# Differential Equations Rate At Which Chemical Amount Changes

1. Feb 7, 2014

### embphysics

1. The problem statement, all variables and given/known data
Suppose that a large mixing tank initially holds 300 gallons of water in which 50 pounds of salt have been dissolved. Pure water is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at the same rate. Determine a differential equation for the amount of salt A(t) in the tank at time t > 0.What is A(0)?

2. Relevant equations

3. The attempt at a solution

The answer to this question is $\displaystyle \frac{dA}{dt} = - \frac{1}{100} A(t)$ Why is is so? Why is the differential equation proportional to the amount of salt at time t? Why couldn't to be, say, inversely proportional? What would the arguments be to show that it is in fact proportional, as opposed to any other alternatives?

I tried to reason through this myself. I said, suppose the rate at which the salt content changes is inversely proportional to the amount at time t. If initially there is a large amount of salt, then the rate at which the amount changes is initially small. Even though the rate is small, A(t) will still decrease, just slowly. So, as time passes, A(t) slowly diminishes. As A(t) decreases, the rate at which it continues to decrease becomes larger.

Through theses arguments I was hoping to show the unreasonableness of the differential equation being inversely proportional to the function. In spite of this, I am not really convinced.

I was hoping someone could provide me with some arguments as to why the differential equation is proportional to the function, as opposed to inversely, or proportional to the square, or the cube, etc.

2. Feb 7, 2014

### jackarms

It's hard to reason through it, but essentially it's directly proportional since when you remove solution, more of the chemical will be removed if it's more concentrated. If there's hardly any in the container, then when you remove some you're not removing much chemical. But if it's really concentrated, then you'll remove a lot of chemical. You can reason through it through the differential equation too:

The change is going to be the concentration of what's coming in times the amount coming in, minus the concentration of what's going out times what's going out. The concentration of the chemical coming in is zero, since it's pure water, so you can ignore that. The concentration of what's going out is going to be, since it's well mixed, the amount that's in there (that's $A$) divided by the amount of solvent (300), then multiply that by 3, since that's the number of gallons going out. So that's where the $-\frac{1}{100}A(t)$ comes from.

3. Feb 7, 2014

### LCKurtz

If $A(t)$ is the amount of salt in a 300 gallon tank at time $t$, then if the tank is well stirred, each gallon will have $\frac{A(t)}{300}$ pounds of salt. If $3$ gallons/minute are pouring out that would give $3\cdot \frac{A(t)}{300}$ gallons per minute leaving the tank.

[Edit:] I see Jackarms beat me to it.