WEBVTT
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Welcome back to recitation.
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In this video I'd like us to
consider the following problem.
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So the question is for what
simple closed curve C, which
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is oriented positively around
the region it encloses,
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does the integral over
C of minus the quantity
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x squared y plus 3x minus 2y dx
plus 4 y squared x minus 2x dy
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achieve its minimum value?
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So again, the thing
we want to find here,
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the point we want to make,
is that we have this integral
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and we're allowed
to vary C. So we're
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allowed to change the curve,
the simple closed curve.
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And we want to know for what
curve C does this integral
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achieve its minimum value?
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So why don't you work
on that, think about it,
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pause the video, and
when you're feeling
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like you're ready to see how I
do it, bring the video back up.
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Welcome back.
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So again, what we would
like to do for this problem
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is we want to take
this quantity, which
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is varying in C, and
we want to figure out
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a way to minimize it.
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To find its minimum value.
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And actually what's
interesting to think about,
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before we proceed
any further, is
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you might think you want to take
the smallest possible simple
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closed curve you can.
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In other words, you might want
to shrink it down to nothing.
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But then this integral would
be 0 and the question is,
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can we do better?
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Because we could have a
minimum value, of course,
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that's negative.
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And so we could do
better by having a larger
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curve put in the right place.
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So I just want to point that
out, if you were thinking
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"I'll just make C
not actually a curve,
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I just shrink it
down to nothing."
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That won't be the
best we can do.
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We can actually
find a curve that
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has an integral that is not
0, but is in fact negative.
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And then we want to make it
the most negative we can.
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So that's the idea.
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So what we're
going to do here is
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we're going to use Green's
theorem to help us.
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And so what we want to remember
is that if we have the integral
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over C of M*dx plus N*dy we want
to write that as the integral
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over R of N sub x
minus M sub y dx dy.
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So what I want to
point out, I'm just
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going to write down
what these are.
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I'm not going to take
the derivatives for you.
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I'm just going to show
you what they are.
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So in this case, with
the M and N that we have,
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we get exactly this
value for the integral.
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We get N sub x minus M sub
y is equal to x squared
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plus 4 y squared minus 4 dx dy.
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You can obviously
compute that yourself.
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Just take the derivative
of N with respect to x,
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subtract the derivative
of M with respect to y,
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you get a little cancellation
and you end up with this.
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And so now, instead of
thinking about trying
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to minimize this quantity
over here in terms of a curve,
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now we can think about trying
to minimize the quantity here
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in terms of a region.
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And the goal here
is to make the sum
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of all of this over
the whole region--
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we want to make it as negative
as we can possibly make it.
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So essentially
what we want to do
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is we want-- on the
boundary of this region,
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we would like the
value here to be 0,
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and on the inside of the region
we'd like it to be negative.
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So let me point that
out again and just
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make sure we understand this.
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To make this quantity
as small as possible,
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what we would like--
let me actually
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just draw a little region.
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So say this is the region.
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To make this integral as small
as possible, what we want
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is that x squared plus
4 y squared minus 4 is
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negative inside the region.
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So if this whole quantity is
less than 0 inside the region,
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and we want it to, on the
boundary of the region,
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equal 0.
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And why do we want it to
equal 0 on the boundary?
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Well, that's because then we've
gotten all the negative we
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could get and we haven't
added in any positive
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and brought the value up.
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So that's really why we want
the boundary of the region
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to be exactly where
this quantity equals 0.
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And so let's think about,
geometrically, what
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describes R-- oops,
that should have an s.
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What describes R,
where x squared
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plus 4 y squared minus 4
is less than 0 inside R,
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and that's the same thing as--
in what we're interested in-- x
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squared plus 4y squared minus 4
equals 0 on the boundary of R.
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And if you look at this,
this is really the expression
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that will probably help you.
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This expression,
if you rewrite it,
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you rewrite it as x squared
plus 4y squared is equal to 4.
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And you see that this is
actually the equation that
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describes an ellipse.
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Maybe you see it more often
if you divide everything by 4,
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and so on the right-hand
side you have a 1,
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and your coefficients are
fractional, potentially, there.
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But this is exactly the
equation for an ellipse.
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And so the boundary of R
is an ellipse described
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by this equation,
but the boundary of R
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is actually just C.
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So C we now know is
exactly the curve
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that is carved out by this
equation on the plane,
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on the xy-plane.
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That's an ellipse.
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So just to remind you
what we were trying to do.
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If you come back
here, we were trying
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to figure out a way to
minimize the certain integral
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over a path.
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And what we did was
instead of trying
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to look at a bunch of
paths and figure out
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what would minimize
that, we tried
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to see if Green's
theorem would help us.
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So Green's theorem allowed
us to take something
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that was an integral
over a path and change it
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to an integral over a region.
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And then when we look at
what we ended up with,
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we realize that we could
make this the most minimum
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if we let this be on the
region where it was negative
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everywhere.
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So we were looking
for a region where
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this quantity was everywhere
negative on the inside and 0
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on the boundary.
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And that's exactly what we did.
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And then we see that
we get to a point
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where the boundary
has this equation,
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x squared plus 4y
squared equals 4.
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We see then the
boundary's an ellipse,
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and C is indeed the
boundary of the region.
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So we see that C is the ellipse
described by this equation.
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So that's where I'll stop.