00:01
Here we have a rate situation.
00:03
These salt tank problems are fairly common.
00:07
Usually what you do is you assign a variable.
00:10
I'm assigning why the variable for the amount of salt in the tank as a function of time.
00:19
And v will use for the variable for the volume of the brine in the tank as a function of time.
00:37
And typically what you do is you're trying to find the amount of salt in the tank.
00:44
And you write the rate of change of the salt as a difference between the rate in minus the rate out.
00:59
And usually there's enough information to determine what those are.
01:03
If we look at the units for the salt, we see that the rate of change should be given in pounds per.
01:12
Minute.
01:15
So pounds per minute.
01:21
And usually you can see what that pounds per minute is by looking at the inlet.
01:28
So here we have two pounds per gallon dissolved in the liquid.
01:32
And the liquid's coming in at five gallons per minute.
01:36
So the rate is simply two pounds per gallon times five gallons per minute.
01:46
We can see that the units will come out to give us pounds per minute, and that's just 10.
01:54
We'll drop the units.
01:56
The rate out sometimes is a little bit more difficult because what we're seeing is we're multiplying the concentration in the brine times the flow rate.
02:09
So the gallons per minute is the flow rate.
02:13
And we don't know the concentration.
02:14
The concentration in the tank is changing, but what we do know is it is y over the volume of the tank, both functions of time, times the flow rate out, which is simply five, four.
02:30
Sorry, four.
02:32
Let's get that right.
02:34
Now the y in this equation, as an unknown, is not a problem because our rate, equation is trying to solve for y.
02:45
So it's perfectly appropriate for y to be on the right -hand side.
02:50
But what we need to do is figure out the volume as a function of time.
02:56
And usually what you do with that is exactly the same thing that you did with the salt.
03:02
The volume is changing according to the rate in minus the rate out.
03:13
And here we recognize that the five gallons per minute is a rate in, and the four gallons per minute is a rate out for the volume.
03:24
So we have a fairly simple differential equation for the volume.
03:30
It is equal to the time rate of change of the volume, time is just one.
03:36
And we can certainly separate and integrate that to give us t plus volume zero, or the volume is t plus 100.
03:53
So the volume is going up linearly in time.
03:58
And our final differential equation for the brine turns out to give us d, y by dt, is equal to 10 minus 4 over t plus 100, with a y there.
04:26
And unfortunately, it is not separable.
04:30
So the common method of separate and integrate needs a little bit of help.
04:35
It's not a huge amount of help, but there is a trick.
04:41
What is making it inseparable is this function of time that's sitting in front of why.
04:49
And the way to handle this is to use the technique called the integrating factor.
04:59
A nice little tool.
05:03
That integrating factor is exponential.
05:07
Raised to the integral of that function for t plus 4 over t plus 100 in the denominator d t.
05:18
And this is nicely a logarithm function.
05:28
If we do the thing where the exponent comes out as a factor, but put it back together in the logarithm, we can actually cancel the logarithm function.
05:44
Which is the inverse of the exponential.
05:48
So our integrating factor turns out to be t plus 100 to the fourth.
05:59
Now what do you do with that integrating factor? what you do with it is you multiply every single thing in the differential equation by that factor.
06:14
Simply multiply it out every single part.
06:20
You can do that to an equation.
06:21
As long as you do it to both sides and every term.
06:26
And i'm actually going to put the 4 .y on the other side.
06:31
You'll see y in a minute.
06:34
And one of the powers cancels...