Gas is being pumped into a spherical balloon at the rate of $30{\mathrm{ft}}^3/\min$ Then the rate at which the radius increases when is reaches the value $15 ft,$ is

Solution:

Let $r$be the radius of the spherical balloon and $V$

be the volume at any time $t$.

It is given that  $\frac{dV}{dt}=30f{t}^3/\min$

Now, $V=\frac{4}{3}\pi r^3$

$\begin{array}{l}\Rightarrow \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\\ \Rightarrow {\left(\frac{dV}{dt}\right)}_{r=15}=4\pi \times {15}^2\times {\left(\frac{dr}{dt}\right)}_{r=15}\\ \Rightarrow 30=900\pi {\left(\frac{dr}{dt}\right)}_{r=15}\\ \Rightarrow {\left(\frac{dr}{dt}\right)}_{r=15}=\frac{1}{30\pi }\mathrm{ft}/\min \end{array}$