Question

"if a snowball melts so that its surface area decreases at a rate of 1 cm 2 min, find the rate at which the diameter decreases when the diameter is 10 cm." (stewart 249) stewart, james. single variable calculus, 8th edition. cengage learning, 20150101. vitalbook file.

Answer 1

We need to find the rate of the diameter, which we can denote as d(d)/dt.

$\frac{dA}{dt} = \frac{dA}{dx} \cdot \frac{dr}{dt}$
$\frac{dA}{dt} = -1$, since it is decreasing.

$-1 = \frac{dA}{dr} \cdot \frac{dr}{dt}$

$A = 4\pi \cdot r^{2}$
$\frac{dA}{dr} = 8\pi \cdot r$

At r = 5:
$\frac{dA}{dr} = 40 \pi$

$\frac{dr}{dt} = -\frac{1}{40 \pi}$

Since the diameter is twice the radius and this is simply the rate at which the radius is decreasing, then the diameter will be decreasing twice as fast:

$\frac{d(d)}{dt} = -\frac{1}{20\pi}$

Thus, the diameter is decreasing at a rate of 1/(20pi) cm/min.