Question

Assume that a spherical raindrop evaporates at a rate proportional to its surface area. If its radius

originally is 3 mm, and one-half hour later has been reduced to 2mm, find an expression for the

radius of the raindrop at any time.

Answer 1

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

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

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

So, $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=k*4\pi r^2$

or $\frac{dr}{dt}=k$

Integrating, we get: $r=kt+C$

To find k and C we use two conditions: $r(0)=3,\;\; r(0.5)=2$

We have:

$3=k*0+C, \\ \;2=k*0.5+C\;\;or\;\;C=3, k=-2$

And finally, $r=-2t+3$ .