Torricelli's law (2)

• Sep 5th 2011, 05:35 AM
alexmahone
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?
• Sep 5th 2011, 07:13 AM
skeeter
Re: Torricelli's law (2)
Quote:

Originally Posted by alexmahone
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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