Welcome back.
Hopefully you have a little intuition now
of what the curl is.
Now let's actually compute it because if your sole goal is to
pass a test and not understand the nature of the universe,
which I think would be sad, but if that is your goal you at
least need to know how to calculate these things.
But it's even more fun when you have the intuition.
And then you'll hopefully never forget it.
We'll take the curl of a fairly fancy vector field.
One that I have trouble visualizing but that we can
mathematically chug through.
So let's say our vector field-- and I'll do a three dimensional
vector field just to do a fairly complicated example; I'm
just going to make it up on the fly --so let's say in the
x-direction the magnitude of the field is, I don't know,
let's say it's x squared, y, sine, z, in the x-direction,
plus-- I don't know --let's make it x, y squared, z in
the j-direction, or the y-direction.
And in the z-direction, I don't know, let's make it cosine of
x times cosine of y, in the z-direction.
Now we said that you can view the curl of this vector field--
and I have no intuition of what this vector field looks
like; I just made this up.
Maybe we'll graph it for fun just to see how messed up it
looks --but we said this curl, you could view it as a cross
product of our del operator and the vector field.
Well, when you were using this engineering notation, when you
have a vector broken down into it x, y, and z components, or
it's i, j, and k components, you can take the determinant of
that matrix-- when I showed you how to compute the cross
product --to figure out the cross product.
So how do we do this?
So the cross product is going to be equal to-- I didn't have
to draw a straight line --so how you take the cross product
of this vector field and the gradient operator?
Well, you write i, j, k on top like you're taking the cross
product of any two three dimensional vectors, and then
you take the first vector-- but it's really a vector operator,
but it's this del operator.
And what are the components of the del operator?
It's the partial derivative with respect to x, the partial
derivative with respect to y, partial derivative with
respect to z, right?
Let me just rewrite the del operator.
You can view it as being equal to the partial with respect to
x, i plus the partial with respect to y, j plus the
partial with respect to z, k.
So its x, y, and z components are the partials with respect
to x, with respect to y, with respect to z.
And then the second, where we're taking this operator
cross the vector field.
So what are the components of the vector field?
I'll probably run out of space, but it's x squared, y, sine, z.
Then here it's xy squared, z-- I should written all of this
bigger --and then the third column, the z component is
cosine of x times cosine of y.
Just the x, y, and z components.
And now we are ready to take the determinant, which will
probably, well, it'll probably get pretty
messy, but let's try it.
So this is equal to the i-unit vector-- let me use a more
vibrant color --the i-unit vector times it's
subdeterminant.
So you cross out it's row and column, and so you take the
determinant of this expression, so it's going to be times--
this times this, but it's really the partial.
If you multiply the partial with respect to y operator
times that expression, you're really just taking, since it's
an operator and not an expression, you're really just
going to take the partial of this with respect to y,
but I'll write it down.
So it's going to be the partial with respect to y of cosine x,
cosine y minus the partial with respect to z times xy squared,
z-- and now we're on to our j component --plus j.
So now what's the magnitude of our curl in the j-direction?
Let's cross out the row in the column of j.
So it's a partial with respect to x of this.
So this is maybe messier than I originally intended.
Cosine of x, cosine of y, cross these columns there, minus
the partial with respect to z of x squared, y sine of z.
And then finally our k component.
Oh, and sorry, when you take the determinant, you use that--
and this is all kind of a bit of voodoo --but you put a plus
here, a minus here, a plus here, so it's kind of
this checkered pattern.
To this is plus i, this should actually be minus j.
Don't want to make that mistake; this is minus j.
This is just kind of the algorithm of how do you
take a determinant.
OK, then finally we have plus k times the determinant
of its submatrix.
So the partial with respect to x of this, sorry, we take out
it's row and column, so xy squared, z minus the partial
with respect to y of this.
Why don't we take this row and columns, and this is the
submatrix of x squared, y, sine of z.
All right, now let me try to simplify it, and I'll
have to get some space.
Hopefully you understood what I did here and now we got this.
And now I think I can erase all of this and just so I
can have some room to simplify things in.
No, that's not what I wanted to do; I wanted to do
it in a darker color.
That's what I wanted to do.
Erase that.
Now we just have to simplify it, taking a bunch of
partial derivatives.
What's the partial derivative of this with respect to y?
Well, x is just a constant, so it's going to be-- well, we can
just put the i out front, but eventually we want to write our
magnitude before the vector --so it's i times the partial
of this with respect to y with our constant.
Cosine of x is just a constant.
And then what's the derivative of this with respect to y?
It's minus sine of y.
So we'll write sine of y, and let's put the minus out front;
these are what's multiplied.
OK, and then we have minus, now we stick the partial with
respect-- sorry, actually, I forgot to do this part.
Let me start over, actually.
So let me just take this expression and I'll
multiply it by i.
The partial of this with respect to y is cosine of x
times minus sine of y-- let's put the minus out front
--minus sine of y.
Now minus the partial of this with respect to z.
Well the partial of this with respect to z, xy squared
is just a constant, right?
So the partial of this with respect to z
is just xy squared.
So minus xy squared, and then we're going to have all of
that; that's the magnitude in our i-direction.
And now we have minus-- because minus in the j-direction
--what's the partial derivative of this with respect to x?
Well, the partial of cosine of x with respect to
x is minus sine of x.
And cosine of y is just a constant, so it just
carries over, cosine of y.
And then that should be-- oh, there we go --minus this
expression, the partial of this with respect to z.
Well, the derivative of sine of z with respect
to z is cosine of z.
This is just a constant, so it's minus x squared,
y, cosine of z.
And that's the magnitude in the j-direction.
We're almost there.
And now finally, plus, what's the partial of
this with respect to x?
Well, these are just constants, so it's y squared, z minus--
once again, we just have a y term; everything else is a
constant --so the partial with respect to y is
x squared, sine of z.
And that's the magnitude in our k-direction.
And we're pretty much done.
I mean we can simplify it a little bit just
to make it clean.
And essentially we could just, I don't have to rewrite it,
we can just multiply this by negative 1.
So then this becomes plus, plus, plus.
Yeah, that's pretty much it.
This is the curl of the vector field v at any
point x, y, and z.
So that's how you calculate it.
You just literally take the cross product of that del
operator and your vector field.
And you'll get something fairly hairy, although this was, I
think, a hairier than average problem.
In the next video, we'll do a little bit of this, but I think
it'll give you more intuition and less of just the algorithm
and the computation of how do you do it.
So you in the next video.