WEBVTT

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CHRISTINE BREINER: Welcome
back to recitation.

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In this video, I'd like us to
work on the following problem.

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We're going to let F be the
vector field that's defined

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by r to the n, times the
quantity x*i plus y*j.

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And r in this case is x squared
plus y squared to the 1/2,

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as it usually is.

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The square root of x
squared plus y squared.

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And then I'd like us
to do the following.

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Use extended Green's
theorem to show

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that F is conservative
for all integers n.

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And then find a
potential function.

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So there are two parts.

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The first part is that you want
to show that F is conservative.

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And then once you know
it's conservative,

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you can find a
potential function.

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So why don't you take a
little while to work on that.

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And then when you're feeling
good about your answer,

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bring the video back up, and
I'll show you what I did.

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OK, welcome back.

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So again, what was the
point of this video?

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We want to do two things.

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We want to work on two problems.

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The first is to show
that this vector field F

00:01:07.100 --> 00:01:08.750
I've given you is conservative.

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And then we want to find
a potential function.

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And we want to be
able to show it's

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conservative for all integers n.

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And what I want to point out is
that for certain integer values

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of n, we're going to
run into some problems

00:01:19.700 --> 00:01:22.220
with differentiability
at the origin.

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OK?

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So we're going to try and
deal with all of it at once,

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and simultaneously deal
with all of the integers,

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by allowing ourselves to
show that F is conservative

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even if we don't include
the origin in our region.

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OK.

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So I want to point out
a few things first.

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And the first thing I want to
point out is if we denote F

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as we usually do in two
dimensions as [M, N],

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then the curl of F is going
to be N sub x minus M sub y.

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OK.

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I actually calculated
these earlier,

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but I want to point out that
M sub y is actually equal to n

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times r to the n
minus 2, times x*y.

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Let me make sure I
wrote that correctly.

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Yes.

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But that is also exactly
equal to N sub x.

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And so what does that give us?

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Since N sub x minus M
sub y is the curl of F,

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when we have this vector [M, N],
we know that the curl of F is

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equal to 0 by this work.

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OK, now if our vector
field was defined

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on a simply-connected
region, then that's

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enough to show that
F is conservative.

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OK?

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We just use Green's
theorem right away.

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Right?

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But the problem is that
we are not necessarily

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on a simply-connected
region because we could

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have problems at the origin.

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And so I'm going
to deal with this

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in a slightly different way.

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To show that F is conservative,
what do I want to show?

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I want to show that
when I take the line

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integral F dot dr over any
closed loop that I get 0.

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That's ultimately what
I'm trying to show.

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So there are fundamentally
two types of curves

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that I'm concerned with.

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Two closed curves in R^2
that I'm concerned with,

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and I'm going to draw a picture
of those two types of curves.

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So in R^2, I'm going to have
curves that miss the origin--

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some curve like this,
which I'll call C_1.

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And then I'm going
to have curves

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that go around the origin,
and I'll call this C_2.

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OK?

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Fundamentally,
there's a difference

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between this curve
and this curve,

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because this curve contains
the region where F is defined

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and differentiable, right?

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Every point on the
interior of this curve, F

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is defined and differentiable
and therefore, I

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can apply regular old
Green's theorem here.

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OK?

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So I know by Green's theorem,
the integral over the closed

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curve C_1 of F dot
dr is equal to 0,

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and that's simply because
the curl of F is equal to 0.

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Right?

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We can immediately
use Green's theorem

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because we know that the
integral over this loop C_1

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is equal to the integral over
this region of the curl of F.

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That's just Green's theorem.

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So I can apply
Green's theorem here.

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Now the problem here is I
can't apply Green's theorem

00:04:20.460 --> 00:04:23.010
because this origin
is a trouble spot.

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Right?

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I'm not necessarily
differentiable there,

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so I have to be a
little more careful.

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OK, and so what
I do is I'm going

00:04:29.295 --> 00:04:32.950
to explain why, immediately,
I can get the integral over C2

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is actually also 0.

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And what I'm going
to do is I'm actually

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going to draw, hopefully,
a circle that contains C_2.

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So I'm going to draw a circle.

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It's a lot of
curves, but this is

00:04:45.290 --> 00:04:47.292
supposed to look like a circle.

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Sorry about that.

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It's a little big on the
low side, but it's a circle.

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OK.

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This is a circle.

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And I'm going to call this C_3.

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Now, I can tell you right
away that the integral

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over the curve C_3 of
F dot dr is equal to 0,

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and let me explain why.

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OK?

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F is a normal vector field
relative to a circle.

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Let's look at this again.

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It's radial, and that's
why we know this.

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F is a radial vector field.

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It's really the vector
field [x, y] times a scalar,

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depending on the radius.

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So if I look at this
picture right here,

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then F is going
to-- let me draw it

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in color-- F is going
to, at any given point,

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be in the radial direction.

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But that is exactly normal
to the tangent direction

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of this curve.

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So this is the
direction F points,

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and this is the
direction the tangent

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vector points to the curve.

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But remember, F dot dr
is the same as F dotted

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with the tangent vector ds.

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OK?

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And so that is why
for this circle,

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it's immediately obvious
that F dot dr is equal to 0.

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Because at any given
point on this circle,

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I'm taking a vector
field, I'm dotting it

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with a vector field that's
orthogonal to it, so I get 0,

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and when I integrate 0 I get 0.

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OK?

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So that's why this is 0.

00:06:10.332 --> 00:06:12.540
And now where the extended
version of Green's theorem

00:06:12.540 --> 00:06:18.655
comes in, is the fact that,
if I look in this region,

00:06:18.655 --> 00:06:21.460
F is defined and differentiable.

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Right?

00:06:21.960 --> 00:06:24.234
F is defined and differentiable
in this entire region

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that I've just shaded.

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Which is the region between
my circle and my curve C_2.

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And what that tells me is that
because this one is 0-- when

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I integrate along
this curve it's

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0-- the integral along this
curve also has to be 0, right?

00:06:38.671 --> 00:06:40.420
That's what you actually
have seen already

00:06:40.420 --> 00:06:42.140
when you talked about the
extended version of Green's

00:06:42.140 --> 00:06:42.970
theorem.

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You can compare the
integral along this curve

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to the integral along this
curve because in the region

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between them, F is everywhere
defined and differentiable.

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So you can apply
Green's theorem there.

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It just now has two
boundary components,

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instead of in this
case where it just

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has one boundary component.

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And so since the integral
on this curve is 0,

00:07:02.770 --> 00:07:06.770
and the curl of F is 0, and F
is defined and differentiable

00:07:06.770 --> 00:07:09.030
everywhere in this
region, that tells you

00:07:09.030 --> 00:07:12.804
that the integral on
the curve C_2 is also 0.

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Let me say that
one more time, OK?

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I'm going to label it in
blue so you can see it.

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I'm going to call this
region that's shaded R.

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So Green's theorem says
that the double integral

00:07:27.620 --> 00:07:32.500
in R of the curl of F
is equal to the integral

00:07:32.500 --> 00:07:34.040
around this curve.

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And then I come in and I
go around this direction

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and I come back out,
and that gives me

00:07:40.560 --> 00:07:45.730
the entire integral of the
curl of F on this region.

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Right?

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The curl of F is 0
everywhere in this region,

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so that integral is 0.

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And so the sum of the integral
on C_3 minus the integral

00:07:55.060 --> 00:07:56.760
on C_2 has to be 0.

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Since this one is
0, that one is 0.

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So you've seen this before.

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I just want to remind you
about where that's coming from.

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All right.

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So now we have to
do one other thing,

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and that's we have to
find a potential function.

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OK, so let's talk about how
to find a potential function.

00:08:18.400 --> 00:08:21.920
I'm going to do this by one of
the methods we saw in lecture.

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have--

00:08:25.730 --> 00:08:29.280
I'm in R^2, and I'm going
to start at a certain point

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and I'm going to
integrate up to (x_1,

00:08:33.860 --> 00:08:37.025
y_1) from this certain point.

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And then I'm going to figure out
what the function is that way.

00:08:39.650 --> 00:08:41.191
So what I'm going
to do-- again, I'll

00:08:41.191 --> 00:08:44.290
write it this way-- I'm going
to figure out f of (x_1, y_1)

00:08:44.290 --> 00:08:49.060
by integrating along a
certain curve, F dot dr.

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Now I can't do exactly
what I did previously,

00:08:51.460 --> 00:08:54.820
because for certain
values of n, I

00:08:54.820 --> 00:08:58.569
run into trouble with
integrating F from the origin.

00:08:58.569 --> 00:09:00.610
So what I'm going to do
is instead of integrating

00:09:00.610 --> 00:09:03.230
from the origin, I'm
going to integrate

00:09:03.230 --> 00:09:05.940
from the point (1, 1).

00:09:05.940 --> 00:09:06.440
OK?

00:09:06.440 --> 00:09:09.420
So I'm going to start
at the point 1 comma 1,

00:09:09.420 --> 00:09:11.890
and I'm going to integrate
in the y-direction,

00:09:11.890 --> 00:09:14.320
and then I'm going to
integrate in the x-direction.

00:09:14.320 --> 00:09:16.400
So this will be my
first curve and this

00:09:16.400 --> 00:09:18.440
will be my second curve.

00:09:18.440 --> 00:09:22.425
And I will land
at x_1 comma y_1.

00:09:22.425 --> 00:09:24.050
So again, this is
one of the strategies

00:09:24.050 --> 00:09:25.630
we've seen previously.

00:09:25.630 --> 00:09:27.780
This is the idea that
I'm going to integrate

00:09:27.780 --> 00:09:32.650
in the y-direction, from y
equals 1 to y equals y_1.

00:09:32.650 --> 00:09:37.055
So this will be the point 1
comma y_1, so x is fixed there.

00:09:37.055 --> 00:09:40.340
And I'm going to integrate
in the x-direction,

00:09:40.340 --> 00:09:45.070
from x equals 1 to x equals
x_1, when y is equal to y_1.

00:09:45.070 --> 00:09:46.710
So let's break this down.

00:09:46.710 --> 00:09:50.800
And let me remind you, also,
the integral along this curve C

00:09:50.800 --> 00:09:57.050
of F dot dr should
be P*dx plus Q*dy.

00:09:57.050 --> 00:09:57.760
Right?

00:09:57.760 --> 00:10:02.270
And so I'm going to look at what
P*dx is and what Q*dy is on C_1

00:10:02.270 --> 00:10:04.271
and on C_2.

00:10:04.271 --> 00:10:04.770
All right.

00:10:04.770 --> 00:10:05.820
So let's do that.

00:10:09.350 --> 00:10:13.240
OK, so I have to remind
myself what P and Q actually

00:10:13.240 --> 00:10:14.830
are in order to do this.

00:10:14.830 --> 00:10:17.670
So let me write that down,
because this will be helpful:

00:10:17.670 --> 00:10:19.960
[P, Q].

00:10:19.960 --> 00:10:26.540
P is r to the n, x,
and Q is r to the n, y.

00:10:26.540 --> 00:10:27.490
All right?

00:10:27.490 --> 00:10:30.760
So that's what we're
dealing with here.

00:10:30.760 --> 00:10:32.580
I'm going to come
back to this picture,

00:10:32.580 --> 00:10:33.730
and then I'm going to
come back and forth

00:10:33.730 --> 00:10:35.120
a little bit at this point.

00:10:35.120 --> 00:10:41.000
So if I want to integrate P*dx
plus Q*dy on the curve C_1,

00:10:41.000 --> 00:10:46.220
what I need to observe first
is that x is fixed, so dx is 0.

00:10:46.220 --> 00:10:49.076
So I'm actually just
going to integrate Q*dy.

00:10:49.076 --> 00:10:49.720
All right.

00:10:49.720 --> 00:10:52.420
So the first integral
along C_1 is just

00:10:52.420 --> 00:10:54.090
a parameterization in y.

00:10:54.090 --> 00:10:57.080
So it's the integral
from 0 to y_1

00:10:57.080 --> 00:11:04.860
of Q evaluated at x equal 1,
and y going from 1 to y_1.

00:11:04.860 --> 00:11:06.090
SPEAKER 1: 1 to y_1.

00:11:06.090 --> 00:11:08.060
CHRISTINE BREINER: y
going from 1 to y1.

00:11:08.060 --> 00:11:08.630
OK?

00:11:08.630 --> 00:11:09.129
Sorry.

00:11:09.129 --> 00:11:10.650
Yes. y going from 1 to y_1.

00:11:10.650 --> 00:11:11.470
Sorry about that.

00:11:11.470 --> 00:11:11.970
Right?

00:11:11.970 --> 00:11:13.540
I was avoiding the
origin, so I'd better not

00:11:13.540 --> 00:11:15.206
put a 0 down there,
because that's where

00:11:15.206 --> 00:11:16.570
I was running into problems.

00:11:16.570 --> 00:11:17.670
OK.

00:11:17.670 --> 00:11:20.780
So Q is r to the n, y.

00:11:20.780 --> 00:11:22.680
So I have to
remember what r is. r

00:11:22.680 --> 00:11:28.960
is x squared plus y
squared to the 1/2.

00:11:28.960 --> 00:11:34.985
So in this case, Q is-- x
is 1, and then I square it

00:11:34.985 --> 00:11:37.880
and I get 1, and then I
have y squared, and then

00:11:37.880 --> 00:11:40.690
to the n over 2--
so this is my r

00:11:40.690 --> 00:11:46.070
to the n part along the curve
C_1-- and then I multiply by y,

00:11:46.070 --> 00:11:47.504
and then I take dy.

00:11:47.504 --> 00:11:48.920
So there are a lot
of pieces here,

00:11:48.920 --> 00:11:51.253
so let me just make sure we
understand what's happening.

00:11:51.253 --> 00:11:53.850
I am interested in
this entire thing,

00:11:53.850 --> 00:11:58.110
P*dx plus Q*dy
along the curve C_1.

00:11:58.110 --> 00:12:02.140
dx is 0 along that
curve. x is 1.

00:12:02.140 --> 00:12:04.601
And y is going from 1 to y_1.

00:12:04.601 --> 00:12:06.850
So if I come back over here,
I see I'm only interested

00:12:06.850 --> 00:12:08.340
in the Q*dy part.

00:12:08.340 --> 00:12:10.740
y is going from 1 to y1.

00:12:10.740 --> 00:12:17.230
And then this is r to the
n, when x is 1 and y is y.

00:12:17.230 --> 00:12:18.340
And this is the y part.

00:12:18.340 --> 00:12:22.000
So this is exactly
Q*dy on the curve C_1.

00:12:22.000 --> 00:12:24.094
Now let's look at what
happens on the curve C_2.

00:12:24.094 --> 00:12:25.510
So if I come back
over here again,

00:12:25.510 --> 00:12:29.870
I want to have P*dx plus
Q*dy on the curve C_2.

00:12:29.870 --> 00:12:34.160
Notice y is fixed at
y1 there, so dy is 0.

00:12:34.160 --> 00:12:36.760
And so I'm only interested
in the P*dx part.

00:12:36.760 --> 00:12:38.650
Everything is going
to be in terms of x.

00:12:38.650 --> 00:12:40.920
And let's see if we can
do the same kind of thing.

00:12:40.920 --> 00:12:44.370
I'm going to be
integrating from 1 to x_1.

00:12:44.370 --> 00:12:48.590
Now r is going to be of the
form x plus y_1 squared,

00:12:48.590 --> 00:12:50.360
to the n over 2.

00:12:50.360 --> 00:12:56.450
And then-- P has an x here
and not a y-- times x dx.

00:12:56.450 --> 00:12:59.690
So again, P is r
to the n times x,

00:12:59.690 --> 00:13:03.790
so this is r to the n times
x exactly on the curve C_2.

00:13:03.790 --> 00:13:06.240
Because on C_2, y
is fixed at y_1,

00:13:06.240 --> 00:13:08.750
so that's why I actually
substituted in a y_1 here.

00:13:08.750 --> 00:13:11.790
It's the same reason
I substituted in a 1

00:13:11.790 --> 00:13:16.130
here for x, because x was
fixed at 1 on the curve C_1.

00:13:16.130 --> 00:13:20.027
So now I have to integrate
these two things.

00:13:20.027 --> 00:13:22.360
I'm going to just write down
what you get in both cases,

00:13:22.360 --> 00:13:24.350
because it's really
single-variable calculus

00:13:24.350 --> 00:13:27.120
at this point in both cases.

00:13:27.120 --> 00:13:29.317
The easiest way to do
this, probably, in my mind,

00:13:29.317 --> 00:13:30.400
is to do a u-substitution.

00:13:33.340 --> 00:13:35.070
Oops, I made a mistake.

00:13:35.070 --> 00:13:36.320
This should be an x squared.

00:13:36.320 --> 00:13:36.990
I apologize.

00:13:36.990 --> 00:13:38.290
This should be an x
squared, because this is

00:13:38.290 --> 00:13:39.720
supposed to be a radius, right?

00:13:39.720 --> 00:13:42.990
It's x squared plus whatever
y is squared, to the n over 2.

00:13:42.990 --> 00:13:45.682
So if you didn't see the squared
here, and you got nervous,

00:13:45.682 --> 00:13:46.390
you were correct.

00:13:46.390 --> 00:13:47.696
There should be a squared here.

00:13:47.696 --> 00:13:49.320
So anyway, I'm going
to go back to what

00:13:49.320 --> 00:13:50.319
I was saying previously.

00:13:50.319 --> 00:13:56.020
To integrate these things,
the easiest thing to do

00:13:56.020 --> 00:13:57.850
is to take what is
inside the parentheses

00:13:57.850 --> 00:14:00.880
and set it equal to u, and then
do a u-substitution from there.

00:14:00.880 --> 00:14:03.076
So again, I'm not going to
actually do that for you,

00:14:03.076 --> 00:14:04.700
but I'm going to tell
you what you get.

00:14:04.700 --> 00:14:06.980
Now, there are two
different situations.

00:14:06.980 --> 00:14:09.550
And the situations
follow when n is

00:14:09.550 --> 00:14:14.290
any integer except negative 2,
and then when n is negative 2.

00:14:14.290 --> 00:14:16.280
And the reason is because
when n is negative 2,

00:14:16.280 --> 00:14:18.550
this exponent is a minus 1.

00:14:18.550 --> 00:14:21.560
So when you integrate, you
end up with a natural log.

00:14:21.560 --> 00:14:24.100
So let me just point
out the two things

00:14:24.100 --> 00:14:26.110
that you get in each
case, and then we'll

00:14:26.110 --> 00:14:28.501
evaluate and see what the
solutions are in each case.

00:14:28.501 --> 00:14:30.000
So I'm just going
to, at this point,

00:14:30.000 --> 00:14:32.220
write down what I
got, because this is

00:14:32.220 --> 00:14:33.920
your single-variable calculus.

00:14:33.920 --> 00:14:40.850
OK, so what I got when n
was not equal to minus 2,

00:14:40.850 --> 00:14:42.490
you get the following thing.

00:14:42.490 --> 00:14:50.030
You get 1 plus y squared,
evaluated at n plus 2,

00:14:50.030 --> 00:14:53.420
over 2, over n plus 2.

00:14:53.420 --> 00:14:56.480
And this is evaluated
from 1 to y_1.

00:14:56.480 --> 00:14:58.570
And then this one you get
a similar thing there,

00:14:58.570 --> 00:15:00.250
but now the y_1 is fixed here.

00:15:00.250 --> 00:15:05.860
So you get an x squared plus
y_1 squared, to the n plus 2,

00:15:05.860 --> 00:15:11.840
over 2, over n plus 2,
evaluated from 1 to x_1.

00:15:11.840 --> 00:15:13.300
So here, the y_1
is fixed and it's

00:15:13.300 --> 00:15:15.210
the x-values that are
changing, and here

00:15:15.210 --> 00:15:16.730
the y-values are changing.

00:15:16.730 --> 00:15:19.260
So when n is not equal to 2,
I get exactly this quantity

00:15:19.260 --> 00:15:21.020
when I integrate
these two terms.

00:15:21.020 --> 00:15:23.280
And so now, let's
see what happens.

00:15:23.280 --> 00:15:23.960
OK?

00:15:23.960 --> 00:15:26.170
Exactly what happens
is the following.

00:15:26.170 --> 00:15:29.760
Notice that when I put in
y_1 here, I get a 1 plus y_1

00:15:29.760 --> 00:15:32.590
squared, to the n plus
2 over 2, over n plus 2.

00:15:32.590 --> 00:15:33.139
Right?

00:15:33.139 --> 00:15:35.680
I'm not going to write it down,
because I'm going to show you

00:15:35.680 --> 00:15:37.270
it gets killed off immediately.

00:15:37.270 --> 00:15:39.000
Where does it get killed off?

00:15:39.000 --> 00:15:42.960
It gets killed off when
I evaluate this one at 1.

00:15:42.960 --> 00:15:43.460
OK?

00:15:43.460 --> 00:15:47.090
So the upper bound here is the
same as the lower bound here.

00:15:47.090 --> 00:15:49.200
When I put in a 1
here, I get 1 plus y_1

00:15:49.200 --> 00:15:52.940
squared to the n plus
2 over 2 over n plus 2.

00:15:52.940 --> 00:15:54.450
It's a lot of n's and 2's.

00:15:54.450 --> 00:15:57.740
But the point is that when
I evaluate this one at y_1

00:15:57.740 --> 00:16:00.440
and I evaluate this one at 1,
I get exactly the same thing,

00:16:00.440 --> 00:16:05.004
but the signs are opposite
and so they subtract off.

00:16:05.004 --> 00:16:07.420
In the final answer, I'm not
going to see this upper bound

00:16:07.420 --> 00:16:08.920
and I'm not going to
see this lower bound,

00:16:08.920 --> 00:16:10.503
because they're going
to subtract off.

00:16:10.503 --> 00:16:12.870
And what I'm actually left
with is just two terms.

00:16:12.870 --> 00:16:15.120
And those two terms I'm
going to write up here.

00:16:15.120 --> 00:16:18.810
Those two terms are
going to be x_1 squared

00:16:18.810 --> 00:16:25.650
plus y_1 squared to the n
plus 2, over 2, over n plus 2.

00:16:25.650 --> 00:16:29.730
Minus, 1 plus 1-- which is
just 2-- to the n plus 2,

00:16:29.730 --> 00:16:32.450
over 2, over n plus 2.

00:16:32.450 --> 00:16:33.570
What it this really?

00:16:33.570 --> 00:16:39.880
This is just r to the n plus 2,
over n plus 2, plus a constant.

00:16:39.880 --> 00:16:42.425
Because this is just
a constant for any n.

00:16:42.425 --> 00:16:46.456
And notice n is not equal
to minus 2-- negative 2.

00:16:46.456 --> 00:16:49.080
That was the place we were going
to run into trouble otherwise.

00:16:49.080 --> 00:16:50.970
And so when n is not
equal to negative 2--

00:16:50.970 --> 00:16:52.480
when you do all
the integration--

00:16:52.480 --> 00:16:55.890
you should arrive at this
as your potential function.

00:16:55.890 --> 00:16:57.000
OK?

00:16:57.000 --> 00:16:59.002
And again, what I
did was I evaluated

00:16:59.002 --> 00:17:00.835
to make it simpler on
ourselves so we didn't

00:17:00.835 --> 00:17:02.140
have to write everything out.

00:17:02.140 --> 00:17:05.370
I noticed that if I evaluate
this at the two bounds,

00:17:05.370 --> 00:17:06.870
and evaluate this
at the two bounds,

00:17:06.870 --> 00:17:09.940
and I add them together,
that the evaluation here

00:17:09.940 --> 00:17:13.720
plus the evaluation here are the
same numerically but opposite

00:17:13.720 --> 00:17:16.240
in sign, and so
they subtract off.

00:17:16.240 --> 00:17:19.300
And then I just have to evaluate
at this one and this one.

00:17:19.300 --> 00:17:26.000
So that's n not
equal to negative 2.

00:17:26.000 --> 00:17:28.790
Now let's do the n equal
to negative 2 case.

00:17:28.790 --> 00:17:31.730
OK, so now I'm integrating
this exact same thing

00:17:31.730 --> 00:17:34.090
in the n equal to
negative 2 case.

00:17:34.090 --> 00:17:37.080
And I'll just write down again
what I get by the substitution.

00:17:37.080 --> 00:17:39.810
And what I get is
natural log of 1

00:17:39.810 --> 00:17:45.740
plus y squared, over 2,
evaluated from 1 to y_1.

00:17:45.740 --> 00:17:52.420
Plus, natural log of x squared
plus y_1 squared, over 2,

00:17:52.420 --> 00:17:54.140
evaluated from 1 to x_1.

00:17:54.140 --> 00:17:55.640
Let me make sure
I have that right.

00:17:55.640 --> 00:17:56.550
Yes.

00:17:56.550 --> 00:17:58.220
And the same kind
of thing is going

00:17:58.220 --> 00:18:01.800
to happen that happened before,
in terms of when I put y_1

00:18:01.800 --> 00:18:07.287
in here, and I put 1 in
here, I get the same thing

00:18:07.287 --> 00:18:08.370
but with an opposite sign.

00:18:08.370 --> 00:18:09.980
Here it's a positive.

00:18:09.980 --> 00:18:12.540
It's natural log 1 plus
y_1 squared over 2.

00:18:12.540 --> 00:18:15.630
And here it's natural log
1 plus y_1 squared over 2,

00:18:15.630 --> 00:18:18.240
but because it's the lower
bound, it's a negative sign.

00:18:18.240 --> 00:18:22.080
So whatever I get here and
what I get here subtract off.

00:18:22.080 --> 00:18:23.750
And then in the end,
I wind up getting

00:18:23.750 --> 00:18:25.850
just the following two terms.

00:18:25.850 --> 00:18:32.520
I get x_1 squared plus
y_1 squared over 2,

00:18:32.520 --> 00:18:36.500
minus natural log of 2 over 2.

00:18:36.500 --> 00:18:39.580
So this term comes from
evaluating this at x_1.

00:18:39.580 --> 00:18:42.690
And this term comes
from evaluating this one

00:18:42.690 --> 00:18:44.260
at y equal 1.

00:18:44.260 --> 00:18:45.900
And if you notice, what is this?

00:18:45.900 --> 00:18:50.110
This is exactly natural
log of r plus a constant.

00:18:50.110 --> 00:18:53.830
So let me step to the other
side so we can see it clearly.

00:18:53.830 --> 00:18:57.110
So this is natural
log of r squared,

00:18:57.110 --> 00:19:00.650
but by log rules, that's really
2 times natural log of r,

00:19:00.650 --> 00:19:04.500
so it divides by 2 and I'm just
left with natural log of r,

00:19:04.500 --> 00:19:06.220
and this is just a constant.

00:19:06.220 --> 00:19:09.910
And so my potential
function in that case

00:19:09.910 --> 00:19:12.777
is exactly natural log
of r plus a constant.

00:19:12.777 --> 00:19:14.235
All right, this
was a long problem,

00:19:14.235 --> 00:19:16.320
so I'm just going to remind
us where we came from

00:19:16.320 --> 00:19:17.100
and what we were doing.

00:19:17.100 --> 00:19:18.516
So let's go back
to the beginning.

00:19:21.400 --> 00:19:23.960
So what we did initially, was
we had this vector field F.

00:19:23.960 --> 00:19:28.782
It was a radial vector field.
r to the n times x*i plus y*j.

00:19:28.782 --> 00:19:30.240
And we wanted to
first show that it

00:19:30.240 --> 00:19:33.110
was conservative for
any integer value of n,

00:19:33.110 --> 00:19:35.350
and then to find its
potential function.

00:19:35.350 --> 00:19:36.980
And obviously we do
it in that order,

00:19:36.980 --> 00:19:38.355
because if it's
not conservative,

00:19:38.355 --> 00:19:41.880
we're not going to find
a potential function.

00:19:41.880 --> 00:19:45.740
In this case, what I observed
first was that the curl of F

00:19:45.740 --> 00:19:47.860
was 0.

00:19:47.860 --> 00:19:52.660
And so in places where I
had a closed curve that

00:19:52.660 --> 00:19:54.544
didn't contain
the origin, I knew

00:19:54.544 --> 00:19:56.460
that the integral all
around that closed curve

00:19:56.460 --> 00:19:59.230
was 0 just by Green's theorem.

00:19:59.230 --> 00:20:02.400
But if I had a closed curve that
contained the origin, because F

00:20:02.400 --> 00:20:04.950
is not differentiable for
all the n-values there,

00:20:04.950 --> 00:20:06.790
I have to be a little careful.

00:20:06.790 --> 00:20:08.550
It's actually even 0, right?

00:20:08.550 --> 00:20:12.020
When x is 0 and y is 0, I'm
going to get something 0 there.

00:20:12.020 --> 00:20:16.360
So I need to figure out a
way to determine the line

00:20:16.360 --> 00:20:17.810
integral on C_2.

00:20:17.810 --> 00:20:18.310
Right?

00:20:18.310 --> 00:20:19.184
And that was my goal.

00:20:19.184 --> 00:20:20.760
For any C_2 that
contains the origin,

00:20:20.760 --> 00:20:23.100
how do I figure out F dot dr.

00:20:23.100 --> 00:20:26.530
And so I just
compared it to what

00:20:26.530 --> 00:20:29.260
I get when I take F
dot dr around a circle.

00:20:29.260 --> 00:20:32.050
Because I know that I can
always find a circle bigger,

00:20:32.050 --> 00:20:34.540
and then I can say I've
got this region here-- in

00:20:34.540 --> 00:20:36.480
between-- on which F
is defined everywhere,

00:20:36.480 --> 00:20:39.950
so I can apply Green's
theorem to that inside region.

00:20:39.950 --> 00:20:43.940
And I know that the curl of
F on the inside region is 0,

00:20:43.940 --> 00:20:48.391
and so the integral on C_2 and
C_3 is going to agree, right?

00:20:48.391 --> 00:20:49.890
Because the integral
on C_3 I showed

00:20:49.890 --> 00:20:52.240
was 0 just geometrically.

00:20:52.240 --> 00:20:56.096
And then the integral
on C_2 then has to be 0.

00:20:56.096 --> 00:20:56.595
All right?

00:20:56.595 --> 00:21:00.010
And so that was just when you
were using the extended version

00:21:00.010 --> 00:21:01.460
of Green's theorem.

00:21:01.460 --> 00:21:05.740
And then to find a potential
function, we came over here.

00:21:05.740 --> 00:21:07.952
And we had to avoid
the origin because

00:21:07.952 --> 00:21:09.910
of the differentiability
problem at the origin.

00:21:09.910 --> 00:21:12.140
So we started-- instead
of where we usually start,

00:21:12.140 --> 00:21:15.630
which is from (0, 0)-- we
started from the point (1, 1).

00:21:15.630 --> 00:21:18.050
And we just determined
the potential function

00:21:18.050 --> 00:21:22.460
going from the point (1,
1) to the point (x_1, y_1)

00:21:22.460 --> 00:21:25.039
along a curve that went
straight up, so x was fixed,

00:21:25.039 --> 00:21:27.080
and then along the curve
that went straight over,

00:21:27.080 --> 00:21:28.640
so y was fixed.

00:21:28.640 --> 00:21:30.925
And so then we were able to
break up this thing where

00:21:30.925 --> 00:21:35.370
I'm integrating over C P*dx plus
Q*dy into two separate pieces,

00:21:35.370 --> 00:21:38.300
and each of them was fairly
simple to write down.

00:21:38.300 --> 00:21:41.930
So let's look at what they were.

00:21:41.930 --> 00:21:45.180
This first one was
where we were moving up.

00:21:45.180 --> 00:21:48.440
And there was no dx.
x was just fixed at 1.

00:21:48.440 --> 00:21:51.180
And y was going from 1 to y_1.

00:21:51.180 --> 00:21:51.980
Right?

00:21:51.980 --> 00:21:54.162
And so x is fixed at
1, so I put a 1 there.

00:21:54.162 --> 00:21:55.370
And y is going from 1 to y_1.

00:21:55.370 --> 00:21:59.240
So I evaluate Q*dy
on that curve.

00:21:59.240 --> 00:22:01.740
And then the next one was P*dx
on the curve where I'm moving

00:22:01.740 --> 00:22:03.620
straight across.

00:22:03.620 --> 00:22:06.970
Right? dy is 0 there, so
I just pick up the P*dx.

00:22:06.970 --> 00:22:10.035
And my y-value was
fixed at y_1, and x

00:22:10.035 --> 00:22:12.320
was varying from 1 to x_1.

00:22:12.320 --> 00:22:15.260
And so then I just had to
be a little bit careful.

00:22:15.260 --> 00:22:17.110
I didn't show you exactly
how you integrate,

00:22:17.110 --> 00:22:20.630
but using a substitution trick--
single-variable calculus--

00:22:20.630 --> 00:22:22.970
shouldn't be too bad
for you at this point.

00:22:22.970 --> 00:22:25.770
We distinguished between when
n was not equal to negative 2

00:22:25.770 --> 00:22:27.460
and when n was
equal to negative 2.

00:22:27.460 --> 00:22:30.420
In the case n not
equal to negative 2,

00:22:30.420 --> 00:22:34.167
we determined the
integral, we simplified,

00:22:34.167 --> 00:22:36.750
and we got to a place where the
potential function was exactly

00:22:36.750 --> 00:22:42.160
equal to r to the n plus 2 over
n plus 2, plus some constant.

00:22:42.160 --> 00:22:45.952
Then in the case where n
was equal to negative 2,

00:22:45.952 --> 00:22:48.410
when you do the substitution,
you get a different integral.

00:22:48.410 --> 00:22:50.430
And in that case, you
get the natural log.

00:22:50.430 --> 00:22:52.630
And so again, we just
had the natural log.

00:22:52.630 --> 00:22:54.237
We have these
different functions.

00:22:54.237 --> 00:22:56.820
We're evaluating the natural log
of these different functions.

00:22:56.820 --> 00:22:58.080
We have the bounds.

00:22:58.080 --> 00:23:00.850
We simplify everything, and
we get exactly to the place

00:23:00.850 --> 00:23:03.650
where you have natural
log of r plus a constant.

00:23:03.650 --> 00:23:06.070
And so we found our potential
function in the case n

00:23:06.070 --> 00:23:10.190
is equal to negative 2,
and then any other n-value.

00:23:10.190 --> 00:23:12.160
So, a very long problem.

00:23:12.160 --> 00:23:13.700
I hope you got
something out of it.

00:23:13.700 --> 00:23:16.160
And this is where I will stop.