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Hi.

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Welcome back to recitation.

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In lecture you've been
learning about vector

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calculus, Stokes' theorem,
all sorts of cool

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stuff like that, curl.

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So I have a nice problem
here for you.

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So let F be the following
vector field.

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So it's the vector field whose
direction is xi hat plus yj

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hat plus zk hat, but in addition
I want to multiply it

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by rho to the n.

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So this is your usual rho from
spherical coordinates.

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This is the square root
of x squared plus y

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squared plus z squared.

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And n can be any number.

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So it might be positive, it
might be negative, it might be

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0, doesn't have to
be an integer.

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Some number, though,
it's some constant.

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So what I'd like you to do is
to show that for this field,

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regardless of what the value of
n happens to be, that it is

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a gradient field.

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So it is the gradient
of some function.

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So why don't you pause the
video, have a go at that, come

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back and we can work
on it together.

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So recall that for something to
be a gradient field, what

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that means--

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well, first of all, that means
that there is some function

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that has it as the gradient,
and we know a lot of other

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characterizations of it.

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And one of them that we know
involves the curls.

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So let's talk about that one.

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So in order to look at the curl
of this field, I'm going

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to want to take partial
derivatives of its components.

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And so I'm going to want
to take partial

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derivatives of rho.

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So let's just remember or
recall that partial rho

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partial x equals z over rho.

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And similarly, partial rho
partial y is y over rho.

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And partial z partial--

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sorry, partial rho partial
z is z over rho.

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So you can just check that using
the fact that you know

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what rho is.

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Et cetera.

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I'm going to write et cetera
because it's the same

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for the other two.

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So let's call-- for
shorthand--

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let's call F, Mi plus
Nj plus P times k.

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So M, N, and P are
its components.

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If F is this, then we know
that curl F is equal to--

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what is it?

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So it's P, the y partial
derivative of P, minus the z

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partial derivative of
N, i hat, plus--

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OK, so it's the z partial
derivative of M minus the x

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partial derivative of
P, j hat, plus--

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it's going to be, what's
the last one?--

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the x partial derivative of N,
minus the y partial derivative

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of M k hat.

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So if this is our formula for
F, then this is our formula

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for the curl of F. And OK, so
now we just have to compute

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these various different partial
derivatives in order

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to see what the curl is.

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And then hopefully
that'll tell us

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something about this field.

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So let's see.

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So Py.

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In our case, so M is equal to
rho to the little n times x.

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Big N is equal to rho to
the little n times y.

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And big P is equal to rho
to the little n times z.

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So let's look at our
components here.

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So P sub y--

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well, z is a constant
with respect to y.

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So this is just rho to the n.

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Well, z times the y partial
of rho to the n.

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So that's by the chain rule, so
we got n rho to the n minus

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1 times y over rho times z.

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So we can rewrite this as nyz
rho to the n minus 2.

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And similarly, so that
was P sub y, so let's

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look at N sub z.

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So N is rho to the n times y.

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So you take the z partial

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derivative, so y is a constant.

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So we need to look
at the z partial

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derivative of P to the n.

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So this is, again, it's n--

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sorry, I think I said
P to the n.

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But of course this isn't
a P, this is a rho.

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So it's n rho to the n minus 1
times partial rho partial z.

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So that's z over rho times y.

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And again, this is equal to
nyz rho to the n minus 2.

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OK, so P sub y, the y partial
of P is nyz rho

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to the n minus 2.

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And the z partial of n is nyz
rho to the n minus 2.

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And they're the same.

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So what does that mean?

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So that means the first
component of curl of F is 0.

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So if you do a little bit more
arithmetic of exactly the same

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sort, you have two more
components to check.

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What you're going to
find is that the

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other ones are 0 also.

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I'm not going to do out all
those partial derivatives for

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you, but I trust that you can
compute the similar looking

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partial derivatives that appear
in these other two

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components--

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this j should have had a hat--

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the other partial derivatives
that appear in these

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components and show that they're
all also equal to 0.

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So in our case, curl of F is
just equal to 0 vector.

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OK, great.

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So what does that mean?

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Well, we want to show that
something is a gradient field.

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So we know that we
can do that.

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We know that that happens
precisely when its curl is 0

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if it's defined on a simply
connected domain.

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On a simply connected region.

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So we know that a function is
a gradient field if it's

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defined on a simply connected
domain and has curl 0.

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Well, let's see.

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So where is this
thing defined?

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Where is F defined?

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Well, if n is bigger than
or equal to 0--

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well, I guess I want strictly
bigger than 0 just to be on

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the safe side.

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If n is positive,
this is defined

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everywhere and we're great.

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If n is negative, then we
have a problem at 0.

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Because then we have
division by rho.

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So we don't want
to divide by 0.

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So if n is negative, there's
a problem at the origin.

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So in the worst case,
F is defined

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everywhere except the origin.

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But that's simply connected.

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Because we're in space.

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If we were in the plane, this
would be a different story.

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But in space, just removing a
point doesn't destroy simply

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connectedness.

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So this field F is defined
everywhere, except the origin

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perhaps, so we're in a simply
connected region.

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So when you have curl F in a
simply connected region, your

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field is, in fact,
a gradient field.

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Now I leave it as an exercise
to you to come up with the

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actual function, or one of the
actual functions, of which

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this is a gradient.

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But we've just shown that
because its curl is 0, and

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because it's defined in a simply
connected region of

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space, that this field really
is the gradient

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field of some function.

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So I'll stop there.

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