CURVES ANDRÉ NEVES. that is, the curve α has finite length. v = p q p q. a i.e., the curve of smallest length connecting p to q is a straight line.


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1 CURVES ANDRÉ NEVES 1. Problems (1) (Ex 1 of 1.3 of Do Crmo) Show tht the tngent line to the curve α(t) (3t, 3t 2, 2t 3 ) mkes constnt ngle with the line z x, y. (2) (Ex 6 of 1.3 of Do Crmo) Let α(t) (e bt cos t, e bt sin t) be curve in R 2 with >, b <. Show tht s t +, the curve α(t) tens to the origin spirling roun it. Show tht α (t) (, ) s t + n lim α (s) s is finite, t t tht is, the curve α hs finite length. (3) (Ex 1 of 1.3 of Do Crmo) Let α : [, b] R 3 be curve with α() p, α(b) q. Show tht for ny vector v Set n show tht (p q). v v p q p q α() α(b) α (t). vt. α (t) t, i.e., the curve of smllest length connecting p to q is stright line. (4) Show tht the length of curve is well efine, i.e., oes not epen on the prmetriztion chosen. (5) Let f : R R be smooth function. Consier the curve c : R R, c(t) (t, f(t)). Compute the geoesic curvture k(t). 1
2 2 ANDRÉ NEVES HINT: The curve c is not prmetrizes by rc length n so we cnnot just compute c. You coul procee s follows. Let t t(s) be chnge of prmeter so tht (s) c(t(s)) is prmetrize by rclength. We cn ssume tht t (s) >. First show tht curvture of is given by k(s) (t (s)) 2 2 c (t) 2 (t(s)).ν(s). where ν is the norml vector to the curve. Then show tht k(t) ċ(t) 2 c(t).ν(t), where ν is the norml vector to the curve c. Finlly use this formul to solve the exercise. (6) Consier close curve c : [, L] R 3 prmetrize by rclength n ssume the existence of φ : [, L] [, ε] R 3 so tht φ(t, ) c(t); For every s fixe, c s (t) φ(t, s) is prmetrize curve. Show tht length (c s ) s s φ k(t). (t, )t. s The formul bove tells you tht if you wnt to ecrese the length of curve in the most effective wy, then you shoul eform it in the irection of its geoesic curvture. (7) Show tht if the geoesic curvture of curve is zero, then the curve prmetrizes stright line 2. Solutions (1) The tngent line to the curve α(t) when t t is spnne by α (t ) (3, 6t, 6t 2 ). The line given in the problem is spnne by (1,, 1). The cosine of the ngle between these to curves is given by cos θ α (t ).(1,, 1) α (t ) 3 + 6t t t 4 n so the ngle oes not epen on t. 1 2
3 CURVES 3 (2) To keep it simple I will just use 1, b 1. To solve ) we just note tht n this implies lim t e t lim cos t te t lim sin te t. t Hence the curve tens to the origin when t. Why spirling? Becuse for every line L θ {x(cos θ, sin θ) x R} mking n ngle θ with the xxis, the curve α crosses it infinitely mny times (whenever t θ + kπ for ll k Z) s t tens to infinity. For b) we hve α (t) t 2 e t t 2(e t e t ) t t n so 2 lim α (t) t e t. t t (3) For the first ientity we write the curve α(t) (x 1 (t), x 2 (t), x 3 (t)) n pply the funmentl theorem of clculus to obtin (p q). v 3 (p i q i )v i i1 3 i1 x i(t)t v i 3 i1 x i(t)v i t α (t). vt. The secon inequlity comes from the fct tht if v is unit vector then for every other vector u we hve v. u u n so α (t). vt We now prove b). Using v p q hve p q α (t) t. α() α(b) p q (p q). v in the formul erive in ) we α (t) t. (4) Let α : [, b] R 3 be curve in spce n choose ny smooth bijection φ : [c, ] [, b]. We cn ssume tht φ (s) for ll s (it is either tht or φ (s) for ll s). Consier the new curve β α φ : [c, ] R 3. The curve β is just reprmetriztion of α n the thing we nee to check is tht they hve the sme length. The prmeter of the curve α is enote
4 4 ANDRÉ NEVES by t n the one of the curve β is enote by s. length (β) c φ() φ(c) β (s) s c α (t) t length (α). α (φ(s)) φ (s)s φ() φ(c) α (t) t (5) Consier φ : R R so tht α c φ becomes prmetrize by rclength. We cn ssume tht φ (s) for ll s. The generl formul is, with α (s) (x (s), y (s)), k(s) α (s).ν(s) where ν(s) ( y (s), x (s)). On the other hn we hve, with t φ(s), α (s) c (t)φ (s) n α (s) c (t)(φ (s)) 2 + c (t)φ (s). Now c (t).ν(s) (why?) n so k(s) (φ (s)) 2.c (t).ν(s). Moreover 1 α (s) c (t) φ (s) n so φ (s) c (t) 1. Thus k(t) c (t) 2 c (t).ν(t) where ν(t) is the norml vector to the curve c (which is nothing but the norml vector to the curve α becuse the curve re the sme!). We hve c (t) (, f (t)) n ν(t) (f (t)) 2 ( f (t), 1). Therefore k(t) f (t) (1 + (f (t)) 2 ) 3/2. (6) Shoul hve been more explicit but it goes without sying tht c s () c s (L) n c s() c s(l) for ll s becuse c s is close curve. In prticulr this implies tht s φ(l, s) s φ(, s) for ll s. This ientities will be use in the lst line of this exercise. We hve, becuse of rclength prmetriztion s tφ(s, t) s 2 stφ(t, ). t φ(t, ) 2 t φ(t, ) stφ(t, ). t φ(t, ).
5 Thus s length(c s) s s 2 stφ(t, ). t φ(t, )t CURVES 5 φ/ t(t, s) t [ s φ(t, ). t φ(t, )] L s tφ(s, t) s t t ( s φ(t, ). t φ(t, )) s φ(t, ). 2 ttφ(t, )t s φ(t, ). 2 ttφ(t, )t φ k(t). (t, )t. s (7) Assume the curve c is prmetrize by rclength. Then c (t).c (t) 1 for ll t implies, fter ifferentition tht c (t).c (t) for ll t. We re ssuming tht k(t) c (t).ν(t) is zero for ll t n so we must hve c (t) for ll t. Why? Well, {c (t), ν(t)} is n orthonorml bsis of R 2 n so the two previous ientities imply inee tht c (t) for ll t. Integrtion in ech component implies tht c (t) v is constnt for ll t n thus c(t) tv + for some vector. This finishes the exercise
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