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Math Help - Rate of Change of a Cone

  1. #1
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    Rate of Change of a Cone

    I'm having a hard time figuring out exactly what to do:

    "A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep."

    Thanks in advance
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  2. #2
    Senior Member
    Joined
    Dec 2008
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    We first write the water volume as a function of the depth, h:

    V=\frac{1}{3}\pi r^2 h=\frac{1}{3}\pi\left(5\cdot\frac{h}{12}\right)^2 h=\frac{5^2}{3\cdot 12^2}\pi h^3.

    Now we may differentiate both sides using the Chain Rule and solve for the rate of depth increase \frac{dh}{dt}.
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