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Let's take a look on the Archimedean spiral. The parametric equation is:

$$c : \mathbb R \to \mathbb R^{2} \,;\, c(t) := (t\cos(t), t\sin(t))$$

The goal of the exercise is to compute the curvature of the spiral in polar coordinates. What I've tried? I convert the parametric equation into polar coordinates:

$$x(\varphi) = r(\varphi)\cos(\varphi)$$

$$y(\varphi) = r(\varphi)\sin(\varphi)$$ Is that correct so far?

I know how to compute the curvature with the parametric equation, that's not the point. But how I can do it with the polar coordinates?

Thanks in advance

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  • $\begingroup$ I don't know how to calculate the derivative of the polar coordinates. Can anybody give me a hint? $\endgroup$
    – Prime Cuts
    Nov 16, 2013 at 0:41
  • $\begingroup$ Did you find what is $r(t)$ and $\theta(t)$? $\endgroup$
    – user99914
    Nov 16, 2013 at 2:50
  • $\begingroup$ Search "polar" in this. Can it help ? $\endgroup$ Nov 16, 2013 at 12:07
  • $\begingroup$ @John: $r(t)$ is the radius, $\theta(t)$ is the angle in dependence of the parameter t, right? How I can find it? $\endgroup$
    – Prime Cuts
    Nov 16, 2013 at 19:53
  • $\begingroup$ @cvis: $r(t)$ is the radius, also the length $|c(t)|$, while $\theta(t)$ is the angle made by $c(t)$ and the $x$-axis. You can check that it is just given by the answer by Ross Millikan below. $\endgroup$
    – user99914
    Nov 16, 2013 at 20:06

3 Answers 3

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Let $r$, $\varphi\colon [a,b]\to\mathbb R$ be $C^{\infty}$. Applying the definition of the curvature to $\gamma=r\cdot(\cos\mathrel\circ\varphi,\sin\mathrel\circ\varphi)$, a lengthy calculation shows that $$\kappa=\frac{r^2\dot\varphi^3+2\dot r^2\dot\varphi-r\ddot r\dot\varphi+r\dot r\ddot\varphi}{(\dot r^2+r^2\dot\varphi^2)^{3/2}}$$

In your case: $r(t)=\varphi(t)=t$, hence $\kappa(t)=\frac{t^2+2}{\sqrt{1+t^2}^3}$.

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  • $\begingroup$ How I can find the $r$ and $\varphi$ in the formula? Can I just follow Ross Millikan's post and say $r=t, \varphi=t$. Is that all? $\endgroup$
    – Prime Cuts
    Nov 16, 2013 at 20:36
  • $\begingroup$ Every time when it looks very simple I'm cunfused a little bit $\endgroup$
    – Prime Cuts
    Nov 16, 2013 at 20:37
  • $\begingroup$ @cvis I've enhanced my answer. $\endgroup$ Nov 17, 2013 at 14:52
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Your parametric equation in polar is $r=t, \theta=t$

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  • $\begingroup$ That is right. And there is not such a "lengthy calculation" using the equation posted by Michael Hoppe. It is immediate after this parametrization. The question is why, when $t \to 0$ the curvature is not infinity. Isn't the spiral supposed to collapse to a point (which as infinite curvature? ) $\endgroup$ Apr 26, 2019 at 13:58
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enter image description here

$$r=θ \\ dr/dθ=1 \\ ds=ρdθ \\ (rdθ)2+(dr)2=(ds)2 \\ (rdθ)2+(dr)2=(ρdθ)2 \\ (r)2+(dr/dθ)2=(ρ)2 \\ (r)2+(1)2=(ρ)2 \\ ρ=\sqrt{r2+1}=\sqrt{θ2+1}$$

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    $\begingroup$ Can you add some further detail or comment? Is $\theta$ the angle $\varphi$ mentioned in the question? $\endgroup$ Apr 29, 2014 at 10:23
  • $\begingroup$ the explicit form of the curve is r=teta and the figure shown is at a point wit r and teta and an adjacent point with r+dr and teta+d(teta) distanced away by ds. $\endgroup$
    – Reza
    Apr 29, 2014 at 13:20

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