Differential equation for draining a pool

Click For Summary
SUMMARY

The discussion centers on solving a differential equation related to the draining of a pool, specifically the equation $$\frac{dD}{dt} = \frac{k - \sqrt{D}}{100}$$. The equilibrium solution is identified as $$D = k^2$$, which is stable. The participants conclude that for the pool to overflow, the condition is $$k > 2$$, while for it to empty, $$k = 0$$ is required. The relationship between the inflow rate and the outflow rate, which varies with the square root of the depth, is crucial for understanding the dynamics of the pool's water level.

PREREQUISITES
  • Understanding of differential equations, specifically first-order equations.
  • Familiarity with equilibrium points in dynamic systems.
  • Knowledge of the square root function and its implications in physical models.
  • Basic concepts of flow rates in fluid dynamics.
NEXT STEPS
  • Study the stability of equilibrium points in differential equations.
  • Explore the implications of varying inflow and outflow rates in fluid dynamics.
  • Learn about the applications of differential equations in real-world scenarios, such as fluid drainage systems.
  • Investigate the use of numerical methods for solving differential equations when analytical solutions are complex.
USEFUL FOR

Mathematicians, engineers, and students studying differential equations, particularly those interested in fluid dynamics and dynamic systems modeling.

Vishak95
Messages
19
Reaction score
0
Hi MHB. Can someone help me with this one please? I don't understand what the question is really saying...in particular part (c).

I tried to set up dD/dt = k - D^1/2 but it doesn't seem correct. Thanks.

View attachment 1589
 

Attachments

  • Differential equation.jpg
    Differential equation.jpg
    45.9 KB · Views: 118
Physics news on Phys.org
I have moved this topic to our Differential Equations sub-forum, as it is a better fit.
 
Your answer to part (a) is not quite correct. We know:

(1) $$\frac{dV}{dt}=\text{flow in}-\text{flow out}$$

We are given:

$$V=100D,\,\text{flow in}=k,\,\text{flow out}=\sqrt{D}$$

So, substituting these given values into (1), what do we obtain?
 
Ok, now I've got this:

$$\frac{dV}{dD}= 100$$

$$\frac{dV}{dt} = \frac{dV}{dD}\cdot \frac{dD}{dt}$$

Leading to:

$$\frac{dD}{dt}= \frac{k-\sqrt{D}}{100}$$
 
Looks good. :D
 
Okay, thanks :)

So now for part (b) I've got the only equilibrium solution:

$$D = k^{2}$$

Which is stable.

But I'm completely lost looking at part (c) :(
 
Your equilibrium point is correct, and is stable since:

i) For $$D<k^2$$ we have:

$$\frac{dD}{dt}>0$$

ii) For $$D>k^2$$ we have:

$$\frac{dD}{dt}<0$$

which shows that for all values of $D$, we must have:

$$\lim_{t\to\infty}D(t)=k^2$$

So, for part (c) you want to look at what values of $k$ cause the equilibrium point to satisfy the condition that the pool overflows or empties. Where would these equilibrium points be?
 
Ok, thanks.

The thing that's still tripping me up is this part:

$$D_{0} \in (0,4)$$

And the fact that right now D0 isn't in the equation...
 
Vishak said:
Ok, thanks.

The thing that's still tripping me up is this part:

$$D_{0} \in (0,4)$$

And the fact that right now D0 isn't in the equation...

$$D_0=D(0)$$

and we are given that:

$$0<D_0<4$$

Can you now state what values of $k$ will cause the pool to empty and overflow?
 
  • #10
I'm guessing that for k > 2, it will overflow and for 0 <or= k < 2 it will empty?
 
  • #11
Vishak said:
I'm guessing that for k > 2, it will overflow and for 0 <or= k < 2 it will empty?

You are correct that for $2<k$ the pool will overflow, since the equilibrium point will be greater than the depth of the pool.

In order for the pool to empty, we require the equilibrium point to be $D=0$, so we require $k=0$.

In other words, in order for the pool to completely empty, there can be no water flowing into the pool, given that the initial amount of water is greater than zero. If there is any water flowing into the pool, no matter how slowly, then the equilibrium point will be greater than zero. Because the rate at which the water leaks out varies as the square root of the depth, as the level of the water decreases, the rate at which it leaks out will decreases as well, until at some point the rate at which it leaks approaches the rate at which it is being pumped in, and we approach equilibrium.
 
  • #12
Thank you so much for that explanation! I understand it now :)
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 52 ·
2
Replies
52
Views
8K
  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K