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### Salt Concentration

```
Date: 01/23/97 at 13:11:47
From: Ruth Carver
Subject: calculus question

Dr. Math,

We're preparing for the calculus BC exam and are having difficulty
with this question.

Tank A contains 100 gallons of fresh water. Tank B contains 100
gallons of a brine solution with a salt concentration of 2 lb/gal.
Both tanks are stirred to keep their contents uniform. The contents of
tank A are pumped into tank B at 3 gal/min and the contents of tank B
are pumped into tank A at the same rate. Find the concentration of
salt in each tank at any time. Find the equilibrium concentration.

Calc 2BC, Mr. Oelker's Class, Germantown Academy
```

```
Date: 01/24/97 at 09:05:41
From: Doctor Jerry
Subject: Re: calculus question

Hello Mr. Oelker's Class,

If A and B denote the pounds of salt in tanks A and B at any time t,
then A(0) = 0 and B(0) = 200.  Moreover, A+B = 200 for all t, since
salt is conserved. At any time t, the rates A' and B' (dA/dt and
dB/dt) are:

A' = -(A/100)*3 + (B/100)*3
B' =  (A/100)*3 - (B/100)*3

We need just one of these equations since we know A+B = 200. Suppose
we look at the first (after replacing B by 200-A). We have:

100A' = 6(100-A)
A(0) = 0

This is a variables-separable differential equation.  Its solution is:

A = 100(1-e^(3t/50))

-Doctor Jerry,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 03/11/97 at 11:06:05
From: Ruth Carver

I think that your answer to the problem is incorrect because you can't
find the concentration in each tank at any time since each is
dependent upon the other.  The amount of salt in tank A depends upon
the amount of salt in tank B which again depends upon the amount of
salt in tank A.

Hope you can help us,
The GA Calc 2BC class.
```

```
Date: 03/11/97 at 13:15:50
From: Doctor Ken

Hi there -

Actually, I think Dr. Jerry's answer is correct.  To understand why,
there are a few things you have to understand: How is it possible to
solve a problem like this?  Why is this solution correct?  What does
it mean?

The first question is how an answer to this problem is possible.
You say that it shouldn't be possible because the concentrations in
the two tanks depend on each other in a complex way.  But here's
something that might convince you that it _is_ possible: you can set
up two tanks, give them the concentrations described in the problem,
stir them a lot, pump liquid between them, and at no point will the
universe explode. In fact, the concentrations in the tanks will
follow a quite precise course of action: the salt concentration in
tank A will go up, and the salt concentration in tank B will go down.
It may be _difficult_ to find an equation for the concentrations, but
it's not _impossible_, because Dr. Jerry did it.

In fact, here's a principle you should keep in mind when doing math.
If somebody's shown you a result or a proof, you can't say it's
incorrect unless you have a specific counterexample or you can point
out a specific place in their work that's incorrect. Anything less
than that doesn't constitute a refutation of their work.

This is NOT to say that you shouldn't question other people's work!
That's the way that you learn things. But it's quite another thing to
say that their work is incorrect, and you should be careful when you
do that.

Anyway, back to the problem! The solution happens through the magic
of differential equations. They are just the ticket in situations
like these, where it can be hard or impossible to solve the problem
using more conventional methods.  Basically, the essence of the
problem is in this statement:

A' = -(A/100)*3 + (B/100)*3

Let's understand this.  A' is the change in the salt concentration of
tank A.  Does it make sense that the change in the salt concentration
is equal to the salt coming in minus the salt going out?  That's what
the right side of the equation says.  The salt going out (it's
negative) is -(A/100)*3, and the salt coming in is (B/100)*3.

Then, since A+B = 200, we can substitute B by 200-A as Dr. Jerry said,
and then we have:

A' = -3A/100 + 6 - 3A/100
A' = -3A/50  + 6

Note that this is an equation.  If we can solve this equation for A,
then we've solved the problem.  That's what Dr. Jerry did.  Why is his
solution correct?  You can just plug things into the above equation
and verify it for yourself.

Let us know if you have any more questions.

-Doctor Ken,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 03/12/97 at 15:30:07
From: Calc BC2 class at Germantown Academy
Subject: Incorrect Response

In your response, you said (as well as that the universe is not going
to explode today) that A' = -(A/100)*3+(B/100)*3, but it's not that
easy. You're saying that A' is the salt coming in minus the salt going
out, assuming that the amount of salt in tank B at any time is
constant. However, the solution in tank B is being diluted by the
fresh water from tank A that enters tank B. This water is not always
fresh, either, as salt is entering it from tank B. Therefore, the
equation for the concentration of salt in each tank at any time is a
bit more complex than that which you derived. We believe, following
this logic, that the concentration in Tank A = (B-200e^(-3t/100))/100
lb/gal and the concentration in Tank B = (A+200e^(-3t/100))/100
lb/gal, where t = time.

Evan Rattner
Karthik Bala
Kate Lockhart
```

```
Date: 03/12/97 at 16:24:38
From: Doctor Ken
Subject: Re: Incorrect Response

Hi there -

completely right that the amount of salt in tank B is not a constant.
In fact, that's exactly what makes the problem hard!  The fact that
A and B are changing is what necessitates differential equations in
the solution.

But you're still mistaken about something: nobody here at Dr. Math
ever assumed that either of A or B _was_ a constant.

Perhaps it'll be a little more clear if we write the equation like
this:

A'(t) = [3*B(t)/100] - [3*A(t)/100]
DSolve[A'[t] == (3*(200-A[t])/100) - (3*A[t]/100), A[t], t]

Now, let me ask you something.  Do you agree that this equation
adequately describes what's happening in the tanks?  Here's what it
says: the _instantaneous_ change in the amount of salt in tank A is
equal to the instantaneous amount of salt coming in from tank B minus
the instantaneous amount of salt going out to tank B.

Another question you might ask yourself: is [3*B(t)/100] really the
instantaneous amount of salt coming in from tank B at time t?  Here's
why it is: you've got B(t) pounds of salt in tank B at time t.  It's
diluted by 100 gallons of water.  So B(t)/100 lbs/gal is the
concentration of salt in tank B.  Since brine is flowing from B to A
at the rate of 3 gallons per minute, the expression 3*B(t)/100 lbs/min
represents the rate at which salt is flowing from B to A.

Now, if you accept that this equation correctly models the situation,
then all you have to do is solve it!  I'm sort of kidding, because
that's not necessarily easy.  Solving equations like these is the
whole reason the study of differential equations was invented.  It's
also why a lot of equation-solving computer programs were written;
plugging it into one I see that:

A(t) = 100 + Constant/E^(3t/50)

We can figure out what the constant is by plugging in the known value
A(0) = 0:

We get 0 = 100 + Constant/E^0
0 = 100 + Constant
-100 = Constant

So our equation is A(t) = 100 - 100/E^(3t/50).

But there are two questions you might be asking, and I'm not sure
which you really _are_ asking: you might be asking why the equation
A'(t) = [3*B(t)/100] - [3*A(t)/100]  is true, or you might be asking
how to solve that for A(t).  If you tell us which, we might be able to
help you better.  Also, if you could write back and explain the
equations that you included in your last message, we might be able to
better understand why you still think Dr. Jerry's answer to this
question is incorrect.

Whew!  I'm glad I was right about the universe not exploding today!

-Doctor Ken,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus

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