$\mathbf{T}\=\frac{1}{\sqrt{\mathrm{\alpha }}}\left[\begin{array}{c}1\\ 6 p\\ 3 p^2\end{array}\right]$

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

$\mathrm{\ρ}\=\sqrt{\mathrm{\α}}$

$\mathrm{\κ}\=\frac{6\sqrt{\mathrm{\beta }}}{{\mathrm{\alpha }}^{3\/2}}$

Table 2.8.6(a)   Items from the Frenet formalism, where $\mathrm{\α}\=9p^4\+36p^2\+1$, $\mathrm{\β}\=9p^4\+p^2\+1$

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

$\=\frac{d}{\mathrm{dp}}\left(\sqrt{\mathrm{\α}}\right) \mathbf{T}\+\left(\frac{6\sqrt{\mathrm{\beta }}}{{\mathrm{\alpha }}^{3\/2}}\right) {\left(\sqrt{\mathrm{\α}}\right)}^2 \mathbf{N}$

$\=\frac{18p\left(p^2\+2\right)}{\sqrt{\mathrm{\α}}} \left(\frac{1}{\sqrt{\mathrm{\alpha }}}\left[\begin{array}{c}1\\ 6 p\\ 3 p^2\end{array}\right]\right)\+\frac{6\sqrt{\mathrm{\β}}}{\sqrt{\mathrm{\α}}}\left(\frac{1}{\sqrt{\mathrm{\α}}\sqrt{\mathrm{\β}}}\left[\begin{array}{c}-3 p\(p^2\+2\)\\ -\(9 p^4-1\)\\ p \(18 p^2\+1\)\end{array}\right]\right)$

$\=\frac{6}{\mathrm{\alpha }}\left(3 p \left(p^2\+2\right)\left[\begin{array}{c}1\\ 6 p\\ 3 p^2\end{array}\right]\+\left[\begin{array}{c}-3 p\(p^2\+2\)\\ -\(9 p^4-1\)\\ p \(18 p^2\+1\)\end{array}\right]\right)$

$\=\frac{6}{\mathrm{\α}} \left[\begin{array}{c}0\\ 9p^4\+36p^2\+1\\ p\left(9p^4\+36p^2\+1\right)\end{array}\right]$

$\=\frac{6}{\mathrm{\α}} \left[\begin{array}{c}0\\ \mathrm{\α}\\ p \mathrm{\α}\end{array}\right]$

$\=\left[\begin{array}{c}0\\ 6\\ 6 p\end{array}\right]$

$\mathbf{R}″\mathbf{·}\mathbf{N}$

$\=\frac{\left[\begin{array}{c}0\\ 6\\ 6 p\end{array}\right]·\left[\begin{array}{c}-3 p\(p^2\+2\)\\ -\(9 p^4-1\)\\ p \(18 p^2\+1\)\end{array}\right]}{\sqrt{\mathrm{\alpha }}\sqrt{\mathrm{\beta }}}$

$\=\frac{6\left(9p^4\+p^2\+1\right)}{\sqrt{\mathrm{\alpha }}\sqrt{\mathrm{\β}}}$

$\=\frac{6 \mathrm{\β}}{\sqrt{\mathrm{\α}}\sqrt{\mathrm{\β}}}$

$\=6\frac{\sqrt{\mathrm{\β}}}{\sqrt{\mathrm{\α}}}$

$\=\left(\frac{6\sqrt{\mathrm{\beta }}}{{\mathrm{\alpha }}^{3\/2}}\right){\left(\sqrt{\mathrm{\α}}\right)}^2\=\mathrm{\κ} {\mathrm{\ρ}}^2$

Initialize

Loading Student:-VectorCalculus

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

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

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

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

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{36p}{\sqrt{9p^4\+36p^2\+1}}\+\frac{18p^3}{\sqrt{9p^4\+36p^2\+1}}$$\stackrel{\text{simplify}}{\=}$$\frac{18p\left(p^2\+2\right)}{\sqrt{9p^4\+36p^2\+1}}$

$\frac{\ⅆ}{\ⅆ p} \mathrm{\ρ}$ = $\frac{1}{2}\frac{36p^3\+72p}{\sqrt{9p^4\+36p^2\+1}}$$\stackrel{\text{assuming real}}{\to }$$\frac{18p\left(p^2\+2\right)}{\sqrt{9p^4\+36p^2\+1}}$

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

$\mathbf{R}″·\mathbf{N}$ = $-\frac{6\left(9p^4-1\right)}{\sqrt{9p^4\+36p^2\+1}\sqrt{9p^4\+p^2\+1}}\+\frac{6p^2\left(18p^2\+1\right)}{\sqrt{9p^4\+36p^2\+1}\sqrt{9p^4\+p^2\+1}}$$\stackrel{\text{simplify}}{\=}$$\frac{6\sqrt{9p^4\+p^2\+1}}{\sqrt{9p^4\+36p^2\+1}}$$$$$

$\mathrm{\κ} {\mathrm{\ρ}}^2$ = $\frac{6\sqrt{9p^4\+p^2\+1}}{\sqrt{9p^4\+36p^2\+1}}$$\stackrel{\text{simplify}}{\=}$$\frac{6\sqrt{9p^4\+p^2\+1}}{\sqrt{9p^4\+36p^2\+1}}$

Initialize

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

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

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

$\mathbf{R}≔⟨p\,3 p^2\,p^3⟩\:$

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

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

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

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

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

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 $\mathrm{\ρ}\prime$ 

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

$\frac{18p\left(p^2\+2\right)}{\sqrt{9p^4\+36p^2\+1}}$

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

$\frac{18p\left(p^2\+2\right)}{\sqrt{9p^4\+36p^2\+1}}$

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

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

$\frac{6\sqrt{9p^4\+p^2\+1}}{\sqrt{9p^4\+36p^2\+1}}$

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

$\frac{6\sqrt{9p^4\+p^2\+1}}{\sqrt{9p^4\+36p^2\+1}}$