DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

SELECTED ANSWERS 1151.2.23 a. cont. Alternatively, let the equation of the plane through P , Q, and R beA s,t · x =0(where we choose A s,t to vary continuously with length 1). We wantto determine A = lim s,t→0 A s,t . For fixed s and t, consider the function F s,t (u) =A s,t · α(u). Then F s,t (0) = F s,t (s) =F s,t (t) =0, so, by the mean value theorem,there are ξ 1 and ξ 2 so that F s,t(ξ ′ 1 )=F s,t(ξ ′ 2 )=0,hence η so that F s,t(η) ′′ =0. NowF s,t(0) ′ = A s,t · T(0) and F s,t(0) ′′ = A s,t · κ 0 N(0). Since ξ i → 0 and η → 0ass, t → 0,we obtain A · T(0) = A · N(0) = 0, so A = ±B(0), as desired.1.3.4 Let L = length(C). Then by Theorem 3.5 we have 2π = ∫ L0 κ(s) ds ≤ ∫ L0cds = cL,so L ≥ 2π/c.2.1.2 a. E = a 2 , F =0,G = a 2 sin 2 u; d. E = G = a 2 cosh 2 u, F =02.1.4 Say all the normal lines pass through the origin. Then there is a function λ so thatx = λn. Differentiating, we have x u = λn u + λ u n and x v = λn v + λ v n. Dotting withn, weget 0 = λ u = λ v . Therefore, λ is a constant and so ‖x‖ = const. Alternatively,from the statement x = λn we proceed as follows. Since n · x u = n · x v =0,wehavex · x u = x · x v =0. Therefore, (x · x) u =(x · x) v =0,so‖x‖ 2 is constant.2.1.6 We check that E = G =4/(1 + u 2 + v 2 ) 2 and F =0,sothe result follows fromExercise 5.2.1.7 b. One of these is: x(u, v) =(cos u + v sin u, sin u − v cos u, v).2.1.12 a. If a cosh(1/a) =R, the area is 2π ( a + R √ R 2 − a 2) .2.2.1 If u- and v-curves are lines of curvature, then F =0(because principal directionsare orthogonal) and m = S(x u ) · x v = k 1 x u · x v =0. Conversely, setting S P (x u )=ax u + bx v ,weinfer that if F = m =0,then 0 = S P (x u ) · x v = Fa+ Gb = Gb, andso b =0. Therefore, x u (and, similarly, x v )isaneigenvector for S P . Moreover, ifS P (x u )=k 1 x u and S P (x v )=k 2 x v ,wedot with x u and x v , respectively, to obtainl = Ek 1 and n = Gk 2 .[]1/b 02.2.3 b. l = b, m = 0, n = cos u(a + b cos u), S P =,0 cos u/(a + b cos u)H = 1 12(b + cos u )a+b cos u , K =cos ub(a+b cos u) ; d. l = −a, m = 0, n = a, S P =[]−(1/a)sech 2 u 00 (1/a)sech 2 , H =0,K = −(1/a) 2 sech 4 u.u2.2.5 We know from Example 1 of Chapter 1, Section 2 that the principal normal of thehelix points along the ruling and is therefore orthogonal to n. Aswemove along aruling, n twists in a plane orthogonal to the ruling, so its directional derivative in thedirection of the ruling is orthogonal to the ruling.2.2.6 E = tanh 2 u, F =0,G = sech 2 u, −l = sech u tanh u = n, m =02.3.5 d. Γ v uv =Γ vvu = f ′ (u)/f(u), Γ uvv = −f(u)f ′ (u), all others 0.2.4.4 κ g = cot u 0 ;wecan also deduce this from Figure 3.1, as the curvature vector κN =(1/ sin u 0 )N has tangential component −(1/ sin u 0 ) cos u 0 x u = cot u 0 (n × T).