Learn about the Curl 2 Video

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Learn about the Curl 2

Learn about the Curl 2 Welcome back, hope 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 the test do 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 its even more when you have the intuition and then you’ll hopefully never forget it. But we’ll take the curl on a fairly fancy vector field. One that I have traveled visualizing but then we can make it mathematically chug through. So let’s say our vector field and I’ll do three dimensional vector fields 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 V = (x²ysinz)^x + xy²zj + cos(x)cos(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 what how mess up it looks but we said that 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 your vector broken down into its xy and z components or it i j and k components you can take the determinant of that matrix when I showed you how to compete 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—so how do you take the cross product of this vector field and the gradient operator or you write I j k on top like your taking the cross product of any two three dimensional vectors and then you take the first vector but in this its really a vector operator but its this del operator and what are the components of the del operator, is the partial derivative with respect to x, the partial derivative with respect to y, the partial derivative with respect to z. V = a/ax[+a/ay J + a/az K right. So its x y and z components of the partial respect to x, which respect to y, which respect to z. And then the second were taking this operator cross the vector fields. So what are the components of the vector field? X²ysinz xy²z cos(x)cos(y) right just the xy and z components. And now we are ready to take the determinant which will probably get pretty messy but let’s try it. So this is equal to the I unit vector times its sub determines, you cross out its row in column and so you’re taking the determine of this expression so its going to be times—well this times this but its really the partial—if you multiply the partial with respect to y operator times an expression you really just taking since its an operator and now an expression you 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 cos x cos y minus the partial with respect to z times xy²z and now were on to our J component plus J. So what’s the magnitude of our curl in the J direction? Let’s cross out the row in the column of J. So the partial with respect to x of this, cos x cos y minus the partial with respect to z, minus the partial with respect to z of x²ysn z and then finally our k component—when you take the determinant you use that the kind of you—and you know this all kind of a bit of voodoo but you put a plus here a minus here a plus here so its kind of this checkered pattern. So this is a 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. Okay and then finally we have plus K times the determinant of its sub matrix so the partial with respect to xy²z minus the partial with respect to y of this. It might get thrown in column so this is the sub matrix of x²ysinz. All right, now let me try to simply and then 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. Now we just have to simplify it. Taking a bunch of partial, what’s the partial derivative of this with respect to y? Well x is just a constant so its going to be—well we can just put the I out from but eventually we want to write our magnitude before the vector. So I times the partial of this with respect to y, what’s our constant cos x is just a constant and then what’s the derivative of this with respect to y? It’s minus sin of y. All right so I write sin of y and let’s put the minus out front. These are what are multiplied. Okay and then we have minus—now we have to take 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 multiply it by I. The partial of this with respect to y is cos x – sin y now minus the partial of this with respect to z. Well the partial of this with respect to z is xy²z just a constant right. So the partial of this with respect to z is just xy². So minus xy² and then we’re going to have all of that that’s the magnitude in our I direction and now we have minus right because minus in the J direction. What’s the partial derivative of this with respect to x? Well partial of cos x with respect to x is minus sign of x so its -x and cos y is just the constant so it’s just carries over. Cos y and then that should be—oh yeah there we go minus this expression the partial of this with respect to z. Well the derivative of sin z with respect to z is cos z, this is just the constant so it’s minus x²y cos 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 its x²z minus, once again we just have a y term everything else is a constant so the partial with respect to y is x² sin of z and that’s the magnitude in our K direction and we’re pretty much done. I mean we can simplify a little bit just to make it clean and essentially we can just multiply by this negative one so this becomes plus, plus, plus and 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 here. Well this is I think a period than average problem. In the next video will do a little bit of this but I think it will give you a more intuition and less of just the algorithm in the computation of how do you do it. See you in the next video.

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Learn about the Curl 2

Welcome back, hope 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 the test do 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 its even more when you have the intuition and then you’ll hopefully never forget it. But we’ll take the curl on a fairly fancy vector field. One that I have traveled visualizing but then we can make it mathematically chug through.

So let’s say our vector field and I’ll do three dimensional vector fields 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 V = (x²ysinz)^x + xy²zj + cos(x)cos(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 what how mess up it looks but we said that 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 your vector broken down into its xy and z components or it i j and k components you can take the determinant of that matrix when I showed you how to compete 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—so how do you take the cross product of this vector field and the gradient operator or you write I j k on top like your taking the cross product of any two three dimensional vectors and then you take the first vector but in this its really a vector operator but its this del operator and what are the components of the del operator, is the partial derivative with respect to x, the partial derivative with respect to y, the partial derivative with respect to z. V = a/ax[+a/ay J + a/az K right. So its x y and z components of the partial respect to x, which respect to y, which respect to z. And then the second were taking this operator cross the vector fields. So what are the components of the vector field? X²ysinz xy²z cos(x)cos(y) right just the xy and z components.

And now we are ready to take the determinant which will probably get pretty messy but let’s try it. So this is equal to the I unit vector times its sub determines, you cross out its row in column and so you’re taking the determine of this expression so its going to be times—well this times this but its really the partial—if you multiply the partial with respect to y operator times an expression you really just taking since its an operator and now an expression you 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 cos x cos y minus the partial with respect to z times xy²z and now were on to our J component plus J. So what’s the magnitude of our curl in the J direction? Let’s cross out the row in the column of J.

So the partial with respect to x of this, cos x cos y minus the partial with respect to z, minus the partial with respect to z of x²ysn z and then finally our k component—when you take the determinant you use that the kind of you—and you know this all kind of a bit of voodoo but you put a plus here a minus here a plus here so its kind of this checkered pattern. So this is a 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. Okay and then finally we have plus K times the determinant of its sub matrix so the partial with respect to xy²z minus the partial with respect to y of this. It might get thrown in column so this is the sub matrix of x²ysinz.

All right, now let me try to simply and then 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. Now we just have to simplify it. Taking a bunch of partial, what’s the partial derivative of this with respect to y? Well x is just a constant so its going to be—well we can just put the I out from but eventually we want to write our magnitude before the vector. So I times the partial of this with respect to y, what’s our constant cos x is just a constant and then what’s the derivative of this with respect to y? It’s minus sin of y. All right so I write sin of y and let’s put the minus out front. These are what are multiplied. Okay and then we have minus—now we have to take 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 multiply it by I. The partial of this with respect to y is cos x – sin y now minus the partial of this with respect to z. Well the partial of this with respect to z is xy²z just a constant right. So the partial of this with respect to z is just xy². So minus xy² and then we’re going to have all of that that’s the magnitude in our I direction and now we have minus right because minus in the J direction. What’s the partial derivative of this with respect to x? Well partial of cos x with respect to x is minus sign of x so its -x and cos y is just the constant so it’s just carries over. Cos y and then that should be—oh yeah there we go minus this expression the partial of this with respect to z.

Well the derivative of sin z with respect to z is cos z, this is just the constant so it’s minus x²y cos 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 its x²z minus, once again we just have a y term everything else is a constant so the partial with respect to y is x² sin of z and that’s the magnitude in our K direction and we’re pretty much done. I mean we can simplify a little bit just to make it clean and essentially we can just multiply by this negative one so this becomes plus, plus, plus and 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 here. Well this is I think a period than average problem.

In the next video will do a little bit of this but I think it will give you a more intuition and less of just the algorithm in the computation of how do you do it. See you in the next video.

Transcription by: Scribe4you Transcription Services