# Related Rates Problems

• Nov 29th 2007, 07:20 AM
haku
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.
• Nov 29th 2007, 07:41 AM
janvdl
Quote:

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.

$\displaystyle \pi$ is a constant
• Nov 29th 2007, 08:21 AM
haku
Does this mean the derivative would be: 2rhttp://www.mathhelpforum.com/math-he...7a9dac6d-1.gif
Is this correct, what happens to the height?
• Nov 29th 2007, 08:36 AM
haku
Okay, so it should be 2rhttp://www.mathhelpforum.com/math-he...7a9dac6d-1.gifh assuming http://www.mathhelpforum.com/math-he...7a9dac6d-1.gif and the height are constants. Or do you get rid of them when you differentiate, to give 2r. Thanks
• Nov 29th 2007, 09:00 AM
haku
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.
• Nov 29th 2007, 09:15 AM
janvdl
I got it! Use implicit differentiation!

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

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

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

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

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

But $\displaystyle r = 5$

$\displaystyle h' = \frac{1}{25} = 0,04$
• Nov 29th 2007, 09:20 AM
haku
Great! Thanks, that has helped me a lot in understanding these problems. :)
• Nov 29th 2007, 09:21 AM
janvdl
Quote:

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 :D
• Nov 29th 2007, 09:27 AM
galactus
r and Pi are constants, as stated by Janvdl.

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

But we need dh/dt.

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

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

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

$\displaystyle \frac{dh}{dt}=\frac{1}{25} \;\ cm/min$
• Nov 29th 2007, 09:36 AM
janvdl
Quote:

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

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

But we need dh/dt.

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

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

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

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

Just one more post and you become an MHF Contributor Galactus! ;)