Mixing
Jacobian of implicit functions
$begingroup$
Question Statement:-
If $u^3+v^3=x+y$ and $u^2+v^2=x^3+y^3$, show that $$frac{partial(u,v)}{partial(x,y)}=frac{1}{2}frac{y^2-x^2}{uv(u-v)}$$
My Solution:-
As the relation that are given between $u,v,x ;& ; y$ do not seemed to me to be separated so as to get the $u$ and $v$ in the terms of $x;&;y$, so I went with the following approach.
We are given with the following relations:-
$$u^3+v^3=x+ytag{1}$$
$u^2+v^2=x^3+y^3tag{2}$
On partially differentiating the above given relations we have,
$$3u^2frac{partial u}{partial{x}}+3v^2frac{partial{v}}{partial{x}}=1tag{3}$$
$$2ufrac{partial u}{partial{x}}+2vfrac{partial{v}}{partial{x}}=3x^2tag{4}$$
The equations $(3)$ and $(4)$ can be written as a matrix as follows
$$begin{bmatrix}
3u^2 & 3v^2\
2u & 2v
end{bmatrix}
begin{bmatrix}
dfrac{partial{u}}{partial{x}}\
dfrac{partial{v}}{partial{x}}
end{bmatrix}
=
begin{bmatrix}
1 \ 3x^2
end{bmatrix} \
implies
begin{bmatrix}
dfrac{partial{u}}{partial{x}}\
dfrac{partial{v}}{partial{x}}
end{bmatrix}=
dfrac{1}{6uv(u-v)}begin{bmatrix}
2v & -2u\
-3v^2 & 3u^2
end{bmatrix}
begin{bmatrix}
1 \ 3x^2
end{bmatrix}\
implies
begin{bmatrix}
dfrac{partial{u}}{partial{x}}\
dfrac{partial{v}}{partial{x}}
end{bmatrix}=
dfrac{1}{6uv(u-v)}begin{bmatrix}
2v-6ux^2\
9u^2x^2-3v^2
end{bmatrix}
$$
So, we get
$$dfrac{partial{u}}{partial{x}}=dfrac{2v-6ux^2}{6uv(u-v)}\
&\
dfrac{partial{v}}{partial{x}}=dfrac{9u^2x^2-3v^2}{6uv(u-v)}$$
And due to the symmetry in the equations $(1)$ and $(2)$, we get
$$dfrac{partial{u}}{partial{y}}=dfrac{2v-6uy^2}{6uv(u-v)}\
&\
dfrac{partial{v}}{partial{y}}=dfrac{9u^2y^2-3v^2}{6uv(u-v)}$$
So, we get the Jacobian determinant as
$$frac{partial(u,v)}{partial(x,y)}=
begin{vmatrix}
dfrac{partial{u}}{partial{x}} & dfrac{partial{u}}{partial{y}}\
dfrac{partial{v}}{partial{x}} & dfrac{partial{v}}{partial{y}}
end{vmatrix}=
begin{vmatrix}
dfrac{2v-6ux^2}{6uv(u-v)} & dfrac{2v-6uy^2}{6uv(u-v)}\
dfrac{9u^2x^2-3v^2}{6uv(u-v)} & dfrac{9u^2y^2-3v^2}{6uv(u-v)}
end{vmatrix}=frac{1}{2}frac{y^2-x^2}{uv(u-v)}$$
My deal with the question:-
This method seemed a bit too long, can you suggest a shorter method. If the method teaches me something new that would be good.
multivariable-calculus jacobian implicit-function
asked Oct 1 '17 at 16:54
user350331user350331
1,28821232
$endgroup$
add a comment |
$begingroup$
Question Statement:-
If $u^3+v^3=x+y$ and $u^2+v^2=x^3+y^3$, show that $$frac{partial(u,v)}{partial(x,y)}=frac{1}{2}frac{y^2-x^2}{uv(u-v)}$$
My Solution:-
As the relation that are given between $u,v,x ;& ; y$ do not seemed to me to be separated so as to get the $u$ and $v$ in the terms of $x;&;y$, so I went with the following approach.
We are given with the following relations:-
$$u^3+v^3=x+ytag{1}$$
$u^2+v^2=x^3+y^3tag{2}$
On partially differentiating the above given relations we have,
$$3u^2frac{partial u}{partial{x}}+3v^2frac{partial{v}}{partial{x}}=1tag{3}$$
$$2ufrac{partial u}{partial{x}}+2vfrac{partial{v}}{partial{x}}=3x^2tag{4}$$
The equations $(3)$ and $(4)$ can be written as a matrix as follows
$$begin{bmatrix}
3u^2 & 3v^2\
2u & 2v
end{bmatrix}
begin{bmatrix}
dfrac{partial{u}}{partial{x}}\
dfrac{partial{v}}{partial{x}}
end{bmatrix}
=
begin{bmatrix}
1 \ 3x^2
end{bmatrix} \
implies
begin{bmatrix}
dfrac{partial{u}}{partial{x}}\
dfrac{partial{v}}{partial{x}}
end{bmatrix}=
dfrac{1}{6uv(u-v)}begin{bmatrix}
2v & -2u\
-3v^2 & 3u^2
end{bmatrix}
begin{bmatrix}
1 \ 3x^2
end{bmatrix}\
implies
begin{bmatrix}
dfrac{partial{u}}{partial{x}}\
dfrac{partial{v}}{partial{x}}
end{bmatrix}=
dfrac{1}{6uv(u-v)}begin{bmatrix}
2v-6ux^2\
9u^2x^2-3v^2
end{bmatrix}
$$
So, we get
$$dfrac{partial{u}}{partial{x}}=dfrac{2v-6ux^2}{6uv(u-v)}\
&\
dfrac{partial{v}}{partial{x}}=dfrac{9u^2x^2-3v^2}{6uv(u-v)}$$
And due to the symmetry in the equations $(1)$ and $(2)$, we get
$$dfrac{partial{u}}{partial{y}}=dfrac{2v-6uy^2}{6uv(u-v)}\
&\
dfrac{partial{v}}{partial{y}}=dfrac{9u^2y^2-3v^2}{6uv(u-v)}$$
So, we get the Jacobian determinant as
$$frac{partial(u,v)}{partial(x,y)}=
begin{vmatrix}
dfrac{partial{u}}{partial{x}} & dfrac{partial{u}}{partial{y}}\
dfrac{partial{v}}{partial{x}} & dfrac{partial{v}}{partial{y}}
end{vmatrix}=
begin{vmatrix}
dfrac{2v-6ux^2}{6uv(u-v)} & dfrac{2v-6uy^2}{6uv(u-v)}\
dfrac{9u^2x^2-3v^2}{6uv(u-v)} & dfrac{9u^2y^2-3v^2}{6uv(u-v)}
end{vmatrix}=frac{1}{2}frac{y^2-x^2}{uv(u-v)}$$
My deal with the question:-
This method seemed a bit too long, can you suggest a shorter method. If the method teaches me something new that would be good.
multivariable-calculus jacobian implicit-function
asked Oct 1 '17 at 16:54
user350331user350331
1,28821232
$endgroup$
1
$begingroup$
Since you’re only interested in the Jacobian determinant, there’s no need to completely solve for $u_x$ &c. Combine your systems of equations into a single matrix equation of the form $AJ=B$. Then $det J=det B/det A$.
$endgroup$
– amd
Oct 1 '17 at 18:09
add a comment |
$begingroup$
Question Statement:-
If $u^3+v^3=x+y$ and $u^2+v^2=x^3+y^3$, show that $$frac{partial(u,v)}{partial(x,y)}=frac{1}{2}frac{y^2-x^2}{uv(u-v)}$$
My Solution:-
As the relation that are given between $u,v,x ;& ; y$ do not seemed to me to be separated so as to get the $u$ and $v$ in the terms of $x;&;y$, so I went with the following approach.
We are given with the following relations:-
$$u^3+v^3=x+ytag{1}$$
$u^2+v^2=x^3+y^3tag{2}$
On partially differentiating the above given relations we have,
$$3u^2frac{partial u}{partial{x}}+3v^2frac{partial{v}}{partial{x}}=1tag{3}$$
$$2ufrac{partial u}{partial{x}}+2vfrac{partial{v}}{partial{x}}=3x^2tag{4}$$
The equations $(3)$ and $(4)$ can be written as a matrix as follows
$$begin{bmatrix}
3u^2 & 3v^2\
2u & 2v
end{bmatrix}
begin{bmatrix}
dfrac{partial{u}}{partial{x}}\
dfrac{partial{v}}{partial{x}}
end{bmatrix}
=
begin{bmatrix}
1 \ 3x^2
end{bmatrix} \
implies
begin{bmatrix}
dfrac{partial{u}}{partial{x}}\
dfrac{partial{v}}{partial{x}}
end{bmatrix}=
dfrac{1}{6uv(u-v)}begin{bmatrix}
2v & -2u\
-3v^2 & 3u^2
end{bmatrix}
begin{bmatrix}
1 \ 3x^2
end{bmatrix}\
implies
begin{bmatrix}
dfrac{partial{u}}{partial{x}}\
dfrac{partial{v}}{partial{x}}
end{bmatrix}=
dfrac{1}{6uv(u-v)}begin{bmatrix}
2v-6ux^2\
9u^2x^2-3v^2
end{bmatrix}
$$
So, we get
$$dfrac{partial{u}}{partial{x}}=dfrac{2v-6ux^2}{6uv(u-v)}\
&\
dfrac{partial{v}}{partial{x}}=dfrac{9u^2x^2-3v^2}{6uv(u-v)}$$
And due to the symmetry in the equations $(1)$ and $(2)$, we get
$$dfrac{partial{u}}{partial{y}}=dfrac{2v-6uy^2}{6uv(u-v)}\
&\
dfrac{partial{v}}{partial{y}}=dfrac{9u^2y^2-3v^2}{6uv(u-v)}$$
So, we get the Jacobian determinant as
$$frac{partial(u,v)}{partial(x,y)}=
begin{vmatrix}
dfrac{partial{u}}{partial{x}} & dfrac{partial{u}}{partial{y}}\
dfrac{partial{v}}{partial{x}} & dfrac{partial{v}}{partial{y}}
end{vmatrix}=
begin{vmatrix}
dfrac{2v-6ux^2}{6uv(u-v)} & dfrac{2v-6uy^2}{6uv(u-v)}\
dfrac{9u^2x^2-3v^2}{6uv(u-v)} & dfrac{9u^2y^2-3v^2}{6uv(u-v)}
end{vmatrix}=frac{1}{2}frac{y^2-x^2}{uv(u-v)}$$
My deal with the question:-
This method seemed a bit too long, can you suggest a shorter method. If the method teaches me something new that would be good.
multivariable-calculus jacobian implicit-function
asked Oct 1 '17 at 16:54
user350331user350331
1,28821232
$endgroup$
Question Statement:-
If $u^3+v^3=x+y$ and $u^2+v^2=x^3+y^3$, show that $$frac{partial(u,v)}{partial(x,y)}=frac{1}{2}frac{y^2-x^2}{uv(u-v)}$$
My Solution:-
As the relation that are given between $u,v,x ;& ; y$ do not seemed to me to be separated so as to get the $u$ and $v$ in the terms of $x;&;y$, so I went with the following approach.
We are given with the following relations:-
$$u^3+v^3=x+ytag{1}$$
$u^2+v^2=x^3+y^3tag{2}$
On partially differentiating the above given relations we have,
$$3u^2frac{partial u}{partial{x}}+3v^2frac{partial{v}}{partial{x}}=1tag{3}$$
$$2ufrac{partial u}{partial{x}}+2vfrac{partial{v}}{partial{x}}=3x^2tag{4}$$
The equations $(3)$ and $(4)$ can be written as a matrix as follows
$$begin{bmatrix}
3u^2 & 3v^2\
2u & 2v
end{bmatrix}
begin{bmatrix}
dfrac{partial{u}}{partial{x}}\
dfrac{partial{v}}{partial{x}}
end{bmatrix}
=
begin{bmatrix}
1 \ 3x^2
end{bmatrix} \
implies
begin{bmatrix}
dfrac{partial{u}}{partial{x}}\
dfrac{partial{v}}{partial{x}}
end{bmatrix}=
dfrac{1}{6uv(u-v)}begin{bmatrix}
2v & -2u\
-3v^2 & 3u^2
end{bmatrix}
begin{bmatrix}
1 \ 3x^2
end{bmatrix}\
implies
begin{bmatrix}
dfrac{partial{u}}{partial{x}}\
dfrac{partial{v}}{partial{x}}
end{bmatrix}=
dfrac{1}{6uv(u-v)}begin{bmatrix}
2v-6ux^2\
9u^2x^2-3v^2
end{bmatrix}
$$
So, we get
$$dfrac{partial{u}}{partial{x}}=dfrac{2v-6ux^2}{6uv(u-v)}\
&\
dfrac{partial{v}}{partial{x}}=dfrac{9u^2x^2-3v^2}{6uv(u-v)}$$
And due to the symmetry in the equations $(1)$ and $(2)$, we get
$$dfrac{partial{u}}{partial{y}}=dfrac{2v-6uy^2}{6uv(u-v)}\
&\
dfrac{partial{v}}{partial{y}}=dfrac{9u^2y^2-3v^2}{6uv(u-v)}$$
So, we get the Jacobian determinant as
$$frac{partial(u,v)}{partial(x,y)}=
begin{vmatrix}
dfrac{partial{u}}{partial{x}} & dfrac{partial{u}}{partial{y}}\
dfrac{partial{v}}{partial{x}} & dfrac{partial{v}}{partial{y}}
end{vmatrix}=
begin{vmatrix}
dfrac{2v-6ux^2}{6uv(u-v)} & dfrac{2v-6uy^2}{6uv(u-v)}\
dfrac{9u^2x^2-3v^2}{6uv(u-v)} & dfrac{9u^2y^2-3v^2}{6uv(u-v)}
end{vmatrix}=frac{1}{2}frac{y^2-x^2}{uv(u-v)}$$
My deal with the question:-
This method seemed a bit too long, can you suggest a shorter method. If the method teaches me something new that would be good.
multivariable-calculus jacobian implicit-function
multivariable-calculus jacobian implicit-function
asked Oct 1 '17 at 16:54
user350331user350331
1,28821232
asked Oct 1 '17 at 16:54
user350331user350331
1,28821232
asked Oct 1 '17 at 16:54
user350331user350331
1,28821232
asked Oct 1 '17 at 16:54
user350331user350331
1,28821232
asked Oct 1 '17 at 16:54
user350331user350331
1,28821232
1,28821232
1
$begingroup$
Since you’re only interested in the Jacobian determinant, there’s no need to completely solve for $u_x$ &c. Combine your systems of equations into a single matrix equation of the form $AJ=B$. Then $det J=det B/det A$.
$endgroup$
– amd
Oct 1 '17 at 18:09
add a comment |
1
$begingroup$
Since you’re only interested in the Jacobian determinant, there’s no need to completely solve for $u_x$ &c. Combine your systems of equations into a single matrix equation of the form $AJ=B$. Then $det J=det B/det A$.
$endgroup$
– amd
Oct 1 '17 at 18:09
1
1
$begingroup$
Since you’re only interested in the Jacobian determinant, there’s no need to completely solve for $u_x$ &c. Combine your systems of equations into a single matrix equation of the form $AJ=B$. Then $det J=det B/det A$.
$endgroup$
– amd
Oct 1 '17 at 18:09
$begingroup$
Since you’re only interested in the Jacobian determinant, there’s no need to completely solve for $u_x$ &c. Combine your systems of equations into a single matrix equation of the form $AJ=B$. Then $det J=det B/det A$.
$endgroup$
– amd
Oct 1 '17 at 18:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
let $f_{1}=u^{3}+v^{3}-x-y$
and $f_{2}=u^{2}+v^{2}-x^3-y^3$
we know, by property of jacobians
$dfrac{partial(u,v)}{partial(x,y)}=(-1)^2dfrac{dfrac{partial(f_{1},f_{2})}{partial(x,y)}}{dfrac{partial(f_{1},f_{2})}{partial(u,v)}}tag{1}$
${dfrac{partial(f_{1},f_{2})}{partial(x,y)}}=3(y^2-x^2)tag{2}$
${dfrac{partial(f_{1},f_{2})}{partial(u,v)}}=6 u v (u-v)tag{3}$
put equation $(2)$ and $(3)$ in $(1)$ ...you will have your result
answered Sep 24 '18 at 21:48
Faraday PathakFaraday Pathak
1,165316
$endgroup$
add a comment |
$begingroup$
This answer blatantly steals the comment by user @amd. Take your first matrix equation and add a column for the derivatives with respect to $y$:
$$begin{bmatrix}
3u^2 & 3v^2\
2u & 2v
end{bmatrix}
begin{bmatrix}
dfrac{partial{u}}{partial{x}} & dfrac{partial{u}}{partial{y}}\
dfrac{partial{v}}{partial{x}} & dfrac{partial{v}}{partial{y}}
end{bmatrix}
=
begin{bmatrix}
1 & 1 \ 3x^2 & 3y^2
end{bmatrix}. \
$$
Now we can take the determinant of both sides:
$$
(6u^2v - 6v^2u)frac{partial(u,v)}{partial(x,y)} = (3y^2 - 3x^2)
$$
which implies the result.
answered Jan 25 at 7:17
hunterhunter
15.3k32640
$endgroup$
$begingroup$
(Explicitly, the labor savings is that, by taking det on both sides before inverting, we wind up inverting a number instead of a matrix, which is easier.) I myself used a different type of labor-savings in writing up this answer: in addition to stealing @amd's suggestion, I stole your LaTeX!
$endgroup$
– hunter
Jan 25 at 7:19
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
let $f_{1}=u^{3}+v^{3}-x-y$
and $f_{2}=u^{2}+v^{2}-x^3-y^3$
we know, by property of jacobians
$dfrac{partial(u,v)}{partial(x,y)}=(-1)^2dfrac{dfrac{partial(f_{1},f_{2})}{partial(x,y)}}{dfrac{partial(f_{1},f_{2})}{partial(u,v)}}tag{1}$
${dfrac{partial(f_{1},f_{2})}{partial(x,y)}}=3(y^2-x^2)tag{2}$
${dfrac{partial(f_{1},f_{2})}{partial(u,v)}}=6 u v (u-v)tag{3}$
put equation $(2)$ and $(3)$ in $(1)$ ...you will have your result
answered Sep 24 '18 at 21:48
Faraday PathakFaraday Pathak
1,165316
$endgroup$
add a comment |
$begingroup$
let $f_{1}=u^{3}+v^{3}-x-y$
and $f_{2}=u^{2}+v^{2}-x^3-y^3$
we know, by property of jacobians
$dfrac{partial(u,v)}{partial(x,y)}=(-1)^2dfrac{dfrac{partial(f_{1},f_{2})}{partial(x,y)}}{dfrac{partial(f_{1},f_{2})}{partial(u,v)}}tag{1}$
${dfrac{partial(f_{1},f_{2})}{partial(x,y)}}=3(y^2-x^2)tag{2}$
${dfrac{partial(f_{1},f_{2})}{partial(u,v)}}=6 u v (u-v)tag{3}$
put equation $(2)$ and $(3)$ in $(1)$ ...you will have your result
answered Sep 24 '18 at 21:48
Faraday PathakFaraday Pathak
1,165316
$endgroup$
add a comment |
$begingroup$
let $f_{1}=u^{3}+v^{3}-x-y$
and $f_{2}=u^{2}+v^{2}-x^3-y^3$
we know, by property of jacobians
$dfrac{partial(u,v)}{partial(x,y)}=(-1)^2dfrac{dfrac{partial(f_{1},f_{2})}{partial(x,y)}}{dfrac{partial(f_{1},f_{2})}{partial(u,v)}}tag{1}$
${dfrac{partial(f_{1},f_{2})}{partial(x,y)}}=3(y^2-x^2)tag{2}$
${dfrac{partial(f_{1},f_{2})}{partial(u,v)}}=6 u v (u-v)tag{3}$
put equation $(2)$ and $(3)$ in $(1)$ ...you will have your result
answered Sep 24 '18 at 21:48
Faraday PathakFaraday Pathak
1,165316
$endgroup$
let $f_{1}=u^{3}+v^{3}-x-y$
and $f_{2}=u^{2}+v^{2}-x^3-y^3$
we know, by property of jacobians
$dfrac{partial(u,v)}{partial(x,y)}=(-1)^2dfrac{dfrac{partial(f_{1},f_{2})}{partial(x,y)}}{dfrac{partial(f_{1},f_{2})}{partial(u,v)}}tag{1}$
${dfrac{partial(f_{1},f_{2})}{partial(x,y)}}=3(y^2-x^2)tag{2}$
${dfrac{partial(f_{1},f_{2})}{partial(u,v)}}=6 u v (u-v)tag{3}$
put equation $(2)$ and $(3)$ in $(1)$ ...you will have your result
answered Sep 24 '18 at 21:48
Faraday PathakFaraday Pathak
1,165316
answered Sep 24 '18 at 21:48
Faraday PathakFaraday Pathak
1,165316
answered Sep 24 '18 at 21:48
Faraday PathakFaraday Pathak
1,165316
answered Sep 24 '18 at 21:48
Faraday PathakFaraday Pathak
1,165316
1,165316
add a comment |
add a comment |
$begingroup$
This answer blatantly steals the comment by user @amd. Take your first matrix equation and add a column for the derivatives with respect to $y$:
$$begin{bmatrix}
3u^2 & 3v^2\
2u & 2v
end{bmatrix}
begin{bmatrix}
dfrac{partial{u}}{partial{x}} & dfrac{partial{u}}{partial{y}}\
dfrac{partial{v}}{partial{x}} & dfrac{partial{v}}{partial{y}}
end{bmatrix}
=
begin{bmatrix}
1 & 1 \ 3x^2 & 3y^2
end{bmatrix}. \
$$
Now we can take the determinant of both sides:
$$
(6u^2v - 6v^2u)frac{partial(u,v)}{partial(x,y)} = (3y^2 - 3x^2)
$$
which implies the result.
answered Jan 25 at 7:17
hunterhunter
15.3k32640
$endgroup$
$begingroup$
(Explicitly, the labor savings is that, by taking det on both sides before inverting, we wind up inverting a number instead of a matrix, which is easier.) I myself used a different type of labor-savings in writing up this answer: in addition to stealing @amd's suggestion, I stole your LaTeX!
$endgroup$
– hunter
Jan 25 at 7:19
add a comment |
$begingroup$
This answer blatantly steals the comment by user @amd. Take your first matrix equation and add a column for the derivatives with respect to $y$:
$$begin{bmatrix}
3u^2 & 3v^2\
2u & 2v
end{bmatrix}
begin{bmatrix}
dfrac{partial{u}}{partial{x}} & dfrac{partial{u}}{partial{y}}\
dfrac{partial{v}}{partial{x}} & dfrac{partial{v}}{partial{y}}
end{bmatrix}
=
begin{bmatrix}
1 & 1 \ 3x^2 & 3y^2
end{bmatrix}. \
$$
Now we can take the determinant of both sides:
$$
(6u^2v - 6v^2u)frac{partial(u,v)}{partial(x,y)} = (3y^2 - 3x^2)
$$
which implies the result.
answered Jan 25 at 7:17
hunterhunter
15.3k32640
$endgroup$
$begingroup$
(Explicitly, the labor savings is that, by taking det on both sides before inverting, we wind up inverting a number instead of a matrix, which is easier.) I myself used a different type of labor-savings in writing up this answer: in addition to stealing @amd's suggestion, I stole your LaTeX!
$endgroup$
– hunter
Jan 25 at 7:19
add a comment |
$begingroup$
This answer blatantly steals the comment by user @amd. Take your first matrix equation and add a column for the derivatives with respect to $y$:
$$begin{bmatrix}
3u^2 & 3v^2\
2u & 2v
end{bmatrix}
begin{bmatrix}
dfrac{partial{u}}{partial{x}} & dfrac{partial{u}}{partial{y}}\
dfrac{partial{v}}{partial{x}} & dfrac{partial{v}}{partial{y}}
end{bmatrix}
=
begin{bmatrix}
1 & 1 \ 3x^2 & 3y^2
end{bmatrix}. \
$$
Now we can take the determinant of both sides:
$$
(6u^2v - 6v^2u)frac{partial(u,v)}{partial(x,y)} = (3y^2 - 3x^2)
$$
which implies the result.
answered Jan 25 at 7:17
hunterhunter
15.3k32640
$endgroup$
This answer blatantly steals the comment by user @amd. Take your first matrix equation and add a column for the derivatives with respect to $y$:
$$begin{bmatrix}
3u^2 & 3v^2\
2u & 2v
end{bmatrix}
begin{bmatrix}
dfrac{partial{u}}{partial{x}} & dfrac{partial{u}}{partial{y}}\
dfrac{partial{v}}{partial{x}} & dfrac{partial{v}}{partial{y}}
end{bmatrix}
=
begin{bmatrix}
1 & 1 \ 3x^2 & 3y^2
end{bmatrix}. \
$$
Now we can take the determinant of both sides:
$$
(6u^2v - 6v^2u)frac{partial(u,v)}{partial(x,y)} = (3y^2 - 3x^2)
$$
which implies the result.
answered Jan 25 at 7:17
hunterhunter
15.3k32640
answered Jan 25 at 7:17
hunterhunter
15.3k32640
answered Jan 25 at 7:17
hunterhunter
15.3k32640
answered Jan 25 at 7:17
hunterhunter
15.3k32640
15.3k32640
$begingroup$
(Explicitly, the labor savings is that, by taking det on both sides before inverting, we wind up inverting a number instead of a matrix, which is easier.) I myself used a different type of labor-savings in writing up this answer: in addition to stealing @amd's suggestion, I stole your LaTeX!
$endgroup$
– hunter
Jan 25 at 7:19
add a comment |
$begingroup$
(Explicitly, the labor savings is that, by taking det on both sides before inverting, we wind up inverting a number instead of a matrix, which is easier.) I myself used a different type of labor-savings in writing up this answer: in addition to stealing @amd's suggestion, I stole your LaTeX!
$endgroup$
– hunter
Jan 25 at 7:19
$begingroup$
(Explicitly, the labor savings is that, by taking det on both sides before inverting, we wind up inverting a number instead of a matrix, which is easier.) I myself used a different type of labor-savings in writing up this answer: in addition to stealing @amd's suggestion, I stole your LaTeX!
$endgroup$
– hunter
Jan 25 at 7:19
$begingroup$
(Explicitly, the labor savings is that, by taking det on both sides before inverting, we wind up inverting a number instead of a matrix, which is easier.) I myself used a different type of labor-savings in writing up this answer: in addition to stealing @amd's suggestion, I stole your LaTeX!
$endgroup$
– hunter
Jan 25 at 7:19
add a comment |
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$begingroup$
Since you’re only interested in the Jacobian determinant, there’s no need to completely solve for $u_x$ &c. Combine your systems of equations into a single matrix equation of the form $AJ=B$. Then $det J=det B/det A$.
$endgroup$
– amd
Oct 1 '17 at 18:09