The Jacobian and area differential

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SUMMARY

The discussion centers on the definition of differentiability in the context of coordinate transformations, specifically the notation used for vector norms. The transformation T(u) is defined in terms of vectors u and p, with the total derivative represented by the matrix J(p). The notation ||x|| indicates the vector norm, which is calculated as the square root of the sum of the squares of its components, specifically for two-dimensional vectors as ||x|| = √(x² + y²).

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  • Understanding of vector notation and operations
  • Familiarity with the concept of differentiability in calculus
  • Knowledge of matrix representation of derivatives
  • Basic understanding of norms in vector spaces
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  • Investigate the Jacobian matrix and its role in transformations
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Mathematicians, students of multivariable calculus, and anyone studying coordinate transformations and their differentiability properties will benefit from this discussion.

WMDhamnekar
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I don't understand the following definition. If we let $u=\langle u,v \rangle$ , $p=\langle p,q\rangle,$ $x=\langle x,y \rangle$,then (x,y)=T(u,v) is given in vector notation by

x=T(u). A coordinate transformation T(u) is differentiable at a point p , if there exists a matrix J(p) for which $\lim_{u\to p}\frac {||T(u)-T(p)-J(p)(u-p)||}{||u-p||}=0.$when it exists, J(p) is the total derivative of T(u). What does this symbol || || indicate?
 
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Dhamnekar Winod said:
I don't understand the following definition. If we let $u=\langle u,v \rangle$ , $p=\langle p,q\rangle,$ $x=\langle x,y \rangle$,then (x,y)=T(u,v) is given in vector notation by

x=T(u). A coordinate transformation T(u) is differentiable at a point p , if there exists a matrix J(p) for which $\lim_{u\to p}\frac {||T(u)-T(p)-J(p)(u-p)||}{||u-p||}=0.$when it exists, J(p) is the total derivative of T(u). What does this symbol || || indicate?


Hi Dhamnekar Winod,

The notation $\|\mathbf x\|$ means the vector norm of $\mathbf x$.
If nothing else is specified it is the usual length of the vector and:
$$\|\mathbf x\| = \sqrt{x^2+y^2}$$
 

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