# Hemispherical water tank

• September 5th 2011, 10:17 PM
alexmahone
Hemispherical water tank
Consider an initially full hemispherical water tank of radius 4 ft, where the radius r of its circular bottom hole is now unknown. At 1 P.M. the bottom hole is opened and at 1:30 P.M. the depth of water in the tank is 2ft.

(a) Use Torricelli's law in the form $\frac{dV}{dt}=-(0.6)\pi r^2\sqrt{2gy}$ (taking constriction into account) to determine when the tank will be empty.

(b) What is the radius of the bottom hole?

My working:

$\pi(8y-y^2)\frac{dy}{dt}=-(0.6)\pi r^2\sqrt{2gy}$

$\int(8y^{1/2}-y^{3/2})dy=-\int{0.6*\sqrt{2g}r^2}dt$

$\frac{16}{3}y^{3/2}-\frac{2}{5}y^{5/2}=-0.6*\sqrt{2g}r^2t+C$

$y(0)=4$

$C=\frac{16}{3}*8-\frac{2}{5}*32=\frac{448}{15}$

$\frac{16}{3}y^{3/2}-\frac{2}{5}y^{5/2}=-0.6*\sqrt{2g}r^2t+\frac{448}{15}$

$y(1800)=2$

$\frac{16}{3}*2^{3/2}-\frac{2}{5}*2^{5/2}=-0.6*8*r^2*1800+\frac{448}{15}$

$8640r^2=17.0445$

$r\approx 0.04442\ ft$