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