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Topic:
How to Define Derivative of a Vector Field in this Case ( Curve in R^3)?
Replies:
3
Last Post:
Jan 18, 2013 4:30 AM




Derivative of a Vector Field
Posted:
Jan 18, 2013 3:48 AM


On Thu, 17 Jan 2013, cku wrote: > Let C:I>R^3 be a smooth curve, and let Z(s) be a vector field along > the curve, parametrized by arclength. > > We define the derivative of a vector field Z along the curve to be the quotient: > > Lim_ds>0 [Z(s+ds)Z(s)]/ds > > Now, I don't know how to make sense of the difference in the numerator:
Z(s + ds) and Z(s) are two vectors in R^3, as is their difference.
For example, the tangent vectors to C. Then it seems, the limit would be the radius of curvature of C at s from Z(s) to the center of curvature.
> The two vectors Z(s+ds) and Z(s) , are in different tangent spaces  > tangent space at s+ds and s respectively  and , AFAIK, the difference > of vectors in different tangent spaces is not defined, except for cases > where there is a natural isomorphism between the tangent spaces, as in > the case where the tangent spaces are those in R^n itself. Any > suggestions,



