Question

The volume of a sphere increases at the rate of 20 cm³/sec. Find the rate of change of its surface area, when its radius is 5 cm

Answer 1

Let r be the radius, s be the surface area and V be the volume of the sphere at any time t.

Then s = 4πr² and V = `(4)/(3)pi"r"^3`

Differentiating w.r.t. t, we get

`"ds"/"dt" = 4pi xx 2"r""dr"/"dt"`

= `8pi"r""dr"/"dt"`

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

= `4pi"r"^2"dr"/"dt"`      ...(1)

From (1), `"dr"/"dt" = (1)/(4pi"r"^2) "dV"/"dt"`

∴ `"ds"/"dt" = 8pi"r" xx (1)/(4pi"r"^2) "dV"/"dt"`

∴ `"ds"/"dt" = (2)/"r"."dV"/"dt"`                     ...(2)

Now, `"dV"/"dt" = (20"cm"^3)/sec` and r = 5 cm

∴ (2) gives, `"ds"/"dt" = (2)/(5) xx 20` = 8

Hence, the surface area of the sphere is changing at the rate 8 cm²/sec.