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