Related Rates Problems

**haku** · November 29th 2007, 07:20 AM

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.

**janvdl** · November 29th 2007, 07:41 AM

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

**haku** · November 29th 2007, 08:21 AM

Does this mean the derivative would be: 2r

Is this correct, what happens to the height?

**haku** · November 29th 2007, 08:36 AM

Okay, so it should be 2r

h assuming

and the height are constants. Or do you get rid of them when you differentiate, to give 2r. Thanks

**haku** · November 29th 2007, 09:00 AM

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.

**janvdl** · November 29th 2007, 09:15 AM

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 $r = 5$

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

**haku** · November 29th 2007, 09:20 AM

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

**janvdl** · November 29th 2007, 09:21 AM

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

**galactus** · November 29th 2007, 09:27 AM

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.

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

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

**janvdl** · November 29th 2007, 09:36 AM

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!

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