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

Given that the volume is increasing at a constant rate, we can set ( \frac{dV}{dt} = k ), where ( k ) is a constant. We can rewrite this as:

$k = 4 \pi r^2 \frac{dr}{dt}$

Now, to find ( \frac{dr}{dt} ), we rearrange this equation:

$\frac{dr}{dt} = \frac{k}{4 \pi r^2}$

This equation shows that the rate of increase of the radius ( \frac{dr}{dt} ) is inversely proportional to ( r^2 ). Therefore, we can conclude that:

$\frac{dr}{dt} \propto \frac{1}{r^2}$

This completes the proof that the rate at which the radius increases is inversely related to the square of its radius.