Derivatives- Application

Question

Assume that spherical raindrop evaporates at a rate proportional to its surface area. If its original radius is 3mm and one hour later, it reduces to 2mm, find an expression for the radius of the raindrop at any time.

Answer 1

$$\dfrac{dV}{dt} = k \cdot 4\pi r^2$$ ($k$ is constant of proportionality)

$V= \dfrac{4}{3} \pi r^3$ for a sphere, so $$\frac{dV}{dt}= 4\pi r^2 \frac{dr}{dt}$$
Compare both $\dfrac{dV}{dt}$. You get $\dfrac{dr}{dt}=k$.

Integrating, you get $r=kt+c$. For $t=0, r=3$ and for $t=1, r=2$ ($r$ is in mm and $t$ is in hours). Solving for above values of $r$ and $t$, you get $r=3-t$.