Lecture Notes for 120 - UCLA Department of Mathematics

i.e., d dv
and X are parallel. In particular,
0 = d dv · dX
dv
=
5.5. RULED SURFACES 126
✓ d↵ dX
+ u (v)
dv dv
= d↵
dv · dX dX
+ u (v)
dv dv
◆
du (v)
+
dv
X · dX
dv
So again we obtain that u (v) corresponds exactly to s =0, which forces to be
c. ⇤
Theorem 5.5.5. (Monge, 1775) A developable surface is a generalized cylinder,
generalized cone, or a tangent developable at almost all points of the surface.
Proof. We can assume that the surface is given by
where dc
dv ? dX
dv
If dX
dv
q (s, v) =c (v)+sX (v)
. The Gauss curvature vanishes precisely when
✓
X ⇥ dc ◆
· dX
dv dv =0.
=0on an interval, then the surface is a generalized cylinder. So we can
assume that dX
dX
dv
6=0. This implies that X and
dv
are linearly independent as they
are orthogonal. The condition ✓
X ⇥ dc ◆
· dX
dv dv =0
2
on the other hand implies that the three vectors are linearly dependent. We already
know that dc
dv ? dX
dv ,sothisforces
✓ ◆
dc dc
dv = dv · X X
6=0then X is tangent to c and so we have a tangent developable. On the
other hand, if dc
ds
=0on an interval, then the surface must be a generalized cone
on that interval.
Thus the surface is divided into regions each of which can be identified with
our three basic types of ruled surfaces and then glued together along lines that go
When dc
dv
through parameter values where either dX
dv
dc
=0or
dv =0.
There is also a similar and very interesting result for ruled minimal surfaces.
Theorem 5.5.6. (Catalan) Any ruled surface that is minimal is planar or a
helicoid at almost all points of the surface.
Proof. Assume that we have a parametrization q (s, v) =c (v)+sX (v) where
dc
dv · dX
dv
=0. In case the surface also has vanishing Gauss curvature it follows that
it is planar as the second fundamental form vanishes. Therefore, we can assume
that both c and X are regular curves and additionally that dc
dv
is not parallel to X.
The mean curvature is given by the general formula
H = L ssg vv 2L sv g sv + L vv g ss
2(g ss g vv gsv)
2
⇤