Properties of geodesics on ruled surfaces 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!
 A: 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.
A: 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.
