Question

Suppose that a glass tank has the shape of a cone with circular cross section as shown in Figure 3.2.10. As in part (a) of Problem 33, assume that $\mathbf{h}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{ft}$corresponds to water filled to the top of the tank, a hole in the bottom is circular with radius $\frac{\mathbf{1}}{\mathbf{32}}$in., $\mathbf{g}\mathbf{=}\mathbf{32}\mathbf{ft}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$, and $\mathbf{c}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{6}$. Use the differential equation in Problem 12 to find the height h(t) of the water.

(b) Can this water clock measure 12 time intervals of length equal to 1 hour? Explain using sound mathematics.

Short answer

(a) So, the height of the water is $h=(-0.0003255t+4\sqrt{2}{)}^5$.

(b) Therefore, $17378.96851$seconds is equivalent to 4.83 hours. The conical container cannot measure time up to 12 hours.

Step-by-step solution

Definition of differential equation

A differential equation is an equation that contains one or more functions with its derivatives.

Find the cross- sectional area(a)

The differential equation that models water draining out of the conical container is

$\frac{dh}{dt}=-e_{a_w}^{A_h}\sqrt{2gh}$

We are given$c=0.6,g$ is the force of gravity, $A_{\text{h}}$and A are the cross-sectional areas of the hole and the container, respectively.

If the water level of the container is at height h then the radius of the circle defined by the surface of the water is $r=h_{/}2$.

Therefore, the cross-sectional area of the container is

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

Differentiation of equation

Substituting the given values and combining terms yields,

$\frac{dh}{dt}=-0.00013021h^{-\frac{3}{2}}$

We can solve the differential equation by separation of variables.

$\begin{array}{rcl}{h}^3& =& -0.00013021d^3\\ h^{\frac{3}{2}}dh& =& \int -0.00013021d\\ \frac{2}{5}{h}^{\frac{5}{2}}& =& -000013021t+k\\ h& =& {\left(-0.0003255t+k_2\right)}^{\frac{2}{3}}\end{array}$

Use initial condition

We use the initial condition $h\left(0\right)=2$to solve for the constant $k_{\mathbf{2}}$.

$$\begin{gathered} 2={\left(-0.0003255\left(0\right)+k_2\right)}^2 \\ 2=k_2^{\frac{2}{3}} \\ 4\sqrt{2}=k_2 \end{gathered}$$

The resulting solution to the differential equation is $h=(-0.0003255t+4\sqrt{2}{)}^5$.

(b) Step 1: Find the value of 

In order to find if the tank of water can last for at least 12 hours we can concentrate on finding at what time the water runs out, in other words when the height of the water is zero.

$\begin{array}{rcl}0& =& (-0.0003255t+4\sqrt{2}{)}^{\frac{5}{2}}\\ 0.0003255t& =& 4\sqrt{2}\\ t& =& 1737896851\end{array}$

Recall that the solution to the differential equation is in terms of seconds and feet.