CHAPTER 1 FUNCTIONS
1.1
FUNCTIONS AND THEIR GRAPHS
1. domain (, ); range [1, )
2. domain [0, ); range (, 1]
3. domain [2, ); y in range and y 5 x 10 0 y can be any nonnegative real number range [0, ).
4. domain (, 0] [3, ); y in range and y x 2 3 x 0 y can be any nonnegative real number
range [0, ).
5. domain (, 3) (3, ); y in range and y 3 4 t , now if t 3 3 t 0 3 4 t 0, or if t 3
3 t 0 3 4 t 0 y can be any nonzero real number range (, 0) (0, ).
6. domain (, 4) ( 4, 4) (4, ); y in range and y 2 2
2 2
4 t 4 16 t 2 16 0 16
2
t 16
, now if t 4 t 2 16 0 2 2
t 16
2
t 16
, or if t 4 t 16 0 2 2
t 16
0, or if
0 y can be any nonzero
real number range (, 18 ] (0, ).
7. (a) Not the graph of a function of x since it fails the vertical line test.
(b) Is the graph of a function of x since any vertical line intersects the graph at most once.
8. (a) Not the graph of a function of x since it fails the vertical line test.
(b) Not the graph of a function of x since it fails the vertical line test.
9. base x; (height)2
2x x2 height 23 x; area is a( x) 12 (base)(height) 12 ( x) 23 x 43 x2 ;
2
perimeter is p ( x) x x x 3 x.
10. s side length s 2 s 2 d 2 s d ; and area is a s 2 a 12 d 2
2
11. Let D diagonal length of a face of the cube and the length of an edge. Then 2 D 2 d 2 and
3
x
2
2
D 2 2 2 3 2 d 2 d . The surface area is 6 2 6 d3 2d 2 and the volume is 3 d3
3/2
d3 .
3 3
12. The coordinates of P are x, x so the slope of the line joining P to the origin is m xx 1 ( x 0).
Thus, x, x
1 , 1 .
m2 m
25
13. 2 x 4 y 5 y 12 x 54 ; L ( x 0)2 ( y 0)2 x 2 ( 12 x 54 )2 x 2 14 x 2 54 x 16
5 x 2 5 x 25
4
4
16
20 x 2 20 x 25
16
20 x 2 20 x 25
4
14. y x 3 y 2 3 x; L ( x 4) 2 ( y 0) 2 ( y 2 3 4)2 y 2 ( y 2 1)2 y 2
y4 2 y2 1 y2
y4 y2 1
Copyright 2016 Pearson Education, Ltd.
1
2
Chapter 1 Functions
15. The domain is (, ).
16. The domain is (, ).
17. The domain is (, ).
18. The domain is (, 0].
19. The domain is (, 0) (0, ).
20. The domain is (, 0) (0, ).
21. The domain is (, 5) (5, 3] [3, 5) (5, ) 22. The range is [2, 3).
23. Neither graph passes the vertical line test
(a)
(b)
Copyright 2016 Pearson Education, Ltd.
Section 1.1 Functions and Their Graphs
24. Neither graph passes the vertical line test
(a)
(b)
x y 1
y 1 x
x y 1
or
or
x y 1
y 1 x
25.
x 0 1 2
y 0 1 0
26.
4 x 2 , x 1
27. F ( x)
2
x 2 x, x 1
x 0 1 2
y 1 0 0
1 , x 0
28. G ( x) x
x, 0 x
29. (a) Line through (0, 0) and (1, 1): y x; Line through (1, 1) and (2, 0): y x 2
x, 0 x 1
f ( x)
x 2, 1 x 2
2,
0,
(b) f ( x)
2,
0,
0 x 1
1 x 2
2 x3
3 x 4
30. (a) Line through (0, 2) and (2, 0): y x 2
0 1
Line through (2, 1) and (5, 0): m 5 2 31 13 , so y 13 ( x 2) 1 13 x 53
x 2, 0 x 2
f ( x) 1
5
3 x 3 , 2 x 5
Copyright 2016 Pearson Education, Ltd.
3
4
Chapter 1 Functions
3 0
(b) Line through (1, 0) and (0, 3): m 0 ( 1) 3, so y 3x 3
1 3
Line through (0, 3) and (2, 1) : m 2 0 24 2, so y 2 x 3
3x 3, 1 x 0
f ( x)
2 x 3, 0 x 2
31. (a) Line through (1, 1) and (0, 0): y x
Line through (0, 1) and (1, 1): y 1
0 1
Line through (1, 1) and (3, 0): m 3 1 21 12 , so y 12 ( x 1) 1 12 x 32
x
1 x 0
f ( x) 1
0 x 1
1
3
1 x 3
2 x 2
(b) Line through (2, 1) and (0, 0): y 12 x
Line through (0, 2) and (1, 0): y 2 x 2
Line through (1, 1) and (3, 1): y 1
10
1x
2 x 0
2
f ( x) 2 x 2 0 x 1
1
1 x 3
32. (a) Line through T2 , 0 and (T, 1): m T (T /2) T2 , so y T2 x T2 0 T2 x 1
0, 0 x T2
f ( x)
2
T
T x 1, 2 x T
A, 0 x T
2
A, T x T
2
(b) f ( x)
3T
A, T x 2
A, 32T x 2T
33. (a) x 0 for x [0, 1)
(b) x 0 for x (1, 0]
34. x x only when x is an integer.
35. For any real number x, n x n 1, where n is an integer. Now: n x n 1 (n 1) x n.
By definition: x n and x n x n. So x x for all real x.
36. To find f(x) you delete the decimal or
fractional portion of x, leaving only
the integer part.
Copyright 2016 Pearson Education, Ltd.
Section 1.1 Functions and Their Graphs
37. Symmetric about the origin
Dec: x
Inc: nowhere
38. Symmetric about the y-axis
Dec: x 0
Inc: 0 x
39. Symmetric about the origin
Dec: nowhere
Inc: x 0
0 x
40. Symmetric about the y-axis
Dec: 0 x
Inc: x 0
41. Symmetric about the y-axis
Dec: x 0
Inc: 0 x
42. No symmetry
Dec: x 0
Inc: nowhere
Copyright 2016 Pearson Education, Ltd.
5
6
Chapter 1 Functions
43. Symmetric about the origin
Dec: nowhere
Inc: x
44. No symmetry
Dec: 0 x
Inc: nowhere
45. No symmetry
Dec: 0 x
Inc: nowhere
46. Symmetric about the y-axis
Dec: x 0
Inc: 0 x
47. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the
origin, the function is even.
48. f ( x) x 5 15 and f ( x) ( x) 5
x
f ( x). Thus the function is odd.
1 1
( x )5
x5
49. Since f ( x) x 2 1 ( x) 2 1 f ( x). The function is even.
50. Since [ f ( x) x 2 x] [ f ( x) ( x) 2 x] and [ f ( x) x 2 x ] [ f ( x) ( x) 2 x] the function is neither
even nor odd.
51. Since g ( x) x3 x, g ( x) x3 x ( x3 x) g ( x). So the function is odd.
52. g ( x) x 4 3 x 2 1 ( x) 4 3( x) 2 1 g ( x), thus the function is even.
53. g ( x)
1
1
g ( x). Thus the function is even.
x 2 1 ( x )2 1
54. g ( x)
x ; g ( x ) x g ( x ). So the function is odd.
x2 1
x2 1
55. h(t ) t 1 1 ; h(t ) t 1 1 ; h(t ) 1 1 t . Since h(t ) h(t ) and h(t ) h(t ), the function is neither even nor odd.
Copyright 2016 Pearson Education, Ltd.
Section 1.1 Functions and Their Graphs
56. Since |t 3 | |(t )3 |, h(t ) h(t ) and the function is even.
57. h(t ) 2t 1, h(t ) 2t 1. So h(t ) h(t ). h(t ) 2t 1, so h(t ) h(t ). The function is neither even
nor odd.
58. h(t ) 2| t | 1 and h(t ) 2| t | 1 2| t | 1. So h(t ) h(t ) and the function is even.
59. s kt 25 k (75) k 13 s 13 t ; 60 13 t t 180
60. K c v 2 12960 c(18)2 c 40 K 40v 2 ; K 40(10) 2 4000 joules
61. r ks 6 k4 k 24 r 24
; 10 24
s 12
s
s
5
k k 14700 P 14700 ; 23.4 14700 V 24500 628.2 cm3
62. P Vk 14.7 1000
V
V
39
63. V f ( x ) x (14 2 x )(22 2 x ) 4 x 3 72 x 2 308 x; 0 x 7.
AB 22 AB 2. So,
64. (a) Let h height of the triangle. Since the triangle is isosceles, AB
2
2
2
h 2 12 2 h 1 B is at (0, 1) slope of AB 1 The equation of AB is
y f ( x) x 1; x [0, 1].
(b) A( x) 2 xy 2 x( x 1) 2 x 2 2 x; x [0, 1].
65. (a) Graph h because it is an even function and rises less rapidly than does Graph g.
(b) Graph f because it is an odd function.
(c) Graph g because it is an even function and rises more rapidly than does Graph h.
66. (a) Graph f because it is linear.
(b) Graph g because it contains (0, 1).
(c) Graph h because it is a nonlinear odd function.
67. (a) From the graph, 2x 1 4x x (2, 0) (4, )
(b) 2x 1 4x 2x 1 4x 0
x2 2 x 8
( x 4)( x 2)
x 0: 2x 1 4x 0
0
0
2x
2x
x 4 since x is positive;
x2 2 x 8
x 0: 2x 1 4x 0
0
2x
x 2 since x is negative;
sign of ( x 4)( x 2)
( x 4)( x 2)
0
2x
Solution interval: (2, 0) (4, )
Copyright 2016 Pearson Education, Ltd.
7
8
Chapter 1 Functions
68. (a) From the graph, x 3 1 x 2 1 x (, 5) ( 1, 1)
3( x 1)
(b) Case x 1: x 3 1 x 2 1 x 1 2
3 x 3 2 x 2 x 5.
Thus, x (, 5) solves the inequality.
3( x 1)
Case 1 x 1: x 3 1 x 2 1 x 1 2
3 x 3 2 x 2 x 5 which
is true if x 1. Thus, x (1, 1)
solves the inequality.
Case 1 x : x 3 1 x 2 1 3 x 3 2 x 2 x 5
which is never true if 1 x,
so no solution here.
In conclusion, x (, 5) (1, 1).
69. A curve symmetric about the x-axis will not pass the vertical line test because the points (x, y) and ( x, y ) lie
on the same vertical line. The graph of the function y f ( x) 0 is the x-axis, a horizontal line for which there
is a single y-value, 0, for any x.
70. price 40 5 x, quantity 300 25x R( x) (40 5 x)(300 25 x)
71. x 2 x 2 h 2 x h
2
2h
; cost 5(2 x) 10h C (h) 10
2
72. (a) Note that 2 km 2, 000 m, so there are
10h 5h 2 2
2h
2
2502 x 2 meters of river cable at $180 per meter and
(2, 000 x) meters of land cable at $100 per meter. The cost is C ( x) 180 2502 x 2 100(2,000 - x).
(b) C (0) $245, 000
C (100) $238,466
C (200) $237,628
C (300) $240,292
C (400) $244,906
C (500) $250,623
C (600) $257,000
Values beyond this are all larger. It would appear that the least expensive location is less than 300 m from
the point P.
1.2
COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS
1. D f : x , Dg : x 1 D f g D fg : x 1. R f : y , Rg : y 0, R f g : y 1, R fg : y 0
2. D f : x 1 0 x 1, Dg : x 1 0 x 1. Therefore D f g D fg : x 1.
R f Rg : y 0, R f g : y 2, R fg : y 0
3. D f : x , Dg : x , D f /g : x , Dg /f : x , R f : y 2, Rg : y 1, R f /g : 0 y 2,
Rg /f : 12 y
Copyright 2016 Pearson Education, Ltd.
Section 1.2 Combining Functions; Shifting and Scaling Graphs
4. D f : x , Dg : x 0, D f /g : x 0, Dg /f : x 0; R f : y 1, Rg : y 1, R f /g : 0 y 1, Rg /f : 1 y
5. (a) 2
(d) ( x 5)2 3 x 2 10 x 22
(g) x 10
(b) 22
(e) 5
(h) ( x 2 3)2 3 x 4 6 x 2 6
(c) x 2 2
(f ) 2
6. (a) 13
(b) 2
(c)
(d)
(e) 0
1
x
(g) x 2
(h)
(f )
1
1
x 1
x 1
x2 x 2
1
1
1 1 x
x 1
x 1
3
4
x 1
7. ( f g h)( x) f ( g (h( x))) f ( g (4 x)) f (3(4 x)) f (12 3 x) (12 3 x) 1 13 3x
8. ( f g h)( x) f ( g (h( x))) f ( g ( x 2 )) f (2( x 2 ) 1) f (2 x 2 1) 3(2 x 2 1) 4 6 x 2 1
f 1 4 f 1 x4 x 1 x4 x 1 15x 4 x1
9. ( f g h)( x) f ( g (h( x))) f g 1x
1
x
2 x 2
f
f
2 x 2 1
2x
3 x
2x
2
3 x
2x
3 3x
8 3x
10. ( f g h)( x) f ( g (h( x))) f g
2 x
11. (a) ( f g )( x)
(d) ( j j )( x)
(b) ( j g )( x)
(e) ( g h f )( x)
(c) ( g g )( x)
(f ) (h j f )( x)
12. (a) ( f j )( x)
(d) ( f f )( x)
(b) ( g h)( x)
(e) ( j g f )( x)
(c) (hh)( x)
(f ) ( g f h)( x)
g(x)
f (x)
( f g )( x )
(a) x 7
x
x7
(b) x 2
3x
13.
(c) x 2
7 2x
3( x 2) 3 x 6
x5
x2 5
(d)
x
x 1
x
x 1
x
x 1
x
1
x 1
(e)
1
x 1
1 1x
x
(f )
1
x
1
x
x
x (xx 1) x
1 .
x 1
g ( x) 1
( f g )( x) g ( x ) x x 1 1 g (1x ) x x 1 1 x x 1 g (1x ) x 1 1 g (1x ) , so g ( x ) x 1.
2
14. (a) ( f g )( x) |g ( x)|
(b)
(c) Since ( f g )( x) g ( x) | x |, g ( x) x .
(d) Since ( f g )( x) f x | x |, f ( x) x 2 . (Note that the domain of the composite is [0, ).)
Copyright 2016 Pearson Education, Ltd.
9
10
Chapter 1 Functions
The completed table is shown. Note that the absolute value sign in part (d) is optional.
g(x)
f(x)
( f g )(x)
1
x 1
| x|
1
x 1
x 1
x 1
x
x
x 1
x2
x
| x|
x
x2
| x|
15. (a) f ( g (1)) f (1) 1
(d) g ( g (2)) g (0) 0
16. (a)
(b)
(c)
(d)
(e)
(b) g ( f (0)) g (2) 2
(e) g ( f (2)) g (1) 1
(c) f ( f (1)) f (0) 2
(f) f ( g (1)) f (1) 0
f ( g (0)) f (1) 2 (1) 3, where g (0) 0 1 1
g ( f (3)) g (1) (1) 1, where f (3) 2 3 1
g ( g (1)) g (1) 1 1 0, where g (1) (1) 1
f ( f (2)) f (0) 2 0 2, where f (2) 2 2 0
g ( f (0)) g (2) 2 1 1, where f (0) 2 0 2
f 12 2 12 52 , where g 12 12 1 12
(f ) f g 12
17. (a) ( f g )( x) f ( g ( x))
( g f )( x) g ( f ( x))
1 1
x
1
x 1
1 x
x
(b) Domain ( f g ): (, 1] (0, ), domain ( g f ): (1, )
(c) Range ( f g ): (1, ), range ( g f ): (0, )
18. (a) ( f g )( x) f ( g ( x)) 1 2 x x
( g f )( x) g ( f ( x)) 1 | x |
(b) Domain ( f g ): [0, ), domain ( g f ): (, )
(c) Range ( f g ): (0, ), range ( g f ): (, 1]
g ( x)
19. ( f g )( x) x f ( g ( x)) x g ( x ) 2 x g ( x) ( g ( x) 2) x x g ( x) 2 x
g ( x) x g ( x) 2 x g ( x) 1 2xx x2x 1
20. ( f g )( x ) x 2 f ( g ( x)) x 2 2( g ( x))3 4 x 2 ( g ( x ))3
21. (a) y ( x 7) 2
(b) y ( x 4)2
22. (a) y x 2 3
(b) y x 2 5
x6
x6
g ( x) 3 2
2
23. (a) Position 4
(b) Position 1
(c) Position 2
(d) Position 3
24. (a) y ( x 1)2 4
(b) y ( x 2) 2 3
(c) y ( x 4) 2 1
(d) y ( x 2)2
Copyright 2016 Pearson Education, Ltd.
Section 1.2 Combining Functions; Shifting and Scaling Graphs
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
Copyright 2016 Pearson Education, Ltd.
11
12
Chapter 1 Functions
35.
36.
37.
38.
39.
40.
41.
42.
43.
44.
45.
46.
Copyright 2016 Pearson Education, Ltd.
Section 1.2 Combining Functions; Shifting and Scaling Graphs
47.
48.
49.
50.
51.
52.
53.
54.
55. (a) domain: [0, 2]; range: [2, 3]
(b) domain: [0, 2]; range: [–1, 0]
Copyright 2016 Pearson Education, Ltd.
13
14
Chapter 1 Functions
(c) domain: [0, 2]; range: [0, 2]
(d) domain: [0, 2]; range: [–1, 0]
(e) domain: [–2, 0]; range: [0, 1]
(f ) domain: [1, 3]; range: [0,1]
(g) domain: [–2, 0]; range: [0, 1]
(h) domain: [–1, 1]; range: [0, 1]
56. (a) domain: [0, 4]; range: [–3, 0]
(b) domain: [–4, 0]; range: [0, 3]
(c) domain: [–4, 0]; range: [0, 3]
(d) domain: [–4, 0]; range: [1, 4]
Copyright 2016 Pearson Education, Ltd.
Section 1.2 Combining Functions; Shifting and Scaling Graphs
(e) domain: [2, 4]; range: [–3, 0]
(f ) domain: [–2, 2]; range: [–3, 0]
(g) domain: [1, 5]; range: [–3, 0]
(h) domain: [0, 4]; range: [0, 3]
57. y 3 x 2 3
58. y (2 x) 2 1 4 x 2 1
1 1 9
( x /3) 2
x2
59. y 12 1 12 12 1 2
60. y 1
61. y 4 x 1
62. y 3 x 1
x
2x
12 16 x2
63. y 4 2x
64. y 13 4 x 2
2
1 x8
65. y 1 (3 x )3 1 27 x3
66. y 1 2x
3
67. Let y 2 x 1 f ( x) and let g ( x) x1/2 ,
, i( x) 2 x 12 , and
1/2
j ( x) 2 x 12 f ( x). The graph of h( x)
h( x) x 12
1/2
1/2
is the graph of g ( x) shifted left 12 unit; the graph
of i ( x) is the graph of h( x) stretched vertically by
a factor of 2; and the graph of j ( x) f ( x) is the
graph of i ( x) reflected across the x-axis.
68. Let y 1 2x f ( x). Let g ( x) ( x)1/2 ,
h( x) ( x 2)1/2 , and i ( x ) 1 ( x 2)1/2
2
1 2x f ( x ). The graph of g ( x) is the graph
of y x reflected across the x-axis. The graph
of h( x) is the graph of g ( x) shifted right two units.
And the graph of i ( x) is the graph of h( x)
compressed vertically by a factor of 2 .
Copyright 2016 Pearson Education, Ltd.
3
15
16
Chapter 1 Functions
69. y f ( x) x3 . Shift f ( x) one unit right followed by
a shift two units up to get g ( x) ( x 1)3 2 .
70.
y (1 x)3 2 [( x 1)3 (2)] f ( x).
Let g ( x) x3 , h( x) ( x 1)3 ,
i( x) ( x 1)3 (2),
and j ( x) [( x 1)3 (2)]. The graph of h( x) is the
graph of g ( x) shifted right one unit; the graph of i ( x)
is the graph of h( x) shifted down two units; and the
graph of f ( x) is the graph of i ( x) reflected across
the x-axis.
71. Compress the graph of f ( x) 1x horizontally by a
factor of 2 to get g ( x) 21x . Then shift g ( x)
vertically down 1 unit to get h( x) 21x 1.
72. Let f ( x) 12 and g ( x) 22 1
x
x
1
x/ 2
2
1
1 1
x2
2
1
1. Since
2
1/ 2 x
2 1.4, we see
that the graph of f ( x ) stretched horizontally by
a factor of 1.4 and shifted up 1 unit is the graph
of g ( x).
73. Reflect the graph of y f ( x) 3 x across the x-axis
to get g ( x ) 3 x .
Copyright 2016 Pearson Education, Ltd.
Section 1.2 Combining Functions; Shifting and Scaling Graphs
74.
y f ( x) (2 x) 2/3 [( 1)(2) x]2/3 ( 1) 2/3 (2 x) 2/3
(2 x)2/3 . So the graph of f ( x) is the graph of
g ( x) x 2/3 compressed horizontally by a factor of 2.
75.
76.
77. (a) ( fg )( x) f ( x) g ( x) f ( x)( g ( x)) ( fg )( x), odd
( x)
(c) ( x)
(b)
( x), odd
( x), odd
f
g
f ( x)
f ( x)
f
g ( x) g
g ( x)
g
f
g ( x)
g ( x)
f ( x)
f ( x)
g
f
(d) f 2 ( x) f ( x) f ( x) f ( x) f ( x) f 2 ( x), even
(e) g 2 ( x) ( g ( x)) 2 ( g ( x))2 g 2 ( x ), even
(f ) ( f g )( x) f ( g ( x)) f ( g ( x)) f ( g ( x)) ( f g )( x), even
(g) ( g f )( x) g ( f ( x)) g ( f ( x)) ( g f )( x), even
(h) ( f f )( x) f ( f ( x)) f ( f ( x)) ( f f )( x), even
(i) ( g g )( x) g ( g ( x )) g ( g ( x)) g ( g ( x )) ( g g )( x ), odd
78. Yes, f ( x) 0 is both even and odd since f ( x) 0 f ( x) and f ( x) 0 f ( x).
79. (a)
(b)
Copyright 2016 Pearson Education, Ltd.
17
18
Chapter 1 Functions
(c)
(d)
80.
1.3
TRIGONOMETRIC FUNCTIONS
1. (a)
s r (10) 45 8 m
(b)
55 m
s r (10)(110) 180
110
18
9
225
2. rs 108 54 radians and 54 180
3. 80 80 180
49 s (6) 49 8.4 cm. (since the diameter 12 cm. radius 6 cm.)
4. d 1 meter r 50 cm rs 30
0.6 rad or 0.6
50
23
0
sin
0
23
0
1
cos
1
0
12
1
0
3
0
und.
5.
tan
cot
und.
sec
1
csc
und.
1
3
2
2
3
2
und.
0
1
und.
und.
1
3
4
1
2
1
2
1
1
2
2
180 34
6.
32
3
6
4
5
6
sin
1
23
12
1
2
1
2
cos
0
1
2
3
2
1
2
23
tan
und.
3
cot
0
sec
csc
1
3
1
3
1
3
1
2
2
und.
2
2
3
1
2
3
2
Copyright 2016 Pearson Education, Ltd.
1
3
3
2
3
2
Section 1.3 Trigonometric Functions
7. cos x 54 , tan x 34
8. sin x 2 , cos x 1
9. sin x 38 , tan x 8
, tan x 12
10. sin x 12
13
5
11. sin x 1 , cos x 2
12. cos x 23 , tan x 1
5
5
5
3
14.
13.
period 4
period
16.
15.
period 4
period 2
18.
17.
period 1
period 6
20.
19.
period 2
5
period 2
Copyright 2016 Pearson Education, Ltd.
19
20
Chapter 1 Functions
21.
22.
period 2
period 2
23. period 2 , symmetric about the origin
24. period 1, symmetric about the origin
s
3
2
s = tan t
1
2
1
1
0
2
t
1
2
3
26. period 4 , symmetric about the origin
25. period 4, symmetric about the s-axis
27. (a) Cos x and sec x are positive for x in the interval
2 , 2 ; and cos x and sec x are negative for x
in the intervals 32 , 2 and 2 , 32 . Sec x is
undefined when cos x is 0. The range of sec x is
(, 1] [1,); the range of cos x is [1, 1].
Copyright 2016 Pearson Education, Ltd.
Section 1.3 Trigonometric Functions
21
(b) Sin x and csc x are positive for x in the intervals
32 , and (0, ); and sin x and csc x are
negative for x in the intervals ( , 0) and
, 32 . Csc x is undefined when sin x is 0. The
range of csc x is (, 1] [1, ); the range of
sin x is [1, 1].
28. Since cot x tan1 x , cot x is undefined when tan x 0
and is zero when tan x is undefined. As tan x
approaches zero through positive values, cot x
approaches infinity. Also, cot x approaches negative
infinity as tan x approaches zero through negative
values.
29. D : x ; R : y 1, 0, 1
30. D : x ; R : y 1, 0, 1
31. cos x 2 cos x cos 2 sin x sin 2 (cos x)(0) (sin x)(1) sin x
32. cos x 2 cos x cos 2 sin x sin 2 (cos x)(0) (sin x)(1) sin x
33. sin x 2 sin x cos 2 cos x sin 2 (sin x)(0) (cos x)(1) cos x
34. sin x 2 sin x cos 2 cos x sin 2 (sin x)(0) (cos x)( 1) cos x
35. cos( A B ) cos( A ( B)) cos A cos( B ) sin A sin( B) cos A cos B sin A( sin B)
cos A cos B sin A sin B
36. sin( A B) sin( A ( B)) sin A cos( B ) cos A sin( B ) sin A cos B cos A( sin B )
sin A cos B cos A sin B
37. If B A, A B 0 cos( A B) cos 0 1. Also cos( A B ) cos( A A) cos A cos A sin A sin A
cos 2 A sin 2 A. Therefore, cos 2 A sin 2 A 1.
38. If B 2 , then cos( A 2 ) cos A cos 2 sin A sin 2 (cos A)(1) (sin A)(0) cos A and
sin( A 2 ) sin A cos 2 cos A sin 2 (sin A)(1) (cos A)(0) sin A . The result agrees with the fact that the
cosine and sine functions have period 2 .
39. cos( x ) cos cos x sin sin x ( 1)(cos x ) (0)(sin x) cos x
Copyright 2016 Pearson Education, Ltd.
22
Chapter 1 Functions
40. sin(2 x) sin 2 cos( x) cos(2 ) sin( x) (0)(cos( x)) (1)(sin( x)) sin x
41. sin 32 x sin 32 cos( x) cos 32 sin( x) ( 1)(cos x ) (0)(sin( x )) cos x
42. cos 32 x cos 32 cos x sin 32 sin x (0)(cos x ) (1)(sin x) sin x
43. sin 712 sin 4 3 sin 4 cos 3 cos 4 sin 3
2
2
cos 2 cos cos 2 sin sin 2
44. cos 11
12
4
3
4
3
4
3
2
2
1
2
6
4
3
2
2
2
2
1
2
3
2
2
2
2 6
4
12 22 23 22 12 23
cos cos cos sin sin
45. cos 12
3 4
3
4
3
4
23 22 12 22 12 23
46. sin 512 sin 23 4 sin 23 cos 4 cos 23 sin 4
47.
1 cos
cos 2 8
2
49. sin 2 12
28 1 22 2 2
48.
1 23 2 3
50. sin 2 38
2
4
1 cos 212
2
2
4
1 cos
cos 2 512
2
1012 1 23 2 3
2
4
1 22 2 2
1 cos 68
2
2
4
51. sin 2 34 sin 23 3 , 23 , 43 , 53
2
2
cos
cos
52. sin 2 cos 2 sin 2 cos 2 tan 2 1 tan 1 4 , 34 , 54 , 74
53. sin 2 cos 0 2sin cos cos 0 cos (2sin 1) 0 cos 0 or 2sin 1 0
cos 0 or sin 12 2 , 32 , or 6 , 56 6 , 2 , 56 , 32
54. cos 2 cos 0 2 cos 2 1 cos 0 2 cos 2 cos 1 0 (cos 1)(2 cos 1) 0
cos 1 0 or 2 cos 1 0 cos 1 or cos 12 or 3 , 53 3 , , 53
sin( A B )
sin A cos B cos A cos B
sin A cos B
A cos B
55. tan( A B ) cos( A B ) cos A cos B sin A sin B cos
cos A cos B
cos A sin B
cos
A cos B
sin A sin B
cos
cos A cos B
A cos B
sin( A B )
sin A cos B cos A cos B
sin A cos B
cos A cos B
56. tan( A B ) cos( A B ) cos A cos B sin A sin B cos
A cos B
cos A sin B
cos
A cos B
sin A sin B
cos
cos A cos B
A cos B
tan A tan B
1 tan A tan B
tan A tan B
1 tan A tan B
57. According to the figure in the text, we have the following: By the law of cosines, c 2 a 2 b 2 2ab cos
12 12 2 cos( A B ) 2 2cos( A B) . By distance formula, c 2 (cos A cos B )2 (sin A sin B )2
cos 2 A 2 cos A cos B cos 2 B sin 2 A 2sin A sin B sin 2 B 2 2(cos A cos B sin A sin B ) . Thus
c 2 2 2 cos( A B ) 2 2(cos A cos B sin A sin B ) cos( A B ) cos A cos B sin A sin B .
Copyright 2016 Pearson Education, Ltd.
Section 1.3 Trigonometric Functions
58. (a) cos( A B ) cos A cos B sin A sin B
sin cos 2 and cos sin 2
Let A B
sin( A B ) cos 2 ( A B) cos 2 A B cos 2 A cos B sin 2 A sin B
sin A cos B cos A sin B
(b) cos( A B) cos A cos B sin A sin B
cos( A ( B )) cos A cos( B ) sin A sin( B )
cos( A B) cos A cos( B ) sin A sin( B) cos A cos B sin A( sin B) cos A cos B sin A sin B
Because the cosine function is even and the sine functions is odd.
59. c 2 a 2 b 2 2ab cos C 22 32 2(2)(3) cos(60) 4 9 12 cos(60) 13 12 12 7.
Thus, c 7 2.65.
60. c 2 a 2 b 2 2ab cos C 22 32 2(2)(3) cos(40) 13 12 cos(40). Thus, c 13 12 cos 40° 1.951.
61. From the figures in the text, we see that sin B hc . If C is an acute angle, then sin C bh . On the other hand,
if C is obtuse (as in the figure on the right in the text), then sin C sin( C ) bh . Thus, in either case,
h b sin C c sin B ah ab sin C ac sin B.
By the law of cosines, cos C
a 2 b2 c2
a 2 c 2 b2
and cos B
. Moreover, since the sum of the interior
2 ab
2 ac
angles of triangle is , we have sin A sin( ( B C )) sin( B C ) sin B cos C cos B sin C
h (2a 2 b 2 c 2 c 2 b 2 ) ah ah bc sin A.
hc a 2bab c a 2cac b bh 2abc
bc
2
2
2
2
2
2
Combining our results we have ah ab sin C, ah ac sin B, and ah bc sin A. Dividing by abc gives
h sin A sin C sin B .
bc
a
c
b
law of sines
62. By the law of sines, sin2 A sin3 B
3/2
. By Exercise 59 we know that c
c
7. Thus sin B 3 3 0.982.
2 7
63. From the figure at the right and the law of cosines,
b 2 a 2 22 2(2a) cos B
a 2 4 4a 12 a 2 2a 4.
Applying the law of sines to the figure,
3/2
b 32 a. Thus, combining results,
b
a 2 2a 4 b 2 32 a 2 0 12 a 2 2a 4 0 a 2 4a 8 . From the quadratic formula and the fact that
4 42 4(1)( 8)
4 34
a 0, we have a
2 1.464.
2
2/2
a
sin A sin B
b
a
64. (a) The graphs of y sin x and y x nearly coincide when x is near the origin (when the calculator is in
radians mode).
(b) In degree mode, when x is near zero degrees the sine of x is much closer to zero than x itself. The curves
look like intersecting straight lines near the origin when the calculator is in degree mode.
Copyright 2016 Pearson Education, Ltd.
23
24
Chapter 1 Functions
65. A 2, B 2 , C , D 1
66. A 12 , B 2, C 1, D 12
67. A 2 , B 4, C 0, D 1
68.
A 2L , B L, C 0, D 0
69–72.
Example CAS commands:
Maple:
f : x - A*sin((2*Pi/B)*(x-C))D1;
A:3; C: 0; D1: 0;
f_list : [seq(f(x), B[1,3,2*Pi,5*Pi])];
plot(f_list, x -4*Pi..4*Pi, scaling constrained,
color [red,blue,green,cyan], linestyle[1,3,4,7],
legend ["B1", "B3","B2*Pi","B3*Pi"],
title "#69 (Section 1.3)");
Mathematica:
Clear[a, b, c, d, f, x]
f[x_]: a Sin[2/b (x c)] d
Plot[f[x]/.{a 3, b 1, c 0, d 0}, {x, 4, 4 }]
Copyright 2016 Pearson Education, Ltd.
Section 1.3 Trigonometric Functions
69. (a) The graph stretches horizontally.
(b) The period remains the same: period | B |. The graph has a horizontal shift of 12 period.
70. (a) The graph is shifted right C units.
(b) The graph is shifted left C units.
(c) A shift of one period will produce no apparent shift. | C | 6
71. (a) The graph shifts upwards | D | units for D 0
(b) The graph shifts down | D | units for D 0.
72. (a) The graph stretches | A| units.
(b) For A 0, the graph is inverted.
Copyright 2016 Pearson Education, Ltd.
25
26
Chapter 1 Functions
1.4
GRAPHING WITH SOFTWARE
1–4.
The most appropriate viewing window displays the maxima, minima, intercepts, and end behavior of the
graphs and has little unused space.
1. d.
2. c.
3. d.
4. b.
5–30. For any display there are many appropriate display widows. The graphs given as answers in Exercises 5–30
are not unique in appearance.
5. [ 2, 5] by [ 15, 40]
6. [ 4, 4] by [ 4, 4]
7. [ 2, 6] by [ 250, 50]
8. [ 1, 5] by [ 5, 30]
Copyright 2016 Pearson Education, Ltd.
Section 1.4 Graphing with Software
9. [ 4, 4] by [ 5, 5]
10. [ 2, 2] by [ 2, 8]
11. [ 2, 6] by [ 5, 4]
12. [ 4, 4] by [ 8, 8]
13. [ 1, 6] by [ 1, 4]
14. [ 1, 6] by [ 1, 5]
15. [ 3, 3] by [0, 10]
16. [ 1, 2] by [0, 1]
Copyright 2016 Pearson Education, Ltd.
27
28
Chapter 1 Functions
17. [ 5, 1] by [ 5, 5]
18. [ 5, 1] by [ 2, 4]
19. [ 4, 4] by [0, 3]
20. [ 5, 5] by [ 2, 2]
21. [ 10, 10] by [ 6, 6]
22. [ 5, 5] by [ 2, 2]
23. [ 6, 10] by [ 6, 6]
24. [ 3, 5] by [ 2, 10]
25. [0.03, 0.03] by [1.25, 1.25]
26. [0.1, 0.1] by [3, 3]
Copyright 2016 Pearson Education, Ltd.
Section 1.4 Graphing with Software
27. [300, 300] by [1.25, 1.25]
28. [50, 50] by [0.1, 0.1]
29. [0.25, 0.25] by[0.3, 0.3]
30. [0.15, 0.15] by [0.02, 0.05]
31. x 2 2 x 4 4 y y 2 y 2 x 2 2 x 8.
The lower half is produced by graphing
y 2 x 2 2 x 8.
32. y 2 16 x 2 1 y 1 16 x 2 . The upper branch
is produced by graphing y 1 16 x 2 .
33.
34.
Copyright 2016 Pearson Education, Ltd.
29
30
Chapter 1 Functions
35.
36.
37.
38.
8
6
4
2
0
1970 1980 1990 2000 2010 2020
39.
40.
(in thousands)
300
26
225
22
R
18
T
150
14
75
10
6
1972 1980 1988 1996 2004 2012
41.
2000 2002 2004 2006 2008
42.
1
600
450
0.5
300
1955
1935
1975
1995
2015
150
0
0.5
0
2
Copyright 2016 Pearson Education, Ltd.
4
6
8
10
Chapter 1 Practice Exercises
31
CHAPTER 1 PRACTICE EXERCISES
1. The area is A r 2 and the circumference is C 2 r. Thus, r 2C A
2C C4 .
2
2
4S . The volume is V 43 r 3 r 3 34V . Substitution into the formula
2/3
for surface area gives S 4 r 2 4 34V .
1/2
2. The surface area is S 4 r 2 r
3. The coordinates of a point on the parabola are (x, x2). The angle of inclination joining this point to the origin
2
satisfies the equation tan xx x. Thus the point has coordinates ( x, x 2 ) (tan , tan 2 ).
h h 500 tan m .
4. tan rise
500
run
6.
5.
Symmetric about the origin.
Symmetric about the y-axis.
8.
7.
Neither
Symmetric about the y-axis.
9. y ( x) ( x )2 1 x 2 1 y ( x). Even.
10. y ( x) ( x)5 ( x)3 ( x) x5 x3 x y ( x). Odd.
11. y ( x) 1 cos( x) 1 cos x y ( x). Even.
12. y ( x) sec( x) tan( x)
13. y ( x)
sin x
cos 2 x
sin2 x sec x tan x y ( x). Odd.
cos x
x 4 1
x4 1
x4 1
3
3
y ( x). Odd.
3
x 2x
x 2 x x 2 x
14. y ( x) ( x) sin( x) ( x) sin x ( x sin x) y ( x). Odd.
Copyright 2016 Pearson Education, Ltd.
32
Chapter 1 Functions
15. y ( x) x cos( x) x cos x. Neither even nor odd.
16. y ( x) ( x) cos( x) x cos x y ( x). Odd.
17. Since f and g are odd f ( x) f ( x) and g ( x ) g ( x).
(a) ( f g )( x) f ( x) g ( x) [ f ( x)] [ g ( x)] f ( x) g ( x) ( f g )( x) f g is even.
(b) f 3 ( x ) f ( x) f ( x) f ( x) [ f ( x)] [ f x ] [ f ( x)] f ( x) f ( x) f ( x) f 3 ( x) f 3 is odd.
(c) f (sin( x)) f (sin( x)) f (sin( x )) f (sin( x)) is odd.
(d) g (sec( x)) g (sec( x)) g (sec( x)) is even.
(e) | g ( x)| | g ( x)| | g ( x) | | g | is even.
18. Let f (a x) f (a x) and define g ( x) f ( x a). Then g ( x) f (( x) a) f (a x) f (a x)
f ( x a ) g ( x) g ( x) f ( x a) is even.
19. (a) The function is defined for all values of x, so the domain is (, ).
(b) Since | x | attains all nonnegative values, the range is [2, ).
20. (a) Since the square root requires 1 x 0, the domain is (,1].
(b) Since 1 x attains all nonnegative values, the range is [2, ).
21. (a) Since the square root requires 16 x 2 0, the domain is [4, 4].
(b) For values of x in the domain, 0 16 x 2 16, so 0 16 x 2 4. The range is [0, 4].
22. (a) The function is defined for all values of x, so the domain is (, ).
(b) Since 32 x attains all positive values, the range is (1, ) .
23. (a) The function is defined for all values of x, so the domain is (, ).
(b) Since 2e x attains all positive values, the range is (3, ) .
24. (a) The function is equivalent to y tan 2 x, so we require 2 x k2 for odd integers k. The domain is given by
x k4 for odd integers k.
(b) Since the tangent function attains all values, the range is (, ).
25. (a) The function is defined for all values of x, so the domain is (, ).
(b) The sine function attains values from –1 to 1, so 2 2sin (3 x ) 2 and hence 3 2 sin (3x ) 1 1.
The range is [3, 1].
26. (a) The function is defined for all values of x, so the domain is (, ).
5
(b) The function is equivalent to y x 2 , which attains all nonnegative values. The range is [0, ) .
27. (a) The logarithm requires x 3 0, so the domain is (3, ).
(b) The logarithm attains all real values, so the range is (, ).
28. (a) The function is defined for all values of x, so the domain is (, ).
(b) The cube root attains all real values, so the range is (, ).
Copyright 2016 Pearson Education, Ltd.
Chapter 1 Practice Exercises
29. (a) Increasing because volume increases as radius increases.
(b) Neither, since the greatest integer function is composed of horizontal (constant) line segments.
(c) Decreasing because as the height increases, the atmospheric pressure decreases.
(d) Increasing because the kinetic (motion) energy increases as the particles velocity increases.
30. (a) Increasing on [2, )
(c) Increasing on (, )
(b) Increasing on [1, )
(d) Increasing on 12 ,
31. (a) The function is defined for 4 x 4, so the domain is [4, 4].
(b) The function is equivalent to y | x |, 4 x 4, which attains values from 0 to 2 for x in the domain.
The range is [0, 2].
32. (a) The function is defined for 2 x 2, so the domain is [2, 2].
(b) The range is [1, 1].
0 1
33. First piece: Line through (0, 1) and (1, 0). m 1 0 11 1 y x 1 1 x
0 1
Second piece: Line through (1, 1) and (2, 0). m 2 1 11 1 y ( x 1) 1 x 2 2 x
1 x, 0 x 1
f ( x)
2 x, 1 x 2
50
34. First piece: Line through (0, 0) and (2, 5). m 2 0 52 y 52 x
05
Second piece: Line through (2, 5) and (4, 0). m 4 2 25 52 y 52 ( x 2) 5 52 x 10 10 52x
5 x, 0 x 2
2
f ( x)
(Note: x 2 can be included on either piece.)
5x
10 2 , 2 x 4
35. (a) ( f g )(1) f ( g (1)) f 1 f (1) 11 1
1 2
(b) ( g f )(2) g ( f (2)) g 12 11 1 or 52
2 2.5
(c) ( f f )( x) f ( f ( x)) f 1x 1/1x x, x 0
2
(d) ( g g )( x) g ( g ( x)) g 1
x
2
1
1 2
x2
4x2
1 2 x 2
36. (a) ( f g )(1) f ( g (1)) f 3 1 1 f (0) 2 0 2
(b) ( g f )(2) f ( g (2)) g (2 2) g (0) 3 0 1 1
(c) ( f f )( x) f ( f ( x)) f (2 x) 2 (2 x) x
(d) ( g g )( x) g ( g ( x)) g 3 x 1 3 3 x 1 1
37. (a) ( f g )( x) f ( g ( x)) f
x 2 2 x 2 x, x 2.
2
( g f )( x) g ( f ( x)) g (2 x 2 )
(b) Domain of f g : [2, ).
Domain of g f : [2, 2].
2 x2 2 4 x2
(c) Range of f g : (, 2].
Range of g f : [0, 2].
Copyright 2016 Pearson Education, Ltd.
33
34
Chapter 1 Functions
1 x 1 x 4 1 x.
( g f )( x) g ( f ( x)) g x 1 x
38. (a) ( f g )( x) f ( g ( x)) f
(b) Domain of f g : (, 1].
Domain of g f : [0, 1].
39.
(c) Range of f g : [0, ).
Range of g f : [0, 1].
y ( f f )( x)
y f ( x)
40.
42.
41.
The graph of f 2 ( x) f1 (| x |) is the same as the
graph of f1 ( x ) to the right of the y-axis. The graph
of f 2 ( x) to the left of the y-axis is the reflection of
y f1 ( x), x 0 across the y-axis.
It does not change the graph.
Copyright 2016 Pearson Education, Ltd.
Chapter 1 Practice Exercises
43.
35
44.
Whenever g1 ( x) is positive, the graph of y
g 2 ( x) | g1 ( x)| is the same as the graph of y g1 ( x).
When g1 ( x) is negative, the graph of y g 2 ( x) is
the reflection of the graph of y g1 ( x) across the
x-axis.
Whenever g1 ( x) is positive, the graph of y
g 2 ( x) g1 ( x) is the same as the graph of y
g1 ( x). When g1 ( x) is negative, the graph of y
g 2 ( x) is the reflection of the graph of y g1 ( x)
across the x-axis.
46.
45.
The graph of f 2 ( x) f1 (| x |) is the same as the
graph of f1 ( x ) to the right of the y-axis. The graph
of f 2 ( x) to the left of the y-axis is the reflection of
y f1 ( x ), x 0 across the y-axis.
Whenever g1 ( x) is positive, the graph of
y g 2 ( x) | g1 ( x)| is the same as graph of
y g1 ( x ). When g1 ( x) is negative, the graph of
y g 2 ( x) is the reflection of the graph of
y g1 ( x) across the x-axis.
48.
47.
The graph of f 2 ( x) f1 (| x |) is the same as the
graph of f1 ( x) to the right of the y-axis. The graph
of f 2 ( x) to the left of the y-axis is the reflection of
y f1 ( x), x 0 across the y-axis.
49. (a) y g ( x 3) 12
(c) y g ( x)
(e) y 5 g ( x)
The graph of f 2 ( x) f1 (| x |) is the same as the
graph of f1 ( x) to the right of the y-axis. The graph
of f 2 ( x) to the left of the y-axis is the reflection of
y f1 ( x), x 0 across the y-axis.
(b) y g x 23 2
(d) y g ( x)
(f ) y g (5 x)
Copyright 2016 Pearson Education, Ltd.
36
Chapter 1 Functions
50. (a) Shift the graph of f right 5 units
(b) Horizontally compress the graph of f by a factor of 4
(c) Horizontally compress the graph of f by a factor of 3 and then reflect the graph about the y-axis
(d) Horizontally compress the graph of f by a factor of 2 and then shift the graph left 12 unit.
(e) Horizontally stretch the graph of f by a factor of 3 and then shift the graph down 4 units.
(f ) Vertically stretch the graph of f by a factor of 3, then reflect the graph about the x-axis, and finally shift the
graph up 14 unit.
51. Reflection of the graph of y x about the x-axis
followed by a horizontal compression by a factor of
1 then a shift left 2 units.
2
52. Reflect the graph of y x about the x-axis, followed
by a vertical compression of the graph by a factor
of 3, then shift the graph up 1 unit.
53. Vertical compression of the graph of y 12 by a
factor of 2, then shift the graph up 1 unit.
x
54. Reflect the graph of y x1/3about the y-axis, then
compress the graph horizontally by a factor of 5.
Copyright 2016 Pearson Education, Ltd.
Chapter 1 Practice Exercises
55.
56.
period 4
period
57.
58.
period 4
period 2
60.
59.
period 2
period 2
61. (a) sin B sin 3 bc b2 b 2sin 3 2
3. By the theorem of Pythagoras,
3
2
a 2 b 2 c 2 a c 2 b 2 4 3 1.
(b) sin B sin 3 bc 2c c
2
sin 3
2
3
2
4 . Thus, a c 2 b 2
3
(2)
4
3
2
62. (a) sin A ac a c sin A
(b) tan A ba a b tan A
63. (a) tan B ba a tanb B
(b) sin A ac c sina A
64. (a) sin A ac
(b) sin A ac
c 2 b2
c
Copyright 2016 Pearson Education, Ltd.
2
4 2 .
3
3
37
38
Chapter 1 Functions
65. Let h height of vertical pole, and let b and c denote
the distances of points B and C from the base of the
pole, measured along the flat ground, respectively.
Then, tan 50 hc , tan 35 bh , and b c 10.
Thus, h c tan 50and h b tan 35 (c 10) tan 35
c tan 50 (c 10) tan 35
c(tan 50 tan 35) 10 tan 35
c 10 tan 35 h c tan 50
tan 50 tan 35
tan 35 tan 50 16.98 m.
10
tan 50 tan 35
66. Let h height of balloon above ground. From the
figure at the right, tan 40 ah , tan 70 bh , and
a b 2. Thus, h b tan 70 h (2 a ) tan 70
and h a tan 40 (2 a) tan 70 a tan 40
a (tan 40 tan 70) 2 tan 70
2 tan 70
h a tan 40
a tan 40
tan 70
2 tan 70 tan 40 1.3 km.
tan
40 tan 70
67. (a)
(b) The period appears to be 4 .
(c) f ( x 4 ) sin( x 4 ) cos
x 4
2
sin( x 2 ) cos 2 sin x cos
x
2
x
2
since the period of sine and cosine is 2 . Thus, f(x) has period 4 .
68. (a)
(b) D ( ,0) (0, ); R [ 1, 1]
21 kp f 21 sin 2 0 for all integers k.
1
Choose k so large that 21 kp 1 0
. But then f 21 kp sin (1/(21)) kp 0
1/(2 ) kp
(c) f is not periodic. For suppose f has period p. Then f
which is a contradiction. Thus f has no period, as claimed.
Copyright 2016 Pearson Education, Ltd.
Chapter 1 Additional and Advanced Exercises
39
CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES
1. There are (infinitely) many such function pairs. For example, f ( x) 3 x and g ( x) 4 x satisfy
f ( g ( x)) f (4 x) 3(4 x) 12 x 4(3 x) g (3 x) g ( f ( x)).
2. Yes, there are many such function pairs. For example, if g ( x) (2 x 3)3 and f ( x) x1/3, then
( f g )( x) f ( g ( x)) f ((2 x 3)3 ) ((2 x 3)3 )1/3 2 x 3.
3. If f is odd and defined at x, then f ( x) f ( x). Thus g ( x) f ( x) 2 f ( x) 2 whereas
g ( x) ( f ( x) 2) f ( x) 2. Then g cannot be odd because g ( x) g ( x) f ( x) 2 f ( x) 2
4 0, which is a contradiction. Also, g ( x) is not even unless f ( x ) 0 for all x. On the other hand, if f is
even, then g ( x) f ( x) 2 is also even: g ( x) f ( x) 2 f ( x ) 2 g ( x).
4. If g is odd and g(0) is defined, then g (0) g ( 0) g (0). Therefore, 2 g (0) 0 g (0) 0.
5. For (x, y) in the 1st quadrant, | x | | y | 1 x
x y 1 x y 1. For (x, y) in the 2nd
quadrant, | x | | y | x 1 x y x 1
y 2 x 1. In the 3rd quadrant, | x | | y | x 1
x y x 1 y 2 x 1. In the 4th
quadrant, | x | | y | x 1 x ( y ) x 1
y 1. The graph is given at the right.
6. We use reasoning similar to Exercise 5.
(1) 1st quadrant: y | y | x | x |
2 y 2 x y x.
(2) 2nd quadrant: y | y | x | x |
2 y x ( x) 0 y 0.
(3) 3rd quadrant: y | y | x | x |
y ( y ) x ( x) 0 0
all points in the 3rd quadrant
satisfy the equation.
(4) 4th quadrant: y | y | x | x |
y ( y ) 2 x 0 x. Combining
these results we have the graph given at the right:
sin 2 x
7. (a) sin 2 x cos 2 x 1 sin 2 x 1 cos 2 x (1 cos x) (1 cos x) 1 cos x 1 cos x
(b) Using the definition of the tangent function and the double angle formulas, we have
1 cos 2 x
2
x
.
cos2 2x 1 cos2 2 2x 11 cos
cos x
tan 2 2x
sin 2 2x
2
8. The angles labeled in the accompanying figure are
equal since both angles subtend arc CD. Similarly, the
two angles labeled α are equal since they both subtend
arc AB. Thus, triangles AED and BEC are similar which
ac
2 a cos b
implies b a c
(a c)(a c) b(2a cos b)
a 2 c 2 2ab cos b 2
c 2 a 2 b 2 2ab cos .
Copyright 2016 Pearson Education, Ltd.
1 cos x
x
1 sincos
sin x
x
40
Chapter 1 Functions
9. As in the proof of the law of sines of Section 1.3, Exercise 61, ah bc sin A ab sin C ac sin B
the area of ABC 12 (base)(height) 12 ah 12 bc sin A 12 ab sin C 12 ac sin B .
10. As in Section 1.3, Exercise 61, (Area of ABC ) 2 14 (base) 2 (height)2 14 a 2 h 2 14 a 2b 2 sin 2 C
a 2 b2 c 2
. Thus,
2 ab
2
(a2 b2 c2 )
14 a 2 b 2 (1 cos 2 C ) . By the law of cosines, c 2 a 2 b 2 2ab cos C cos C
a 2 b2 c 2 2
a 2b 2
(area of ABC )2 14 a 2b 2 (1 cos2 C ) 14 a 2b 2 1
4 1
2 ab
4 a 2b 2
1 4a 2 b 2 ( a 2 b 2 c 2 ) 2 1 [(2ab ( a 2 b 2 c 2 )) (2ab ( a 2 b 2 c 2 ))]
16
16
1 [((a b) 2 c 2 )(c 2 ( a b) 2 )] 1 [(( a b) c)(( a b) c)(c ( a b))(c (a b))]
16
16
a b c a b c a b c a b c
abc
s ( s a)( s b)( s c), where s
.
2
2
2
2
2
Therefore, the area of ABC equals s( s a )( s b)( s c) .
11. If f is even and odd, then f ( x) f ( x) and f ( x) f ( x) f ( x) f ( x) for all x in the domain of f.
Thus 2 f ( x ) 0 f ( x) 0.
f ( x) f ( x)
f ( x ) f ( ( x ))
f ( x) f ( x)
E ( x)
E ( x) E is an even
2
2
2
f ( x) f ( x)
f ( x) f ( x)
f ( x) f ( x)
. Then O( x)
function. Define O ( x) f ( x) E ( x ) f ( x)
2
2
2
f ( x) f ( x)
f ( x) f ( x)
O( x) O is an odd function f ( x) E ( x ) O ( x) is the sum of an even
2
2
12. (a) As suggested, let E ( x)
and an odd function.
(b) Part (a) shows that f ( x) E ( x) O( x) is the sum of an even and an odd function. If also
f ( x) E1 ( x ) O1 ( x), where E1 is even and O1 is odd, then f ( x) f ( x) 0
( E1 ( x) O1 ( x)) ( E ( x) O( x)) . Thus, E ( x) E1 ( x) O1 ( x) O ( x) for all x in the domain of f (which is
the same as the domain of E E1 and O O1). Now ( E E1 )( x) E ( x) E1 ( x) E ( x) E1 ( x) (since E
and E1 are even) ( E E1 )( x) E E1 is even. Likewise, (O1 O)( x) O1 ( x ) O( x)
O1 ( x) (O( x)) (since O and O1 are odd) (O1 ( x) O ( x)) (O1 O ) ( x) O1 O is odd.
Therefore, E E1 and O1 O are both even and odd so they must be zero at each x in the domain of f by
Exercise 11. That is, E1 E and O1 O, so the decomposition of f found in part (a) is unique.
2
2
13. y ax 2 bx c a x 2 ba x b 2 4ba c a x 2ba
4a
4ba c
2
2
(a) If a 0 the graph is a parabola that opens upward. Increasing a causes a vertical stretching and a shift of
the vertex toward the y-axis and upward. If a 0 the graph is a parabola that opens downward. Decreasing
a causes a vertical stretching and a shift of the vertex toward the y-axis and downward.
(b) If a 0 the graph is a parabola that opens upward. If also b 0, then increasing b causes a shift of the graph
downward to the left; if b 0, then decreasing b causes a shift of the graph downward and to the right.
If a 0 the graph is a parabola that opens downward. If b 0, increasing b shifts the graph upward to the
right. If b 0, decreasing b shifts the graph upward to the left.
(c) Changing c (for fixed a and b) by c shifts the graph upward c units if c 0, and downward c units
if c 0.
14. (a) If a 0, the graph rises to the right of the vertical line x b and falls to the left. If a < 0, the graph falls
to the right of the line x b and rises to the left. If a 0, the graph reduces to the horizontal line y c.
As | a | increases, the slope at any given point x x0 increases in magnitude and the graph becomes steeper. As
| a | decreases, the slope at x0 decreases in magnitude and the graph rises or falls more gradually.
(b) Increasing b shifts the graph to the left; decreasing b shifts it to the right.
(c) Increasing c shifts the graph upward; decreasing c shifts it downward.
Copyright 2016 Pearson Education, Ltd.
Chapter 1 Additional and Advanced Exercises
41
15. Each of the triangles pictured has the same base
b vt v (1 s) . Moreover, the height of each
triangle is the same value h. Thus 12 (base)(height)
12 bh A1 A2 A3 … . In conclusion,
the object sweeps out equal areas in each one
second interval.
16. (a) Using the midpoint formula, the coordinates of P are
y
a0 b0
, 2
2
, . Thus the slope
a b
2 2
of OP x ba /2
ba .
/2
b0
(b) The slope of AB 0 a ba . The line segments AB and OP are perpendicular when the product of their
slopes is 1
ba ba ba . Thus, b2 a2 a b (since both are positive). Therefore, AB is
2
2
perpendicular to OP when a b.
17. From the figure we see that 0 2 and AB AD 1. From trigonometry we have the following:
sin . We can see that:
sin EB
EB, cos AE
AE , tan CD
CD, and tan EB
cos
AB
AB
AD
AE
area ADC 1 ( AE )( EB) 1 ( AD)2 1 ( AD) (CD)
area AEB area sector DB
2
2
sin
12 sin cos 12 (1) 2 12 (1)(tan ) 12 sin cos 12 12 cos
2
18. ( f g )( x) f ( g ( x)) a (cx d ) b acx ad b and ( g f )( x) g ( f ( x)) c (ax b) d acx cb d
Thus ( f g )( x) ( g f )( x) acx ad b acx bc d ad b bc d . Note that f (d ) ad b and
g (b) cb d , thus ( f g )( x) ( g f )( x) if f (d ) g (b).
Copyright 2016 Pearson Education, Ltd.
CHAPTER 2
2.1
LIMITS AND CONTINUITY
RATES OF CHANGE AND TANGENTS TO CURVES
1. (a)
f
f (3) f (2)
3 2 2819 19
x
(b)
f
f (1) f ( 1)
1( 1) 22 0 1
x
2. (a)
g (3) g (1)
( )
g
3 1 3 21 2
x
(b)
g (4) g ( 2)
g
4 ( 2) 8 6 8 0
x
3. (a)
h
t
(b)
h
t
4. (a)
5.
6.
11 4
h 34 h 4
3
4 4
2
g
g ( ) g (0) (2 1) (2 1)
0
2
t
0
R R (2) R (0)
20
(b)
0 3 3 3
h 2 h 6
2
6
3
g
g ( ) g ( )
(2 1) (2 1)
( )
0
t
2
81 1 31
2 1
2
P P (2) P (1) (816 10) (1 4 5) 2 2 0
21
1
7. (a)
(b)
8. (a)
2
y
((2 h )2 5) (22 5) 4 4 h h 2 51
4h h h 4 h. As h 0, 4 h 4 at P (2, 1) the slope is 4.
x
h
h
y ( 1) 4( x 2) y 1 4 x 8 y 4 x 9
2
y (7(2 h )2 ) (7 22 ) 744 h h 2 3
4 hhh 4 h. As h 0, 4 h
x
h
h
4 at P (2, 3) the slope
is 4.
(b) y 3 ( 4)( x 2) y 3 4 x 8 y 4 x 11
9. (a)
2
y
((2 h )2 2(2 h) 3) (22 2(2) 3) 4 4 h h 2 4 2 h 3( 3)
2h h h 2 h. As h 0, 2 h 2 at
x
h
h
P(2, 3) the slope is 2.
(b) y (3) 2( x 2) y 3 2 x 4 y 2 x 7.
10. (a)
2
y
((1 h ) 2 4(1 h )) (12 4(1)) 1 2 h h 2 4 4 h ( 3)
h h 2h h 2. As h 0, h 2 2 at P (1, 3) the
x
h
h
slope is 2.
(b) y ( 3) ( 2)( x 1) y 3 2 x 2 y 2 x 1.
11. (a)
2
3
2
3
y
(2 h )3 23
812h 4hh h 8 12 h 4hh h 12 4h h 2 . As h 0, 12 4h h 2 12, at P (2, 8)
x
h
the slope is 12.
(b) y 8 12( x 2) y 8 12 x 24 y 12 x 16.
12. (a)
2
3
2
3
y
2 (1 h )3 (213 )
213h h3h h 1 3h 3hh h 3 3h h 2 . As h 0, 3 3h h 2 3, at
x
h
P(1, 1) the slope is 3.
(b) y 1 ( 3)( x 1) y 1 3 x 3 y 3x 4.
Copyright 2016 Pearson Education, Ltd.
43
44
Chapter 2 Limits and Continuity
13. (a)
2
3
y
(1 h )3 12(1 h) (13 12(1)) 13h 3h 2 h3 1212h ( 11)
9h 3hh h 9 3h h 2 .
h
x
h
2
As h 0, 9 3h h 9 at P(1, 11) the slope is 9.
(b) y (11) (9)( x 1) y 11 9 x 9 y 9 x 2.
14. (a)
y
(2 h )3 3(2 h ) 2 4 (23 3(2)2 4) 812 h 6 h 2 h3 12 12 h 3h 2 4 0 3h 2 h3
h 3h h 2 .
x
h
h
2
As h 0, 3h h 0 at P (2, 0) the slope is 0.
(b) y 0 0( x 2) y 0.
15. (a)
Q
Slope of PQ
p
t
Q1 (10, 225)
650 225 42.5 m/s
20 10
Q2 (14,375)
650375 45.83 m/s
20 14
Q3 (16.5, 475)
650 475 50.00 m/s
20 16.5
Q4 (18,550)
650550 50.00 m/s
20 18
(b) At t 20, the sportscar was traveling approximately 50 m/s or 180 km/h.
16. (a)
Slope of PQ
Q
p
t
Q1 (5, 20)
80 20 12 m/s
10 5
Q2 (7, 39)
80 39 13.7 m/s
10 7
Q3 (8.5,58)
80 58 14.7 m/s
10 8.5
Q4 (9.5, 72)
80 72 16 m/s
10 9.5
(b) Approximately 16 m/s
p
17. (a)
Profit (1000s)
200
160
120
80
40
0
(b)
(c)
2010 2011 2012 2013 2014
Ye ar
t
p
174 62 112 56 thousand dollars per year
2014
t
2012
2
p
62 27 35 thousand dollars per year.
The average rate of change from 2011 to 2012 is t 2012
2011
p
111
62 49 thousand dollars per year.
The average rate of change from 2012 to 2013 is t 20132012
So, the rate at which profits were changing in 2012 is approximately 12 (35 49) 42 thousand dollars
per year.
18. (a) F ( x) ( x 2)/( x 2)
x
1.2
1.1
F ( x) 4.0
3.4
F 4.0 ( 3) 5.0;
x
1.2 1
F 3.04 ( 3) 4.04;
x
1.011
1.01
3.04
1.001
3.004
1.0001
3.0004
F 3.4 ( 3) 4.4;
x
1.11
F 3.004 ( 3) 4.004;
x
1.0011
Copyright 2016 Pearson Education, Ltd.
1
3
Section 2.1 Rates of Change and Tangents to Curves
45
F 3.0004 ( 3) 4.0004;
x
1.00011
(b) The rate of change of F ( x) at x 1 is 4.
19. (a)
g
g (2) g (1)
21 2211 0.414213
x
g
g (1 h ) g (1)
(1 h ) 1 1hh 1
x
g
g (1.5) g (1)
1
1.51 1.5
x
0.5
0.449489
(b) g ( x) x
1 h
1.1
1.01
1.001
1.0001
1.00001
1.000001
1 h
1.04880
1.004987
1.0004998
1.0000499
1.000005
1.0000005
0.4880
0.4987
0.4998
0.499
0.5
0.5
1 h 1 /h
(c) The rate of change of g ( x) at x 1 is 0.5.
(d) The calculator gives lim
h 0
20. (a) i)
ii)
1 h 1 1
2.
h
11
1
f (3) f (2)
3 2
6
16
3 2
1
1
1 1
2 T
f (T ) f (2)
T
TT 22 2TT 22T 2T2(TT 2) 2T2(2
21T , T 2
T 2
T )
(b) T
f (T )
( f (T ) f (2))/(T 2)
2.1
2.01
2.001
0.476190
0.497512
0.499750
0.2381
0.2488
0.2500
(c) The table indicates the rate of change is 0.25 at t 2.
(d) lim 21T 14
T 2
2.0001
0.4999750
0.2500
2.00001
0.499997
0.2500
2.000001
0.499999
0.2500
NOTE: Answers will vary in Exercises 21 and 22.
21. (a) [0, 1]:
s 150 15 km/h; [1, 2.5]: s 20 15 10 km/h; [2.5, 3.5]: s 30 20 10 km/h
t
10
t
2.51
3
t
3.5 2.5
(b) At P 12 , 7.5 : Since the portion of the graph from t 0 to t 1 is nearly linear, the instantaneous rate of
change will be almost the same as the average rate of change, thus the instantaneous speed at t 12 is
15 7.5 15 km/h. At P (2, 20): Since the portion of the graph from t 2 to t 2.5 is nearly linear, the
10.5
20 0 km/h.
instantaneous rate of change will be nearly the same as the average rate of change, thus v 20
2.5 2
For values of t less than 2, we have
Q
Q1 (1, 15)
Q2 (1.5, 19)
Q3 (1.9, 19.9)
Slope of PQ
s
t
15 20 5 km/h
1 2
19 20 2 km/h
1.5 2
19.9 20 1 km/h
1.9 2
Thus, it appears that the instantaneous speed at t 2 is 0 km/h.
At P(3, 22):
Slope of PQ st
Q
Q
35 22 13 km/h
Q1 (4, 35)
Q1 (2, 20)
4 3
Q2 (3.5, 30)
Q3 (3.1, 23)
30 22 16 km/h
3.53
23 22 10 km/h
3.13
Q2 (2.5, 20)
Q3 (2.9, 21.6)
Thus, it appears that the instantaneous speed at t 3 is about 7 km/h.
Copyright 2016 Pearson Education, Ltd.
Slope of PQ
s
t
20 22 2 km/h
2 3
20 22 4 km/h
2.53
21.6 22 4 km/h
2.9 3
46
Chapter 2 Limits and Continuity
(c) It appears that the curve is increasing the fastest at t 3.5. Thus for P (3.5, 30)
Slope of PQ st
Slope of PQ st
Q
Q
3530 10 km/h
Q1 (4, 35)
22 30 16 km/h
Q1 (3, 22)
4 3.5
33.5
34 30 16 km/h
Q2 (3.75, 34)
2530 20 km/h
Q2 (3.25, 25)
3.753.5
3.253.5
32 30 20 km/h
Q3 (3.6, 32)
2830 20 km/h
Q3 (3.4, 28)
3.6 3.5
3.4 3.5
Thus, it appears that the instantaneous speed at t 3.5 is about 20 km/h.
22. (a) [0, 3]:
A 10 15
t
3 0
A 3.9 15
t
5 0
L ; [0, 5]:
1.67 day
L ; [7, 10]:
2.2 day
A 0 1.4
t
10 7
L
0.5 day
(b) At P(1, 14):
Q
Q1 (2, 12.2)
Q2 (1.5, 13.2)
Q3 (1.1, 13.85)
Slope of PQ
A
t
12.2 14 1.8 L/day
2 1
13.2 14 1.6 L/day
1.51
13.8514 1.5 L/day
1.11
Q
Q1 (0, 15)
Q2 (0.5, 14.6)
Q3 (0.9, 14.86)
Slope of PQ
A
t
1514 1 L/day
0 1
14.6 14 1.2 L/day
0.51
14.86 14 1.4 L/day
0.9 1
Thus, it appears that the instantaneous rate of consumption at t 1 is about 1.45 L/day.
At P(4, 6):
Slope of PQ At
Q
Slope of PQ At
Q
10 6 4 L/day
Q1 (3, 10)
3.9 6 2.1 L/day
Q1 (5, 3.9)
3 4
Q2 (4.5, 4.8)
Q3 (4.1, 5.7)
5 4
4.86 2.4 L/day
4.5 4
5.7 6 3 L/day
4.1 4
Q2 (3.5, 7.8)
Q3 (3.9, 6.3)
7.86 3.6 L/day
3.5 4
6.36 3 L/day
3.9 4
Thus, it appears that the instantaneous rate of consumption at t 1 is 3 L/day.
At P(8, 1):
Slope of PQ At
A
Q
Slope
of
PQ
Q
t
1.4 1 0.6 L/day
Q1 (7, 1.4)
0.51 0.5 L/day
Q1 (9, 0.5)
7 8
9 8
1.31 0.6 L/day
Q2 (7.5, 1.3)
0.7 1 0.6 L/day
Q2 (8.5, 0.7)
7.58
8.58
1.04 1 0.6 L/day
Q3 (7.9, 1.04)
0.951 0.5 L/day
Q3 (8.1, 0.95)
7.9 8
8.18
Thus, it appears that the instantaneous rate of consumption at t 1 is 0.55 L/day.
(c) It appears that the curve (the consumption) is decreasing the fastest at t 3.5. Thus for P(3.5, 7.8)
Slope of PQ st
Q
Slope of PQ At
Q
11.2 7.8 3.4 L/day
Q1 (2.5, 11.2)
4.87.8 3 L/day
Q1 (4.5, 4.8)
2.53.5
Q2 (4, 6)
Q3 (3.6, 7.4)
4.53.5
6 7.8 3.6 L/day
4 3.5
7.4 7.8 4 L/day
3.6 3.5
Q2 (3, 10)
Q3 (3.4, 8.2)
10 7.8 4.4 L/day
33.5
8.2 7.8 4 L/day
3.4 3.5
Thus, it appears that the rate of consumption at t 3.5 is about 4 L/day.
2.2
LIMIT OF A FUNCTION AND LIMIT LAWS
1. (a) Does not exist. As x approaches 1 from the right, g ( x) approaches 0. As x approaches 1 from the left, g ( x)
approaches 1. There is no single number L that all the values g ( x) get arbitrarily close to as x 1.
(b) 1
(c) 0
(d) 0.5
2. (a) 0
(b) 1
Copyright 2016 Pearson Education, Ltd.
Section 2.2 Limit of a Function and Limit Laws
47
(c) Does not exist. As t approaches 0 from the left, f (t ) approaches 1. As t approaches 0 from the right,
f (t ) approaches 1. There is no single number L that f (t ) gets arbitrarily close to as t 0.
(d) 1
3. (a) True
(d) False
(g) True
(b) True
(e) False
(c) False
(f) True
4. (a) False
(d) True
(b) False
(e) True
(c) True
5. lim | xx | does not exist because | xx | xx 1 if x 0 and | xx | xx 1 if x 0. As x approaches 0 from the left, | xx |
x0
approaches 1. As x approaches 0 from the right, | xx | approaches 1. There is no single number L that all the
function values get arbitrarily close to as x 0.
1 become increasingly large and negative. As x approaches 1
6. As x approaches 1 from the left, the values of x
1
from the right, the values become increasingly large and positive. There is no number L that all the function
values get arbitrarily close to as x 1, so lim x11 does not exist.
x 1
7. Nothing can be said about f ( x) because the existence of a limit as x x0 does not depend on how the function
is defined at x0 . In order for a limit to exist, f ( x) must be arbitrarily close to a single real number L when x is
close enough to x0 . That is, the existence of a limit depends on the values of f ( x) for x near x0 , not on the
definition of f ( x) at x0 itself.
8. Nothing can be said. In order for lim f ( x) to exist, f ( x) must close to a single value for x near 0 regardless of
x 0
the value f (0) itself.
9. No, the definition does not require that f be defined at x 1 in order for a limiting value to exist there. If f (1) is
defined, it can be any real number, so we can conclude nothing about f (1) from lim f ( x) 5.
x 1
10. No, because the existence of a limit depends on the values of f ( x) when x is near 1, not on f (1) itself. If
lim f ( x) exists, its value may be some number other than f (1) 5. We can conclude nothing about lim f ( x),
x 1
whether it exists or what its value is if it does exist, from knowing the value of f (1) alone.
11.
lim ( x 2 13) ( 3)2 13 9 13 4
x3
12. lim ( x 2 5 x 2) (2)2 5(2) 2 4 10 2 4
x 2
13. lim 8(t 5)(t 7) 8(6 5)(6 7) 8
t 6
14.
15.
lim ( x3 2 x 2 4 x 8) (2)3 2(2)2 4( 2) 8 8 8 8 8 16
x 2
2(2) 5
lim 2 x 53
11(2)3
x2 11 x
93 3
Copyright 2016 Pearson Education, Ltd.
x 1
48
16.
17.
Chapter 2 Limits and Continuity
(2) 23 4 (2) 25 252
lim 4 x (3x 4)2 4 12 3 12 4
x 1/2
y2
18. lim
2
y 2 y 5 y 6
19.
20.
2 23 1 (8 2) 43 1 (6) 13 2
lim (8 3s )(2 s 1) 8 5 23
t 2/3
2
2
2 2
4
4 1
410
20
6
5
(2) 2 5(2) 6
y 3
lim
z 2 10 42 10 16 10 6
lim (5 y )4/3 [5 (3)]4/3 (8)4/3 (8)1/3
z 4
21. lim
3
3h 1 1
22. lim
5h 4 2
lim
h
h 0
h 0
h 0
2
24 16
4
3
3 32
3(0) 1 1
1 1
(5h 4) 4
5 h 4 2 5h 4 2
lim
h
5 h 4 2 h 0 h 5 h 4 2
lim
h 0 h
5h
5h 4 2
x 5
1
23. lim 2x 5 lim ( x 5)(
lim x 15 51 5 10
x 5)
x 5 x 25
x 5
x 5
24.
x 3
lim 2 x 3 lim
lim 1 1 12
x 3 x 4 x 3 x 3 ( x 3)( x 1) x 3 x 1 31
25.
2
( x 5)( x 2)
lim x x3x510 lim
lim ( x 2) 5 2 7
x 5
x 5
x 5
x 5
2
26. lim x x7x210 lim
x 2
x2
( x 5)( x 2)
lim ( x 5) 2 5 3
x2
x 2
(t 2)(t 1)
2
27. lim t 2 t 2 lim (t 1)(t 1) lim tt12 1112 32
t 1 t 1
t 1
t 1
28.
2
(t 2)(t 1)
lim t 23t 2 lim (t 2)(t 1) lim tt 22 11 22 13
t 1 t t 2
t 1
t 1
29.
2( x 2)
lim 32 x 42 lim 2
lim 22 42 12
x 2 x 2 x
x 2 x ( x 2) x 2 x
30. lim
5 y 3 8 y 2
4
y 0 3 y 16 y
31.
32.
2
lim
y 2 (5 y 8)
2
2
y 0 y (3 y 16)
1 x
1
lim
5 y 8
2
y 0 3 y 16
816 12
x1
lim xx 11 lim x x 1 lim 1xx x11 lim 1x 1
x1
x 1
1 1
lim x 1 x x 1 lim
x 0
x 0
x1
( x 1) ( x 1)
( x 1)( x 1)
x
2x
lim ( x 1)(
1 lim ( x 1)(2 x 1) 21 2
x 1) x
x 0
x 0
Copyright 2016 Pearson Education, Ltd.
lim
h0
5
5h 4 2
5 5
4
4 2
Section 2.2 Limit of a Function and Limit Laws
4
33. lim u 3 1 lim
u 1 u 1
(u 2 1)(u 1)(u 1)
u 1 (u u 1)(u 1)
v 2 v 16
(u 2 1)(u 1)
u 2 u 1
u 1
(v 2)( v 2 2v 4)
3
34. lim v4 8 lim
35.
lim
2
v 2 2v 4 4 4 4 12 3
2
(4)(8)
32 8
v 2 (v 2)( v 4)
lim
2
v 2 (v 2)(v 2)( v 4)
x 3
lim x 3 lim
lim
x 9 x 9
x 9 ( x 3)( x 3) x 9
2
(11)(11)
111 43
1
x 3
1 1
9 3 6
x 4 2 x
x 3 2 lim ( x1) x3 2 lim x 3 2 4 2 4
x 1
x 3 2 x 3 2 x1 ( x 3)4
x 1 lim
x 3 2 x 1
38.
x 2 8 3
lim
x 1
x 1
lim
x 1
x 8 3 x 8 3 lim ( x 8)9 lim ( x1)( x1)
x 1 ( x 1) x 83 x 1 ( x 1) x 8 3
( x 1) x 8 3
2
x 1
x 1
x 2
x 2 12 4
lim
x 2
x 2
2
2
2
lim
39. lim
x 4(2 2) 16
( x 1)
37. lim
x 1
x (4 x )
x (2 x )(2 x )
lim
lim x 2
2 x
x 4 2 x
x 4
x 4
36. lim 4 x x lim
x 2 8 3
2
2
323 13
x 12 4 x 12 4 lim ( x 12)16 lim ( x2)( x2)
x 2 ( x 2) x 12 4 x 2 ( x 2) x 12 4
( x 2) x 12 4
2
2
2
2
x2
lim
x 2 12 4
x2
2
2
4
12
16 4
x 5 3 lim ( x2) x 5 3 lim ( x2) x 5 3
40. lim
lim
( x 5) 9
x 2 ( x 2)( x 2)
x 2 x 5 3 x 2 x 5 3 x 5 3 x 2
2
2
x 2
2
2
2
x 2 5 3
x 2
lim
41.
2
2
( x 2)
x2
9 3
32
4
2 x 5 2 x 5 lim 4( x 5) lim 9 x
x 3 ( x 3) 2 x 5 x 3 ( x 3) 2 x 5
( x 3) 2 x 5
(3 x )(3 x )
3 x 6 3
lim
lim
x 3 ( x 3) 2 x 5 x 3 2 x 5 2 4 2
2
2
lim 2 x x3 5 lim
x
3
x 3
2
2
2
2
2
2
2
2
lim (4 x)5 x 9 lim (4 x)5 x 9
25( x 9)
16 x
x4
x 4 5 x 9 x 4 5 x 9 5 x 9 x 4
(4 x ) 5 x 9
lim
lim 5 x 9 5 25 5
42. lim
4 x
2
lim
2
(4 x ) 5 x 2 9
2
2
2
2
x 4
2
2
2
(4 x )(4 x )
x 4
4 x
8
4
2
43. lim (2sin x 1) 2sin 0 1 0 1 1
44.
45. lim sec x lim cos1 x cos1 0 11 1
46.
x 0
x 0
x 0
lim sin 2 x lim sin x (sin 0)2 02 0
x 0
x 0
sin x sin 0 0 0
lim tan x lim cos
x
cos 0 1
x 0
x 0
Copyright 2016 Pearson Education, Ltd.
49
50
Chapter 2 Limits and Continuity
x sin x 1 0 sin 0 1 0 0 1
47. lim 13cos
x
3cos 0
3
3
x 0
48. lim ( x 2 1)(2 cos x) (02 1)(2 cos 0) (1)(2 1) (1)(1) 1
x 0
49.
lim
x
x 4 cos( x ) lim
x 4 lim cos( x ) 4 cos 0 4 1 4
x
50. lim 7 sec2 x
x 0
x
lim (7 sec2 x) 7 lim sec2 x 7 sec2 0 7 (1) 2 2 2
x 0
x 0
51. (a) quotient rule
(c) sum and constant multiple rules
(b) difference and power rules
52. (a) quotient rule
(c) difference and constant multiple rules
(b)
power and product rules
53. (a) lim f ( x) g ( x) lim f ( x) lim g ( x) (5)(2) 10
x c
x c
x c
(b) lim 2 f ( x) g ( x) 2 lim f ( x) lim g ( x) 2(5)(2) 20
x c
x c
x c
(c) lim [ f ( x) 3 g ( x)] lim f ( x) 3 lim g ( x) 5 3(2) 1
x c
(d)
x c
x c
lim f ( x )
f ( x)
xc
lim
lim f ( x ) lim g ( x ) 5(52) 75
x c f ( x ) g ( x )
x c
x c
54. (a) lim [ g ( x) 3] lim g ( x) lim 3 3 3 0
x 4
x4
x 4
(b) lim xf ( x ) lim x lim f ( x) (4)(0) 0
x 4
x4
x4
2
(c) lim [ g ( x)]2 lim g ( x) [3]2 9
x 4
x 4
g ( x)
lim g ( x )
(d) lim f ( x ) 1 limx f (4x ) lim 1 031 3
x 4
x4
x 4
55. (a) lim [ f ( x) g ( x )] lim f ( x) lim g ( x ) 7 ( 3) 4
x b
x b
x b
(b) lim f ( x) g ( x) lim f ( x) lim g ( x) (7)(3) 21
x b
x b
xb
(c) lim 4 g ( x) lim 4 lim g ( x) (4)(3) 12
x b
xb xb
(d) lim f ( x)/g ( x) lim f ( x)/ lim g ( x) 73 73
x b
56. (a)
x b
x b
lim [ p( x) r ( x) s( x)] lim p ( x) lim r ( x) lim s ( x ) 4 0 (3) 1
x 2
x 2
x 2
x 2
(b) lim p( x) r ( x) s ( x) lim p ( x) lim r ( x) lim s ( x) (4)(0)(3) 0
x 2
x 2
x2
x 2
(c) lim [4 p ( x) 5r ( x)]/s ( x) 4 lim p ( x) 5 lim r ( x ) lim s ( x) [4(4) 5(0)]/ 3 16
3
x 2
x 2
x 2
x 2
2
(1 h ) 2 12
h (2 h )
lim 1 2 h h h 1 lim h lim (2 h) 2
h
h 0
h 0
h 0
h 0
57. lim
Copyright 2016 Pearson Education, Ltd.
Section 2.2 Limit of a Function and Limit Laws
51
2
h ( h 4)
( 2 h ) 2 ( 2)2
lim 4 4h h h 4 lim h lim (h 4) 4
h
h 0
h 0
h 0
h 0
58. lim
59. lim
h 0
60. lim
h 0
61. lim
h 0
[3(2 h ) 4][3(2) 4]
lim 3hh 3
h
h 0
21 h 12 lim 22 h 1 lim 2(2 h) lim
h 0 2 h
h
7h 7
lim
h
h 0
h0 2 h ( 2 h )
7 h 7 7 h 7 lim (7 h)7 lim
h
lim
h 7 h 7
h 0 h 7 h 7 h 0 h 7 h 7 h0
3(0 h ) 1 3(0) 1
lim
h
h 0
h 0
62. lim
63. lim
x 0
h
14
h 0 h (4 2 h )
1
1
7h 7
2 7
3h 1 1 3h 1 1 lim (3h1)1 lim 3h lim 3 3
h 3h 1 1
h 0 h 3h 1 1 h 0 h 3h 1 1 h0 3h11 2
5 2 x 2 5 2(0)2 5 and lim
x0
5 x 2 5 (0) 2 5; by the sandwich theorem, lim f ( x) 5
x 0
64. lim (2 x 2 ) 2 0 2 and lim 2 cos x 2(1) 2; by the sandwich theorem, lim g ( x) 2
x 0
x 0
2
x 0
x 1
65. (a) lim 1 x6 1 60 1 and lim 1 1; by the sandwich theorem, lim 2x2sin
cos x
x0
x 0
x0
(b) For x 0, y ( x sin x)/(2 2 cos x) lies
between the other two graphs in the figure,
and the graphs converge as x 0.
lim lim
1
x 0 2
66. (a) lim
x2
24
1
x 0 2
x 2 1 0 1 and lim 1 1 ; by the sandwich theorem, lim 1cos x 1 .
2
24
2
2
2
x 0
x0 2 2
x 0 x
2
(b) For all x 0, the graph of f ( x) (1 cos x)/x
lies between the line y 12 and the parabola
y 12 x 2 /24, and the graphs converge as
x 0.
67. (a) f ( x) ( x 2 9)/( x 3)
x
3.1
3.01
3.001
3.0001
3.00001
3.000001
f ( x)
6.1
6.01
6.001
6.0001
6.00001
6.000001
x
2.9
2.99
2.999
2.9999
2.99999
2.999999
5.9
5.99
The estimate is lim f ( x) 6.
5.999
5.9999
5.99999
5.999999
f ( x)
x 3
Copyright 2016 Pearson Education, Ltd.
52
Chapter 2 Limits and Continuity
(b)
2
(c) f ( x) xx 39
( x 3)( x 3)
x 3 if x 3, and lim ( x 3) 3 3 6.
x 3
x 3
68. (a) g ( x) ( x 2 2)/ x 2
x
g ( x)
1.4
2.81421
1.41
2.82421
1.414
2.82821
1.4142
2.828413
1.41421
2.828423
1.414213
2.828426
(b)
2
(c) g ( x) x 2
x 2
x 2 x 2 x 2 if x 2, and lim x 2 2 2 2 2.
x 2
x 2
69. (a) G ( x) ( x 6)/( x 2 4 x 12)
x
5.9
5.99
G ( x) .126582 .1251564
x
G ( x)
6.1
.123456
6.01
.124843
5.999
.1250156
5.9999
.1250015
5.99999 5.999999
.1250001 .1250000
6.001
.124984
6.0001
.124998
6.00001
.124999
6.000001
.124999
(b)
(c) G ( x)
x 6
x6
( x 6)(
x 1 2 if x 6, and lim x 1 2 61 2 81 0.125.
x 2)
( x 2 4 x 12)
x 6
Copyright 2016 Pearson Education, Ltd.
Section 2.2 Limit of a Function and Limit Laws
70. (a) h( x) ( x 2 2 x 3)/( x 2 4 x 3)
x
h( x )
2.9
2.052631
2.99
2.005025
2.999
2.000500
2.9999
2.000050
2.99999
2.000005
2.999999
2.0000005
x
3.1
h( x) 1.952380
3.01
1.995024
3.001
1.999500
3.0001
1.999950
3.00001
1.999995
3.000001
1.999999
(b)
2
( x 3)( x 1)
(c) h( x) x2 2 x 3 ( x 3)( x 1) xx 11 if x 3, and lim xx 11 3311 42 2.
x 4 x 3
x 3
71. (a) f ( x) ( x 2 1)/(| x | 1)
x
f ( x)
1.1
2.1
1.01
2.01
1.001
2.001
1.0001
2.0001
1.00001
2.00001
1.000001
2.000001
x
f ( x)
.9
1.9
.99
1.99
.999
1.999
.9999
1.9999
.99999
1.99999
.999999
1.999999
(b)
(c)
( x 1)( x 1) x 1, x 0 and x 1
2
x 1
x
1
f ( x)
, and lim (1 x ) 1 (1) 2.
x 1 ( x 1)( x 1)
x 1
( x 1) 1 x, x 0 and x 1
72. (a) F ( x) ( x 2 3 x 2)/(2 | x |)
x
F ( x)
2.1
1.1
2.01
1.01
2.001
1.001
2.0001
1.0001
2.00001
1.00001
2.000001
1.000001
x
F ( x)
1.9
.9
1.99
.99
1.999
.999
1.9999
.9999
1.99999
.99999
1.999999
.999999
Copyright 2016 Pearson Education, Ltd.
53
54
Chapter 2 Limits and Continuity
(b)
( x 2)( x 1) ,
x0
2
2 x
(c) F ( x ) x 3 x 2
, and lim ( x 1) 2 1 1.
2 x
( x 2)( x 1)
x 2
x
1,
x
0
and
x
2
2 x
73. (a) g ( ) (sin )/
g ( )
.1
.998334
.01
.999983
.1
g ( )
.998334
lim g( ) 1
.001
.999999
.0001
.999999
.00001
.999999
.000001
.999999
.01
.001
.999983 .999999
.0001
.999999
.00001
.999999
.000001
.999999
0
(b)
74. (a) G (t ) (1 cos t )/t 2
t
G (t )
.1
.499583
.01
.499995
.001
.499999
.0001
.5
.00001
.5
.000001
.5
t
.1
G (t ) .499583
lim G (t ) 0.5
.01
.499995
.001
.499999
.0001 .00001 .000001
.5
.5
.5
t 0
(b)
Copyright 2016 Pearson Education, Ltd.
Section 2.2 Limit of a Function and Limit Laws
55
75. lim f ( x) exists at those points c where lim x 4 lim x 2 . Thus, c 4 c 2 c 2 (1 c 2 ) 0 c 0, 1, or 1.
x c
x c
x c
Moreover, lim f ( x) lim x 2 0 and lim f ( x ) lim f ( x) 1.
x 0
x 0
x 1
x 1
76. Nothing can be concluded about the values of f , g , and h at x 2. Yes, f (2) could be 0. Since the conditions
of the sandwich theorem are satisfied, lim f ( x ) 5 0.
x 2
lim f ( x ) lim 5
lim f ( x ) 5
f ( x ) 5
4
x4
x lim
x 4 4 2
lim f ( x) 5 2(1) lim f ( x) 2 5 7.
x
2
x
lim
2
x 4
x 4
x4
77. 1 lim
x 4
78. (a) 1 lim
f ( x)
x 2 x
(b) 1 lim
2
lim f ( x )
x 2
lim x
x 2
f ( x)
2
x 2 x
x4
lim
x 2
2
lim f ( x )
x 24
f ( x)
x
lim f ( x) 4.
x 2
lim 1 lim f ( x )
x 2 x x 2 x
f ( x)
2.
12 xlim
2 x
f ( x ) 5
f ( x ) 5
79. (a) 0 3 0 lim x 2 lim ( x 2) lim x 2 ( x 2) lim [ f ( x) 5]
x2
x 2
x2
x 2
lim f ( x) 5 lim f ( x) 5.
x2
x 2
f ( x ) 5
(b) 0 4 0 lim x 2 lim ( x 2) lim f ( x) 5 as in part (a).
x 2
x 2
x 2
2
f ( x)
f ( x)
f ( x)
80. (a) 0 1 0 lim 2 lim x lim 2 lim x 2 lim 2 x 2 lim f ( x).
x
x
x 0
x 0
x 0
x0
x 0 x 0 x
That is, lim f ( x) 0.
x 0
f ( x)
f ( x)
f ( x)
f ( x)
(b) 0 1 0 lim 2 lim x lim 2 x lim x . That is, lim x 0.
x0
x0
x 0 x x 0 x0 x
81. (a) lim x sin 1x 0
x 0
(b) 1 sin 1x 1 for x 0:
x 0 x x sin 1x x lim x sin 1x 0 by the sandwich theorem;
x 0
x 0 x x sin 1x x lim x sin 1x 0 by the sandwich theorem.
x 0
82. (a) lim x 2 cos
x 0
0
1
x3
Copyright 2016 Pearson Education, Ltd.
56
Chapter 2 Limits and Continuity
(b) 1 cos
1 for x 0 x x cos x lim x cos 0 by the sandwich theorem since
1
x3
2
2
lim x 2 0.
1
x3
2
2
x 0
1
x3
x 0
83-88. Example CAS commands:
Maple:
f : x - (x^4 16)/(x 2);
x0 : 2;
plot( f (x), x x0-1..x0 1, color black,
title "Section 2.2, #83(a)" );
limit( f (x), x x 0 );
In Exercise 83, note that the standard cube root, x^(1/3), is not defined for x<0 in many CASs. This can be
overcome in Maple by entering the function as f : x - (surd(x 1, 3) 1)/x.
Mathematica: (assigned function and values for x0 and h may vary)
Clear[f , x]
f[x _]: (x 3 x 2 5x 3)/(x 1) 2
x0 1; h 0.1;
Plot[f[x],{x, x0 h, x0 h}]
Limit[f[x], x x0]
2.3
THE PRECISE DEFINITION OF A LIMIT
1.
Step 1: x 5 x 5 5 x 5
Step 2: 5 7 2,or 5 1 4.
The value of δ which assures x 5 1 x 7 is the smaller value, 2.
2.
Step 1:
Step 2:
x 2 x 2 2 x 2
2 1 1, or 2 7 5.
The value of which assures x 2 1 x 7 is the smaller value, 1.
Step 1:
Step 2:
x (3) x 3 3 x 3
3 72 12 , or 3 12 52 .
3.
The value of which assures x (3) 72 x 12 is the smaller value, 12 .
Copyright 2016 Pearson Education, Ltd.
Section 2.3 The Precise Definition of a Limit
4.
Step 1:
Step 2:
x 32 x 32 32 x 32
32 72 2, or 32 12 1.
The value of which assures x 32 72 x 12 is the smaller value, 1.
5.
Step 1:
Step 2:
x 12 x 12 12 x 12
1 , or 1 4 1 .
12 94 18
2 7
14
1.
The value of which assures x 12 94 x 74 is the smaller value, 18
6.
Step 1:
Step 2:
7. Step 1:
Step 2:
x 3 x 3 3 x 3
3 2.7591 0.2409, or 3 3.2391 0.2391.
The value of which assures x 3 2.7591 x 3.2391 is the smaller value, 0.2391.
x 5 x 5 5 x 5
From the graph, 5 4.9 0.1, or 5 5.1 0.1; thus 0.1 in either case.
8. Step 1:
Step 2:
x (3) x 3 3 x 3
From the graph, 3 3.1 0.1, or 3 2.9 0.1; thus 0.1.
9. Step 1:
Step 2:
x 1 x 1 1 x 1
9 7 , or 1 25 9 ; thus 7 .
From the graph, 1 16
16
16
16
16
10. Step 1:
Step 2:
x 3 x 3 3 x 3
From the graph, 3 2.61 0.39, or 3 3.41 0.41; thus 0.39.
11. Step 1:
Step 2:
x 2 x 2 2 x 2
From the graph, 2 3 2 3 0.2679, or 2 5 5 2
thus 5 2.
12. Step 1:
x (1) x 1 1 x 1
Step 2:
From the graph, 1 25
thus
13. Step 1:
Step 2:
5 2
.
2
5 2
2
0.118 or 1 23 22 3
0.2361;
0.1340;
x (1) x 1 1 x 1
9 0.36; thus 9 0.36.
From the graph, 1 16
97 0.77, or 1 16
25
9
25
25
Copyright 2016 Pearson Education, Ltd.
57
58
Chapter 2 Limits and Continuity
14. Step 1:
Step 2:
x 12 x 12 12 x 12
1 1 1
From the graph, 12 2.01
2 2.01
thus 0.00248.
1 1 1
0.00248, or 12 1.99
1.99 2
15. Step 1:
Step 2:
( x 1) 5 0.01 x 4 0.01 0.01 x 4 0.01 3.99 x 4.01
x 4 x 4 4 x 4 0.01.
16. Step 1:
(2 x 2) (6) 0.02 2 x 4 0.02 0.02 2 x 4 0.02
4.02 2 x 3.98 2.01 x 1.99
x (2) x 2 2 x 2 0.01.
Step 2:
17. Step 1:
x 1 1 0.1 0.1 x 1 1 0.1 0.9 x 1 1.1 0.81 x 1 1.21
Step 2:
0.19 x 0.21
x 0 x . Then, 0.19 0.19 or 0.21; thus, 0.19.
18. Step 1:
Step 2:
0.00251;
x 12 0.1 0.1 x 12 0.1 0.4 x 0.6 0.16 x 0.36
x 14 x 14 14 x 14 .
en 14 0.16 0.09 or 14 0.36 0.11; thus 0.09.
19. Step 1:
Step 2:
20. Step 1:
Step 2:
21. Step 1:
Step 2:
22. Step 1:
Step 2:
19 x 3 1 1 19 x 3 1 2 19 x 4 4 19 x 16
4 x 19 16 15 x 3 or 3 x 15
x 10 x 10 10 x 10.
Then 10 3 7, or 10 15 5; thus 5.
x 7 4 1 1 x 7 4 1 3 x 7 5 9 x 7 25 16 x 32
x 23 x 23 23 x 23.
Then 23 16 7, or 23 32 9; thus 7.
1 1 0.05 0.05 1 1 0.05 0.2 1 0.3 10 x 10 or 10 x 5.
x 4
x 4
x
2
3
3
x 4 x 4 4 x 4.
Then 4 10
or 23 , or 4 5 or 1; thus 23 .
3
x 2 3 0.1 0.1 x 2 3 0.1 2.9 x 2 3.1 2.9 x 3.1
x 3 x 3 3 x 3.
en 3 2.9 3 2.9
thus 0.0286
23. Step 1:
Step 2:
0.0291, or 3 3.1 3.1 3
0.0286;
x 2 4 0.5 0.5 x 2 4 0.5 3.5 x 2 4.5 3.5 x 4.5 4.5 x 3.5,
for x near 2.
x (2) x 2 2 x 2.
Then 2 4.5 4.5 2 0.1213, or 2 3.5 2 3.5
thus 4.5 2 0.12.
Copyright 2016 Pearson Education, Ltd.
0.1292;
Section 2.3 The Precise Definition of a Limit
24. Step 1:
Step 2:
25. Step 1:
Step 2:
26. Step 1:
Step 2:
59
1 ( 1) 0.1 0.1 1 1 0.1 11 1 9 10 x 10 or 10 x 10 .
x
x
10
x
10
11
9
9
11
x (1) x 1 1 x 1.
1 ; thus 1 .
Then 1 10
91 , or 1 10
11
9
11
11
( x 2 5) 11 1 x 2 16 1 1 x 2 16 1 15 x 2 17 15 x 17.
x 4 x 4 4 x 4.
Then 4 15 4 15 0.1270, or 4 17 17 4
17 4 0.12.
0.1231; thus
120 5 1 1 120 5 1 4 120 6 1 x 1 30 x 20 or 20 x 30.
x
x
x
4 120 6
x 24 x 24 24 x 24.
Then 24 20 4, or 24 30 6; thus 4.
27. Step 1:
Step 2:
mx 2m 0.03 0.03 mx 2m 0.03 0.03 2m mx 0.03 2m 2 0.03
x 2 0.03
.
m
m
x 2 x 2 2 x 2.
Then 2 2 0.03
0.03
, or 2 2 0.03
0.03
. In either case, 0.03
.
m
m
m
m
m
28. Step 1:
Step 2:
mx 3m c c mx 3m c c 3m mx c 3m 3 mc x 3 mc
x 3 x 3 3 x 3.
Then 3 3 mc mc , or 3 3 mc mc . In either case, mc .
29. Step 1:
(mx b) m2 b c c mx m2 c c m2 mx c m2 12 mc x 12 mc .
Step 2:
x 12 x 12 12 x 12 .
Then 12 12 mc mc , or 12 12 mc mc . In either case, mc .
30. Step 1:
(mx b) (m b) 0.05 0.05 mx m 0.05 0.05 m mx 0.05 m
Step 2:
1 0.05
x 1 0.05
.
m
m
x 1 x 1 1 x 1.
Then 1 1 0.05
0.05
, or 1 1 0.05
0.05
. In either case, 0.05
.
m
m
m
m
m
31. lim (3 2 x) 3 2(3) 3
x 3
Step 1:
Step 2:
32.
(3 2 x) (3) 0.02 0.02 6 2 x 0.02 6.02 2 x 5.98 3.01 x 2.99 or
2.99 x 3.01.
0 x 3 x 3 3 x 3.
Then 3 2.99 0.01, or 3 3.01 0.01; thus 0.01.
lim (3 x 2) (3)(1) 2 1
x 1
Step 1:
Step 2:
(3 x 2) 1 0.03 0.03 3 x 3 0.03 0.01 x 1 0.01 1.01 x 0.99.
x (1) x 1 1 x 1.
Then 1 1.01 0.01, or 1 0.99 0.01; thus 0.01.
2
33. lim xx 24 lim
x 2
x 2
( x 2)( x 2)
lim ( x 2) 2 2 4, x 2
( x 2)
x 2
Copyright 2016 Pearson Education, Ltd.
60
Chapter 2 Limits and Continuity
Step 1:
Step 2:
34.
4 0.05 0.05
x2 4
x 2
1.95 x 2.05, x 2.
x 2 x 2 2 x 2.
Then 2 1.95 0.05, or 2 2.05 0.05; thus 0.05.
x 2 6x 5 lim (x 5)(x 1) lim ( x 1) 4, x 5.
x 5
(x 5)
x 5
x 5
x 5
2
( x 5)( x 1)
x
6
x
5
(4) 0.05 0.05 ( x 5) 4 0.05 4.05 x 1 3.95,
Step 1:
x 5
lim
Step 2:
35.
( x 2)( x 2)
4 0.05 3.95 x 2 4.05, x 2
( x 2)
lim
x 3
Step 1:
Step 2:
x 5
5.05 x 4.95, x 5.
x (5) x 5 5 x 5.
Then 5 5.05 0.05, or 5 4.95 0.05; thus 0.05.
1 5 x 1 5(3) 16 4
1 5 x 4 0.5 0.5 1 5 x 4 0.5 3.5 1 5 x 4.5 12.25 1 5 x 20.25
11.25 5 x 19.25 3.85 x 2.25.
x (3) x 3 3 x 3.
Then 3 3.85 0.85, or 3 2.25 0.75; thus 0.75.
36. lim 4x 24 2
x2
Step 1:
Step 2:
4 2 0.4 0.4 4 2 0.4 1.6 4 2.4 10 x 10 10 x 10 or 5 x 5 .
x
x
x
16
4
24
4
6
3
2
x 2 x 2 2 x 2.
Then 2 53 13 , or 2 52 12 ; thus 13 .
37. Step 1:
Step 2:
(9 x) 5 4 x 4 x 4 4 x 4 4 x 4 .
x 4 x 4 4 x 4.
Then 4 4 , or 4 4 . Thus choose .
38. Step 1:
Step 2:
(3x 7) 2 3x 9 9 3 x 9 3 3 x 3 3 .
x 3 x 3 3 x 3.
Then 3 3 3 3 , or 3 3 3 3 . Thus choose 3 .
39. Step 1:
Step 2:
x 5 2 x 5 2 2 x 5 2 (2 ) 2 x 5 (2 )2
(2 ) 2 5 x (2 ) 2 5.
x 9 x 9 9 x 9.
Then 9 2 4 9 4 2 , or 9 2 4 9 4 2 . Thus choose the smaller
distance, 4 2 .
40. Step 1:
Step 2:
4 x 2 4 x 2 2 4 x 2 (2 ) 2 4 x (2 ) 2
(2 ) 2 x 4 (2 )2 (2 )2 4 x (2 )2 4.
x 0 x .
Then (2 ) 2 4 2 4 4 2 , or (2 ) 2 4 4 2 . Thus choose the
smaller distance, 4 2 .
Copyright 2016 Pearson Education, Ltd.
Section 2.3 The Precise Definition of a Limit
41. Step 1:
For x 1, x 2 1 x 2 1 1 x 2 1 1 x 1
Step 2:
1 x 1 near x 1.
x 1 x 1 1 x 1.
Then 1 1 1 1 , or 1 1 1 1. Choose
61
min 1 1 , 1 1 , that is, the smaller of the two distances.
42. Step 1:
Step 2:
For x 2, x 2 4 x 2 4 4 x 2 4 4 x 4 4 x
4 near x 2.
x (2) x 2 2 x 2.
Then 2 4 4 2, or 2 4 2 4 . Choose
min
43. Step 1:
Step 2:
44. Step 1:
Step 2:
4 2, 2 4 .
1 1 1 1 1 1 1 1 x 1 .
x
x
x
1
1
x 1 x 1 1 x 1 .
Then 1 11 1 11 1 , or 1 11 11 1 1 .
Choose 1 , the smaller of the two distances.
1 1 1 1 1 1 1 13 1 1 3
3
3
3
3
x2 3
x2 3
x2
x2
2
133 x 133 133 x 133 , or 133 x 133 for x near
x 3 x 3 3 x 3 .
3 3
3 , or
13
13
Choose min 3 133 , 133 3 .
Then 3
45. Step 1:
Step 2:
46. Step 1:
Step 2:
47. Step 1:
Step 2:
48. Step 1:
Step 2:
3.
3
3
13
3
13
3.
(6) ( x 3) 6 , x 3 x 3 3 x 3.
x 2 9
x 3
x (3) x 3 3 x 3.
Then 3 3 , or 3 3 . Choose .
2 (x 1) 2 , x 1 1 x 1 .
x 2 1
x 1
x 1 x 1 1 x 1 .
Then 1 1 , or 1 1 . Choose .
x 1: (4 2 x) 2 0 2 2 x since x 1. Thus, 1 2 x 0;
x 1: (6 x 4) 2 0 6 x 6 since x 1. Thus, 1 x 1 6 .
x 1 x 1 1 x 1 .
Then 1 1 2 2 , or 1 1 6 6 . Choose 6 .
x 0: 2 x 0 2 x 0 2 x 0;
x 0: 2x 0 0 x 2.
x 0 x .
Then 2 2 , or 2 2. Choose 2 .
Copyright 2016 Pearson Education, Ltd.
62
Chapter 2 Limits and Continuity
49. By the figure, x x sin 1x x for all x 0 and x x sin 1x x for x 0. Since lim ( x) lim x 0, then by
x 0
the sandwich theorem, in either case, lim x sin 1x 0.
x 0
x 0
50. By the figure, x 2 x 2 sin 1x x 2 for all x except possibly at x 0. Since lim ( x 2 ) lim x 2 0, then by the
x 0
sandwich theorem, lim x 2 sin 1x 0.
x 0
x 0
51. As x approaches the value 0, the values of g ( x) approach k. Thus for every number 0, there exists a 0
such that 0 x 0 g ( x) k .
52. Write x h c. Then 0 x c x c , x c (h c) c , h c c h ,
h 0 0 h 0 .
Thus, lim f ( x) L for any 0, there exists 0 such that f ( x) L whenever 0 x c
x c
f (h c) L whenever 0 h 0 lim f (h c) L.
h 0
53. Let f ( x) x 2 . The function values do get closer to 1 as x approaches 0, but lim f ( x) 0, not 1. The
x 0
function f ( x) x 2 never gets arbitrarily close to 1 for x near 0.
54. Let f ( x) sin x, L 12 , and x0 0. There exists a value of x (namely x 6 ) for which sin x 12 for any
given 0. However, lim sin x 0, not 12 . The wrong statement does not require x to be arbitrarily close to x0 .
x 0
As another example, let g ( x) sin 1x , L 12 , and x0 0. We can choose infinitely many values of x near 0 such
that sin 1x 12 as you can see from the accompanying figure. However, lim sin 1x fails to exist. The wrong
x 0
statement does not require all values of x arbitrarily close to x0 0 to lie within 0 of L 12 . Again you can
see from the figure that there are also infinitely many values of x near 0 such that sin 1x 0. If we choose 14
we cannot satisfy the inequality sin 1x 12 for all values of x sufficiently near x0 0.
55.
2
πx
A 60 0.1 0.1 2x 60 0.1 59.9 4 60.1 π4 (59.9) x 2 π4 (60.1)
2
59.9 x 2 60.1
2
or 8.7331 x 8.7476. To be safe, the left endpoint was rounded up and
the right endpoint was rounded down.
56. V RI VR I VR 5 0.1 0.1 120
5 0.1 4.9 120
5.1 10
R 10
R
R
49 120 51
(120)(10)
(120)(10)
R 49 23.53 R 24.48.
51
To be safe, the left endpoint was rounded up and the right endpoint was rounded down.
Copyright 2016 Pearson Education, Ltd.
Section 2.3 The Precise Definition of a Limit
63
57. (a) x 1 0 1 x 1 f ( x) x. Then f ( x) 2 x 2 2 x 2 1 1. That is,
f ( x) 2 1 12 no matter how small is taken when 1 x 1 lim f ( x) 2.
x 1
(b) 0 x 1 1 x 1 f ( x) x 1. Then f ( x) 1 ( x 1) 1 x x 1. That is, f ( x) 1 1
no matter how small is taken when 1 x 1 lim f ( x) 1.
x 1
(c) x 1 0 1 x 1 f ( x) x. Then f ( x) 1.5 x 1.5 1.5 x 1.5 1 0.5.
Also, 0 x 1 1 x 1 f ( x) x 1. Then f ( x) 1.5 ( x 1) 1.5 x 0.5
x 0.5 1 0.5 0.5. Thus, no matter how small is taken, there exists a value of x such that
x 1 but f ( x) 1.5 12 lim f ( x) 1.5.
x 1
58. (a) For 2 x 2 h( x) 2 h( x) 4 2. Thus for 2, h( x) 4 whenever 2 x 2 no matter
how small we choose 0 lim h( x) 4.
x2
(b) For 2 x 2 h( x) 2 h( x) 3 1. Thus for 1, h( x) 3 whenever 2 x 2 no matter
how small we choose 0 lim h( x) 3.
x2
(c) For 2 x 2 h( x) x 2 so h( x) 2 x 2 2 . No matter how small 0 is chosen, x 2 is close to 4
when x is near 2 and to the left on the real line x 2 2 will be close to 2. Thus if 1, h( x) 2
whenever 2 x 2 no matter how small we choose 0 lim h( x) 2.
x2
59. (a) For 3 x 3 f ( x) 4.8 f ( x) 4 0.8. Thus for 0.8, f ( x) 4 whenever 3 x 3 no
matter how small we choose 0 lim f ( x) 4.
x 3
(b) For 3 x 3 f ( x) 3 f ( x) 4.8 1.8. Thus for 1.8, f ( x) 4.8 whenever 3 x 3
no matter how small we choose 0 lim f ( x) 4.8.
x 3
(c) For 3 x 3 f ( x) 4.8 f ( x) 3 1.8. Again, for 1.8, f ( x) 3 whenever 3 x 3 no
matter how small we choose 0 lim f ( x) 3.
x 3
60. (a) No matter how small we choose 0, for x near 1 satisfying 1 x 1 , the values of g ( x) are
near 1 g ( x) 2 is near 1. Then, for 12 we have g ( x) 2 12 for some x satisfying 1 x 1 ,
or 0 x 1 lim g ( x) 2.
x 1
(b) Yes, lim g ( x) 1 because from the graph we can find a 0 such that g ( x) 1 if 0 x (1) .
x 1
Copyright 2016 Pearson Education, Ltd.
64
Chapter 2 Limits and Continuity
61-66. Example CAS commands (values of del may vary for a specified eps):
Maple:
f : x - (x^4-81)/(x-3); x0 : 3;
.
plot( f (x), x x0-1..x0 1, color black,
title "Section 2.3, #61(a)" );
# (a)
L : limit( f (x), x x0 );
# (b)
epsilon : 0.2;
# (c)
plot( [f (x), L-epsilon,L epsilon], x x0-0.01..x0 0.01,
color black, linestyle [1,3,3], title "Section 2.3, #61(c)" );
q : fsolve( abs( f (x)-L ) epsilon, x x0-1..x0 1 );
# (d)
delta : abs(x0-q);
plot( [f (x), L-epsilon, L epsilon], x x0-delta..x0 delta, color black, title "Section 2.3, #61(d)" );
for eps in [0.1, 0.005, 0.001 ] do
# (e)
q : fsolve( abs( f (x)-L ) eps, x x0-1..x0 1 );
delta : abs(x0-q);
head : sprintf ("Section 2.3, #61(e)\n epsilon %5f , delta %5f \n", eps, delta );
print(plot( [f (x), L-eps, L eps], x x0-delta..x0 delta,
color black, linestyle [1,3,3], title head ));
end do:
Mathematica (assigned function and values for x0, eps and del may vary):
Clear[f , x]
y1: L eps; y2: L eps; x0 1;
f[x _]: (3x 2 (7x 1)Sqrt[x] 5)/(x 1)
Plot[f [x], {x, x0 0.2, x0 0.2}]
L: Limit[f [x], x x0]
eps 0.1; del 0.2;
Plot[{f [x], y1, y2}, {x, x0 del, x0 del}, PlotRange {L 2eps, L 2eps}]
Copyright 2016 Pearson Education, Ltd.
Section 2.4 One-Sided Limits
2.4
65
ONE-SIDED LIMITS
1. (a) True
(e) True
(i) False
(b) True
(f) True
(j) False
(c) False
(g) False
(k) True
(d) True
(h) False
(l) False
2. (a) True
(e) True
(i) True
(b) False
(f) True
(j) False
(c) False
(g) True
(k) True
(d) True
(h) True
3. (a)
lim f ( x) 22 1 2, lim f ( x) 3 2 1
x 2
x 2
(b) No, lim f ( x) does not exist because lim f ( x) lim f ( x)
x 2
x 2
(c)
x 2
lim f ( x ) 42 1 3, lim f ( x) 42 1 3
x 4
x 4
(d) Yes, lim f ( x) 3 because 3 lim f ( x) lim f ( x)
x 4
x 4
4. (a)
x 4
lim f ( x) 22 1, lim f ( x) 3 2 1, f (2) 2
x 2
x 2
(b) Yes, lim f ( x) 1 because 1 lim f ( x) lim f ( x )
x 2
x 2
(c)
lim
x 1
f ( x) 3 (1) 4, lim
x 1
x 2
f ( x) 3 (1) 4
(d) Yes, lim f ( x ) 4 because 4 lim
f ( x) lim
x 1
x 1
x 1
f ( x)
5. (a) No, lim f ( x) does not exist since sin 1x does not approach any single value as x approaches 0
(b)
x 0
lim f ( x) lim 0 0
x 0
x 0
(c) lim f ( x) does not exist because lim f ( x) does not exist
x 0
x 0
6. (a) Yes, lim g ( x) 0 by the sandwich theorem since x g ( x) x when x 0
x 0
(b) No, lim g ( x) does not exist since x is not defined for x 0
x 0
(c) No, lim g ( x) does not exist since lim g ( x ) does not exist
x 0
7. (a)
x 0
(b)
lim f ( x) 1 lim f ( x )
x 1
x 1
(c) Yes, lim f ( x) 1 since the right-hand and left-hand
x 1
limits exist and equal 1
Copyright 2016 Pearson Education, Ltd.
66
Chapter 2 Limits and Continuity
8. (a)
(b)
lim f ( x) 0 lim f ( x )
x 1
x 1
(c) Yes, lim f ( x) 0 since the right-hand and left-hand
x 1
limits exist and equal 0
9. (a) domain: 0 x 2
range: 0 y 1 and y 2
(b) lim f ( x) exists for c belonging to (0, 1) (1, 2)
x c
(c) x 2
(d) x 0
10. (a) domain: x
range: 1 y 1
(b) lim f ( x) exists for c belonging to
x c
(, 1) ( 1, 1) (1, )
(c) none
(d) none
11.
13.
14.
15.
lim
x 0.5
x2
x 1
0.5 2
0.51
3/2
1/2
lim
x 1
x 1
x2
11
1 2
0 0
2( 2) 5
x
2 x 5 2
(2) 12 1
x 1 x x 2 1 ( 2) ( 2)
x 2
lim
lim
x 1
lim
h 0
2
2
x11 xx 6 37 x 111 116 371 12 71 72 1
h2 4h 5 5
h
lim
h 0
h 0 h
lim
h 0
6 5h 2 11h 6
h
( h 4 h 5) 5
h 2 4 h 5 5 h 2 4h 5 5
2
lim
h
4
5
5
h
h
h
0
h
h2 4h 5 5
2
h ( h 4)
lim
16.
12.
3
h 4h 5 5
2
0 4 2
5 5
5
2
2
lim 6 5hh 11h 6 6 5h2 11h 6
h 0
6 5h 11h 6
lim
h 0 h
6 (5h 2 11h 6)
6 5h 2 11h 6
lim
h 0 h
h (5h 11)
6 5h 2 11h 6
Copyright 2016 Pearson Education, Ltd.
(0 11)
11
6 6
2 6
Section 2.4 One-Sided Limits
x2
( x 2)
lim ( x 3) x 2 lim ( x 3) ( x 2)
17. (a)
x 2
(|x 2| ( x 2) for x 2)
x 2
lim ( x 3) ((2) 3) 1
x 2
x2
( x 2)
lim ( x 3) x 2 lim ( x 3) ( x 2)
x 2
x 2
(b)
(|x 2| ( x 2) for x 2)
lim ( x 3)(1) (2 3) 1
x 2
18. (a)
lim
x 1
2 x ( x 1)
lim
x 1
x 1
lim
x 1
(b)
2 x ( x 1)
( x 1)
lim
x 1
2 x ( x 1)
lim
x 1
x 1
(|x 1| x 1 for x 1)
2x 2
2 x ( x 1)
( x 1)
(|x 1| ( x 1) for x 1)
lim 2 x 2
x 1
19. (a)
20. (a)
21.
lim 33 1
(b)
lim (t t ) 4 4 0
(b)
3
t 4
lim sin 2 lim sinx x 1
0
2
lim 23
3
lim (t t ) 4 3 1
t 4
(where x 2 )
x 0
kt lim k sin k lim sin k 1 k
22. lim sint kt lim k sin
kt
(where kt )
sin 3 y
3sin 3 y
sin 3 y
14 lim 3 y 34 lim 3 y 43 lim sin 43
4
y
0
y 0
y 0
y 0
(where 3y )
t 0
0
t 0
0
23. lim
24.
h 0
t 0
x
x 0
2t 2 lim
26. lim tan
t
t 0
sin 3 h
3h
(where 3h)
sin 2 x
cos
2x
lim sin 2 x lim
25. lim tanx2 x lim
x 0
1 1 1
sin
3
3
0
1
1
1
13 sin3h3h 13 hlim
3 lim
h 0
0
lim sinh3h lim
t
sin t
cos t
1 lim 2sin 2 x 1 2 2
cos
x 0 2 x x 0 2 x
x 0 x cos 2 x
cos t 2 lim cos t
1
2 lim tsin
lim sin t 2 1 1 2
t
t 0
t 0 t
t 0
x 1
12 lim sin2 x2 x lim cos15 x 12 1 (1) 12
x
x
sin
2
cos
5
x 0
x 0
x 0
csc 2 x lim
27. lim xcos
5x
x 0
2
6 x cos x lim 3cos x x 2 x
28. lim 6 x 2 (cot x)(csc 2 x) lim sin
3 11 3
x sin 2 x
sin x sin 2 x
x 0
x 0
x x cos x lim
29. lim sin
x cos x
x 0
x 0
x
x cos x
x 0 sin x cos x sin x cos x
x
xlim
x 1 xlim
0 sin x cos x
0 sin x
lim sin1 x lim cos1 x lim sin1 x (1)(1) 1 2
x 0 x x 0
x 0 x
Copyright 2016 Pearson Education, Ltd.
67
68
Chapter 2 Limits and Continuity
2
30. lim x x2xsin x lim
0 12 12 (1) 0
x 1 1 sin x
x
x 0 2 2 2
x 0
(1cos )(1 cos )
2
2
cos lim
cos
sin
31. lim 1sin
lim (2sin 1cos
lim (2sin cos
2
(2sin cos )(1 cos )
)(1 cos )
)(1 cos )
0
0
0
0
0 0
lim (2 cos sin
(2)(2)
)(1 cos )
0
32.
x (1 cos x )
lim x x 2cos x lim
lim
2
x 0 sin 3 x
x 0 sin 3 x
x 0
x (1 cos x )
9 x2
lim
sin 2 3 x
9 x2
x 0
1 cos x
9x
sin3 x3 x
2
sin(1 cos t )
lim sin 1 since 1 cos t 0 as t 0
1cos t
0
34. lim
sin(sin h )
lim sin 1 since sin h 0 as h 0
sin h
0
h 0
sin lim
35. lim sin
2
sin 5 x lim
36. lim sin
4x
lim
x 0
33. lim
t 0
1 lim 1 cos x
9 x0
x
sin3 x3 x
1 (0)
92 0
2
1
sin 2 1 lim sin 2
12 1 1 12
2 0
sin 2
0 sin 2 2
0
sin 5 x 4 x 5
x 0 sin 4 x 5 x 4
x 0
54 xlim
sin5 x5x sin4 x4 x 54 11 54
0
37. lim cos 0 1 0
0
2 lim sin cos 2
2 1
38. lim sin cot 2 lim sin cos
lim 2cos
sin 2
2sin cos
cos
2
0
0
0
0
xlim
sin 3x 1 8 x 3
0 cos 3 x sin 8 x 3 x 8
83 lim cos13 x sin3 x3 x sin8 x8 x 83 1 1 1 83
x0
3 x lim
39. lim tan
sin 8 x
sin 3 x 1
x 0 cos 3 x sin 8 x
x 0
1 1 1 1
sin 3 y cot 5 y
sin 3 y sin 4 y cos 5 y
sin 3 y
lim y cos 4 y sin 5 y lim
y
cot
4
y
y
y 0
y 0
y 0
40. lim
sin 3 y
3y
y 0
lim
sin 4 y
4y
5y
sin 5 y
sin 4 y
cos 4 y
cos 5 y
cos 4 y
sin
34
5
tan lim cos lim sin sin 3 lim sin
2
2
3
0 cot 3 0 cos
0 cos cos 3 0
sin 3
41. lim
2
cos 5 y
sin 5 y
345 y
345 y
12
5
12
5
sin33 cos 3cos 3 (1)(1) 113 3
4
2
cos
cos 4 (4sin 2 cos 2 )
cos 4 (2sin cos )2
cot 4 lim
sin 4
lim 2cos 42sin 2 lim
lim
2
2
2
2
2
2
2
0 sin cos 2 sin 4
0 sin cot 2 0 sin 2 cos 2 0 sin cos 2 sin 4 0 sin cos 2 sin 4
42. lim
sin 2 2
111 1
cos 4 cos2 1
cos 4 cos2
lim sin44
lim sin14
2
2
0 cos 2 sin 4
0
cos 2 0 4 cos 2 1
lim
4 cos 4 cos 2
2
2
2
43. Yes. If lim f ( x) L lim f ( x), then lim f ( x) L. If lim f ( x) lim f ( x), then lim f ( x) does not
exist.
x a
xa
x a
x a
x a
Copyright 2016 Pearson Education, Ltd.
x a
Section 2.4 One-Sided Limits
69
44. Since lim f ( x) L if and only if lim f ( x) L and lim f ( x) L, then lim f ( x) can be found by
x c
x c
calculating lim f ( x).
x c
x c
x c
45. If f is an odd function of x, then f ( x) f ( x). Given lim f ( x) 3, then lim f ( x) 3.
x 0
x 0
46. If f is an even function of x, then f ( x) f ( x). Given lim f ( x) 7 then lim
can be said about lim
x 2
x 2
f ( x) because we don’t know lim f ( x).
x 2
f ( x) 7. However, nothing
x 2
47. I (5, 5 ) 5 x 5 . Also, x 5 x 5 2 x 5 2 . Choose 2 lim
x 5 0.
48. I (4 , 4) 4 x 4. Also, 4 x 4 x 2 x 4 2 . Choose 2 lim
4 x 0.
x 5
x 4
49. As x 0 the number x is always negative. Thus, x (1) xx 1 0 which is always true
x
x 1.
x 0 x
independent of the value of x. Hence we can choose any 0 with x 0 lim
50. Since x 2 we have x 2 and x 2 x 2. Then, x 2 1 xx 22 1 0 which is always true so
x2
long as x 2. Hence we can choose any 0, and thus 2 x 2 x 2 1 . Thus, lim
x 2 1.
x 2 x 2
x2
51. (a)
(b)
lim
x 400
x 400. Just observe that if 400 x 401, then x 400. Thus if we choose 1, we have for
any number 0 that 400 x 400 x 400 400 400 0 .
lim x 399. Just observe that if 399 x 400 then x 399. Thus if we choose 1, we have for
x 400
any number 0 that 400 x 400 x 399 399 399 0 .
(c) Since lim x lim x we conclude that lim x does not exist.
x 400
52. (a)
x 400
lim f ( x) lim
x 0
x 0
x 0 0;
x 400
x 0 x 0 x 2 for x positive. Choose 2
lim f ( x) 0.
x 0
(b)
lim f ( x) lim x 2 sin 1x 0 by the sandwich theorem since x 2 x 2 sin 1x x 2 for all x 0.
x 0
x 0
Since x 2 0 x 2 0 x 2 whenever x , we choose and obtain x 2 sin 1x 0
if x 0.
(c) The function f has limit 0 at x0 0 since both the right-hand and left-hand limits exist and equal 0.
Copyright 2016 Pearson Education, Ltd.
70
2.5
Chapter 2 Limits and Continuity
CONTINUITY
1. No, discontinuous at x 2, not defined at x 2
2. No, discontinuous at x 3, 1 lim g ( x) g (3) 1.5
x 3
3. Continuous on [1, 3]
4. No, discontinuous at x 1, 1.5 lim k ( x) lim k ( x) 0
x 1
x 1
5. (a) Yes
(b) Yes, lim
(c) Yes
(d) Yes
6. (a) Yes, f (1) 1
x 1
f ( x) 0
(b) Yes, lim f ( x) 2
(c) No
(d) No
7. (a) No
(b) No
x 1
8. [1, 0) (0, 1) (1, 2) (2, 3)
9. f (2) 0, since lim f ( x) 2(2) 4 0 lim f ( x)
x 2
x 2
10. f (1) should be changed to 2 lim f ( x )
x 1
11. Nonremovable discontinuity at x 1 because lim f ( x) fails to exist ( lim f ( x) 1 and lim f ( x) 0).
x 1
x 1
x 1
Removable discontinuity at x 0 by assigning the number lim f ( x) 0 to be the value of f (0) rather
x 0
than f (0) 1.
12. Nonremovable discontinuity at x 1 because lim f ( x) fails to exist ( lim f ( x) 2 and lim f ( x) 1).
x 1
x 1
x 1
Removable discontinuity at x 2 by assigning the number lim f ( x) 1 to be the value of f (2) rather than
x 2
f (2) 2.
13. Discontinuous only when x 2 0 x 2
14. Discontinuous only when ( x 2)2 0 x 2
15. Discontinuous only when x 2 4 x 3 0 ( x 3)( x 1) 0 x 3 or x 1
16. Discontinuous only when x 2 3 x 10 0 ( x 5)( x 2) 0 x 5 or x 2
17. Continuous everywhere. (|x 1| sin x defined for all x; limits exist and are equal to function values.)
18. Continuous everywhere. (|x| 1 0 for all x; limits exist and are equal to function values.)
Copyright 2016 Pearson Education, Ltd.
Section 2.5 Continuity
71
19. Discontinuous only at x 0
20. Discontinuous at odd integer multiples of 2 , i.e., x (2n 1) 2 , n an integer, but continuous at all other x.
21. Discontinuous when 2x is an integer multiple of , i.e., 2 x n , n an integer x n2 , n an integer, but
continuous at all other x.
22. Discontinuous when 2x is an odd integer multiple of 2 , i.e., 2x (2n 1) 2 , n an integer x 2n 1, n an
integer (i.e., x is an odd integer). Continuous everywhere else.
23. Discontinuous at odd integer multiples of 2 , i.e., x (2n 1) 2 , n an integer, but continuous at all other x.
24. Continuous everywhere since x 4 1 1 and 1 sin x 1 0 sin 2 x 1 1 sin 2 x 1; limits exist and are
equal to the function values.
25. Discontinuous when 2 x 3 0 or x 32 continuous on the interval 32 , .
26. Discontinuous when 3 x 1 0 or x 13 continuous on the interval 13 , .
27. Continuous everywhere: (2 x 1)1/3 is defined for all x; limits exist and are equal to function values.
28. Continuous everywhere: (2 x)1/5 is defined for all x; limits exist and are equal to function values.
2
( x 3)( x 2)
lim ( x 2) 5 g (3)
x 3
x 3
x 3
29. Continuous everywhere since lim x xx36 lim
x 3
30. Discontinuous at x 2 since lim f ( x) does not exist while f (2) 4.
x 2
31. lim sin( x sin x) sin( sin ) sin( 0) sin 0, and function continuous at x .
x
32. lim sin( 2 cos(tan t )) sin( 2 cos(tan(0))) sin 2 cos(0) sin 2 1, and function continuous at t 0.
t 0
33. lim sec ( y sec2 y tan 2 y 1) lim sec ( y sec2 y sec2 y ) lim sec (( y 1) sec2 y ) sec ((1 1)sec 2 1)
y 1
y 1
y 1
sec 0 1, and function continuous at y 1.
34. lim tan 4 cos(sin x1/3 ) tan 4 cos(sin(0)) tan 4 cos(0) tan 4 1, and function continuous at x 0.
x 0
35. lim cos
cos
cos cos 4 22 , and function continuous at t 0.
19
3
sec
2
t
19
3
sec
0
16
t 0
36. lim
x 6
csc2 x 5 3 tan x csc2 6 5 3 tan 6 4 5 3
9 3, and function continuous
1
3
at x 6 .
Copyright 2016 Pearson Education, Ltd.
72
37.
Chapter 2 Limits and Continuity
cos2 x cos x
cos x 1
sin lim (cos x )
x
x
x 0
lim sin
x 0
cos x 1
sin lim cos x lim
x 0
x
x 0
sin(1 0) 0 (See Example 5 in Section 2.4.)
The function is not defined at x 0 and thus is not continuous there.
38.
sin 2 x
sin x
2 1
(sin 2 x sin x )
lim sec
lim
sec lim
sec 3 3 sec 3 2
3x
x 0 3x
x 0 3x
x 0
(See Theorem 7 in Section 2.4.)
The function is not defined at x 0 and thus is not continuous there.
2
39. g ( x) xx 39
( x 3)( x 3)
x 3, x 3 g (3) lim ( x 3) 6
( x 3)
x 3
2
40. h(t ) t t3t210
3
41. f ( s ) s3 1
s 1
(t 5)(t 2)
t 5, t 2 h(2) lim (t 5) 7
t 2
t 2
2
2
( s 2 s 1)( s 1)
s ss11 , s 1 f (1) lim s ss11
( s 1)( s 1)
s 1
3
2
2
( x 4)( x 4)
42. g ( x) 2x 16 ( x 4)( x 1) xx14 , x 4 g (4) lim xx14 85
x 3 x 4
x 4
43. As defined, lim f ( x ) (3)2 1 8 and lim (2a )(3) 6a. For f ( x) to be continuous we must have
x 3
6a 8 a 43 .
x 3
44. As defined, lim g ( x) 2 and lim g ( x ) b(2)2 4b. For g ( x) to be continuous we must have
x 2
4b 2 b 12 .
x 2
45. As defined, lim f ( x) 12 and lim f ( x ) a 2 (2) 2a 2a 2 2a. For f ( x ) to be continuous we must have
2
x 2
x 2
12 2a 2a a 3 or a 2.
46. As defined, lim g ( x) 0bb1 bb1 and lim g ( x) (0) 2 b b. For g ( x) to be continuous we must have
x 0
x 0
b b b 0 or b 2.
b 1
47. As defined, lim
x 1
f ( x) 2 and lim
x 1
f ( x) a (1) b a b, and lim f ( x) a (1) b a b and
x 1
lim f ( x) 3. For f ( x) to be continuous we must have 2 a b and a b 3 a 52 and b 12 .
x 1
48. As defined, lim g ( x) a (0) 2b 2b and lim g ( x) (0)2 3a b 3a b, and lim g ( x) (2)2 3a b
x 0
x 0
x 2
4 3a b and lim g ( x) 3(2) 5 1. For g ( x) to be continuous we must have 2b 3a b and 4 3a b 1
x 0
a 32 and b 32 .
Copyright 2016 Pearson Education, Ltd.
Section 2.5 Continuity
49. The function can be extended: f (0)
73
50. The function cannot be extended to be continuous at
x 0. If f (0) 2.3, it will be continuous from the
right. Or if f (0) 2.3, it will be continuous from
the left.
2.3.
51. The function cannot be extended to be continuous
at x 0. If f (0) 1, it will be continuous from the
right. Or if f (0) 1, it will be continuous from
the left.
52. The function can be extended: f (0)
7.39.
53. f ( x) is continuous on [0, 1] and f (0) 0, f (1) 0
by the Intermediate Value Theorem f ( x) takes on
every value between f (0) and f (1) the equation
f ( x ) 0 has at least one solution between x 0
and x 1.
54. cos x x (cos x) x 0. If x 2 , cos 2 2 0. If x 2 , cos 2 2 0. Thus cos x x 0 for
some x between 2 and 2 according to the Intermediate Value Theorem, since the function cos x x is
continuous.
55. Let f ( x) x3 15 x 1, which is continuous on [4, 4]. Then f (4) 3, f (1) 15, f (1) 13, and f (4) 5.
By the Intermediate Value Theorem, f ( x ) 0 for some x in each of the intervals 4 x 1, 1 x 1, and
1 x 4. That is, x3 15 x 1 0 has three solutions in [4, 4]. Since a polynomial of degree 3 can have at
most 3 solutions, these are the only solutions.
56. Without loss of generality, assume that a b. Then F ( x) ( x a )2 ( x b)2 x is continuous for all values of x,
so it is continuous on the interval [a, b]. Moreover F (a ) a and F (b) b. By the Intermediate Value Theorem,
since a a 2 b b, there is a number c between a and b such that F ( x) a 2 b .
Copyright 2016 Pearson Education, Ltd.
74
Chapter 2 Limits and Continuity
57. Answers may vary. Note that f is continuous for every value of x.
(a) f (0) 10, f (1) 13 8(1) 10 3. Since 3 10, by the Intermediate Value Theorem, there exists a c so
that 0 c 1 and f (c) .
(b) f (0) 10, f (4) (4)3 8(4) 10 22. Since 22 3 10, by the Intermediate Value Theorem,
there exists a c so that 4 c 0 and f (c) 3.
(c) f (0) 10, f (1000) (1000)3 8(1000) 10 999, 992, 010. Since 10 5, 000, 000 999,992, 010, by the
Intermediate Value Theorem, there exists a c so that 0 c 1000 and f (c) 5, 000, 000.
58. All five statements ask for the same information because of the intermediate value property of continuous
functions.
(a) A root of f ( x ) x3 3 x 1 is a point c where f (c) 0.
(b) The point where y x3 crosses y 3x 1 have the same y-coordinate, or y x3 3 x 1 f ( x)
x3 3 x 1 0.
(c) x3 3x 1 x3 3 x 1 0. The solutions to the equation are the roots of f ( x ) x3 3 x 1.
(d) The points where y x3 3 x crosses y 1 have common y-coordinates, or y x3 3 x 1 f ( x)
x3 3 x 1 0.
(e) The solutions of x3 3x 1 0 are those points where f ( x ) x3 3 x 1 has value 0.
sin( x 2)
59. Answers may vary. For example, f ( x) x 2 is discontinuous at x 2 because it is not defined there.
However, the discontinuity can be removed because f has a limit (namely 1) as x 2.
60. Answers may vary. For example, g ( x) x11 has a discontinuity at x 1 because lim g ( x) does not exist.
x 1
lim g ( x) and lim g ( x) .
x 1
x 1
61. (a) Suppose x0 is rational f ( x0 ) 1. Choose 12 . For any 0 there is an irrational number x (actually
infinitely many) in the interval ( x0 , x0 ) f ( x) 0. Then 0 |x x0 | but | f ( x) f ( x0 )|
1 12 , so lim f ( x) fails to exist f is discontinuous at x0 rational.
x x0
On the other hand, x0 irrational f ( x0 ) 0 and there is a rational number x in ( x0 , x0 ) f ( x) 1.
Again lim f ( x) fails to exist f is discontinuous at x0 irrational. That is, f is discontinuous at every point.
x x0
(b) f is neither right-continuous nor left-continuous at any point x0 because in every interval ( x0 , x0 ) or
( x0 , x0 ) there exist both rational and irrational real numbers. Thus neither limits lim f ( x) and
x x0
lim f ( x) exist by the same arguments used in part (a).
x x0
f ( x)
62. Yes. Both f ( x) x and g ( x) x 12 are continuous on [0, 1]. However g ( x ) is undefined at x 12 since
f ( x)
g 12 0 g ( x ) is discontinuous at x 12 .
63. No. For instance, if f ( x) 0, g ( x ) x , then h( x) 0 x 0 is continuous at x 0 and g ( x) is not.
64. Let f ( x ) x11 and g ( x) x 1. Both functions are continuous at x 0. The composition f g f ( g ( x))
1
1 is discontinuous at x 0, since it is not defined there. Theorem 10 requires that f ( x ) be continuous
( x 1) 1 x
at g (0), which is not the case here since g (0) 1 and f is undefined at 1.
Copyright 2016 Pearson Education, Ltd.
Section 2.5 Continuity
75
65. Yes, because of the Intermediate Value Theorem. If f (a ) and f (b) did have different signs then f would have
to equal zero at some point between a and b since f is continuous on [a, b].
66. Let f ( x ) be the new position of point x and let d ( x) f ( x) x. The displacement function d is negative if x is
the left-hand point of the rubber band and positive if x is the right-hand point of the rubber band. By the
Intermediate Value Theorem, d ( x) 0 for some point in between. That is, f ( x) x for some point x,
which is then in its original position.
67. If f (0) 0 or f (1) 1, we are done (i.e., c 0 or c 1 in those cases). Then let f (0) a 0 and f (1) b 1
because 0 f ( x) 1. Define g ( x) f ( x ) x g is continuous on [0, 1]. Moreover, g (0) f (0) 0 a 0 and
g (1) f (1) 1 b 1 0 by the Intermediate Value Theorem there is a number c in (0, 1) such that
g (c) 0 f (c) c 0 or f (c) c.
f (c )
68. Let 2 0. Since f is continuous at x c there is a 0 such that x c f ( x) f (c)
f (c) f ( x) f (c) .
If f (c) 0, then 12 f (c ) 12 f (c) f ( x) 32 f (c) f ( x) 0 on the interval (c , c ).
If f (c) 0, then 12 f (c) 32 f (c) f ( x) 12 f (c) f ( x) 0 on the interval (c , c ).
69. By Exercise 52 in Section 2.3, we have lim f ( x ) L lim f (c h ) L.
x c
h0
Thus, f ( x ) is continuous at x c lim f ( x ) f ( c ) lim f ( c h ) f ( c ).
x c
h 0
70. By Exercise 69, it suffices to show that lim sin(c h) sin c and lim cos(c h) cos c.
h 0
h 0
Now lim sin(c h) lim (sin c)(cos h) (cos c )(sin h) (sin c ) lim cos h (cos c ) lim sin h .
h 0
h 0
h 0
h 0
By Example 11 Section 2.2, lim cos h 1 and lim sin h 0. So lim sin(c h) sin c and thus f ( x) sin x is
h 0
continuous at x c. Similarly,
h 0
h 0
lim cos(c h) lim (cos c)(cos h) (sin c)(sin h) (cos c) lim cos h (sin c) lim sin h cos c. Thus,
h 0
h 0
h 0
g ( x) cos x is continuous at x c.
h 0
71. x 1.8794, 1.5321, 0.3473
72. x 1.4516, 0.8547, 0.4030
73. x 1.7549
74. x
3.5156
75. x
76. x
1.8955, 0, 1.8955
0.7391
Copyright 2016 Pearson Education, Ltd.
76
2.6
Chapter 2 Limits and Continuity
LIMITS INVOLVING INFINITY; ASYMPTOTES OF GRAPHS
1. (a) lim f ( x) 0
(c)
(e)
x 2
lim
x 3
(b)
f ( x) 2
lim f ( x) 1
(f)
x 0
(h)
x 0
lim f ( x) 0
x
2. (a) lim f ( x) 2
(c)
(e)
(g)
(i)
(b)
x 4
lim f ( x) 1
x 2
(f)
lim f ( x)
(h)
lim f ( x)
(j)
x 3
x 3
lim f ( x)
x 0
lim f ( x) 1
x
lim f ( x ) 3
x 2
(d) lim f ( x) does not exist
f ( x)
lim
f ( x) 2
(d) lim f ( x) does not exist
(g) lim f ( x) does not exist
(i)
lim
x 3
x 3
x 0
(k) lim f ( x) 0
(l)
x
Note: In these exercises we use the result lim
x x
1
m/ n
Theorem 8 and the power rule in Theorem 1: lim
x 2
lim
x 3
f ( x)
lim f ( x)
x 0
lim f ( x) does not exist
x 0
lim f ( x) 1
x
0 whenever mn 0. This result follows immediately from
1
m/n
x x
1
x x
lim
3. (a) 3
(b) 3
4. (a)
(b)
5. (a) 12
(b) 12
6. (a) 18
(b) 18
7. (a) 53
(b) 53
8. (a) 34
(b) 34
m/ n
lim 1x
x
9. 1x sinx2 x 1x lim sinx2 x 0 by the Sandwich Theorem
x
1 lim cos 0 by the Sandwich Theorem
10. 31 cos
3
3
3
t
r
12. lim 2 r r 7 sin5sin
lim
r
r
010 1
10
2 1 sin t
t
cos t
t 1 t
sin t lim t
11. lim 2tt cos
t
lim 10 1
r 200 2
1 sinr r
7
sin r
r 2 r 5 r
Copyright 2016 Pearson Education, Ltd.
m/ n
0m / n 0.
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
13. (a)
14. (a)
lim 52xx73 lim
2 3x
7
x 5 x
x
52
2 73
3
2
x
7
lim 3 2
lim 1 1x 7 2
x x x x 7 x 1 x 2 3
(b) 2 (same process as part (a))
15. (a)
2 (same process as part (a))
5
(b)
x
x
1 1
x 1 lim x x 2 0
2
3
x x 3 x 1 2
lim
(b) 0 (same process as part (a))
x
16. (a)
17. (a)
lim 3 x2 7 lim
3 7
x x2
0
2
x 1 2
x x 2
(b) 0 (same process as part (a))
x
7 x3
lim 37 9 7
2
x x 3 x 6 x x 1 x 2
(b) 7 (same process as part (a))
9 13
9 x4 x
x
lim
92
4
2
5
6
1
x 2 x 5 x x 6 x 2 2 3 4
(b)
lim
3
x
18. (a)
lim
x
19. (a)
20. (a)
(b)
21. (a)
(b)
22. (a)
(b)
23.
5
10 1 31
x x 2 x6
4
lim 10 x 6x 31 lim
x
x
x
x
0
1
x
9 (same process as part (a))
2
(b) 0 (same process as part (a))
1
lim x 2 7 x 2 lim x 71 2 x 2 , since x n 0 and x 7 .
3
x
2
x x 1
x 1 x x
3
2
1
lim x 2 7 x 2 lim x 71 2 x 2 , since x n 0 and x 7 .
x x 1
x
x 1 x x
1
3
lim 3 x 35 x 1 lim 3 x 5x2 x3 , since x n 0 and 3x 4 .
7
2
4
x 6 x 7 x 3
x 67 x 3 x
7
2
4
1
3
lim 3 x 35 x 1 lim 3 x 5x2 x3 , since x n 0 and 3x 4 .
x 6 x 7 x 3
8
x 67 x 3 x
3
2
3
lim 5 x 2 x 5 9 lim 5 x 25 x 49 x
x 3 x 4 x
5
3 x x 4
x
, since x n 0, 5x 3 , and the denominator 4.
8
3
3
2
5
lim 5 x 2 x 5 9 lim 5 x 25 x 49 x , since x n 0, 5x 3 , and the denominator 4.
3
x
4
x
3
x
x
4
x
x
8 x 2 3 lim
2 x2 x
x
lim
x
24.
2
lim x 2 x 1
8
x 3
x
25.
3
lim 12 x
x x 7 x
1/3
5
8 32
x
2 1x
8 32
x
1
lim
x 2 x
1/3
1 1x 12
x
lim
3
x 8 2
x
5
8 0
2 0
42
1/3
1 1x 12
x
lim
x 8 32
x
5
18000
1/3
18
1 x
12 x
5
2
lim x 7 lim x 7 010
x 1 x
x 1 x
Copyright 2016 Pearson Education, Ltd.
1/3
12
77
78
Chapter 2 Limits and Continuity
x 2 5 x lim
x3 x 2
x
26. lim
x
27.
1 5
x x2
1 12 23
x
1 5
x x2
lim
1
2
x
x
x 1 2 3
x
00
1 0 0
2 1
1/ 2 2
2 x x 1
lim 3 x 7 lim x 7 x 0
3
x
x
28.
x
29.
2 1
1/ 2
2 x
lim
lim x 1
x 2 x
x 2 1
x1/ 2
1 21
x 5 x
x /15
1 x (1/5) (1/3) lim
lim
1
(1/5) (1/3)
3
5
x x x
x 1 x
x 1 1
x 2 /15
3
lim
1
4
30. lim x2 x 3 lim
5/3
x 12
1/3
31. lim 2 x8/5 x
7 lim
3 x x
x
x x
x
1
x 1 x
x x x
32.
0 0
1 7
2 x1/15 19/15
8/5
x
x
3
1
1 3/5
11/10
x
x
1 5 3
x
x 5 x 3
x 2/3
lim
52
2/3
1
4
x 2 x x 4 x 2 1/3 x
3
lim
x
33.
34.
35.
36.
37.
39.
41.
2
2
x 2 1
lim x 1/ x2 lim
x 1
x ( x 1)/ x
x
lim
x
( x 2 1)/ x 2
1/ x 2
lim (111/
(110)0 1
( x 1)/ x
x)
x
lim
( x 2 1)/ x 2
x 2 1
x 2 1/ x 2
11/ x 2
lim
lim
lim
( 1100) 1
x 1
x ( x 1)/ x 2
x ( x 1)/( x )
x ( 11/ x )
lim
x 3
x
2
lim
x
4 x 25
lim
4 3 x 3
6
x
x 9
x
lim
x
( x 3)/ x 2
2
4 x 25 / x
2
6
x 9 / x
lim
6
x
43. lim
45. (a)
lim
2
1/3
x 0 3 x
6
( x 9)/ x
6
(13/ x )
x
lim
x
4 25/ x
19/ x
negative
positive
42.
positive
positive
44. lim
(b)
2
( 4/ x3 3)
40.
4
2
x 7 ( x 7)
x 8
(43 x3 )/( x3 )
lim
2 x
x 8
(4 x 25)/ x
2
positive
negative
lim
2
38.
3
x 2
lim
( x 3)/ x
positive
positive
x 2
x
(4 3 x3 )/ x 6
lim 31x
x 0
lim
6
(10)
12
4 0
(0 3)
3
1 0
lim
5
2x
positive
negative
lim
1
x 3
positive
positive
3x
2 x 10
negative
negative
1
2
x 0 x ( x1)
negative
positivepositive
x 0
x 3
lim
x 5
lim
2
1/3
x 0 3 x
Copyright 2016 Pearson Education, Ltd.
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
46. (a)
4
47. lim
x 0 x
49.
51.
52.
2
lim
1/5
x 0 x
2/5
(b)
4
lim
1/5 2
x 0 ( x
)
1/5
48. lim
2/3
1
x 0 x
lim tan x
x 2
2
lim
x 0 x
50.
lim
x 2
lim
1
1/3 2
x 0 ( x
)
sec x
lim (1 csc )
0
lim (2 cot ) and lim (2 cot ) , so the limit does not exist
0
53. (a)
(b)
(c)
(d)
54. (a)
(b)
(c)
(d)
0
1
lim
2
lim
2
x 2 x 4
1
x 2 x 4
lim
1
( x 2)( x 2)
lim
1
( x 2)( x 2)
x 2
1
lim
2
lim
2
x 2 x 4
1
x 2 x 4
x 2
lim
1
( x 2)( x 2)
lim
1
( x 2)( x 2)
x 2
x 2
lim 2x lim ( x 1)(x x 1)
x 1 x 1 x 1
lim 2x lim ( x 1)(x x 1)
x 1 x 1 x 1
x lim
x
2
( x 1)( x 1)
x 1 x 1 x 1
lim 2x lim ( x 1)(x x 1)
x 1 x 1 x 1
lim
x 0
x2
2
1
x
x 0
x2
2
1
x
x2
2
1
x
22/3
2
x2
x 1 2
1
x
1
2
lim
3
x 2
56. (a)
(c)
(d)
lim
x 2
x 0
x 2 1
2 x4
2
lim 2xx 14 lim
x 1
x 0
x 1
2
x
1
lim 2 x 4 41
x 0
1
1
1
positivenegative
1
negativenegative
positive
positivenegative
negative
positivenegative
negative
negativenegative
1
x
1
negative
1
x
1
positive
1
21/3
1
positivenegative
positive
positivepositive
0 lim
(b) lim 0 lim
(c) lim
2
2
(d) lim
55. (a)
1
positivepositive
1/3
1/3
0
3
2
positive
positive
( x 1)( x 1)
2204 0
2 x4
(b)
lim
x 2
x 2 1
2 x4
Copyright 2016 Pearson Education, Ltd.
positive
negative
79
80
Chapter 2 Limits and Continuity
57. (a)
lim
x 0
x 2 3 x 2 lim ( x 2)( x1)
2
x3 2 x 2
x 0 x ( x 2)
negativenegative
positivenegative
(b)
x 2 3 x 2 lim ( x 2)( x 1) lim x 1 1 , x 2
3
2
2
2
4
x 2 x 2 x
x 2 x ( x 2)
x 2 x
(c)
lim
lim
x 2
x 2 3 x 2 lim ( x 2)( x 1) lim x 1 1 , x 2
2
2
4
x3 2 x 2
x 2 x ( x 2)
x 2 x
(d) lim x 33 x 22 lim
( x 2)( x 1)
x 2
x 2 ( x 2)
2
(e) lim x 33 x 22 lim
( x 2)( x 1)
x 0 x 2 x
x 2 ( x 2)
2
x 2 x 2 x
58. (a)
lim
x 2
x 0
lim x 21 14 , x 2
x 2 x
negativenegative
positivenegative
( x 2)( x 1)
( x 1)
x 2 3 x 2 lim
lim x ( x 2)
3
x ( x 2)( x 2)
x 2 x 4 x
x 2
x 2
(c)
lim
(e)
x 2 3 x 2 lim ( x 2)( x 1) lim ( x 1) 1 1
x ( x 2)( x 2)
x ( x 2)
2(4) 8
x3 4 x
x 2
x 2
(b)
(d)
lim
x 0
x 2 3 x 2 lim ( x 2)( x 1) lim ( x 1)
x ( x 2)( x 2)
x ( x 2)
x3 4 x
x 0
x 0
( x 2)( x 1)
2
negative
negativepositive
negative
negativepositive
( x 1)
0 0
lim x 3 3 x 2 lim x ( x 2)( x 2) lim x ( x 2) (1)(3)
x 1
x 4 x
x 1
lim x (xx12)
x 0
and lim
x 0
x 1
x ( x 2)
x 1
negative
positivepositive
negative
negativepositive
so the function has no limit as x 0.
59. (a)
60. (a)
61. (a)
(c)
62. (a)
(c)
3
lim 2 1/3
t
(b)
1 7
lim 3/5
t
t 0
(b)
1
lim 2/3
(b)
t 0
x 0 x
2
( x 1) 2/3
1
lim 2/3
2 2/3
x
(
x
1)
x 1
(d)
1
lim 1/3
(b)
x 0 x
1
( x 1)4/3
1
lim 1/3
1 4/3
x
(
x
1)
x 1
63. y x11
(d)
3
lim 2 1/3
t
t 0
1 7
lim 3/5
t
t 0
1
lim 2/3
x 0 x
2
( x 1)2/3
1
lim 2/3
2 2/3
x
(
x
1)
x 1
1
lim 1/3
x 0 x
1
( x 1) 4/3
1
lim 1/3
1 4/3
x
(
x
1)
x 1
64. y x11
Copyright 2016 Pearson Education, Ltd.
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
65. y 2 x1 4
66. y x33
67. y xx32 1 x 1 2
68. y x2x1 2 x21
69. Here is one possibility.
70. Here is one possibility.
71. Here is one possibility.
72. Here is one possibility.
Copyright 2016 Pearson Education, Ltd.
81
82
Chapter 2 Limits and Continuity
73. Here is one possibility.
74. Here is one possibility.
75. Here is one possibility.
76. Here is one possibility.
f ( x)
f ( x)
2 as well.
g
x ( x )
77. Yes. If lim g ( x ) 2 then the ratio the polynomials’ leading coefficients is 2, so lim
x
78. Yes, it can have a horizontal or oblique asymptote.
f ( x)
79. At most 1 horizontal asymptote: If lim g ( x ) L, then the ratio of the polynomials’ leading coefficients is L,
x
f ( x)
L as well.
x g ( x )
so lim
80.
lim
x
x 9 x 4 x 9 x 4 lim ( x 9) ( x 4)
x 9 x 4 xlim
x 9 x 4 x x 9 x 4
5
lim
x
x
5
lim
101 0
x 9 x 4
x 1 9x 1 4x
2
2
( x 2 25) ( x 2 1)
81. lim x 2 25 x 2 1 lim x 2 25 x 2 1 x 2 25 x 2 1 lim
x
x 25 x 1 x x 2 25 x 2 1
x
26
lim
x 2 25 x 2 1
x
82.
lim x 2 3 x lim x 2 3 x
x
x
x2
x 1 32 x
x
x2
x 1 252 1 12
x
x 2 3 x
lim
x 2 3 x x
3
lim
26
x
lim
lim
3x
x 1 32 1
101 0
x
( x 2 3) ( x 2 )
x 2 3 x
101 0
x
Copyright 2016 Pearson Education, Ltd.
lim
x
3
x 2 3 x
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
83.
83
2
(4 x 2 ) (4 x 2 3 x 2)
lim 2 x 4 x 2 3 x 2 lim 2 x 4 x 2 3x 2 2 x 4 x 2 3 x 2 lim
2 x 4 x 3 x 2 x 2 x 4 x 2 3 x 2
x
x
3 x 2
lim
x 2 x 4 x 2 3 x 2
3 2x
lim
x 2 4 3x 22
lim
x
3 x 2
x2
2x
x2
4 3x 22
x
lim
x 2 xx
3 x 2
x
4 3x 22
x
3202 43
x
84.
2
(9 x 2 x ) (9 x 2 )
x
lim 9 x 2 x 3 x lim 9 x 2 x 3 x 9 x 2 x 3 x lim
lim
x
9 x x 3 x x 9 x 2 x 3 x
x
x 9 x 2 x 3 x
lim
x
85.
xx
9 x2
x 3xx
x2 x2
x
1
1 16
9 1x 3 3 3
2
2
( x 2 3 x ) ( x 2 2 x )
lim x 2 3 x x 2 2 x lim x 2 3x x 2 2 x x 2 3 x x 2 2 x lim
x
x 3 x x 2 x x x 2 3 x x 2 2 x
x
5
x
5
lim
lim
151 52
2
2
3
x
86.
lim
x 3 x x 2 x
x
1 x 1 2x
2
2
( x 2 x ) ( x 2 x )
2x
x 2 x x 2 x lim x 2 x x 2 x x 2 x x 2 x lim
lim
x x x x x x 2 x x 2 x x x 2 x x 2 x
x
x
2
lim
121 1
1
1
lim
x 1 x 1 x
87. For any 0, take N 1. Then for all x N we have that f ( x) k k k 0 .
88. For any 0, take N 1. Then for all y N we have that f ( x) k k k 0 .
89. For every real number B 0, we must find a 0 such that for all x, 0 x 0 21 B.
x
Now, 12 B 0 12 B 0 x 2 B1 x 1 . Choose 1 , then 0 x x 1
x
x
21 B so that lim 12 .
x
x0
B
B
B
x
90. For every real number B 0, we must find a 0 such that for all x, 0 x 0 1 B. Now,
x
1 B 0 x 1 . Choose 1 . Then 0 x 0 x 1 1 B so that lim 1 .
B
B
B
x
x
x0 x
91. For every real number B 0, we must find a 0 such that for all x, 0 x 3
( x 3)2
2 B 0
2
B 0 2 B1 ( x 3)2 B2 0 X 3
2
2
( x 3)
( x 3)
0 x 3 2 2 B 0 so that lim 2 2 .
( x 3)
x 3 ( x 3)
2 B. Now,
( x 3) 2
2 . Choose
B
92. For every real number B 0, we must find a 0 such that for all x, 0 x (5)
1
B.
( x 5) 2
1
B 0 ( x 5)2 B1 x 5 1 . Choose 1 . Then 0 x (5)
B
B
( x 5)2
1
x 5 1 1 2 B so that lim
.
2
B
( x 5)
x 5 ( x 5)
Now,
Copyright 2016 Pearson Education, Ltd.
2 , then
B
84
Chapter 2 Limits and Continuity
93. (a) We say that f ( x) approaches infinity as x approaches x0 from the left, and write lim f ( x) ,
x x0
if for every positive number B, there exists a corresponding number 0 such that for all x,
x0 x x0 f ( x) B.
(b) We say that f ( x) approaches minus infinity as x approaches x0 from the right, and write lim f ( x) ,
x x0
if for every positive number B (or negative number B) there exists a corresponding number 0 such
that for all x, x0 x x0 f ( x ) B.
(c) We say that f ( x) approaches minus infinity as x approaches x0 from the left, and write lim f ( x) , if
x x0
for every positive number B (or negative number B) there exists a corresponding number 0 such that
for all x, x0 x x0 f ( x) B.
94. For B 0, 1x B 0 x B1 . Choose B1 . Then 0 x 0 x B1 1x B so that lim 1x .
x 0
95. For B 0, 1x B 0 1x B 0 x B1 B1 x. Choose B1 . Then x 0 B1 x
1x B so that lim 1x .
x 0
96. For B 0, x 1 2 B x 1 2 B ( x 2) B1 x 2 B1 x 2 B1 . Choose B1 .
Then 2 x 2 x 2 0 B1 x 2 0 x 1 2 B 0 so that lim x 1 2 .
x 2
97. For B 0, x 1 2 B 0 x 2 B1 . Choose B1 . Then 2 x 2 0 x 2 0 x 2 B1
x 1 2 B 0 so that lim
98.
1 .
x 2
x 2
For B 0 and 0 x 1, 1 2 B 1 x 2 B1 (1 x)(1 x) B1 . Now 12x 1 since x 1. Choose 21B .
1 x
Then 1 x 1 x 1 0 1 x 21B (1 x )(1 x ) B1 12x B1 1 2 B for 0 x 1 and
1 x
x near 1 lim 1 2 .
x 1 1 x
2
99. y xx1 x 1 x11
2
100. y xx 11 x 1 x21
Copyright 2016 Pearson Education, Ltd.
Section 2.6 Limits Involving Infinity; Asymptotes of Graphs
2
102. y 2xx 14 12 x 1 2 x3 4
2
104. y x 21 x 12
101. y xx 14 x 1 x31
103. y x x1 x 1x
105. y
x
4 x 2
2
3
x
106. y
x
1
4 x 2
Copyright 2016 Pearson Education, Ltd.
85
86
Chapter 2 Limits and Continuity
1
107. y x 2/3 1/3
108. y sin
x
x 2 1
109. (a) y (see accompanying graph)
(b) y (see accompanying graph)
(c) cusps at x 1 (see accompanying graph)
110. (a) y 0 and a cusp at x 0 (see the
accompanying graph)
(b) y 32 (see accompanying graph)
(c) a vertical asymptote at x 1 and contains the
point 1,
CHAPTER 2
3
23 4
(see accompanying graph)
PRACTICE EXERCISES
1. At x 1:
lim f ( x) lim f ( x) 1
x 1
x 1
lim f ( x) 1 f ( 1)
x 1
f is continuous at x 1.
At x 0 :
lim f ( x) lim f ( x) 0
x 0
lim f ( x) 0.
x 0
x 0
But f (0) 1 lim f ( x)
x 0
f is discontinuous at x 0.
If we define f (0) 0, then the discontinuity at
x 0 is removable.
At x 1:
lim f ( x) 1 and lim f ( x) 1
x 1
x 1
lim f ( x ) does not exist
x 1
f is discontinuous at x 1.
Copyright 2016 Pearson Education, Ltd.
Chapter 2 Practice Exercises
2. At x 1:
87
lim f ( x) 0 and lim f ( x) 1
x 1
lim f ( x) does not exist
x 1
x 1
f is discontinuous at x 1.
At x 0 :
lim f ( x ) and lim f ( x)
x 0
x 0
lim f ( x) does not exist
x 0
f is discontinuous at x 0.
At x 1:
lim f ( x) lim f ( x) 1 lim f ( x) 1.
x 1
x 1
x 1
But f (1) 0 lim f ( x)
x 1
f is discontinuous at x 1.
If we define f (1) 1, then the discontinuity at
x 1 is removable.
3. (a) lim (3 f (t )) 3 lim f (t ) 3( 7) 21
t t0
t t0
2
(b) lim ( f (t )) 2 lim f (t ) (7) 2 49
t t0
t t0
(c) lim ( f (t ) g (t )) lim f (t ) lim g (t ) ( 7)(0) 0
t t0
t t0
lim f (t )
f (t )
t t0
lim f (t )
(d) lim g (t ) 7 lim (0g (t ) 7) lim g (0t ) lim 7 077 1
t t0
t t
t t
t t0
t t0
t t0
(e) lim cos ( g (t )) cos lim g (t ) cos 0 1
t t0
t t0
(f)
lim | f (t )| lim f (t ) | 7| 7
t t0
t t0
(g) lim ( f (t ) g (t )) lim f (t ) lim g (t ) 7 0 7
t t0
(h) lim
t t0
1
f (t )
t t0
t t0
1
lim f (t ) 17 17
t t0
4. (a) lim g ( x) lim g ( x) 2
x 0
x 0
(b) lim ( g ( x) f ( x)) lim g ( x) lim f ( x)
x 0
x 0
x 0
2 12 22
(c) lim ( f ( x) g ( x)) lim f ( x) lim g ( x) 12 2
x 0
x 0
(d) lim f 1( x ) lim 1f ( x ) 11 2
x 0
2
x 0
x 0
(e) lim ( x f ( x)) lim x lim f ( x) 0 12 12
x 0
(f)
lim
x 0
x 0
x 0
1 (1)
lim f ( x ) lim cos x
f ( x )cos x
2
x 0
x 0
12
x 1
lim x lim 1
0 1
x0
x 0
5. Since lim x 0 we must have that lim (4 g ( x)) 0. Otherwise, if lim (4 g ( x)) is a finite positive number,
x 0
x 0
x 0
4 g ( x )
4 g ( x )
we would have lim x and lim x so the limit could not equal 1 as x 0. Similar
x 0
x 0
reasoning holds if lim (4 g ( x)) is a finite negative number. We conclude that lim g ( x) 4.
x 0
x 0
Copyright 2016 Pearson Education, Ltd.
88
Chapter 2 Limits and Continuity
6. 2 lim x lim g ( x) lim x lim lim g ( x ) 4 lim lim g ( x) 4 lim g ( x) (since lim g ( x) is a
x0
x 4 x 0
x 4 x 0
x 0
x 4 x 4 x 0
2
1
constant) lim g ( x ) 4 2 .
x 0
7. (a) lim f ( x) lim x1/3 c1/3 f (c) for every real number c f is continuous on ( , ).
x c
x c
x c
x c
x c
x c
1 h(c ) for every nonzero real number c h is continuous on ( , 0) and
c 2/3
x c
x c
c
(b) lim g ( x) lim x3/4 c3/4 g (c) for every nonnegative real number c g is continuous on [0, ).
(c) lim h( x) lim x 2/3
(, ).
(d) lim k ( x) lim x 1/6 1/1 6 k (c ) for every positive real number c k is continuous on (0, )
8. (a)
n I
n 12 , n 12 , where I the set of all integers.
(n , (n 1) ), where I the set of all integers.
(b)
n I
(c) (, ) ( , )
(d) (, 0) (0, )
x2 4 x 4
( x 2)( x 2)
lim
lim x 2 , x 2; the limit does not exist because
2
x 0 x 5 x 14 x x 0 x ( x 7)( x 2) x 0 x ( x 7)
lim x (xx27) and lim x (xx27)
x 0
x 0
x2 4 x 4
( x 2)( x 2)
0 0
lim 3 2
lim x ( x 7)( x 2) lim x (xx27) , x 2, and lim x (xx27) 2(9)
x
5
x
14
x
x 2
x2
x 2
x 2
9. (a) lim
(b)
3
x ( x 1)
x2 x
lim 3 2
lim 2 x 1
lim 2 1 , x 0 and x 1.
5
4
3
x 0 x 2 x x
x 0 x ( x 2 x 1) x 0 x ( x 1)( x 1) x 0 x ( x 1)
2
Now lim 2 1 and lim 2 1 lim 5 x 4x 3 .
x ( x 1)
x ( x 1)
x
2
x x
x 0
x 0
x 0
2
x ( x 1)
x
x
1
lim 5 4 3 lim 3 2
lim 2
, x 0 and x 1. The limit does not exist because
x 1 x 2 x x
x 1 x ( x 2 x 1) x 1 x ( x 1)
lim 2 1 and lim 2 1 .
x 1 x ( x 1)
x 1 x ( x 1)
10. (a) lim
(b)
11. lim 11 xx lim
1 x
lim 1 12
x 1 (1 x )(1 x ) x 1 1 x
x 1
2
( x2 a2 )
2
12. lim x 4 a 4 lim
x a x a
2
2
2
2
x a ( x a )( x a )
1
12
2
2
2a
xa x a
lim
( x h)2 x 2
( x 2 2 hx h 2 ) x 2
lim
lim (2 x h) 2 x
h
h
h 0
h 0
h 0
13. lim
( x h)2 x 2
( x 2 2 hx h 2 ) x 2
lim
lim (2 x h) h
h
h
x 0
x 0
x 0
14. lim
15.
1
1
2 (2 x )
lim 2 xx 2 lim 2 x (2 x ) lim 421 x 14
x 0
x 0
x 0
(2 x )3 8
( x3 6 x 2 12 x 8) 8
lim
lim ( x 2 6 x 12) 12
x
x
x 0
x 0
x 0
16. lim
Copyright 2016 Pearson Education, Ltd.
Chapter 2 Practice Exercises
1/3
2/3
1/3
( x 1)( x 1)
1
1 lim ( x 1) ( x x 1)( x 1) lim
lim 2/3 x 1/3
11 2
2/3
1/3
2/3
1/3
x 1
x 1 ( x 1) ( x 1)( x x 1) x 1 ( x 1)( x x 1) x 1 x x 1 111 3
1/3
17. lim x
x 1
18.
2/3
( x1/3 4)( x1/3 4)
( x1/3 4)( x1/3 4) ( x 2/3 4 x1/3 16)( x 8)
lim x 16 lim
lim
2/3
1/3
x 8
x 64
x 8
x 64
( x 64) ( x1/3 4) ( x 8)
x 8
( x 8)( x 4 x 16)
( x1/3 4) ( x 8)
(4 4) (88)
lim
lim 2/3 1/3
161616 83
2/3
1/3
x 64 ( x 64) ( x 4 x 16) x 64 x 4 x 16
x 64
cos x
tan 2 x
sin 2 x
19. lim tan x lim cos
lim
2 x sin x
x 0
20.
x 0
x 0
cos x
x
2 x 1 1 1 2 2
sin2 x2 x cos
2 x sin x x
lim csc x lim sin1 x
x
x
21. lim sin 2x sin x sin 2 sin sin 2 1
x
22. lim cos 2 ( x tan x) cos 2 ( tan ) cos 2 ( ) (1)2 1
x
23. lim 3sin8 xx x lim
8
8 4
sin x
x 0 3 x 1 3(1) 1
x 0
2 x 1 lim
cos 2 x 1 lim
sin 2 x
cossin2xx1 cos
cos 2 x 1 x 0 sin x (cos 2 x 1) x 0 sin x (cos 2 x 1)
x 0
2
24. lim cossin2 xx 1 lim
x 0
2
4sin x cos 2 x
4(0)(1) 2
0
11
x 0 cos 2 x 1
lim
1/3
25.
26.
lim [4 g ( x)]1/3 2 lim 4 g ( x)
x 0
x 0
1
lim
2 lim 4 g ( x) 8, since 23 8. Then lim g ( x ) 2.
x 0
x 0
2 lim ( x g ( x)) 12 5 lim g ( x) 12 lim g ( x) 12 5
x 5 x g ( x)
x 5
x 5
x 5
2
27. lim 3gx( x)1 lim g ( x) 0 since lim (3 x 2 1) 4
x 1
28.
x 1
x 1
2
lim 5 x 0 lim g ( x) since lim (5 x 2 ) 1
x 2
g ( x)
29. At x 1:
lim
x 2
lim
lim
x 1
x ( x 2 1)
f ( x) lim
x 1
x ( x 2 1)
x 1
x 2
x 2 1
2
x 1 | x 1|
lim x 1, and
x 1
f ( x) lim
x ( x 2 1)
2
x 1 | x 1|
lim
lim
x 1
2
x 1 ( x 1)
lim ( x) (1) 1. Since lim
x 1
x ( x 2 1)
x 1
f ( x)
f ( x) lim f ( x) does not exist, the
x1
function f cannot be extended to a continuous
function at x 1.
Copyright 2016 Pearson Education, Ltd.
89
90
Chapter 2 Limits and Continuity
At x 1:
lim f ( x) lim
x 1
x 1
lim
x 1
x ( x 2 1)
2
| x 1|
x ( x 2 1)
x 2 1
lim
x ( x 2 1)
2
x 1 ( x 1)
lim ( x) 1, and lim f ( x) lim
x 1
x 1
x 1
x ( x 2 1)
| x 2 1|
lim x 1.
x 1
Again lim f ( x) does not exist so f cannot be extended to a continuous function at x 1 either.
x 1
30. The discontinuity at x 0 of f ( x) sin
lim sin 1x does not exist.
1x is nonremovable because x
0
31. Yes, f does have a continuous extension at a 1:
define f (1) lim x 41 43 .
x 1 x x
32. Yes, g does have a continuous extension at a 2 :
5 cos
g 2 lim 4 2 54 .
2
33. From the graph we see that lim h(t ) lim h(t )
t 0
t 0
so h cannot be extended to a continuous function
at a 0.
34. From the graph we see that lim k ( x) lim k ( x)
x 0
x 0
so k cannot be extended to a continuous function at
a 0.
35. (a) f (1) 1 and f (2) 5 f has a root between 1 and 2 by the Intermediate Value Theorem.
(b), (c) root is 1.32471795724
36. (a) f (2) 2 and f (0) 2 f has a root between 2 and 0 by the Intermediate Value Theorem.
(b), (c) root is -1.76929235424
Copyright 2016 Pearson Education, Ltd.
Chapter 2 Practice Exercises
2 3x
37. lim 52xx73 lim
38.
x 5 x
x
39.
5200 52
7
2
lim x 43x 8 lim
x
1 4 8
2
3 x3
x 3 x 3 x
3x
2 32
2
lim 2 x2 3 lim
x 5 x 7
20
5 0 52
x
x 5
7
x2
000 0
1
40.
2
lim 2 1
lim 7x 1 100 0 0
1
x
7
x
1
x
x
x
2
x
41.
x 2 7 x lim x 7
1
x x 1
x 1
42.
lim
x
43.
44.
45.
46.
x 4 x3 lim
x 1
3
128
x 12 x 128 x 12 3
lim
x
lim sin x lim 1 0 since
x x
x x
x as x lim sinx x 0.
x
lim cos 1 lim 2 0 lim cos 1 0.
x2 x
lim x sin
lim
x sin x
x
1 sinx x 2
x
x
1 sinx x
1100 0 1
x 2/3 x 1 lim 1 x 5/3 1 0 1
2/3
2
2
x x cos x x 1 cos x 1 0
lim
x 2/3
2
2
2
47. (a) y xx 34 is undefined at x 3 : lim xx 34 and lim xx 34 , thus x 3 is a vertical asymptote.
x 3
x 3
x 2 x 2 and lim x 2 x 2 , thus x 1 is a vertical
2
2
x 1 x 2 x 1
x 1 x 2 x 1
2
(b) y x2 x 2 is undefined at x 1: lim
x 2 x 1
asymptote.
2
2
(c) y x2 x 6 is undefined at x 2 and 4: lim x2 x 6 lim xx43 65 ; lim
x 2 x 6 lim x 3
2
x 2 x 8
x
2
x
8
x
2 x 8 x 4 x 4
x 2
x2
x 4
2
x
x
6
x
3
lim 2
lim x 4 . Thus x 4 is a vertical asymptote.
x 4 x 2 x 8 x 4
48. (a)
2
1 1
2
2
x 1 x x 1
x 1 2
x x 1
2
y 12 x : lim 12 x lim x 1 11 1 and lim 12 x lim
x
1 1
x2
1
x 1 2
11 1, thus y 1 is a
x
horizontal asymptote.
y
x 4
:
x4
(c) y
x2 4
:
x
(b)
lim
1 4
x 4
x
lim
1 0
x4
1 0
x 1 4x
lim
x2 4
lim
x
x
x
x
lim
x
1 42
x
1
1 42
x
x
x
1, thus y 1 is a horizontal asymptote.
1 0
1 and lim
1
x
lim
x
1 42
x
1
x2 4
lim
x
x
1 42
x
x
x2
1 0
11 1,
1
thus y 1 and y 1 are horizontal asymptotes.
(d) y
x 2 9 :
9 x 2 1
lim
x
x 2 9 lim
9 x 2 1 x
1 92
x
9 12
x
1 0 1 and lim
9 0
3
x
x 2 9 lim
9 x 2 1 x
thus y 13 is a horizontal asymptote.
Copyright 2016 Pearson Education, Ltd.
1 92
x
9 12
x
1 0 1 ,
90
3
91
92
Chapter 2 Limits and Continuity
CHAPTER 2
1. (a) x
ADDITIONAL AND ADVANCED EXERCISES
0.1
xx
0.01
0.001
0.0001 0.00001
0.7943 0.9550 0.9931 0.9991 0.9999
Apparently, lim x x 1
x 0
(b)
2. (a) x
10
1x
1/(ln x )
3.
1000
0.3679 0.3679 0.3679
1
x x
Apparently, lim
(b)
100
1/(ln x )
2
lim L lim L0 1 v 2 L0
c
v c
v c
0.3678 1e
1
lim v 2
v c
2
c
2
L0 1 c 2 0
c
The left-hand limit was needed because the function L is undefined if v c (the rocket cannot move faster than
the speed of light).
4. (a)
x
1 0.2 0.2 2x 1 0.2 0.8 2x 1.2 1.6
2
x 2.4 2.56 x 5.76.
(b)
x
1 0.1 0.1 2x 1 0.1 0.9 2x 1.1 1.8
2
x 2.2 3.24 x 4.84.
Copyright 2016 Pearson Education, Ltd.
Chapter 2 Additional and Advanced Exercises
93
5. |10 2(t 20) 104 10| 0.0005 | 2(t 20) 104 | 0.0005 0.0005 2(t 20) 104 0.0005
2.5 t 20 2.5 17.5 t 22.5 Within 2.5 C.
6. We want to know in what interval to hold values of h to make V satisfy the inequality
|V 1000| |36 h 1000| 10. To find out, we solve the inequality:
990 h 1010 8.8 h 8.9
|36 h 1000| 10 10 36 h 1000 10 990 36 h 1010 36
36
where 8.8 was rounded up, to be safe, and 8.9 was rounded down, to be safe.
The interval in which we should hold h is about 8.9 8.8 0.1 cm wide (1 mm). With stripes 1 mm wide, we can
expect to measure a liter of water with an accuracy of 1%, which is more than enough accuracy for cooking.
7. Show lim f ( x) lim ( x 2 7) 6 f (1).
x 1
x 1
Step 1: |( x 2 7) 6| x 2 1 1 x 2 1 1 x 1 .
Step 2: | x 1| x 1 1 x 1.
Then 1 1 or 1 1 . Choose min 1 1 , 1 1 , then 0 | x 1|
2
|( x 7) 6| and lim f ( x) 6. By the continuity text, f ( x ) is continuous at x 1.
x 1
1 2g 1 .
4
x 14
x 14 2 x
Step 1: 21x 2 21x 2 2 21x 2 412 x 412 .
Step 2: X 14 x 14 14 x 14 .
Then 14 412 14 412 4(2 ) , or 14 412 412 14 4(2 ) .
Choose 4(2 ) , the smaller of the two values. Then 0 x 14 21x 2 and lim 21x 2.
x 14
By the continuity test, g ( x) is continuous at x 14 .
8. Show lim g ( x) lim
9. Show lim h( x) lim 2 x 3 1 h(2).
x 2
Step 1:
x2
2 x 3 1 2 x 3 1 1 2 x 3 1
Step 2: | x 2| x 2 or 2 x 2.
(1 )2 3
(1 )2 3
x
.
2
2
2
2
2
(1 ) 2 3
(1 ) 2 3 1(1 )2
2 , or 2 (1 ) 3 (1 ) 3 2 (1 ) 1
2
2
2
2
2
2
2
2
2
2
2 . Choose 2 , the smaller of the two values. Then, 0 | x 2| 2 x 3 1 ,
Then 2
so lim 2 x 3 1. By the continuity test, h( x) is continuous at x 2.
x2
10. Show lim F ( x) lim 9 x 2 F (5).
x 5
Step 1:
x 5
9 x 2 9 x 2 9 (2 )2 x 9 (2 ) 2 .
Step 2: 0 | x 5| x 5 5 x 5.
Then 5 9 (2 )2 (2 )2 4 2 2, or 5 9 (2 ) 2 4 (2 ) 2 2 2.
Choose 2 2, the smaller of the two values. Then, 0 | x 5| 9 x 2 , so lim 9 x 2.
By the continuity test, F ( x) is continuous at x 5.
x 5
11. Suppose L1 and L2 are two different limits. Without loss of generality assume L2 L1. Let 13 ( L2 L1 ). Since
lim f ( x) L1 there is a 1 0 such that 0 | x x0 | 1 | f ( x ) L1 | f ( x ) L1
x x0
Copyright 2016 Pearson Education, Ltd.
94
Chapter 2 Limits and Continuity
13 ( L2 L1 ) L1 f ( x) 13 ( L2 L1 ) L1 4 L1 L2 3 f ( x) 2 L1 L2 . Likewise, lim f ( x) L2 so
x x0
there is a 2
such that 0 | x x0 | 2 | f ( x ) L2 | f ( x) L2 13 ( L2 L1 ) L2 f ( x) 13 ( L2 L1 ) L2
2 L2 L1 3 f ( x) 4 L2 L1 L1 4 L2 3 f ( x) 2 L2 L1. If min{1 , 2 } both inequalities must
4 L L 3 f ( x) 2 L1 L2
hold for 0 | x x0 | : 1 2
5( L1 L2 ) 0 L1 L2 . That is, L1 L2 0 and
L1 4 L2 3 f ( x) 2 L2 L1
L1 L2 0, a contradiction.
12. Suppose lim f ( x) L. If k 0, then lim k f ( x) lim 0 0 0 lim f ( x) and we are done. If k 0, then given
x c
x c
x c
x c
any 0, there is a 0 so that 0 | x c | | f ( x) L | | k | | k || f ( x) L | | k ( f ( x) L)|
|(kf ( x)) (kL)| . Thus lim k f ( x) kL k lim f ( x) .
x c
x c
13. (a) Since x 0 , 0 x3 x 1 ( x3 x) 0 lim f ( x3 x) lim f ( y ) B where y x3 x.
x 0
y 0
(b) Since x 0 , 1 x x 0 ( x x) 0 lim f ( x x) lim f ( y ) A where y x3 x.
3
3
x 0
3
y 0
(c) Since x 0 , 0 x x 1 ( x x ) 0 lim f ( x x ) lim f ( y ) A where y x 2 x 4 .
4
2
2
4
x 0
4
2
4
y 0
2
(d) Since x 0 , 1 x 0 0 x x 1 ( x x ) 0 lim f ( x x 4 ) A as in part (c).
4
2
2
x 0
14. (a) True, because if lim ( f ( x) g ( x)) exists then lim ( f ( x) g ( x)) lim f ( x) lim [( f ( x) g ( x)) f ( x)]
x a
lim g ( x) exists, contrary to assumption.
x a
(b) False; for example take f ( x)
11
x 0 x x
lim ( f ( x) g ( x)) lim
x 0
x a
x a
x a
1 and g ( x ) 1 . Then neither lim f ( x ) nor lim g ( x ) exists, but
x
x
x 0
x 0
0 0 exists.
xlim
0
(c) True, because g ( x) | x | is continuous g ( f ( x)) | f ( x)| is continuous (it is the composite of
continuous functions).
1, x 0
(d) False; for example let f ( x)
f ( x) is discontinuous at x 0. However | f ( x )| 1 is
1, x 0
continuous at x 0.
2
( x 1)( x 1)
2, x 1.
x 1 ( x 1)
15. Show lim f ( x ) lim xx 11 lim
x 1
x 1
x 2 1 , x 1
Define the continuous extension of f ( x) as F ( x ) x 1
. We now prove the limit of f ( x) as x 1
2 , x 1
exists and has the correct value.
2
Step 1: xx 11 (2)
( x 1)( x 1)
2 ( x 1) 2 , x 1 1 x 1.
( x 1)
Step 2: | x (1)| x 1 1 x 1.
Then 1 1 , or 1 1 . Choose . Then 0 | x (1)|
2
xx 11 (2) lim F ( x) 2. Since the conditions of the continuity test are met by F ( x), then f ( x) has
x 1
a continuous extension to F ( x) at x 1.
Copyright 2016 Pearson Education, Ltd.
Chapter 2 Additional and Advanced Exercises
2
16. Show lim g ( x) lim x 2x2x63 lim
x 3
x 3
x 3
95
( x 3)( x 1)
2, x 3.
2( x 3)
x 2 2 x 3 , x 3
Define the continuous extension of g ( x) as G ( x) 2 x 6
. We now prove the limit of g ( x) as x 3
,
x3
2
exists and has the correct value.
2
Step 1: x 2x2x63 2
( x 3)( x 1)
2 x21 2 , x 3 3 2 x 3 2.
2( x 3)
Step 2: | x 3| x 3 3 x 3.
Then, 3 3 2 2, or 3 3 2 2. Choose 2. Then 0 | x 3|
2
x 2x2x63 2 lim
x 3
( x 3)( x 1)
2. Since the conditions of the continuity test hold for G ( x), g ( x) can be
2( x 3)
continuously extended to G ( x ) at x 3.
17. (a) Let 0 be given. If x is rational, then f ( x) x | f ( x) 0| | x 0| | x 0| ; i.e., choose .
Then | x 0| | f ( x) 0| for x rational. If x is irrational, then f ( x) 0 | f ( x) 0| 0
which is true no matter how close irrational x is to 0, so again we can choose . In either case, given
0 there is a 0 such that 0 | x 0| | f ( x) 0| . Therefore, f is continuous at x 0.
(b) Choose x c 0. Then within any interval (c , c ) there are both rational and irrational numbers. If c
is rational, pick 2c . No matter how small we choose 0 there is an irrational number x in
(c , c ) | f ( x) f (c)| |0 c | c 2c . That is, f is not continuous at any rational c 0. On the
other hand, suppose c is irrational f (c) 0. Again pick 2c . No matter how small we choose 0
there is a rational number x in (c , c ) with | x c | 2c 2c x 32c . Then | f ( x) f (c)| | x 0|
| x | 2c f is not continuous at any irrational c 0.
|c|
If x c 0, repeat the argument picking 2 2c . Therefore f fails to be continuous at any nonzero
value x c.
18. (a) Let c mn be a rational number in [0, 1] reduced to lowest terms f (c) 1n . Pick 21n . No matter
how small 0 is taken, there is an irrational number x in the interval (c , c ) | f ( x) f (c)|
0 1n 1n 21n . Therefore f is discontinuous at x c, a rational number.
(b) Now suppose c is an irrational number f (c) 0. Let 0 be given. Notice that 12 is the only rational
number reduced to lowest terms with denominator 2 and belonging to [0, 1]; 13 and 23 the only rationals with
denominator 3 belonging to [0, 1]; 14 and 34 with denominator 4 in [0, 1]; 15 , 52 , 53 and 54 with denominator 5
in [0, 1]; etc. In general, choose N so that N1 there exist only finitely many rationals in [0, 1] having
denominator N , say r1 , r2 , , rp . Let min {| c ri |: i 1, , p}. Then the interval (c , c )
contains no rational numbers with denominator N . Thus, 0 | x c | | f ( x) f (c)| | f ( x ) 0|
| f ( x)| N1 f is continuous at x c irrational.
Copyright 2016 Pearson Education, Ltd.
96
Chapter 2 Limits and Continuity
(c) The graph looks like the markings on a typical
ruler when the points ( x, f ( x)) on the graph of
f ( x ) are connected to the x-axis with vertical
lines.
19. Yes. Let R be the radius of the equator (earth) and suppose at a fixed instant of time we label noon as the zero
point, 0, on the equator 0 R represents the midnight point (at the same exact time). Suppose x1 is a point
on the equator “just after” noon x1 R is simultaneously “just after” midnight. It seems reasonable that the
temperature T at a point just after noon is hotter than it would be at the diametrically opposite point just after
midnight: That is, T ( x1 ) T ( x1 R) 0. At exactly the same moment in time pick x2 to be a point just before
midnight x2 R is just before noon. Then T ( x2 ) T ( x2 R ) 0. Assuming the temperature function T is
continuous along the equator (which is reasonable), the Intermediate Value Theorem says there is a point c
between 0 (noon) and R (simultaneously midnight) such that T (c) T (c R ) 0; i.e., there is always a pair
of antipodal points on the earth’s equator where the temperatures are the same.
2
2
20. lim f ( x ) g ( x) lim 14 ( f ( x) g ( x))2 ( f ( x) g ( x))2 14 lim ( f ( x ) g ( x)) lim ( f ( x) g ( x))
x c
x c
x c
xc
2
2
1
4 (3 (1) ) 2.
1(1 a )
1 1 a 1 1 a
lim
1 12
a
1
1
a
a
(
1 1 a )
1 1 0
a 0
a 0
a 0
a 0
1(1 a )
a
1
At x 1: lim r (a ) lim
lim
1
1 0
a 1
a 1 a ( 1 1 a ) a 1 a ( 1 1 a )
21. (a) At x 0: lim r (a ) lim 1 a1 a lim
a 0
1 (1 a )
1 1 a
a
lim 1 a1 a 1 1 a lim
lim
a
1 1 a
a 0
a 0 a ( 1 1 a ) a 0 a ( 1 1 a )
1
lim
(because the denominator is always negative); lim r (a)
a 0 1 1 a
a 0
1
lim
(because the denominator is always positive).
a 0 1 1 a
(b) At x 0: lim r (a ) lim
a 0
Therefore, lim r (a ) does not exist.
a 0
At x 1: lim r (a ) lim
a 1
a 1
1 1 a
a
lim
1
a 1 1 1 a
1
Copyright 2016 Pearson Education, Ltd.
Chapter 2 Additional and Advanced Exercises
97
(c)
(d)
22. f ( x) x 2 cos x f (0) 0 2 cos 0 2 0 and f ( ) 2 cos( ) 2 0. Since f ( x ) is
continuous on [ , 0], by the Intermediate Value Theorem, f ( x ) must take on every value between [ 2, 2].
Thus there is some number c in [ , 0] such that f (c) 0; i.e., c is a solution to x 2 cos x 0.
23. (a) The function f is bounded on D if f ( x) M and f ( x) N for all x in D. This means M f ( x) N for
all x in D. Choose B to be max {| M |, | N |}. Then | f ( x)| B. On the other hand, if | f ( x)| B, then
B f ( x) B f ( x ) B and f ( x) B f ( x) is bounded on D with N B an upper bound and
M B a lower bound.
(b) Assume f ( x) N for all x and that L N . Let L 2N . Since lim f ( x) L there is a 0 such that
x x0
0 | x x0 | | f ( x) L | L f ( x ) L L L 2N f ( x) L L 2N L 2N f ( x)
3 L2 N . But L N L 2N N N f ( x) contrary to the boundedness assumption f ( x) N . This
contradiction proves L N .
(c) Assume M f ( x) for all x and that L M . Let M2 L . As in part (b), 0 | x x0 | L M2 L
f ( x) L M2 L 3L 2 M f ( x) M2 L M , a contradiction.
| a b | a b a b
2 2 22a a.
2
|a b|
If a b, then a b 0 | a b | (a b) b a max {a, b} a 2 b 2 a 2 b b 2 a 22b b.
| a b |
Let min {a, b} a 2 b 2 .
24. (a) If a b, then a b 0 | a b | a b max {a, b} a 2 b
(b)
25.
lim
x 0
sin(1cos x )
sin(1cos x )
x 1 cos x lim sin(1cos x ) lim 1cos 2 x
lim 1cos x 1cos
x
x
1 cos x x 0 1cos x
x 0
x 0 x (1 cos x )
2
x lim sin x sin x 1 0 0.
1 lim x (1sin
cos x )
x
1 cos x
2
x 0
x 0
Copyright 2016 Pearson Education, Ltd.
98
Chapter 2 Limits and Continuity
26.
lim sin x lim sinx x x x 1 lim
sin x
x
x 0 sin x
x 0
x 0
27. lim
x 0
1
sin x
x
lim
x 0
x 1 1 0 0.
sin(sin x )
sin(sin x )
sin(sin x )
lim sin x sinx x lim sin x lim sinx x 1 1 1.
x
x 0
x 0
x 0
sin( x 2 x )
sin( x 2 x )
sin( x 2 x )
lim
( x 1) lim
lim ( x 1) 1 1 1.
2
2
x
x 0
x 0 x x
x 0 x x
x 0
28. lim
sin( x 2 4)
sin( x 2 4)
sin( x 2 4)
lim
(
x
2)
lim
lim ( x 2) 1 4 4.
2
2
x2
x 2
x 2 x 4
x2 x 4
x2
29. lim
sin( x 3)
sin( x 3)
sin( x 3)
lim
1 lim
lim
x
9
3
3
x
x
x 3
x 9
x 9
x 9
x 9
30. lim
1 1 1 1 .
6 6
x 3
31. Since the highest power of x in the numerator is 1 more than the highest power of x in the denominator, there is
3/ 2
an oblique asymptote. y 2 x 2 x 3 2 x 3 , thus the oblique asymptote is y 2 x.
x 1
x 1
x; thus
32. As x , 1x 0 sin 1x 0 1 sin 1x 1, thus as x , y x x sin 1x x 1 sin 1x
the oblique asymptote is y x.
33. As x , x 2 1 x 2 x 2 1 x 2 ; as x , x 2 x, and as x , x 2 x; thus the oblique
asymptotes are y x and y x.
34. As x , x 2 x x 2 2 x x( x 2) x 2 ; as x , x 2 x, and as x , x 2 x;
asymptotes are y x and y x.
Copyright 2016 Pearson Education, Ltd.
CHAPTER 3
3.1
DERIVATIVES
TANGENTS AND THE DERIVATIVE AT A POINT
1. P1: m1 1, P2 : m2 5
2. P1: m1 2, P2 : m2 0
3. P1: m1 52 , P2 : m2 12
4. P1: m1 3, P2 : m2 3
[4 ( 1 h ) 2 ](4 ( 1)2 )
(1 2 h h 2 ) 1
lim
h
h
h 0
h 0
h (2 h )
lim h 2; at (1, 3): y 3 2( x ( 1))
h 0
5. m lim
y 2 x 5, tangent line
2
[(1 h 1) 2 1] [(1 1) 2 1]
lim hh lim h 0;
h
h 0
h 0
h 0
6. m lim
at (1,1) : y 1 0( x 1) y 1, tangent line
7. m lim 2 1 hh 2 1 lim 2 1hh 2 2 1 h 2
h 0
4(1 h ) 4
lim
2 1 h 2
h 0
h0 2 h 1 h 1
2
1;
h 0 1 h 1
lim
at (1, 2): y 2 1( x 1) y x 1, tangent line
1
8. m lim
h 0
lim
( 1 h )2
h
2
( 2 h h )
2
h0 h ( 1 h )
1
( 1)2
lim
1( 1 h )2
h0 h ( 1 h )
2 h 2;
2
h0 ( 1 h )
lim
2
at (1,1):
y 1 2( x (1)) y 2 x 3, tangent line
Copyright 2016 Pearson Education, Ltd.
99
100
Chapter 3 Derivatives
2
3
( 2 h )3 ( 2)3
lim 812 h 6hh h 8
h
h 0
h 0
2
9. m lim
lim (12 6h h ) 12;
h 0
at (2, 8): y 8 12( x (2)) y 12 x 16,
tangent line
1
10. m lim
( 2 h )3
1
( 2)3
h
h 0
(12 h 6 h 2 h3 )
lim
lim
8 ( 2 h )3
h 0 8 h ( 2 h )
2
lim 126h h3
h 0 8( 2 h )
3
3
h 0 8h ( 2 h )
3
12
8( 8) 16 ;
3 ( x ( 2))
at 2, 18 : y 18 16
3 x 1 , tangent line
y 16
2
[(2 h ) 2 1]5
(5 4 h h 2 ) 5
h (4 h )
lim
lim
4; at (2,5): y 5 4( x 2), tangent line
h
h
h 0
h 0
h 0 h
11. m lim
h ( 3 2 h )
[(1 h ) 2(1 h )2 ]( 1)
(1 h 2 4 h 2 h 2 ) 1
lim
lim
3; at (1, 1) : y 1 3( x 1), tangent line
h
h
h
h 0
h 0
h 0
12. m lim
3 h
13. m lim (3 h )h 2
3
h 0
8
14. m lim
h 0
(2 h )2
h
2
lim
h 0
lim
(3 h ) 3( h 1)
lim h(h2h1) 2; at (3,3): y 3 2( x 3), tangent line
h ( h 1)
h 0
8 2(2 h )2
h 0 h (2 h )
2
lim
h 0
8 2(4 4 h h 2 )
h (2 h )2
lim
2 h (4 h )
h 0 h (2 h )
2
48 2; at (2, 2): y 2 2( x 2)
(2 h )3 8
(812 h 6 h 2 h3 ) 8
h (12 6 h h 2 )
lim
lim
12; at (2,8): y 8 12(t 2), tangent line
h
h
h
h 0
h 0
h 0
15. m lim
[(1 h )3 3(1 h )] 4
(1 3h 3h 2 h3 33h ) 4
h (6 3h h 2 )
lim
lim
6; at (1, 4): y 4 6(t 1), tangent line
h
h
h
h 0
h 0
h 0
16. m lim
(4 h ) 4
4 h 2
lim 4hh 2 4 h 2 lim
h
4 h 2 h 0 h 4 h 2
h 0
at (4, 2): y 2 14 ( x 4), tangent line
17. m lim
h 0
lim
(8 h ) 1 3
(9 h ) 9
lim 9hh 3 9 h 3 lim
h
9 h 3 h 0 h 9 h 3
h 0
h 0
at (8,3): y 3 16 ( x 8), tangent line
18. m lim
h 0 h
h
4 h 2
lim
h 0 h
h
9 h 3
1 1;
4 2 4
1 1;
9 3 6
5h ( 2 h )
5( 1 h )2 5
5(1 2 h h 2 ) 5
lim
lim
10, slope
h
h
h
h 0
h 0
h 0
19. At x 1, y 5 m lim
Copyright 2016 Pearson Education, Ltd.
Section 3.1 Tangents and the Derivative at a Point
101
h (4 h )
[1(2 h )2 ]( 3)
(1 4 4 h h 2 ) 3
lim
lim
4, slope
h
h
h
h 0
h 0
h 0
20. At x 2, y 3 m lim
1
1
2 (2 h )
h
21. At x 3, y 12 m lim (3 h )h1 2 lim 2h (2 h) lim 2h(2
14 , slope
h)
h 0
h 0
h 0
h 1 ( 1)
22. At x 0, y 1 m lim h 1 h
h 0
lim
h 0
( h 1) ( h 1)
lim h (2hh1) 2
h ( h 1)
h 0
23. (a) It is the rate of change of the number of cells when t 5. The units are the number of cells per hour.
(b) P (3) because the slope of the curve is greater there.
(c)
6.10(5 h ) 2 9.28(5 h ) 16.43 [6.10(5) 2 9.28(5) 16.43]
61.0h 6.10h 2 9.28h
lim
h 0
h 0
h
h
lim 51.72 6.10h 51.72 52 cells/h.
P (5) lim
h 0
24. (a) From t 0 to t 3, the derivative is positive.
(b) At t 3, the derivative appears to be 0. From t 2 to t 3, the derivative is positive but decreasing.
[( x h )2 4( x h ) 1]( x 2 4 x 1)
h
h 0
25. At a horizontal tangent the slope m 0 0 m lim
( x 2 2 xh h 2 4 x 4 h 1) ( x 2 4 x 1)
(2 xh h 2 4 h )
lim
lim (2 x h 4) 2 x 4; 2 x 4 0
h
h
h 0
h 0
h 0
lim
x 2. Then f (2) 4 8 1 5 (2, 5) is the point on the graph where there is
a horizontal tangent.
2
2
3
[( x h )3 3( x h )]( x3 3 x )
( x3 3 x 2 h 3 xh 2 h3 3 x 3h ) ( x3 3 x )
lim
lim 3 x h 3 xhh h 3h
h
h
h 0
h 0
h 0
26. 0 m lim
lim (3x 2 3xh h 2 3) 3 x 2 3; 3x 2 3 0 x 1 or x 1. Then f (1) 2 and f (1) 2 (1, 2)
h 0
and (1, 2) are the points on the graph where a horizontal tangent exists.
1
1
( x 1) ( x h 1)
27. 1 m lim ( x h ) h1 x 1 lim h ( x 1)( x h 1) lim h( x 1)( hx h 1)
h 0
h 0
h 0
1 ( x 1) 2 1 x 2 2 x 0
( x 1)2
x ( x 2) 0 x 0 or x 2. If x 0, then y 1 and m 1 y 1 ( x 0) ( x 1). If x 2,
then y 1 and m 1 y 1 ( x 2) ( x 3).
( x h) x
h
lim
1 .
28. 14 m lim x hh x lim x hh x x h x lim
x h x h 0 h x h x h 0 h x h x
2 x
h 0
h 0
Thus, 14 1 x 2 x 4 y 2. The tangent line is y 2 14 ( x 4) 4x 1.
2 x
f (2 h ) f (2)
(100 4.9(2 h ) 2 ) (100 4.9(2) 2 )
4.9(4 4h h 2 ) 4.9(4)
lim
lim
lim (19.6 4.9h) 19.6.
h
h
h
h 0
h 0
h 0
h 0
29. lim
The minus sign indicates the object is falling downward at a speed of 19.6 m/s.
f (10 h ) f (10)
3(10 h )2 3(10) 2
lim
h
h
h 0
h 0
30. lim
3(20 h h 2 )
60 m/s.
h
h 0
lim
f (3 h ) f (3)
(3 h )2 (3)2 )
[9 6 h h 2 9]
lim
lim
lim (6 h) 6
h
h
h
h 0
h 0
h 0
h 0
31. lim
Copyright 2016 Pearson Education, Ltd.
102
Chapter 3 Derivatives
4
3
4
(2 h ) 3 (2)
f (2 h ) f (2)
lim 3
h
h
h 0
h 0
32. lim
3
4
lim 3
h 0
[12 h 6 h 2 h3 ]
lim 43 [12 6h h 2 ] 16
h
h 0
33. At ( x0 , mx0 b) the slope of the tangent line is lim
h 0
( m ( x0 h ) b ) ( mx0 b )
mh
lim h lim m m.
( x0 h ) x0
h 0
h 0
The equation of the tangent line is y (mx0 b) m( x x0 ) y mx b.
41 h 12 2 4 h
2 4 h
lim
lim 2 4 h 2 4 h
h
lim
h
4
2
4
h
2
h
4
h
h 0
h 0 2 h 4 h 2 4 h
h 0
h0
4 (4 h )
h
1
1
1
16
lim
lim
lim
2 4 2 4
h 0 2 h 4 h 2 4 h h 0 2 h 4 h 2 4 h h 0 2 4 h 2 4 h
34. At x 4, y 1 12 and m lim
1 1
4 h 2
1 lim h sin 1 0 yes, f ( x) does have a tangent at the
2
h sin h
f (0 h ) f (0)
lim
h
h
h 0
h 0
35. Slope at origin lim
origin with slope 0.
1 lim sin 1 . Since lim sin 1 does not exist, f ( x) has no tangent at the origin.
h sin
g (0 h ) g (0)
lim h h
h
h 0
h 0
36. lim
37.
h
h0
lim
h 0
f (0 h ) f (0)
h
lim
h 0
h
h 0
1 0 , and
h
h0
h
f (0 h ) f (0)
h
lim
h 0
yes, the graph of f has a vertical tangent at the origin.
38.
lim
h 0
U (0 h ) U (0)
h
lim 0h1 , and lim
h 0
h 0
U (0 h ) U (0)
h
vertical tangent at (0, 1) because the limit does not exist.
lim 1h0 . Therefore, lim
h 0
h 0
f (0 h ) f (0)
h
lim 1h1 0 no, the graph of f does not have a
h 0
39. (a) The graph appears to have a cusp at x 0.
(b)
lim
h 0
2/5
f (0 h ) f (0)
1 and lim 1 limit does not exist the graph
lim h h 0 lim 3/5
3/5
h
h
h 0 h
h 0
h 0
2/5
of y x
does not have a vertical tangent at x 0.
40. (a) The graph appears to have a cusp at x 0.
(b)
lim
h 0
f (0 h ) f (0)
h
lim
h 0
h 4/5 0
h
lim
does not have a vertical tangent at x 0.
1
1/5
h 0 h
and lim
1
1/5
h 0 h
limit does not exist y x 4/5
Copyright 2016 Pearson Education, Ltd.
Section 3.1 Tangents and the Derivative at a Point
103
41. (a) The graph appears to have a vertical tangent
at x 0.
1/5
f (0 h ) f (0)
1 y x1/5 has a vertical tangent at x 0.
lim h h0 lim 4/5
h
h 0
h 0 h
(b) lim
h 0
42. (a) The graph appears to have a vertical tangent
at x 0.
h3/5 0
f (0 h ) f (0)
lim
lim 21 5 the graph of y x3/5 has a vertical tangent at x 0.
h
h
h 0
h 0
h 0 h
(b) lim
43. (a) The graph appears to have a cusp at x 0.
(b)
lim
h 0
f (0 h ) f (0)
h
lim
h 0
2/5
the graph of y 4 x
4 h 2/5 2 h
h
lim
h 0
4 2
h3/5
and lim
h 0
4 2
h3/5
limit does not exist
2 x does not have a vertical tangent at x 0.
44. (a) The graph appears to have a cusp at x 0.
(b)
5/3
2/3
f (0 h ) f (0)
lim h h5h lim
h
h 0
h 0
h 0
lim
h 0 lim
2/3
5
h1/3
5
1/3
h 0 h
does not exist the graph of
y x5/3 5 x 2/3 does not have a vertical tangent at x 0.
Copyright 2016 Pearson Education, Ltd.
104
Chapter 3 Derivatives
45. (a) The graph appears to have a vertical tangent
at x 1 and a cusp at x 0.
(b) x 1:
(1 h )2/3 (1 h 1)1/3 1
(1 h ) 2/3 h1/3 1
lim
y x 2/3 ( x 1)1/3 has a vertical tangent
h
h
h 0
h 0
lim
at x 1;
x 0:
f (0 h ) f (0)
h 2/3 ( h 1)1/3 ( 1)1/3
1 ( h 1)1/3 1
lim
lim
h h does not exist
1/3
h
h
h 0
h 0
h 0 h
2/3
1/3
lim
yx
( x 1)
does not have a vertical tangent at x 0.
46. (a) The graph appears to have vertical tangents
at x 0 and x 1.
(b) x 0:
x 1:
f (0 h ) f (0)
h1/3 ( h 1)1/3 ( 1)1/3
lim
y x1/3 ( x 1)1/3 has a vertical tangent at x 0;
h
h
h 0
h 0
lim
f (1 h ) f (1)
(1 h )1/3 (1 h 1)1/3 1
lim
y x1/3 ( x 1)1/3 has a vertical tangent at x 1.
h
h
h 0
h 0
lim
47. (a) The graph appears to have a vertical tangent
at x 0.
(b)
|h| 0
|h|
f (0 h ) f (0)
f (0 h ) f (0)
lim hh0 lim 1 ; lim
lim
lim |h|
h
h
h
h
h 0
h 0
h 0
h 0
x 0
h 0
lim 1 y has a vertical tangent at x 0.
h 0 |h|
lim
Copyright 2016 Pearson Education, Ltd.
Section 3.2 The Derivative as a Function
48. (a) The graph appears to have a cusp at x 4.
(b)
49-52.
|4 (4 h )| 0
|4 (4 h )|
|h|
f (4 h ) f (4)
lim
lim h lim 1 ; lim
lim
h
h
h
h 0
h 0 h
h 0
h 0
h 0
|h|
lim | h | lim 1 y 4 x does not have a vertical tangent at x 4.
h 0 | h |
h 0
lim
h 0
f (4 h ) f (4)
h
Example CAS commands:
Maple:
f : x - x^3 2*x;x0 : 0;
plot( f (x), x x0-1/2..x0 3, color black,
# part (a)
title "Section 3.1, #49(a)" );
q : unapply( (f (x0 h)-f (x0))/h, h );
# part (b)
L : limit( q(h), h 0 );
sec_lines : seq( f(x0) q(h)*(x-x0), h 1..3 );
tan_ line : f(x0) L*(x-x0);
plot( [f(x),tan_line,sec_lines], x x0-1/2..x0 3, color black,
# part (c)
# part (d)
linestyle [1,2,5,6,7], title "Section 3.1, #49(d)",
legend ["y f(x)","Tangent line at x 0","Secant line (h 1)",
"Secant line (h 2)","Secant line (h 3)"] );
Mathematica: (function and value for x0 may change)
Clear[f , m, x, h]
x0 p;
f[x_ ]: Cos[x] 4Sin[2x]
Plot[f [x],{x, x0 1, x0 3}]
dq[h_ ]: (f [x0 h] f [x0])/h
m Limit[dq[h], h 0]
ytan: f [x0] m(x x0)
y1: f [x0] dq[1](x x0)
y2: f [x0] dq[2](x x0)
y3: f [x0] dq[3](x x0)
Plot[{f [x], ytan, y1, y2, y3}, {x, x0 1, x0 3}]
3.2
THE DERIVATIVE AS A FUNCTION
1. Step 1:
Step 2:
Step 3:
f ( x) 4 x 2 and f ( x h) 4 ( x h) 2
2
f ( x h ) f ( x ) [4 ( x h )2 ](4 x 2 )
(4 x 2 2 xh h 2 ) 4 x 2
h ( 2 x h )
2 xhh h
2 x h
h
h
h
h
f ( x) lim (2 x h) 2 x; f (3) 6, f (0) 0, f (1) 2
h 0
Copyright 2016 Pearson Education, Ltd.
105
106
Chapter 3 Derivatives
[( x h 1)2 1][( x 1)2 1]
h
h 0
2. F ( x) ( x 1)2 1 and F ( x h) ( x h 1)2 1 F ( x ) lim
2
( x 2 2 xh h 2 2 x 2h 11) ( x 2 2 x 11)
lim 2 xh hh 2h lim (2 x h 2) 2( x 1);
h
h0
h 0
h 0
lim
F (1) 4, F (0) 2, F (2) 2
g (t ) 12 and g (t h)
1
(t h ) 2
t 2 ( t h )2
1 1
( t h )2 t 2
2
2
2
g (t h ) g ( t )
t 2 (t 2 2th h2 )
h ( 2t h )
(t h )h t h
2th 2 h2
2t 2h2
2 2
h
(t h ) t h
(t h ) t h
(t h ) 2 t 2 h
(t h ) t
g (t ) lim 2t 2h 2 2 2t2 32 ; g (1) 2, g (2) 14 , g 3 2
3 3
t t
t
h 0 ( t h ) t
3. Step 1:
t
Step 2:
Step 3:
4.
1 ( z h )
1 z
2
2( z h ) 2 z
1( z h )
(1 z h ) z (1 z )( z h )
zh z h z 2 zh
k ( z ) 12zz and k ( z h) 2( z h) k ( z ) lim
lim
lim z z 2(
h
2(
z
h
)
zh
z h ) zh
h 0
h 0
h 0
h
1
1
1
1
1
lim 2( z h) zh lim 2( z h ) z 2 ; k (1) 2 , k (1) 2 , k
2 4
2z
h 0
h 0
p ( ) 3 and p ( h) 3( h)
5. Step 1:
3( h ) 3
h
p ( h ) p ( )
h
Step 2:
3h
h
3
3 3h 3
3 3h 3
3
h0 3 3h 3
p ( ) lim
Step 3:
3 3h 3 3 3h 3 (3 3h)3
h
3 3h 3 h 3 3h 3
2 32
3
3 ; p (1) 3 , p (3) 12 , p 23
3 3
2 3
2 3
6. r ( s ) 2 s 1 and r ( s h) 2( s h) 1 r ( s ) lim
h 0
2 s 2 h 1 2 s 1
h
2s h1 2 s 1 2s 2h1 2s 1 lim (2 s 2h1)(2s 1)
h
h 0
2s 2h1 2s 1 h0 h 2s 2h1 2s 1
lim
lim
h 0 h
2h
2s 2h1 2s 1
2
h 0 2 s 2 h 1 2 s 1
lim
2
2
2 s 1 2 s 1
2 2 s 1
1 ;
2 s 1
r (0) 1, r (1) 1 , r 12 1
3
2
dy
2( x h )3 2 x3
2( x3 3 x 2 h 3 xh 2 h3 ) 2 x3
lim
h
h
h 0
h 0
7. y f ( x) 2 x3 and f ( x h) 2( x h)3 dx lim
2
2
3
h (6 x 2 6 xh 2 h 2 )
lim (6 x 2 6 xh 2h 2 ) 6 x 2
h
h 0
h 0
lim 6 x h 6hxh 2 h lim
h 0
3
2
2
3
2
2
3
2
(( s h )3 2( s h )2 3) ( s 3 2 s 2 3)
lim s 3s h 3sh h 2 s h 4 sh 2h 3 s 2 s 3
h
h 0
h 0
8. r s 3 2 s 2 3 dr
lim
ds
2
2
3
2
h (3s 2 3sh h 2 4 s 2 h )
lim (3s 2 3sh h 2 4 s 2h) 3s 2 2s
h
h 0
h 0
lim 3s h 3sh hh 4 sh 2h lim
h 0
Copyright 2016 Pearson Education, Ltd.
Section 3.2 The Derivative as a Function
lim
9. s r (t ) 2tt1 and r (t h) 2(tthh) 1 ds
dt
lim
t h
2tt1
2( t h ) 1
( t h )(2 t 1) t (2t 2 h 1)
(2t 2 h 1)(2 t 1)
h
h
107
h 0
h 0
2
2
(t h )(2t 1) t (2t 2 h 1)
2
t
t
2
ht
h
2
t
2
ht
t
lim (2t 2h 1)(2t 1) h lim
lim (2t 2h h1)(2t 1) h lim (2t 2h 11)(2t 1)
(2t 2 h 1)(2t 1) h
h 0
h 0
h 0
h 0
1
1 2
(2t 1)(2
t 1)
(2t1)
h (t h )t t (t h )
(t h )t
10.
(t h ) 1 (t 1 )
h t 1h 1t
t h
t
dv lim
lim
lim
h
h
dt
h 0
h 0
h0
11.
dp
( q h )3/2 q 3/2
( q h )( q h )1/2 qq1/2
lim
lim
lim
dq h0
h
h
h 0
h 0
1/2
1/2
1/2
1/2
lim ht h t h lim t ht 1 t 1 1 1
2
2
h 0 h (t h )t
h
q [( q h )1/2 q1/2 ]
h
1/2
h ( qhh )
2
2
h 0 (t h )t
t2
t2
1/2
][( q h ) q ]
lim q[( qh )h[(qqh )1/2
( q h )1/2 lim h[(qq[(qh)1/2h )qq1/2] ] ( q h )1/2 lim ( qh )1/2q q1/2 ( q h )1/2 q2 q1/2 23 q1/2
q1/2 ]
h 0
12.
dz lim
dw h0
1
( w h )2 1
h
lim
w2 1 ( w h )2 1
2
w2 1 ( w2 2 wh h 2 1)
lim
h0 h ( w h )2 1 w2 1
2
2
2
2
2
lim
2
2
2
2
2
2
2 w h
lim
w 1 ( wh) 1 h0 h ( wh) 1 w 1 w 1 ( wh) 1
2
2
13. f ( x) x 9x and f ( x h) ( x h) ( x 9 h)
3
h 0
w 1 ( wh) 1 w 1 ( wh) 1
h0 h ( w h ) 1 w 1
h 0 h ( w h ) 1 w 1 w 1 ( w h ) 1
1
w2 1
h 0
3
2
2
2
2
2
2
w
( w2 1)3/2
( x h ) 9 x 9
( xh)
x
f ( x h) f ( x)
x ( x h )2 9 x x 2 ( x h ) 9( x h )
h
h
x( x h)h
2
9 h
x 2 x h xh x( 9x x h)xh x h 9 x 9h x xh(xxh
h) h
2
h ( x 2 xh 9)
9 ;
xx (xxh
x( x h)h
h)
2
2
9 x 9 1 9 ;
f ( x) lim xx (xxh
2
2
h)
x
h 0
x
m f (3) 0
k ( x h) k ( x )
lim
h
h 0
h 0
h
1
1 ; k (2) 1
lim h (2 x )(2
lim
x h ) h 0 (2 x )(2 x h )
16
(2 x ) 2
h 0
14. k ( x) 21 x and k ( x h) 2 ( 1x h) k ( x) lim
15.
3
2
2
3
2
2
3
2
2
3
ds lim [(t h ) (t h ) ](t t ) lim (t 3t h 3th h ) (t 2th h ) t t
dt
h
h
h 0
h 0
h 0 h (2 x )(2 x h )
h
2
2
2
3
lim 3t h 3th hh 2th h
2
h 0
h (3t 2 3th h 2 2t h )
2
2
2
ds
lim
lim (3t 3th h 2t h) 3t 2t ; m dt
5
h
t1
h 0
h 0
( x h) 3
16.
3
2 1x h 21 x lim (2 x)(2 x h)
x 3
dy
1 ( x h ) 1 x
lim
lim
dx h0
h
h 0
lim (1 x h4)(1 x )
h 0
17. f ( x)
( x h 3)(1 x ) ( x 3)(1 x h )
(1 x h )(1 x )
h
2
2
3 x x 3 x 3 x xh 3h lim
4h
lim x h 3 x xh
h (1 x h )(1 x )
h(1 x h)(1 x )
h 0
h 0
4 ; dy
4 2 94
dx x2
(1 x )2
(3)
8 and f ( x h)
x2
8[( x 2) ( x h 2)]
f ( x h) f ( x )
8
h
( x h) 2
8
8
( x h)2
x 2
h
x2 x h2 x2 x h2
h x h2 x 2
x 2 x h2
8
8h
8
f ( x) lim
h x h2 x 2 x 2 x h2
h x h2 x2 x 2 x h2
h 0 x h 2 x 2 x 2 x h 2
8
4
; m f (6) 4 12 the equation of the tangent line at (6, 4) is
( x 2) x 2
4 4
x2 x 2 x 2 x 2
1
1
1
y 4 2 ( x 6) y 2 x 3 4 y 2 x 7.
Copyright 2016 Pearson Education, Ltd.
108
Chapter 3 Derivatives
18. g ( z ) lim
(1 4 ( z h ) ) 1 4 z
h
h 0
h
lim 4 z h 4 z 4 z h 4 z lim (4 z h)(4 z )
h
h 0
4 z h 4 z h 0 h 4 z h 4 z
1 ; m g (3) 1 1 the equation
2
2 4 z
2 4 3
of the tangent line at (3, 2) is w 2 12 ( z 3) w 12 z 23 2 w 12 z 72 .
lim
h 0 h
4 z h 4 z
lim
h 0
1
4 z h 4 z
f (t h ) f (t )
h
19. s f (t ) 1 3t 2 and f (t h) 1 3(t h)2 1 3t 2 6th 3h 2 ds
lim
dt
h 0
(13t 2 6th 3h 2 ) (13t 2 )
lim (6t 3h) 6t ds
6
h
dt t 1
h 0
h 0
lim
dy
lim 1x x 1 h
1 x 1 h 1 1x
f ( x h) f ( x)
lim
h
h
h 0
h 0
20. y f ( x) 1 1x and f ( x h) 1 x 1 h dx lim
h 0
dy
lim x ( x h h) h lim x ( x1 h) 12 dx
13
x
h 0
h 0
x 3
h
2
2
f ( h ) f ( )
4 h
4
2 and f ( h)
2
ddr lim
lim
lim 2 4 2 4 h
h
h
4
4 ( h )
h 0
h 0 h 4 4 h
h 0
2
4
2
4
h
4(4 ) 4(4 h )
lim 2 4 2 4 h
lim
2 4 2 4 h
h0 h 4 4 h
h 0 2 h 4 4 h 4 4 h
21. r f ( )
lim
h 0 4 4 h
2
2
4 4 h
(4 ) 2 4
22. w f ( z ) z z and f ( z h) ( z h)
1
ddr
1
(4 ) 4
0 8
z h z h ( z z )
f ( z h) f ( z )
lim
h
h
h 0
h 0
z h dw
lim
dz
z h z 1 lim ( z h) z 1 lim 1 1 1
lim h z hh z lim 1 z hh z
2 z
h 0 h z h z
h 0 z h z
h 0
h 0
z h z
dw
54
dz
z 4
1 1
f ( z ) f ( x)
( x 2) ( z 2)
z2 x2
z
1
lim
lim ( z x )( z 2)( x 2) lim ( z x )( zx2)(
lim
1 2
zx
x 2) z x ( z 2)( x 2)
( x 2)
zx z x
zx
zx
zx
23. f ( x) lim
2
2
2
2
f ( z ) f ( x)
( z x )( z x ) 3( z x )
( z 2 3 z 4) ( x 2 3 x 4)
lim
lim z 3 zz xx 3 x lim z x z 3x z 3 x lim
z
x
z
x
zx
zx
zx
zx
zx
zx
( z x ) ( z x ) 3
lim
lim ( z x) 3 2 x 3
zx
zx
zx
24. f ( x) lim
z x
g ( z ) g ( x)
z ( x 1) x ( z 1)
z 1 x 1
x
1
lim
lim ( z x )( z 1)( x 1) lim ( z x )(zz1)(
lim
1 2
zx
x 1) z x ( z 1)( x 1)
( x 1)
zx
zx zx
zx
zx
25. g ( x) lim
26. g ( x) lim
zx
g ( z ) g ( x)
(1 z ) (1 x )
lim
lim
zx
zx
zx
zx
z x
zx
z x lim
lim
zx
z x
z x ( z x )( z x )
zx
1
1
z x
2 x
27. Note that as x increases, the slope of the tangent line to the curve is first negative, then zero (when x 0), then
positive the slope is always increasing which matches (b).
28. Note that the slope of the tangent line is never negative. For x negative, f 2 ( x) is positive but decreasing as
x increases. When x 0, the slope of the tangent line to x is 0. For x 0, f 2 ( x) is positive and increasing. This
graph matches (a).
Copyright 2016 Pearson Education, Ltd.
Section 3.2 The Derivative as a Function
109
29. f3 ( x) is an oscillating function like the cosine. Everywhere that the graph of f3 has a horizontal tangent we
expect f3 to be zero, and (d) matches this condition.
30. The graph matches with (c).
31. (a) f is not defined at x 0, 1, 4. At these points, the left-hand and right-hand derivatives do not agree. For
f ( x ) f (0)
f ( x ) f (0)
example, lim
slope of line joining (4, 0) and (0, 2) 12 but lim
slope of line
x 0
x 0
x 0
x 0
f ( x ) f (0)
joining (0, 2) and (1, 2) 4. Since these values are not equal, f (0) lim
does not exist.
x 0
x 0
(b)
(b) Shift the graph in (a) down 3 units
32. (a)
33.
y’
2
1
6
7
8
9
10
11
x
1
2
3
4
5
34. (a)
(b) The fastest is between the 20th and 30th days;
slowest is between the 40th and 50th days.
35. Answers may vary. In each case, draw a tangent line and estimate its slope.
(a) i) slope 0.77 dT
0.77 °C
ii) slope 1.43 dT
1.43 °C
dt
h
dt
h
iii) slope 0 dT
0 °C
dt
h
iv) slope 1.88 dT
1.88 °C
dt
h
Copyright 2016 Pearson Education, Ltd.
110
Chapter 3 Derivatives
(b) The tangent with the steepest positive slope appears to occur at t 6 12 p.m. and slope 3.63
dT
3.63 °C
. The tangent with the steepest negative slope appears to occur at t 12 6 p.m. and
dt
h
slope 4.00 dT
4.00 °C
dt
h
(c)
36. Answers may vary. In each case, draw a tangent line and estimate the slope.
kg
kg
(a) i) slope 10.4 dW
10.4 month
dt
ii) slope 17.5 dW
17.5 month
dt
kg
iii) slope 3.1 dW
3.1 month
dt
(b) The tangent with the steepest negative slope appears to occur at t 2.7 months. and slope 27
kg
dW
27 month
dt
(c)
f (0 h ) f (0)
h
f (0 h) f (0)
h
37. Left-hand derivative: For h 0, f (0 h) f (h) h 2 (using y x 2 curve) lim
h 0
2
lim h h0 lim h 0;
h 0
h 0
Right-hand derivative: For h 0, f (0 h) f (h) h (using y x curve) lim
f (0 h ) f (0)
h
lim h h 0 lim 1 1; Then lim
h 0
h 0
h 0
h 0
f (0 h ) f (0)
the
derivative
f (0) does not exist.
h
lim
h 0
f (1 h ) f (1)
h
lim 2h 2 lim 0 0;
h 0
h 0
f (1 h ) f (1)
(2 2 h ) 2
Right-hand derivative: When h 0, 1 h 1 f (1 h) 2(1 h) 2 2h lim
lim
h
h
h 0
h 0
2
h
lim h lim 2 2;
h 0
h 0
f (1 h ) f (1)
f (1 h ) f (1)
Then lim
lim
the derivative f (1) does not exist.
h
h
h 0
h 0
38. Left-hand derivative: When h 0, 1 h 1 f (1 h) 2 lim
h 0
39. Left-hand derivative: When h 0,1 h 1 f (1 h) 1 h lim
1 h 1 1 h 1 lim (1 h)1 lim 1 1 ;
h
1 h 1 h0 h 1 h 1 h0 1 h 1 2
h 0
h 0
f (1 h ) f (1)
h
lim
Copyright 2016 Pearson Education, Ltd.
lim
h 0
1 h 1
h
Section 3.2 The Derivative as a Function
Right-hand derivative: When h 0,1 h 1 f (1 h) 2(1 h) 1 2h 1 lim
(2 h 1) 1
lim 2 2;
h
h 0
h 0
f (1 h ) f (1)
f (1 h ) f (1)
lim
the derivative f (1) does not exist.
Then lim
h
h
h 0
h 0
h 0
f (1 h ) f (1)
h
lim
40. Left-hand derivative: lim
h 0
f (1 h ) f (1)
h
lim
h 0
(1 h ) 1
h
lim 1 1;
h 0
1 (1 h )
1 1
1 h
f (1 h ) f (1)
1 h
Right-hand derivative: lim
lim
lim
lim h(1hh ) lim 11h 1;
h
h
h
h 0
h 0
h 0
h 0
h 0
f (1 h ) f (1)
f (1 h ) f (1)
Then lim
lim
the derivative f (1) does not exist.
h
h
h 0
h 0
41. f is not continuous at x 0 since lim f ( x) does not exist and f (0) 1
x 0
1/3
g ( h ) g (0)
1 ;
lim h h0 lim 2/3
h
h
h 0
h 0
h 0
g ( h ) g (0)
h 2/3 0 lim 1 ;
Right-hand derivative: lim
lim
1/3
h
h
h 0 h
h 0
h 0
g ( h ) g (0)
g ( h ) g (0)
lim
the derivative g (0) does not exist.
Then lim
h
h
h 0
h 0
42. Left-hand derivative: lim
43. (a) The function is differentiable on its domain 3 x 2 (it is smooth)
(b) none
(c) none
44. (a) The function is differentiable on its domain 2 x 3 (it is smooth)
(b) none
(c) none
45. (a) The function is differentiable on 3 x 0 and 0 x 3
(b) none
(c) The function is neither continuous nor differentiable at x 0 since lim f ( x ) lim f ( x)
h 0
h 0
46. (a) f is differentiable on 2 x 1, 1 x 0, 0 x 2, and 2 x 3
(b) f is continuous but not differentiable at x 1: lim f ( x) 0 exists but there is a corner at x 1 since
x 1
f ( 1 h ) f ( 1)
f ( 1 h ) f ( 1)
lim
3 and lim
3 f (1) does not exist
h
h
h 0
h 0
(c) f is neither continuous nor differentiable at x 0 and x 2:
at x 0, lim f ( x) 3 but lim f ( x) 0 lim f ( x ) does not exist;
x 0
x 0
x 2
x 2
x 0
at x 2, lim f ( x ) exists but lim f ( x ) f (2)
47. (a) f is differentiable on 1 x 0 and 0 x 2
(b) f is continuous but not differentiable at x 0: lim f ( x) 0 exists but there is a cusp at x 0,
f (0 h ) f (0)
so f (0) lim
does not exist
h
(c) none
x 0
h 0
Copyright 2016 Pearson Education, Ltd.
111
112
Chapter 3 Derivatives
48. (a) f is differentiable on 3 x 2, 2 x 2, and 2 x 3
(b) f is continuous but not differentiable at x 2 and x 2: there are corners at those points
(c) none
f ( x h) f ( x )
( x h)2 ( x 2 )
x 2 2 xh h 2 x 2 lim ( 2 x h) 2x
lim
lim
h
h
h
h 0
h 0
h 0
h 0
49. (a) f ( x) lim
(b)
(c) y 2 x is positive for x 0, y is zero when x 0, y is negative when x 0
(d) y x 2 is increasing for x 0 and decreasing for 0 x ; the function is increasing on intervals
where y 0 and decreasing on intervals where y 0
f ( x h) f ( x )
lim
h
h 0
h 0
50. (a) f ( x) lim
(b)
x1h x1 lim x( x h) lim
h
h 0 x ( x h ) h
1
12
x
h 0 x ( x h )
(c) y is positive for all x 0, y is never 0, y is never negative
(d) y 1x is increasing for x 0 and 0 x
51. (a)
z 3 x3
3 3
3
3
f ( z ) f ( x)
lim z x lim 3(z zxx )
Using the alternate formula for calculating derivatives: f ( x) lim
z
x
zx
zx
zx
2
( z x )( z 2 zx x 2 )
x 2 x 2 f ( x ) x 2
lim z zx
3
3(
z
x
)
zx
zx
lim
(b)
(c) y is positive for all x 0, and y 0 when x 0; y is never negative
3
(d) y x3 is increasing for all x 0 (the graph is horizontal at x 0 ) because y is increasing where y 0; y is
never decreasing
Copyright 2016 Pearson Education, Ltd.
Section 3.2 The Derivative as a Function
52. (a)
113
z 4 x4
4 4
f ( z ) f ( x)
Using the alternate form for calculating derivatives: f ( x) lim
lim z x
z
x
zx
zx
4
4
3
2
2
3
( z x )( z 3 xz 2 x 2 z x3 )
lim z xz 4 x z x x3 f ( x) x3
4(
z
x
)
zx
zx
z x lim
lim 4(
z x)
zx
(b)
(c) y is positive for x 0, y is zero for x 0, y is negative for x 0
4
(d) y x4 is increasing on 0 x and decreasing on x 0
2
2
2
2
(2( x h )2 13( x h ) 5) (2 x 2 13 x 5)
lim 2 x 4 xh 2h 13 xh13h 5 2 x 13 x 5 lim 4 xh 2hh 13h
h
h 0
h 0
h 0
53. y lim
lim (4 x 2h 13) 4 x 13, slope at x. The slope is 1 when 4 x 13 1 4 x 12 x 3
h 0
y 2 32 13 3 5 16. Thus the tangent line is y 16 (1)( x 3) y x 13 and the point of
tangency is (3, 16).
x h x x h x lim ( x h) x lim 1 1 .
h
h 0
x h x h0 x h x h h0 x h x 2 x
54. For the curve y x , we have y lim
Suppose a, a is the point of tangency of such a line and (1, 0) is the point on the line where it crosses the
0
x-axis. Then the slope of the line is a a( 1)
a a1 which must also equal 1 ; using the derivative formula at
2 a
x a a a1 1 2a a 1 a 1. Thus such a line does exist: its point of tangency is (1, 1), its slope is
1
2 a
2 a
12 ; and an equation of the line is y 1 12 ( x 1) y 12 x 12 .
55. Yes; the derivative of f is f so that f ( x0 ) exists f ( x0 ) exists as well.
56. Yes; the derivative of 3g is 3 g so that g (7) exists 3 g (7) exists as well.
g (t )
57. Yes, lim h(t ) can exist but it need not equal zero. For example, let g (t ) mt and h(t ) t. Then g (0) h(0) 0,
t0
g (t )
but lim h (t ) lim mt
lim m m, which need not be zero.
t0
t 0
t0 t
58. (a) Suppose | f ( x)| x 2 for 1 x 1. Then | f (0)| 02 f (0) 0. Then f (0) lim
h 0
lim
h 0
f (0 h ) f (0)
h
f (h)
f ( h)
f (h)0
f ( h)
lim h . For | h | 1, h 2 f (h) h 2 h h h f (0) lim h 0 by the
h
h0
h 0
Sandwich Theorem for limits.
(b) Note that for x 0,
| f ( x)| | x 2 sin 1x | | x 2 ||sin 1x | | x 2 | 1 x 2 (since 1 sin x 1). x 2 (since 1 sin x 1). By part (a), f is
differentiable at x 0 and f (0) 0.
Copyright 2016 Pearson Education, Ltd.
114
Chapter 3 Derivatives
1 is the derivative of the function y x so
2 x
xh x
gets closer to y 1 as h gets smaller and
h
2 x
59. The graphs are shown below for h 1, 0.5, 0.1 The function y
that
1 lim x h x . The graphs reveal that y
h
2 x
h 0
smaller.
60. The graphs are shown below for h 2,1, 0.5. The function y 3 x 2 is the derivative of the function y x3 so that
( x h )3 x 3
( x h )3 x 3
.
The
graphs
reveal
that
y
gets closer to y 3 x 2 as h gets smaller and smaller.
h
h
h 0
3 x 2 lim
61. The graphs are the same. So we know that for
| x|
f ( x) | x |, we have f ( x) x .
62. Weierstrass’s nowhere differentiable continuous function.
Copyright 2016 Pearson Education, Ltd.
Section 3.3 Differentiation Rules
63-68. Example CAS commands:
Maple:
f : x -> x^3 x^2 - x;
x0 : 1;
plot( f(x), x x0-5..x0 2, color black,
title "Section 3.2, #63(a)" );
q : unapply( f(x h)-f(x))/h, (x,h) );
# (b)
L : limit( q(x,h), h 0 );
# (c)
m : eval( L, x x0 );
tan_line : f(x0) m*(x-x0);
plot( [f(x),tan_line], x x0-2..x0+3, color black,
linestyle [1, 7], title "Section 3.2 #63(d)",
legend ["y f(x)","Tangent line at x 1"] );
Xvals : sort( [x0 2^(-k) $ k 0..5, x0-2^(-k) $ k 0..5 ] ):
# (e)
Yvals : map( f, Xvals ):
evalf[4]( convert(Xvals,Matrix) , convert(Yvals,Matrix) >);
plot( L, x x0-5..x0 3, color black, title "Section 3.2 #63(f )" );
Mathematica: (functions and x0 may vary) (see section 2.5 re. RealOnly ):
Miscellaneous`RealOnly`
Clear[f, m, x, y, h]
x0 π/4;
f[x_ ]: x 2 Cos[x]
Plot[f[x], {x, x0 3, x0 3}]
q[x_,h_ ]: (f[x h] f[x])/h
m[x_ ]: Limit[q[x, h], h 0]
ytan: f[x0] m[x0] (x x0)
Plot[{f[x], ytan},{x, x0 3, x0 3}]
m[x0 1]//N
m[x0 1]//N
Plot[{f[x], m[x]},{x, x0 3, x0 3}]
3.3
DIFFERENTIATION RULES
dy
d ( x 2 ) d (3) 2 x 0 2 x
1. y x 2 3 dx dx
dx
dy
2. y x 2 x 8 dx 2 x 1 0 2 x 1
d2y
dx 2
d2y
dx 2
2
2
2
d (5t 3 ) d (3t 5 ) 15t 2 15t 4 d s d (15t 2 ) d (15t 4 ) 30t 60t 3
3. s 5t 3 3t 5 ds
dt
2
dt
dt
dt
dt
dt
2
4. w 3z 7 7 z 3 21z 2 dw
21z 6 21z 2 42 z d w
126 z 5 42 z 42
2
dz
dz
dy
5. y 43 x3 x dx 4 x 2 1
d2y
dx 2
8x
Copyright 2016 Pearson Education, Ltd.
115
116
Chapter 3 Derivatives
6.
y x3 x2 4x dx x 2 x 14
3
2
dy
d2y
2x 1
dx 2
2
7. w 3 z 2 z 1 dw
6 z 3 z 2 36 12 d w
18 z 4 2 z 3 184 23
2
dz
z
dz
z
z
z
2
8. s 2t 1 4t 2 ds
2t 2 8t 3 22 83 d 2s 4t 3 24t 4 34 244
dt
t
t
dt
t
9. y 6 x 2 10 x 5 x 2 dx 12 x 10 10 x 3 12 x 10 103
dy
x
dy
10. y 4 2 x x 3 dx 2 3x 4 2 34
x
d2y
dx 2
d2y
dx 2
t
12 0 30x 4 12 304
x
0 12 x 5 12
5
x
2
11. r 13 s 2 52 s 1 dr
23 s 3 52 s 2 23 52 d 2r 2s 4 5s 3 24 53
ds
3s
2s
s
ds
s
2
12. r 12 1 4 3 4 ddr 12 2 12 4 4 5 12
124 45 d r2 24 3 48 5 20 6
2
243 485 206
d
d ( x3 x 1) ( x3 x 1) d (3 x 2 )
13. (a) y (3 x 2 ) ( x3 x 1) y (3 x 2 ) dx
dx
(3 x 2 ) (3 x 2 1) ( x3 x 1) (2 x) 5 x 4 12 x 2 2 x 3
(b) y x5 4 x3 x 2 3x 3 y 5 x 4 12 x 2 2 x 3
14. (a)
(b)
y (2 x 3)(5 x 2 4 x) y (2 x 3)(10 x 4) (5 x 2 4 x)(2) 30 x 2 14 x 12
y (2 x 3)(5 x 2 4 x) 10 x3 7 x 2 12 x y 30 x 2 14 x 12
d ( x 5 1 ) ( x 5 1 ) d ( x 2 1)
15. (a) y ( x 2 1) ( x 5 1x ) y ( x 2 1) dx
x
x dx
( x 2 1) (1 x 2 ) ( x 5 x 1 ) (2 x) ( x 2 1 1 x 2 ) (2 x 2 10 x 2) 3 x 2 10 x 2 12
x
(b) y x3 5 x 2 2 x 5 1x y 3x 2 10 x 2 12
x
16. y (1 x 2 )( x3/4 x 3 )
(a) y (1 x 2 ) 34 x 1/4 3 x 4 ( x3/4 x 3 )(2 x)
(b)
3 3 11 x 7/4 1
4
x2
4 x1/ 4 x 4
7/4
1
y x3/4 x 3 x11/4 x 1 y 31/ 4 34 11
x
4
4x
x
x2
17. y 32xx25 ; use the quotient rule: u 2 x 5 and v 3 x 2 u 2 and v 3 y vu2uv
6 x 46 x 215
(3 x 2)
(3 x 2)(2) (2 x 5)(3)
(3 x 2)2
v
19
(3 x 2) 2
18. y 423 x ; use the quotient rule: u 4 3 x and v 3x 2 x u 3 and v 6 x 1 y vu2uv
3x x
(3 x 2 x )( 3) (4 3 x )(6 x 1)
(3 x 2 x )2
2
2
2
v
9 x 3 x 218 x 2 21x 4 9 x 224 x 2 4
(3 x x )
(3 x x )
2
4 ; use the quotient rule: u x 2 4 and v x 0.5 u 2 x and v 1 g ( x) vuuv
19. g ( x) xx 0.5
2
( x 0.5)(2 x ) ( x 2 4)(1)
( x 0.5) 2
2
2
2
2 x x x 2 4 x x 42
( x 0.5)
( x 0.5)
Copyright 2016 Pearson Education, Ltd.
v
Section 3.3 Differentiation Rules
(t 2)(1) (t 1)(1) t 2 t 1
(t 1)(t 1)
20. f (t ) 2t 1 (t 2)(t 1) tt12 , t 1 f (t )
2
2
2
t t 2
(t 2)
(1 t 2 )( 1) (1t )(2t )
21. v (1 t ) (1 t 2 ) 1 1t2 dv
dt
(1 t 2 )2
1t
s 1
f ( s )
s 1
23. f ( s )
d (
ds
NOTE:
( s 1)
17
(2 x 7)2
( s 1) ( s 1)( s 1)
1
2 s
1
2 s
( s 1)2
(2 x )(5) (5 x 1)
4x
25. v 1 x x4 x v
27. y
(2 x 7)
2
(1t )
1
s ( s 1) 2
2 s ( s 1) 2
2 s
2 x
2
s ) 1 from Example 2 in Section 3.2
24. u 5 x 1 du
dx
26. r 2
2
1
(t 2) 2
1t 22 t 2 2t t 22t 21
(1t )
(2 x 7)(1) ( x 5)(2)
22. w 2xx57 w
2 x 7 2 x 2 10
2
(2 x 7)
(t 2)
5 x1
1
x
4 x3/ 2
x 1 2 (1 x 4 x )
x
x2
2 x21
x
(0) 1 1
2
1 1 1
r 2
2
3/ 2 1/ 2
1
1
; use the quotient rule: u 1 and v ( x 2 1) ( x 2 x 1) u 0 and
( x 2 1)( x 2 x 1)
2
2
3
2
3
2
3
2
dy
v ( x 1)(2 x 1) ( x x 1)(2 x) 2 x x 2 x 1 2 x 2 x 2 x 4 x 3 x 1 dx vu2uv
v
0 1(4 x3 3 x 2 1)
3
2
2 2 2
24 x2 32x 1 2
2
( x 1) ( x x 1)
( x 1) ( x x 1)
( x 1)( x 2)
2
28. y ( x 1)( x 2) x 2 3 x 2 y
x 3 x 2
( x 2 3 x 2)(2 x 3) ( x 2 3 x 2)(2 x 3)
2
( x 1) ( x 2)
2
2
6 x 2 12 6( x 2)
2
2
2
( x 1) ( x 2)
( x 1) ( x 2) 2
29.
y 12 x 4 32 x 2 x y 2 x3 3x 1 y 6 x 2 3 y 12 x y (4) 12 y ( n) 0 for all n 5
30.
1 x5 y 1 x 4 y 1 x3 y 1 x 2 y (4) x y (5) 1 y ( n ) 0 for all n 6
y 120
24
6
2
31.
y ( x 1)( x 2)( x 3) y ( x 2)( x 3) ( x 1)( x 3) ( x 1)( x 2) x 2 5 x 6 x 2 2 x 3
x 2 x 2 3x 2 8 x 1 y 6 x 8 y 6 y ( n ) 0 for n 4.
32.
y (4 x 3 3x )(2 x ) x 4 x 3 8 x 2 3 x 6 x 4 x 4 8 x 3 3 x 2 6 x y 16 x 3 24 x 2 6 x 6
y 48 x 2 48 x 6 y 96 x 48 y iv 96 y ( n ) 0 for n 5
3
33. y x x 7 x 2 7 x 1 dx 2 x 7 x 2 2 x 72
dy
x
d2y
dx 2
2 14 x 3 2 143
x
Copyright 2016 Pearson Education, Ltd.
117
118
Chapter 3 Derivatives
2
34. s t 52t 1 1 5t 12 1 5t 1 t 2 ds
0 5t 2 2t 3 5t 2 2t 3 25 23
dt
2
t
t
t
d 2s 10t 3 6t 4 103 64
dt
35. r
t
( 1)( 2 1)
3
t
t
3
2
31 1 13 1 3 ddr 0 3 4 3 4 34 d r2 12 5 12
5
( x 2 x )( x 2 x 1)
x ( x 1)( x 2 x 1)
d
3
4
x ( x 1)
x 4 x 1 x4 1 x 3
x4
x4
x
x
2
4
4
5 12
du
3
d
u
dx 0 3x 3 x 4 2 12 x 5
x
dx
x
36. u
x4
37. w 133z z (3 z ) 13 z 1 1 (3 z ) z 1 13 3 z z 1 83 z dw
z 2 0 1 z 2 1 21 1
dz
2
d w
2z
dz 2
3
0 2z
3
z
23
z
dw
d 2w
4 z 3 2 12 z 2
dz
dz
38.
w ( z 1)( z 1)( z 2 1) ( z 2 1)( z 2 1) z 4 1
39.
q2 3 q4 1 q6 3q4 q 2 3 q 2 1 q 2 q 4
p
12 4 12
4
12q 4
12q q3
40. p
dp q q 3
d 2 p 1 q 4
q 5 2
5q 6
dq 6
6
6
2
dq
q 2 3
3
3
q 2 3
3
2
3
2
( q 1) ( q 1)
( q 3q 3q 1) ( q 3q 3q 1)
d2p
3
1
2 q 3
dq
q
q 2 3
3
2q 6q
q 2 3
2 q ( q 2 3)
21q 12 q 1 dq 12 q 2 1 2
dp
2q
41. u (0) 5, u (0) 3, v(0) 1, v (0) 2
d (uv ) uv vu d (uv )
(a) dx
u (0)v (0) v(0)u (0) 5 2 (1)(3) 13
dx
x 0
u (0) v(0) ( 1)( 3) (5)(2)
7
v
x 0 v(0)u(0)
(v (0))
( 1)
u (0) v(0) v (0)u (0)
(5)(2) ( 1)( 3)
d v uvvu d v
7
(c) dx
25
u u
dx u x 0
(u (0))
(5)
(b)
d u
dx v
d u
vu2uv dx
v
2
2
(d)
2
2
2
d (7v 2u ) 7v 2u d (7v 2u ) |
x 0 7v (0) 2u (0) 7 2 2( 3) 20
dx
dx
42. u (1) 2, u (1) 0, v(1) 5, v (1) 1
d (uv ) |
(a) dx
x 1 u (1)v (1) v (1)u (1) 2 ( 1) 5 0 2
v (1)u(1)
2( 1) 50
d v
(c) dx
12
u x 1 u (1)v((1)u(1))
(2)
(b)
v (1)u (1) u (1)v(1) 50 2( 1)
d u
2
25
dx v x 1
( v (1))2
(5) 2
2
(d)
2
d (7v 2u ) |
x 1 7v (1) 2u (1) 7 ( 1) 2 0 7
dx
43. y x3 4 x 1. Note that (2, 1) is on the curve: 1 23 4(2) 1
(a) Slope of the tangent at ( x, y ) is y 3x 2 4 slope of the tangent at (2, 1) is y (2) 3(2)2 4 8. Thus
the slope of the line perpendicular to the tangent at (2, 1) is 18 the equation of the line perpendicular to
the tangent line at (2, 1) is y 1 18 ( x 2) or y 8x 54 .
(b) The slope of the curve at x is m 3x 2 4 and the smallest value for m is 4 when x 0 and y 1.
Copyright 2016 Pearson Education, Ltd.
Section 3.3 Differentiation Rules
119
(c) We want the slope of the curve to be 8 y 8 3 x 2 4 8 3x 2 12 x 2 4 x 2. When
x 2, y 1 and the tangent line has equation y 1 8( x 2) or y 8 x 15; When x 2,
y (2)3 4(2) 1 1, and the tangent line has equation y 1 8( x 2) or y 8 x 17.
44. (a) y x3 3 x 2 y 3 x 2 3. For the tangent to be horizontal, we need m y 0 0 3x 2 3
3x 2 3 x 1. When x 1, y 0 the tangent line has equation y 0. The line perpendicular to
this line at (1, 0) is x 1. When x 1, y 4 the tangent line has equation y 4. The line
perpendicular to this line at (1, 4) is x 1.
(b) The smallest value of y is 3, and this occurs when x 0 and y 2. The tangent to the curve at (0, 2)
has slope 3 the line perpendicular to the tangent at (0, 2) has slope 13 y 2 13 ( x 0) or y 13 x 2
is an equation of the perpendicular line.
dy
45. y 42 x dx
( x 2 1)(4) (4 x )(2 x )
( x 2 1)2
x 1
2
2
4 x 2 482x
( x 1)
4( x 2 1)
( x 2 1) 2
. When x 0, y 0 and y
4(0 1)
4, so the tangent
1
to the curve at (0, 0) is the line y 4 x. When x 1, y 2 y 0, so the tangent to the curve at (1, 2) is the
line y 2.
2
( x 4)(0) 8(2 x )
16(2)
46. y 28 y
216 x 2 . When x 2, y 1 and y 2 2 12 , so the tangent line to the
2
2
( x 4)
curve at (2, 1) has the equation y 1 12 ( x 2), or y 2x 2.
x 4
( x 4)
(2 4)
47. y ax 2 bx c passes through (0, 0) 0 a (0) b(0) c c 0; y ax 2 bx passes through (1, 2)
2 a b; y 2ax b and since the curve is tangent to y x at the origin, its slope is 1 at x 0 y 1
when x 0 1 2a (0) b b 1. Then a b 2 a 1. In summary a b 1 and c 0 so the curve is
y x 2 x.
48. y cx x 2 passes through (1, 0) 0 c(1) 1 c 1 the curve is y x x 2 . For this curve, y 1 2 x
and x 1 y 1. Since y x x 2 and y x 2 ax b have common tangents at x 1, y x 2 ax b must
also have slope 1 at x 1. Thus y 2 x a 1 2 1 a a 3 y x 2 3 x b. Since this last curve
passes through (1, 0), we have 0 1 3 b b 2. In summary, a 3, b 2 and c 1 so the curves are
y x 2 3x 2 and y x x 2 .
49. y 8 x 5 m 8; f ( x) 3 x 2 4 x f ( x) 6 x 4;6 x 4 8 x 2 f (2) 3(2) 2 4(2) 4 (2, 4)
50. 8 x 2 y 1 y 4 x 12 m 4; g ( x) 13 x3 23 x 2 1 g ( x) x 2 3 x; x 2 3 x 4 x 4 or x 1
g (4) 13 (4)3 23 (4)2 1 53 , g (1) 13 (1)3 23 ( 1)2 1 56 4, 53 or 1, 56
( x 2)(1) x (1)
2 ; 2 1 4 ( x 2) 2
2
( x 2)2 ( x 2)2
2 x 2 x 4 or x 0 if x 4, y 44 2 2, and if x 0, y 00 2 0 (4, 2) or (0, 0).
51. y 2 x 3 m 2 m 12 ; y x x 2 y
y 8
( x 2) 2
y 8
2
52. m x 3 ; f ( x ) x 2 f ( x) 2 x; m f ( x) x 3 2 x xx 38 2 x x 2 8 2 x 2 6 x x 2 6 x 8 0
x 4 or x 2 f (4) 42 16, f (2) 22 4 (4, 16) or (2, 4).
53. (a) y x3 x y 3 x 2 1. When x 1, y 0 and y 2 the tangent line to the curve at (1, 0) is
y 2( x 1) or y 2 x 2.
Copyright 2016 Pearson Education, Ltd.
120
Chapter 3 Derivatives
(b)
y x3 x
y 2 x 2
(c)
x x 2x 2 x 3x 2 ( x 2)(x 1) 0 x 2 or x 1. Since y 2(2) 2 6; the
3
3
2
other intersection point is (2, 6)
54. (a) y x3 6 x 2 5 x y 3x 2 12 x 5. When x 0, y 0 and y 5 the tangent line to the curve at
(0, 0) is y 5 x.
(b)
y x3 6 x 2 5 x
y 5 x
(c)
x 6x 5x 5x x 6x 0 x ( x 6) 0 x 0 or x 6. Since y 5(6) 30,
3
2
3
2
2
the other intersection point is (6, 30).
50
55. lim xx 11 50 x 49
x 1
56.
x 1
50(1) 49 50
2/9
lim x x 11 92 x 7/9
x 1
x 1
2
92
9( 1)7/9
x 0
x 0 , since g is differentiable at x 0
x 1
x 1 , since f is differentiable at x 1
57. g ( x) 2a x 3
58. f ( x) a2bx
lim (2 x 3) 3 and lim a a a 3
x 0
x 0
lim a a and lim (2bx) 2b a 2b, and
x 1
2
x 1
since f is continuous at x 1 lim (ax b) a b and lim (bx 3) b 3 a b b 3
a 3 3 2b b 32 .
x 1
x 1
59. P( x) an x n an 1 x n 1 a2 x 2 a1 x a0 P ( x) nan x n 1 (n 1)an 1 x n 2 2a2 x a1
dR CM M 2
60. R M 2 C2 M3 C2 M 2 13 M 3 , where C is a constant dM
d (u c ) u dc c du u 0 c du c du . Thus when one of the functions is a
61. Let c be a constant dc
0 dx
dx
dx
dx
dx
dx
constant, the Product Rule is just the Constant Multiple Rule the Constant Multiple Rule is a special case of
the Product Rule.
Copyright 2016 Pearson Education, Ltd.
Section 3.4 The Derivative as a Rate of Change
d 1
62. (a) We use the Quotient rule to derive the Reciprocal Rule (with u 1): dx
v
v0 1 dv
dx
v
2
(b) Now, using the Reciprocal Rule and the Product Rule, we’ll derive the Quotient Rule:
1 dv
dx
v2
121
12 dv
.
dx
v
dxd u 1v u dxd 1v 1v dudx (Product Rule) u v1 dvdx 1v dudx (Reciprocal Rule)
d u u v v u , the Quotient Rule.
dx
v v
v
d u
dx v
2
dv
dx
du
dx
du
dx
2
63. (a)
dv
dx
2
d (uvw) d ((uv ) w) (uv ) dw w d (uv ) uv dw w u dv v du
dx
dx
dx
dx
dx
dx
dx
uvw uv w u vw
(b)
d (u u u u ) d
dx 1 2 3 4
dx
u1u2u3 u4 u1u2u3 dudx4 u4 dxd u1u2u3
du
du
du
du
d u u u u u u u
dx
1 2 3 4 1 2 3 dx4 u4 u1u2 dx3 u3u1 dx2 u3u2 dx1
du
du
du
uv dwdx wu dxdv wv dudx
(using (a) above)
du
d u u u u u u u
dx
1 2 3 4 1 2 3 dx4 u1u2u4 dx3 u1u3u4 dx2 u2u3u4 dx1
u1u2u3u4 u1u2u3 u4 u1u2 u3u4 u1u2u3u4
d (u u ) u u u
(c) Generalizing (a) and (b) above, dx
1
n
1 2
n 1un u1u2 un 2 un 1un u1u2 un
64.
d ( x m ) d 1
dx
dx x m
x m 0 1( m x m 1 )
m 2
(x )
m 1
m2xm
x
m x m 1 2 m m x m 1
2
65. P VnRT
an2 . We are holding T constant, and a, b, n, R are also constant so their derivatives are zero
nb
dP
dV
V
(V nb )0 ( nRT )(1)
(V nb )2
V 2 (0) ( an 2 )(2V )
(V 2 )2
2
nRT 2 2an3
(V nb )
V
kmq h2 ddt A 2(km)q3 2qkm
66.
( km ) q 2 h2
A( q) km
cm 2 ( km ) q 1 cm h2 q dA
dq
q
3.4
THE DERIVATIVE AS A RATE OF CHANGE
hq
2
2
2
3
1. s t 2 3t 2, 0 t 2
(a) displacement s s (2) s (0) 0 m 2 m 2 m, vav st 22 1 m/s
2
(b) v ds
2t 3 | v(0)| | 3| 3 m/s and | v(2)| 1 m/s; a d 2s 2 a (0) 2 m/s 2 and a(2) 2 m/s 2
dt
(c)
dt
v 0 2t 3 0 t 32 . v is negative in the interval 0 t 32 and v is positive when 32 t 2 the
body changes direction at t 32 .
2. s 6t t 2 , 0 t 6
(a) displacement s s (6) s (0) 0 m, vav st 06 0 m/ s
2
(b) v ds
6 2t | v(0)| |6| 6 m/ s and | v(6)| | 6| 6 m/ s; a d 2s 2 a (0) 2 m/ s 2 and
dt
dt
a(6) 2 m/ s 2
(c) v 0 6 2t 0 t 3. v is positive in the interval 0 t 3 and v is negative when 3 t 6 the
body changes direction at t 3.
Copyright 2016 Pearson Education, Ltd.
122
3.
Chapter 3 Derivatives
s t 3 3t 2 3t , 0 t 3
(a) displacement s s (3) s (0) 9 m, vav st 39 3 m/ s
2
(b) v ds
3t 2 6t 3 | v(0)| | 3| 3 m/ s and | v(3)| | 12| 12 m/ s; a d 2s 6t 6
dt
dt
a(0) 6 m/ s 2 and a (3) 12 m/ s 2
(c) v 0 3t 2 6t 3 0 t 2 2t 1 0 (t 1) 2 0 t 1. For all other values of t in the interval
the velocity v is negative (the graph of v 3t 2 6t 3 is a parabola with vertex at t 1 which opens
downward the body never changes direction).
4
4. s t4 t 3 t 2 , 0 t 3
9
(a) s s (3) s (0) 94 m, vav st 34 43 m/ s
(b) v t 3 3t 2 2t | v(0)| 0 m/ s and | v(3)| 6 m/s; a 3t 2 6t 2 a(0) 2 m/ s 2 and
a(3) 11 m/ s 2
(c) v 0 t 3 3t 2 2t 0 t (t 2)(t 1) 0 t 0, 1, 2 v t (t 2)(t 1) is positive in the interval for
0 t 1 and v is negative for 1 t 2 and v is positive for 2 t 3 the body changes direction at t 1
and at t 2.
5. s 252 5t , 1 t 5
t
(a) s s (5) s (1) 20 m, vav 420 5 m/ s
4 m/s 2
(b) v 50
103 a(1) 140 m/ s 2 and a(5) 25
52 | v(1)| 45 m/s and | v(5)| 15 m/ s; a 150
3
4
t
t
t
t
(c) v 0 5035t 0 50 5t 0 t 10 the body does not change direction in the interval
t
6. s t25
, 4t 0
5
(a) s s (0) s (4) 20 m, vav 20
5 m/s
4
25 | v ( 4)| 25 m/ s and | v (0)| 1 m/ s; a 50
a(4) 50 m/ s 2 and a(0) 52 m/ s 2
(t 5)3
(t 5)2
v 0 25 2 0 v is never 0 the body never changes direction
(t 5)
(b) v
(c)
7. s t 3 6t 2 9t and let the positive direction be to the right on the s -axis.
(a) v 3t 2 12t 9 so that v 0 t 2 4t 3 (t 3)(t 1) 0 t 1 or 3; a 6t 12 a(1) 6 m/ s 2 and
a(3) 6 m/ s 2 . Thus the body is motionless but being accelerated left when t 1, and motionless
but being accelerated right when t 3.
(b) a 0 6t 12 0 t 2 with speed | v(2)| |12 24 9| 3 m/s
(c) The body moves to the right or forward on 0 t 1, and to the left or backward on 1 t 2. The positions
are s (0) 0, s (1) 4 and s (2) 2 total distance | s (1) s (0)| | s (2) s (1)| | 4| | 2| 6 m.
8. v t 2 4t 3 a 2t 4
(a) v 0 t 2 4t 3 0 t 1 or 3 a (1) 2 m/s 2 and a(3) 2 m/s 2
(b) v 0 (t 3) (t 1) 0 0 t 1 or t 3 and the body is moving forward; v 0 (t 3)(t 1) 0
1 t 3 and the body is moving backward
(c) velocity increasing a 0 2t 4 0 t 2; velocity decreasing a 0 2t 4 0 0 t 2
9. sm 1.86t 2 vm 3.72t and solving 3.72t 27.8 t 7.5 s on Mars; s j 11.44t 2 v j 22.88t and
solving 22.88t 27.8 t 1.2 s on Jupiter.
Copyright 2016 Pearson Education, Ltd.
Section 3.4 The Derivative as a Rate of Change
123
10 . (a) v(t ) s (t ) 24 1.6t m/s, and a(t ) v (t ) s (t ) 1.6 m/s 2
(b) Solve v(t ) 0 24 1.6t 0 t 15s
(c) s (15) 24(15) .8(15)2 180 m
2
(d) Solve s (t ) 90 24t .8t 2 90 t 3015
4.39 s going up and 25.6 s going down
2
(e) Twice the time it took to reach its highest point or 30 s
11. s 15t 12 g s t 2 v 15 g s t so that v 0 15 g s t 0 g s 15
. Therefore g s 15
43 0.75 m/s 2
t
20
12. Solving sm 250t 0.8t 2 0 t (250 0.8t ) 0 t 0 or 312.5 312.5 s on the moon;
solving se 250t 4.9t 2 0 t (250 4.9t ) 0 t 0 or 51.02 51.02 s on the earth. Also,
vm 250 1.6t 0 t 156.25 and sm (156.25) 19530.75 m, the height it reaches above the moon’s surface;
ve 250 9.8t 0 t 25.51 and se (25.51) 0.05 m, the height it reaches above the earth’s surface.
13. (a) s 56 4.9t 2 v 9.8t speed | v | 9.8t m/s and a 9.8 m/s 2
(b) s 0 56 4.9t 2 0 t
(c) When t
14. (a)
56 3.4 s
4.9
56 , v 9.8 56 32.8 m/s
4.9
4.9
lim v lim 9.8(sin )t 9.8t so we expect v 9.8t m/s in free fall
2
2
(b) a dv
9.8 m/s 2
dt
(b) between 3 and 6 seconds: 3 t 6
(d)
15. (a) at 2 and 7 seconds
(c)
16. (a) P is moving to the left when 2 t 3 or 5 t 6; P is moving to the right when 0 t 1; P is standing
still when 1 t 2 or 3 t 5
(b)
17. (a)
(c)
(e)
(f)
57 m/s
at 8 s, 0 m/s
From t 8 until t 10.8 s, a total of 2.8 s
Greatest acceleration happens 2 s after launch
(g) From t 2 to t 10.8 s; during this period, a
(b) 2 s
(d) 10.8 s, 27 m/s
v (10.8) v (2)
32 m/s 2
10.8 2
Copyright 2016 Pearson Education, Ltd.
124
Chapter 3 Derivatives
18. (a) Forward: 0 t 1 and 5 t 7; Backward: 1 t 5; Speeds up: 1 t 2 and 5 t 6;
Slows down: 0 t 1, 3 t 5, and 6 t 7
(b) Positive: 3 t 6; negative: 0 t 2 and 6 t 7; zero: 2 t 3 and 7 t 9
(c) t 0 and 2 t 3
(d) 7 t 9
19.
s 490t 2 v 980t a 980
(a) Solving 160 490t 2 t 74 s. The average velocity was
s (4/7) s (0)
280 cm/s.
4/7
(b) At the 160 cm mark the balls are falling at v(4/7) 560 cm/s. The acceleration at the 160 cm mark
was 980 cm/s2.
17 29.75 flashes per second.
(c) The light was flashing at a rate of 4/7
20. (a)
(b)
21. C position, A velocity, and B acceleration. Neither A nor C can be the derivative of B because B’s
derivative is constant. Graph C cannot be the derivative of A either, because A has some negative slopes
while C has only positive values. So, C (being the derivative of neither A nor B) must be the graph of position.
Curve C has both positive and negative slopes, so its derivative, the velocity, must be A and not B. That leaves
B for acceleration.
22. C position, B velocity, and A acceleration. Curve C cannot be the derivative of either A or B because C
has only negative values while both A and B have some positive slopes. So, C represents position. Curve C
has no positive slopes, so its derivative, the velocity, must be B. That leaves A for acceleration. Indeed, A is
negative where B has negative slopes and positive where B has positive slopes.
23. (a) c(100) 11, 000 cav 11,000
$110
100
(b) c( x) 2000 100 x .1x 2 c ( x) 100 .2 x. Marginal cost c ( x) the marginal cost of producing
100 machines is c(100) $80
(c) The cost of producing the 101st machine is c(101) c(100) 100 201
$79.90
10
x
x
, which is marginal revenue. r (100) 20000
24. (a) r ( x) 20000 1 1x r ( x) 20000
$2.
2
2
x
100
(b) r (101) $1.96.
(c) lim r ( x) lim 20000
0. The increase in revenue as the number of items increases without bound will
2
x
approach zero.
Copyright 2016 Pearson Education, Ltd.
Section 3.4 The Derivative as a Rate of Change
125
25. b(t ) 106 104 t 103 t 2 b(t ) 104 (2)(103 t ) 103 (10 2t )
(b) b(5) 0 bacteria/h
(a) b(0) 104 bacteria/h
(c) b(10) 104 bacteria/h
1
26. S ( w) 120
180
w
27. (a) y 6 1 12t
1
80 w
; S increases more rapidly at lower weights where the derivative is greater.
t
6 1 6t 144
dydt 12t 1
2
2
dy
(b) The largest value of dt is 0 m/h when t 12 and the fluid level is falling the slowest at that time.
dy
The smallest value of dt is 1 m/h, when t 0, and the fluid level is falling the fastest at that time.
dy
(c) In this situation, dt 0 the graph of y is
dy
always decreasing. As dt increases in value,
the slope of the graph of y increases from 1
to 0 over the interval 0 t 12.
28. Q(t ) 200(30 t )2 200(900 60t t 2 ) Q (t ) 200(60 2t ) Q (10) 8, 000 litres/min is the rate the
Q (10) Q (0)
water is running at the end of 10 min. Then
10, 000 litres/min is the average rate the water flows
10
during the first 10 min. The negative signs indicate water is leaving the tank.
29. s (v) 0.21 0.01272v; s (50) 0.846, s (100) 1.482. The units of ds / dv are m/km/h; ds / dv gives,
roughly, the number of additional meters required to stop the car if its speed increases by 1 km/h.
4 r 2 dV
30. (a) V 43 r 3 dV
dr
dr
r 2
4 (2) 2 16 m3 /m
(b) When r 2, dV
16 so that when r changes by 1 unit, we expect V to change by approximately 16 .
dr
Therefore when r changes by 0.2 units V changes by approximately (16 )(0.2) 3.2 10.05 m3 .
Note that V (2.2) V (2) 11.09 m3 .
31. 200 km/h 55 95 m/s 500
m/s, and D 10
t 2 V 20
t. Thus V 500
20
t 500
t 25s. When
9
9
9
9
9
9
t 25, D 10
(25)2 6250
m
9
9
v
v
v2
v2
0
0
0
0
; 580 v0t 4.9t 2 so that t 9.8
32. s v0t 4.9t 2 v v0 9.8t ; v 0 t 9.8
580 9.8
19.6
v0 (19.6)(580) 106.6 m/s and, finally,
106.6 60 s 60 min 1 km
1 min 1 h 1000 m 384 km/h.
s
33.
(a) v 0 when t 6.12 s
(b) v 0 when 0 t 6.12 body moves right (up); v 0 when 6.12 t 12.24 body moves left (down)
Copyright 2016 Pearson Education, Ltd.
126
Chapter 3 Derivatives
(c) body changes direction at t 6.12 s
(d) body speeds up on (6.12, 12.24] and slows down on [0, 6.25)
(e) The body is moving fastest at the endpoints t 0 and t 12.24 when it is traveling 60 m/s. It’s moving
slowest at t 6.12 when the speed is 0.
(f ) When t 6.12 the body is s 183.6 m from the origin and farthest away.
34.
(a) v 0 when t 32 s
(b) v 0 when 0 t 1.5 body moves left (down); v 0 when 1.5 t 5 body moves right (up)
(c) body changes direction at t 32 s
(d) body speeds up on 32 , 5 and slows down on 0, 32
(e) body is moving fastest at t 5 when the speed | v(5)| 7 units/s; it is moving slowest at t 32 when
the speed is 0
(f ) When t 5 the body is s 12 units from the origin and farthest away.
35.
(a) v 0 when t 6 3 15 s
(b) v 0 when 6 3 15 t 6 3 15 body moves left (down); v 0 when 0 t 6 3 15 or 6 3 15 t 4
body moves right (up)
(c) body changes direction at t 6 3 15 s
(d) body speeds up on 6 3 15 , 2 6 3 15 , 4 and slows down on 0, 6 3 15 2, 6 3 15 .
(e) The body is moving fastest at t 0 and t 4 when it is moving 7 units/sec and slowest at t 6 3 15 s
(f ) When t 6 3 15 the body is at position s 6.303 units and farthest from the origin.
Copyright 2016 Pearson Education, Ltd.
Section 3.5 Derivatives of Trigonometric Functions
36.
(a) v 0 when t 6 3 15
(b) v 0 when 0 t 6 3 15 or 6 3 15 t 4 body is moving left (down); v 0 when 6 3 15 t 6 3 15
body is moving right (up)
(c) body changes direction at t 6 3 15 s
(d) body speeds up on 6 3 15 , 2 6 3 15 , 4 and slows down on 0, 6 3 15 2, 6 3 15
(e) The body is moving fastest at 7 units/s when t 0 and t 4; it is moving slowest and stationary
at t 6 3 15
(f ) When t 6 3 15 the position is s 10.303 units and the body is farthest from the origin.
3.5
DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
dy
d (cos x ) 10 3sin x
1. y 10 x 3cos x dx 10 3 dx
dy
d (sin x ) 3 5cos x
2. y 3x 5sin x dx 23 5 dx
2
x
x
dy
3. y x 2 cos x dx x 2 ( sin x) 2 x cos x x 2 sin x 2 x cos x
dy
4. y x sec x 3 dx x sec x tan x sec x 0
2 x
5.
x sec x tan x sec x
2 x
dy
y csc x 4 x 7 dx csc x cot x 4
2 x
dy
d (cot x ) cot x d ( x 2 ) 2 x 2 csc 2 x (cot x )(2 x ) 2
6. y x 2 cot x 12 dx x 2 dx
3
3
dx
x
x 2 csc 2 x 2 x cot x 23
x
x
x
sin x sin x(sec 2 x 1)
7. f ( x) sin x tan x f ( x) sin x sec2 x cos x tan x sin x sec 2 x cos x cos
x
8.
x csc x cot x g ( x ) csc x ( csc 2 x) ( csc x cot x ) cot x csc3 x csc x cot 2 x
g ( x ) cos2 x sin1 x cos
sin x
sin x
csc x(csc2 x cot 2 x)
Copyright 2016 Pearson Education, Ltd.
127
128
9.
Chapter 3 Derivatives
y x sec x
1
dy d
d
1
1
( x ) sec x x (sec x ) 2 sec x x sec x tan x 2
x
dx dx
dx
x
x
dy
d (sec x ) sec x d (sin x cos x)
10. y (sin x cos x) sec x dx (sin x cos x) dx
dx
(sin x cos x)(sec x tan x) (sec x)(cos x sin x)
2
2
sin x cos x sin x 2cos x cos x sin x
cos x
(sin x cos x ) sin x
cos 2 x
1 sec 2 x
cos 2 x
x sin x
coscos
x
Note also that y sin x sec x cos x sec x tan x 1 sec x.
dy
x
11. y 1cot
cot x
dx
2
d (cot x ) (cot x ) d (1 cot x )
(1 cot x ) dx
dx
(1 cot x )
2
2
csc x csc x cot x csc x cot x
(1 cot x ) 2
dy
x
12. y 1cos
sin x
dx
sin x 12
(1 sin x )
dy
dx
2
(1 sin x ) 2
(1 cot x )( csc2 x ) (cot x )( csc2 x )
(1 cot x )2
2
csc x2
(1 cot x )
d (cos x ) (cos x ) d (1 sin x )
(1sin x ) dx
dx
(1sin x )
2
(1sin x ) 2
(1sin x )( sin x ) (cos x )(cos x )
(1sin x ) 2
2
2
sin x sin x 2cos x
(1sin x )
1
1sin
x
dy
13. y cos4 x tan1 x 4sec x cot x dx 4sec x tan x csc2 x
dy
14. y cosx x cosx x dx
x ( sin x ) (cos x )(1)
x2
(cos x )(1) x ( sin x )
cos 2 x
x sin x2 cos x cos x 2x sin x
cos x
x
dy
d (sec x tan x ) (sec x tan x ) d (sec x tan x )
15. y (sec x tan x) (sec x tan x) dx (sec x tan x) dx
dx
(sec x tan x)(sec x tan x sec2 x) (sec x tan x) (sec x tan x sec2 x)
(sec 2 x tan x sec x tan 2 x sec3 x sec2 x tan x) (sec2 x tan x sec x tan 2 x sec3 x tan x sec2 x ) 0.
Note also that y sec x tan x (tan x 1) tan x 1 0.
2
2
2
2
dy
dx
dy
16. y x 2 cos x 2 x sin x 2 cos x dx ( x 2 ( sin x) (cos x)(2 x)) (2 x cos x (sin x)(2)) 2( sin x)
x 2 sin x 2 x cos x 2 x cos x 2sin x 2sin x x 2 sin x
17. f ( x) x3 sin x cos x f ( x) x3 sin x( sin x) x3 cos x(cos x) 3 x 2 sin x cos x
x3 sin 2 x x3 cos 2 x 3 x 2 sin x cos x
18. g ( x ) (2 x) tan 2 x g ( x) (2 x) (2 tan x sec 2 x) (1) tan 2 x 2(2 x) tan x sec 2 x tan 2 x
2(2 x) tan x (sec 2 x tan x)
19. s tan t t ds
sec 2 t 1
dt
20. s t 2 sec t 1 ds
2t sec t tan t
dt
csc t ds
21. s 11csc
t
dt
(1csc t )( csc t cot t ) (1 csc t )(csc t cot t )
t ds
22. s 1sin
cos t
dt
(1cos t )(cos t ) (sin t )(sin t )
(1csc t )
(1cos t )
2
2
2
2
csc t cot t csc t cot t csc2 t cot t csc t cot t 2 csc t cot2 t
(1 csc t )
(1csc t )
2
2
1
cos t cos t 2sin t cos t 1 2 1cos
cos1t 1
t
(1cos t )
(1cos t )
23. r 4 2 sin ddr 2 dd (sin ) (sin )(2 ) ( 2 cos 2 sin ) ( cos 2sin )
Copyright 2016 Pearson Education, Ltd.
Section 3.5 Derivatives of Trigonometric Functions
24. r sin cos ddr ( cos (sin )(1)) sin cos
25. r sec csc ddr (sec )( csc cot ) (csc )(sec tan )
21
sin
1 sec 2 csc 2
cos 2
1
sin
1
cos1 sin1 cos
sin sin cos cos
26. r (1 sec ) sin ddr (1 sec ) cos (sin ) (sec tan ) (cos 1) tan 2 cos sec2
dp
27. p 5 cot1 q 5 tan q dq sec 2 q
dp
28. p (1 csc q ) cos q dq (1 csc q )( sin q ) (cos q )( csc q cot q ) ( sin q 1) cot 2 q sin q csc2 q
29. p
(cos q )(cos q sin q ) (sin q cos q )( sin q )
sin q cos q
dp
cos 2 q cos q sin q sin 2 q cos q sin q
12 sec 2 q
2
cos q
dq
cos q
cos 2 q
cos q
tan q
(1 tan q )(sec2 q ) (tan q )(sec 2 q )
dp
30. p 1 tan q dq
31. p
2
dp
dq
sec 2 q tan q sec2 q tan q sec2 q
( q 2 1)( q cos q sin q (1)) ( q sin q )(2 q )
q 1
q 3 cos q q 2 sin q q cos q sin q
2
( q 1)
2
(1 tan q )2
sec2 q
(1 tan q )2
q3 cos q q 2 sin q q cos q sin q 2 q 2 sin q
( q 2 1) 2
( q 2 1)2
32. p
q sin q
(1 tan q ) 2
3q tan q
dp
dq
q sec q
3
( q sec q )(3 sec2 q ) (3q tan q )( q sec q tan q sec q (1))
( q sec q ) 2
2
3q sec q q sec q (3q sec q tan q 3q sec q q sec q tan 2 q sec q tan q )
( q sec q )2
q sec q 3q sec q tan q q sec q tan 2 q sec q tan q
3
2
( q sec q )2
y csc x y csc x cot x y ((csc x)( csc2 x) (cot x )( csc x cot x)) csc3 x csc x cot 2 x
(csc x)(csc2 x cot 2 x ) (csc x )(csc2 x csc2 x 1) 2 csc3 x csc x
(b) y sec x y sec x tan x y (sec x)(sec2 x) (tan x )(sec x tan x) sec3 x sec x tan 2 x
(sec x )(sec 2 x tan 2 x) (sec x)(sec 2 x sec 2 x 1) 2sec3 x sec x
33. (a)
34. (a) y 2 sin x y 2 cos x y 2( sin x) 2sin x y 2 cos x y (4) 2 sin x
(b) y 9 cos x y 9sin x y 9 cos x y 9( sin x) 9sin x y (4) 9 cos x
35. y sin x y cos x slope of tangent at x is
y ( ) cos ( ) 1; slope of tangent at x 0 is
y (0) cos (0) 1; and slope of tangent at x 32 is
y ( 32 ) cos 32 0. The tangent at ( , 0) is
y 0 1( x ), or y x ; the tangent at (0, 0) is
y 0 1 ( x 0), or y x; and the tangent at
3 , 1 is y 1.
2
Copyright 2016 Pearson Education, Ltd.
129
130
Chapter 3 Derivatives
36. y tan x y sec2 x slope of tangent at x 3 is
sec 2 3 4; slope of tangent at x 0 is sec 2 (0) 1; and
slope of tangent at x 3 is sec 2 3 4. The tangent
3 , 3 is y 3 4 x 3 ; the
at 3 , tan 3
tangent at (0, 0) is y x; and the tangent at
3 , tan 3 3 , 3 is y 3 4 x 3 .
37. y sec x y sec x tan x slope of tangent at
x 3 is sec 3 tan 3 2 3; slope of tangent
4 4 2. The tangent at the point
4
3 , sec 3 3 , 2 is y 2 2 3 x 3 ; the
tangent at the point 4 , sec 4 4 , 2 is
y 2 2 x 4 .
at x is sec tan
38. y 1 cos x y sin x slope of tangent at x 3 is
sin 3 23 ; slope of tangent at x 32 is sin 3π
1.
2
The tangent at the point 3 , 1 cos 3 3 , 23
x 3 ; the tangent at the point
32 ,1 cos 32 32 , 1 is y 1 x 32
is y 32
3
2
39. Yes, y x sin x y 1 cos x; horizontal tangent occurs where 1 cos x 0 cos x 1 x
40. No, y 2 x sin x y 2 cos x; horizontal tangent occurs where 2 cos x 0 cos x 2. But there are
no x-values for which cos x 2.
41. No, y x cot x y 1 csc2 x; horizontal tangent occurs where 1 csc2 x 0 csc2 x 1. But there are no
x-values for which csc 2 x 1.
42. Yes, y x 2 cos x y 1 2 sin x; horizontal tangent occurs where 1 2 sin x 0 1 2sin x
12 sin x x 6 or x 56
43. We want all points on the curve where the tangent
line has slope 2. Thus, y tan x y sec 2 x so that
y 2 sec 2 x 2 sec x 2 x 4 . Then the
tangent line at 4 , 1 has equation y 1 2 x 4 ; the
tangent line at 4 , 1 has equation y 1 2 x 4 .
Copyright 2016 Pearson Education, Ltd.
Section 3.5 Derivatives of Trigonometric Functions
131
44. We want all points on the curve y cot x where the tangent
line has slope 1. Thus y cot x y csc 2 x so that
y 1 csc2 x 1 csc 2 x 1 csc x 1 x 2 .
The tangent line at 2 , 0 is y x 2 .
1sin2 cosx x
45. y 4 cot x 2 csc x y csc 2 x 2 csc x cot x sin1 x
(a) When x 2 , then y 1; the tangent line is y x 2 2.
(b) To find the location of the horizontal tangent set y 0 1 2 cos x 0 x 3 radians. When x 3 ,
then y 4 3 is the horizontal tangent.
2sincosxx1
46. y 1 2 csc x cot x y 2 csc x cot x csc 2 x sin1 x
(a) If x 4 , then y 4; the tangent line is y 4 x 4.
(b) To find the location of the horizontal tangent set y 0 2 cos x 1 0 x
When x 34 , then y 2 is the horizontal tangent.
47. lim sin
1x 12 sin 12 12 sin 0 0
48.
1 cos( csc x) 1 cos( csc( 6 )) 1 cos( ( 2)) 2
x 2
49.
lim
x 6
lim
sin 12
6 6
dd (sin ) cos
6
50. lim tan 1 dd (tan ) sec2
4 4
4
51. lim sec cos x tan
x 0
52. lim sin
x 0
54.
4
cos 6 23
sec2 4 2
1 sec tan 4 sec 1
4 sec x 1 sec 1 tan 4 sec
0
tan x2tansecx x sin tan0tan2 sec0 0 sin 2 1
53. lim tan 1
t 0
6
sin t
t
tan 1 lim
sin t
tan (1 1) 0
t 0 t
1
cos 1 1
cos lim cos
lim cos sin
1
sin
0
0 sin
lim
0
3 radians.
4
Copyright 2016 Pearson Education, Ltd.
132
Chapter 3 Derivatives
55. s 2 2 sin t v ds
2 cos t a dv
2 sin t j da
2 cos t. Therefore, velocity v 4
dt
dt
dt
2 m/s; speed | v 4 | 2 m/s; acceleration a 4 2 m/s 2 ; jerk j 4 2 m/s3 .
56. s sin t cos t v ds
cos t sin t a dv
sin t cos t j da
cos t sin t. Therefore velocity
dt
dt
dt
v 4 0 m/s; speed v 4 0 m/s; acceleration a 4 2 m/s 2 ; jerk j 4 0 m/s3 .
57. lim f ( x) lim
x 0
58.
x 0
sin 2 3 x lim 9 sin 3 x
3x
x2
x 0
f ( x) f (0) 9 c.
sin3x3x 9 so that f is continuous at x 0 xlim
0
lim g ( x) lim ( x b) b and lim g ( x) lim cos x 1 so that g is continuous at x 0 lim g ( x)
x 0
x 0
x 0
x 0
x 0
d
lim g ( x) b 1. Now g is not differentiable at x 0: At x 0, the left-hand derivative is dx ( x b)|x 0 1,
x 0
d (cos x )|
but the right-hand derivative is dx
x 0 sin 0 0. The left- and right-hand derivatives can never agree
at x 0, so g is not differentiable at x 0 for any value of b (including b 1).
59.
d 999 (cos x ) sin x because d 4 (cos x ) cos x the derivative of cos x any number of times that is a
dx999
dx 4
3
249 4
999
multiple of 4 is cos x. Thus, dividing 999 by 4 gives 999 249 4 3 d 999 (cos x ) d 3 d 2494 (cos x)
dx dx
dx
3
d 3 (cos x) sin x.
dx
(cos x )(0) (1)( sin x )
dy
(cos x ) 2
(sin x )(0) (1)(cos x )
(b) y csc x sin1 x dx
(c)
sin x sec x tan x d (sec x ) sec x tan x
cos
dx
x
d (csc x ) csc x cot x
cos
x
cos
x
1
sin x sin x csc x cot x dx
sin x
dy
60. (a) y sec x cos1 x dx
sin2x cos1 x
cos x
2
(sin x )2
2
2
dy
(sin x )( sin x ) (cos x )(cos x )
d (cot x ) csc2 x
cos
x
y cot x sin x dx
sin x 2 cos x 21 csc2 x dx
2
(sin x )
sin x
sin x
cm
3
5 3 s ; t 4 v 10sin 34 5 2 cms
61. (a) t 0 x 10 cos(0) 10 cm; t 3 x 10 cos 3 5 cm; t 34 x 10 cos 34 5 2 cm
(b)
t 0 v 10sin(0) 0 cm
; t 3 v 10sin 3
s
62. (a) t 0 x 3cos(0) 4sin(0) 3 m; t 2 x 3cos 2 4sin 2 4 m;
t x 3cos( ) 4 sin( ) 3 m
(b) t 0 v 3sin(0) 4 cos(0) 4 ms ; t 2 v 3sin 2 4 cos 2 3 ms ;
t v 3 sin( ) 4 cos( ) 4 ms
63.
As h takes on the values of 1, 0.5, 0.3 and 0.1 the corresponding dashed curves of y
sin( x h ) sin x
cos x. The same is true as h takes
h
h 0
d (sin x ) lim
and closer to the black curve y cos x because dx
on the values of 1, 0.5, 0.3 and 0.1.
sin( x h ) sin x
get closer
h
Copyright 2016 Pearson Education, Ltd.
Section 3.5 Derivatives of Trigonometric Functions
133
64.
As h takes on the values of 1, 0.5, 0.3, and 0.1 the corresponding dashed curves of y
d (cos x ) lim
and closer to the black curve y sin x because dx
takes on the values of 1, 0.5, 0.3, and 0.1.
h 0
cos( x h ) cos x
get closer
h
cos( x h ) cos x
sin x. The same is true as h
h
65. (a)
sin( x h ) sin( x h )
The dashed curves of y
are closer to the black curve y cos x than the corresponding
2h
dashed curves in Exercise 63 illustrating that the centered difference quotient is a better approximation of
the derivative of this function.
(b)
cos( x h ) cos( x h )
The dashed curves of y
are closer to the black curve y sin x than the corresponding
2h
dashed curves in Exercise 64 illustrating that the centered difference quotient is a better approximation of
the derivative of this function.
66. lim
h 0
|0 h||0 h|
|h||h|
lim 2 h lim 0 0 the limits of the centered difference quotient exists even though the
2h
x 0
h 0
derivative of f ( x) | x | does not exist at x 0.
67. y tan x y sec 2 x, so the smallest value
y sec2 x takes on is y 1 when x 0; y has no
maximum value since sec 2 x has no largest value
on 2 , 2 ; y is never negative since sec 2 x 1.
68. y cot x y csc2 x so y has no smallest
value since csc2 x has no minimum value on
(0, ); the largest value of y is 1, when x 2 ;
the slope is never positive since the largest value
y csc 2 x takes on is 1.
Copyright 2016 Pearson Education, Ltd.
134
Chapter 3 Derivatives
69. y sinx x appears to cross the y -axis at y 1, since
lim sinx x 1; y sinx2 x appears to cross the y -axis at
x 0
y 2, since lim sinx2 x 2; y sinx4 x appears to
x 0
cross the y -axis at y 4, since lim sinx4 x 4.
x 0
However, none of these graphs actually cross the
y -axis since x 0 is not in the domain of the
sin( 3 x )
functions. Also, lim sinx5 x 5, lim
3,
x
x 0
x 0
and lim sinxkx k the graphs of y sinx5 x ,
x 0
sin( 3 x )
sin kx
y
, and y x approach 5, 3, and k,
x
respectively, as x 0. However, the graphs do not
actually cross the y -axis.
sin h
h
sin h
h
70. (a) h
1
0.01
0.001
0.0001
.017452406
.017453292
.017453292
.017453292
sin h . 180
sin h
lim
h
h
h 0
x 0
lim
180
.99994923
1
1
1
lim 180 sin h 180 lim 180 sin
.h
h 0
0
180
(converting to radians)
180
h 180
cos h 1
h
(b) h
1
0.01
0.001
0.0001
lim
h 0
0.0001523
0.0000015
0.0000001
0
cos h 1
0, whether h is measured in degrees or radians.
h
(sin x cos h cos x sin h ) sin x
sin( x h ) sin x
lim
h
h
h 0
h 0
cos h 1
sin h
cos h 1
sin h
lim sin x h
lim cos x h (sin x ) lim
(cos x) lim h
h
h 0
h 0
h 0
h 0
d (sin x ) lim
(c) In degrees, dx
180
(sin x)(0) (cos x) 180
cos x
(cos x cos h sin x sin h) cos x
cos( x h ) cos x
lim
h
h
h 0
h 0
(cos x )(cos h 1) sin x sin h
cos h 1
sin h
lim
lim cos x h
lim sin x h
h
h 0
h 0
h 0
cos h 1
sin h
sin x
(cos x) lim
(sin x ) lim h (cos x)(0) (sin x) 180
h
180
h 0
h 0
d (cos x ) lim
(d) In degrees, dx
(e)
180 sin x; dxd (sin x) dxd 180 sin x 180 cos x;
d (cos x ) d sin x 2 cos x; d (cos x ) d 2 cos x 3 sin x
180
180
dx 180
dx 180
dx
dx
d 2 (sin x ) d cos x
dx 180
dx 2
2
2
3
3
2
3
2
3
Copyright 2016 Pearson Education, Ltd.
3
Section 3.6 The Chain Rule
3.6
THE CHAIN RULE
1. f (u ) 6u 9 f (u ) 6 f ( g ( x)) 6; g ( x) 12 x 4 g ( x) 2 x3 ;
dy
therefore dx f ( g ( x)) g ( x) 6 2 x3 12 x3
2. f (u ) 2u 3 f (u ) 6u 2 f ( g ( x)) 6(8 x 1) 2 ; g ( x) 8 x 1 g ( x) 8;
dy
therefore f ( g ( x)) g ( x) 6(8 x 1) 2 8 48(8 x 1)2
dx
3. f (u ) sin u f (u ) cos u f ( g ( x)) cos(3 x 1); g ( x) 3 x 1 g ( x) 3;
dy
therefore f ( g ( x)) g ( x ) (cos(3 x 1))(3) 3cos(3 x 1)
dx
4.
f (u ) cos u
f (u ) sin u
f ( g ( x )) sin( x /3); g ( x ) x /3
g ( x ) 1/3; therefore,
dy
dx f ( g ( x )) g ( x ) sin( x /3)( 1/3) (1/3)sin( x /3)
5.
f (u ) u
f (u ) 2 1u
1 ; g ( x ) sin x
f ( g ( x )) 2 sin
x
g ( x ) cos x; therefore,
dy
cos x
dx f ( g ( x )) g ( x ) 2 sin x
6. f (u ) sin u f (u ) cos u f ( g ( x)) cos( x cos x); g ( x) x cos x g ( x) 1 sin x;
dy
therefore f g ( x )) g ( x) (cos( x cos x))(1 sin x)
dx
7.
f (u ) tan u f (u ) sec 2 u f ( g ( x )) sec2 ( x 2 ); g ( x ) x 2 g ( x ) 2 x;
dy
therefore dx f ( g ( x )) g ( x ) sec2 ( x 2 )(2 x ) 2 x sec2 ( x 2 )
8.
f (u ) sec u f (u ) sec u tan u f ( g ( x )) sec 1x 7 x tan( 1x 7 x ); g ( x ) 1x 7 x
dy
g ( x ) 12 7; therefore, dx f ( g ( x )) g ( x ) 12 7 sec( 1x 7 x ) tan( 1x 7 x )
x
dy
dy
dy
dy
x
9. With u (2 x 1), y u 5 : dx du du
5u 4 2 10(2 x 1)4
dx
10. With u (4 3x), y u 9 : dx du du
9u 8 (3) 27(4 3 x)8
dx
7u 8 17 1 7x
11. With u 1 7x , y u 7 : dx du du
dx
dy
dy
12. With u 2x 1, y u 10: dx du du
10u 11
dx
dy
13. With u
dy
x , y u :
x2
8
1
x
4 dy
dx
8
1
1
1
4 x
4 x
dy du
4u 3 4x 1 12
du dx
x
dy
11
4 x 1
12 u 1/2 (6 x 4)
14. With u 3 x 2 4 x 6, y u1/2 : dx du du
dx
dy
x
2
x2
8
1
x
3x2
3x2 4 x 6
Copyright 2016 Pearson Education, Ltd.
3
x
4
1
x2
135
136
Chapter 3 Derivatives
dy
dy
dy
dy
15. With u tan x, y sec u: dx du du
(sec u tan u )(sec2 x) (sec(tan x ) tan(tan x )) sec2 x
dx
16. With u 1x , y cot u: dx du du
( csc2 u )
dx
dy
csc
1
x2
2
1
x2
1
x
dy
17. With u tan x, y u 3: dx du du
3u 2 sec2 x 3tan 2 x sec2 x
dx
18. With u cos x, y 5u 4 : dx du du
(20u 5 )( sin x) 20(cos 5 x)(sin x)
dx
dy
dy
d (3 t ) 1 (3 t ) 1/2
19. p 3 t (3 t )1/2 dt 12 (3 t ) 1/2 dt
2
dp
3
1
2 3 t
d (2r r 2 ) 1 (2r r 2 ) 2/3 (2 2r )
20. q 2r r 2 (2r r 2 )1/3 dr 13 (2r r 2 )2/3 dr
3
dq
2 2r
3(2 r r 2 ) 2/3
d (3t ) 4 ( sin 5t ) d (5t ) 4 cos 3t 4 sin 5t 4 (cos 3t sin 5t )
21. s 34 sin 3t 54 cos 5t ds
34 cos 3t dt
dt
5
dt
3
3
t
2 cos 2 sin 32 t
32 t sin 32 t dtd 32 t 32 cos 32 t 32 sin 32 t
d
22. s sin 32 t cos 32 t ds
cos 32 t dt
dt
2
23. r (csc cot )1 ddr (csc cot )2 dd (csc cot ) csc cot csc2
(csc cot )
csc (cot csc )
(csc cot ) 2
csccsc
cot
24. r 6(sec tan )3/2 ddr 6 32 (sec tan )1/2 dd (sec tan ) 9 sec tan (sec tan sec2 )
d (sin 4 x ) sin 4 x d ( x 2 ) x d (cos 2 x ) cos 2 x d ( x )
25. y x 2 sin 4 x x cos 2 x dx x 2 dx
dx
dx
dx
dy
d (sin x)) 2 x sin 4 x x( 2 cos 3 x d (cos x)) cos 2 x
x 2 (4sin 3 x dx
dx
x 2 (4sin 3 x cos x) 2 x sin 4 x x((2 cos 3 x) ( sin x)) cos 2 x
4 x 2 sin 3 x cos x 2 x sin 4 x 2 x sin x cos 3 x cos 2 x
d (sin 5 x) sin 5 x d 1 x d (cos3 x ) cos3 x d ( x )
26. y 1x sin 5 x 3x cos3 x y 1x dx
dx x
3 dx
dx 3
1x (5sin 6 x cos x) (sin 5 x) 12 3x ((3cos 2 x)( sin x)) (cos3 x) 13
5x sin
27.
6
x cos x 12 sin
5
1
x
x
2
3
x x cos x sin x 13 cos x
(3x 2) (3x 2) (1) 4 4
(3x 2) 3 ( 1) 4
(3x 2)
1 (3 x 2)6 4 1
y 18
2
6
18
2x
5
1
2 x2
5
1
x3
1
2 x2
1
x 3 (4 12 )2
2x
4
3
2
1
x
2
5 d
dx
6
18
dydx 3(5 2 x)4 (2) 84 2x 1 x2
28. y (5 2 x) 3 18 2x 1
6(5 2 x)4 12
dy
dx
2
x
3
6
(5 2 x )4
2 1
x
2
3
x
Copyright 2016 Pearson Education, Ltd.
2
d
dx
1
2 x2
Section 3.6 The Chain Rule
d ( x 1) ( x 1) 3 (4)(4 x 3)3 d (4 x 3)
29. y (4 x 3) 4 ( x 1) 3 dx (4 x 3)4 (3)( x 1) 4 dx
dx
dy
(4 x 3) 4 (3)( x 1)4 (1) ( x 1)3 (4)(4 x 3)3 (4) 3(4 x 3) 4 ( x 1) 4 16(4 x 3)3 ( x 1) 3
(4 x 3)3
( x 1)4
[3(4 x 3) 16( x 1)]
(4 x 3)3 (4 x 7)
( x 1)4
30. y (2 x 5) 1 ( x 2 5 x)6 dx (2 x 5)1 (6)( x 2 5 x)5 (2 x 5) ( x 2 5 x)6 (1)(2 x 5) 2 (2)
dy
6 ( x 2 5 x )5
2( x 2 5 x )6
(2 x 5) 2
d (tan (2 x1/2 )) tan (2 x1/2 ) d ( x ) 0
31. h( x) x tan 2 x 7 h( x) x dx
dx
2
1/2
x sec (2 x
d (2 x1/2 ) tan(2 x1/2 ) x sec 2
) dx
2 x 1x tan 2 x x sec2 2 x tan 2 x
1
1
1
1
1
1
x sec tan x 2 x sec x 2 x sec x sec x tan x
x
d sec 1 sec 1 d ( x 2 ) x 2 sec 1 tan 1 d 1 2 x sec 1
32. k ( x) x 2 sec 1x k ( x) x 2 dx
x
x dx
x
x dx x
x
2
1
x
2
33. f ( x) 7 x sec x f ( x) 12 (7 x sec x)1/2 ( x (sec x tan x) (sec x) 1) x sec x tan x sec x
2 7 x sec x
34. g ( x) tan 3 x4 g ( x)
( x 7)4 (sec2 3 x3) (tan 3 x )4( x 7)3 .1
4 2
( x 7)
35. f ( ) 1sin
cos
[( x 7) ]
f ( ) 2
2
(2sin )(cos cos 2 sin 2 )
(1 cos )3
3t
36. g (t ) 13sin
2t
sin
1 cos
( x 7)3 (3( x 7) sec2 3 x 4 tan 3 x
8
( x 7)
(3( x 7) sec2 3 x 4 tan 3 x )
( x 7)5
) (sin )( sin )
dd 1sincos 12sincos (1cos )(cos(1 cos
)
2
(2 sin )(cos 1)
(1 cos )3
2 sin
(1 cos )2
13sin2t3t g (t ) (1sin 3t )((12)sin(33t )2t )(3cos 3t ) 22sin 3(1t 9sincos3t3)t 6t cos 3t
1
2
37. r sin( 2 ) cos(2 ) ddr sin( 2 )( sin 2 ) dd (2 ) cos(2 ) (cos( 2 )) dd ( 2 )
sin( 2 )( sin 2 )(2) (cos 2 )(cos ( 2 ))(2 ) 2sin( 2 )sin(2 ) 2 cos(2 ) cos( 2 )
sec2 1 1 tan 1 (sec tan ) 2 1
tan tan sec
1 sec sec2 1 1 tan 1 sec tan sec
2
2
38. r sec tan 1 ddr sec
2
1
2
t
t 1
t
t 1
dq
cos
dt cos
2(t 1) t
2(t 1)3/ 2
t
t 1
d
dt
t 2
2(t 1)3/ 2
t
t 1
cos
t
t 1
2
cos
39. q sin
2 1
t 1 cos
2
t 1
d
t 1(1) t . dt
1
t 1
t 1
t
t 1
sint t csc2 sint t t costt sin t
dq
d
40. q cot( sint t ) dt csc 2 sint t dt
2
dy
d sin( t 2) 2sin( t 2) cos( t 2) d ( t 2)
41. y sin 2 ( t 2) dt 2sin( t 2) dt
dt
2 sin( t 2) cos( t 2)
Copyright 2016 Pearson Education, Ltd.
t
2 t 1
t 1
137
138
Chapter 3 Derivatives
dy
d (sec t ) (2sec t )(sec t tan t ) d ( t ) 2 sec2 t tan t
42. y sec 2 t dt (2sec t ) dt
dt
d (1 cos 2t ) 4(1 cos 2t ) 5 ( sin 2t ) d (2t )
43. y (1 cos 2t ) 4 dt 4(1 cos 2t ) 5 dt
dt
dy
8sin 2t
(1 cos 2t )5
dydt 2 1 cot 2t dtd 1 cot 2t 2 1 cot 2t csc2 2t dtd 2t 1cot
2
44. y 1 cot 2t
3
3
dy
csc2 2t
2t
45. y (t tan t )10 dt 10 (t tan t )9 t sec 2 t 1 tan t 10t 9 tan 9 t (t sec2 t tan t )
10t10 tan 9 t sec2 t 10t 9 tan10 t
46. y (t 3/4 sin t ) 4/3 t 1 (sin t ) 4/3 dt t 1 43 (sin t )1/3 cos t t 2 (sin t ) 4/3
dy
47. y
(sin t )1/3 (4t cos t 3cos t )
4(sin t )1/3 cos t (sin t ) 4/3
3t
t2
3t 2
3
t2
t 4t
3
48. y 53tt 42
t2
t 4t
dy
dt
3
2
(t 3 4t )(2t ) t 2 (3t 2 4)
3
3
( t 4t )
2
4
4
2
2
2
3t 4 2t 4 8t 2 3t 4 4t 2 3t ( t 4t ) 3t (t 4)
( t 4t ) 2
( t 3 4t ) 2
t 4 ( t 2 4t ) 4
(t 2 4) 4
3
t 2)
dydt 5 53tt 42 (5t 2)(53t 2)(3t 4)5 5 53tt 42 15t (56t152)t 20 5 (5(3tt 4)2) (5t262) 130(5
(3t 4)
5
6
6
2
6
2
6
2
6
dy
d cos (2t 5) cos(cos (2t 5)) ( sin (2t 5)) d (2t 5)
49. y sin (cos (2t 5)) dt cos(cos (2t 5)) dt
dt
2 cos(cos(2t 5))(sin(2t 5))
dydt sin 5sin 3t dtd 5sin 3t sin 5sin 3t 5cos 3t dtd 3t
53 sin 5sin 3t cos 3t
50. y cos 5sin 3t
3
2
2
t dy 3 1 tan 4 t d 1 tan 4 t 3 1 tan 4 t 4 tan 3 t d tan t
51. y 1 tan 4 12
12 dt
12
12
12 dt
12
dt
2
2
4
3
2
4
3
2
t
t
t
t
t
t
1
12 1 tan 12 tan 12 sec 12 12 1 tan 12 tan 12 sec 12
3
2
dy
52. y 16 1 cos 2 (7t ) dt 63 [1 cos 2 (7t )]2 2 cos(7t )( sin(7t ))(7) 7 1 cos 2 (7t ) (cos(7t ) sin(7t ))
dy
d (1 cos (t 2 )) 1 (1 cos (t 2 )) 1/2 sin(t 2 ) d (t 2 )
53. y (1 cos (t 2 ))1/2 dt 12 (1 cos (t 2 ))1/2 dt
2
dt
2
12 (1 cos (t ))
54. y 4sin
1/2
2
(sin (t )) 2t
1 cos (t 2 )
1 t 4 cos 1 t 1 t 4 cos 1 t
2 cos 1 t
1 t 2 t
dy
dt
cos 1 t
t sin (t 2 )
d
dt
1
d 1
2 1 t dt
t
t t t
dy
55. y tan 2 (sin 3 t ) dt 2 tan(sin 3 t ) sec 2 (sin 3 t ) (3sin 2 t (cos t )) 6 tan(sin 3 t ) sec 2 (sin 3 t ) sin 2 t cos t
Copyright 2016 Pearson Education, Ltd.
4
3
Section 3.6 The Chain Rule
56.
dy
y cos 4 (sec 2 3t ) dt 4 cos3 sec 2 (3t )
139
sin(sec2 (3t ) 2 sec(3t ) sec(3t ) tan(3t ) 3
24 cos3 sec2 (3t ) sin sec2 (3t ) sec 2 (3t ) tan(3t )
dy
57. y 3t (2t 2 5) 4 dt 3t 4 (2t 2 5)3 (4t ) 3 (2t 2 5)4 3(2t 2 5)3 [16t 2 2t 2 5] 3(2t 2 5)3 (18t 2 5)
dy
58. y 3t 2 1 t dt 12 3t 2 1 t
3 2 1 t
1/2
1
2
1/2
1 (1 t ) 1/2 ( 1)
2
12 1t 2 1t 1
1
1
1
3
2
1
t
2 2 1t
2 3t 2 1t 4 1t 2 1t
2 3t 2 1t
1
y 3 1 1x x1 x3 1 1x y x3 dxd 1 1x 1 1x dxd x3
2
2
3 2 1 1x 1 6 1 1x 6 1 1x 6 1 1x 6 1 1x 1x 1 1x 6 1 1x 1 2x
x
x
x
x
x
x
x
2
3
59. y 1 1x
2
2
2
2
2
2
3
2
2
4
3
2
3
3
y 1 x 12 x1/2 12 1 x x1/2
2
3
y 12 1 x 12 x 3/2 x 1/2 (2) 1 x 12 x 1/2
2
3
3
1 3/2
1
1/2
1
1
1
1
2 2 x
1 x x 1 x 2 x 1 x 2 x 1 x 1
3
3
21x 1 x 1 12 1 21x 1 x 32 1
2 x
2 x
1
60. y 1 x
2
2
d csc(3 x 1))
61. y 19 cot (3x 1) y 19 csc 2 (3 x 1)(3) 13 csc2 (3x 1) y 23 (csc(3 x 1) dx
d (3 x 1)) 2 csc 2 (3 x 1) cot(3 x 1)
23 csc(3 x 1)( csc(3x 1) cot(3x 1) dx
3x 13 3sec2 3x y 3 2sec 3x sec 3x tan 3x 13 2sec2 3x tan 3x
62. y 9 tan 3x y 9 sec 2
63. y x(2 x 1)4 y x 4(2 x 1)3 (2) 1 (2 x 1)4 (2 x 1)3 (8 x (2 x 1)) (2 x 1)3 (10 x 1)
y (2 x 1)3 (10) 3(2 x 1)2 (2)(10 x 1) 2(2 x 1)2 (5(2 x 1) 3(10 x 1)) 2(2 x 1) 2 (40 x 8)
16(2 x 1)2 (5 x 1)
64. y x 2 ( x3 1)5 y x 2 5( x3 1) 4 (3 x 2 ) 2 x( x3 1)5 x ( x3 1) 4 [15 x3 2 ( x3 1)] ( x3 1) 4 (17 x 4 2 x)
y ( x3 1) 4 (68 x3 2) 4 ( x3 1)3 (3 x 2 ) (17 x 4 2 x) 2 ( x3 1)3 [( x3 1) (34 x3 1) 6 x 2 (17 x 4 2 x)]
136 x6 47 x3 1
2 x3 1
3
1 g (1) 1 and g (1) 1 ; f (u ) u 5 1 f (u ) 5u 4 f ( g (1)) f (1) 5;
2
2 x
5
1
therefore, ( f g )(1) f ( g (1)) g (1) 5 2 2
65. g ( x) x g ( x)
66. g ( x) (1 x )1 g ( x) (1 x) 2 (1)
f ( g (1)) f
1 g ( 1) 1 and g ( 1) 1 ; f (u ) 1 1 f (u ) 1
2
4
u
u2
(1 x ) 2
1 4; therefore, ( f g )( 1) f ( g ( 1)) g ( 1) 4 1 1
2
4
Copyright 2016 Pearson Education, Ltd.
140
Chapter 3 Derivatives
10 csc2 10u
5 g(1) 5 and g (1) 5 ; f (u ) cot u f (u ) csc 2 u
2
10
10
10
2 x
csc 2 ; therefore, ( f g )(1) f ( g (1)) g (1) 5
f ( g (1)) f (5) 10
2
10
10 2
4
67. g ( x) 5 x g ( x)
1 2sec u tan u f g f 4 1 2sec2 4 tan 4 5; therefore, ( f g ) 14 f g 14 g 14 5
68. g ( x) x g ( x) g 14 4 and g 14 ; f (u ) u sec 2 u f (u ) 1 2 sec u sec u tan u
2
1
4
69. g ( x) 10 x 2 x 1 g ( x) 20 x 1 g (0) 1 and g (0) 1; f (u ) 22 u f (u )
u 1
2
u 2 1(2)(2u )(2u )
2
u 2 1
22u 22 f ( g (0)) f (1) 0; therefore, ( f g )(0) f ( g (0)) g (0) 0 1 0
(u 1)
uu 11 f (u) 2 uu 11 dud uu 11
2
70. g ( x) 12 1 g ( x) 23 g (1) 0 and g (1) 2; f (u )
x
x
4(u 1)
f ( g (1)) f (0) 4; therefore,
(u 1)(1)(u 1)(u 1)(1) 2((uu1)(2)
1)
(u 1)
2 uu 11
2
3
3
( f g )(1) f ( g (1)) g (1) (4)(2) 8
71. y f ( g ( x)), f (3) 1, g (2) 5, g (2) 3 y f ( g ( x)) g ( x) y x 2 f ( g (2)) g (2) f (3) 5
(1) 5 5
72. r sin( f (t )), f (0) 3 , f (0) 4 dr
cos( f (t )) f (t ) dr
dt
dt
dy
dy
73. (a) y 2 f ( x) dx 2 f ( x) dx
dy
x2
dy
x 3
f (3) g (3) 2 5
dy
dy
3 5 (4)(2 ) 15 8
g ( x ) f ( x ) f ( x ) g ( x )
dy
2
[ g ( x )]
dy
dy
dx
x2
dy
(e) y f ( g ( x)) dx f ( g ( x)) g ( x) dx
x2
x 3
f (3) g (3) g (3) f (3)
g (2) f (2) f (2) g (2)
2
[ g (2)]
(2) 13 (8)( 3)
dy
1
dy
74. (a) y 5 f ( x) g ( x) dx 5 f ( x) g ( x) dx
(b)
37
6
f ( x )
dy
f (2)
dx
3 1 1 242
2 f ( x)
2 8 6 8 12 2
x 2 2 f (2)
dy
dy
2
3
3
5
2( g (3)) g (3) 2(4) 3 5 32
y ( g ( x)) dx 2( g ( x)) g ( x ) dx
x 3
dy
dy
y (( f ( x )) 2 ( g ( x)) 2 )1/2 dx 12 (( f ( x)) 2 ( g ( x))2 ) 1/2 (2 f ( x) f ( x) 2 g ( x) g ( x)) dx
x2
12 (( f (2)) 2 ( g (2)) 2 )1/2 (2 f (2) f (2) 2 g (2) g (2)) 12 (82 22 )1/2 (2 8 13 2 2 (3)) 5
3 17
dy
(h)
22
f ( g (2)) g (2) f (2)(3) 13 (3) 1
(f ) y ( f ( x))1/2 dx 12 ( f ( x)) 1/2 f ( x)
(g)
(c) y f ( x) g ( x ) dx f ( x) g ( x) g ( x) f ( x) dx
f ( x)
cos( f (0)) f (0) cos 3 4 12 4 2
2 f (2) 2 13 32
(b) y f ( x ) g ( x) dx f ( x) g ( x) dx
(d) y g ( x ) dx
t 0
x 1
5 f (1) g (1) 5 13 38 1
dy
dy
y f ( x)( g ( x))3 dx f ( x)(3( g ( x)) 2 g ( x)) ( g ( x))3 f ( x) dx
3
dy
( g ( x ) 1) f ( x ) f ( x ) g ( x )
3 f (0)( g (0)) g (0) ( g (0))
f ( x)
(1) (5) 6
2
(c) y g ( x ) 1 dx
f (0) 3(1)(1) 2 13
( g ( x ) 1)
2
dy
dx
x 1
x 0
3
( g (1) 1) f (1) f (1) g (1)
( g (1) 1)
2
Copyright 2016 Pearson Education, Ltd.
1
( 41) 13 (3) 83
( 41)2
Section 3.6 The Chain Rule
141
13 19
dy
dy
(e) y g ( f ( x )) dx g ( f ( x)) f ( x) dx
g ( f (0)) f (0) g (1)(5) 83 (5) 40
3
x 0
dy
dy
(d) y f ( g ( x)) dx f ( g ( x)) g ( x) dx
x 0
f ( g (0)) g (0) f (1) 13 13
(f ) y ( x11 f ( x))2 dx 2( x11 f ( x))3 11x10 f ( x) dx
dy
4 323 13
dy
2(1 3) 3 11 13 23
dy
dy
(g) y f ( x g ( x)) dx f ( x g ( x))(1 g ( x)) dx
43 94
13
75.
2(1 f (1))3 (11 f (1))
f (0 g (0))(1 g (0)) f (1) 1 13
1 so that dsdt dds ddt 1 5 5
ds ds d : s cos ds sin ds
sin 32
dt
d dt
d
d 3
2
76.
x 0
x 1
dy
dy
dy
dy
dy
dy
dx dx
: y x 2 7 x 5 dx 2 x 7 dx
9 so that dt dx dx
9 13 3
dt
dt
dt
x 1
dy
77. With y x, we should get dx 1 for both (a) and (b):
dy
dy
dy
(a) y u5 7 du 15 ; u 5 x 35 du
5; therefore, dx du du
15 5 1, as expected
dx
dx
(b) y 1 u1 du 12 ; u ( x 1)1 du
( x 1) 2 (1)
dx
dy
u
1
( x 1)
1 2
1 2 ( x 1)2
( x 1)
1 ; therefore dy dy du 1 1
dx
du dx
u 2 ( x 1)2
( x 1)2
1
1, again as expected
( x 1)2
dy
78. With y x3/2 , we should get dx 32 x1/2 for both (a) and (b):
dy
(a) y u 3 du 3u 2 ; u x du
dx
as expected.
dy
(b) y u du
1 ; therefore, dy dy du 3u 2 1 3(
dx
du dx
2 x
2 x
x )2
1 3
2
2 x
x,
1 ; u x3 du 3 x 2 ; therefore, dy dy du 1 3 x 2 1 3 x 2 3 x1/2 ,
dx
du dx
dx
2
2 u
2 u
2 x3
again as expected.
79. y
xx11 and x 0 y 0011 (1)2 1. y 2 xx11 ( x1)( x11)( x1)1 2 (( xx1)1) ( x21) 4(( xx1)1)
2
2
2
y x 0
4(0 1)
(0 1)3
2
34 4 y 1 4( x 0) y 4 x 1
1
80. y x 2 x 7 and x 2 y (2) 2 (2) 7 9 3. y 12 x 2 x 7
y x 2
2(2) 1
2 (2) 2 (2) 7
63 12 y 3 12 ( x 2) y 12 x 2
dy
81. y 2 tan 4x dx 2sec2 4x
(a)
3
4 2 sec2 4x
1/2
(2 x 1)
2 x 1
2 x2 x7
2 and y(1) tangent line is
dy
2 sec 2 ( 4 ) slope of tangent is ; thus, y (1) 2 tan 4
dx x 1
given by y 2 ( x 1) y x 2
(b) y 2 sec 2 4x and the smallest value the secant function can have in 2 x 2 is 1 the minimum
value of y is 2 and that occurs when 2 2 sec2 4x 1 sec 2 4x 1 sec 4x x 0.
Copyright 2016 Pearson Education, Ltd.
142
Chapter 3 Derivatives
82. (a) y sin 2 x y 2 cos 2 x y (0) 2 cos(0) 2 tangent to y sin 2 x at the origin is y 2 x;
y sin 2x y 12 cos 2x y (0) 12 cos 0 12 tangent to y sin 2x at the origin is
y 12 x. The tangents are perpendicular to each other at the origin since the product of their slopes is 1.
(b) y sin( mx ) y m cos( mx ) y (0) m cos 0 m; y sin mx y m1 cos mx
y (0) m1 cos(0) m1 . Since m m1 1, the tangent lines are perpendicular at the origin.
(c) y sin( mx ) y m cos(mx). The largest value cos( mx) can attain is 1 at x 0 the largest value y can
attain is | m | because y m cos (mx) m cos mx m 1 m . Also, y sin mx y m1 cos mx
y m1 cos
m1 cos m1 the largest value y can attain is m1 .
x
m
x
m
(d) y sin(mx) y m cos(mx) y (0) m slope of curve at the origin is m. Also, sin(mx) completes m
periods on [0, 2 ]. Therefore the slope of the curve y sin(mx) at the origin is the same as the number of
periods it completes on [0, 2 ]. In particular, for large m, we can think of “compressing” the graph of
y sin x horizontally which gives more periods completed on [0, 2 ], but also increases the slope of the
graph at the origin.
83. s A cos(2 bt ) v ds
A sin(2 bt )(2 b) 2 bA sin(2 bt ). If we replace b with 2b to double the
dt
frequency, the velocity formula gives v 4 bA sin(4 bt ) doubling the frequency causes the velocity to
4 2b 2 A cos(2 bt ). If we replace b with 2b in the acceleration
double. Also v 2 bA sin(2 bt ) a dv
dt
formula, we get a 16 2 b 2 A cos(4 bt ) doubling the frequency causes the acceleration to quadruple.
8 3b3 A sin(2 bt ). If we replace b with 2b in the jerk formula,
Finally, a 4 2b 2 A cos(2 bt ) j da
dt
we get j 64 3b3 A sin(4 bt ) doubling the frequency multiplies the jerk by a factor of 8.
84. (a)
2 ( x 101) 4 y 20 cos 2 ( x 101) 2 40 cos 2 ( x 101) . The
y 20sin 365
365
365
365
365
2 ( x 101)
temperature is increasing the fastest when y is as large as possible. The largest value of cos 365
2 ( x 101) 0 x 101 on day 101 of the year ( April 11), the temperature is
is l and occurs when 365
increasing the fastest.
cos 2 (101 101) 40 cos(0) 40 0.34 C/day
(b) y (101) 40
365
365
365
365
85. s (1 4t )1/2 v ds
12 (1 4t )1/2 (4) 2(1 4t )1/2 v(6) 2(1 4 6)1/2 52 m/s; v 2(1 4t )1/2
dt
4 m/s 2
a dv
12 2(1 4t ) 3/2 (4) 4(1 4t ) 3/2 a (6) 4(1 4 6)3/2 125
dt
2 k s a dvds dsdt dvds v
d k s
86. We need to show a dv
is constant: a dv
dv
dv and dv
ds
dt
dt
ds dt
ds
2
k k
2 s
s k2 which is a constant.
87. v proportional to 1 v k for some constant k dv
ds
s
2
k2
s
k Thus, a dv dv ds dv v k k
dt
ds dt
ds
s
2 s 3/ 2
2 s 3/ 2
2
acceleration is a constant times so a is inversely proportional to s .
1
s2
1
s2
d dx f ( x ) d ( f ( x )) f ( x ) f ( x ) f ( x ), as required.
88. Let dx
f ( x). Then, a dv
dv
dx dv
f ( x ) dx
dt
dt
dt
dx dt
dx
dx
89. T 2
L dT 2 1 1 Therefore, dT dT dL kL k L 1 2 k
g
dL
du
dL du
2
gL
gL
g
2 L g
g L
g
g
required.
Copyright 2016 Pearson Education, Ltd.
L kT , as
g
2
Section 3.6 The Chain Rule
90. No. The chain rule says that when g is differentiable at 0 and f is differentiable at g (0), then f o g is
differentiable at 0. But the chain rule says nothing about what happens when g is not differentiable at 0
so there is no contradiction.
sin 2( x h ) sin 2 x
91. As h 0, the graph of y
h
approaches the graph of y 2 cos 2 x because
sin 2( x h ) sin 2 x
d (sin 2 x ) 2 cos 2 x.
lim
dx
h
h 0
92. As h 0, the graph of y
cos[( x h ) 2 ]cos( x 2 )
h
2
approaches the graph of y 2 x sin ( x ) because
cos[( x h )2 ] cos( x 2 )
d [cos ( x 2 )] 2 x sin ( x 2 ).
dx
h
h 0
lim
93. From the power rule, with y x1/4 , we get dx 14 x 3/4 . From the chain rule, y
dy
dy
dx
1
2
x
d
dx
x
x 2 1 x 2 1 x 14 x3/4 , in agreement.
94. From the power rule, with y x3/4 , we get dx 34 x 1/4 . From the chain rule, y x x
dy
dx
dy
1
d
2 x x dx
dy
1
2 x x
dx
x x
x
1
2 x
x
1
3
2 x x 2
4xx
x 3 x
3 x
4 x
x
34 x 1/4 , in agreement.
95. (a)
(b)
df
1.27324sin 2t 0.42444sin 6t 0.2546sin10t 0.18186sin14t
dt
Copyright 2016 Pearson Education, Ltd.
143
144
Chapter 3 Derivatives
df
dg
(c) The curve of y dt approximates y dt
the best when t is not , 2 , 0, 2 , nor .
96. (a)
(b)
(c)
dh 2.5464 cos(2t ) 2.5464 cos (6t ) 2.5465cos (10t ) 2.54646 cos(14t ) 2.54646 cos (18t )
dt
dh/dt
10
2
0
t
2
10
3.7
IMPLICIT DIFFERENTIATION
1. x 2 y xy 2 6 :
Step 1:
Step 2:
Step 3:
Step 4:
x
dy
2 dy
y 2 x x 2 y dx y 2 1 0
dx
dy
dy
x 2 dx 2 xy dx 2 xy y 2
dy 2
( x 2 xy ) 2xy y 2
dx
dy
2 xy y 2
2
dx
x 2 xy
dy
dy
dy
dy
2. x3 y 3 18 xy 3 x 2 3 y 2 dx 18 y 18 x dx (3 y 2 18 x) dx 18 y 3 x 2 dx
3. 2 xy y 2 x y :
Step 1:
Step 2:
Step 3:
Step 4:
2x 2 y 2 y 1
dy
dy
dy
dx
dx
dx
dy
dy dy
2 x dx 2 y dx dx 1 2 y
dy
(2 x 2 y 1) 1 2 y
dx
dy
1 2 y
2 x 2 y 1
dx
Copyright 2016 Pearson Education, Ltd.
6 y x2
y 2 6 x
Section 3.7 Implicit Differentiation
dy
dy
dy
dy
4. x3 xy y 3 1 3 x 2 y x dx 3 y 2 dx 0 (3 y 2 x) dx y 3x 2 dx
5. x 2 ( x y ) 2 x 2 y 2 :
dy
dy
Step 1: x 2 2( x y ) 1 dx ( x y ) 2 (2 x ) 2 x 2 y dx
dy
dy
Step 2: 2 x 2 ( x y ) dx 2 y dx 2 x 2 x 2 ( x y ) 2 x( x y ) 2
dy
Step 3: dx 2 x 2 ( x y ) 2 y 2 x [1 x( x y ) ( x y )2 ]
y 3 x 2
3 y2 x
x 1 x 2 xy x 2 2 xy y 2
2 x 1 x ( x y ) ( x y )2
x 1 x ( x y ) ( x y ) 2
dy
dx
2 x 2 ( x y ) 2 y
y x2 ( x y )
x 2 y x3 y
Step 4:
145
dy
dy
dy
x2 x 3 x y xy
3
2
2
x 2 y x3 y
dy
6. (3xy 7)2 6 y 2(3 xy 7) 3x dx 3 y 6 dx 2(3 xy 7)(3 x) dx 6 dx 6 y (3xy 7)
dy
y (3 xy 7)
dy
dx [6 x(3xy 7) 6] 6 y (3xy 7) dx x (3 xy 7) 1
dy
7. y 2 xx 11 2 y dx
( x 1) ( x 1)
( x 1)2
3 xy 2 7 y
13 x 2 y 7 x
2 dy
1
dx
( x1)2
y ( x1)2
2 x y
8. x3 x 3 y x 4 3x3 y 2 x y 4 x3 9 x 2 y 3x3 y 2 y (3 x3 1) y 2 4 x3 9 x 2 y
y
2 4 x 3 9 x 2 y
3 x3 1
dy
dy
9. x tan y 1 (sec 2 y ) dx dx
1 cos 2 y
sec2 y
dy
dy
dy
dy
10. xy cot( xy ) x dx y csc2 ( xy ) x dx y x dx x csc 2 ( xy ) dx y csc2 ( xy ) y
dy
dy
y csc2 ( xy ) 1
y
dx x x csc 2 ( xy ) y csc2 ( xy ) 1 dx
x
x 1 csc2 ( xy )
dy
dy
dy
11. x tan( xy ) 0 1 sec 2 ( xy ) y x dx 0 x sec 2 ( xy ) dx 1 y sec 2 ( xy ) dx
2
1 y sec2 ( xy )
x sec 2 ( xy )
2
y
cos ( xy ) y
cos ( xy ) y
1
x
x
x
x sec2 ( xy ) x
dy
dy
dy
dy
12. x 4 sin y x3 y 2 4 x3 (cos y ) dx 3x 2 y 2 x3 2 y dx (cos y 2 x3 y ) dx 3 x 2 y 2 4 x3 dx
3 x 2 y 2 4 x3
cos y 2 x3 y
dy
dy
dy
dy
13. y sin 1y 1 xy y cos 1y (1) 12 dx sin 1y dx x dx y dx 1y cos 1y sin 1y x y
y
dy
dx
y
sin x
1y cos 1y
1
y
y2
y sin
cos xy
1
y
1
y
14. x cos(2 x 3 y ) y sin x x sin(2 x 3 y )(2 3 y ) cos(2 x 3 y ) y cos x y sin x
2 x sin(2 x 3 y ) 3 xy sin(2 x 3 y ) cos(2 x 3 y ) y cos x y sin x
cos(2 x 3 y ) 2 x sin(2 x 3 y ) y cos x (sin x 3x sin(2 x 3 y )) y
cos(2 x 3 y ) 2 x sin(2 x 3 y ) y cos x
y
sin x 3 x sin(2 x 3 y )
15. 1/2 r1/2 1 12 1/2 12 r 1/2 ddr 0 ddr 1 1 ddr 2 r r
2 r 2
2
Copyright 2016 Pearson Education, Ltd.
146
Chapter 3 Derivatives
16. r 2 32 2/3 34 3/4 ddr 1/2 1/3 1/4 ddr 1/2 1/3 1/4
r cos( r )
17. sin(r ) 12 cos(r ) r ddr 0 ddr [ cos(r )] r cos(r ) ddr cos( r ) r , cos(r ) 0
csc
18. cos r cot r ( sin r ) ddr csc2 r ddr ddr sin r r csc2 ddr rsin
r
2
2
dy
d y d
d x
( y ) dx
19. x 2 y 2 1 2 x 2 yy 0 2 yy 2 x dx y xy ; now to find 2 , dx
y
dx
y
y ( 1) xy
y
2
since y x d y y y x y (1 y ) 1
y x xy
2
y
2
y
dx
2
2
y
2
2
2
3
y
3
y3
;
y 1/3
1/3
20. x 2/3 y 2/3 1 23 x 1/3 23 y 1/3 dx 0 dx 23 y 1/3 23 x 1/3 y dx x 1/3 x
y
dy
Differentiating again, y
dy
dy
x1/3 ( 13 y 2/3 ) y y1/3 ( 13 x 2/3 )
2/3
x
y1/3
2/3 1/3 1 1/3 4/3
1
2 3x
y
3y x
4/3 1/31 2/3
dx
3x
3y x
y1/3
x1/3 13 y 2/3 1/3 y1/3 13 x 2/3
x
x 2/3
d2y
y ( x 1) y
21. y 2 x 2 2 x 2 yy 2 x 2 y 22x y 2 x y1 ; then y
y2
d 2 y y y 2 ( x 1)2
y ( x 1) x y1
y2
dx 2
y3
22. y 2 2 x 1 2 y 2 y y 2 2 y y (2 y 2) 2 y y11 ( y 1) 1; then y ( y 1)2 y
d2y
( y 1) 2 ( y 1) 1
dx 2
y
1
( y 1)3
y
1
; we can differentiate the
y 1
y 1/ 2 1
y 1/2 1 y 0 y 1/2 1 y 12 [ y ]2 y 3/2
23. 2 y x y y 1/2 y 1 y y y 1/2 1 1 dx y
dy
equation y y 1/2 1 1 again to find y : y 12 y 3/2 y
2
3/ 2
1
1
y
2 y 1/ 2 1
d2y
1
2 y
3/ 2 11/ 2 3
1/ 2
3
dx
(y
1)
2y (y
1)
2 1 y
y
24. xy y 2 1 xy y 2 yy 0 xy 2 yy y y ( x 2 y ) y y ( x 2 y ) ;
d2y
y
2
( x 2 y ) y y (1 2 y)
2 y 2 2 xy
2 y( x y)
dx
( x 2 y )3
( x2 y)
2
y
y
( x 2 y ) ( x 2 y ) y 1 2 ( x 2 y )
( x 2 y )
2
1
( x2 y)
[ y ( x 2 y ) y ( x 2 y )2 y 2 ]
( x 2 y )
2
2 y ( x 2 y )2 y 2
( x 2 y )3
( x 2 y )3
2
25. x3 y 3 16 3 x 2 3 y 2 y 0 3y 2 y 3 x 2 y x 2 ; we differentiate y 2 y x 2 to find y :
y
2
2 x 2 y x 2
2
2
2
y
y y y [2 y y ] 2x y y 2 x 2 y[ y ] y
y2
d2y
2
333232 2
dx (2,2)
2
4
2 x 2 x3
Copyright 2016 Pearson Education, Ltd.
y
y
2
2 xy 3 2 x 4
y5
Section 3.7 Implicit Differentiation
147
y
( x 2 y )( y ) ( y )(1 2 y )
26. xy y 2 1 xy y 2 yy 0 y ( x 2 y ) y y ( x 2 y ) y
; since
2
y (0,1) 12 we obtain y
(0, 1)
( x2 y)
( 2) 12 ( 1)(0)
4
14
dy
dy
dy
dy
27. y 2 x 2 y 4 2 x at (2, 1) and (2, 1) 2 y dx 2 x 4 y 3 dx 2 2 y dx 4 y 3 dx 2 2 x
dy
dy
dy
dy
dx (2 y 4 y 3 ) 2 2 x dx x31 dx
1 and dx
1
2y y
( 2, 1)
( 2, 1)
dy
dy
28. ( x 2 y 2 ) 2 ( x y )2 at (1, 0) and (1, 1) 2 ( x 2 y 2 ) 2 x 2 y dx 2( x y ) 1 dx
dy
dy
dx [2 y ( x 2 y 2 ) ( x y )] 2 x ( x 2 y 2 ) ( x y ) dx
dy
and dx
(1, 1)
2
2
2 x ( x y ) ( x y )
2 y ( x2 y 2 )( x y )
dy
dx
(1,0)
1
1
2 x y
29. x 2 xy y 2 1 2 x y xy 2 yy 0 ( x 2 y ) y 2 x y y 2 y x ;
(a) the slope of the tangent line m y (2, 3) 74 the tangent line is y 3 74 ( x 2) y 74 x 12
(b) the normal line is y 3 74 ( x 2) y 74 x 29
7
30. x 2 y 2 25 2 x 2 yy 0 y xy ;
m y (3, 4) xy
(a)
(3, 4)
34
y 4 34 ( x 3) y 34 x 25
4
y 4 43 ( x 3) y 43 x
y
31. x 2 y 2 9 2 xy 2 2 x 2 yy 0 x 2 yy xy 2 y x ;
y
(a) the slope of the tangent line m y ( 1, 3) x
(b)
( 1, 3)
the normal line is y 3 13 ( x 1) y 13 x 83
3 the tangent line is y 3 3( x 1) y 3x 6
32. y 2 2 x 4 y 1 0 2 yy 2 4 y 0 2( y 2) y 2 y y 1 2 ;
(a) the slope of the tangent line m y ( 2, 1) 1 the tangent line is y 1 1( x 2) y x 1
(b) the normal line is y 1 1( x 2) y x 3
33. 6 x 2 3xy 2 y 2 17 y 6 0 12 x 3 y 3xy 4 yy 17 y 0 y (3x 4 y 17) 12 x 3 y
12 x 3 y
y 3 x 4 y 17 ;
12 x 3 y
(a) the slope of the tangent line m y ( 1, 0) 3 x 4 y 17
( 1, 0)
76 the tangent line is y 0 76 ( x 1)
y 76 x 76
(b) the normal line is y 0 76 ( x 1) y 76 x 76
Copyright 2016 Pearson Education, Ltd.
148
Chapter 3 Derivatives
34. x 2 3 xy 2 y 2 5 2 x 3xy 3 y 4 yy 0 y 4 y 3x 3 y 2 x y
(a) the slope of the tangent line m y
( 3, 2)
3 y 2 x
;
4 y 3x
3 y 2 x
0 the tangent line is y 2
4 y 3 x ( 3, 2)
(b) the normal line is x 3
2 y
35. 2 xy sin y 2 2 xy 2 y (cos y ) y 0 y (2 x cos y ) 2 y y 2 x cos y ;
(a) the slope of the tangent line m y
1,
2
2 y
2 x cos y
y x
2
1,
2 the tangent line is y 2 2 ( x 1)
2
(b) the normal line is y 2 2 ( x 1) y 2 x 2 2
36. x sin 2 y y cos 2 x x(cos 2 y )2 y sin 2 y 2 y sin 2 x y cos 2 x y (2 x cos 2 y cos 2 x)
sin 2 y 2 y sin 2 x
sin 2 y 2 y sin 2 x y cos 2 x 2 x cos 2 y ;
(a) the slope of the tangent line m y
y 2 2 x 4 y 2 x
4 2
,
sin 2 y 2 y sin 2 x
cos 2 x 2 x cos 2 y
4 2
,
2 the tangent line is
2
(b) the normal line is y 2 12 x 4 y 12 x 58
2 cos( x y )
37. y 2sin( x y ) y 2[cos( x y )] ( y ) y [1 2 cos( x y )] 2 cos ( x y ) y 1 2 cos( x y ) ;
2 cos( x y )
(a) the slope of the tangent line m y (1, 0) 1 2 cos( x y )
y 2 x 2
(b) the normal line is y 0 21 ( x 1) y 2x 21
(1, 0)
2 the tangent line is y 0 2 ( x 1)
38. x 2 cos 2 y sin y 0 x 2 (2 cos y )( sin y ) y 2 x cos 2 y y cos y 0 y [ 2 x 2 cos y sin y cos y ]
2 x cos 2 y y
2 x cos 2 y
2
2 x cos y sin y cos y
;
(a) the slope of the tangent line m y (0, )
2 x cos 2 y
2
2 x cos y sin y cos y (0, )
0 the tangent line is y
(b) the normal line is x 0
39. Solving x 2 xy y 2 7 and y 0 x 2 7 x 7 7,0 and
2
7,0 are the points where the curve
2 x y
2
crosses the x-axis. Now x xy y 7 2 x y xy 2 yy 0 ( x 2 y ) y 2x y y x 2 y
2 x y
m x 2 y the slope at 7,0 is m 2 7 2 and the slope at
7
slope is 2 in each case, the corresponding tangents must be parallel.
dy
dy
dy
7, 0 is m 2 77 2. Since the
y2
40. xy 2 x y 0 x dx y 2 dx 0 dx 1 x ; the slope of the line 2 x y 0 is 2. In order to be
parallel, the normal lines must also have slope of 2. Since a normal is perpendicular to a tangent, the slope
y2
of the tangent is 12 . Therefore, 1 x 12 2 y 4 1 x x 3 2 y. Substituting in the original equation,
y (3 2 y ) 2(3 2 y ) y 0 y 2 4 y 3 0 y 3 or y 1. If y 3, then x 3 and
y 3 2( x 3) y 2 x 3. If y 1, then x 1 and y 1 2( x 1) y 2 x 3.
Copyright 2016 Pearson Education, Ltd.
Section 3.7 Implicit Differentiation
41. y 4 y 2 x 2 4 y 3 y 2 yy 2 x 2(2 y 3 y ) y 2 x y
3
, 23
4
x
y 2 y3
x
y 2 y3
is
3 1
,
4 2
3
4
12
2 8
3
, 23
4
3
4
3 6 3
8
2
x ; the slope of the tangent line at
y 2 y3
1
1 4 3 213 1; the slope of the tangent line at
2
149
4
, is
3 1
4 2
24 32 3
y 2 3 x 2
y 2 3 x 2
42. y 2 (2 x) x3 2 yy (2 x) y 2 (1) 3x 2 y 2 y (2 x ) ; the slope of the tangent line is m 2 y (2 x )
(1, 1)
4
1
1
2 2 the tangent line is y 1 2( x 1) y 2 x 1; the normal line is y 1 2 ( x 1) y 2 x 32
3
3
43. y 4 4 y 2 x 4 9 x 2 4 y 3 y 8 yy 4 x3 18 x y (4 y 3 8 y ) 4 x3 18 x y 4 x 318 x 2 x3 9 x
4 y 8 y
2 y 4 y
( 3)(189)
27
27
27
27
m; (3, 2): m 2(8 4) 8 ;(3, 2): m 8 ;(3, 2): m 8 ;(3, 2): m 8
y (2 y 2 4)
x (2 x 2 9)
44. x3 y 3 9 xy 0 3x 2 3 y 2 y 9 xy 9 y 0 y (3 y 2 9 x) 9 y 3 x 2 y
(a) y (4, 2) 54 and y (2, 4) 54 ;
(b) y 0
3 y x2
y 2 3 x
9 y 3 x 2
2
3 y 9 x
3 y x2
y 2 3 x
9x 0 x 54x 0
2
2
0 3 y x 2 0 y x3 x3 x3
3
x2
3
6
3
x3 ( x3 54) 0 x 0 or x 3 54 33 2 there is a horizontal tangent at x 33 2. To find the
corresponding y -value, we will use part (c).
(c)
2
dx 0 y 3 x 0 y 2 3 x 0 y 3 x ; y
dy
3 y x2
3/2 3/2
3/2
3/2
x
x
6 3 0 or x
0 or x
3 x x3
3x 9 x 3x 0 x3 6 3x3/2 0
3
6 3 x 0 or x 3 108 33 4. Since the equation
x3 y 3 9 xy 0 is symmetric in x and y, the graph is symmetric about the line y x. That is, if ( a, b) is
a point on the folium, then so is (b, a ). Moreover, if y ( a, b) m, then y ( a , b ) m1 . Thus, if the folium has
a horizontal tangent at (a, b), it has a vertical tangent at (b, a ) so one might expect that with a horizontal
tangent at x 3 54 and a vertical tangent at x 33 4, the points of tangency are 3 54, 33 4 and
33 4, 3 54 , respectively. One can check that these points do satisfy the equation x3 y 3 9 xy 0.
x y
45. x 2 2 xy 3 y 2 0 2 x 2 xy 2 y 6 yy 0 y (2 x 6 y ) 2 x 2 y y 3 y x the slope of
x y
the tangent line m y (1, 1) 3 y x
(1, 1)
1 the equation of the normal line at (1, 1) is y 1 1( x 1)
y x 2. To find where the normal line intersects the curve we substitute into its equation:
x 2 2 x(2 x) 3(2 x) 2 0 x 2 4 x 2 x 2 3(4 4 x x 2 ) 0 4 x 2 16 x 12 0 x 2 4 x 3 0
( x 3)( x 1) 0 x 3 and y x 2 1. Therefore, the normal to the curve at (1, 1) intersects the
curve at the point (3, 1). Note that it also intersects the curve at (1, 1).
q
46. Let p and q be integers with q 0 and suppose that y x p x p /q . Then y q x p Since p and q are integers
d ( y q ) d ( x p ) qy q 1
and assuming y is a differentiable function of x, dx
px p 1 dx
dx
dx
dy
p 1
p 1
p
p
p
p
x
q xp p / q q . x p 1( p p /q ) q x( p /q ) 1
q ( x p /q ) q 1
x
Copyright 2016 Pearson Education, Ltd.
dy
px p 1
qy
q 1
p 1
p
q x q 1
y
150
Chapter 3 Derivatives
y 0
dy
47. y 2 x dx 21y If a normal is drawn from (a, 0) to ( x1 , y1 ) on the curve its slope satisfies x1 a 2 y1
1
y1 2 y1 ( x1 a ) or a x1 12 Since x1 0 on the curve, we must have that a 12 . By symmetry, the two
x x
points on the parabola are x1 , x1 and x1 , x1 . For the normal to be perpendicular, x 1a a x1 1
1 1
2
x1
1 x1 ( a x1 ) 2 x1 x1 12 x1 x1 14 and y1 12 Therefore, 14 , 12 and a 34 .
2
( a x1 )
48. 2 x 2 3 y 2 5 4 x 6 yy 0 y 32 yx y (1, 1) 32 yx
2
2
y 2 x3 2 yy 3x 2 y 32xy y (1, 1) 32xy
(1, 1)
23 and y (1,1) 32 yx
2
(1, 1)
23 and y (1, 1) 32xy
(1, 1)
(1, 1)
23 ; also,
23 . Therefore the
tangents to the curves are perpendicular at (1, 1) and (1, 1) (i.e., the curves are orthogonal at these two points of
intersection).
49. (a) x 2 y 2 4, x 2 3 y 2 (3 y 2 ) y 2 4 y 2 1 y 1. If y 1 x 2 (1) 2 4 x 2 3 x 3.
If y 1 x 2 (1)2 4 x 2 3 x 3.
dy
dy
dy
dy
x 2 y 2 4 2 x 2 y dx 0 m1 dx xy and x 2 3 y 2 2 x 6 y dx m2 dx 3xy
3, 1 : m1 dydx 13 3 and m2 dydx 3(1)3 33 m1 m2 3 33 1
dy
dy
At 3, 1 : m1 dx ( 1)3 3 and m2 dx 3( 31) 33 m1 m2 3 33 1
3
dy
dy
At 3, 1 : m1 dx 1 3 and m2 dx 3(1)3 33 m1 m2 3 33 1
3
3
dy
dy
At 3, 1 : m1 dx ( 1) 3 and m2 dx 3( 1) 33 m1 m2 3 33 1
At
(b) x 1 y 2 , x 13 y 2 , 13 y 2 1 y 2 y 2 34 y 23 . If y 23 x 1
. x 1 y 1 2 y m
If y 23 x 1 23
dy
2
dy
dx
2
1
4
1
dy
dx
1
2y
3
2
2
1
4
and x 13 y 2
dy
1 23 y dx m2 dx 23y
At , : m
dy
At 14 , 23 : m1 dx
1
4
.
3
2
1
dy
dx
dy
3 1
1
1 and m2 dx 3 3 m1 m2 1
2( 3 /2)
3
2( 3 /2)
3
3
3
dy
3
3
1
1
1
and m2 dx
m1 m2
3 1
2( 3 /2)
3
2( 3/2)
3
3
3
32xy 1 x2 y x2 x3
(0)
x4 x3 x 4 4 x3 0 x3 ( x 4) 0 x 0 or x 4. If x 0 y 2 0 and 13 32xy 1 is
dy
dy
2
dy
50. y 13 x b, y 2 x3 dx 13 and 2 y dx 3x 2 dx 32xy 13
2
2
2
4
indeterminate at (0, 0). If x 4 y
dy
2
2
2
(4)2
8. At (4, 8), y 13 x b 8 13 (4) b b 28
.
2
3
dy
dy
dy
51. xy 3 x 2 y 6 x 3 y 2 dx y 3 x 2 dx 2 xy 0 dx 3xy 2 x 2 y 3 2 xy dx
y 3 2 xy
2
3 xy x
2
dx x 2 y (2 x dx ) 0 dx ( y 3 2 xy ) 3 xy 2 x 2 dx
also, xy 3 x 2 y 6 x (3 y 2 ) y 3 dy
dy
dy
dy
y 3 2 xy
3 xy 2 x 2
3 xy 2 x 2
y 3 2 xy
;
;
dx appears to equal 1 The two different treatments view the graphs as functions symmetric across the
thus dy
dy
dx
line y x, so their slopes are reciprocals of one another at the corresponding points (a, b) and (b, a).
Copyright 2016 Pearson Education, Ltd.
Section 3.7 Implicit Differentiation
dy
dy
dy
dy
151
2
3 x
52. x3 y 2 sin 2 y 3 x 2 2 y dx (2sin y )(cos y ) dx dx (2 y 2sin y cos y ) 3x 2 dx 2 y 2sin
y cos y
2
x
dx 2 y 2 sin y cos y dx
2 sin y3cos
; also, x3 y 2 sin 2 y 3 x 2 dy
y 2 y
dy
2 sin y cos y 2 y
3x2
dx appears to
; thus dy
1 The two different treatments view the graphs as functions symmetric across the line y x so their
equal dy
dx
slopes are reciprocals of one another at the corresponding points (a, b) and (b, a).
53-60.
Example CAS commands:
Maple:
q1: x^3-x*y y^3 7;
pt : [x 2, y 1];
p1: implicitplot( q1, x -3..3, y -3..3 ):
p1;
eval( q1, pt );
q2 : implicitdiff( q1, y, x );
m : eval( q2, pt );
tan_line : y 1 m*(x-2);
p2 : implicitplot( tan_line, x -5..5, y -5..5, color green ):
p3 : pointplot( eval([x, y], pt), color blue):
display( [p1,p2,p3], "Section 3.7 #53(c)" );
Mathematica: (functions and x0 may vary):
Note use of double equal sign (logic statement) in definition of eqn and tanline.
<<Graphics`ImplicitPlot`
Clear[x, y]
{x0, y0}{1, /4};
eqn x Tan[y/x] 2;
ImplicitPlot[eqn,{x, x0 3, x0 3},{y, y0 3, y0 3}]
eqn/.{x x0, y y0}
eqn/.{ y y[x]}
D[%, x]
Solve[%, y'[ x]]
slope y '[x]/.First[%]
m slope/.{x x0, y[x] y0}
tanline y y0 m (x x0)
ImplicitPlot[{eqn, tanline}, {x, x0 3, x0 3},{y, y0 3, y0 3}]
Copyright 2016 Pearson Education, Ltd.
152
3.8
Chapter 3 Derivatives
RELATED RATES
1. A r 2 dA
2 r dr
dt
dt
2. S 4 r 2 dS
8 r dr
dt
dt
dy
dy
2 dt 5 dx
dt 5(2) 10
3. y 5 x, dx
dt
dt
dy
dy
4. 2 x 3 y 12, dt 2 2 dx
3 dt 0 2 dx
3(2) 0 dx
3
dt
dt
dt
dy
dy
5. y x 2 , dx
3 dt 2 x dx
; when x 1 dt 2(1)(3) 6
dt
dt
dy
dy
dy
6. x y 3 y, dt 5 dx
3 y 2 dt dt ; when y 2 dx
3(2) 2 (5) (5) 55
dt
dt
dy
dy
dy
7. x 2 y 2 25, dx
2 2 x dx
2 y dt 0; when x 3 and y 4 2(3)(2) 2( 4) dt 0 dt 32
dt
dt
dy
dy
4 ,
4 y 1.
8. x 2 y3 27
12 3 x 2 y 2 dt 2 xy 3 dx
0; when x 2 (2)2 y 3 27
dt
dt
3
12 2(2) 13 dxdt 0 dxdt 92
Thus 3(2) 2 13
2
3
dy
9. L x 2 y 2 , dx
1, dt 3 dL
dt
dt
dL
dt
(5)( 1) (12)(3)
2
(5) (12)
2
1
2 x2 y 2
2x 2 y
dy
dt
dx
dt
dy
x dx
y dt
dt
x2 y2
; when x 5 and y 12
31
13
10. r s 2 v3 12, dr
4, ds
3 dr
2 s ds
3v 2 dv
0; when r 3 and s 1 (3) (1) 2 v3 12 v 2
dt
dt
dt
dt
dt
4 2(1)(3) 3(2)2 dv
0 dv
16
dt
dt
11. (a)
2
m dS 12 x dx ; when x 3 dS 12(3)( 5) 180 m
S 6 x 2 , dx
5 min
dt
min
dt
dt
dt
3
m dV 3 x 2 dx ; when x 3 dV 3(3) 2 ( 5) 135 m
(b) V x3 , dx
5 min
dt
dt
dt
dt
min
2
12. S 6 x 2 , dS
72 cms dS
12 x dx
72 12(3) dx
dx
2 cm
; V x3 dV
3 x 2 dx
; when x 3
dt
dt
dt
dt
dt
dt
dt
s
dV
3(3) 2 (2) 54 cms
dt
3
13. (a) V r 2 h dV
r 2 dh
dt
dt
(b) V r 2 h dV
2 rh dr
dt
dt
14. (a) V 13 r 2 h dV
13 r 2 dh
dt
dt
(b) V 13 r 2 h dV
23 rh dr
dt
dt
(c) V r 2 h dV
r 2 dh
2 rh dr
dt
dt
dt
(c) dV
13 r 2 dh
2 rh dr
dt
dt 3
dt
15. (a)
(c)
dV 1 volt/s
dt
dV R dl I dR
dt
dt
dt
dl 1 amp/s
dt
3
1 dV R dI dR 1 dV V dI
dR
dt
I dt
dt
dt
I dt
I dt
(b)
Copyright 2016 Pearson Education, Ltd.
Section 3.8 Related Rates
(d)
16. (a)
dR 1 1 12
dt
2
2
13 12 (3) 23 ohms/s, R is increasing
P RI 2 dP
I 2 dR
2 RI dI
dt
dt
dt
dI
(b) P RI 2 0 dP
I 2 dR
2 RI dI
dR
2 RI
2 dt
dt
dt
dt
dt
I
17. (a) s x 2 y 2 ( x 2 y 2 )1/2 ds
dt
(b) s x 2 y 2 ( x 2 y 2 )1/2 ds
dt
(c) s
153
x
dI 2 P dI
2 PI
dt
I2
I 3 dt
dx
x 2 y 2 dt
x
dx
x 2 y 2 dt
y
dy
x 2 y 2 dt
dy
dy
y dy
x 2 y 2 s 2 x 2 y 2 2s ds
2 x dx
2 y dt 2 s 0 2 x dx
2 y dt dx
x dt
dt
dt
dt
dt
dy
18. (a) s x 2 y 2 z 2 s 2 x 2 y 2 z 2 2 s ds
2 x dx
2 y dt 2 z dz
dt
dt
dt
ds
dt
(b)
x
y
dx
dy
dz
z
x 2 y 2 z 2 dt
x 2 y 2 z 2 dt
x 2 y 2 z 2 dt
y
dy
From part (a) with dx
0 ds
2 2 2 dt 2 z 2 2 dz
dt
dt
x y z
x y z dt
dy
y dy
(c) From part (a) with ds
0 0 2 x dx
2 y dt 2 z dz
dx
x dt xz dz
0
dt
dt
dt
dt
dt
19. (a) A 12 ab sin dA
12 ab cos ddt
dt
(b) A 12 ab sin dA
12 ab cos ddt 12 b sin da
dt
dt
(c) A 12 ab sin dA
12 ab cos ddt 12 b sin da
12 a sin db
dt
dt
dt
1 cm 2 /min.
20. Given A r 2 , dr
0.01 cm/s, and r 50 cm. Since dA
2 r dr
, then dA
2 (50) 100
dt
dt
dt r 50
dt
21. Given ddt 2 cm/s, dw
2 cm/s, 12 cm and w 5 cm.
dt
(a) A w dA
dw
w ddt dA
12(2) 5(2) 14 cm 2 /s, increasing
dt
dt
dt
(b) P 2 2 w dP
2 ddt 2 dw
2(2) 2(2) 0 cm/s, constant
dt
dt
(c) D w2 2 ( w2 2 )1/2 dD
12 w2 2
dt
2)
2w dwdt 2 ddt dDdt w w (5)(2)25(12)(
144
1/2
d
dt
2
dw
dt
2
14
cm/s, decreasing
13
dy
yz dx
xz dt xy dz
dV
(3)(2)(1) (4)(2)(2) (4)(3)(1) 2 m3 /s
22. (a) V xyz dV
dt
dt
dt
dt (4, 3, 2)
dy
(b) S 2 xy 2 xz 2 yz dS
(2 y 2 z ) dx
(2 x 2 z ) dt (2 x 2 y ) dz
dt
dt
dt
dS
dt
(4, 3, 2)
(10)(1) (12)(2) (14)(1) 0 m 2 /s
(c) x 2 y 2 z 2 ( x 2 y 2 z 2 )1/2 ddt
ddt (4, 3, 2)
dx
x y z dt
2
2
(1) (2) (1) 0 m/s
3
29
4
29
y
x
2
2
2
x y z
2
dy
dt
z
dz
x 2 y 2 z 2 dt
2
29
23. Given: dx
1.5 m/s, the ladder is 3.9 m long, and x 3.6, y 1.5 at the instant of time
dt
dy
3.6 (1.5) 3.6 m/s, the ladder is sliding down the wall
(a) Since x 2 y 2 15.21 dt xy dx
1.5
dt
x dydt y dxdt . The area is
(b) The area of the triangle formed by the ladder and walls is A 12 xy dA
12
dt
changing at 12 [3.6(3.6) 1.5(1.5)] 10.71
5.355 m 2 /s.
2
x sin d 1 dx d
1 (1.5) 1 rad / s
1
(c) cos 3.9
dx 1.5
dt
3.9 dt
dt
3.9sin dt
Copyright 2016 Pearson Education, Ltd.
154
Chapter 3 Derivatives
dy
dy
24. s 2 y 2 x 2 2 s ds
2 x dx
2 y dt ds
1s x dx
y dt ds
dt
dt
dt
dt
dt
1 [5( 442) 12( 481)] 614 knots
169
25. Let s represent the distance between the girl and the kite and x represents the horizontal distance between the
girl and kite s 2 (90) 2 x 2 ds
xs dx
dt
dt
120(7.5)
6 m/s.
150
1 cm/min. Also V 15 r 2 dV 30 r dr
3000
26. When the diameter is 10 cm, the radius is 5 cm and dr
dt
dt
dt
1
dV
30 (5) 3000
0.05 . The volume is changing at about 0.157 cm3/min.
dt
27. V 13 r 2 h, h 83 (2r ) 34r r 43h V 13
(a)
(b)
43h h 1627 h dVdt 169h dhdt
2
3
2
dh
9
90 0.1119 m/s 11.19 cm/s
(10) 256
dt h 4
16 42
4
h
dr
dh
90
15 0.1492 m/s 14.92 cm/s
4
4
r 3 dt 3 dt 3 256 32
h 754 h dVdt 2254 h dhdt dhdt h5 2254(50)
(5)
28. (a) V 13 r 2 h and r 152h V 13 152h
(b)
2
8 0.0113 m/min 1.13 cm/min
225
dh dr
8
r 152h dr
15
15
dt
2 dt
2
225
dt h 5
2
3
2
154 0.0849 m/s 8.49 cm/s
dy
dy
29. (a) V 3 y 2 (3R y ) dV
3 [2 y (3R y ) y 2 ( 1)] dt dt 3 (6 Ry 3 y 2 )
dt
1
dV at R 13 and y 8
dt
dy
1 m/min
we have dt 1441 ( 6) 24
(b) The hemisphere is one the circle r 2 (13 y )2 169 r 26 y y 2 m
12 (26 y y 2 ) 1/2 (26 2 y ) dt dr
(c) r (26 y y 2 )1/2 dr
dt
dt
dy
dr
dt
y 8
2885 m/min
13 y
dy
26 y y 2 dt
138
1
268 64 24
30. If V 43 r 3 , S 4 r 2 , and dV
kS 4k r 2 , then dV
4 r 2 dr
4k r 2 4 r 2 dr
dr
k , a constant.
dt
dt
dt
dt
dt
Therefore, the radius is increasing at a constant rate.
100 m3 /min, then dV
4 r 2 dr
dr
1 m/min. Then S 4 r 2
31. If V 43 r 3 , r 5, and dV
dt
dt
dt
dt
dS
8 r dr
8 (5)(1) 40 m 2 /min, the rate at which the surface area is increasing.
dt
dt
32. Let s represent the length of the rope and x the horizontal distance of the boat from the dock.
(a) We have s 2 x 2 36 dx
xs ds
2s ds
. Therefore, the boat is approaching the dock at
dt
dt
dt
(b)
s 36
dx
10
(2) 2.5 m/s.
dt s 10
102 36
8
cos 6r sin ddt 62 dr
ddt 2 6 dr
. Thus, r 10, x 8, and sin 10
r dt
r sin dt
3 rad/s
ddt 26 8 (2) 20
10 10
Copyright 2016 Pearson Education, Ltd.
Section 3.8 Related Rates
155
33. Let s represent the distance between the bicycle and balloon, h the height of the balloon and x the horizontal
distance between the balloon and the bicycle. The relationship between the variables is s 2 h 2 x 2
1 [68(1) 51(17)] 11 m/s.
ds
1s h dh
x dx
ds
85
dt
dt
dt
dt
34. (a) Let h be the height of the coffee in the pot. Since the radius of the pot is 7.5, the volume of the coffee is
dV 160 cm/min.
1
V 56.25 h dV
56.25 dh
the rate the coffee is rising is dh
56.25
dt
dt
dt
dt
56.25
3
(b) Let h the height of the coffee in the pot. From the figure, the radius of the filter r h2 V 13 r 2 h 12h ,
the volume of the filter. The rate the coffee is falling is dh
dt
4 dV
h 2 dt
4 ( 160) 80 cm/min.
144
18
1 (0) 233 ( 2) 466 L/min increasing about 0.2772 L/min
41
35. y QD 1 dt D 1 dt QD 2 dD
2
dt
1681
dy
dQ
(41)
36. Let P( x, y ) represent a point on the curve y x 2 and the angle of inclination of a line containing P and the
2
y
origin. Consequently, tan x tan xx x sec2 ddt dx
ddt cos 2 dx
. Since dx
10 m/s and
dt
dt
dt
cos 2 x 3
x 2 32 1 , we have d
1 rad/s.
10
dt x 3
y 2 x2
92 32
37. The distance from the origin is s x 2 y 2 and we wish to find
ds
12 ( x 2 y 2 ) 1/2
dt (5, 12)
2x 2 y
dx
dt
dy
dt
(5, 12)
(5)( 1) (12)( 5)
5m/s
25144
s
38. Let s distance of the car from the foot of perpendicular in the textbook diagram tan 40
2
1 ds d cos ds ;
sec 2 ddt 40
dt
dt
40 dt
ds 80 and 0 d 2 rad/sec. A half second later the car has
dt
dt
80 (since s increases)
traveled 40 m right of the perpendicular | | 4 , cos 2 12 , and ds
dt
1
(2)
ddt 40
(80) 1 rad/s.
39. Let s 4.9t 2 represent the distance the ball has
fallen, h the distance between the ball and the
ground, and I the distance between the shadow and
the point directly beneath the ball. Accordingly,
s h 15 and since the triangle LOQ and triangle
PRQ are similar we have I 159hh h 15 4.9t 2
and I
9(15 4.9t 2 )
15(15 4.9t 2 )
1352 9 dI
2703
dt
4.9t
4.9t
dI
441 m/s.
dt t 1
2
sec 2 ddt 242 dx
dx
40. When x represents the length of the shadow, then tan 24
dt
dt
x
x
3 rad/ min . At x 18, cos 3 dx
We are given that ddt 0.27 2000
5
dt
9 m/min 0.1767 m/min 17.7 cm/min.
160
Copyright 2016 Pearson Education, Ltd.
x 2 sec2 d
24
dt
x 2 sec2 d .
24
dt
3
and sec 53
ddt 2000
156
Chapter 3 Derivatives
41. The volume of the ice is V 43 r 3 43 43 dV
4 r 2 dr
dr
5 cm/min when dV
dt
dt
dt
dt r 6 72
10 cm3 /min, the thickness of the ice is decreasing at 725 cm/min. The surface area is
S 4 r 2 ds
8 r dr
dS
dt
dt
dt
48
r 6
725 103 cm2 /min, the outer surface area of the ice is decreasing
at 10
cm 2 /min.
3
42. Let s represent the horizontal distance between the car and plane while r is the line-of-sight distance between
r dr
ds
5 (160) 200 km/h speed of plane
the car and plane 9 s 2 r 2 ds
dt
dt
dt
2
r 5
r 9
16
speed of car 200 km/h the speed of the car is 80 km/h.
43. Let x represent distance of the player from second base and s the distance to third base. Then dx
5 m/s
dt
(a)
2 x dx
ds
xs dx
. When the player is 9 m from first base, x 18 s 9 13
s 2 x 2 729 2 s ds
dt
dt
dt
dt
and ds
dt
18 ( 5) 10 2.774 m/s
9 13
13
d
d
(b) sin 1 27
cos 1 dt1 272 ds
dt1 2 27
dt
s
d
dt1
27
10
13
9 13 (18)
5 rad/s; cos 27 sin d 2
2
2 dt
39
s
d1
2 27
dt
s cos
1
lim 2 27
x 0
x 729
d
ds
dt
27
s 2 xs
1
6
x
s
d
27
13 dt2
9 13 (18)
dx
dt
27
s2
d 2
dt
27
ds
s 2 sin 2 dt
dx
dt
s
10
x
s
ds
s sin 2 dt
5
39
13
27
dx
x 2 729 dt
27
s 2 xs
13
272 ds
dt2 2 27
dt
rad / s.
lim
(4.5) rad/s;
27 ds . Therefore, x 18 and s 9
sk dt
(c)
ds s27
ds . Therefore, x 18 and s 9
k dt
s cos 1 dt
s
dx
dt
d1
x 0 dt
27
s2
dx
dt
27
dx lim d 2 1 rad/s
6
x 2 729 dt
x 0 dt
44. Let a represent the distance between point O and ship A, b the distance between point O and ship B, and D the
distance between the ships. By the Law of Cosines, D 2 a 2 b 2 2ab cos120 dD
21D 2a da
2b db
dt
dt
dt
1 2a da 2b db a db b da . When a 5, da 14, b 3, and db 21, then dD 413 where D 7. The
2D
dt
dt
dt
dt
dt
dt
dt
2D
dD
ships are moving dt 29.5 knots apart.
45. The hour hand moves clockwise from 4 at 30/h 0.5/min. The minute hand, starting at 12, chases
the hour hand at 360 /h 6 /min. Thus, the angle between them is decreasing and is changing at 0.5/min
6/min
5.5/min.
46. The volume of the slick in cubic meters is V 43
a2 b2 , where a is the length of the major axis and b is the
3
a d b
b d a
3
db
da
length of the minor axis. dV
dt 4 2 dt 2 2 dt 2 16 a dt b dt . Convert all measurements to meters and
substitute:
dV
dt
316 2(1000)(3) 34 (1000)(9) 316 (12,750) 7,510 m3 /h
Copyright 2016 Pearson Education, Ltd.
3.9 Linearization and Differentials
3.9
157
LINEARIZATION AND DIFFERENTIALS
1. f ( x) x3 2 x 3 f ( x) 3x 2 2 L( x ) f (2)( x 2) f (2) 10( x 2) 7 L( x) 10 x 13 at x 2
2. f ( x) x 2 9 ( x 2 9)1/2 f ( x)
12 ( x2 9)1/2 (2 x) xx9 L( x) f (4)( x 4) f (4)
2
54 ( x 4) 5 L( x) 54 x 95 at x 4
3. f ( x) x 1x f ( x) 1 x 2 L( x) f (1) f (1)( x 1) 2 0( x 1) 2
4. f ( x) x1/3 f ( x)
1 L ( x ) f ( 8)( x ( 8)) f (8) 1 ( x 8) 2 L ( x ) 1 x 4
12
12
3
3 x 2/3
5. f ( x) tan x f ( x) sec2 x L( x) f ( ) f ( )( x ) 0 1( x ) x
6. (a) f ( x) sin x f ( x) cos x L( x ) f (0) f (0)( x 0) x L( x) x
(b) f ( x) cos x f ( x) sin x L( x) f (0) f (0)( x 0) 1 L( x) 1
(c) f ( x) tan x f ( x) sec 2 x L( x) f (0) f (0)( x 0) x L( x) x
7. f ( x) x 2 2 x f ( x) 2 x 2 L( x) f (0)( x 0) f (0) 2( x 0) 0 L( x) 2 x at x 0
8. f ( x) x 1 f ( x) x 2 L( x) f (1)( x 1) f (1) (1)( x 1) 1 L( x) x 2 at x 1
9. f ( x) 2 x 2 4 x 3 f ( x) 4 x 4 L( x) f (1)( x 1) f (1) 0( x 1) (5) L( x) 5 at x 1
10.
f ( x) 1 x f ( x) 1 L( x) f (8)( x 8) f (8) 1( x 8) 9 L( x) x 1 at x 8
11.
1 ( x 8) 2 L( x) 1 x 4 at x 8
f ( x) 3 x x1/3 f ( x) 13 x 2/3 L( x) f (8)( x 8) f (8) 12
12
3
12.
f ( x) xx1 f ( x)
(1)( x 1) (1)( x )
( x 1) 2
1 L( x ) f (1)( x 1) f (1) 1 ( x 1) 1 L( x) 1 x 1
4
2
4
4
( x 1)2
at x 1
13. f ( x) k (1 x) k 1. We have f (0) 1 and f (0) k . L( x) f (0) f (0)( x 0) 1 k ( x 0) 1 kx
14. (a) f ( x) (1 x)6 [1 ( x)]6 1 6( x) 1 6 x
1
(b) f ( x) 12x 2 1 ( x) 2[1 (1)( x)] 2 2 x
(c) f ( x) 1 x
1/2
1 12 x 1 2x
2
(d) f ( x) 2 x 2 2 1 x2
2 1 2 1
1/2
(e) f ( x ) (4 3x )1 3 41 3 1 34x
(f)
x
f ( x ) 1
2
x
2/3
13
1 x2
2 2
x2
4
41 3 1 13 34x 41 3 1 4x
1 2x x
2/3
x
2x
1 23
1 6 3 x
2
x
15. (a) (1.0002)50 (1 0.0002)50 1 50(0.0002) 1 .01 1.01
(b) 3 1.009 (1 0.009)1/3 1 13 (0.009) 1 0.003 1.003
Copyright 2016 Pearson Education, Ltd.
158
Chapter 3 Derivatives
12 ( x 1)1/2 cos x L f ( x) f (0)( x 0) f (0)
32 ( x 0) 1 L f ( x ) 32 x 1, the linearization of f ( x); g ( x) x 1 ( x 1)1/2 g ( x ) 12 ( x 1)1/2
16. f ( x) x 1 sin x ( x 1)1/2 sin x f ( x)
Lg ( x) g (0)( x 0) g (0) 12 ( x 0) 1 Lg ( x) 12 x 1, the linearization of g ( x); h( x) sin x
h( x) cos x Lh ( x) h(0)( x 0) h(0) (1)( x 0) 0 Lh ( x) x, the linearization of h( x).
L f ( x) Lg ( x) Lh ( x) implies that the linearization of a sum is equal to the sum of the linearizations.
17. y x3 3 x x3 3 x1/2 dy 3x 2 32 x 1/2 dx dy 3 x 2
3
2 x
dx
18. y x 1 x 2 x (1 x 2 )1/2 dy (1) (1 x 2 )1/2 ( x) 12 (1 x 2 ) 1/2 (2 x) dx
1 x2
1/2
dx
(1 x 2 ) x 2 dx
2
1 x
1 2 x 2
19. y
2 x dy (2)(1 x ) (2 x )(2 x ) dx 2 2 x 2 dx
1 x 2
(1 x 2 )2
(1 x 2 ) 2
20. y
2 x
3 1 x
2
x 1/ 2 31 x1/ 2 2 x1/ 2 3 x 1/ 2
1/ 2
1 2
2
1
dx 3 x 33 dx dy
dx
2 x 1/ 2 dy
1/ 2 2
2
1/ 2
9(1
x
)
3
x
(1
x )2
31 x
91 x
21. 2 y 3/2 xy x 0 3 y1/2 dy y dx x dy dx 0 (3 y1/2 x) dy (1 y ) dx dy
1 y
dx
3 y x
22. xy 2 4 x3/2 y 0 y 2 dx 2 xy dy 6 x1/2 dx dy 0 (2 xy 1) dy (6 x1/2 y 2 )dx dy
23. y sin (5 x ) sin (5 x1/2 ) dy (cos (5 x1/2 ))
52 x1/2 dx dy
5cos 5 x
2 x
dx
24. y cos ( x 2 ) dy [ sin ( x 2 )](2 x)dx 2 x sin ( x 2 )dx
( x ) dx dy 4x sec dx
3
25. y 4 tan x3 dy 4 sec 2
x3
3
2
2
2 x3
3
26. y sec x 2 1 dy [sec ( x 2 1) tan ( x 2 1)](2 x) dx 2 x [sec ( x 2 1) tan ( x 2 1)]dx
27. y 3csc (1 2 x ) 3csc (1 2 x1/2 ) dy 3( csc (1 2 x1/2 )) cot (1 2 x1/2 ) ( x 1/2 ) dx
dy 3 csc (1 2 x ) cot (1 2 x ) dx
x
28. y 2 cot
2 cot x dy 2 csc ( x
1
x
1/2
2
1/2
) 12 ( x 3/2 ) dx dy 1 3 csc2
29. f ( x ) x 2 2 x, x0 1, dx 0.1 f ( x) 2 x 2
(a) f f ( x0 dx) f ( x0 ) f (1.1) f (1) 3.41 3 0.41
(b) df f ( x0 ) dx [2(1) 2](0.1) 0.4
(c) | f df | |0.41 0.4| 0.01
Copyright 2016 Pearson Education, Ltd.
x
dx
1
x
6 x y2
dx
2 xy 1
3.9 Linearization and Differentials
159
30. f ( x ) 2 x 2 4 x 3, x0 1, dx 0.1 f ( x) 4 x 4
(a) f f ( x0 dx) f ( x0 ) f (.9) f (1) .02
(b) df f ( x0 ) dx [4(1) 4](.1) 0
(c) | f df | |.02 0| .02
31. f ( x ) x3 x, x0 1, dx 0.1 f ( x) 3x 2 1
(a) f f ( x0 dx) f ( x0 ) f (1.1) f (1) .231
(b) df f ( x0 )dx [3(1) 2 1](.1) .2
(c) | f df | |.231 .2| .031
32. f ( x) x 4 , x0 1, dx 0.1 f ( x) 4 x3
(a) f f ( x0 dx) f ( x0 ) f (1.1) f (1) .4641
(b) df f ( x0 )dx 4(1)3 (.1) .4
(c) | f df | |.4641 .4| .0641
33. f ( x) x 1 , x0 0.5, dx 0.1 f ( x) x 2
(a) f f ( x0 dx) f ( x0 ) f (.6) f (.5) 13
1 2
(b) df f ( x0 )dx (4) 10
5
(c)
1
| f df | | 13 52 | 15
34. f ( x) x3 2 x 3, x0 2, dx 0.1 f ( x) 3x 2 2
(a) f f ( x0 dx) f ( x0 ) f (2.1) f (2) 1.061
(b) df f ( x0 )dx (10)(0.10) 1
(c) | f df | |1.061 1| .061
35. V 43 r 3 dV 4 r02 dr
36. V x3 dV 3 x02 dx
37. S 6 x 2 dS 12 x0 dx
38. S r r 2 h 2 r (r 2 h 2 )1/2 , h constant dS
(r 2 h 2 )1/2 r r (r 2 h 2 )1/2 dS
dr
dr
dS
2 r02 h 2
r02 h 2
dr , h constant
39. V r 2 h, height constant dV 2 r0 h dr
40. S 2 rh dS 2 r dh
41. Given r 2 m, dr .02 m
(a) A r 2 dA 2 r dr 2 (2)(.02) .08 m 2
(100%) 2%
(b) .08
4
42. C 2 r and dC 2 cm dC 2 dr dr 1 the diameter grew about 2 cm; A r 2
dA 2 r dr 2 (5) 1 10 cm 2
Copyright 2016 Pearson Education, Ltd.
r 2 h 2 r 2
2
r h2
160
Chapter 3 Derivatives
43. The volume of a cylinder is V r 2 h. When h is held fixed, we have dV
2 rh, and so dV 2 rh dr .
dr
For h 30 cm, r 6 cm, and dr 0.5 cm, the volume of the material in the shell is approximately
dV 2 rh dr 2 (6)(30)(0.5) 180 565.5 cm3 .
44. Let angle of elevation and h height of building. Then h 30 tan , so dh 30 sec2 d . We want
| d | 0.04sin cos
| dh | 0.04h, which gives: |30sec2 d | 0.04 |30 tan | 12 | d | 0.04sin
cos
cos
5
5
| d | 0.04sin 12 cos 12 0.01 radian. The angle should be measured with an error of less than
0.01 radian (or approximately 0.57 degrees), which is a percentage error of approximately 0.76%.
45. The percentage error in the radius is
(a)
drdt 100 2%.
r
dC 100
Since C 2 r dC
2 dr
. The percentage error in calculating the circle’s circumference is Cdt
dt
dt
dr
dr
2
dt
2 r 100 dtr 100 2%.
(b) Since A r 2 dA
2 r dr
. The percentage error in calculating the circle’s area is given by
dt
dt
2 r drdt 100 2 drdt 100 2(2%) 4%.
dAdt 100
A
r
r2
46. The percentage error in the edge of the cube is
dxdt 100 0.5%.
x
(a) Since S 6 x 2 dS
12 x dx
. The percentage error in the cube’s surface area is
dt
dt
100 2(0.5%) 1%
dSdt 100 12 x dxdt 100
S
6 x2
dx
dt
2 x
dV 100 3x2 dxdt 100
(b) Since V x3 dV
3 x 2 dx
. The percentage error in the cube’s volume is Vdt
dt
dt
100 3(0.5%) 1.5%
dx
dt
x3
3 x
47. V h3 dV 3 h 2 dh ; recall that V dV. Then | V | (1%)(V )
|3 h 2 dh |
of h is 13 %.
(1)( h3 )
(1)( h3 )
| dV | 100
100
(1)( h3 )
1 h 1 % h. Therefore the greatest tolerated error in the measurement
| dh | 300
100
3
48. (a) Let Di represent the interior diameter. Then V r 2 h
h
Di 2
2
Di2 h
4
and h 10 V
5 Di2
2
2
2
1 5 Di Di
dV 5 Di dDi . Recall that V dV . We want | V | (1%)(V ) | dV | 100
2 40
Di2
dD
5 Di dDi 40 D i 200. The inside diameter must be measured to within 0.5%.
i
(b) Let De represent the exterior diameter, h the height and S the area of the painted surface. S De h
dD
dS hdDe dS
D e . Thus for small changes in exterior diameter, the approximate percentage
s
e
change in the exterior diameter is equal to the approximate percentage change in the area painted, and to
estimate
the amount of paint required to within 5%, the tank’s exterior diameter must be measured to within 5%.
Copyright 2016 Pearson Education, Ltd.
3.9 Linearization and Differentials
49. Given D 100 cm, dD 1 cm, V 43
161
D2 D6 dV 2 D2 dD 2 (100)2 (1) 102 . Then dVV (100%)
3
3
4
104
104
106
26 102 % 26 % 3% 26
10
106
106
6
50. V 43 r 3 43
3
D2 D6 dV D2 dD; recall that V dV . Then V (3%)V 1003 D6 200D
3
3
2
3
D3 D 2 dD D3 dD D (1%) D the allowable percentage error in measuring the
dV 200
2
200
100
diameter is 1%.
51.
b dg
dWmoon
b dg
(1.6)2
1
2
b
W a g a bg dW bg dg 2 dW
b dg
g
earth
(9.8)2
9.8
1.6
37.52, so a change of gravity
2
on the moon has about 38 times the effect that a change of the same magnitude has on Earth.
3
52. C (t ) (14 t83t)2 0.06, where t is measured in hours. When the time changes from 20 min to 30 min, t in hours
changes from 13 to 12 , so the differential estimate for the change in C is
C 13 12 13 16 C 13 0.584 mg/mL.
3
53. The relative change in V is estimated by dVV/dr r 4krkr4
1.1r and
r 4 r r . If the radius increases by 10%, r changes to
r 0.1r. The approximate relative increase in V is thus
54. (a) T 2
dT 2 L g dg Lg
L
g
1/2
1
2
3/2
3/2
4(0.1r )
0.4 or 40%.
r
dg
(b) If g increases, then dg 0 dT 0. The period T decreases and the clock ticks more frequently. Both the
pendulum speed and clock speed increase.
(c) 0.001 100(9803/2 ) dg dg 0.977 cm/s 2 the new g 979 cm/s 2
Q(a ) f (a ) implies that b0 f (a).
Since Q ( x) b1 2b2 ( x a ), Q(a ) f (a ) implies that b1 f (a ).
f ( a )
iii. Since Q ( x) 2b2 , Q (a ) f (a ) implies that b2 2 .
55. (a) i.
ii.
In summary, b0 f (a ), b1 f (a ), and b2
f ( a )
.
2
2
(b) f ( x) (1 x)1 ; f ( x) 1(1 x)2 (1) (1 x) ; f ( x) 2(1 x)3 (1) 2(1 x)3 Since
f (0) 1, f (0) 1, and f (0) 2, the coefficients are b0 1, b1 1, b2 22 1. The quadratic
approximation is Q ( x) 1 x x 2 .
(c)
As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(d) g ( x) x 1; g ( x) 1x 2 ; g ( x) 2 x 3
Copyright 2016 Pearson Education, Ltd.
162
Chapter 3 Derivatives
Since g (1) 1, g (1) 1, and g (1) 2, the coefficients are b0 1, b1 1, b2 22 1. The quadratic
approximation is Q ( x) 1 ( x 1) ( x 1) 2 .
As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(e) h( x) (1 x)1/2 ; h( x) 12 (1 x) 1/2 ; h( x) 14 (1 x) 3/2
1
Since h(0) 1, h(0) 12 , and h(0) 14 , the coefficients are b0 1, b1 12 , b2 24 18 . The quadratic
2
approximation is Q( x) 1 2x x8 .
As one zooms in, the two graphs quickly become
indistinguishable. They appear to be identical.
(f ) The linearization of any differentiable function u ( x) at x a is L( x) u ( a ) u ( a )( x a ) b0 b1 ( x a ),
where b0 and b1 are the coefficients of the constant and linear terms of the quadratic approximation. Thus,
the linearization for f ( x) at x 0 is 1 x; the linearization for g ( x) at x 1 is 1 ( x 1) or 2 x; and the
linearization for h( x) at x 0 is 1 2x .
56. E ( x) f ( x) g ( x) E ( x ) f ( x) m( x a ) c. Then E (a ) 0 f ( a ) m( a a ) c 0 c f ( a ).
E ( x)
f ( x ) m( x a ) c
f ( x) f (a )
Next we calculate m: lim x a 0 lim
0 lim x a m 0 (since c f (a ))
x
a
x a
x a
x a
f (a ) m 0 m f (a). Therefore, g ( x) m( x a ) c f (a )( x a ) f (a) is the linear approximation,
as claimed.
5760. Example CAS commands:
Maple:
with(plots):
a : 1: f : x -> x^3 x^2 2*x;
plot(f(x), x 1..2);
diff (f(x), x);
fp : unapply (, x);
L: x ->f(a) fp(a)*(x a);
plot({f(x), L(x)}, x 1..2);
err: x -> abs(f(x) L(x));
plot(err(x), x 1..2, title #absolute error function#);
err(1);
Mathematica: (function, x1, x2, and a may vary):
Clear[f , x]
{x1, x2} {1, 2};a 1;
f[x_ ]: x 3 x 2 2x
Plot [f[x], {x, x1, x2}]
lin[x_ ] f[a] f [a](x a)
Plot[{f[x], lin[x]},{x, x1, x2}]
err[x_ ] Abs [f[x] lin[x]]
Plot[err[x], {x, x1, x2}]
err//N
Copyright 2016 Pearson Education, Ltd.
Chapter 3 Practice Exercises
163
After reviewing the error function, plot the error function and epsilon for differing values of epsilon (eps) and
delta (del)
eps 0.5; del 0.4
Plot[{err[x], eps}, {x, a del, a del}]
CHAPTER 3
PRACTICE EXERCISES
dy
1. y x5 0.125 x 2 0.25 x dx 5 x 4 0.25 x 0.25
dy
2. y 3 0.7 x3 0.3 x7 dx 2.1x 2 2.1x6
dy
3. y x3 3( x 2 2 ) dx 3 x 2 3(2 x 0) 3 x 2 6 x 3 x( x 2)
dy
4. y x7 7 x 11 dx 7 x 6 7
dy
5. y ( x 1) 2 ( x 2 2 x) dx ( x 1) 2 (2 x 2) ( x 2 2 x)(2( x 1)) 2( x 1)[( x 1)2 x( x 2)]
2( x 1)(2 x 2 4 x 1)
6. y (2 x 5)(4 x)1 dx (2 x 5)(1)(4 x)2 (1) (4 x)1 (2) (4 x) 2 [(2 x 5) 2(4 x)] 3(4 x)2
dy
dy
7. y ( 2 sec 1)3 d 3( 2 sec 1)2 (2 sec tan )
2
8. y 1 csc2 4
9. s
10. s
t
1 t
ds
dt
1 ds
dt
t 1
2
2
dy
2 1 csc2 4
d
csc cot
2
2
1 csc2 4 (csc cot )
2
1 t 21 t t 21 t 1 t t
1
2
2
2
2 t 1 t
2 t 1 t
1 t
( t 1) (0) 1
t 1
2
1
2 t
1
2 t
t 1
2
dy
11. y 2 tan 2 x sec 2 x dx (4 tan x)(sec 2 x) (2sec x)(sec x tan x) 2sec 2 x tan x
12. y
1 2 csc2 x 2 csc x dy (2 csc x )( csc x cot x ) 2( csc x cot x ) (2 csc x cot x )(1 csc x )
dx
sin 2 x sin x
13. s cos 4 (1 2t ) ds
4 cos3 (1 2t )( sin(1 2t ))(2) 8 cos3 (1 2t ) sin(1 2t )
dt
14. s cot 3 2t ds
3cot 2
dt
2t csc2 2t t 2 t6 cot 2 2t csc2 2t
2
2
Copyright 2016 Pearson Education, Ltd.
164
Chapter 3 Derivatives
15. s (sec t tan t )5 ds
5(sec t tan t ) 4 sec t tan t sec2 t 5(sec t )(sec t tan t )5
dt
16. s csc5 (1 t 3t 2 ) ds
5csc4 (1 t 3t 2 ) ( csc (1 t 3t 2 ) cot (1 t 3t 2 )) (1 6t )
dt
5(6t 1) csc5 (1 t 3t 2 ) cot (1 t 3t 2 )
17. r 2 sin (2 sin )1/2 ddr 12 (2 sin )1/2 (2 cos 2sin ) cos sin
2 sin
12 (cos )1/2 ( sin ) 2(cos )1/2 cossin 2 cos 2 coscossin
18. r 2 cos 2 (cos )1/2 ddr 2
19. r sin 2 sin(2 )1/2 ddr cos(2 )1/2 12 (2 )1/2 (2) cos 2
20. r sin 1 ddr cos 1 1
dy
21. y 12 x 2 csc 2x dx 12 x 2 csc 2x cot 2x
2 1 1
cos
2 1
1
x2 csc 2x 12 2 x csc 2x cot 2x x csc 2x
2
2 1 x sin x 2 2 x cos x sin x x
dy
1
2 1
2
22. y 2 x sin x dx 2 x cos x
23. y x 1/2 sec (2 x) 2 dx x 1/2 sec (2 x) 2 tan(2 x)2 (2(2 x) 2) sec (2 x)2 12 x 3/2
dy
1/2
8x
or
2
sec (2 x) tan (2 x)
2
12 x 3/2 sec (2 x) 2 12 x1/2 sec (2 x) 2 16 tan (2 x) 2 x 2
1 sec (2 x ) 2 16 x 2 tan(2 x ) 2 1
2 x3/ 2
24. y x csc( x 1)3 x1/2 csc ( x 1)3 dx x1/2 ( csc ( x 1)3 cot ( x 1)3 ) (3( x 1)2 ) csc ( x 1)3 12 x 1 2
dy
2
3
3 x ( x 1) csc ( x 1) cot ( x 1)
3
csc( x 1)3
12
2 x
3
or 1 csc ( x 1)3 1 6 x( x 1)2 cot ( x 1)
2 x
x csc ( x 1)3 1x 6( x 1) 2 cot ( x 1)3 or 1 csc ( x 1)3
2 x
dy
25. y 5cot x 2 dx 5( csc2 x 2 )(2 x) 10 x csc2 ( x 2 )
dy
26. y x 2 cot 5 x dx x 2 ( csc2 5 x)(5) (cot 5 x)(2 x) 5 x 2 csc2 5 x 2 x cot 5 x
27.
dy
y x 2 sin 2 (2 x 2 ) dx x 2 (2sin (2 x 2 )) (cos (2 x 2 ))(4 x) sin 2 (2 x 2 )(2 x)
8 x3 sin(2 x 2 ) cos(2 x 2 ) 2 x sin 2 (2 x 2 )
28. y x 2 sin 2 ( x3 ) dx x 2 (2sin ( x3 )) (cos ( x3 ))(3x 2 ) sin 2 ( x3 )(2 x 3 ) 6sin ( x3 ) cos ( x3 ) 2 x 3 sin 2 ( x3 )
dy
(t 1)
(4t )(1)
4t
4
dsdt 2 t4t1 (t 1)(4)
2 t 1 (t 1) 8t
(t 1)
29. s t4t1
30. s
2
3
3
2
2
3
3
1
1 (15t 1) 3 ds 1 ( 3)(15t 1) 4 (15)
15
dt
15
15(15t 1)3
(15t 1) 4
Copyright 2016 Pearson Education, Ltd.
Chapter 3 Practice Exercises
31. y
32. y
x
x 1
2
2 x
2 x 1
dy
( x 1)
dx 2 x x1
2
1
2 x
( x 1)3
( x 1)2
1/2
x 2 x 1 1 1/2 dy 1 1 1
x
dx
2
x
x2
33. y
( x 1)3
(2 x 1) 1 2 x 1 4 x 1
x
x
x
4
2 x 1
(2 x 1) 2
(2 x 1)3 (2 x 1)3
dy
dx 2 2 x
x (1) ( x1)2 x 1 x
1
x2
1
2 x 2 1 1x
34. y 4 x x x 4 x( x x1/2 )1/2 dx 4 x 12 ( x x1/2 ) 1/2 1 12 x 1/2 ( x x1/2 )1/2 (4)
dy
( x x ) 1 2 2 x 1 1 4( x x ) ( x x )1 2 (2 x x 4 x 4 x ) 6 x 5 x
2 x
x x
35. r
)( sin )
sin cos cos sin
cossin1 ddr 2 cossin1 (cos 1)(cos(cos )(sin
2 cos 1
1)
(cos 1)
2
(2sin )(1 cos )
(cos 1)3
1
36. r 1sin
cos
2
2
2
2
2sin
(cos 1)2
1)
(cos cos2 sin 2 sin )
ddr 2 1sincos 1 (1 cos )(cos(1cos) (sin) 1)(sin ) (12(sin
cos )
2
3
2
2(sin 1)(cos sin 1)
(1cos )3
dy
37. y (2 x 1) 2 x 1 (2 x 1)3/2 dx 32 (2 x 1)1/2 (2) 3 2 x 1
1 (3 x 4) 19/20 (3)
38. y 20(3x 4)1/4 (3 x 4)1/5 20(3 x 4)1/20 dx 20 20
dy
39. y 3(5 x 2 sin 2 x)3/2 dx 3 32 (5 x 2 sin 2 x)5/2 [10 x (cos 2 x)(2)]
dy
3
(3 x 4)19/ 20
9(5 x cos 2 x )
5 x2 sin 2 x
5/ 2
2
40. y (3 cos3 3 x)1/3 dx 13 (3 cos3 3 x) 4/3 (3cos 2 3x)( sin 3 x)(3) 3cos 33x sin 34/3x
dy
(3 cos 3 x )
y2
41. xy 2 x 3 y 1 ( xy y ) 2 3 y 0 xy 3 y 2 y y ( x 3) 2 y y x 3
dy
dy
dy
dy
dy
42. x 2 xy y 2 5 x 2 2 x x dx y 2 y dx 5 0 x dx 2 y dx 5 2 x y dx ( x 2 y ) 5 2 x y
dy
dx
5 2 x y
x2 y
dy
dy
dy
dy
43. x3 4 xy 3 y 4/3 2 x 3 x 2 4 x dx 4 y 4 y1/3 dx 2 4 x dx 4 y1/3 dx 2 3x 2 4 y
dy
dy
dx (4 x 4 y1/3 ) 2 3x 2 4 y dx
2 3 x 2 4 y
4 x 4 y1/3
44. 5 x 4/5 10 y 6/5 15 4 x 1/5 12 y1/5 dx 0 12 y1/5 dx 4 x 1/5 dx 13 x 1/5 y 1/5
dy
dy
dy
Copyright 2016 Pearson Education, Ltd.
1
3( xy )1/5
165
166
Chapter 3 Derivatives
45. ( xy )1/2 1 12 ( xy ) 1/2 x dx y 0 x1/2 y 1/2 dx x 1/2 y1/2 dx x 1 y dx x
dy
dy
dy
dy
( x 1)(1) ( x )(1)
dy
dy
dy
dy
y
y
46. x 2 y 2 1 x 2 2 y dx y 2 (2 x) 0 2 x 2 y dx 2 xy 2 dx x
47. y 2 xx1 2 y dx
1/2
48. y 2 11 xx
( x 1) 2
dy
dx
(1 x )(1) (1 x )( 1)
dy
y 4 11 xx 4 y 3 dx
dy
dx
(1 x )2
dp
1
2 y ( x 1) 2
dp
1
2 y 3 (1 x ) 2
dp
dp
dp
49. p3 4 pq 3q 2 2 3 p 2 dq 4 p q dq 6q 0 3 p 2 dq 4q dp 6q 4 p dq (3 p 2 4q ) 6q 4 p
dp
dq
6q 4 p
3 p 2 4q
50. q (5 p 2 2 p )3/2 1 32 (5 p 2 2 p )5/2 10 p dq 2 dq 23 (5 p 2 2 p)5/2 dq (10 p 2)
2
5/ 2
dp
dp
dp
5p 2p
dp
dq 3(5 p 1)
51. r cos 2 s sin 2 s r ( sin 2 s )(2) (cos 2 s ) dr
(cos 2 s ) 2r sin 2 s 2sin s cos s
2 sin s cos s 0 dr
ds
ds
2 r sin 2 s sin 2 s
(2 r 1)(sin 2 s )
dr
(2r 1)(tan 2s )
ds
cos 2 s
cos 2 s
52. 2rs r s s 2 3 2 r s dr
dr
1 2s 0 dr
(2 s 1) 1 2 s 2r dr
ds
ds
ds
ds
dy
2
dy
53. (a) x3 y 3 1 3 x 2 3 y 2 dx 0 dx x 2
(b)
d2y
2
2 xy (2 yx
2
) x 2
y
y
2
2
2 xy
2 x4
y
dy
y 2 ( 2 x ) ( x 2 ) 2 y dx
dx
4
2
y
2 xy 3 2 x 4
y
y5
y
dy
dy
dy
dy
d2y
y 2 1 2x 2 y dx 22 dx 12 dx ( yx 2 )1 2 ( yx 2 )2 y (2 x) x 2 dx
x
yx
dx
2 1
2 xy x 2
d2y
2 xy 2 1
yx
dx
2
dx 2
4
y2 x4
4
d2y
1 2 s 2r
2 s 1
y3 x4
dy
dy
dy
54. (a) x 2 y 2 1 2 x 2 y dx 0 2 y dx 2 x dx xy
y x 1 (since y 2 x2 1)
x
yx y
2
dy
x d y y (1) x dx
dx
y
dx 2
y2
y2
dy
(b)
2
2
y3
y3
55. (a) Let h( x) 6 f ( x) g ( x) h( x) 6 f ( x) g ( x) h(1) 6 f (1) g (1) 6 12 (4) 7
(b) Let h( x) f ( x) g 2 ( x) h( x) f ( x) (2 g ( x)) g ( x) g 2 ( x) f ( x) h(0) 2 f (0) g (0) g (0) g 2 (0) f (0)
2(1)(1) 12 (1) 2 (3) 2
f ( x)
(c) Let h( x) g ( x ) 1 h( x)
( g ( x ) 1) f ( x ) f ( x ) g ( x )
( g ( x ) 1) 2
h(1)
( g (1) 1) f (1) f (1) g (1)
( g (1) 1) 2
(51) 12 3( 4)
(51) 2
1 1
2
4
(d) Let h( x) f ( g ( x)) h( x) f ( g ( x)) g ( x) h(0) f ( g (0)) g (0) f (1) 12 12
(e) Let h( x) g ( f ( x)) h( x) g ( f ( x)) f ( x) h(0) g ( f (0)) f (0) g (1) f (0) ( 4)(3) 12
Copyright 2016 Pearson Education, Ltd.
5
12
Chapter 3 Practice Exercises
(f ) Let h( x) ( x f ( x))3/2 h( x) 32 ( x f ( x))1/2 (1 f ( x)) h(1) 32 (1 f (1))1/2 (1 f (1))
32 (1 3)1/2 1 12 92
(g) Let h( x) f ( x g ( x)) h( x) f ( x g ( x))(1 g ( x)) h(0) f ( g (0))(1 g (0))
f (1) 1 12 12 23 43
56. (a) Let h( x) x f ( x) h( x) x f ( x) f ( x)
1 h(1)
2 x
13
1 f (1) f (1) 1 15 ( 3) 12 10
2 1
Let h( x) ( f ( x)) h( x) 12 ( f ( x))1/2 ( f ( x)) h(0) 12 ( f (0))1/2 f (0) 12 (9)1/2 (2) 13
1
Let h( x) f x h( x) f x 1 h(1) f 1 1 15 12 10
2 x
2 1
2
2
1/2
(b)
(c)
(d) Let h( x) f (1 5 tan x) h ( x) f (1 5 tan x)(5sec x) h (0) f (1 5 tan 0)(5sec 0)
f (1)(5) 15 (5) 1
f ( x)
(e) Let h( x) 2 cos x h( x)
(2 cos x ) f ( x ) f ( x )( sin x )
(2 cos x ) 2
h(0)
(21) f (0) f (0)(0)
(2 1) 2
3( 2)
23
9
2
h(1) 10 sin 2 (2 f (1) f (1)) f 2 (1) 10 cos 2 2 20( 3) 15 0 12
(f ) Let h( x) 10sin 2x f 2 ( x) h( x) 10sin 2x (2 f ( x) f ( x)) f 2 ( x) 10 cos 2x
dy
2t ; y 3sin 2 x dx 3(cos 2 x )(2) 6 cos 2 x 6 cos(2t 2 2 ) 6 cos (2t 2 );
57. x t 2 dx
dt
dy
dy
dy
thus, dt dx dx
6 cos (2t 2 ) 2t dt
dt
t 0
6 cos(0) 0 0
dt 1 (u 2 2u ) 2/3 (2u 2) 2 (u 2 2u ) 2/3 (u 1); s t 2 5t ds 2t 5
58. t (u 2 2u )1/3 du
3
3
dt
ds ds dt [2(u 2 2u )1/3 5] 2 (u 2 2u ) 2/3 (u 1)
2(u 2 2u )1/3 5; thus du
dt du
3
(2 2(2))
ds
du
[2(22 2(2))1/3 5] 23
u 2
2
2/3
(2 1) 2(2 81/3 5)(82/3 ) 2(2 2 5) 14 92
dw dw dr
ds
dr ds
cos
8cos s dw
6
ds
2 8sin s 6
s 0
8sin s 6 2
60. 2 t 1 2 t 2 ddt
cos 8sin s 6 2
6
6
2
3
2
6
6
8sin
2 4
ddt 0 ddt (2 t 1) 2 ddt 2t 1 ; r ( 2 7)1/3
2
ddr 13 ( 2 7)2/3 (2 ) 23 ( 2 7) 2/3 ; now t 0 and 2t 1 1 so that ddt
and ddr
1
23 (1 7) 2/3 16 dr
dt
dy
t0
ddr
dy
(1) 16
dy
d2y
dx 2
(3 y
dy
1)( 2 cos x ) ( 2sin x ) 6 y dx
2
2
(3 y 1)
dy
dy
3 y 1
d y
2
2
dx (0, 1)
(31)( 2 cos 0)( 2sin 0)(60)
d2y
dx
2
x
2/3
dy
23 y 1/3 dx
y
x
2/3 2
2/3
2 x 1/3
3
11 1
dy
y 2/3
d y
2
dx
2
x
dy
8 8
2/3
(8, 8)
dy
(8, 8)
2
3
1; dx
1/3
(0, 1)
2 sin(0)
0;
31
12
(31)2
62. x1/3 y1/3 4 13 x 2/3 13 y 2/3 dx 0 dx 2/3 dx
dy
t 0, 1
dr
16
t 0 d t 0
x
61. y 3 y 2 cos x 3 y 2 dx dx 2sin x dx (3 y 2 1) 2sin x dx 2sin
2
dx
2
r 2 2 1 r 2 8sin s ; thus,
cos 8sin 2 8cos
(cos 0)(8)
3
59. r 8sin s 6 dr
8cos s 6 ; w sin ( r 2) dw
cos
dr
ds
y 2/3
x 2/3
8 13 13 32 1
( 1) 82/3
84/3
Copyright 2016 Pearson Education, Ltd.
2
3
1/3
82/3
4
6
167
168
Chapter 3 Derivatives
1
1
f (t h ) f ( t )
2t 1(2t 2 h 1)
2( t h ) 1 2 t 1
h
(2t 2h 1)(2t 1) h (2t 2h21)(2
h
h
t 1) h
f ( t h ) f (t )
2
2
lim
(2t 2h21)(2t 1) f (t ) lim
2
h
h 0
h 0 (2t 2 h 1)(2t 1) (2t 1)
63. f (t ) 2t11 and f (t h) 2(t 1h) 1
64. g ( x) 2 x 2 1 and g ( x h) 2( x h)2 1 2 x 2 4 xh 2h 2 1
g ( x h ) g ( x ) (2 x 2 4 xh 2 h 2 1) (2 x 2 1)
h
h
4 xh 2 h 2
g ( x h) g ( x)
4 x 2h g ( x) lim
lim (4 x 2h) 4 x
h
h
h 0
h 0
65. (a)
(b)
lim f ( x) lim x 2 0 and lim f ( x) lim x 2 0 lim f ( x) 0. Since lim f ( x) 0 f (0) it
x 0
x 0
x 0
x 0
x 0
x 0
x 0
x 0
follows that f is continuous at x 0.
(c) lim f ( x ) lim (2 x) 0 and lim f ( x) lim (2 x) 0 lim f ( x) 0. Since this limit exists, it
x 0
x 0
follows that f is differentiable at x 0.
x 0
66. (a)
(b)
lim f ( x) lim x 0 and lim f ( x ) lim tan x 0 lim f ( x) 0. Since lim f ( x) 0 f (0), it
x 0
x 0
x 0
x 0
x 0
x 0
x 0
x 0
x 0
x 0
x 0
follows that f is continuous at x 0.
(c) lim f ( x) lim 1 1 and lim f ( x) lim sec 2 x 1 lim f ( x) 1. Since this limit exists it follows
that f is differentiable at x 0.
67. (a)
(b)
lim f ( x) lim x 1 and lim f ( x ) lim (2 x) 1 lim f ( x) 1. Since lim f ( x ) 1 f (1), it
x 1
x 1
x 1
x 1
x 1
x 1
x 1
x 1
x 1
x 1
x 1
follows that f is continuous at x 1.
(c) lim f ( x ) lim 1 1 and lim f ( x) lim 1 1 lim f ( x ) lim f ( x ), so lim f ( x) does not
exist f is not differentiable at x 1.
x 1
Copyright 2016 Pearson Education, Ltd.
x 1
Chapter 3 Practice Exercises
68. (a)
169
lim f ( x ) lim sin 2 x 0 and lim f ( x) lim mx 0 lim f ( x) 0, independent of m; since
x 0
x 0
x 0
x 0
x 0
f (0) 0 lim f ( x ) it follows that f is continuous at x 0 for all values of m.
(b)
x 0
lim f ( x) lim (sin 2 x) lim 2 cos 2 x 2 and lim f ( x) lim (mx) lim m m f is
x 0
x 0
x 0
x 0
x 0
x 0
differentiable at x 0 provided that lim f ( x) lim f ( x) m 2.
x 0
x 0
69. y 2x 2 x1 4 12 x (2 x 4) 1 dx 12 2(2 x 4)2 ; the slope of the tangent is 32 32 12 2(2 x 4) 2
dy
2 2(2 x 4) 2 1
1
(2 x 4)2 1 4 x 2 16 x 16 1 4 x 2 16 x 15 0
(2 x 4) 2
(2 x 5)(2 x 3) 0 x 52 or x 32 52 , 95 and 32 , 14 are points on the curve where the slope is 32 .
70.
y x
1 dy
1
1
;
1 2 . The derivative is equal to 2 when x
, so the points where the slope is 2 are
2 x dx
2
2x
1
1
, 0 and
, 0 .
2
2
dy
dy
71. y 2 x3 3 x 2 12 x 20 dx 6 x 2 6 x 12; the tangent is parallel to the x-axis when dx 0
6 x 2 6 x 12 0 x 2 x 2 0 ( x 2)( x 1) 0 x 2 or x 1 (2, 0) and (1, 27) are points
on the curve where the tangent is parallel to the x-axis.
dy
dy
72. y x3 dx 3 x 2 dx
( 2, 8)
12; an equation of the tangent line at (2, 8) is y 8 12( x 2)
y 12 x 16; x-intercept: 0 12 x 16 x 43 43 , 0 ; y -intercept : y 12(0) 16 16 (0, 16)
dy
73. y 2 x3 3 x 2 12 x 20 dx 6 x 2 6 x 12
x when dy 1 24; 6 x 2 6 x 12 24
(a) The tangent is perpendicular to the line y 1 24
1
dx
24
2
2
x x 2 4 x x 6 0 ( x 3)( x 2) 0 x 2 or x 3 (2, 16) and (3, 11) are points
x .
where the tangent is perpendicular to y 1 24
dy
(b) The tangent is parallel to the line y 2 12 x when dx 12 6 x 2 6 x 12 12 x 2 x 0
x( x 1) 0 x 0 or x 1 (0, 20) and (1, 7) are points where the tangent is parallel to y 2 12 x.
dy
x
74. y sin
x
dx
x ( cos x ) ( sin x )(1)
x
2
2
dy
m1 dx
x
the tangents intersect at right angles.
dy
75. y tan x, 2 x 2 dx sec 2 x; now the slope
of y 2x is 12 the normal line is parallel to
dy
y 2x when dx 2. Thus, sec2 x 2
1 2
cos 2 x
cos 2 x 12 cos x 1 x 4 and x 4
2
dy
2 1 and m2 dx
for 2 x 2 4 , 1 and 4 , 1 are points
where the normal is parallel to y 2x .
Copyright 2016 Pearson Education, Ltd.
2 1. Since m 1
1
2
m2
x
170
Chapter 3 Derivatives
dy
dy
76. y 1 cos x dx sin x dx
2 , 1
1
the tangent at 2 , 1 is the line y 1 x 2
2
y x 1; the normal at , 1 is
2
y 1 (1) x y x 1
2
2
dy
dy
77. y x 2 C dx 2 x and y x dx 1; the parabola is tangent to y x when 2 x 1 x 12 y 12 ; thus,
1 1 2 C C 1
2
2
4
dy
dy
78. y x3 dx 3 x 2 dx
3a 2 the tangent line at (a, a3 ) is y a3 3a 2 ( x a ). The tangent line
xa
3
3
intersects y x3 when x a 3a 2 ( x a ) ( x a ) ( x 2 xa a 2 ) 3a 2 ( x a ) ( x a )( x 2 xa 2a 2 ) 0
dy
( x a ) 2 ( x 2a ) 0 x a or x 2a. Now dx
x 2 a
3( 2a ) 2 12a 2 4 (3a 2 ), so the slope at x 2a
is 4 times as large as the slope at (a, a3 ) where x a.
3 ( 2)
1 the line through (0, 3) and (5, 2) is
0 5
dy
dy
y x 3; y xc1 dx c 2 , so the curve is tangent to y x 3 dx 1 c 2
( x 1)
( x 1)
2
c
c
( x 1) c, x 1. Moreover, y x 1 intersects y x 3 x 1 x 3, x 1
2
79. The line through (0, 3) and (5, 2) has slope m
c ( x 1)( x 3), x 1. Thus c c ( x 1) ( x 1)( x 3) ( x 1)[ x 1 ( x 3)] 0,
x 1 ( x 1)(2 x 2) 0 x 1 (since x 1 ) c 4.
80. Let b, a 2 b 2 be a point on the circle x 2 y 2 a 2 . Then x 2 y 2 a 2 2 x 2 y dx 0 dx xy
2
2
dy
b
dx
normal line through b, a 2 b 2 has slope ab b normal line is
2
2
x a
a b
2
2
2
2
a 2 b2
2
y a b 2 ab b ( x b) y
a 2 b2
x
a 2 b 2 y a bb x which passes
b
through the origin.
dy
dy
dy
dy
81. x 2 2 y 2 9 2 x 4 y dx 0 dx 2xy dx
(1, 2)
dy
14 the tangent line is y 2 14 ( x 1) 14 x 94
and the normal line is y 2 4( x 1) 4 x 2.
82.
x3 y 2 2
d 3
dy
dy
3x 2
3
dy
x y 2 0 3x 2 2 y 0
, so the
At the point (1, 1) ,
dx
dx
dx
2y .
dx
2
3
3
5
tangent line has the equation ( y 1) ( x 1) or y x and the normal line has the equation
2
2
2
2
2
1
( y 1) ( x 1) or y x .
3
3
3
dy y 2
dy
dx
2 the tangent
dx x 5
(3, 2)
line is y 2 2( x 3) 2 x 4 and the normal line is y 2 21 ( x 3) 12 x 72 .
dy
dy
dy
83. xy 2 x 5 y 2 x dx y 2 5 dx 0 dx ( x 5) y 2
Copyright 2016 Pearson Education, Ltd.
Chapter 3 Practice Exercises
dy
dy
dy
84. ( y x) 2 2 x 4 2( y x) dx 1 2 ( y x) dx 1 ( y x) dx
171
1 y x
dy
dx
34 the tangent
yx
(6, 2)
line is y 2 34 ( x 6) 34 x 52 and the normal line is y 2 43 ( x 6) 43 x 10.
2 xy y
dy
dy
dy
dy
1
x dx y 0 x dx y 2 xy dx
dx
45 the tangent line
x
2 xy
(4, 1)
is y 1 54 ( x 4) 54 x 6 and the normal line is y 1 54 ( x 4) 54 x 11
.
5
85. x xy 6 1
dy
dy
1/ 2
dy
86. x3/2 2 y 3/2 17 32 x1/2 3 y1/2. dx 0 dx x1/ 2 dx
2y
(1, 4)
14 the tangent line is
y 4 14 ( x 1) 14 x 17
and the normal line is y 4 4( x 1) 4 x.
4
dy
dy
dy
dy
dy dy
87. x3 y 3 y 2 x y x3 3 y 2 dx y 3 (3x 2 ) 2 y dx 1 dx 3 x3 y 2 dx 2 y dx dx 1 3 x 2 y 3
1 3 x2 y3
dy
dy
dy
dy
dx (3 x3 y 2 2 y 1) 1 3 x 2 y 3 dx 3 2
is undefined. Therefore, the
dx
24 , but dx
3 x y 2 y 1
(1, 1)
1
curve has slope 2 at (1, 1) but the slope is undefined at (1, 1).
(1, 1)
dy
88. y sin( x sin x) dx [cos( x sin x)](1 cos x ); y 0 sin( x sin x) 0 x sin x k , k 2, 1, 0, 1, 2
dy
(for our interval) cos( x sin x) cos(k ) 1. Therefore, dx 0 and y 0 when 1 cos x 0 and x k .
For 2 x 2 , these equations hold when k 2, 0, and 2(since cos( ) cos 1.) Thus the curve has
horizontal tangents at the x-axis for the x-values 2 , 0, and 2 (which are even integer multiples of ) the
curve has an infinite number of horizontal tangents.
89. B graph of f , A graph of f . Curve B cannot be the derivative of A because A has only negative slopes
while some of B’s values are positive.
90. A graph of f , B graph of f . Curve A cannot be the derivative of B because B has only negative slopes
while A has positive values for x 0.
91.
92.
93. (a) 0, 0
(b) largest 1700, smallest about 1400
94. rabbits/day and foxes/day
95. lim sin2 x lim sinx x (2 x11) (1) 11 1
x 0 2 x x x 0
x 0
sin r lim
97. lim tan
2r
r0
lim cos17 x sin7 x7 x 1 32 1 1 72 2
3x sin 7 x 32 x
x0 2 x 2 x cos 7 x
0
7 x lim
96. lim 3 x 2tan
x
r0
2
7
lim cos 2r 12 (1) 11 12
sinr r tan2r2r 12 12 (1) r
0
sin 2 r
2r
Copyright 2016 Pearson Education, Ltd.
172
Chapter 3 Derivatives
sin(sin )
lim
sin (sin )
. Let x sin . Then x 0 as 0
0 sin
sin(sin ) sin
sin
0
0
sin(sin )
sin
x
lim sin lim x 1
0
x 0
98. lim
99.
lim
4 tan 2 tan 1
tan 2 5
lim
2
4 1
tan
lim
2
1
tan 2
1 5
tan 2
(4 0 0)
(1 0) 4
1 2
2
2
(0 2)
1
2
cot
100. lim
lim cot (500) 52
2
5cot 7 cot 8
7
8
0
0 5 cot 2
cot
x lim
x sin x lim
101. lim 2x2sin
cos x
2(1 cos x )
x 0
102. lim
0
x 0
1 cos
2
lim
0
xx
x
x
lim 22 2 x sinx x lim 2 x 2 x sinx x (1)(1)(1) 1
sin
sin
sin
2
x0 2
x0 2
x sin x
2
x0 2 2sin 2x
lim sin 2 sin 2 1 (1)(1) 1 1
2 2
2
0 2
2 2
2sin 2 2
1 sin x 1; let tan x 0 as x 0 lim g ( x ) lim tan(tan x ) lim tan 1.
cos
x
x
x0
x 0
x0 tan x
0
103. lim tanx x lim
x0
Therefore, to make g continuous at the origin, define g (0) 1.
tan(tan x )
tan(tan x )
sin x 1 1 lim sin x (using the result of # 103); let
104. lim f ( x) lim sin(sin x ) lim tan x sin(sin
x ) cos x
x 0
x0
x0
x0 sin(sin x )
sin
x
sin x 0 as x 0 lim sin(sin x ) lim sin 1. Therefore, to make f continuous at the origin,
define f (0) 1.
0
x0
105. (a) S 2 r 2 2 rh and h constant dS
4 r dr
2 h dr
(4 r 2 h) dr
dt
dt
dt
dt
(b) S 2 r 2 2 rh and r constant dS
2 r dh
dt
dt
(c) S 2 r 2 2 rh dS
4 r dr
2 r dh
h dr
(4 r 2 h) dr
2 r dh
dt
dt
dt
dt
dt
dt
(d) S constant dS
0 0 (4 r 2 h) dr
2 r dh
(2r h) dr
r dh
dr
2 rr h dh
dt
dt
dt
dt
dt
dt
dt
106. S r r 2 h 2 dS
r
dt
(a)
r drdt h dhdt r 2 h2 dr ;
r 2 h2
2 dr
dS r dt
h constant dh
0
dt
dt
r 2 h2
(b) r constant dr
0 dS
dt
dt
rh
dh
dt
2
r 2 h 2 dr
r 2 h 2 2r 2 dr
dt
r h dt
r 2 h 2 dt
2
(c) In general, dS
r 2 h 2 2r 2 dr
2rh 2 dh
dt
dt
r h dt
r
h
107. A r 2 dA
2 r dr
; so r 10 and dr
2 m /sec dA
(2 )(10) 2 40 m 2 /s
dt
dt
dt
dt
108. V s3 dV
3s 2 ds
ds
12 dV
; so s 20 and dV
1200 cm3 /min ds
dt
dt
dt
dt
dt
dt
3s
Copyright 2016 Pearson Education, Ltd.
1
(1200) 1 cm/min
30(20)2
Chapter 3 Practice Exercises
dR
dR
dR
173
dR
109. dt1 1 ohm/sec, dt2 0.5 ohm/s; and R1 R1 R1 12 dR
12 dt1 12 dt2 Also, R1 75 ohms
dt
1
2
R
R1
R2
1 1 R 30 ohms. Therefore, from the derivative equation,
and R2 50 ohms R1 75
50
1 dR 1 ( 1) 1 (0.5)
1 1
5625 5000
(30) 2 dt
(75)2
(50)2
5625
9(625)
1 0.02 ohm/s.
50(5625) 50
dRdt (900) 5000
56255000
110. dR
3 ohms/s and dX
2 ohms/s; Z R 2 X 2 dZ
dt
dt
dt
X 20 ohms dZ
dt
(10)(3) (20)( 2)
R dR
X dX
dt
dt
R2 X 2
so that R 10 ohms and
1 0.45 ohm/s.
5
102 202
dy
111. Given dx
10 m/s and dt 5 m/s, let D be the distance from the origin D 2 x 2 y 2
dt
dy
dy
2 D dD
2 x dx
2 y dt D dD
x dx
y dt . When ( x, y ) (3, 4), D 32 (4) 2 5 and
dt
dt
dt
dt
5 dD
(3)(10) (4)(5) dD
10
2. Therefore, the particle is moving away from the origin at 2 m/s
dt
dt
5
(because the distance D is increasing).
112. Let D be the distance from the origin. We are given that dD
11 units/s. Then D 2 x 2 y 2 x 2 ( x3/2 ) 2
dt
x 2 x3 2 D dD
2 x dx
3x 2 dx
x(2 3 x) dx
; x 3 D 32 33 6 and substitution in the derivative
dt
dt
dt
dt
equation gives (2)(6)(11) (3)(2 9) dx
dx
4 units/s.
dt
dt
113. (a) From the diagram we have h3 1.2
r 1.2
h r 0.4 h.
r
3
(b) V 13 r 2 h 13
0.45 h h 0.163 h dVdt 0.16 h2 dhdt , so dVdt 0.2 and h 2 dhdt 265 m/min.
2
3
114. From the sketch in the text, s r ds
r ddt dr
. Also r 0.4 is constant dr
0 ds
dt
dt
dt
dt
r ddt (0.4) ddt . Therefore, ds
2 m/s and r 0.4 m ddt 5 rad/s
dt
115. (a) From the sketch in the text, ddt 0.6 rad/s and x tan . Also x tan dx
sec2 ddt ; at point A,
dt
(sec 2 0)(0.6) 0.6. Therefore the speed of the light is 0.6 53 km/s when it
x 0 0 dx
dt
reaches point A.
(b)
(3/5) rad 1 rev 60s
2 rad min 18
revs/min
s
b a
116. From the figure, ar BC
r
b
b2 r 2
. We are given
that r is constant. Differentiation gives,
b r (b)
2
1 da
r dt
2
db
dt
2
b r
2
b
db
dt
b2 r 2
. Then, b 2r and
2 r ( 0.3r )
2
2
(2r ) r ( 0.3r ) (2 r ) (2 r )2 r 2
db 0.3r da r
dt
dt
(2 r ) 2 r 2
3r 2 ( 0.3r )
3r
4 r 2 (0.3 r )
3r 2
(3r 2 )( 0.3r ) (4 r 2 )(0.3r )
3 3r 2
0.3r r
3 3
10 3
m/s. Since da
is positive, the distance OA is
dt
increasing when OB 2r , and B is moving toward O at the rate of 0.3r m/s.
Copyright 2016 Pearson Education, Ltd.
174
Chapter 3 Derivatives
117. (a) If f ( x ) tan x and x 4 , then f ( x) sec 2 x,
f 4 1 and f 4 2. The linearization of
f ( x ) is L( x) 2 x 4 ( 1) 2 x 2 2 .
(b) if f ( x) sec x and x 4 , then f ( x ) sec x tan x,
f 4 2 and f 4 2. The linearization of
f ( x) is L( x) 2 x 4 2 2x
1
118. f ( x) 1 tan
f ( x)
x
2(4 )
.
4
sec2 x . The linearization at x 0 is L ( x) f (0)( x 0) f (0) 1 x.
(1 tan x ) 2
119. f ( x) x 1 sin x 0.5 ( x 1)1/2 sin x 0.5 f ( x) 12 ( x 1) 1/2 cos x
L( x) f (0)( x 0) f (0) 1.5( x 0) 0.5 L( x) 1.5 x 0.5 , the linearization of f ( x ) .
2
1 x 3.1 2(1 x) 1 ( x 1)1/2 3.1 f ( x) 2(1 x)2 (1) 12 ( x 1)1/2
1 x
2
1
L( x) f (0)( x 0) f (0) 2.5 x 0.1 , the linearization of f ( x) .
2 2 1 x
(1 x)
120. f ( x)
121. S r r 2 h 2 , r constant dS r 12 (r 2 h 2 ) 1/2 2h dh
h0 to h0 dh dS
r h0 ( dh )
rh
r 2 h2
dh. Height changes from
r 2 h02
2
r | dr | r . The measurement of the
122. (a) S 6r 2 dS 12r dr. We want | dS | (2%) S |12r dr | 12
100
100
edge r must have an error less than 1%.
2
(b) When V r 3 , then dV 3r 2 dr. The accuracy of the volume is dV
(100%) 3r 3dr (100%)
V
r (100%) 3%
3r (dr )(100%) 3r 100
3
2
r
123. C 2 r r 2C , S 4 r 2 C , and V 43 r 3 C 2 . It also follows that dr 21 dC , dS 2C
dC and
2
6
dV C 2 dC. Recall that C 10 cm and dC 0.4 cm.
2
(a) dr 0.4
0.2
cm
2
drr (100%) 0.2 210 (100%) (.04)(100%) 4%
Copyright 2016 Pearson Education, Ltd.
Chapter 3 Additional and Advanced Exercises
175
dSS (100%) 8 100 (100%) 8%
6 (100%) 12%
(c) dV 10 (0.4) 20 cm dV
(100%) 20 1000
V
2
(b) dS 20
(0.4) 8 cm
2
2
2
2
2
4.5 h 4.2 m. The same triangles imply that 6 a a h 10.8a 1 1.8
124. Similar triangles yield 10.5
1.8
h
h
1.8
1
dh 10.8a 2 da 10.8
da 10.8
100
2
2
a
CHAPTER 3
a
0.53 centimeters.
10.8
1001 0.533
100
4.5
2
ADDITIONAL AND ADVANCED EXERCISES
1. (a) sin 2 2sin cos dd (sin 2 ) dd (2sin cos ) 2 cos 2 2[(sin )( sin ) (cos )(cos )]
cos 2 cos 2 sin 2
(b) cos 2 cos 2 sin 2 dd (cos 2 ) dd (cos 2 sin 2 )
2sin 2 (2 cos )( sin ) (2sin )(cos )
sin 2 cos sin sin cos sin 2 2sin cos
2. The derivative of sin ( x a ) sin x cos a cos x sin a with respect to x is cos( x a ) cos x cos a sin x sin a,
which is also an identity. This principle does not apply to the equation x 2 2 x 8 0, since x 2 2 x 8 0 is
not an identity: it holds for 2 values of x ( 2 and 4), but not for all x.
3. (a) f ( x) cos x f ( x) sin x f ( x) cos x, and g ( x) a bx cx 2 g ( x) b 2cx g ( x) 2c;
also, f (0) g (0) cos(0) a a 1; f (0) g (0) sin(0) b b 0; f (0) g (0)
cos(0) 2c c 12 . Therefore, g ( x) 1 12 x 2 .
(b) f ( x ) sin( x a) f ( x) cos( x a ), and g ( x) b sin x c cos x g ( x) b cos x c sin x; also
f (0) g (0) sin(a ) b sin(0) c cos(0) c sin a; f (0) g (0) cos( a ) b cos(0) c sin(0)
b cos a. Therefore, g ( x ) sin x cos a cos x sin a.
(c) When f ( x ) cos x, f ( x) sin x and f (4) ( x) cos x; when g ( x) 1 12 x 2 , g ( x) 0 and g (4) ( x) 0.
Thus f (0) 0 g (0) so the third derivatives agree at x 0 . However, the fourth derivatives do not
agree since f (4) (0) 1 but g (4) (0) 0. In case (b), when f ( x) sin( x a ) and
g ( x) sin x cos a cos x sin a , notice that f ( x) g ( x) for all x, not just x 0. Since this is an identity, we
have f ( n) ( x) g ( n ) ( x) for any x and any positive integer n.
4. (a) y sin x y cos x y sin x y y sin x sin x 0; y cos x y sin x
y cos x y y cos x cos x 0; y a cos x b sin x y a sin x b cos x
y a cos x b sin x y y ( a cos x b sin x) ( a cos x b sin x) 0
(b) y sin(2 x) y 2 cos(2 x) y 4 sin(2 x) y 4 y 4 sin(2 x ) 4 sin(2 x) 0. Similarly,
y cos(2 x) and y a cos(2 x) b sin(2 x) satisfy the differential equation y 4 y 0. In general,
y cos( mx), y sin( mx) and y a cos (mx ) b sin (mx ) satisfy the differential equation y m 2 y 0.
5. If the circle ( x h) 2 ( y k ) 2 a 2 and y x 2 1 are tangent at (1, 2), then the slope of this tangent is
m 2 x (1, 2) 2 and the tangent line is y 2 x. The line containing (h, k) and (1, 2) is perpendicular to
y 2 x kh12 12 h 5 2k the location of the center is (5 2k , k ). Also, ( x h)2 ( y k )2 a 2
1 ( y)2
x h ( y k ) y 0 1 ( y ) 2 ( y k ) y 0 y k y . At the point (1, 2) we know y 2 from the
Copyright 2016 Pearson Education, Ltd.
176
Chapter 3 Derivatives
tangent line and that y 2 from the parabola. Since the second derivatives are equal at (1, 2) we obtain
1 (2) 2
2 k 2 k 92 . Then h 5 2k 4 the circle is ( x 4)2 y 92
we have that a 5 25 .
a2 . Since (1, 2) lies on the circle
2
, where 0 x 60.
2
dr 3 x
The marginal revenue is dx
40 2 x 3 40x 401 dxdr 3 40x 3 40x 240x 3 3 40x 1 40x .
x
6. The total revenue is the number of people times the price of the fare: r ( x) xp x 3 40
2
dr 0 x 40 (since x 120 does not belong to the domain). When 40 people are on the bus the
Then dx
x
marginal revenue is zero and the fare is p (40) 3 40
( x40) $4.00.
2
dy
(0.04u )v u (0.05v) 0.09uv 0.09 y the rate of growth of the total
7. (a) y uv dt du
v u dv
dt
dt
production is 9% per year.
dy
(b) If du
0.02u and dv
0.03v, then dt (0.02u )v (0.03v)u 0.01uv 0.01 y, increasing at 1% per year.
dt
dt
8. When x 2 y 2 25, then y xy . The tangent line
to the balloon at (4, 3) is y 3 43 ( x 4)
y 43 x 25
. The top of the gondola is 5 + 2.5 =
3
7.5 m below the center of the balloon. The
intersection of y 7.5 and y 43 x 25
is at the
3
far right edge of the gondola 7.5 43 x 25
3
x 85 . Thus the gondola is 2 x 1.25 m wide.
9. Answers will vary. Here is one possibility.
(a) s(0) 10 cos 4 10
2
2
10. s (t ) 10 cos t 4 v(t ) ds
10sin t 4 a(t ) dv
d 2s 10 cos t 4
dt
dt
dt
(b) Left : 10, Right:10
(c) Solving 10 cos t 4 10 cos t 4 1 t 34 when the particle is farthest to the left. Solving
10 cos t 4 10 cos t 4 1 t 4 , but t 0 t 2 4 74 when the particle is farthest to
0, v 0, a 34 10, and a 74 10.
(d) Solving 10 cos t 4 0 t 4 v 4 10, v 4 10 and a 4 0.
the right. Thus, v
3
4
7
4
11. (a) s (t ) 19.6t 4.9t 2 v(t ) ds
19.6 9.8t 9.8(2 t ). The maximum height is reached when v(t ) 0
dt
t 2 s. The velocity when it leaves the hand is v(0) 19.6 m/s.
(b) s (t ) 19.6t 0.8t 2 v(t ) ds
19.6 1.6t. The maximum height is reached when
dt
v(t ) 0 t = 12.25s. The maximum height is about s (12.25) 120 m.
Copyright 2016 Pearson Education, Ltd.
Chapter 3 Additional and Advanced Exercises
177
12. s1 3t 3 12t 2 18t 5 and s2 t 3 9t 2 12t v1 9t 2 24t 18 and v2 3t 2 18t 12; v1 v2
9t 2 24t 18 3t 2 18t 12 2t 2 7t 5 0 (t 1)(2t 5) 0 t 1 s and t 2.5 s.
13. m v 2 v02 k x02 x 2 m 2v dv
k 2 x dx
m dv
k 22vx dx
m dv
kx 1v dx
. Then substituting
dt
dt
dt
dt
dt
dt
dx v m dv kx, as claimed.
dt
dt
14. (a) x At 2 Bt C on [t1 , t2 ] v dx
2 At B v
dt
2 A B A(t t ) B is
t1 t2
2
t1 t2
2
1
2
the instantaneous velocity at the midpoint. The average velocity over the time interval is
At22 Bt2 C At12 Bt1 C t2 t1 [ At2 t1 B] A(t t ) B.
vav xt
2 1
t t
t t
2
2
1
1
(b) On the graph of the parabola x At 2 Bt C , the slope of the curve at the midpoint of the interval [t1 , t2 ]
is the same as the average slope of the curve over the interval.
15. (a) To be continuous at x requires that lim sin x lim (mx b) 0 m b m b ;
(b) If y
cos x, x
m, x
x
x
is differentiable at x , then lim cos x m m 1 and b .
x
f ( x ) f (0)
lim
x 0
x 0
x 0
x 0 f (0). f (0) lim
16. f ( x) is continuous at 0 because lim 1cos
x
x
lim 1cos
2
x 0
x
x 0
1 cos x lim sin x 2
1
1 cos x
1 cos x
x 0 x
1 cos x 0
x
x
12 . Therefore f (0) exists with value 12 .
17. (a) For all a, b and for all x 2, f is differentiable at x. Next, f differentiable at x 2 f continuous at
x 2 lim f ( x) f (2) 2a 4a 2b 3 2a 2b 3 0. Also, f differentiable at x 2
x 2
f ( x)
a, x 2
. In order that f (2) exist we must have a 2a (2) b a 4a b 3a b.
2ax b, x 2
Then 2a 2b 3 0 and 3a b a 34 and b 94 .
(b) For x 2, the graph of f is a straight line having a slope of 34 and passing through the origin; for x 2, the
graph of f is a parabola. At x 2, the value of the y -coordinate on the parabola is 32 which matches the
y -coordinate of the point on the straight line at x 2. In addition, the slope of the parabola at the match up
point is 34 which is equal to the slope of the straight line. Therefore, since the graph is differentiable at the
match up point, the graph is smooth there.
18. (a) For any a, b and for any x 1, g is differentiable at x. Next, g differentiable at x 1 g continuous
at x 1 lim g ( x) g (1) a 1 2b a b b 1. Also, g differentiable at x 1
x 1
a , x 1
g ( x)
. In order that g (1) exist we must have a 3a(1)2 1 a 3a 1 a 12 .
3ax 2 1, x 1
For x 1, the graph of g is a straight line having a slope of 12 and a y -intercept of 1. For x 1, the
graph of g is a cubic. At x 1, the value of the y -coordinate on the cubic is 32 which matches the
(b)
y -coordinate of the point on the straight line at x 1. In addition, the slope of the cubic at the match up
point is 12 which is equal to the slope of the straight line. Therefore, since the graph is differentiable at
the match up point, the graph is smooth there.
d ( f ( x)) d ( f ( x)) f ( x)( 1) f ( x ) f ( x ) f ( x ) f is even.
19. f odd f ( x) f ( x) dx
dx
Copyright 2016 Pearson Education, Ltd.
178
Chapter 3 Derivatives
d ( f ( x )) d ( f ( x)) f ( x)( 1) f ( x) f ( x) f ( x ) f is odd.
20. f even f ( x) f ( x) dx
dx
h ( x ) h ( x0 )
f ( x ) g ( x ) f ( x0 ) g ( x0 )
lim
x x0
x x0
x x0
x x0
f ( x ) g ( x ) f ( x ) g ( x0 ) f ( x ) g ( x0 ) f ( x0 ) g ( x0 )
g ( x) g ( x )
f ( x) f ( x )
lim
lim f ( x) x x 0 lim g ( x0 ) x x 0
x
x
0
0
0
x x0
x x0
x x0
g
(
x
)
g
(
x
)
g
(
x
)
g
(
x
)
f ( x0 ) lim x x 0 g ( x0 ) f ( x0 ) 0 lim x x 0 g ( x0 ) f ( x0 ) g ( x0 ) f ( x0 ), if g is
0
0
x x
x x
21. Let h( x) ( fg )( x) f ( x) g ( x) h( x) lim
0
0
continuous at x0 . Therefore ( fg ) ( x) is differentiable at x0 if f ( x0 ) 0, and ( fg )( x0 ) g ( x0 ) f ( x0 ).
22. From Exercise 21 we have that fg is differentiable at 0 if f is differentiable at 0, f (0) 0 and g is continuous at 0.
(a) If f ( x) sin x and g ( x) | x |, then | x | sin x is differentiable because f (0) cos(0) 1, f (0) sin (0) 0
and g ( x) | x | is continuous at x 0.
(b) If f ( x) sin x and g ( x) x 2/3 , then x 2/3 sin x is differentiable because
f (0) cos (0) 1, f (0) sin (0) 0 and g ( x) x 2/3 is continuous at x 0.
(c) If f ( x) 1 cos x and g ( x) 3 x, then 3 x (1 cos x) is differentiable because f (0) sin (0) 0,
f (0) 1 cos (0) 0 and g ( x) x1/3 is continuous at x 0.
(d) If f ( x) x and g ( x) x sin 12 , then x 2 sin 1x is differentiable because f (0) 1, f (0) 0 and
sin t 0 (so g is continuous at x 0 ).
x0 xlim
t
lim x sin 1x lim
x 0
sin
1
x
1
x
23. If f ( x) x and g ( x) x sin 1x , then x 2 sin 1x is differentiable at x 0 because f (0) 1, f (0) 0 and
lim x sin 1x lim
x 0
lim sin t 0 (so g is continuous at x 0 ). In fact, from Exercise 21,
sin 1x
1
x 0
x
t
t
h(0) g (0) f (0) 0. However, for x 0, h( x) x 2 cos 1x 12 2 x sin 1x . But
x
lim h( x) lim cos 1x 2 x sin 1x does not exist because cos 1x has no limit as x 0. Therefore,
x 0
x 0
the derivative is not continuous at x 0 because it has no limit there.
24. From the given conditions we have f ( x h) f ( x) f (h), f (h) 1 hg (h) and lim g (h) 1. Therefore,
h 0
f ( x h) f ( x )
f ( x ) f ( h) f ( x )
f ( h) 1 f ( x) lim g (h) f ( x) 1 f ( x)
lim
lim
f
(
x
)
h
h
h
h 0
h 0
h 0
h0
f ( x) lim
f ( x) f ( x) and f ( x) exists at every value of x.
25. Step 1:
Step 2:
dy
du
du
The formula holds for n 2 (a single product) since y u1u2 dx dx1 u2 u1 dx2
Assume the formula holds for n k :
du
du
dy
y u1u2 uk dx dx1 u2u3 uk u1 dx2 u3
uk ... u1u2
du
uk 1 dxk .
d (u1u2 uk )
du
uk 1 u1u2 uk dxk 1
dx
du
du
du
du
dx1 u2 u3 uk u1 dx2 u3 uk u1u2 uk 1 dxk uk 1 u1u2 uk dxk 1
du
du
du
du
dx1 u2u3 uk 1 u1 dx2 u3 uk 1 u1u2 uk 1 dxk uk 1 u1u2 uk dxk 1 .
If y u1u2
uk uk 1 u1u2
uk uk 1, then dx
dy
Thus the original formula holds for n (k 1) whenever it holds for n k .
Copyright 2016 Pearson Education, Ltd.
Chapter 3 Additional and Advanced Exercises
mk k !(mm! k )! . Then 1m 1!(mm!1)! m and mk km1 k !(mm! k )! (k 1)!(mm!k 1)! m!((kk 1)1)!(mm!(mk )!k )
m !( m 1)
( m 1)!
1
( k 1)!( m k )! ( k 1)!(( m 1) ( k 1))! m
k 1 . Now, we prove Leibniz’s rule by mathematical induction.
26. Recall
Step 1:
Step 2:
d (uv )
dv v du . Assume that the statement is true for n k , that is:
u dx
dx
dx
k
k
k 1
d (uv )
k d k 2u d 2 v
d
u
d
u
dv
d k 1v u d k v .
k v k k 1 dx 2
... kk1 du
dv dx k 1
dx k
dx
dx
dx k 2 dx 2
dx k
If n 1, then
d d (uv ) d k 1u v d k u dv k d k u dv k d k 1u d 2 v
dx
dx k dx k 1
dx k dx dx k dx
dx k 1 dx 2
k
1
2
k
2
3
2
k
1
k
d u v
k2 d k u1 d 2v 2k d k u2 d v3 ... kk1 d u2 d k v1 kk1 du
dx dx k
dx
dx
dx
dx
dx dx
d k v u d k 1u d k 1u v ( k 1) d k u dv k k d k 1u d 2v
du
1
2 k 1
dx dx k
dx k 1 dx k 1
dx k dx
dx
dx 2
k
k
1
k
1
k
k
k du d v
k 1 d k 1u d 2v
d
v
d
u
d
u
dv
k 1 k dx k u k 1 k 1 v (k 1) k dx 2
...
dx
dx
dk
dx
dx k 1 dx 2
k
k
1
d v u d v.
kk 1 du
k 1
dx k
d
If n k 1, then
k 1
(uv )
k
dx k 1
dx
dx
Therefore the formula (c) holds for n (k 1) whenever it holds for n k .
2
27. (a) T 2 4g L L
T 2g
4
2
2
(b) T 2 4g L T 2
g
L
(1s 2 )(9.8 m/s 2 )
4 2
L 0.248 m
L; dT 2 1 dL dL; dT
g 2 L
Lg
(0.248 m)(9.8 m/s 2 )
(0.01 m) 0.02015 s.
(c) Since there are 86, 400 s in a day, we have we have (0.02015 s)(86, 400 s/day) 1700 s/day, or
28.3 min/day; the clock will lose about 28.3 min/day.
28. v s 3 dv
3s 2 ds
k (6 s 2 ) ds
2k . If s0 the initial length of the cube’s side, then s1 s0 2k
dt
dt
dt
s
s
2k s0 s1. Let t the time it will take the ice cube to melt. Now, t 20k s 0 s
0
1
1 34
1/3
11 h.
Copyright 2016 Pearson Education, Ltd.
1
( v0 )1/3
(v0 )1/3 43 v0
1/3
179
CHAPTER 4
4.1
APPLICATIONS OF DERIVATIVES
EXTREME VALUES OF FUNCTIONS
1. An absolute minimum at x c2 , an absolute maximum at x b. Theorem 1 guarantees the existence of such
extreme values because h is continuous on [a, b].
2. An absolute minimum at x b, an absolute maximum at x c. Theorem 1 guarantees the existence of such
extreme values because f is continuous on [a, b].
3. No absolute minimum. An absolute maximum at x c. Since the function’s domain is an open interval, the
function does not satisfy the hypotheses of Theorem 1 and need not have absolute extreme values.
4. No absolute extrema. The function is neither continuous nor defined on a closed interval, so it need not fulfill
the conclusions of Theorem 1.
5. An absolute minimum at x a and an absolute maximum at x c. Note that y g ( x) is not continuous but still
has extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when the
hypothesis is not satisfied, absolute extrema may or may not occur.
6. Absolute minimum at x c and an absolute maximum at x a. Note that y g ( x) is not continuous but still has
absolute extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when the
hypothesis is not satisfied, absolute extrema may or may not occur.
7. Local minimum at (1, 0), local maximum at (1, 0).
8. Minima at (2, 0) and (2, 0), maximum at (0, 2).
9. Maximum at (0, 5). Note that there is no minimum since the endpoint (2, 0) is excluded from the graph.
10. Local maximum at (3, 0), local minimum at (2, 0), maximum at (1, 2), minimum at (0, 1).
11. Graph (c), since this is the only graph that has positive slope at c.
12. Graph (b), since this is the only graph that represents a differentiable function at a and b and has negative
slope at c.
13. Graph (d), since this is the only graph representing a function that is differentiable at b but not at a.
14. Graph (a), since this is the only graph that represents a function that is not differentiable at a or b.
15. f has an absolute min at x 0 but does not have
an absolute max. Since the interval on which f is
defined, 1 x 2, is an open interval, we do not
meet the conditions of Theorem 1.
Copyright 2016 Pearson Education, Ltd.
181
182
Chapter 4 Applications of Derivatives
16. f has an absolute max at x 0 but does not have an
absolute min. Since the interval on which f is defined,
1 x 1, is an open interval, we do not meet the
conditions of Theorem 1.
17. f has an absolute max at x 2 but does not have an
absolute min. Since the function is not continuous at
x 1, we do not meet the conditions of Theorem 1.
18. f has an absolute max at x 4 but does not have an
absolute min. Since the function is not continuous at
x 0, we do not meet the conditions of Theorem 1.
19. f has an absolute max at x 2 and an absolute min at
x 32 . Since the interval on which f is defined,
0 x 2 , is an open interval we do not meet the
conditions of Theorem 1.
20. f has an absolute max at x 0 and an absolute min
at x 2 and x 1 but does not have an absolute
y
(0, 1)
maximum. Since f is defined on a union of halfopen intervals, we do not meet the conditions of
Theorem 1.
y f ( x)
1
0
21. f ( x) 23 x 5 f ( x) 23 no critical points;
f (2) 19
, f (3) 3 the absolute maximum
3
is 3 at x 3 and the absolute minimum is 19
3
at x 2
Copyright 2016 Pearson Education, Ltd.
2
x
Section 4.1 Extreme Values of Functions
22. f ( x) x 4 f ( x) 1 no critical points;
f ( 4) 0, f (1) 5 the absolute maximum is 0
at x 4 and the absolute minimum is 5 at x 1
23. f ( x ) x 2 1 f ( x) 2 x a critical point at
x 0; f (1) 0, f (0) 1, f (2) 3 the absolute
maximum is 3 at x 2 and the absolute minimum is
1 at x 0
24.
f ( x ) 4 x 3 f ( x ) 3x 2 a critical point at
x 0; f ( 2) 12, f (0) 4, f (1) 3 the absolute
maximum is 12 at x 2 and the absolute
minimum is 3 at x 1
y
(2, 12)
10
5
f ( x ) 4 x3
(1, 3)
2
1
0
25. F ( x) 12 x 2 F ( x) 2 x 3 23 , however
x
x
x 0 is not a critical point since 0 is not in the domain;
F (0.5) 4, F (2) 0.25 the absolute maximum
is 0.25 at x 2 and the absolute minimum is 4 at
x 0.5
26. F ( x) 1x x 1 F ( x) x 2 12 , however
x
x 0 is not a critical point since 0 is not in the
domain; F (2) 12 , F (1) 1 the absolute
maximum is 1 at x 1 and the absolute minimum
is 12 at x 2
Copyright 2016 Pearson Education, Ltd.
1
x
183
184
Chapter 4 Applications of Derivatives
27. h( x) 3 x x1/3 h( x) 13 x 2/3 a critical point
at x 0; h(1) 1, h(0) 0, h(8) 2 the
absolute maximum is 2 at x 8 and the absolute
minimum is 1 at x 1
28. h( x) 3 x 2/3 h( x) 2 x 1/3 a critical point at
x 0; h(1) 3, h(0) 0, h(1) 3 the absolute
maximum is 0 at x 0 and the absolute minimum is
3 at x 1 and x 1
29. g ( x) 4 x 2 (4 x 2 )1/2
g ( x) 12 (4 x 2 ) 1/2 ( 2 x)
x
4 x 2
critical
points at x 2 and x 0, but not at x 2 because 2
is not in the domain;
g (2) 0, g (0) 2, g (1) 3 the absolute
maximum is 2 at x 0 and the absolute minimum is
0 at x 2
30.
g ( x) 5 x 2 (5 x 2 )1/2
g ( x) 12 (5 x 2 )1/2 (2 x)
x
5 x 2
critical points at x 5 and x 0, but not at
x 5 because 5 is not in the domain;
f 5 0, f (0) 5
the absolute maximum is 0 at x 5 and the
absolute minimum is 5 at x 0
31. f ( ) sin f ( ) cos 2 is a critical
point, but 2π is not a critical point because 2 is
not interior to the domain; f
2 1, f 2 1,
f 56 12 the absolute maximum is 1 at 2
and the absolute minimum is 1 at 2
32. f ( ) tan f ( ) sec2 f has no critical
points in 3 , 4 . The extreme values therefore
occur at the endpoints: f
3 3 and f 4 1
the absolute maximum is 1 at 4 and
the absolute minimum is 3 at 3
Copyright 2016 Pearson Education, Ltd.
Section 4.1 Extreme Values of Functions
185
33. g ( x) csc x g ( x) (csc x)(cot x) a critical
point at x 2 ; g 3 2 , g π2 1, g 23 2
3
the absolute maximum is
3
2 at x and x 2 ,
3
3
3
and the absolute minimum is 1 at x 2
34. g ( x) sec x g ( x) (sec x)(tan x) a critical
point at x 0; g 3 2, g (0) 1, g 6 2 the
3
absolute maximum is 2 at x and the absolute
minimum is 1 at x 0
35.
3
f (t ) 2 | t | 2 t 2 2 (t 2 )1/2
f (t ) 12 (t 2 ) 1/2 (2t ) t 2 |tt| a critical
t
point at t 0; f (1) 1, f (0) 2, f (3) 1 the
absolute maximum is 2 at t 0 and the absolute
minimum is 1 at t 3
36. f (t ) | t 5| (t 5) 2 ((t 5)2 )1/2
f (t ) 12 ((t 5)2 ) 1/2 (2(t 5))
t 5
(t 5)2
5 a critical point at t 5; f (4) 1, f (5) 0,
| tt 5|
f (7) 2 the absolute maximum is 2 at t 7 and
the absolute minimum is 0 at t 5
37. f ( x) x 4/3 f ( x) 43 x1/3 a critical point at x 0; f (1) 1, f (0) 0, f (8) 16 the absolute
maximum is 16 at x 8 and the absolute minimum is 0 at x 0
38. f ( x) x5/3 f ( x) 53 x 2/3 a critical point at x 0; f (1) 1, f (0) 0, f (8) 32 the absolute
maximum is 32 at x 8 and the absolute minimum is 1 at x 1
39. g ( ) 3/5 g ( ) 53 2/5 a critical point at 0; g (32) 8, g (0) 0, g (1) 1 the absolute
maximum is 1 at 1 and the absolute minimum is 8 at 32
40. h( ) 3 2/3 h( ) 2 1/3 a critical point at 0; h(27) 27, h(0) 0, h(8) 12 the absolute
maximum is 27 at 27 and the absolute minimum is 0 at 0
41. y x 2 6 x 7 y 2 x 6 2 x 6 0 x 3. The critical point is x 3.
42. f ( x) 6 x 2 x3 f ( x) 12 x 3 x 2 12 x 3 x 2 0 3 x(4 x) 0 x 0 or x 4. The critical points are
x 0 and x 4.
Copyright 2016 Pearson Education, Ltd.
186
Chapter 4 Applications of Derivatives
43. f ( x) x(4 x)3 f ( x) x[3(4 x) 2 (1)] (4 x)3 (4 x)2 [3 x (4 x )] (4 x)2 (4 4 x)
4(4 x) 2 (1 x) 4(4 x)2 (1 x) 0 x 1 or x 4. The critical points are x 1 and x 4.
44. g ( x) ( x 1)2 ( x 3)2 g ( x ) ( x 1)2 2( x 3)(1) 2( x 1)(1) ( x 3)2 2( x 3) ( x 1)[( x 1) ( x 3)]
4( x 3)( x 1) ( x 2) 4( x 3)( x 1)( x 2) 0 x 3 or x 1 or x 2. The critical points are x 1, x 2,
and x 3.
3
3
3
x
x
x
45. y x 2 2x y 2 x 22 2 x 2 2 2 x 2 2 0 2 x3 2 0 x 1; 2 x 2 2 undefined x 2 0 x 0.
x
The domain of the function is (, 0) (0, ), thus x 0 is not the domain, so the only critical point is x 1.
2
46. f ( x ) xx 2 f ( x)
( x 2)2 x x 2 (1)
( x 2)2
2
2
2
( x 2)
( x 2)
( x 2)
x 4 x2 x 4 x2 0 x 2 4 x 0 x 0 or x 4; x 4 x2 undefined
( x 2) 2 0 x 2. The domain of the function is (, 2) (2, ), thus x 2 is not the domain, so the only
critical points are x 0 and x 4
47. y x 2 32 x y 2 x 16 2 x
x
3/2
16 2 x3/2 16 0 2 x3/2 16 0 x 4; 2 x3/2 16 undefined
x
x
x
x 0 x 0. The critical points are x 4 and x 0.
48. g ( x) 2 x x 2 g ( x)
1 x
2 x x2
1 x
2 x x2
0 1 x 0 x 1;
1 x
2 x x2
2 x x 2 0 x 0 or x 2. The critical points are x 0, x 1, and x 2.
49. Minimum value is 1 at x 2.
50. To find the exact values, note that y 3x 2 2,
which is zero when x 23 . Local maximum at
, 4 (0.816, 5.089); local minimum
at , 4
(0.816, 2.911)
2
3
4 6
9
2
3
4 6
9
Copyright 2016 Pearson Education, Ltd.
undefined 2 x x 2 0
Section 4.1 Extreme Values of Functions
51. To find the exact values, note that y 3 x 2 2 x 8
(3x 4)( x 2), which is zero when x 2 or x 43 .
Local maximum at (2, 17); local minimum at
4 , 41
3
27
52. Note that y 5 x 2 ( x 5)( x 3), which is zero at
x 0, x 3, and x 5. Local maximum at (3, 108);
local minimum at (5, 0); (0, 0) is neither a maximum
nor a minimum.
53. Minimum value is 0 when x 1 or x 1.
54. Note that y
x 2
, which is zero at x 4 and is
x
undefined when x 0. Local maximum at (0, 0);
absolute minimum at (4, 4)
55. The actual graph of the function has asymptotes
at x 1, so there are no extrema near these values.
(This is an example of grapher failure.) There is
a local minimum at (0, 1).
Copyright 2016 Pearson Education, Ltd.
187
188
Chapter 4 Applications of Derivatives
56. Maximum value is 2 at x 1;
minimum value is 0 at x 1 and x 3.
57. Maximum value is 12 at x 1;
minimum value is 12 at x 1.
58. Maximum value is 12 at x 0;
minimum value is 12 as x 2.
59. y x 2/3 (1) 23 x 1/3 ( x 2)
5x 4
33 x
crit. pt.
derivative
extremum
value
x 54
0
local max
12 101/3 1.034
25
x0
undefined
local min
0
Copyright 2016 Pearson Education, Ltd.
Section 4.1 Extreme Values of Functions
2
60. y x 2/3 (2 x) 23 x 1/3 ( x 2 4) 8 x3 8
3 x
crit. pt.
derivative
extremum
value
x 1
0
minimum
3
x0
undefined
local max
0
x 1
0
minimum
3
61. y x
1
2 4 x
2
(2 x ) (1) 4 x 2
x 2 (4 x 2 )
4 x
2
crit. pt.
derivative
extremum
value
x 2
undefined
local max
0
x 2
0
minimum
x 2
0
maximum
2
2
x2
undefined
local min
0
62. y x 2
1 ( 1) 2 x
2 3 x
5 x 2 12 x
2 3 x
3 x
2
4 2 x 2
4 x
x 2 (4 x )(3 x )
2 3 x
crit. pt.
derivative
extremum
value
x0
0
minimum
0
x 12
5
0
local max
x3
undefined
minimum
0
144 151/2 4.462
125
2, x 1
63. y
1, x 1
crit. pt.
derivative
extremum
value
x 1
undefined
minimum
2
1, x 0
64. y
2 2 x, x 0
crit. pt.
derivative
extremum
value
x0
undefined
local min
3
x 1
0
local mix
4
Copyright 2016 Pearson Education, Ltd.
189
190
Chapter 4 Applications of Derivatives
2 x 2, x 1
65. y
2 x 6, x 1
crit. pt.
derivative
extremum
value
x 1
0
maximum
5
x 1
undefined
local min
1
x3
0
maximum
5
1 x 2 1 x 15 , x 1
2
4
66. We begin by determining whether f ( x) is defined at x 1, where f ( x) 4
x3 6 x 2 8 x,
x 1
Clearly, f ( x) 12 x 12 if x 1, and lim f (1 h) 1. Also, f ( x) 3 x 2 12 x 8 if x 1, and
h 0
lim f (1 h) 1. Since f is continuous at x 1, we have that f (1) 1.
h 0
12 x 12 , x 1
Thus, f ( x)
3 x 2 12 x 8, x 1
Note that 12 x 12 0 when x 1, and 3 x 2 12 x 8 0 when x
12 122 4(3)(8) 12 48
6 2 2 33 .
2(3)
But 2 2 3 3 0.845 1, so the critical points occur at x 1 and x 2 2 3 3 3.155.
crit. pt.
derivative
extremum
value
x 1
0
local max
4
x 3.155
0
local min
3.079
67. (a) No, since f ( x) 23 ( x 2) 1/3 , which is undefined at x 2.
(b) The derivative is defined and nonzero for all x 2. Also, f (2) 0 and f ( x ) 0 for all x 2.
(c) No, f ( x) need not have a global maximum because its domain is all real numbers. Any restriction of f to a
closed interval of the form [a, b] would have both a maximum value and minimum value on the interval.
(d) The answers are the same as (a) and (b) with 2 replaced by a.
x3 9 x, x 3 or 0 x 3
3 x3 9, x 3 or 0 x 3
68. Note that f ( x)
. Therefore, f ( x)
.
3
3
x 9 x, 3 x 0 or x 3
3 x 9, 3 x 0 or x 3
(a) No, since the left- and right-hand derivatives at x 0, are 9 and 9, respectively.
(b) No, since the left- and right-hand derivatives at x 3, are 18 and 18, respectively.
(c) No, since the left- and right-hand derivatives at x 3, are 18 and 18, respectively.
Copyright 2016 Pearson Education, Ltd.
Section 4.1 Extreme Values of Functions
191
(d) The critical points occur when f ( x) 0 (at x 3) and when f ( x) is undefined (at x 0 and x 3).
The minimum value is 0 at x 3, at x 0, and at x 3; local maxima occur at 3, 6 3 and 3, 6 3 .
69. Yes, since f ( x ) | x | x 2 ( x 2 )1/2 f ( x) 12 ( x 2 ) 1/2 (2 x)
x
| xx | is not defined at x 0. Thus it is
( x 2 )1/ 2
not required that f be zero at a local extreme point since f may be undefined there.
70. If f (c) is a local maximum value of f, then f ( x) f (c) for all x in some open interval (a, b) containing c. Since
f is even, f ( x) f ( x ) f (c) f (c) for all x in the open interval (b, a ) containing c. That is, f assumes
a local maximum at the point c. This is also clear from the graph of f because the graph of an even function is
symmetric about the y -axis.
71. If g (c) is a local minimum value of g, then g ( x ) g (c) for all x in some open interval (a, b) containing c.
Since g is odd, g ( x) g ( x) g (c) g ( c) for all x in the open interval (b, a ) containing c. That is,
g assumes a local maximum at the point c. This is also clear from the graph of g because the graph of an odd
function is symmetric about the origin.
72. If there are no boundary points or critical points the function will have no extreme values in its domain. Such
functions do indeed exist, for example f ( x) x for x . (Any other linear function f ( x) mx b with
m 0 will do as well.)
73. (a) V ( x) 160 x 52 x 2 4 x3
V ( x) 160 104 x 12 x 2 4( x 2)(3 x 20)
The only critical point in the interval (0, 5) is at x 2. The maximum value of V ( x ) is 144 at x 2.
(b) The largest possible volume of the box is 144 cubic units, and it occurs when x 2 units.
74. (a) f ( x) 3ax 2 2bx c is a quadratic, so it can have 0, 1, or 2 zeros, which would be the critical points of f.
The function f ( x ) x3 3 x has two critical points at x 1 and x 1. The function f ( x ) x3 1 has one
critical point at x 0. The function f ( x) x3 x has no critical points.
(b) The function can have either two local extreme values or no extreme values. (If there is only one critical
point, the cubic function has no extreme values.)
v
gt
2v
75. s 12 gt 2 v0 t s0 ds
gt v0 0 t g0 . Now s (t ) s0 t 2 v0 0 t 0 or t g0 .
dt
v
v s s s is the maximum height over the interval 0 t
v
Thus s g0 12 g g0
2
0
v0
g
0
v02
2g
0
0
2v0
.
g
76. dI
2sin t 2 cos t , solving dI
0 tan t 1 t 4 n where n is a nonnegative integer (in this exercise
dt
dt
t is never negative) the peak current is 2 2 amps.
Copyright 2016 Pearson Education, Ltd.
192
Chapter 4 Applications of Derivatives
77. Maximum value is 11 at x 5; minimum value is 5
on the interval [3, 2]; local maximum at (5, 9)
78. Maximum value is 4 on the interval [5, 7];
minimum value is 4 on the interval [2, 1].
79. Maximum value is 5 on the interval [3, );
minimum value is 5 on the interval (, 2].
80. Minimum value is 4 on the interval [1, 3]
81-86.
Example CAS commands:
Maple:
with(student):
f : x - x^4 -8*x^2 4*x 2;
domain : x -20/25..64/25;
plot( f(x), domain, color black, title "Section 4.1 #81(a)" );
Df : D(f );
plot( Df(x), domain, color black, title "Section 4.1 #81(b)" )
StatPt : fsolve( Df(x) 0, domain )
SingPt : NULL;
EndPt : op(rhs(domain));
Copyright 2016 Pearson Education, Ltd.
Section 4.2 The Mean Value Theorem
193
Pts : evalf ([EndPt,StatPt,SingPt]);
Values : [seq( f(x), x Pts )];
Maximum value is 2.7608 and occurs at x 2.56 (right endpoint).
Minimum value is -6.2680 and occurs at x1.86081 (singular point).
Mathematica: (functions may vary):
<<Miscellaneous `RealOnly`
Clear[f,x]
a 1; b 10/3;
f[x_ ] 2 2x 3 x 2/3
f '[ x]
Plot[{f[x], f '[x]}, {x, a, b}]
NSolve[f '[x] 0, x]
{f[a], f[0], f[x]/.%, f[b]}//N
In more complicated expressions, NSolve may not yield results. In this case, an approximate solution
(say 1.1 here) is observed from the graph and the following command is used:
FindRoot[f '[x] 0, {x, 1.1}]
4.2
THE MEAN VALUE THEOREM
1.
When f ( x) x 2 2 x 1 for 0 x 1, then
2. When f ( x ) x 2/3 for 0 x 1, then
f (1) f (0)
f (c) 3 2c 2 c 12 .
10
c1/3 c 278 .
f (1) f (0)
f (c ) 1 23
1 0
3. When f ( x ) x 1x for 12 x 2, then
f (2) f (1/2)
f (c ) 0 1 12 c 1.
2 1/2
c
4. When f ( x) x 1 for 1 x 3, then
f (3) f (1)
f (c ) 22 1 c 32 .
31
2 c 1
5. When f ( x ) x3 x 2 for 1 x 2, then
f (2) f ( 1)
f (c) 2 3c 2 2c c 13 7 .
2 ( 1)
1 7
1.22 and 13 7 0.549 are both in the interval 1 x 2.
3
x3 2 x 0
g (2) g ( 2)
6. When g ( x)
, then 2( 2) g (c) 3 g (c). If 2 x 0, then g ( x) 3 x 2 3 g (c)
2
x 0 x 2
2
3c 3 c 1. Only c 1 is in the interval. If 0 x 2, then g ( x) 2 x 3 g (c ) 2c 3 c 32 .
7. Does not; f ( x) is not differentiable at x 0 in (1, 8).
8. Does; f ( x) is continuous for every point of [0, 1] and differentiable for every point in (0, 1).
9. Does; f ( x) is continuous for every point of [0, 1] and differentiable for every point in (0, 1).
10. Does not; f ( x) is not continuous at x 0 because lim f ( x) 1 0 f (0).
x 0
11. Does not; f is not differentiable at x 1 in (2, 0).
Copyright 2016 Pearson Education, Ltd.
194
Chapter 4 Applications of Derivatives
12. Does; f ( x) is continuous for every point of [0, 3] and differentiable for every point in (0, 3).
13. Since f ( x) is not continuous on 0 x 1, Rolle’s Theorem does not apply: lim f ( x) lim x 1 0 f (1).
x 1
x 1
14. Since f ( x) must be continuous at x 0 and x 1 we have lim f ( x) a f (0) a 3 and
x 0
lim f ( x) lim f ( x) 1 3 a m b 5 m b. Since f ( x) must also be differentiable at x 1
x 1
x 1
we have lim f ( x) lim f ( x ) 2 x 3| x 1 m |x 1 1 m. Therefore, a 3, m 1 and b 4.
x 1
x 1
15. (a)
(b) Let r1 and r2 be zeros of the polynomial P( x) x n an 1 x n 1 a1 x a0 , then P (r1 ) P (r2 ) 0.
Since polynomials are everywhere continuous and differentiable, by Rolle’s Theorem P (r ) 0 for some r
between r1 and r2 , where P ( x) nx n 1 (n 1)an 1 x n 2 a1.
16. With f both differentiable and continuous on [a, b] and f (r1 ) f (r2 ) f (r3 ) 0 where r1 , r2 and r3 are in [a, b],
then by Rolle’s Theorem there exists a c1 between r1 and r2 such that f (c1 ) 0 and a c2 between r2 and r3 such
that f (c2 ) 0. Since f is both differentiable and continuous on [a, b], Rolle’s Theorem again applies and we
have a c3 between c1 and c2 such that f (c3 ) 0. To generalize, if f has n 1 zeros in [a, b] and f ( n) is continuous
on [a, b], then f ( n) has at least one zero between a and b.
17. Since f exists throughout [a, b] the derivative function f is continuous there. If f has more than one zero
in [a, b], say f (r1 ) f (r2 ) 0 for r1 r2 , then by Rolle’s Theorem there is a c between r1 and r2 such that
f (c ) 0, contrary to f 0 throughout [a, b]. Therefore f has at most one zero in [a, b]. The same argument
holds if f 0 throughout [a, b].
18. If f ( x ) is a cubic polynomial with four or more zeros, then by Rolle’s Theorem f ( x) has three or more zeros,
f ( x) has 2 or more zeros and f ( x) has at least one zero. This is a contradiction since f ( x) is a non-zero
constant when f ( x) is a cubic polynomial.
19. With f (2) 11 0 and f (1) 1 0 we conclude from the Intermediate Value Theorem that
f ( x) x 4 3x 1 has at least one zero between 2 and 1. Then 2 x 1 8 x3 1 32 4 x3 4
29 4 x3 3 1 f ( x) 0 for 2 x 1 f ( x) is decreasing on [2, 1] f ( x) 0 has exactly one
solution in the interval (2, 1).
20. f ( x) x3 42 7 f ( x) 3x 2 83 0 on (, 0) f ( x ) is increasing on (, 0). Also, f ( x) 0 if x 2
x
x
and f ( x ) 0 if 2 x 0 f ( x ) has exactly one zero in (, 0).
21. g (t ) t t 1 4 g (t )
1 1 0 g (t ) is increasing for t in (0, ); g (3)
2 t 2 t 1
3 2 0 and
g (15) 15 0 g (t ) has exactly one zero in (0, ).
22. g (t ) 11 t 1 t 3.1 g (t )
1 1 0 g (t ) is increasing for t in ( 1, 1); g ( 0.99) 2.5 and
2 1t
(1t )2
g (0.99) 98.3 g (t ) has exactly one zero in (1, 1).
Copyright 2016 Pearson Education, Ltd.
Section 4.2 The Mean Value Theorem
195
23. r ( ) sin 2 3 8 r ( ) 1 23 sin 3 cos 3 1 13 sin 23 0 on (, ) r ( ) is increasing on
(, ); r (0) 8 and r (8) sin 2 83 0 r ( ) has exactly one zero in (, ).
24. r ( ) 2 cos 2 2 r ( ) 2 2sin cos 2 sin 2 0 on (, ) r ( ) is increasing on
(, ); r (2 ) 4 cos(2 ) 2 4 1 2 0 and r (2 ) 4 1 2 0 r ( ) has exactly one
zero in (, ).
2
25. r ( ) sec 13 5 r ( ) (sec )(tan ) 34 0 on 0, 2 r ( ) is increasing on 0, 2 ; r (0.1) 994
and r (1.57) 1260.5 r ( ) has exactly one zero in 0, .
2
26. r ( ) tan cot r ( ) sec 2 csc2 1 sec2 cot 2 0 on 0, 2 r ( ) is increasing on
2 4
0, ; r 0 and r (1.57) 1254.2 r ( ) has exactly one zero in 0, .
4
27. By Corollary 1, f ( x ) 0 for all x f ( x) C , where C is a constant. Since f (1) 3 we have
C 3 f ( x) 3 for all x.
28. g ( x) 2 x 5 g ( x) 2 f ( x) for all x. By Corollary 2, f ( x ) g ( x) C for some constant C. Then
f (0) g (0) C 5 5 C C 0 f ( x ) g ( x) 2 x 5 for all x.
29. g ( x) x 2 g ( x) 2 x f ( x) for all x. By Corollary 2, f ( x) g ( x) C.
(a) f (0) 0 0 g (0) C 0 C C 0 f ( x) x 2 f (2) 4
(b) f (1) 0 0 g (1) C 1 C C 1 f ( x) x 2 1 f (2) 3
(c) f (2) 3 3 g (2) C 3 4 C C 1 f ( x) x 2 1 f (2) 3
30. g ( x) mx g ( x) m, a constant. If f ( x) m, then by Corollary 2, f ( x) g ( x ) b mx b where b is a
constant. Therefore all functions whose derivatives are constant can be graphed as straight lines y mx b.
2
3
4
31. (a) y x2 C
(b) y x3 C
(c) y x4 C
32. (a) y x 2 C
(b) y x 2 x C
(c) y x3 x 2 x C
33. (a) y x 2 y 1x C
(b) y x 1x C
(c) y 5 x 1x C
34. (a) y 12 x 1/2 y x1/2 C y x C
(b) y 2 x C
35. (a) y 12 cos 2t C
(c) y 12 cos 2t 2 sin 2t C
(b) y 2sin 2t C
2
(c) y 2 x 2 x C
36. (a) y tan C
(b) y 1/2 y 23 3/2 C
(c)
37. f ( x) x 2 x C ; 0 f (0) 02 0 C C 0 f ( x) x 2 x
38. g ( x) 1x x 2 C ; 1 g (1) 11 (1)2 C C 1 g ( x) 1x x 2 1
Copyright 2016 Pearson Education, Ltd.
y 23 3/2 tan C
196
Chapter 4 Applications of Derivatives
39. r ( ) 8 cot C ; 0 r 4 8 4 cot 4 C 0 2 1 C C 2 1
r ( ) 8 cot 2 1
40. r (t ) sec t t C ; 0 r (0) sec(0) 0 C C 1 r (t ) sec t t 1
9.8t 5 s 4.9t 2 5t C ; at s 10 and t 0 we have C 10 s 4.9t 2 5t 10
41. v ds
dt
32t 2 s 16t 2 2t C; at s 4 and t 12 we have C 1 s 16t 2 2t 1
42. v ds
dt
43. v ds
sin( t ) s 1 cos( t ) C ; at s 0 and t 0 we have C 1 s
dt
1cos( t )
44. v ds
2 cos 2t s sin 2t C ; at s 1 and t 2 we have C 1 s sin 2t 1
dt
dv
32 v 32t C; at v = 20 and t = 0 we have C 20 v 32t 20
dt
ds
v
32t 20 s 16t 2 20t C; at s = 5 and t = 0 we have C 5 s 16t 2 20t 5
dt
45. a
46. a 9.8 v 9.8t C1; at v 3 and t 0 we have C1 3 v 9.8t 3 s 4.9t 2 3t C2 ; at s 0 and
t 0 we have C2 0 s 4.9t 2 3t
47. a 4sin(2t ) v 2 cos(2t ) C1; at v 2 and t 0 we have C1 0 v 2 cos(2t ) s sin(2t ) C2 ; at
s 3 and t 0 we have C2 3 s sin(2t ) 3
48. a 92 cos 3t v 3 sin 3t C1; at v 0 and t 0 we have C1 0 v 3 sin 3t s cos 3t C2 ; at
3t
s 1 and t 0 we have C2 0 s cos
49. If T (t ) is the temperature of the thermometer at time t, then T (0) 19 C and T (14) 100 C. From the Mean
Value Theorem there exists a 0 t0 14 such that
T (14) T (0)
8.5 C / s T (t0 ), the rate at which the
14 0
temperature was changing at t t0 as measured by the rising mercury on the thermometer.
50. Because the trucker's average speed was 115 km/h, by the Mean Value Theorem, the trucker must have been
going that speed at least once during the trip.
51. Because its average speed was approximately 7.667 knots, and by the Mean Value Theorem, it must have been
going that speed at least once during the trip.
52. The runner’s average speed for the marathon was approximately 19.1 km/h. Therefore, by the Mean Value
Theorem, the runner must have been going that speed at least once during the marathon. Since the initial speed
and final speed are both 0 mph and the runner’s speed is continuous, by the Intermediate Value Theorem, the
runner’s speed must have been 18 km/h at least twice.
53. Let d (t ) represent the distance the automobile traveled in time t. The average speed over 0 t 2 is
The Mean Value Theorem says that for some 0 t0 2, d ( t0 )
automobile at time t0 (which is read on the speedometer).
d (2) d (0)
.
20
d (2) d (0)
. The value d ( t0 ) is the speed of the
20
Copyright 2016 Pearson Education, Ltd.
Section 4.2 The Mean Value Theorem
197
54. a (t ) v (t ) 1.6 v(t ) 1.6t C ; at (0, 0) we have C 0 (t ) 1.6t. When t 30, then v(30) 48 m/s.
11
55. The conclusion of the Mean Value Theorem yields bb aa 12 c 2
c
2
aabb a b c ab.
2
56. The conclusion of the Mean Value Theorem yields bb aa 2c c a 2 b .
57. f ( x) [cos x sin( x 2) sin x cos( x 2)] 2sin( x 1) cos( x 1) sin( x x 2) sin 2( x 1)
sin(2 x 2) sin (2 x 2) 0. Therefore, the function has the constant value f (0) sin 2 1 0.7081
which explains why the graph is a horizontal line.
58. (a) f ( x) ( x 2)( x 1) x( x 1)( x 2) x5 5 x3 4x is one possibility.
(b) Graphing f ( x) x5 5 x3 4 x and f ( x) 5 x 4 15 x 2 4 on [3, 3] by [ 7, 7] we see that each
x-intercept of f ( x) lies between a pair of x-intercepts of f ( x), as expected by Rolle’s Theorem.
(c) Yes, since sin is continuous and differentiable on (, ).
59. f ( x) must be zero at least once between a and b by the Intermediate Value Theorem. Now suppose that f ( x ) is
zero twice between a and b. Then by the Mean Value Theorem, f ( x) would have to be zero at least once
between the two zeros of f ( x), but this can’t be true since we are given that f ( x) 0 on this interval.
Therefore, f ( x) is zero once and only once between a and b.
60. Consider the function k ( x) f ( x) g ( x). k ( x) is
continuous and differentiable on [a, b], and since
k (a) f (a) g (a ) and k (b) f (b) g (b), by the
Mean Value Theorem, there must be a point c in
(a, b) where k (c) 0. But since k (c) f (c ) g (c),
this means that f (c ) g (c), and c is a point where
the graphs of f and g have tangent lines with the
same slope, so these lines are either parallel or are
the same line.
61. f ( x) 1 for 1 x 4 f ( x) is differentiable on 1 x 4 f is continuous on 1 x 4 f satisfies the
f (4) f (1)
f (4) f (1)
conditions of the Mean Value Theorem 41 f (c) for some c in 1 x 4 f (c) 1
1
3
f (4) f (1) 3
Copyright 2016 Pearson Education, Ltd.
198
Chapter 4 Applications of Derivatives
62. 0 f ( x) 12 for all x f ( x) exists for all x, thus f is differentiable on (1, 1) f is continuous on [1, 1]
f satisfies the conditions of the Mean Value Theorem
f (1) f ( 1)
f (1) f ( 1)
f (c ) for some c in [1, 1]
1 ( 1)
0
12 0 f (1) f (1) 1. Since f (1) f (1) 1 f (1) 1 f (1) 2 f ( 1), and
2
since 0 f (1) f (1) we have f (1) f (1). Together we have f (1) f (1) 2 f (1).
63. Let f (t ) cos t and consider the interval [0, x] where x is a real number. f is continuous on [0, x] and f is
differentiable on (0, x) since f (t ) sin t f satisfies the conditions of the Mean Value Theorem
f ( x ) f (0)
x (0) f (c) for some c in [0, x ] cosxx 1 sin c. Since 1 sin c 1 1 sin c 1
1 cosxx 1 1. If x 0, 1 cosxx 1 1 x cos x 1 x |cos x 1| x | x | . If x 0, 1 cosxx 1 1
x cos x 1 x x cos x 1 x ( x) cos x 1 x |cos x 1| x | x | . Thus, in both cases,
we have |cos x 1| | x | . If x 0, then |cos 0 1| |1 1| |0| |0|, thus |cos x 1| | x | is true for all x.
64. Let f ( x) sin x for a x b. From the Mean Value Theorem there exists a c between a and b such that
sin b sin a
sin b sin a
sin b sin a
cos c 1 b a 1 b a 1 |sin b sin a | | b a | .
ba
65. Yes. By Corollary 2 we have f ( x) g ( x) c since f ( x) g ( x). If the graphs start at the same point x a,
then f (a ) g (a ) c 0 f ( x ) g ( x).
66. Assume f is differentiable and | f ( w) f ( x )| | w x | for all values of w and x. Since f is differentiable,
f ( w) f ( x )
f ( x) exists and f ( x) lim
using the alternative formula for the derivative. Let g ( x) x ,
w x
w x
f ( w) f ( x )
lim
which is continuous for all x. By Theorem 10 from Chapter 2, | f ( x)| lim
w x
f ( w) f ( x )
w x
w x
w x
| f ( w) f ( x )|
| f ( w) f ( x )|
lim
. Since f ( w) f ( x) w x for allw and x |w x| 1 as long as w x. By Theorem 5
|w x|
w x
| f ( w) f ( x )|
from Chapter 2, f ( x) lim
lim 1 1 f ( x) 1 1 f ( x) 1.
|w x|
w x
w x
f (b ) f ( a )
67. By the Mean Value Theorem we have b a
f (c) for some point c between a and b. Since b a 0 and
f (b) f (a), we have f (b) f (a) 0 f (c ) 0.
68. The condition is that f should be continuous over [a, b]. The Mean Value Theorem then guarantees the
f (b ) f ( a )
existence of a point c in (a, b) such that b a
f (c). If f is continuous, then it has a minimum and
maximum value on [a, b], and min f f (c) max f , as required.
69.
f ( x) (1 x 4 cos x) 1 f ( x) (1 x 4 cos x)2 (4 x3 cos x x 4 sin x)
x3 (1 x 4 cos x) 2 (4 cos x x sin x) 0 for 0 x 0.1 f ( x) is decreasing when 0 x 0.1
min f 0.9999 and max f 1. Now we have 0.9999
f (0.1) 1
1 0.09999 f (0.1) 1 0.1
0.1
1.09999 f (0.1) 1.1.
4 x3 0 for 0 x 0.1 f ( x) is increasing when
(1 x 4 )3
f (0.1) 2
1.0001
0 x 0.1 min f 1 and max f 1.0001. Now we have 1
0.1
70. f ( x) (1 x 4 ) 1 f ( x) (1 x 4 ) 2 (4 x3 )
0.1 f (0.1) 2 0.10001 2.1 f (0.1) 2.10001.
Copyright 2016 Pearson Education, Ltd.
Section 4.3 Monotonic Functions and the First Derivative Test
71. (a) Suppose x 1, then by the Mean Value Theorem
199
f ( x ) f (1)
0 f ( x ) f (1). Suppose x 1, then by the
x 1
f ( x ) f (1)
0 f ( x) f (1). Therefore f ( x) 1 for all x since f (1) 1.
x 1
f ( x ) f (1)
f ( x ) f (1)
Yes. From part (a), lim
0 and lim
0. Since f (1) exists, these two one-sided limits
x 1
x 1
x 1
x 1
Mean Value Theorem
(b)
are equal and have the value f (1) f (1) 0 and f (1) 0 f (1) 0.
72. From the Mean Value Theorem we have
q
f (b ) f ( a )
f (c) where c is between a and b. But f (c ) 2 pc q 0
ba
has only one solution c 2 p . (Note: p 0 since f is a quadratic function.)
4.3
MONOTONIC FUNCTIONS AND THE FIRST DERIVATIVE TEST
1. (a) f ( x) x( x 1) critical points at 0 and 1
(b) f | | increasing on ( , 0) and (1, ), decreasing on (0, 1)
0
1
(c) Local maximum at x 0 and a local minimum at x 1
2. (a) f ( x) ( x 1)( x 2) critical points at 2 and 1
(b) f | | increasing on (, 2) and (1, ), decreasing on (2, 1)
2
1
(c) Local maximum at x 2 and a local minimum at x 1
3. (a) f ( x) ( x 1) 2 ( x 2) critical points at 2 and 1
(b) f | | increasing on (2, 1) and (1, ), decreasing on (, 2)
2
1
(c) No local maximum and a local minimum at x 2
4. (a) f ( x) ( x 1) 2 ( x 2)2 critical points at 2 and 1
(b) f | | increasing on (, 2) (2, 1) (1, ), never decreasing
2
(c) No local extrema
5. (a)
(b)
1
f ( x ) ( x 1)( x 2)( x 3) critical points at 2, 1, 3
f | | | increasing on ( 2, 1) and (3, ), decreasing on ( , 2) and (1, 3)
2
1
3
(c) Local maximum at x 1, local minima at x 2 and x 3
6. (a) f ( x) ( x 7)( x 1)( x 5) critical points at 5, 1 and 7
(b) f | | | increasing on ( 5, 1) and (7, ), decreasing on ( , 5) and (1, 7)
5
1
7
(c) Local maximum at x 1, local minima at x 5 and x 7
x 2 ( x 1)
7. (a) f ( x) ( x 2) critical points at x 0, x 1 and x 2
(b) f )( | | increasing on (, 2) and (1, ), decreasing on (2, 0) and (0, 1)
2
0
(c) Local minimum at x 1
1
Copyright 2016 Pearson Education, Ltd.
200
Chapter 4 Applications of Derivatives
( x 2)( x 4)
8. (a) f ( x) ( x 1)( x 3) critical points at x 2, x 4, x 1, and x 3
(b) f | )( | )( increasing on ( , 4), (1, 2) and (3, ), decreasing on
4
1
2
3
( 4, 1) and (2, 3)
(c) Local maximum at x 4 and x 2
2
9. (a) f ( x) 1 42 x 2 4 critical points at x 2, x 2 and x 0.
x
x
2
0
(b) f | )( | increasing on (, 2) and (2, ), decreasing on (2, 0) and (0, 2)
2
(c) Local maximum at x 2, local minimum at x 2
10. (a) f ( x) 3 6 3 x 6 critical points at x 4 and x 0
x
x
(b) f ( | increasing on (4, ), decreasing on (0, 4)
4
0
(c) Local minimum at x 4
11. (a) f ( x) x 1/3 ( x 2) critical points at x 2 and x 0
(b) f | )( increasing on (, 2) and (0, ), decreasing on (2, 0)
2
0
(c) Local maximum at x 2, local minimum at x 0
12. (a) f ( x) x 1/2 ( x 3) critical points at x 0 and x 3
(b) f ( | increasing on (3, ), decreasing on (0, 3)
3
0
(c) No local maximum and a local minimum at x 3
13. (a) f ( x) (sin x 1)(2 cos x 1), 0 x 2 critical points at x 2 , x 23 , and x 43
(b) f [ | | | ] increasing on 23 , 43 , decreasing on 0, 2 ,
0
2
3
2
4
3
2 , 23 and 43 , 2
2
and x 2
(c) Local maximum at x 43 and x 0, local minimum at x 2π
3
14. (a) f ( x) (sin x cos x)(sin x cos x), 0 x 2 critical points at x 4 , x 34 , x 54 , and x 74
(b) f [ | | | | ] increasing on 4 , 34 and
0
3 , 5
4
4
4
3
4
and 74 , 2
5
4
7
4
2
54 , 74 , decreasing on 0, 4 ,
(c) Local maximum at x 0, x 34 and x 74 , local minimum at x 4 , x 54 and x 2
15. (a) Increasing on (2, 0) and (2, 4), decreasing on (4, 2) and (0, 2)
(b) Absolute maximum at (4, 2), local maximum at (0, 1) and (4, 1); Absolute minimum at (2, 3), local
minimum at (2, 0)
16. (a) Increasing on (4, 3.25), ( 1.5, 1), and (2, 4), decreasing on (3.25, 1.5) and (1, 2)
(b) Absolute maximum at (4, 2), local maximum at ( 3.25, 1) and (1, 1); Absolute minimum at ( 1.5, 1), local
minimum at ( 4, 0) and (2, 0)
17. (a) Increasing on ( 4, 1), (0.5, 2), and (2, 4), decreasing on ( 1, 0.5)
(b) Absolute maximum at (4, 3), local maximum at (1, 2) and (2, 1); No absolute minimum, local minimum
at (4, 1) and (0.5, 1)
Copyright 2016 Pearson Education, Ltd.
Section 4.3 Monotonic Functions and the First Derivative Test
201
18. (a) Increasing on (4, 2.5), (1, 1), and (3, 4), decreasing on (2.5, 1) and (1, 3)
(b) No absolute maximum, local maximum at ( 2.5, 1), (1, 2) and (4, 2); No absolute minimum, local
minimum at ( 1, 0) and (3, 1)
19. (a) g (t ) t 2 3t 3 g (t ) 2t 3 a critical point at t 32 ; g | , increasing on
3/2
, 32 , decreasing on 32 ,
(b) local maximum value of g 32 21
at t 32 , absolute maximum is 21
at t 23
4
4
20. (a) g (t ) 3t 2 9t 5 g (t ) 6t 9 a critical point at t 32 ; g | , increasing on , 32 ,
decreasing on 32 ,
3/2
(b) local maximum value of g 32 47
at t 23 , absolute maximum is 47
at t 32
4
4
21. (a) h( x) x3 2 x 2 h( x) 3 x 2 4 x x(4 3 x) critical points at x 0, 43 h | | ,
increasing on 0, 43 , decreasing on ( , 0) and 43 ,
0
4/3
(b) local maximum value of h 43 32
at x 34 ; local minimum value of h(0) 0 at x 0, no absolute
27
extrema
22. (a) h( x) 2 x3 18 x h( x) 6 x 2 18 6 x 3 x 3 critical points at x 3
3, , decreasing on 3, 3
(b) a local maximum is h 3 12 3 at x 3; local minimum is h 3 12 3 at x 3, no absolute
h | | , increasing on , 3 and
3
3
extrema
23. (a) f ( ) 3 2 4 3 f ( ) 6 12 2 6 (1 2 ) critical points at 0, 12
f | | , increasing on 0, 12 , decreasing on (, 0) and
0
1/2
(b) a local maximum is f
12 ,
12 14 at 12 , a local minimum is f (0) 0 at 0, no absolute extrema
2 2 critical points at 2
f | | , increasing on 2, 2 , decreasing on , 2 and 2,
2
2
(b) a local maximum is f 2 4 2 at 2, a local minimum is f 2 4 2 at 2, no
24. (a) f ( ) 6 3 f ( ) 6 3 2 3
absolute extrema
25. (a) f (r ) 3r 3 16r f (r ) 9r 2 16 no critical points f , increasing on (, ), never
decreasing
(b) no local extrema, no absolute extrema
26. (a) h(r ) (r 7)3 h(r ) 3(r 7) 2 a critical point at r 7 h | , increasing on
( , 7) ( 7, ), never decreasing
(b) no local extrema, no absolute extrema
Copyright 2016 Pearson Education, Ltd.
7
202
Chapter 4 Applications of Derivatives
27. (a) f ( x) x 4 8 x 2 16 f ( x) 4 x3 16 x 4 x( x 2)( x 2) critical points at x 0 and x 2
f | | | , increasing on ( 2, 0) and (2, ), decreasing on ( , 2) and (0, 2)
2
0
2
(b) a local maximum is f (0) 16 at x 0, local minima are f (2) 0 at x 2, no absolute maximum;
absolute minimum is 0 at x 2
28. (a) g ( x) x 4 4 x3 4 x 2 g ( x) 4 x3 12 x 2 8 x 4 x ( x 2)( x 1) critical points at x 0, 1, 2
g | | | , increasing on (0, 1) and (2, ), decreasing on (, 0) and (1, 2)
0
1
2
(b) a local maximum is g (1) 1 at x 1, local minima are g (0) 0 at x 0 and g (2) 0 at x 2, no absolute
maximum; absolute minimum is 0 at x 0, 2
29. (a) H (t ) 32 t 4 t 6 H (t ) 6t 3 6t 5 6t 3 (1 t )(1 t ) critical points at t 0, 1
H | | | , increasing on (, 1) and (0, 1), decreasing on (1, 0) and (1, )
1
0
1
(b) the local maxima are H (1) 12 at t 1 and H (1) 12 at t 1, the local minimum is H (0) 0 at t 0,
absolute maximum is 12 at t 1; no absolute minimum
30. (a) K (t ) 15t 3 t 5 K (t ) 45t 2 5t 4 5t 2 (3 t )(3 t ) critical points at t 0, 3
K | | | , increasing on (3, 0) (0, 3), decreasing on (, 3) and (3, )
3
0
3
(b) a local maximum is K (3) 162 at t 3, a local minimum is K (3) 162 at t 3, no absolute extrema
x 1 3
critical points at x 1 and x 10 f ( | ,
x 1
10
1
3
x 1
31. (a) f ( x) x 6 x 1 f ( x) 1
increasing on (10, ), decreasing on (1, 10)
(b) a local minimum is f (10) 8, a local and absolute maximum is f (1) 1, absolute minimum of 8 at x 10
32. (a) g ( x) 4 x x 2 3 g ( x) 2 2 x 2 2 x
x
3/ 2
x
critical points at x 1 and x 0 g ( | ,
increasing on (0, 1), decreasing on (1, )
(b) a local minimum is f (0) 3, a local maximum is f (1) 6, absolute maximum of 6 at x 1
33. (a) g ( x) x 8 x 2 x(8 x 2 )1/2 g ( x) (8 x 2 )1/2 x 12 (8 x 2 )1/2 (2 x)
critical points at x 2, 2 2 g
on 2 2, 2 and 2, 2 2
1
2(2 x )(2 x )
2 2 x 2 2 x
( | | ) , increasing on (2, 2), decreasing
2
2 2
0
2
2 2
(b) local maxima are g (2) 4 at x 2 and g 2 2 0 at x 2 2, local minima are g (2) 4 at
x 2 and g 2 2 0 at x 2 2, absolute maximum is 4 at x 2; absolute minimum is 4 at x 2
34. (a) g ( x) x 2 5 x x 2 (5 x)1/2 g ( x) 2 x(5 x )1/2 x 2 12 (5 x) 1/2 ( 1)
5 x (4 x )
critical points
2 5 x
at x 0, 4 and 5 g | | ), increasing on (0, 4), decreasing on (, 0) and (4, 5)
0
4
5
(b) a local maximum is g (4) 16 at x 4, a local minimum is 0 at x 0 and x 5, no absolute maximum;
absolute minimum is 0 at x 0, 5
Copyright 2016 Pearson Education, Ltd.
Section 4.3 Monotonic Functions and the First Derivative Test
2
35. (a) f ( x) xx 23 f ( x)
2 x ( x 2) ( x 2 3)(1)
( x 2)2
( x 3)( x 1)
( x 2)2
203
critical points at x 1, 3
f | )( | , increasing on (, 1) and (3, ), decreasing on (1, 2) and (2, 3),
1
3
2
discontinuous at x 2
(b) a local maximum is f (1) 2 at x 1, a local minimum is f (3) 6 at x 3, no absolute extrema
36. (a) f ( x )
2
2
3
2 2
x3 f ( x ) 3 x (3 x 1) x (6 x ) 3 x ( x 1) a critical point at x 0 f | ,
2
2
2
3x 1
(3 x 1)
(3 x 2 1)2
0
increasing on (, 0) (0, ), and never decreasing
(b) no local extrema, no absolute extrema
37. (a) f ( x) x1/3 ( x 8) x 4/3 8 x1/3 f ( x) 43 x1/3 83 x 2/3
4( x 2)
3 x 2/3
critical points at x 0, 2
f | )( , increasing on (2, 0) (0, ), decreasing on (, 2)
2
0
(b) no local maximum, a local minimum is f (2) 6 3 2 7.56 at x 2, no absolute maximum; absolute
minimum is 6 3 2 at x 2
38. (a) g ( x) x 2/3 ( x 5) x5/3 5 x 2/3 g ( x) 53 x 2/3 10
x 1/3
3
5( x 2)
5( x 2)
critical points at
3
33 x
3 x
x 2 and x 0 g | )( , increasing on (, 2) and (0, ), decreasing on (2, 0)
2
0
3
(b) local maximum is g (2) 3 4 4.762 at x 2, a local minimum is g (0) 0 at x 0, no absolute
extrema
39. (a) h( x) x1/3 ( x 2 4) x7/3 4 x1/3 h( x) 73 x 4/3 34 x 2/3
x 0, 2 h
7
, 0 and 0,
2
7
2
7
3 3 x2
0
2/ 7
7
3
(b) local maximum is h 2 247/62 3.12 at x 2 , the local minimum is h
7
absolute extrema
, , decreasing on
)( | , increasing on , 2 and
|
2/ 7
7 x 2 7 x 2 critical points at
7
7
40. (a) k ( x) x 2/3 ( x 2 4) x8/3 4 x 2/3 k ( x) 83 x5/3 83 x 1/3
2
7
2
7
24 3 2
3.12, no
77/ 6
8( x 1)( x 1)
critical points at x 0, 1
33 x
k | )( | , increasing on ( 1, 0) and (1, ), decreasing on ( , 1) and (0, 1)
1
0
1
(b) local maximum is k (0) 0 at x 0, local minima are k ( 1) 3 at x 1, no absolute maximum;
absolute minimum is 3 at x 1
41. (a) f ( x) 2 x x 2 f ( x) 2 2 x a critical point at x 1 f | ] and f (1) 1 and f (2) 0
a local maximum is 1 at x 1, a local minimum is 0 at x 2.
(b) There is an absolute maximum of 1 at x 1; no absolute minimum.
(c)
Copyright 2016 Pearson Education, Ltd.
1
2
204
Chapter 4 Applications of Derivatives
42. (a) f ( x) ( x 1) 2 f ( x) 2( x 1) a critical point at x 1 f | ] and
1
0
f ( 1) 0, f (0) 1 a local maximum is 1 at x 0, a local minimum is 0 at x 1
(b) no absolute maximum; absolute minimum is 0 at x 1
(c)
43. (a) g ( x) x 2 4 x 4 g ( x) 2 x 4 2( x 2) a critical point at x 2 g [ | and
1
2
g (1) 1, g (2) 0 a local maximum is 1 at x 1, a local minimum is g (2) 0 at x 2
(b) no absolute maximum; absolute minimum is 0 at x 2
(c)
44. (a) g ( x) x 2 6 x 9 g ( x) 2 x 6 2( x 3) a critical point at x 3 g [ | and
4
g (4) 1, g (3) 0 a local maximum is 0 at x 3, a local minimum is 1 at x 4
(b) absolute maximum is 0 at x 3; no absolute minimum
(c)
3
45. (a) f (t ) 12t t 3 f (t ) 12 3t 2 3(2 t )(2 t ) critical points at t 2 f [ | |
3
2
2
and f (3) 9, f (2) 16, f (2) 16 local maxima are 9 at t 3 and 16 at t 2, a local minimum
is 16 at t 2
(b) absolute maximum is 16 at t 2; no absolute minimum
(c)
Copyright 2016 Pearson Education, Ltd.
Section 4.3 Monotonic Functions and the First Derivative Test
205
46. (a) f (t ) t 3 3t 2 f (t ) 3t 2 6t 3t (t 2) critical points at t 0 and t 2 f | | ]
0
2
3
and f (0) 0, f (2) 4, f (3) 0 a local maximum is 0 at t 0 and t 3, a local minimum is 4 at t 2
(b) absolute maximum is 0 at t 0, 3; no absolute minimum
(c)
3
47. (a) h( x) x3 2 x 2 4 x h( x) x 2 4 x 4 ( x 2) 2 a critical point at x 2 h [ | and
h(0) 0 no local maximum, a local minimum is 0 at x 0
(b) no absolute maximum; absolute minimum is 0 at x 0
(c)
0
2
48. (a) k ( x) x3 3 x 2 3 x 1 k ( x) 3 x 2 6 x 3 3( x 1) 2 a critical point at x 1 k | ]
1
and k (1) 0, k (0) 1 a local maximum is 1 at x 0, no local minimum
(b) absolute maximum is 1 at x 0; no absolute minimum
(c)
49. (a) f ( x) 25 x 2 f ( x)
x
25 x 2
0
critical points at x 0, x 5, and x 5 f ( | ),
5
0
f (5) 0, f (0) 5, f (5) 0 local maximum is 5 at x 0; local minimum of 0 at x 5 and x 5
(b) absolute maximum is 5 at x 0; absolute minimum of 0 at x 5 and x 5
(c)
Copyright 2016 Pearson Education, Ltd.
5
206
Chapter 4 Applications of Derivatives
50. (a) f ( x) x 2 2 x 3,3 x f ( x)
2 x 2
x 2 2 x 3
only critical point in 3 x is at x 3
f [ , f (3) 0 local minimum of 0 at x 3, no local maximum
3
(b) absolute minimum of 0 at x 3, no absolute maximum
(c)
2
51. (a) g ( x) x2 2 , 0 x 1 g ( x) x 2 4 x21 only critical point in 0 x 1 is x 2 3 0.268
x 1
( x 1)
4 336 1.866 local minimum of 4 336 at x 2 3, local
g [ | ), g 2 3
0
0.268
1
maximum at x 0.
(b) absolute minimum of
(c)
52. (a) g ( x)
3
at x 2
4 3 6
3, no absolute maximum
8x
x 2 , 2 x 1 g ( x)
only critical point in 2 x 1 is x 0
(4 x 2 ) 2
4 x 2
g ( | ], g (0) 0 local minimum of 0 at x 0, local maximum of 13 at x 1.
2
0
1
(b) absolute minimum of 0 at x 0, no absolute maximum
(c)
53. (a) f ( x) sin 2 x, 0 x f ( x) 2 cos 2 x, f ( x) 0 cos 2 x 0 critical points are x 4 and x 34
f [ | | ] , f (0) 0, f 4 1, f
0
4
3
4
34 1, f ( ) 0 local maxima are 1 at x 4
and 0 at x , and local minima are 1 at x 34 and 0 at x 0.
(b) The graph of f rises when f 0, falls when f 0, and has local
extreme values where f 0. The function f has a local minimum
value at x 0 and x 34 , where the values
of f change from negative to positive. The function f has a local
maximum value at x and x 4 , where the values of f change
from positive to negative.
Copyright 2016 Pearson Education, Ltd.
Section 4.3 Monotonic Functions and the First Derivative Test
207
54. (a) f ( x ) sin x cos x, 0 x 2 f ( x) cos x sin x, f ( x) 0 tan x 1 critical points are x 34
and x 74 f [ | | ] , f (0) 1, f
0
3
4
7
4
2
34 2, f 74 2, f (2 ) 1 local
maxima are 2 at x 34 and 1 at x 2 , and local minima are 2 at x 74 and 1 at x 0.
(b) The graph of f rises when f 0, falls when
f 0, and has local extreme values where
f 0. The function f has a local minimum
value at x 0 and x 74 , where the values
of f change from negative to positive. The
function f has a local maximum value at x 2
and x 34 , where the values of f change from
positive to negative.
55. (a) f ( x) 3 cos x sin x, 0 x 2 f ( x) 3 sin x cos x, f ( x) 0 tan x 1 critical points
0
7
6
6
2
3
2, f (2 ) 3
are x 6 and x 76 f [ | | ] , f (0) 3, f 6 2, f 76
local maxima are 2 at x 6 and 3 at x 2 , and local minima are 2 at x 76 and 3 at x 0.
(b) The graph of f rises when f 0, falls when
f 0, and has local extreme values where
f 0. The function f has a local minimum
value at x 0 and x 76 , where the values
of f change from negative to positive. The
function f has a local maximum value at
x 2 and x 6 , where the values of f
change from positive to negative.
56. (a) f ( x) 2 x tan x, 2 x 2 f ( x) 2 sec 2 x, f ( x) 0 sec2 x 2 critical points are
x 4 and x 4 f ( | | ) , f 4 2 1, f 4 1 2 local maximum
2
4
4
2
is 2 1 at x 4 , and local minimum is 1 2 at x 4 .
(b) The graph of f rises when f 0, falls when
f 0, and has local extreme values where
f 0. The function f has a local minimum
value at x 4 , where the values of f change
from negative to positive. The function f has a
local maximum value at x 4 , where the
values of f change from positive to negative.
57. (a) f ( x) 2x 2sin 2x f ( x) 12 cos 2x , f ( x) 0 cos 2x 12 a critical point at x 23
f [ | ] and f (0) 0, f
0
2 /3
2
3
2
3
3, f (2 ) local maxima are
0 at x 0 and at x 2 , a local minimum is 3 3 at x 23
Copyright 2016 Pearson Education, Ltd.
208
Chapter 4 Applications of Derivatives
(b) The graph of f rises when f 0, falls when
f 0, and has a local minimum value at the
point where f changes from negative to
positive.
58. (a) f ( x ) 2 cos x cos 2 x f ( x) 2 sin x 2 cos x sin x 2(sin x )(1 cos x) critical points at
x , 0, f [ | ] and f ( ) 1, f (0) 3, f ( ) 1 a local maximum is
0
1 at x , a local minimum is 3 at x 0
(b) The graph of f rises when f 0, falls when
f 0, and has local extreme values where
f 0. The function f has a local minimum
value at x 0, where the values of f change
from negative to positive.
59. (a) f ( x) csc 2 x 2 cot x f ( x) 2(csc x)( csc x)(cot x) 2( csc2 x) 2(csc2 x) (cot x 1) a critical
point at x 4 f ( | ) and f 4 0 no local maximum, a local minimum is 0 at x 4
0
/4
(b) The graph of f rises when f 0, falls when
f 0, and has a local minimum value at the
point where f 0 and the values of f change
from negative to positive. The graph of f
steepens as f ( x) .
60. (a) f ( x) sec2 x 2 tan x f ( x) 2(sec x)(sec x)(tan x) 2sec2 x (2sec 2 x) (tan x 1) a critical point
at x 4 f ( | ) and f 4 0 no local maximum, a local minimum is 0 at x 4
/2
/4
/2
(b) The graph of f rises when f 0, falls when
f 0, and has a local minimum value where
f 0 and the values of f change from
negative to positive.
Copyright 2016 Pearson Education, Ltd.
Section 4.3 Monotonic Functions and the First Derivative Test
209
61. h( ) 3cos 2 h( ) 32 sin 2 h [ ] , (0, 3) and (2 , 3) a local maximum is 3 at 0,
2
0
a local minimum is 3 at 2
62. h( ) 5sin 2 h( ) 52 cos 2 h [ ], (0, 0) and ( , 5) a local maximum is 5 at , a local
minimum is 0 at 0
63. (a)
(b)
0
(c)
64. (a)
(b)
(c)
(d)
65. (a)
(b)
66. (a)
(b)
Copyright 2016 Pearson Education, Ltd.
(d)
210
Chapter 4 Applications of Derivatives
67. The function f ( x) x sin 1x has an infinite number of local maxima and minima on its domain, which is
( , 0) (0, ). The function sin x has the following properties: a) it is continuous on (, ); b) it is
periodic; and c) its range is [1, 1]. Also, for a 0, the function 1x has a range of (, a ] [a, )
on a1 , 0 0, a1 . In particular, if a 1, then 1x 1 or 1x 1 when x is in [ 1, 0) (0, 1]. This means sin 1x
1 , 3 , 5 ,.
2
2
2
x
takes on the values of 1 and 1 infinitely many times on [ 1, 0) (0, 1], namely at
x 2 , 32 , 52 , . Thus sin 1x has infinitely many local maxima and minima in [ 1, 0) (0, 1]. On the
1 and since x 0 we have x x sin 1x x. On the interval
[ 1, 0), 1 sin 1x 1 and since x 0 we have x x sin 1x x. Thus f ( x) is bounded by the lines
y x and y x. Since sin 1x oscillates between 1 and 1 infinitely many times on [ 1, 0) (0, 1] then f will
1
x
interval (0, 1], 1 sin
oscillate between y x and y x infinitely many times. Thus f has infinitely many local maxima and minima.
We can see from the graph (and verify later in Chapter 7) that lim x sin 1x 1 and lim x sin 1x 1. The
x
x
graph of f does not have any absolute maxima, but it does have two absolute minima.
b 4a4ac , a parabola whose
4a
vertex is at x 2ba . Thus when a 0, f is increasing on 2ab , and decreasing on , 2ab ; when a 0, f is
increasing on , 2ab and decreasing on 2ab , . Also note that f ( x) 2ax b 2a x 2ba for
2
2
68. f ( x) ax 2 bx c a x 2 ba x c a x 2 ba x b 2 4ba c a x 2ba
a 0, f
| ; for a 0, f
b /2 a
2
2
| .
b /2 a
69. f ( x) ax 2 bx f ( x) 2a x b, f (1) 2 a b 2, f (1) 0 2a b 0 a 2, b 4
f ( x) 2 x 2 4 x
70. f ( x) ax3 bx 2 cx d f ( x) 3ax 2 2bx c, f (0) 0 d 0, f (1) 1 a b c d 1,
f (0) 0 c 0, f (1) 0 3a 2b c 0 a 2, b 3, c 0, d 0 f ( x) 2 x3 3 x 2
4.4
CONCAVITY AND CURVE SKETCHING
3
2
1. y x3 x2 2 x 13 y x 2 x 2 ( x 2)( x 1) y 2 x 1 2 x 12 . The graph is rising on
and concave down on , 12 . Consequently,
a local maximum is 32 at x 1, a local minimum is 3 at x 2, and 12 , 34 is a point of inflection.
(, 1) and (2, ), falling on (1, 2), concave up on
1,
2
4
2. y x4 2 x 2 4 y x3 4 x x( x 2 4) x( x 2)( x 2) y 3 x 2 4
3x 2 3x 2 . The graph
, and
is rising on (2, 0) and (2, ), falling on (, 2) and (0, 2), concave up on , 2 and
3
2
3
concave down on 2 , 2 . Consequently, a local maximum is 4 at x 0, local minima are 0 at x 2, and
2 , 16
3 9
and
2 , 16
3 9
3
3
are points of inflection.
Copyright 2016 Pearson Education, Ltd.
Section 4.4 Concavity and Curve Sketching
3. y 34 ( x 2 1)2/3 y
211
34 23 ( x2 1)1/3 (2 x) x( x2 1)1/3 , y )1 ( 0| 1)( the
graph is rising on (1, 0) and (1, ), falling on (, 1) and (0, 1) a local maximum is 34 at x 0, local
minima are 0 at x 1; y ( x 2 1)1/3 ( x) 13 ( x 2 1)4/3 (2 x)
x 2 3
3 3 ( x 2 1)4
,
y | ) ( )( | the graph is concave up on , 3 and
3
1
1
3
3
down on 3, 3 points of inflection at 3, 3 44
3, , concave
9 x1/3 ( x 2 7) y 3 x 2/3 ( x 2 7) 9 x1/3 (2 x) 3 x 2/3 ( x 2 1), y | )( | the
4. y 14
14
14
2
1
1
0
graph is rising on (, 1) and (1, ), falling on (1, 1) a local maximum is 27
at x 1, a local minimum is
7
27
at x 1; y x 5/3 ( x 2 1) 3 x1/3 2 x1/3 x 5/3 x 5/3 (2 x 2 1), y )( the graph is
7
concave up on (0, ), concave down on (, 0) a point of inflection at (0, 0).
0
5. y x sin 2 x y 1 2 cos 2 x, y [ | | ] the graph is rising on 3 , 3 ,
2 /3
/3
/3
2 /3
falling on 23 , 3 and 3 , 23 local maxima are 23 23 at x 23 and 3 23 at x 3 , local minima
are 3 at x and 2 3 at x 2 ; y 4sin 2 x, y
3
2
3
3
2
the graph is concave up on 2 , 0 and
2
2 2
at , , (0, 0), and ,
2
[ | | | ]
3
2 /3
/2
, 2 , concave down on 2 , and
2 3
3
2
0
/2
2 /3
0, 2 points of inflection
6. y tan x 4 x y sec2 x 4, y ( | | ) the graph is rising on 2 , 3 and
/3
/2
π , π , falling on π , π a local maximum is 3 4 at x , a local minimum is
3 2
/3
/2
3 3
3
3
3 43 at x 3 ;
y 2(sec x)(sec x)(tan x) 2(sec2 x)(tan x), y ( | ) the graph is concave up on 0, 2 ,
/2
0
concave down on 2 , 0 a point of inflection at (0, 0)
/2
7. If x 0, sin x sin x and if x 0, sin x sin( x)
sin x. From the sketch the graph is rising on
32 , 2 , 0, 2 and 32 , 2 , falling on
3
3
2 , 2 , 2 , 0 and 2 , 2 ; local minima
are 1 at x 32 and 0 at x 0; local maxima are
1 at x 2 and 0 at x 2 ; concave up on
(2 , ) and ( , 2 ), and concave down on
( , 0) and (0, ) points of inflection are
( , 0) and ( , 0)
8. y 2 cos x 2 x y 2 sin x 2, y [
34 , 4 and 54 , 32 , falling on , 34 and
2
2
4
|
| | ] rising on
3 /4
4 , 54
/4
5 /4
3 /2
local maxima are 2 2 at x ,
at x and 3 2 at x 3 , and local minima are 2 3 2 at x 3 and
4
2
2
4
Copyright 2016 Pearson Education, Ltd.
4
2 54 2
212
Chapter 4 Applications of Derivatives
at x 54 ; y 2 cos x, y [ | | ] concave up on , 2 and 2 , 32 ,
/2
/2
concave down on 2 , 2 points of inflection at
3 /2
2 , 22
and ,
2
9. When y x 2 4 x 3, then y 2 x 4 2( x 2)
and y 2. The curve rises on (2, ) and falls on
(, 2). At x 2 there is a minimum. Since y 0,
the curve is concave up for all x.
10. When y 6 2 x x 2 , then y 2 2 x 2(1 x)
and y 2. The curve rises on (, 1) and falls on
(1, ). At x 1 there is a maximum. Since y 0,
the curve is concave down for all x.
11. When y x3 3 x 3, then y 3 x 2 3
3( x 1)( x 1) and y 6 x. The curve rises on
( , 1) (1, ) and falls on ( 1, 1). At x 1 there is
a local maximum and at x 1 a local minimum. The
curve is concave down on (, 0) and concave up on
(0, ). There is a point on inflection at x 0.
12. When y x (6 2 x)2 , then
y 4 x(6 2 x) (6 2 x) 2 12(3 x)(1 x) and
y 12(3 x) 12(1 x) 24( x 2). The curve
rises on ( , 1) (3, ) and falls on (1, 3). The curve
is concave down on ( , 2) and concave up on
(2, ). At x 2 there is a point of inflection.
13. When y 2 x3 6 x 2 3, then y 6 x 2 12 x
6 x( x 2) and y 12 x 12 12( x 1). The
curve rises on (0, 2) and falls on (, 0) and (2, ).
At x 0 there is a local minimum and at x 2 a local
maximum. The curve is concave up on ( , 1) and
concave down on (1, ). At x 1 there is a point of
inflection.
Copyright 2016 Pearson Education, Ltd.
2
2
Section 4.4 Concavity and Curve Sketching
14. When y 1 9 x 6 x 2 x3 , then y 9 12 x 3 x 2
3( x 3)( x 1) and y 12 6 x 6( x 2). The
curve rises on ( 3, 1) and falls on ( , 3) and
( 1, ). At x 1 there is a local maximum and at
x 3 a local minimum. The curve is concave up on
(, 2) and concave down on (2, ). At x 2
there is a point of inflection.
15. When y ( x 2)3 1, then y 3( x 2)2 and
y 6( x 2). The curve never falls and there are no
local extrema. The curve is concave down on ( , 2)
and concave up on (2, ). At x 2 there is a point of
inflection.
16. When y 1 ( x 1)3 , then y 3( x 1)2 and
y 6( x 1). The curve never rises and there are no
local extrema. The curve is concave up on ( , 1)
and concave down on ( 1, ). At x 1 there is a
point of inflection.
17. When y x 4 2 x 2 , then y 4 x3 4 x
4 x( x 1)( x 1) and y 12 x 2 4
12 x 1
3
x . The curve rises on (1, 0)
1
3
and (1, ) and falls on (, 1) and (0, 1). At x 1
there are local minima and at x 0 a local maximum.
,
The curve is concave up on , 1 and
3
and concave down on 1 , 1 . At x
points of inflection.
3
3
1
3
1
there are
3
18. When y x 4 6 x 2 4, then y 4 x3 12 x
4 x x 3 x 3 and y 12 x 2 12
and 0, 3 , and falls on 3, 0 and 3, . At
12( x 1)( x 1). The curve rises on , 3
x 3 there are local maxima and at x 0 a local
minimum. The curve is concave up on ( 1,1) and
concave down on ( , 1) and (1, ). At x 1 there
are points of inflection.
Copyright 2016 Pearson Education, Ltd.
213
214
Chapter 4 Applications of Derivatives
19. When y 4 x3 x 4 , then
y 12 x 2 4 x3 4 x 2 (3 x) and y 24 x 12 x 2
12 x(2 x). The curve rises on (, 3) and falls on
(3, ). At x 3 there is a local maximum, but there is
no local minimum. The graph is concave up on (0, 2)
and concave down on (, 0) and (2, ). There are
inflection points at x 0 and x 2.
20. When y x 4 2 x3 , then y 4 x3 6 x 2 2 x 2 (2 x 3)
and y 12 x 2 12 x 12 x( x 1). The curve rises on
32 , and falls on , 32 . There is a local
minimum at x 32 , but no local maximum. The
curve is concave up on (, 1) and (0, ), and
concave down on (1, 0). At x 1 and x 0 there
are points of inflection.
21. When y x5 5 x 4 , then
y 5 x 4 20 x3 5 x3 ( x 4) and y 20 x3 60 x 2
20 x 2 ( x 3). The curve rises on ( , 0) and (4, ),
and falls on (0, 4). There is a local maximum at x 0,
and a local minimum at x 4. The curve is concave
down on (, 3) and concave up on (3, ). At x 3
there is a point of inflection.
4
22. When y x 2x 5 , then
2
3
and y 3 2x 5 12 52x 5 2x 5 52
2
5 2x 5 ( x 4). The curve is rising on (, 2) and
4
3
3
y 2x 5 x(4) 2x 5 12 2x 5 52x 5 ,
(10, ), and falling on (2, 10). There is a local
maximum at x 2 and a local minimum at x 10.
The curve is concave down on ( , 4) and concave
up on (4, ). At x 4 there is a point of inflection.
23. When y x sin x, then y 1 cos x and y sin x.
The curve rises on (0, 2 ). At x 0 there is a local
and absolute minimum and at x 2 there is a local
and absolute maximum. The curve is concave down
on (0, ) and concave up on ( , 2 ). At x there is
a point of inflection.
Copyright 2016 Pearson Education, Ltd.
Section 4.4 Concavity and Curve Sketching
24. When y x sin x, then y 1 cos x and y sin x.
The curve rises on (0, 2 ). At x 0 there is a local
and absolute minimum and at x 2 there is a local
and absolute maximum. The curve is concave up on
(0, ) and concave down on ( , 2 ). At x there is
a point of inflection.
25. When y 3x 2 cos x, then y 3 2sin x and
y 2 cos x. The curve is increasing on 0, 43 and
, and decreasing on
5 , 2
3
4 , 5
3
3
. At x 0 there
is a local and absolute minimum, at x 43 there is a
local maximum, at x 53 there is a local minimum,
and at x 2 there is a local and absolute maximum.
The curve is concave up on 0, 2 and 32 , 2 , and
2 2
2
is concave down on , 3 . At x and x 3
2
there are points of inflection.
26. When y 43 x tan x, then y 43 sec2 x and
y 2sec 2 x tan x. The curve is increasing on
6 , 6 , and decreasing on 2 , 6 and 6 , 2 .
At x there is a local minimum, at x there is
6
6
a local maximum, there are no absolute maxima or
absolute minima. The curve is concave up on
2 , 0 , and is concave down on 0, 2 . At x 0
there is a point of inflection.
27. When y sin x cos x, then y sin 2 x cos 2 x
cos 2 x and y 2sin 2 x. The curve is increasing
on 0, 4 and 34 , , and decreasing on 4 , 34 . At
x 0 there is a local minimum, at x there is
4
a local and absolute maximum, at x 34 there is a
local and absolute minimum, and at x there is
a local maximum. The curve is concave down on
0, 2 , and is concave up on 2 , . At x 2 there is
a point of inflection.
28. When y cos x 3 sin x, then y sin x 3 cos x
and y cos x 3 sin x. The curve is increasing on
0, 3 and 43 , 2 , and decreasing on 3 , 43 . At
x 0 there is a local minimum, at x 3 there is
a local and absolute maximum, at x 43 there is a
local and absolute minimum, and at x 2 there is
a local maximum. The curve is concave down on
Copyright 2016 Pearson Education, Ltd.
215
216
Chapter 4 Applications of Derivatives
0, 56 and 116 , 2 , and is concave up on
56 , 116 . At x 56 and x 116 there are points
of inflection.
4 x 9/5 .
29. When y x1/5 , then y 15 x 4/5 and y 25
The curve rises on (, ) and there are no extrema.
The curve is concave up on ( , 0) and concave
down on (0, ). At x 0 there is a point of inflection.
6 x 8/5 .
30. When y x 2/5 , then y 52 x 3/5 and y 25
The curve is rising on (0, ) and falling on (, 0).
At x 0 there is a local and absolute minimum.
There is no local or absolute maximum. The curve is
concave down on ( , 0) and (0, ). There are no
points of inflection, but a cusp exists at x 0.
31. When y
y
x
x 2 1
, then y
1
and
( x 2 1)3/ 2
3 x . The curve is increasing on ( , ).
( x 2 1)5/ 2
There are no local or absolute extrema. The curve is
concave up on ( , 0) and concave down on (0, ).
At x 0 there is a point of inflection.
2
32. When y 21xx1 , then y
( x 2)
(2 x 1) 2 1 x 2
and
3
2
y 4 x 312 x 2 3/7 2 . The curve is decreasing on
(2 x 1) (1 x )
1, 12 and 12 , 1 . There are no absolute extrema,
0.92, 12
there is a local maximum at x 1 and a local
minimum at x 1. The curve is concave up on
(1, 0.92) and 12 , 0.69 , and concave down on
and (0.69, 1). At x 0.92 and x 0.69
there are points of inflection.
33. When y 2 x 3 x 2/3 , then y 2 2 x 1/3 and
y 23 x 4/3 . The curve is rising on (, 0) and (1, ),
and falling on (0, 1). There is a local maximum at
x 0 and a local minimum at x 1. The curve is
concave up on ( , 0) and (0, ). There are no points
of inflection, but a cusp exists at x 0.
Copyright 2016 Pearson Education, Ltd.
Section 4.4 Concavity and Curve Sketching
34. When y 5 x 2/5 2 x, then y 2 x 3/5 2
2 x 3/5 1 and y 65 x 8/5 . The curve is rising
on (0, 1) and falling on ( , 0) and (1, ). There is
a local minimum at x 0 and a local maximum at
x 1. The curve is concave down on (, 0) and
(0, ). There are no points of inflection, but a cusp
exists at x 0.
35. When y x 2/3 52 x 52 x 2/3 x5/3 , then
y 53 x 1/3 53 x 2/3 53 x 1/3 (1 x) and
y 95 x 4/3 10
x 1/3 95 x 4/3 (1 2 x). The curve
9
is rising on (0, 1) and falling on (, 0) and (1, ).
There is a local minimum at x 0 and a local
maximum at x 1. The curve is concave up on
, 12 and concave down on 12 , 0 and (0, ).
There is a point of inflection at x 12 and a cusp
at x 0.
36. When y x 2/3 ( x 5) x5/3 5 x 2/3 , then
y 53 x 2/3 10
x 1/3 53 x 1/3 ( x 2) and
3
y 10
x 1/3 10
x 4/3 10
x 4/3 ( x 1). The curve
9
9
9
is rising on ( , 0) and (2, ), and falling on (0, 2).
There is a local minimum at x 2 and a local
maximum at x 0. The curve is concave up on
(1, 0) and (0, ), and concave down on (, 1).
There is a point of inflection at x 1 and a cusp
at x 0.
37. When y x 8 x 2 x(8 x 2 )1/2 , then
y (8 x 2 )1/2 ( x ) 12 (8 x 2 ) 1/2 ( 2 x)
2 1/2
(8 x )
(8 2 x 2 )
2(2 x )(2 x )
2 2x 2 2x
3
and
1
y 12 (8 x 2 ) 2 (2 x)(8 2 x 2 ) (8 x 2 ) 2 ( 4 x)
2 x ( x 2 12)
(8 x 2 )3
. The curve is rising on (2, 2), and falling
on 2 2, 2 and 2, 2 2 . There are local minima
x 2 and x 2 2, and local maxima at x 2 2
and x 2. The curve is concave up on 2 2, 0 and
concave down on 0, 2 2 . There is
a point of inflection at x 0.
Copyright 2016 Pearson Education, Ltd.
217
218
Chapter 4 Applications of Derivatives
32 (2 x2 )1/2 (2 x)
3 x 2 x 2 3x 2 x 2 x and
y (3)(2 x 2 )1/2 (3x) 12 (2 x 2 )1/2 (2 x )
6(1 x )(1 x )
. The curve is rising on 2, 0 and
2 x 2 x
falling on 0, 2 . There is a local maximum at x 0,
38. When y (2 x 2 )3/2 , then y
and local minima at x 2. The curve is concave
down on ( 1, 1) and concave up on 2, 1 and
1, 2 . There are points of inflection at x 1.
x
39. When y 16 x 2 , then y
y
16 x 2
and
16
. The curve is rising on (4, 0) and
(16 x 2 )3/ 2
falling on (0, 4). There is a local and absolute
maximum at x 0 and local and absolute minima at
x 4 and x 4. The curve is concave down on
(4, 4). There are no points of inflection.
3
40. When y x 2 2x , then y 2 x 22 2 x 2 2 and
x
3
x
y 2 43 2 x 3 4 . The curve is falling on (, 0) and
x
x
(0, 1), and rising on (1, ). There is a local minimum at
x 1. There are no absolute maxima or absolute minima.
The curve is concave up on , 3 2 and (0, ), and
concave down on 3 2, 0 . There is a point of inflection
3
at x 2.
2
41. When y xx 23 , then y
and y
2 x ( x 2) ( x 2 3)(1)
( x 2) 2
(2 x 4)( x 2) 2 ( x 2 4 x 3)2( x 2)
( x 2) 4
( x 3)( x 1)
( x 2)2
2 . The curve
( x 2)3
is rising on ( , 1) and (3, ), and falling on (1, 2) and
(2, 3). There is a local maximum at x 1 and a local
minimum at x 3. The curve is concave down on
(, 2) and concave up on (2, ). There are no points
of inflection because x 2 is not in the domain.
3
42. When y x3 1, then y
x2
and y 3 2 x 5/3 .
( x 1)
( x 1) 2/3
3
The curve is rising on (, 1), (1, 0), and (0, ). There
are no local or absolute extrema. The curve is concave up
on ( , 1) and (0, ), and concave down on ( 1, 0).
There are points of inflection at x 1 and x 0.
Copyright 2016 Pearson Education, Ltd.
Section 4.4 Concavity and Curve Sketching
43. When y
2
2
8 x , then y 8( x 4) and y 16 x ( x 12) .
2
2
2
2
x 4
( x 4)
( x 4)3
The curve is falling on (, 2) and (2, ), and is rising
on ( 2, 2). There is a local and absolute minimum at
x 2, and a local and absolute maximum at x 2. The
curve is concave down on , 2 3 and 0, 2 3 , and
concave up on 2 3, 0 and 2 3, . There are points
of inflection at x 2 3, x 0, and x 2 3. y 0 is a
horizontal asymptote.
44. When y
2
4
5 , then y 20 x3 and y 100 x ( x 3) .
x 5
( x 4 5)2
( x 4 5)3
4
The curve is rising on (, 0), and is falling on (0, ).
There is a local and absolute maximum at x 0, and there
is no local or absolute minimum. The curve is concave up
on , 4 3 and 4 3, , and concave down on 4 3, 0
4
4
and 0, 3 . There are points of inflection at x 3 and
4
x 3. There is a horizontal asymptote of y 0.
x 2 1, | x | 1
2 x, | x | 1
45. When y | x 2 1|
, then y
2
2 x, | x | 1
1 x , | x | 1
2, | x | 1
and y
. The curve rises on (1, 0) and (1, )
2, | x | 1
and falls on (, 1) and (0, 1). There is a local maximum
at x 0 and local minima at x 1. The curve is concave
up on ( , 1) and (1, ), and concave down on ( 1, 1).
There are no points of inflection because y is not
differentiable at x 1 (so there is no tangent line at
those points).
x 2 2 x, x 0
46. When y | x 2 2 x | 2 x x 2 , 0 x 2,
2
x 2 x, x 2
2 x 2, x 0
2, x 0
then y 2 2 x, 0 x 2, and y 2, 0 x 2 .
2 x 2, x 2
2, x 2
The curve is rising on (0, 1) and (2, ), and falling on
( , 0) and (1, 2). There is a local maximum at x 1 and
local minima at x 0 and x 2. The curve is concave up
on (, 0) and (2, ), and concave down on (0, 2). There
are no points of inflection because y is not differentiable
at x 0 and x 2 (so there is no tangent at those points).
1 , x0
x , x 0
2 x
, then y
47. When y | x|
1
x , x 0
2 x , x 0
Copyright 2016 Pearson Education, Ltd.
219
220
Chapter 4 Applications of Derivatives
x 3/ 2 ,
x0
4
and y
.
3/ 2
( x ) , x 0
4
Since lim y and lim y there is a cusp at
x 0
x 0
x 0. There is a local minimum at x 0, but no local
maximum. The curve is concave down on (, 0) and
(0, ). There are no points of inflection.
x 4, x 4
48. When y | x 4 |
, then
4 x, x 4
( x 4)3/ 2
1 ,x4
,x4
2 x4
4
and y
.
y
3/ 2
1
2 4 x , x 4
(4 x )
,
x
4
4
Since lim y and lim y there is a cusp at
x 4
x 4
x 4. There is a local minimum at x 4, but no local
maximum. The curve is concave down on (, 4) and
(4, ). There are no points of inflection.
49. y 2 x x 2 (1 x)(2 x ), y | |
1
2
rising on ( 1, 2), falling on ( , 1) and (2, )
there is a local maximum at x 2 and a local
minimum at x 1; y 1 2 x, y |
1/2
concave up on , 12 , concave down on 12 ,
a point of inflection at x 12
50. y x 2 x 6 ( x 3)( x 2), y | |
2
3
rising on (, 2) and (3, ), falling on (2, 3)
there is a local maximum at x 2 and a local minimum at
x 3; y 2 x 1, y |
1/2
concave up on 12 , , concave down on , 12
a point of inflection at x 12
51. y x( x 3)2 , y | | rising on (0, ), falling
0
3
on ( , 0) no local maximum, but there is a local minimum at
x 0; y ( x 3)2 x(2) ( x 3) 3( x 3)( x 1), y
| | concave up on (, 1) and (3, ), concave
1
3
down on (1, 3) points of inflection at x 1 and x 3
52. y x 2 (2 x), y | | rising on (, 2), falling
0
2
on (2, ) there is a local maximum at x 2, but no local
minimum; y 2 x(2 x) x 2 (1) x(4 3 x), y
| | concave up on 0, 43 , concave down on , 0
0
4/3
and 43 , points of inflection at x 0 and x 43
Copyright 2016 Pearson Education, Ltd.
Section 4.4 Concavity and Curve Sketching
53. y x( x 2 12) x x 2 3 x 2 3 ,
y | | | rising on 2 3, 0 and
2 3
0
2 3
2 3, , falling on , 2 3 and 0, 2 3 a local
maximum at x 0, local minima at
x 2 3; y 1 ( x 2 12) x(2 x ) 3( x 2)( x 2),
y | | concave up on ( , 2) and (2, ),
2
2
concave down on ( 2, 2) points of inflection
at x 2
54. y ( x 1)2 (2 x 3), y | | rising on
3/2
1
32 , , falling on , 32 no local maximum,
a local minimum at x 32 ;
y 2( x 1)(2 x 3) ( x 1)2 (2) 2( x 1)(3 x 2),
y | | concave up on , 23 and (1, ),
2/3
1
concave down on 23 , 1 points of inflection at x 23 and
x 1
55. y (8 x 5 x 2 )(4 x) 2 x(8 5 x)(4 x) 2 ,
y | | | rising on 0, 85 , falling on
0
8/5
4
( , 0) and 85 , a local maximum at x 85 ,
a local minimum at x 0;
y (8 10 x)(4 x) 2 (8 x 5 x 2 )(2)(4 x)(1)
4(4 x)(5 x 2 16 x 8), y |
8 2 6
5
concave
| |
8 2 6
5
4
up on , 8 25 6 and 8 25 6 , 4 , concave down on
8 2 6 8 2 6
, 5
5
and (4, ) points of inflection at x
8 2 6
and
5
x4
56. y ( x 2 2 x)( x 5)2 x( x 2)( x 5) 2 ,
y | | | rising on ( , 0) and (2, ),
0
2
5
falling on (0, 2) a local maximum at x 0,
a local minimum at x 2;
y (2 x 2)( x 5)2 2( x 2 2 x)( x 5)
2( x 5)(2 x 2 8 x 5), y | | |
4 6
2
4 6
2
5
concave up on 42 6 , 42 6 and (5, ), concave down on
, 42 6 and 42 6 , 5 points of inflection at x 42 6 and
x5
Copyright 2016 Pearson Education, Ltd.
221
222
Chapter 4 Applications of Derivatives
57. y sec2 x, y ( ) rising on 2 , 2 , never falling
/2
/2
no local extrema;
y 2(sec x)(sec x)(tan x) 2 (sec 2 x) (tan x),
y ( | ) concave up on 0, 2 , concave down
/2
/2
0
on 2 , 0 , 0 is a point of inflection.
58. y tan x, y ( | ) rising on 0, 2 , falling on
/2
/2
0
2 , 0 no local maximum, a local minimum at
x 0; y sec 2 x, y ( )
/2
/2
concave up on 2 , 2 no points of inflection
59. y cot 2 , y ( | ) rising on (0, ) , falling on
0
2
( , 2 ) a local maximum at , no
local minimum; y 12 csc2 2 , y ( ) never concave
2
0
up, concave down on (0, 2 ) no points of inflection
60. y csc2 2 , y ( ) rising on (0, 2 ) , never falling
0
no local extrema;
y 2 csc csc
2
2
2
cot 2 12
csc2 2 cot 2 , y ( | )
0
2
concave up on ( , 2 ), concave down on (0, )
a point of inflection at
61. y tan 2 1 (tan 1)(tan 1),
y ( | | ) rising on 2 , 4 and
/4
/4
/2
, , falling on , a local maximum at , a
/2
4 4
4 2
4
local minimum at 4 ; y 2 tan sec2 ,
y ( | ) concave up on 0, 2 , concave down
/2
/2
0
on 2 , 0 a point of inflection at 0
62. y 1 cot 2 (1 cot )(1 cot ),
y ( | | ) rising on 4 , 34 , falling on
/4
3 /4
0, 4 and 34 , a local maximum
at 3 , a local minimum at ;
0
4
4
y 2(cot )( csc2 ), y ( | )
0
/2
Copyright 2016 Pearson Education, Ltd.
Section 4.4 Concavity and Curve Sketching
concave up on 0, 2 , concave down on 2 ,
a point of infection at 2
63. y cos t , y [ | | ] rising on 0, 2 and
/2
3 /2
2
3 , 2 , falling on , 3 local maxima at t and t 2 ,
2
2 2
2
local minima at t 0 and t 3 ; y sin t , y [ | ]
0
2
0
2
concave up on ( , 2 ), concave down on (0, ) a point of
inflection at t
64. y sin t , y [ | ] rising on (0, ), falling on
0
2
( , 2 ) a local maximum at t ,
local minima at t 0 and t 2 ; y cos t ,
y [ | | ] concave up on 0, 2 and
0
/2
3 /2
2
32 , 2 , concave down on 2 , 32
points of inflection at t 2 and t 32
65. y ( x 1) 2/3 , y ) ( rising on ( , ), never
1
falling no local extrema; y 23 ( x 1) 5/3 , y ) (
1
concave up on ( , 1), concave down on ( 1, ) a point of
inflection and vertical tangent at x 1
66. y ( x 2)1/3 , y )( rising on (2, ), falling on
2
(, 2) no local maximum, but a local minimum at
x 2; y 13 ( x 2)4/3 , y )( concave down on
2
(, 2) and (2, ) no points of inflection, but there is a cusp at
x2
67. y x 2/3 ( x 1), y )( | rising on (1, ),
0
1
falling on (, 1) no local maximum, but
a local minimum at x 1; y 13 x 2/3 23 x 5/3
13 x 5/3 ( x 2), y | )( concave up on
2
0
(, 2) and (0, ), concave down on (2, 0) points of
inflection at x 2 and x 0, and a vertical tangent at x 0
68. y x 4/5 ( x 1), y | ) ( rising on (1, 0) and
1
0
(0, ), falling on ( , 1) no
local maximum, but a local minimum at x 1;
y 15 x 4/5 54 x 9/5 15 x 9/5 ( x 4), y )( |
0
4
concave up on ( , 0) and (4, ), concave down on (0, 4)
points of inflection at x 0 and x 4, and a vertical tangent
at x 0
Copyright 2016 Pearson Education, Ltd.
223
224
Chapter 4 Applications of Derivatives
2 x, x 0
69. y
, y | rising on (, ) no
2 x, x 0
0
2, x 0
local extrema; y
, y )( concave up
2, x 0
0
on (0, ), concave down on (, 0) a point of inflection at
x0
x 2 , x 0
70. y
, y | rising on (0, ), falling on
2
0
x , x 0
(, 0) no local maximum,
2 x, x 0
but a local minimum at x 0; y
,
2 x, x 0
y | concave up on (, )
0
no point of inflection
71. The graph of y f ( x) the graph of y f ( x) is concave up on
(0, ), concave down on (, 0) a point of inflection at x 0;
the graph of y f ( x) y | | the graph
y f ( x) has both a local maximum and a local minimum
72. The graph of y f ( x) y | the graph of
y f ( x) has a point of inflection, the graph of
y f ( x) y | | the graph of y f ( x) has
both a local maximum and a local minimum
73. The graph of y f ( x) y | |
the graph of y f ( x) has two points of inflection, the graph of
y f ( x) y | the graph of y f ( x) has a local
minimum
Copyright 2016 Pearson Education, Ltd.
Section 4.4 Concavity and Curve Sketching
74. The graph of y f ( x) y | the graph of
y f ( x) has a point of inflection; the graph of
y f ( x) y | | the graph of y f ( x) has
both a local maximum and a local minimum
2
75. y 2 x 2 x 1
x 1
Since 1 and 1 are roots of the denominator, the domain is
( , 1) ( 1, 1) (1, ).
1
2
y
; y
( x 1)
( x 1)2
( x 1)3
There are no critical points. The function is decreasing on its
domain. There are no inflection points. The function is concave
down on ( , 1) ( 1, 1) and concave up on (1, ). The
numerator and denominator share a factor of x 1. Dividing out
this common factor gives y 2xx11 ( x 1), which shows that
x 1 is a vertical asymptote. Now dividing numerator and
2(1/ x )
denominator by x gives y 1(1/ x ) , which shows that y 2 is a
horizontal asymptote. The graph will have a hole at x 1,
2( 1) 1
y 1( 1)1 23 . The x-intercept is 12 .
76.
y
x 2 49
x 2 5 x 14
Since 7 and 2 are roots of the denominator, the domain is
( , 7) ( 7, 2) (2, ).
10
5
y
; y
( x 7)
2
( x 2)
( x 1)3
There are no critical points. The function is increasing on its
domain. There are no inflection points. The function is concave
up on ( , 7) ( 7, 2) and concave down on (2, ). The
numerator and denominator share a factor of x 7. Dividing out
this common factor gives y xx 72 ( x 7), which shows that
x 1 is a vertical asymptote. Now dividing numerator and
1(7/ x )
denominator by x gives y 1(2/ x ) , which shows that y 1 is a
horizontal asymptote. The graph will have a hole at x 7,
( 1) 7
y ( 7)2 14
. The x-intercept is 72 .
9
Copyright 2016 Pearson Education, Ltd.
225
226
77.
Chapter 4 Applications of Derivatives
4
y x 21
x
Since 0 is a root of the denominator, the domain is
( , 0) (0, ).
y
2 x4 2
3
6
; y 2
x
x4
There are critical points at x 1. The function is increasing on
( 1, 0) (1, ) and decreasing on ( , 1) (0, 1). There are no
inflection points. The function is concave up on its domain. The
y-axis is a vertical asymptote. Dividing numerator and
2
2
denominator by x 2 gives y x 11/ x , which shows that there
are no horizontal asymptotes. For large x , the graph is close to
the graph of y x 2 .
78.
2
y x2x 4
Since 0 is a root of the denominator, the domain is
( , 0) (0, ).
2
y x 24 ; y 43
2x
x
There are no critical points at x 2. The function is increasing
on ( , 2) (2, ) and decreasing on ( 2, 0) (0, 2). There
are no inflection points. The function is concave down on ( , 0)
and concave up on (0, ). The y-axis is a vertical asymptote.
Dividing numerator and denominator by x gives y x 24/ x , which
shows that the line y 2x is an asymptote.
79. y
1
x 2 1
Since 1 and 1 are roots of the denominator, the domain is
( , 1) ( 1, 1) (1, ).
y
2x ;
( x 2 1)2
2
y 6 x2 23
( x 1)
There is a critical point at x 0, where the function has a local
maximum. The function is increasing on ( , 1) ( 1, 0) and
decreasing on (0, 1) (1, ). The function is concave up on
( , 1) (1, ) and concave down on ( 1, 1). The lines x 1
and x 1 are vertical asymptotes. The x-axis is a horizontal
asymptote.
Copyright 2016 Pearson Education, Ltd.
Section 4.4 Concavity and Curve Sketching
80.
y
x2
x 2 1
Since 1 and 1 are roots of the denominator, the domain is
( , 1) ( 1, 1) (1, ).
y
2
y 6 x2 23
2x ;
( x 2 1)2
( x 1)
There is a critical point at x 0, where the function has a local
maximum. The function is increasing on ( , 1) ( 1, 0) and
decreasing on (0, 1) (1, ). There are no inflection points. The
function is concave up on ( , 1) (1, ) and concave down on
( 1, 1). The lines x 1 and x 1 are vertical asymptotes.
Dividing numerator and denominator by x 2 gives y
1
1(1/ x 2 )
which shows that the line y 1 is a horizontal asymptote. The xintercept is 0 and the y-intercept is 0.
81.
2
y x 22
x 1
Since 1 and 1 are roots of the denominator, the domain is
( , 1) ( 1, 1) (1, ).
y
2
y 6 x2 23
2x ;
( x 2 1)2
( x 1)
There is a critical point at x 0, where the function has a local
maximum. The function is increasing on ( , 1) ( 1, 0) and
decreasing on (0, 1) (1, ). There are no inflection points. The
function is concave up on ( , 1) (1, ) and concave down on
( 1, 1). The lines x 1 and x 1 are vertical asymptotes.
Dividing numerator and denominator by x 2 gives y
1(2/ x 2 )
1(1/ x 2 )
which shows that the line y 1 is a horizontal asymptote. The
x-intercepts are 2 and the y-intercept is 2 .
82.
2
y x 2 4
x 2
2 and 2 are roots of the denominator, the domain is
Since
, 2 2, 2 2, .
y
4x ;
( x 2)2
2
y
4(3 x 2 2)
( x 2 2)3
There is a critical point at x 0, where the function has a local
2, and
minimum. The function is increasing on 0, 2
decreasing on , 2 2, 0 . There are no inflection
points. The function is concave up on 2, 2 and concave
2, . The lines x 2 and x 2
down on , 2
are vertical asymptotes. Dividing numerator and denominator by
Copyright 2016 Pearson Education, Ltd.
227
228
Chapter 4 Applications of Derivatives
x 2 gives y
1(4/ x 2 )
which shows that the line y 1 is a
1(2/ x 2 )
horizontal asymptote. The x-intercepts are 2 and the yintercept is 2 .
83.
2
y xx1
Since 1 is a root of the denominator, the domain is
( , 1) ( 1, ).
2
y x 2 x2 ; y
( x 1)
2
( x 1)3
There is a critical point at x 0, where the function has a local
minimum, and a critical point at x 2 where the functions has a
local maximum. The function is increasing on ( , 2) (0, )
and decreasing on ( 2, 1) ( 1, 0). There are no inflection
points. The function is concave up on ( 1, ) and concave down
on ( , 1) . The line x 1 is a vertical asymptote. Dividing
numerator by denominator gives y x 1 x11 , which shows
that the line y x 1 is an oblique asymptote. (See Section 2.6.)
The x-intercept is 0 and the y-intercept is 0.
84.
2
y xx 14
Since 1 is a root of the denominator, the domain is
( , 1) ( 1, ).
2
y x 2 x 2 4 ; y
( x 1)
6
( x 1)3
There are no critical points. The function is decreasing on its
domain. There are no inflection points. The function is concave up
on ( 1, ) and concave down on ( , 1) . The line x 1 is a
vertical asymptote. Dividing numerator by denominator gives
y 1 x x31 , which shows that the line y 1 x is an oblique
asymptote. (See Section 2.6.) The x-intercepts are 2 and the yintercept is 4.
85.
2
y x xx11
Since 1 is a root of the denominator, the domain is
( , 1) (1, ).
2
y x 2 x2 ; y 2 3
( x 1)
x 1
There is a critical point at x 0, where the function has a local
maximum, and a critical point at x 2 where the function has a
local minimum. The function is increasing on ( , 0) (2, )
and decreasing on (0, 1) (1, 2). There are no inflection points.
The function is concave up on (1, ) and concave down on
( , 1). The line x 1 is a vertical asymptote. Dividing
Copyright 2016 Pearson Education, Ltd.
Section 4.4 Concavity and Curve Sketching
numerator by denominator gives y x x11 which shows that
the line y x is an oblique asymptote. (See Section 2.6.) The yintercept is 1.
2
86. y x xx11
Since 1 is a root of the denominator, the domain is
( , 1) (1, ).
2
y 2 x x 2 ; y 2 3
( x 1)
x 1
There is a critical point at x 0, where the function has a local
minimum, and a critical point at x 2 where the function has a
local maximum. The function is increasing on (0, 1) (1, 2) and
decreasing on ( , 0) (2, ). There are no inflection points.
The function is concave up on ( , 1) and concave down on
(1, ). The line x 1 is a vertical asymptote. Dividing numerator
by denominator gives y x x11 which shows that the line
y x is an oblique asymptote. (See Section 2.6.) The yintercept is 1.
87.
3
2
( x 1)3
y x 32x 3 x 1 ( x 1)( x 2)
x x 2
Since 1 and 2 are roots of the denominator, the domain is
( , 2) ( 2, 1) (1, ).
y
( x 1)( x 5)
, x 1; y 18 3 , x 1
( x 2)2
( x 2)
Since 1 is not in the domain, the only critical point is at x 5,
where the function has a local maximum. The function is
increasing on ( , 5) (1, ) and decreasing on
( 5, 2) ( 2, 1). There are no inflection points. The function is
concave up on ( 2, 1) (1, ) and concave down on ( , 2).
The line x 2 is a vertical asymptote. Dividing numerator by
the denominator gives y x 4 x 9 2 which shows that the line
y x 4 is an oblique asymptote. (See Section 2.6.) The y-
intercept is 12 . The graph has a hole at the point (1, 0).
Copyright 2016 Pearson Education, Ltd.
229
230
88.
Chapter 4 Applications of Derivatives
3
y x x 2 2
x x
( x 1)( x 2 x 2)
( x 1)( x )
Since 1 and 0 are roots of the denominator, the domain is
( , 0) (0, 1) (1, ).
2
y x 2 2 , x 1; y 42 , x 1
x
x
There is a critical point at x 2 where the function has a local
minimum, and a critical point at x 2 where the function has a
local maximum. The function is increasing on 2, 0 0, 2
2, . There are no inflection
and decreasing on , 2
points. The function is concave up on ( , 0) and concave down
on (0, 1) (1, ). The y-axis is a vertical asymptote. Dividing
numerator by denominator gives y x 1 2x which shows that
the line y x 1 is an oblique asymptote. (See Section 2.6.)
The graph has a hole at the point (1, 4).
89. y
x
x 2 1
Since 1 and 1 are roots of the denominator, the domain is
( , 1) ( 1, 1) (1, ).
y
x 2 1 ;
( x 2 1)2
3
y 2 x2 6 x3
( x 1)
There are no critical points. The function is decreasing on its
domain. There is an inflection point at x 0. The function is
concave up on ( 1, 0) (1, ) and concave down on
( , 1) (0, 1). The lines x 1 and x 1 are vertical
asymptotes. Dividing numerator and denominator by x 2 gives
y 1/ x 2 which show that the x-axis is a horizontal asymptote.
1(1/ x )
The x-intercept is 0 and the y-intercept is 0.
90. y
x 1
x 2 ( x 2)
Since 0 and 2 are roots of the denominator, the domain is
( , 0) (0, 2) (2, ).
2
3
2
y 2 x3 5 x 24 ; y 6 x 244 x 403x 24
x ( x 2)
x ( x 2)
There are no critical points. The function is increasing on ( , 0)
and decreasing on (0, 2) (2, ). There is an inflection point at
approximately x 1.223. The function is concave up on
( , 0) (0, 1.223) (2, ) and concave down on (1.223, 2).
The lines x 0 (the y-axis) and x 2 are vertical asymptotes.
Dividing numerator and denominator by x 3 gives
y
(1/ x 2 ) (1/ x 3 )
1 (2/ x )
which shows that the x-axis is a horizontal
asymptote. The x-intercept is 1.
Copyright 2016 Pearson Education, Ltd.
Section 4.4 Concavity and Curve Sketching
91. y
8
x2 4
The domain is ( , ).
2
x ; y 16(3 x 4)
y 16
2
2
2
3
( x 4)
( x 4)
There is a critical point at x 0, where the function has a local
maximum. The function is increasing on ( , 0) and decreasing
on (0, ). There are inflection points at x 2 / 3 and at
x 2 / 3. The function is concave up on
, 2 / 3 2 / 3, and concave down on
2 / 3, 2 / 3 . Dividing numerator and denominator by x2
gives y
8/ x 2
which shows that the x-axis is a horizontal
1(4/ x 2 )
asymptote. The y-intercept is 2.
4x
The domain is ( , ).
x2 4
2
4( x 4)
8 x ( x 2 12)
y 2 2 ; y 2 3
( x 4)
( x 4)
92. y
There is a critical point at x 2, where the function has a local
minimum, and at x 2, where the function has a local maximum.
The function is increasing on ( 2, 2) and decreasing on
( , 2) (2, ). There are inflection points at
x 2 3, x 0, and x 2 3. The function is concave up on
2 3, 0 2 3, and concave down on
, 2 3 0, 2 3 . Dividing numerator and denominator by
x 2 gives y
4/ x
which shows that the x-axis is a horizontal
1 (4/ x 2 )
asymptote. The x-intercept is 0 and the y-intercept is 0.
93. Point
P
Q
R
S
T
94.
y
y
0
0
95.
Copyright 2016 Pearson Education, Ltd.
231
232
Chapter 4 Applications of Derivatives
96.
97. Graphs printed in color can shift during a press run, so your values may differ somewhat from those given here.
(a) The body is moving away from the origin when |displacement| is increasing as t increases, 0 t 2 and
6 t 9.5; the body is moving toward the origin when |displacement| is decreasing as t increases, 2 t 6
and 9.5 t 15.
(b) The velocity will be zero when the slope of the tangent line for y s (t ) is horizontal. The velocity is zero
when t is approximately 2, 6, or 9.5 s.
(c) The acceleration will be zero at those values of t where the curve y s (t ) has points of inflection. The
acceleration is zero when t is approximately 4, 7.5, or 12.5 s.
(d) The acceleration is positive when the concavity is up, 4 t 7.5 and 12.5 t 15; the acceleration is
negative when the concavity is down, 0 t 4 and 7.5 t 12.5.
98. (a) The body is moving away from the origin when |displacement| is increasing as t increases, 1.5 t 4,
10 t 12 and 13.5 t 16; the body is moving toward the origin when |displacement| is decreasing as
t increases, 0 t 1.5, 4 t 10 and 12 t 13.5.
(b) The velocity will be zero when the slope of the tangent line for y s (t ) is horizontal. The velocity is zero
when t is approximately 0, 4, 12 or 16 s.
(c) The acceleration will be zero at those values of t where the curve y s (t ) has points of inflection. The
acceleration is zero when t is approximately 1.5, 6, 8, 10.5, or 13.5 s.
(d) The acceleration is positive when the concavity is up, 0 t 1.5, 6 t 8 and 10 t 13.5, the
acceleration is negative when the concavity is down, 1.5 t 6, 8 t 10 and 13.5 t 16.
2
99. The marginal cost is dc
which changes from decreasing to increasing when its derivative d 2c is zero. This is a
dx
dx
point of inflection of the cost curve and occurs when the production level x is approximately 60 thousand units.
dy
100. The marginal revenue is dx and it is increasing when its derivative
0 t 2 and 5 t 9; marginal revenue is decreasing when
2 t 5 and 9 t 12.
2
d y
dx 2
d2y
dx 2
is positive the curve is concave up
0 the curve is concave down
101. When y ( x 1)2 ( x 2), then y 2( x 1)( x 2) ( x 1)2 . The curve falls on (, 2) and rises on (2, ).
At x 2 there is a local minimum. There is no local maximum. The curve is concave upward on (, 1) and
5 , , and concave downward on 1, 5 . At x 1 or x 5 there are inflection points.
3
3
3
102. When y ( x 1) 2 ( x 2)( x 4), then y 2( x 1)( x 2)( x 4) ( x 1) 2 ( x 4) ( x 1)2 ( x 2)
( x 1)[2( x 2 6 x 8) ( x 2 5 x 4) ( x 2 3x 2)] 2( x 1)(2 x 2 10 x 11). The curve rises on (, 2) and
(4, ) and falls on (2, 4). At x 2 there is a local maximum and at x 4 a local minimum. The curve is concave
downward on (, 1) and 52 3 , 52 3 and concave upward on 1, 52 3 and 52 3 , . At x 1, 52 3 and
5 3
there are inflection points.
2
Copyright 2016 Pearson Education, Ltd.
Section 4.4 Concavity and Curve Sketching
233
103. The graph must be concave down for x 0 because
f ( x) 12 0.
x
104. The second derivative, being continuous and never zero, cannot change sign. Therefore the graph will always
be concave up or concave down so it will have no inflection points and no cusps or corners.
105. The curve will have a point of inflection at x 1 if 1 is a solution of y 0; y x3 bx 2 cx d
y 3 x 2 2bx c y 6 x 2b and 6(1) 2b 0 b 3.
2
2
106. (a) f ( x) ax 2 bx c a x 2 ba x c a x 2 ba x b 2 4ba c a x 2ba
4a
vertex is at x 2ba the coordinates of the vertex are
2ba ,
b 4a4ac a parabola whose
2
2
b 2 4ac
4a
(b) The second derivative, f ( x) 2a, describes concavity when a 0 the parabola is concave up and
when a 0 the parabola is concave down.
107. A quadratic curve never has an inflection point. If y ax 2 bx c where a 0, then y 2ax b and y 2a.
Since 2a is a constant, it is not possible for y to change signs.
108. A cubic curve always has exactly one inflection point. If y ax3 bx 2 cx d where a 0, then
y 3ax 2 2bx c and y 6ax 2b. Since 3ab is a solution of y 0, we have that y changes its sign at
x 3ba and y exists everywhere (so there is a tangent at x 3ba ). Thus the curve has an inflection point at
x b . There are no other inflection points because y changes sign only at this zero.
3a
109. y ( x 1)( x 2), when y 0 x 1 or x 2; y | | points of inflection at x 1
and x 2
1
2
110. y x 2 ( x 2)3 ( x 3), when y 0 x 3, x 0 or x 2; y | | | points of
inflection at x 3 and x 2
3
Copyright 2016 Pearson Education, Ltd.
0
2
234
Chapter 4 Applications of Derivatives
111. y a x3 bx 2 cx y 3a x 2 2bx c and y 6a x 2b; local maximum at x 3
3a (3) 2 2b(3) c 0 27 a 6b c 0; local minimum at x 1 3a (1)2 2b(1) c 0
3a 2b c 0; point of inflection at (1, 11) a (1)3 b(1)2 c(1) 11 a b c 11 and
6a (1) 2b 0 6a 2b 0. Solving 27 a 6b c 0, 3a 2b c 0, a b c 11, and 6a 2b 0
a 1, b 3, and c 9 y x3 3 x 2 9 x
2
2
112. y xbx ca y bx 2cx 2ab ; local maximum at x 3
(bx c )
minimum at (1, 2)
b ( 1)2 2c ( 1) a b
(b ( 1) c )
2
b (3)2 2c (3) ab
(b (3) c ) 2
0 9b 6c ab 0; local
( 1) 2 a
0 b 2c a b 0 and b( 1) c 2 a 2b 2c 1.
2
Solving 9b 6c ab 0, b 2c a b 0, and a 2b 2c 1 a 3, b 1, and c 1 y xx 13 .
113. If y x5 5 x 4 240, then y 5 x3 ( x 4) and
y 20 x 2 ( x 3). The zeros of y' are extrema, and
there is a point of inflection at x 3.
114. If y x3 12 x 2 then y 3 x( x 8) and y 6( x 4).
The zeros of y and y are extrema, and points of
inflection, respectively.
115. If y 54 x5 16 x 2 25, then y 4 x( x3 8) and
y 16( x3 2). The zeros of y and y are extrema,
and points of inflection, respectively.
4
3
116. If y x4 x3 4 x 2 12 x 20, then
y x3 x 2 8 x 12 ( x 3)( x 2)2 . So y has a
local minimum at x 3 as its only extreme value.
Also y 3 x 2 2 x 8 (3 x 4)( x 2) and there
are inflection points at both zeros, 43 and 2, of y .
Copyright 2016 Pearson Education, Ltd.
Section 4.5 Applied Optimization
235
117. The graph of f falls where f 0, rises where f 0,
and has horizontal tangents where f 0. It has
local minima at points where f changes from
negative to positive and local maxima where f
changes from positive to negative. The graph of f is
concave down where f 0 and concave up where
f 0. It has an inflection point each time f
changes sign, provided a tangent line exists there.
118. The graph f is concave down where f 0, and
concave up where f 0. It has an inflection point
each time f changes sign, provided a tangent line
exists there.
4.5
APPLIED OPTIMIZATION
1. Let and w represent the length and width of the rectangle, respectively. With an area of 16 cm.2 , we have that
2( 2 16)
( )( w) 16 w 16 1 the perimeter is P 2 2w 2 32 1 and P ( ) 2 322
. Solving
2
P( ) 0
2( 4)( 4)
2
0 4, 4. Since 0 for the length of a rectangle, must be 4 and w 4 the
perimeter is 16 in., a minimum since P ( ) 163 0.
2. Let x represent the length of the rectangle in meters (0 x 4). Then the width is 4 x and the area is
A( x) x(4 x) 4 x x 2 . Since A( x) 4 2 x, the critical point occurs at x 2. Since, A( x) 0 for 0 x 2
and A( x) 0 for 2 x 4, this critical point corresponds to the maximum area. The rectangle with the largest
area measures 2 m by 4 2 2 m, so it is a square.
Graphical Support:
3. (a) The line containing point P also contains the points (0, 1) and (1, 0) the line containing P is y 1 x
a general point on that line is ( x, 1 x).
(b) The area A( x) 2 x(1 x), where 0 x 1.
(c) When A( x) 2x 2 x 2 , then A( x) 0 2 4 x 0 x 12 . Since A(0) 0 and A(1) 0, we conclude
that A 12 12 sq units is the largest area. The dimensions are 1 unit by 12 unit.
Copyright 2016 Pearson Education, Ltd.
236
Chapter 4 Applications of Derivatives
4. The area of the rectangle is A 2 xy 2 x (12 x 2 ), where
0 x 12. Solving A( x) 0 24 6 x2 0 x 2 or
2. Now 2 is not in the domain, and since A(0) 0 and
A 12 0, we conclude that A(2) 32 square units is the
maximum area. The dimensions are 4 units by 8 units.
5. The volume of the box is V ( x ) x(45 2 x)(24 2 x)
1080 x 138 x 2 4 x3 , where 0 x 12. Solving
V ( x) 0 1080 276 x 12 x 2 12( x 5)( x 18) 0
x 5 or 18, but 18 is not in the domain. Since
V (0) V (4) 0, V 5 2450 cm3 must be the maximum
volume of the box with dimensions 14 35 5 cm.
6. The area of the triangle is A 12 ba b2 400 b 2 , where
0 b 20. Then dA
12 400 b 2
db
200 b 2
400 b 2
b2
2 400 b 2
0 the interior critical point is b 10 2.
When b 0 or 20, the area is zero A 10 2 is the
2
2
maximum area. When a b 400 and b 10 2, the
value of a is also 10 2 the maximum area occurs when
a b.
7. The area is A( x) x(800 2 x), where 0 x 400. Solving
A( x) 800 4 x 0 x 200. With A(0) A(400) 0,
the maximum area is A(200) 80, 000 m 2 . The
dimensions are 200 m by 400 m.
8. The area is 2 xy 216 y 108
. The amount of fence
x
needed is P 4 x 3 y 4 x 324 x 1 , where 0 x;
dP 4 324 0 x 2 81 0 the critical points are 0
2
dx
x
and 9, but 0 and 9 are not in the domain. Then
P (9) 0 at x 9 there is a minimum the dimensions
of the outer rectangle are 18 m by 12 m 72 meters of
fence will be needed.
9. (a) We minimize the weight tS where S is the surface area, and t is the thickness of the steel walls of the
tank. The surface area is S x 2 4 xy where x is the length of a side of the square base of the tank, and y
is its depth. The volume of the tank must be 4 m3 y 42 . Therefore, the weight of the tank is
w( x) t x
2
x
16
. Treating the thickness as a constant gives w( x) t
x
2x . The critical value is
16
x2
at x 2. Since w(2) t 2 323 0, there is a minimum at x 2. Therefore, the optimum dimensions
2
of the tank are 20 m on the base edges and 1 m deep.
(b) Minimizing the surface area of the tank minimizes its weight for a given wall thickness. The thickness of
the steel walls would likely be determined by other considerations such as structural requirements.
Copyright 2016 Pearson Education, Ltd.
Section 4.5 Applied Optimization
237
10. (a) The volume of the tank being 20 m3 , we have that yx 2 20 y 202 . The cost of building the
x
tank is c( x) 5 x 2 30 x 202 , where 0 x. Then c( x) 10 x 600
0 the critical points are 0 and 3 60 ,
2
x
x
but 0 is not in the domain. Thus, c 3 60 0 at x 3 60 we have a minimum. The values of
x 3 60 m and y
3
60
m will minimize the cost.
3
2
(b) The cost function c 5( x 4 xy ) 10 xy, can be separated into two items: (1) the cost of the materials and
labor to fabricate the tank, and (2) the cost for the excavation. Since the area of the sides and bottom of
the tanks is ( x 2 4 xy ), it can be deduced that the unit cost to fabricate the tanks is $5/m 2 . Normally,
excavation costs are per unit volume of excavated material. Consequently, the total excavation cost can be
2
taken as 10 xy 10
( x 2 y ). This suggests that the unit cost of excavation is $10/m
where x is the length of
x
x
a side of the square base of the tank in meters. For the least expensive tank, the unit cost for the excavation
2
is $10/m
$0.67
. The total cost of the least expensive tank is $230, which is the sum of $179 for
3
15 m
m
fabrication and $51 for the excavation.
11. The area of the printing is ( y 10)( x 20) 312.5.
Consequently, y 312.5
10. The area of the paper
x 20
is A( x) x 312.5
10 , where 20 x. Then
x 20
A( x) 312.5
10 x
x 20
312.5
( x 20) 2
10( x 20)2 6250
( x 20)2
0
the critical points are 5 and 45, but 5 is not in
the domain. Thus A(45) 0 at x 45 we have a
minimum. Therefore the dimensions 45 by 22.5 cm
minimize the amount of paper.
12. The volume of the cone is V 13 r 2 h, where r x 9 y 2 and h y 3 (from the figure in the text). Thus,
V ( y ) 3 (9 y 2 )( y 3) 3 (27 9 y 32 y 2 y 3 ) V ( y ) 3 (9 6 y 3 y 2 ) (1 y )(3 y ). The critical
points are 3 and 1, but 3 is not in the domain. Thus V (1) (6 6(1)) 0 at y 1 we have a maximum
volume of V (1) 3 (8)(4) 323 cubic units.
13. The area of the triangle is A( )
Solving A( ) 0
Since A( )
3
ab sin
, where 0 .
2
ab cos
0 2 .
2
0, there is a maximum at 2 .
ab sin
A 2
2
14. A volume V r 2 h 100 h 1000
. The amount of material is the
2
r
surface area given by the sides and bottom of the can
S 2 rh r 2 2000
r 2 , 0 r. Then
r
dS 2000 2 r 0 r 3 1000 0. The critical points are 0 and 10 ,
3
dr
r2
r2
2
but 0 is not in the domain. Since d 2s 4000
2
0,
we
have
a
dr
r3
10
minimum surface area when r 3 cm and h 1000
310 cm.
r2
Comparing this result to the result found in Example 2, if we include
both ends of the can, then we have a minimum surface area when the
can is shorter—specifically, when the height of the can is the same as
its diameter.
Copyright 2016 Pearson Education, Ltd.
238
Chapter 4 Applications of Derivatives
15. With a volume of 1000 cm3 and V r 2 h, then h 1000
. The amount of aluminum used per can is
2
r
3
A 8r 2 2 rh 8r 2 2000
. Then A(r ) 16r 2000
0 8r 1000
0 the critical points are 0 and 5, but
2
2
r
r
r
r 0 results in no can. Since A(r ) 16 1000
and h:r 8: .
0 we have a minimum at r 5 h 40
3
r
16. (a) The base measures 3 2 x cm by 4522 x cm, so the volume formula is V ( x)
x (30 2 x )(45 2 x )
2
2 x3 75 x 2 675 x.
(b) We require x 0, 2 x 30, and 2 x 45. Combining these requirements, the domain is the interval
(0, 15).
(c) The maximum volume is approximately 1782.5 cm3 when x 5.89 cm.
(d) V ( x) 6 x 2 150 x 675. The critical point occurs when V ( x) 0, at
x
150 ( 150)2 4(6)(675) 150 6300
2(6)
12
2525 7 , that is, x 5.89 or x 19.11. We discard the larger
value because it is not in the domain. Since V ( x) 12 x 150, which is negative when x 5.89, the
critical point corresponds to the maximum volume. The maximum volume occurs when
x 2525 7 5.89, which confirms the result in (c).
17. (a) The “sides” of the suitcase will measure 60 2 x cm by 45 2 x cm and will be 2x cm apart, so the volume
formula is V ( x) 2 x(60 2 x)(45 2x ) 8 x3 420 x 2 5400 x.
(b) We require x 0, 2 x 45, and 2 x 60. Combining these requirements, the domain is the interval (0, 9).
(c) The maximum volume is approximately 20.68 cm3 when x 8.49 cm.
(d) V ( x) 24 x 2 840 x 5400 24( x 2 35 x 225). The critical point is at
x
35 ( 35)2 4(1)(225)
35 2 325 3552 13 , that is, x 8.49 or x 26.51. We discard the larger value
2(1)
because it is not in the domain. Since V ( x) 24(2 x 35) which is negative when x 8.49, the critical
point corresponds to the maximum volume. The maximum value occurs at x 3552 13 8.49, which
confirms the results in (c).
(e) 8 x3 468 x 2 5400 x 17500 4(2 x3 105 x 2 1350 x 4375) 0 4(2 x 25)( x 5)( x 35) 0.
Since 35 is not in the domain, the possible values of x are x 12.5 cm or x 5 cm.
Copyright 2016 Pearson Education, Ltd.
Section 4.5 Applied Optimization
239
(f ) The dimensions of the resulting box are 2x cm, (60 2 x) cm, and (45 2 x) cm. Each of these
measurements must be positive, so that gives the domain of (0, 22.5).
18. If the upper right vertex of the rectangle is located at ( x, 4 cos 0.5 x) for 0 x , then the rectangle has width
2x and height 4 cos 0.5x, so the area is A( x) 8 x cos 0.5 x.. Solving A( x) 0 graphically for 0 x , we find
that x 2.214. Evaluating 2x and 4 cos 0.5x for x 2.214, the dimensions of the rectangle are approximately
4.43 (width) by 1.79 (height), and the maximum area is approximately 7.923.
19. Let the radius of the cylinder be r cm, 0 r 10. Then the height is 2 100 r 2 and the volume is
V (r ) 2 r 2 100 r 2 cm3 . Then, V (r ) 2 r 2
2 r (200 3r 2 )
100 r 2
1
2 100 r
2
(2r ) 2 100 r 2 (2r )
. The critical point for 0 r 10 occurs at r
100 r 2
200 10 2 . Since V ( r ) 0 for 0 r 10 2
3
3
3
2 r 10, the critical point corresponds to the maximum volume. The dimensions are
3
2 8.16 cm and h 20 11.55 cm, and the volume is 4000 2418.40 cm3 .
3
3 3
3
and V (r ) 0 for 10
r 10
2 r 3 4 r (100 r 2 )
20. (a) From the diagram we have 4 x 276 and
V x 2 . The volume of the box is
V ( x) x 2 (276 4 x), where 0 x 69. Then
V ( x) 552 x 12 x 2 12 x(46 x) 0 the
critical points are 0 and 46, but x 0 results in
no box. Since V ( x) 552 24 x 0 at x 46
we have a maximum. The dimensions of the
box are 46 46 92 cm.
(b) In terms of length, V ( ) x2
2764 . The
2
graph indicates that the maximum volume
occurs near 92, which is consistent with the
result of part (a).
21. (a) From the diagram we have 3h 2w 276 and
Then V (h) 276h 92 h 2 92 h 184
h 0
3
V h 2 w V (h) h 2 138 32 h 138h 2 32 h3
h 0 or h 184
, but h 0 results in no box.
3
Since V (h) 276 9h 0 at h 184
, we
3
have a maximum volume at h 184
and
3
w 138 32 h 46.
(b)
Copyright 2016 Pearson Education, Ltd.
240
Chapter 4 Applications of Derivatives
22. From the diagram the perimeter is P 2r 2h r ,
where r is the radius of the semicircle and h is the
height of the rectangle. The amount of light
transmitted proportional to A 2rh 14 r 2
r ( P 2r r ) 14 r 2 rP 2r 2 43 r 2 . Then
dA P 4r 3 r 0 r 2 P
dr
2
83
(4 ) P
2
P
4
P
2h P 83 83 83 . Therefore,
2 r 8 gives the proportions that admit the most
h
4
2
light since d 2A 4 32 0.
dr
23. The fixed volume is V r 2 h 23 r 3 h V 2 23r , where h is the height of the cylinder and r is the radius
r
of the hemisphere. To minimize the cost we must minimize surface area of the cylinder added to twice the
surface area of the hemisphere. Thus, we minimize C 2 rh 4 r 2 2 r
V
r2
23r 4 r 2 2rV 83 r 2 .
1/3
Then dC
2V2 16
r 0 V 83 r 3 r 83V
. From the volume equation, h V 2 23r
dr
3
r
r
1/3
1/3
1/3 1/3
1/3
1/3
1/3 1/3
2
4
V
2
3
V
3
2
4
V
2
3
V
3
V
d
C
4
V
16
1/3 2/3
. Since 2 3 3 0, these dimensions do minimize
3
32 1/3
32 1/3
dr
r
the cost.
24. The volume of the trough is maximized when the area of the cross section is maximized. From the diagram
the area of the cross section is A( ) cos sin cos , 0 2 . Then A( ) sin cos 2 sin 2
(2sin 2 sin 1) (2sin 1)(sin 1) so A( ) 0 sin 12 or sin 1 6 because
sin 1 when 0 2 . Also, A( ) 0 for 0 6 and A( ) 0 for 6 2 . Therefore, at 6 there
is a maximum.
25. (a) From the diagram we have: AP x, RA L x 2 , PB 21.6 x,
CH DR 28 RA 28 L x 2 , QB x 2 (21.6 x)2 ,
HQ 28 CH QB 28 28 L x 2 x 2 (21.6 x)2
2
2
L x 2 x 2 (21.6 x) 2 , RQ RH HQ
2
(21.6) 2
L x 2 x 2 (21.6 x 2 )
L2 x 2
L2 x 2 x 2 ( x 21.6) 2
2
2
2
. It follows that RP PQ RQ
2
2
(21.6) 2
L2 x 2 L2 x 2 2 L2 x 2 43.2 x (21.6) 2 43.2 x (21.6) 2 (21.6)2
43.22 x 2 4( L2 x 2 )(43.2 x (21.6) 2 ) L2 x 2
43.22 x 2
4[43.2 x (21.6)2 ]
3
3
43.2 x3
43.2 x3
4 x4x43.2 2 x2x21.6 .
2
2
43.2 x (21.6)
43.2 x 43.2
2
3
(b) If f ( x) 4 x4x43.2 is minimized, then L2 is minimized. Now f ( x)
4 x 2 (8 x 129.6)
(4 x 43.2)2
x 129.6
and f ( x) 0 when x 129.6
. Thus L2 is minimized when x 129.6
.
8
8
8
Copyright 2016 Pearson Education, Ltd.
f ( x ) 0 when
Section 4.5 Applied Optimization
241
(c) When x 16.2, then L 28 cm.
26. (a) From the figure in the text we have P 2 x 2 y y P2 x. If P 36, then y 18 x. When the
cylinder is formed, x 2 r r 2x and h y h 18 x. The volume of the cylinder is
2
3 x (12 x )
0 x 0 or 12; but when x 0 there is no
4
3
V r 2 h V ( x ) 18 x4 x . Solving V ( x)
cylinder. Then V ( x) 3 3 2x V (12) 0 there is a maximum at x 12. The values of x 12 cm
and y 6 cm give the largest volume.
(b) In this case V ( x) x 2 (18 x). Solving V ( x) 3 x(12 x) 0 x 0 or 12; but x 0 would result in
no cylinder. Then V ( x) 6 (6 x) V (12) 0 there is a maximum at x 12. The values of
x 12 cm and y 6 cm give the largest volume.
27. Note that h 2 r 2 3 and so r 3 h 2 . Then the volume is given by V 3 r 2 h 3 (3 h 2 )h h 3 h3 for
0 h 3, and so dV
r 2 (1 r 2 ). The critical point (for h 0 ) occurs at h 1. Since dV
0 for
dh
dh
0 h 1, and dV
0 for 1 h 3, the critical point corresponds to the maximum volume. The cone of
dh
greatest volume has radius 2 m, height 1 m, and volume 23 m3 .
y
28. Let d ( x 0) 2 ( y 0)2 x 2 y 2 and ax b 1 y ba x b. We can minimize d by minimizing
x2 y2
D
2
x 2 ba x b
D 2 x 2 ba x b ba 2 x 2ab x 2ab . D 0
2
2
2
2
2
2 x b 2 x ba 0 x 2ab 2 is the critical point y ba
D
a
ab 2
a 2 b 2
ab 2
a b2
2
a b
2
2
b
2 2 0 the critical point is a local minimum
a
x y 1 that is closest to the origin.
a b
2
2
x
x
2
2
b
a 2b . D 2 2b 2
a b2
a2
2
2
ab , a b is the point on the line
a 2 b2 a 2 b 2
2
29. Let S ( x) x 1x , x 0 S ( x) 1 12 x 21 . S ( x) 0 x 21 0 x 2 1 0 x 1. Since x 0, we
x
only consider x 1. S ( x) 23 S (1) 23 0 local minimum when x 1
x
1
3
3
30. Let S ( x) 1x 4 x 2 , x 0 S ( x) 12 8 x 8 x 21 . S ( x) 0 8 x 21 0 8 x3 1 0 x 12 .
S ( x) 23 8 S
x
x
1 2 8 0 local minimum when x 1 .
2
2
(1/2)3
x
x
31. The length of the wire b perimeter of the triangle circumference of the circle. Let x length of a side of the
equilateral triangle P 3 x, and let r radius of the circle C 2 r. Thus b 3 x 2 r r b23 x .
The area of the circle is r 2 and the area of an equilateral triangle whose sides are x is 12 ( x)
Thus, the total area is given by A 43 x 2 r 2 43 x 2
b23x 43 x2 b43x
2
A 23 x 23 (b 3x) 23 x 23b 29 x . A 0 23 x 23b 29 x 0 x
Copyright 2016 Pearson Education, Ltd.
2
3b .
3 9
x x .
3
2
3 2
4
242
Chapter 4 Applications of Derivatives
A 23 29 0 local minimum at the critical point. P 3
triangular segment and C 2
b23x b 3x b 39b9
3b
9b m is the length of the
3 9
3 9
3 b
m is the length of the circular segment.
3 9
32. The length of the wire b perimeter of the triangle circumference of the circle. Let x length of a side of the
square P 4 x, and let r radius of the circle C 2 r. Thus b 4 x 2 r r b24 x . The area of the
circle is r 2 and the area of a square whose sides are x is x 2 . Thus, the total area is given by A x 2 r 2
2
b4 x
x 2 b24 x x 2 4
A 2 x 24 (b 4 x) 2 x 2b 8 x, A 0 2 x 2b 8 x 0
x b . A 2 8 0 local minimum at the critical point. P 4 b 4b m is the length of the
2
4
square segment and C 2
b4 x
2
4
4
b 4 x b 44b 4b m is the length of the circular segment.
33. Let ( x, y ) x, 43 x be the coordinates of the corner that intersects the line. Then base 3 x and height
y 43 x, thus the area of the rectangle is given by A (3 x) 43 x 4 x 43 x 2 , 0 x 3. A 4 83 x, A 0
x 32 . A 34 A 32 0 local maximum at the critical point. The base 3 32 32 and the height
43 32 2.
34. Let ( x, y ) x, 9 x 2 be the coordinates of the corner that intersects the semicircle. Then base 2x and
height y 9 x 2 , thus the area of the inscribed rectangle is given by A (2 x) 9 x 2 , 0 x 3. Then
2
2
2
2(9 x ) 2 x
A 2 9 x 2 (2 x) x 2
18 4 x2 , A 0 18 4 x 2 0 x 3 2 2 , only x 3 22 lies in
2
9 x
9 x
4 x
0 x 3. A is continuous on the closed interval 0 x 3 A has an absolute maxima and absolute minima.
9 absolute maxima. Base of rectangle is 3 2 and height
A(0) 0, A(3) 0, and A 3 2 2 3 2
3 2
2
is 3 22 .
35. (a) f ( x) x 2 ax f ( x) x 2 (2 x3 a ), so that f ( x ) 0 when x 2 implies a 16
(b) f ( x) x 2 ax f ( x) 2 x 3 ( x3 a), so that f ( x) 0 when x 1 implies a 1
36. If f ( x) x3 ax 2 bx, then f ( x) 3 x 2 2ax b and f ( x) 6 x 2a.
(a) A local maximum at x 1 and local minimum at x 3 f ( 1) 0 and f (3) 0 3 2a b 0 and
27 6a b 0 a 3 and b 9.
(b) A local minimum at x 4 and a point inflection at x 1 f (4) 0 and f (1) 0 48 8a b 0 and
6 2a 0 a 3 and b 24.
37. (a)
s (t ) 4.9t 2 29.4t 34.3 v(t ) s (t ) 9.8t 29.4. At t 0, the velocity is v(0) 29.4 m/s.
(b) The maximum height occurs when v(t ) 0, when t 3. The maximum height is s (3) 78.4 m and it
occurs at t 3 s.
(c) Note that s (t ) 4.9t 2 29.4t 34.3 4.9(t 1)(t 7), so s 0 at t 1 or t 7. Choosing the positive
value of t, the velocity when s 0 is v(7) 39.2 m/s.
38.
Copyright 2016 Pearson Education, Ltd.
Section 4.5 Applied Optimization
243
Let x be the distance from the point on the shoreline nearest Jane’s boat to the point where she lands her boat.
Then she needs to row 4 x 2 km at 2 km/h and walk 6 x km at 5 km/h. The total amount of time to reach
4 x 2
65 x hours (0 x 6). Then
2
the village is f ( x)
f ( x) 0. we have:
x
2 4 x2
f ( x) 12
1
(2 x) 15
2 4 x2
x
2 4 x2
15 . Solving
15 5 x 2 4 x 2 25 x 2 4 4 x 2 21x 2 16 x 4 . We discard
21
the negative value of x because it is not in the domain. Checking the endpoints and critical point, we have
f (0) 2.2, f
2.12, and f (6) 3.16. Jane should land her boat
4
21
4 0.87 kilometers down the
21
shoreline from the point nearest her boat.
39.
2 h h 2 10 and L( x )
x
x 5
x
h 2 ( x 5) 2
2 10x ( x 5)2 when x 0. Note that L( x) is
2
minimized when f ( x) 2 10
( x 5)2 is
x
2
minimized. If f ( x) 0, then
x
2 2 10
102 2( x 5) 0 ( x 5) 1 203 0
x
x
x 5 (not acceptable since distance is never
negative) or x 2.71 . Then
L(2.71) 91.82 9.58 m.
40. (a) s1 s2 sin t sin t 3 sin t sin t cos 3 sin 3 cos t sin t 12 sin t 23 cos t tan t 3
t 3 or 43
(b) The distance between the particles is s (t ) | s1 s2 | sin t sin t 3 12 sin t 3 cos t
s (t )
sin t 3 cos t cos t 3 sin t
2 sin t 3 cos t
since d | x | x critical times and endpoints are
dx
| x|
0, 3 , 56 , 43 , 116 , 2 ; then s (0) 23 , s 3 0, s 56 1, s 43 0, s 116 1, s (2 ) 23 the
greatest distance between the particles is 1.
(c) Since s (t )
sin t 3 cos t cos t 3 sin t we can conclude that at t and 4 , s(t ) has cusps and the
3
2 sin t 3 cos t
3
distance between the particles is changing the fastest near these points.
41. I k2 , let x distance the point is from the stronger light source 6 x distance the point is from the other
d
light source. The intensity of illumination at the point from the stronger light is I1
k2
illumination at the point from the weaker light is I 2
the intensity of the second light k1 8k2 . I1
I
16 k2
x3
x 4 m. I
2 k2
(6 x )3
48k2
x
4
16(6 x )3 k2 2 x3k2
x3 (6 x )3
6 k2
(6 x )
4
I (4)
(6 x ) 2
8 k2
x
2
and I 0
48k2
4
4
6 k2
(6 4)4
k1
x2
, and intensity of
. Since the intensity of the first light is eight times
. The total intensity is given by I I1 I 2
16(6 x )3 k2 2 x3k2
x3 (6 x )3
8 k2
x
2
k2
(6 x ) 2
0 16(6 x)3 k2 2 x3 k2 0
0 local minimum. The point should be 4 m from the
stronger light source.
Copyright 2016 Pearson Education, Ltd.
244
Chapter 4 Applications of Derivatives
v2
42. R g0 sin 2 ddR
2
d R2
d
2v02
cos 2
g
4v 2
g0 sin 2 4
and ddR 0
2v02
cos 2
g
0
4v 2
2
4 . d R2 g0 sin 2
d
4v02
g 0 local maximum. Thus, the firing angle of
4 45
4
will maximize the range R.
43. (a) From the diagram we have d 2 4r 2 w2 . The strength of the beam is S kwd 2 kw (4r 2 w2 ).
When r 15, then S 900kw kw3 . Also, S ( w) 900k 3kw2 3k (300 w2 ) so
S ( w) 0 w 10 3; S 10 3 0 and 10 3 is not acceptable. Therefore S 10 3 is the
(b)
maximum strength. The dimensions of the strongest beam are 10 3 by 10 6 centimeters.
(c)
Both graphs indicate the same maximum value and are consistent with each other. Changing k does not
change the dimensions that give the strongest beam (i.e., do not change the values of w and d that produce
the strongest beam).
44. (a) From the situation we have w2 900 d 2 . The stiffness of the beam is S kwd 3 kd 3 (900 d 2 )1/2 ,
where 0 d 30. Also, S (d )
(b)
4 kd 2 (675 d 2 )
900 d 2
critical points at 0, 30, and 15 3. Both d 0 and
d 30 cause S 0. The maximum occurs at d 15 3. The dimensions are 15 by 15 3 centimeters.
(c)
Both graphs indicate the same maximum value and are consistent with each other. The changing of k has
no effect.
45. (a) s 10 cos( t ) v 10 sin( t ) speed |10 sin( t )| 10 |sin( t ) | the maximum speed is
10 31.42 cm/s since the maximum value of |sin ( t )| is 1; the cart is moving the fastest at t 0.5 s, 1.5
s, 2.5 s and 3.5 s when |sin ( t )| is 1. At these times the distance is s 10 cos 2 0 cm and
2
2
a 10 cos ( t ) | a | 10 |cos ( t )| | a | 0 cm/s
2
(b) | a | 10 2 |cos ( t )| is greatest at t 0.0 s, 1.0 s, 2.0 s, 3.0 s, and 4.0 s, and at these times the magnitude
of the cart’s position is | s | 10 cm from the rest position and the speed is 0 cm/s.
46. (a) 2sin t sin 2t 2 sin t 2 sin t cos t 0 (2 sin t )(1 cos t ) 0 t k where k is a positive integer
(b) The vertical distance between the masses is s(t ) | s1 s2 | ( s1 s2 ) 2
((sin 2t 2sin t )2 )1/2
1/2
s (t ) 12 ((sin 2t 2sin t )2 )1/2 (2)(sin 2t 2sin t )(2 cos 2t 2 cos t)
2(cos 2t 2 cos t )(sin 2t 2sin t )
|sin 2t 2sin t |
4(2 cos t 1)(cos t 1)(sin t )(cos t 1)
critical times at 0, 23 , , 43 , 2 ; then s (0) 0,
|sin 2t 2sin t |
Copyright 2016 Pearson Education, Ltd.
Section 4.5 Applied Optimization
245
3 23 , s( ) 0, s 43 sin 83 2sin 43 3 23 , s(2 ) 0
s 23 sin 43 2sin 23
the greatest distance is 3 23 at t 23 and 43
47. (a) s (12 12t ) 2 (8t ) 2 ((12 12t )2 64t 2 )1/2
(b) ds
12 ((12 12t )2 64t 2 ) 1/2 [2(12 12t )(12) 128t ]
dt
208t 144
(12 12t 2 ) 64t 2
ds
dt
t 0
12 knots and
ds
8 knots
dt t1
(d) The graph supports the conclusions in parts (b)
and (c).
(c) The graph indicates that the ships did not see
each other because s (t ) 5 for all values of t.
(e)
lim ds
t dt
208 144t 2082 208 4 13 which equals the square
lim
lim
2
2
2
144 64
t 144(1t ) 64t
t 144 1 1 64
t
2
(208t 144)2
root of the sums of the squares of the individual speeds.
48. The distance OT TB is minimized when OB is a
straight line. Hence
1 2 .
49. If v kax kx 2 , then v ka 2kx and v 2k , so v 0 x a2 . At x a2 there is a maximum since
2
v a2 2k 0. The maximum value of v is ka4 .
50. (a) According to the graph, y (0) 0.
(b) According to the graph, y ( L) 0.
(c) y (0) 0, so d 0. Now y ( x) 3ax 2 2bx c, so y (0) 0 implies that c 0. Therefore, y ( x) ax3 bx 2
and y ( x) 3ax 2 2bx. then y ( L) aL3 bL2 H and y ( L) 3aL2 2bL 0, so we have two linear
equations in two unknowns a and b. The second equation gives b 3aL
. Substituting into the first
2
3
3
equation, we have aL3 3aL
H , or aL2 H , so a 2 H3 . Therefore, b 3 H2 and the equation for y is
2
y ( x) 2 H3 x3 3 H2 x 2 , or y ( x) H
L
L
2
.
x 3 3 x 2
L
L
L
L
51. The profit is p nx nc n( x c) [a( x c) 1 b(100 x)]( x c) a b(100 x)( x c)
a (bc 100b) x 100bc bx 2 . Then p ( x) bc 100b 2bx and p ( x) 2b. Solving
p ( x) 0 x 2c 50. At x 2c 50 there is a maximum profit since p ( x) 2b 0 for all x.
Copyright 2016 Pearson Education, Ltd.
246
Chapter 4 Applications of Derivatives
52. Let x represent the number of people over 50. The profit is p( x) (50 x)(200 2 x) 32(50 x) 6000
2 x 2 68 x 2400. Then p ( x) 4 x 68 and p 4. Solving p ( x) 0 x 17. At x 17 there is a
maximum since p (17) 0. It would take 67 people to maximize the profit.
53. (a) A(q ) kmq 1 cm h2 q, where q 0 A(q) kmq 2 h2
points are
2 km , 0, and
h
2 km , but only
h
hq 2 2 km
2q2
and A( q ) 2kmq 3 . The critical
2 km is in the domain. Then A
h
2 km
h
0 at q
2 km there
h
is a minimum average weekly cost.
( k bq ) m
cm h2 q kmq 1 bm cm h2 q, where q 0 A(q ) 0 at q 2km
as in (a). Also
h
q
3
A(q) 2kmq 0 so the most economical quantity to order is still q 2 km
which minimizes the
h
(b) A(q )
average weekly cost.
c( x)
54. We start with c( x) the cost of producing x items, x 0, and x the average cost of producing x items,
assumed to be differentiable. If the average cost can be minimized, it will be at a production level at which
d c ( x ) 0 xc( x ) c ( x ) 0 (by the quotient rule) xc( x) c( x) 0 (multiply both sides by x 2 )
2
dx
x
c( x)
x
c( x) x where c ( x ) is the marginal cost. This concludes the proof. (Note: The theorem does not assure a
production level that will give a minimum cost, but rather, it indicates where to look to see if there is one. Find
the production levels where the average cost equals the marginal cost, then check to see if any of them give a
minimum.)
55. The profit p ( x) r ( x) c( x) 6 x ( x3 6 x 2 15 x) x3 6 x 2 9 x, where x 0. Then p ( x) 3x 2 12 x 9
3( x 3)( x 1) and p ( x) 6 x 12. The critical points are 1 and 3. Thus p (1) 6 0 at x 1 there is a
local minimum, and p (3) 6 0 at x 3 there is a local maximum. But p (3) 0 the best you can do is
break even.
c( x)
56. The average cost of producing x items is c ( x) x x 2 20 x 20, 000 c( x) 2 x 20 0 x 10, the
only critical value. The average cost is c (10) $19,900 per item is a minimum cost because c(10) 2 0.
57. Let x the length of a side of the square base of the box and h the height of the box. V x 2 h 6 h 62 .
x
The total cost is given by
3
C 60 x 2 40(4 xh) 60 x 2 160 x 62 60 x 2 960
, x 0 C 120 x 960
120 x 2960
2
x
x
3
C 0 12 x 296 0
x
x
x
3
12 x 96 0 x 4; C 120 1920
C (4) 120 1920
0 local
x2
42
minimum. x 2 h 62 3 / 2 and C (2) 60(2)2 960
720 the box is 2 m 2 m 3 / 2 m, with a
2
2
minimum cost of $720.
58. Let x the number of $10 increases in the charge per room, then price per room 50 10 x, and the number of
rooms filled each night 800 40x the total revenue is R ( x) (50 10 x)(800 40 x)
400 x 2 6000 x 40000, 0 x 20 R ( x) 800 x 6000; R ( x) 0 800 x 6000 0
x 15
; R ( x) 800 R 15
800 0 local maximum. The price per room is 50 10 15
$125.
2
2
2
2
3
dM
dM
dR CM M 2 . Solving d R C 2 M 0 M C . Also. d R 2 0 at M C there is a
59. We have dM
2
3
2
2
maximum.
Copyright 2016 Pearson Education, Ltd.
Section 4.5 Applied Optimization
247
60 . (a) If v cr0 r 2 cr 3 , then v 2cr0 r 3cr 2 cr 2r0 3r and v 2cr0 6cr 2c r0 3r . The solution of
2r
2r
2r
2r
v 0 is r 0 or 30 , but 0 is not in the domain. Also, v 0 for r 30 and v 0 for r 30 at r 30
there is a maximum.
(b) The graph confirms the findings in (a).
2
2
61. If x 0, then x 1 0 x 2 1 2 x x x1 2. In particular if a, b, c and d are positive integers,
16.
2
then a a1
b 2 1
b
c 2 1
c
d 2 1
d
a x x a x
62. (a) f ( x)
f ( x)
a x
a x
2
x
2
b2 d x
b d x
2
2 3/ 2
g ( x)
2
2 1/ 2
2
2
2
2
a x 3/x 2
a x
2
b 2 d x
d x
b 2
2
2
2
function of x
(b) g ( x)
2 1/ 2
2
2
d x b d x
2 1/ 2
2
b2 d x
2
0 f ( x) is an increasing
a x
2 1/ 2
2
a2
2
2 3/ 2
b2 d x d x
2
2
b d x
2
2 3/ 2
0 g ( x) is a decreasing function of x
dt is an increasing function of x (from part (a)) minus a decreasing function
(c) Since c1 , c2 0, the derivative dx
2
dt 1 f ( x) 1 g ( x ) d t 1 f ( x) 1 g ( x) 0 since f ( x) 0 and
of x (from part (b)): dx
2
c
c
c
c
1
2
dt is an increasing function of x.
g ( x) 0 dx
dx
1
2
63. At x c , the tangents to the curves are parallel. Justification: The vertical distance between the curves is
D( x) f ( x) g ( x), so D( x) f ( x) g ( x). The maximum value of D will occur at a point c where D 0. At
such a point, f (c) g (c) 0, or f (c) g (c).
64. (a) f ( x ) 3 4 cos x cos 2 x is a periodic function with period 2
(b) No, f ( x) 3 4 cos x cos 2 x 3 4 cos x (2 cos 2 x 1) 2(1 2 cos x cos 2 x) 2(1 cos x)2 0
f ( x) is never negative.
65 . (a) If y cot x 2 csc x where 0 x , then y (csc x)
2 cot x csc x . Solving y 0 cos x 12
x 4 . For 0 x 4 we have y 0 and y 0 when 4 x . Therefore, at x 4 there is a maximum
value of y 1.
(b)
Copyright 2016 Pearson Education, Ltd.
248
Chapter 4 Applications of Derivatives
The graph confirms the findings in (a).
66. (a) If y tan x 3 cot x where 0 x 2x , then y sec2 x 3csc2 x. Solving y 0 tan x 3 x 3 ,
but 3 is not in the domain. Also, y 2sec2 x tan x 6 csc 2 x cot x 0 for all 0 x 2 . Therefore at
(b)
x 3 there is a minimum value of y 2 3 .
The graph confirms the findings in (a).
2
2
67 . (a) The square of the distance is D( x) x 32 x 0 x 2 2 x 94 , so D ( x) 2 x 2 and the critical
point occurs at x 1. Since D ( x) 0 for x 1 and D ( x) 0 for x 1, the critical point corresponds to the
(b)
minimum distance. The minimum distance is D(1) 25 .
The minimum distance is from the point 32 , 0 to the point (1, 1) on the graph of y x, and this occurs at
the value x 1 where D( x) , the distance squared, has its minimum value.
68. (a)
Calculus Method:
The square of the distance from the point 1, 3 to x, 16 x 2 is given by
D( x) ( x 1) 2
16 x 2 3
Then D( x) 2 12
2
483 x 2
2
2
x 2 2 x 1 16 x 2 2 48 3 x 2 3 2 x 20 2 48 3 x 2 .
(6 x) 2
6x
483 x 2
2
. Solving D( x) 0 we have:
6 x 2 48 3x 2 36 x 4(48 3 x 2 ) 9 x 48 3 x 2 12 x 2 48 x 2 We discard x 2 as
an extraneous solution, leaving x 2. Since D( x) 0 for 4 x 2 and D( x) 0 for 2 x 4, the critical
point corresponds to the minimum distance. The minimum distance is D(2) 2 .
Geometry Method:
The semicircle is centered at the origin and has radius 4. The distance from the origin to 1, 3 is
3 2. The shortest distance from the point to the semicircle is the distance along the radius
containing the point 1, 3 . That distance is 4 2 2.
12
2
Copyright 2016 Pearson Education, Ltd.
Section 4.6 Newton’s Method
249
(b)
The minimum distance is from the point 1, 3 to the point 2, 2 3 on the graph of y 16 x 2 , and
this occurs at the value x 2 where D ( x), the distance squared, has its minimum value.
4.6
NEWTON’S METHOD
1.
y x 2 x 1 y 2 x 1 xn 1 xn n2 x n 1 ; x0 1 x1 1 12111 32 x2 23 9 4 3
4 2 1
x 2 x 1
n
6 9 2 1 13 .61905;
x2 23 412
9
3 21 21
3
1
x0 1 x1 1 12111 2 x2 2 44211 53 1.66667
x3 3 x 1
2. y x3 3x 1 y 3 x 2 3 xn 1 xn n 2 n
3 xn 3
; x0 0 x1 0 13 13 x2 13
1 11
27
1 3
3
1 29 0.32222
13 90
90
x 4 x 3
3. y x 4 x 3 y 4 x3 1 xn 1 xn n 3 n
4 xn 1
1296
6 3
; x0 1 x1 1 14113 65 x2 65 625864 5
125
1
750 1875 6 171 5763 1.16542; x 1 x 1 113 2 x 2 16 2 3
65 1296
0
1
2
4320 625
5 4945 4945
41
32 1
51 1.64516
2 11
31
31
4. y 2 x x 2 1 y 2 2 x xn 1 xn
2 xn xn2 1
;
2 2 xn
x0 0 x1 0 02001 12 x2 12
1 5 .41667; x 2 x 2 4 41 5 x 5
12 12
0
1
2
12
2 4
2
2
29 2.41667
12
1 14 1
2 1
5 25
1 5 20 25 4
1
4
2 12 52 12
2 5
x4 2
625 2
4 xn
16
512
5. y x 4 2 y 4 x3 xn 1 xn n 3 ; x0 1 x1 1 142 54 x2 54 256125 54 625
2000
113 2500 113 2387 1.1935
54 2000
2000
2000
x4 2
625
4 xn
16
2
6. From Exercise 5, xn 1 xn n 3 ; x0 1 x1 1 142 1 14 54 x2 54 256125
512 5 113 1.1935
54 625
2000
4 2000
f (x )
7. f ( x0 ) 0 and f ( x0 ) 0 xn 1 xn f ( xn ) gives x1 x0 x2 x0 xn x0 for all n 0. That is all, of
n
the approximations in Newton’s method will be the root of f ( x ) 0.
Copyright 2016 Pearson Education, Ltd.
250
Chapter 4 Applications of Derivatives
8. It does matter. If you start too far away from x 2 , the calculated values may approach some other root.
Starting with x0 0.5, for instance, leads to x 2 as the root, not x 2 .
f (x )
f (h)
9. If x0 h 0 x1 x0 f ( x0 ) h f ( h)
0
h
h
1
h
h 2 h h;
2 h
f ( h)
f (x )
if x0 h 0 x1 x0 f ( x0 ) h f ( h)
0
h
10.
h
1
2 h
h
h 2 h h.
f ( x) x1/3 f ( x) 13 x 2/3
xn 1 xn
x1/3
n
1 x 2/3
3 n
2 xn ; x0 1
x1 2, x2 4, x3 8, and x4 16 and so
forth. Since xn 2 xn 1 we may conclude that
n xn .
11. i)
ii)
iii)
iv)
is equivalent to solving x3 3 x 1 0 .
is equivalent to solving x3 3 x 1 0 .
is equivalent to solving x3 3 x 1 0 .
is equivalent to solving x3 3 x 1 0 .
All four equations are equivalent.
x 10.5sin x
12. f ( x) x 1 0.5sin x f ( x) 1 0.5cos x xn 1 xn n10.5cos x n ; if x0 1.5, then x1 1.49870
n
13. f ( x) tan x 2 x f ( x) sec2 x 2 xn 1 xn
tan( xn ) 2 xn
sec2 xn
; x0 1 x1 1.2920445
x2 1.155327774 x16 x17 1.165561185
x 4 2 x3 x 2 2 xn 2
14. f ( x ) x 4 2 x3 x 2 2 x 2 f ( x) 4 x3 6 x 2 2 x 2 xn 1 xn n 3 n 2 n
x4 0.630115396; if x0 2.5, then x4 2.57327196
15. (a) The graph of f ( x) sin 3x 0.99 x 2 in the
window 2 x 2, 2 y 3 suggests three
roots. However, when you zoom in on the
x-axis near x 1.2, you can see that the graph
lies above the axis there. There are only two
roots, one near x 1, the other near x 0.4.
(b) f ( x) sin 3 x 0.99 x 2
f ( x) 3cos 3 x 2 x
xn 1 xn
sin 3 xn 0.99 xn2
3cos 3 xn 2 xn
and the solutions are approximately
0.35003501505249 and –1.0261731615301
Copyright 2016 Pearson Education, Ltd.
4 xn 6 xn 2 xn 2
; if x0 0.5, then
Section 4.6 Newton’s Method
251
16. (a) Yes, three times as indicted by the graphs
(b) f ( x) cos 3 x x f ( x) 3sin 3 x 1
cos(3 x ) x
xn 1 xn 3sin(3nx ) n1 ; at approximately
n
0.979367, 0.887726, and 0.39004 we have
cos 3 x x
17. f ( x) 2 x 4 4 x 2 1 f ( x) 8 x3 8 x xn 1 xn
2 xn4 4 xn2 1
8 xn3 8 xn
; if x0 2, then x6 1.30656296; if
x0 0.5, then x3 0.5411961; the roots are approximately 0.5411961 and 1.30656296 because f ( x ) is an
even function.
18. f ( x) tan x f ( x) sec 2 x xn 1 xn
approximate to be 3.14159.
tan( xn )
sec2 ( xn )
; x0 3 x1 3.13971 x2 3.14159 and we
19. From the graph we let x0 0.5 and f ( x) cos x 2 x
cos( x ) 2 x
xn 1 xn sin(nx ) 2n x1 .45063
n
x2 .45018 at x 0.45 we have cos x 2 x.
20. From the graph we let x0 0.7 and
x cos( x )
f ( x) cos x x xn 1 xn 1nsin( x n)
n
x1 .73944 x2 .73908 at x 0.74
we have cos x x.
21. The x-coordinate of the point of intersection of y x 2 ( x 1) and y 1x is the solution of x 2 ( x 1) 1x
x3 x 2 1x 0 The x-coordinate is the root of f ( x) x3 x 2 1x f ( x) 3 x 2 2 x 12 . Let x0 1
xn 1 xn
xn3 xn2 x1
n
3 xn2 2 xn 12
xn
x
x1 0.83333 x2 0.81924 x3 0.81917 x7 0.81917 r 0.8192
Copyright 2016 Pearson Education, Ltd.
252
Chapter 4 Applications of Derivatives
22. The x-coordinate of the point of intersection of y x and y 3 x 2 is the solution of x 3 x 2
x 3 x 2 0 The x-coordinate is the root of f ( x ) x 3 x 2 f ( x) 1 2 x . Let x0 1
xn 1 xn
xn 3 xn 2
1 2 x
n
2 xn
2 x
x1 1.4 x2 1.35556 x3 1.35498 x7 1.35498 r 1.3550
23. If f ( x) x3 2 x 4, then f (1) 1 0 and f (2) 8 0 by the Intermediate Value Theorem the equation
x3 2 xn 4
x3 2 x 4 0 has a solution between 1 and 2. Consequently, f ( x) 3 x 2 2 and xn 1 xn n
3 xn2 2
. Then
x0 1 x1 1.2 x2 1.17975 x3 1.179509 x4 1.1795090 the root is approximately 1.17951.
24. We wish to solve 8 x 4 14 x3 9 x 2 11x 1 0. Let f x 8 x 4 14 x3 9 x 2 11x 1, then
f ( x) 32 x3 42 x 2 18 x 11 xn 1 xn
x0
32 xn3 42 xn2 18 xn 11
.
approximation of corresponding root
–1.0
0.1
0.6
2.0
25.
8 xn4 14 xn3 9 xn2 11xn 1
–0.976823589
0.100363332
0.642746671
1.983713587
f (x )
xi3 xi
i
4 xi2 2
f ( x) 4 x 4 4 x 2 f ( x) 16 x3 8 x xi 1 xi f ( xi ) xi
. Iterations are performed using the
procedure in problem 13 in this section.
(a) For x0 2 or x0 0.8, xi 1 as i gets large.
(b) For x0 0.5 or x0 0.25, xi 0 as i gets large.
(c) For x0 0.8 or x0 2, xi 1 as i gets large.
(d) (If your calculator has a CAS, put it in exact mode, otherwise approximate the radicals with a decimal
value.) For x0 721 or x0
21
, Newton’s method does not converge. The values of xi alternate
7
between x0 721 or x0
as i increases.
21
7
26. (a) The distance can be represented by
, where x 0. The distance
2
D( x) is minimized when f ( x) ( x 2) 2 x 2 12 is
2
minimized. If f ( x) ( x 2) 2 x 2 12 , then
D( x) ( x 2) 2 x 2 12
2
f ( x) 4( x3 x 1) and f ( x) 4 (3 x 2 1) 0. Now
f ( x) 0 x3 x 1 0 x( x 2 1) 1 x 21 .
x 1
1
(b)
Let g ( x) 21 x ( x 2 1)1 x g ( x) ( x 2 1) 2 (2 x) 1 2 2 x 2 1 xn 1 xn
( x 1)
x 1
xn2 1
x0 1 x4 0.68233 to five decimal places.
Copyright 2016 Pearson Education, Ltd.
xn
2 xn
2
xn2 1 1
;
Section 4.7 Antiderivatives
( xn 1)40
27. f ( x) ( x 1) 40 f ( x) 40( x 1)39 xn 1 xn
x87 x88 x89
40( xn 1)39
253
39 xn 1
. With x0 2, our computer gave
40
x200 1.11051, coming within 0.11051 of the root x 1.
28. Since s r 3 r 3r . Bisect the angle to obtain a right triangle with hypotenuse r and opposite side
3
of length 1. Then sin 2 1r sin 2r 1r sin 23r 1r sin 23r 1r 0. Thus the solution r is a root of
r
f (r ) sin 23r 1r f (r ) 32 cos 23r 12 ; r0 1 rn 1 rn
2r
sin 23r r1
n
32 cos 23r
n
2 rn
n
1
r1 1.00280
rn2
3
r2 1.00282 r3 1.00282 r 1.0028 1.00282
2.9916
4.7
ANTIDERIVATIVES
1. (a) x 2
(b)
x3
3
(c)
x3 x 2 x
3
2. (a) 3x 2
(b)
x8
8
(c)
x8 3 x 2 8 x
8
3. (a) x 3
(b)
x3
4. (a) x 2
(b)
x4 x3
(c)
3
3
(c) x3 x 2 3 x
2
3
x 2 x 2 x
2
2
5. (a)
1
x
(b)
5
x
(c) 2 x 5x
6. (a)
1
x2
(b)
1
4x 2
(c)
x4 1
4
2 x2
x3
(b)
x
(c)
2
3
8. (a) x 4/3
(b)
1 x 2/3
2
(c)
3 x 4/3 3 x 2/3
4
2
9. (a) x 2/3
(b)
x1/3
(c) x 1/3
7. (a)
10. (a)
1 x 3 1
3 1
(b)
1
1
x 1
(c)
11. (a) cos ( x )
(b)
3cos x
(c)
12. (a) sin ( x)
(b)
sin 2x
(c)
1 tan x
2
(b)
2 tan 3x
14. (a) cot x
(b)
cot 32x
13. (a)
x3 2 x
1 x 2
2
cos ( x )
cos (3 x)
2 sin 2x sin x
(c) 23 tan 32x
(c) x 4 cot (2 x)
Copyright 2016 Pearson Education, Ltd.
254
Chapter 4 Applications of Derivatives
15. (a) csc x
(b)
1 csc(5 x)
5
(c) 2 csc 2x
16. (a) sec x
(b)
4 sec(3 x)
3
(c) 2 sec 2x
2
17.
( x 1) dx x2 x C
18.
(5 6 x ) dx 5 x 3 x 2 C
19.
3t 2 2t dt t3 t4 C
20.
4t dt t C
21.
(2 x3 5 x 7)dx 12 x 4 52 x 2 7 x C
22.
(1 x 2 3x5 )dx x 13 x3 12 x 6 C
23.
x dx x x dx
24.
2x dx 2x 2x dx x
25.
x 1/3 dx x 2 C 32 x 2/3 C
2
2
1
x2
1
5
2
1
3
2
x3
2
1
3
t3
6
3
2 x 2
2
1
5
2/3
2 x2 C x 1 x2 C
2
5 x2
26.
3
1/ 4
x 5/4 dx x 1 C 44 C
4
27.
x 3 x dx x1/2 x1/3 dx x x C 23 x3/2 34 x4/3 C
28.
dx x 2x dx
x
2
3/ 2
4/3
3
2
4
3
1/2
1 1/2
2
2
x
1
2
x3/ 2
2
8 y 1/2 4 dy
8 y 2 y 1/4 dy 82y 2 y
30.
1 1
7 y 5/ 4
17 y5/4 dy 17 y y
31.
2 x 1 x 3 dx
32.
x 3 ( x 1) dx
33.
t t t
dt
t2
34.
4 t
dt
t3
dt 4t t dt 4
35.
37.
dy
x
1/ 2
2 x 1 C 13 x3/2 4 x1/2 C
3
2
29.
2
y
4
x 1 x3 1 x C 1 x3 x C
1
3
3
x
3
3
3
1
5
t2
2
1/ 4
1
4
3/ 4
3
4
C 4 y 2 83 y3/4 C
y
C 7 1/4 4 C
y
2 x 2 x2 dx 22x 2 x1 C x2 2x C
1
2
x2 x3 dx x1 x2 C 1x 21x C
1
2
2
t 3/ 2 t1/ 2
t2
t2
dt t
1/2
1/ 2
1/ 2
t 3/2 dt t 1 t 1 C 2 t 2 C
2
2
t
t 2
2
t 3/ 2
23
2 cos t dt 2sin t C
36.
5sin t dt 5 cos t C
7 sin 3 d 21cos 3 C
38.
3cos 5 d 53 sin 5 C
4
t3
t1/ 2
t3
3
5/2
C 22 23/ 2 C
t
Copyright 2016 Pearson Education, Ltd.
3t
Section 4.7 Antiderivatives
2
39.
3csc2 x dx 3cot x C
40.
sec3 x dx tan3 x C
41.
csc cot d 1 csc C
2
2
42.
2 sec tan d 2 sec C
5
5
43.
(4sec x tan x 2sec2 x) dx 4sec x 2 tan x C
44.
1 (csc 2 x csc x cot x ) dx 1 cot x 1 csc x C
2
2
2
45.
(sin 2 x csc2 x) dx 12 cos 2 x cot x C
46.
(2 cos 2 x 3sin 3x) dx sin 2 x cos 3x C
47.
1 cos 4t dt
2
12 12 cos 4t dt 12 t 12 sin44t C 2t sin84t C
48.
1 cos 6t dt
2
12 12 cos 6t dt 12 t 12 sin66t C 2t sin126t C
49.
3 x 3 dx 3 x
51.
(1 tan 2 ) d sec2 d tan C
52.
(2 tan 2 ) d (1 1 tan 2 ) d (1 sec2 ) d tan C
53.
cot 2 x dx (csc2 x 1) dx cot x x C
54.
(1 cot 2 x) dx (1 (csc 2 x 1)) dx (2 csc 2 x) dx 2 x cot x C
55.
cos (tan sec ) d (sin 1) d cos C
56.
csc
d
csc sin
3 1
3 1
C
x
2 1 dx x 2 C
2
sin d
1
1
d
d sec2 d tan C
csccsc
sin sin
1sin
cos
2
4
d (7 x 2) C
dx
28
58.
d
dx
59.
d 1 tan (5 x 1) C
dx 5
60.
d
dx
61.
d 1 C
dx x 1
(3 x 5) 1
C
3
2
4(7 x 2)3 (7)
(7 x 2)3
28
57.
50.
(3 x 5)2 (3)
3
(3 x 5)2
15 (sec2 (5x 1))(5) sec2 (5x 1)
3cot x31 C 3 csc2 x31 13 csc2 x31
(1)(1)( x 1)2 ( x11)
d
63. (a) Wrong: dx
62.
2
d
x C
dx x 1
x (1)
1
( x(1)(1)
x 1)
( x 1)
sin x C sin x cos x x sin x cos x x sin x
x2
2
2x
2
x2
2
x2
2
Copyright 2016 Pearson Education, Ltd.
2
2
255
256
Chapter 4 Applications of Derivatives
d ( x cos x C ) cos x x sin x x sin x
(b) Wrong: dx
d ( x cos x sin x C ) cos x x sin x cos x x sin x
(c) Right: dx
(b)
(c)
sec3 C 3sec2 (sec tan ) sec3 tan tan sec 2
3
3
2
d
1
1
Right: d 2 tan C 2 (2 tan ) sec2 tan sec2
Right: dd 12 sec2 C 12 (2sec ) sec tan tan sec2
64. (a) Wrong: dd
(2 x 1)3
C
3
3
d
65. (a) Wrong: dx
3(2 x1)2 (2)
2(2 x 1) 2 (2 x 1)2
3
2
2
d ((2 x 1) C ) 3(2 x 1) (2) 6(2 x 1) 3(2 x 1) 2
(b) Wrong: dx
d ((2 x 1)3 C ) 6(2 x 1) 2
(c) Right: dx
d ( x 2 x C )1/2 1 ( x 2 x C ) 1/2 (2 x 1)
2 x 1
66. (a) Wrong: dx
2x 1
2
2 x 2 x C
d ( x 2 x)1/2 C 1 ( x 2 x) 1/2 (2 x 1)
(b) Wrong: dx
2
d
(c) Right: dx
1
3
2 x 1
2 x2 x
2x 1
2 x 1 C dxd 13 (2 x 1)3/2 C 63 (2 x 1)1/2 (2) 2 x 1
3
x 3)
xx23 C 3 xx23 ( x2)( x12)( x3)1 3 (( xx2)3) ( x52) 15(
( x 2)
2
3
d
67. Right: dx
2
2
d
68. Wrong: dx
sin( x 2 )
C
x
xcos( x 2 )(2 x ) sin( x 2 )1
x
2
2
2
2 x 2 cos( x 2 ) sin( x 2 )
x
2
2
4
x cos( x 2 ) sin( x 2 )
x2
dy
69. Graph (b), because dx 2 x y x 2 C. Then y (1) 4 C 3.
dy
70. Graph (b), because dx x y 12 x 2 C. Then y (1) 1 C 32 .
71.
dy
2 x 7 y x 2 7 x C ; at x 2 and y 0 we have 0 22 7(2) C C 10 y x 2 7 x 10
dx
72.
2
2
2
dy
10 x y 10 x x2 C ; at x 0 and y 1 we have 1 10(0) 02 C C 1 y 10 x x2 1
dx
73.
74.
2
2
dy
12 x x 2 x y x 1 x2 C ; at x 2 and y 1 we have 1 21 22 C C 12
dx
x
2
2
y x 1 x2 12 or y 1x x2 12
dy
9 x 2 4 x 5 y 3 x3 2 x 2 5 x C ; at x 1 and y 0 we have 0 3( 1)3 2( 1) 2 5( 1) C
dx
3
2
C 10 y 3 x 2 x 5 x 10
75.
1/3
dy
3 x 2/3 y 3 x1 C 9 y 9 x1/3 C ; at x 1 and y 5 we have 5 9(1)1/3 C C 4
dx
3
1/3
y 9x
76.
4
dy
1 12 x 1/2 y x1/2 C ; at x 4 and y 0 we have 0 41/2 C C 2 y x1/2 2
dx
2 x
Copyright 2016 Pearson Education, Ltd.
Section 4.7 Antiderivatives
77.
ds 1 cos t s t sin t C ; at t 0 and s 4 we have 4 0 sin 0 C C 4 s t sin t 4
dt
78.
ds cos t sin t s sin t cos t C ; at t and s 1 we have 1 sin cos C C 0
dt
257
s sin t cos t
79.
dr sin r cos ( ) C ; at r 0 and 0 we have 0 cos ( 0) C C 1 r cos ( ) 1
d
80.
dr cos r 1 sin ( ) C ; at r 1 and 0 we have 1 1 sin( 0) C C 1 r 1 sin ( ) 1
d
81.
dv 1 sec t tan t v 1 sec t C ; at v 1 and t 0 we have 1 1 sec (0) C C 1 v 1 sec t 1
dt
2
2
2
2
2
2
82.
C C 7 2
dv 8t csc 2 t v 4t 2 cot t C ; at v 7 and t we have 7 4 2 cot
dt
2
2
2
2
2
v 4t cot t 7
83.
d2y
dx 2
dy
dy
2 6 x dx 2 x 3 x 2 C1; at dx 4 and x 0 we have 4 2(0) 3(0)2 C1 C1 4
dy
dx 2 x 3 x 2 4 y x 2 x3 4 x C2 ; at y 1 and x 0 we have 1 02 03 4(0) C2 C2 1
y x 2 x3 4 x 1
84.
d2y
dx 2
dy
dy
dy
0 dx C1; at dx 2 and x 0 we have C1 2 dx 2 y 2 x C2 ; at y 0 and x 0 we have
0 2(0) C2 C2 0 y 2 x
85.
d 2 r 2 2t 3 dr t 2 C ; at dr 1 and t 1 we have 1 (1) 2 C C 2 dr t 2 2
1
1
1
dt
dt
dt
dt 2
t3
1
1
1
r t 2t C2 ; at r 1 and t 1 we have 1 1 2(1) C2 C2 2 r t 2t 2 or r 1t 2t 2
86.
d 2 s 3t ds 3t 2 C ; at ds 3 and t 4 we have 3 3(4)
1
8
dt
16
dt
16
dt 2
2
2
3
t s t C ; at
C1 C1 0 ds
316
2
dt
16
3
3
4 C C 0 s t
s 4 and t 4 we have 4 16
2
2
16
87.
88.
d3y
6
d2y
dx
2
2
6 x C1; at
d2y
2
8 and x 0 we have 8 6(0) C1 C1 8
d2y
6x 8
dx 2
dy
dy
dy
dx 3 x 8 x C2 ; at dx 0 and x 0 we have 0 3(0) 2 8(0) C2 C2 0 dx 3 x 2 8 x
y x3 4 x 2 C3 ; at y 5 and x 0 we have 5 03 4(0)2 C3 C3 5 y x3 4 x 2 5
dx
3
dx
d 3 0 d 2 C ; at d 2 2 and t 0 we have d 2 2 d 2t C ; at d 1 and t 0 we have
2
1
dt
dt
2
dt 3
dt 2
dt 2
dt 2
2 1
d
1
1
1
2 2(0) C2 C2 2 dt 2t 2 t 2 t C3 ; at 2 and t 0 we have
2 02 12 (0) C3 C3 2 t 2 12 t 2
89. y (4) sin t cos t y cos t sin t C1; at y 7 and t 0 we have 7 cos (0) sin (0) C1 C1 6
y cos t sin t 6 y sin t cos t 6t C2 ; at y 1 and t 0 we have
1 sin (0) cos (0) 6(0) C2 C2 0 y sin t cos t 6t y cos t sin t 3t 2 C3 ; at
Copyright 2016 Pearson Education, Ltd.
258
Chapter 4 Applications of Derivatives
y 1 and t 0 we have 1 cos (0) sin (0) 3(0)2 C3 C3 0 y cos t sin t 3t 2
y sin t cos t t 3 C4 ; at y 0 and t 0 we have 0 sin (0) cos (0) 03 C4
C4 1 y sin t cos t t 3 1
90.
y (4) cos x 8sin(2 x) y sin x 4 cos (2 x) C1; at y 0 and x 0 we have
0 sin(0) 4 cos(2(0)) C1 C1 4 y sin x 4 cos(2 x) 4 y cos x 2sin(2 x) 4 x C2 ; at
y 1 and x 0 we have 1 cos(0) 2sin(2(0)) 4(0) C2 C2 0 y cos x 2sin(2 x) 4 x
y sin x cos(2 x) 2 x 2 C3 ; at y 1 and x 0 we have 1 sin(0) cos(2(0)) 2(0) 2 C3 C3 0
y sin x cos(2 x) 2 x 2 y cos x 12 sin(2 x) 32 x3 C4 ; at y 3 and x 0 we have
3 cos(0) 12 sin(2(0)) 23 (0)3 C4 C4 4 y cos x 12 sin(2 x) 32 x3 4
91. m y 3 x 3 x1/2 y 2 x3/2 C ; at (9, 4) we have 4 2(9)3/2 C C 50 y 2 x3/2 50
92. (a)
d2y
dx 2
dy
dy
6 x dx 3 x 2 C1; at y 0 and x = 0 we have 0 3(0) 2 C1 C1 0 dx 3 x 2
y x3 C2 ; at y = 1 and x = 0 we have C2 1 y x3 1
(b) One, because any other possible function would differ from x3 1 by a constant that must be zero because
of the initial conditions
93.
dy
1 43 x1/3 y
dx
1 43 x1/3 dx x x4/3 C; at (1, 0.5) on the curve we have
0.5 1 14/3 C C 0.5 x x 4/3 12
94.
dy
x 1 y
dx
1
95.
2
( x 1) dx x2 x C ; at (1, 1) on the curve we have
2
( 1)2
(1) C C 12 y x2 x 12
2
dy
sin x cos x y
dx
(sin x cos x) dx cos x sin x C ; at (, 1) on the curve we have
1 = cos() sin() + C C = 2 y = cos x sin x 2
96.
dy
1 sin x 12 x 1/2 sin x y
dx
2 x
12 x1/2 sin x dx x1/2 cos x C; at (1, 2) on the curve
we have 2 11/2 cos (1) C C 0 y x cos x
97. (a)
ds 9.8t 3 s 4.9t 2 3t C ; (i) at s = 5 and t = 0 we have C = 5 s 4.9t 2 3t 5;
dt
displacement = s(3) s(1) = ((4.9)(9) 9 + 5) (4.9 3 + 5) = 33.2 units; (ii) at s = 2 and t = 0 we have
C = 2 s 4.9t 2 3t 2; displacement = s(3) s(1) = ((4.9)(9) 9 2) (4.9 3 2) = 33.2 units;
(iii) at s s0 and t = 0 we have C s0 s 4.9t 2 3t s0 ;
displacement s (3) s (1) ((4.9)(9) 9 s0 ) (4.9 3 s0 ) 33.2 units
(b) True. Given an antiderivative f(t) of the velocity function, we know that the body’s position function is
s = f(t) + C for some constant C. Therefore, the displacement from t = a to t = b is
(f(b) + C) (f(a) + C) = f(b) f(a). Thus we can find the displacement from any antiderivative f as the
numerical difference f(b) f(a) without knowing the exact values of C and s.
Copyright 2016 Pearson Education, Ltd.
Section 4.7 Antiderivatives
259
98. a(t ) v (t ) 20 v(t) = 20t + C; at (0, 0) we have C = 0 v(t) = 20t. When t = 60, then
v(60) = 20(60) = 1200 m/s.
2
99. Step 1: d 2s k ds
kt C1 ; at ds
30 and t = 0 we have
dt
dt
dt
2
2
C1 30 ds
kt 30 s k t2 30t C2 ; at s = 0 and t = 0 we have C2 0 s kt2 30t
dt
0 0 kt 30 t 30
Step 2: ds
dt
k
Step 3: 75
30 30 75 (30)2 (30)2 75 (30)2 k 6
k 30
k
2
k
2
2k
k
2k
2
100. d 2s k ds
k dt kt C ; at ds
13.3 when t = 0 we have
dt
dt
dt
2
kt 13.3 s kt2 13.3t C1 ; at s = 0 when t = 0 we have
13.3 = k(0) + C C = 13.3 ds
dt
k (0)2
2
0 2 13.3(0) C1 C1 0 s kt2 13.3t. Then ds
0 kt 13.3 0 t 13.3
and
dt
k
s 13.3
k
13.3 13.3 13.7 88.445 176.89 13.7 k 88.445 6.46 m .
k 13.3
k
2
k
2
k
k
13.7
s2
101. (a) v a dt (15t1/2 3t 1/2 )dt 10t 3/2 6t1/2 C ;
ds (1) 4 4 10(1)3/2 6(1)1/2 C C 0 v 10t 3/2 6t1/2
dt
(b) s v dt (10t 3/2 6t1/2 )dt 4t 5/2 4t 3/2 C ;
s (1) 0 0 4(1)5/2 4(1)3/2 C C 0 s 4t 5/2 4t 3/2
2
1.6t C1 ; at ds
0 and t = 0 we have C1 0 ds
1.6t s 0.8t 2 C2 ; at s = 1.2
102 . d 2s 1.6 ds
dt
dt
dt
dt
and t = 0 we have C2 1.2 s 0.8t 2 1.2. Then s 0 0 0.8t 2 1.2 t
1.2 1.22 s, since t > 0
0.8
2
2
103. d 2s a ds
a dt at C ; ds
v0 when t = 0 C v0 ds
at v0 s at2 v0t C1; s s0
dt
dt
dt
dt
when t = 0 s0
2
a (0)2
v0 (0) C1 C1 s0 s at2 v0t s0
2
2
104. The appropriate initial value problem is: Differential Equation: d 2s g with Initial Conditions: ds
v0 and
dt
dt
s s0 when t = 0. Thus ds
g dt gt C1; ds
(0) v0 v0 ( g )(0) C1 C1 v0 ds
gt v0 .
dt
dt
dt
Thus s ( gt v0 )dt 12 gt 2 v0t C2 ; s (0) s0 12 ( g )(0)2 v0 (0) C2 C2 s0
Thus s 12 gt 2 v0t s0 .
f ( x) dx 1 x C1 x C
(b)
g ( x) dx x 2 C1 x C
(c)
f ( x ) dx 1 x C1 x C
(d)
g ( x) dx ( x 2) C1 x C
(e)
[ f ( x) g ( x)]dx 1 x ( x 2) C1 x x C
105 (a)
(f)
[ f ( x) g ( x)]dx 1 x ( x 2) C1 x x C
Copyright 2016 Pearson Education, Ltd.
260
Chapter 4 Applications of Derivatives
106. Yes. If F ( x) and G ( x) both solve the initial value problem on an interval I then they both have the same
first derivative. Therefore, by Corollary 2 of the Mean Value Theorem there is a constant C such that
F ( x) G ( x) C for all x. In particular, F ( x0 ) G ( x0 ) C , so C F ( x0 ) G ( x0 ) 0. Hence F ( x) G ( x)
for all x.
107110. Example CAS commands:
Maple:
with(student):
f : x - cos(x)^2 sin(x);
ic : [x Pi,y 1];
F : unapply( int( f(x), x ) C, x );
eq : eval( y F(x), ic );
solnC : solve( eq, {C} );
Y : unapply( eval( F(x), solnC ), x );
DEplot( diff(y(x),x) f(x), y(x), x 0..2*Pi, [[y(Pi) 1]],
color black, linecolor black, stepsize 0.05, title "Section 4.7 #107" );
Mathematica: (functions and values may vary)
The following commands use the definite integral and the Fundamental Theorem of calculus to construct the
solution of the initial value problems for Exercises 107 -110.
Clear x, y, yprime
yprime[x_] Cos[x]2 Sin[x];
initxvalue π; inityvalue 1;
y[x_] Integrate[yprime[t], {t, initxvalue, x}] inityvalue
If the solution satisfies the differential equation and initial condition, the following yield True
yprime[x] D[y[x], x] //Simplify
y[initxvalue]inityvalue
Since exercise 110 is a second order differential equation, two integrations will be required.
Clear[x, y, yprime]
y2prime[x_] 3 Exp[x/2] 1;
initxval 0; inityval 4; inityprimeval 1;
yprime[x_] Integrate[y2prime[t],{t, initxval, x}] inityprimeval
y[x_] Integrate[yprime[t], {t, initxval, x}] inityval
Verify that y[x] solves the differential equation and initial condition and plot the solution (red) and its derivative
(blue).
y2prime[x] D[y[x], {x, 2}]//Simplify
y[initxval]inityval
yprime[initxval]inityprimeval
Plot[{y[x], yprime[x]}, {x, initxval 3, initxval 3}, PlotStyle {RGBColor[1,0,0], RGBColor[0,0,1]}]
Copyright 2016 Pearson Education, Ltd.
Chapter 4 Practice Exercises
CHAPTER 4
261
PRACTICE EXERCISES
1. No, since f ( x) x3 2 x tan x f ( x) 3 x 2 2 sec2 x 0 f ( x) is always increasing on its domain
2. No, since g ( x) csc x 2 cot x g ( x) csc x cot x 2 csc 2 x cos2 x
sin x
g ( x) is always decreasing on its domain
2 1 (cos x 2) 0
sin 2 x
sin 2 x
3. No absolute minimum because lim (7 x)(11 3 x)1/3 . Next f ( x) (11 3 x)1/3 (7 x)(11 3 x) 2/3
(113 x ) (7 x )
(113 x )2/3
x
4(1 x )
(113 x )
x 1 and x 11
are critical points. Since f 0 if x 1 and f 0
2/3
3
if x 1, f (1) 16 is the absolute maximum.
4. f ( x) ax2b f ( x)
x 1
a ( x 2 1) 2 x ( ax b )
2
( x 1)
2
( ax 2 2bx a )
( x 2 1)2
1 (9a 6b a ) 0 5a 3b 0. We
; f (3) 0 64
require also that f (3) 1. Thus 1 3a8b 3a b 8. Solving both equations yields a 6 and b 10. Now,
f ( x)
2(3 x 1)( x 3)
( x 2 1) 2
so that f | | | | . Thus f changes sign at x 3 from
1
1/3
1
3
positive to negative so there is a local maximum at x 3 which has a value f (3) 1.
5. Yes, because at each point of [0, 1) except x 0, the function’s value is a local minimum value as well as a
local maximum value. At x 0 the function’s value, 0, is not a local minimum value because each open
interval around x 0 on the x-axis contains points to the left of 0 where f equals 1.
6. (a) The first derivative of the function f ( x) x3 is zero at x 0 even though f has no local extreme value at
x 0.
(b) Theorem 2 says only that if f is differentiable and f has a local extreme at x c then f (c) 0. It does not
assert the (false) reverse implication f (c) 0 f has a local extreme at x c.
7. No, because the interval 0 x 1 fails to be closed. The Extreme Value Theorem says that if the function is
continuous throughout a finite closed interval a x b then the existence of absolute extrema is guaranteed on
that interval.
8. The absolute maximum is | 1| 1 and the absolute minimum is |0| 0. This is not inconsistent with the
Extreme Value Theorem for continuous functions, which says a continuous function on a closed interval attains
its extreme values on that interval. The theorem says nothing about the behavior of a continuous function on an
interval which is half open and half closed, such as [1, 1), so there is nothing to contradict.
9. (a) There appear to be local minima at x 1.75
and 1.8. Points of inflection are indicated at
approximately x 0 and x 1.
(b) f ( x) x 7 3x5 5 x 4 15 x 2 x 2 ( x 2 3)( x3 5). The pattern y | | | |
3
indicates a local maximum at x 5 and local minima at x 3.
Copyright 2016 Pearson Education, Ltd.
3
0
3
5
3
262
Chapter 4 Applications of Derivatives
(c)
10. (a) The graph does not indicate any local
extremum. Points of inflection are indicated
at approximately x 34 and x 1.
(b) f ( x) x7 2 x 4 5 103 x 3 ( x3 2)( x 7 5). The pattern f )( | | indicates a
x
7
3
0
7
5
3
2
local maximum at x 5 and a local minimum at x 2.
(c)
11. (a) g (t ) sin 2 t 3t g (t ) 2sin t cos t 3 sin(2t ) 3 g 0 g (t ) is always falling and hence must
decrease on every interval in its domain.
(b) One, since sin 2 t 3t 5 0 and sin 2 t 3t 5 have the same solutions: f (t ) sin 2 t 3t 5 has the same
derivative as g (t ) in part (a) and is always decreasing with f ( 3) 0 and f (0) 0. The Intermediate Value
Theorem guarantees the continuous function f has a root in [3, 0].
dy
12. (a) y tan d sec2 0 y tan is always rising on its domain y tan increases on every
interval in its domain
(b) The interval 4 , is not in the tangent’s domain because tan is undefined at 2 . Thus the tangent
need not increase on this interval.
13. (a) f ( x ) x 4 2 x 2 2 f ( x) 4 x3 4 x. Since f (0) 2 0, f (1) 1 0 and f ( x) 0 for 0 x 1, we
may conclude from the Intermediate Value Theorem that f ( x) has exactly one solution when 0 x 1.
(b) x 2 2 2 48 0 x 2 3 1 and x 0 x .7320508076 .8555996772
Copyright 2016 Pearson Education, Ltd.
Chapter 4 Practice Exercises
14. (a) y xx1 y
263
1
0, for all x in the domain of xx1 y xx1 is increasing in every interval in its
( x 1)2
domain.
(b) y x3 2 x y 3 x 2 2 0 for all x the graph of y x3 2 x is always increasing and can never
have a local maximum or minimum
15. Let V (t ) represent the volume of the water in the reservoir at time t, in minutes, let V (0) a0 be the initial amount
and V (1440) a0 1,000,000,000 liters be the amount of water contained in the reservoir after the rain, where
24 h 1440 min. Assume that V (t ) is continuous on [0, 1440] and differentiable on (0, 1440). The Mean Value
V (1400) V (0)
Theorem says that for some t0 in (0, 1440) we have V (t0 ) 14400
a0 1,000,000,000 a0
1440
L
1,000,000,000
694,444 L/min. Therefore at t0 the reservoir’s volume was increasing at a rate in excess of
1440 min
500,000 L/min.
16. Yes, all differentiable functions g ( x) having 3 as a derivative differ by only a constant. Consequently, the
d (3 x ). Thus g ( x ) 3 x K , the same form as F ( x ).
difference 3 x g ( x) is a constant K because g ( x) 3 dx
17. No, xx1 1 x11 xx1 differs from x11 by the constant 1. Both functions have the same derivative
( x(x1)1)x(1) ( x11) dxd x11 .
d
x
dx x 1
18. f ( x) g ( x)
2
2
2 x f ( x ) g ( x ) C for some constant C the graphs differ by a vertical shift.
( x 2 1) 2
19. The global minimum value of 12 occurs at x 2.
20. (a) The function is increasing on the intervals [3, 2] and [1, 2].
(b) The function is decreasing on the intervals [2, 0) and (0, 1].
(c) The local maximum values occur only at x 2, and at x 2; local minimum values occur at x 3 and
at x 1 provided f is continuous at x 0.
21. (a) t 0, 6, 12
(b)
t 3, 9
(c)
6 t 12
(d)
0 t 6, 12 t 14
22. (a) t 4
(b)
at no time
(c)
0t 4
(d)
4t 8
23.
24.
Copyright 2016 Pearson Education, Ltd.
264
Chapter 4 Applications of Derivatives
25.
26.
27.
28.
29.
30.
31.
32.
33. (a) y 16 x 2 y | | the curve is rising on ( 4, 4), falling on ( , 4) and (4, )
4
4
a local maximum at x 4 and a local minimum at x 4; y 2 x y | the curve is
concave up on (, 0), concave down on (0, ) a point of inflection at x 0
(b)
Copyright 2016 Pearson Education, Ltd.
0
Chapter 4 Practice Exercises
265
34. (a) y x 2 x 6 ( x 3)( x 2) y | | the curve is rising on (, 2) and (3, ),
2
3
falling on ( 2, 3) local maximum at x 2 and a local minimum at x 3; y 2 x 1 y |
1/2
concave up on 12 , , concave down on , 12 a point of inflection at x 12
(b)
35. (a) y 6 x ( x 1)( x 2) 6 x3 6 x 2 12 x y | | | the graph is rising on (1, 0)
1
0
2
and (2, ), falling on (, 1) and (0, 2) a local maximum at x 0, local minima at x 1 and
x y | |
and , , concave down on , points of
x 2; y 18 x 2 12 x 12 6 (3 x 2 2 x 2) 6 x 13 7
the curve is concave up on , 13 7
1 7
3
1 7
3
1 7
3
1 7
3
1 7 1 7
3
3
inflection at x 13 7
(b)
36 . (a) y x 2 (6 4 x) 6 x 2 4 x3 y | | the curve is rising on , 32 , falling on
0
3/2
32 , a local maximum at x 32 ; y 12 x 12 x2 12 x(1 x) y 0| 1| concave
up on (0, 1), concave down on ( , 0) and (1, ) points of inflection at x 0 and x 1
(b)
37. (a) y x 4 2 x 2 x 2 ( x 2 2) y | | | the curve is rising on , 2 and
2
0
2
2, , falling on 2, 2 a local maximum at x 2 and a local minimum at x 2;
y 4 x3 4 x 4 x( x 1)( x 1) y | | | concave up on ( 1, 0) and (1, ),
1
0
1
concave down on ( , 1) and (0, 1) points of inflection at x 0 and x 1
Copyright 2016 Pearson Education, Ltd.
266
Chapter 4 Applications of Derivatives
(b)
38. (a) y 4 x 2 x 4 x 2 (4 x 2 ) y | | | the curve is rising on (2, 0) and (0, 2),
2
0
2
falling on (, 2) and (2, ) a local maximum at x 2, a local minimum at x 2; y 8 x 4 x3
4 x (2 x 2 ) y | | | concave up on , 2 and 0, 2 , concave
down on 2, 0 and
2
0
2
2, points of inflection at x 0 and x 2
(b)
39. The values of the first derivative indicate that the curve is rising on (0, ) and falling on (, 0). The slope of
the curve approaches as x 0 , and approaches as x 0 and x 1. The curve should therefore have a
cusp and local minimum at x 0, and a vertical tangent at x 1.
40. The values of the first derivative indicate that the curve is rising on 0, 12 and (1, ), and falling on ( , 0) and
1 , 1 . The derivative changes from positive to negative at x 1 , indicating a local maximum there. The slope
2
2
of the curve approaches as x 0 and x 1 , and approaches as x 0 and as x 1 , indicating
cusps and local minima at both x 0 and x 1.
Copyright 2016 Pearson Education, Ltd.
Chapter 4 Practice Exercises
267
41. The values of the first derivative indicate that the curve is always rising. The slope of the curve approaches
as x 0 and as x 1, indicating vertical tangents at both x 0 and x 1.
42. The graph of the first derivative indicates that the curve is rising on 0, 17 16 33 and 17 16 33 , , falling on
(, 0) and 17 16 33 , 17 16 33 a local maximum at x 17 16 33 , a local minimum at x 17 16 33 . The derivative
approaches as x 0 and x 1, and approaches as x 0 , indicating a cusp and local minimum at
x 0 and a vertical tangent at x 1.
43. y xx13 1 x 43
2
45. y x x1 x 1x
44. y x2x5 2 x10
5
2
46. y x xx 1 x 1 1x
Copyright 2016 Pearson Education, Ltd.
268
Chapter 4 Applications of Derivatives
3
4
2
48. y x 21 x 2 12
47. y x 2x 2 x2 1x
x
2
49. y x 2 4 1 21
x 3
50. y
x 3
x
x2 1 4
x 4
x2 4
2
51. (a) Maximize f ( x) x 36 x x1/2 (36 x)1/2 where 0 x 36
f ( x) 12 x 1/2 12 (36 x)1/2 (1)
36 x x
derivative fails to exist at 0 and 36; f (0) 6, and
2 x 36 x
f (36) 6 the numbers are 0 and 36
(b) Maximize g ( x) x 36 x x1/2 (36 x)1/2 where 0 x 36 g ( x) 12 x 1/2 12 (36 x) 1/2 (1)
36 x x
critical points at 0, 18 and 36; g (0) 6, g (18) 2
2 x 36 x
18 6 2 and g (36) 6 the numbers
are 18 and 18
52. (a) Maximize f ( x) x (20 x) 20 x1/2 x3/2 where 0 x 20 f ( x) 10 x 1/2 32 x1/2 203 x 0
are critical points; f (0) f (20) 0 and f
x 0 and x 20
3
are 20
and 40
.
3
3
1/2
(b) Maximize g ( x) x 20 x x (20 x)
20
3
20
3
20 20
3
2 x
40 20
the numbers
3 3
where 0 x 20 g ( x) 2 20 x 1 0 20 x 12
2 20 x
x 79
. The critical points are x 79
and x 20. Since g 79
81
and g (20) 20, the numbers must be
4
4
4
4
79 and 1 .
4
4
53. A( x) 12 (2 x)(27 x 2 ) for 0 x 27
A( x) 3(3 x)(3 x) and A( x) 6 x. The
critical points are 3 and 3, but 3 is not in the
domain. Since A(3) 18 0 and A 27 0, the
maximum occurs at x 3 the largest area
is A(3) 54 sq units.
Copyright 2016 Pearson Education, Ltd.
Chapter 4 Practice Exercises
269
54. The volume is V x 2 h 1 h 12 . The surface
2
area is S ( x) x 4 x
S ( x)
3
x
2( x 2)2( x3 2)
x2
1
x2
x
2
4x , where x 0
the critical points are 0 and
2 , but 0 is not in the domain. Now
S ( 3 2) 2 83 0 at x 3 2 there is a
2
minimum. The dimensions 3 2 m by 3 2 m by
3
2 / 2 m minimize the surface area.
r2 3
55. From the diagram we have h2
2
2
2
r 2 124h . The volume of the cylinder is
V r 2 h 124h
2
h (12h h ), where
3
4
3. Then V (h) 34 (2 h)(2 h) the
0h2
critical points are 2 and 2, but 2 is not in the
domain. At h 2 there is a maximum since
V (2) 3 0. The dimensions of the largest
cylinder are radius 2 and height 2.
56. From the diagram we have x radius and y height
12 2x and V ( x) 13 x 2 (12 2 x), where 0 x 6
V ( x) 2 x(4 x) and V (4) 8 . The critical
points are 0 and 4; V (0) V (6) 0 x 4 gives the
maximum. Thus the values of r 4 and h 4 yield
the largest volume for the smaller cone.
x , where p is the profit on grade B tires and 0 x 4. Thus
57. The profit P 2 px py 2 px p 40510
x
P ( x)
2p
(5 x ) 2
2
( x 10 x 20) the critical points are 5 5 , 5, and 5 5 , but only 5 5 is in the
domain. Now P ( x) 0 for 0 x 5 5 and P ( x) 0 for 5 5 x 4 at x 5 5 there is a local
maximum. Also P(0) 8 p, P 5 5 4 p 5 5 11 p, and P(4) 8 p at x 5 5 there is an
absolute maximum. The maximum occurs when x 5 5 and y 2 5 5 , the units are hundreds of tires,
i.e., x 276 tires and y 553 tires.
58. (a) The distance between the particles is | f (t )| where f (t ) cos t cos t 4 . Then,
f (t ) sin t sin t . Solving f (t ) 0 graphically, we obtain t 1.178, t 4.320, and so on.
4
Copyright 2016 Pearson Education, Ltd.
270
Chapter 4 Applications of Derivatives
Alternatively, f (t ) 0 may be solved analytically as follows. f (t ) sin t 8 8 sin t 8 8
critical points occur when cos t 8 0, or t 38 k . At each of these values, f (t ) cos 38 0.765
sin t 8 cos 8 cos t 8 sin 8 sin t 8 cos 8 cos t 8 sin 8 2sin 8 cos t 8 so the
units, so the maximum distance between the particles is 0.765 units.
(b) Solving cos t cos t 4 graphically, we obtain t 2.749, t 5.890, and so on.
Alternatively, this problem can be solved analytically as follows.
cos t cos t 4
cos 8 8 cos t 8 8
cos t 8 cos 8 sin t 8 sin 8 cos t 8 cos 8 sin t 8 sin 8
2sin t 8 sin 8 0
sin t 8 0; t 78 k
t
The particles collide when t 78 2.749. (Plus multiples of if they keep going.)
59. The dimensions will be x cm by 25 2 x cm by 40 2 x cm, so V ( x) x(25 2 x)(40 2 x)
4 x3 130 x 2 1000 x for 0 x 12. Then V ( x) 12 x 2 260 x 1000 4( x 5)(3x 50), so the critical
point in the correct domain is x 5. This critical point corresponds to the maximum possible volume because
V ( x) 0 for 0 x 5 and V ( x) 0 for 5 x 12.5. The box of largest volume has a height of 5 cm and a
base measuring 15 cm by 30 cm, and its volume is 2250 cm3.
Graphical support:
60. The length of the ladder is
d1 d 2 2.4sec 1.8csc . We wish to maximize
I ( ) 2.4sec 1.8csc
I ( ) 2.4sec tan 1.8csc cot .
Then I ( ) 0
3
2.4sin 3 1.8cos3 0 tan 26
d1 3.243 and d 2 2.677 the length of the
ladder is about 5.92 m.
Copyright 2016 Pearson Education, Ltd.
Chapter 4 Practice Exercises
271
61. g ( x) 3 x x3 4 g (2) 2 0 and g (3) 14 0 g ( x) 0 in the interval [2, 3] by the Intermediate
Value Theorem. Then g ( x) 3 3x 2 xn 1 xn
3 xn xn3 4
33 xn2
forth to x5 2.195823345.
; x0 2 x1 2.22 x2 2.196215, and so
62. g ( x) x 4 x3 75 g (3) 21 0 and g (4) 117 0 g ( x) 0 in the interval [3, 4] by the Intermediate
x 4 x3 75
Value Theorem. Then g ( x) 4 x3 3 x 2 xn 1 xn n 3 n
4 xn 3 xn2
; x0 3 x1 3.259259 x2 3.229050,
and so forth to x5 3.22857729.
6
3
63.
( x5 4 x 2 9) dx x5 dx 4 x 2 dx 9 1dx x6 43x 9 x C
64.
6t t dt 6t dt tdt 6
65.
66.
67.
dr
( r 3)3
68.
5 dr
69.
t3
4
5
t 53
t
t
5
4
t t t3
r 3
2
t6
6
t3
4
5
1 t4 t2 C t6 t4 t2 C
4 4
2
16 2
3
dt t tdt
dt 5t
3/2
t
3/2
dt 5t
2
3
3 1
t 2
3
dt 4t dt 5
(r 3) 3 dr
5 r 3
5 dt
t3
1
31
5/ 2
2
dt t3 5 t31 C t 5 52 C 52 t 5/2 52 C
2t
2t
1
2
2
31
32 1
4 t31 C 10t 1 2 22 C
t
( r 3) 31
( r 3) 2
C 2 C 1 2 C
31
2( r 3)
r 3
dr 5
2 1
21
(1) C
5
r 3
C
9 2 3 1 d ------- (1)
Put 3 1 t 3 2 d dt 2 d dt3
(1) becomes 9 2 3 1 d
70.
x
5 x 2
9
3
1 1
2
t dt 3 t1
2
1
3
dx ------- (1)
Put 5 x 2 t 2 xdx dt xdx dt2
(1) becomes
x
5 x
1 1
2
dx dt /2 12 t 1
2
t
2 1
C
C 3 23 t 2 C 2 3 1
1
2
12 t1 C 5 x 2 C.
2
Copyright 2016 Pearson Education, Ltd.
32
272
71.
Chapter 4 Applications of Derivatives
x 4 1 x5
1/5
dx ------- (1)
Put 1 x5 t 5 x 4 dx dt x 4 dx dt5
(1) becomes
72.
1 1
5
(t )1/5 dt5 15 t 1
4
5
5 1
4
15 t4 C 15 54 t 5 C 14 5 x 2
5
4/5
C.
(5 x)5/7 dx ------- (1)
Put 5 x t dx dt
(1) becomes
5 1
7
(5 x)5/7 dx t 5/7 (dt ) t5
1
7
7 (5 x )12/7 C.
C 12
s C. Differentiate the solution to check:
73. Our trial solution based on the chain rule is 10 tan 10
s C sec 2 s . Thus sec 2 s ds 10 tan s C.
10 tan 10
10
10
10
d
ds
74.
csc2 es ds ------- (1)
Put es t eds dt ds dte
(1) becomes
75.
csc 3 cot 3 d ------- (1)
Put
3 t 3d dt d dt
3
(1) becomes
76.
csc2 es ds 1e ( cot es ) C cotees C.
csc 3 cot 3 d 1 csc t cot t dt 1 ( csc t ) C 1 csc t 3 C.
3
3
3
sec 5t tan 5t dt ------- (1)
Put 5t p dt 5dp
(1) becomes sec 5t tan 5t dt sec p tan p dp 5sec p C 5sec 5t C.
77. Using the hint cos 2 6x
1cos 3x
1 cos 2 6x
2
2
cos 2 6x dx 12 1 cos 3x dx 12 x 3sin 3x C.
78. Put 2x t dx 2dt
(1) becomes sec 2 2x dx sec 2 t dt 2 tan t C.
x 2 1 dx
x2
y x 1x 1
79. y
80. y
(1 x 2 ) dx x x 1 C x 1x C ; y 1 when x 1 1 11 C 1 C 1
x 1x dx ( x2 2 x1 ) dx ( x2 2 x2 ) dx x3 2 x x1 C x3 2 x 1x C;
2
3
2
3
y 1 when x 1 13 2 11 C 1 C 13 y x3 2 x 1x 13
Copyright 2016 Pearson Education, Ltd.
3
Chapter 4 Additional and Advanced Exercises
81.
dr
dt
15 t dt (15t 3t
1/2
3
t
1/2
273
) dt 10t 3/2 6t1/2 C ; dr
8 when t 1 10(1)3/2 6(1)1/2 C 8
dt
10t 3/2 6t1/2 8 r (10t 3/2 6t1/2 8) dt 4t 5/2 4t 3/2 8t C ; r 0 when t 1
C 8. Thus dr
dt
4(1)5/2 4(1)3/2 8(1) C1 0 C1 0. Therefore, r 4t 5/2 4t 3/2 8t
82.
d 2 r cos t dt sin t C ; r 0 when t 0 sin 0 C 0 C 0. Thus, d 2 r sin t
dt 2
dt 2
dr
sin t dt cos t C1; r 0 when t 0 1 C1 0 C1 1. Then
dt
dr cos t 1 r (cos t 1) dt sin t t C ; r 1 when t 0 0 0 C 1 C 1. Therefore,
2
2
2
dt
r sin t t 1
CHAPTER 4
ADDITIONAL AND ADVANCED EXERCISES
1. If M and m are the maximum and minimum values, respectively, then m f ( x) M for all x
then f is constant on I.
I . If m M
2 x 0
has an absolute minimum value of 0 at x 2 and an absolute
2. No, the function f ( x) 3 x 6,
2
9x ,0x2
maximum value of 9 at x 0, but it is discontinuous at x 0.
3. On an open interval the extreme values of a continuous function (if any) must occur at an interior critical point.
On a half-open interval the extreme values of a continuous function may be at a critical point or at the closed
endpoint. Extreme values occur only where f 0, f does not exist, or at the endpoints of the interval. Thus
the extreme points will not be at the ends of an open interval.
4. The pattern f | | | | indicates a local maximum at x 1 and a local minimum
1
at x 3.
2
3
4
5. (a) If y 6( x 1)( x 2) 2 , then y 0 for x 1 and y 0 for x 1. The sign pattern is
f | | f has a local minimum at x 1. Also y 6( x 2) 2 12( x 1)( x 2)
1
2
6( x 2)(3x) y 0 for x 0 or x 2, while y 0 for 0 x 2. Therefore f has points of inflection at
x 0 and x 2. There is no local maximum.
(b) If y 6 x ( x 1)( x 2), then y 0 for x 1 and 0 x 2; y 0 for 1 x 0 and x 2. The sign pattern
is y | | | . Therefore f has a local maximum at x 0 and local minima at x 1 and
1
x 2. Also, y 18
0
2
1 7
x 3
x , so y 0 for
1 7
3
1 7
x 13 7 and y 0 for all other x f
3
has points of inflection at x 13 7 .
f (6) f (0)
6. The Mean Value Theorem indicates that 60
indicates the most that f can increase is 12.
f (c) 2 for some c in (0, 6). Then f (6) f (0) 12
7. If f is continuous on [ a, c) and f ( x ) 0 on [ a, c), then by the Mean Value Theorem for all x [a, c ) we have
f (c) f ( x)
0 f (c) f ( x) 0 f ( x) f (c). Also if f is continuous on (c, b] and f ( x) 0 on (c, b], then
c x
for all x (c, b] we have
x [a, b].
f ( x ) f (c )
0 f ( x) f (c) 0 f ( x) f (c). Therefore f ( x) f (c) for all
x c
Copyright 2016 Pearson Education, Ltd.
274
Chapter 4 Applications of Derivatives
8. (a) For all x, ( x 1) 2 0 ( x 1) 2 (1 x 2 ) 2 x (1 x 2 ) 12
(b) There exists c ( a, b) such that
| f (b) f (a)| 12 |b a | .
c f (b ) f ( a ) f (b ) f ( a ) c
ba
ba
1 c 2
1 c 2
x 1.
2
1 x 2
1
2 , from part (a)
9. No. Corollary 1 requires that f ( x) 0 for all x in some interval I, not f ( x ) 0 at a single point in I.
10. (a) h( x) f ( x) g ( x) h( x) f ( x) g ( x) f ( x) g ( x) which changes signs at x a since f ( x), g ( x) 0
when x a, f ( x), g ( x) 0 when x a and f ( x), g ( x) 0 for all x. Therefore h( x) does have a local
maximum at x a.
(b) No, let f ( x) g ( x ) x3 which have points of inflection at x 0, but h( x ) x 6 has no point of inflection
(it has a local minimum at x 0).
11. From (ii), f (1) b1ca2 0 a 1; from (iii), either 1 lim f ( x) or 1 lim f ( x). In either case,
x
1 1x
x 1
b
1,
and
c
1.
For
if
then
lim
1
b
0
lim
0 and if
2
2
2
x bx cx 2 x bx c x
x x c x
lim f ( x) lim
x
c 0, then lim
1 1x
lim
2
x bx x
12.
x
1 1x
1 1x
x
2
x
. Thus a 1, b 0, and c 1.
dy
2
2 k 4 k 2 36
3
x
2
kx
3
0
x
x has only one value when 4k 2 36 0 k 2 9 or k 3.
dx
6
13. The area of the ABC is A( x) 12 (2) 1 x 2
(1 x 2 )1/2 , where 0 x 1. Thus A( x) x 2
1 x
0 and 1 are critical points. Also A ( 1) 0 so
A(0) 1 is the maximum. When x 0 the ABC is
isosceles since AC BC 2.
14.
f ( c h ) f ( c )
f (c ) for
h
h 0
lim
12 | f (c) | 0 there exists a
0 such that 0 | h |
f ( c h ) f ( c )
f ( c h )
f (c) 12 | f (c) |
f (c) 12 | f (c) | . Then f (c) 0 12 | f (c) |
h
h
f ( c h )
f (c) 12 | f (c) | h f (c) 12 | f (c) | . If f (c) 0, then | f (c) | f (c)
f ( c h )
f ( c h )
32 f (c) h 12 f (c) 0; likewise if f (c ) 0, then 0 12 f (c ) h 23 f (c).
(a) If f (c) 0, then h 0 f (c h) 0 and 0 h
maximum.
(b) If f (c) 0, then h 0 f (c h) 0 and 0 h
minimum.
f (c h) 0. Therefore, f (c) is a local
f (c h) 0. Therefore, f (c) is a local
15. The time it would take the water to hit the ground from height y is
2y
, where g is the acceleration of gravity.
g
The product of time and exit velocity (rate) yields the distance the water travels:
D( y )
2y
g
64(h y ) 8
2 ( hy y 2 )1/2 ,
g
are critical points. Now D(0) 0, D h2 8
0 y h D ( y ) 4
2
g
h h2 h2
2 1/2
2 ( hy y 2 ) 1/2 ( h 2 y ) 0, h
2
g
and h
4h g2 and D( h) 0 the best place to
drill the hole is at y h2 .
Copyright 2016 Pearson Education, Ltd.
Chapter 4 Additional and Advanced Exercises
275
tan tan
; and tan ah . These equations give
tan
16. From the figure in the text, tan( ) b n a ; tan( ) 1 tan
a
b a tan h
a
h
1 tan
h
h tan a
h a tan . Solving for tan
gives tan
bh
or (h 2 a(b a )) tan bh.
h 2 a (b a )
d
(h 2 a (b a )) sec 2 dh b. Then
2
2
Differentiating both sides with respect to h gives 2h tan
d
0 2h tan
dh
bh
h 2 a (b a )
b 2h
b 2bh 2 bh ab(b a ) h a (b a ) h a(a b).
17. The surface area of the cylinder is S 2 r 2 2 rh.
From the diagram we have Rr HH h h RHR rH
and S ( r ) 2 r (r h) 2 r r H r HR
2 1 HR
r 2 Hr, where 0 r R.
2
Case 1: H R S (r ) is a quadratic equation containing the origin and concave upward S ( r ) is maximum
at r R.
Case 2: H R S (r ) is a linear equation containing the origin with a positive slope S (r ) is maximum
at r R.
Case 3: H R S (r ) is a quadratic equation containing the origin and concave downward.
RH . For simplification
Then dS
4 1 HR r 2 H and dS
0 4 1 HR r 2 H 0 r 2( H
dr
dr
R)
RH .
we let r* 2( H
R)
RH
(a) If R H 2 R, then 0 H 2 R H 2( H R) r* 2( H
R. Therefore, the maximum occurs at
R)
the right endpoint R of the interval 0 r R because S (r ) is an increasing function of r.
2
(b) If H 2 R, then r* 22RR R S (r ) is maximum at r R.
RH
(c) If H 2 R, then 2 R H 2 H H 2( H R ) 2( HH R ) 1 2( H
R r* R. Therefore, S (r ) is
R)
RH .
a maximum at r r* 2( H
R)
Conclusion: If H (0, 2 R], then the maximum surface area is at r R. If H
RH .
at r r* 2( H
R)
(2 R, ), then the maximum is
18. f ( x) mx 1 1x f ( x) m 12 and f ( x) 23 0 when x 0. Then f ( x) 0 x
x
If f
0, then m 1 m 2
1
m
1 yields a minimum.
m
m 1 0 m 14 . Thus the smallest acceptable value for m is 14 .
x
19. (a) The profit function is P( x) (c ex) x (a bx) ex 2 (c b) x a. P ( x ) 2ex c b 0
x c2eb . P ( x) 2e 0 if e 0 so that the profit function is maximized at x c2eb .
(b) The price therefore that corresponds to a production level yielding a maximum profit is
p x c b c e c2eb c 2b dollars.
2e
(c b) c2eb a (c4be ) a.
(c) The weekly profit at this production level is P( x) e c2eb
2
2
(d) The tax increases cost to the new profit function is F ( x) (c ex) x (a bx tx) ex 2 (c b t ) x a.
Now F ( x) 2ex c b t 0 when x t b2e c c 2bet . Since F ( x) 2e 0 if e 0, F is maximized
when x c 2bet units per week. Thus the price per unit is p c e c 2bet c 2b t dollars. Thus, such a tax
increases the cost per unit by c 2b t c 2b 2t dollars if units are priced to maximize profit.
Copyright 2016 Pearson Education, Ltd.
276
Chapter 4 Applications of Derivatives
20. (a)
The x-intercept occurs when 1x 3 0 1x 3 x 13 .
1 3
f (x )
x
(b) By Newton’s method, xn 1 xn f ( xn ) . Here f ( xn ) xn2 21 . So xn 1 xn n1 xn x1 3 xn2
xn
n
xn2
xn xn 3 xn2 2 xn 3 xn2 xn (2 3 xn ).
xq a
f (x )
21. x1 x0 f ( x0 ) x0 0 q 1
qx0q x0q a
qx0q 1
qx0
0
x0q ( q 1) a
q 1
and qa1 with weights m0 q and m1 1q .
x
qx0q 1
x0
n
so that x is a weighted average of x
q 1
q
a
1
x0q 1 q
1
0
0
In the case where x0
q 1
a we have x q a and x a
1
0
x0q 1
x0q 1 q
a
x0q 1
1
qa1
x0
q
dy
q 1 1
q
q
dy
22. We have that ( x h) 2 ( y h)2 r 2 and so 2( x h) 2( y h) dx 0 and 2 2 dx 2( y h)
dy
dy
2 x 2 y dx 2h 2h dx , by the former. Solving for h, we obtain h
dy
equation yields 2 2 dx 2 y
d2y
dx
2
dy
2
x y dx
dy
1 dx
dy
x y dx
dy
1 dx
a .
x0q 1
d2y
dx 2
0 hold. Thus
. Substituting this into the second
dy
0. Dividing by 2 results in 1 dx y
d2y
dx
2
dy
x y dx
dy
1 dx
0.
23. (a) a (t ) s (t ) k (k 0) s (t ) kt C1 , where s (0) 30 C1 30 s (t ) kt 30. So
2
2
2
s (t ) kt2 30t C2 where s (0) 0 C2 0 so s (t ) kt2 30t. Now s (t ) 30 when kt2 30t 30.
2
Solving for t we obtain t 30 30k 60k . At such t we want s (t ) 0, thus k
30 302 60 k
k
30 0 or
2
2
k 30 30k 60k 30 0. In either case we obtain 302 60k 0 so that k 30
15 m/s 2 .
60
2
(b) The initial condition that s (0) 15 m/s implies that s (t ) kt 15 and s (t ) kt2 15t where k is as
above. The car is stopped at a time t such that s (t ) kt 15 0 t 15
. At this time the car has
k
112.5 60 7.5 m. Thus halving the initial
15 15k 152k 112.5
k
30
traveled a distance s 15
2k 15
k
k
2
2
2
velocity quarters stopping distance.
24. h( x ) f 2 ( x) g 2 ( x) h( x) 2 f ( x) f ( x ) 2 g ( x) g ( x) 2[ f ( x) f ( x) g ( x) g ( x)]
2[ f ( x) g ( x) g ( x)( f ( x ))] 2 0 0. Thus h( x) c, a constant. Since h(0) 5, h( x ) 5 for all x in the
domain of h. Thus h(10) 5.
dy
25. Yes. The curve y x satisfies all three conditions since dx 1 everywhere, when x 0, y 0, and
everywhere.
Copyright 2016 Pearson Education, Ltd.
d2y
dx 2
0
Chapter 4 Additional and Advanced Exercises
277
26. y 3 x 2 2 for all x y x3 2 x C where 1 13 2 1 C C 4 y x3 2 x 4.
3
27. s (t ) a t 2 v s (t ) 3t C. We seek v0 s (0) C. We know that s (t*) b for some t* and s is at a
4
4
maximum for this t*. Since s (t ) 12t Ct k and s (0) 0 we have that s (t ) 12t Ct and also s (t*) 0 so
b 31/3 C 4/3 43b
[ (3C )1/3 ]4
C (3C )1/3 b (3C )1/3 (C 312C ) b (3C )1/3 34C
12
(4b )3/ 4
(4b )3/ 4
C 3 . Thus v0 s (0) 3 2 3 2 b3/4 .
that t* (3C )1/3 . So
28. (a) s (t ) t1/2 t 1/2 v(t ) s (t ) 23 t 3/2 2t1/2 k where v(0) k 43 v(t ) 23 t 3/2 2t1/2 43 .
4 t 5/2 4 t 3/2 4 t k where s (0) k 4 . Thus s (t ) 4 t 5/2 4 t 3/2 4 t 4 .
(b) s (t ) 15
2
2
15
3
3
15
3
3
15
29. The graph of f ( x) ax 2 bx c with a 0 is a parabola opening upwards. Thus f ( x ) 0 for all x if f ( x ) 0
for at most one real value of x. The solutions to f ( x) 0 are, by the quadratic equation
we require (2b) 2 4ac 0 b 2 ac 0.
2b (2b ) 2 4 ac
. Thus
2a
30. (a) Clearly f ( x) (a1 x b1 )2 (an x bn )2 0 for all x. Expanding we see
a12 a22 an2 x 2 2 a1b1 a2b2 an bn x b12 b22 bn2 0. Thus
a1b1 a2b2 anbn 2 a12 a22 an2 b12 b22 bn2 0 by Exercise 29. Thus
a1b1 a2b2 an bn 2 a12 a22 an2 b12 b22 bn2 .
f ( x) a12 x 2 2a1b1 x b12 an2 x 2 2an bn x bn2
(b) Referring to Exercise 29: It is clear that f ( x) 0 for some real x b 2 4ac 0, by quadratic formula.
Now notice that this implies that f ( x) (a1 x b1 )2 (an x bn ) 2
a12 a22 an2 x 2 2 a1b1 a2 b2 an bn x b12 b22 bn2 0
2
a1b1 a2b2 an bn a12 a22 an2 b12 b22 bn2 But now f ( x) 0 ai x bi 0
a1b1 a2 b2 an bn a12 a22 an2 b12 b22 bn2 0
2
for all i 1, 2, , n ai x bi 0 for all i 1, 2, , n.
Copyright 2016 Pearson Education, Ltd.
CHAPTER 5 INTEGRALS
5.1
AREA AND ESTIMATING WITH FINITE SUMS
1. f ( x) x 2
(a) x
(b)
Since f is increasing on [0, 1], we use left endpoints to
obtain lower sums and right endpoints to obtain upper
sums.
1 0
12 and xi ix 2i a lower sum is
2
1 0
x 4 14 and xi ix 4i a lower sum is
(c) x
(d) x
1
2
i0
3
2
i 0
2
2
i 1
4
2
2i 12 12 02 12 81
1 0 1
2 and xi ix 2i an upper sum is
2
1 0
14 and xi ix 4i an upper sum is
4
2
2
2
2i 12 12 12 12 85
2
30 15
32
4i 14 14 14 12 34 12 14 16
2
2
2
i 1
Since f is increasing on [0, 1], we use left endpoints to
obtain lower sums and right endpoints to obtain upper
sums.
1 0 1
2 and xi ix 2i a lower sum is
2
1
3
i0
3
3
i0
2
3
i 1
4
3
2i 12 12 03 12 161
3
36 9
4i 14 14 03 14 12 34 256
64
(b)
1 0
x 4 14 and xi i x 4i a lower sum is
(c)
1 0
x 2 12 and xi ix 2i an upper sum is
(d) x
2
4i 14 14 02 14 12 34 14 87 327
2. f ( x) x3
(a) x
1 0 1
4 and xi ix 4i an upper sum is
4
3
3
3
2i 12 12 12 13 12 89 169
3
25
4i 14 14 14 12 34 13 100
256 64
3
3
3
i 1
Copyright 2016 Pearson Education, Ltd.
279
280
Chapter 5 Integrals
3. f ( x) 1x
Since f is decreasing on [1, 5], we use left endpoints to
obtain upper sums and right endpoints to obtain lower
sums.
2
(a) x 521 2 and xi 1 i x 1 2i a lower sum is x1 2 2 13 15 16
15
i
(b)
(c)
(d)
i 1
4
5
1
1 1 1 1 1 1 1 77
x 4 1 and xi 1 ix 1 i a lower sum is
xi
60
2 3 4 5
i1
1
5 1
1 2 2 1 1 8
x 2 2 and xi 1 i x 1 2i an upper sum is
xi
3
3
i0
3
5 1
1 1 1 1 1 1 1 25
x 4 1 and xi 1 ix 1 i an upper sum is
2 3 4
xi
12
i0
Since f is increasing on [2, 0] and decreasing on
[0, 2], we use left endpoints on [2, 0] and right
endpoints on [0, 2] to obtain lower sums and use right
endpoints on [2, 0] and left endpoints on [0, 2] to
obtain upper sums.
4. f ( x) 4 x 2
2 ( 2)
2 and xi 2 ix 2 2i a lower sum is 2 (4 (2)2 ) 2 (4 22 ) 0
2
1
4
2 ( 2)
x 4 1 and xi 2 i x 2 i a lower sum is (4 ( xi ) 2 ) 1 (4 ( xi ) 2 ) 1
i 0
i 3
2
2
2
2
(a) x
(b)
1((4 (2) ) (4 (1) ) (4 1 ) (4 2 )) 6
2 ( 2)
(c) x 2 2 and xi 2 ix 2 2i an upper sum is 2 (4 (0) 2 ) 2 (4 02 ) 16
(d) x
2 ( 2)
1 and xi 2 ix 2 i an upper sum is
4
1((4 (1)2 ) (4 02 ) (4 02 ) (4 12 )) 14
5. f ( x) x 2
2
3
i 1
i 2
(4 ( xi )2 ) 1 (4 ( xi )2 ) 1
1 0 1
2
2
3 2
1 2
Using 2 rectangles x
12 4 4 1032 165
12 f 14 f 43
1 0 1
4
4
5 f 7
8
8
Using 4 rectangles x
14 f 18 f 83 f
2
2
2
2
21
14 18 83 85 78 64
Copyright 2016 Pearson Education, Ltd.
Section 5.1 Area and Estimating with Finite Sums
6. f ( x) x3
281
1 0 1
2
2
3 3
1 3
Using 2 rectangles x
12 4 4 22864 327
12 f 14 f 43
1 0
14
4
5 f 7
8
8
Using 4 rectangles x
14 f 81 f 83 f
3
3
3
14 1 3 35 7
7. f ( x) 1x
8
3
496 124 31
128
4 83
83
Using 2 rectangles x 521 2 2( f (2) f (4))
2 12 14 32
Using 4 rectangles Δx 541 1
32 f 52 f 72 f 92 1 23 52 72 92
1 f
496
3 1488
496 315
57 9 57 9
8. f ( x) 4 x 2
Using 2 rectangles x
2 ( 2)
2
2
2( f (1) f (1)) 2(3 3) 12
Using 4 rectangles x
2 ( 2)
1
4
1 f 3
2
2
1 f 32 f 12 f
2
2
2
2
1 4 32 4 12 4 12 4 32
16 94 2 14 2 16 10
11
2
9. (a) D (0)(1) (12)(1) (22)(1) (10)(1) (5)(1) (13)(1) (11)(1) (6)(1) (2)(1) (6)(1) 87 centimeters
(b) D (12)(1) (22)(1) (10)(1) (5)(1) (13)(1) (11)(1) (6)(1) (2)(1) (6)(1) (0)(1) 87 centimeters
10. (a) D (1)(300) (1.2)(300) (1.7)(300) (2.0)(300) (1.8)(300) (1.6)(300) (1.4)(300) (1.2)(300)
(1.0)(300) (1.8)(300) (1.5)(300) (1.2)(300) 5220 meters (NOTE: 5 minutes 300 seconds)
(b) D (1.2)(300) (1.7)(300) (2.0)(300) (1.8)(300) (1.6)(300) (1.4)(300) (1.2)(300) (1.0)(300)
(1.8)(300) (1.5)(300) (1.2)(300) (0)(300) 4920 meters (NOTE: 5 minutes 300 seconds)
11. (a)
D (0)(10) (15)(10) (5)(10) (12)(10) (10)(10) (15)(10) (12)(10) (5)(10) (7)(10)
(12)(10) (15)(10) (10)(10) 1180 m 1.18 km
(b) D (15)(10) (5)(10) (12)(10) (10)(10) (15)(10) (12)(10) (5)(10) (7)(10) (12)(10)
(15)(10) (10)(10) (12)(10) 1300 m 1.3 km
12. (a) The distance traveled will be the area under the curve. We will use the approximate velocities at the
midpoints of each time interval to approximate this area using rectangles. Thus,
D (32)(0.001) (82)(0.001) (116)(0.001) (143)(0.001) (164)(0.001) (181)(0.001) (194)(0.001)
(207)(0.001) (216)(0.001) (224)(0.001) 1.56 km
Copyright 2016 Pearson Education, Ltd.
282
Chapter 5 Integrals
(b) Roughly, after 0.0063 hours, the car would have gone 0.78 km, where 0.0060 hours 22.7 s.
At 22.7 s, the velocity was approximately 192 km/h.
13. (a) Because the acceleration is decreasing, an upper estimate is obtained using left endpoints in summing
acceleration t. Thus, t 1 and speed [9.8 + 5.944 + 3.605 + 2.187 + 1.326](1) = 22.862 m/s
(b) Using right endpoints we obtain a lower estimate: speed [5.944 + 3.605 + 2.187 + 1.326 + 0.805] =
13.867 m/s
(c) Upper estimates for the speed at each second are:
t 0 1
2
3
4
5
v 0 9.8 15.744 19.349 21.536 22.862
Thus, the distance fallen when t 3 seconds is s [9.8 + 15.744 + 19.349](1) 44.893 m.
14. (a) The speed is a decreasing function of time right endpoints give a lower estimate for the height
(distance) attained. Also
t
0
1
2
3
4
5
v 122.5 112.7 102.9 93.1 83.3 73.5
gives the time-velocity table by subtracting the constant g 9.8 from the speed at each time increment
t 1s. Thus, the speed 73.5 m/s after 5 seconds.
(b) A lower estimate for height attained is h [112.7 102.9 93.1 83.3 73.5](1) 465.5 m.
15. Partition [0, 2] into the four subintervals [0, 0.5], [0.5, 1], [1, 1.5], and [1.5, 2]. The midpoints of these
subintervals are m1 0.25, m2 0.75, m3 1.25, and m4 1.75. The heights of the four approximating
1 , f ( m ) (0.75)3 27 , f ( m ) (1.25)3 125 , and f (m ) (1.75)3 343
rectangles are f (m1 ) (0.25)3 64
2
3
4
64
64
64
3
3
3
3
1 3
1 5
1 7
1 31
Notice that the average value is approximated by 12 14
2
4
2
4
2
4
2 16
approximate area under
length 1of [0,2]
. We use this observation in solving the next several exercises.
curve f ( x) x3
16. Partition [1,9] into the four subintervals [1, 3], [3, 5], [5, 7], and [7, 9]. The midpoints of these subintervals are
m1 2, m2 4, m3 6, and m4 8. The heights of the four approximating rectangles are f (m1 ) 12 ,
f (m2 ) 14 , f (m3 ) 16 , and f (m4 ) 18 . The width of each rectangle is x 2. Thus,
25
25 average value
area
25 .
Area 2 12 2 14 2 16 2 18 12
128 96
length of [1,9]
17. Partition [0, 2] into the four subintervals [0, 0.5], [0.5, 1], [1, 1.5], and [1.5, 2]. The midpoints of the
subintervals are m1 0.25, m2 0.75, m3 1.25, and m4 1.75. The heights of the four approximating
rectangles are f (m1 ) 12 sin 2 4 12 12 1, f (m2 ) 12 sin 2 34 12 12 1, f (m3 ) 12 sin 2 54
1, and f (m ) sin
12 1
2
2
1
2
1
2
4
1
2
2 7
12
4
1. The width of each rectangle is x .
1
2
2
Thus, Area (1 1 1 1) 12 2 average value lengtharea
2 1.
of [0, 2] 2
18. Partition [0, 4] into the four subintervals [0, 1], [1, 2], [2, 3], and [3, 4]. The midpoints of the subintervals
are m1 12 , m2 23 , m3 52 , and m4 72 . The heights of the four approximating rectangles are
Copyright 2016 Pearson Education, Ltd.
1
2
Section 5.1 Area and Estimating with Finite Sums
4
1
f (m1 ) 1 cos 42 1 cos 8
1 cos
3
8
4
4
3
0.27145 (to 5 decimal places), f (m2 ) 1 cos 42
4
283
4
4
5
0.97855, f (m3 ) 1 cos 42 1 cos 58
0.97855, and
4
4
7
f (m4 ) 1 cos 42 1 cos 78
0.27145. The width of each rectangle is x 1. Thus,
Area (0.27145)(1) (0.97855)(1) (0.97855)(1) (0.27145)(1) 2.5 average value lengtharea
2.5
85
of [0,4]
4
19. Since the leakage is increasing, an upper estimate uses right endpoints and a lower estimate uses left endpoints:
(a) upper estimate (70)(1) (97)(1) (136)(1) (190)(1) (265)(1) 758 liters,
lower estimate (50)(1) (70)(1) (97)(1) (136)(1) (190)(1) 543 liters.
(b) upper estimate (70 97 136 190 265 369 516 720) 2363 liters,
lower estimate (50 70 97 136 190 265 369 516) 1693 liters.
(c) worst case: 2363 720t 25, 000 t 31.4 hours;
best case: 1693 720t 25, 000 t 32.4 hours
20. Since the pollutant release increases over time, an upper estimate uses right endpoints and a lower estimate
uses left endpoints;
(a) upper estimate (0.2)(30) (0.25)(30) (0.27)(30) (0.34)(30) (0.45)(30) (0.52)(30) 60.9 tons
lower estimate (0.05)(30) (0.2)(30) (0.25)(30) (0.27)(30) (0.34)(30) (0.45)(30) 46.8 tons
(b) Using the lower (best case) estimate: 46.8 (0.52)(30) (0.63)(30) (0.70)(30) (0.81)(30) 126.6 tons,
so near the end of September 125 tons of pollutants will have been released.
21. (a) The diagonal of the square has length 2, so the side length is 2. Area
2 2
2
(b) Think of the octagon as a collection of 16 right triangles with a hypotenuse of length 1 and an acute angle
measuring 216 8 .
Area 16 12 sin 8
cos 8 4 sin 4 2 2 2.828
(c) Think of the 16-gon as a collection of 32 right triangles with a hypotenuse of length 1 and an acute angle
.
measuring 232 16
Area /
(d) Each area is less than the area of the circle, . As n increase, the area approaches .
22. (a) Each of the isosceles triangles is made up of two right triangles having hypotenuse 1 and an acute angle
measuring 22n n The area of each isosceles triangle is AT 2 12 sin n cos n 12 sin 2n .
(b)
sin 2n
The area of the polygon is AP nAT n2 sin 2n , so lim n2 sin 2n lim
2
n
n
n
2
(c) Multiply each area by r .
AT 12 r 2 sin 2n
AP n2 r 2 sin 2n
lim AP r 2
n
23-26.
Example CAS commands:
Maple:
Copyright 2016 Pearson Education, Ltd.
284
Chapter 5 Integrals
with( Student[Calculus 1] );
f := x -> sin(x);
a := 0;
b := Pi;
Plot( f (x), x a..b, title "#23(a) (Section 5.1)" );
N : [ 100, 200, 1000 ];
# (b)
for n in N do
Xlist : [ a+1.*(b-a)/n*i $ i 0..n ];
Ylist : map( f, Xlist );
end do:
for n in N do
Avg[n] : evalf(add(y,y Ylist)/nops(Ylist));
# (c)
end do;
avg : FunctionAverage( f (x), x a..b, output value );
evalf( avg );
FunctionAverage(f(x),x a..b, output plot);
# (d)
fsolve( f(x) avg, x 0.5 );
fsolve( f(x) avg, x 2.5 );
fsolve( f(x) Avg[1000], x 0.5 );
fsolve( f(x) Avg[1000], x 2.5 );
Mathematica: (assigned function and values for a and b may vary):
Symbols for π, , powers, roots, fractions, etc. are available in Palettes.
Never insert a space between the name of a function and its argument.
Clear[x]
f[x_] : x Sin[1/x]
{a, b}{π/4, π}
Plot[f[x],{x, a, b}]
The following code computes the value of the function for each interval midpoint and then finds the
average. Each sequence of commands for a different value of n (number of subdivisions) should be
placed in a separate cell.
n 100; dx (b a) /n;
values Table[N[f[x]],{x, a dx/2, b, dx}]
average Sum[values[[i]],{i, 1, Length[values]}] / n
n 200; dx (b a) /n;
values Table[N[f[x]],{x, a dx/2, b, dx}]
average Sum[values[[i]],{i, 1, Length[values]}] / n
n 1000; dx (b a) /n;
values Table[N[f[x]],{x, a dx/2, b, dx}]
average Sum[values[[i]],{i, 1, Length[values]}] / n
FindRoot[f[x] average,{x, a}]
5.2
SIGMA NOTATION AND LIMITS OF FINITE SUMS
1.
k6k1 11 21 62 12
7
3
2
6(1)
6(2)
k 1
Copyright 2016 Pearson Education, Ltd.
Section 5.2 Sigma Notation and Limits of Finite Sums
3
2.
k 1
285
k 1 11 2 1 3 1
1 2 3 0 12 32 76
k
4
3. cos k cos(1 ) cos(2 ) cos(3 ) cos(4 ) 1 1 1 1 0
k 1
4.
5
sin k sin(1 ) sin(2 ) sin(3 ) sin(4 ) sin(5 ) 0 0 0 0 0 0
k 1
3
5. (1)k 1 sin k (1)11 sin 1 (1) 21 sin 2 (1)31 sin 3 0 1 23
k 1
3 2
2
4
6. (1)k cos k (1)1 cos(1 ) (1) 2 cos(2 ) ( 1)3 cos(3 ) (1)4 cos(4 ) ( 1) 1 ( 1) 1 4
k 1
7. (a)
(b)
(c)
6
2k 1 211 221 231 241 251 261 1 2 4 8 16 32
k 1
5
2k 20 21 22 23 24 25 1 2 4 8 16 32
k 0
4
2k 1 211 201 211 221 231 241 1 2 4 8 16 32
k 1
All of them represent 1 2 4 8 16 32
8. (a)
(b)
(c)
6
(2)k 1 (2)11 (2)21 ( 2)31 ( 2) 41 ( 2)51 ( 2)6 1 1 2 4 8 16 32
k 1
5
(1)k 2k (1)0 20 (1)1 21 (1) 2 22 (1)3 23 (1) 4 24 (1)5 25 1 2 4 8 16 32
k 0
3
(1)k 1 2k 2 (1)21 22 2 (1) 11 21 2 (1)01 20 2 (1)11 21 2 (1)21 22 2 (1)31 23 2
k 2
1 2 4 8 16 32;
(a) and (b) represent 1 2 4 8 16 32; (c) is not equivalent to the other two
9. (a)
(b)
(c)
4
k 2
2
( 1) k 1
( 1) 2 1 ( 1)31 ( 1) 4 1
21 31 41 1 12 13
k 1
( 1)k
( 1)0
( 1)1
( 1)2
k 1 01 11 21 1 12 13
k 0
1
k 1
( 1) k
( 1) 1 ( 1)0 ( 1)1
1 2 0 2 1 2 1 12 13
k 2
(a) and (c) are equivalent; (b) is not equivalent to the other two.
10. (a)
(b)
(c)
4
(k 1)2 (1 1)2 (2 1)2 (3 1)2 (4 1)2 0 1 4 9
k 1
3
(k 1) 2 (1 1) 2 (0 1) 2 (1 1)2 (2 1) 2 (3 1) 2 0 1 4 9 16
k 1
1
k 2 (3)2 (2) 2 (1) 2 9 4 1
k 3
(a) and (c) are equivalent to each other; (b) is not equivalent to the other two.
Copyright 2016 Pearson Education, Ltd.
286
Chapter 5 Integrals
6
4
k 1
13. 1k
k 1 2
k 1
5
5
5
15. (1) k 1 k1
14. 2k
k 1
17. (a)
4
12. k 2
11. k
16. (1) k k5
k 1
n
n
k 1
n b
k 1
n
k 1
n
k 1
k 1
3ak 3 ak 3(5) 15
(b) 6k 16 bk 16 (6) 1
(c)
n
n
k 1
n
k 1
n
k 1
n
k 1
n
k 1
n
k 1
k 1
k 1
(ak bk ) ak bk 5 6 1
(d) (ak bk ) ak bk 5 6 11
(e)
18. (a)
(c)
19. (a)
n
(bk 2ak ) bk 2 ak 6 2(5) 16
k 1
n
n
k 1
n
k 1
n
n
k 1
k 1
k 1
10
10(10 1)
55
2
8ak 8 ak 8(0) 0
(b)
(ak 1) ak 1 0 n n
k
k 1
10
(d)
(b)
2
n
n
250bk 250 bk 250(1) 250
k 1
n
n
k 1
k 1
k 1
n
(bk 1) bk 1 1 n
10
k 1
k2
10(101)(2(10) 1)
385
6
13
13(131)(2(13) 1)
819
6
k 1
10(10 1)
(c) k 2 552 3025
k 1
20. (a)
(c)
13
3
k
k 1
13
13(131)
91
2
(b)
2
k2
k 1
13(131)
k 3 2 912 8281
k 1
56
7
7
k 1
k 1
6
6
6
k 1
k 1
k 1
6
6
6
k 1
k 1
k 1
5
5
5
5
k 1
k 1
k 1
k 1
7
7
7
7
k 1
k 1
k 1
k 1
21. 2k 2 k 2
7(7 1)
2
23. (3 k 2 ) 3 k 2 3(6)
24. (k 2 5) k 2 5
5
5
k 1
k 1
5(51)
2
240
k
22. 15k 15
15
5(51)
2
6(6 1)(2(6) 1)
73
6
6(6 1)(2(6) 1)
5(6) 61
6
25. k (3k 5) (3k 2 5k ) 3 k 2 5 k 3
26. k (2k 1) (2k 2 k ) 2 k 2 k 2
5(51)(2(5) 1)
6
7(7 1)(2(7) 1)
6
5
7(7 1)
308
2
Copyright 2016 Pearson Education, Ltd.
Section 5.2 Sigma Notation and Limits of Finite Sums
3
3
287
5
5
5
5
2
3
k 3 k 1 k 3 k 1 5(51) 5(51) 3376
27. 225
225
225
2
2
k 1
k 1
k 1
k 1
2
2
7 3 7
7
7
28. k k4 k 14 k 3
k 1
k 1 k 1
k 1
29. (a)
7(7 1) 2 1 7(7 1) 2
4
588
2
2
7
500
3 3(7) 21
(b) 7 7(500) 3500
k 1
k 1
264
262
k 3
j1
(c) Let j k 2 k j 2; if k 3 j 1 and if k 264 j 262 10 10 10(262) 2620
36
28
28
28
k 9
j 1
j 1
j 1
17
15
k 3
j 1
30. (a) Let j k 8 k j 8; if k 9 j 1 and if k 36 j 28 k ( j 8) j 8
28(281)
8(28) 630
2
(b) Let j k 2 k j 2; if k 3 j 1 and if k 17 j 15 k 2 ( j 2) 2
15
15
15
15
j 1
j 1
j 1
j 1
( j 2 4 j 4) j 2 4 j 4
15(151)(2(15) 1)
15(151)
4 2 4(15) 1240 480 60 1780
6
71
(c) Let j k 17 k j 17; if k 18 j 1 and if k 71 j 54 k (k 1)
54
54
2
54
2
54
k 3
54
( j 17)(( j 17) 1) ( j 33 j 272) j 33 j 272
j 1
j 1
j 1
j 1
j 1
54(54 1)(2(54) 1)
54(54 1)
33 2
272(54) 53955 49005 14688 117648
6
31. (a)
(c)
32. (a)
(c)
33. (a)
n
4 4n
k 1
n
(b)
n
n
n
c cn
k 1
2
n ( n 1)
(k 1) k 1 2 n n 2 n
k 1
k 1
k 1
n
1n 2n 1n 2n n 1 2n2
k 1
n
(b)
n
nc nc n c
k 1
n ( n 1)
k2 12 2 n2n1
n
n
k 1
(b)
Copyright 2016 Pearson Education, Ltd.
(c)
288
Chapter 5 Integrals
34. (a)
(b)
(c)
35. (a)
(b)
(c)
36. (a)
(b)
(c)
37. | x1 x0 | |1.2 0| 1.2, | x2 x1 | |1.5 1.2| 0.3, | x3 x2 | 2.3 1.5 0.8, | x4 x3 | 2.6 2.3 0.3,
and | x5 x4 | |3 2.6| 0.4; the largest is || P || 1.2.
38. | x1 x0 | | 1.6 (2)| 0.4,| x2 x1 | | 0.5 ( 1.6) | 1.1,| x3 x2 | | 0 (0.5) | 0.5,
| x4 x3 | |0.8 0| 0.8, and | x5 x4 | |1 0.8| 0.2; the largest is || P || 1.1.
39. f ( x) 1 x 2
Let x 1n0 1n and ci ix ni . The right-hand sum is
i 1
n
1 ci2
1n 1n i11 ni n1 i1 n2 i2
n
3
n
3
n3 13 i 2 1
n
1
n i 1
2 n3 12
6
n
2
n
n ( n 1)(2 n 1)
. Thus,
6n
3
2
6n
n
lim 1 ci2 1n
n i 1
2 n3 12
lim 1 6 n 1 13 23
n
Copyright 2016 Pearson Education, Ltd.
3
1 2 n 3n3 n
Section 5.2 Sigma Notation and Limits of Finite Sums
40. f ( x ) 2 x
289
Let x 3n0 n3 and ci ix 3ni . The right-hand sum
n
n
n
n3 i1 6ni n3 18n i1i 18n n(n21) 9n n9n .
i 1
n
Thus, lim 6ni n3 lim 9n 9n lim 9 9n 9.
n
n i 1
n n
is 2ci
2
2
2
2
2
2
41. f ( x ) x 2 1
Let x 3n0 n3 and ci ix 3ni . The right-hand sum
3
n
n
n
2
2
is ci2 1 n3 3ni 1 n3 n3 9i2 1
n
i1
i 1
i 1
n
27
i 2 n3 n 273
n
n
i 1
18 27
92
n
n
2
n ( n 1)(2 n 1)
6
n
9(2 n3 3n 2 n )
2 n3
3
3. Thus, lim ci2 1 n3
n i 1
18 27n 92
n
9 3 12.
lim
3
2
n
42. f ( x) 3x 2
Let x 1n0 1n and ci ix ni . The right-hand sum is
n
3ci2
i 1
1n i13 ni 1n n3 i1i2 n3 n(n1)(26 n1)
n
n
2
3
3
2
2 n 3n3 n
2 3n 12
n
2
2n
n
. Thus, lim 3ci2
n i 1
2 3n 12
lim 2 n 22 1.
n
43. f ( x ) x x 2 x(1 x)
3
1n
Let x 1n0 1n and ci ix ni . The right-hand sum is
i 1
n
ci ci2
1n i1 ni ni 1n n1 i1i n1 i1i2
n
n
2
n
2
3
2
3
2
n ( n 1)
n ( n 1)(2 n 1)
13
n 2n 2n 3n3 n
2
6
n
n
2n
6n
3 1
n
1 1n 2 n n2
2 6 . Thus, lim ci ci2 1n
n i 1
1 1 2 3n 12
lim 2n 6 n 12 62 65 .
n
12
44. f ( x) 3 x 2 x 2
Let x 1n0 1n and ci ix ni . The right-hand sum is
n
n
n
n
2
3ci 2ci2 1n 3ni 2 ni 1n 32 i 23 i 2
n i 1
n i 1
i 1
i 1
2
2
n ( n 1)
n ( n 1)(2 n 1)
3
3
n
3
n
2
n
3
n 1
2
2
3
2
2
2
6
n
n
2n
3n
3 1
n
3 3n 2 n n2
2 1
2 3 . Thus, lim 3ci 2ci n
n i 1
Copyright 2016 Pearson Education, Ltd.
290
Chapter 5 Integrals
3 3 2 n3 12
lim 2 n 3 n 32 32 13
.
6
n
Let x 1n0 1n and ci ix ni . The right-hand sum is
45. f ( x) 2 x3
n
n
n
3
2ci3 1n 2 ni 1n 24 i3 24
n i 1
n
i 1
i 1
2 n 2 ( n 2 2 n 1)
4n
lim
n
4
2
n 22n 1
2n
1 2n 12
n
2
1 2n 12
n
12 .
2
2
2 5n2n 5 4n 62n 2 n 22n 1 2
3n
5 5
lim 2 2 n
n
4n
4 6n 22
n
3
1 n2 12
n
4
5.3
THE DEFINITE INTEGRAL
1.
0 x dx
4.
1 1x dx
7.
/4 (sec x) dx
2 2
4
0
5 5n
2
4 6n 22
n
3
0
1 n2 12
n
4
1 2x dx
5.
2 11x dx
8.
0
3.
7 ( x 3x) dx
6.
0 4 x dx
5
1
2
2
(tan x) dx
(b)
1
5
5 g ( x) dx 1 g ( x) dx 8
2
5
n i 1
2 f ( x) dx 1 f ( x) dx 1 f ( x) dx 6 (4) 10
5
n
3
/4
. Thus, lim ci2 ci3 1n
3
2.
2
5
7.
2 52 34 14 12
2 g ( x) dx 0
2
2
(c) 3 f ( x ) dx 3 f ( x) dx 3(4) 12
1
1
5
1 [ f ( x) g ( x)] dx 1 f ( x) dx 1 g ( x) dx 6 8 2
5
5
5
(f ) [4 f ( x) g ( x)] dx 4 f ( x) dx g ( x) dx 4(6) 8 16
1
1
1
(e)
n i 1
Let x
5
n
. Thus, lim 2ci3 1n
2
(d)
n ( n 1) 2
2
0 ( 1)
1n and ci 1 ix 1 ni .
n
n
The right-hand sum is ci2 ci3 1n
i 1
n
n
2
3
2
3
i
1 n 1 ni 1n 2 5ni 4i2 i 3 1n
n
n
i 1
i 1
n
n
n
n
n
2
3
n2 52i 4i3 i 4 n2 52 i 43 i 2 14 i3
n
n
n
n i 1
n i 1
n i 1
i 1
i 1
n ( n 1)
n ( n 1)(2 n 1)
n ( n 1) 2
5
2
4
1
n ( n) 2
3
4
2
6
2
n
n
n
46. f ( x ) x 2 x3
9. (a)
Copyright 2016 Pearson Education, Ltd.
Section 5.3 The Definite Integral
9
9
1 2 f ( x) dx 2 1 f ( x) dx 2(1) 2
9
9
9
(b) [ f ( x) h( x)] dx f ( x) dx h( x) dx 5 4 9
7
7
7
10. (a)
(c)
9
9
9
7 [2 f ( x) 3h( x)] dx 2 7 f ( x) dx 3 7 h( x) dx 2(5) 3(4) 2
1
9
2
2
9 f ( x) dx 1 f ( x) dx (1) 1
7
9
9
(e) f ( x) dx f ( x) dx f ( x) dx 1 5 6
1
1
7
7
9
9
9
(f ) [h( x) f ( x)] dx [ f ( x) h( x)] dx f ( x) dx h( x) dx 5 4 1
9
7
7
7
(d)
1 f (u ) du 1 f ( x) dx 5
1
2
(c) f (t ) dt f (t ) dt 5
2
1
11. (a)
3
0
0 g (t ) dt 3 g (t ) dt 2
0
0
(c) [ g ( x)] dx g ( x) dx 2
3
3
12. (a)
4
4
3
3
2
0
3
1
1 h(r ) dr 1 h(r ) dr 1 h(r ) dr 6 0 6
3
1
3
(b) h(u ) du h(u ) du h(u ) du 6
3
1
1
14. (a)
15. The area of the trapezoid is A 12 ( B b)h
12 (5 2)(6) 21
4 x
3
2 2
dx 21 square units
16. The area of the trapezoid is A 12 ( B b)h
12 (3 1)(1) 2
3/2
1/2
0
3 g (u ) du 3 g (t ) dt 2
0 g (r )
0
(d)
dr 1 g (t ) dt 1 ( 2) 1
3 2
2 3
2
(b)
3 f ( z ) dz 0 f ( z ) dz 0 f ( z ) dz 7 3 4
3
4
(b) f (t ) dt f (t ) dt 4
4
3
13. (a)
2
1 3 f ( z ) dz 3 1 f ( z ) dz 5 3
2
2
(d) [ f ( x)] dx f ( x) dx 5
1
1
(b)
(2 x 4) dx 2 square units
Copyright 2016 Pearson Education, Ltd.
291
292
Chapter 5 Integrals
17. The area of the semicircle is A 12 r 2 12 (3) 2
92
3
3
9 x 2 dx 92 square units
18. The graph of the quarter circle is A 14 r 2 14 (4) 2
4
0
4
16 x 2 dx 4 square units
19. The area of the triangle on the left is A 12 bh
12 (2)(2) 2. The area of the triangle on the right is
A 12 bh 12 (1)(1) 12 . Then, the total area is 2.5
1
2
| x| dx 2.5 square units
20. The area of the triangle is A 12 bh 12 (2)(1) 1
1
(1 | x|) dx 1 square unit
1
21. The area of the triangular peak is A 12 bh 12 (2)(1) 1.
The area of the rectangular base is S w (2)(1) 2.
1
Then the total area is 3 (2 | x|) dx 3 square
1
units
Copyright 2016 Pearson Education, Ltd.
Section 5.3 The Definite Integral
22. y 1 1 x 2 y 1 1 x 2 ( y 1) 2 1 x 2
x 2 ( y 1)2 1, a circle with center (0, 1) and radius
of 1 y 1 1 x 2 is the upper semicircle. The area
of this semicircle is A 12 r 2 12 (1) 2 2 . The area
of the rectangular base is A w (2)(1) 2. Then the
1
total area is 2 2 1 1 x 2 dx 2 2
1
square units
b
2
23.
0 2x dx 12 (b)( b2 ) b4
25.
a 2s ds 12 b(2b) 12 a(2a) b a
b
2
24.
0 4 x dx 12 b(4b) 2b
2
26.
a 3t d t 12 b(3b) 12 a(3a) 32 (b a )
27. (a)
2 4 x dx 12 [ (2) ] 2
(b)
0 4 x dx 14 [ (2) ]
28. (a)
1 3x 1 x dx 13x dx 1 1 x dx 12 [(1)(3)] 14 [ (1) ] 4 32
b
2
2
(b)
2
2
0
2
0
0
1
2
0
1
2
2
2
2
2
2
2
1
1 3x 1 x dx 13x dx 0 3x dx 1 1 x dx 12 [(1)(3)] 12 [(1)(3)] 12 [ (1) ] 2
2 (1)2 1
2
29.
1
31.
d
2
2
b
x dx
2
2
30.
2
2
(2 )2 2
2 32
2
2
2
32.
2.5
0.5 x d x
5 2
2
(2.5) 2 (0.5) 2
2 3
2
5 2 2 24
2
r dr
Copyright 2016 Pearson Education, Ltd.
2
2
2
293
294
33.
Chapter 5 Integrals
3
0
3 7 7
x dx
3
7 2
3
34.
3
1
t dt 2 1
3
1/2 2
35.
0
37.
a x dx 2 a2 3a2
2a
3
24
(2 a )2
2
36.
(0.3)3
0.009
3
0.3 2
0 s d s
3
d 2
3
/2 2
0
2
3
3a
24
3a a2 a 2
2
38.
a
39.
0
3 b b
x dx
40.
0 x dx 3 9b
41.
3 7 dx 7(1 3) 14
42.
0 5 x dx 5 0 x dx 5 22 02 10
43.
0 (2t 3) dt 21 t dt 0 3 dt 2 22 02 3(2 0) 4 6 2
44.
0
45.
2 1 2z dz 21 dz 2 2z dz 21 dz 12 1 z dz 1[1 2] 12 22 12 1 12 23 74
46.
3 (2 z 3) dz 3 2 z dz 3 3 dz 2 0 z dz 3 3 dz 2 32 02 3[0 3] 9 9 0
47.
1 3u du 31 u du 3 0 u du 0 u du 3 23 03 13 03 3 23 13 3 73 7
3
3
b 2
3
3
1
2
1
2
t 2 dt
1
2
2
0
1
0
1
2
(3b )3
3
2
2
2
2
2 2
2
2 dt 2 02 2 2 0 1 2 1
1
0
2
2
3
2 2
2 2
2
0
1
0
2
2
3b 2
2
2
2
t dt
x dx
0
1 2
2
3
3
2
2
3
3
3
3
3 1 3
7
1
1/2
u 2 du 24 u 2 du u 2 du 24 13 23 24 38 7
1/2
0
0
24u 2 du 24
1
48.
1/2
49.
0 (3x x 5) dx 3 0 x dx 0 x dx 0 5 dx 3 23 03 22 02 5[2 0] (8 2) 10 0
50.
1 (3x x 5) dx 0 (3x x 5) dx 3 0 x dx 0 x dx 0 5 dx 3 13 03 12 02 5(1 0)
2
2 2
2
0
1
2
2
2
2
3
3
1 2
1
2
1
2
32 5 72
Copyright 2016 Pearson Education, Ltd.
3
3
2
2
Section 5.3 The Definite Integral
51. Let x b n 0 bn and let x0 0, x1 x, x2 2x, ,
xn 1 (n 1)x, xn nx b. Let the ck 's be the right
endpoints of the subintervals c1 x1 , c2 x2 , and so on.
The rectangles defined have areas:
f (c1 )x f (x)x 3(x)2 x 3( x)3
f (c2 )x f (2x)x 3(2x) 2 x 3(2)2 (x)3
f (c3 )x f (3x)x 3(3x )2 x 3(3)2 (x)3
f (cn )x f (nx)x 3(nx)2 x 3(n)2 (x)3
n
n
Then Sn f (ck )x 3k 2 (x)3
n
3
k 1
3(x) k
3
2
k 1
3
3 b3
n
k 1
n ( n 1)(2 n 1)
6
b
3
b2 2 n3 12 3 x 2 dx lim b2 2 n3 12 b3 .
0
n
n
n
52. Let x b n 0 bn and let x0 0, x1 x, x2 2x, . . . ,
xn 1 (n 1)x, xn nx b. Let the ck 's be the right
endpoints of the subintervals c1 x1 , c2 x2 , and so on.
The rectangles defined have areas:
f (c1 ) x f (x ) x (x)2 x (x)3
f (c2 ) x f (2x)x (2x) 2 x (2)2 (x )3
f (c3 ) x f (3x) x (3x)2 x (3) 2 (x)3
f (cn )x f (nx) x (nx) 2 x ( n) 2 (x)3
n
n
n
k 1
k 1
Then Sn f (ck )x k 2 (x)3 (x)3 k 2
2
x dx lim
2 .
n
n ( n 1)(2 n 1)
6
b
2
3
b3
b
3
b3
0
3
n
6
n
3
n
6
1
n2
k 1
1
n2
b3
3
53. Let x b n 0 bn and let x0 0, x1 x, x2 2x, ,
xn 1 (n 1)x, xn nx b. Let the ck 's be the right
endpoints of the subintervals c1 x1 , c2 x2 , and so on.
The rectangles defined have areas:
f (c1 )x f ( x) x 2( x )( x) 2( x)2
f (c2 )x f (2x)x 2(2Δx)(Δx) 2(2)(Δx) 2
f (c3 )x f (3x)x 2(3x)(x) 2(3)(x)2
f (cn ) x f ( nx) x 2(nx)( x) 2( n)( x) 2
n
n
Then Sn f (ck )x 2k (x)2
k 1
n
k 1
n ( n1)
2
2
2(x)2 k 2 b 2
k 1
n
b
b 1
2
1
n
2 x dx lim b 2 1 1n b 2 .
0
n
Copyright 2016 Pearson Education, Ltd.
295
296
Chapter 5 Integrals
54. Let x b n 0 bn and let x0 0, x1 x, x2 2x, ,
xn 1 (n 1)x, xn nx b. Let the ck 's be the right
endpoints of the subintervals c1 x1 , c2 x2 , and so on.
The rectangles defined have areas:
f (c1 )x f (x)x 2x 1 (x) 12 (x)2 x
f (c3 )x f (3x)x
f (c2 )x f (2x)x
f (cn )x f (nx)x
2 x 1 ( x ) 1 (2)( x ) 2 x
2
2
3x 1 (x) 1 (3)( x) 2 x
2
2
n2 x 1 (x) 12 (n)(x)2 x
n ( n 1)
1 k ( x) 2 x 1 ( x) 2 k x 1 1 b
b ( n)
2
2
2 n 2 n
k 1
k 1
k 1
k 1
b
14 b 2 1 1n b 2x 1 dx lim 14 b 2 1 1n b 14 b 2 b.
0
n
n
n
n
Then Sn f (ck )x
55. av( f )
n
2
(x 1) dx x dx 1 d x
1
3 0
3
2
0
3 3
1 3 1
3
3
3 2
1
3 0
3
1
3 0
3 0 1 1 0.
03 x2 dx 13 12 03 x2 dx
56. av( f ) 31 0
2
3
16 33 32 .
01
1
1
0
0
57. av( f ) 110 (3x 2 1) dx 3 x 2 dx 1 dx
3
3 13 (1 0) 2.
01
1
1
0
0
58. av( f ) 110 (3 x 2 3) dx 3 x 2 dx 3 dx
2
3
3 13 3(1 0) 2.
Copyright 2016 Pearson Education, Ltd.
Section 5.3 The Definite Integral
03
3
3
3
0
0
0
59. av( f ) 31 0 (t 1)2 dt 13 t 2 dt 23 t dt 13 1 dt
3
2
2
13 33 23 32 02 13 (3 0) 1.
t t dt t dt t dt
60. av( f ) 1(12)
1
1 1 2
3 2
2
2
1 1
3 2
1
2 2
2 ( 2)2
13 t 2 dt 13
t dt 13 12 2
0
0
3
( 2)3
13 13 13 3 12 32 .
1
61. (a) av( g ) 1(11) (| x| 1)dx
1
0
1
1
0
12 ( x 1) dx 12 ( x 1) dx
0
0
1
1
1
1
0
0
12 x dx 12 1 dx 12 x dx 12 1 dx
2
2
2 ( 1)2
12 02 2 12 (0 (1)) 12 12 02 12 (1 0)
12 .
13
3
(b) av( g ) 311 (| x | 1) dx 12 ( x 1) dx
3
3
1
1
1
2
2
12 x dx 12 1 dx 12 32 12 12 (3 1) 1.
3
(c) av( g ) 31(11) (| x | 1) dx
1
1
3
14
(| x | 1) dx 14 (| x | 1) dx
1
1
14 (1 2) 14 (see parts (a) and (b) above).
Copyright 2016 Pearson Education, Ltd.
297
298
Chapter 5 Integrals
0
0
1
1
62. (a) av(h) 0(11) | x | dx ( x) dx
0
2
x dx 02
1
( 1)2
12 .
2
01
1
(b) av(h) 110 | x | dx x dx
2
2
12 02
0
12 .
1
(c) av(h) 1(11) | x | dx
1
0
1
12 | x | dx | x | dx
1
0
1
1
1
1
2 2 2 2 (see parts (a) and (b) above).
63. Consider the partition P that subdivides the interval [a, b] into n subintervals of width x b n a and let ck be
the right endpoint of each subinterval. So the partition is P a, a b n a , a
ck a
k (b a )
. We get the Riemann sum
n
n
n
k 1
k 1
f (ck )x c bna
c (b a )
n
2(b a )
n (b a )
, ..., a n
n
n
1
k 1
and
c (b a )
n c(b a ).
n
b
As n and P 0 this expression remains c(b a ). Thus, c dx c(b a ) .
a
64. Consider the partition P that subdivides the interval [0, 2] into n subintervals of width x 2n 0 n2 and let ck
be the right endpoint of each subinterval. So the partition is P 0, n2 , 2 n2 , . . . , n n2 2 and ck k n2 2nk .
We get the Riemann sum
n
n
k 1
k 1
n
n
n
n ( n 1)
4( n 1)
f (ck )x 2 2nk 1 n2 n2 4nk 1 n82 k n2 1 n82 2 n2 n n 2 . As n and
k 1
k 1
k 1
2
4( n 1)
(2 x 1) dx 6.
P 0 the expression n 2 has the value 4 2 6. Thus,
0
Copyright 2016 Pearson Education, Ltd.
Section 5.3 The Definite Integral
299
65. Consider the partition P that subdivides the interval [ a, b] into n subintervals of width x b n a and let ck be
2(b a )
n (b a )
, ..., a n
n
n
n
n
k (b a )
k (b a ) 2
ck a n . We get the Riemann sum
f (ck ) x ck2 b n a b n a
a n
k 1
k 1
k 1
n
n
n
n
(b a ) 2
2 2ak (b a ) k 2 (b a )2 b a
2 2 a (b a )
k 2
k2
b n a
a
n a n
2
n
n
n
k 1
k 1
k 1
k 1
the right endpoint of each subinterval. So the partition is P a, a b n a , a
b n a na 2
(b a )a
1
1 1 (b a )3 2 n 2
a(b a) 2 1 n 6 1 n
(b a)a 2 a (b a )2 1
b 2
3
3
2 a (b a )2 n ( n 1) (b a )3 n ( n 1)(2 n 1)
2
2 n 1 (b a ) ( n 1)(2 n 1)
2
(
b
a
)
a
a
(
b
a
)
2
3
6
6
n
n
n
n2
3
2
and
(b a )
6
3
As n and P 0 this expression has value
3
3
2 ba 2 a3 ab 2 2a 2 b a3 13 (b3 3b 2 a 3ba 2 a3 ) b3 a3 . Thus,
3
a x dx b3 a3 .
0 ( 1)
1n and
n
let ck be the right endpoint of each subinterval. So the partition is P 1, 1 1n , 1 2 1n , , 1 n 1n 0
n
n
k
k 2 1
and ck 1 k 1n 1 kn . We get the Riemann sum
f (ck )x
1 n 1 n n
k 1
k 1
n
n
n
n
2
n ( n 1)
n ( n 1)(2 n 1)
k
2k
k
3
2
1n
k 13 k 2 n2 n 32 2 13
1 n 1 n n n 1 n2
6
n
n
n
k 1
k 1
k 1
k 1
3( n 1) ( n 1)(2 n 1)
2 2 n
. As n and || P || 0 this expression has value 2 32 13 65 .
6n2
0
( x x 2 ) dx 56 .
Thus,
1
66. Consider the partition P that subdivides the interval [1, 0] into n subintervals of width x
2 ( 1)
n3 and
n
let ck be the right endpoint of each subinterval. So the partition is P 1, 1 n3 , 1 2 n3 , , 1 n n3 2
n
n
3
3k 2
3k
f (ck )x
and ck 1 k n3 1 3nk . We get the Riemann sum
3 1 n 2 1 n 1 n
k 1
k 1
n
n
n
n
2
n ( n 1)
n ( n 1)(2 n 1)
n3
3 18nk 27 k2 2 6nk 1 18
1 722
k 813 k 2 18
n 722 2 813
6
n
n
n
n
n
n
n
k 1
k 1
k 1
k 1
36( n 1) 27( n 1)(2 n 1)
18 n
. As n and || P || 0 this expression has value 18 36 27 9.
2n2
2
2
67. Consider the partition P that subdivides the interval [1, 2] into n subintervals of width x
Thus, (3 x 2 x 1)dx 9.
1
1( 1)
n2 and let
n
ck be the right endpoint of each subinterval. So the partition is P 1, 1 n2 , 1 2 n2 , , 1 n n2 1 and
n
n
n
3
ck 1 k n2 1 2nk . We get the Riemann sum
f (ck )x
ck3 n2 n2
1 2nk
k 1
k 1
k 1
n
n
n
n
n
2
3
n2
1 6nk 12k2 8k3 n2 1 6n
k 122 k 2 83
k3
n
n
n
n
k 1
k 1
k 1
k 1
k 1
1 1
n ( n 1)
n ( n 1)(2 n 1) 16 n ( n 1) 2
( n 1)(2 n 1)
( n1)2
n2 n 122 2 243
4 2
2 6 nn1 4
4 2 2 6 1 n
2
6
n
n
n
n
n
68. Consider the partition P that subdivides the interval [1, 1] into n subintervals of width x
Copyright 2016 Pearson Education, Ltd.
300
Chapter 5 Integrals
1 1
2 6 1 n 4
2 3n 12
n
1
4
1 n2 12
n
1
. As n and || P || 0 this expression has value 2 6 8 4 0.
1
Thus, x3 dx 0.
1
69. Consider the partition P that subdivides the interval [a, b] into n subintervals of width x b n a and let ck be
2(b a )
n (b a )
, , a n b and
n
3
n
n
n
k (b a )
k (b a )
3 ba
b
a
ck a n . We get the Riemann sum
f (ck )x
ck n n
a n
k 1
k 1
k 1
n
n
n
n
n
3a 2 ( b a )
3a ( b a ) 2
(b a )3
3 3a 2 k (b a ) 3ak 2 (b a )2 k 3 (b a )3 b a
b n a
n a3
k
k2 3
k3
a
2
3
2
n
n
n
n
n
n
k 1
k 1
k 1
k 1
k 1
2
2
2
3
4
3a (b a ) n ( n 1) 3a (b a ) n ( n 1)(2 n 1) (b a )
n ( n 1)
b n a na3
2
4 2
6
n2
n3
n
the right endpoint of each subinterval. So the partition is P a, a b n a , a
(b a )a3
(b a )a3
3a ( b a )
2
2
nn1
a (b a )
2
1
3
3a 2 (b a )2 1 n a (b a )3
1 2
2
value (b a) a3
2
3a ( b a )
2
2
( n 1)(2 n 1)
n2
2 n3 12
n
a (b a)3
1
(b a )
4
4
(b a )
4
(b a ) 4
4
4
4
( n 1)
2
2
n2
1 n2 12
n
1
. As n and || P || 0 this expression has
b
4
4
4
b4 a4 . Thus, x3 dx b4 a4 .
a
70. Consider the partition P that subdivides the interval [0, 1] into n subintervals of width x 1n0 1n and let ck be
n
n
n
3
ck 0 k 1n kn . We get the Riemann sum f (ck )x 3ck ck3 1n 1n 3 kn kn
the right endpoint of each subinterval. So the partition is P 0, 0 1n , 0 2 1n , , 0 n 1n 1 and
k 1
n ( n 1)
1n n3 k 13 k 3 32 2 14
n
n
n
k 1
k 1
n
n
k 1
k 1
1
1 2 1
2
n ( n 1) 2
3 n 1 1 ( n 1) 3 1 n 1 n n2 . As n and
2
2 n
4
2 1
4
1
n2
1
|| P || 0 this expression has value 32 14 54 . Thus, (3 x x3 ) dx 54 .
0
71. To find where x x 2 0, let x x 2 0 x(1 x ) 0 x 0 or x 1. If 0 x 1, then 0 x x 2 a 0 and
b 1 maximize the integral.
72. To find where x 4 2 x 2 0, let x 4 2 x 2 0 x 2 ( x 2 2) 0 x 0 or x 2. By the sign graph,
++++++ 0 0 0 +++++++, we can see that x 4 2 x 2 0 on 2, 2 a 2 and b 2
2
0
minimize the integral.
73.
2
f ( x) 1 2 is decreasing on [0, 1] maximum value of f occurs at 0 max f f (0) 1; minimum value of
1 x
f occurs at 1 min f f (1)
12
1 1 . Therefore, (1 0) min
2
112
f
1 1
dx (1 0)
0 1 x 2
1 1
dx 1. That is, an upper bound 1 and a lower bound 12 .
0 1 x 2
Copyright 2016 Pearson Education, Ltd.
max f
Section 5.3 The Definite Integral
301
1
1 1, min f
0.8. Therefore
1 02
1 (0.5)2
0.5
0.5 1
f ( x) dx (0.5 0) max f 52
dx 12 . On [0.5, 1], max f 1 2 0.8
0
0 1 x 2
1 (0.5)
74. See Exercise 73 above. On [0, 0.5], max f
(0.5 0) min f
and min f
f
1 0.5. Therefore (1 0.5) min
112
Then 14 52
1
1 dx (1 0.5) max
0.5 1 x 2
f 14
1
1 dx 2 .
5
0.5 1 x 2
0.5 1
1
1 dx 1 2 13 1 1 dx 9 .
dx
2 5
20
10
0.5 1 x 2
0 1 x 2
1 x 2
0
1
1
1
0
0
75. 1 sin x 2 1 for all x (1 0)(1) sin x 2 dx (1 0)(1) or sin x 2 dx 1 sin x 2 dx cannot
0
equal 2.
76. f ( x) x 8 is increasing on [0, 1] max f f (1) 1 8 3 and min f f (0) 0 8 2 2. Therefore,
(1 0) min f
1
x 8 dx (1 0) max f 2 2
0
1
0
x 8 dx 3.
b
77. If f ( x) 0 on [a, b], then min f 0 and max f 0 on [a, b]. Now, (b a ) min f f ( x) dx (b a ) max f .
a
b
Then b a b a 0 (b a) min f 0 f ( x) dx 0.
a
b
78. If f ( x) 0 on [ a, b], then min f 0 and max f 0. Now, (b a ) min f f ( x) dx (b a ) max f . Then
a
b
b a b a 0 (b a ) max f 0 f ( x) dx 0.
a
1
1
1
0
0
0
79. sin x x for x 0 sin x x 0 for x 0 (sin x x) dx 0 (see Exercise 78) sin x dx x dx 0
1
1
1
0
0
0
2
2
1
sin x dx x dx sin x dx 12 02 sin x dx 12 . Thus an upper bound is 12 .
0
2
2
2
1
80. sec x 1 x2 on 2 , 2 sec x 1 x2 0 on 2 , 2 sec x 1 x2 dx 0 (see Exercise 77)
0
sec x dx 1 dx x dx sec x dx (1 0) sec x dx . Thus a lower bound is .
1
1
0
0
2
1
1
0
0
2
since [0, 1] is contained in 2 , 2 sec x dx 1 x2 dx 0 sec x dx 1 x2 dx
1
1
0
0
1 1 2
2 0
1
1 13
2 3
0
1
7
6
7
6
0
b
81. Yes, for the following reasons: av( f ) b 1 a f ( x) dx is a constant K. Thus
a
b
b
b
b
b
a av( f ) dx a K dx K (b a) a av( f ) dx (b a) K (b a) b1a a f ( x) dx a f ( x) dx.
82. All three rules hold. The reasons: On any interval [ a, b] on which f and g are integrable, we have:
b
b
b
b
b
(a) av( f g ) b 1 a [ f ( x) g ( x)]dx b 1 a f ( x) dx g ( x ) dx b 1 a f ( x) dx b 1 a g ( x) dx
a
a
a
a
a
av( f ) av( g )
b
b
b
(b) av(kf ) b 1 a kf ( x) dx b 1 a k f ( x)dx k b 1 a f ( x) dx k av( f )
a
a
a
Copyright 2016 Pearson Education, Ltd.
302
Chapter 5 Integrals
b
b
b
a
a
a
(c) av( f ) b 1 a f ( x) dx b 1 a g ( x) dx since f ( x) g ( x) on [ a, b], and b 1 a g ( x) dx av( g ).
Therefore, av( f ) av( g ).
83. (a) U max1 x max 2 x max n x where max1 f ( x1 ), max 2 f ( x2 ) , , max n f ( xn ) since f is
increasing on [ a, b]; L min1 x min 2 x min n x where min1 f ( x0 ), min 2 f ( x1 ) , ,
min n f ( xn 1 ) since f is increasing on [a, b]. Therefore
U L (max1 min1 )x (max 2 min 2 )x (max n min n )x
( f ( x1 ) f ( x0 ))x ( f ( x2 ) f ( x1 ))x ( f ( xn ) f ( xn 1 ))x ( f ( xn ) f ( x0 )) x
( f (b) f (a)) x.
(b) U max1 x1 max 2 x2 max n xn where max1 f ( x1 ), max 2 f ( x2 ) , , max n f ( xn ) since f
is increasing on [ a, b]; L min1 x1 min 2 x2 ... min n xn where min1 f ( x0 ), min 2 f ( x1 ), ,
min n f ( xn 1 ) since f is increasing on [ a, b]. Therefore
U L (max1 min1 ) x1 (max 2 min 2 )x2 (max n min n ) xn
( f ( x1 ) f ( x0 )) x1 ( f ( x2 ) f ( x1 ))x2 ( f ( xn ) f ( xn 1 )) xn
( f ( x1 ) f ( x0 )) xmax ( f ( x2 ) f ( x1 ))xmax ( f ( xn ) f ( xn 1 )) xmax . Then
U L ( f ( xn ) f ( x0 )) xmax ( f (b) f ( a )) xmax f (b) f ( a ) xmax since f (b) f ( a ). Thus
lim (U L) lim ( f (b) f (a )) xmax 0, since xmax P .
P 0
P 0
84. (a) U max1 x max 2 x max n x where
max1 f ( x0 ), max 2 f ( x1 ), , max n f ( xn 1 )
since f is decreasing on [ a, b];
L min1 x min 2 x min n x where
min1 f ( x1 ), min 2 f ( x2 ), , min n f ( xn )
since f is decreasing on [a, b]. Therefore
U L (max1 min1 )x (max 2 min 2 ) x
... (max n min n ) x
( f ( x0 ) f ( x1 )) x ( f ( x1 ) f ( x2 )) x
... ( f ( xn 1 ) f ( xn )) x
( f ( x0 ) f ( xn )) x ( f (a) f (b)) x.
(b) U max1 x1 max 2 x 2 ... max n xn where max1 f ( x0 ), max 2 f ( x1 ), , max n f ( xn 1 )
since f is decreasing on [a, b]; L min1 x1 min 2 x2 min n xn where
min1 f ( x1 ), min 2 f ( x2 ), , min n f ( xn ) since f is decreasing on [a, b]. Therefore
U L (max1 min1 )x1 (max 2 min 2 )x2 (max n min n )xn
( f ( x0 ) f ( x1 ))x1 ( f ( x1 ) f ( x2 ))x2 ( f ( xn 1 ) f ( xn ))xn ( f ( x0 ) f ( xn ))xmax
( f (a ) f (b)xmax f (b) f (a) xmax since f (b) f (a). Thus
lim (U L) lim f (b) f (a) xmax 0, since xmax P .
P 0
P 0
with points x 0, x x,
85. (a) Partition 0, 2 into n subintervals, each of length x 2n
0
1
x2 2x,... , xn nx 2 . Since sin x is increasing on 0, 2 , the upper sum U is the sum of the areas
of the circumscribed rectangles of areas f ( x1 )x (sin x)x, f ( x2 )x (sin 2x)x,... ,
f ( xn )x (sin nx)x.
Copyright 2016 Pearson Education, Ltd.
Section 5.3 The Definite Integral
303
cos x cos n 12 x
cos 4n cos n 12 2n
Then U (sin x sin 2x ... sin nx)x 2
x
2n
2sin 2x
2sin 4n
cos 4n cos 2 4n
4 n sin 4n
(b) The area is
/2
0
cos 4n cos 2 4n
sin
4n
4n
sin x dx lim
cos 4n cos 2 4n
sin
n
1cos 2 1.
1
4n
4n
n
86. (a) The area of the shaded region is xi mi which is equal to L.
i 1
n
(b) The area of the shaded region is xi M i which is equal to U.
i 1
(c) The area of the shaded region is the difference in the areas of the shaded regions shown in the second part
of the figure and the first part of the figure. Thus this area is U L.
n
n
i 1
i 1
87. By Exercise 86, U L xi M i xi mi where M i max { f ( x) on the ith subinterval} and
n
n
i 1
i 1
mi min { f ( x) on ith subinterval}. Thus U L ( M i mi )xi xi provided xi for each
n
n
i 1
i 1
i 1, , n. Since xi xi (b a) the result, U L (b a ) follows.
88. The car drove the first 240 kilometers in 5 hours and the
second 240 kilometers in 3 hours, which means it drove
480 kilometers in 8 hours, for an average value of
480 km/h 60 km/h. In terms of average value of
8
functions, the function whose average value we seek is
48, 0 t 5
v(t ) 80,
5 t 8 , and the average value is
(48)(5) (80)(3)
60.
8
89-94.
Example CAS commands:
Maple:
with( plots );
with( Student[Calculus1] );
f : x -> 1-x;
a : 0;
b : 1;
N :[4, 10, 20, 50];
P : [seq( RiemannSum( f(x), x a..b, partition n, method random, output plot ), n N )]:
display( P, insequence true);
Copyright 2016 Pearson Education, Ltd.
304
Chapter 5 Integrals
95-98.
Example CAS commands:
Maple:
with( Student[Calculus1] );
f : x - sin(x);
a : 0;
b : Pi;
plot( f(x), x a..b, title "#95(a) (Section 5.3)" );
N : [ 100, 200, 1000 ];
# (b)
for n in N do
Xlist : [ a 1.*(b-a)/n*i $ i 0..n ];
Ylist : map( f, Xlist );
end do:
for n in N do
Avg[n] : evalf(add(y,y Ylist)/nops(Ylist));
# (c)
end do;
avg : FunctionAverage( f(x), x a..b, output value );
evalf( avg );
FunctionAverage(f(x),x a..b, output plot);
fsolve( f(x) avg, x 0.5 );
fsolve( f(x) avg, x 2.5 );
fsolve( f(x) Avg[1000], x 0.5 );
fsolve( f(x) Avg[1000], x 2.5 );
95-98.
# (d)
Example CAS commands:
Mathematica: (assigned function and values for a, b, and n may vary)
Sums of rectangles evaluated at left-hand endpoints can be represented and evaluated by this set of commands
Clear[x, f, a, b, n]
{a, b}{0, π}; n 10; dx (b a)/n;
f Sin[x]2 ;
xvals Table[N[x],{x, a, b dx, dx}];
yvals f /.x xvals;
boxes MapThread[Line[{{#1, 0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals, xvals dx, yvals}];
Plot[f, {x, a, b}, Epilog boxes];
Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N
Sums of rectangles evaluated at right-hand endpoints can be represented and evaluated by this set of
commands.
Clear[x, f, a, b, n]
{a, b}{0, π}; n 10; dx (b a)/n;
f Sin[x]2 ;
xvals Table[N[x], {x, a dx, b, dx}];
yvals f /.x xvals;
boxes MapThread[Line[{{#1, 0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals, dx,xvals, yvals}];
Plot[f, {x, a, b}, Epilog boxes];
Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N
Copyright 2016 Pearson Education, Ltd.
Section 5.4 The Fundamental Theorem of Calculus
Sums of rectangles evaluated at midpoints can be represented and evaluated by this set of commands.
Clear[x, f, a, b, n]
{a, b}{0, π}; n 10; dx (b a)/n;
f Sin[x]2 ;
xvals Table[N[x], {x, a dx/2, b dx/2, dx}];
yvals f /.x xvals;
boxes MapThread[Line[{{#1, 0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals, dx/2, xvals dx/2, yvals}];
Plot[f, {x, a, b},Epilog boxes];
Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N
5.4
THE FUNDAMENTAL THEOREM OF CALCULUS
1.
2
3
2
3
2
2 2
x3 3 x 2 (2) 3(2) (0) 3(0) 10
x
(
x
3)
dx
(
x
3
x
)
dx
0
0
2 0 3
2 3
2
3
3
2.
(1)
( 1)
2
2
2
2
1 x 2 x 3 dx x3 x 3x 1 3 (1) 3(1) 3 (1) 3(1) 203
2
1
2
3.
4.
3
2 ( x 3)
1
1 x
299
dx
4
x 300
dx
300
3
3
2
1
1
1
124
3 3 1
3
125 125
( x 3) 2 (5) (1)
1
1
1
1
1
(1)300 ( 1)300
1 1 0
300
300
4
5.
1
3
4
3 44 3 14
2 x3
3 x4
1 753
3x dx x 4 1 64 16 1
4
16
16
16
16
16
1
1
4
x4
6. x 2 x 3 dx x 2 3x
2
4
1
3
3
34
( 2)4
81
105
32 3(3)
( 2)2 3( 2) 6
4
4
4
4
7.
0 x x dx x3 32 x
8.
1 x
9.
0
1
2
32 6/5
/3
3
32
1
3/2
0
13 23 0 1
dx 5 x 1/5 52 (5) 52
1
(2 tan 0) 2 3 0 2 3
2sec2 x dx [2 tan x]0 /3 2 tan 3
Copyright 2016 Pearson Education, Ltd.
305
306
Chapter 5 Integrals
10.
0 (1 cos x) dx [ x sin x]0 ( sin ) (0 sin 0)
11.
/4 csc cot d [ csc ] /4 csc 34 csc 4 2 2 0
3 /4
/3
12.
4
0
3 /4
sin u
2
cos u
/3
4
cos u 0
du
4
4
4
(1/2)
1
13.
2 1cos2 2t dt /2 12 12 cos 2t dt 12 t 14 sin 2t 2 12 (0) 14 sin 2(0) 12 2 14 sin 2 2 4
14.
/3 sin t dt Use the double angle formula cos 2t 1 2sin t which implies that sin t
0
0
0
/3
/3
2
2
/3
sin 2 t dt
1 cos(2t )
.
2
/3
/3 1 cos 2t
t sin 2t
dt
4 /3
2
2
/3
2
1 3 1
3
3
6 4 2 6 4 2 3 4
/4
/4
(sec2 x 1) dx [tan x x]0 /4 tan 4 4 (tan(0) 0) 1 4
15.
0
tan 2 x dx
16.
0
/6
(sec x tan x)2 dx
0
/6
0
(sec2 x 2sec x tan x tan 2 x ) dx
/6
0
(2sec2 x 2sec x tan x 1) dx
(2 tan 0 2sec 0 0) 2 3 6 2
12 cos 2(0) 24 2
[2 tan x 2sec x x]0 /6 2 tan 6 2sec 6 6
/8
/8
12 cos 2
17.
0
18.
3 4sec t t dt /3 (4sec t t ) dt 4 tan t t 3
sin 2 x dx 12 cos 2 x
4
0
4
2
4
2
2
2
4 tan 4 4 tan 3 (4(1) 4) 4 3 3 4 3 3
4
3
19.
20.
1 (r 1) dr 1 (r 2r 1) dr r3 r r 1 3 (1) (1) 13 1 1 83
1
3
3
2
1
3
2
(t 1)(t 2 4) dt
1
2
( 1)3
2
3
3
3
4
3
(t 3 t 2 4t 4) dt t4 t3 2t 2 4t
3
3
3
3 4 3 3
3 4
3
2
2
2( 3) 4( 3) 10 3
4 3 2( 3) 4 3 4
3
Copyright 2016 Pearson Education, Ltd.
2
Section 5.4 The Fundamental Theorem of Calculus
21.
1 y 5 2 y
dy
1
u7 1
2 2
u5
22.
3 y
23.
1
24.
3
du
u 7 u 5
2 2
1
8
18
u8 1
1 ( 2)
1
3
du 16
4
4
4u 4 2 16 4(1) 4 16
4 2
1
y3
( 1)3
( 3)3
( y 2 2 y 2 ) dy 3 2 y 1 3 ( 21) 3 ( 23) 22
3
3
3
1
2
2 s2 s
2
ds
(1 s 3 2 ) ds s 2
2
1
s 1
s
1/3
2/3
8 x 1 2 x
1
1
2 2 1 2 2 23/4 1 2 4 8 1
1
2
3
8 1/3 2 x 2/3
8
dx 2 x x1/3
dx (2 x 2/3 2 x 1/3 x1/3 ) dx 2 x 53 x5/3 3 x 2/3 34 x 4/3
1
1
1
x1/3
x
2(8) 53 (8)5/3 3(8)2/3 34 (8) 4/3 2(1) 53 (1)5/3 3(1) 2/3 34 (1)4/3 137
20
25.
sin 2 x dx
/2 2sin
/2 2sin2sinx cosx x dx /2 cos x dx sin x
x
26.
0
/3
(cos x sec x)2 dx
/3
0
1
(sin( )) sin 2
/2
(cos 2 x 2 sec 2 x) dx
/3 cos 2 x 1
0
2
2 sec2 x dx
2 cos 2 x 52 sec2 x dx 14 sin 2 x 52 x tan x 0
14 sin 2 3 52 3 tan 3 14 sin 2(0) 52 (0) tan(0) 56 9 8 3
/3
/3 1
0
27.
4 | x | dx 4 | x | dx 0 | x | dx 4 x dx 0 x dx x2 4 x2 0 02 2 42 02 16
28.
0 12 cos x cos x dx 0
4
0
4
0
/2 1
2
4
(cos x cos x ) dx
sin 2 sin 0 1
29.
30.
31.
32.
0
/2
1
x cos x 2 dx sin x 2
2
0
2 sin
1
5
x
x
dx 2cos x
2
1
/2
5
5
2
2
/3
/3
sin 2 x cos x dx
0
2
4
1 (cos x cos x ) dx
/2 2
2
( 4) 2
/2
cos x dx [sin x]0 /2
0
2 1 cos 1 2 2 cos 1
2 1 x
0
0
1
1
sin sin 0
2
2
2
dx x(1 x 2 )1/2 dx 1 x 2
2
x
2
26 5
(sin x)2 cos x dx 13 (sin x)3
/3
0
13 sin 3 3 13 sin 3 (0) 83
Copyright 2016 Pearson Education, Ltd.
2
2
307
308
Chapter 5 Integrals
33. (a)
0 cos t dt [sin t ]0
x
x
d
sin x sin 0 sin x dx
0 cos t dt
x
d (sin x ) cos x 1 x 1/2 cos x
dx
2
2 x
x
d
d
1 1/2 cos x
(b) dx
0 cos t dt (cos x ) dx ( x ) (cos x ) 2 x
2 x
d sin x 3t 2 dt d (sin 3 x 1) 3sin 2 x cos x
3t 2 dt [t 3 ]1sin x sin 3 x 1 dx
1
dx
sin
x
2
2
2
d
d (sin x ) 3sin x cos x
3t dt (3sin x) dx
dx 1
sin x
1
34. (a)
(b)
4
0
4
d t
dt 0
u du t 4
(b)
4
t
t4
d
d t
u du u1/2 du 23 u 3/2 23 (t 4 )3/2 0 23 t 6 dt
u du dt
0
0
0
t4
35. (a)
36. (a)
dtd (t 4 ) t 2 (4t3 ) 4t5
tan
tan
sec2 y dy [tan y ]0tan tan (tan ) 0 tan (tan ) dd
0
2
2
dd (tan(tan )) (sec (tan ))sec
0
d tan sec 2 y dy (sec 2 (tan )) d (tan )
d 0
d
(b)
x 3 2/3
0
37. (a)
t
dt 3t1/3
x3
0
sec2 y dy
(sec2 (tan )) sec2
3
d x t 2/3 dt d (3 x ) 3
3( x 0) 3 x dx
0
dx
x3 dxd x3 x 2 3x2 3
3
d x t 2/3 dt
dx 0
(b)
23 t 6 4t 5
2/3
t t
x dx
x dx t t t t t t
(b)
x dx t t t t t t
t
38. (a)
3
x3
t
d
dt 0
4
1
t
d
dt 1
39. y
x
42.
y x
t
d 1 5/2
dt 5
3
x3
2
3
x3
3
t 3/2
3 1
2
1 5/2
5
3 1
2
d
dt
13
10
13
10
1 5 3/2
5 2
2
3
t 3/2
x
sin t 2 dt
x
0
3 2
2
1 1/2
2
1 3/2
2
1 3/2
2
3 2
2
3 2
2
x
dy
0
y
4
3
2 x2 1
1 t 2 dt dx 1 x 2
0
41.
x5
5
4
dy
40. y 1t dt dx 1x , x
1
dy
sin t 2 dt dx sin( x )2
0
dxd ( x ) (sin x) 12 x1/2 sin2 xx
2
2
2
dy
d x sin t 3 dt 1 x sin t 3 dt x sin ( x 2 )3 d ( x 2 ) x sin t 3 dt
sin t 3 dt dx x dx
2
2
2
dx
2
x2
2 x 2 sin x6
x2
2
sin t 3 dt
Copyright 2016 Pearson Education, Ltd.
Section 5.4 The Fundamental Theorem of Calculus
2
2
x t2
x t2
dy
dt
dt dx 2x 2x 0
1 t 2 4
3 t 2 4
x 4 x 4
43.
y
44.
x
x
dy
y (t 3 1)10 dt dx 3 (t 3 1)10 dt
0
0
45.
y
0
46.
y
tan x dt
dy
1
dx
0
1t 2
1 tan 2 x
3
sin x
1t
2
2
d x (t 3 1)10 dt 3( x3 1)10 x (t 3 1)10 dt
dx 0
0
dy
dt
2
, x 2 dx
d (sin x )
1
1sin 2 x dx
x cos x 1 since x
cos1 x (cos x) cos
cos x
2
cos x
2
(tan x) sec x 1
d
dx
1
sec2 x
2
47. x 2 2 x 0 x( x 2) 0 x 0 or x 2;
Area
2
3
( x 2 2 x)dx
0
( x 2 2 x)dx
2
2
( x 2 2 x)dx
0
2
0
2
3
3
x3 x 2 x3 x 2 x3 x 2
3
2
0
3
( 2)3
( 3)
3 (2) 2 3 (3)2
3
3
( 2)
03 02 3 (2) 2
3
2 0
23
3
2
03
3
2
28
3
48. 3 x 2 3 0 x 2 1 x 1;
because of symmetry about the y -axis,
1
2
Area 2 (3 x 2 3)dx (3 x 2 3)dx
0
1
2 [ x3 3 x]10 [ x3 3 x]12
2[((13 3(1)) (03 3(0))) ((23 3(2)) (13 3(1))] 2(6) 12
49. x3 3x 2 2 x 0 x( x 2 3x 2) 0
x( x 2)( x 1) 0 x 0, 1, or 2;
1
2
0
1
Area ( x3 3 x 2 2 x)dx ( x3 3 x 2 2 x)dx
1
2
4
4
4
4
x4 x3 x 2 x4 x3 x 2 14 13 12 04 03 02
0
1
4
4
24 23 22 14 13 12 12
Copyright 2016 Pearson Education, Ltd.
309
310
50.
Chapter 5 Integrals
x1/3 x 0 x1/3 1 x 2/3 0 x1/3 0 or 1 x 2/3 0 x 0 or
1 x 2/3 x 0 or 1 x 2 x 0 or x 1;
0
1
8
1
0
1
Area ( x1/3 x) dx ( x1/3 x) dx ( x1/3 x) dx
0
1
8
2
2
2
34 x 4/3 x2 43 x 4/3 x2 43 x 4/3 x2
1
0
1
2
3 4/3 02 3
4/3 ( 1)
4 (0) 2 4 (1) 2
4/3 12
4/3 02
3
3
4 (1) 2 4 (0) 2
(8) (1)
4/3
3
4
82
2
3
4
4/3
12
2
14 14 (20 34 12 ) 83
4
51. The area of the rectangle bounded by the lines y 2, y 0, x , and x 0 is 2 . The area under the curve
y 1 cos x on [0, ] is (1 cos x) dx [ x sin x]0 ( sin ) (0 sin 0) . Therefore the area of the
0
shaded region is 2 .
52. The area of the rectangle bounded by the lines by the lines x 6 , x 56 , y sin 6 12 sin 56 , and y 0 is
3 . The area under the curve y sin x on 6 , 56 is 5/6/6 sin x dx [ cos x]5/6/6
cos 56 cos 6 23 23 3. Therefore the area of the shaded region is 3 3 .
1 5
2 6
6
53. On 4 , 0 : The area of the rectangle bounded by the lines y 2, y 0, 0, and 4 is 2 4
4 2 . The area between the curve y sec tan and y 0 is
0
0
sec tan d sec /4
/4
2 1. Therefore the area of the shaded region on 4 , 0 is 4 2 ( 2 1).
On 0, 4 : The area of the rectangle bounded by 4 , 0, y 2, and y 0 is 2 4 4 2 . The area
( sec 0) sec 4
under the curve y sec tan is
/4
0
/4
sec tan d sec 0
sec 4 sec 0 2 1. Therefore the area of
the shaded region on 0, 4 is 4 2 ( 2 1). Thus, the area of the total shaded region is
2
4
2 1 4 2 2 1 2 2 .
2 2 . The area
54. The area of the rectangle bounded by the lines y 2, y 0, t 4 , and t 1 is 2 1 4
under the curve y sec 2 t on 4 , 0 is
0
4
sec2 t dt [tan t ]0 4 tan 0 tan 4 1. The area under the
1
2
3
3
1
curve y 1 t 2 on [0, 1] is (1 t 2 ) dt t t3 1 13 0 03 32 . Thus, the total area under the curves
0
0
on 4 , 1 is 1 23 53 . Therefore the area of the shaded region is 2 2 53 13 2 .
55.
x
dy
y 1t dt 3 dx 1x and y ( ) 1t dt 3 0 3 3 (d) is a solution to this problem.
Copyright 2016 Pearson Education, Ltd.
Section 5.4 The Fundamental Theorem of Calculus
x
1
dy
56.
y sec t dt 4 dx sec x and y (1)
57.
y sec t dt 4 dx sec x and y (0) sec t dt 4 0 4 4 (b) is a solution to this problem.
58.
x
1
dy
y 1t dt 3 dx 1x and y (1) 1t dt 3 0 3 3 (a) is a solution to this problem.
59.
y sec t dt 3
1
x
1
sec t dt 4 0 4 4 (c) is a solution to this problem.
0
dy
0
0
1
1
x
60.
2
61. Area
311
b /2
b /2
h x dx hx
4h
b2
2
y
x
1
1 t 2 dt 2
b /2
4 hx3
2
3b b /2
2
2
4 h b2
4 h b2
b
b
h 2
h 2
3b 2
3b 2
bh bh bh bh bh 2 bh
bh
2
6
2
6
3
3
0 one arch of y sin kx will occur over the interval 0, k the area
62. k
1k cos k k
63.
1k cos (0)
/k
0
sin kx dx k1 cos kx
k2
dc 1 1 x 1/2 c x 1 t 1/2 dt [t1/2 ]x
0
2
dx
0 2
2 x
x; c(100) c(1) 100 1 $9.00
3
3
3
64. r 2 2 2 dx 2 1 1 2 dx 2 x x11 2 3 (311) 0 (011)
0
0
0
( x 1)
( x 1)
1
1
2 3 4 1 2 2 4 4.5 or $4500
65. (a) t 0 T 30 2 25 0 20 C; t 16 T 30 2 25 16 24 C;
t 25 T 30 2 25 25 30 C
(b) average temperature 2510
25
0
30 2 25 t dt 251 30t 4 3(25 t )3/2 0
25
1 30(25) 4 3(25 25)3/2 1 30(0) 4 3(25 0)3/2 23.33 C
25
25
66. (a) t 0 H 0 1 5(0)1/3 1 m; t 4 H 4 1 5(4)1/3 5 53 4 10.17 m;
t 8 H 8 1 5(8)1/3 13 m
t 1 5t1/3 dt 81 23 (t 1)3/2 15
t 4/3
4
0
0
3/2
4/3
3/2
4/3
18 23 (8 1) 15
(8) 18 23 (0 1) 15
(0) 29
9.67 m
4
4
3
(b) average height 81 0
67.
x
2
8
8
x
2
1 f (t ) dt x 2 x 1 f ( x) dxd 1 f (t ) dt dxd ( x 2 x 1) 2 x 2
Copyright 2016 Pearson Education, Ltd.
/k
0
312
Chapter 5 Integrals
x
x
68.
0 f (t ) dt x cos x f ( x) dxd 0 f (t ) dt cos x x sin x f (4) cos (4) (4) sin (4) 1
69.
f ( x) 2
x 1 9
11 9
dt f ( x) 1 (9x 1) x92 f (1) 3; f (1) 2
dt 2 0 2;
1t
2 1t
2
L( x) 3( x 1) f (1) 3( x 1) 2 3 x 5
70.
g ( x) 3
x2
1
g (1) 3
sec(t 1) dt g ( x ) (sec( x 2 1))(2 x) 2 x sec( x 2 1) g (1) 2(1) sec ((1)2 1) 2;
( 1) 2
1
1
sec(t 1) dt 3 sec(t 1) dt 3 0 3;
1
L( x) 2( x (1)) g (1) 2( x 1) 3 2 x 1
71. (a) True: since f is continuous, g is differentiable by Part 1 of the Fundamental Theorem of Calculus.
(b) True: g is continuous because it is differentiable.
(c) True: since g (1) f (1) 0.
(d) False, since g (1) f (1) 0.
(e) True, since g (1) 0 and g (1) f (1) 0.
(f ) False: g ( x) f ( x) 0, so g never changes sign.
(g) True, since g (1) f (1) 0 and g ( x) f ( x) is an increasing function of x (because f ( x) 0).
72. Let a x0 x1 x2 " xn b be any partition of [a, b] and left F be any antiderivative of f.
n
(a) [ F ( xi ) F ( xi 1 )]
i 1
[ F ( x1 ) F ( x0 )] [ F ( x2 ) F ( x1 )] [ F ( x3 ) F ( x2 )] " [ F ( xn 1 ) F ( xn 2 )] [ F ( xn ) F ( xn 1 )]
F ( x0 ) F ( x1 ) F ( x1 ) F ( x2 ) F ( x2 ) " F ( xn 1 ) F ( xn 1 ) F ( xn )
F ( xn ) F ( x0 ) F (b) F (a )
(b) Since F is any antiderivative of f on [a, b] F is differentiable of [a, b] F is continuous on [a, b].
Consider any subinterval [ xi 1, xi ] in [a, b], then by the Mean Value Theorem there is at least one
number ci in ( xi 1, xi ) such that [ F ( xi ) F ( xi 1 )] F (ci )( xi xi 1 ) f (ci )( xi xi 1 ) f (ci )xi .
n
n
Thus F (b) F (a) [ F ( xi ) F ( xi 1 )] f (ci )xi .
i 1
i 1
n
(c) Taking the limit of F (b) F (a) f (ci )xi we obtain lim ( F (b) F (a)) lim f (ci )xi
P 0
P 0 i 1
i 1
n
b
F (b) F (a) f ( x) dx
a
t
d
73. (a) v ds
dt
f ( x) dx f (t ) v(5) f (5) 2 m/s
dt
0
df
(b) a dt is negative since the slope of the tangent line at t = 5 is negative
3
(c) s f ( x) dx 12 (3)(3) 92 m since the integral is the area of the triangle formed by y = f(x), the x-axis
0
and x = 3
(d) t = 6 since from t = 6 to t = 9, the region lies below the x-axis
(e) At t = 4 and t = 7, since there are horizontal tangents there
(f) Toward the origin between t = 6 and t = 9 since the velocity is negative on this interval. Away from the
origin between t = 0 and t = 6 since the velocity is positive there.
(g) Right or positive side, because the integral of f from 0 to 9 is positive, there being more area above the
x-axis than below it.
Copyright 2016 Pearson Education, Ltd.
Section 5.4 The Fundamental Theorem of Calculus
74. If the marginal cost is
x2
x
115, by the net change theorem the production cost is
1000 2
x
p( x )
t2
t
1 3 1 2
115 dt
x x 115 x. Thus the average cost per unit for 600 units is
3000
4
0 1000 2
p(600)
85.
600
7578.
Example CAS commands:
Maple:
with( plots );
f : x - x^3-4*x^2 3*x;
a : 0;
b : 4;
F : unapply( int(f(t),t a..x), x );
# (a)
p1: plot( [f(x),F(x)], x a..b, legend ["y f(x)","y F(x)"], title "#75(a) (Section 5.4)" ):
p1;
dF : D(F);
# (b)
q1: solve( dF(x) 0, x );
pts1: [ seq( [x,f(x)], x remove(has,evalf([q1]),I) ) ];
p2 : plot( pts1, style point, color blue, symbolsize 18, symbol diamond, legend "(x,f(x))
where F'(x) 0" ):
display( [p1, p2], title "75(b) (Section 5.4)" );
incr : solve( dF(x)>0, x );
decr : solve( dF(x)<0, x );
# (c)
df : D(f );
# (d)
p3 : plot( [df(x),F(x)], x a..b, legend ["y f '(x)","y F(x)"], title "#75(d) (Section 5.4)" ):
p3;
q2 : solve( df(x) 0, x );
pts2 : [ seq( [x,F(x)], x remove(has,evalf([q2]),I) ) ];
p4 : plot( pts2, style point, color blue, symbolsize 18, symbol diamond, legend "(x,f(x))
where f '(x) 0" ):
display( [p3,p4], title "75(d) (Section 5.4)" );
79-82.
Example CAS commands:
Maple:
a : 1;
u : x - x^2;
f : x - sqrt(1-x^2);
F : unapply( int( f(t),t a..u(x) ), x );
dF : D(F);
cp : solve( dF(x) 0, x );
solve( dF(x)>0, x );
solve( dF(x)<0, x );
# (b)
Copyright 2016 Pearson Education, Ltd.
313
314
Chapter 5 Integrals
d2F : D(dF);
solve( d2F(x) 0, x );
# (c)
plot( F(x), x -1..1, title "#79(d) (Section 5.4)" );
83.
Example CAS commands:
Maple:
f : `f `;
q1: Diff( Int( f(t), t a..u(x) ), x );
d1: value( q1 );
84.
Example CAS commands:
Maple:
f : `f `;
q2 : Diff( Int( f(t), t a..u(x) ), x,x );
value( q2 );
75-84.
Example CAS commands:
Mathematica: (assigned function and values for a, and b may vary)
For transcendental functions the FindRoot is needed instead of the Solve command.
The Map command executes FindRoot over a set of initial guesses
Initial guesses will vary as the functions vary.
Clear[x, f, F]
{a, b}{0, 2π}; f[x_] Sin[2x] Cos[x/3]
F[x_] Integrate[f[t],{t, a, x}]
Plot[{f[x], F[x]},{x, a, b}]
x/.Map[FindRoot[F'[x] 0, {x, #}] &, {2, 3, 5, 6}]
x/.Map[FindRoot[f '[x]0, {x, #}] &, {1, 2, 4, 5, 6}]
Slightly alter above commands for 79 84.
Clear[x, f, F, u]
a 0; f[x_] x 2 2x 3
u[x_] 1 x 2
F[x_] Integrate[f[t], {t, a, u(x)}]
x/.Map[FindRoot[F'[x] 0, {x, #}] &, {1, 2, 3, 4}]
x/.Map[FindRoot[F"[x] 0, {x, #}] &, {1, 2, 3, 4}]
After determining an appropriate value for b, the following can be entered
b 4;
Plot[{F[x],{x, a, b}]
5.5
INDEFINITE INTEGRALS AND THE SUBSTITUTION METHOD
1. Let u 2 x 4 du 2 dx 12 du dx
5
5
5
6
6
2(2 x 4) dx 2u 12 du u du 16 u C 16 (2 x 4) C
Copyright 2016 Pearson Education, Ltd.
Section 5.5 Indefinite Integrals and the Substitution Method
315
2. Let u 7 x 1 du 7 dx 17 du dx
1/2
dx 7u1/2 17 du u1/2 du 23 u 3/2 C 23 (7 x 1)3/2 C
7 7 x 1 dx 7(7 x 1)
3. Let u x 2 5 du 2 x dx 12 du x dx
4
2
2 x( x 5) dx 2u
4 1
du
2
4
u du 13 u
4. Let u x 4 1 du 4 x3 dx 14 du x3 dx
3
3
2
4
( x4 x1) dx 4 x ( x 1) dx 4u
4
2
2 1
du
4
3
C 13 ( x 2 5)3 C
2
u du u
1
C
1 C
x 4 1
5. Let u 3 x 2 4 x du (6 x 4)dx 2(3x 2)dx 12 du (3x 2)dx
2
4
4
4
5
2
1/3
4/3
5
(3x 2)(3x 4 x) dx u 12 du 12 u du 101 u C 101 (3x 4 x) C
6. Let u 1 x du
(1 x )1/3
dx
x
(1
1 dx 2 du 1 dx
2 x
x
1/3 1
1/3
x)
dx u 2 du 2
x
u du 2 34 u
C 32 (1 x ) 4/3 C
7. Let u 3 x du 3 dx 13 du dx
sin 3x dx 13 sin u du 13 cos u C 13 cos 3x C
8. Let u 2 x 2 du 4 x dx 14 du x dx
2
2
x sin (2 x ) dx 14 sin u du 14 cos u C 14 cos 2 x C
9. Let u 2t du 2 dt 12 du dt
sec 2t tan 2t dt 12 sec u tan u du 12 sec u C 12 sec 2t C
10. Let u 1 cos 2t du 12 sin 2t dt 2 du sin 2t dt
1 cos 2t sin 2t dt 2u du 32 u C 32 1 cos 2t C
2
2
3
3
11. Let u 1 r 3 du 3r 2 dr 3du 9r 2 dr
2
1/2
1/2
3 1/2
9r dr3 3u du 3(2)u C 6(1 r ) C
1 r
12. Let u y 4 4 y 2 1 du (4 y 3 8 y ) dy 3 du 12 ( y 3 2 y ) dy 12( y 4 4 y 2 1) 2 ( y 3 2 y ) dy
3u 2 du u 3 C ( y 4 4 y 2 1)3 C
13. Let u x3/2 1 du 32 x1/2 dx 32 du x dx
2
x sin ( x
32
1) dx 23 sin 2 u du 23 u2 14 sin 2u C 13 ( x3/2 1) 16 sin (2 x3/2 2) C
14. Let u 1x du 12 dx
x
dx cos2 (u) du cos2 (u) du u2 14 sin 2u C 21x 14 sin 2x C 21x 14 sin 2x C
1 cos 2 1
x
x2
Copyright 2016 Pearson Education, Ltd.
316
Chapter 5 Integrals
15. (a) Let u cot 2 du 2 csc2 2 d 12 du csc2 2 d
csc 2 cot 2 d 12 u du 12 u2 C u4 C 14 cot 2 C
2
2
2
2
(b) Let u csc 2 du 2 csc 2 cot 2 d 12 du csc 2 cot 2 d
csc 2 cot 2 d 12 u du 12 u2 C u4 C 14 csc 2 C
2
2
2
2
16. (a) Let u 5 x 8 du 5 dx 15 du dx
5dxx8 15 1u du 15 u
1/2
du 15 (2u1/2 ) C 52 u1/2 C 52 5 x 8 C
(b) Let u 5 x 8 du 12 (5 x 8) 1/2 (5) dx 52 du
5dxx8 52 du 52 u C 52 5 x 8 C
17. Let u 3 2 s du 2 ds 12 du ds
3 2s ds u 12 du 12 u
1/2
1/2
32 u3/2 C 13 (3 2s)3/2 C
du 12
18. Let u 5s 4 du 5 ds 15 du ds
51s 4 ds 1u 15 du 15 u
dx
5 x 8
du 15 (2u1/2 ) C 52 5s 4 C
19. Let u 1 2 du 2 d 12 du d
1 d u 12 du 12 u
4
2
1/4
4
54 u5/4 C 52 (1 2 )5/4 C
du 12
20. Let u 7 3 y 2 du 6 y dy 12 du 3 y dy
3 y 7 3 y dy u 12 du 12 u
2
1/2
32 u3/2 C 13 (7 3 y 2 )3/2 C
du 12
21. Let u 1 x du 1 dx 2 du 1 dx
1
dx
x (1 x )2
2 du
u2
2 x
x
2
2
u C
C
1 x
22. Let u sin x du cos x dx
sin x 1 sin x cos x dx u
2
1/2
u5/2 du 23 u 3/2 72 u 7/2 C 23 sin 3/2 x 72 sin 7/2 x C
23. Let u 3 x 2 du 3dx 13 du dx
sec (3x 2) dx (sec u ) 13 du 13 sec u du 13 tan u C 13 tan(3x 2) C
2
2
2
24. Let u tan x du sec2 x dx
2
2
2
3
3
tan x sec x dx u du 13 u C 13 tan x C
5
6
x
1
cos 3 dx u (3 du) 3 6 u C 12 sin 6 3x C
25. Let u sin 3x du 13 cos 3x dx 3 du cos 3x dx
sin 5 3x
Copyright 2016 Pearson Education, Ltd.
Section 5.5 Indefinite Integrals and the Substitution Method
2 x
7
8
1
1
sec 2 dx u (2 du) 2 8 u C 4 tan8 2x C
26. Let u tan 2x du 12 sec2 2x dx 2 du sec2 2x dx
tan 7 2x
3
2
r 1 du r dr 6 du r 2 dr
27. Let u 18
6
r 18r 1 dr u (6 du ) 6 u du 6 u6 C 18r 1 C
5
3
2
5
6
5
6
3
5
r du 1 r 4 dr 2 du r 4 dr
28. Let u 7 10
2
r 7 10r dr u (2 du ) 2 u du 2 u4 C 12 7 10r C
3
5
4
3
4
3
5
4
29. Let u x3/2 1 du 32 x1/2 dx 32 du x1 2 dx
1/2
x
sin( x3/2 1) dx (sin u ) 23 du 23 sin u du 23 ( cos u ) C 23 cos( x3/2 1) C
v
csc cot 2 dv 2du 2u C 2 csc v 2 C
30. Let u csc v 2 du 12 csc v 2 cot v 2 dv 2du csc v 2 cot v 2 dv
v
2
31. Let u cos(2t 1) du 2sin(2t 1) dt 12 du sin(2t 1) dt
sin(2t 1)
cos2 (2t 1) dt 12 udu2 21u C 2 cos(21 t 1) C
32. Let u sec z du sec z tan z dz
1/2
1/2
sec z tan z dz 1 du u du 2u C 2 sec z C
sec z
u
33. Let u 1t 1 t 1 1 du t 2 dt du 12 dt
1 cos 1 1 dt
t
t2
t
(cos u )(du ) cos u du sin u C sin 1t 1 C
34. Let u t 3 t1/2 3 du 12 t 1/2 dt 2du 1 dt
t
1 cos(
t
t 3) dt (cos u )(2 du ) 2 cos u du 2sin u C 2sin( t 3) C
35. Let u sin 1 du cos 1
2
1 sin 1 cos 1 d
1 d du 1 cos 1 d
2
2
u du 12 u 2 C 12 sin 2 1 C
36. Let u csc du csc cot
cos
2 1 d 2du 1 cot csc d
sin 2 d 1 cot csc d 2 du 2u C 2 csc C sin2 C
37. Let u 1 x x u 1 dx du
1x x dx u u1 du u
1/2
u 1/2 du 23 u 3/2 2u1/2 C 23 (1 x )3/2 2(1 x )1/2 C
38. Let u 1 1x du 12 dx
x 1 dx
x5
1
x2
x
x 1 dx
x
1/2
x12 1 1x dx u du u
du 32 u 3/2 C 32 1 1x
Copyright 2016 Pearson Education, Ltd.
3/2
C
317
318
Chapter 5 Integrals
39. Let u 2 1x du 12 dx
x
2 1x dx u du u1/2 du 32 u 3/2 C 32 2 1x
1
x2
3/2
C
40. Let u 1 12 du 23 dx 12 du 13 dx
x
x 2 1 dx
x2
1
x3
x
1
x3
1 12 dx
x
x
u 12 du 12
1/2
3/2
u du 13 u C 13 1 12
x
C
3/2
41. Let u 1 33 du 94 dx 19 du 14 dx
x
x
3
x
3
xx113 dx x14 x x3 3 dx x14 1 x33 dx u 19 du 19 u
1/2
2 u 3/2 C 2 1 3
du 27
3
27
x
C
3/2
42. Let u x3 1 du 3x 2 dx 13 du x 2 dx
4
2
xx 1 dx xx 1 dx 1u 13 du 13 u
3
3
1/2
du 23 u1/2 C 23 ( x3 1)3/2 C
43. Let u x 1. Then du dx and x u 1. Thus x( x 1)10 dx (u 1) u10 du (u11 u10 )du
1 u12 1 u11 C 1 ( x 1)12 1 ( x 1)11 C
12
11
12
11
44. Let u 4 x. Then du 1 dx and (1)du dx and x 4 u. Thus x 4 xdx (4 u ) u (1) du
(4 u )(u1/2 ) du (u 3/2 4u1/2 ) du 52 u 5/2 83 u 3/2 C 52 (4 x)5/2 83 (4 x)3/2 C
45. Let u 1 x. Then du 1 dx and (1)du dx and x 1 u. Thus ( x 1)2 (1 x)5 dx
(2 u )2 u 5 (1) du (u 7 4u 6 4u 5 ) du 18 u8 74 u 7 23 u 6 C 18 (1 x)8 74 (1 x)7 23 (1 x )6 C
46. Let u x 5. Then du dx and x u 5. Thus ( x 5)( x 5)1/3 dx (u 10)u1/3 du (u 4/3 10u1/3 ) du
73 u 7/3 15
u 4/3 C 73 ( x 5)7/3 15
( x 5)4/3 C
2
2
47. Let u x 2 1. Then du 2 x dx and 12 du x dx and x 2 u 1. Thus x3 x 2 1 dx (u 1) 12 u du
12 (u 3/2 u1/2 ) du 12 52 u 5/2 32 u 3/2 C 15 u 5/2 13 u 3/2 C 15 ( x 2 1)5/2 13 ( x 2 1)3/2 C
48. Let u x3 1 du 3x 2 dx and x3 u 1. So 3x5 x3 1 dx (u 1) u du (u 3/2 u1/2 ) du
52 u 5/2 23 u 3/2 C 52 ( x3 1)5/2 23 ( x3 1)3/2 C
49. Let u x 2 4 du 2 x dx and 12 du x dx. Thus
14 u 2 C 14 ( x 2 4) 2 C
50. Let u 2 x 1 x 12 (u 1) dx 12 du. Thus
x
dx
( x 2 4)3
x
(2 x 1)2/3
2
3
( x 4) x dx u
1 ( u 1)
2
2/3
u
3 1
du 12
2
12 du 14 u1/3 u 2/3 du
3 (2 x 1) 4/3 3 (2 x 1)1/3 C
14 43 u 4/3 3u1/3 C 16
4
Copyright 2016 Pearson Education, Ltd.
3
u du
Section 5.5 Indefinite Integrals and the Substitution Method
319
51. (a) Let u tan x du sec2 x dx; v u 3 dv 3u 2 du 6dv 18u 2 du; w 2 v dw dv
2
6 dv
2
2
2
3 2
6 dw
2
18(2tan tanx secx) x dx (218uu ) du (2v) w 6 w dw 6w
3
6 3 C
2u
2
1
C 26 v C
2
6
C
2 tan 3 x
2
(b) Let u tan 3 x du 3 tan x sec2 x dx 6 du 18 tan 2 x sec2 x dx; v 2 u dv du
2
6 du
2
6 dv
18(2tan tanx secx) x dx (2u ) v 6v C 26u C 2 tan6 x C
3
2
2
2
3
(c) Let u 2 tan 3 x du 3 tan 2 x sec2 x dx 6 du 18 tan 2 x sec2 x dx
2
2
6 du
18(2tan tanx3secx)2 x dx u 2 u6 C 2 tan6 3 x C
52. (a) Let u x 1 du dx; v sin u dv cos u du; w 1 v 2 dw 2v dv 12 dw v dv
2
2
2
1 sin ( x 1) sin( x 1) cos( x 1) dx 1 sin u sin u cos u du v 1 v dv
12 w dw 13 w3/2 C 13 (1 v 2 )3/2 C 13 (1 sin 2 u )3/2 C 13 (1 sin 2 ( x 1))3/2 C
(b) Let u sin( x 1) du cos( x 1) dx; v 1 u 2 dv 2u du 12 dv u du
2
2
1/2
1 sin ( x 1) sin( x 1) cos( x 1) dx u 1 u du 12 v dv 12 v
dv
3/2
(c)
12 23 v
C 13 v3/2 C 13 (1 u 2 )3/2 C 13 (1 sin 2 ( x 1))3/2 C
Let u 1 sin 2 ( x 1) du 2sin( x 1) cos( x 1) dx 12 du sin( x 1) cos( x 1) dx
1 sin 2 ( x 1) sin( x 1) cos( x 1) dx 12 u du 12 u1/2 du 12 23 u 3/2 C 13 (1 sin 2 ( x 1))3/2 C
1 du (2r 1) dr ; v u dv 1 du 1 dv 1 du
53. Let u 3(2r 1) 2 6 du 6(2r 1)(2) dr 12
6
(2 r 1) cos 3(2r 1) 2 6
3(2 r 1) 2 6
dr
2 u
cos u
u
12 u
du (cos v) dv sin v C sin u C
1
12
1
6
1
6
1
6
16 sin 3(2r 1)2 6 C
54. Let u cos du sin
sin
sin
2 1 d 2du sin d
cos3 d cos3 d u23/du2 2 u
3/2
du 2(2u 1/2 ) C 4 C
u
55. Let u 3t 2 1 du 6t dt 2du 12t dt
s 12t (3t 2 1)3 dt u 3 (2 du ) 2 14 u 4 C 12 u 4 C 12 (3t 2 1)4 C ;
s 3 when t 1 3 12 (3 1)4 C 3 8 C C 5 s 12 (3t 2 1)4 5
56. Let u x 2 8 du 2 x dx 2 du 4 x dx
y 4 x( x 2 8) 1/3 dx u 1/3 (2du ) 2 32 u 2/3 C 3u 2/3 C 3( x 2 8)2/3 C ;
y 0 when x 0 0 3(8)
du dt
57. Let u t 12
2/3
C C 12 y 3( x 2 8) 2/3 12
dt 8sin 2 u du 8 u 1 sin 2u C 4 t 2sin 2t C ;
s 8sin 2 t 12
2 4
12
6
Copyright 2016 Pearson Education, Ltd.
4
C
cos
320
Chapter 5 Integrals
2sin C C 8 1 9
s 8 when t 0 8 4 12
6
3
3
s4 t
12
2sin 2t 9 4t 2sin 2t 9
6
3
58. Let u 4 du d
6
r 3cos 2 4 d 3cos 2u du 3 u2 14 sin 2u C 23 4 43 sin 2 2 C ;
r 8 when 0 8 38 34 sin 2 C C 2 43 r 23 4 43 sin 2 2 2 43
r 32 34 sin 2 2 8 43 r 23 43 cos 2 8 43
59. Let u 2t 2 du 2 dt 2 du 4 dt
ds
dt
4sin 2t 2 dt (sin u )(2 du ) 2 cos u C1 2 cos 2t 2 C1;
100 we have 100 2 cos 2 C1 C1 100 ds
2 cos 2t 2 100
at t 0 and ds
dt
dt
s 2 cos 2t 2 100 dt (cos u 50) du sin u 50u C2 sin 2t 2 50 2t 2 C2 ;
at t 0 and s 0 we have 0 sin 2 50 2 C2 C2 1 25
s sin 2t 2 100t 25 (1 25 ) s sin 2t 2 100t 1
60. Let u tan 2 x du 2sec2 2 x dx 2du 4sec2 2 x dx; v 2 x dv 2dx 12 dv dx
dy
dx
2
2
2
4sec 2 x tan 2 x dx u (2 du ) u C1 tan 2 x C1;
dy
dy
at x 0 and dx 4 we have 4 0 C1 C1 4 dx tan 2 2 x 4 (sec 2 2 x 1) 4 sec2 2 x 3
y (sec 2 2 x 3) dx (sec2 v 3) 12 dv 12 tan v 32 v C2 12 tan 2 x 3x C2 ;
at x 0 and y 1 we have 1 12 (0) 0 C2 C2 1 y 12 tan 2 x 3x 1
61. Let u 2t du 2 dt 3 du 6dt
s 6 sin 2t dt (sin u )(3 du ) 3 cos u C 3 cos 2t C ;
at t 0 and s 0 we have 0 3cos 0 C C 3 s 3 3cos 2t s 2 3 3cos( ) 6 m
62. Let u t du dt du 2 dt
v 2 cos t dt (cos u )( du ) sin u C1 sin( t ) C1;
at t 0 and v 8 we have 8 (0) C1 C1 8 v ds
sin( t ) 8 s ( sin( t ) 8) dt
dt
sin u du 8t C2 cos( t ) 8t C2 ; at t 0 and s 0 we have 0 1 C2 C2 1
s 8t cos ( t ) 1 s (1) 8 cos 1 10 m
5.6
DEFINITE INTEGRAL SUBSTITUTIONS AND THE AREA BETWEEN CURVES
1. (a) Let u y 1 du dy; y 0 u 1, y 3 u 4
3
4 1/2
0 y 1 dy 1 u
4
du 23 u 3/2 23 (4)3/2 23 (1)3/2 23 (8) 23 (1) 14
3
1
(b) Use the same substitution for u as in part (a); y 1 u 0, y 0 u 1
0
1 1/2
1 y 1 dy 0 u
1
du 23 u 3/2
0
23 (1)3/2 0 23
Copyright 2016 Pearson Education, Ltd.
Section 5.6 Definite Integral Substitutions and the Area Between Curves
2. (a) Let u 1 r 2 du 2r dr 12 du r dr ; r 0 u 1, r 1 u 0
1
0
0 r 1 r dr 1 12 u du 13 u
2
(1)3/2 13
3/2 0
0 13
1
(b) Use the same substitution for u as in part (a); r 1 u 0, r 1 u 0
1
0
2
1 r 1 r dr 0 12 u du 0
3. (a) Let u tan x du sec2 x dx; x 0 u 0, x 4 u 1
/4
0
1
1
tan x sec2 x dx u du u2 12 0 12
0
0
2
2
(b) Use the same substitution as in part (a); x 4 u 1, x 0 u 0
0
0
0
2
/4 tan x sec x dx 1u du u2 1 0 12 12
2
4. (a) Let u cos x du sin x dx du sin x dx; x 0 u 1, x u 1
1
2
3 1
2
3
3
0 3cos x sin x dx 1 3u du [u ] 1 (1) ((1) ) 2
(b) Use the same substitution as in part (a); x 2 u 1, x 3 u 1
3
1
2
2
2 3cos x sin x dx 1 3u du 2
5. (a) u 1 t 4 du 4t 3dt 14 du t 3dt ; t 0 u 1, t 1 u 2
2
1 3
2
4 3
3
u 4 24 14 15
0 t (1 t ) dt 1 14 u du 16
1 16 16 16
(b) Use the same substitution as in part (a); t 1 u 2, t 1 u 2
1 3
2
4 3
3
1t (1 t ) dt 2 14 u du 0
6. (a) Let u t 2 1 du 2t dt 12 du t dt ; t 0 u 1, t 7 u 8
7
2
1/3
0 t (t 1)
34 u 4/3 1 83 (8)4/3 83 (1)4/3 458
8
8
dt 12 u1/3 du 12
1
(b) Use the same substitution as in part (a); t 7 u 8, t 0 u 1
0
2
1
1/3
1/3
7 t (t 1) dt 8 12 u
8
du 12 u1/3 du 45
8
1
7. (a) Let u 4 r 2 du 2r dr 12 du rdr ; r 1 u 5, r 1 u 5
1
5
1 (45rr2 )2 dr 55 12 u
2
du 0
(b) Use the same substitution as in part (a); r 0 u 4, r 1 u 5
1
5
0 (45rr2 )2 dr 5 4 12 u
2
5
du 5 12 u 1 5 12 (5) 1 5 12 (4)1 18
4
8. (a) Let u 1 v3/2 du 32 v1/2 dv 20
du 10 v dv; v 0 u 1, v 1 u 2
3
0 (1v3/ 2 )2 dv 1 u12 203 du 203 1 u
1 10 v
2
2 2
1
du 20
3 u
2
1
1 1 10
20
3 2 1
3
Copyright 2016 Pearson Education, Ltd.
321
322
Chapter 5 Integrals
(b) Use the same substitution as in part (a); v 1 u 2, v 4 u 1 43/2 9
70
1 (1v3/ 2 )2 dv 2 u12 203 du 203 u1 2 203 19 12 203 187 27
4 10 v
9
9
9. (a) Let u x 2 1 du 2 x dx 2 du 4 x dx; x 0 u 1, x 3 u 4
0
3
4x
2
x 1
dx
4 2
1
4
du 2u 1/2 du [4u1/2 ] 14 4(4)1/2 4(1)1/2 4
1
u
(b) Use the same substitution as in part (a); x 3 u 4, x 3 u 4
3
4
3 x42x1 dx 4 2u du 0
10. (a) Let u x 4 9 du 4 x3 dx 14 du x3 dx; x 0 u 9, x 1 u 10
1
10
3
0 xx4 9 dx 9 14 u
1/2
10
du 14 (2)u1/2 12 (10)1/2 12 (9)1/2
9
10 3
2
(b) Use the same substitution as in part (a); x 1 u 10, x 0 u 9
0
3
9
1 xx4 9 dx 10 14 u
1/2
du
10 1 1/2
u
du 3 210
9 4
1
1
11. (a) Let u 4 5t t (u 4), dt du; t 0 u 4, t 1 u 9.
5
5
0 t 4 5t dt 25 4 (u 4) u du 25 4 u
1
1
1
9
9
3/2
4u1/2 du
9
1 2 5/2 8 3/2
1 2
8
8
2
506
u u (243) (27) (32) 8)
25 5
3
25
5
3
5
3
4
375
(b) Use the same substitution as in (a); t 1 u 9, t 9 u 49.
9
1
t 4 5t dt
49
1 49 3/2
1 2
8
u 4u1/2 du u5/2 u 3/2
9
25
25 5
3
9
1 2
8
8
2
86,744
(16,807) (343) (243) 27
25 5
3
3
375
5
12. (a) Let u 1 cos 3t du 3sin 3t dt 13 du sin 3t dt ; t 0 u 0, t 6 u 1 cos 2 1
/6
1
2
2
1
2
2
0 (1 cos 3t ) sin 3t dt 0 13 u du 13 u2 0 16 (1) 16 (0) 16
(b) Use the same substitution as in part (a); t 6 u 1, t 3 u 1 cos 2
/3
2
2
2
2
/6 (1 cos 3t )sin 3t dt 1 13 u du 13 u2 1 16 (2) 16 (1) 12
13. (a) Let u 4 3sin z du 3cos z dz 13 du cos z dz; z 0 u 4, z 2 u 4
2
0
4 1 1
cos z
dz
du
4
4 3sin z
u 3
0
(b) Use the same substitution as in part (a); z u 4 3sin( ) 4, z u 4
z
dz 1 13 du 0
4cos
4 u
3sin z
4
Copyright 2016 Pearson Education, Ltd.
Section 5.6 Definite Integral Substitutions and the Area Between Curves
323
14. (a) Let u 2 tan 2t du 12 sec2 2t dt 2 du sec2 2t dt ; t 2 u 2 tan 4 1, t 0 u 2
/2 2 tan 2t sec 2t dt 1 u (2 du ) [u ] 1 2 1 3
0
2
2
2 2
2
2
(b) Use the same substitution as in part (a); t 2 u 1, t 2 u 3
/2
3
2
2 3
2
2
/2 (2 tan 2t ) sec 2t dt 21 u du [u ] 1 3 1 8
15. Let u t 5 2t du (5t 4 2) dt ; t 0 u 0, t 1 u 3
1
5
3 1/2
4
0 t 2t (5t 2) dt 0 u
3
du 23 u 3/2 23 (3)3/2 23 (0)3/2 2 3
0
dy
; y 1 u 2, y 4 u 3
2 y
3 1
3
du u 2 du [u 1 ]32 13
2 u2
2
16. Let u 1 y du
4
dy
1 2 y (1 y )2
12 16
17. Let u cos 2 du 2sin 2 d 12 du sin 2 d ; 0 u 1, 6 u cos 2 6 12
/6
0
cos 3 2 sin 2 d
1/2 3
u
1
12 du 12 11/2 u 3 du 12 u2 1 4 1 4(1)1 34
2
1/2
1
2
2
2
1
1
4
34
3 4 3 4 12
u 5 (6 du ) 6 u4
1/ 3
2(1)
1/ 3 2u 1/ 3
2 1
3
18. Let u tan 6 du 16 sec 2 6 d 6 du sec 2 6 d ; u tan 6 1 , 32 u tan 4 1
3 /2
cot 5 6 sec 2 6 d
3
1
19. Let u 5 4 cos t du 4sin t dt 14 du sin t dt ; t 0 u 5 4 cos 0 1, t u 5 4 cos 9
9
9 1/4
9 1/4
1/4
5/4
5/4
5/2
0 5(5 4 cos t ) sin t dt 1 5u 14 du 54 1 u du 54 54 u 1 9 1 3 1
20. Let u 1 sin 2t du 2 cos 2t dt 12 du cos 2t dt ; t 0 u 1, t 4 u 0
/4
0
0
0
(1 sin 2t )3/2 cos 2t dt 12 u 3/2 du 12 52 u 5/2 15 (0)5/2 15 (1)5/2 15
1
1
21. Let u 4 y y 2 4 y 3 1 du (4 2 y 12 y 2 ) dy; y 0 u 1, y 1 u 4(1) (1)2 4(1)3 1 8
1
2
3
0 (4 y y 4 y 1)
2/3
8
(12 y 2 2 y 4) dy u 2/3 du [3u1/3 ] 18 3(8)1/3 3(1)1/3 3
1
22. Let u y 3 6 y 2 12 y 9 du (3 y 2 12 y 12) dy 13 du ( y 2 4 y 4) dy; y 0 u 9, y 1 u 4
1
3
2
0 ( y 6 y 12 y 9)
1/2
( y 2 4 y 4) dy
4
4 1 1/2
u
du 13 (2u1/2 ) 23 (4)1/2 23 (9)1/2 23 (2 3) 23
3
9
9
3
23. Let u 3/2 du 32 1/2 d 23 du d ; 0 u 0, 2 u
3
0
2
0
0 23 2 14 sin 2 23 (0) 3
cos 2 ( 3/2 ) d cos 2 u 23 du 23 u2 14 sin 2u
Copyright 2016 Pearson Education, Ltd.
324
Chapter 5 Integrals
24. Let u 1 1t du t 2 dt ; t 1 u 0, t 12 u 1
1/2 2
1 t
sin 2 1 1t dt
1
0
1
sin 2 u du u2 14 sin 2u 12 14 sin(2) 02 14 sin 0
0
12 14 sin 2
25. Let u 4 x 2 du 2 x dx 12 du x dx; x 2 u 0, x 0 u 4, x 2 u 0
A
0
2
2
4
0
0
0
4
x 4 x 2 dx x 4 x 2 dx 12 u1/2 du 12 u1/2 du 2
4 1 1/2
4
u du u1/2 du
0 2
0
4
23 u 3/2 23 (4)3/2 23 (0)3/2 16
3
0
26. Let u 1 cos x du sin x dx; x 0 u 0, x u 2
2
2
0 (1 cos x) sin x dx 0 u du u2 0 22 02 2
2
2
2
27. Let u 1 cos x du sin x dx du sin x dx; x u 1 cos ( ) 0, x 0 u 1 cos 0 2
A
0
2
2
2
3(sin x) 1 cos x dx 3u1/2 (du ) 3 u1/2 du 2u 3/2 2(2)3/2 2(0)3/2 25/2
0
0
0
28. Let u sin x du cos x dx 1 du cos x dx; x 2 u sin 2 0, x 0 u
Because of symmetry about x 2 , A 2
(cos x )(sin( sin x )) dx 2 (sin u ) 1 du
0 2
0
/2 2
sin u du [ cos u ]0 ( cos ) ( cos 0) 2
0
29. For the sketch given, a 0, b ; f ( x) g ( x) 1 cos 2 x sin 2 x
A
(1 cos 2 x )
0
2
1 cos 2 x
;
2
sin 2 x
dx 12 (1 cos 2 x) dx 12 x 2 12 [( 0) (0 0)] 2
0
0
30. For the sketch given, a 3 , b 3 ; f (t ) g (t ) 12 sec 2 t (4 sin 2 t ) 12 sec2 t 4sin 2 t ;
A
12
/3
/3 (1cos 2t )
1 sec 2 t 4sin 2 t dt 1 /3 sec 2 t dt 4
sin 2t dt 12
sec2 t dt 4
dt
2
2
2
/3
/3
/3
/3
/3
/3
/3
sec2t dt 2
/3
/3
sin 2t
(1 cos 2t ) dt 12 [tan t ]/3/3 2 t 2
3 4 3 3 43
/3
/3
31. For the sketch given, a 2, b 2; f ( x) g ( x) 2 x 2 ( x 4 2 x 2 ) 4 x 2 x 4 ;
A
2
2
2
5
64 64 320192 128
(4 x 2 x 4 ) dx 43x x5 32
32
32
32
3
5
5
3
5
15
15
3
2
2
32. For the sketch given, c 0, d 1; f ( y ) g ( y ) y 2 y 3 ;
1
1
1
1
1
(1 0) (10)
y3 y 4
1
A ( y 2 y 3 ) dy y 2 dy y 3 dy 3 4 3 4 13 14 12
0
0
0
0 0
Copyright 2016 Pearson Education, Ltd.
Section 5.6 Definite Integral Substitutions and the Area Between Curves
325
33. For the sketch given, c 0, d 1; f ( y ) g ( y ) (12 y 2 12 y 3 ) (2 y 2 2 y ) 10 y 2 12 y3 2 y;
1
1
1
1
1
1
1
A (10 y 2 12 y 3 2 y ) dy 10 y 2 dy 12 y 3 dy 2 y dy 10
y 3 12 y 4 22 y 2
3 0 4
0
0
0
0
0
0
10
0 (3 0) (1 0) 43
3
34. For the sketch given, a 1, b 1; f ( x) g ( x) x 2 ( 2 x 4 ) x 2 2 x 4 ;
1
1
12 22
A ( x 2 2 x 4 ) dx x3 25x 13 52 13 52 32 54 1015
15
1
1
3
5
2
35. We want the area between the line y 1, 0 x 2, and the curve y x4 , minus the area of a triangle
(formed by y x and y 1) with base 1 and height 1. Thus, A
2
0
1 dx (1)(1) x
x2
4
2
x3
12 0
1
2
1
2
8 1 2 2 1 5
2 12
2
3 2 6
36. We want the area between the x-axis and the curve y x 2 , 0 x 1 plus the area of a triangle (formed by
1
1
x 1, x y 2, and the x-axis) with base 1 and height 1. Thus, A x 2 dx 12 (1)(1) x3 12 13 12 65
0
0
3
37. AREA A1 A2
A1: For the sketch given, a 3 and we find b by solving the equations y x 2 4 and y x 2 2 x
simultaneously for x: x 2 4 x 2 2 x 2 x 2 2 x 4 0 2( x 2)( x 1) x 2 or x 1 so
b 2: f ( x) g ( x) ( x 2 4) ( x 2 2 x) 2 x 2 2 x 4 A1
2
2
3
(2 x 2 2 x 4) dx
3
2
23x 22x 4 x 16
4 8 (18 9 12) 9 16
11
;
3
3
3
3
2
A2: For the sketch given, a 2 and b 1: f ( x) g ( x) ( x 2 x) ( x 2 4) 2 x 2 2 x 4
1
1
3
(2 x 2 2 x 4) dx 23x x 2 4 x 32 1 4 16
48
3
2
2
23 1 4 16
4 8 9;
3
A2
Therefore, AREA A1 A2 11
9 38
3
3
38. AREA A1 A2
A1: For the sketch given, a 2 and b 0: f ( x) g ( x) (2 x3 x 2 5 x) ( x 2 3x ) 2 x3 8 x
A1
0
0
(2 x3 8 x) dx 24x 8 2x 0 (8 16) 8;
2
2
4
2
A2: For the sketch given, a 0 and b 2: f ( x) g ( x) ( x 2 3 x) (2 x3 x 2 5 x) 8 x 2 x3
2
2
A2 (8 x 2 x3 ) dx 8 2x 24x (16 8) 8;
0
0
2
4
Therefore, AREA A1 A2 16
39. AREA A1 A2 A3
A1: For the sketch given, a 2 and b 1: f ( x) g ( x) ( x 2) (4 x 2 ) x 2 x 2
A1
1
1
3
2
( x 2 x 2) dx x3 x2 2 x 13 12 2 83 42 4 73 12 1463 11
;
6
2
2
Copyright 2016 Pearson Education, Ltd.
326
Chapter 5 Integrals
A2: For the sketch given, a 1 and b 2: f ( x) g ( x ) (4 x 2 ) ( x 2) ( x 2 x 2)
2
2
A2 ( x 2 x 2) dx x3 x2 2 x 83 24 4 13 12 2 3 8 12 92 ;
1
1
3
2
A3: For the sketch given, a 2 and b 3: f ( x) g ( x) ( x 2) (4 x 2 ) x 2 x 2
3
3
2
3
A3 ( x 2 x 2) dx x3 x2 2 x
2
2
273 92 6 83 24 4 9 92 83 ;
Therefore, AREA A1 A2 A3 11
92 9 92 83 9 56 49
6
6
40. AREA A1 A2 A3
A1: For the sketch given, a 2 and b 0: f ( x) g ( x)
A1 13
0
x x ( x 4 x)
x3
3
x3
3
x
3
4
3
1
3
3
0
( x3 4 x) dx 13 x4 2 x 2 0 13 (4 8) 43 ;
2
2
4
3
A2: For the sketch given, a 0 and we find b by solving the equations y x3 x and y 3x simultaneously
3
3
for x: x3 x 3x x3 34 x 0 3x ( x 2)( x 2) 0 x 2, x 0, or x 2 so b 2: f ( x) g ( x)
2
3
4
2
2
3x x3 x 13 ( x3 4 x) A2 13 ( x3 4 x) dx 13 (4 x x3 ) 13 2 x 2 x4 13 (8 4) 43 ;
0
0
0
A3: For the sketch given, a 2 and b 3: f ( x) g ( x)
3
x ( x 4 x)
x3
3
x
3
1
3
3
3
25 ;
A3 13 ( x3 4 x) dx 13 x4 2 x 2 13 81
2 9 16
8 13 81
14 12
4
4
4
2
2
4
25 32 25 19
Therefore, AREA A1 A2 A3 43 43 12
12
4
41. a 2, b 2;
f ( x ) g ( x) 2 ( x 2 2) 4 x 2
A
2
2
3
(4 x 2 ) dx 4 x x3 8 83 8 83
2
2
2 24
83 32
3
3
42. a 1, b 3;
f ( x) g ( x) (2 x x 2 ) (3) 2 x x 2 3
3
3
3
A (2 x x 2 3)dx x 2 x3 3x
1
1
9 27
9 1 13 3 11 13 32
3
3
43. a 0, b 2;
2
f ( x) g ( x) 8 x x 4 A (8 x x 4 )dx
0
2
8 2x x5 16 32
80532 48
5
5
0
2
5
Copyright 2016 Pearson Education, Ltd.
Section 5.6 Definite Integral Substitutions and the Area Between Curves
327
44. Limits of integration: x 2 2 x x x 2 3x 0
x( x 3) 0 a 0 and b 3;
f ( x) g ( x) x ( x 2 2 x) 3x x 2
3
3
A (3x x 2 )dx 32x x3 27
9 27218 92
2
0
0
2
3
45. Limits of integration: x 2 x 2 4 x 2 x 2 4 x 0
2 x( x 2) 0 a 0 and b 2;
f ( x ) g ( x ) ( x 2 4 x ) x 2 2 x 2 4 x
2
2
A (2 x 2 4 x)dx 23x 42x
0
0
16
16
32
48
8
3 2 6 3
3
2
46. Limits of integration: 7 2 x 2 x 2 4 3 x 2 3 0
3( x 1)( x 1) 0 a 1 and b 1;
f ( x) g ( x) (7 2 x 2 ) ( x 2 4) 3 3 x 2
1
1
A (3 3x 2 )dx 3 x x3 3 1 13 1 13
1
1
3
6 23 4
47. Limits of integration: x 4 4 x 2 4 x 2 x 4 5 x 2 4 0
( x 2 4)( x 2 1) 0 ( x 2)( x 2)( x 1)( x 1) 0
x 2, 1,1, 2; f ( x) g ( x ) ( x 4 4 x 2 4) x 2
x 4 5 x 2 4 and
g (x) f ( x) x 2 ( x 4 4 x 2 4) x 4 5 x 2 4
A
1
2
x4 5x2 4 dx 11( x4 5x2 4) dx
2
( x 4 5 x 2 4) dx
1
1
1
2
5
3
5
3
5
3
x5 53x 4 x x5 53x 4 x 5x 53x 4 x
2
1
1
180 8
15 53 4 32
40
8 15 53 4 15 53 4 32
40
8 15 53 4 60
60
30015
5
3
5
3
5
3
48. Limits of integration: x a 2 x 2 0 x 0 or
a 2 x 2 0 x 0 or a 2 x 2 0 x a, 0, a;
A
0
a
a
x a 2 x 2 dx x a 2 x 2 dx
0
0
a
12 32 (a 2 x 2 )3/2 12 32 (a 2 x 2 )3/2
a
0
2 3/2 1 2 3/2 2 a3
1
3 (a ) 3 (a )
3
Copyright 2016 Pearson Education, Ltd.
328
Chapter 5 Integrals
49. Limits of integration: y
x
x, x 0
and
x, x 0
5 y x 6 or y 5x 56 ; for x 0; x 5x 56
5 x x 6 25( x) x 2 12 x 36
x 2 37 x 36 0 ( x 1)( x 36) 0
x 1, 36 (but x 36 is not a solution);
for x 0 : 5 x x 6 25 x x 2 12 x 36
x 2 13 x 36 0 ( x 4)( x 9) 0
x 4,9; there are three intersection points and
A
0
1
x5 6 x dx 04 x5 6 x dx 49 x x5 6 dx
0
4
9
3/2
( x 6)2
( x 6)2
( x 6)2
10 32 x 10 32 x3/2 23 x3/2 10
1
0
4
36 25 2 100 2 43/2 36 0 2 93/2 225 2 43/2 100 50 20 5
10
10 3
10
3
10
3
10
3
10
10
3
3
50. Limits of integration: y | x 2 4|
x 2 4, x 2 or x 2
4 x2 , 2 x 2
2
for x 2 and x 2 : x 2 4 x2 4
2 x 2 8 x 2 8 x 2 16 x 4; for 2 x 2 :
2
4 x 2 x2 4 8 2 x 2 x 2 8 x 2 0 x 0; by
symmetry of the graph,
2
2
A 2 x2 4 (4 x 2 ) dx
0
2
4
2
3
3
4
2 x2 4 x 2 4 dx 2 x2 2 8 x x6
0
2
2
2 82 0 2 32 64
16 86 40 56
64
6
3
3
51. Limits of integration: c 0 and d 3;
f ( y) g ( y) 2 y 2 0 2 y 2
3
3
2 y3
A 2 y 2 dy 3 2 9 18
0
0
52. Limits of integration: y 2 y 2 ( y 1)( y 2) 0
c 1 and d 2; f ( y ) g ( y ) ( y 2) y 2
2
2
y3
y2
A ( y 2 y 2 ) dy 2 2 y 3
1
1
42 4 83 12 2 13 6 83 12 2 13 92
Copyright 2016 Pearson Education, Ltd.
Section 5.6 Definite Integral Substitutions and the Area Between Curves
53. Limits of integration: 4 x y 2 4 and
4 x 16 y y 2 4 16 y y 2 y 20 0
( y 5)( y 4) 0 c 4 and d 5;
16 y
y 2 4 y 2 y 20
f ( y) g ( y) 4 4
4
5
5
y3 y 2
A 14 ( y 2 y 20) dy 14 3 2 20 y
4
4
125
25
64
16
1
1
4 3 2 100 4 3 2 80
14
189
9
3 2 180 2438
54. Limits of integration: x y 2 and x 3 2 y 2
y 2 3 2 y 2 3 y 2 3 0 3( y 1)( y 1) 0
c 1 and d 1; f ( y ) g ( y ) (3 2 y 2 ) y 2
3 3 y 2 3(1 y 2 ) A 3
1
1
1
(1 y 2 ) dy
y3
3 y 3 3 1 13 3 1 13 3 2 1 13 4
1
55. Limits of integration: x y 2 and x 2 3 y 2
y
y2 2 3 y2 2 y2 2 0
2( y 1)( y 1) 0 c 1 and d 1;
f ( y ) g ( y ) (2 3 y 2 ) ( y 2 ) 2 2 y 2 2(1 y 2 )
1
x y 0
1
1
y3
A 2 (1 y 2 ) dy 2 y 3
1
1
2 1 13 2 1 13 4 23 83
56. Limits of integration: x y 2/3 and
x 2 y 4 y 2/3 2 y 4 c 1 and d 1;
1
f ( y ) g ( y ) (2 y 4 ) y 2/3 A (2 y 4 y 2/3 ) dy
1
1
0
1
y5
2 y 5 53 y 5/3 2 15 53 2 15 53
1
2 2 15 53 12
5
x 3y2 2
1
2
57. Limits of integration: x y 2 1 and x | y | 1 y 2
y 2 1 | y | 1 y 2 y 4 2 y 2 1 y 2 (1 y 2 )
y4 2 y2 1 y2 y4 2 y4 3 y2 1 0
(2 y 2 1)( y 2 1) 0 2 y 2 1 0 or y 2 1 0 y 2 12
or y 2 1 y 22 or y 1.
Substitution shows that 2 2 are not solutions y 1;
for 1 y 0, f ( x) g ( x) y 1 y 2 ( y 2 1)
1 y 2 y (1 y 2 )1/2 , and by symmetry of the graph,
Copyright 2016 Pearson Education, Ltd.
1
2
x
329
330
Chapter 5 Integrals
0
0
0
A 2 1 y 2 y (1 y 2 )1/2 dy 2 (1 y 2 ) dy 2 y (1 y 2 )1/2 dy
1
1
1
0
0
y3
2(1 y 2 )3/ 2
1
2
2 y 3 2 12
2 (0 0) 1 3 3 0 2
3
1
1
58. AREA A1 A2
Limits of integration: x 2 y and
x y3 y 2 y3 y 2 2 y 0
y ( y 2 y 2) y ( y 1)( y 2) 0 y 1, 0, 2:
for 1 y 0, f ( y ) g ( y ) y 3 y 2 2 y
0
0
y 4 y3
A1 ( y3 y 2 2 y ) dy 4 3 y 2
1
1
5 ;
0 14 13 1 12
for 0 y 2, f ( y ) g ( y ) 2 y y 3 y 2
2
2
y4
y3
A2 (2 y y 3 y 2 ) dy y 2 4 3
0
0
5 8 37
4 16
83 0 83 ; Therefore, A1 A2 12
4
3 12
59. Limits of integration: y 4 x 2 4 and y x 4 1
x 4 1 4 x 2 4 x 4 4 x 2 5 0
( x 2 5)( x 1)( x 1) 0 a 1 and b 1;
f ( x ) g ( x ) 4 x 2 4 x 4 1 4 x 2 x 4 5
A
1
1
3
5
(4 x 2 x 4 5) dx 43x x5 5 x
1
1
43 15 5 43 15 5 2 43 15 5 104
15
60. Limits of integration: y x3 and y 3 x 2 4
x3 3 x 2 4 0 ( x 2 x 2)( x 2) 0
( x 1)( x 2) 2 0 a 1 and b 2;
f ( x) g ( x) x3 (3 x 2 4) x3 3 x 2 4
A
2
2
4
3
( x3 3x 2 4) dx x4 33x 4 x
1
1
16
24
8 14 1 4 27
4
3
4
Copyright 2016 Pearson Education, Ltd.
Section 5.6 Definite Integral Substitutions and the Area Between Curves
61. Limits of integration: x 4 4 y 2 and x 1 y 4
4 4 y2 1 y4 y4 4 y2 3 0
y 3 y 3 ( y 1)( y 1) 0 c 1 and d 1
since x 0; f ( y ) g ( y ) (4 4 y 2 ) (1 y 4 )
1
3 4 y 2 y 4 A (3 4 y 2 y 4 ) dy
1
1
4y
y
56
3 y 3 5 2 3 34 15 15
1
3
5
y2
62. Limits of integration: x 3 y 2 and x 4
y2
3 y2
3 0 34 ( y 2)( y 2) 0
4
y2
c 2 and d 2; f ( y ) g ( y ) (3 y 2 ) 4
2
2
3 2
2
y
y
y
3 1 4 A 3 1 4 dy 3 y 12
2
2
8
8
16
3 2 12 2 12 3 4 12 12 4 8
3 y2 4
63. a 0, b ; f ( x) g ( x) 2sin x sin 2 x
0
0
A (2sin x sin 2 x )dx 2 cos x cos22 x
2(1) 12 2 1 12 4
64. a 3 , b 3 ; f ( x) g ( x) 8cos x sec2 x
A
/3
/3
(8cos x sec 2 x)dx [8 sin x tan x]/3/3
8 23 3 8 23 3 6 3
65. a 1, b 1; f ( x ) g ( x) (1 x 2 ) cos 2x
1
x x3 2 sin 2x 1 13 2 1 13 2
1
2 23 2 43 4
1
A 1 x 2 cos 2x dx
1
3
Copyright 2016 Pearson Education, Ltd.
331
332
Chapter 5 Integrals
66. A A1 A2
a1 1, b1 0 and a2 0, b2 1;
f1 ( x) g1 ( x) x sin 2x and f 2 ( x) g 2 ( x)
sin 2x x by symmetry about the origin,
1
A1 A2 2 A1 A 2 sin 2x x dx
0
1
2
2 2 cos 2x x2 2 2 0 12 2 1 0
0
4
4
2
1
2 2 2 2
67. a 4 , b 4 ; f ( x) g ( x) sec2 x tan 2 x
A
/4
/4
/4
/4
/4
/4
(sec2 x tan 2 x) dx
[sec 2 x (sec2 x 1)] dx
1 dx [ x]/4/4 4 4 2
68. c 4 , d 4 ; f ( y ) g ( y )
tan 2 y ( tan 2 y ) 2 tan 2 y 2(sec2 y 1)
A
/4
/4
2 sec 2 y 1 dy 2[(tan y y )]/4/4
2 1 4 1 4 4 1 4 4
69. c 0, d 2 ; f ( y ) g ( y )
3 sin y cos y 0 3sin y cos y
/2
/2
sin y cos y dy 3 23 (cos y )3/2
0
0
2(0 1) 2
A 3
Copyright 2016 Pearson Education, Ltd.
Section 5.6 Definite Integral Substitutions and the Area Between Curves
70. a 1, b 1; f ( x ) g ( x) sec2 3x x1/3
1
A sec 2 3x x1/3 dx
1
1
3 tan 3x 43 x 4/3
1
3
3
3
3 4 3 34 6 3
71. A A1 A2
Limits of integration: x y 3 and x y y y3
y 3 y 0 y ( y 1)( y 1) 0 c1 1, d1 0
and c2 0, d 2 1; f1 ( y ) g1 ( y ) y 3 y and
f 2 ( y ) g 2 ( y ) y y 3 by symmetry
about the origin, A1 A2 2 A2 A
1
1
y2 y4
2 ( y y 3 ) dy 2 2 4 2 12 14 12
0
0
72. A A1 A2
Limits of integration: y x3 and y x5 x3 x5
x5 x3 0 x3 ( x 1)( x 1) 0 a1 1, b1 0
and a2 0, b2 1; f1 ( x) g1 ( x ) x3 x5 and
f 2 ( x) g 2 ( x) x5 x3 by symmetry about the
1
origin, A1 A2 2 A2 A 2 ( x3 x5 ) dx
0
1
2 x4 x6 2 14 16 16
0
4
6
73. A A1 A2
Limits of integration: y x and y 12 x 12 , x
x
3
x 1 x 1, f1 ( x) g1 ( x) x 0 x
1
x
0
1
A1 x dx x2 12 ; f 2 ( x) g 2 ( x) 12 0
0
x
0
2
2
2
1
1
x 2 A2 x 2 dx x1 12 1 12 ;
A A1 A2 12 12 1
Copyright 2016 Pearson Education, Ltd.
333
334
Chapter 5 Integrals
74. Limits of integration: sin x cos x x 4 a 0 and
b 4 ; f ( x) g ( x) cos x sin x
/4
A
0
2
2
2
2
(cos x sin x) dx [sin x cos x ]0 /4
(0 1) 2 1
75. (a) The coordinates of the points of intersection of the
line and parabola are c x 2 x c and y c
(b) f ( y ) g ( y ) y y 2 y the area of
c
the lower section is, AL [ f ( y ) g ( y )] dy
0
2
c
0
c
y dy 2 23 y 3/2 43 c3/2 . The area of
0
the entire shaded region can be found by setting c 4 : A
43 43/2 438 323 . Since we want c to divide
the region into subsections of equal area we have A 2 AL 32
2 34 c3/2 c 42/3
3
c
c
c
3/ 2
3
(c x 2 ) dx cx x3
2 c3/2 c 3
c
c
c
3/2
32
4
3 c . Again, the area of the whole shaded region can be found by setting c 4 A 3 . From the
(c) f ( x) g ( x) c x 2 AL
[ f ( x) g ( x)] dx
condition A 2 AL , we get 43 c3/2 32
c 42/3 as in part (b).
3
76. (a) Limits of integration: y 3 x 2 and y 1
3 x 2 1 x 2 4 a 2 and b 2;
f ( x) g ( x) (3 x 2 ) (1) 4 x 2
A
2
2
3
(4 x 2 ) dx 4 x x3
2
2
8 83 8 83 16 16
32
3
3
(b) Limits of integration: let x 0 in y 3 x 2
y 3; f ( y ) g ( y ) 3 y 3 y
3
3
3
2(3 y )3/ 2
2(3 y )1/2 A 2 (3 y )1/2 dy 2 (3 y )1/2 (1) dy (2)
3
1
1
1
3/2
32
4
4
3 0 (3 1)
3 (8) 3
Copyright 2016 Pearson Education, Ltd.
Section 5.6 Definite Integral Substitutions and the Area Between Curves
335
77. Limits of integration: y 1 x and y 2
1 x 2 , x
x
x
0 x x 2 x (2 x) 2
x 4 4 x x2 x2 5x 4 0
( x 4)( x 1) 0 x 1, 4 (but x 4 does not
satisfy the equation); y 2 and y 4x 2 4x
x
x
8 x x 64 x3 x 4. Therefore,
AREA A1 A2 : f1 ( x) g1 ( x) 1 x1/2 x4
4
4
A2 2 x 1/2 4x dx 4 x1/2 x8 4 2 16
4 81 4 15
17
; Therefore,
8
8
8
1
1
1
2
1
; f 2 ( x) g 2 ( x) 2 x 1/2 4x
A1 1 x1/2 4x dx x 23 x3/2 x8 1 23 18 0 37
24
0
0
2
51 88 11
AREA A1 A2 37
17
3724
24
8
24
3
78. Limits of integration: ( y 1) 2 3 y
y2 2 y 1 3 y y2 y 2 0
( y 2)( y 1) 0 y 2 since y 0; also,
2 y 3 y 4 y 9 6 y y 2 y 2 10 y 9 0
( y 9)( y 1) 0 y 1 since y 9 does not satisfy
the equation;
AREA A1 A2
1
1
2 y 3/ 2
f1 ( y ) g1 ( y ) 2 y 0 2 y1/2 A1 2 y1/2 dy 2 3 34 ;
0
0
2
2
f 2 ( y ) g 2 ( y ) (3 y ) ( y 1)2 A2 [3 y ( y 1)2 ] dy 3 y 12 y 2 13 ( y 1)3
1
1
6 2 13 3 12 0 1 13 12 76 . Therefore, A1 A2 43 76 15
25
6
a
a
3
3
79. Area between parabola and y a 2 : A 2 (a 2 x 2 ) dx 2 a 2 x 13 x3 2 a3 a3 0 43a ;
0
0
Area of triangle AOC: 12 (2a)(a 2 ) a3 ; limit of ratio lim
a3
34 which is independent of a.
3
a0 4 a
3
80.
b
b
b
b
b
a
a
a
a
a
A 2 f ( x) dx f ( x) dx 2 f ( x) dx f ( x) dx f ( x) dx 4
81. Neither one; they are both zero. Neither integral takes into account the changes in the formulas for the region’s
upper and lower bounding curves at x 0. The area of the shaded region is actually
A
0
1
1
0
1
0
1
0
x ( x ) dx x ( x ) dx 2 x dx 2 x dx 2.
82. It is sometimes true. It is true if f ( x) g ( x) for all x between a and b. Otherwise it is false. If the graph of f lies
below the graph of g for a portion of the interval of integration, the integral over that portion will be negative
and the integral over [a, b] will be less than the area between the curves (see Exercise 71).
Copyright 2016 Pearson Education, Ltd.
336
Chapter 5 Integrals
83. Let u 2 x du 2 dx 12 du dx; x 1 u 2, x 3 u 6
1 sinx2 x dx 2 sin u2 u 12 du 2 sinu u du [ F (u )]2 F (6) F (2)
6
3
6
6
84. Let u 1 x du dx du dx; x 0 u 1, x 1 u 0
1
0
0
1
1
0 f (1 x) dx 1 f (u )(du) 1 f (u) du 0 f (u) du 0 f ( x) dx
85. (a) Let u x du dx; x 1 u 1, x 0 u 0
f odd f ( x) f ( x). Then
0
1
0
0
0
1
1
1
f ( x)dx f (u )(du ) f (u ) ( du ) f (u ) du
1
f (u ) du 3
0
(b) Let u x du dx; x 1 u 1, x 0 u 0
f even f ( x) f ( x). Then
86. (a) Consider
0
a
0
1
0
0
1
1
1
0
f ( x)dx f (u )(du ) f (u )du f (u ) du 3
f ( x) dx when f is odd. Let u x du dx du dx and x a u a and
x 0 u 0. Thus
a
0
a
0
0
a
a
a
a
0
0
f ( x) dx f (u ) du f (u ) du f (u ) du f ( x) dx. Thus
a
0
a
a
a f ( x) dx a f ( x) dx 0 f ( x) dx 0 f ( x) dx 0 f ( x) dx 0.
(b)
/2 sin x dx [ cos x] /2 cos 2 cos 2 0 0 0.
/2
/2
87. Let u a x du dx; x 0 u a, x a u 0
I
a
f ( x ) dx
0 f ( x) f (a x)
0
a
f ( a u )
a
a f ( a x ) dx
f ( x ) dx
f ( x) f (a x)
0 f ( x) f (a x)
0
Therefore, 2 I a I a2 .
I I
xy
a
f ( a u ) du
f ( a x ) dx
a f (a u ) f (u ) (du ) 0 f (u ) f (a u ) 0 f ( x) f (a x)
a f ( x) f (a x)
a
dx dx [ x]0a a 0 a.
0 f ( x) f (a x)
0
xy
t du 1 dt 1 du 1 dt ; t x u y, t xy u 1. Therefore,
88. Let u t du 2 dt xy
t
u
t
t
xy
1
y
1
y
x 1t dt y u1 du y u1 du 1 u1 du 1 1t dt
89. Let u x c du dx; x a c u a, x b c u b
b c
b
b
a c f ( x c) dx a f (u ) du a f ( x) dx
Copyright 2016 Pearson Education, Ltd.
Section 5.6 Definite Integral Substitutions and the Area Between Curves
90. (a)
91-94.
(b)
(c)
Example CAS commands:
Maple:
f : x - x^3/3-x^2/2-2*x 1/3;
g : x - x-1;
plot( [f(x),g(x)], x -5..5, legend ["y f(x)","y g(x)"], title "#91(a) (Section 5.6)" );
# (b)
q1: [ -5, -2, 1, 4 ];
q2 : [seq( fsolve( f(x) g(x), x q1[i]..q1[i 1] ), i 1..nops(q1)-1 )];
for i from 1 to nops(q2)-1 do
# (c)
area[i] : int( abs(f(x)-g(x)),x q2[i]..q2[i 1] );
end do;
add( area[i], i 1..nops(q2)-1 );
# (d)
Mathematica: (assigned functions may vary)
Clear[x, f, g]
f[x_] x 2 Cos[x]
g[x_] x 3 x
Plot[{f[x], g[x]}, {x, 2, 2}]
After examining the plots, the initial guesses for FindRoot can be determined.
pts x/.Map[FindRoot[f[x] g[x],{x, #}]&, { 1, 0, 1}]
i1 NIntegrate[f[x] g[x], {x, pts[[1]], pts[[2]]}]
i2 NIntegrate [f[x] g[x], {x, pts[[2]], pts[[3]]}]
i1 i2
Copyright 2016 Pearson Education, Ltd.
337
338
Chapter 5 Integrals
CHAPTER 5
PRACTICE EXERCISES
1. (a) Each time subinterval is of length t 0.4 s. The distance traveled over each subinterval, using the
midpoint rule, is h 12 (vi vi 1 )t , where vi is the velocity at the left endpoint and vi 1 the velocity at
the right endpoint of the subinterval. We then add h to the height attained so far at the left endpoint vi to
arrive at the height associated with velocity vi 1 at the right endpoint. Using this methodology we build the
following table based on the figure in the text:
t (s)
0 0.4
v (m/s) 0 10
h (m) 0 2
0.8
25
9
1.2
55
25
1.6
100
56
2.0
190
114
2.4
180
188
t (s)
v (m/s)
h (m)
6.8
37
660.6
7.2
25
672
7.6
12
679.4
8.0
0
681.8
6.4
50
643.2
2.8
165
257
3.2
150
320
3.6
140
378
4.0
130
432
4.4
115
481
4.8
105
525
5.2
90
564
5.6
76
592
6.0
65
620.2
NOTE: Your table values may vary slightly from ours depending on the v-values you read from the graph.
Remember that some shifting of the graph occurs in the printing process.
The total height attained is about 680 m.
(b) The graph is based on the table in part (a).
2. (a) Each time subinterval is of length t 1 s. The distance traveled over each subinterval, using the
midpoint rule, is s 12 (vi vi 1 ) t , where vi is the velocity at the left, and vi 1 the velocity at the right,
endpoint of the subinterval. We then add s to the distance attained so far at the left endpoint vi to arrive at
the distance associated with velocity vi 1 at the right endpoint. Using this methodology we build the table
given below based on the figure in the text, obtaining approximately 26 m for the total distance traveled:
t (s)
v (m/s)
s (m)
0
0
0
1
0.5
0.25
2
1.2
1.1
3
2
2.7
4
3.4
5.4
5
4.5
9.35
6
4.8
14
7
4.5
18.65
8
3.5
22.65
(b) The graph shows the distance traveled by the
moving body as a function of time for 0 t 10.
Copyright 2016 Pearson Education, Ltd.
9
2
25.4
10
0
26.4
Chapter 5 Practice Exercises
10
3. (a)
10
a
4k 14 ak 14 (2) 12
k 1
10
(b)
k 1
k 1
10
(c)
(d)
4. (a)
(b)
k 1
10
k 1
10
10
k 1
k 1
k 1
10
10
k 1
k 1
k 1
10
52 bk 52 bk 52 (10) 25 0
20
20
k 1
20
k 1
k 1
3ak 3 ak 3(0) 0
20
20
(ak bk ) ak bk 0 7 7
k 1
20
(d)
10
(ak bk 1) ak bk 1 2 25 (1)(10) 13
k 1
20
(c)
10
(bk 3ak ) bk 3 ak 25 3(2) 31
1 2bk
2
7
k 1
k 1
20
20
12 72 bk 12 (20) 72 (7) 8
k 1
k 1
20
20
(ak 2) ak 2 0 2(20) 40
k 1
k 1
k 1
5. Let u 2 x 1 du 2 dx 12 du dx; x 1 u 1, x 5 u 9
5
1 (2 x 1)
1/2
9
9
dx u 1/2 12 du u1/2 3 1 2
1
1
6. Let u x 2 1 du 2 x dx 12 du x dx; x 1 u 0, x 3 u 8
3
2
1/3
1 x( x 1)
8
8
dx u1/3 12 du 83 u 4/3 83 (16 0) 6
0
0
7. Let u 2x 2 du dx; x u 2 , x 0 u 0
cos 2x dx /2 (cos u )(2 du ) [2sin u ] /2 2 sin 0 2sin 2 2(0 (1)) 2
0
0
0
8. Let u sin x du cos x dx; x 0 u 0, x 2 u 1
/2
1
1
0
(sin x)(cos x) dx u du u2 12
0
0
9. (a)
2 f ( x) dx 13 2 3 f ( x) dx 13 (12) 4
2
2
2
5
5
2
(b)
2 f ( x) dx 2 f ( x) dx 2 f ( x) dx 6 4 2
(c)
5 g ( x) dx 2 g ( x) dx 2
(d)
2 ( g ( x)) dx 2 g ( x) dx (2) 2
(e)
2
5
5
2
5
5
f ( x) g ( x)
5
dx f ( x) dx g ( x) dx (6) (2)
1 5
5 2
1 5
5 2
1
5
1
5
8
5
Copyright 2016 Pearson Education, Ltd.
339
340
Chapter 5 Integrals
10. (a)
0 g ( x) dx 17 0 7 g ( x) dx 17 (7) 1
2
2
2
2
1
(b)
1 g ( x) dx 0 g ( x) dx 0 g ( x) dx 1 2 1
(c)
2 f ( x) dx 0 f ( x) dx
(d)
0 2 f ( x) dx 2 0 f ( x) dx 2( ) 2
(e)
0 [ g ( x) 3 f ( x)] dx 0 g ( x) dx 3 0 f ( x) dx 1 3
0
2
2
2
2
2
2
11. x 2 4 x 3 0 ( x 3)( x 1) 0 x 3 or x 1;
1
3
0
1
Area ( x 2 4 x 3) dx ( x 2 4 x 3) dx
1
3
3
3
x3 2 x 2 3x x3 2 x 2 3x
0
1
3
2
1
3 2(1) 3(1) 0
3
3
33 2(32 ) 3(3) 13 2(1) 2 3(1)
13 1 0 13 1 83
2
12. 1 x4 0 4 x 2 0 x 2;
Area
2
2
3 2
1 dx 1 dx
3
x2
4
x2
4
2
3
x x x3
x 12
2 12 2
3
3
2 2 ( 2) 3 33 2 23
2 12
12
12
12
3
13
43 43 4 43 4
13. 5 5 x 2/3 0 1 x 2/3 0 x 1;
1
8
Area (5 5 x 2/3 ) dx (5 5 x 2/3 ) dx
1
1
1
8
5/3
5/3
5 x 3 x 5 x 3 x
1
1
5/3
5/3
5(1) 3(1)
5(1) 3(1)
5/3
5/3
5(8) 3(8)
5(1) 3(1)
[2 (2)] [(40 96) 2] 62
Copyright 2016 Pearson Education, Ltd.
Chapter 5 Practice Exercises
14. 1 x 0 x 1;
1
4
0
1
Area (1 x ) dx (1 x ) dx
1
4
x 23 x3/2 x 23 x3/2
0
1
3/2
2
2
1 3 (1)
0 4 3 (4)3/2 1 23 (1)3/2
16
1
1
3 4 3 3 2
15.
f ( x) x, g ( x) 12 , a 1, b 2
x
b
A [ f ( x) g ( x)] dx
a
2
1
16.
x dx 1 1
x2
2
1
x2
2
1
x 1
4
2
1
2
1
2
b
f ( x) x, g ( x) 1 , a 1, b 2 [ f ( x) g ( x)] dx
a
x
A
2
1
4 2
2
x dx 2 x
1
x
x2
2
2
1
2 12 2 7 42 2
b
1
1
a
0
0
17. f ( x) (1 x )2 , g ( x) 0, a 0, b 1 A [ f ( x) g ( x)] dx (1 x ) 2 dx (1 2 x x) dx
1
1
(1 2 x1/2 x) dx x 43 x3/2 x2 1 43 12 16 (6 8 3) 16
0
0
2
b
1
1
a
0
0
18. f ( x) (1 x3 )2 , g ( x) 0, a 0, b 1 A [ f ( x) g ( x)] dx (1 x3 ) 2 dx (1 2 x3 x 6 ) dx
1
4
7
9
x x2 x7 1 12 71 14
0
19. f ( y ) 2 y 2 , g ( y ) 0, c 0, d 3
d
3
c
0
A [ f ( y ) g ( y )] dy (2 y 2 0) dy
3
2 y 2 dy 23 [ y 3 ]30 18
0
Copyright 2016 Pearson Education, Ltd.
341
342
Chapter 5 Integrals
20. f ( y ) 4 y 2 , g ( y ) 0, c 2, d 2
d
2
c
2
A [ f ( y ) g ( y )] dy
2
(4 y 2 ) dy
y3
4 y 3 2 8 83 32
3
2
y2
y 2
21. Let us find the intersection points: 4 4
y 2 y 2 0 ( y 2)( y 1) 0 y 1 or
y2
y2
y 2 c 1, d 2; f ( y ) 4 , g ( y ) 4
d
2 y2 y2
A [ f ( y ) g ( y )] dy 4 4 dy
c
1
2
2
y3
y2
14 ( y 2 y 2 ) dy 14 2 2 y 3
1
1
14 24 4 83 12 2 13 89
y 2 4
y 16
22. Let us find the intersection points: 4 4
y 2 y 20 0 ( y 5)( y 4) 0 y 4 or
y 2 4
y 16
y 5 c 4, d 5; f ( y ) 4 , g ( y ) 4
d
5 y 16 y 2 4
A [ f ( y ) g ( y )] dy 4 4 dy
c
4
2
3 5
5
y
y
14 ( y 20 y 2 ) dy 14 2 20 y 3
4
4
25
125
16
64
1
4 2 100 3 2 80 3
14 92 180 63 14 92 117 81 (9 234) 243
8
23. f ( x) x, g ( x) sin x, a 0, b 4
b
/4
a
0
A [ f ( x) g ( x)] dx
/4
x2 cos x
0
2
2
( x sin x) dx
32 22 1
24. f ( x) 1, g ( x) |sin x |, a 2 , b 2
b
/2
a
/2
A [ f ( x) g ( x)] dx
0
/2
2
(1 sin x) dx
/2
0
/2
0
(1 |sin x |) dx
(1 sin x) dx
(1 sin x) dx 2[ x cos x]0 /2
2 2 1 2
Copyright 2016 Pearson Education, Ltd.
Chapter 5 Practice Exercises
25. a 0, b , f ( x) g ( x) 2sin x sin 2 x
0
0
A (2sin x sin 2 x) dx 2 cos x cos22 x
2 (1) 12 2 1 12 4
26. a 3 , b 3 , f ( x) g ( x) 8cos x sec2 x
A
/3
/3
(8cos x sec2 x) dx [8sin x tan x]/3/3
8 23 3 8 23 3 6 3
27. f ( y )
y , g ( y ) 2 y, c 1, d 2
d
2
c
1
A [ f ( y ) g ( y )] dy [ y (2 y )] dy
2
y 2 y dy 23 y3/2 2 y y2 1
1
43 2 4 2 23 2 12 34 2 76 8 26 7
2
2
28. f ( y ) 6 y, g ( y ) y 2 , c 1, d 2
d
2
c
1
A [ f ( y ) g ( y )] dy (6 y y 2 ) dy
2
y2
y3
6 y 2 3 12 2 83 6 12 13
1
3 13
4 73 12 2414
6
6
29. f ( x) x3 3 x 2 x 2 ( x 3) f ( x) 3 x 2 6 x 3 x( x 2) f | | f (0) 0 is a
0
3
2
3
maximum and f (2) 4 is a minimum. A ( x3 3x 2 ) dx x4 x3 81
27 27
4
4
0
0
30.
4
a
2
2
a
a
A (a1/2 x1/2 )2 dx (a 2 ax1/2 x) dx ax 43 ax3/2 x2 a 2 43 a a a a2
0
0
0
2
2
a 2 1 43 12 a6 (6 8 3) a6
Copyright 2016 Pearson Education, Ltd.
343
344
Chapter 5 Integrals
1
31. The area above the x-axis is A1 ( y 2/3 y ) dy
0
1
y
3y
1 ; the area below the x-axis is
5 2 10
0
0
0
3 y5/3 y 2
11 the
A2 ( y 2/3 y ) dy 5 2 10
1
1
total area is A1 A2 65
5/3
32.
A
2
/4
0
(cos x sin x) dx
5 /4
3 /2
5 /4
/4
(sin x cos x ) dx
(cos x sin x) dx
[sin x cos x]0 /4 [ cos x sin x]5/4/4
[sin x cos x]35 /2
/4
22 22 (0 1) 22 22 22 22
(1 0) 22 22 8 22 2 4 2 2
x
d2y
dy
1
33.
y x 2 1t dt dx 2 x 1x 2 2 12 ; y (1) 1 1t 1 and y (1) 2 1 3
1
1
dx
x
34.
y (1 2 sec t ) dt dx 1 2 sec x
x
dy
0
d2y
dx 2
0
2 12 (sec x) 1/2 (sec x tan x) sec x (tan x);
dy
x 0 y (1 2 sec t ) dt 0 and x 0 dx 1 2 sec 0 3
0
x sin t
5
dy
dt 3 dx sinx x ; x 5 y sint t dt 3 3
t
5
35.
y
5
36.
y
1
x
dy
2 sin 2 t dt 2 so that dx 2 sin 2 x; x 1 y
1
1
2 sin 2 t dt 2 2
37. Let t sin x dt cos x dx dt cos x dx
t 32 1
3/2
3/2
3/2
8(sin
x
)
cos
x
dx
8
t
(
dt
)
8
t
dt
8
3 C 16t 1/2 C 16(sin x)1/2 C
2 1
38. Let t tan x dt sec2 x dx
5 1
2
5/2
5/2
2
(tan x) sec x dx t dt t 5
2 1
39.
C 23 t 3/2 C 23 tan 3/2 x C
2
[6 1 4sin(4 1)]d 6 d 1d 4 sin(4 1)d 6 2 4
3 2 cos(4 1) C
Copyright 2016 Pearson Education, Ltd.
cos(4 1)
C
4
Chapter 5 Practice Exercises
40. Let t 3 dt 3d 13 dt d
2
3csc2 (3 ) d
3
12 1
2 3csc 2 t 13 dt 13 (2t 1/2 3csc2 t ) dt 23 t 1 cot t C
2 1
t
43 t cot t C 43 3 cot t C
41.
dx x dx 25 x dx x3 25 x21 C x3 25 1x C x3 25x C
x 5x x 5x dx x 25
x2
42.
2
(t 2) 2 2
t
5
2 1
3
2
2
3
dt t 4t5 4 2 dt t 45t 6 dt 13 44 65 dt t 3 dt 4 t 4 dt 6 t 5 dt
t
t
t
t
t
2
3 1
4 1
2
5 1
2
3
4
t31 4 t41 6 t51 C t2 4 t3 6 t4 C 43 34 12 C
3t
3
2t
1
2t
dp
43. Let 6t 3/2 p 6 32 t 2 dt dp 9t1/2 dt dp tdt 9
t cos 6t
3/2
dt cos dp9 91 sin p C 19 sin(6t3/2 ) C
44. Let 1 csc t ( csc cot )d dt (csc cot )d dt
1 1
2
C 23 t 3/2 C 23 (1 csc )3/2 C
csc cot 1 csc d tdt t1
1
2
45.
2
2
2 (4 x 2 x 9) dx x4 2 x2 9 x 2 [((2) (2) 9(2)) ((2) (2) 9(2))]
4
3
2
4
2
4
2
[(16 4 18) (16 4 18)] [(30 (6)] 36
46.
3
3
6
3
5
2
6
3
0 (6s 9s 7) ds 6 s6 9 s3 7s 0 [((3) 3(3) 7(3)) (0)] 729 81 21 669
3 9
3
3
47.
31
2
2
2
1 x3 dx 9 x31 1 9 x2 1 92 [(3) (1) ] 92 91 1 92 98 4
48.
8
8
4 1
1
8 4/3
3
3
x
x
x
dx 4 1 3[(8)1/3 (1)1/3 ] 3 12 1 3 12
1
3 1
1 3 1
49.
9 dx
4 x x
1
9 3/2
2
x
dx x 3
4
2 1
3
9
4
1
2
x1
2
9
4
23
2[(9)1/2 (4)1/2 ] 2 13 12 2 16 13
Copyright 2016 Pearson Education, Ltd.
3
345
346
Chapter 5 Integrals
50. Let 5 x t
(5 x )1/3
dx
x
1 dx dt 1 dx 2dt
2 x
x
1 1
3
1/3
t (2dt ) 2 t1
3
1
4/3
2 t 4 32 (5 x )4/3
3
32 2 3 2 32 44/3 32 2 3 2 32 4 3 4 3 2 3 4 3 2
1/3
9 (5 x )
1
x
dx 32 (5 x )
4/3 9
72 dx
(4 x3)
51. Let 4 x 3 t 4dx dt dx dt4
3
72 dt4
(t )
3
31
t 3 dt 18 t31 9t 2 9(4 x 3)2
1
1 72 dx
1 1 9 1 40
9(4 x 3)2 9 (4(1) 3)2 (4(0) 3)2 9 (7)2 (3)2 9 49
9
49
49
0
0 (4 x 3)3
52. Let 9 7r t 7dr dt dr dt7
(97r ) t
dt /7
dr
3 2
2
3
t
2/3
2 1
1/3
3
dt 17 t 2 17 t 1 73 t1/3 C 73 (9 7 r )1/3 C
1
3
3
2
dr
73 (9 7r )1/3 73 (23)1/3 (9)1/3
2
3
0 (9 7 r )
0
2
3
1
1
53. Let t 1 x3/4 dt 34 x 4 dx 34 dt x 4 dx
1/4
3/4
x 1 x
1
1/2 x
1/4
1/3
1 1
3
dx t1/3 34 dt 34 t1
3
1
t 4/3 1 x3/4
1
3/4 4/3
4/3
C
4/3
1 x dx 1 x 1/2 0 1 12 3/4 1 12 3/4
3/4 1/3
4/3
dt x 4 dx
54. Let t 1 5 x5 dt 5 5 x 4 dx 25
5 5/2
4
x (1 5 x )
55.
56.
3 (5 9)4/3 3 (5 1)4/3 3 (2)4/3 3 (4) 4/3
2
2
1 2
2
dx
dt 1
t 5/2 25
25
t
5/2
5 1
1 t 2 2 t 3/2 2 (1 5 x5 ) 3/2 C
25
75
75
5 1
2
3/2
5 3/2
3/2
3/2 4
3/2
2 (1 5 x5 ) 3/2
2 1 5 3
2 32
x (1 5 x5 )5/2 dx 75
75
(1)
1
75
2
1247
0
0
1 cos 2(8r )
2
0 cos 8r dr 0
/4
0
sin
2
4
2t dt
2
2
dt 1 t sin 4t 2
2
4
/4
0
sin 4 4 2
sin 2
12 4
0
4
4
sin 3
12 4 4 2 14 12 4 14 14 12 4 2 8 2
r 1 sin16 0
dr 12 (1 cos16r )dr 12 r sin16
16
16
2
0 2
0
/4 1cos 2 2t 4
0
Copyright 2016 Pearson Education, Ltd.
Chapter 5 Practice Exercises
57.
58.
/2
0
cos 2 d
5 /6
/3
/2 1 cos 2
2
0
sin 2 d
d 12 sin22
/2
0
sin 2
12 2 2 2 (0 0) 12 2 4
5
5 /6
5 /6 1cos 2
1 sin 2
1 5 sin 2 6 sin 2 3
d
2
2
2
2 6
2 3
2
/3
/3
12 56 43 3 43 12 2 23 4 43
59.
2
tan 2 6x dx
2
2
tan x
sec2 6x 1 dx 1 6 x 6 tan 26 2 6 tan 6
6
6 3 6 12 3 4 3
3
60.
cot 2 3 d
3
61.
0
/3
3
cot
cosec2 3 1 d 1 3 3cot 3 3cot 3 ( )
3
2 3 2
3
3
1
3
sin x cos x dx
sin 2 x dx 1 cos 2 x 0
0
12 cos
2
2
2
2
/3
/3
0
cos 2 3
2
1 1 1 1 1 1
12 21 1/2
2 2 2
4
2 4
8
62.
5 /6
5 /6
/3 sec z tan z dz [ sec z ] /3 sec 56 sec 3 23 23 43
63. Let cos x t sin x dx dt
5 1
2
5/2
5/2
7(cos x) sin x dx 7t dt 7 t5
2
/2
0
1
/2
7(cos x)5/2 sin x dx 2(cos x)7/2
0
7
7 72 t 2 2(cos x)7/2 C
2 cos 2
7/2
2(cos 0)7/2 2
64. Let cos 4 x t sin 4 x 4 dx dt
2
3
2
3
12 cos 4 x sin 4 x dx 3t (dt ) 3 t3 cos 4 x
/3
/312 cos 4 x sin 4 x dx cos 4 x /3 cos 43 cos 4 3 12 12 0
/3
2
3
3
3
3
3
65. Let 4 5cos 2 x t 10 cos x sin x dx dt 5cos x sin x dx dt2
5cos x sin x dx 1
2
4 5cos 2 x
/2 5cos x sin x
0
1 1
dt 1 t 2 1 t1/ 2
2 1 1
2 1/2
t
2
/2
dx 4 5cos 2 x
0
4 5 cos x
2
4 5cos 2 x
4 5cos 2 2 4 5cos 2 0 4 9 (2 3) 1
Copyright 2016 Pearson Education, Ltd.
347
348
Chapter 5 Integrals
dt
66. Let 8 19 tan x t 19sec2 x dx dt sec2 x dx 19
sec 2 x
1
dx 19
(819 tan x ) 4/3
/4
0
4 1
3
4/3
t dx 191 t 4
3 1
1/3
3 (8 19 tan x) 1/3
1 t
19
19
1/3
/4
sec2 x
3 (8 19 tan x ) 1/3
dx 19
0
(819 tan x )4/3
3 8 19 tan 1/3 (8 19 tan 0)1/3
19
4
3 1 1 3 1 1 1
19
19 3 2 38
3 27 3 8
1
b(1) 12 (2b) b
k
2
2
2
k
m
(
k
)
m
(
k
)
(b) av( f ) k (1 k ) (mx b) dx 21k mx2 bx 21k 2 b(k ) 2 b(k )
k
k
21k (2bk ) b
67. (a) av( f ) 1(11)
68. (a)
(b)
69.
1
m(1)2
m( 1)2
1 (mx b) dx 12 mx2 bx 1 12 2 b(1)
2
2
3
3
3
1
3x dx 13 3 x1/2 dx 33 23 x3/2 33 32 (3)3/2 32 (0)3/2 33 (2 3) 2
0
0
30 0
a
a
a
1
yav
ax dx 1a a x1/2 dx aa 23 x3/2 aa 23 (a )3/2 23 (0)3/2 aa 23 a a 23 a
0
0
0
a0
yav
b
f av b 1 a f ( x) dx b 1 a [ f ( x)]ba b 1 a [ f (b) f (a )]
a
f (b ) f ( a )
so the average value of f
ba
over [a, b] is
the slope of the secant line joining the points (a, f (a )) and (b, f (b)), which is the average rate of change
of f over [a, b].
b
70. Yes, because the average value of f on [a, b] is b 1 a f ( x) dx. If the length of the interval is 2, then b a 2
a
b
and the average value of the function is 12 f ( x) dx.
a
71. We want to evaluate
365
1
1
f ( x) dx 365
365 0 0
2 ( x 101) 4 dx 20
2 ( x 101) dx 4
sin 365
0 20sin 365
365 0
365 0
365
365
365
dx
2 ( x 101) is 2 365 and that we are integrating this function over an
Notice that the period of y sin 365
2
365
20 365 sin 2 ( x 101) dx 4 365 dx is 20 0 4 365 4.
interval of length 365. Thus the value of 365
365
365
365 0
365
0
72.
675
1
(8.27 105 (26T 1.87T 2 )) dT
675 20 20
675
1 8.27T 26T 1.87T
655
2105
310
5 20
2
3
2
3
2
3
1 8.27(675) 26(675) 1.87(675) 8.27(20) 26(20) 1.87(20)
655
5
5
5
5
210
310
210
310
1 (3724.44 165.40) 5.43 the average value of C on [20, 675]. To find the temperature T at
655
v
which Cv 5.43, solve 5.43 8.27 105 (26T 1.87T 2 ) for T. We obtain 1.87T 2 26T 284000 0
T
26 (26) 2 4(1.87)( 284000)
2124996
26 3.74
. So T 382.82 or T 396.72. Only T 396.72 lies in the
2(1.87)
interval [20, 675], so T 396.72 C.
Copyright 2016 Pearson Education, Ltd.
Chapter 5 Practice Exercises
73.
dy
dx
75.
dy
d x 6 dt 6
dx
1 3t 4
dx
3 x 4
76.
dy
d 2
d sec x 1
d
sec x tan x
1
1
dx
sec x t 2 1 dt dx 2 t 2 1 dt sec2 x 1 dx (sec x) 1sec2 x
dx
349
dy
2 cos3 x
d (7 x 2 ) 14 x 2 cos3 (7 x 2 )
74. dx 2 cos3 (7 x 2 ). dx
77. Yes. The function f, being differentiable on [a, b], is then continuous on [a, b]. The Fundamental Theorem of
Calculus says that every continuous function on [a, b] is the derivative of a function on [a, b].
78. The second part of the Fundamental Theorem of Calculus states that if F ( x) is an antiderivative of f ( x )
b
on [a, b], then f ( x) dx F (b) F (a). In particular, if F ( x) is an antiderivative of 1 x 4 on [0, 1], then
a
1
4
0 1 x dx F (1) F (0).
1
x
1 t 2 dt
dy
2
x
2
x
2
y
x
80.
y
0
1 dt cos x 1 dt dy d cos x 1 dt d cos x 1 dt
dx
dx 0
dx 0
cos x 1t 2
0
1t 2
1t 2
1t 2
1
1cos 2 x
1 1 t dt dx dxd 1 1 t dt dxd 1 1 t dt 1 x
79.
(cos x) ( sin x)
d
dx
1
sin 2 x
2
1 csc x
sin x
81. We estimate the area A using midpoints of the vertical intervals, and we will estimate the width of
the parking lot on each interval by averaging the widths at top and bottom. This gives the estimate
A 5 0212 12218 18217 17 216.5 16.5218 182 21 21222 22214 657.5 m 2 . The cost is
Area ($21/m 2 ) (657.5 m 2 )($21/m 2 ) $13,897.50 the job cannot be done for $10,000.
82. (a) Before the chute opens for A, a 9.8 m/s 2 . Since the helicopter is hovering v0 0 m/s
v 9.8 dt 9.8t v0 9.8t. Then s0 2000 m s 9.8t dt 4.9t 2 s0 4.9t 2 2000.
At t 4s, s 4.9(4)2 2000 1921.6 m when A’s chute opens;
(b) For B, s0 2200 m, v0 0, a 9.8 m/s 2 v 9.8 dt 9.8t v0 9.8t s
2
2
2
9.8t dt 4.9t s0 4.9t 2200. At t 13 s, s 4.9(13) 2200 1371.9 m when B’s chute
opens;
(c) After the chutes open, v 4.9 m/s s 4.9 dt 4.9t s0 . For A, s0 1921.6 m and for B,
s0 1371.9 m. Therefore, for A, s 4.9t 1921.6 and for B, s 4.9t 1371.9. When they hit the
ground, s 0 for A, 0 4.9t 1921.6 t 1921.6
392 seconds, and for B,
4.9
0 4.9t 1371.9 t 1371.9
280 seconds to hit the ground after the chutes open, Since B’s chutes
4.9
opens 58 seconds after A’s opens B hits the ground first.
Copyright 2016 Pearson Education, Ltd.
350
Chapter 5 Integrals
CHAPTER 5
ADDITIONAL AND ADVANCED EXERCISES
1
1
0
0
1. (a) Yes, because f ( x) dx 17 7 f ( x) dx 17 (7) 1
1
(b) No. For example, 8 x dx [4 x 2 ]10 4, but
0
2
5
5
2
1
3/ 2
8 x dx 2 2 x 3 4 3 2 13/2 03/2 4 3 2
0
2 0
1
4
2. (a) True: f ( x) dx f ( x) dx 3
5
(b) True:
2
[ f ( x) g ( x)] dx
5
2
f ( x) dx
5
2
g ( x) dx
2
2
5
5
2
2
f ( x) dx f ( x) dx
g ( x) dx
43 2 9
(c) False:
5
2
2
f ( x)dx 4 3 7
5
2
g ( x) dx
5
2
the other hand, f ( x) g ( x) [ g ( x) f ( x)] 0
3.
[ f ( x ) g ( x)] dx
5
2
0
5
2
[ g ( x) f ( x)] dx 0. On
[ g ( x) f ( x)] dx 0 which is a contradiction.
x
x
x
0
0
0
y 1a f (t ) sin a ( x t ) dt 1a f (t ) sin ax cos at dt 1a f (t ) cos ax sin at dt
sin ax x
cos ax x
dy
a
f (t ) cos at dt a
f (t ) sin at dt dx
0
0
x
sin ax d x
cos ax d x
cos ax f (t ) cos at dt a dx
f (t ) cos at dt sin ax f (t ) sin at dt a dx
f (t ) sin at dt
0
0
0
0
x
x
x
0
0
cos ax f (t ) cos at dt sinaax ( f ( x) cos ax) sin ax f (t ) sin at dt cosaax ( f ( x) sin ax)
dy
x
x
d2y
0
0
dx 2
dx cos ax f (t ) cos at dt sin ax f (t ) sin at dt. Next,
x
d x f (t ) cos at dt a cos ax x f (t ) sin at dt (sin ax ) d x f (t ) sin at dt
a sin ax f (t ) cos at dt (cos ax) dx
dx 0
0
0
0
x
x
0
0
a sin ax f (t ) cos at dt (cos ax) f ( x) cos ax a cos ax f (t ) sin at dt (sin ax) f ( x) sin ax
x
x
a sin ax f (t ) cos at dt a cos ax f (t ) sin at dt f ( x). Therefore, y a 2 y
0
0
x
cos ax x
a cos ax f (t ) sin at dt a sin ax f (t ) cos at dt f ( x) a 2 sinaax f (t ) cos at dt a f (t ) sin at dt
0
0
0
0
f ( x). Note also that y (0) y (0) 0.
x
4.
x
y
x
d ( x) d
dt dx
dx
1 4 y
2
1 4 y . Then
12 1 4 y 2
0
1
1 4t
1
dy
dx
1/2
dy
dx
dy
(8 y ) dx
dy
d y
1
dt dy
0 1 4t 2 dt dx from the chain rule
0 1 4t
y
2
1
1
2
2
d 1 4 y 2 d 1 4 y 2 dy
dx
dy
dx
d2y
dx 2
4 y 14 y 4 y. Thus d y 4 y, and the constant of proportionality
2
dy
4 y dx
1 4 y
2
1 4 y
2
2
dx 2
is 4.
Copyright 2016 Pearson Education, Ltd.
Chapter 5 Additional and Advanced Exercises
x2
0
5. (a)
d
f (t ) dt x cos x dx
x2
0
f (t ) dt cos x x sin x f ( x 2 )(2 x) cos x x sin x
f ( x 2 ) cos x 2xx sin x . Thus, x 2 f (4)
f ( x) 2
0
(b)
f ( x)
t dt t3
0
3
351
3
cos 2 2 sin 2
14
4
3
3
13 f ( x) 13 f ( x) x cos x f ( x) 3 x cos x f ( x) 3 3 x cos x
f (4) 3 3(4) cos 4 3 12
6.
a
a
2
2
0 f ( x) dx a2 a2 sin a 2 cos a. Let F (a) 0 f (t ) dt f (a) F (a). Now F (a) a2 a2 sin a 2 cos a
f (a) F (a ) a 12 sin a a2 cos a 2 sin a f 2 2 12 sin 2 22 cos 2 2 sin 2 2 12 2 12
7.
b
b
2
2
1 f ( x) dx b 1 2 f (b) dbd 1 f ( x) dx 12 (b 1)
1/2
(2b)
b
b 2 1
f ( x)
x
x 2 1
d x u f (t ) dt du x f (t ) dt ; the derivative of the
8. The derivative of the left side of the equation is: dx
0 0
0
d x f (u )( x u ) du d x f (u ) x du d x u f (u ) du
right side of the equation is: dx
dx 0
0
dx 0
d x x f (u ) du d x u f (u ) du x f (u ) du x d x f (u ) du x f ( x) x f (u ) du x f ( x) x f ( x )
dx
0
0
0
dx 0
dx 0
x
f (u ) du. Since each side has the same derivative, they differ by a constant, and since both sides equal 0
0
x u
x
when x 0, the constant must be 0. Therefore, f (t ) dt du f (u )( x u ) du.
0 0
0
9.
dy
3x2 2 y
dx
3
2
3
3
(3x 2) dx x 2 x C. Then (1, 1) lies on the curve 1 2(1) C 1
C 4 y x 2 x 4
10. The acceleration due to gravity downward is 9.8 m/s 2 v 9.8 dt 9.8t v0 , where v0 is the initial
velocity v 9.8t 9.8 s (9.8t 9.8) dt 4.9t 2 9.8t C. If the release point, at t 0, is s 0,
then C 0 s 4.9t 2 9.8t. Then s 5.2 5.2 4.9t 2 9.8t 4.9t 2 9.8t 5.2 0. The discriminant
of this quadratic equation is 5.88 which says there is no real time when s 5.2 m. You had better duck.
11.
3
0
8 f ( x) dx 8 x
3
dx 4 dx
0
0
5/3
3
3
5 x [4 x]0
8
0 53 (8)5/3 (4(3) 0) 96
12 36
5
5
12.
2/3
3
0
3
2
4 f ( x) dx 4 x dx 0 ( x 4) dx
3
0
3
23 ( x)3/2 x3 4 x
0
4
3
0 23 (4)3/2 33 4(3) 0 16
3 73
3
Copyright 2016 Pearson Education, Ltd.
352
Chapter 5 Integrals
13.
0 g (t ) dt 0 t dt 1 sin t dt
2
1
2
1
2
t2 1 cos t
0
1
2
12 0 1 cos 2 1 cos 12 2
14.
2
1
2
0 h( z ) dz 0 1 z dz 1 (7 z 6)
1
1/3
dz
2
3 (7 z 6) 2/3
23 (1 z )3/2 14
0
1
3/2
3/2
2
2
3 (1 1) 3 (1 0)
2/3
3 (7(2) 6)
3 (7(1) 6) 2/3
14
14
6
3
55
2
3 7 14 42
15.
1
2
1
2
2
2 f ( x) dx 2 dx 1 (1 x ) dx 1 2 dx
1
3
[ x]12 x x3 [2 x]12
1
3
( 1)3
(1 (2)) 1 13 1 3 2(2) 2(1)
1 23 23 4 2 13
3
16.
2
0
1
2
2
1 h(r ) dr 1 r dr 0 (1 r ) dr 1 dr
0
1
2
3
r2 r r3 [r ]12
1
0
3
( 1)2
0 2 1 13 0 (2 1) 12 32 1 76
2
2 1
2
1
2
b
2
17. Ave. value b 1 a f ( x) dx 21 0 f ( x) dx 12 x dx ( x 1) dx 12 x2 12 x2 x
a
0
1
0
1
0
2
2
2
12 12 0 22 2 12 1 12
1
2
3
b
3
18. Ave. value b 1 a f ( x) dx 31 0 f ( x) dx 13 dx 0 dx dx 13 [1 0 0 3 2] 32
1
2
a
0
0
19. Let f ( x) x5 on [0, 1]. Partition [0, 1] into n subintervals with x 1n0 1n . Then 1n , n2 , , nn are the right-hand
is the upper sum for f ( x) x on
j 5 1
n
endpoints of the subintervals. Since f is increasing on [0, 1], U n
j 1
1
1 1 2
lim 1 2 n x5 dx x6 16
1n nlim
n
n
n
n
0
0
n
n
j 5
[0, 1] lim n
n j 1
5
5
5
n 5
5
5
5
6
Copyright 2016 Pearson Education, Ltd.
6
1
Chapter 5 Additional and Advanced Exercises
353
20. Let f ( x) x3 on [0, 1]. Partition [0, 1] into n subintervals with x 1n0 1n . Then 1n , n2 , , nn are the right-hand
is the upper sum for f ( x) x on
j 3 1
n
endpoints of the subintervals. Since f is increasing on [0, 1], U n
[0, 1] lim
n j 1
j 3 1
n
n
j 1
3
1
3
3
3
3
3
n3 1 x3 dx x 4 1
lim 1n 1n n2 nn lim 1 2
n4
0
4 0 4
n
n
21. Let y f ( x) on [0, 1]. Partition [0, 1] into n subintervals with x 1n0 1n . Then 1n , n2 , , nn are the right-hand
is a Riemann sum of y f ( x) on
endpoints of the subintervals. Since f is continuous on [0, 1], f
j 1
[0, 1] lim f
n j 1
22. (a)
j
n
1
n
lim f f f f ( x) dx
j
n
1
n
1
n n
1
n
1
n
n
2
n
0
1
lim 12 [2 4 6 2n] lim 1n n2 n4 n6 2nn 2 x dx [ x 2 ]10 1, where f ( x) 2 x on [0, 1]
0
n n
n
15
15
15
1 [115 215 n15 ] lim 1 1
n2 nn
16
n
n
n
n
n
15
(b) lim
f ( x) x
(c)
1 15
x
0
1
1 , where
dx x16 16
0
16
on [0, 1]
1
1
lim 1 sin n sin 2n sin nn sin n dx 1 cos x 1 cos 1 cos 0 2 , where
0
0
n n
f ( x) sin x on [0, 1]
0
1 115 215 n15 lim 1 lim 1 [115 215 n15 ] lim 1 1 x15 dx 0 1
17
16
n n n n16
n
n
n n 0
(d) lim
(e)
(see part (b) above)
lim 115 115 215 n15 lim n16 [115 215 n15 ]
n n
n n
1
15
lim n lim 116 [1 215 n15 ] lim n x15 dx (see part (b) above)
0
n
n n
n
23. (a) Let the polygon be inscribed in a circle of radius r. If we draw a radius from the center of the circle (and
the polygon) to each vertex of the polygon, we have n isosceles triangles formed (the equal sides are equal
to r, the radius of the circle) and a vertex angle of n where n 2n . The area of each triangle is
2
2
An 12 r 2 sin n the area of the polygon is A nAn nr2 sin n nr2 sin 2n .
(b)
2
n r 2 sin 2 lim
n
n 2
n
lim A lim nr2 sin 2n lim
n
n
r2
r 2 r 2 2lim
/ n
sin 2n
sin 2n
2
n
2
n
24. Partition [0, 1] into n subintervals, each of length x 1n with the points x0 0, x1 1n , x2 n2 , , xn nn 1.
The inscribed rectangles so determined have areas f ( x0 ) x (0)2 x, f ( x1 )x
1n x,
2
n2 x,, f ( xn1 ) nn1 x. The sum of these areas is
2
2
2
( n 1) 1
( n 1)
1
2
Sn 02 1n n2 nn1 x 1 2
n n n n . Then
n
n
n
f ( x2 ) x
2
2
2
2
2
2
2
2
2
2
3
3
2
3
1
( n 1)2
2
lim Sn lim 1 3 23 3 x 2 dx 13 13 .
0
n
n
n
n n
Copyright 2016 Pearson Education, Ltd.
2
3
354
Chapter 5 Integrals
25. (a)
g (1) f (t ) dt 0
1
1
3
(b) g (3) f (t ) dt 12 (2)(1) 1
1
(c)
g (1)
1
1
f (t ) dt
1
f (t ) dt 14 ( 22 )
1
(d) g ( x) f ( x) 0 x 3, 1, 3 and the sign chart for g ( x) f ( x) is | | | . So g has a
3
relative maximum at x 1.
(e) g (1) f (1) 2 is the slope and g (1)
1
1
3
f (t )dt , by (c). Thus the equation is y 2( x 1)
1
y 2x 2 .
(f ) g ( x) f ( x) 0 at x 1 and g ( x) f ( x) is negative on (3, 1) and positive on (1, 1) so there is an
inflection point for g at x 1. We notice that g ( x) f ( x) 0 for x on (1, 2) and g ( x) f ( x) 0 for x
on (2, 4), even though g (2) does not exist, g has a tangent line at x 2, so there is an inflection point at
x 2.
(g) g is continuous on [3, 4] and so it attains its absolute maximum and minimum values on this interval. We
saw in (d) that g ( x) 0 x 3, 1, 3. We have that g (3)
3
1
1
3
g (1) f (t ) dt 0
1
3
2
f (t ) dt 22 2
4
g (3) f (t ) dt 1
1
f (t ) dt
g (4) f (t ) dt 1 12 1 1 12
1
1
Thus, the absolute minimum is 2 and the absolute maximum is 0. Thus, the range is [2 , 0].
26.
x
x
y sin x cos 2t dt 1 sin x cos 2t dt 1 y cos x cos(2 x); when x we have
y cos cos(2 ) 1 1 2. And y sin x 2sin(2 x); when x , y sin
x cos 2t dt 1 0 0 1 1.
1 dxd 1x 1x x x1 1x 1x 2x
x 1
dx
dt f ( x) 1x dx
1/ x t
27.
f ( x)
28.
f ( x)
sin x 1
1
dt f ( x)
cos x 1t 2
1sin 2 x
29.
g ( y)
2 y
30.
f ( x)
2
1
x
y
x 3
x
(sin x)
d
dx
1
1cos 2 x
2 d
sin t 2 dt g ( y ) sin 2 y dy
2 y
t (5 t ) dt f ( x) ( x 3)(5 ( x 3))
(cos x)
d
dx
cos x sin x 1 1
cos x sin x
cos 2 x sin 2 x
y
sin y dyd y sin y4 y sin
2 y
2
dxd ( x 3) x(5 x) dxdx ( x 3)(2 x) x(5 x)
6 x x 2 5 x x 2 6 6 x. Thus f ( x) 0 6 6 x 0 x 1. Also, f ( x) 6 0 x 1 gives
a maximum.
Copyright 2016 Pearson Education, Ltd.
CHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS
6.1
VOLUMES USING CROSS-SECTIONS
1.
A( x)
x x 2 x; a 0, b 4; V b A( x) dx 4 2 x dx x 2 4 16
2
2.
A( x )
(diagonal)2
2
(diameter) 2
4
a
2
2 x 2 x 2
2
4
0
1 2 x2 x4 ; a 1, b 1;
2 1 x 2
0
4
2
1
5
b
1
V A( x) dx 1 2 x 2 x 4 dx x 23 x3 x5 2 1 23 15 16
15
1
a
1
2
3.
2
A( x ) (edge)2 1 x 2 1 x 2 2 1 x 2 4 1 x 2 ; a 1, b 1;
b
1
1
V A( x) dx 4 1 x 2 dx 4 x x3 8 1 13 16
3
1
a
1
3
2 2 1 x 2 1 x2 ; a 1, b 1;
4.
2
1
b
1
V A( x) dx 2 1 x 2 dx 2 x x3 4 1 13 83
a
1
1
1 x 2 1 x 2
(diagonal) 2
A( x )
2
2
2
2
2
3
5. (a) STEP 1) A( x ) 12 (side) (side) sin 3 12 2 sin x 2 sin x sin 3 3 sin x
STEP 2) a 0, b
b
a
0
0
STEP 3) V A( x) dx 3 sin x dx 3 cos x 3(1 1) 2 3
2
(b) STEP 1) A( x ) (side) 2 sin x
2 sin x 4 sin x
STEP 2) a 0, b
b
a
0
STEP 3) V A( x) dx 4 sin x dx 4 cos x 0 8
6. (a) STEP 1) A( x )
(diameter) 2
4 (sec x tan x) 2 4 sec 2 x tan 2 x 2sec x tan x
4
4 sec 2 x sec2 x 1 2 sin2x
cos x
STEP 2) a 3 , b 3
b
STEP 3) V A( x) dx
a
/3
/3 4
2sec x 1
2
2sin x
cos 2 x
dx 2 tan x x 2
4
4 2 3 3 2 11 2 3 3 2 11 4 4 3 23
2
2
Copyright 2016 Pearson Education, Ltd.
/3
1
cos x /3
355
356
Chapter 6 Applications of Definite Integrals
(b) STEP 1) A( x) (edge)2 (sec x tan x) 2 2sec2 x 1 2 sin2x
STEP 2) a 3 , b 3
b
/3
a
/3
STEP 3) V A( x) dx
2sec x 1
2
cos x
2 sin x
cos 2 x
dx 2 2 3 4 3
3
2
3
7. (a) STEP 1) A( x ) (length) (height) (6 3x ) (10) 60 30 x
STEP 2) a 0, b 2
b
2
2
STEP 3) V A( x) dx (60 30 x) dx 60 x 15 x 2 (120 60) 0 60
0
0
a
(b) STEP 1) A( x ) (length) (height) (6 3 x)
20 2(6 3 x )
2
(6 3x)(4 3x) 24 6 x 9 x
2
STEP 2) a 0, b 2
b
2
a
0
STEP 3) V A( x)dx
24 6 x 9 x2 dx 24 x 3x2 3x3 0 (48 12 24) 0 36
2
8. (a) STEP 1) A( x ) 12 (base) (height)
x 2x (6) 6 x 3x
STEP 2) a 0, b 4
6 x1/2 3 x dx 4 x3/2 32 x 2 (32 24) 0 8
0
0
b
STEP 3) V A( x) dx
a
(b) STEP 1) A( x) 12
4
4
2
diameter
12 x2 2 x x 4 x 8 x x3/2 14 x2
2
x
2
2
3/ 2
1 2
4
STEP 2) a 0, b 4
1 x3 8 64 16 (0)
x x3/2 14 x 2 dx 12 x 2 52 x5/2 12
5
3 8
15
0 8
0
b
STEP 3) V A( x) dx 8
a
9.
A( y ) 4 (diameter) 2 4
4
4
5 y 2 0 54 y 4 ;
2
d
c 0, d 2; V A( y ) dy
c
2 5 4
y dy 54
0 4
2
y5
5
5 4 2 0 8
0
2
10.
2
A( y ) 12 (leg)(leg) 12 1 y 2 1 y 2 12 2 1 y 2 2 1 y 2 ; c 1, d 1;
1
d
1
y3
V A( y ) dy 2 1 y 2 dy 2 y 3 4 1 13 83
c
1
1
Copyright 2016 Pearson Education, Ltd.
Section 6.1 Volumes Using Cross-Sections
357
11. The slices perpendicular to the edge labeled 5 are triangles, and by similar triangles we have bh 43 h 34 b.
The equation of the line through (5, 0) and (0, 4) is y 54 x 4, thus the length of the base 54 x 4 and
the height 34 54 x 4 53 x 3. Thus A( x ) 12 (base) (height) 12 54 x 4 53 x 3
b
5 6 2 12
x 5 x6
0 25
6 x 2 12 x 6 and V
25
A( x) dx
5
a
dx 252 x3 65 x2 6 x 0 (10 30 30) 0 10
5
12. The slices parallel to the base are squares. The cross section of the pyramid is a triangle, and by similar
259 y 2 V cd A( y) dy 05 259 y 2 dy
triangles we have bh 53 b 53 h. Thus A( y ) (base) 2 53 y
2
5
3 y 3 15 0 15
25
0
13. (a) It follows from Cavalieri’s Principle that the volume of a column is the same as the volume of a right
prism with a square base of side length s and altitude h. Thus,
STEP 1) A( x ) (sidelength) 2 s 2 ;
STEP 2) a 0, b h;
b
h
a
0
STEP 3) V A( x) dx s 2 dx s 2 h
(b) From Cavalieri’s Principle we conclude that the volume of the column is the same as the volume of the
prism described above, regardless of the number of turns V s 2 h
14. 1)
The solid and the cone have the same altitude
of 12.
The cross sections of the solid are disks of
2)
diameter x 2x 2x . If we place the vertex of
the cone at the origin of the coordinate system
and make its axis or symmetry coincide with
the x-axis then the cone’s cross sections will
be circular disks of diameter 4x 4x 2x
(see accompanying figure).
The solid and the cone have equal altitudes and
identical parallel cross sections. From
Cavalier’s Principle we conclude that the solid
and the cone have the same volume.
3)
2
2
15. R( x) y 1 2x V R( x) dx 1 2x
2
0
0
dx 02 1 x x4 dx x x2 12x 0
2
2
2
3
2
8 2
2 42 12
3
16.
2
R ( y ) x 2 V R ( y ) dy
3y
2
0
2
2 3y 2
2
dy 94 y 2 dy 43 y 3 43 8 6
2
0
0
0
17. R ( y ) tan 4 y ; u 4 y du 4 dy 4 du dy; y 0 u 0, y 1 u 4 ;
2
1
1
/4
/4
2
/4
V R ( y ) dy tan 4 y dy 4
tan 2 u du 4
1 sec 2 u du 4 u tan u 0
0
0
0
0
4 4 1 0 4
Copyright 2016 Pearson Education, Ltd.
358
18.
Chapter 6 Applications of Definite Integrals
R ( x) sin x cos x; R( x) 0 a 0 and b 2 are the limits of integration;
V
/2
0
R( x) dx
2
/2
0
(sin x cos x)2 dx
/2 (sin 2 x )2
0
4
dx; u 2 x du 2 dx du
dx
;
8
4
2
x 0 u 0, x 2 u V 18 sin 2 u du 8 u2 14 sin 2u 8 2 0 0 16
0
0
2
19. R ( x) x 2 V R( x) dx
2
0
2
0
x2 dx 02 x4 dx x5 0 325
2
2
5
2
20. R ( x) x3 V R ( x) dx
2
0
2
x3 dx x6 dx x7 128
7
0
0
0
2
2
2
7
3
21. R ( x) 9 x 2 V R( x) dx
2
3
9 x 2 dx 9 x x3
3
3
3
3
3
2 π 18 36π
2 9(3) 27
3
22.
1
R ( x) x x 2 V R ( x) dx
2
0
0
1
2
1
x x 2 dx x 2 2 x3 x 4 dx
0
1
3
4
5
x3 24x x5 13 12 15
0
(10 15 6)
30
30
23. R ( x) cos x V
/2
0
/2
0
R( x) dx
2
/2
cos x dx sin x 0
(1 0)
Copyright 2016 Pearson Education, Ltd.
Section 6.1 Volumes Using Cross-Sections
24. R ( x) sec x V
/4
/4
/4
2
/4
/4
25.
R( x) dx
sec 2 x dx tan x /4 [1 (1)] 2
R ( x) 2 sec x tan x V
/4
0
/4
/4
0
0
R( x) dx
2
2 sec x tan x dx
2
2 2 2 sec x tan x sec2 x tan 2 x dx
/4
/4
2 dx 2 2 sec x tan x dx
0
0
/4
(tan x) 2 sec2 x dx
0
/4
3
/4
/4
2 x 0 2 2 sec x 0 tan3 x
0
2 0 2 2 2 1 13 (13 0)
2 2 2 11
3
26. R ( x) 2 2sin x 2(1 sin x) V
/2
0
/2
0
4(1 sin x) 2 dx 4
/2
0
4
/2
4
/2 3 cos 2 x
2sin x
2
1 sin 2 x 2sin x dx
1 1 (1 cos 2 x) 2sin x dx
2
0
0
R( x) dx
2
2
4 32 x sin42 x 2 cos x
/2
0
4 34 0 0 (0 0 2) (3 8)
1
1
27. R ( y ) 5 y 2 V R( y ) dy 5 y 4 dy
1
2
1
1
y 5 [1 (1)] 2
1
Copyright 2016 Pearson Education, Ltd.
359
360
Chapter 6 Applications of Definite Integrals
2
2
28. R ( y ) y 3/2 V R( y ) dy y 3 dy
2
0
0
2
y4
4 4
0
29. R ( y ) 2sin 2 y V
/2
0
/2
0
R( y ) dy
2
/2
2sin 2 y dy cos 2 y 0
[1 (1)] 2
30.
0
y
R ( y ) cos 4 V R ( y ) dy
2
2
0
0
y
y
cos 4 dy 4 sin 4 4[0 ( 1)] 4
2
2
y 2/3 y 6 dy 53 y 5/3 71 y 7
1635
1
y
31. R1 ( y ) y 3 , R2 ( y ) y1/3 V R2 2 R12 dy
0
1
0
1
x y3
1
0
x y1/ 3
0
32. R ( y )
2y
y 2 1
0
1
x
1
V R( y ) dy
2
0
dy; [u y 2 1 du 2 y dy;
1
2 y y2 1
2
y 0 u 1, y 1 u 2]
2
2
1
1
V u 2 du u1 12 (1) 2
b
33. For the sketch given, a 2 , b 2 ; R( x) 1, r ( x) cos x; V R ( x) r ( x)
a
/2
/2
(1 cos x) dx 2
/2
0
/2
(1 cos x) dx 2 x sin x 0
2
2 2 1 2 2
Copyright 2016 Pearson Education, Ltd.
2
dx
Section 6.1 Volumes Using Cross-Sections
d
34. For the sketch given, c 0, d 4 ; R( y ) 1, r ( y ) tan y; V R( y ) r ( y )
c
/4
0
2
1
0
1
2
2
dx
1 x 2 dx x x3 1 13 0 23
0
0
3
1
36. r ( x) 2 x and R ( x) 2 V R ( x) r ( x)
0
1
1
2
2
dx
(4 4 x) dx 4 x x2 4 1 12 2
0
0
2
37. r ( x) x 2 1 and R( x) x 3
2
V R( x) r ( x)
1
2
2
dx
2
2
( x 3)2 x 2 1 dx
1
2
x 2 6 x 9 x 4 2 x 2 1 dx
1
x 4 x 2 6 x 8 dx x5 x3 62x 8 x
1
1
2
5
3
2
2
33
3 28 3 8 530533 1175
5
32
8 24 16 15 13 62 8
5 3 2
38. r ( x) 2 x and R ( x) 4 x 2
2
V R( x) r ( x)
1
2
2
dx
2
16 8 x 2 x 4 4 4 x x 2 dx
1
2
2
12 4 x 9 x 2 x 4 dx 12 x 2 x 2 3x3 x5
1
1
2
2
4 x 2 (2 x)2 dx
1
5
2
dy
1 tan 2 y dy 0 /4 2 sec2 y dy 2 y tan y 0 /4 2 1 2
35. r ( x) x and R ( x) 1 V R( x) r ( x)
1
2
24 8 24 32
12 2 3 15 15 33
108
5
5
5
Copyright 2016 Pearson Education, Ltd.
361
362
Chapter 6 Applications of Definite Integrals
39. r ( x) sec x and R ( x ) 2
V
/4
/4
/4
/4
R ( x) r ( x)
2
2
dx
2 sec2 x dx 2 x tan x/4/4
2 1 2 1 ( 2)
40.
1
R ( x ) sec x and r ( x) tan x V R ( x) r ( x)
0
0
1
1
2
2
dx
sec2 x tan 2 x dx 1 dx x 0
1
0
1
2
dy
2
dy
2
dy
41. r ( y ) 1 and R ( y ) 1 y V R ( y ) r ( y )
0
1
1
2
(1 y ) 2 1 dy 1 2 y y 2 1 dy
0
0
1
1
y3
2 y y 2 dy y 2 3 1 13 43
0
0
42.
1
R ( y ) 1 and r ( y ) 1 y V R ( y ) r ( y )
0
1
2
1
1 (1 y )2 dy 1 1 2 y y 2 dy
0
0
1
1
y3
2 y y 2 dy y 2 3 1 13 23
0
0
43.
4
y V R( y) r ( y )
R ( y ) 2 and r ( y )
0
2
4
4
y2
(4 y ) dy 4 y 2 (16 8) 8
0
0
44.
R ( y ) 3 and r ( y ) 3 y 2
V
0
3
R( y ) r ( y )
2
2
dy
3
3
3 3 y 2 dy 3 y 2 dy y 3
0
3
0
0
3
Copyright 2016 Pearson Education, Ltd.
Section 6.1 Volumes Using Cross-Sections
45.
1
dy
R ( y ) 2 and r ( y ) 1 y V R ( y ) r ( y )
0
2
2
2
1
4 1 y dy 1 2 y y dy
0
1
3 43 12 18683 76
1
y2
3 2 y y dy 3 y 43 y 3/2 2
0
0
46.
1
R ( y ) 2 y1/3 and r ( y ) 1 V R( y ) r ( y )
0
2
2
dy
2
1
1
2 y1/3 1 dy 4 4 y1/3 y 2/3 1 dy
0
0
1
1
3 y 5/3
3 4 y1/3 y 2/3 dy 3 y 3 y 4/3 5
0
0
3 3 53 35
47. (a) r ( x) x and R ( x) 2
4
V R ( x) r ( x)
0
2
2
4
dx
4
(4 x) dx 4 x x2 (16 8) 8
0
0
2
dy y dy
(c) r ( x) 0 and R ( x) 2 x V R ( x) r ( x) dx 2 x dx
2
(b) r ( y ) 0 and R ( y ) y 2 V R( y ) r ( y )
0
2
4
2
2
2
0
2 4
0
y5
5
4
2
2
0
32
5
0
4
4 4 x x dx 4 x 8x3 x2 0 16 643 162 83
0
4
3/ 2
2
2
(d) r ( y ) 4 y 2 and R ( y ) 4 V R( y ) r ( y )
0
2
2
dy 16 4 y dy
2 2
2
0
2
2
2
y5
16 16 8 y 2 y 4 dy 8 y 2 y 4 dy 83 y 3 5 64
32
224
3
5
15
0
0
0
y
48. (a) r ( y ) 0 and R ( y ) 1 2
dy
1 dy 1 y dy
2
V R( y ) r ( y)
2
2
0
2
0
y 2
2
2
0
2
y2
4
y2
y3
8 2
y 2 12 2 24 12
3
0
Copyright 2016 Pearson Education, Ltd.
363
364
Chapter 6 Applications of Definite Integrals
y
(b) r ( y ) 1 and R ( y ) 2 2
2
dy 2 1 dy 4 2 y 1 dy
0
2
y 2
2
2
V R( y ) r ( y )
2
0
2
y2
4
2
0
2
y2
y3
8 2 2 8
3 2 y 4 dy 3 y y 2 12 6 4 12
3
3
0
0
1
49. (a) r ( x) 0 and R ( x ) 1 x 2 V R( x) r ( x)
1
2
2
dx
1
1 x 2 dx 1 2 x 2 x 4 dx
1
1
2
1
1
3
5
10 3 16
x 23x x5 2 1 32 15 2 1515
15
1
1
(b) r ( x) 1 and R ( x) 2 x 2 V
1
R( x) r ( x) dx 2 x 1 dx
2
2 2
1
2
1
1
4 4 x 2 x 4 1 dx 3 4 x 2 x 4 dx 3 x 43 x3 x5 2 3 34 15
1
1
1
1
5
215 (45 20 3) 56
15
1
(c) r ( x) 1 x 2 and R( x) 2 V
1
1
R( x) r ( x) dx 4 1 x dx
2
2 2
1
2
1
1
4 1 2 x 2 x 4 dx 3 2 x 2 x 4 dx 3 x 23 x3 x5 2 3 23 15
1
1
1
1
5
1
215 (45 10 3) 64
15
50. (a) r ( x) 0 and R ( x) bh x h
b
V R ( x) r ( x)
0
2
2
dx
b
bh x h dx h x 2 2bh x h 2 dx
0
0 b
2
b
2
2
2
b
3
2
2
h 2 x 2 xb x h 2 b3 b b h3 b
3b
0
h
y
(b) r ( y ) 0 and R( y ) b 1 h V
0
R( y) r( y) dy b 1 dy
2
h
2 h
0
2
2
h
2 y y2
y2
y3
b 2 1 h 2 dy b 2 y h 2 b 2 h h h3 b3 h
0
h
3h 0
51.
R ( y ) b a 2 y 2 and r ( y ) b a 2 y 2
a
V R( y ) r ( y )
a
2
2
dy
2
2
b a 2 y 2 b a 2 y 2 dy
a
a
a
a
4b a 2 y 2 dy 4b
a
a
a 2 y 2 dy
2
4b area of semicircle of radius a 4b 2a 2a 2 b 2
Copyright 2016 Pearson Education, Ltd.
y 2
h
Section 6.1 Volumes Using Cross-Sections
365
5
5
52. (a) A cross section has radius r 2 y and area r 2 2 y. The volume is 2 ydy y 2 25 .
0
0
A(h). Therefore dV
dV
dh A(h) dh
, so dh
A(1h ) dV
.
(b) V (h) A(h)dh, so dV
dh
dt
dh dt
dt
dt
dt
3
3
81 3 units
83 units
.
For h 4, the area is 2 (4) 8 , so dh
dt
s
s
53. (a)
R( y ) a 2 y 2 V
ha
a
ha
3
y3
( h a )3
a 2 y 2 dy a 2 y 3
a 2 h a3 3 a3 a3
a
2
3
3
h (3a h )
a 2 h 13 h3 3h 2 a 3ha 2 a3 a3 a 2 h h3 h 2 a ha 2
3
(b) Given dV
0.2 m3 /s and a 5 m, find dh
dt
dt
. From part (a), V (h)
3
h 2 (15 h )
5 h 2 3h
3
h 4
2
dV
dV
dV
dh
0.2
dh 10 h h dt dh dt h(10 h) dh
dh
4 (10
(201)(6) 1201 m/s.
dt
dt h 4
4)
54. Suppose the solid is produced by revolving
y 2 x about the y -axis. Cast a shadow of
the solid on a plane parallel to the xy -plane.
Use an approximation such as the Trapezoid Rule,
2
n
dˆ
b
2
to estimate R( y ) dy 2k y.
a
k 1
55. The cross section of a solid right circular cylinder with a cone removed is a disk with radius R from which a
disk of radius h has been removed. Thus its area is A1 R 2 h 2 ( R 2 h 2 ). The cross section of the
2
R 2 h 2 . Therefore its area is A2 R 2 h 2 R 2 h 2 .
We can see that A1 A2 . The altitudes of both solids are R. Applying Cavalieri’s Principle we find
hemisphere is a disk of radius
Volume of Hemisphere (Volume of Cylinder) (Volume of Cone) R 2 R 13 R 2 R 32 R3 .
56.
6
6
60536 365 cm3.
6
2
4
x 36 x 2 V
x
R ( x ) 12
36 x 2 dx 144
R( x) dx 144
36 x x dx
6
2
0
2
0
12 x3 x 12 63 6 6 12 36 196
144
5 0 144
5
144
5
144
5
5
3
0
The plumb bob will weigh about W (8.5) 365 192 g, to the nearest gram.
57.
R ( y ) 256 y 2 V
7
16
2
7
y
256 y 2 dy 256 y 3
16
16
R( y ) dy
7
3
3
3
3
3
(256)(7) 73 (256)(16) 163 73 256(16 7) 163 1053 cm3 3308 cm3
Copyright 2016 Pearson Education, Ltd.
366
Chapter 6 Applications of Definite Integrals
58. (a)
R ( x ) | c sin x |, so V
0
R( x)2 dx 0 (c sin x)2 dx 0 c 2 2c sin x sin 2 x dx
2 x dx c 2 1 2c sin x cos 2 x dx c 2 1 x 2c cos x sin 2 x
c 2 2c sin x 1cos
0 2
2
2
2
4 0
0
c 2 2 2c 0 (0 2c 0) c 2 2 4c . Let V (c) c 2 2 4c . We find the
2 4 2 4; Evaluate V at the endpoints: V (0) 2 and V (1) 32 4 2 (4 ) .
extreme values of V (c) : dV
(2c 4) 0 c 2 is a critical point, and V 2 4 2 8
dc
2
2
2
2
Now we see that the function’s absolute minimum value is 2 4, taken on at the critical point c 2 .
(See also the accompanying graph.)
2
(b) From the discussion in part (a) we conclude that the function’s absolute maximum value is 2 , taken on
at the endpoint c 0.
(c) The graph of the solid’s volume as a function
of c for 0 c 1 is given at the right. As c
moves away from [0, 1] the volume of the solid
increases without bound. If we approximate the
solid as a set of solid disks, we can see that the
radius of a typical disk increases without
bounds as c moves away from [0, 1].
59. Volume of the solid generated by rotating the region bounded by the x -axis and y f ( x) from x a to
b
x b about the x-axis is V [ f ( x )]2 dx 4 , and the volume of the solid generated by rotating the same
a
b
b
a
a
b
region about the line y 1 is V [ f ( x) 1]2 dx 8 . Thus f ( x) 1 dx f ( x) dx 8 4
b
a
2
2
a
f ( x) 2 f ( x) 1 f ( x) dx 4 (2 f ( x) 1) dx 4 2 f ( x) dx dx 4
2
b
b
b
a
a
a
2
b
b
a
a
f ( x) dx 12 (b a ) 2 f ( x) dx 4b2 a
60. Volume of the solid generated by rotating the region bounded by the x-axis and y f ( x) from x a to x b
b
about the x-axis is V f ( x) dx 6 , and the volume of the solid generated by rotating the same
2
a
b
region about the line y 2 is V f ( x) 2 dx 10 . Thus
2
a
a f ( x) 2 dx a f ( x) dx 10 6 a f ( x) 4 f ( x) 4 f ( x) dx 4
b
b
2
b
2
2
2
b
b
b
b
b
a
a
a
a
a
(4 f ( x) 4) dx 4 4 f ( x) dx 4 dx 4 f ( x) dx (b a) 1 f ( x) dx 1 b a
6.2
VOLUMES USING CYLINDRICAL SHELLS
1. For the sketch given, a 0, b 2;
b
shell
V 2 radius
a
shell
height
dx 2 x 1 dx 2 x dx 2 2
2
0
x2
4
2
0
x3
4
2 3 6
Copyright 2016 Pearson Education, Ltd.
x2
2
2
x4
16 0
4
2
16
16
Section 6.2 Volumes Using Cylindrical Shells
2. For the sketch given, a 0, b 2;
b
shell
V 2 radius
a
shell
height
dx 2 x 2 dx 2 2x dx 2 x 2 4 1 6
2
2
x2
4
0
x3
4
0
2
2
x4
16 0
3. For the sketch given, c 0, d 2;
d
shell
V 2 radius
c
shell
height
2
dy 2 y y dy 2 y dy 2 2
2
2 3
2
0
0
y4
4
0
4. For the sketch given, c 0, d 3;
d
shell
V 2 radius
c
shell
height
3
5. For the sketch given, a 0, b 3;
b
shell
V 2 radius
a
shell
height
dx 2 x x 1 dx;
3
2
0
u x 2 1 du 2 x dx; x 0 u 1, x 3 u 4
4
4
V u1/2 du 23 u 3/2 23 43/2 1 23 (8 1) 143
1
1
6. For the sketch given, a 0, b 3;
b
shell
V 2 radius
a
shell
height
dx 2 x
3
0
dx;
x 9
9x
3
[u x3 9 du 3 x 2 dx 3 du 9 x 2 dx; x 0 u 9, x 3 u 36]
36
36
V 2 3u 1/2 du 6 2u1/2 12
9
9
36 9 36
7. a 0, b 2;
b
shell
V 2 radius
a
2
shell
height
dx 2 x x dx
2
2
shell
height
dx 2 x 2 x dx
2
x
2
0
2 x 2 32 dx 3x 2 dx x3 8
0
0
0
8. a 0, b 1;
b
shell
V 2 radius
a
1
1
0
dx 3x dx x
2 32x
0
2
1
0
2
3
3
y4
2 y 3 3 y 2 dy 2 y 3 dy 2 4 92
0
0
0
dy
x
2
3 1
0
Copyright 2016 Pearson Education, Ltd.
367
368
Chapter 6 Applications of Definite Integrals
9. a 0, b 1;
b
shell
V 2 radius
a
1
shell
height
dx 2 x (2 x) x dx
1
2
0
1
2 2 x x 2 x3 dx 2 x 2 x3 x4
0
0
3
4
5
2 1 13 14 2 121243 10
12
6
10. a 0, b 1;
dx 2 x 2 x x dx
2 x 2 2 x dx 4 x x dx
b
shell
V 2 radius
a
1
1
shell
height
1
2
0
2
2
0
3
0
1
2
4
4 x2 x4 4 12 14
0
11. a 0, b 1;
b
shell
V 2 radius
a
1
shell
height
dx 2 x x (2 x 1) dx
1
0
1
2 x3/2 2 x 2 x dx 2 52 x5/2 23 x3 12 x 2
0
0
20 15 7
2 52 23 12 2 1230
15
12. a 1, b 4;
dx 2 x x dx
3 x dx 3 x 2 4 1
b
shell
V 2 radius
a
4
shell
height
4 1/2
2
3
1
3
2
1
3/2 4
1
1/2
3/2
2 (8 1) 14
13. (a)
x sin x , 0 x
sin x, 0 x
x
x f ( x)
x f ( x)
; since sin 0 0 we have
x0
0,
x0
x,
sin x, 0 x
x f ( x)
x f ( x) sin x, 0 x
x0
sin x,
b
(b) V 2
a
shell
radius
shell
height
dx 2 x f ( x) dx and x f ( x) sin x, 0 x by part (a)
0
V 2 sin x dx 2 cos x 0 2 ( cos cos 0) 4
0
Copyright 2016 Pearson Education, Ltd.
Section 6.2 Volumes Using Cylindrical Shells
14. (a)
2
x tanx x , 0 x 4
x g ( x)
x0
x 0,
tan 2 x, 0 x /4
x g ( x)
; since tan 0 0 we have
x0
0,
tan 2 x, 0 x /4
x g ( x)
x g ( x) tan 2 x, 0 x /4
2
x0
tan x,
dx 2 x g ( x) dx and x g ( x) tan x, 0 x /4 by part (a)
V 2
tan x dx 2 sec x 1 dx 2 tan x x
2 1
b
(b) V 2
a
shell
radius
/4
2
/4
shell
height
2
0
/4
0
/4
2
0
0
15. c 0, d 2;
dy 2 y y ( y) dy
2 y y dy 2
d
shell
V 2 radius
c
2
3/2
2
shell
height
0
5/ 2
y3
3
2y
5
2
0
2
0
2 23 2 8 52 83 16 52 13
3 2 5
16
15
2 52
5
3
16. c 0, d 2;
dy 2 y y ( y) dy
2 y y dy 2 16
d
shell
V 2 radius
c
2
3
2
shell
height
y4
4
2
0
2
0
y3
3
2
2
4
0
1
3
16 56 403
17. c 0, d 2;
dy 2 y 2 y y dy
2 2 y y dy 2
2
d
shell
V 2 radius
c
2
2
2
shell
height
2 y3
3
3
0
2
0
y4
4
2
0
16
3
16
4
8
32 13 14 32
12
3
18. c 0, d 1;
dy 2 y 2 y y y dy
2 y y y dy 2 y y dy
d
shell
V 2 radius
c
1
1
shell
height
1
2
0
2
0
2
3
0
1
y3 y 4
2 3 4 2 13 14 6
0
Copyright 2016 Pearson Education, Ltd.
4
4
2
2
369
370
Chapter 6 Applications of Definite Integrals
19. c 0, d 1;
d
shell
V 2 radius
c
shell
height
dy 2 y y ( y) dy
1
0
1
1
2 2 y 2 dy 43 y3 43
0
0
20. c 0, d 2;
d
shell
V 2 radius
c
dy 2 y y dy
2
shell
height
y
2
0
2
2
2y
2 2 dy 3 y 3 83
0
0
21. c 0, d 2;
d
shell
V 2 radius
c
dy 2 y (2 y) y dy
2
shell
height
2
0
2
y
y
2 y y 2 y 3 dy 2 y 2 3 4
0
0
2
3
2
4
2 4 83 16
6 (48 32 48) 163
4
22. c 0, d 1;
d
shell
V 2 radius
c
shell
height
dy 2 y (2 y) y dy
1
2
0
1
1
y3 y 4
2 2 y y 2 y 3 dy 2 y 2 3 4
0
0
2 1 13 14 6 (12 4 3) 56
(b) V 2
b
shell
radius
23. (a) V 2
a
b
a
shell
radius
shell
height
dx 2 x (3x)dx 6 x dx 2 x 16
dx 2 (4 x) (3x)dx 6 4x x dx 6 2x x
shell
height
dx 2 ( x 1) (3x)dx 6 x x dx 6 x x
shell
height
dy 2 y 2 y dy 2 2 y y dy 2 y y
shell
height
2
3 2
0
2 2
0
0
2
2
0
0
2
2
2
1 3
3
0
6 8 83 32
b
(c) V 2
a
shell
radius
2
2
0
0
2
2
1 2
2
0
1 3
3
6 83 2 28
d
(d) V 2
c
shell
radius
6
0
1
3
6
0
1 2
3
2 (36 24) 24
Copyright 2016 Pearson Education, Ltd.
2
6
1 3
9
0
Section 6.2 Volumes Using Cylindrical Shells
d
(e) V 2
c
shell
radius
dy 2 (7 y) 2 y dy 2 14 y y dy
6
shell
height
6
1
3
0
1 2
3
13
3
0
6
2 14 y 13
y 2 19 y3 2 (84 78 24) 60
6
0
d
(f ) V 2
c
shell
radius
dy 2 ( y 2) 2 y dy 2 4 y y dy
6
shell
height
6
1
3
0
1 2
3
4
3
0
6
2 4 y 23 y 2 19 y3 2 (24 24 24) 48
0
b
24. (a) V 2
a
shell
radius
2 16 32
5
b
(b) V 2
a
dx 2 x 8 x dx 2 8x x dx 2 4x x
2
shell
height
0
965
shell
radius
2
3
4
2
1 5
5
0
2
0
dx 2 (3 x) 8 x dx 2 24 8x 3x x dx
2
shell
height
2
3
0
3
4
0
2
2 24 x 4 x 2 34 x 4 15 x5 2 48 16 12 32
264
5
5
0
b
(c) V 2
a
shell
radius
dx 2 ( x 2) 8 x dx 2 16 8x 2x x dx
2
shell
height
2
3
0
3
4
0
2
2 16 x 4 x 2 12 x 4 15 x5 2 32 16 8 32
336
5
5
0
(e) V 2
d
shell
radius
(d) V 2
c
d
shell
radius
c
dy 2 y y dy 2 y dy y (128)
dy 2 (8 y) y dy 2 8 y y dy 2 6 y y
8
shell
height
1/3
0
8
shell
height
8 4/3
0
6
7
8
1/3
1/3
0
7/3 8
0
6
7
768
7
4/3
4/3
0
3
7
7/3 8
0
2 96 384
576
7
7
d
(f ) V 2
c
shell
radius
dy 2 ( y 1) y dx 2 y y dy 2 y y
8
shell
height
8
1/3
0
4/3
1/3
3 7/3
7
0
3
4
4/3 8
0
2π 384
12 936π
7
7
b
25. (a) V 2
a
shell
radius
shell
height
dx 2 (2 x) x 2 x dx 2 4 3x x dx
2
2
2
1
2
3
1
2
2 4 x x3 14 x 4 2 (8 8 4) 2 4 1 14 272
1
b
(b) V 2
a
shell
radius
dx 2 ( x 1) x 2 x dx 2 2 3x x dx
2
shell
height
2
2
1
3
1
2
2 2 x 32 x 2 14 x 4 2 (4 6 4) 2 2 23 14 272
1
dy 2 y y y dy 2 y y ( y 2) dy
4 y dy 2 y y 2 y dy y 2 y y y
d
(c) V 2
c
1 3/2
0
shell
radius
1
shell
height
4
0
3/2
1
4
1
2
8
5
5/2 1
0
2 5/2
5
85 (1) 2 64
64
16 2 52 13 1 725
5
3
c
1
0
shell
radius
2 4
1
dy 2 (4 y) y y dy 2 (4 y) y ( y 2) dy
4 4 y y dy 2 y y 6 y 4 y 8 dy
d
(d) V 2
1 3
3
3/2
shell
height
4
1
4
0
1
2
3/2
1
Copyright 2016 Pearson Education, Ltd.
371
372
Chapter 6 Applications of Definite Integrals
1
4
4 83 y3/2 52 y 5/2 2 13 y 3 52 y 5/2 3 y 2 83 y 3/2 8 y
0
1
shell
radius
4 83 52 2 64
64
48 64
32 2 13 52 3 83 8 1085 .
3
5
3
b
26. (a) V 2
a
shell
height
dx 2 (1 x) 4 3x x dx 2 x x 3x 3x 4 x 4 dx
1
2
1
4
1
5
4
3
2
1
1
2 16 x6 15 x5 34 x 4 x3 2 x 2 4 x 2 16 15 43 1 2 4 2 16 15 34 1 2 4 565
1
d
(b) V 2
c
shell
radius
shell
height
1
4
4 y5/4 dy 4
y
0
3 1
1
dy 2 y y y dy 2 y
1
4
0
4 y
4 y
3 dy
3
4
4
1
4 ydy [u 4 y y 4 u du du; y 1 u 3, y 4 u 0]
0
3
3
169 y 9/4 4 (4 u ) u du 169 (1) 4 4 u u 3/2 du 169 4 83 u 3/2 52 u 5/2
0
0
3 3
3 0
3
3
27. (a) V 2
169 4 8 3 18
3 169 885 872
5
45
d
c
shell
radius
shell
height
6
24 14 15 24
20
5
d
(b) V 2
c
shell
radius
shell
height
1
1
1
y 4 y5
dy 2 y 12 y 2 y 3 dy 24 y 3 y 4 dy 24 4 5
0
0
0
dy 2 (1 y) 12 y y dy 24 (1 y) y y dy
1
2
1
3
0
2
3
0
1
1
y3 y 4 y5
1 4
24 y 2 2 y 3 y 4 dy 24 3 2 5 24 13 12 15 24 30
5
0
0
dy 2 y 12 y y dy 24 y y y dy
24 y y y dy 24 y y 24
d
(c) V 2
c
shell
radius
1 8 2
0 5
1
shell
height
13 3
5
0
8
5
4
8 3
15
13
20
2
3
4
y5
5
1 8
0 5
1
8
15
0
2
13
20
3
1
5
(32 39 12) 24 2
24
60
12
(d) V 2
dy 2 y 12 y y dy 24 y y y dy
24 y y y y dy 24 y y y dy 24 y y
1
3
d
c
shell
radius
4
2
5
0
24
d
c
shell
radius
2
2
2
5
0
1 2
0 5
2 3
5
shell
height
2
1
3
0
(8 9 12) 24 2
24
60
12
2 3 1
15 20 5
28. (a) V 2
1
shell
height
2
3 3
5
4
2
5
2
2 3
15
3
20
3
y5
5
4
1
0
dy 2 y dy 2 y y dy 2 y dy
y2
2
2
0
y4
4
y2
2
2
2
0
y4
4
2
3
0
4
y4 y6
26 32 1 4 32 1 1 32 2 8
2 4 24 2 24 24
4 24
4 6
24
3
0
d
(b) V 2
c
shell
radius
shell
height
dy 2 (2 y) dy 2 (2 y) y dy
2
0
y2
2
y4
4
y2
2
2
2
2
0
y4
4
2
y4
y5
2 y3 y5 y 4 y 6
2 2 y 2 2 y 3 4 dy 2 3 10 4 24 2 16
32 16
64
85
3 10
4
24
0
0
Copyright 2016 Pearson Education, Ltd.
y5
4
Section 6.2 Volumes Using Cylindrical Shells
d
(c) V 2
c
shell
radius
373
dy 2 (5 y) dy 2 (5 y) y dy
y2
2
2
shell
height
0
y4
4
y2
2
0
y4
4
2
2
2
2
y5
5 y3 5 y5 y 4 y 6
2 5 y 2 54 y 4 y 3 4 dy 2 3 20 4 24 2 40
160
16
64
8
3
20
4
24
0
0
d
(d) V 2
c
shell
radius
dy 2 y dy 2 y y dy
2
shell
height
y2
2
5
8
0
y4
4
y2
2
2
0
2
y4
4
2
5
8
4
6
3
5
2
y5
5 y 4 dy 2 y y 5 y 5 y 2 16 64 40 160 4
2 y 3 4 85 y 2 32
4
24
24
160
4
24 24 160
0
0
dy
2 y y y dy 2 y y dy
d
29. (a) About x-axis: V 2
c
shell
height
3/2
2
shell
radius
1
1
0
0
shell
radius
1
2 52 y 5/2 13 y 3 2 52 13 215
0
b
About y -axis: V 2
1
a
1
shell
height
dx
2 x x x 2 dx 2 x 2 x3 dx
0
0
1
3
4
2 x3 x4 2 13 14 6
0
b
(b) About x-axis: R ( x) x and r ( x) x 2 V
a
1
R( x) r ( x) dx x x dx
2
2
4
0
3
5
x3 x5 13 15 215
0
About y -axis: R( y )
1
1
2
d
y and r ( y ) y V
c
R( y) r ( y) dy y y dy
2
1
2
2
0
y 2 y3
2 3 12 13 6
0
b
30. (a) V
a
R( x) r( x) dx 2 x dx
2
4
2
2
x
2
0
2
3 x 2 2 x 4 dx x4 x 2 4 x
0 4
0
4
4
3
16 16 16 16
b
(b) V 2
a
shell
radius
shell
height
4
dx 2 x 2 x dx 2 x 2 dx 2 2x dx
4
0
4
x
2
x
2
0
4
0
x2
2
2 x 2 x6 2 16 64
323
6
0
3
b
(c) V 2
a
shell
radius
shell
height
4
dx 2 (4 x) 2 x dx 2 (4 x) 2 dx 2 8 4x dx
4
x
2
0
4
0
2 8 x 2 x 2 x6 2 32 32 64
643
6
0
3
Copyright 2016 Pearson Education, Ltd.
x
2
4
0
x2
2
374
Chapter 6 Applications of Definite Integrals
b
(d) V R ( x) r ( x)
a
2
2
4 3 2
x 10 x 28
0 4
d
31. (a) V 2
c
shell
radius
dx (8 x) 6 dx 64 16 x x 36 6 x dx
4
x 2
2
2
0
4
x2
4
2
0
dx x4 5x2 28x 0 [16 (5) (16) (7) (16)] (3) (16) 48
4
3
shell
height
dy 2 y( y 1) dy
2
1
2
y
y
y 2 y dy 2 3 2
1
1
3
2
2
2
2 73 2 12 3 (14 12 3) 53
2 83 42 13 12
b
(b) V 2
a
shell
radius
shell
height
dx 2 x(2 x) dx 2 2x x dx 2 x
2
2
1
1
2
2
2
x3
3 1
b
shell
shell
(c) V 2 radius
height
dx 12 2 103 x (2 x) dx 2 12 203 163 x x2 dx
a
2
8 x 2 1 x3 2 40 32 8 20 8 1 2 3 2
2 20
x
3
3
3
3
1
3 3 3 3 3 3
2 4 83 1 13 2 1238 331 2 34 32 43
d
(d) V 2
c
d
32. (a) V 2
c
shell
radius
shell
radius
2
shell
height
shell
height
dy 2 y y 0 dy
2
2
( y 1)3
dy 2 ( y 1)( y 1) dy 2 ( y 1)2 2 3 23
1
1
1
2
2
0
2
4
2
y4
2 y3 dy 2 4 2 24 8
0
0
b
(b) V 2
a
shell
radius
4
shell
height
dx
2 x 2 x dx 2
0
4
4
0
2 x x3/2 dx
5
2 x 2 52 x5/2 2 16 252
0
2 16 64
25 (80 64) 325
5
b
(c) V 2
a
shell
radius
shell
height
dx 2 (4 x) 2 x dx 2 8 4x 2x x dx
4
0
4
2 8 x 83 x3/2 x 2 52 x5/2 2
0
c
16 16
3
4
d
(d) V 2
2
shell
radius
shell
height
4
1/2
3/2
0
32 643 16 645 215 (240 320 192) 215 (112) 22415
dy 2 (2 y) y dy 2 2 y y dy 2 y
2
0
2
2
2
3
0
(4 3) 8
32
12
3
Copyright 2016 Pearson Education, Ltd.
2 3
3
y4
4
2
0
Section 6.2 Volumes Using Cylindrical Shells
dy 2 y y y dy
2 y y dy 2 2
(b) V 2
dy 2 (1 y) y y dy
2 y y y y dy 2
d
33. (a) V 2
c
1
shell
radius
2
4
375
1
shell
height
y3
3
0
d
c
shell
radius
1
2
y5
5
2
d
34. (a) V 2
c
shell
radius
1
3
0
4
15
1
5
3
0
y2
2
4
0
1111
2 3 4 5
1
1
shell
height
3
3
0
y3
3
y4
4
y5
5
1
0
260 (30 20 15 12) 730
shell
height
dy 2 y 1 y y dy
1
3
0
1
1
y 2 y3 y5
2 y y 2 y 4 dy 2 2 3 5
0
0
2 12 13 15 230 (15 10 6) 11
15
(b) Use the washer method:
d
V R( y ) r ( y )
c
2
2
dy 1 y y dy 1 y y 2 y dy
1
3 2
2
1
0
1
2
6
4
0
y3 y7 2 y5
(105 35 15 42) 97
y 3 7 5 1 13 71 52 105
105
0
(c) Use the washer method:
2
2
d
1
1
2
2
V R( y ) r ( y ) dy 1 y y 3 0 dy 1 2 y y 3 y y 3 dy
c
0
0
1
1
y3
y7
y 4 2 y5
1 y 2 y 6 2 y 2 y 3 2 y 4 dy y 3 7 y 2 2 5
0
0
(70 30 105 2 42) 121
1 13 71 1 12 52 210
210
dy 2 (1 y) 1 y y dy 2 (1 y) 1 y y dy
2 1 y y y y y dy 2 1 2 y y y y dy 2 y y
d
(d) V 2
c
shell
radius
1
1
shell
height
3
1
3
0
2
1
4
0
2
3
4
0
2 1 1 13 14 15
d
35. (a) V 2
c
shell
radius
260 (20 15 12) 23
30
shell
height
dy 2 y 8 y y dy
1
2
0
2
y4
2 2 y 3/2 y 3 dy 2 4 5 2 y 5/2 4
0
0
2
2
4 2 2 5
24 2 423 44
2
5
4
5
4
3
0
2 4 85 1 85 (8 5) 245
Copyright 2016 Pearson Education, Ltd.
2
y3
3
y4
4
y5
5
1
0
376
Chapter 6 Applications of Definite Integrals
b
(b) V 2
a
shell
radius
shell
height
dx 2 x x dx 2 x dx 2 x
4
4
x2
8
0
3/2
0
x3
8
4
x4
32 0
2 5/2
5
5
44 2 26 28 27 (32 20) 29 3 24 3 48
2 252 32
5
32
160
160
5
5
dx 2 x 2 x x x dx
2 x x x dx 2 x x dx
b
36. (a) V 2
a
1
shell
radius
1
shell
height
2
0
1
2
0
2
3
0
1
3
4
2 x3 x4 2 13 14 6
0
b
(b) V 2
a
shell
radius
shell
height
dx 2 1 x 2 x x x dx 2 1 x x x dx
1
1
2
0
2
0
1
2
4
1
2 x 2 x 2 x3 dx 2 x2 32 x3 x4 2 12 23 14 212 (6 8 3) 6
0
0
b
37. (a) V R( x) r ( x)
a
2
2
dx x
1/2
1
1/16
1
1 dx
1
2 x1/2 x
(2 1) 2 14 16
1/16
7 9
1 16
16
d
(b) V 2
c
shell
radius
shell
height
dy 2 y dy
1
1
y4
0
1
16
2
y
y2
y 3 16 dy 2 12 y 2 32
1
1
2
2
1 2 1
2 18 12 32
32
2 (8 1) 9
d
38. (a) V R ( y ) r ( y )
c
2
2
2
dy dy
2
1
1
y4
1
16
y
1 1 1 1
13 y 3 16 24
8
3 16
1
( 2 6 16 3) 11
48
48
b
(b) V 2
a
shell
radius
shell
height
dx 2 1 dx 2 x x dx 2 x
1
1/4
1
2 23 12 23 18 32
1
x
1
1/2
1/4
34 1 16 161 48 (4 16 48 8 3) 1148
Copyright 2016 Pearson Education, Ltd.
2 3/2
3
1
x2
2 1/4
Section 6.2 Volumes Using Cylindrical Shells
377
39. (a) Disk: V V1 V2
b
b
V1 1 R1 ( x) dx and V2 2 R2 ( x) dx with R1 ( x)
2
a1
2
a2
x2
3
and R2 ( x) x ,
a1 2, b1 1; a2 0, b2 1 two integrals are required
(b) Washer: V V1 V2
dx with R ( x)
V R ( x) r ( x) dx with R ( x)
b
V1 1 R1 ( x) r1 ( x)
2
2
1
a1
b2
2
2
2
a2
2
2
two integrals are required
d
(c) Shell: V 2
c
shell
radius
shell
height
x2
3
x 2
3
2
dy 2 y
d
c
and r1 ( x) 0; a1 2 and b1 0;
shell
height
and r2 ( x) x ; a2 0 and b2 1
dy where shell height y 3 y 2 2 2 y ;
2
2
2
c 0 and d 1. Only one integral is required. It is, therefore preferable to use the shell method.
However, whichever method you use, you will get V .
40. (a) Disk: V V1 V2 V3
d
Vi i Ri ( y ) dy, i 1, 2, 3 with R1 ( y ) 1 and c1 1, d1 1; R2 ( y )
2
ci
y and c2 0 and d 2 1;
R3 ( y ) ( y )1/4 and c3 1, d3 0 three integrals are required
(b) Washer: V V1 V2
d
Vi i Ri ( y ) ri ( y )
ci
2
2
dy, i 1, 2 with R ( y) 1, r ( y) y , c 0 and d 1;
1
1
1
1
R2 ( y ) 1, r2 ( y ) ( y )1/4 , c2 1 and d 2 0 two integrals are required
b
(c) Shell: V 2
a
shell
radius
shell
height
dx 2 x
b
a
shell
height
dx , where shell height x x x x , a 0
2
4
2
4
and b 1 only one integral is required. It is, therefore preferable to use the shell method.
However, whichever method you use, you will get V 56 .
b
41. (a) V R ( x) r ( x)
a
2
4
2
2
4
4
4
dx 25 x 2 (3)2 dx 25 x 2 9 dx 16 x 2 dx
4
4
4
16 x 13 x3 64 64
64 64
256
3
3
3
4
Volume of portion removed 500 256 244
(b) Volume of sphere 43 (5)3 500
3
3
3
3
b
42. V 2
a
shell
radius
shell
height
dx 2 x sin x 1 dx; [u x 1 du 2x dx;
1
2
2
1
x 1 u 0, x 1 u ] sin u du cos u 0 ( 1 1) 2
0
Copyright 2016 Pearson Education, Ltd.
378
Chapter 6 Applications of Definite Integrals
b
43. V 2
a
shell
radius
2
2
shell
height
dx 2 x x h dx 2 x h x dx 2 x x
r
r
h
r
0
0
h 2
r
h 3
3r
r
h 2
2
0
2
2 r 3h r 2h 13 r h
d
44. V 2
c
shell
radius
shell
height
dy 2 y r y r y dy 4 y r y dy
r
2
2
2
r
2
0
2
2
0
0
r2
r2
[u r 2 y 2 du 2 y dy; y 0 u r 2 , y r u 0] 2 2 u du 2 u1/2 du 43 u 3 2
0
r
0
43 r 3
6.3
1.
ARC LENGTH
x2 2 2 x x2 2 x
3
3
L 1 x 2 2 x 2 dx 1 2 x 2 x 4 dx
0
0
3
2
3
3
1 x 2 dx 1 x 2 dx x x3
0
0
0
1/2
dy 1 3
32
dx
3
3 27
12
3
2.
dy
32
dx
xL
4
1 94 x dx;
0
u 1 9 x du 9 dx 4 du dx;
4
4
9
x 0 u 1; x 4 u 10]
10
10
8 10 10 1
L u1/2 94 du 94 23 u 3/2 27
1
1
3.
y
dx y 2 1 dx
dy
dy
4 y2
L
3
1
3
1
4
1
2
1
16 y 4
3
1 y 4 12 1 4 dy y 4 12 1 4 dy
16 y
1
16 y
2
3 2
2
1
1
y 4 y 2 dy 1 y 4 y 2 dy
3
y 3 y 1
3 4
1
9
2
273 121 13 14 9 121 13 14
( 1 4 3)
( 2)
9 12 53
12
6
Copyright 2016 Pearson Education, Ltd.
Section 6.3 Arc Length
4.
y2
L 1 y 2 dy
y 2 dy
2
dx 1 y1/2 1 y 1/2 dx
dy
2
2
dy
9
1
4
1
1
4
1
y
9
1
y
1
2
1
4
1
y
9
9
12 y 1 dy 12 y1/2 y 1/2 dy
1
1
y
9
9 y 3/ 2
12 32 y 3/2 2 y1/2 3 y1/2
1
1
3
33 3 13 1 11 13 32
3
5.
y
dx y 3 1 dx
dy
dy
4 y3
L
2
1
2
1
2
1 y 6 12
6
1
2
1
16 y 6
1 dy 2
1
16 y 6
y 6 12
1 dy
16 y 6
2
2
2 3 y 3
3 y 3
y 4 y 2
y 4 dy 1 y 4 dy 4 8
1
1
1 11
16
(16)(2)
14 18 4 32
4
4 8
18 4 123
12832
32
6.
y 2 y
2
dx y 1 dx
dy
2
dy
2 y2
L
3
2
2
4
4
1
4
1 14 y 4 2 y 4 dy
y4 2 y 4 dy
2
3
3
12 y 2 y 2 dy 12 y 2 y 2 dy
2
2
3
2
1
4
3
y3
12 3 y 1 12 27
1 83 12
3 3
2
12 26
83 12 12 6 12 13
3
4
7.
2/3
dy
dy 2
1/3 1 1/3
x
x
x 2/3 12 x16
dx
4
dx
L
8
1
8
1
2/3
1 x 2/3 12 x16 dx
2/3
x 2/3 12 x16 dx
8
1
x1/3 14 x1/3 dx
2
x1/3 14 x1/3 dx 34 x4/3 83 x2/3 1
8
83 2 x 4/3 x 2/3 83 2 24 22 (2 1)
1
8
8
1
83 (32 4 3) 99
8
Copyright 2016 Pearson Education, Ltd.
379
380
8.
Chapter 6 Applications of Definite Integrals
dy
x 2 2 x 1 4 2 x 2 2 x 1 14 1 2
dx
(4 x 4)
(1 x )
(1 x)2 14
L
2
0
2
2
1 dy
1
(1 x) 4 12
dx
(1 x )2
16(1 x )4
1 (1 x)4 12
(1 x) 4 12
0
(1 x ) 4
dx
16
(1 x ) 4
dx
16
2
2
2 (1 x )
(1
x
)
dx
4
0
2
2
3
(1 x )2
(1 x )2 4 dx; [u 1 x du dx; x 0 u 1, x 2 u 3] L u 2 14 u 2 du
0
1
3
1 1 1 1081 4 3 106 53
u3 14 u 1 9 12
3 4
12
12
6
1
3
9.
x x
dy
dy 2
x 2 1 2 dx
dx
4x
L
3
1
1 x
4
2
1
4 x2
2
4
1
2
1
16 x 4
3
4
1
10.
3
3
L
1
1/2
1 x
8
4
1
4 x4
2
8
1
2
12 1 8 dx
16 x
1
8
4
1/2 x 41x dx x5 121x 1/2
sec4 y 1
L
/4
/4
y
x5
5
1
12 x 3
0.5
3
0
373
15 121 1601 23 480
dx
dy
x
3
1
5
4
11.
2
y
2
4
4
1
1
8
1
0
1
16 x8
1/2 x 12 161x dx 1/2 x 41x dx
1
1
4x
2
1 1 1 53
9 12
3 3
6
x x
dy
dy 2
x 4 1 4 dx
dx
4x
x3
3
4
2
x 2 1 2 dx x3 41x
4x
1
y
6
2
2
4
3
10
8
12 1 4 dx
16 x
1 x 12 161x dx 1 x 41x dx
3
y
0.5
sec y 1
dx
dy
2
4
1 sec 4 y 1 dy
/4
/4
sec 2 y dy
/4
tan y /4 1 (1) 2
Copyright 2016 Pearson Education, Ltd.
1
1.5
x
Section 6.3 Arc Length
12.
dy
dx
3x 1
dy 2
3x 4 1 dx
1
L
4
1 3x 4 1 dx
2
1
2
3 x 2 dx
1
3 x3 33 1 (2)3 33 (1 8) 7 3 3
2
3
13. (a)
L 1 dx
dy 2
dx
2
1
2
1
(c)
14. (a)
1 4 x 2 dx
L 6.13
dy
dy 2
sec2 x dx sec4 x
dx
0
L
/3
(c)
15. (a)
cos y
2
(b)
2
1 cos 2 y dy
L 3.82
dx y dx
dy
dy
1 y 2
L
1/2
1/2
1/2
1/2
(c)
1 sec 4 x dx
dx cos y dx
dy
dy
0
16. (a)
(b)
L 2.06
L
(c)
(b)
dy
dy 2
2
x
4 x2
dx
dx
1
1 y 2
y
2
2
1 y 2
1/2
(b)
y2
1 y
2
dy
1/2
1/2
1 dy
1 y 2
dy
L 1.05
Copyright 2016 Pearson Education, Ltd.
381
382
Chapter 6 Applications of Definite Integrals
dx
17. (a) 2 y 2 2 dy
3
L
18. (a)
(b)
(b)
dy
dy 2
cos x cos x x sin x dx x 2 sin 2 x
dx
0
19. (a)
2
L 9.29
L
(c)
2
dx
dy
1 ( y 1) 2 dy
1
(c)
( y 1)
1 x 2 sin 2 x dx
L 4.70
dy
dy 2
tan
x
tan 2 x
dx
dx
L
/6
0
(b)
1 tan 2 x dx
/6
0
sin 2 x cos 2 x dx
cos 2 x
/6 dx
/6
sec x dx
0 cos x
0
(c)
20. (a)
L 0.55
dx
dy
sec y 1
dx
sec2 y 1 dy
L
/4
/3
/4
/3
(c)
21. (a)
2
(b)
2
1 sec 2 y 1 dy
/4
/3
| sec y | dy
sec y dy
L 2.20
corresponds to
dy 2
dx
1 here, so take dy as 1 . Then y
4x
dx
2 x
x C and since (1, 1) lies on the curve,
C 0. So y x from (1, 1) to (4, 2).
(b) Only one. We know the derivative of the function and the value of the function at one value of x.
22. (a)
corresponds to
dx
dy
2
1
y4
dy
here, so take dx as
1 . Then x 1 C and, since (0, 1) lies on the curve,
y
y2
C 1. So y 11x .
(b) Only one. We know the derivative of the function and the value of the function at one value of x.
Copyright 2016 Pearson Education, Ltd.
Section 6.3 Arc Length
23.
x
y
dy
cos 2t dt dx cos 2 x L
0
/4
/4
2 cos x dx 2 sin x 0
0
24.
2/3 3/2
1
1 1 x2/3 dx
y 1 x
,
2
1 cos 2 x dx
2 /4
x
1
2/3
3
2
3 1
2 2
1
x
2/3
1 dx
32 x 1/3
1
x
1 x
2/3
2 cos 2 x dx
2/3
1
2
1/ 2
L
x1/3
dx
1
2 /4
3
4
1
2 /4
1 x 2/3 1/ 2
dx
1
x1/3
1
1 dx 1
x 1/3 dx 32 x 2/3
2 /4
2 /4 x1/3
2 /4
3
4
dy
y 3 2 x, 0 x 2 dx 2 L
2
0
2
/4
0
total length 8 6
2
4
1 cos 2 x dx
2 sin 4 2 sin(0) 1
1
/4
0
1/2
dy
2
x 1 dx 23 1 x 2/3
4
2/3
2 /4
32 (1) 2/3 32
25.
/4
0
383
1 (2) 2 dx
2
0
2
5 dx 5 x 2 5.
0
2
d (2 0) (3 ( 1)) 2 5
26. Consider the circle x 2 y 2 r 2 , we will find the length of the portion in the first quadrant, and multiply our
result by 4.
dy
y r 2 x 2 , 0 x r dx
r
r
0
r 2 x2
4
r
dx
0
r 2 x2
dx 4r
x
r2 x
2
2
r
r
r 2 dx
1 x dx 4 1 2x 2 dx 4
2
2
2
0
0
0 r x2
r x
r x
L 4
2
r
d 9 x 2 d y ( y 3) 2 18 x dx 2 y ( y 3) ( y 3)2 3( y 3)( y 1)
27. 9 x 2 y ( y 3)2 dy
dy
dy
dx
dy
( y 3)( y 1)
( y 3)( y 1)
dx
dy;
6x
6x
( y 3) 2 ( y 1)2
4 y ( y 3)
2
2
2
2
( y 3)( y 1)
( y 3) ( y 1)
ds 2 dx 2 dy 2
dy dy 2
dy 2 dy 2
6x
36 x 2
y 2 2 y 1 4 y
( y 1) 2
( y 1)2
dy 2 dy 2 4 y 1 dy 2
dy 2 4 y dy 2
4
y
d 4 x 2 y 2 d 64 8 x 2 y dy 0 dy 4 x dy 4 x dx;
28. 4 x 2 y 2 64 dx
dx
dx
y
y
dx
2
2
2
2
2
y 2 16 x 2
ds 2 dx 2 dy 2 dx 2 4yx dx dx 2 16 x2 dx 2 1 16 x2 dx 2
dx 2 4 x 64216 x dx 2
2
y
y
y
y
2
20 x 264 dx 2 42 (5 x 2 16) dx 2
y
29.
y
2x
x
0
1
dt, x 0 2 1 1 y f ( x) x C where C is any real
dy 2
dt
dy 2
dx
dy
dx
number.
Copyright 2016 Pearson Education, Ltd.
384
Chapter 6 Applications of Definite Integrals
30. (a) From the accompanying figure and definition of the
differential (change along the tangent line) we see
that dy f ( xk 1 ) xk length of kth tangent fin is
xk 2 (dy )2 xk 2 f ( xk 1 ) xk 2 .
n
n
n k 1
n k 1
xk 2 f ( xk 1 ) xk 2
(b) Length of curve lim (length of kth tangent fin) lim
n
lim 1 f ( xk 1 ) xk
2
b
a
n k 1
1 f ( x) dx
2
4
31.
x 2 y 2 1 y 1 x 2 ; P 0, 14 , 12 , 43 , 1 L
k 1
xi xi 1 2 yi yi 1 2
14 0 415 1 12 14 23 415 43 12 47 23 1 34 0 47
2
2
2
2
2
2
2
2
1.55225
y y
2
x
x
x1
1
L 2 1 m 2 dx 1 m 2 x x2 1 m 2 x2 x1 1
x2 x1 2 y2 y1 2
x2 x1 2 y2 y1 2
x
x
x2 x1
2
1
x2 x1
x2 x1 2
33.
dy
y 2 x3/2 dx 3 x1/2 ; L( x)
x
0
1 3t1/2
3
dy
x
0
x
0
2
1
2
19 x
19 x
2 u 3/2
u du 27
1
2 (1 9 x)3/2 2 ;
27
27
2(10 10 1)
27
y x3 x 2 x 4 x1 4 dx x 2 2 x 1
L( x)
x x
y2 y1 2
x2 x1
x2 x1 2 y2 y1 2 .
1
34.
1
dt 0x 1 9t dt;
[u 1 9t du 9dt ; t 0 u 1, t x u 1 9 x] 19
2 (10)3 2 2
L(1) 27
27
dy
x1 , lie on y mx b, where m x2 x1 , then dx m
32. Let ( x1 , y1 ) and ( x2 , y2 ), with x2
1
( x 1)2 1 2 ;
4( x 1)2
4( x 1)
2
2
x
x
[4(t 1)4 1]2
4(t 1)4 1
1 (t 1)2 1 2 dt 1
dt
1
dt
2
0
0
4(t 1)
16(t 1)4
4(t 1)
16(t 1)4 16(t 1)8 8(t 1) 4 1
16(t 1)
4
dt
x
0
16(t 1)8 8(t 1)4 1
16(t 1)
4
dt
x
0
[4(t 1)4 1]2
16(t 1)
4
dt
x 4(t 1)4 1
0 4(t 1) 2
dt
x
x 1
(t 1)2 1 2 dt ; [u t 1 du dt ; t 0 u 1, t x u x 1] u 2 14 u 2 du
1
0
4(t 1)
x 1
13 u 3 14 u 1
1
1 ; L(1) 8 1 1 59
13 ( x 1)3 4( x11) 13 14 13 ( x 1)3 4( x11) 12
3 8 12
24
Copyright 2016 Pearson Education, Ltd.
Section 6.4 Areas of Surfaces of Revolution
35-40.
Example CAS commands:
Maple:
with( plots );
with( Student[Calculus1] );
with( student );
f : x - sqrt(1-x^2);a : -1;
b : 1;
N : [2, 4, 8];
for n in N do
xx : [seq( a i*(b-a)/n, i 0..n )];
pts : [seq([x, f (x)], x xx)];
L : simplify(add( distance(pts[i 1], pts[i]), i 1..n ));
T : sprintf("#35(a) (Section 6.3)\nn %3d L %8.5f \n", n, L );
P[n] : plot( [f (x), pts], x a..b, title T ):
end do:
display( [seq(P[n], n N)], insequence true, scaling constrained );
L : ArcLength( f(x), x a..b, output integral ):
L evalf ( L );
# (b)
# (a)
# (c)
Mathematica: (assigned function and values for a, b, and n may vary)
Clear[x, f ]
{a, b} {1, 1}; f[x_ ] Sqrt[1 x 2 ]
p1 Plot[f[x], {x, a, b}]
n 8;
pts Table[{xn, f[xn]}, {xn, a, b, (b a)/n}]/ / N
Show[p1,Graphics[{Line[pts]}]}]
Sum[ Sqrt[ (pts[[i 1, 1]] pts[[i, 1]]) 2 (pts[[i 1, 2]] pts[[i, 2]]) 2 ], {i, 1, n}]
NIntegrate[Sqrt[ 1 f '[ x]2 ], {x, a, b}]
6.4
AREAS OF SURFACES OF REVOLUTION
1. (a)
dy
dy 2
sec2 x dx sec 4 x
dx
S 2
/4
0
(b)
(tan x) 1 sec 4 x dx
(c) S 3.84
Copyright 2016 Pearson Education, Ltd.
385
386
Chapter 6 Applications of Definite Integrals
2. (a)
dy
dy 2
2
x
4 x2
dx
dx
(b)
2
S 2 x 2 1 4 x 2 dx
0
(c) S 53.23
3. (a)
dx 1
xy 1 x 1y dy
2
dx
dy
y
21
1 y
S 2
2
1
y4
(b)
1 y 4 dy
(c) S 5.02
4. (a)
cos y
2
dx cos y dx
dy
dy
(b)
2
S 2 (sin y ) 1 cos2 y dy
0
(c) S 14.42
5. (a)
x1/2 y1/2 3 y 3 x1/2
(b)
2
12 x1/2
2
dy 2
dx 1 3x 1/2
2
2
4
S 2 3 x1/2 1 1 3x 1/2 dx
1
dy
dx 2 3 x1/2
(c) S 63.37
6. (a)
1 y
dx 1 y 1/2 dx
dy
dy
2
(b)
1/2 2
y 2 y 1 1 y 1/2 dx
1
S 2
2
2
(c) S 51.33
Copyright 2016 Pearson Education, Ltd.
Section 6.4 Areas of Surfaces of Revolution
7. (a)
tan y
2
dx tan y dx
dy
dy
(b)
2
y
/3 y
2
0 tan t dt 1 tan y dy
/3 y
2 tan t dt sec y dy
0 0
(c) S 2.08
S 2
1
0
8. (a)
dy
dx
x2 1
2
5 x
1
(c) S 8.55
1
0.2
3
1
x
0.8
2
t 2 1 dt
1
2
dx S 2 1 dx
5 4
4
y 2x dx 12 ; S 2 y 1 dx
a
0.6
1
dy 2
b
x
y
t 2 1 dt x dx
dy
0.4
tan t dt
y
0
9.
y
0
(b)
t 2 1 dt 1 x 2 1 dx
1
1
0
2
5 x
S 2
x
0.5
x 1
dy 2
dx
387
x
2
0
1
4
3
x
4
x dx 2 x2 4 5;
2 0
0
5
2
Geometry formula: base circumference 2 (2), slant height 42 22 2 5
Lateral surface area 12 (4 ) 2 5 4 5 in agreement with the integral value
10.
dy 2 2 y 1 2 dy 4 5 y dy 2 5 y
d
dx 2; S
y 2x x 2 y dy
2 x 1 dydx
c
2
2
2
0
2 2
0
2
0
2 5 4 8 5; Geometry formula: base circumference 2 (4), slant height 42 22 2 5
Lateral surface area 12 (8 ) 2 5 8 5 in agreement with the integral value
11.
dx 1 ;
dy
2
dx 2
dy 2
b
3
S 2 y 1 dx
a
1
( x 1)
2
3
dx 2 5 13 ( x 1) dx 2 5 x2 x 1
1 12
2
2
2 5 92 3 12 1 2 5 (4 2) 3 5; Geometry formula: r1 12 12 1, r2 23 12 2, slant height
(2 1)2 (3 1) 2 5 Frustum surface area r1 r2
slant height (1 2) 5 3 5 in
agreement with the integral value
12.
d
dy 2 (2 y 1) 1 4 dy 2 5 (2 y 1) dy
dx 2; S
y 2x 12 x 2 y 1 dy
2 x 1 dydx
c
2
2
2
1
1
2
2 5 y 2 y 2 5 (4 2) (1 1) 4 5; Geometry formula: r1 1, r2 3,
1
slant height (2 1)2 (3 1) 2 5 Frustum surface area (1 3) 5 4 5 in agreement with
the integral value
Copyright 2016 Pearson Education, Ltd.
388
13.
Chapter 6 Applications of Definite Integrals
2
4
3
2
dy
x 2 dy
x9 S 29x
dx
3
dx
0
4
1 x9 dx;
u 1 x 4 du 4 x3 dx 1 du x3 dx;
9
9
4
9
25/9 1/2 1
u 4 du
1
S 2
x 0 u 1, x 2 u 25
9
25/9
2 23 u 3/2
1
14.
3 125
1 3 12527 27 98
27
81
dy
dy 2
12 x 1/2 dx 41x
dx
S
15/4
3/4
2 x 1 41x dx 2
15/4
15/4
3/4
x 14 dx
3/2
3/2
3/2
2 23 x 14
43 15
14
43 14
4
3/4
3
43 42 1 43 (8 1) 283
15.
dy
(2 2 x )
12
dx
2 x x2
S
1.5
0.5
2
1.5
2
1.5
16.
dy 2
dx
2 x x2
2 2 x x 2 1
(1 x ) 2
2 x x2
(1 x ) 2
2 x x2
dx
2
2
2 x x 2 2 x x 1 22 x x dx
0.5
0.5
1 x
2 x x
dx 2
1.5
x 0.5 2
dy
dy 2
1 dx 4( x11)
dx
2 x 1
5
5
1
1
S 2 x 1 1 4( x11) dx 2
2
5
1
( x 1) 14 dx
5
3/2
x 54 dx 2 23 x 54
1
3/2
3/2 4 25 3/2
3/2
43 5 54
1 54
3 4
94
3
3
43 53 33 6 (125 27) 986 493
2
2
17.
y S
dx y 2 dx
dy
dy
2
1 2 y 3
0 3
4
1 y 4 dy;
u 1 y 4 du 4 y3 dy 1 du y 3 dy;
4
2
y 0 u 1, y 1 u 2 S 2 13 u1/2 14 du
1
2
2
6 u1 2 du 6 32 u 3/2 9 ( 8 1)
1
1
Copyright 2016 Pearson Education, Ltd.
Section 6.4 Areas of Surfaces of Revolution
18.
x 13 y3/2 y1/2 0, when 1 y 3. To get positive
area, we take x 13 y 3/2 y1/2
14 y 2 y 1
3
S 2 13 y 3/2 y1/2 1 14 y 2 y 1 dy
1
3
2 13 y 3/2 y1/2 14 y 2 y 1 dy
1
dx 1 y1/2 y 1/2 dx
dy
2
dy
2
y y
2 dy 13 y1/2 13 y 1 y1/2 y1 dy 13 13 y 1 ( y 1) dy
1/ 2
3 1 3/2
2
y y1/2
1 3
1/ 2 2
1/ 2
3
3 1 2 2
y3 y3
1 1
27 9
y
y
1
dy
y
9 3
9 3 3 9 3 1
3
1 3
1
9 (18 1 3) 169
19.
dx
dy
1 dx
dy
4 y
4
15/4
0
2
1 S 15/4 2 2
4 y
0
15/4
5 y dy 4 23 (5 y )3/2
0
15/4
4 y 1 41 y dy 4
0
3 19 13 1
(4 y ) 1 dy
3/2
3/2 3/2
83 5 15
53/2 83 54
5
4
83 5 5 5 8 5 83 40 585 5 353 5
20.
dx
dy
dx
1
dy
2 y 1
2
1 S 1 2
2 y 1
5/8
2 y 1 1 2 y11 dy 2
1
5/8
(2 y 1) 1 dy 2
1
5/8
1
3/2 4 2
5 5
4 2 82 2 5 5
16 2 5 5
2 2 23 y3/2 43 2 13/2 85
12
3 1 8 8 3
5/8
82 2
21.
1
dy
dy 2
x
x 2 S 2 x
dx
dx
0
22.
y 13 x 2 2
2
0
2
x
3/2
1 x 2 dx
1
2 y1/2 dy
3/2
2
2
1 x2
2 2 1
3
3
0
dy x x 2 2 dx ds 1 2 x 2 x 4 dx S 2
2
x 1 2 x 2 x 4 dx
0
x2 1 dx 2 0 2 x x2 1 dx 2 0 2 x3 x dx 2 x4 x2 0 2 44 22 4
2
4
2
2
2
23. ds dx 2 dy 2 y3 1 3 1 dy y 6 12 1 6 1 dy y 6 12 1 6 dy
y
y
y
4
16
16
2
2
2
2
y 3 1 3 dy y3 1 3 dy; S 2 y ds 2 y y3 1 3 dy 2π y 4 14 y 2 dy
1
1
1
4y
4y
4y
2
y5
2 5 14 y 1 2 32
1 15 14 2 31
18 240 (8 31 5) 253
5
20
5 8
1
Copyright 2016 Pearson Education, Ltd.
389
390
Chapter 6 Applications of Definite Integrals
sin x S 2
dy 2
dy
24.
y cos x dx sin x dx
25.
y a 2 x 2 dx 12 a 2 x 2
S 2
a
1/2
dy 2
dx
a2 x2
x2
a2 x2
a2 x2 x2 dx 2 aa a dx 2 a xaa
a
a x
(cos x) 1 sin 2 x dx
x
(2 x)
2
a
a 2 x 2 1 2x 2 dx 2
a
/2
/2
dy
2
2 a [a (a)] (2 a )(2a ) 4 a 2
26.
S 2 x 1 dx 2 x
h r
h r r h r
dy 2
dy
22r
2
h
h
x2
2 0
2
hr
0 h
r2
h2
y hr x dx hr dx
2 r
h2
2 h2
2
2
hr
0 h
r2
h2
2
dy. Now, x y 16 x 16 y
d
c
2
7
16
y
2
16 y
2
dx
dy
2
y2
2
16 y
; S
2
7
16
2
2
dx
27. The area of the surface of one wok is S 2 x 1 dy
dx
dy
h 2 r 2 h x dx
0
h2
h 2 r 2 dx 2 r
h
h2
2
2 162 y 2 1
y2
2
16 y
2
dy
2
7
16
2
2
2
2
162 y 2 y 2 dy
16 dy 32 9 288 904.78 cm 2 . The enamel needed to cover one surface of one wok is
V S 0.5 mm S 0.05 cm (904.78)(0.05) cm3 45.24 cm3 . For 5000 woks, we need
5000 V 5000 45.24 cm3 (5)(45.24) L 226.2 L 226.2 liters of each color are needed.
28.
dy
y r 2 x 2 dx 12
2
2
2
dy
r x
2
x2 ;
r 2 x2
S 2
ah
2
r 2 x 2 1 2x 2 dx
r x
a
2x
2
R x
2
x
2
R x
2
dx
dy
2
x2 ;
R x2
2
S 2
ah
a
R2 x2 1
R2 x2 x2 dx 2 R aah dx 2 Rh
ah
a
30. (a)
r x
2
dx
dy
2
2
y R 2 x 2 dx 12
2
2
x
2
r x x dx 2 r aah dx 2 rh, which is independent of a.
ah
a
29.
2x
dx
x 2 y 2 152 x 152 y 2 dy
S
15
7.5
2 152 y 2 1
y2
2
15 y
2
y
152 y
dx
dy
2
2
y2
2
15 y 2
;
15
152 y 2 y 2 dy 2 15
dy
7.5
7.5
dy 2
15
(2 )(15)(22.5) 675 square meters
(b) 2121 square meters
Copyright 2016 Pearson Education, Ltd.
x2
dx
R x2
2
Section 6.5 Work and Fluid Forces
391
31. (a) An equation of the tangent line segment is
(see figure) y f (mk ) f (mk ) x mk . When
x xk 1 we have
r1 f (mk ) f (mk )( xk 1 mk )
x
x
f (mk ) f (mk ) 2k f (mk ) f (mk ) 2k ;
when x xk we have
r2 f (mk ) f (mk ) xk mk
x
f (mk ) f (mk ) 2k ;
2
(b) L2k xk r2 r1 xk f (mk ) 2k f (mk ) 2k xk f (mk )xk
2
x
2
2
x
2
2
xk 2 f (mk )xk 2 , as claimed
Lk
(c) From geometry it is a fact that the lateral surface area of the frustum obtained by revolving the tangent line
segment about the x-axis is given by Sk r1 r2 Lk 2 f (mk )
xk 2 f (mk )xk 2 using
parts (a) and (b) above. Thus, Sk 2 f (mk ) 1 f (mk ) xk .
2
n
n
b
(d) S lim Sk lim 2 f (mk ) 1 f (mk ) xk 2 f ( x) 1 f ( x) dx
32.
n k 1
n k 1
y 1 x
2/3 3/2
1
0
2/3 3/2
1
2
a
1/2
dy
dx 32 1 x 2/3
S 2 2 1 x
2
1
x
2/3
32 x 1/3
1 x
x 2/3 dx 4 1 x 2/3
1/ 2
2/3
1/3
x
1
1 dx 4 1 x 2/3
0
3/2
dy 2
dx
1 x 2/3 1 1
x 2/3
x 2/3
1
0
u 1 x du 23 x 1/3 dx 32 du x 1/3 dx; x 0 u 1, x 1 u 0
0
0
S 4 u 3/2 32 du 6 52 u 5/2 6 0 52 125
1
1
2/3
6.5
x1/3 dx;
3/2
WORK AND FLUID FORCES
1. The force required to stretch the spring from its natural length of 2 m to a length of 5 m is F ( x ) kx.
3
3
3
The work done by F is W F ( x) dx k x dx k2 x 2 92k . This work is equal to 1800 J
0
0
0
92 k 1800 k 400 N/m
2. (a) We find the force constant from Hooke’s Law: F kx k Fx k 800
200 N/cm.
4
2
2
0
0
(b) The work done to stretch the spring 2 cm beyond its natural length is W kx dx 200 x dx
2
200 x2 200(2 0) 400 cm N 4 J
0
2
(c) We substitute F 1600 into the equation F 200 x to find 1600 200 x x 8 cm.
Copyright 2016 Pearson Education, Ltd.
392
Chapter 6 Applications of Definite Integrals
3. We find the force constant from Hooke’s law: F kx. A force of 2 N stretches the spring to 0.02 m
N . The force of 4 N will stretch the rubber band y m, where F ky y F
2 k (0.02) k 100 m
k
0.04
4N
y 0.04 m 4 cm. The work done to stretch the rubber band 0.04 m is W
kx dx
N
0
100 m
y
100
0.04
0
0.04
x dx 100 x2
0
2
(100)(0.04) 2
0.08 J
2
N . The work done to
4. We find the force constant from Hooke’s law: F kx k Fx k 90
k 90 m
1
5
5
5
stretch the spring 5 m beyond its natural length is W kx dx 90 x dx 90 x2 (90) 25
1125 J
2
0
0
0
2
N
5. (a) We find the spring’s constant from Hooke’s law: F kx k Fx 96,000
96,000
k 12,000 cm
2012
8
1
1
0
0
(b) The work done to compress the assembly the first centimeter is W kx dx 12, 000 x dx
1
(1)2
12, 000 x2 (12, 000) 2
0
2
(12,000)(1)
6, 000 N cm. The work done to compress the assembly the
2
second centimeter is:
2
2
2
W kx dx 12, 000 x dx 12, 000 x2 12,000
4 1
2
1
1
1
2
(12,000)(3)
18, 000 N cm
2
(70)(9.8)
N . If someone
457.3 mm
1.5
6. First, we find the force constant from Hooke’s law: F kx k Fx
compresses the scale x 3 mm, he/she must weigh F kx 457.3(3) 1372 N (140 kg). The work done to
3
3
compress the scale this far is W kx dx 457.3 x2 (457.3)(9) 4116 N mm 4.116 J
0
0
2
7. The force required to haul up the rope is equal to the rope’s weight, which varies steadily and is proportional
to x, the length of the rope still hanging: F ( x ) 0.624 x. The work done is: W
50
0
F ( x) dx
50
0
0.624x dx
50
0.624 x2 780 J
0
2
8. The weight of sand decreases steadily by 300 N over the 6 m, at 50 N/m. So the weight of sand when the
b
6
a
0
bag is x m off the ground is F ( x ) 600 50 x. The work done is: W F ( x)dx (600 50 x) dx
6
600 x 25 x 2 2700 J
0
9. The force required to lift the cable is equal to the weight of the cable paid out: F ( x ) (60)(60 x)
where x is the position of the car off the first floor. The work done is: W
60
2
2
2
60 60 x x2 60 602 602 60260 108,000 J
0
Copyright 2016 Pearson Education, Ltd.
60
0
F ( x) dx 60
60
0
(60 x) dx
Section 6.5 Work and Fluid Forces
393
10. Since the force is acting toward the origin, it acts opposite to the positive x-direction. Thus F ( x) k2 .
x
b
b
b
The work done is W k2 dx k 12 dx k 1x k
a
a
k
a
x
1 1 k ( a b )
b a
ab
11. Let r the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a
constant rate, the amount of water in the bucket is proportional to (6 x), the distance the bucket is being
raised. The leakage rate of the water is 13 N/m raised and the weight of the water in the bucket is
6
6
F 13(6 x). So: W 13 (6 x) dx 13 6 x x2 234 J.
0
0
2
12. Let r the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a
constant rate, the amount of water in the bucket is proportional to (6 x), the distance the bucket is being
raised. The leakage rate of the water is 32.5 N/m raised and the weight of the water in the bucket is
6
2
6
F 32.5(6 x). So: W 32.5(6 x) dx 32.5 6 x x2 585 J.
0
0
Note that since the force in Exercise 12 is 2.5 times the force in Exercise 11 at each elevation, the total work is
also 2.5 times as great.
13. We will use the coordinate system given.
(a) The typical slab between the planes at y and y y
has a volume of V (10)(12)y 120y m3. The
force F required to lift the slab is equal to its weight:
F 9800 V 9800 120y N. The distance
through which F must act is about y m, so the work
done lifting the slab is about W force distance
9800 120 y y J. The work it takes to lift all the
water is approximately
20
20
0
0
W W 9800 120 y y J.
This is a Riemann sum for the function 9800 120 y over the interval 0 y 20. The work of pumping the
tank empty is the limit of these sums:
20
y2
9800 120 y dy (9800)(120) 2 (9800)(120) 400
(9800)(120)(200) 235,200,000 J
2
0
0
W
20
J
W
(b) The time t it takes to empty the full tank with 5 hp motor is t 3678
235,200,000
63,948 s
J/s
3678 J/s
17.763 h t 17 h and 46 min
(c) Following all the steps of part (a), we find that the work it takes to lower the water level 10 m is
10
10
y2
W 9800 120 y dy (9800)(120) 2 (9800)(120) 100
58,800,000 J and the time is
2
0
0
W
t 3678
15987 s 266 min
J/s
(d) In a location where water weighs 9780 N3 :
m
a) W (9780)(24, 000) 234,720,000 J .
b) t 234,720,000
63817 s 17.727 h t 17 h and 44 min
3678
Copyright 2016 Pearson Education, Ltd.
394
Chapter 6 Applications of Definite Integrals
In a location where water weighs 9820 N3
m
a) W (9820)(24, 000) 235,680,000 J
b) t 235,680,000
64,078 s 17.80 h t 17 h and 48 min
3678
14. We will use the coordinate system given.
(a) The typical slab between the planes at y and y y has
a volume of V (6)(4) y 24y m3 . The force F
required to lift the slab is equal to its weight:
F 9800V 9800 24y N. The distance through
which F must act is about y m, so the work done lifting
the slab is about W force distance
6
9800 24 y y J. The work it takes to lift all the water is approximately W W
3
6
9800 24 y y J. This is a Riemann sum for the function 9800 24 y over the interval 3 y 6. The
3
work it takes to empty the cistern is the limit of these sums:
6
6
y2
W 9800 24 y dy (9800)(24) 2 (9800)(24)(18 4.5) (9800)(24)(13.5) 3,175,200 J
3
3
(b) t 370WJ/s 3,175,200
8581.6 s 2.38 hours 2 h and 23 min
370
(c) Following all the steps of part (a), we find that the work it takes to empty the tank halfway is
W
4.5
3
4.5
y2
9800 24 y dy (9800)(24) 2 (9800)(24) 9800
92 (9800)(24) 11.25
1,323,000 J.
2
2
3
W 1,323,000 3575.68 s 59.5 min
Then the time is t 370
370
(d) In a location where water weighs 9780 N3 :
m
a) W (9780)(24)(13.5) 3,168,720 J.
b) t 3,168,720
8564.11 s 2.379 hours 2 h and 22.7 min
370
c) W (9780)(24) 11.25
1,320,300 J; t 1,320,300
3568.38 s 0.99 hours 59.5 min
2
370
In a location where water weighs 9820 N3 :
m
a) W (9820)(24)(13.5) 3,181,680 J.
b) t 3,181,680
8599.14 s 2.389 hours 2 h and 23.3 min
370
c) W (9820)(24) 11.25
1,325,700 J; t 1,325,700
3583 s 0.995 hours 59.7 min
2
370
, thickness y, and height below the top of the tank (10 y). So the
work to pump the oil in this slab, W , is 8820 (10 y ) π y. The work to pump all the oil to top of the
15. The slab is a disk of area x 2
y 2
2
y
2
10 8820
4
0
tank is W
10
y
10 y
4 1,837,500 J 5,772,676.5 J
10 y2 y3 dy 8820π
4 3
0
3
4
Copyright 2016 Pearson Education, Ltd.
Section 6.5 Work and Fluid Forces
395
and
y 2
16. Each slab of oil is to be pumped to a height of 11 m. So the work to pump a slab is 8820(11 y )( ) 2
m3 , half the
since the tank is half full and the volume of the original cone is V 13 r 2 h 13 52 (10) 250
3
y2
1
volume 250π
m3 , and with half the volume the cone is filled to a height y, 250
y y 3 500 m.
6
3
4
6
3
500 8820
4
0
So W
11y y
2
3
11 y
dy 8820π
4 3
3
y4
4
3
500
1,843,840 J.
0
17. The typical slab between the planes at y and y y has a volume of V (radius)2 (thickness)
62 y 9 y m3. The force F required to lift the slab is equal to its weight:
2
F 7840V 7840 9 y N F 70,560 y N. The distance through which F must act is about
9
9
0
0
(9 y ) m. The work it takes to lift all the kerosene is approximately W W 70,560 (9 y )y J
which is a Riemann sum. The work to pump the tank dry is the limit of these sums:
9
9
y2
W 70,560 (9 y ) dy 70,560 9 y 2 70,560
0
0
812 (70,560)(40.5 ) 8,977,666 J
18. (a) Follow all the steps of Example 5 but make the substitution of 10,100 N3 for 8820 N3 . Then,
m
m
8
8
y
10,100 10 y
10,100 1083 84
10,100
10,100 83
2
3 10
W 10,100
(10
y
)
y
dy
8
2
3
4
4
4
4
3
4
4
3
3
0
0
5,415,268 J
(b) Exactly as done in Example 5 but change the distance through which F acts to distance (11 y ) m.
3
4
8
3
4
8
(11 y ) y 2 dy 8820 11 y y 8820
Then W 8820
4 3
4
4
0 4
0
1183 84
3
4
8820
4
83 113 2
3
8820348 5 (2940 )(82 )(5)(2) 5,911,220.7 J
19. The typical slab between the planes at y and y y has a volume of about V (radius)2 (thickness)
y y m3. The force F ( y) required to lift this slab is equal to its weight: F ( y) 11,500 V
2
11,500 y y 11,500 y y m. The distance through which F ( y ) must act to lift the slab to the top of
2
the reservoir is about (4 y ) m3 , so the work done is approximately W 11,500 y (4 y )y J. The work
n
done lifting all the slabs from y 0 m to y 4 m is approximately W 11,500 yk 4 yk y J. Taking
k 0
4
4
0
0
the limit of these Riemann sums as n , we get W 11,500 y (4 y ) dy 11,500
4
4 y y2 dy
11,500 2 y 2 13 y 3 11,500 32 64
368,000
J 385,369 J.
3
3
0
20. The typical slab between the planes at y and y y has volume of about V (length)(width)(thickness)
2 2.25 y 2 (3)y m3 . The force F ( y ) required to lift this slab is equal to its weight:
Copyright 2016 Pearson Education, Ltd.
396
Chapter 6 Applications of Definite Integrals
F ( y ) 8300 V 8300 2 2.25 y 2 (3) y 49,800 2.25 y 2 y N. The distance through which F ( y )
must act to lift the slab to the level of 4.5 m above the top of the reservoir is about (6 y ) m, so the work done
is approximately W 49,800 2.25 y 2 (6 y ) y J. The work done lifting all the slabs from y 1.5 m to
n
y 1.5 m is approximately W 49,800 2.25 yk2 6 yk y J. Taking the limit of these Riemann sums
k 0
as n , we get W
1.5
1.5
49,800 2.25 y 2 (6 y )dy 49,800
1.5
1.5
(6 y ) 2.25 y 2 dy
1.5
1.5
49,800
6 2.25 y 2 dy
y 2.25 y 2 dy . To evaluate the first integral, we use we can interpret
1.5
1.5
1.5
1.5
2
2
1.5 2.25 y dy as the area of the semicircle whose radius is 1.5, thus 1.5 6 2.25 y dy
6
1.5
1.5
2.25 y 2 dy 6 12 (1.5)2 6.75 . To evaluate the second integral let
u 2.25 y 2 du 2 y dy; y 1.5 u 0, y 1.5 u 0, thus
1.5
1.5
0
2
2
1.5 y 2.25 y dy 12 0 u du 0. Thus, 49,800 1.5 6 2.25 y dy
1.5
1.5
49,800
6 2.25 y 2 dy
y 2.25 y 2 dy 49,800(6.75 0) 336,150 1,056,046 J
1.5
1.5
21. The typical slab between the planes at y and y y has a volume of about V (radius)2 (thickness)
2
25 y 2 y m3 . The force F ( y ) required to lift this slab is equal to its weight:
2
F ( y ) 9800 V 9800 25 y 2 y 9800 25 y 2 y N. The distance through which F ( y ) must
act to lift the slab to the level of 4 m above the top of the reservoir is about (4 y ) m, so the work done is
approximately W 9800 25 y 2 (4 y )y N m. The work done lifting all the slabs from y 5 m to
0
y 0 m is approximately W 9800 25 y 2 (4 y )y N m. Taking the limit of these Riemann sums,
0
5
we get W 9800 25 y 2 (4 y ) dy 9800
5
0
5
0
100 25 y 4 y 2 y3 dy
y4
9800 100 y 25
y 2 43 y 3 4 9800 500 25225 43 125 625
15, 073, 099.75 J
2
4
5
22. The typical slab between the planes at y and y y has a volume of about V (radius)2 (thickness)
2
9 y 2 y 9 y 2 y m3 . The force is F ( y ) 88003 N V 8800 9 y 2 y N. The distance
m
through which F ( y ) must act to lift the slab to the level of 0.6 m above the top of the tank is about (3.6 y )
3
to y 3 m is approximately W 8800 9 y 2 (3.6 y )y J. Taking the limit of these
0
m, so the work done is W 8800 9 y 2 (3.6 y )y J. The work done lifting all the slabs from y 0 m
Copyright 2016 Pearson Education, Ltd.
Section 6.5 Work and Fluid Forces
3
3
397
Riemann sums, we get W 8800 9 y 2 (3.6 y ) dy 8800 9 y 2 (3.6 y ) dy
0
0
3
3
9 y 2 3.6 y 3
y4
8800 32.4 9 y 3.6 y 2 y 3 dy 8800 32.4 y 2 3 4
0
0
8800 97.2 81
1.2 27 81
(8800 )(44.55) 1,231,630 J. It would cost
2
4
(0.4)(1,231,630) 492,652¢ = $4926.52. Yes, you can afford to hire the firm.
x
dv by the chain rule W x2 mv dv dx m x2 v dv dx m 1 v 2 ( x ) 2
23. F m dv
mv dx
x1 dx
x1 dx
dt
2
x1
12 m v 2 ( x2 ) v 2 ( x1 ) 12 mv22 12 mv12 , as claimed.
24. W = (1/2)(0.06 kg)(50 m/s)2 75 J
25. 144 km/h
90 km
1h
1 min 1000 m
60 min 60 s 1 km 40 m/s;
1h
m 0.15 kg;
W 12 (0.15 kg)(40 m/s) 2 120 J
26. W = (1/2)(0.05 kg)(84 m/s)2 176.4 J
27. m 0.06 kg; W
1
(0.06 kg)(68 m/s)2 = 138.72 J
2
28. m 0.2 kg; W 12 (0.2 kg)(40 m/s)2 160 J
29. We imagine the milkshake divided into thin slabs by planes perpendicular to the y -axis at the points of a
partition of the interval [0, 18]. The typical slab between the planes at y and y y has a volume of about
y cm . The force F ( y) required to lift this slab is equal to its
V
y N. The distance through which F ( y) must act to lift this slab
V (radius) 2 (thickness)
9.8
weight: F ( y ) 0.8 1000
y 36 2
12
7.84
1000
3
y 36 2
12
to the level of 3 cm above the top is about (21 y ) cm. The work done lifting the slab is about
W 7.84
1000
( y1236) (21100 y) y J. The work done lifting all the slabs from y 0 to y 18 is approximately
2
2
18
( y 36) 2 (21 y ) y J which is a Riemann sum. The work is the limit of these sums as the norm
W 7.84
5
0
10 12
of the partition goes to zero:
W
18 7.84
5
0 10 12
( y 36)2 (21 y ) dy 7.84π
5
18
10 12 0
18
27,216 216 y 51y2 y3 dy
y4
184 17 183 108 182 27,216 18 2.175 J
7.84π
17 y 3 108 y 2 27,216y 7.84π
5
0 105 12 4
10 12 4
30. We fill the pipe and the tank. To find the work required to fill the tank note that radius = 3 m, then
V 9y m3 . The force required will be F = 9800 V = 9800 9 y = 88,200 y N. The distance
through which F must act is y so the work done lifting the slab is about W1 88,200 y y N. The work it
Copyright 2016 Pearson Education, Ltd.
398
Chapter 6 Applications of Definite Integrals
115.5
115.5
108
115.5
108
takes to lift all the water into the tank is: W1 W1 88,200 y y J. Taking the limit we end up
with W1
115.5
108
y2
88,200 y dy 88,200 2
88,200
[115.52 1082 ] 232,234,776 J
2
108
1 m. Then
To find the work required to fill the pipe, do as above, but take the radius to be 5 cm 20
1 y m3 and F 9800 V 9800 y. Also take different limits of summation and integration:
V 400
400
2 108
108 9800
9800 y
y
dy
9800
400
400 2
400
0
0
108
W2 W2 W2
0
1082 448,883 J
2
The total work is W W1 W2 232,234,776 448,883 232,683,659 J. The time it takes to fill the tank and
W 232,683,659 116,342 s 32 h
the pipe is Time 2000
2000
31. Work
35,780,000 1000 MG
r2
6,370,000
35,780,000
35,780,000 dr
1000 MG 1r
6,370,000 r 2
6,370,000
dr 1000 MG
(1000) 5.975 104 6.672 1011
1
1
35,780,000
6,370,000
5.144 1010 J
be the x-coordinate of the second electron. Then r 2 ( 1)2
32. (a) Let
0
W F( ) d
1
0 (23 1029 )
1 ( 1)
2
0
29
(23 1029 ) 1 1 11.5 1029
23 10
1 1
2
d
(b) W W1 W2 where W1 is the work done against the field of the first electron and W2 is the work done
against the field of the second electron. Let
be the x-coordinate of the third electron. Then r12 ( 1)2
and r22 ( 1) 2
5
W1 23 102
3
r1
5
W2 23 102
3
29
29
r2
d
d
5
29
23 10 2 d
3 ( 1)
5 23 1029
3 ( 1)
23 1029.
12
Therefore W W1 W2
2
d
5
23 1029 11 (23 1029 ) 14 12 23
1029 , and
4
3
5
10
23 1029 11 (23 1029 ) 16 14 23 12
3
29
(3 2)
234 1029 1223 1029 233 1029 7.67 1029 J
33. To find the width of the plate at a typical depth y, we first find an equation for the line of the plate’s right-hand
edge: y x 5. If we let x denote the width of the right-hand half of the triangle at depth y, then x 5 y and
the total width is L( y ) 2 x 2(5 y ). The depth of the strip is ( y ). The force exerted by the water against
one side of the plate is therefore F
0.6
1.6
w( y ) L( y ) dy
0.6
1.6
9800 ( y ) 2(1.6 y ) dy
1.6 y y 2 dy 19,600 0.8 y 2 13 y3
1.6
1.6
19,600
0.6
0.6
19,600 0.8 0.36 13 0.216 0.8 2.56 13 4.096 (19,600)(0.6826667 0.216)
(19,600)(0.466667) 9,146.7 N
Copyright 2016 Pearson Education, Ltd.
Section 6.5 Work and Fluid Forces
399
34. An equation for the line of the plate’s right-hand edge is y x x y. Thus the total width is
L( y ) 2 x 2 y. The depth of the strip is (0.6 y ). The force exerted by the water is
0
0
0
1
1
0
1
F w(0.6 y ) L( y ) dy 9800 (0.6 y ) 2 y dy 19,600
0.6 0.4 y y2 dy
y3
19,600 0.6 y 0.2 y 2 3 ( 19,600) 0.6 0.2 13 ( 19,600)(0.46667) 9,146.7 N
1
b
strip
35. (a) The width of the strip is L( y ) 1.2, the depth of the strip is (3 y ) F w depth
F ( y ) dy
a
0.9
0
9800(3 y )(1.2)dy 39,200
0.9
0
0.9
y2
(3 y ) dy 39,200 3 y 2 39,200 2.7 0.81
89,964 N
2
0
b
strip
(b) The width of the strip is L( y ) 0.9, the depth of the strip is (3 y ) F w depth
F ( y ) dy
a
1.2
0
9800(3 y )(0.9) dy 29,400
1.2
0
1.2
y2
(3 y )dy 29,400 3 y 2 29,400(3.6 0.72) 84,672 N
0
b
strip
36. The width of the strip is L( y ) 2 25 y 2 , the depth of the strip is (6 y ) F w depth
F ( y ) dy
a
5
5
5
5
9800 (6 y ) 2 25 y 2 dy 19,600 (6 y ) 25 y 2 dy 19,600 6 25 y 2 dy y 25 y 2 dy
0
0
0
0
To evaluate the first integral, we use we can interpret
5
0
5
5
0
0
radius is 5, thus 6 25 y 2 dy 6
25 y 2 dy as the area of a quarter circle whose
25 y 2 dy 6 14 (5)2 752 . To evaluate the second integral let
5
0
0
25
u 25 y 2 du 2 y dy; y 0. u 25, y 5 u 0, thus y 25 y 2 dy 12
25
5
25 1/2
u du 13 u 3/2 125
. Thus, 19,600 6
3
0
0
0
12
1,492,404 N.
u du
5
25 y 2 dy y 25 y 2 dy 19,600 752 125
3
0
37. Using the coordinate system of Exercise 32, we find the equation for the line of the plate’s right-hand edge to
be y 2 x 4 x
(a)
y4
2
and L( y ) 2 x y 4. The depth of the strip is (1 y ).
0
0
4
4
(9800) (4)(4)
0
3y
y
4 3 y y 2 dy 9800 4 y 2 3
4
4
F w(1 y ) L( y ) dy 9800 (1 y )( y 4) dy 9800
0
(3)(16) 64
( 9800)( 120 64)
3 (9800)(16 24 64
)
182,933 N
2
3
3
(3)(16)
( 10,050)( 120 64) 187,600 N
(b) F (10,050) (4)(4) 2 64
3
3
38. Using the coordinate system given, we find an equation for the
line of the plate’s right-hand edge to be y 2 x 4 x
4 y
2
and L( y ) 2 x 4 y. The depth of the strip is (1 y )
1
1
0
0
F w (1 y )(4 y ) dy 9800 y 2 5 y 4 dy
Copyright 2016 Pearson Education, Ltd.
2
3
400
Chapter 6 Applications of Definite Integrals
1
y3 5 y 2
9800 3 2 4 y (9800) 13 52 4 (9800) 2156 24
0
(9800)(11)
17,967 N
6
39. Using the coordinate system given in the accompanying figure,
we see that the total width is L( y ) 1.6 and the depth of the strip
is (0.85 y ) F
0.84
0
0.84
0
w(0.85 y ) L( y ) dy
10,050(0.85 y ) 1.6 dy (10,050)(1.6)
0.84
y2
(10,050)(1.6) 0.85 y 2
0
0.84
0
(0.85 y ) dy
16,080 (0.85)(0.84) 0.84
2
(16,080)(0.84)(1.7 0.84)
5808 N
(2)
40. Using the coordinate system given in the accompanying figure,
we see that the right-hand edge is x 1 y 2 so the total width
is L( y ) 2 x 2 1 y 2 and the depth of the strip is ( y ). The
force exerted by the water is therefore
0
F w ( y ) 2 1 y 2 dy
1
0
3/2
1 y 2 (2 y ) dy 9800 23 1 y 2
1
1
9800
0
(9800) 23 (1 0) 6533 N
41. (a)
F 9800 N3 (2 m) 1 m 2 19,600 N
m
b
strip
(b) The width of the strip is L( y ) 1, the depth of the strip is (2 y ) F w depth
F ( y ) dy
a
1
1
1
y2
9800(2 y )(1) dy 9800 (2 y ) dy 9800 2 y 2 9800 2 12 14,700 N
0
0
0
(c) The width of the strip is L( y ) 1, the depth of the strip is (2 y ), the height of the strip is
b
strip
F w depth
F ( y ) dy
a
1/ 2
0
1/ 2
y2
9800 2 2 y 2
0
9800 2
9800 (2 y )(1) 2 dy 9800 2
1/ 2
0
2 dy
(2 y ) dy
16,135 N
2
2
1
4
42. The width of the strip is L( y ) 34 2 3 y , the depth of the strip is (6 y ), the height of the strip is
b
strip
F w depth
F ( y ) dy
a
2 3
0
9800(6 y ) 34 2 3 y
2 3
y3
14,700 12 y 3 3 y 2 y 2 3 3
3
0
2 3
23 dy 14,700
12 3 6 y 2 y 3 y2 dy
3 0
14,700 72 36 12 3 8 3 246,734 N
3
2 dy
3
Copyright 2016 Pearson Education, Ltd.
Section 6.5 Work and Fluid Forces
43. The coordinate system is given in the text. The right-hand edge is x
401
y and the total width is
L( y ) 2 x 2 y .
1
(a) The depth of the strip is (2 y ) so the force exerted by the liquid on the gate is F w(2 y ) L( y ) dy
0
1
1
1
1
8,000(2 y ) 2 y dy 16,000 (2 y ) y dy 16,000 2 y1/2 y3/2 dy 16,000 43 y3/2 52 y 5/2
0
0
0
0
16,000 43 52 16,000
(20 6) 14,933 N
15
1
(b) We need to solve 25, 000 w( H y ) 2 y dy for h. 25, 000 16, 000 23H 52 H 2.94 m.
0
44. Suppose that h is the maximum height. Using the coordinate system given in the text, we find an equation for
the line of the end plate’s right-hand edge is y 3x x 13 y. The total width is L( y ) 2 x 23 y and the
h
depth of the typical horizontal strip at level y is (h y ). Then the force is F w(h y ) L( y ) dy Fmax ,
0
where Fmax 25,000 N. Hence,
h
hy 2 y3
hy y 2 dy (9800) 23 2 3
0
0
h
Fmax w (h y ) 23 y dy (9800) 23
0
h
F
25,000
h3 h 3 9 9800 3 9 9800 2.842 m. The volume of
h2 h3 (9800) 32 16 h3 9800
9
3
(9800) 23
3
max
water which the tank can hold is V 12 (Base)(Height) 10, where Height h and
1
2
(Base) 13 h V 13 h 2 (10) 10
h 2 10
(2.842) 2 26.92 m3 .
3
3
45. The pressure at level y is p( y ) w y the average
b
b
pressure is p b1 p ( y ) dy b1 w y dy
0
b
y
b1 w 2
0
2
0
wb b2 wb2 . This is the pressure at
2
level b2 , which is the pressure at the middle of the
plate.
b
b
b
b
y2
46. The force exerted by the fluid is F w(depth)(length) dy w y a dy ( w a) y dy ( w a) 2
0
0
0
0
(ab) p Area, where p is the average value of the pressure.
w ab2
2
wb
2
Copyright 2016 Pearson Education, Ltd.
402
Chapter 6 Applications of Definite Integrals
47. When the water reaches the top of the tank the force on the movable side is (9800) 2 1 y 2 ( y )dy
1
0
0
3/2
2
(9800) 1 y
(2 y ) dy (9800) 23 1 y 2
(9800) 3 (1) 6,533 J. The force compressing
1
1
the spring is F 3, 000 x so when the tank is full we have 6,533 3,000x x 2.18 m. Therefore the
movable end does not reach the required 2.5 m to allow drainage the tank will overflow.
0
2 1/2
48. (a) Using the given coordinate system we see that the total
width L( y ) 1 and the depth of the strip is (1 y ).
1
1
0
0
Thus, F w(1 y )L( y ) dy (9800)(1 y ) dy
1
1
y2
(9800) (1 y ) dy (9800) y 2
0
0
(9800) 1 12 (9800) 12 4900 N
(b) Find a new water level Y such that FY (0.75)(4900 N) 3675 N. The new depth of the strip is (Y y )
Y
Y
and Y is the new upper limit of integration. Thus, FY w(Y y )L( y ) dy 9800 (Y y ) dy
0
Y
0
Y
y
(9800) (Y y ) dy (9800) Yy 2 (9800) Y 2 Y2 (9800) Y2 . Therefore,
0
0
2
Y
6.6
2 FY
(9800)
7350
9800
2
2
0.75 0.866 m. So, Y 1 Y 1 0.866 0.134 m 13.4 cm
MOMENTS AND CENTERS OF MASS
1. Since the plate is symmetric about the y -axis and its density is
constant, the distribution of mass is symmetric about the y -axis
and the center of mass lies on the y -axis. This means that x 0.
M
It remains to find y Mx . We model the distribution of mass
with vertical strips. The typical strip has center of mass:
2
( x , y ) x, x 2 4 , length: 4 x 2 width: dx,
area: dA 4 x 2 dx, mass: dm dA
4 x2 dx
4 x dx 16 x dx. The moment of the
16 x dx 16 x 16 2 16 2
The moment of the strip about the x-axis is y dm
plate about the x-axis is M x y dm
2
2 2
x2 4
2
2
4
2
4
2
2
x5
5 2
25
5
2
2
3
22 32 32
128
. The mass of the plate is M (4 x 2 ) dx 4 x x3 2
5
5
2
M
Therefore y Mx
1285 12 The plate’s center of mass is the point x , y 0, 12 .
5
323 5
Copyright 2016 Pearson Education, Ltd.
8 83 323 .
25
5
Section 6.6 Moments and Centers of Mass
403
2. Applying the symmetry argument analogous to the one
in Exercise 1, we find x 0. To find y Mx
, we use the
M
vertical strips technique. The typical strip has center of
2
mass: ( x , y ) x, 252 x , length: 25 x 2 , width: dx,
area: dA 25 x 2 dx, mass: dm
25 x2 dx.
dA
The moment of the strip about the x-axis is
2
y dm 252 x
25 x dx 25 x dx.
2 2
2
2
The moment of the plate about the x-axis is M x y dm
5 2
5
25 x2 dx 2 55 625 50 x2 x4 dx
2
5
5
5
2 625 x 50
x3 x5 2 2 625 5 50
53 55 625 5 10
1 625 83 . The mass of the plate
3
3
3
5
10.
M
25 x 2 dx 25 x x3 2 53 53 34 53. Therefore y M
5
5
5
is M dm
5
3
5
3
x
54 83
3
4
3
The plate’s center of mass is the point ( x , y ) (0, 10).
3. Intersection points: x x 2 x 2 x x 2 0
x(2 x) 0 x 0 or x 2. The typical vertical
x x2 ( x)
strip has center of mass: ( x , y ) x,
2
2
x, x2 , length: x x 2 ( x) 2 x x 2 ,
width: dx, area: dA 2 x x 2 dx, mass: dm
dA
2 x x2 dx. The moment of the strip about the x-axis is y dm x2 2 x x2 dx; about the y-axis
2
2
it is x dm x (2 x x 2 ) dx. Thus, M x y dm 2 x 2 2 x x 2 dx 2 2 x3 x 4 dx
0
0
2
2
2 x2 x5 2 23 25 2 23 1 54 45 ; M y x dm x 2 x x 2 dx
0
0
2
2
2
2 4 ; M dm 2
2 x 2 x3 dx 23 x3 x4 23 23 24 12
0 2 x x dx
3
0
0
2
2
M
M
2 x x 2 dx x 2 x3 4 83 43 . Therefore, x M 43 43 1 and y M
0
0
2
4
5
5
4
4
4
3
45
y
43 53 ( x , y ) 1, 53 is the center of mass.
4. Intersection points: x 2 3 2 x 2 3 x 2 3 0
3( x 1)( x 1) 0 x 1. or x 1 Applying the
symmetry argument analogous to the one in Exercise 1,
we find x 0 The typical vertical strip has center of mass:
2 x 2 x 2 3
2
( x , y ) x,
x, x2 3 , length: 2 x 2 x 2 3
2
3 1 x 2 , width: dx, area: dA 3 1 x 2 dx,
Copyright 2016 Pearson Education, Ltd.
x
404
Chapter 6 Applications of Definite Integrals
1 x2 dx. The moment of the strip about the x-axis is
y dm 32 x 2 31 x 2 dx 32 x 4 3 x 2 x 2 3 dx 32 x 4 2 x 2 3 dx;
1
1
45 32 ;
M x y dm 32 x 4 2 x 2 3 dx 32 x5 23x 3x 32 2 15 32 3 3 310
5
15
1
1
1
1
M
M dm 3 1 x 2 dx 3 x x3 3 2 1 13 4 . Therefore, y M 532
85
4
1
1
mass: dm
dA 3
5
3
3
x
( x , y ) 0, 85 is the center of mass.
y y3
5. The typical horizontal strip has center of mass: ( x , y ) 2 ,
length: y y 3 , width: dy, area: dA y y 3 dy,
y y3 dy. The moment of the strip about the
2
y y
y -axis is x dm 2 y y 3 dy 2 y y3 dy
2 y 2 2 y 4 y 6 dy; the moment about the x-axis is y dm y y y3 dy y 2 y 4 dy. Thus,
1
1
1
y
y
M x y dm y 2 y 4 dy 3 5 13 15 215 ; M y x dm 2 y 2 2 y 4 y 6 dy
0
0
0
mass: dm dA
3
3
1
5
y3 2 y5 y 7
15 4 ; M dm
2 3 5 7 2 13 52 71 2 35342
57
105
0
1
3
( y y) dy
0
1
y2 y4
2 4
0
4
16 and y M 2
16 , 8 is the
4 105
12 14 4 . Therefore, x MM 105
15 4 158 ( x , y ) 105
M
15
y
x
center of mass.
6. Intersection points: y y 2 y y 2 2 y 0
y ( y 2) 0 y 0 or y 2 The typical horizontal
y 2 y y y 2
strip has center of mass: ( x , y )
, y 2 , y ,
2
area: dA 2 y y 2 dy, mass: dm dA 2 y y 2 dy.
The moment about the y -axis is x dm 2 y 2 2 y y 2 dy 2 2 y3 y 4 dy; the moment about the x-axis
2
2
y
2y
is y dm y 2 y y 2 dy 2 y 2 y3 dy. Thus, M x y dm 2 y 2 y 3 dy 3 4
0
0
length: y y 2 y 2 y y 2 , width: dy,
3
2
163 164 1612 (4 3) 43 ; M y x dm 02 2 2 y3 y 4 dy 2 y2 y5 0 2 8 325
4
2
2 405 32 45 ; M dm
0
2 y y 2 dy
2
2 y3
y 3
0
5
4 83 43 . Therefore,
43 53 and y MM 43 43 1 ( x , y ) 53 , 1 is the center of mass.
M
x My 45
x
Copyright 2016 Pearson Education, Ltd.
4
Section 6.6 Moments and Centers of Mass
405
7. Applying the symmetry argument analogous to the one used in
Exercise 1, we find x 0. The typical vertical strip has center
cos x
2
of mass: ( x , y ) x,
, length: cos x, width: dx,
area: dA cos x dx, mass: dm dA cos x dx. The moment
of the strip about the x-axis is y dm cos2 x cos x dx
2 cos 2 x dx 2
M x y dm
1 cos 2 x
2
dx (1 cos 2x) dx; thus,
4
/2
/2
sin 2 x
(1 cos 2 x ) dx 4 x 2
0 2 64 ; M dm
/2 4 2
/2 4
/2 cos x dx sin x /2 2 . Therefore, y Mx 42 8 ( x , y ) 0, 8 is the center of mass.
/2
/2
M
8. Applying the symmetry argument analogous to the one used in
Exercise 1, we find x 0. The typical vertical strip has center
sec2 x
of mass: ( x , y ) x, 2 , length: sec2 x , width: dx,
area: dA sec2 x dx, mass: dm
dA
sec2 x dx. The
2
moment about the x-axis is y dm sec2 x ( sec2 x) dx
2 sec 4 x dx. M x
/4
/4
y dm 2
/4
/4
sec 4 x dx 2
/4
/4
tan 2 x 1sec2 x dx
tan x 2 sec2 x dx 2 /4 sec2 x dx 2 (tan3x)
/4
2
/4
/4
3
2 13 13 2 1 (1) 3
M
Therefore, y Mx
43 ; M dm
/4
/4
/4
/4
2 tan x /4
/4
/4 sec x dx tan x /4 1 (1) 2 .
2
43 21 23 ( x , y ) 0, 32 is the center of mass.
9. By symmetry, x 1.
y
Mx
21
(2 x x 2 )2 (2 x 2 4 x )2 dx
0 2
M
2
23 x 4 6 x 3 6 x 2
0
1
0
2
8
3 x5 3 x 4 2 x 3
10
2
5
0
1
2
2
2
2
0 (2 x x ) (2 x 4 x ) dx
0 3x 4 x 2 dx
x 2x 2x 4
2
2
3
2
2
0
8
y
dx
Mx 5
M
4
y 2x x2
2
5
Copyright 2016 Pearson Education, Ltd.
1,
2
5
2
y 2x2
x
4x
406
Chapter 6 Applications of Definite Integrals
10. (a) Since the plate is symmetric about the line x y and its
density is constant, the distribution of mass is symmetric
about this line. This means that x y The typical vertical
strip has center of mass: ( x , y ) x,
9 x 2
,
2
9 x 2 width: dx, area: dA 9 x 2 dx,
length:
mass: dm
9 x 2 dx. The moment about the
dA
x-axis is y dm
9 x 2
2
9 x 2 dx 2 9 x 2 dx
9 x 2 dx 2 9 x x3 2 (27 9) 9 ; M dm dA dA
0 2
0
Thus, M x y dm
3
(Area of a quarter of a circle of radius 3)
3
3
(9 ) 94 4
94 94 . Therefore, y Mx
M
( x , y ) 4 , 4 is the center of mass.
(b) Applying the symmetry argument analogous to the
one used in Exercise 1, we find that x 0. The
typical vertical strip has the same parameters as in
part (a). Thus, M x y dm
3
3 2
2
3
0 2
9 x2 dx
9 x2 dx 2(9 ) 18 ; M dm dA
dA
(Area of a semi-circle of radius 3)
92 92 . Therefore, y MM (18 ) 92 4 , the
x
same y as in part (a) ( x , y ) 0, 4 is the center of mass.
11. By symmetry, x y .
My
3
0
3
0
y
x 3 9 x 2 dx
3
2
3x x 9 x 2 dx
1
3
1
9
3
x 2 (9 x 2 )3/2
2
0 2
3
To compute M, we multiply the area of the “triangle” by
9
obtaining 9 .
4
yx
My
M
9
2
9 94
4
2
y 9 x2
0
,
2
.
4
Copyright 2016 Pearson Education, Ltd.
1
2
3
x
, 42
Section 6.6 Moments and Centers of Mass
407
12. Applying the symmetry argument analogous to the one used in Exercise
1, we find that y 0. The typical vertical strip has center of mass:
13 13
( x , y ) x, x 2 x ( x, 0), length: 13 13 23 , width: dx ,
x
x
x
area: dA 23 dx, mass: dm dA 2 3 dx. The moment about the y -axis is x dm x 2 3 dx 2 2 dx. Thus,
x
x
M y x dm
a
a2
1 x2
( a 2 1)
a
2
dx 2 1x 2
1
M
. Therefore, x My
2
2 x 2
1 2
13. M x y dm
x
x
1a 1 2 (aa1) ; M dm 1 x dx x1 1 a1 1
2 ( a 1)
a
a2
a
3
2
2
a 2 2 a ( x , y ) 2 a , 0 . Also, lim x 2.
a 1
( a 2 1) a 1
a
dx x dx
2 1
1 x2
2
x2
2
2
x2
(1) 2 12 1;
2
2 2
2
dx 2 x 2 dx 2 x 1 2 12
1
1 x2
1
2
M y x dm x
1
dx
2
x2
2 2
2 dx 2 2 x dx 2 x 2 2 1 4 1 3; M dm 2
2
1
1 x2 dx 1 x x2 dx
1 x
2 1
2
2
2
x x2
2
2
2
2
M
M
2
2 dx 2 x 1 2(2 1) 2. So x My 32 and y Mx 12 ( x , y ) 23 , 12 is the center of mass.
1
14. We use the vertical strip approach:
1 x x
0 2
M x y dm
0
1
2
x x2 dx
1
12 x 2 x 4 12 x dx 6 x3 x5 dx
0
1
6 x4 x6 6 14 16 64 1 12 ;
0
4
6
1
1
1
1
M y x dm x x x 2 dx x 2 x3 12 x dx 12 x3 x 4 dx 12 x4 x5 12 14 15
0
0
0
0
1
4
5
1220 53 ;
3
4
1
1
12 1. So x M y 3 and
M dm x x 2 dx 12 x 2 x3 dx 12 x3 x4 12 13 14 12
M
5
0
0
0
M
y Mx 12 53 , 12 is the center of mass.
b
15. (a) We use the shell method: V 2 shell
radius
a
4
4 x
4
4
shell
height dx 1 2 x x x dx 16 1 x dx
4
4
16 x1/2 dx 16 23 x3/2 16 23 8 23
1
1
323 (8 1) 2243
(b) Since the plate is symmetric about the x-axis and its density
( x) 1x is a function of x alone, the
distribution of its mass is symmetric about the x-axis. This means that y 0. We use the vertical strip
4
4
4
4
approach to find x : M y x dm x 4 4 dx x 8 1x dx 8 x 1/2 dx 8 2 x1/2
1
1
1
1
x
x
x
Copyright 2016 Pearson Education, Ltd.
408
Chapter 6 Applications of Definite Integrals
dx 8 x
4
4
8(2 2 2) 16; M dm 4 4 dx 8 1
1 x
1
x
x
4 3/2
1
x
1
4
dx 8 2 x 1/2
1
M
8 1 (2) 8. So x My 16
2 ( x , y ) (2, 0) is the center of mass.
8
(c)
b
4
16. (a) We use the disk method: V R ( x) dx
2
1
a
[1 4] 3
dx 4 x dx 4 4 (1)
4
1
x 1
4 2
4
x2
1
(b) We model the distribution of mass with vertical strips: M x y dm
4
1
4
2x 2 dx 4 2 x dx
1 2
x
1 x
2
4
4
3/ 2
4
4
4
2 x 3/2 dx 2 2 2 1 (2) 2; M y x dm x 2x dx 2 x1/2 dx 2 2 x3
1
1
1
1
x 1
4
4
4
4
2 16
23 28
; M dm 2x dx 2 xx dx 2 x 1/2 dx 2 2 x1/2 2(4 2) 4.
3
3
1
1
1
1
M
So x My
283 7 and y M x 2 1 ( x , y ) 7 , 1 is the center of mass.
4
3
M
4
2
3 2
(c)
17. The mass of a horizontal strip is dm
dA L dy, where L is the width of the triangle at a distance of y above
its base on the x-axis as shown in the figure in the text. Also, by similar triangles we have Lb
L bh (h y ). Thus, M x y dm
h
0
2
h
bh 2 12 13 bh
; M dm
6
M
y Mx
0
3
3
hy 2 y3
hy y 2 dy hb 2 3 hb h2 h3
0
0
y bh (h y ) dy hb
b
h
h
h y
h
h
h
2
h
y2
(h y ) dy hb (h y ) dy hb hy 2 hb h 2 h2 2bh . So
0
0
the center of mass lies above the base of the triangle one-third of the way toward
bh 2
6
2
bh
h
3
the opposite vertex. Similarly the other two sides of the triangle can be placed on the x-axis and the same results
will occur. Therefore the centroid does lie at the intersection of the medians, as claimed.
Copyright 2016 Pearson Education, Ltd.
Section 6.6 Moments and Centers of Mass
18. From the symmetry about the y -axis it follows that x 0. It also follows
that the line through the points (0, 0) and (0, 3) is a median
y 13 (3 0) 1 ( x , y ) (0, 1).
19. From the symmetry about the line x y it follows that x y . It also
follows that the line through the points (0, 0) and
y x 23 12 0
13 ( x , y ) 13 , 13 .
12 , 12 is a median
20. From the symmetry about the line x y it follows that x y . It also
a2 , a2 is a median
follows that the line through the point (0, 0) and
y x 23 a2 0
13 a ( x , y ) a3 , a3 .
21. The point of intersection of the median from the vertex (0, b) to the
x a2 0 23 a3 ( x , y ) a3 , b3 .
opposite side has coordinates 0, a2 y (b 0) 13 b3 and
22. From the symmetry about the line x a2 it follows that x a2 . It also
follows that the line through the points
a2 , 0 and a2 , b is a median
y 13 (b 0) b3 ( x , y ) a2 , b3 .
23.
y x1/2 dy 12 x 1/2 dx ds (dx)2 (dy )2
1 41x dx;
3/2 2
0 x 1 41x dx 0 x 14 dx 23 x 14
2
Mx
2
3/2
3/2 2 9 3/2
3/2 2
23 2 14
14
3 4
14
3
24.
y x3 dy 3 x 2 dx dx (dx)2 3 x 2 dx
Mx
1 3
0 x
0
278 81 136
1 9 x4 dx;
2
1 9 x 4 dx;
1 du x3 dx;
[u 1 9 x 4 du 36 x3 dx 36
x 0 u 1, x 1 u 10] M x
10
1/2
1 361 u
10
du 36 32 u 3/2 54 103/2 1
1
Copyright 2016 Pearson Education, Ltd.
409
410
Chapter 6 Applications of Definite Integrals
0
0
2
25. From Example 4 we have M x a(a sin )(k sin )d a 2 k sin 2 d a2k (1 cos 2 ) d
0
a2k sin22 a 2k ; M y a(a cos )(k sin ) d a 2 k sin cos d a2k sin 2 0;
0
0
0
0
2
2
2
0 2ak. Therefore, x My 0 and y MMx a 2k 21ak a4 0, a4 is
M
M ak sin d ak cos
0
2
the center of mass.
26. M x y dm (a sin ) a d
0
a2
/2
a2
/2
0
0
(sin )(1 k cos ) d a 2
/
sin d a 2 k
/2
2
a 0 (1) a k
2
/2
/2
a2
/2
0
0
0
a 2 sin
a
2
(cos )(1 k cos ) d a 2
/2
2
0 a2k sin22
2
0
0 12 a2 a2k a2 a2k 2a2 a2 k a2 (2 k );
2
2
a2 cos 1 k cos d
(cos )(1 k cos )d
d a2 /2 cos d a2 k /2 1cos2 2 d
/2 1 cos 2
0
sin cos d
0
1 0 a 2 ( 1) 0 a 2 k
2
cos d a 2 k
2
a 2 cos /2 a 2 k sin2
/2
M y x dm (a cos ) a d
0
(sin )(1 k cos ) d
sin cos d a 2 / 2 sin d a 2 k
0 /2 a 2 k sin2
2
a2 sin 1 k cos d
/2
0
a 2 cos
a2
0
a2k sin22
a 2 sin
2
2
2
2
2
(1 0) a2k 2 0 (0 0) a 2 (0 1) a2k ( 0) 2 0 a 2 a 4k a 2 a 4k 0;
M
0
/2
1 k cos d a 0
0
a d a
a k sin
(1 k cos ) d a
/2
(1 k cos ) d
0 /2 a k sin /2 2 k 0 a ( 0) 2 k a2 ak a 2 k a 2ak
M
M
a 2 (2 k )
a (2 k )
a ( 2k ). So x My 0 and y Mx a ( 2 k ) 2k 0, 2a2ka
is the center of mass.
k
27.
f ( x) x 6, g ( x) x 2 , f ( x) g ( x) x 6 x 2
x 2 x 6 0 x 3, x 2;
1
3
3
M ( x 6) x 2 dx 12 x 2 6 x 13 x3
2
2
92 18 9 2 12 83 125
6
3
3
2
6 3 x 2 6 x x3 dx 6 1 x3 3 x 2 1 x 4
1
x 125/6
2 x( x 6) x dx 125
2
125 3
4
2
6 9 27 81 6 8 12 4 1 ;
125
4
125
3
2
Copyright 2016 Pearson Education, Ltd.
Section 6.6 Moments and Centers of Mass
3 1
( x 6)2
2 2
1
y 125/6
411
x2 dx 1253 32 x2 12 x 36 x4 dx 1253 13 x3 6 x2 36 x 15 x5 2
2
3
3 9 54 108 243 3 8 24 72 32 4 1 , 4 is the center of mass.
125
5
125
3
5
2
28.
f ( x) 2, g ( x) x 2 ( x 1), f ( x) g ( x) 2 x 2 ( x 1)
x3 x 2 2 0 x 1; 1
1
1
M 2 x 2 ( x 1) dx 2 x3 x 2 dx
0
0
1
2 x 14 x 4 13 x3 2 14 13 0 17
;
12
0
1
2
1
12 1 2 x x 4 x3 dx
x 17/12
0 x 2 x ( x 1) dx 17
0
1
12 x 2 1 x5 1 x 4 12 1 1 1 0 33 ;
17
5
4
5 4
85
0 17
2
1 2
1
2
6
5
4
7
6
5 1
1
y 17/12
0 12 2 x ( x 1) dx 176 0 4 x 2 x x dx 176 4 x 17 x 13 x 15 x 0
6 4 1 1 1 0 698 33 , 698 is the center of mass.
17
7 3 5
595
85 595
29.
f ( x) x 2 , g ( x) x 2 ( x 1), f ( x) g ( x)
x 2 x 2 ( x 1) x3 2 x 2 0 x 0, x 2;
1
2
2
M x 2 x 2 ( x 1) dx 2 x 2 x3 dx
0
0
2
23 x3 14 x 4 16
4 0 43 ;
3
0
1 2 x x 2 x 2 ( x 1) dx 3 2 2 x3 x 4 dx
x 4/3
0
4 0
2
34 12 x 4 15 x5 34 8 32
0 65 ;
5
0
2
2
2
2
1 2 1 x2
y 4/3
x 2 ( x 1) dx 83 2 x5 x6 dx 83 13 x6 71 x7 83 64
128
0 87 56 , 78 is the
0 2
3
7
0
0
center of mass.
30.
f ( x) 2 sin x, g ( x) 0, x 0, x 2 ;
M
2
0
x 41
2 sin x dx 2 x cos x 02 (4 1) (0 1) 4 ;
2
0
41
2
0
1;
x 2 sin x 0 dx 41
2
0
2
2
2
x sin x dx 41 x 2 41 sin x x cos x 0
0
0
2 x dx 41
2 x x sin x dx
41 4 2 0 41 (0 2 ) 0 221 ;
y 41
2 1
(2 sin x)2 (0) 2 dx
2
0
Copyright 2016 Pearson Education, Ltd.
412
Chapter 6 Applications of Definite Integrals
2
2
2
81 4 4sin x sin 2 x dx 81 4 4 sin x dx 81 sin 2 x dx
0
0
0
81
2
0
2
2
2
4 4 sin x dx 81 0 1cos2 2 x dx 81 4 x 4 cos x 02 161 0 dx 161 0 cos 2 x dx
2
4
2
[u 2 x du 2dx, x 0 u 0, x 2 u 4 ] 81 4 x 4 cos x 0 161 x 0 321 cos u du
0
2
2
4
81 4 x 4 cos x 0 161 x 0 321 sin u 0 81 (8 4) 81 (0 4) 161 (2 ) 0 0 89 221 , 98 is the
center of mass.
31. Consider the curve as an infinite number of line segments joined together. From the derivation of arc length we
have that the length of a particular segment is ds (dx)2 (dy )2 . This implies that M x y ds, M y x ds
M
and M ds. If
is constant, then x My
M
x ds
x ds
y ds
y ds
length and y Mx
length .
ds
ds
32. Applying the symmetry argument analogous to the one used in Exercise 1, we find that x 0. The typical vertical
a x2
2
2
strip has center of mass: ( x , y ) x, 24 p , length: a 4x p , width: dx, area : dA a 4x p dx,
a dx. Thus, M y dm
x2
4p
mass: dm dA
4
5
2 pa 2
a x 2 dx 2 a 2 x x 2
2 pa
16 p
80 p
2
2a 2
2 pa
8a
pa
3
x2
4p
2 pa
5
2 2 a2 x x 2
80 p 0
2 pa
2 pa
a a dx
2 pa
16 2a 2
pa 8080
pa 1 16
2a 2
80
3
3
ax 12x p
2 ax 12x p
2 pa
0
2 pa 1
2 pa 2
x
x2
4p
25 p 2 a 2 pa
2a 2 pa
80 p 2
8a 5 pa ; M dm 22 papa a 4x p dx
64
pa 80
2
23 pa pa
2 2a pa 12 p 4a
2
4 4a
pa 1 12
pa 1212 4
M
8a 2 pa 3 3
. So y Mx
8a pa 5 a, as claimed
5
33. The centroid of the square is located at (2, 2). The volume is V (2 )( y )( A) (2 )(2)(8) 32 and the surface
area is S (2 )( y )( L) (2 )(2) 4 8 32 2 (where
8 is the length of a side).
34. The midpoint of the hypotenuse of the triangle is 32 , 3 y 2 x is an
equation of the median the line y 2 x contains the centroid. The point
32 , 3 is 3 25 units from the origin the x-coordinate of the centroid
solves the equation
x 32 (2 x 3)2 25
2
x 2 3 x 94 4 x 2 12 x 9 54
5 x 2 15 x 9 1 x 2 3x 2 ( x 2)( x 1) 0 x 1 since the centroid must lie inside the triangle
y 2. By the Theorem of Pappus, the volume is V (distance traveled by the centroid)(area of the region)
2 (5 x ) 12 (3)(6) (2 )(4)(9) 72
35. The centroid is located at (2, 0) V (2 )( x )( A) (2 )(2)( ) 4 2
Copyright 2016 Pearson Education, Ltd.
Section 6.6 Moments and Centers of Mass
413
36. We create the cone by revolving the triangle with
vertices (0, 0), (h, r ) and (h, 0) about the x-axis (see the accompanying
figure).Thus, the cone has height h and
base radius r. By Theorem of Pappus, the lateral surface area swept out by the
h2 r 2 r r 2 h2 . To
hypotenuse L is given by S 2 yL 2 2r
calculate the volume we need the position of the centroid of the triangle.
From the diagram we see that the centroid lies on the
line y 2rh x. The x-coordinate of the centroid solves the equation
( x h)2 2rh x 2r
13 h2 r4
2
2
2 r 4 h
2
2
2
2
2
4h 2r x 2 4 h2h r x r4
0 x 23h or 43h x 23h , since the centroid must lie inside the
9
4h
2
2
triangle y 2rh x 3r . By the Theorem of Pappus, V 2
3r 12 hr 13 r 2 h.
37. S 2 y L 4 a 2 (2 y )( a ) y 2a , and by symmetry x 0
38. S 2 L 2 a 2a ( a ) 2 a 2 ( 2)
39. V 2 y A 43 ab 2 (2 y ) 2ab y 34b and by symmetry x 0
40. V 2 A V 2 a 34a 2a
2
a3 (3 4)
3
41. V 2 A (2 ) (area of the region) (distance from the centroid to the line y x a). We must find the
distance from 0, 34a to y x a. The line containing the centroid and perpendicular to y x a has slope
the point 4a 63a , 4a63a . Thus, the distance from the centroid to the line y x a is
2
2
4a63a 34a 64a 36a 2(46a3a ) V (2 ) 2(46a3a ) 2a 2 a 6(43 )
1 and contains the point 0, 34a . This line is y x 34a . The intersection of y x a and y x 34a is
3
2
42. The line perpendicular to y x a and passing through the centroid 0, 2a has equation y x 2a . The
intersection of the two perpendicular lines occurs when x a x 2a x 2a2a x 2 a2a
y 2a2a . Thus the distance from the centroid to the line y x a is
2a2 a 0 2a2 a 22a
2
a (2 )
a (2 )
. Therefore, by the Theorem of Pappus the surface area is S 2
( a )
2
2
43. If we revolve the region about the y -axis: r a, h b A 12 ab,V 13 a 2b, and
of Pappus: 13 a 2b 2 x 12 ab x a3 ; If we revolve the region about the x-axis:
r b, h a A 12 ab, V 13 b 2 a, and
1 b 2 a 2 y 1 ab
3
2
y . By the Theorem of Pappus:
y b3 a3 , b3 is the center
of mass.
Copyright 2016 Pearson Education, Ltd.
2
2 a 2 (2 ).
x . By the Theorem
414
Chapter 6 Applications of Definite Integrals
44. Let O(0, 0), P(a, c), and Q(a, b) be the vertices of the given triangle. If we revolve the region about the
x-axis: Let R be the point R (a, 0). The volume is given by the volume of the outer cone, radius RP c,
minus the volume of the inner cone, radius RQ b, thus V 13 c 2 a 13 b 2 a 13 a c 2 b 2 , the area is
given by the area of triangle OPR minus area of triangle OQR,
A 12 ac 12 ab 12 a (c b), and
y . By
the Theorem of Pappus: 13 a c 2 b 2 2 y 12 a (c b) y c 3b ; If we revolve the region about the
y -axis: Let S and T be the points S (0, c) and T (0, b), respectively. Then the volume is the volume of the
cylinder with radius OR a and height RP c, minus the sum of the volumes of the cone with radius
SP a and height OS c and the portion of the cylinder with height OT b and radius TQ a
with a cone of height OT b and radius TQ a removed. Thus
V a 2 c 13 a 2 c a 2b 13 a 2 b 23 a 2 c 23 a 2 b 23 a 2 (a b). The area of the triangle is the
same as before, A 12 ac 12 ab 12 a(c b), and
2 a 2 ( a b) 2 x 1 a (c b) x 2 a ( a b )
2
3
3(c b )
CHAPTER 6
1.
x . By the Theorem of Pappus:
2 a ( a b )
3(c b ) , c 2b is the center of mass.
PRACTICE EXERCISES
A( x) 4 (diameter) 2 4
x x2
2
4 x 2 x x 2 x 4 ; a 0, b 1
b
1
V A( x)dx 4 x 2 x5/2 x 4 dx
a
0
1
2
5
4 x2 74 x 7/2 x5 4 12 74 15
0
(35 40 14) 9
470
2.
280
A( x) 12 (side)2 sin 3 43 2 x x
2
43 4 x 4 x x x 2 ; a 0, b 4
b
4
a
0
3 4
V A( x)dx 43
4 x 4 x3/2 x2 dx
43 2 x 2 85 x5/2 x3 43 32 8532 64
3
0
324 3 1 85 32 8153 (15 24 10) 8153
Copyright 2016 Pearson Education, Ltd.
Chapter 6 Practice Exercises
3.
415
A( x) 4 (diameter) 2 4 (2sin x 2 cos x)2
4 4 sin 2 x 2sin x cos x cos 2 x (1 sin 2 x);
b
a 4 , b 54 V A( x) dx
a
5 /4
/4
(1 sin 2 x) dx x cos22 x
5 /4
/4
cos 5 x
cos
54 2 2 4 2 2 2
4.
A( x ) (edge)2
b
6
a
0
A( x ) dx
2
6 x 0 6 x 36 24 6 x 36 x 4 6 x3/2 x2 ; a 0, b 6 V
2
4
36 24 6 x 36x 4 6 x3/2 x2 dx 36 x 24 6 23 x3/2 18x2 4 6 52 x5/2 x3 0
3
3
216 16 6 6 6 18 62 85 6 6 62 63 216 576 648 1728
72 360 1728
180051728 72
5
5
5
5.
2
A( x ) 4 (diameter)2 4 2 x x4
4
4
0
4x x ; a 0, b 4 V A( x) dx
2
5/2
4
4x x dx 2x x
5/2
x4
16
2
4
2 7/2
7
b
x4
16
4
x5
516 0
4
a
32 32 78 52 32 324 1 87 52
(35 40 14) 72
835
35
6.
A( x) 12 (edge)2 sin 3 43 2 x 2 x
2
4 3x; a 0, b 1
2
43 4 x
1
b
1
V A( x) dx 4 3x dx 2 3 x 2 2 3
0
a
0
7. (a) disk method:
b
1
1
2
V R ( x) dx 3x 4 dx
2
a
1
1
9 x8 dx x9 2
1
1
(b) shell method:
b
V 2
a
1
1 5
shell
4
x
shell
radius height dx 0 2 x 3 x dx 2 30 x dx 2 3 6
0
6
1
Note: The lower limit of integration is 0 rather than 1.
(c) shell method:
b
V 2
a
1
shell
4
3 1 12
3x x
3 1
shell
radius height dx 2 1 (1 x) 3 x dx 2 5 2 2 5 2 5 2 5
1
3
6
1
Copyright 2016 Pearson Education, Ltd.
6
416
Chapter 6 Applications of Definite Integrals
(d) washer method:
b
R( x) 3, r ( x) 3 3x 4 3 1 x 4 V R( x) r ( x)
a
2
2
dx 9 9 1 x dx
4 2
1
1
1
5
9
1
1
9 1 1 2 x 4 x8 dx 9 2 x 4 x8 dx 9 25x x9 18 52 19 2513 265
1
1
1
8. (a) washer method:
b
R ( x ) 43 , r ( x) 12 V R ( x) r ( x)
a
x
2
2
dx dx x
2
4
x3
1
2
1 2
2
16 5
5
2
x
4 1
1 1 16 1 ( 2 10 64 5) 57
516
1 16
14 10
5
2
5
4
20
20
32 2
(b) shell method:
2
V 2 x
1
dx 2 4x 2 1 4 2
4
x3
(c) shell method:
b shell
a radius
V 2
2
x2
4 1
1
1
2
4
2
5
4
1
4
5
2
2
2 8
4
1
4 1 x dx
shell
height dx 2 1 (2 x ) 2 dx 2 1
2
x
x
x
3
2
3
2
2
2 42 4x x x4 2 (1 2 2 1) 4 4 1 14 32
x
1
(d) washer method:
b
V R ( x) r ( x)
a
2
2
dx
2
2
2
72 4 43 dx
1
x
2
494 16 1 2x 3 x 6 dx
1
2
5
494 16 x x 2 x5
1
494 16 2 14 5132 1 1 15 494 16
14 1601 15 49 16 (40 1 32) 494 7110
103
20
9. (a) disk method:
5
x 1 dx 15 ( x 1) dx x2 x 1 252 5 12 1 242 4 8
1
V
5
2
2
(b) washer method:
d
R( y ) 5, r ( y ) y 2 1 V R( y ) r ( y )
c
2
2
dy 25 y 1 dy
2
2
2
2
2
2
y
25 y 4 2 y 2 1 dy 24 y 4 2 y 2 dy 24 y 5 32 y 3
2
2
2
5
2
(45 6 5) 1088
2 24 2 32
23 8 32 3 52 13 32
5
15
15
Copyright 2016 Pearson Education, Ltd.
Chapter 6 Practice Exercises
(c) disk method:
417
R( y ) 5 y 2 1 4 y 2
d
2
2
V R ( y ) dy 4 y 2
2
c
dy
2
2
8y
y
16 8 y 2 y 4 dy 16 y 3 5
2
2
3
2
2 32 64
32
64 1 23 15
3
5
5
(15 10 3) 512
64
15
15
10. (a) shell method:
d
V 2
c
shell
shell
radius height dy
4
4
y2
y3
2 y y 4 dy 2 y 2 4 dy
0
0
4
y3 y 4
2 3 16 2
0
(b) shell method:
b
V 2
a
2
643 644 212 64 323
4
4
shell
3/2
x 2 dx 2 54 x5/2 x3
shell
radius height dx 0 2 x 2 x x dx 2 0 2 x
0
3
4
54 32 643 12815
(c) shell method:
b
V 2
a
4
4
shell
1/2
4 x 2 x3/2 x 2 dx
shell
radius height dx 0 2 (4 x ) 2 x x dx 2 0 8 x
4
3
2 16
x3/2 2 x 2 54 x5/2 x3 2 16
8 32 54 32 64
64 43 1 54 23 64 1 54 645
3
3
3
0
(d) shell method:
3
d
4
4
y2
shell
2
2 y
V 2 shell
radius height dy 0 2 (4 y ) y 4 dy 2 0 4 y y y 4 dy
c
4
y
y
4 y 2 y 2 4 dy 2 2 y 2 23 y 3 16 2 32 32 64 16 32 2 83 1 323
0
0
2
3
4
4
11. disk method:
R( x) tan x, a 0, b 3 V
/3
0
12. disk method:
0
0
V (2 sin x)2 dx
tan 2 x dx
/3
0
sec2 x 1 dx tan x x0 /3 3
3 3
4 4sin x sin 2 x dx 0 4 4sin x 1cos2 2 x dx
4 x 4 cos x 2x sin42 x 4 4 2 0 (0 4 0 0)
0
92 8 2 (9 16)
13. (a) disk method:
2
x 2 2 x dx x 4 4 x3 4 x 2 dx x5 x 4 34 x3 32
16 32
5
3
0
0
0
V
2
2
5
2
(6 15 10) 16
16
15
15
Copyright 2016 Pearson Education, Ltd.
418
Chapter 6 Applications of Definite Integrals
(b) washer method:
2
2
2
2
2
x 15
V 12 x 2 2 x 1 dx dx ( x 1)4 dx 2 5 2 52 85
0
0
0
0
(c) shell method:
2
2
shell
2
2
shell
radius height dx 2 0 (2 x) x 2 x dx 2 0 (2 x) 2 x x dx
2
2
2
2 4 x 2 x 2 2 x 2 x3 dx 2 x3 4 x 2 4 x dx 2 x4 43 x3 2 x 2 2 4 32
8
3
0
0
0
b
V 2
a
4
23 (36 32) 83
(d) washer method:
2
2
4 4 x 2 8 x x 4 4 x3 4 x 2 dx 8 x 4 4 x3 8 x 4 dx 8
0
0
2
2
2
2
2
V 2 x 2 2 x dx 22 dx 4 4 x 2 2 x x 2 2 x dx 8
0
0
0
2
5
x5 x 4 4 x 2 4 x 8
0
14. disk method:
V 2
/4
0
4 tan 2 x dx 8
/4
0
325 16 16 8 8 5 (32 40) 8 725 405 325
sec2 x 1 dx 8 tan x x0 /4 2 (4 )
15. The material removed from the sphere consists of a cylinder
and two “caps.” From the diagram, the height of the cylinder
3 22 , i.e. h 1. Thus
2
Vcy1 (2h) 3 6 m3 . To get the volume of a cap,
is 2h, where h 2
2
2
use the disk method and x 2 y 2 22 : Vcap x 2 dy
1
2
2
y3
4 y 2 dy 4 y 3 8 83 4 13
1
1
53 m3 . Therefore, Vremoved Vcy1 2Vcap 6 103
283 m3 .
2
x
16. We rotate the region enclosed by the curve y 12 1 4121
and the x-axis around the x-axis. To find the
b
volume we use the disk method: V R ( x) dx
a
12
11/2
11/2
1 dx 12 x
4 x2
121
2
11/2
11/2
2
2
x
12 1 4121
dx
11/2
11/2
2
x dx
12 1 4121
112 132 1 3634 114 132 1 13
11/2
4 x3
4
24 11
363
363 11/2
2
3
88 276 cm3
264
3
Copyright 2016 Pearson Education, Ltd.
2
Chapter 6 Practice Exercises
17.
3/ 2
419
2 x L 1 2 x dx
dy 2
y x1/2 x 3 dx 12 x 1/2 12 x1/2 dx
dy
4
1 1
4 x
1 1
4 x
1
dx 14 14 x1/2 x1/2 dx 14 12 x1/2 x1/2 dx 12 2 x1/2 23 x3/2 1
1
12 4 23 8 2 23 12 2 14
10
3
3
L
18.
4
2
1 1 2 x
4 x
dx 2 y 1/3
x y 2/3 dy
3
13
8
dx
dy
2
4
8
4 y 2/3
L
9
1
dy 1
dx
1 dy
2
8
1
2/3
4 dy 8 9 y 4 dy
2/3
1
9y
3 y1/3
9 y 2/3 4 y 1/3 dy; [u 9 y 2/3 4 du 6 y 1/3dy; y 1 u 13, y 8 u 40]
1
40
40 1/2
1 2 u 3/2
u du 18
1 403/2 133/2 7.634
3
13 27
13
1
L 18
19.
dx x x dx x x 2858
32
dy 2
32 1 1/5
2
1 dx
1
20.
2
dy
dy 2
12 x1/5 12 x 1/5 1 dx 14 x 2/5 21 41 x 2/5 12 x1/5 12 x 1/5
dx
1 1/5
2
1
y L 1 y dy
1 y 3 1 dx 1 y 2 1
x 12
2
y
dy
4
dx
dy
y
2
1 y 4 1 1 dy 2
16
2 y4
1
1
32
5 4/5
8
1
5 6/5
12
2
4
1
16
1
2
2
1
y4
1
16
1
4
1
2
1
y4
2
2
2 1 2
1 2 1
1
1 3 1
4 y y 2 dy 1 4 y y 2 dy 12 y y
1
8 1 1 1 7 1 13
12
2
12
12 2 12
dx;
dy 2
b
21. S 2 y 1 dx
a
2
3
2x 1
0
2 x 2 dx 2
2 x 1
dx;
dy 2
b
22. S 2 y 1 dx
a
1
3
dy 2
1
dx 2 x11 S 2
0
2 x 1
dy
dx
2
3
0
3
x 1 dx 2 2 32 ( x 1)3/2 2 2 32 (8 1) 283 2
0
3
1
dy
dy 2
x 2 dx x 4 S 2 x3
dx
0
2 x 1 1 2 x11 dx
1 x 4 dx 6
1
1 x 4 4 x3 dx
0
3/2
6 23 1 x 4
9 2 2 1
0
dy;
d
dx
23. S 2 x 1 dy
c
2
2
S 2 4 y y 2
1
d
c
2
12 (42 y )
4 y y2
2 y
4 y y
dx
1 dy
2
2
4 y y 2 44 y y 2
4 y y2
4
4 y y2
2
4
dy 4 dx 4
1
4 y y2
dy;
dx
24. S 2 x 1 dy
dx
dy
1
dx 1 1 dx
dy
dy
2 y
2
1
4y
6
4 y 1
S 2
4y
2
6
4 32 (4 y 1)3/2 6 (125 27) 6 (98) 493
2
Copyright 2016 Pearson Education, Ltd.
y
4 y 1
4y
dy
6
2
4 y 1 dy
420
Chapter 6 Applications of Definite Integrals
25. The equipment alone: the force required to lift the equipment is equal to its weight F1 ( x) 100 N . The
b
40
a
0
work done is W1 F1 ( x) dx 100 dx 100 x 0 4000 J; the rope alone: the force required to lift the
40
rope is equal to the weight of the rope paid out at elevation x F2 ( x) 0.8(40 x). The work done is
b
40
40
2
2
(0.8)(1600)
0.8(40 x) dx 0.8 40 x x2 0.8 402 402
640 J; the total work
2
0
0
W2 F2 ( x) dx
a
is W W1 W2 4000 640 4640 J
26. The force required to lift the water is equal to the water’s weight, which varies steadily from 9.8 4000 N
to 9.8 2000 N over the 1500 m elevation. When the truck is x m off the base of Mt. Washington, the
x (39,200) 1 x
water weight is F ( x ) 9.8 4000 221500
N. The work done is
1500
3000
b
1500
a
0
W F ( x) dx
1500
x2
39,200 x 23000
0
x
39,200 1 3000
dx
2
(39,200)(1500) 44,100,000 J
39,200 1500 41500
1500
3
4
27. Using a proportionality constant of 1, the work in lifting the weight of w N from r a to a is
r
r
2
2
2
r a wt dt w t2 r a w2 r ( r a ) w2 (2ar a ).
2
28. Force constant: F kx 200 k (0.8) k 250 N/m; the 300 N force stretches the spring
x Fk 300
1.2 m; the work required to stretch the spring that far is then W
250
1.2
0
1.2
0
F ( x) dx
1.2
0
250 x dx
1.2
250 x dx 125 x 2 125(1.2) 2 180 J
0
29. We imagine the water divided into thin slabs by planes
perpendicular to the y -axis at the points of a partition
of the interval [0,8]. The typical slab between the planes at
y and y y has a volume of about
V (radius) 2 (thickness)
54 y y
2
y 2 y m3 . The force F ( y ) required to lift this slab is
25
16
equal to its weight: F ( y ) 9800V
(9800)(25)
y 2 y N. The distance through which F ( y ) must act to lift this slab to the level 6 m above the
16
(9800)(25)
top is about (6 8 y ) m, so the work done lifting the slab is about W
y 2 (14 y )y J. The
16
work done lifting all the slabs from y 0 to y 8 to the level 6 m above the top is approximately
8
W
0
(9800)(25)
y 2 (14 y )y J so the work to pump the water is the limit of these Riemann sums as the
16
8 (9800)(25)
(9800)(25) 8
y 2 (14 y ) dy
14 y 2 y3
(16)
16
0
0
norm of the partition goes to zero: W
8
143 83 84 65,680,230 J
4
14 y 3 y (9800) 25
(9800) 25
16 3
4
16
0
4
Copyright 2016 Pearson Education, Ltd.
dy
Chapter 6 Practice Exercises
421
30. The same as in Exercise 29, but change the distance through which F ( y ) must act to (8 y ) rather than
(6 8 y ). Also change the upper limit of integration from 8 to 5. The integral is:
83 53 54 8,518,707 J
5
05 8 y 2 y3 dy (9800) 2516 83 y3 y4 0
4
5 (9800)(25) 2
y (8 y ) dy (9800) 25
16
16
0
W
4
(9800) 25
16
y
5 y . A typical
31. The tank’s cross section looks like the figure in Exercise 29 with right edge given by x 10
2
horizontal slab has volume V (radius) 2 (thickness)
y y y. The force required to lift this
y 2
2
2
4
slab is its weight: F ( y ) 9000 y 2 y. The distance through which F ( y ) must act is (2 10 y ) m, so the
10
10
y2
12 y3 y 4
work to pump the liquid is W 9000 (12 y ) 4 dy 2250 3 4 3,375,000 J; the time
0
0
needed to empty the tank is
3,375,000 J
257 s.
41,250 J
32. A typical horizontal slab has volume about V (6)(2 x)y (6) 2 1.5625 y 2 y and the force required
2
to lift this slab is its weight F ( y ) (8950)(6) 2 1.5625 y y. The distance through which F ( y ) must
act is (2 1.25 y ) m, so the work to pump the olive oil from the half-full tank is
W 8950
0
107,400
0
1.25
1.25
(3.25 y )(6) 2 1.5625 y 2 dy
1.5625 y 2 ( 2 y ) dy 349,050
1.25
3.25 1.5625 y 2 dy 53,700
1/2
0
0
3/2
(area of a quarter circle having radius 1.25) 23 (53,700) 1.5625 y 2
1.25
(349,050)(0.390625 ) 69,922 498,270 J
33. Intersection points: 3 x 2 2 x 2 3x 2 3 0
3( x 1)( x 1) 0 x 1 or x 1. Symmetry
suggests that x 0. The typical vertical strip has center of
2 x 2 3 x 2
x 2 3 ,
mass: ( x , y ) x,
x
,
2
2
length: 3 x 2 2 x 2 3 1 x 2 , width: dx,
1
y dm 32 x 2 31 x 2 dx 32 x 4 2 x 2 3 dx M x y dm 32 x 4 2 x 2 3 dx
1
1
1
32 x5 23x 3 x 3 15 23 3 315 ( 3 10 45) 325 ; M dm 3 1 x 2 dx
1
1
area: dA 3 1 x 2 dx, and mass: dm dA 3 1 x 2 dx the moment about the x-axis is
5
3
1
3
3 x x3 6
1
1 13 4 y MM 5324 85 . Therefore, the centroid is ( x , y ) 0, 85 .
x
Copyright 2016 Pearson Education, Ltd.
422
Chapter 6 Applications of Definite Integrals
34. Symmetry suggests that x 0. The typical vertical strip
2
has center of mass: ( x , y ) x, x2 , length: x 2 , width: dx,
area: dA x 2 dx, mass: dm dA x 2 dx the
moment about the x-axis is y dm 2 x 2 x 2 dx
2
2
2 x 4 dx M x y dm 2 x 4 dx 10 x5
2
2
35. The typical vertical strip has: center of mass: ( x , y )
4 x2
2
x, 2 4 , length: 4 x4 , width: dx,
4 dx the moment about the x-axis is
2
area: dA 4 x4 dx, mass: dm dA
x2
4
4 x2
4
2
4 dx 16 dx; moment about: x dm 4 x dx 4 x dx.
64
Thus, M y dm 16 dx 16 x
; M x dm
4 x dx 2 x (32 16) 16 ; M dm 4 dx 4 x
y dm
x2
4
4
x
x3
4
0
x4
16
2 0
4
x4
16
2
x2
4
4
x5
516 0
2
64
5
2
4
x4
16 0
2
128
5
4
0
x3
4
y
x2
4
4
x3
12 0
3 12 . Centroid is ( x , y ) 3 , 12 .
16 1264 323 x MM 1632 3 32 and y MM 128
2 5
532
5
y
x
36. A typical horizontal strip has: center of mass:
y2 2 y
( x , y ) 2 , y , length: 2 y y 2 , width: dy,
area: dA 2 y y 2 dy, mass: dm dA
2 y y 2 dy; the moment about the x-axis
is y dm y 2 y y 2 dy 2 y 2 y 3 dy;
y 2 y 2 y y 2 dy
2
the moment about the y -axis is x dm
2
2
4 y 2 y 4 dy M x y dm
2
0 2 y y dy 23 y 4 0 23 8 164 163 164 1216 43 ; M y x dm
2
2
3
3
y4
2
2
2
y
y
4 y 2 y 4 dy 2 34 y 3 5 2 438 32
32
; M dm 2 y y 2 dy y 2 3
5
15
0
0
0
0
2
5
2
3
4 83 43 x MM 1532 43 85 and y MM 3443 1. Therefore, the centroid is ( x , y ) 85 , 1 .
y
x
Copyright 2016 Pearson Education, Ltd.
Chapter 6 Practice Exercises
423
37. A typical horizontal strip has: center of mass:
y2 2 y
( x , y ) 2 , y , length: 2 y y 2 , width: dy,
area: dA 2 y y 2 dy, mass: dm dA
(1 y ) 2 y y 2 dy the moment about the
x-axis is y dm y (1 y ) 2 y y 2 dy
y 2 y
x dm 2 (1 y ) 2 y y 2 dy 12 4 y 2 y 4 (1 y ) dy 12 4 y 2 4 y 3 y 4 y 5 dy M x y dm
2 y 2 2 y 3 y 3 y 4 dy 2 y 2 y 3 y 4 dy; the moment about the y -axis is
2
2
y
y
4 (11) 44 ;
2 y 2 y 3 y4 dy 23 y 3 4 5 16
16
32
16 13 14 52 16
(20 15 24) 15
3
4
5
60
15
0
0
4
2
5
2
3
5
6
2
y5
y6
M y x dm 12 4 y 2 4 y 3 y 4 y 5 dy 12 43 y 3 y 4 5 6 12 432 24 25 26
0
0
2
2
4 43 2 54 86 4 2 54 24
; M dm (1 y ) 2 y y 2 dy 2 y y 2 y3 dy
5
0
0
2
83 95 and y MM 1544 83 4440 1011 . Therefore,
M
y3 y 4
y 2 3 4 4 83 16
83 x My 24
4
5
0
x
11 .
the center of mass is ( x , y ) 95 , 10
9 3
dx 6
1 x3/ 2
M
x 1/2 4 x My 12 3 and y M x 9 5
M
4
M
4
9
1
20
9
dx 4; M x dx 2x 52; M x dx 6 x
9x 9
1 2 x3
(b) M x
3
2 x3/ 2
, length:
3 , width: dx, area: dA 3 dx,
x3/ 2
x3/ 2
3
3
3
9
mass: dm dA 3/ 2 dx the moment about the x-axis is y dm 3/ 2 3/ 2 dx 3 dx; the moment
x
2x
x
2x
3
3
about the y -axis is x dm x 3/ 2 dx 1/ 2 dx.
x
x
9
9
91 9
9
9 x 2 20 ; M
3 dx 3 2 x1/2 12 ;
(a) M x
dx
x
y
3
3/
2
2
2 1
9
1
1 2 x
1
x
38. A typical vertical strip has: center of mass: ( x , y ) x,
9
2
9
1
x 1
M
y
9 2
1
3
x3/ 2
3/2 9
1
9
1
1/2 9
1
3
x3/ 2
M
12 x My 13
and y Mx 13
3
b
2
a
0
2
y
2 y y 2 dy 39,200 y 2 3
0
0
39. F W strip
depth L( y ) dy F 2 (9800)(2 y )(2 y ) dy 39,200
2
(39,200) 4 83 (39,200) 43 52,267 J
Copyright 2016 Pearson Education, Ltd.
3
424
Chapter 6 Applications of Definite Integrals
b
40. F W strip
depth L( y ) dy F
a
11,800
0.5
0
11,800 0.5 y (2 y 4) dy 11,800
y 2 2 y 2 4 y dy
0.5
75 250 11,800 (3000 75 15 250)
(11,800) 1 200
31000
3000
41.
0.5
0
2 3 y 2 y 2 dy 11,800 103 y 23 y 2 23 y3 0 (11,800) 1 23 0.25 23 0.125
0.5
0
(11,800)(1625)
6392 N.
3000
b
4
4
y
F W strip
L( y ) dy F 9800 (9 y ) 2 2 dy 9800 9 y1/2 3 y 3/2 dy
depth
0
0
a
4
9800 6 y 3/2 52 y 5/2 (9800) 6 8 52 32 9800
(48 5 64)
5
0
h
(9800)(176)
344,960 N
5
42. Place the origin at the bottom of the tank. Then F W strip
depth L ( y ) dy , h the height of the mercury
0
column, strip depth
h
h
h
y2
h y, L( y ) 0.09 F 133,350(h y ) (0.09) dy 12,001.5 ( h y ) dy 12,001.5 h y 2
0
0
0
2
12,001.5 h 2 h2
12,001.5 2
12,001.5 2
h 150,000 to get h 5 m. The volume of the mercury
h . Now solve
2
2
is s 2 h 0.32 5 0.45 m3 .
CHAPTER 6
ADDITIONAL AND ADVANCED EXERCISES
b
x
1. V f ( x) dx b 2 ab f (t ) dt x 2 ax for all x
2
a
a f ( x) 2 x a
2
2
a
2 x a
f ( x)
a
x
2. V f ( x) dx a 2 a f (t ) dt x 2 x for all x
2
0
2
a
3. s ( x) Cx
x
x
0
2
2 x 1
1 f (t ) dt Cx 1 f ( x) C f ( x) C 2 1 for C 1
2
0
f ( x)
a f ( x) 2 x 1 f ( x)
2
C 2 1 dt k . Then f (0) a a 0 k f ( x)
x
0
C 2 1 dt a f ( x) x C 2 1 a,
where C 1.
4. (a) The graph of f ( x) sin x traces out a path from (0, 0) to ( , sin ) whose length is
L
0
1 cos 2
d . The line segment from (0, 0) to ( , sin ) has length
( 0)2 (sin 0) 2 2 sin 2 . Since the shortest distance between two points is the length of
the straight line segment joining them, we have immediately that
0 1 cos
2
d
2
sin 2
if 0
2 .
(b) In general, if y f ( x) is continuously differentiable and f (0) 0, then
0 1 f (t ) dt
2
2
f 2 ( ) for
0.
Copyright 2016 Pearson Education, Ltd.
Chapter 6 Additional and Advanced Exercises
425
5. We can find the centroid and then use Pappus’ Theorem to calculate the volume. f ( x) x, g ( x) x 2 ,
f ( x) g ( x) x x 2 x 2 x 0 x 0, x 1;
1
1
1; M x x 2 dx 12 x 2 13 x3
0
0
1
1 1 x x x 2 dx 6 1 x 2 x3 dx 6 1 x3 1 x 4 6 1 1 0 1 ;
12 13 0 16 ; x 1/6
0
0
4
3 4
2
3
0
2
1
1 1 1 x 2 x 2 dx 3 1 x 2 x 4 dx 3 1 x3 1 x5 3 1 1 0 2 The centroid is 1 , 2 .
y 1/6
0 2
3
5
3 5
3
2 5
0
0
12 , 52 to the axis of rotation, y x. To calculate this distance we must find the point
on y x that also lies on the line perpendicular to y x that passes through 12 , 52 . The equation of this line
9 . The point of intersection of the lines x y 9 and y x is
is y 52 1 x 12 x y 10
209 , 209 .
10
is the distance from
Thus,
109 12 209 52 101 2 . Thus V 2 101 2 16 30π 2 .
2
2
6. Since the slice is made at an angle of 45 , the volume of the wedge is half the volume of the cylinder of radius
1
2
7.
and height 1. Thus, V 12
12 (1) 8 .
2
1 1 dx A 3 2
x
0
y 2 x ds
3
x 1x 1 dx 43 (1 x)3/2 28
0 3
8. This surface is a triangle having a base of 2 a and a height of 2 ak . Therefore the surface area is
1 (2 a )(2 ak ) 2 2 a 2 k .
2
9.
2
2
2
3
4
F ma t 2 d 2 a tm v dx
3tm C ; v 0 when t 0 C 0 dx
3tm x 12t m C1 ;
dt
dt
dt
4
x 0 when t 0 C1 0 x 12t m . Then x h t (12 mh)1/4 . The work done is
W F dx
(12 mh )1/ 4
0
1/ 4
6 (12 mh )
(12 mh )1/ 4 2 t 3
dx
t
1
F (t ) dt dt
t 3m dt 3m 6
181m
0
0
(12mh)6/4 (1218mhm)
3/ 2
12mh18 m12 mh 23h 2 3mh 43h 3mh
10. Converting to grams and centimeters, 3.6 N/cm 360 N/m. Thus, F 360 x W
0.15
0
0.15
180 x 2
0
360 x dx
4.05 J. Since W 12 mv02 12 mv12 , where W 4.05 J, m 0.05 kg
and v1 0 m/s, we have 4.05
12 201 v02 v02 4.05 40. For the projectile height, s 9.8t 2 v0t (since
v
0
s 0 at t 0 ) ds
and the height is
v 9.8t v0 . At the top of the ball’s path, v 0 t 9.8
dt
v
v
0
s 4.9 9.8
2
v0
0 9.8
v02
40 8.26 m.
4.05
19.6
19.6
M
11. From the symmetry of y 1 x n , n even, about the y -axis for 1 x 1, we have x 0. To find y Mx ,
n
we use the vertical strips technique. The typical strip has center of mass: ( x , y ) x, 12x , length: 1 x n ,
Copyright 2016 Pearson Education, Ltd.
426
Chapter 6 Applications of Definite Integrals
width: dx, area: dA 1 x n dx, mass: dm 1 dA 1 x n dx. The moment of the strip about the x-axis is
1 x dx M 1 1 x dx 2 1 1 1 2 xn x2n dx x 2 xn 1 x2n 1 1 1 2 1
n 1 2n 1 0 n 1 2n 1
x
1 2
0 2
2
n 2
y dm
n 2
2
1
1
( n 1)(2 n 1) 2(2 n 1) ( n 1)
1 4 n 2 n 1
2n2
2n (3nn1)(2
. Also, M
dA
1 x n dx
( n 1)(2 n 1)
n 1)
( n 1)(2 n 1)
1
1
n 1 1
1
M
( n 1)
2n2
2 1 x n dx 2 x xn 1 2 1 n11 n2n1 . Therefore, y Mx ( n 1)(2
2nn1
n 1) 2 n
0
0
0, 2nn1
is the location of the centroid. As n , y 12 so the limiting position of the centroid is 0, 12 .
12. Align the telephone pole along the x-axis as shown
in the accompanying figure. The slope of the top
10052 10033 1 1 (52 33)
length of pole is
100 12
12
33 19 x
10019
. Thus, y 100
12
100 12
19 x is an equation of the line
1001 33 12
representing the top of the pole. Then
b
12
2
12
19 x dx
19 x
1
x 1001 33 12
x 33 12
10,000 0
0
M y x y 2 dx
a
2
dx;
2
2
b
12
12
19 x dx
19 x dx. Thus, x M y 151,596 6.88 (using
1
M y 2 dx 1001 33 12
33 12
10,000 0
M
22,036
a
0
a calculator to compute the integrals). By symmetry about the x-axis, y 0 so the center of mass is about 6.9
meters from the top of the pole.
13. (a) Consider a single vertical strip with center of mass ( x, y ). If the plate lies to the right of the line, then the
moment of this strip about the line x b is ( x b) dm ( x b) dA the plate’s first moment about
x b is the integral ( x b) dA x dA b dA M y b A.
(b) If the plate lies to the left of the line, the moment of a vertical strip about the line x b is (b x ) dm
dA the plate’s first moment about x b is (b x) dA b dA x dA b A M y .
(b x )
14. (a) By symmetry of the plate about the x-axis, y 0. A typical vertical strip has center of mass: ( x , y )
( x, 0), length: 4 ax , width: dx, area: 4 ax dx, mass: dm
dA kx 4 ax dx, for some
a
proportionality constant k. The moment of the strip about the y -axis is M y x dm 4k x 2 ax dx
0
4k
a
a x5/2 dx 4k
0
a
4k a x3/2 dx 4k
0
a
a
8k a 4
a 72 x 7/2 4k a1/2 72 a 7/2 7 . Also, M dm 4k x ax dx
0
0
a
M
8k a 4
8k a 3
a 52 x5/2 4k a1/2 52 a5/2 5 . Thus, x My 7 5 3 75 a
0
8k a
( x , y ) 57a , 0 is the center of mass.
y2 a
y2
y 2 4a 2
(b) A typical horizontal strip has center of mass: ( x , y ) 4 a2 , y 8a , y , length: a 4a ,
y2
width: dy, area: a 4 a dy, mass: dm
y2
dA y a 4a dy. Thus,
Copyright 2016 Pearson Education, Ltd.
Chapter 6 Additional and Advanced Exercises
M x y dm
2a
2 a
427
0
2a
y2
y2
y2
y y a 4a dy y 2 a 4a dy y 2 a 4 a dy
2
a
0
0
2a
4
2a
y4
y4
y5
y5
a5 8a 4 32 a 5 0;
ay 2 4a dy ay 2 4a dy a3 y 3 20 a
a3 y 3 20a 8a3 32
a
20
3
20 a
2 a
0
2a
0
0
2a y 2 4a 2
2a
y2
y a 4a dy 81a
y
8
a
2 a
2 a
M y x dm
2a
4a 2 y 2
y 2 4a 2 4a dy 1 2 y 16a 4 y 4 dy
32 a
2 a
0
2a
2a
y6
y6
16a 4 y y5 dy 1 2 16a 4 y y 5 dy 1 2 8a 4 y 2 6
1 2 8a 4 y 2 6
2 a
32 a 0
32 a
2 a 32a
0
0
12
32 a
2a
2a
2a
4a y
2
3
2
3
1 0
M dm
y
dy 41a
y 4a 2 y 2 dy 4a
2a 4a y y dy 41a 0 4a y y dy
2 a 4 a
2 a
6
6
6
1 2 8a 4 4a 2 646a 1 2 8a 4 4a 2 646a 1 2 32a 6 323a 1 2 32 32a 6 43 a 4 ;
32 a
16 a
32a
16a
2
2
0
2a
4
y4
y4
41a 2a 2 y 2 4
41a 2a 2 y 2 4 2 41a 2a 2 4a 2 164a 21a 8a 4 4a 4 2a3 . Therefore,
2 a
0
M
x My
43 a4 21a 23a and y MM 0 is the center of mass.
x
3
15. (a) On [0, a ] a typical vertical strip has center of mass: ( x , y ) x,
b2 x2 a 2 x2
,
2
b 2 x 2 a 2 x 2 , width: dx, area: dA b 2 x 2 a 2 x 2 dx, mass: dm
length:
dA
2
2
b 2 x 2 a 2 x 2 dx. On [a, b] a typical vertical strip has center of mass: ( x , y ) x, b 2 x ,
b 2 x 2 , width: dx, area: dA b 2 x 2 dx, mass: dm
length:
dA
b 2 x 2 dx. Thus,
a
b
M x y dm 12 b 2 x 2 a 2 x 2 b 2 x 2 a 2 x 2 dx 12 b 2 x 2
0
a
b
a
2 b 2 a 2 x 2 b 2 x x3 2 b 2 a 2 a 2 b3 b3 b 2 a a3
a
0
2 ab 2 a3 2 32 b3 ab 2 a3 3b 3a b 3 a ;
b 2 x 2 dx
a
b
a
b
2 b 2 x 2 a 2 x 2 dx 2 b 2 x 2 dx 2 b 2 a 2 dx 2 b 2 x 2 dx
0
a
0
a
3
3
3
3
3
3
3
a
b
M y x dm x b 2 x 2 a 2 x 2 dx x
0
a
0 x b x
a
2
2 1/2
dx
a
0 x a x
a
2 1/2
2
dx
a
b 2 x 2 dx
a x b x
b
3
2
2 1/2
dx
b
2 b2 x 2 3/ 2
2 a 2 x 2 3/ 2
2 b2 x 2 3/ 2
3
2
3
2
3
0
0
a
2
3/2
3/2
2 3/2
2
2 3/2
b3
a3
3 b2 a 2
b2
3 0 a
3 0 b a
3 3
Copyright 2016 Pearson Education, Ltd.
b3 a 3 M ;
3
x
428
Chapter 6 Applications of Definite Integrals
We calculate the mass geometrically: M A
b a
3
3
3
b2
a2
4
4
2
4
2
My
M
(b a ) a ab b
4 a ab b
4 a ab b
3
3
M
34 b2 a 2 34 (b a )(b a ) 3 ( a b ) ; likewise y Mx 3 ( a b ) .
2
2
b a
b a
4
2
b 2
lim 34 a aab
b
ba
(b)
b a . Thus, x
4
3
2
a2 a2 a2
aa
2
2
2
2
2
( x , y ) , is the limiting position of the
4
3
3a 2
2a
2a
2a 2a
centroid as b a. This is the centroid of a circle of radius a (and the two circles coincide when b a ).
16. Since the area of the triangle is 400, the diagram may be
labeled as shown at the right. The centroid of
the triangle is
. The shaded portion is
a3 , 400
3a
1600 400 1200. Write ( x y ) for the centroid of the
remaining region. The centroid of the whole square is
obviously (20, 20). Think of the square as a sheet of
uniform density, so that the centroid of the square
is the average of the centroids of the two regions,
weighted by area: 20
400 a3 1200( x )
1600
and
400 400
1200( y )
3a
which we solve to get x 80
9a
1600
3
80/3( a 5/3)
and y
. Set x 22 cm (Given). It follows that a 42, whence y 4840
25.6 cm. The distances
a
189
20
of the centroid ( x , y ) from the other sides are easily computed. (If we set y 22 cm above, we will find
x 25.6 cm.)
17. The submerged triangular plate is depicted in the
figure at the right. The hypotenuse of the triangle
has slope 1 y (2) ( x 0) x ( y 2)
is an equation of the hypotenuse. Using a typical
horizontal strip, the fluid pressure is
strip
F (9800) strip
depth length dy
2
6
(9800)( y ) ( y 2) dy
2
y3
y 2 2 y dy 62.4 3 y 2
6
6
9800
2
(9800)(112)
(9800) 208
32
365,867 N
3
3
(9800) 83 4 216
36
3
18. Consider a rectangular plate of length and width
w. The length is parallel with the surface of the fluid
of weight density . The force on one side of the
0
2
y2
w ( y)() dy 2 w 2w .
The average force on one side of the plate is
plate is F
0
0
2
y2
( y ) dy w 2 2w . Therefore the force 2w
w
w
(the average pressure up and down ) · (the area of the plate).
Fav w
0
2w (w)
Copyright 2016 Pearson Education, Ltd.
CHAPTER 7 TRANSCENDENTAL FUNCTIONS
7.1
INVERSE FUNCTIONS AND THEIR DERIVATIVES
1. Yes one-to-one, the graph passes the horizontal line test.
2. Not one-to-one, the graph fails the horizontal line test.
3. Not one-to-one since (for example) the horizontal line y 2 intersects the graph twice.
4. Not one-to-one, the graph fails the horizontal line test.
5. Yes one-to-one, the graph passes the horizontal line test.
6. Yes one-to-one, the graph passes the horizontal line test.
7. Not one-to one since the horizontal line y 3 intersects the graph an infinite number of times.
8. Yes one-to-one, the graph passes the horizontal line test.
9. Yes one-to-one, the graph passes the horizontal line test.
10. Not one-to one since (for example) the horizontal line y 1 intersects the graph twice.
11. Domain: 0 x 1, Range: 0 y
12. Domain: x 1, Range: y 0
13. Domain: 1 x 1, Range: 2 y 2
14. Domain: x , Range: 2 y 2
Copyright 2016 Pearson Education, Ltd.
429
430
Chapter 7 Transcendental Functions
15. Domain: 0 x 6, Range: 0 y 3
16. Domain: 2 x 1, Range: 1 y 3
17. The graph is symmetric about y x.
(b)
y 1 x 2 y 2 1 x 2 x 2 1 y 2 x 1 y 2 y 1 x 2 f 1 ( x)
18. The graph is symmetric about y x.
y 1x x 1y y 1x f 1 ( x)
19. Step 1:
Step 2:
20. Step 1:
Step 2:
21. Step 1:
Step 2:
22. Step 1:
y x2 1 x2 y 1 x
y x 1 f 1 ( x)
y x 2 x y , since x 0.
y x f 1 ( x)
y x3 1 x3 y 1 x ( y 1)1/3
y 3 x 1 f 1 ( x)
y x 2 2 x 1 y ( x 1)2
Step 2:
y 1 x f
23. Step 1:
y ( x 1) 2
Step 2:
y 1
1
y x 1, since x 1 x 1 y
( x)
y x 1, since x 1 x
y 1
y x 1 f 1 ( x)
Copyright 2016 Pearson Education, Ltd.
Section 7.1 Inverse Functions and Their Derivatives
24. Step 1:
y x 2/3 x y 3/2
Step 2:
y x3/2 f 1 ( x)
25. Step 1:
y x5 x y1/5
Step 2:
y 5 x f 1 ( x);
x and f 1 f ( x) x5 x
5
Domain and Range of f 1 : all reals; f f 1 ( x) x1/5
26. Step 1:
y x 4 x y1/4
Step 2:
y 4 x f 1 ( x);
1/5
x and f 1 f ( x) x4 x
4
Domain of f 1 : x 0, Range of f 1 : y 0; f f 1 ( x) x1/4
27. Step 1:
Step 2:
431
1/4
y x3 1 x3 y 1 x ( y 1)1/3
y 3 x 1 f 1 ( x);
Domain and Range of f 1 : all reals;
f f 1 ( x) ( x 1)1/3
28. Step 1:
Step 2:
1 ( x 1) 1 x and f 1 f ( x) x3 1 1 x3 x
1/3
3
1/3
y 12 x 72 12 x y 72 x 2 y 7
y 2 x 7 f 1 ( x);
Domain and Range of f 1 : all reals;
f f 1 ( x) 12 (2 x 7) 72 x 72 72 x and f 1 f ( x) 2 12 x 72 7 ( x 7) 7 x
29. Step 1:
Step 2:
y 12 x 2 1y x 1
y
x
y
1 f 1 ( x )
x
x 1 x1 x and
Domain of f 1 : x 0, Range of f 1 : y 0; f f 1 ( x)
1
1
x
2
11 x and f 1 f ( x)
x
since x 0
30. Step 1:
y 13 x3 1y x
Step 2:
1 3 1 f 1 ( x );
y 1/3
x
x
Domain of f
f 1 f ( x)
31. Step 1:
x
1
1
y1/3
: x 0, Range of f 1 : y 0; f f 1 ( x)
1
x3
1/3
1/3 3
1
x
1x
1
2 y 3
y xx23 y ( x 2) x 3 xy 2 y x 3 xy x 2 y 3 x y 1
Copyright 2016 Pearson Education, Ltd.
1
1
x2
11 x
x
432
Chapter 7 Transcendental Functions
Step 2:
y 2xx13 f 1 ( x);
(2 x 3)3( x 1) 5 x x and
5
2 (2 x 3) 2( x 1)
x 1
2 x 3 3
Domain of f 1 : x 1, Range of f 1 : y 2; f f 1 ( x) 2xx13
2( x 3)3( x 2) 5 x x
xx32 1 ( x 3)( x2) 5
2 xx 32 3
f 1 f ( x)
32. Step 1:
Step 2:
x
y x 3
x 3
1
3x 2
y
x y x 3 y x y x x 3 y x y3y1
2
f ( x);
y x 1
Domain of f
1
: (, 0] (1, ), Range of f
1
:[0, 9) (9, ); f f
x3x1
( x)
; If x 1 or
2
x3x1 3
2
1
3 x
3x
x3x1
x 1
3x
3 x x and f 1 f ( x ) x 3
9x
3x
2
3
2
x
3 3 x 3( x 1)
3x
1
x
1
x
x 3
x 1 3
x 3
2
2
x 0 x3x1 0
99x x
33. Step 1:
Step 2:
y x 2 2 x, x 1 y 1 ( x 1)2 , x 1 y 1 x 1, x 1 x 1 y 1
y 1 x 1 f 1 ( x);
Domain of f 1 : [1, ), Range of f 1 : (, 1];
f f 1 ( x) 1 x 1
2 1 x 1 1 2 x 1 x 1 2 2 x 1 x and
2
f 1 f ( x) 1
x2 2 x 1, x 1 1 ( x 1)2 , x 1 1 | x 1| 1 (1 x) x
y5 2 x3 1 y5 1 2 x3 y 21 x3 x 3 y 21
34. Step 1:
Step 2:
y 2 x3 1
5
5
1
: (, ), Range of f
1
1/5
5
5 1/5
1
y 2x 3 2x y 3
: (, ); f f
x 1 1 x x and f f ( x)
35. (a)
1
5
x 1
(b)
12
Copyright 2016 Pearson Education, Ltd.
5
2 x 21 1
3 1/5
3 2 x 1 1
(2 x3 1) 1 3 2 x3
3
2 x
2
2
y
df
df 1
2, dx
dx x 1
1/5
5 3
( x) 2 3 x 21 1
x 2 23 f 1 ( x) 2x 23
(c)
5
y 3 x 21 f 1 ( x);
Domain of f
1/5
1/5
Section 7.1 Inverse Functions and Their Derivatives
36. (a)
(b)
y 15 x 7 15 x y 7
x 5 y 35 f 1 ( x) 5 x 35
(c)
37. (a)
df
df 1
1,
dx x 1 5 dx
5
x 34/5
y 5 4x 4x 5 y
(b)
y
x 54 4 f 1 ( x) 54 4x
(c)
38. (a)
df
df 1
4,
dx x 1/2
dx
x
2
df
4 x x 5 20,
dx x 5
df 1
dx
(c)
(b)
y f 1 ( x)
2
39. (a)
14
y 2 x 2 x 2 12 y
x 1
(c)
x 3
1 x 1/2
2 2
x 50
1
20
x 50
x, g ( f ( x)) 3 x3 x
(b)
3
f ( g ( x)) 3 x
f ( x ) 3x 2 f (1) 3, f (1) 3;
g ( x) 13 x 2/3 g (1) 13 , g (1) 13
(d) The line y 0 is tangent to f ( x) x3 at
(0, 0); the line x 0 is tangent to g ( x) 3 x at
(0, 0)
x,
(b)
1/3 3
40. (a) h(k ( x)) 14 4 x
x
3
k (h( x)) 4 x4
1/3
2
(c) h( x) 34x h(2) 3, h(2) 3;
k ( x) 43 (4 x)2/3 k (2) 13 , k (2) 13
3
(d) The line y 0 is tangent to h( x) x4 at (0, 0);
the line x 0 is tangent to k ( x ) (4 x)1/3 at
(0, 0)
41.
df
df 1
3 x 2 6 x dx
dx
x f (3)
df1
dx
x 3
91
42.
df
df 1
2 x 4 dx
dx
Copyright 2016 Pearson Education, Ltd.
x f (5)
df1
dx x 5
16
433
434
43.
Chapter 7 Transcendental Functions
df 1
dx
45. (a)
df 1
x 4
dx
x f (2)
df1
dx x 2
11 3
dg 1
dx
44.
3
x 0
dg 1
dx
x f (0)
dg1
dx
x 0
12
y mx x m1 y f 1 ( x) m1 x
(b) The graph of y f 1 ( x) is a line through the origin with slope m1 .
46.
b f 1 ( x) 1 x b ; the graph of f 1 ( x) is a line with slope 1 and y -intercept
y mx b x m m
m
m
m
y
b.
m
47. (a)
y x 1 x y 1 f 1 ( x) x 1
(b) y x b x y b f 1 ( x) x b
(c) Their graphs will be parallel to one another and
lie on opposite sides of the line y x
equidistant from that line.
48. (a)
y x 1 x y 1 f 1 ( x) 1 x; the
lines intersect at a right angle
y x b x y b f 1 ( x) b x; the
lines intersect at a right angle
(c) Such a function is its own inverse.
(b)
49. Let x1 x2 be two numbers in the domain of an increasing function f. Then, either x1 x2 or x1 x2 which
implies f ( x1 ) f ( x2 ) or f ( x1 ) f ( x2 ), since f ( x) is increasing. In either case, f ( x1 ) f ( x2 ) and f is oneto-one. Similar arguments hold if f is decreasing.
df
df 1
50.
f ( x) is increasing since x2 x1 13 x2 56 13 x1 65 ; dx 13 dx 11 3
51.
f ( x) is increasing since x2 x1 27 x23 27 x13 ; y 27 x3 x 13 y1/3 f 1 ( x) 13 x1/3 ;
3
df
df 1
2
81
x
12
12/3 91 x 2/3
dx
dx
81x 1 x1/3
9x
3
52.
f ( x) is decreasing since x2 x1 1 8 x23 1 8 x13 ; y 1 8 x3 x 12 (1 y )1/3 f 1 ( x) 12 (1 x)1/3 ;
df
df 1
24 x 2 dx 1 2
1 2/3 16 (1 x)2/3
dx
6(1 x )
24 x 1 (1 x )1/3
2
53.
f ( x) is decreasing since x2 x1 1 x2 1 x1 ; y (1 x)3 x 1 y1/3 f 1 ( x) 1 x1/3 ;
3
df
df 1
2
1
3(1
x
)
dx
dx
3(1 x )2
1/3
1 x
3
1 1 x 2/3
2/3
3
3x
Copyright 2016 Pearson Education, Ltd.
Section 7.1 Inverse Functions and Their Derivatives
54.
435
f ( x) is increasing since x2 x1 x25/3 x15/3 ; y x5/3 x y 3/5 f 1 ( x) x3/5 ;
df
df 1
53 x 2/3 dx 5 12/3
dx
x
3
x
3/5
3 3 x 2/5
5
5 x 2/5
55. The function g ( x) is also one-to-one. The reasoning: f ( x) is one-to-one means that if x1 x2 then
f ( x1 ) f ( x2 ), so f ( x1 ) f ( x2 ) and therefore g ( x1 ) g ( x2 ). Therefore g ( x) is one-to-one as well.
56. The function h( x) is also one-to-one. The reasoning: f ( x) is one-to-one means that if x1 x2 then
f ( x1 ) f ( x2 ), so f (1x ) f (1x ) , and therefore h( x1 ) h( x2 ).
1
2
57. The composite is one-to-one also. The reasoning: If x1 x2 then g ( x1 ) g ( x2 ) because g is one-to-one. Since
g ( x1 ) g ( x2 ), we also have f g ( x1 ) f g ( x2 ) because f is one-to-one. Thus, f g is one-to-one
because x1 x2 f g ( x1 ) f g ( x2 ) .
58. Yes, g must be one-to-one. If g were not one-to-one, there would exist numbers x1 x2 in the domain of g
with g ( x1 ) g ( x2 ). For these numbers we would also have f g ( x1 ) f g ( x2 ) , contradicting the
assumption that f g is one-to-one.
59. ( g f )( x) x g f ( x) x g f ( x) f ( x) 1
60. W (a )
f ( a )
f (a)
f 1 ( y )
a2 dy 0 aa 2 x f (a) f ( x) dx S (a);
2
2
W (t ) f 1 f (t ) a 2 f (t ) t 2 a 2 f (t ); also
t
t
t
S (t ) 2 f (t ) x dx 2 x f ( x) dx f (t )t 2 f (t )a 2 2 x f ( x) dx
a
a
a
S (t ) t 2 f (t ) 2 t f (t ) a 2 f (t ) 2 t f (t ) t 2 a 2 f (t ) W (t ) S (t ). Therefore, W (t ) S (t )
for all t [a, b].
61-68. Example CAS commands:
Maple:
with(plots); # 61
f : x - sqrt(3* x-2);
domain: 2/3..4;
x0: 3;
Df : D(f);
# (a)
plot( [f (x), Df (x)], x domain, color [red,blue], linestyle [1,3], legend [" y f(x)"," y f '(x)"],
title "# 61(a) (Section 7.1)");
q1: solve( y f(x), x );
# (b)
g : unapply( q1, y );
m1: Df(x0);
# (c)
Copyright 2016 Pearson Education, Ltd.
436
Chapter 7 Transcendental Functions
t1: f(x0)+m1*(x -x0);
y t1;
m2 : 1/Df(x0);
# (d)
t2 : g(f(x0)) m2*(x-f(x0));
y t2;
domaing : map(f , domain);
# (e)
p1: plot( [f(x), x], x domain, color [pink, green], linestyle [1,9], thickness [3, 0] ):
p2 : plot( g(x), x domaing, color cyan, linestyle 3, thickness 4 ):
p3 : plot( t1, x x0-1..x0+1, color red, linestyle 4, thickness 0 ) :
p4 : plot( t2, x f x0 -1..f x0 1, color blue, linestyle 7, thickness 1) :
p5 : plot([[x0, f (x0)], [f(x0),x0]], color green ) :
display( [p1, p2, p3, p4, p5], scaling constrained, title "# 61(e) Section 7.1 " );
Mathematica: (assigned function and values for a, b, and x0 may vary)
If a function requires the odd root of a negative number, begin by loading the RealOnly package that allows
Mathematica to do this.
<<Miscellaneous `RealOnly`
Clear [x, y]
{a,b} {2, 1}; x0 1/2;
f[x _ ] (3x 2) / (2 x 11)
Plot [{f[x], f'[ x]}, {x, a, b}]
sol x Solve[y f[x], x]
g[y _ ] x / . sol x[[1]]
y0 f[x0]
f tan[x _ ] y0 f'[x0] (x-x0)
g tan[y _ ] x0 1 / f'[x0] (y y0)
Plot [{f[x], f tan[x], g[x], gtan[x], Identity[x]},{x, a, b},
Epilog Line[{{x0, y0},{y0, x0}}], PlotRange {{a,b},{a,b}}, AspectRatio Automatic]
69-70. Example CAS commands:
Maple:
with(plots);
eq : cos(y) x^(1/5);
domain: 0..1;
x0: 1/2;
f : unapply( solve( eq, y ), x);
# (a)
Df : D(f);
plot( [f (x), Df (x), x domain, color [red, blue], linestyle [1,3], legend [" y f(x)"," y f '(x)"],
title "# 70(a) (Section 7.1)" );
q1: solve( eq, x );
# (b)
g : unapply( q1, y );
Copyright 2016 Pearson Education, Ltd.
Section 7.2 Natural Logarithms
m1: Df(x0);
437
# (c)
t1: f(x0)+m1*(x -x0);
y t1;
m2 : 1/Df(x0);
# (d)
t2 : g(f(x0)) m2*(x-f(x0));
y t2;
domaing : map(f , domain);
# (e)
p1: plot( [f(x), x], x domain, color [pink, green], linestyle [1,9], thickness [3, 0] ):
p2 : plot( g(x), x domaing, color cyan, linestyle 3, thickness 4 ) :
p3 : plot( t1, x x0-1..x0+1, color red, linestyle 4, thickness 0 ) :
p4 : plot( t2, x f x0 -1..f x0 1, color blue, linestyle 7, thickness 1 ) :
p5 : plot( [[x0, f (x0)], [f(x0),x0]], color green ) :
display( [p1, p2, p3, p4, p5], scaling constrained, title "# 70(e) Section 7.1 " );
Mathematica: (assigned function and values for a, b, and x0 may vary)
For problems 69 and 70, the code is just slightly altered. At times, different "parts" of solutions need to be used, as
in the definitions of f[x] and g[y]
Clear [x, y]
{a,b} {0, 1}; x0 1/ 2 ;
eqn Cos[y] x1/5
soly Solve[eqn, y]
f[x _ ] y / . soly[[2]]
Plot[{f[x], f '[x]}, {x, a, b}]
solx Solve[eqn, x]
g[y _] x / . sol x[[1]]
y0 f[x0]
ftan[x _ ] y0 f'[x0] (x x0)
gtan[y _ ] x0 1/ f'[x0] (y y0)
Plot [{f[x], ftan[x], g[x], gtan[x], Idenity[x]},{x, a, b},
Epilog Line[{{x0, y0},{y0, x0}}], PlotRange {{a, b},{a, b}}, AspectRatio Automatic]
7.2
NATURAL LOGARITHMS
1. (a) ln 0.75 ln 34 ln 3 ln 4 ln 3 ln 22 ln 3 2 ln 2
(b) ln 94 ln 4 ln 9 ln 22 ln 32 2 ln 2 2 ln 3
(d) ln 3 9 13 ln 9 13 ln 32 32 ln 3
(c) ln 12 ln1 ln 2 ln 2
(e) ln 3 2 ln 3 ln 21/2 ln 3 12 ln 2
(f ) ln 13.5 12 ln 13.5 12 ln 27
12 ln 33 ln 2 12 (3 ln 3 ln 2)
2
1 ln 1 3 ln 5 3 ln 5
2. (a) ln 125
(b) ln 9.8 ln 49
ln 7 2 ln 5 2 ln 7 ln 5
5
Copyright 2016 Pearson Education, Ltd.
438
Chapter 7 Transcendental Functions
(d) ln 1225 ln 352 2 ln 35 2 ln 5 2 ln 7
(c) ln 7 7 ln 73/2 32 ln 7
7 ln 7 ln 53 ln 7 3 ln 5
ln 0.056 ln 125
(e)
3. (a) ln sin ln
1 ln
2
sin5 ln sin ln 5
sin
5
ln
2
(b) ln 3x 9 x ln
(c)
(f )
1
7 ln 5 ln 7 ln 7 1
ln 25
2 ln 5
2
ln 35 ln
1
3x
3 x 2 9 x
3x
ln ( x 3)
4t 4 ln 2 ln 4t 4 ln 2 ln 2t 2 ln 2 ln 22t ln t 2
2
4. (a) ln sec ln cos ln (sec )(cos ) ln 1 0
8x44 ln (2 x 1)
1/3
3
(t 1)(t 1)
(c) 3 ln t 2 1 ln (t 1) 3 ln t 2 1 ln (t 1) 3 13 ln t 2 1 ln (t 1) ln (t 1) ln (t 1)
(b) ln (8 x 4) ln 22 ln (8 x 4) ln 4 ln
6.
1 (k ) 1
y ln kx y kx
x
8.
y ln t 3/2 dt 3/12
t
12.
y ln (2 2) d
14.
d (ln x)
y (ln x)3 dx 3(ln x) 2 dx
5.
y ln 3x y 31x (3) 1x
7.
y ln t 2 dt 12 (2t ) 2t
t
9.
y ln 3x ln 3x 1 dx
10.
y ln 10
ln 10 x 1 dx
x
11.
y ln ( 1) d 11 (1) 11
13.
y ln x3 dx
15.
d (ln t ) (ln t ) 2
y t (ln t )2 dx (ln t ) 2 2t (ln t ) dt
16.
d (ln t ) (ln t )1/2
y t ln t t (ln t )1/2 dt (ln t )1/2 12 t(ln t ) 1/2 dt
17.
x
y x4 ln x 16
x3 ln x x4 1x 416x x3 ln x
dx
18.
y x 2 ln x
19.
y t dt
dy
dy
3x
dy
2
1
10 x 1
3x
1
x3
2
3
x
dy
4
4
dy
3 1/2
2
3
2t
1
x
dy
4
t
dy
1
x
(10x )
dy
dy
2
1
3 x 1
dy
dy
212 (2) 11
3(ln x ) 2
x
2t ln t
(ln t )2 2 ln t
t
t (ln t )1/ 2
1
(ln t )1/2
2t
2(ln t )1/ 2
3
dydx 4 x2 ln x x2 1x 2x ln x 4 x6 (ln x)3 ( x 2 x ln x ) 4 x7 (ln x)3 8x7 (ln x)4
ln t
dy
4
3
t 1t (ln t )(1)
t2
1ln t
t2
Copyright 2016 Pearson Education, Ltd.
Section 7.2 Natural Logarithms
1
t (1 ln t )(1) 11ln t
1 ln t
ln t
dy
dt t
2
t
t2
t2
t
20.
y
21.
y 1 ln x y
22.
y 1 ln x y
23.
y ln (ln x) y ln1 x
24.
1
1
y ln ln (ln x) y ln (ln
d ln (ln x ) ln (ln
1 d (ln x ) x (ln x )1ln (ln x )
x ) dx
x ) ln x dx
25.
dy
y sin (ln ) cos (ln ) d sin (ln ) cos (ln ) cos (ln ) 1 sin (ln ) 1
ln x
x ln x
1x lnx x lnx x
(1 ln x ) 1x (ln x ) 1x
(1 ln x )2
(1 ln x )2
1
x (1 ln x )2
(1ln x)2 ln x 1
(1 ln x ) ln x x 1x ( x ln x ) 1x
(1 ln x ) 2
1
x
(1 ln x )2
ln x
(1 ln x ) 2
1
x ln x
sin (ln ) cos (ln ) cos (ln ) sin (ln ) 2 cos (ln )
2
dy
sec (tan sec )
sec
tan sec
26.
tan sec
y ln (sec tan ) d secsec
tan
27.
y ln
28.
1 x 1
y 12 ln 11 xx 12 ln (1 x) ln (1 x) y 12 11x 11x (1) 12 (11 xx)(1
x ) 1 x 2
29.
ln t
y 11ln
t
dt
30.
y ln t ln t1/2
1
ln x 12 ln ( x 1) y 1x 12
x x 1
x11 2(2 xx(x1)1)x 23x(xx21)
dy
(1ln t )2
1/2 1/2
12 ln t
1t lnt t 1t lnt t
(1ln t ) 1t (1 ln t ) t1
(1ln t ) 2
dydt 12 ln t1/2
1/2
1/12 12 t 1/2
t
dy
1/2
d ln t1/2 1 ln t1/2
dt
2
1/2
d t1/2
1/12 dt
t
1
4t ln t
31.
1
y ln sec (ln ) d sec (ln
d sec (ln )
) d
32.
y ln 1 2 ln
sin cos
2
t (1ln t ) 2
sec (ln ) tan (ln ) d
tan (ln )
d (ln )
sec (ln )
dy
2
sin
12 (ln sin ln cos ) ln (1 2 ln ) d 12 cos
sin
cos
1 2 ln
12 cot tan (1 24 ln )
33.
x 2 15
5 ln x 2 1 1 ln (1 x) y 52 x 1 1 (1) 10 x 1
y ln
2
x 2 1 2 1 x
x 2 1 2(1 x )
1 x
34.
y ln
( x 1)5
( x 2) 20
( x 2) 4( x 1)
x2
12 5 ln ( x 1) 20 ln ( x 2) y 12 x51 x20
52 ( x 1)( x 2) 52 ( x 31)(
2
x 2)
Copyright 2016 Pearson Education, Ltd.
439
440
Chapter 7 Transcendental Functions
x2
35.
y 2
36.
y
x /2
3
x
x
ln 3 x
33 x2
dy
d x 2 ln x 2 d x 2 2 x ln | x | x ln | x |
ln t dt dx ln x 2 dx
2 dx 2
2
x ln 3 x 13 x2/3 ln x 12 x1/2
dy
d 3 x ln
ln t dt dx ln 3 x dx
d
x dx
ln x
2 x
2
2
0
0
38.
1 3x32 dx ln 3x 2 1 ln 2 ln 5 ln 52
40.
4r82r5 dr ln 4r 5 C
37.
3 1x dx ln x 3 ln 2 ln 3 ln 23
39.
y 25 dy ln y 25 C
41.
0 2cos t dt ln |2 cos t |0 ln 3 ln1 ln 3; or let u 2 cos t du sin t dt with t 0 u 1 and
2y
2
2
sin t
2
sin t
3
3
dt u1 du ln | u | 1 ln 3 ln 1 ln 3
0 2 cos t
1
t u 3
42.
/3 4 sin
0
1 4 cos
d [ln |1 4 cos |]0 /3 ln |1 2| ln 3 ln 13 ; or let u 1 4 cos du 4 sin d with
0 u 3 and 3 u 1
/3 4 sin
0
1 4 cos
1 1
du [ln | u |]13 ln 3 ln 13
3 u
d
2 2 ln x
ln 2
dx
2u du [u 2 ]0ln 2 (ln 2)2
x
1
0
43. Let u ln x du 1x dx; x 1 u 0 and x 2 u ln 2;
44. Let u ln x du 1x dx; x 2 u ln 2 and x 4 u ln 4;
2 x dxln x ln 2 u1 du ln u ln 2 ln (ln 4) ln (ln 2) ln ln 2 ln ln 2 ln ln 2 ln 2
4
ln 4
ln 4
ln 4
ln 22
2 ln 2
45. Let u ln x du 1x dx; x 2 u ln 2 and x 4 u ln 4;
4
ln 4 2
2 x(lndxx)2 ln 2 u
du u1
ln 4
ln 2
ln14 ln12
1 1 1 1 1 1
2 ln 2 ln 2
2 ln 2 ln 4
ln 22 ln 2
46. Let u ln x du 1x dx; x 2 u ln 2 and x 16 u ln16;
16
ln 16 1/2
2 2 x dxln x 12 ln 2 u
ln 16
du u1/2
ln 16 ln 2 4 ln 2 ln 2 2 ln 2 ln 2 ln 2
ln 2
2
t dt du ln | u | C ln |6 3 tan t | C
47. Let u 6 3 tan t du 3 sec2 t dt ; 63sec
u
3 tan t
sec y tan y
48. Let u 2 sec y du sec y tan y dy; 2sec y dy du
ln | u | C ln |2 sec y | C
u
49. Let u cos 2x du 12 sin 2x dx 2 du sin 2x dx; x 0 u 1 and x 2 u 1 ;
2
/2
0
/2 sin 2x
0
cos 2x
tan 2x dx
dx 2
1/ 2 du
u
1
2 ln | u |1/1 2 2 ln 12 2 ln 2 ln 2
Copyright 2016 Pearson Education, Ltd.
Section 7.2 Natural Logarithms
50. Let u sin t du cos t dt ; t 4 u 1 and t 2 u 1;
2
/2 cos t
1
du ln | u | 1
cot t dt
dt
ln 1 ln
1/ 2
/4
/4 sin t
1/ 2 u
2
/2
2
51. Let u sin 3 du 13 cos 3 d 6 du 2 cos 3 d ; 2 u 12 and u 23 ;
/2
2 cot 3 d
2 cos 3
/2 sin 3
3/2 du
3/2
6 ln | u | 1/2 6 ln 23 ln 12
u
d 6
1/2
6 ln 3 ln 27
u 1 ;
52. Let u cos 3 x du 3sin 3 x dx 2du 6 sin 3x dx; x 0 u 1 and x 12
2
/12
0
53.
6 tan 3 x dx
/12 6 sin 3 x
1/ 2 du
1/ 2
dx 2
2 ln | u | 1
2 ln 1 ln 1 2 ln
cos 3 x
u
1
2
0
2 ln 2
2 xdx 2 x 2 x dx1 x ; let u 1 x du 2 1 x dx; 2 x dx1 x duu ln | u | C
ln 1 x C ln 1 x C
54. Let u sec x tan x du sec x tan x sec2 x dx (sec x)(tan x sec x) dx sec x dx du
;
u
sec x dx
ln (sec x tan x) u duln u (ln u )
55.
12
y x( x 1) x( x 1)
1 2 1
u du 2(ln u )1 2 C 2
ln (sec x tan x) C
ln y 12 ln x( x 1) 2 ln y ln ( x) ln ( x 1)
2 y 1
x x11
y
x( x 1) 1x x11 x(2xx(1)x(21)x1) 2 2xx(x11)
y 12
56.
y
x2 1 ( x 1)2 ln y 12 ln x2 1 2 ln ( x 1) yy 12 x2 x1 x21
y
57.
2
x 2 1 ( x 1) 2
x 1
x 2 1 x 1
2
2 x 2 x 1 | x 1|
2
x 2 1 ( x 1) 2 x 2 x x 1
x 2 1 ( x 1)
x 1( x 1)
ln y 12 ln t ln (t 1) 1y dydt 12 1t t 11
dy
1
dt 12 t t 1 1t t 11 12 t t 1 t (t11)
2 t (t 1)
y t t 1 t t 1
1/2
3/ 2
58.
1/2
y t (t11) t (t 1)
dy
dt 12
59.
ln y 12 ln t ln(t 1) 1y dt 12 1t t 11
1 2t 1
t ( t 1) t (t 1)
dy
2t 1
2 t 2 t
3/ 2
dy
y 3 (sin ) ( 3)1/2 sin ln y 12 ln ( 3) ln (sin ) 1y d 2(13) cos
sin
dy
d 3 (sin ) 2(13) cot
Copyright 2016 Pearson Education, Ltd.
441
442
60.
Chapter 7 Transcendental Functions
221
2
dy
1
y (tan ) 2 1 (tan )(2 1)1/2 ln y ln (tan ) 12 ln (2 1) 1y d sec
tan
2
dy
2
1
d (tan ) 2 1 sec
(sec2 ) 2 1 tan
tan
2 1
61.
2 1
dy
dy
y t (t 1)(t 2) ln y ln t ln (t 1) ln (t 2) 1y dt 1t t 11 t 12 dt t (t 1)(t 2) 1t t 11 t 12
(t 1)(t 2) t (t 2) t (t 1)
t (t 1)(t 2)
3t 2 6t 2
t (t 1)(t 2)
62.
dy
1
y t (t 1)(
ln y ln1 ln t ln (t 1) ln (t 2) 1y dt 1t t 11 t 12
t 2)
dy
1
1
(t 1)(t 2) t (t 2) t (t 1) 3t 2 6t 2
1 1 1
dt t (t 1)(
2
t 2) t t 1 t 2 t (t 1)(t 2)
t (t 1)(t 2)
t 3 3t 2 2t
dy
dy
63.
5 ln y ln ( 5) ln ln (cos ) 1
sin
5
y cos
15 1 cos
cos
y d
d
64.
y sin ln y ln ln (sin ) 12 ln (sec ) 1y d 1 cos
sin
sec
dy
dy
d sin 1 cot 12 tan
sec
65.
2
15 1 tan
(sec )(tan )
2 sec
y
1
y x x 2/3
ln y ln x 12 ln x 2 1 23 ln ( x 1) y 1x 2x 3( x21)
( x 1)
x 1
1 1
y x x 2/3
x 2
x x 2 1 3( x 1)
( x 1)
2
( x 1)10
66.
y
67.
y3
(2 x 1)5
x ( x 2)
x 2 1
y 13 3
68.
y3
69. (a)
y
ln y 13 ln x ln( x 2) ln x 2 1 y 13
x ( x 2)
x 2 1
x ( x 1)( x 2)
x 1(2 x 3)
2
y 13 3
( x 1)10
y
ln y 12 10 ln ( x 1) 5 ln (2 x 1) y x51 2 x51 y
1
x
1 2x
x 2 x 2 1
1
x
1 2x
x 2 x 2 1
x51 2 x51
ln y 13 ln x ln ( x 1) ln ( x 2) ln x 2 1 ln (2 x 3)
x 1(2 x 3) x
x ( x 1)( x 2)
2
(2 x 1)5
1 1 1 2x 2
x 1 x 2 x 2 1 2 x 3
sin x tan x 0 x 0; f ( x ) 0 for x 0 and f ( x ) 0 for
f ( x) ln (cos x) f ( x) cos
x
4
0 x 3 there is a relative maximum at x 0 with f (0) ln (cos 0) ln 1 0; f 4
ln 12 12 ln 2 and f 3 ln cos 3 ln 12 ln 2. Therefore, the absolute
ln cos 4
minimum occurs at x 3 with f 3 ln 2 and the absolute maximum occurs at x 0 with f (0) 0.
Copyright 2016 Pearson Education, Ltd.
Section 7.2 Natural Logarithms
f ( x) cos (ln x) f ( x)
(b)
sin (ln x )
0 x 1; f ( x) 0 for 12 x 1 and
x
there is a relative maximum at x 1 with f (1) cos (ln 1) cos 0 1; f
443
f ( x) 0 for 1 x 2
12 cos ln 12
cos ( ln 2) cos (ln 2) and f (2) cos (ln 2). Therefore, the absolute minimum occurs at x 12 and
12 f (2) cos (ln 2), and the absolute maximum occurs at x 1 with f (1) 1.
x 2 with f
f ( x) x ln x f ( x) 1 1x ; if x 1, then f ( x) 0 which means that f ( x) is increasing
70. (a)
f (1) 1 ln1 1 f ( x) x ln x 0, if x 1 by part (a) x ln x if x 1
(b)
5
5
5
71.
1 ln 2 x ln x dx 1 ln x ln 2 ln x dx (ln 2)1 dx (ln 2)(5 1) ln 2 ln 16
72.
A
/3
0
/4
tan x dx 0
4
0
/3
sin x dx /3 sin x dx ln |cos x | 0
ln |cos x | 0
/4
cos x
/4 cos x
0
tan x dx
ln 1 ln 1 ln 12 ln1 ln 2 ln 2 23 ln 2
2
3
73. V
0
74. V
2
y 1
2
6
75. V 2
2
3 1
dy 4
0 y 1
dy 4
cot x dx
2 cos x
2
x
dx 2
2 1
dx 2
12x
1
x2
2
dx ln (sin x) 6 ln 1 ln 12 ln 2
6 sin x
1/2
ln | y 1|30 4 ln 4 ln1 4 ln 4
ln | x |12 2 2 ln 2 ln 12 2 2 ln 2 ln 24 ln 16
2
3
2
3
3
76. V 93x dx 27 33x dx 27 ln x3 9 27 ln 36 ln 9 27 ln 4 ln 9 ln 9
0 x 9
0 x 9
0
27 ln 4 54 ln 2
77. (a)
2
2
y x8 ln x 1 y 1 4x 1x
8 x2 4
8 x 1
dx
4
x
4
4 4 x
(b) x
L
12
4
dx
dy
2
2
2
2
8
y
dy dy 2 ln y 9 2 ln 12 1 2 ln 4
dx 2 1 dx
dy
8
y
dy
2
2
2
dx x8 ln | x | 4 8 ln 8 2 ln 4 6 ln 2
1 dy
y 2
y
2 ln 4
4
1 x4x4 x4x 4 L 48 1 y2 dx
12 y 2 16
8y
4
2
y
1 8 2y
12 y
4 8
2
y
2
2
y 2 16
y 2 16
1 8y 8y
y2
16
2
12
4
8 2 ln 3 8 ln 9
78. L
2
1
dy
1 12 dx dx 1x y ln | x | C ln x C since x 0 0 ln 1 C C 0 y ln x
x
Copyright 2016 Pearson Education, Ltd.
444
Chapter 7 Transcendental Functions
1
1 1x dx
1 2x
2
79. (a) M y x 1x dx 1, M x
12
2 1
1 x
2
(b)
2
2
21
dx
1 x
dx 21x 14 , M
1
M
2
ln | x | 1 ln 2 x My ln12 1.44 and
1
M
y Mx ln42 0.36
80. (a) M y
16
1
x
dx x dx x 42; M dx
16 1/2
1
1
x
12 ln | x |1 ln 4; M
16
3/2 16
1
2
3
16
x
16
16 1
1
2 x
1
1 16 1 dx
2 1 x
1
x
M
M
dx 2 x1 2 6 x My 7 and y Mx ln64
1
x
1
dx 4 dx 60, M dx 2 x dx
3; M
4 x
dx 4 dx 4 ln | x | 4 ln 16 x
(b) M y
16
1
1
x
x
1 2 16
1
16
4
x
16
x
1
16
1
1
x
1
1
2 x
1
x
16 1
x
4
x
16 3 2
4
x
1
My
M
16
1
1
15
ln 16
and
M
y Mx 4 ln3 16
81.
2
f ( x) ln x3 1 , domain of f: (1, ) f ( x) 33x ; f ( x) 0 3 x 2 0 x 0, not in the domain:
x 1
f ( x) undefined x3 1 0 x 1, not a domain. On (1, ), f ( x) 0 f is increasing on (1, ) f
is one-to-one
82.
g ( x) x 2 ln x , domain of g: x 0.652919 g ( x)
2 x 1x
2
2 x ln x
2 x 2 1
2 x x 2 ln x
; g ( x) 0 2 x 2 1 0
no real solutions; g ( x) undefined 2 x x 2 ln x 0 x 0 or x 0.652919, neither in domain. On
x 0.652919, g ( x) 0 g is increasing for x 0.652919 g is one-to-one
83.
84.
dy
1 1x
dx
d2y
dx 2
at (1, 3) y x ln | x | C ; y 3 at x 1 C 2 y x ln | x | 2
dy
dy
sec 2 x dx tan x C and 1 tan 0 C dx tan x 1 y (tan x 1) dx ln |sec x | x C1 and
0 ln |sec 0| 0 C1 C1 0 y ln |sec x | x
85. (a)
L( x) f (0) f (0) x, and f ( x) ln (1 x) f ( x) x 0 11x
(b) Let f ( x) ln ( x 1). Since f ( x )
x 0
1 L( x) ln1 1 x L( x) x
1
0 on [0, 0.1], the graph of f is concave down on this
( x 1) 2
interval and the largest error in the linear approximation will occur when x 0.1. This error is
0.1 ln(1.1) 0.00469 to five decimal places.
Copyright 2016 Pearson Education, Ltd.
Section 7.3 Exponential Functions
445
(c) The approximation y x for ln (1 x) is best
for smaller positive values of x; in particular
for 0 x 0.1 in the graph. As x increases, so
does the error x ln (1 x). From the graph an
upper bound for the error is 0.5 ln (1 0.5)
0.095; i.e., | E ( x)| 0.095 for 0 x 0.5.
Note from the graph that 0.1 ln(1 0.1)
0.00469 estimates the error in replacing
ln (1 x) by x over 0 x 0.1. This is
consistent with the estimate given in part (b)
above.
d ln a 1 a 1 and d ln a ln x 0 1 1 . Since in a and
86. For all positive values of x, dx
2
x
dx
x
x
x
x a
x
x
ln a ln x have the same derivative, then ln ax ln a ln x C for some constant C. Since this equation holds
for all positive values of x, it must be true for x 1 ln a1 ln a ln 1 C ln a 0 C ln a1 ln a C.
Thus ln a ln a C C 0 ln ax ln a ln x.
87. (a)
x . Since |sin x | and |cos x | are less than
y acos
sin x
(b)
or equal to 1, we have for a 1
1 y 1 for all x.
a 1
a 1
Thus, lim y 0 for all x the graph of y looks
a
more and more horizontal as a .
88. (a) The graph of y x ln x appears to be
concave upward for all x 0.
(b)
y x ln x y
1 1 y 1 1 1
2 x x
4 x3 2 x 2
x2
1 0 x 4 x 16. Thus y 0 if
x
4
0 x 16 and y 0 if x 16 so a point of inflection exists at x 16. The graph of y x ln x
closely resembles a straight line x 10 and it is impossible to discuss the point of inflection visually from
the graph.
7.3
EXPONENTIAL FUNCTIONS
1. (a) e 0.3t 27 ln e 0.3t ln 33 (0.3t ) ln e 3 ln 3 0.3t 3 ln 3 t 10 ln 3
(b) e kt 12 ln ekt ln 21 kt ln e ln 2 t lnk2
(c) e(ln 0.2)t 0.4 eln 0.2
0.4 0.2t 0.4 ln 0.2t ln 0.4 t ln 0.2 ln 0.4 t lnln 0.4
0.2
t
Copyright 2016 Pearson Education, Ltd.
446
Chapter 7 Transcendental Functions
2. (a) e 0.01t 1000 ln e0.01t ln 1000 ( 0.01t ) ln e ln1000 0.01t ln1000 t 100 ln 1000
1 ln e kt ln101 kt ln e ln10 kt ln 10 t ln10
(b) e kt 10
k
21 2t 21 t 1
(c) e(ln 2)t 12 eln 2
t
3. e t x 2 ln e t ln x 2 t 2 ln x t 4(ln x) 2
2
2
2
4. e x e2 x 1 et e x 2 x 1 et ln e x 2 x 1 ln et t x 2 2 x 1
5.
d ( 5 x ) y 5e 5 x
y e 5 x y e 5 x dx
6.
d
y e 2 x 3 y e2 x 3 dx
7.
d (5 7 x ) y 7e57 x
y e57 x y e57 x dx
8.
ye
9.
y xe x e x y e x xe x e x xe x
10.
d ( 2 x ) y 2e 2 x 2(1 2 x )e 2 x 4 xe 2 x
y (1 2 x)e 2 x y 2e 2 x (1 2 x)e 2 x dx
11.
y x 2 2 x 2 e x y (2 x 2)e x x 2 2 x 2 e x x 2 e x
12.
d (3 x )
y 9 x 2 6 x 2 e3 x y (18 x 6)e3 x 9 x 2 6 x 2 e3 x dx
23x y 32 e2 x 3
4 x x y e 4 x x d 4 x x2 y 2 2 x e 4 x x
2
2
dx
x
2
y (18 x 6)e3 x 3 9 x 2 6 x 2 e3 x 27 x 2 e3 x
13.
y e (sin cos ) y e (sin cos ) e (cos sin ) 2e cos
14.
y ln 3 e ln 3 ln ln e ln 3 ln d 1 1
15.
y cos e
16.
y 3e2 cos 5 d 3 2 e 2 cos 5 3 cos 5 e2 dd (2 ) 5(sin 5 ) 3e2
dy
sin e e sin e e ( ) 2 e sin e
2
2
dy
d
dy
d
d
2
2
2
d
d
2 e2 3cos 5 2 cos 5 5 sin 5
17.
y ln 3te t ln 3 ln t ln e t ln 3 ln t t dt 1t 1 1t t
dy
Copyright 2016 Pearson Education, Ltd.
2
2
2
Section 7.3 Exponential Functions
447
18.
dy
d (sin t ) 1 cos t cos t sin t
y ln 2et sin t ln 2 ln et ln sin t ln 2 t ln sin t dt 1 sin1 t dt
sin t
sin t
19.
y ln e ln e ln 1 e ln 1 e d 1 1 dd 1 e 1 e 1
1 e
1 e
1 e
1 e
20.
y ln
1
dy
ln ln 1 d
1
1
1
d
d
1
2
1
1
2
1
1
2 1
2 1
d
d
1
1
1
dy
1
2 1 1/ 2
21.
d (cos t ) (1 t sin t )ecos t
y e(cos t ln t ) ecos t eln t tecos t dt ecos t tecos t dt
22.
dy
y esin t ln t 2 1 dt esin t (cos t ) ln t 2 1 2t esin t esin t ln t 2 1 (cos t ) 2t
23.
0 sin e dt y sin e
24.
d e 2 x ln e 4 x d e4 x (2 x) 2e2 x 4 x
y 4 x ln t dt y ln e2 x dx
dx
dy
ln x
t
2
x
2x
4xe 8e
4 xe2 x 4 xe4 x
25. ln y e y sin x
dxd (ln x) sinx x
e2 x
e
ln x
e4 x dxd 4 x
4 x
y ye (sin x) e cos x y e sin x e cos x
y
1
y
y
y
1
y
y
ye y cos x
1 ye y sin x
y
y
e cos x y 1 ye y sin x
y
26. ln xy e x y ln x ln y e x y 1x 1y y 1 y e x y y 1y e x y e x y 1x
y xe
1
x y
1 ye x y
y y xe x 1 y
x y
x1 ye
x y
2x
2x
27. e 2 x sin ( x 3 y ) 2e 2 x 1 3 y cos ( x 3 y ) 1 3 y cos(2ex 3 y ) 3 y cos(2ex 3 y ) 1
y
2e2 x cos ( x 3 y )
3 cos ( x 3 y )
28. tan y e x ln x sec 2 y y e x 1x y
29.
e
31.
ln 2 e dx e ln 2 e
3x
3x
5e x dx e3 5e x C
ln 3 x
x ln 3
ln 3
eln 2 3 2 1
xe 1 cos y
2
x
x
30.
2e 3e
32.
ln 2 e
x
0
x
2 x
dx 2e x 32 e2 x C
0
dx e x
e0 eln 2 1 2 1
ln 2
Copyright 2016 Pearson Education, Ltd.
448
Chapter 7 Transcendental Functions
( x 1)
dx 8e( x 1) C
34.
ln 9
dx e(2 x 1) C
8e
35.
ln 4 e
ln 9 x 2
dx 2e x 2
2 e(ln 9) 2 e(ln 4) 2 2 eln 3 eln 2 2(3 2) 2
ln 4
36.
0
ln16 x 4
dx 4e x /4
0
ln16
e
2e
(2 x 1)
33.
4 e(ln16) 4 e0 4 eln 2 1 4(2 1) 4
37. Let u r1 2 du 12 r 1 2 dr 2 du r 1 2 dr ;
r
e r dr e
r1 2
r 1 2 dr 2 eu du 2eu C 2e r
12
C 2e r C
38. Let u r1 2 du 12 r 1 2 dr 2 du r 1 2 dr ;
r
e r dr e
r1 2
r 1 2 dr 2 eu du 2e r
12
C 2e r C
2
2
39. Let u t 2 du 2t dt du 2t dt ; 2tet dt eu du eu C e t C
4
4
40. Let u t 4 du 4t 3 dt 14 du t 3 dt ; t 3et dt 14 eu du 14 et C
1/ x
41. Let u 1x du 12 dx du 12 dx; e 2 dx eu du eu C e1 x C
x
x
x
42. Let u x 2 du 2 x 3 dx 12 du x 3 dx;
1 x 2
e x3 dx e
x 2
x 3 dx 12 eu du 12 eu C 12 e x
2
2
C 12 e 1/ x C
43. Let u tan du sec 2 d ; 0 u 0, 4 u 1;
4
0
1 etan sec2 d 0 4 sec2 d 01eu du tan 0 4 eu 0 tan 4 tan(0) e1 e0
1
(1 0) (e 1) e
44. Let u cot du csc 2 d ; 4 u 1, 2 u 0;
4 1 e
2
cot
csc2 d 42 csc2 d 10 eu du cot 24 eu 1 cot 2 cot 4 e0 e1
0
(0 1) (1 e) e
45. Let u sec t du sec t tan t dt du
sec t tan t dt ;
e
sec ( t )
u
sec( t )
sec ( t ) tan ( t ) dt 1 eu du e C e
C
46. Let u csc ( t ) du csc ( t ) cot ( t ) dt ;
e
csc( t )
csc( t ) cot ( t ) dt eu du eu C ecsc ( t ) C
Copyright 2016 Pearson Education, Ltd.
Section 7.3 Exponential Functions
47. Let u ev du ev dv 2 du 2ev dv; v ln 6 u 6 , v ln 2 u 2 ;
ln ( 6) 2e cos e dv 2 6 cos u du 2 sin u 6 2 sin 2 sin 6 2 1 12 1
ln ( 2)
v
2
v
2
2
2
48. Let u e x du 2 xe x dx; x 0 u 1, x ln u eln ;
0
ln
dx cos u du sin u sin ( ) sin (1) sin (1) 0.84147
2
2 xe x cos e x
2
1
1
r
49. Let u 1 e r du e r dr ; e r dr u1 du ln | u | C ln 1 er C
1 e
50.
x
11e x dx ee x 1 dx; let u e
x
1 du e x dx du e x dx;
ee x 1 dx u1 du ln | u | C ln e
x
51.
dy
et sin
dt
x
1 C
et 2 y et sin et 2 dt;
let u et 2 du et dt y sin u du cos u C cos et 2 C ; y (ln 2) 0
cos eln 2 2 C 0 cos (2 2) C 0 C cos 0 1; thus, y 1 cos et 2
52.
dy
e t sec 2
dt
et y et sec2 et dt;
let u et du e t dt 1 du e t dt y 1 sec 2 u du 1 tan u C
1 tan et C ; y (ln 4) 2 1 tan e ln 4 C 2 1 tan 1 C 2
1 (1) C 2 C 3 ; thus, y 3 1 tan e t
53.
d2y
dx 2
2e x dx 2e x C ; x 0 and dx 0 0 2e0 C C 2; thus dx 2e x 2
dy
dy
dy
y 2e x 2 x C1 ; x 0 and y 1 1 2e0 C1 C1 1 y 2e x 2 x 1 2 e x x 1
54.
d2y
dy
dy
1 e 2t dt t 12 e 2t C ; t 1 and dt 0 0 1 12 e 2 C C 12 e 2 1; thus
dt 2
dy
t 12 e 2t 12 e 2 1 y 12 t 2 14 e 2t 12 e 2 1 t C1; t 1 and y 1 1 12 14 e2 12 e 2 1 C1
dt
C1 12 14 e2 y 12 t 2 14 e 2t 12 e 2 1 t 12 14 e2
55.
y 2 x y 2 x ln 2
57.
y 5 s ds 5 s (ln 5) 12 s 1 2
58.
y 2s ds 2 s (ln 2)2 s ln 22
2
dy
2ln 5s 5 s
dy
2
y 3 x y 3 x (ln 3)(1) 3 x ln 3
56.
s2s (ln 4)s2s
2
2
Copyright 2016 Pearson Education, Ltd.
449
450
Chapter 7 Transcendental Functions
59.
y x y x ( 1)
61.
y (cos ) 2 d 2(cos )
62.
y (ln ) d (ln )( 1) 1
63.
y 7sec ln 7 d 7sec ln 7 (ln 7)(sec tan ) 7sec (ln 7)2 (sec tan )
64.
y 3tan ln 3 d 3tan ln 3 (ln 3) sec2 3tan (ln 3)2 sec2
65.
y 2sin 3t dt 2sin 3t ln 2 (cos 3t )(3) (3 cos 3t ) 2sin 3t (ln 2)
66.
y 5 cos 2t dt 5 cos 2t ln 5 (sin 2t )(2) (2sin 2t ) 5 cos 2t (ln 5)
67.
y log 2 5 ln 2 d
68.
y log3 (1 ln 3)
69.
x ln x ln x 2 ln x 3 ln x y 3
y ln
ln 4
ln 4
ln 4
ln 4
ln 4
x ln 4
70.
x ln e ln x x ln x
y ln
25 2 ln 5 2 ln 5 2 ln 5
71.
x 1 x3 ln x y 1
y x3 log10 x x3 lnln10
x3 1x 3x 2 ln x ln110 x 2 3x 2 ln 10
ln 10
ln 10
dy
2 1 (sin )
dy
(ln )
dy
dy
dy
dy
y t1e dt (1 e)t e
60.
( 1)
dy
ln 5
dy
ln12 51 (5) ln1 2
11ln 3 (ln 3) 11ln 3
ln(1 ln 3)
dy
d ln13
ln 3
2
2 ln1 5 ( x ln x) y 2 ln1 5 1 1x 2 xxln15
ln x
1 x 2 3 x 2 log x
ln10
10
1
(2 ln r ) 1 2 ln r
lnln 3r lnln 9r (lnln3)(lnr 9) dydr (ln 3)(ln
9)
r
r (ln 3)(ln 9)
2
72.
y log3 r log 9 r
73.
ln 3 ln x 1
y log3 xx 11
ln 3
74.
x 1 ln 3
y log5
7x
3x2
ln 5
log5 3 x7x 2
(ln 3) ln
xx 11 ln x 1 ln ( x 1) ln ( x 1) dy 1 1
x1
ln 3
(ln 5) 2
ln 3 7x x 2
dx
(ln 5) 2
ln 5
x 1
x 1
2
( x 1)( x 1)
ln25 ln 5 12 ln 37xx2
ln 3 7x x 2
dy
(3 x 2) 3 x
12 ln 7 x 12 ln (3 x 2) dx 277 x 2(33x 2) 2 x (3 x 2) x (3 x1 2)
75.
dy sin ln cos ln
1
y sin log 7 sin ln
sin log 7 ln17 cos log 7
ln 7
d
ln 7
ln 7 ln 7
Copyright 2016 Pearson Education, Ltd.
Section 7.3 Exponential Functions
76.
y log 7
sin cos
ln (sin ) ln(cos ) ln 2
ln(sin ) ln(cos ) ln e ln 2
ln 7
ln 7
e 2
dy
d (sincos
sin
1 ln 2 ln17
)(ln 7) (cos )(ln 7) ln 7 ln 7
(cot tan 1 ln 2)
x
77.
e x y 1
y log10 e x ln
ln10
ln10
ln10
78.
5
y 2log
5l n y
79.
dy
y 3log 2 t 3(ln t ) (ln 2) dt 3(ln t ) (ln 2) (ln 3)
80.
y 3log8 log 2 t
81.
y log 2 8t ln 2
82.
83.
5
2 l n 5
2 lnln5 5 ln 55 (1) 5 ln1 5 5 ln 5 2log5 ( ln 51)5
2
2
ln 5 2 log5
2 lnln5
ln t dy
3ln log 2 t 3ln ln 2
ln 8
ln 8
dt
t ln1 2 1t log2 3 3log t
2
3
1
1
1
ln38 (ln t )/(ln
2) t ln 2 t (ln t )(ln 8) t (ln t )(ln 2)
3 ln 2(ln 2)(ln t ) 3 ln t dy 1
ln 8 ln t ln 2
ln 2
ln 2
sin t
t ln eln 3
t ln 3sin t
y ln 3
ln 3
dt
t
t (sin t )(ln 3) t sin t dy sin t t cos t
dt
ln 3
x
x
5 dx ln5 5 C
x
ln 33x
3 3
ln 3
84. Let u 3 3x du 3x ln 3 dx ln13 du 3x dx; 3 x dx ln13 u1 du ln13 ln | u | C
85.
86.
451
1
1
1
1
1 1
2
1 21
2 d
d
0
0 2
ln 2
ln 2
0
0
2 5
d
0
1
2 5
11
ln 2
1
2
ln 12
2(ln11ln 2) 2 ln1 2
0
2
1
1
5
5
1
d 1 1 1 11 (1 25) ln124ln 5 ln245
ln 5 ln 5
ln 5
ln 5
2
87. Let u x 2 du 2 x dx 12 du x dx; x 1 u 1, x 2 u 2;
1
2
x dx 2 1 2u du 1 2 2 1 22 21 1
2 ln 2
ln 2
1 2
2 ln 2 1
2
u
x2
88. Let u x1/2 du 12 x 1/2 dx 2 du dx ; x 1 u 1, x 4 u 2;
x
( u 1) 2
42 x
4 x1/ 2
2
dx 2
x 1/2 dx 2 2u du 2ln 2 ln12
1
1
1
x
1
23 22 ln42
Copyright 2016 Pearson Education, Ltd.
C
452
Chapter 7 Transcendental Functions
89. Let u cos t du sin t dt du sin t dt ; t 0 u 1, t 2 u 0;
2 cos t
0
7
0
70 7 ln67
u
0
sin t dt 7u du ln7 7 ln17
1
1
90. Let u tan t du sec2 t dt ; t 0 u 0, t 4 u 1;
0 3
4 1 tan t
1
1u
1 1 u
3
du 1
0 3
ln 3
0
2
sec t dt
ln13 13 13 3 ln2 3
1
0
91. Let u x 2 x ln u 2 x ln x u1 du
2 ln x (2 x ) 1x du
2u (ln x 1) 12 du x 2 x (1 ln x) dx;
dx
dx
4
8
x 2 u 2 16, x 4 u 4 65,536;
4 2x
65,536
2 x (1 ln x) dx 12 16
du 12 u 16
65,536
12 (65,536 16) 65,520
32, 760
2
2
2
92. Let u 1 2 x du 2 x 2 (2 x) ln 2dx 2 ln1 2 du 2 x x dx
x2
1x22 dx 2 ln1 2
x2
3 1
2
ln 1 2 x
1 du 1 ln | u | C
C
u
2 ln 2
2 ln 2
93.
3x
3
95.
0
2 1
2 1
2 1 x 2 dx x
3
0
97.
3
dx 3 x
3 1
C
log10 x
dx
x
x
96.
1 x
3
e
dx xln 2 e ln21
1
ln 2
ln 2
ln 2
2 1 1
ln
2 ln 2
x)
1
1
C
1x dx ln10
12 u 2 C (ln2 ln10
u du ln10
2
4 log 2 x
4 ln x 1
dx
dx; u ln x du 1x dx; x 1 u 0, x 4 u ln 4
x
x
1 ln 2
ln 4
4 ln x 1
ln 4 1
(ln 4) 2
(ln 4)2
dx
u du ln12 12 u 2
ln12 12 (ln 4)2 2 ln 2 ln 4 ln 4
x
ln 2
0
1 ln 2
0
1
99.
e (ln 2) 1
2
ln x
1x dx; u ln x du 1x dx
ln10
ln x
ln10
98.
2 1 dx x 2 C
94.
2
2
2
2
lnln 2x dx 1 lnxx dx 12 (ln x) 1 12 [(ln 4) (ln1) ] 12 (ln 4)
4 ln 2 log 2 x
4 ln 2
dx
x
x
1
1
4
4
12 (2 ln 2) 2 2(ln 2) 2
e 2 ln10 log10 x
e (ln 10)(2 ln x ) 1
dx
(ln 10)
x
x
1
1
100.
dx (ln x)2 1 (ln e)2 (ln 1)2 1
e
Copyright 2016 Pearson Education, Ltd.
Section 7.3 Exponential Functions
2 log 2 ( x 2)
2
dx ln12
ln ( x 2)
x2
0
0
101.
2
453
x 1 2 dx ln12 (ln( x2 2)) ln12 (ln24) (ln22)
2
2
0
2
4(ln 2) (ln 2)
ln12 2 2 23 ln 2
2
2
10 log10 (10 x )
10 10 ln (10 x )
dx ln10
x
1 10
1 10
102.
1
10 x
dx
10
ln10
10
2
2
ln(10 x ) 2
10 (ln 100) (ln1)
ln10
20
20
2
1 10
2
10 4(ln 10) 2 ln 10
ln10
20
2
dx
dx
105. x log
x
10
2
2
0
3 2 log 2 ( x 1)
3
dx ln22 ln( x 1) x11
x 1
2
2
104.
9
(ln 1)
2
2 ln10
dx ln210 ln( x21) ln10
(ln 10)
2
9 2 log10 ( x 1)
9
2
dx ln10
ln ( x 1) x11
x
1
0
0
103.
2
ln 2
3
2
2
ln ( x 1) 2
2 (ln 2) (ln1) ln 2
2
ln 2 2
2
2
dx (ln10) dx; u ln x du dx
ln 10
ln x
1
x
1
ln x
1
x
1
x
1x dx (ln10) u1 du (ln10) ln | u | C (ln 10) ln |ln x | C
(ln 10) ln1x
2
(ln 8) 2
(ln x ) 1
2 (ln x )
dx (ln 8)2 1 C ln x C
x
106.
dx
2
x log8 x
x lndx 2 (ln 8)
107.
ln x 1
ln x
dt ln | t | 1 ln |ln x | ln1 ln (ln x ), x 1
t
x
ln 8
1
ex 1
ex
dt
ln
|
t
|
ln e x ln1 x ln e x
1
1 t
108.
109.
1/ x 1
1/ x
dt ln | t | 1 ln 1x ln1
t
1
x
x
1
1
ln1 ln | x | ln1 ln x, x 0
ln x ln1 log x, x 0
110. ln1a 1t dt ln1a ln | t | ln
a
a ln a
y
111. y ( x 1) x ln y ln( x 1) x x ln( x 1) y ln ( x 1) x ( x11) y ( x 1) x xx1 ln ( x 1)
112. y x 2 x 2 x y x 2 x 2 x ln y x 2 ln x 2 x 2 x ln x
y 2 x y x 2 (2 2 ln x) y
113. y
1
y x2
y 2 x 2 x 1x 2 ln x 2 2 ln x
x x x (2 2 ln x) 2 x 2 x x x ln x
2
2x
2
2x
2x
t t1 2 t1/2 ln y ln t1/2 2t ln t 1y dydt 12 (ln t ) 2t 1t ln2t 12 dydt t ln2t 12
t
t
t
Copyright 2016 Pearson Education, Ltd.
454
Chapter 7 Transcendental Functions
114. y t t t
t ln y ln t t t1/2 (ln t ) 1 dy 1 t 1/2 (ln t ) t1/2 1 ln t 2 dy ln t 2 t t
1/ 2
1/ 2
y dt
2
y
t
dt
2 t
2 t
x y (sin x) x ln (sin x) x cot x
115. y (sin x) x ln y ln (sin x) x x ln (sin x) y ln (sin x) x cos
sin x
sin x x(lnx x)(cos x)
y
116. y xsin x ln y ln xsin x (sin x)(ln x) y (cos x)(ln x) (sin x) 1x
sin x x (ln x )(cos x )
y xsin x
x
d x x ; if u x x ln u ln x x x ln x u x 1 1 ln x 1 ln x
117. y sin x x y cos x x dx
u
x
x
u x (1 ln x) y cos x x x x (1 ln x ) x x cos x x (1 ln x)
y
118. y (ln x)ln x ln y (ln x) ln (ln x ) y
y
ln(ln x ) 1
x
(ln x)
1x ln (ln x) (ln x) ln1x dxd (ln x) ln(lnx x) 1x
ln x
119. f ( x) e x 2 x f ( x) e x 2; f ( x) 0 e x 2 x ln 2; f (0) 1, the absolute maximum;
f (ln 2) 2 2 ln 2 0.613706, the absolute minimum; f (1) e 2 0.71828, a relative or local maximum
since f ( x) e x is always positive.
120. The function f ( x) 2esin( x /2) has a maximum whenever sin 2x 1 and a minimum whenever sin 2x 1.
Therefore the maximums occur at x 2k (2 ) and the minimums occur at x 3 2k (2 ), where k is
any integer. The maximum is 2e 5.43656 and the minimum is 2e 0.73576.
121. f ( x) xe x f ( x) xe x (1) e x e x xe x f ( x) e x xe x (1) e x xe x 2e x
f ( x) 0 e x xe x e x (1 x) 0 e x 0 or 1 x 0 x 1, f (1) (1)e1 1e ; using second
(a)
derivative test, f (1) (1)e1 2e1 1e 0 absolute maximum at 1, 1e
f ( x) 0 xe x 2e x e x ( x 2) 0 e x 0 or x 2 0 x 2, f (2) (2)e2 22 ; since
(b)
f (1) 0 and f (3) e
122.
x
f ( x) e 2 x f ( x)
1 e
x
2x
e 16e e
1e
4x
3
(3 2) 13 0 point of inflection at
e
e
2, 22
e
1e2 e e 2e2 e e3 f ( x) 1e2 e 3e3 e e3 21e2 2e2
2
2
2
2 2
1e2
1e2
1 e
x
x
x
x
x
x
x
x 2
x
x
x
Copyright 2016 Pearson Education, Ltd.
x
x
2x 3
x
x
x
Section 7.3 Exponential Functions
f ( x) 0 e x e3 x 0 e x 1 e2 x 0 e2 x 1 x 0; f (0)
(a)
1 e2 x
f (0)
2
e0 16e2(0) e4(0)
1e
4 0 absolute maximum at 0, 1
2
8
2(0) 3
x
e0
12 ; f ( x) undefined
1 e2(0)
0 e2 x 1 no real solutions. Using the second derivative test,
f ( x) 0 e x 1 6e2 x e4 x e x 0 or 1 6e2 x e4 x 0 e2 x
(b)
455
ln 3 2 2
2,
or x ln 32 2 f ln 3 2 2 3 2 2 and f ln32 2 32 2 ; since
2
2
( 6) 36 4
3 2
2
2
4 2 2
2
4 2 2
ln 3 2 2 3 2 2
f (1) 0, f (0) 0, and f (1) 0 points of inflection at
,
and
2
4 2 2
ln 3 2 2 3 2 2
,
.
2
4 2 2
123. f ( x) x 2 ln 1x f ( x) 2 x ln 1x x 2 11 x 2 2 x ln 1x x x(2 ln x 1); f ( x) 0 x 0 or
x
ln x 12 . Since x 0 is not in the domain of f, x e1/2 1 . Also, f ( x) 0 for 0 x 1 and
1 . Therefore,
e
f ( x) 0 for x
f
1
e
e
1e ln
e
e 1e ln e1/2 21e ln e 21e
is the absolute maximum value of
f assumed at x 1 .
e
124. f ( x) ( x 3)2 e x f ( x) 2( x 3)e x ( x 3) 2 e x
( x 3)e x (2 x 3) ( x 1)( x 3)e x ; thus
f ( x) 0 for x 1 or x 3, and f ( x) 0 for
1 x 3 f (1) 4e 10.87 is a local maximum
and f (3) 0 is a local minimum. Since f ( x) 0 for
all x, f (3) 0 is also an absolute minimum.
e2 x e x dx e2 e x
e 2 eln 3 e2 e0 92 3 12 1 82 2 2
0
0
125.
ln 3
126.
2 ln 2
ln 3
2x
0
127. L
2 ln 3
0
e x /2 e x /2 dx 2e x/2 2e x/2 0
2 ln 2
1
x
dy
x/2
1 e4 dx dx e 2 y e x /2 C ; y (0) 0 0 e0 C C 1 y e x /2 1
0
1 dy 2 1 e 2 e dy
dy 2 dy e 2 e dy
ln 2 e y e y
2
0
128. S 2
2
2eln 2 2e ln 2 2e0 2e0 (4 1) (2 2) 5 4 1
ln 2 e y e y
2
0
e y e y
2
e y e y
2
2
2
ln 2 e y e y
2
0
ln 2 e y e y 2
2
0
1
4
2y
ln 2
2 0
2y
Copyright 2016 Pearson Education, Ltd.
2 y
2 y
456
Chapter 7 Transcendental Functions
ln 2
2 12 e2 y 2 y 12 e2 y
0
2
12 4 2 ln 2 12 14 2 2 81 2 ln 2 1615 ln 2
129. y 12 e x e x dx 12 e x e x ; L
1
dy
1
1 12 e x e x
0
dx 01 1 e4 12 e 4 dx
2
2 x
2x
e2 x 1 e2 x dx 1
4
2
4
0
e e dx e e dx e e e 0
dy
0
ln 3
ln 2
x
e 1
e4 x 2e2 x 1 4e2 x dx ln 3
2
ln 2
e2 x 1
ln 3 e x e x
dx;
ln 2 e x e x
ex
x
e 1
11
02
x
1
2
x
ln 3
e 1
ln 2
x
1 22 xe
x
22 xe ; L
dx
2
e 1
ln 3
ln 2
1
2
1
1
e
4e2 x
e2 x 1
2
e2 1
2e
dx
e2 x 1
2
x
x 1
0
x
e2 1 dx ln 3 e2 1 dx ln 3
2
ln 2 e2 1
ln 2 2 1 dx
e2 1
e4 x 2e2 x 1 dx ln 3
2
ln 2
e2 x 1
2
x
x
1
2
130. y ln e x 1 ln(e x 1) dx ex
2 12 e2 ln 2 2 ln 2 12 e 2 ln 2 12 0 12
x
ex
x
x
e x
ex
let u e x e x du e x e x dx, x ln 2 u eln 2 e ln 2 2 12 32 ,
ln 23 ln 169
8 31
83
du ln | u | 3 2 ln 83
32u
x ln 3 u eln 3 e ln 3 3 13 83
dy
sin x tan x; L
131. y ln cos x dx cos
x
4
0
4
0
4
sec x dx ln |sec x tan x |0
dy
132. y ln csc x dx
4
6
ln
133. (a)
(b)
2
1 tan x dx
4
1 tan 2 x dx
0
(0) ln 2 1
4
sec2 x dx
0
ln sec 4 tan 4
4
cos x cot x
cot x; L
csc x
6
2
1 cot x dx
4
4
6
1 cot 2 x dx
ln csc 6 cot 6
4
6
csc 2 x dx
csc x dx ln |csc x cot x | 6 ln csc 4 cot 4
2 1 ln 2 3 ln 22 31
d ( x ln x x C ) x 1 ln x 1 0 ln x
dx
x
e
e
average value e11 ln x dx e11 x ln x x 1 e11
1
(e ln e e) (1ln1 1) e11 (e e 1) e11
21
2
dx ln | x | 1 ln 2 ln1 ln 2
1 x
134. average value 211
135. (a)
(b)
f ( x) e x f ( x) e x ; L( x) f (0) f (0)( x 0) L( x ) 1 x
f (0) 1 and L(0) 1 error 0; f (0.2) e0.2 1.22140 and L(0.2) 1.2 error 0.02140
Copyright 2016 Pearson Education, Ltd.
Section 7.3 Exponential Functions
457
(c) Since y e x 0, the tangent line approximation always lies below the curve y e x . Thus
L( x) x 1 never overestimates e x .
136. (a)
y e x y e x 0 for all x the graph of y e x is always concave upward
ln b x
e dx area of the trapezoid AEFD
ln a
(b) area of the trapezoid ABCD
12 ( AB CD)(ln b ln a )
ln b x
ln a
e dx e
ln a
eln b
2
(ln b ln a). Now ( AB CD) is the height of the
1
2
midpoint M e(ln a ln b) 2 since the curve containing the points B and C is linear
e(ln a ln b) 2 (ln b ln a )
ln b x
ln a
(c)
ln b x
x ln b
ln a e dx e ln a e
ln b
e dx
eln a eln b
2
(ln b ln a)
eln a b a, so part (b) implies that
e(ln a ln b ) 2 (ln b ln a) b a e
ln a
eln b
2
(ln b ln a) e
(ln a ln b ) 2
a a b
ln bb ln
a
2
a a b eln a eln b b a a b ab b a a b
eln a 2 eln b 2 ln bb ln
a
2
ln b ln a
2
ln b ln a
2
2 2x
2 2x
dx 2
dx; [u 1 x 2 du 2 x dx; x 0 u 1, x 2 u 5]
2 1 x 2
0 1 x 2
5
5
A 2 u1 du 2 ln | u | 1 2(ln 5 ln1) 2 ln 5
1
137. A
1
138. A 2
1
(1 x )
1
1 x
1 x dx 2 2
dx 2
1
1 2
ln
1
2
ln22 12 2 ln22 32 ln32
1
139. From zooming in on the graph at the right, we
estimate the third root to be x 0.76666
140. The functions f ( x) x ln 2 and g ( x) 2ln x appear
to have identical graphs for x 0. This is no
accident, because x ln 2 eln 2ln x eln 2
141. (a)
ln x
2ln x.
f ( x) 2 x f ( x) 2 x ln 2; L( x) 20 ln 2 x 20 x ln 2 1 0.69 x 1
Copyright 2016 Pearson Education, Ltd.
458
Chapter 7 Transcendental Functions
(b)
142. (a)
1 , and f (3) ln 3 L( x) 1 ( x 3) ln 3 x 1 1 0.30 x 0.09
f ( x) log3 x f ( x) x ln
3
ln 3
3ln 3
ln 3 3ln 3 ln 3
(b)
dy
143. (a) The point of tangency is ( p, ln p) and mtangent 1p since dx 1x . The tangent line passes through (0, 0)
the equation of the tangent line is y 1p x. The tangent line also passes through ( p, ln p )
ln p 1p p 1 p e, and the tangent line equation is y 1e x.
(b)
d2y
dx 2
12 for x 0 y ln x is concave downward over its domain. Therefore, y ln x lies below the
x
graph of y 1e x for all x 0, x e, and ln x ex for x 0, x e.
(c) Multiplying by e, e ln x x or ln x e x.
e
(d) Exponentiating both sides of ln x e x, we have eln x e x , or x e e x for all positive x e.
(e) Let x to see that e e . Therefore, e is bigger.
144. Using Newton’s Method: f ( x) ln( x) 1 f ( x) 1x xn 1 xn
ln xn 1
1
xn
xn 1 xn 2 ln xn . Then,
x1 2, x2 2.61370564, x3 2.71624393, and x5 2.71828183. Many other methods may be used. For
example, graph y ln x 1 and determine the zero of y.
7.4
EXPONENTIAL CHANGE AND SEPARABLE DIFFERENTIAL EQUATIONS
1. (a)
y e x y e x 2 y 3 y 2 e x 3e x e x
(c) y e x Ce3 x 2 y e x 32 Ce 3 x 2 2 y 3 y 2 e x 32 Ce3 x 2 3 e x Ce3 x 2 e x
(b)
y e x e3 x 2 y e x 32 e3 x 2 2 y 3 y 2 e x 32 e3 x 2 3 e x e3 x 2 e x
Copyright 2016 Pearson Education, Ltd.
Section 7.4 Exponential Change and Separable Differential Equations
2. (a)
3.
459
y2
y 1x y 12 1x
x
2
1
( x13)
( x 3)2
2
1
( x 1C )
( x C )2
2
(b)
y x 13 y
(c)
y x 1C y
y2
ex x2 y 1x et dt e x x 1x 1x et dt e x xy e x
x et
x t
dt y 12 et dt 1x
1 t
x 1
y 1x
y2
x
t
t
x 2 y xy e x
4.
y
x
1 x 4 1
y
5.
1
1 t
2 x3
1 x 4
4
x 1 t 4 dt
3 1
4
1 x
dt y 12
4 x3
1
4
1 x
1 x 4
1 t dt 1 y 12xx4 y 1 y 12xx4 y 1
1 x 1
x
1
4
3
3
4
y e x tan 1 2e x y e x tan 1 2e x e x 1 2 2e x e x tan 1 2e x 2 2 x
1 4e
1 2e x
y y
2
y y 2 2 x ; y ( ln 2) e( ln 2) tan 1
1 4e2 x
1 4e
2
2
2
2e ln 2 2 tan 1 1 2 4 2
( x 2) y e 2xy; y(2) (2 2)e 0
x2
22
6.
y ( x 2)e x y e x 2 xe x
7.
y cosx x y x sin x2cos x y sinx x 1x cosx x y sinx x x xy sin x y
x
y
cos(( 2)2) 0
xy y sin x; y 2
8.
y lnxx y
y 1
ln x x 1x
(ln x )
2
ln x
1 x 2 y x 2 x 2 x 2 y xy y 2 ;
ln x (ln x ) 2
(ln x )2
y (e) lnee e.
9. 2 xy dx 1 2 x1/2 y1/2 dy dx 2 y1/2 dy x 1/2 dx 2 y1/2 dy x 1/2 dx
dy
2 23 y 3/2 2 x1/2 C1 23 y 3/2 x1/2 C , where C 12 C1
3
y dy x 2 y1/2 dx y 1/2 dy x 2 dx y 1/2 dy x 2 dx 2 y1/2 x3 C 2 y1/2 13 x3 C
10.
dy
x2
dx
11.
dy
e x y dy e x e y dx e y dy e x dx
dx
12.
dy
3x 2 e y dy 3 x 2 e y dx e y dy 3 x 2 dx
dx
y
x
y
x
y
x
e dy e dx e e C e e C
y
2
y
3
y
3
e dy 3x dx e x C e x C
Copyright 2016 Pearson Education, Ltd.
460
13.
Chapter 7 Transcendental Functions
y cos
dy
dx
y cos2 y dx sec y y dy dx sec y y dy dx. In the integral on the left2
y dy
hand side, substitute u
y du
2
1 dy 2 du 1 dy, and we have
2 y
y
2
sec u du dx 2 tan u x C x 2 tan y C
14.
dy
2
15.
1 dx
2 xy
2 xy dx 1 dy
y
32
3
2
2 ydy 1 dx 2 y1 2 dy x 1 2 dx 2 y1 2 dy x 1 2 dx
x
12
dy x 1 C1 2 y 3 2 3 x 32 C1 2
2
y
x
x dx e y x dx e e
dy
dy
x
y
u
y
x
x
right-hand side, substitute u
e dy 2 e du e
y
dy e e
y 3 x C, where C 32 C1
3
x
dx e y dy e
dx e y dy e
x
1
x du
dx 2 du 1 dx, and we have
2 x
x
x
dx. In the integral on the
x
2eu C1 e y 2e x C , where C C1
16. (sec x) dx e y sin x dx e y sin x cos x dy e y esin x cos x dx e y dy esin x cos x dx
dy
e
17.
y
dy
dy e
dy
2x
dx
sin x
cos x dx e
sin x
e
1 y 2 dy 2 x 1 y 2 dx
2
| y | 1 y sin x C
18.
y
C1 e
dy
1 y
2
y
sin x
e
2 x dx
C , where C C1
dy
1 y
2
2 x dx sin 1 y x 2 C since
2 x y
2x y
2x y
x
dy
e x y dy e x y dx dy e xe y dx e2 y dx e2 y dy e x dx
dx
e
e
e e
e
2y
e
2y
dy e x dx e2 e x C1
y2
dy 3 x 2 dx
e 2 y 2e x C where C 2C1
19.
dy
y2
y 2 dx 3 x 2 y 3 6 x 2 y 2 dy 3 x 2 y 3 2 dx
3
y 2
dy 3x 2 dx 3
y 2
13 ln y 3 2 x3 C
20.
dy
xy 3x 2 y 6 ( y 3)( x 2) y13 dy ( x 2)dx
dx
y13 dy ( x 2)dx
ln | y 3| 12 x 2 2 x C
21.
1 dy ye x
x dx
22.
2
2
2 ye x e x
1
dy
y y 2
x2
2
y 2 y y 21 y dy xe x dx y 21 y dy xe x dx
2
xe dx 2 ln
2
y 2 12 e x C 4 ln
2
2
y 2 e x C 4 ln
y 2 ex C
e y 1 e x 1 e 11 dy e x 1 dx e 11 dy e x 1 dx
e dy e x 1 dx ln 1 e y e x x C ln 1 e y e x x C
1 e
dy
e x y e x e y 1
dx
y
y
y
y
Copyright 2016 Pearson Education, Ltd.
2
Section 7.4 Exponential Change and Separable Differential Equations
23. (a)
461
0.99 0.00001
y y0 e kt 0.99 y0 y0 e1000 k k ln1000
ln (0.9)
(b) 0.9 e( 0.00001)t (0.00001)t ln (0.9) t 0.00001 10,536 years
(c)
24. (a)
(b)
y y0 e(20,000) k y0 e0.2 y0 (0.82) 82%
ln (90) ln(1013)
dp
kp p p0 ekh where p0 1013; 90 1013e20k k
0.121
dh
20
6.05
p 1013e
2.389 hectopascals
900 h
(c) 900 1013e( 0.121) h 0.121h ln 1013
ln(1013) ln(900)
0.9777 km
0.121
25.
dy
0.6 y y y0 e 0.6t ; y0 100 y 100e0.6t y 100e0.6 54.88 grams when t 1 h
dt
26.
A A0 e kt 800 1000e10k k
ln (0.8)
A 1000e(ln (0.8) 10)t , where A represents the amount of sugar
10
(ln (0.8) 10)24
that remains after time t. Thus after another 14 hours, A 1000e
585.35 kg
L
27. L( x) L0 e kx 20 L0 e6k ln 12 6k k ln62 0.1155 L( x) L0 e 0.1155 x ; when the intensity is
L
one-tenth of the surface value, 100 L0 e 0.1155 x ln10 0.1155 x x 19.9 m
28. V (t ) V0 et 40 0.1V0 V0 et 40 when the voltage is 10% of its original value t 40 ln (0.1) 92.1 s
29.
ln 2
y y0 e kt and y0 1 y ekt at y 2 and t 0.5 we have 2 e0.5k ln 2 0.5k k 0.5 ln 4.
Therefore, y e(ln 4)t y e 24 ln 4 424 2.81474978 1014 at the end of 24 hours
30.
y y0 e kt and y (3) 10, 000 10, 000 y0 e3k ; also y (5) 40, 000 y0 e5k . Therefore
y0 e5k 4 y0 e3k e5k 4e3k e2k 4 k ln 2. Thus, y y0 e(ln 2)t 10, 000 y0 e3ln 2 y0 eln 8
10, 000 8 y0 y0 10,000
1250
8
31. (a) 10, 000ek (1) 7500 ek 0.75 k ln 0.75 and y 10, 000e(ln 0.75)t . Now 1000 10, 000e(ln 0.75)t
0.1 8.00 years (to the nearest hundredth of a year)
ln 0.1 (ln 0.75)t t lnln0.75
(b) 1 10, 000e(ln 0.75)t ln 0.0001 (ln 0.75)t t lnln0.0001
32.02 years (to the nearest hundredth
0.75
of a year)
dz
dy
k
k ( r ky ) kz. The equation dz / dt kz has solution z ce kt , so
dt
dt
1
r ky ce kt and y r ce kt .
k
1
(a) Since y (0) y0 , we have y0 ( r c ) and thus c r ky0 . So
k
1
r
r
kt
y r [ r ky0 ]e
y0 e kt .
k
k
k
32. Let z r ky. Then
Copyright 2016 Pearson Education, Ltd.
462
Chapter 7 Transcendental Functions
r
r r
(b) Since k 0, lim y0 e kt .
k
k k
t
y
y r/k
y y0
t
33. Let y ( t ) be the population at time t , so t (0) 1147 and we are interested in t (20). If the population
continues to decline at 39% per year, the population in 20 years would be 1147 (0.61)20 0.06 1, so the
species would be extinct.
34. (a) We will ignore leap years. There are (60)(60)(24)(365) 31,536,000 seconds in a year. Thus, assuming
exponential growth, P 314,419,198e kt , with t in years, and
31,536,000 314,419,199
ln
0.0083583.
12
314,419,198
(You don’t really need to compute that logarithm: it will be very nearly equal to 1 over the denominator
of the fraction.)
314,419,199 314,419,198e12 k /31,536,000 k
(b) In seven years, P 314,419,198e(0.0083583)(7) 333,664,000 . (We certainly can’t estimate this
population to better than six significant digits.)
35.
0.9 P0 P0 ek k ln 0.9; when the well’s output falls to one-fifth of its present value P 0.2 P0
ln 0.2
0.2 P0 P0 e(ln 0.9)t 0.2 e(ln 0.9)t ln (0.2) (ln 0.9)t t ln 0.9 15.28 years
36. (a)
dp
1 p dp 1 dx ln p 1 x C p e( 0.01x C ) eC e 0.01x C e 0.01x ;
100
1
dx
p
100
100
p(100) 20.09 20.09 C1e( 0.01)(100) C1 20.09e 54.61 p( x) 54.61e 0.01x (in dollars)
(b) p(10) 54.61e( 0.01)(10) $49.41, and p(90) 54.61e( 0.01)(90) $22.20
(c) r ( x) xp ( x) r ( x) p ( x) xp ( x);
p ( x ) .5461e 0.01x
r ( x) (54.61 .5461x)e 0.01x . Thus,
r ( x) 0 54.61 .5461x x 100. Since
r 0 for any x 100 and r 0 for x 100,
then r ( x) must be a maximum at x 100.
37.
ln (0.5)
A A0 e kt and A0 10 A 10ekt , 5 10ek (24360) k 24360 0.000028454 A 10e0.000028454t ,
ln 0.2
then 0.2(10) 10e0.000028454t t 0.000028454
56563 years
38.
A A0 e kt and 12 A0 A0 e139k 12 e139k k
ln(0.5)
0.00499; then
139
ln 0.05 600 days
0.05 A0 A0 e0.00499t t 0.00499
Copyright 2016 Pearson Education, Ltd.
Section 7.4 Exponential Change and Separable Differential Equations
39.
y y0 e kt y0 e( k )(3 k ) y0 e 3
40. (a)
(b)
y0
e3
463
y
200 (0.05)( y0 ) after three mean lifetimes less than 5% remains
ln 2 0.262
A A0 e kt 12 e 2.645k k 2.645
1 3.816 years
k
ln 2 t ln 20 ln 2 t t 2.645 ln 20 11.431 years
(c) (0.05) A A exp 2.645
2.645
ln 2
41. T Ts T0 Ts e kt , T0 90C, Ts 20C, T 60C 60 20 70e 10k 74 e10k
k
0.05596
ln 74
10
(a) 35 20 70e0.05596t t 27.5 min is the total time it will take 27.5 10 17.5 minutes longer to
reach 35C
(b) T Ts T0 Ts e kt , T0 90C, Ts 15C 35 15 105e0.05596t t 13.26 min
42. T 18 T0 18 e kt 2 18 T0 18 e10k and 10 18 T0 18 e 20k . Solving
16 T0 18 e10k and 8 T0 18 e20k simultaneously T0 18 e10 k 2 T0 18 e20 k
ln 2
T 18
e10k 2 k ln102 and 16 0 10 k 16 e10 10 T0 18 T0 18 16 eln 2
e
18 32 14
39 T
43. T Ts To Ts e kt 39 Ts 46 Ts e10k and 33 Ts 46 Ts e 20k 46Ts e 10 k and
33Ts
e20k
46 Ts
e 3346TT
10 k 2
s
s
s
39 Ts 2
46 Ts
33 Ts 46 Ts 39 Ts 2
1518 79Ts Ts2 1521 78Ts Ts2 Ts 3 Ts 3C
44. Let x represent how far above room temperature the silver will be 15 min from now, y how far above room
temperature the silver will be 120 min from now, and t0 the time the silver will be 10°C above room
temperature. We then have the following time-temperature table:
time in min. 0
20 (Now) 35
140
t0
temperature Ts 70 Ts 60 Ts x Ts y Ts 10
1 ln 6 0.00771
T Ts T0 Ts e kt 60 Ts Ts 70 Ts Ts e20k 60 70e 20 k k 20
7
(a) T Ts T0 Ts e 0.00771t Ts x Ts 70 Ts Ts e(0.00771)(35) x 70e0.26985 53.44C
(b) T Ts T0 Ts e 0.00771t Ts y Ts 70 Ts Ts e (0.00771)(140)
y 70e1.0794 23.79C
(c) T Ts T0 Ts e 0.00771t Ts 10 Ts 70 Ts Ts e (0.00771)t0 10 70e0.00771t0
1
ln 17 0.00771t0 t0 0.00771
ln 71 252.39 252.39 20 232 minutes from now the
silver will be 10°C above room temperature
Copyright 2016 Pearson Education, Ltd.
464
Chapter 7 Transcendental Functions
ln 2 0.0001216
45. From Example 4, the half-life of carbon-14 is 5700 yr 12 c0 c0 e k (5700) k 5700
c c0 e 0.0001216t (0.445)c0 c0 e0.0001216t t 0.0001216 6659 years
ln(0.445)
46. From Exercise 45, k 0.0001216 for carbon-14.
(a) c c0 e 0.0001216t (0.17)c0 c0 e0.0001216t t 14,571.44 years 12,571 BC
(b) (0.18)c0 c0 e0.0001216t t 14,101.41 years 12,101 BC
(c) (0.16)c0 c0 e 0.0001216t t 15, 069.98 years 13, 070 BC
47. From Exercise 45, k 0.0001216 for carbon- 14 y y0 e 0.0001216t . When t 5000
y y0 e0.0001216(5000) 0.5444 y0 y 0.5444 approximately 54.44% remains
y
0
48. From Exercise 45, k 0.0001216 for carbon-14. Thus, c c0 e 0.0001216t (0.995)c0 c0 e0.0001216t
ln(0.995)
t 0.0001216 41 years old
49. e (ln 2/5730)t 0.15
ln 2
5730ln(0.15)
t ln(0.15) t
15,683 years
5730
ln 2
50. (a) e (ln 2/5730)(500) 0.94131, or about 94%.
(b) We’ll assume that the error could be 1% of the original amount. If the percentage of carbon-14 remaining
5730ln(0.93131)
were 0.93131, the Ice Maiden’s actual age would be
588 years.
ln 2
7.5
INDETERMINATE FORMS AND L’HÔPITAL’S RULE
x 2 1
2
2x
x2 x 4
1. lHôpital: lim
x2
x2
14 or lim x2 2 lim ( x 2)(
lim x 1 2 14
x 2)
x2 x 4
x2
x2
5x
2. lHôpital: lim sinx5 x 5cos
5 or lim sinx5 x 5 lim sin5 x5 x 5 1 5
x 0
1
x 0
x 0
5 x 0
2
2
10 5 or lim 5 x 3 x lim
3. lHôpital: lim 5 x 23 x lim 1014x x 3 lim 14
2
7
x 7 x 1
x
x
x 7 x 1
5 3x
1
x 7 2
75
x
lim x x 1 3
x1 4 x 4 x 3 11
( x 1) x 2 x 1
x3 1 lim 3 x 2 3 or lim x3 1 lim
3
2
3
2
x 1 4 x x 3 x 1 ( x 1) 4 x 4 x 3
x 1 4 x x 3 x 1 12 x 1 11
4. lHôpital: lim
x lim sin x lim cos x 1 or lim 1 cos x lim
5. lHôpital: lim 1cos
2
2
2
x 0 x
x 0 2 x
x 0 2
x 0 x
x 0
2
sinx x 1cos1 x 12
(1 cos x ) 1 cos x
1 cos x
x2
sin 2 x
lim sin x
x 0 x (1 cos x ) x 0 x
lim
2
2
Copyright 2016 Pearson Education, Ltd.
Section 7.5 Indeterminate Forms and L’Hôpital’s Rule
2
2
6. lHôpital: lim 2 3x 3 x lim 4 x23 lim 64x 0 or lim 2 3x 3 x lim
x x x 1
x 3 x 1
x
7.
x 2 lim 1 1
2
4
x2 x 4 x 2 2 x
9.
lim t 2 4t 15 lim 32t t 14 2( 3) 1 23
7
11.
lim
3
8.
3( 3)2 4
2
t 3 t t 12
t 3
3
1
x
x
10 0
2
lim x x 525 lim 21x 10
x 5
x 5
10.
3
2
lim 33t 3 lim 9t2 9
t 1 4t t 3 t 1 12t 1 11
14.
5t 5
lim sin 5t lim 5cos
2
t 0 2t
t 0 2
2
x 7 x 3
x 21x
x
x
x 8 x 2 lim 116 x lim 16 2
2
3
x 12 x 5 x x 24 x 5 x 24
13.
lim sint t lim
16.
1
x 1 2 3
x lim 30 5
lim 5 x 3 2 x lim 15 x 2 2 lim 30
42 x
42 7
12.
15.
x x x 1
2 3
x x2
lim
2
t 0
t 0
cos t (2t ) 0
2
1
2
8x
x lim 16 16 16
lim cos
lim 16
x 1
sin x
cos x
1
x 0
x 0
x 0
x lim cos x 1
lim sin x3 x lim cos x21 lim sin
6x
6
6
x 0
x
x 0 3 x
x 0
x 0
17.
2
lim 2 lim
23 2
2 cos(2 ) 2 sin(2 ) sin 2
18.
3
lim 3 lim
3
3 sin 3
3 cos 3
19.
sin lim cos lim sin
1
lim 11cos
14
2
2sin 2
4 cos 2
( 4)( 1)
2
2
2
20.
x 1
lim
lim 1 1
1
x 1 ln x sin( x ) x 1 x cos( x ) 1
21.
2
lim x
lim
x 0 ln(sec x ) x 0
22.
lim
2x
secsecx tanx x
ln(csc x )
x 2 x
2
lim
2
2 x lim
lim tan
x
x 0
2 2 2
2
2
x 0 sec x 1
lim cot x lim csc2 x 12 1
2 x 2 2 x 2 x 2 2 2 2
x cot x
csccsc
x
x 2 x
2
23.
t (1 cos t )
(1cos t ) t (sin t )
sin t (sin t t cos t )
t cos t t sin t 1110 3
lim
lim
lim
lim cos t coscos
1cos t
sin t
t
1
t 0 t sin t
t 0
t 0
t 0
24.
t cos t lim cos t (cos t t sin t ) 1 (1 0) 2
lim t sin t lim sin tsin
t
cos t
1
t 0 1cos t t 0
t 0
Copyright 2016 Pearson Education, Ltd.
465
466
25.
26.
27.
Chapter 7 Transcendental Functions
x 2 sec x lim cos x lim sin1 x 11 1
x 2
lim
x 2
x 2
29.
30.
31.
32.
33.
34.
x 2
2 x tan x lim cot x lim csc1 x lim sin 2 x 1
2
x 2
x 2
x 2
0
0
12 1 lim ln 12 12 ln 1 ln1 ln 2 ln 2
x 2x
lim
x
x 0 2 1
lim
12 0 1
(1) 2 x ( x ) (ln 2) 2 x
x
0
ln 2
(ln 2)20
(ln 2) 2 x
x 0
x
2
1
0
0
ln 3
lim 3 x 1 lim 3x ln 3 30 ln 3 ln
2
x 0 2 1
ln( x 1)
ln( x 1)
2
ln x
ln 2
lim log x lim
x
x
log x
lim
x 0
x 0
ln( x 3)
ln 3
x
lim
ln x
lim
lnln 2x ln 3 lim
3
ln x 2 2 x
x 0
lim
ln e x 1
ln x
x 0
x11 (ln 2) lim x (ln 2) lim 1 ln 2
1
x x
x x 1
x 1
(ln 2) lim
lim log ( 2x 3) lim
x
2 ln 2
x 0 2 ln 2
2 x2
lim 2 x2 2 x lim 42 xx 22 lim 22 1
ex
x 0
lim
x 0
x 2 x
xe x
e x 1
36.
ay a
ay a 2 a
lim
y
y
y 0
y 0
lim
lim
2
1x ln 3 lim x 3 ln 3 lim 1 ln 3
x13 ln 2 x x ln 2 x 1 ln 2
2
x2 2 x
1
x
e x 1
1
x
ln x ln 3 lim
ln 2 x ln( x 3)
ln 2 x
5 y 25 5
(5 y 25)1/ 2 5
lim
lim
y
y
y 0
y 0
y 0
38.
x 2
35.
37.
2
30 (ln 3)(1)
3sin (ln 3)(cos )
ln 3
1
1
0
sin
lim 3 1 lim
lim
x
lim
28.
x 0
x
x
lim e xxe 110 1
x 0
e
12 (5 y 25)1/ 2 (5) lim
1
2
y 0
5
12
y 0 2 5 y 25
1
a lim ay a
1/ 2
x 0
2
1
1/ 2
(a)
a
12 , a 0
y 0 2 ay a 2
lim
lim ln 2 x ln( x 1) lim ln x2x1 ln lim x2x1 ln lim 12 ln 2
x
x
x
x
lim (ln x ln sin x) lim ln sinx x ln lim sinx x ln lim cos1 x ln1 0
x 0
x 0
x 0
x 0
Copyright 2016 Pearson Education, Ltd.
Section 7.5 Indeterminate Forms and L’Hôpital’s Rule
39.
40.
x 0
1
(ln x 1) 1
sin x cos 2 x sin 2 x
cos x
x 0
lim
t
49.
50.
h
h
lim e2h1 lim e2 12
h 0
h 0
t
2
t
t
lim e tt lim e t 2t lim e t 2 lim et 1
t e 1
t
e
t e
t e
2
lim x 2 e x lim xx lim 2 xx lim 2x 0
x e
x
x e
x e
sin x
lim x sin x lim 12cos x lim
02 0
2
2
x 0 x tan x
x 0 x sec x tan x x 0 2 x sec x tan x 2 sec x
e 1 lim 2 e 1e
lim
x
48.
0 10 1
1
44.
h2
1 lim sin lim cos 1
lim cos
0 e 1 0 e 1 0 e
h 0
1
x
x cos x lim (1cos x ) (sin x )(cos x )
sin1 x cos
x 0
sin x
sin x
43.
eh (1 h )
1 1
x
1 1
(01) 1
2
lim (csc x cot x cos x) lim
x 0
x 0
47.
( x 1)
lim
lim ( x ln1x)x x 1
x11 ln1x xlim
(lnxx1)(ln
x)
x 1
1
x 1 (ln x ) ( x 1) x 1
lim
lim
46.
x
33 (1)(0)
62 3
110
x 1
45.
cos x
x 0
x cos x
x 0
(3 x 1)(sin x ) x
3sin x (3 x 1)(cos x ) 1
3cos x 3cos x (3 x 1)( sin x )
3 x 1 1
lim
lim
lim
x
sin x
x sin x
sin x x cos x
cos x cos x x sin x
x 0
x 0
x 0
x 0
lim
42.
cos x
sin x
x 0
lim
41.
lim 2(ln x)(sin x) lim 2(ln x) sin x 1
2(ln x ) 1x
(ln x ) 2
lim ln(sin x ) lim
2
x 0 x sin x
x
x
x 0 x cos x sin x
2x
2x
x
x
e 2e lim
4e 2e
lim x 2cos
22 1
x sin x
x sin x 2 cos x
x 0
2
2
x 0
2
sin cos lim 1sin cos lim 2sin lim 2 cos 2 2
lim tan
2
2
0
0
sec 1
0 tan
0
2
3 x 3 2 x
9sin 3 x 2
lim sin 3 x 3 x x lim 2sin x3cos
lim 3cos 3 x 3 2 x lim
cos 2 x cos x sin 2 x x 0 sin x cos 2 x sin 3 x x 0 2sin x sin 2 x cos x cos 2 x 3cos 3 x
x 0 sin x sin 2 x
x 0
24 12
51. The limit leads to the indeterminate form 1. Let f ( x) x1/(1 x ) ln f ( x) ln x1/(1 x ) 1ln xx . Now
1
lim ln f ( x) lim 1ln xx lim x1 1. Therefore lim x1/(1 x ) lim f ( x) lim eln f ( x ) e 1 1e
x 1
x 1
x 1
x 1
x 1
Copyright 2016 Pearson Education, Ltd.
x 1
467
468
Chapter 7 Transcendental Functions
x . Now
52. The limit leads to the indeterminate form 1. Let f ( x) x1/( x 1) ln f ( x) ln x1/( x 1) ln
x 1
x lim
lim ln f ( x) lim ln
x 1
x 1
x 1
x 1
1x 1. Therefore lim x1/( x1) lim f ( x) lim eln f ( x) e1 e
1
x 1
x 1
x 1
53. The limit leads to the indeterminate form 0 . Let f ( x) (ln x)1/ x ln f ( x) ln(ln x)1/ x
ln(ln x )
lim
x
x
x
lim ln f ( x ) lim
x
x ln1 x 0. Therefore lim (ln x)1/ x lim f ( x) lim eln f ( x) e0 1
1
x
x
x
54. The limit leads to the indeterminate form 1. Let f ( x) (ln x)1/( x e) ln f ( x)
ln(ln x )
x e
lim
x e
lim
x e
ln(ln x )
x e
x ln1 x 1 . Therefore (ln x)1/( x e) lim f ( x) lim eln f ( x) e1/ e
1
ln(ln x )
. Now
x
e
x e
lim ln f ( x)
x e
x e
x 1. Therefore
55. The limit leads to the indeterminate form 00. Let f ( x) x 1/ln x ln f ( x) ln
ln x
lim x 1/ln x lim f ( x) lim eln f ( x ) e 1 1e
x 0
x 0
x 0
x 1. Therefore
56. The limit leads to the indeterminate form 0 . Let f ( x) x1/ln x ln f ( x) ln
ln x
lim x1/ln x lim f ( x ) lim e1n f ( x ) e1 e
x
x
x
57. The limit leads to the indeterminate form 0 . Let f ( x) (1 2 x)1/(2 ln x ) ln f ( x)
lim ln f ( x) lim
x
x
1/2
ln(1 2 x )
2 ln x
ln(1 2 x )
lim 1x2 x lim 12 12 . Therefore lim (1 2 x)1/(2 ln x ) lim f ( x)
2 ln x
x
x
x
x
lim eln f ( x ) e
x
58. The limit leads to the indeterminate form 1. Let f ( x) e x x
ln e x x
lim ln f ( x) lim
x 0
x 0
x
1/ x
ln f ( x)
ln e x x
x
lim e 1 2. Therefore lim e x x 1/ x lim f ( x) lim eln f ( x) e2
x
x
x 0 e x
x 0
x 0
x 0
59. The limit leads to the indeterminate form 00. Let f ( x) x x ln f ( x) x ln x ln f ( x) ln1 x
x
1x lim ( x) 0. Therefore lim x x lim f ( x) lim eln f ( x)
lim ln f ( x) lim ln1 x lim
x 0
x 0
x
x 0 1
x2
x 0
x 0
x 0
x 0
e0 1
60. The limit leads to the indeterminate form 0 . Let f ( x) 1 1x
x 2
1
lim 1 x2 lim
x 0
x
x 0
1
1 x 1
x 0
ln 1 x
x
ln f ( x)
ln f ( x) x xlim
0
lim xx1 0. Therefore lim 1 1x
x 0
1
1
f ( x) lim eln f ( x ) e0 1
xlim
0
x 0
x
Copyright 2016 Pearson Education, Ltd.
Section 7.5 Indeterminate Forms and L’Hôpital’s Rule
61. The limit leads to the indeterminate form 1. Let f ( x)
469
xx12 ln f ( x) ln xx12 x ln xx12
x
x
3
1 1
( x 2)(
ln x 2
ln( x 2) ln( x 1)
x 1)
x 2 x 1
lim ln f ( x) lim x ln xx12 lim 1x 1 lim
lim
lim
1
1
1
x
x
x
x
x
x
x
x2
x2
x
3x2
(
x
2)( x 1)
x
lim
lim lim 3. Therefore, lim lim f ( x) lim e
6x
x 2 x 1
x2 x
x x 1
6
x 2
62. The limit leads to the indeterminate form 0 . Let f ( x)
2
lim ln f ( x) lim 1x ln xx 21
x
x
ln f ( x )
x
e3
ln f ( x) ln ln
x 2 1
x2
1/ x
x 2 1
x2
1/ x
1
x
x 2 1
x2
2
2x
ln xx 21
1
ln x 2 1 ln( x 2)
2
x 2 1 x 2
lim
lim
lim
lim x2 4 x 1
x
x
1
x
x
x
x ( x 1)( x 2)
lim f ( x) lim e
x 2 4 x 1 lim
2 x 4 lim 2 0. Therefore, lim x 2 1
2
2
x x 2
x x 2 x x 2 x 3 x 4 x 1 x 6 x 4
lim
x
3
1/ x
x
ln f ( x )
x
e0 1
63.
1
3
2
lim x 2 ln x lim ln1 x lim x2 lim 2x x lim 32x 0
x 0
x 0 x 2 x 0 x3 x 0
x 0
64.
2(ln x ) 1
2
(ln x )2
x
lim 2 ln1 x lim 1x lim
lim x(ln x)2 lim 1 lim
1
x 0
x 0 x x 0 x2 x 0 x x 0 x 2 x 0
65.
1
x
lim x tan 2 x lim
lim 2 1
1 1
cot x
x 0
x 0
2
x 0 csc 2 x
66.
lim sin x ln x lim
x 0
68.
69.
70.
lim 9 x 1
x x 1
x
sin x
lim
lim
1
x
x 0
x 2
1
lim sinx x
x 0
sec x
tan x
lim
x 2
lim 9
x 1
2
9 3
1 1
1
x lim
1 1
cos1 x cos
sin x
sin x
x 2
x
cos
sin x
lim cos x 1
1
x 0 sin x
x 0
cot x lim
lim csc
x
x 0
23 1 0
x
x 1 4
3
x
71.
x 0
ln x lim
sin x tan x lim sin x sec x cos x tan x 0 0
x 0
csc x cot x lim
1
csc x
x
1
x 0
x 0
x 0
9 x 1
x 1
x
2 x2
x
67. lim
lim 2x 0
x
x
lim 2x 3x lim
x 3 4
Copyright 2016 Pearson Education, Ltd.
470
72.
73.
74.
Chapter 7 Transcendental Functions
x
x
lim 1 2 x 10 1
x
x
x 5 1 x 5 1 0 1
2
2
1 42
lim 2 x 4x lim
x 5 2
x2
x2 x
lim e x lim e x
x xe
e
x 0
lim
x
lim e 1 lim
1/ x
e x ( x 1) (2 x 1)
1
x
x ( x 1)
lim e x
1/ x
x
lim
x 0
x
x
x 0
x
lim e1/ x
e1/ x 12
x
12
x
x 0
75. Part (b) is correct because part (a) is neither in the 00 nor
form and so lHôpital’s rule may not be used.
2 lim
2 is not an
2
76. Part (b) is correct; the step lim 2 x2xcos
in part (a) is false because lim 2 x2xcos
x
2 sin x
x
x 0
x 0
x 0
indeterminate quotient form.
77. Part (d) is correct, the other parts are indeterminate forms and cannot be calculated by the incorrect arithmetic
f ( c )
f (0) f ( 2)
f ( c )
f (b ) f ( a )
f ( c )
f (3) f (0)
78. (a) We seek c in 2, 0 so that g (c ) g (0) g ( 2) 00 42 12 . Since f (c) 1 and g (c) 2c we have that
1 1 c 1.
2c
2
(b) We seek c in (a, b) so that g (c ) g (b) g ( a )
b a 1 . Since
b a
b2 a 2
f (c) 1 and g (c) 2c we have that
1 1 c ba .
2c b a
2
(c) We seek c in (0, 3) so that g (c ) g (3) g (0) 9300 13 . Since f (c) c 2 4 and g (c) 2c we have
that
c 2 4 1 c 1 37 c 1 37 .
2c
3
3
3
3 x lim 9 9 cos 3 x
79. If f ( x ) is to be continuous at x 0, then lim f ( x ) f (0) c f (0) lim 9 x 3sin
3
2
x 0
x 0
5x
x 0
15 x
sin 3 x lim 81cos 3 x 27 .
lim 2730
x
10
x 0
x 0 30
80.
tan 2 x a sin bx
x
x3
x2
x 0
lim
lim
tan 2 x ax x 2 sin bx
x3
x 0
lim
2 sec2 2 x a bx 2 cos bx 2 x sin bx
3 x2
x 0
8sec2 2 x tan 2 x b 2 x 2 sin bx 4bx cos bx 2 sin bx
6x
x 0
1666b 0 16 6b 0 b 83
lim
lim
0
0
2sec2 2 x 2 bx 2 cos bx 2 x sin bx
3x2
x 0
lim (2sec2 2 x a bx 2 cos bx 2 x sin bx) a 2 0 a 2; lim
x 0
will be in form if
32sec2 2 x tan 2 2 x 16sec4 2 x b3 x 2 cos bx 6b2 x sin bx 6b cos bx
6
x 0
81. (a)
Copyright 2016 Pearson Education, Ltd.
Section 7.5 Indeterminate Forms and L’Hôpital’s Rule
471
(b) The limit leads to the indeterminate form :
x 2 x 2 x
2
x
lim x x 2 x lim x x 2 x x x 2 x lim
lim
x
x x x x x x 2 x x x x 2 x
x
1 1 1
2
1 1 0
x 1 1 1x
lim
82.
lim x 2 1 x lim x
x
x
x 2 1
xx lim x
x
x
x lim x
x 2 x
x 2 1
x2
1
1
x2
1
x
83. The graph indicates a limit near 1. The limit leads
2 x 2 (3 x 1) x 2
x 1
x 1
to the indeterminate form 00 : lim
2
3/ 2
1/ 2
lim 2 x 3 xx 1 x
x 1
9 1/ 2
1 1/ 2
2 lim 4 x 2 x 2 x
1
x 1
4 92 12
415 1
1
84. (a) The limit leads to the indeterminate form 1. Let f ( x) 1 1x
ln f ( x)
ln f ( x) x ln 1 1x xlim
x
x 2
lim ln1 x1 lim 1 x1 lim 1 1 1 lim 1 1 x lim f ( x)
lim
x x
2
x 1
x 1x
x
x x
x 1 1x 1 0
x
ln 1 1x
lim eln f ( x ) e1 e
x
(b)
x
1 1x x
10
100
1000
10,000
100,000
2.5937424601
2.70481382942
2.71692393224
2.71814592683
2.71826823717
Both functions have limits as x approaches
infinity. The function f has a maximum but no
minimum while g has no extrema. The limit of
f ( x ) leads to the indeterminate form 1.
(c) Let f ( x) 1 12
x
ln f (x) x ln 1 x
x
lim ln f ( x) lim
x
ln 1 x 2
x
lim
2 x 3
1 x 2
2
lim
lim f ( x) lim e
ln f ( x )
x
x
Therefore lim 1 12
x
2
1
x x
x
x
x
2 x 2 lim 4 x lim 4 0.
3
2
x x x x 3 x 1 x 6 x
e0 1
Copyright 2016 Pearson Education, Ltd.
472
Chapter 7 Transcendental Functions
85. Let
f (k ) 1 kr
k
ln f (k )
ln 1 rk 1
k
86. (a)
y
y x1/ x ln y lnxx y
1
k
k
y
k k
1x ( x)ln x y 1ln x
r
lim krk r
1
k 1 rk
k
lim
x . The sign pattern is y | |
x2
y x1/ x ln y ln2x y
1x x2 2 x ln x
x4
x
1/ x
x2
which indicates a maximum value of y e
2
k
1/e
(b)
1
rk 2
1 rk 1
2
f (k ) lim eln f ( k ) e r .
klim
k
lim 1r r. Therefore lim 1 kr
k
k
1
lim ln1 rk lim
0
e
when x e
y 1 2 3ln x
x
x . The sign pattern is
1/ x 2
y | | which indicates a maximum of y e1/(2e) when x e
0
(c)
e
n
y x1/ x ln y lnnx
1x xn (ln x) nxn 1
x
x
2n
y
x n 1 (1 n ln x )
x
2n
n
x1/ x . The sign pattern is
y | | which indicates a maximum of y e1/( ne) when x n e
n
0
(d)
87. (a)
n
e
x
lim x1/ x lim eln x
x
x
1/ x n
n
lim e(ln x ) x exp lim lnnx exp lim 1n e0 1
x
x x
x nx
y x tan 1x , lim x tan 1x
sec2 1 1
tan 1
x x2
lim 1 x lim
lim sec2 1x 1; lim x tan 1x
x
1
x x x
x
x2
sec2 1 1
tan 1x
x x2
lim 1 lim
lim sec2
1
x x x
x
x2
x and as x .
(b)
88.
2x
3 x e2 x
3x
x 2 x e
2 x e
lim
3 2e2 x
3x
x 2 3e
lim lim 0; lim
2
4e2 x
3x
x 9e
1/ h
1/ h
f (0 h ) f (0)
f (0) lim
lim e h 0 lim e h
h
h 0
h 0
h 0
2
e 0
lim
3 e 2 x
ex
x 2 3
1
1h
h2
lim 2 lim
lim h 2
2
1/
h
1/
h
1/
h
2
h 0 e
h 0 e 3 h 0 2 e
h
2
h 1/ h
2
h 0
lim
1x 1 the horizontal asymptote is y 1 as
3 x e2 x
4
x
3x
x 9e
x 2 x e
32 the horizontal asymptotes are y 0 as x and y 32 as x .
y 3 x e 3 x , lim
89. (a) We should assign the value 1 to
f ( x) (sin x) x to make it continuous at x 0.
Copyright 2016 Pearson Education, Ltd.
Section 7.6 Inverse Trigonometric Functions
(b) ln f ( x) x ln(sin x)
ln(sin x )
1
x
lim ln f ( x) lim
x 0
ln(sin x )
x 0
1
x
lim
x 0
sin1 x (cos x) lim x2
1
2
x
x 0 tan x
lim 22x 0 lim f ( x) e0 1
x 0 sec x
x 0
(c) The maximum value of f ( x) is close to 1 near the point x 1.55 (see the graph in part (a)).
(d) The root in question is near 1.57.
90. (a) When sin x 0 there are gaps in the sketch.
The width of each gap is .
(b) Let f ( x) (sin x) tan x
ln f ( x) (tan x) ln(sin x) lim ln f ( x)
x 2
ln(sin x )
cot x
lim
x 2
lim
x 2
sin1 x (cos x)
csc2 x
x 0 lim f ( x ) e0 1.
lim ( cos
csc x )
x 2
Similarly,
x 2
lim
2
x
f ( x) e0 1. Therefore,
lim f ( x) 1.
x 2
(c) From the graph in part (b) we have a minimum of about 0.665 at x 0.47 and the maximum is about
1.491 at x 2.66.
7.6
INVERSE TRIGONOMETRIC FUNCTIONS
1. (a) 4
(b) 3
(c) 6
2. (a) 4
(b) 3
(c) 6
3. (a) 6
(b) 4
(c) 3
4. (a) 6
(b) 4
(c) 3
5. (a) 3
(b) 34
(c) 6
6. (a) 4
(b) 3
(c) 6
3
4
(b) 6
(c)
(b) 6
(c)
7. (a)
2
3
8. (a)
3
4
Copyright 2016 Pearson Education, Ltd.
2
3
473
474
Chapter 7 Transcendental Functions
2
15.
17.
19.
21.
14.
lim tan 1 x 2
16.
lim sec1 x 2
18.
x 1
x
x
lim csc1 x lim sin 1 1x 0
x
y cos 1 x 2 dx
dy
23. y sin 1 2t dt
1 x
dy
2
2
1
25. y sec 1 (2 s 1) ds
2t
2
dy
26. y sec1 5s ds
dy
27.
dy
2
y csc1 x 2 1 dx
x
2
|2 s 1| 4 s 2 4 s
x 1
x
1
|2 s 1| s 2 s
1
x 1 1
2
2
2 x
x 1 x 2 x
2
4
2
2t
12 1 2
2
2
| x| x 2 4
| 2x | 2x 1 | x| x 4 4
dy
29. y sec1 1t cos 1 t dt 1 2
1t
2
dy
30. y sin 1 32 csc1 t3 dt
t
2
31.
t
3
y cot 1 t cot 1 t1/2 dt
dy
23t
2
t2
3 1
12 t 1/ 2
1 t
lim csc1 x lim sin 1 1x 0
x
|s| 25 s 2 1
2x
2
x
dy
dy
1 2t 2
lim sec 1 x lim cos 1 1x 2
24. y sin 1 (1 t ) dt
2
dy
28. y csc1 2x dx
lim tan 1 x 2
x
1 x 4
2
5
x 1
22. y cos 1 1x sec 1 x dx
2 x
|2 s 1| (2 s 1) 2 1
|5 s| (5 s )2 1
lim cos 1 x
20.
2x
12. cot sin 1 23 cot 3 1
3
lim sin 1 x 2
x
tan 6 13
11. tan sin 1 12
13.
10. sec cos 1 12 sec 3 2
9. sin cos 1 22 sin 4 1
1/ 2 2
t
2
t 4 9
9
6
t t 4 9
1
2 t (1t )
Copyright 2016 Pearson Education, Ltd.
1
1 (1t )2
1
| x| x 2 1
1
2t t 2
Section 7.6 Inverse Trigonometric Functions
32.
y cot 1 t 1 cot 1 (t 1)1/2 dt
dy
1
y ln tan
34.
y tan 1 (ln x) dx
35.
dy
et
y csc1 et dt
2
t
t
1x
dy
1 (ln x )2
1
x 1 (ln x ) 2
e t
y cos 1 et dt
37.
y s 1 s 2 cos 1 s s 1 s 2
38.
39.
dy
1 e t
s2
1 s
2
y tan
2
e t
1e2t
cos1 s dyds 1 s2 s 12 1 s2
1/2
1/2
2
2
2
1 s
1 s
sec1 s dydx 12 s2 1
1
x tan
x 1 csc
2
1/2
1
1
(2 s )
1 s 2
1 s
1/2
1
1/2
2
1 s 2 s 12 1 s s 2 1 2 s 2
1 s
x 1 csc
2
1
y s 2 1 sec1 s s 2 1
1
1
1
2 t 1(1t 1)
2t t 1
e2t 1
2
2
1
e 1
|e |
36.
1 s2
1 (t 1)1/ 2
1
2
dy
1
dx 1 x1
tan x
tan 1 x 1 x 2
33.
x
12 (t 1)1/ 2
475
dy
x dx
1/2
(2s )
12 x2 1
1
|s| s 2 1
1/ 2
(2 x )
1/ 2
1 x 2 1
2
s
s 2 1
1
| x| x 2 1
1
|s| s 2 1
1
x x 2 1
s|s|1
|s| s 2 1
1
| x| x 2 1
for x 1
40.
2
dy
y cot 1 1x tan 1 x 2 tan 1 x 1 tan 1 x dx 0 x 2 1 2 21 1 2 0
41.
y x sin 1 x 1 x 2 x sin 1 x 1 x 2
sin 1 x
42.
43.
x
1 x
2
x
1 x
2
1 x
x 1
1 x
dydx sin 1 x x 11x 12 1 x2
1/2
1/2
2
(2 x)
sin 1 x
y ln x 2 4 x tan 1 2x dx
91 x2 dx sin
1 x 1
dy
x 1 x2x 4 tan 1 2x 42 xx tan 1 2x
2 x tan 1 x
2
x 4
2
1
2
x 2
2
2
C
1 x
3
Copyright 2016 Pearson Education, Ltd.
2
0,
476
44.
Chapter 7 Transcendental Functions
114 x dx 12 1(22 x) dx 12 1duu , where u 2 x and du 2dx
2
2
2
12 sin 1 u C 12 sin 1 (2 x) C
1 x
C
17
45.
171 x2 dx 17 12 x2 dx 117 tan
46.
913x2 dx 13 3 12 x2 dx 3 13 tan
47.
, where u 5 x and du 5 dx
x 25dxx2 2 u du
u 2 2
1
C tan C
3
9
x
3
1
x
3
1 sec1 u C 1 sec 1 5 x C
2
48.
2
2
2
x 5dxx2 4 u udu2 4 , where u 5 x and du 5 dx
12 sec 1 u2 C 12 sec1 25 x C
1
49.
0 44dss 4sin
50.
0
2
3 2 /4
ds
9 4 s 2
1 s 1
4 sin 1 12 sin 1 0
2 0
3 2 /4
du
0
9 u 2
3 2 /2
12
, where u 2s and du 2ds; s 0 u 0, s 3 42 u 3 2 2
12 sin 1 u3
0
51.
52.
2
12 sin 1 22 sin 1 0 12 4 0 8
2 2 du
, where u 2t and du 2dt ; t 0 u 0, t 2 u 2 2
8 u 2
2 2
1 1 tan 1 u
14 tan 1 2 2 tan 1 0 14 tan 1 1 tan 1 0 14 4 0
8 0
8
2 8
0 8dt2t 12 0
2
4 6 0 23
2
16
2 3
2 4dt3t 2 13 2 3 4duu 2 , where u 3t and du 3dt; t 2 u 2 3, t 2 u 2 3
2 3
1 12 tan 1 u2
1 tan 1 3 tan 1 3 1 3 3
2 3
3 3
3
2 3 2 3
53.
2 /2
1
dy
2
y 4 y 1
2
2
du
u u 2 1
, where u 2 y and du 2dy; y 1 u 2, y 22 u 2
2
sec1 | u |
sec1 2 sec 1 | 2 | 4 3 12
2
54.
2 /3
2/3
dy
2
y 9 y 1
2
2
du
u u 2 1
, where u 3 y and du 3dy; y 23 u 2, y 32 u 2
2
sec1 | u |
sec1 2 sec 1 | 2 | 4 3 12
2
Copyright 2016 Pearson Education, Ltd.
Section 7.6 Inverse Trigonometric Functions
55.
14(3drr 1)2 32 1duu 2 , where u 2(r 1) and du 2dr
32 sin 1 u C 32 sin 1 2(r 1) C
56.
46(drr 1)2 6 4duu 2 , where u r 1 and du dr
6sin 1 u2 C 6sin 1 r 21 C
57.
2(dxx 1) 2duu , where u x 1 and du dx
2
2
1 tan 1 u C 1 tan 1 x 1 C
2
58.
2
2
2
1(3dxx1) 13 1duu , where u 3x 1 and du 3dx
2
2
13 tan 1 u C 13 tan 1 (3x 1) C
59.
(2 x 1) (2dxx1) 4 12 u udu 4 , where u 2 x 1 and du 2dx
2
2
12 12 sec1 u2 C 14 sec 1 2 x21 C
60.
( x 3) (dxx3) 25 u udu25 , where u x 3 and du dx
2
2
15 sec1 u5 C 15 sec1 x 53 C
61.
/2
1
d 2
/2 12cos
1 1duu , where u sin and du cos d ; 2 u 1, 2 u 1
(sin )
2
2
1
2 tan 1 u 2 tan 1 1 tan 1 (1) 2 4 4
1
62.
/4
1
2
xdx
/6 1csc
3 1duu , where u cot x and du csc x dx; x 6 u 3, x 4 u 1
(cot x )
2
2
2
1
tan 1 u tan 1 1 tan 1 3 4 3 12
3
63.
ln 3 e x dx
3 du
, where u e x and du e x dx;
2x
1
1 e
1u 2
3
1
1
1
0
tan
64.
e / 4
1
4 dt
t 1 ln 2 t
u
1
4
/4 du
0
1u 2
, where u ln t and du 1t dt ; t 1 u 0, t e /4 u 4
/4
ydy
3 tan 1 3 4 12
tan
4 tan 1 u
0
65.
x 0 u 1, x ln 3 u 3
4 tan 1 4 tan 1 0 4 tan 1 4
2
1 y 12 1duu , where u y and du 2 y dy
4
2
12 sin 1 u C 12 sin 1 y 2 C
Copyright 2016 Pearson Education, Ltd.
477
478
66.
Chapter 7 Transcendental Functions
sec2 y dy
2
1 tan y 1duu , where u tan y and du sec y dy
2
2
sin 1 u C sin 1 (tan y ) C
1
67.
x dx 4 x3 1 x dx4 x 4 1(dxx2) sin ( x 2) C
68.
2 dxx x 1 x dx2 x 1 1(dxx 1) sin ( x 1) C
69.
1 362dtt t 2 61 4t 2dt 2t 1 61 22 dt(t 1)2 6 sin
70.
1/2 364dtt 4t 2 31/2 4 42t 2dt4t 1 31/2 22 2(2dtt 1)2 3 sin
2
2
2
1
2
0
2
2
0
0
1
1
sin 1 0 6 6 0
1 t 1 0
6 sin 1 12
2 1
1
sin 1 0
1 2t 1 1
3 sin 1 12
2 1/2
3 6 0 2
dy
dy
dy
C
1 y 1
2
71.
y 2 2 y 5 4 y 2 2 y 1 22 ( y 1)2 12 tan
72.
y 6 y 10 1 y 6 y 9 1( y 3) tan ( y 3) C
73.
1 x 82dxx 2 81 1 x dx2 x 1 81 1( dxx 1) 8 tan ( x 1) 1 8 tan 1 tan
74.
2 x 26dxx 10 22 1 x dx6 x9 22 1( xdx3) 2 tan ( x 3) 2 2 tan 1 tan (1) 2 4 4
75.
xx244dx x2x 4dx x24 4dx; x2x 4dx 12 u1du where u x 4 du 2 xdx 12 du xdx
dy
dy
2
1
dy
2
2
2
2
2
2
4
4
4
2
2
1
1
1
1
4
1
0 8 4 0 2
2
2
1
2
2
2
x2 4 dx 12 ln x 2 4 2 tan 1 2x C
x 4
76.
t t6t210dt (t t3)2 1dt Let w t 3 w 3 t dw dt ww11 dw w w1dw w 11dw;
2
w w1dw 12 u1du where u w 1 du 2w dw 12 du w dw w w1dw w 11dw
2
2
2
2
77.
2
2
1
2
2
12 ln w 1 tan ( w) C 12 ln (t 3) 1 tan (t 3) C 12 ln t 6t 10 tan 1 (t 3) C
2
1
2
2
x x 2x91 dx 1 2xx 109 dx dx x2x 9 dx 10 x 19 dx; x22x 9 dx u1 du where
2
2
2
2
2
u x 2 9 du 2 xdx dx 22 x dx 10 21 dx x ln x 2 9 10
tan 1 3x C
3
x 9
x 9
Copyright 2016 Pearson Education, Ltd.
Section 7.6 Inverse Trigonometric Functions
78.
t 2tt 13t 4dt t 2 2t t 12 dt t 2 dt t 2t 1dt 2 t 11dt; t 22t 1 dt u1 du where
3
2
2
2
2
2
u t 2 1 du 2t dt (t 2)dt 22t dt 2 21 dt 12 t 2 2t ln t 2 1 2 tan 1 (t ) C
t 1
79.
t 1
( x 1) dxx2 2 x ( x 1) xdx2 2 x11 ( x 1) dx( x 1)2 1 u duu 2 1 , where u x 1 and du dx
sec1 | u | C sec1 | x 1| C
80.
( x 2) dxx2 4 x 3 ( x 2) xdx2 4 x 41 ( x 2) dx( x2)2 1 u u12 1du, where u x 2 and du dx
sec1 | u | C sec 1 | x 2| C
81.
sin 1 x
u
e 1 x dx e du, where u sin
eu C esin
82.
cos 1 x
1
dx
1 x 2
C
u
1
cos 1 x
x and du dx2
1 x
C
sin 1 x dx u 2 du, where u sin 1 x and du dx
1 x2
1 x 2
2
3
u3 C
tan 1 x
1/2
1 x2 dx u
sin x C
1
3
3
du , where u tan 1 x and du dx 2
1 x
23 u 3/2 C 23 tan 1 x
85.
x and du
2
e C e
84.
x
e 1 x dx e du, where u cos
u
83.
1
2
tan 1 y11 y 2 dy
1
1 y 2
dy
tan 1 y
3/2
C 23
tan 1 x C
u1 du, where u tan
3
1
y and du
dy
1 y 2
ln | u | C ln tan 1 y C
86.
sin y1 1 y dy
1
2
1
1 y 2
dy
sin 1 y
u1 du, where u sin
1
y and du
dy
1 y 2
ln | u | C ln sin 1 y C
87.
dx /3 sec2u du, where u sec1 x and du dx ; x 2 u , x 2 u
/4
4
3
x x 1
x x 2 1
2
1
2 sec sec x
2
2
/3
tan u /4 tan 3 tan 4 3 1
Copyright 2016 Pearson Education, Ltd.
479
480
88.
Chapter 7 Transcendental Functions
/3
1
2/ 3 x x 1 dx /6 cos u du, where u sec x and du x dxx2 1 ; x 23 u 6 , x 2 u 3
2
cos sec1 x
2
/3
sin u /6 sin 3 sin 6
89.
x ( x 1) tan1 1 x 2 9 dx 2 u 219 du where u tan
1
23 tan 1 tan 3
90.
x
1 x
e dx u du where u sin
e sin
1e 2 x
12 sin
91.
3 1
2
1
x du
1
1
1 dx 2du
1
dx
2 x
(1 x ) x
C
x
1 x
1
e du
1e
2x
e x dx
C
1 x 2
e
5
2
lim sin x 5 x lim 1125 x 5
x 0
x 0
1
12 x2 1
1/ 2
1/ 2
92.
x 2 1
2
lim x 11 lim
1
x 1 sec x x 1 sec x
93.
lim x tan 1 2x
x
94.
12 x
1
2
1 9 x4
2
tan
3
x
6
lim
lim
lim
4
7 x2
x 0
x 0 14 x
x 0 7 19 x
95.
2 3 x4 1
2(0 1)
2x
4 2
1 2
2
1 x
tan
x
1 x 4
lim
lim 2
10 2 22 1
lim
1
1
1
x 2
sin x x 0
x 0 x sin x x 0 x
1 x2 3/ 2 (1 0)3/ 2
1 x 2
xlim
lim
x 1
tan 1 2 x 1
x
1
(2 x )
1
| x| x2 1
lim
x
lim x | x | 1
x 1
2 x 2
1 4 x 2
2
x
2
2
2
x 1 4 x
lim
76
2x
96.
x
2
x
lim e tan
2x
x
1 x
e
e x
x
lim
x
e tan
1 x
e 2x
e
2x
e x tan 1 e x 2e x
e2 x
2e 1
1
lim
x
e
4e
1
2x
2 e2 x
lim
2
e2 x 1
e x tan 1 e x
x
1 x
e2 x 3 lim tan 1 e x 13e2 x 0 0 0
lim tan xe
2
x
2
x 4e
4 e2 x 1 x 4e
4 e x e x
Copyright 2016 Pearson Education, Ltd.
4e
e2 x e 2 x 3
2x
2
e2 x 1
Section 7.6 Inverse Trigonometric Functions
97.
tan 1 x
lim
x x 1
x 0
2
lim
tan 1
x x (11 x )
x
2 x 1
x 0
x 1
tan 1
lim
x 0
x
x (1 x )
3x2
2 x 1
1
2 tan 1 x
lim
lim 2x (1 x )
x 0 (3 x 2) x x 1 x 0 12 x 13 x 2
2 x x 1
2
lim
2 1
12 x 2 13 x 2
x 0
x1 2
98.
lim
lim
x 0 sin x
x 0 2 sin x
2x
sin 1 x 2
2
1
1 x 4
1
x
1
lim
lim
1
x 0 sin 1 x 1 x 2 x 0 sin 1 x x 1 1 x 2
1 x 2
1 x 2
1 x 2
2
2
lim 2 1 x 12 x 1 11 1
x 0 1 x x 1 x sin x
x tan 1 x
2
1
dx
99. If y ln x 12 ln 1 x 2 tanx x C , then dy 1x x 2 1 x 2
1 x
x
x1 x 2 x3 x tan 1 x 1 x 2
1
1
1x x 2 1 2 tan 2 x dx
dx tan 2 x dx, which verifies the formula
2
2
1 x
x
x
x1 x
x 1 x
4
x4
100. If y x4 cos 1 5 x 54
1 25 x 2
dx, then
4
4
dy x3 cos 1 5 x x4 5 2 54 x 2 dx x3 cos 1 5 x dx, which verifies the formula
1 25 x
1 25 x
101. If y x sin 1 x
2 x 2 1 x2 sin 1 x C, then
2
2 2 x sin 1 x
2
dy sin 1 x
2 2 x2 sin 1 x 2 1 x 2 1 2 dx sin 1 x dx, which verifies the
2
1 x
1 x
1 x
formula
2
102. If y x ln a 2 x 2 2 x 2a tan 1 ax C , then dy ln a 2 x 2 22 x 2 2 2 2 dx
a x
1 x 2
a
2
2
ln a 2 x 2 2 a 2 x 2 2 dx ln a 2 x 2 dx, which verifies the formula
a x
dy
103. dx
dy
104. dx
1
1 x
2
dy
dx
1 x
2
y sin 1 x C ; x 0 and y 0 0 sin 1 0 C C 0 y sin 1 x
1 1 dy
1 1 dx y tan 1 ( x) x C ; x 0 and y 1 1 tan 1 0 0 C C 1
x 2 1
1 x 2
1
y tan ( x) x 1
dy
105. dx
1
dy
dx
y sec1 | x | C ; x 2 and y sec 1 2 C C sec 1 2
x x 2 1
x x 2 1
3 23 y sec 1 ( x) 23 , x 1
Copyright 2016 Pearson Education, Ltd.
481
482
Chapter 7 Transcendental Functions
dy
106. dx
1 2 dy 1 2 dx y tan 1 x 2sin 1 x C ; x 0 and y 2
1 x 2
1 x 2
1 x 2
1 x 2
1
1
1
1
2 tan
0 2sin
107. (a) The angle
0 C C 2 y tan
x 2sin
x2
is the large angle between the wall and the right end of the blackboard minus the small angle
cot 1 5x cot 1 x .
between the left end of the blackboard and the wall
d
dt
(b)
1
5
x 2
1 5
20 4 x 2
1
5 2 12
; ddt 0 20 4 x 2 0 x
2
25 x
1 x
25 x 2 1 x 2
1 x
5. Since x 0,
consider only x 5
5 cot 1 55 cot 1 5 0.729728 41.8103. Using the
first derivative test, ddt
16
0 and ddt
52
x 1
x 10
380 0 local maximum of 41.8103° when
1375
x 5 2.24 m.
108. V
/3
0
22 (sec y ) 2 dy 4 y tan y /3 4 3
0
3
109. V 13 r 2 h 13 (3sin ) 2 (3cos ) 9 cos cos3 , where 0 2
dV
9 (sin ) 1 3cos 2 0 sin 0 or cos 1 the critical points are: 0, cos 1 1 , and
d
1
3
3
cos 1 ; but cos 1 1 is not in the domain. When 0, we have a minimum and when
3
3
1 1
cos
54.7, we have a maximum volume.
3
110. 65 90
90 180
65
21 65 22.78 42.22
65 tan 1 50
111. Take each square as a unit square. From the diagram we have the following: the smallest angle has a tangent
of 1 tan 1 1; the middle angle has a tangent of 2 tan 1 2; and the largest angle has a tangent of
3
tan 1 3. The sum of these three angles is
tan 1 1 tan 1 2 tan 1 3 .
112. (a) From the symmetry of the diagram, we see that sec 1 x is the vertical distance from the graph of
y sec1 x to the line y and this distance is the same as the height of y sec1 x above the x-axis at x;
i.e., sec1 x sec1 ( x).
(b) cos 1 ( x) cos 1 x, where 1 x 1 cos 1 1x cos 1 1x , where x 1 or x 1
1
sec ( x) sec
1
x
113. sin 1 (1) cos 1 (1) 2 0 2 ;sin 1 (0) cos 1 (0) 0 2 2 ; and sin 1 (1) cos 1 (1) 2 2 . If
x (1, 0) and x a, then sin 1 ( x) cos 1 ( x) sin 1 ( a) cos 1 ( a ) sin 1 a cos 1 a
sin 1 a cos 1 a 2 2 from Equations (3) and (4) in the text.
Copyright 2016 Pearson Education, Ltd.
Section 7.6 Inverse Trigonometric Functions
483
114.
d csc1 u d sec1 u 0
115. csc1 u 2 sec1 u dx
dx 2
du
dx
2
|u| u 1
du
dx
2
, |u | 1
|u | u 1
d (tan y ) d ( x )
116. y tan 1 x tan y x dx
dx
dy
dy
sec2 y dx 1 dx
1
sec2 y
1
1 x
2
2
1 , as
1 x 2
indicated by the triangle
df 1
117. f ( x) sec x f ( x) sec x tan x dx
x b
1
df
dx x f 1 ( b )
1
sec sec 1 b tan sec1 b
d sec 1 x
of sec 1 x is always positive, we obtain the right sign by writing dx
d cot 1 u d tan 1 u 0
118. cot 1 u 2 tan 1 u dx
dx 2
du
dx
1u
2
119. The function f and g have the same derivative (for x 0), namely
1
b b 2 1
1
| x| x 2 1
. Since the slope
.
du
dx
1u 2
1
. The functions therefore differ by a
x ( x1)
constant. To identify the constant we can set x equal to 0 in the equation f ( x) g ( x) C , obtaining
sin 1 (1) 2 tan 1 (0) C 2 0 C C 2 . For x 0, we have sin 1 xx 11 2 tan 1 x 2 .
120. The functions f and g have the same derivative for x 0, namely
1 . The functions therefore differ by a
1 x 2
constant for x 0. To identify the constant we can set x equal to 1 in the equation f ( x) g ( x) C , obtaining
sin 1
tan 1 C C C 0. For x 0, we have sin
1
1
2
4
1
1
4
x 2 1
tan 1 1x .
2
3
3
1
1 dx tan 1 x
dx
tan 1 3 tan 1 33
3/3
3/3 1 x 2
3/3 1 x 2
2
3 6 2
121. V
3
dy
122. Consider y r 2 x 2 dx
dy
x
2
r x
2
; Since dx is undefined at x r and x r , we will find the length from
x 0 to x r (in other words, the length of 18 of a circle) L
0
2
r/ 2
0
r sin 1
r/ 2
2
1 2x 2 dx
r x
r/ 2
r 2 dx r / 2
0
r x2
2
0
1
r/ 2
dx r sin 1 rx
0
r x
r
2
2
2
dx
2
2
r x
x
r sin 1 r / r 2 r sin 1 (0)
0 r . The total circumference of the circle is C 8L 8 2 r.
1
2
4
r
r
4
4
Copyright 2016 Pearson Education, Ltd.
484
Chapter 7 Transcendental Functions
123. (a)
1
b
1
A( x ) 4 (diameter)2 4 1 2 1 2 2 V A( x) dx dx2 tan 1 x
1
a
1 1 x
1 x
1 x 1 x
2
2
( )(2) 4 2
2
(b)
1
b
1
A( x) (edge)2 1 2 1 2 4 2 V A( x) dx 4dx2 4 tan 1 x
1
a
1 1 x
1 x
1 x 1 x
4[tan 1 (1) tan 1 (1)] 4 4 ( 4 ) 2
2
124. (a)
A( x) 4 (diameter)2 4 4 2 2 0 4 4 2
1 x
1 x
2 /2
sin 1 x
sin 1
2 /2
(b)
A( x )
1 x
2
V A( x) dx
1
2
2
2
2
1 x
2
2
4
b
4
2
2
2 /2
2 /2
2
2 4 2
1 0.84107
125. (a) sec1 1.5 cos 1 1.5
1 0.72973
(b) csc1 (1.5) sin 1 1.5
(c) cot 1 2 2 tan 1 2 0.46365
126. (a) sec1 (3) cos 1 13 1.91063
dx
dx 2 sin 1 x
2 /2
2 /2 1 x 2
V A( x) dx
2
a
2 /2
2 /2 1 x 2
a
sin
(diagonal)2
12 4 2 2 0
2
1 x
b
1 0.62887
(b) csc1 1.7 sin 1 1.7
(c) cot 1 (2) 2 tan 1 (2) 2.67795
127. (a) Domain: all real numbers except those having
the form 2 k where k is an integer.
Range: 2 y 2
(b) Domain: x ; Range: y
The graph of y tan 1 (tan x ) is periodic, the
graph of y tan(tan 1 x) x for x .
128. (a) Domain: x ; Range: 2 y 2
Copyright 2016 Pearson Education, Ltd.
Section 7.6 Inverse Trigonometric Functions
(b) Domain: 1 x 1; Range: 1 y 1
The graph of y sin 1 (sin x) is periodic; the
graph of y sin (sin 1 x ) x for 1 x 1.
129. (a) Domain: x ; Range: 0 y
(b) Domain: 1 x 1; Range: 1 y 1
The graph of y cos 1 (cos x) is periodic; the
graph of y cos (cos 1 x ) x for 1 x 1.
130. Since the domain of sec1 x is (, 1] [1, ), we
have sec (sec1 x) x for | x | 1. The graph of
y sec(sec 1 x) is the line y x with the open line
segment from (1, 1) to (1, 1) removed.
131. The graphs are identical for y 2sin 2 tan 1 x
4 sin tan 1 x cos tan 1 x
4
x
1
4x
x 2 1 from the triangle
x 2 1 x 2 1
132. The graphs are identical for y cos 2sec1 x
x
2
cos 2 sec 1 x sin 2 sec 1 x 12 x 21 22x
x
2
x
from the triangle
Copyright 2016 Pearson Education, Ltd.
485
486
Chapter 7 Transcendental Functions
133. The values of f increase over the interval [1, 1]
because f 0, and the graph of f steepens as the
values of f increase toward the ends of the interval.
The graph of f is concave down to the left of the
origin where f 0, and concave up to the right of
the origin where f 0, There is an inflection point
at x 0 where f 0 and f has a local minimum
value.
134. The values of f increase throughout the interval
(, ) because f 0, and they increase most
rapidly near the origin where the values of f are
relatively large. The graph of f is concave up to the
left of the origin where f 0, and concave down to
the right of the origin where f 0. There is an
inflection point at x 0 where f 0 and f has a
local maximum value.
7.7
HYPERBOLIC FUNCTIONS
sinh x 3 ,
1 169 1625 54 , tanh x cosh
x
5
1. sinh x 34 cosh x 1 sinh 2 x 1 34
34
2
5
4
1 5 , sech x 1 4 , and csch x 1 4
coth x tanh
x
3
cosh x
5
sinh x
3
2. sinh x 43 cosh x 1 sinh 2 x 1 16
9
25 5 , tanh x sinh x
9
3
cosh x
43 4 , coth x 1 5 ,
tanh x
4
53 5
1 3 , and csch x 1 3
sech x cosh
sinh x
4
x
5
3. cosh x 17
, x 0 sinh x cosh 2 x 1
15
1 17 ,
coth x tanh
x
8
17 2 1
15
289 1
225
64 8 ,
225 15
169 1
25
144 12 ,
25
5
ln x
e ln x
2
e
lnx
12
5
sinh x 5 12 ,
tanh x cosh
13
x
13
1 13 , sech x 1 5 , and csch x 1 5
coth x tanh
x 12
cosh x 13
sinh x 12
15
1 15 , and csch x 1 15
sech x cosh
x 17
sinh x
8
4. cosh x 13
, x 0 sinh x cosh 2 x 1
5
5. 2 cosh (ln x) 2 e
8
sinh x 15 8 ,
tanh x cosh
x
17 17
1 x 1
x
eln x
Copyright 2016 Pearson Education, Ltd.
Section 7.7 Hyperbolic Functions
6.
2
2 ln x
2ln x
ln x
ln x
sinh (2 ln x) e 2e
e 2e
2
x2 1
4
2
2 x x 21
2x
7. cosh 5 x sinh 5 x
e5 x e 5 x e5 x e5 x
e5 x
2
2
9. (sinh x cosh x) 4
e x e x e x e x
2
2
487
8. cosh 3 x sinh 3 x e
e e
4
x 4
3x
e 3 x e3 x e 3 x e 3 x
2
2
4x
10. ln(cosh x sinh x) ln(cosh x sinh x) ln cosh 2 x sinh 2 x ln1 0
11. (a) sinh 2 x sinh( x x) sinh x cosh x cosh x sinh x 2sinh x cosh x
(b) cosh 2 x cosh( x x) cosh x cosh x sinh x sin x cosh 2 x sinh 2 x
12. cosh 2 x sinh 2 x
e x e x
2
2
e x e x
2
2e x 14 4e0 14 (4) 1
e e e e e e e e
2
x
1
4
x
x
x
x
x
x
x
14 2e x
dy
13 2 cosh 3x
13.
y 6sinh 3x dx 6 cosh 3x
14.
y 12 sinh 2 x 1 dx 12 cosh(2 x 1) (2) cosh(2 x 1)
15.
dy
y 2 t tanh t 2t1/2 tanh t1/2 dt sech 2 t1/2
16.
y t 2 tanh 1t t 2 tanh t 1 dt sech 2 t 1 t 2 t 2 2t tanh t 1 sech 2 1t 2t tanh 1t
dy
12 t 1/2 2t1/2 tanh t1/2 t 1/2 sech 2 t tanht t
dy
dy
z coth z
17. y ln(sinh z ) dz cosh
sinh z
dy
sinh z tanh z
18. y ln(cosh z ) dz cosh
z
19.
dy
y (sech )(1 ln sech ) d
sech tanh
sech
sech sech tanh 1 ln sech
sech tanh sech tanh (1 ln sech ) (sech tanh ) 1 1 ln sech
sech tanh ln sech
20.
dy
y csch 1 ln csch d csch
csch coth
csch
1 ln csch csch coth
csch coth 1 ln csch csch coth csch coth 1 1 ln csch
csch coth ln csch
Copyright 2016 Pearson Education, Ltd.
488
21.
Chapter 7 Transcendental Functions
sinh v 1 2 tanh v sech 2v tanh v tanh v sech 2 v
y ln cosh v 12 tanh 2 v dv cosh
2
v
dy
(tanh v ) 1 sech 2 v (tanh v) tanh 2 v tanh 3 v
22.
dy
v 1 2 coth v csch 2 v coth v coth v csch 2 v
y ln sinh v 12 coth 2 v dv cosh
sinh v
2
(coth v ) 1 csch 2 v (coth v) coth 2 v coth 3 v
e 2e x2 1 x2x x2 1 x2 x1 2 x dydx 2
23.
y x 2 1 sech ln x x 2 1
24.
y 4 x 2 1 csch ln 2x 4 x 2 1
25.
y sinh 1 x sinh 1 x1/2 dx
ln x
ln x
1
2
e 2e 4 x2 1 2 x(22 x) 4 x2 1 4 x4 x1 4 x dydx 4
ln 2 x
ln 2 x
12 x1/ 2
dy
1/ 2 2
1 x
1
2
1
1
2 x 1 x
2 x (1 x )
(2) 12 ( x 1)1/ 2
26.
y cosh 1 2 x 1 cosh 1 2 x 1
27.
y (1 ) tanh 1 d (1 )
28.
dy
y ( 2 2 ) tanh 1 ( 1) d 2 2 1 2 (2 2) tanh 1 ( 1)
1( 1)
1/2
dy
dy
dx
2
2( x 1)1/ 2 1
(1) tanh
1
1
1 2
1
x 1 4 x 3
1
4 x 2 7 x 3
1 tanh 1
1
2
2 2 (2 2) tanh 1 ( 1) (2 2) tanh 1 ( 1) 1
2
29.
1 t 1/ 2
dy
(1) coth 1 t1/2 1 coth 1 t
y (1 t ) coth 1 t (1 t ) coth 1 t1/2 dt (1 t ) 2
2 t
1 t1/ 2 2
30.
y 1 t 2 coth 1 t dt 1 t 2
11t 2t coth 1 t 1 2t coth 1 t
31.
y cos 1 x x sech 1 x dx
1
32.
dy
2
dy
1 x
2
x 1 2 (1) sech 1 x
x 1 x
1
1 x 2
1
1 x 2
sech 1 x sech 1 x
sech 1x
1/2
1/2
dy
dx 1x 1 x 2 1 12 1 x 2
2 x sech 1 x 1x 1x x sech 1 x x sech 1 x
1 x
1 x
x 1 x
y ln x 1 x 2 sech 1 x ln x 1 x 2
1/2
2
2
Copyright 2016 Pearson Education, Ltd.
2
Section 7.7 Hyperbolic Functions
33.
34.
dy
y csch 1 12 d
y csch 1 2 d
1
2
1 12
(ln 2)2
dy
2
35.
y sinh 1 (tan x) dx
36.
y cosh 1 (sec x) dx
dy
dy
12
ln 1
2
1 (tan x )
ln(1) ln(2)
1 12
2
2
2
sec x
(sec x )(tan x )
2
sec x 1
1 12
x
sec 2x sec
|sec x|
2
ln 2
2
1 22
1 2
sec2 x
ln 2
2
2
489
(sec x )(tan x )
37. (a) If y tan 1 (sinh x) C , then dx
dy
(b) If y sin 1 (tanh x) C , then dx
dy
2
tan x
|sec x||sec x|
| sec x |
|sec x|
(sec x )(tan x )
sec x,
|tan x|
0 x 2
cosh x cosh x sech x, which verifies the formula
1 sinh 2 x
cosh 2 x
sech 2 x
2
1 tanh x
2
x sech x, which verifies the formula
sech
sech x
2
2
dy
38. If y x2 sech 1 x 12 1 x 2 C , then dx x sech 1 x x2 1 2 2 x 2 x sech 1 x, which verifies the
x
1
x
4 1 x
formula
2
x coth x, which verifies the formula
2
39. If y x 21 coth 1 x 2x C , then dx x coth 1 x x 21
dy
40. If y x tanh 1 x 12 ln 1 x 2 C , then dx tanh 1 x x
dy
1
1 x 2
1
2
tanh x, which verifies the
1
1 x 2
1 2 x
2 1 x 2
formula
41.
sinh 2 x dx 12 sinh u du, where u 2 x and du 2 dx
42.
cosh u
cosh 2 x
C 2 C
2
sinh 5x dx 5 sinh u du, where u 5x and du 15 dx
5cosh u C 5cosh 5x C
43.
6 cosh 2x ln 3 dx 12 cosh u du, where u 2x ln 3 and du 12 dx
12 sinh u C 12sinh 2x ln 3 C
44.
1
4 cosh (3x ln 2) dx 43 cosh u du, where u 3x ln 2 and du 3 dx
43 sinh u C 43 sinh(3 x ln 2) C
Copyright 2016 Pearson Education, Ltd.
1
490
45.
Chapter 7 Transcendental Functions
sinh u
tanh 7x dx 7 cosh u du, where u 7x and du 17 dx
7 ln | cosh u | C1 7 ln cosh 7x C1 7 ln e
x/7
e x /7 C 7 ln e x /7 e x /7 7 ln 2 C
1
1
2
7 ln e x /7 e x /7 C
46.
u du , where u and du d
coth 3 d 3 cosh
sinh u
3
3
/ 3
3 ln sinh u C1 3 ln sinh C1 3 ln e
3
e / 3 C
1
2
3 ln e / 3 e / 3 3 ln 2 C1 3 ln e / 3 e / 3 C
47.
sech x 12 dx sech u du, where u x 12 and du dx
2
2
tanh u C tanh x 12 C
48.
2
2
csch (5 x)dx csch u du, where u (5 x) and du dx
( coth u ) C coth u C coth (5 x) C
49.
sech t tanh t
dt 2
t
1/2
sech u tanh u du, where u t t
and du dt
2 t
2( sech u ) C 2 sech t C
50.
csch ( ln t ) coth (ln t )
dt
t
csch u coth u du, where u ln t and du dtt
csch u C csch(ln t ) C
51.
ln 4
ln 4
15/8
x dx
ln 2 coth x dx ln 2 cosh
3/4 u1 du where u sinh x, du cosh x dx;
sinh x
x ln 2 u sinh(ln 2) e
ln 2
e ln 2
2
3 , x ln 4 u sinh(ln 4) eln 4 e ln 4 4 14 15
2 12
2
4
15/8
15
3
ln | u | 3/4 ln 8 ln 4 ln 15
. 4 ln 52
8 3
52.
ln 2
0
2
tanh 2 x dx
ln 2 sinh 2 x
17/8 1
dx 12
du where u cosh 2 x, du 2sinh (2 x ) dx,
u
cosh 2 x
1
0
x 0 u cosh 0 1, x ln 2 u cosh (2 ln 2) cosh (ln 4) e
17/8
12 ln | u | 1 12 ln 17
8
53.
ln2
ln4
e 2
2 ln 4
ln 2 e 2
ln 2
ln 4
e ln 4
2
17
4 14
2
ln1 12 ln 17
8
ln 2
2
ln 2 2
2e e 2e d
e 1 d e 2
ln 4
ln 4
ln 4
2e cosh d
2 ln 2
2
1 ln 4 3 ln 2 2 ln 2 3 ln 2
ln 4 81 ln 2 32
32
32
Copyright 2016 Pearson Education, Ltd.
8
8
Section 7.7 Hyperbolic Functions
54.
ln 2
0 4e
sinh d
0
ln 2
2
ln 2
4e e 2e d 2
1 e2 d 2 e 2
0
0
0 2 ln 2 2 ln 2 1 ln 4
2 ln 2 e 2
55.
ln 2
2 ln 2
e0
2
/4
491
1
8
1
2
1
4
1
3
4
/4 cosh(tan ) sec d 1 cosh u du where u tan , du sec d , x 4 u 1, x 4 u 1,
2
2
1 1
1 1
1 1
1
sinh u 1 sinh(1) sinh(1) e 2e e 2e e e 2 e e e e 1
56.
/2
0
1
2sinh(sin ) cos d 2 sinh u du where u sin , du cos d , x 0 u 0, x 2 u 1
0
1
2 cosh u 0 2(cosh1 cosh 0) 2 e 2e 1 e e1 2
1
57.
2 cosh(ln t )
ln 2
dt
cosh u du where u ln t , du 1t dt ,
t
0
1
sinh u 0
ln 2
58.
sinh(ln 2) sinh(0) e
4 8 cosh x
2
dx 16 cosh u du where u
1
x
1
ln 2
x 1 u 0, x 2 u ln 2
1
e ln 2 0 2 2 3
2
2
4
x x1/2 , du 12 x 1/2 dx dx , x 1 u 1, x 4 u 2
2 x
16 sinh u 12 16(sinh 2 sinh1) 16 e 2e
59.
2
2
8 e e e e
ee
2
1
2
2
1
ln 2 cosh 2x dx ln 2 cosh2x 1 dx 12 ln 2 (cosh x 1)dx 12 sinh x x ln 2
0
0
2
0
0
12 (sinh 0 0) (sinh( ln 2) ln 2) 12 (0 0) e
eln 2 ln 2 1
2
2
ln 2
12 2 ln 2 1 1 1 ln 2
2
2
4
83 12 ln 2 83 ln 2
60.
ln10
0
ln10
ln10
ln10
4sinh 2 2x dx
4 cosh2 x 1 dx 2
(cosh x 1)dx 2 sinh x x 0
0
2 sinh(ln 10) ln 10 (sinh 0 0) e
5 ln 5
61. sinh 1 12
12
25 1
144
1(1/2)
0
ln10
ln
2
3
2 3
2 3
dx
4 x
2
sinh 1 2x
0
(b) sinh 1 3 ln
25 1
9
ln 3
64. coth 1 54 12 ln (1/4) 12 ln 9 ln 3
1 1(9/25)
65. sech 1 53 ln (3/5) ln 3
0
62. cosh 1 53 ln 53
63. tanh 1 12 12 ln 1 (1/2) ln33
67. (a)
1 2 ln10 9.9 2 ln10
e ln10 2 ln10 10 10
(9/4)
66. csch 1 1 ln 3 4/3 ln 3 2
3
1/ 3
sinh 1 3 sinh 0 sinh 1 3
3 3 1 ln 3 2
Copyright 2016 Pearson Education, Ltd.
492
Chapter 7 Transcendental Functions
68. (a)
0
1/3
6 dx
1 9 x
2
1 dx
, where u 3 x, du 3 dx, a 1
0 a 2 u 2
1 1
1
1
2
2sinh u 2 sinh 1 sinh
0
1
2
(b) 2sinh 1 2 ln 1 1 1 2 ln 1 2
69. (a)
2
5/4 11x dx coth
1
2
0 2sinh 1 1
2
x
coth 1 2 coth 1 54
5/4
1 ln 1
(b) coth 1 2 coth 1 54 12 ln 3 ln 9/4
1/4 2
3
1/2
1/2
1
x tanh 1 12 tanh 1 0 tanh 1 12
0
1
1/2
1 ln 3
(b) tanh 1 12 12 ln
11/2 2
70. (a)
0 11x dx tanh
71. (a)
1/5 x 1dx16 x 4/5 u aduu , u 4 x, du 4 dx, a 1
2
3/13
12/13
2
2
2
12/13
sech 1u
sech 1 12
sech 1 54
13
4/5
1 1(12/13)2
1 1(4/5)2
1 4
13 169 144
16
sech
ln
ln
ln 5 25
(b) sech 1 12
(4/5) ln
13
5
(12/13)
12
4
ln 54 3 ln 13125 ln 2 ln 23 ln 2 23 ln 43
72. (a)
(b)
73. (a)
2
1 x 4dx x2 12 csch
1
2
1 x 2
12
2 1
csch 11 csch 1 12 12 csch1 12 csch 11
5/4
csch 1 12 csch 11 12 ln 2 (1/2)
ln 1 2 12 ln 12 25
0
cos x
0 1sin x dx 0 11u du where u sin x, du cos x dx;
2
2
0
sinh 1 u sinh 1 0 sinh 1 0 0
0
(b) sinh 1 0 sinh 1 0 ln 0 0 1 ln 0 0 1 0
74. (a)
e
1
, where u ln x, du 1x dx, a 1
1 x 1dx(ln x)2 0 adu
2
u 2
1
sinh 1 u sinh 1 1 sinh 1 0 sinh 1 1
0
(b) sinh 1 1 sinh 1 0 ln 1 12 1 ln 0 02 1 ln 1 2
f ( x) f ( x)
f ( x) f ( x)
f ( x) f ( x)
f ( x) f ( x)
2 f ( x)
and O( x)
. Then E ( x) O( x )
2 f ( x).
2
2
2
2
f x f ( x )
f ( x) f ( x)
f ( x ) f ( ( x ))
x
E ( x) E ( x) is even, and O( x)
2
2
2
75. Let E (x)
Also, E
Copyright 2016 Pearson Education, Ltd.
Section 7.7 Hyperbolic Functions
493
f ( x) f ( x)
O( x) O( x) is odd. Consequently, f ( x) can be written as a sum of an even and an odd
2
f ( x) f ( x)
f ( x) f ( x)
f ( x) f ( x)
because
0 if f is even, and f ( x)
because
function. f ( x)
2
2
2
f ( x) f ( x)
2 f ( x)
2 f ( x)
0 if f is odd. Thus, if f is even f ( x) 2 0 and if f is odd, f ( x) 0 2
2
76.
y
y sinh 1 x x sinh y x e 2e
y
2 x e y 1y 2 xe y e 2 y 1 e 2 y 2 xe y 1 0
e
e y 2 x 24 x 4 e y x x 2 1 sinh 1 x y ln x x 2 1 Since e y 0, we cannot choose
2
e y x x 2 1 because x x 2 1 0.
mg
tanh
k
77. (a) v
lim v lim
t
t
735
0.235
(c)
gk
t mg 1 tanh 2
m
m dv
mg sech 2
dt
when t 0.
(b)
mg
sech 2
k
gk
t dv
m
dt
mg
tanh
k
kg
t
m
gk
t
m
gk
g sech 2
m
gk
t . Thus
m
gk
t mg kv 2 . Also, since tanh x 0 when x 0, v 0
m
mg
lim tanh
k t
kg
t
m
mg
(1)
k
mg
k
735,000
55.93 s
235
2
78. (a) s (t ) a cos kt b sin kt ds
ak sin kt bk cos kt d 2s ak 2 cos kt bk 2 sin kt
dt
dt
2
2
k (a cos kt b sin kt ) k s (t ) acceleration is proportional to s. The negative constant k 2
implies that the acceleration is directed toward the origin.
2
ak sinh kt bk cosh kt d 2s ak 2 cosh kt bk 2 sinh kt
(b) s (t ) a cosh kt b sinh kt ds
dt
dt
k 2 (a cosh kt b sinh kt ) k 2 s (t ) acceleration is proportional to s. The positive constant k2 implies
that the acceleration is directed away from the origin.
2
cosh 2 x sinh 2 x dx 1 dx 2
0
0
79. V
2
80. V 2
ln 3
0
81.
sech 2 x dx 2 tanh x 0
ln 3
y 12 cosh 2 x y sinh 2 x L
ln 5
0
12 e
82. (a)
(b)
2x
e 2 x
2
ln 5
0
3 1/ 3
2
3 1/ 3
1 (sinh 2 x)2 dx
ln 5
0
cosh 2 x dx 12 sinh 2 x
14 5 15 56
ex 1 1
e x 1x
1 21x
x
x
ex ex
e
e
e
lim tanh x lim x x lim x 1 lim
1 lim e 1 1100 1
x
x e e
x e x
x e x 1
x 1 2 x
x
e
e
ex e
x
x
lim tanh x lim e x e x lim
x
x e e
e x 1x
e
x 1
x e x
e
ex 1
x
2x
ex e
x lim e2 x 1 0011 1
x e x 1 e
x e 1
ex
lim
Copyright 2016 Pearson Education, Ltd.
ln 5
0
494
Chapter 7 Transcendental Functions
(c)
(d)
(e)
x
lim sinh x lim e 2e
x
x
x
x
lim sinh x lim e 2e
x
0
lim
0
1
x
e x ex
lim e2 21ex
2
x
x
lim
x
x
e x
2
ex
x 2
1
2
e
e
x
x
2
lim x 2 1 e1 lim e 1 100 0
x
x
1
x e e
x e x x
x
2x
lim sech x lim
x
e
(f )
x 1
e 1x
1 21x
e x 1x
x
x
e e
e
e
e
lim coth x lim x x lim x 1 lim
1 lim e 1 1100 1
x
x e e
x e x
x e x 1 x
x 1 2 x
e
e
e
ex
(g)
e x 1x
x
x
x
2x
e
e
lim coth x lim x x lim x e1 e x lim e2 x 1
e e
e
e
x 0
x 0
x 0
x 0 e 1
x
x
e
(h)
(i)
83. (a)
x
x
x
lim coth x lim e x e x lim
x 0
x 0
e
e
e 1x
2x
x
e x lim e2 x 1
e
x 1
x 0 e x
e
e
x 0
e
1
x
x
2
lim x 2 1 . e x lim 22 xe 001 0
x
x
e
e
e
e
e
1
x
x
x
x
lim csch x lim
x
e
dy
w x tan
yH
cosh H
w
dx
Hw Hw sinh Hw x sinh Hw x
(b) The tension at P is given by T cos H T H sec H 1 tan 2
H 1 sinh Hw x
2
w x wy
H cosh Hw x w H
cosh H
w
84. s 1a sinh ax sinh ax as ax sinh 1 as x 1a sinh 1 as; y 1a cosh ax 1a cosh 2 ax
1a sinh 2 ax 1 1a a 2 s 2 1 s 2 12
a
85. To find the length of the curve: y a1 cosh ax y sinh ax L
b
0
b
1 (sinh ax) 2 dx L cosh ax dx
0
b
b
b
1a sinh ax 1a sinh ab. The area under the curve is A a1 cosh ax dx 12 sinh ax 12 sinh ab
0
0
a
0 a
1a sinh ab which is the area of the rectangle of height 1a and length L as claimed, and which is illustrated
1a
below.
Copyright 2016 Pearson Education, Ltd.
Section 7.8 Relative Rates of Growth
495
86. (a) Let the point located at (cosh u , 0) be called T. Then A(u ) area of the triangle OTP minus the area
under the curve y x 2 1 from A to T A(u ) 12 cosh u sinh u
(b)
12 cosh 2 u 12 sinh 2 u sinh 2 u 12 cosh 2 u sinh 2 u 12 (1) 12
A(u ) 12 cosh u sinh u
cosh u
1
x 2 1 dx.
x 2 1 dx A(u ) 12 cosh 2 u sinh 2 u cosh 2 u 1 sinh u
cosh u
1
(c) A(u ) 12 A(u ) u2 C , and from part (a) we have A(0) 0 C 0 A(u ) u2 u 2 A
7.8
RELATIVE RATES OF GROWTH
1. (a) slower, lim x x3 lim 1x 0
x e
x e
3
3 x 2 2sin x cos x
2
x lim
(b) slower, lim x sin
x
2 x lim 6 4sin 2 x 0 by the Sandwich
lim 6 x 2 cos
ex
ex
ex
x
x
x
2 x 10 for all reals, and lim 2 0 lim 10
Theorem because 2x 6 4sin
x
x
e
ex
ex
x e
x e
e
x
1/ 2
x
(c) slower, lim
x e
lim x x lim
x
x e
x e
x
32 lim
x
3 x 0 since 3 1
2e
x 2e
x
x e
x/2
(f ) slower, lim e x lim
x e
(g)
1
0
x
x 2 xe
ex
4 since 4e 1
x e
x
(d) faster, lim 4x lim
(e) slower, lim
x
12 x1/ 2 lim
1
x e
x/2
0
ex
2
same, lim x lim 12 12
x e
x
(h) slower, lim
log10 x
e
x
x
1
ln x lim
x
1
lim
0
x
x
x
x (ln10) e
x (ln10) e
x (ln10) xe
lim
4
3
2
x 1 lim 40 x 30 lim 120 x lim 240 x lim 240 0
2. (a) slower, lim 10 x 30
x
x
x
x
x
e
x
(b) slower, lim
x ln x x
ex
x
lim
x
x (ln x 1)
4
lim 12xx
x e
5
(d) slower, lim 2 lim
x
x
x e
x
ex
x
1 x 4
ex
(c) slower, lim
e
x
e
x
lim
x
x e
e
4 x3
2x
2
x e
lim
x e
lim ln x 11 lim ln x lim 1x lim 1 0
ln x 1 x 1x
x
x
2
ex
lim 12 x2 x
x 4e
x e
x
lim 242xx
x 8e
x e
x
lim 242 x
x 16
x xe
0 0
5 0 since 25e 1
x 2e
x
(e) slower, lim e x lim
x e
1 0
2x
x e
x
(f ) faster, lim xex lim x
x e
x
1
cos x
e
e
1
(g) slower, since for all reals we have 1 cos x 1 e 1 ecos x e1 e x e x ex and also
1
1
cos x
lim e x 0 lim ex , so by the Sandwich Theorem we conclude that lim e x 0
x e
x e
x e
Copyright 2016 Pearson Education, Ltd.
e
x
496
Chapter 7 Transcendental Functions
x 1
(h) same, lim e x lim
x e
1
x e
lim 1e 1e
( x x 1)
x
2
3. (a) same, lim x 24 x lim 2 2x x 4 lim 22 1
x
x
x
5
2
x
(b) slower, lim x 2x lim x 1
x
x
x 4 x3
x2
(c) same, lim
x
(d) same, lim
x
( x 3)2
x
x
x ln x
(e) slower, lim
4
3
lim x 4x
lim
x2
x
3
lim 1 1x 1 1
x
x
2 x 3
lim 22 1
2x
x
1
ln x
lim x lim 1x 0
x
x
2
x x
(ln 2)2 x
(ln 2) 2 2 x
lim
2x
2
x
x
x
(f ) slower, lim 22 lim
x x
3 x
(g) slower, lim x e2 lim
x lim 1 0
x
x
x e x e
x x
2
(h) same, lim 8 x2 lim 8 8
x x
x
2
4. (a) same, lim x 2 x lim 1 3/1 2 1
x
x
x
x
2
(b) same, lim 10 x2 lim 10 10
x x
x
2 x
(c) slower, lim x e2 lim
(d) slower, lim
1 0
x
x e
x
x
log10 x 2
x2
x
3
ln x2
1
ln10
1 lim 2 ln x 2 lim x 1 lim 1 0
lim 2 ln10
2
ln10
2
ln10 x x 2
x x
x x
x x
2
(e) faster, lim x 2.x lim ( x 1)
x
x
x
101 lim
x
(f ) slower, lim
2
x x
(g) faster, lim
(1.1) x
x x
2
1 0
x 2
x 10 x
(ln1.1)(1.1) x
(ln1.1)2 (1.1) x
lim
2
x
2
x
x
lim
2
x lim 1 100 1
(h) same, lim x 100
2
x
x
x
x
ln x lim 1 1
log3 x
lim lnln 3x
x ln x
x
5. (a) same, lim
x ln 3
ln 3
x 2
2
1
x
x
1 ln x lim 1 1
(c) same, lim ln x lim 2
(b) same, lim lnln2xx lim
x ln x
2
2x
1
x
x ln x
1/ 2
12 x1/ 2 lim x lim x
1
x x
x 2 x
x 2
(d) faster, lim ln xx lim xln x lim
x
x
Copyright 2016 Pearson Education, Ltd.
Section 7.8 Relative Rates of Growth
(e) faster, lim lnxx lim
x
x
1
1x
497
lim x
x
x lim 5 5
(f ) same, lim 5ln
ln x
x
x
1
x
(g) slower, lim ln x lim x ln1 x 0
x
x
x
x
faster, lim lne x lim e1
x
x
(h)
x
lim xe x
x
ln x 2
ln 2
2
log 2 x 2
same, lim ln x lim ln x ln12 lim lnlnxx ln12 lim 2lnlnxx ln12 lim 2 ln22
x
x
x
x
x
6. (a)
ln10 x 1
log10 10 x
ln10
lim ln
x
x ln x
x
(b) same, lim
10 1 lim 1 1
x ln10 x ln10
1 lim 10 x
lim ln10 x ln10
1
ln10 x ln x
x
1
(c) slower, lim lnxx lim
x
x
1
x (ln x)
0
1
2
slower, lim lnx x lim 2 1 0
x
x x ln x
(d)
x 2 lim x 2 lim 1 2 lim x 2
x ln x x x
ln
x
x
x lim
(e) faster, lim x ln2 ln
x
x
x
1
x
e lim
(f ) slower, lim ln
x
x
1 0
x
x e ln x
1/ln xx lim 1 0
1x x ln x
2
ln(2 x 5)
(h) same, lim ln x lim 2 x1 5 lim 2 2x x 5 lim 22 lim 1 1
x
x x
x
x
x
ln(ln x )
lim
x ln x
x
(g) slower, lim
e x lim e x /2 e x grows faster then e x /2 ; since for x ee we have ln x e and
x/2
e
x
x
7. lim
lim
x
ln x x
ex
lim
x
lnex (ln x) x grows faster then e x ; since x ln x for all x 0 and
x
x x lim
x x x x grows faster then (ln x ) x . Therefore, slowest to fastest are:
x
ln
x
x (ln x )
x
x /2 x
x x
lim
e
8.
so the order is d, a, c, b
, e , (ln x) , x
lim
x
(ln 2) x
x
2
lim
ln(ln 2) (ln 2) x
x
2
2x
lim
x
ln(ln 2) 2 (ln 2) x
2
ln(ln 2) 2
2
lim (ln 2) x 0 (ln 2) x grows slower than
x
x
2 x lim
2
0 x 2 grows slower than 2 x ; lim 2 xx lim 2e 0 2 x
x
2 x
x (ln 2)2
x (ln 2) 2
x e
x
x 2 x
x
x
x 2 ; lim x x lim
x 2
grows slower than e . Therefore, the slowest to the fastest is: ln 2 , x , 2 and e so the order is c, b, a, d
9. (a) false; lim xx 1
x
(b) false; lim x x 5 11 1
x
Copyright 2016 Pearson Education, Ltd.
498
Chapter 7 Transcendental Functions
(c) true; x x 5 x x 5 1 if x 1 (or sufficiently large)
(d) true; x 2 x 2xx 1 if x 1 (or sufficiently large)
x
(e) true; lim e2 x lim 1x 0
x e
x e
(f ) true; x xln x 1 lnxx 1 xx 1 1 2 if x 1 (or sufficiently large)
x
lim 1 1
x x
(g) false; lim lnln2xx lim
x
x 2 5
x
(h) true;
10. (a) true;
1
x
2
2x
( x 5)2
x x 5 1 5x 6 if x 1 (or sufficiently large)
x
x 13 x 1 if x 1 (or sufficiently large)
1x x3
1 1
x 2
x
(b) true;
1x
(c) false; lim
x
1 1x 2 if x 1 (or sufficiently large)
1 1
x 2
x
1x
lim 1 1x 1
x
x 3 if x is sufficiently large
(d) true; 2 cos x 3 2 cos
2
2
x
(e) true; e x x 1 xx and
e
e
x 0 as x 1 x 2 if x is sufficiently large
ex
ex
(f ) true; lim x ln2 x lim lnxx lim
(g)
1x 0
x x
x
x 1
ln(ln x ) ln x
true; ln x ln x 1 if x is sufficiently large
1
2
x
ln x lim
lim x 21 lim 12 1 2
2
2x
x ln( x 1) x 2 x x 2 x
x
x 2 1
(h) false; lim
1
2
f ( x)
g ( x)
f ( x)
11. If f ( x ) and g ( x) grow at the same rate, then lim g ( x ) L 0 lim f ( x ) L1 0. Then g ( x ) L 1 if x is
f ( x)
x
f ( x)
x
sufficiently large L 1 g ( x ) L 1 g ( x ) L 1 if x is sufficiently large f O( g ). Similarly,
g ( x)
L1 1 g O( f ).
f ( x)
f ( x)
12. When the degree of f is less than the degree of g since in that case lim g ( x ) 0.
x
f ( x)
13. When the degree of f is less than or equal to the degree of g since lim g ( x ) 0 when the degree of f is smaller
f ( x)
x
than the degree of g, and lim g ( x ) ba (the ratio of the leading coefficients) when the degrees are the same.
x
14. Polynomials of a greater degree grow at a greater rate than polynomials of a lesser degree. Polynomials of the
same degree grow at the same rate.
15.
ln( x 1)
lim
x ln x
x
lim
x11 lim x lim 1 1 and lim ln( x 999) lim x 1999 lim x 1
1
ln x
x
x x
x x 999
1x x x1 x 1
Copyright 2016 Pearson Education, Ltd.
Section 7.8 Relative Rates of Growth
16.
17.
x 1 a lim x lim 1 1. Therefore, the relative rates are the same.
1x x x a x 1
ln( x a )
lim
x ln x
x
lim
10 x 1
x
lim
x
x 1
x
lim 10 xx 1 10 and lim
x
x
lim x 1
x x
1 1. Since the growth rate is transitive, we
conclude that 10 x 1 and x 1 have the same growth rate (that of
18.
x4 x
x2
lim
x
499
4
lim x 4 x 1 and lim
x x
4
x
3
4
x 4 x3
x2
4
x ).
3
lim x 4x 1. Since the growth rate is transitive, we
x
x
conclude that x x and x x have the same growth rate (that of x 2).
19.
n 1
n
x e
x e
x e
20. If p( x) an x n an 1 x n 1
lim
p( x)
x e
x
a1 x a0 , then
n 1
n
an lim xx an 1 lim xex
x e
p( x)
19). Therefore, lim
x
x e
21. (a)
lim nx! 0 x n o e x for any non-negative integer n
lim x x lim nx x
1/ n
lim xln x lim
x
x
x
x
a1 lim xx a0 lim 1x where each limit is zero (from Exercise
x e
x e
0 e grows faster than any polynomial.
x (1 n )/ n
n 1n
1n lim x1/ n ln x o x1/ n for any positive integer n
x
(b) ln e17,000,000 17, 000, 000 e1710
6
1/106
e17 24,154,952.75
(c) x 3.430631121 1015
(d) In the interval 3.41 1015 ,3.45 1015 we have
ln x 10 ln(ln x). The graphs cross at about
3.4306311 1015.
22.
lim ln x
lim 1/nx1
nx
x x n
ln
x
x
1
lim
lim
n
n 1
a0
a1
n
an
an 1
x an x an 1x a1x a0
x an nx
lim an x n 1 n
x
x
x
than any non-constant polynomial (n 1)
23. (a)
lim
n log 2 n
n n (log 2 n )
2
lim log1 n 0 n log 2 n
n
(b)
2
grows slower then n(log 2 n) 2 ;
lim
n
n log 2 n
n
3/2
lnln n2 1 lim 1n
1/ 2
ln 2 n 1 n 1/ 2
n n
2
lim
ln22 lim
1 0 n log n grows slower
2
1/2
n n
than n3/2 . Therefore, n log 2 n grows at the
slowest rate the algorithm that takes
O(n log 2 n) steps is the most efficient in the
long run.
Copyright 2016 Pearson Education, Ltd.
0 ln x grows slower
500
Chapter 7 Transcendental Functions
24. (a)
lim
log 2 n 2
n
n
2(ln n ) 1n
ln n
(ln n )2
lim ln 2 lim
lim
1 0
2
n
n
n n (ln 2)
2
2
log 2 n grows slower then n; lim
ln 2 2
n
log 2 n 2
n n log 2 n
lim
log 2 n
n
1n 2 lim 1 0
1 1/ 2
ln 2 n n1/ 2
x 2 n
ln12 lim
2
lim ln n 2 lim n
ln 2 2 n n (ln 2)2 n 1
n
lim
1 lim ln n
ln n
ln 2
1/ 2
n n
ln 2 n n1/ 2
(b)
2
log 2 n grows slower than n log 2 n.
2
Therefore log 2 n grows at the slowest rate
the algorithm that takes O log 2 n
2
steps
is the most efficient in the long run.
25. It could take one million steps for a sequential search, but at most 20 steps for a binary search because
219 524, 288 1, 000, 000 1, 048,576 220.
26. It could take 450,000 steps for a sequential search, but at most 19 steps for a binary search because
218 262,144 450, 000 524, 288 219.
CHAPTER 7
PRACTICE EXERCISES
2 2 e 2 x 2e 2 x
1.
y 10e x /5 dx (10) 15 e x /5 2e x /5
3.
1 e 4 x dy 1 x 4e 4 x e 4 x (1) 1 4e4 x xe4 x 1 e 4 x 1 e 4 x xe 4 x
y 14 xe 4 x 16
dx
4
4
4
16
4.
y x 2 e 2/ x x 2 e 2 x
5.
y ln sin 2 d
dy
dy
dy
1
6. y ln sec2 d
dy
2.
y 2e 2 x dx
1
1
1
dy
dx x 2 2 x 2 e 2 x e 2 x 2 x 2 2 x e 2 x 2e 2/ x 1 x
2(sin )(cos )
cos 2 cot
2sin
sin 2
2(sec )(sec tan )
sec2
2 tan
ln x2
dy
2
x
1
ln 2 dx ln 2
x2 ln 2 x
2
7.
2
y log 2 x2
8.
y log5 (3x 7)
2
3x37 (ln 5)(33 x7)
ln(3 x 7)
dy
dx ln15
ln 5
9. y 8t dt 8t (ln 8)( 1) 8t (ln 8)
dy
Copyright 2016 Pearson Education, Ltd.
Chapter 7 Practice Exercises
dy
10. y 92t dt 92t (ln 9)(2) 92t (2 ln 9)
dy
11. y 5 x3.6 dx 5(3.6) x 2.6 18 x 2.6
12. y 2 x 2 dx
dy
13.
2 2 x 2 1 2 x 2 1
y
y ( x 2) x 2 ln y ln( x 2) x 2 ( x 2) ln( x 2) y ( x 2) x 1 2 (1) ln( x 2)
dy
dx ( x 2) x 2 ln( x 2) 1
14.
1
y
y 2(ln x) x /2 ln y ln 2(ln x) x /2 ln(2) 2x ln(ln x ) y 0 2x lnx x ln(ln x) 12
y 2 ln1 x 12 ln(ln x) 2(ln x) x /2 (ln x) x /2 ln(ln x) ln1x
15.
y sin
16.
1
u
u 1u 2
2
1 u sin
1
y sin 1
1u 2
1
1 u
1
1
1/2
12 v 3/ 2
dy
dv
1 v
y ln cos
18.
y z cos 1 z 1 z 2 z cos 1 z 1 z 2
cos 1 z
z
1 z
1/ 2
2
1 z
2
y z sec1 z z 2 1 z sec1 z z 2 1
dy
| z| z 1
z
z 1
sec1 z
1
1t 2
dy
2
1
2v3/ 2 1v 1
tan t
21.
u
|u| 1u 2
1
3/2 v 1
2v v 1
2v
v 1
2v3/ 2 v v1
1/2
( 2 z )
cos 1 z
z
y 1 t 2 cot 1 2t dx 2t cot 1 2t 1 t 2
2
2
2
20.
z
1/2
y t tan 1 t 12 ln t dt tan 1 t t
dydz cos1 z 1z z 12 1 z 2
19.
1/ 2
1
2
1
y 1x1
cos x
1 x 2 cos 1 x
17.
x
,0 u 1
1
v
2
sin v
1
1u
( 2u )
dy
u
du 2
1/ 2 2
1u 2 1 1u 2
2
1 1u
2 1/2
1 z
2
z 1
1
2
1
t
1
t 1
1t 2 2t
142t
2
dydz z |z| 1z 1 sec1 z (1) 12 z 2 1
1/2
2
sec 1 z , z 1
Copyright 2016 Pearson Education, Ltd.
1/2
2z
501
502
22.
Chapter 7 Transcendental Functions
1 1/ 2
1/2
1/2 x
dy
y 2 x 1sec 1 x 2( x 1)1/2 sec1 x1/2 dx 2 12 x 1
sec1 x1/2 x 1 2
x
x
1
1
x
21x
2 x 1
2 sec
sec1 x 1
x
x 1
23.
y csc1 (sec ) d
24.
y 1 x 2 e tan
25.
y
dy
x
sec tan
|sec | sec2 1
y 2 xe tan
1
x
tan 1, 0
|tan
|
2
1
1
tan 1 x
1 x 2 e 2 2 xe tan x e tan x
1 x
ln y ln 2 x2 1 ln(2) ln x2 1 1 ln(cos 2 x) y 0 2 x 1 2sin 2 x
2 x 2 1
cos 2 x
y
26.
1
2 x tan 2 x
x 2 1
cos 2 x
y
2 x 2 1
2
2 x tan 2 x
cos 2 x x 2 1
y
y
1 ln(3 x 4) ln(2 x 4)
y 10 32 xx44 ln y ln 10 32 xx44 10
y 101 3 x3 4 2 x2 4
2 cos 2 x
x 2 1
3x34 x1 2
3 1 y 10 3 x 4 1
1
y 10
3x4 x2
2 x 4 10
27.
5
(t 1)(t 1)
y (t 2)(t 3) ln y 5 ln(t 1) ln(t 1) ln(t 2) ln(t 3)
dy
(t 1)(t 1) 5
dt 5 (t 2)(t 3) t 11 t 11 t 11 t 13
28.
1
y
u 1
u
1 ln 2 u
u 2 1
u 1 u
d (sin )
dy
cot
ln(sin )
2
1
y
dy
d
2
cos
sin
1 1/2 ln(sin )
2
ln1x ln(ln x) yy ln1x ln1x 1x ln(ln x) (lnx1) 1x y ln x 1/ln x 1x(lnln(lnx)x)
y (ln x)1/ln x ln y
31.
e cos e dx cos t dt , where e t and e dx dt
x
2
x
x
sin t C sin e x C
e sin 5e 7 dx 15 sin t dt, where 5e 7 t and e dx dt5
x
30.
x
1 1 1
t 1 t 2 t 3
1y dudy u1 ln 2 12 u2u1
y (sin ) ln y ln y (sin )
1
t 1
u
dy
32.
dy
dt
y 2u22 ln y ln 2 ln u u ln 2 12 ln u 2 1
du 2u22
29.
5
x
x
x
cos t C 15 cos 5e x 7 C
Copyright 2016 Pearson Education, Ltd.
2
Chapter 7 Practice Exercises
33.
503
e sec e 7 dx sec u du, where u e 7 and du e dx
x
2
x
x
2
x
tan u C tan e x 7 C
34.
e csc e 1 cot e 1 dy csc u cot u du, where u e 1 and du e dy
y
y
y
y
y
csc u C csc e y 1 C
35.
sec x e
2
tan x
dx eu du, where u tan x and du sec2 x dx
eu C e tan x C
36.
csc x e
2
cot x
dx eu du , where u cot x and du csc2 x dx
eu C ecot x C
37.
1
1
1 3x14 dx 13 7 u1 du, where u 3x 4, du 3 dx; x 1 u 7, x 1 u 1
13 ln u
38.
1
13 ln | 1| ln | 7 | 13 0 ln 7
7
e ln x
1
dx u1/2 du , where u ln x, du 1x dx;
x
0
1
23 u 3/2 23 13/2 23 03/2 23
0
1
ln37
x 1 u 0, x e u 1
3 dx 3 1/2 1 du, where u cos x , du 1 sin x dx; x 0 u 1, x u 1
3
3
1 u
3
2
3
1/2
3 ln | u |1 3 ln 12 ln |1 | 3ln 12 ln 23 ln 8
tan 3x dx
sin x
39.
0
40.
x dx 2
1 du , where u sin x, du cos x dx;
1/6 2 cot x dx 21/6 cos
sin x
1/2 u
0 cos x
1/4
1/4
1/ 2
x 16 u 12 , x 14 u 1
2
1/ 2
2x ln | u | 1/2 2 ln 1 ln 12 2 ln1 12 ln 2 ln1 ln 2 2 12 ln 2 ln 2
2
41.
4
9
2
0 t 2 2t25 dt 25 u1 du, where u t 25, du 2t dt; t 0 u 25, t 4 u 9
9
9
ln | u |25 ln | 9 | ln | 25 | ln 9 ln 25 ln 25
42.
/6
1/2
/2 1cossint t dt 2 u1 du, where u 1 sin t , du cos t dt; t 2 u 2, t 6 u 12
ln | u |2 ln 12 ln 2 ln1 ln 2 ln 2 2 ln 2 ln 4
1/2
Copyright 2016 Pearson Education, Ltd.
504
43.
Chapter 7 Transcendental Functions
tan(ln v )
dv
v
sin u
tan u du cos u du, u ln v and du 1v dv
ln | cos u | C ln | cos(ln v ) | C
44.
v ln1 v dv u1 du, where u ln v and du 1v dv
ln | u | C ln | ln v | C
45.
(ln x )3
dx
x
3
u du, where u ln x and du 1x dx
2
u2 C 12 (ln x)2 C
46.
ln( x 9)
x 9 dx p dp, where ln( x 9) p and x 19 dx dp
p2
2 C 12 ln( x 9) C
47.
2
1r csc 1 ln r dr csc u du, where u 1 ln r and du 1r dr
2
2
cot u C cot 1 ln r C
48.
49.
sin 2 ln x
dx
x
sin t dt , where 2 ln x t and 1x dx dt
cos t C cos 2 ln x C
x2
2
u
x3 dx 12 3 du, where u x and du 2 x dx
C
1 3u C 1 3 x
2 ln
3
2 ln 3
50.
2
tan x
2
sec 2 x dx 2u du , where u tan x and du sec2 x dx
tan x
ln12 2u C 2ln 2 C
9
51.
1 5x dx 5ln x1 5(ln 9 ln1) 5ln 9
52.
1 41x dx 14 ln x 1 14 ln 81 ln1 14 ln 3 0 14 4 ln 3 ln 3
53.
15 1 ln 4
1 8x 21x dx 12 1 14 x 1x dx 12 18 x ln | x |1 12 168 ln 4 18 ln1 16
2
81
4
9
81
4
4
2
4
15 ln 4 15 ln 2
16
16
54.
1 32x x82 dx 32 1 1x 12 x
dx 23 ln | x | 12 x1 1 32 ln 8 128 (ln1 12)
23 ln 8 32 12 23 ln 8 21
2 ln 8 7 ln 82/3 7 ln 4 7
2 3
8
8
2
8
Copyright 2016 Pearson Education, Ltd.
Chapter 7 Practice Exercises
55.
1 ( x 1)
2 e
0
dx eu du, where u ( x 1), du dx; x 2 u 1, x 1 u 0
1
0
eu e0 e1 e 1
1
56.
0
ln 2 e
2w
dw 12
0
ln(1/4)
eu du, where u 2 w, du 2dw; w ln 2 u ln 14 , w 0 u 0
0
12 eu
1 e0 eln(1/4) 12 1 14 83
ln(1/4) 2
57.
ln 5 r
1
e 3er 1
3/2
16
dr 13 u 3/2 du, where u 3er 1, du 3e r dr ; r 0 u 4, r ln 5 u 16
4
16
14 12 23 14 16
23 u 1/2 23 161/2 41/2 23
4
58.
0 e e 1
ln 9
1/2
8
d u1/2 du, where u e 1, du e d ; 0 u 0, ln 9 u 8
0
8
11/ 2
23 u 3/2 23 83/2 03/2 23 29/2 0 2 3 323 2
0
59.
e
1 1x (1 7 ln x)
1/3
8
dx 17 u 1/3 du, where u 1 7 ln x, du 7x dx; x 1 u 1, x e u 8
1
8
3 u 2/3 3 82/3 12/3 3 (4 1) 9
14
14
14
1 14
60.
1/ 2
x 5ln x dx 5t dt 5 t1/2 10 t C 10 ln x C , where ln x t , 1x dx dt;
e3
e2 x 5ln x dx 10 ln x e2 10 ln e 10 ln e 10 3 2
e3
61.
2
3
2
3 ln v 1
3
ln 4 2
2
dv
ln(v 1) v11 dv
u du ,
v 1
1
1
ln 2
where u ln(v 1), du v11 dv;
v 1 u ln 2, v 3 u ln 4
ln 4
(ln 2)3
13 u 3
13 (ln 4)3 (ln 2)3 13 (2 ln 2)3 (ln 2)3 3 (8 1) 73 (ln 2)3
ln 2
62.
p2
(1 ln t )(t ln t )dt pdp 2 C
9
(9 ln 9)2 (3ln 3)2
(t ln t )2
2 12 (18ln 3)2 (3ln 3)2 12 315(ln 3) 2
2
2
3
9
3 (1 ln t )(t ln t )dt
63.
8 log 4
1
8
d ln14 (ln ) 1 d ln14
1
ln 8
2 ln1 4 u 2
0
64.
e 8(ln 3) log3
1
ln t dt dp, 1 ln t dt dp;
(t ln t )2
C where t ln t p, (t ) 1t
2
d
ln 8
0
u du, where u ln , du 1 d ; 1 u 0, 8 u ln 8
2
1 (ln 8) 2 02 (3ln 2) 9 ln 2
ln16
4 ln 2
4
d 801u du, where u ln , du 1 d
e 8(ln 3)(ln )
e
d 8 (ln ) 1
(ln 3)
1
1
Copyright 2016 Pearson Education, Ltd.
505
506
Chapter 7 Transcendental Functions
1 u 0, e u 1
1
4 u 2 4 12 02 4
0
65.
3/4
3/4
3/2
3/4 964 x2 dx 33/4 32 2(2 x)2 dx 33/2 321u 2 du, where u 2 x, du 2 dx;
x 34 u 32 , x 34 u 32
3/2
3 sin 1 u3
3 sin 1 12 sin 1 12 3 6 6 3 3
3/2
66.
1/5
1/5
1
1/5 4625 x2 dx 65 1/5 22 5(5 x)2 dx 65 1 221u 2 du, where u 5 x, du 5dx;
x 15 u 1, x 15 u 1
1
65 sin 1 u2 56 sin 1 12 sin 1 12 56 6 6 65 3 25
1
67.
2
2
2 3
3
2 433t 2 dt 3 2 22 3 t 2 dt 3 2 3 22 1u 2 du, where u 3 t , du 3dt;
t 2 u 2 3, t 2 u 2 3
2 3
3 12 tan 1 u2
23 tan 1
2 3
68.
69.
3
3
3 31t 2 dt 3 3 12 t 2 dt 13 tan
1
3 tan 1 3 23 3 3 3
tan
3
t
3
3
1
3
1
3
3 tan 1 1 1 3 4 36
3
y 41y 2 1 dy (2 y ) (22 y )2 1 dy u u12 1 du where u 2 y and du 2 dy
sec1 | u | C sec1 2 y C
70.
y y2416 dy 24 y y1 4 dy 24 12 sec
71.
2 /3 | y | 91y 2 1 dy 2 /3 |3 y | (33 y )2 1 dy 2 | u | 1u 2 1 du, where u 3 y, du 3 dy;
2
2
2/3
2
1 y
4
2/3
C 6sec
1 y
C
4
2
y 32 u 2, y 32 u 2
2
sec1 u sec1 2 sec1 2 3
2
72.
6/ 5
2 5
1
2
| y| 5 y 3
6/ 5
5
2/ 5
5 y 3
dy
5
2
2
dy
6
2
1
2
u u
3
2
du , where u 5 y, du 5dy;
y 2 u 2, y
5
Copyright 2016 Pearson Education, Ltd.
6
u
5
6
Chapter 7 Practice Exercises
6
507
1 sec1 u
1 sec 1 2 sec 1 2 1 4 6 1 312 212 363
3 2
3
3
3
3
12 3
3
73.
21x x2 dx 1 x21 2 x 1 dx 1(1x 1)2 dx 11u 2 du, where u x 1 and du dx
sin 1 u C sin 1 ( x 1) C
74.
x21 4 x 1 dx 3 x214 x 4 dx
sin 1
75.
1
1
2
3 x 2
2
dx
C sin C
1
2
du where u x 2 and du dx
3 u 2
1 x 2
3
u
3
1
1
1
2 v2 24v 5 dv 22 1 v2 14v 4 dv 22 1(v1 2)2 dv 20 11u 2 du, where u v 2, du dv;
v 2 u 0, v 1 u 1
1
2 tan 1 u 2 tan 1 1 tan 1 0 2 4 0 2
0
76.
1
1
1
3/2
1 4v2 34v 4 dv 34 1 34 v21v 14 dv 34 1 3 2 1 v 1 2 dv 34 1/2 3 12 u 2 du where u v 12 , du dv;
2
34 2 tan 1
3
77.
2u
3
3/2
1/2
2
2
v 1 u 12 , v 1 u 32
23 tan 1 3 tan 1 1 23 3 6 23
3
26 6 23 2 43
(t 1) t12 2t 8 dt (t 1) t 21 2t 19 dt (t 1) (t11)2 32 dt u u12 32 du, where u t 1 and du dt
13 sec1 u3 C 13 sec1 t 31 C
78.
(3t 1) 19t 2 6t dt (3t 1) 9t12 6t 11 dt (3t 1) (31t 1)2 12 dt 13 u u12 1 du, where u 3t 1 and du 3dt
13 sec 1 u C 13 sec1 3t 1 C
79. 3 y 2 y 1 ln 3 y ln 2 y 1 y (ln 3) ( y 1) ln 2 (ln 3 ln 2) y ln 2 ln 32 y ln 2 y ln 23
ln 2
80. 4 y 3 y 2 ln 4 y ln 3 y 2 y ln 4 ( y 2) ln 3 2 ln 3 (ln 3 ln 4) y
ln 9
(ln12) y 2 ln 3 y ln12
2
2
2
2
81. 9e 2 y x 2 e2 y x9 ln e2 y ln x9 2 y (ln e) ln x9 y 12 ln x9 ln
82. 3 y 3ln x ln 3 y ln(3ln x) y ln 3 ln(3ln x) y
ln(3ln x )
ln 3 ln(ln x )
ln 3
ln 3
Copyright 2016 Pearson Education, Ltd.
x 2 ln x ln | x | ln 3
9
3
508
Chapter 7 Transcendental Functions
83. ln( y 1) x ln y eln( y 1) e( x ln y ) e x eln y y 1 ye x y ye x 1 y 1 e x 1 y
84. ln(10 ln y ) ln 5 x eln(10 ln y ) eln 5 x 10 ln y 5 x ln y 2x eln y e x /2 y e x /2
2
a 1
a
85. lim x x3x1 4 lim 2 x13 5
86. lim x b 1 lim axb 1 ba
87. lim tanx x tan 0
x lim sec x 1 1
88. lim x tan
sin x
1 cos x 11 2
x 1
x 1 x 1
x 1
2
x
89.
x 0
2
sin(2 x )
lim sin 2x lim 2sin x2cos2x lim
x 0 tan x
x 0 2 x sec x
sin( mx )
x 1 bx
2
x 0 2 x sec x
2
lim
x 0
2 cos(2 x )
2
tan x2 2 x 2sec2 x2
x 0 2 x 2sec x
2
0221 1
m cos( mx )
90. lim sin( nx ) lim n cos( nx ) mn
x 0
91.
92.
x 0
lim sec(7 x) cos(3x) lim
x 2
x 2
cos(3 x )
cos(7 x )
lim
3sin(3 x )
x 2
7 sin(7 x )
73
x sec x lim cosxx 10 0
lim
x 0
x 0
cos x lim sin x 0 0
93. lim (csc x cot x) lim 1sin
x
cos x
1
x 0
x 0
x 0
lim lim 1 x lim 1 x lim 1
2
95.
lim x 2 x 1 x 2 x lim x 2 x 1 x 2 x
x
x
x 0
1
x4
2
1
4
x 0 x
lim
1
x2
1 x 2
4
x 0 x
94.
x 0
1
4
x 0 x
x 2 x 1 x 2 x
x 2 x 1 x 2 x
lim
x
2 x 1
x 2 x 1 x 2 x
Notice that x x 2 for x 0 so this is equivalent to
lim
x
96.
2 x 1
x
x 2 x 1
x2
x2 x
x2
x3 x3
2
2
x x 1 x 1
lim
2 1x
lim
x 1 1x 12 1 1x
2
1
1 1
x
lim 2 x lim 6 x lim 12 x lim 12 lim 1 0
x
x 1 x 1 x x 1 x 4 x x 12 x x 24 x x 2 x
lim
x3 x 2 1 x3 x 2 1
2
3
2
4
2
x
3
2
(ln10)10 x
ln10
1
x 0
97. The limit leads to the indeterminate form 00 : lim 10 x1 lim
x 0
(ln 3)3
ln 3
0 1
98. The limit leads to the indeterminate form 00 : lim 3 1 lim
0
sin x
99. The limit leads to the indeterminate form 00 : lim 2 x 1 lim
x 0 e 1
sin x
ex
x 0
1 lim 2
e 1
x 0
100. The limit leads to the indeterminate form 00 : lim 2 x
x 0
2sin x (ln 2)(cos x )
sin x
ln 2
(ln 2)( cos x )
ex
Copyright 2016 Pearson Education, Ltd.
ln 2
1
1 e x
Chapter 7 Practice Exercises
509
x lim 5cos x 5
101. The limit leads to the indeterminate form 00 : lim 5x5cos x lim 5sin
x
x
x 0 e x 1
x 0 e 1
2
e
x 0
2
2
2
2
2
x lim 2 x cos x sin x
102. The limit leads to the indeterminate form 00 : lim x sin3 x lim 2 x cos2 x sin
2
4
2
x 0 tan x
3 tan x sec x
x 0
2
x 0 3 tan x 3 tan x
6 8 x 4 cos x 2 24 x 2 sin x 2
6 x cos x 4 x sin x
6 x cos x 4 x sin x
lim
lim
66 1
3
2
2
5
3
4
2
2
2
2
x 0 12 tan x sec x 6 tan x sec x x 0 12 tan x 18 tan x 6 tan x x 0 60 tan x sec x 54 tan x sec x 6sec x
2
lim
2
2
2
3
2
103. The limit leads to the indeterminate form 00 : lim
t ln(1 2t )
104. The limit leads to the indeterminate form 00 : lim
sin 2 ( x )
t 0
t
x 4 e
lim
2 2 cos(2 x )
ex 4
x4
x4
2
1122t
lim
2t
t 0
3 x
lim
2 (sin x )(cos x )
e x 4 1
x 4
lim
sin(2 x )
x4
e x 4 1
2 2
105. The limit leads to the indeterminate form 00 : lim
t 0
lim lim 1
et
t
1
t
t 0
et 1
t
106. The limit leads to the indeterminate form
: lim e1/ y ln y lim
y 0
t 0
ln y
y 0 e
y 1
et
1
lim
y 0 e
y 1
y 1
y
2
y
lim 1 0
y
0
ey
ln f ( x) ln x ln lim ln f ( x) lim ln x ln ; this limit is currently of the
e x 1
e x 1
ln x
e x 1
e x 1
x
x
x
x/2
x/2
e
1
form 0 . Before we put in one of the indeterminate forms, we rewrite x e x / 2 e x / 2 coth 2x ; the limit is
e 1 e e
ln coth 2x
ln coth 2x
x
0 : lim
lim ln x ln coth 2 lim
;
the
limit
leads
to
the
indeterminate
form
1
1
0 x
x
x
ln x
ln x
107. Let f ( x)
e x 1
e x 1
csch 2 2x 1
coth x 2
2 lim x(ln x)2 lim x(ln x)2 lim 2 x (ln x) 1x (ln x)2 lim 2 ln x (ln x)2
lim
cosh x
1
cosh x
x
1 x 2 sinh 2x cosh 2x x sinh x x
x
(ln x )2 x
2
ln x
2 1 2(ln x ) 1
x
e x 1
2 ln x lim
2
lim x sinh x x lim 2xsinh
lim
0
lim
2
x
x
x
x x cosh x sinh x x x cosh x x sinh x
x e 1
x
lim eln f ( x ) e0 1
x
108. Let f ( x) 1 3x
ln 1 3 x
x
ln f ( x) lim
; the limit leads to the
ln f ( x) x ln 1 3x xlim
x
0
x 0
1
3 x 2
1
x 0
indeterminate form
: lim 1 3 x2 lim x3x3 0 lim 1 3x
x 0
109. (a)
log x
x
x 0
1
eln f ( x ) e0 1
xlim
0
x
lnln 2x lim ln 3 ln 3 same rate
ln x
x ln 3 x ln 2 ln 2
lim log2 x lim
x
3
Copyright 2016 Pearson Education, Ltd.
510
Chapter 7 Transcendental Functions
x 2 lim 2 x lim 1 1 same rate
2
x x 1 x 2 x x
x
x x 1x
lim
(b) lim
(c)
lim xe x lim e x faster
lim
x
100
x
x 100 x x 100
x
faster
1
x tan x
x xe
(d) lim
(e)
(f )
110. (a)
(b)
(c)
(d)
(e)
(f )
111. (a)
(b)
(c)
(d)
(e)
(f )
112. (a)
x
lim
1
sin 1 x 1
x
x 1
lim csc1 x lim
x
lim sinhx x lim
x e
x
x 2
2
1 x 1
x 2
x
e e lim 1e
x
x
2e
x
x
2 x
2
x
x
lim 3 x lim 23 0 slower
1
1 same rate
x 1 1
2
x
lim
12 same rate
x 2
x
2 ln x lim ln 2 1 1 same rate
lim ln 2 2x lim ln2(ln
x)
2
x ln x
x
x 2 ln x 2
3
2
2
10
x
2
x
30
x
4
x
60
x
4
lim
lim
lim
lim 60x 0 slower
x
ex
ex
x
x
x e
x e
x 2
1 1
1 1
2
tan x
tan x
lim
lim
lim 1 x2 lim 1 1 1 same rate
1
1
x
x
x
x x
x 1 2
x
lim
x
lim sin 1 x1
sin 1 1x
x 1
x2
lim
sech x
x
x e
x
x
2
2
x 2
2
1 x1
x
lim
lim
3
x 2 x
x 2 1 1
x x
lim e ex lim
e
x
faster
x2
x e
x
2
e e
x
x
2
2 x
x 1 e
lim
2 same rate
11
2 4
x x
1 12 2 for x sufficiently large true
1
x
2
x
11
2 4
x x
x 2 1 M for any positive integer M whenever x
1
4
x
lim x xln x lim 1 1 1 the same growth rate false
x
x 1 x
M false
1/ x
ln x
ln(ln x )
lim ln x lim 1 lim ln1x 0 grows slower true
x
x
x
x
tan 1 x for all x true
1
2
cosh x 1 1 e 2 x 1 (1 1) 1 if x 0 true
2
2
ex
1
4
x
11
2 4
x x
21 1 if x 0 true
x 1
Copyright 2016 Pearson Education, Ltd.
Chapter 7 Practice Exercises
(b)
1
4
x
lim
x 1 1
x2 x4
0 true
1
2
x x 1
lim
1
(c)
lim ln x lim x 0 true
x x 1 x 1
(d)
ln 2 x ln 2 1 1 1 2 if x 2 true
ln x
ln x
(e)
sec1 x
1
(f )
2 if x 1 true
cos 1 1x
1
1
sinh x 1 1 e
2
ex
2 x
2
12 if x 0 true
df
df 1
df 1
113. dx e x 1 dx
x f (ln 2)
dx
x ln 2
df 1
1
dx
1 1
x f (ln 2) e x 1 x ln 2 21 3
114. y f ( x) y 1 1x 1x y 1 x y11 f 1 ( x) x11 ; f 1 f ( x)
f f 1 ( x) 1
1
df
1
x 1
1 ( x 1) x;
1
f ( x) 12 dx
x
f ( x)
df 1
dx
f ( x)
1
1 x and
1 1x 1 1x
1
1
x2 ;
2
( x 1) 2 f ( x )
1 1 1
x
f 1(x )
115. y x ln 2 x x y x 22x ln(2 x) 1 ln 2 x;
solving y 0 x 12 ; y 0 for x 12 and y 0 for
x 12 relative minimum of 12 at x 12 ;
1 1 and f e 0 absolute minimum is
f 2e
e
2
12 at x 12 and the absolute maximum is 0 at x 2e
116. y 10 x(2 ln x) y 10(2 ln x) 10 x 1x
20 10 ln x 10 10(1 ln x); solving y 0
x e; y 0 for x e and y 0 for x e
relative maximum at x e of 10e; y 0 on (0, e2 ] and
y e 2 10e2 (2 2 ln e) 0 absolute minimum is
0 at x e2 and the absolute maximum is 10e at x e
e
1
1
117. A 2 lnx x dx 2u du u 2 1, where u ln x and du 1x dx; x 1 u 0, x e u 1
0
1
0
118. (a)
20 1
20
20 ln 2, and A 2 1 dx ln | x | 2 ln 2 ln1 ln 2
dx ln | x | 10 ln 20 ln10 ln 10
2
1
1 x
10 x
A1
Copyright 2016 Pearson Education, Ltd.
511
512
Chapter 7 Transcendental Functions
(b)
kb 1
kb
kb ln b ln b ln a, and A b 1 dx ln | x | b ln b ln a
dx ln | x | ka ln kb ln ka ln ka
2
a
x
a
ka
a x
A1
dy
119. y ln x dx 1x ;
x 1x dydt e 1e m/s
dy
dy
dy
dx dx
dt 1x
dt
dt
2
1 3 y
dx ( dy / dt ) dx 4
;
dt
( dy / dx )
dt
e x /3
120. y 3e x /3 dx e x /3 ;
dy
x 3 y 3e 1 dx
dt
x 3
14 3 3e
1e
14 e e 1 0.54 m/s
2
2
2
2
121. A xy xe x dA
e x ( x)(2 x)e x e x 1 2 x 2 . Solving dA
0 1 2 x2 0 x 1 ;
dx
dx
dA 0 for x 1 and dA 0 for 0 x 1 absolute maximum of 1 e 1/2
dx
dx
2
2
2
1/2
1
ye
units high.
e
2
1 at x 1 units long by
2e
2
122. A xy x ln2x lnxx dA
12 ln2x 1ln2 x . Solving dA
0 1 ln x 0 x e; dA
0 for x e and
dx
dx
dx
x
x
x
x
dA 0 for x e absolute maximum of ln e 1 at x e units long and y 1 units high.
dx
e
e
e2
123. (a)
y ln x y 1 ln3/x2 2ln x
x
x x
2x
2x x
y 34 x 5/2 (2 ln x) 12 x 5/2 x 5/2 34 ln x 2 ;
2
2
solving y 0 ln x 2 x e ; y 0 for x e and y 0
for x e 2 a maximum of 2e ; y 0 ln x 83 x e8/3 ;
the curve is concave down on 0, e8/3 and concave up on e8/3 , ; so there is an inflection point at
e
8/3
, 84/3
3e
x2
.
2
2
y 2 xe x y 2e x 4 x 2 e x
(b) y e
2
2
4 x 2 2 e x ; solving y 0 x 0; y 0 for x 0 and
y 0 for x 0 a maximum at x 0 of e0 1; there are
points of inflection at x 1 ; the curve is concave down
2
for 1 x 1 and concave up otherwise.
2
(c) y (1 x)e
x
2
y e x (1 x)e x xe x
y e x xe x ( x 1)e x ; solving
y 0 xe x 0 x 0; y 0 for x 0 and y 0 for
x 0 a maximum at x 0 of (1 0)e0 1; there is a point of
inflection at x 1 and the curve is concave up for x 1 and
concave down for x 1.
Copyright 2016 Pearson Education, Ltd.
Chapter 7 Practice Exercises
513
124. y x ln x y ln x x 1x ln x 1; solving y 0 ln x 1 0
ln x 1 x e ; y 0 for x e 1 and y 0 for x e1
1
a minimum of e1 ln e1 1e at x e1. This minimum is an
absolute minimum since y 1x is positive for all x 0.
dy
y cos 2
125. dx
126. y
y
dy
y cos
2
y
dx 2 tan y x C y tan 1 x 2C
2
3 y ( x 1) 2
( y 1)
y dy 3( x 1) 2 dx y ln y ( x 1)3 C
y 1
127. yy sec y 2 sec2 x
ydy
sec2 x dx
sec y
2
128. y cos 2 ( x) dy sin x dx 0 y dy
tan x C sin y 2 2 tan x C
sin y 2
2
1
2
sin x dx y 1 C y
2 C
1
2
cos( x )
cos( x )
cos 2 ( x )
129. dx e x y 2 e y dy e ( x 2) dx e y e( x 2) C. We have y (0) 2,so e 2 e 2 C C 2e2 and
dy
e y e( x 2) 2e 2 y ln e ( x 2) 2e2
dy
130. dx
y ln y
1 x
2
tan
dy
y ln y dx 2 ln(ln y ) tan 1 ( x) C y ee
1 ( x ) C
1 x
. We have y (0) e 2 e2 ee
1
tan
e tan (0) C 2 tan 1 (0) C ln 2 0 C ln 2 C ln 2 y ee
tan 1 (0) C
1 ( x ) ln 2
y 1 ln x C. We have y(1) 1 2 ln 1 1 ln1 C
2 ln 2 C ln 22 ln 4. So 2 ln y 1 ln x ln 4 ln(4 x) ln y 1 12 ln(4 x) ln(4 x)1/2
2
ln y 1
e
eln(4 x ) y 1 2 x y 2 x 1
131. x dy y y dx 0
dy
y y
dx
2 ln
x
1/ 2
x
dx e
132. y 2 dy
2x
e 1
2x
dy
e
2
e x1 dx
y
3
y3
(1)3
3 e x e x C. We have y (0) 1 3 e0 e0 C C 13 . So
y
e x e x 13 y 3 3 e x e x
3
1 y 3 e x e x 1
1/3
A
133. Since the half life is 5700 years and A(t ) A0 e kt we have 20 A0 e5700k 12 e5700 k ln(0.5) 5700k
ln(0.5)
ln(0.5)
t
ln(0.5)
k 5700 . With 10% of the original carbon-14 remaining we have 0.1A0 A0 e 5700 0.1 e 5700
ln(0.5)
ln(0.1) 5700 t t
(5700) ln(0.1)
18,935 years (rounded to the nearest year).
ln(0.5)
Copyright 2016 Pearson Education, Ltd.
t
514
Chapter 7 Transcendental Functions
134. T Ts To Ts e kt 82 5 (104 5)e k /4 , time in hours, k 4 ln 79 4 ln 97
21 5 (104 5)e
4 ln(9/7)t
t ln 69 1.78 h 107 min, the total time the time it took to cool from
4 ln 7
82 C to 21 C was 107 15 92 min
x cot 1 5 x , 0 x 50 d
135. cot 1 60
dx
3 30
601 301 30 2
1
; solving
2
2
2
2
2
2
x
50 x
60
x
30
(50
x
)
1 60
1 30
d 0 x 2 200 x 3200 0 x 100 20 17, but 100 20 17 is not in the domain; d 0 for
dx
dx
x 20 5 17 and ddx 0 for 20 5 17 x 50 x 20 5 17 17.54 m maximizes
136. v x 2 ln 1x x 2 (ln1 ln x) x 2 ln x dv
2 x ln x x 2
dx
1x x(2 ln x 1); solving
dv 0 2 ln x 1 0 ln x 1 x e 1/2 ; dv 0 for x e 1/2 and dv 0 for x e 1/2 a relative
dx
dx
dx
2
1/2
1/2 r
maximum at x e
; h x and r 1 h e e 1.65 cm
CHAPTER 7
1.
2.
lim
b 1
b
ADDITIONAL AND ADVANCED EXERCISES
2
x
lim 1x tan 1t dt lim
0
x
b
dx lim sin 1 x lim sin 1 b sin 1 0 lim sin 1 b 0 lim sin 1 b 2
0 b1
1 x
b 1
b 1
b 1
1
0
x
1
0 tan t dt
x
x
,
form
1
lim tan1 x 2
x
3.
y cos x
4.
1/ x
1 x 1/ 2 sec 2
1
2
1 x 1/ 2
2
x 0
2
lim
y x ex
2/ x
lim x e
x
ln cos x
ln y 1x ln cos x and lim
x
12 lim cos x
x 0
ln y
x 2/ x
x 0
2 ln x e x
x
x
1/ x
lim
x 0
2
sin x
21 lim tan x
x cos x
x
x 0
e1/2 1
e
lim ln y lim 21e lim 2e lim 2e 2
x
x x e
x
x
x
x
x
x 1 e
x
x e
lim e y e 2
x
1 1 1n 1 ! 1n 1 which can be interpreted as a
1 1 ! 21n xlim
x n 1 n 2
n 1
1 2
1 n
5. lim
Riemann sum with partitioning
1
n
1
n
1
n
1 1
1
dx ln 1 x ln 2
0
0 1 x
x 1n lim n11 n 1 2 ! 21n
x
Copyright 2016 Pearson Education, Ltd.
Chapter 7 Additional and Advanced Exercises
6. lim 1n [e1/ n e 2/ n ! e] lim 1n e(1/ n ) 1n e2(1/ n) ! 1n en(1/ n) which can be interpreted as a
x
x
1
1
x 1n lim 1n [e1/ n e2/ n ! e] e x dx e x e 1
0
0
x
Riemann sum with partitioning
7.
t
t
t
t
A(t ) e x dx e x 1 et ,V (t ) e2 x dx 2 e2 x 2 1 e2t
0
0
0
0
lim A(t ) lim 1 et 1
(a)
t
t
t
1 e 2 t
V (t )
lim A(t ) lim 2
(b)
1e t
t
t 0
1 e 2t
V (t )
lim A(t ) lim 2
(c)
t 0
1e t
2
lim 1e 1e lim 1 et
2
t 0
t 0
1e
t
t
2
t
ln 2 0;
lim log a 2 lim ln
a
8. (a)
a 0
(b)
a 0
ln 2 ;
lim log a 2 lim ln
a
a 1
a 1
ln 2 ;
lim log a 2 lim ln
a
a 1
a 1
ln 2 0
lim log a 2 lim ln
a
a
9.
a
2 e
e 2 log 2 x
2 e ln x dx (ln x ) 1 ;
dx
ln 2
ln 2 1 x
ln 2
x
1
1
A1
2 e
e 2 log 4 x
2 e ln x dx (ln x ) 1
dx
2 ln 2
4
ln 4 1 x
2 ln 2
1
1
A2
A1 : A2 2 :1
10.
1
2
1
y tan
x
2
1 1
1 x
x2
1 2 1 2 0 tan 1 x tan 1 1x is a constant
1 x
1 x
1
x tan 1 1x
y
and the constant is for x 0; it is for x 0
2
2
1
1
lim tan x tan 1x 0 2 2 and lim tan 1 x tan 1 1x 0 2 2
x 0
x 0
since tan 1 x tan 1 1x is odd. Next the
x
(xx )
x x
11. ln x ( x ) x x ln x and ln( x x ) x x ln x x x 2 ln x; then, x x ln x x 2 ln x x x x 2 ln x 0 x x x 2 or
x
2
ln x 0. ln x 0 x 1; x x x ln x 2 ln x x 2. Therefore, x
Copyright 2016 Pearson Education, Ltd.
( x ) when x 2 or x 1.
515
516
Chapter 7 Transcendental Functions
12. In the interval x 2 the function
sin x 0 (sin x)sin x is not defined for all values in
that interval or its translation by 2 .
13. f ( x) e g ( x ) f ( x) e g ( x ) g ( x), where g ( x)
14. (a)
(c)
172
x f (2) e0
2
116
1 x 4
1
x
df
2 lnxe e x 2 x
dx
e
t dt 0
(b) f (0) 2 ln
t
1
df
2 x f ( x) x 2 C ; f (0) 0 C 0 f ( x) x 2 the graph of f ( x) is a parabola
dx
15. (a) g ( x) h( x) 0 g ( x ) h( x); also g ( x) h( x) 0 g ( x) h( x) 0 g ( x) h( x ) 0
g ( x ) h( x); therefore h( x) h( x) h( x) 0 g ( x) 0
(b)
f ( x ) fO ( x ) f E ( x ) fO ( x )
f ( x ) f ( x ) f E ( x ) fO ( x ) f E ( x ) fO ( x )
E
f E ( x);
2
2
2
f ( x ) fO ( x ) f E ( x ) fO ( x )
f ( x ) f ( x ) f E ( x ) fO ( x ) f E ( x ) fO ( x )
E
fO ( x )
2
2
2
(c) Part b such a decomposition is unique.
16. (a)
g (0) g (0)
g (0 0) 1 g (0) g (0) 1 g 2 (0) g (0) 2 g (0) g (0) g 3 (0) 2 g (0) g 3 (0) g (0) 0
g (0) g 2 (0) 1 0 g (0) 0
g ( x ) g ( h) g ( x)
1 g ( x ) g ( h )
g ( x h) g ( x )
(b) g ( x) lim
lim
h
h
h 0
h 0
g ( x ) g ( h) g ( x ) g 2 ( x ) g ( h)
h1 g ( x ) g ( h )
h 0
lim
2
g ( h ) 1 g 2 ( x )
lim h 1 g ( x ) g ( h ) 1 1 g 2 ( x) 1 g 2 ( x) 1 g ( x)
h 0
(c)
dy
dy
1 y2
dx tan 1 y x C tan 1
dx
1 y 2
g ( x) x C; g (0) 0 tan 1 0 0 C
C 0 tan 1 g ( x) x g ( x) tan x
1
1
1 2
1 2x
G M
dx 2 tan 1 x 2 and M y
dx ln 1 x 2 ln 2 x My ln 2 ln 4 ;
2
2
0
0 1 x
0 1 x
0
2
17. M
symmetry
Copyright 2016 Pearson Education, Ltd.
G
y 0 by
517
dx dx ln | x | ln 4 ln ln16 ln 2 ln 2
(b) M x
dx x dx x ;
M
dx dx ln | x | ln16 ln 2;
18. (a) V
y
x
M
4
2
1
1/4 2 x
4
1
1/4
2 x
4 1 1
1/4 2 2 x
4 1
4 1
4 1/4 x
4
1 4 1/2
2 1/4
1
1 4 1
8 1/4 x
2 x
1
4
4
8
3
1
24
64 1
24
1/4
1
8
1
2
4
4
4
4
63
24
32 1221 74 and
2 1 3 ; therefore, x M y 63
M
24
2 2
1/4
23 ln32
4
1 3/2
3
1/4
1
8
4
1/2
4 1 1/2
x
dx x
1/4 2
dx
1/4 2 x
Mx
y M 12 ln 2
4
1/4
2
b csc dL k b csc b csc cot ; solving dL 0
19. (a) L k a b cot
d
d
R4
r4
R4
r4
r 4 b csc2 bR 4 csc cot 0 (b csc ) r 4 csc R 4 cot 0; but b csc 0 since
4
4
2 r 4 csc R 4 cot 0 cos r 4 cos 1 r 4 , the critical value of
(b)
R
4
cos 1 56 cos 1 (0.48225) 61
R
20. In order to maximize the amount of sunlight, we need to maximize the angle formed by extending the two red
line segments to their vertex. The angle between the two lines is given by 1 ( 2 ) . From trig we
105 tan 1 105
have tan 1 135
and tan 2 60
2 tan 1 60
1
135 x
x
x
x
105 tan 1 60
(1 ( 2 )) tan 1 135
x
x
ddx
1
105
1 135
x
2
105
1
602
x
(135 x )2 1 60 2
x
105
60
(135 x ) 2 11,025 x 2 3600
d 0
105
60 0 60 (135 x)2 11,025
dx
(135 x ) 2 11,025 x 2 3600
105( x2 3600)
x 2 360 x 30,600 0 x 180 30 70. Since x 0, consider only x 180 30 70. Using the first
derivative test, ddx
x 30
9 0 and d
1050
dx
x 120
9 0 local max when x 180 30 70 71 m.
1500
Copyright 2016 Pearson Education, Ltd.
CHAPTER 8
8.1
TECHNIQUES OF INTEGRATION
USING BASIC INTEGRATION FORMULAS
1
16 x
dx
1.
0 8 x 2 2
u 8 x 2 2 du 16 x dx
u 2 when x 0, u 10 when x 1
1
10
16 x dx 1 du ln u 10
2
2 u
0 8 x 2 2
ln10 ln 2 ln 5
x2
2. 2
dx
x 1
1
.
Use long division to write the integrand as 1 2
x 1
x2
1
dx 1 dx 2
dx
2
x 1
x 1
x tan 1 x C
3.
sec x tan x dx
2
Expand the integrand: sec x tan x sec2 x 2sec x tan x tan 2 x
2
sec2 x 2sec x tan x (sec2 x 1)
2sec2 x 2sec x tan x 1
sec x tan x dx 2 sec x dx 2 sec x tan x dx 1 dx
2
2
2 tan x 2sec x x C
We have used Formulas 8 and 10 from Table 8.1.
/3
1
dx
4.
2
/4 cos x tan x
u tan x du sec2 x dx
1
dx
cos2 x
u 1 when x 4, u 3 when x / 3
/3
3
1
1 du ln u 3
dx
1
1 u
/4 cos2 x tan x
1
ln 3 ln1 ln 3
2
Copyright 2016 Pearson Education, Ltd.
519
520
Chapter 8 Techniques of Integration
1 x
dx
5.
1 x2
Write as the sum of two integrals:
1 x
1
x
dx
dx
dx
2
2
1 x
1 x
1 x2
For the first integral use Formula 18 in Table 8.1 with a 1.
For the second:
u 1 x2
du 2 x dx
x
1
1
dx
1/2 du
2u
1 x2
u 1 x2
1 x
dx sin 1 x 1 x 2 C
So
2
1 x
1
6.
dx
x x
u x 1 du
1
2 x
1
1
dx 2
du
u
x x
dx
2ln u C 2ln
x 1 C
du csc2 z dz
1
e cot z
7.
dz
sin 2 z
u cot z
sin 2 z
dz
e cot z
dz e u du
sin 2 z
e u C e cot z C
3
2ln z
8.
dz
16 z
u ln z 3 3ln z
du
3
dz
z
Using Formula 5 in Table 8.1,
3
2ln z
1
u
16 z dz 48 2 du
3
2u
2ln z
C
C
48ln 2
48ln 2
Copyright 2016 Pearson Education, Ltd.
Section 8.1 Using Basic Integration Formulas
1
9.
dz
z
e e z
Multiply the integrand by
ez
ez
.
ez
1
dz
dx
z
2z
e e z
e 1
u ez
du e z du
ez
1
dx 2
du
2z
u 1
e 1
tan 1 u C tan 1 e z C
2
8
10.
dx
2
1 x 2 x 2
u x 1 du dx
u 0 when x 1, u 1 when x 2
2
1
8
1
dx
8
du
2
2
1 x 2 x 2
0 u 1
1
8 tan 1 u 8 0 2
0
4
0
4
11.
dx
1 1 (2 x 1)2
u 2 x 1 du 2dx
u 1 when x 1, u 1 when x 0
0
1
4
1
dx
2
du
2
1 1 u 2
1 1 (2 x 1)
1
2 tan 1 u 2
1
4 4
3
2
4x 7
12.
dx
1 2 x 3
Use long division to write the integrand as 2 x 3
2
.
2x 3
3
2
3
3
3
2
4x 7
dx 2 x dx 3 dx
dx
1
1
1 2 x 3
1 2 x 3
3
3
2 3
1 2 x dx 13 dx x 1 3x 1 8 12 4
3
For the last integral,
u 2 x 3 du 2 dx
u 1 when x 1, u 9 when x 3
Copyright 2016 Pearson Education, Ltd.
521
522
Chapter 8 Techniques of Integration
3
9
2
1
dx
du
1 2 x 3
1 u
9
1
ln u ln 9 ln1 2 ln 3
3
2
4x 7
dx 4 2ln 3
So
1 2 x 3
1
dt
13.
1 sec t
Multiply the integrand by
1 sec t
.
1 sec t
1
1 sec t 1 sec t
cos t
cos t
cot 2 t 2 1 csc2 t 2
2
1 sec t 1 sec t
tan t
sin t
sin t
1
cos
t
dt 1 dt csc2 t dt 2 dt
1 sec t
sin t
t cot t csc t C
Here we have used Formula 9 in Table 8.1 for the second integral, and the substitution u sin t , du cos t dt
1
1
1
for the third integral, which gives it the form
.
2 du
u
sin t
u
14.
csc t sin 3t dt
Write sin 3t as sin(2t t ) and expand.
csc t sin 3t
cos 2t sin t (2sin t cos t )cos t
sin t
cos 2t 2cos2 t 2cos 2t 1
csc t sin 3t dt 2cos 2t dt 1 dt
sin 2t t C
/4
15.
0
1 sin
cos2
d
Split into two integrals.
/4
0
/4
1 sin
d
0
cos2
/4
0
/4
1
d
0
cos2
/4
sec2 d
0
/4
tan sec 0
sin
cos2
sin
cos 2
d
d
(1 2) (0 1) 2
The second integral is evaluated with the substitution u cos
du sin d , which gives
sin d 1 du 1 1 .
2
u cos
cos2
u
Copyright 2016 Pearson Education, Ltd.
Section 8.1 Using Basic Integration Formulas
1
d
16.
2 2
Write the integrand as
1
1 ( 1)2
. With u 1, du d ,
1
1
d
d
2 2
1 ( 1)2
1
du sin 1 u C sin 1 ( 1) C
1 u2
We have used Formula 18 in Table 8.1 with a 1.
ln y
dy
17.
y 4 ln 2 y
Write the integrand as
u 1 4 ln 2 y
du
ln y
1
.
y 1 4 ln 2 y
8 ln y
dy
y
ln y
1
ln y
dy
dy
y 4ln 2 y
y 1 4ln 2 y
1 1
1
1
2
du ln u C ln(1 4 ln y ) C
8u
8
8
Note that the argument of the logarithm is positive, so we don’t need absolute value bars.
2 y
18.
dy
2 y
u
y
du
1
dy
2 y
Using Formula 5 in Table 8.1,
2 y
dy 2u du
2 y
1 u
1
2 C
2 y C
ln 2
ln 2
1
d
19.
sec tan
Multiply the integrand by
cos
.
cos
1
cos
cos
d
d
sec tan cos
1 sin
u 1 sin
du cos d
Copyright 2016 Pearson Education, Ltd.
523
524
Chapter 8 Techniques of Integration
cos d 1 du ln u C
1 sin
u
ln (1 sin ) C
We can discard the absolute value because 1 sin is never negative.
1
dt
20.
t 3 t2
Use Formula 5 in Table 7.10, with a 3.
1
1
t
dt
csch 1
C
2
3
3
t 3 t
4t 3 t 2 16t
21.
dt
t2 4
4
.
Use long division to write the integrand as 4t 1 2
t 4
4t 3 t 2 16t
1
dt 4t dt 1 dt 4 2
dt
2
t 4
t 4
t
2t 2 t 2 tan 1 C
2
To evaluate the third integral we used Formula 19 in Table 8.1 with a 2.
x 2 x 1
22.
dx
2x x 1
Split into two integrals.
1
x 2 x 1
1
dx
dx
dx
x
2 x 1
2x x 1
x 1 ln x C
For the first integral we used u x 1, du
23.
/2
0
1 cos d
Multiply the integrand by
/2
0
1
dx, du u C
2 x 1
/2
1 cos d
0
1 cos
.
1 cos
/2
1 cos2
sin
d
d .
0
1 cos
1 cos
(Note that when 0 / 2, sin 0 so sin 2 sin . )
u 1 cos du sin d
u 2 when 0, u 1 when / 2
/2
0
1
2
sin
1 du 1 du 2 u 2 2 2 2
d
1
2 u
1 u
1 cos
Copyright 2016 Pearson Education, Ltd.
Section 8.1 Using Basic Integration Formulas
24.
2
(sec t cot t ) dt
Expand the integrand:
sec t cot t 2 sec2 t 2sec t cot t cot 2 t
sec2 t 2sec t cot t csc 2 t 1
2
2
2
(sec t cot t ) dt sec t dt 2 csc t dt csc t dt 1 dt
tan t 2ln csc t cot t cot t t C
We have used Formulas 8, 9 and 15 from Table 8.1.
1
dy
25.
e2 y 1
Multiply the integrand by
ey
ey
.
1
ey
dy
dy ; u e y
2y
y
2
y
e 1
e e 1
du e y dy
ey
1
dy
du
y 2y
u u2 1
e e 1
sec 1 u C sec 1 e y C
We have used Formula 20 in Table 8.1.
6
dy
26.
y 1 y
du
1
u
y
6
1
dy 12
du
1 u2
y 1 y
2 y
dy
12 tan 1 y C
2
dx
27.
x 1 4 ln 2 x
u 2 ln x
du
2
dx
x
1
2
dx
du
2
1 u2
x 1 4 ln x
sin 1 u C sin 1 2 ln x C
Copyright 2016 Pearson Education, Ltd.
525
526
Chapter 8 Techniques of Integration
1
28.
dx
( x 2) x 2 4 x 3
u x2
du dx
1
1
dx
du
2
u u2 1
( x 2) x 4 x 3
sec1 u C sec 1 x 2 C
We have used Formula 20 in Table 8.1 with a 1.
29.
(csc x sec x )(sin x cos x ) dx
Expand the integrand and separate into two integrals.
(csc x sec x )(sin x cos x ) 1 cot x tan x 1 cot x tan x
(csc x sec x )(sin x cos x ) dx cot x dx tan x dx
ln sin x ln sec x C ln sin x ln cos x C
We have used Formulas 12 and 13 from Table 8.1.
x
30. 3sinh ln 5 dx
2
u
x
ln 5
2
1
du dx
2
x
3sinh ln 5 dx 6 sinh u du
2
x
6cosh u C 6cosh ln 5 C
2
3
2 x3
31.
dx
2 x2 1
2x
Use long division to write the integrand as 2 x 2 .
x 1
3
3
3
3
2 x3
2x
2x
dx
2
x
dx
2
x
dx
dx
2
2
2
2
2 x 1
x 1
2
2 x 1
For the second integral we use u x 2 , du 2 x dx.
3
3
2x
2 3
2
2 2 x dx 2 x 2 1 dx x 2 ln x 1 2
3
(9 2) (ln 8 ln1)
7 ln 8 9.079
Copyright 2016 Pearson Education, Ltd.
Section 8.1 Using Basic Integration Formulas
32.
1
2
1 1 x sin x dx is the integral of an odd function over an interval symmetric to 0, so its value is 0.
0
1 y
33.
dy
1 1 y
1 y
Multiply the integrand by
1 y
and split the indefinite integral into a sum.
1 y
1 y
y
1
dy
dy
dy
1 y dy
1 y2
1 y2
1 y2
sin 1 y 1 y 2 C
The first integral is Formula 18 in Section 8.1, and for the second we use the substitution
u 1 y 2 , du 2 y dy . So
0
0
1 y
dy sin 1 y 1 y 2
1
1 1 y
(0 1) 0 1
2
2
34.
e
z ez
dz
z
Write the integrand as e z ee and use the substitution u e z , du e z dz.
e
z ez
z
dz e z ee dz eu du
z
eu C e e C
7
35.
dx
( x 1) x 2 2 x 48
u x 1,
du dx;
x 2 2 x 48 u 2 72
We use Formula 20 in Table 8.1.
7
7
dx
du
2
2
u u 72
( x 1) x 2 x 48
1
u
x 1
7sec 1 C sec 1
C
7
7
7
Copyright 2016 Pearson Education, Ltd.
527
528
Chapter 8 Techniques of Integration
1
36.
dx
(2 x 1) 4 x 4 x 2
u 2 x 1,
du 2dx;
4 x 4 x 2 u 2 12
We use Formula 20 in Table 8.1.
1
1
1
dx
du
2 u u 2 12
(2 x 1) 4 x 4 x 2
1
1
sec 1 u C sec 1 2 x 1 C
2
2
3
2
2 7 7
37.
d
2 5
Use long division to write the integrand as 2 1
5
.
2 5
3
2
5
2 7 7
d 2 d d 1 d
d
2 5
2 5
1
1
5
3 2 ln 2 5 C
3
2
2
In the last integral we have used the substitution u 2 5, du 2 d .
1
d
38.
cos 1
Multiply the integrand by
cos 1
.
cos 1
cos 1 cos 1
1 cos
csc2 csc cot
2
cos 1 cos 1 cos 1
sin 2
1
csc csc cot d csc d csc cot d
2
2
cot csc C
We have used Formulas 9 and 11 from Table 8.1.
1
39.
dx
1 ex
Use one step of long division to write the integrand as 1
ex
1 ex
.
ex
1
dx
1
dx
dx x ln 1 e x C
1 ex
1 ex
For the second integral we have used the substitution u 1 e x , du e x dx. Note that 1 e x is always positive.
Copyright 2016 Pearson Education, Ltd.
Section 8.1 Using Basic Integration Formulas
x
40.
dx
1 x3
u x 3/2 ,
du
3 1/2
x dx
2
2 1
x
dx
du
3
3 1 u2
1 x
2
2
tan 1 u C tan 1 x 3/2 C
3
3
41. The area is
/4
/4
/4
2 cos x sec x dx 2sin x ln sec x tan x /4
2 ln 2 1 2 ln 2 1
2 1
2 2 ln
2 2 ln 3 2 2 1.066
2 1
42. The volume using the washer method is
/4
/4
4 cos x sec x dx.
2
2
Split into two integrals; for the first write 4cos2 x as 2 1 cos 2x and for the second use Formula 8 in
Table 8.1.
/4
4 cos x sec x dx 4 cos x dx sec x dx
2
/4
/4
2
/4
/4
/4
/4
2
2
/4
2 1 cos 2 x dx
/4
sec2 x dx
/4
/4
/4
2 x sin 2 x /4 tan x /4
1 1 1 ( 1) 2
2 2
43. For y ln (cos x ) , dy / dx tan x. The arc length is given by
/3
1 tan x dx
2
0
/3
0
/3
0
sec x dx since sec x is positive on the interval of integration.
sec x dx ln sec x tan x
/3
0
ln 2 3 ln 1 0 ln 2 3
44. For y ln (sec x ) , dy / dx tan x. The arc length is given by
/4
0
/4
0
1 tan x dx
2
/4
0
sec x dx since sec x is positive on the interval of integration.
sec x dx ln sec x tan x
ln
/4
0
2 1 ln 1 0 ln 2 1
Copyright 2016 Pearson Education, Ltd.
529
530
Chapter 8 Techniques of Integration
45. Since secant is an even function and the domain is symmetric to 0, x 0.
For the y-coordinate:
1 /4
2
/4 sec x dx
2
y
/4
/4 sec x dx
1 tan x /4
2
/4
/4
ln sec x tan x /4
1
ln 2 1 ln 2 1
1
1
0.567
2 1 ln 3 2 2
ln
2 1
46. Since both cosecant and the domain are symmetric around / 2, x / 2.
1 5 /6 2
/6 csc x dx
y 2
5 /6
/6 csc x dx
47.
12
12 cot x
5 /6
/6
ln csc x cot x
3 3
ln 2 3 ln 2 3
5 /6
/6
3
3
0.658
2 3 ln 7 4 3
ln
2 3
1 3x e dx xe
3
x3
x3
C
1
48.
dx
1 sin 2 x
Multiply the integrand by
sec2 x
sec2 x
.
sec2 x
1
sec2 x
dx
dx
dx
1 sin 2 x
sec 2 x tan 2 x
1 2 tan 2 x
u tan x,
du sec2 x dx
sec2 x
1
dx
du
2
1 2u 2
1 2 tan x
v 2u,
dv 2 du
1 du 1 1 dv
1 2u 2
2 1 v2
1
tan 1 v C
2
1
tan 1 2 tan x C
2
Copyright 2016 Pearson Education, Ltd.
Section 8.2 Integration by Parts
49.
x
7
x 4 1 dx
u x 4 1, du 4 x 3dx;
x
x 7dx
u 1
du
4
1
(u 1) u du
4
1
1
u 3/2 du u1/2 du
4
4
1 5/2 1 3/2
u u C
10
6
3/2
1 3/2
1 4
u 3u 5 C
x 1
3x 4 2 C
30
30
7
x 4 1 dx
50.
( x 1)( x 1)
2
2/3
dx
The easiest substitution to use is probably u
x 1
2
, du
dx.
x 1
(1 x )2
The integral can be written as
1
1
dx u 2/3 du
x 1 2/3
2
( x 1)2
x 1
3
3 x 1
u1/3 C
2
2 x 1
8.2
1/3
C
INTEGRATION BY PARTS
1. u x, du dx; dv sin 2x dx, v 2 cos 2x ;
x sin 2x dx 2 x cos 2x 2 cos 2x dx 2 x cos 2x 4sin 2x C
2. u , du d ; dv cos d , v 1 sin ;
cos d sin 1 sin d sin 1 cos C
2
3.
cos t
2
()
t
sin t
()
2t
cos t
()
2
sin t
0
2
2
t cos t dt t sin t 2t cos t 2sin t C
Copyright 2016 Pearson Education, Ltd.
531
532
Chapter 8 Techniques of Integration
4.
sin x
()
2
x
cos x
()
2 x
sin x
()
2
cos x
2
2
x sin x dx x cos x 2 x sin x 2 cos x C
0
2
5. u ln x, du dx
; dv x dx, v x2 ;
x
2
2
2
2
1 x ln x dx x2 ln x 1 1 x2 dxx 2 ln 2 x4 1 2 ln 2 43 ln 4 34
2
2
2
4
6. u ln x, du dx
; dv x3 dx, v x4 ;
x
e 3
e
e
e
1 x ln x dx x4 ln x 1 1 x4 dxx e4 16x 1
4
4
4
4
3e 4 1
16
7. u x, du dx; dv e x dx, v e x ;
x
x
x
x
x
x e dx xe e dx xe e C
8. u x, du dx; dv e3 x dx, v 13 e3 x ;
3x
x e dx 3x e
3x
13 e3 x dx 3x e3 x 91 e3 x C
e x
9.
()
x 2
e x
()
2 x
e x
()
2
e x
2 x
x e
0
dx x 2 e x 2 x e x 2e x C
e2 x
10.
()
x 2 2 x 1
12 e 2x
()
2x 2
14 e2 x
2
18 e2 x
()
x 2 x 1 e dx 12 x 2 x 1 e
2
0
2x
2
2x
14 (2 x 2)e2 x 14 e2 x C
12 x 2 23 x 54 e2 x C
11. u tan 1 y, du
1
dy
1 y 2
tan y dy y tan
1
; dv dy , v y;
y
y dy
1 y
2
y tan 1 y 12 ln 1 y 2 C y tan 1 y ln 1 y 2 C
Copyright 2016 Pearson Education, Ltd.
Section 8.2 Integration by Parts
12. u sin 1 y, du
sin
1
dy
1 y 2
; dv dy, v y;
y dy y sin 1 y
y dy
1 y 2
y sin 1 y 1 y 2 C
13. u x, du dx; dv sec2 x dx, v tan x;
2
x sec x dx x tan x tan x dx x tan x ln |cos x | C
14.
2
2
4 x sec 2 x dx; [ y 2 x, dy 2dx] y sec y dy y tan y tan y dy y tan y ln | sec y | C
2 x tan 2 x ln |sec 2 x | C
ex
15.
()
x3
ex
( )
3x 2
ex
()
6 x
ex
6
()
ex
x e dx x e 3x e 6 xe 6e C x 3x 6 x 6 e C
3 x
0
3 x
2 x
x
3
x
2
x
e p
16.
p4
()
e p
()
4 p3
e p
()
12 p 2
e p
( )
24 p
e p
24
()
e p
4 p
p e
0
dp p 4 e p 4 p3e p 12 p 2 e p 24 pe p 24e p C
p 4 4 p3 12 p 2 24 p 24 e p C
ex
17.
()
x 2 5 x
ex
( )
2 x 5
ex
2
0
()
ex
x 5 x e dx x 5 x e (2 x 5)e 2e C x e 7 xe 7e C
2
x
2
x
x
x
x2 7 x 7 e x C
Copyright 2016 Pearson Education, Ltd.
2 x
x
x
533
534
Chapter 8 Techniques of Integration
er
18.
()
r 2 r 1
er
( )
2r 1
er
2
er
()
0
r r 1 e dr r r 1 e (2r 1)e 2e C
2
r
5 x
5 x
2
r
r
r
r 2 r 1 (2r 1) 2 e r C r 2 r 2 e r C
ex
19.
()
x5
ex
5x4
ex
()
()
20 x3
ex
( )
60 x 2
ex
()
120 x
ex
120
( )
ex
4 x
3 x
2 x
x
x
x e dx x e 5 x e 20 x e 60 x e 120 xe 120e C
0
x5 5 x 4 20 x3 60 x 2 120 x 120 e x C
e 4t
20.
()
t 2
14 e4t
()
1 e 4t
2t
16
()
1 e 4t
2
64
0
2 4t
t e
2
2
2
2 t e 4 t 2 e 4 t C t e 4 t t e 4 t 1 e 4t C
dt t4 e4t 16
64
4
8
32
1 e 4t C
t4 8t 32
21.
I e sin d ; [u sin , du cos d ; dv e d , v e ] I e sin e cos d ;
[u cos , du sin d ; dv e d , v e ] I e sin e cos e sin d
e sin e cos I C 2 I e sin e cos C I 12 e sin e cos C , where C C2
is another arbitrary constant
22.
I e y cos y dy;[u cos y, du sin y dy; dv e y dy, v e y ]
I e y cos y e y ( sin y ) dy e y cos y e y sin y dy;
[u sin y, du cos y dy; dv e y dy, v e y ] I e y cos y e y sin y e y cos y dy
Copyright 2016 Pearson Education, Ltd.
Section 8.2 Integration by Parts
535
e y cos y e y sin y I C 2 I e y sin y cos y C I 12 e y sin y e y cos y C , where
C C2 is another arbitrary constant
23. I e 2 x cos 3 x dx; [u cos 3 x; du 3 sin 3x dx, dv e2 x dx; v 12 e2 x ]
I 12 e 2 x cos 3 x 32 e 2 x sin 3 x dx; [u sin 3 x, du 3cos 3 x, dv e2 x dx; v 12 e2 x ]
I 12 e 2 x cos 3 x 23 12 e 2 x sin 3 x 32 e2 x cos 3x dx 12 e2 x cos 3 x 34 e 2 x sin 3 x 94 I C
2x
4 C
13
I 12 e 2 x cos 3 x 34 e2 x sin 3x C I e13 3 sin 3x 2 cos 3 x C , where C 13
4
24.
e
2 x
sin 2 x dx; [ y 2 x, du 2dx] 12 e y sin y dy I ; [u sin y, du cos y dy; dv e y dy, v e y ]
I 12 e y sin y e y cos y dy [u cos y, du sin y ; dv e y dy, v e y ]
I 12 e y sin y 12 e y cos y e y ( sin y ) dy 12 e y sin y cos y I C
2 x
2 I 12 e y sin y cos y C I 14 e y sin y cos y C e 4 sin 2 x cos 2 x C , where
C C2
25.
3s 9 x 2
3s 9
e x 2 x dx 2 xe x dx; [u x, du dx; dv e x dx, v e x ];
e
ds
;
3
3
ds 2 x dx
3
2
3
xe dx 23 xe e dx 23 xe e C 23 3s 9e
x
x
x
x
x
3s 9
e 3s 9 C
26. u x, du dx; dv 1 x dx, v 23 (1 x)3 ;
1
3
1
1
0 x 1 x dx 23 x (1 x) 0 23 0 (1 x) dx 0 23 52 (1 x)
3
2
5 2 1
4
0 15
2
27. u x, du dx; dv tan 2 x dx, v tan 2 x dx sin 2 x dx 1cos2 x dx
cos x
3
3
3
0
x tan 2 x dx x tan x x
3
3 3 ln 12 18 3 3 ln 2 18
0
2
0
(tan x x ) dx 3
cos x
dx
cos 2 x
dx tan x x;
3
3 3 ln |cos x | x2 0
2
2
(2xx1)x dx ; dv dx, v x; ln x x2 dx x ln x x2 x2( xx11) x dx
(2 x 1) dx
x ln x x 2 x 1 x ln x x 2 2 x11 dx x ln x x 2 2 x ln | x 1| C
28. u ln x x 2 , du
2
u ln x
29. sin (ln x) dx; du 1x dx (sin u ) eu du. From Exercise 21, (sin u ) eu du eu
dx eu du
12 x cos (ln x) x sin (ln x) C
Copyright 2016 Pearson Education, Ltd.
sin u cos u
2
C
536
Chapter 8 Techniques of Integration
u ln z
30. z (ln z ) 2 dz ; du 1z dz eu u 2 eu du e2u u 2 du ;
dz eu du
e 2u
()
u 2
12 e2u
()
2u
14 e2u
()
2
18 e2u
2 2u
u e
0
2u
2
du u2 e2u u2 e2u 14 e 2u C e4 2u 2 2u 1 C
2
z4 2(ln z ) 2 2 ln z 1 C
31.
x sec x dx Let u x , du 2 x dx 12 du x dx x sec x dx 12 sec u du 12 ln |sec u tan u | C
2
2
2
12 ln |sec x 2 tan x 2 | C
32.
cos x
dx Let u
x
x , du
1 dx 2du 1 dx
2 x
x
cos x
dx 2
x
cos u du 2 sin u C 2 sin x C
u ln x
33. x(ln x)2 dx; du 1x dx eu u 2 eu du e2u u 2 du ;
dx eu du
e 2u
()
u 2
12 e2u
( )
2u
14 e2u
()
2
18 e2u
2 2u
u e
0
2
2u
du u2 e2u u2 e2u 14 e 2u C e4 2u 2 2u 1 C
2
2
2
x4 2 ln x 2 ln x 1 C x2 ln x x2 ln x x4 C
2
34.
2
2
x(ln1 x)2 dx Let u ln x, du 1x dx x(ln1 x)2 dx u12 du u1 C ln1 x C
35. u ln x, du 1x dx; dv 12 dx, v 1x ;
x
ln x
x
36.
2
ln x
dx x
ln x
x1 dx x 1x C
2
(ln x )3
dx Let u ln x, du 1x dx
x
(ln x )3
dx
x
3
4
4
u du 14 u C 14 (ln x) C
Copyright 2016 Pearson Education, Ltd.
Section 8.2 Integration by Parts
37.
3 x4
x e
4
4
dx Let u x 4 , du 4 x3 dx 14 du x3 dx x3e x dx 14 eu du 14 eu C 14 e x C
3
3
38. u x3 , du 3x 2 dx; dv x 2 e x dx, v 13 e x ;
5 x3
3 x3 2
3 x3
x e dx x e x dx 13 x e
3
3
3
13 e x 3 x 2 dx 13 x3e x 13 e x C
;
32
32
32
52
3
2
2 2
2
2 2
2
x x 1 dx 13 x x 1 13 x 1 2 x dx 13 x x 1 152 x 1 C
39. u x 2 , du 2 x dx; dv x 2 1 x dx, v 13 x 2 1
40.
32
x sin x dx Let u x , du 3x dx 13 du x dx x sin x dx 13 sin u du 13 cos u C
2
3
3
2
2
2
3
13 cos x3 C
41. u sin 3x, du 3cos 3 x dx; dv cos 2 x dx, v 12 sin 2 x;
sin 3x cos 2 x dx 12 sin 3x sin 2 x 23 cos 3x sin 2 x dx
u cos 3 x, du 3sin 3 x dx; dv sin 2 x dx, v 12 cos 2 x;
sin 3x cos 2 x dx 12 sin 3x sin 2 x 23 12 cos 3x cos 2 x 23 sin 3x cos 2 x dx
12 sin 3x sin 2 x 34 cos 3x cos 2 x 94 sin 3 x cos 2 x dx
54 sin 3x cos 2 x dx 12 sin 3x sin 2 x 43 cos 3 x cos 2 x
sin 3 x cos 2 x dx 52 sin 3x sin 2 x 53 cos 3x cos 2 x C
42. u sin 2 x, du 2 cos 2 x dx; dv cos 4 x dx, v 14 sin 4 x;
sin 2 x cos 4 x dx 14 sin 2 x sin 4 x 12 cos 2 x sin 4 x dx
u cos 2 x, du 2sin 2 x dx; dv sin 4 x dx, v 14 cos 4 x ;
sin 2 x cos 4 x dx 14 sin 2 x sin 4 x 12 14 cos 2 x cos 4 x 12 sin 2 x cos 4 x dx
14 sin 2 x sin 4 x 81 cos 2 x cos 4 x 14 sin 2 x cos 4 x dx
34 sin 2 x cos 4 x dx 14 sin 2 x sin 4 x 18 cos 2 x cos 4 x
sin 2 x cos 4 x d
0
advertisement
Related documents
Download
advertisement
Add this document to collection(s)
You can add this document to your study collection(s)
Sign in Available only to authorized usersAdd this document to saved
You can add this document to your saved list
Sign in Available only to authorized users