### DIFFERENTIAL GEOMETRY COURSE

19. DEVELOPMENT

Definition 167 : We call a surface developable if it is a bent of a plane, so if it is isometric to a plane.

Proposition 168 : Any ruled surface whose tangent plane is constant along each rule, is developable.

Proof: Let z(s) be an orthogonal trajectory of the rules, where the parameter s is arc length.
Let v(s) be a unit normal in the direction of the rule, so that y(s,u) = z(s) + u v(s) is a parametrization of the ruled surface.
Because the tangent plane is constant along each rule, so for each s the same plane for all u, the three vectors v ' , z ' and v all are lying in the plane through the origin parallel to the tangent plane.
Since ||v|| = 1, v ' is perpendicular to v, and so is the direction vector z ' of the orthogonal trajectory.
So v ' (s) = λ(s) z ' (s). We can now check by simple calculation that a1 1 = (1+uλ)2, a1 2 = 0, and a2 2 = 1.
Now draw a planar curve x(s) with κ(s) = -λ(s), and let n(s) be the principal normal vector of x(s). According to Frenet we have n ' = -κ x ' = λ x ' .
Check by calculation that y(s,u) → x(s) + u n(s) is an isometry from the ruled surface to the plane wherein x(s) is lying. To do this, you only have to calculate the first fundamental form of this plane with this parametrization.

Explanation 169 : We call the curve x(s) trace curve of z(s). We get it as a trace of z(s) if we draw z(s) with wet ink on the ruled surface and roll the surface over the plane. Since the tangent plane is constant along the rule, this rolling also gives the isometry (development).

Remark 170 : Proposition 168 is convertible. Also see 163 and 164.

Problem 171 : Check the following assertions in the context of the proof of 168.
i) For the second fundamental form we find h1 1 = (1+uλ) z ". N , h1 2 = v ' . N = 0, h2 2 = 0 . N = 0.
ii) According to 152, z(s) and the rules are curvature lines.
iii) According to 141, the rules are asymptotic lines. The total curvature is 0 everywhere (see 151). Each point is parabolic (see 134 2) ).