Answer Happy

1. 1. For parametrised curves C : (x(t), y(t)), a <t < b with image in H = {(x,y) R2: > 0} we define their hyperbolic length by Vi(1)2 + ()? Lhyp (C) = dt X(t) a) Lyp (C) is not changed by regular reparametrisation, meaning that Lhyp(C) = Lup(©) if C is parametrised by by (x(t(s)), y(t(s))) where t(s) is a one-variable func- tion satisfying either t(e) = a, t(d) = b, t'(s) > 0 (orientation preserving) or t(C) = b, (d) = a, t'(8) < 0 (orientation reversing). (3p) b) Express Lhyp (C) as an integral I = F(x, y, y') dx for curves C parame trised as graphs of the form (x, y(x)), SuSx2. (2p) c) Derive the Euler equation for the integral I from (b). (3p) d) Solve the Euler equation. (4p) e) Show that every solution parametrises a subset of either a half circle of the form {.? + (y -b)2 = R',>0} or of a half line {r > 0, y =b}. (3p) f) Show that the half lines and half circles in (e) have infinite length. (3p) g) Show that there are infinitely many half circles as in (e) avoiding the half line {r > 0, y =b}. A good way to do so is to describe this family in terms of centers and radii. (2p) Remark: If one views those half lines and half circles as the lines of a geometry on H, one can show that all axioms of euclidean geometry are valid except for the following axiom of parallels: For a line l and a point P in its complement, there is a unique line lp through P and disjoint from l. Historically, this discovery, due independently to Bolyai and Lobachevsky, was very influential. For a while, it had been unknown whether the axiom of parallels is a consequence of the other axioms. Finding the above alternative geometry showed its independence. (in total 20p)