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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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So the problem is as follows.

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For which of the following
vector fields is the domain

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where each vector field is
defined and continuously

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differentiable a simply
connected region.

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So there's a lot there.

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I'm going to break that down
first, and then show you the

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vector fields.

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So we're starting with some
different vector fields.

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And we want to determine first
the domain for each vector

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field where it is both defined
and continuously

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differentiable.

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And then once you've determined
that domain, the

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next object is to determine
whether or not that region is

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simply connected.

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So there are two parts for
each of these problems.

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So again, the first thing is you
want to determine all of

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the values for which the vector
field is both defined

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and continuously
differentiable.

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You want to look at that region
that contains all those

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values, and you want to
determine if that region is

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simply connected.

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So there are four different
vector fields, and I'll just

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point them out here.

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They're all in the plane.

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So the first one is root
x i plus root y j.

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The second one is i plus j,
divided by the square root of

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1 minus x squared
minus y squared.

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The third one looks fairly
similar to the second one, but

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it's i plus j, divided by the
square root of the quantity x

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squared plus y squared
minus 1.

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And the fourth one is i plus j
times the quantity natural log

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of r squared-- natural log of
x squared plus y squared.

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So there are four different
vector fields here.

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You have do two things
for each one.

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So find the domain where it's
defined and continuously

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differentiable, and then
determine if that domain is

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simply connected.

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So why don't you pause the
video, work on these, and then

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when you're ready to see what
I did, you can bring

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the video back up.

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OK, welcome back.

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So we're going to do
this problem one

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vector field at a time.

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We're going to take each vector
field, I'm going to

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determine where it's defined and
differentiable, and then

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I'm going to determine if that
region is simply connected.

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So we're going to do each
one separately.

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So I'm going to start off
with a, which was root x

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i plus root y j.

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And I want to point out first,
for the function f of x is

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equal to square root of x, it
is defined for all x greater

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than or equal to 0, and it
is differentiable for x

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greater than 0.

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I left out the T. There we go.

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So it is both defined and
differentiable when x is

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greater than 0.

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And now if I replace this with
a y, the same thing is true

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for y greater than or equal to
0, and y greater than 0.

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So we know the region where
this vector field is both

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defined and differentiable is
when x is greater than 0 and y

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is greater than 0.

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So we need the region--

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let me draw it this way--

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the region that is the first
quadrant in the xy plane.

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And it's not including the
x-axis or the y-axis.

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So it's this fully
shaded region.

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OK, that is the region where
it's defined and

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differentiable.

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If I was going to write that
precisely, I would say

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something like, all (x, y) with
x greater than 0 and y

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greater than 0.

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Right?

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You need both.

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So it's exactly the (x, y) pairs
with x positive and y

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positive, and that's
the region.

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And now the question is, is that
region simply connected?

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Well, the way we think about
simply connectedness from what

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you've seen in class, is you
want to show that if you take

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a closed curve that's contained
in the region, that

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everything on the interior of
that closed curve is also in

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the region.

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And you notice that is in fact
true for this first quadrant.

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Any closed curve I draw that's
in the region, all the points

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on the interior of the curve
are also in the region.

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So this domain where it's
defined and differentiable is

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simply connected.

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So the first one is
simply connected.

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All right.

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So that's part a.

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Part b--

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let me rewrite that one so we
don't have to zoom over to the

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other side--

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this was i plus j divided by the
function square root of 1

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minus x squared minus
y squared.

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Well, we already know that the
square root function is

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defined as long as the inside
function is greater than or

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equal to 0.

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Because it's in the denominator,
we actually need

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this function 1 minus x squared
minus y squared to be

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greater than 0.

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And that's also where it's going
to be differentiable.

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The differentiable and the
defined regions are exactly

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the same, and they're both where
1 minus x squared minus

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y squared is greater than 0.

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Right?

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The function is only defined
as long as this quantity is

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greater than 0, and that's
exactly where it's

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differentiable as well.

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And so what does this
correspond to?

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Well, if you think about it,
this is actually 1 is greater

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than x squared plus y squared.

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And what are the points
that look like this?

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Well, the x- and y-points that
satisfy this inequality are

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the x- and y-values
that are on the

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interior of the unit circle.

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So if I draw a picture
of that.

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Let me try and dot
the unit circle.

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It's not containing the
boundary, but it's all the

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points that are on the interior
of the unit circle.

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Every point here, when I take
the ordered pair (x, y) and

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it's on the interior of the unit
circle, it satisfies this

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inequality.

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Those are the only points
that do that.

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So this is the region that has
this vector field both

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differentiable and defined.

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Right?

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Now, is this region
simply connected?

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It is, again for the
same reason.

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Because if you take any closed
curve here, and you look at

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the interior of that closed
curve, every point on the

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interior of that closed curve
is also in the region.

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Right?

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So it is also simply
connected.

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OK.

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So we had two so far that
were simply connected.

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They were different looking
regions, but ultimately they

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both had any closed curve-- the
interior of it-- was all

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contained in the region that
we were interested in.

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So now the third one we have is
somewhat similar-looking to

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part b, except that what's
in the square

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root is a little different.

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So now we can use exactly
the same logic as what

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we did in part b.

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And what we see is by the exact
same logic that this

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vector field will be defined and
differentiable as long as

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x squared plus y squared
minus 1 is positive.

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Because that's where the
square root function is

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differentiable, and that's also
where 1 divided by the

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square root of this
thing is defined.

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So they correspond to exactly
the same regions.

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And that is x squared plus
y squared greater than 1.

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So if you think about that, what
we did previously was we

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had x squared plus y squared
less than 1.

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So obviously if you want x
squared plus y squared greater

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than 1, we're taking all the (x,
y) pairs that are outside

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the unit circle.

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So again, we take
the unit circle.

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We don't include the unit
circle, because that's where x

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squared plus y squared
equals 1.

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And then we want all of the
values outside of that region.

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So this extends off
to infinity.

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All of the values outside
of that region.

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Now, is this region
simply connected?

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It is not.

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And the point is that while you
do have some curves, that

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if I take a closed curve, all of
the points on the interior

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of that closed curve
are in the region.

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There are some curves for
which that's not true.

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For example, if I take the
circle of radius--

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what does that look like--

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2, 1 1/2, something like that?

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If I look at all of
the points on the

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interior of this curve--

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I'm going to try and shade
it without getting rid of

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everything, so you can see
still what's behind--

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if I look at all those points,
notice in particular, there

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are a bunch of points-- for
instance, this one here, this

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one here, and this one here--
all the ones inside the unit

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circle, are on the interior of
this curve, but they're not in

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the region.

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Right?

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I'll shade it extra dark.

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All the points that are in here,
that are inside the unit

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circle, are actually still on
the interior of this curve,

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but they're not in the region.

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And while there are some curves
for which everything on

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the inside is in the region, but
there are some curves for

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which it's not true, and that
is what we know about

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not-simply connectedness.

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So we know this one is
not simply connected.

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00:09:46,425 --> 00:09:52,530

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OK.

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So now we have one left.

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And the last one was i plus j,
times natural log of x squared

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plus y squared.

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00:10:07,800 --> 00:10:10,100

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So in this vector field, what
I'm really interested in is

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the behavior of this function
natural log of x

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squared plus y squared.

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And the point I want to make is
that natural log is defined

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as long as the input value
is positive, and it is

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differentiable everywhere
it's defined.

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And so this function will
be both defined and

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differentiable as long as x
squared plus y squared is

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greater than 0.

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So it's defined for all of these
values, and natural log

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is differentiable everywhere
it's defined.

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So it's also differentiable
for all of these values.

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And so we see that this vector
field is defined and

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differentiable everywhere
except at one point.

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And so let me draw--

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this is zooming in showing that
it's missing that point--

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I want to make it extra large.

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But it's really only
missing one point.

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And what point is that?

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That point is the origin.

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So everywhere except
the origin.

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Maybe I should make it smaller,
because maybe it

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looks like it's missing
a whole circle.

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It's just missing the point.

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It's just missing the origin.

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But every other point on the xy
plane is a place where this

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vector field is differentiable
and defined.

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Right?

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So it's only missing that one
point, but that still gives us

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the fact that this region
is not simply connected.

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And again, it's exactly the
same type of logic as the

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previous problem, that I could
draw curves, where every point

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on the interior is contained
in the region.

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But there are curves
that also fail.

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Right?

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If I draw a curve that contains
the origin, every

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point on the interior of this
region except the origin, is

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contained in the domain
of interest, right?

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But because the interior of the
curve contains the origin,

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I know that this region is, in
fact, not simply connected.

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So if I look at all of R2--

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so if I look at all the x- and
y-values except x equals 0 and

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y equals 0, except
the origin--

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I get a region that's not
simply connected.

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And this one is maybe a little
tricky, so I'm going to say it

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one more time.

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OK?

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So while there are some curves
that we can see some portions

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of this region behave like
simply connected regions, when

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you're around the origin, any
curve you take around the

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origin is going to contain the
origin on its interior.

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Right?

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But the origin is not in our
domain of interest. And

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therefore, there are curves that
we can take that their

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interior contains a point that's
not in the region.

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And that's what it means to
be not simply connected.

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OK?

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So let me go back to the
beginning and just remind you

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again real quickly
what we did here.

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We had these four
vector fields.

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We wanted to do two things
with all of them.

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We wanted to first find the
regions where they were

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defined and differentiable, and
then determine if those

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regions were simply connected.

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And so we had two examples
where the

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regions were simply connected.

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And then I gave you two examples
where the regions

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were not simply connected.

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And so hopefully this was
informative for how we can

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understand that vector fields
are not necessarily always

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defined everywhere, but also to
understand what this simply

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connected region term
actually means.

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And I guess that's
where I'll stop.

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