For surfaces in $\mathbb R^3$, does $ ds^2 = du^2 + G(u)^2 dv^2$ implies $|\frac{\partial G(u)}{\partial u}| ≤1 $?

55 Views Asked by At

Let we have a regular surface in $\mathbb R^3$, parameterized by $ \vec r(u,v)$. Suppose that the first fundamental form (metric) of this surface is given by:

$ ds^2 = du^2 + G(u)^2 dv^2$ (G(u) only depends on u, it's a particular case of the geodesic form).

What can we say about $\frac{\partial G(u)}{\partial u}$?

This question appeared when I was studying surfaces of revolution. In this case we have $ \vec r(w,v) = (f(w)*cos(v),f(w)*sin(v), g(w))$. And:

$ ds^2 =(f'(w)^2 + g'(w)^2) dw^2 + f(w)^2 dv^2$

And: $|\frac{\partial G(u)}{\partial u}|= |\frac{f'(w)}{\sqrt{f'(w)^2 + g'(w)^2}}| ≤1 $

I was wondering if this bound is a general property of surfaces in $\mathbb R^3$ or it's possible to find some surfaces where $\frac{\partial G(u)}{\partial u}$ is unbounded.