Thomas Calculus 13th [Solutions]

142 Chapter 3 Derivatives
25.
6 4 3 2
4 3
b( t) 10 10 t 10 t b ( t) 10 (2)(10 t)
10 (10 2 t)
(a) b (0)
4
10 bacteria/hr (b) b (5) 0
(c) b (10)
4
10 bacteria/hr
3
bacteria/hr
S ( w ) ; S increases more rapidly at lower weights where the derivative is greater.
1 180 1
26.
120 w 80w
27. (a)
(b)
y
6 1 t
12
The largest value of dy
dt
The smallest value of dy
2
2
6 1 t t dy t
dt 12
6 144
dt
dy
(c) In this situation, 0 the graph of y is
dt
always decreasing. As dy increases in value,
dt
the slope of the graph of y increases from 1
to 0 over the interval 0 t 12.
1
is 0 m/h when t 12 and the fluid level is falling the slowest at that time.
is 1 m/h, when t 0, and the fluid level is falling the fastest at that time.
28.
Q( t) 200(30 t)
2
2
200(900 60 t t ) Q ( t) 200( 60 2 t ) Q (10) 8,000 gallons/min is the rate
Q(10) Q(0)
10,000 gallons/min is the average rate the water
the water is running at the end of 10 min. Then
10
flows during the first 10 min. The negative signs indicate water is leaving the tank.
29. s ( v) 1.1 0.108 v; s (35) 4.88, s (70) 8.66. The units of ds/
dv are ft/mph; ds/
dv gives, roughly, the
number of additional feet required to stop the car if its speed increases by 1 mph.
3 2
30. (a) V
4
r dV 4 r dV
2 3
4 (2) 16 ft /ft
3 dr
dr r 2
(b) When r 2, dV 16 so that when r changes by 1 unit, we expect V to change by approximately 16 .
dr
3
Therefore when r changes by 0.2 units V changes by approximately (16 )(0.2) 3.2 10.05 ft .
3
Note that V (2.2) V (2) 11.09 ft .
200 km/hr 55 m/sec m/sec, and D 10 t V 20
t Thus V
500 20
t
500
31.
5 500
9 9
2
t 25, D
10
(25)
6250
m
9 9
2
9 9 .
9 9
9
t 25sec. When
2 v0
2
32. s v t 16t v v 32 t; v 0 t ;1900 v t 16t so that t
33.
0 0 32
0
v0 (64)(1900) 80 19 ft/sec and, finally,
80 19 ft 60 sec
sec 1 min
60 min 1 mi
1 hr 5280 ft
2 2
v0 v0 v0
1900
32 32 64
238 mph.
(a) v 0 when t 6.25sec
(b) v 0 when 0 t 6.25 body moves right (up); v 0 when 6.25 t 12.5 body moves left (down)
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