Why is a surface in the $R^3$ approximated by a quadratic equation?

Question

On the above, if there is a tangent plane, then is it best to approximate the surface $S$ by a linear equation? Why and how does one approximate the equation for $S$ by a quadratic one $Z=1/2(AX^2+BY^2)$? I cannot understand the motivations...

Answer 1

Take a surface $S\subset{\mathbb R}^3$ and a point ${\bf p}\in S$. Then you can consider he tangent plane $T$ at ${\bf p}$ as $(x,y)$-plane and the surface normal at ${\bf p}$ as $z$-axis. In this way in the neighborhood of ${\bf p}$ the surface appears as graph $z=f(x,y)$ for a certain function $f$. Things are set up in such a way that $$f(0,0)=0\>, \quad f_x(0,0)=0\>, f_y(0,0)=0\ .$$ These equations simply express that the point ${\bf p}$ has coordinates $(0,0,0)$ and that $T$ is the tangent plane to $S$ at ${\bf p}$. If we look at the Taylor expansion of $f$ at $(0,0)$ it therefore has the form $$f(x,y)=ax^2 + 2bxy + cy^2+\ {\rm higher\ terms}\ .\tag{1}$$ Linear algebra then tells us that we can rotate the coordinate axes in the $(x,y)$-plane such that in $(1)$ no mixed term $2bxy$ appears. As a consequence the second Taylor approximation of $f$ can be written (in the new coordinates) as $$z=Ax^2+By^2$$ with certain real constants $A$, $B$. Now that these preliminaries are out of the way the interesting differential geometry begins $\ldots$