Question

The radius of the soap bubble is increasing at the rate of $0.2$ cm/sec. If its radius is $5$ cm, find the rate of increasing of its volume.

Answer 1

Radius of soap bubble $=r$

$\frac{dr}{dt}=0.2cm/sec$

Volume of bubble $\frac{4}{3}\pi r^3$

$\therefore \frac{dr}{dt}=\frac{4}{3}\pi \frac{d}{dt}{(r}^3)=\frac{4}{3}\pi {(3r}^3)\frac{dr}{dt}=4\pi r^2\frac{dr}{dt}whenr=5cm,\frac{dv}{dt}=4\pi {\left(5\right)}^2\left(0.2\right)=100\times 0.2\pi =20\pi {cm}^3/sec$