Answer to Question #280920 in Calculus for Kelvin lee Boacon

Question

5. A spherical snowball with an outer layer of ice melts, so that the radius of the snowball

decreases at the rate of 1/5 cm/sec. Find the rate at which the volume decreases when the

diameter is 50 cm.

Answer 1

The formula for volume of a sphere is $V=\frac{4}{3} r^{3} \pi.$

Differentiating with respect to t, time.

$\frac{d V}{d t}=4 r^{2}\left(\frac{d r}{d t}\right)$

The rate of change of the snowball is given by $\frac{d V}{d t}.$

We know $\frac{d r}{d t}=-\frac{1}{5}.$

We want to find the rate of change when $d=50\Rightarrow r=\frac{50}{2}=25.$

Hence,

$\begin{aligned} &\frac{d V}{d t}=4(25)^{2}(-\frac{1}{5}) \\ &\frac{d V}{d t}=-500 \end{aligned}$

Thus, the volume of the snowball is decreasing at a rate of $500 \frac{\mathrm{cm}^{3}}{\mathrm{sec}}.$