![]() ![]() Also, k γ (t) k2 = k γ (u) k2, which gives 5 4 cos t = 5 4 cos u, hence cos t = cos u. Hence, the unique equation γ (t) = γ (−t) implies t(tt2 1 √ √ self-intersection is at γ ( 3) = γ (− 3) = 0.ġ.4.8 If γ (t) = γ (u) then sin t = sin u. This shows that γ is not periodic and that for a self-intersection we must have t = −u. As k is the largest such divisor, k ′ = 1, so T0 = T0′. Then, ki Ti /k = k ′ ki Ti so kk ′ divides each ki. By Exercise 1.4.4, T0 = k ′ T0′ for some integer k ′ and then there are integers ki′ such that T0′ = ki′ Ti for all i such that γi is not constant. (ii) The sequence is as in (i) and Tr → 0 as r → ∞, then by the mean value theorem 0 = (f (t Tr ) − f (t))/Tr = f˙(t λTr ) for some 0 T or T ′ ∈ P). Iterating this argument gives the desired sequence. 1.4.5 (i) Choose T1 > 0 such that γ is T1 -periodic then T1 is not the smallest positive number with this property, so there is a T2 > 0 such that γ is T2 -periodic. Thus, when t is retricted to the interval −π 0, or on −k if k 0 this contradicts the definition of T0. ![]() Hence, γ˙ (t) = 0 ⇐⇒ t is an odd multiple of π. (ii) γ˙ (t) = (− sin t − sin 2t, cos t cos 2t) so k γ˙ (t) k = 2 cos 2t. 1.3.4 (i) If γ˜ (t) = γ (ϕ(t)), let ψ be the inverse of the reparametrization map ϕ. Using the fact that dt/dt˜ 6= 0, it is easy to see that d2γ˜ /dt˜2 and d3γ˜ /dt˜3 are linearly independent when t = t0. (iii) Let γ˜ (t˜) be a reparametrization of γ (t), and suppose γ has an ordinary cusp at t = t0. γ = (0, 6) at t = 0 so the origin is an ordinary cusp. and (3, 2) and then γ¨ and γ are easily seen to 3be linearly independent at t = 0. So there are four cases: if (m, n) = (2, 2) or (3, 3) then either γ¨ or γ is zero at t = 0, so the only possibilities for an ordinary cusp are (m, n) = (2, 3). ![]() γ¨ and γ are both 0 at t = 0 so γ¨ and γ¨ are linearly dependent at t = 0 similarly. (ii) γ is regular since γ˙ 6= 0 for 0 3 the first components of. ![]() (If b and c are parallel there are infinitely-many such planes.) 1.3.1 (i) γ˙ = sin 2t(−1, 1) vanishes when t is an integer multiple of π/2, so γ is not regular. This implies 1.2.9 γ = 0 =⇒ γ = a tb that γ is contained in the plane passing through the point with position vector a and parallel to the vectors b and c (i.e. b t2 c, where a, b, c are constant vectors. We have k γ˙ (t) k2 = 2 cosh2 t − 2 cosh t so the arc-length is Z ap Z ap 2 2 cosh t − 2 cosh t dt = 2 2 cosh2 t − 2 cosh t dt. Solutions to the Exercises in Elementary Differential Geometry Chapter 1 1.1.1 It is a parametrization of the part of the parabola with x ≥ 0. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |