Gas is leaking out of a spherical balloon at the rate of 1800 cubic cm per sec. when the radius of the balloon is 720cm, the rate at which the surface area is sharinking is

Let $r$ be the radius of the spherical baloon, and $V$ be the volume of it

Given $\frac{dV}{dt}=1800 cc per sec$

We know that $V=\frac{4}{3}{\mathrm{\pi r}}^3$

Differentiate both sides with respect to time

$\frac{dV}{dt}=\frac{4}{3}\mathrm{\pi }\left(3{\mathrm{r}}^2\frac{\mathrm{dr}}{\mathrm{dt}}\right)$

Substitute the appropriate values in the above equaiton

$1800=4\mathrm{\pi }{\left(720\right)}^2\frac{\mathrm{dr}}{\mathrm{dt}}$

We know that $S=4{\mathrm{\pi r}}^2$

Differentiate both sides

$\frac{ds}{dt}=8\mathrm{\pi r}\left(\frac{\mathrm{dr}}{\mathrm{dt}}\right)$

$\begin{array}{rcl}\frac{ds}{dt}& =& 8\mathrm{\pi }\left(720\right)\left(\frac{1800}{4\mathrm{\pi }\left(720\right)\left(720\right)}\right)\\ & =& \frac{360}{72}\\ & =& 5 \end{array}$

Therefore, the rate of change in the surface area is