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2.2. The Shape of a Surface

Consider a parametrized surface X(u1,u2) such that X(0,0) = P. By Taylor Theorem, we have
X(u1,u2) = X(0,0) + Xi(0,0)ui + (1/2)Xij(0,0)uiuj + o(r3)
where r2 = (u1)2 + (u2)2. Then the distance from X(u1,u2) to the tangent plane at X(0,0) is given by
(X(u1,u2) - X(0,0)) (0,0) = (1/2) hij(0,0)uiuj + o(r3).
This expression is dominated by the quadratic term, so the shape of the surface will be similar to the shape of the graph of the quadratic function f(x,y) = h11x2 + 2h12xy + h22y2. This shape is determined by the expression h = h11h22 - (h12)2. When h > 0, the graph of the quadratic function intersects the horizontal plane only at the origin, while if h < 0, the intersection is a pair of intersecting lines.
The point P of a parametrized surface is called elliptic, hyperbolic, or parabolic according as h is positive, negative, or zero at P. At an elliptic point P, the tangent plane meets the surface only at P locally, i.e. there is a neighborhood of P that meets the tangent plane only at P. On the other hand, if P is hyperbolic, the intersection of the tangent plane and a small neighborhood of P will consist of two curves with distinct tangent lines, so arbitrarily close to P, there are points of the surface on both sides of the tangent plane. At a parabolic point, the behavior of the surface can be quite a bit more complicated (see the Exercises at the end of this section for some examples.)
The behavior of a parametrized surface in a neighborhood of P can also be described in terms of the normal sections at P, i.e. the intersections of the surface with planes through the normal line at P. If the curvature kN of such a normal section is non-zero, then the center of the osculating circle to this curve is P + (1/kN). At an elliptic point, all normal curvatures will have the same sign and the centers of osculating circles for all normal sections lie on the same side of the tangent plane. At a hyperbolic point, the normal curvature takes on both positive and negative values, and the centers of some osculating circles lie on one side of the tangent plane and some will lie on the other. At a parabolic point, at least one normal section has normal curvature 0.

Exercises

  1. For all surfaces in the exercises 1. -6. of section 1.1, which points are elliptic, which are hyperbolic and which are parabolic?
  1. Let X(u1,u2) = ( u1, u2, (u1)3 - 3u1(u2)2 ). Show that the origin is the only parabolic point and that all of the metric coefficients hij are xero at this point. (Such a point is called planar.) What is the intersection of the surface and the tangent plane at this point? (This point is called a "Monkey saddle". A saddle for a person riding a bicycle has two dips, but a saddle for a monkey requires a third depression for the tail.)
  1. Let X(u1,u2) = ( u1, u2, (u1)4 + (u2)4). Show that the origin is the only parabolic point and that all of the metric coefficients hij are zero at this point. (Such a point is called planar.) What is the intersection of the surface and the tangent plane at this point?
  1. Let X(u1,u2) = ( u1, u2, (u1)4 + c(u2)2) for a constant c. Show that the origin is the only parabolic point. What is the intersection of the surface and the tangent plane at this point?
  1. Let X(u1,u2) = ( u1, u2, ((u1)2 - u2)((u1)2 - cu2)) for a constant c. Show that the origin is a parabolic point. What is the intersection of the surface and the tangent plane at this point? Which points are parabolic, which are ellipic, and which are hyperbolic?
  1. Show that all point on a hyperboloid of one sheet are hyperbolic:
    2. (x1)2/(a1)2 + (x2)2/(a2)2 - (x3)2/(a3)2 = 1
  1. For the torus of revolution of problem 7 in section 1.1, which points are parabolic, which are elliptic, and which are hyperbolic?
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