Question

The volume of a sphere is increasing at the rate of 3 cubic centimetre per second. Find the rate of its surface area, when the radius is 2cm.

Answer 1

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

differentiating on both sides, we get,

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{1}{4\pi r^2}\frac{dV}{dt}$

$r=2cm,\frac{dV}{dt}=3c{m}^2/sec$

$\therefore \frac{dr}{dt}=\frac{1}{4\pi 2^2}\left(3\right)=\frac{3}{16\pi }cm/sec$

$S=4\pi r^2$

$\frac{dS}{dt}=8\pi r\frac{dr}{dt}$

$=8\pi \left(2\right)\times \frac{3}{16\pi }cm/sec$

$\therefore \frac{dS}{dt}=3c{m}^2/sec$