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?