Let $\mathbf{R}\prime \=\frac{d}{\mathrm{ds}}\mathbf{R}\left(s\right)$ and $\stackrel{\.}{\mathbf{R}}\=\frac{d}{\mathrm{dp}}\mathbf{R}\left(p\right)$. Start with the formulas $\mathrm{\tau }\=\left[\mathbf{R}\prime \mathbf{R}″\mathbf{R}‴\right]\/{\mathrm{\kappa }}^2$ and $\mathrm{\κ}\=∥\stackrel{\.}{\mathbf{R}}\times \stackrel{..}{\mathbf{R}}∥\/{\mathrm{\ρ}}^3$. The intermediate calculations, heavily dependent on the chain rule for differentiation of composite functions, are summarized in Table 2.6.8(a).

The last row in Table 2.6.8(a) is obtained by observing that in the box product, only rows that are independent will "survive." This becomes clear in the following concluding steps where the additive properties of a determinant are essential.

If $\left|\left(A\right)\left(B\right)\left(C\right)\right|$ represents the determinant of matrix whose columns are the vectors A, B, and C, then

$\left|\left(U\+V\right)\left(W\right)\left(Q\right)\right|$ = $\left|\left(U\right)\left(W\right)\left(Q\right)\right|\+\left|\left(V\right)\left(W\right)\left(Q\right)\right|$ 

In other words, if a column is the sum of two other column vectors, then the determinant splits into the sum of two determinants. And what is true for columns is true for rows, because the determinant of a matrix has the same value as the determinant of the transpose of the matrix.

Write $\left[\mathbf{R}\prime \mathbf{R}″\mathbf{R}‴\right]$ as the determinant

$\|\left(\frac{\stackrel{\.}{\mathbf{R}}}{\mathrm{\ρ}}\right)\(\left(-\frac{\stackrel{\.}{\mathrm{\ρ}}}{{\mathrm{\ρ}}^3}\stackrel{\.}{\mathbf{R}}\+\frac{\stackrel{..}{\mathbf{R}}}{{\mathrm{\ρ}}^2}\right)\left(\mathrm{\alpha } \stackrel{\.}{\mathbf{R}}\+\mathrm{\beta } \stackrel{..}{\mathbf{R}}\+\frac{1}{{\mathrm{\rho }}^3}\stackrel{..\.}{\mathbf{R}}\right)\|$ 

By the additivity property of determinants, this splits into a sum of determinants, but any of the resulting determinants that have two columns proportional will be zero. Hence, the "surviving" determinant is the box product

$\left[\frac{\stackrel{\.}{\mathbf{R}}}{\mathrm{\ρ}} \frac{\stackrel{..}{\mathbf{R}}}{{\mathrm{\ρ}}^2} \frac{\stackrel{..\.}{\mathbf{R}}}{{\mathrm{\ρ}}^3}\right]\=\frac{1}{{\mathrm{\ρ}}^6}\left[\stackrel{\.}{\mathbf{R}}\stackrel{..}{\mathbf{R}}\stackrel{..\.}{\mathbf{R}}\right]$

Consequently, $\mathrm{\tau }\=\left[\mathbf{R}\prime \mathbf{R}″\mathbf{R}‴\right]\/{\mathrm{\kappa }}^2$ becomes

$\frac{\left[\stackrel{\.}{\mathbf{R}}\stackrel{..}{\mathbf{R}}\stackrel{..\.}{\mathbf{R}}\right]\/{\mathrm{\ρ}}^6}{\(∥\stackrel{\.}{\mathbf{R}}\times \stackrel{..}{\mathbf{R}}∥\/{\mathrm{\rho }}^3{\)}^2}$ = $\frac{\left[\stackrel{\.}{\mathbf{R}}\stackrel{..}{\mathbf{R}}\stackrel{..\.}{\mathbf{R}}\right]}{{∥\stackrel{\.}{\mathbf{R}}\times \stackrel{..}{\mathbf{R}}∥}^2}$