Question

A spherical ball of ice melts in such a way that the rate at which its volume decreases at any instant is directly proportional to its surface area. Prove that the radius of the ice ball decreases at a constant rate.

Answer 1

Since Ball of ice is spherical.

Let r be the radius of the bubble and V be the volume of the bubble.

Given that the rate at which its volume decreases at any instant is directly proportional to its surface area.

So, we write

`(dV)/dt ∝ S`

`(dV)/dt` = −kS  .....(1)

Negative because volume is decreasing and k is some constant.

We need to prove that the radius of the ice ball decreases at a constant rate.

i.e., we need to find `(dr)/dt`.

To find a, we find the surface area and rate of change of volume:

Surface Area of Sphere (S) = 4πr²   ...(2)

Rate of change of volume of sphere `(dV)/(dt)`

Volume of sphere (V) = `4/3 πr^3`

`(dV)/(dt) = (d(4/3 π r^3))/(dt)`

`(dV)/(dt) = 4/3 π (d(r^3))/(dt)`

`(dV)/dt = 4/3 π * (d(r^3))/(dt) * (dr)/dt`

`(dV)/dt = 4/3 π  3r^2 * (dr)/dt`

`(dV)/dt = 4πr^2 * (dr)/dt`   ...(3)

Putting `(dV)/dt` = −kS in (3)

⇒ −kS = `4πr^2 * (dr)/dt`

Putting S = 4πr² from (2):

`-k xx 4πr^2 = 4πr^2 xx (dr)/dt`

`-k xx (4πr^2)/(4πr^2) = (dr)/dt`

`-k = (dr)/dt`

`(dr)/dt = -k`

Thus, the radius of the ice ball decreases at a constant rate.

Hence proved.