# what is a nice way to show that developable surfaces must have a principal curvature=0?

So a developable surface can be parametrized as

$x(s, t)=\alpha(s)+t \beta(s)$

I can see that $\beta(s)$ is the direction of the principal curvature plane with k=0, but why is it the minimum or maximum curvature plane cutting through that point? Is $\alpha(s)$ a plane curve on the other principal curvature plane?

• What is definition of developable surface that you use? Wikipedia defines it precisely as surface with zero curvature – Blazej Sep 26 '15 at 8:18
• I'm using this definition: x(s,t)=α(s)+tβ(s) – UXkQEZ7 Sep 26 '15 at 8:36

The vector you have given $x(s, t)=\alpha(s)+t \beta(s)$ is a ruled surface with generator $\beta(s)$.
It is developable if $(T, \beta(s),\beta{'}(s)) = 0$
$(T, \beta(s),\beta{'}(s)) \ne 0.$
$K = k_1\cdot k_2 = 0$ is necessary and sufficient condition. When parametric lines of principal curvature $k_1=0$ for $K=0$ then that parameter defines the straight edge or regression line of a developable surface.