In my notes, it says to consider an arc length parameterization r(s). Then we can show that B' points in the direction of N such that we can write B'=-N. I understand why ||B'||= but am unsure how to show why B' is parallel to N? Also, is the minus sign just for convention or is there a reason it is there?
Any help would be greatly appreciated!
$\def\vt{\mathbf{T}} \def\vn{\mathbf{N}} \def\vb{\mathbf{B}} \def\vr{\mathbf{r}}$In what follows differentiation with respect to arc length is indicated with a prime. We have \begin{align*} \vt &= \vr' \\ \vn &= \vt'/\|\vt'\| \\ \vb &= \vt\times\vn. \end{align*} Then \begin{align*} \vb' &= \vt'\times\vn+\vt\times\vn' \\ &= \|\vt'\|\underbrace{\vn\times\vn}_{\mathbf{0}} + \vt\times\vn' \\ &= \vt\times\vn'. \end{align*} Note that $\vb$ is orthogonal to $\vt$ by definition of the cross product. Since $\|\vb\|=1$, $\vb'$ is also orthogonal to $\vb$. (Proof: $(\vb\cdot\vb)' = 2\vb\cdot\vb' = 0$.) Since $\vt,\vn,\vb$ form an orthogonal basis, it must be the case that $\vb'$ is proportional to $\vn$. We take $\vb' = -\tau\vn$. The minus sign is a convention. It is a version of the right-hand rule. Note that $\tau>0$ indicates a twisting of the coordinate frame about the $\vt$ axis in a right-handed manner. (Take your right hand and point your thumb in the $\vt$ direction. Twist your hand about this axis in the direction in which your fingers are pointing. This twist corresponds to $\tau>0$.)