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Math Help - word problem using implicit differentiation

  1. #1
    Junior Member LexiRae's Avatar
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    word problem using implicit differentiation

    fluid is entering a spherical tank with a 6 inch radius at a constant rate of 30 pi cubic inches per hour. when the water is h feet deep, the volume of the water in the tank is given by:
    V = [(pi)(h^2)/3] (18-h)
    what is the rate at which the depth of the water in the tank is increasing at the moment when the water is 2 inches deep?

    thanks so much i really appreciate it! im so confused on this one!
    Last edited by LexiRae; November 28th 2008 at 12:32 PM. Reason: wrote the wrong thing
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by LexiRae View Post
    fluid is entering a spherical tank with a 6 inch radius at a constant rate of 30 pi cubic inches per hour. when the water is h feet deep, the volume of the water in the tank is given by:
    V = [(pi)(h^2)/3] (18-h)
    what is the rate at which the depth of the water in the tank is increasing at the moment when the water is 2 inches deep?

    thanks so much i really appreciate it! im so confused on this one!
    First lets simplfy the equation a bit

    V=6\pi h^2-\frac{\pi h^3}{3}

    Now taking the derviative with respect to time gives

    \frac{dV}{dt}=12 \pi h \frac{dh}{dt}-\pi h^2 \frac{dh}{dt}=\left(12 \pi h-\pi h^2 \right) \frac{dh}{dt}

    Now so we can solve for \frac{dh}{dt} and plug in what we know

    \frac{dh}{dt}=\frac{\frac{dV}{dt}}{12 \pi h-\pi h^2}=\frac{30}{12\pi (2)-\pi (2)^2 }=\frac{30}{20 \pi}=\frac{3}{2\pi}
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