# How do I find the instantaneous rate of change of the volume of a cylinder as the radius varies while the surface area is held fixed.

I have a question in my Calculus 1 homework that I'm not sure where to begin with.

I need to calculate the instantaneous rate of change of the volume of a cylinder as the radius varies while the surface area is held fixed.

I know that volume $V=\pi r^2 h$ and surface area $S=2\pi rh+2\pir^2$ however I'm not sure how to relate them in an equation.

• My error in V was a typo but I don't understand why you have changed S. Surely $S=2\pi rh$ doesn't include the two ends? – Alex Modell Oct 12 '16 at 19:42

Hint:

Find $h$ from the equation of the surface: $$h=\frac{S}{2\pi r}-r$$ and substitute in the volume: $$V=\pi r^2\left(\frac{S}{2\pi r}-r \right)$$ This is the equation that gives the volume as a function o f $r$ for a given total surface $S$.

• i thought about lateral surface – hamam_Abdallah Oct 12 '16 at 19:52
• I've used the total surface... But the spirit is the same.... :) – Emilio Novati Oct 12 '16 at 19:54

Hint:

using the former result and logarithmic derivative, we get the rate of change

$\frac{dV}{V}=\frac{dr}{r}-3\frac{dr}{r}=-2\frac{dr}{r}$

the rate of change of the volume is twice the rate of change of the radius.

• Could you explain why $S=2\pi rh$ please? – Alex Modell Oct 12 '16 at 19:50
• sorry i made the correction – hamam_Abdallah Oct 12 '16 at 20:12