DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

50 Chapter 2. Surfaces: Local TheorynN1tanh uθFigure 2.7Meusnier’s Formula, Proposition 2.5, we have k 2 = κ n = κ cos φ = (cosh u)(− tanh u) =− sinh u.Amazingly, then, we have K = k 1 k 2 =(1/ sinh u)(− sinh u) =−1. ▽Example 7. Let’s now consider the case of a general surface of revolution, parametrized as inExample 2 of Section 1, byx(u, v) = ( f(u) cos v, f(u) sin v, g(u) ) ,where f ′ (u) 2 +g ′ (u) 2 =1. Recall that the u-curves are called meridians and the v-curves are calledparallels. Thenand so we havex u = ( f ′ (u) cos v, f ′ (u) sin v, g ′ (u) )x v = ( −f(u) sin v, f(u) cos v, 0 )n = ( −g ′ (u) cos v, −g ′ (u) sin v, f ′ (u) )x uu = ( f ′′ (u) cos v, f ′′ (u) sin v, g ′′ (u) )x uv = ( −f ′ (u) sin v, f ′ (u) cos v, 0 )x vv = ( −f(u) cos v, −f(u) sin v, 0 ) ,E =1, F =0, G = f(u) 2 and l = f ′ (u)g ′′ (u) − f ′′ (u)g ′ (u), m =0, n = f(u)g ′ (u).By Exercise 2.2.1, then k 1 = f ′ (u)g ′′ (u) − f ′′ (u)g ′ (u) and k 2 = g ′ (u)/f(u). Thus,K = k 1 k 2 = ( f ′ (u)g ′′ (u) − f ′′ (u)g ′ (u) ) g ′ (u)f(u) = −f′′ (u)f(u) ,since from f ′ (u) 2 + g ′ (u) 2 =1we deduce that f ′ (u)f ′′ (u)+g ′ (u)g ′′ (u) =0,and sof ′ (u)g ′ (u)g ′′ (u) − f ′′ (u)g ′ (u) 2 = −(f ′ (u) 2 + g ′ (u) 2 )f ′′ (u) =−f ′′ (u).Note, as we observed in the special case of Example 6, that on every surface of revolution, themeridians and the parallels are lines of curvature. ▽