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Curves in 3 dimensions

  1. Dec 16, 2008 #1
    This is my last annoying post here, probably. :)

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

    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]?


    2. Relevant equations



    3. 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 :)
     
  2. jcsd
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