Question

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

Answer 1

You are trying to show that dr/dt is constant, i.e. does not depend on the value of any variable.

V = 4/3*pi*r^3
S =4*pi*r^2

Taking the derivative of V with respect to t gives:
dV/dt = 4*pi*r^2 * dr/dt

But dV/dt = k*S, so now you have
k*S = 4*pi*r^2 * dr/dt

Replace S with 4*pi*r^2 to get
k * 4 * pi * r^2 = 4 * pi * r^2 * dr/dt.

Divide both sides by 4*pi*r^2 to get

k = dr/dt.

Answer 2

Thank you so much

Answer 3

Wait so how does this verify that k is an independent constant?