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.
00:01:04.763 --> 00:01:07.100
The first is to show
that this vector field F
00:01:07.100 --> 00:01:08.750
I've given you is conservative.
00:01:08.750 --> 00:01:11.260
And then we want to find
a potential function.
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And we want to be
able to show it's
00:01:12.790 --> 00:01:14.360
conservative for all integers n.
00:01:14.360 --> 00:01:17.909
And what I want to point out is
that for certain integer values
00:01:17.909 --> 00:01:19.700
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?
00:01:22.720 --> 00:01:25.290
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
00:01:32.750 --> 00:01:36.130
even if we don't include
the origin in our region.
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OK.
00:01:37.370 --> 00:01:41.010
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.
00:03:10.460 --> 00:03:17.110
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?
00:03:51.300 --> 00:03:55.000
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?
00:04:03.470 --> 00:04:05.500
We can immediately
use Green's theorem
00:04:05.500 --> 00:04:09.260
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.
00:04:15.260 --> 00:04:17.900
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.
00:04:27.660 --> 00:04:29.295
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
00:04:32.950 --> 00:04:35.170
is actually also 0.
00:04:35.170 --> 00:04:37.060
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,
00:05:05.220 --> 00:05:07.770
and let me explain why.
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OK?
00:05:08.550 --> 00:05:13.680
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
00:05:50.980 --> 00:05:53.670
with the tangent vector ds.
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OK?
00:05:54.170 --> 00:05:55.896
And so that is why
for this circle,
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it's immediately obvious
that F dot dr is equal to 0.
00:05:58.940 --> 00:06:01.170
Because at any given
point on this circle,
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I'm taking a vector
field, I'm dotting it
00:06:03.260 --> 00:06:05.810
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?
00:06:08.670 --> 00:06:10.332
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.
00:06:21.460 --> 00:06:21.960
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.
00:06:25.150 --> 00:06:29.650
Which is the region between
my circle and my curve C_2.
00:06:29.650 --> 00:06:33.486
And what that tells me is that
because this one is 0-- when
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I integrate along
this curve it's
00:06:34.860 --> 00:06:38.671
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.
00:06:42.970 --> 00:06:45.580
You can compare the
integral along this curve
00:06:45.580 --> 00:06:48.580
to the integral along this
curve because in the region
00:06:48.580 --> 00:06:51.220
between them, F is everywhere
defined and differentiable.
00:06:51.220 --> 00:06:53.320
So you can apply
Green's theorem there.
00:06:53.320 --> 00:06:55.100
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.
00:06:58.920 --> 00:07:02.770
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.
00:07:15.470 --> 00:07:18.470
Let me say that
one more time, OK?
00:07:18.470 --> 00:07:20.470
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.
00:07:23.750 --> 00:07:27.620
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.
00:07:34.040 --> 00:07:36.580
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.
00:07:45.730 --> 00:07:46.230
Right?
00:07:46.230 --> 00:07:49.450
The curl of F is 0
everywhere in this region,
00:07:49.450 --> 00:07:51.280
so that integral is 0.
00:07:51.280 --> 00:07:55.060
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.
00:07:56.760 --> 00:07:58.230
Since this one is
0, that one is 0.
00:07:58.230 --> 00:07:59.354
So you've seen this before.
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I just want to remind you
about where that's coming from.
00:08:04.840 --> 00:08:05.780
All right.
00:08:05.780 --> 00:08:08.020
So now we have to
do one other thing,
00:08:08.020 --> 00:08:11.455
and that's we have to
find a potential function.
00:08:11.455 --> 00:08:13.830
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.
00:08:24.510 --> 00:08:25.730
have--
00:08:25.730 --> 00:08:29.280
I'm in R^2, and I'm going
to start at a certain point
00:08:29.280 --> 00:08:33.860
and I'm going to
integrate up to (x_1,
00:08:33.860 --> 00:08:37.025
y_1) from this certain point.
00:08:37.025 --> 00:08:39.650
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.
00:08:49.060 --> 00:08:51.460
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.