Question:
The radius of a sphere is increasing at the rate of 0.2 cm/sec. The rate at which the volume of the sphere increase when radius is 15 cm, is
(a) $12 \pi \mathrm{cm}^{3} / \mathrm{sec}$
(b) $180 \pi \mathrm{cm}^{3} / \mathrm{sec}$
(c) $225 \pi \mathrm{cm}^{3} / \mathrm{sec}$
(d) $3 \pi \mathrm{cm}^{3} / \mathrm{sec}$
Solution:
(b) $180 \pi \mathrm{cm}^{3} / \mathrm{sec}$
Let $r$ be the radius and $V$ be the volume of the sphere at any time $t .$ Then,
$\mathrm{V}=\frac{4}{3} \pi r^{3}$
$\Rightarrow \frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t}$
$\Rightarrow \frac{d V}{d t}=4 \pi(15)^{2} \times 0.2$
$\Rightarrow \frac{d V}{d t}=180 \pi \mathrm{cm}^{3} / \mathrm{sec}$