Chapter 10.1, Problem 53E

To determine

We need to show that

a. The water level at time t is $y\left(t\right)={\left(y_0{}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{3aB\sqrt{2g}}{2\pi }t\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$ where $y_0$ is the water level at time $t=0$

b. If the total volume of water in the tank has volume V at time $t=0$, then $y_0=\sqrt{\raisebox{1ex}{$2aV$}\!\left/ \!\raisebox{-1ex}{$\pi $}\right.}$.

c. The tank is empty at time $t_e=\left(\frac{2}{3B\sqrt{g}}\right){\left(\frac{2\pi V^3}{a}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$.