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Find a parameterization of a curve in $\mathbb{R}^2$ with signed curvature is given by $κ_\sigma(s) = \frac{1}{1+s^2}$.

I have tried to to use this explanation, but I think I don't have enough information or I am missing something. I have also tried to use the method given by the proof of the fundamental theorem of plane curves. But the resulting integral is not easy, so I think that's not a good way to solve the problem

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Use the expression $$ \beta(s) = \left(\int_{s_0}^s \cos \theta(u)\,du, \int_{s_0}^s \sin \theta(u)\,du\right) \quad \text{with $\theta(u) = \int_{s_0}^u \kappa(t)\,dt$.} $$ The curve $\beta$ has tangent vector $\beta'(s) = (\cos \theta(s), \sin(\theta))$, so it is unit speed parametrized, and its curvature is $\|\beta''(s)\| = \theta'(s) = \kappa(s)$.

The equation gives an explicit expression of the curve in terms of $\kappa$, and mathematically it is equivalent with the method explained in the link you added (solving the two systems of differential equations amounts to taking an integral twice.)

Here we can work out the integrals. Plugging $\kappa$ in the formula and integrating (with $s_0=0$) gives $\theta(s) = \arctan(s)$. Now note that $$ \cos(\arctan(s)) = \frac{1}{\sqrt{1+s^2}}, \quad \sin(\arctan(s)) = \frac{s}{\sqrt{1+s^2}}. $$ Integrating once again then gives $$ \beta(s) = (\mathrm{arsinh}(s), \sqrt{1+s^2}). $$

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