Calculating the Gauss Curvature of a Surface with Curves in 3 Dimensions

[mooshasta]

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

Homework Statement

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

Given the surface [tex]\psi (s,t) = \alpha (t) + (s-t) v(t)[/tex], where [tex]v(t) = \frac{d \alpha}{dt}[/tex], t is in I, and s > t, I want to calculate [tex]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}[/tex] in terms of t, s, and [itex]\kappa (t)[/itex]

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 [itex]\Re ^2[/itex]?

Homework Equations

The Attempt at a Solution

I calculated that [tex]\frac{\delta \psi}{\delta s} = v(t)[/tex] and [tex]\frac{\delta \psi}{\delta t} = (s-t) v'(t)[/tex], 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 [tex]G = (s-t)^2 ( v'(t) \cdot v'(t)) = (s-t)^2 \kappa ^2 (t)[/tex], since (from what I understand) [itex]\kappa (t) = ||v'(t)||[/itex], 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 :)

 

[mighty2000]

Thank you for your question. I can see that you have put a lot of effort into understanding this problem and your calculations seem to be on the right track.

To calculate the Gaussian curvature, we can use the following formula:

K = \frac{EG-F^2}{(EG-F^2)^2}

Substituting the values you have calculated, we get:

K = \frac{(s-t)^2 \kappa^2 (t)}{(s-t)^2}

= \kappa^2 (t)

Since \kappa (t) is non-zero for all points t, we can see that the Gaussian curvature is always 0. This means that the surface is locally flat, or isometric to \Re^2. This makes sense since we are adding a linear term (s-t) to the parametrization of the curve, which does not change the curvature of the curve.

I hope this helps you understand the problem better. Keep up the good work and don't hesitate to ask for further clarification if needed.Scientist