yego.me
💡 Stop wasting time. Read Youtube instead of watch. Download Chrome Extension

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.

More Articles

View All
Impedance of simple networks
Let’s talk about the idea of the impedance of some simple networks. Now, what I’ve shown here is a very simple network. It has two impedances in it, Z1 and Z2, and inside these boxes are one of our favorite passive components, either an R, an L, or a C. T…
The presidential incumbency advantage | US government and civics | Khan Academy
What we’re going to do in this video is talk about the incumbent advantage. This is the idea that the person who is already in power, the incumbent, has an advantage in elections. In particular, we’re going to focus on presidential elections, although thi…
What EVERYONE Needs To Do With Their Money ASAP
What’s up guys, it’s Graham here. So I want to start this video off on a bit of a serious note because if economic growth begins to slow down, inflation persists, and stocks head for a bear market, it’s more important than ever right now to make sure you’…
Types of mixtures | Intermolecular forces and properties | AP Chemistry | Khan Academy
I suspect that you might already be familiar with the term “mixture.” It really does mean what you think it means. If you take two or more substances and you were to mix them together, you are dealing with a mixture, and it could be a solid, a liquid, or …
Determining whether real world model is linear or exponential
The table represents the cost of buying a small piece of land in a remote village since the year 1990. Which kind of function best models this relationship? I’m using this as an example from the Khan Academy exercises, and we’re really trying to pick bet…
Why Scientists Are Puzzled By This Virus
Very recently, scientists discovered that your body is teeming with trillions of the most bizarre viruses. These viruses are not your enemies but critical to your health, protecting you from disease, maybe even killing cancer. A new frontier of science, s…