If the rate of change of volume of a sphere is equal to the rate of change of its radius, then its radius is equal to

Solution:

Let $r$ be the radius and $V$ be the volume of the

sphere. Then,

$\begin{array}{l} V=\frac{4}{3}\pi r^3\\ \Rightarrow \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\end{array}$

$\Rightarrow \frac{dr}{dt}=4\pi r^2\frac{dr}{dt}$                         $\left[\because \frac{dV}{dt}=\frac{dr}{dt}\text{ (given) }\right]$

$\Rightarrow 4\pi r^2=1\Rightarrow r=\frac{1}{2\sqrt{\pi }}$