Re: jacobian
- From: smn <smnewberger@xxxxxxxxxxx>
- Date: 18 May 2007 00:46:33 -0700
On May 17, 11:18 pm, holde...@xxxxxxxxx wrote:
Suppose u = F(x,y) and v = G(x,y), and J = d(f,g)/d(u,v) (i.e.,
determinant of jacobian of (f(u,v), g(u,v)).
Show dF/dx = (1/J) * dy / dv [The 'd' s are all "partial"
symbols.]
There are a few problems like this for me to do. Can someone show me
how to do this one, so I can (hopefully) do the others. I don't see
how 1/J gets there. I don't see how this works--is it some sort of
chain rule?
The 2x2 matrix j at u,v whose determinant is J has an inverse matrix
which is just the Jacobian matrix of (F,G) (the inverse of the map
(f,g) ) at the corresponding x,y whose entries are dF/dx,dG/dx (first
column) dF/dy,dG/dy (2nd column).The entries of j are df/dx ... so you
just need the formula for the inverse of a 2x2 matrix , or for its
entries ,often called Cramers rule .
Regards,smn
.
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- From: holdenf2
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