Air is being pumped into balloons at the rate off 8 cm³/sec. How fast does balloons' surface enlarge when the radius of the balloon is 10 cm³?

Let the radius of the balloon be r, the volume be v and the surface area is s.

The volume of the balloon is v = 4πr³/3. The derivative of the volume with respect to time is

The rate that air is being pumped into the balloon is 8 cm³/sec, so

     dv/dr = 4πr²dr/dt = 8

Therefore,

     dr/dt = 2/πr²

The surface area of the balloon is

     s = 4πr²

Thus the surface enlarge speed of the balloon is

Substitude dr/dt = 2/πr² into ds/dr,

When the radius r is 10 cm,