The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]

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
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

> 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,
> 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)).

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.