00:01
Here we have an example of a rate problem.
00:06
These are typically done the same way in the sense that there's usually a rate of input and a rate of output.
00:16
And here we're going to call the variable y is equal to the amount of the chemical.
00:24
In this case, it looks like salt.
00:26
We'll just call it salt in the tank in pounds.
00:33
Whereas v is the volume of fluid solution in the tank in gallons.
00:46
We're also going to subject this to the limit.
00:50
There is a maximum volume.
00:52
There's usually a situation where the input and output flow rates in the valve system are equal, in which case the volume remains constant.
01:03
Here we see there's more inlet four gallons per minute compared.
01:08
Or to the half a gallon per minute.
01:10
So at some point, the tank will become full, whereas if the flow rates were opposite, the tank could become empty.
01:19
But we'll call the volume max.
01:22
We'll call that equal to 50 gallons.
01:25
And that will set a limit to the time period that we're allowed to continue this.
01:32
But our goal is to figure out the amount of salt in the tank.
01:38
Usually these rate problems you set up with the rate of salt going into the tank is the difference between the rate in and the rate out.
02:04
We can write a similar equation for the volume that it is changing as well.
02:11
It has a rate in, which is four gallons per minute, and we also have a rate out of a half a gallon per minute.
02:26
Notice that that is a fairly easy equation to solve for the volume as a function of time.
02:34
It is separable.
02:38
We have dv is 3 .5 d t, and we can simply integrate that to get v is 3 .5 t plus v0, which is 10.
02:54
We can figure out at what time we can get a limit on this, when is volume equal to v max? and we get that that time max is simply 50 minus 10, which is 40, divided by 3 .5.
03:17
That unit should be minutes, so we'll leave it there.
03:22
And so we're assuming that from t equals 0 to 40 over 3 .5 minutes is a good time period.
03:32
Now let's work on the salt that's going into the tank.
03:38
What we need is the concentration times the flow rate.
03:43
So one pound per gallon times four gallons per minute gives us the unit that we want, which is pounds per minute, the amount of salt per unit time.
03:57
And we get one times four for the rate in.
04:02
On the rate out we need the concentration in the tank and that is going to depend on the y and the volume times the rate of output 0 .5 rate of output flow in gallons per minute now it's a good thing that we solved for the volume as a function of time we're going to need that in our y equation and we'll clean it up a little bit but de y by d t will simply be equal to four minus one half.
04:43
We'll get that combined a little bit times 3 .5t plus 10 times y.
04:56
I'm going to rewrite this, but we can see that this equation is not separable.
05:07
So we can't use just the simple separable technique separate and integrate.
05:15
For the equation, but there is a neat little trick that we can use.
05:22
And it is follows.
05:25
Let me get the y and the time stuff together and simplify the denominator just a tad bit.
05:35
So no dangling one -haps and whatnot.
05:39
So what makes this non -separable is this function of time that's sitting in front of the y.
05:46
And the way you solve this is with the integrating factor.
06:00
So that integrating factor is calculated from e raised to the integral of dt times that function of time.
06:11
There's really a one up there.
06:13
So it's 1 over 7t plus 20.
06:21
And that is going to give us a logarithm.
06:24
We can do a substitution u is equal to 70 plus 20.
06:29
And du is equal to 7d .t.
06:37
And what we get then is e raised to the 1 seventh logarithm of 7t plus 20.
06:52
Now that fraction in front can be placed with the argument to the logarithm.
07:01
Remember, exponents of any type, whether negative, positive, fractional, you name it, can go inside the logarithm as raising the argument to some power.
07:15
The logarithm pulls out that power as a factor.
07:21
And so what's nice about that is now the exponential and the logarithm cancel each other.
07:31
And you have a nice integrating factor.
07:34
Now, it does have this fractional power, but everything works the way it should that what you want to do with that is now multiply everything in that equation by the integrating factor.
07:52
It might not seem like that buys you anything, but we'll see what it buys you.
08:14
Okay, and that second term, we have one -seventh minus seven -sevenths, or minus six -sevents.
08:25
And don't forget the other side gets multiplied, you can do anything you want to an equation as long as you treat everyone in the equation fairly.
08:37
Okay, now what did this buy you? it turns out that these two terms can be combined into an exact differential that includes the integrating factor, the 7t plus 1 20 to the 1 7 times y.
08:59
Those can be all be grouped together into a single exact differential...