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### hw9.soln

Course: MATH 131A, Fall 2006
School: San Jose State
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Word Count: 442

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113, MATH SPRING 2010 HOMEWORK 9 SOLUTIONS 7.6 (Pressley): Recall that the principal curvatures 1 2 of a surface S at a point p are the minimum and maximum of the normal curvature n (v ) taken over all directions v Tp S . Recall also that n (v ) = (0) N, where is a unit speed curve on the surface with (0) = v and N is the unit normal to the surface. Rigid motions map the unit normal to the unit normal and...

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113, MATH SPRING 2010 HOMEWORK 9 SOLUTIONS 7.6 (Pressley): Recall that the principal curvatures 1 2 of a surface S at a point p are the minimum and maximum of the normal curvature n (v ) taken over all directions v Tp S . Recall also that n (v ) = (0) N, where is a unit speed curve on the surface with (0) = v and N is the unit normal to the surface. Rigid motions map the unit normal to the unit normal and preserve the dot product. It follows that the principal curvatures are preserved by rigid motions. Therefore, the Gaussian curvature K = 1 2 and the mean curvature H = (1 + 2 )/2 are also preserved. A dilation Da : (x, y, z ) (ax, ay, az ) (a > 0) multiplies the curvature of any curve by a1 . By the same argument as above, Da therefore multiplies the normal curvatures of a surface by a1 , hence it multiplies the Gaussian curvature by (a1 )2 and the mean curvature by a1 . 7.11 (Pressley): We will use the parametrization (u, v ) = (f (u) cos v, f (u) sin v, g (u)), where f (u) = eu and g (u) = 1 e2u cosh1 (eu ), u < 0, 0 v < 2 . (i) Each parallel of the pseudosphere P is a circle with radius equal to the distance from the axis of revolution, which is f (u) = eu , u < 0. Thus its length is 2eu . (ii) By Example 7.2, E = 1, F = 0 and G = f (u)2 , so if U = {(u, v : ) u < 0, 0 v 2 }, then E G F 2 dudv area(P ) = U 2 0 eu du dv = 0 = 2. (iii) Also by Example 7.2, the principal curvatures are 1 = f g f g = and 2 = K = 1 1 f = 2u 1 g e e2u 1, since K = 1. (iv) Since 1 < 0 and 2 > 0, all points of P are hyperbolic. 8.1 (Pressley): The xz -plane is orthogonal to the hyperboloid (which we denote by H ), therefore its intersection with H (which is a meridian) is a geodesic, being a normal section, by Proposition 8.4. Similarly, the xy -plane is orthogonal to H , so for the same reason the unit circle in the xy -plane is also a geodesic. It was shown in exercise 4.4 that for every the straight line (x z ) cos = (1 y ) sin , (x + z ) sin = (1 + y ) cos is contained in H . Taking (x, y, z ) = (1, 0, 0) and solving the above equations for , we obtain sin = cos . This implies that = /4 or = 5/4. The corresponding lines are geodesics by 1 2 Proposition 8.3. This describes four geodesics passing through (1, 0, 0). 8.14 (Pressley): Suppose that every parallel of a surface of revolution S is a geodesic. Let (t) = (f (t), 0, g (t)) be the prole curve of S . Then f (t) = 0, for all t, by Proposition 8.5. Therefore, f is constant and since f (t) is the distance from the axis of revolution, it follows that S is a cylinder.
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