Yesterday I needed to do some calculations with circles and "ventured" to calculate the arc length via the $\int{\sqrt{1+\left(f'(x)\right)^2}}$ formula and was baffled to see that in the case of unit half-circles that amounts to $\ ds := \sqrt{1+\left(f'(x)\right)^2} = \frac{1}{f(x)}$


Stuffing that differential equation in WA yields the four solutions $f(x) = \pm\sqrt{-2 c\_1 x - c\_1^2 - x^2 + 1}\ $and $f(x) = \pm\sqrt{2 c\_1 x - c\_1^2 - x^2 + 1}\ $  

  




> 
> **Questions:**
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> * has that special property been noted before?
> * are there other functions, whose arclength differential satisfies $\sqrt{1+\left(f'(x)\right)^2} = g\left(f(x)\right)$, when $g()$ is an ordinary, explicit function?
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> 
> 



Has this Peculiar Property of Unit Circles Already been Noticed?
You are asking whether there are functions $f$ such that $(f')^2=h(f)$. (I denote $h=g^2-1$). The answer is yes: these functions are solutions of the differential equation
$f'=\sqrt{h(f)}$. The general solution of this equation is: $$\int\frac{df}{\sqrt{h(f)}}=x+c.$$
In your case, $h$ is a quadratic rational function. Taking other quadratic functions you obtain various integrals which can be evaluated in terms of elementary functions. If $h$ is a polynomial of degree $3$ or $4$, the integral cannot be expressed in terms of elementary functions but it
is an elliptic integral. It is actually one of the earliest problems where an elliptic integral was encountered: the arc length of ellipse and Bernoulli lemniscate.


See, for example, [MR1469740](http://www.ams.org/mathscinet-getitem?mr=1469740) Prasolov, Viktor; Solovyev, Yuri, Elliptic functions and elliptic integrals. Translated from the Russian manuscript by D. Leites. American Mathematical Society, Providence, RI, 1997.


