The radius of a spherical button is increasing at the rate `10 cm//sec`. Find the rate of change of its surface area when its radius is 4 cm.

Explanation

Let r be the radius of the spherical balloon.
Then, surface area of the balloon, `S=4pir^2`
`therefore`Rate of change of surface area of the balloon, `(dS)/dt=8pir(dr)/dt`
When r = 4cm and `(dr)/dt=10 cm//sec`(given), then rate of change of surface area of the balloon, `(dS)/dt=8pi. 4. 10= 320pi cm^2//sec`.