00:01
Here we're going to look at a salt water mixing and time rate of change problem, where the volume in a tank is changing due to stuff in, mainly salt water, and the mixture coming out.
00:20
So we're going to call y equal to the number of kilograms of salt in the water, and that is the variable that we would like to find in time, and it's starting with an initial condition of no salt.
00:46
So this is going to be a problem where we set it up, where the rate of change of y in the tank is d .y by dt, and that is equal to the difference in the rate of salt in minus the rate of salt.
01:08
So there is a valve letting things out and some pipe that's letting things in.
01:14
The rate in is no problem.
01:16
The rate in is basically we're taking 7 liters per minute times 0 .7 kilograms per liter.
01:27
And that just gives us 4 .9 kilograms per minute.
01:32
And that is what we mean by the rate of salt in kilograms per minute is the unit that we're using.
01:42
The rate out is a little bit trickier, and the reason being is because the volume is also changing.
01:52
We do not have the same rate in for water as we have out, and so the volume of water will be changing.
02:02
So we're going to write the rate out as why the amount of salt in the tank divided by the volume, which is a function of time, and then the 5 liters per minute is the time rate of change of the volume.
02:26
Okay, so volume will be in liters, and multiplying by 5 liters per minute will give us the rate of salt coming out in kilograms per minute.
02:40
Okay, so what we need is a, for volume as a function of time.
02:54
And we're going to write it as, again, a difference in the volume in and the volume out.
03:02
So we'll write a little equation for it.
03:04
It's not too hard to work it out.
03:07
But the rate of change of the volume is 7 liters per minute in minus 5 liters per minute out.
03:18
So again, we're doing a rate in and a rate out.
03:22
And that is positive 2 liters per minute.
03:28
Okay.
03:30
And so we can simply integrate this really quickly, and we get that volume is plus two, plus arbitrary integration constant.
03:44
Sorry, plus 2t.
03:46
Yeah, let's integrate that in time.
03:49
That would be helpful.
03:52
And i could show that integration, but it's pretty fairly straightforward.
03:57
And here we have to use volume at time to equal zero should be equal to 100 liters.
04:06
And so we can see that that is the constant, c, the arbitrary integration constant.
04:13
So finally, what we have is that the volume as a function of time is 100 plus 2t.
04:25
And we can finally then write our equation, differential equation, for y.
04:32
And think about how to solve it.
04:37
Goody gum drops.
04:40
So d .y by dt is equal to positive 4 .9 is in kilograms per minute.
04:50
And then we are writing our rate out as y times 5, 5 times y, divided by the volume, which is 100 plus 2t.
05:11
And we can see that we just can't separate and integrate this.
05:16
So we're going to have to use a little trick.
05:20
And the little trick is to combine the y stuff on one side.
05:27
And the stuff, again, that just depends on time on the other side.
05:33
And it's just a constant, 4 .9.
05:36
Okay.
05:38
So we've got all the y stuff together.
05:40
And the trick is to use an integrating factor.
05:47
So remember what you're doing with an integrating factor is you're trying to get all the y stuff written as the differential of some function of time.
06:01
And then you can separate that and integrate it.
06:05
That's why it's called an integrating factor.
06:07
But the integrating factor, if, is going to look like the function sitting in front of y, function of time.
06:18
Doesn't matter what that function is as long as you can integrate it.
06:22
You're in good shape.
06:24
So let's go ahead and figure out that integrating factor.
06:30
We're going to take e and integrate all the things in front of y.
06:36
And let's see we get 100 plus 2t dt and i'll just go ahead and write that out.
06:49
It's a one -over type thing.
06:51
Yeah, you might have to work on that a little bit, but we basically get a natural logarithm because it is one over that 100 plus 2t.
07:05
And i will combine the five -haves as a power so i can do an...
07:12
Getting rid of the logarithm.
07:16
Yeah, that's kind of nice to do.
07:24
Okay, so our integrating factors is 100 plus 2 t raised to the five halves.
07:35
Okay.
07:36
Now watch what we do with it.
07:38
We multiply the entire equation through by this integrating factor.
07:47
So multiply the whole differential equation and we'll see what this buys us.
08:00
Okay.
08:02
So equation one with the integrating factor will become yucky looking, but we'll see how this made life better.
08:11
100 plus 2t to the five halves, d .y by dt plus 5, 100 plus 2t to the five halves, y over 100 plus 2t is equal to 100 plus 2t to the five - halves 4 .9 times that.
08:59
Okay, now what is good about this is we can now lump the two y parts together and say that we have basically used the multiplication rule to split those out into the two parts.
09:20
We're kind of merging them into one function, 100 plus 2t to the 5 halves times why all of that taking its differential is equal to the other side.
09:41
100 plus 2t.
09:43
We're not going to mess around with that side.
09:50
Okay.
09:51
And you can kind of check it out...
