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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. –  Arturo Magidin Feb 15 '11 at 21:20
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2 Answers 2

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 $\frac{dS}{dt} = -\frac{S}{20}$

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Why is it $-\frac{S}{20}$? Can you put it in words? –  bobobobo Feb 15 '11 at 17:33
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Ross's answer was good, but here's a bit more clarity.

The question is "how much salt is lost after 20 minutes (of above described loss scenario)?"

In $y = y_0 e^{kt}$, we have to find $k$.

So how much salt is lost?

Well,

  • You lose $\frac{1}{2}$ gallons of water per minute
  • The concentration of salt per gallon is $\frac{S}{10}$, because there is S salt in the 10 gallons of water (at any instant of time)
  • Thus the net rate of change of salt in the vat is $ -\frac{1}{2} $ [gallons/min] $ \frac{S}{10} $ [salt/gallon] $ = -\frac{S}{20} $ [salt/min]

  • Thus the differential equation is

$ \frac{dS}{dt} = -\frac{S}{20} $

Matching with

$ \frac{dy}{dt} = ky $

for a differential equation, this means k = $-\frac{1}{20}$.

Then

$y = 5 e^{-\frac{1}{20}t}$

$t = 20$ so

$y = 5 e^{-1}$

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