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So, I'm quite in a pickle here to calculate the curvature of the Archimedes spiral, and everything I searched for, people just say the curvature is

$\kappa = \dfrac{2+t^2}{a(1+t^2)^{3/2}}$

without ever enlightening how to reach this result. What I know is this:

the spiral is given by $\mathbf{x}(t) = (x(t), y(t)) = (rt\cos(t), rt\sin(t))$, and we have the natural coordinate being $\mathbf{x}_{\text{nat}}(s) = \mathbf{x}(t(s))$ with relation to the arc lenght $s$, where $t(s)$ is the inverse function of $s(t) = \int_{0}^{t} |\mathbf{\dot{x}}(\tau)|d\tau$.

First, I need to find said natural coordinate $\mathbf{x}_{\text{nat}}(s) = \mathbf{x}(t(s))$ by using the relation between $t(s)$ and $s(t)$ given by

$t'(s) = \dfrac{1}{\dot{s}(t(s))} = \dfrac{1}{r\sqrt{1+t^2}}$ ,

and afterwords, I need to find the curvature of said spiral by the relation

$\kappa(s) = |\mathbf{x}''_{\text{nat}}(s)|$ ,

and I don't know what to do. If we calculate the integral of $s(t)$, we obtain a not invertible function for $t(s)$, and then we are stuck with just $t'(s)$. But the natural coordinate uses $t(s)$, so I'm running in circles trying to carry on this exercise for days and nothing helped me until now. Is someone could give me some light on this I would appreciate greatly.

Daniel Simões Aristone
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  • Borrowing from Handbook of Mathematics: The equation of the Archimedean spiral in polar coordinates is $\rho = a\varphi, , a>0$. The radius of curvature is $r=a\frac{(\varphi^2+1)^{3/2}}{\varphi^2+2}$. – Anton Vrdoljak Oct 16 '23 at 18:13
  • Borrowing from Wolfram MathWorld: If the two-dimensional curve is in polar coordinates, then $\kappa = \frac{r^2+2r_{\varphi}^2-rr_{\varphi \varphi}}{(r^2+r_{\varphi}^2)^{3/2}}$. – Anton Vrdoljak Oct 16 '23 at 18:21
  • This does not help me, I'm afraid. This is information about known formulae, and I must calculate, derive such expression for the curvature, I cannot simply say "it is known to be this", I must use the relation of $\kappa(s) = |\mathbf{x}''_{\text{nat}}(s)| $ and obtain such curvature myself. – Daniel Simões Aristone Oct 16 '23 at 18:25
  • Ok, at lest you may check some references on a given link in previous comment. Next, you may check too this old post: https://math.stackexchange.com/questions/568752/curvature-of-the-archimedean-spiral-in-polar-coordinates – Anton Vrdoljak Oct 16 '23 at 18:48
  • But I did. I checked this sources before, and none of them gave me the help I actually need, unfortunately, because they do not tell what calculations we must do in order to find the curvature expression. Thanks for your time, though. – Daniel Simões Aristone Oct 16 '23 at 19:44

1 Answers1

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This answer might not be what you were hoping for, but it is in general impossible to calculate the curvature via the formula $\kappa(s) = |\mathbf{x_{\mathrm{nat}}''}(s)|$.

This approach rarely works, because it is rarely possible to get a unit-speed parametrisation. First, you might not be able to work out the integral for $s(t)$. And second, it might be impossible to invert this expression. This is exactly the problem you are facing here.

What do you need to do then? Well, you need to use a formula that gives the curvature $\kappa(t)$ of a curve (that is not necessarily parametrized by the natural coordinate). This formula is $$ \kappa(t) = \frac{\det(\mathbf{x}'(t)\;\mathbf{x}''(t))}{\|\mathbf{x}'(t)\|^{3/2}} = \frac{x'(t)y''(t) - x''(t)y'(t)}{[x'(t)^2 + y'(t)^2]^{3/2}}. \tag{1} $$

Plug in $x(t) = rt\cos t$ and $ y(t) = r t \sin t$ and you should find the expression for $\kappa(t)$.

For completeness, I briefly derive the formula. Let $\mathbf{x}$ be a parametrization of a plane curve. Deriving once (and omitting the argument $t$) gives $\mathbf{x}' = v T$, where $v$ is the speed. If you use the Frenet-Serret formula $T'=v\kappa N$ for plane curves, you find after a second derivation $$ \mathbf{x}'' = v' T + v^2\kappa N. $$ Then $\det(\mathbf{x}'\; \mathbf{x}'') = v^3 \kappa \det(T\; N) = v^3\kappa$, which gives equation $(1)$.

Ernie060
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