$\stackrel{\.}{\mathbf{R}}\=\left[\begin{array}{c}-3p^2\+3\\ 6p\\ 3p^2\+3\end{array}\right]$

$\mathbf{T}\prime \=\stackrel{\.}{\mathbf{T}}\/\mathrm{\ρ}\=\left[\begin{array}{c}-\frac{2}{3}\frac{p}{{\left(p^2\+1\right)}^3}\\ -\frac{1}{3}\frac{p^2-1}{{\left(p^2\+1\right)}^3}\\ 0\end{array}\right]$

$\mathrm{\ρ}\=∥\stackrel{\.}{\mathbf{R}}∥\=3\sqrt{2}\left(p^2\+1\right)$

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

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

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

$\mathbf{B}\=\mathbf{T}\times \mathbf{N}\=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ -\frac{1}{2}\frac{\sqrt{2}\left(p^2-1\right)}{p^2\+1}& \frac{\sqrt{2}p}{p^2\+1}& 1\/\sqrt{2}\\ -\frac{2p}{p^2\+1}& -\frac{p^2-1}{p^2\+1}& 0\end{array}\right|$ =  $\left[\begin{array}{c}\frac{1}{2}\frac{\sqrt{2}\left(p^2-1\right)}{p^2\+1}\\ -\frac{\sqrt{2}p}{p^2\+1}\\ \frac{1}{2}\sqrt{2}\end{array}\right]$

Table 2.6.4(a)   Manual calculation of the TNB-frame

$\mathrm{\tau }\= -\left(\stackrel{\.}{\mathbf{B}}\/\mathrm{\rho }\right)·\mathbf{N}$

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

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

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

$\stackrel{\.}{\mathbf{T}}\left(p\right)·\stackrel{\.}{\mathbf{B}}\left(p\right)$

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

$\=-\frac{8p^2}{{\left(p^2\+1\right)}^4}-\frac{2{\left(p^2-1\right)}^2}{{\left(p^2\+1\right)}^4}$

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

$-\mathrm{\kappa } \mathrm{\tau } {\mathrm{\rho }}^2$

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

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

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 }$

$\mathbf{R}$  = $\stackrel{\text{TNB frame}}{\to }$ $\stackrel{\text{assuming real}}{\to }$ $\stackrel{\text{assign to a name}}{\to }$$\mathrm{TNB}$

Obtain the torsion via the Context Panel system

$\mathbf{R}$ = $\stackrel{\text{torsion}}{\to }$$\frac{1}{3}\frac{\mathrm{csgn}\left(p^2\+1\right)\sqrt{2}}{\sqrt{\frac{{\mathrm{csgn}\left(1\,p^2\+1\right)}^2{p}^4\+2{\mathrm{csgn}\left(1\,p^2\+1\right)}^2{p}^2\+{\mathrm{csgn}\left(1\,p^2\+1\right)}^2\+2}{{\left(p^2\+1\right)}^2}}{\left(p^2\+1\right)}^3}$$\stackrel{\text{assuming real}}{\to }$$\frac{1}{3{\left(p^2\+1\right)}^2}$

Obtain the torsion by the upper-right formula in Table 2.6.1

$∥\frac{\ⅆ}{\ⅆ p} \mathbf{R}∥$ = $3\sqrt{2}\sqrt{{\left(p^2\+1\right)}^2}$$\stackrel{\text{assuming real}}{\to }$$3\sqrt{2}\left(p^2\+1\right)$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{\ρ}$ 

$-\left(\frac{\ⅆ}{\ⅆ p} {\mathbf{TNB}}_3\right)·{\mathbf{TNB}}_2\/\mathrm{\ρ}$ = $-\frac{1}{6}\frac{\left(-\frac{2\left(\frac{\sqrt{2}p}{p^2\+1}-\frac{\sqrt{2}\left(p^2-1\right)p}{{\left(p^2\+1\right)}^2}\right)p}{p^2\+1}-\frac{\left(\frac{2\sqrt{2}{p}^2}{{\left(p^2\+1\right)}^2}-\frac{\sqrt{2}}{p^2\+1}\right)\left(p^2-1\right)}{p^2\+1}\right)\sqrt{2}}{p^2\+1}$$\stackrel{\text{simplify}}{\=}$$\frac{1}{3{\left(p^2\+1\right)}^2}$

Obtain the torsion by the lower-right formula in Table 2.6.1

$\stackrel{\.}{\mathbf{R}}\=\frac{\ⅆ}{\ⅆ p} R$$\stackrel{\text{assign}}{\to }$

$\stackrel{..}{\mathbf{R}}\=\frac{{\ⅆ}^2}{{\ⅆ}^{}{p}^2} R$$\stackrel{\text{assign}}{\to }$

$\stackrel{..\.}{\mathbf{R}}\=\frac{{\ⅆ}^3}{{\ⅆ}^{}{p}^3} R$$\stackrel{\text{assign}}{\to }$

$\frac{\stackrel{\.}{\mathbf{R}}·\left(\stackrel{..}{\mathbf{R}}\times \stackrel{..\.}{\mathbf{R}}\right)}{{\∥\stackrel{\.}{\mathbf{R}}\times \stackrel{..}{\mathbf{R}}\∥}^2}$ = $\frac{1}{3{\left(p^2\+1\right)}^2}$$\stackrel{\text{simplify}}{\=}$$\frac{1}{3{\left(p^2\+1\right)}^2}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{\τ}$$$

Evaluate the right-hand side of the given identity

$\mathbf{R}$ = $\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}$

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

Evaluate the left-hand side of the given identity

$\left(\frac{\ⅆ}{\ⅆ p} {\mathrm{TNB}}_1\right)·\left(\frac{\ⅆ}{\ⅆ p} {\mathrm{TNB}}_3\right)$ = $\left(-\frac{\sqrt{2}p}{p^2\+1}\+\frac{\sqrt{2}\left(p^2-1\right)p}{{\left(p^2\+1\right)}^2}\right)\left(\frac{\sqrt{2}p}{p^2\+1}-\frac{\sqrt{2}\left(p^2-1\right)p}{{\left(p^2\+1\right)}^2}\right)\+\left(-\frac{2\sqrt{2}{p}^2}{{\left(p^2\+1\right)}^2}\+\frac{\sqrt{2}}{p^2\+1}\right)\left(\frac{2\sqrt{2}{p}^2}{{\left(p^2\+1\right)}^2}-\frac{\sqrt{2}}{p^2\+1}\right)$$\stackrel{\text{simplify}}{\=}$$-\frac{2}{{\left(p^2\+1\right)}^2}$$$

Evaluate the TNB-frame at $p\=-1$ 

$\genfrac{}{}{0ex}{}{\mathrm{TNB}}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{p\=-1}$ = $\stackrel{\text{assign to a name}}{\to }$$\mathrm{TNBp}$

Graph each vector in the evaluated TNB-frame

$\mathrm{TNBp}\left[1\right]$ = $\stackrel{\text{to free Vector}}{\to }$ $\stackrel{\text{plot arrow}}{\to }$

$\mathrm{TNBp}\left[2\right]$ = $\stackrel{\text{to free Vector}}{\to }$ $\stackrel{\text{plot arrow}}{\to }$ $$

$\mathrm{TNBp}\left[3\right]$ = $\stackrel{\text{to free Vector}}{\to }$ $\stackrel{\text{plot arrow}}{\to }$ $$

Graph R and add the vectors of the TNB-frame

$R$ = $\stackrel{\text{to list}}{\to }$$\left[-p^3\+3p\,3p^2\,p^3\+3p\right]$$\to$   $$

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

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

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

$\mathrm{TNB}≔\mathrm{simplify}\left(\left[\mathrm{TNBFrame}\left(\mathbf{R}\right)\right]\right) assuming p\colon\colon \mathrm{real}$

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

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

Obtain the torsion by the upper-right formula in Table 2.6.1

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

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

$\mathrm{simplify}\left(-\frac{1}{\mathrm{\ρ}} \mathrm{DotProduct}\left(\mathrm{diff}\left({\mathbf{TNB}}_3\,p\right)\,{\mathbf{TNB}}_2\right)\right) assuming p\colon\colon \mathrm{real}$ = $\frac{1}{3{\left(p^2\+1\right)}^2}$$$

Obtain the torsion by the lower-right formula in Table 2.6.1

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

$\stackrel{..}{\mathbf{R}}≔\mathrm{diff}\left(\mathbf{R}\,p\$2\right)\:$

$\stackrel{..\.}{\mathbf{R}}≔\mathrm{diff}\left(\mathbf{R}\,p\$3\right)\:$

Apply the DotProduct, CrossProduct, Norm, and simplify commands as appropriate.

$\mathrm{simplify}\left(\frac{\mathrm{DotProduct}\left(\stackrel{\.}{\mathbf{R}}\,\mathrm{CrossProduct}\left(\stackrel{..}{\mathbf{R}}\,\stackrel{..\.}{\mathbf{R}}\right)\right)}{\mathrm{Norm}{\left(\mathrm{CrossProduct}\left(\stackrel{\.}{\mathbf{R}}\,\stackrel{..}{\mathbf{R}}\right)\right)}^2}\right)$ = $\frac{1}{3{\left(p^2\+1\right)}^2}$ 

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

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

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

$\mathrm{simplify}\left(\mathrm{DotProduct}\left(\mathrm{diff}\left({\mathrm{TNB}}_1\,p\right)\,\mathrm{diff}\left({\mathrm{TNB}}_3\,p\right)\right)\right)$ = $-\frac{2}{{\left(p^2\+1\right)}^2}$$$

$\mathrm{TNBFrame}\left(\mathbf{R}\,\mathrm{range}\=-2..0\,\mathrm{output}\=\mathrm{plot}\,\mathrm{caption}\=''''\,\mathrm{frames}\=1\,\mathrm{curveoptions}\=\left[\mathrm{scaling}\=\mathrm{constrained}\,\mathrm{axes}\=\mathrm{frame}\,\mathrm{color}\=\mathrm{blue}\,\mathrm{labels}\=\left[x\,y\,z\right]\,\mathrm{orientation}\=\left[-65\,70\,0\right]\right]\,\mathrm{tangentoptions}\=\left[\mathrm{color}\=\mathrm{black}\,\mathrm{width}\=.2\right]\,\mathrm{normaloptions}\=\left[\mathrm{color}\=\mathrm{red}\,\mathrm{width}\=.2\right]\,\mathrm{binormaloptions}\=\left[\mathrm{color}\=\mathrm{green}\,\mathrm{width}\=.2\right]\right)$

$\mathrm{TNBFrame}\left(\mathbf{R}\,\mathrm{range}\=-2..0\,\mathrm{output}\=\mathrm{animation}\,\mathrm{caption}\=''''\,\mathrm{frames}\=30\,\mathrm{curveoptions}\=\left[\mathrm{scaling}\=\mathrm{constrained}\,\mathrm{axes}\=\mathrm{frame}\,\mathrm{color}\=\mathrm{blue}\,\mathrm{labels}\=\left[z\,y\,z\right]\,\mathrm{orientation}\=\left[-65\,70\,0\right]\right]\,\mathrm{tangentoptions}\=\left[\mathrm{color}\=\mathrm{black}\,\mathrm{width}\=.2\right]\,\mathrm{normaloptions}\=\left[\mathrm{color}\=\mathrm{red}\,\mathrm{width}\=.2\right]\,\mathrm{binormaloptions}\=\left[\mathrm{color}\=\mathrm{green}\,\mathrm{width}\=.2\right]\right)$

$$\begin{gathered} V≔\mathrm{map}\left(\mathrm{ConvertVector}\,\mathrm{eval}\left(\mathrm{TNB}\,p\=-1\right)\,\mathrm{rooted}\,\left[-2\,3\,-4\right]\right)\: \\ {p}_1≔\mathrm{PlotVector}\left(V\,\mathrm{color}\=\left[\mathrm{black}\,\mathrm{red}\,\mathrm{green}\right]\,\mathrm{width}\=.2\right)\: \\ {p}_2≔\mathrm{PlotPositionVector}\left(\mathrm{ConvertVector}\left(\mathbf{R}\,\mathrm{position}\right)\,p\=-1.5..0\,\mathrm{curveoptions}\=\left[\mathrm{color}\=\mathrm{blue}\right]\right)\: \\ \mathrm{plots}:-\mathrm{display}\left(p_1\,p_2\,\mathrm{scaling}\=\mathrm{constrained}\,\mathrm{axes}\=\mathrm{frame}\,\mathrm{labels}\=\left[x\,y\,z\right]\,\mathrm{tickmarks}\=\left[2\,2\,8\right]\,\mathrm{orientation}\=\left[-65\,70\,0\right]\right) \end{gathered}$$