# 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, - Mathematics and Statistics

Sum

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 = 4πr2

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 = 4πr2

∴ "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.

Concept: Application of Differential Equations
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