Question

The volume of a sphere is increasing at the rate of $8c{m}^3/sec$. Find the rate at which its surface area is increasing when the radius of the sphere is $12cm$.

Answer 1

Let $r$ be the radius of sphere, $V$ be the volume and $S$ be the surface area of the sphere.

Then,

$V$ = $\frac{4}{3}\pi r^{3}andS=4\pi r^2$
$\therefore \frac{dV}{dt}=\frac{4}{3}\pi \left(3r^2\right)\frac{dr}{dt}=4\pi r^2\frac{dr}{dt}Also,\frac{dV}{dt}=8c{m}^3/sec\Rightarrow 8=4\pi r^2\frac{dr}{dt}\Rightarrow \frac{dr}{dt}=\frac{2}{\pi r^2}...\left(i\right)$
Now,
$\frac{dS}{dt}=4\pi \left(2r\right)\frac{dr}{dt}=8\pi r\frac{dr}{dt}=8\pi r\left(\frac{2}{\pi r^2}\right)\left[\text{using}\right(i\left)\right]=\frac{16}{r}c{m}^2/secWhenr=12cm,\frac{dS}{dt}=\frac{16}{r}=\frac{16}{12}=\frac{4}{3}=1\frac{1}{3}c{m}^2/s$