Tangent, Normal and Binormal Vectors — Question 9

Question 9 – Solution

Strategy Let $\theta=t^3$ and separate angular position from the positive rate $d\theta/dt$.

Frame Because $3t^2>0$, $$\mathbf T=\left\langle-\sin(t^3),\cos(t^3),0\right\rangle,\quad \mathbf N=\left\langle-\cos(t^3),-\sin(t^3),0\right\rangle,\quad \mathbf B=\left\langle 0,0,1\right\rangle.$$

Dependence The frame at a geometric point depends on position and orientation, not on how fast that point is reached. The speed $\|\vec r'\|=6t^2$ varies, but normalization removes it. At $t=0$ the parametrization stops, so the usual Frenet construction from velocity is not regular there.