The volume of a sphere is increasing
Question:

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) $\frac{1}{24 \pi} \mathrm{cm} / \mathrm{sec}$

Solution:

(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,

$\mathrm{V}=\frac{4}{3} \pi \mathrm{r}^{3}$

$\Rightarrow \frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t}$

$\Rightarrow \frac{d r}{d t}=\frac{1}{4 \pi r^{2}} \frac{d V}{d t}$

$\Rightarrow \frac{d r}{d t}=\frac{3}{4 \pi(2)^{2}}$

$\Rightarrow \frac{d r}{d t}=\frac{3}{16 \pi} \mathrm{cm} / \mathrm{sec}$

Administrator

Leave a comment

Please enter comment.
Please enter your name.