WEBVTT
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CHRISTINE BREINER: Welcome
back to recitation.
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In this video, I want us to work
on the following problem, which
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is to show that this vector
field is not conservative.
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And the vector field
is minus y*i plus x*j,
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all divided by x
squared plus y squared.
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So you can think about this
in two separate components,
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if you need to, as minus y
divided by x squared plus y
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squared i plus x over x
squared plus y squared j.
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So that's really
exactly the same thing.
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So your object is to show
that this vector field is not
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conservative.
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And why don't you work on that
for awhile, pause the video,
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and then when you're ready
to see my solution you
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can bring the video back up.
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So welcome back.
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Well, maybe some of you thought
I had a typo in this problem
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initially, and I
wanted you to show
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it, in fact, was conservative,
but it actually is not
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a conservative vector field.
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And let me explain how we can
show it is not conservative
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and why probably what
you did initially to try
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and show it was not didn't work.
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So maybe that wording
was a little confusing,
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but let me take you through it.
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So the first thing
I would imagine
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you tried is you looked
at M sub y and N sub x.
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So M in this case is negative y
over x squared plus y squared.
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And N in this case,
capital N in this case,
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is x divided by x
squared plus y squared.
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So if you worked that
out, you probably did
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or maybe you didn't,
and I'll just show you.
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M sub y-- let me just
double check-- is y
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squared minus x squared over
x squared plus y squared,
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I think with an
extra squared on it.
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Yeah.
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And that's also
equal to N sub x.
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Right?
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So what you know so far,
what you might have thought
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immediately, was well, N sub x
minus M sub y is the curl of F
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and that's equal to 0, and
therefore this vector field
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is conservative.
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But the problem is the
theorem you were thinking
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about referencing doesn't hold.
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And the reason is
because there are
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two hypotheses in that theorem.
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And one is that if I
define this vector field,
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if I call it capital F, the
vector field-- or the theorem
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is that capital F defined
everywhere, and curl
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of F equal to 0,
implies F conservative.
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OK.
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So that's the theorem you
might have been trying to use.
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You see from this the
curl of F equals 0,
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but the problem is the first
part of this statement,
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that F being defined
everywhere, is not true.
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In fact, there's one place in
R^2 where this vector field is
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not defined, and that is
when x is 0 and y is 0.
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Because at that point,
obviously, the denominator
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is zero and we run into trouble.
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So you cannot use this theorem
to say F is conservative
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because it's not
defined everywhere.
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Or I should be careful
how I say that.
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There is somewhere
that it is not defined.
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So even though the
curl of F equals 0,
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the first part of the
statement is not true.
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So you cannot get anything
out of this theorem.
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So knowing the curl of F equals
0 doesn't tell you whether it's
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conservative or not.
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OK?
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So now what I'm
going to do is I'm
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going to show-- I
told you we want
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to show it's not conservative.
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I'm going to show you
how we can show that.
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And what we're going to do is
we're going to find a loop,
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a closed loop, so a
closed curve in R^2,
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that when I integrate this
vector field over that closed
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curve, I don't get zero.
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And then we would know that
the vector field is not
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conservative.
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So that's what
we're going to do.
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So let me write
it out explicitly
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and then we'll figure out the
curve we want and then we'll
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parameterize the
curve appropriately.
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So I'm going to show,
for some closed curve--
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I'm going to pick
my curve and I'm
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going to integrate
over the closed curve
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and I'm going to integrate this.
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Minus y over x squared
plus y squared dx plus x
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over x squared
plus y squared dy.
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And I'm going to show that
if I pick a certain curve,
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I'm going to get something
that's not equal to zero, OK?
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And the curve I'm going to
pick is the unit circle.
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So we're going to let
C be the unit circle.
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Let C equal-- I'll just
write the unit circle.
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But how can I parameterize
the unit circle easily?
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I can parameterize
the unit circle easily
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by x equal to cosine theta
and y equal to sine theta.
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So let me do that.
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And why am I picking
the unit circle?
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We'll see why that
is in a second.
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So we're going to let x equal
cosine theta and y equal sine
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theta.
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And so now we know
what goes here
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and we know what goes here.
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By the way, what is x
squared plus y squared?
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It's cosine squared
theta plus sine squared
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theta, which is equal to 1.
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This is exactly the
square of the radius
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and since we're on the
unit circle, that's 1.
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That's why I picked
the unit circle
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because I wanted the
denominator to be very simple.
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So I've got the x's and the y's.
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Now what's dx?
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dx is going to be equal to
minus sine theta d theta.
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And what's dy?
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I'll just write it
right underneath here.
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dy is going to equal
cosine theta d theta.
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So let me just point out
again what we were doing here.
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We want to parameterize
the unit circle.
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I chose to parameterize it in
theta, which I haven't told you
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what my bounds are yet, but
I've done everything else.
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I needed to know
what x and y were
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and also what dx and dy were.
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And so now I can
start substituting in.
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So let's figure out what I get
when I start substituting in.
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I'm integrating now.
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Again, I said I didn't
mention the bounds.
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What are the bounds on theta
to get the whole unit circle?
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I'm just going to
integrate from 0 to 2*pi.
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So I integrate from 0 to 2*pi.
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That takes me all the
way around the circle.
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Minus y is negative sine theta.
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This part is 1 as I
mentioned earlier.
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And then dx is minus
sine theta d theta.
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So I have a minus sine theta
times a minus sine theta.
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That's going to give me a
sine squared theta d theta.
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And then this term,
the second term,
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when I integrate
over the curve, I'm
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going to just rewrite
another one here separately
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momentarily.
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x is cosine theta.
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x squared plus y squared is 1.
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And dy is cosine theta d theta.
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So I get cosine
squared theta d theta.
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All right?
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Now here we are.
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If I tried to integrate
both of these separately,
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it would take potentially
a very long time
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and be kind of annoying.
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But if you notice, because
I can add these integrals,
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I can add over-- they
have the same bounds,
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so I can put them back together.
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Sine squared theta plus cosine
squared theta always equals 1.
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That's a trig identity
that's good to know.
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So this in fact is equal to the
integral from 0 to 2*pi of d
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theta.
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So let me come
back one more time
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and make sure we understand.
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We're integrating from 0 to 2*pi
sine squared theta d theta plus
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cosine squared theta d theta.
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That's just sine squared
theta plus cosine squared
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theta d theta.
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So that's 1.
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So 1 times d theta.
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But what is this?
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Well, this integral from 0
to 2 pi of d theta is theta
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evaluated at 2*pi and 0.
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I actually get 2*pi, which is
in particular not equal to 0,
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right?
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So that actually shows you
that this vector field is not
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conservative.
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Now why does this make sense?
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This makes sense
because if I really
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think about what I'm
doing-- actually,
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this is the place where maybe
it'll ultimately make the most
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sense-- what you're
doing is you're
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looking at how theta
changes as you go all
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the way around the origin once.
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And theta changes by 2*pi.
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If you go all the
way around one time,
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the theta value starts at
0 and then goes up to 2*pi.
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And so that's exactly where
this 2*pi is coming from.
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So that actually
is ultimately how
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you were going to be able to
show that this vector field is
00:08:27.460 --> 00:08:28.880
not conservative.
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So let me go back and just
remind you where we came from.
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We started off
with a vector field
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and we wanted to know if
it was not conservative.
00:08:38.284 --> 00:08:40.450
We wanted to show-- sorry--
it was not conservative.
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So the first thing you
might want to check,
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which maybe you did, was
if the curl was zero.
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And in fact the curl is zero.
00:08:48.920 --> 00:08:50.470
And so maybe you
thought, well, she
00:08:50.470 --> 00:08:51.685
might have done something
wrong, or she might
00:08:51.685 --> 00:08:52.893
have written something wrong.
00:08:52.893 --> 00:08:56.510
But we can't actually appeal to
the theorem you want to appeal
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to to make any conclusion
about the vector field,
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because the vector field in
our example is not defined
00:09:02.390 --> 00:09:04.570
for every value of (x,y).
00:09:04.570 --> 00:09:08.090
So for every value in the
xy-plane, we cannot define F.
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There's one value for
which we cannot define F.
00:09:10.890 --> 00:09:13.050
And so we cannot say
that if the curl's 0,
00:09:13.050 --> 00:09:15.540
then the vector field
is conservative.
00:09:15.540 --> 00:09:18.130
We can't draw any conclusions
from this theorem.
00:09:18.130 --> 00:09:19.890
So then we had to
actually find a way
00:09:19.890 --> 00:09:22.820
to show it was not conservative
without looking at the curl.
00:09:22.820 --> 00:09:25.120
And that amounts to showing
there is a closed curve
00:09:25.120 --> 00:09:27.274
that when I integrate
over that closed curve--
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when I look at what the vector
field does over that closed
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curve-- I get
something non-zero.
00:09:32.300 --> 00:09:34.610
And we picked an easy example.
00:09:34.610 --> 00:09:36.270
This is actually
what the integral
00:09:36.270 --> 00:09:39.240
will look like over the
closed curve in x and y.
00:09:39.240 --> 00:09:41.080
We let our closed curve
be the unit circle,
00:09:41.080 --> 00:09:43.070
then we parameterized in theta.
00:09:43.070 --> 00:09:46.020
And we see that actually what
this vector field is doing
00:09:46.020 --> 00:09:48.630
is it's looking at
what is d theta?
00:09:48.630 --> 00:09:50.440
It's finding out
what d theta is.
00:09:50.440 --> 00:09:53.890
And so we find out the integral
from 0 to 2*pi of d theta is
00:09:53.890 --> 00:09:55.385
obviously not zero.
00:09:55.385 --> 00:09:57.260
And that gives us that
the vector field's not
00:09:57.260 --> 00:09:58.420
conservative.
00:09:58.420 --> 00:09:59.900
So that's where I'll stop.