Question

A spherical soap bubble is expanding so that its radius is increasing at the rate of 0.02 cm/sec. At what rate is the surface area is increasing, when its radius is 5 cm?

Answer 1

Let r be the radius and S be the surface area of the soap bubble at any time t.

Then S = 4πr²

Differentiating w.r.t. t, we get

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

∴ `"dS"/"dt" = 8pir "dr"/"dt"`           ...(1)

Now, `"dr"/"dt" = (0.02cm)/sec` and r = 5 cm

∴ (1) gives, `"dS"/"dt" = 8π(5)(0.02)` = 0.8π

Hence, the surface area of the soap bubble is increasing at the rate if `(0.8picm^2)/sec`.