Question

Suppose that $\mathbf{r}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{h}\mathbf{\right)}$defines the shape of a water clock for which the time marks are equally spaced. Use the differential equation in Problem 12 to find f(h) and sketch a typical graph of as a function of r. Assume that the cross-sectional area ${\mathbf{A}}_{\mathbf{h}}$of the hole is constant. [Hint: In this situation $\frac{\mathbf{dh}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\mathbf{a}$, where $\mathbf{a}\mathbf{>}\mathbf{0}$is a constant.] (reference equation in problem 12)

$\frac{\mathbf{dh}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\mathbf{c}\frac{{\mathbf{A}}_{\mathbf{h}}}{{\mathbf{A}}_{\mathbf{w}}}\sqrt{\mathbf{2}\mathbf{gh}}\mathbf{,}$

Short answer

So, the graph is shown as below:

Step-by-step solution

Definition of cross – section

Cross – section is a cutting or piece of something cut off at right angles to an axis also: a representation of such a cutting.

 Find the area of cross - section

The differential equation that models water draining out of a container is

$\frac{dh}{dt}=c_{A_w}^{A_h}\sqrt{2gh}$

The values $A_{\mathbf{h}}$and $A_{\mathbf{w}}$are the cross-sectional areas of the hole and the container, respectively.

Let $r=f\left(h\right)$define the radius of the container when the water is at height h. Then the cross-sectional area of the container is

$A=\pi \left(f\right(h){)}^2$

The radius of the hole is constant $r_{\mathbf{h}}$.

Then the cross-sectional area of the hole is $A_{\text{r}}=\pi {\left(r_{\text{h}}\right)}^2$.

Substitution

Substituting the expressions for the cross-sectional areas and the value of gravity into the differential equation yields,

$\frac{dh}{dt}=-8c\frac{r_h^2}{{\left(f\right(h\left)\right)}^2}\sqrt{h}$

In order for the water level to decline at a constant rate it means we must have the differential equation

$\frac{d\lambda }{dx}=-k$

Where k is a constant.

Simplify the expression

Therefore, we have

$$\begin{gathered} k=8c\frac{r_h^2}{{\left(f\right(h\left)\right)}^2}\sqrt{h} \\ \left(f\right(h){)}^2=\frac{8c{r}_h^2}{k}\sqrt{h} \\ f\left(h\right)=\sqrt{\frac{8c{r}_h^2}{k}\sqrt{h}} \end{gathered}$$

Simplifying the expression we get the shape of the container as

$f\left(h\right)=\frac{2r_{\text{h}}\sqrt{2c}}{\sqrt{k}}{h}^{\frac{1}{4}}$

Find the value of h 

Recall that $r=f\left(h\right)$.

Therefore,

$$\begin{gathered} r=\frac{2r_h\sqrt{2c}}{\sqrt{k}}{h}^4 \\ {h}^{-4}=\frac{2r_4\sqrt{2c}}{r\sqrt{k}} \\ h={\left(\frac{r\sqrt{k}}{2r_{\text{h}}\sqrt{2c}}\right)}^4 \\ h=\frac{r^4{k}^2}{64r_{\text{h}}^4{c}^2} \end{gathered}$$

Plot the graph

Let $k=\frac{2}{43200}$so that the water runs out at exactly 12 hours.

Let $c=0.6$and $r_{\mathbf{h}}=\frac{1}{32}$inches.

$h=2023r^4$

The resulting plot is shown below.