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If curl is $xy hati + xy hatj + -(x+y)z hatk$ what is the vector function?
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If curl of a vector is $xy hati + xy hatj + -(x+y)z hatk$ what is the vector ? How to approach the problem. Is there specific method to get the vector function ? Only thing I get is the usual curl matrix or curl formula. How would I get the the vector by comparing $(partial A_z/partial y - partial A_x/partial z) $ = xy $(partial A_z/partial x - partial A_x /partial z)$ = xy and $(partial A_y/partial x - partial A_x/partial y )$ = -(x+y)z .
vector-fields
asked Jul 31 at 9:10
user187604
876
 |Â
show 12 more comments
up vote
1
down vote
favorite
If curl of a vector is $xy hati + xy hatj + -(x+y)z hatk$ what is the vector ? How to approach the problem. Is there specific method to get the vector function ? Only thing I get is the usual curl matrix or curl formula. How would I get the the vector by comparing $(partial A_z/partial y - partial A_x/partial z) $ = xy $(partial A_z/partial x - partial A_x /partial z)$ = xy and $(partial A_y/partial x - partial A_x/partial y )$ = -(x+y)z .
vector-fields
asked Jul 31 at 9:10
user187604
876
Have you tried using the usual curl matrix or curl formula? I suspect you should be able to obtain something from the 3 equations
– Calvin Khor
Jul 31 at 9:13
@CalvinKhor $(partial A_z/partial y - partial A_x/partial z) $ = xy $(partial A_z/partial x - partial A_x /partial z)$ = xy and $(partial A_y/partial x - partial A_x/partial y )$ = -(x+y)z . Would it help me too much?
– user187604
Jul 31 at 9:24
I guess not. $$nabla times (nabla times u) = nabla(nablacdot u) - Delta u$$ may be useful?
– Calvin Khor
Jul 31 at 9:32
@CalvinKhor please do the problem with the formula you mentioned. It'll help me a lot.
– user187604
Jul 31 at 9:34
I don't know if it can be done with it. I'm hoping it leads to an answer.
– Calvin Khor
Jul 31 at 9:36
 |Â
show 12 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If curl of a vector is $xy hati + xy hatj + -(x+y)z hatk$ what is the vector ? How to approach the problem. Is there specific method to get the vector function ? Only thing I get is the usual curl matrix or curl formula. How would I get the the vector by comparing $(partial A_z/partial y - partial A_x/partial z) $ = xy $(partial A_z/partial x - partial A_x /partial z)$ = xy and $(partial A_y/partial x - partial A_x/partial y )$ = -(x+y)z .
vector-fields
asked Jul 31 at 9:10
user187604
876
If curl of a vector is $xy hati + xy hatj + -(x+y)z hatk$ what is the vector ? How to approach the problem. Is there specific method to get the vector function ? Only thing I get is the usual curl matrix or curl formula. How would I get the the vector by comparing $(partial A_z/partial y - partial A_x/partial z) $ = xy $(partial A_z/partial x - partial A_x /partial z)$ = xy and $(partial A_y/partial x - partial A_x/partial y )$ = -(x+y)z .
vector-fields
asked Jul 31 at 9:10
user187604
876
edited Jul 31 at 9:29
asked Jul 31 at 9:10
user187604
876
asked Jul 31 at 9:10
user187604
876
asked Jul 31 at 9:10
user187604
876
876
Have you tried using the usual curl matrix or curl formula? I suspect you should be able to obtain something from the 3 equations
– Calvin Khor
Jul 31 at 9:13
@CalvinKhor $(partial A_z/partial y - partial A_x/partial z) $ = xy $(partial A_z/partial x - partial A_x /partial z)$ = xy and $(partial A_y/partial x - partial A_x/partial y )$ = -(x+y)z . Would it help me too much?
– user187604
Jul 31 at 9:24
I guess not. $$nabla times (nabla times u) = nabla(nablacdot u) - Delta u$$ may be useful?
– Calvin Khor
Jul 31 at 9:32
@CalvinKhor please do the problem with the formula you mentioned. It'll help me a lot.
– user187604
Jul 31 at 9:34
I don't know if it can be done with it. I'm hoping it leads to an answer.
– Calvin Khor
Jul 31 at 9:36
 |Â
show 12 more comments
Have you tried using the usual curl matrix or curl formula? I suspect you should be able to obtain something from the 3 equations
– Calvin Khor
Jul 31 at 9:13
@CalvinKhor $(partial A_z/partial y - partial A_x/partial z) $ = xy $(partial A_z/partial x - partial A_x /partial z)$ = xy and $(partial A_y/partial x - partial A_x/partial y )$ = -(x+y)z . Would it help me too much?
– user187604
Jul 31 at 9:24
I guess not. $$nabla times (nabla times u) = nabla(nablacdot u) - Delta u$$ may be useful?
– Calvin Khor
Jul 31 at 9:32
@CalvinKhor please do the problem with the formula you mentioned. It'll help me a lot.
– user187604
Jul 31 at 9:34
I don't know if it can be done with it. I'm hoping it leads to an answer.
– Calvin Khor
Jul 31 at 9:36
Have you tried using the usual curl matrix or curl formula? I suspect you should be able to obtain something from the 3 equations
– Calvin Khor
Jul 31 at 9:13
Have you tried using the usual curl matrix or curl formula? I suspect you should be able to obtain something from the 3 equations
– Calvin Khor
Jul 31 at 9:13
@CalvinKhor $(partial A_z/partial y - partial A_x/partial z) $ = xy $(partial A_z/partial x - partial A_x /partial z)$ = xy and $(partial A_y/partial x - partial A_x/partial y )$ = -(x+y)z . Would it help me too much?
– user187604
Jul 31 at 9:24
@CalvinKhor $(partial A_z/partial y - partial A_x/partial z) $ = xy $(partial A_z/partial x - partial A_x /partial z)$ = xy and $(partial A_y/partial x - partial A_x/partial y )$ = -(x+y)z . Would it help me too much?
– user187604
Jul 31 at 9:24
I guess not. $$nabla times (nabla times u) = nabla(nablacdot u) - Delta u$$ may be useful?
– Calvin Khor
Jul 31 at 9:32
I guess not. $$nabla times (nabla times u) = nabla(nablacdot u) - Delta u$$ may be useful?
– Calvin Khor
Jul 31 at 9:32
@CalvinKhor please do the problem with the formula you mentioned. It'll help me a lot.
– user187604
Jul 31 at 9:34
@CalvinKhor please do the problem with the formula you mentioned. It'll help me a lot.
– user187604
Jul 31 at 9:34
I don't know if it can be done with it. I'm hoping it leads to an answer.
– Calvin Khor
Jul 31 at 9:36
I don't know if it can be done with it. I'm hoping it leads to an answer.
– Calvin Khor
Jul 31 at 9:36
 |Â
show 12 more comments
1 Answer
1
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oldest
votes
up vote
1
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accepted
One way to approach this is to convert to an equivalent differential form and then apply the algorithm described in this answer to find an antiderivative of it. In detail, apply the Hodge star operator to the 1-form $A,dx+B,dy+C,dz$ to get $eta = A,dywedge dz + B,dzwedge dx + C,dxwedge dy$. Steps 1 and 2 of the algorithm produce the form $$left[tzB(tx,ty,tz)-tyC(tx,ty,tz)right],dtwedge dx + left[txC(tx,ty,tz)-tzA(tx,ty,tz)right],dtwedge dy + left[tyA(tx,ty,tz)-txB(tx,ty,tz)right],dtwedge dz$$ and so a vector field with curl $(A,B,C)$ is $$int_0^1 left[zB(tx,ty,tz)-yC(tx,ty,tz),xC(tx,ty,tz)-zA(tx,ty,tz),yA(tx,ty,tz)-xB(tx,ty,tz)right],t,dt.$$
In this problem, we have $A:(x,y,z)mapsto xy$, $B:(x,y,z)mapsto xy$ and $C:(x,y,z)mapsto-(x+y)z$, so, after some simplification, $$mathbf F(x,y,z) = left[(2x+y)yz,mathbf i - (x+2y)xz,mathbf j - (x-y)xy,mathbf kright]int_0^1 t^3,dt \ = frac14left[(2x+y)yz,mathbf i - (x+2y)xz,mathbf j - (x-y)xy,mathbf kright].$$
There are, of course, other antiderivatives. Just as you can add an arbitrary constant to the antiderivative of a single-variable scalar function to get another antiderivative, you can add any irrotational vector field to $mathbf F$ to get another vector field with the same curl.
answered Jul 31 at 22:02
amd
25.7k2943
As an example of the “constant of integration,†if you compute the difference between this answer and the one obtained by trial and error ($xyz,mathbf i - xyz,mathbf j$) you’ll find that its curl does indeed vanish.
– amd
Aug 3 at 1:48
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
One way to approach this is to convert to an equivalent differential form and then apply the algorithm described in this answer to find an antiderivative of it. In detail, apply the Hodge star operator to the 1-form $A,dx+B,dy+C,dz$ to get $eta = A,dywedge dz + B,dzwedge dx + C,dxwedge dy$. Steps 1 and 2 of the algorithm produce the form $$left[tzB(tx,ty,tz)-tyC(tx,ty,tz)right],dtwedge dx + left[txC(tx,ty,tz)-tzA(tx,ty,tz)right],dtwedge dy + left[tyA(tx,ty,tz)-txB(tx,ty,tz)right],dtwedge dz$$ and so a vector field with curl $(A,B,C)$ is $$int_0^1 left[zB(tx,ty,tz)-yC(tx,ty,tz),xC(tx,ty,tz)-zA(tx,ty,tz),yA(tx,ty,tz)-xB(tx,ty,tz)right],t,dt.$$
In this problem, we have $A:(x,y,z)mapsto xy$, $B:(x,y,z)mapsto xy$ and $C:(x,y,z)mapsto-(x+y)z$, so, after some simplification, $$mathbf F(x,y,z) = left[(2x+y)yz,mathbf i - (x+2y)xz,mathbf j - (x-y)xy,mathbf kright]int_0^1 t^3,dt \ = frac14left[(2x+y)yz,mathbf i - (x+2y)xz,mathbf j - (x-y)xy,mathbf kright].$$
There are, of course, other antiderivatives. Just as you can add an arbitrary constant to the antiderivative of a single-variable scalar function to get another antiderivative, you can add any irrotational vector field to $mathbf F$ to get another vector field with the same curl.
answered Jul 31 at 22:02
amd
25.7k2943
As an example of the “constant of integration,†if you compute the difference between this answer and the one obtained by trial and error ($xyz,mathbf i - xyz,mathbf j$) you’ll find that its curl does indeed vanish.
– amd
Aug 3 at 1:48
add a comment |Â
up vote
1
down vote
accepted
One way to approach this is to convert to an equivalent differential form and then apply the algorithm described in this answer to find an antiderivative of it. In detail, apply the Hodge star operator to the 1-form $A,dx+B,dy+C,dz$ to get $eta = A,dywedge dz + B,dzwedge dx + C,dxwedge dy$. Steps 1 and 2 of the algorithm produce the form $$left[tzB(tx,ty,tz)-tyC(tx,ty,tz)right],dtwedge dx + left[txC(tx,ty,tz)-tzA(tx,ty,tz)right],dtwedge dy + left[tyA(tx,ty,tz)-txB(tx,ty,tz)right],dtwedge dz$$ and so a vector field with curl $(A,B,C)$ is $$int_0^1 left[zB(tx,ty,tz)-yC(tx,ty,tz),xC(tx,ty,tz)-zA(tx,ty,tz),yA(tx,ty,tz)-xB(tx,ty,tz)right],t,dt.$$
In this problem, we have $A:(x,y,z)mapsto xy$, $B:(x,y,z)mapsto xy$ and $C:(x,y,z)mapsto-(x+y)z$, so, after some simplification, $$mathbf F(x,y,z) = left[(2x+y)yz,mathbf i - (x+2y)xz,mathbf j - (x-y)xy,mathbf kright]int_0^1 t^3,dt \ = frac14left[(2x+y)yz,mathbf i - (x+2y)xz,mathbf j - (x-y)xy,mathbf kright].$$
There are, of course, other antiderivatives. Just as you can add an arbitrary constant to the antiderivative of a single-variable scalar function to get another antiderivative, you can add any irrotational vector field to $mathbf F$ to get another vector field with the same curl.
answered Jul 31 at 22:02
amd
25.7k2943
As an example of the “constant of integration,†if you compute the difference between this answer and the one obtained by trial and error ($xyz,mathbf i - xyz,mathbf j$) you’ll find that its curl does indeed vanish.
– amd
Aug 3 at 1:48
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
One way to approach this is to convert to an equivalent differential form and then apply the algorithm described in this answer to find an antiderivative of it. In detail, apply the Hodge star operator to the 1-form $A,dx+B,dy+C,dz$ to get $eta = A,dywedge dz + B,dzwedge dx + C,dxwedge dy$. Steps 1 and 2 of the algorithm produce the form $$left[tzB(tx,ty,tz)-tyC(tx,ty,tz)right],dtwedge dx + left[txC(tx,ty,tz)-tzA(tx,ty,tz)right],dtwedge dy + left[tyA(tx,ty,tz)-txB(tx,ty,tz)right],dtwedge dz$$ and so a vector field with curl $(A,B,C)$ is $$int_0^1 left[zB(tx,ty,tz)-yC(tx,ty,tz),xC(tx,ty,tz)-zA(tx,ty,tz),yA(tx,ty,tz)-xB(tx,ty,tz)right],t,dt.$$
In this problem, we have $A:(x,y,z)mapsto xy$, $B:(x,y,z)mapsto xy$ and $C:(x,y,z)mapsto-(x+y)z$, so, after some simplification, $$mathbf F(x,y,z) = left[(2x+y)yz,mathbf i - (x+2y)xz,mathbf j - (x-y)xy,mathbf kright]int_0^1 t^3,dt \ = frac14left[(2x+y)yz,mathbf i - (x+2y)xz,mathbf j - (x-y)xy,mathbf kright].$$
There are, of course, other antiderivatives. Just as you can add an arbitrary constant to the antiderivative of a single-variable scalar function to get another antiderivative, you can add any irrotational vector field to $mathbf F$ to get another vector field with the same curl.
answered Jul 31 at 22:02
amd
25.7k2943
One way to approach this is to convert to an equivalent differential form and then apply the algorithm described in this answer to find an antiderivative of it. In detail, apply the Hodge star operator to the 1-form $A,dx+B,dy+C,dz$ to get $eta = A,dywedge dz + B,dzwedge dx + C,dxwedge dy$. Steps 1 and 2 of the algorithm produce the form $$left[tzB(tx,ty,tz)-tyC(tx,ty,tz)right],dtwedge dx + left[txC(tx,ty,tz)-tzA(tx,ty,tz)right],dtwedge dy + left[tyA(tx,ty,tz)-txB(tx,ty,tz)right],dtwedge dz$$ and so a vector field with curl $(A,B,C)$ is $$int_0^1 left[zB(tx,ty,tz)-yC(tx,ty,tz),xC(tx,ty,tz)-zA(tx,ty,tz),yA(tx,ty,tz)-xB(tx,ty,tz)right],t,dt.$$
In this problem, we have $A:(x,y,z)mapsto xy$, $B:(x,y,z)mapsto xy$ and $C:(x,y,z)mapsto-(x+y)z$, so, after some simplification, $$mathbf F(x,y,z) = left[(2x+y)yz,mathbf i - (x+2y)xz,mathbf j - (x-y)xy,mathbf kright]int_0^1 t^3,dt \ = frac14left[(2x+y)yz,mathbf i - (x+2y)xz,mathbf j - (x-y)xy,mathbf kright].$$
There are, of course, other antiderivatives. Just as you can add an arbitrary constant to the antiderivative of a single-variable scalar function to get another antiderivative, you can add any irrotational vector field to $mathbf F$ to get another vector field with the same curl.
answered Jul 31 at 22:02
amd
25.7k2943
edited Jul 31 at 22:16
answered Jul 31 at 22:02
amd
25.7k2943
answered Jul 31 at 22:02
amd
25.7k2943
answered Jul 31 at 22:02
amd
25.7k2943
25.7k2943
As an example of the “constant of integration,†if you compute the difference between this answer and the one obtained by trial and error ($xyz,mathbf i - xyz,mathbf j$) you’ll find that its curl does indeed vanish.
– amd
Aug 3 at 1:48
add a comment |Â
As an example of the “constant of integration,†if you compute the difference between this answer and the one obtained by trial and error ($xyz,mathbf i - xyz,mathbf j$) you’ll find that its curl does indeed vanish.
– amd
Aug 3 at 1:48
As an example of the “constant of integration,†if you compute the difference between this answer and the one obtained by trial and error ($xyz,mathbf i - xyz,mathbf j$) you’ll find that its curl does indeed vanish.
– amd
Aug 3 at 1:48
As an example of the “constant of integration,†if you compute the difference between this answer and the one obtained by trial and error ($xyz,mathbf i - xyz,mathbf j$) you’ll find that its curl does indeed vanish.
– amd
Aug 3 at 1:48
add a comment |Â
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Have you tried using the usual curl matrix or curl formula? I suspect you should be able to obtain something from the 3 equations
– Calvin Khor
Jul 31 at 9:13
@CalvinKhor $(partial A_z/partial y - partial A_x/partial z) $ = xy $(partial A_z/partial x - partial A_x /partial z)$ = xy and $(partial A_y/partial x - partial A_x/partial y )$ = -(x+y)z . Would it help me too much?
– user187604
Jul 31 at 9:24
I guess not. $$nabla times (nabla times u) = nabla(nablacdot u) - Delta u$$ may be useful?
– Calvin Khor
Jul 31 at 9:32
@CalvinKhor please do the problem with the formula you mentioned. It'll help me a lot.
– user187604
Jul 31 at 9:34
I don't know if it can be done with it. I'm hoping it leads to an answer.
– Calvin Khor
Jul 31 at 9:36