The radius of a sphere is increasing at a constant rate of 2cm/sec. At the instant when the volume of the sphere is 36π cm^3, what is the rate that the surface area is increasing? Surface area=4π r^2 Volume=4/3π r^3

Question

The radius of a sphere is increasing at a constant rate of 2cm/sec. At the instant when the volume of the sphere is 36π cm^3, what is the rate that the surface area is increasing? Surface area=4π r^2   Volume=4/3π r^3

anonymous · Answer

$$S(r(t))=4\pi r^2$$
$$S'(r(t))=8\pi r r'$$ and you are told that 
$$r'=2$$ so this becomes 
$$S'(r(t))=16\pi r$$ now your only job is to find r and substitute

anonymous · Answer

How do I find r?