Question
1. Let F(x,y) = -3yi + 3xj Sketch some vectors in the vector field F(x,y) Decide whether the vector field F(x,y) is conservative.
1. Let F(x,y) = -3yi + 3xj Sketch some vectors in the vector field F(x,y) Decide whether the vector field F(x,y) is conservative.


Answers
Determine whether or not the vector field is conservative. If it is conservative, find a function $f$ such that $\mathbf{F}=\nabla f$ $$\boldsymbol{F}(x, y, z)=2 x y \mathbf{i}+\left(x^{2}+2 y z\right) \mathbf{j}+y^{2} \mathbf{k}$$
Determine whether the vector fields conservative. So here, computer curl and see if it is a serial factor. One. Sci fi Why co science See? So I component should beat the ripped about this way's respect too. Why so Cose Isay minus derivative of this with respect to see which is also co sci fi Jake Hambone is derivative off. This was respectful C which is zero minus directive off. This with respect to X, which is also zero Okay, components of the roof. This was respect Tio X, which is zero minus or if the revolver one with respect or why, which is also zero. So this is indeed a zero factor. So it is conservative. If it's conservative, we can find a potential function. So we want the reality ofthe f with respect to X Toby won derivative of F with respect or why should be, uh, Sisay on derivative off half. With respect, Izzie should be why cosign so f equals X close and the function of Boise has from the first question or why sign Z plus any function ofthe taxi if you hose anti. The Ruthie with respect with these are y sizes. Yeah, plus any function of X Y. So we're in order for these. Three questions will be true. F should be X plus y sign Z plus any constant.
For this problem we are given the vector field F equals Y. Coast of X, Y I plus X. Cause of X Y j minus sign of Ziad K. And we are asked to determine if F is a conservative vector field. So we can apply the theorem that I have a reminder of over to the side. First requirement is that fs continuous partial derivatives. And since we just have the product of variables with co sign and sign, you can see that that trivially is going to be true, then we need to check if the curl of F is going to equal the zero vector. So we'll have Dell cross F equals the determinant of the matrix. I Then we'd have dx or for the partial derivative of something with respect to X. And we have Y coast of X Y. Then we'd have next that J dy and X coast of X Y. And then we'd have K and disease. And one moment here then we would have negative signs said as our last component, their negative signs said. So first term is going to be D by D X. Negative sign said minus D by d zed x coast of xy subtly zero minus zero in the I direction. Then we'd have plus D by D X negative signs, E G minus D by DZ dy coast of X Y. So I'll be plus zero minus zero in the J direction. And then we'd have for the third component D by D X S X coast of X Y. So that would be we'd have to apply product rule. So it'll be coast of X. Yeah. Cause of X. Y plus uh Xy coast of X. Y minus D by D. Y. Why? Cossacks? Why? So we'd have minus coast of X. Y minus X. Y. Coast of X. Y. Which we can see that will go to zero as well. So we can see that. Yes our curl does equal the zero vector. I did not mean to scroll down that far. So the second part of the problem then is to find a function little F. Such that the gradient of F equals our vector field given. So we need that the partial derivative of F. With respect to X is going to equal our original X components. So that's going to be why coast of X. Y. And that with respect to why it would need to be explosive. Xy with respect to Zed would need to be negative signs said. So integrating that first equation we would get that F of X must equal. So we would integrate, let's see here we'd make the substitution U equals X. Y. And that would divide out with the Y out front. So let's see here or actually know the Y is constant. Then we yeah we just integrate this. Okay. I'm sorry. Mostly talking to myself but the Y will cancel So we'll have just a sign of X. Y plus some unknown function G of Y and Z. Then we differentiate that so we'll have fx is going to be equal to why cause of X. Y. Or actually sorry we differentiate that with respect to why I realized I was going backwards there. So differentiating with respect to why we would get X coast of X. Y plus the partial derivative of G. With respect to why. And we can see that from what we have in that second term there that has to equal our F. Y. So we can see then that G. Y must equal zero or the partial derivative of the with respect to why must equal zero which indicates then that G must be equal to some function of ZD plus a constant. So we now have that FX is going to equal sine of X. Y plus H of Z plus K. Lastly we differentiate this with respect to Zed which will give us each prime of Z. Which we see must equal negative sign of said. Which then in turn means that H. Of said is going to be positive co sign of said Keep in mind derivative of coast is negative sign. So that would give us positive co sign of said. And we already are accounting for there being some sort of constant. So that means that our F of X is going to be F of X equals sine of X. Y plus Coast of Zed plus KK
Determinative. It is conservative. So we hump. Here's a curl. So the first guy component should be the repeatable, this ex co sign Why? With respect to why which gives us negative exile. Why mind the store a rebuttable thiss with respect to see which is also website on Why so we actually don't have to compete with the other two. If this company is not if the first come parents already none zero about this Chuck again derivative of this was respectful wise. Next have access sign why and the root tip off Exley. Someone always respect too. So minus the ripped him off This boy's respect. You see, it's also exciting. Why? So we already we did get a down zero answer in the first part. So this is now conservative and if it is not conservative, then we cannot find a We cannot find the potential function
For this problem, we are given the vector field F equals X Y Z squared I plus x squared Y Z squared J plus x squared y squared Z. K. And we are asked to determine if F is conservative of conservative. So the way that we go about doing this is we can apply the theorem that I have written over to the side here that if F has continuous partial derivatives and the curl of F equals zero, then f is a conservative vector field. And we can see that F has shown here trivially. It must have continuous partial derivatives. So our next step then is to find the curl of F which is going to be the determinant of I then dx then our X component X, Y Z squared, then jay, partial why are why component two X squared. Why is it squared? Okay. And partial zed or Z component X squared Y squared zed. So we can see then that our curl would have to be D by D, Y x squared Y squared said. So that would be two X Y or two X squared Y zed minus D by D zed X word, Y, Z squared. So it'll be minus two X squared Y Z in the I direction. Then we would have minus because we have that plus minus pattern when taking a determinant like this. We could have minus D by D X X word Y squared said. So that will give us two X y squared Z minus D by disease to our a minus divided, said X Y Z squared. So that would give us minus X Y or a minus uh, two X Y zed. And at this point we can actually stop because we can see that that component is not going to equal zero. The first component of the curl does, but the second does not. So if the curl does not equal the zero vector, then that tells us that it is not conservative.