Question

The surface area of a spherical balloon is increasing at the rate of $2c{m}^2/sec$. At what rate is the volume of the balloon is increasing when the radius of the balloon is $6cm$?

Answer 1

Let $r,V$ and $S$ be respectively the radius, volume and surface area of the spherical balloon at time $t$.
$\therefore V=\frac{4}{3}\pi r^3$
and $S=4\pi r^2$
Rate of change of surface area with respect to $t$
$\Rightarrow \frac{dS}{dt}=2$
$\Rightarrow \frac{d}{dt}\left(4\pi r^2\right)=2$
$\Rightarrow 8\pi r\frac{dr}{dt}=24$
$\Rightarrow \frac{dr}{dt}=\frac{1}{4\pi r}$ .... (i)
Rate of change of volume of the balloon with respect to $t$
$=\frac{dV}{dt}$
$=\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)$
$=\frac{4}{3}\pi \cdot 3r^2\frac{dr}{dt}$
$=4\pi r^2\cdot \frac{dr}{dt}$
$=4\pi r^2\cdot \frac{1}{4\pi r}$ ....[From (i)]
$=r$
$=6$
Hence, the volume of the spherical balloon is increasing at the rate of $6c{m}^3/sec$.