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

We are given:

\(\displaystyle V=100D,\,\text{flow in}=k,\,\text{flow out}=\sqrt{D}\)

So, substituting these given values into (1), what do we obtain?

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\(\displaystyle \frac{dD}{dt}>0\)

\(\displaystyle \frac{dD}{dt}<0\)

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

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

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?

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\(\displaystyle D_0=D(0)\)Ok, thanks.

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

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

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

and we are given that:

\(\displaystyle 0<D_0<4\)

Can you now state what values of $k$ will cause the pool to empty and overflow?

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You are correct that for $2<k$ the pool will overflow, since the equilibrium point will be greater than the depth of the pool.I'm guessing that for k > 2, it will overflow and for 0 <or= k < 2 it will empty?

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.

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