Thomas Calculus 13th [Solutions]

236 Chapter 4 Applications of Derivatives
44.
2
v
ds
32t 2 s 16t 2 t C ; at s 4 and t 1 we have C 1
dt
2
s
2
16t
2t
1
45. v
ds
sin( t) s 1 cos( t) C ; at s 0 and t 0 we have C
dt
1 1 cos( t)
s
46.
ds 2 cos 2t sin 2t
2
v s C ; at s 1 and t we have C 1 s sin 2t
1
dt
dv
dt
t
t
47. a e v e C ; at v = 20 and t = 0 we have C = 19 v e 19
ds
dt
t
t
v e 19 s e 19 t C ; at s = 5 and t = 0 we have C = 4 s e 19t
4
t
t
48. a 9.8 v 9.8 t C 1;
at v 3 and t 0 we have C1 3 v 9.8t
3
2
t 0 we have C2 0 s 4.9t 3t
2
s 4.9t 3 t C 2;
at s 0 and
49. a 4sin(2 t) v 2cos(2 t) C 1;
at v 2 and t 0 we have C 1 0 v 2cos(2 t)
s sin(2 t) C 2;
at
s 3 and t 0 we have C2 3 s sin(2 t) 3
50. a 9 cos 3t
v 3 sin 3t
C
2
1;
at v 0 and t 0 we have C 3 3
1 0 v sin t s cos 3t
C 2;
at
s 1 and t 0 we have C
3
2 0 s cos
t
51. 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
T (14) T (0)
Value Theorem there exists a 0 t 0 14 such that
8.5 C/ sec T ( t
14 0
0),
the rate at which the
temperature was changing at t t 0 as measured by the rising mercury on the thermometer.
52. Because the trucker's average speed was 79.5 mph, by the Mean Value Theorem, the trucker must have been
going that speed at least once during the trip.
53. 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.
54. The runners average speed for the marathon was approximately 11.909 mph. Therefore, by the Mean Value
Theorem, the runner must have been going that speed at least once during the marathon. Since the initial speed
and final speed are both 0 mph and the runners speed is continuous, by the Intermediate Value Theorem, the
runners speed must have been 11 mph at least twice.
d(2) d (0)
55. Let d( t ) represent the distance the automobile traveled in time t. The average speed over 0 t 2 is .
2 0
d (2) d (0)
The Mean Value Theorem says that for some 0 t0 2, d ( t 0) . The value d ( t
2 0
0)
is the speed of the
automobile at time t 0 (which is read on the speedometer).
56. a( t) v ( t) 1.6 v( t) 1.6 t C ; at (0, 0) we have C 0 ( t) 1.6 t . When t 30, then v(30) 48 m/sec.
57. The conclusion of the Mean Value Theorem yields
1 1
b a 2 a b
b a 2
c ab
58. The conclusion of the Mean Value Theorem yields 2 2 2 c c .
b a
2
1 c a b c ab.
b a a b
59. f ( x) [cos xsin( x 2) sin x cos( x 2)] 2sin( x 1) cos( x 1) sin( x x 2) sin 2( x 1)
2
sin(2x 2) sin (2x 2) 0. Therefore, the function has the constant value f (0) sin 1 0.7081
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