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I am trying to solve the following problem:

Show that a unit-speed curve $\gamma$ with nowhere vanishing curvature is a geodesic on the ruled surface $\sigma(u,v)=\gamma(u)+v\delta(u)$, where $\gamma$ is a smooth function of $u$, if and only if $\delta$ is perpendicular to the principal normal of $\gamma$ at $\gamma(u)$ for all values of $u$.

Edit (rather large): My professor wrote the question down wrong. I fixed it on here. Sadly, even with it right, I can't get either direction.

Any help would be appreciated. Thanks!

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A unit-speed curve $\gamma(u)$ (i.e. parametrized by arc-length) is a geodesic in a surface $S$ iff $\gamma''(u)$ is perpendicular to $S$.

The normal to $\sigma(u,v)=\gamma(u)+v\delta(u)$ is parallel to $$ \frac{\partial\sigma}{\partial u}\times\frac{\partial\sigma}{\partial v} =(\gamma'+v\delta')\times\delta\tag{1} $$ On $\gamma$, $v=0$. Thus, $\gamma$ is a geodesic iff $\gamma''\times(\gamma'\times\delta)=0$. Using Lagrange's formula and the fact that $\gamma''\cdot\gamma'=0$, we get $$ \gamma''\cdot\delta\gamma'-\gamma''\cdot\gamma'\delta=0 \Leftrightarrow\gamma''\cdot\delta=0\tag{2} $$ Thus, $\gamma$ is a geodesic iff $\gamma''\cdot\delta=0$, where $\gamma''$ is the parallel to the principal normal since $\gamma$ is unit-speed.

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How did you get the line "Thus, $\gamma$ is a geodesic iff $\gamma'' \times (\gamma'\times \delta)=0$"? Are you just setting $v=0$, and if so, why? –  Samuel Reid Dec 5 '12 at 5:25
    
As I mentioned, on $\gamma$, the coordinate $v=0$. Thus, the normal to the surface at a point on $\gamma$ is $\gamma'\times\delta$. Thus, $\gamma''$ is perpendicular to the surface when $\gamma''\times(\gamma'\times\delta)=0$. –  robjohn Dec 5 '12 at 5:32
    
Oh, that makes perfect sense. I never thought of that. Thank you! –  Samuel Reid Dec 5 '12 at 5:48
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are you enrolled in UofCalgary PMAT 423? I have the same question as you.

(=>) this is what I was thinking as well. Just remember that we're supposed to show that γ is perpendicular to the principal normal of γ, not to γ". Use γ" = kn (not N which is the standard unit normal of the surface).

(<=) here we have to use the idea that the principal normal of γ is perpendicular to γ at every point on γ. Since kn = t' = γ", where k is the curvature of γ (not the surface), this implies γ" is perpendicular to γ at every point which then implies γ" is perpendicular to γ'. A property of the cross product is that if A x B = C then C is perpendicular to A and to B so (N x γ') is perpendicular to γ'. Since Y' is perpendicular to γ" and γ" is parallel to N (because γ lies on the surface) then (N x γ') must be perpendicular to γ" as well. From this we get γ" • (N x γ') = 0 = kg. Since the geodesic curvature equals 0, then the surface must be geodesic.

I think this is the correct way to do this question but it both directions just seems too simple. If you have any ideas I'd love to here it.

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Yeah I am! You're right about the principal normal thing, thanks. I'm not sure about your claim $\gamma"$ is parallel to $N$ (because $\gamma$ lies on the surface). Why would that necessarily be true? All we know is that $\gamma'' \perp \gamma'$. I guess if we wanted to show $\gamma'' \parallel N$, we could show $\gamma''\perp \sigma_u, \sigma_v$, but for that we need $\sigma_u\cdot\gamma''=0=v\gamma''\cdot\delta'$, and we know nothing about $\delta$. Likewise with $\sigma_v$. So that seems problematic to me... –  Samuel Reid Dec 2 '12 at 22:31
    
Also, if you're in my class anyway, did you get 7.3.9? –  Samuel Reid Dec 2 '12 at 22:37
    
IThat was the one point I wasn't entirely sure about. My reasoning behind it is that if γ lies on the boundary of the surface then γ and S would have parallel normals and γ lies on the ruled surface because it is a ruling. I included that part because I think it's possible that γ" and (N x γ') could be parallel without that but I could be wrong –  Kevin Dec 2 '12 at 22:56
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As for 7.3.9, we have γ(t), σ(u,v) and γ lies on the surface. Using u=v=t, set γ(t) = σ(t,t). Kn = γ" • N where N is the standard unit normal of σ. From there it's just a calculation question. When I hit it with a math stick I got -2/sqrt(5). –  Kevin Dec 2 '12 at 23:09
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let us continue this discussion in chat –  Samuel Reid Dec 2 '12 at 23:20
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