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Re: Urgent Help needed
*To*: mathgroup at smc.vnet.net
*Subject*: [mg20523] Re: [mg20390] Urgent Help needed
*From*: "Richard I. Pelletier" <bitbucket at home.com>
*Date*: Sat, 30 Oct 1999 00:13:46 -0400
*Organization*: @Home Network
*References*: <7v3bei$5t8@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
> Vladimir Tsyrlin wrote:
> >
> > Given the implcit form of a curve, i.e. F(x,y,z) = 0, do you know how to
> > find the curvature of F at a point in 3D space? All the references I have
> > assume F is in parametric form and take the standard differential geometry
> > approach.
> >
> > --
> > **************************************************
> > *************Vladimir Tsyrlin *******************
> > vtsyrlin at cs.rmit.edu.au vtsyrlin at ozemail.com.au
> > ******************************************************
I have seen one post suggesting that you _meant_ the curvature of a
curve in 3-space, and therefore, need 2 equations. Let me assume you
_meant_ one equation, and therefore, the curvatures of a surface in
3-space.
And rather than try to show you the gory details, let me give you the
name of the answer, and one reference.
What you want is called _the shape operator_. It is the covariant
derivative of a unit normal along a tangent to the surface. The nice
thing about the equation F(x,y,z)=0 is that the gradient of F _is_
normal to the surface, so you just need to compute a covariant
derivative.
Once you have the shape operator, in this case a 2x2 matrix, the
principal curvatures are its eigenvalues, so the mean and Gaussian
curvatures are their average and product resp.
Chapter V of Barrett O'Neill Elementary Differential Geometry (1966
ed.) is entitled _Shape Operators_, and pp. 216-219 work out the
details for precisely this problem.
It does seem very common to focus on the case where you have a
parametric representation, but you don't need one.
Vale,
Rip
--
Multiplication is not commutative before breakfast.
Richard I. Pelletier
NB eddress: r i p 1 [at] h o m e [dot] c o m
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