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**Topics covered:** Continuation; Applications to Temperature, Mixing, RC-circuit, Decay, and Growth Models.

**Instructor/speaker:** Prof. Arthur Mattuck

Lecture 8: Continuation

I usually like to start a lecture with something new, but this time I'm going to make an exception, and start with the finishing up on Friday because it involves a little more practice with complex numbers.

I think that's what a large number of you are still fairly weak in. So, to briefly remind you, it will be sort of self-contained, but still, it will use complex numbers.

And, I think it's a good way to start today.

So, remember, the basic problem was to solve something with, where the input was sinusoidal in particular. The k was on both sides, and the input looked like cosine omega t.

And, the plan of the solution consisted of transporting the problem to the complex domain. So, you look for a complex solution, and you complexify the right hand side of the equation, as well. So, cosine omega t becomes the real part of this complex function.

The reason for doing that, remember, was because it's easier to handle when you solve linear equations.

It's much easier to handle exponentials on the right-hand side than it is to handle sines and cosines because exponentials are so easy to integrate when you multiply them by other exponentials. So, the result was, after doing that, y tilda turned out to be one, after I scale the coefficient, one over one plus omega over k And then, the rest was e to the i times (omega t minus phi), where phi had a certain meaning.

It was the arc tangent of a, it was a phase lag.

And, this was then, I had to take the real part of this to get the final answer, which came out to be something like one over the square root of one plus the amplitude one omega / k squared, and then the rest was cosine omega t plus minus phi.

It's easier to see that that part is the real part of this; the problem is, of course you have to convert this. Sorry, this should be i omega t, in which case you don't need the parentheses, either. So, the problem was to use the polar representation of this complex number to convert it into something whose amplitude was this, and whose angle was minus phi. Now, that's what we call the polar method, going polar.

I'd like, now, for the first few minutes of the period, to talk about the other method, the Cartesian method. I think for a long while, many of you will be more comfortable with that anyway.

Although, one of the objects of the course should be to get you equally comfortable with the polar representation of complex numbers. So, if we try to do the same thing going Cartesian, what's going to happen?

Well, I guess the same point here.

So, the starting point is still y tilde equals one over, sorry, this should have an i here, one plus i times omega over k, e to the i omega t.

But now, what you're going to do is turn this into its Cartesian, turn both of these into their Cartesian representations as a plus ib.

So, if you do that Cartesianly, of course, what you have to do is the standard thing about dividing complex numbers or taking the reciprocals that I told you at the very beginning of complex numbers.

You multiply the top and bottom by the complex conjugate of this in order to make the bottom real.

So, what does this become? This becomes one minus i times omega over k divided by the product of this in its complex conjugate, which is the real number, one plus omega over k squared So, I've now converted this to the a plus bi form.

I have also to convert the right-hand side to the a plus bi form. So, it will look like cosine omega t plus i sine omega t.

Having done that, I take the last step, which is to take the real part of that. Remember, the reason I want the real part is because this input was the real part of the complex input. So, once you've got the complex solution, you have to take its real part to go back into the domain you started with, of real numbers, from the domain of complex numbers.

So, I want the real part is going to be, the real part of that is, first of all, there's a factor out in front.

That's entirely real. Let's put that out in front, so doesn't bother us particularly.

And now, I need the product of this complex number and that complex number. But, I only want the real part of it. So, I'm not going to multiply it out and get four terms. I'm just going to look at the two terms that I do want. I don't want the others.

All right, the real part is cosine omega t, from the product of this and that.

And, the rest of the real part will be the product of the two i terms. But, it's i times negative i, which makes one. And therefore, it's omega over k times sine omega t.

Now, that's the answer.

And that's the answer, too; they must be equal, unless there's a contradiction in mathematics.

But, it's extremely important. And that's the other reason why I'm giving you this, that you learn in this course to be able to convert quickly and automatically things that look like this into things that look like that.

And, that's done by means of a basic formula, which occurs at the end of the notes for reference, as I optimistically say, although I think for a lot of you will not be referenced, stuff in the category of, yeah, I think I've vaguely seen that somewhere.

But, well, we never used it for anything.

Okay, you're going to use it all term.

So, the formula is, the famous trigonometric identity, which is, so, the problem is to convert this into the other guy. And, the thing which is going to do that, enable one to combine the sine and the cosine terms, is the famous formula that a times the cosine, I'm going to use theta to make it as neutral as possible, -- -- so, theta you can think of as being replaced by omega t in this particular application of the formula. But, I'll just use a general angle theta, which doesn't suggest anything in particular.

So, the problem is, you have something which is a combination with real coefficients of cosine and sine, and the important thing is that these numbers be the same.

In practice, that means that the omega t, you're not allowed to have omega one t here, and some other frequency, omega two t here.

That would correspond to using theta one here, and theta two here. And, though there is a formula for combining that, nobody remembers it, and it's, in general, less universally useful than the first. If you're going to memorize a formula, and learn this one, it's best to start with the ones where the two are equal. That's the basic formula.

The others are variations of it, but there is a sizable variations. All right, so the answer is that this is equal to some other constant, real constant, times the cosine of theta minus phi.

Of course, most people remember this vaguely.

What they don't remember is what the c and the phi are, how to calculate them. I don't suggest you memorize the formulas for them. You can if you wish.

Instead, memorize the picture, which is much easier.

Memorize that a and b are the two sides of a right triangle.

Phi is the angle opposite the b side, and c is the length of the hypotenuse. Okay, that's worth putting up.

I think that's a pink formula. It's even worth two of those, but I will thrift. Now, let's apply it to this case to see that it gives the right answer.

So, to use this formula, how I use it?

Well, I should take, I will reproduce the left-hand side. So that part, I just copy. And, how about the right?

Well, the amplitude, it's combined into a single cosine term whose amplitude is, well, the two sides of the right triangle are one, and omega over k.

The hypotenuse in that case is going to be, well, why don't we write it here? So, we have one, and omega over k. And, here's phi.

So, the hypotenuse is going to be the square root of one plus omega over k squared.

And, that's going to be multiplied by the cosine of omega t minus this phase lag angle phi.

You can write, if you wish, phi equals the arc tangent, but you are not learning a lot by that.

Phi is the arc tangent of omega over k.

That's okay, but it's true. But, notice there's cancellation now. This over that is equal to what? Well, it's equal to this.

And, so when we get in this way, by combining these two factors, one gets exactly the same formula that we got before.

So, as you can see, in some sense, there's not, if you can remember this trigonometric identity, there's not a lot of difference between the two methods except that this one requires this extra step. The answer will come out in this form, and you then, to see what it really looks like, really have to convert it to this form, the form in which you can see what the phase lag and the amplitude is. It's amazing how many people who should know, this includes working mathematicians, theoretical mathematicians, includes even possibly the authors of your textbooks.

I'm not sure, but I've caught them in this, too, who in this form, everybody remembers that it's something like that. Unfortunately, when it occurs as the answer in an answer book, the numbers are some colossal mess here plus some colossal mess here. And theta is, again, a real mess, involving roots and some cube roots, and whatnot. The only thing is, these two are the same real mess.

That amounts to just another pure oscillation with the same frequency as the old guy, and with the amplitude changed, and with a phase shift, move to the right or left.

So, this is no more general than that.

Notice they both have two parameters in them, these two coefficients. This one has the two parameters in an altered form. Okay, well, I wanted, because of the importance of this formula, I wanted to take a couple of minutes out for a proof of the formula, -- -- just to give you chance to stare at it a little more now.

There are three proofs I know. I'm sure there are 27.

The Pythagorean theorem now has several hundred.

But, there are three basic proofs.

There is the one I will not give you, I'll call the high school proof, which is the only one one normally finds in books, physics textbooks or other textbooks. The high school proof takes the right-hand side, applies the formula for the cosine of the difference of two angles, which it assumes you had in trigonometry, and then converts it into this.

It shows you that once you've done that, that a turns out to be c cosine phi and b, the number b is c sine phi, and therefore it identifies the two sides. Now, the thing that's of course correct and it's the simplest possible argument, the thing that's no good about it is that the direction at which it goes is from here to here.

Well, everybody knew that. If I gave you this and told you, write it out in terms of cosine and sine, I would assume it dearly hope that practically all of you can do that. Unfortunately, when you want to use the formula, it's this way you want to use it in the opposite direction.

You are starting with this, and want to convert it to that.

Now, the proof, therefore, will not be of much help. It requires you to go in the backwards direction, and match up coefficients.

It's much better to go forwards.

Now, there are two proofs that go forwards.

There's the 18.02 proof. Since I didn't teach most of you 18.02, I can't be sure you had it.

So, I'll spend one minute giving it to you.

What is the 18.02 proof? It is the following picture.

I think this requires deep colored chalk.

This is going to be pretty heavy.

All right, first of all, the a and the b are the given.

So, I'm going to put in that vector.

So, there is the vector whose sides are, whose components are a and b. I'll write it without the i and j. I hope you had from Jerison that form for the vector, if you don't like that, write ai plus bj, okay?

Now, there's another vector lurking around.

It's the unit vector whose, I'll write it this way, u because it's a unit vector, and theta to indicate that it's angle is theta. Now, the reason for doing that is because you see that the left-hand side is a dot product of two vectors. The left-hand side of the identity is the dot product of the vector b> with the vector whose components are cosine theta and sine theta.

That's what I'm calling this unit vector.

It's a unit vector because cosine squared plus sine squared is one.

All right, well, you should, that was 18.02.

There must be an 18.03 proof also. Yes.

There are different ways of explaining why I want to put the minus i there instead of i.

But, the simplest is because I want, when I take the real part, to get the left-hand side. I will.

What's the polar representation of this guy?

Well, if has the angle theta, then a negative b, a minus bi goes down below. It has the angle minus phi.

So, it's e to the minus i phi.

That's the first guy. And, how about the second guy?

Well, the second guy is e to the i theta.

And now, what do I want? The real part of this, and I want the real part of this.

So, let's just say take the real parts of both sides.

If I take the real part of the left-hand side, I get a cosine theta plus b sine theta.

As I mentioned in the notes, it took mathematicians 300 or 400 years to get used to complex numbers.

So, if it takes you three or four weeks, that's not too bad.

So, we are talking about basic linear equations.

And finally, there is the most general case, where you allow k to be non-constant.

Now, there's one other thing, which I want to talk about.

I will do all these in a certain order.

But, from the beginning, you should keep in mind that there's another between the first two cases.

So, it's understood when I write these, that k is positive.

I want to talk about that, too.

It applies to other things, too.

Let me give you a mixing problem.

The typical mixing problem gives another example.

You have a tank, a room, I don't know, where everything's getting mixed in.

It has a certain volume, which I will call v.

Something is flowing in, a gas or a liquid.

And, r will be the flow rate, in some units.

But, I do have to say what flows in.

So, this is the data. r is part of the data.

r is the flow rate. v is the volume.

I think I won't bother writing that down, and the problem is to determine what happens to x of t.

Well, because it's something you can physically think about.

How about the rate of the salt outflow?

So, I'm going to convert this to concentrations.

How about the x? How do I convert that?

Well, what's the relation between x?

Now, that's not in standard form.

And, you can see it falls exactly in this category.

That means that I can talk about it.

I said concentration. I mean, that concentration has nothing to do with this concentration.

Now, why is that, is the basic parameter.

What is this? Well, r is the rate of outflow, and the rate of inflow, what's r over v?

r over v is the fractional rate of outflow.

What are the units? This is volume per minute.

In other words, it's like cosine omega t.

I'm asking, how closely does the concentration in the tank follow C sub e?

Now, what would that depend upon?

We know that from the temperature thing.

So, this means, closely means, that the phase lag is, big or small?

Remember, you saw it here first.

r, yeah, we had that. Okay, see, I had it in high school too. That's the capacitance.

That's the voltage drop across resistance.

And, that has to be the voltage drop.

And then, there is some sign convention.

It's the second equation that really falls into the category.

Okay, what's the differential equation, which is going to be, it's going to govern this situation?

What I want to know is how much B there is at any given time.

So, I want a differential equation for the quantity of the radioactive product at any given time.

Well, what's it going to look like?

But, I'm interested in the rate at which it's coming in to B.

Okay, so how does the differential equation look like?

It looks like B prime plus k2 B equals an exponential, k1 A zero e to the negative k1 t.

But, there's no reason to expect that that constant really has anything to do with k2.

The technique of solving the equation is identical.

But, you cannot interpret. So, the technique is the same, and therefore it's worth learning.

So, the constant here is negative.

Then, when I solve, my k, in other words, is now properly written as negative a.

And therefore, this formula should now become not this, but the negative k is a t.

And, here it's negative a t. And, here it is positive a t.

Now, why is it, if this is going to be the solution, why are all those things totally irrelevant?

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