For a implicit surface, $F(x,y,z)=0$. My main goal is obtaining
\begin{align}
K = – \frac{\begin{vmatrix}
H(F) & \nabla F^T \\
\nabla F & 0
\end{vmatrix}}{|\nabla F|^4}
= – \frac{\begin{vmatrix}
F_{xx} & F_{xy} & F_{xz} & F_x \\
F_{xy} & F_{yy} & F_{yz} & F_y \\
F_{xz} & F_{yz} & F_{zz} & F_z \\
F_x & F_y & F_z & 0
\end{vmatrix}}{|\nabla F|^4}
\end{align}
to do that as a first I want to know its metric, i.e., First fundamental form for given $F(x,y,z)=0$.
What I know is from $dF=0$, so its normal vector is proportional to gradient of F, this implies that tge unit normal vector is given by $N = \frac{\nabla F}{|\nabla F|}$. But How about its tangent vector?
This problem can be equivalent to find tangent vector of implicit surface.
For a implicit curve, $F(x,y)=0$.
I can easily compute
\begin{align}
\frac{dF}{ds} = F_x \frac{dx}{ds} + F_y \frac{dy}{ds} = \nabla F \cdot T =0
\end{align}
This has
and find unit tangent vector
\begin{align}
T = \left( \frac{dx}{ds}, \frac{dy}{ds} \right) = \pm \frac{(F_y, -F_x)}{\sqrt{F_x^2+ F_y^2}}
\end{align}
But for a implicit surface $F(x,y,z)=0$
\begin{align}
\frac{dF}{ds} = F_x \frac{dx}{ds} + F_y \frac{dy}{ds}
+ F_z \frac{dz}{ds} = \nabla F \cdot T =0
\end{align}
I am problem of finding $T$….
Best Answer
From the link Question on calculating curvature of a surface given implicitly, I notice that implicit function theorem gives $z=h(x,y)$ for $F(x,y,z)=0$. If $F_z\neq 0$, we have $X_u = (1,0, -\frac{F_x}{F_z} )$, $X_v = (0, 1, -\frac{F_y}{F_z})$.
Then now from the definition of Gaussian curvature \begin{align} K = \frac{eg-f^2}{EG-F^2} = \frac{\det(X_{uu} X_{u} X_v) \det(X_{vv} X_u X_v) - \det(X_{uv} X_u X_v)^2 }{(|X_u|^2 |X_v|^2 - (X_u \cdot X_v)^2)^2} \end{align} I obtain the desired result.