A spherical balloon is filled with$4500 \pi$ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of $72\pi$ cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases $49$ minutes after the leakage began is :

Let $r$be the radius of the balloon and $V$ be its volume. It is given that

           $\begin{array}{l}\frac{dV}{dt}=-72\pi \\ \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=-72\pi \Rightarrow \frac{4}{3}\pi r^3=-72\pi t+C\end{array}$

When          $t=0,\frac{4}{3}\pi r^3=V=4500\pi$

$\therefore \frac{4}{3}\pi r^3=4500\pi -72\pi t$

When $t=49,\frac{4}{3}\pi r^3=4500\pi -\left(72\pi \right)\left(49\right)=972\pi$

$\Rightarrow r^3=729$ so $r=9$

Also $\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)=-72\pi \Rightarrow 4\pi r^2\frac{dr}{dt}=-72\pi$

$\Rightarrow \frac{dr}{dt}=-\frac{72\pi }{4\pi r^2}=-\frac{18}{r^2}\text{.}$So ${\left.\frac{dr}{dt}\right|}_{r=9}=-\frac{2}{9}\text{. }$