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 \(2π cm^3 /s.\)