The volume of a spherical bubble is increasing at a constant rate - AQA - A-Level Maths Mechanics - Question 10 - 2019 - Paper 1

Since the volume is increasing at a constant rate, we can denote that constant rate as $k$, such that: $\frac{dV}{dt} = k.$

Substituting this into our previous equation results in: $k = 4 \pi r^2 \frac{dr}{dt}.$

Next, we isolate $\frac{dr}{dt}$: $\frac{dr}{dt} = \frac{k}{4 \pi r^2}.$

From this expression, it is clear that the rate of increase of the radius $\frac{dr}{dt}$ is inversely proportional to $r^2$, which we can write as: $\frac{dr}{dt} \propto \frac{1}{r^2}.$

This concludes the proof.