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Hi need to compute the curvature of a curve defined by:

\begin{cases} r_1(s) = a \cdot \cos\left[{\frac{s}{a} \cos\alpha}\right] \\ r_2(s) = a \cdot \sin\left[{\frac{s}{a} \cos\alpha}\right] \\ r_3(s) = s \cdot \sin\left({\alpha}\right) \end{cases}

With $a \in \mathbb{R}; a > 0$ and $\alpha \in \mathbb{R}$

but to do this first I need to test if the curve is parametrized by the arc complement then I cant compute the curvature as:

$k=||\frac{dT}{ds}||$

with $T(s)=\frac{dr}{ds}$

P. Lawrence
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efirvida
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1 Answers1

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If we start with the curve $$x=a\cos(\frac{t\cos \alpha}{a}),y=a\sin(\frac{t\cos \alpha}{a}),z=t\sin\alpha$$ we calculate $ \mathbf v=[\dot x,\dot y,\dot z] \text { and }\dot s=\vert \vert \mathbf v \vert \vert=1$ so the curve really is parametrized by the arc length and there is no harm in taking $s=t.$ Since $\dot s=1,$ the formula for the curvature is just$$\kappa=\vert \vert \mathbf v \times \mathbf a\vert \vert$$. We already worked out $\mathbf v$. Differentiate its components to find $\mathbf a$ and then take the cross-product and calculate its magnitude.The result is $$\kappa=\frac{\cos^2\alpha}{\vert a \vert}.$$

P. Lawrence
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  • So if I understand well, If the norm of the speed is 1, then the curve is parametrized by the arc length? – efirvida Jan 07 '23 at 20:36
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    @efirvida if the speed, which is the norm of the vector, is 1, then $\dot s=t$ and thus $s+c=t$. Before, you were measuring arc length as the arc distance from some point $P_0$ on the curve. Take a new point $P_$ on the curve such that the arc length from $P_$ to $P_0$ is $c$. Now measure the arc length from $P_*$ and then $s=t$ so the curve is parametrized by its arc length. – P. Lawrence Jan 08 '23 at 05:16
  • @p-lawrence, thank you for the explanation, its clear for me now! – efirvida Jan 08 '23 at 18:43