Question

The radius of an air bubble is increasing at the rate of $\frac{1}{2}cm/s$. At what rate is the volume of the bubble increasing when the radius is $1cm$?

Answer 1

The air bubble is in the shape of a sphere.
Now, the volume of an air bubble $\left(V\right)$ with radius $\left(r\right)$ is given by,
$V=\frac{4}{3}\pi r^3$
The rate of change of volume $\left(V\right)$ with respect to time $\left(t\right)$ is given by,
$\frac{dV}{dt}=\frac{4}{3}\pi \frac{d}{dr}\left(r^3\right)\cdot \frac{dr}{dt}\frac{4}{3}\pi \left(3r^2\right)\cdot \frac{dr}{dt}=4\pi r^2\frac{dr}{dt}$
It is given that $\frac{dr}{dt}=\frac{1}{2}$ cm/s
Therefore, when $r=1$ cm,
$\frac{dV}{dt}=4\pi (1{)}^2\left(\frac{1}{2}\right)=2\pi c{m}^3/s$
Hence, the rate at which the volume of the bubble increases is $2\pi c{m}^3/s$.