4.3.2

Instructor: Example 3. A tank containing 100 liters of a brine solution initially has 4 kilograms of salt dissolved in the solution. At time t equals 0, another brine solution flows into the tank at a rate of 2 liters per minute. This brine solution contains a concentration of 0.5 kilograms per liter of salt. At the same time, a stopcock is opened at the bottom of the tank, allowing the combined solution flow out at a rate of 2 liters per minute, as we see here, so that the level of liquid in the tank remains constant. We're adding to liters, taking out two liters. 

Find the amount of salt in the tank as a function of time measured in minutes. And find the limiting amount of salt in the tank, assuming that the solution is well mixed at all times. Now we've got this diagram already showing all the information that we need, but let's first define some variables. 

First, we want to know how much salt is in the tank. So I'm going to define a function that's u of t. That is the amount of salt. Amount of salt in the tank, at time t, which means that du dt, the derivative of that with respect to time, is the rate of change of the amount of salt. 

With this additional fact that the initial amount, u of 0 equals 4, the amount of salt initially, is 4 kilograms. Now we can determine what du dt is. We can determine what du dt is, because that is the inflow rate minus the outflow. Inflow minus the outflow. 

Now we know that the amount going in is 2 liters per minute multiplied by 1/2 a kilogram of salt per liter. So this is 2 times 0.5. And the outflow rate is however much salt-- so we know how much is being added in, so that's the inflow. That one's just a fixed amount over our time. But the outflow rate is a percentage of what is going out. 

So we know that we know what the amount is, but OK, but that divided by 100-- because there's a concentration there, OK. We're talking amounts over time, so we have out of the liters times 2, because it's 2 liters per minute. So that this right here is what sets up the whole rest of the problem. So be sure you understand what's going on there. 

We have the amount of salt, the amount of salt per liter, because these are all salt per minute, salt per liter, things like that. OK. And we have a rate of 2 liters per minute. So let me just write out what those units are. 

That would be-- this is in kilograms, isn't it. Kilograms, yes, all right. So this is kilograms per liter. And this is 2 liters per minute. So this is kilograms per minute. Whereas, also, this is the 2, is the liters per minute multiplied by our kilograms per liter. So that is also kilograms per minute. So we're adjusting for units, really. OK. 

Well, if we determine what that is, then du dt is in fact, 1 minus u over our u of t, and then over 50, reducing that. So let's just say that's 1/50 u of t. And in fact, we might just want to drop the, of t. We know it's a function. 

Now, we have this along with the fact that u of 0, the initial amount is 4. So this is an initial value problem that we're going to solve. 

So let's go ahead and rearrange some things. Let's see. This would be du dt equals 50 minus u over 50. See, we need to get our functions in terms of u on the left side. We don't actually have anything in terms of t, but just our u. 

So let's go ahead and rearrange that as we have in the past, to say that that is 50 over, 50 minus u is going to be a denominator. So we have du, 50 minus u. And that is dt, and that'd be actually, 1/50. Try like this. 

And now we are going to take the antiderivative of both sides, which will lead us to 1/50 t on the right side plus c. And, left side that will be minus natural log absolute value of 50 minus u. And as in the previous problems, the absolute values will be useless. Or they will be-- well, because of our constant, the size of our constant, they're not useless, but we can eliminate them that way. 

All right. Now exponentiating-- actually first, we want to remove that negative sign. So I'm going to divide by negative 1, and that will make this negative 1/50 t. And I'm going to leave plus c because I don't know the sign of that. 

So that would be-- well, we'll go ahead and do that. I'm going to make this minus that. Now we've got natural log, absolute value of 50 minus u equals negative 1/50 t minus c exponentiating. 

We've now got 50 minus u equals e to the negative 1/50 t minus c. OK, let's go ahead and write that as a constant, the constant as a coefficient. Make that c1. And we'll say, c1 over here equals e to the c. 

Now, solving for u, let's bring this over here. Solving for u, that should be u of t. In fact, let's do it all the way up here. u of t equals-- we've got our c1. Let's see. Subtracting 50 dividing by negative 1, that'll be 50 minus c1 e to the negative 1/50 t. 

Now let's take our initial value, u of 0 is 4, evaluating this in here. Well, putting in 0, substituting 0 as our t-value will make this entire term 1. So this is going to be 50 minus c1. We have 4, so that would be 46. c1 equals 46, making u of t our particular solution. 50 minus 46 e to the negative 1 over 50 t. 

Now, the question of what is the long-term behavior, that's the other part of this. Find the limiting amount in the tank. Well, if we applied a limit-- and let's make this the limit as t goes to infinity of u of t. Let's make that the limit of 50 minus 46e to the negative 1/50 t. 

We know that negative exponent, that term is going to 0. So the long-term behavior is going to be 50 pounds. That is the limiting amount. 

[ENDS AT 9:41]