Prove constant gaussian curvature zero Prove that a tangent developable has constant Gaussian curvature zero. Also compute its mean curvature.
I have a tangent developable as let $\gamma:(a,b)\rightarrow R^3$ be a regular space curve, and s>0. Then $\sigma(s,t) = \gamma(t)+s\gamma'(t)$ is called tangent developable.
I'm using the equation for Gaussian curvature, $K=\frac{eg-f^2}{EG-F^2}$.
I have that $e=<\sigma_{ss},N>=0$, because $\sigma_{ss}=0$ But when looking to calculate f or F, i'm having trouble evaluating the cross products or the normal without knowing an expression for $\gamma$.
Any direction would be great. thanks
 A: Here's another proof: if $\sigma(s,t) = \gamma(t)+s\gamma'(t)$, then $$\sigma_s(s,t) = \gamma'(t) \quad\mbox{and}\quad \sigma_t(s,t) = \gamma'(t)+s\gamma''(t),$$so that $$N(\sigma(s,t)) = \frac{\sigma_s(s,t)\times \sigma_t(s,t)}{\|\sigma_s(s,t)\times \sigma_t(s,t)\|} = \frac{\gamma'(t)\times \gamma''(t)}{\|\gamma'(t)\times \gamma''(t)\|}.$$Then $${\rm d}N_{\sigma(s,t)}\left(\sigma_s(s,t)\right) = (N\circ \sigma)_s(s,t) = 0$$and $\sigma_s(s,t) \neq 0$. So ${\rm d}N_{\sigma(s,t)}$ is singular, and $K(\sigma(s,t)) = \det {\rm d}N_{\sigma(s,t)} = 0$. No need to waste time computing $e,f$ and $g$.
A: You are not given the expression for $\gamma$, so everything will depends on $\gamma$. 
So you can get 
$$ \sigma_t = \gamma'(t) + s\gamma''(t), \ \ \sigma_s = \gamma'(t),$$
so 
$$ N = \frac{\sigma_s \times \sigma_t}{\| \sigma_s \times \sigma_t\|} = \frac{\gamma'' \times \gamma'}{\|\gamma'' \times \gamma'\|}.$$
This should be sufficient for your to show $f= 0$, since you get represent $f$ in terms of $\gamma$ and compute. 
