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Math Help - another related rates

  1. #1
    Eater of Worlds
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    another related rates

    Here's another related rates problem if anyone would like to chew on it.

    A cylindrical beaker has radius r and height h. Water is pouring out as the beaker is tilted. Find a formula that relates the rate of change of the volume of water in the beaker and the rate of change of the angle the beaker makes with the vertical.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by galactus View Post
    Here's another related rates problem if anyone would like to chew on it.

    A cylindrical beaker has radius r and height h. Water is pouring out as the beaker is tilted. Find a formula that relates the rate of change of the volume of water in the beaker and the rate of change of the angle the beaker makes with the vertical.
    ill give it a try.. is it..

    V = \pi r^2 h - \pi r^3 \tan \theta, where \theta is the angle with respect to the vertical.. so
     dV = -\pi r^3 \sec^2 \theta d\theta
    Last edited by kalagota; November 10th 2007 at 04:38 AM.
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  3. #3
    Eater of Worlds
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    You're certainly on the right track, k.

    Except, r is constant, so we differentiate wrt theta.

    If we let, say, d = the depth of the water at the axis of the cylinder, then

    d=h-rtan{\theta}

    Then we have:

    V={\pi}r^{2}d

    V={\pi}r^{2}(h-rtan{\theta})

    V={\pi}r^{2}h-{\pi}r^{3}tan{\theta}

    r and h are constant, so differentiating wrt theta:

    \frac{dV}{dt}=-{\pi}r^{3}sec^{2}{\theta}\frac{d{\theta}}{dt}
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by kalagota View Post
    ill give it a try.. is it..

    V = \pi r^2 h - \pi r^3 \tan \theta, where \theta is the angle with respect to the vertical.. so
     dV = -3\pi r^2 \sec^2 \theta d\theta
    ah yeah, a typo error on my scratch, i didnt notice it while typing here.. i've edited it..
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