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?
Best Answer
The differential equation is $\frac{dS}{dt} = -\frac{S}{20}$, which has solution $S=Ce^{\frac{-t}{20}}$. S(0) is given as 5, so $S(t)=5e^{\frac{-t}{20}}$ At t=20, this is $5e^{-1}$
Added: I just took the equation from your post. In a small amount of time, $\delta t$ you have $\frac{1}{2}\delta t$ pure water added and the same quantity of salty water removed. This represents $\frac{1}{20}\delta t$ of the volume of the salty water, so you should remove that much salt. The change in salt is $-\frac{1}{20}S\delta t$. This gives $\frac{dS}{dt} = -\frac{S}{20}$