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MathGroup Archive 2005

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Re: Normal Disappear Problem

  • To: mathgroup at
  • Subject: [mg55335] Re: [mg55302] Normal Disappear Problem
  • From: Daniel Lichtblau <danl at>
  • Date: Sat, 19 Mar 2005 04:46:17 -0500 (EST)
  • References: <>
  • Sender: owner-wri-mathgroup at

gouqizi.lvcha at wrote:
> Hi, All:
> I have the following parametric equation for an unit sphere:
> x = cos(u)sin(v)
> y = sin(u)sin(v)
> z = cos(v)
> 0<=u<2*Pi ; 0<=v<=Pi
> Then I use
> normal = (Dx/Du, Dy/Du, Dz/Du) CROSS (Dx/Dv, Dy/Dv, Dz/Dv) to get the
> normal vector.
> I get the follwoing after calculation (with normalization):
> normal =  [sin(v) ^2 cos(u), sin(v)^2  sin(u), cos(u)^2  cos(v) sin(v)
> +  sin(u)^2  cos(v) sin(v)]
> Now when u=0, v=0 , Normal = (0,0,0)! How can it be? We know the fact
> that a sphere should have normal everywhere.
> Rick

Your tangent vectors are:

In[2]:= {x,y,z} = {Cos[u]*Sin[v], Sin[u]*Sin[v], Cos[v]};

In[3]:= tanvex = {{D[x,u],D[y,u],D[z,u]},{D[x,v],D[y,v],D[z,v]}}
Out[3]= {{-(Sin[u] Sin[v]), Cos[u] Sin[v], 0},
  {Cos[u] Cos[v], Cos[v] Sin[u], -Sin[v]}}

The theorem indicates each must vanish somewhere, hence their cross 
product must vanish.

Upshot: the sphere has everywhere a unit normal, but it cannot be 
obtained as a cross product from an everywhere smooth tangential frame.

Daniel Lichtblau
Wolfram Research

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