Question 3 Find the Frenet frame for rβ(t)=β¨t,t2,t3β©\vec r(t)=\left\langle t,t^2,t^3\right\rangle at t=1t=1. Use the cross-product method for π©\mathbf B and verify the orientation. Show solutionHide solution+Question 3 β Solution Strategy Use π»=rββ²/β₯rββ²β₯\mathbf T=\vec r'/\|\vec r'\|, π©=(rββ²Γrββ³)/β₯rββ²Γrββ³β₯\mathbf B=(\vec r'\times\vec r'')/\|\vec r'\times\vec r''\|, and π΅=π©Γπ»\mathbf N=\mathbf B\times\mathbf T. See the diagram in the original worksheet below. Derivatives At t=1t=1, rββ²=β¨1,2,3β©\vec r'=\left\langle 1,2,3\right\rangle and rββ³=β¨0,2,6β©\vec r''=\left\langle 0,2,6\right\rangle. Thus π»=114β¨1,2,3β©,π©=119β¨3,β3,1β©.\mathbf T=\frac 1{\sqrt{14}}\left\langle 1,2,3\right\rangle,\qquad \mathbf B=\frac 1{\sqrt{19}}\left\langle 3,-3,1\right\rangle. Normal Therefore π΅=π©Γπ»=1266β¨β11,β8,9β©.\boxed{\mathbf N=\mathbf B\times\mathbf T=\frac 1{\sqrt{266}}\left\langle-11,-8,9\right\rangle}. Verification Direct cross multiplication gives π»Γπ΅=π©\mathbf T\times\mathbf N=\mathbf B and all norms equal 1.