Chapter 8.2, Problem 64ES

Suppose that a tank containing a liquid is vented to the air at the top and has an outlet at the bottom through which the liquid can drain. It follows from Torricelli's law in physics that if the outlet is opened at time $t=0,$ then at each instant the depth of the liquid $h\left(t\right)$ and the area $A\left(h\right)$ of the liquid's surface are related by

$A\left(h\right)\frac{dh}{dt}=-k\sqrt{h}$

where k is a positive constant that depends on such factors as the viscosity of the liquid and the cross-sectional area of the outlet. Use this result in these exercises, assuming that h is in feet, $A\left(h\right)$ is in square feet, and t is in seconds.

Follow the directions of Exercise 63 for the cylindrical tank in the accompanying figure, assuming that the tank is filled to a depth of 4 feet at time $t=0$ and that the constant in Torricelli's law is $k=\mathrm{0.025.}$