Question

If the rate at which water vapor condenses onto a spherical raindrop is proportional to the surface area of the raindrop, show that the radius of the raindrop will increase at a constant rate?

Answer 1

Let us setup the following variables:

# { (r, "Radius of raindrop at time t","(cm)"), (S, "Surface area of raindrop at time t", "(cm"^3")"), (V, "Volume of raindrop at time t", "(cm"^2")"), (t, "time", "(sec)") :} #

The standard formula for Surface Area of a sphere, and the volume are:

`S = 4pir^2 \ \`, and `\ \ V= 4/3 pi r^3`

Differentiating wrt `r`, we have:

`(dV)/(dr)= 4 pi r^2 = S`

We are given that the rate at which water vapor condenses onto a spherical raindrop is proportional to the surface area of the raindrop, Thus:

`(dV)/(dt) prop S => (dV)/(dt) =k S \ \`.....[A], say for some constant `k`

And os applying the chain rule, we have:

`(dV)/(dr) * (dr)/(dt) =k S`

`:. S (dr)/(dt) =k S`

`:. (dr)/(dt) =k`, a constant, QED