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?

The vector you have given $x(s, t)=\alpha(s)+t \beta(s)$ is a ruled surface with generator $ \beta(s)$.

It may _or may not be developable_ depending on tangent vector triple product.

It is developable if $ (T, \beta(s),\beta{'}(s)) = 0 $

and skew ( twisted with negative Gauss curvature K) if

$(T, \beta(s),\beta{'}(s)) \
e 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.