$\stackrel{\.}{\mathbf{R}}\=\left[\begin{array}{c}-3p^2\+3\\ 6p\\ 3p^2\+3\end{array}\right]$

$\mathbf{T}\prime \=\stackrel{\.}{\mathbf{T}}\/\mathrm{\ρ}\=\left[\begin{array}{c}-\frac{2}{3}\frac{p}{{\left(p^2\+1\right)}^3}\\ -\frac{1}{3}\frac{p^2-1}{{\left(p^2\+1\right)}^3}\\ 0\end{array}\right]$

$\mathrm{\ρ}\=∥\stackrel{\.}{\mathbf{R}}∥\=3\sqrt{2}\left(p^2\+1\right)$

$\mathrm{\κ}\=\parallel \mathbf{T}\prime \parallel$ = $\frac{1}{3{\left(p^2\+1\right)}^2}$

$\mathbf{T}\=\stackrel{\.}{\mathbf{R}}\/\mathrm{\ρ}\=\left[\begin{array}{c}-\frac{1}{2}\frac{\sqrt{2}\left(p^2-1\right)}{p^2\+1}\\ \frac{\sqrt{2}p}{p^2\+1}\\ \frac{1}{2}\sqrt{2}\end{array}\right]$

$\mathbf{N}\=\mathbf{T}\prime \/\mathrm{\κ}\=\left[\begin{array}{c}-\frac{2p}{p^2\+1}\\ -\frac{p^2-1}{p^2\+1}\\ 0\end{array}\right]$

$\mathbf{B}\=\mathbf{T}\times \mathbf{N}\=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ -\frac{1}{2}\frac{\sqrt{2}\left(p^2-1\right)}{p^2\+1}& \frac{\sqrt{2}p}{p^2\+1}& 1\/\sqrt{2}\\ -\frac{2p}{p^2\+1}& -\frac{p^2-1}{p^2\+1}& 0\end{array}\right|$ =  $\left[\begin{array}{c}\frac{1}{2}\frac{\sqrt{2}\left(p^2-1\right)}{p^2\+1}\\ -\frac{\sqrt{2}p}{p^2\+1}\\ \frac{1}{2}\sqrt{2}\end{array}\right]$

Table 2.6.4(a)   Manual calculation of the TNB-frame

$\mathrm{\tau }\= -\left(\stackrel{\.}{\mathbf{B}}\/\mathrm{\rho }\right)·\mathbf{N}$

$\=-\left[\begin{array}{c}\frac{2}{3}\frac{p}{{\left(p^2\+1\right)}^3}\\ \frac{1}{3}\frac{p^2-1}{{\left(p^2\+1\right)}^3}\\ 0\end{array}\right]·\left[\left[\begin{array}{c}-\frac{2p}{p^2\+1}\\ -\frac{p^2-1}{p^2\+1}\\ 0\end{array}\right]\right]$

$$  $\=-\left(-\frac{4}{3}\frac{p^2}{{\left(p^2\+1\right)}^4}-\frac{1}{3}\frac{{\left(p^2-1\right)}^2}{{\left(p^2\+1\right)}^4}\right)$

$\=\frac{1}{3{\left(p^2\+1\right)}^2}$

$\stackrel{\.}{\mathbf{T}}\left(p\right)·\stackrel{\.}{\mathbf{B}}\left(p\right)$

$\=\left[\begin{array}{c}-\frac{2\sqrt{2}p}{{\left(p^2\+1\right)}^2}\\ -\frac{\sqrt{2}\left(p^2-1\right)}{{\left(p^2\+1\right)}^2}\\ 0\end{array}\right]·\left[\begin{array}{c}\frac{2\sqrt{2}p}{{\left(p^2\+1\right)}^2}\\ \frac{\sqrt{2}\left(p^2-1\right)}{{\left(p^2\+1\right)}^2}\\ 0\end{array}\right]$

$\=-\frac{8p^2}{{\left(p^2\+1\right)}^4}-\frac{2{\left(p^2-1\right)}^2}{{\left(p^2\+1\right)}^4}$

$\=-\frac{2}{{\left(p^2\+1\right)}^2}$

$-\mathrm{\kappa } \mathrm{\tau } {\mathrm{\rho }}^2$

$\=-\frac{1}{3{\left(p^2\+1\right)}^2} \frac{1}{3{\left(p^2\+1\right)}^2} {\left(3\sqrt{2}\left(p^2\+1\right)\right)}^2$

$\=-\frac{2}{{\left(p^2\+1\right)}^2}$

Loading Student:-VectorCalculus

$\mathrm{BasisFormat}\left(\mathrm{false}\right)\:$

$\mathbf{R}\=⟨3 p-p^3\,3 p^2\,3 p\+ p^3⟩$$\stackrel{\text{assign}}{\to }$

$\mathbf{R}$$\left[\begin{array}{c}-p^3\+3p\\ 3p^2\\ p^3\+3p\end{array}\right]$ = $\stackrel{\text{TNB frame}}{\to }$$\left[\begin{array}{c}-\frac{1}{2}\frac{\sqrt{2}\left(p^2-1\right)\mathrm{csgn}\left(p^2\+1\right)}{p^2\+1}\\ \frac{\sqrt{2}\mathrm{csgn}\left(p^2\+1\right)p}{p^2\+1}\\ \frac{1}{2}\sqrt{2}\mathrm{csgn}\left(p^2\+1\right)\end{array}\right]\,\left[\begin{array}{c}-\frac{1}{2}\frac{\sqrt{2}\left(\mathrm{csgn}\left(1\,p^2\+1\right)p^4\+4\mathrm{csgn}\left(p^2\+1\right)p-\mathrm{csgn}\left(1\,p^2\+1\right)\right)}{\sqrt{\frac{{\mathrm{csgn}\left(1\,p^2\+1\right)}^2{p}^4\+2{\mathrm{csgn}\left(1\,p^2\+1\right)}^2{p}^2\+{\mathrm{csgn}\left(1\,p^2\+1\right)}^2\+2}{{\left(p^2\+1\right)}^2}}{\left(p^2\+1\right)}^2}\\ \frac{\sqrt{2}\left(\mathrm{csgn}\left(1\,p^2\+1\right)p^3-\mathrm{csgn}\left(p^2\+1\right)p^2\+\mathrm{csgn}\left(1\,p^2\+1\right)p\+\mathrm{csgn}\left(p^2\+1\right)\right)}{\sqrt{\frac{{\mathrm{csgn}\left(1\,p^2\+1\right)}^2{p}^4\+2{\mathrm{csgn}\left(1\,p^2\+1\right)}^2{p}^2\+{\mathrm{csgn}\left(1\,p^2\+1\right)}^2\+2}{{\left(p^2\+1\right)}^2}}{\left(p^2\+1\right)}^2}\\ \frac{1}{2}\frac{\sqrt{2}\mathrm{csgn}\left(1\,p^2\+1\right)}{\sqrt{\frac{{\mathrm{csgn}\left(1\,p^2\+1\right)}^2{p}^4\+2{\mathrm{csgn}\left(1\,p^2\+1\right)}^2{p}^2\+{\mathrm{csgn}\left(1\,p^2\+1\right)}^2\+2}{{\left(p^2\+1\right)}^2}}}\end{array}\right]\,\left[\begin{array}{c}\frac{p^2-1}{\sqrt{\frac{{\mathrm{csgn}\left(1\,p^2\+1\right)}^2{p}^4\+2{\mathrm{csgn}\left(1\,p^2\+1\right)}^2{p}^2\+{\mathrm{csgn}\left(1\,p^2\+1\right)}^2\+2}{{\left(p^2\+1\right)}^2}}{\left(p^2\+1\right)}^2}\\ -\frac{2p}{\sqrt{\frac{{\mathrm{csgn}\left(1\,p^2\+1\right)}^2{p}^4\+2{\mathrm{csgn}\left(1\,p^2\+1\right)}^2{p}^2\+{\mathrm{csgn}\left(1\,p^2\+1\right)}^2\+2}{{\left(p^2\+1\right)}^2}}{\left(p^2\+1\right)}^2}\\ \frac{1}{\sqrt{\frac{{\mathrm{csgn}\left(1\,p^2\+1\right)}^2{p}^4\+2{\mathrm{csgn}\left(1\,p^2\+1\right)}^2{p}^2\+{\mathrm{csgn}\left(1\,p^2\+1\right)}^2\+2}{{\left(p^2\+1\right)}^2}}\left(p^2\+1\right)}\end{array}\right]$$\stackrel{\text{assuming real}}{\to }$$\left[\begin{array}{c}-\frac{1}{2}\frac{\sqrt{2}\left(p^2-1\right)}{p^2\+1}\\ \frac{\sqrt{2}p}{p^2\+1}\\ \frac{1}{2}\sqrt{2}\end{array}\right]\,\left[\begin{array}{c}-\frac{2p}{p^2\+1}\\ -\frac{p^2-1}{p^2\+1}\\ 0\end{array}\right]\,\left[\begin{array}{c}\frac{1}{2}\frac{\sqrt{2}\left(p^2-1\right)}{p^2\+1}\\ -\frac{\sqrt{2}p}{p^2\+1}\\ \frac{1}{2}\sqrt{2}\end{array}\right]$$\stackrel{}{\to }$