Question

Suppose that a raindrop evaporates in such a way that it maintains a spherical shape. Given that the volume of a pshere of radius r is V=(4/3)pi(r)^3 and its surface area is A = 4pi(r)^2, if the radius changes in time, show that V' = Ar'. If the rate of evaporation (V') is porportional to the surface area, show that the radius changes as a constant rate.

Answer 1

V=(4/3)πr³, V'=4πr²r'=Ar' when we substitute A=4πr². If V'=kA where k is a constant, then kA=Ar' and r'=k so rate of change of radius is constant.