Analytic Surfaces

$S (u, v)$
$S' (u, v) = (\frac{∂S}{∂u}, \frac{∂S}{∂v}, \frac{∂S}{∂u} \times \frac{∂S}{∂v})$

Perhaps the most important representation is the basic derivative.

$\frac{df}{dX} = (\frac{∂f}{∂x}, \frac{∂f}{∂y}, \frac{∂f}{∂z})$

Example

$z = g (x, y)$
$f = - g (x, y) + z = 0$
$\frac{df}{dX} = (- \frac{\partial g}{∂x}, - \frac{\partial g}{∂y}, 1)$

Alternatively using vector calculus.

$f = (x, y, g (x, y))$
$f_{x} = (1, 0, g_{x})$
$f_{y} = (0, 1, g_{y})$
$N = f_{x} \times f_{y}$

$N = |\begin{matrix} i & j & k \\ 1 & 0 & g_{x} \\ 0 & 1 & g_{y} \end{matrix}| = (- g_{x}, - g_{y}, 1)$

What just happened?

In maths there are mysteries. You may think that you have seen it all but no. The surface was completely algebraically described by z and g. By rearanging the equation to zero and differentiating I got the surface normal. Zero is often another dimension. I thought that a derivative of zero is always zero, hence it will generate a normal.