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I've got a curve in $\mathbb R^2 : (x(s) , y(s)) $ ($s \in (0,L)$).

For the tangent, it holds: $t(s) = \left( \matrix{x'(s) \\ y'(s)} \right) $. Since we have arc length, it holds also $||t|| = 1$.

Let $\psi(s)$ s.t. $cos(\psi(s)) = x'(s)$ and $sin(\psi(s)) = y'(s)$.

The curvature is defined by $K(s) := \psi'(s)$

Let $f: (-R, R) \rightarrow \mathbb R, f(x) = \sqrt{R^2-x^2} $

Now (finally) the question: what's the parametrization of the graph of this function? (in arc length)

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  • $\begingroup$ Why do you need parametrization in arc length? $\endgroup$ – Michael Hoppe May 15 '18 at 21:25
  • $\begingroup$ Because I want, in a second step, calculate the curvature of the curve. $\endgroup$ – DMan May 15 '18 at 21:29

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