00:01
Here we have a rate problem in which the amount of salt inside of a solution in a tank is going to be changing.
00:11
So some variable definitions.
00:14
I'm going to use y, which will be a function of time, is the amount of salt in the tank in pounds.
00:27
That's the unit we'll be using.
00:30
Whereas v is the volume of salt in gallons.
00:42
We see that the inlet rate for the fluid is equal to the outlet rate.
00:48
So we are going to then have a simplified problem in which the volume is not a function of time.
00:56
It is constant.
00:59
That will simplify the answer quite a bit.
01:03
But the way you set up these rate problems is a rate is the derivative of the quantity with respect to time.
01:12
So, d .y by d .t is the difference between the rate in minus the rate out.
01:25
So we have to determine those rates in and out, which come from the inlet and the outlet side to set up our differential equations.
01:35
And it's not too hard if we look at the units we expect for d .y by dt.
01:43
It should be in units of pounds per minute.
01:50
So we see that if we look at the inlet and multiply the concentration in the solution by the volume flow rate, let's give these things names, but just simply take those and multiply them together.
02:14
We will get the gallons to cancel, and we will have pounds per minute.
02:22
And that's going to be our rate in two times three is the concentration times the flow rate.
02:36
Now on the outlet side, all we have is the flow rate.
02:41
It's still three gallons per minute.
02:43
But we need the concentration coming out, and that is going to be the y over the volume in the tank times three.
02:59
So again we have concentration times flow rate.
03:07
And our volume is not changing.
03:10
So that makes life a little bit easier.
03:13
We have y over 300 times three.
03:20
So we now have a differential equation that we can conceivably solve.
03:27
One of the things i like to do with this, but it does turn out that this is separable...