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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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 trj Posts: 95 From: Korea Registered: 2/23/12
Re: How to Define Derivative of a Vector Field in this Case ( Curve in R^3)?
Posted: Jan 18, 2013 1:04 AM

> Hi, All:
>
> Let C:I-->R^3 be a smooth curve, and let Z(s) be a
> a vector field along the curve,
> parametrized by arc-length.
>
> We define the derivative of a vector field Z along
> g 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
> he difference in the numerator:
>
> The two vectors Z(s+ds) and Z(s) , are in different
> t tangent spaces --
>
> tangent space at s+ds and s respectively -- and ,
> , AFAIK, the difference
>
> of vectors in different tangent spaces is not
> t defined, except for cases
>
> where there is a natural isomorphism between the
> e tangent spaces, as in the
>
> case where the tangent spaces are those in R^n
> n itself. Any suggestions,
>
>

That is not quite how you are supposed to do it.
Instead you express Z using Euclidean coordinate functions:

Z(s) = (z1(s),z2(s),z3(s))

The derivative of Z is now the vector field

Z'(s) = (z1'(s),z2'(s),z3'(s)).

Date Subject Author
1/18/13 Bacle H
1/18/13 trj
1/18/13 William Elliot
1/18/13 trj