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Math Help - Related Rates Problems

  1. #1
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    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.
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  2. #2
    Bar0n janvdl's Avatar
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    Quote Originally Posted by haku View Post
    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
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  3. #3
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    Does this mean the derivative would be: 2r
    Is this correct, what happens to the height?
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  4. #4
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    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
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  5. #5
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    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.
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  6. #6
    Bar0n janvdl's Avatar
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    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
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  7. #7
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    Great! Thanks, that has helped me a lot in understanding these problems.
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  8. #8
    Bar0n janvdl's Avatar
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    Quote Originally Posted by haku View Post
    Great! Thanks, that has helped me a lot in understanding these problems.
    Thank you too, you got me to remember Calculus again
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  9. #9
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    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
    Last edited by galactus; November 29th 2007 at 09:28 AM. Reason: Janvdl beat me to it :)
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  10. #10
    Bar0n janvdl's Avatar
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    Quote Originally Posted by galactus View Post
    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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