Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 3 / 14G (Geometry)

At a regular point, choose the unit normal $\mathbf{n}=({\sigma }_x\times {\sigma }_y)/|{\sigma }_x\times {\sigma }_y|$. The first fundamental form is the induced squared lengthThe second fundamental form isThe Gaussian curvature is the determinant of the shape operator, $K=(eg-f^2)/(EG-F^2)$, independent of the normal orientation.

For the given graph, ${\sigma }_x=(1,0,y)$ and ${\sigma }_y=(0,1,x)$, so with $S=1+x^2+y^2$,ThereforeThe surface is regular everywhere. The printed description “hyperboloid” is a naming error: this graph is a hyperbolic paraboloid, and the computation uses the actual parametrization.

Assume the parametrization is $C^2$ and regular, so its curvature is defined. Write $\mathbf{v}={\mathbf{a}}^{\prime }+y{\mathbf{b}}^{\prime }$ and $W=|\mathbf{v}\times \mathbf{b}|>0$. Since ${\sigma }_x=\mathbf{v}$ and ${\sigma }_y=\mathbf{b}$, its first fundamental form isFor the unit normal $\mathbf{n}=(\mathbf{v}\times \mathbf{b})/W$, the mixed coefficient of the second fundamental form iswhere the term containing ${\mathbf{b}}^{\prime }$ twice vanishes. Also $g=\mathbf{n}\cdot {\sigma }_{yy}=0$, while $EG-F^2=W^2$. ThusThe Gaussian curvature vanishes exactly when the displayed scalar triple product is zero. This is the Gaussian curvature of a ruled surface criterion; at a singular point with $W=0$, curvature and the claimed equivalence are not defined.