Past exam of the geometry course of the University of Oxford / 2008 / A / 2 / iii

Put $b={\gamma }^{\prime }+v{a}^{\prime }$. Then $r_u=b$ and $r_v=a$, soThis is the induced metric of the ruled surface wherever $b\times a\ne 0$.

For the ruled surface, $r_{vv}=0$, so $N=0$. AlsoThusThe denominator is positive at every regular point, proving that zero Gaussian curvature is equivalent to the vanishing of the stated scalar triple product. In particular the Gaussian curvature of a regular ruled surface is always nonpositive.

For the helicoid,Hence $E=1+v^2$, $F=0$, $G=1$. Since $r_{uu}=(-v\cos u,-v\sin u,0)$ and $r_{vv}=0$, both $L$ and $N$ vanish. The mean-curvature formula therefore givesFor comparison, $M=1/\sqrt{1+v^2}$, so $K=-1/(1+v^2{)}^2$; zero mean curvature does not mean zero Gaussian curvature.