$\stackrel{\.}{\mathbf{R}}\=\left[\begin{array}{c}1\\ 6 p\\ 3 p^2\end{array}\right]$

$\mathbf{T}\prime \=\stackrel{\.}{\mathbf{T}}\/\mathrm{\ρ}\=\left[\begin{array}{c}-\frac{18p\left(p^2\+2\right)}{{\left(9p^4\+36p^2\+1\right)}^2}\\ -\frac{6\left(9p^4-1\right)}{{\left(9p^4\+36p^2\+1\right)}^2}\\ \frac{6p\left(18p^2\+1\right)}{{\left(9p^4\+36p^2\+1\right)}^2}\end{array}\right]$

$\mathrm{\ρ}\=∥\stackrel{\.}{\mathbf{R}}∥\=1\/\sqrt{9p^4\+36p^2\+1}$

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

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

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

$\mathbf{B}\=\mathbf{T}\times \mathbf{N}\=\frac{\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 1& 6 p& 3 p^2\\ -3p\left(p^2\+2\right)& -\left(9p^4-1\right)& p\left(18p^2\+1\right)\end{array}\right|}{\left(9p^4\+36p^2\+1\right)\sqrt{9p^4\+p^2\+1}}$ =  $\left[\begin{array}{c}\frac{3p^2}{\sqrt{9p^4\+p^2\+1}}\\ -\frac{p}{\sqrt{9p^4\+p^2\+1}}\\ \frac{1}{\sqrt{9p^4\+p^2\+1}}\end{array}\right]$

Table 2.6.2(a)   Manual calculation of the TNB-frame

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

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

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

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

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

$\=\left[\begin{array}{c}-\frac{18p\left(p^2\+2\right)}{{\left(9p^4\+36p^2\+1\right)}^{3\/2}}\\ -\frac{6\left(9p^4-1\right)}{{\left(9p^4\+36p^2\+1\right)}^{3\/2}}\\ \frac{6p\left(18p^2\+1\right)}{{\left(9p^4\+36p^2\+1\right)}^{3\/2}}\end{array}\right]·\left[\begin{array}{c}\frac{3p\left(p^2\+2\right)}{{\left(9p^4\+p^2\+1\right)}^{3\/2}}\\ \frac{9p^4-1}{{\left(9p^4\+p^2\+1\right)}^{3\/2}}\\ -\frac{p\left(18p^2\+1\right)}{{\left(9p^4\+p^2\+1\right)}^{3\/2}}\end{array}\right]$

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

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

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

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