Question

1) For this problem use the following space curve: Ft) =< 1,3 sin(t), 3 cos(t) > a) Determine the unit tangent vector: T. b) Determine the unit normal vector: N. c) Determine the curvature of this space curve at the point: (0,0,3). d) Determine the arc length of the curve between t O and t=1

Answer 1

T(t) = < t, 3sin Lt, 31056) > (a) determine the unit tangent vectores 8(t) = tît 3 Sint) 9 + 3 cos Lt) Å Tangent vector drit) dt ł + 3 cost ] -3 sincta =< t, 3 cosct), -3 Sinits) Unit Tangent vector 119 CDI = V(4) - + (3 cos (t))+(-3 sinu) =r1+9 = To unit Tangent rectus 1 + 3 ws(t] -3 sincti Vio <io 3 cos(t) - 3sinct T (t) = vios VTO

(6) determine the unit Normal Vector N NCH = TILE I T' TCH = Cut 3 cast Vio -3 sint > Vio dT dt - Los Lo -3 sint -3 sint 3 cost Vio > Vio -3 cost = vgsn't +gost 3 11 g Nit = 40, -3sint Vio -3cast Vio 3 Vio Nit) = (0, -sint, -cost>

() determine the cunature of this space wwe at the point: (0,0,3) Curvature = 11 T'All 11 WWII K(+) = 11 r'(t) x 8 "(H11 118'(+) 113 =co, -35nt, -3 Bost> rict 8 (4) = 1 3 cost -3 sint -3 sint - cost e = ? (-9cost - 9 st) -] (-360st) + < (-35nt) = 1 (-9) +3] (cost) - 3 sont Ê --gî + 3custj-sint Ê 2-9, 30st, -3 sint> 1181 H x5 11(H) = √ 81 + gusht tasn't V 81+9 3 vo V 90 =3 2 KH)=1/50 Vio &'(t) 1 = V10

@ determine the arre length of the curve bet tro and t=1 8(t) = (t 3 sint, 3 st) x = t y=3 sint 2=3 cost dy JE = 3 cost dz -3 sint dt = 1 dt 1 1 (anc length) L = (45) +4*47) VI + (40387 + 1-24mt2 dt o 1 1 ttg dt N W to NTG [ VO (1-0) L