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Solution: | Solution: | ||
Note that | Note that | ||
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| − | </math> | + | <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> | ||
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| + | <math>y=u*\sin v \longrightarrow \frac{\partial y}{\partial u}= \sin v \; \frac{\partial y}{\partial v} = u*\cos v</math> | ||
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| + | </font> | ||
Therefore the Jacobian matrix is | Therefore the Jacobian matrix is | ||
Revision as of 09:32, 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 $
$ 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
