00:01
So we have a mixing problem where salt is going into a tank of water.
00:08
Here we are going to call the variables, why is the amount of salt in the tank, whereas volume is, of course, the volume of water.
00:23
And it's good to be cognizant of units.
00:27
So the units for y will be in kilograms.
00:32
What we're using for volume is liters.
00:39
And i'm being a little bit looser with the volume because i see that we have water getting poured into the tank at the same rate.
00:49
It's coming out.
00:51
Or solution anyway may not be pure water.
00:55
And so the volume will remain constant.
01:02
And that makes these problems usually a little bit easier.
01:06
But the way you set up the problem is with a rate equation.
01:11
Where after y is a function of time.
01:18
And it is going to come from the solution of a rate equation where d .y by dt, which will be in kilograms per minute in this problem, is the difference between the rate at which it is coming into the tank minus the rate it's leaving the tank.
01:43
And what we mean by those rates is, again, need something in kilograms per minute.
01:49
So what we see is if we take the concentration of the salt in the incoming solution, the 0 .4 kilograms per liter, and we multiply it by the rate the fluid is coming in, we have the correct units of kilograms per minute.
02:13
So the liters will cancel and we'll wind up with kilograms per minute, the rate in.
02:20
The rate out is a little bit trickier.
02:23
It's not too hard.
02:25
But our concentration leaving is going to be y over the volume of the tank, and again times the 5 liters per minute.
02:39
And this is, of course, the same y that we're solving for by solving the differential equation.
02:48
So putting it all together, we have a d -y by d -t is equal to, we'll drop the units, those are kind of awkward.
02:57
And the rate out is 5 over 100, that's 0 .05 times y.
03:07
Now usually i like to write the side, the right -hand side, a little bit differently.
03:15
I like to have the leading term of y, so i'll pull out a minus point of, and then we're left with y minus 40.
03:30
And you'll see why in a little bit that that may be helpful.
03:33
But mostly why i'm doing that is because i see that the point 0 .05 has units of time per time per minute, whereas y has units of kilograms.
03:53
And so i'm going to want to separate the things with time.
03:57
And that is indeed the technique that we're going to be using to solve the equation is to separate and integrate.
04:11
By separate, we simply mean cross -multiply to get the time stuff on one side and the y quantities on the other.
04:27
And that little trick i did with a minus 0 .05 is going to make the integration a little bit easier.
04:34
And maybe the solution a little bit quicker to arrive at.
04:42
Okay, so now is the integrate...