Salt ConcentrationDate: 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 Subject: Your answer 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 Subject: Re: Your answer 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 - You've made a couple of astute points about the situation. You're 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/ |
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