Answer
See explanations.
Work Step by Step
Step 1. We are given that the drop picks up moisture at a rate proportional to its surface area; we have $\frac{dV}{dt}=k(4\pi r^2)$ where $k$ is a constant and $r$ is the radius.
Step 2. Using the formula $V=\frac{4}{3}\pi r^3$, we have $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$
Step 3. Comparing steps 1 and 2, we have $4\pi r^2\frac{dr}{dt}=k(4\pi r^2)$, which gives $\frac{dr}{dt}=k$. This means that the drop’s radius increases at a constant rate.
