Question:
The volume of a sphere is increasing at the rate of $4 \pi \mathrm{cm}^{3} / \mathrm{sec}$. The rate of increase of the radius when the volume is $288 \pi \mathrm{cm}^{3}$, is
(a) $1 / 4$
(b) $1 / 12$
(c) $1 / 36$
(d) $1 / 9$
Solution:
(c) $1 / 36$
Let $r$ be the radius and $V$ be the volume of the sphere at any time $t$. Then,
$V=\frac{4}{3} \pi r^{3}$
$\Rightarrow \frac{4}{3} \pi r^{3}=288 \pi$
$\Rightarrow r^{3}=\frac{288 \times 3}{4}$
$\Rightarrow r^{3}=216$
$\Rightarrow r=6$
$\Rightarrow \frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t}$
$\Rightarrow \frac{d V}{d t}=4 \pi(6)^{2} \frac{d r}{d t}$
$\Rightarrow 4 \pi=144 \pi \frac{d r}{d t}$
$\Rightarrow \frac{d r}{d t}=\frac{1}{36}$