WEBVTT

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This is also written in the
form, it's the k that's on the

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right hand side.
Actually, I found that source

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is of considerable difficulty.
And, in general,

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it is.
For these, the temperature

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concentration model,
it's natural to have the k on

00:01:01.000 --> 00:01:07.000
the right-hand side,
and to separate out the (q)e as

00:01:05.000 --> 00:01:11.000
part of it.
Another model for which that's

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true is mixing,
as I think I will show you on

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Monday.
On the other hand,

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there are some common
first-order models for which

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it's not a natural way to
separate things out.

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Examples would be the RC
circuit, radioactive decay,

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stuff like that.
So, this is not a universal

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utility.
But I thought that that form of

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writing it was a sufficient
utility to make a special case,

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and I emphasize it very heavily
in the nodes.

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Let's look at the equation.
And, this form will be good

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enough, the y prime.
When you solve it,

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let me remind you how the
solutions look,

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because that explains the
terminology.

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The solution looks like,
after you have done the

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integrating factor and
multiplied through,

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and integrated both sides,
in short, what you're supposed

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to do, the solution looks like y
equals, there's the term e to

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the negative k out
front times an integral which

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you can either make definite or
indefinite, according to your

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preference.
q of t times e to the kt inside

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dt,
it will help you to remember

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the opposite signs if you think
that when q is a constant,

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one, for example,
you want these two guys to

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cancel out and produce a
constant solution.

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That's a good way to remember
that the signs have to be

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opposite.
But, I don't encourage you to

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remember the formula at all.
It's just a convenient thing

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for me to be able to use right
now.

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And then, there's the other
term, which comes by putting up

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the arbitrary constant
explicitly, c e to the negative

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kt.
So, you could either write it

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this way, where this is somewhat
vague, or you could make it

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definite by putting a zero here
and a t there,

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and change the dummy variable
inside according to the way the

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notes tell you to do it.
Now, when you do this,

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and if k is positive,
that's absolutely essential,

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only when that is so,
then this term,

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as I told you a week or so ago,
this term goes to zero because

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k is positive as t goes to
infinity.

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So, this goes to zero as t
goes, and it doesn't matter what

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c is, as t goes to infinity.
This term stays some sort of

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function.
And so, this term is called the

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steady-state or long-term
solution, or it's called both,

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a long-term solution.
And this, which disappears,

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gets smaller and smaller as
time goes on,

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is therefore called the
transient because it disappears

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at the time increases to
infinity.

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So, this part uses the initial
condition, uses the initial

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value.
Let's call it y of zero,

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assuming that you
started the initial value,

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t, when t was equal to zero,
which is a common thing to do,

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although of course not
necessary.

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The starting value appears in
this term.

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This one is just some function.
Now, the general picture or the

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way that looks is,
the steady-state solution will

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be some solution like,
I don't know,

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like that, let's say.
So, that's a steady-state

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solution, the SSS.
Well, what do the other guys

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look like?
Well, the steady-state solution

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has this starting point.
Other solutions can have any of

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these other starting points.
So, in the beginning,

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they won't look like the
steady-state solution.

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But, we know that as time goes
on, they must approach it

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because this term represents the
difference between the solution

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and the steady-state solution.
So, this term is going to zero.

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And therefore,
whatever these guys do to start

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out with, after a while they
must follow the steady-state

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solution more and more closely.
They must, in short,

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be asymptotic to it.
So, the solutions to any

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equation of that form will look
like this.

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Up here, maybe it started at
127.

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That's okay.
After a while,

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it's going to start approaching
that green curve.

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Of course, they won't cross
each other.

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That's the rock star,
and these are the groupies

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trying to get close to it.
Now, but something follows from

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that picture.
Which is the steady-state

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solution?
What, in short,

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is so special about this green
curve?

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All these other white solution
curves have that same property,

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the same property that all the
other white curves and the green

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curve, too, are trying to get
close to them.

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In other words,
there is nothing special about

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the green curve.
It's just that they all want to

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get close to each other.
And therefore,

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though you can write a formula
like this, there isn't one

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steady-state solution.
There are many.

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Now, this produces vagueness.
You talk about the steady-state

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solution; which one are you
talking about?

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I have no answer to that;
the usual answer is whichever

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one looks simplest.
Normally, the one that will

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look simplest is the one where c
is zero.

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But, if this is a peculiar
function, it might be that for

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some other value of c,
you get an even simpler

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expression.
So, the steady-state solution:

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about the best I can see,
either you integrate that,

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don't use an arbitrary
constant, and use what you get,

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or pick the simplest.
Pick the value of c,

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which gives you the simplest
answer.

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Pick the simplest function,
and that's what usually called

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the steady-state solution.
Now, from that point of view,

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what I'm calling the input in
this input response point of

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view, which we are going to be
using, by the way,

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constantly, well,
pretty much all term long,

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but certainly for the next
month or so, I'm constantly

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going to be coming back to it.
The input is the q of t.

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In other words,

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it seems rather peculiar.
But the input is the right-hand

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side of the equation of the
differential equation.

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And the reason is because I'm
always thinking of the

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temperature model.
The external water bath at

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temperature T external,
the internal thing here,

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the problem is,
given this function,

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the external water bath
temperature is driving,

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so to speak,
the temperature of the inside.

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And therefore,
the input is the temperature of

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the water bath.
I don't like the word output,

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although it would be the
natural thing because this

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temperature doesn't look like an
output.

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Anyone might be willing to say,
yeah, you are inputting the

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value of the temperature here.
This, it's more likely,

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the normal term is response.
This thing, this plus the water

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bath, is a little system.
And the response of the system,

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i.e.
the change in the internal

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temperature is governed by the
driving external temperature.

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So, the input is q of t,
and the response of

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the system is the solution to
the differential equation.

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Now, if the thing is special,
as it's going to be for most of

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this period, it has that special
form, then I'm going to,

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I really want to call q sub e
the input.

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I want to call q sub e the
input, and there is no standard

00:10:02.000 --> 00:10:08.000
way of doing that,
although there's a most common

00:10:06.000 --> 00:10:12.000
way.
So, I'm just calling it the

00:10:10.000 --> 00:10:16.000
physical input,
in other words,

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the temperature input,
or the concentration input.

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And, that will be my (q)e of t,
and by the

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subscript e, you will understand
that I'm writing it in that form

00:10:26.000 --> 00:10:32.000
and thinking of this model,
or concentration model,

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or mixing model as I will show
you on Monday.

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By the way, this is often
handled, I mean,

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how would you handle this to
get rid of a k?

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Well, divide through by k.
So, this equation is often,

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in the literature,
written this way:

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one over k times y prime plus y
is equal to, well,

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now they call it q of t,
not (q)e of t because they've

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gotten rid of this funny factor.
But

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I will continue to call it (q)e.
So, in other words,

00:11:04.000 --> 00:11:10.000
and this part this is just,
frankly, called the input.

00:11:09.000 --> 00:11:15.000
It doesn't say physical or
anything.

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And, this is the solution,
it's then the response,

00:11:16.000 --> 00:11:22.000
and this funny coefficient of y
prime,

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that's not in standard linear
form, is it, anymore?

00:11:23.000 --> 00:11:29.000
But, it's a standard form if
you want to do this input

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response analysis.
So, this is also a way of

00:11:31.000 --> 00:11:37.000
writing the equation.
I'm not going to use it because

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how many standard forms could
this poor little course absorb?

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I'll stick to that one.
Okay, you have,

00:11:47.000 --> 00:11:53.000
then, the superposition
principle, which I don't think

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I'm going to-- the solution,
which solution?

00:11:57.000 --> 00:12:03.000
Well, normally it means any
solution, or in other words,

00:12:02.000 --> 00:12:08.000
the steady-state solution.
Now, notice that terminology

00:12:07.000 --> 00:12:13.000
only makes sense if k is
positive.

00:12:10.000 --> 00:12:16.000
And, in fact,
there is nothing like the

00:12:13.000 --> 00:12:19.000
picture, the picture doesn't
look at all like this if k is

00:12:17.000 --> 00:12:23.000
negative, and therefore,
the terms would steady state,

00:12:20.000 --> 00:12:26.000
transient would be totally
inappropriate if k were

00:12:24.000 --> 00:12:30.000
negative.
So, this assumes definitely

00:12:26.000 --> 00:12:32.000
that k has to be greater than
zero.

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Otherwise, no.
So, I'll call this the physical

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input.
And then, you have the

00:12:35.000 --> 00:12:41.000
superposition principle,
which I really can't improve

00:12:40.000 --> 00:12:46.000
upon what's written in the
notes, this superposition of

00:12:44.000 --> 00:12:50.000
inputs.
Whether they are physical

00:12:47.000 --> 00:12:53.000
inputs or nonphysical inputs,
if the input q of t produces

00:12:51.000 --> 00:12:57.000
the response, y of t,

00:12:54.000 --> 00:13:00.000
and q two of t produces the
response, y two of t,

00:13:02.000 --> 00:13:08.000
-- then a simple calculation
with the differential equation

00:13:07.000 --> 00:13:13.000
shows you that by,
so to speak,

00:13:10.000 --> 00:13:16.000
adding, that the sum of these
two, I stated it very generally

00:13:15.000 --> 00:13:21.000
in the notes but it corresponds,
we will have as the response

00:13:21.000 --> 00:13:27.000
y1, the steady-state response y1
plus y2,

00:13:25.000 --> 00:13:31.000
and a constant times y1.
That's an expression,

00:13:30.000 --> 00:13:36.000
essentially,
of the linear,

00:13:32.000 --> 00:13:38.000
it uses the fact that the
special form of the equation,

00:13:35.000 --> 00:13:41.000
and we will have a very
efficient and elegant way of

00:13:38.000 --> 00:13:44.000
seeing this when we study higher
order equations.

00:13:41.000 --> 00:13:47.000
For now, I will just,
the little calculation that's

00:13:45.000 --> 00:13:51.000
done in the notes will suffice
for first-order equations.

00:13:48.000 --> 00:13:54.000
If you don't have a complicated
equation, there's no point in

00:13:52.000 --> 00:13:58.000
making a fuss over proofs using
it.

00:13:54.000 --> 00:14:00.000
But essentially,
it uses the fact that the

00:13:57.000 --> 00:14:03.000
equation is linear.
Or, that's bad,

00:14:01.000 --> 00:14:07.000
so linearity of the ODE.
In other words,

00:14:04.000 --> 00:14:10.000
it's a consequence of the fact
that the equation looks the way

00:14:09.000 --> 00:14:15.000
it does.
And, something like this would

00:14:12.000 --> 00:14:18.000
not, in any sense,
be true if the equation,

00:14:15.000 --> 00:14:21.000
for example,
had here a y squared

00:14:18.000 --> 00:14:24.000
instead of t.
Everything I'm saying this

00:14:21.000 --> 00:14:27.000
period would be total nonsense
and totally inapplicable.

00:14:27.000 --> 00:14:33.000
Now, today, what I wanted to
discuss was, what's in the notes

00:14:32.000 --> 00:14:38.000
that I gave you today,
which is, what happens when the

00:14:36.000 --> 00:14:42.000
physical input is trigonometric?
For certain reasons,

00:14:41.000 --> 00:14:47.000
that's the most important case
there is.

00:14:44.000 --> 00:14:50.000
It's because of the existence
of what are called Fourier

00:14:49.000 --> 00:14:55.000
series, and there are a couple
of words about them.

00:14:53.000 --> 00:14:59.000
That's something we will be
studying in about three weeks or

00:14:58.000 --> 00:15:04.000
so.
What's going on,

00:15:01.000 --> 00:15:07.000
roughly, is that,
so I'm going to take the

00:15:06.000 --> 00:15:12.000
equation in the form y prime
plus ky equals k times

00:15:12.000 --> 00:15:18.000
(q)e of t,
and the input that I'm

00:15:17.000 --> 00:15:23.000
interested in is when this is a
simple one that you use on the

00:15:23.000 --> 00:15:29.000
visual that you did about two
points worth of work for handing

00:15:30.000 --> 00:15:36.000
in today, cosine omega t.

00:15:36.000 --> 00:15:42.000
So, if you like,
k here.

00:15:37.000 --> 00:15:43.000
So, the (q)e is cosine omega t.
That was the physical input.

00:15:41.000 --> 00:15:47.000
And, omega, as you know,
is, you have to be careful when

00:15:44.000 --> 00:15:50.000
you use the word frequency.
I assume you got this from

00:15:48.000 --> 00:15:54.000
physics class all last semester.
But anyway, just to remind you,

00:15:52.000 --> 00:15:58.000
there's a whole yoga of five or
six terms that go whenever

00:15:56.000 --> 00:16:02.000
you're talking about
trigonometric functions.

00:16:00.000 --> 00:16:06.000
Instead of giving a long
explanation, the end of the

00:16:03.000 --> 00:16:09.000
second page of the notes just
gives you a reference list of

00:16:08.000 --> 00:16:14.000
what you are expected to know
for 18.03 and physics as well,

00:16:12.000 --> 00:16:18.000
with a brief one or two line
description of what each of

00:16:17.000 --> 00:16:23.000
those means.
So, think of it as something to

00:16:20.000 --> 00:16:26.000
refer back to if you have
forgotten.

00:16:23.000 --> 00:16:29.000
But, omega is what's called the
angular frequency or the

00:16:27.000 --> 00:16:33.000
circular frequency.
It's somewhat misleading to

00:16:31.000 --> 00:16:37.000
call it the frequency,
although I probably will.

00:16:36.000 --> 00:16:42.000
It's the angular frequency.
It's, in other words,

00:16:39.000 --> 00:16:45.000
it's the number of complete
oscillations.

00:16:42.000 --> 00:16:48.000
This cosine omega t
is going up and down right?

00:16:47.000 --> 00:16:53.000
So, a complete oscillation as
it goes down and then returns to

00:16:51.000 --> 00:16:57.000
where it started.
That's a complete oscillation.

00:16:55.000 --> 00:17:01.000
This is only half an
oscillation because you didn't

00:16:58.000 --> 00:17:04.000
give it a chance to get back.
Okay, so the number of complete

00:17:03.000 --> 00:17:09.000
oscillations in how much time,
well, in two pi,

00:17:06.000 --> 00:17:12.000
in the distance,
two pi on the t-axis in the

00:17:09.000 --> 00:17:15.000
interval of length two pi
because, for example,

00:17:13.000 --> 00:17:19.000
if omega is one,
cosine t takes two

00:17:16.000 --> 00:17:22.000
pi to repeat itself,
right?

00:17:20.000 --> 00:17:26.000
If omega were two,
it would repeat itself.

00:17:23.000 --> 00:17:29.000
It would make two complete
oscillations in the interval,

00:17:28.000 --> 00:17:34.000
two pi.
So, it's what happens to the

00:17:31.000 --> 00:17:37.000
interval, two pi,
not what happens in the time

00:17:35.000 --> 00:17:41.000
interval, one,
which is the natural meaning of

00:17:39.000 --> 00:17:45.000
the word frequency.
There's always this factor of

00:17:43.000 --> 00:17:49.000
two pi that floats around to
make all of your formulas and

00:17:48.000 --> 00:17:54.000
solutions incorrect.
Okay, now, so,

00:17:51.000 --> 00:17:57.000
what I'm out to do is,
the problem is for the physical

00:17:55.000 --> 00:18:01.000
input, (q)e cosine omega t,

00:17:59.000 --> 00:18:05.000
find the response.
In other words,

00:18:02.000 --> 00:18:08.000
solve the differential
equation.

00:18:07.000 --> 00:18:13.000
In short, for the visual that
you looked at,

00:18:11.000 --> 00:18:17.000
I think I've forgot the colors
now.

00:18:14.000 --> 00:18:20.000
The input was in green,
maybe, but I do remember that

00:18:19.000 --> 00:18:25.000
the response was in yellow.
I think I remember that.

00:18:24.000 --> 00:18:30.000
So, find the response,
yellow, and the input was,

00:18:28.000 --> 00:18:34.000
what color was it,
green?

00:18:30.000 --> 00:18:36.000
Blue, blue.
Light blue.

00:18:34.000 --> 00:18:40.000
Okay, so we've got to solve the
differential equation.

00:18:39.000 --> 00:18:45.000
Now, it's a question of how I'm
going to solve the differential

00:18:46.000 --> 00:18:52.000
equation.
I'm going to use complex

00:18:49.000 --> 00:18:55.000
numbers throughout,
A because that's the way it's

00:18:54.000 --> 00:19:00.000
usually done.
B, to give you practice using

00:18:59.000 --> 00:19:05.000
complex numbers,
and I don't think I need any

00:19:04.000 --> 00:19:10.000
other reasons.
So, I'm going to use complex

00:19:09.000 --> 00:19:15.000
numbers.
I'm going to complexify.

00:19:13.000 --> 00:19:19.000
To use complex numbers,
what you do is complexification

00:19:18.000 --> 00:19:24.000
of the problem.
So, I'm going to complexify the

00:19:23.000 --> 00:19:29.000
problem, turn it into the domain
of complex numbers.

00:19:29.000 --> 00:19:35.000
So, take the differential
equation, turn it into a

00:19:33.000 --> 00:19:39.000
differential equation involving
complex numbers,

00:19:37.000 --> 00:19:43.000
solve that, and then go back to
the real domain to get the

00:19:42.000 --> 00:19:48.000
answer, since it's easier to
integrate exponentials.

00:19:46.000 --> 00:19:52.000
And therefore,
try to introduce,

00:19:49.000 --> 00:19:55.000
try to change the trigonometric
functions into complex

00:19:53.000 --> 00:19:59.000
exponentials,
simply because the work will be

00:19:57.000 --> 00:20:03.000
easier to do.
All right, so let's do it.

00:20:02.000 --> 00:20:08.000
To change this differential
equation, remember,

00:20:05.000 --> 00:20:11.000
I've got cosine omega t here.

00:20:09.000 --> 00:20:15.000
I'm going to use the fact that
e to the i omega t,

00:20:14.000 --> 00:20:20.000
Euler's formula,
that the real part of it is

00:20:17.000 --> 00:20:23.000
cosine omega t.
So, I'm going to view this as

00:20:22.000 --> 00:20:28.000
the real part of this complex
function.

00:20:25.000 --> 00:20:31.000
But, I will throw at the
imaginary part,

00:20:28.000 --> 00:20:34.000
too, since at one point we will
need it.

00:20:31.000 --> 00:20:37.000
Now, what is the equation,
then, that it's going to turn

00:20:36.000 --> 00:20:42.000
into?
The complexified equation is

00:20:41.000 --> 00:20:47.000
going to be y prime plus ky
equals, and now,

00:20:46.000 --> 00:20:52.000
instead of the right hand side,
k times cosine omega t,

00:20:53.000 --> 00:20:59.000
I'll use the whole complex
exponential, e i omega t.

00:21:00.000 --> 00:21:06.000
Now, I have a problem because

00:21:06.000 --> 00:21:12.000
y, here, in this equation,
y means the real function which

00:21:09.000 --> 00:21:15.000
solves that problem.
I therefore cannot continue to

00:21:13.000 --> 00:21:19.000
call this y because I want y to
be a real function.

00:21:16.000 --> 00:21:22.000
I have to change its name.
Since this is complex function

00:21:20.000 --> 00:21:26.000
on the right-hand side,
I will have to expect a complex

00:21:24.000 --> 00:21:30.000
solution to the differential
equation.

00:21:28.000 --> 00:21:34.000
I'm going to call that complex
solution y tilda.

00:21:32.000 --> 00:21:38.000
Now, that's what I would also
use as the designation for the

00:21:38.000 --> 00:21:44.000
variable.
So, y tilda is the complex

00:21:42.000 --> 00:21:48.000
solution.
And, it's going to have the

00:21:46.000 --> 00:21:52.000
form y1 plus i times y2.

00:21:49.000 --> 00:21:55.000
It's going to be the complex
solution.

00:21:53.000 --> 00:21:59.000
And now, what I say is,
so, solve it.

00:21:57.000 --> 00:22:03.000
Find this complex solution.
So, find the program is to find

00:22:03.000 --> 00:22:09.000
y tilde, --
-- that's the complex solution.

00:22:08.000 --> 00:22:14.000
And then I say,
all you have to do is take the

00:22:12.000 --> 00:22:18.000
real part of that,
and that will answer the

00:22:16.000 --> 00:22:22.000
original problem.
Then, y1, that's the real part

00:22:20.000 --> 00:22:26.000
of it, right?
It's a function,

00:22:23.000 --> 00:22:29.000
you know, like this is cosine
plus sine, as it was over here,

00:22:28.000 --> 00:22:34.000
it will naturally be something
different.

00:22:31.000 --> 00:22:37.000
It will be something different,
but that part of it,

00:22:36.000 --> 00:22:42.000
the real part will solve the
original problem,

00:22:40.000 --> 00:22:46.000
the original,
real, ODE.

00:22:44.000 --> 00:22:50.000
Now, you will say,
you expect us to believe that?

00:22:47.000 --> 00:22:53.000
Well, yes, in fact.
I think we've got a lot to do,

00:22:51.000 --> 00:22:57.000
so since the argument for this
is given in the nodes,

00:22:54.000 --> 00:23:00.000
so, read this in the notes.
It only takes a line or two of

00:22:58.000 --> 00:23:04.000
standard work with
differentiation.

00:23:02.000 --> 00:23:08.000
So, read in the notes the
argument for that,

00:23:05.000 --> 00:23:11.000
why that's so.
It just amounts to separating

00:23:09.000 --> 00:23:15.000
real and imaginary parts.
Okay, so let's,

00:23:13.000 --> 00:23:19.000
now, solve this.
Since that's our program,

00:23:17.000 --> 00:23:23.000
all we have to find is the
solution.

00:23:20.000 --> 00:23:26.000
Well, just use integrating
factors and just do it.

00:23:25.000 --> 00:23:31.000
So, the integrating factor will
be, what, e to the,

00:23:29.000 --> 00:23:35.000
I don't want to use that
formula.

00:23:34.000 --> 00:23:40.000
So, the integrating factor will
be e to the kt is the

00:23:38.000 --> 00:23:44.000
integrating factor.
If I multiply through both

00:23:42.000 --> 00:23:48.000
sides by the integrating factor,
then the left-hand side will

00:23:46.000 --> 00:23:52.000
become y e to the kt,
the way it always does,

00:23:50.000 --> 00:23:56.000
prime, Y tilde, sorry,

00:23:53.000 --> 00:23:59.000
and the right-hand side will
be, now I'm going to start

00:23:57.000 --> 00:24:03.000
combining exponentials.
It will be k times e to the

00:24:03.000 --> 00:24:09.000
power i times omega t plus k.

00:24:11.000 --> 00:24:17.000
I'm going to write that k plus
omega t.

00:24:31.000 --> 00:24:37.000
i omega t plus k.

00:24:36.000 --> 00:24:42.000
Thank you.
i omega t plus k,

00:24:40.000 --> 00:24:46.000
or k plus i omega t.

00:24:47.000 --> 00:24:53.000
kt?
Sorry.

00:24:48.000 --> 00:24:54.000
So, it's k times e to the i
omega t times e to the kt.

00:24:57.000 --> 00:25:03.000
So, that's (k plus i omega)

00:25:06.000 --> 00:25:12.000
times t. Sorry.

00:25:10.000 --> 00:25:16.000
So, y tilda e to the kt
is k divided by,

00:25:17.000 --> 00:25:23.000
now I integrate this,
so it essentially reproduces

00:25:23.000 --> 00:25:29.000
itself, except you have to put
down on the bottom k plus i

00:25:30.000 --> 00:25:36.000
omega.
I'll take the final step.

00:25:35.000 --> 00:25:41.000
What's y tilda equals,
see, when you do it this way,

00:25:38.000 --> 00:25:44.000
then you don't get a messy
looking formula that you

00:25:42.000 --> 00:25:48.000
substitute into and that is
scary looking.

00:25:44.000 --> 00:25:50.000
This is never scary.
Now, I'm going to do two things

00:25:48.000 --> 00:25:54.000
simultaneously.
First of all,

00:25:49.000 --> 00:25:55.000
here, if I multiply,
after I get the answer,

00:25:52.000 --> 00:25:58.000
I'm going to want to multiply
it by e to the negative kt,

00:25:56.000 --> 00:26:02.000
right,
to solve for y tilda.

00:26:00.000 --> 00:26:06.000
If I multiply this by e to the
negative kt, then that just gets

00:26:04.000 --> 00:26:10.000
rid of the k that I put in,
and left back with e to the i

00:26:08.000 --> 00:26:14.000
omega t.
So, that side is easy.

00:26:10.000 --> 00:26:16.000
All that is left is e to the i
omega t.

00:26:14.000 --> 00:26:20.000
Now, what's interesting is this
thing out here,

00:26:18.000 --> 00:26:24.000
k plus i omega.
I'm going to take a typical

00:26:22.000 --> 00:26:28.000
step of scaling it.
And you scale it.

00:26:24.000 --> 00:26:30.000
I'm going to divide the top and
bottom by k.

00:26:29.000 --> 00:26:35.000
And, what does that produce?
One divided by one plus i times

00:26:35.000 --> 00:26:41.000
omega over k.

00:26:40.000 --> 00:26:46.000
What I've done is take these
two separate constants,

00:26:45.000 --> 00:26:51.000
and shown that the critical
thing is their ratio.

00:26:51.000 --> 00:26:57.000
Okay, now, what I have to do
now is take the real part.

00:26:57.000 --> 00:27:03.000
Now, there are two ways to do
this.

00:27:01.000 --> 00:27:07.000
There are two ways to do this.
Both are instructive.

00:27:08.000 --> 00:27:14.000
So, there are two methods.
I have a multiplication.

00:27:13.000 --> 00:27:19.000
The problem is,
of course, that these two

00:27:17.000 --> 00:27:23.000
things are multiplied together.
And, one of them is,

00:27:23.000 --> 00:27:29.000
essentially,
in Cartesian form,

00:27:26.000 --> 00:27:32.000
and the other is,
essentially,

00:27:29.000 --> 00:27:35.000
in polar form.
You have to make a decision.

00:27:35.000 --> 00:27:41.000
Either go polar,
it sounds like go postal,

00:27:40.000 --> 00:27:46.000
doesn't it, or worse,
like a bear,

00:27:45.000 --> 00:27:51.000
savage, attack it savagely,
which that's a very good,

00:27:52.000 --> 00:27:58.000
aggressive attitude to have
when doing a problem,

00:27:58.000 --> 00:28:04.000
or we can go Cartesian.
Going polar is a little faster,

00:28:05.000 --> 00:28:11.000
and I think it's what's done in
the nodes.

00:28:08.000 --> 00:28:14.000
The notes to do both of these.
They just do the first.

00:28:11.000 --> 00:28:17.000
On the other hand,
they give you a formula,

00:28:14.000 --> 00:28:20.000
which is the critical thing
that you will need to go

00:28:18.000 --> 00:28:24.000
Cartesian.
I hope I can do both of them if

00:28:21.000 --> 00:28:27.000
we sort of hurry along.
How do I go polar?

00:28:24.000 --> 00:28:30.000
To go polar,
what you want to do is express

00:28:27.000 --> 00:28:33.000
this thing in polar form.
Now, one of the things I didn't

00:28:32.000 --> 00:28:38.000
emphasize enough,
probably, when I talked to you

00:28:35.000 --> 00:28:41.000
about complex numbers last time
is, so I will remind you,

00:28:40.000 --> 00:28:46.000
which saves my conscience and
doesn't hurt yours,

00:28:43.000 --> 00:28:49.000
suppose you have alpha as a
complex number.

00:28:47.000 --> 00:28:53.000
See, this complex number is a
reciprocal.

00:28:50.000 --> 00:28:56.000
The good number is what's down
below.

00:28:52.000 --> 00:28:58.000
Unfortunately,
it's downstairs.

00:28:55.000 --> 00:29:01.000
You should know,
like you know the back of your

00:28:58.000 --> 00:29:04.000
hand, which nobody knows,
one over alpha.

00:29:03.000 --> 00:29:09.000
So that's the form.
The number I'm interested in,

00:29:05.000 --> 00:29:11.000
that coefficient,
it is of the form one over

00:29:08.000 --> 00:29:14.000
alpha.
One over alpha times alpha is

00:29:10.000 --> 00:29:16.000
equal to one.

00:29:13.000 --> 00:29:19.000
And, from that,
it follows, first of all,

00:29:15.000 --> 00:29:21.000
if I take absolute values,
if the absolute value of one

00:29:19.000 --> 00:29:25.000
over alpha times the absolute
value of this is equal to one,

00:29:22.000 --> 00:29:28.000
so, this is equal to one over
the absolute value of alpha.

00:29:26.000 --> 00:29:32.000
I think you all knew that.
I'm a little less certain you

00:29:29.000 --> 00:29:35.000
knew how to take care of the
angles.

00:29:33.000 --> 00:29:39.000
How about the argument?
Well, the argument of the

00:29:36.000 --> 00:29:42.000
angle, in other words,
the angle of one over alpha

00:29:40.000 --> 00:29:46.000
plus, because when you multiply,
angles add.

00:29:44.000 --> 00:29:50.000
Remember that.
Plus, the angle associated with

00:29:48.000 --> 00:29:54.000
alpha has to be the angle
associated with one.

00:29:51.000 --> 00:29:57.000
But what's that?
One is out here.

00:29:54.000 --> 00:30:00.000
What's the angle of one?
Zero.

00:30:06.000 --> 00:30:12.000
Therefore, the argument,
the absolute value of this

00:30:10.000 --> 00:30:16.000
thing is want over the absolute
value.

00:30:14.000 --> 00:30:20.000
That's easy.
And, you should know that the

00:30:18.000 --> 00:30:24.000
argument of want over alpha is
equal to minus the argument of

00:30:23.000 --> 00:30:29.000
alpha.
So, when you take reciprocal,

00:30:27.000 --> 00:30:33.000
the angle turns into its
negative.

00:30:30.000 --> 00:30:36.000
Okay, I'm going to use that
now, because my aim is to turn

00:30:35.000 --> 00:30:41.000
this into polar form.
So, let's do that someplace,

00:30:40.000 --> 00:30:46.000
I guess here.
So, I want the polar form for

00:30:48.000 --> 00:30:54.000
one over one plus i times omega
over k.

00:31:00.000 --> 00:31:06.000
Okay, I will draw a picture.

00:31:04.000 --> 00:31:10.000
Here's one.
Here is omega over k.

00:31:09.000 --> 00:31:15.000
Let's call this angle phi.

00:31:12.000 --> 00:31:18.000
It's a natural thing to call
it.

00:31:15.000 --> 00:31:21.000
It's a right triangle,
of course.

00:31:18.000 --> 00:31:24.000
Okay, now, this is going to be
a complex number times e to an

00:31:24.000 --> 00:31:30.000
angle.
Now, what's the angle going to

00:31:28.000 --> 00:31:34.000
be?
Well, this is a complex number,

00:31:32.000 --> 00:31:38.000
the angle for the complex
number.

00:31:35.000 --> 00:31:41.000
So, the argument of the complex
number, one plus i times omega

00:31:40.000 --> 00:31:46.000
over k is how much?

00:31:43.000 --> 00:31:49.000
Well, there's the complex
number one plus i over one plus

00:31:48.000 --> 00:31:54.000
i times omega over k.

00:31:53.000 --> 00:31:59.000
Its angle is phi.
So, the argument of this is

00:31:57.000 --> 00:32:03.000
phi, and therefore,
the argument of its reciprocal

00:32:01.000 --> 00:32:07.000
is negative phi.
So, it's e to the minus i phi.

00:32:06.000 --> 00:32:12.000
And, what's A?

00:32:09.000 --> 00:32:15.000
A is one over the absolute
value of that complex number.

00:32:14.000 --> 00:32:20.000
Well, the absolute value of
this complex number is one plus

00:32:20.000 --> 00:32:26.000
omega over k squared.

00:32:24.000 --> 00:32:30.000
So, the A is going to be one
over that, the square root of

00:32:29.000 --> 00:32:35.000
one plus omega over k,
the quantity squared,

00:32:33.000 --> 00:32:39.000
times e to the minus i phi.

00:32:39.000 --> 00:32:45.000
See, I did that.

00:32:43.000 --> 00:32:49.000
That's a critical step.
You must turn that coefficient.

00:32:46.000 --> 00:32:52.000
If you want to go polar,
you must turn is that

00:32:49.000 --> 00:32:55.000
coefficient, write that
coefficient in the polar form.

00:32:52.000 --> 00:32:58.000
And for that,
you need these basic facts

00:32:54.000 --> 00:33:00.000
about, draw the complex number,
draw its angle,

00:32:57.000 --> 00:33:03.000
and so on and so forth.
And now, what's there for the

00:33:02.000 --> 00:33:08.000
solution?
Once you've done that,

00:33:06.000 --> 00:33:12.000
the work is over.
What's the complex solution?

00:33:10.000 --> 00:33:16.000
The complex solution is this.
I've just found the polar form

00:33:16.000 --> 00:33:22.000
for this.
So, I multiply it by e to the i

00:33:20.000 --> 00:33:26.000
omega t,
which means these things add.

00:33:25.000 --> 00:33:31.000
So, it's equal to A,
this A, times e to the i omega

00:33:30.000 --> 00:33:36.000
t minus i times phi.

00:33:37.000 --> 00:33:43.000
Or, in other words,
the coefficient is one over,

00:33:42.000 --> 00:33:48.000
this is a real number,
now, square root of one plus

00:33:47.000 --> 00:33:53.000
omega over k squared.

00:33:53.000 --> 00:33:59.000
And, this is e to the,
see if I get it right,

00:33:58.000 --> 00:34:04.000
now.
And finally,

00:34:00.000 --> 00:34:06.000
now, what's the answer to our
real problem?

00:34:05.000 --> 00:34:11.000
y1: the real answer.
I mean: the really real answer.

00:34:11.000 --> 00:34:17.000
What is it?
Well, this is a real number.

00:34:13.000 --> 00:34:19.000
So, I simply reproduce that as
the coefficient out front.

00:34:17.000 --> 00:34:23.000
And for the other part,
I want the real part of that.

00:34:20.000 --> 00:34:26.000
But you can write that down
instantly.

00:34:23.000 --> 00:34:29.000
So, let's recopy the
coefficient.

00:34:25.000 --> 00:34:31.000
And then, I want just the real
part of this.

00:34:28.000 --> 00:34:34.000
Well, this is e to the i times
some crazy angle.

00:34:32.000 --> 00:34:38.000
So, the real part is the cosine
of that crazy angle.

00:34:36.000 --> 00:34:42.000
So, it's the cosine of omega t
minus phi.

00:34:40.000 --> 00:34:46.000
And, if somebody says,

00:34:42.000 --> 00:34:48.000
yeah, well, okay,
I got the omega k,

00:34:45.000 --> 00:34:51.000
I know what that is.
That came from the problem,

00:34:49.000 --> 00:34:55.000
the driving frequency,
driving angular frequency.

00:34:52.000 --> 00:34:58.000
That was omega,
and k, I guess,

00:34:55.000 --> 00:35:01.000
k was the conductivity,
the thing which told you how

00:34:59.000 --> 00:35:05.000
quickly the heat that penetrated
the walls of the little inner

00:35:03.000 --> 00:35:09.000
chamber.
So, that's okay,

00:35:07.000 --> 00:35:13.000
but what's this phi?
Well, the best way to get phi

00:35:11.000 --> 00:35:17.000
is just to draw that picture,
but if you want a formula for

00:35:15.000 --> 00:35:21.000
phi, phi will be,
well, I guess from the picture,

00:35:19.000 --> 00:35:25.000
it's the arc tangent of omega,
k, divided by k,

00:35:23.000 --> 00:35:29.000
over one,
which I don't have to put

00:35:28.000 --> 00:35:34.000
in.
So, it's this number,

00:35:30.000 --> 00:35:36.000
phi, in reference to this
function.

00:35:34.000 --> 00:35:40.000
See, if the phi weren't there,
this would be cosine omega t,

00:35:39.000 --> 00:35:45.000
and we all know what that looks

00:35:44.000 --> 00:35:50.000
like.
The phi is called the phase lag

00:35:48.000 --> 00:35:54.000
or phase delay,
something like that,

00:35:51.000 --> 00:35:57.000
the phase lag of the function.
What does it represent?

00:35:56.000 --> 00:36:02.000
It represents,
let me draw you a picture.

00:36:02.000 --> 00:36:08.000
Let's draw the picture like
this.

00:36:05.000 --> 00:36:11.000
Here's cosine omega t.

00:36:08.000 --> 00:36:14.000
Now, regular cosine would look
sort of like that.

00:36:13.000 --> 00:36:19.000
But, I will indicate that the
angular frequency is not one by

00:36:18.000 --> 00:36:24.000
making my cosine squinchy up a
little too much.

00:36:23.000 --> 00:36:29.000
Everybody can tell that that's
the cosine on a limp axis,

00:36:28.000 --> 00:36:34.000
something for Salvador Dali,
okay.

00:36:31.000 --> 00:36:37.000
So, there's cosine of
something.

00:36:36.000 --> 00:36:42.000
So, what was it?
Blue?

00:36:37.000 --> 00:36:43.000
I don't have blue.
Yes, I have blue.

00:36:41.000 --> 00:36:47.000
Okay, so now you will know what
I'm talking about because this

00:36:46.000 --> 00:36:52.000
looks just like the screen on
your computer when you put in

00:36:52.000 --> 00:36:58.000
the visual for this.
Frequency: your response order

00:36:56.000 --> 00:37:02.000
one.
So, this is cosine omega t.

00:36:59.000 --> 00:37:05.000
Now, how will cosine omega t

00:37:04.000 --> 00:37:10.000
minus phi look?

00:37:07.000 --> 00:37:13.000
Well, it'll be moved over.
Let's, for example,

00:37:10.000 --> 00:37:16.000
suppose phi were pi over two.
Now, where's pi over two on the

00:37:15.000 --> 00:37:21.000
picture?
Well, what I do is cosine omega

00:37:18.000 --> 00:37:24.000
t minus this.
I move it over by one,

00:37:22.000 --> 00:37:28.000
so that this point becomes that
one, and it looks like,

00:37:26.000 --> 00:37:32.000
the site will look like this.
In other words,

00:37:30.000 --> 00:37:36.000
I shove it over by,
so this is the point where

00:37:33.000 --> 00:37:39.000
omega t equals pi over two.

00:37:39.000 --> 00:37:45.000
It's not the value of t.
It's not the value of t.

00:37:43.000 --> 00:37:49.000
It's the value of omega t.

00:37:46.000 --> 00:37:52.000
And, when I do that,
then the blue curve has been

00:37:50.000 --> 00:37:56.000
shoved over one quarter of its
total cycle, and that turns it,

00:37:55.000 --> 00:38:01.000
of course, into the sine curve,
which I hope I can draw.

00:38:01.000 --> 00:38:07.000
So, this goes up to there,
and then, it's got to get back

00:38:05.000 --> 00:38:11.000
through.
Let me stop there while I'm

00:38:08.000 --> 00:38:14.000
ahead.
So, this is sine omega t,

00:38:11.000 --> 00:38:17.000
the yellow thing,

00:38:13.000 --> 00:38:19.000
but that's also,
in another life,

00:38:16.000 --> 00:38:22.000
cosine of omega t minus pi over
two.

00:38:21.000 --> 00:38:27.000
The main thing is you don't
subtract, the pi over two is not

00:38:26.000 --> 00:38:32.000
being subtracted from the t.
It's being subtracted from the

00:38:32.000 --> 00:38:38.000
whole expression,
and this whole expression

00:38:35.000 --> 00:38:41.000
represents an angle,
which tells you where you are

00:38:39.000 --> 00:38:45.000
in the travel,
a long cosine to this.

00:38:41.000 --> 00:38:47.000
What this quantity gets to be
two pi, you're back where you

00:38:46.000 --> 00:38:52.000
started.
That's not the distance on the

00:38:49.000 --> 00:38:55.000
t axis.
It's the angle through which

00:38:51.000 --> 00:38:57.000
you go through.
In other words,

00:38:54.000 --> 00:39:00.000
does number describes where you
are on the cosine cycle.

00:38:58.000 --> 00:39:04.000
It doesn't tell you,
it's not aiming at telling you

00:39:01.000 --> 00:39:07.000
exactly where you are on the t
axis.

00:39:04.000 --> 00:39:10.000
The response function looks
like one over the square root of

00:39:09.000 --> 00:39:15.000
one plus omega over k squared
times cosine omega t minus phi.

00:39:19.000 --> 00:39:25.000
And, I asked you on the problem
set, if k goes up,

00:39:24.000 --> 00:39:30.000
in other words,
if the conductivity rises,

00:39:28.000 --> 00:39:34.000
if heat can get more rapidly
from the outside to the inside,

00:39:34.000 --> 00:39:40.000
for example,
how does that affect the

00:39:38.000 --> 00:39:44.000
amplitude?
This is the amplitude,

00:39:42.000 --> 00:39:48.000
A, and the phase lag.
In other words,

00:39:47.000 --> 00:39:53.000
how does this affect the
response?

00:39:51.000 --> 00:39:57.000
And now, you can see.
If k goes up,

00:39:55.000 --> 00:40:01.000
this fraction is becoming
smaller.

00:39:59.000 --> 00:40:05.000
That means the denominator is
becoming smaller,

00:40:05.000 --> 00:40:11.000
and therefore,
the amplitude is going up.

00:40:12.000 --> 00:40:18.000
What's happening to the phase
lag?

00:40:14.000 --> 00:40:20.000
Well, the phase lag looks like
this: phi one omega over k.

00:40:20.000 --> 00:40:26.000
If k is going up,

00:40:23.000 --> 00:40:29.000
then the size of this side is
going down, and the angle is

00:40:28.000 --> 00:40:34.000
going down.
Now, that part is intuitive.

00:40:32.000 --> 00:40:38.000
I would have expected everybody
to get that.

00:40:36.000 --> 00:40:42.000
It's the heat gets in quickly,
more quickly,

00:40:40.000 --> 00:40:46.000
then the amplitude will match
more quickly.

00:40:44.000 --> 00:40:50.000
This will rise,
and get fairly close to one,

00:40:47.000 --> 00:40:53.000
in fact, and there should be
very little lag in the way the

00:40:53.000 --> 00:40:59.000
response follows input.
But how about the other one?

00:40:57.000 --> 00:41:03.000
Okay, I give you two minutes.
The other one,

00:41:01.000 --> 00:41:07.000
you will figure out yourself.