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:

means that the determinant of the second fundamental tensor must be zero .

We can rewrite this

We obtain the following requirement for zero gaussian curvature: .

Now we have two possibilities:

1. or

2. is collinear with , 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 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 -curve be curved?

Thanks in advance!

Re: Question about proof that all surfaces with zero Gaussian curvature are developab

Make that .