# The volume of a spherical balloon is increasing at the rate

Question:

The volume of a spherical balloon is increasing at the rate of $25 \mathrm{~cm}^{3} / \mathrm{sec}$. Find the rate of change of its surface area at the instant when radius is $5 \mathrm{~cm}$.

Solution:

Let $r$ be the radius and $V$ be the volume of the sphere at any time $t$. Then,

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

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

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

$\Rightarrow \frac{d r}{d t}=\frac{25}{4 \pi(5)^{2}}$     $\left[\because r=5 \mathrm{~cm}\right.$ and $\left.\frac{d V}{d t}=25 \mathrm{~cm}^{3} / \mathrm{sec}\right]$

$\Rightarrow \frac{d r}{d t}=\frac{1}{4 \pi} \mathrm{cm} / \mathrm{sec}$

Now, let $S$ be the surface area of the sphere at any time $t$. Then,

$S=4 \pi r^{2}$

$\Rightarrow \frac{d S}{d t}=8 \pi r \frac{d r}{d t}$

$\Rightarrow \frac{d S}{d t}=8 \pi(5) \times \frac{1}{4 \pi}$         $\left[\because r=5 \mathrm{~cm}\right.$ and $\left.\frac{d r}{d t}=\frac{1}{4 \pi} \mathrm{cm} / \mathrm{sec}\right]$

$\Rightarrow \frac{d S}{d t}=10 \mathrm{~cm}^{2} / \mathrm{sec}$