Final Review II: Examples

Transcription

1 Final Review II: Examles We go through last year's nal. This should not be treated as a samle nal. Question. Let (s) cos s; cos s; sin s. a) Prove that s is the arc length arameter. b) Calculate ; ; T ; N ; B. Solution. a) We calculate b) We have which gives We further have Thus T _(s) sin s (s) ; sin s ; cos s ) k_(s)k : () cos s ; N ; cos s ; sin s ; () cos s ; cos s ; sin s : (3) sin s (s) ; sin s ; cos s : (4) (_(s) (s)) (s) (s) : (5)

2 Dierential Geometry of Curves & Surfaces Question. (5 ts) Let C be a regular curve with curvature / at C. Let C~ be the orthogonal rojection of C onto the osculating lane (the lane sanned by T and N) at. Prove that C and C~ have the same curvature at. Proof. Let C be arametrized as (s) where s is the arc length arameter. Wlog x(). Denote T (); N(); B(), (), () by T ; N ; B, ;. Then C~ is given by ~(s) (s) (((s) ()) B ) B : (6) Note that s in general is not the arc length arameter of ~. Taking derivative we have Consequently ~_() T. Now by the Frenet-Serret equations Therefore ~_(s) _(s) (_(s) B ) B T (s) (T (s) B ) B : ~(s) (s) [N(s) (N(s) B ) B ] ) ~() N (7) ~() k~ _() ~()k k~_()k 3 (): (8) Thus ends the roof.

3 Question 3. Let S be a surface with rst fundamental form du + (u + ) dv. Calculate i. The angle between the curves u + v and u v ; ii. The area of the curvilinear triangle bounded by u v; u v; v. Solution. i. The intersection of the two curves is at u v, where we have E ; G ; F. We arametrize: u + v : u(t) t; v(t) t; (9) u v : U(t) V (t) t: () Thus E u_ U _ + F u_ V_ + U_ v_ + G v_ V_ cos E u_ + F u_ v_ + G v_ E U_ + F U_ V_ + G V_ : () Thus /. ii. The region bounded by u v; u v; v is f(u; v)j 6 v 6 ; v 6 u 6 vg f(u; v)j 6 u 6 ; 6 v 6 jujg: () We have Area E G F du dv (u + ) du dv v v 7 6 : v 3 (u + ) du 3 + v dv dv 3

4 Dierential Geometry of Curves & Surfaces Question 4. ( ts) Consider the surface atch (u; v) (a (u + v); b (u v); u v) where a; b > are constants. Calcualte at (; ; ) its rincial curvatures, rincial vectors, mean curvature, and Gaussian curvature. Solution. We see that (; ). Now calculate which leads to Next we have and u (a; b; v); v (a; b; u) (3) E a + b + 4 v ; F a b + 4 u v; G a + b + 4 u : (4) N u v k u v b (u + v) + a (u v) + a b At (; ) these become b (u + v) a (v u) a b A; (5) uu (; ; ); uv (; ; ); vv (; ; ): (6) E a + b G; F a b ; (7) N (; ; ); uu (; ; ); uv (; ; ); vv (; ; ): (8) Consequently at (; ) L N; M : (9) We solve " det a + b a b a b a + b!# () which gives b ; a ; H a b For the rincial vectors, we solve c ; c. a b ; K a b : ()! a + b a b i i a b a + b i which gives t c [ u v ] c@ b A; t c ( u + v ) c@ a A: () 4

5 Question 5. Consider the surface atch (u; v) (u cos v; u sin v; v). a) Calculate the Christoel symbols k ij, i; j; k ;. b) Is u a geodesic? Justify your claim. Solution. a) We calculate thus Furthermore u (cos v; sin v; ); v ( u sin v; u cos v; ); (3) u v (sin v; cos v; u): (4) uu (; ; ); uv ( sin v; cos v; ); vv ( u cos v; u sin v; ): (5) Thus clearly. Furthermore ; u u + ; (6) u; : (7) b) We arametrize u by arc length as u(s) ; v(s) v(s). To nd v(s) we calculate Thus v(s) satises E ; F ; G + u : (8) along the curve. Consequently we can take v(s) s/. Now along the curve u we have G v_ v_ (9) ; ; (3) and other k ij 's are all zero. The geodesic equations now read u v_ ; (3) v+ u_ v_ : (3) We see that (3) is not satised. So u is not a geodesic. 5

6 Dierential Geometry of Curves & Surfaces Question 6. Let S be a surface with non-ositive Gaussian curvature. Prove that there cannot be a smooth closed geodesic that bounds a simly connected region. Proof. Assume there is such a curve. Denote it by C and the region enclosed by. Gauss- Bonnet now gives K + g 6 g : (33) C C Contradiction. 6

7 Question 7. Let be a sace curve. Let S be the ruled surface generated by and its binormals. Prove that is a geodesic on S. Proof. (Short) The tangent lane of S is sanft ; Bg thus T_ k N?T S. Consequently r T. Proof. (Long) We arametrize by its arc length u. Then S is given by the surface atch We have and (u; v) (u) + v B(u): (34) u T v N ; v B (35) uu N v ( T + B) v T + N v B; uv N ; vv : (36) Next we have Thus We then solve which gives so Similarly u v N v T ) N S N v T : (37) + v L + v ; M ; N : (38) + v v T + N v B uu (T v N) + B + (N + v T ) (39) v + v; (4) v + ; (4) v ; (4) ; v: (43) N (T v N) + B N + v T + v ) v + v ; ; (44) and. The unit tangent vector T is given by u_(t) u + v_(t) v with u(t) t; v(t). Note that by our setu t is the arc length here. We check u + (u_ ; v_) u_ ij v_ + ; v+ (u_ ; v_) u_ ij v_ + : (45) as v along the curve u(t) t; v(t). 7