# Math Help - Related Rates Problems

1. ## Related Rates Problems

The volume, V, of a cylinder of radius r and height h is given by the formula, V=Pi *(r^2) *h.
Tomato soup is leaking out the bottom of a cylindrical can of diameter 10cm at a rate of Pi cm^3/sec. How quickly is the surface of the soup falling.

How do you solve this question? I have tried differentiating V but the Pi confuses me. Please help.

Thanks.

2. Originally Posted by haku
The volume, V, of a cylinder of radius r and height h is given by the formula, V=Pi *(r^2) *h.
Tomato soup is leaking out the bottom of a cylindrical can of diameter 10cm at a rate of Pi cm^3/sec. How quickly is the surface of the soup falling.

How do you solve this question? I have tried differentiating V but the Pi confuses me. Please help.

Thanks.
$\pi$ is a constant

3. Does this mean the derivative would be: 2r
Is this correct, what happens to the height?

4. Okay, so it should be 2rh assuming and the height are constants. Or do you get rid of them when you differentiate, to give 2r. Thanks

5. The answer should be 0.04, but I can't get to it. Everything you've done seems okay. Any ideas? Thanks for your help with this problem.

6. I got it! Use implicit differentiation!

$\frac{dV}{dh} = \pi r^2 (h')$

But we are told that $\frac{dV}{dh} = \pi$

$\pi = \pi r^2 (h')$

$h' = \frac{\pi}{\pi r^2}$

$h' = \frac{1}{r^2}$

But $r = 5$

$h' = \frac{1}{25} = 0,04$

7. Great! Thanks, that has helped me a lot in understanding these problems.

8. Originally Posted by haku
Great! Thanks, that has helped me a lot in understanding these problems.
Thank you too, you got me to remember Calculus again

9. r and Pi are constants, as stated by Janvdl.

$V={\pi}r^{2}h$

But we need dh/dt.

$\frac{dV}{dt}={\pi}\left(r^{2}\frac{dh}{dt}+2rh\fr ac{dr}{dt}\right)$

But dr/dt=0, because it remains constant.

${\pi}={\pi}\left(25\frac{dh}{dt}+0\right)$

$\frac{dh}{dt}=\frac{1}{25} \;\ cm/min$

10. Originally Posted by galactus
r and Pi are constants, as stated by Janvdl.

$V={\pi}r^{2}h$

But we need dh/dt.

$\frac{dV}{dt}={\pi}\left(r^{2}\frac{dh}{dt}+2rh\fr ac{dr}{dt}\right)$

But dr/dt=0, because it remains constant.

${\pi}={\pi}\left(25\frac{dh}{dt}+0\right)$

$\frac{dh}{dt}=\frac{1}{25} \;\ cm/min$
Just one more post and you become an MHF Contributor Galactus!