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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

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is the domain where
each vector field

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is defined and
continuously differentiable

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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

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show you the 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,

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the 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

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to determine all of the values
for which the vector field is

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both defined and
continuously differentiable.

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You want to look at that region
that contains all those values,

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and you want to determine
if that region is simply

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connected.

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So there are four
different vector fields,

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and I'll just 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 1

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minus x squared minus y squared.

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The third one looks fairly
similar to the second one,

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but it's i plus j,
divided by the square root

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of the quantity x 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

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if that domain is
simply connected.

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

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and then when you're
ready to see what I did,

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you can bring the video back up.

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OK, welcome back.

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So we're going to do
this problem one vector

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field at a time.

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We're going to take
each vector field,

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I'm going to determine
where it's defined

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and differentiable,
and then I'm going

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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

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was root x i plus root y j.

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And I want to point out
first, for the function

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

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greater 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

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when x is greater than 0.

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And now if I replace
this with a y,

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the same thing is true for y
greater than or equal to 0,

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and y greater than 0.

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So we know the region
where this vector field is

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both defined and differentiable
is when x is greater

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than 0 and y is greater than 0.

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So we need the region-- let me
draw it this way-- the region

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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 full shaded region.

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OK, that is the
region where it's

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defined and differentiable.

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If I was going to
write that precisely,

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I would say something like, all
(x, y) with x greater than 0

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and y 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 positive,

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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

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connectedness, from what
you've seen in class, is you

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want to show that if you
take a closed curve that's

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contained in the
region, that everything

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on the interior of that closed
curve is also in 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

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the points on the
interior of the curve

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are also in the region.

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So this domain where it's
defined and differentiable

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is simply connected.

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So the first one, it
is simply connected.

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All right.

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So that's part a.

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Part b-- let me
rewrite that one so

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we don't have to zoom
over to the 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

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is defined as long as the
inside function is greater

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than or equal to 0.

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Because it's in the
denominator, we actually

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need this function 1
minus x squared minus y

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squared to be 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

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are exactly the
same, and they're

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both where 1 minus x
squared minus y squared

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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

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it's 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

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is actually 1 is greater 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

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are the x- and y-values that
are on the interior of the unit

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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,

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but it's all the points that
are on the interior of the unit

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circle.

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

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and it's on the interior
of the unit circle,

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it satisfies this 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

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field both 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,

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and you look at the interior
of that closed curve,

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every point on the interior
of that closed curve

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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,

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but ultimately they both
had any closed curve,

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the interior of it was all
contained in the region

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that we were interested in.

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So now the third one
we have is somewhat

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similar-looking to
part b, except that

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what's in the square root
is a little different.

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So now we can use
exactly the same logic

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as what we did in part b.

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And what we see is by
the exact same logic

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that this vector field will
be defined and differentiable

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as long as 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

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1 divided by the 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

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was we had x squared plus
y squared less than 1.

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So obviously if
you want x squared

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plus y squared greater than
1, we're taking all the (x, y)

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pairs that are outside
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,

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because that's where x 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,

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that if I take a
closed curve, all

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of the points on the
interior of that closed curve

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are in the region, there are
some curves for which that's

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not true.

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For example, if
I take the circle

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of radius-- what does that
look like-- 2, 1 and 1/2,

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something like that?

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If I look at all of the points
on the interior of this curve--

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

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of everything, so
you can see still

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what's behind-- if I look
at all those points, notice

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in particular, there
are a bunch of points--

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for instance, this one
here, this one here,

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

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are on the interior
of this curve,

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but they're not in 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

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are inside the unit
circle, are actually still

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on the interior of this curve,
but they're not in the region.

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And while there are some
curves for which everything

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on the inside is in
the region, but there

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are some curves for
which it's not true,

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and that is what we know about
not-simply connectedness.

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So we know this one is
not simply connected.

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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

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of x squared plus y squared.

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

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is the behavior of this function
natural log of x squared

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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,

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and it is differentiable
everywhere it's defined.

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And so this function will be
both defined and differentiable

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as long as x squared plus y
squared is greater than 0.

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So it's defined for
all of these values,

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and natural log is
differentiable everywhere it's

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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

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is defined and differentiable
everywhere except at one point.

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And so let me draw--
this is a zooming

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in of that it's
missing that point-- I

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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

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is a place where
this vector field

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is differentiable and defined.

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Right?

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So it's only missing
that one point,

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but that still gives us the
fact that this region is not

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simply connected.

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And again, it's
exactly the same type

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of logic as the
previous problem,

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that I could draw curves, where
every point on the interior

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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 point

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on the interior of this
region except the origin,

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is 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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00:12:04,570 --> 00:12:08,080
I know that this region is,
in fact, not simply connected.

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00:12:08,080 --> 00:12:12,177
So if I look at all of R^2--
so if I look at all the (x,

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00:12:12,177 --> 00:12:16,120
y)-values except x
equals 0 and y equals 0--

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except the origin-- I get
a region that's not simply

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connected.

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00:12:22,847 --> 00:12:24,430
And this one is maybe
a little tricky,

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so I'm going to say
it one more time.

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OK?

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00:12:27,740 --> 00:12:30,190
So while there are
some curves that we

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00:12:30,190 --> 00:12:32,600
can see some portions
of this region

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00:12:32,600 --> 00:12:36,850
behave like simply connected
regions, when you're

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00:12:36,850 --> 00:12:39,990
around the origin, any curve
you take around the origin

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00:12:39,990 --> 00:12:42,650
is going to contain the
origin on its interior.

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00:12:42,650 --> 00:12:43,350
Right?

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00:12:43,350 --> 00:12:47,190
But the origin is not in
our domain of interest.

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00:12:47,190 --> 00:12:49,610
And therefore, there
are curves that we

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00:12:49,610 --> 00:12:52,330
can take that their interior
contains a point that's

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00:12:52,330 --> 00:12:53,480
not in the region.

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00:12:53,480 --> 00:12:56,541
And that's what it means
to be not simply connected.

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00:12:56,541 --> 00:12:57,040
OK?

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00:12:57,040 --> 00:12:58,340
So let me go back
to the beginning

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00:12:58,340 --> 00:13:00,673
and just remind you again
real quickly what we did here.

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00:13:03,440 --> 00:13:05,530
We had these four vector fields.

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00:13:05,530 --> 00:13:07,610
We wanted to do two
things with all of them.

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00:13:07,610 --> 00:13:11,010
We wanted to first find the
regions where they were defined

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00:13:11,010 --> 00:13:14,560
and differentiable, and then
determine if those regions were

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00:13:14,560 --> 00:13:15,890
simply connected.

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00:13:15,890 --> 00:13:18,210
And so we had two
examples where the regions

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00:13:18,210 --> 00:13:19,670
were simply connected.

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00:13:19,670 --> 00:13:22,300
And then I gave you two examples
where the regions were not

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00:13:22,300 --> 00:13:23,720
simply connected.

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00:13:23,720 --> 00:13:25,900
And so hopefully this
was informative for how

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00:13:25,900 --> 00:13:29,080
we can understand that vector
fields are not necessarily

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00:13:29,080 --> 00:13:31,330
always defined
everywhere, but also

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00:13:31,330 --> 00:13:35,730
to understand what this simply
connected region term actually

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00:13:35,730 --> 00:13:37,120
means.

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00:13:37,120 --> 00:13:39,600
And I guess that's
where I'll stop.