Let $c$: $I \to \Bbb{R}^3$ be an arc length parametrized curve with curvature $k(t) \ne$ 0 for all $t \in I$. Show that the torsion of $c$ is given by

$\tau(t)$ = $⟨c'(t) \space \times \space c''(t), \space \space c'''(t)⟩ \over {(k(t))^2}$

From the Frenet equations, I know that:

$\tau(t)$ = ⟨$n'(t)$, $b(t)$⟩

$\kappa(t)$ = || $c'(t)$||

$b(t)$ = $c'(t)n(t)$

$n(t)$ = $\frac{c''(t)}{k}$

However, I can't just plug these in to verify that the equation in the question is true. It does work out if I plug in $\kappa(t)$ instead of $\kappa$ into $n(t)$ = $\frac{c''(t)}{k}$ , but I think it works out incorrectly... despite that I got the answer, $\kappa(t)$ is a function, not a constant (like in the Frenet equations).

I can't just simply write the following, right?:

$\tau(t)$ = ⟨$n'(t)$, $b(t)$⟩

= ⟨$\frac{c''(t)}{k(t)}$, $c'(t)n(t)$⟩

How would I go about solving this equation and taking care of the $\kappa$?

  • $\begingroup$ To clarify: $k$ and $\kappa$ are both the curvature (function), right? Also, there are several "it"s whose meaning is not entirely clear (to me); is your question that you tried to substitute the formula for $n(t)$ into the formula for the torsion but the expected expression doesn't seem to come out? $\endgroup$ – Andrew D. Hwang May 9 '17 at 23:17
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    $\begingroup$ $b(t) \ne c'(t)n(t)$, particularly as $c'(t)$ and $n(t)$ are both vectors. Instead $b(t) = c'(t) \times n(t)$ $\endgroup$ – Paul Sinclair May 10 '17 at 0:28

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