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Given a surface $ F: D \in R^2\longrightarrow S \in R^3 $ with smooth parameteric representation: $F(u,v) = (x(u,v),y(u,v), z(u,v)) $ .

Denote by $N = F_u \times F_v $ , how can one prove that $N$ at $p$ is orthogonal to any curve which lies in the surface and passes through p?

I never saw such a proof, and I can't find it in Calculus books... (it bother me because always people refer to $N$ as the normal, but never saw such a proof)

Thanks

homogenity
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1 Answers1

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Let $\gamma : I \to S$ be such a curve, so that $\gamma(t) = F(u(t),v(t))$ for some functions $(u,v) : I \to D$. Then $\gamma'(t) = F_u u'(t) + F_v v'(t)$ by the chain rule. Since $N$ is orthogonal to $F_u$ and $F_v$, it is orthogonal to everything in their span, and in particular to $\gamma'(t)$.

fuglede
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  • I think what the OP needs in fact is the following easy theorem in three-dimensional analytic geometry: a straight line $;l;$ intersecting a plane $;\pi;$ is perpendicular to it (i.e., $;l\perp\pi;$) iff $;l;$ is perpendicular to two straight lines contained in $;\pi;$ passing through the line's foot at $;\pi;$ , with foot = the intersection point of $;l,,,\pi;$ . With this we're done since $;N\perp F_u;,;N\perp F_v;$ . – DonAntonio Dec 28 '13 at 11:39
  • Hi, thanks a lot ! A couple of questions, if I may:

    1. Why does if $\gamma$ lies on our surface, then $\gamma(t)=F(u(t),v(t)) $ for some functions $u,v$?

    2. I'm a little confused about the notations-

    $\gamma ' (t)$ is the tangent vector to $\gamma(t)$ .. But, since $\gamma(t)$ is a path, $\gamma'(t) $ must also be a path.. Why are we looking at it as a vector?

    Thanks !

     – homogenity Dec 28 '13 at 16:56
  • @homogenity: 1) By definition, the surface is the image of $F$, so every point $\gamma(t) \in S$ is of the form $F(u(t),v(t))$, for some $(u(t),v(t)) \in D$, and taking all of the values of $t$ together, we may view these $u(t)$ and $v(t)$ as functions on $I$ in their own. 2) Right, $\gamma'(t)$ you could view as a $t$-dependent path of tangent vectors; we're only really interested in $\gamma'(t)$ at the value of $t$ for which $\gamma(t) = p$. – fuglede Jan 02 '14 at 09:40
  • Great !!! Thanks ! – homogenity Jan 03 '14 at 10:56