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Math Help - Related Rates problem

  1. #1
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    Related Rates problem

    Hey everybody. I need help understanding this problem

    A solution is passing through a conical filter 24 cm deep and 16 cm across the top into a cylindrical vessel of diameter 12 cm.
    At what rate is the level of the solution in the cylinder rising if when the depth of the solution in the conical filter is 12 cm

    Its level is falling at the rate of 1 cm/min in the filter
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  2. #2
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    Quote Originally Posted by JohnShade123 View Post
    Hey everybody. I need help understanding this problem

    A solution is passing through a conical filter 24 cm deep and 16 cm across the top into a cylindrical vessel of diameter 12 cm.
    At what rate is the level of the solution in the cylinder rising if when the depth of the solution in the conical filter is 12 cm

    Its level is falling at the rate of 1 cm/min in the filter
    The volume(v) of solution in the conical filter is

    v=\frac{1}{3}\pi r^2h

    \frac{r}{h}=\frac{8}{24}

    r=\frac{h}{3}

    v=\frac{1}{3}\pi (\frac{h}{3})^2h=\frac{1}{27}\pi h^3

    \frac{dv}{dt}=\frac{1}{9}\pi h^2 \frac{dh}{dt}

    The level(h) of the solution in the conical filter is falling at the rate of 1 cm/min so

    \frac{dh}{dt}=-1

    When h=12

    \frac{dv}{dt}=\frac{1}{9}\pi (12)^2(-1)=-16\pi

    The volume(V) of solution in the cylinder is

    V=\pi(6)^2H=36\pi H

    \frac{dV}{dt}=36\pi \frac{dH}{dt}

    When h=12, \frac{dv}{dt}=-16\pi

    So \frac{dV}{dt}=16\pi

    16\pi=36\pi \frac{dH}{dt}

    Solve for \frac{dH}{dt}
    Last edited by ione; April 19th 2010 at 11:51 PM.
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  3. #3
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    Thanks for the help. It is actually much easier once you understand it
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