Water depth in conical pond with evaporation

[DivGradCurl]

I just need some help with this problem. Thank you.

"A pond forms as water collects in a conical depression of radius $$a$$ and depth $$h$$. Suppose that water flows in at a constant rate $$k$$ and is lost through evaporation at a rate proportional to the surface area.

(a) Show that the volume $$V(t)$$ of water in the pond at time $$t$$ satisfies the differential equation

$$\frac{dV}{dt} = k - \alpha \pi \left( \frac{3a}{\pi h} \right) ^{2/3} V ^{2/3}$$

where $$\alpha$$ is the coefficient of evaporation.

(b) Find the equilibrium depth of water in the pond. Is the equilibrium asymptotically stable?

(c) Find a condition that must be satisfied if the pond is not to overflow."

My work:

(a)

Consider the following

$$\frac{dV}{dt} = k - \alpha \pi \underbrace{\left( \frac{3a}{\pi h} \right) ^{2/3} V ^{2/3}} _{r^2}$$

where $$r$$ is the radius of the pond. Thus, we have

$$r^2 = \left( \frac{3a}{\pi h} \right) ^{2/3} V ^{2/3}$$

$$r^3 = \frac{3aV}{\pi h}$$

$$V = \frac{1}{3} \pi r^2 \left( \frac{hr}{a} \right) = \frac{1}{3} \pi r^2 L$$

which satisfies the differential equation.

(b)

$$\frac{dV}{dt} = 0$$

$$k - \alpha \pi r^2 = 0$$

$$r = \sqrt{\frac{k}{\alpha \pi}}$$

Then, the equilibrium depth of water in the pond is

$$L = \frac{h}{a} \sqrt{\frac{k}{\alpha \pi}}$$

In this particular case, I don't know how to show whether or not it is asymptotically stable.

(c)

$$L = \frac{h}{a} \sqrt{\frac{k}{\alpha \pi}} \leq h$$


Any help is highly appreciated.

 

[DivGradCurl]

An equilibrium depth of water in the pond implies that there is an equilibrium volume. Thus, a direction field can show whether or not it is asymptotically stable, which is what I have at

http://mygraph.cjb.net/

Here are the (random) values that I used to plot it:

$$\left\{ \begin{array}{ll} k = 1 \\ \alpha = 0.6 \\ a = 0.3 \\ h = 0.5 \end{array} \right.$$

Based on this information, I'd say the equilibrium is asymptotically stable.

Thanks anyway

 

[thepatientmental]

Thank you.



Great work so far! For part (b), we can determine the stability of the equilibrium by analyzing the sign of the derivative of the right-hand side of the differential equation at the equilibrium point. In this case, we have

\frac{d}{dV} \left( k - \alpha \pi \left( \frac{3a}{\pi h} \right) ^{2/3} V ^{2/3} \right) = -\frac{2}{3} \alpha \pi \left( \frac{3a}{\pi h} \right) ^{2/3} V ^{-1/3}

Substituting in the equilibrium point r = \sqrt{\frac{k}{\alpha \pi}}, we get

\frac{d}{dV} \left( k - \alpha \pi r^2 \right) = -\frac{2}{3} \alpha \pi \left( \frac{3a}{\pi h} \right) ^{2/3} \left( \frac{k}{\alpha \pi} \right) ^{-1/3} = -2\left( \frac{a}{h} \right) ^{2/3}

Since \left( \frac{a}{h} \right) ^{2/3} is always positive, we can see that the derivative is always negative at the equilibrium point. This means that the equilibrium point is stable, but we cannot determine if it is asymptotically stable without further information about the initial conditions.

For part (c), you are correct in stating that the condition for the pond to not overflow is

L = \frac{h}{a} \sqrt{\frac{k}{\alpha \pi}} \leq h

This means that the equilibrium depth of water must be less than or equal to the depth of the cone. If the equilibrium depth is greater than the cone's depth, then the pond will overflow.