Question

Assume that a rain drop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the rain drop.

Answer 1

Let the surface area of the raindrop be $A$.
Thus, the rate of evaporation will be given by $\frac{dV}{dt}$.
As per the given condition,
$$\begin{gathered} \frac{dV}{dt}\propto A \\ \Rightarrow \frac{dV}{dt}=-kA \end{gathered}$$
Here, $k$ is a constant. Also, the negative sign appears when V decreases and $t$ increases.
Now,
$V=\frac{4}{3}\mathrm{\pi }{r}^3$
Here, $r$ is the radius of the spherical drop.
$$\begin{gathered} \therefore \frac{d}{dt}\left(\frac{4}{3}{\mathrm{\pi r}}^3\right)=-k\times 4{\mathrm{\pi r}}^2 \\ \Rightarrow \frac{4}{3}\times 3\pi r^2\frac{dr}{dt}=-k\times 4\pi r^2 \\ \Rightarrow \frac{dr}{dt}=-k \\ \mathrm{It}\mathrm{is}\mathrm{the}\text{required differential equation.} \end{gathered}$$