Posted by **Mishaka** on Tuesday, January 17, 2012 at 7:37pm.

The radius, r, of a sphere is increasing at a constant rate of 0.05 meters per second.

A. At the time when the radius of the sphere is 12 meters, what is the rate of increase in its volume?

B. At the time when the volume of the sphere is 36pi cubic meters, what is the rate of increase in its surface area?

C. Express the rate at which the volume of the sphere changes with respect to the surface area of the sphere (as a function of r).

Parts A and B I have figured out (A. = 28.8pi cubic meters per second; B. = 1.2pi squared meters per second) But I can't quite figure out C. How do I relate surface area and volume together into one equation???

- Calculus (Parts A and B done, just help with C) -
**Damon**, Tuesday, January 17, 2012 at 7:53pm
V = (4/3)pi r^3

dV/dr = 4 pi r^2 which happens to be the area

so

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

the differential change of volume is the surface area * dr

PS

I wonder how you did parts a and b without knowing that

- Calculus (Parts A and B done, just help with C) -
**Mishaka**, Wednesday, January 18, 2012 at 3:32pm
I did notice this when I took the derivative for part A of the volume. But it seemed too simple to just plug in this value, I wanted to make sure that I was certain about the placement. Thank you!!!

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