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

$\mathbf{N}\=\frac{\left[\begin{array}{c}3p^6\+5p^4\+2p^2-2\\ {\mathrm{\λ}}^2 p\\ \left(p^6-p^4-5p^2-5\right)p\end{array}\right]}{\mathrm{\lambda } \sqrt{{\mathrm{\lambda }}^2\+1} \sqrt{{\mathrm{\lambda }}^3\+1}}$

$\mathrm{\ρ}\=\frac{\sqrt{{\mathrm{\lambda }}^2\+1}}{\mathrm{\λ}}$

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

Table 2.8.4(a)   Items from the Frenet formalism, where $\mathrm{\λ}\=1\+p^2$

$\mathrm{\rho }\prime \mathbf{T}\+\mathrm{\kappa } {\mathrm{\rho }}^2\mathbf{N}$

$\=\frac{d}{\mathrm{dp}}\left(\frac{\sqrt{{\mathrm{\lambda }}^2\+1}}{\mathrm{\λ}}\right)\mathbf{T}\+\left(\frac{2\sqrt{{\mathrm{\lambda }}^3\+1}}{{\left({\mathrm{\lambda }}^2\+1\right)}^{3\/2}}\right){\left(\frac{\sqrt{{\mathrm{\lambda }}^2\+1}}{\mathrm{\λ}}\right)}^2\mathbf{N}$

$\=-\frac{2 p}{{\mathrm{\lambda }}^2\sqrt{{\mathrm{\lambda }}^2\+1}} \left(\frac{\left[\begin{array}{c}-2 p\\ {\mathrm{\lambda }}^2\\ 1-p^2\end{array}\right]}{\mathrm{\lambda } \sqrt{{\mathrm{\λ}}^2\+1}}\right)\+\frac{2\sqrt{{\mathrm{\lambda }}^3\+1}}{{\mathrm{\λ}}^2\sqrt{{\mathrm{\lambda }}^2\+1}}\left(\frac{\left[\begin{array}{c}3p^6\+5p^4\+2p^2-2\\ {\mathrm{\lambda }}^2 p\\ \left(p^6-p^4-5p^2-5\right)p\end{array}\right]}{\mathrm{\lambda } \sqrt{{\mathrm{\lambda }}^2\+1} \sqrt{{\mathrm{\lambda }}^3\+1}}\right)$

$\=\frac{2}{{\mathrm{\lambda }}^3 \left({\mathrm{\lambda }}^2\+1\right)}\left(-p \left[\begin{array}{c}-2 p\\ {\mathrm{\lambda }}^2\\ 1-p^2\end{array}\right]\+\left[\begin{array}{c}3p^6\+5p^4\+2p^2-2\\ {\mathrm{\lambda }}^2 p\\ \left(p^6-p^4-5p^2-5\right)p\end{array}\right]\right)$

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

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

$\mathbf{R}″\mathbf{·}\mathbf{T}$

$\=\frac{2}{{\mathrm{\lambda }}^4\sqrt{{\mathrm{\lambda }}^2\+1}}\left[\begin{array}{c}3 p^2-1\\ 0\\ p \(p^2-3\)\end{array}\right]·\left[\begin{array}{c}-2 p\\ {\mathrm{\lambda }}^2\\ 1-p^2\end{array}\right]$

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

$\=\frac{-2 p}{{\mathrm{\lambda }}^4\sqrt{{\mathrm{\lambda }}^2\+1}}\left({\mathrm{\λ}}^2\right)$

$\= -\frac{2 p}{{\mathrm{\λ}}^2\sqrt{{\mathrm{\λ}}^2\+1}}\=\mathrm{\ρ}\prime$

$\mathbf{R}″\mathbf{·}\mathbf{N}$

$\=\frac{2}{{\mathrm{\lambda }}^4 \sqrt{{\mathrm{\lambda }}^2\+1} \sqrt{{\mathrm{\lambda }}^3\+1}}\left[\begin{array}{c}3 p^2-1\\ 0\\ p \(p^2-3\)\end{array}\right]·\left[\begin{array}{c}3p^6\+5p^4\+2p^2-2\\ {\mathrm{\lambda }}^2 p\\ \left(p^6-p^4-5p^2-5\right)p\end{array}\right]$

$\=\frac{2}{{\mathrm{\lambda }}^4 \sqrt{{\mathrm{\lambda }}^2\+1} \sqrt{{\mathrm{\lambda }}^3\+1}}\left(p^6\+3p^4\+3p^2\+2\right) {\left(p^2\+1\right)}^2$

$\=\frac{2}{{\mathrm{\lambda }}^4 \sqrt{{\mathrm{\lambda }}^2\+1} \sqrt{{\mathrm{\lambda }}^3\+1}} \left({\mathrm{\lambda }}^3\+1\right) {\mathrm{\lambda }}^2$

$\=\frac{2\sqrt{{\mathrm{\λ}}^3\+1}}{{\mathrm{\λ}}^2 \sqrt{{\mathrm{\λ}}^2\+1}}$

$\=\left(\frac{2\sqrt{{\mathrm{\lambda }}^3\+1}}{{\left({\mathrm{\lambda }}^2\+1\right)}^{3\/2}}\right){\left(\frac{\sqrt{{\mathrm{\lambda }}^2\+1}}{\mathrm{\lambda }}\right)}^2\=\mathrm{\kappa } {\mathrm{\rho }}^2$

Initialize

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

$\mathbf{R}\=⟨\frac{1}{1\+p^2}\,p\,\frac{p}{1\+p^2}⟩$$\stackrel{\text{assign}}{\to }$

Obtain $\mathrm{\ρ}\=∥\mathbf{R}\prime \left(p\right)∥$ and $\mathbf{R}″\left(p\right)$ 

$∥\frac{\ⅆ}{\ⅆ p} \mathbf{R}∥$ = $\sqrt{\frac{p^4\+2p^2\+2}{{\left(p^2\+1\right)}^2}}$$\stackrel{\text{assuming positive}}{\to }$$\frac{\sqrt{p^4\+2p^2\+2}}{p^2\+1}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{\ρ}$

$\frac{{\ⅆ}^2}{{\ⅆ}^{}{p}^2} \mathbf{R}$ = $\left[\begin{array}{c}\frac{8p^2}{{\left(p^2\+1\right)}^3}-\frac{2}{{\left(p^2\+1\right)}^2}\\ 0\\ -\frac{6p}{{\left(p^2\+1\right)}^2}\+\frac{8p^3}{{\left(p^2\+1\right)}^3}\end{array}\right]$$\stackrel{\text{simplify}}{\=}$$\left[\begin{array}{c}\frac{2\left(3p^2-1\right)}{{\left(p^2\+1\right)}^3}\\ 0\\ \frac{2p\left(p^2-3\right)}{{\left(p^2\+1\right)}^3}\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{Temp}$$\mathbf{R}″\=\mathrm{Temp}$$\stackrel{\text{assign}}{\to }$

Obtain T 

$\mathbf{R}$ = $\left[\begin{array}{c}\frac{1}{p^2\+1}\\ p\\ \frac{p}{p^2\+1}\end{array}\right]$$\stackrel{\text{tangent vector}}{\to }$$\left[\begin{array}{c}-\frac{2p}{{\left(p^2\+1\right)}^2}\\ 1\\ -\frac{p^2-1}{{\left(p^2\+1\right)}^2}\end{array}\right]$$\stackrel{\text{Euclidean-normalize}}{\to }$$\left[\begin{array}{c}-\frac{2p}{\sqrt{\frac{p^4\+2p^2\+2}{{\left(p^2\+1\right)}^2}}{\left(p^2\+1\right)}^2}\\ \frac{1}{\sqrt{\frac{p^4\+2p^2\+2}{{\left(p^2\+1\right)}^2}}}\\ -\frac{p^2-1}{\sqrt{\frac{p^4\+2p^2\+2}{{\left(p^2\+1\right)}^2}}{\left(p^2\+1\right)}^2}\end{array}\right]$$\stackrel{\text{assuming real}}{\to }$$\left[\begin{array}{c}-\frac{2p}{\left(p^2\+1\right)\sqrt{p^4\+2p^2\+2}}\\ \frac{p^2\+1}{\sqrt{p^4\+2p^2\+2}}\\ -\frac{p^2-1}{\left(p^2\+1\right)\sqrt{p^4\+2p^2\+2}}\end{array}\right]$$\stackrel{\text{assign to a name}}{\to }$$T$

Obtain N 

$\mathbf{R}$ = $\left[\begin{array}{c}\frac{1}{p^2\+1}\\ p\\ \frac{p}{p^2\+1}\end{array}\right]$$\stackrel{\text{principal normal}}{\to }$$\left[\begin{array}{c}\frac{2\left(3p^6\+5p^4\+2p^2-2\right)}{\left(p^4\+2p^2\+2\right){\left(p^2\+1\right)}^3\sqrt{\frac{p^4\+2p^2\+2}{{\left(p^2\+1\right)}^2}}}\\ \frac{2p}{\left(p^4\+2p^2\+2\right)\left(p^2\+1\right)\sqrt{\frac{p^4\+2p^2\+2}{{\left(p^2\+1\right)}^2}}}\\ \frac{2\left(p^6-p^4-5p^2-5\right)p}{\left(p^4\+2p^2\+2\right){\left(p^2\+1\right)}^3\sqrt{\frac{p^4\+2p^2\+2}{{\left(p^2\+1\right)}^2}}}\end{array}\right]$$\stackrel{\text{2-normalize}}{\to }$$\left[\begin{array}{c}\frac{3p^6\+5p^4\+2p^2-2}{\sqrt{\frac{p^6\+3p^4\+3p^2\+2}{{\left(p^2\+1\right)}^2{\left(p^4\+2p^2\+2\right)}^2}}\left(p^4\+2p^2\+2\right){\left(p^2\+1\right)}^3\sqrt{\frac{p^4\+2p^2\+2}{{\left(p^2\+1\right)}^2}}}\\ \frac{}{}\end{array}\right]$