# Let $\beta$ be a unit speed curve with $\kappa \gt 0$. Show that $(\beta''\times \beta''')\cdot \beta^{4}=\kappa^5\frac{d}{ds}(\frac{\tau}{\kappa})$

Let $\beta$ be a unit speed curve with $\kappa \gt 0$. Show that

$$(\beta''\times \beta''')\cdot \beta^{4}=\kappa^5\frac{d}{ds}(\frac{\tau}{\kappa})$$

Simple calculation seems too frustrating. I'm lost on how to show this equation. What ideas may I use? I would greatly appreciate any solutions or suggestions.

• How do you define $\beta^4$ for a curve $\beta(s)$? – Robert Lewis Apr 17 '15 at 15:45
• Or do you mean $\beta^4 = \beta^{(4)} = (d^4\beta / ds^4)$? – Robert Lewis Apr 17 '15 at 18:11

## 1 Answer

It's really not too bad provided we exploit the formulas of Frenet-Serret, the definition of the binormal vector $B = T \times N$, and a few standard vector identities:

First, since $\beta(s)$ is a unit speed curve, we have the unit tangent vector

$T = \beta´(s); \tag{1}$

second, from (1), we have

$\beta''(s) = T' = \kappa N, \tag{2}$

where $N$ is the unit normal field to $\beta(s)$; this is basically the first Frenet-Serret equation. Third, from (2) we compute

$\beta'''(s) = T'' = \kappa' N + \kappa N'; \tag{3}$

fourth, we deploy the second Frenet-Serret equation,

$N' = -\kappa T + \tau B \tag{4}$

by inserting its right-hand side for $N'$ in (3). yielding

$\beta''' = \kappa' N - \kappa^2 T + \kappa \tau B; \tag{5}$

fifth, we calculate $\beta'' \times \beta'''$ via (2) and (5):

$\beta'' \times \beta''' = \kappa N \times (\kappa' N - \kappa^2 T + \kappa \tau B)$ $= \kappa \kappa' N \times N- \kappa^3 N \times T + \kappa^2 \tau N \times B = \kappa^3 B + \kappa^2 \tau T, \tag{6}$

where we have used $N \times N = 0$, $N \times T = -B$ (from $B = T \times N$), and $T = N \times B$ (also from $B = T \times N$, by cyclic permutation) in completing the derivation of (6); sixth, and this is the "grungiest" computation we will perform, we exploit (5) to find

$\beta^{(4)} = \dfrac{d^4\beta}{ds^4} = \kappa'' N + \kappa' N´ - 2 \kappa \kappa' T - \kappa^2 T' + \kappa' \tau B + \kappa \tau' B + \kappa \tau B'; \tag{7}$

we use (2), (4), and the third Frenet-Serret equation,

$B' = -\tau N \tag{8}$

to remove the derivatives $T'$, $N'$, $B'$ from (7):

$\beta^{(4)} = \kappa'' N + \kappa' (-\kappa T + \tau B) -2\kappa \kappa' T - \kappa^3 N + \kappa' \tau B + \kappa \tau' B -\kappa \tau^2 N; \tag{9}$

we are now in position to perform the lucky seventh and final step, taking the dot product of $\beta'' \times \beta'''$ as in (6) with $\beta^{(4)}$ as in (9); before committing equations to paper (well, actually to digital display), we make a few observations which will not only serve to make our calculations more efficient, but will also spare Yours Truly just a tad of $\LaTeX$, and every little tad saved helps, as I'm sure my colleagues here on MSE will agree. We observe: $T$, $N$, $B$ form an orthonormal frame in $\Bbb R^3$ at each point of $\beta(s)$, thus $T \cdot T = N \cdot N = B \cdot B = 1$ whilst $T \cdot N = T \cdot B = N \cdot B = 0$; since $\beta'' \times \beta'''$ is a linear combination of $T$ and $B$, all the terms of $\beta^{(4)}$ which contain $N$ will vanish when the dot product is taken, so they can be a priori neglected; we further observe: $-\kappa T + \tau B$ is normal to $\beta'' \times \beta'''$, as my easily be checked by direct calculation, so the term $\kappa'(-\kappa T + \tau B)$ may also be ignored; we are then left with

$(\beta'' \times \beta''') \cdot \beta^{(4)} = (\kappa^3 B + \kappa^2 \tau T) \cdot (-2 \kappa \kappa' T + \kappa' \tau B + \kappa \tau' B)$ $= \kappa^3 \kappa' \tau + \kappa^4 \tau' - 2 \kappa^3 \kappa' \tau = \kappa^4 \tau' - \kappa^3 \kappa' \tau = \kappa^3 (\tau' \kappa - \tau \kappa')$ $= \kappa^5 \dfrac{\tau' \kappa - \tau \kappa'}{\kappa^2} = \kappa^5 \dfrac{d}{ds} (\dfrac{\tau}{\kappa}) \tag{10}$

as per request.