Math 137 Winter 2010
Assignment 6
Due Friday, March 5
All solutions must be clearly stated and fully justified.
Use the format given on UWAce under
Content, in the folder Assignments; it is the file
Math 137 Assignment Templates
.
Text problems:
Section 3.7:
10, 12, 18, 34
Section 3.8:
4, 8a,b,c, 14, 20
Section 3.9:
12, 14, 22, 30
Section 3.10: 12, 18, 24, 34
Section 3.7:
10. A ball is thrown upward with height in feet after t seconds s(t) = 80t – 16t
2
.
a) What is the maximum height reached by the ball?
At maximum height v(t) = s
′
(t) = 80 – 32 t = 0
⇒
t = 80/32 = 5/2.
Max height is s(5/2) = 80(5/2)
– 16(5/2)
2
= 200 – 100 = 100 ft.
b)
What is the velocity of the ball when it is 96 ft above the ground on its way up?
When it is
96 ft above the ground on its way down?
s(t) = 80t – 16t
2
= 96
⇒
16t
2
– 80 t + 96 = 0
⇒
16(t – 3)(t – 2) = 0
⇒
the ball is 96 ft above the
ground at t = 2 sec (on the way up) and also at t = 3 sec (on the way down).
So v(2) = 80 – 32(2)
= 16 ft/sec on the way up, and v(3) = 80 – 32(3) = – 16 ft/sec, or 16 ft/sec on the way down.
12. Sodium Chlorate crystals are cubic.
a) Let V be the volume of a cube with side length x; calculate dV/dx when x = 3 mm and explain
its meaning.
The volume of a cube of side x is V(x) = x
3
; dV/dx = 3x
2
, and at x = 3 this is 27 mm
3
/mm.
This
is the rate at which the volume is increasing as x increases past 3 mm.
b)
Show that the rate of change of the volume of a cube with respect to its edge length is equal
to half the surface area of the cube.
The surface area of a cube is S(x) = 6x
2
= 2(3x
2
) = 2dV/dx, so dV/dx = ½ S(x).
18.
A tank holds 5000 gallons of water, which drains completely in 40 minutes.
The volume V
of water in the remaining in the tank after t minutes is V = 5000(1 – t/40)
2
, 0
≤
t
≤
40.
Find the
rate at which the water is draining from the tank after a) 5 min, b) 10 min, c) 20 min, and d) 40
min.
At what time is the water flowing out the fastest?
The slowest?
Summarize your findings.
V
′
(t) = 5000(1 – t/40)( – 1/40) = –250(1 – t/40)
a)
V
′
(5) =
–250(1 – 5/40) = –218.75 gal/min
b)
V
′
(10) =
–250(1 – 10/40) = –187.5 gal/min
c)
V
′
(20) =
–250(1 – 20/40) = –125 gal/min
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V
′
(40) =
–250(1 – 40/40) = 0 gal/min
The outflow is fastest at the beginning, when t = 0, for V
′
(0) = –250 gal/min.
The outflow is
slowest at the end, when t = 40 min.
As the tank empties, the water flows out more slowly.
34.
A population of fish is introduced into a pond and harvested regularly.
A model for the rate
of change of fish population is given by dP/dt = r
0
(1 – P(t)/P
c
)P(t) –
β
P(t), where r
0
is the birth
rate of the fish, P
c
is the carrying capacity of the pond, and
β
is the percentage of the population
that is harvested.
a)
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