# Curves in 3 dimensions

1. Dec 16, 2008

### mooshasta

This is my last annoying post here, probably. :)

1. The problem statement, all variables and given/known data

I have a curve $\alpha: I \rightarrow \Re^3$ parametrized by arclength. $\kappa(t) \neq 0$ for all I.

Given the surface $$\psi (s,t) = \alpha (t) + (s-t) v(t)$$, where $$v(t) = \frac{d \alpha}{dt}$$, t is in I, and s > t, I want to calculate $$E = \frac{\delta \psi}{\delta s} \cdot \frac{\delta \psi}{\delta s}, F = \frac{\delta \psi}{\delta s} \cdot \frac{\delta \psi}{\delta t}, G = \frac{\delta \psi}{\delta t} \cdot \frac{\delta \psi}{\delta t}$$ in terms of t, s, and $\kappa (t)$

From here I want to show that the Gauss curvature of the surface is 0 at all points t and s > t. This seems pretty intuitive to me but I think I'm supposed to use the quantities above to get to this conclusion. Finally, is the surface locally isometric to $\Re ^2$?

2. Relevant equations

3. The attempt at a solution

I calculated that $$\frac{\delta \psi}{\delta s} = v(t)$$ and $$\frac{\delta \psi}{\delta t} = (s-t) v'(t)$$, so E = 1 (from unit speed), F = 0 (also from unit speed, because velocity and acceleration must be perpendicular to maintain a certain speed), and that $$G = (s-t)^2 ( v'(t) \cdot v'(t)) = (s-t)^2 \kappa ^2 (t)$$, since (from what I understand) $\kappa (t) = ||v'(t)||$, from Frenet formulas. \

I'm not sure these are correct, and I'm also not sure how to calculate Gauss curvature from these quantities, however..

Any help or hints is appreciated :)