Rate of change in a radius and surface area...?

The volume of a spherical balloon is increasing at the constant rate 8 cubic feet per minute. How fast is the radius increasing when the radius exactly 10 feet? How fast is the surface area increasing at that time? Explain your work. Thanks :)

Rate of change in a radius and surface area...?

Let the volume be v ft^3, the surface area be s ft^2, and the radius be r ft, and time t be measured in minutes.

Then:

v = (4/3)pi r^3

dv/dt = (4/3)pi 3r^2 dr/dt

dv/dt = 4pi r^2 dr/dt

Putting dv/dt = 8 and r = 10:

8 = 4 pi 10^2 dr/dt

dr/dt = 8 / 400pi

dr/dt = 1 / 50pi ft/min.

s = 4pi r^2

ds/dt = 8pi r dr/dt

= 8 pi 10 / (50 pi)

= (8 / 5) ft^2/min.