No. 09-0 ☆ Consider the ellipsoid in R³ defined in the Cartesian coordinates (x, y, z) by: x = 2 sin(u) cos(v), y = 2 sin(u) sin(v), z = 3 cos(u), where u and v are parameters. The infinitesimal arc-length ds on the surface is given by ds² = dx²+dy² + dz². == In terms of the parameters u, v, ds = (E(u, v)du² + 2F(u, v)du dv + G(u, v) dv²) == where: E(u, v) 5 sin² (u) + 4, = F(u, v) = 0, G(u, v) = 4 sin² (u). A path on the surface parametrised by arc-length s with so < s < $1, from a point with parameters u₁ = u(so) and vo v(so) to a point with parameters u₁ = u(s1) and 201 = = v(s1) (where uo, vo, u1, V₁ are constants) is a geodesic if it satisfies the Euler- Lagrange equations for the distance functional S=81 L[u, v] == ds, S=80 which are given by: d (a) ds d 1 (Eù + Fv) — —½ (ù² E„ + 2úvF« + ¿²G„) = 0; 2 1 น (b) (Gv + Fù) — —½ (ù² E₁ + 2ù¿F¸ + i²G,) = 0. ds - - 2 (Here the notation E means the partial derivative и du derivative and ❞ means ds d² u ds² and so on.) E and so on. The notation u means the |მu Calculate the Euler-Lagrange equations by evaluating and simplifying the expressions in (a) and (b) above. You do not need to solve the differential equations. No. First, d ds d ds (Eù + Fv) = E₁ü² + Еúv + Еü + F₁v² + Fчüv + Fü, and (Gv + Fu) = G₁v² + G₁üv + G + F₁u² + F₁uv + Fü. Now: Eu = 10 cos(u) sin(u); E₁₂ = 0; Fu = 0; F₁₁ = 0; G₁ = 8 cos(u) sin(u); Gv = 0. Thus: Can you Show more details?? d ds ·(Eù + Fv) = ù² (10 cos(u) sin(u)) + úv((0) + (0)) + ü(5 sin² (u) + 4) + ü(0) + v² (0) = = ü (5 sin² (u) + 4) + 10 u² cos(u) sin(u). d ds (Gv + Fù) = ú² (0) + uv ((8 cos(u) sin(u)) + (0)) + ü(0) + (4 sin² (u)) + v² (0) = 4 sin² (u)+8uv cos(u) sin(u). So the Euler-Lagrange equations are given by: (a) 1 ü (5 sin²(u) + 4) + 10 ú² cos(u) sin(u) – ½-½ (80² cos(u) sin(u) + 10 û² cos(u) sin(u)) = 0 or 5 ü sin² (u) + (5 ü² – 4v²) cos(u) sin(u) +4ü = 0; 1 (b) 4 ʊ sin² (u) + 8 ú v cos(u) sin(u) – ±±(0) - or 4 ü sin² (u) + 8 ú v cos(u) sin(u) = 0. (0) = 0,