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?
Best Answer
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.