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3d curl computation example


4m read
·Nov 11, 2024

So let's go ahead and work through an actual curl computation.

Let's say our vector-valued function V, which is a function of x, y, and z, this is going to be three-dimensional, is defined by the functions, uh, and I don't know, let's say the first component is x * y, the second one is cosine of z, and then the last component is z^2 + y.

So, if you take this guy, how do you compute the curl of that vector-valued function?

So what you do, as I mentioned in the last video, is you imagine taking this Del operator and taking the cross product between that and your vector-valued function.

What that means, when you expand it, is that the Del operator you just kind of fill it with partial differential operators, you could say. But really, it's just the symbol ∂/∂x, ∂/∂y, ∂/∂z.

Uh, and these are things that are just waiting to take in some kind of function. So, we're going to take the cross product between that and the function that we have defined here.

So let me just actually copy it over here a little residue, and to compute this cross product, we take a certain determinant.

So I'm going to write over here, determinant. It's going to be of a 3x3 matrix, but really it's kind of like a "quote unquote matrix" because each component has something funky.

So, the top row, just like we would have with any other cross product that we're computing, is going to have i, j, and k, these unit vectors in three-dimensional space.

And the second row here is going to have all of these partial differential operators since that's the first vector in our cross product. So that's ∂/∂x, ∂/∂y, and again all of these are just kind of waiting to be given a function that they can take the derivative of.

And then that third row is going to be the functions that we have. So the first component here is xy, the second component is cosine of z, and then that final component is z^2 + y.

Um, so I'll give some room here, maybe make it more visible.

So this is the determinant we need to compute, and this is going to be broken up into three different parts.

Uh, the first one we take this top part i and multiply it by the determinant of this submatrix.

So when we do that, um, this subdeterminant, we're taking the partial derivative with respect to y of z^2 + y.

Now, as far as y is concerned, z looks like a constant, so z^2 is a constant, and the partial derivative of this entire guy is just 1.

So that'll look like 1, and then we're subtracting off the partial derivative with respect to z of cosine of z, and that just looks the same as, you know, a derivative of cosine z, which is negative sin(z).

So that's -sin(z).

So that's the first part. And then as the next part, we're going to take j, but we're subtracting because you're always kind of thinking plus minus plus when you're doing these determinants.

So, we're going to subtract off j multiplied by its own little subdeterminant, and this time the subdeterminant is going to involve the two columns that it's not part of.

So you're imagining this first column and this second column as being part of a matrix.

So the first thing you do is take this partial derivative with respect to x of z^2 + y. Well, no x shows up there, right? That's z^2, and y.

Um, each looks like constants as far as x is concerned, so that's 0.

Then we take the partial with respect to z of x * y, and again, there's no z that shows up there, so that's also 0.

So we're kind of subtracting off 0.

And then finally, we're adding this last component.

So, we're going to add that last component k multiplied by the determinant of this submatrix of the columns that it's not part of.

So this involves the partial derivative with respect to x of cosine z. Well, no x shows up there, so that's just 0.

So that's just a 0.

And then we're subtracting off the partial with respect to y of x * y. Well, x looks like a constant, y looks like the variable, so that partial derivative is just x.

So we're subtracting off x, which means if we simplify this.

So the curl of our vector field, the curl of our vector field as a whole, as this function of x, y, and z is equal to, and that first component, the i component, we've got 1 - (-sin(z)).

So - (-sin(z)) that's 1 + sin(z).

And then the j component, we're subtracting off, but it's 0.

Usually, if you were subtracting off, you'd have to make sure to remember to flip those, but both of those are 0, so the entire j component here, or the y component of the output is 0.

And then finally, we're adding, uh, the k component is 0 - x, so that entire thing is just -x.

And that's the curl of the function, and in general, that's how you do it.

You would, um, you would take a look at the way that your function is defined in each component there and imagine taking the cross product between this Del symbol, this ∂/∂x, ∂/∂y, ∂/∂z, and you take the cross product between that and your function.

And, uh, it involves taking six different partial derivatives, and you're just mainly, it's a matter of bookkeeping to make sure you do it right, and you'll end up with something like this.

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