Curl 2 The mechanics of calculating curl.
- 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
- 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.
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