WEBVTT
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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 calculus,
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Stokes' theorem, all sorts of
cool 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
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is x i hat plus y
j hat plus z k hat,
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but in addition I want to
multiply it 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
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plus y 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,
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it might be 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,
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that it is 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,
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come back and we can
work on it together.
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So recall that for something to
be a gradient field, what that
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means-- well, first
of all, that means
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that there is some function
that has it as the gradient,
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and we know a lot of other
characterizations of it.
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And one of them that we
know involves the curl.
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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,
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I'm going to want to
take partial derivatives
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of its components.
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And so I'm going to want to
take partial derivatives of rho.
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So let's just remember or
recall that partial rho
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partial x equals x over rho.
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And similarly, partial rho
partial y is y over rho.
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And partial z partial--
sorry, partial rho partial z
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is z over rho.
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So you can just check that using
the fact that you know what rho
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is.
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Et cetera.
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I'm going to write et
cetera because it's
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the same for the other two.
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So let's call-- for
shorthand-- let's call F,
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M*i plus N*j 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
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is equal to-- what is it?
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So it's P, the y-th
partial derivative of P,
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minus the z-th partial
derivative of N,
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i hat, plus-- OK, so it's the
z-th partial derivative of M
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minus the x-th 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
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is our formula for
the curl of F. And OK,
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so now we just have to compute
these various different partial
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derivatives in order to
see what the curl is.
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And then hopefully that'll tell
us something about this field.
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So let's see.
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So P_y.
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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-- 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-th
partial of rho to the n.
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So that's by the
chain rule, so we
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got n rho to the n minus 1
times y over rho times z.
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So we can rewrite this as
n*y*z rho to the n minus 2.
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And similarly-- so
that was P sub y,
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so let's 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-th partial
derivative, so y is a constant.
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So we need to look at the
z-th partial derivative of P
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to the n.
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So this is, again,
it's n-- sorry,
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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
n*y*z rho to the n minus 2.
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OK, so P sub y,
the y partial of P,
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is n*y*z rho to the n minus 2.
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And the z partial of n is
n*y*z 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
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of exactly the same
sort-- you have two more
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components to check--
what you're going to find
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is that the other
ones are 0 also.
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I'm not going to do out all
those partial derivatives
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for you, but I trust that you
can compute the similar looking
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partial derivatives that appear
in these other two components--
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this j should have had a hat--
the other partial derivatives
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that appear in these components
and show that they're all also
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equal to 0.
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So in our case, curl of F is
just equal to the zero 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
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if it's defined on a
simply connected domain
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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-- well,
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I guess I want strictly
bigger than 0 just
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to be on the safe side.
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If n is positive, this
is defined everywhere
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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 everywhere
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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
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doesn't destroy
simply connectedness.
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So this field F is
defined everywhere,
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except the origin
perhaps, so we're
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in a simply connected region.
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So when you have curl F in
a simply connected region,
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your field is, in
fact, a gradient field.
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Now I leave it as
an exercise to you
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to come up with the
actual function,
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or one of the actual functions,
of which this is a gradient.
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But we've just shown
that because its curl
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is 0, and because it's defined
in a simply connected region
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of space, that this field
really is the gradient field
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of some function.
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So I'll stop there.