Line 35: Line 35:
  
 
x=u*\cos v \longrightarrow \frac{\partial x}{\partial u}= \cos v \; \frac{\partial x}{\partial v} = -u*\sin v
 
x=u*\cos v \longrightarrow \frac{\partial x}{\partial u}= \cos v \; \frac{\partial x}{\partial v} = -u*\sin v
 +
\smallskip
 
y=u*\sin v \longrightarrow \frac{\partial y}{\partial u}= \sin v \; \frac{\partial y}{\partial v} = u*\cos v
 
y=u*\sin v \longrightarrow \frac{\partial y}{\partial u}= \sin v \; \frac{\partial y}{\partial v} = u*\cos v
  

Revision as of 08:28, 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 function (which is in the form of a vector. Let F be a function such that

$ F(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} $

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: Note that $ x=u*\cos v \longrightarrow \frac{\partial x}{\partial u}= \cos v \; \frac{\partial x}{\partial v} = -u*\sin v \smallskip 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} $

Now after doing



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