WEBVTT
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Welcome back to recitation.
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In this video what
I'd like you to do
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is work on proving
the following product
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rule for the del operator.
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So we're going to let
capital F be a vector field
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and u be a scalar function.
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And we want to show the
product rule for the del
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operator which--
it's in quotes but it
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should remind you of the product
rule we have for functions.
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And it is that del dot
the quantity u times
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F-- so u is the scalar function
and F is the vector field-- is
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actually equal to the gradient
of u dotted with F plus u
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times del dot F. Where F
again is the vector field.
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So why don't you take a
moment to prove this fact
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and you can pause the
video while doing that.
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And then when you're ready
to check if your solution is
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correct, bring the video back
up and I'll show you what I did.
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OK, welcome back.
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Again what we wanted
to do is prove
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this sort of pseudo product
rule for the del operator.
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And what we're doing
is we're trying
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to see what happens if
you have a vector field
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and you multiply it
by a scalar function
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and you apply the
del operator to that.
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So we're going to see
if we actually come up
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with what we should on the
right-hand side, which,
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since I'm calling it a
rule, we really hope we do.
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In fact, we will.
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So let me start off.
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What I'm going to do is I'm
going to write symbolically
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what the left-hand side means.
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And then we're going to
break it up into pieces
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and show you get, in
fact, what you would
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get on the right-hand side.
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So symbolically,
what do we have?
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Oh actually, first
I'm going to call F--
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the components of F,
as is usually done
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in lecture--
capitals P, Q, and R.
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So those will be
the components of F.
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That's how we've been
denoting this, usually.
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And notice that if we wanted
F to be a vector field in two
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dimensions, we'd just
make R 0, and then we'd
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have a vector field
in two dimensions.
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So we can certainly
do that if we want,
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but we're going to prove
it in a little more--
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in the three-dimensional case.
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So remember that del dotted with
any vector field is supposed
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to be-- symbolically, what was
written was you should think
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about this as del, del x, comma
del, del y, comma del, del z,
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dotted with this
vector field, u*F.
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Now what is this vector field?
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Because u is a scalar, when I
multiply u by the vector field
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F, the components are going to
be u capital P comma u capital
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Q comma u capital R. So
those are my components.
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Now symbolically,
this is what we've
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seen when we're looking at the
del operator acting on a vector
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field.
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So what do we actually do?
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Well, what we actually
do, of course,
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is we take the x derivative
of the first component,
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we take the y derivative
of the second component,
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and we take the z derivative
of the third component.
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And then we add those together.
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So it's really a symbolic
idea of a dot product.
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It's not a true
dot product here.
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But let's actually write
down what we get there.
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We get del del x of
the quantity u*P,
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plus del del y of the
quantity u*Q plus del del z
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of the quantity u*R. That's
exactly what this symbolically
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means is this, what I've
written in the next line down.
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So maybe I should
write equals again.
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This is another equals.
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The top thing equals
the next line down,
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equals the next line down.
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And just to have it nice and
even I'll put the equals there,
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so when we look back
it's easy to see.
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Now how do I deal with this?
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Well, notice that u is a
function and P is a function.
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Why is that?
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Because P was a component
of a vector field.
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So it is just a function that
is in the first component
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of the vector field.
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P, Q, and R are all
individually functions
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that depend on x, y, and z.
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So here, at this
step, I can actually
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take the regular product
rule we have for functions.
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And I can do it in this
one, this one, and this one.
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And what I'm going to do, so I
don't have to write del del x
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and del del y all
over the place,
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I'm going to use the subscripts
notation for derivatives.
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So I'm going to write,
what this actually
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equals is u sub x
times P plus u P sub x.
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So here, the del del x is now
corresponding to a subscript.
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So notice that I've just used
the product rule on functions
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at this step, because u is a
function and P is a function.
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I can do the same thing for
the y derivative of u*Q.
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I get u sub y Q plus u Q sub y.
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And then the last component,
I do the same thing for z.
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I get u sub z R plus u R sub z.
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Now if I made a mistake it
will become very apparent
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in the next moment.
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But I don't think
I made a mistake.
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I want to remind us
where we want to go,
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and then we'll see if we
have the pieces to get there.
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So I'm going to go back over
to the side we have over here.
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And I want to remind
you, we started off
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with a del operator acting on
this vector field u times F,
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capital F. And what
we'd like to see
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is if we can get this to
equal the gradient of u
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dotted with the vector
field plus u times
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the del operator acting on F.
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So let's see if
we can first find
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components of the vector
field and components
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of the gradient of u.
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And let's see where those are.
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They're actually very
nicely placed here.
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Notice that this
underlined component
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is the first component
of gradient u
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and the first component of F.
And this underlined component
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is the second component
of gradient of u
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and the second component of
F. And this third underlined
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component here is the
third component of grad u
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and the third component of F.
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So if I take those three
components together--
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I'm going to write
the equals up here--
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I get exactly grad
u dotted with F.
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That's the first thing I get.
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And then I'm going to
see what else I get.
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But let me just make sure you
understand, look at these three
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underlined components together.
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You get u sub x times P plus
u sub y times Q plus u sub z
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times R. Well, [u x, u y, u z]
is the gradient vector field
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for u, and [P, Q, R] is
F. So when I dot those,
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I get exactly u_x*P
plus u_y*Q plus u_z*R.
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You notice these two, or
the dot product of this,
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gives you those
three components.
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And now there are three
components remaining.
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Notice what they
all have in common.
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They all have a u
in the first spot.
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And then it's
multiplied by P sub x,
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and then here it's
multiplied by Q sub y,
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and here it's
multiplied by R sub z.
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But that is exactly-- P sub
x plus Q sub y plus R sub
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z is exactly the del operator
acting on this vector field F.
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So this is something
you've seen previously.
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So I'm going to do
these as squiggles.
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That's exactly equal to u times
the del operator of this vector
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field F.
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So again let me remind
you, del dot F actually
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is going to give you P sub
x plus Q sub y plus R sub z.
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And then if I multiply that
by a u, I get this u in front.
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So what I have done is
for an arbitrary function
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u-- I guess I've assumed
that function had
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first derivatives so that
I could do all this stuff--
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and for a vector field
that had first derivatives,
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I showed that if I take
the del operator of u times
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the vector field, I actually
get the gradient of u
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dotted with the vector
field plus u times
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the del operator of
the vector field.
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So that is what
I wanted to show.
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If you remember,
what I wanted to show
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was exactly that sort of pseudo
product rule for this del
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operator.
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So I think that that
is where I'll stop.
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I'm going to step
off to the side,
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so you can see it all
again, for a moment.
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But that's it.
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So that's where I'll stop.