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Math Help - Error In Problem Creation?

  1. #1
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    Related Rates Problem

    The problem is as follows:

    A conical tank with its vertex down is 12 ft high and 12 ft in diameter at the top. Water is being pumped in at the rae of 4 cubic feet per minute. Find the rate at which the water level is rising: a) when the water is 2 ft deep; b) and 8 ft deep.

    Since they did not give me the rate at which the radius changes, I used the relationship 3r = h, giving me h/3 = r; then I squared each side to and substituted into the equation--which is given-- v = (\pi/3)(h^3/9) I took the derivative with respect to time and got \frac{dh}{dt} = \frac{dV}{dt}\frac{9}{\pi h^2}

    For a) I got 9/pi, but the answer is 4/pi. Did I set the equations up wrong, how about my derivative?
    Last edited by Bashyboy; May 13th 2012 at 12:16 PM.
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  2. #2
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    Re: Error In Problem Creation?

    r = 6 , h = 12

    \frac{r}{h} = \frac{6}{12} = \frac{1}{2}

    so ... h = 2r \implies r = \frac{h}{2}

    V = \frac{\pi}{3} r^2 h

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

    V = \frac{\pi}{12} h^3

    try again
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