Section 3.4
Selected Solutions
#1, #8, #12, #21
1. (a)
Write the volume V of a cube as a function of the side length s.
V(s) = s3
(b)
Find the (instantaneous) rate of change of the volume V with respect to a side s.
V'(s) = 3s2
(c)
Evaluate the rate of change of V at s = 1 and s = 5
V'(1) = 3(1)2 = 3
V'(5) = 3(5)2 = 75
(d)
If s is measured in inches and V is measured in cubic inches, what units would
dV
be appropriate for
?
ds
Consider the graph:
cu. in
35
30
25
20
15
10
5
1
2
3
The rate of change (slope or
inches
rise
cubic inches
) is measured in
.
run
inch
Therefore, the appropriate label for
dV
is cubic inches per inch.
ds
8. Draining a Tank. The number of gallons of water in a tank t minutes after the tank
has started to drain is Q(t) = 200(30 - t) 2. How fast is the water running out at the end of
10 minutes?
Since t = minutes and Q = gallons, the rate of change is measured in gallons
per minute.
Q(t) = 200(900 - 60t + t2) = 180000 - 12000t + 200t2
Q'(t) = -12000 + 400t
Q'(10) = -12000 + 400(10) = -12000 + 4000 = 8000
Therefore the water is draining at a rate of 8000 gal/min
What is the average rate at which the water flows out during the first 10 minutes.
When t = 0, Q(0) = 200(30 - 0)2 = 200(30)2 = 180,000 gallons
When t = 10, Q(10) = 200(30 - 10) 2 = 200(20)2 = 80,000 gallons
This means that in 10 minutes, the water decreased by 100,000 gallons
100,000 gal
= 10,000 gallons per minute.
10 min
Graphical Support:
gallons
1.8e+05
1.6e+05
1.4e+05
1.2e+05
1e+05
(10, 80000)
80000
60000
40000
20000
1
2
3
4
5
6
7
8
9
10 11 12 minutes
12. Thoroughbred Racing. A racehorse is running a 10-furlong race. As the horse
passes each furlong marker (F), a steward records the time elapsed (t seconds) since
the beginning of the race, as shown in the table below:
F 0
t 0
1
20
2
33
3
46
4
59
5
73
6
86
7
100
8
112
9
124
10
135
(a) How long does it take the horse to finish the race?
135 seconds
(b) What is the average speed of the horse over the first 5 furlongs?
5 d 73 = 0.068 furlongs per second
(c) What is the approximate speed of the horse as it passes the 3-furlong marker?
2
4-2
=
= 0.077 furlongs per second
59 - 33 26
(d) During which portion of the race is the horse running the fastest?
Between 9 and 10 seconds because it only took 11 seconds.
Graphical Support:
12 Furlongs
11
10
9
8
7
6
5
4
3
2
1
20
40
60
80
100 120 140 160 Seconds
(e)
F
0
F/Sec 0
1
2
3
4
5
6
7
8
9
10
0.050 0.077 0.077 0.077 0.071 0.077 0.071 0.083 0.083 0.091
The acceleration (chane in velocity) was increasing fastest between 0 and 1 second.
12 Furlongs
11
10
9
8
7
6
5
4
3
2
1
20
40
60
80
100 120 140 160 Seconds
21. Particle Motion. A particle moves along a line so that its position at any time t ≥ 0 is
given by the funtion s(t) = (t - 2) 2(t - 4) where s is measured in meters and t is measured
in seconds.
POSITION Ò
s(t)
VELOCITY Ò
v(t) = (t2 - 4t + 4)(1) + (t - 4)(2t - 4)
= t2 - 4t + 4 + 2t2 - 4t - 8t + 16
= 3t2 - 16t + 20
ACCELERATION Ò
a(t) = 6t - 16
(a)
= (t2 - 4t + 4)(t - 4)
Find the instantaneous velocity at time t.
v(t) = 3t2 - 16t + 20
(b) Find the acceleration of the particle at any time t.
a(t) = 6t - 16
(c) When is the particle at rest?
The particle is at rest when the velocity = 0
3t2 - 16t + 20 = 0
(3t - 10)(t - 2) = 0
3t - 10 = 0
or
t-2=0
t = 10/3
or
t=2
When t = 2, s(2) = ((2)2 - 4(2) + 4)((2) - 4) = 0
When t = 10/3, s(10/3) = ( (10/3)2 - 4(10/3) + 4)((10/3) - 4) = - 1.185
(d) Describe the motion of the particle. At what values of t does the particle change
directions?
At t = 0, the particle is at s = -16. It moves to the right until t = 2 seconds and s = 0. Then
1
it moves back -1.185 meters to the left until 3 seconds. It then begins moving to the
3
right and continues to the right from there on.
Graphical Support:
20 Meters
15
10
5
Meters
2
1
2
3
4
5
Seconds
1
-5
1
-10
-1
-15
-2
-20
2
3
4
5
Seconds