Question

7.(16 points) Consider the curve F(t) = 4 cos(t)ī + 4 sin(t); +3tk. (a) Find the unit tangent vector T(t) and the unit normal vector function Ñ (t) at the point (-4,0,37). (b) Compute the curvature k at the point (-4,0,31).

Answer 1

The given anum ģ(t): 4 COSL+ C + 4 sint) (b) tzt 0- ) differentiate (i with respect we have. (+) usint ê + 4 cost Ĵ + 2* to 't' Now 11 7' (t)) = [-u Bint)2+ e cost)2+(332 t J 16 sint+ 16 Coszt + 9 11 ( Then 16+9 25 5 cmit Tomgent vector is &lt) Tita 11710 - $+270.5 costi 5 mis

know that so Filt) Now We mit Noomal vector 16 given by Nit) T'It) 11710) differentüte @ with respect to t. - cost -cost 2-4 sint tok [ta - 4 cost. C- y sint ġ. 11 7'ct)) lys cost)2+(-45 Sint)? [cos²z+8in²t] IT'(tl). 5 51825 oll Then amit Normal vector is. ÑUE Filt) - cost î – sint il IF(t)/1 (X,9,2)-(-4,0,3) Notu. te → t= (-4,0,31 ) For X = 4 Grest Y=usint 2= 35, + 4COS I Ĵ + 3 so Fit) (zit TOT) sinta - } + 3 5 AB Mit) lt == Ñ LT) ,-.COS(IT) Î – Sinctij o Ñ (T)= ti?

(b) we know that curvature of curve is given by. K= 118' (t) x 8"(t) | Il rit)113 si gilt)= -u Bint + 4costſ + 3R gla -4 9 + 3 -4 cost îs usint Now. ģ"(t) = t=17' (X,9,2)=(-4,034 "(t) = - 4 Then 7'000) x (T) = n 2 ری O: -4. 3 4 0 ů (0-o] – } [o-123 +R [016] ÖLIT)X?"60 129 +1GR. Nowe ll 31(c)xä"(t) 11 5 (123P+ (16)2 J 144+ 256 = 400 20 11 glut) x7"190)11 Nowe 11 lut)ll: J64)2 +63)2.</25 - 5 Then curvatures 20 KE (503 KE 0.16. K= 145 =)