Question

A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1lb of salt per gallon is entering at a rat of 3 gal/min, and the misture is allowed to flow out of the tank at a rate of 2 gal/min. Find the amount of salt in the tank at any time prior to the instant when the solution begins to overflow. Find the concentration (in pounds per gallon) of salt in the tank when it is on the point of overflowing. Compare this concentration with the theoretical limiting concentration if the tank had infinite capacity.

How on earth...

Answer 1

If \(A(t)\) models the amount of salt in the solution, then \(A'(t)\) models the rate of change of \(A\), which is given by the equation
\[A'(t)=\Bigg(\text{rate in}\Bigg)\Bigg(\text{concentration in}\Bigg)-\Bigg(\text{rate out}\Bigg)\Bigg(\text{concentration out}\Bigg)\]
Here, you're given that the solution is coming in at a rate of 3 gal/min with a concentration of 1 lb/gal, and flowing out at a rate of 2 gal/min:
\[A'(t)=3-2\Bigg(\text{concentration out}\Bigg)\]
The concentration of the solution flowing out is given by
\[\frac{\text{amount of salt in tank at time }t}{\text{volume of water in tank at time }t}=\frac{A(t)}{200+(3-2)t}\]
So you have the linear ODE
\[A'(t)=3-\frac{2}{200+t}A(t)\\
A'(t)+\frac{2}{200+t}A(t)=3\]

Answer 2

The tank will overflow when the volume exceeds the capacity of the tank, 500 gal. This occurs when
\[500<200+t~~\Rightarrow~~t>300\]

Answer 3

So first, solve for \(A(t)\) for values of \(t\le300\) ("prior to overflow")

Answer 4

The concentration of salt in the tank at the point of overflow is thus given by \(A(300)\).

For the infinite capacity part, you have to find the limiting value of \(A(t)\), and simply compare to \(A(300)\).

Make sense?

Answer 5

Refer to this link if you're still having trouble: http://tutorial.math.lamar.edu/Classes/DE/Modeling.aspx

Answer 6

I am at this point.


\(\large{\dfrac{dx}{dt} +\dfrac{2}{t+200} x=3}\)

\(\large{\frac{dx}{dt}((t+200)^2)=3(t+200)^2}\)

Answer 7

Have you learned how to solve linear ODE's? You have to find an integrating factor.

Answer 8

That is what I did. The integrating factor was,

\(\mu(t)=(t+200)^2\)

Answer 9

Okay, looks right:
\[\large\mu(t)=e^{\int\frac{2}{200+t}~dt}=e^{2\ln|200+t|}=(200+t)^2\]
Multiplying both sides of the equation by this, you have
\[(200+t)^2x'+2(200+t)x=3(200+t)^2\]
Note that the left side is a derivative:
\[\frac{d}{dt}\left[(200+t)^2x\right]=3(200+t)^2\\
(200+t)^2x=3\int(200+t)^2~dt\]

Answer 10

\(x=t+200\)?

Answer 11

+C

Answer 12

Let's see...
\[(200+t)^2x=(200+t)^3+C\\
x=(200+t)+\frac{C}{(200+t)^2}\]
Close!

Answer 13

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
Now, you're given an IVP; initially, the concentration of the solution in the tank is \(A(0)=\color{red}{100}\). Use this to solve for \(C\).
\(\color{blue}{\text{End of Quote}}\)