Thread: Rate of Change - Volume and Surface Area of Cylinder

We have a special type of cylinder whose height, h, is always twice its radius r. What is the rate of change of its volume V with respect to r in terms of the total surface area, A?

I figured V= (pi)(r^2)(h) and since h=2r , V = (pi)(r^2)(2r)
Surface Area = 2(pi)(r^2)+(pi)(r)(h) = 2(pi)(r^2)+(pi)(2r^2)

I'm really not sure what to do after that though, thanks.

- Adrian

Originally Posted by Adrian

We have a special type of cylinder whose height, h, is always twice its radius r. What is the rate of change of its volume V with respect to r in terms of the total surface area, A?

I figured V= (pi)(r^2)(h) and since h=2r , V = (pi)(r^2)(2r)
Surface Area = 2(pi)(r^2)+(pi)(r)(h) = 2(pi)(r^2)+(pi)(2r^2)

I'm really not sure what to do after that though, thanks.

- Adrian

I'm afraid your total surface area, A, is lacking.

The problems asks us to find dV/dr in terms of A.

V = pi(r^2)h = pi(r^2)(2r) = 2pi(r^3)

A = 2[pi(r^2)] +2pi(r)*h
A = 2pi(r^2) +2pi(r)(2r)
A = 2pi(r^2) +4pi(r^2)
A = 6pi(r^2)

So,
V = 2pi(r^3)
= [pi(r^2)](2r)
= [A/6](2r)
= (1/3)A*r

Differentiate that with respect to r,
dV/dr = (1/3)A ----------------------------answer.

That was very helpful, and generally the path I was trying to take, I think my problem was the surface area, which you helped with a lot, thanks again.