WEBVTT
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The topic for today is how to
change variables.
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So, we're talking about
substitutions and differential
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equations, or changing
variables.
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That might seem like a sort of
fussy thing to talk about in the
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third or fourth lecture,
but the reason is that so far,
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you know how to solve two kinds
of differential equations,
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two kinds of first-order
differential equations,
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one where you can separate
variables, and the linear
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equation that we talked about
last time.
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Now, the sad fact is that in
some sense, those are the only
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two general methods there are,
that those are the only two
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kinds of equations that can
always be solved.
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Well, what about all the
others?
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The answer is that to a great
extent, all the other equations
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that can be solved,
the solution can be done by
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changing the variables in the
equation to reduce it to one of
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the cases that we can already
do.
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Now, I'm going to give you two
examples of that,
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two significant examples of
that today.
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But, ultimately,
as you will see,
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the way the equations are
solved is by changing them into
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a linear equation,
or an equation where the
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variables are separable.
However, that's for a few
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minutes.
The first change of variables
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that I want to talk about is an
almost trivial one.
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But it's the most common kind
there is, and you've already had
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it in physics class.
But I think it's so important
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in the science and engineering
subjects that it's a good idea,
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even in 18.03,
to call attention to it
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explicitly.
So, in that sense,
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the most common change of
variables is the one simple one
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called scaling.
So, again, the kind of equation
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I'm talking about is a general
first-order equation.
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And, scaling simply means to
change the coordinates,
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in effect, or axes,
to change the coordinates on
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the axes to scale the axes,
to either stretch them or
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contract them.
So, what does the change of
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variable actually look like?
Well, it means you introduce
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new variables,
where x1 is equal to x times
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something or times a constant.
I'll write it as divided by a
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constant, since that tends to be
a little bit more the way people
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think of it.
And y, the same.
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So, the new variable y1 is
related to the old one by an
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equation of that form.
So, a, b are constants.
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So, those are the equations.
Why does one do this?
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Well, for a lot of reasons.
But, maybe we can list them.
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You, for example,
could be changing units.
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That's a common reason in
physics.
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Changing the units that he
used, you would have to make a
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change of coordinates of this
form.
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Perhaps the even more important
reason is to,
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sometimes it's used to make the
variables dimensionless.
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In other words,
so that the variables become
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pure numbers,
with no units attached to them.
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Since you are well aware of the
tortures involved in dealing
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with units in physics,
the point of making variables,
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I'm sorry, dimensionless,
I don't have to sell that.
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Dimensionless,
i.e.
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no units, without any units
attached.
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It just represents the number
three, not three seconds,
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or three grahams,
or anything like that.
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And, the third reason is to
reduce or simplify the
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constants: reduce the number or
simplify the constants in the
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equation.
Reduce their number is self
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explanatory.
Simplify means make them less,
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either dimensionless also,
or if you can't do that,
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at least less dependent upon
the critical units than the old
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ones were.
Let me give you a very simple
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example which will illustrate
most of these things.
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It's the equation.
It's a version of the cooling
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law, which applies at very high
temperatures,
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and it runs.
So, it's like Newton's cooling
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laws, except it's the internal
and external temperatures vary,
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what's important is not the
first power as in Newton's Law,
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but the fourth power.
So, it's a constant.
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And, the difference is,
now, it's the external
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temperature, which,
just so there won't be so many
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capital T's in the equation,
I'm going to call M,
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to the forth power minus T to
the forth power.
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So, T is the internal
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temperature, the thing we are
interested in.
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And, M is the external
constant, which I'll assume,
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now, is a constant external
temperature.
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So, this is valid if big
temperature differences,
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Newton's Law,
breaks down and one needs a
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different one.
Now, you are free to solve that
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equation just as it stands,
if you can.
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There are difficulties
connected with it because you're
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dealing with the fourth powers,
of course.
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But, before you do that,
one should scale.
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How should I scale?
Well, I'm going to scale by
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relating T to M.
So, that is very likely to use
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is T1 equals T divided by M.
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This is now dimensionless
because M, of course,
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has the units of temperature,
degrees Celsius,
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degrees absolute,
whatever it is,
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as does T.
And therefore,
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by taking the ratio of the two,
there are no units attached to
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it.
So, this is dimensionless.
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Now, how actually do I change
the variable in the equation?
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Well, watch this.
It's an utterly trivial idea,
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and utterly important.
Don't slog around doing it this
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way, trying to stuff it in,
and divide first.
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Instead, do the inverse.
In other words,
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write it instead as T equals
MT1, the reason being that it's
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T that's facing you in that
equation, and therefore T you
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want to substitute for.
So, let's do it.
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The new equation will be what?
Well, dT-- Since this is a
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constant, the left-hand side
becomes dT1 / dt times M equals
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k times M to the forth minus M
to the forth T1 to the forth,
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so I'm going to factor
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out that M to the forth,
and make it one minus T1 to the
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forth, okay?
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Now, I could divide through by
M and get rid of one of those,
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and so, the new equation,
now, is dT1 / dt,
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d time, is equal to-- Now,
I have k M cubed out
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front here.
I'm going to just give that a
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new name, k1.
Essentially,
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it's the same equation.
It's no harder to solve and no
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easier to solve than the
original one.
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But it's been simplified.
For one, I think it looks
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better.
So, to compare the two,
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I'll put this one up in green,
and this one in green,
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too, just to convince you it's
the same, but indicate that it's
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the same equation.
Notice, so, T1 has been
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rendered, is now dimensionless.
So, I don't have to even ask
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when I solve this equation,
oh, please tell me what the
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units of temperature are.
How you are measuring
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temperature makes no difference
to this equation.
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k1 still has units.
What units does it have?
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It's been simplified because it
now has the units of,
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since this is dimensionless and
this is dimensionless,
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it has the units of inverse
time.
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So, k1, whereas it had units
involving both degrees and
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seconds before,
now it has inverse time as its
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units.
And, moreover,
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there's one less constant.
So, one less constant in the
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equation.
It just looks better.
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This business,
I think you know that k1,
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the process of forming k1 out
of k M cubed is
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called lumping constants.
I think they use standard
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terminology in physics and
engineering courses.
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Try to get all the constants
together like this.
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And then you lump them there.
They are lumped for you,
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and then you just give the lump
a new name.
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So, that's an example of
scaling.
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Watch out for when you can use
that.
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For example,
it would have probably been a
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good thing to use in the first
problem set when you were
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handling this problem of drug
elimination and hormone
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elimination production inside of
the thing.
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You could lump constants,
and as was done to some extent
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on the solutions to get a neater
looking answer,
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one without so many constants
in it.
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Okay, let's now go to serious
stuff, where we are actually
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going to make changes of
variables which we hope will
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render unsolvable equations
suddenly solvable.
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Now, I'm going to do that by
making substitutions.
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But, it's, I think,
quite important to watch up
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there are two kinds of
substitutions.
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There are direct substitutions.
That's where you introduce a
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new variable.
I don't know how to write this
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on the board.
I'll just write it
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schematically.
So, it's one which says that
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the new variable is equal to
some combination of the old
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variables.
The other kind of substitution
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is inverse.
It's just the reverse.
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Here, you say that the old
variables are some combination
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of the new.
Now, often you'll have to stick
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in a few old variables,
too.
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But the basic,
it's what appears on the
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left-hand side.
Is it a new variable that
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appears on the left-hand side by
itself, or is it the old
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variable that appears on the
left-hand side?
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Now, right here,
we have an example.
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If I did it as a direct
substitution,
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I would have written T1 equals
T over M.
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That's the way I define the new
variable, which of course you
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have to do if you're introducing
it.
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But when I actually did the
substitution,
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I did the inverse substitution.
Namely, I used T equals T1,
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M times T1. And,
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the reason for doing that was
because it was the capital T's
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that faced me in the equation
and I had to have something to
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replace them with.
Now, you see this already in
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calculus, this distinction.
But that might have been a year
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and a half ago.
Just let me remind you,
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typically in calculus,
for example,
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when you want to do this kind
of integral, let's say,
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x times the square root of one
minus x squared dx,
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the substitution you'd use for
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that is u equals one minus x
squared,
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right?
And then, you calculate,
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and then you would observe that
this, the x dx,
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more or less makes up du,
apart from a constant factor,
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there.
So, this would be an example of
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direct substitution.
You put it in and convert the
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integral into an integral of u.
What would be an example of
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inverse substitution?
Well, if I take away the x and
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ask you, instead,
to do this integral,
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then you know that the right
thing to do is not to start with
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u, but to start with the x and
write x equals sine or cosine u.
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So, this is a direct
substitution in that integral,
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but this integral calls for an
inverse substitution in order to
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be able to do it.
And notice, they would look
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practically the same.
But, of course,
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as you know from your
experience, they are not.
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They're very different.
Okay, so I'm going to watch for
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that distinction as I do these
examples.
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The first one I want to do is
an example as a direct
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substitution.
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So, it applies to the equation
of the form y prime equals,
00:14:52.000 --> 00:14:58.000
there are two kinds of terms on
the right-hand side.
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Let's use p of x,
p of x times y plus q of x
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times any power
whatsoever of y.
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Well, notice,
for example,
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if n were zero,
what kind of equation would
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this be?
y to the n would be
00:15:14.000 --> 00:15:20.000
one, and this would be a linear
equation, which you know how to
00:15:20.000 --> 00:15:26.000
solve.
So, n equals zero we already
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know how to do.
So, let's assume that n is not
00:15:27.000 --> 00:15:33.000
zero, so that we're in new
territory.
00:15:33.000 --> 00:15:39.000
Well, if n were equal to one,
you could separate variables.
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So, that too is not exciting.
But, nonetheless,
00:15:42.000 --> 00:15:48.000
it will be included in what I'm
going to say now.
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If n is two or three,
or n could be one half.
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So anything:
even zero is all right.
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It's just silly.
Any number: it could be
00:15:57.000 --> 00:16:03.000
negative.
n equals minus five.
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That would be fine also.
This kind of equation,
00:16:04.000 --> 00:16:10.000
to give it its name,
is called the Bernoulli
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equation, named after which
Bernoulli, I haven't the
00:16:11.000 --> 00:16:17.000
faintest idea.
There were, I think,
00:16:14.000 --> 00:16:20.000
three or four of them.
And, they fought with each
00:16:18.000 --> 00:16:24.000
other.
But, they were all smart.
00:16:20.000 --> 00:16:26.000
Now, the key trick,
if you like,
00:16:23.000 --> 00:16:29.000
method, to solving any
Bernoulli equation,
00:16:26.000 --> 00:16:32.000
let me call another thing.
Most important is what's
00:16:30.000 --> 00:16:36.000
missing.
It must not have a pure x term
00:16:34.000 --> 00:16:40.000
in it.
And that goes for a constant
00:16:37.000 --> 00:16:43.000
term.
In other words,
00:16:38.000 --> 00:16:44.000
it must look exactly like this.
Everything multiplied by y,
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or a power of y,
two terms.
00:16:45.000 --> 00:16:51.000
So, for example,
if I add one to this,
00:16:48.000 --> 00:16:54.000
the equation becomes
non-doable.
00:16:51.000 --> 00:16:57.000
Right, it's very easy to
contaminate it into an equation
00:16:55.000 --> 00:17:01.000
that's unsolvable.
It's got to look just like
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that.
Now, you've got one on your
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homework.
You've got several.
00:17:05.000 --> 00:17:11.000
Both part one and part two have
Bernoulli equations on them.
00:17:10.000 --> 00:17:16.000
So, this is practical,
in some sense.
00:17:13.000 --> 00:17:19.000
What do we got?
The idea is to divide by y to
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the n.
Ignore all formulas that you're
00:17:20.000 --> 00:17:26.000
given.
Just remember that when you see
00:17:23.000 --> 00:17:29.000
something that looks like this,
or something that you can turn
00:17:28.000 --> 00:17:34.000
into something that looks like
this, divide through by y to the
00:17:34.000 --> 00:17:40.000
nth power, no matter what n is.
All right, so y prime over y to
00:17:40.000 --> 00:17:46.000
the n is equal to p of x times
one over y to the n minus one,
00:17:44.000 --> 00:17:50.000
right, plus q of x.
00:17:49.000 --> 00:17:55.000
Well, that certainly doesn't
look any better than what I
00:17:53.000 --> 00:17:59.000
started with.
And, in your terms,
00:17:55.000 --> 00:18:01.000
it probably looks somewhat
worse because it's got all those
00:17:59.000 --> 00:18:05.000
Y's at the denominator,
and who wants to see them
00:18:03.000 --> 00:18:09.000
there?
But, look at it.
00:18:06.000 --> 00:18:12.000
In this very slightly
transformed Bernoulli equation
00:18:10.000 --> 00:18:16.000
is a linear equation struggling
to be free.
00:18:14.000 --> 00:18:20.000
Where is it?
Why is it trying to be a linear
00:18:17.000 --> 00:18:23.000
equation?
Make a new variable,
00:18:20.000 --> 00:18:26.000
call this hunk of it in new
variable.
00:18:23.000 --> 00:18:29.000
Let's call it V.
So, V is equal to one over y to
00:18:27.000 --> 00:18:33.000
the n minus one.
00:18:30.000 --> 00:18:36.000
Or, if you like,
you can think of that as y to
00:18:34.000 --> 00:18:40.000
the one minus n.
What's V prime?
00:18:39.000 --> 00:18:45.000
So, this is the direct
substitution I am going to use,
00:18:44.000 --> 00:18:50.000
but of course,
the problem is,
00:18:46.000 --> 00:18:52.000
what am I going to use on this?
Well, the little miracle
00:18:51.000 --> 00:18:57.000
happens.
What's the derivative of this?
00:18:54.000 --> 00:19:00.000
It is one minus n times y to
the negative n times y prime
00:18:59.000 --> 00:19:05.000
In other words,
00:19:04.000 --> 00:19:10.000
up to a constant,
this constant factor,
00:19:07.000 --> 00:19:13.000
one minus n,
it's exactly the left-hand side
00:19:11.000 --> 00:19:17.000
of the equation.
Well, let's make N not equal
00:19:15.000 --> 00:19:21.000
one either.
As I said, you could separate
00:19:18.000 --> 00:19:24.000
variables if n equals one.
What's the equation,
00:19:22.000 --> 00:19:28.000
then, turned into?
A Bernoulli equation,
00:19:28.000 --> 00:19:34.000
divided through in this way,
is then turned into the
00:19:36.000 --> 00:19:42.000
equation one minus n,
sorry, V prime divided by one
00:19:44.000 --> 00:19:50.000
minus n is equal to p of x times
V plus q of x.
00:19:55.000 --> 00:20:01.000
It's linear.
00:20:01.000 --> 00:20:07.000
And now, solve it as a linear
equation.
00:20:03.000 --> 00:20:09.000
Solve it as a linear equation.
You notice, it's not in
00:20:06.000 --> 00:20:12.000
standard form,
not in standard linear form.
00:20:09.000 --> 00:20:15.000
To do that, you're going to
have to put the p on the other
00:20:13.000 --> 00:20:19.000
side.
That's okay,
00:20:14.000 --> 00:20:20.000
that term, on the other side,
solve it, and at the end,
00:20:17.000 --> 00:20:23.000
don't forget that you put in
the V.
00:20:19.000 --> 00:20:25.000
It wasn't in the original
problem.
00:20:22.000 --> 00:20:28.000
So, you have to convert the
problem, the answer,
00:20:25.000 --> 00:20:31.000
back in terms of y.
It'll come out in terms of V,
00:20:28.000 --> 00:20:34.000
but you must put it back in
terms of y.
00:20:32.000 --> 00:20:38.000
Let's do a really simple
example just to illustrate the
00:20:38.000 --> 00:20:44.000
method, and to illustrate the
fact that I don't want you to
00:20:45.000 --> 00:20:51.000
memorize formulas.
Learn methods,
00:20:49.000 --> 00:20:55.000
not final formulas.
So, suppose the equation is,
00:20:54.000 --> 00:21:00.000
let's say, y prime equals y
over x minus y squared.
00:21:01.000 --> 00:21:07.000
That's a Bernoulli equation.
00:21:06.000 --> 00:21:12.000
I could, of course,
have concealed it by writing xy
00:21:09.000 --> 00:21:15.000
prime plus xy prime minus xy
equals negative y squared.
00:21:13.000 --> 00:21:19.000
Then, it wouldn't look
instantly like a Bernoulli
00:21:16.000 --> 00:21:22.000
equation.
You would have to stare at it a
00:21:19.000 --> 00:21:25.000
while and say,
hey, that's a Bernoulli
00:21:22.000 --> 00:21:28.000
equation.
Okay, but so I'm handing it to
00:21:25.000 --> 00:21:31.000
you a silver platter,
as it were.
00:21:27.000 --> 00:21:33.000
So, what do we do?
Divide through by y squared.
00:21:32.000 --> 00:21:38.000
So, it's y prime over y squared
equals one over x times one over
00:21:38.000 --> 00:21:44.000
y minus one.
00:21:41.000 --> 00:21:47.000
And now, the substitution,
then, I'm going to make,
00:21:46.000 --> 00:21:52.000
is for this thing.
V equals one over y.
00:21:51.000 --> 00:21:57.000
It's a direct substitution.
00:21:53.000 --> 00:21:59.000
V prime is going to be negative
one over y squared
00:21:59.000 --> 00:22:05.000
times y prime.
Don't forget to use the chain
00:22:05.000 --> 00:22:11.000
rule when you differentiate with
respect-- because the
00:22:08.000 --> 00:22:14.000
differentiation is with respect
to x, of course,
00:22:12.000 --> 00:22:18.000
not with respect to y.
Okay, so what's this thing?
00:22:15.000 --> 00:22:21.000
That's the left-hand side.
The only thing is it's got a
00:22:19.000 --> 00:22:25.000
negative sign.
So, this is minus V prime
00:22:22.000 --> 00:22:28.000
equals, one over x stays one
over x, one over y.
00:22:26.000 --> 00:22:32.000
So, it's V over x minus one.
00:22:30.000 --> 00:22:36.000
So, let's put that in standard
form.
00:22:32.000 --> 00:22:38.000
In standard form,
it will look like,
00:22:35.000 --> 00:22:41.000
first imagine multiplying it
through by negative one,
00:22:39.000 --> 00:22:45.000
and then putting this term on
the other side.
00:22:42.000 --> 00:22:48.000
And, it will turn into V prime
plus V over X is equal to one.
00:22:47.000 --> 00:22:53.000
So, that's the linear equation
00:22:51.000 --> 00:22:57.000
in standard linear form that we
are asked to solve.
00:22:54.000 --> 00:23:00.000
And, the solution isn't very
hard.
00:22:57.000 --> 00:23:03.000
The integrating factor is,
well, I integrate one over x.
00:23:03.000 --> 00:23:09.000
That makes log x.
And, e to the log x,
00:23:05.000 --> 00:23:11.000
so, it's e to the log x,
which is, of course,
00:23:09.000 --> 00:23:15.000
just x itself.
So, I should multiply this
00:23:12.000 --> 00:23:18.000
through by x squared,
be able to integrate it.
00:23:15.000 --> 00:23:21.000
Now, some of you,
I would hope,
00:23:17.000 --> 00:23:23.000
just can see that right away,
that if you multiply this
00:23:21.000 --> 00:23:27.000
through by x,
it's going to look good.
00:23:24.000 --> 00:23:30.000
So, after we multiply through
by x, which I get?
00:23:27.000 --> 00:23:33.000
(xV) prime for the-- maybe I
shouldn't skip a step.
00:23:33.000 --> 00:23:39.000
You are still learning this
stuff, so let's not skip a step.
00:23:38.000 --> 00:23:44.000
So, it becomes x V prime plus V
equals x,
00:23:44.000 --> 00:23:50.000
okay?
After I multiplied through by
00:23:47.000 --> 00:23:53.000
the integrating factor,
this now says this is xV prime,
00:23:53.000 --> 00:23:59.000
and I quickly check that that,
in fact, is what it's equal to,
00:23:59.000 --> 00:24:05.000
equals x, and therefore xV is
equal to one half x squared plus
00:24:05.000 --> 00:24:11.000
a constant. And,
00:24:08.000 --> 00:24:14.000
therefore, V is equal to one
half x plus C over x.
00:24:14.000 --> 00:24:20.000
You can leave it at that form,
00:24:19.000 --> 00:24:25.000
or you can combine terms.
It doesn't matter much.
00:24:23.000 --> 00:24:29.000
Am I done?
The answer is,
00:24:25.000 --> 00:24:31.000
no I am not done,
because nobody reading this
00:24:29.000 --> 00:24:35.000
answer would know what V was.
V wasn't in the original
00:24:33.000 --> 00:24:39.000
problem.
It was y that was in the
00:24:35.000 --> 00:24:41.000
original problem.
And therefore,
00:24:37.000 --> 00:24:43.000
the relation is,
one is the reciprocal of the
00:24:40.000 --> 00:24:46.000
other.
And therefore,
00:24:41.000 --> 00:24:47.000
I have to turn this expression
upside down.
00:24:44.000 --> 00:24:50.000
Well, if you're going to have
to turn it upside down,
00:24:47.000 --> 00:24:53.000
you probably should make it
look a little better.
00:24:50.000 --> 00:24:56.000
Let's rewrite it as x squared
plus 2c,
00:24:53.000 --> 00:24:59.000
combining fractions,
I think they call it in high
00:24:56.000 --> 00:25:02.000
school or elementary school,
plus 2c.
00:25:00.000 --> 00:25:06.000
How's that? x squared
plus 2c divided by 2x.
00:25:03.000 --> 00:25:09.000
Now, 2c, you don't call it
00:25:07.000 --> 00:25:13.000
constant 2c because this is just
as arbitrary to call it c1.
00:25:12.000 --> 00:25:18.000
So, I'll call that,
so, my answer will be y equals
00:25:16.000 --> 00:25:22.000
2x divided by x squared plus an
arbitrary constant.
00:25:20.000 --> 00:25:26.000
But, to indicate it's different
from that one,
00:25:23.000 --> 00:25:29.000
I'll call it C1. C1 is
00:25:27.000 --> 00:25:33.000
two times the old one,
but that doesn't really matter.
00:25:31.000 --> 00:25:37.000
So, there's the solution.
It has an arbitrary constant in
00:25:37.000 --> 00:25:43.000
it, but you note it's not an
additive arbitrary constant.
00:25:40.000 --> 00:25:46.000
The arbitrary constant is
tucked into the solution.
00:25:44.000 --> 00:25:50.000
If you had to satisfy an
initial condition,
00:25:47.000 --> 00:25:53.000
you would take this form,
and starting from this form,
00:25:50.000 --> 00:25:56.000
figure out what C1 was in order
to satisfy that initial
00:25:54.000 --> 00:26:00.000
condition.
Thus, Bernoulli equation is
00:25:57.000 --> 00:26:03.000
solved.
Our first Bernoulli equation:
00:25:59.000 --> 00:26:05.000
isn't that exciting?
So, here was the equation,
00:26:05.000 --> 00:26:11.000
and there is its solution.
Now, the one I'm asking you to
00:26:11.000 --> 00:26:17.000
solve on the problem set in part
two is no harder than this,
00:26:18.000 --> 00:26:24.000
except I ask you some hard
questions about it,
00:26:24.000 --> 00:26:30.000
not very hard,
but a little hard about it.
00:26:30.000 --> 00:26:36.000
I hope you will find them
interesting questions.
00:26:33.000 --> 00:26:39.000
You already have the
experimental evidence from the
00:26:37.000 --> 00:26:43.000
first problem set,
and the problem is to explain
00:26:40.000 --> 00:26:46.000
the experimental evidence by
actually solving the equation in
00:26:45.000 --> 00:26:51.000
the scene.
I think you'll find it
00:26:47.000 --> 00:26:53.000
interesting.
But, maybe that's just a pious
00:26:51.000 --> 00:26:57.000
hope.
Okay, I like,
00:26:52.000 --> 00:26:58.000
now, to turn to the second
method, where a second class of
00:26:56.000 --> 00:27:02.000
equations which require inverse
substitution,
00:27:00.000 --> 00:27:06.000
and those are equations,
which are called homogeneous,
00:27:04.000 --> 00:27:10.000
a highly overworked word in
differential equations,
00:27:08.000 --> 00:27:14.000
and in mathematics in general.
But, it's unfortunately just
00:27:14.000 --> 00:27:20.000
the right word to describe them.
So, these are homogeneous,
00:27:19.000 --> 00:27:25.000
first-order ODE's.
Now, I already used the word in
00:27:23.000 --> 00:27:29.000
one context in talking about the
linear equations when zero is
00:27:28.000 --> 00:27:34.000
the right hand side.
This is different,
00:27:32.000 --> 00:27:38.000
but nonetheless,
the two uses of the word have
00:27:35.000 --> 00:27:41.000
the same common source.
The homogeneous differential
00:27:39.000 --> 00:27:45.000
equation, homogeneous newspeak,
is y prime equals,
00:27:43.000 --> 00:27:49.000
it's a question of what the
right hand side looks like.
00:27:47.000 --> 00:27:53.000
And, now, the supposed way to
say it is, you should be able to
00:27:52.000 --> 00:27:58.000
write the right-hand side as a
function of a combined variable,
00:27:57.000 --> 00:28:03.000
y divided by x.
In other words,
00:28:01.000 --> 00:28:07.000
after fooling around with the
right hand side a little bit,
00:28:06.000 --> 00:28:12.000
you should be able to write it
so that every time a variable
00:28:11.000 --> 00:28:17.000
appears, it's always in the
combination y over x.
00:28:15.000 --> 00:28:21.000
Let me give some examples.
For example,
00:28:19.000 --> 00:28:25.000
suppose y prime were,
let's say, x squared y divided
00:28:23.000 --> 00:28:29.000
by x cubed plus y cubed.
00:28:29.000 --> 00:28:35.000
Well, that doesn't look in that
form.
00:28:31.000 --> 00:28:37.000
Well, yes it is.
Imagine dividing the top and
00:28:34.000 --> 00:28:40.000
bottom by x cubed.
What would you get?
00:28:37.000 --> 00:28:43.000
The top would be y over x,
if you divided it by x
00:28:40.000 --> 00:28:46.000
cubed.
And, if I divide the bottom by
00:28:43.000 --> 00:28:49.000
x cubed, also,
which, of course,
00:28:45.000 --> 00:28:51.000
doesn't change the value of the
fraction, as they say in
00:28:49.000 --> 00:28:55.000
elementary school,
one plus (y over x) cubed.
00:28:52.000 --> 00:28:58.000
So, this is the way it started
00:28:55.000 --> 00:29:01.000
out looking, but you just said
ah-ha, that was a homogeneous
00:28:59.000 --> 00:29:05.000
equation because I could see it
could be written that way.
00:29:05.000 --> 00:29:11.000
How about another homogeneous
equation?
00:29:10.000 --> 00:29:16.000
How about x y prime?
Is that a homogeneous equation?
00:29:18.000 --> 00:29:24.000
Of course it is:
otherwise, why would I be
00:29:24.000 --> 00:29:30.000
talking about it?
If you divide through by x,
00:29:29.000 --> 00:29:35.000
you can tuck it inside the
radical, the square root,
00:29:33.000 --> 00:29:39.000
if you remember to square it
when you do that.
00:29:36.000 --> 00:29:42.000
And, it becomes the square root
of x squared over x squared,
00:29:41.000 --> 00:29:47.000
which is one,
plus y squared over x squared.
00:29:44.000 --> 00:29:50.000
It's homogeneous.
00:29:47.000 --> 00:29:53.000
Now, you might say,
hey, this looks like you had to
00:29:50.000 --> 00:29:56.000
be rather clever to figure out
if an equation is homogeneous.
00:29:55.000 --> 00:30:01.000
Is there some other way?
Yeah, there is another way,
00:29:58.000 --> 00:30:04.000
and it's the other way which
explains why it's called
00:30:02.000 --> 00:30:08.000
homogeneous.
You can think of it this way.
00:30:07.000 --> 00:30:13.000
It's an equation which is,
in modern speak,
00:30:12.000 --> 00:30:18.000
invariant, invariant under the
operation zoom.
00:30:18.000 --> 00:30:24.000
What is zoom?
Zoom is, you increase the scale
00:30:23.000 --> 00:30:29.000
equally on both axes.
So, the zoom operation is the
00:30:30.000 --> 00:30:36.000
one which sends x into
a times x,
00:30:36.000 --> 00:30:42.000
and y into a times y.
00:30:42.000 --> 00:30:48.000
In other words,
you change the scale on both
00:30:46.000 --> 00:30:52.000
axes by the same factor,
a.
00:30:48.000 --> 00:30:54.000
Now, what I say is,
gee, maybe you shouldn't write
00:30:53.000 --> 00:30:59.000
it like this.
Why don't we say,
00:30:56.000 --> 00:31:02.000
we introduce,
how about this?
00:31:00.000 --> 00:31:06.000
So, think of it as a change of
variables.
00:31:02.000 --> 00:31:08.000
We will write it like that.
So, you can put here an equals
00:31:06.000 --> 00:31:12.000
sign, if you don't know what
this meaningless arrow means.
00:31:10.000 --> 00:31:16.000
So, I'm making this change of
variables, and I'm describing it
00:31:14.000 --> 00:31:20.000
as an inverse substitution.
But of course,
00:31:16.000 --> 00:31:22.000
it wouldn't make any
difference.
00:31:19.000 --> 00:31:25.000
It's exactly the same as the
direct substitution I started
00:31:22.000 --> 00:31:28.000
out with underscaling.
The only difference is,
00:31:25.000 --> 00:31:31.000
I'm not using different scales
on both axes.
00:31:28.000 --> 00:31:34.000
I'm expanding them both
equally.
00:31:32.000 --> 00:31:38.000
That's what I mean by zoom.
Now, what happens to the
00:31:36.000 --> 00:31:42.000
equation?
Well, what happens to dy over
00:31:40.000 --> 00:31:46.000
dx?
Well, dx is a dx1.
00:31:43.000 --> 00:31:49.000
dy is a dy1.
00:31:47.000 --> 00:31:53.000
And therefore,
the ratio, dy by dx is the same
00:31:51.000 --> 00:31:57.000
as dy1 over dx1.
00:31:54.000 --> 00:32:00.000
So, the left-hand side becomes
dy1 over dx1,
00:31:58.000 --> 00:32:04.000
and the right-hand side becomes
F of, well, y over x is the same
00:32:04.000 --> 00:32:10.000
as y over, since I've scaled
them equally,
00:32:08.000 --> 00:32:14.000
this is the same as y1 over x1.
00:32:14.000 --> 00:32:20.000
So, it's y1 over x1,
and the net effect is I simply,
00:32:18.000 --> 00:32:24.000
everywhere I have an x,
I change it to x1,
00:32:22.000 --> 00:32:28.000
and wherever I have a y,
I change it to y1,
00:32:25.000 --> 00:32:31.000
which, what's in a name?
It's the identical equation.
00:32:31.000 --> 00:32:37.000
So, I haven't changed the
equation at all via zoom
00:32:35.000 --> 00:32:41.000
transformation.
And, that's what makes it
00:32:38.000 --> 00:32:44.000
homogeneous.
That's not an uncommon use of
00:32:42.000 --> 00:32:48.000
the word homogeneous.
When you say space is
00:32:45.000 --> 00:32:51.000
homogeneous, every direction,
well, that means,
00:32:49.000 --> 00:32:55.000
I don't know.
It means, okay,
00:32:51.000 --> 00:32:57.000
I'm getting into trouble there.
I'll let it go since I can't
00:32:56.000 --> 00:33:02.000
prepare any better,
I haven't prepared any better
00:33:00.000 --> 00:33:06.000
explanation, but this is a
pretty good one.
00:33:06.000 --> 00:33:12.000
Okay, so, suppose we've got a
homogeneous equation.
00:33:13.000 --> 00:33:19.000
How do we solve it?
So, here's our equation,
00:33:20.000 --> 00:33:26.000
F of y over x.
Well, what substitution would
00:33:29.000 --> 00:33:35.000
you like to make?
Obviously, we should make a
00:33:34.000 --> 00:33:40.000
direct substitution,
z equals y over x.
00:33:38.000 --> 00:33:44.000
So, why did he say that this
was going to be an example of
00:33:42.000 --> 00:33:48.000
inverse substitution?
Because I wanted to confuse
00:33:45.000 --> 00:33:51.000
you.
But look, that's fine.
00:33:47.000 --> 00:33:53.000
If you write it in that form,
you'll know exactly what to do
00:33:51.000 --> 00:33:57.000
with the right-hand side.
And, this is why everybody
00:33:55.000 --> 00:34:01.000
loves to do that.
But for Charlie,
00:33:57.000 --> 00:34:03.000
you have to substitute into the
left-hand side as well.
00:34:03.000 --> 00:34:09.000
And, I can testify,
for many years of looking with
00:34:06.000 --> 00:34:12.000
sinking heart at examination
papers, what happens if you try
00:34:10.000 --> 00:34:16.000
to make a direct substitution
like this?
00:34:13.000 --> 00:34:19.000
You say, oh,
I need z prime.
00:34:15.000 --> 00:34:21.000
z prime equals,
well, I better use the quotient
00:34:18.000 --> 00:34:24.000
rule for differentiating that.
And, it comes out this long,
00:34:22.000 --> 00:34:28.000
and then either a long pause,
what do I do now?
00:34:26.000 --> 00:34:32.000
Because it's not at all obvious
what to do at that point.
00:34:30.000 --> 00:34:36.000
Or, much worse,
two pages of frantic
00:34:32.000 --> 00:34:38.000
calculations,
and giving up in total despair.
00:34:37.000 --> 00:34:43.000
Now, the reason for that is
because you tried to do it
00:34:40.000 --> 00:34:46.000
making a direct substitution.
All you have to do instead is
00:34:45.000 --> 00:34:51.000
use it, treat it as an inverse
substitution,
00:34:48.000 --> 00:34:54.000
write y equals zx.
What's the motivation for doing
00:34:51.000 --> 00:34:57.000
that?
It's clear from the equation.
00:34:54.000 --> 00:35:00.000
This goes through all of
mathematics.
00:34:57.000 --> 00:35:03.000
Whenever you have to change a
variable, excuse me,
00:35:00.000 --> 00:35:06.000
whenever you have to change a
variable, look at what you have
00:35:05.000 --> 00:35:11.000
to substitute for,
and focus your attention on
00:35:08.000 --> 00:35:14.000
that.
I need to know what y prime is.
00:35:12.000 --> 00:35:18.000
Okay, well, then I better know
what y is.
00:35:15.000 --> 00:35:21.000
If I know what y is,
do I know what y prime is?
00:35:19.000 --> 00:35:25.000
Oh, of course.
y prime is z prime x plus z
00:35:22.000 --> 00:35:28.000
times the derivative of this
factor, which is one.
00:35:26.000 --> 00:35:32.000
And now, I turned with that one
00:35:31.000 --> 00:35:37.000
stroke, the equation has now
become z prime x plus z is equal
00:35:36.000 --> 00:35:42.000
to F of z.
Well, I don't know.
00:35:40.000 --> 00:35:46.000
Can I solve that?
Sure.
00:35:42.000 --> 00:35:48.000
That can be solved because this
is x times dz / dx.
00:35:48.000 --> 00:35:54.000
Just put the z on the other
side, it's F of z minus z.
00:35:53.000 --> 00:35:59.000
And now, this side is just a
00:35:55.000 --> 00:36:01.000
function of z.
Separate variables.
00:36:00.000 --> 00:36:06.000
And, the only thing to watch
out for is, at the end,
00:36:03.000 --> 00:36:09.000
the z was your business.
You've got to put the answer
00:36:07.000 --> 00:36:13.000
back in terms of z and y.
Okay, let's work an example of
00:36:11.000 --> 00:36:17.000
this.
Since I haven't done any
00:36:13.000 --> 00:36:19.000
modeling yet this period,
let's make a little model,
00:36:17.000 --> 00:36:23.000
differential equations model.
It's a physical situation,
00:36:21.000 --> 00:36:27.000
which will be solved by an
equation.
00:36:24.000 --> 00:36:30.000
And, guess what?
The equation will turn out to
00:36:27.000 --> 00:36:33.000
be homogeneous.
Okay, so the situation is as
00:36:32.000 --> 00:36:38.000
follows.
We are in the Caribbean
00:36:34.000 --> 00:36:40.000
somewhere, a little isolated
island somewhere with a little
00:36:39.000 --> 00:36:45.000
lighthouse on it at the origin,
and a beam of light shines from
00:36:44.000 --> 00:36:50.000
the lighthouse.
The beam of light can rotate
00:36:48.000 --> 00:36:54.000
the way the lighthouse beams.
But, this particular beam is
00:36:53.000 --> 00:36:59.000
being controlled by a guy in the
lighthouse who can aim it
00:36:57.000 --> 00:37:03.000
wherever he wants.
And, the reason he's interested
00:37:01.000 --> 00:37:07.000
in aiming it wherever he wants
is there's a drug boat here,
00:37:06.000 --> 00:37:12.000
[LAUGHTER] which has just been
caught in the beam of light.
00:37:13.000 --> 00:37:19.000
So, the drug boat,
which has just been caught in a
00:37:16.000 --> 00:37:22.000
beam of light,
and feels it'd a better escape.
00:37:20.000 --> 00:37:26.000
Now, the lighthouse keeper
wants to keep the drug boat;
00:37:24.000 --> 00:37:30.000
the light is shining on it so
that the U.S.
00:37:27.000 --> 00:37:33.000
Coast Guard helicopters can
zoom over it and do whatever
00:37:31.000 --> 00:37:37.000
they do to drug boats,
--
00:37:34.000 --> 00:37:40.000
-- I don't know.
So, the drug boat immediately
00:37:37.000 --> 00:37:43.000
has to follow an escape
strategy.
00:37:39.000 --> 00:37:45.000
And, the only one that occurs
to him is, well,
00:37:42.000 --> 00:37:48.000
he wants to go further away,
of course, from the lighthouse.
00:37:47.000 --> 00:37:53.000
On the other hand,
it doesn't seem sensible to do
00:37:50.000 --> 00:37:56.000
it in a straight line because
the beam will keep shining on
00:37:54.000 --> 00:38:00.000
him.
So, he fixes the boat at some
00:37:57.000 --> 00:38:03.000
angle, let's say,
and goes off so that the angle
00:38:00.000 --> 00:38:06.000
stays 45 degrees.
So, it goes so that the angle
00:38:05.000 --> 00:38:11.000
between the beam and maybe,
draw the beam a little less
00:38:11.000 --> 00:38:17.000
like a 45 degree angle.
So, the angle between the beam
00:38:16.000 --> 00:38:22.000
and the boat,
the boat's path is always 45
00:38:20.000 --> 00:38:26.000
degrees, goes at a constant 45
degree angle to the beam,
00:38:26.000 --> 00:38:32.000
hoping thereby to escape.
On the other hand,
00:38:30.000 --> 00:38:36.000
of course, the lighthouse guy
keeps the beam always on the
00:38:36.000 --> 00:38:42.000
boat.
So, it's not clear it's a good
00:38:40.000 --> 00:38:46.000
strategy, but this is a
differential equations class.
00:38:44.000 --> 00:38:50.000
The question is,
what's the path of the boat?
00:38:48.000 --> 00:38:54.000
What's the boat's path?
Now, an obvious question is,
00:38:52.000 --> 00:38:58.000
why is this a problem in
differential equations at all?
00:38:57.000 --> 00:39:03.000
In other words,
looking at this,
00:38:59.000 --> 00:39:05.000
you might scratch your head and
try to think of different ways
00:39:04.000 --> 00:39:10.000
to solve it.
But, what suggests that it's
00:39:09.000 --> 00:39:15.000
going to be a problem in
differential equations?
00:39:13.000 --> 00:39:19.000
The answer is,
you're looking for a path.
00:39:17.000 --> 00:39:23.000
The answer is going to be a
curve.
00:39:20.000 --> 00:39:26.000
A curve means a function.
We are looking for an unknown
00:39:24.000 --> 00:39:30.000
function, in other words.
And, what type of information
00:39:29.000 --> 00:39:35.000
do we have about the function?
The only information we have
00:39:34.000 --> 00:39:40.000
about the function is something
about its slope,
00:39:38.000 --> 00:39:44.000
that its slope makes a constant
45° angle with the lighthouse
00:39:44.000 --> 00:39:50.000
beam.
Its slope makes a constant
00:39:53.000 --> 00:39:59.000
known angle to a known angle.
Well, if you are trying to find
00:40:04.000 --> 00:40:10.000
a function, and all you know is
something about its slope,
00:40:09.000 --> 00:40:15.000
that is a problem in
differential equations.
00:40:13.000 --> 00:40:19.000
Well, let's try to solve it.
Well, let's see.
00:40:16.000 --> 00:40:22.000
Well, let me draw just a little
bit.
00:40:19.000 --> 00:40:25.000
So, here's the horizontal.
Let's introduce the
00:40:23.000 --> 00:40:29.000
coordinates.
In other words,
00:40:25.000 --> 00:40:31.000
there's the horizontal and
here's the boat to indicate
00:40:30.000 --> 00:40:36.000
where I am with respect to the
picture.
00:40:35.000 --> 00:40:41.000
So, here's the boat.
Here's the beam,
00:40:38.000 --> 00:40:44.000
and the path of the boat is
going to make a 45° angle with
00:40:44.000 --> 00:40:50.000
it.
So, this is the path that we
00:40:47.000 --> 00:40:53.000
are talking about.
And now, let's label what I
00:40:51.000 --> 00:40:57.000
know.
Well, this angle is 45°.
00:40:54.000 --> 00:41:00.000
This angle, I don't know,
but of course I can calculate
00:41:00.000 --> 00:41:06.000
it easily enough because it has
to do with, if I know the
00:41:05.000 --> 00:41:11.000
coordinates of this point,
(x, y), then of course that
00:41:11.000 --> 00:41:17.000
horizontal angle,
I know the slope of this line,
00:41:15.000 --> 00:41:21.000
and that angle will be related
to the slope.
00:41:22.000 --> 00:41:28.000
So, let's call this alpha.
And now, what I want to know is
00:41:29.000 --> 00:41:35.000
what the slope of the whole path
is.
00:41:35.000 --> 00:41:41.000
So, y prime-- let's call y
equals y of x,
00:41:42.000 --> 00:41:48.000
the unknown function whose
path, whose graph is going to be
00:41:50.000 --> 00:41:56.000
the boat's path,
unknown graph.
00:41:54.000 --> 00:42:00.000
What's its slope?
Well, its slope is the tangent
00:42:00.000 --> 00:42:06.000
of the sum of these two angles,
alpha plus 45°.
00:42:08.000 --> 00:42:14.000
Now, what do I know?
Well, I know that the tangent
00:42:11.000 --> 00:42:17.000
of alpha is how much?
That's y over x.
00:42:14.000 --> 00:42:20.000
In other words,
00:42:16.000 --> 00:42:22.000
if this was the point,
x over y, this is the angle it
00:42:19.000 --> 00:42:25.000
makes with a horizontal,
if you think of it over here.
00:42:23.000 --> 00:42:29.000
So, this angle is the same as
that one, and it's y over,
00:42:26.000 --> 00:42:32.000
its slope of that line is y
over x.
00:42:30.000 --> 00:42:36.000
So, the tangent of the angle is
y over x.
00:42:32.000 --> 00:42:38.000
How about the tangent of 45°?
That's one, and there's a
00:42:36.000 --> 00:42:42.000
formula.
This is the hard part.
00:42:38.000 --> 00:42:44.000
All you have to know is that
the formula exists,
00:42:41.000 --> 00:42:47.000
and then you will look it up if
you have forgotten it,
00:42:44.000 --> 00:42:50.000
relating the tangent or giving
you the tangent of the sum of
00:42:48.000 --> 00:42:54.000
two angles, and you can,
if you like,
00:42:50.000 --> 00:42:56.000
clever, derive it from the
formula for the sign and cosine
00:42:54.000 --> 00:43:00.000
of the sum of two angles.
But, one peak is worth a
00:42:57.000 --> 00:43:03.000
thousand finesses.
So, it is the tangent of alpha
00:43:02.000 --> 00:43:08.000
plus the tangent of 45°.
Let me read it out in all its
00:43:06.000 --> 00:43:12.000
gory details,
divided by one,
00:43:08.000 --> 00:43:14.000
so you'll at least learn the
formula, one minus tangent alpha
00:43:12.000 --> 00:43:18.000
times tangent 45°.
00:43:15.000 --> 00:43:21.000
This would work for the tangent
of the sum of any two angles.
00:43:20.000 --> 00:43:26.000
That's the formula.
So, what do I get then?
00:43:23.000 --> 00:43:29.000
y prime is equal to the tangent
of alpha, which is y over x,
00:43:27.000 --> 00:43:33.000
oh, I like that combination,
plus one, divided by (one minus
00:43:32.000 --> 00:43:38.000
y over x times one).
00:43:37.000 --> 00:43:43.000
Now, there is no reason for
doing anything to it,
00:43:40.000 --> 00:43:46.000
but let's make it look a little
prettier, and thereby,
00:43:44.000 --> 00:43:50.000
make it less obvious that it's
a homogeneous equation.
00:43:48.000 --> 00:43:54.000
If I multiply top and bottom by
x, it looks prettier.
00:43:52.000 --> 00:43:58.000
x plus y over x minus y equals
y prime.
00:43:55.000 --> 00:44:01.000
That's our
differential equation.
00:44:00.000 --> 00:44:06.000
But, notice,
that let step to make it look
00:44:02.000 --> 00:44:08.000
pretty has undone the good work.
It's fine if you immediately
00:44:06.000 --> 00:44:12.000
recognize this as being a
homogeneous equation because you
00:44:10.000 --> 00:44:16.000
can divide the top and bottom by
x.
00:44:12.000 --> 00:44:18.000
But here, it's a lot clearer
that it's a homogeneous equation
00:44:16.000 --> 00:44:22.000
because it's already been
written in the right form.
00:44:20.000 --> 00:44:26.000
Okay, let's solve it now,
since we know what to do.
00:44:23.000 --> 00:44:29.000
We're going to use as the new
variable, z equals y over x.
00:44:27.000 --> 00:44:33.000
And, as I wrote up there for y
00:44:33.000 --> 00:44:39.000
prime, we'll substitute z prime
x plus z.
00:44:40.000 --> 00:44:46.000
And, with that,
let's solve.
00:44:44.000 --> 00:44:50.000
Let's solve it.
The equation becomes z prime x
00:44:50.000 --> 00:44:56.000
plus z is equal to z plus one
over one minus z.
00:44:57.000 --> 00:45:03.000
We want to separate variables,
00:45:04.000 --> 00:45:10.000
so you have to put all the z's
on one side.
00:45:07.000 --> 00:45:13.000
So, this is going to be x,
dz / dx equals this thing minus
00:45:11.000 --> 00:45:17.000
z, which is (z plus one) over
(one minus z) minus z.
00:45:14.000 --> 00:45:20.000
And now, as you realize,
00:45:18.000 --> 00:45:24.000
putting it on the other side,
I'm going to have to turn it
00:45:22.000 --> 00:45:28.000
upside down.
Just as before,
00:45:24.000 --> 00:45:30.000
if you have to turn something
upside down, it's better to
00:45:28.000 --> 00:45:34.000
combine the terms,
and make it one tiny little
00:45:31.000 --> 00:45:37.000
fraction.
Otherwise, you are in for quite
00:45:35.000 --> 00:45:41.000
a lot of mess if you don't do
this nicely.
00:45:39.000 --> 00:45:45.000
So, z plus one minus z,
that gets rid of the z's.
00:45:43.000 --> 00:45:49.000
The numerator is one minus z
squared over one minus z,
00:45:48.000 --> 00:45:54.000
I hope, one,
is that right,
00:45:50.000 --> 00:45:56.000
(one plus z squared) over (one
minus z).
00:45:56.000 --> 00:46:02.000
And so, the question is dz,
00:45:58.000 --> 00:46:04.000
and put this on the other side
and turn it upside down.
00:46:05.000 --> 00:46:11.000
So, that will be (one minus z)
over (one plus z squared) on the
00:46:10.000 --> 00:46:16.000
left-hand side and on the
right-hand side,
00:46:14.000 --> 00:46:20.000
dx over x.
Well, that's ready to be
00:46:17.000 --> 00:46:23.000
integrated just as it stands.
The right-hand side integrates
00:46:23.000 --> 00:46:29.000
to be log x.
The left-hand side is the sum
00:46:26.000 --> 00:46:32.000
of two terms.
The integral of one over one
00:46:30.000 --> 00:46:36.000
plus z squared is the arc
tangent of z,
00:46:34.000 --> 00:46:40.000
maybe?
The derivative of this is one
00:46:37.000 --> 00:46:43.000
over one plus z squared.
00:46:40.000 --> 00:46:46.000
How about the term z over one
plus z squared?
00:46:44.000 --> 00:46:50.000
Well, that integrates to be a
00:46:46.000 --> 00:46:52.000
logarithm.
It is more or less the
00:46:48.000 --> 00:46:54.000
logarithm of one plus
z squared.
00:46:51.000 --> 00:46:57.000
If I differentiate this,
I get one over one plus z^2
00:46:54.000 --> 00:47:00.000
times 2z,
but I wish I had negative z
00:46:58.000 --> 00:47:04.000
there instead.
Therefore, I should put a minus
00:47:01.000 --> 00:47:07.000
sign, and I should multiply that
by half to make it come out
00:47:05.000 --> 00:47:11.000
right.
And, this is log x on the right
00:47:09.000 --> 00:47:15.000
hand side plus,
put in that arbitrary constant.
00:47:13.000 --> 00:47:19.000
And now what?
Well, let's now fool around
00:47:16.000 --> 00:47:22.000
with it a little bit.
The arc tangent,
00:47:19.000 --> 00:47:25.000
I'm going to simultaneously,
no, two steps.
00:47:22.000 --> 00:47:28.000
I have to remember your
innocence, although probably a
00:47:26.000 --> 00:47:32.000
lot of you are better
calculators than I am.
00:47:31.000 --> 00:47:37.000
I'm going to change this,
use as many laws of logarithms
00:47:35.000 --> 00:47:41.000
as possible.
I'm going to put this in the
00:47:38.000 --> 00:47:44.000
exponent, and put this on the
other side.
00:47:41.000 --> 00:47:47.000
That's going to turn it into
the log of the square root of
00:47:45.000 --> 00:47:51.000
one plus z squared.
00:47:48.000 --> 00:47:54.000
And, this is going to be plus
the log of x plus c.
00:47:53.000 --> 00:47:59.000
And, now I'm going to make,
00:47:55.000 --> 00:48:01.000
go back and remember that z
equals y over x.
00:48:00.000 --> 00:48:06.000
So, this becomes the arc
tangent of y over x equals.
00:48:04.000 --> 00:48:10.000
Now, I combine the logarithms.
00:48:09.000 --> 00:48:15.000
This is the log of x times this
square root, right,
00:48:12.000 --> 00:48:18.000
make one logarithm out of it,
and then put z equals y over z.
00:48:16.000 --> 00:48:22.000
And, you see that if you do
00:48:19.000 --> 00:48:25.000
that, it'll be the log of x
times the square root of one
00:48:23.000 --> 00:48:29.000
plus (y over x) squared,
00:48:27.000 --> 00:48:33.000
and what is that?
Well, if I put this over x
00:48:30.000 --> 00:48:36.000
squared and take it out,
it cancels that.
00:48:34.000 --> 00:48:40.000
And, what you are left with is
the log of the square root of x
00:48:38.000 --> 00:48:44.000
squared plus y squared plus a
constant.
00:48:42.000 --> 00:48:48.000
Now, technically,
00:48:43.000 --> 00:48:49.000
you have solved the equation,
but not morally because,
00:48:47.000 --> 00:48:53.000
I mean, my God,
what a mess!
00:48:49.000 --> 00:48:55.000
Incredible path.
It tells me absolutely nothing.
00:48:52.000 --> 00:48:58.000
Wow, what is the screaming?
Change me to polar coordinates.
00:48:56.000 --> 00:49:02.000
What's the arc tangent of y
over x?
00:49:00.000 --> 00:49:06.000
Theta.
In polar coordinates it's
00:49:02.000 --> 00:49:08.000
theta.
This is r.
00:49:04.000 --> 00:49:10.000
So, the curve is theta equals
the log of r plus a constant.
00:49:09.000 --> 00:49:15.000
And, I can make even that
little better if I exponentiate
00:49:14.000 --> 00:49:20.000
everything, exponentiate both
sides, combine this in the usual
00:49:19.000 --> 00:49:25.000
way, the and what you get is
that r is equal to some other
00:49:24.000 --> 00:49:30.000
constant times e to the theta.
00:49:30.000 --> 00:49:36.000
That's the curve.
It's called an exponential
00:49:33.000 --> 00:49:39.000
spiral, and that's what our
little boat goes in.
00:49:37.000 --> 00:49:43.000
And notice, probably if I had
set up the problem in polar
00:49:42.000 --> 00:49:48.000
coordinates from the beginning,
nobody would have been able to
00:49:48.000 --> 00:49:54.000
solve it.
But, anyone who did would have
00:49:51.000 --> 00:49:57.000
gotten that answer immediately.
Thanks.