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Thread: 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 $\displaystyle x^2=by$ 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) $\displaystyle A(y)\frac{dy}{dt}=-k\sqrt{y}$

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

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

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

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

    $\displaystyle y(0)=4$

    $\displaystyle C=8$

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

    $\displaystyle y(1)=1$

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

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

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

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

    (b) $\displaystyle y=0$

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

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

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

    (c) $\displaystyle 2^2=b*4$

    $\displaystyle b=1$

    $\displaystyle k=7\pi b=7\pi$

    $\displaystyle k=a\sqrt{2g}$

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

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

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

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

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

    My textbook says that my answers for (a) and (b) are correct. But for (c), it has $\displaystyle r=\frac{1}{60}\sqrt{\frac{7}{12}}\approx0.15\ in$. Where have I gone wrong?
    Last edited by alexmahone; Sep 5th 2011 at 06:29 AM.
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  2. #2
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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 $\displaystyle x^2=by$ 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) $\displaystyle A(y)\frac{dy}{dt}=-k\sqrt{y}$

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

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

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

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