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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

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William Elliot

Posts: 1,238
Registered: 1/8/12
Derivative of a Vector Field
Posted: Jan 18, 2013 3:48 AM
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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 arc-length.
>
> 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,




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