If we define the developable ruled surface like $\varphi(u,v)=\alpha(u) + v\cdot w(u)$, how can we proof the statement above?
Some books I've looked for has it as a definition rather than a proposition. I've tried calculating the tangent plane of a point of the generator (which I've seen defined as $L_u=\varphi(u,t)$), but didn't get any results, as the matrix I ended up with was the following: \begin{pmatrix} a_1'(u)+tw_1'(u) & a_2'(u)+tw_2'(u) & a_3'(u)+tw_3'(u) \\ w_1(u) & w_2(u) & w_3(u) \end{pmatrix} and I don't really know how to continue.
Another approach was to say that as the gaussian curvature is zero in all of the points of the generator, one of the principal curvatures has the direction of the generator, so the tangent plane touches the generator in all of its points. Would this be correct?
Could anyone help, please? Thanks in advance!
You have
$$\begin{cases} \frac{\partial \phi}{\partial u}(u,v) = \alpha^\prime(u) + v w^\prime(u)\\ \frac{\partial \phi}{\partial v}(u,v) = w(u) \end{cases}$$ hence the normal vector
$$N(u,v) = \alpha^\prime(u) \times w(u) +v(w^\prime(u) \times w(u)).$$
This has to be independent in direction from $v$ by hypothesis and implies that $\alpha^\prime(u) \times w(u)$ and $w^\prime(u) \times w(u)$ are colinear vectors. Which implies that $\det(\alpha^\prime(u), w(u), w^\prime(u))=0$ as desired.