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

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

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

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

Table 2.8.8(a)   Items from the Frenet formalism

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

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

$\=6\sqrt{2}p \left(\left[\begin{array}{c}\frac{1- p^2}{\sqrt{2}\(p^2\+1\)}\\ \frac{\sqrt{2}p}{p^2\+1}\\ 1\/\sqrt{2} \end{array}\right]\right)\+6\left(\left[\begin{array}{c}-\frac{2p}{p^2\+1}\\ -\frac{p^2-1}{p^2\+1}\\ 0\end{array}\right]\right)$

$\=\left[\begin{array}{c}\frac{6p\left(-p^2\+1\right)}{p^2\+1}-\frac{12p}{p^2\+1}\\ \frac{12p^2}{p^2\+1}-\frac{6\left(p^2-1\right)}{p^2\+1}\\ 6p\end{array}\right]$

$\=\left[\begin{array}{c}-6p\\ 6\\ 6p\end{array}\right]$

$\mathbf{R}″·\mathbf{T}$

$\=\left[\begin{array}{c}-6 p\\ 6\\ 6 p\end{array}\right]·\left[\begin{array}{c}\frac{1- p^2}{\sqrt{2}\(p^2\+1\)}\\ \frac{\sqrt{2}p}{p^2\+1}\\ 1\/\sqrt{2} \end{array}\right]$

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

$\=6\sqrt{2}p\=\mathrm{\ρ}\prime$

$\mathbf{R}″\mathbf{·}\mathbf{N}$

$\=\left[\begin{array}{c}-6 p\\ 6\\ 6 p\end{array}\right]·\left[\begin{array}{c}-\frac{2p}{p^2\+1}\\ -\frac{p^2-1}{p^2\+1}\\ 0\end{array}\right]$

$\=\frac{12p^2}{p^2\+1}-\frac{6\left(p^2-1\right)}{p^2\+1}$

$\=6$

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

Initialize

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 }$

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

$∥\frac{\ⅆ}{\ⅆ p} \mathbf{R}∥$ = $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{\ρ}$

$\frac{{\ⅆ}^2}{{\ⅆ}^{}{p}^2} \mathbf{R}$ = $\stackrel{\text{simplify}}{\=}$ $\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{a}\text{ssign to a name}}{\to }$$N$$$

Obtain the curvature $\mathrm{\κ}$ 

$\mathbf{R}$  = $\stackrel{\text{curvature}}{\to }$$\frac{1}{6}\frac{\sqrt{{\left(-\frac{1}{6}\frac{\sqrt{2}\left(-3p^2\+3\right)\mathrm{csgn}\left(1\,p^2\+1\right)}{{\mathrm{csgn}\left(p^2\+1\right)}^2\left(p^2\+1\right)}-\frac{1}{3}\frac{\sqrt{2}\left(-3p^2\+3\right)p}{\mathrm{csgn}\left(p^2\+1\right){\left(p^2\+1\right)}^2}-\frac{\sqrt{2}p}{\mathrm{csgn}\left(p^2\+1\right)\left(p^2\+1\right)}\right)}^2\+{\left(-\frac{\sqrt{2}p\mathrm{csgn}\left(1\,p^2\+1\right)}{{\mathrm{csgn}\left(p^2\+1\right)}^2\left(p^2\+1\right)}-\frac{2\sqrt{2}{p}^2}{\mathrm{csgn}\left(p^2\+1\right){\left(p^2\+1\right)}^2}\+\frac{\sqrt{2}}{\mathrm{csgn}\left(p^2\+1\right)\left(p^2\+1\right)}\right)}^2\+{\left(-\frac{1}{6}\frac{\sqrt{2}\left(3p^2\+3\right)\mathrm{csgn}\left(1\,p^2\+1\right)}{{\mathrm{csgn}\left(p^2\+1\right)}^2\left(p^2\+1\right)}-\frac{1}{3}\frac{\sqrt{2}\left(3p^2\+3\right)p}{\mathrm{csgn}\left(p^2\+1\right){\left(p^2\+1\right)}^2}\+\frac{\sqrt{2}p}{\mathrm{csgn}\left(p^2\+1\right)\left(p^2\+1\right)}\right)}^2}\sqrt{2}}{\mathrm{csgn}\left(p^2\+1\right)\left(p^2\+1\right)}$$\stackrel{\text{assuming real}}{\to }$$\frac{1}{3{\left(p^2\+1\right)}^2}$$\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{3p\sqrt{2}\left(p^2-1\right)}{p^2\+1}\+\frac{6\sqrt{2}p}{p^2\+1}\+3\sqrt{2}p$$\stackrel{\text{simplify}}{\=}$$6\sqrt{2}p$

$\frac{\ⅆ}{\ⅆ p} \mathrm{\ρ}$ = $6\sqrt{2}p$

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

$\mathbf{R}″·\mathbf{N}$ = $\frac{12p^2}{p^2\+1}-\frac{6\left(p^2-1\right)}{p^2\+1}$$\stackrel{\text{simplify}}{\=}$$6$$$$$

$\mathrm{\κ} {\mathrm{\ρ}}^2$ = $6$

Initialize

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

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

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

$\mathbf{R}≔⟨3 p-p^3\,3 p^2\,3 p\+p^3⟩\:$

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

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

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

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

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

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 $\mathrm{\ρ}\prime$ 

$\mathrm{simplify}\left(\mathrm{DotProduct}\left(\mathbf{R}″\,\mathbf{T}\right)\right)$ = $6\sqrt{2}p$$$

$\mathrm{diff}\left(\mathrm{\ρ}\,p\right)$ = $6\sqrt{2}p$$$

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

$\mathrm{simplify}\left(\mathrm{DotProduct}\left(\mathbf{R}″\,\mathbf{N}\right)\right)$ = $6$$$

$\mathrm{\κ} {\mathrm{\ρ}}^2$ = $6$$$