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
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
= 9t − 3t + 3t − 3t + t − t + 3t − t + 1
= 9t 4 − 6t 3 + 7t 2 − 2t + 1
2
y
+
6 y − 2 9 y2 −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)
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
Instructor’s Resource Manual
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
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 . Let n
be a natural number that satisfies
1 / n < b − a . Let S = {k : k n > 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
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
2
e.
2
2
2
2
triangle, then a + b ≠ c .
72. a.
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
a + b = c . If a triangle is not a right
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
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
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
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
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
Instructor’s Resource Manual
–3 < 1 – 6 x ≤ 4
9. –4 < –6 x ≤ 3
e.
2
1 ⎡ 1 2⎞
> x ≥ – ; ⎢– , ⎟
3
2 ⎣ 2 3⎠
f.
2. a.
c.
b.
(2, 7)
(−∞, −2]
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
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.
Instructor’s Resource Manual
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 < b 2 . Also, a 2 < ab .
Thus, a 2 < ab < b 2 and a 2 < b 2 . ⇐ Let
a 2 < b 2 , so a ≠ b Then
0 < ( a − b ) = a 2 − 2ab + b 2
2
x4 − 2 x2 ≥ 8
b.
< b 2 − 2ab + b 2 = 2b ( b − a )
x4 − 2 x2 − 8 ≥ 0
Since b > 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 < b ⇔ <1⇔ < .
b
b a
(−∞, −2] ∪ [2, ∞)
c.
[( x 2 + 1) − 5][( x 2 + 1) − 2] < 0
30. (b) and (c) are true.
(a) is false: Take a = −1, b = 1 .
(d) is false: if a ≤ b , then −a ≥ −b .
31. a.
3x + 7 > 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, ∞)
Instructor’s Resource Manual
Section 0.2
9
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
Instructor’s Resource Manual
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
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
9
x +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 +
Instructor’s Resource Manual
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
Problem Set 0.3
5. d1 = (5 + 2) 2 + (3 – 4)2 = 49 + 1 = 50
1.
d 2 = (5 − 10)2 + (3 − 8)2 = 25 + 25 = 50
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
Instructor’s Resource Manual
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
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.
37. 6 – 2 y = 10 x – 2
–2 y = 10 x – 8
y = –5 x + 4;
slope = –5; y-intercept = 4
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
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
36. −4 y = 5 x − 6
5
3
y = − x+
4
2
5
3
slope = − ; y -intercept =
4
2
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
–1 – 2
3
=– ;
3 +1
4
3
y + 3 = – ( x – 3)
4
3
3
y=– x–
4
4
e.
m=
f.
x=3
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.
Instructor’s Resource Manual
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
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.
Instructor’s Resource Manual
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⎞
2
⎜⎜ 1 −
⎟⎟
1−
2
⎝
⎠
2 R=
r=
R
⎛
⎞⎛
2
2
2⎞
1+
⎜⎜ 1 +
⎟⎜ 1 −
⎟
2
2 ⎟⎜
2 ⎟⎠
⎝
⎠⎝
1
1− 2 +
2R
r=
1
1−
2
r = (3 − 2 2) R ≈ 0.1716 R
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
Instructor’s Resource Manual
Section 0.3
19
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 .
Instructor’s Resource Manual
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
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 ) .
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
77. x 2 + ( y – 6) 2 = 25; passes through (3, 2)
tangent line: 3x – 4y = 1
The dirt hits the wall at y = 8.
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
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
Instructor’s Resource Manual
Section 0.4
23
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
Instructor’s Resource Manual
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
25
21. y =
1
2
; y-intercept = 1
x +1
Symmetric with respect to the y-axis
22. y =
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
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
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
Instructor’s Resource Manual
Section 0.4
27
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
Instructor’s Resource Manual
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
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=
Instructor’s Resource Manual
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⎭
Instructor’s Resource Manual
Section 0.5
31
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
Instructor’s Resource Manual
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
⎧⎪ – 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
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
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
⎧180 − 180 − 10t
⎪
⎪
⎪
s = ⎨ 902 + (10t − 90) 2
⎪
2
2
⎪ 90 + (270 − 10t )
⎪
⎪⎩
if 0 ≤ t ≤ 9
if 9 < t ≤ 18
if 0 ≤ t ≤ 9
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
Instructor’s Resource Manual
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
decreased from ∞ to
≈ 2.8889 . On
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
3. 2; left
4. a quotient of two polynomial functions
b.
c.
2. f(g(x))
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
2
d.
(f
⎛ 2 ⎞ ⎛1⎞ 1 3
g )(1) = f ⎜
⎟=⎜ ⎟ + =
⎝1+ 3 ⎠ ⎝ 2 ⎠ 2 4
( g f )(1) = g (12 + 1) =
2
2
=
2+3 5
f.
⎛ 2 ⎞
( g g )(3) = g ⎜
⎟=
⎝ 3+3⎠
2
2 3
= 10 =
1
+3 3 5
3
)(t ) = t 3 + 1 +
(Φ
1
⎛1⎞ ⎛1⎞
)(r ) = Φ⎜ ⎟ = ⎜ ⎟ + 1 = 3 + 1
r
r
r
⎝ ⎠ ⎝ ⎠
Φ )(r ) =
(r + 1) =
c.
(
d.
Φ 3 ( z ) = ( z 3 + 1) 3
e.
(Φ –
f.
((Φ –
1
6. g 3 (x) = (x 2 +1) 3 = (x 4 + 2x 2 + 1)(x 2 + 1)
6
)
= 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
9. ⎡ g 2 (π ) − g (π ) ⎤
⎣
⎦
≈ 4.789
1
5t
⎛1⎞
)⎜ ⎟
⎝t⎠
11. a.
b.
12. a.
b.
2 x2 – 1
x
Domain: (– ∞, – 1] ∪ [1, ∞)
d.
38
g ( x) = x , f ( x) = x + 7
g (x) = x15 , f (x) = x 2 + x
f ( x) =
( 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]
(f
( g f )( x) = g ⎛⎜ x 2 – 1 ⎞⎟ =
⎝
⎠
1/ 3
2
= ⎡⎢(11 − 7π ) − 11 − 7π ⎤⎥
⎣
⎦
10. [ g 3 (π) – g (π)]1/ 3 = [(6π – 11)3 – (6π – 11)]1/ 3
≈ 7.807
3
1 1
⎛ 1⎞
= ⎜ ⎟ + 1– 1 = 3 + 1 – t
⎝ t⎠
t
t
4. a.
2
1/ 3
1
5t
)(t ) = (Φ –
4
= x + 3x + 3x + 1
( g g g )( x) = ( g g )( x 2 + 1)
8. g(2.03) ≈ 0.000205
r 3 +1
)(5t) = [(5t) 3 +1] –
= 125t3 + 1 –
f ) ( x) = g ⎛⎜ x 2 − 4 ⎞⎟ = 1 + x 2 − 4
⎝
⎠
= 1 + x2 – 4
1
t
(Φ +
3
2
= x2 + 2 x – 3
3
b.
g ) ( x) = f ( 1 + x ) = 1 + x − 4
(f
(g
e.
3. a.
5.
4
f ( x) =
13. p = f
2
x
3
, g ( x) = x 2 + x + 1
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, ∞ )
Section 0.6
Instructor’s Resource Manual
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
Instructor’s Resource Manual
Section 0.6
39
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
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
=
;
1− x −1
x
=
x
= x;
x − ( x − 1)
1
x −1
=
= 1 – x;
x
1 – x –1 x − 1 − x
Section 0.6
41
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
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 = = –
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
43
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°
cos
b.
5.34 tan 21.3°
≈ 0.8845
sin 3.1°+ cot 23.5°
c.
tan (0.452) ≈ 0.4855
d.
sin (–0.361) ≈ –0.3532
6. a.
b.
7. a.
= cos
π
=
and by the Pythagorean Identity, sin
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
9. a.
2
3
⎛1⎞
= 1− ⎜ ⎟ = .
6
4
⎝2⎠
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
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
3
b.
π
⎛π ⎞
sin ⎜ ⎟
⎛π⎞
⎝6⎠ = 3
tan ⎜ ⎟ =
3
⎝ 6 ⎠ cos ⎛ π ⎞
⎜ ⎟
6
⎝ ⎠
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)
Instructor’s Resource Manual
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 cos2 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
= cos 2 z =
1
sec z
(sec t – 1)(sec t + 1) = sec 2 t – 1 = tan 2 t
c.
sec t – sin t tan t =
d.
12. a.
(1 + cos )(1 − cos ) = 1 − cos 2
14. a.
= sin 2
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 – csc2 t
csc2 t
=–
= – cos 2 t = –
2
b.
=
sin 4 x = sin[2(2 x)] = 2sin 2 x cos 2 x
cot 2 t
csc2 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
sec v
=
sin 2 t
cos 2 t
1
cos 2 t
= sin 2 t
= sin 2 v + cos 2 v = 1
Instructor’s Resource Manual
Section 0.7
45
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
Instructor’s Resource Manual
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
Instructor’s Resource Manual
Section 0.7
47
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.
2
sin (– t ) = sin t ; even
c.
1
csc(– t ) =
= – csc t; odd
sin(– t )
( π)
=
e.
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
27. cos 2
28. sin 2
2
2
π ⎛
π ⎞ ⎛1⎞
1
= ⎜ cos ⎟ = ⎜ ⎟ =
3 ⎝
3⎠
4
⎝2⎠
2
2
π ⎛ π⎞
1
⎛1⎞
= ⎜ sin ⎟ = ⎜ ⎟ =
6 ⎝
6⎠
4
⎝2⎠
3
3
π
2
32. a.
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 ) =
=
sin(−t ) = – sin t = sin t ; even
d.
π
1 – cos 4 1 – 2
π 1 – cos 2 8
31. sin
=
=
=
8
2
2
2
2
2
b.
(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 and Δx = cos
Δy sin
m=
=
= tan
Δx cos
3
1
π⎞ ⎛1⎞
3 π ⎛
29. sin 6 = ⎜ sin 6 ⎟ = ⎜ 2 ⎟ = 8
⎝
⎠ ⎝ ⎠
48
Section 0.7
Instructor’s Resource Manual
39. a.
b.
= 3
π
=
3
tan
3x + 3 y = 6
3 y = – 3x + 6
3
3
x + 2; m = –
3
3
y=–
tan
3
3
=–
=
5π
6
40. m1 = tan
1
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).
and m2 = tan 2
tan 2 + tan(− 1 )
tan = tan( 2 − 1 ) =
1 − tan 2 tan(− 1 )
=
41. a.
b.
tan 2 − tan 1
m − m1
= 2
1 + tan 2 tan 1 1 + m1m2
3–2
1
=
1 + 3(2 ) 7
≈ 0.1419
tan
=
tan =
–1 – 12
1+
( 12 ) (–1)
= –3
≈ 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; ≈ 1.7127
tan =
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 =
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
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
Instructor’s Resource Manual
Section 0.7
49
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 ⎦
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.
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
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)
Instructor’s Resource Manual
51.
a.
C sin( t + φ ) = (C cos φ )sin t + (C sin φ ) cos 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( t + φ1 ) + A2 sin( t + φ 2 ) + A3 (sin t + φ 3 )
= A1 (sin t cos φ1 + cos t sin φ1 )
+ A2 (sin t cos φ 2 + cos t sin φ 2 )
+ A3 (sin t cos φ 3 + cos t sin φ 3 )
= ( A1 cos φ1 + A2 cos φ 2 + A3 cos φ 3 ) sin t
+ ( A1 sin φ1 + A2 sin φ 2 + A3 sin φ 3 ) cos t
= C sin ( t + 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.
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
53. a.
b.
c.
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
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.
(
)
(
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
p and q must be integers.
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 =
Instructor’s Resource Manual
10. True:
y − x = −( x − y ) so
( x − y )( y − x) = ( x − y )(−1)( x − y )
20. True:
since 1 + r ≥ 1 − r ,
2
= (−1)( x − y ) .
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
1 1
> 1; <
b
b a
a < b < 0; a < b;
[ a, b] and [b, c ]
If r < −1, r = −r and 1 − r = 1 + r ,
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
4
If either x = 0 or y = 0, the
inequality is easily seen to be true.
4
4
22. True:
4
4
x = x and y = y , so x < y
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
y satisfies
= 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⎞ ⎛
1⎞
a2 + 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)
2
For every real number y, whether it
is positive, zero, or negative, the
cube root x = 3 y satisfies
x3 =
also, –1 < r < 1.
If 0 < r < 1, then r = r and
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,
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
Instructor’s Resource Manual
Section 0.8
53
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
Instructor’s Resource Manual
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
2 3
f ( x) ⋅ g ( x) = x x = x
59. False:
60. True:
61. True:
62. False:
63. False:
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 + ⎟ = ;
n ⎠ ⎝ 1⎠
2⎠
4
⎝
⎝
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 ⎠
−1
cot x =
b.
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
=
The domain of the tangent function
π
excludes all nπ + where n is an
2
integer.
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
55
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
⎛ 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
44 ± 442 – 4(21)(15) 44 ± 26 3 5
=
= ,
2(21)
42
7 3
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
Instructor’s Resource Manual
21. |t – 5| = |–(5 – t)| = |5 – t|
If |5 – t| = 5 – t, then 5 − t ≥ 0.
t ≤5
⎛ 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
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
28. x 2 + y 2 − 8 x + 6 y = 0
23. If x ≤ 2, then
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
2
( AB) + ( BC )2 = ( AC ) 2 , so ΔABC is a right
triangle.
⎛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
57
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.
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
35.
Instructor’s Resource Manual
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
a
1
b−0
b
=− =− ;
0−a
4
a
a = 4b
b=
⎛ 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
59
44. a.
b.
f (– x) =
3(– x)
2
(– x) + 1
=–
3x
2
x +1
; 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
2
f (x) = x – 1
; even
47. V(x) = x(32 – 2x)(24 – 2x)
Domain [0, 12]
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
Instructor’s Resource Manual
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
Review and Preview
61
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
Review and Preview
Instructor’s Resource Manual
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
lim ( x 2 + 2t − 1) = (−2) 2 + 2t − 1 = 3 + 2t
x →−2
(
t →−1
t →−1
(
= –t – t = –2t
12.
) ( ( −1) − 1) = 0
2
) ( ( −1)
6. lim t 2 − x 2 =
2
)
x2 – 4
( x – 2)( x + 2)
= lim
x→2 x – 2
x→2
x–2
= lim( x + 2)
=3+3=6
13.
x→2
lim
(t + 4)(t − 2) 4
(3t − 6) 2
t →2
= lim
=2+2=4
8.
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
x2 – t 2
( x + t )( x – t )
= lim
x→–t x + t
x→ – t
x+t
= lim ( x – t )
lim
x→ –t
x →−2
5. lim t 2 − 1 =
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
Instructor’s Resource Manual
Section 1.1
63
© 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. 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
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
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
18.
( x + h) 2 − x 2
x 2 + 2 xh + h 2 − x 2
= lim
h→0
h →0
h
h
lim
lim
sin x
2x
=0
2
x
(1 − cos x ) / x
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
lim
(1 – cos x) 2
x2
x →0
2
t
23.
=0
(t − 1) /(sin(t − 1))
2.
3.56519
1.1
2.1035
1.01
2.01003
1.001
2.001
0.229849
0
1.1884
0.0249792
0.9
1.90317
0.01
0.00249998
0.99
1.99003
0.001
0.00024999998
0.999
1.999
sin x
= 0.5
x →0 2 x
1− cos t
2t
t
20.
1.
0.1
–1.
–0.229849
–0.1
–0.0249792
–0.01
–0.00249998
–0.001
–0.00024999998
Section 1.1
2
1.
1.
lim
64
x2
22.
h 2 + 2 xh
= lim(h + 2 x) = 2 x
h →0
h →0
h
x
( x – sin x) 2
x →0
= lim
19.
( x − sin x) 2 / x 2
x
21.
x →3
= 36
16. lim
1 − cos t
=0
2t
2
lim
t −1
=2
− 1)
t →1 sin(t
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.
x −sin( x − 3) − 3
x −3
x
24.
4.
1. + π4
0.1 +
0.158529
3.1
3.01
3.001
2.
0.001 +
lim
4
0.1 + π
0.0500
0.01 + π
0.0050
0.001 + π
0.0005
–0.001 + π
–0.0005
x−π
x →π
=0
−0.1 +
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
–0.01
f.
–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
0.0000210862
2.12072 × 10−7
π
−0.001 +
0.1
1
t
0.00199339
−0.01 +
0.357907
t →0
0.11921
0.001 +
1.
1 – cot t
= 0.25
(2 − 2sin u ) / 3u
0.01 + π2
(1 − cot t ) /(1 / t )
lim
(tan x − 1)2
0.1 + π2
t
26.
(x − )
0.2505
1. + π2
–0.0500
–0.0050
0.255008
u
28.
–0.4597
1 + sin ( x − 32π )
0.300668
π 2
4
−1. + π2
–0.01 + π
lim
0.674117
−0.001 + π4
x→ π
0.4597
–0.1 + π
0.2495
4
−0.01 + π4
1. + π
–1. + π
0.245009
π
−0.1 + π4
(1 + sin( x − 3π / 2)) /( x − π )
x
4
−1. + π4
x – sin( x – 3) – 3
=0
lim
x →3
x–3
25.
0.201002
π
0.0000166666
1.66667 × 10−7
0.0000166666
1.66667 × 10−7
2.999
4
0.01 +
0.00166583
2.99
0.0320244
π
0.00166583
0.158529
2.9
( 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
© 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.
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.
lim f ( x) = 2
c.
lim f ( x) = 4
d.
x → –3+
lim f ( x) does not exist.
x →1
b. g(1) does not exist.
x → –3−
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
33.
a.
b.
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
© 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.
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.
lim
−
x →1
lim
c.
x →0
does not exist.
x −1
x −1
x −1
= −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−
x2 − 1
x2 − 1
=2
= −2 and lim
x −1
x →1+ x − 1
38. lim
x →0
= lim
x →0
c.
( x + 2 − 2)( x + 2 + 2)
d.
x( x + 2 + 2)
x →0
b.
lim
x →0
= lim
x →0
1
x− x =0
x
1
does not exist.
x
lim x(−1)
1/ x
lim x (−1)
1
2
=
=
=
4
0+2 + 2 2 2
x+2+ 2
46. a) Does not exist
c)
lim f ( x) does not exist.
=0
b) 0
−1
c)
=0
1/ x
x →0+
45. a) 1
x( x + 2 + 2)
1
+
x → 0+
x( x + 2 + 2)
x+2−2
= lim
39. a.
b.
x+2− 2
x
x →0
= lim
lim
x →1+
1
d)
−1
b) 0
d) 0.556
x →1
lim f ( x) = 0
47. lim x does not exist since
x →0
for x < 0.
x →0
40.
x is not defined
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
Instructor’s Resource Manual
Section 1.2
67
© 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.
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
3.
ε
10. If x is within 0.001 of 2, then
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
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.
12. 0 < x + 21 < δ ⇒ (3x – 1) – (–64) < ε
3 x – 1 + 64 < ε ⇔ 3x + 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
2 x – 11x + 5
(2 x – 1)( x – 5)
–9 =
–9
x–5
x–5
= 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 )
⇔
2x + 2
2
x – 25
( x – 5)( x + 5)
– 10 =
– 10 = x + 5 – 10
x–5
x–5
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
© 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.
5
⇔ x−4 ⋅
x − 3( 2 x − 1 + 7( x − 3))
5
To bound
x – 3( 2 x – 1 + 7( x – 3))
1
2
1
2
δ ≤ . If δ ≤ , then
x – 3( 2 x – 1 + 7( x – 3))
hence x − 4 ⋅
19. 0 < x – 1 < δ ⇒
, agree that
7
9
< x < , so
2
2
5
0.65 <
<ε
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
⇔ x –1 <
δ=
ε
14
(10 x – 6)( x – 1)2
( 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 ⎭
18. 0 < x – 1 < δ ⇒
( x –1)2
–4 <ε
⇔ 10( x – 1) < ε
5
ε
⇔ x–4 <
( x – 1) 2
10 x3 – 26 x 2 + 22 x – 6
⇔
< 1.65 and
10 x3 – 26 x 2 + 22 x – 6
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
(2 x + 1) – 3 = 2 x 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
© 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.
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 ⇒
2
( x – 2 x – 1) – 2 = x 2 – 2 x – 3
= 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) < B for
x − a < δ or x − a < δ ⇒ f ( x) < B. Thus,
x − a < δ ⇒ f ( x) g ( x) − 0 = f ( x) g ( x)
= f ( x) g ( x) < B ⋅
ε
B
= ε so lim f ( x)g(x) = 0.
x→ a
x→c
can find δ > 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
Instructor’s Resource Manual
Section 1.2
71
© 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. Choose ε > 0. Since lim f ( x) = L, there is a
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
1.3 Concepts Review
1. 48
2. 4
3. – 8; – 4 + 5c
4. 0
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
Problem Set 1.3
1. lim (2 x + 1)
= lim 2 x + lim 1
3
= 2 lim x + lim 1
2,1
x→1
30. Suppose that L > M. Then L – M =
> 0. Now
and δ = min{δ1 , δ 2} where
2
0 < x – a < δ1 ⇒ f ( x) – L < ε and
2.
Thus, for 0 < x – a < δ ,
8
x+6
3
2
x→–1
2, 1
2
3. lim [(2 x +1)( x – 3)]
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
= lim (2 x 2 + 1) ⋅ lim (7 x 2 + 13)
x→ 2
+ 1 has
1
, then 2.75 < x < 3
4
or 3 < x < 3.25 and by graphing
If δ ≤
y = g ( x) =
= 3 lim x 2 – lim 1
= 3(–1) – 1 = 2
4.
x 4 – 4x 3 + x 2 + x + 6
x – 4x + x 2 + x + 6
an asymptote at x ≈ 3.49.
c.
3
x→–1
⎛
⎞
= 3⎜ lim x ⎟ – lim 1
⎝ 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 .
x 3 – x 2 – 2x – 4
5
= lim 3x 2 – lim 1
x→ –1
31. (b) and (c) are equivalent to the definition of
limit.
4
lim (3x 2 – 1)
x→ –1
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 = > 2ε
which is a contradiction.
Thus L ≤ M .
b. No, because
x→1
= 2(1) + 1 = 3
0 < x – a < δ 2 ⇒ g ( x) – M < ε .
g(x) =
x→1
x →1
take ε <
33. a.
4
x→1
L − ε < f ( x) < L + ε and f ( x) < M .
x→ 2
6
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
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.
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
Instructor’s Resource Manual
Section 1.3
73
© 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.
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
=
13.
3
1
144
lim
x →2
=
x2 − 4
2
x +4
=
(
lim ( x
2
) 4−4
=
=0
+ 4) 4 + 4
lim x − 4
x→2
2
x→2
x→2
( x − 3)( x + 1)
x2 − 2 x − 3
= lim
x →−1
x →−1
x +1
( x + 1)
lim
= lim ( x − 3) = −4
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
= lim ( x − 3) = −1
16.
2
2(5)4 − 9(5)3 + 19
( x − 3)( x − 2 )
x2 − 5x + 6
14. lim
= lim
x→2
x→2
x−2
( x − 2)
15.
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
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.
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
Instructor’s Resource Manual
Section 1.3
75
© 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.
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
45.
x→c
2
⎡
⎤
38. lim f (x) = 0 ⇔ ⎢ lim f (x) ⎥ = 0
⎣
⎦
x→c
x→c
46.
2
⇔ lim f ( x) = 0
lim
x →1–
lim
x → 2+
lim f 2 ( x) = 0
47.
x →c
⇔ lim
x →c
f 2 ( x) = 0
48.
⇔ lim f ( x) = 0
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
(3 x − 1)
=
2
π3 + (– π)3
=0
–π
(22 + 1) 2
(3 ⋅ 2 − 1)
2
lim ( x − x ) = lim x − lim
x →3−
x →3−
=
x
= –1
x
lim
x 2 + 2 x = 32 + 2 ⋅ 3 = 15
x →0 –
x →3+
f ( x) g ( x) = 1; g ( x) =
1
f ( x)
lim g ( x) = 0 ⇔ lim
1
=0
f ( x)
x →a
5⋅ 2
5
2
=
2
5
x = 3− 2 =1
lim
⇔
If f ( x) =
x→c
( x 2 + 1) x
x →a
= ⎛⎜ lim x ⎞⎟ = c 2 = c
⎝ x →c ⎠
40. a.
49.
x2 – 9
x+3
1+ x
1+1
2
=
=
4 + 4 x 4 + 4(1)
8
x →3−
x→c
⇔
x2 – 9
x →3+
32 – 9
=0
3+3
=
x →c
x→c
⇔ lim [ f (x) – L] = 0
( 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
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
© 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.
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
3 x tan x
3 x (sin x / cos x)
3x
= lim
= lim
x →0
x → 0 cos x
sin x
sin x
0
=0
1
5. lim
sin x 1
sin x 1
1
= lim
= ⋅1 =
2x
2 x →0 x
2
2
sin 3
3 sin 3
3
sin 3
= lim ⋅
= lim
3
2 →0 3
→0 2
→0 2
3
3
= ⋅1 =
2
2
6. lim
sin 3
sin 3
cos sin 3
= lim sin = lim
→ 0 tan
→0
→0
sin
cos
7. lim
1 + x2 + x + x2 + x – 2 x + 1
x → 0+
=
x →0
2
MP = ( x – 1)2 + ( y – 0) 2 =
x →0
1 + x2 + x + x2 – x + 1
=1
x
1
(1)( x) =
2
2
1
x
Area of ΔMOP = (1)( y ) =
2
2
⎡
sin 3
1 ⎤
= lim ⎢cos ⋅ 3 ⋅
⋅ sin ⎥
→0⎢
3
⎥⎦
⎣
⎡
sin 3
1
= 3 lim ⎢cos ⋅
⋅ sin
→0 ⎢
3
⎣
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
sin 5
tan 5
sin 5
= lim cos 5 = lim
→ 0 sin 2
→ 0 sin 2
→ 0 cos 5 sin 2
sin 5 1 2 ⎤
⎡ 1
= lim ⎢
⋅5⋅
⋅ ⋅
5
2 sin 2 ⎥⎦
→0 ⎣ cos 5
5
sin 5
2 ⎤
⎡ 1
= lim ⎢
⋅
⋅
2 →0 ⎣ cos 5
5
sin 2 ⎥⎦
5
5
= ⋅1⋅1⋅1 =
2
2
8. lim
cot π sin
→0
2 sec
9. lim
= lim
→0
cos π
sin π
sin
2
cos
cos π sin cos
2sin π
⎡ cos π cos sin 1 π
= lim ⎢
⋅
⋅ ⋅
π sin π
2
→0 ⎣
1
sin
π
⎡
=
⋅
lim ⎢cos π cos ⋅
2 π →0 ⎣
sin π
1
1
⋅1⋅1⋅1⋅1 =
=
2π
2π
= lim
2. 1
→0
3. the denominator is 0 when t = 0 .
4. 1
Problem Set 1.4
cos x 1
= =1
x →0 x + 1 1
lim
→π / 2
cos =
π
2
⎤
⎥
⎦
⎤
⎥
⎦
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 ⋅1 ⋅1 ⋅1 = 3
⎥⎦
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
© 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.
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.
lim
→0
sin 2
2
= lim
→0
= lim
→0
sin
× lim
→0
sin
sin
sin
= 1× 1 = 1
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 ≤ lim
≤1
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
© 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.
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 →∞
n
b0 x + b1 x
x →∞
0
=0
1+ 0
x →∞
=
20.
21.
2
1
x2 + 3
= lim
2 + 1x
n +1
x →∞
x 2 +3
x
2 + 1x
= lim
x →∞
1+
2
3
x2
lim
x →∞
26. lim
lim ⎛⎜ 2 x 2 + 3 – 2 x 2 – 5 ⎞⎟
⎠
x →∞ ⎝
= lim
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 →∞
80
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
+
bn –1
x n –1
n →∞
+
an
xn
+
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
+
t2 – 9
(t + 3)(t – 3)
= lim
+
+
t +3
t → –3 t + 3
t → –3
= lim (t – 3) = –6
lim
=0
+
+ bn –1 x + bn
an –1
x n –1
lim
lim
= lim
+
+
+ an –1 x + an
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
x →∞
n –1
= –∞
+
+ 12
2x +1
x
= lim
= lim
=0
x+4
x →∞ 1 + 4
x →∞ 1 + 4
x
x
= lim
y→–∞ 1 –
n2
∞
n
n3/ 2
= lim
= =∞
3
n →∞
1
2
1
n + 2n + 1
1+ 2 + 3
n
n
n →∞
2
x
= lim
2
=2
2 x +1
x2
b0 +
b1
x
2
19. For x > 0, x = x 2 .
2x + 1
a0 +
a1
x
n
25. lim
n →∞
lim
y2 – 2 y + 2
a0 x n + a1 x n –1 +
= 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
−
→ π+ , 2 → π2 while sin → 0 .
2
lim
→π+
sin
= −∞
x
Section 1.5
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.
−
33. As x → 3− , x3 → 27, while x − 3 → 0 .
43.
3
lim
x →3−
x
= −∞
x−3
π+
π2
,π →
while cos → 0 – .
2
2
π
= –∞
lim
π + cos
→
34. As
→
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
x
1
thus lim
= ∞.
−
x
x
x
x →0
1
(Since x < 0, – > 0. )
x
=−
39. For x < 0, x = – x, thus
lim
x →0 –
x
x
= lim
x →0 –
45.
–x
= –1
x
40. For x > 0, x = x, thus lim
x →0 +
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
Instructor’s Resource Manual
Section 1.5
81
© 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.
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
82
Section 1.5
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.
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
Let u =
1
1⎞
⎛
→ ∞, so lim sin ⎜ x + ⎟
x
x⎠
x →∞
⎝
does not exist. (See part a.)
As x → ∞, x +
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
=∞
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−
=1
x
⎛ 1⎞
63. lim ⎜1 + ⎟
x⎠
x →∞ ⎝
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+
Section 1.5
83
© 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.
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
→0
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
Define g(0) = 1
8. lim t – 2 = 1 = g (3); continuous
9. h(3) does not exist, so h(t) is not continuous at 3.
sin( )
= 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
Section 1.6
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.
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
= 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
Instructor’s Resource Manual
Section 1.6
85
© 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.
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
Section 1.6
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.
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( ) is
⎣
⎦
the length of the side of the rectangle which
makes angle 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
π
such that f ( 0 ) = 0.
2
Hence, D can be circumscribed by a square.
0
between 0 and
65. Yes, g is continuous at R .
lim g ( r ) =
r →R−
GMm
= lim g ( r )
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
© 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.
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.
4. True.
By definition, where c = 0, L = 0.
5. False:
If f(c) is not defined, lim f ( x ) might
exist; e.g., f ( x) =
x →c
x2 – 4
= −4.
x →−2 x + 2
6. True:
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 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.
x –4
.
x+2
f(–2) does not exist, but lim
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
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
© 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.
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:
lim may not exist
x →0 −
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
It could be the case where
lim f ( x ) = 2
x→2
26. False:
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
24. True:
lim g( x ) exist; e.g., f ( x) =
x→c
g ( x) =
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:
That f(x) ≠ g(x) for all x does not
imply that lim f ( x) ≠ lim g ( x). For
x →c
x →c
example, if f ( x) =
2
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.
g ( x) =
x→2
30. False:
x→ 2
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
x+7
for c = −2 .
x+2
Squeeze Theorem
x →c
x
= 1 = f ( 2.3)
2
x–3
and
x+2
27. True:
x →c
lim
x→c
x→c
continuous at x = c.
23. True:
That lim [ f ( x ) + g ( x )] exists does
not imply that lim f ( x ) and
x →−∞
19. False:
Choose ε = 0. 001 f (2) then since
lim f ( x ) = f (2), there is some δ
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
15. False:
18. False.
25. True:
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
© 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.
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)
= lim
t+2
( t − 2 )2
20.
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
x →0 –
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 > 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 > 0. We need to find
a > 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
© 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.
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
Instructor’s Resource Manual
Section 1.7
91
© 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.
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 →−∞
Horizontal: y = 0, because
lim
x →∞
sin x
x2
= lim
sin x
x →−∞
x2
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
= lim
h →0
h →0
= lim
h →0
92
−1
a2
F ( 2 ) = 2 ≈ 1.414
= 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
1. a.
3. a.
=
−h
a (a + h)
−h
a (a + h)
h
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
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.
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
Instructor’s Resource Manual
Review and Preview
93
© 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.
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
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.
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 →0
3
8. y = x – 1
a., b.
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
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
Instructor’s Resource Manual
Section 2.1
95
© 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.
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
1
e.
14. a.
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
−
= lim 2+ h 2
h
h →0
− 2(2h+ h)
= lim
h
h →0
1
= lim −
h→0 2(2 + h)
h
h −1
h
1
= lim
h →0 h − 1
= −1
y + 1 = –1(x – 0); y = –x – 1
h →0
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
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. a.
f ( + h) – f ( )
h
v = lim
h →0
2( + h) + 1 – 2 + 1
h
h →0
= lim
2 + 2h + 1 – 2 + 1
h
= lim
h →0
= lim
( 2 + 2h + 1 – 2 + 1)( 2 + 2h + 1 + 2 + 1)
h( 2 + 2h + 1 + 2 + 1)
h →0
2h
= lim
2 + 2h + 1 + 2 + 1)
h →0 h(
2
=
2 +1 + 2 +1
1
b.
2 +1
=
=
1
2 +1
ft/s
1
2
2 +1 = 2
3
2
The object reaches a velocity of 1 ft/s when t = 3 .
2
2
2 + 1= 4;
=
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
Section 2.1
97
© 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.
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.
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.
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.
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
1.
f (c + h) – f (c) f (t ) – f (c)
;
h
t –c
2.
f (c )
3. continuous; f ( x) = x
4.
=
dy
dx
Problem Set 2.2
1.
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 =
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
9
π inch 3 / sec
16
3.
m tan = 7
b. m tan = 0
c.
m tan = –1
d. m tan = 17. 92
h →0
5h + h 2
= lim (5 + h) = 5
h
h →0
h →0
= lim
4.
f (4) = lim
h →0
2
32. y = f ( x) = sin x sin 2 x
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
99
© 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.
6.
f ( x + h) – f ( x )
h
[ ( x + h) + ] – ( x + )
= lim
h
h →0
h
= lim
=
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
= lim
8.
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
2 xh + h + h
= lim (2 x + h + 1) = 2 x + 1
h
h →0
f ( x) = lim
h →0
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
h( x + h) – h( x )
h
h →0
⎡⎛ 2
2 ⎞ 1⎤
= lim ⎢⎜
– ⎟⋅ ⎥
h → 0 ⎣⎝ x + h x ⎠ h ⎦
13. h ( x) = lim
⎡ –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
h →0 ( x 2
f ( x + h) – f ( x )
f ( x) = lim
h
h →0
[( 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
2
2
2
= lim (3x + 3hx + h + 4 x + 2h) = 3 x + 4 x
h →0
g ( x + h) – g ( x )
h
–12 x – 6h
2
2
+ 1)( x + 2hx + h + 1)
=−
12 x
2
( x + 1)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
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.
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 )( 3 x + 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
3 x + 3h + 3 x
=
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)( 3 x + 3 x + 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
101
© 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.
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
2
f (t ) – f ( x)
t–x
2
(t − 3t ) – ( x – 3 x)
t–x
t→x
= lim
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
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.
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 .
Instructor’s Resource Manual
Section 2.2
103
© 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.
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
2
x + x Δx
×
(
) ⎤⎥ × 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
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.
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
© 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.
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
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
f (−u ) − f ( x0 )
−u + x0
u → x0
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
= lim
t →− 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.
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
© 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.
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 )
2
1. Dx (2 x ) = 2 Dx ( x ) = 2 ⋅ 2 x = 4 x
= 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
3
2
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)
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
⎛ ⎞
8. Dx ⎜ ⎟ = Dx ( x –3 ) = (–3x –4 ) = –3 x –4
⎝ x3 ⎠
3
=–
x4
⎛ 100 ⎞
9. Dx ⎜
= 100 Dx ( x –5 ) = 100(–5 x –6 )
5 ⎟
⎝ x ⎠
500
= –500 x –6 = –
x6
⎛ 3 ⎞ 3
3
10. Dx ⎜
Dx ( x –5 ) =
=
(–5 x –6 )
5⎟
4
⎝ 4x ⎠ 4
15 –6
15
=–
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
Instructor’s Resource Manual
Section 2.3
107
© 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.
⎛ 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)
2
2
= x(2 x) + ( x + 1)(1) = 3 x + 1
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
2
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
= 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)
2
(3 x + 1)
2
=–
6x
2
(3x + 1) 2
⎛ 2 ⎞ (5 x 2 –1) Dx (2) – (2) Dx (5 x 2 –1)
34. Dx ⎜
⎟=
(5 x 2 – 1) 2
⎝ 5x2 – 1 ⎠
=
108
(5 x 2 –1)(0) – 2(10 x )
2
(5 x –1)
Section 2.3
2
=–
20 x
(5 x 2 –1)2
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.
⎛
⎞ (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)
2
(4 x – 3 x + 9)
−8 x + 3
=–
2
8x − 3
2
(4 x – 3x + 9)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
Instructor’s Resource Manual
Section 2.3
109
© 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.
⎛ 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) =
=
46. a.
g (0) f (0) – f (0) g (0)
g 2 (0)
–3(–1) – 4(5)
(–3)
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
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.
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) ]
2
2
⎛ 1 ⎞ ( x + 4) Dx (1) – (1) Dx ( x + 4)
50. Dx ⎜
⎟=
( x 2 + 4)2
⎝ x2 + 4 ⎠
=
( x 2 + 4)(0) – (2 x)
( x 2 + 4)2
At x = 1: mtan = −
=–
2x
( x 2 + 4) 2
2(1)
2
2
=–
2
25
(1 + 4)
1
2
Tangent line: y – = – ( x –1)
5
25
2
7
y = – x+
25
25
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
⎝
⎠
2
mtan = x + 2 x –1 = 1
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.
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
= Dx f ( x) − Dx g ( x)
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) .
Instructor’s Resource Manual
Section 2.4
111
© 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.
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 =
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.)
4 3
πr ; the volume
3
2.4 Concepts Review
2
59. Dx (7 – x ) = –2 x
The line through (4, 0) and ( x0 , y0 ) has
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
© 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 ⎞
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
= sec2 x
=
2
− sin x – cos x
2
=
2
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
1
cos 2 x
tan x(cos x – sin x) – sec2 x(sin x + cos x)
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 2 x sin x
1 ⎞⎛ cos 2 x ⎞
= ⎜ sin x −
−
−
⎟⎜
⎟
2
cos x cos x cos x ⎠⎝ sin 2 x ⎠
⎝
cos 2 x
1
cos x
− cos x −
−
sin x
sin x sin 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
=
=
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
⎛ sin x + cos x ⎞
Dx ⎜
⎟
tan x
⎝
⎠
tan x Dx (sin x + cos x) − (sin x + cos x) Dx (tan x )
=
tan 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
12. Dx ( sin x tan x ) = sin xDx [ tan x ] + tan xDx [sin 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
Instructor’s Resource Manual
Section 2.4
113
© 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.
b. Dt(20 sin t) = 20 cos t
π
π
At t = : rate = 20 cos = 10 3 ≈ 17. 32 ft/s
6
6
y = tan 2 x = (tan x)(tan x)
17.
Dx y = (tan x)(sec 2 x) + (tan x)(sec 2 x)
= 2 tan x sec 2 x
3
y = sec x = (sec x)(sec x)
18.
Dx y = (sec2 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.
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
= 2 tan x sec 2 x
Now, sec2 x is never 0, but tan x = 0 at
x = kπ where k is an integer.
= 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 ⎦⎤
2
x=
2
= −2sin x + 2 cos x
2
2
22. Dx cos 2 x = Dx (2 cos x − 1) = 2 Dx cos x − Dx 1
= −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)
27.
π
At x = : mtan = –2;
4
y=1
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.
2
28.
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
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.
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
Instructor’s Resource Manual
Section 2.4
115
© 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.
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
= (–5u –6 )(1) = –5( x + 3) –6 = –
–9
x +1
x –1
Dx y = Du y ⋅ Dx u
13. y = u 3 and u =
5
( x + 3)6
= (3u 2 )
( x –1) 2
2
8. y = u and u = 3x + x – 3
Dx y = Du y ⋅ Dx u
⎛ 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 ) ⋅
2
9. y = sin u and u = x + x
Dx y = Du y ⋅ Dx u
= (cos u)(2x + 1)
( x − π) Dx ( x − 2) − ( x − 2) Dx ( x − π)
⎛ x−2⎞
= −3 ⎜
⎟
⎝ x−π⎠
2
= (2 x + 1) cos( x + x)
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⎟
( x + 2)2
( x + 2)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
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.
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 ( x7 + 3)3 ] = (2 – 3 x 2 )4 Dx ( x 7 + 3)3 + ( x 7 + 3)3 Dx (2 – 3 x 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 ( x7 + 3) 2 (29 x7 – 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 ) = 3sin 2 cos
d
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
Instructor’s Resource Manual
Section 2.5
117
© 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.
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 + 3 x) = 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. D [cos 4 (sin
2
3
= –4 cos (sin
)] = 4 cos3 (sin
2
) sin(sin
2
2
)(cos
) D cos(sin
2
)D (
2
2
) = 4 cos3 (sin
3
) = –8 cos (sin
2
2
)[– sin(sin
) sin(sin
2
2
)]D (sin
)(cos
2
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
Section 2.5
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.
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
43. ( fg ) (2 ) = ( fg + gf
44. ( f g ) (2) =
≈
d
d
F ( cos x ) = F ( cos x ) ( cos x )
dx
dx
= − sin xF ( cos x )
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 ) ⎤⎦
)(2) = 2(0) + 1(1) = 1
= sec 2 ( 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
56.
=1
d
d
⎡⎣ g ( tan 2 x ) ⎤⎦ = g ' ( tan 2 x ) ⋅ tan 2 x
dx
dx
(
45. ( f g ) (6) = f ( g (6)) g (6)
= f (2) g (6) ≈ (1)(−1) = –1
= 2 g ' ( tan 2 x ) sec 2 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 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.
(
52.
) (
= F ( x ) × 2sin F ( x ) × cos ( F ( x ) ) × Dx ⎣⎡ F ( x ) ⎦⎤
)
(
)
+ F ( x ) sin 2 F ( x )
= 2 xF x 2 + 1
[
51.
(
+ F ( x ) sin 2 F ( x )
Dx F x 2 +1 = F x 2 +1 Dx x 2 +1
49. Dt (F (t ))
50.
)
−2
] = −2(F (t ))
−3
= 2 F ( x ) F ( x ) sin F ( x ) cos F ( x )
F (t )
+ F ( x ) sin 2 F ( x )
d ⎡ 1 ⎤
−3
⎢
⎥ = −2(F (z )) F (z )
dz ⎣⎢ (F (z ))2 ⎦⎥
58. Dx ⎡⎣sec3 F ( x ) ⎤⎦ = 3sec 2 ⎡⎣ F ( x ) ⎤⎦ Dx ⎡⎣sec F ( x ) ⎤⎦
= 3sec 2 ⎡⎣ 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
⎥
⎦
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
( x ) sin f ( x )
g ( 0 ) = − f ( 0 ) sin f ( 0 ) = −2sin1 ≈ −1.683
−1 ⎤
( )
= 2y −
( F ( y ))
⎞
⎟
2 ⎟
⎟
⎠
)
= g ' ( tan 2 x ) sec 2 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
© 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.
61. F ( x ) = − f ( x ) g ( x ) sin g ( x ) + f
F (1) = − f (1) g (1) sin g (1) + f
( x ) cos g ( x )
(1) cos g (1)
c. Dt L =
= −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)
−3
x-axis at x = 3 / 2 .
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
Section 2.5
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
⎞
⎟
⎟
⎠
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
2
16 cos 2 2t + 49sin 2 2t
⎛
sin 2t
= 2 cos 2t ⎜1 +
⎜
25 – cos 2 2t
⎝
y−
68. a.
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
a.
(cos 2π t ,sin 2π t )
b.
(0 – cos 2πt ) 2 + ( y – sin 2πt )2 = 52 , so
y = sin 2πt + 25 – cos 2 2πt
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
© 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.
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.
⋅
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 is the angle between
tan = 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
3
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 =
( )
1 + 12 (–1)
=
– 23
1
2
= –3,
so ≈ –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
© 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.
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
2
78. Dx x – 1 =
=
x2 – 1
2
x –1
80. a.
b.
sin x
sin x
x2 – 1
(2 x) =
79. Dx sin x =
=
2x
⋅ 2x =
=
2
x
x
=
x
x
82. a.
Dx ( x 2 –1)
2 x x 2 –1
sin x
sin x
b.
2
x –1
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)
=
=
D x f ( x ) − f ( x ) D x g ( x ) g ( x ) D x f ( x ) − f ( x ) D x 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 ( 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
1
1
= f
= ⎡⎣ f
g ( x2 ) = f
= f
= ⎡⎣ f
122
1
( f ( f ( x2 ) ) ) f ( f ( x2 ) ) f
( x1 )⎤⎦
2
⎡⎣ f
( x2 )⎤⎦
1
1
( x2 ) f ( x1 ) = f ( f ( x1 ) ) f ( x1 ) f ( x2 ) f ( x1 )
2
( f ( f ( f ( x )))) f ( f ( f ( x ))) f
( f ( x2 ) ) f ( x2 )
( f ( f ( x1 ) ) ) f ( f ( x1 ) ) f ( x1 ) f ( x2 ) = f ( f ( x2 ) ) f ( x2 ) f ( x1 ) f ( x2 )
2
( x1 )⎤⎦
Section 2.5
2
⎡⎣ f
( x2 )⎤⎦
2
2
= g ( x1 )
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.
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
© 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.
9.
10.
f ( x) = 2 x; f ( x) = 2; f (2) = 2
12.
f ( x) = 15 x 2 + 4 x + 1
f (u ) =
f ( x) = 30 x + 4
f (2) = 64
11.
f (u ) =
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 ( ) = –2(cos π) –3 (– sin π)π = 2 π(cos π) –3 (sin π)
f ( ) = 2π[(cos π) –3 (π)(cos π) + (sin π)(–3)(cos π) –4 (– sin π)(π)] = 2π2 [(cos π)−2 + 3sin 2 π(cos π) −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) =
f (2) =
( 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)
8
13
4
( x –1)(2 x – 2) – ( x 2 – 2 x – 3)(2)
( x –1)
3
=
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
Dxn −1 ( x n ) = n(n –1)(n – 2)(n – 3)...(2) x
Dxn ( x n ) = n(n –1)(n – 2)(n – 3)...2(1) x 0 = n!
124
=
Section 2.6
18. Let k < n.
Dxn ( x k ) = Dxn − k [ Dxk ( x k )] = Dx (k !) = 0
so Dxn [an x n –1 +
19. a.
+ a1 x + a0 ] = 0
Dx4 (3x3 + 2 x –19) = 0
b.
11
10
D12
x (100 x − 79 x ) = 0
c.
2
5
D11
x ( x – 3) = 0
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.
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 ) =
d 2s
dt 2
= –4
e.
a(t ) =
dt 2
= 6t – 18
3t 2 –18t + 24 > 0
3(t – 2)(t – 4) > 0
(−∞, 2) ∪ (4, ∞)
c.
3t 2 –18t + 24 < 0
(2, 4)
d. 6t – 18 < 0
6t < 18
t < 3; (−∞,3)
ds
= 12 – 4t
dt
12 – 4t < 0
t > 3; (3, ∞)
v(t ) =
d 2s
e.
d. a(t) = –4 < 0 for all t
24. a.
ds
= 3t 2 –18t + 24
dt
b.
26. a.
b. 12 – 4t > 0
4t < 12
t < 3; ( −∞,3)
c.
v(t ) =
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. 12t < 0
t<0
The acceleration is negative for negative t.
= 6t –12
Instructor’s Resource Manual
Section 2.6
125
© 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.
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 ) =
b.
dt
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 ) =
b.
1–
4
t2
t2 – 4
t2
c.
d.
1–
8
3
4
t
2
dt
2
=
< 0; (0, 2)
< 0; The acceleration is not negative for
t
any positive t.
e.
2
=
1
(12t 2 – 84t + 120)
10
ds1
= 4 – 6t
dt
ds
v2 (t ) = 2 = 2t – 2
dt
31. v1 (t ) =
b.
4 – 6t = 2t – 2
8t = 6
3
t = sec
4
4 – 6t = 2t – 2 ; 4 – 6t = –2t + 2
t=
>0
> 0; (2, ∞)
d 2s
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
8
t3
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
126
Section 2.6
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.
33. a.
v(t) = –32t + 48
initial velocity = v0 = 48 ft/sec
b. –32t + 48 = 0
3
t = sec
2
c.
d.
s = –16(1.5) 2 + 48(1.5) + 256 = 292 ft
2
–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.
48 – 32t = 0
t = 1.5
b. v(1) = 16 ft/sec upward
2
48t –16t = 0
–16t(–3 + t) = 0
t = 3 sec
35. v(t ) = v0 – 32t
v0 – 32t = 0
t=
(t – 4)(t + 2) (6t – 6)
(t – 4)(t + 2)
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
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.
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)
v (t ) a(t )
34. v(t) = 48 –32t
a.
37. v(t ) = 3t 2 – 6t – 24
< 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
n n
⎛ ⎞ n−k
k
Dxn (uv) =
⎜ ⎟ Dx (u ) Dx (v)
k
⎝
⎠
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
Instructor’s Resource Manual
Section 2.6
127
© 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. a.
4. 2 x + 2 2 yDx y = 0
Dx y = –
2x
2
2
y
=–
x
2
y
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 –3
3. x(2 y )
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
x 2 Dx y – 2 xyDx y = y 2 – 2 xy
9
dy
dx
12 x 2 + 7 y 2
8. x 2 Dx y + 2 xy = y 2 + x(2 y ) Dx y
3
2. 3 y 2
4.
Dx y =
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
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.
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 – 2y = 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 =
3
2
2
At (1, –1), y =
1 + y sin( xy )
–2 xy sin( xy 2 ) – 2 y
2
3
y ( x3 + 3xy 2 ) = –3 x 2 y – y 3
2
–3 x y – y
3
18.
x3 + 3 xy 2
36
9
=–
28
7
9
Tangent line: y – 3 = – ( x – 1)
7
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
23.
y 2 sin( xy 2 ) – 6 x
6
= –6
1
Tangent line: y – 0 = –6(x – 1)
y + 2 xyy + y 2 = 0
19.
y [1 – 2 xy sin( xy 2 )] = y 2 sin( xy 2 ) – 6 x
1 – 2 xy sin( xy 2 )
1
( x –1)
2
Tangent line: y –1 = –
16. y + [– sin( xy 2 )][2 xyy + y 2 ] + 6 x = 0
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
⎟
⎝
⎠
At (1, 3), y = –
–2 xy 2 – 4 y
2
3
4
3
Tangent line: y + 1 =
2
13. x y + 3 x y + y + 3xy y = 0
y =
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
Instructor’s Resource Manual
Section 2.7
129
© 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.
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.
31.
y
(x + 2)2 + y2 = 1
5 x
−
−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
x 2 cos x + 2 x sin x
2
33 ( x sin 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
4
3
2 [1 + cos( x + 2 x )]
dy (tan 2 x + sin 2 x) –1/ 2 (2 tan x sec 2 x + 2 sin x cos x)
=
dx
2
=
tan x sec 2 x + sin x cos x
33. s 2 + 2 st
2
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
5
2 x 2 cos x
=−
32.
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
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 =
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.
xy + y + 3 y 2 y = 0
37. a.
y (6 y 2 – x 2 ) =
y (x + 3y2 ) = – y
y =–
y
y =
x + 3y2
⎛ –y
xy + ⎜
⎜ x + 3y2
⎝
b.
⎞ ⎛ –y ⎞
2
⎟⎟ + ⎜⎜
⎟ + 3y y
2⎟
⎠ ⎝ x + 3y ⎠
⎛ –y
+6 y ⎜
⎜ x + 3 y2
⎝
⎞
⎟⎟ = 0
⎠
2y
6 y3
xy + 3 y 2 y –
+
x + 3 y2
2y
y (x + 3y2 ) =
y (x + 3y2 ) =
y =
2
x + 3y2
2 xy
–
( x + 3 y 2 )2
6 y3
( x + 3 y 2 )2
2 2
(x + 3y )
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 y 5
(6 y 2 – x 2 ) 2
Instructor’s Resource Manual
Section 2.7
131
© 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.
43. Implicitly differentiate the first equation.
4 x + 2 yy = 0
45. x 2 – x(2 x) + 2(2 x) 2 = 28
7 x 2 = 28
2x
y =–
y
Implicitly differentiate the second equation.
2 yy = 4
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
2
y
Solve for the points of intersection.
y =
y2 =
2 x2 + 4 x = 6
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
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
⇒
x=
1
2
x
y
Implicitly differentiate the second equation.
2( x –1) + 2 yy = 0
tan =
0–2
= –2; = 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
( )( )
=
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– x
y
⎛1 3⎞
1
1
At ⎜⎜ ,
⎟⎟ : m1 = – 3 , m2 = 3
2
2
⎝
⎠
1 + 1
2
3
3
tan =
= 3 = 3 →
2
1+ 1 − 1
3
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 =–
3
At (2, 4): m1 = 2, m2 = 0
47. x 2 – xy + y 2 = 16 , when y = 0,
⎛1 3⎞
⎛1
3⎞
Points of intersection: ⎜⎜ ,
⎟⎟ and ⎜⎜ , –
⎟
2 ⎟⎠
⎝2 2 ⎠
⎝2
Implicitly differentiate the first equation.
2 x + 2 yy = 0
y =
y – 2x
4y – x
π
3
3
⎛1
3⎞
1
1
, m2 = –
At ⎜⎜ , –
⎟⎟ : m1 =
2 ⎠
3
3
⎝2
1
1
2
− −
−
3
3
3
tan =
=
=− 3
2
1
1
1+
–
3
( )( )
3
=
132
3
2π
3
Section 2.7
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.
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
© 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.
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
© 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. Let x be the distance from the beam to the point
opposite the lighthouse and be the angle
between the beam and the line from the
lighthouse to the point opposite.
x d
tan = ;
= 2(2π) = 4π rad/min,
1 dt
d
dx
sec2
=
dt dt
1
1
5
At x = , = tan –1 and sec2 = .
2
2
4
dx 5
= (4π) ≈ 15.71 km/min
dt 4
4000
x
d
4000 dx
sec2
=−
dt
x 2 dt
1 d
1
4000
and x =
=
≈ 7322.
When = ,
2 dt 10
tan 12
16. tan =
dx
≈ sec 2
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 be the
angle that the light makes with the ground at
the tip of Chris' shadow.
6
d
6 ds
tan = so sec2
=–
and
s
dt
s 2 dt
6 cos 2
d
=–
dt
s2
When s = 6,
6
( )
1
2
2
2
1
d
⎛1⎞
=–
⎜ ⎟=– .
2
24
dt
⎝ ⎠
6
Chris must lift his head at the rate of
1
rad/s.
24
18. Let 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
2
and the
height of the triangle is a cos .
2
1⎛
⎞⎛
⎞ 1
A = ⎜ 2a sin ⎟ ⎜ a cos ⎟ = a 2 sin
2⎝
2 ⎠⎝
2⎠ 2
d
1
1
A = (100)2 sin = 5000sin ;
=
dt 10
2
dA
d
= 5000 cos
dt
dt
π dA
π ⎞⎛ 1 ⎞
⎛
When = ,
= 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
ds ds 1
= ft/s
.
dt dt 2
π
= , so
4
Instructor’s Resource Manual
Section 2.8
135
© 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
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 ;
d
π 11π
rad/h
= 2π – =
a = 5, b = 4,
dt
6
6
s 2 = 41 – 40 cos
ds
d
2s = 40sin
dt
dt
π
At 3:00, = 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
By similar triangles,
(for t > 1), so s =
48
64 – 16t
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
© 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.
27. a.
dx
= 2 ft/s. Let y
dt
y
18
216
=
be the height of the opposite end of the ladder. By similar triangles,
, so y =
.
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
πh3 π
–
3
6
dV
dh
dh
= 16πh
– πh 2
dt
dt
dt
V=
4 3
πr
3
dV
dr
= 4πr 2
dt
dt
k (4πr 2 ) = 4πr 2
dh
is
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
dV
= k (4πr 2 )
dt
a.
= 8πh 2 –
This is the same as in Problem 21, so
29.
dr
dt
dr
=k
dt
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
© 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.
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
8. dy = 2( x10 + sin 2 x )[10 x9 +
1
2 sin 2 x
⋅ (cos 2 x)(2)]dx
⎛
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 + csc 2 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 = –
b.
dy = –
3. Δx is small.
4. larger ; smaller
138
Section 2.9
dx
x
2
dx
x
2
=–
=–
0.5
(1)2
= –0.5
0.75
(–2)2
= –0.1875
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.
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≈
2
Δy = [(2.88) – 3] – [(3) – 3] = –0.7056
dy = 2xdx = 2(3)(–0.12) = –0.72
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
© 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.
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
= 1512 + 1512 – 2(151)(151) cos 0.53 ≈ 79.097
s = 45, 602 – 45, 602 cos
ds =
1
2 45, 602 – 45, 602 cos
22,801sin
=
45, 602 – 45, 602 cos
d
22,801sin 0.53
=
⋅ 45, 602sin d
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 = (151)(151) sin 0.53 ≈ 5763.33
2
2
22,801
A=
sin ; = 0.53, d = 0.005
2
22,801
dA =
(cos )d
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.
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.
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
2
2
38. g ( x) = x cos x; g '( x) = − x sin x + 2 x cos x
a =π /2
The linear approximation is then
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
2
π ⎛
π⎞
⎜x− ⎟
4 ⎝
2⎠
x+
π3
8
Instructor’s Resource Manual
Section 2.9
141
© 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.
45.
f ( x ) = 1 − x2 ;
(
)
−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 ) , f (x ) = (1 + x ) −1 , a = 0
The linear approximation is then
L(x ) = 1 + (x ) = x + 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
= −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
= −1
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.
y
49. a. lim ε ( h ) = lim ( f ( x + h ) − f ( x ) − f
h→0
= f ( x) − f ( x) − f
5
−5
5
x
−5
( x) h)
( x) 0 = 0
⎡ f ( x + h) − f ( x)
= lim ⎢
−f
h→0 h
h
⎣
= f ( x) − f ( x) = 0
b. lim
ε (h)
⎤
( x )⎥
⎦
= −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.
x
−5
=0
y
5
−5
5
x
−5
= 0.5
y
8. False:
5
f ( x) = ax 2 + b, g ( x) = ax 2 + c,
b ≠ c . Then f ( x) = 2ax = g ( x), but
f(x) ≠ g(x).
−5
5
9. True:
x
−5
=1
5
5
−5
Instructor’s Resource Manual
Dx f ( g ( x)) = f ( g ( x)) g ( x); since
g(x) = x, g ( x) = 1, so
Dx f ( g ( x)) = f ( g ( x)).
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
h →0
x
=2
Section 2.10
143
© 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. True:
29. True:
If f ( x) = x3 g ( x), then
Dx2 (sin x ) = – sin x;
Dx f ( x) = x3 g ( x) + 3x 2 g ( x)
Dx3 (sin x) = – cos x;
= x 2 [ xg ( x) + 3 g ( x)].
16. False:
Dx4 (sin x) = sin x;
Dx y = 3 x 2 ; At (1, 1):
Dx5 (sin x) = cos x
2
mtan = 3(1) = 3
Tangent line: y – 1 = 3(x – 1)
17. False:
30. False:
Dx3 (cos x) = sin x;
Dx4 (cos x ) = Dx [ Dx3 (cos x)] = Dx (sin x)
+ g ( x) f ( x) + f ( x) g ( x)
= f ( x) g ( x) + 2 f ( x) g ( x) + f ( x) g ( x)
3
Dxn +3 (cos x ) = Dxn (sin x).
The degree of y = ( x + x) is 24, so
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:
Since D1x+3 (cos x) = D1x (sin x),
8
Dx25 y = 0.
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:
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
( π2 )
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)
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 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 .
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.
Sample Test Problems
3( x + h)3 – 3x3
9 x 2 h + 9 xh 2 + 3h3
= lim
= lim (9 x 2 + 9 xh + 3h 2 ) = 9 x 2
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
h →0
⎡
–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
h →0
3h
= lim
h →0 h(
f.
g.
h →0
3
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 3x – 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 3 x)(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.
3 x + 3h + 3 x )
= lim
f ( x) = lim
= lim
h.
3( x + h) – 3 x
( 3x + 3h – 3x )( 3x + 3h + 3 x )
= lim
h
h →0
h( 3 x + 3h + 3x )
2 xh + h 2
h ⎛⎜ ( x + h) 2 + 5 + x 2 + 5 ⎞⎟
⎝
⎠
= lim
h →0
2x + h
2
2
( x + h) + 5 + x + 5
=
2x
2
2 x +5
=
x
2
x +5
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
Instructor’s Resource Manual
Section 2.10
145
© 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.
b.
(t 3 + t ) – ( x3 + x)
t–x
t→x
g ( x) = lim
= lim
t→x
2
g.
2
(t – x)(t + tx + x ) + (t – x)
t–x
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
t → x (t 2
+ 1)( x 2 + 1)(t – x)
–( x + t )
2x
= lim
=–
2
t → x (t 2 + 1)( x 2 + 1)
( x + 1)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
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.
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
2
(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 – 3[sin (2)(cos )(– sin ) + cos3 ]
= – sin + 6sin 2 cos – 3cos3
16.
−3 x 2 + 10 x + 3
( 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 ⎠
−24t 2 + 60t + 10
(6t 2 + 2t ) 2
10. Dx (3x + 2) 2 / 3 =
2t + 6
2 2t + 6
D 2 (sin + cos3 )
2
⎛ 3 x – 5 ⎞ ( x 2 + 1)(3) – (3 x – 5)(2 x)
8. Dx ⎜
⎟=
( x 2 + 1)2
⎝ x2 + 1 ⎠
=
t
1
= cos – 3sin cos 2
7. Dz ( z + 4 z + 2 z ) = 3z + 8 z + 2
=
( x3 + x ) 2
15. D (sin + cos3 ) = cos + 3cos 2 (– sin )
= 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
17. D [sin(
18.
2
)] = cos(
2
)(2 ) = 2 cos(
2
)
d
(cos3 5 x) = (3cos2 5 x )(– sin 5 x )(5)
dx
= –15cos 2 5 x sin 5 x
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
Instructor’s Resource Manual
Section 2.10
147
© 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.
19.
d
[sin 2 (sin(π ))] = 2sin(sin(π )) cos(sin(π ))(cos(π ))(π) = 2π sin(sin(π )) cos(sin(π )) cos(π )
d
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. D tan 3 = (sec2 3 )(3) = 3sec2 3
22.
23.
d ⎛ sin 3 x ⎞ (cos 5 x 2 )(cos 3x)(3) – (sin 3x)(– 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
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.
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
39. s = t 3 – 6t 2 + 9t
ds
v(t ) =
= 3t 2 – 12t + 9
dt
a(t ) =
35. V =
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
dt 2
= 6t –12
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, ∞)
4 3
πr
3
dV
= 4πr 2
dr
d 2s
a.
2
1 ⎛7 1 ⎞
⎛7
⎞
y = ⎜ – 2⎟ = ; ⎜ , ⎟
16 ⎝ 4 16 ⎠
⎝4
⎠
128t – 16t 2 = 0
–16t(t – 8) = 0
The object hits the ground when t = 8s
v = 128 – 32(8) = –128 ft/s
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
Section 2.10
149
© 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.
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
2
dy
tan( xy ) + xy sec ( xy )
=–
dx
x 2 sec2 ( xy )
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) + 3 g 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⎠
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.
c.
Dx3 x = Dx ( Dx2 x ) = Dx (0) = 0
d.
Dx2 (
49. a.
x ( x − 1)( x − 2 ) = 0
2
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 ) ,
x ) = Dx (2 x) = 2
=
D sin
b.
3. x ( x − 1)( x − 2 ) ≤ 0
=
D cos
sin
sin
cos
cos
cos = cot
sin
(− sin ) = − tan
( 0,1) , (1, 2 ) , and ( 2, ∞ ) . The solution set is
{ x | x ≤ 0 or 1 ≤ x ≤ 2} , or ( −∞, 0] ∪ [1, 2] .
cos
1
( x + 1)−1/ 2 ; a = 3
2
L( x) = f (3) + f '(3)( x − 3)
4.
(
and ( 3, ∞ ) . The solution set is { x | 2 < x < 3} or
( 2,3) .
−
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, ∞ ) .
−
−
−
x ( x − 2)
5.
x2 − 4
x ( x − 2)
( x − 2 )( x + 2 )
−
≥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
−
x2 − x − 6 > 0
2.
)
x ( x + 1)( x + 2 ) = 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) ,
−
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
( x − 2 )( x − 3) < 0
( x − 2 )( x − 3) = 0
−
x3 + 3x 2 + 2 x ≥ 0
Review and Preview Problems
1.
−
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.
−
−
f ( x) = x + 1; f '( x) = −
50. a.
( 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, ∞ ) .
−
−
−
−
−
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, ∞ ) .
−
−
−
−
−
Instructor’s Resource Manual
Review and Preview
151
© 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.
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, ∞ ) .
−
−
−
−
−
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
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
.
1
−1/ 2
=
x
cos x
2 x
1
cos 2 x
−1/ 2
f ' ( x ) = ( sin 2 x )
⋅ cos 2 x ⋅ 2 =
2
sin 2 x
4− x
His distance swimming will be
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
2
19. Consider the diagram:
1 + sin 2 x
f ' ( x ) = cos
π
= x ( 9 − 2 x )( 24 − 2 x )
( 2sin x )( cos 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.)
14.
an integer.
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.
Review and Preview
x2 + 1 4 − x
+
hours.
4
10
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.
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.
Instructor’s Resource Manual
Review and Preview
153
© 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.
CHAPTER
Applications of the
Derivative
3
7.
3.1 Concepts Review
1. continuous; closed and bounded
2. extreme
3. endpoints; stationary points; singular points
4.
3
( x) = 2 x + 3; 2x + 3 = 0 when x = – .
2
3
Critical points: –2, – , 1
2
9
⎛ 3⎞
(–2) = –2, ⎜ – ⎟ = – , (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
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.
11. h (r ) = −
1
2
; h (r ) is never 0; h (r ) is not defined
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−
15. g ( x) = −
; −
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.)
x → 0+
2x
2 2
(1 + x )
Critical point: 0
g(0) = 1
No maximum value, no minimum value.
12. g ( x) = −
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 ( ) = cos ; cos = 0 when
=
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
155
© 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.
x –1
; a ( x) does not exist when x = 1.
x –1
19. a ( x) =
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 ' (
sec
y
3(3s – 2)
; f ( s) does not exist when s = 2 .
3
3s – 2
1
( )
21. g ( x) =
2
22. s (t ) =
− π4
; 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
3
π π 5π
,
1
when
2
3 3 3
π
π π 5π
, , , 2π
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
Section 3.1
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
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
Maximum value: 16
; Minimum value: −1
5
,
Critical points: −2π , −
π
h ' ( t ) is undefined when t = −2 and h ' ( t ) = 0
Maximum value: 1
Minimum value: −1
,−
, 0,
2
Maximum value:
H ( 6π ) = 1 ; H ( 7π ) = −1 ; H ( 8π ) = 1
π
4
⎛ 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 =
π
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
156
( sec tan ) + 2 sec
= sec ( tan + 2 )
( tan + 2 ) = 0 when = 0 .
2
Consider the graph:
Critical points: −1, 2 , 4
3
2
= 0, f(4) = 10
f(–1) = 5, f
3
Maximum value = 10, minimum value = 0
x=−
)=
3
,−
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
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.
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
© 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.
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
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.
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 <
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.
π
and
2
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 ( ) = –2 cos sin ; R ( ) > 0 when
and
3π
<
2
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.
<π
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).
< 2π.
⎡ π ⎤ ⎡ 3π
⎤
R( ) 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).
Instructor’s Resource Manual
Section 3.2
159
© 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.
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].
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) = 36 x 2 – 24 x = 12 x(3 x – 2); g ( x) > 0
G ( x) = 60 x3 – 30 x = 30 x(2 x 2 – 1);
2
when x < 0 or x > . g(x) is concave up on
3
⎛2 ⎞
⎛ 2⎞
(– ∞, 0) ∪ ⎜ , ∞ ⎟ and concave down on ⎜ 0, ⎟ .
3
⎝
⎠
⎝ 3⎠
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
15
⎛ 1 ⎞ 15
⎛1⎞
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
,
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.
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) =
f ( x) =
–2(5 x + 1)
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) = –
27.
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) = , 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).
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 ⎠
Instructor’s Resource Manual
Section 3.2
161
© 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.
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
30.
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
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
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.
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.
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.
b.
f ( x) < 0 : (2.0, 4.7) ∪ (9.9, 10]
[ f ( x ) ⋅ g ( x)] = [ f ( x) g ( x) + f ( x) g ( x)]
c.
f ( x) < 0 : [0, 3.4) ∪ (7.6, 10]
= 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.
c.
46. a.
[ f ( g ( x))] = [ f ( g ( x)) g ( x)]
⎡ 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)
Instructor’s Resource Manual
Section 3.2
163
© 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.
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
e.
t
ds
d 2s
are approaching zero.
and
dt
dt 2
s
>0
s
Concave down.
Concave up.
164
Section 3.2
t
t
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.
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
© 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.
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)
d
d2
> 0,
> 0 , where is the angle that
dt
dt 2
the tower makes with the vertical.
Concave up.
( 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
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.
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 = 3x + 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
Instructor’s Resource Manual
Section 3.2
167
© 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.
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 ( ) = 2 cos 2 ; 2 cos 2 ≠ 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
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.
( ) = 2sin cos
5.
−
π
<
<
9. h ' ( y ) = 2 y +
π
2
2
Critical point: 0
⎛ π ⎞
( ) < 0 on ⎜ − , 0 ⎟ ,
⎝ 2 ⎠
( ) = 2 cos 2 – 2sin 2 ;
Local minimum at x = 0
⎛ π⎞
( ) > 0 on ⎜ 0, ⎟ ,
⎝ 2⎠
(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 )
f '' ( x ) =
(
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
3
g '' ( 0 ) = 2
Local minima at z = 0 .
Instructor’s Resource Manual
4
2
⎛ 34⎞
2
h⎜ −
⎟ = 2− 3
⎝ 2 ⎠
− 24
( )
Local minima at −
10.
f '( x) =
(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.
3
3
⎛
4⎞
h ' ( y ) < 0 on ⎜ −∞, −
⎟
2 ⎠
⎝
⎛ 34 ⎞
, 0 ⎟ and ( 0, ∞ )
h ' ( y ) > 0 on ⎜ −
⎝ 2
⎠
2
h '' ( y ) = 2 − 3
y
r ( x ) = 12 x 2 ; r (0) = 0; the Second Derivative
Test fails.
Local minimum at z = 0; no local maxima
f '( x) =
y2
Critical point: −
6. r ( z ) = 4 z 3
Critical point: 0
r ( z ) < 0 on (−∞, 0);
r ( z ) > 0 on (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
© 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.
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
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 ⎠
⎝
⎠
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)
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.
19.
f ( x) > 0 on (−∞, 2), f ( x) > 0 on (2, ∞)
No local minimum or maximum values
20.
2
; 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 .
⎛ 2⎞
s = –⎜ ⎟
⎝ 15 ⎠
2
5s 3 / 5
=
15s3 / 5 + 2
5s 3 / 5
; r ( s ) = 0 when
5/3
, r ( s ) does not exist at s = 0.
⎛ 2⎞
Critical points: – ⎜ ⎟
⎝ 15 ⎠
170
Section 3.3
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
;
1 + sin
( ) does not exist at
=
f ( x) = 20( x – 2)3 ; f (2) = 0
16. r ( s ) = 3 +
f (t ) = 1 +
3π
, but ( ) does not exist at that point
2
either.
No critical points
No local minimum or maximum values
f ( x) = 5( x – 2) 4
Critical point: 2
15. g (t ) = –
6
r (s) = –
5/3
,0
g( )=
sin cos
; g ( ) = 0 when
sin
=
π 3π
;
,
2 2
g ( ) 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 ⎠
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.
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
Instructor’s Resource Manual
Section 3.3
171
© 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.
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
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.
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
Instructor’s Resource Manual
Section 3.3
173
© 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.
⎛ − B − B 2 − 3 AC
f '' ⎜
⎜
3A
⎝
Problem Set 3.4
⎞
⎟
⎟
⎠
⎛ − B − B 2 − 3 AC
= 6A⎜
⎜
3A
⎝
⎞
⎟ + 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 < ∞
200
2. 2x +
x
3. S =
n
i =1
( yi − bxi )2
4. marginal revenue; marginal cost
174
Section 3.4
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, ∞) .
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.
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
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.
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
© 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.
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
© 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.
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
© 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.
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) =
– ; T ( x) = 0 has no solution.
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
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.
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 = ±
dh
4
3
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
© 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.
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
dr
3 ⎠ ⎝ r2
3 ⎠
⎝ 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
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.
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
tan =
cos =
r
⎛
x = h⎜
⎜
⎝
dr
r3
> 0, so this minimizes the perimeter.
34. The distance from the fence to the base of the
h
.
ladder is
tan
The length of the ladder is x.
h +w
h
cos = tan
; x cos =
+ w;
x
tan
h
w
+
x=
sin
cos
dx
h cos
w sin w sin 3 − h cos3
=−
+
;
d
sin 2
cos 2
sin 2 cos2
h
when tan 3 =
w
= tan −1 3
3
h
+ w2 / 3
,
w
2/3
+ w2 / 3
⎛ h 2 / 3 + w2 / 3
h 2 / 3 + w2 / 3 ⎞⎟
+ w⎜
3
3
⎟
⎜
h
w
⎠
⎝
⎞
⎟
⎟
⎠
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.
A = 2 xy = 2 x r 2 − x 2
2A
= 2r +
Q = 2r + r = 2r +
2
r
r
dQ
2A
= 2−
; Q = 0 when r = A
dr
r2
2A
=
=2
( A )2
4A
h
2/3
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
=
w
35. x is limited by 0 ≤ x ≤ 12 .
2 Ar
2
3
3
; sin =
= ( h 2 / 3 + w 2 / 3 )3 / 2
2A
; =
2
r2
The perimeter is
d 2Q
h
h
⎛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.
33. A =
3
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
.
=0
h
w
Instructor’s Resource Manual
Section 3.4
181
© 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.
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
r 2 – h2
1
r
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 . The
area is
⎛1⎞
A = 3(3sin ) + 2 ⎜ ⎟ (3cos )(3sin )
⎝2⎠
= 9sin + 9 cos sin .
dA
= 9 cos + 9(− sin ) sin + 9 cos cos
d
= 9(2 cos 2 + cos − 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 − sin 2 + cos 2 )
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 + cos − 1 = 0; cos = −1, ; = π,
2
3
π
Since 0 ≤ ≤ , the critical points are
2
π
π
0, , and .
3
2
When = 0 , A = 0.
When
When
π
27 3
,A=
≈ 11.7.
3
4
π
= , A = 9.
2
=
The carrying capacity is maximized when
=
π
.
3
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. The circumference of the top of the tank is the
circumference of the circular sheet minus the arc
length of the sector,
20π − 10 meters. The radius of the top of the
20π − 10
5
= (2π − ) meters. The
tank is r =
2π
π
slant height of the tank is 10 meters, so the height
of the tank is
2
5 ⎞
5
⎛
h = 102 − ⎜10 − ⎟ =
4π −
π ⎠
π
⎝
2
1
1 ⎡5
⎤
V = πr 2 h = π ⎢ (2π − ) ⎥
3
3 ⎣π
⎦
125
(2π − )2 4π − 2
=
3π2
dV 125 ⎛
=
⎜ 2(2π − )(−1) 4π
d
3π2 ⎝
+
=
(2π − )2
125(2π − )
3π
2
4π −
125(2π − )
3π
2
4π −
2
2
(3
(3
2π − = 0 or 3
2
2
2
2
meters.
⎡5
⎢ π 4π −
⎣
−
( 12 ) (4π − 2
4π −
2
2⎤
⎥
⎦
2
)⎞
⎟
⎟
⎠
− 12π + 4π2 ) ;
− 12π + 4π2 ) = 0
− 12π + 4π2 = 0
2 6
2 6
π, = 2π +
π
3
3
< 2π, the only critical point is
= 2π, = 2π −
Since 0 <
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
© 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.
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.
1
⎛ 1
⎞ a ⎛
⎛π
⎞
⎞
x = a cos ⎜ − ⎟ = a ⎜
cos +
sin ⎟ =
⎜ cos + sin ⎟
2
2⎠
2
2⎠
2
2⎝
⎝4 2⎠
⎝ 2
1
⎛ 1
⎞ a ⎛
⎛π
⎞
⎞
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 ⎛
⎡ a ⎛
⎞⎤ ⎡ a ⎛
⎞⎤
⎞⎤
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⎝
⎛
⎞
⎛
⎞
= 4a 2 ⎜ cos 2 − sin 2 ⎟ − 2a 2 ⎜1 − 2 cos sin ⎟ = 4a 2 cos − 2a 2 (1 − sin )
2
2⎠
2
2⎠
⎝
⎝
dA
1
= −4a 2 sin + 2a 2 cos ; −4a 2 sin + 2a 2 cos = 0 when tan = ;
d
2
1
2
sin =
, cos =
5
5
d2A
d
2
< 0 when tan =
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
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.
47. a.
L ( ) = 15(9 + 25 − 30 cos )−1/ 2 sin = 15(34 − 30 cos )−1/ 2 sin
L ( )=−
15
(34 − 30 cos )−3 / 2 (30sin ) sin + 15(34 − 30 cos )−1/ 2 cos
2
= −225(34 − 30 cos ) −3 / 2 sin 2 + 15(34 − 30 cos ) −1/ 2 cos
= 15(34 − 30 cos )−3 / 2 [−15sin 2 + (34 − 30 cos ) cos ]
= 15(34 − 30 cos ) −3 / 2 [−15sin 2 + 34 cos − 30 cos 2 ]
= 15(34 − 30 cos ) −3 / 2 [−15 + 34 cos − 15cos 2 ]
= −15(34 − 30 cos )−3 / 2 [15cos 2 − 34 cos + 15]
L = 0 when cos =
34 ± (34)2 − 4(15)(15) 5 3
= ,
2(15)
3 5
⎛3⎞
= 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 ( ) = 65(25 + 169 − 130 cos )−1/ 2 sin = 65(194 − 130 cos )−1/ 2 sin
L ( )=−
65
(194 − 130 cos )−3 / 2 (130sin ) sin + 65(194 − 130 cos ) −1/ 2 cos
2
= −4225(194 − 130 cos )−3 / 2 sin 2 + 65(194 − 130 cos ) −1/ 2 cos
= 65(194 − 130 cos )−3 / 2 [−65sin 2 + (194 − 130 cos ) cos ]
= 65(194 − 130 cos )−3 / 2 [−65sin 2 + 194 cos − 130 cos 2 ]
= 65(194 − 130 cos )−3 / 2 [−65cos 2 + 194 cos − 65]
= −65(194 − 130 cos )−3 / 2 [65cos 2 − 194 cos + 65]
L = 0 when cos =
194 ± (194)2 − 4(65)(65) 13 5
= ,
2(65)
5 13
⎛5⎞
= cos −1 ⎜ ⎟
⎝ 13 ⎠
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
Instructor’s Resource Manual
Section 3.4
185
© 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.
d.
L ( ) = hm(h 2 + m2 − 2hm cos )−1/ 2 sin
L ( ) = −h 2 m2 (h 2 + m 2 − 2hm cos ) −3 / 2 sin 2 + hm(h 2 + m 2 − 2hm cos )−1/ 2 cos
= hm(h 2 + m 2 − 2hm cos )−3 / 2 [− hm sin 2 + (h 2 + m 2 ) cos − 2hm cos 2 ]
= hm(h 2 + m 2 − 2hm cos ) −3 / 2 [− hm cos 2 + (h 2 + m 2 ) cos − hm]
= −hm(h 2 + m2 − 2hm cos )−3 / 2 [hm cos 2 − (h 2 + m2 ) cos + hm]
L = 0 when hm cos 2 − (h 2 + m2 ) cos + hm = 0
(h cos − m)(m cos − h) = 0
cos =
m h
,
h m
Since h < m, cos =
h
so
m
⎛h⎞
= cos −1 ⎜ ⎟ .
⎝m⎠
⎛
⎛
⎛ h ⎞⎞
⎛ h ⎞⎞
L ⎜ cos −1 ⎜ ⎟ ⎟ = hm ⎜ h 2 + m 2 − 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
© 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.
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
© 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.
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
=
[ yi − (5 + bxi )]2
db db i =1
54. C(x) = 7000 + 100x
55. n = 100 + 10
R (n) = np (n) = 300n –
d
=
[ yi − (5 + bxi )]2
db
i =1
=
= 300n –
⎡n
= 2⎢
− xi yi + 5 xi + bxi 2
⎢⎣ i =1
(
n
= −2
i =1
Setting
0 = −2
n
i =1
n
b
i =1
b=
i =1
xi yi + 10
xi yi + 5
xi 2 =
n
n
i =1
xi + 2b
n
i =1
n
i =1
i =1
i =1
n
i =1
xi + 2b
xi + b
xi yi − 5
xi yi − 5
n
n
xi 2
n
i =1
n
i =1
xi
xi 2
n
i =1
xi
xi 2
b. Using the formula from a., we get that
(2037) − 5(52)
b=
≈ 3.0119
590
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
58.
xi
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.
c.
Estimate n ≈ 200
P (n) = 200 – n; 200 – n = 0 when n = 200.
P (n) = –1, so profit is maximum at n = 200.
2
n
i =1
n2
2
57.
⎤
)⎥⎥⎦
dS
= 0 gives
db
i =1
n
0=−
xi yi + 10
n2
– (7000 + 100n)
2
= −7000 + 200n –
2( yi − 5 − bxi )(− xi )
i =1
n2
2
56. P (n) = R (n) – C (n)
n
n
250 – p (n)
n
so p(n) = 300 –
5
2
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
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.
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
© 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.
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
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.
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
⎧⎪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
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.
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
Instructor’s Resource Manual
Section 3.5
191
© 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.
4. Domain: (– ∞ , ∞ ); range: (– ∞ , ∞ )
Neither an even nor an odd function
y-intercept: –1; x-intercept: 1
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⎦
⎝
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)
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
⎝
⎞
1 ⎞ ⎛ 1
, ∞⎟
⎟∪⎜
6⎠ ⎝ 6
⎠
1 ⎞
⎟ ; inflection
6⎠
⎛ 1
5 ⎞
⎛ 1 5 ⎞
points H ⎜ –
, − ⎟ and H ⎜
, ⎟
36
6
⎝
⎠
⎝ 6 36 ⎠
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 ⎠
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
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.
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
© 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.
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)
2
(– ) 2
=
= ( ); even
(– ) =
2
+1
(– )2 + 1
function; symmetric with respect to the y-axis.
y-intercept: 0; -intercept: 0
2
( )=
; ( ) = 0 when = 0
2
( + 1)2
Critical point: 0
( ) > 0 when > 0
( ) is increasing on [0, ∞ ) and
decreasing on (– ∞ , 0].
Global minimum (0) = 0; no local maxima
2(1 – 3 2 )
( )=
; ( ) > 0 when
( 2 + 1)3
1
1
< <
–
3
3
⎛ 1 1 ⎞
( ) is concave up on ⎜ –
,
⎟ and
3 3⎠
⎝
1 ⎞ ⎛ 1
⎛
⎞
concave down on ⎜ – ∞, –
, ∞ ⎟;
⎟∪⎜
3⎠ ⎝ 3 ⎠
⎝
⎛ 1 1⎞ ⎛ 1 1⎞
inflection points ⎜ –
, ⎟, ⎜
, ⎟
3 4⎠ ⎝ 3 4⎠
⎝
2
lim
→∞
2
+1
= lim
2
lim
→–∞
2
+1
1
→∞ 1 + 1
2
= 1;
1
= 1;
→–∞ 1 + 1
2
= lim
y = 1 is a horizontal asymptote. No vertical
asymptotes
= 0;
y = 0 is a horizontal asymptote.
194
Section 3.5
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.
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
© 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.
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
Section 3.5
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.
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
6
2
if x > 0
+
⎩
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
197
© 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.
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
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.
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
Instructor’s Resource Manual
Section 3.5
199
© 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.
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
Section 3.5
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.
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
f ( x) = 12ax 2 + 6bx + 2c = 2(6ax 2 + 3bx + c)
37.
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.
Instructor’s Resource Manual
Section 3.5
201
© 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.
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
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.
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
− π − π
π
−π
π
π 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
− π − π
(
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
π
π 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 .
Instructor’s Resource Manual
Section 3.5
203
© 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.
y
Justification:
f (1) = g (1) = 1
c=0
− π − π
π
−π
π
π 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)
π
π x
y
− π − π
'( 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
− π − π
'( x) = g '( x 4 )4 x3
c=2
π
π
π 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
Section 3.5
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.
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
Section 3.5
205
© 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.
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 3 x + 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
Section 3.5
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
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.
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.
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.
Instructor’s Resource Manual
Section 3.5
207
© 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.
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
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.
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
209
© 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.
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 ( ) = cos
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 ( ) is not continuous at = −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 ( ) is not continuous at = .
2
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.
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) = 2 x +
f (b) – f (a )
1
[ (b 2 – a 2 ) + (b – a )]
=
b–a
b–a
= ( a + b) +
2 c+
= ( a + b) +
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.
Instructor’s Resource Manual
Section 3.6
211
© 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.
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 < b and both either positive or negative, the
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 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
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.
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
213
© 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.
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
So, f '
50.
≤M y−x .
0 → f = constant .
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
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.
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
n
hn
mn
f (mn )
1
0.5
1.5
0.375
2
0.25
1.25
–1.546875
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
Instructor’s Resource Manual
Section 3.7
215
© 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.
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
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.
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
Instructor’s Resource Manual
Section 3.7
217
© 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. Let f ( x ) = 2 x 2 − sin x .
16. Let f ( x) = x 4 – 47 .
y
f ( x ) = 4 x3
2
−2
x
2
−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
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.
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
© 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.
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
⎡ 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
f (in )
(1 + in ) 24 (20in − 1) + 1
= in −
f (in )
(1 + in )23 (500in − 4)
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
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.
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
© 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.
c.
33. a.
x = 5( x – x 2 )
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
. Then x satisfies
c.
(1 + 5 ) / 2 ≈ 1.618034 .
x1 = 0
x2 = 5 ≈ 2.236068
1
x
x4 = 5 + 5 + 5 ≈ 2.7730839
x5 = 5 + 5 + 5 + 5 ≈ 2.7880251
x = 5 + x , and x must satisfy x ≥ 0
x2 = 5 + 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
b.
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 –
Suppose f(r) = 0. Then r –
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,
f (r )
.
f (r )
f (r )
= 0, so f(r) = 0.
f (r )
x2 − x − 5 = 0
(
8
= 1.6
5
x2 − x − 1 = 0
x=
Let x = 1 + 1 + 1 +
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
© 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.
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
y
b
2
D
(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
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 and b =
= 6.2 csc
a=
cos
sin
Note that the angle determines the position of
the rod. The total length of the rod is then
L = a + b = 8.6sec + 6.2 csc
⎛ π⎞
The domain for is the open interval ⎜ 0, ⎟ .
⎝ 2⎠
Section 3.7
223
© 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.
The derivative of L is
8.6sin 3 − 6.2 cos3
L '( ) =
sin 2 ⋅ cos 2
Thus, L ' ( ) = 0 provided
8.6sin 3 − 6.2 cos3
8.6sin
=0
3
= 6.2 cos3
sin 3
6.2
8.6
6.2
=
8.6
=
cos3
tan 3
tan =
3
Note that the angle determines the position of
the rod. The total length of the rod is then
L = a + b = 8csc + 8csc ( 75 − )
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 − 3
8.6
1
=
π
4
As suggested in the diagram, let a and b represent
the lengths of the segments AB and BC, and let
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 and
a=
sin
8
b=
= 8csc ( 75 − )
sin ( 75 − )
The domain for
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
n
1 π4 ≈ 0.78540
2
3
0.73373
0.73098
4
0.73097
L '(
)=
(
8 sin 2 ⋅ cos ( − 75 ) − cos ⋅ sin 2 ( − 75 )
)
sin ⋅ sin ( − 75 )
We will use Newton’s method to solve
L ' ( ) = 0 starting with 1 = 40 .
5
0.73097
Note that ≈ 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
n
1
40
2
2 37.54338
3 37.50000
4
37.5
Note that = 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 −
B
105
o
a
A
224
8 feet
Section 3.7
D
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.
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 4 + 3x 2 + 1)8 (4 x3 + 6 x )dx = u8 du
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
=
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.
( x 2 + π)dx = x 2 dx + π 1dx =
4.
( 3x2 + 3 ) dx = 3
=3
5.
6.
7.
x3
+ πx + C
3
x 2 dx + 3 1dx
x3
+ 3 x + C = x3 + 3 x + C
3
x5 / 4 dx =
x9 / 4
9
4
+C =
4 9/4
x
+C
9
⎛ x5 / 3
⎞
3x 2 / 3 dx = 3 x 2 / 3 dx = 3 ⎜
+ C1 ⎟
⎜ 5
⎟
⎝ 3
⎠
9
= x5 / 3 + C
5
1
3 2
dx = x −2 / 3 dx = 3 x1/ 3 + C = 33 x + C
x
8.
7 x −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
© 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.
(3 x 2 − πx)dx = 3 x 2 dx − π x dx
10.
4 x6 + 3x 4
17.
x
dx = (4 x3 + 3 x) dx
⎛ x3
⎞
⎛ x2
⎞
= 3 ⎜ + C1 ⎟ − π ⎜
+ C2 ⎟
⎜ 3
⎟
⎜ 2
⎟
⎝
⎠
⎝
⎠
= 4 x3 dx + 3 x dx
πx 2
=x −
+C
2
= x4 +
3
(4 x5 − x3 )dx = 4 x5 dx − x3 dx
11.
=
=
x
2 x6 x 4
−
+C
3
4
=
x101 x100
+
+C
101 100
=
6
4
20.
( x3 + x ) dx =
=
(
)
=
( x5 + 5 x 4 − 3 x3 +
)
3 x 2 dx
=3 x
dx − 2 x
−1
(
16.
=
2 x1/ 2
1
2
= 2 2x −
+
dx
=
(
4x
4
+C
Section 3.8
2
2
( )
2
1 + 2 ) z3
(
2
+C
z dz =
2
dz = ⎡ 1 + 2 z ⎤ dz
⎣
⎦
3
dz =
z4 + 2z2 + 1
z
dz
2 9/ 2 4 5/ 2
z
+ z
+ 2 z1/ 2 + C
9
5
s( s + 1)2
24.
3x −4
+C
−4
3
z
)
= z 7 / 2 dz + 2 z 3 / 2 + z −1/ 2 dz
−2
⎛ 2x 3 ⎞
+ ⎟⎟ dx =
⎜⎜
x5 ⎠
⎝ x
)
( z 2 + 1)2
23.
3x
2x
−
+C
−1
−2
3 1
=− +
+C
x x2
=
2z
= 1+ 2
⎛ 3
2 ⎞
−2
−3
⎜ 2 − 3 ⎟ dx = (3 x − 2 x ) dx
x ⎠
⎝x
−3
x 4 x3 / 2
x 4 2 x3
+
+C =
+
+C
3
4
4
3
(z +
22.
x6
3x4
3 x3
=
+ x5 −
+
+C
6
4
3
−2
x3 dx + x1/ 2 dx
2
= x5 dx + 5 x 4 dx −3 x3 dx + 3 x 2 dx
15.
x3 x 2
+
+C
3
2
21. Let u = x + 1; then du = dx.
u3
( x + 1)3
( x + 1)2 dx = u 2 du =
+C =
+C
3
3
⎡ x 2 x3 + 5 x 2 − 3x + 3 ⎤ dx
⎥⎦
⎣⎢
14.
x4 1
+ +C
4 x
2
27 x
x
45 x
2x
+
−
+
+C
8
2
4
2
x 4 x −1
−
+C
4
−1
( x 2 + x) dx = x 2 dx + x dx =
= 27 x7 dx + 3 x5 dx − 45 x3 dx + 2 x dx
8
dx = ( x3 − x −2 ) dx
19.
(27 x7 + 3x5 − 45 x3 + 2 x)dx
13.
3
= x3 dx − x −2 dx =
( x100 + x99 )dx = x100 dx + x99 dx
12.
3x 2
+C
2
x6 − x
18.
⎛ x6
⎞ ⎛ x4
⎞
= 4 ⎜ + C1 ⎟ − ⎜
+ C2 ⎟
⎜ 6
⎟ ⎜ 4
⎟
⎝
⎠ ⎝
⎠
226
3
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 − cos )d = sin d − cos d
= − cos − sin + C
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.
(t 2 − 2 cos t )dt = t 2 dt − 2 cos t dt
26.
=
3y
t3
− 2sin t + C
3
2
2y + 5y
27. Let g ( x) = 2 x + 1 ; then g ( x) = 2 .
(
)
2 x +1
3
[ g ( x)]3 g ( x)dx
2 dx =
[ g ( x)]4 + C = (
=
)
2 x +1
4
4
4
+C
28. Let g ( x) = πx3 + 1 ; then g ( x) = 3πx 2 .
(πx3 + 1)4 3πx 2 dx =
=
[ g ( x)]4 g ( x) dx
[ 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 6
1 ⎛ u7
u du = ⎜
=
+ C1 ⎟
⎟
3
3 ⎜⎝ 7
⎠
=
=
32. Let u = 2 y 2 + 5; then du = 4 y dy
(5 x3 + 3x − 8)7
+C
21
30. Let u = 5 x3 + 3x − 2; then du = (15 x 2 + 3)dx .
(5 x 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
(4t )(2t 2 − 11)1/ 3 dt
4
3 1/ 3
3⎛3
⎞
u du = ⎜ u 4 / 3 + C1 ⎟
=
4
4⎝4
⎠
9
= (2t 2 − 11) 4 / 3 + C
16
9
= 3 (2t 2 − 11) 4 + C
16
3
(4 y )(2 y 2 + 5) −1/ 2 dy
4
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
3x x + 4 dx
x 2 x3 + 4 dx =
3
1
1
=
u du =
u1/ 2 du
3
3
1⎛ 2
⎞
= ⎜ u 3/ 2 + 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
⎞
= ⎜ u 3/ 2 + C1 ⎟
4⎝3
⎠
3/
2
1 4
= x + 2 x2
+C
6
=
(
)
35. Let u = 1 + cos x ; then du = − sin x dx .
sin x (1 + cos x ) dx = − − sin x (1 + cos x ) dx
4
4
⎛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 .
sin x cos x 1 + sin 2 x dx
3
3t 2t 2 − 11 dt =
Instructor’s Resource Manual
dy =
1
⋅ 2sin x cos x 1 + sin 2 x dx
2
1
1
=
u du =
u1/ 2 du
2
2
1⎛2
⎞
= ⎜ u 3/ 2 + C1 ⎟
2⎝3
⎠
3/
2
1
= 1 + sin 2 x
+C
3
=
(
)
Section 3.8
227
© 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.
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.
=
228
=−
( x) g ( x) ] dx = f ( x) g ( x) + C
1
(2 x + 5)3 / 2
⎡
− x3
3x2 ⎤
+
⎢
⎥ dx
3/ 2
2 x + 5 ⎦⎥
⎣⎢ (2 x + 5)
=
[ 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
Section 3.8
[ f ( x) g ( x) + f
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
⎠
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.
dx =
2 x −1
⎡ x2
⎤
+ 2 x x − 1 ⎥ dx
⎢
⎢⎣ 2 x − 1
⎦⎥
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
g ( x)
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
2
45. Let f ( x) = x 2 , g ( x) = x − 1 .
f ( x) = (− x 2 + 3 x + C1 )dx
39.
g ( x) f ( x) − f ( x) g ( x)
=
x3
2x + 5
+C
f ( x)dx =
47.
d
f ( x)dx = f ( x) + C
dx
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
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.
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
= f m ( x )[ g n ( x)] + [ f m ( x)] g n ( x)
= −2 cos x − x sin x + C1 x + C2
= 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,
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 −1 ( x) g n −1 ( x)[nf ( x) g ( x) + mg ( x) f ( x)]dx
= f m ( x) g n ( x) + C .
50. Let u = sin[( x 2 + 1)4 ];
b.
then du = cos ⎡( x 2 + 1)4 ⎤ 4( x 2 + 1)3 (2 x)dx .
⎣
⎦
F16 ( x) = x sin x + 16 cos x +
Cn x16− n
n =1 (16 − n)!
16
du = 8 x cos ⎡ ( x 2 + 1)4 ⎤ ( x 2 + 1)3 dx
⎣
⎦
sin 3 ⎡( x 2 + 1)4 ⎤ cos ⎡( x 2 + 1) 4 ⎤ ( x 2 + 1)3 x dx
⎣
⎦
⎣
⎦
4
⎛
⎞
1
1 3
1 u
= u 3 ⋅ du =
u du = ⎜
+ C1 ⎟
⎟
8
8
8 ⎜⎝ 4
⎠
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
x +C .
2
1
x dx = − x 2 + C .
2
x dx =
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
sin 2 x dx =
dx = x − sin 2 x + C .
2
2
4
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.
1
⎛ x⎞
⎛x⎞ 9
⎛ x⎞
sin 3 ⎜ ⎟ dx = cos ⎜ ⎟ − cos ⎜ ⎟ + C
6
2
2
2
⎝ ⎠
⎝ ⎠
⎝6⎠
c.
( x 2 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
© 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.
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
x3
y + C1 =
+ x + C2
3
8.
y=
y=−
1
2
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:
4 = (1 + C )2 / 3 ; C = 7
y = ( x3 / 2 + 7)2 / 3
dy
= x −3 + 2
dx
y + C1 = −
Section 3.9
x dx
y 3 / 2 = x3 / 2 + C
x
1
+x−
3
3
dy = ( x −3 + 2) dx
x
y
2 3/ 2
2
y
+ C1 = x3 / 2 + C2
3
3
3
dy = ( x −3 + 2) dx
dy
=
dx
y dy =
x3
y=
+ x+C
3
At x = 1, y = 1:
1
1
1 = + 1 + C; C = −
3
3
6.
dy x
=
dx y
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
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.
10.
dy
= y4
dt
13.
y −4 dy = dt
−
1
3 y3
y=−
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
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 2 + 4t − 1) dt
s + C1 =
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
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
u −3 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
y
2
5
y=
du
= u 3 (t 3 − t )
dt
−
(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
© 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.
16.
dy
= 3y2
dx
19. v = (2t + 1)1/ 3 dt =
y −2 dy = 3 dx
−
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
(2t + 1) 4 / 3 dt − 1dt
s=
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
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
(3t + 1) −2 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
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.
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.
Instructor’s Resource Manual
Section 3.9
233
© 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.
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).
3. True:
4. False:
increasing for all x, but f ' ( x ) does
f ( x) = 18 x5 + 16 x3 + 4 x;
4
2
6. False:
When f ( x) > 0, f ( x) is increasing.
8. False:
If f (c) = 0 , c is a candidate, but not
necessarily an inflection point. For
=
16. True:
17. True:
18. False:
10. True:
f ( x) = ax 2 + bx + c;
f ( x) = 2ax + b; f ( x) = 2a
12. True:
23. False:
x →−∞
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
The function is differentiable on
(0, 2).
f ( x) =
lim (2 x3 + x + tan x) = ∞ while
+
Instructor's Resource Manual
3 3
,
.
3 3
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 .
−
At x = 3 there is a removable
discontinuity.
x
so f (0) does not exist.
x
dy
d2y
= cos x;
= − sin x; –sin x = 0
dx
dx 2
has infinitely many solutions.
P = 2x +
lim (2 x3 + x + tan x) = −∞.
14. False:
1
= –1.
–1
For example if f ( x) = x 4 ,
f (0) = f (0) = 0 but f has a
minimum at x = 0.
x →∞
x →− π
2
x
–1
21. False:
lim (2 x3 + x) = ∞ while
x→ π
2
x→ – ∞ 1
x2
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) = −∞
13. True:
= lim
1 + 12
20. True:
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
There are two points: x = −
22. True:
11. False:
x →∞ 1
x2
19. False:
example, if f ( x) = x 4 , P (0) = 0 but
x = 0 is not an inflection point.
9. True:
x2 + 1
For example, f ( x) = x is increasing
on [–1, 1] but f (0) = 0.
7. True:
= lim
1
x2
1
= –1 and
–1
lim
f ( x) = 90 x + 48 x + 4 , which is
greater than zero for all x.
3
1+
x→ – ∞ 1 – x 2
not exist at x = 0.
5. True:
x →∞ 1 – x 2
=
For example, let f(x) = sin x.
f ( x) = x1/ 3 is continuous and
lim
x2 + 1
24. True:
By the Mean Value Theorem, the
derivative must be zero between each
pair of distinct x-intercepts.
Section 3.10
235
© 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.
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:
30. True:
31. False:
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
However, if the constant is 0, the
functions are the same.
a ≤ g ( x ) ≤ b whenever a ≤ x ≤ b ,
then there is at least one fixed point
on [ a, b ] . The given conditions satisfy
x
For example, let f ( x) = e .
these criteria.
lim e x = 0, so y = 0 is a horizontal
x →−∞
32. True:
33. True:
f ( xn )
= –2 xn .
f ( xn )
asymptote.
40. True:
If f(c) is a global maximum then f(c)
is the maximum value of f on
S where (a, b) is any interval
(a, b)
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) = 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.
f ( x) dx, f ( x) is a
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
Section 3.10
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.
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
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,
3(u – 2)
f (2) does not exist.
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
Critical points: –1, 0,
1
Global minimum f (–2) = ;
4
⎛ 1⎞
global maximum f ⎜ – ⎟ = 4.
⎝ 2⎠
f ( x) = –
s
; f ( s ) does not exist when s = 0.
s
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.
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
Section 3.10
237
© 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.
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
11.
f ( ) = cos ; f ( ) = 0 when
15.
16.
1
x>− .
2
f(x) is increasing on [–2, 1] and concave down on
⎛ 1 ⎞
⎜− , ∞⎟ .
⎝ 2 ⎠
Critical points:
17.
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.
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 ( ) = 2sin cos – cos = cos (2sin – 1);
=
f ( x) = 4 x3 – 20 x 4 = 4 x3 (1 – 5 x); f ( x) > 0
when 0 < x <
3
⎛ 4π ⎞
f ⎜ ⎟=–
≈ –0.87
2
⎝ 3 ⎠
⎛ 4π ⎞
Global minimum f ⎜ ⎟ ≈ –0.87;
⎝ 3 ⎠
⎛π⎞
global maximum f ⎜ ⎟ = 1
⎝2⎠
f ( ) = 0 when
f ( x) = −6 x 2 − 6 x + 12 = –6(x + 2)(x – 1);
f ( x) = −12 x − 6 = –6(2x + 1); f ( x) < 0 when
⎡ π 4π ⎤
⎢4 , 3 ⎥
⎣
⎦
12.
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) > 0 when –2 < x < 1.
π
= in
2
π π 4π
, ,
4 2 3
⎛π⎞ 1
⎛π⎞
f ⎜ ⎟=
≈ 0.71, f ⎜ ⎟ = 1,
4
2
⎝ ⎠
⎝2⎠
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
Section 3.10
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.
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
2
( x + 1)2
; f ( x) = 0 when x = 0.
8(3 x 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 >
Instructor's Resource Manual
Section 3.10
239
© 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.
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(3x – 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.
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
f ( x) = 1 +
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
Section 3.10
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.
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
Section 3.10
241
© 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.
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.
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.
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
© 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.
⎧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
=
−2
( x − 1)
( x − 1) 2
g (3) − g (2) 2 − 3
=
= −1
3− 2
1
−2
= –1; c = 1 ± 2
(c − 1)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.
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.
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 – sec 2 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
Instructor's Resource Manual
Section 3.10
245
© 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.
( x − 3x + 3 x ) dx
( x − 3x + 3x ) dx
3
53.
=
58. Let u = cos x ; then du = − sin x dx or
−du = sin x dx .
2
3
2
1/ 2
1 4
2
x − x3 + 3 ⋅ x3/ 2 + C
4
3
1
= x 4 − x3 + 2 x3/ 2 + C
4
=
2 x 4 − 3x 2 + 1
54.
x2
=
= − u 4 du
1
= − u5 + C
5
1
= − cos5 x + C
5
dx
( 2 x2 − 3 + x−2 ) dx
2 3
x − 3x − x −1 + C
3
2 x3
1
=
− 3x − + C or
x
3
59. u = tan(3 x 2 + 6 x), du = (6 x + 6) sec 2 (3x 2 + 6 x)
2 x4 − 9 x2 − 3
+C
3x
y 3 − 9 y sin y + 26 y −1
dy
y
=
( y 2 − 9sin y + 26 ) dy
=
1 3
y + 9 cos y + 26 y + C
3
( x + 1) tan 2 ( 3x 2 + 6 x ) sec2 ( 3x 2 + 6 x ) dx
1 2
1
u du = u 3 + C
6
18
1
= tan 3 3 x 2 + 6 x + C
18
=
(
56. Let u = y 2 − 4 ; then du = 2 ydy or
1
du = ydy .
2
)
60. u = t 4 + 9, du = 4t 3 dt
1
du
t3
dt = 4
u
t4 + 9
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
=
(
(
)
1/ 3
dz =
61. Let u = t 5 + 5 ; then du = 5t 4 dt or
)
57. Let u = 2 z 2 − 3 ; then du = 4 zdz or
z 2z2 − 3
1
du = zdz .
4
1
u1/ 3 du
4
1 3
= ⋅ u4/3 + C
4 4
4/3
3
=
+C
2z2 − 3
16
(
Section 3.10
(
t 4 t5 + 5
1
u1/ 3 ⋅ du
4
=
246
u 4 ⋅ −du
=
=
55.
( cos x )4 sin x dx
cos 4 x sin x dx =
)
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
=
(
)
)
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.
62. Let u = x 2 + 4 ; then du = 2 x dx or
x
1
dx =
2
2
x +4
1
du = xdx .
2
66. Let u = y 3 − 3 y ; then
(
)
(
y2 −1
u
(y
1
u −1/ 2 du
2
1
= ⋅ 2u1/ 2 + C
2
=
3
− 3y
)
2
dy =
x2
1
dx =
3
3
x +9
1
du = x 2 dx .
3
=−
u
1
u −1/ 2 du
3
1
= ⋅ 2u1/ 2 + C
3
2 3
=
x +9 +C
3
( y + 1)
dy =
=
1
u2
du
( 2 y − 1)
3
dy =
=
du
u3
1
dy =
69.
x +1
dx
y = 2 x +1 + C
y = 2 x + 1 + 14
sin y dy = dx
70.
− cos y = x + C
x = –1 – cos y
dy =
71.
2t − 1 dt
1
y = (2t − 1)3 2 + C
3
1
y = (2t − 1)3 2 − 1
3
y −4 dy = t 2 dt
72.
−
−
1
3 y3
1
3 y3
y=3
Instructor's Resource Manual
+C
y = − cos x + C
y = –cos x + 3
u −3 du
1
= − u −2 + C
2
1
=−
+C
2
2 ( 2 y − 1)
3y − 9 y
dy = sin x dx
68.
65. Let u = 2 y − 1 ; then du = 2dy .
2
1
3
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
u2
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 .
2
du
1
u −2 du
3
1
= ⋅ −u −1 + C
3
1
1
=− ⋅ 3
+C
3 y − 3y
du
=
1
1
3
=
= 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 .
du
=
t3
+C
3
=
t3 2
−
3 3
1
2 − t3
Section 3.10
247
© 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.
7. Aregion = 0.5 (1 + 1.5 + 2 + 2.5) = 3.5
2 y dy = (6 x − x3 )dx
73.
1 4
x +C
4
1
y 2 = 3x 2 − x 4 + 9
4
1
y = 3x 2 − x 4 + 9
4
y 2 = 3x 2 −
8. Aregion = 0.5 (1.5 + 2 + 2.5 + 3) = 4.5
9. Aregion = Arect + Atri = 1x +
1
1
1
10. Aregion = bh = x ⋅ xt = x 2t
2
2
2
cos y dy = x dx
74.
1
1
x ⋅ x = x2 + x
2
2
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 36
2
4
⎝
⎠
5
= a 2 cot 36
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
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.
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
P −1/ 3 dP = C1dt ; P 2 / 3 = C1t + C2
2
3
P(0) = 1000: C2 = ⋅10002 / 3 = 150
2
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.
b.
for 0 ≤ t < 1
⎧−32t
v(t ) = ⎨
32(
1)
24
for
1 < t ≤ 2.5
−
t
−
+
⎩
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
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.
4
CHAPTER
4.1 Concepts Review
1. 2 ⋅
The Definite Integral
8
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
6
(k − 1) =
1.
k =1
6
k =1
k−
6
1
k =1
i =1
i2 =
(−1)5 25 (−1)6 26 (−1)7 27
+
+
6
7
8
1154
=−
105
+
6(7)(13)
= 91
6
6
7
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
7.
8
(l + 1)2 = 42 + 52 + 62 + 7 2 + 82 + 92 = 271
6
n cos(nπ) =
n =1
n =1
( −1)n ⋅ n
= –1 + 2 – 3 + 4 – 5 + 6
=3
+
4.
(−1)3 23 (−1) 4 24
+
4
5
=
6(7)
− 6(1)
2
= 15
6
(−1)k 2k
k =3 ( k + 1)
7
6.
=
2.
(−1)m 2m − 2
5.
6
8.
⎛ kπ ⎞
k sin ⎜ ⎟
⎝ 2 ⎠
k =−1
⎛ 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
l =3
9. 1 + 2 + 3 +
+ 41 =
41
i
i =1
10. 2 + 4 + 6 + 8 +
+ 50 =
25
2i
i =1
11. 1 +
1 1
+ +
2 3
+
1 100 1
=
100 i =1 i
Instructor’s Resource Manual
Section 4.1
249
© 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 1 1
12. 1 − + − +
2 3 4
1 100 (−1)i +1
=
100 i =1 i
−
13. a1 + a3 + a5 + a7 +
14.
+ a99 =
f ( w1 )Δx + f ( w2 )Δx +
n
=
10
20.
=
50
i =1
a2i −1
=4
+ f ( wn )Δx
i =1
10
=
i =1
i =1
bi
10
10
k =1
n =1
an + 2
17.
p =0
p =1
ap −
10
n =1
bn
n
10
q =1
bp
=
q =1
aq −
q =1
= 40 − 50 −
bq −
19.
25.
(3i − 2)
i =1
100
=3
i =1
i−
100
2
i =1
= 3(5050) − 2(100)
= 14,950
i+
i =1
4n3 − 3n 2 − n
6
(2i − 3) 2 =
n
i 2 − 12
n
n
1
i =1
=
i =1
10(11)
2
n
2n3 + 3n 2 + n 3n 2 + 3n
−
+n
3
2
10
q
i2 − 3
=
=4
q =1
n
i =1
i =1
= −65
100
k =1
2n(n + 1)(2n + 1) 3n(n + 1)
−
+n
6
2
n
10
k2
=
24.
(aq − bq − q)
10
k 3 + 20
10
i =1
= 40 − 50
= –10
18.
(5k 3 + 20k 2 )
(2i 2 − 3i + 1) = 2
23.
10
p =1
k =1
k =1
10
k =1
(a p +1 − b p +1 )
10
=
k2
= 5(3025) + 20(385)
= 22,825
= 3(40) + 2(50)
= 220
9
10
k3 −
10
5k 2 (k + 4) =
22.
(3an + 2bn )
=3
10
k =1
=5
n =1
3
i =1
= 3025 − 385
= 2640
= 90
16.
i =1
k =1
= 40 + 50
10
10
i−
(k 3 − k 2 ) =
21.
10
ai +
10
= 4(385) – 55 – 3(10)
= 1455
(ai + bi )
15.
i2 −
i =1
10
10
(4i 2 − i − 3)
i =1
10
f ( wi )Δx
i =1
[(i − 1)(4i + 3)]
i =1
10
(4i 2 − 12i + 9)
i =1
n
i+
i =1
n
9
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) +
2S = n(n + 1)
n(n + 1)
S=
2
+ (n + 1) + (n + 1) + (n + 1)
250
Section 4.1
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.
26. S − rS = a + ar + ar 2 +
+ ar n
27. a.
+ ar n + ar n +1 )
− (ar + ar 2 +
= a − ar n +1
(1)
1− 2
⎛1⎞
⎜ ⎟ =
1
k =0 ⎝ 2 ⎠
2
10
k
10
k
11
10
⎛1⎞
⎛1⎞
⎜ ⎟ = 1− ⎜ ⎟
2
⎝
⎠
⎝2⎠
k =1
n +1
a − ar
= S (1 − r ); S =
1− r
10
2k =
b.
k =0
10
10
⎛1⎞
= 2 − ⎜ ⎟ , so
⎝2⎠
=
1023
.
1024
1 − 211
= 211 − 1, so
−1
2k = 211 − 2 = 2046 .
k =1
28.
+ [ a + (n − 2)d ] + [ a + (n − 1)d ] + (a + nd )
S = a + ( a + d ) + ( a + 2d ) +
+ S = (a + nd ) + [ a + (n − 1)d ] + [ a + (n − 2)d ] +
2S = (2a + nd ) + (2a + nd ) + (2a + nd ) +
2S = (n + 1)(2a + nd)
(n + 1)(2a + nd )
S=
2
+ ( a + 2d ) + ( a + d ) + a
+ (2a + nd ) + (2a + nd ) + (2a + nd )
( i + 1)3 − i3 = 3i 2 + 3i + 1
29.
n
⎡( i + 1)3 − i 3 ⎤ =
⎣
⎦
i =1
n
i =1
( 3i
n
( n + 1)3 − 13 = 3
2
i2 + 3
i =1
n
n3 + 3n 2 + 3n = 3
i2 + 3
n
n
i+
i =1
i =1
2n3 + 6n 2 + 6n = 6
)
+ 3i + 1
n
1
i =1
n ( n + 1)
2
+n
i 2 + 3n 2 + 3n + 2n
i =1
3
2
2n + 3n + n
=
6
n ( n + 1)( 2n + 1)
6
=
n
i2
i =1
n
i2
i =1
Instructor’s Resource Manual
Section 4.1
251
© 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.
( i + 1)4 − i 4 = 4i3 + 6i 2 + 4i + 1
30.
n
i =1
⎡(i + 1)4 − i 4 ⎤ =
⎣
⎦
n
4i3 + 6i 2 + 4i + 1)
(
i =1
n
(n + 1)4 − 14 = 4
n 4 + 4 n3 + 6 n 2 + 4 n = 4
i =1
n
i3 + 6
i3 + 6
Solving for
n
i2 + 4
i =1
i =1
n
n
n
i+
i =1
1
i =1
n(n + 1)(2n + 1)
n(n + 1)
+4
+n
6
2
i3 gives
i =1
n
4
i =1
n
4
(
) (
)
i3 = n 4 + 4n3 + 6n 2 + 4n − 2n3 + 3n 2 + n − 2n 2 + 2n − n
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
⎡( i + 1)5 − i5 ⎤ =5
⎣⎢
⎦⎥
n
i 4 + 10
i =1
( n + 1)5 − 15 = 5
n5 + 5n 4 + 10n3 + 10n 2 + 5n = 5
n
i =1
n
i =1
n
Solving for
n
i =1
i 4 + 10
n
i3 + 10
i =1
2
n 2 ( n + 1)
4
i 4 + 52 n 2 ( n + 1)
2
n
i2 + 5
i+
i =1
+ 10
n
1
i =1
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
i 4 yields
i =1
n
i =1
n(n + 1)(2n + 1)(3n 2 + 3n − 1)
i 4 = 15 ⎡ n5 + 52 n 4 + 53 n3 − 16 n ⎤ =
⎣
⎦
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 +
also half the area of the grid or
n(n + 1)
. Thus, 1 + 2 +
2
33. x =
2
⎡ n(n + 1) ⎤
3
3
+ n3 or ⎢
⎥ . Thus, 1 + 2 +
⎣ 2 ⎦
n(n + 1)
.
2
n(n + 1)
+n=
. From the diagram the area is
2
+n =
Suppose we have a square grid with sides of length 1 + 2 +
13 + 23 +
+ n . The shaded area is
2
⎡ n(n + 1) ⎤
+ n3 = ⎢
⎥ .
⎣ 2 ⎦
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
Section 4.1
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.
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
⎣
⎦
n
35. a.
i =1
b.
( xi − x ) =
n
i =1
xi −
n
x = nx − nx = 0
i =1
1 n
1 n 2
( xi − x )2 =
( xi − 2 x xi + x 2 )
n i =1
n i =1
1 n 2 2x n
1 n 2
=
xi −
xi +
x
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 2⎞
⎛1 n 2⎞
xi ⎟ − 2 x 2 + x 2 = ⎜
xi ⎟ − 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
⎣
⎦
37. Let S (c) =
n
i =1
S '(c) =
=
=
( xi − c)2 . Then
d n
( xi − c )2
dc i =1
n
d
( xi − c )2
dc
i =1
n
i =1
= −2
2( xi − c)(−1)
n
i =1
xi + 2nc
S ''(c) = 2n
Set S '(c) = 0 and solve for c :
−2
c
n
xi + 2nc
i =1
n
= 1n xi = x
i =1
=0
Since S ''( x) = 2n > 0 we know that x minimizes S (c) .
Instructor’s Resource Manual
Section 4.1
253
© 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.
i
The number of gifts given on the nth day is
38. a.
m=
m =1
i (i + 1)
.
2
12
The total number of gifts is
i (i + 1)
= 364 .
i =1 2
n
b. For n days, the total number of gifts is
i (i + 1)
.
i =1 2
i (i + 1) n i 2 n i 1 n 2 1 n
1 ⎡ n(n + 1)(2n + 1) ⎤ 1 ⎡ n(n + 1) ⎤
=
+
=
i +
i= ⎢
⎥+ 2⎢ 2 ⎥
2
2
2
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
i ( n − m + i ) = ( n − m)
i =1
m
i =1
i+
m
i2
i =1
m(m + 1) m(m + 1)(2m + 1)
= ( n − m)
+
2
6
m(m + 1)(3n − m + 1)
=
6
42.
1
1
1
+
+
+
1⋅ 2 2 ⋅ 3 3 ⋅ 4
+
1
n(n + 1)
⎛ 1⎞ ⎛ 1 1⎞ ⎛1 1⎞
= ⎜1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ +
⎝ 2⎠ ⎝ 2 3⎠ ⎝3 4⎠
1
= 1−
n +1
1 ⎞
⎛1
+⎜ −
⎟
n
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
Section 4.1
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.
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
⎛1 3 2 ⎞ 1
1 ⎡⎛ 1 ⎛ 1 ⎞
1 9 3 17
31
⎢⎜ ⋅ ⎜ ⎟ + 1⎟ + ⎛⎜ ⋅12 + 1⎞⎟ + ⎜ ⋅ ⎜⎛ ⎟⎞ + 1⎟ + ⎜⎛ ⋅ 22 + 1⎟⎞ ⎥ = ⎛⎜ + + + 3 ⎞⎟ =
⎜
⎟
⎜
⎟
2⎢ 2 ⎝2⎠
2
2
2⎝8 2 8
⎠ ⎥⎦
⎠ ⎝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=
Instructor’s Resource Manual
Section 4.1
255
© 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.
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 ⎞
+
+
+
+
A( Sn ) = ⎢⎜
2 n⎟ ⎜ 2 n⎟
⎠ ⎝n
⎠
⎣⎝ n
53. Δx =
⎛ n 2 ⎞⎤
1
+⎜
+
=
(1 + 2 + 3 +
2 n ⎟⎥
⎝n
⎠ ⎦ n2
+ n) + 2 =
n(n + 1)
2n
2
+2 =
1 5
+
2n 2
⎛ 1 5⎞ 5
lim A( Sn ) = lim ⎜ + ⎟ =
n →∞
n→∞ ⎝ 2n 2 ⎠ 2
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
+ ⎟+⎜
+ ⎟+ +⎜
+ ⎟⎥ =
A( Sn ) = ⎢⎜
(1 + 22 + 32 + + n 2 ) + 1
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
Section 4.1
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.
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 ⎞ i3
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.
⎛8 4
4 ⎞ 8
lim A( Sn ) = lim ⎜ + +
⎟= .
n →∞
n→∞ ⎝ 3 n 3n 2 ⎠ 3
⎛ 8 ⎞ 16
By symmetry, A = 2 ⎜ ⎟ =
.
⎝3⎠ 3
60.
1
i
, xi =
n
n
3
=
n
4
(13 + 23 +
+ n3 ) =
⎤
( n3 ) ⎥
n
⎦
1
4
1 ⎡ n(n + 1) ⎤
⎢
⎥
n4 ⎣ 2 ⎦
(1 + 2 +
+ n)
2
i 2
⎡i
⎤1
+
f (ti )Δt = ⎢ + 2 ⎥ =
⎣n
⎦ n n2 n
n
⎛1 1 ⎞
=⎜ + ⎟+2
⎝ 2 2n ⎠
1
5
lim A( Sn ) = + 2 =
2
2
n →∞
1
The object traveled 2 ft.
2
4 ⎡ 2n3 + 3n 2 + n ⎤ 8 4
4
= ⎢
⎥= + + 2
3
n
3 ⎣⎢
3
n
3n
⎦⎥
1
n2
n
2
⎛ i 2⎞ 1 n
⎜ 2 + ⎟= 2 i+
n ⎠ n i =1 i =1 n
i =1 ⎝ n
1 ⎡ n(n + 1) ⎤
=
⎢
⎥+2
n2 ⎣ 2 ⎦
⎡ n2 + n ⎤
=⎢
⎥+2
2
⎣⎢ 2n ⎦⎥
⎛ 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(n + 1) ⎤
1 ⎡ n(n + 1) ⎤
⎥ + 2⎢ 2 ⎥
4 ⎢⎣
2
⎦
⎦
n
n ⎣
A( Sn ) =
2
+
+ n3 ) +
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
3
⎛ i ⎞ ⎛1⎞ i
f ( xi )Δx = ⎜ ⎟ ⎜ ⎟ =
⎝ n ⎠ ⎝ n ⎠ n4
⎡1
1 3
A( Sn ) = ⎢ (13 ) +
(2 ) +
4
n4
⎣n
(13 + 23 +
=
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 →∞
Instructor’s Resource Manual
Section 4.1
257
© 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.
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 n
i2 =
b3 ⎡ n(n + 1)(2n + 1) ⎤
⎢
⎥
6
⎦
n3 ⎣
n3 i =1
b3 ⎡
3 1 ⎤
=
2+ + ⎥
⎢
6 ⎣
n n2 ⎦
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
Section 4.1
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.
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
3. RP =
5
i =1
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)
= 2(0.75) + 3(0.5) + 3.75(1.25) + 5(0.5) + 5.5(1) = 15.6875
4. RP =
4
i =1
f ( xi )Δxi = f (−2)(−1.3 + 3) + f (−0.5)(0 + 1.3) + f (0)(0.9 − 0) + f (2)(2 − 0.9)
= 4(1.7) + 3.25(1.3) + 3(0.9) + 2(1.1) = 15.925
5. RP =
8
i =1
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)
= [–0.21875 – 0.46875 – 0.46875 – 0.21875 + 0.28125 + 1.03125 + 2.03125 + 3.28125](0.5) = 2.625
6. RP =
6
i =1
f ( xi )Δxi = [ f (0.5) + f (1) + f (1.5) + f (2) + f (2.5) + f (3) ] (0.5)
= [1.5 + 5 + 14.5 + 33 + 63.5 + 109](0.5) = 113.25
7.
8.
9.
10.
3
1
2
0
1
x3 dx
( x + 1)3 dx
x
π
dx
i =1
=
(sin x)2 dx
2
2i
, xi =
n
n
f ( xi ) = xi + 1 =
n
2
−1 1 + x
0
11. Δx =
f ( xi )Δx =
2i
+1
n
n
⎡
⎛ 2 ⎞⎤ 2
⎢1 + i ⎜ n ⎟ ⎥ n
⎝ ⎠⎦
i =1 ⎣
2 n
4 n
2
4 ⎡ n(n + 1) ⎤
i = (n) +
1+
⎢
⎥
n i =1 n 2 i =1
n
n2 ⎣ 2 ⎦
⎛ 1⎞
= 2 + 2 ⎜1 + ⎟
⎝ n⎠
2
0
⎡
⎛ 1 ⎞⎤
( x + 1)dx = lim ⎢ 2 + 2 ⎜1 + ⎟ ⎥ = 4
n →∞ ⎣
⎝ n ⎠⎦
Instructor’s Resource Manual
Section 4.2
259
© 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.
12. Δx =
2
2i
, xi =
n
n
14. Δx =
2
2
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
3i ⎞
6i
⎛
f ( xi ) = 2 ⎜ −2 + ⎟ + π = π − 4 +
n⎠
n
⎝
13. Δx =
i =1
=
f ( xi )Δx =
2
3i ⎞
36i 27i 2
⎛
f ( xi ) = 3 ⎜ −2 + ⎟ + 2 = 14 −
+
n⎠
n
⎝
n2
n
n ⎡
⎛ 36 ⎞ ⎛ 27 ⎞ ⎤ 3
f ( xi )Δx = ⎢14 − ⎜ ⎟ i + ⎜ ⎟ i 2 ⎥
⎝ n ⎠ ⎝ n2 ⎠ ⎦ n
i =1
i =1 ⎣
=
3 n
108 n
81 n 2
i+
i
14 −
n i =1
n 2 i =1 n3 i =1
= 42 −
⎡
4⎛
3 1 ⎞ ⎤ 14
( x 2 + 1) dx = lim ⎢ 2 + ⎜ 2 + +
⎟⎥ =
n n2 ⎠ ⎦ 3
3⎝
n →∞ ⎣
n
n
6i ⎤ 3
⎡
⎢π − 4 + n ⎥ n
⎦
i =1 ⎣
3 n
18 n
18 ⎡ n(n + 1) ⎤
i = 3(π − 4) +
(π − 4) +
⎢
⎥
2
n i =1
n i =1
n2 ⎣ 2 ⎦
⎛ 1⎞
= 3π − 12 + 9 ⎜ 1 + ⎟
⎝ n⎠
⎡
⎛ 1 ⎞⎤
(2 x + π) dx = lim ⎢3π − 12 + 9 ⎜ 1 + ⎟ ⎥
−2
n →∞ ⎣
⎝ n ⎠⎦
= 3π − 3
108 ⎡ n(n + 1) ⎤ 81 ⎡ n(n + 1)(2n + 1) ⎤
⎢
⎥+ ⎢
⎥
6
⎦
n 2 ⎣ 2 ⎦ n3 ⎣
3 1 ⎞
⎛ 1 ⎞ 27 ⎛
= 42 − 54 ⎜1 + ⎟ + ⎜ 2 + +
⎟
n n2 ⎠
⎝ n⎠ 2 ⎝
1
−2
(3 x 2 + 2) dx
⎡
3 1 ⎞⎤
⎛ 1 ⎞ 27 ⎛
= lim ⎢ 42 − 54 ⎜1 + ⎟ + ⎜ 2 + +
⎟ ⎥ = 15
n n2 ⎠ ⎦
n→∞ ⎣
⎝ n⎠ 2 ⎝
5
5i
, xi =
n
n
5i
f ( xi ) = 1 +
n
15. Δx =
n
i =1
1
=
f ( xi )Δx =
n
⎡
⎛ 5 ⎞⎤ 5
⎢1 + i ⎜ n ⎟ ⎥ n
⎝ ⎠⎦
i =1 ⎣
5 n
25 n
25 ⎡ n(n + 1) ⎤
i = 5+
1+
⎢
⎥
n i =1 n 2 i =1
n2 ⎣ 2 ⎦
= 5+
5
0
16. Δx =
3
3i
, xi = −2 +
n
n
25 ⎛ 1 ⎞
⎜1 + ⎟
2 ⎝ n⎠
⎡ 25 ⎛ 1 ⎞ ⎤ 35
( x + 1) dx = lim ⎢5 + ⎜1 + ⎟ ⎥ =
2 ⎝ n ⎠⎦ 2
n →∞ ⎣
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 ( xi )Δx = ⎢90 − i ⎜
i
=
−
i
+
i
90
+
⎜ 2 ⎟⎥
⎟
n i =1
⎝ n ⎠
n 2 i =1
n3 i =1
⎝ n ⎠⎦ n
i =1
i =1 ⎣
= 1800 −
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 ⎛
2+ +
( x 2 + x) dx = lim ⎢1800 − 3800 ⎜1 + ⎟ +
⎜
⎟⎥ =
−10
n n2 ⎠ ⎦
3 ⎝
3
n→∞ ⎣
⎝ n⎠
10
260
Section 4.2
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.
21. The area under the curve is equal to the area of a
17.
semi-circle:
5
0
1
1
27
(1)(2) + 1(2) + 3(2) + (3)(3) =
2
2
2
=
18.
f ( x) dx
A
−A
A2 − x 2 dx = 12 π A2 .
22. The area under the curve is equal to the area of a
triangle:
y
4
2
4
−4
2
2
0
f ( x) dx =
4 x
1
1
9
(1)( 3) + (1)( 2 ) + (1)( 2 ) =
2
2
2
⎛1⎞
f ( x ) dx = 2 ⎜ ⎟ 4 ⋅ 4 = 16
⎝2⎠
23. s ( 4 ) =
24. s ( 4 ) =
19.
25. s ( 4 ) =
26. s ( 4 ) =
27.
2
0
f ( x) dx =
1
1
1 π
(π ⋅12 ) + (1)(1) = +
4
2
2 4
20.
4
0
4
0
4
0
4
0
v ( t ) dt =
1 ⎛ 4 ⎞ 2
4⎜ ⎟ =
2 ⎝ 60 ⎠ 15
1
v ( t ) dt = 4 + 4 ( 9 − 1) = 20
2
v ( t ) dt =
1
2 (1) + 2 (1) = 3
2
1
2
v ( t ) dt = π ( 2 ) + 0 = π
4
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
Instructor’s Resource Manual
Section 4.2
261
© 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.
t
28.
s(t)
20
10
40
40
60
90
80
160
100
250
120
360
3
e.
3
f.
g.
2
30.
31. a.
b.
t
s(t)
20
20
40
80
60
160
80
240
100
320
120
400
s(t)
20
20
−1
2
d.
0
120
-100
0
x dx +
2
1
x dx
x 2 x dx = −
0
−1
x 2 dx + 0
1 2
0
x dx
2 2
1
x dx
13 ⎛ 23 13 ⎞
+⎜ − ⎟ = 2
3 ⎜⎝ 3 3 ⎟⎠
f ( x) dx = 0 because this is an odd
−1
1
−1
1
−1
g ( x ) dx = 3 + 3 = 6
f ( x) dx = 3 + 3 = 6
[ − g ( x)] dx = −3 + (−3) = −6
xg ( x) dx = 0 because xg(x) is an odd
function.
1
f.
−1
f 3 ( x) g ( x) dx = 0 because f 3 ( x) g ( x)
is an odd function.
33. RP =
x dx = (−3 − 2 − 1 + 0 + 1 + 2)(1) = −3
2
(x −
(x −
−3
3
1
e.
100
3
1
−1
d.
60
−3
−1
c.
⎡1
⎤
x ) dx = 6 ⎢ (1)(1) ⎥ = 3
⎣2
⎦
x
)
2
1
dx = 6 x 2 dx = 6 ⋅
0
=
1 n
( xi + xi −1 )( xi − xi −1 )
2 i =1
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
+(−1) 2 + 0 + 1 + 4](1) = 19
c.
1
b.
80
−3
1
function.
80
x
−1
32. a.
60
3
−1
x dx + 0
+
60
−3
0
1
1
= − (1)(1) + 1(1) + (1)(1) = 1
2
2
40
3
(−3)3 (3)3
+
=0
3
3
x x dx = −
=−
t
1
1
(3)(3) + (3)(3) = 9
2
2
x x dx =
−3
h.
29.
x dx =
−3
13
=2
3
262
Section 4.2
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.
(
)
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
⎦
Rp =
n
= xi .
xi2 Δxi
i =1
n
1 2
=
( xi + xi −1 xi + xi2−1 )( xi − xi −1 )
3
i =1
=
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
( x3 + 1) dx = 5.24
2
Right:
0
Midpoint:
36. Left:
Right:
1
0
( x3 + 1) dx = 6.84
2
0
1
37. Left:
0
cos x dx ≈ 0.8638
1
Right:
cos x dx ≈ 0.8178
0
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
f ( xi )Δxi = lim
If xi =
i 1
+ , f ( xi ) = 0 .
n πn
P →0 i =1
n
lim
P →0 i =1
n
1
=1
n →∞ i =1 n
lim
f ( xi )Δxi = lim
n
n →∞ i =1
0=0
Thus f is not integrable on [0, 1].
( x3 + 1) dx = 5.98
tan x dx ≈ 0.5398
1
0
Midpoint:
tan x dx ≈ 0.6955
1
0
tan x dx ≈ 0.6146
Instructor’s Resource Manual
Section 4.2
263
© 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.
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
Section 4.3
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.
2
9.
10.
2 f ( x) dx = 2
1
2
0
2 f ( x) dx = 2
= 2 ⎡⎢
⎣
11.
2
0
1
0
f ( x) dx +
2
f ( x ) dx = 2(3) = 6
1
2
0
f ( x) dx
2
1
f ( x)dx ⎤⎥ = 2(2 + 3) = 10
⎦
2
[ 2 f ( x) + g ( x)] dx = 2
f ( x) dx +
0
1
2
= 2 ⎡⎢ f ( x) dx +
f ( x) dx ⎤⎥ +
0
1
⎣
⎦
= 2(2 + 3) + 4 = 14
12.
1
1
0
0
[2 f ( s ) + g ( s )] ds = 2
2
0
f ( s ) ds +
2
0
g ( x) dx
g ( x) dx
1
0
1
2
[2 f ( s ) + 5 g ( s )] ds = −2
2
= −2(3) − 5 ⎡⎢ g ( s ) ds −
⎣ 0
= –6 – 5[4 + 1] = –31
14.
15.
1
1
2
0
2
1
1
0
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
⎝
⎠
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)
f ( s ) ds − 5
2
1
g ( s ) ds
⎡
23. G ( x ) = Dx ⎢
⎢⎣
⎡
24. G ( x ) = Dx ⎢
⎣⎢
g ( s ) ds ⎤⎥
⎦
2
0
[3 f ( x) + 2 g ( x)] dx = 0
25.
=
dt
x
1
=
2t dt ⎤⎥ = 2 x
⎦
1
18. G ( x) = Dx ⎡⎢ 2t dt ⎤⎥ = Dx ⎡⎢ −
x
⎣
⎦
⎣
x
1
2t dt ⎤⎥ = −2 x
⎦
( 2t 2 + t ) dt ⎤⎥⎦ = 2x2 +
19. G ( x) = Dx ⎡⎢
⎣
0
20. G ( x) = Dx ⎡⎢
⎣
cos3 (2t ) tan(t ) dt ⎤⎥
1
⎦
x
x
= cos3 (2 x) tan( x)
t2
1+ t2
t2
0
− x2 1 + t 2
dt
dt +
t2
− x2
1+ t2
0
x
t2
0
1+ t2
dt +
dt
x
t2
0
1+ t2
dt
2
= 3 (2 + 3) + 2(4) + 2π = 5 3 + 4 2 + 2π
17. G ( x) = Dx ⎡⎢
⎣
1
− x2 )
(
x2
G '( x) = −
−2 x ) +
(
2
1 + x2
1 + ( − x2 )
1
2
2
= 3 ⎡⎢ f (t ) dt +
f (t ) dt ⎤⎥ + 2
g (t ) dt
0
1
0
⎣
⎦
2
− x2
=−
⎡ 3 f (t ) + 2 g (t ) + π ⎤ dt
⎣
⎦
0
⎤
2 z + sin z dz ⎥
⎦⎥
x2 + x
x
G ( x) =
[3 f (t ) + 2 g (t )] dt
+π
1
= (2 x + 1) 2( x 2 + x) + sin( x 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
16.
⎤
sin t dt ⎥ = 2 x sin( x 2 )
⎥⎦
x2
x
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, ∞ ).
Instructor’s Resource Manual
Section 4.3
265
© 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.
28.
35.
1+ x
f
( x) =
f
(1 + x ) − (1 + x ) 2 x = − x + 2 x − 1
( x) =
( x + 1)
( x + 1)
1+ x
2
2
2
2
2
2
2
So, f(x) is increasing on [0, ∞ ) and concave up on
( 0, −1 + 2 ) .
29.
( x ) = cos x;
f
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.
( x ) = x + sin x;
f
f
f ( x) dx =
0
2
0
(2 − x)dx +
4
2
( x − 2) dx
= 2+2 = 4
36.
( 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.
( x) =
f
32. f(x) is increasing on x ≥ 0 and concave up on
( 0,1) , ( 2,3) ,...
4
0
=
=
37. a.
33.
( 3 + x − 3 ) dx
3
4
( 3 + x − 3 ) dx + 3 ( 3 + x − 3 ) dx
0
3
0
( 6 − x ) dx +
4
3
x dx =
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
f ( x) dx =
2
0
4
2 dx +
2
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
f ( x) dx =
1
0
dx +
2
1
x dx +
4
2
(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
266
Section 4.3
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.
d.
39.
a.
b.
f.
0
F (0) =
0
( t 4 + 1) dt = 0
41.
1 ≤ 1 + x4 ≤ 1 + x4 .
y = F ( x)
1
dy
= F '( x) = x 4 + 1
dx
0
(
1
dx ≤
dy = x + 1 dx
y = 15 x5 + x + C
c.
42.
0
0 = 15 05 + 0 + C
1
40.
a.
G ( x) =
G (0) =
0
0
0
G (2π ) =
b.
1
0
( x4 + 1) dx = F (1) = 15 15 + 1 = 65 .
x
1
0
1
0
4 + x 2 dx ≤
1
sin t dt
43.
sin t dt = 0
0
1 + x 4 dx ≤
6
5
1
0
( 4 + x2 ) dx
4 + x 2 dx ≤
( 4 + x4 ) dx = 01( 3 + 1 + x4 ) dx
1
1
= 3 dx + (1 + x 4 ) dx
0
0
= 3 + 65 =
2π
(1 + x 4 ) dx
21
0
5
Here, we have used the result from problem 39:
Thus y = F ( x) = 15 x5 + x
0
2 dx ≤
2≤
C=0
d.
0
On the interval [0,1], 2 ≤ 4 + x 4 ≤ 4 + x 4 .
Thus
1
Now apply the initial condition y (0) = 0 :
1
1 + x 4 dx ≤
0
By problem 39d, 1 ≤
)
4
t ≤ t . Since 1 + x 4 ≥ 1 for all x,
For t ≥ 1 ,
5 ≤ f ( x) ≤ 69 so
4⋅5 ≤
sin t dt = 0
21
5
20 ≤
( 5 + x3 ) dx ≤ 4 ⋅ 69
5 + x3 ) dx ≤ 276
0(
4
0
4
Let y = G ( x) . Then
dy
= G '( x) = sin x .
dx
dy = sin x dx
y = − cos x + C
c.
d.
Apply the initial condition
0 = y (0) = − cos 0 + C . Thus, C = 1 ,
and hence y = G ( x) = 1 − cos x .
π
0
e.
sin x dx = G (π ) = 1 − cos π = 2
44.
5
On [2,4], 85 ≤ ( x + 6 ) ≤ 105 . Thus,
2 ⋅ 85 ≤
4
2
65,536 ≤
( x + 6 )5 dx ≤ 2 ⋅105
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
Instructor’s Resource Manual
Section 4.3
267
© 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.
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⎠
12.7628 ≤
20 ⎛
49.
lim
On [ 4π ,8π ]
1 sin 2 x ≤ 5 + 1
5 ≤ 5 + 20
20
( 4π ) (5) ≤
20π ≤
(5 + 201 sin 2 x ) dx ≤ ( 4π ) (5 + 201 )
8π
1 sin 2 x dx ≤ 101 π
5 + 20
) 5
4π (
8π
4π
51.
lim
x 1+ t
0
x 1+ t
1
x →0 x
5
1⎞
1 + ⎟ dx ≤ 16.1051
⎜
10 ⎝
x⎠
50.
47.
Let F ( x) =
0
1
2+t
2+t
dt . Then
F ( x) − F (0)
x−0
x →0
1+ 0 1
= F '(0) =
=
2+0 2
dt = lim
x 1+ t
dt
2+t
1 ⎡ x 1+ t
= lim
dt −
⎢
x →1 x − 1 ⎣ 0 2 + t
F ( x) − F (1)
= lim
x −1
1
x→
1+1 2
= F '(1) =
=
2 +1 3
x →1 x − 1 1
x
1
1 1+ t
0 2+t
⎤
dt ⎥
⎦
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
Section 4.3
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.
52.
x
0
59.
f (t ) dt = x 2
=
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.
61.
1 x3
3
Differentiate both sides with respect to x:
d x2
d 1 3
f (t ) dt =
x
0
dx
dx 3
( )
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.
not identically zero.
⎧ t
⎪ 5 du,
⎪ 0
⎪⎪ 100
5 du +
a. s ( t ) = ⎨
0
⎪
⎪ 100
5 du +
⎪
⎩⎪ 0
2
0
f ( x)dx −
b
a g ( x ) dx
( x) − g ( x )]dx
⎧⎪2 + ( t − 2 ) , t ≤ 2
v (t ) = ⎨
⎪⎩ 2 − ( t − 2 ) , t > 2
t≤2
⎧ t,
=⎨
−
>2
4
,
t
t
⎩
t
0
v ( u ) du
⎧ 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 ⎥⎦
⎩ ⎣
⎧
t2
,
⎪
⎪
2
=⎨
t2
⎪
⎪⎩−4 + 4t − 2
0≤t ≤2
t>2
t2
− 4t + 4 = 0; t = 4 + 2 2 ≈ 6.83
2
False; A counterexample is f ( x ) = 0 for all x,
except f (1) = 1 . Thus,
b
a
False. a = 0, b = 1, f(x) = 0, g(x) = –1 is a
counterexample.
x
2
54.
58.
b
a[ f
s (t ) =
( ) ( 2x) = x
x
f ( x2 ) =
2
f x
2
True.
f ( x ) dx = 0 , but f is
0 ≤ t ≤ 100
u ⎞
⎛
6−
du
⎜
100 ⎝
100 ⎟⎠
700 ⎛
u ⎞
6−
⎟ du +
100 ⎜
⎝ 100 ⎠
t
100 < t ≤ 700
t
700
( −1) du,
t > 700
⎧
⎪
⎪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
⎩
Instructor’s Resource Manual
Section 4.3
269
© 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.
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 .
3.
c. s ( t ) = 0 = 2400 − t ; t = 2400
− f ( x ) ≤ f ( x) ≤ f ( x) , so
63.
b
a
b
− f ( x) dx ≤
b
a
a
f ( x ) dx ≥ −
b
a
4.
f ( x) dx ⇒
f ( x) dx
and combining this with
b
a
f ( x ) dx ≥
b
a
b
f ( x) dx ≤
If x > a ,
64.
x
a
b
x
(3 x 2 − 2 x + 3) dx = ⎡ x3 − x 2 + 3 x ⎤
⎣
⎦ −1
−1
= (8 – 4 + 6) – (–1 –1 – 3) = 15
2
2
(4 x3 + 7) dx = ⎡ x 4 + 7 x ⎤
⎣
⎦1
1
= (16 + 14) – (1 + 7) = 22
6.
⎡ 1⎤
8
⎛ 1⎞
dt = ⎢ − ⎥ = ⎜ − ⎟ − (−1) =
2
1 t3
9
⎝ 9⎠
⎣ t ⎦1
1
4
f ( x ) dx ≤ M ( x − a) by the
7.
x
a
f ( x) dx ≥ − M ( x − a ) by
8.
x
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.
3
2
3
4
16
⎡2
⎤
⎛2 ⎞
t dt = ⎢ t 3 / 2 ⎥ = ⎜ ⋅ 8 ⎟ − 0 =
3
⎣3
⎦0 ⎝ 3 ⎠
4
0
8 3
1
the Boundedness Property. Thus
a
4
3
⎡ 1⎤
⎛ 1⎞
dw = ⎢ − ⎥ = ⎜ − ⎟ − (−1) =
1 w2
4
⎝ 4⎠
⎣ w ⎦1
f ( x) dx
a
f ( x) dx = −
2
2
5.
Boundedness Property. If x < a ,
a
2
⎡ x5 ⎤
32 1 33
+ =
x 4 dx = ⎢ ⎥ =
−1
5
⎣⎢ ⎦⎥ −1 5 5 5
2
f ( x) dx,
we can conclude that
a
2.
8
⎡3
⎤
⎛3
⎞ ⎛ 3 ⎞ 45
w dw = ⎢ w4 / 3 ⎥ = ⎜ ⋅16 ⎟ − ⎜ ⋅1⎟ =
⎝4
⎠ ⎝4 ⎠ 4
⎣4
⎦1
−2
⎡ y3
⎛ 2 1 ⎞
1 ⎤
⎜⎜ y + 3 ⎟⎟ dy = ⎢ − 2 ⎥
−4
y ⎠
⎝
⎣⎢ 3 2 y ⎦⎥ −4
⎛ 8 1 ⎞ ⎛ 64 1 ⎞ 1783
= ⎜− − ⎟−⎜− − ⎟ =
96
⎝ 3 8 ⎠ ⎝ 3 32 ⎠
−2
4
s4 − 8
1
s2
ds =
4
1
4
2
( s − 8s
−2
⎡ s3 8 ⎤
) ds = ⎢ + ⎥
⎢⎣ 3 s ⎥⎦1
⎛ 64
⎞ ⎛1
⎞
= ⎜ + 2 ⎟ − ⎜ + 8 ⎟ = 15
⎝ 3
⎠ ⎝3 ⎠
4.4 Concepts Review
11.
1. antiderivative; F(b) – F(a)
12.
2. F(b) – F(a)
3. F (d ) − F (c )
4.
2
1
13.
1 4
u du
3
Problem Set 4.4
1.
14.
2
⎡ x4 ⎤
x dx = ⎢ ⎥ = 4 − 0 = 4
0
⎣⎢ 4 ⎦⎥ 0
2
3
π/2
0
π/2
π/6
π/2
cos x dx = [sin x ]0
=1–0=1
π/2
2sin t dt = [ −2 cos t ]π / 6 = 0 + 3 = 3
1
⎡2
⎤
(2 x 4 − 3x 2 + 5) dx = ⎢ x5 − x3 + 5 x ⎥
0
⎣5
⎦0
22
⎛2
⎞
= ⎜ −1+ 5⎟ − 0 =
5
5
⎝
⎠
1
1
3
⎡3
⎤
( x 4 / 3 − 2 x1/ 3 ) dx = ⎢ x 7 / 3 − x 4 / 3 ⎥
0
2
⎣7
⎦0
15
⎛3 3⎞
= ⎜ − ⎟−0 = −
7
2
14
⎝
⎠
1
270
Section 4.4
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. u = 3x + 2, du = 3 dx
1
2
2
u ⋅ du = u 3 / 2 + C = (3x + 2)3 / 2 + C
3
9
9
16. u = 2x – 4, du = 2 dx
1
3
3
u1/ 3 ⋅ du = u 4 / 3 + C = (2 x − 4)4 / 3 + C
2
8
8
25. u = x 2 + 4, du = 2 x dx
1
1
sin(u ) ⋅ du = − cos u + C
2
2
1
= − cos( x 2 + 4) + C
2
26. u = x3 + 5, du = 3 x 2 dx
17. u = 3x + 2, du = 3 dx
1
1
1
cos(u ) ⋅ du = sin u + C = sin(3 x + 2) + C
3
3
3
18. u = 2x – 4, du = 2 dx
1
1
sin u ⋅ du = − cos u + C
2
2
1
= − cos(2 x − 4) + C
2
19. u = 6x – 7, du = 6dx
1
1
sin u ⋅ du = − cos u + C
6
6
1
= − cos(6 x − 7) + C
6
20. u = πv − 7, du = π dv
1
1
1
cos u ⋅ du = sin u + C = sin(πv − 7) + C
π
π
π
1
1
1
cos u ⋅ du = sin u + C = sin( x3 + 5) + C
3
3
3
27. u = x 2 + 4, du =
1
1
1
u ⋅ du = u 3 / 2 + C = ( x 2 + 4)3 / 2 + C
2
3
3
22. u = x3 + 5, du = 3 x 2 dx
1
1
1
u ⋅ du = u10 + C = ( x3 + 5)10 + C
3
30
30
9
23. u = x + 3, du = 2 x dx
1
7
u −12 / 7 ⋅ du = − u −5 / 7 + C
2
10
7 2
= − ( x + 3)−5 / 7 + C
10
2
24. u = 3 v + π, du = 2 3v dv
u7 / 8 ⋅
=
4
15
2 3
(
3
du =
4
15 3
3 v2 + π
)
dx
3
28. u = z 2 + 3, du =
2z
dz
2
3 2
⎛
⎞
3⎜ z + 3 ⎟
⎝
⎠
3
3
3
3
cos u ⋅ du = sin u + C = sin z 2 + 3 + C
2
2
2
29. u = ( x3 + 5)9 ,
du = 9( x3 + 5)8 (3x 2 )dx = 27 x 2 ( x3 + 5)8 dx
cos u ⋅
1
1
du =
sin u + C
27
27
1
sin ⎡( x3 + 5)9 ⎤ + C
⎣
⎦
27
30. u = (7 x 7 + π)9 , du = 441x 6 (7 x 7 + π)8 dx
sin u ⋅
=−
1
1
du = −
cos u + C
441
441
1
cos(7 x 7 + π)9 + C
441
31. u = sin( x 2 + 4), du = 2 x cos( x 2 + 4) dx
2
1
x +4
sin u du = − cos u + C = − cos x 2 + 4 + C
=
21. u = x 2 + 4, du = 2 x dx
x
2
15 / 8
u15 / 8 + C
+C
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)
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
Instructor’s Resource Manual
Section 4.4
271
© 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.
33.
u = cos( x3 + 5), du = −3 x 2 sin( x3 + 5) dx
41.
1
⎛ 1⎞
u 9 ⋅ ⎜ − ⎟ du = − u10 + C
3
30
⎝
⎠
1
= − cos10 ( x3 + 5) + C
30
34.
3
42.
u = x 2 + 1, du = 2 x dx
0
( x 2 + 1)10 (2 x)dx =
2 10
u du
1
25
2
⎡ u11 ⎤
=⎢
⎥
⎣⎢ 11 ⎦⎥1
3
x2 + 1
1
x3 + 3 x
1
3
16 −1/ 2
u
du
4
43.
−1
0
1
0
1
u = t + 2, du = dt
3
−1
1
2
5 −2
u du
1
dt =
(t + 2)
4
⎡ 1⎤
= ⎢ − ⎥ − [ −1] =
5
⎣ 5⎦
38.
44.
5
⎡ 1⎤
= ⎢− ⎥
⎣ u ⎦1
39.
9
π/2
=
dx
36
cos 2 x sin x dx = −
π /2
0
cos 2 x ( − sin x ) dx
0
sin 2 3 x cos 3x dx
1 π /2 2
1
sin 3 x ( 3cos 3 x ) dx =
3 0
3
−1 2
u du
0
−1
⎡ u3 ⎤
1
⎛ 1⎞
= ⎢ ⎥ = ⎜− ⎟−0 = −
9
⎝ 9⎠
⎢⎣ 9 ⎥⎦ 0
9
u = 3x + 1, du = 3 dx
8
1 8
1
3x + 1 dx =
3 x + 1 ⋅ 3dx =
5
5
3
3
45.
u = x 2 + 2 x, du = (2 x + 2) dx = 2( x + 1) dx
1
( x + 1)( x 2 + 2 x)2 dx
0
11
=
25
16
u du
=
25
⎡2
⎤
⎡2
⎤ ⎡2
⎤ 122
= ⎢ u 3 / 2 ⎥ = ⎢ (125) ⎥ − ⎢ (64) ⎥ =
9
9
9
9
⎣
⎦16 ⎣
⎦ ⎣
⎦
40.
x3 + 3 x
u = sin 3 x, du = 3cos 3 x dx
0
⎡2
⎤
udu = ⎢ u 3 / 2 ⎥
2
1
3
⎣
⎦1
2
2
52
⎡
⎤ ⎡
⎤
= ⎢ (27) ⎥ − ⎢ (1) ⎥ =
⎣3
⎦ ⎣3 ⎦ 3
y − 1 dy =
1
⎡2
⎤
= ⎢ u1/ 2 ⎥
⎣3
⎦4
0 2
u du
1
u = y – 1, du = dy
10
3x2 + 3
3
⎡ u3 ⎤
=−
= ⎢− ⎥
⎢⎣ 3 ⎥⎦1
⎛ 1⎞ 1
= 0−⎜− ⎟ =
⎝ 3⎠ 3
⎡2
⎤
udu = ⎢ u 3 / 2 ⎥
3
⎣
⎦0
⎛2
⎞ ⎛2 ⎞ 2
= ⎜ ⋅13 / 2 ⎟ − ⎜ ⋅ 0 ⎟ =
⎝3
⎠ ⎝3 ⎠ 3
37.
1
3
u = cos x, du = − sin x dx
π/2
x3 + 1 (3x 2 ) dx =
dx =
⎛2 ⎞ ⎛2 ⎞ 8
= ⎜ ⋅6⎟ − ⎜ ⋅ 2⎟ =
⎝3 ⎠ ⎝3 ⎠ 3
u = x3 + 1, du = 3 x 2 dx
0
⎡4
⎤
u du = ⎢ u 3 / 2 ⎥
3
⎣
⎦ 25
u = x3 + 3x, du = (3 x 2 + 3) dx
=
2047
⎡1
⎤ ⎡1
⎤
= ⎢ (2)11 ⎥ − ⎢ (1)11 ⎥ =
11
11
11
⎣
⎦ ⎣
⎦
36.
7 + 2t 2 ⋅ ( 4t ) dt
−3
25
⎡4
⎤ ⎡4
⎤
= ⎢ (125) ⎥ − ⎢ (125) ⎥ = 0
⎣3
⎦ ⎣3
⎦
u = tan( x −3 + 1) , du = −3 x −4 sec2 ( x −3 + 1) dx
1
25
=2
5
⎛ 1⎞
u ⋅ ⎜ − ⎟ du = − u 6 / 5 + C
3
18
⎝
⎠
6/5
5 ⎡
=−
tan( x −3 + 1) ⎤
+C
⎣
⎦
18
3
7 + 2t 2 (8t ) dt = 2
−3
5
35.
u = 7 + 2t 2 , du = 4t dt
u = 2x + 2, du = 2 dx
7
1
1 7
2
dx =
dx
1 2x + 2
2 1 2x + 2
16
1 16 −1/ 2
u
du = ⎡⎣ u ⎤⎦ = 4 − 2 = 2
=
4
2 4
46.
02
1
2
( x 2 + 2 x)2 2( x + 1) dx
3 2
u
0
3
⎡ u3 ⎤
9
du = ⎢ ⎥ =
6
⎣⎢ ⎦⎥ 0 2
u = x − 1, du =
4
1
=2
( x − 1)3
x
1 3
u du
0
1
2 x
dx = 2
4
1
dx
( x − 1)3
2 x
dx
1
⎡u4 ⎤
1
= 2⎢ ⎥ =
⎢⎣ 4 ⎥⎦ 0 2
272
Section 4.4
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.
47.
u = sin , du = cos d
1/ 2 3
u du
0
48.
50.
51.
⎡u
=⎢ ⎥
⎣⎢ 4 ⎦⎥ 0
=
56.
3/2
1
1 π
1
π
cos u du = [ sin u ]−π = 0
π −π
π
57.
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
=
58.
cos1 3
u du
1
1 − cos 1
8
3
2sin 3 ⎛⎜ π8 ⎞⎟
⎝ ⎠
=
9
59.
a. Between 0 and 3, f ( x) > 0 . Thus,
3
3
0
c.
3
0
−
1
f ''( x) dx = f '(3) − f '(0)
= −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,
4
0
f ( x) dx > 0 .
b. Since f is an antiderivative of f ' ,
4
u = 3 x, du = 3dx; v = 5 x, dv = 5dx
0
f '( x) dx = f (3) − f (0)
= 0 − 2 = −2 < 0
2 π5
u = cos x, du = − sin x dx
f ( x) dx > 0 .
b. Since f is an antiderivative of f ' ,
0
55.
1
cos 4 1 1
+
8
8
3
u = 2 x5 , du = 10 x 4 dx
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
=−
1 sin( π3 / 8) 2
1 ⎡ 3 ⎤ sin( π / 8)
3 / 8) u du = ⎣u ⎦
sin(
−
π
− sin( π3 / 8)
3
9
1
2 π5
cos u du = [sin u ]0
0
10
1
1
= (sin(2π5 ) − 0) = sin(2π5 )
10
10
54.
cos1
4
0
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 ⎡u4 ⎤
=− ⎢ ⎥
2 ⎣⎢ 4 ⎦⎥
u = sin( x3 ), du = 3x 2 cos( x3 )dx
u = πx 2 , du = 2πx dx
1
10
53.
u = cos( x 2 ), du = −2 x sin( x 2 )dx
1
−
2
3/2
1
1⎛4 ⎞ 1
u −3 du = ⎡u −2 ⎤
= ⎜ − 1⎟ =
⎣
⎦
1
2
2⎝3 ⎠ 6
1 π
1
1
π
sin u du = − [ cos u ]0 = − (−1 − 1)
2π 0
2π
2π
1
=
π
52.
u = π sin , du = π cos d
1
1
−0 =
64
64
u = cos , du = − sin d
−
49.
4 ⎤1/ 2
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
Instructor’s Resource Manual
Section 4.4
273
© 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.
61.
1 2
t +C
2
V ( 0 ) = C = 0 since no water has leaked out at
V (t ) =
V (t ) =
( 20 − t ) dt = 20t −
67.
a.
b n
bn
x dx + n n
a
a
b.
⎡
t2 ⎤
219
V ( t ) dt = ⎢t −
⎥ =
0
220
⎣⎢
⎦⎥ 0 220
V (10 ) − V ( 9 ) =
t ⎞
T
⎜1 − 110 ⎟ dt = T − 220
⎝
⎠
=
T ≈ 110 hrs
63.
Use a midpoint Riemann sum with n = 12
partitions.
V=
12
=
≈ 1(5.4 + 6.3 + 6.4 + 6.5 + 6.9 + 7.5 + 8.4
+ 8.4 + 8.0 + 7.5 + 7.0 + 6.5)
= 84.8
64.
10
i =1
65.
68.
12
i =0
P ( ti ) Δti
mass =
2
0
(
bn
⎡ n ( n +1) / n ⎤
y dy = ⎢
y
⎥ n
⎣ n +1
⎦a
)
x
Let y = G ( x) =
a
f (t ) dt . Then
G ( x ) = F ( x) − F (a) . Now choose x = b to
obtain
b
a
δ ( x) = m ( x) = 1+
bn
n
an
G ( a ) = 0 . Thus, C = − F (a) and
≈ 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.
1 ⎡ n +1 ⎤ b
x
⎦a
n +1 ⎣
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
Use a midpoint Riemann sum with n = 12
partitions.
E=
x n dx =
n
b n +1 − a n +1
n +1
n
(b n +1 − a n +1 ) = An
nBn =
n +1
=
f ( xi ) Δxi
⎛ 6200 + 6300 + 6500 + 6500 + 6600 ⎞
≈ 1⎜
⎟
⎝ + 6700 + 6800 + 7000 + 7200 + 7200 ⎠
= 67, 000
b
a
1
(b n +1 − a n +1 )
n +1
An =
Use a midpoint Riemann sum with n = 10
partitions.
V=
n
(n + 1)b n +1 − (n + 1)a n +1
= b n +1 − a n +1
n +1
c. Bn =
f ( xi ) Δxi
i =1
y dy
⎛ b n +1 a n +1 ⎞ ⎛ n n +1
n n +1 ⎞
b −
a ⎟
=⎜
−
⎟+
⎜ n + 1 n + 1 ⎟ ⎜⎝ n + 1
n +1
⎠
⎝
⎠
2
T⎛
0
a
bn
n
an
x n dx +
b
t ⎞
201
⎜1 − 110 ⎟ dt = 220
⎝
⎠
55 = V (T ) − V ( 0 ) =
b
y dy = b n +1 − a n +1
b
⎡ x n +1 ⎤
⎡ n ( n +1) / n ⎤
y
=⎢
⎥ +⎢
⎥ n
⎦a
⎢⎣ n + 1 ⎥⎦ a ⎣ n + 1
10 ⎛
9
y dy = An
Bn + An = b n +1 − a n +1 . Thus
1
1
bn
n
an
(a )(a n ) + An + Bn = (b)(b n ) or
1
Time to drain: 20t − t 2 = 200; t = 20 hours.
2
V (1) − V ( 0 ) =
a
x n dx = Bn ;
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
x
4
δ ( x ) dx = m ( 2 ) =
5
2
f (t ) dt = G ( b ) = F (b) − F (a)
3
69.
3 2
x dx
0
⎡ x3 ⎤
= ⎢ ⎥ = 9−0 = 9
⎣⎢ 3 ⎦⎥ 0
70.
2 3
x dx
0
⎡ x4 ⎤
= ⎢ ⎥ = 4−0 = 4
⎢⎣ 4 ⎥⎦ 0
2
274
Section 4.4
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.
71.
72.
73.
π
0
77.
π
sin x dx = [ − cos x ]0 = 1 + 1 = 2
First Fundamental Theorem of Calculus.
Since G is differentiable at c , G is
continuous there. Now suppose c = a .
2
1
1 ⎤
⎡
(1 + x + x 2 ) dx = ⎢ x + x 2 + x3 ⎥
0
2
3 ⎦0
⎣
8⎞
20
⎛
= ⎜2+ 2+ ⎟−0 =
3⎠
3
⎝
2
Then lim G ( x) = lim
2
74
4
(2
−2
1
1
⎡1 ⎤
= ⎢ x3 ⎥ = = 0.333
⎣ 3 ⎦0 3
x − 3 x ) dx = 2
4
−2
x dx − 3
= 2 [ (−2 − 1 + 0 + 1 + 2 + 3)(1) ]
−2
x dx
b
1
⎡1
⎤
x dx = ⎢ x x ⎥ = ( b b − a a )
a
2
⎣
⎦a 2
76.
For b > 0, if b is an integer,
b
=
i =1
i=
(b − 1)b
.
2
If b is not an integer, let n = b . Then
0
x dx = 0 + 1 + 2 + ⋅⋅⋅ + (n − 1) + n(b − n)
(n − 1)n
+ n(b − n)
2
( b − 1) b
=
+ b (b − b ) .
2
=
x
a
f (m) dt ≤
x
a
f (t ) dt ≤
x
a
f ( M ) dt
( x − a ) f ( m) ≤ G ( x ) ≤ ( x − a ) f ( M )
By the Squeeze Theorem
lim ( x − a ) f (m) ≤ lim G ( x )
x →a +
x →a +
Thus,
a
lim G ( x) = 0 =
x→a+
a
f (t ) dt = G (a)
Therefore G is right-continuous at x = a .
Now, suppose c = b . Then
b
lim G ( x) = lim
x →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 )
x dx = 0 + 1 + 2 + ⋅⋅⋅ + (b − 1)
0
b −1
b
Then
≤ lim ( x − a ) f ( M )
4
d ⎛1
⎞ 1 ⎛ x⎞ x
⎜ x x ⎟ = x⎜ ⎟ + = x
dx ⎝ 2
⎠ 2 ⎝ x⎠ 2
b
Min-Max Existence Theorem) m and M such
that f (m) ≤ f ( x) ≤ f ( M ) for all x in [ a, b ] .
x→a+
1
⎡1
⎤
−3 ⎢ (2)(2) + (4)(4) ⎥
2
⎣2
⎦
= –24
75.
f (t ) dt . Since f is
continuous on [ a, b ] , there exist (by the
n
⎛ 1− 0 ⎞ ⎛ 1 ⎞ 1
0
i
i 2 , which for
+
=
⎜
⎟ ⎜ ⎟
3
n
n
⎝
⎠
⎝
⎠
n i =1
i =1
77
= 0.385 .
n = 10 equals
200
1 2
x dx
0
x
x→a a
x →c
The right-endpoint Riemann sum is
n
a. Let c be in ( a, b ) . Then G '(c) = f (c) by the
x →b −
Thus
lim G ( x) = 0 =
x →b −
b
b
f (t ) dt = G (b)
Therefore, G is left-continuous at x = b .
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 ] .
Instructor’s Resource Manual
Section 4.4
275
© 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.
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
x
F ( x) =
f (t ) dt
−1
0
=
−1
6.
x
0 dt +
0
1 dt
= 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.
1
b−a
b
a
8.
0; 2
π
0
f ( x ) dx
f ( x ) dx
10.
2.
1
2−0
1
=
3
5.
4 x3 dx
(
3
0
11.
= 40
0
3
x + 16
2
24 − 4 =
3
)
(
1
[sin π − sin 0] = 0
π
x cos x 2 dx =
12.
(
1 ⎡2 3
⎤
x + 16 ⎥
⎢
2 ⎣3
⎦0
6 −2
π/2
0
sin 2 x cos x dx
2 ⎡1 3 ⎤
sin x ⎥
π ⎢⎣ 3
⎦0
(
2
y 1 + y2
1
2
3π
=
)
3
(
⎡1
dy = ⎢ 1 + y 2
⎣8
) ⎥⎦1
625
609
−2 =
= 76.125
8
8
2
π −4
π /4
0
tan x sec2 x =
(1 − 0 ) =
π /4
1 ⎡1
⎤
tan 2 x ⎥
π / 4 − 1 ⎢⎣ 2
⎦0
2
π −4
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
)
4 ⎤2
π /2
13.
)
1
17
−2(−2) + 12 (−2)2 + 2 + 12 =
3
6
π
(0 − 0) = 0
1
π / 4 −1
=
1 ⎛1
2⎞
⎜ sin x ⎟
π ⎝2
⎠0
2
dx =
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⎩
=
π
1
2 −1
=
3
1
1
dx = ⎡ x 2 + 16 ⎤ =
⎢
⎥0 3
2
⎣
⎦
3
x + 16
x2
[sin x ]π0
π
π/ 2
x
2
1
0
1
π /2−0
=
4
1 4 2
1 ⎡5 ⎤
5 x dx = ⎢ x3 ⎥ = 35
4 −1 1
3 ⎣ 3 ⎦1
1
3.
3−0
4.
3
1
3
1
= ⎡ x4 ⎤
2 ⎣ ⎦1
π
π −0
Problem Set 4.5
1
3 −1
1
2 x dx ⎞⎟
⎠
π
1
9.
4. f ( x + p ) = f ( x ) ; period
1.
cos x dx =
2
0
1 π
1
π
sin x dx = ( − cos x )0
π −0 0
π
1
2
= − ( −1 − 1) =
=
2
0
π
2. f ( c )
3.
1 π
=
4.5 Concepts Review
1.
1 2
( x + x ) dx
2 + 3 −3
1 0
= ⎛⎜ ( − x + x ) dx +
5 ⎝ −3
2
1
4
= ⎡ x2 ⎤ =
⎣
⎦
0
5
5
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
)
276
Section 4.5
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.
3
x + 1 dx = c + 1 ( 3 − 0 )
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.
4
1
( ax + b ) dx = ( ac + b )( 4 − 1)
4
16.
1
−1
5
⎡a 2
⎤
⎢ 2 x + bx ⎥ = 3ac + 3b; c = 2
⎣
⎦1
x 2 dx = c 2 (1 − ( −1) )
1
3
⎡1 3 ⎤
2
⎢ 3 x ⎥ = 2c ; c = ± 3 ≈ ±0.58
⎣
⎦ −1
17.
3
(1 − x ) dx = (1 − c ) (3 + 4)
2
−4
26.
2
b 2
y dy
0
c=
b
3
3
⎡ 1 3⎤
2
⎢ x − 3 x ⎥ = 7 − 7c
⎣
⎦ −4
B
39
c=±
≈ ±2.08
3
18.
1
0
A
27.
B
1
20.
21.
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
π
−π
⎡x x ⎤
x dx = c ( 2 + 2 ) ; ⎢
⎥ = 4 c ; c = −1,1
⎣ 2 ⎦ −2
sin z dz = sin c (π + π )
[ − cos z ]π−π
22.
π
0
28.
2
2
= f (c)
⎡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
⎡ − x 2 ( 2 x − 3) ⎤
⎢
⎥ = c − c2
6
⎢⎣
⎥⎦ 0
19.
( ax + b ) dx
B−A
x (1 − x ) dx = c (1 − c )(1 − 0 )
c=
b
⎡1 ⎤
= c 2 ( b − 0 ) ; ⎢ y 3 ⎥ = bc 2
⎣ 3 ⎦0
b
⎡1
⎤
ay 2 dy = ac 2 ( b − 0 ) ; ⎢ ay 3 ⎥ = abc 2
0
⎣3
⎦0
b
c=
b 3
3
29.. Using c = π yields 2π (5) 4 = 1250π ≈ 3927
= 2π sin c; c = 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
Instructor’s Resource Manual
Section 4.5
277
© 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.
31.
Using c = 0.5 yields 2
2
1 + 0.5
2
= 3.2
38.
3π
π
(
π
=
32.
−π
π
=
5
)
(sin 2 x + 2sin x cos x + cos 2 x) dx
(1 + 2sin x cos x) dx =
−π
π
=2
40.
x cos( x3 ) dx
(sin x + cos x)2 dx
−π
⎛ 16 ⎞
Using c = 15 yields ⎜ ⎟ (20 − 10) ≈ 13.8 .
⎝ 15 ⎠
3π 2
0
3π
2⎡
2
= sin 3 3 π3
sin( x3 ) ⎤
⎣
⎦
0
3
3
=
39.
x 2 cos( x3 ) dx = 2
− 3π
0
π/2
π
−π
dx +
π
−π
sin 2 x dx
dx + 0 = 2[ x]0π = 2π
z sin 2 ( z 3 ) cos( z 3 )dz = 0 , since
−π / 2
(− 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
(sin x + cos x) dx =
π
−π
sin x dx + 2
π
0
= − z sin 2 ( z 3 ) cos( z 3 )
41.
36.
37.
−1
x
3
2 4
(1 + x )
odd.
π/2
−π / 2
1
dx +
−1
1
x dx +
−1
x 2 dx +
1
−1
x3 dx
1
= 2 [ x]
42.
100
⎡ x3 ⎤
8
+0+ 2⎢ ⎥ +0 =
3
3
⎣⎢ ⎦⎥ 0
(v + sin v + v cos v + sin 3 v)5 dv = 0
−100
since (−v + sin(−v) − v cos(−v) + sin 3 (−v))5
= (−v − sin v − v cos v − sin 3 v)5
= −(v + sin v + v cos v + sin 3 v)5
43.
(x
1
−1
3
)
+ x3 dx = 2
1
0
x3 dx +
1
−1
x3 dx
1
⎡ x4 ⎤
1
= 2⎢ ⎥ +0 =
4
2
⎣⎢ ⎦⎥ 0
44.
( x sin
π/4
−π / 4
5
)
2
x + x tan x dx = 0
2
dx = 0 , since the integrand is
sin x
dx = 0 , since the integrand is odd.
1 + cos x
1
−1
1
0
π
1
(1 + x + x 2 + x3 ) dx
=
cos x dx
= 0 + 2 [sin x ]0 = 0
1
−1
since − x sin 5 (− x) + − x tan(− x)
2
= − x sin 5 x − x tan x
45.
−a
f ( x) dx =
−b
−a
−b
b
a
f ( x) dx when f is even.
f ( x) dx = −
b
a
f ( x) dx when f is odd.
278
Section 4.5
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.
46. u = − x, du = − dx
b
a
=
f ( − x ) dx = −
−a
−b
b. ku =
−b
−a
−a
f (u ) du =
−b
f (u ) du
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
π/2
cos x dx = 8
0
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
0
π2
⎡ 1
⎤
= 8 ⎢ − cos 2 x ⎥
⎣ 2
⎦0
49.
1+π
1
π2
sin x dx =
π
0
54.
a. V = 0 by periodicity.
b. V = 0 by periodicity.
sin 2 x dx
2
c. Vrms
=
= –4(–1 – 1) = 8
sin x dx =
k b
1 b
u dx =
ku dx = ku
a
b−a
b−a a
π
0
=
φ +1
Vˆ 2 sin 2 (120π t + φ ) dt
φ
1
0
Vˆ 2 sin 2 (120π t ) dt
by periodicity.
u = 120π t , du = 120π dt
sin x dx
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
=
51.
1+π
1
sin 2 x dx =
cos x dx =
= 2 [sin x ]0
53.
55.
sin 2 x dx
π
0
cos x dx = 2
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
1
[ − cos 2 x ]0π / 2 = 1
2
π /2
52.
π/2
0
π /2
0
f (m) ≤ f ( x) ≤ f ( M ) for all x in [ a, b ] . Thus
b
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
fdx = f
dx =
f ( x)dx ⋅ dx
a
a
a
a
b−a
b
1 b
=
f ( x)dx ⋅ (b − a ) =
f ( x) dx
a
a
b−a
a
(b − a) f (m) ≤
b
a
b
a
f ( x) dx ≤
b
a
f ( M ) dx
f ( x) dx ≤ (b − a ) f ( M )
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
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
Instructor’s Resource Manual
Section 4.5
279
© 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.
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.
a
Note that
π +a
f ( x ) dx = 0 since f is odd, and
−a
f ( x ) dx = 0 since
π −a
f (π + x ) = − f (π − x ) .
π /2
0
f ( x ) dx =
π
J1 (1) ≈ 0.69 (Bessel
2
function)
π /2
f ( x ) dx = 0
−π / 2
3π / 2
c. 2π =
=2
=
57.
2π
0
2π
0
2π
0
2π
2
cos x dx +
0
cos 2 x dx, thus
sin x dx
0
3π / 2
2π
cos 2 x dx
−3π / 2
13π / 6
0
π /6
4π / 3
2
sin x dx = π
π /6
10π / 3
a. Even
c. On [ 0, π ] , sin x = sin x .
59.
u = cos x , du = − sin x dx
f ( x ) dx =
=−
sin u du = cos u + C
f ( x ) dx = − cos ( cos x ) + C
−π / 2
π /2
f ( x ) dx = 2
3π / 2
0
0
f ( x ) dx
π
0
f ( x ) dx +
3π / 2
π
−3π / 2
0
f ( x ) dx = 2
3π / 2
0
f ( x ) dx
f ( x ) dx
= 2 ( cos1 − 1) ≈ −0.92
4π / 3
10π / 3
13π / 6
f ( x ) dx =
4π / 3
π /6
a
a
g ( x) dx =
0
a ⎛c ⎞
f ⎜ x ⎟ dx
c ⎝a ⎠
a2 c
f ( x ) dx ≈ −0.44
0
f ( x) dx
⎛c ⎞
f ⎜ x ⎟ dx
⎝b ⎠
a
0
a2 c
c2 0
g ( x ) dx +
f ( x) dx +
a 2 + b2 c
c
2
0
b
h( x) dx
0
2 c
b
c2 0
f ( x) dx =
f ( x) dx
c
0
f ( x) dx since
a 2 + b 2 = c 2 from the triangle.
60.
If f is odd, then f (− x ) = − f ( x) and we can
write
−a
⎛ 3⎞
⎛1⎞
f ( x ) dx = 2 cos1 − cos ⎜⎜
⎟⎟ + cos ⎜ ⎟
⎝2⎠
⎝ 2 ⎠
≈ −0.44
f ( x ) dx ≈ 1.055
b
b
b2 c
f ( x) dx =
f ( x) dx
0 c
c
c2 0
=
0
f ( x ) dx = 0
π /6
π /6
c
=
= 2 (1 − cos1) ≈ 0.92
f ( x ) dx =
4π / 3
f ( x ) dx =
0
Thus,
= cos1 − 1 ≈ −0.46
3π / 2
2π
f ( x ) dx ≈ 1.055 (numeric integration)
a
a
f ( x) dx =
0 c
c
=
f ( x ) dx = 1 − cos1 ≈ 0.46
π /2
f ( x ) dx = 0
f ( x ) dx = 0
0
c2
b
b b
B=
h( x) dx =
0
0 c
Likewise, on [π , 2π ] ,
π /2
2π
f ( x ) dx =
c
=
= cos ( cos x ) + C
0
2π
0
a. Written response.
b. A =
sin x ⋅ sin ( cos x ) dx
f ( x ) dx ≈ 0.69
0
f ( x ) dx = 0 ;
13π / 6
b. 2π
π /2
f ( x ) dx =
2
=−
f ( x) dx =
a
0
0
−a
[ − f (− x)] dx =
f (u ) du = −
a
0
0
a
f (u ) du
f ( x) dx
On the second line, we have made the
substitution u = − x .
280
Section 4.5
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.
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
x
x0 = 1.00
;h =
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
Right Riemann Sum:
dx ≈ 0.25[ f ( x0 ) + f ( x1 ) +
3
1
1
x2
3
1
2
dx ≈ 0.25[ f ( x1 ) + f ( x2 ) + ... + f ( x8 )] ≈ 0.5655
0.25
[ f ( x0 ) + 2 f ( x1 ) + + 2 f ( x7 ) + f ( x8 )] ≈ 0.6766
2
x
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:
1
+ f ( x7 )] ≈ 0.7877
dx ≈
Fundamental Theorem of Calculus:
2.
f ( x) =
1
3
x
x0 = 1.00
;h =
3
1
2
⎡ 1⎤
dx = ⎢ – ⎥ = – + 1 = ≈ 0.6667
1 x2
3
3
⎣ x ⎦1
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
1
1
x3
Right Riemann Sum:
Trapezoidal Rule:
Parabolic Rule:
3
1
1
x3
3
1
1
3
3
1
1
3
x
x
dx ≈ 0.25[ f ( x0 ) + f ( x1 ) +
dx ≈ 0.25[ f ( x1 ) + f ( x2 ) + ... + f ( x8 )] ≈ 0.3392
dx ≈
dx ≈
+ f ( x7 )] ≈ 0.5799
0.25
[ f ( x0 ) + 2 f ( x1 ) +
2
+ 2 f ( x7 ) + f ( x8 )] ≈ 0.4596
0.25
[ f ( x0 ) + 4 f ( x1 ) + 2 f ( x2 ) +
3
Fundamental Theorem of Calculus:
3
1
+ 4 f ( x7 ) + f ( x8 )] ≈ 0.4455
3
4
⎡ 1 ⎤
dx = ⎢ −
= ≈ 0.4444
3
2⎥
9
x
⎣ 2 x ⎦1
1
Instructor's Resource Manual
Section 4.6
281
© 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.
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
2
Left Riemann Sum:
Trapezoidal Rule:
Parabolic Rule:
x dx ≈ 0.25[ f ( x0 ) + f ( x1 ) +
0
2
Right Riemann Sum:
0
f ( x5 ) ≈ 1.1180
f ( x6 ) ≈ 1.2247
f ( x7 ) ≈ 1.3229
f ( x8 ) ≈ 1.4142
+ f ( x7 )] ≈ 1.6847
0.25
[ f ( x0 ) + 2 f ( x1 ) + + 2 f ( x7 ) + f ( x8 )] ≈ 1.8615
2
2
0.25
xdx ≈
[ f ( x1 ) + 4 f ( x2 ) + 2 f ( x3 ) + + 4 f ( x7 ) + f ( x8 )] ≈ 1.8755
0
3
2
xdx ≈
0
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 =
3
Left Riemann Sum:
1
Trapezoidal Rule:
Parabolic Rule:
x x 2 + 1 dx ≈ 0.25[ f ( x0 ) + f ( x1 ) +
3
Right Riemann Sum:
1
(
)
5
0.25
[ f ( x0 ) + 2 f ( x1 ) + + 2 f ( x7 ) + f ( x8 )] ≈ 9.6184
2
3
0.25
x x 2 + 1dx ≈
[ f ( x0 ) + 4 f ( x1 ) + 2 f ( x2 ) + + 4 f ( x7 ) + f ( x8 )] ≈ 9.5981
1
3
3
1
f ( x) = x x 2 + 1 ; h =
x0 = 0.00
x1 = 0.125
x2 = 0.250
x3 = 0.375
x4 = 0.500
+ f ( x7 )] ≈ 8.6093
x x 2 + 1 dx ≈ 0.25[ f ( x1 ) + f ( x2 ) + ... + f ( x8 )] ≈ 10.6274
x x 2 + 1 dx ≈
Fundamental Theorem of Calculus:
5.
= 1.25
= 1.50
= 1.75
= 2.00
x dx ≈ 0.25[ f ( x1 ) + f ( x2 ) + ... + f ( x8 )] ≈ 2.0383
Fundamental Theorem of Calculus:
4.
x5
x6
x7
x8
3
(
)
1
⎡1
⎤
x x 2 + 1dx = ⎢ ( x 2 + 1)3 / 2 ⎥ = 10 10 – 2 2 ≈ 9.5981
1
⎣3
⎦1 3
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
Section 4.6
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.
)
5
x x 2 + 1 dx ≈ 0.125[ f ( x0 ) + f ( x1 ) +
0
1
Right Riemann Sum:
Trapezoidal Rule:
(
1
Left Riemann Sum:
0
(
)
5
x x 2 + 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
x x 2 + 1 dx ≈
[ f ( x0 ) + 4 f ( x1 ) + 2 f ( x2 ) + + 4 f ( x7 ) + f ( x8 )] ≈ 5.2580
0
3
1
0
x x 2 + 1 dx ≈
(
Parabolic Rule:
)
Fundamental Theorem of Calculus:
6.
+ f ( x7 )] ≈ 3.4966
(
)
(
)
1
5
6⎤
⎡1 2
x x 2 + 1 dx = ⎢
x + 1 ⎥ = 5.25
0
⎣12
⎦0
1
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
3/ 2
f ( x) = ( x + 1)
;h =
4
Left Riemann Sum:
1
4
Right Riemann Sum:
Trapezoidal Rule:
Parabolic Rule:
( x + 1)3 / 2 dx ≈ 0.375[ f ( x0 ) + f ( x1 ) +
1
+ f ( x7 )] ≈ 18.5464
( x + 1)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
[ f ( x0 ) + 4 f ( x1 ) + 2 f ( x2 ) + + 4 f ( x7 ) + f ( x8 )] ≈ 20.0979
( x + 1)3 / 2 dx ≈
1
3
4
1
( x + 1)3 / 2 dx ≈
Fundamental Theorem of Calculus:
4
1
3/ 2
( x + 1)
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.
Instructor’s Resource Manual
Section 4.6
283
© 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.
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.
f
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
( x) = −
1
x
2
( x) =
; f
The largest that f
(c)
2
x3
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
dx ≈
[ f ( x0 ) + 2 f ( x1 ) + + 2 f ( x11 ) + f ( x12 )]
1 x
2
≈ 1.1007
12.
f
( x) = −
1
(1 + x )
2
The largest that f
; f
(c)
( x) =
2
(1 + x )3
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
dx ≈
[ f ( x0 ) + 2 f ( x1 ) + + 2 f ( x4 ) + f ( x5 )]
1 1+ x
2
≈ 0.6956
2
13.
f
( x) =
1
2 x
; f
The largest that f
( x) = −
(c)
1
4 x3 / 2
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
[ f ( x0 ) + 2 f ( x1 ) + + 2 f ( x7 ) + f ( x8 )]
x dx ≈
1
2
≈ 4.6637
284
Section 4.6
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.
14.
f
1
( x) =
2 x +1
; f
The largest that f
( 3 − 1)3 ⎛
12n
15.
f
2
4
f ( ) ( x) =
1
2
x
24
(c)
1
4 ( x + 1)
3/ 2
can be on [1,3] occurs when c = 1 , and f (1) =
1
4 × 23 / 2
.
100
⎞
Round up: n = 3
⎟ ≤ 0.01; n ≥
12
2
⎠
3
0.667
[ f ( x0 ) + 2 f ( x1 ) + 2 f ( x2 ) + f ( x3 )]
x + 1 dx ≈
1
2
≈ 3.4439
1
⎜
⎝ 4 × 23 / 2
( x) = −
( x) = −
( x) =
; f
2
x
3
; f
6
( 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
dx ≈
[ f ( x0 ) + 4 f ( x1 ) + ... + 4 f ( x5 ) + f ( x6 )]
1 x
3
≈ 1.0989
16.
f
( x) =
f
( x) =
1
2 x +1
; f
3
8 ( x + 1)
5/ 2
( x) = −
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
x + 1 dx ≈ ⎡⎣ f ( x0 ) + 4 f ( x1 ) + f ( x2 ) ⎤⎦ ≈ 10.5464
4
3
4
Instructor’s Resource Manual
Section 4.6
285
© 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.
17.
m+ h
b
⎡a
⎤
(ax 2 + bx + c)dx = ⎢ x3 + x 2 + cx ⎥
m–h
2
⎣3
⎦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
f ( x) dx = [ f (m − h) + 4 f (m) + f (m + h)] = 2am3 h + 2amh3 + 2bm2 h + bh3 + 2chm + 2dh
m−h
3
3
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 = −
2
However, f ( x) = 3ax + 2bx + c, f ( x) = 6ax + 2b, f
19. The left Riemann sum will be smaller than
b
a
(3)
(l − k )5
180n
4
f (4) (m) for some m between l and k.
( x) = 6a, and f (4) ( x) = 0 , so En = 0.
f ( x ) dx .
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
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.
20. The right Riemann sum will be larger than
b
a
f ( x ) dx .
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
If f is concave down, then f
b
a
f ( x ) dx .
( c ) < 0 for any c ∈ ( a, b ) . Thus, the error En =
( b − a )3
24n 2
(c) < 0 .
f
Since the error
is negative, then the Riemann sum must be greater than the integral.
22. The Trapezoidal Rule approximation will be smaller than
If f is concave down, then f
(c) < 0
b
a
f ( x ) dx .
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
a
x k dx ≈ [– a k + 2 ⋅ 0 + a k ] = 0
2
a
1
1
⎡ 1 k +1 ⎤
[a k +1 – a k +1 ] = 0
x k dx = ⎢
x ⎥ =
[a k +1 – (– a) k +1 ] =
–a
1
k
+
k
k
1
1
+
+
⎣
⎦ –a
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–
Instructor’s Resource Manual
Section 4.6
287
© 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.
b. T ≈ 1.9886; f ( x) = cos x
T–
( )
π
[cos π – cos 0] 12
2
≈ 1.999987
12
The correct value is 2.
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,
24
Distance =
0
v(t ) dt ≈
8
i =1
4. False:
=
g ( x) = x 2 + 7 x − 5 are a
counterexample.
v(ti ) Δt
= ( 31 + 54 + 53 + 52 + 35 + 31 + 28 )
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 =
≈
10
i =1
120
0
F (t ) dt
F (ti ) Δt = 12(71 + 68 +
f ( x) = x 2 + 2 x + 1 and
+ + an −1 + an − 2 + an + an −1
= a0 + 2a1 + 2a2 + + 2an −1 + an
+ 148)
= 13, 740 gallons
100
8. True:
100
(2i − 1) = 2
i =1
4.7 Chapter Review
=
Concepts Test
Theorem 4.3.D
9. True:
i =1
2. True:
Obtained by integrating both sides of
the Product Rule
3. True:
If F ( x) =
288
Section 4.7
1
i =1
(ai + 1) 2 =
10
i =1
ai2 + 2
10
i =1
ai +
100
1
i =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.
f ( x) dx, f ( x) is a
derivative of F(x).
100
2(100)(100 + 1)
− 100 = 10, 000
2
10
1. True:
i =1
i−
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.
12. False:
13. False:
1
−1 x dx
is a counterexample.
A counterexample is
⎧0, x ≠ 0
f ( x) = ⎨
⎩1, x = 0
with
1
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 ) =
2
⎡ f ( x ) ⎤⎦ dx = 0 .
−1 ⎣
If f ( x ) is continuous, then
2
2
[ f ( x)] ≥ 0 , and if [ f ( x)] is greater
than 0 on [a, b], the integral will be
also.
14. False:
Dx ⎡⎢
f ( z ) dz ⎤⎥ = f ( x)
⎣ a
⎦
15. True:
sin x + cos x has period 2π , so
x
x + 2π
x
32. False:
33. False:
34. False:
lim kf ( x) = k lim f ( x) and
x →a
x→a
x →a
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 = 2
a
0
f ( x) dx because f
z (t ) = t 2 is a counterexample.
b
0
f ( x ) dx = F (b) − F (0)
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.
limits exist.
17. True:
a
−a
G ( x ) dx = G (b) − G (a )
Odd-exponent terms cancel
themselves out over the interval, since
they are odd.
lim [ f ( x) + g ( x) ]
x→a
a
35. True:
x→a
= lim f ( x) + lim g ( x ) when all the
b
F ( x ) dx
is even.
(sin x + cos x) dx
is independent of x.
16. True:
=
b
a
38. True:
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 ) =
x
0
and use Theorem 4.3.B.
f ( t ) dt is even,
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.
Instructor’s Resource Manual
( )
cos x 2 dx
Section 4.7
289
© 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.
Sample Test Problems
12. u = 2 y 3 + 3 y 2 + 6 y, du = (6 y 2 + 6 y + 6) dy
1
( y2 + y +1)
5
5
⎡1
⎤
1. ⎢ x 4 − x3 + 2 x3 / 2 ⎥ =
⎣4
⎦0 4
dy =
1 5 3
2 y +3 y 2 + 6 y
355
=
2
1⎤
13
⎡2
2. ⎢ x3 − 3 x − ⎥ =
3
x
⎣
⎦1 6
1 355 −1/ 5
u
du
6 11
(
1 ⎡5 4/5 ⎤
5
3554 / 5 − 114 / 5
u ⎥
=
⎢
6 ⎣4
24
⎦11
13.
⎡⎛ i ⎞ 2 ⎤ ⎛ 1 ⎞ 7
⎢⎜ ⎟ − 1⎥ ⎜ ⎟ =
⎥⎦ ⎝ 2 ⎠ 4
i =1 ⎢⎣⎝ 2 ⎠
14.
f ( x) =
)
4
π
3
⎡1
26 ⎤
50 26 π
3. ⎢ y 3 + 9 cos y − ⎥ =
− +
− 9 cos1
3
y
3 π
3
⎣
⎦1
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 (3 x 2 + 6 x)
1 2
1
u du = u 3 + C
6
18
π
1 ⎡ 3 2
1
tan (3 x + 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⎣
15.
11.
16.
)
(
(
290
Section 4.7
1 ⎞
5 − ⎟ dx =
2 ⎜⎝
x2 ⎠
4
1⎤
39
⎡
⎢5 x + x ⎥ = 4
⎣
⎦2
(3i − 3i −1 )
i =1
= (3 − 1) + (32 − 3) + (33 − 32 ) +
)
)
)
x + 1 dx
4⎛
18.
( x + 1) sin x 2 + 2 x + 3 dx
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+5−4
5
1 5 2 3
1 ⎡2
⎤
3x x − 4 dx = ⎢ ( x3 − 4)3 / 2 ⎥
5−2 2
3 ⎣3
⎦2
= 294
n
(
0
3
17.
3
3
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
(2 − x + 1) 2 dx
=
2
9.
3
0
1
1
, f (7) =
x+3
10
+ (3n − 3n−1 )
= 3n − 1
10
19.
i =1
(6i 2 − 8i ) = 6
10
i =1
i2 − 8
10
i
i =1
⎡10(11)(21) ⎤
⎡10(11) ⎤
= 6⎢
⎥ − 8 ⎢ 2 ⎥ = 1870
6
⎣
⎦
⎣
⎦
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.
20. a.
1 1 1 13
+ + =
2 3 4 12
e.
b. 1 + 0 + (–1) + (–2) + (–3) + (–4) = –9
c.
−2
0
2
f (− x) dx = −
f ( x) dx = −2
0
24. a.
2
2
+0−
−1 = 0
2
2
1+
78
21. a.
1
n=2 n
50
nx 2n
b.
4
n =1
x − 1 dx =
0
22.
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
c.
⎧
9⎡
3 1 ⎤⎫
= lim ⎨48 − ⎢ 2 + + ⎥ ⎬
2⎣
n n2 ⎦ ⎭
n→∞ ⎩
= 48 – 9 = 39
f ( x) dx =
1
0
f ( x) dx +
1
b.
c.
d.
f ( x) dx = −
1
2
2
0
0
3 f (u ) du = 3
0
2
2
0
0
2
0
g ( x) − 3
2
0
f ( x) dx = 2
f ( x ) dx = −
x dx −
2
4
0
x dx
f ( x) dx = 2(−4) = −8
0
f (u ) du = 3(2) = 6
2
−2
2
= −2
f ( x) dx
d.
2
2
−2
−2
f ( x) dx = −4
[ 2 g ( x) − 3 f ( x)] dx
=2
25. a.
c.
1
0
b. Since f ( x ) ≤ 0 , f ( x ) = − f ( x ) and
= –4 + 2 = –2
0
4
( x − x ) dx =
4
⎧
27 ⎡ n(n + 1)(2n + 1) ⎤ ⎫
= lim ⎨48 − ⎢
⎥⎬
6
n→∞ ⎩
⎦⎭
n3 ⎣
2
4
0
⎡1 2 ⎤
⎢2 x ⎥ − 6 = 8 – 6 = 2
⎣
⎦0
⎧⎪ 48 n
27 n 2 ⎫⎪
1−
i ⎬
= lim ⎨
3
n→∞ ⎩
⎪ n i =1 n i =1 ⎭⎪
23. a.
x dx = 1 + 2 + 3 = 6
0
2
−2
2
−2
=2
0
f ( x ) dx
f ( x ) dx = 8
g ( x) dx = 0
[ f ( x) + f (− x)] dx
2
0
f ( x) dx + 2
2
0
f ( x) dx
= 4(–4) = –16
f ( x) dx
= 2(–3) – 3(2) = –12
Instructor’s Resource Manual
Section 4.7
291
© 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.
2
e.
0
[ 2 g ( x) + 3 f ( x)] dx
=2
2
0
2
g ( x) dx + 3
f ( x) dx
0
= 2(5) + 3(–4) = –2
0
f.
26.
27.
−2
100
−100
−1
−4
g ( x) dx = −
2
g ( x) dx = −5
0
c.
G ( x) = −
d.
G ( x) =
e.
G ( x) =
G ( x) =
f.
=−
−1
⎡ x3 ⎤ = 9c 2
⎣ ⎦ −4
G ( x) =
G ( x) =
c.
g ( x ) dg (u )
0
du = [ g (u )]0g ( x )
x
0
−x
0
f (−t ) dt =
x
0
f (u )(− du )
f (u ) du
4
0
30. a.
x dx =
1
2
x +1
3 2
1 x dx
b.
2 ⎡ 3 / 2 ⎤ 4 16
=
x
⎦0 3
3⎣
3
1
26
= ⎡ x3 ⎤ =
⎣
⎦
1
3
3
2x
4
x +1
3x2
6
x +1
31.
−
2x 1
dt −
dt
1 t
t
1
1
⋅5 − ⋅2 = 0
f (x) =
5x
2x
f ( x) =
1
2
x +1
5x 1
2x
t
dt =
5x 1
1
G ( x ) = f ( x + 1) − f ( x )
b.
32. Left Riemann Sum:
2
1
1
1 + x4
Right Riemann Sum:
2
1
1+ x
4
f ''(c) =
En = −
dx ≈
dx ≈ 0.125[ f ( x0 ) + f ( x1 ) +
2
1
1
1 + x4
Midpoint Riemann Sum:
1
f (t ) dt
G ( x ) = sin 2 x
29. a.
33.
1
f ( x)
x
G ( x) = − f ( x)
c2 = 7
c = − 7 ≈ −2.65
b.
x
0
f ( z ) dz +
G ( x) = g ( g ( x)) g ( x)
3 x 2 dx = 3c 2 (−1 + 4)
G ( x) =
x
x2 0
du
= g ( g ( x)) − g (0)
( x3 + sin 5 x) dx = 0
28. a.
1
dx ≈ 0.125[ f ( x1 ) + f ( x2 ) +
2
1
1
1 + x4
(1 + c 4 )3
(2 − 1)3
(12)82
≤
f ''(c) =
+ f ( x8 )] ≈ 0.1767
dx ≈ 0.125[ f ( x0.5 ) + f ( x1.5 ) +
0.125
[ f ( x0 ) + 2 f ( x1 ) +
2
4c 2 (5c 4 − 3)
+ f ( x7 )] ≈ 0.2319
+ 2 f ( x7 ) + f ( x8 )] ≈ 0.2043
(
(4)(22 ) (5)(24 ) − 3
(1 + 1 )
4
+ f ( x7.5 )] ≈ 0.2026
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
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.
34.
4
0
1
0.5
dx ≈
[ f ( x0 ) + 4 f ( x1 ) + 2 f ( x2 ) +
1 + 2x
3
4
f ( ) (c) =
En = −
35.
n2 >
(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 = −
384
+ 4 f ( x7 ) + f ( x8 )] ≈ 1.1050
4 3
(1 + c )
(2 − 1)3
12n
2
(
(4)(22 ) (5)(24 ) + 3
≤
f ''(c) =
(1 + 1 )
4
1
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
2
1.
1 ⎛1⎞
1 1 1
−⎜ ⎟ = − =
2 ⎝2⎠
2 4 4
2. x − x 2
6.
(
( 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
Instructor’s Resource Manual
Review and Preview
293
© 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.
11.
=
12.
13.
2
⎡ x5 x 4
⎤
x − 2 x + 2 dx = ⎢ −
+ 2x⎥
−1
2
⎢⎣ 5
⎥⎦ −1
2
(
4
)
3
12 ⎛ 27 ⎞ 51
−⎜− ⎟ =
5 ⎝ 10 ⎠ 10
3 23
y dy
0
3
3
3
= ⋅ y 5 3 = ⋅ 35 3 ≈ 3.74
0
5
5
2
2⎛
⎡
x2 x4 ⎞
x3 x5 ⎤
16
+
⎜1 −
⎟ dx = ⎢ x − + ⎥ =
⎟
0⎜
2
16
6
80
⎝
⎠
⎣⎢
⎦⎥ 0 15
14. Let u = 1 + 94 x; then du = 94 dx and
1+
9
4
x dx =
4
9
8 ⎛ 9 ⎞
=
⎜1 + x ⎟
27 ⎝ 4 ⎠
3
2
u du =
4 2 32
u +C
93
+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 ⎟⎠
Review and Preview
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.
5
CHAPTER
Applications of the Integral
6. To find the intersection points, solve
5.1 Concepts Review
1.
b
f ( x)dx; −
a
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.
d
c
[ q( y) − p( y )] dy
A=
Problem Set 5.1
1. Slice vertically.
9
⎛
⎞ ⎛8
⎞ 125
⎜ −9 + + 18 ⎟ − ⎜ + 2 − 12 ⎟ =
2
6
⎝
⎠ ⎝3
⎠
ΔA ≈ ( x 2 + 1)Δx
A=
2
⎡1
⎤
( x 2 + 1)dx = ⎢ x3 + x ⎥ = 6
–1
3
⎣
⎦ −1
2
7. Solve x3 − x 2 − 6 x = 0 .
2. Slice vertically.
ΔA ≈ ( x3 − x + 2)Δx
A=
2
33
1
⎡1
⎤
( x3 − x + 2)dx = ⎢ x 4 − x 2 + 2 x ⎥ =
−1
2
⎣4
⎦ −1 4
2
A=
−2
=
2
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 + 3 x ⎥ =
−3
⎣ 3
⎦ −3 3
5. To find the intersection points, solve 2 – x 2 = x .
2
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
1
Instructor’s Resource Manual
3
0
(− x3 + x 2 + 6 x)dx
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
⎣
⎦
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
⎣
⎦
A=
( x3 − x 2 − 6 x)dx +
0
4. Slice vertically.
1
0
−2
1
1
⎡ 1
⎤
⎡1
⎤
+ ⎢ − x 4 + x3 + 3 x 2 ⎥
= ⎢ x 4 − x3 − 3 x 2 ⎥
4
3
4
3
⎣
⎦ −2 ⎣
⎦0
⎛8
⎞ ⎛ 8
⎞ 40
= ⎜ + 2 + 4⎟ − ⎜ − + 2 − 4⎟ =
3
3
⎝
⎠ ⎝
⎠ 3
A=
ΔA1 ≈ ( x3 − x 2 − 6 x)Δx
A = A1 + A2
1
⎡1
⎤
( x + x + 2)dx = ⎢ x3 + x 2 + 2 x ⎥
2
⎣3
⎦ −2
2
x( x 2 − x − 6) = 0
x(x + 2)(x – 3) = 0
x = –2, 0, 3
Slice vertically.
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
3
1
⎡ 1
⎤
(− x 2 + x + 6) dx = ⎢ − x3 + x 2 + 6 x ⎥
−2
2
⎣ 3
⎦ −2
3
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
Section 5.1
295
© 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.
9. To find the intersection points, solve
13.
2
y +1 = 3 – y .
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
(– 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
A=
1
ΔA ≈ −( x − 4)( x + 2)Δx = (− x 2 + 2 x + 8)Δx
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
⎣
⎦
A=
2
0
3
⎡ 1
⎤
(− 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.
A=
3
14.
2
22
1
⎡ 1
⎤
( − y 2 − y + 6)dy = ⎢ − y 3 − y 2 + 6 y ⎥ =
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=
15.
1 ⎞
⎛
ΔA ≈ ⎜ 3 − x 2 ⎟ Δx
3 ⎠
⎝
A=
4
3
1 ⎤
⎛ 1 2⎞
⎡
3 − x ⎟ dx = ⎢3 x − x3 ⎥ = 9 − 3 = 6
0 ⎜⎝
3 ⎠
9 ⎦0
⎣
3
Estimate the area to be (3)(2) = 6.
12.
1
ΔA ≈ − ( x 2 − 7)Δx
4
A=
ΔA ≈ (5 x − x 2 )Δx
3
1 ⎤
⎡5
(5 x − x 2 )dx = ⎢ x 2 − x3 ⎥ ≈ 11.33
1
3 ⎦1
⎣2
⎛ 1⎞
Estimate the area to be (2) ⎜ 5 ⎟ = 11 .
⎝ 2⎠
A=
296
3
Section 5.1
2
1
1 ⎡1
⎤
− ( x 2 − 7)dx = − ⎢ x3 − 7 x ⎥
0
4
4 ⎣3
⎦0
2
1⎛8
⎞ 17
= − ⎜ − 14 ⎟ =
≈ 2.83
4⎝3
⎠ 6
⎛ 1⎞
Estimate the area to be (2) ⎜ 1 ⎟ = 3 .
⎝ 2⎠
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.
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).
3
ΔA2 ≈ x Δx
A = A1 + A2 =
0
−3
0
− x3 dx +
3
0
x3 dx
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
3
ΔA1 ≈ − x Δx
Estimate the area to be
ΔA2 ≈ 3 x Δx
A = A1 + A2 =
0
−2
− 3 x dx +
0
2 3
0
x dx
1
(4)(4) = 8 .
2
20.
2
⎛ 3 3 2 ⎞ ⎛ 33 2 ⎞
⎡ 3
⎤
⎡3
⎤
= ⎢− x4 / 3 ⎥ + ⎢ x4 / 3 ⎥ = ⎜
⎟+⎜
⎟
⎣ 4
⎦ −2 ⎣ 4
⎦ 0 ⎜⎝ 2 ⎟⎠ ⎜⎝ 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
2
⎡
⎤
(10 − x ) dx = ⎢10 x − x3 2 ⎥
0
3
⎣
⎦0
= 90 – 18 = 72
Estimate the area to be 9 · 8 =72.
A=
9
Instructor’s Resource Manual
9 ± 81 − 64
2
9 ± 17
x=
2
⎛
9 − 17
9 + 17 ⎞
is extraneous so x =
. ⎟⎟
⎜⎜ x =
2
2
⎝
⎠
x=
Section 5.1
297
© 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.
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
+
(− x 2 − 5 x − 6)dx
−2
23
17
−
≈ 15.92
4
4
Estimate the area to be
−2
−3
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 ⎠
3/ 2
2 ⎛ 9 + 17 ⎞
= ⎜⎜
⎟
3 ⎝ 2 ⎟⎠
A=
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
1
⎡ 2
⎤
(−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
A=
1
8
1 ⎤
⎡
(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.
A=
8
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
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.
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 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.
−1
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 ⎞ 64
⎛ 8⎞ ⎛
= ⎜ 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
Instructor’s Resource Manual
Section 5.1
299
© 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.
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⎞
⎤
− x + − (−2 x + 2) ⎥ dx =
⎜ x + ⎟ dx .
−1 ⎢⎣ 2
−
1
2
2⎠
⎝2
⎦
The right-hand part of the triangle has area
5⎡ 1
5⎛ 3
7
15 ⎞
⎤
− x + − ( x − 4) ⎥ dx = ⎜ − x + ⎟ dx .
⎢
2⎣ 2
2 ⎝ 2
2
2⎠
⎦
The triangle has area
2 ⎛3
5⎛ 3
3⎞
15 ⎞
x + ⎟ dx + ⎜ − x + ⎟ dx
⎜
−1 ⎝ 2
2⎝ 2
2⎠
2⎠
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
0
0
⎡( x + 6) − x3 ⎤ dx =
⎣
⎦
A( R ) = A( R1 ) + A( R2 )
A( R2 ) =
0
=
−4
⎛3
⎞
⎜ x + 6 ⎟ dx +
⎝2
⎠
2
0
(− x3 + x + 6)dx
(− x3 + x + 6)dx
0
2
31.
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
9
(3t 2 − 24t + 36)dt = ⎡t 3 − 12t 2 + 36t ⎤ = (729 – 972 + 324) – (–1 – 12 – 36) = 130
⎣
⎦ −1
−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
2
−1
(3t 2 − 24t + 36) dt +
6
6
2
(−3t 2 + 24t − 36) dt +
9
6
(3t 2 − 24t + 36) dt
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 ⎞ ⎛
+ sin 2t ⎟ dt = ⎢ t − cos 2t ⎥
= ⎜ + ⎟ −⎜0 − ⎟ =
+1
⎜
0
2⎠ 4
⎝2
⎠
⎣2 2
⎦0
⎝ 4 2⎠ ⎝
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
Section 5.1
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.
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
⎞
⎞
+ sin 2t dt =
⎜ + sin 2t ⎟ dt + 7 π /12 ⎜ − − sin 2t ⎟ dt +
0
0
2
⎝2
⎠
⎝ 2
⎠
7 π /12
⎡1 1
⎤
= ⎢ t − cos 2t ⎥
⎣2 2
⎦0
11π /12
3π / 2
11π /12
⎛1
⎞
⎜ + sin 2t ⎟ dt
⎝2
⎠
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
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 =
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.
A=
6 −2
x dx
1
b. Find c so that
6
1
5
⎡ 1⎤
= ⎢− ⎥ = − + 1 =
6
6
⎣ x ⎦1
c −2
1
x dx =
5
.
12
1
⎡ 1⎤
x dx = ⎢ − ⎥ = 1 −
c
⎣ x ⎦1
1 5
12
1− = , c =
c 12
7
12
The line x =
bisects the area.
7
c −2
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
A( A) =
c
1
5
;
12
d=
x 2 − 4 x + 4 = 8 when x = 2 + 2 2, so the object
34. a.
1
36
and y = 1, so we find d such that
1
1 1
5 1 1
dy = ;
dy = ⎡⎣ 2 y ⎤⎦
d y
d
12 d y
2
2
Slicing the region horizontally, the area is
1
1
5
5
⎛ 1 ⎞
<
dy + ⎜ ⎟ (5) . Since
the
1/ 36 y
36 12
⎝ 36 ⎠
=
0
–3
0
–3
[9 – (– x + 6)]dx +
(3 + x)dx +
3
0
3
0
[9 – ( x + 6)]dx
(3 – x)dx
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
+
=
–2
–3
0
–2
[(– x + 6) – ( x + 6)]dx
(– x 2 – x + 6)dx +
0
–2
(–2 x)dx
−2
0
1
37
⎡ 1
⎤
= ⎢ – x3 – x 2 + 6 x ⎥ + ⎡ – x 2 ⎤ =
⎣
⎦
−2
2
6
⎣ 3
⎦ −3
Instructor’s Resource Manual
Section 5.1
301
© 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.
A(C ) = A( B) =
A( D) =
0
–2
π/6
37
(by symmetry)
6
[( x + 6) – x 2 ]dx +
2
0
[(– 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
A( A + B + C + D) =
3
⎡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
⎝
1 ⎤
⎡
(9 – x 2 )dx = ⎢9 x – x3 ⎥
–3
3 ⎦ −3
⎣
3
= 36
5.2 Concepts Review
36. Let f(x) be the width of region 1 at every x.
ΔA1 ≈ f ( x)Δx, so A1 =
b
a
f ( x)dx .
Let g(x) be the width of region 2 at every x.
ΔA2 ≈ g ( x)Δx, so A2 =
b
a
g ( x)dx .
Since f(x) = g(x) at every x in [a, b],
A1 =
b
a
f ( x)dx =
b
a
g ( x)dx = A2 .
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.
38. Sketch the graph.
1. πr 2 h
2. π( R 2 − r 2 )h
3. πx 4 Δx
4. π[( x 2 + 2)2 − 4]Δx
Problem Set 5.2
1. Slice vertically.
ΔV ≈ π( x 2 + 1) 2 Δx = π( x 4 + 2 x 2 + 1)Δx
V =π
2
0
( x 4 + 2 x 2 + 1)dx
2
2
⎡1
⎤
⎛ 32 16
⎞ 206π
= π ⎢ x5 + x3 + x ⎥ = π ⎜ + + 2 ⎟ =
3
15
⎣5
⎦0
⎝ 5 3
⎠
≈ 43.14
2. Slice vertically.
ΔV ≈ π(− x 2 + 4 x)2 Δx = π( x 4 − 8 x3 + 16 x 2 )Δx
V =π
3
0
( x 4 − 8 x3 + 16 x 2 )dx
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⎞
⎞
⎜ − sin x ⎟ dx + π / 6 ⎜ sin x − ⎟ dx
0
2
2⎠
⎝
⎠
⎝
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
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.
3. a.
Slice vertically.
6.
2 2
2
4
ΔV ≈ π(4 − x ) Δx = π(16 − 8 x + x )Δx
V =π
2
0
(16 – 8 x 2 + x 4 )dx
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
V =π
4
1 ⎤
⎡
(4 – y )dy = π ⎢ 4 y – y 2 ⎥
0
2 ⎦0
⎣
= π(16 − 8) = 8π ≈ 25.13
4
V =π
4. a.
3
0
3
2187π
⎡1 ⎤
x 6 dx = π ⎢ x7 ⎥ =
≈ 981.52
7
7
⎣
⎦0
7.
Slice vertically.
ΔV ≈ π(4 − 2 x)2 Δx
0≤ x≤2
V =π
=
2
⎡ 1
⎤
(4 − 2 x)2 dx = π ⎢ − (4 − 2 x)3 ⎥
0
⎣ 6
⎦0
2
32π
≈ 33.51
3
2
⎛ 1 ⎞
⎛1⎞
ΔV ≈ π ⎜ ⎟ Δx = π ⎜ ⎟ Δx
⎝x⎠
⎝ x2 ⎠
b. Slice vertically.
y
x = 2−
2
V =π
2
4
2
≈ 0.79
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
V =π
=
⎛ x2
ΔV ≈ π ⎜
⎜ π
⎝
V=
4
0
3
2
3
⎡1 ⎤
⎛ 81 16 ⎞
x3 dx = π ⎢ x 4 ⎥ = π ⎜ − ⎟
⎣ 4 ⎦2
⎝ 4 4⎠
65π
≈ 51.05
4
2
⎞
x4
Δx
⎟ Δx =
⎟
π
⎠
4
x4
1 ⎡1 ⎤
1024
dx = ⎢ x5 ⎥ =
≈ 65.19
π
π ⎣ 5 ⎦0
5π
Instructor’s Resource Manual
Section 5.2
303
© 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.
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
⎡
8 ⎞ ⎤ 100π
⎛
= π ⎢(27 − 9) − ⎜ −18 + ⎟ ⎥ =
≈ 104.72
3 ⎠⎦
3
⎝
⎣
10.
=
6
2
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
x
4/3
(
ΔV ≈ π 2 y
27
⎡3
⎤
⎛ 6561 3 ⎞
dx = π ⎢ x 7 / 3 ⎥ = π ⎜
− ⎟
7
7⎠
⎣
⎦1
⎝ 7
6558π
≈ 2943.22
7
V = 4π
4
0
)
2
Δy = 4πy Δy
4
⎡1 ⎤
y dy = 4π ⎢ y 2 ⎥ = 32π ≈ 100.53
⎣ 2 ⎦0
14.
11.
ΔV ≈ π( y 2 ) 2 Δy = πy 4 Δy
V =π
304
3 4
y dy
0
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
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.
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
( x 2 − x 4 )dx = 36π ⎢ x3 − x5 ⎥
0
3
5 ⎦0
⎣
⎛ 1 1 ⎞ 24π
= 36π ⎜ − ⎟ =
≈ 15.08
5
⎝3 5⎠
1
V = 36π
19. Sketch the region.
2
ΔV ≈ π ⎛⎜ 4 − y 2 ⎞⎟ Δy = π(4 − y 2 )Δy
⎝
⎠
V =π
2
1 ⎤
⎡
(4 − y 2 )dy = π ⎢ 4 y − y 3 ⎥
−2
3 ⎦ −2
⎣
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 ⎢⎣⎜⎝
=
Instructor’s Resource Manual
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
Section 5.2
305
© 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.
20. Sketch the region.
V =π
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
⎣
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
=π
−1
1
2
x
V =π
=
4⎛
⎡⎛ 1 4 3 2
⎞
⎢⎜ 256 x − 8 x + 9 ⎟
⎠
⎣⎝
⎛ 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
0
4
1 5⎤
1
⎡
⎞
x
8 − x 4 ⎟ dx = π ⎢8 x −
0 ⎜⎝
32 ⎠
160 ⎥⎦ 0
⎣
32 ⎞ 128π
⎛
= π ⎜ 32 − ⎟ =
≈ 80.42
5 ⎠
5
⎝
=π
4⎛
23.
4
⎡ y 2 y3 ⎤
y y2 ⎞
⎜ −
⎟ 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.
V=
2
⎛ 2 4 − x 2 ⎞ dx = 2 4(4 − x 2 )dx
⎜
⎟
−2 ⎝
−2
⎠
2
2
⎡
x3 ⎤
⎡⎛ 8 ⎞ ⎛
8 ⎞ ⎤ 128
= 4 ⎢ 4 x − ⎥ = 4 ⎢ ⎜ 8 − ⎟ − ⎜ −8 + ⎟ ⎥ =
3
3
3 ⎠⎦
3
⎠ ⎝
⎣⎝
⎣⎢
⎦⎥ −2
≈ 42.67
306
Section 5.2
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.
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
V=
2
−2
4 4 − x 2 dx = 4(2π) = 8π ≈ 25.13
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 .
V=
1
−1
( x8 − 2 x6 + x 4 )dx
1
16
2
1 ⎤
⎡1
= ⎢ x9 − x 7 + x5 ⎥ =
≈ 0.051
7
5 ⎦ −1 315
⎣9
27. The square at x has sides of length 1 − x 2 .
V=
⎛2 ⎞
= 2(π r 2 L) − 8 ⎜ r 3 ⎟
⎝3 ⎠
16
= 2π r 2 L − r 3
3
4 − x 2 dx = 2π .
25. The square at x has sides of length
V=
Thus, the volume inside the “+” for two cylinders
of radius r and length L is
V = vol. of cylinders - vol. of common region
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
⎛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
⎡
x
2
(1 − x 2 )dx = ⎢ x − ⎥ =
≈ 0.67
0
3 ⎦⎥
3
⎣⎢
0
1
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
2
≈ 229.99 in 3
2
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
(r − y 2 )dy = r 2 y − y 2 |0r
0
3
2
3 1 3
= r − r = r3
3
3
Thus, the volume of the “+” is
V = vol. of cylinders - vol. of common region
⎛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
29. Using the result from Problem 28, the volume of
one octant of the common region in the “+” is
r 2
1
(r − y 2 )dy = r 2 y − y 2 |0r
0
3
1
2
= r3 − r3 = r3
3
3
Instructor’s Resource Manual
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 4
V=
( x − 2 x5 / 2 + x)dx
8 0
=
1
=
π ⎡1 5 4 7 / 2 1 2 ⎤
9π
x − x
+ x ⎥ =
≈ 0.050
⎢
8 ⎣5
7
2 ⎦ 0 560
Section 5.2
307
© 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.
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 .
V =π
a.
Revolving about the line x = 4, the radius of
=π
the disk at y is 4 − 3 y 2 = 4 − y 2 / 3 .
V =π
=π
8
0
8
0
8
b. Revolving about the line y = 8, the inner
radius of the disk at x is 8 − x3 = 8 − x3 / 2 .
=π
4
0
0
(64 − 16 x3 / 2 + x3 )dx
4
(16 − 8 y 2 / 3 + y 4 / 3 )dy
4
(8 − x3 / 2 )2 dx
32
1 ⎤
⎡
= π ⎢64 x − x5 / 2 + x 4 ⎥
5
4 ⎦0
⎣
1024
⎡
⎤ 576π
= π ⎢ 256 −
+ 64 ⎥ =
≈ 361.91
5
5
⎣
⎦
(4 − y 2 / 3 ) 2 dy
24 5 / 3 3 7 / 3 ⎤
⎡
= π ⎢16 y −
y
+ y ⎥
5
7
⎣
⎦0
768 384 ⎞
⎛
= π ⎜ 128 −
+
⎟
5
7 ⎠
⎝
1024π
=
≈ 91.91
35
V =π
4
0
4
0
⎡82 − (8 − x3 / 2 )2 ⎤ dx
⎣
⎦
(16 x3 / 2 − x3 )dx
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
4 − y 2 dy +
2
0
(4 − y 2 )dy
2
1 ⎤
⎡
⎛ 8⎞
= 2 π + ⎢ 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 .
V =π
=π
308
8
0
8⎡ 2
4
0 ⎢⎣
(
− 4− y
) ⎥⎦ dy
2/3 2 ⎤
(8 y 2 / 3 − y 4 / 3 )dy
Section 5.2
at x is a rectangle with base of length 2 r 2 − x 2
and height x tan .
V=
r
0
2 x tan
r 2 − x 2 dx
r
⎡ 2
⎤
= ⎢ − tan (r 2 − x 2 )3 / 2 ⎥
⎣ 3
⎦0
2
= r 3 tan
3
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.
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
=
2π
h
0
2
⎛ 1 ⎞
h = r2 − ⎜
r⎟ =
⎝ 3 ⎠
h
y
π ⎡ 2 3/ 2 ⎤
dy =
y ⎥
k
k ⎢⎣ 3
⎦0
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
⎛ z⎞
ΔV ≈ Az Δz = A ⎜1 − ⎟ Δz V = A
⎝ h⎠
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,
V=
2 2
r =
3
h⎛
2
z⎞
1 − ⎟ dz
⎜
0⎝
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
Instructor’s Resource Manual
Section 5.2
309
© 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.
5.3 Concepts Review
3. a, b.
1. 2πx f ( x)Δx
2. 2π
3. 2π
4. 2π
2
0
2
0
2
0
x 2 dx; π
2
0
(4 − y 2 )dy
(1 + x) x dx
(1 + y )(2 − y ) dy
c. ΔV ≈ 2πx x Δx = 2πx3 / 2 Δx
d, e. V = 2π
Problem Set 5.3
=
1. a, b.
3 3/ 2
x dx
0
3
⎡2
⎤
= 2π ⎢ x 5 / 2 ⎥
⎣5
⎦0
36 3
π ≈ 39.18
5
4. a,b.
⎛1⎞
c. ΔV ≈ 2πx ⎜ ⎟ Δx = 2πΔx
⎝x⎠
d,e. V = 2π
4
1
c. ΔV ≈ 2πx(9 − x 2 )Δx = 2π(9 x − x3 )Δx
dx = 2π [ x ]1 = 6π ≈ 18.85
4
3
1 ⎤
⎡9
(9 x − x3 )dx = 2π ⎢ x 2 − x 4 ⎥
0
4 ⎦0
⎣2
81
81
81
π
⎛
⎞
= 2π ⎜ − ⎟ =
≈ 127.23
2
⎝ 2 4⎠
d, e. V = 2π
2. a, b.
3
5. a, b.
c. ΔV ≈ 2πx ( x 2 )Δx = 2πx3Δx
d, e. V = 2π
1 3
x dx
0
1
π
⎡1 ⎤
= 2π ⎢ x 4 ⎥ = ≈ 1.57
⎣ 4 ⎦0 2
c.
ΔV ≈ 2π(5 − x) x Δx
= 2π(5 x1/ 2 − x3 / 2 )Δx
d, e. V = 2π
5
0
(5 x1/ 2 − x3 / 2 )dx
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
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.
6. a, b.
c. ΔV ≈ 2πx (3 x − x 2 )Δx = 2π(3 x 2 − x3 )Δx
3
1 ⎤
⎡
(3 x 2 − x3 )dx = 2π ⎢ x3 − x 4 ⎥
0
4 ⎦0
⎣
81 ⎞ 27π
⎛
= 2π ⎜ 27 − ⎟ =
≈ 42.41
4⎠
2
⎝
d, e. V = 2π
3
9. a, b.
c. ΔV ≈ 2π(3 − x)(9 − x 2 )Δx
= 2π(27 − 9 x − 3 x 2 + x3 )Δx
d, e. V = 2π
3
0
(27 − 9 x − 3 x 2 + x3 )dx
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.
d, e. V = 2π
1
0
1
π
⎡1 ⎤
y 3 dy = 2π ⎢ y 4 ⎥ = ≈ 1.57
⎣ 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
⎠
d, e. V = 2π
1⎛1 4
x
0 ⎜⎝ 4
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
d, e. V = 2π
4
0
(
)
y + 1 Δ y = 2 π( y 3 / 2 + y ) Δ y
( y 3 / 2 + y ) dy
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.
Instructor’s Resource Manual
Section 5.3
311
© 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.
c. ΔV ≈ 2π(2 − y ) y 2 Δy = 2π(2 y 2 − y 3 )Δy
2
1 ⎤
⎡2
(2 y 2 − y 3 )dy = 2π ⎢ y 3 − y 4 ⎥
0
4 ⎦0
⎣3
⎛ 16
⎞ 8π
= 2π ⎜ − 4 ⎟ =
≈ 8.38
3
⎝
⎠ 3
d, e. V = 2π
2
12. a, b.
c. ΔV ≈ 2π(3 − y )
(
(
)
2 y + 1 Δy
V = 2π
2
0
(3 + 3
V = 2π
c.
V =π
d.
V = 2π
a.
A=
b.
V = 2π
c.
V =π
)
2 y1/ 2 − y − 2 y 3 / 2 dy
2
16 ⎞ 88π
⎛
= 2π ⎜ 6 + 8 − 2 − ⎟ =
≈ 55.29
5⎠
5
⎝
13. a.
π
b
2π
c.
2π
d.
2π
14. a.
b.
c.
π
b
a
b
a
b
a
d
2π
3
1
1
x2
dx
2
⎡⎛ 1
⎤
⎞
⎢⎜ + 1⎟ − (−1)2 ⎥ dx
1 ⎢⎝ x 3
⎥⎦
⎠
⎣
3⎛ 1
2 ⎞
=π ⎜
+ ⎟ dx
1 ⎝ x6
x3 ⎠
3
⎛ 1 ⎞
(4 − x) ⎜ ⎟ dx
⎝ x3 ⎠
3⎛ 4
1 ⎞
= 2π ⎜ − ⎟ dx
1 ⎝ x3 x 2 ⎠
3
1
16.
x [ f ( x) − g ( x) ] dx
( x − a ) [ f ( x) − g ( x) ] dx
(b − x) [ f ( x) − g ( x) ] dx
d
c
d
c
y [ f ( y ) − g ( y ) ] dy
=π
(3 − y ) [ f ( y ) − g ( y ) ] dy
d.
2
Section 5.3
( x3 + 1) dx
0
2
0
2
0
2
0
x( x3 + 1)dx = 2π
2
0
( x 4 + x)dx
⎡ ( x3 + 2)2 − (−1)2 ⎤
⎣
⎦
( x 6 + 4 x3 + 3)dx
V = 2π
= 2π
312
⎛ 1 ⎞
x ⎜ ⎟ dx = 2π
⎝ x3 ⎠
2⎤
⎡ f ( y ) 2 − g ( y ) 2 ⎤ dy
⎣
⎦
c
2π
3
1
dx
⎡ f ( x ) − g ( x) dx
⎣
⎦
2
a
b.
x3
b.
)
⎡
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.
2
0
2
0
(4 − x)( x3 + 1)dx
(− x 4 + 4 x3 − x + 4)dx
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.
17. To find the intersection point, solve
y=
y=
y3
.
32
π
4
π
x=
2
x2 =
6
y − 1024 y = 0
y ( y 5 − 1024) = 0
y = 0, 4
4 ⎛
y3 ⎞
V = 2π y ⎜ y − ⎟ dy
0 ⎜
32 ⎟⎠
⎝
4
0
sin( x 2 ) = cos( x 2 ) .
tan( x 2 ) = 1
y6
1024
= 2π
21. To find the intersection point, solve
= 2π
⎛ 3/ 2 y ⎞
−
⎜y
⎟ dy
⎜
32 ⎟⎠
⎝
4
4
= 2π
= 2π
23. a.
20. y = ± a 2 − x 2 , − a ≤ x ≤ a
−a
a 2 − x 2 dx − 4π
⎞ dx
⎟
⎠
a
−a
x sin x dx
2π
The curves intersect when x = 0 and x = 1.
1
0
[ x 2 − ( x 2 )2 ] dx = π
1
0
( x 2 − x 4 )dx
1
b.
V = 2π
1
0
x( x − x 2 )dx = 2π
1
0
( 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
a 2 − x 2 dx = πa 2 .)
−a
2
Instructor’s Resource Manual
2π
0
1 ⎤
⎡1
⎛ 1 1 ⎞ 2π
= π ⎢ x3 − x5 ⎥ = π ⎜ − ⎟ =
≈ 0.42
5 ⎦0
⎣3
⎝ 3 5 ⎠ 15
b
b
a
2 x dx + 2π
V =π
2
⎡ 1
⎤
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
= 4πb
0
(2 x + x sin x)dx
= 2π(4π2 ) + 2π(−2π) = 4π 2 (2π − 1) ≈ 208.57
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 ⎝
⎠
(b − x) ⎛⎜ 2 a 2 − x 2
⎝
2π
)
2 − 1 ≈ 1.30
x(2 + sin x)dx
0
0
2π
(
= 2π ⎡ x 2 ⎤ + 2π [sin x − x cos x ]0
⎣ ⎦0
19. Let R be the region bounded by y = b − x ,
a
2π
π/2
2π
4
⎡8
2
y 4 y5 ⎤
= 2π ⎢ y 3 / 2 − y 5 / 2 −
+
⎥
5
32 160 ⎥⎦
⎣⎢ 3
0
32 ⎞ 208π
⎛ 64 64
= 2π ⎜ − − 8 + ⎟ =
≈ 43.56
5
5 ⎠
15
⎝ 3
2
⎡ x cos( x 2 ) − x sin( x 2 ) ⎤ dx
⎣
⎦
0
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 ⎟⎠
⎝
−a
π /2
x ⎡ cos( x 2 ) − sin( x 2 ) ⎤ dx
⎣
⎦
⎡⎛ 1
1 ⎞ 1⎤
= 2π ⎢⎜
+
⎟− ⎥ = π
⎣⎝ 2 2 2 2 ⎠ 2 ⎦
4
V = 2π
0
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
= 4π
π/2
V = 2π
x 2 + x − 2a = 0
−1 ± 1 + 8a
2
−1 + 1 + 8a
x=
2
x=
Section 5.3
313
© 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.
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
V = 4π
2
V=
2
⎡ 1 + 4a − 1 + 8a ⎤
+ ⎢a −
⎥
2
⎣
⎦
⎡ 2a + 1 − 1 + 8a ⎤
=⎢
⎥
2
⎣
⎦
r2
S Δx
S
r
r2
0
x 2 dx =
r
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 ≈
r
4
⎡1 ⎤
dx = 4π ⎢ x3 ⎥ = πr 3
⎣ 3 ⎦0 3
r 2
x
0
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
b
[f
a
(t )] + [ g (t ) ] dt
2
2
4. Mean Value Theorem (for derivatives)
2
ΔV ≈ πr Δa
V =π
1
(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
−π
1
0
1.
L=
2a 1 + 8a da
=
1
0
5π
−π
2
0
To integrate
2a 1 + 8a da
1
0
2a 1 + 8a da , use the
substitution u = 1 + 8a.
1
9 1
1
2a 1 + 8a da =
(u − 1) u du
0
1 4
8
1 9 3 / 2 1/ 2
=
(u
− u )du
32 1
5
1 + (6 x1/ 2 ) 2 dx =
1/ 3
1 + 36 x dx
5
(
2a 1 + 8a da
1
5
1/ 3
⎡1 2
⎤
= ⎢ ⋅ (1 + 36 x)3 / 2 ⎥
⎣ 36 3
⎦1/ 3
1
=
181 181 − 13 13 ≈ 44.23
54
⎡⎛ 2
9 ⎞ ⎛ 1 ⎞⎤
= π ⎢⎜ + 3 + 1 − ⎟ − ⎜ − ⎟ ⎥
4 ⎠ ⎝ 12 ⎠ ⎦
⎣⎝ 3
−π
f ( x) = 4 x3 / 2 , f ( x ) = 6 x1/ 2
2.
f ( x) =
L=
=
2
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
⎣
⎦
4 x 4 + 4 x 2 + 1 dx =
2
1
(2 x 2 + 1)dx
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
© 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.
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
8
1 + ⎡ − x −1/ 3 (4 − x 2 / 3 )1/ 2 ⎤ dx
⎣
⎦
1
8
4x
1
−2 / 3
dx =
8
1
2x
−1/ 3
L=
⎡3
⎤
= 2 ⎢ x 2 / 3 ⎥ = 3(4 − 1) = 9
2
⎣
⎦1
=
L=
=
1
2
1
3
1
⎛ x2
1
+
⎜
⎜ 2 2 x2
⎝
1
2
⎛ y4
3 ⎞
+
⎜
⎟ dy
⎜ 6 2 y4 ⎟
⎝
⎠
3
1
3
3
2
dx 2 dy
=t ,
=t
dt
dt
L=
y4
1
y3 1
5. g ( y ) =
+
, g ( y) =
−
16 2 y 2
4 y3
=
y8 1
9
+ +
dy =
36 2 4 y8
3
3
3
L=
2
⎞
⎟ dy
⎟
⎠
⎞
⎟ dx
⎟
⎠
⎛ x2
⎡ x3 1 ⎤
1 ⎞
dx = ⎢ − ⎥
+
⎜
⎟
1 ⎜ 2
2 x 2 ⎟⎠
⎝
⎣⎢ 6 2 x ⎦⎥1
⎛ 9 1 ⎞ ⎛ 1 1 ⎞ 14
= ⎜ − ⎟−⎜ − ⎟ =
≈ 4.67
⎝2 6⎠ ⎝6 2⎠ 3
=
⎛ y4
3
1+ ⎜
−
⎜ 6 2 y4
⎝
7.
⎞
⎟ dx
⎟
⎠
x4 1
1
+ +
dx =
4 2 4 x4
3
=
x2
1
−
2 2 x2
⎛ x2
1
1+ ⎜
−
⎜ 2 2 x2
⎝
3
1
y5
1
y4
3
+
, g ( y) =
−
3
30 2 y
6 2 y4
⎛ y4
⎡ y5
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) =
3
dx
8
4.
y5
1
+
30 2 y 3
6. x =
−2
−3
−2
−3
2
−2
−3
(t 2 )2 + (t )2 dt =
1
0
t 4 + t 2 dt
1
(
)
1
⎡1
⎤
t t 2 + 1 dt = ⎢ (t 2 + 1)3 / 2 ⎥ = 2 2 − 1
0
⎣3
⎦0 3
≈ 0.61
=
⎛ y3 1 ⎞
1+ ⎜
−
⎟ dy
⎜ 4 y3 ⎟
⎝
⎠
y6 1 1
+ +
dy =
16 2 y 6
1
0
2
⎛ y3 1 ⎞
+
⎜
⎟ dy
⎜ 4 y3 ⎟
⎝
⎠
1
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
(
4
1
(6t ) 2 + (6t 2 ) 2 dt =
4
1
36t 2 + 36t 4 dt
4
6t 1 + t 2 dt = ⎡ 2(1 + t 2 )3 / 2 ⎤
⎣
⎦1
)
= 2 17 17 − 2 2 ≈ 134.53
Instructor’s Resource Manual
Section 5.4
315
© 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.
9.
12. x = y +
3
2
3
g ( y) = y + , g ( y) = 1
2
L=
3
1
1 + (1)2 = 2
3
1
dy = 2 2
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
16 cos 2 t + 16sin 2 t dt =
0
2
⎛9 5⎞
d = ⎜ − ⎟ + (3 − 1) 2 = 8 = 2 2
⎝2 2⎠
(4 cos t ) 2 + (−4sin t )2 dt
π
0
4dt
13.
= 4π ≈ 12.57
dx
dy
= 1,
= 2t
dt
dt
10.
L=
−
−2
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
=
0
(2
5 cos 2t
) + ( −2
2
5 sin 2t
20 cos 2 2t + 20sin 2 2t dt =
5π
=
≈ 3.51
2
11.
3
1
1 + (2)2 dx = 5
3
1
dx = 2 5
At x = 1, y = 2(1) + 3 = 5.
At x = 3, y = 2(3) + 3 = 9.
2
)
2
π/ 4
0
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=
1 + 4t 2 dt
with n = 8,
−
π/4
0
Let f (t ) = 1 + 4t 2 . Using the Parabolic Rule
−2
0
π/ 4
2
12 + (2t )2 dt =
−1
−1
L=
2
0
2
d = (3 − 1) + (9 − 5) = 20 = 2 5
dx
dy
1
= 2t ,
=
dt
dt 2 t
L≈
4
1
2
⎛ 1 ⎞
(2t ) 2 + ⎜
⎟ dt =
⎝2 t ⎠
Let f (t ) = 4t 2 +
4
1
4t 2 +
1
dt
4t
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
cos 2 t + 4sin 2 2t dt
⎛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
L≈
16.
π 2−0 ⎡
dx
dy
= 1,
= sec 2 t
dt
dt
L=
π 4
0
π 4
0
π 4−0 ⎡
⎛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 ⎠⎦
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
π/ 2
=
3a
2
π/ 2
⎡1
⎤
3a ⎢ sin 2 t ⎥
with the same result.)
⎣2
⎦0
The total length is 6a.
18. a.
( )
OT = length PT = a
b. From Figure 18 of the text,
PQ PQ
QC QC
sin =
=
and cos =
=
.
a
a
PC
PC
Therefore PQ = a sin and QC = a cos .
x = OT − PQ = a − a sin = a ( − sin )
y = CT − CQ = a − a cos = a (1 − cos )
19. From Problem 18,
x = a( − sin ), y = a (1 − cos )
dx
dy
= a(1 − cos ),
= a sin
d
d
2
so
2
⎛ dx ⎞ ⎛ dy ⎞
2
⎜
⎟ +⎜
⎟ = [ a (1 − cos ) ] + [ a sin
⎝d ⎠ ⎝d ⎠
]2
= a 2 − 2a 2 cos + a 2 cos 2 + a 2 sin 2
= 2a 2 − 2a 2 cos = 2a 2 (1 − cos )
1 − cos
⎛ ⎞
= 4a 2 sin 2 ⎜ ⎟ .
2
⎝2⎠
The length of one arch of the cycloid is
= 4a 2
2π
0
17.
9a 2 cos 2 t sin 2 t (sin 2 t + cos 2 t ) dt
π/ 2
1 + sec 4 t dt
Let f (t ) = 1 + sec4 t . Using the Parabolic
≈
0
c.
12 + (sec 2 t )2 dt =
Rule with n = 8, L ≈
π/ 2
9a 2 cos 2 t sin 4 t + 9a 2 sin 2 t cos 4 t dt
⎡ 1
⎤
3a cos t sin tdt = 3a ⎢ − cos 2 t ⎥
0
2
⎣
⎦0
(The integral can also be evaluated as
=
Let f (t ) = cos 2 t + 4sin 2 2t . Using the
Parabolic Rule with n = 8,
π/ 2
0
⎛ ⎞
4a 2 sin 2 ⎜ ⎟ d =
⎝2⎠
2π
0
⎛ ⎞
2a sin ⎜ ⎟ d
⎝2⎠
2π
⎡
⎤
= 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
Instructor’s Resource Manual
Section 5.4
317
© 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.
20. a.
= t , the point P is at x = a t − a sin( t ), y = a − a cos( t ) at time t.
Using
dx
= a − a cos( t ) = a (1 − cos( t ))
dt
dy
= a sin( t )
dt
2
ds
⎡ dy ⎤ ⎡ dx ⎤
= ⎢ ⎥ +⎢ ⎥
dt
⎣ dt ⎦ ⎣ dt ⎦
2
= a2
2
sin 2 ( t ) + a 2
2
− 2a 2
= 2a
1
(1 − cos( t )) = 2a
2
2
cos( t ) + a 2
sin 2
t
= 2a
2
2
sin
cos 2 ( t ) = 2a 2
2
− 2a 2
2
cos( t )
t
2
π
t
= 1, which occurs when t = (2k + 1). The speed is a minimum when
2
2k π
t
.
sin
= 0, which occurs when t =
2
b. The speed is a maximum when sin
c.
21. a.
From Problem 18a, the distance traveled by the wheel is a , so at time t, the wheel has gone a = a t miles.
Since the car is going 60 miles per hour, the wheel has gone 60t miles at time t. Thus, a = 60 and the
maximum speed of the bug on the wheel is 2a = 2(60) = 120 miles per hour.
dy
= x3 − 1
dx
L=
2
1 + x3 − 1 dx =
1
2
2 3/ 2
x dx
1
(
)
2
⎡2
⎤
= ⎢ x5 / 2 ⎥ = 4 2 − 1 ≈ 1.86
⎣5
⎦1 5
b.
L=
0
2 − 2 cos t dt =
4π
0
L=
π/6
2
23.
318
Section 5.4
0
at dt −
0
−1
at dt
0
f ( x ) = 6 x, f ( x ) = 6
1
1
0
0
A = 2π 6 x 1 + 36 dx = 12 37 π
x dx
1
24.
π/3
x
f ( x) = 25 − x 2 , f ( x) = −
A = 2π
4
1 + 64sin cos x − 1 dx
π/3
1
⎡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
=
at dt =
1
2π
dy
= 64sin 2 x cos 4 x − 1
dx
π/3
−1
a 2 t 2 cos 2 t + a 2t 2 sin 2 tdt
a a
⎡a ⎤ ⎡a ⎤
= ⎢ t2 ⎥ − ⎢ t2 ⎥ = + = a
⎣ 2 ⎦ 0 ⎣ 2 ⎦ −1 2 2
⎛t⎞
2 sin ⎜ ⎟ dt
⎝2⎠
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
L=
f (t ) = 1 − cos t , g (t ) = sin t
4π
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
b.
= 2π
= 2π
3
−2
3
−2
3
−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
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.
25.
f ( x) =
A = 2π
x3
, f ( x) = x 2
3
7
1
3
x
3
26.
f '( x ) = − x(r 2 − x 2 )−1/ 2
1 + x 4 dx
⎡1
⎤
= 2π ⎢ (1 + x 4 )3 / 2 ⎥
18
⎣
⎦1
=
29. y = f ( x) = r 2 − x 2
7
π
= 250 2 − 2 2
9
(
)
f ( x) =
A = 2π
x6 + 2
8x
3
1
2
=
x4
1
x3
1
+
, f ( x) =
−
8 4 x2
2 2 x3
⎛ x4
⎛ x3
1 ⎞
1 ⎞
+
1+ ⎜ −
⎜
⎟
⎟
⎜ 8 4x2 ⎟
⎜ 2 2 x3 ⎟
⎝
⎠
⎝
⎠
2
3
r
−r
r
= 2π
−r
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 2 dx = 2π
r
−r
rdx = 2π rx |−r r = 4π r 2
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
= 2π
= 2π
= 2π
3
π
0
π
0
0
r sin t (−r sin t ) 2 + (r cos t )2 dt
r sin t r 2 sin 2 t + r 2 cos 2 tdt
r sin t r 2 dt
π 2
0
2
r sin tdt = −2π r 2 cos t |π0
= −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 = l. Therefore,
2πr
=
.
l
b. The area of the sector is equal to the lateral
surface area. Therefore, the lateral surface
1
1 ⎛ 2πr ⎞
area is l 2 = 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
= 2π
⎡ x8 3 x 2
1 ⎤
= 2π ⎢
+
−
⎥
4
⎣⎢128 32 32 x ⎦⎥1
27.
r
−r
r
= 2π
⎛ x4
1 ⎞ x6 1
1
+
+ +
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 ⎟
⎝
⎠
= 2π
−r
= 2π
248π 2
≈ 122.43
9
2
r
A = 2π
(
)
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,
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 ⎦
Instructor’s Resource Manual
Section 5.4
319
© 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.
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
a − ( x − a)2
A = 2π
= 2π
= 2π
b.
2
b+h
a 2 − ( x − a)2 1 +
b
b+h
b
b+ h
( x − a)2
a 2 − ( x − a)2
dx
a 2 − ( x − a )2 + ( x − a) 2 dx
a dx = 2πa[ x ]bb + h = 2 π ah
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
b.
2π
0
(1 − cos t )3 / 2 dt
e.
⎛t⎞
1 − cos t = 2sin 2 ⎜ ⎟ , so
⎝2⎠
⎛t⎞
sin 3 ⎜ ⎟ dt
⎝2⎠
2π
t
t
⎛ ⎞
⎛ ⎞
sin ⎜ ⎟ sin 2 ⎜ ⎟ dt
0
⎝2⎠
⎝2⎠
2π
⎛ t ⎞⎡
⎛ t ⎞⎤
sin ⎜ ⎟ ⎢1 − cos 2 ⎜ ⎟ ⎥ dt
0
⎝ 2⎠⎣
⎝ 2 ⎠⎦
A = 2 2πa 2
= 8πa 2
= 8πa 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
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.
f (t ) = −3sin t , g (t ) = 3cos t
36. a.
L=
2π
0
2π
=
0
9sin 2 t + 9 cos 2 tdt
3dt = 3[t ]02 π = 6π ≈ 18.850
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
2
2
5.5 Concepts Review
1. F ⋅ (b − a);
b
a
F ( x) dx
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.
W=
2
0
2
12 x dx = ⎡6 x 2 ⎤ = 24 ft-lb
⎣
⎦0
3. F(0.01) = 0.6; k = 60
F(x) = 60x
W=
2
sin t + π cos πt dt ≈ 86.58
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 =
y = x 4 , y = 4 x3 , L =
10
1
1 + 4 x 2 dx ≈ 1.47894
0
1
0
1 + 16 x6 dx ≈ 1.60023
9
y = x , y = 10 x ,
L=
1
y=x
L=
1 + 10018 dx ≈ 1.75441
0
100
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
© 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.
11. A slab of thickness Δy at height y has width
d
⎡1
⎤
kxdx = ⎢ kx 2 ⎥
0
2
⎣
⎦0
1
1
= k (d 2 − 0) = kd 2
2
2
5. W =
d
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=
7. W =
27
2
⎡1 ⎤
9 s ds = 9 ⎢ s 2 ⎥ = 18 ft-lb
0
⎣ 2 ⎦0
2
3
1
3
1
6 s ds = ⎡3s 2 ⎤ + ⎡3s 2 ⎤
⎣
⎦2 ⎣
⎦2
2
2
= 3(9 – 4) + 3(1 – 4) = 6 ft-lb
6 s ds +
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 ⎠
⎝
5
0
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
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
W=
= 20δ
3
0
20δ 6 y − y 2 (8 − y )dy
3
0
6 y − y 2 (3 − y ) dy
+20δ
3
0
6 y − y 2 (5)dy
3
⎡1
⎤
= 20δ ⎢ (6 y − y 2 )3 / 2 ⎥ +100δ
⎣3
⎦0
= 8δ ( y 2 − 15 y + 50)Δy
W=
4
=
=
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.
W=
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
8δ ( y 2 − 15 y + 50) dy
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
3
0
6 y − y 2 dy
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
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.
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
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
A = 1; p(v) = cv −1.4
f ( x) = cx −1.4
W=
21.
16
2
= 16, x2 = = 2
1
1
16
cx −1.4 dx
2
1.4
= 40(16) (−2.5)(16
≈ 2075.83 in.-lb
40002
−2
W=
)
f ( x) = c(2 x) −1.4
1
c(2 x)
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
; f (4000) = 5000
= 5000 , k = 80,000,000,000
4200 80, 000, 000, 000
4000
22. F ( x) =
16
2
x1 =
= 8, x2 = = 1
2
2
W =2
x2
x2
dx
4200
A = 2; p (v) = cv −1.4
−1.4
k
⎡ 1⎤
= 80, 000, 000, 000 ⎢ − ⎥
⎣ x ⎦ 4000
20, 000, 000
=
≈ 952,381 mi-lb
21
17. c = 40(16)1.4
8
f ( x) =
k
16
= c ⎡ −2.5 x −0.4 ⎤
⎣
⎦2
−0.4
−0.4
10
1 ⎤
⎡
(10 − y )dy = ⎢10 y − y 2 ⎥
0
2 ⎦0
⎣
= 100 – 50 = 50 ft-lb
W = W1 + W2 = 250 ft-lb
10
W2 =
f ( x)dx .
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
500
W2 =
)
k
x2
where x is the distance between the
charges. F (2) = 10;
W=
k
= 10, k = 40
4
5
⎡ 40 ⎤
dx = ⎢ − ⎥ = 32 ergs
1 x2
⎣ x ⎦1
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
W=
4
4
11,520v −1.4 dv = ⎡ −28,800v −0.4 ⎤
⎣
⎦1
1
= −28,800(4−0.4 − 1−0.4 ) ≈ 12,259 ft-lb
Instructor’s Resource Manual
Section 5.5
323
© 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.
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
3 ⎞
3 ⎤
⎡
600 − y ⎟ dy = ⎢600 y − y 2 ⎥
0 ⎜⎝
2 ⎠
4 ⎦0
⎣
= 48,000 – 4800 = 43,200 ft-lb
W=
80 ⎛
24. The total work is equal to the work W1 needed to
fill the pipe plus the work W2 needed to fill the
tank.
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,
F=
3
0
⎣⎢
= 6 ⋅ 62.4 ⋅10.5 = 3931.2 pounds
W1 =
30 δπy
0
4
dy =
( 62.4 ) π ⎡ 1
4
(
0
−3
⎞⎞
+ 3 ⎟ ⎟ dy
⎠⎠
⎝ ⎝ 3
δ (− y ) ⎜ 2 ⎜
⎛ y2
⎞
+ 3 y ⎟ dy
⎜
⎜
⎟
3
⎝ 3
⎠
0
r = 102 − (40 − y )2 = − y 2 + 80 y − 1500 .
= 1684.8 pounds
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
0
⎡
δ (3 − y )(6) dy = 6δ ⎢3 y −
⎢⎣
= 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
F=
+ 3.
3
⎛ ⎛ y
0
−3 3
(30 ≤ y ≤ 50) is π r 2 where
50
y
⎛ ⎛ y
⎞⎞
ΔF ≈ δ (− y ) ⎜ 2 ⎜
+ 3 ⎟ ⎟ Δy and
⎠⎠
⎝ ⎝ 3
⎡ y3 3 y 2 ⎤
= −2δ ⎢
+
⎥
2 ⎥⎦
⎢⎣ 3 3
−3
W2 =
324
)
y= 3 ⋅ x − 3 3 or x =
= −2δ
ΔW2 = δπr 2 Δy y = δπ(− y3 + 80 y 2 − 1500 y )Δy
0
The equation of the line in Quadrant I is
30
≈ 22, 054 ft-lb
The cross sectional area at height y feet
3
(−3, 0), (3, 0) and 0, −3 3 .
F=
2⎤
⎢2 y ⎥
⎣
⎦0
y2 ⎤
⎥
2 ⎦⎥
27. Place the equilateral triangle in the coordinate
system such that the vertices are
2
δπy
⎛1⎞
ΔW1 = δπ ⎜ ⎟ Δy ( y ) =
Δy
2
4
⎝ ⎠
⎡
δ (5 − y )(6) dy = 6δ ⎢5 y −
2 ⎤3
y
⎥
2 ⎥⎦
0
⎡
3 y 2 y3 ⎤
= δ ⎢9 y −
− ⎥ = 62.4 ⋅ 26
8
4 ⎥⎦
⎣⎢
−4
= 1622.4 pounds
29. ΔF ≈ δ (1 − y )
=δ
1
0
( y ) Δy; F =
( y1 2 − y3 2 ) dy
1
0
δ (1 − y )
( y ) dy
1
2
⎡2
⎤
⎛ 4⎞
= δ ⎢ y3 2 − y5 2 ⎥ = 62.4 ⎜ ⎟
5
⎣3
⎦0
⎝ 15 ⎠
= 16.64 pounds
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.
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 ⎠
⎛
0 ⎝
b
δ ⎜ ab − 2ay +
x
From the diagram, we let h = the depth of an
arbitrary slice along the width of the bottom of
the pool.
20
Using the Pythagorean Theorem, we can find that
the length of the bottom of the pool is
b
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.
⎡ ay 2 ay 3 ⎤
−
⎢
⎥
3b ⎥⎦
⎢⎣ 2
0
h−4
b
4
x
4 26
a 2⎞
y ⎟ dy
b ⎠
2
⎡
⎛ 2
ay 3 ⎤
2 ab
= δ ⎢ aby − ay 2 +
⎥ = δ ⎜⎜ ab − ab +
3b ⎦⎥
3
⎣⎢
⎝
0
⎞
⎟
⎟
⎠
2
ab
3
The total force on one half of the dam is twice the
=δ
10
4
⎛ 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
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).
ΔF ≈ δ (102 − y )(2)Δy; F =
2 ⎤2
2
0
2δ (102 − y ) dy
⎡
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
© 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.
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
W1 =
4
0
δ 40(10 − y )dy
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 ) − 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
© 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.
6. M y = (−3) ⋅ 5 + (−2) ⋅ 6 + 3 ⋅ 2 + 4 ⋅ 7 + 7 ⋅1 = 14
5.6 Concepts Review
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
1. right;
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 =
0
7
0
7
4. x =
0
7
0
(
=
49
2
x x dx
=
x dx
7
⎡ 2 x3 / 2 ⎤
⎣3
⎦0
x(1 + x3 )dx
(1 + x3 )dx
+ 16,807
5
(7 +
7
⎡ 2 x5 / 2 ⎤
⎣5
⎦0
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
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
f ( x) = 2 − x; g ( x) = 0
2
x=
x[(2 − x) − 0]dx
0
2
0
2
=
0
[(2 − x) − 0]dx
[2 x − x 2 ]dx
2
0
Instructor’s Resource Manual
[2 − x]dx
Section 5.6
327
© 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.
10.
2
⎛ 2 1 3⎞
⎜x − x ⎟
3 ⎠0
⎝
8
3
=
=
2
4
−
2
1 2⎞
⎛
⎜ 2x − x ⎟
2 ⎠0
⎝
2
=
3
1
y= 2
2
2
=
0
[(2 − x )2 − 02 ]dx
0
2
0
4−
[(2 − x ) − 0]dx
4
x=
[4 − 4 x + x 2 ]dx
4
2
⎛
2 1 3⎞
8
⎜ 4x − 2x + x ⎟
3 ⎠0 8 − 8 + 3
⎝
=
=
4
4
2
=
3
=
x
( 13 x2 ) dx = 13 04 x3dx
4 1 2
x dx
0 3
4
1 ⎡ 1 x4 ⎤
64
3 ⎣4
⎦0
= 3
4
64
1 ⎡ 1 x3 ⎤
9
3 ⎣3
⎦0
y=
9.
=
0
( )
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
2
=
1
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
Section 5.6
0
1 3
x dx
0
2
1
y= 2
2
+ 15 x5 ⎤
⎦−
8 2
3
x=
2
2
=
32 2
15
8 2
3
4
=
5
=
x( x3 )dx
1
14
1
4
=
1 3 2
( x ) dx
0
1 3
x dx
0
=
1 4
x
0
dx
1
⎡ 1 x4 ⎤
⎣ 4 ⎦0
1
= 2
1
=
⎡ 1 x5 ⎤
⎣ 5 ⎦0
1
4
x dx
1
4
1
5
1
4
=
4
5
1
1 6
0
=
=
⎡ 1 x7 ⎤
⎣ 14 ⎦ 0
1
4
2
7
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.
12.
y=
(
⎡
⎢⎣ 2 x
1 4
2 1
4
=
4
1
⎤
− (2 x − 4)2 ⎥ dx
⎦
2
⎡ 2 x − ( 2 x − 4 ) ⎤ dx
⎣
⎦
1
2
)
(− x 2 + 5 x − 4)dx
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
(
=
2
− 18
−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 .
13.
x2 − x − 3 = 0
1 ± 13
2
x=
(1+ 13 )
x( x + 3 − x 2 )dx
2
(1− 13 )
x=
2
(1+ 13 )
2
(1− 13 )
To find the intersection point, solve
2x − 4 = 2 x .
x−2 = x
x2 − 4 x + 4 = x
4
x=
=
=
1
4
2
(1+ 13 )
( x 2 + 3x − x3 )dx
2
(1− 13 )
2
=
x2 − 5x + 4 = 0
(x – 4)(x – 1) = 0
x = 4 (x = 1 is extraneous.)
( x + 3 − x 2 )dx
(1+ 13 )
⎡ 1 x 2 + 3 x − 1 x3 ⎤ 2
3
⎣2
⎦ (1− 13 )
2
x ⎡⎣ 2 x − (2 x − 4) ⎤⎦ dx
(1+ 13 )
⎡ 1 x3 + 3 x 2 − 1 x 4 ⎤ 2
2
4
⎣3
⎦ (1− 13 )
⎡ 2 x − (2 x − 4) ⎤ dx
⎦
1 ⎣
4 3/ 2
2
2 (x
− x + 2 x)dx
1
4 1/ 2
2 ( x − x + 2)dx
1
4
2 ⎡ 52 x5 / 2 − 13 x3 + x 2 ⎤
⎣
⎦1
4
2 ⎡ 23 x3 / 2 − 12 x 2 + 2 x ⎤
⎣
⎦1
2
=
13 13
6
=
(1+ 13 )
=
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
Instructor’s Resource Manual
Section 5.6
329
© 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+ 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+ 5 (− y 4
2 1− 5
1+ 5
1− 5
13 13
6
143 13
30
13 13
6
=
11
=
5
1 ⎡ − 1 y5
2⎣ 5
1− 5
1+ 5
1− 5
1+ 5
=
=
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
2 ± 20
2
Section 5.6
(4 − y 4 )dy
⎡2 y − 1 y3 ⎤
3
⎣
⎦−
y = 0 (by symmetry)
330
1+ 5
+ 23 y 4 − 12 y 2 − 16 y ⎤
⎦1−
5
=
−20 5
5
20 5
3
= −3
y=
y=
(− y 2 + 2 y + 4)dy
1+ 5
1+ 5
=
+ 6 y 3 − 24 y − 16)dy
⎡ − 1 y3 + y 2 + 4 y ⎤
⎣ 3
⎦1−
15.
x=
− ( y 2 − 3 y − 4) 2 ⎤ dy
⎦
⎡ (− y ) − ( y 2 − 3 y − 4) ⎤ dy
⎣
⎦
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
1− 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
1 2
xdx
x |
0
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
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.
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
19. m( R1 ) = δ
a
c
m( R2 ) = δ
M x ( R1 ) =
b
2
a
c
2
b
b
M y ( R2 ) = δ
=δ
c
a
b
a
M x ( R1 ) =
a
c
2
2
2
2
=
δ
δ
2 a
δ
c
M y ( R3 ) = δ
=δ
+δ
c
a
b
a
c
b
x( g ( x) − f ( x))dx
x( g ( x) − f ( x))dx
x( g ( x) − f ( x))dx
= M y ( R1 ) + M y ( R2 )
b
m( R3 ) = δ
=δ
=δ
b
a
b
a
b
a
a
b
(( g ( x))2 − ( f ( x)) 2 )dx
x(h( x) − g ( x))dx
x( g ( x) − f ( x))dx
a
(h( x) − f ( x))dx
(h( x) − g ( x) + g ( x) − f ( x))dx
(h( x) − g ( x))dx +δ
=
=
δ
b
2
a
δ
b
2
a
δ
b
2
a
δ
=δ
c
b
=δ
( g ( x) − f ( x))dx
(( g ( x)) 2 − ( f ( x))2 )dx
b
2
M y ( R3 ) = δ
(( g ( x))2 − ( f ( x)) 2 )dx
2
= M x ( R1 ) + M x ( R2 )
+
b
a
b
a
( g ( x) − f ( x))dx
((h( x))2 − ( f ( x))2 )dx
((h( x))2 − ( g ( x)) 2 + ( g ( x))2 − ( f ( x)) 2 )dx
((h( x))2 − ( g ( x)) 2 )dx
b
(( g ( x))2 − ( f ( x )) 2 )dx
2
= M x ( R1 ) + M x ( R2 )
(( g ( x)) 2 − ( f ( x))2 )dx
b
δ
((h( x))2 − ( g ( x)) 2 )dx
Now,
( g ( x) − f ( x))dx
c
a
+
x( g ( x) − f ( x))dx
2 a
2
M y ( R2 ) = δ
a
b
a
b
a
b
a
x(h( x) − f ( x))dx
x(h( x) − g ( x) + g ( x) − f ( x))dx
x(h( x) − g ( x))dx +δ
b
a
x( g ( x) − f ( x))dx
= M y ( R1 ) + M y ( R2 )
= m( R1 ) + m( R2 )
M x ( R3 ) =
b
M y ( R1 ) = δ
(( g ( x)) − ( f ( x)) )dx
( g ( x) − f ( x))dx + δ
( g ( x) − f ( x))dx
δ
M x ( R2 ) =
x( g ( x) − f ( x))dx
b
b
M x ( R3 ) =
Now,
m( R3 ) = δ
a
(( g ( x)) − ( f ( x)) )dx
δ
M y ( R1 ) = δ
m( R2 ) = δ
(h( x) − g ( x))dx
= m( R1 ) + m( R2 )
( g ( x) − f ( x))dx
b
M x ( R2 ) =
a
( g ( x) − f ( x))dx
δ
b
20. m( R1 ) = 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
© 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.
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 ⎤
= ⎢ x4 ⎥ =
4
⎣
⎦0 4
4
From Problem 11, x = .
5
1⎛
4 ⎞ 2π
V = A(2πx ) = ⎜ 2π ⋅ ⎟ =
4⎝
5⎠ 5
Using cylindrical shells:
25. A =
1 3
x dx
0
V = 2π
1
0
x ⋅ x3 dx = 2π
1 4
x dx
0
1
2π
⎡1 ⎤
= 2π ⎢ x 5 ⎥ =
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.
ΔV = 2π xh( x)Δx , so V = 2π
Δm ≈ h( x)Δx , so m =
b
a
b
a
xh( x)dx .
h( x)dx = A .
b
ΔM y ≈ xh( x)Δx so M y = xh( x)dx .
a
,
x=
My
b
=
a
xh( x)dx
m
A
The distance traveled by the centroid is 2πx .
(2πx ) A = 2π
b
a
xh( x)dx
Therefore, V = 2πxA .
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.
29. a.
ΔV ≈ 2π( K − y ) w( y )Δy
V = 2π
b.
d
c
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
( K − y ) w( y )dy
Δm ≈ w( y )Δy , so m =
d
c
w( y )dy = A .
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
⎝
⎠
= 2π
d
c
32. a.
π
since
2
π
f (sin x) = f ( sin(π − x ) ) . Thus x = .
2
π
x=
0
x f (sin x) dx
π
0
=
f (sin x)dx
π
2
Therefore
π
A=
Instructor’s Resource Manual
The graph of f (sin x) on [0, π ] is
symmetric about the line x =
h
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.
b 3⎤
1
⎡1
= ⎢ by 2 −
y ⎥ = bh 2
3h ⎦ 0 6
⎣2
M
h
y= x =
m
3
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
The length of a segment at y is b −
b.
sin 2πn
n →∞
( 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
Therefore, V = 2π( K − y ) A .
30. a.
lim 4πr 3 n sin
0
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
0
x sin x cos 4 x dx =
π
=
π π
sin x cos 4 x dx
2 0
π⎡ 1
π
⎤
− cos5 x ⎥ =
2 ⎢⎣ 5
⎦0 5
Section 5.6
333
© 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.
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
2
g ( x) dx
S
yS ≥ y R
x≈
≥
1 1
2 0
x≈
=
i =1
=
8
i =1
=
6.5 + 8 +
+ (35)(10) + (40)(8)
+ 10 + 8
hi
=
( −25)(6.5) + ( −15)(8) +
6.5 + 8 +
+ (5)(10) + (10)(8)
+ 10 + 8
y≈
1 8
((hi − 4) 2 − (−4) 2 )
2 i =1
8
i =1
2
hi
(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.
1695
≈ 23.38
72.5
y≈
hi
−480
=
≈ −6.62
72.5
=
(5)(6.5) + (10)(8) +
1 8
(hi ) 2
2 i =1
=
R
hi
xi hi
i =1
f ( x)dx
8
xi hi
i =1
8
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
i =1
8
8
R
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
(1/ 2)(6.52 + 82 + + 102 + 82 )
(6.5 + 8 + + 10 + 8)
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
© 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.
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
x≈
xi ( f ( xi ) − g ( xi ))
i =1
19
i =1
a.
P ( X ≥ 2) = P (2) + P(3) + P(4)
= 0.05 + 0.05 + 0.05 = 0.15
b.
E( X ) =
(20)(145 − 13) + (40)(149 − 10) + (380)(85 − 85)
(145 − 13) + (149 − 19) + (85 − 85)
482,860
=
≈ 173.69
2780
i =1
( f ( xi ) − g ( xi ))
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.
+ 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
4.
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)
=0
5.
a.
b.
2.
sum, integral
3.
4.
5
0
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
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 ) = −0.1(0.98) + 100(0.018)
+ 1000(0.002)
= 3.702
7.
a.
P ( X ≥ 2) = P (2) + P(3) + P(4)
=
f ( x) dx
xi pi
= 0(0.7) + 1(0.15) + 2(0.05)
1 19
[( f ( xi )) 2 − ( g ( xi )) 2 ]
2 i =1
19
5
i =1
( f ( xi ) − g ( xi ))
=
y≈
2.
b.
3 2 1
6
+ + =
= 0.6
10 10 10 10
E ( X ) = 1(0.4) + 2(0.3) + 3(0.2) + 4(0.1) = 2
cumulative distribution function
Problem Set 5.7
1.
a.
P ( X ≥ 2) = P (2) + P(3) = 0.05 + 0.05 = 0.1
b.
E( X ) =
4
i =1
xi pi
= 0(0.8) + 1(0.1) + 2(0.05) + 3(0.05)
= 0.35
Instructor’s Resource Manual
Section 5.7
335
© 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.
8.
a.
P ( X ≥ 2) = P (2) + P(3) + P(4)
=
2
2
b.
2
0
(−1)
(−2)
5
+
+
=
= 0.5
10
10
10
10
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
20
20
0
x⋅
3
x (20 − x) dx
4000
( 20x2 − x3 ) dx
=
3
4000
=
3 ⎡ 20 x3 x 4 ⎤
− ⎥ = 10
⎢
4000 ⎣⎢ 3
4 ⎥⎦
0
20
0
20
1
1
dx =
⋅18 = 0.9
20
20
a.
E( X ) =
c.
20
= 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.
11.
a.
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
3
=
64
For −20 ≤ x ≤ 20 ,
x 1
1
1
1
( x + 20) =
F ( x) =
dt =
x+
−20 40
40
40
2
P ( X ≥ 2) =
8
2
c.
b.
8
0
x⋅
3
x(8 − x) dx
256
14.
a.
3
256
=
3 ⎡ 8 x3 x 4 ⎤
− ⎥ =4
⎢
256 ⎣⎢ 3
4 ⎥⎦
0
b.
For 0 ≤ x ≤ 8
12.
a.
=
336
20
2
3
x(20 − x) dx
4000
3 ⎤ 20
3 ⎡ 2 x
⎢10 x − ⎥ = 0.972
4000 ⎢⎣
3 ⎥⎦
2
Section 5.7
)
4
3 ⎡ 4 x5 ⎤
dx =
⎢ x − ⎥ = 2.4
64 ⎣⎢
5 ⎦⎥
0
For 0 ≤ x ≤ 4
x
x
P ( X ≥ 2) =
8
2
1
(8 − x) dx
32
E( X ) =
8
0
x⋅
1
(8 − x) dx
32
8
x
P ( X ≥ 2) =
−x
4
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
F ( x) =
(4x
3
3 2
x (4 − x) dx
64
1 ⎡
x2 ⎤
9
=
⎢8 x − ⎥ =
32 ⎢⎣
2 ⎥⎦
16
2
8
8
c.
x⋅
8
8 x 2 − x3 ) dx
0(
=
0
F ( x) =
3 ⎡ 2 x3 ⎤
3
27
⋅ 72 =
⎢4 x − ⎥ =
256 ⎣⎢
3 ⎦⎥
256
32
2
E( X ) =
4
4
0
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
8
=
E( X ) =
0
c.
For 0 ≤ x ≤ 8
F ( x) =
=
x
0
x
1
1 ⎡
t2 ⎤
(8 − t ) dt =
⎢8t − ⎥
32
32 ⎣⎢
2 ⎦⎥
0
1
1
x − x2
4
64
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.
a.
4π
P ( X ≥ 2) =
⎛πx ⎞
sin ⎜
⎟ dx
8
⎝ 4 ⎠
2
9
b.
E( X ) =
c.
For 1 ≤ x ≤ 9
1
4
=
π⎡ 4
π x⎤
1
1
− cos ⎥ = − (−1 − 0) =
⎢
8⎣ π
4 ⎦2
2
2
c.
⎛πx ⎞
sin ⎜
⎟ dx
8
⎝ 4 ⎠
Using integration by parts or a CAS,
E(X ) = 2 .
E( X ) =
4
0
π
x⋅
=−
19.
For 0 ≤ x ≤ 4
x
16.
a.
xπ
F ( B) =
4
c.
⎛π x ⎞
cos ⎜
⎟ dx
8
⎝ 8 ⎠
Using a CAS, E ( X ) ≈ 1.4535
E( X ) =
0
20.
a.
a.
⎡ ⎛ π t ⎞⎤
⎛ πt ⎞
cos ⎜ ⎟ dt = ⎢sin ⎜ ⎟ ⎥
8
⎝ 8 ⎠
⎣ ⎝ 8 ⎠⎦0
b.
E( X ) =
=
c.
4
P ( X ≥ 2) =
4
1
2
4
a.
x
c.
1
P ( X ≥ 2) =
=
4
2
9
2
9
⎡ 81 ⎤
dx = ⎢ −
⎥
3
40 x
⎣ 80 x 2 ⎦ 2
81
77
≈ 0.24
320
Instructor’s Resource Manual
E( X ) =
=
4
ln 4 ≈ 1.85
3
1
f ( x) dx = F (b) − F (a) due to
a+b
.
2
a +b
2
a
4
−4 4
⎡ 4⎤
+
dt = ⎢ − ⎥ =
t
3
⎣
⎦1 3x 3
3t
4x − 4
=
3x
18.
b.
⎡4
⎤
x⋅
dx = ⎢ ln x ⎥
2
⎣3
⎦1
3x
4
b
a
1
1 ⎛ a+b
⎞
dx =
− a⎟
⎜
b−a
b−a⎝ 2
⎠
1 b−a 1
=
⋅
=
b−a 2
2
For 1 ≤ x ≤ 4
F ( x) =
f ( x) dx = 1
A
The midpoint of the interval [a,b] is
=
1
⎡ 4⎤
dx = ⎢ − ⎥ =
2
⎣ 3x ⎦ 2 3
3x
4
f ( x) dx = 0;
a+b⎞
a+b⎞
⎛
⎛
P⎜ X <
⎟ = P⎜ X ≤
⎟
2 ⎠
2 ⎠
⎝
⎝
x
⎛π x ⎞
= sin ⎜
⎟
⎝ 8 ⎠
17.
f (t ) dt. By the First
the Second Fundamental Theorem of Calculus.
xπ
0
A
A
B
P ( a ≤ X ≤ b) =
For 0 ≤ x ≤ 4
F ( x) =
x
A
Proof of P (a ≤ X ≤ b) = F (b) − F (a ) :
π
x⋅
81 81x 2 − 81
=
80
80 x 2
Proof of F ( x) = f ( x) :
F ( A) =
⎛πx⎞
P ( X ≥ 2) =
cos ⎜
⎟ dx
2 8
⎝ 8 ⎠
⎡ ⎛ π x ⎞⎤
π
π
1
= ⎢sin ⎜
⎟ ⎥ = sin − sin = 1 −
2
4
2
⎣ ⎝ 8 ⎠⎦ 2
b.
80 x 2
+
Fundamental Theorem of Calculus,
F ( x) = f ( x).
Proof of F ( A) = 0 and F ( B) = 1:
4π
4
81
By definition, F ( 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
x⋅
21.
F ( x) =
b
a
2
x⋅
1
1 ⎡ x2 ⎤
dx =
⎢ ⎥
b−a
b − a ⎣⎢ 2 ⎦⎥
b
a
2
b −a
a+b
=
2(b − a )
2
x
a
1
1
x−a
dt =
( x − a) =
b−a
b−a
b−a
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
© 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.
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.
5
=
5
⎡ 5kx 2 kx3 ⎤
−
⎢
⎥ =1
3 ⎥⎦
⎢⎣ 2
0
125k 125k
−
=1
2
3
375k − 250k = 6
24.
Solve
k
5
0
5
0
d.
6
125
4
e.
k ( 2 − x − 2 ) dx = 1
Due to the symmetry about the line x = 2, the
solution can be found by solving
2
2
0
kx dx = 1
k ⋅ x2
2
0
=1
3
4
338
41
3
4
( 2 − x − 2 ) dx
(2 − ( x − 2)) dx =
4
=
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
P (3 ≤ X ≤ 4) =
41
0
2
4
y
1
k=
4
=
21
0
⎡ t2 ⎤
y2
=⎢
⎥ =
⎣⎢ 28,800 ⎦⎥ 0 28,800
4k = 1
b.
0
x
+ x −1
8
0
⎧
⎪
x2
⎪
⎪⎪
8
F ( x) = ⎨
2
⎪ x
⎪− 8 + x − 1
⎪
⎩⎪1
=−
( 25x2 − 10x3 + x4 ) dx = 1
0
2
x
⎡t2 ⎤
x2
t dt = ⎢ ⎥ =
4
8
⎣⎢ 8 ⎦⎥ 0
x1
If 0 ≤ x ≤ 2, F ( x) =
x2
=
8
kx 2 (5 − x) 2 dx = 1
Solve
1 3 2 1 ⎡ 2 x3 ⎤
2 4
x + ⎢2 x − ⎥ = + = 2
12 0 4 ⎣⎢
3 ⎦⎥
3 3
If 2 < x ≤ 4, F ( x) =
5
a.
2
0
2
⎡ 25 x3 5 x 4 x5 ⎤
k⎢
−
+ ⎥ =1
2
5 ⎥⎦
⎢⎣ 3
0
625
k =1
6
6
k=
625
25.
1
x ⋅ (2 − x − 2 ) dx
4
4
kx(5 − x) dx = 1.
k=
4
0
4
1
1
x ⋅ (2 + ( x − 2)) dx + x ⋅ (2 − ( x − 2)) dx
2
4
4
1 2 2
1 4
(4 x − x 2 ) dx
=
x dx +
4 0
4 2
=
Since the PDF must integrate to one, solve
0
E( X ) =
1
4
4
3
1
1 ⎡
t2 ⎤
= +
⎢ 240t − ⎥
2 14, 400 ⎣⎢
2 ⎦⎥
y
120
(4 − x) dx
=
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
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.
26.
a.
180
Solve
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
FY ( y ) ≈ (7.54731× 1027 ) y 7 ( y8 − 0.202475 y 7
150
180
0
x⋅
8
⎛ 3
⎞
− t ⎟ dt
k ⋅ t6 ⎜
0
⎝ 127 ⎠
Using a CAS,
y
FY ( y ) =
1
x 2 (180 − x) dx
87, 480, 000
E( 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
k
0.6
6
= 108
0
28.
a.
0.6 6
x (0.6 − x)8 dx
0
=1
= 1− k
P ( X ≥ 100) = k
c.
x (0.6 − x)8 dx
d.
0.6
E( X ) =
=k
0
x ⋅ kx 6 (0.6 − x)8 dx
F ( x) =
0
x
F ( x) =
x ⋅ x 2 (200 − x)8 dx
0
(2.417 × 10−23 )t 2 (200 − t )8 dx
+(1.76 × 1011 ) x 4 − (3.2853 × 1013 ) x3
+(3.942 × 1015 ) x 2 − (2.816 × 1017 ) x
95,802, 719t 6 (0.6 − t )8 dt
+9.39 × 1018 )
Using a CAS,
F ( x ) ≈ 6,386,850 x 7 ( x8 − 5.14286 x 7
e.
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.
0
( x8 − 1760 x 7 + 136889 x 6 − (6.16 × 108 ) x5
≈ 0.2625
d.
200
E( X ) = k
Using a CAS, F(x) ≈ (2.19727 ×10−24 ) x3 ⋅
0.6 7
x (0.6 − x)8 dx
0
x
x (200 − x)8 dx
= 50 using a CAS
≈ 0.884 using a CAS
c.
200 2
100
≈ 0.0327 using a CAS.
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
8
kx (0.6 − x) dx = 1.
0
Solve
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
© 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.
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
0.2
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
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
x
P ( X = x) =
x
expressions, P( a < X < b), P( a ≤ X ≤ b),
P (a < X ≤ b) and P (a ≤ X < b), are
if 1 ≤ x < 2
if 2 ≤ x < 3
if 3 ≤ x < 4
if x ≥ 4
0.8
0.6
0.4
0.2
0
35.
1
b.
P (0.5 < Y < 0.6) = F (0.6) − F (0.5)
c.
f ( y) = F ( y) =
d.
E (Y ) =
Since P( S ) = 1, P( A ∪ A ) = 1.
c
P ( A) + P( A ) = 1 and P( A ) = 1 − P ( A).
32.
5
x
1.2 1
1
−
=
1.6 1.5 12
c
c
4
P (Y < 2) = P(Y ≤ 2) = F (2) = 1
By the defintion of a complement of a set,
Since P( A ∪ Ac ) = P( A) + P( Ac ),
3
2
a.
=
A ∪ Ac = S , where S denotes the sample space.
if 0 ≤ x < 1
1.0
equivalent.
31.
if x < 0
⎧ 0
⎪ 0.7
⎪
⎪⎪0.85
F ( x) = ⎨
⎪ 0.9
⎪0.95
⎪
⎪⎩1
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
g ( y) =
and G ( y ) = y 2 − 1 .
2
y −1
30.
1
1
0
y⋅
2
( y + 1)2
2
( 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
© 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.
36.
a.
P ( Z > 1) = 1 − P ( Z ≤ 1) = 1 − F (1)
= 1−
b.
38.
Concepts Test
P (1 < Z < 2) = P (1 ≤ Z ≤ 2) = F (2) − F (1)
=
37.
1 8
=
9 9
4 1 1
− =
9 9 3
f ( z) = F ( z) =
d.
E (Z ) =
0
2z
,0≤ z≤3
9
3
z⋅
⎡ 2 z3 ⎤
2z
dz = ⎢
⎥ =2
9
⎣⎢ 27 ⎦⎥
0
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
E( X ) =
E( X 2 ) =
3
E( X ) =
4
0
x⋅
8 2 3
x ⋅
0
256
x(8 − x) dx = 19.2 and
8 3 3
x ⋅
0
256
x (8 − x) dx = 102.4
using a CAS
39.
40.
41.
V ( X ) = E ⎡( X − ) 2 ⎤ , where = E ( X ) = 2
⎣
⎦
4
15 2
4
V ( X ) = ( x − 2) 2 ⋅
x (4 − x) 2 dx =
0
512
7
3
x ( 8 − x ) dx = 4
256
8
3
16
V ( X ) = ( x − 4) 2 ⋅
x(8 − x) dx =
0
256
5
8
= E(X ) =
E ⎡⎢( X −
⎣
0
x⋅
)2 ⎤⎥⎦ = E ( X 2 − 2 X
= E( X 2 ) − E ( 2 X
) + E(
= E( X 2 ) − 2 ⋅ E( X ) +
2
2
= E( X ) − 2
= E( X 2 ) −
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
+
2
2
+ 2)
)
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. 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,
V = 2π
1
0
x(− x 2 + x)dx . To use the
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.
2
since E ( X ) =
2
For Problem 37, V ( X ) = E ( X 2 ) − 2 and
32 2 4
using previous results, V ( X ) =
−2 =
7
7
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
© 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.
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
20. True: The computation of E(X) would be the
same as the computation for the center of
mass of the wire.
x
A
1
1
4. V = π ⎡( x − x 2 + 2)2 − (2)2 ⎤ dx
⎦
0⎣
=π
f (t ) dt , then F ( x) = f ( x)
P ( X = 1) = P (1 ≤ X ≤ 1) =
1
1
f ( x) dx = 0
Sample Test Problems
1. A =
342
1
1 ⎤
1
⎡1
( x − x 2 )dx = ⎢ x 2 − x3 ⎥ =
0
3 ⎦0 6
⎣2
1
Section 5.8
1
0
( x 4 − 2 x3 − 3x 2 + 4 x) dx
1
1
7π
⎡1
⎤
= π ⎢ x5 − x 4 − x3 + 2 x 2 ⎥ =
2
⎣5
⎦ 0 10
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
6. x =
x( x − x 2 )dx
0
1
0
y=
( x − x )dx
=
1 1 ( x − x 2 ) 2 dx
2 0
1
2
0
=
2
1
( x − x )dx
⎡ 1 x3 − 1 x 4 ⎤
4
⎣3
⎦0
1
⎡ 1 x 2 − 1 x3 ⎤
3
⎣2
⎦0
=
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:
( x 2 − x3 )dx
1
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, …}.
If F ( x) =
1
0
1 ⎤
π
⎡1
= 2π ⎢ x3 − x 4 ⎥ =
4 ⎦0 6
⎣3
18. True: See Problem 30 in Section 5.6.
22. True:
x( x − x 2 )dx = 2π
0
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.
E ( X ) = 5 ⋅1 = 5
0
5. V = 2π (3 − x)( x − x 2 )dx
16. True: At slice y, ΔA ≈ (9 − y 2 )Δy .
21. True:
1
3. V = 2π
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.
8. 8 = F(8) = 8k, k = 1
a.
4
12. x =
8
1
⎡1 ⎤
x dx = ⎢ x 2 ⎥ = (64 − 4)
2
2
⎣
⎦2 2
= 30 in.-lb
8
W=
6
9. W =
0
4
=
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 = yΔy
5
5
11. a.
4x = x 2 .
x2 − 4 x = 0
x(x – 4) = 0
x = 0, 4
0
b. To find the intersection points, solve
y
= y.
4
y2
=y
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
=2
− ( x 2 )2 ⎤ dx
⎦
(4 x − x 2 )dx
(16 x 2 − x 4 ) dx
1 ⎡ 16 x3
2⎣3
4
− 15 x5 ⎤
⎦0
32
3
4
0
4
=
1024
15
32
3
=
32
5
⎡ (4 x) 2 − ( x 2 ) 2 ⎤ dx
⎣
⎦
(16 x 2 − x 4 ) dx
0
(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
64
3
32
3
32
3
32
3
13. V = π
14. a.
4
1 ⎤
⎡
(4 x − x 2 )dx = ⎢ 2 x 2 − x3 ⎥
3 ⎦0
⎣
64 ⎞ 32
⎛
= ⎜ 32 − ⎟ =
3 ⎠ 3
⎝
A=
=
(4 x 2 − x3 ) dx
4
100
4
=
=
4
0
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 ⎠
To find the intersection points, solve
A=
=
1 4
2 0
=π
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
0
6
1 ⎤
⎡
(10 − y )dy = 1560π ⎢10 y − y 2 ⎥
0
2 ⎦0
⎣
= 65,520π ≈ 205,837 ft-lb
W2 =
⎡ 4 x3 − 1 x 4 ⎤
4
⎣3
⎦0
y=
6
(4 x − x 2 ) dx
4
(62.4)(52 )π(10 − y )dy
= 1560π
x(4 x − x 2 )dx
0
⎡1 ⎤
x dx = ⎢ x 2 ⎥ = 8 in.-lb
0
⎣ 2 ⎦0
−4
W=
b.
0
4
3
0
δ (3 − y )(16π ) dy =16πδ
3
0
(3 − y ) dy
3
⎡
y2 ⎤
= 16πδ ⎢3 y − ⎥ = 16πδ (4.5) ≈ 14,114.55 lbs.
2 ⎦⎥
⎣⎢
0
Section 5.8
343
© 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.
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
1
1
x + +
dx =
2 16 x 4
3
3
4
1
L2 =
1
⎛ 2
1
⎜x + 2
4x
⎝
⎞
⎟ dx
⎠
3
3
0
b.
=
3
17. V =
2
⎛ 9 − x 2 ⎞ dx = 3 (9 − x 2 ) dx
⎜
⎟
−3 ⎝
−3
⎠
3
18. A =
b
a
19. V = π
⎡ f 2 ( x) − g 2 ( x) ⎤ dx
⎦
a⎣
20. V = 2π
b
a
( x − a ) [ f ( x) − g ( x) ] dx
2
1
d.
(
)
17
≈ 0.354
48
0.5
0
b.
(
)
85
≈ 0.332
256
2
0
x⋅
21. M y = δ
Mx =
344
a
x [ f ( x) − g ( x) ] dx
δ
b
⎡ f 2 ( x) − g 2 ( x ) ⎤ dx
⎦
2 a⎣
Section 5.8
2
1
1 ⎡
x5 ⎤
8 − x3 dx = ⎢ 4 x 2 − ⎥
12
12 ⎣⎢
5 ⎦⎥
0
(
)
x
1
1 ⎡
t4 ⎤
8 − t 3 dt = ⎢8t − ⎥
0 12
12 ⎢⎣
4 ⎦⎥
0
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 ⎡
x4 ⎤
8 − x3 dx = ⎢8 x − ⎥
12
12 ⎢⎣
4 ⎥⎦
1
F ( x) =
=
25. a.
2
1
1 ⎡
x4 ⎤
8 − x3 dx = ⎢8 x − ⎥
12
12 ⎢⎣
4 ⎥⎦
1
= 0.8
[ f ( x) − g ( x)] dx
b
2
c. E ( X ) =
3
1 ⎤
⎡
= ⎢9 x − x3 ⎥ = (27 − 9) − (−27 + 9) = 36
3 ⎦ −3
⎣
g ( x) 1 + [ g ( x) ] dx
P (0 ≤ X < 0.5) = P(0 ≤ X ≤ 0.5)
=
(t 2 + 1)dt
⎡1
⎤
= 2 ⎢ t3 + t ⎥ = 4 3
⎣3
⎦0
2
P ( X ≥ 1) = P(1 ≤ X ≤ 2)
=
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
a
f ( x) 1 + [ f ( x) ] dx
A4 = π ⎡ f 2 (b) − g 2 (b) ⎤
⎣
⎦
Total surface area = A1 + A2 + A3 + A4 .
=
t 4 + 2t 2 + 1 dt = 2
b
a
b
A3 = π ⎡ f 2 (a) − g 2 (a) ⎤
⎣
⎦
24. a.
3
2
A2 = 2π
16.
0
1 + [ g ( x) ] dx
a
23. A1 = 2π
1⎤
1 ⎞ ⎛ 1 1 ⎞ 53
⎡1
⎛
= ⎢ x3 − ⎥ = ⎜ 9 − ⎟ − ⎜ − ⎟ =
3
4
x
12
⎣
⎦1 ⎝
⎠ ⎝3 4⎠ 6
L=2
2
L4 = f (b) − g (b)
Total length = L1 + L2 + L3 + L4
⎛
1 ⎞
1 + ⎜ x2 −
⎟ dx
4 x2 ⎠
⎝
3
1 + [ f ( x) ] dx
a
b
L3 = f (a) − g (a)
dy
1
15.
= x2 −
dx
4 x2
L=
b
22. L1 =
c. E ( X ) =
6
0
(6 − 3) 2 3
=
36
4
1
6− x
,
(6 − x) =
36
18
⎛6− x⎞
x ⋅⎜
⎟ dx
⎝ 18 ⎠
6
1 ⎡
x3 ⎤
= ⎢3 x 2 − ⎥ = 2
18 ⎢⎣
3 ⎥⎦
0
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.
Review and Preview Problems
1. By the Power Rule
1
x −2 +1 x −1
1
−2
=
=
=
= − +C
dx
x
dx
2
−2 + 1 −1
x
x
2. By the Power Rule
1
2
x −1.5+1
x −0.5
−1.5
=
=
=
=−
+C
dx
x
dx
1.5
−1.5 + 1 −0.5
x
x
3. By the Power Rule
1
x −1.01+1
x −0.01
100
−1.01
=
=
=
= − 0.01 + C
dx
x
dx
1.01
−1.01 + 1 −0.01
x
x
4. By the Power Rule
1
x −0.99 +1
x 0.01
−0.99
=
=
=
= 100 x 0.01 + C
dx
x
dx
−0.99 + 1 0.01
x 0.99
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
5. F (1) =
1
t
2
b. (1 +
dt = 0
6. By the First Fundamental Theorem of Calculus
d x1
1
F ( x) =
dt =
1
dx t
x
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
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
2
⎞
⎟ (2 x) = x
⎠
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⎠
c. (1 +
11
≈ 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
or x =
π
+ 2kπ =
6
6
where k is any integer.
≈ 2.704814
Review and Preview
345
© 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.
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) =
1
, sec( x) is never 0 .
cos( x)
17. In the triangle, relative to
,
1
x2 − 1
sin =
cos =
tan = x 2 − 1
x
x
x
1
cot =
sec = x csc =
x2 − 1
x2 − 1
,
opp = x , adj = 1 − x 2 , hypot = 1 so that
x
sin = x cos = 1 − x 2 tan =
1 − x2
cot =
1 − x2
x
1
sec =
1− x
19. In the triangle, relative to
1
y2
dy = xdx
dy
y2
2
csc =
=
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 ' =
opp = x 2 − 1 , adj = 1 , hypot = x so that
18. In the triangle, relative to
y ' = xy 2 → dy = xy 2 dx
21.
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
,
opp = 1 , adj = x , hypot = 1 + x 2 so that
sin =
1
1+ x
2
x
cos =
cot = x sec =
1+ x
1 + x2
x
20. In the triangle, relative to
2
tan =
1
x
csc = 1 + x 2
,
opp = 1 − x 2 , adj = x , hypot = 1 so that
sin = 1 − x 2
cot =
346
x
1 − x2
cos = x tan =
sec =
1 − x2
x
1
1
csc =
x
1 − x2
Review and Preview
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.
6
CHAPTER
Transcendental
Functions
6.1 Concepts Review
1.
2.
3.
x1
1
t
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
dt ; (0, ∞); (– ∞, ∞)
1
x
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⎠
4
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
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
2
x + 3x + π
⋅ Dx ( x 2 + 3 x + π ) =
=
2x + 3
2
x + 3x + π
13.
4. Dx ln(3x3 + 2 x) =
=
1
3
3x + 2 x
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
14.
⎡ 1 2
⎤
1 + ( x + 1) –1/ 2 ⋅ 2 x ⎥
⎢
⎦
x + x2 + 1 ⎣ 2
1
1
x2 + 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
⎡ 1 2
⎤
–1/ 2
⋅ 2x⎥
⎢1 + 2 ( x – 1)
⎣
⎦
x + x –1
1
2
1
2
x −1
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
© 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. Let u = 2x + 1 so du = 2 dx.
1
1 1
dx =
du
2x +1
2 u
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
17. Let u = 3v 2 + 9v so du = 6v + 9.
6v + 9
1
dv =
du = ln u + C
2
u
3v + 9v
18. Let u = 2 z 2 + 8 so du = 4z dz.
1 1
z
dz =
du
2
4 u
2z + 8
1
1
= ln u + C = ln 2 z 2 + 8 + C
4
4
23. By long division,
so
2
1
dx
x
1
dx .
x
dx = – u –2 du
24. By long division,
1 1
dx =
du
5
10
u
2x + π
1
1
= ln u + C = ln 2 x5 + π + C
10
10
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
so
3
1
x2 3
dx
+ x+
4 4
4 2x −1
Let u = 2 x − 1 ; then du = 2dx . Hence
1
1 1
1
dx =
du = ln u + C
2x −1
2 u
2
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
so
x4
dx =
x+4
x3 dx − 4 x 2 dx + 16 xdx − 64dx + 256
x4
0
x2 + x x 3
3
= + +
2 x − 1 2 4 4(2 x − 1)
3
3
x2 + x
x
dx =
dx + dx +
dx
2x −1
2
4
4(2 x − 1)
21. Let u = 2 x5 + π so du = 10 x 4 dx .
x4
x2
1
dx = x dx + 1 dx +
dx
x −1
x −1
x2
=
+ x + ln x − 1 + C
2
25. By long division,
x4
256
= x3 − 4 x 2 + 16 x − 64 +
x+4
x+4
x(ln x)
1
1
= +C =
+C
u
ln x
3
x2
1
= x +1+
x −1
x −1
=
= u 2 + C = (ln x) 2 + C
–1
1
1
9
1
ln 9 – ln 3 = ln 4 = ln 4 3 = ln 3
4
4
3
4
=
)
2 ln x
dx = 2 udu
x
20. Let u = ln x, so du =
1
t +1
⎡1
⎤
dt = ⎢ ln 2t 2 + 4t + 3 ⎥
0 2t 2 + 4t + 3
4
⎣
⎦0
1
= ln 3v 2 + 9v + C
19. Let u = ln x so du =
1 1
du
4 u
2t + 4t + 3
1
1
= ln u + C = ln 2t 2 + 4t + 3 + C
4
4
16. Let u = 1 – 2x so du = –2dx.
1
1 1
dx = –
du
1 – 2x
2 u
1
1
= – ln u + C = – ln 1 – 2 x + C
2
2
(
dt =
2
=
1
dx
x+4
x 4 4 x3
−
+ 8 x 2 − 64 x + 256 ln x + 4 + C
4
3
26. By long division,
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
© 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.
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
Section 6.1
349
© 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.
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
1/ 3 t
1
1
dt = 2
t
dt
x1
dt = 2
t
1 1
x1
dt =
dt
1/ 3 t
1 t
1/ 3 1
x1
–
dt =
dt
1 t
1 t
1
– ln = ln x
3
ln 3 = ln x
x=3
1/ 3 t
46. a.
dt +
x1
1
1
x1
1
t
1 1
<
for t > 1,
t
t
x1
so ln x =
dt <
1 t
1
x
dt
1
t
dt =
x –1/ 2
1
t
dt
x
= ⎡⎣ 2 t ⎤⎦ = 2( x –1)
1
so ln x < 2( x – 1)
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.
b. If x > 1, 0 < ln x < 2( x – 1) ,
x →∞
and lim
x →∞
π
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
1
dx =
du = ln u + C
1 + sin x
u
= ln 1 + sin x + C = ln(1 + sin x) + C
(since 1 + sin x ≥ 0 for all x ).
53. V = 2π
4
2πx
1
2
x + x –1
2 ⋅1 + 1
f (1) = cos [ln(1 + 1 –1)] ⋅
2
1 + 1 –1
= 3cos 2 (0) = 3
2πx
x2 + 4
dx
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
4
1
4
dx = ⎡⎢ π ln x 2 + 4 ⎤⎥
⎣
⎦
1
x +4
= π ln 20 − π ln 5 = π ln 4 ≈ 4.355
du
dx
2x + 1
xf ( x)dx =
= π ln x 2 + 4 + C
L=
2
4
1
Let u = x 2 + 4 so du = 2x dx.
2πx
1
dx = π du = π ln u + C
2
u
x +4
=1
= cos 2 [ln( x 2 + x –1)] ⋅
2
4
= ⎡−
⎣ 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
⎜
⎟⋅ =
= lim
dx = ln 2 ≈ 0.693
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
1
⎝ dx ⎠
⎝ 2 2x ⎠
2
2
1
2
2⎛ x
1 ⎞
⎛x 1 ⎞
⎜ + ⎟ dx = 1 ⎜ + ⎟ dx
⎝ 2 2x ⎠
⎝ 2 2x ⎠
2
⎤
1 ⎡ x2
⎡
⎛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
π
3
= ⎡−
⎣ ln cos x ⎤⎦ 0
= ln cos 0 − ln cos π 3
⎛ 1 ⎞
= ln(1) − ln(0.5) = ln ⎜
⎟
⎝ 0.5 ⎠
= ln 2 ≈ 0.69315
Instructor’s Resource Manual
Section 6.1
351
© 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.
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
dt
t
y–x
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) = −
c.
20
0.1
cos(ln x)dx ≈ −8.37
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
⎝
⎠
a.
1⎡
0 ⎢⎣
5
⎛1⎞
⎛ 1 ⎞⎤
x ln ⎜ ⎟ − x 2 ln ⎜ ⎟ ⎥ dx =
≈ 0.139
x
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
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.
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
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.
10.
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.
<π
f ( ) is decreasing at = 0 because f(0) = 1 and
f( ) < 1 for 0 < < π . f( ) is decreasing at
= π because f( π ) = –1 and f( ) > –1 for
0 < < π . Therefore f( ) is strictly decreasing
on 0 ≤ ≤ π and so it has an inverse.
f ( ) = – sin < 0 for 0 <
f ( x) = – csc 2 x < 0 for 0 < x <
f(x) is decreasing on 0 < x <
14.
f (r ) =
1
r
cos 4 tdt = –
r
1
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.
Instructor’s Resource Manual
Section 6.2
353
© 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.
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 +
2
( x) = x – 1, x ≥ 0
Step 3: f –1 ( x) = 2 +
2
( f ( x)) = ( x + 1) – 1 = ( x + 1) – 1 = x
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 +
Step 2: f
( y) = 1 – y , y ≤ 0
Step 3: f
Check:
–1
( x) = 1 – x 2 , x ≤ 0
f ( f –1 ( x)) = – 1 – (1 – x 2 ) = – x 2 = – x
= –(–x) = x
1
x–3
1
x–3= –
y
1
y
)
1
⎛2+
⎜
⎝
1
( x 1–2 )
1 ⎞–2
⎟
x2 ⎠
=
1
= x2
⎛ 1 ⎞
⎜ 2⎟
⎝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
f –1 ( f ( x)) = –
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
1
x –2
= 2+
Step 2: f –1 ( y ) = −
y=–
f(f
(
2
y
2
x
Step 3: f –1 ( x) = −
2
Check:
19. Step 1:
–1
1
2
f –1 ( f ( x)) = 1 – (– 1 – x ) 2 = 1 – (1 – x) = x
Step 2: f
,x>0
x2
= x =x
–1
–1
1
= 2 + (x – 2) = x
1 – x = (– y ) 2 = y 2
x = 3–
,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
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.
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
3
( f ( x)) = 1 + 3 ( x –1) = 1 + ( x –1) = x
3
3
3
3
( x)) = [(1 + x ) –1] = ( x ) = x
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
=
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
1+
1–
1+ x
1– x
1+ x
1– x
2– y
y –1
1/ 3
⎛2– y⎞
x=⎜
⎟
⎝ y –1 ⎠
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:
1/ 3
=
x + 1 + x –1 2 x
=
=x
x +1 – x +1 2
–1
1 + x –1 + x 2 x
=
=
=x
+1 1+ x +1 – x 2
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=
( )
1/ 3
⎥⎦
3⎤
⎡
1 – ⎢ xx +–11 ⎥
⎣
⎦
x +1 + x – 1 2x
=
=
=x
x +1 – x +1 2
x3 =
1+ y
( y) =
1− y
( x)) =
3 ⎤1/ 3
x3 y – x3 = 2 – y
Check:
f(f
( xx+–11 )
x3 y + y = x3 + 2
Step 3: f –1 ( x) =
–1
⎡
1+ ⎢
⎣
f –1 ( f ( x)) =
3
f –1 ( f ( x)) = ( x5 / 2 ) 2 / 5 = x
f –1 ( f ( x)) =
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
© 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.
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
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.
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) =
33.
37.
f ( x) = 15 x 4 + 1 and y = 2 corresponds to x = 1,
so ( f −1 ) (2) =
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
=
1
) (3) ≈
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 f –1 )(h( x)) = ( g –1 f –1 )( f ( g ( x)))
= g –1 [ f –1 ( f ( g ( x)))] = g –1 [ g ( x)] = x
Similarly,
h(( g –1 f –1 )( x)) = f ( g (( g –1 f –1 )( x)))
= f ( g ( g –1 ( f –1 ( x )))) = f ( f –1 ( x)) = x
Thus h –1 = g –1 f –1
35.
( f −1 ) (3) ≈ −
1
3
Instructor’s Resource Manual
Section 6.2
357
© 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. Find f −1 ( x) :
y=
ax + b
cx + d
cxy + dy = ax + b
(cy – a)x = b – dy
b − dy
dy − b
x=
=−
cy − a
cy − a
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
f −1 ( x) = −
dx − b
cx − a
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
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 ⎠
( )
acx 2 + (bc − a 2 ) x − ab + dcx 2
+(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
⎛ 1 −2⎞
1
1
⎟=
h(h −1 ( x)) = h ⎜ x
=
=x
1
⎜ 3 ⎟ ⎡ 1 − 2⎤ + 2
x
⎝
⎠ ⎣ x
⎦
( )
( )
43. f has an inverse because it is monotonic
(increasing):
If a = d, then f = f −1 requires b = 0 and
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 ) =
c.
dy − b
cy − a
b. If bc – ad = 0, then f(x) is either a constant
function or undefined.
h( x ) = f ( g ( x)) = f (3x + 2) =
=
f −1 ( y ) = −
1
f
( )
π
2
1
f
=
( 56π )
=
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
© 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.
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].
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
b
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
increasing for p > 1, ab ≤
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)
a p –1
b q –1
0
0
x
dx +
y
dy
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 ≤
a
0
f ( x) dx +
b
0
f −1 ( y ) dy with equality
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
© 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.
5. ln ecos x = cos x
19.
6. ln e –2 x –3 = –2 x – 3
ex
ex
x
=
eln x
2
9. eln 3+ 2 ln x = eln 3 ⋅ e2 ln x = 3 ⋅ eln x = 3 x 2
10. 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
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
x
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 )
=
⎡
20. Dx ⎢e1
⎢⎣
=−
⎞
⎟
⎝ x ⎠
15. Dx e2 ln x = Dx e
2
= x ex +
= ex
2
– 1
⎛ 1
2
= e x Dx ⎜ –
2
– 1
2
=e x
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
= ex
(4 x –1)
x+2
13. Dx e
=e
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
© 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.
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 ⇒ – a > – b ⇒ e
increasing function.
25.
f ( x) = e
2x
–a
>e
–b
x
, since e is an
8
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.
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
© 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.
28.
f ( x) = e x + x
Domain = (−∞, ∞)
f ( x) = e x + 1 ,
30.
f ( x) = e x
f ( x) = ln(2 x − 1) . Since 2 x − 1 > 0 if and only if
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.
1
2
1
2
, domain = ( , ∞)
f ( x) =
2
,
2x −1
f ( x) =
−4
(2 x − 1) 2
Since f ( x) > 0 for all domain values, f is
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) =
x
f
f
2x
,
x +1
2
f ( x) =
−5
2
−2( x − 1)
( x 2 + 1) 2
( −∞ , −1) −1 ( −1,0) 0 (0,1) 1 (1,∞ )
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, 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
Section 6.3
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.
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 (
f has points of
2
2
x
4
2
, e)
y
−3
34.
f ( x) = e x − e− x
x
f ( x) = e + e
x
f
f
−x
Domain = (−∞, ∞)
,
f ( x) = e x − e− x
(−∞, 0) 0 (0, ∞)
+
+
+
−
+
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
−3
3
y
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=
x
f
f
4± 2
≈ 2 ± 0.707
2
( −∞ ,1.293) ≈1.293 (1.293,2) 2 (2,2.707) ≈ 2.707 (2.707,∞ )
+
+
+
0
−
−
−
+
−
−
−
+
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
Section 6.3
363
© 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.
35.
f ( x) =
x − t2
0
e
f ( x) = e− x ,
2
37. Let u = 3x + 1, so du = 3dx.
1 3 x +1
1 u
1
3dx =
e3 x +1dx =
e
e du = eu + C
3
3
3
1 3 x +1
= e
+C
3
Domain = (−∞, ∞)
dt
f ( x) = −2 xe− x
2
(−∞, 0) 0 (0, ∞)
+
+
+
+
0
−
x
f
f
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)
y
39. Let u = x 2 + 6 x , so du = (2x + 6)dx.
2
1 u
1
e du = eu + C
( x + 3)e x + 6 x dx =
2
2
1 x2 +6 x
= e
+C
2
3
3
−3
38. Let u = x 2 − 3, so du = 2x dx.
2
1 x 2 −3
1 u
xe x −3 dx =
e
e du
2 x dx =
2
2
1
1 2
= eu + C = e x −3 + C
2
2
x
40. Let u = e x − 1, so du = e x dx .
ex
1
dx =
du = ln u + C = ln e x − 1 + C
u
e −1
−3
36.
f ( x) =
x
0
te− t dt
f ( x) = xe
−x
,
( −∞ ,0)
−
+
x
f
f
x
Domain = (−∞, ∞)
f ( x) = (1 − x)e
0
0
+
(0,1)
+
+
1
1
41. Let u = − , so du =
dx .
x
x2
e−1/ x
dx = eu du = eu + C = e−1/ x + C
x2
−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
0
x
43. Let u = 2x + 3, so du = 2dx
1 u
1
1
e2 x +3 dx =
e du = eu + C = e 2 x +3 + C
2
2
2
=
−2
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 =
9
x
e x ⋅ ee dx = eu du = eu + C = ee + C
1 2 x +3
e
dx
0
4
−3
x
Let u = e x , so du = e x dx.
te−t dt ) .
Note: It can be shown with techniques in
2
1
Chapter 7 that 0 te− t dt = 1 − ≈ 0.264
e
y
x
e x + e dx = e x ⋅ ee dx
42.
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
1
⎡ 1
⎤
dx = ⎢ – e3 / x ⎥ = – e3 / 2 + e3 ≈ 5.2
1 x2
3
3
3
⎣
⎦1
364
Section 6.3
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.
ln 3
45. V = π
0
(e x )2 dx = π
ln 3
⎡1
⎤
= π ⎢ e2 x ⎥
⎣2
⎦0
46. V =
ln 3 2 x
e dx
0
1 ⎞
⎛1
= π ⎜ e2 ln 3 − e0 ⎟ = 4π ≈ 12.57
2 ⎠
⎝2
2
1
2πxe− x dx .
0
y = et cos t , so dy = (et cos t − et sin t )dt
2
ds = dx 2 + dy 2
2
2πxe− x dx = −π e− x (−2 x)dx = −π eu 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
−1
= π(1 − e ) ≈ 1.99
0
1
1− e
1− e
( x − 0);
−1 =
⇒ y −1 =
1− 0 e
e
e
1− e
y=
x +1
e
=
53. a.
1
1 ⎡⎛ 1 − e
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 − 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
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
Let u = − x 2 , so du = –2x dx.
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
2
1+1
=
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
Instructor’s Resource Manual
Section 6.3
365
© 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.
x2
F ( x) =
c.
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
2 2
1 + [ln( e ) ]
⋅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
a.
b.
1
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
55. a.
−3
3
⎞
⎛ 1 ⎞
⎟ dx = 2 0 exp ⎜ − 2 ⎟ dx ≈ 3.11
⎠
⎝ x ⎠
8π −0.1x
e
sin x dx
0
b.
≈ 0.910
lim (1 + x)1 x = e ≈ 2.72
56. a.
x →0
lim (1 + x)−1 x =
b.
57.
⎛ 1
exp ⎜ −
⎝ x2
x →0
f ( x) = e
1
≈ 0.368
e
a.
− x2
f ( x) = −2 xe− x
b.
2
2
2
2
366
c.
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
© 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.
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
4
4. x = 64
x = 4 64 = 2 2
⎛ x⎞ 1
5. log9 ⎜ ⎟ =
⎝3⎠ 2
x
= 91/ 2 = 3
3
x=
1
2x
1
3
2⋅4
=
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
14. x ln 5 = ln 13
ln13
x=
≈ 1.5937
ln 5
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
ln12 = ln 4
–1
ln12
= –1
ln 4
ln12
= 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
© 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
20. Dx log10 ( x3 + 9) =
3x
=
21.
3
( x + 9) ln10
⋅ Dx ( x3 + 9)
26.
1
(1) + ln( z + 5) ⋅ 3z ln 3
z +5
⎡ 1
⎤
= 3z ⎢
+ ln( z + 5) ln 3⎥
⎣z +5
⎦
(
=D
=
=
2
)=D
2
2 –1
ln 3
ln10
2
(
2
–
=–
2
2
–
– ) –1/ 2 (2 –1)
27.
1 u
1 2u
2 du = ⋅
+C
2
2 ln 2
2
=
5
x
x
=
1
2 x
28.
dx.
5u
+C
ln 5
2⋅5 x
+C
ln 5
4
⎡5 x ⎤
25
5 ⎞
⎥ = 2 ⎛⎜
dx = 2 ⎢
−
⎟
1
ln
5
ln
5
ln
5⎠
⎢
⎥
x
⎝
⎣
⎦1
40
=
≈ 24.85
ln 5
368
Section 6.4
2
2
d ( x2 )
d
10
= 10( x ) ln10 x 2 = 10( x ) 2 x ln10
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
105 x –1
+C
5ln10
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
24. Let u = 5x – 1, so du = 5 dx.
1
1 10u
105 x –1 dx = 10u du = ⋅
+C
5
5 ln10
1
⎡103 x –10 –3 x ⎤
Thus, (10 + 10 )dx = ⎢
⎥
0
⎢⎣ 3ln10 ⎥⎦ 0
1 ⎛
1 ⎞ 999,999
=
⎜1000 –
⎟=
3ln10 ⎝
1000 ⎠ 3000 ln10
≈ 144.76
23. Let u = x 2 so du = 2xdx.
x ⋅ 2 x dx =
10 –3 x
+C
3ln10
1
– ) log10 3
– ) ln 3
ln 3
=
⋅D
ln10
ln10
ln 3 1
⋅ (
ln10 2
2
1
0
103 x dx + 10 –3 x dx
=
= 3z ⋅
2–
1
0
103 x
+C
3ln10
Now let u = –3x, so du = –3dx.
1
1 10u
10 –3 x dx = – 10u du = – ⋅
+C
3
3 ln10
Dz [3z ln( z + 5)]
log10 (3
(103 x + 10 –3 x )dx =
Let u = 3x, so du = 3dx.
1
1 10u
103 x dx = 10u du = ⋅
+C
3
3 ln10
2
( x3 + 9) ln10
22. D
1
0
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)
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.
30.
x
x
d (e x )
d
2
= 2(e ) ln 2 e x = 2(e ) e x ln 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
x
2 +1)
(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
9
−3
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.
f ( x) = (ln 2) 2 2− x
4
⎛ ln( x 2 + 1) 2 x ln x ⎞
= ( x 2 + 1)ln x ⎜
+
⎟
⎜
x
x 2 + 1 ⎟⎠
⎝
dy
=e
dx
,
Domain = (−∞, ∞)
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
32. y = (ln x 2 ) 2 x +3 = e(2 x +3) ln(ln x
−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.
= 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
8
−2
x
g (e) = πe π−1 ≈ 26.74
g (e) is larger than f (e) .
−3
Instructor’s Resource Manual
Section 6.4
369
© 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.
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
f
−
−
−
0
+
+
+
f
−
0
+
+
+
0
−
y
−5
39.
f ( x) =
2
dt
f ( x) = 2− x ,
Domain = (−∞, ∞)
f ( x) = −2(ln 2) x 2− x
2
2
(−∞, 0) 0 (0, ∞)
x
f
+
+
+
f
+
0
−
(0,
−5
5
x
0 − t2
1
2
dt ) ≈ (0, −0.81)
y
5
−5
f ( x) = x log3 ( x 2 + 1) =
x ln( x 2 + 1)
. Since
ln 3
x 2 + 1 > 0 for all x, domain = (−∞, ∞)
f ( x) =
x − t2
1
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
5
x
x
(−∞, −1) −1 (−1, 0) 0 (0,1) 1 (1, ∞)
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)
38.
5
⎤
1 ⎡ 2 x2
2 ⎡ x3 + 3 x ⎤
⎢
⎢
⎥
+ ln( x 2 +1) ⎥ , f ( x ) =
ln 3 ⎢ x 2 +1
ln 3 ⎢ x 2 +1 ⎥
⎥⎦
⎣
⎣
⎦
−5
5
x
−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
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.
40.
f ( x) =
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
44. 115 = 20 log10 (121.3P )
log10 (121.3P ) = 5.75
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
+
0
+
frequency of C, r = 21/12 ≈ 1.0595
f
−
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)
2 (p times) and has only powers of 2
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
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.
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 =
p
where p and q are integers,
q
q ≠ 0 . Then 2 p q = 3 or 2 p = 3q. But
2p = 2⋅2
y
−5
46. Assume log 2 3 =
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.
Instructor’s Resource Manual
Section 6.4
371
© 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.
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
d 2
( x ln x)
dx
2
1⎞
⎛
= e x ln x ⎜ 2 x ln x + x 2 ⋅ ⎟
x⎠
⎝
f ( x) = e x
2 ln 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 x ln x d
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,
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.
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.
1
x
1
+
−
x →0
x2
= 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
Since fu ( x ) > 0 on (0, u) and fu ( x ) < 0 on
(u, ∞ ), fu ( x) attains its maximum at x0 = u.
fu (u ) > fu (u + 1) means
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
b.
1
x
Therefore, lim x x = e0 = 1 .
Therefore, when x ≠ e , ln x < ln e , which
52. a.
x →0 +
= 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.
u u e −u > (u + 1)u e −(u +1) .
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
x
dx
0
≈ 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 ⎠
Instructor’s Resource Manual
Section 6.4
373
© 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.
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
r
(7, 30) and (10, 70). Thus, 30 = C 7 and
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
r
r
(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
0.47
.
Hence, the last equation is y = 2 x
The given answers are only approximate.
Student answers may also vary.
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
Section 6.5
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.
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.
Instructor’s Resource Manual
Section 6.5
375
© 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.
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
110 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) = 81.6 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 20 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) = 83.7 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
= k dt or ln T(t)-T1 = kt + C
T − T1
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 →∞
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.
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
12t
=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.
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
(
)
Instructor’s Resource Manual
Section 6.5
377
© 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.
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.
y = 15e0.012⋅25 − 5 ≈ 15.25. The population in 2010
will be about 15.25 million.
=0
n→∞ (1 + ε ) n
x →0 −
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
dN = k dt
L−N
–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 ) .
1/ x
= lim (1 + 2 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
= a dt
y + ba
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
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.
43.
e.
f ( x)
= k > 0 can be written as
f ( x)
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
© 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.
6.6 Concepts Review
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)
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
(
)
P( x)dx
1 d ⎛
; ⎜
x dx ⎝
)
y⎞
2
⎟ = 1; x + Cx
x⎠
4. particular
h →0
kx
(
Problem Set 6.6
kx
=e .
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
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.
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 ⎤
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
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 ] = xe2 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.
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
Integrating factor: e
3dx
y (t ) = 400 – 350e – t / 50 goes through (0, 50).
= e3 x
y (40) = 400 – 350e –0.8 ≈ 242.735 lb of salt
D[ ye3 x ] = e5 x
Then ye3 x =
ye3 x =
dy
y
⎛ y ⎞
= (2)(4) – ⎜
=8
⎟ (4) or y +
dt
50
⎝ 200 ⎠
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).
Section 6.6
381
© 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.
18.
dy
–2 y
2
=
y = 0.
or y +
dt 50 + t
50 + t
Integrating factor:
2
⎛
⎞
exp ⎜
dt ⎟ = e2 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 )
b. Let A be the number of gallons of pure alcohol
drained away.
200
(100 – A) + 0.25A = 50 ⇒ A =
3
It took
I (t ) = 10 –6 + C exp(–106 t )
( )
I (t ) = 10 –6 [1 – exp(–106 t )] goes through (0, 0).
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
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 ⎠
c would need to satisfy
200
3
5
c>
d.
+
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 .
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 )
20. 3.5I = 120sin 377t
dy
⎛ 5 ⎞
= 5(0.25) – ⎜
⎟ y = 1.25 – 0.05 y
dt
⎝ 100 ⎠
y =
(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
Section 6.6
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.
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
P ( x ) dx +C
= Q ( x )e
P ( x ) dx +C
d ⎛ P ( x ) dx +C ⎞
y ⎟ = Q ( x )e
⎜e
⎟
dx ⎜⎝
⎠
ye
P ( x ) dx +C
y=e
−
=
P ( x ) dx
+ C2 e
Instructor’s Resource Manual
P ( x ) dx C
e dx + C1
Q ( x )e
Q ( x )e
−
P ( x ) dx
dx
P ( x ) dx
Section 6.6
383
© 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.
6.7 Concepts Review
5.
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
Section 6.7
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.
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
The integrating factor is e
1dx
= ex .
e x y '+ ye x = e x ( x + 2)
( )
d x
e y = ( x + 2) e x
dx
x
Note: Solutions to Problems 22-28 are given along with
the corresponding solutions to 11-16.
11., 22.
e y = ( x + 2) e dx
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
x
xn
x
x
x
e y = ( x + 2)e − e + C
−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
.
10.
Instructor’s Resource Manual
12., 23.
Euler's
Method yn
3.0
Improved Euler
Method yn
3.0
0.2
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
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
© 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., 24.
xn
0.0
Euler's
Method yn
0.0
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
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
14., 25.
xn
= (1 + h) yn −1 = (1 + h) n y0 = (1 + h )
n
0.2
0.0
0.004
0.4
0.008
0.024
0.6
0.040
0.076
(1 + 1/ N ) N
0.8
0.112
0.176
we know that lim (1 + 1/ N )
1.0
xn
1.0
16., 27.
Improved Euler
Method yn
0.0
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,
0.0
15., 26.
Euler's
Method yn
0.0
0.240
Euler's
Method yn
1.0
Improved Euler
Method yn
1.0
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
1.0
Euler's
Method yn
2.0
Improved Euler
Method yn
2.0
≈ e for large N. From Chapter 7,
N
N →∞
0.340
1.2
xn
b.
=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 )
= h [ f ( x0 ) + f ( x1 ) + f ( x2 ) ] = h
3−1
i =0
f ( xi )
At the nth step of Euler's method,
yn = yn −1 + hf ( xn −1 ) = h
19. a.
x1
x0
y '( x)dx =
n −1
i =0
f ( xi )
x1
sin x 2 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
Section 6.7
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.
c.
x3
x0
x3
sin x 2 dx
x0
y '( x)dx =
x3
x0
c.
+ ( 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
y ( x3 ) ≈ 0.004999
Continuing in this fashion, we have
xn
x0
y '( x)dx =
y ( xn ) ≈ h
xn
x0
xn
sin x 2 dx
x0
n −1
y ( xn ) − y ( x0 ) ≈
i =0
( xi +1 − xi ) sin xi2
y ( xn ) ≈ h
n −1
i =0
f ( xi −1 )
The result y ( xn ) ≈ h
b.
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
xn
0
f ( x) dx ,
c.
are equivalent.
x1
x0
y '( x)dx =
x1
x0
y ( x1 ) − 0 ≈ 0.1 0 + 1
y ( x1 ) ≈ 0.1
x2
x0
x + 1 dx
y ( x2 ) − y ( x0 ) ≈ ( x1 − x0 ) x0 + 1
+ ( x2 − x1 ) x1 + 1
y ( x2 ) − y (0) = h x0 + 1 + h x1 + 1
y ( x2 ) − 0 ≈ 0.1 0 + 1 + 0.1 0.1 + 1
y ( x2 ) ≈ 0.204881
xi −1 + 1
2. yn −1 + hf ( xn −1 , yn −1 )
x + 1 dx
y ( x1 ) − y (0) = h x0 + 1
y '( x)dx =
i =0
( xi +1 − xi ) xi −1 + 1
h
y1 − y0 = [ f ( x0 , y0 ) + f ( x1 + yˆ1 )] ⇒
2
h
y1 = y0 + [ f ( x0 , y0 ) + f ( x1 + yˆ1 )]
2
1. xn −1 + h
h
3. yn −1 + [ f ( xn −1 , yn −1 ) + f ( xn , yˆ n )]
2
y ( x1 ) − y ( x0 ) ≈ ( x1 − x0 ) x0 + 1
x2
x0
n −1
i =0
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.
f ( xi −1 ) is the same as
Riemann sum for approximating
xn
x + 1 dx
x0
n −1
When n = 10 , this becomes
y ( x10 ) = y (1) ≈ 1.198119
n −1
i =0
y '( x)dx =
y ( xn ) − y ( x0 ) ≈
When n = 10 , this becomes
y ( x10 ) = y (1) ≈ 0.269097
b.
x + 1 dx
+ ( x2 − x1 ) x1 + 1 + ( x3 − x2 ) x2 + 1
+ 0.1sin 0.22
20. a.
x3
x0
y ( x3 ) − y ( x0 ) ≈ ( x1 − x0 ) x0 + 1
y ( x3 ) − y ( x0 ) ≈ ( x1 − x0 ) sin x02
d.
y '( x)dx =
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
0.01
0.013468
0.000045
0.005
0.006765
0.000011
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
Instructor’s Resource Manual
Section 6.7
387
© 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.
6.8 Concepts Review
⎡ π π⎤
1. ⎢ – , ⎥ ; arcsin
⎣ 2 2⎦
1 ⎞
⎛
13. cos(arccot 3.212) = cos ⎜ arctan
⎟
3.212 ⎠
⎝
≈ cos 0.3018 ≈ 0.9548
14. sec(arccos 0.5111) =
⎛ π π⎞
2. ⎜ – , ⎟ ; arctan
⎝ 2 2⎠
3. 1
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 ⎜⎜ –
⎟⎟ = – since sin ⎜ – ⎟ = –
2
3
3
2
⎝
⎠
⎝
⎠
⎛
2⎞
π
2
⎛ π⎞
4. sin –1 ⎜⎜ –
⎟⎟ = – 4 since sin ⎜⎝ – 4 ⎟⎠ = – 2
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
cos(arccos 0.5111)
Section 6.8
17. cos(sin(tan −1 2.001)) ≈ 0.6259
18. sin 2 (ln(cos 0.5555)) ≈ 0.02632
19.
= sin −1
x
8
20.
= tan −1
x
6
21.
= sin −1
5
x
22.
= cos −1
9
or
x
= sec−1
x
9
23. Let
1
be the angle opposite the side of length 3,
and
2
=
and tan
24. Let
and
1
2
1
= .
x
1
= tan –1
–
2.
Then tan
1
=
3
x
3
1
– tan –1 .
x
x
be the angle opposite the side of length 5,
2 = 1 − , and y the length of the unlabeled
1
=
side. Then
tan
=
– , so
1
=
5
=
y
= tan −1
1− 2
5
2
and y = x 2 − 25.
, tan
2
=
2
=
y
2
2
x − 25
x − 25
5
2
− tan −1
2
2
x − 25
x − 25
,
⎡
⎡
⎛ 2 ⎞⎤
⎛ 2 ⎞⎤
25. cos ⎢ 2sin –1 ⎜ – ⎟ ⎥ = 1 – 2sin 2 ⎢sin –1 ⎜ – ⎟ ⎥
⎝ 3 ⎠⎦
⎝ 3 ⎠⎦
⎣
⎣
2
1
⎛ 2⎞
= 1– 2⎜ – ⎟ =
9
⎝ 3⎠
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.
()
()
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
=
29. tan(sin –1 x) =
30. sin(tan –1 x) =
=
2
4
16
⎛ 12 ⎞
⎛ 4 ⎞ 12
⋅ 1– ⎜ ⎟ – 1– ⎜ ⎟ ⋅ = –
5
13
5
13
65
⎝ ⎠
⎝ ⎠
sin(sin –1 x)
cos(sin –1 x)
1
csc(tan
1
1+
1
tan 2 (tan –1 x )
=
–1
x)
1
x2
34. a.
1 – x2
1
=
1
1+
x
=
1 + cot 2 (tan –1 x)
=
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
since
2
lim
→π / 2−
π
lim tan x = – since
2
x→ – ∞
lim tan = −∞
–1
→−π / 2+
=
35. a.
⎛1⎞
lim sec –1 x = lim cos –1 ⎜ ⎟
x →∞
x →∞
⎝x⎠
π
= lim cos –1 z =
2
z → 0+
⎛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
x →1−
1 – x2
sin(sin −1 x) = x,
tan = ∞
Thus, since sin is continuous, the Composite
lim sin(sin −1 x) = lim x = 1 .
x →1−
x →1−
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 .
⎣
⎦
Instructor’s Resource Manual
Section 6.8
389
© 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.
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 + sec 2 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
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.
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
⎡
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
=
1
2
3(1 + sin –1 x)2
(
1 – x2
x +4
Chain Rule:
dy dy du
1
du
=
=
⋅
=
2 dx
dx du dx
1− u
⎛ −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 ⎥⎦
(
; then y = sin −1 ( u ( x) ) so by the
)
⎤
⎥
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 =
)
⎤
⎥
2
⎥ + 1 ⋅ arcsec(x + 1)
2
2
( x + 1) − 1 ⎥
⎦
2x
55.
)
cos 3x dx
Let u = 3 x, du = 3dx ; then
1
cos 3 x (3dx) =
3
1
1
1
cos u du = sin u + C = sin 3 x + C
3
3
3
cos 3 x dx =
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 =
sin 2 x(2 cos 2 x)dx
2
1
=
u du
2
u2
1
=
+ C = sin 2 2 x + C
4
4
58. Let u = cos x, so du = − sin x dx .
tan x dx =
sin x
1
(− sin x)dx
dx = −
cos x
cos x
1
du = − ln u + C = − ln cos x + C
u
= ln sec x + C
=−
Instructor’s Resource Manual
Section 6.8
391
© 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.
65. Let u = 2x, so du = 2 dx.
1
1
1
dx =
2dx
2
2 1 + (2 x)2
1+ 4x
59. Let u = e2 x , so du = 2e2 x dx .
1
cos(e2 x )(2e2 x )dx
e2 x cos(e2 x )dx =
2
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
⎤
cos(e2 x ) dx = ⎢ sin(e2 x ) ⎥
⎣2
⎦0
1
⎡1
⎤
= ⎢ sin(e2 ) − sin(e0 ) ⎥
2
⎣2
⎦
1 2x
e
0
66. Let u = e x , so du = e x dx .
ex
2x
60. Let u = sin x, so du = cos x dx.
u3
sin 3 x
sin 2 x cos x dx = u 2 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
Let u =
1
=
dx
2
2
−1
x2 − 1
x
= ⎡sec−1 x ⎤
⎣
⎦
2⎞ π π π
⎜⎜
⎟⎟ = − =
⎝ 2 ⎠ 3 4 12
1
−1
d =−
dx
⎛ 3 ⎞
12⎜ 1− x 2 ⎟
⎝ 4 ⎠
1
1
2 3
⎛ 3 ⎞
x ⎟⎟
1−⎜⎜
⎝ 2 ⎠
2
dx
3
3
x, du =
dx ; then
2
2
1
1 ⎛ 2 ⎞
dx =
⎜
⎟
2
2 3⎝ 3⎠
⎛ 3 ⎞
x
12 − 9 x
1
1− u2
du
2
dx . Let u = 12 − 9 x 2 , du = −18 x dx;
dx = −
1
18
1
(−18 dx)
12 − 9 x 2
12 − 9 x 2
1 1
⎛ 1⎞
=−
du = ⎜ − ⎟ (2 u ) + C
18 u
⎝ 18 ⎠
=−
1 + cos
1 + cos 2
1
du = − tan −1 u + C
=−
2
1+ u
1
dx =
1−⎜⎜
x ⎟⎟
⎝ 2 ⎠
x
x ⎤ = tan 1 − tan (−1)
⎦ −1
1
2
du
then
−1
64. Let u = cos , so du = − sin d .
2
2
−1 ⎛
dx = ⎡ tan
⎣
sin
2
2
−1
1
dx =
⎛ 3 ⎞
1
1
x ⎟⎟ + C
= sin −1 u + C = sin −1 ⎜⎜
3
3
⎝ 2 ⎠
68.
π ⎛ π⎞ π
−⎜− ⎟ =
4 ⎝ 4⎠ 2
=
12 − 9 x
2 3
⎜ ⎟ − cos
⎝2⎠
−1 1 + x 2
1
=
dx = [arcsin x]0 2 / 2
−1 ⎛ 1 ⎞
1
1
=
67.
π
2
– arcsin 0 =
2
4
= arcsin
62.
2
x 2
1+ e
1 + (e )
1 + u2
= arctan u + C = arctan ex + C
sin e2 − sin1
≈ 0.0262
2
=
ex
dx =
12 − 9 x 2
+C
9
(− sin )d
= − tan −1 (cos ) + C
π/2
0
sin
1 + cos 2
= − tan −1 0 + tan −1 1 = −0 +
392
π/2
d = ⎡ − tan −1 (cos ) ⎤
⎣
⎦0
Section 6.8
π π
=
4 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.
1
69.
2
=
x − 6 x + 13
1
1
dx =
2
( x − 6 x + 9) + 4
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 1 be the angle between the viewer’s
line of sight to the top of the picture and the
horizontal. Then call 2 = 1 − , so = 1 − 2 .
dx
dx
( x − 3)2 + 4
Let u = x − 3, du = dx, a = 2; then
1
⎛u⎞
dx =
du = tan −1 ⎜ ⎟ + C
2
2
2
a
⎝a⎠
( x − 3) + 4
u +a
1
=
1
1
⎛ x−3⎞
tan −1 ⎜
⎟+C
2
⎝ 2 ⎠
1
70.
2
2 x + 8 x + 25
1
2
dx =
1
⎛ 17 ⎞
⎟
⎝ 2 ⎠
( x + 2) 2 + ⎜
2
1
2
2( x + 4 x + 4 +
17
)
2
dx =
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
x 4x − 9
1
du =
u u 2 − a2
1 −1 ⎛ 2 x
sec ⎜
3
⎝ 3
72.
x +1
dx =
1
dx =
2
2 x 4 x2 − 9
(2 dx) =
1 −1 ⎛ u ⎞
sec ⎜ ⎟ + C =
a
⎝a⎠
⎞
⎟+C
⎠
x
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
x +1
7.6
2.6
; tan 2 =
;
b
b
7.6
2.6
= tan −1
− tan −1
b
b
If b = 12.9, ≈ 0.3335 or 19.1° .
tan 1 =
⎛ 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 ⎠
Instructor’s Resource Manual
Section 6.8
393
© 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
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
( )⎤⎦ + 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
–
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.
( )
( )
⎡
⎛1⎞
⎛ 5 ⎞⎤
tan ⎢3 tan –1 ⎜ ⎟ + tan –1 ⎜ ⎟ ⎥
4
⎝ ⎠
⎝ 99 ⎠ ⎦
⎣
=
⎡
⎡
⎛ 1 ⎞⎤
⎛ 1 ⎞⎤
tan ⎢ 4 tan –1 ⎜ ⎟ ⎥ – tan ⎢ tan –1 ⎜
⎟⎥
⎝ 5 ⎠⎦
⎝ 239 ⎠ ⎦
⎣
⎣
=
⎡
⎡
⎛ 1 ⎞⎤
⎛ 1 ⎞⎤
1 + tan ⎢ 4 tan –1 ⎜ ⎟ ⎥ tan ⎢ tan –1 ⎜
⎟⎥
5
⎝
⎠
⎝ 239 ⎠ ⎦
⎣
⎦
⎣
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
represent ∠DAB, then ∠CAB is
Let
2
. Since
b
ΔABC is isosceles, AE =
b
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
= 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
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.
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
–1 < x < 1
Proof: Let = arcsin x, so x = sin
x
sin
sin
Then
=
=
cos
1 – x2
1 – sin 2
x
so = arctan
.
1 – x2
for
dx
81.
a2 – x2
=
= tan
dx
=
( )
2⎤
⎡
a 2 ⎢1 – ax ⎥
⎣
⎦
1
dx
1
dx
⋅
=
⋅
2
a
a
1 – ax
1 – ax
( )
( )
2
since a > 0
x
1
, so du = dx.
a
a
1
dx
du
⋅
=
= sin −1 u + C
2
2
a
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 = – arcsin x
2
⎛π
⎞
Then x = sin ⎜ − ⎟ = cos
2
⎝
⎠
so = 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
=
dx
2
2
2
a 1+ x a
a +x
83. Let u =
(a)
1
1
1
du = tan −1 u + C
a 1+ u2
a
1
⎛ x⎞
= tan −1 ⎜ ⎟ + C
a
⎝a⎠
=
Instructor’s Resource Manual
Section 6.8
395
© 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.
x
, so du = (1/ a ) dx. Since a > 0,
a
dx
1
1
1
=
dx
2
2
2
a
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
be the angle subtended by viewer’s eye.
⎛ 12 ⎞
⎛2⎞
= tan −1 ⎜ ⎟ − tan −1 ⎜ ⎟
b
⎝ ⎠
⎝b⎠
⎛ 12 ⎞
d
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
=
a − x2 +
(−2 x)
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)
d
Since
> 0 for b in ⎡⎣0, 2 6
db
d
and
< 0 for b in 2 6, ∞ , the angle is
db
)
1 2
1 − x2 + a2
a − x2 +
= a2 − 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⎞
⎛ x⎞
= cos −1 ⎜ ⎟ − cos −1 ⎜ ⎟
⎝b⎠
⎝a⎠
⎛
⎞
⎛
⎜
⎟ ⎛ 1 ⎞⎛ dx ⎞ ⎜
−1
−1
d
=⎜
⎟ ⎜ ⎟⎜ ⎟ − ⎜
2 ⎝ b ⎠⎝ dt ⎠ ⎜
dt ⎜
⎜ 1 − bx ⎟⎟
⎜ 1 − ax
⎝
⎠
⎝
( )
b.
( )
⎛ a+x
= tan −1 ⎜
⎜ 2
2
⎝ b −x
⎛
⎜
⎜
d
1
=⎜
dt ⎜ ⎛
a+ x
⎜⎜ 1 + ⎜ 2 2
⎝ ⎝ b −x
⎞
⎟
⎠
2
⎞
⎟ ⎛ b2 − x2 + (a + x) x
⎟⎜
b2 − x2
⎟⎜
2
2
b −x
⎟⎜
⎟⎟ ⎜⎝
⎠
⎞
⎛
⎞
⎟ ⎛ 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
⎣
Section 6.8
⎞ dx
⎟
⎟ dt
⎠
⎞
⎛ x⎞
⎟ − sin −1 ⎜ ⎟
⎟
⎝b⎠
⎠
⎡⎛
b2 − x2
= ⎢⎜
2
2
⎜ 2
⎣⎢⎝ b − x + (a + x)
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
© 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.
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 (t ) denote the angle of elevation.
⎛ 15t ⎞
−1 ⎛ t ⎞
Then h(t) = 15t, so (t ) = tan −1 ⎜
⎟ = tan ⎜ ⎟ .
⎝ 60 ⎠
⎝4⎠
d
1 ⎛1⎞
4
=
⎜ ⎟=
dt 1 + t 2 ⎝ 4 ⎠ 16 + t 2
(4)
d
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 (t ) = tan −1 ⎜
⎟ = tan ⎜
x
(
t
)
⎝
⎝
⎠
⎛
−2
1
d
so
=
⎜⎜
2
2
dt
⎝ ( 5 − 10t )
1 + 2 / 5 − 10t
( (
=
20
( 5 − 10t ) 2 + 4
When t = 0,
91.
))
⎞
⎟,
5 − 10t ⎠
⎞
⎟⎟ (−10)
⎠
2
.
d
20
=
≈ 2.22 radians per minute.
dt
9
Let x represent the position on the shoreline and let
represent the angle of the beam (x = 0 and = 0
when the light is pointed at P). Then
d
1
1 dx
2 dx
⎛x⎞
= tan −1 ⎜ ⎟ , so
=
=
2
dt 1 + x 2 dt 4 + x 2 dt
⎝2⎠
92 Let x represent the length of the rope and let
represent the angle of depression of the rope.
⎛8⎞
Then = sin −1 ⎜ ⎟ , so
⎝x⎠
d
1
8 dx
8
dx
=
−
=−
.
2 dt
2
2
dt
dt
x
8
−
x
x
64
1− x
( )
dx
= −5 , we obtain
dt
d
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 represent the angle subtended by the
⎛ 6376 ⎞
earth. Then = 2sin −1 ⎜
⎟ , so
⎝ x ⎠
d
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
d
≈ 3.96 × 10−4 radians
these values, we obtain
dt
per second.
=−
12, 752
(2)
When x = 1,
dx
d
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.
Instructor’s Resource Manual
Section 6.8
397
© 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.
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 e 2 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 e 2 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
Section 6.9
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.
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 3x ⋅ 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
Instructor’s Resource Manual
Section 6.9
399
© 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
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(3 x + 2)dx =
sinh u du = cosh u + C
3
3
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
=
1
1 – (coth x)2
(–csch 2 x) =
–csch 2 x
–csch 2 x
=1
⎡1
⎤ ln 3
cosh 2 xdx = ⎢ sinh 2 x ⎥
0
⎣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
= Dx tanh (coth x)
ln 3
1
cosh u du
2π
1
2 z
dz .
dz = 2 cosh u du = 2sinh u + C
= 2sinh z + C
41. Let u = 2 z1/ 4 , so du =
sinh(2 z1/ 4 )
4 3
1
1
⋅ 2 z –3 / 4 dz =
dz.
4
4
2 z3
dz = 2 sinh u du = 2 cosh u + C
z
= 2 cosh(2 z1/ 4 ) + C
1 ln 9 ln 19
1⎛
1 ⎞ 20
(e − e ) = ⎜ 9 − ⎟ =
4
4⎝
9⎠
9
Section 6.9
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.
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
1
1 u2
u du = ⋅
+C
2
2 2
1
= [ln(sinh x 2 )]2 + C
4
46. Area =
ln 5
– ln 5
cosh 2 x dx = 2
0
47. Note that the graphs of y = sinh x and y = 0
intersect at the origin.
=
ln 2
tanh x dx
8
tanh x dx = 2
1
0
0
π cosh 2 x dx =
π 1
(1 + cosh 2 x)dx
2 0
π ⎛ sinh 2
⎞
− 0⎟
⎜1 +
2⎝
2
⎠
π π sinh 2
= +
≈ 4.42
2
4
50. Volume =
cosh 2 x dx
⎡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
0
sinh 2 x ⎤
π⎡
x+
⎢
2⎣
2 ⎥⎦ 0
=π
ln 5
ln 5
Area =
8
=
⋅ cosh x 2 ⋅ 2 xdx = 2 x coth x 2 dx .
x coth x 2 ln(sinh x 2 )dx =
8
0
49. Volume =
=
45. Let u = ln(sinh x 2 ) , so
1
−8
(− tanh x) dx +
1
1
= [ln(cosh x)]2 + C
2
du =
0
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
=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 =
2
sinh x dx = [cosh x]ln
0
eln 2 + e− ln 2 e0 + e0 1 ⎛
1⎞
1
−
= ⎜ 2 + ⎟ −1 =
2⎝
2⎠
4
2
2
=π
ln10
0
ln10 ⎛ e x
0
⎜
⎜
⎝
ln10 e 2 x
0
π sinh 2 xdx
− e− x
2
2
⎞
⎟ dx
⎟
⎠
– 2 + e –2 x
π
dx =
4
4
ln10
0
(e 2 x – 2 + e –2 x )dx
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
1
0
1
0
1
0
( ) dx
dy 2
2πy 1 + dx
2π cosh x 1 + sinh 2 x dx
2π cosh x cosh x dx
π(1 + cosh 2 x)dx
1
π
π
⎡
⎤
= ⎢ πx + sinh 2 x ⎥ = π + sinh 2 ≈ 8.84
2
2
⎣
⎦0
Instructor’s Resource Manual
Section 6.9
401
© 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.
2
1
⎛ dy ⎞
2πy 1 + ⎜ ⎟ dx = 2π sinh x 1 + cosh 2 xdx
0
0
⎝ dx ⎠
Let u = cosh x, so du = sinh x dx
1
⎡u
⎤
2π sinh x 1 + cosh 2 xdx = 2π 1 + u 2 du = 2π ⎢ 1 + u 2 + ln u + 1 + u 2 + C ⎥
2
⎣2
⎦
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
⎡
⎤
2π sinh x 1 + cosh 2 xdx = π ⎢cosh x 1 + cosh 2 x + ln cosh x + 1 + cosh 2 x ⎥
0
⎣
⎦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
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.
c.
Length of the curve is
24
−24
2
⎛ dy ⎞
1 + ⎜ ⎟ dx =
⎝ dx ⎠
24
−24
2⎛
x ⎞
1 + sinh ⎜ ⎟dx =
⎝ 24 ⎠
24
⎡
⎛ x ⎞
⎛ x ⎞⎤
cosh ⎜ ⎟ dx = ⎢ 24sinh ⎜ ⎟ ⎥
= 48sinh1 ≈ 56.4
−24
⎝ 24 ⎠ ⎦ −24
⎝ 24 ⎠
⎣
24
Surface area ≈ (56.4)(100) = 5640 ft 2
56. Area =
1
cosh t sinh t −
2
cosh t
1
x 2 − 1 dx =
cosh t
1
1
⎡1
⎤
cosh t sinh t − ⎢ x x 2 − 1 − ln x + x 2 − 1 ⎥
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
=
so y =
t
0
sech u du by the Fundamental
Theorem of Calculus.
(0, gd(0)) = (0, tan −1 0) = (0, 0).
Instructor’s Resource Manual
Section 6.9
403
© 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.
59. Area =
x
0
cosh t dt = [sinh t ]0x = sinh x
61.
Arc length =
x
1 + [ Dt cosh t ]2 dt =
0
=
x
0
x
0
1 + sinh 2 tdt
cosh t dt = [ sinh t ]0 = sinh x
x
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:
3. True:
ln 0 is undefined.
d2y
dx
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
Section 6.10
x
) – ln(2e ) = ln
= ln e = 1
404
f ( g ( x)) = 4 + eln( x − 4)
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
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.
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
Section 6.10
405
© 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.
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 ⎠
⎝
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
11.
4.
x2 −4 x
d
d 5
1
log10 ( x5 − 1) =
( x − 1)
5
dx
( x − 1) ln10 dx
=
12.
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
2
1−
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
2
= sec x
sec 2 x
d
1
⎛x⎞
ln sin 2 ⎜ ⎟ =
2
2
dx
⎝ ⎠ sin
=
5x4
5.
sec 2 x
=
d
sec−1 e x =
dx
ex
=
2
d x2 −4 x
d 2
e
= e x −4 x
( x − 4 x)
dx
dx
= (2 x − 4)e
d
tan x
dx
tan x + 1
2
d
2sin −1 3 x =
dx
=
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
3.
tan 2 x + 1
1
≈ 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.
46. False:
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
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.
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
3dx =
e3 x −1dx =
e
e du
3
3
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
26. Let u = sin 3x, so du = 3 cos 3x dx.
1
1
6 cot 3x dx = 2
3cos 3x dx = 2 du
sin 3 x
u
= 2 ln u + C = 2 ln sin 3 x + C
d
4 tan 5 x sec5 x
dx
27. Let u = e x , so du = e x dx .
= 20sec 2 5 x sec 5 x + 20 tan 5 x sec 5 x tan 5 x
= 20sec 5 x(sec2 5 x + tan 2 5 x)
= 20sec 5 x(2sec 2 5 x − 1)
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 ⎠⎦
⎣
⎛ x2
d
tan −1 ⎜
⎜ 2
dx
⎝
⎞
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
e 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
⎞ 4x
⎟+ 4
⎟ x +4
⎠
2
(
dx = 3
1
(2 x + 1)dx
x + x −5
x + x −5
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
2
⎛ x2 ⎞⎤
⎛ x2 ⎞
⎛ 4x ⎞
d ⎡
⎢ x tan −1 ⎜ ⎟ ⎥ = (1) tan −1 ⎜ ⎟ + ( x) ⎜ 4
⎟
⎜
⎟
⎜
⎟
dx ⎢⎣
⎝ x +4⎠
⎝ 2 ⎠ ⎥⎦
⎝ 2 ⎠
2
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
2
)
Section 6.10
407
© 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. Let u = e x +3 + 1, so du = e x +3 dx .
e
x+2
33. Let u = ln x, so du =
1
1
1 1
e x +3 dx =
du
3
x
+
e e
e u
+1
+1
e
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
⎠
dx =
x +3
=
e x +3e−1
e x +3 + 1
=
−1
2
=
e x +3 + 1
2
= tanh( x − 3) + C
Check:
d
d
[tanh( x − 3)] = sech 2 ( x − 3) ( x − 3)
dx
dx
= 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
35.
π
4
f ( x) > 0 when cos x > sin x which occurs 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
π
π
≤x< .
2
4
f ( x) = – sin x – cos x; f ( x) = 0 when
–
32. Let u = sin x, so du = cos x dx.
cos x
1
dx =
du = tan −1 u + C
2
1 + sin x
1+ u2
408
f ( x) = cos x – sin x; f ( x) = 0 when tan x = 1,
x=
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
x + x(ln x) 2
sech 2 ( x − 3)dx = sech 2 u du = tanh u + C
= 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
4 x cos x dx = 2 (cos x )2 x dx = 2 cos u du
1
1
⋅ dx
1 + (ln x) x
2
x + x(ln x)
1
=−
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+ 2
30. Let u = x 2 , so du = 2x dx.
dx = −
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 –
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.
⎛ 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 =
, so = tan −1
.
x
x
d
1
⎛ 500 ⎞ dx
=
⎜−
⎟
2
dt 1 + 500 ⎝ x 2 ⎠ dt
(x)
=−
500
dx
x + 250, 000 dt
2
When
= 30°, x =
500
= 500 3 and
tan 30°
d
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.
Instructor’s Resource Manual
Section 6.10
409
© 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. (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.
Integrating factor: e
2
2
2
Therefore, y = 1 + 2e – x .
46. Integrating factor is e – ax .
D[ ye – ax ] = 1; y = e ax ( x + C )
47. Integrating factor is e –2 x .
48. a.
Q (t ) = 3 – 0.02Q
b.
Q (t ) + 0.02Q = 3
y (20) = 10, 000e
≈ 19, 601
The population will be about 19,600.
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
u = 2x
du = 2 dx
3.
e3t dt =
u = 3t
du = 3 dt
sin t
cos t
5.
1
1
sin u du = − cos u + C =
2
2
ln
1 u
1
1
e du = eu + C = e3t + C
3
3
3
1
1
x sin x dx =
sin u du = − cos u + C =
2
2
2
u=x
2
2
u = 3 x2
du = 6 x dx
dx =
sin 2 x cos x dx = u 2 du =
7.
x x 2 + 2 dx =
u = sin x
du = cos x dx
u = x2 + 2
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
6.
1
− cos x 2 + C
2
xe3 x
1
du = − ln u + C =
u
dt = −
u = cos t
du = −sin t dt
du = 2 x dx
4.
Q → 150 g, as t → ∞ .
c.
1
− cos 2 x + C
2
2.
2
y = 1 + Ce – x
If x = 0, y = 3, then 3 = 1 + C, so C = 2.
(0.03365)(20)
sin 2 x dx =
2
2
D[ ye x ] = 2 xe x ; ye x = e x + C ;
y = 10, 000e0.03365t
1.
2
= ex
D[ ye –2 x ] = e – x ; y = – e x + Ce2 x
14, 000 = 10, 000e10 k
ln(1.4)
k=
≈ 0.03365
10
43. Integrating factor is x . D [ yx ] = 0; y = Cx
2 xdx
8.
(
1 2
x +2
3
x
2
x +1
u = x 2 +1
du = 2 x dx
1
2
u3
sin 3 x
+C =
+C
3
3
1 3
u du = u 2 + C
3
3
) 2 +C
dx =
1 1
1
du = ln u + C
2 u
2
= ln u + C = ln x 2 + 1 + C
410
Review and Preview
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.
⎡ ⎛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)
x
= 2e sin 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
e
0
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
Instructor’s Resource Manual
Review and Preview
411
© 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.
7
CHAPTER
Techniques of Integration
7.1 Concepts Review
5.
1. elementary function
x +4
ex
x
1
⎛x⎞
tan –1 ⎜ ⎟ + C
2
⎝2⎠
dx =
2+e
= ln u + C
3. e x
4.
=
6. u = 2 + e x , du = e x dx
u 5 du
2.
dx
2
2 3
u du
1
du
u
= ln 2 + e x + C
= ln(2 + e x ) + C
Problem Set 7.1
1
( x – 2)5 dx = ( x – 2)6 + C
6
1.
3 x dx =
2.
1
3
3x ⋅ 3dx =
2
(3x)3 / 2 + C
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
1 5 5
=
u 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 1/ 2
=−
u du
2 1
1 1 1/ 2
=
u du
2 0
8.
2t 2
2t 2 + 1
= dt –
5
⎡ u6 ⎤
56 − 16
=⎢ ⎥ =
12
⎣⎢ 12 ⎦⎥1
=
7. u = x 2 + 4, du = 2x dx
x
1 du
dx =
2
2 u
x +4
1
= ln u + C
2
1
= ln x 2 + 4 + C
2
1
= ln( x 2 + 4) + C
2
1
2
1
0
1 − x 2 (−2 x dx)
dt =
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
6 z 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
Section 7.1
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.
10. u = 2t + 1, du = 2dt
5
5 du
dt =
2
2t + 1
u
16. u = 1 – x , du = –
3 / 4 sin
= 5 u +C
= 5 2t + 1 + C
tan z
11.
2
cos z
0
tan z sec2 z dz = u du
17.
12. u = cos z, du = –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
–1
2
= sin ( x ) + C
15. u = sin x, du = cos x dx
π / 4 cos x
2 / 2 du
dx =
0 1 + sin 2 x
0
1 + u2
= [tan −1 u ]0 2 / 2
2
2
≈ 0.6155
= tan −1
1/ 2
1
sin u du
sin u du
1⎞
⎛
= −2 ⎜ cos1 − cos ⎟
2⎠
⎝
≈ 0.6746
u = tan z, du = sec2 z dz
ecos z sin z dz = – ecos z (– sin z dz )
1
1/ 2
dx
= [−2 cos u ]11/ 2
dz = tan z sec2 z dz
1 2
u +C
2
1
= tan 2 z + C
2
dx = –2
1– x
=2
=
1– x
1
2 1– x
19.
3x2 + 2 x
1
dx = (3x – 1)dx +
dx
x +1
x +1
3
= x 2 – x + ln x + 1 + C
2
1
x3 + 7 x
dx = ( x 2 + x + 8)dx + 8
dx
x –1
x –1
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
dx =
sin u du
x
2
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
6e x
1 − e2 x
dx = 6
du
1− u2
du
= 6sin −1 u + C
= 6sin −1 (e x ) + C
Instructor’s Resource Manual
Section 7.1
413
© 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.
22. u = x 2 , du = 2x dx
x
1
du
dx =
4
2
4 + u2
x +4
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 = –2e2 x dx
3e2 x
1– e
2x
dx = –
3 du
2
u
= –3 u + C
= –3 1 – e2 x + C
24.
25.
x3
dx =
1 4 x3
dx
4 x4 + 4
x4 + 4
1
= ln x 4 + 4 + C
4
1
= ln( x 4 + 4) + C
4
1
0
2
t 3t dt =
26.
29. u = e x , du = e x dx
e x sec e x dx = sec u du
= ln sec e x + tan e x + C
1 1 t2
2t 3 dt
2 0
30. u = e x , du = e x dx
e x sec 2 (e x )dx = sec 2 u du = tan u + C
1
π / 6 cos x
2
sin x dx
0
cos x π / 6
⎡ 2
= ⎢–
⎣⎢ ln 2
=–
28. u = cos(4t – 1), du = –4 sin(4t – 1)dt
sin(4t − 1)
sin(4t − 1)
dt =
dt
2
1 − sin (4t − 1)
cos 2 (4t − 1)
1 1
=−
du
4 u2
1
1
= u −1 + C = sec(4t − 1) + C
4
4
= 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
=–
⎤
⎥
⎦⎥ 0
1
(2 3 / 2 – 2)
ln 2
2−2 3/2
ln 2
≈ 0.2559
=
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
dx = x −
u
sin x
= x − ln u + C
π / 6 cos x
2
(– sin x dx)
0
= 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
414
Section 7.1
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.
33. u = t 3 − 2 , du = 3t 2 dt
39. u = 3 y 2 , du = 6 y dy
t 2 cos(t 3 − 2)
y
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)
2 3
34.
1 + cos 2 x
2
sin 2 x
dt =
1
dx =
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
2
1
u
du
e tan
−1
⎛ 3y2
1
= sin −1 ⎜
⎜ 4
6
⎝
1 + 4t 2
dt
2t
1 u
dt =
e du
2
2
1 + 4t
−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
2
1
(t + 1)e−t − 2t −5 = − eu du
2
1 u
1 − t 2 − 2t − 5
= − e +C = − e
+C
2
2
Instructor’s Resource Manual
1
42 − u 2
du
⎞
⎟+C
⎟
⎠
40. u = 3x, du = 3 dx
cosh 3 x 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
sinh u du
x 2 sinh x3 dx =
3
1
= cosh u + C
3
1
= cosh x3 + C
3
42. u = 2x, du = 2 dx
5
5
dx =
2
9 − 4 x2
1
32 − u 2
du
5
⎛u⎞
= sin −1 ⎜ ⎟ + C
2
⎝3⎠
5 −1 ⎛ 2 x ⎞
= sin ⎜ ⎟ + C
2
⎝ 3 ⎠
43. u = e3t , du = 3e3t dt
4−e
2
1
6
1
⎛u⎞
= sin −1 ⎜ ⎟ + C
6
⎝4⎠
e3t
=− u +C
= − 1 + cot 2t + C
37. u = tan −1 2t , du =
16 − 9 y 4
dy =
6t
dt =
1
3
1
2
2 − u2
du
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
Section 7.1
415
© 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 = cos x, du = –sin x dx
π/2
0
sin x
1
dx = −
du
2
0 16 + cos x
1 16 + u 2
1
1
=
du
0 16 + u 2
51.
= 2(e2 x − e −2 x )dx
1 e4 + 1 1
= ln
− ln 2
2
2
e2
1
2
=
=
x + 2x + 5
1
( x + 1) 2 + 22
2
=
=
x – 4x + 9
1
( x – 2)2 + ( 5)2
1
x – 4x + 4 + 5
=
Section 7.1
dx =
1
18 x + 18
dx
2
18 9 x + 18 x + 10
)
3– x
52.
16 + 6 x – x
2
dx =
1
2
6 – 2x
16 + 6 x – x 2
dx
= 16 + 6 x – x 2 + C
tan x
2
dx =
sec x – 4
sin x
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
55. The length is given by
L=
dx
2
9 x + 18 x + 9 + 1
(3x + 3)2 + 12
u = 3x + 3, du = 3 dx
dx
1
du
=
2
2
2
3 u + 12
(3 x + 3) + 1
416
dx
d ( x – 2)
1
= tan –1 (3x + 3) + C
3
52 – ( x – 3)2
9 x + 18 x + 10
1
= ln 9 x 2 + 18 x + 10 + C
18
1
= ln 9 x 2 + 18 x + 10 + C
18
54.
⎛ x–2⎞
tan –1 ⎜
⎟+C
5
⎝ 5 ⎠
9 x + 18 x + 10
dx
x +1
=
2
dx
=
–( x – 3)2 + 52
dx
1
2
=
x + 2x + 1+ 4
dx
2
d ( x + 1)
dx =
dx
49.
1
2
–( x – 6 x + 9 – 25)
⎛ 2t ⎞
1
⎟+C
= sec –1 ⎜
⎜ 3 ⎟
3
⎝
⎠
1
⎛ x +1⎞
tan –1 ⎜
⎟+C
2
⎝ 2 ⎠
1
48.
dx =
2
53. u = 2t , du = 2dt
dt
du
=
2
t 2t – 9
u u 2 – 32
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.
dx
=
(
− e−2 x
1 e2 + e−2 1
dx =
du
0 e 2 x + e −2 x
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
50.
=
=
=
2
⎛ dy ⎞
1 + ⎜ ⎟ dx
a
⎝ dx ⎠
b
π/ 4
0
π/ 4
0
π/ 4
0
= ln
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
2 + 1 − ln 1 = ln
2 + 1 ≈ 0.881
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.
56. sec x =
=
57. u = x – π , du = dx
2 π x sin x
π (u + π) sin(u + π)
dx =
du
0 1 + cos 2 x
– π 1 + cos 2 (u + π)
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
du = 0 by symmetry.
– π 1 + cos 2 u
π π sin u
π π sin u
du = 2
du
2
– π 1 + cos u
0 1 + cos 2 u
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 ⎠⎦
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 = ⎜
+
⎟dx
⎝ cos x 1 + sin x ⎠
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
dx +
dx = –
+
cos x
1 + sin x
u
v
= – ln u + ln v + C
=
= – ln cos x + ln 1 + sin x + C
= ln
⎛π⎞
= 2π ⎜ ⎟ = π2
⎝2⎠
1 + sin x
+C
cos x
= ln sec x + tan x + C
58.
V = 2π
3π
4 ⎛x+
– π ⎜⎝
4
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–
= 2π
π
2 ⎛u +
– π ⎜⎝
2
π⎞ 2
2
2
2
sin u +
cos u –
cos u +
sin u du
⎟
2⎠ 2
2
2
2
π
π
2 ⎛ u + π ⎞ 2 sin u du = 2 2π 2 u sin u du +
⎜
⎟
−π ⎝
−π
2⎠
2
2
π
2π 2π u sin u du = 0 by symmetry. Therefore,
−
2
π
π
= 2π2 2 2 sin u du = 2 2π2 [− cos u ] 2 = 2 2π2
= 2π
2
V
0
Instructor’s Resource Manual
2π 2
π
2
−π
2
sin u du
0
Section 7.1
417
© 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.
6. u = x
dv = sin 2x dx
1
du = dx
v = – cos 2 x
2
1
1
x sin 2 x dx = – x cos 2 x – – cos 2 x dx
2
2
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
1. u = x
v = ex
du = dx
x
= (t – 3) sin (t – 3) + cos (t – 3) + C
dv = e x dx
x
x
x
x
xe dx = xe − e dx = xe − e + C
dv = e3 x dx
1
du = dx
v = e3 x
3
1
1
xe3 x dx = xe3 x − e3 x dx
3
3
1 3x 1 3x
= xe − e + C
3
9
2. u = x
dv = e5t +π dt
1
du = dt
v = e5t +π
5
1
1 5t +π
te5t +π dt = te5t +π –
e
dt
5
5
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
(t + 7)e2t +3 dt = (t + 7)e2t +3 –
e
dt
2
2
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
du = dt
v=
2
(t + 1)3 / 2
3
2
2
t 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
dv = 3 2t + 7dt
3
v = (2t + 7)4 / 3
du = dt
8
3
3
t 3 2t + 7dt = t (2t + 7) 4 / 3 – (2t + 7)4 / 3 dt
8
8
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 3 x − x dx = x ln 3 x − x + C
x
12. u = ln(7 x5 )
du =
5
dx
x
dv = dx
v=x
5
ln(7 x5 )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
17. u = ln t
1
du = dt
t
v=x
arctan x = x arctan x −
x
1+ x
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
2
dx
1
e
v=x
5x
1 + 25 x 2
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
dv =
15. u = ln x
du =
2
dv =
x
2
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
1
3
5
⎡1
⎤
2 x ln x3 dx = ⎢ (2 x)3 2 ln x3 ⎥ −
⎣3
⎦1
5 32
1
2
x dx
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
z 3 ln z dz = z 4 ln z −
z ⋅ dz
4
4
z
1 4
1 3
= z ln z −
z dz
4
4
1
1
= z 4 ln z − z 4 + C
4
16
19. u = ln z
1
du = dz
z
3
1
2
x2
dx
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
5
dv = 2 xdx
1
v = (2 x)3 / 2
3
5
dx
⎡ 1
⎤
dx = ⎢ − ln 2 x5 ⎥ + 5
⎣ x
⎦2
18. u = ln x3
3
du = dx
x
1
25 / 2 3 / 2 ⎤
= ⎡ (2 x)3 / 2 ln x3 −
x
3
⎢⎣ 3
⎥⎦1
dx
x2
1
v=−
x
5
dx
x
3 ln 2 x5
=
1
x2
1
1
du = dx
v=–
x
x
ln x
ln x
1⎛1⎞
dx = –
– – ⎜ ⎟ dx
x
x⎝ x⎠
x2
ln x 1
=–
– +C
x
x
16. u = ln 2 x5
2
v = t3 / 2
3
e
e 2 1/ 2
⎡2
⎤
t ln t dt = ⎢ t 3 / 2 ln t ⎥ –
t dt
⎣3
⎦1 1 3
e
dv = dx
arctan 5 x dx = x arctan 5 x –
dv = t dt
20. u = arctan t
1
du =
dt
1+ t2
dv = t dt
1
v = t2
2
1
1 t2
t arctan t dt = t 2 arctan t –
dt
2
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⎞
dt
arctan ⎜ ⎟ dt = t arctan ⎜ ⎟ +
⎝t ⎠
⎝ t ⎠ 1+ t2
⎛1⎞ 1
= t arctan ⎜ ⎟ + ln(1 + t 2 ) + C
⎝t ⎠ 2
π/2
dv = t 5 dt
7
dt
t
1
v = t6
6
du =
1
7 5
t 5 ln(t 7 )dt = t 6 ln(t 7 ) –
t dt
6
6
1
7
= t 6 ln(t 7 ) – t 6 + C
6
36
dv = csc2 x dx
v = − cot x
23. u = x
du = dx
π/6
22. u = ln(t 7 )
π/ 2
x csc2 x dx = [ − x cot x ]π / 6 +
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
π6
=
π4
π4
x sec2 x dx = [ x tan x ]π 6 −
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
4
x5 x3 + 4dx = x3 ( x3 + 4)3 / 2 –
x ( x + 4)3 / 2 dx = x3 ( x3 + 4)3 / 2 – ( x3 + 4)5 / 2 + C
9
3
9
45
26. u = x7
dv = x 6 x 7 + 1 dx
du = 7 x 6 dx
v=
x13 x7 + 1dx =
2 7 7
2 6 7
2
4 7
( x + 1)5 / 2 + C
x ( x + 1)3 / 2 –
x ( x + 1)3 / 2 dx = x 7 ( x 7 + 1)3 / 2 –
21
3
21
105
4
27. u = t
dv =
du = 4t 3 dt
t7
(7 – 3t 4 )3 / 2
420
2 7
( x + 1)3 / 2
21
v=
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
© 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.
28. u = x 2
dv = x 4 – x 2 dx
1
du = 2x dx
v = – (4 – x 2 )3 / 2
3
1
2
1
2
x3 4 – x 2 dx = – x 2 (4 – x 2 )3 / 2 +
x(4 – x 2 )3 / 2 dx = – x 2 (4 – x 2 )3 / 2 – (4 – x 2 )5 / 2 + C
3
3
3
15
29. u = z 4
dv =
du = 4 z 3 dz
z7
4 2
(4 – z )
v=
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
dx = 2 x ln x – 2
dx = 2 x ln x – 4 x + C
1/
x
x 2
33. u = x
dv = (3 x + 10)49 dx
1
(3x + 10)50
150
x
1
x
1
(3 x + 10)50 –
(3 x + 10)50 dx =
x(3x + 10) 49 dx =
(3x + 10)50 –
(3 x + 10)51 + C
150
150
150
22,950
du = dx
34. u = t
du = dt
v=
dv = ( t − 1) dt
12
v=
1
( t − 1)13
13
1
1 1
⎡t
13 ⎤
t (t − 1)12 dt = ⎢ ( t − 1) ⎥ −
( t − 1)13 dt
0
⎣13
⎦ 0 13 0
1
1
1
1
⎡t
13
= ⎢ ( t − 1) −
( t − 1)14 ⎤⎥ =
13
182
182
⎣
⎦0
x
dv = 2 dx
1 x
du = dx
v=
2
ln 2
x x
1
x 2 x dx =
2 –
2 x dx
ln 2
ln 2
x x
1
=
2 –
2x + C
2
ln 2
(ln 2)
35. u = x
Instructor's Resource Manual
dv = a z dz
1 z
du = dz
v=
a
ln a
z z
1
za z dz =
a –
a z dz
ln a
ln a
z z
1
a –
az + C
=
2
ln a
(ln a)
36. u = z
37. u = x 2
du = 2x dx
dv = e x dx
v = ex
x 2 e x dx = x 2 e x − 2 xe x dx
u=x
dv = e x dx
du = dx
v = ex
(
x 2 e x dx = x 2 e x − 2 xe x − e x dx
)
= x 2 e x − 2 xe x + 2e x + C
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
38. u = x 4
dv = xe x dx
1 2
v = ex
2
du = 4 x3 dx
du = et dt
e cos t dt = e sin t − et sin t dt
2
2
1
x e dx = x 4 e x – 2 x3e x dx
2
du = e dt
v = –cos t
e cos t dt = e sin t − ⎡ −et cos t + et cos t dt ⎤
⎣
⎦
t
2
v = ex
2
2 ⎛
2
2
1
⎞
x5 e x dx = x 4 e x – ⎜ x 2 e x − 2 xe x dx ⎟
2
⎝
⎠
2
2
1 4 x2
= x e – x2e x + e x + C
2
2 et cos t dt = et sin t + et cos t + C
et cos t dt =
v=z
du = ae dt
at
du = ae dt
at
v = sin t
(
e sin t dt = – e cos t + a eat sin t – a e at sin t dt
)
at
)
eat sin t dt = – e at cos t + aeat sin t – a 2 e at sin t dt
= z ln 2 z – 2 z ln z + 2 z + C
(1 + a 2 ) e at sin t dt = – e at cos t + ae at sin t + C
dv = dx
eat sin t dt =
v=x
ln 2 x 20 dx = x ln 2 x 20 – 40 ln x 20 dx
u = ln x 20
20
du =
dx
x
dv = cos t dt
at
(
40 ln x 20
dx
x
at
u = e at
ln z dz = z ln z – 2 z ln z – dz
du =
v = –cos t
e sin t dt = – e cos t + a e at cos t dt
v=z
40. u = ln 2 x 20
dv = sin t dt
at
dv = dz
2
1 t
e (sin t + cos t ) + C
2
42. u = e at
ln 2 z dz = z ln 2 z – 2 ln z dz
2
t
et cos t dt = et sin t + et cos t − et cos t dt
dv = dz
u = ln z
1
du = dz
z
dv = sin t dt
t
2
39. u = ln 2 z
2 ln z
du =
dz
z
t
u = et
dv = 2 xe x dx
du = 2x dx
v = sin t
t
5 x2
u = x2
dv = cos t dt
43. u = x 2
du = 2 x dx
dv = dx
v=x
– e at cos t
a2 + 1
+
aeat sin t
a2 + 1
+C
dv = cos x dx
v = sin x
2
(
ln 2 x 20 dx = x ln 2 x 20 – 40 x ln x 20 – 20 dx
)
= x ln 2 x 20 – 40 x ln x 20 + 800 x + C
x cos x dx = x 2 sin x − 2 x sin x dx
u = 2x
du = 2dx
dv = sin x dx
v = − cos x
(
x 2 cos x dx = x 2 sin x − −2 x cos x + 2 cos x dx
)
2
= x sin x + 2 x cos x − 2sin x + C
44. u = r 2
du = 2r dr
dv = sin r dr
v = –cos r
r 2 sin r dr = – r 2 cos r + 2 r cos r dr
u=r
du = dr
dv = cos r dr
v = sin r
(
)
r 2 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
422
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 = [sin(ln x) − cos(ln x)] + C
2
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 = [cos(ln x ) + sin(ln x)] + C
2
47. u = (ln x)3
du =
3ln 2 x
dx
x
dv = dx
v=x
(ln x)3 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
dv = dx
v=x
(ln x) 4 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(3x)
3
1
1
sin x sin(3 x) dx = – sin x cos(3 x) + cos x cos(3 x) dx
3
3
u = cos x
dv = cos(3x)dx
1
v = sin(3 x)
du = –sin x dx
3
Instructor's Resource Manual
Section 7.2
423
© 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
1 ⎡1
1
⎤
sin x sin(3 x) dx = – sin x cos(3 x) + ⎢ cos x sin(3 x) + sin x sin(3 x) dx ⎥
3
3 ⎣3
3
⎦
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(3 x) dx = – sin x cos(3x) + cos x sin(3x) + C
8
8
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 = – cos(5 x ) cos(7 x) –
sin(5 x ) cos(7 x )dx
7
7
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 = – cos(5 x) cos(7 x) – ⎢ sin(5 x ) sin(7 x ) –
cos(5 x ) sin(7 x )dx ⎥
7
7 ⎣7
7
⎦
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 = – cos(5 x) cos(7 x) – sin(5 x) sin(7 x) + C
24
24
51. u = e z
du = e z dz
dv = sin z dz
1
v = – cos z
1
e z sin z dz = – e z cos z +
u=e z
du = e z dz
dv = cos z dz
1
v = sin z
1
e z sin z dz = − e z cos z +
=–
2
1
e z cos z +
+ 2
2
2
e z sin z –
⎡1 z
⎢ e sin z −
⎣
2
2
Section 7.2
–
2
+ 2
e z cos z +
⎤
e z sin z dz ⎥
⎦
e z sin z dz
1
e z sin z dz = – e z cos z +
e z sin z dz =
424
e z cos z dz
2
e z sin z + C
e z ( sin z – cos z )
e z sin z + C =
+C
2
2
+ 2
+ 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.
z
52. u = e
dv = cos z dz
du = e z dz
1
v=
sin z
1
e z cos z dz = e z sin z –
u=e z
dv = sin z dz
1
v = – cos z
du = e z dz
z
e
=
1
2
cos z dz =
e z sin z +
+ 2
2
e z sin z dz
1
e
2
z
sin z −
⎡ 1
⎢− e
⎣
2
e z cos z –
e z cos z dz =
2
2
z
cos z +
e
z
⎤
cos z dz ⎥
⎦
e z cos z dz
1
e z cos z + e z sin z + C
e z ( cos z + sin z )
e z cos z dz =
+C
2
+ 2
dv = x dx
53. u = ln x
du =
1
dx
x
x ln x dx =
v=
≠ –1
dv = x dx
2 ln x
dx
x
x (ln x)2 dx =
=
≠ –1
x +1
x +1
x +1
1
ln x –
x dx =
ln x –
+ C,
+1
+1
+1
( + 1)2
54. u = (ln x)2
du =
x +1
,
+1
v=
x +1
,
+1
≠ –1
x +1
2
x +1
x +1 ⎤
2 ⎡ x +1
(ln x)2 –
x ln x dx =
(ln x)2 −
ln x −
⎢
⎥+C
+1
+1
+1
+ 1 ⎣⎢ + 1
( + 1) 2 ⎦⎥
x +1
x +1
x +1
(ln x) 2 – 2
ln x + 2
+ C,
+1
( + 1)2
( + 1)3
Problem 53 was used for
55. u = x
du = x –1dx
x ln x dx.
dv = e x dx
v=
1
x e x
x e x dx =
–
56. u = x
du = x –1dx
x sin x dx = –
≠ –1
e x
x –1e x dx
dv = sin x dx
1
v = – cos x
x cos x
Instructor's Resource Manual
+
x –1 cos x dx
Section 7.2
425
© 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.
57. u = x
dv = cos x dx
1
v = sin x
du = x –1dx
x sin x
x cos x dx =
x –1 sin x dx
–
58. u = (ln x)
du =
dv = dx
–1
(ln x)
x
v=x
dx
(ln x) –1 dx
(ln x) dx = x(ln x) –
59. u = (a 2 – x 2 )
dv = dx
2
2
du = –2 x(a – x )
–1
v=x
dx
(a 2 – x 2 ) dx = x (a 2 – x 2 ) + 2
60. u = cos
–1
x 2 ( a 2 – x 2 ) –1 dx
dv = cos x dx
x
du = –( – 1) cos
–2
v = sin x
x sin x dx
cos x dx = cos –1 x sin x + ( – 1) cos –2 x sin 2 x dx
= cos −1 x sin x + ( − 1) cos − 2 x(1 − cos 2 x) dx = cos –1 x sin x + ( – 1) cos –2 x dx – ( – 1) cos x dx
−1
cos x dx = cos
cos
cos x dx =
61. u = cos
–1
–1
x sin x
=
=
cos –1 x sin x
x=
+ ( – 1) cos –2
1
sin x
x sin 2
+ ( − 1) cos − 2
x(1 − cos 2
+ ( – 1) cos –2
x dx – ( – 1) cos
cos −1 x sin x
x dx =
v=
+ ( − 1) cos − 2
cos –1 x sin x
+
–1
cos –2
x dx
x) dx
x dx
x dx
x dx
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 3x ⎤
⎤
– x 3 e3 x + ⎢ x 2 e3 x –
xe3 x dx ⎥ = x 4 e3 x – x3e3 x + x 2 e3 x – ⎢ xe3 x –
e 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
x 4 e3 x dx =
62.
1 4 3x
x e
3
1
= x 4 e3 x
3
=
426
x dx
cos –2 x dx
x sin x dx
cos –1 x sin x
cos −1 x sin x
cos
–1
−2
dv = cos x dx
x dx =
cos
+
x
du = – ( – 1) cos –2
cos
x sin x + ( − 1) cos
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.
63.
64.
1 4
4 3
1
4⎡ 1
⎤
x sin 3 x –
x 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 3x +
cos 2 3 x dx ⎥
18
6 ⎣12
4
18
6
⎦
1
5
5
1
1
⎡
⎤
dx ⎥
= cos5 3 x sin 3 x + cos3 3 x sin 3 x + ⎢ cos 3 x sin 3 x +
18
72
8 ⎣6
2
⎦
x
1
5
5
5
= cos5 3x sin 3x + cos3 3x sin 3x + cos 3x sin 3x + + C
18
72
48
16
cos6 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
e
1
dv = dx
v=x
ln x dx = [ x ln x ]1 −
e
66. V =
e
1
e
1
dx = [ x ln x − x]1e = (e – e) – (1 · 0 – 1) = 1
π(ln x) 2 dx
u = (ln x)2
du =
2 ln x
dx
x
dv = dx
v=x
e
e
e
⎛
⎞
(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
π
e
Instructor's Resource Manual
Section 7.2
427
© 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.
67.
9
0
9 –x /3 ⎛
3e – x / 3 dx = –9
68. V =
9
0
0
e
1 ⎞
9
–x /3 9
]0 = – + 9 ≈ 8.55
⎜ – dx ⎟ = –9[e
3
⎝
⎠
e3
π(3e – x / 3 ) 2 dx = 9π
9 –2 x / 3
e
dx
0
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
cos x dx ⎞⎟
sin x dx ⎟ − ⎛⎜ [ − x cos x ]0 +
= ⎜ ⎣⎡ x sin x ⎦⎤ 0 −
0
0
⎠
⎝
⎠ ⎝
2π
–1 ≈ 0.11
4
x sin x dx and x cos x dx.
= [ x sin x + cos x + x cos x – sin x]0π / 4 =
Use Problems 60 and 61 for
⎛ 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
71.
e
1
2π
0
ln x 2 dx = 2
u = ln x
1
du = dx
x
e
2
1
e
1
e
1
ln x dx
dv = dx
v=x
⎛
⎝
ln x dx = 2 ⎜ [ x ln x]1e −
x ln x 2 dx = 2
e
1
e
1
⎞
⎠
(
)
dx ⎟ = 2 e − [ x]1e = 2
x ln x dx
u = ln x
1
du = dx
x
dv = x dx
1
v = x2
2
e
⎛
e
⎡1
⎤
2 x ln x dx = 2 ⎜ ⎢ x 2 ln x ⎥ –
1
⎜ ⎣2
⎦1
⎝
428
Section 7.2
e
⎞
⎛1
⎡1 ⎤ ⎞ 1
x dx ⎟ = 2 ⎜ e2 – ⎢ x 2 ⎥ ⎟ = (e2 + 1)
1 2
⎟
⎜2
⎣ 4 ⎦1 ⎟⎠ 2
⎠
⎝
e1
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.
1 e
(ln x)2 dx
2 1
u = (ln x)2
du =
1
2
e
1
x=
y=
72. a.
dv = dx
2 ln x
dx
x
v=x
1
(ln x)2 dx = ⎛⎜ [ x(ln x) 2 ]1e – 2
2⎝
1 (e 2
2
+ 1)
=
2
1 (e – 2)
2
2
=
e
1
1
ln x dx ⎞⎟ = (e – 2)
⎠ 2
e2 + 1
4
e–2
4
u = cot x
dv = csc2 x dx
du = – csc2 x dx
v = –cot x
cot x csc2 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
dv = cot x csc x dx
v = –csc x
cot x csc2 x dx = − csc2 x − cot x csc2 x dx
2
cot x csc2 x dx = − csc2 x + C
1
cot x csc2 x dx = − csc 2 x + C
2
c.
73. a.
1
1
1
1
– cot 2 x = – (csc2 x –1) = – csc2 x +
2
2
2
2
p ( x ) = x3 − 2 x
g ( x) = e x
All antiderivatives of g ( x) = e x
( x3 − 2 x)e x dx = ( x3 − 2 x)e x − (3x 2 − 2)e x + 6 xe x − 6e x + C
b.
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 − 3 x + 1)(− cos x) − (2 x − 3)(− sin x) + 2 cos x + C
Instructor's Resource Manual
Section 7.2
429
© 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.
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)
u=x
x sin x dx . 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
an =
π
⎤
1⎡ ⎡ 1
1 π
⎤
cos(
)
(
)
cos(nx) f ( x)dx ⎥
nx
f
x
−
+
⎢
⎢
⎥
π⎣ ⎣ n
⎦ −π n −π
⎦
Term 2
Term 1
Term 1 =
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 ⎡
cos(nx) f ( x) dx ⎥⎤ = 0.
± ( f (−π) − f (π)) +
⎢
−π
n→∞ πn ⎣
⎦
lim an = lim
n →∞
Gn [(n + 1)(n + 2) ⋅⋅⋅ (n + n)]1 n ⎡⎛ 1 ⎞⎛ 2 ⎞
=
76.
= ⎢⎜1 + ⎟⎜ 1 + ⎟
n
[n n ]1 n
⎣⎝ n ⎠⎝ n ⎠
⎛ G ⎞ 1 ⎡⎛ 1 ⎞⎛ 2 ⎞
ln ⎜ n ⎟ = ln ⎢⎜1 + ⎟⎜ 1 + ⎟
⎝ n ⎠ n ⎣⎝ n ⎠⎝ n ⎠
=
⎛ n ⎞⎤
⎜1 + ⎟ ⎥
⎝ 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
1/ n
⎛ n ⎞⎤
⎜1 + ⎟ ⎥
⎝ n ⎠⎦
2
⎞
⎟ = 1 ln x dx = 2 ln 2 – 1
⎠
4
⎞ 2 ln 2–1
= 4e –1 =
⎟=e
e
⎠
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.
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
(1 t ) dt = (1 t ) dt + 1
functions will differ by a constant, so it is perfectly correct (notationally) to write
78.
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 –
b
b
a
xf ( x)dx
Starting with the same integral,
u = f(x)
dv = dx
du = f ( x)dx
v=x–a
b
a
b
f ( x) dx = [ ( x – a) f ( x) ]a –
b
80. u = f ( x)
du = f ( x)dx
f (b) – f (a ) =
a
( x – a) f ( x)dx
dv = dx
v=x–a
b
a
f ( x)dx = [ ( x – a) f ( x) ]a –
b
b
a
( x – a ) f ( x)dx = f (b)(b – a ) –
b
a
( x – a ) f ( x)dx
Starting with the same integral,
u = f ( x)
dv = dx
du = f ( x)dx
v=x–b
f (b) − f (a) =
b
a
f ( x)dx = [ ( x – b) f ( x) ]a –
b
b
a
( x – b) f ( x)dx = f (a)(b − a) –
b
a
( x – b) f ( x)dx
81. Use proof by induction.
n = 1: f (a) + f (a )(t – a) +
t
a
(t – x) f ( x )dx = f (a ) + f (a )(t – a ) + [ f ( x)(t – x)]ta +
t
a
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)
( x)dx by parts.
f
a
n!
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 = ⎢ –
( x) ⎥ +
( x)dx
f
f
f
a
a ( n + 1)!
n!
⎣⎢ (n + 1)!
⎦⎥
t
a
Instructor's Resource Manual
Section 7.2
431
© 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.
=
(t – a )n +1 ( n +1)
(a) +
f
(n + 1)!
(t – x)n +1 ( n + 2)
( x)dx
f
a (n + 1)!
t
f (i ) ( a )
(t – a) n +1 ( n +1)
(t – a)i +
(a) +
f
(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
Thus f (t ) = f (a ) +
n
(t – x)n +1 ( n + 2)
( x)dx
f
a ( n + 1)!
t
Thus, the statement is true for n + 1.
82. a.
B( , ) =
1
0
x −1 (1 − x)
−1
≥ 1,
dx where
≥1
x = 1 – u, dx = –du
1
0
x −1 (1 − x)
−1
dx =
0
1
1
(1 − u ) −1 (u ) −1 (−du ) = (1 − u ) −1 u
−1
du = B ( , )
0
Thus, B( , ) = B( , ).
b.
B( , ) =
u=x
1
0
x −1 (1 − x)
−1
dx
−1
du = ( − 1) x
−2
v=−
dx
1
B( ,
−1
dv = (1 − x)
⎡ 1
⎤
) = ⎢ − x −1 (1 − x) ⎥ +
⎣
⎦0
−1
=
B ( − 1, + 1)
1
dx
(1 − x)
−1
1
0
x
−2
−1
(1 − x) dx =
1
0
x
−2
(1 − x) dx
(*)
Similarly,
B( ,
)=
u = (1 − x)
1
x
−1
0
−1
(1 − x)
c.
dx
−2
⎡1
) = ⎢ x (1 − x)
⎣
v=
dx
1
−1 ⎤
⎥ +
⎦0
1
−1
dx
x
1
0
x (1 − x)
−2
−1
dx =
1
0
x (1 − x)
−2
dx =
−1
B ( + 1,
− 1)
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)
m(m + 1) ( m + 2 )
B(1, m + n − 1) =
Thus, B (n, m) =
1
0
Section 7.2
⋅ 2 ⋅1
(m + n − 2)
(1 − x)m+ n − 2 dx = −
B(1, m + n − 1).
1
1
[(1 − x)m+ n −1 ]10 =
m + n −1
m + n −1
( n − 1) (n − 2) ( n − 3)
m(m + 1) ( m + 2 )
If n > m, then B (n, m) = B (m, n) =
432
−1
dv = x
du = − ( − 1) (1 − x)
B( ,
−1
⋅ 2 ⋅1
(m + n − 2) ( 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
© 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.
dv = f (t )dt
83. u = f(t)
du = f (t )dt
b
a
v = f (t )
b
= f (b) f (b) − f (a) f (a) −
b
[ f (t )]2 ≥ 0, so −
84.
b
f (t ) f (t )dt = [ f (t ) f (t ) ]a –
a
b
a
a
[ f (t )]2 dt
[ f (t )]2 dt = −
b
a
[ f (t )]2 dt
[ f (t )]2 ≤ 0 .
x⎛ t
0
⎞
⎜ 0 f ( z )dz ⎟ dt
⎝
⎠
u=
t
0
f ( z )dz dv = dt
du = f(t)dt
v=t
x
x⎛ t
x
x
x
⎞
⎡ t
⎤
⎜ f ( z )dz ⎟ dt = ⎢t 0 f ( z )dz ⎥ – 0 t f (t )dt = 0 x f ( z )dz – 0 t f (t )dt
0⎝ 0
⎠
⎣
⎦0
x
By letting z = t,
0
x
x f ( z )dz =
0
x f (t )dt , so
x⎛ t
0
x
x
x
⎞
⎜ 0 f ( z )dz ⎟ dt = 0 x f (t )dt – 0 t f (t )dt = 0 ( x – t ) f (t )dt
⎝
⎠
85. Let I =
x t1
t
⋅⋅⋅ n −1
0 0
0
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.
I=
x t1 t2
t
⋅⋅⋅ n −1
0 0 0
0
where F ( tn −1 ) =
f (tn ) dtn ...dt3 dt2 dt1 =
tn −1
0
t1 t2
t
⋅⋅⋅ n − 2
0 0
0
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 ) =
t1
0
f ( tn ) dtn ,
du = f ( t1 ) dt1 ,
I=
dv = ( x − t1 )
v=−
1 ⎧⎪ ⎡ 1
( x − t1 )n −1
⎨⎢−
( n − 2 )! ⎩⎪ ⎣ n − 1
t1
0
n−2
1
( x − t1 )n −1
n −1
t =x
1
⎤1
+
f ( tn ) dtn ⎥
⎦ t1 =0 n − 1
x
0
f ( t1 )( x − t1 )
n −1
⎪⎫
dt1 ⎬ .
⎭⎪
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)
=
Instructor’s Resource Manual
Section 7.3
433
© 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.
86. Proof by induction.
n = 1:
dv = e x dx
u = P1 ( x)
du =
dP1 ( x)
dx
dx
v = ex
e x P1 ( x)dx = e x P1 ( x) – e x
Note that
dP1 ( x)
dP ( x ) x
dP ( x)
dx = e x P1 ( x) – 1
e 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,
dPn +1 ( x)
dx
dx
e x Pn +1 ( x)dx = e x Pn +1 ( x) – e x
Note that
dPn +1 ( x)
is a polynomial of degree n, so
dx
⎡
e x Pn +1 ( x)dx = e x Pn +1 ( x) − ⎢ e x
⎢⎣
= e x Pn +1 ( x) + e x
n +1
(−1) j
(−1) j
j =0
d j Pn +1 ( x )
j =1
(3 x 4 + 2 x 2 )e x dx = e x
87.
n
4
dx j
(–1) j
= ex
d j ⎛ dPn +1 ( x) ⎞ ⎤
x
x
⎟ ⎥ = e Pn +1 ( x ) − e
j ⎜
dx
⎥
⎠⎦
dx ⎝
n +1
(−1) j
j =0
n
( −1) j
j =0
d j +1Pn +1 ( x )
dx j +1
d j Pn +1 ( x)
dx j
d j (3 x 4 + 2 x 2 )
j =0
dx j
= 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 + cos 2 x
dx
2
1.
2
2.
(1 – sin 2 x) cos x dx
3.
sin 2 x(1 – sin 2 x) cos x dx
4. cos mx cos nx =
1
[cos(m + n) x + cos(m − n) x ]
2
Problem Set 7.3
sin 2 x dx =
1.
2. u = 6x, du = 6 dx
1
sin 4 6 x dx =
sin 4 u du
6
1 – cos 2 x
dx
2
1
1
dx –
cos 2 x dx
2
2
1
1
= x – sin 2 x + C
2
4
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
=
=
434
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.
sin 3 x dx = sin x(1 − cos 2 x)dx
3.
5.
= sin x dx − sin x cos 2 x dx
=
1
= − cos x + cos3 x + C
3
cos5 d =
π/ 2
0
π/2
0
(1 – sin 2 )2 cos d
(1 – 2sin 2 + sin 4 ) cos d
π/2
2
1
⎡
⎤
= ⎢sin – sin 3 + sin 5 ⎥
3
5
⎣
⎦0
2
1
8
⎛
⎞
= ⎜1 – + ⎟ – 0 =
15
⎝ 3 5⎠
cos3 x dx =
4.
π/2
0
= cos x (1 − sin 2 x)dx
= cos x dx − cos x sin 2 x dx
1
= sin x − sin 3 x + C
3
6.
π/2
0
sin 6 d =
π / 2 ⎛ 1 – cos 2
0
⎜
⎝
2
3
⎞
⎟ d
⎠
1 π/ 2
(1 – 3cos 2 + 3cos 2 2 – cos3 2 )d
0
8
1 π/ 2
3 π/2
3 π/ 2 2
1 π/2 3
=
d –
2 cos 2 d +
cos 2 –
cos 2 d
0
0
0
8
16
8
8 0
1
3
3 π / 2 ⎛ 1 + cos 4 ⎞
1 π/2
2
= [ ]0π / 2 – [sin 2 ]0π / 2 +
⎜
⎟ d – 0 (1 – sin 2 ) cos 2 d
0
8
16
8
2
8
⎝
⎠
1 π 3 π/ 2
3 π/2
1 π/2
1 π/2 2
d +
4 cos 4 d –
2 cos 2 d +
sin 2 ⋅ 2 cos 2 d
= ⋅ +
0
0
0
8 2 16
64
16
16 0
π 3π 3
1
1
5π
= + + [sin 4 ]0π / 2 – [sin 2 ]0π / 2 + [sin 3 2 ]0π / 2 =
16 32 64
16
48
32
=
sin 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
7.
=–
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 = –
8.
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
cos3 3 sin –2 3 d = (1 – sin 2 3 ) sin –2 3 cos 3 d
9.
=
1
(sin −2 3 − 1)3cos 3 d
3
1
1
= – csc 3 – sin 3 + C
3
3
sin1/ 2 2 z cos3 2 z dz = (1 – sin 2 2 z ) sin1/ 2 2 z cos 2 z dz
10.
=
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
© 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.
2
11.
2
1
⎛ 1 – cos 6t ⎞ ⎛ 1 + cos 6t ⎞
(1 – 2 cos 2 6t + cos 4 6t )dt
sin 4 3t cos 4 3t dt = ⎜
⎟ ⎜
⎟ dt =
2
2
16
⎝
⎠ ⎝
⎠
1 ⎡
1
1
1
⎤
=
cos12t dt +
(1 + 2 cos12t + cos 2 12t )dt
1 – (1 + cos12t ) + (1 + cos12t )2 ⎥ 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.
1
⎛ 1 + cos 2 ⎞ ⎛ 1 – cos 2 ⎞
cos6 sin 2 d = ⎜
(1 + 2 cos 2 – 2 cos3 2 – cos 4 2 )d
⎟ ⎜
⎟d =
2
2
16
⎝
⎠ ⎝
⎠
1
1
1
1
=
d +
2 cos 2 d – (1 – sin 2 2 ) cos 2 d –
(1 + cos 4 )2 d
16
16
8
64
1
1
1
1
1
=
d +
2 cos 2 d –
2 cos 2 d +
2sin 2 2 cos 2 d –
(1 + 2 cos 4 + cos 2 4 )d
16
16
16
16
64
1
1
1
1
1
sin 2 2 ⋅ 2 cos 2 d –
4 cos 4 d –
(1 + cos8 )d
=
d +
d –
16
16
64
128
128
1
1
1
1
1
1
+ sin 3 2 −
−
−
sin 4 −
sin 8 + C
=
16
48
64
128
128
1024
5
1
1
1
=
+ sin 3 2 –
sin 4 –
sin 8 + C
128
48
128
1024
sin 4 y cos 5 y dy =
13.
=
14.
1
2
[sin 9 y + sin(− y )] dy =
1
(sin 9 y − sin y ) dy
2
1⎛ 1
1
1
⎞
⎜ − cos 9 y + cos y ⎟ + C = cos y − cos 9 y + C
2⎝ 9
2
18
⎠
cos y cos 4 y dy =
1
1
1
1
1
[cos 5 y + cos(−3 y )]dy = sin 5 y − sin(−3 y) + C = sin 5 y + sin 3 y + C
10
6
10
6
2
2
15.
16.
1
⎛ w⎞
⎛ w⎞
⎛ 1 – cos w ⎞ ⎛ 1 + cos w ⎞
(1 – cos w – cos 2 w + cos3 w)dw
sin 4 ⎜ ⎟ cos 2 ⎜ ⎟ dw = ⎜
⎟ ⎜
⎟ dw =
8
2
2
⎝2⎠
⎝2⎠
⎝
⎠ ⎝
⎠
1 ⎡
1
1 ⎡1 1
⎤
⎤
2
=
1 – cos w – (1 + cos 2w) + (1 – sin 2 w) cos w⎥ dw =
⎢ 2 – 2 cos 2 w – sin w cos w⎥dw
8 ⎢⎣
2
8
⎦
⎣
⎦
1
1
1
= w – sin 2 w – sin 3 w + C
16
32
24
sin 3t sin t dt = −
(
1
[cos 4t − cos 2t ] dt
2
)
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
© 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.
17.
x cos 2 x sin x dx
u=x
tan 4 x dx =
19.
du = 1 dx
=
2
dv = cos x sin x dx
=
1
v = − (cos x) 2 (− sin x) dx = − cos3 x
3
t =cos x
Thus
= tan 2 x sec2 x dx − (sec2 x − 1)dx
1
= tan 3 x − tan x + x + C
3
x cos 2 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⎣
⎤
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
⎦
18.
3
cot 4 x dx =
20.
=
=
= cot 2 x csc2 x dx − (csc2 x − 1)dx
1
= − cot 3 x + cot x + x + C
3
21. tan 3 x =
du = 1 dx
dv = sin 3 x cos x dx
v = (sin x)3 (cos x) dx =
t =sin x
1 4
sin x
4
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
⎤
(1 − cos 2 x) 2 dx ⎥ =
x sin 4 x −
4 ⎢⎣
4
⎦
1⎡
1
⎤
x sin 4 x −
(1 − 2 cos 2 x + cos 2 2 x) dx ⎥ =
4 ⎢⎣
4
⎦
( cot 2 x )( cot 2 x ) dx
( cot 2 x ) (csc2 x − 1) dx
( cot 2 x csc2 x − cot 2 x ) dx
x sin x cos x dx
u=x
( tan 2 x )( tan 2 x ) dx
( tan 2 x ) (sec2 x − 1) dx
( tan 2 x sec2 x − tan 2 x ) dx
( tan x ) ( tan 2 x ) dx
=
( tan x ) ( sec2 x − 1) dx
=
1
tan 2 x + ln cos x + C
2
cot 3 2t dt =
22.
=
( cot 2t ) ( cot 2 2t ) dt
( cot 2t ) ( csc2 2t − 1)dt
= cot 2t csc2 2t dt − cot 2t dt
1
1
= − cot 2 2t − ln sin 2t + C
4
2
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
⎦
Instructor’s Resource Manual
Section 7.3
437
© 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.
23.
⎛ ⎞
tan 5 ⎜ ⎟ d
⎝2⎠
d
⎛ ⎞
u = ⎜ ⎟ ; du =
2
2
⎝ ⎠
⎛ ⎞
tan 5 ⎜ ⎟ d = 2 tan 5 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.
1
⎛ ⎞
⎛ ⎞
tan 4 ⎜ ⎟ − tan 2 ⎜ ⎟ − 2 ln cos + C
2
2
⎝2⎠
⎝2⎠
cot 5 2t dt
u = 2t ; du = 2dt
1
cot 5 2t dt =
cot 5 u du
2
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
=
cot 3 u csc2 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 x sec4 xdx =
=
)
)
)
)
(
)(
)
(
(
)
)
( tan −3 x )(sec2 x )(sec2 x ) dx
( tan −3 x )(1 + tan 2 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 x sec4 x dx =
=
( tan −3 / 2 x )(sec2 x )(sec2 x )
( tan −3 / 2 x )(1 + tan 2 x )(sec2 x ) dx
= tan −3 / 2 x sec2 x dx + tan1/ 2 x sec2 x dx
2
= −2 tan −1/ 2 x + tan 3 / 2 x + C
3
438
Section 7.3
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.
tan 3 x sec2 x dx
27.
Let u = tan x . Then du = sec 2 x dx .
tan 3 x sec2 x dx =
u 3 du =
1 4
1
u + C = tan 4 x + C
4
4
tan 3 x sec−1/ 2 x dx = tan 2 x sec−3 / 2 x(sec x tan x)dx
28.
(sec2 x − 1)(sec−3 / 2 x ) (sec x tan x ) dx
=
= sec1/ 2 x ( sec x tan x ) dx − sec−3 / 2 x ( sec x tan x ) dx
=
29.
30.
31.
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
dx . Making the substitution and changing the limits as necessary, we get
L
L
L
mπ x
nπ x
L π
dx =
cos
cos
cos mu cos nu du = 0 (See Problem 29
−L
L
L
π −π
π
0
π( x + sin x)2 dx = π
π
0
( x 2 + 2 x sin x + sin 2 x) dx = π
π
π 2
x
0
dx + 2π
π
0
x sin x dx +
π
2
π
0
(1 − cos 2 x)dx
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
π/2
V =π
33. a.
0
1
π
π
−π
sin 2 u du = π
π / 2 1 – cos 2u
0
f ( x) sin(mx)dx =
2
1
π
π/ 2
1
⎡1
⎤
du = π ⎢ u – sin 2u ⎥
4
⎣2
⎦0
=
π2
≈ 2.4674
4
⎛ N
⎞
1 N
an
⎜⎜ an sin(nx) ⎟⎟ sin(mx)dx =
−π
π n =1
⎝ n =1
⎠
π
π
−π
sin(nx) sin(mx) dx
From Example 6,
⎧0 if n ≠ m
so every term in the sum is 0 except for when n = m.
sin(nx) sin(mx)dx = ⎨
⎩π if n = m
If m > N, there is no term where n = m, while if m ≤ N, then n = m occurs. When n = m
π
−π
an
1
π
π
−π
π
−π
sin(nx) sin(mx) dx = am π so when m ≤ N,
f ( x) sin(mx) dx =
Instructor’s Resource Manual
1
⋅ am ⋅ π = am .
π
Section 7.3
439
© 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.
N
⎛ N
⎞⎛ N
⎞
π
1 N
sin(nx) sin(mx) dx
am sin(mx) ⎟ dx =
an
am
⎜⎜ an sin(nx) ⎟⎟ ⎜⎜
⎟
−π
−π
−π
π n =1 m =1
⎝ n =1
⎠ ⎝ m =1
⎠
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
1
π
b.
π
f 2 ( x)dx =
1
π
π
Proof by induction
x
x
n = 1: cos = cos
2
2
Assume true for k ≤ n.
34. a.
x
x
cos cos
2
4
cos
x
2n
⋅ cos
x
2n +1
⎡
1
3
= ⎢ cos
x + cos
x+
n
2
2n
⎢⎣
+ cos
2n –1 ⎤ 1
x
x⎥
cos
n
n –1
n +1
2
2
⎥⎦ 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⎣
⎦
n
⎡
⎡
1
3
2 –1 ⎤ ⎛
1 ⎞ 1
1
3
⎢cos n x + cos n x + + cos n x ⎥ ⎜ cos n +1 x ⎟ n –1 = ⎢cos n +1 x + cos n +1 x +
⎠2
2
2
2
2
2
2
⎣⎢
⎦⎥ ⎝
⎣⎢
⎡
1
3
lim ⎢ cos
x + cos
x+
n
n →∞ ⎢
2
2n
⎣
b.
=
1
x
c.
1
x
x
0
x
0
+ cos
⎡
2n –1 ⎤ 1
1
1
3
x⎥
x + cos
x+
= lim ⎢cos
n
n –1
n
x
n
→∞
2
2
2n
⎦⎥ 2
⎣⎢
+ cos
+ cos
2n +1 –1 ⎤ 1
x⎥
2n +1 ⎦⎥ 2n
2n –1 ⎤ x
x⎥
2n ⎦⎥ 2n –1
cos t dt
1
sin x
cos t dt = [sin t ]0x =
x
x
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
, etc.
2
2 2+ 2 2+ 2+ 2
⋅
⋅
2
2
2
⎛π⎞
⎛π⎞
= lim cos ⎜ 2 ⎟ cos ⎜ 2 ⎟
⎜2⎟
⎜4⎟
n→∞
⎝ ⎠
⎝ ⎠
440
2+ 2
,
2
Section 7.3
⎛ π
cos ⎜ 2
⎜ 2n
⎝
⎛π⎞
⎛π⎞
⎛π⎞
= cos ⎜ 2 ⎟ cos ⎜ 2 ⎟ cos ⎜ 2 ⎟
⎜2⎟
⎜4⎟
⎜8⎟
⎝ ⎠
⎝ ⎠
⎝ ⎠
( )
⎞ sin π2
2
⎟=
=
π
⎟
π
2
⎠
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.
36. Since (k − sin x)2 = (sin x − k ) 2 , the volume of S is
= πk 2
π
0
dx − 2k π
π
0
sin x dx +
π
0
π(k − sin x)2 = π
π
0
(k 2 − 2k sin x + sin 2 x) dx
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) =
a.
π2
≈ 4.93,
2
2
2
π
π
⎛2⎞
f ⎜ ⎟ = 4−8+
≈ 0.93, f (1) = π2 − 4π +
≈ 2.24
2
2
⎝π⎠
S has minimum volume when k =
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
u
x+4
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
=
=2
1 t +e
1 u+e
1
2
2 e
=2
du – 2
du
1
1 u+e
=
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 3 x + πdx = (u 3 – π)u (3u 2 du )
3 7 3π 4
u – u +C
7
4
2
2
= 2[u ]1 2 – 2e ⎡⎣ln u + e ⎤⎦
2. u = 3 x + π , u 3 = x + π, 3u 2 du = dx
= (3u 6 – 3πu 3 )du =
dx =
u = t , u 2 = t , 2u du = dt
6.
1
t
0 t +1
=2
1
1
dt =
u2
0 u2
+1
u
0 u2
+1
du = 2
1
1
1
0
0 u2
= 2 du – 2
(2u du )
1 u2
+1−1
0
2
u +1
du
du = 2[u ]10 – 2[tan –1 u ]10
+1
π
= 2 – 2 tan 1 = 2 – ≈ 0.4292
2
–1
Section 7.4
441
© 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.
7. u = (3t + 2)1/ 2 , u 2 = 3t + 2, 2u du = 3dt
1 2
⎛2
⎞
(u – 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 dt =
8. u = (1 – x)1/ 3 , u 3 = 1 – x, 3u 2 du = – dx
x(1 – x)
2/3
3
2
2
dx = (1 – u )u (–3u )du
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
12. t = sec x, dt = sec x tan x dx
π
Note that 0 ≤ x < .
2
t 2 – 1 = tan x = tan x
2 2
=
= 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⎠
2
11. x = 2 tan t , dx = 2sec t dt
dx
2sec 2 t dt
1
=
=
cos t dt
2
3/ 2
2 3/ 2
4
( x + 4)
(4sec t )
=
442
1
x
sin t + C =
+C
4
4 x2 + 4
Section 7.4
=
2
t
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
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
dt
3
–3
t2 –1
–2
3
=
t
sec –1 (–3)
2π / 3
dt =
sec –1 (–3)
– tan x
2π / 3
sec3 x
– 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
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
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.
16. x = π tan t, dx = π sec 2 t dt
πx – 1
2
x +π
=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
π
πx − 1
0
x 2 + π2
1 2
x
x + π2 + + C
π
π
dx
⎡
= ⎢ π x 2 + π2 – ln
⎢
⎣
= 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
=
x2 + 2 x + 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
2
= sec t dt = ln sec t + tan t + C
x + 4x + 5
= ln
= 3 u 2 + 4 – 3ln u 2 + 4 + u + C
= 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
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
x2 + 4 x + 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 ⎠
Instructor’s Resource Manual
Section 7.4
443
© 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.
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
dx =
du
2
x + 2x + 2
u2 + 1
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
π
8
π
=
8
=
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
dx =
du
2
x – 6 x + 18
u2 + 9
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
–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
0 x2
+ 2x + 5
1
x
= 2π
dx
0 ( x + 1) 2 + 4
= 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/4
1
π π/ 4
cos 2 t dt
dt =
–1
–1
2
tan (1/ 2) sec t
8 tan (1/ 2)
π/ 4
⎛1 1
⎞
+ cos 2t ⎟ dt
tan –1 (1/ 2) ⎜⎝ 2 2
⎠
π/4
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 = 2sec2 t dt
u = x 2 + 9, du = 2 x dx
x dx
1 du 1
= ln u + C
2
x +9 2 u
1
1
= ln x 2 + 9 + C = ln x 2 + 9 + C
2
2
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
)
(
)
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.
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
x3 dx
0
9 + x2
=
x2
3
0
x dx =
9 + x2
3 2
3
3 2
⎡ u3
⎤
(u 2 − 9) du = ⎢ – 9u ⎥
3
⎣⎢ 3
⎦⎥ 3
≈ 5.272
=
31. a.
u2 − 9
udu
u
3 2
= 18 – 9 2
4 − x2
x2
=
u 2 du
4 – u2
1
du = −4
du + du
4 − u2
4 − u2
1 u+2
= −4 ⋅ ln
+u+C
4 u−2
= − ln
4 − x2 + 2
2
+ 4− x +C
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–a
b 2 – ( x + a) 2 dx
0
x + a = b sin t, dx = b cos t dt
x dx = –
−4 + 4 − u 2
and y = b 2 – ( x + a )2 dx .
A=4
u = 4 – x 2 , u 2 = 4 – x 2 , 2u du = –2x dx
4 – x2
dx =
x
( 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,
sin –1 ( a / b )
b 2 cos 2 t dt
π/2
= 2b 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
π/ 2
A=4
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 ln
2
4 − x2
−
+ 4 − x2 + C
x
x
2− 4− x
x
+ 4 − x2 + C
− ln
= ln
= ln
4− x +2
4 − x2 − 2
= ln
2
4− x −2
4 − x2 + 2
(2 − 4 − x )
2
2
4− x −4
⎛ 2 − 4 − x2
= ln ⎜
⎜
x
⎝
= ln
a
For
⎛ 2a 2 – x 2 – a ⎞ dx
⎜
⎟
⎠
a
2a 2 – x 2 dx + 2a 2
–a
a 2 – x 2 dx is the area of 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
a 2 – x 2 dx =
.
–a
2
2
( 4 − x − 2)
a 2 – x 2 dx –
–a
Note that
To reconcile the answers, note that
2
a 2 – x 2 dx –
–a
a
=
2
a
A=
a
2a 2 – x 2 dx =
(2 − 4 − x )
–a
− x2
= a2
2 2
2
2
⎞
⎟ = 2 ln 2 − 4 − x
⎟
x
⎠
=
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
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.
Instructor’s Resource Manual
Section 7.4
445
© 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.
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.
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
dx =
dx –
dx
x( x + 1)
x
x +1
= ln x – ln x + 1 + C
2.
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
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
dx =
dx –
dx
2
3
x
3 x+3
x + 3x
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
dx = –
dx +
dx
2
2
x
+
1
2 x –1
x –1
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
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.
4.
5.
5x
3
2
=
5x
2
=
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
=
dx –
dx
3
2
x
6
6 x+3
2x + 6x
5
5
= ln x – ln x + 3 + C
6
6
x − 11
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
dx = 3
dx − 2
dx
2
+
−1
x
x
4
x + 3x − 4
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
dx = –
dx +
dx
2
7 x–4
7 x+3
x – x – 12
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
dx −
dx
dx = 4
2
x+5
x−2
x + 3x − 10
= 4 ln x + 5 − ln x − 2 + C
x+π
2
2
=
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
dx =
dx –
dx
2
2
π
π
–
2
–
x
x
x – 3πx + 2π
= 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
dx =
dx –
dx
2
2
x
–
1
x
+5
2x + 9x – 5
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
dx =
dx +
dx = ln 3 x – 2 + 4 ln x + 1 + C
2
3
x
–
2
x
+
1
3
3x + x – 2
17 x – 3
2
=
Instructor’s Resource Manual
Section 7.5
447
© 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.
12.
13.
5– x
=
2
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
dx =
dx +
dx =
ln x – π +
ln x – 4 + C
2
π
–
4
x
–
π
4
–
π
x
–
4
π
–
4
4
–
π
x − x(π + 4) + 4π
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
dx =
dx −
dx +
dx = 2 ln x − ln x + 1 + ln x − 2 + C
3
2
x
x +1
x−2
x − x − 2x
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)(3x + 2)
A=
1
1
6
, B = – ,C =
35
7
5
7 x2 + 2 x – 3
1
1
1
1
6
1
dx =
dx –
dx +
dx
(2 x – 1)(3 x + 2)( x – 3)
35 2 x – 1
7 3x + 2
5 x–3
=
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
dx =
dx −
dx +
dx = ln 2 x − 1 − ln x + 3 + 3ln x − 2 + C
2
2x −1
x+3
x−2
(2 x − 1)( x + x − 6)
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
3
2
4 x – 28 x + 56 x – 32
448
Section 7.5
dx =
1
1
1
1
1
dx +
dx = x + ln x – 4 + C
4
4 x–4
4
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.
17.
x3
2
x +x–2
3x – 2
2
x +x–2
19.
2
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
x3
18.
3x – 2
= x –1 +
x3 + x 2
2
dx = ( x − 1) dx +
= x – 4+
8
1
1 1
1
8
1
dx +
dx = x 2 – x + ln x + 2 + ln x – 1 + C
3 x+2
3 x −1
2
3
3
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
x3 + x 2
dx –
dx = x 2 − 4 x + 18ln x + 3 – 4 ln x + 2 + C
dx = ( x − 4) dx +
2
x+3
x+2
2
x + 5x + 6
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
x4 + 8x2 + 8
dx = x dx – 2 dx + 7
dx + 7
dx = x 2 – 2 ln x + 7 ln x + 2 + 7 ln x – 2 + C
3
2
–
2
x
x
x
+
2
x – 4x
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
dx –
dx +
dx
dx = ( x3 + 4 x 2 + 16 x + 68) dx –
2
3
2
x−4
4 x
4
x
x – 4x
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
2
=
A
B
+
x − 3 ( x − 3)2
( x − 3)
x + 1 = A(x – 3) + B
A = 1, B = 4
x +1
1
4
4
+C
dx =
dx +
dx = ln x − 3 −
2
2
x−3
x−3
( x − 3)
( x − 3)
Instructor’s Resource Manual
Section 7.5
449
© 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.
22.
23.
5x + 7
5x + 7
=
2
2
=
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
dx =
dx −
dx = 5ln x + 2 +
+C
2
2
+
x
x
+
2
2
x + 4x + 4
( x + 2)
3x + 2
3
2
x + 3x + 3x + 1
3x + 2
=
( x + 1)
3
=
A
B
C
+
+
x + 1 ( x + 1)2 ( x + 1)3
2
3x + 2 = A( x + 1) + B ( x + 1) + C
A = 0, B = 3, C = –1
3x + 2
3
1
3
1
dx =
dx −
dx = −
+
+C
3
2
2
3
x + 1 2( x + 1) 2
x + 3x + 3x + 1
( x + 1)
( x + 1)
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 ⎤
dx = ⎢
–
+
–
+
−
+
⎥ dx
2
5
2
2
3
4
1 – x (1 – x)
( x – 2) (1 – x)
(1 – x)
(1 – x)
(1 – x)5 ⎦⎥
⎣⎢ x – 2 ( x – 2)
2
=
5
= 128ln x – 2 +
25.
3 x 2 − 21x + 32
3
2
x − 8 x + 16 x
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
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
+C
dx =
dx +
dx −
dx = 2 ln x + ln x − 4 +
3
2
2
x
−
x
x
−
4
4
x − 8 x + 16
( x − 4)
26.
27.
450
x 2 + 19 x + 10
4
3
=
x 2 + 19 x + 10
3
=
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
+ ln 2 x + 5 + C
dx = ⎜ – +
+
+
dx = – ln x – –
4
3
2
3 2x + 5 ⎟
x x2
x
2 x + 5x
x
x
⎝
⎠
2 x2 + x – 8
3
=
2x2 + x – 8
2
=
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
dx = –2 dx +
dx = −2 dx + 2
dx +
dx
2
2
3
2
x
x
x + 4x
x +4
x +4
x +4
1
⎛x⎞
= –2 ln x + 2 ln x 2 + 4 + tan –1 ⎜ ⎟ + C
2
⎝2⎠
Section 7.5
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.
28.
3x + 2
3x + 2
=
2
2
x( x + 2) + 16 x x( x + 4 x + 20)
1
1
13
A = , B = – ,C =
10
10
5
=
A
Bx + C
+
2
x x + 4 x + 20
1 x + 13
– 10
1 1
1 1
14
1
1
2x + 4
5
dx +
dx −
dx +
dx =
dx
2
2
2
2
10
x
20
10
x
5
x + 4 x + 20
x + 4 x + 20
( x + 2) + 16
x( x + 2) + 16 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 =
2 x 2 – 3x – 36
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
dx = –4
dx +
dx = –2 ln 2 x – 1 + ln x 2 + 9 + C
2
2
2
x
–
1
2
(2 x – 1)( x + 9)
x +9
2
1
=
1
=
4
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
⎛x⎞
dx =
dx –
dx −
dx = ln x – 2 – ln x + 2 – tan –1 ⎜ ⎟ + C
4
2
32 x – 2
32 x + 2
8 x +4
32
32
16
⎝2⎠
x – 16
1
2
( x – 1) ( x + 4)
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
dx = –
dx +
dx +
dx +
dx
2
2
2
125
x
–1
25
125
x
+
4
25
( x –1) ( x + 4)
( x –1)
( x + 4)2
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
2
( x + 3)( x – 4 x + 5)
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 = ⎢1 – ⎜
dx
⎟+
⎢⎣ 13 ⎝ x + 3 ⎠ x 2 – 4 x + 5 ⎥⎦
( x + 3)( x 2 – 4 x + 5)
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
Instructor’s Resource Manual
Section 7.5
451
© 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.
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
dt = sin t – ln sin t + 3 – tan –1 (sin t – 2) – ln sin 2 t – 4sin t + 5 + C
2
13
13
26
(sin t + 3)(sin t – 4sin t + 5)
= x−
34. x = sin t, dx = cos t dt
cos t
1
1
1
1
⎛x⎞
dt =
dx = ln x − 2 − ln x + 2 − tan −1 ⎜ ⎟ + C
4
4
32
32
16
⎝2⎠
sin t − 16
x − 16
which is the result of Problem 30.
cos t
1
1
1
⎛ sin t ⎞
dt = ln sin t – 2 – ln sin t + 2 – tan –1 ⎜
⎟+C
4
32
32
16
⎝ 2 ⎠
sin t – 16
35.
x3 – 4 x
2
=
2
Ax + B
+
Cx + D
( x + 1)
x + 1 ( x 2 + 1) 2
A = 1, B = 0, C = –5, D = 0
x3 – 4 x
2
( x + 1)
2
2
x
dx =
2
x +1
dx − 5
36. x = cos t, dx = –sin t dt
(sin t )(4 cos 2 t –1)
(cos t )(1 + 2 cos 2 t + cos 4 t )
4 x2 − 1
x
2
( x + 1)
dt = –
4 x2 − 1
2
dx =
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
2 x3 + 5 x 2 + 16 x
5
3
=
2
=
=
2
x(2 x 2 + 5 x + 16)
2
2
dx =
2
2 x 2 + 5 x + 16
16sec2
4
d
2
2
=
Ax + B
2
+
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
dx =
dx +
dx =
dx +
dx +
dx
5
3
2
2
2
2
2
2
2
x + 8 x + 16 x
x +4
( x + 4)
x +4
( x + 4)
( x + 4)2
8
dx, let x = 2 tan , dx = 2sec2 d .
To integrate
2
2
( x + 4)
8
4
=
⎛1 1
⎞
= cos 2 d = ⎜ + cos 2 ⎟ d
2
2
⎝
⎠
( x + 4)
16sec
x
x
1
1
1
1
1
+C
=
+ sin 2 + C =
+ sin cos + C = tan –1 +
2
2 x2 + 4
2
4
2
2
2 x3 + 5 x 2 + 16 x
x
5
1
x
x
3
x
2x – 5
dx = tan –1 –
+ tan –1 +
+ C = tan –1 +
+C
5
3
2
2
2 2( x + 4) 2
2 x +4
2
2 2( x 2 + 4)
x + 8 x + 16 x
452
Section 7.5
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.
38.
x –17
x –17
A
B
=
+
( x + 4)( x – 3) x + 4 x – 3
=
2
x + x –12
A = 3, B = –2
6
x –17
3
2 ⎞
6
–
⎜
⎟ dx = ⎡⎣3ln x + 4 – 2 ln x – 3 ⎤⎦ 4 = (3ln10 – 2 ln 3) – (3ln 8 – 2 ln1)
⎝ x +4 x –3⎠
x + x –12
= 3ln10 – 2 ln 3 – 3ln 8 ≈ –1.53
39. u = sin
4
du = cos
cos
π/4
0
6⎛
dx =
2
4
(1 – sin
2
)(sin
1
2
2
d
2
=
2
+ 1)
2
d =
1
1/ 2
0
2
2
(1 – u )(u + 1)
2
du =
1/ 2
0
1
1
du +
1+ u
4
1/ 2
⎡ 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
41.
1
1/ 2
0
2
u +1
du +
1
2
1
1/ 2
0
2
(u + 1) 2
du
1/ 2
⎡1 1+ u 1
⎤
u
= ⎢ ln
+ tan –1 u +
⎥
2
4(u + 1) ⎥⎦ 0
⎢⎣ 8 1 − u 2
1
1
1
+ tan –1
+
≈ 0.65
2 −1 2
2 6 2
3x + 13
2
1
2
(u + 1) 2
=
x + 4x + 3
A = –2, B = 5
5 3 x + 13
1
du
2 +1
(To integrate
40.
(1 – u )(1 + u )(u 2 + 1)2
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
du =
du +
2
2
2
0
8 0
1− u
8
(1 − u )(u + 1)
1
= ln
8
1
1/ 2
0
2
x + 4x + 3
du, let u = tan t.)
A
B
3 x + 13
=
+
( x + 3)( x + 1) x + 3 x + 1
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
dy = 1 dt = t + C1
y (1 − y )
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
Instructor’s Resource Manual
Section 7.5
453
© 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.
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
dy =
dt = t + C1
y (12 − y )
10
10
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 )
b. y (3) =
Thus:
y
= e1.2t +12C1 = Ce1.2t or
12 − y
(C =e12C1 )
y (t ) =
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
dy = 0.0003 dt = 0.0003t + C1
y (8000 − y )
Section 7.5
8000e2.4t
7+e2.4t
⇒ A=
⎛1
⎞
1
1
⎜ +
⎟ dy =
4000 ⎝ y (4000 − y ) ⎠
⎛
⎞
1
1
y
ln ⎜
[ln y − ln(4000 − y)] =
⎟
4000
4000 ⎝ 4000 − y ⎠
so that
y
= e4t + 4000C1 = Ce4t
or
4000 − y
(C =e4000C1 )
0.001t + C1 =
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
= 0.001y (4000 − y ) so that
dt
1
dy = 0.001 dt = 0.001t + C1
y (4000 − y )
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
⎛ 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 ⎠
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
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.
45.
dy
= ky ( L − y ) so that
dt
1
dy = k dt = kt + C1
y( L − y)
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 ⎥⎦
dy
= kdt
y (16 – y )
1 ⎛1
1 ⎞
⎜ +
⎟ dy =
L ⎝ y L− y⎠
1
1 ⎛ y ⎞
[ln y − ln( L − y )] = ln ⎜
⎟ so that
L
L ⎝ L− y⎠
Thus: kt + C1 =
1 ⎛1
1 ⎞
⎜ +
⎟ dy = kt + C
16 ⎝ y 16 – 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
C
y0
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
© 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.
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
A
B
=
+
(a − x )(b − x) a − x b − x
1 ⎛1
1 ⎞
⎜ +
⎟ dy = kdt
10 ⎝ y 10 – y ⎠
y
ln
= 10kt + C
10 – y
1
1
,B=
a −b
a −b
dx
(a − x)(b − x)
A=−
y
= Ce10kt
10 – y
1
1 ⎞
⎛
+
⎜−
⎟ dx = k dt
⎝ a−x b−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
a −b
ln a − x
=
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
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.
d. If a = b, the differential equation is, after
separating variables
dx
= k dt
(a − x) 2
dx
= k dt
(a − x) 2
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
=
+
⎜
⎟ dy
( y − m)( M − y ) M − m ⎝ y − m M − y ⎠
= k dt
A=
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 ⎞
=
+
⎜
⎟ dy
( A − y )( B + y ) A + B ⎝ A − y B + y ⎠
= k dt
C=
− 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
dx = 1
du
2
2
π / 6 sin x(sin 2 x + 1) 2
2 u (u + 1)
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
du
1
2
2
2 u (u + 1)
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
Instructor’s Resource Manual
Section 7.5
457
© 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.
7.6 Concepts Review
5. Trig identity cos 2 u =
1. substitution
substitution.
2
⎛ 1 + cos 4 x ⎞
cos 4 2 x dx = ⎜
⎟ dx =
2
⎝
⎠
2. 53
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
xe −5 x dx = − xe−5 x − − e−5 x dx
5
5
1 −5 x 1 −5 x
= − xe
− e
+C
5
25
1
1⎞
⎛
= − e −5 x ⎜ x + ⎟ + C
5
5⎠
⎝
du = 1 dx
sin 3 x cos x dx = u 3 du =
u = sin x
du = cos x dx
(
)
u = x 2 +9
du = 2 x dx
3. Substitution
2
ln x
dx =
x
ln 2
u = ln x
du = 1 dx
0
ln 2
⎡ u2 ⎤
u du = ⎢ ⎥
⎢⎣ 2 ⎥⎦ 0
=
( ln 2 )2
2
≈ 0.2402
x
4. Partial fractions
x
x
dx =
dx
2
( x − 3)( x − 2)
x − 5x + 6
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
2
x − 5x + 6
dx =
Section 7.6
u4
sin 4 x
+C =
+C
4
4
7. Partial fractions
1
1
dx =
dx
2
( x + 4)( x + 2)
x + 6x + 8
1
A
B
=
+
=
( 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 ⎞
=
−
dx
1 x 2 + 6 x + 8 2 1 ⎜⎝ x + 2 x + 4 ⎟⎠
=
1
1 ⎡ ( x + 2)
2
⎡ ln x + 2 − ln x + 4 ⎤⎦ = ⎢ ln
1
2⎣
2 ⎣ ( x + 4)
2
⎤
⎥
⎦1
1⎛ 4
3 ⎞ 1 10
= ⎜ ln − ln ⎟ = ln ≈ 0.0527
2⎝ 6
5⎠ 2 9
3
2
−
dx =
( x − 3) ( x − 2)
3ln x − 3 − 2 ln x − 2 = ln
458
⎡
⎤
⎢
⎥
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
6. Substitution
2. Substitution
x
1 1
dx =
du = ln u + C = ln x 2 + 9 + C
2
2 u
x +9
1
1 + cos 2u
and
2
( x − 3)3
( x − 2) 2
+C
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.
8. Partial fractions
1
1
dt =
dt
2
(1
−
t
)(1
+ t)
1− t
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
2 1 dt = 1
2 ⎛ 1 + 1 ⎞ dx
⎜
⎟
0 1− t2
2 0 ⎝ 1− t 1+ t ⎠
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
0
7
x x + 2 dx =
2
u = x+2
u2 = x+2
2u du = dx
(u 2 − 2)(u )2u du =
2 ⎡ 5
3u − 10u 3 ⎤
⎣
⎦
15
=
2
10. Substitution
4
1
dt =
3 t − 2t
e x 3e x + 1 dx = u 3u + 1 du
u = e x , du = e x dx
(
)(
2 ⎣⎡ln u − 2 ⎤⎦
8
6
=
2
3
) 2 +C
2
9e x − 2 3e x + 1
135
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
4
π
0
π
0
4
sin 2 x dx =
= 4 [ − cos u ]
0
2
ex
=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
=−
=−
2
2
5
5 x − 11
11 − 5 x
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
15. a. Substitution, Formula 18
dx
1
du
=
=
2
4 9 − u 2 F18
9 − 16 x
9 − 16e
cos 2 x sin x dx = 0
sin 2 x dx = 8
1
(2 cos t + 1)(3 − 4 cos t ) 2 + C
20
≈ 1.223
6 −2
12. Use of symmetry; substitution
0
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
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 ⎣
8
u = 2t , u 2 = 2t
u du = dt
+C
b. Substitution; Formula 96
cos t 3 − 4 cos t sin t dt = − u 3 − 4u du =
⎡ u 5 2u 3 ⎤
7
2u 4 − 4u 2 du =2 ⎢ −
⎥
2
3 ⎥⎦
⎢⎣ 5
7
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
© 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.
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
2
5 x + 11
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
u 9 − u 2 du
sin 2 x cos x 9 − 2sin 2 x dx =
4
u = 2 sin x
du = 2 cos x dx
a =3
81 2 −1 ⎛ 2 sin x ⎞
sin ⎜⎜
⎟⎟ + C
32
3
⎝
⎠
2
16 − u
du =
u
F 55
a=4
u = 3t
du = 3 dt
16 − 3t 2 − 4ln
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
16 − u 2
du =
u
F 55
a=4
⎧
1⎪
4 + 16 − 3t 6
6
⎨ 16 − 3t − 4 ln
3⎪
3 t3
⎩
460
Section 7.6
a= 5
3
ln 3 x 2 + 5 + 3 x 4 + C
6
20. a. Substitution; Formula 48
5 2
t 2 3 + 5t 2 dt =
u 3 + u 2 du
25
u= 5t
=
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
1
3
u = 3x 2
du = 2 3 x dx
=
F 45
{
18. a. Substitution, Formula 55
16 − 3t
dt =
t
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
x 2 9 − 2 x 2 dx =
u 9 − u 2 du =
4
F 57
u= 2 x
+
=
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 ⎪
⎩
⎭
(
)
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.
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
=
du
3sin t + 5
3u + 5
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
24. a. Formula 100a
dz
5
=
ln
z 5 − 4 z F100 a 5
b. Complete the square; substitution;
Formula 45.
dt
dt
=
=
3 2 29
2
t + 3t − 5
(t + ) −
2
u = t+ 32
du = dt
du
u2 −
29
4
=
F 45
a = 29 2
ln (t +
u = cos x
du = −sin x dx
2
t + 3t − 5 + C
22. a. Complete the square; substitution;
Formula 47.
( x + 1) − 4
dx =
x +1
u = x +1
du = dx
⎛ x +1⎞
x 2 + 2 x − 3 − 2sec−1 ⎜
⎟+C
⎝ 2 ⎠
b. Complete the square; substitution;
Formula 47.
( x − 2) 2 − 4
dx =
x−2
u = x−2
du = dx
u2 − 4
du =
F 47
u
23. a. Formula 98
y
dy
3y + 5
25. Substitution; Formula 84
1
sinh 2 3t dt = sinh 2 u du =
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
dx = 2 sech u du =
F 82
x
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
dt = −
du =
2 cos t + 1
2u + 1 F 98
a=2
x − 4 x − 2sec
5 − 4 cos x − 5
+C =
du = 3 dt
a=2
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
u2 − 4
du =
F 47
u
x2 − 4 x
dx =
x−2
+C
5 − 4z + 5
b. Substitution, Formula 100a
sin x
du
dx = −
cos x 5 − 4 cos x
u 5 − 4u
−
x2 + 2 x − 3
dx =
x +1
5 − 4z − 5
a = −4
b =5
4
3
)+
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 =
F 96
a=4
b = −1
u = cos t
du = −sin t dt
−
Instructor’s Resource Manual
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 = −
du =
cos t + 1
u + 1 F 99
n=2
a =1
b =1
u = cos t
du = −sin t dt
2⎡
− ⎢u 2 u + 1 − 2
5⎣
⎤
du ⎥ =
u + 1 ⎦ F 98
u
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
dx = π − 2 ≈ 1.14159
0 1 + sin x
32. Using a CAS, we obtain:
0
3
x dx ≈ 0.76803
33. Using a CAS, we obtain:
π /2
231π
sin12 x dx =
≈ 0.35435
0
2048
34. Using a CAS, we obtain:
π
x
3π
cos 4 dx =
≈ 1.17810
0
2
8
35. Using a CAS, we obtain:
4
t
dt ≈ 0.11083
1 1 + t8
36. Using a CAS, we obtain:
3 4 −x / 2
0
x e
39. Using a CAS, we obtain:
2
3 x + 2x −1
dx = 4 ln ( 2 ) + 2 ≈ 4.77259
2 x2 − 2 x + 1
40. Using a CAS, we obtain:
3
π
du
= 2 tan −1 5 − ≈ 0.72973
1 u 2u − 1
2
41.
1 ⎡
x
dx ⎤
+3
⎢
⎥ =
2
2
36 ⎢⎣ (9 + x )
(9 + x 2 )2 ⎥⎦ F 95
sech
38. Using a CAS, we obtain:
π /4
x3
dx ≈ −0.00921
−π / 4 4 + tan x
( )
30. Formula 95, Formula 17
1
dx =
(9 + x 2 )3 F 95
1
37. Using a CAS, we obtain:
π /2
1
dx ≈ 1.10577
0
1 + 2 cos5 x
dx = 768 − 3378e −3/ 2 ≈ 14.26632
1
c
dx = ⎡ln x + 1 ⎤⎦ = ln(c + 1)
0
x +1 F3 ⎣
ln(c + 1) = 1 ⇒ c + 1 = e ⇒
c
0
c = e − 1 ≈ 1.71828
42. Formula 17
c
c 2
dx = ⎡ 2 tan −1 x ⎤ = 2 tan −1 c
⎦0
0 x2 + 1
F 17 ⎣
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
c
0
ln( x + 1) dx =( x + 1) [ ln( x + 1) − 1]0 =
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
© 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. 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( x3 + 1) dx = x ln( x3 + 1) − 3
u = ln( x3 +1)
du =
3x
x3
x3 + 1
2
ln( x3 + 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) − 3x + 3 ⎜
dx
⎜ ( x + 1)( x 2 − x + 1) ⎟⎟
⎝
⎠
1
2
( x + 1)( x − x + 1)
=
( 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
dx −
dx =
2
x +1
x − x +1
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
0
3
4
5
2.0000 1.6976 1.6621 1.6615 1.6615
ln( x3 + 1) dx = 1 ⇒ c ≈ 1.6615
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
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
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
)
(
1 ⎞
⎛
x ln( x3 + 1) − 3 ⎜1 −
⎟ dx =
3
⎝ x +1⎠
ln x + 1 −
c
0
⎧
⎛
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
b
a
− x3
0
( f ( x) − g ( x)) dx =
8
0
v = −3e
c +1 8 − (c + 1) x
−x
−c
⎡
3 ( x + 3) ⎤ = −18e 3 (c + 3) + 54
⎢ −18e
⎥
⎣
⎦0
b.
c
0
c.
( f ( x) − g ( x)) dx =
cx − x 2 dx =
⎛ c3 ⎞ ⎛ 2 ⎞ c
x = ⎜ ⎟⎜ ⎟ =
⎜ 6 ⎟ ⎝ c2 ⎠ 3
⎝ ⎠
x =2⇒c =6
0
6e
−x
3
dx =
For notational convenience, let
u
1
c=
e
c
0
−c
−1 ⇒
3
1
=e
c +1
−c
3
Let
(c − x) dx =
h (c ) =
c
⎡
x2 ⎤
c2
⎢cx − ⎥ =
2 ⎥⎦
2
⎣⎢
0
c.
c
−c
c
a
a
( f ( x) − g ( x)) dx =
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+
⎡ cx 2 x3 ⎤
c3
− ⎥ =
⎢
3 ⎥⎦
6
⎢⎣ 2
0
b.
b
⎛ −c
⎞
−18 ⎜ e 3 − 1⎟
⎝
⎠
52. f ( x) = c g ( x) = x a = 0 b = c
b
3
c
dx =
⎛ 256 ⎞ ⎛ c + 1 ⎞
8
c. x = ⎜
=
⎜ 3(c + 1)2 ⎟⎟ ⎝⎜ 32 ⎠⎟ 3(c + 1)
⎝
⎠
8
1
x =2⇒
=2⇒c=
3(c + 1)
3
x( f ( x) − g ( x)) dx =
−x
c
32
(c + 1)
b
dx =
c −x
−x ⎤
⎡
6 ⎢ −3 xe 3 ⎥ + 18 e 3 dx =
0
⎣
⎦0
8
a
− x3
−x
⎡
64
32
⎛ c + 1 ⎞ 2 ⎤ c +1
=
−
=
⎢8 x − ⎜ 2 ⎟ x ⎥
(c + 1) (c + 1)
⎝
⎠ ⎦0
⎣
a.
xe
0 u=x
dv = e 3
du = dx
3(c + 1)2
c +1
c
x( f ( x) − g ( x)) dx = 6
⎡ 2 ⎛ c + 1 ⎞ 3 ⎤ c +1
256
512
=
−
=
⎢4x − ⎜ 3 ⎟ x ⎥
2
⎝
⎠ ⎦0
(c + 1)
3(c + 1)2
⎣
256
b.
g ( x) = 0 a = 0 b = c
−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
© 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.
b
a.
a
x( f ( x) − g ( x )) dx =
⎛πx ⎞ 2⎤
⎢ cx sin ⎜ 2c ⎟ − x ⎥ dx
⎝
⎠
⎣
⎦
c⎡
0
⎛ πt2 ⎞
cos ⎜
⎟ dt
⎜ 2 ⎟
0
⎝
⎠
⎛ π x2 ⎞
d
C ( x) = cos ⎜
⎟
⎜ 2 ⎟
dx
⎝
⎠
b. C ( x) =
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 ⎠
4c3
π
π
3⎤
⎡c
⎢ ⎥
⎢⎣ 3 ⎥⎦
⎡ c 3 ⎤ 4c 3 c 3
⎥= 2 −
3
⎢⎣ 3 ⎥⎦ π
b.
b
a
⎛πx ⎞ ⎤
⎢ c sin ⎜ 2c ⎟ − x ⎥ dx =
⎝
⎠ ⎦
⎣
c⎡
0
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
55. a. erf ( x) =
x −t 2
e
0
π
dt
d
2 − x2
erf ( x) =
e
dx
π
x sin t
0
t
dt
sin x
d
Si ( x) =
dx
x
⎛ π t2 ⎞
sin ⎜
⎟ dt
⎜ 2 ⎟
0
⎝
⎠
⎛ π x2 ⎞
d
S ( x) = sin ⎜
⎟
⎜ 2 ⎟
dx
⎝
⎠
56. a. S ( x) =
b. erf ( x) =
x
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.
( f ( x) − g ( x)) dx =
b. Si ( x) =
57. a. (See problem 55 a.) . Since erf ( x) > 0 for all
x , erf ( x) is increasing on (0, ∞) .
[sin u − u cos u ]0 2 − ⎢
2
⎛ 4 1⎞
= c3 ⎜
− ⎟
⎝π2 3⎠
x
(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 ⎠
π
x 2 < 2π . Thus C ( x) is concave
2
up on ( 2, 2) .
when π <
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
© 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.
17. False:
7.7 Chapter Review
x2
2
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:
Use the substitution u = x 2 – 3 x + 5,
du = (2x – 3)dx.
= 1+
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
2
x( x + 1)
=
2
–x
+
x x2 + 1
x+2
2
x ( x 2 − 1)
1 2
3
1
=− −
+
−
x x 2 2( x − 1) 2( x + 1)
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
x 2 25 − 4 x 2 dx = u 2 25 − u 2 du
8
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:
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
x 25 − 4 x 2 dx = −
u = 1 + e x , du = e x dx
11. False:
−
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
1
8
u du =
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
© 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.
Sample Test Problems
1.
4
t
dt = ⎡ 9 + t 2 ⎤ = 5 − 3 = 2
⎢⎣
⎥⎦ 0
2
9+t
4
0
cot 2 (2 )d =
2.
10.
cos 2 2
sin 2 2
1 − sin 2 2
=
= (cos x + csc x − sin x)dx
d
= sin x + ln csc x − cot x + cos x + C
2
d = (csc 2 − 1)d
sin 2 2
1
= − cot 2 − + C
2
3.
4.
π / 2 cos x
e
sin x dx
0
=
⎛
cos 2 x ⎞
⎜ cos x +
⎟ dx
⎜
sin x ⎟⎠
⎝
⎛
1 − sin 2 x ⎞
= ⎜ cos x +
⎟ dx
⎜
sin x ⎟⎠
⎝
sin x + cos x
dx =
tan x
π/2
⎡ −ecos x ⎤
⎣
⎦0
(Use Formula 15 for csc x dx .)
11.
= e − 1 ≈ 1.718
⎡ sin 2 x x
⎤
x sin 2 x dx = ⎢
− cos 2 x ⎥
0
2
⎣ 4
⎦0
(Use integration by parts with u = x,
dv = sin 2 x dx .)
2
=
1
4
x 2 e x dx = e x (2 − 2 x + x 2 ) + C
Use integration by parts twice.
2
tan t , dy =
3
13. y =
dy
3
⎛
y +y
2 ⎞
5.
dy = ⎜ y 2 − y + 2 −
⎟ dy
y +1
1+ y ⎠
⎝
1
1
= y 3 − y 2 + 2 y − 2 ln 1 + y + C
3
2
6.
2 + 3y2
=
sin 3 (2t ) dt = [1 – cos 2 (2t )]sin(2t )dt
=
1
1
= – cos(2t ) + cos3 (2t ) + C
2
6
=
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.
9.
dy
3/ 2
2 y +1
0
e 2t
t
e −2
= ⎡⎣ 2 y + 1 ⎤⎦
3/ 2
0
=
14.
(Use the substitution u = et − 2 ,
du = et dt
which gives the integral
Instructor’s Resource Manual
u+2
⎞
du. ⎟
u
⎠
1
1
3
1
3
1
3
ln
ln
ln
15.
sec2 t
2 sec t
sec t dt =
3
2
sec2 t dt
3
2
3
=
y 2 + 23
2
3
1
3
+
2
3
y2 +
dt
ln sec t + tan t + C1
y 2 + 23 + y
y
2
3
+ C1
+ C1
2
+ y +C
3
Note that tan t =
= 2 −1 = 1
dt = et + 2 ln et − 2 + C
⎛ x −1 ⎞
sin –1 ⎜
⎟+C
2
⎝ 3 ⎠
1
=
16 + 4 x – 2 x
(Complete the square.)
12.
π/4
π/4
dx
y
2
3
, so sec t =
y 2 + 23
2
3
.
w3
1
1
dw = – w2 – ln 1 – w2 + C
2
2
1– w
Divide the numerator by the denominator.
2
tan x
dx = – ln ln cos x + C
ln cos x
Use the substitution u = ln cos x .
Section 7.7
467
© 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
t+2
dt −
dt
2
t −1
–1
t + t +1
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
3dt
16.
t
=
=
=
3
=
(
22. u = ln( y 2 + 9)
du =
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
dv = cot 2 x dx
v = –cot x – x
x cot 2 x dx = – x cot x – x 2 – (– cot x – x)dx
1
= – x cot x – x 2 + ln sin x + C
2
25.
Use cot 2 x = csc2 x − 1 for cot 2 x dx.
20. u = x , du =
sin x
x
1 −1/ 2
x
dx
2
= −2 cos x + C
2
dt
t
ln t 2
[ln(t 2 )]2
dt =
+C
t
4
v=y
2 y2
y2 + 9
dy
−3et / 3 (9 cos 3t − sin 3t )
+C
82
Use integration by parts twice.
et / 3 sin 3t dt =
t +9
1
–t + 1
dt +
dt
t
t + 9t
t2 + 9
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⎠
3
dt =
3x
x
cos x cos 2 x
−
+C
cos dx = −
2
2
2
4
Use a product identity.
sin
2
26.
dx = 2 sin u du
21. u = ln t 2 , du =
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
ln( y 2 + 9)dy = y ln( y 2 + 9) –
1
⎛ 2t + 1 ⎞
= ln t − 1 − ln t 2 + t + 1 − 3 tan −1 ⎜
⎟+C
2
⎝ 3 ⎠
17.
2y
2
dv = dy
27.
⎛x⎞
⎛ 1 + cos x ⎞
cos 4 ⎜ ⎟ 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
tan 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
2x ⎛ 1
⎜
1+ x
1+ x ⎝ 2 x
(u + 1)(u − 1) + 1
=2
du = 2
u +1
x
dx =
u2
⎞
du
dx ⎟ = 2
1+ u
⎠
1 ⎞
⎛
⎜ u −1+
⎟ du
u +1⎠
⎝
⎛ u2
⎞
= 2⎜
− u + ln u + 1 ⎟ + C
⎜ 2
⎟
⎝
⎠
(
)
= x − 2 x + 2 ln 1 + x + C
468
Section 7.7
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.
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
29.
=
2 5/ 2
2
tan
x + tan 9 / 2 x + C
5
9
30. u = t1/ 6 + 1, (u – 1)6 = t , 6(u − 1)5 du = dt
dt
1/ 6
t (t
+ 1)
6(u − 1)5 du
=
6
(u − 1) u
=
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
e2 y
9−e
32.
2y
dy = −
1 −1/ 2
u
du = − u + C = − 9 − e2 y + C
2
cos5 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.
eln(3cos x ) dx = 3cos x dx = 3sin x + C
x2 + a2
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
2
+ 9− y +C
y
3
, so csc t = and
3
y
2
9– y
.
y
35. u = e4x , du = 4e4 x dx
e4 x
8x
1+ e
dx =
36. x = a tan t, dx = a sec 2 t dt
1 du
1
= tan −1 (e 4 x ) + C
4 1+ u2 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
dx =
4
1 ( x 2 + a 2 )3 / 2
3a 2
x3
Note that tan t =
+C
x
, so csc t =
a
x2 + a2
.
x
37. u = w + 5, u 2 = w + 5, 2u du = dw
w
2
dw = 2 (u 2 – 5)du = u 3 – 10u + C
3
w+5
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
=–
= –2 1 + cos t + C
1 + cos t
u
39. u = cos 2 y, du = –2 cos y sin y dy
sin y cos y
4
9 + cos y
dy = –
1
du
2 9 + u2
⎛ cos 2 y ⎞
1
= – tan –1 ⎜
⎟+C
⎜ 3 ⎟
6
⎝
⎠
Instructor’s Resource Manual
Section 7.7
469
© 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.
dx
40.
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
dx =
dx + 2
dx +
dx
2 2
2
x
x ( x + 3)
x
x2 + 3
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 ⎞
tan –1 ⎜
= ln x – – ln x 2 + 3 +
⎟+C
x 2
3
⎝ 3⎠
=
42. x = 4 tan t, dx = 4sec2 t dt
=
⎞
1
1
1⎛
x
x
⎟+C =
cos t dt = sin t + C = ⎜
+C
16
16
16 ⎜⎝ x 2 + 16 ⎟⎠
16 x 2 + 16
=
A
B
C
+
+
2 x + 1 (2 x + 1) 2 (2 x + 1)3
dx
(16 + x 2 )3 / 2
43. a.
b.
c.
d.
e.
f.
44. a.
470
3 – 4 x2
(2 x + 1)
3
7 x – 41
2
( x – 1) (2 – x)
3x + 1
2
( x + x + 10)
=
3
=
2
A
B
C
D
E
+
+
+
+
2
2
2 – x (2 – x)
x – 1 ( x – 1)
(2 – x)3
Ax + B
2
x + x + 10
( x + 1) 2
2
2
2 2
( x – x + 10) (1 – x )
x5
4
2
( x + 3) ( x + 2 x + 10)
(3 x 2 + 2 x –1)2
2
(2 x + x + 10)
2⎡
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
Cx + D
2
2 x + x + 10
+
Cx + D
2
(2 x + x + 10)
2
+
Ex + F
2
(2 x + x + 10)3
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
© 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.
b.
V = 2π
x
2
1
3x – x
dx = −π
2
2
= −π ⎡ 2 3 x − x 2 ⎤ + 3π
⎢⎣
⎥⎦1
2 −2 x + 3 − 3
1
3x − x
2
1
9
4
2
1
3 − 2x
3x − x
2
dx + 3π
2
1
1
3x − x2
dx
2
1
2
dx = – π
(
− 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 =
L=
x2
x
,y =
16
8
2
⎛x⎞
1 + ⎜ ⎟ dx =
⎝8⎠
4
0
4
0
1+
x2
dx
64
x = 8 tan t, dx = 8sec2 t
L=
tan –1 1
2 sec t ⋅ 8sec 2 t dt
0
=8
tan –1 1
2 sec3 t dt
0
= 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
tan –1 1
2
0
Note: Use Formula 28 for sec3 t dt.
1
3
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
2
0
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
⎝
48. V = 2π
0
2
0
u=2–x
x=2–u
V = 2π
0
2
2
4 x 2 2 – xdx
du = –dx
dx = –du
4(2 – u )2 u (– du ) = 8π
2
8
2
⎡8
⎤
(4u1/ 2 – 4u 3 / 2 + u 5 / 2 )du = 8π ⎢ u 3 / 2 – u 5 / 2 + u 7 / 2 ⎥
0
5
7
⎣3
⎦0
2
⎛ 16 2 32 2 16 2 ⎞
⎛ 128 2 ⎞ 1024 2π
= 8π ⎜⎜
–
+
≈ 43.3287
⎟⎟ = 8π ⎜⎜
⎟⎟ =
5
7 ⎠
105
⎝ 3
⎝ 105 ⎠
Instructor’s Resource Manual
Section 7.7
471
© 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.
ln 3
49. V = 2π
0
2(e x –1)(ln 3 – x)dx = 4π
ln 3
0
x
[(ln 3)e x – xe x – ln 3 + x]dx
xe x dx = xe x – e x dx = xe – e x + C by using integration by parts.
Note that
ln 3
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
⎣
⎦
50. A =
18
3 3
3
x
2
x2 + 9
⎡⎛
1
⎤
⎞
= 4π ⎢⎜ 3ln 3 – 3ln 3 + 3 – (ln 3) 2 + (ln 3) 2 ⎟ – (ln 3 + 1) ⎥
2
⎠
⎣⎝
⎦
dx
x = 3 tan t, dx = 3sec2 t dt
A=
π/3
18
π/6
27 tan 2 t sec t
51. A = –
0
3sec2 t dt = 2
π/3
1 ⎞
⎛ 2
⎞
⎛
⎡ 1 ⎤
= 2⎜ –
+ 2 ⎟ = 4 ⎜1 –
dt = 2 ⎢ –
⎟ ≈ 1.6906
⎥
π / 6 sin 2 t
3
3⎠
⎣ sin t ⎦ π / 6
⎝
⎠
⎝
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.
–1 ⎛
2
–1
⎞
36
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
V =π
6
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+
2
3
−
cos 2 x
2
sin x
dx =
π/3
π/6
sin 2 x + cos 2 x
2
sin x
dx =
π/3
π/6
1
dx =
sin x
π/3
π/6
csc x dx
⎛ 2 3 +3⎞
⎛ 1 ⎞
− ln 2 − 3 = ln ⎜
⎟⎟ ≈ 0.768
⎟ − ln(2 − 3) = ln ⎜⎜
3
⎝ 3⎠
⎝ 3 ⎠
1
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.
81 − 4 x 2
dx =
x
54. a. First substitute u = 2 x, du = 2 dx to obtain
81 − u 2
du , then use Formula 55:
u
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
(
e x 9 − e2 x
x
3
) 2 dx = e8 ( 45 − 2e2 x )
9 − e2 x +
3
3
) 2 dx = (9 − u 2 ) 2 du , then use Formula 62:
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 =
cos x sin 2 x + 4 dx =
sin x
sin 2 x + 4 + 2 ln sin x + sin 2 x + 4 + C
2
1
b. First substitute u = 2 x , du = 2dx to obtain
Then use Formula 18:
u 2 + 4 du , then use Formula 44:
2
dx =
1 − 4x
1
1 2x +1
dx = ln
+C .
4 2x −1
1 − 4 x2
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
⎩
1
2
du
1− u2
.
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
3
3 ⎣ x +1
⎥
1 + x3
x2 − x + 1 ⎦ 3 ⎢
( x − )2 +
2
4
⎢
⎥
1
u = x − , du = dx
⎢⎣
⎥⎦
2
3
⎡
⎤
⎡
⎤
u−
x +1
1
−1 2
2
⎢
⎥ = 1 ⎢ln
⎥
= ln x + 1 −
du
+
3
tan
(
x
−
)
3
2 ⎥
3
⎢
⎥ F17 3 ⎢
2
u2 +
x
x
1
−
+
⎣
⎦
4
⎣
⎦
)
(
so
1⎡
dx = ⎢ln
0 1 + x3
3⎢
⎣
c
1
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
(
(
2
(c − 1 )
2
3
2
(c − 1 )
2
3
1
0 1 + x3
⎤
) + π6 ⎤⎥⎦ ⎥⎥⎦ .
⎤
) + π6 ⎤⎥⎦ ⎥⎥⎦ − 0.5 and G (c) = 1 +1c
3
we apply Newton’s
dx = 0.5 :
n
1
2
3
4
5
6
an 1.0000 0.3287 0.5090 0.5165 0.5165 0.5165
Thus c ≈ 0.5165 .
Instructor’s Resource Manual
Section 7.7
473
© 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.
Review and Preview Problems
1. lim
x2 + 1
x →2 x 2
−1
=
22 + 1
=
2
2 −1
5
3
2 x + 1 2(3) + 1 7
=
=
3+5
8
x →3 x + 5
2. lim
14. Note that, if = sec−1 x, then
1
1
se c = x ⇒ cos = ⇒ = 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+
2
x +1
2
1
5
x 2 = 1 + 0 = 1 or:
7. lim
= lim
1
x →∞ x 2 − 1 x →∞
1− 2 1− 0
x
lim
x2 + 1
x →∞ x 2
−1
= lim 1 +
x →∞
2
2
x −1
1
2x + 1
x = 2+0 = 2
8. lim
= lim
5 1+ 0
x →∞ x + 5
x →∞
1+
x
9.
10.
11.
12.
lim e− x = lim
lim e
− x2
2x
lim e
x →∞
17.
f ( x ) = x3e− x
y
5
5
10
x
=0
e
x →∞
x →− ∞
1
= lim
We would conjecture lim x 2 e− x = 0 .
=0
x →∞ x 2
x →∞
lim e
1
x →∞ e x
x →∞
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
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.
18.
f ( x ) = x 4 e− x
23.
y
x
a
0
1+ x
dx =
2
u = x2
du = 2 x dx
2
a
a
1⎡
ln(1 + x 2 ) ⎤ = ln ⎛⎜ 1 + a 2 ⎞⎟
⎢
⎥
⎦0
⎝
⎠
2⎣
1
2
4
8
16
⎛
⎞
ln ⎜ 1+ a 2 ⎟ 0.3466 0.8047 1.4166 2.0872 2.7745
⎝
⎠
5
10
x
24.
−2
a
0
1
a
dx = [ ln(1 + x) ]0 = ln (1 + a )
1+ x
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 →∞
25.
10 − x
19. y = x e
y
a
1
⎡ 1⎤
dx = ⎢ − ⎥ = 1 −
1 x2
a
⎣ x ⎦1
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 ⎤
dx = ⎢ −
= ⎢1 − ⎥
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.
4
1
a
20. Based on the results from problems 15-19, we
would conjecture
lim x n e− x = 0
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
=
u =− x 2
du =−2 x dx
a
16
1
2
4
8
16
1⎡ 2⎤
e− a
− ⎢e− x ⎥ = 1 −
2⎣
2
⎦
1−
dx = [ ln x ]a = ln
4
x
a
8
0.632 0.865 0.982 0.9997 0.9999+
dx
41
a
ln
1
1
2
4
a
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 )
Instructor’s Resource Manual
1
Review and Preview
475
© 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.
8
CHAPTER
Indeterminate Forms and
Improper Integrals
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
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.
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
e – ln(1 + x) –1
x
x
e +
= lim
x →0
2
1
(1+ x )2
2
= lim
x →0
=
e
x
– 1+1x
= lim
x →0
20. The limit is of the form
–1
= 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
x →0
8 x3
1
1+ x 2
24 x 2
1
1
= lim –
=–
24
x →0 24(1 + x 2 ) 2
Rule twice.)
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
x →0
0
1 + sin t dt
x
ex – e– x
= – ∞.
0
.
0
= lim 1 + sin x = 1
x →0
2x
1+1
=1
2
Instructor’s Resource Manual
Section 8.1
477
© 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.
24. The limit is of the form
x
0
lim
x →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
=
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.
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’H pital’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’H pital’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
Instructor’s Resource Manual
Section 8.1
479
© 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.
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.
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.
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.
40.
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
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.
Instructor’s Resource Manual
Section 8.1
481
© 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.
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
x ⋅ 2 ln 2(1 + x ln 2)
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 = ∞.
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.
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
x →0
x
= lim
1
x
x →0 +
–
x →0
x →0
1
x2
2
13. lim (csc 2 x – cot 2 x) = lim
x →0
14.
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
lim (cos x)csc x = lim eln y = 1
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 +
lim csc x(ln(cos x)) = lim
=1
x
+
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⎠
⎝
⎠
Instructor's Resource Manual
Section 8.2
483
© 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.
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
2
∞
22. The limit is of the form ∞ , which is not an
indeterminate form.
lim x x = ∞
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
indeterminate form.
x→ π
+ , which is not an
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
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.
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
© 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 + e−t > 1 for all t, so
39.
x
x
1 + e−t dt >
1
1
d.
dt = x − 1 .
∞
.
The limit is of the form
∞
x
1 + e−t dt
1
lim
x
x →∞
x →∞
n →∞
−x
n →∞
=1
n
lim
lim
1
+
x →1
41. a.
sin t dt
x −1
= lim
+
x →1
= lim
sin x
= sin(1)
1
lim
a = lim e
n →∞
n →∞
42. a.
c.
x →0 +
n →∞
since lim
n →∞
lim
n
n →∞
= lim
n →∞
486
a −1
1
n
n
n→∞
a ln a = ln a
Section 8.2
x →0
ln x
+
1
n2
ln x
1
x
1
x
1
x
1
x →0 + – 2
x
ln y
= lim
lim x x = lim e
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 +
Let y = ( x x ) x , then ln y = x ln( x x ).
lim ( x x ) x = lim eln y = 1
x →0 +
−
−
x →0+
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
1
n2
lim x ln( x x ) = 0
0
This limit is of the form ,
0
n
( ) (1 − ln n)
n (ln n − 1) = ∞
x →0 +
n = lim eln y = 1
( n a − 1) = nlim
→∞
0
,
0
n →∞
+
lim
n →∞
lim n
n
The limit is of the form
1
n →∞
1
n
The limit is of the form 00.
x →0
ln n
= lim n = 0
n →∞ n
n→∞ 1
n
= lim
lim x ln x = lim
lim
lim
n −1
Let y = x x , then ln y = x ln x.
=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
n→∞
1
Let y = a , then ln y = ln a.
n
1
lim ln a = 0
n →∞ n
ln y
n
n = 1 by part b.
n
n −1
1
n
n →∞
n
n
n
since lim
0
40. This limit is of the form .
0
x
( n n − 1) = nlim
→∞
This limit is of the form
1+ e
1
= lim
lim n
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.
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.
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 +
x x x
x x
Let y = (( x ) ) , then ln y = x ln(( 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 )
= lim
+
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 +
)
x →0 +
c.
x
( x( x ) )2
2 ln x
x
1
x →0+ – 2
x
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 →0+ – x ( x x ) ⎡ x x (ln x +1) ln x + x ⎤
⎢
x ⎥⎦
⎣
= lim
44. a.
–∞
.
∞
x x(ln x) + x x ln x + x
0
=
=0
1⋅ 0 + 1⋅ 0 + 1
x →0
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
+
= lim
x
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
+
1 ln x
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 =
Instructor's Resource Manual
Section 8.2
487
© 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.
1k + 2k +
lim
1/ t
+ nk
c.
n k +1
n →∞
1 ⎡⎛ 1 ⎞ ⎛ 2 ⎞
⎢⎜ ⎟ + ⎜ ⎟ +
n →∞ n ⎢⎝ n ⎠
⎝n⎠
⎣
k
k
⎛n⎞
⎜ ⎟
⎝n⎠
= lim
k
⎤
⎥
⎥⎦
9 ⎞
⎛1
lim ⎜ 2t + 5t ⎟
+ ⎝ 10
10 ⎠
t →0
= 10 2 ⋅
10 9
5 ≈ 4.562
48. a.
k
n
1 ⎛i⎞
⋅⎜ ⎟
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
= lim
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 +
lim
+ nk
2
n
1 ⎛i⎞
= lim
⋅⎜ ⎟
n →∞
i =1 n ⎝ n ⎠
n k +1
n →∞
lim
k
1/ t
⎛ n
⎞
46. Let y = ⎜ ci xit ⎟
⎜
⎟
⎝ i =1
⎠
0
⎞
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
i =1
ci = 1.
=
i =1
ci ln xi =
n
=
ei =1
ln xi ci
ci xit
i =1
ci xit ln xi
ln xi ci
= lim eln y
t →0 +
= x1c1 x2c2
xncn =
n
i =1
xi ci
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
, so the limit is of the form
= lim
n →∞
lim
n →∞
c.
2nx
xe
nx
∞
.
∞
2nx
xenx
= lim
2x
n →∞ x 2 enx
∞
.
∞
=0
1
2
xe− x dx = ⎡ − xe− x − e− x ⎤ = 1 −
⎣
⎦0
0
e
1
1
3
4 xe−2 x dx = ⎡ −2 xe−2 x − e−2 x ⎤ = 1 −
⎣
⎦0
0
e2
1
1
0
1
4
9 xe−3 x dx = ⎡ −3xe−3 x − e−3 x ⎤ = 1 −
⎣
⎦0
e3
1
5
16 xe−4 x dx = ⎡ −4 xe−4 x − e−4 x ⎤ = 1 −
⎣
⎦0
e4
1
0
1
6
25 xe−5 x = ⎡ −5 xe−5 x − e−5 x ⎤ = 1 −
⎣
⎦0
0
e5
1
1
7
36e−6 x dx = ⎡ −6 xe−6 x − e−6 x ⎤ = 1 −
⎣
⎦0
0
e6
d. Guess: lim
1 2
n →∞ 0
n xe− nx dx = 1
1
1 2 − nx
n xe dx
0
i =1
1/ t
⎛ n
⎞
lim ⎜ ci xi t ⎟
⎜
⎟
t →0+ ⎝ i =1
⎠
n
n
1
i =1
n
e
nx
1
n
n
n2 x
This limit is of the form
1
1 k
n x
n →∞ e nx
⎡ 1 k +1 ⎤
x dx = ⎢
x ⎥
⎣ k +1
⎦0
1
=
k +1
=
n 2 xe− nx =
= 2 5 ≈ 3.162
5
= 5 2 ⋅ 54 ≈ 4.163
= ⎡ − 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
1 2 − nx
n xe dx
n →∞ 0
lim
=1.
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.
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
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
2.
3.
b
lim
b →∞ 0
0
–∞
7.
∞
f ( x)dx
11.
Problem Set 8.3
In this section and the chapter review, it is understood
∞
a
means lim [ g ( x)]
b →∞
b
a
2.
3.
4.
12.
and likewise for
= ∞ – e100 = ∞
e x dx = ⎡ e x ⎤
⎣ ⎦100
100
The integral diverges.
5
dx
–∞ x4
∞
1
⎡ 1 ⎤
1
1
= ⎢–
=–
–0=
3⎥
3(–125)
375
⎣ 3x ⎦ – ∞
1
x2
dx
b
14.
∞
1
xe – x dx
u = x, du = dx
1
dv = e – x dx, v = – e – x
∞
1
∞
x dx
b
2
ln 2 + 1
⎡ ln x 1 ⎤
= lim ⎢ −
− ⎥ =
b →∞ ⎣
x
x ⎦2
2
∞
2
2
1
2 xe – x dx = ⎡⎢ – e – x ⎤⎥ = 0 – (– e –1 ) =
e
⎣
⎦1
∞
⎡ x⎤
2
= ⎢2
=∞
⎥ =∞–
1
πx ⎣ π ⎦1
π
The integral diverges.
dx
∞
1
⎡1
⎤
dx = ⎢ (ln x)2 ⎥ = ∞ – = ∞
e x
2
2
⎣
⎦e
The integral diverges.
∞ ln x
b
1
1
⎡1
⎤
e4 x dx = ⎢ e4 x ⎥ = e4 – 0 = e4
–∞
4
⎣4
⎦ –∞ 4
∞
1
dx = [ln(ln x)]e∞ = ∞ – 0 = ∞
x ln x
The integral diverges.
∞
e
⎡ ln x ⎤
= lim ⎢ −
+ lim
b →∞ ⎣
x ⎥⎦ 2 b →∞
= ⎡ 1 + x 2 ⎤ = ∞ – 82 = ∞
5.
⎢⎣
⎥⎦ 9
9
2
1+ x
The integral diverges.
6.
∞
⎡
⎤
1
dx = ⎢ –
⎥
2
1 (1 + x 2 ) 2
⎣⎢ 2(1 + x ) ⎦⎥1
x
∞
–5
1
∞
∞
dx
1
1
1
dx, dv =
dx, v = − .
2
x
x
x
∞ ln x
b ln x
dx = lim
dx
2 x2
b →∞ 2 x 2
∞
∞
dx =
13. Let u = ln x, du =
similar expressions.
1.
x
⎛ 1⎞ 1
= 0–⎜– ⎟ =
⎝ 4⎠ 4
4. p > 1
that [ g ( x)]
∞
1⎡
ln(1 + x 2 ) ⎤
⎣
⎦
10
2
1
= ∞ – ln 101 = ∞
2
The integral diverges.
∞
⎡ x 0.00001 ⎤
9.
=⎢
⎥ = ∞ – 100, 000 = ∞
1 x 0.99999
⎣⎢ 0.00001 ⎦⎥1
The integral diverges.
10.
0
dx
10 1 + x 2
∞
cos x dx
f ( x)dx;
∞
⎡
1
⎤
= –
1 x1.00001 ⎢⎣ 0.00001x 0.00001 ⎥⎦
1
1
1
⎛
⎞
= 0–⎜–
= 100, 000
⎟=
⎝ 0.00001 ⎠ 0.00001
∞
Instructor’s Resource Manual
∞
xe – x d = ⎡ – xe – x ⎤ +
⎣
⎦1
∞ –x
e dx
1
∞
2
= ⎡ – xe – x – e – x ⎤ = 0 – 0 – (– e –1 – e –1 ) =
⎣
⎦1
e
15.
1
⎡
⎤
1
= ⎢–
⎥
3
2
– ∞ (2 x – 3)
⎣⎢ 4(2 x – 3) ⎦⎥ – ∞
1
dx
=–
1
1
– (–0) = −
4
4
Section 8.3
489
© 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.
16.
4
1/ 3 ⎤ ∞
dx
∞
(π − x )
= ⎡ −3 (π − x )
⎣
2/3
= ∞ + 33 π − 4 = ∞
⎦4
The integral diverges.
17.
x
∞
–∞
x2 + 9
dx =
x
0
–∞
x2 + 9
The integral diverges since both
18.
dx
∞
–∞ ( x2
dx
0
=
–∞ ( x2
2
0
dx = ⎡ x 2 + 9 ⎤ + ⎡ x 2 + 9 ⎤ = (3 – ∞) + (∞ – 3)
2
⎣⎢
⎦⎥ – ∞ ⎢⎣
⎦⎥ 0
x +9
0
x
–∞
2
x +9
dx and
∞
x
0
2
x +9
dx diverge.
dx
∞
+
∞
0
x
+ 16)
( x + 16) 2
dx
1
x
x
by using the substitution x = 4 tan .
=
tan –1 +
2
2
2
128
4 32( x + 16)
( x + 16)
+ 16)
2
∞
dx +
2
0
0
⎡ 1
⎤
x
x
⎡ 1 ⎛ π⎞ ⎤
π
=0–⎢
=⎢
tan –1 +
⎥
⎜ – ⎟ + 0⎥ =
2
– ∞ ( x 2 + 16) 2
4 32( x + 16) ⎥⎦
⎣128 ⎝ 2 ⎠ ⎦ 256
⎢⎣128
–∞
dx
0
∞
⎡ 1
⎤
x
x
1 ⎛ π⎞
π
=⎢
tan –1 +
⎥ =
⎜ ⎟ + 0 – (0) =
2
0 ( x 2 + 16) 2
128 ⎝ 2 ⎠
256
4 32( x + 16) ⎥⎦
⎢⎣128
0
∞
dx
π
π
π
=
+
=
– ∞ ( x 2 + 16) 2
256 256 128
dx
∞
19.
1
∞
dx =
–∞ x2
1
∞
dx =
0
1
dx +
2
0
x + 1⎤
1
⎡1
–1 1 1 ⎛ π ⎞
dx = ⎢ tan –1
⎥ = 3 tan 3 – 3 ⎜ – 2 ⎟ =
– ∞ ( x + 1) 2 + 9
3
3
⎣
⎦ –∞
⎝
⎠
1
0
∞
0
1⎛
–1 1 ⎞
⎜ π + 2 tan
⎟
6⎝
3⎠
∞
1⎛
1⎞ 1⎛
1⎞ π
dx = ⎜ π + 2 tan –1 ⎟ + ⎜ π – 2 tan –1 ⎟ =
6⎝
3⎠ 6 ⎝
3⎠ 3
+ 2 x + 10
1
∞
x
∞
–∞ 2 x
dx =
e
For
dx
x + 1⎤
1⎛ π⎞ 1
1 1⎛
1⎞
⎡1
= ⎜ ⎟ – tan –1 = ⎜ π – 2 tan –1 ⎟
dx = ⎢ tan –1
⎥
2
3 ⎦0 3 ⎝ 2 ⎠ 3
3 6⎝
3⎠
⎣3
( x + 1) + 9
1
–∞ x2
20.
1
∞
– ∞ ( x + 1) + 9
– ∞ ( x + 1) + 9
0 ( x + 1) 2 + 9
+ 2 x + 10
1
1
x +1
dx = tan –1
by using the substitution x + 1 = 3 tan
2
3
3
( x + 1) + 9
2
0
x
– ∞ e –2 x
0
x
– ∞ e –2 x
dx =
0
–∞
dx +
∞
x
0
2x
e
dx
xe2 x dx, use u = x, du = dx, dv = e2 x dx, v =
0
1 2x
e .
2
0
1 0 2x
1
1
1
⎡1
⎤
⎡1
⎤
xe2 x dx = ⎢ xe2 x ⎥ –
e dx = ⎢ xe2 x – e2 x ⎥ = 0 – – (0) = –
–∞
4
4
4
⎣2
⎦ –∞ 2 –∞
⎣2
⎦ –∞
∞ x
∞ –2 x
1
For
dx =
xe dx, use u = x, du = dx, dv = e –2 x dx, v = – e –2 x .
0 e2 x
0
2
0
∞
1
⎡ 1
⎤
xe –2 x dx = ⎢ – xe –2 x ⎥ +
0
⎣ 2
⎦0 2
∞ x
1 1
dx = – + = 0
–∞ 2 x
4 4
e
∞
490
Section 8.3
∞ –2 x
e dx
0
∞
1
1⎞ 1
⎡ 1
⎤
⎛
= ⎢ – xe –2 x – e –2 x ⎥ = 0 – ⎜ 0 – ⎟ =
4
4⎠ 4
⎣ 2
⎦0
⎝
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.
21.
∞
–∞
0
sech x dx =
–∞
–1
0
= [tan (sinh x)]– ∞
∞
sech x dx =
0
25. The area is given by
∞
∞⎛ 1
2
1 ⎞
dx = ⎜
–
dx
2
1 4x −1
1 ⎝ 2 x –1 2 x + 1 ⎟⎠
sech x dx
+ [tan –1 (sinh x)]∞
0
∞
⎡ ⎛ π ⎞⎤ ⎡ π
⎤
= ⎢0 – ⎜ – ⎟ ⎥ + ⎢ – 0 ⎥ = π
2
2
⎝
⎠
⎣
⎦
⎣
⎦
22.
∞
1
=
csch x dx =
2e x
∞
e2 x –1
1
∞
1
1
dx =
sinh x
=
2
∞
ex – e– x
1
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 ⎠
dx
dx
Let u = e x , du = e x dx .
∞
2e x
1
2x
e
–1
dx =
2
∞
e
2
u –1
= [ln(u –1) – ln(u + 1)]∞
e
du =
∞⎛
e
1
1 ⎞
–
⎜
⎟ du
⎝ u –1 u + 1 ⎠
26. The area is
∞
1
dx =
1 x2 + x
∞
⎡ u –1 ⎤
= ⎢ln
⎥
⎣ u + 1⎦ e
24.
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.
∞
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 ⎞
⎜ –
⎟ dx
⎝ x x +1⎠
∞
⎡ 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
∞⎛ 1
1
x ⎤
1
∞ ⎡
= ⎡⎣ln x − ln x + 1 ⎤⎦ = ⎢ln
= 0 − ln = ln 2
⎥
1
2
⎣ x + 1 ⎦1
.
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.
1
1 ⎡ 2x −1 ⎤
∞
⎡ ln 2 x − 1 − ln 2 x + 1 ⎤⎦ = ⎢ ln
1
2⎣
2 ⎣ 2 x + 1 ⎥⎦1
29. FP =
∞ − rt
0
e
f (t ) dt =
∞
0
100, 000e−0.08t
∞
⎡ 1
⎤
= ⎢−
100, 000e−0.08t ⎥ = 1,250,000
⎣ 0.08
⎦0
The present value is $1,250,000.
30. FP =
∞ −0.08t
0
e
(100, 000 + 1000t )dt
∞
= ⎡ −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.
∞
−∞
f ( x) dx =
= 0+
a
−∞
0 dx +
b
a
1
dx +
b−a
∞
b
0 dx
1
1
(b − a )
[ x ]b + 0 =
b−a a
b−a
Instructor’s Resource Manual
Section 8.3
491
© 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.
b.
∞
=
−∞
a
=
−∞
x f ( x) dx
b
x ⋅ 0 dx +
a
x
1
dx +
b−a
∞
b
x ⋅ 0 dx
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
2
∞
=
−∞
a
=
−∞
( x − ) 2 dx
1 ⎡( x −
⎢
= 0+
b−a ⎢
3
⎣
=
2
=
c.
a.
( x − )2 ⋅ 0 dx
⎥
⎦a
+0
2
+ 3b
2
1
( b − a )3
12 ( b − a )
( b − a )2
12
0
=
32.
b
1
⎡ 1 b3 − 3 b 2 a + 3 ba 2 − 1 a3 ⎤
4
4
4
⎦
3(b − a ) ⎣ 4
=
P ( X < 2) =
=
∞
b
− a3 + 3a 2 − 3a
3
= (a + b) / 2 to obtain
1 b − 3b
b−a
Next, substitute
=
1
dx +
b−a
3
3
=
)3 ⎤⎥
a
( x − )2
)3 − ( a − )3
1 (b −
b−a
2
b
( x − )2 ⋅ 0 dx +
−∞
0 dx +
2
−∞
2
0
f ( x) dx
1
dx
10 − 0
2 1
=
10 5
∞
−∞
f ( x) dx =
0
−∞
0 dx +
∞
0
(x)
−1 −( x / )
e
dx
In the second integral, let u = ( x / ) . Then,
du = ( / )(t / ) −1 dt . When x = 0, u = 0 and when
x → ∞, u → ∞ . Thus,
∞
−∞
=
492
f ( x) dx =
∞ −u
0
e
∞
0
(x)
−1 − ( x / )
e
dx
∞
du = ⎡ −e−u ⎤ = −0 + e0 = 1
⎣
⎦0
Section 8.3
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.
b.
∞
=
−∞
2
3
=
2
0
xf ( x) dx =
−∞
∞ 2 − ( x / 3) 2
x e
0
∞
=
−∞
−1
⎛ x⎞
x⎜ ⎟
⎝ ⎠
∞
x ⋅ 0 dx +
0
e−( x /
)
dx
3
π
2
dx =
( x − )2 f ( x) dx =
0
−∞
( x − )2 ⋅ 0 dx +
2
9
∞
0
( x − )2 xe−( x
2
/ 9)
dx
3
3
3
π− =
π − π =0
2
2
2
The probability of being less than 2 is
=
c.
2
−∞
f ( x ) dx =
0
−∞
0 dx +
2
0
(x)
−1 − ( x / )
e
dx = 0 + ⎡⎢ −e −( x /
⎣
)
2
⎤
⎥⎦
0
2
= 1 − e−(2 / ) = 1 − e−(2 / 3) ≈ 0.359
33.
f ( x) = –
x–
3
2π
1
f ( x) = –
3
2π
2
e –( x – ) / 2
2
2
e –( x – ) / 2
2
+
( x – )2 –( x – )2 / 2
e
5
2π
2
⎛ ( x – )2
1 ⎞ –( x – )2 / 2 2
–
=⎜
=
⎟e
3
⎜ 5 2π
2π ⎟⎠
⎝
2
2
1
[( x − )2 − 2 ]e –( x – ) / 2
5
2π
f ( x) = 0 when ( x – )2 = 2 so x =
34.
a.
b.
∞
–∞
=
f ( x)dx =
∞
–∞
±
CM k
and the distance from
⎡ 1
dx = CM k ⎢ –
M x k +1
⎣ kx k
∞
xf ( x)dx =
∞
M
x
kM k
k +1
dx = kM k
to each inflection point is
∞
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, = 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
Section 8.3
493
© 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.
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 ∞
2
=
=
+
kM
x
–
x
dx
( x – ) 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 –
∞
2
∞
⎡
⎤
⎤
1
2k 2 M k +1 ⎡
1
k 3M k +2
= kM k ⎢ –
+
–
–
⎥
⎢
⎥
k –2
k –1 ⎣⎢ (k –1) x k –1 ⎦⎥
(k –1)2
⎣⎢ (k – 2) x
⎦⎥ M
M
∞
⎡ 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
=
(34 b.) and since, in our problem,
k −1
= 20, 000 and
k =3, we have
20000 =
3
4 × 104
M or M =
.
2
3
2
b. By 34 c.,
2
c.
∞
105
=
kM 2
(k − 2)(k − 1)
4 ⎞2
2
so that
36. u = Ar
37. a.
25
$100,000.
494
of one percent earn over
Section 8.3
−∞
sin x dx =
0
sin x dx +
−∞
a
+ lim
0 a →−∞
∞
0
sin x dx
[ − cos x ]0a
Both do not converge since –cos x is
oscillating between –1 and 1, so the integral
diverges.
b.
lim
a
a →∞ − a
sin x dx = lim [− cos x]−a a
a →∞
= lim [− cos a + cos(−a)]
⎛ 4 × 10
⎡1⎤
−⎜
⎟ lim
⎜ 3 ⎟ t →∞ ⎢⎣ x3 ⎥⎦ 5
10
⎝
⎠
Thus 6
∞
a →∞
t
⎛ 4 × 10
1⎤
64
⎡ 1
=⎜
−
=
⎟ lim
⎜ 3 ⎟ t →∞ ⎣⎢1015 t 3 ⎦⎥ 27 × 103
⎝
⎠
≈ 0.0024
∞
= lim [ − cos x ]
3
4 ⎞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 .
3 ⎛ 4 × 10
4 × 10
= ⎜
⎟ =
⎜
⎟
4⎝ 3 ⎠
3
4 ⎞3
2
A⎡
x
= ⎢
r ⎣⎢ r 2 + x 2
8
⎛ 4 × 104 ⎞
t
3
f ( x) dx = ⎜
dx =
⎟ lim
⎜ 3 ⎟ t →∞ 105 x 4
⎝
⎠
dx
∞
a
a →∞
= lim [− cos a + cos a] = lim 0 = 0
a →∞
38. a.
b.
a→∞
The total mass of the wire is
∞ 1
π
dx = from Example 4.
0 1 + x2
2
∞
⎡1
⎤
dx = ⎢ ln 1 + x 2 ⎥ which
0 1 + x2
⎣2
⎦0
diverges. Thus, the wire does not have a
center of mass.
∞
x
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.
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
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
N
f ( x )dx =
0
f ( x)dx +
∞
N
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 n2 2 2 1 x2
=
1 1 ⎡ 1⎤
1 1
+ –
= + (–0 + 1) = 1
2 2 ⎢⎣ x ⎥⎦1
2 2
∞
∞
(By viewing
for
1
x2
Thus,
1
n=2 n
2
as a lower Riemann sum
)
∞
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
Instructor’s Resource Manual
Section 8.3
495
© 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.
1
100
1
1.1
x
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
2
0
dx =
≈ 4.71
1 ⎡ −1 ⎤10
tan x
⎦0
π⎣
π(1 + x )
1.4711
≈
≈ 0.468
π
50
50
1
1
dx = ⎡ tan −1 x ⎤
⎦0
0 π(1 + x 2 )
π⎣
1.5508
≈
≈ 0.494
π
100
100
1
1
dx = ⎡ tan −1 x ⎤
2
⎦0
0 π(1 + x )
π⎣
1.5608
≈
≈ 0.497
π
43.
1
1
0
2π
1
2
2π
3 1
0
2π
4 1
0
2π
0
3.
dx
10
= lim ⎡ 2 x – 3 ⎤⎦
b
x – 3 b→3+ ⎣
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.
6.
1
b
dx
= lim ⎡sin –1 x ⎤
⎣
⎦0
0
2
b →1–
1– x
π
π
= lim sin –1 b – sin –1 0 = – 0 =
–
2
2
b →1
b
x
∞
dx = lim ⎡ 1 + x 2 ⎤
⎥⎦100
100
2
b →∞ ⎢⎣
1+ x
exp(–0.5 x 2 )dx ≈ 0.5000
b →∞
b
1
4– x
4. p < 1
496
10
3
The integral diverges.
2. 2
b→4
3
= lim 1 + b 2 + 10, 001 = ∞
1. unbounded
– 0
dx
exp(–0.5 x 2 )dx ≈ 0.4987
8.4 Concepts Review
lim
3
= 2 7 – lim 2 b – 3 = 2 7
7.
3.
33 2
3(b – 1) 2 / 3
3
3
2 – lim
=
–0=
3
3
2
2
2
2
b→1+
⎡
⎤
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.
exp(–0.5 x 2 )dx ≈ 0.3413
exp(–0.5 x 2 )dx ≈ 0.4772
3
⎡ 3( x – 1) 2 / 3 ⎤
= lim ⎢
⎥
1 ( x – 1)1/ 3 b →1+
2
⎣⎢
⎦⎥ b
dx
3
=
≈ 4.50
dx = [ln x]100
1 = ln100 ≈ 4.61
1
100
≈ 3.69
1.
100
1
1
42.
1 ⎤
⎡
dx = ⎢ –
⎥
⎣ 0.1x 0.1 ⎦1
1.01
1
= 0.99
100
1
100
Problem Set 8.4
100
⎡ 1⎤
dx = ⎢ – ⎥
2
⎣ x ⎦1
x
1
100
Section 8.4
dx
3
1
–1 x3
b
dx = lim
b →0
1
– –1 x3
b
3
1
+ b
x3
dx + lim
b→0
dx
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.
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.
8.
–5
1
5
2/3
x
b →0 + 5
1/ 3 b
= lim ⎡3x
⎣
b →0+
1
b
dx = lim
x
2/3
b→0– b
1/ 3 –5
⎤ + lim ⎡3 x
⎦ 5 b→0 – ⎣
1
–5
dx + lim
x
2/3
–1
x
= lim
⎤
⎦b
dx
b
b →0 – –1
128 –5 / 7
x –5 / 7 dx + lim
b →0 + b
x
b
dx
128
⎡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
128 –5 / 7
9.
dx
b →0 –
3
= 0 – 3 5 + 33 5 – 0 = 33 −5 − 3 5 = −6 3 5
1
10.
03
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
11.
4
dx
0
(2 – 3x)1/ 3
b→ 2
dx
b
= lim
– 0
3
(2 – 3 x)1/ 3
+ lim
b→ 2
+ b
3
b
dx
4
(2 – 3 x)1/ 3
4
⎡ 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
13.
x
8
12.
5
2 2/3
(16 − 2 x )
x
–4
0
16 – 2 x
2
dx =
⎡ 3
⎤
dx = lim ⎢ − (16 − 2 x 2 )1/ 3 ⎥
−⎣ 4
⎦
b→ 8
lim
b→ – 8
x
b
+ 0
16 – 2 x
2
dx +
x
–4
– b
16 – 2 x 2
dx
⎡ 1
⎤
⎡ 1
⎤
– ln 16 – 2 x 2 ⎥ + lim ⎢ – ln 16 – 2 x 2 ⎥
+⎢
–
4
4
⎣
⎦
⎣
⎦b
0 b→ – 8
8
lim
b→ –
=
b→ – 8
5
3
3
3
= lim − (16 − 2b 2 )1/ 3 + 3 6 = 3 6
− 4
4
4
b→ 8
–4
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
b
x
dx = lim ⎡ – 9 – x 2 ⎤ = lim − 9 – b 2 + 9 = 3
⎢
⎥⎦ 0 b→3–
2
b →3 – ⎣
9– x
–1
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.
Instructor’s Resource Manual
Section 8.4
497
© 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
dx
1 ⎤
= ⎢
−
⎥ dx by using a partial fraction decomposition.
(
x
−
1)(
x
+
2)
3(
x
−
1)
3(
x
+ 2) ⎦
x + x−2
⎣
3
b
3
dx
dx
dx
= lim
+ lim
0 x 2 + x – 2 b →1– 0 x 2 + x – 2 b→1+ b x 2 + x – 2
dx
16. Note that
=
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
2
x − x − x +1
3
dx
= lim
0 x3 – x 2 – x + 1 b →1–
1
=
2
2( x − 1)
b
dx
0
3
−
1
1
+
4( x − 1) 4( x + 1)
2
x – x – x +1
+ lim
dx
3
+ b
b→1
3
x – x2 – x + 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.
18. Note that
x1/ 3
x
2/3
−9
=
1
1/ 3
x
+
9
1/ 3
x
( x 2 / 3 − 9)
.
b
x1/ 3
27
27
27
⎡3
⎤
⎛3
⎞ ⎛
⎞
dx = lim ⎢ x 2 / 3 + ln x 2 / 3 – 9 ⎥ = lim ⎜ b 2 / 3 + ln b 2 / 3 – 9 ⎟ – ⎜ 0 + ln 9 ⎟
0 x2 / 3 – 9
– ⎣2
–
2
2
2
2
⎦ 0 b→27 ⎝
⎠ ⎝
⎠
b →27
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
© 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.
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
⎤
dx = lim ⎢ sin 2 / 3 x ⎥
3
0
+
2
⎦b
sin x
b →0 ⎣
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
26.
⎡1
⎤
tan x sec x dx = lim ⎢ tan 3 x ⎥
–
3
⎦0
b→ π ⎣
2
π
dx
x⎤
⎡
= lim cot ⎥
0 cos x – 1 b →0+ ⎢⎣
2 ⎦b
π
2
–1
–3
dx
b
= lim ⎡ 2 ln(– x) ⎤⎦
–3
x 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.
24.
π/4
0
27.
b
sec2 x
1 ⎤
⎡
dx = lim ⎢ –
2
–
tan x – 1 ⎥⎦ 0
(tan x – 1)
b→ π ⎣
ln 3
0
–
b→ π
4
ln 3
= 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 −
e x dx
1
1
+
= –(– ∞) – 1
tan b – 1 0 – 1
The integral diverges.
28. Note that
4
2
29.
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
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
dx
= lim [ln(ln x)]be = ln(ln e) – lim ln(ln b) = ln 1 – ln 0 = 0 +
x ln x b→1+
b →1+
The integral diverges.
e
1
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
dx
4c
⎡
⎤
= lim ⎢ln x + x 2 − 4c 2 ⎥ = ln ⎡⎣ (4 + 2 3)c ⎤⎦ − lim ln b + b 2 − 4c 2
2c
+⎣
2
2
⎦b
b → 2c +
b → 2c
x − 4c
= ln ⎡⎣ (4 + 2 3)c ⎤⎦ − ln 2c = ln(2 + 3)
4c
Instructor’s Resource Manual
Section 8.4
499
© 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.
32.
x dx
2c
c
2
x + xc – 2c
2
x dx
2c
=
(x + )
c
c 2
2
=
– 94 c 2
2c
c
( x + 2c ) dx − c
2
( x + 2c ) − 94 c2 2
⎡
c
c
= lim ⎢ x 2 + xc – 2c 2 − ln x + + x 2 + xc – 2c 2
+
2
2
b →c ⎣
dx
2c
0
( x + 2c )
2
− 94 c 2
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
dx = ⎢
= –
+2
=1–
+2
⎥ +2 c
2
2
c x (1 + x)
c
c
2 1+ c
1+ c
(1 + x)
(1 + x)2
(1 + x)
⎣1 + x ⎦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.
1
1
Thus, lim
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
dx
1
dx
1
Thus,
0
x(1 + x)
= lim
⎡
⎤
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
1
c →0 c
This is a proper integral.
35.
3
x
–3
9 – x2
dx =
= – 9 + lim
+
b → –3
36.
500
0
x
–3
9 – x2
dx +
9 – b 2 – lim
b →3
0
x
b
dx = lim ⎡ – 9 – x 2 ⎤ + lim ⎡ – 9 – x 2 ⎤
⎢
⎥⎦ b b→3– ⎢⎣
⎥⎦ 0
0
2
b→ –3+ ⎣
9– x
3
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 − x 2
−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
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.
37.
1
4
–4 16 – x 2
1
0
dx =
–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
1
4
dx +
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
− ln x
1
−1 2
dx =
−1
x − ln x
dx +
1
0
−1 2
x − ln x
−1 2
dx +
1
12
0
x − ln x
b
= lim ⎡ −2 − ln x ⎤
⎣
⎦b
b →−1+
dx +
1
1
12
x − ln x
dx
12
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.
∞
1
0
xp
dx =
1
1
0
xp
1
∞
dx +
1
xp
1
= −3π + 3π lim b −1/ 3
x →0 +
0
40.
0
0
1
0
1
dx and
x
∞1
1
x
dx diverge.
1
ln x dx +
c →0 − c
1
lim x ln x − x c +
+
c →0
[
b
1
ln x dx
] [ x ln x − x ]1b
= −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
c
+ b
where 1 < c < ∞.
b→1
f ( x)dx + lim
b
8
b
b →∞ c
( x − 8) −2 / 3 dx = lim ⎡3( x − 8)1/ 3 ⎤
⎣
⎦0
0
b →8−
= 3(0) – 3(–2)= 6
42.
=
ln x dx = lim
c →0+
= lim
41.
b
dx = ∞ .
If p = 1, both
∞
The limit tends to infinity as b → 0, so the
volume is infinite.
44. Since ln x < 0 for 0 < x < 1, b > 1
x →∞
If p = 0,
b →0
∞
⎡ 1
⎤
0,
dx = ⎢
x − p +1 ⎥
1 xp
⎣ − p +1
⎦1
1
∞
diverges since lim x − p +1 = ∞ .
∞
1
1
V = π x −4 / 3 dx = lim π ⎡ −3x −1/ 3 ⎤
⎣
⎦b
0
b →0 +
b.
since lim x − p +1 = ∞ .
If p < 1 and p
1
= lim ⎡3x1/ 3 ⎤ = 3
⎣
⎦b
b →0 +
1
⎡ 1
⎤
dx = ⎢
x − p +1 ⎥ diverges
0 xp
⎣ − p +1
⎦0
1
If p > 1,
1 −2 / 3
x
dx
0
43. a.
dx
f ( x)dx
45.
1 sin x
dx is not an improper integral since
x
sin x
is bounded in the interval 0 x 1.
x
0
46. For x
1⎛ 1
1 ⎞
dx
−
⎜
3
0 ⎝ x x + 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 ⎤
dx = lim ⎢ –
+
⎥ = – blim
1 x4
→∞ 3b3 3
b →∞ ⎣ 3 x3 ⎦1
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
= lim
1,
1
Section 8.4
501
© 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, x 2 ≥ x so – x 2 ≤ – x, thus
47. For x
From Example 2 of Section 8.2, lim
51. a.
2
= –0 +
= lim [– e – x ]1b = – lim
1
b →∞ eb
b→∞
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
∞ – x2
e dx
1
Thus, by the Comparison Test,
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 = lim
dx
2
2
b →∞
x+2
x+2
∞
= lim ⎡⎣ 2 x + 2 ⎤⎦ = lim 2
2
b →∞
b →∞
(
)
∞ n –1 – x
b+2 −2 = ∞
1
≤
1
2
. Since
∞
1
of
a
b
a
x →b
g ( x)dx implies the convergence of
implies the divergence of
b
a
b
a
g ( x)dx.
M
∞
1
1
x2
∞ n –1 – x
M
x
e dx
dx
x n –1e – x dx
1
∞ n –1 – x
1
x
e dx
1
1
= ⎡ – e – x ⎤ = – e –1 + 1 = 1 – , so the
⎣
⎦0
e
integral converges when n = 1. For 0 x 1,
1 –x
e dx
0
0 ≤ x n –1 ≤ 1 for n > 1. Thus,
e
x→a
f ( x)dx and the divergence of
52.
x n –1
lim f ( x) = lim g ( x) = ∞, then the convergence
x →b
b
x n –1e – x dx +
x n –1e – x dx +
converges.
2
50. If 0 f(x) g(x) on [a, b] and either
lim f ( x) = lim g ( x) = ∞ or
x→a
M
1
1
integral is finite, so
∞
1
e dx =
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
( )
≤
M
x
= 1+
49. Since x 2 ln ( x + 1) ≥ x 2 , we know that
1
b
1 1
⎡ 1⎤
dx = lim ⎢ – ⎥ = – lim +
2
1 x
b →∞ ⎣ x ⎦1
b →∞ b 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
x →∞ e x
e– x ≤ e– x .
∞ –x
e dx
1
xa
x
= x n –1e – x ≤ e – x . By the comparison test
from Problem 50,
53. a.
(1) =
f ( x)dx
b.
1 n −1 – x
0
x
∞ 0 −x
x e dx
0
e dx converges.
∞
= ⎡ −e − x ⎤ = 1
⎣
⎦0
∞ n −x
x e dx
0
n
−x
(n + 1) =
Let u = x , dv = e dx,
du = nx n −1dx, v = −e− x .
(n + 1) = [− x n e− x ]∞
0 +
= 0+n
c.
502
Section 8.4
∞
0
nx n −1e− x dx
∞ n −1 − x
0
x
e dx = n (n)
From parts a and b,
(1) = 1, (2) = 1 ⋅ (1) = 1,
(3) = 2 ⋅ (2) = 2 ⋅1 = 2! .
Suppose (n) = (n − 1)!, then by part b,
(n + 1) = n (n) = n[(n − 1)!] = n ! .
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.
∞ –x
54. n = 1,
0
∞
xe – x dx = 1 = 1! = (2 –1)!
0
∞ 2 –x
x e dx = 2 = 2! = (3 –1)!
0
∞ 3 –x
n = 2,
n = 3,
x e dx = 6 = 3! = (4 –1)!
n = 4,
0
∞ 4 –x
x e dx
0
n = 5,
55. a.
e dx = 1 = 0! = (1 –1)!
∞
–∞
= 24 = 4! = (5 –1)!
∞
f ( x)dx =
0
Cx –1e –
y
Let y = x, so x =
∞
0
–1 – x
Cx
e
C –
–∞
∞
dx =
0
∞
xf ( x)dx =
Let y = x, so x =
dx
1
and dx =
0
y
–1
⎛ y⎞
C⎜ ⎟
⎝ ⎠
x
dy .
–
( )
1
e– y
1
( ) = 1 when C =
∞
=
b.
x
( )
=
( )
x –1e –
1
and dx =
x
C
dy =
∞
y –1e – y dy = C –
0
.
∞
dx =
x e–
0
( )
∞
y ⎞ –y 1
1
dy =
y e – y dy =
⎜ ⎟ e
0
( ) ⎝ ⎠
( )
(Recall that ( + 1) = ( ) for > 0.)
2
c.
=
–∞
( x – ) f ( x)dx =
∞
=
∞⎛
2
( ) 0
x +1e –
x
dx –
0
⎞
⎟
⎠
⎜x–
⎝
–1 ∞
2
0
( )
=
=
=
∞⎛
y⎞
⎜ ⎟
( ) 0 ⎝ ⎠
=
1
2
∞
( )
1
2
2
( )
+
2
–
0
y
+1
+1 – y
e
( + 2) –
2 2
2
+
2
2
Instructor’s Resource Manual
e– y
1
dy –
2
2
2
=
( )
( )
y
–1 – x
dx +
e
dx =
2
–2
( )
and dx =
1
( )
∞
∞⎛ 2
⎜x
0 ⎜
x –1e –
0
⎝
x
–
2
( )=
x+
2
2
⎞ –1 –
⎟x e
⎟
⎠
x
dx
dx
dy .
–1
2
–2 ∞
⎛ y⎞
y ⎞ –y 1
1
e
dy
+
e – y dy
0 ⎜ ⎟
0 ⎜ ⎟
(
)
⎝ ⎠
⎝ ⎠
( )
∞
( )
x
x
1
( )
–1 ∞
⎛
2
2
dy –
2
x e–
In all three integrals, let y = x, so x =
2
dx
1
( + 1) =
( )
0
∞
x
dy.
∞⎛
=
( )
0
y e – y dy +
( + 1) +
2
2
( )
2
2
∞
( )
0
( )=
y –1e – y dy
1
2
( )
( + 1)
( )–
2
2
( )
( )+
2
2
2
Section 8.4
503
© 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.
56. a.
∞
L{t }( s ) =
0
t e – st dt
x
1
, so dt = dx, then
s
s
Let t =
∞
0
∞ 1
x ⎞ –x 1
( + 1)
e
dx =
x e – x dx =
.
⎜
⎟
0 ⎝s⎠
0 s +1
s
s +1
∞⎛
t e – st dt =
If s ≤ 0 , t e− st → ∞ as t → ∞, so the integral does not converge. Thus, the transform is defined only when
s > 0.
b.
L{e t }( s ) =
∞
0
e t e – st dt =
∞ ( – s )t
e
dt
0
⎧∞ if > s
lim e( – s )b = ⎨
b →∞
⎩0 if s >
−1
1
=
Thus, L{e t }( s ) =
–s s−
c.
L{sin( t )}( s ) =
Let I =
∞
0
∞
0
∞
⎡ 1 ( – s )t ⎤
=⎢
e
⎥ =
⎣ –s
⎦0
when s > . (When s ≤
1 ⎡
lim e( – s )b – 1⎤⎥
– s ⎢⎣b→∞
⎦
, the integral does not converge.)
sin( t )e – st dt
sin( t )e – st dt and use integration by parts with u = sin( t), du =
cos( t)dt,
1
dv = e – st dt , and v = – e – st .
s
∞
∞
⎡ 1
⎤
cos( t )e – st dt
Then I = ⎢ – sin( t )e – st ⎥ +
0
s
s
⎣
⎦0
Use integration by parts on this integral with
1
u = cos( t), du = – sin( t)dt, dv = e – st dt , and v = – e – st .
s
∞
∞
⎛
⎞
∞
⎡ 1
⎤
⎡ 1
⎤
I = ⎢ – sin( t )e – st ⎥ + ⎜ ⎢ – cos( t )e – st ⎥ –
sin( t )e – st dt ⎟
⎟
⎣ s
⎦ 0 s ⎜⎝ ⎣ s
⎦0 s 0
⎠
∞
2
1⎡
⎛
⎞⎤
I
= – ⎢e – st ⎜ sin( t ) + cos( t ) ⎟ ⎥ –
s⎣
s
⎝
⎠⎦0 s 2
Thus,
∞
2⎞
⎛
1 ⎡ – st ⎛
⎞⎤
I ⎜1 +
⎟ = – ⎢e ⎜ sin( t ) + cos( t ) ⎟ ⎥
⎜
s⎣
s
⎝
⎠⎦0
s 2 ⎟⎠
⎝
I=–
(
∞
1
s 1+
2
s2
)
s
⎡ – st ⎛
⎞⎤
⎢ e ⎜ sin( t ) + s cos( t ) ⎟ ⎥ = – 2
⎝
⎠⎦0
s +
⎣
⎡
⎤
⎛
⎞
lim e – sb ⎜ sin( b) + cos( b) ⎟ – ⎥
s
s
⎝
⎠
⎦
2 ⎢b →∞
⎣
⎛
⎞ ⎧0 if s > 0
lim e – sb ⎜ sin( b) + cos( b) ⎟ = ⎨
s
b →∞
⎝
⎠ ⎩∞ if s ≤ 0
Thus, I =
504
s2 +
Section 8.4
2
when s > 0.
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.
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
dx =
dx +
dx
0 x p + xq
0 x p + xq
1 x p + xq
Both of these integrals must converge.
1
1 1
1
1 1 1
dx >
dx =
dx which
0 x p + xq
0 2x p
2 0 xp
converges if and only if p < 1.
∞
∞ 1
1
1 ∞ 1
dx >
dx =
dx which
1 x p + xq
1 2xq
2 1 xq
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 ⎞
dy
⎜1 – 2
⎟ dy =
⎟
0 ⎜
0
y +1⎠
y2 + 1
⎝
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.
Instructor’s Resource Manual
Section 8.5
505
© 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.
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
x→a
lim [ f ( x) g ( x)] = 1.
Consider f ( x) = 3x 2 and
x →∞
f ( x)
6x + 1
1
= lim
= while
2
g ( x) x →0 12 x + 2 2
lim
f ( x)
3x 2 + x + 1 1
= lim
=
g ( x ) x →0 4 x 3 + 2 x + 3 3
x →0
g ( x) = x 2 + 1, then
lim
lim
x →0
x→a
11. False:
Consider f ( x) = 3x 2 + x + 1 and
18. False:
f ( x)
3x2
= lim
g ( x) x→∞ x 2 + 1
19. True:
3
= lim
= 3, but
x →∞ 1 + 1
2
p > 1. See Example 4 of Section 8.4.
1
x
1
∞
0
x
1
xp
∞ 1
0
lim [ f ( x) – 3 g ( x)]
x →∞
= lim [3x 2 – 3( x 2 + 1)]
12. True:
20. False:
13. True:
14. True:
21. True:
See Example 7 of Section 8.2.
1/ f ( x )
Let y = [1 + f ( x)]
ln[1 + f ( x)]
= lim
f ( x)
x→a
x →a
lim
∞
–∞
0
.
0
22. False:
f ( x)
f ( x)
lim [1 + f ( x)]1/ f ( x ) = lim eln y = e1 = e
1 and
1.
1
dx .
x +1
0
−∞
f ( x)dx +
∞
f ( x )dx =
0
∞
0
f ( x)dx
f ( x)dx.
x →a
f ( x )dx converge so their sum
See Problem 37 of Section 8.3.
∞
0
f ( x)dx = lim
b
b →∞ 0
= lim [ f ( x)]b0 = lim
b →∞
b →∞
24. True:
∞
0
f ( x )dx ≤
f ( x )dx
f (b) – f (0)
∞ –x
0
Use repeated applications of
l'Hôpital's Rule.
= lim – e
e0 = 1 and p(0) is the constant term.
must converge.
b →∞
25. False:
Section 8.5
dx;
= 0 – f(0) = –f(0).
f(0) must exist and be finite since
f ( x) is continuous on [0, ).
1
=1
x →a 1 + f ( x)
506
xp
1
converges.
23. True:
1
1+ f ( x )
x
1
∞
dx +
Thus, both integrals making up
= lim
x→a
–∞
–∞
1
ln[1 + f ( x)]
lim
ln[1 + f ( x)] = lim
f ( x)
x→a f ( x)
x →a
16. True:
0
f ( x)dx =
0
, then
This limit is of the form
∞
p
If f is an even function, then
f(–x) = f(x) so
1
ln y =
ln[1 + f ( x)].
f ( x)
15. True:
∞
Consider
As x → a, f ( x) → 2 while
1
→ ∞.
g ( x)
1
dx diverges for p
= lim [–3] = –3
x →∞
1
0
dx diverges for p
xp
1
x →∞
dx =
p
–b
e dx = lim [– e – x ]b0
+ 1 = 1, so
b →∞
∞
0
f ( x)dx
The integrand is bounded on the
⎡ π⎤
interval ⎢0, ⎥ .
⎣ 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.
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
x
4x
4
= lim
=4
x →0 tan x x →0 sec 2 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 = e2
x →0
x →0
=0
Section 8.5
507
© 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.
x ln x = lim
lim
x →0 +
1
x
x →0+
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
x→ π
2
= lim
x→ π
2
508
cos 2 3 x
x sin x – π2
cos x
x→ π
2
19.
20.
21.
22.
= lim
x→ π
2
0
.
0
sin x + x cos x 1
= =1
sin x
1
∞
⎡ 1 ⎤
= ⎢–
⎥ = 0 +1 = 1
0 ( x + 1) 2
⎣ x + 1⎦0
∞
∞
0
dx
∞
π
π
= ⎡ tan –1 x ⎤ = – 0 =
⎣
⎦
0
2
2
1+ x
dx
2
1
1
1
⎡1
⎤
e2 x dx = ⎢ e2 x ⎥ = e2 – 0 = e2
–∞
2
⎣2
⎦ –∞ 2
1
dx
= lim[– ln(1 – x)]b–1
x b→1
= – lim ln(1 – b) + ln 2 = ∞
1
–1 1 –
b→1
cos x sin x
π
cos
3 x sin 3 x
x→
= lim
The integral diverges.
2
2
2
2
2
cos x − sin x
3(cos 3 x − sin 3 x)
Section 8.5
x sin x – π2
π
⎛
⎞
lim ⎜ x tan x – sec x ⎟ = lim
2
⎠ x → π2 cos x
x→ π ⎝
2
lim
2
3cos x
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
= lim
x→ π
2
1
2
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
=−
1
1
=
3(0 − 1) 3
23.
dx
= [ln( x + 1)]0∞ = ∞ – 0 = ∞
x +1
The integral diverges.
∞
0
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.
24.
2
dx
1
2
x(ln x)1/ 5
= lim
b
dx
1
x(ln x)1/ 5
b →1– 2
b
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
25.
∞
dx
x2 + x4
1
=
∞⎛
1
⎜ 2
⎝x
1
−
2
⎡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
dx
2
+ lim
4/5
5
= [(ln 2)4 / 5 – (– ln 2)4 / 5 ]
4
∞
π
π
π π
⎞
⎡ 1
⎤
dx = ⎢ – – tan –1 x ⎥ = 0 – + 1 + tan –1 1 = 1 + – = 1 –
2⎟
x
2
4
2
4
⎣
⎦
1+ x ⎠
1
1
1
26.
1
⎡ 1 ⎤
=⎢
= −0 =1
⎥
– ∞ (2 – x ) 2
⎣ 2 – x ⎦ –∞ 1
27.
b
0
b
0 dx
dx
dx
⎡1
⎤
⎡1
⎤
= lim
+ lim
= lim ⎢ ln 2 x + 3 ⎥ + lim ⎢ ln 2 x + 3 ⎥
–2 2 x + 3
– –2 2 x + 3
+ b 2x + 3
– 2
⎦ –2 b→ – 3 + ⎣ 2
⎦b
b→ – 3
b→ – 3
b→ – 3 ⎣
2
2
2
2
dx
1
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.
28.
29.
30.
31.
4
dx
1
x –1
∞
2
b →1+
b →1+
∞
1
1
⎡ 1 ⎤
= ⎢–
= –0 +
=
⎥
2
ln 2 ln 2
⎣ ln x ⎦ 2
x(ln x)
dx
∞
∞
dx
0
ex / 2
⎡
2 ⎤
2
= ⎢–
= –0 + = 2
x/2 ⎥
1
⎣ e
⎦0
dx
5
3
= lim [2 x – 1]b4 = 2 3 – lim 2 x – 1 = 2 3 – 0 = 2 3
(4 – x)
2/3
dx
b
= lim
b→4– 3
(4 – x)
2/3
b → 4+ b
b
dx
5
+ lim
(4 – x)
2/3
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–
32.
33.
∞
2
∞
∞
2
2⎤
1
1
⎡ 1
xe – x dx = ⎢ – e – x ⎥ = 0 + e –4 = e –4
2
2
2
⎣
⎦2
x
–∞ x2
+1
dx =
0
x
– ∞ x2
+1
dx +
∞
0
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 + x4
–∞ 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
© 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.
35.
ex
=
e2 x + 1
ex
(e x ) 2 + 1
Let u = e x , du = e x dx
ex
∞
0
e
2x
∞
dx =
+1
∞
π
π π π
du = ⎡ tan –1 u ⎤ = – tan –1 1 = – =
⎣
⎦
1
2
2 4 4
u +1
1
2
1
36. Let u = x3 , du = 3 x 2 dx
∞ 2 – x3
∞ 1 –u
1
x e dx =
e du =
–∞
–∞ 3
3
The integral diverges.
37.
0
–∞
e – u du +
0
∞
1
1
1
1
e du = ⎡ – e – u ⎤ + ⎡ – e – u ⎤ = (–1 + ∞) + (–0 + 1)
⎣
⎦
⎣
⎦
–∞ 3
0
3
3
3
∞ –u
0
x
3
dx = 0
9 − x2
See Problem 35 in Section 8.4.
−3
38. let u = ln(cos x), then du =
π
2
π
3
tan x
(ln cos x) 2
dx =
39. For p ≠ 1, p ≠ 0,
1
lim
b →∞ b p –1
1
∞
xp
b →0 b
1
p –1
∞1
1
xp
1
∞
⎡
⎤
1
1
1
dx = ⎢ –
+
⎥ = lim
p –1
p
p
–1
1 x
p
–1
b
→∞
(1 – p )b
⎣⎢ ( p – 1) x
⎦⎥1
∞
1
1
= ∞ when p < 1, p ≠ 0.
dx = [ln x]1∞ = ∞ – 0 . The integral diverges.
x
∞
1dx = [ x]1∞ = ∞ – 1 . The integral diverges.
1
1
⎡
⎤
1
1
1
dx = ⎢ –
+ lim
⎥ =
p –1
0 xp
1
–
p
b
→
0
( p – 1)b p –1
⎣⎢ ( p – 1) x
⎦⎥ 0
1
1
converges when p – 1 < 0 or p < 1.
When p = 0,
1
ln
1
1
⎡ 1⎤ 2
du = ⎢ – ⎥
=–
+0 =
2
1
u
ln
2
ln 2
⎣ ⎦ –∞
u
1
dx converges when p > 1 and diverges when p ≤ 1.
When p = 1,
0
du =
1
2
–∞
ln
1
40. For p ≠ 1, p ≠ 0,
lim
–∞
1
1–
2
ln
2 u
b →∞ b p −1
When p = 0,
1
1
⋅ – sin x dx = – tan x dx
cos x
= 0 when p – 1 > 0 or p > 1, and lim
When p = 1,
11
0
x
dx = [ln x]10 = 0 – lim ln b = ∞ . The integral diverges.
b →0+
1
0
1dx = [ x]10 = 1 – 0 = 1
dx 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
1
3
∞
1
x6 + x
1
x6 + x
<
1
x
3
. Hence,
∞
1
1
x6 + x
dx <
∞
1
1
x3
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
ln x
ln x
e
x
< 1 and
1
=
< .
e 2 x (e x ) 2 e x
Hence,
∞ ln x
∞
1
dx < e – x dx = [– e – x ]1∞ = –0 + e –1 = .
x
2
1 e
1
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⎤
dx <
dx = ⎢ – ⎥ = –0 + 1 = 1.
1 x3
1 x2
⎣ x ⎦1
∞ ln x
dx converges.
Thus,
1 x3
∞ ln x
Contrapositive: f ( x ) ≠ 2 x ⇒ f ( x) ≠ x 2 (AT)
8. Original: a < b ⇒ a 2 < b 2
Converse: a 2 < b 2 ⇒ a < b
Contrapositive: a 2 ≥ b 2 ⇒ a ≥ b
Review and Preview Problems
1. Original: If x > 0 , then x 2 > 0 (AT)
Converse: If x 2 > 0 , then x > 0
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.
Contrapositive: If x ≤ 0 , then x 2 ≤ 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
1 1 4 2 1 7
+ = + + =
2 4 4 4 4 4
1 1 1 1 1 12 + 6 + 4 + 3 25
= + + + =
=
12
12
i =1 i 1 2 3 4
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
Review and Preview
511
© 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.
⎛∞⎞
14. By L’Hopital’s Rule ⎜ ⎟ twice:
⎝∞⎠
20.
x
∞
1
2
x +1
t →∞ 1
n2
2n 2 1
lim
= lim
= =
2
4 2
n →∞ 2n + 1 n→∞ 4n
1
lim
2 t →∞
⎛∞⎞
15. By L’Hopital’s Rule ⎜ ⎟ twice:
⎝∞⎠
lim
x →∞
x
2
e
x
= lim
x →∞
2x
e
= lim
x
x →∞
2
ex
=0
lim
en
n →∞
17.
∞1
1
x
= lim
n →∞
2n
= lim
en
t
dx = lim
n →∞
1
2
en
=0
lim [ ln x ]
dx =
21.
∞
1
1
x2
t
1
t →∞ 1
x2
dx =
u = x 2 +1
du = 2 x dx
du = ∞
t
dx = lim
x
dx =
t →∞ 1 x 2 + 1
x2 + 1
Integral does not converge.
1
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
∞
lim
Integral does not converge.
18.
u
x +1
u = ln x
du = 1 x dx
t →∞ 1 x
t
= lim ln t
1 t →∞
t →∞
2
2
Integral does not converge (see problem 17).
⎛∞⎞
16. By L’Hopital’s Rule ⎜ ⎟ twice:
⎝∞⎠
n2
t 2 +1 1
x
t
dx = lim
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
Review and Preview
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.
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 →∞
n
n→∞ 1 + 3
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 =
n3 + 3n 2 + 3n
n →∞ n3
1 + n3 + 32
= lim
3. bounded above
= lim
=1
+ 32 + 13
n
+ 3n 2 + 3n + 1
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
4.
a1 = 5, a2 =
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
14
29
50
77
, a3 =
, a4 = , a5 =
3
5
7
9
2
3n + n
3n 2 + 2
= lim
= ∞;
n →∞ 2n –1
n→∞ 2 – 1
n
n →∞ 2n – 1
= 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 this
material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1
1
1
1
9. a1 = –1, a2 = , a3 = – , a4 = , a5 = –
2
3
4
5
n
1 cos(nπ) 1
cos ( nπ ) = ( −1) , so – ≤
≤ .
n
n
n
1
1
lim – = lim = 0, so by the Squeeze
n →∞ n n→∞ n
Theorem, the sequence converges to 0.
10. a1 = e –1 sin1 ≈ 0.3096, a2 = e –2 sin 2 ≈ 0.1231,
a3 = e
–3
sin 3 ≈ 0.0070, a4 = e
–4
sin 4 ≈ –0.0139,
a5 = e –5 sin 5 ≈ –0.0065
–1 ≤ sin n ≤ 1 for all n, so
– e – n ≤ e – n sin n ≤ e – n .
lim – e – n = lim e – n = 0, so by the Squeeze
n →∞
n →∞
Theorem, the sequence converges to 0.
11. a1 =
a3 =
e2
e4
≈ 2.4630, a2 =
≈ 6.0665,
3
9
e6
e8
≈ 23.7311, a4 =
≈ 110.4059,
17
27
e10
≈ 564.7812
39
Consider
e2 x
2e2 x
4e2 x
= lim
= lim
=∞
lim
x →∞ x 2 + 3 x –1 x →∞ 2 x + 3 x →∞ 2
by using l’Hôpital’s Rule twice. The sequence
diverges.
a5 =
e2
e4
12. a1 =
≈ 1.8473, a2 =
≈ 3.4124,
4
16
6
8
e
e
a3 =
≈ 6.3036, a4 =
≈ 11.6444,
64
256
a5 =
e
2n
4n
10
e
≈ 21.510
1024
⎛ e2
=⎜
⎜ 4
⎝
n
⎞ e2
> 1 so the sequence diverges.
⎟ ,
⎟ 4
⎠
π
π2
13. a1 = – ≈ –0.6283, a2 =
≈ 0.3948,
5
25
a3 = –
π3
π4
≈ –0.2481, a4 =
≈ 0.1559,
125
625
a5 = –
π5
≈ –0.0979
3125
(– π)n
n
1
1
+ 3 ≈ 1.9821, a2 = + 3 = 3.0625,
4
16
1
1
a3 =
+ 3 3 ≈ 5.2118, a4 =
+ 9 ≈ 9.0039,
64
256
1
a5 =
+ 9 3 ≈ 15.589
1024
14. a1 =
n
1
⎛1⎞
⎜ ⎟ converges to 0 since –1 < < 1.
4
4
⎝ ⎠
3n / 2 =
( 3)
n
3 ≈ 1.732 > 1 .
diverges since
Thus, the sum diverges.
15. a1 = 2.99, a2 = 2.9801, a3 ≈ 2.9703,
a4 ≈ 2.9606, a5 ≈ 2.9510
(0.99)n converges to 0 since –1 < 0.99 < 1, thus
2 + (0.99)n converges to 2.
16. a1 =
a3 =
a5 =
1
2100
≈ 0.3679, a2 =
≈ 1.72 × 1029 ,
2
e
e
3100
≈ 2.57 × 1046 , a4 =
3
e
5100
4100
e
4
≈ 2.94 × 1058 ,
≈ 5.32 × 1067
e5
Consider lim
x →∞
x100
ex
. By Example 2 of
x100
Section 8.2, lim
ex
x →∞
= 0 . Thus, lim
n100
en
n →∞
= 0;
converges
17. a1 =
a3 =
a5 =
ln1
1
ln 3
3
ln 5
5
= 0, a2 =
ln 2
2
≈ 0.4901,
≈ 0.6343, a4 =
ln 4
≈ 0.6931 ,
2
≈ 0.7198
Consider lim
x →∞
ln x
x
= lim
x →∞
1
x
1
= lim
2 x
x →∞
using l’Hôpital’s Rule. Thus, lim
n →∞
ln n
n
2
x
= 0 by
= 0;
converges.
π
⎛ π⎞
= ⎜ – ⎟ , – 1 < – < 1, thus the sequence
n
5
⎝ 5⎠
5
converges to 0.
514
Section 9.1
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.
18. a1 =
a3 =
a5 =
ln1
2
= 0, a2 =
ln 13
6
ln 15
10
lim
ln 1n
n →∞
2n
2
ln 14
2 2
a3 = 61/ 6 ≈ 1.3480, a4 = 81/ 8 = 23 / 8 ≈ 1.2968,
≈ –0.4901,
a5 = 101/10 ≈ 1.2589
Consider lim (2 x)1/ 2 x . This limit is of the form
x →∞
≈ –0.5089
ln 1x
x →∞
x →∞
20. a1 = 21/ 2 ≈ 1.4142, a2 = 41/ 4 = 21/ 2 ≈ 1.4142,
≈ –0.3466,
≈ –0.4485, a4 =
Consider lim
= lim −
ln 12
2
2x
∞ 0 . Let y = (2 x)1/ 2 x , then ln y =
= lim
− ln x
x →∞
2x
= lim
x →∞
− 1x
1
2x
= 0 by using l’Hôpital’s Rule. Thus,
x
ln 2 x
2x
∞
This limit is of the form .
∞
lim
1
1
ln 2 x.
2x
x →∞ 2 x
ln 2 x = lim
x →∞
1
ln 2 x
1
= lim x = lim
=0
x →∞ 2 x
x →∞ 2 x →∞ 2 x
lim
= 0; converges
lim (2 x)1/ 2 x = lim eln y = 1
x →∞
1/ 2
⎛ 2⎞
19. a1 = ⎜1 + ⎟
⎝ 1⎠
= 3 ≈ 1.7321,
⎛ 2⎞
a2 = ⎜ 1 + ⎟
⎝ 2⎠
⎛ 2⎞
a3 = ⎜ 1 + ⎟
⎝ 3⎠
2/ 2
= 2,
3/ 2
⎛ 2⎞
a4 = ⎜1 + ⎟
⎝ 4⎠
4/2
5/ 2
⎛5⎞
=⎜ ⎟
⎝3⎠
x →∞
1/ 2 n
= 1; converges
Thus lim (2n)
n →∞
n
1
or an = 1 −
;
n +1
n +1
1 ⎞
1
⎛
lim ⎜1 −
= 1; converges
⎟ = 1 − lim
n +1⎠
n →∞ ⎝
n→∞ n + 1
21. an =
3/ 2
≈ 2.1517,
2
9
⎛3⎞
=⎜ ⎟ = ,
2
4
⎝ ⎠
22. an =
5/ 2
⎛ 2⎞
⎛7⎞
a5 = ⎜ 1 + ⎟
=⎜ ⎟
≈ 2.3191
5
⎝
⎠
⎝5⎠
2
Let = h, then as n → ∞, h → 0 and
n
n/2
⎛ 2⎞
lim ⎜1 + ⎟
= lim (1 + h)1/ h = e by
n⎠
h→0
Theorem 6.5A; converges
n →∞ ⎝
n
2
n +1
Consider
x
2
x
. Now, lim
x
= lim
x →∞ 2 x
1
x →∞ 2 x
by l’Hôpital’s Rule. Thus, lim
n
n →∞ 2n +1
ln 2
=0
= 0;
converges
n
n
; lim
2n − 1 n→∞ 2n − 1
1
1
= lim
= , but due to (−1)n , the terms of
2
n →∞ 2 − 1
23. an = (−1)n
n
the sequence alternate between positive and
negative, so the sequence diverges.
24. an =
1
1 − nn−1
= n;
lim n = ∞ ; diverges
n →∞
25. an =
n
2
2
=
n
2
2
=
n – (n –1)
n – (n – 2n + 1)
n
1
1
lim
= lim
= ; converges
2
n →∞ 2n –1 n→∞ 2 – 1
n
;
2n –1
n
Instructor’s Resource Manual
Section 9.1
515
© 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.
26. an =
n
n(n + 1)
n2 + n
=
=
;
(n + 1) − n1+1 (n + 1)2 − 1 n 2 + 2n
2
lim
n +n
n →∞ n 2
+ 2n
= lim
1 + 1n
n→∞ 1 + 2
n
= 1; converges
1
n
sin x
= 1; converges
lim
x →0 x
n2
n2
3n
;
n →∞ 3n
29. an =
lim
2n
n2
2n
n →∞ n 2
= lim
2n
n →∞ 3n
= lim
2
=0
ln 3
by using l’Hôpital’s Rule twice; converges
lim
n→∞ 3n (ln 3) 2
;
2n ln 2
2n (ln 2) 2
= lim
= ∞;
2
n →∞ 2n
n→∞
= lim
diverges
30. an =
an =
2n 2 − 1
n2 + n
1
1
n +1 – n
1
–
;
=
=
n n + 1 n(n + 1) n(n + 1)
to a limit L ≤ 2.
1
5
9
13
31. a1 = , a2 = , a3 = , a4 =
2
4
8
16
an is positive for all n, and an +1 < an for all
4n − 7
converges to a limit L ≥ 0.
2n +1
, so {an }
n + 2n
, so {an } converges
⎛
1 ⎞
1
< 1, so
an +1 = an ⎜ 1 −
and 1 −
⎜ (n + 1) 2 ⎟⎟
(n + 1)2
⎝
⎠
{an } converges to a limit L ≥ 0.
3
5
41
34. a1 = 1; a2 = ; a3 = ; a4 =
2
3
24
an < 2 for all n since
1
1
1
1
1
+ + +
1+ + + ≤
0
1
n
n! 2
2!
2
2 +1
<
1
= 0; converges
n →∞ n( n + 1)
2
2
3
⎛ 3 ⎞⎛ 8 ⎞ 2
33. a2 = ; a3 = ⎜ ⎟ ⎜ ⎟ = ;
4
⎝ 4 ⎠⎝ 9 ⎠ 3
⎛ 3 ⎞ ⎛ 8 ⎞⎛ 15 ⎞ 5
a4 = ⎜ ⎟ ⎜ ⎟⎜ ⎟ = ;
⎝ 4 ⎠ ⎝ 9 ⎠⎝ 16 ⎠ 8
⎛ 3 ⎞⎛ 8 ⎞ ⎛ 15 ⎞ ⎛ 24 ⎞ 3
a5 = ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ =
⎝ 4 ⎠⎝ 9 ⎠ ⎝ 16 ⎠ ⎝ 25 ⎠ 5
an > 0 for all n and an +1 < an since
lim
n ≥ 2 since an +1 − an = −
< 2 for all n, and an < an +1 for all
n since an +1 − an =
sin n
1
1
27. an = n sin ; lim n sin = lim
= 1 since
n n→∞
n n→∞ 1
28. an = (–1)n
1
7
17
31
32. a1 = ; a2 = ; a3 = ; a4 =
2
6
12
20
∞
k
⎛1⎞
⎜ ⎟ =2
k =0 ⎝ 2 ⎠
the sum never reaches 2. an < an +1 since each
term is the previous term plus a positive quantity,
so {an } converges to a limit L ≤ 2.
1
3
1⎛3⎞ 7
35. a1 = 1, a2 = 1 + (1) = , a3 = 1 + ⎜ ⎟ = ,
2
2
2⎝2⎠ 4
1 ⎛ 7 ⎞ 15
a4 = 1 + ⎜ ⎟ =
2⎝4⎠ 8
1 1
Suppose that 1 < an < 2, then < an < 1, so
2 2
3
1
3
< 1 + an < 2, or < an +1 < 2. Thus, since
2
2
2
1 < a2 < 2, every subsequent term is between
3
2
and 2.
1
1
an < 1, so an < 1 + an = an +1
2
2
and the sequence is nondecreasing, so {an }
converges to a limit L ≤ 2.
an < 2 thus
516
Section 9.1
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.
1⎛
2⎞
⎜2+ ⎟ =
2⎝
2⎠
1 ⎛ 3 4 ⎞ 17
a3 = ⎜ + ⎟ = , a4
2 ⎝ 2 3 ⎠ 12
36. a1 = 2, a2 =
un +1 = 3 + un > un if 3 + un > un 2 or
3
,
2
1 ⎛ 17 24 ⎞ 577
= ⎜ + ⎟=
2 ⎝ 12 17 ⎠ 408
un 2 – un – 3 < 0. un 2 – un – 3 = 0 when
Suppose an > 2 , and consider
(
1⎛
2 ⎞
2
>2 2⇔
⎜ an + ⎟ > 2 ⇔ an +
an ⎠
an
2⎝
2
> 0 , which is always true. Hence,
an > 2 for all n. Also,
n
1
2
3
4
5
6
7
8
9
10
11
lim un ≈ 2.3028
u=
41.
(
)
1
1 + 13 , then
2
(
(
)
)
(
)
Instructor’s Resource Manual
)
)
)
)
(
n
1
2
3
4
5
6
7
8
lim un ≈ 1.1118
un
0
1
1.1
1.11053
1.11165
1.11177
1.11178
1.11178
n →∞
)
(
(
(
1
1 ± 13 so
2
1⎛
2 ⎞
40. If a = lim an where an +1 = ⎜ an + ⎟ , then
an ⎠
2⎝
n →∞
1⎛
2⎞
a = ⎜ a + ⎟ or 2a 2 = a 2 + 2; a 2 = 2 when
a⎠
2⎝
a = ± 2, so a = 2, since a > 0.
1
3 < 3 + un < 7 + 13 and
2
1
1
3 < 3 + un = un +1 <
7 + 13 = 1 + 13
2
2
⎛ 1
1
7 + 13 = 1 + 13 can be seen by
⎜⎜
2
⎝ 2
squaring both sides of the equality and noting
that both sides are positive.) Hence, since
1
0 < u1 = 3 ≈ 1.73 < 1 + 13 ≈ 2.3028,
2
1
3 < un < 1 + 13 for all n; {un } is bounded
2
above.
)
)
1
1 + 13 ≈ 2.3028 since u > 0 and
2
(
n →∞
(
)
1
1 – 13 < 0.
2
un
1.73205
2.17533
2.27493
2.29672
2.30146
2.30249
2.30271
2.30276
2.30277
2.30278
2.30278
(
) (
u 2 – u – 3 = 0 when u =
2 < an +1 ≤ an and the
38. Suppose that 0 < un <
(
n →∞
series converges to a limit L ≥ 2.
37.
)
39. If u = lim un , then u = 3 + u or u 2 = 3 + u;
1⎛
2 ⎞
an +1 ≤ an ⇔ ⎜ an + ⎟ ≤ an
⎜
2⎝
an ⎟⎠
1 1
⇔
≤ an ⇔ 2 ≤ an
an 2
which is true. Hence,
)
(
an 2 + 2 > 2 2an ⇔ an 2 − 2 2an + 2 > 0 ⇔
( an − 2 )
(
1
1 ± 13 , thus un +1 > un if
2
1
1
1
1 – 13 < un < 1 + 13 ,
1 – 13 < 0
2
2
2
1
and 0 < un < 1 + 13 for all n, as shown
2
above, so {un } is increasing. Hence, by Theorem
D, {un } converges.
un =
)
42. Since 1.1 > 1, 1.1a > 1.1b if a > b. Thus, since
u3 = 1.1 > 1 = u2 , u4 = 1.11.1 > 1.11 = u3 .
Suppose that un < un +1 for all n ≤ N. Then
u N +1 = 1.1u N > 1.1u N –1 = u N , since u N > u N –1
by the induction hypothesis. Thus, un is
increasing.
1.1un < 2 if and only if un ln1.1 < ln 2;
un <
ln 2
≈ 7.3. Thus, unless un > 7.3,
ln1.1
un +1 = 1.1un < 2. This means that {un } is
bounded above by 2, since u1 = 0.
Section 9.1
517
© 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
k
→ 0; using Δx = , an equivalent
n
n
definite integral is
43. As n → ∞,
1
0
sin x dx = [– cos x]10 = – cos1 + cos 0 = 1 – cos1
≈ 0.4597
1
k
→ 0; using Δx = , an equivalent
n
n
definite integral is
1 1
π
dx = [tan –1 x]10 = tan –1 1 – tan –1 0 =
0 1 + x2
4
44. As n → ∞,
45.
n
n − (n + 1)
−1
1
−1 =
=
=
;
n +1
n +1
n +1 n +1
1
1
< ε is the same as < n + 1. For any given
ε
n +1
1
ε > 0 , choose N > − 1 then
ε
n≥N ⇒
46. For n > 0,
n
n +1
=
n
2
.
n
2
n +1 n +1
< ε is the
ε
n
n +1
and L is a lower bound for {an }. Then {– an } is
a nondecreasing sequence and –L is an upper
bound for {– an }. By what was just proven,
{– an } converges to a limit A ≤ –L, so {an }
converges to a limit B = –A ≥ L.
49. If {bn } is bounded, there are numbers N and M
with N ≤ bn ≤ M for all n. Then
an N ≤ an bn ≤ an M .
lim an N = N lim an = 0 and
n→∞
n→∞
n →∞
by the Squeeze Theorem, and by Theorem C,
lim an bn = 0.
n →∞
50. Suppose {an + bn } converges. Then, by
Theorem A
lim [(an + bn ) – an ] = lim (an + bn ) – lim an .
n →∞
n→∞
n→∞
But since (an + bn ) – an = bn , this would mean
that {bn } converges. Thus {an + bn } diverges.
{an } and {bn }
48. Suppose that {an } is a nondecreasing sequence,
and U is an upper bound for {an }, so
S = {an : n ∈ } is bounded above. By the
completeness property, S has a least upper bound,
which we call A. Then A ≤ U by definition and
an ≤ A for all n. Suppose that lim an ≠ A, i.e.,
n →∞
that {an } either does not converge, or does not
converge to A. Then there is some ε > 0 such that
Section 9.1
< A for all n,
2
which contradicts A being the least upper bound
for the set S. For the second part of Theorem D,
suppose that {an } is a nonincreasing sequence,
51. No. Consider an = (−1)n and bn = (−1)n +1 . Both
< ε.
47. Recall that every rational number can be written
as either a terminating or a repeating decimal.
Thus if the sequence 1, 1.4, 1.41, 1.414, … has a
limit within the rational numbers, the terms of the
sequence would eventually either repeat or
terminate, which they do not since they are the
decimal approximations to 2, which is
irrational. Within the real numbers, the least
upper bound is 2.
518
ε
lim an M = M lim an = 0, so lim an bn = 0
n2 + 1
1 1
= n+ > .
n
n ε
1
1
Since n + > n , it suffices to take n > . So for
n
ε
1
any given ε > 0 , choose N > , then
2
A – an > ε for all n, an < A –
n →∞
same as
n≥N ⇒
A – an ≤ ε for n ≥ N since {an } is
nondecreasing and an ≤ A for all n. However, if
n →∞
n
−1 < ε.
n +1
2
A – an > ε for all n, since if A – a N ≤ ε ,
diverge, but
an + bn = (−1) + (−1)n +1 = (−1) n (1 + (−1)) = 0 so
n
{an + bn }
52. a.
converges.
f3 = 2, f 4 = 3, f5 = 5, f 6 = 8,
f 7 = 13, f8 = 21, f9 = 34, f10 = 55
b. Using the formula,
1 ⎡1 + 5 1 − 5 ⎤ 1 ⎡ 2 5 ⎤
f1 =
−
⎢
⎥=
⎢
⎥ =1
2 ⎦
5⎣ 2
5⎣ 2 ⎦
2
2⎤
⎡
1 ⎢⎛ 1 + 5 ⎞ ⎛ 1 − 5 ⎞ ⎥
f2 =
⎜⎜
⎟⎟ − ⎜⎜
⎟⎟
5 ⎢⎝ 2 ⎠ ⎝ 2 ⎠ ⎥
⎣
⎦
⎡
⎤
1 1 + 2 5 + 5 − (1 − 2 5 + 5)
=
⎢
⎥
4
5⎣
⎦
1 ⎡4 5 ⎤
=
⎢
⎥ = 1.
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.
f
φ n +1 − (−1)n +1φ − n −1
lim n +1 = lim
n →∞ f n
n→∞
φ n − (−1) n φ − n
n +1
= lim
φ n +1 − ( −1)n +1
φ
n→∞
φn −
φ
φ − ( −1)
2 n +1
φ
n→∞
n
(
x
n +1
= lim
( −1)n
x
⎛ 1⎞
54. Let f ( x) = ⎜1 + ⎟ .
⎝ x⎠
)
1−
2
( −1) n
φ
=φ
n
⎛ 1⎞
lim ⎜1 + ⎟ = e .
n⎠
n →∞ ⎝
2n
(
)
1
⎡1
⎤
1+ 5 ⎥ − 1+ 5 −1
2
⎣2
⎦
⎛3
5⎞ ⎛1
5⎞
= ⎜⎜ +
⎟⎟ − ⎜⎜ +
⎟⎟ − 1 = 0
⎝2 2 ⎠ ⎝2 2 ⎠
φ 2 − φ −1 = ⎢
c.
⎛ 1⎞
lim ⎜1 + ⎟ = lim (1 + x )1/ x = e, so
x⎠
x →∞ ⎝
x →0 +
Therefore φ satisfies x 2 − x − 1 = 0 .
Using the Quadratic Formula on
x 2 − x − 1 = 0 yields
1± 1+ 4 1± 5
=
.
2
2
1+ 5
;
φ=
2
1
2
2(1 − 5) 1 − 5
− =−
=−
=
1− 5
2
φ
1+ 5
x=
53.
x
1 ⎞
⎛
55. Let f ( x) = ⎜1 + ⎟ .
⎝ 2x ⎠
x
1/ x
1 ⎞
⎛
⎛ x⎞
lim ⎜1 + ⎟ = lim ⎜ 1 + ⎟
+
2x ⎠
2⎠
x →∞ ⎝
x →0 ⎝
1/ 2
⎡⎛ x ⎞ 2 / x ⎤
= lim ⎢⎜ 1 + ⎟ ⎥
2⎠ ⎥
x →0+ ⎢⎝
⎣
⎦
= e1/ 2 , so
n
1 ⎞
⎛
lim ⎜ 1 + ⎟ = e1/ 2 .
2n ⎠
n →∞ ⎝
x
1 ⎞
⎛
56. Let f ( x) = ⎜1 + ⎟ .
⎝ x2 ⎠
1
x
⎛ ⎛ 1 ⎞ 2 ⎞1/ x
1 ⎞
⎛
lim ⎜ 1 + 2 ⎟ = lim ⎜ 1 + ⎜ ⎟ ⎟
x →∞ ⎝
x →∞ ⎜
x ⎟
x ⎠
⎝ ⎝ ⎠ ⎠
⎛1⎞
Using the fact that lim f ( x ) = lim+ f ⎜ ⎟ , we
x →∞
x →0
⎝ x⎠
can write
1
⎛ ⎛ 1 ⎞ 2 ⎞1/ x
lim ⎜ 1 + ⎜ ⎟ ⎟ = lim+ 1 + x 2
x →∞ ⎜
x →0
x ⎟
⎝ ⎝ ⎠ ⎠
(
From the figure shown, the sides of the triangle
have length n – 1 + 2x. The small right triangles
3
marked are 30-60-90 right triangles, so x =
;
2
thus the sides of the large triangle have lengths
2
3
n − 1 + 3 and Bn =
n −1+ 3
4
3 2
=
n + 2 3n − 2n − 2 3 + 4 while
4
(
)
n(n + 1) ⎛ 1 ⎞
π
π ⎜ ⎟ = ( n 2 + n)
2
8
⎝2⎠
An
= lim
n →∞ Bn
n →∞
lim
= lim
n→∞ 2
(
π (n2
8
+ 2 3n – 2n – 2 3 + 4)
π 1+
1
n
3 1 + 2 n 3 – n2 – 2
3
n2
+
4
n2
1/ x
)
=
π
2 3
which leads
. Then,
(
ln y = ln 1 + x 2
ln y =
)
1/ x
(
1
ln 1 + x 2
x
lim+ ln y = lim+
)
(
ln 1 + x 2
x
x→0
x→0
+ n)
)
)
= lim+
3 2
(n
4
(
(
Let y = 1 + x 2
x →0
2
An =
1/ x
to the indeterminate form 1∞ .
)
(
)
2x
1 + x2
) = lim 1+2xx
2
x →0
1
+
=0
This gives us
lim+ ln y = 0
x →0
ln ⎛⎜ lim+ y ⎞⎟ = 0
⎝ x →0 ⎠
lim+ y = e0 = 1 or
x →0
(
lim+ 1 + x 2
x →0
)
1/ x
=1
n
1 ⎞
⎛
Thus, lim ⎜1 + 2 ⎟ = 1 .
n →∞ ⎝
n ⎠
Instructor’s Resource Manual
Section 9.1
519
© 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.
x
x
⎛ x −1 ⎞
57. Let f ( x ) = ⎜
⎟ .
⎝ x +1⎠
⎛1⎞
Using the fact that lim f ( x ) = lim+ f ⎜ ⎟ , we
x →∞
x →0
⎝x⎠
can write
1/ x
x
⎛ 1x − 1 ⎞
⎛ x −1 ⎞
=
lim ⎜
lim
⎜
⎟
⎟
x →∞ ⎝ x + 1 ⎠
x → 0+ ⎜ 1 + 1 ⎟
⎝x ⎠
1/ x
⎛ 1− x ⎞
= lim+ ⎜ 1+xx ⎟
⎟
x →0 ⎜
⎝ x ⎠
1/ x
⎛ 1− x ⎞
= lim+ ⎜
⎟ which leads to the
x →0 ⎝ 1 + x ⎠
indeterminate form 1∞ .
1/ x
⎛1− x ⎞
Let y = ⎜
⎟
⎝ 1+ x ⎠
lim+ ln y = lim+
x →0
y ⎤⎥ = lim+
⎦ x →0
= lim+
x →0
1 ⎛1− x ⎞
ln ⎜
⎟
x ⎝ 1+ x ⎠
ln
lim+ ln y = lim+
x →0
1− x
1+ x
x
−2
1 − x2
n
. Then,
⎛ n −1 ⎞
−2
Thus, lim ⎜
⎟ =e .
n →∞ ⎝ n + 1 ⎠
x→0
1 ⎛ 2 x2 + 1 ⎞
ln ⎜
⎟
x ⎜⎝ 3 x 2 + 1 ⎟⎠
2
(l'Hopital's Rule)
⎛ 1− x ⎞
lim+ ⎜
⎟
x →0 ⎝ 1 + x ⎠
or
Section 9.1
which leads
( )
( )
1/ x
520
1/ x
⎛ 2 x2 + 1 ⎞
= lim+ ⎜ 2
⎟
x →0 ⎜ 3 x + 1 ⎟
⎝
⎠
1/ x
−2
This gives us,
ln ⎡⎢ lim+ y ⎤⎥ = −2
⎣ x →0 ⎦
x →0
1/ x
⎛ 2 x22+1 ⎞
= lim+ ⎜ x2 ⎟
x →0 ⎜ 3 x +1 ⎟
⎝ x2 ⎠
1/ x
⎞
⎟
⎟
⎠
⎛ 2 x2 + 1 ⎞
ln y = ln ⎜ 2
⎜ 3 x + 1 ⎟⎟
⎝
⎠
1 ⎛1− x ⎞
ln ⎜
⎟
x ⎝ 1+ x ⎠
lim+ y = e −2
x
⎛ 2 + 12
⎞
x
⎟ = lim+ ⎜
⎟
⎜ 3 + 12
0
x
→
⎠
x
⎝
1/ x
1/ x
ln ⎡⎢ lim+
⎣ x →0
⎛ 2 + x2
lim ⎜
x →∞ ⎜⎝ 3 + x 2
⎛ 2 x2 + 1 ⎞
Let y = ⎜ 2
⎜ 3 x + 1 ⎟⎟
⎝
⎠
⎛ 1− x ⎞
ln y = ln ⎜
⎟
⎝ 1+ x ⎠
x →0
⎛1⎞
Using the fact that lim f ( x ) = lim+ f ⎜ ⎟ , we
x →∞
x →0
⎝x⎠
can write
to the indeterminate form 1∞ .
. Then,
1/ x
ln y =
⎛ 2 + x2 ⎞
58. Let f ( x) = ⎜
⎟ .
⎜ 3 + x2 ⎟
⎝
⎠
= e−2
ln 2 x2 +1
3 x +1
⎡
⎤
ln ⎢ lim+ y ⎥ = lim+
x
⎣ x →0 ⎦ x →0
6x ⎤
⎡ 4x
= lim+ ⎢ 2
−
⎥ (l'Hopital's Rule)
x →0 ⎣ 2 x + 1 3 x 2 + 1 ⎦
=0
This gives us,
ln ⎡⎢ lim+ y ⎤⎥ = 0
⎣ x →0 ⎦
1/ x
lim+ y = e0 = 1 or
x →0
⎛ 1− x ⎞
lim+ ⎜
⎟
x →0 ⎝ 1 + x ⎠
=1
Thus,
⎛ 2 + n2
lim ⎜
n →∞ ⎜ 3 + n 2
⎝
n
⎞
⎟⎟ = 1 .
⎠
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.
⎛ 2 + x2
59. Let f ( x ) = ⎜
⎜ 3 + x2
⎝
Problem Set 9.2
x2
⎞
⎟⎟
⎠
∞
⎛1⎞
Using the fact that lim f ( x ) = lim+ f ⎜ ⎟ , we
x →∞
x →0
⎝ x⎠
can write
⎛ 2 + x2
lim ⎜
x →∞ ⎜⎝ 3 + x 2
⎛
= lim+ ⎜
x →0 ⎜
⎝
⎞
⎟
⎟
⎠
2 x 2 +1
x2
3 x 2 +1
x2
x2
1/ x
⎞
⎟
⎟
⎠
2
∞
1/ x 2
3.
1/ x 2
)
⎤
⎥ (l'Hopital's Rule)
⎥
⎦⎥
⎛ 1− x ⎞
lim ⎜
⎟
x → 0+ ⎝ 1 + x ⎠
Thus,
⎛ 2 + n2
lim ⎜
n →∞ ⎜ 3 + n 2
⎝
⎞
⎟⎟
⎠
n2
= e −1 .
9.2 Concepts Review
1. an infinite series
2. a1 + a2 +
3.
r < 1;
1
6
⎛ 1⎞
+⎜– ⎟
⎝ 4⎠
–5
+
; a geometric series
2
k
; a geometric
+ an
a
1– r
2
k
∞
4.
1/ x 2
x →0
–4
=
= e −1
; a geometric
1
series with a = 3, r = – ;
5
3
3 5
S=
= =
⎛ 1⎞ 6 2
1– ⎜ – ⎟ 5
⎝ 5⎠
Thus, by Theorem B,
k
k
∞ ⎡
⎛1⎞
⎛ 1 ⎞ ⎤ 8 5 31
⎢2 ⎜ ⎟ + 3 ⎜ – ⎟ ⎥ = + =
⎝ 5 ⎠ ⎥⎦ 3 2 6
k =0 ⎢⎣ ⎝ 4 ⎠
2 x 2 +1
3 x 2 +1
2
= −1
This gives us,
ln ⎡⎢ lim+ y ⎤⎥ = −1
⎣ x →0 ⎦
or
⎛ 1⎞
+⎜– ⎟
⎝ 4⎠
1 ⎛1⎞
⎛ 1⎞
3⎜ – ⎟ = 3 – 3 ⋅ + 3⎜ ⎟ −
5
5 ⎝5⎠
⎝
⎠
k =0
( )
x
⎡
−1
= lim+ ⎢
⎢
2
x →0
2 x + 1 3x 2 + 1
⎣⎢
lim+ y = e −1
–3
1
⎛1⎞
⎛1⎞
2⎜ ⎟ = 2 + 2⋅ + 2⎜ ⎟ +
4
4
⎝4⎠
k =0 ⎝ ⎠
∞
⎛ 2 x2 + 1 ⎞
lim+ ln y = lim+ 2 ln ⎜ 2
⎜ 3x + 1 ⎟⎟
x →0
x →0 x
⎝
⎠
)(
⎛ 1⎞
=⎜– ⎟
⎝ 4⎠
1
7
6
7
1
2
2 8
= = .
series with a = 2, r = ; S =
4
1 – 14 43 3
1
(
– k –2
k
∞
. Then,
⎛ 2 x2 + 1 ⎞
ln y = ln ⎜ 2
⎜ 3 x + 1 ⎟⎟
⎝
⎠
y ⎤⎥ = lim+
⎦ x →0
; a geometric
a = (–4)3 , r = –4; r = 4 > 1 so the series diverges.
1/ x 2
ln ⎡⎢ lim+
⎣ x →0
⎛ 1⎞
⎜– ⎟
k =1 ⎝ 4 ⎠
= (–4)3 + (–4)4 + (–4)5 +
with
which
leads to the indeterminate form 1∞ .
ln
2
1
2.
⎛ 2x2 + 1 ⎞
= lim+ ⎜ 2
⎟
x →0 ⎜ 3 x + 1 ⎟
⎝
⎠
⎛ 2 x2 + 1 ⎞
Let y = ⎜ 2
⎜ 3x + 1 ⎟⎟
⎝
⎠
k
1 1 1 1⎛1⎞
⎛1⎞
⎜ ⎟ = + ⋅ + ⎜ ⎟ +
7 7 7 7⎝7⎠
k =1 ⎝ 7 ⎠
1
1
series with a = , r = ; S = 7 =
7
7
1 – 17
1/ x 2
⎛ 2 + 12
x
= lim ⎜
x →0+ ⎜ 3 + 12
x
⎝
⎞
⎟
⎟
⎠
1.
2
5 5 1 5⎛1⎞
⎛1⎞
5⎜ ⎟ = + ⋅ + ⎜ ⎟ +
2 2 2 2⎝2⎠
k =1 ⎝ 2 ⎠
; a geometric
5
5
1
series with a = , r = ; S = 2 =
2
2
1 – 12
∞
⎛1⎞
3⎜ ⎟
k =1 ⎝ 7 ⎠
k +1
3
49
1 – 17
=
= 5.
2
=
3
3 1 3 ⎛1⎞
+ ⋅ + ⎜ ⎟ +
49 49 7 49 ⎝ 7 ⎠
geometric series with a =
S=
5
2
1
2
3
49
6
7
=
; a
3
1
,r = ;
49
7
1
14
Thus, by Theorem B,
k
k +1
∞ ⎡
1 69
⎛1⎞
⎛1⎞ ⎤
.
⎢2 ⎜ ⎟ – 3 ⎜ ⎟ ⎥ = 5 – =
4
7
14
14
⎝
⎠
⎝
⎠
k =1 ⎣⎢
⎦⎥
4. diverges
Instructor’s Resource Manual
Section 9.2
521
© 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.
∞
5.
k –5
4 3 2 1
1 2
= – – – – +0+ + +
k
+
2
3
4
5
6
8
9
k =1
∞
;
6.
1 – k5
k –5
= lim
= 1 ≠ 0; the series
k →∞ k + 2 k →∞ 1 + 2
k
2
9 9 9 9⎛9⎞
⎛9⎞
⎜ ⎟ = + ⋅ + ⎜ ⎟ +
8
8 8 8 8⎝8⎠
k =1 ⎝ ⎠
; a geometric
9
9 9
series with a = , r = ; > 1, so the series
8
8 8
diverges.
lim
k
diverges.
∞
7.
1 ⎞ ⎛ 1 1⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞
⎛1
⎜ –
⎟ = ⎜ – ⎟+⎜ – ⎟+⎜ – ⎟+
–1⎠ ⎝ 2 1⎠ ⎝ 3 2 ⎠ ⎝ 4 3 ⎠
k
k
k =2 ⎝
⎛1 ⎞ ⎛1 1⎞
S n = ⎜ – 1⎟ + ⎜ – ⎟ +
⎝2 ⎠ ⎝3 2⎠
lim Sn = lim –1 +
n →∞
n→∞
;
1 ⎞ ⎛1
1 ⎞
1
⎛ 1
+⎜
–
⎟+⎜ –
⎟ = –1 + ;
n
–
1
n
–
2
n
n
–
1
n
⎝
⎠ ⎝
⎠
∞
1
1 ⎞
⎛1
= –1, so
⎜ –
⎟ = –1
n
k =2 ⎝ k k – 1 ⎠
∞
8.
∞
∞
3
1
1
=3
which diverges since
diverges.
k =1 k
k =1 k
k =1 k
∞
9.
k!
k =1 100
1
2
6
+
+
+
100 10, 000 1, 000, 000
=
k
n +1
1
an , a1 =
. an > 0 for all n, and for n >99, an +1 > an , so the
100
100
sequence is eventually an increasing sequence, hence lim an ≠ 0. The sequence can also be described by
Consider {an }, where an +1 =
n →∞
an =
n!
100
n
∞
, hence
k!
k =1 100
k
∞
10.
2
2 2 2
= + + +
3 8 15
k =1 ( k + 2) k
diverges.
=
∞
1 ⎞ ⎛1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞
⎛1
⎜ –
⎟ = ⎜ – ⎟+⎜ – ⎟+⎜ – ⎟+⎜ – ⎟+
⎝1 3 ⎠ ⎝ 2 4 ⎠ ⎝ 3 5 ⎠ ⎝ 4 6 ⎠
k =1 ⎝ k k + 2 ⎠
1 ⎞ ⎛1
1 ⎞
⎛ 1⎞ ⎛1 1⎞ ⎛1 1⎞
⎛ 1
Sn = ⎜1 – ⎟ + ⎜ – ⎟ + ⎜ – ⎟ + + ⎜
–
⎟+⎜ –
⎟
3
2
4
3
5
n
–
1
n
+
1
n
n
+
2⎠
⎝
⎠ ⎝
⎠ ⎝
⎠
⎝
⎠ ⎝
1
1
1
3
2n + 3
3
2n + 3
–
= 1+ –
= –
= –
2
2 n + 1 n + 2 2 (n + 1)(n + 2) 2 n + 3n + 2
2+ 3
∞
2
3
3
2n + 3
3
3
n n2
= – lim
= , so
lim Sn = – lim
= .
2
3
2
2 n→∞ n + 3n + 2 2 n→∞ 1 + + 2 2
2
n →∞
k =1 ( k + 2) k
n n
∞
11.
⎛e⎞
⎜ ⎟
k =1 ⎝ π ⎠
k +1
2
e
e
(
(
π)
π)
S=
=
2
∞
12.
522
4k +1
k =1 7
k –1
2
π–e
π
1 – πe
2
2
2
⎛e⎞ ⎛e⎞ e ⎛e⎞ ⎛e⎞
= ⎜ ⎟ +⎜ ⎟ ⋅ +⎜ ⎟ ⎜ ⎟ +
⎝π⎠ ⎝π⎠ π ⎝π⎠ ⎝π⎠
=
e2
≈ 5.5562
π(π – e)
2
=
16
4
⎛4⎞
+ 16 ⋅ + 16 ⎜ ⎟ +
1
7
⎝7⎠
Section 9.2
2
e
⎛e⎞
; a geometric series with a = ⎜ ⎟ , r = < 1;
π
⎝π⎠
; a geometric series with a = 16, r =
4
16
16 112
< 1; S =
=
=
3
7
3
1 – 74
7
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.
13.
∞
⎛ 3
3
–
⎜⎜
2
k2
k = 2 ⎝ ( k – 1)
⎞ ⎛3 3⎞ ⎛ 3 3⎞ ⎛3 3 ⎞
⎟⎟ = ⎜ – ⎟ + ⎜ – ⎟ + ⎜ – ⎟ + ;
⎠ ⎝ 1 4 ⎠ ⎝ 4 9 ⎠ ⎝ 9 16 ⎠
⎛
3⎞ ⎛ 3 1⎞ ⎛1 3 ⎞
3
3
⎛
Sn = ⎜ 3 – ⎟ + ⎜ – ⎟ + ⎜ – ⎟ + + ⎜
–
⎜ (n – 2)2 (n – 1)2
4 ⎠ ⎝ 4 3 ⎠ ⎝ 3 16 ⎠
⎝
⎝
3
3
= 3–
; lim S = 3 – lim
= 3, so
2 n →∞ n
→∞
n
n
n2
∞ ⎛
3
3 ⎞
–
⎜⎜
⎟ = 3.
2
k 2 ⎟⎠
k = 2 ⎝ (k – 1)
∞
14.
2
2 2 2 2
= + + + +
k
–
5
1
2 3 4
k =6
=2
∞
1
1
which diverges since
diverges.
k
k =1
k =1 k
=
2
10
1
1 – 10
∞
2⎛1⎞
⎜ ⎟
10
⎝ 10 ⎠
k =1
21
= 100
1
1 – 100
100
21. Let s = 1 – r, so r = 1 – s. Since 0 < r < 2,
–1 < 1 – r < 1, so
∞
21 ⎛ 1 ⎞
⎜
⎟
k =1 100 ⎝ 100 ⎠
∞
=
k –1
∞
k =0
(1 − s ) s k
k =0
(1 – s ) s k –1 =
(−1) k x k =
22.
∞
r (1 − r )k =
k =1
21 7
=
=
99 33
13
= 1000
1
1 − 1000
∞
k =0
=
k –1
36 10,000
727
+
=
100 1 − 1
1980
s < 1, and
17. 0.013013013... =
∞
1− s
=1
1− s
(− x) k =
k =0
∞
(– x)k –1;
k =1
if –1 < x < 1 then
–1 < –x < 1 so x < 1 ;
∞
13 ⎛ 1 ⎞
⎜
⎟
k =1 1000 ⎝ 1000 ⎠
k –1
∞
k =1
(– x) k –1 =
1
1
=
1 − (− x) 1 + x
13
=
999
18. 0.125125125... =
=
k –1
36 ∞ 71 ⎛ 1 ⎞
+
⎜
⎟
100 k =1 10, 000 ⎝ 100 ⎠
71
=
2
=
9
16. 0.21212121
⎞
⎟⎟
⎠
20. 0.36717171... =
∞
15. 0.22222
=
⎞ ⎛ 3
3
–
⎟⎟ + ⎜⎜
2
n2
⎠ ⎝ (n – 1)
125
1000
1
1 − 1000
=
19. 0.4999... =
∞
125 ⎛ 1 ⎞
⎜
⎟
1000
⎝ 1000 ⎠
k =1
k –1
125
999
4 ∞ 9 ⎛1⎞
+
⎜ ⎟
10 k =1 100 ⎝ 10 ⎠
k –1
9
=
4
1
+ 100 =
10 1 − 1 2
10
Instructor’s Resource Manual
Section 9.2
523
© 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.
k
= ln k – ln(k + 1)
k +1
Sn = (ln1 – ln 2) + (ln 2 – ln 3) + (ln 3 – ln 4) +
23. ln
lim Sn = lim – ln(n + 1) = – ∞, thus
n →∞
n →∞
∞
ln
k =1
+ (ln( n – 1) – ln n) + (ln n – ln( n + 1)) = ln1 – ln(n + 1) = – ln(n + 1)
k
diverges.
k +1
⎛
1 ⎞
k2 –1
24. ln ⎜ 1 –
= ln
= ln(k 2 – 1) – ln k 2 = ln[(k + 1)(k – 1)] – ln k 2 = ln(k + 1) + ln(k – 1) – 2 ln k
⎟
2
2
k
⎝ k ⎠
Sn = (ln 3 + ln1 – 2 ln 2) + (ln 4 + ln 2 – 2 ln 3) + (ln 5 + ln 3 – 2 ln 4) +
+(ln n + ln(n – 2) – 2 ln(n – 1)) + (ln(n + 1) + ln(n – 1) – 2 ln n)
= –ln 2 + ln(n + 1) – ln n = – ln 2 + ln
lim Sn = – ln 2 + lim ln
n →∞
n →∞
n +1
n
n +1
n +1⎞
⎛
= – ln 2 + ln ⎜ lim
⎟ = – ln 2 + ln1 = – ln 2
n
⎝ n→∞ n ⎠
2
⎛2⎞
⎛2⎞
⎛2⎞
25. The ball drops 100 feet, rebounds up 100 ⎜ ⎟ feet, drops 100 ⎜ ⎟ feet, rebounds up 100 ⎜ ⎟ feet, drops
⎝3⎠
⎝3⎠
⎝3⎠
2
⎛2⎞
100 ⎜ ⎟ , etc. The total distance it travels is
⎝3⎠
2
3
2
3
⎛2⎞
⎛2⎞
⎛2⎞
⎛2⎞
⎛2⎞
⎛2⎞
100 + 200 ⎜ ⎟ + 200 ⎜ ⎟ + 200 ⎜ ⎟ + ... = −100 + 200 + 200 ⎜ ⎟ + 200 ⎜ ⎟ + 200 ⎜ ⎟ + ...
⎝3⎠
⎝3⎠
⎝3⎠
⎝3⎠
⎝3⎠
⎝3⎠
= −100 +
26. Each gets
∞
⎛2⎞
200 ⎜ ⎟
⎝3⎠
k =1
k –1
= −100 +
200
= 500 feet
1 − 23
∞
1 1 1 1⎛1 1⎞
1⎛1⎞
+ ⋅ + ⎜ ⋅ ⎟ + ... =
⎜ ⎟
4 4 4 4⎝4 4⎠
k =1 4 ⎝ 4 ⎠
1
4
k –1
=
1−
1
4
=
1
3
(This can be seen intuitively, since the size of the leftover piece is approaching 0, and each person gets the same
amount.)
∞
27. $1 billion + 75% of $1 billion + 75% of 75% of $1 billion + ... =
k =1
∞
28.
$1 billion (0.90) k –1 =
k =1
($1 billion)0.75k –1 =
$1 billion
= $4 billion
1 − 0.75
$1 billion
= $10 billion
1 − 0.90
29. As the midpoints of the sides of a square are connected, a new square is formed. The new square has sides
1
2
1
times the sides of the old square. Thus, the new square has area the area of the old square. Then in the next step,
2
1
of each new square is shaded.
8
Area =
∞
1
1 1 1 1
1⎛1⎞
⋅1 + ⋅ + ⋅ + ... =
⎜ ⎟
8
8 2 8 4
k =1 8 ⎝ 2 ⎠
The area will be
524
Section 9.2
k –1
=
1
8
1−
1
2
=
1
4
1
.
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.
∞
1 1⎛8⎞ 1⎛8 8⎞
+ ⎜ ⎟+ ⎜ ⋅ ⎟+
30.
9 9⎝9⎠ 9⎝9 9⎠
31.
1⎛8⎞
=
⎜ ⎟
k =1 9 ⎝ 9 ⎠
k –1
=
1
9
= 1; the whole square will be painted.
1 − 89
∞
3 3 ⎛ 1 1 ⎞ 3 ⎛ 1 1 ⎞⎛ 1 1 ⎞
3⎛ 1 ⎞
+ ⎜ ⋅ ⎟ + ⎜ ⋅ ⎟⎜ ⋅ ⎟ + ... =
⎜ ⎟
4 4 ⎝ 4 4 ⎠ 4 ⎝ 4 4 ⎠⎝ 4 4 ⎠
k =1 4 ⎝ 16 ⎠
k –1
=
3
4
1
1 − 16
=
4
5
The original does not need to be equilateral since each smaller triangle will have
1
area of the previous larger
4
triangle.
∞
π
32. Ratio of inscribed circle to triangle is
3 3
, so
π
3⎛1⎞
⋅ ⎜ ⎟
k =1 3 3 4 ⎝ 4 ⎠
k –1
=
( )=
π
4 3
1 − 14
π
3 3
(This can be seen intuitively, since every small triangle has a circle inscribed in it.)
33. a. We first note that, at each stage, the number of sides is four times the number in the previous stage and the
length of each side is one-third the length in the previous stage. Summarizing:
Stage
# of sides
length/
side (in.)
perimeter
0
3
9
27
pn
()
36
( )
⎛4⎞
27 ⎜ ⎟
⎝3⎠
1
3(4)
9
n
3(4n )
9
1
3
1
3n
n
n
4
⎛4⎞
The perimeter of the Koch snowflake is lim pn = lim 27 ⎜ ⎟ which is infinite since > 1 .
3
n →∞
n →∞
⎝3⎠
b. We note the following:
3 2
s
4
2. The number of new triangles added at each stage is equal to the number of sides the figure had at the
previous stage and
1. The area of an equilateral triangle of side s is
3. the area of each new triangle at a given stage is
3
(side length at that stage)2 . Using results from part a. we
4
can summarize:
Stage Additional triangles Area of each new Δ Additional area, An
(col 2, part a.)
(see col 3, part a.)
( )
3 2
9
4
( )
36
0
original
3 2
9
4
1
3
3 2
3
4
n
3 4n −1
(
)
Instructor’s Resource Manual
3⎛ 9 ⎞
⎜ ⎟
4 ⎝ 3n ⎠
2
( )
⎛4⎞
3 3⎜ ⎟
⎝9⎠
n−2
Section 9.2
525
© 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.
Thus the area of the Koch snowflake is
∞
n=0
An =
81 3 27 3 ∞
⎛4⎞
+
+ 3 3⎜ ⎟
4
4
⎝9⎠
n =1
n −1
=
81 3 27 3 ⎛ 3 3 ⎞
⎟
+
+⎜
⎜ (1 − 4 ) ⎟
4
4
9 ⎠
⎝
=
81 3 1 ⎛ 81 3 ⎞ 4 ⎛ 81 3 ⎞ 8 ⎛ 81 3 ⎞
+ ⎜⎜
⎟+ ⎜
⎟= ⎜
⎟
4
3 ⎝ 4 ⎟⎠ 15 ⎜⎝ 4 ⎟⎠ 5 ⎜⎝ 4 ⎟⎠
Note: By generalizing the above argument it can be shown that, no matter what the size of the original
equilateral triangle, the area of the Koch snowflake constructed from it will be
8
5
times the area of the original
triangle.
34. We note the following:
1. Each triangle contains the angles 90, ,90 −
succeeding triangle. Summarizing:
# triangle
base
height
1
h cos
h sin
2
h sin cos
h sin 2
n
h (sin n −1 ) cos
h sin n
2. The height of each triangle will be the hypotenuse of the
area An
1 2
h sin cos
2
1 2 3
h sin cos
2
1 2 2 n −1
cos
h sin
2
∞
Thus the total area of the small triangles is A =
n =1
∞
Now consider the infinite geometric series S =
An =
h 2 ⎛ cos ⎞ ∞
2 n −1
⎜
⎟ (sin )
2 ⎝ sin ⎠n = 2
(sin 2 )n −1 =
n =1
∞
then:
(sin 2 )n −1 = S − 1 =
n=2
In
1
cos 2
−1 =
sin 2
cos 2
1
1 − sin
=
1
cos 2
A=
Therefore:
ABC , height = h and base = h tan ; thus the area of
2
ABC =
h 2 ⎛ cos ⎞ ⎛ sin 2
⎜
⎟⎜
2 ⎝ sin ⎠ ⎜⎝ cos 2
⎞ h2
tan
⎟=
⎟ 2
⎠
1
h2
(h tan )h =
tan , the same as A .
2
2
35. Both Achilles and the tortoise will have moved.
100 + 10 + 1 +
=
∞
1
1
⎛1⎞
+
+ ... = 100 ⎜ ⎟
10 100
⎝ 10 ⎠
k =1
k –1
100
1
= 111 yards
1
9
1 − 10
Also, one can see this by the following reasoning. In the time it takes the tortoise to run
run d yards. Solve d = 100 +
526
Section 9.2
d
yards, Achilles will
10
d
1000
1
.d =
= 111 yards
10
9
9
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.
36. a.
Say Trot and Tom start from the left, Joel
from the right. Trot and Joel run towards
each other at 30 mph. Since they are 60
miles apart they will meet in 2 hours. Trot
will have run 40 miles and Tom will have
run 20 miles, so they will be 20 miles apart.
Trot and Tom will now be approaching each
other at 30 mph, so they will meet after
2/3 hour. Trot will have run another
40/3 miles and will be 80/3 miles from the
left. Joel will have run another 20/3 miles
and will be at 100/3 miles from the left, so
they will be 20/3 miles apart. They will meet
after 2/9 hour, during which Trot will have
run 40/9 miles, etc. So Trot runs
40 +
40 40
+
+
3
9
=
∞
⎛1⎞
40 ⎜ ⎟
k =1 ⎝ 3 ⎠
k –1
=
40
1 − 13
39. Let X = number of rolls needed to get first 6 For
X to equal n , two things must occur:
1. Mary must get a non-6 (probability =
2. Mary rolls a 6 (probability =
⎛5⎞
Pr( X = n) = ⎜ ⎟
⎝6⎠
) on the nth
n −1
⎛1⎞
⎜ ⎟ and
⎝6⎠
1
n
−
1
∞
⎛5⎞ ⎛1⎞
6 =1
=
⎜ ⎟ ⎜ ⎟
6
6
1
−
⎝ ⎠ ( 56 )
n =1 ⎝ ⎠
∞
40. EV ( X ) =
n ⋅ Pr ( X = n ) =
n =1
∞
1 ⎛5⎞
n⋅ ⋅⎜ ⎟
6 ⎝6⎠
n =1
n −1
1⎛ 1 ⎞ ∞
1 ⎛ 6 ⎞⎛ 5/ 6
= ⎜ ⎟ n ⋅ pn = ⎜ ⎟ ⎜
6 ⎝ p ⎠ n =1
6 ⎝ 5 ⎠ ⎜ (1 − 5 / 6 )2
⎝
1 ⎛ 6 ⎞ ⎛ 5 ⎞ ⎛ 36 ⎞
= ⎜ ⎟⎜ ⎟⎜ ⎟ = 6
6 ⎝ 5 ⎠⎝ 6 ⎠⎝ 1 ⎠
37. Note that:
⎞
⎟
⎟
⎠
∞
1. If we let tn be the probability that Peter wins
on his nth flip, then the total probability that
Peter wins is T =
n −1
and T =
∞
⎛ 1 ⎞⎛ 4 ⎞
⎜ ⎟⎜ ⎟
n =1 ⎝ 3 ⎠ ⎝ 9 ⎠
⎛ 13 ⎞ 1 9 3
⎜
⎟= ⋅ =
⎜ (1 − 4 ) ⎟ 3 5 5
9
⎝
⎠
In this case (see problem 37),
∞
n −1
2
p ⎡(1 − p ) ⎤
⎣
⎦
n =1
p
(2 p − p )
2
=
converges, and c ≠ 0. Then
∞
3. The probability that Peter wins on his nth flip
requires that (i) he gets a head on the nth
flip, and (ii) neither he nor Paul gets a head
on their previous n-1 flips. Thus:
2
tn = p ⎡(1 − p ) ⎤
⎣
⎦
k =1
tn
2. The probability that neither man wins in their
k
k
⎛2 2⎞
⎛4 ⎞
first k flips is ⎜ ⋅ ⎟ = ⎜ ⎟ .
⎝3 3⎠
⎝9 ⎠
⎛ 1 ⎞⎛ 4 ⎞
tn = ⎜ ⎟ ⎜ ⎟
⎝ 3 ⎠⎝ 9 ⎠
41. (Proof by contradiction) Assume
∞
n =1
=
1
6
roll. Thus,
b. Tom and Joel are approaching each other at
20 mph. They are 60 miles apart, so they will
meet in 3 hours. Trot is running at 20 mph
during that entire time,