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Math Help - Find Rate of Change of Water Depth

  1. #1
    Super Member nycmath's Avatar
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    Find Rate of Change of Water Depth

    A hemispherical water tank with radius 6 meters is filled to a depth of h meters. The volume of the tank is given to be V = (1/3)pih(108-h^2), where 0 < h < 6. If water is being bumped into the tank at a rate of 3 cubic meters per minute, find the rate of change of the depth of the water when h = 2 meters.

    My Work:

    dr/dt = 3
    I need dv/dt when h is 2.
    I then replaced h by r/3 and proceeded to substituted
    r/3 for h in the given formula before taking the implicit derivative.

    I simplified and found the new volume equation to be V = 12pi - (pi/81)*r^3.

    I differentiated my new volume formula implicitly, substituted r = 6 or the given radius of the water tank and simplified.

    My Answer is dv/dt = 32pi meters per minute.

    Book's answer is (3/32pi) meters per minute.

    How do I find the right answer? What are the steps for this question?
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  2. #2
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    Re: Find Rate of Change of Water Depth

    I haven't looked at your problem very hard yet but
    "If water is being bumped into the tank at a rate of 3 cubic meters per minute"
    This is talking about cubic metres which is a volume
    So i think dv/dt=3
    and you want to find
    dh/dt when h=2

    I got the same answer as the book.
    Last edited by Melody2; November 19th 2013 at 06:13 PM. Reason: I worked out the answer
    Thanks from nycmath
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  3. #3
    Super Member nycmath's Avatar
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    Re: Find Rate of Change of Water Depth

    It was the wrong set up on my part. It's hard knowing what we have and what we need to find from related rates applications. I will need to practice more questions.
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    Re: Find Rate of Change of Water Depth

    Do you need any more help?
    Once you get the hang of these questions they are quite easy.
    Always use the units to help you know what you have been given and what you need to find.
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