Answer
Result See proof.
Work Step by Step
Step 1 Given: \[ \begin{align*} \mathbf{T}' &= \mathbf{N} \kappa \\ \mathbf{N}' &= -\mathbf{T}\kappa + \mathbf{B}\tau \\ \mathbf{B}' &= \tau \mathbf{N} \end{align*} \] Step 2 Since \[ \mathbf{r}''(s) = \frac{d\mathbf{T}}{ds} = \kappa\mathbf{N} \] Then \[ \begin{align*} \mathbf{r}'''(s) &= \mathbf{B}\kappa - \kappa^2\mathbf{T} + \frac{d\kappa}{ds}\mathbf{N} \\ &= \kappa(-\kappa\mathbf{T} + \tau\mathbf{B}) + \frac{d\kappa}{ds}\mathbf{N} \end{align*} \] Since \[ \mathbf{r}'(s) \times \mathbf{r}''(s) = \mathbf{T} \times (\kappa\mathbf{N}) = \kappa\mathbf{T} \times \mathbf{N} = \kappa\mathbf{B} \] Step 3 Hence \[ [\mathbf{r}'(s) \times \mathbf{r}''(s)] \cdot \mathbf{r}'''(s) = -\kappa^3\mathbf{B} \cdot \mathbf{T} + \kappa\frac{d\kappa}{ds}\mathbf{B} \cdot \mathbf{N} + \kappa^2\tau\mathbf{B} \cdot \mathbf{B} = \kappa^2\tau \] It follows \[ \tau = \frac{[\mathbf{r}'(s) \times \mathbf{r}''(s)] \cdot \mathbf{r}'''(s)}{\|\mathbf{r}''(s)\|^2} \] and \[ \mathbf{B} = \frac{\mathbf{r}'(s) \times \mathbf{r}''(s)}{\|\mathbf{r}''(s)\|} \]
