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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,

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which is to show that this
vector field is not

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conservative.

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And the vector field is minus
yi plus xj all divided by x

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squared plus y squared.

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So you can think about this in
two separate components, if

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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 can bring

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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

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problem initially, and I wanted
you to show it , in

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fact, was conservative, but it
actually is not a conservative

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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--

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didn't work.

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So maybe that wording was a
little confusing, but let me

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take you through it.

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So the first thing I would
imagine you tried is you

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looked at m sub y and n sub x.

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So m in this case is
negative y over x

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squared plus y squared.

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And N in this case, capital N in
this case, is x divided by

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x squared plus y squared.

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So if you worked that out, you
probably did or maybe you

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didn't, and I'll
just show you.

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M sub y-- let me just
double check--

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is y 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 is

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conservative.

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But the problem is the theorem
you were thinking about

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referencing doesn't hold.

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And the reason is because there
are two hypotheses in

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that theorem.

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And one is that if I define this
vector field, if I call

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it capital F, the
vector field--

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or the theorem 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, but the problem is

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the first part of this
statement, that F being

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defined everywhere,
is not true.

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In fact, there's one place in r2
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 is

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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, the first part of

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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

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whether it's conservative
or not.

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OK?

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So now what I'm going to do
is I'm going to show--

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I told you we want 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, a

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closed loop.

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So a closed curve in R2 that
when I integrate this vector

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field over that closed 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 and then we'll

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figure out the curve we want
and then we'll parameterize

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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 going to integrate over

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the closed curve and I'm going
to integrate this.

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Minus y over x squared plus
y squared dx plus x over x

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squared plus y squared dy.

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And I'm going to show that if
I pick a certain curve, I'm

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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--

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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 by x equal to

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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

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y equal sine theta.

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And so now we know what
goes here and we

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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 theta, which is

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equal to 1.

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This is exactly the square of
the radius and since we're on

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the unit circle, that's 1.

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That's why I picked the unit
circle because I wanted the

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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

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you what my bounds are yet, but
I've done everything else.

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I needed to know what x
and y were and also

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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, when I integrate over

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the curve, I'm going to just
rewrite another one here

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separately 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, it would

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take potentially a very long
time and be kind of annoying.

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But if you notice, because I can
add these integrals, I can

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add over they have the
same bounds, so I can

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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

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to 2 pi of d theta.

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So let me come back one more
time and make sure we

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understand.

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We're integrating from 0 to 2 pi
sine squared theta d theta

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plus cosine squared
theta d theta.

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That's just sine squared
theta plus cosine

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squared 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

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not equal to 0, 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 think

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about what I'm doing--

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actually, this is the place
where maybe it'll ultimately

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make the most sense--

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what you're doing is you're
looking at how theta changes

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as you go all 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, the theta value

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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 you were going to be able

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to show that this vector field
is 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 and we wanted to know if

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it was not conservative.

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We wanted to show-- sorry--
it was not conservative.

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So the first thing you might
want to check, which maybe you

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did, was if the curl was zero.

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And in fact the curl is zero.

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And so maybe you thought, well,
she might have done

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something wrong, or she might
have written something wrong.

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But we can't actually appeal
to the theorem you want to

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appeal to to make any conclusion
about the vector

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field because the vector field
in our example is not defined

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for every value of (x,y).

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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. And so we

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cannot say that if the curl's
0, then the vector field is

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conservative.

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We can't draw any conclusions
from this theorem.

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So then we had to actually find
a way to show it was not

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conservative without looking
at the curl.

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And that amounts to showing
there is a closed curve that

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when I integrate over that
closed curve-- when I look at

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what the vector field does
over that closed curve--

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I get something non-zero.

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And we picked an easy example.

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This is actually what the
integral will look like over

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the closed curve in x and y.

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We let our closed curve be
the unit circle, then we

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parameterized in theta.

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And we see that actually what
this vector field is doing is

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it's looking at what
is d theta?

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It's finding out what
d theta is.

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And so we find out the integral
from 0 to 2 pi of d

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theta is obviously not zero.

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And that gives us that the
vector field's not

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conservative.

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So that's where I'll stop.

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