surface normalSurfaceNormal2007-07-08 13:09:352008-06-13 16:29:01Definitiontangent plane (elementary)parametre planeparameter planeparametre curveparameter curveGaussian coordinates% this is the default PlanetMath preamble. as your knowledge
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Let $S$ be a smooth surface in $\mathbb{R}^3$. The {\em surface normal} of $S$ at a point $P$ of $S$ is the line passing through $P$ and perpendicular to the tangent plane $\tau$ of $S$ at the point $P$, i.e. perpendicular to all lines in $\tau$.
If the surface $S$ is given in a parametric form
$$x = x(u,\,v),\quad y = y(u,\,v),\quad z = z(u,\,v),$$
it is useful to interpret the parameters $u$ and $v$ as the rectangular coordinates of a point in a plane, the so-called {\em parameter plane}. We can consider on $S$ the so-called {\em parameter curves}, namely the $u$-{\em curves} which correspond the lines parallel to the $u$-axis and the $v$-{\em curves} which correspond the lines parallel to the $v$-axis in the parameter plane. One $u$-curve and one $v$-curve passes through every point on the surface (the values of $u$ and $v$ in a point of $S$ are the {\em Gaussian coordinates} of this point). The surface normal at any point of $S$ is perpendicular to both parameter curves, and thus its direction cosines $a$, $b$, $c$ satisfy the equations
\begin{align*}
\begin{cases}
\displaystyle{a\frac{\partial x}{\partial u}+b\frac{\partial y}{\partial u}+c\frac{\partial z}{\partial u}= 0},\\
\\
\displaystyle{a\frac{\partial x}{\partial v}+b\frac{\partial y}{\partial v}+c\frac{\partial z}{\partial v}= 0.}
\end{cases}
\end{align*}
This homogeneous pair of linear equations determines the ratio of the direction cosines
$$a:b:c = \frac{\partial(y,\,z)}{\partial(u,\,v)}:\frac{\partial(z,\,x)}{\partial(u,\,v)}:\frac{\partial(x,\,y)}{\partial(u,\,v)}$$
via the Jacobians.\\
\textbf{Example.} Determine the direction cosines of the normal of the helicoid
$$x = u\cos{v},\quad y = u\sin{v},\quad z = cv.$$
We have the Jacobians
$$\left| \begin{matrix}
\frac{\partial y}{\partial u} & \frac{\partial z}{\partial u}\\
\frac{\partial y}{\partial v} & \frac{\partial z}{\partial v}
\end{matrix}\right| =
\left|\begin{matrix}
\sin{v} & 0 \\
u\cos{v} & c
\end{matrix}\right| = c\sin{v},\;\;
\left|\begin{matrix}
\frac{\partial z}{\partial u} & \frac{\partial x}{\partial u}\\
\frac{\partial z}{\partial v} & \frac{\partial x}{\partial v}
\end{matrix}\right| =
\left|\begin{matrix}0 & \cos{v} \\
c & -u\sin{v}\end{matrix}\right| = -c\cos{v},\;\;
\left|\begin{matrix}
\frac{\partial x}{\partial u} & \frac{\partial y}{\partial u}\\
\frac{\partial x}{\partial v} & \frac{\partial y}{\partial v}
\end{matrix}\right| =
\left|\begin{matrix}\cos{v} & \sin{v}\\
-u\sin{v} & u\cos{v}\end{matrix}\right| = u.$$
These are the components of the normal vector of the helicoid surface in the point with the Gaussian coordinates $u$ and $v$.\, The length of the vector is\, $\sqrt{(c\sin{v})^2+(-c\cos{v})^2+u^2} = \sqrt{u^2+c^2}$.\, If we \PMlinkname{divide}{Division} the vector by its length, we obtain a unit vector, the components of which are the direction cosines of the surface normal:
$$\frac{c\sin{v}}{\sqrt{u^2+c^2}},\;\;-\frac{c\cos{v}}{\sqrt{u^2+c^2}},\;\;\frac{u}{\sqrt{u^2+c^2}}.$$