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- Newsgroups: sci.math
- Path: sparky!uunet!stanford.edu!ames!news.hawaii.edu!uhunix.uhcc.Hawaii.Edu!lady
- From: lady@uhunix.uhcc.Hawaii.Edu (Lee Lady)
- Subject: Re: Curvature of a Line in Space
- Message-ID: <1992Nov13.062155.2710@news.Hawaii.Edu>
- Followup-To: sci.math
- Summary: The curvature is the rate at which is tangent vector is turning.
- Keywords: curvature
- Sender: root@news.Hawaii.Edu (News Service)
- Nntp-Posting-Host: uhunix.uhcc.hawaii.edu
- Organization: University of Hawaii (Mathematics Dept)
- References: <1992Oct28.210821.2790@TorreyPinesCA.ncr.com> <1992Nov3.002252.8053@shell.shell.com> <1992Nov5.014717.9834@nas.nasa.gov>
- Date: Fri, 13 Nov 1992 06:21:55 GMT
- Lines: 32
-
- In article <1992Nov5.014717.9834@nas.nasa.gov> asimov@wk223.nas.nasa.gov (Daniel A. Asimov) writes:
- > .... The curvature vector can be defined as the second
- >derivative of the position vector *with respect to arclength*.
- >Then, the magnitude of this vector is the curvature.
- >With respect to an arbitrary parametrization, you will not in general
- >get the same thing.
-
- More intuitively, the curvature is the rate at which the tangent vector
- is turning as one moves along the curve at a speed of 1. In other
- words, if we take the tangent vector at two values s and s+h of the
- arc length and denote the angle between them by delta_theta, then the
- curvature is the limit of delta_theta/h as h goes to 0.
-
- For a curve in the plane, this is easy to see. In that case, we can let
- theta denote the angle between the unit tangent vector and the
- horizontal. (The unit tangent vector is the first derivative of the
- position vector with respect to arc lenth.) Since the unit tangent
- vector has length 1 it then has components (cos theta, sin theta).
- Differentiating with respect to s using the chain rule shows that the
- magnitude of its derivative equals the derivative of theta, i.e. the
- rate at which the tangent vector is turning.
-
- I believe that you can adapt this proof for a curve in three space by
- projecting the curve onto its osculating place at the point in question.
- Not being a differential geometer, though, I'm not prepared to offer
- details of a proof.
-
- --
- It is a poor sort of skepticism which merely delights in challenging
- those claims which conflict with one's own belief system.
- --Bogus quote
- lady@uhunix.uhcc.hawaii.edu lady@uhunix.bitnet
-