The volume of a sphere is increasing at $3 \mathrm{~cm}^{3} / \mathrm{sec}$. The rate at which the radius increases when radius is $2 \mathrm{~cm}$, is
(a) $\frac{3}{32 \pi} \mathrm{cm} / \mathrm{sec}$
(b) $\frac{3}{16 \pi} \mathrm{cm} / \mathrm{sec}$
(c) $\frac{3}{48 \pi} \mathrm{cm} / \mathrm{sec}$
(d) $\frac{1}{24 \pi} \mathrm{cm} / \mathrm{sec}$
(b) $\frac{3}{16 \pi} \mathrm{cm} / \mathrm{sec}$
Let $r$ be the radius and $V$ be the volume of the sphere at any time $t$. Then,
$\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}$
$\Rightarrow \frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t}$
$\Rightarrow \frac{d r}{d t}=\frac{1}{4 \pi r^{2}} \frac{d V}{d t}$
$\Rightarrow \frac{d r}{d t}=\frac{3}{4 \pi(2)^{2}}$
$\Rightarrow \frac{d r}{d t}=\frac{3}{16 \pi} \mathrm{cm} / \mathrm{sec}$
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