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===Basic Definition===
 
===Basic Definition===
  
The Jacobian Matrix is just a matrix that takes the partial derivatives of each element of a function (which is in the form of a vector. Let F be a function such that
+
The Jacobian Matrix is just a matrix that takes the partial derivatives of each element of a transformation (which is in the form of a vector. Let T be a transformation such that
  
<font size=4><math>F(u,v)=<x,y> </math></font>
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<font size=4><math>T(u,v)=<x,y> </math></font>
  
 
then the Jacobian matrix of this function would look like this:
 
then the Jacobian matrix of this function would look like this:
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\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}  \end{bmatrix}</math>
 
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}  \end{bmatrix}</math>
  
To help illustrate this, let's do an example:  
+
In general, the Jacobian Matrix of a Transformation F, looks like this:
  
Example #1: Let's take the Transformation:
+
[[Image:Example.jpg]]
  
<font size=4><math>T(u,v) = <u * \cos v,r * \sin v> </math> </font>.
+
To help illustrate this, let's do an example:
 +
 
 +
===Example #1:=== Let's take the Transformation:      <font size=4><math>T(u,v) = <u * \cos v,r * \sin v> </math> </font>.
  
 
What would be the Jacobian Matrix of this Transformation?  
 
What would be the Jacobian Matrix of this Transformation?  
  
Solution:
+
===Solution:===
Note that
+
 
+
  
 
<font size = 4>
 
<font size = 4>
<math>x=u*\cos v \longrightarrow \frac{\partial x}{\partial u}= \cos v \; \frac{\partial x}{\partial v} = -u*\sin v</math>
+
<math>x=u*\cos v \longrightarrow \frac{\partial x}{\partial u}= \cos v , \; \frac{\partial x}{\partial v} = -u*\sin v</math>
  
<math>y=u*\sin v \longrightarrow \frac{\partial y}{\partial u}= \sin v \; \frac{\partial y}{\partial v} = u*\cos v</math>
+
<math>y=u*\sin v \longrightarrow \frac{\partial y}{\partial u}= \sin v , \; \frac{\partial y}{\partial v} = u*\cos v</math>
  
 
</font>
 
</font>
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</math>
 
</math>
  
Now after doing
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This example actually showcased the transformation "T" as the change from polar coordinates into Cartesian coordinates.
 +
 
 +
Let's do another example.
 +
 
 +
===Example #2:=== Let
 +
 
 +
 
 +
==Application: Jacaobian Determinants==
 +
 
 +
The determinant of Example #1 gives:
 +
 
 +
<font size=5> <math> \left|\begin{matrix}
 +
\cos v & -u * \sin v \\
 +
\sin v &  u * \cos v \end{matrix}\right|=~~ u \cos^2 v + u \sin^2 v =~~ u </math></font>
 +
 
 +
Notice that, in an integral when changing from cartesian coordinates (dxdy) to polar coordinates <math> (drd\theta)</math>, the equation is as such:
 +
 
 +
<font size=4><math> dxdy=r*drd\theta </math></font>
 +
 
 +
in this case, since <math>u =r</math> and <math> v = \theta</math>, then
 +
 
 +
<font size=4><math> dxdy=u*dudv </math></font>
 +
 
 +
It is easy to extrapolate, then, that the transformation from one set of coordinates to another set is merely
 +
 
 +
<font size=4><math> dC1=det(J(T))dC2 </math></font>
 +
 
 +
where C1 is the first set of coordinates, det(J(C1)) is the determinant of the Jacobian matrix made from the Transformation T, T is the Transformation from C1 to C2 and C2 is the second set of coordinates.
 +
 
 +
It is important to notice several aspects: first, the determinant is assumed to exist and be non-zero, and therefore the Jacobian matrix must be square and invertible.
 +
 
 +
For integrals, changing variables is quite useful. The most obvious case is that of u-substitution. However, for larger dimensions, this gets slightly trickier. Suppose we wanted to change
  
  
  
 
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Revision as of 09:52, 8 May 2013


Jacobians and their applications

by Joseph Ruan


Basic Definition

The Jacobian Matrix is just a matrix that takes the partial derivatives of each element of a transformation (which is in the form of a vector. Let T be a transformation such that

$ T(u,v)=<x,y> $

then the Jacobian matrix of this function would look like this:

$ J(u,v)=\begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} $

In general, the Jacobian Matrix of a Transformation F, looks like this:

Example.jpg

To help illustrate this, let's do an example:

===Example #1:=== Let's take the Transformation: $ T(u,v) = <u * \cos v,r * \sin v> $ .

What would be the Jacobian Matrix of this Transformation?

Solution:

$ x=u*\cos v \longrightarrow \frac{\partial x}{\partial u}= \cos v , \; \frac{\partial x}{\partial v} = -u*\sin v $

$ y=u*\sin v \longrightarrow \frac{\partial y}{\partial u}= \sin v , \; \frac{\partial y}{\partial v} = u*\cos v $

Therefore the Jacobian matrix is

$ \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix}= \begin{bmatrix} \cos v & -u*\sin v \\ \sin v & u*\cos v \end{bmatrix} $

This example actually showcased the transformation "T" as the change from polar coordinates into Cartesian coordinates.

Let's do another example.

===Example #2:=== Let


Application: Jacaobian Determinants

The determinant of Example #1 gives:

$ \left|\begin{matrix} \cos v & -u * \sin v \\ \sin v & u * \cos v \end{matrix}\right|=~~ u \cos^2 v + u \sin^2 v =~~ u $

Notice that, in an integral when changing from cartesian coordinates (dxdy) to polar coordinates $ (drd\theta) $, the equation is as such:

$ dxdy=r*drd\theta $

in this case, since $ u =r $ and $ v = \theta $, then

$ dxdy=u*dudv $

It is easy to extrapolate, then, that the transformation from one set of coordinates to another set is merely

$ dC1=det(J(T))dC2 $

where C1 is the first set of coordinates, det(J(C1)) is the determinant of the Jacobian matrix made from the Transformation T, T is the Transformation from C1 to C2 and C2 is the second set of coordinates.

It is important to notice several aspects: first, the determinant is assumed to exist and be non-zero, and therefore the Jacobian matrix must be square and invertible.

For integrals, changing variables is quite useful. The most obvious case is that of u-substitution. However, for larger dimensions, this gets slightly trickier. Suppose we wanted to change



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