Compute Unit Normal Vector to Surface: Difference b/w N & n

[Lee33]

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?

 

[SteamKing]

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.

 

[Lee33]

I don't have a calculus text. This class is differential geometry.

But will it just be $n = \frac{x_u(u_0,v_0) \times x_v(u_0,v_0)}{||x_u(u_0,v_0) \times x_v(u_0,v_0)||}?$

 

[vanhees71]

That's correct. The tangent vectors of the coordinate lines are \partial_u \vec{x} and \partial_v \vec{x} if \vec{x}(u,v) 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
\mathrm{d}^2 \vec{F}=\mathrm{d} u \, \mathrm{d} v \; \partial_{u} \vec{x} \times \partial_{v} \vec{x}.

You can simplify this a bit for your case, where
\vec{x}=u \vec{e}_x + v \vec{e}_y + h(u,v) \vec{e}_z.

 

[Lee33]

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?

 

[vanhees71]

I get the opposite sign.

 

[Lee33]

Thank you!