Question

The surface area of a spherical balloon is increasing at the rate of 2cm²/sec. At what rate the volume of the balloon is increasing when radius of the balloon is 6 cm?

Answer 1

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

Then S = 4πr² and V = `(4)/(3)πr^3`

Differentiating w.r.t. t, we get

`(dS)/(dt) = 4π xx 2r (dr)/dt = 8πr (dr)/dt`     ...(1)

and `(dV)/dt = 4/3π xx 3r^2 (dr)/dt = 4πr^2 (dr)/dt`

From (1), `(dr)/dt = 1/(8πr)*(dS)/dt`

∴ `(dV)/dt = 4πr^2 xx 1/(8πr) (dS)/dt`

∴ `(dV)/dt = r/2*(dS)/dt`          ...(2)

Now, `(dS)/dt` = 2 cm²/sec and r = 6 cm

∴ (2) gives, `(dV)/dt = 6/2 xx 2` = 6

Hence, the volume of the spherical balloon is increasing at the rate of 6 cm³/sec.