Since $\mathbf{R}\prime \left(s\right)\=\mathbf{T}$, it then follows that $\mathbf{R}″\=\mathbf{T}\prime \=\mathrm{\κ} \mathbf{N}$ (via the first equation on the right in Table 2.7.1). So,

Note the use, in the fourth line, of the third equation on the right in Table 2.7.1. The last line simplifies the notation in anticipation of the additional calculations needed to evaluate the box product. However, it is first necessary to obtain $\mathbf{R}⁗$.

By invoking the additive property of determinants, $\left[\mathbf{x}\mathbf{ }\mathbf{y} \left(\mathbf{z}\+\mathbf{w}\right)\right]\=\left[\mathbf{x}\mathbf{ }\mathbf{y}\mathbf{ }\mathbf{z}\right]\+\left[\mathbf{x}\mathbf{ }\mathbf{y}\mathbf{ }\mathbf{w}\right]$, and by writing $\mathbf{R}″\=\mathrm{\κ} \mathbf{N}$ as $b_2\mathbf{N}$, the box product becomes

$\left[\mathbf{R}″\mathbf{R}‴\mathbf{R}⁗\right]\=$$b_2\left(a_4{c}_3-a_3{c}_4\right)\left[\mathbf{NBT}\right]$ = $b_2\left(a_4{c}_3-a_3{c}_4\right)\left[\mathbf{BTN}\right]$ = $b_2\left(a_4{c}_3-a_3{c}_4\right)$

where the minus signs introduced by row interchanges in the determinant cancel, and

$\left[\mathbf{BTN}\right]\=\mathbf{B}·\left(\mathbf{T}\times \mathbf{N}\right)\=\mathbf{B}·\mathbf{B}\=1$ 

Making the substitutions $b_2\=\mathrm{\kappa }\,a_3\=-{\mathrm{\kappa }}^2\,a_4\=-3 \mathrm{\kappa } \mathrm{\kappa }\prime \,c_3\=\mathrm{\kappa } \mathrm{\tau }\,c_4\=2 \mathrm{\kappa }\prime \mathrm{\tau }\+\mathrm{\kappa } \mathrm{\tau }\prime$, into $b_2\left(a_4{c}_3-a_3{c}_4\right)$ leads to

The only place in Maple where symbolic calculations can be made with vectors is in the Physics:-Vectors package. Unfortunately, linear input for this package works best, although the output reflects notation greatly desired in the realm of physics.

In the following calculations, the basis vectors $\left\{\mathbf{i}\,\mathbf{j}\,\mathbf{k}\right\}$ take the place of the vectors T, N, and B, respectively. Within the package, a left-underscore for the letters i, j, or k indicates a unit basis vector ; a right-underscore, a symbolic vector.

Write this final result as ${\mathrm{\κ}}^3\left(\mathrm{\κ} \mathrm{\τ}\prime -\mathrm{\κ}\prime \mathrm{\τ}\right)\={\mathrm{\κ}}^5\left(\frac{\mathrm{\κ} \mathrm{\τ}\prime -\mathrm{\κ}\prime \mathrm{\τ}}{{\mathrm{\κ}}^2}\right)\={\mathrm{\κ}}^5$ $\frac{d}{\mathrm{ds}}\left(\frac{\mathrm{\τ}}{\mathrm{\κ}}\right)$.

Slightly more insight into the details of these calculations can be obtained if, instead of using the basis vectors $\left\{\mathbf{i}\,\mathbf{j}\,\mathbf{k}\right\}$, the derivatives of R are written in terms of T, N, and B.

The simplification of the box product to a new product involving just the vectors of the TNB-frame, and not linear combinations of these vectors, is the reward for this extra effort.