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A very large tank initially contains 15 gallons of saltwater containing 6 pounds of salt. Saltwater containing 1 pound of salt per gallon is pumped in to the top of the tank a rate of 2 gallons per minute while a well mixed solution leaves the bottom of the tank at a rate of 1 gallon per minute. What does the solution to the differential equation predict about the concentration of salt in the tank in the long run.

I assumed this was an infinite tank since we are talking about an infinite time period.

Here is my differential equation where $V$ is the volume and $S$ is the weight of salt. $$V=15+2t-t=15+t$$

$$\frac{dS}{dt}=2-\frac{S}{15+t}$$

Solving it gives $$S=\frac{t^2+30t+C}{t+15}$$

We know concentration is going to be the ratio of amount of salt $S$ and the volume $V$.

$$\lim_{t \rightarrow\infty}{\frac{S}{V}}= \lim_{t \rightarrow\infty}\frac{t^2+30t+90}{(t+15)^2} = 1$$ So I believe that the concentration will go to $1\frac{lbs}{gal}$ in the long run. This seems intuitive to me since the concentration coming in is $1$ and will eventually dilute the rest.

However, my teacher says that eventually, the tank will be full of salt. She says the limit of $1$ corresponds to the percentage of the tank full of salt?? Who is wrong, and why?

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  • $\begingroup$ You have not defined $S$, and have not explained the ODE ${dS\over dt}=\ldots$. But otherwise your explanations are correct. In particular your limit $1$ means that in the end we shall have $1$ pound of salt per gallon of fluid. $\endgroup$ – Christian Blatter Feb 27 at 19:40
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    $\begingroup$ @ChristianBlatter I should clarify. $S$ is the lbs of salt $\endgroup$ – Jac Frall Feb 27 at 19:45
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You are right and your teacher is wrong.

$$\lim_{t \rightarrow\infty}{\frac{S}{V}}= 1$$

means that in the long run the concentration approaches 1 lb per gallon. The units of $\frac{S}{V}$ are $[lb/gal]$ and not, as your teacher proposed, a fraction.

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