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Math Help - related rates help

  1. #1
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    related rates help

    The Question is:
    Water drains from the conical tank shown in the figure at the rate of 5ft^3/min.
    How fast is the water level dropping from when h=6ft?

    ok the relation between the variables h and r is: 5r/2=h

    And i know the answer to the question is .276 ft/min, but i don't know how you get that.

    What i did was do the derivative of the cone volume formula: (1/3)pi(r^2)h using the product rule and i differentiate the h as well. Plus i even substitute r with 2h/5 (since i want to know r in terms of h) after that i replace everything with the appropriate numbers and my answer is different. Can you guys show me how to do this step by step? Thank you.

    EDIT: there is an upside down cone with a base radius of 4in and height of 10 in and there is an r and h which represents the height of the water and the base radius of it as well. Also the question before that says: What is the relation between the variables h and r? And that is every part of the question.
    Last edited by maximade; November 23rd 2009 at 04:45 PM. Reason: clearer
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  2. #2
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    PLease, post all the information you were given, in order to help you better.
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  3. #3
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    Quote Originally Posted by maximade View Post
    The Question is:
    Water drains from the conical tank shown in the figure at the rate of 5ft^3/min.
    How fast is the water level dropping from when h=6ft?

    ok the relation between the variables h and r is: 5r/2=h

    And i know the answer to the question is .276 ft/min, but i don't know how you get that.

    What i did was do the derivative of the cone volume formula: (1/3)pi(r^2)h using the product rule and i differentiate the h as well. Plus i even substitute r with 2h/5 (since i want to know r in terms of h) after that i replace everything with the appropriate numbers and my answer is different. Can you guys show me how to do this step by step? Thank you.
    since h = \frac{5r}{2} , r = \frac{2h}{5}

    V = \frac{\pi}{3} \left(\frac{2h}{5}\right)^2 h

    V = \frac{4\pi}{75} h^3

    you know \frac{dV}{dt} = -5 \, ft^3/min ... take the time derivative of the above equation and determine \frac{dh}{dt} when h = 6
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