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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:

  • 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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    $\begingroup$ Are you asking whether is has ever been noted before that $\sqrt{1+(\tan(\theta))^2}$ equals $\sec(\theta)$? $\endgroup$
    – Jason Starr
    Commented Jul 15, 2017 at 7:38
  • $\begingroup$ @JasonStarr no, Iam aware, that the arclength differential can often be expressed by simple functions; what I am looking for, are arclength differentials, that are explicit functions, where the variable has been replaced by the original function and $sec(\theta)$ doesn't qualify as $g(tan(\theta))$, or do I miss something? $\endgroup$
    – Manfred Weis
    Commented Jul 15, 2017 at 7:59
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    $\begingroup$ Let me clarify my comment. The answer to your first question, "has that special property been noted before" is: yes, for circles this is one of the elementary identities of trigonometry equivalent to the identity $\sqrt{1+(\tan(\theta))^2} = \sec(\theta)$. $\endgroup$
    – Jason Starr
    Commented Jul 15, 2017 at 8:09
  • $\begingroup$ @JasonStarr write down an explicit function $g(x)$ with some text editor and then replace '$x$' with '$f(\theta)$'; if you then have $\sqrt{1+(f'(\theta))^2} = g(\theta)$, then that specific $f(\theta)$ is an example of what i am looking for. $\endgroup$
    – Manfred Weis
    Commented Jul 15, 2017 at 8:10
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    $\begingroup$ My favorite example of a plane curve whose arclength is explicitly computable is the (rational) cubic $9y^2-x(x-3)^2=0$. When one parametrizes it by $(x,y) = (t^2,\,t^3/3{-}t)$, one has $\mathrm{d}s = (t^2+1)\,\mathrm{d}t$, so $s_1-s_0 = (t_1-t_0)((t_1-t_0)^2+3)/3$, which can be expressed rationally in terms of $(x_1,y_1)$ and $(x_0,y_0)$, since $t_i = 3y_i/(x_i-3)$ (as long as $x_i\not=3$). $\endgroup$
    – Robert Bryant
    Commented Jul 15, 2017 at 14:04

2 Answers 2

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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 Prasolov, Viktor; Solovyev, Yuri, Elliptic functions and elliptic integrals. Translated from the Russian manuscript by D. Leites. American Mathematical Society, Providence, RI, 1997.

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    $\begingroup$ In particular, for the astroid, $h(f)$ equals $(1-f^{2/3})^{-1}-1=f^{2/3}/(1-f^{2/3}).$ I had a superfluous minus sign in my comment above. In this case $df/\sqrt{h(f)}$ equals $f^{-1/3}\sqrt{1-f^{2/3}}df$. Substituting $e=1-f^{2/3}$, $de = -(2/3)f^{-1/3}df$, the differential is $-(3/2)e^{1/2}de = d(-e^{3/2})$. So the astroid is one more case where the integral above can be evaluated in terms of elementary functions. $\endgroup$
    – Jason Starr
    Commented Jul 15, 2017 at 8:38
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    $\begingroup$ I believe that all cases when $h$ is an algebraic function have been in principle classified: J. Davenport, On the integration of algebraic functions. Springer-Verlag, Berlin-New York, 1981. $\endgroup$
    – Alexandre Eremenko
    Commented Jul 15, 2017 at 8:44
  • $\begingroup$ thank you all for the nice feedback; maybe those examples are useful for class rooms.. $\endgroup$
    – Manfred Weis
    Commented Jul 15, 2017 at 9:52
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This is related to the fact that the circle has constant curvature. Indeed the curvature of a curve obtained plotting $ y=f(x) $ is $\mathcal{C}_{f}(x)=f''(x)(1+(f'(x))^{2})^{-3/2} $, which comes from the definition of the curvature as the derivative of the angle in function of the distance $d\alpha/ds$. Note that's there no other function in $C^{2}(\mathbb{R})$ than the constant function $f:x\mapsto 0$ such that for all $x$ one has $\mathcal{C}_{f}(x)=f(x)$ (I asked the question some years ago here or on MSE).

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