# Partial Derivatives problem.

• Apr 20th 2008, 09:01 PM
Rombie
Partial Derivatives problem.
I'm having trouble with the last part of a larger problem so I'll go through what I've done in case it's tripped me up.

Here's the problem.

There is a magical growing cylinder with these properties. The height grows at a constant rate $\alpha > 0$ (i.e. $\frac{dh}{dt}=\alpha$). The radius grows proportionately to the difference between the height and the radius (i.e. $\frac{dr}{dt}=h-r$)

1) Find the formula for the volume. $V=\pi r^2h$
2) Express the change in volume with time $\frac{dV}{dt}$, in terms of r and h.
$\frac{dV}{dt}=\frac{\partial V}{\partial r}\frac{dr}{dt}+\frac{\partial V}{\partial h}\frac{dh}{dt}$
$=2\pi rh(h-r) + \pi r^2\alpha$
$=2\pi rh^2 - 2\pi r^2h + \pi r^2\alpha$

3) Find the rate at which the volume of the plant is changing when $r = 2$, $h = 3$ and $\alpha = 2$

bung them into the equation

$\frac{dV}{dt}=2\pi\times 2\times 3^2 - 2\pi\times2^2\times 3 + \pi\times 2^2\times2$
$\frac{dV}{dt}=36\pi - 24\pi +8\pi = 20\pi$ Do I need to put in any units?

4) Find a non-trivial expression for the radius of the cylinder in terms of it's height when V isn't changing i.e. r = f(h);

I set $0 = 2\pi rh^2 - 2\pi r^2 h + \alpha\pi r^2$
$2\pi r^2 h - 2\pi rh^2 = \alpha\pi r^2$
$2\pi r(rh - h^2) = \alpha\pi r^2$
$rh - h^2 = \frac{\alpha}{2} r$
$rh - \frac{\alpha}{2} r = h^2$
$rh - \frac{\alpha}{2} r = h^2$
$r(h - \frac{\alpha}{2} ) = h^2$
$r = \frac{h^2}{h - \frac{\alpha}{2}}$

Hopefully all that is okay. The last little bit says.

If $r > f(h)$ and $h > \frac{\alpha}{2}$ is the volume increasing or decreasing?

What I've been trying to do is use that fact to make deductions about what in this equation $=2\pi rh^2 - 2\pi r^2h + \pi r^2\alpha$ is positive or negative and whats larger than what. I think it should turn out something like they have to add up to be greater or less than zero I'm just having trouble finding the right way to show it. That's if it is all correct to this point.

Thank you for any help.