$\mathbf{T}\=\left[\begin{array}{c}\frac{1}{\sqrt{9p^4\+36p^2\+1}}\\ \frac{6p}{\sqrt{9p^4\+36p^2\+1}}\\ \frac{3p^2}{\sqrt{9p^4\+36p^2\+1}}\end{array}\right]$

$\mathbf{N}\=\left[\begin{array}{c}-\frac{3p\left(p^2\+2\right)}{\sqrt{9p^4\+36p^2\+1}\sqrt{9p^4\+p^2\+1}}\\ -\frac{9p^4-1}{\sqrt{9p^4\+36p^2\+1}\sqrt{9p^4\+p^2\+1}}\\ \frac{p\left(18p^2\+1\right)}{\sqrt{9p^4\+36p^2\+1}\sqrt{9p^4\+p^2\+1}}\end{array}\right]$

$\mathbf{B}\=\left[\begin{array}{c}\frac{3p^2}{\sqrt{9p^4\+p^2\+1}}\\ -\frac{p}{\sqrt{9p^4\+p^2\+1}}\\ \frac{1}{\sqrt{9p^4\+p^2\+1}}\end{array}\right]$

$\mathrm{\ρ}\=\sqrt{9p^4\+36p^2\+1}$

$\mathrm{\κ}\=\frac{6\sqrt{9p^4\+p^2\+1}}{{\left(9p^4\+36p^2\+1\right)}^{3\/2}}$

$\mathrm{\τ}\=\frac{1}{9p^4\+p^2\+1}$

Table 2.7.13(a)   Frenet formalism

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

$\=\frac{1}{9p^4\+p^2\+1}\left[\begin{array}{c}\frac{1}{\sqrt{9p^4\+36p^2\+1}}\\ \frac{6p}{\sqrt{9p^4\+36p^2\+1}}\\ \frac{3p^2}{\sqrt{9p^4\+36p^2\+1}}\end{array}\right]\+\frac{6\sqrt{9p^4\+p^2\+1}}{{\left(9p^4\+36p^2\+1\right)}^{3\/2}}\left[\begin{array}{c}\frac{3p^2}{\sqrt{9p^4\+p^2\+1}}\\ -\frac{p}{\sqrt{9p^4\+p^2\+1}}\\ \frac{1}{\sqrt{9p^4\+p^2\+1}}\end{array}\right]$

$\=\left[\begin{array}{c}\frac{162p^6\+27p^4\+54p^2\+1}{\left(9p^4\+p^2\+1\right){\left(9p^4\+36p^2\+1\right)}^{3\/2}}\\ \frac{210p^3}{\left(9p^4\+p^2\+1\right){\left(9p^4\+36p^2\+1\right)}^{3\/2}}\\ \frac{3\left(9p^6\+54p^4\+3p^2\+2\right)}{\left(9p^4\+p^2\+1\right){\left(9p^4\+36p^2\+1\right)}^{3\/2}}\end{array}\right]$

$\frac{d\mathbf{T}}{\mathrm{dp}}\frac{1}{\mathrm{\ρ}}\=\frac{6}{9p^4\+36p^2\+1}\left[\begin{array}{c}-3 p \(p^2\+2\)\\ 1-9 p^4\\ p \(18 p^2\+1\)\end{array}\right]$ = $\mathbf{d}\times \mathbf{T}$ 

$\frac{d\mathbf{N}}{\mathrm{dp}}\frac{1}{\mathrm{\ρ}}$ = $\left[\begin{array}{c}\frac{3\left(81p^{10}\+486p^8\+1278p^6\+34p^4-3p^2-2\right)}{{\left(9p^4\+36p^2\+1\right)}^2{\left(9p^4\+p^2\+1\right)}^{3\/2}}\\ -\frac{p\left(2997p^8\+1296p^6\+1998p^4\+144p^2\+37\right)}{{\left(9p^4\+36p^2\+1\right)}^2{\left(9p^4\+p^2\+1\right)}^{3\/2}}\\ -\frac{1458p^{10}\+243p^8-306p^6-1278p^4-54p^2-1}{{\left(9p^4\+36p^2\+1\right)}^2{\left(9p^4\+p^2\+1\right)}^{3\/2}}\end{array}\right]$ = $\mathbf{d}\times \mathbf{N}$ 

$\frac{d\mathbf{B}}{\mathrm{dp}}\frac{1}{\mathrm{\ρ}}$ = $\frac{1}{\sqrt{9p^4\+36p^2\+1}{\left(9p^4\+p^2\+1\right)}^{3\/2}}\left[\begin{array}{c}3 p \(p^2\+2\)\\ 9 p^4-1\\ -p \(18 p^2\+1\)\end{array}\right]$ = $\mathbf{d}\times \mathbf{B}$

$\frac{d\mathbf{T}}{\mathrm{dp}}\frac{1}{\mathrm{\rho }}\times \frac{d^2\mathbf{T}}{{\mathrm{dp}}^2}\frac{1}{{\mathrm{\rho }}^2}$= $\frac{36}{{\left(9p^4\+36p^2\+1\right)}^{9\/2}}\left[\begin{array}{c}162 p^6\+27p^4\+54p^2\+1\\ 210 p^3\\ 3\left(9p^6\+54p^4\+3p^2\+2\right)\end{array}\right]$ = ${\mathrm{\kappa }}^2 \mathbf{d}$

Initialize

Loading Student:-VectorCalculus

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

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

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

$∥\frac{\ⅆ}{\ⅆ p} \mathbf{R}∥$ = $\sqrt{9p^4\+36p^2\+1}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{\ρ}$$$

$\mathbf{R}$ = $\stackrel{\text{c}\text{urvature}}{\to }$$\frac{1}{2}\frac{\sqrt{\frac{{\left(36p^3\+72p\right)}^2}{{\left(9p^4\+36p^2\+1\right)}^3}\+4{\left(-\frac{3p\left(36p^3\+72p\right)}{{\left(9p^4\+36p^2\+1\right)}^{3\/2}}\+\frac{6}{\sqrt{9p^4\+36p^2\+1}}\right)}^2\+4{\left(-\frac{3}{2}\frac{p^2\left(36p^3\+72p\right)}{{\left(9p^4\+36p^2\+1\right)}^{3\/2}}\+\frac{6p}{\sqrt{9p^4\+36p^2\+1}}\right)}^2}}{\sqrt{9p^4\+36p^2\+1}}$$\stackrel{\text{assuming positive}}{\to }$$\frac{6\sqrt{9p^4\+p^2\+1}}{{\left(9p^4\+36p^2\+1\right)}^{3\/2}}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{\κ}$

$\mathbf{R}$  = $\stackrel{\text{torsion}}{\to }$$\frac{1}{\sqrt{\frac{9p^4\+p^2\+1}{{\left(9p^4\+36p^2\+1\right)}^2}}{\left(9p^4\+36p^2\+1\right)}^{3\/2}\sqrt{\frac{9p^4\+p^2\+1}{9p^4\+36p^2\+1}}}$$\stackrel{\text{assuming positive}}{\to }$$\frac{1}{9p^4\+p^2\+1}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{\τ}$$$$$

$\mathbf{R}$

$\stackrel{\text{TNB frame}}{\to }$

$\stackrel{\text{assuming positive}}{\to }$

$\stackrel{\text{assign to a name}}{\to }$

$Q$

$\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 the Darboux vector

$\mathrm{\τ} \mathbf{T}\+\mathrm{\kappa } \mathbf{B}$  = $\stackrel{\text{assuming positive}}{\to }$ $\stackrel{\text{assign to a name}}{\to }$$d$$$

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

$\left(\frac{\ⅆ}{\ⅆ p} \mathbf{T}\right)\/\mathrm{\ρ}$  = $\stackrel{\text{assuming positive}}{\to }$  

$\mathbf{d}\times \mathbf{T}$ = $\stackrel{\text{assuming positive}}{\to }$ $$

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

$\left(\frac{\ⅆ}{\ⅆ p} \mathbf{N}\right)\/\mathrm{\ρ}$ = $\stackrel{\text{assuming positive}}{\to }$ $$

$\mathbf{d}\times \mathbf{N}$ = $\stackrel{\text{assuming positive}}{\to }$ $$

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

$\left(\frac{\ⅆ}{\ⅆ p} \mathbf{B}\right)\/\mathrm{\ρ}$ = $\stackrel{\text{assuming positive}}{\to }$ $$

$\mathbf{d}\times \mathbf{B}$ = $\stackrel{\text{assuming positive}}{\to }$ $$

$\frac{\left(\frac{\ⅆ}{\ⅆ p} \mathbf{T}\right)}{\mathrm{\rho }}\times \frac{\left(\frac{{\ⅆ}^2}{{\ⅆ}^{}{p}^2} \mathbf{T}\right)}{{\mathrm{\rho }}^2}$

$\stackrel{\text{assuming positive}}{\to }$

${\mathrm{\κ}}^2 \mathbf{d}$ = $\left[\begin{array}{c}\frac{36\left(162p^6\+27p^4\+54p^2\+1\right)}{{\left(9p^4\+36p^2\+1\right)}^{9\/2}}\\ \frac{7560p^3}{{\left(9p^4\+36p^2\+1\right)}^{9\/2}}\\ \frac{108\left(9p^6\+54p^4\+3p^2\+2\right)}{{\left(9p^4\+36p^2\+1\right)}^{9\/2}}\end{array}\right]$$$

Initialize

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

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

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

$\mathbf{R}≔⟨p\,3 p^2\,p^3⟩\:$

$\mathrm{\ρ}≔\mathrm{Norm}\left(\mathrm{diff}\left(\mathbf{R}\,p\right)\right)$ = $\sqrt{9p^4\+36p^2\+1}$$$

$\mathrm{\κ}≔\mathrm{simplify}\left(\mathrm{Curvature}\left(\mathbf{R}\right)\right) assuming p\>0$

$\frac{6\sqrt{9p^4\+p^2\+1}}{{\left(9p^4\+36p^2\+1\right)}^{3\/2}}$

$\mathrm{\τ}≔\mathrm{simplify}\left(\mathrm{Torsion}\left(\mathbf{R}\right)\right) assuming p\>0$

$\frac{1}{9p^4\+p^2\+1}$

Obtain and display the vectors of the TNB-frame

$$\begin{gathered} \mathbf{Temp}≔\mathrm{TNBFrame}\left(\mathbf{R}\right)\: \\ \mathbf{Q}≔\mathrm{map}\left(\mathrm{simplify}\,\left[\mathbf{Temp}\right]\right) assuming p\>0\: \\ \mathbf{T}\,\mathbf{N}\,\mathbf{B}\mathbf{≔}{\mathbf{Q}}_1\,{\mathbf{Q}}_2\,{\mathbf{Q}}_3 \end{gathered}$$

Obtain the Darboux vector

$$\begin{gathered} \mathbf{Temp}≔\mathrm{\τ} \mathbf{T}\+\mathrm{\κ} \mathbf{B}\: \\ \mathbf{d}≔\mathrm{simplify}\left(\mathbf{Temp}\right) assuming p\>0 \end{gathered}$$

$\mathrm{simplify}\left(\mathrm{diff}\left(\mathbf{T}\,p\right)\/\mathrm{\ρ}\right) assuming p\>0$

$\mathrm{simplify}\left(\mathrm{CrossProduct}\left(\mathbf{d}\,\mathbf{T}\right)\right) assuming p\>0$

$\mathrm{simplify}\left(\frac{\mathrm{diff}\left(\mathbf{N}\,p\right)}{\mathrm{\rho }}\right) assuming p\>0$

$\mathrm{simplify}\left(\mathrm{CrossProduct}\left(\mathbf{d}\mathbf{\,}\mathbf{N}\right)\right) assuming p\>0$

$\mathrm{simplify}\left(\mathrm{diff}\left(\mathbf{B}\,p\right)\/\mathrm{\ρ}\right) assuming p\>0$

$\mathrm{simplify}\left(\mathrm{CrossProduct}\left(\mathbf{d}\,\mathbf{B}\right)\right) assuming p\>0$

$\mathrm{simplify}\left(\mathrm{CrossProduct}\left(\frac{\mathrm{diff}\left(\mathbf{T}\,p\right)}{\mathrm{\rho }}\,\frac{\mathrm{diff}\left(\mathbf{T}\,p\,p\right)}{{\mathrm{\rho }}^2}\right)\right)$

${\mathrm{\κ}}^2 \mathbf{d}$