Question

Suppose that a drop of mist is a perfect sphere and that, through condensation, the drop picks up moisture at a rate proportional to its surface area. Show that under these circumstances the drop's radius increases at a constant rate.

Answer 1

The volume of a sphere is given by

$V= \frac{4}{3} \pi r^3$

The area of a sphere is given by

$A=4\pi r^2$

$\frac{dV}{dt} \propto A \\ \\ \frac{dV}{dt} = kA$

But

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

By substitution,

$4\pi r^2 \frac{dr}{dt}=kA \\ \\ \Rightarrow4\pi r^2 \frac{dr}{dt}=4k\pi r^2 \\ \\ \frac{dr}{dt} =k$

Thus, the drop's radius increases at a constant rate.