
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 $\displaystyle \frac{dV}{dt}=(0.6)\pi r^2\sqrt{2gy}$ (taking constriction into account) to determine when the tank will be empty.
(b) What is the radius of the bottom hole?
My working:
$\displaystyle \pi(8yy^2)\frac{dy}{dt}=(0.6)\pi r^2\sqrt{2gy}$
$\displaystyle \int(8y^{1/2}y^{3/2})dy=\int{0.6*\sqrt{2g}r^2}dt$
$\displaystyle \frac{16}{3}y^{3/2}\frac{2}{5}y^{5/2}=0.6*\sqrt{2g}r^2t+C$
$\displaystyle y(0)=4$
$\displaystyle C=\frac{16}{3}*8\frac{2}{5}*32=\frac{448}{15}$
$\displaystyle \frac{16}{3}y^{3/2}\frac{2}{5}y^{5/2}=0.6*\sqrt{2g}r^2t+\frac{448}{15}$
$\displaystyle y(1800)=2$
$\displaystyle \frac{16}{3}*2^{3/2}\frac{2}{5}*2^{5/2}=0.6*8*r^2*1800+\frac{448}{15}$
$\displaystyle 8640r^2=17.0445$
$\displaystyle r\approx 0.04442\ ft$