Question 2.7.2: For a cylindrical tank of height 2m and radius 0.3 m, filled...

In this situation, the cross sectional area A(h) of the tank at height h is constant because each is a circle of radius 0.3, so that A(h) = 0.09π. In addition, the area of the hole in square meters is a = π(0.02)^{2} = 0.0004π, and the gravitational constant is g =9.8 m/s^{2}. Since we have already established that A(h)dh/dt =−a \sqrt {2gh}, we therefore conclude that h satisfies the equation

0.09π \frac {dh}{dt}=−0.0004π \sqrt {19.6h}

Simplifying, it follows that

\frac {dh}{dt}=−0.019676 \sqrt {h}

Separating variables, we have

h^{−1/2}dh=−0.019676dt

and upon integrating, it follows that

2h^{1/2} =−0.019676t +C

Thus,

h(t ) = (C0 −0.009838t )^{2}

Because h(0) = 2, C_{0} =\sqrt {2}. Furthermore, with h(t ) = (\sqrt {2}−0.009838t )^{2}, we can see that h(t ) = 0 when t = 143.75 sec, at which time the tank is empty. A plot of h(t ) confirms precisely the behavior observed in the direction field in figure 2.12.