Partial Derivatives, Gradient, Divergence, Curl - Multivariable Calculus Video Lecture - Engineering Mathematics

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.
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.