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Here's the question, from my $10 mathbook that everyone is growing to love and hate:

A saltwater solution initially contains 5lb of salt in 10 gal of fluid. If water flows in at 0.5 gal/min and the mixture flows out at the same rate, how much salt is present after 20 mins?

I get that this is a differential equation problem, and that the quantity we are observing is the concentration of the salt in the water.

The rate of change of concentration ($\frac{dSolute}{dt}$) is proportional to: the current concentration, and the rate at which the fluid is (being replaced) by fresh water.

And so I of course have $y =y_0 e^{kt}$.

How do I think about this problem?

There is a solution given, which is

$ \frac{dS}{dt} = -\frac{1}{2}(\frac{S}{10})$

At $t=20$, $S=5e^{-1}$

I can guess where the constants came from ($-\frac{1}{2}$, 10) but I don't see what's going on here, especially with the $S=5e^{-1}$ quantity.

Working backwards, it says

$$\frac{\mbox{change in concentration}}{\mbox{unit of time}} = \mbox{(rate at which we are losing solution)}\left(\frac{\mbox{current salt concentration}}{\mbox{volume of water}}\right).$$

Why is that the formula?

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    Hope you don't mind the edit; this way the entire formula fits (at least in my window) without the need for a horizontal scrollbar.2011-02-15

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