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Welcome back to recitation.

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In this video what I'd like you
to do is work on proving

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the following product rule
for the del operator.

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So we're going to let capital F
be a vector field and u be a

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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 should remind you of

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the product rule we have
for functions.

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And it is that del dot the
quantity u times F--

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so u is the scalar function and
F is the vector field--

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is actually equal to the
gradient of u dotted with F

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plus u 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 and you can

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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

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show you what I did.

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OK, welcome back.

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Again what we wanted to do is
prove this sort of pseudo

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product rule for the
del operator.

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And what we're doing is we're
trying to see what happens if

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you have a vector field and you
multiply it by a scalar

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function and you apply the
del operator to that.

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So we're going to see if we
actually come up with what we

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should on the right hand side,
which, since I'm calling it a

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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 and show you

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get, in fact, what you would
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-- the components of

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F, as is usually done
in lecture--

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capitals P, Q, and R. 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 have a vector

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field in two dimensions.

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So we can certainly do that if
we want, but we're going to

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prove it in a little more-- in
the three dimensional case.

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So remember that del dotted
with any vector field is

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supposed to be symbolically--

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what was written was you should
think about this as

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del, del x, comma del, del y,
comma del, del z, dotted with

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this vector field.

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uF.

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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

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capital Q, comma u capital R.
So those are my components.

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Now symbolically this is what
we've seen when we're looking

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at the del operator acting
on a vector field.

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So what do we actually do?

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Well, what we actually do, of
course, is we take the x

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derivative of the first
component, we take the y

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derivative of the second
component, and we take the z

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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
uP, plus del del y of

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the quantity uQ, plus del del
z of the quantity uR.

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That's exactly what this
symbolically means is this,

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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, equals

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the next line down.

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And just to have it nice and
even I'll put the equals

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there, 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 of

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the vector field.

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P, Q, and R are all individually
functions that

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depend on x, y, and z.

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So here at this step I can
actually take the regular

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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, I'm going to use the

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subscripts notation
for derivatives.

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So I'm going to write what this
actually equals is u sub

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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, I get

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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 in

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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 and then we'll see

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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 with a del

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operator acting on this vector
field u times F. Capital F.

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And what we'd like to see is
if we can get this to equal

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the gradient of u 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 components of the vector

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field and components 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 is the first

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component of gradient u and the
first component of F. And

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this underlined component is
the second component of

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gradient of u and the second
component of F. And this third

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underlined 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. That's the

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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

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three 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, ux, uy, uz is the
gradient vector field for

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u, and P, Q, R is F. So when I
dot those, I get exactly uxP

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plus uyQ plus uzR.

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You notice these two, or the dot
product of this, gives you

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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, and then here it's

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multiplied by Q sub
y, and here it's

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multiplied by R sub z.

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But that is exactly--

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P sub x plus Q sub y plus R
sub z is exactly the del

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operator acting on this vector
field F. So this is something

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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

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vector field F.

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So again let me remind you, del
dot F actually is going to

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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 u--

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I guess I've assumed that
function had first derivatives

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so that I could do all this
stuff, and for a vector field

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that had first derivatives--

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I showed that if I take the del
operator of u times the

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vector field, I actually get the
gradient of u dotted with

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the vector field plus u
times the del operator

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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 was exactly that sort

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of pseudo product rule for
this del 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 so you can see it all

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again for a moment.

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But that's it.

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So that's where I'll stop.

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