Thread: Question about proof that all surfaces with zero Gaussian curvature are developable

1. Question about proof that all surfaces with zero Gaussian curvature are developable

Hey guys, I need your help in understanding a proof given by Struijk on the claim that all surfaces with zero Gaussian curvature are developable surfaces. I am an engineering student rather than a mathematician, so excuse me if this is a complete beginners-question Basically, the part of the proof I do understand is:
$\displaystyle K = 0$ means that the determinant of the second fundamental tensor must be zero $\displaystyle ef-g^2 = 0$.
We can rewrite this $\displaystyle ef - g^2 = (X_u \cdot N_u)(X_v \cdot N_v)-(X_v \cdot N_u)(X_u \cdot N_v) = (X_u \times X_v) \cdot (N_u \times N_v)$

We obtain the following requirement for zero gaussian curvature: $\displaystyle N \cdot (N_u \times N_v) = 0$.

Now we have two possibilities:
1. $\displaystyle N_u = 0$ or $\displaystyle N_v = 0$
2. $\displaystyle N_u$ is collinear with $\displaystyle N_v$, which basically implies that case 1 will be true if we change to different coordinates

So far so good, but now the part that I don't understand. If we look at case 1 only, how does for example $\displaystyle N_u = 0$ imply developability, meaning that the normal vector does not change along a straight line on the surface (generator)? Why must this be true and why can't the $\displaystyle u = constant$-curve be curved?

Make that $\displaystyle eg-f^2=0$.
curvature, developable, gaussian, proof, question, surfaces 