Assume that a spherical raindrop evaporates at a rate proportional to its surface area. If its radius originally is 3 mm and 1 hour later has been reduced to 2 mm, find an expression for the radius of the raindrop at any time t.

#### Solution

Let r be the radius, V be the volume and S be the surface area of the spherical raindrop at time t.

Then V = `4/3 pi"r"^3` and S = 4πr^{2}

The rate at which the raindrop evaporates is `"dV"/"dt"` which is proportional to the surface area.

∴ `"dV"/"dt" prop "S"`

∴ `"dV"/"dt"` = - kS, where k > 0 ...(1)

Now, V = `4/3pi"r"^3` and S = 4πr^{2}

∴ `"dV"/"dt" = (4pi)/3 xx 3"r"^2 "dr"/"dt" = 4 pi "r"^2 "dr"/"dt"`

∴ (1) becomes, `4 pi "r"^2 "dr"/"dt" = - "k"(4 pi "r"^2)`

∴ `"dr"/"dt"` = - k

∴ dr = - k dt

On integrating, we get

`int "dr" = - "k" int "dt" + "c"`

∴ r = - kt + c

Initially, i.e. when t = 0, r = 3

∴ 3 = - k × 0 + c ∴ c = 3

∴ r = - kt + 3

When t = 1, r = 2

∴ 2 = - k × 1 + 3

∴ k = 1

∴ r = - t + 3

∴ r = 3 - t, where 0 ≤ t ≤ 3.

This is the required expression for the radius of the raindrop at any time t.