0
$\begingroup$

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.

Thanks for your help in advance!

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Alex Modell Oct 12 '16 at 19:42
0
$\begingroup$

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 $S$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ i thought about lateral surface $\endgroup$ – hamam_Abdallah Oct 12 '16 at 19:52
  • $\begingroup$ I've used the total surface... But the spirit is the same.... :) $\endgroup$ – Emilio Novati Oct 12 '16 at 19:54
0
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.