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Confused about the Jacobian matrix


Jacobian matrix normalizationEvaluating $int_0^infty int_0^infty fracx^2 + y^21 + (x^2 - y^2)^2 e^-2xy :mathrmdx :mathrmdy$Surface Integral do I use a jacobian?How to set the limits for Jacobian IntegrationComputing Jacobian of matrix-valued map $X mapsto X^TAX$Jacobian Matrix - unknown functionOn the relationship between the volume element in curvilinear coordinates and the JacobianWhat is the Jacobian in this transformationHow to do this surface integral?Geometric Meaning of the Jacobian of a Linear Transformation













2












$begingroup$



Find $$int_0^infty int_0^infty e^-2xy , mathrm d x mathrm dy$$ using $u = x^2 - y^2$ and $v=2xy$.




I have tried using the Jacobian matrix to obtain the Jacobian of the transformation. However, confusion arises since I do not know what should be kept constant. Do I directly differentiate $u$ with respect to $x$ while keeping $y$ constant, or do I substitute $y$ from $v=2xy$, and then differentiate $u$ with respect to $x$ while keeping $v$ constant?










share|cite|improve this question









New contributor




user205891 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    When finding $fracpartial upartial x$ and $fracpartial upartial y$, just differentiate $u$ as a function of $(x,y)$, ignore the $v$. Similarly, when differentiating $v$, don't worry about $u$. So for example, $fracpartial upartial x = 2x$.
    $endgroup$
    – Minus One-Twelfth
    2 days ago











  • $begingroup$
    However, the Jacobian matrix says that we should derive x with respect to u, x wrt v, y wrt u, and y wrt v. Therefore, it is not deriving u wrt to x and y, but it's the other way around.
    $endgroup$
    – user205891
    2 days ago






  • 1




    $begingroup$
    There is a trick that you can use if you want: the Jacobian determinant of the inverse map is the inverse (reciprocal) of the Jacobian determinant of the forward map. Therefore, you can just do it the way I described, get the Jacobian determinant of that, and then take the reciprocal. (This will get you something in terms of $x$ and $y$, which you will need to express in terms of $u$ and $v$ for the integral.)
    $endgroup$
    – Minus One-Twelfth
    2 days ago
















2












$begingroup$



Find $$int_0^infty int_0^infty e^-2xy , mathrm d x mathrm dy$$ using $u = x^2 - y^2$ and $v=2xy$.




I have tried using the Jacobian matrix to obtain the Jacobian of the transformation. However, confusion arises since I do not know what should be kept constant. Do I directly differentiate $u$ with respect to $x$ while keeping $y$ constant, or do I substitute $y$ from $v=2xy$, and then differentiate $u$ with respect to $x$ while keeping $v$ constant?










share|cite|improve this question









New contributor




user205891 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    When finding $fracpartial upartial x$ and $fracpartial upartial y$, just differentiate $u$ as a function of $(x,y)$, ignore the $v$. Similarly, when differentiating $v$, don't worry about $u$. So for example, $fracpartial upartial x = 2x$.
    $endgroup$
    – Minus One-Twelfth
    2 days ago











  • $begingroup$
    However, the Jacobian matrix says that we should derive x with respect to u, x wrt v, y wrt u, and y wrt v. Therefore, it is not deriving u wrt to x and y, but it's the other way around.
    $endgroup$
    – user205891
    2 days ago






  • 1




    $begingroup$
    There is a trick that you can use if you want: the Jacobian determinant of the inverse map is the inverse (reciprocal) of the Jacobian determinant of the forward map. Therefore, you can just do it the way I described, get the Jacobian determinant of that, and then take the reciprocal. (This will get you something in terms of $x$ and $y$, which you will need to express in terms of $u$ and $v$ for the integral.)
    $endgroup$
    – Minus One-Twelfth
    2 days ago














2












2








2





$begingroup$



Find $$int_0^infty int_0^infty e^-2xy , mathrm d x mathrm dy$$ using $u = x^2 - y^2$ and $v=2xy$.




I have tried using the Jacobian matrix to obtain the Jacobian of the transformation. However, confusion arises since I do not know what should be kept constant. Do I directly differentiate $u$ with respect to $x$ while keeping $y$ constant, or do I substitute $y$ from $v=2xy$, and then differentiate $u$ with respect to $x$ while keeping $v$ constant?










share|cite|improve this question









New contributor




user205891 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$





Find $$int_0^infty int_0^infty e^-2xy , mathrm d x mathrm dy$$ using $u = x^2 - y^2$ and $v=2xy$.




I have tried using the Jacobian matrix to obtain the Jacobian of the transformation. However, confusion arises since I do not know what should be kept constant. Do I directly differentiate $u$ with respect to $x$ while keeping $y$ constant, or do I substitute $y$ from $v=2xy$, and then differentiate $u$ with respect to $x$ while keeping $v$ constant?







calculus multivariable-calculus partial-derivative jacobian






share|cite|improve this question









New contributor




user205891 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




user205891 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Rodrigo de Azevedo

13k41960




13k41960






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asked 2 days ago









user205891user205891

112




112




New contributor




user205891 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





user205891 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






user205891 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • $begingroup$
    When finding $fracpartial upartial x$ and $fracpartial upartial y$, just differentiate $u$ as a function of $(x,y)$, ignore the $v$. Similarly, when differentiating $v$, don't worry about $u$. So for example, $fracpartial upartial x = 2x$.
    $endgroup$
    – Minus One-Twelfth
    2 days ago











  • $begingroup$
    However, the Jacobian matrix says that we should derive x with respect to u, x wrt v, y wrt u, and y wrt v. Therefore, it is not deriving u wrt to x and y, but it's the other way around.
    $endgroup$
    – user205891
    2 days ago






  • 1




    $begingroup$
    There is a trick that you can use if you want: the Jacobian determinant of the inverse map is the inverse (reciprocal) of the Jacobian determinant of the forward map. Therefore, you can just do it the way I described, get the Jacobian determinant of that, and then take the reciprocal. (This will get you something in terms of $x$ and $y$, which you will need to express in terms of $u$ and $v$ for the integral.)
    $endgroup$
    – Minus One-Twelfth
    2 days ago

















  • $begingroup$
    When finding $fracpartial upartial x$ and $fracpartial upartial y$, just differentiate $u$ as a function of $(x,y)$, ignore the $v$. Similarly, when differentiating $v$, don't worry about $u$. So for example, $fracpartial upartial x = 2x$.
    $endgroup$
    – Minus One-Twelfth
    2 days ago











  • $begingroup$
    However, the Jacobian matrix says that we should derive x with respect to u, x wrt v, y wrt u, and y wrt v. Therefore, it is not deriving u wrt to x and y, but it's the other way around.
    $endgroup$
    – user205891
    2 days ago






  • 1




    $begingroup$
    There is a trick that you can use if you want: the Jacobian determinant of the inverse map is the inverse (reciprocal) of the Jacobian determinant of the forward map. Therefore, you can just do it the way I described, get the Jacobian determinant of that, and then take the reciprocal. (This will get you something in terms of $x$ and $y$, which you will need to express in terms of $u$ and $v$ for the integral.)
    $endgroup$
    – Minus One-Twelfth
    2 days ago
















$begingroup$
When finding $fracpartial upartial x$ and $fracpartial upartial y$, just differentiate $u$ as a function of $(x,y)$, ignore the $v$. Similarly, when differentiating $v$, don't worry about $u$. So for example, $fracpartial upartial x = 2x$.
$endgroup$
– Minus One-Twelfth
2 days ago





$begingroup$
When finding $fracpartial upartial x$ and $fracpartial upartial y$, just differentiate $u$ as a function of $(x,y)$, ignore the $v$. Similarly, when differentiating $v$, don't worry about $u$. So for example, $fracpartial upartial x = 2x$.
$endgroup$
– Minus One-Twelfth
2 days ago













$begingroup$
However, the Jacobian matrix says that we should derive x with respect to u, x wrt v, y wrt u, and y wrt v. Therefore, it is not deriving u wrt to x and y, but it's the other way around.
$endgroup$
– user205891
2 days ago




$begingroup$
However, the Jacobian matrix says that we should derive x with respect to u, x wrt v, y wrt u, and y wrt v. Therefore, it is not deriving u wrt to x and y, but it's the other way around.
$endgroup$
– user205891
2 days ago




1




1




$begingroup$
There is a trick that you can use if you want: the Jacobian determinant of the inverse map is the inverse (reciprocal) of the Jacobian determinant of the forward map. Therefore, you can just do it the way I described, get the Jacobian determinant of that, and then take the reciprocal. (This will get you something in terms of $x$ and $y$, which you will need to express in terms of $u$ and $v$ for the integral.)
$endgroup$
– Minus One-Twelfth
2 days ago





$begingroup$
There is a trick that you can use if you want: the Jacobian determinant of the inverse map is the inverse (reciprocal) of the Jacobian determinant of the forward map. Therefore, you can just do it the way I described, get the Jacobian determinant of that, and then take the reciprocal. (This will get you something in terms of $x$ and $y$, which you will need to express in terms of $u$ and $v$ for the integral.)
$endgroup$
– Minus One-Twelfth
2 days ago











1 Answer
1






active

oldest

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2












$begingroup$

It is always the old, original, variables with respect the new ones...but you can do it the other way around and take the inverse matrix's determinant!:



$$J=Jfrac(x,y)(u,v)=beginvmatrixfracpartial upartial x&fracpartial upartial y\fracpartial vpartial x&fracpartial vpartial yendvmatrix^-1=beginvmatrix2x&-2y\2y&2xendvmatrix^-1=frac14(x^2+y^2)$$



We took the inverse determinant since we differentiated the new variables wrt the original ones...and there exists that relation.






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    2












    $begingroup$

    It is always the old, original, variables with respect the new ones...but you can do it the other way around and take the inverse matrix's determinant!:



    $$J=Jfrac(x,y)(u,v)=beginvmatrixfracpartial upartial x&fracpartial upartial y\fracpartial vpartial x&fracpartial vpartial yendvmatrix^-1=beginvmatrix2x&-2y\2y&2xendvmatrix^-1=frac14(x^2+y^2)$$



    We took the inverse determinant since we differentiated the new variables wrt the original ones...and there exists that relation.






    share|cite|improve this answer









    $endgroup$

















      2












      $begingroup$

      It is always the old, original, variables with respect the new ones...but you can do it the other way around and take the inverse matrix's determinant!:



      $$J=Jfrac(x,y)(u,v)=beginvmatrixfracpartial upartial x&fracpartial upartial y\fracpartial vpartial x&fracpartial vpartial yendvmatrix^-1=beginvmatrix2x&-2y\2y&2xendvmatrix^-1=frac14(x^2+y^2)$$



      We took the inverse determinant since we differentiated the new variables wrt the original ones...and there exists that relation.






      share|cite|improve this answer









      $endgroup$















        2












        2








        2





        $begingroup$

        It is always the old, original, variables with respect the new ones...but you can do it the other way around and take the inverse matrix's determinant!:



        $$J=Jfrac(x,y)(u,v)=beginvmatrixfracpartial upartial x&fracpartial upartial y\fracpartial vpartial x&fracpartial vpartial yendvmatrix^-1=beginvmatrix2x&-2y\2y&2xendvmatrix^-1=frac14(x^2+y^2)$$



        We took the inverse determinant since we differentiated the new variables wrt the original ones...and there exists that relation.






        share|cite|improve this answer









        $endgroup$



        It is always the old, original, variables with respect the new ones...but you can do it the other way around and take the inverse matrix's determinant!:



        $$J=Jfrac(x,y)(u,v)=beginvmatrixfracpartial upartial x&fracpartial upartial y\fracpartial vpartial x&fracpartial vpartial yendvmatrix^-1=beginvmatrix2x&-2y\2y&2xendvmatrix^-1=frac14(x^2+y^2)$$



        We took the inverse determinant since we differentiated the new variables wrt the original ones...and there exists that relation.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        DonAntonioDonAntonio

        179k1494233




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