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The radius, r, of a sphere is increasing at 0.03 meters per second. The rate of increase is constant.

AWhat is the rate of increase in its volume when the radius of the sphere is 18 meters?

BWhat is the rate of increase in the sphere’s surface area when the volume is 288 cubic meters?

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

Please help, I don't get any of this. An explanation would be great, thanks!

Verified Answer :

LouisRemiRichard

A)
Given: dr/dt = .03 m/s and r = 18 m
Volume of a sphere = (4/3)PIr^3
Take the derivative of both sides, so
dv/dt = 4PIr^2dr/dt
Plug in the givens and you have the rate of increase of the volume
B)
Given: V = 288 m^3 and I'm going to assume that r and dr/dt are the same as in part A
Surface Area = 4PIr^2
Take the derivative on both sides, so
dA/dt = 8PIrdr/dt
Again, plug in the givens and you'll have the rate of increase in the surface area
C)
This part, it's been awhile since I've done related rates, so it may take me awhile, but perhaps the next person can answer it before I finish.

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