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 $1 cm$?

Answer 1

The correct answer is $2π cm^3 /s.$
The air bubble is in the shape of a sphere. Now, the volume of an air bubble $(V)$ with radius $(r)$ is given by,
$v=\frac{4}{3}πr^3$
The rate of change of volume $(V)$ with respect to time $(t)$ is given by,
$\frac{dv}{dt}=\frac{4}{3}π\frac{d}{dr}(r^3).\frac{dr}{dt}$ [Bychain rule]
$=\frac{4}{3}π(3r^2)\frac{dr}{dt}$
$4πr^2 \frac{dr}{dt}$
It is given that $\frac{dr}{dt}=\frac{1}{2} cm/s$
Therefore, when $r=1 cm$,
$\frac{dv}{dt}=4π(1)^2(\frac{1}{2})=2π cm^3/s$
Hence, the rate at which the volume of the bubble increases is $2π cm^3 /s.$