Question

A spherical balloon loses its volume due to escape of air from it in such a way that decrease of volume at any instant is proportional to its surface area. Show that the radius is decreasing at a constant rate.

Answer 1

For a sphere with radius r:

Volume (V) = `4/3 pir^3`

Surface Area (S) = `4pir^2`

The problem states that the rate of decrease of volume `(-(dV)/dt)` is proportional to the surface area (S):

`- (dV)/(dt)` ∝ S

Using a constant of proportionality k (where k > 0):

`(dV)/(dt) = -k.S`

Using the chain rule to find `(dV)/(dt)` in terms of the radius r:

`(dV)/(dt) = (dV)/(dr) . (dr)/(dt)`

Since `(dV)/(dr) = d/(dr)(4/3pir^3) = 4pir^2`, we have:

`(dV)/(dt) = 4pir^2 . (dr)/(dt)`

Substitute the expressions for `(dV)/(dt) and S` into the original proportionality equation:

`4pir^2 . (dr)/(dt) = -k(4pir^2)`

Divide both sides by 4πr² (assuming r ≠ 0):

`(dr)/(dt) = -k`

Since k is a constant, `(dr)/(dt)` is constant. This shows that the radius of the balloon decreases at a constant rate over time.