Curvartura

Curvatura
La curvatura de la curva dada por la función vectorial r
r't× r''(t) r't3
Ej 1)
Halle la curvatura de la cubica alabeadart=(t,t2,t3) en un punto general y en (0,0,0)
r't=(1,2t,3t)
r''t=(1,2,3)
r't× r''(t)= ijk12t3t2026t
=i12t2-6t2-j6t-0+k2-0=6t2,-6t,2
kt=(6t2)2+(-6t)2+22(1)2+(2t)2+(3t2)23
kt=36t4+36t2+41+4t2+9t43
kt=kt=4(9t4+9t2+1)1+4t2+9t432
kt=29t4+9t2+11+4t2+9t432
Evaluando t enel punto (0,0,0)
k0=29(0)4+9(0)2+11+4(0)2+9(0)432
kt=21132=2(1)1=21=2
Resultado = 2
Ej 2) rt=(etcost,etsent,t) en el punto (1,0,0)rt=r't× r'' r't3
r't=(-etsent+etcost,etcost+etsent,1)
r''t=-etcost-etsent-etsent+etcost,-etsent+etcost+etcost+etsent,0
r''t=-2etsent,2etcost,0r't× r''t= ijk-etsent+etcostetcost+etsent1-2etsent2etcost0
=i-2etcost-j2etsent+k-2e2tsentcost+-2e2tcos2t+2e2tsentcost +2e2tsen2t=-2etcost,-2etsent ,2e2t cos2t+sen2t
cos2t+sen2t=1
kt=-2etcost2+-2etsent2+2e2t2-etsent+etcost2+etcost+etsent2+132kt=4e2tcos2t+4e2tsen2t+4e4te2tsen2t-2e2tsent cost+e2tcos2t+e2tcos2t+2e2tsent cost+e2tsen2t+132
kt=4e2t(cos2t+sen2t)+4e4te2tsen2t-2e2tsent cost+e2tcos2t+e2tcos2t+2e2tsentcost+e2tsen2t+132
kt=4e2t(1+e2t)e2tsen2t-2e2tsent cost+e2tcos2t+e2tcos2t+2e2tsent cost+e2tsen2t+132
kt=2et1+e2t2e2tsen2t+2e2tcos2t+132kt=2et1+e2t2e2t(sen2t+cos2t)+132
kt=2et1+e2t2e2t+132

Evaluando t en 0
k0=2e01+e2(0)2e2(0)+132
kt=22332=2.8284275.196152=0.544336
Resultado = 0.544336