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  4. If the rate of change of volume of a sphere is equal to the rate of change of its radius then its radius =

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Q. If the rate of change of volume of a sphere is equal to the rate of change of its radius then its radius =

Application of Derivatives

Solution:

$V=\frac{4}{3} \pi r^{3}$
$\frac{dV}{dt}=\frac{4}{3} \pi\cdot3r^{2} \frac{dr}{dt}$
$=4\,\pi r^{2} \frac{dr}{dt}$
Since $\frac{dV}{dt}=\frac{dr}{dt}$
$\therefore 1=4\pi r^{2}$
$\Rightarrow r^{2}=\frac{1}{4\pi}$
$\Rightarrow r=\frac{1}{2\sqrt{\pi}}$