Chapter 1.5, Problem 43PS

Solution

To find: the time required to drain the pool when it is completely filled.

  $t=\frac{lw\sqrt{3h_0}}{6\pi d^2}$ .

Given:

For a swimming pool with rectangular base, the height h of water in the pool t seconds after it begins draining is given by $h={\left(\sqrt{h_0}-\frac{2\pi d^2\sqrt{3}}{lw}t\right)}^2$ , where l and w are the pool’s length and width, d is the diameter of the drain, and $h_0$ is the water’s initial height.

Calculation:

The height h of water in the pool t seconds after it begins draining is given by: $h={\left(\sqrt{h_0}-\frac{2\pi d^2\sqrt{3}}{lw}t\right)}^2$

Now, when the pool is completely drained, the height of water in the pool becomes 0, that is $h=0$ .

So, substituting $h=5$ in the given equation,

  $0={\left(\sqrt{h_0}-\frac{2\pi d^2\sqrt{3}}{lw}t\right)}^2$

Take square root of both sides,

  $0=\sqrt{h_0}-\frac{2\pi d^2\sqrt{3}}{lw}t$

This is equivalent to:

  $\sqrt{h_0}=\frac{2\pi d^2\sqrt{3}}{lw}t$

Solving for t,

  $t=\frac{lw\sqrt{h_0}}{2\pi d^2\sqrt{3}}$

Now to rationalize the denominator, multiply and divide the right-hand side by $\sqrt{3}$ .

  $\begin{array}{c}t=\frac{lw\sqrt{h_0}}{2\pi d^2\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}\\ =\frac{lw\sqrt{3h_0}}{2\pi d^2\cdot 3}\\ =\frac{lw\sqrt{3h_0}}{6\pi d^2}\end{array}$