln ⎜
⎟ +⎪
⎜ 2
⎟ ⎪
⎪
c
c
1
−
+
⎝
⎠ ⎬
⎨
⎪
⎪
3π
−1 2
(c − 1 ) +
⎪ 3 tan
⎪
2
3
6
⎩
⎭
Using Newton’s Method , with
⎧
⎛
c +1 ⎞ ⎫
⎪c(ln(c3 + 1) − 3) + ln ⎜
⎟ +⎪
⎜ 2
⎟ ⎪
⎪
−
+
c
c
1
⎝
⎠ ⎬
G (c ) = ⎨
⎪
⎪
3π
−1 2
−1
(c − 1 ) +
⎪ 3 tan
⎪
2
3
6
⎩
⎭
dx =
x3 +1
dv = dx, v = x
b.
c. Summarizing
2
( x− 1 )
2
3
)
49. There is no antiderivative that can be expressed
in terms of elementary functions; an
approximation for the integral, as well as a
process such as Newton’s Method, must be used.
Several approaches are possible. c ≈ 9.2365
50. There is no antiderivative that can be expressed
in terms of elementary functions; an
approximation for the integral, as well as a
process such as Newton’s Method, must be used.
Several approaches are possible. c ≈ 1.96
Instructor’s Resource Manual
Section 7.6
463
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
51.
f ( x) = 8 − x g ( x) = cx a = 0 b =
a.
b
∫a
x( f ( x) − g ( x)) dx = ∫
8
0
8
c +1
53.
c +1 8 x − (c + 1) x 2
dx =
f ( x ) = 6e
a.
8
− x3
g ( x) = 0 a = 0 b = c
b
c
∫a x( f ( x) − g ( x)) dx = 6∫0 u =xex
⎡ 2 ⎛ c + 1 ⎞ 3 ⎤ c +1
256
512
=
−
=
⎢4x − ⎜ 3 ⎟ x ⎥
2
⎝
⎠ ⎦0
(c + 1)
3(c + 1)2
⎣
256
c +1
∫0
−x
v = −3e
( f ( x) − g ( x)) dx = ∫
8
0
c +1 8 − (c + 1) x
dx =
c
−x
−c
⎡
3 ( x + 3) ⎤ = −18e 3 (c + 3) + 54
⎢ −18e
⎥
⎣
⎦0
b.
32
(c + 1)
c.
b
c
2
dx =
c
3
dx =
For notational convenience, let
u
1
c=
e
−c
−1 ⇒
3
1
=e
c +1
−c
3
Let
c
∫a ( f ( x) − g ( x)) dx = ∫0 (c − x) dx =
h (c ) =
c
⎡
x2 ⎤
c2
⎢ cx − ⎥ =
2 ⎥⎦
2
⎣⎢
0
⎛ c3 ⎞ ⎛ 2 ⎞ c
x = ⎜ ⎟⎜ ⎟ =
⎜ 6 ⎟ ⎝ c2 ⎠ 3
⎝ ⎠
x =2⇒c =6
−x
−c
⎡ cx 2 x3 ⎤
c3
− ⎥ =
⎢
3 ⎥⎦
6
⎢⎣ 2
0
b
c
u = −18e 3 ; then
u (c + 3) + 54
cu
3(u + 18)
=
+
=
x=
u + 18
u + 18
u + 18
cu
+3
u + 18
cu
c
= −1 ⇒
= −1 ⇒
x =2⇒
18
u + 18
1+
52. f ( x) = c g ( x) = x a = 0 b = c
∫a x( f ( x) − g ( x)) dx = ∫0 cx − x
b
∫a ( f ( x) − g ( x)) dx = ∫0 6e
⎛ −c
⎞
−18 ⎜ e 3 − 1⎟
⎝
⎠
⎛ 256 ⎞ ⎛ c + 1 ⎞
8
c. x = ⎜
=
⎜ 3(c + 1)2 ⎟⎟ ⎝⎜ 32 ⎠⎟ 3(c + 1)
⎝
⎠
8
1
x =2⇒
=2⇒c=
3(c + 1)
3
c.
3
c
8
b.
−x
c −x
−x ⎤
⎡
6 ⎢ −3 xe 3 ⎥ + 18∫ e 3 dx =
0
⎣
⎦0
⎡
64
32
⎛ c + 1 ⎞ 2 ⎤ c +1
=
−
=
⎢8 x − ⎜ 2 ⎟ x ⎥
(c + 1) (c + 1)
⎝
⎠ ⎦0
⎣
a.
dx =
dv = e 3
du = dx
3(c + 1)2
b.
− x3
−c
1
−e 3 ,
c +1
1 −c
1
h′(c ) = e 3 −
3
(c + 1)2
and apply Newton’s Method
n
an
1
2
3
4
5
6
2.0000 5.0000 5.6313 5.7103 5.7114 5.7114
c ≈ 5.7114
54.
⎛πx ⎞
f ( x) = c sin ⎜
⎟ g ( x) = x a = 0 b = c
⎝ 2c ⎠
(Note: the value for b is obtained by setting
x
⎛πx ⎞
c sin ⎜
⎟ = x This requires that be a zero for
c
⎝ 2c ⎠
⎛π ⎞
the function h(u ) = u − sin ⎜ u ⎟ . Applying
⎝2 ⎠
Newton’s Method to h we discover that the zeros
of h are -1, 0, and 1. Since we are dealing with
x
positive values, we conclude that =1 or x = c.)
c
464
Section 7.6
Instructor’s Resource Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
a.
⎛πx⎞
c⎡
b
∫a x( f ( x) − g ( x)) dx = ∫0 ⎢⎣cx sin ⎜⎝ 2c ⎟⎠ − x
2⎤
⎥ dx
⎦
c
⎡ x3 ⎤
⎛πx ⎞
= ∫ cx sin ⎜
−
dx
⎢ ⎥
⎟
0
⎝ 2c ⎠
⎢⎣ 3 ⎦⎥ 0
π
π
c
u=
=∫
π
0
=
F 40
2c
x , du =
2c
dx
2 c ⎛ 2c u ⎞ sin u ⎛ 2c ⎞ du −
⎜
⎝π
⎟
⎠
⎜ ⎟
⎝π ⎠
π
4c3
π
57. a. (See problem 55 a.) . Since erf ′( x) > 0 for all
x , erf ( x) is increasing on (0, ∞) .
⎡ c3 ⎤
⎢ ⎥
⎢⎣ 3 ⎥⎦
⎡ c 3 ⎤ 4c 3 c 3
⎥= 2 −
3
⎢⎣ 3 ⎥⎦ π
c⎡
b
⎛πx⎞
⎤
2 c
⎡ 2c 2
2c 2 c 2
⎛πx ⎞ x ⎤
−
−
−
=
cos
⎢
⎥ =
⎜
⎟
π
2
⎝ 2c ⎠ 2 ⎦⎥ 0
⎣⎢ π
⎛ 2 1⎞
c2 ⎜ − ⎟
⎝π 2⎠
⎛ 12 − π 2 ⎞
c3 ⎜
⎜ 3π 2 ⎟⎟
2
⎝
⎠ = c ⎡ 2(12 − π ) ⎤
c. x =
⎢
⎥
⎛ 4 −π ⎞
⎢⎣ 3π (4 − π ) ⎥⎦
c2 ⎜
⎟
⎝ 2π ⎠
3π (4 − π )
≈ 3.798
x =2⇒c=
12 − π 2
2
dt
d
2 − x2
erf ( x) =
e
dx
π
b. Si ( x) = ∫
x sin t
0
∴
x −t 2
e
π ∫0
55. a. erf ( x) =
t
dt
sin x
d
Si ( x) =
dx
x
⎛ π t2 ⎞
x
56. a. S ( x) = ∫ sin ⎜
⎟ dt
⎜ 2 ⎟
0
⎝
⎠
⎛ π x2 ⎞
d
∴ S ( x) = sin ⎜
⎟
⎜ 2 ⎟
dx
⎝
⎠
Instructor’s Resource Manual
−4 x
π
2
e− x which is negative on
(0, ∞) , so erf ( x) is not concave up
anywhere on the interval.
58. a.
∫a ( f ( x) − g ( x)) dx = ∫0 ⎢⎣c sin ⎜⎝ 2c ⎟⎠ − x ⎥⎦ dx =
∴
b. erf ′′( x) =
[sin u − u cos u ]0 2 − ⎢
2
⎛ 4 1⎞
= c3 ⎜
− ⎟
⎝π2 3⎠
b.
⎛ πt2 ⎞
x
b. C ( x) = ∫ cos ⎜
⎟ dt
⎜ 2 ⎟
0
⎝
⎠
⎛ π x2 ⎞
d
∴ C ( x) = cos ⎜
⎟
⎜ 2 ⎟
dx
⎝
⎠
(See problem 56 a.) Since
⎛π
⎞
S ′( x) = sin ⎜ x 2 ⎟ , S ′( x) > 0 when
2
⎝
⎠
0<
π
2
x 2 < π or 0 < x 2 < 2; thus
S ( x) is increasing on
( 0, 2 ) .
⎛π
⎞
b. Since S ′′( x) = π x cos ⎜ x 2 ⎟ , S ′′( x) > 0
⎝2 ⎠
when
π
π
3π π 2
< x < 2π ,
0 < x2 <
and
2
2
2
2
or 0 < x 2 < 1 and 3 < x 2 < 4.
Thus S ( x) is concave up on
(0,1) ∪ ( 3, 2) .
59. a. (See problem 56 b.) Since
⎛π
⎞
C ′( x) = cos ⎜ x 2 ⎟ , C ′( x) > 0 when
⎝2 ⎠
π
π
3π π 2
0 < x2 <
or
< x < 2π ; thus
2
2
2
2
C ( x) is increasing on (0,1) ∪ ( 3, 2) .
⎛π
⎞
b. Since C ′′( x) = −π x sin ⎜ x 2 ⎟ , C ′′( x) > 0
⎝2 ⎠
when π <
π
x 2 < 2π . Thus C ( x) is concave
2
up on ( 2, 2) .
60. From problem 58 we know that S ( x) is concave
up on (0,1) and concave down on (1, 3) so the
first point of inflection occurs at x = 1 . Now
1
⎛π ⎞
S (1) = ∫ sin ⎜ t 2 ⎟ dt . Since the integral cannot
0
⎝2 ⎠
be integrated directly, we must use some
approximation method. Methods may vary but
the result will be S (1) ≈ 0.43826 . Thus the first
point of inflection is (1, 0.43826)
Section 7.6
465
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
x2
17. False:
7.7 Chapter Review
x –1
Concepts Test
1. True:
The resulting integrand will be of the
form sin u.
2. True:
The resulting integrand will be of the
1
form
.
2
a + u2
3. False:
18. True:
19. True:
20. False:
Try the substitution
u = x 4 , du = 4 x3 dx
4. False:
= 1+
2
Use the substitution u = x 2 – 3 x + 5,
du = (2x – 3)dx.
1
1
−
2( x − 1) 2( x + 1)
x2 + 2
2
3
3
=− +
+
x 2( x + 1) 2( x − 1)
x( x − 1)
2
x2 + 2
x( x + 1)
2
=
2
–x
+
x x2 + 1
x+2
x ( x 2 − 1)
1 2
3
1
=− −
+
−
x x 2 2( x − 1) 2( x + 1)
2
b2
.
4a
21. False:
To complete the square, add
22. False:
The resulting integrand will be of the
1
form
.
2
a − x2
Polynomials can be factored into
products of linear and quadratic
polynomials with real coefficients.
23. True:
Polynomials with the same values for
all x will have identical coefficients
for like degree terms.
7. True:
This integral is most easily solved
with a partial fraction decomposition.
24. True:
8. False:
This improper fraction should be
reduced first, then a partial fraction
decomposition can be used.
Let u = 2 x ; then du = 2dx and
1 2
2
2
2
∫ x 25 − 4 x dx = 8 ∫ u 25 − u du
which can be evaluated using
Formula 57.
9. True:
Because both exponents are even
positive integers, half-angle formulas
are used.
25. False:
It can, however, be solved by the
5. True:
6. True:
10. False:
11. False:
The resulting integrand will be of the
1
form
.
2
a + u2
substitution u = 25 − 4 x 2 ; then
du = −8 x dx and
Use the substitution
u = 1 + e x , du = e x dx
∫x
Use the substitution
−
u = – x 2 – 4 x, du = (−2 x − 4)dx
26. True:
12. True:
13. True:
14. True:
This substitution eliminates the
radical.
Then expand and use the substitution
u = sin x, du = cos x dx
The trigonometric substitution
x = 3sin t will eliminate the radical.
dv = x dx
1
v = x3
3
15. True:
Let u = ln x
1
du = dx
x
16. False:
Use a product identity.
466
Section 7.7
2
25 − 4 x 2 dx = −
1
u du =
8∫
3
1
(25 − 4 x 2 ) 2 + C
12
Since (see Section 7.6, prob 55 a.)
2 − x2
erf ′( x) =
e
> 0 for all x ,
π
erf ( x) is an increasing function.
27. True:
by the First Fundamental Theorem of
Calculus.
28. False:
Since (see Section 7.6, prob 55 b.)
sin x
Si ′( x) =
, which is negative on,
x
say, (π , 2π ) , Si ( x) will be decreasing
on that same interval.
Instructor’s Resource Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Sample Test Problems
10.
4
dt = ⎡ 9 + t 2 ⎤ = 5 − 3 = 2
⎢⎣
⎥⎦ 0
2
9+t
t
4
1.
∫0
2.
2
∫ cot (2θ )dθ = ∫
=∫
cos 2 2θ
sin 2 2θ
1 − sin 2 2θ
= ∫ (cos x + csc x − sin x)dx
dθ
= sin x + ln csc x − cot x + cos x + C
dθ = ∫ (csc 2θ − 1)dθ
sin 2 2θ
1
= − cot 2θ − θ + C
2
3.
4.
∫
=
(Use Formula 15 for ∫ csc x dx .)
π/2
⎡ −ecos x ⎤
⎣
⎦0
11.
= e − 1 ≈ 1.718
12.
π/4
⎡ sin 2 x x
⎤
x sin 2 x dx = ⎢
− cos 2 x ⎥
2
⎣ 4
⎦0
(Use integration by parts with u = x,
dv = sin 2 x dx .)
π/4
∫0
=
1
4
∫x
∫
3
2
∫ sin (2t )dt = ∫ [1 – cos (2t )]sin(2t )dt
y–2
1
2y – 4
dy = ∫
dy
7. ∫
2
2
2 y – 4y + 2
y – 4y + 2
1
= ln y 2 – 4 y + 2 + C
2
8.
∫0
9.
∫ et − 2 dt = e
2y +1
e 2t
= ⎡⎣ 2 y + 1 ⎤⎦
t
3/ 2
0
14.
(Use the substitution u = et − 2 ,
du = et dt
u+2
⎞
du. ⎟
which gives the integral ∫
u
⎠
Instructor’s Resource Manual
2
tan t , dy =
3
dy
2 + 3y2
1
=
1
=
1
=
1
3
3
3
ln
ln
15.
w3
sec2 t
y 2 + 23
2
3
1
3
ln sec t + tan t + C1
y
+
2
3
y 2 + 23 + y
2
3
y2 +
dt
2 sec t
∫ sec t dt =
3
ln
2
sec2 t dt
3
2
3
=∫
+ C1
+ C1
2
+ y +C
3
Note that tan t =
= 2 −1 = 1
+ 2 ln et − 2 + C
e dx = e x (2 − 2 x + x 2 ) + C
2 x
=
1
1
= – cos(2t ) + cos3 (2t ) + C
2
6
dy
⎛ x −1 ⎞
sin –1 ⎜
⎟+C
2
⎝ 3 ⎠
1
=
16 + 4 x – 2 x
(Complete the square.)
2
13. y =
⎛
y +y
2 ⎞
5. ∫
dy = ∫ ⎜ y 2 − y + 2 −
⎟ dy
1+ y ⎠
y +1
⎝
1
1
= y 3 − y 2 + 2 y − 2 ln 1 + y + C
3
2
3/ 2
dx
∫
Use integration by parts twice.
3
6.
⎛
sin x + cos x
cos 2 x ⎞
dx = ∫ ⎜ cos x +
⎟ dx
⎜
tan x
sin x ⎟⎠
⎝
⎛
1 − sin 2 x ⎞
= ∫ ⎜ cos x +
⎟ dx
⎜
sin x ⎟⎠
⎝
2
π / 2 cos x
e
sin x dx
0
∫
1
y
2
3
, so sec t =
y 2 + 23
2
3
.
1
– ln 1 – w2 + C
2
Divide the numerator by the denominator.
∫ 1 – w2 dw = – 2 w
tan x
2
∫ ln cos x dx = – ln ln cos x
+C
Use the substitution u = ln cos x .
Section 7.7
467
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16.
3dt
t+2
1
22. u = ln( y 2 + 9)
∫ t 3 – 1 = ∫ t − 1 dt − ∫ t 2 + t + 1 dt
(
18. u = ln y, du =
∫
1
dy
y
5
23.
24.
(ln y )
1
dy = ∫ u 5 du = (ln y )6 + C
y
6
19. u = x
du = dx
∫ x cot
2
1
= – x cot x – x 2 + ln sin x + C
2
25.
Use cot 2 x = csc2 x − 1 for ∫ cot 2 x dx.
26.
20. u = x , du =
∫
sin x
x
1 −1/ 2
x
dx
2
dx = 2 ∫ sin u du
= −2 cos x + C
21. u = ln t 2 , du =
2
dt
t
ln t 2
[ln(t 2 )]2
dt
=
+C
∫ t
4
v=y
+ 9)dy = y ln( y 2 + 9) – ∫
2 y2
y2 + 9
dy
−3et / 3 (9 cos 3t − sin 3t )
+C
82
Use integration by parts twice.
t /3
∫ e sin 3t dt =
t +9
–t + 1
1
∫ t 3 + 9t dt = ∫ t dt + ∫ t 2 + 9 dt
1
t
1
dt + ∫
dt
= ∫ dt − ∫
2
2
t
t +9
t +9
1
1
⎛t⎞
= ln t – ln t 2 + 9 + tan –1 ⎜ ⎟ + C
2
3
⎝3⎠
dv = cot 2 x dx
v = –cot x – x
x dx = – x cot x – x 2 – ∫ (– cot x – x)dx
2
dy
⎛
18 ⎞
= y ln( y 2 + 9) − ∫ ⎜ 2 −
⎟ dy
2
⎜
y + 9 ⎟⎠
⎝
⎛ y⎞
= y ln( y 2 + 9) – 2 y + 6 tan –1 ⎜ ⎟ + C
⎝3⎠
)
∫ sinh x dx = cosh x + C
y +9
2
∫ ln( y
1
⎛ 2t + 1 ⎞
= ln t − 1 − ln t 2 + t + 1 − 3 tan −1 ⎜
⎟+C
2
⎝ 3 ⎠
17.
2y
du =
1
1 2t + 4
dt − ∫
dt
=∫
2 t2 + t +1
t −1
1
1 2t + 1 + 3
dt − ∫
dt
=∫
2 t2 + t +1
t −1
1
1 2t + 1
3
1
dt − ∫
dt − ∫
dt
=∫
t −1
2 t2 + t +1
2 t+1 2+3
2
4
dv = dy
27.
3x
x
cos x cos 2 x
−
+C
cos dx = −
2
2
2
4
Use a product identity.
∫ sin
2
x⎞
⎛ 1 + cos x ⎞
⎜ ⎟ dx = ∫ ⎜
⎟ dx
2
⎝2⎠
⎝
⎠
1
1
1
= ∫ dx + ∫ 2 cos x dx + ∫ cos 2 x dx
4
4
4
1
1
1
= ∫ dx + ∫ cos x dx + ∫ (1 + cos 2 x)dx
4
2
8
3
1
1
= x + sin x + sin 2 x + C
8
2
16
∫ cos
∫ tan
4⎛
3
2 x sec 2 x dx =
1
(sec2 2 x – 1) d (sec 2 x)
2∫
1
1
= sec3 (2 x) – sec(2 x) + C
6
2
28. u = x , du =
1
2 x
dx
u2
2x ⎛ 1
⎞
=
2
dx
⎜
⎟
∫ 1+ x
∫ 1 + u du
1+ x ⎝ 2 x ⎠
(u + 1)(u − 1) + 1
1 ⎞
⎛
= 2∫
du = 2 ∫ ⎜ u − 1 +
⎟ du
u +1
u +1⎠
⎝
x
dx = ∫
⎛ u2
⎞
= 2⎜
− u + ln u + 1 ⎟ + C
⎜ 2
⎟
⎝
⎠
(
)
= x − 2 x + 2 ln 1 + x + C
468
Section 7.7
Instructor’s Resource Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
29.
∫ tan
=
3/ 2
x sec4 x dx = ∫ tan 3 / 2 x(1 + tan 2 x) sec2 x dx = ∫ tan 3 / 2 x sec2 x dx + ∫ tan 7 / 2 x sec2 x dx
2 5/ 2
2
x + tan 9 / 2 x + C
tan
5
9
30. u = t1/ 6 + 1, (u – 1)6 = t , 6(u − 1)5 du = dt
dt
∫ t (t1/ 6 + 1)
=∫
6(u − 1)5 du
(u − 1) u
6
=∫
6du
1
1
= –6∫ du + 6 ∫
du = –6 ln t1/ 6 + 1 + 6 ln t1/ 6 + C
u
u −1
u (u – 1)
31. u = 9 − e2 y , du = −2e 2 y dy
∫
32.
e2 y
9−e
∫ cos
5
2y
dy = −
1 −1/ 2
u
du = − u + C = − 9 − e2 y + C
2∫
x sin xdx = ∫ (1 – sin 2 x) 2 (sin1/ 2 x) cos x dx = ∫ sin1/ 2 x cos x dx – 2∫ sin 5 / 2 x cos x dx + ∫ sin 9 / 2 x cos x dx
2
4
2
= sin 3 / 2 x – sin 7 / 2 x + sin11/ 2 x + C
3
7
11
33.
∫e
ln(3cos x )
dx = ∫ 3cos x dx = 3sin x + C
34. y = 3 sin t, dy = 3 cos t dt
∫
9 − y2
3cos t
dy = ∫
⋅ 3cos t dt
y
3sin t
1 − sin 2 t
= 3∫ (csc t − sin t ) dt
sin t
= 3 ⎡⎣ln csc t − cot t + cos t ⎤⎦ + C
= 3∫
9− y
3
= 3ln −
y
y
Note that sin t =
cot t =
2
+ 9− y +C
2
y
3
, so csc t = and
3
y
2
9– y
.
y
1
x2 + a2
∫
dx = ∫
du
∫ 1 + e8 x dx = 4 ∫ 1 + u 2
=
1
tan −1 (e 4 x ) + C
4
a sec t
a sec2 t dt
a tan 4 t
1 sec3 t
1 cos t
=
dt =
dt
∫
2 ∫
4
a tan t
a 2 sin 4 t
1 ⎛ 1 1 ⎞
1
=
+C = –
–
csc3 t + C
2⎜ 3
3 ⎟
2
3a
a ⎝
sin t ⎠
x
=–
4
4
1 ( x 2 + a 2 )3 / 2
3a 2
x3
Note that tan t =
35. u = e4x , du = 4e4 x dx
e4 x
36. x = a tan t, dx = a sec 2 t dt
+C
x
, so csc t =
a
x2 + a2
.
x
37. u = w + 5, u 2 = w + 5, 2u du = dw
w
2 3
2
∫ w + 5 dw = 2∫ (u – 5)du = 3 u – 10u + C
2
= ( w + 5)3 / 2 –10( w + 5)1/ 2 + C
3
38. u = 1 + cos t, du = –sin t dt
sin t dt
du
∫ 1 + cos t = – ∫ u = –2 1 + cos t + C
39. u = cos 2 y, du = –2 cos y sin y dy
sin y cos y
1
du
∫ 9 + cos4 y dy = – 2 ∫ 9 + u 2
⎛ cos 2 y ⎞
1
= – tan –1 ⎜
⎟+C
⎜ 3 ⎟
6
⎝
⎠
Instructor’s Resource Manual
Section 7.7
469
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
40.
∫
dx
1 – 6x – x
dx
=∫
2
10 – ( x + 3)2
⎛ x+3⎞
= sin –1 ⎜
⎟+C
⎝ 10 ⎠
41.
4 x2 + 3x + 6
A B Cx + D
+
+
x x2 x2 + 3
x ( x + 3)
A = 1, B = 2, C = –1, D = 2
4 x2 + 3x + 6
1
1
–x + 2
∫ x2 ( x2 + 3) dx = ∫ x dx + 2∫ x2 dx + ∫ x2 + 3 dx
2
2
=
1
1
1
2x
1
dx + 2∫
dx
= ∫ dx + 2 ∫ dx − ∫
2
2
2
2 x +3
x
x
x +3
2 1
2
⎛ x ⎞
– ln x 2 + 3 +
tan –1 ⎜
⎟+C
x 2
3
⎝ 3⎠
= ln x –
42. x = 4 tan t, dx = 4sec 2 t dt
1
dx
1
∫ (16 + x2 )3 / 2 = 16 ∫ cos t dt = 16 sin t + C
43. a.
3 – 4 x2
(2 x + 1)
3
=
2
( x – 1) (2 – x)
c.
d.
e.
f.
44. a.
470
( x + x + 10)
=
3
3x + 1
2
=
2
A
B
C
D
E
+
+
+
+
2
2
2 – x (2 – x)
x – 1 ( x – 1)
(2 – x)3
Ax + B
x + x + 10
2
( x + 1) 2
( x – x + 10) (1 – x )
2
2
2 2
x5
( x + 3) ( x + 2 x + 10)
4
2
(3 x 2 + 2 x –1)2
(2 x + x + 10)
2
2⎡
⎞
1⎛
x
x
⎜
⎟+C =
+C
16 ⎜⎝ x 2 + 16 ⎟⎠
16 x 2 + 16
A
B
C
+
+
2 x + 1 (2 x + 1) 2 (2 x + 1)3
7 x – 41
b.
=
3
=
2
+
( x + x + 10)2
=
A
B
C
D
Ex + F
Gx + H
+
+
+
+
+
2
2
2
2
1 – x (1 – x)
1 + x (1 + x)
x – x + 10 ( x – x + 10)2
=
A
B
C
D
Ex + F
Gx + H
+
+
+
+
+
x + 3 ( x + 3)2 ( x + 3)3 ( x + 3)4 x 2 + 2 x + 10 ( x 2 + 2 x + 10) 2
Ax + B
2 x + x + 10
2
Cx + D
2
+
Cx + D
(2 x + x + 10)
2
2
+
Ex + F
(2 x + x + 10)3
2
2
⎤
2
1
V = π∫ ⎢
dx
⎥ dx = π∫
1 ⎢
1 3x – x2
2⎥
⎣ 3x – x ⎦
1
A
B
= +
2
x
3
–x
3x – x
1
1
A= ,B=
3
3
21⎛1
π
2π
π
1 ⎞
2
ln 2 ≈ 1.4517
V = π∫ ⎜ +
dx = ⎡⎣ln x – ln 3 – x ⎤⎦ = (ln 2 + ln 2) =
1
1 3 ⎝ x 3 – x ⎟⎠
3
3
3
Section 7.7
1
Instructor’s Resource Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
b.
V = 2π ∫
x
2
1
3x – x
2
dx = −π ∫
2 −2 x + 3 − 3
1
3x − x
2
2
= −π ⎡ 2 3 x − x 2 ⎤ + 3π ∫
⎢⎣
⎥⎦1
1
2
3 − 2x
3x − x
2
2
1
1
3x − x2
dx + 3π ∫
dx
2
1
9
4
dx = – π ∫
1
2
(
− x − 23
)
2
⎡
⎛ 2 x − 3 ⎞⎤
dx = ⎢ −2π 3x − x 2 + 3π sin −1 ⎜
⎟⎥
⎝ 3 ⎠ ⎦1
⎣
1
1
⎛ 1⎞
= −2π 2 + 3π sin −1 + 2π 2 − 3π sin −1 ⎜ − ⎟ = 6π sin −1 ≈ 6.4058
3
3
⎝ 3⎠
45. y =
x2
x
, y′ =
16
8
L=∫
2
4
x2
⎛ x⎞
1 + ⎜ ⎟ dx = ∫ 1 + dx
0
64
⎝8⎠
4
0
x = 8 tan t, dx = 8sec2 t
tan –1 1
2 sec t ⋅ 8sec 2 t dt
0
L=∫
tan –1 1
2 sec3 t dt
0
= 8∫
= 4 ⎡⎣sec t tan t + ln sec t + tan t ⎤⎦
⎡⎛ 5 ⎞ ⎛ 1 ⎞
⎛ 1+ 5 ⎞
1
5⎤
= 4 ⎢⎜⎜
⎥ = 5 + 4 ln ⎜⎜
⎟⎟ ⎜ ⎟ + ln +
⎟⎟ ≈ 4.1609
2 2 ⎥⎦
⎝ 2 ⎠
⎣⎢⎝ 2 ⎠ ⎝ 2 ⎠
46. V = π∫
1
3
dx = π∫
3
tan –1 1
2
0
Note: Use Formula 28 for ∫ sec3 t dt.
1
dx
( x + 5 x + 6)
( x + 3) ( x + 2) 2
1
A
B
C
D
=
+
+
+
2
2
2
3
2
x
x
+
+
( x + 3) ( x + 2)
( x + 3)
( x + 2)2
2
0
2
0
2
A = 2, B = 1, C = –2, D = 1
3
3⎡ 2
1
2
1 ⎤
1
1 ⎤
⎡
–
– 2 ln x + 2 –
V = π∫ ⎢
dx = π ⎢ 2 ln x + 3 –
+
+
⎥
0 x + 3 ( x + 3) 2
x + 2 ( x + 2)2 ⎥⎦
x+3
x + 2 ⎥⎦ 0
⎣
⎢⎣
⎡⎛
1
1⎞ ⎛
1
1 ⎞⎤
4⎞
⎛7
= π ⎢⎜ 2 ln 6 – – 2 ln 5 – ⎟ – ⎜ 2 ln 3 – – 2 ln 2 – ⎟ ⎥ = π ⎜ + 2 ln ⎟ ≈ 0.06402
5⎠
6
5⎠ ⎝
3
2 ⎠⎦
⎝ 15
⎣⎝
47. V = 2π∫
x
3
dx
x + 5x + 6
x
A
B
=
+
2
2
x
x
+
+3
x + 5x + 6
A = –2, B = 3
3⎡
2
3 ⎤
3
V = 2π ∫ ⎢ –
dx = 2π [ –2 ln( x + 2) + 3ln( x + 3) ]0
+
0 ⎣ x + 2 x + 3 ⎥⎦
2⎞
32
⎛
= 2π[(–2 ln 5 + 3ln 6) – (–2 ln 2 + 3ln 3)] = 2π ⎜ 3ln 2 + 2 ln ⎟ = 2π ln
≈ 1.5511
5⎠
25
⎝
0
2
2
48. V = 2π∫ 4 x 2 2 – xdx
0
u=2–x
x=2–u
du = –dx
dx = –du
2
0
2
8
2
⎡8
⎤
V = 2π ∫ 4(2 – u )2 u (– du ) = 8π∫ (4u1/ 2 – 4u 3 / 2 + u 5 / 2 )du = 8π ⎢ u 3 / 2 – u 5 / 2 + u 7 / 2 ⎥
0
2
5
7
⎣3
⎦0
⎛ 16 2 32 2 16 2 ⎞
⎛ 128 2 ⎞ 1024 2π
= 8π ⎜⎜
–
+
≈ 43.3287
⎟⎟ = 8π ⎜⎜
⎟⎟ =
5
7 ⎠
105
⎝ 3
⎝ 105 ⎠
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49. V = 2π∫
ln 3
0
Note that
2(e x –1)(ln 3 – x)dx = 4π∫
∫ xe
x
ln 3
0
x
[(ln 3)e x – xe x – ln 3 + x]dx
dx = xe x – ∫ e x dx = xe – e x + C by using integration by parts.
ln 3
⎡⎛
1
⎤
⎞
= 4π ⎢⎜ 3ln 3 – 3ln 3 + 3 – (ln 3) 2 + (ln 3) 2 ⎟ – (ln 3 + 1) ⎥
2
⎠
⎣⎝
⎦
1 ⎤
⎡
V = 4π ⎢ (ln 3)e x – xe x + e x – (ln 3) x + x 2 ⎥
2 ⎦0
⎣
1
⎡
⎤
= 4π ⎢ 2 – ln 3 – (ln 3) 2 ⎥ ≈ 3.7437
2
⎣
⎦
18
3 3
50. A = ∫
3
x
x2 + 9
2
dx
x = 3 tan t, dx = 3sec2 t dt
π/3
18
π/6
27 tan 2 t sec t
A=∫
51. A = – ∫
0
π/3
1 ⎞
⎛ 2
⎞
⎛
⎡ 1 ⎤
= 2⎜ –
+ 2 ⎟ = 4 ⎜1 –
dt = 2 ⎢ –
⎟ ≈ 1.6906
⎥
π / 6 sin 2 t
3
3⎠
⎣ sin t ⎦ π / 6
⎝
⎠
⎝
3sec2 t dt = 2 ∫
π/3
cos t
t
dt
–1)2
t
A
B
=
+
2
(t –1) (t –1)2
(t –1)
A = 1, B = 1
0
0 ⎡ 1
1 ⎤
1 ⎞⎤
6
⎡
1 ⎤
⎛
⎡
A = –∫ ⎢
+
dt = – ⎢ ln t –1 –
⎥
⎥ = – ⎢(0 + 1) – ⎜ ln 7 + 7 ⎟ ⎥ = ln 7 – 7 ≈ 1.0888
–6 t –1 (t –1) 2
t
–1
⎣
⎦
⎝
⎠
⎣
⎦
–6
⎣⎢
⎦⎥
–6 (t
52.
2
–1 ⎛
–1
⎞
6
36
V = π∫ ⎜
dx = π∫
dx
–3 ⎝ x x + 4 ⎟⎠
–3 x 2 ( x + 4)
36
A B
C
= +
+
2
2
x+4
x ( x + 4) x x
9
9
A = – , B = 9, C =
4
4
–1
9
9
9 ⎤
9π –1 ⎛ 1 4
1 ⎞
9π ⎡
4
⎤
–
dx =
– +
+
+
+
⎟ dx =
∫
⎢ – ln x – x + ln x + 4 ⎥
–3 ⎢ 4 x x 2 4( x + 4) ⎥
–3 ⎜⎝ x x 2
4
x
4
4
+
⎣
⎦ –3
⎠
⎣
⎦
9π ⎡
4
⎤
9
π
8
3
π
⎛
⎞
⎛
⎞
=
(4 + ln 3) – ⎜ – ln 3 + ⎟ ⎥ =
(4 + 3ln 3) ≈ 34.3808
⎜ + 2 ln 3 ⎟ =
4 ⎝3
4 ⎢⎣
3 ⎠⎦
⎝
⎠ 2
V = π∫
–1 ⎡
53. The length is given by
π/3
∫π / 6
1 + [ f ′( x)]2 dx = ∫
= ⎡⎣ln csc x − cot x ⎤⎦
472
Section 7.7
π/3
π/6
π/3
π/6
= ln
1+
cos 2 x
2
sin x
2
3
−
dx = ∫
π/3
π/6
sin 2 x + cos 2 x
2
sin x
dx = ∫
π/3
π/6
π/3
1
dx = ∫
csc x dx
/6
π
sin x
⎛ 2 3 +3⎞
⎛ 1 ⎞
− ln 2 − 3 = ln ⎜
⎟⎟ ≈ 0.768
⎟ − ln(2 − 3) = ln ⎜⎜
3
⎝ 3⎠
⎝ 3 ⎠
1
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∫
81 − 4 x 2
81 − u 2
dx = ∫
du , then use Formula 55:
x
u
∫
54. a. First substitute u = 2 x, du = 2 dx to obtain
81 − 4 x 2
9 + 81 − 4 x 2
dx = 81 − 4 x 2 − 9 ln
+C
x
2x
(
b. First substitute u = e x , du = e x dx to obtain ∫ e x 9 − e2 x
(
2x
x
∫e 9−e
) 2 dx = e8 ( 45 − 2e2 x )
x
3
9 − e2 x +
) 2 dx = ∫ (9 − u 2 ) 2 du , then use Formula 62:
3
3
243 −1 ⎛ e x
sin ⎜
⎜ 3
8
⎝
⎞
⎟+C
⎟
⎠
55. a. First substitute u = sin x, du = cos x dx to obtain ∫ cos x sin 2 x + 4 dx = ∫ u 2 + 4 du , then use Formula 44:
∫ cos x
sin 2 x + 4 dx =
sin x
sin 2 x + 4 + 2 ln sin x + sin 2 x + 4 + C
2
b. First substitute u = 2 x , du = 2dx to obtain
Then use Formula 18:
∫
∫
1
dx =
1 − 4x
1 2x +1
dx = ln
+C .
4 2x −1
1 − 4 x2
2
1
du
.
∫
2 1− u2
1
56. By the First Fundamental Theorem of Calculus,
⎧ x cos x − sin x
⎧ sin x
x≠0
⎪
⎪
Si ′( x) = ⎨ x
Si ′′( x) = ⎨
x2
⎪⎩ 1
⎪
x=0
0
⎩
for x ≠ 0
for x = 0
57. Using partial fractions (see Section 7.6, prob 46 b.):
1
1
A
Bx + C
( A + B) x 2 + ( B + C − A) x + ( A + C )
=
=
+
=
⇒
1 + x3 ( x + 1)( x 2 − x + 1) x + 1 x 2 − x + 1
( x + 1)( x 2 − x + 1)
A + C = 1 B + C = A A = −B ⇒ A =
1
1
2
B=−
C= .
3
3
3
Therefore:
⎡
⎤
⎢
⎥
⎥
1
1⎡ 1
x−2
x−2
⎤ 1⎢
⎢
dx
dx
dx
ln
x
1
dx
=
−
=
+
−
∫ 1 + x3
∫ x 2 − x + 1 ⎥⎦ 3 ⎢
∫ ( x − 1 )2 + 3 ⎥⎥
3 ⎢⎣ ∫ x + 1
2
4
⎢
⎥
1
u = x − , du = dx
⎢⎣
⎥⎦
2
3
⎡
⎤
⎡
⎤
u−
x +1
1
−1 2
2
⎢
⎥ = 1 ⎢ ln
⎥
du
3
tan
(
x
)
= ln x + 1 − ∫
+
−
3
2 ⎥
3
⎢
⎥ F17 3 ⎢
2
u2 +
x
x
1
−
+
⎣
⎦
4
⎣
⎦
)
(
so
c
1
1⎡
∫0 1 + x3 dx = 3 ⎢⎢ln
⎣
1⎡
Letting G (c) = ⎢ ln
3⎢
⎣
c +1
⎡
+ 3 ⎢ tan −1
2
⎣
c − c +1
c +1
⎡
+ 3 ⎢ tan −1
2
⎣
c − c +1
Method to find the value of c such that
c
(
(
1
2
(c − 1 )
2
3
2
(c − 1 )
2
3
⎤
) + π6 ⎤⎥⎦ ⎥⎥⎦ .
⎤
) + π6 ⎤⎥⎦ ⎥⎥⎦ − 0.5 and G′(c) = 1 +1c
∫0 1 + x3 dx = 0.5
3
we apply Newton’s
:
n
1
2
3
4
5
6
an 1.0000 0.3287 0.5090 0.5165 0.5165 0.5165
Thus c ≈ 0.5165 .
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Review and Preview Problems
1. lim
x2 + 1
−1
x →2 x 2
=
22 + 1
=
2 −1
2
5
3
2 x + 1 2(3) + 1 7
=
=
3+5
8
x →3 x + 5
2. lim
14. Note that, if θ = sec−1 x, then
1
1
se c θ = x ⇒ cos θ = ⇒ θ = cos −1 . Hence
x
x
1
lim sec−1 x = lim cos −1 = 1
x
x →∞
x →∞
15.
f ( x ) = xe− x
y
x2 − 9
( x + 3)( x − 3)
= lim
=
x
−
3
x−3
x →3
x →3
lim ( x + 3) = 3 + 3 = 6
3. lim
2
x →3
x2 − 5x + 6
( x − 2)( x − 3)
= lim
=
x−2
x−2
x→2
x →2
lim ( x − 3) = 2 − 3 = −1
4. lim
5
x→2
10
x
−2
sin 2 x
2sin x cos x
5. lim
= lim
=
x
x →0 x
x →0
⎛ sin x ⎞
lim 2 ⎜
⎟ cos x = 2(1)(1) = 2
x →0 ⎝ x ⎠
We would conjecture lim xe− x = 0 .
x →∞
16.
f ( x ) = x 2 e− x
y
tan 3x
⎛ sin 3 x ⎞ ⎛ 3 ⎞
= lim ⎜
⎟⎜ ⎟ =
x →0 x
x →0 ⎝ cos 3 x ⎠ ⎝ 3 x ⎠
6. lim
⎛ sin 3x ⎞ ⎛ 1 ⎞
lim 3 ⎜
⎟⎜
⎟ = 3(1)(1) = 3
x →0 ⎝ 3 x ⎠ ⎝ cos 3 x ⎠
1+
x +1
2
1
5
x 2 = 1 + 0 = 1 or:
= lim
7. lim
1
x →∞ x 2 − 1 x →∞
1− 2 1− 0
x
2
lim
x2 + 1
x →∞ x 2
−1
= lim 1 +
x →∞
2
x −1
2
1
2x + 1
x = 2+0 = 2
8. lim
= lim
5 1+ 0
x →∞ x + 5
x →∞
1+
x
9.
lim e− x = lim
1
We would conjecture lim x 2 e− x = 0 .
x →∞
17.
f ( x ) = x3e− x
y
5
=0
x →∞ e x
x →∞
5
10.
11.
12.
lim e
− x2
x →∞ x 2
x →∞
lim e
2x
lim e
10
x
=0
e
x →∞
x →− ∞
1
= lim
x
−2
= 1+ 0 = 1
2+
10
−5
= ∞ (has no finite value)
We would conjecture lim x3e− x = 0 .
x →∞
− 2x
=
lim e
2u
(u = − x ) u → ∞
=∞
(has no finite value)
13.
474
lim tan −1 x =
x →∞
π
2
Review and Preview
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18.
f ( x ) = x 4 e− x
23.
y
x
a
∫0
1+ x
dx =
2
a
1⎡
ln(1 + x 2 ) ⎤ = ln ⎛⎜ 1 + a 2 ⎞⎟
⎢
⎥
⎦0
⎝
⎠
2⎣
u = x2
du = 2 x dx
2
1
a
2
4
8
16
⎛
⎞
ln ⎜ 1+ a 2 ⎟ 0.3466 0.8047 1.4166 2.0872 2.7745
⎝
⎠
5
10
x
24.
−2
1
∫0 1 + x dx = [ln(1 + x)]0 = ln (1 + a )
a
a
a
1
2
4
8
16
ln (1+ a ) 0.6931 1.0986 1.6094 2.1972 2.8332
We would conjecture lim x10 e − x = 0 .
x →∞
10 − x
19. y = x e
25.
y
a
1
⎡ 1⎤
∫1 x 2 dx = ⎢⎣− x ⎥⎦1 = 1 − a
a
1
a
1
1−
a
480,000
26.
240,000
a
∫1
2
4
8
16
0.5 0.75 0.875 0.9375
a
⎡ 1 ⎤
1⎡
1 ⎤
= ⎢1 − ⎥
dx = ⎢ −
3
2⎥
x
⎣ 2 x ⎦1 2 ⎣ a 2 ⎦
1
a
2
4
8
16
1⎡ 1 ⎤
⎢1− ⎥ 0.375 0.46875 0.4921875 0.498046875
2 ⎣ a2 ⎦
10
20
x
2 −x
=0.
We would conjecture lim x e
x →∞
27.
20. Based on the results from problems 15-19, we
would conjecture
lim x n e− x = 0
4
∫a
1
4
dx = ⎡⎣ 2 x ⎤⎦ = 4 − 2 a
a
x
a
4− 2 a
1
1
2
2 2.58579
1
1
4
8
3 3.29289
1
16
3.5
x →∞
21.
a −x
e
0
∫
a
1
1−e− a
22.
a
∫0 xe
28.
a
dx = ⎡ −e− x ⎤ = 1 − e− a
⎣
⎦0
− x2
2
4
16
=
u =− x 2
du =−2 x dx
a
1
2
4
8
16
1⎡ 2⎤
e− a
− ⎢e− x ⎥ = 1 −
2⎣
2
⎦
1−
4
4
a
8
0.632 0.865 0.982 0.9997 0.9999 +
dx
41
∫a x dx = [ln x ]a = ln a
ln
1
1
2
1
4
1
8
1
16
4
1.38629 2.07944 2.77259 3.46574 4.15888
a
2
1
2
2 ea
0.81606028
0.93233236
0.999999944
1− (8.02×10−29 )
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Indeterminate Forms and
Improper Integrals
8
CHAPTER
8.1 Concepts Review
7. The limit is not of the form
1. lim f ( x); lim g ( x)
x→a
2.
As x → 1– , x 2 – 2 x + 2 → 1, and x 2 – 1 → 0 – so
x →a
f ′( x)
g ′( x)
9. The limit is of the form
0
1. The limit is of the form .
0
2 x – sin x
2 – cos x
= lim
=1
lim
x
1
x →0
x →0
1
0
.
0
cos x
– sin x
lim
= lim
=1
x →π / 2 π / 2 – x x →π / 2 –1
0
.
0
1 – 2 cos 2 x
10. The limit is of the form
sin –1 x
=
=
1– 2
= –1
1
x2 + 6 x + 8
x → –2 x 2
– 3x –10
2
2
=
=–
–7
7
=
3
=3
1
x →0
476
x3 – 3 x 2 + x
x3 – 2 x
Section 8.1
11. The limit is of the form
lim
x →0
2x + 6
x → –2 2 x – 3
=
= lim
+
7
x
2
x
–1
–1
x →0
3x2 + 6 x + 1
3x2 – 2
0
.
0
7 x ln 7
2 x
= lim
x →0+ 2 x ln 2
= lim
x →0
2 x
+
7
x
ln 7
2
x
ln 2
ln 7
≈ 2.81
ln 2
13. The limit is of the form
= lim
0
.
0
1 – 2t
–3
t – t2
3
2 t
= lim
= 2 =–
lim
1
1
2
t →1 ln t
t →1
12. The limit is of the form
0
. (Apply l’Hôpital’s
0
Rule twice.)
0
6. The limit is of the form .
0
lim
0
.
0
t
0
5. The limit is of the form .
0
lim
–1
ex – e– x
ex + e– x 2
= lim
= =1
2
x →0 2sin x
x →0 2 cos x
0
.
0
3
1+ 9 x 2
lim
x →0 1
1– x 2
3sin 2 x cos x
lim
3. The limit is of the form
sec2 x
0
.
0
3
ln(sin x)3
= lim sin x
lim
x →π / 2 π / 2 – x
x →π / 2
0
=
=0
–1
2. The limit is of the form
tan –1 3 x
ln x 2
lim
Problem Set 8.1
4. The limit is of the form
0
.
0
1 2x
2
1
= lim x
= lim
=1
x →1 x 2 – 1 x →1 2 x
x →1 x 2
4. Cauchy’s Mean Value
x – sin 2 x
= lim
x →0 tan x
x →0
= –∞
8. The limit is of the form
x →0
x →0
x2 + 1
x →1–
3. sec2 x; 1; lim cos x ≠ 0
lim
x2 – 2 x + 2
lim
lim
0
.
0
lim
=
1
1
=–
–2
2
ln cos 2 x
x →0
= lim
7x
2
= lim
–2sin 2 x
cos 2 x
x →0
14 x
–4 cos 2 x
x →0 14 cos 2 x – 28 x sin 2 x
= lim
–2sin 2 x
x →0 14 x cos 2 x
=
–4
2
=–
14 – 0
7
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0
.
0
3sin x
3cos x
lim
= lim
1
–x
x →0 –
x →0 – –
14. The limit is of the form
19. The limit is of the form
Rule twice.)
2 –x
lim
x →0 –
0
. (Apply l’Hôpital’s
0
Rule three times.)
tan x – x
sec2 x – 1
lim
= lim
x →0 sin 2 x – 2 x x →0 2 cos 2 x – 2
2sec 2 x tan x
2sec 4 x + 4sec2 x tan 2 x
= lim
–8cos 2 x
x →0 –4sin 2 x
x →0
2+0
1
=
=–
–8
4
= lim
0
16. The limit is of the form . (Apply l’Hôpital’s
0
Rule three times.)
sin x – tan x
cos x – sec2 x
lim
= lim
x →0 x 2 sin x
x →0 2 x sin x + x 2 cos x
– sin x – 2sec2 x tan x
= lim
x →0 2sin x + 4 x cos x – x 2 sin x
– cos x – 2sec4 x – 4sec2 x tan 2 x
= lim
x →0
6 cos x – x 2 cos x – 6 x sin x
–1 – 2 – 0
1
=
=–
6–0–0
2
17. The limit is of the form
0
. (Apply l’Hôpital’s
0
Rule twice.)
x2
2x
2
lim
= lim
= lim
+ sin x – x
+ cos x – 1
+ − sin x
x →0
x →0
x →0
0
This limit is not of the form . As
0
x → 0+ , 2 → 2, and − sin x → 0− , so
2
lim
= −∞.
+ sin x
x →0
18. The limit is of the form
0
. (Apply l’Hôpital’s
0
Rule twice.)
e – ln(1 + x) –1
x →0
x
e +
x
= lim
x →0
8 x3
2
1
(1+ x )2
2
= lim
x →0
=
e
x
– 1+1x
= lim
1
1+ x 2
–1
24 x 2
1
1
= lim –
=–
24
x →0 24(1 + x 2 ) 2
x →0
20. The limit is of the form
= lim
–2 x
(1+ x 2 ) 2
x →0
48 x
0
. (Apply l’Hôpital’s
0
Rule twice.)
cosh x –1
sinh x
cosh x 1
= lim
= lim
=
lim
2`
2
x
2
2
0
0
x →0
x
→
x
→
x
21. The limit is of the form
0
. (Apply l’Hôpital’s
0
Rule twice.)
1 − cos x − x sin x
lim
2
+
x → 0 2 − 2 cos x − sin x
− x cos x
= lim
x → 0+ 2sin x − 2 cos x sin s
x sin x – cos x
= lim
2
2
+
x →0 2 cos x – 2 cos x + 2sin x
0
This limit is not of the form .
0
As x → 0+ , x sin x – cos x → −1 and
2 cos x – 2 cos 2 x + 2sin 2 x → 0+ , so
x sin x – cos x
lim
= –∞
+ 2 cos x – 2 cos 2 x + 2sin 2 x
x →0
22. The limit is of the form
lim
sin x + tan x
ex + e– x – 2
0
.
0
cos x + sec2 x
= lim
ex – e– x
0
This limit is not of the form .
0
x →0 –
x →0 –
As x → 0 – , cos x + sec 2 x → 2, and
e x – e – x → 0 – , so lim
cos x + sec2 x
x →0 –
23. The limit is of the form
x
x
lim
tan –1 x – x
x →0
= lim – 6 – x cos x = 0
15. The limit is of the form
0
. (Apply l’Hôpital’s
0
∫
lim 0
x →0
1 + sin t dt
x
ex – e– x
= – ∞.
0
.
0
= lim 1 + sin x = 1
x →0
2x
1+1
=1
2
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24. The limit is of the form
x
lim
x →0
∫0
t cos t dt
2
+
= lim
x →0
x
cos x
+
2 x
26. Note that sin (1 0 ) is undefined (not zero), so
0
.
0
= lim
x →0
+
l'Hôpital's Rule cannot be used.
1
⎛1⎞
As x → 0, → ∞ and sin ⎜ ⎟ oscillates rapidly
x
⎝ x⎠
between –1 and 1, so
x cos x
2x
=∞
lim
x →0
25. It would not have helped us because we proved
sin x
lim
= 1 in order to find the derivative of
x →0 x
sin x.
( ) ≤ lim
x 2 sin 1x
tan x
x2
.
x →0 tan x
x2
x 2 cos x
=
tan x
sin x
x 2 cos x
⎡⎛ x ⎞
⎤
= lim ⎢⎜
⎟ x cos x ⎥ = 0 .
x →0 sin x
x →0 ⎣⎝ sin x ⎠
⎦
lim
Thus, lim
x 2 sin
( 1x ) = 0 .
x →0 tan x
A table of values or graphing utility confirms
this.
27. a.
OB = cos t , BC = sin t and AB = 1 – cos t , so the area of triangle ABC is
The area of the sector COA is
region ABC is
1
sin t (1 – cos t ).
2
1
1
t while the area of triangle COB is cos t sin t , thus the area of the curved
2
2
1
(t – cos t sin t ).
2
1 sin t (1 – cos t )
area of triangle ABC
= lim 2
1
t →0+ area of curved region ABC t →0+ 2 (t – cos t sin t )
lim
sin t (1 – cos t )
cos t – cos 2 t + sin 2 t
4sin t cos t – sin t
4 cos t – 1 3
= lim
= lim
= lim
=
+ t – cos t sin t
+ 1 – cos 2 t + sin 2 t
+
+
4 cos t sin t
4 cos t
4
t →0
t →0
t →0
t →0
(L’Hôpital’s Rule was applied twice.)
= lim
1
1
1
t cos 2 t , so the area of the curved region BCD is cos t sin t – t cos 2 t.
2
2
2
1 cos t (sin t – t cos t )
area of curved region BCD
= lim 2
lim
1 (t – cos t sin t )
+ area of curved region ABC
t →0
t →0+
2
b. The area of the sector BOD is
cos t (sin t – t cos t )
sin t (2t cos t – sin t )
2t (cos 2 t – sin 2 t )
t (cos 2 t – sin 2 t )
= lim
= lim
= lim
2
2
t – sin t cos t
4 cos t sin t
2 cos t sin t
t →0+
t →0+ 1 – cos t + sin t
t →0+
t →0+
= lim
cos 2 t – 4t cos t sin t – sin 2 t
1– 0 – 0 1
=
2–0
2
2 cos t – 2sin t
t →0
(L’Hôpital’s Rule was applied three times.)
= lim
+
478
Section 8.1
2
2
=
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28. a.
Note that ∠DOE has measure t radians. Thus the coordinates of E are (cost, sint).
Also, slope BC = slope CE . Thus,
0− y
sin t − 0
=
(1 − t ) − 0 cos t − (1 − t )
(1 − t ) sin t
cos t + t − 1
(t − 1) sin t
y=
cos t + t –1
(t – 1) sin t
lim y = lim
+
+ cos t + t – 1
t →0
t →0
0
This limit is of the form .
0
(t – 1) sin t
sin t + (t – 1) cos t 0 + (–1)(1)
=
= –1
lim
= lim
– sin t + 1
–0 + 1
t →0+ cos t + t – 1 t →0+
−y =
b. Slope AF = slope EF . Thus,
t
t − sin t
=
1 − x 1 − cos t
t (1 − cos t )
= 1− x
t − sin t
t (1 + cos t )
x = 1−
t − sin t
t cos t – sin t
x=
t – sin t
t cos t – sin t
lim x = lim
+
+
t – sin t
t →0
t →0
0
The limit is of the form . (Apply l’Hôpital’s Rule three times.)
0
t cos t – sin t
–t sin t
= lim
lim
+
+
t – sin t
t →0
t →0 1 – cos t
– sin t – t cos t
t sin t – 2 cos t 0 – 2
= lim
= lim
=
= –2
+
+
sin t
cos t
1
t →0
t →0
ex −1
ex
⎛0⎞
29. By l’Hộpital’s Rule ⎜ ⎟ , we have lim f ( x) = lim
= lim
= 1 and
x
⎝0⎠
x →0 +
x →0+
x →0+ 1
ex −1
ex
= lim
= 1 so we define f (0) = 1 .
x →0− x
x →0− 1
lim f ( x) = lim
x →0 −
1
ln x
⎛0⎞
30. By l’Hộpital’s Rule ⎜ ⎟ , we have lim f ( x) = lim
= lim x = 1 and
⎝0⎠
x →1+
x →1+ x − 1 x →1+ 1
1
ln x
= lim x = 1 so we define f (1) = 1 .
lim f ( x) = lim
x →1−
x →1− x − 1 x →1− 1
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31. A should approach 4πb 2 , the surface area of a sphere of radius b.
2 2
⎡
2πa 2 b arcsin a a– b
⎢
2
lim ⎢ 2πb +
a →b + ⎢
a 2 – b2
⎣
Focusing on the limit, we have
lim
a →b
a 2 – b2
a
2
a 2 arcsin
+
2
2 2
⎤
a 2 arcsin a a– b
⎥
2
⎥ = 2πb + 2πb lim+
a →b
a 2 – b2
⎥
⎦
2a arcsin
= lim
a →b
a –b
+
a 2 – b2
a
⎛
+ a2 ⎜
⎝a
a
a 2 –b2
b
2
a –b
2
⎞
⎟
2
2
⎛
⎞
⎠ = lim ⎜ 2 a 2 – b 2 arcsin a – b + b ⎟ = b.
⎟
a
a →b + ⎜
⎝
⎠
Thus, lim A = 2πb 2 + 2πb(b) = 4πb 2 .
a →b +
32. In order for l’Hôpital’s Rule to be of any use, a(1)4 + b(1)3 + 1 = 0, so b = –1 – a.
Using l’Hôpital’s Rule,
ax 4 + bx3 + 1
4ax3 + 3bx 2
lim
= lim
x →1 ( x – 1) sin πx
x →1 sin πx + π( x – 1) cos πx
To use l’Hôpital’s Rule here,
4a(1)3 + 3b(1)2 = 0, so 4a + 3b = 0, hence a = 3, b = –4.
36 x 2 – 24 x
12
6
3 x 4 – 4 x3 + 1
12 x3 – 12 x 2
= lim
=
=–
= lim
2
–2π
π
x →1 2π cos πx – π ( x – 1) sin πx
x →1 ( x – 1) sin πx
x →1 sin πx + π( x – 1) cos πx
lim
a = 3, b = –4, c = –
6
π
33. If f ′(a ) and g ′(a ) both exist, then f and g are
both continuous at a. Thus, lim f ( x) = 0 = f (a )
38.
x →a
and lim g ( x ) = 0 = g (a ).
x →a
lim
x→a
f ( x)
f ( x) – f (a )
= lim
g ( x) x→a g ( x) – g (a )
f ( x )– f ( a )
x–a
lim
x → a g ( x )– g ( a )
x–a
=
cos x – 1 +
x2
2
34. lim
x →0
35. lim
x
36. lim
x →0
4
ex – 1 – x –
x →0
f ( x )– f ( a )
x–a
x →a
g ( x )– g ( a )
lim
x–a
x→a
lim
x
x2
2
=
–
4
1 – cos( x 2 )
3
x sin x
=
=
f ′(a)
g ′(a )
1
24
x3
6
=
1
24
1
2
tan x − x
sec2 x − 1
= lim 1
=2
x → 0 arcsin x − x
x →0
−1
2
37. lim
1− x
480
Section 8.1
The slopes are approximately 0.02 / 0.01 = 2 and
0.01/ 0.01 = 1 . The ratio of the slopes is
therefore 2 /1 = 2 , indicating that the limit of the
ratio should be about 2. An application of
l'Hopital's Rule confirms this.
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41.
39.
The slopes are approximately 0.005 / 0.01 = 1/ 2
and 0.01/ 0.01 = 1 . The ratio of the slopes is
therefore 1/ 2 , indicating that the limit of the
ratio should be about 1/ 2 . An application of
l'Hopital's Rule confirms this.
The slopes are approximately 0.01/ 0.01 = 1 and
−0.01/ 0.01 = 1 . The ratio of the slopes is
therefore −1/1 = −1 , indicating that the limit of
the ratio should be about −1 . An application of
l'Hopital's Rule confirms this.
42. If f and g are locally linear at zero, then, since
lim f ( x ) = lim g ( x ) = 0 , f ( x) ≈ px and
40.
x →0
x →0
g ( x) ≈ qx , where p = f '(0) and q = g '(0) .
Then f ( x) / g ( x) ≈ px / px = p / q when x is
near 0.
The slopes are approximately 0.01/ 0.01 = 1 and
0.02 / 0.01 = 2 . The ratio of the slopes is
therefore 1/ 2 , indicating that the limit of the
ratio should be about 1/ 2 . An application of
l'Hopital's Rule confirms this.
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8.2 Concepts Review
1.
f ′( x)
g ′( x )
2. lim
x →a
∞
.
∞
3sec x tan x
5. The limit is of the form
lim
x→ π
2
f ( x)
g ( x)
or lim
1
x →a 1
g ( x)
f ( x)
3sec x + 5
= lim
tan x
x→ π
= lim
x→ π
2
3. ∞ – ∞, 0°, ∞°, 1∞
3 tan x
= lim 3sin x = 3
sec x x→ π
2
ln sin 2 x
= lim
x →0+ 3ln tan x x →0+
lim
x →0+
1. The limit is of the form
∞
.
∞
7. The limit is of the form
1 1000 x999
1000
ln x1000
= lim x
lim
x
1
x →∞
x →∞
1000
= lim
=0
x →∞ x
2. The limit is of the form
∞
. (Apply l’Hôpital’s
∞
Rule twice.)
2x
x →∞
x →∞
= lim
x →∞
x ⋅ 2 ln 2(1 + x ln 2)
x
x
x →∞
2
x →∞ 2 x
e
x
( 1x )
ln 2(1 + x ln 2)
=0
∞
4. The limit is of the form . (Apply l’Hôpital’s
∞
Rule three times.)
3x
3
= lim
lim =
1
x →∞ ln(100 x + e x ) x →∞
(100 + e x )
x
100 x + e
= lim
x →∞
= lim
x →∞
100 + e x
3e x
ex
=3
= lim
x →∞
)
=0
8. The limit is of the form
–∞
. (Apply l’Hôpital’s
∞
ln(4 – 8 x) 2
lim
= lim
–
tan πx
x→ 1
x→ 1
( 2)
1
(4–8 x )2
( 2)
= lim
(2)
–
x→ 1
–
2(4 – 8 x)(–8)
π sec2 πx
–16 cos 2 πx
32π cos πx sin πx
= lim
–
π(4 – 8 x )
–8π
x→ 1
(2)
300 + 3e x
ex
(2)
–
x→ 1
∞
.
∞
cot x
– csc2 x
= lim
1
– ln x x →0+ –
9. The limit is of the form
lim
x →0 +
2 x – ln x
= lim
2 x – ln x
sin 2 x
⎡ 2x
⎤
= lim ⎢
csc x – ln x ⎥ = ∞
+ ⎣ sin x
⎦
x →0
x
since lim
= 1 while lim csc x = ∞ and
+ sin x
x →0 +
x →0
x →0
lim
Section 8.2
x ln x1000
x →∞
x →0 +
482
1000
= lim
(
1
1 1000 x999
ln x1000 x1000
1
x
= lim – 4 cos πx sin πx = 0
= 0 (See Example 2).
300 x + 3e x
ln(ln x1000 )
= lim
lim
ln x
x →∞
x →∞
∞
.
∞
Rule twice.)
2 x ln 2
= lim
x ⋅ 2 x ln 2
2
10000
lim
2(ln x) 1x
x →∞
2 ln x
= lim
3.
= lim
1 2sin x cos x
sin 2 x
3 sec 2 x
tan x
2 cos 2 x 2
=
3
3
= lim
Problem Set 8.2
(ln x)2
–∞
.
–∞
6. The limit is of the form
4. ln x
lim
sec 2 x
2
+
– ln x = ∞.
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10. The limit is of the form
∞
, but the fraction can
∞
15. The limit is of the form 00.
2
Let y = (3 x) x , then ln y = x 2 ln 3x
be simplified.
2 csc2 x
2
2
= lim
= =2
lim
x →0 cot 2 x
x →0 cos 2 x 12
1000
11. lim ( x ln x
x →0
The limit is of the form
lim
= lim
1
x
x →0
x →0 +
x
–
x →0
x →0
1
x2
2
13. lim (csc 2 x – cot 2 x) = lim
x →0
14.
x →0 +
lim csc x(ln(cos x)) = lim
x →0
The limit is of the form
ln(cos x)
sin x
0
.
0
1 (– sin x )
ln(cos x)
= lim cos x
cos x
x →0 sin x
x →0
sin x
0
= lim –
=– =0
1
x →0 cos 2 x
lim
sin 2 x
=1
lim (cos x)csc x = lim eln y = 1
x
x →0
lim (tan x – sec x) = lim
x→ π
2
x→ π
2
sin x – 1
cos x
2
x →0
17. The limit is of the form 0∞ , which is not an
indeterminate form.
0
The limit is of the form .
0
sin x – 1
cos x
0
= lim
=
=0
lim
–1
x → π cos x
x → π – sin x
lim (5cos x) tan x = 0
–
x →(π / 2 )
2
2
2
⎛ x 2 – sin 2 x ⎞
⎛
1 ⎞
⎛ 1
1 ⎞
18. lim ⎜ csc2 x –
= lim ⎜
=
lim
–
⎟
⎟
⎜
⎟
x →0 ⎜ x 2 sin 2 x ⎟
x →0 ⎝
x →0 ⎝ sin 2 x x 2 ⎠
x2 ⎠
⎝
⎠
Consider lim
x 2 – sin 2 x
2
x →0
2
lim
+
x →0
1 – cos 2 x
x →0
x →0 sin 2
x →0 +
x2
=0
2
Let y = (cos x)csc x , then ln y = csc x(ln(cos x))
⎛ x ⎞
12. lim 3 x 2 csc 2 x = lim 3 ⎜
⎟ = 3 since
x →0
x →0 ⎝ sin x ⎠
x
lim
=1
x →0 sin x
sin x
= lim –
16. The limit is of the form 1∞.
x →0
= lim
1
x2
1 ⋅3
3x
2
x →0 + – 3
x
= lim
∞
.
∞
lim (3x) x = lim eln y = 1
= lim – 1000 x = 0
2
1
x2
+
2
1000 x999
1
x →0
ln 3 x
lim
∞
.
∞
1000
+
The limit is of the form
1
x
x →0
ln x1000
x →0
ln x1000
) = lim
ln 3 x
lim x 2 ln 3 x = lim
x sin x
2
x – sin x
x →0
= lim
2
x sin x
x →0 sin 2
= lim
2
. The limit is of the form
2
2 x – 2sin x cos x
= lim
x →0 2 x sin
2
2
x + 2 x sin x cos x
2
2
2
2
x + 4 x sin x cos x + x cos x – x sin x
x →0 12 cos
0
. (Apply l’Hôpital’s Rule four times.)
0
= lim
1 – cos 2 x + sin 2 x
2
x →0
x – sin x cos x
2
x sin x + x 2 sin x cos x
4sin x cos x
= lim
x →0 6 x cos x 2
+ 6 cos x sin x − 4 x 2 cos x sin x − 6 x sin 2 x
4 cos 2 x – 4sin 2 x
2
2
2
2
2
2
x – 4 x cos x – 32 x cos x sin x – 12sin x + 4 x sin x
=
4 1
=
12 3
2
2
⎛ x 2 – sin 2 x ⎞
1
⎛1⎞
=⎜ ⎟ =
Thus, lim ⎜
⎟
2
2
⎜
⎟
9
x →0 x sin x
⎝ 3⎠
⎝
⎠
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19. The limit is of the form 1∞.
24. The limit is of the form 1∞.
Let y = ( x + e x / 3 )3 / x , then ln y =
3
ln( x + e x / 3 ).
x
3
3ln( x + e x / 3 )
ln( x + e x / 3 ) = lim
x
x →0 x
x →0
0
The limit is of the form .
0
(
3 + ex / 3
= lim
x →0
x+e
x/3
=
x →0 x 2
)
x →0
20. The limit is of the form (–1)0 .
The limit does not exist.
21. The limit is of the form 10 , which is not an
indeterminate form.
lim (sin x)cos x = 1
x2
x →0
0
.
0
(Apply l’Hôpital’s rule twice.)
1 (– sin x )
ln(cos x)
− tan x
lim
= lim cos x
= lim
2
2x
x →0
x →0
x →0 2 x
x
− sec 2 x −1
1
=
=−
2
2
2
x →0
2
lim (cos x)1/ x = lim eln y = e−1/ 2 =
x →0
1
e
25. The limit is of the form 0∞ , which is not an
indeterminate form.
lim (tan x) 2 / x = 0
x →0 +
26. The limit is of the form ∞ + ∞, which is not an
indeterminate form.
x→ π
2
∞
22. The limit is of the form ∞ , which is not an
indeterminate form.
lim x x = ∞
lim (e – x – x) = lim (e x + x) = ∞
x→ – ∞
x →∞
27. The limit is of the form 00. Let
y = (sin x) x , then ln y = x ln(sin x).
x →∞
23. The limit is of the form ∞ 0 . Let
1
y = x1/ x , then ln y = ln x.
x
1
ln x
lim ln x = lim
x →∞ x
x →∞ x
–∞
.
The limit is of the form
∞
1
ln x
1
= lim x = lim = 0
lim
x →∞ x
x →∞ 1
x →∞ x
lim x
ln(cos x) .
ln(cos x)
ln(cos x ) = lim
x →0
x →∞
x2
= lim
lim ( x + e x / 3 )3 / x = lim eln y = e 4
1/ x
1
The limit is of the form
4
=4
1
x →0
1
lim
lim
3
1 + 13 e x / 3
3ln( x + e x / 3 )
x +e x / 3
lim
= lim
x
1
x →0
x →0
2
Let y = (cos x)1/ x , then ln y =
= lim e
x →∞
ln y
=1
ln(sin x)
lim x ln(sin x) = lim
x →0
+
x →0
1
x
+
–∞
.
∞
1 cos x
sin x
The limit is of the form
lim
x →0
ln(sin x)
+
1
x
= lim
x →0 +
–
1
x2
⎡ x
⎤
= lim ⎢
(– x cos x) ⎥ = 1 ⋅ 0 = 0
+ ⎣ sin x
⎦
x →0
lim (sin x ) x = lim eln y = 1
x →0 +
x →0+
28. The limit is of the form 1∞. Let
1
ln(cos x – sin x).
x
1
ln(cos x − sin x)
lim ln(cos x − sin x ) = lim
x
x →0 x
x →0
y = (cos x – sin x)1/ x , then ln y =
= lim
1
(− sin x − cos x)
cos x −sin x
1
− sin x − cos x
= lim
= −1
x →0 cos x − sin x
x →0
lim (cos x − sin x )1/ x = lim eln y = e−1
x →0
484
Section 8.2
x →0
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29. The limit is of the form ∞ – ∞.
1⎞
1⎞
x – sin x
⎛
⎛ 1
lim ⎜ csc x – ⎟ = lim ⎜
– ⎟ = lim
x ⎠ x →0 ⎝ sin x x ⎠ x →0 x sin x
x →0 ⎝
0
The limit is of the form . (Apply l’Hôpital’s
0
Rule twice.)
x – sin x
1 – cos x
= lim
lim
x →0 x sin x
x →0 sin x + x cos x
sin x
0
= lim
= =0
2
x →0 2 cos x – x sin x
x
⎛ 1⎞
⎛ 1⎞
Let y = ⎜ 1 + ⎟ , then ln y = x ln ⎜ 1 + ⎟ .
x
⎝
⎠
⎝ x⎠
(
ln 1 + 1x
⎛ 1⎞
lim x ln ⎜ 1 + ⎟ = lim
1
x →∞
⎝ x ⎠ x →∞
x
lim
(
ln 1 + 1x
x →∞
1
x
) = lim
x →∞
)
0
.
0
1
1+ 1
x
(– )
–
1
x2
1
x2
1
=1
x →∞ 1 + 1
x
= lim
x
31. The limit is of the form 3∞ , which is not an
indeterminate form.
lim (1 + 2e )
x →0 +
=∞
32. The limit is of the form ∞ – ∞.
x ⎞
ln x – x 2 + x
⎛ 1
lim ⎜
–
⎟ = lim
x →1 ⎝ x – 1 ln x ⎠ x →1 ( x – 1) ln x
0
.
0
Apply l’Hôpital’s Rule twice.
1 − 2x +1
ln x − x 2 + x
lim
= lim x
x →1 ( x − 1) ln x
x →1 ln x + x −1
x
The limit is of the form
2
1− 2x + x
−4 x + 1 −3
3
= lim
=
=−
2
2
x →1 x ln x + x − 1 x →1 ln x + 2
= lim
1
ln(cos x).
x
1
ln(cos x)
ln(cos x) = lim
x
x
x →0
0
The limit is of the form .
0
lim
x →0
1
(– sin x)
ln(cos x)
sin x
= lim cos x
= lim –
=0
1
x
x →0
x →0
x →0 cos x
lim
lim (cos x)1/ x = lim eln y = 1
x →0
34. The limit is of the form 0 ⋅ – ∞.
ln x
lim ( x1/ 2 ln x) = lim
x →0 +
The limit is of the form
lim
x →0 +
1
x
x →0+
ln x
1
x
= lim
x →0 +
–
–∞
.
∞
1
x
1
2 x3/ 2
= lim – 2 x = 0
x →0+
35. Since cos x oscillates between –1 and 1 as
x → ∞, this limit is not of an indeterminate form
previously seen.
Let y = ecos x , then ln y = (cos x)ln e = cos x
⎛ 1⎞
lim ⎜1 + ⎟ = lim eln y = e1 = e
x⎠
x →∞ ⎝
x →∞
x 1/ x
Let y = (cos x)1/ x , then ln y =
x →0
30. The limit is of the form 1∞.
The limit is of the form
33. The limit is of the form 1∞.
Instructor's Resource Manual
lim cos x does not exist, so lim ecos x does not
x →∞
x →∞
exist.
36. The limit is of the form ∞ – ∞.
lim [ln( x + 1) – ln( x – 1)] = lim ln
x →∞
x →∞
x +1
x –1
1 + 1x
x +1
x +1
= lim
= 1, so lim ln
=0
1
x –1
x →∞ x – 1 x →∞ 1 –
x →∞
lim
x
37. The limit is of the form
0
, which is not an
–∞
indeterminate form.
x
lim
=0
x →0+ ln x
38. The limit is of the form – ∞ ⋅ ∞, which is not an
indeterminate form.
lim (ln x cot x) = – ∞
x →0 +
Section 8.2
485
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1 + e−t > 1 for all t, so
39.
d.
1 + e−t dt > ∫ dt = x − 1 .
x
x
∫1
lim n
n →∞
1 + e−t dt
x
∫1
lim
x →∞
1+ e
1
= lim
x →∞
x
−x
x
lim
∫1
+
x →1
sin t dt
x −1
= lim
+
x →1
n
since lim
n →∞
=1
n
lim
1
n
= lim
sin x
= sin(1)
1
n
n→∞
42. a.
n
n −1
= lim
1
n
n →∞
lim
a = lim e
ln y
n →∞
x →0 +
n →∞
n →∞
lim
n
n →∞
= lim
n →∞
a −1
1
n
n
−
n→∞
a ln a = ln a
Section 8.2
1
x
1
x
1
x
1
x →0 + – 2
x
ln y
= lim
x →0 +
–∞
.
∞
= lim – x = 0
x →0 +
=1
b. The limit is of the form 10 , since
lim x x = 1 by part a.
x →0 +
lim x ln( x x ) = 0
x →0+
lim ( x x ) x = lim eln y = 1
n
1
n
1 n
n2
−
a ln a
1
n2
x →0 +
Note that 10 is not an indeterminate form.
a −1
a = 1 by part a.
= lim
ln x
Let y = ( x x ) x , then ln y = x ln( x x ).
0
This limit is of the form ,
0
since lim
ln x
x →0 +
( n a − 1) = nlim
→∞
+
lim x x = lim e
n = lim eln y = 1
n
x →0
x →0 +
n →∞
lim n
1
n2
n (ln n − 1) = ∞
+
lim
1
n →∞
−
The limit is of the form
ln n
= lim n = 0
n →∞ n
n→∞ 1
n
1
n2
The limit is of the form 00.
x →0
lim
lim
( ) (1 − ln n)
n →∞
lim x ln x = lim
=1
b. The limit is of the form ∞ 0 .
1
Let y = n n , then ln y = ln n .
n
1
ln n
lim ln n = lim
n →∞ n
n →∞ n
∞
.
This limit is of the form
∞
n
Let y = x x , then ln y = x ln x.
1
Let y = a , then ln y = ln a.
n
1
lim ln a = 0
n →∞ n
n
0
,
0
n = 1 by part b.
n
n →∞
486
n −1
This limit is of the form
0
40. This limit is of the form .
0
c.
n
1
∞
.
The limit is of the form
∞
41. a.
( n n − 1) = nlim
→∞
c.
The limit is of the form 01 , since
lim x x = 1 by part a.
x →0 +
x
Let y = x( x ) , then ln y = x x ln x
lim x x ln x = – ∞
x →0+
lim x( x
x →0
+
x
)
= lim eln y = 0
x →0 +
Note that 01 is not an indeterminate form.
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
d. The limit is of the form 10 , since
1
ln x
= lim x = 0, so lim x1/ x = lim eln y = 1
x →∞ x
x →∞ 1
x →∞
x →∞
lim
lim ( x x ) x = 1 by part b.
x →0 +
1 ln x
Let y = (( x ) ) , then ln y = x ln(( x ) ).
x x x
x x
lim x ln(( x x ) x ) = 0
x →0 +
lim (( x x ) x ) x = lim eln y = 1
x →0 +
x →0 +
Note that 10 is not an indeterminate form.
e.
The limit is of the form 00 , since
lim ( x
(xx )
x →0 +
Let y = x( x
) = 0 by part c.
( xx )
)
x →0
x →0
x →0 +
ln x
1
x
x( x )
+
x
x( x )
x
2
x
)
Note: lim x(ln x )2 = lim
x →0 +
lim x( x
x →0 +
( xx )
)
x
(ln x)2
x →0 +
2 ln x
x
1
x →0+ – 2
x
x →0 +
c.
x
x →0+ – x ( x x ) ⎡ x x (ln x +1) ln x + x ⎤
⎢
x ⎥⎦
⎣
= lim
lim (1x + 2 x )1/ x = ∞
lim (1x + 2 x )1/ x = 0
1
x
– x( x
The limit is of the form (1 + 1)∞ = 2∞ , which
is not an indeterminate form.
x →0 –
x x(ln x) + x x ln x + x
0
=
=0
1⋅ 0 + 1⋅ 0 + 1
x →0
44. a.
–∞
.
∞
= lim
x
y ′ < 0 on (e, ∞). When x = e, y = e1/ e .
b. The limit is of the form (1 + 1) – ∞ = 2 – ∞ ,
which is not an indeterminate form.
1
+
x
( x( x ) )2
= lim
y is maximum at x = e since y ′ > 0 on (0, e) and
, then ln y = x( x ) ln x.
The limit is of the form
lim
⎛ 1 ln x ⎞ 1x ln x
y′ = ⎜
−
⎟e
⎝ x2 x2 ⎠
y ′ = 0 when x = e.
x
x
ln x
lim x( x ) ln x = lim
+
y = x1/ x = e x
1
x
= lim – 2 x ln x = 0
x →0 +
= lim eln y = 1
x →0+
The limit is of the form ∞0 .
Let y = (1x + 2 x )1/ x , then
ln y =
1
ln(1x + 2 x )
x
1
ln(1x + 2 x )
ln(1x + 2 x ) = lim
x
x →∞ x
x →∞
∞
The limit is of the form . (Apply
∞
l’Hôpital’s Rule twice.)
1 (1x ln1 + 2 x ln 2)
ln(1x + 2 x )
1x + 2 x
lim
= lim
1
x
x →∞
x →∞
lim
= lim
2 x ln 2
x →∞ 1x
+ 2x
2 x (ln 2)2
= lim
x →∞ 1x
ln1 + 2 x ln 2
= ln 2
lim (1x + 2 x )1/ x = lim eln y = eln 2 = 2
x →∞
x →∞
d. The limit is of the form 10 , since 1x = 1 for
all x. This is not an indeterminate form.
43.
lim (1x + 2 x )1/ x = 1
x →−∞
ln x
x
ln x
= −∞, so lim x1/ x = lim eln y = 0
lim
+ x
x →0
x →0+
x →0+
ln y =
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Section 8.2
487
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
45.
1/ t
1k + 2k + " + n k
lim
c.
n k +1
n →∞
1 ⎡⎛ 1 ⎞ ⎛ 2 ⎞
⎛n⎞
⎢⎜ ⎟ + ⎜ ⎟ + " ⎜ ⎟
n →∞ n ⎢⎝ n ⎠
⎝n⎠
⎝n⎠
⎣
k
k
= lim
k
⎤
⎥
⎥⎦
9 ⎞
⎛1
lim ⎜ 2t + 5t ⎟
+ ⎝ 10
10 ⎠
t →0
= 10 2 ⋅
10 9
5 ≈ 4.562
48. a.
k
n
1 ⎛i⎞
= lim ∑ ⋅ ⎜ ⎟
n →∞
i =1 n ⎝ n ⎠
The summation has the form of a Reimann sum
for f ( x ) = x k on the interval [ 0,1] using a
regular partition and evaluating the function at
1
i
each right endpoint. Thus, Δxi = , xi = , and
n
n
b.
k
⎛i⎞
f ( xi ) = ⎜ ⎟ . Therefore,
⎝n⎠
1k + 2k + " + n k
lim
n k +1
n →∞
e
2
1 ⎛i⎞
= lim ∑ ⋅ ⎜ ⎟
n →∞
i =1 n ⎝ n ⎠
n
lim
k
1
⎡ 1 k +1 ⎤
= ∫ x k dx = ⎢
x ⎥
0
⎣ k +1
⎦0
1
=
k +1
1/ t
⎞
1 ⎛ n
, then ln y = ln ⎜ ∑ ci xit ⎟ .
⎜
⎟
t ⎝ i =1
⎠
⎛ n
⎞
ln ⎜ ∑ ci xit ⎟
⎜
⎟
⎞
1 ⎛ n
⎠
lim ln ⎜ ∑ ci xit ⎟ = lim ⎝ i =1
⎜
⎟
+ t
+
t
t →0
⎝ i =1
⎠ t →0
The limit is of the form
0
, since
0
⎛
⎞
ln ⎜ ∑ ci xit ⎟
⎜
⎟
⎠ = lim
lim ⎝ i =1
t
t →0 +
t →0+
n
∑ ci = 1.
i =1
n
∑ ci xit ln xi
n
∑ ci xit i =1
i =1
n
n
i =1
i =1
1/ t
= ei =1
= lim eln y
t →0 +
n
= x1c1 x2c2 … xncn = ∏ xi ci
i =1
1/ t
47. a.
1 ⎞
⎛1
lim ⎜ 2t + 5t ⎟
+⎝2
2 ⎠
t →0
b.
4 ⎞
⎛1
lim ⎜ 2t + 5t ⎟
+⎝5
5
⎠
t →0
1/ t
488
Section 8.2
= lim
n →∞
lim
n →∞
c.
1
∫0 xe
2nx
xenx
xe
−x
= lim
2x
n →∞ x 2 enx
nx
= 2 5 ≈ 3.162
5
= 5 2 ⋅ 54 ≈ 4.163
∞
.
∞
=0
1
2
dx = ⎡ − xe− x − e− x ⎤ = 1 −
⎣
⎦0
e
1
−2 x
1
3
dx = ⎡ −2 xe−2 x − e−2 x ⎤ = 1 −
⎣
⎦0
e2
1
−3 x
1
4
dx = ⎡ −3xe−3 x − e−3 x ⎤ = 1 −
⎣
⎦0
e3
∫0 4 xe
∫0 9 xe
1
5
dx = ⎡ −4 xe−4 x − e−4 x ⎤ = 1 −
⎣
⎦0
e4
1
−4 x
1
−5 x
∫016 xe
∫0 25 xe
∫0 36e
∞
.
∞
2nx
−6 x
1
6
= ⎡ −5 xe−5 x − e−5 x ⎤ = 1 −
⎣
⎦0
e5
1
7
dx = ⎡ −6 xe−6 x − e−6 x ⎤ = 1 −
⎣
⎦0
e6
d. Guess: lim
∫
1 2
n →∞ 0
n xe− nx dx = 1
1
1 2
n
∑ ln xici
, so the limit is of the form
xe− nx dx = ⎡ − nxe− nx − e− nx ⎤
⎣
⎦0
n +1
= −(n + 1)e− n + 1 = 1 −
en
1
⎛ n +1⎞
lim ∫ n 2 xe− nx dx = lim ⎜1 −
⎟
0
n →∞
n →∞ ⎝
en ⎠
n +1
= 1 − lim
if this last limit exists. The
n →∞ e n
∞
.
limit is of the form
∞
n +1
1
lim
= lim
= 0, so
n →∞ e n
n →∞ en
∫0 n
= ∑ ci ln xi = ∑ ln xi ci
⎛ n
⎞
lim ⎜ ∑ ci xi t ⎟
⎜
⎟
t →0+ ⎝ i =1
⎠
nx
This limit is of the form
1
n
1
n x
n →∞ e nx
1
⎛ n
⎞
46. Let y = ⎜ ∑ ci xit ⎟
⎜
⎟
⎝ i =1
⎠
n2 x
n 2 xe− nx =
1 2
∫n
n →∞ 0
lim
xe− nx dx = 1 .
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
49. Note f(x) > 0 on [0, ∞).
⎛ x 25 x3 ⎛ 2 ⎞ x ⎞
lim f ( x) = lim ⎜
+
+⎜ ⎟ ⎟ = 0
x →∞
x →∞ ⎜ e x
e x ⎝ e ⎠ ⎟⎠
⎝
Therefore there is no absolute minimum.
f ′( x) = (25 x 24 + 3 x 2 + 2 x ln 2)e− x
− ( x 25 + x3 + 2 x )e− x
7.
8.
= (− x 25 + 25 x 24 − x3 + 3 x 2 − 2 x + 2 x ln 2)e− x
Solve for x when f ′( x) = 0 . Using a numerical
method, x ≈ 25.
A graph using a computer algebra system verifies
that an absolute maximum occurs at about x = 25.
8.3 Concepts Review
1. converge
3.
∫– ∞ f ( x)dx; ∫0
∞
0
f ( x)dx
11.
In this section and the chapter review, it is understood
means lim [ g ( x)]
b →∞
b
a
∞
x
∞
3.
∫1
∞
dx
∞
= ⎡ 1 + x 2 ⎤ = ∞ – 82 = ∞
∫9
⎢⎣
⎥⎦ 9
2
1+ x
The integral diverges.
x dx
∞
⎡ x⎤
2
∫1 πx = ⎢2 π ⎥ = ∞ – π = ∞
⎣
⎦1
The integral diverges.
∞
dx
x
1
dx = [ln(ln x)]e∞ = ∞ – 0 = ∞
x ln x
The integral diverges.
∞
∫e
∞
1
⎡1
2⎤
∫e x dx = ⎢⎣ 2 (ln x) ⎥⎦ e = ∞ – 2 = ∞
The integral diverges.
∞ ln x
b
b
ln 2 + 1
⎡ ln x 1 ⎤
= lim ⎢ −
− ⎥ =
b →∞ ⎣
x
x ⎦2
2
14.
∞
∫1
xe – x dx
u = x, du = dx
1
1
1
⎡1
⎤
4. ∫ e4 x dx = ⎢ e4 x ⎥ = e4 – 0 = e4
–∞
4
⎣4
⎦ –∞ 4
6.
∞
⎡
⎤
1
∫1 (1 + x2 )2 dx = ⎢⎢ – 2(1 + x2 ) ⎥⎥
⎣
⎦1
∞
b 1
⎡ ln x ⎤
= lim ⎢ −
+ lim
dx
b →∞ ⎣
x ⎥⎦ 2 b →∞ ∫2 x 2
∞
2
2
1
2 xe – x dx = ⎡⎢ – e – x ⎤⎥ = 0 – (– e –1 ) =
e
⎣
⎦1
1
∞
∞
dx
–5
⎡ 1 ⎤
1
1
∫– ∞ x 4 = ⎢⎣ – 3x3 ⎥⎦ = – 3(–125) – 0 = 375
–∞
5
∞
)⎤
⎦10
1
1
1
dx, dv =
dx, v = − .
2
x
x
x
∞ ln x
b ln x
dx
∫2 x 2 dx = blim
→∞ ∫2 x 2
= ∞ – e100 = ∞
dx = ⎡ e x ⎤
⎣ ⎦100
The integral diverges.
∫100 e
2
13. Let u = ln x, du =
2.
5.
12.
and likewise for
similar expressions.
1.
1
⎛ 1⎞ 1
= 0–⎜– ⎟ =
⎝ 4⎠ 4
Problem Set 8.3
∞
a
x
⎡ x 0.00001 ⎤
9. ∫
=⎢
⎥ = ∞ – 100, 000 = ∞
1 x 0.99999
⎣⎢ 0.00001 ⎦⎥1
The integral diverges.
4. p > 1
that [ g ( x)]
∞
∫10 1 + x2 dx = 2 ⎡⎣ln(1 + x
∞
b
∫ cos x dx
b →∞ 0
lim
dx
1
= ∞ – ln 101 = ∞
2
The integral diverges.
10.
2.
∞
⎡
1
⎤
∫1 x1.00001 = ⎢⎣ – 0.00001x0.00001 ⎥⎦
1
1
1
⎛
⎞
= 0–⎜–
= 100, 000
⎟=
⎝ 0.00001 ⎠ 0.00001
∞
Instructor’s Resource Manual
dv = e – x dx, v = – e – x
∞
∫1
∞
∞
xe – x d = ⎡ – xe – x ⎤ + ∫ e – x dx
⎣
⎦1
1
∞
2
= ⎡ – xe – x – e – x ⎤ = 0 – 0 – (– e –1 – e –1 ) =
⎣
⎦1
e
1
⎡
⎤
1
= ⎢–
15. ∫
⎥
3
2
– ∞ (2 x – 3)
⎣⎢ 4(2 x – 3) ⎦⎥ – ∞
1
dx
=–
1
1
– (–0) = −
4
4
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16.
∞
1/ 3 ⎤ ∞
dx
∫4 (π − x )2 / 3 = ⎡⎣−3 (π − x )
⎦4
= ∞ + 33 π − 4 = ∞
The integral diverges.
17.
∞
x
∫– ∞
x2 + 9
dx = ∫
0
x
–∞
x2 + 9
dx + ∫
0
The integral diverges since both
18.
∞
0
dx
0
1
dx
–1
2
x +9
∞
dx
∞
0
dx = ⎡ x 2 + 9 ⎤ + ⎡ x 2 + 9 ⎤ = (3 – ∞) + (∞ – 3)
2
⎣⎢
⎦⎥ – ∞ ⎢⎣
⎦⎥ 0
x +9
x
x
∫– ∞
∫– ∞ ( x2 + 16)2 = ∫– ∞ ( x2 + 16)2 + ∫0
∫ ( x 2 + 16)2 = 128 tan
∞
dx and
∞
∫0
x
2
x +9
dx diverge.
dx
2
( x + 16) 2
x
x
by using the substitution x = 4 tan θ.
+
2
4 32( x + 16)
0
⎡ 1
⎤
x
⎡ 1 ⎛ π⎞ ⎤
π
–1 x
∫– ∞ ( x2 + 16)2 = ⎢⎢128 tan 4 + 32( x2 + 16) ⎥⎥ = 0 – ⎢⎣128 ⎜⎝ – 2 ⎟⎠ + 0⎥⎦ = 256
⎣
⎦ –∞
0
dx
∞
⎡ 1
⎤
x
1 ⎛ π⎞
π
–1 x
∫0 ( x2 + 16)2 = ⎢⎢128 tan 4 + 32( x2 + 16) ⎥⎥ = 128 ⎜⎝ 2 ⎟⎠ + 0 – (0) = 256
⎣
⎦0
∞
dx
π
π
π
∫– ∞ ( x 2 + 16)2 = 256 + 256 = 128
∞
19.
dx
1
∞
1
∞
0
1
1
1
∫ ( x + 1)2 + 9 dx = 3 tan
–1
0
1
⎡1
–1 1 1 ⎛ π ⎞
–1 x + 1 ⎤
∫– ∞ ( x + 1)2 + 9 dx = ⎢⎣ 3 tan 3 ⎥⎦ – ∞ = 3 tan 3 – 3 ⎜⎝ – 2 ⎟⎠ =
∞
∫0
1⎛
–1 1 ⎞
⎜ π + 2 tan
⎟
6⎝
3⎠
∞
1⎛
1
∞
∞
∫– ∞
For
dx
x + 1⎤
1⎛ π⎞ 1
1 1⎛
1⎞
⎡1
= ⎜ ⎟ – tan –1 = ⎜ π – 2 tan –1 ⎟
dx = ⎢ tan –1
⎥
2
3 6⎝
3⎠
3 ⎦0 3 ⎝ 2 ⎠ 3
⎣3
( x + 1) + 9
1
∫– ∞ x2 + 2 x + 10 dx = 6 ⎜⎝ π + 2 tan
20.
( x + 1)2 + 9
x +1
by using the substitution x + 1 = 3 tan θ.
3
1
0
1
∞
∫– ∞ x 2 + 2 x + 10 dx = ∫– ∞ ( x + 1)2 + 9 dx = ∫– ∞ ( x + 1)2 + 9 dx + ∫0
x
e
2x
0
dx = ∫
0
x
– ∞ e –2 x
x
∞
x
0
2x
dx + ∫
0
∫– ∞ e –2 x dx = ∫– ∞ xe
2x
e
1⎛
–1 1 ⎞ π
⎟ + ⎜ π – 2 tan
⎟=
3⎠ 6 ⎝
3⎠ 3
–1 1 ⎞
dx
dx, use u = x, du = dx, dv = e2 x dx, v =
0
1 2x
e .
2
0
1 0 2x
1
1
⎡ 1 2x ⎤
⎡ 1 2x 1 2x ⎤
2x
∫– ∞ xe dx = ⎢⎣ 2 xe ⎥⎦ – ∞ – 2 ∫– ∞ e dx = ⎢⎣ 2 xe – 4 e ⎥⎦ – ∞ = 0 – 4 – (0) = – 4
∞ x
∞
1
For ∫
dx = ∫ xe –2 x dx, use u = x, du = dx, dv = e –2 x dx, v = – e –2 x .
0 e2 x
0
2
0
∞
∞
1 ∞
1
1⎞ 1
⎡ 1
⎤
⎡ 1
⎤
⎛
xe –2 x dx = ⎢ – xe –2 x ⎥ + ∫ e –2 x dx = ⎢ – xe –2 x – e –2 x ⎥ = 0 – ⎜ 0 – ⎟ =
0
4
4⎠ 4
⎣ 2
⎦0 2
⎣ 2
⎦0
⎝
∞ x
1 1
∫– ∞ 2 x dx = – 4 + 4 = 0
e
∞
∫0
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21.
∞
25. The area is given by
∞
∞⎛ 1
2
1 ⎞
∫1 4 x 2 − 1dx = ∫1 ⎜⎝ 2 x –1 – 2 x + 1 ⎟⎠ dx
∞
0
∫– ∞ sech x dx = ∫– ∞ sech x dx = ∫0 sech x dx
= [tan –1 (sinh x)]0– ∞ + [tan –1 (sinh x)]∞
0
∞
⎡ ⎛ π ⎞⎤ ⎡ π
⎤
= ⎢0 – ⎜ – ⎟ ⎥ + ⎢ – 0 ⎥ = π
2
2
⎝
⎠
⎣
⎦
⎣
⎦
22.
∞
csch x dx = ∫
∫1
∞
1
∞
2e
x
1
2x
–1
=∫
e
=
1⎛
⎛ 1 ⎞⎞ 1
= ⎜ 0 − ln ⎜ ⎟ ⎟ = ln 3
2⎝
⎝ 3 ⎠⎠ 2
2x −1
Note:. lim ln =
= 0 since
2x + 1
x →∞
⎛ 2x −1 ⎞
.
lim ⎜
⎟ =1
x →∞ ⎝ 2 x + 1 ⎠
∞
1
2
dx = ∫
dx
1 ex – e– x
sinh x
dx
Let u = e x , du = e x dx .
2e x
∞
∫1
e
2x
–1
dx = ∫
∞
e
∞⎛ 1
1 ⎞
du = ∫ ⎜
–
⎟ du
e
⎝ u –1 u + 1 ⎠
u –1
2
26. The area is
∞
∞⎛ 1
1
1 ⎞
∫1 x 2 + x dx = ∫1 ⎜⎝ x – x + 1 ⎟⎠ dx
2
∞
⎡ u –1 ⎤
= [ln(u –1) – ln(u + 1)]∞
e = ⎢ln
⎥
⎣ u + 1⎦ e
e –1
= 0 – ln
≈ 0.7719
e +1
b –1
b –1 ⎞
⎛
= 0 since lim
= 1⎟
⎜ lim ln
b
b +1 ⎠
+
1
b
→∞
b
→∞
⎝
23.
24.
∞
x ⎤
1
∞ ⎡
= ⎡⎣ln x − ln x + 1 ⎤⎦ = ⎢ ln
= 0 − ln = ln 2
⎥
1
2
⎣ x + 1 ⎦1
.
27. The integral would take the form
∞ 1
∞
k∫
dx = [ k ln x ]3960 = ∞
3960 x
which would make it impossible to send anything
out of the earth's gravitational field.
∞
⎡ 1
⎤
cos x dx = ⎢
(sin x − cos x) ⎥
∫
x
⎣ 2e
⎦0
1
1
= 0 − (0 − 1) =
2
2
(Use Formula 68 with a = –1 and b = 1.)
∞ −x
e
0
28. At x = 1080 mi, F = 165, so
k = 165(1080) 2 ≈ 1.925 × 108 . So the work done
in mi-lb is
∞
∞
1
1.925 × 108 ∫
dx = 1.925 × 108 ⎡ − x −1 ⎤
⎣
⎦1080
1080 x 2
8
1.925 × 10
=
≈ 1.782 × 105 mi-lb.
1080
∞
⎡ 1
⎤
= ⎢−
(cos x + sin x) ⎥
∫
x
⎣ 2e
⎦0
1
1
= 0 + (1 + 0) =
2
2
(Use Formula 67 with a = –1 and b = 1.)
∞ −x
e sin x dx
0
1
1 ⎡ 2x −1 ⎤
∞
⎡ ln 2 x − 1 − ln 2 x + 1 ⎤⎦ = ⎢ ln
1
2⎣
2 ⎣ 2 x + 1 ⎥⎦1
∞
∞
0
0
29. FP = ∫ e− rt f (t ) dt = ∫ 100, 000e−0.08t
∞
⎡ 1
⎤
= ⎢−
100, 000e−0.08t ⎥ = 1,250,000
⎣ 0.08
⎦0
The present value is $1,250,000.
∞
30. FP = ∫ e−0.08t (100, 000 + 1000t )dt
0
∞
= ⎡ −1, 250, 000e−0.08t − 12,500te−0.08t − 156, 250e−0.08t ⎤ = 1,406,250
⎣
⎦0
The present value is $1,406,250.
31.
a.
∞
a
b
1
∞
∫−∞ f ( x) dx = ∫−∞ 0 dx + ∫a b − a dx + ∫b
= 0+
0 dx
1
1
(b − a )
[ x ]b + 0 =
b−a a
b−a
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b.
μ=∫
∞
−∞
=∫
a
x f ( x) dx
b
−∞
x ⋅ 0 dx + ∫ x
a
∞
1
dx + ∫ x ⋅ 0 dx
b
b−a
b
1 ⎡ x2 ⎤
= 0+
⎢ ⎥ +0
b − a ⎢⎣ 2 ⎥⎦
a
2
2
=
b −a
2(b − a)
=
(b + a)(b − a )
2(b − a)
=
a+b
2
σ2 = ∫
∞
−∞
=∫
a
( x − μ ) 2 dx
b
−∞
= 0+
( x − μ )2 ⋅ 0 dx + ∫ ( x − μ )2
a
3 ⎤b
1 ⎡( x − μ )
⎢
⎥ +0
b−a ⎢
3
⎥
⎣
⎦a
3
=
∞
1
dx + ∫ ( x − μ )2 ⋅ 0 dx
b
b−a
3
1 (b − μ ) − ( a − μ )
b−a
3
1 b3 − 3b 2 μ + 3bμ 2 − a3 + 3a 2 μ − 3a μ 2
b−a
3
Next, substitute μ = (a + b) / 2 to obtain
=
σ2 =
=
=
c.
1
( b − a )3
12 ( b − a )
( b − a )2
12
0
−∞
=
a.
2
P ( X < 2) = ∫
=∫
32.
1
⎡ 1 b3 − 3 b 2 a + 3 ba 2 − 1 a3 ⎤
4
4
4
⎦
3(b − a ) ⎣ 4
−∞
2
0 dx + ∫
0
f ( x) dx
1
dx
10 − 0
2 1
=
10 5
∞
x
θ (θ )
∞β
0
∫−∞ f ( x) dx = ∫−∞ 0 dx + ∫0
β −1 −( x / θ ) β
e
dx
In the second integral, let u = ( x / θ ) β . Then,
du = ( β / θ )(t / θ ) β −1 dt . When x = 0, u = 0 and when
x → ∞, u → ∞ . Thus,
∞
∞β
∫−∞ f ( x) dx = ∫0
∞
(x)
θ θ
β −1 − ( x / θ ) β
e
dx
∞
= ∫ e−u du = ⎡ −e−u ⎤ = −0 + e0 = 1
⎣
⎦0
0
492
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b.
μ=∫
∞
−∞
xf ( x) dx = ∫
0
−∞
x ⋅ 0 dx + ∫
∞
0
β
θ
⎛ x⎞
x⎜ ⎟
⎝θ ⎠
β −1
e−( x / θ ) dx∂
2 ∞ 2 − ( x / 3)2
3
π
x e
dx =
3 ∫0
2
=
σ2 = ∫
∞
−∞
( x − μ )2 f ( x) dx = ∫
0
−∞
( x − μ )2 ⋅ 0 dx +
2
2 ∞
( x − μ )2 xe− ( x / 9) dx
∫
0
9
3
3
3
π −μ =
π − π =0
2
2
2
The probability of being less than 2 is
=
c.
2
f ( x ) dx = ∫
∫−∞
0
2β
( )
0 dx + ∫ θ θx
−∞
0
β
β −1 − ( x / θ ) β
e
2
β
dx = 0 + ⎡⎢ −e −( x / θ ) ⎤⎥
⎣
⎦0
2
= 1 − e−(2 / θ ) = 1 − e−(2 / 3) ≈ 0.359
33.
f ′( x) = –
x–μ
σ
3
2π
2
2
e –( x – μ ) / 2σ
2
2
2
( x – μ ) –( x – μ )2 / 2σ 2
e –( x – μ ) / 2σ +
e
3
σ 2π
σ 5 2π
1
f ′′( x) = –
⎛ ( x – μ )2
1 ⎞ –( x – μ )2 / 2σ 2
–
=⎜
=
⎟e
⎜ σ 5 2π σ 3 2π ⎟
⎝
⎠
2
2
1
[( x − μ )2 − σ 2 ]e –( x – μ ) / 2σ
σ 5 2π
f ′′( x) = 0 when ( x – μ )2 = σ 2 so x = μ ± σ and the distance from μ to each inflection point is σ.
34.
a.
b.
⎡ 1
dx = CM k ⎢ –
M x k +1
⎣ kx k
∞
f ( x)dx = ∫
∫– ∞
μ=∫
∞
–∞
∞
CM k
∞
kM k
M
k +1
xf ( x)dx = ∫ x
dx = kM k ∫
∞
1 ⎞ C
C
⎤
k⎛
= . Thus, = 1 when C = k.
⎥ = CM ⎜ 0 +
k ⎟
k
kM ⎠ k
⎦M
⎝
∞
M
b 1
⎛
⎞
dx = kM k ⎜ lim ∫
dx ⎟
k
M
x
x
⎝ b→∞
⎠
1
k
x
This integral converges when k > 1.
b
⎛
⎡
⎤
1
k⎜
When k > 1, μ = kM
lim ⎢ –
⎥
⎜⎜ b→∞ ⎢ (k –1) x k –1 ⎥
⎣
⎦
M
⎝
⎞
⎛
⎞ kM
1
⎟ = kM k ⎜ –0 +
⎟=
k –1 ⎟ k –1
⎜
⎟⎟
(k –1) M
⎝
⎠
⎠
The mean is finite only when k > 1.
Instructor’s Resource Manual
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493
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c.
Since the mean is finite only when k > 1, the variance is only defined when k > 1.
2
⎛ 2 2kM
∞
∞⎛
kM ⎞ kM k
k 2M 2 ⎞ 1
k ∞
=
+
kM
x
–
x
dx
σ 2 = ∫ ( x – μ ) 2 f ( x)dx = ∫ ⎜ x –
dx
⎜
⎟
⎟
∫M ⎜
–∞
M⎝
k –1
k –1 ⎠ x k +1
(k –1)2 ⎟⎠ x k +1
⎝
2k 2 M k +1 ∞ 1
k 3M k +2 ∞ 1
dx +
dx
∫
∫
M x k –1
M xk
k –1
(k –1) 2 M x k +1
The first integral converges only when k – 1 > 1 or k > 2. The second integral converges only when k > 1,
which is taken care of by requiring k > 2.
= kM k ∫
1
∞
dx –
∞
∞
⎤
⎤
2k 2 M k +1 ⎡
1
k 3M k +2
+
–
–
⎥
⎢
⎥
k –1 ⎣⎢ (k –1) x k –1 ⎦⎥
(k –1)2
⎢⎣ (k – 2) x k –2 ⎦⎥ M
M
⎡
1
σ 2 = kM k ⎢ –
∞
⎡ 1 ⎤
⎢– k ⎥
⎣ kx ⎦ M
⎛
⎞ 2k 2 M k +1 ⎛
⎞ k 3M k +2 ⎛
1
1
1 ⎞
= kM k ⎜ –0 +
+
–
–0 +
–0 +
⎟
⎜
⎟
⎟
k –2 ⎟
k –1 ⎟
2 ⎜
⎜
⎜
k
–1
(k – 2) M
(k –1) M
kM k ⎠
⎝
⎠
⎝
⎠ (k –1) ⎝
=
kM 2 2k 2 M 2 k 2 M 2
+
–
k – 2 (k –1) 2 (k –1) 2
⎛ k 2 – 2k + 1 – k 2 + 2k ⎞
⎛ 1
kM 2
k ⎞
= kM 2 ⎜
= kM 2 ⎜
=
–
⎟
⎟
⎜ k – 2 (k –1)2 ⎟
⎜ (k – 2)(k –1) 2
⎟ (k – 2)(k –1)2
⎝
⎠
⎝
⎠
35. We use the results from problem 34:
a.
To have a probability density function (34 a.)
we need C = k ; so C = 3. Also,
kM
μ=
(34 b.) and since, in our problem,
k −1
μ = 20, 000 and
k =3, we have
20000 =
3
4 × 104
M or M =
.
2
3
b. By 34 c., σ 2 =
kM 2
(k − 2)(k − 1)
4 ⎞2
2
so that
36. u = Ar ∫
c.
∞
∫105
37. a.
Thus 6
25
$100,000.
494
sin x dx
0
a →−∞
Both do not converge since –cos x is
oscillating between –1 and 1, so the integral
diverges.
b.
a
[− cos x]− a
∫ sin x dx = alim
a →∞ − a
→∞
lim
a
= lim [− cos a + cos(−a)]
a →∞
= lim [− cos a + cos a] = lim 0 = 0
a →∞
38. a.
of one percent earn over
Section 8.3
∞
0
a
3
⎛ 4 × 10
1⎤
64
⎡ 1
=⎜
−
=
⎟ lim
⎜ 3 ⎟ t →∞ ⎣⎢1015 t 3 ⎦⎥ 27 × 103
⎝
⎠
≈ 0.0024
∞
∫−∞ sin x dx = ∫−∞ sin x dx + ∫0
a →∞
t
⎞
⎡1⎤
⎟ lim ⎢ 3 ⎥
⎟ t →∞ ⎣ x ⎦ 5
10
⎠
4 ⎞3
∞
= lim [ − cos x ]0 + lim [ − cos x ]a
3
⎛ 4 × 104
−⎜
⎜ 3
⎝
( r + x 2 )3 / 2
⎤
⎞
A⎛
a
⎟
⎥ = ⎜1 −
⎟
r ⎝⎜
⎥⎦ a
r 2 + a2 ⎠
dx
x
=
Note that ∫
by using
2
2 3/ 2
(r + x )
r 2 r 2 + x2
the substitution x = r tan θ .
8
⎛ 4 × 104 ⎞
t
3
f ( x) dx = ⎜
dx =
⎟ lim
⎜ 3 ⎟ t →∞ ∫105 x 4
⎝
⎠
dx
2
A⎡
x
= ⎢
r ⎣⎢ r 2 + x 2
3 ⎛ 4 × 10
4 × 10
⎟ =
⎟
4⎝ 3 ⎠
3
σ2 = ⎜
⎜
∞
a
b.
a→∞
The total mass of the wire is
∞ 1
π
∫0 1 + x2 dx = 2 from Example 4.
∞
⎡1
2 ⎤
∫0 1 + x2 dx = ⎢⎣ 2 ln 1 + x ⎥⎦0 which
diverges. Thus, the wire does not have a
center of mass.
∞
x
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39. For example, the region under the curve y =
⎡
⎤
1 ⎤
1
⎡
, n + 1⎥
⎢ n, n + 2 ⎥ and ⎢ n + 1 –
2
2n ⎦
2(n + 1)
⎣
⎢⎣
⎥⎦
1
1
will never overlap since
≤ and
2
2
2n
1
1
≤ .
2
8
2(n + 1)
1
x
to the right of x = 1.
Rotated about the x-axis the volume is
∞ 1
π∫
dx = π . Rotated about the y-axis, the
1 x2
∞
1
volume is 2π ∫ x ⋅ dx which diverges.
1
x
40. a.
The graph of f consists of a series of isosceles
triangles, each of height 1, vertices at
1
1
⎛
⎞
⎛
⎞
⎜ n – 2 , 0 ⎟ , (n, 1), and ⎜ n + 2 , 0 ⎟ ,
2n
2n
⎝
⎠
⎝
⎠
based on the x-axis, and centered over each
integer n.
lim f ( x) does not exist, since f(x) will be 1
Suppose lim f ( x) = M ≠ 0, so the limit
x →∞
exists but is non-zero. Since lim f ( x) = M ,
x →∞
there is some N > 0 such that when x ≥ N,
M
f ( x) – M ≤
, or
2
M
M
M–
≤ f ( x) ≤ M +
2
2
Since f(x) is nonnegative, M > 0, thus
M
> 0 and
2
∞
∫0
f ( x )dx = ∫
N
0
f ( x)dx + ∫
∞
N
x →∞
at each integer, but 0 between the triangles.
Each triangle has area
1
1⎡
1
1 ⎞⎤
⎛
bh = ⎢ n +
–⎜n –
⎟ ⎥ (1)
2
2
2⎣
2n
2n 2 ⎠ ⎦
⎝
=
f ( x)dx
∞
M
⎡ Mx ⎤
dx = ∫ f ( x)dx + ⎢
⎥ =∞
N 2
0
0
⎣ 2 ⎦N
so the integral diverges. Thus, if the limit
exists, it must be 0.
≥∫
b.
N
f ( x)dx + ∫
∞
N
For example, let f(x) be given by
1
⎧ 2
3
⎪2n x – 2n + 1 if n – 2 ≤ x ≤ n
2n
⎪
1
⎪
f ( x) = ⎨ –2n 2 x + 2n3 + 1 if n < x ≤ n +
2n 2
⎪
⎪0
otherwise
⎪
⎩
for every positive integer n.
⎛
⎛
1 ⎞
1 ⎞
3
f ⎜n –
= 2n 2 ⎜ n –
⎟ – 2n + 1
2⎟
2n ⎠
2n 2 ⎠
⎝
⎝
3
3
= 2n – 1 – 2n + 1 = 0
f ( n ) = 2 n 2 ( n ) – 2 n3 + 1 = 1
1⎛ 1 ⎞
1
⎜ ⎟=
2 ⎝ n 2 ⎠ 2n 2
∞
∫0
f ( x)dx is the area in all of the triangles,
thus
∞
∫0
∞
f ( x)dx = ∑
1
n =1 2n
2
=
1 ∞ 1
∑
2 n =1 n 2
=
1 1 ∞ 1 1 1 ∞ 1
+ ∑
≤ +
dx
2 2 n = 2 n 2 2 2 ∫1 x 2
=
1 1 ⎡ 1⎤
1 1
+ –
= + (–0 + 1) = 1
2 2 ⎢⎣ x ⎥⎦1
2 2
∞
∞
(By viewing
1
∑ n2
as a lower Riemann sum
n=2
for
1
x2
Thus,
)
∞
∫0
f ( x )dx converges, although
lim f ( x) does not exist.
x →∞
lim f (n) = lim (–2n 2 x + 2n3 + 1) = 1 = f (n)
x→n+
x →n+
⎛
⎛
1 ⎞
1 ⎞
3
f ⎜n+
= –2n 2 ⎜ n +
⎟ + 2n + 1
2⎟
2n ⎠
2n 2 ⎠
⎝
⎝
= –2n3 –1 + 2n3 + 1 = 0
Thus, f is continuous at
1
1
n–
, n, and n +
.
2
2n
2n 2
Note that the intervals
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Section 8.3
495
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
41.
∫1
1.1
x
∫1
x
100 1
x
1
⎡
⎤
dx = ⎢ –
0.01 ⎥
⎣ 0.01x
⎦1
100
⎡ x 0.01 ⎤
dx = ⎢
⎥
x 0.99
⎣⎢ 0.01 ⎦⎥1
∫1
1
10
∫0
2
dx =
≈ 4.71
1 ⎡ −1 ⎤10
tan x
⎦0
π⎣
π(1 + x )
1.4711
≈
≈ 0.468
π
50
1
1
−1 50
∫0 π(1 + x2 ) dx = π ⎡⎣ tan x ⎤⎦0
1.5508
≈
≈ 0.494
π
100
1
1
−1 100
∫0 π(1 + x2 ) dx = π ⎡⎣ tan x ⎤⎦ 0
1.5608
≈
≈ 0.497
π
1
1
2
2π
1
∫0
∫0
2π
3 1
∫0
2π
4 1
∫0
2π
dx
33 2
3(b – 1) 2 / 3
3
3
=
2 – lim
–0=
3
3
2
2
2
2
b→1+
3
⎡
⎤
3
= lim ⎢ –
2. ∫
⎥
1/ 3
1 ( x – 1) 4 / 3 b →1+
⎢⎣ ( x – 1) ⎥⎦ b
3
3
3
=–
+ lim
=–
+∞
3
1/ 3
3
+
2 b→1 ( x –1)
2
The integral diverges.
3.
3
dx
10
dx
∫3
10
= lim ⎡ 2 x – 3 ⎤⎦
b
x – 3 b→3+ ⎣
= 2 7 – lim 2 b – 3 = 2 7
b →3+
4.
9
∫0
dx
b
= lim ⎡ −2 9 – x ⎤⎦
0
9 – x b→9 – ⎣
= lim − 2 9 – b + 2 9 = 6
b →9 –
5.
exp(–0.5 x 2 )dx ≈ 0.3413
exp(–0.5 x 2 )dx ≈ 0.4772
3
⎡ 3( x – 1) 2 / 3 ⎤
⎢
⎥
∫1 ( x – 1)1/ 3 = blim
2
→1+ ⎣⎢
⎦⎥ b
3
=
≈ 4.50
dx = [ln x]100
1 = ln100 ≈ 4.61
1
100
≈ 3.69
1.
100
1.01
∫1
43.
1 ⎤
⎡
dx = ⎢ –
⎥
⎣ 0.1x 0.1 ⎦1
1
100
Problem Set 8.4
= 0.99
100
1
100
∫1
42.
100
⎡ 1⎤
dx = ⎢ – ⎥
2
⎣ x ⎦1
x
1
100
6.
1
b
dx
= lim ⎡sin –1 x ⎤
⎣
⎦0
2
b →1–
1– x
π
π
= lim sin –1 b – sin –1 0 = – 0 =
–
2
2
b →1
∫0
∞
b
dx = lim ⎡ 1 + x 2 ⎤
∫100
⎥⎦100
2
b →∞ ⎢⎣
1+ x
x
exp(–0.5 x 2 )dx ≈ 0.4987
= lim 1 + b 2 + 10, 001 = ∞
exp(–0.5 x 2 )dx ≈ 0.5000
The integral diverges.
b →∞
7.
3
1
b
b
8.4 Concepts Review
1. unbounded
2. 2
3.
lim
b→4
∫
b
– 0
1
4– x
4. p < 1
496
Section 8.4
dx
1
3
1
∫–1 x3 dx = blim
∫ 3 dx + blim
∫ 3 dx
→0 – –1 x
→0+ b x
3
⎡ 1 ⎤
⎡ 1 ⎤
= lim ⎢ –
+ lim ⎢ –
2⎥
2⎥
–
+
b →0 ⎣ 2 x ⎦ –1 b →0 ⎣ 2 x ⎦ b
⎛
1
1⎞ ⎛ 1
1 ⎞
= ⎜ lim –
+ + – + lim
2 2 ⎟ ⎜ 18
2⎟
–
+
b →0 2b ⎠
⎝ b→0 2b
⎠ ⎝
1⎞ ⎛ 1
⎛
⎞
= ⎜ −∞ + ⎟ + ⎜ – + ∞ ⎟
2⎠ ⎝ 8
⎝
⎠
The integral diverges.
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
–5
∫5
8.
x
2/3
1
b
1
–5
∫ 2 / 3 dx + blim
∫
b →0 + 5 x
→0– b
dx = lim
x
2/3
9.
dx
= lim
= lim ⎡3x1/ 3 ⎤ + lim ⎡3 x1/ 3 ⎤
⎣
⎦ 5 b→0 – ⎣
⎦b
b →0+
b
dx
–5 / 7
128 –5 / 7
∫
b →0 + b
dx + lim
x
dx
128
b
⎡7
⎤
⎡7
⎤
= lim ⎢ x 2 / 7 ⎥ + lim ⎢ x 2 / 7 ⎥
– ⎣2
+
⎦ –1 b→0 ⎣ 2
⎦b
b →0
7
7
7
7
= lim b 2 / 7 – (–1)2 / 7 + (128) 2 / 7 – lim b 2 / 7
– 2
+ 2
2
2
b →0
b→0
7 7
21
= 0 – + (4) – 0 =
2 2
2
= lim 3b1/ 3 – 33 5 + 33 –5 – lim 3b1/ 3
b →0+
3
x
∫ x
b →0 – –1
–5
b
128 –5 / 7
∫–1
b →0 –
3
= 0 – 3 5 + 33 5 – 0 = 33 −5 − 3 5 = −6 3 5
10.
1
∫0 3
x
1 – x2
dx = lim
∫
x
b
b →1– 0 3
1 – x2
dx
b
⎡ 3
⎤
= lim ⎢ – (1 – x 2 )2 / 3 ⎥
–⎣ 4
⎦0
b →1
3
3
3 3
= lim − (1 – b 2 ) 2 / 3 + = –0 + =
– 4
4
4 4
b →1
4
dx
0
(2 – 3x)1/ 3
11. ∫
= lim
b→ 2
∫
– 0
3
dx
b
(2 – 3 x)1/ 3
+ lim
b→ 2
∫
4
+ b
3
4
b
dx
(2 – 3 x)1/ 3
⎡ 1
⎤
⎡ 1
⎤
= lim ⎢ – (2 – 3 x)2 / 3 ⎥ + lim ⎢ – (2 – 3x ) 2 / 3 ⎥
–
+
⎦ 0 b→ 2 ⎣ 2
⎦b
b→ 2 ⎣ 2
3
3
1
1
1
1
= lim − (2 – 3b) 2 / 3 + (2)2 / 3 – (–10) 2 / 3 + lim (2 – 3b)2 / 3
–
+
2
2
2
b→ 2
b→ 2 2
3
3
1
1
1
= 0 + 22 / 3 − 102 / 3 + 0 = (22 / 3 − 102 / 3 )
2
2
2
12.
13.
8
∫
5
x
2 2/3
(16 − 2 x )
–4
x
∫0
16 – 2 x
=
lim
2
dx =
⎡ 3
⎤
dx = lim ⎢ − (16 − 2 x 2 )1/ 3 ⎥
−⎣ 4
⎦
b→ 8
lim
b→ – 8
b
x
dx +
+ ∫0
16 – 2 x 2
b→ – 8
5
3
3
3
= lim − (16 − 2b 2 )1/ 3 + 3 6 = 3 6
− 4
4
4
b→ 8
–4
– ∫b
x
16 – 2 x 2
dx
–4
b
⎡ 1
⎤
⎡ 1
⎤
– ln 16 – 2 x 2 ⎥ + lim ⎢ – ln 16 – 2 x 2 ⎥
+⎢
–
4
4
⎣
⎦
⎣
⎦b
0 b→ – 8
8
b→ –
=
lim
b
1
1
1
1
lim − ln 16 – 2b 2 + ln16 – ln16 + lim
ln 16 – 2b 2
+ 4
–
4
4
4
b→ – 8
b→ – 8
1
⎡
⎤ ⎡ 1
⎤
= ⎢ –(– ∞) + ln16 ⎥ + ⎢ – ln16 + (– ∞) ⎥
4
⎣
⎦ ⎣ 4
⎦
The integral diverges.
14.
15.
3
∫0
–1
b
dx = lim ⎡ – 9 – x 2 ⎤ = lim − 9 – b 2 + 9 = 3
⎢
⎥⎦ 0 b→3–
2
b →3 – ⎣
9– x
x
dx
∫–2 ( x + 1)4 / 3
b
⎡
⎤
3
3
3
= lim –
+
= –(– ∞) – 3
= lim ⎢ –
⎥
1/
3
1/
3
–
(–1)1/ 3
b → –1– ⎣⎢ ( x + 1)
⎦⎥ –2 b→ –1 (b + 1)
The integral diverges.
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Section 8.4
497
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16. Note that
3
dx
⎡
dx
1
1
⎤
∫ x2 + x − 2 = ∫ ( x − 1)( x + 2) = ∫ ⎢⎣ 3( x − 1) − 3( x + 2) ⎥⎦ dx
dx
3
dx
b
by using a partial fraction decomposition.
dx
+ lim ∫
∫0 x2 + x – 2 = blim
∫ 2
2
→1– 0 x + x – 2 b→1+ b x + x – 2
3
b
1
1
⎡1
⎤
⎡1
⎤
= lim ⎢ ln x –1 – ln x + 2 ⎥ + lim ⎢ ln x –1 – ln x + 2 ⎥
– ⎣3
+
3
3
⎦ 0 b→1 ⎣ 3
⎦b
b →1
3
b
⎡1
x –1 ⎤
⎡1
x –1 ⎤
1 b –1 1 1 1 2
1 b –1
= lim ⎢ ln
⎥ + lim+ ⎢ 3 ln x + 2 ⎥ = lim– 3 ln b + 2 – 3 ln 2 + 3 ln 5 – lim+ 3 ln b + 2
– 3
x
+
2
b →1 ⎣
b →1
⎦ 0 b→1 ⎣
⎦ b b→1
1 1⎞ ⎛1 2
⎛
⎞
= ⎜ – ∞ – ln ⎟ + ⎜ ln + ∞ ⎟
3 2⎠ ⎝3 5
⎝
⎠
The integral diverges.
1
17. Note that
3
1
=
2
2
−
1
1
+
4( x − 1) 4( x + 1)
x − x − x + 1 2( x − 1)
3
b
3
dx
dx
dx
+ lim ∫
∫0 x3 – x2 – x + 1 = blim
– ∫0 x3 – x 2 – x + 1
+ b x3 – x 2 – x + 1
b→1
→1
3
b
⎡
⎤
⎡
⎤
1
1
1
1
1
1
= lim ⎢ –
– ln x − 1 + ln x + 1 ⎥ + lim ⎢ –
– ln x − 1 + ln x + 1 ⎥
–
+
4
4
b →1 ⎣ 2( x –1) 4
⎦ 0 b→1 ⎣ 2( x –1) 4
⎦b
⎡⎛
⎡ 1 1
⎛
1
1 b +1 ⎞ ⎛ 1
1
1 b + 1 ⎞⎤
⎞⎤
lim ⎢⎜ –
+ ln
+ ⎜ – + 0 ⎟ ⎥ + lim ⎢ – + ln 2 – ⎜ –
+ ln
⎟
⎟⎥
b −1 ⎠ ⎝ 2
⎠ ⎦ b→1+ ⎣ 4 4
b →1– ⎣⎝ 2(b –1) 4
⎝ 2(b –1) 4 b − 1 ⎠ ⎦
1⎞ ⎛ 1 1
⎛
⎞
= ⎜ ∞ + ∞ – ⎟ + ⎜ – + ln 2 + ∞ – ∞ ⎟
2⎠ ⎝ 4 4
⎝
⎠
The integral diverges.
x1/ 3
18. Note that
x
2/3
−9
1
=
1/ 3
x
+
9
1/ 3
x
( x 2 / 3 − 9)
x1/ 3
.
b
27
⎡ 3 2 / 3 27
⎤
⎛ 3 2 / 3 27
⎞ ⎛
⎞
2/3
2/3
∫0 x2 / 3 – 9 dx = b→lim27 – ⎢⎣ 2 x + 2 ln x – 9 ⎥⎦0 = b→lim27 – ⎜⎝ 2 b + 2 ln b – 9 ⎟⎠ – ⎜⎝ 0 + 2 ln 9 ⎟⎠
27
27
=
– ∞ – ln 9
2
2
The integral diverges.
27
19.
π/4
∫0
b
⎡ 1
⎤
tan 2 xdx = lim ⎢ – ln cos 2 x ⎥
–
2
⎦0
b→ π ⎣
4
1
1
= lim − ln cos 2b + ln1 = –(–∞) + 0
–
2
2
b→ π
4
The integral diverges.
20.
π/2
∫0
π/2
csc xdx = lim ⎡⎣ln csc x – cot x ⎤⎦
b
+
b →0
= ln 1 – 0 – lim ln csc b – cot b
b →0 +
= 0 – lim ln
b →0
+
1 – cos b
sin b
1 – cos b
0
is of the form .
0
b →0+ sin b
1 – cos b
sin b 0
= lim
= =0
lim
+ sin b
+ cos b
1
b →0
b →0
1 – cos b
Thus, lim ln
= – ∞ and the integral
+
sin b
b →0
diverges.
lim
498
Section 8.4
Instructor's Resource Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
21.
π/2
∫0
1 − cos x
x
= sin 2 ,
2
2
1
1
x
= − csc 2 .
cos x − 1
2
2
sin x
π/ 2
dx = lim ⎡⎣ ln 1 – cos x ⎤⎦
b
+
1 – cos x
b →0
25. Since
= ln1 – lim ln 1 – cos b = 0 – (– ∞)
b →0+
The integral diverges.
22.
23.
π/2
⎡3 2/3 ⎤
sin
x⎥
∫0 3 sin x dx = blim
+⎢
⎦b
→0 ⎣ 2
3 2/3 3 2/3 3
= (1)
– (0)
=
2
2
2
π/2
π/2
∫0
cos x
π
b
– lim cot = 0 – ∞
2 b →0 +
2
The integral diverges.
= cot
b
⎡1
⎤
tan x sec x dx = lim ⎢ tan 3 x ⎥
–
3
⎦0
b→ π ⎣
2
π
dx
x⎤
⎡
cot ⎥
∫0 cos x – 1 = blim
+⎢
2 ⎦b
→0 ⎣
π
2
26.
–1
∫–3 x
dx
b
= lim ⎡ 2 ln(– x) ⎤⎦
–3
ln(– x) b→ –1– ⎣
= lim 2 ln(–b) – 2 ln 3 = 0 – 2 ln 3
2
b →−1–
1
1
= lim tan 3 b – (0)3 = ∞
–
3
3
b→ π
= –2 ln 3
2
The integral diverges.
27.
24.
π/4
∫0
sec2 x
b
1 ⎤
⎡
dx = lim ⎢ –
2
–
tan x – 1 ⎥⎦ 0
(tan x – 1)
b→ π ⎣
ln 3
∫0
–
b→ π
4
= lim ⎡ 2 e x –1 ⎤
⎥⎦ b
+⎢
x
e –1 b→0 ⎣
= 2 3 – 1 – lim 2 eb – 1 = 2 2 – 0 = 2 2
b →0+
4
= lim −
ln 3
e x dx
1
1
+
= –(– ∞) – 1
tan b – 1 0 – 1
The integral diverges.
28. Note that
4
∫2
29.
e
4 x − x 2 = 4 − ( x 2 − 4 x + 4) = 22 − ( x − 2)2 . (by completing the square)
dx
4 x – x2
= lim
∫
dx
b
b→4– 2
dx
4 x – x2
b
x – 2⎤
π
π
⎡
–1 b – 2
– sin –1 0 = – 0 =
= lim ⎢sin –1
⎥ = lim– sin
–⎣
2
2
2
2
⎦ 2 b→4
b→ 4
[ln(ln x)]b = ln(ln e) – lim ln(ln b) = ln 1 – ln 0 = 0 + ∞
∫1 x ln x = blim
→1+
b →1+
e
The integral diverges.
10
⎡
1
⎤
1
1
1
= lim –
30. ∫
=–
+ lim
=–
+∞
99
99
1 x ln100 x b →1+ ⎢⎣ 99 ln 99 x ⎥⎦
+
99 ln 10 b→1 99 ln b
99 ln 99 10
b
The integral diverges.
10
31.
dx
4c
⎡
⎤
= lim ⎢ln x + x 2 − 4c 2 ⎥ = ln ⎡⎣ (4 + 2 3)c ⎤⎦ − lim ln b + b 2 − 4c 2
+⎣
2
2
⎦b
b → 2c +
b → 2c
x − 4c
= ln ⎡⎣ (4 + 2 3)c ⎤⎦ − ln 2c = ln(2 + 3)
4c
∫2c
dx
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499
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32.
2c
x dx
∫c
=∫
x 2 + xc – 2c 2
2c
x dx
( x + 2c )
c
2
– 94 c 2
=∫
2c
c
( x + 2c ) dx − c 2c
dx
∫
0
2
2
2
( x + 2c ) − 94 c2
( x + 2c ) − 94 c2
⎡
c
c
= lim ⎢ x 2 + xc – 2c 2 − ln x + + x 2 + xc – 2c 2
+
2
2
b →c ⎣
2c
⎤
⎥
⎦b
c 5c
⎡
c
c
⎤
= 4c 2 − ln
+ 4c 2 – lim ⎢ b 2 + bc – 2c 2 − ln b + + b 2 + bc – 2c 2 ⎥
+
2
2
2
2
b →c ⎣
⎦
c 9c ⎛
c 3c
⎞
c 9c c 3c
c
= 2c − ln – ⎜ 0 − ln
+ 0 ⎟ = 2c − ln + ln = 2c − ln 3
2
2
2
2
2
2 2 ⎝
2
2
⎠
1
33. For 0 < c < 1,
1
dv =
x
is continuous. Let u =
x (1 + x)
1
1
, du = –
dx .
1+ x
(1 + x) 2
dx, v = 2 x .
1
⎡2 x ⎤
1
1
1
2 2 c
xdx
2 c
xdx
xdx
∫c x (1 + x) dx = ⎢1 + x ⎥ + 2∫c (1 + x)2 = 2 – 1 + c + 2∫c (1 + x)2 = 1 – 1 + c + 2∫c (1 + x)2
⎣
⎦c
1
1
⎡ 2 c
1
1
xdx
xdx ⎤
+ 2∫
dx = lim ⎢1 –
⎥ = 1 – 0 + 2∫0
2
c
1+ c
c →0 ⎢⎣
x (1 + x)
(1 + x) 2
(1 + x) ⎥⎦
This last integral is a proper integral.
Thus, lim ∫
1
1
c →0 c
1
34. Let u =
1+ x
1
dv =
x
, du = –
1
2(1 + x)3 / 2
dx
dx, v = 2 x .
1
⎡ 2 x ⎤
1
1
2 1 2 c
x
x
=⎢
dx =
–
+∫
dx
For 0 < c < 1, ∫
⎥ + ∫c
3
/
2
c x(1 + x)
c
2
1+ c
(1 + x)3 / 2
(1 + x)
⎣ 1 + x ⎦c
1
Thus,
1
∫0
dx
dx
x(1 + x)
= lim ∫
1
c →0 c
⎡
⎤
1
1
x
2 c
x
= lim ⎢ 2 –
+∫
dx ⎥ = 2 – 0 + ∫
dx
3
/
2
c
0
1+ c
x(1 + x) c→0 ⎢⎣
(1 + x)
(1 + x)3 / 2
⎥⎦
dx
This is a proper integral.
35.
3
∫–3
x
9 – x2
dx = ∫
= – 9 + lim
+
b → –3
36.
500
0
x
–3
9 – x2
9 – b 2 – lim
b →3
0
b
dx = lim ⎡ – 9 – x 2 ⎤ + lim ⎡ – 9 – x 2 ⎤
⎢
⎥⎦ b b→3– ⎢⎣
⎥⎦ 0
0
2
b→ –3+ ⎣
9– x
dx + ∫
3
x
9 – b 2 + 9 = –3 + 0 – 0 + 3 = 0
–
0
b
⎡ 1
⎤
⎡ 1
⎤
dx = lim ⎢ − ln 9 − x 2 ⎥ + lim ⎢ − ln 9 − x 2 ⎥
∫−3 9 − x2
−3 9 − x 2
0 9 − x2
+⎣ 2
−
⎦ b b→3 ⎣ 2
⎦0
b →3
1
1
ln 9 − b 2 − lim ln 9 − b 2 + ln 3 = (− ln 3 − ∞) + (∞ + ln 3)
= − ln 3 + lim
+ 2
− 2
b →−3
b →3
The integral diverges.
3
x
dx = ∫
Section 8.4
0
x
dx + ∫
3
x
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37.
1
4
∫–4 16 – x2
dx = ∫
1
0
–4 16 – x 2
0
b
⎡1 x + 4 ⎤
⎡1 x + 4 ⎤
dx = lim ⎢ ln
⎥ + lim– ⎢ 8 ln x – 4 ⎥
0 16 – x 2
+ 8
x
–
4
b → –4 ⎣
⎦ b b →4 ⎣
⎦0
dx + ∫
1
4
1
1 b+4
1 b+4 1
= ln1 – lim ln
+ lim ln
– ln1 = (0 + ∞) + (∞ – 0)
+8
–8
8
b
–
4
b–4 8
b→ –4
b→4
The integral diverges.
38.
1
1
∫−1 x
dx = ∫
−1
− ln x
1
−1 2
x − ln x
dx + ∫
1
0
−1 2
x − ln x
−1 2
dx + ∫
1
12
0
x − ln x
1
12
x − ln x
dx
12
b
= lim ⎡ −2 − ln x ⎤
⎣
⎦b
b →−1+
1
dx + ∫
b
+ lim ⎡ −2 − ln x ⎤
+ lim ⎡ −2 − ln x ⎤ + lim ⎡ −2 − ln x ⎤
⎣
⎦ −1 2 b→0+ ⎣
⎦b
⎣
⎦1 2
b →0 −
b→1−
= (−2 ln 2 + 0) + (−∞ + 2 ln 2) + (−2 ln 2 + ∞ ) + (0 + 2 ln 2)
The integral diverges.
39.
∞
∫0
1
xp
1
1
0
xp
dx = ∫
If p > 1,
dx + ∫
1
∞
1
xp
1
= −3π + 3π lim b −1/ 3
x →0 +
b →0
∞
The limit tends to infinity as b → 0, so the
volume is infinite.
⎡ 1
⎤
dx = ⎢
x − p +1 ⎥
If p < 1 and p ≠ 0, ∫
1 xp
⎣ − p +1
⎦1
1
∞
diverges since lim x − p +1 = ∞ .
44. Since ln x < 0 for 0 < x < 1, b > 1
x →∞
If p = 0,
∫0
∫0
If p = 1, both
40.
∞
∫0
1
dx and
x
∞1
∫1
x
= lim [ x ln x − x ]c + [ x ln x − x ]1
1
dx diverge.
c →0
b
+
= −1 − lim (c ln c − c) + b ln b − b + 1
= b ln b − b
Thus, b ln b – b = 0 when b = e.
f ( x)dx
∫
b
− 0
b →1
f ( x)dx + lim
8
∫0 ( x − 8)
∫
c
+ b
where 1 < c < ∞.
−2 / 3
b→1
f ( x)dx + lim
1⎛ 1
b
dx = lim ⎡3( x − 8)1/ 3 ⎤
⎣
⎦0
b →8−
f ( x)dx
45.
1 sin x
∫0
dx is not an improper integral since
x
sin x
is bounded in the interval 0 ≤ x ≤ 1.
x
⎞
1
x
1
Instructor’s Resource Manual
1
1+ x
4
< 1 so
1
4
4
x (1 + x )
<
1
x4
.
b
1 1
⎡ 1 ⎤
+
⎢–
⎥ = – blim
∫1 x 4 dx = blim
→∞ ⎣ 3 x3 ⎦1
→∞ 3b3 3
1 1
= –0 + =
3 3
∞
1
Thus, by the Comparison Test ∫
dx
1 x 4 (1 + x 4 )
converges.
∞
⎡1
⎤
dx = lim ⎢ ln x 2 + 1 ⎥
− ∫b 2
− ⎣2
⎦b
x +1
b →0
b →0
1
1
1
= ln 2 − lim ln b 2 + 1 = ln 2
2
2
b →0− 2
1
b
46. For x ≥ 1,
∫0 ⎜⎝ x − x3 + x ⎟⎠ dx
= lim
∫
b →∞ c
= 3(0) – 3(–2)= 6
42.
b
c →0+
= lim
41.
1
b
∫0 ln x dx = clim
∫ ln x dx + ∫1 ln x dx
→0 − c
dx = ∞ .
1
1
1
V = π∫ x −4 / 3 dx = lim π ⎡ −3x −1/ 3 ⎤
⎣
⎦b
0
b →0 +
b.
since lim x − p +1 = ∞ .
∞
1
dx = lim ⎡3x1/ 3 ⎤ = 3
⎣
⎦b
b →0 +
1
⎡ 1
− p +1 ⎤
∫0 x p dx = ⎢⎣ − p + 1 x ⎥⎦ diverges
0
1
1 −2 / 3
∫0 x
43. a.
dx
1
Section 8.4
501
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
47. For x ≥ 1, x 2 ≥ x so – x 2 ≤ – x, thus
From Example 2 of Section 8.2, lim
51. a.
2
x →∞ e x
e– x ≤ e– x .
∞ –x
e dx
1
∫
= –0 +
1
= lim [– e – x ]1b = – lim
b →∞ eb
b→∞
∞ – x2
∫1
e
dx
converges.
48. Since x + 2 − 1 ≤ x + 2 we know that
∞
1
1
1
. Consider ∫
≥
dx
0
x + 2 −1
x+2
x+2
∞
b
1
1
dx
∫2 x + 2 dx = blim
∫
2
→∞
x+2
∞
= lim ⎡⎣ 2 x + 2 ⎤⎦ = lim 2
2
b →∞
b →∞
(
)
∞ n –1 – x
b+2 −2 = ∞
∫1
1
2
. Since
∞
∫1
∫a g ( x)dx
of
b
∫a
52.
implies the divergence of
b
b
∫a g ( x)dx.
∞ n –1 – x
∫1
x
e dx
1
1
dx = ⎡ – e – x ⎤ = – e –1 + 1 = 1 – , so the
⎣
⎦0
e
integral converges when n = 1. For 0 ≤ x ≤ 1,
0 ≤ x n –1 ≤ 1 for n > 1. Thus,
x
= x n –1e – x ≤ e – x . By the comparison test
from Problem 50,
53. a.
∫a
dx
1 –x
implies the convergence of
f ( x)dx and the divergence of
x2
∫0 e
e
x→a
x →b
1
converges.
x n –1
lim f ( x) = lim g ( x) = ∞, then the convergence
x →b
b
∞
1
x n –1e – x dx
integral is finite, so
2
50. If 0 ≤ f(x) ≤ g(x) on [a, b] and either
lim f ( x) = lim g ( x) = ∞ or
x→a
M
1
∞
x n –1e – x dx + ∫ x n –1e – x dx
M
x n –1e – x dx + ∫
= 1+ ∫
∞
1
M
1
M
1
by part a and Problem 46. The remaining
⎡ 1⎤
dx = ⎢ − ⎥ = 1
⎣ x ⎦1
x
x ln ( x + 1) x
we can apply the Comparison Test of Problem 46
∞
1
to conclude that ∫
dx converges.
1 x 2 ln x + 1
( )
≤
e dx = ∫
x
≤∫
49. Since x 2 ln ( x + 1) ≥ x 2 , we know that
1
b
1 1
⎡ 1⎤
– ⎥ = – lim +
∫1 x2 dx = blim
⎢
b →∞ b 1
→∞ ⎣ x ⎦1
= –0 + 1 = 1
1
∞
b.
Thus, by the Comparison Test of Problem 46, we
∞
1
conclude that ∫
dx diverges.
0
x+2
2
=0
for a any positive real number.
x n +1
= 0 for any positive real
Thus lim
x →∞ e x
number n, hence there is a number M such
x n +1
that 0 <
≤ 1 for x ≥ M. Divide the
ex
x n –1
1
≤
inequality by x 2 to get that 0 <
x
e
x2
for x ≥ M.
+ e –1
1 1
=
e e
Thus, by the Comparison Test,
xa
f ( x)dx
b.
1 n −1 – x
∫0 x
e dx converges.
∞
∞
Γ(1) = ∫ x0 e− x dx = ⎡ −e− x ⎤ = 1
⎣
⎦0
0
∞
Γ(n + 1) = ∫ x n e− x dx
0
Let u = x , dv = e− x dx,
n
du = nx n −1dx, v = −e− x .
∞
n −1 − x
Γ(n + 1) = [− x n e− x ]∞
e dx
0 + ∫ nx
0
∞ n −1 − x
x e dx
0
= 0 + n∫
c.
502
Section 8.4
= nΓ(n)
From parts a and b,
Γ(1) = 1, Γ(2) = 1 ⋅ Γ(1) = 1,
Γ(3) = 2 ⋅ Γ(2) = 2 ⋅1 = 2! .
Suppose Γ(n) = (n − 1)!, then by part b,
Γ(n + 1) = nΓ(n) = n[(n − 1)!] = n ! .
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
∞ –x
54. n = 1,
∫0
n = 2,
∫0
n = 3,
∫0
n = 4,
∫0
n = 5,
∫0
55. a.
e dx = 1 = 0! = (1 –1)!
∞
xe – x dx = 1 = 1! = (2 –1)!
∞ 2 –x
x e dx = 2 = 2! = (3 –1)!
∞ 3 –x
x e dx = 6 = 3! = (4 –1)!
∞ 4 –x
x e dx = 24 = 4! = (5 –1)!
∞
∞
α –1 – β x
∫– ∞ f ( x)dx = ∫0 Cx
y
Let y = βx, so x =
∞
α –1 – β x
∫0 Cx
e
e
and dx =
β
dx = ∫
∞
0
μ=∫
∞
–∞
α –1
∞
xf ( x)dx = ∫ x
0
y
Let y = βx, so x =
dy .
1
e– y
1
β –α Γ(α )
dy =
β
=
C
∫
∞ α –1 – y
y
βα 0
e
dy = C β –α Γ(α )
βα
.
Γ(α )
β α α –1 – β x
βα ∞ α –β x
x e
dx =
x e
dx
Γ(α )
Γ(α ) ∫0
and dx =
β
1
β
⎛ y⎞
C⎜ ⎟
⎝β⎠
C β –α Γ(α ) = 1 when C =
b.
dx
1
β
dy.
α
∞ α –y
βα ∞⎛ y ⎞ – y 1
1
1
1
α
μ=
dy =
y e dy =
Γ(α + 1) =
αΓ(α ) =
⎜ ⎟ e
∫
∫
0
0
β
βΓ(α )
βΓ(α )
βΓ(α )
β
Γ(α ) ⎝ β ⎠
(Recall that Γ(α + 1) = αΓ(α) for α > 0.)
c.
2
α ⎞ β α α –1 – β x
β α ∞ ⎛ 2 2α
α 2 ⎞ α –1 – β x
+
–
x e
dx =
x
x
dx
σ = ∫ ( x – μ ) f ( x)dx = ∫ ⎜ x – ⎟
⎜
⎟x e
–∞
0
Γ(α ) ∫0 ⎜⎝
β
β ⎠ Γ(α )
β 2 ⎟⎠
⎝
β α ∞ α +1 – β x
2αβ α –1 ∞ α – β x
α 2 β α –2 ∞ α –1 – β x
=
x
e
dx
x
e
dx
+
x e
dx
–
Γ(α ) ∫0
Γ(α ) ∫0
Γ(α ) ∫0
∞
2
∞⎛
2
In all three integrals, let y = βx, so x =
α +1
βα ∞⎛ y ⎞
⎜ ⎟
Γ(α ) ∫0 ⎝ β ⎠
σ2 =
=
=
=
1
β
2
e– y
∞ α +1 – y
y e dy –
0
∫
Γ(α )
1
β 2 Γ(α )
α 2 +α
β
2
Γ(α + 2) –
–
2α 2
β
2
+
α2
β
2
Instructor’s Resource Manual
dy –
β
2α
2
2α
β 2 Γ(α )
=
β
and dx =
1
β
dy .
α
1
β
y
α –1
2αβ α –1 ∞ ⎛ y ⎞ – y 1
α 2 β α –2 ∞ ⎛ y ⎞
e
dy
+
⎜ ⎟
⎜ ⎟
β
Γ(α ) ∫0 ⎝ β ⎠
Γ(α ) ∫0 ⎝ β ⎠
∞ α –y
y e dy +
0
∫
Γ(α )
Γ(α + 1) +
α2
β 2 Γ(α )
α2
β
2
∞ α –1 – y
∫
Γ(α ) 0
Γ(α ) =
y
1
β 2 Γ(α )
e
e– y
1
β
dy
dy
(α + 1)αΓ(α ) –
2α
β 2 Γ(α )
αΓ(α ) +
α2
β2
α
β2
Section 8.4
503
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
56. a.
∞
L{tα }( s ) = ∫ tα e – st dt
0
Let t =
x
1
, so dt = dx, then
s
s
∞ α – st
t e dt
0
∫
α
∞⎛ x ⎞
∞ 1
1
Γ(α + 1)
= ∫ ⎜ ⎟ e – x dx = ∫
xα e – x dx =
.
0 ⎝s⎠
0 sα +1
s
sα +1
If s ≤ 0 , tα e− st → ∞ as t → ∞, so the integral does not converge. Thus, the transform is defined only when
s > 0.
∞
b.
∞
∞
1 ⎡
⎡ 1 (α – s )t ⎤
=
lim e(α – s )b –1⎤⎥
L{eα t }( s ) = ∫ eα t e – st dt = ∫ e(α – s )t dt = ⎢
e
⎥
0
0
⎦
⎣α – s
⎦ 0 α – s ⎢⎣b→∞
⎧∞ if α > s
lim e(α – s )b = ⎨
b →∞
⎩0 if s > α
−1
1
when s > α. (When s ≤ α , the integral does not converge.)
=
Thus, L{eα t }( s ) =
α – s s −α
c.
∞
L{sin(α t )}( s ) = ∫ sin(α t )e – st dt
0
∞
Let I = ∫ sin(α t )e – st dt and use integration by parts with u = sin(α t), du = α cos(α t)dt,
0
dv = e
– st
1
dt , and v = – e – st .
s
∞
α ∞
⎡ 1
⎤
Then I = ⎢ – sin(α t )e – st ⎥ + ∫ cos(α t )e – st dt
s
⎣
⎦0 s 0
Use integration by parts on this integral with
1
u = cos(αt), du = –α sin(αt)dt, dv = e – st dt , and v = – e – st .
s
∞
∞
⎛
⎞
α ⎡ 1
α ∞
⎡ 1
⎤
⎤
I = ⎢ – sin(α t )e – st ⎥ + ⎜ ⎢ – cos(α t )e – st ⎥ – ∫ sin(α t )e – st dt ⎟
⎟
⎣ s
⎦ 0 s ⎜⎝ ⎣ s
⎦0 s 0
⎠
∞
1⎡
α
α2
⎛
⎞⎤
I
= – ⎢e – st ⎜ sin(α t ) + cos(α t ) ⎟ ⎥ –
s⎣
s
⎝
⎠⎦0 s2
Thus,
∞
⎛ α2 ⎞
1 ⎡ – st ⎛
α
⎞⎤
I ⎜1 +
⎟ = – ⎢ e ⎜ sin(α t ) + cos(α t ) ⎟ ⎥
⎜
s⎣
s
⎝
⎠⎦0
s 2 ⎠⎟
⎝
I=–
(
∞
1
2
s 1 + α2
s
)
s
⎡
α
⎡ – st ⎛
α
⎛
⎞ α⎤
⎞⎤
e – sb ⎜ sin(α b) + cos(α b) ⎟ – ⎥
⎢ e ⎜ sin(α t ) + s cos(α t ) ⎟ ⎥ = – 2
⎢blim
2
s
⎝
⎠⎦0
⎝
⎠ s⎦
s + α ⎣ →∞
⎣
α
⎛
⎞ ⎧0 if s > 0
lim e – sb ⎜ sin(α b) + cos(α b) ⎟ = ⎨
s
b →∞
⎝
⎠ ⎩∞ if s ≤ 0
Thus, I =
504
α
s +α 2
Section 8.4
2
when s > 0.
Instructor’s Resource Manual
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
57. a.
The integral is the area between the curve
1– x
y2 =
and the x-axis from x = 0 to x =1.
x
1– x
y2 =
; xy 2 = 1 – x; x( y 2 + 1) = 1
x
1
x=
2
y +1
1– x
→ ∞, while
x
As x → 0, y =
when x = 1, y =
∞
∫0
1
2
y +1
1–1
= 0, thus the area is
1
dy = lim [tan –1 y ]b0
b →∞
= lim tan –1 b – tan –1 0 =
b →∞
π
2
b. The integral is the area between the curve
1+ x
y2 =
and the x-axis from x = –1 to
1– x
x = 1.
1+ x 2
y2 =
; y – xy 2 = 1 + x; y 2 –1 = x( y 2 + 1);
1– x
x=
2
58. For 0 < x < 1, x p > x q so 2 x p > x p + x q and
1
1
. For 1 < x, x q > x p so
>
p
q
p
2x
x +x
1
1
q
p
.
>
2 x > x + x q and
p
q
2 xq
x +x
∞
1
∞
1
1
1
∫0 x p + xq dx = ∫0 x p + xq dx + ∫1 x p + xq dx
Both of these integrals must converge.
1
1 1
1
1 1 1
∫0 x p + xq dx > ∫0 2 x p dx = 2 ∫0 x p dx which
converges if and only if p < 1.
∞
∞ 1
1
1 ∞ 1
∫1 x p + xq dx > ∫1 2 x q dx = 2 ∫1 xq dx which
converges if and only if q > 1. Thus, 0 < p < 1
and 1 < q.
8.5 Chapter Review
Concepts Test
1. True:
See Example 2 of Section 8.2.
2. True:
Use l'Hôpital's Rule.
y –1
y2 + 1
3. False:
1 + (–1)
=
1 – (–1)
When x = –1, y =
0
= 0, while
2
1+ x
→ ∞.
1– x
The area in question is the area to the right of
1+ x
and to the left of the
the curve y =
1– x
line x = 1. Thus, the area is
∞⎛
∞ 2
y2 – 1 ⎞
∫0 ⎜⎜1 – y 2 + 1 ⎟⎟ dy = ∫0 y 2 + 1 dy
⎝
⎠
4. False:
as x → 1, y =
b
= lim ⎡ 2 tan –1 y ⎤
⎦0
b →∞ ⎣
⎛ π⎞
lim 2 tan –1 b – 2 tan –1 0 = 2 ⎜ ⎟ = π
b →∞
⎝2⎠
lim
x →∞
1000 x 4 + 1000
4
0.001x + 1
=
1000
= 106
0.001
lim xe −1/ x = ∞ since e−1/ x → 1 and
x →∞
x → ∞ as x → ∞ .
5. False:
For example, if f(x) = x and
g ( x) = e x ,
lim
x
x →∞ e x
= 0.
6. False:
See Example 7 of Section 8.2.
7. True:
Take the inner limit first.
8. True:
Raising a small number to a large
exponent results in an even smaller
number.
9. True:
Since lim f ( x) = –1 ≠ 0, it serves
x→a
only to affect the sign of the limit of
the product.
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505
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10. False:
Consider f ( x) = ( x – a )2 and
g ( x) =
1
2
17. False:
g ( x) = 4 x3 + 2 x + 3; f ′( x) = 6 x + 1
, then lim f ( x ) = 0
x →a
( x – a)
and lim g ( x) = ∞, while
g ′( x) = 12 x 2 + 2, and so
f ′( x)
6x + 1
1
= lim
= while
lim
2
2
x →0 g ′( x )
x →0 12 x + 2
x→a
lim [ f ( x) g ( x)] = 1.
x→a
11. False:
Consider f ( x) = 3x 2 and
x →∞
x →0
f ( x)
3x2
= lim
g ( x) x→∞ x 2 + 1
3
= lim
= 3, but
x →∞ 1 + 1
2
18. False:
p > 1. See Example 4 of Section 8.4.
19. True:
∫0
∞
1
1
x
1
lim [ f ( x ) – 3 g ( x)]
x →∞
∞
∫1
2
= lim [3x – 3( x + 1)]
x →∞
1
12. True:
13. True:
14. True:
Let y = [1 + f ( x)]
ln[1 + f ( x)]
= lim
f ( x)
x→a
x →a
∞
∞
∫– ∞ f ( x)dx
f ( x)dx
f ( x)dx.
0
.
0
converges.
22. False:
See Problem 37 of Section 8.3.
23. True:
∫0
f ′( x)
∞
b
∫
b →∞ 0
f ′( x)dx = lim
f ′( x )dx
= lim [ f ( x)]b0 = lim f (b) – f (0)
f ′( x)
x →a
converge so their sum
b →∞
b →∞
= 0 – f(0) = –f(0).
f(0) must exist and be finite since
f ′( x) is continuous on [0, ∞).
24. True:
∞
∫0
∞
f ( x )dx ≤ ∫ e – x dx = lim [– e – x ]b0
0
Use repeated applications of
l'Hôpital's Rule.
= lim – e
e0 = 1 and p(0) is the constant term.
must converge.
Section 8.5
∞
0
Thus, both integrals making up
b →∞
25. False:
506
∞
0
lim [1 + f ( x)]1/ f ( x ) = lim eln y = e1 = e
16. True:
1
dx .
x +1
∫– ∞ f ( x)dx = ∫0
1
=1
x →a 1 + f ( x)
15. True:
∞
∫0
∫– ∞ f ( x)dx = ∫−∞ f ( x)dx + ∫0
= lim
x→a
dx;
If f is an even function, then
f(–x) = f(x) so
1
ln[1 + f ( x)]
lim
ln[1 + f ( x)] = lim
f ( x)
x→a f ( x)
x →a
lim
xp
21. True:
1
ln y =
ln[1 + f ( x)].
f ( x)
1
1+ f ( x )
1
∞
1
Consider
, then
This limit is of the form
x
dx + ∫
20. False:
See Example 7 of Section 8.2.
1/ f ( x )
p
diverges for p ≥ 1 and
As x → a, f ( x) → 2 while
1
→ ∞.
g ( x)
1
dx diverges for p ≤ 1.
xp
= lim [–3] = –3
x →∞
1
0
dx = ∫
p
∫0 x p dx
x
2
f ( x)
3x 2 + x + 1 1
= lim
=
g ( x ) x →0 4 x 3 + 2 x + 3 3
lim
g ( x) = x 2 + 1, then
lim
Consider f ( x) = 3x 2 + x + 1 and
–b
+ 1 = 1, so
b →∞
∞
∫0
f ( x)dx
The integrand is bounded on the
⎡ π⎤
interval ⎢ 0, ⎥ .
⎣ 4⎦
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Sample Test Problems
1. The limit is of the form
0
.
0
2 x3
6 x2
= lim
= lim 6 x3 = ∞
x →∞ ln x
x →∞ 1
x →∞
lim
4x
4
= lim
=4
x →0 tan x x →0 sec 2 x
x
lim
2. The limit is of the form
1
→ ∞ . A number
x
less than 1, raised to a large power, is a very
⎛ ⎛ 1 ⎞32
⎞
small number ⎜ ⎜ ⎟ = 2.328 × 10−10 ⎟ so
⎜⎝ 2 ⎠
⎟
⎝
⎠
9. As x → 0, sin x → 0 , and
0
.
0
tan 2 x
2sec2 2 x 2
= lim
=
3
x →0 sin 3 x
x →0 3cos 3 x
lim
0
3. The limit is of the form . (Apply l’Hôpital’s
0
Rule twice.)
sin x − tan x
cos x − sec2 x
lim
= lim
2x
1 x2
x →0
x →0
3
3
= lim
− sin x − 2sec x(sec x tan x)
2
3
x →0
x →0 +
10.
lim x ln x = lim
x →0 +
lim
x →0 +
= ∞ (L’Hôpital’s Rule does not apply
x2
since cos(0) = 1.)
2 x cos x
x →0
x →0 sin x
0
The limit is of the form .
0
2 x cos x
2 cos x – 2 x sin x
lim
= lim
cos x
x →0 sin x
x →0
2–0
=
=2
1
5. lim 2 x cot x = lim
6. The limit is of the form
∞
.
∞
− 1−1x
ln(1 − x)
= lim
2
x →1 cot πx
x →1− −π csc πx
sin 2 πx
= lim
x →1− π(1 − x )
lim
−
The limit is of the form
0
.
0
sin 2 πx
2π sin πx cos πx
lim
= lim
=0
− π(1 – x )
−
−π
x →1
x →1
7. The limit is of the form
ln t
t →∞ t 2
= lim
1
t
t →∞ 2t
∞
.
∞
= lim
1
t →∞ 2t 2
Instructor’s Resource Manual
x →0+
ln x
1
x
∞
.
∞
The limit is of the form
=0
x →0
lim
lim (sin x)1/ x = 0 .
cos x
4. lim
∞
.
∞
8. The limit is of the form
ln x
1
x
1
x
1
x →0 + – 2
x
= lim
= lim – x = 0
x →0 +
11. The limit is of the form 00.
Let y = x x , then ln y = x ln x.
lim x ln x = lim
x →0 +
x →0+
ln x
1
x
∞
.
∞
The limit is of the form
lim
x →0 +
ln x
1
x
1
x
1
x →0 + – 2
x
ln y
= lim
lim x x = lim e
x →0 +
x →0 +
= lim – x = 0
x →0 +
=1
12. The limit is of the form 1∞.
2
ln(1 + sin x).
x
2
2 ln(1 + sin x )
lim ln(1 + sin x) = lim
x
x →0 x
x →0
0
The limit is of the form .
0
Let y = (1 + sin x)2 / x , then ln y =
2
cos x
2 ln(1 + sin x)
= lim 1+sin x
1
x
x →0
x →0
2 cos x 2
= lim
= =2
1
x →0 1 + sin x
lim
lim (1 + sin x)2 / x = lim eln y = e 2
x →0
x →0
=0
Section 8.5
507
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13.
x ln x = lim
lim
x →0 +
x →0+
1
x
The limit is of the form
lim
ln x
1
x
x →0 +
= lim
x →0 +
–
17. The limit is of the form 1∞.
ln x
Let y = (sin x) tan x , then ln y = tan x ln(sin x).
∞
.
∞
1
x
1
2 x3/ 2
lim tan x ln(sin x ) = lim
x→ π
2
= lim – 2 x = 0
x →0+
The limit is of the form
sin x
= lim
x→ π
2
lim (sin x) tan x = lim eln y = 1
18.
lim t1/ t = lim eln y = 1
t →∞
∞
16. The limit is of the form . (Apply l’Hôpital’s
∞
Rule three times.)
tan 3x
3sec2 3 x
= lim
lim
2
x → π tan x
x → π sec x
508
3(cos 3 x − sin 3 x)
Section 8.5
=−
1
1
=
3(0 − 1) 3
x→ π
2
sin x + x cos x 1
= =1
sin x
1
∞
⎡ 1 ⎤
∫0 ( x + 1)2 = ⎢⎣ – x + 1⎥⎦0 = 0 + 1 = 1
20.
∫0 1 + x2 = ⎣⎡ tan
21.
1 2
1 2
⎡ 1 2x ⎤
2x
∫– ∞ e dx = ⎢⎣ 2 e ⎥⎦ – ∞ = 2 e – 0 = 2 e
22.
ln(1 – x)]–1
∫–11 – x = blim[–
→1
∞
∞
dx
dx
∞
π
π
x⎤ = – 0 =
⎦0 2
2
1
1
1
–1
dx
b
= – lim ln(1 – b) + ln 2 = ∞
b→1
cos x sin x
= lim
= lim
2
π
π
cos
3 x sin 3 x
x→
x → cos 3 x
2
2
2
cos x
= lim
0
.
0
19.
3cos x
cos 2 x − sin 2 x
x sin x – π2
x→ π
2
2
2
x sin x – π2
π
⎛
⎞
lim ⎜ x tan x – sec x ⎟ = lim
2
⎠ x → π2 cos x
x→ π ⎝
2
lim
2
x→ π
2
The limit is of the form
1⎞
x – sin x
⎛ 1
lim ⎜
– ⎟ = lim
+ ⎝ sin x
+
x
⎠ x →0 x sin x
x →0
0
The limit is of the form . (Apply l’Hôpital’s
0
Rule twice.)
x – sin x
1 – cos x
= lim
lim
+ x sin x
+ sin x + x cos x
x →0
x →0
sin x
0
= lim
= =0
+ 2 cos x – x sin x
2
x →0
2
x→ π
2
1
= lim
cos x(1 + ln(sin x)) 0
= =0
sin x
1
x→ π
ln t
1
= lim t = lim = 0
t →∞ t
t →∞ 1 t →∞ t
2
0
.
0
lim
lim
15.
sin x ln(sin x)
cos x
cos x ln(sin x) + sin x cos x
sin x ln(sin x)
= lim
cos x
sin x
x→ π
x→ π
2
2
14. The limit is of the form ∞ 0 .
1
Let y = t1/ t , then ln y = ln t.
t
1
ln t
lim ln t = lim
t →∞ t
t →∞ t
∞
The limit is of the form .
∞
t →∞
x→ π
2
The integral diverges.
23.
∞
dx
= [ln( x + 1)]0∞ = ∞ – 0 = ∞
x +1
The integral diverges.
∫0
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
24.
2
dx
∫12 x(ln x)1/ 5
= lim
dx
b
∫1 x(ln x)1/ 5
b →1– 2
2
b
⎡5
⎤
⎡5
⎤
= lim ⎢ (ln x)4 / 5 ⎥ + lim ⎢ (ln x) 4 / 5 ⎥
1/ 5
1
+ ∫b
–
+
⎦
⎦b
x(ln x)
b →1
b →1 ⎣ 4
b →1 ⎣ 4
2
+ lim
dx
2
4/5 ⎞
⎛5
5⎛ 1⎞
5 ⎞ 5
5⎛ 1⎞
⎛5
= ⎜ (0) – ⎜ ln ⎟ ⎟ + ⎜ (ln 2)4 / 5 – (0) ⎟ = (ln 2) 4 / 5 – ⎜ ln ⎟
⎜4
4⎝ 2⎠
4⎝ 2⎠ ⎟ ⎝ 4
4 ⎠ 4
⎝
⎠
5
= [(ln 2) 4 / 5 – (ln 2) 4 / 5 ] = 0
4
4/5
5
= [(ln 2)4 / 5 – (– ln 2)4 / 5 ]
4
∞
25.
∞⎛ 1
π
1 ⎞
π
π π
⎡ 1
–1 ⎤
–1
∫1 x 2 + x4 = ∫1 ⎝⎜ x2 − 1 + x 2 ⎠⎟ dx = ⎢⎣ – x – tan x ⎥⎦1 = 0 – 2 + 1 + tan 1 = 1 + 4 – 2 = 1 – 4
26.
1
⎡ 1 ⎤
∫– ∞ (2 – x)2 = ⎢⎣ 2 – x ⎥⎦ – ∞ = 1 − 0 = 1
27.
b
0 dx
dx
dx
⎡1
⎤
⎡1
⎤
∫–2 2 x + 3 = lim3 – ∫–2 2 x + 3 + lim3 + ∫b 2 x + 3 = lim3 – ⎢⎣ 2 ln 2 x + 3 ⎥⎦ –2 + lim3 + ⎢⎣ 2 ln 2 x + 3 ⎥⎦b
b→ –
b→ –
b→ –
b→ –
2
2
2
2
∞
dx
1
1
dx
0
b
0
⎛
⎞ ⎛
⎞
1
1
1
1
⎛1
⎞
ln 2b + 3 – (0) ⎟ + ⎜ ln 3 – lim
ln 2b + 3 ⎟ = (– ∞) + ⎜ ln 3 + ∞ ⎟
= ⎜ lim
⎜⎜
⎟
⎜
⎟
–
+
2 ⎟ ⎜2
2
3 2
⎝2
⎠
⎟
b→ – 3
2
⎝ b→ – 2
⎠ ⎝
⎠
The integral diverges.
4
28.
∫1
29.
∫2
dx
x –1
∞
∞
= lim [2 x – 1]b4 = 2 3 – lim 2 x – 1 = 2 3 – 0 = 2 3
b →1+
b →1+
∞
1
1
⎡ 1 ⎤
= ⎢–
= –0 +
=
⎥
2
ln 2 ln 2
⎣ ln x ⎦ 2
x(ln x)
dx
dx
∞
⎡
2 ⎤
2
= ⎢–
= –0 + = 2
x/2 ⎥
1
⎣ e
⎦0
30.
∫0
31.
+ lim ∫
∫3 (4 – x)2 / 3 = blim
∫
2/3
2/3
b →4+ b (4 – x )
→ 4 – 3 (4 – x )
5
ex / 2
dx
5
dx
b
dx
b
5
= lim ⎡ –3(4 – x)1/ 3 ⎤ + lim ⎡ –3(4 – x)1/ 3 ⎤
⎣
⎦ 3 b → 4+ ⎣
⎦b
b→4–
= lim − 3(4 – b)1/ 3 + 3(1)1/ 3 – 3(–1)1/ 3 + lim 3(4 – b)1/ 3 = 0 + 3 + 3 + 0 = 6
b→4+
b→4–
∞
∞
2
2⎤
1
1
⎡ 1
xe – x dx = ⎢ – e – x ⎥ = 0 + e –4 = e –4
2
2
2
⎣
⎦2
32.
∫2
33.
∫– ∞ x 2 + 1 dx = ∫– ∞ x2 + 1 dx + ∫0
∞
0
x
∞
x
x
2
x +1
0
∞
1⎡
1
ln( x 2 + 1) ⎤ + ⎡ ln( x 2 + 1) ⎤ =
⎣
⎦
⎣
⎦
–∞ 2
0
2
(0 + ∞) + (∞ – 0)
=
34.
dx
The integral diverges.
0
∞
⎡1
⎤
⎡1
⎤
dx = ⎢ tan –1 x 2 ⎥ + ⎢ tan –1 x 2 ⎥
∫– ∞ 1 + x 4
–∞ 1 + x4
0 1 + x4
2
2
⎣
⎦ –∞ ⎣
⎦0
π
π
1
1
1
1
π
π
⎛
⎞
⎛
⎞
= tan –1 0 – ⎜ ⎟ + ⎜ ⎟ – tan –1 0 = 0 – + – 0 = 0
2
2⎝ 2⎠ 2⎝ 2⎠ 2
4 4
∞
x
dx = ∫
0
x
Instructor’s Resource Manual
dx + ∫
∞
x
Section 8.5
509
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
35.
ex
=
e2 x + 1
ex
(e x ) 2 + 1
Let u = e x , du = e x dx
∫0
∞
∞ 1
π
π π π
dx = ∫
du = ⎡ tan –1 u ⎤ = – tan –1 1 = – =
2
⎣
⎦
1
1
2
2 4 4
u +1
+1
ex
∞
e
2x
36. Let u = x3 , du = 3 x 2 dx
∞ 2 – x3
∞ 1 –u
1 0 –u
1 ∞ –u
1
1
1
1
–u 0
–u ∞
∫– ∞ x e dx = ∫– ∞ 3 e du = 3 ∫– ∞ e du + 3 ∫0 e du = 3 ⎡⎣ –e ⎤⎦ – ∞ + 3 ⎡⎣ – e ⎤⎦ 0 = 3 (–1 + ∞) + 3 (–0 + 1)
The integral diverges.
37.
3
x
∫−3
dx = 0
9 − x2
See Problem 35 in Section 8.4.
38. let u = ln(cos x), then du =
∫
π
2
π
3
tan x
(ln cos x) 2
–∞
1
1–
2
ln
2 u
dx = ∫
39. For p ≠ 1, p ≠ 0,
1
lim
b →∞ b p –1
= 0 when p – 1 > 0 or p > 1, and lim
x
∞
b →0 b
∞
= ∞ – 1 . The integral diverges.
dx converges when p > 1 and diverges when p ≤ 1.
40. For p ≠ 1, p ≠ 0,
lim
1
p –1
1
⎡
⎤
1
1
1
∫0 x p dx = ⎢⎢ – ( p – 1) x p –1 ⎥⎥ = 1 – p + blim
→
0
( p – 1)b p –1
⎣
⎦0
1
1
converges when p – 1 < 0 or p < 1.
11
1
When p = 1,
ln b = ∞ . The integral diverges.
∫0 x dx = [ln x]0 = 0 – blim
→0+
When p = 0,
∫01dx = [ x]0 = 1 – 0 = 1
1
1
∫0 x p dx
1
1
converges when p < 1 and diverges when 1 ≤ p.
41. For x ≥ 1, x 6 + x > x 6 , so
Section 8.5
1
x 6 + x > x 6 = x3 and
converges since 3 > 1 (see Problem 39). Thus
510
= ∞ when p < 1, p ≠ 0.
dx = [ln x]1∞ = ∞ – 0 . The integral diverges.
∫1 1dx = [ x]1
xp
1
b →∞ b p −1
∞1
1
∞
When p = 0,
1
ln
1
1
⎡ 1⎤ 2
du = ⎢ – ⎥
=–
+0 =
2
1
u
ln
2
ln 2
⎣ ⎦ –∞
u
1
1
∫1
∞
du = ∫
1
2
–∞
ln
⎡
⎤
1
1
1
+
∫1 x p dx = ⎢⎢ – ( p – 1) x p –1 ⎥⎥ = blim
p –1
p
–1
→∞
(1 – p )b
⎣
⎦1
∞
When p = 1,
∫1
1
⋅ – sin x dx = – tan x dx
cos x
∞
∫1
x6 + x
1
x6 + x
<
1
x
3
. Hence,
∞
∫1
1
x6 + x
∞
1
1
x3
dx < ∫
dx which
dx converges.
Instructor’s Resource Manual
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of
this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
42. For x > 1, ln x < e x , so
ln x
e
ln x
ln x
x
< 1 and
1
=
< .
e 2 x (e x ) 2 e x
Hence,
∞ ln x
∞ –x
–x ∞
–1 1
∫1 e2 x dx < ∫1 e dx = [– e ]1 = –0 + e = e .
∞ ln x
Thus, ∫
dx converges.
1 e2 x
43. For x > 3, ln x > 1, so
∞ ln x
∫3
x
dx > ∫
∞1
3
x
ln x 1
> . Hence,
x
x
dx = [ln x]3∞ = ∞ – ln 3.
The integral diverges, thus
∞ ln x
∫3
x
dx also
diverges.
44. For x ≥ 1, ln x < x, so
ln x
ln x 1
.
< 1 and
<
x
x3
x2
4. Original:
f continuous at c ⇒ f differentiable at c
Converse:
f differentiable at c ⇒ f continuous at c (AT)
Contrapositive:
f non-differentiable at c ⇒ f discontinuous at c
5. Original:
f right continuous at c ⇒ f continuous at c
Converse:
f continuous at c ⇒ f right continuous at c
(AT)
Contrapositive:
f discontinuous at c ⇒ f not right continuous at c
6. Original: f ′( x) ≡ 0 ⇒ f ( x) = c (AT)
Converse: f ( x) = c ⇒ f ′( x) ≡ 0 (AT)
Contrapositive: f ( x) ≠ c ⇒ f ′( x) ≡ 0 (AT)
7. Original: f ( x) = x 2 ⇒ f ′( x) = 2 x (AT)
Converse: f ′( x) = 2 x ⇒ f ( x) = x 2
Hence,
∞
1
(Could have f ( x) = x 2 + 3 )
∞
⎡ 1⎤
∫1 x3 dx < ∫1 x2 dx = ⎢⎣ – x ⎥⎦1 = –0 + 1 = 1.
∞ ln x
dx converges.
Thus, ∫
1 x3
∞ ln x
Contrapositive: f ′( x ) ≠ 2 x ⇒ f ( x) ≠ x 2 (AT)
8. Original: a 0 , then x 2 > 0 (AT)
Converse: If x 2 > 0 , then x > 0
1 1 4 2 1 7
+ = + + =
2 4 4 4 4 4
9. 1 +
1 1 1 1 1
+ + + +
=
2 4 8 16 32
32 16 8
4
2
1 63
+ + + + +
=
32 32 32 32 32 32 32
10. 1 +
Contrapositive: If x 2 ≤ 0 , then x ≤ 0 (AT)
2. Original: If x 2 > 0 , then x > 0
Converse: If x > 0 , then x 2 > 0 (AT)
4
11.
1
1 1
1
1
∑i = 1+ 2 + 3+ 4 =
i =1
12 + 6 + 4 + 3 25
=
12
12
2
Contrapositive: If x ≤ 0 , then x ≤ 0
3. Original:
f differentiable at c ⇒ f continuous at c (AT)
Converse:
f continuous at c ⇒ f differentiable at c
Contrapositive:
f discontinuous at c ⇒ f non-differentiable at c
(AT)
Instructor’s Resource Manual
4
12.
∑
(−1) k
k
=
−1 1 −1 1
+ + + =
2 4 8 16
2
−8 + 4 − 2 + 1 −5
=
16
16
k =1
⎛∞⎞
13. By L’Hopital’s Rule ⎜ ⎟ :
⎝∞⎠
x
1 1
lim
= lim =
x →∞ 2 x + 1 x →∞ 2 2
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511
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⎛∞⎞
14. By L’Hopital’s Rule ⎜ ⎟ twice:
⎝∞⎠
20.
∞
∫1
x
2
x +1
dx = lim ∫
t →∞ 1
n2
2n 2 1
lim
= lim
= =
2
4 2
n →∞ 2n + 1 n→∞ 4n
x
lim
x →∞
e
x
= lim
x →∞
2x
e
= lim
x
x →∞
2
ex
lim
en
n →∞
17.
∞1
∫1
x
= lim
n →∞
2n
= lim
en
dx = lim ∫
t
n →∞
1
2
=0
21.
en
=0
lim [ ln x ]
dx =
∞
∫1
1
x2
t
1
t →∞ 1
x2
u = x 2 +1
du = 2 x dx
t
dx = lim ∫1
x
dx =
t →∞
x2 + 1
x2 + 1
Integral does not converge.
1
∞
∫2
x (ln x)
2
dx = lim ∫
t
t →∞ 2
(
1
2
ln x +1
2
1
x(ln x) 2
)
∞
=∞
1
dx =
ln t
⎡ 1⎤
du = lim ⎢ − ⎥
=
∫
t →∞ ln 2 u 2
t →∞ ⎣ u ⎦ ln 2
1 ⎤
1
⎡ 1
lim ⎢
−
=
≈ 1.443
⎥
t →∞ ⎣ ln 2 ln t ⎦ ln 2
[ ]=∞
dx = lim ∫
22.
x
∞
∫1
lim
Integral does not converge.
18.
dx =
u = ln x
du = 1 x dx
t →∞ 1 x
t
= lim ln t
1 t →∞
t →∞
x +1
Integral does not converge (see problem 17).
⎛∞⎞
16. By L’Hopital’s Rule ⎜ ⎟ twice:
⎝∞⎠
n2
2
t 2 +1 1
1
lim ∫
du = ∞
2 t →∞ 2 u
⎛∞⎞
15. By L’Hopital’s Rule ⎜ ⎟ twice:
⎝∞⎠
2
x
t
ln t
1
Integral converges.
dx =
t
⎡ 1⎤
⎡ 1⎤
lim ⎢ − ⎥ = lim ⎢1 − ⎥ = 1
t⎦
t →∞ ⎣ x ⎦1 t →∞ ⎣
Integral converges.
19.
1
∞
∫1
1.001
x
dx = lim ∫
t
1
t →∞ 1 x1.001
dx =
t
1000 ⎤
⎡ 1000 ⎤
⎡
lim ⎢ −
= lim ⎢1000 −
⎥
⎥ = 1000
0.001
t →∞ ⎣ x
t 0.001 ⎦
⎦1 t →∞ ⎣
Integral converges.
512
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this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
9
CHAPTER
Infinite Series
7
26
63
124
215
5. a1 = , a2 =
, a3 = , a4 =
, a5 =
8
27
64
125
216
9.1 Concepts Review
1. a sequence
lim
n3 + 3n 2 + 3n
(n + 1)3
n →∞
2.
lim an exists (finite sense)
n →∞
4. –1; 1
6. a1 =
Problem Set 9.1
1
2
3
4
5
1. a1 = , a2 = , a3 = , a4 = , a5 =
2
5
8
11
14
n
1
1
lim
= lim
= ;
3
n →∞ 3n –1 n →∞ 3 – 1
n
converges
5
8
11
14
17
2. a1 = , a2 = , a3 = , a4 = , a5 =
2
3
4
5
6
3 + n2
3n + 2
= lim
= 3;
n →∞ n + 1
n →∞ 1 + 1
lim
n
converges
a4 =
+ 32 + 13
n
+ 3n 2 + 3n + 1
=1
n
n→∞ 1 + 3
n
n3 + 3n 2 + 3n
n→∞ n3
1 + n3 + 32
= lim
3. bounded above
= lim
n
5
14
29
, a2 =
, a3 =
,
3
5
7
50 5 2
77
=
, a5 =
9
9
11
3 + 22
3n 2 + 2
3
n
= lim
=
;
2
n →∞ 2n + 1
n→∞ 2 + 1
n
converges
lim
1
2 1
3
4 2
7. a1 = – , a2 = = , a3 = – , a4 = = ,
3
4 2
5
6 3
5
a5 = –
7
n
1
= lim
= 1, but since it alternates
lim
n →∞ n + 2 n→∞ 1 + 2
n
6
18
38
3. a1 = = 2, a2 =
= 2, a3 = ,
3
9
17
66 22
102 34
a4 =
= , a5 =
=
27 9
39 13
lim
4n 2 + 2
n →∞ n 2
+ 3n – 1
= lim
4+
n →∞ 1 + 3
n
2
n2
–
1
n2
between positive and negative, the sequence
diverges.
= 4;
converges
2
3
4
5
8. a1 = –1, a2 = , a3 = – , a4 = , a5 = –
3
5
7
9
⎧−1 for n odd
cos(nπ) = ⎨
⎩ 1 for n even
lim
n
n →∞ 2n – 1
4.
a1 = 5, a2 =
14
29
50
77
, a3 =
, a4 = , a5 =
3
5
7
9
3n + n
3n 2 + 2
= lim
= ∞;
n →∞ 2n –1
n→∞ 2 – 1
2
= lim
1
n→∞ 2 – 1
n
=
1
, but since cos(n π )
2
alternates between 1 and –1, the sequence
diverges.
lim
n
diverges
Instructor’s Resource Manual
Section 9.1
513
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of th

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