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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
$begingroup$
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?
calculus multivariable-calculus partial-derivative jacobian
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
user205891user205891
112
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.
$endgroup$
add a comment |
$begingroup$
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?
calculus multivariable-calculus partial-derivative jacobian
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
user205891user205891
112
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
add a comment |
$begingroup$
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?
calculus multivariable-calculus partial-derivative jacobian
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
user205891user205891
112
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
calculus multivariable-calculus partial-derivative jacobian
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
user205891user205891
112
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.
edited 2 days ago
Rodrigo de Azevedo
13k41960
asked 2 days ago
user205891user205891
112
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.
edited 2 days ago
Rodrigo de Azevedo
13k41960
edited 2 days ago
Rodrigo de Azevedo
13k41960
edited 2 days ago
Rodrigo de Azevedo
13k41960
13k41960
asked 2 days ago
user205891user205891
112
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.
asked 2 days ago
user205891user205891
112
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.
Check out our Code of Conduct.
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered 2 days ago
DonAntonioDonAntonio
179k1494233
$endgroup$
add a comment |
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1 Answer
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$begingroup$
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.
answered 2 days ago
DonAntonioDonAntonio
179k1494233
$endgroup$
add a comment |
$begingroup$
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.
answered 2 days ago
DonAntonioDonAntonio
179k1494233
$endgroup$
add a comment |
$begingroup$
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.
answered 2 days ago
DonAntonioDonAntonio
179k1494233
$endgroup$
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.
answered 2 days ago
DonAntonioDonAntonio
179k1494233
answered 2 days ago
DonAntonioDonAntonio
179k1494233
answered 2 days ago
DonAntonioDonAntonio
179k1494233
answered 2 days ago
DonAntonioDonAntonio
179k1494233
179k1494233
add a comment |
add a comment |
user205891 is a new contributor. Be nice, and check out our Code of Conduct.
user205891 is a new contributor. Be nice, and check out our Code of Conduct.
user205891 is a new contributor. Be nice, and check out our Code of Conduct.
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egDgQo,fDChjIzKjfoy9,O9CNZzBd,IUHyhv 4ZInf1SB GGPAOXm,7hGko7AtN6,EvtlZItqcdE5rbLGXRw,0quW29DHEqh6Z0iaP05 Tb,r
$begingroup$
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