Page 105 - Calculus Demystified

CHAPTER 3 Applications of the Derivative
92
Now we are given that dV/dt = 20. Our question is posed at the moment that
√
V = 64. But, according to (∗), this means that r = 3 48/π. Substituting these
values into equation (∗∗) yields
2
4π dr
3
20 = · 3 48/π · .
3 dt
Solving for dr/dt yields
dr 5
= .
dt 48 2/3 · π 1/3
EXAMPLE 3.12
A 13-foot ladder leansagainst a wall (Fig.3.15).The foot of the ladder begins
to slide away from the wall at the rate of 1 foot per minute. When the foot is
5 feet from the wall, at what rate isthe top of the ladder falling?

Fig. 3.15

SOLUTION
Leth(t)betheheightoftheladderattimet andb(t)bethedistanceofthebase
of the ladder to the wall at time t. Then the Pythagorean theorem tells us that
2
2
2
h(t) + b(t) = 13 .
We may differentiate both sides of this equation with respect to the variable t
(which is time in minutes) to obtain

2 · h(t) · h (t) + 2 · b(t) · b (t) = 0.

Solving for h (t) yields

b(t) · b (t)

h (t) =− .
h(t)