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

$\mathbf{N}\= -\left[\begin{array}{c}\cos \(p\)\\ \sin \(p\)\\ 0\end{array}\right]$

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

$\mathrm{\ρ}\=\sqrt{2}$ 

$\mathrm{\κ}\=1\/2$ 

$\mathrm{\τ}\=1\/2$

Table 2.7.12(a)   Frenet formalism

Left-hand side

Right-hand side

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

$\mathbf{d}\times \mathbf{T}$ = $-\frac{1}{2}\left[\begin{array}{c}\cos \(p\)\\ \sin \(p\)\\ 0\end{array}\right]$$$

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

$\mathbf{d}\times \mathbf{N}$ = $\frac{1}{\sqrt{2}}\left[\begin{array}{c}\sin \(p\)\\ -\cos \(p\)\\ 0\end{array}\right]$ 

$\frac{d\mathbf{B}}{\mathrm{dp}}\frac{1}{\mathrm{\ρ}}$ = $\frac{1}{2}\left[\begin{array}{c}\cos \(p\)\\ \sin \(p\)\\ 0\end{array}\right]$ 

$\mathbf{d}\times \mathbf{B}$ = $\frac{1}{2}\left[\begin{array}{c}\cos \(p\)\\ \sin \(p\)\\ 0\end{array}\right]$ 

$\frac{d\mathbf{T}}{\mathrm{dp}}\frac{1}{\mathrm{\ρ}}\times \frac{d^2\mathbf{T}}{{\mathrm{dp}}^2}\frac{1}{{\mathrm{\ρ}}^2}\=\left[\begin{array}{c}0\\ 0\\ \sqrt{2}\/8\end{array}\right]$ = ${\mathrm{\kappa }}^2 \mathbf{d}$

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}\=⟨\cos \left(p\right)\,\sin \left(p\right)\,p⟩$$\stackrel{\text{assign}}{\to }$

$∥\frac{\ⅆ}{\ⅆ p} \mathbf{R}∥$ = $\sqrt{2}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{\ρ}$$$

$\mathbf{R}$ = $\stackrel{\text{curvature}}{\to }$$\frac{1}{4}\sqrt{2{\cos \left(p\right)}^2\+2{\sin \left(p\right)}^2}\sqrt{2}$$\stackrel{\text{simplify}}{\=}$$\frac{1}{2}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{\κ}$$$

$\mathbf{R}$ = $\stackrel{\text{torsion}}{\to }$$\frac{1}{2}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{\τ}$$$

$\mathbf{R}$  = $\stackrel{\text{TNB frame}}{\to }$ $\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 and display the Darboux vector

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

$\mathbf{d}$ = $\frac{1}{2}{Q}_1\+\frac{1}{2}{Q}_3$$$

$\left(\frac{\ⅆ}{\ⅆ p} \mathbf{T}\right)\/\mathrm{\ρ}$ = $$

$\mathbf{d}\times \mathbf{T}$ = $$

$\left(\frac{\ⅆ}{\ⅆ p} \mathbf{N}\right)\/\mathrm{\ρ}$ = $$

$\mathbf{d}\times \mathbf{N}$ = $$

$\left(\frac{\ⅆ}{\ⅆ p} \mathbf{B}\right)\/\mathrm{\ρ}$ = $$

$\mathbf{d}\times \mathbf{B}$ = $$

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

$\stackrel{\text{simplify}}{\=}$

${\mathrm{\κ}}^2 \mathbf{d}$ = $$

Initialize

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

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

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

$\mathbf{R}≔⟨\cos \left(p\right)\,\sin \left(p\right)\,p⟩\:$

$\mathrm{\ρ}≔\mathrm{Norm}\left(\mathrm{diff}\left(\mathbf{R}\,p\right)\right)$ = $\sqrt{2}$$$

$\mathrm{\κ}≔\mathrm{simplify}\left(\mathrm{Curvature}\left(\mathbf{R}\right)\right)$ = $\frac{1}{2}$$$

$\mathrm{\τ}≔\mathrm{Torsion}\left(\mathbf{R}\right)$ = $\frac{1}{2}$$$

Obtain and display the vectors of the TNB-frame

$$\begin{gathered} \mathbf{T}≔\mathrm{TangentVector}\left(\mathbf{R}\,\mathrm{normalized}\right)\: \\ \mathbf{N}≔\mathrm{PrincipalNormal}\left(\mathbf{R}\,\mathrm{normalized}\right)\: \\ \mathbf{B}≔\mathrm{Binormal}\left(\mathbf{R}\,\mathrm{normalized}\right)\: \end{gathered}$$

$\mathbf{T}\,\mathbf{N}\,\mathbf{B}$

Obtain the Darboux vector

$\mathbf{d}≔\mathrm{\τ} \mathbf{T}\+\mathrm{\κ} \mathbf{B}$ = $$

$\frac{\mathrm{diff}\left(\mathbf{T}\,p\right)}{\mathrm{\ρ}}$ = $$

$\mathrm{CrossProduct}\left(\mathbf{d}\,\mathbf{T}\right)$ = $$

$\frac{\mathrm{diff}\left(\mathbf{N}\,p\right)}{\mathrm{\ρ}}$ = $$

$\mathrm{CrossProduct}\left(\mathbf{d}\,\mathbf{N}\right)$ = $$

$\frac{\mathrm{diff}\left(\mathbf{B}\,p\right)}{\mathrm{\ρ}}$ = $$

$\mathrm{CrossProduct}\left(\mathbf{d}\,\mathbf{B}\right)$ = $$

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

${\mathrm{\κ}}^2 \mathbf{d}$