00:01
An example of a rate problem usually involves some sort of solution in a tank with some solution coming in and some coming out.
00:11
And you're usually interested in the quantity of time salt in the tank, which is a function of time.
00:25
Now here, if the rate of flow, so r is the volume flow rate, and if it is the same in as it is out, then the volume in the tank remains constant.
00:50
And that's the situation that we're going to consider here.
00:57
The c in the c out, as we'll see, are concentrations.
01:01
And let us pick some units.
01:02
The volume flow rate will be in gallons per minute.
01:09
It's important to make sure all your units are consistent.
01:13
Q is going to be in pounds.
01:16
Volume, of course, is going to be in gallons.
01:18
So what we want those concentrations to be is in pounds per gallon, once per gallon for both of those.
01:36
So let's set up a differential equation for q, which is what we'll solve to get q as a function of time.
01:43
It's equal to the difference between the rate coming in, which will be in pounds per minute, minus the rate out, which will also be in pounds per minute.
02:02
So notice what we're doing there is we're setting up the change of q in time.
02:09
And q, since it has units of pounds, and time has to have units of time, we want pounds per minute on all the quantities on both sides.
02:22
And we can get that by taking the concentrations in and out, multiplying those both by the rate.
02:36
So notice that if you take pounds per gallon and multiply by gallons per minute, you will get pounds per minute.
02:47
That's kind of the thing to remember.
02:50
Units help.
02:53
Okay, now the cn, cn is usually a quantity that's given as well as the flow rate.
03:00
The quantity c -out, actually, there's a little trick that you pull, and that's that the tank is well -mixed, which means there's the same concentration not only throughout the tank, but coming out the output stream.
03:19
So that c -out is the same as q over v -0.
03:23
The amount of salt in the tank, divided by the entire volume.
03:29
And that is a time -dependent quantity.
03:34
If your flow rates were different, you'd have some time dependence in both terms.
03:43
But here we have a simplifying assumption here.
03:48
So dq by dt is equal to c in, and we'll put r in front of everything, minus q over v -0.
04:02
Now this can be solved with the technique of and integrate.
04:14
But i am going to do one last little cleanup thing.
04:18
I am going to get the leading term in front of q to be a positive one.
04:23
And to do that, i will have to bring out the minus 1 over v0.
04:33
And then we have a minus v0 times c in.
04:41
And now we're ready to solve.
04:45
So we simply switch the time and the q quantity.
04:52
That is our separate.
04:57
Sometimes a little bit more work is necessary to separate, but this is fairly straightforward.
05:04
And i'm going to leave things in terms of symbols simply because we're going to be changing some parameters in the numerical.
05:16
So now we'll integrate, and we must add an arbitrary integration constant, which reminder that is found by figuring out, being told, given the initial quantity q0.
05:36
On the left, we have a logarithm, a straight -up logarithm.
05:43
On the right, very simple, integral, just time, and then our constant.
05:53
And we may invert this by raising both sides to the e -base.
06:00
Not raising both sides, but using both sides as an exponent of the e -base.
06:05
And we finally get q.
06:10
We'll bring over the cn, b, not, and then plus.
06:17
I'll call it another constant, but we're really breaking this up into a product here.
06:24
Remember adding exponents is like breaking the base up into two factors.
06:32
And that second factor is another constant.
06:34
That i'll just call a.
06:37
Constant is just a constant.
06:40
Call it anything you want...