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While an explicit function $y(x)$'s arc length $s$ is easily obtained as

$$s = \int \sqrt{1+|y'(x)|^2}\,dx,$$

is there any formula for implicit functions given by $f(x,y) = 0$? One can use the implicit differentiation $y'(x) = -\frac{\partial_y f}{\partial_x f}$ to obtain

$$s = \int\sqrt{1 + |\partial_y f / \partial_x f|^2}\,dx,$$

but that still requires (locally) solving for $y(x)$. Is there any formulation that does not require this, e.g. another implicit equation involving $s$?

Thoughts so far:

One could rewrite $s$ as

$$s = \int |\nabla f|\, |\partial_x f|dx,$$

or symmetrize to

$$s = \int |\nabla f|\, \underbrace{(|\partial_x f|dx + |\partial_y f|dy)}_{(*)}/2$$

where $(*)$ might be strongly related to $|df|$ I guess (though it's not identical due to the $|\cdot|$), but then?

Tobias Kienzler
  • 6,785

1 Answers1

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Consider the divergence theorem on the two-dimensional region $\mathcal R = \{(x,y):f(x,y)\le 0\}$ bounded by the curve $\mathcal C = \partial\mathcal R = \{(x,y):f(x,y)=0\}$, $$\iint_{\mathcal R} \nabla\cdot\mathbf v\,\mathrm dA = \oint_{\mathcal C}\mathbf v\cdot\hat{\mathbf n}\,\mathrm d\ell.$$ If we take $\mathbf v=\hat{\mathbf n}=(\nabla f)/\|\nabla f\|$, we have $\mathbf v\cdot\hat{\mathbf n} = 1$, so $$\iint_{\mathcal R} \nabla\cdot\left(\frac{\nabla f}{\|\nabla f\|}\right)\,\mathrm dA = \oint_{\mathcal C}\mathrm d\ell,$$ which is the arc length of the curve.

I don't know if this formula is useful at all, but it does satisfy your requirements.

  • Technically correct, and interesting. I'll have to think about this a bit – Tobias Kienzler Jan 30 '16 at 20:58
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    @Tobias: By the way, it may appear as though this still requires you to know the bounds of $\mathcal R$, but you can just multiply the integrand by its characteristic function, i.e. $1$ if $f \le 0$ and $0$ otherwise, and integrate over all of $\mathbb R^2$. –  Jan 30 '16 at 21:27