Question

Suppose that a droplet of mist is a perfect sphere and that, through condensation, the droplet picks up moisture at a rate proportional to its surface area. Show that under these circumstances the droplet's radius increases at a constant rate.

Answer 1

If V is volume and A is surface area then dV/dt=k A
Here V is 4/3 pi r^3 and A=4pi r^2
dV/dt becomes 4 pi r^2 dr/dt. Putting the value you get dr/dt=k.
Implies radius changes at constant rate.

Answer 2

so dv/dt=4pir^2(dr/dt), but how did you get the dr/dt=k?

Answer 3

Substitute it in dV/dt = k 4 pi r^2
You get 4 pi r^2 dr/dt= k 4 pi r^2.
4 pi r^2 cancels on both sides. you get dr/dt=k.