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

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

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

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

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

Loading Student:-VectorCalculus

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

$\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}$

$\frac{{\ⅆ}^2}{{\ⅆ}^{}{p}^2} \mathbf{R}$ = $\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 

$\mathbf{R}$ = $\stackrel{\text{tangent vector}}{\to }$ $\stackrel{\text{Euclidean-normalize}}{\to }$ $\stackrel{\text{assuming real}}{\to }$ $\stackrel{\text{assign to a name}}{\to }$$T$

Obtain N 

$\mathbf{R}$ = $\stackrel{\text{principal normal}}{\to }$ $\stackrel{\text{2-normalize}}{\to }$ $\stackrel{\text{assuming real}}{\to }$ $\stackrel{\text{assign to a name}}{\to }$$N$

Obtain the curvature $\mathrm{\κ}$ 

$\mathbf{R}$  = $\stackrel{\text{curvature}}{\to }$$\frac{1}{2}\frac{\sqrt{4{\left(\frac{p\left(\frac{4p^3\+4p}{{\left(p^2\+1\right)}^2}-\frac{4\left(p^4\+2p^2\+2\right)p}{{\left(p^2\+1\right)}^3}\right)}{{\left(\frac{p^4\+2p^2\+2}{{\left(p^2\+1\right)}^2}\right)}^{3\/2}{\left(p^2\+1\right)}^2}\+\frac{8p^2}{\sqrt{\frac{p^4\+2p^2\+2}{{\left(p^2\+1\right)}^2}}{\left(p^2\+1\right)}^3}-\frac{2}{\sqrt{\frac{p^4\+2p^2\+2}{{\left(p^2\+1\right)}^2}}{\left(p^2\+1\right)}^2}\right)}^2\+\frac{{\left(p^2\+1\right)}^6{\left(\frac{4p^3\+4p}{{\left(p^2\+1\right)}^2}-\frac{4\left(p^4\+2p^2\+2\right)p}{{\left(p^2\+1\right)}^3}\right)}^2}{{\left(p^4\+2p^2\+2\right)}^3}\+4{\left(-\frac{1}{2}\frac{\left(\frac{1}{p^2\+1}-\frac{2p^2}{{\left(p^2\+1\right)}^2}\right)\left(\frac{4p^3\+4p}{{\left(p^2\+1\right)}^2}-\frac{4\left(p^4\+2p^2\+2\right)p}{{\left(p^2\+1\right)}^3}\right)}{{\left(\frac{p^4\+2p^2\+2}{{\left(p^2\+1\right)}^2}\right)}^{3\/2}}\+\frac{-\frac{6p}{{\left(p^2\+1\right)}^2}\+\frac{8p^3}{{\left(p^2\+1\right)}^3}}{\sqrt{\frac{p^4\+2p^2\+2}{{\left(p^2\+1\right)}^2}}}\right)}^2}}{\sqrt{\frac{p^4\+2p^2\+2}{{\left(p^2\+1\right)}^2}}}$$\stackrel{\text{assuming real}}{\to }$$\frac{2\sqrt{p^6\+3p^4\+3p^2\+2}}{{\left(p^4\+2p^2\+2\right)}^{3\/2}}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{\κ}$

Compute $\mathrm{\rho }\prime \mathbf{T}\+\mathrm{\kappa } {\mathrm{\rho }}^2\mathbf{N}$ and compare with $\mathbf{R}″$ 

$\left(\frac{\ⅆ}{\ⅆ p} \mathrm{\ρ}\right) \mathbf{T}\+\mathrm{\κ} {\mathrm{\ρ}}^2 \mathbf{N}$ = $\stackrel{\text{assuming real}}{\to }$

$\mathbf{R}″$ =

Compare the scalar projection of $\mathbf{R}″$ on T with $\mathrm{\ρ}\prime$

$\mathbf{R}″·\mathbf{T}$ = $-\frac{4\left(3p^2-1\right)p}{{\left(p^2\+1\right)}^4\sqrt{p^4\+2p^2\+2}}-\frac{2p\left(p^2-3\right)\left(p^2-1\right)}{{\left(p^2\+1\right)}^4\sqrt{p^4\+2p^2\+2}}$$\stackrel{\text{simplify}}{\=}$$-\frac{2p}{\sqrt{p^4\+2p^2\+2}{\left(p^2\+1\right)}^2}$

$\frac{\ⅆ}{\ⅆ p} \mathrm{\ρ}$ = $-\frac{2\sqrt{p^4\+2p^2\+2}p}{{\left(p^2\+1\right)}^2}\+\frac{1}{2}\frac{4p^3\+4p}{\left(p^2\+1\right)\sqrt{p^4\+2p^2\+2}}$$\stackrel{\text{assuming real}}{\to }$$-\frac{2p}{\sqrt{p^4\+2p^2\+2}{\left(p^2\+1\right)}^2}$

Compare the scalar projection of $\mathbf{R}″$ on N with $\mathrm{\κ} {\mathrm{\ρ}}^2$ 

$\mathbf{R}″·\mathbf{N}$ = $\frac{2\left(3p^2-1\right)\left(3p^6\+5p^4\+2p^2-2\right)}{{\left(p^2\+1\right)}^4\sqrt{p^4\+2p^2\+2}\sqrt{p^6\+3p^4\+3p^2\+2}}\+\frac{2p^2\left(p^2-3\right)\left(p^6-p^4-5p^2-5\right)}{{\left(p^2\+1\right)}^4\sqrt{p^4\+2p^2\+2}\sqrt{p^6\+3p^4\+3p^2\+2}}$$\stackrel{\text{simplify}}{\=}$$\frac{2\sqrt{p^6\+3p^4\+3p^2\+2}}{\sqrt{p^4\+2p^2\+2}{\left(p^2\+1\right)}^2}$

$\mathrm{\κ} {\mathrm{\ρ}}^2$ = $\frac{2\sqrt{p^6\+3p^4\+3p^2\+2}}{\sqrt{p^4\+2p^2\+2}{\left(p^2\+1\right)}^2}$$\stackrel{\text{simplify}}{\=}$$\frac{2\sqrt{p^6\+3p^4\+3p^2\+2}}{\sqrt{p^4\+2p^2\+2}{\left(p^2\+1\right)}^2}$

Initialize

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

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

Define R and obtain $\mathbf{R}″\,\mathbf{T}\,\mathbf{N}\,\mathrm{\ρ}\,\mathrm{\κ}$ 

$\mathbf{R}≔⟨1\/\left(1\+p^2\right)\,p\,p\/\left(1\+p^2\right)⟩\:$

$\mathbf{R}″≔\mathrm{simplify}\left(\mathrm{diff}\left(\mathbf{R}\,p\,p\right)\right)\:$

$$\begin{gathered} \mathbf{Temp}\mathbf{≔}\mathrm{TangentVector}\left(\mathbf{R}\,\mathrm{normalized}\right)\: \\ \mathbf{T}≔\mathrm{simplify}\left(\mathbf{Temp}\right) assuming p\colon\colon \mathrm{real}\: \end{gathered}$$

$$\begin{gathered} \mathbf{Temp}\mathbf{≔}\mathrm{PrincipalNormal}\left(\mathbf{R}\,\mathrm{normalized}\right)\: \\ \mathbf{N}≔\mathrm{simplify}\left(\mathbf{Temp}\right) assuming p\colon\colon \mathrm{real}\: \end{gathered}$$

$\mathrm{\ρ}≔\mathrm{simplify}\left(\mathrm{Norm}\left(\mathrm{diff}\left(\mathbf{R}\,p\right)\right)\right) assuming p\colon\colon \mathrm{real}\:$

$\mathrm{\κ}≔\mathrm{simplify}\left(\mathrm{Curvature}\left(\mathbf{R}\right)\right) assuming p\colon\colon \mathrm{real}\:$

Display $\mathbf{R}″\,\mathbf{T}\,\mathbf{N}\,\mathrm{\ρ}\,\mathrm{\κ}$ 

$\mathbf{R}″\,\mathbf{T}\,\mathbf{N}\,\mathrm{\ρ}\,\mathrm{\κ}$

Obtain the right-hand side of the decomposition formula

$\mathrm{simplify}\left(\mathrm{diff}\left(\mathrm{\ρ}\,p\right) \mathbf{T}\+\mathrm{\κ} {\mathrm{\ρ}}^2 \mathbf{N}\right)$

Obtain the scalar projection of $\mathbf{R}″$ on T and compare to $\mathrm{\ρ}\prime$ 

$\mathrm{simplify}\left(\mathrm{DotProduct}\left(\mathbf{R}″\,\mathbf{T}\right)\right)$

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

$\mathrm{simplify}\left(\mathrm{diff}\left(\mathrm{\ρ}\,p\right)\right)$

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

Obtain the scalar projection of $\mathbf{R}″$ on N and compare to $\mathrm{\κ} {\mathrm{\ρ}}^2$ 

$\mathrm{simplify}\left(\mathrm{DotProduct}\left(\mathbf{R}″\,\mathbf{N}\right)\right)$

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

$\mathrm{simplify}\left(\mathrm{\κ} {\mathrm{\ρ}}^2\right)$

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