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

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

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

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

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

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

Table 2.7.15(a)   Frenet formalism

$\mathbf{d}\=\mathrm{\tau } \mathbf{T}\+\mathrm{\kappa } \mathbf{B}$

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

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

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

$\frac{d\mathbf{N}}{\mathrm{dp}}\frac{1}{\mathrm{\ρ}}$ = $\frac{\sqrt{2}}{3{\left(p^2\+1\right)}^3}\left[\begin{array}{c}{p}^2-1\\ -2 p\\ 0\end{array}\right]$ = $\mathbf{d}\times \mathbf{N}$ 

$\frac{\mathrm{dB}}{\mathrm{dp}}\frac{1}{\mathrm{\ρ}}$ = $\frac{1}{3{\left(p^2\+1\right)}^3}\left[\begin{array}{c}2 p\\ p^2-1\\ 0\end{array}\right]$ = $\mathbf{d}\times \mathbf{B}$ 

$\frac{d\mathbf{T}}{\mathrm{dp}}\frac{1}{\mathrm{\rho }}\times \frac{d^2\mathbf{T}}{{\mathrm{dp}}^2}\frac{1}{{\mathrm{\rho }}^2}$= $\frac{\sqrt{2}}{27{\left(p^2\+1\right)}^6}\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$ = ${\mathrm{\kappa }}^2 \mathbf{d}$ 

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

Frenet formalism: $\mathrm{\ρ}\,\mathrm{\κ}\,\mathrm{\τ}\,\mathbf{T}\,\mathbf{N}\,\mathrm{B}$ 

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

$∥\frac{\ⅆ}{\ⅆ p} \mathbf{R}∥$ = $3\sqrt{2}\sqrt{{\left(p^2+1\right)}^2}$$\stackrel{\text{assuming positive}}{\to }$$3\sqrt{2}\left(p^2+1\right)$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{\rho }$$$$$

$\mathbf{R}$ = $\left[\begin{array}{c}-p^3\+3p\\ 3p^2\\ p^3\+3p\end{array}\right]$$\stackrel{\text{curvature}}{\to }$$\frac{\sqrt{{\left(-\frac{\sqrt{2}\left(-3p^2+3\right)\mathrm{csgn}\left(1\,p^2+1\right)}{6{\mathrm{csgn}\left(p^2+1\right)}^2\left(p^2+1\right)}-\frac{\sqrt{2}\left(-3p^2+3\right)p}{3\mathrm{csgn}\left(p^2+1\right){\left(p^2+1\right)}^2}-\frac{\sqrt{2}p}{\mathrm{csgn}\left(p^2+1\right)\left(p^2+1\right)}\right)}^2+{\left(-\frac{\sqrt{2}p\mathrm{csgn}\left(1\,p^2+1\right)}{{\mathrm{csgn}\left(p^2+1\right)}^2\left(p^2+1\right)}-\frac{2\sqrt{2}{p}^2}{\mathrm{csgn}\left(p^2+1\right){\left(p^2+1\right)}^2}+\frac{\sqrt{2}}{\mathrm{csgn}\left(p^2+1\right)\left(p^2+1\right)}\right)}^2+{\left(-\frac{\sqrt{2}\left(3p^2+3\right)\mathrm{csgn}\left(1\,p^2+1\right)}{6{\mathrm{csgn}\left(p^2+1\right)}^2\left(p^2+1\right)}-\frac{\sqrt{2}\left(3p^2+3\right)p}{3\mathrm{csgn}\left(p^2+1\right){\left(p^2+1\right)}^2}+\frac{\sqrt{2}p}{\mathrm{csgn}\left(p^2+1\right)\left(p^2+1\right)}\right)}^2}\sqrt{2}}{6\mathrm{csgn}\left(p^2+1\right)\left(p^2+1\right)}$$\stackrel{\text{assuming positive}}{\to }$$\frac{1}{3{\left(p^2+1\right)}^2}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{\kappa }$

$\mathbf{R}$ = $\left[\begin{array}{c}-p^3\+3p\\ 3p^2\\ p^3\+3p\end{array}\right]$$\stackrel{\text{torsion}}{\to }$$\frac{\mathrm{csgn}\left(p^2+1\right)\sqrt{2}}{3\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)}^3}$$\stackrel{\text{assuming positive}}{\to }$$\frac{1}{3{\left(p^2+1\right)}^2}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{\tau }$

$\mathbf{R}$ = $\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{\sqrt{2}\left(p^2-1\right)\mathrm{csgn}\left(p^2+1\right)}{2\left(p^2+1\right)}\\ \frac{\sqrt{2}\mathrm{csgn}\left(p^2+1\right)p}{p^2+1}\\ \frac{\mathrm{csgn}\left(p^2+1\right)\sqrt{2}}{2}\end{array}\right],\left[\begin{array}{c}-\frac{\sqrt{2}\left(\mathrm{csgn}\left(1\,p^2+1\right)p^4+4p\mathrm{csgn}\left(p^2+1\right)-\mathrm{csgn}\left(1\,p^2+1\right)\right)}{2\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-p^2\mathrm{csgn}\left(p^2+1\right)+\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{\sqrt{2}\mathrm{csgn}\left(1\,p^2+1\right)}{2\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}{\left(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]$$\stackrel{\text{assuming positive}}{\to }$$\left[\begin{array}{c}\frac{\left(-p^2+1\right)\sqrt{2}}{2p^2+2}\\ \frac{\sqrt{2}p}{p^2+1}\\ \frac{\sqrt{2}}{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{\sqrt{2}\left(p^2-1\right)}{2p^2+2}\\ -\frac{\sqrt{2}p}{p^2+1}\\ \frac{\sqrt{2}}{2}\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$Q$

$\mathbf{T}\=Q_1$$\stackrel{\text{assign}}{\to }$

$\mathbf{N}\=Q_2$$\stackrel{\text{assign}}{\to }$

$\mathbf{B}\=Q_3$$\stackrel{\text{assign}}{\to }$

Obtain and display the Darboux vector

$\mathbf{d}\=\mathrm{\τ} \mathbf{T}\+\mathrm{\κ} \mathbf{B}$$\stackrel{\text{assign}}{\to }$

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

$\mathbf{T}\prime \=\mathbf{d}\times \mathbf{T}$ 

$\left(\frac{\ⅆ}{\ⅆ p} \mathbf{T}\right)\/\mathrm{\ρ}$ = $\left[\begin{array}{c}\frac{1}{6}\frac{\sqrt{2}\left(-\frac{2p\sqrt{2}}{2p^2\+2}-\frac{4\left(-p^2\+1\right)\sqrt{2}p}{{\left(2p^2\+2\right)}^2}\right)}{p^2\+1}\\ \frac{1}{6}\frac{\sqrt{2}\left(-\frac{2\sqrt{2}{p}^2}{{\left(p^2\+1\right)}^2}\+\frac{\sqrt{2}}{p^2\+1}\right)}{p^2\+1}\\ 0\end{array}\right]$$\stackrel{\text{assuming positive}}{\to }$$\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]$$$

$\mathbf{d}\times \mathbf{T}$ = $\left[\begin{array}{c}-\frac{2}{3}\frac{p}{{\left(p^2\+1\right)}^3}\\ -\frac{1}{2}\left(\frac{1}{3}\frac{\left(-p^2\+1\right)\sqrt{2}}{{\left(p^2\+1\right)}^2\left(2p^2\+2\right)}\+\frac{1}{3}\frac{\sqrt{2}\left(p^2-1\right)}{{\left(p^2\+1\right)}^2\left(2p^2\+2\right)}\right)\sqrt{2}\+\frac{2}{3}\frac{-p^2\+1}{{\left(p^2\+1\right)}^2\left(2p^2\+2\right)}\\ \frac{\left(\frac{1}{3}\frac{\left(-p^2\+1\right)\sqrt{2}}{{\left(p^2\+1\right)}^2\left(2p^2\+2\right)}\+\frac{1}{3}\frac{\sqrt{2}\left(p^2-1\right)}{{\left(p^2\+1\right)}^2\left(2p^2\+2\right)}\right)\sqrt{2}p}{p^2\+1}\end{array}\right]$ 

$\mathbf{N}\prime \=d\times \mathbf{N}$ 

$\left(\frac{\ⅆ}{\ⅆ p} \mathbf{N}\right)\/\mathrm{\ρ}$ = $\left[\begin{array}{c}\frac{1}{6}\frac{\sqrt{2}\left(\frac{4p^2}{{\left(p^2\+1\right)}^2}-\frac{2}{p^2\+1}\right)}{p^2\+1}\\ \frac{1}{6}\frac{\sqrt{2}\left(-\frac{2p}{p^2\+1}-\frac{2\left(-p^2\+1\right)p}{{\left(p^2\+1\right)}^2}\right)}{p^2\+1}\\ 0\end{array}\right]$ = $\stackrel{\text{assuming positive}}{\to }$$\left[\begin{array}{c}\frac{1}{3}\frac{\sqrt{2}\left(p^2-1\right)}{{\left(p^2\+1\right)}^3}\\ -\frac{2}{3}\frac{\sqrt{2}p}{{\left(p^2\+1\right)}^3}\\ 0\end{array}\right]$

$\mathbf{d}\times \mathbf{N}$ = $\left[\begin{array}{c}-\frac{1}{3}\frac{\sqrt{2}\left(-p^2\+1\right)}{{\left(p^2\+1\right)}^3}\\ -\frac{2}{3}\frac{\sqrt{2}p}{{\left(p^2\+1\right)}^3}\\ \frac{\left(\frac{1}{3}\frac{\left(-p^2\+1\right)\sqrt{2}}{{\left(p^2\+1\right)}^2\left(2p^2\+2\right)}\+\frac{1}{3}\frac{\sqrt{2}\left(p^2-1\right)}{{\left(p^2\+1\right)}^2\left(2p^2\+2\right)}\right)\left(-p^2\+1\right)}{p^2\+1}\end{array}\right]$ 

$\mathbf{B}\prime \=\mathbf{d}\times \mathbf{B}$ 

$\left(\frac{\ⅆ}{\ⅆ p} \mathbf{B}\right)\/\mathrm{\ρ}$ = $\left[\begin{array}{c}\frac{1}{6}\frac{\sqrt{2}\left(\frac{2p\sqrt{2}}{2p^2\+2}-\frac{4\sqrt{2}\left(p^2-1\right)p}{{\left(2p^2\+2\right)}^2}\right)}{p^2\+1}\\ \frac{1}{6}\frac{\sqrt{2}\left(\frac{2\sqrt{2}{p}^2}{{\left(p^2\+1\right)}^2}-\frac{\sqrt{2}}{p^2\+1}\right)}{p^2\+1}\\ 0\end{array}\right]$$\stackrel{\text{assuming positive}}{\to }$$\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]$

$\mathbf{d}\times \mathbf{B}$ = $\left[\begin{array}{c}\frac{2}{3}\frac{p}{{\left(p^2\+1\right)}^3}\\ -\frac{1}{2}\left(\frac{1}{3}\frac{\left(-p^2\+1\right)\sqrt{2}}{{\left(p^2\+1\right)}^2\left(2p^2\+2\right)}\+\frac{1}{3}\frac{\sqrt{2}\left(p^2-1\right)}{{\left(p^2\+1\right)}^2\left(2p^2\+2\right)}\right)\sqrt{2}\+\frac{2}{3}\frac{p^2-1}{{\left(p^2\+1\right)}^2\left(2p^2\+2\right)}\\ -\frac{\left(\frac{1}{3}\frac{\left(-p^2\+1\right)\sqrt{2}}{{\left(p^2\+1\right)}^2\left(2p^2\+2\right)}\+\frac{1}{3}\frac{\sqrt{2}\left(p^2-1\right)}{{\left(p^2\+1\right)}^2\left(2p^2\+2\right)}\right)\sqrt{2}p}{p^2\+1}\end{array}\right]$$$

$\frac{\left(\frac{\ⅆ}{\ⅆ p} \mathbf{T}\right)}{\mathrm{\rho }}\times \frac{\left(\frac{{\ⅆ}^2}{{\ⅆ}^{}{p}^2} \mathbf{T}\right)}{{\mathrm{\rho }}^2}$ = $\left[\begin{array}{c}0\\ 0\\ \frac{1}{108}\frac{\sqrt{2}\left(-\frac{2p\sqrt{2}}{2p^2\+2}-\frac{4\left(-p^2\+1\right)\sqrt{2}p}{{\left(2p^2\+2\right)}^2}\right)\left(\frac{8\sqrt{2}{p}^3}{{\left(p^2\+1\right)}^3}-\frac{6\sqrt{2}p}{{\left(p^2\+1\right)}^2}\right)}{{\left(p^2\+1\right)}^3}-\frac{1}{108}\frac{\sqrt{2}\left(-\frac{2\sqrt{2}{p}^2}{{\left(p^2\+1\right)}^2}\+\frac{\sqrt{2}}{p^2\+1}\right)\left(-\frac{2\sqrt{2}}{2p^2\+2}\+\frac{16p^2\sqrt{2}}{{\left(2p^2\+2\right)}^2}\+\frac{32\left(-p^2\+1\right)\sqrt{2}{p}^2}{{\left(2p^2\+2\right)}^3}-\frac{4\left(-p^2\+1\right)\sqrt{2}}{{\left(2p^2\+2\right)}^2}\right)}{{\left(p^2\+1\right)}^3}\end{array}\right]$$\stackrel{\text{simplify}}{\=}$$\left[\begin{array}{c}0\\ 0\\ \frac{1}{27}\frac{\sqrt{2}}{{\left(p^2\+1\right)}^6}\end{array}\right]$$$

${\mathrm{\κ}}^2 \mathbf{d}$ = $\left[\begin{array}{c}\frac{1}{9}\frac{\frac{1}{3}\frac{\left(-p^2\+1\right)\sqrt{2}}{{\left(p^2\+1\right)}^2\left(2p^2\+2\right)}\+\frac{1}{3}\frac{\sqrt{2}\left(p^2-1\right)}{{\left(p^2\+1\right)}^2\left(2p^2\+2\right)}}{{\left(p^2\+1\right)}^4}\\ 0\\ \frac{1}{27}\frac{\sqrt{2}}{{\left(p^2\+1\right)}^6}\end{array}\right]$$$

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$\mathrm{with}\left(\mathrm{Student}:-\mathrm{VectorCalculus}\right)\:$

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

Define the position vector R and obtain $\mathrm{\ρ}\,k\,\mathrm{\τ}$