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