Find the versors of the Frenet Triad

Question

I have the following curve:

$C(t)=(\frac{t^5}{5}, \frac{t^3}{3}, t ^ 2).$

The problem is to find $T$, $N$ and $B$ of the Frenet Frame.

I know the fact that $\vec{B} = \vec{T} \times \vec{N}$.

I've already looked into this questions's answer. I got everything, besides this fact $\frac{\gamma'(x)}{|\gamma'(x)|}=\frac{1}{\sqrt{3}}(\cos(x)-\sin(x),\sin(x)+\cos(x),1)$.

It's unclear to me what $|\gamma'(x)|$ really is.

So that is, computing $\vec{T}$ and $\vec{B}$ is what I've been struggling. That's the only problem I face.

Thanks for any help.

Answer 1

$\gamma(x)$ there is equivalent to the $C(t)$ here.

$C'(t)=(t^{4},t^{2},2t)$

$|C'(t)|=\sqrt{t^{8}+t^{4}+4t^{2}}=t\sqrt{t^{6}+t^{2}+4}$

$\displaystyle \mathbf{T}=\frac{1}{\sqrt{t^{6}+t^{2}+4}} \left( \begin{array}{c} t^{3} \\ t \\ 2 \end{array} \right)$

With careful manipulation, we have

$\displaystyle \mathbf{B}=\frac{1}{\sqrt{t^{6}+9t^{4}+1}} \left( \begin{array}{c} -1 \\ 3t^{2} \\ -t^{3} \end{array} \right)$

$\displaystyle \mathbf{N}= \frac{1}{\sqrt{(t^{6}+t^{2}+4)(t^{6}+9t^{4}+1)}} \left( \begin{array}{c} t^{2}(t^{2}+6) \\ 2-t^{6} \\ t(3t^{4}+1) \end{array} \right)$

$\displaystyle \kappa= \frac{2\sqrt{t^{6}+9t^{4}+1}}{t(t^{6}+t^{2}+4)^{\frac{3}{2}}}$

$\displaystyle \tau= -\frac{3}{t^{6}+9t^{4}+1}$