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Math Help - Torricelli's law (2)

  1. #1
    MHF Contributor alexmahone's Avatar
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    Torricelli's law (2)

    A water tank has the shape obtained by revolving the parabola x^2=by around the y-axis. The water depth is 4 ft at 12 noon, when a circular plug in the bottom of the tank is removed. At 1 P.M. the depth of the water is 1 ft.

    (a) Find the depth y(t) of water remaining after t hours.

    (b) When will the tank be empty?

    (c) If the initial radius of the top surface of the water is 2 ft, what is the radius of the circular hole in the bottom?

    My working:

    (a) A(y)\frac{dy}{dt}=-k\sqrt{y}

    \pi x^2\frac{dy}{dt}=-k\sqrt{y}

    \pi by \frac{dy}{dt}=-k\sqrt{y}

    \sqrt{y} dy=-\frac{k}{\pi b} dt

    y^{3/2}=-\frac{kt}{\pi b}+C

    y(0)=4

    C=8

    y^{3/2}=-\frac{kt}{\pi b}+8

    y(1)=1

    1=-\frac{k}{\pi b}+8

    \frac{k}{\pi b}=7

    y^{3/2}=-7t+8

    y=(8-7t)^{2/3}

    (b) y=0

    t=\frac{8}{7}\ h

    \frac{1}{7}\ h=(8+\frac{4}{7})\ min=8\ min\ 34\ s

    The tank will be empty at 1:08:34 P.M.

    (c) 2^2=b*4

    b=1

    k=7\pi b=7\pi

    k=a\sqrt{2g}

    7\pi=a\sqrt{2*32*(3600)^2}=8*3600a\ [32\ ft/s^2=32*(3600)^2\ ft/h^2]

    a=\frac{7\pi}{8*3600}

    \pi r^2=\frac{7\pi}{8*3600}

    r^2=\frac{7}{8*3600}

    r=\frac{1}{120}\sqrt{\frac{7}{2}}\ ft=\frac{1}{10}\sqrt\frac{7}{2}\ in

    My textbook says that my answers for (a) and (b) are correct. But for (c), it has r=\frac{1}{60}\sqrt{\frac{7}{12}}\approx0.15\ in. Where have I gone wrong?
    Last edited by alexmahone; September 5th 2011 at 07:29 AM.
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  2. #2
    MHF Contributor
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    Re: Torricelli's law (2)

    Quote Originally Posted by alexmahone View Post
    A water tank has the shape obtained by revolving the parabola x^2=by around the y-axis. The water depth is 4 ft at 12 noon, when a circular plug in the bottom of the tank is removed. At 1 P.M. the depth of the water is 1 ft.

    (a) Find the depth y(t) of water remaining after t hours.

    (b) When will the tank be empty?

    (c) If the initial radius of the top surface of the water is 2 ft, what is the radius of the circular hole in the bottom?

    My working:

    (a) A(y)\frac{dy}{dt}=-k\sqrt{y}

    \pi x^2\frac{dy}{dt}=-k\sqrt{y}

    \pi by \frac{dy}{dt}=-k\sqrt{y}

    \sqrt{y} dy=-\frac{k}{\pi b} dt

    \frac{2}{3} y^{3/2}=-\frac{kt}{\pi b}+C ... this line (corrected)
    ...
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