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Re: Urgent Help needed

  • To: mathgroup at
  • Subject: [mg20523] Re: [mg20390] Urgent Help needed
  • From: "Richard I. Pelletier" <bitbucket at>
  • Date: Sat, 30 Oct 1999 00:13:46 -0400
  • Organization: @Home Network
  • References: <7v3bei$>
  • Sender: owner-wri-mathgroup at

> 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   vtsyrlin at
> > ******************************************************

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

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

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.

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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