Question

A balloon which always remains spherical on inflation, is being inflated pumping in 900 cubic cm of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.

Answer 1

At any instant time t let the radius of the balloon be r and its volume be V, then Volume of balloon $V=\left(\frac{4}{3}\right)\pi r^3$
The balloon is being inflated at 900 cubic cm/s i.e., the rate of change of volume with respect to time is $900c{m}^3/s.$
On differentiating w.r.t. t, we get
Rate of change of volume $\frac{dV}{dt}=\left(\frac{4}{3}\pi \right)\left(3r^2\frac{dr}{dt}\right)$
Given $r=15cm\Rightarrow 900=\left(\frac{4}{3}\pi \right)\left\{3\right(15{)}^2\frac{dr}{dt}\}\Rightarrow \frac{dr}{dt}=\frac{900}{3\times (15{)}^2}\times \frac{3}{4\pi }$
$\Rightarrow$ Rate of change of radius r, $\frac{dr}{dt}=\frac{900}{4\pi \times (15{)}^2}=\frac{225}{\pi \times 225}=\frac{1}{\pi }cm/s$
Hence, the rate at which the radius of the balloon increases when the radius is 15 cm is $\frac{1}{\pi }$ cm/s.