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Unit normal vector

  • Thread starter Lee33
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I have seen different equations for computing the unit normal vector to a surface. What is the difference between the normal vector ##N## of a curve and ##n##. I have seen this formula for a unit normal vector: ##n = \frac{x_1 \times x_2}{|x_1 \times x_2|}##. Which is different from ##N##. For example,

Let ##f(u,v)=(u,v,h(u,v))## be a parametrization of the graph ##T_h## of ##h:\mathbb{R}^2\to \mathbb{R}##. Compute a unit normal vector to ##T_h##.

How can I compute this unit normal vector?
 
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  • #2
SteamKing
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Use the gradient of f(u,v) to determine N. Then divide N by its magnitude to get n. Check your calculus text for a discussion of the gradient and how it is related to the normal vector.
 
  • #3
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I don't have a calculus text. This class is differential geometry.

But will it just be ##n = \frac{x_u(u_0,v_0) \times x_v(u_0,v_0)}{||x_u(u_0,v_0) \times x_v(u_0,v_0)||}?##
 
  • #4
vanhees71
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That's correct. The tangent vectors of the coordinate lines are [itex]\partial_u \vec{x}[/itex] and [itex]\partial_v \vec{x}[/itex] if [itex]\vec{x}(u,v)[/itex] is the parametrization of your surface. The surface normal vector is given by their cross product, and then you normalize it.

BTS: the surface-element vectors are
[tex]\mathrm{d}^2 \vec{F}=\mathrm{d} u \, \mathrm{d} v \; \partial_{u} \vec{x} \times \partial_{v} \vec{x}.[/tex]

You can simplify this a bit for your case, where
[tex]\vec{x}=u \vec{e}_x + v \vec{e}_y + h(u,v) \vec{e}_z.[/tex]
 
  • #5
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Thanks vanhees71.

I got my normal vector to be ##\frac{\langle f_x , f_y, -1 \rangle}{\sqrt{f_x^2+f_y^2+1}}.## Is that correct?
 
  • #6
vanhees71
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I get the opposite sign.
 
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  • #7
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Thank you!
 

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