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Math Help - Torricelli's law (3)

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    MHF Contributor alexmahone's Avatar
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    Torricelli's law (3)

    Suppose that an initially full hemispherical water tank of radius 1 m has its flat side as its bottom. It has a bottom hole of radius 1 cm. If this bottom hole is opened at 1 P.M., when will the tank be empty?

    My working:

    A(y)=\pi r^2=\pi(1-y^2)

    \pi(1-y^2) \frac{dy}{dt}=-\pi\left(\frac{1}{100}\right)^2 \sqrt{2*9.8y}

    \int(y^{-1/2}-y^{3/2})dy=-\int{\frac{\sqrt{19.6}}{10^4} dt

    2y^{1/2}-\frac{2}{5}y^{5/2}=-\frac{\sqrt{19.6}}{10^4} t+C

    y(0)=1

    C=2-\frac{2}{5}=\frac{8}{5}

    2y^{1/2}-\frac{2}{5}y^{5/2}=-\frac{\sqrt{19.6}}{10^4} t+\frac{8}{5}

    When y=0,

    t=\frac{80000}{5*\sqrt{19.6}}\approx 3614\ s

    The tank is empty about 14 seconds after 2:00 P.M.
    Last edited by alexmahone; September 5th 2011 at 02:25 PM. Reason: Figured it out!
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