MA 3972-MA-Book April 11, 2018 14:46
180 STEP 4. Review the Knowledge You Need to Score High
r
4
h
10
Figure 9.1-2
Thus, you can reduce the equation to one variable:
V=
1
3
π
(
2 h
5
) 2
h=
4
75
πh^3.
Step 4: Differentiate both sides of the equation with respect tot.
dV
dt
=
4
75
π(3)h^2
dh
dt
=
4
25
πh^2
dh
dt
Step 5: Substitute known values.
2 =
4
25
πh^2
dh
dt
;
dh
dt
=
(
25
2
)
1
πh^2
m/min
Evaluating
dh
dt
ath=5;
dh
dt
∣∣
∣∣
h= 5
=
(
25
2
)
1
π(5)^2
m/min
=
1
2 π
m/min.
Step 6: Thus, the water level is rising at
1
2 π
m/min when the water is 5 m high.
Shadow Problem
A light on the ground 100 feet from a building is shining at a 6-foot-tall man walking away
from the light and toward the building at the rate of 4 ft/sec. How fast is his shadow on the
building becoming shorter when he is 40 feet from the building? (See Figure 9.1-3.)
100 ft
Building
Light 6 ft
Figure 9.1-3