So Xx times Xy = (-x f'(x) cos y, -x f'(x) sin y, x). If x 0 then X is regular, and for x > 0 the unif normal is
so again a modified Gauss mapping, equivalent to N is given by
The parabolic set occurs when 0 = = -f'(x) f"(x), i.e. at extrema or at inflections of the profile curve. Furthermore, grad = (-(f"(x))2 - f'(x) f'''(x), 0), so Ñ is good if f"(x) = 0 implies f'(x) 0 and f'''(x) 0.
If x0 is a value for which f''(x0) = 0, then the parabolic curve can be parametrized by x(t) = x0, y(t) = t, and we obtain
If f'(x0) 0 and f'''(x0) 0, then the Gauss map has an ordinary fold along the parabolic curve. For example, these conditions are satisfied by the bell surface:
On the other hand, if x0 is a value for which f'(x0) = 0 and f"(x0) 0, then the Gauss map is good, but not excellent, because the parabolic curve is parametrized by x(t) = x0, y(t) = t, and N(t) = (0, 0) for all t. An example is the top of a torus of revolution:
0 < b < a, a - b < x < a + b, with x0 = a.
Let be a regular space curve wit curvature nowhere zero. Define the canal surface about of radius r to be
where P and B are the principal normal and the binormal of the curve . To find the normal of the surface X, we form
where and are the curvatuer and torsion of , and s is the arclength along . Then
so X will be regular if r is a sufficiently small positive number. Moreover,
so the parabolic set occurs when the two vectors
are linearly dependent, i.e. when cos y = 0. This occurs at the curves x(t) = t, y(t) = ±/2, for which we have
A straightforward calculation gives the Gaussian curvature K = cos y/(r (rcos y - 1)) and K/y = sin y/(r (rcos y - 1)2), so K = 0 implies grad K 0, since 0. Therfore the Gauss map is good. Now taking derivatives with respect to t, we obtain
so N' = 0 if and only if = 0 and then N" = 's'P. So the Gauss map N is excellent if ' 0 whenever = 0, and then N(x, y) has a cusp at (t ,±/2) if and only if (t) = 0.
For example consider the warped torus, a canal surface of the space curve
(t) = (cos t, sin t, sin(nt))
where n is an integer, n 2. The curvature of is nowhere zero, since (s')3 = |(' x ")| 1. Furthermore
so = 0 if and only if t = /2n, 3/2n, ..., (4n - 1)/2n, provided that 0. Taking derivatives of 2(s')^6 and n(1 - n2)cos nt shows that tau = 0 implies ' 0, so long as 0. Therfore a canal surface of has an excellent Gauss map, with 4n cusps. For = 0 a canal surface of is a torus of revolution, and each component of the parabolic curve is collapsed to a point by the Gauss map.