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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



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 arclength. > > 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, > > please? > 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)).



