Explicit nontrivial examples of arc length parametrization

Question

For pedagogical purposes (and for some numerical experiments) I was looking for some nontrivial explicit (i.e. closed-form) examples of arclength parametrized curves. I know that arclength parametrization always exists for a regular curve, but it seems that the general consensus is that trying to compute this explicitly is generally madness.

Of course any straight line and circular arc can be explicitly parametrized this way, but I was wondering if there are any other known nontrivial examples (especially ones with nonconstant curvature)?

Answer 1

A nice simple one is Neile's parabola, $$ y = \frac{2}{3}ax^{3/2}. $$ One finds $dy/dx = a\sqrt{x}$, so the arc-length integral is $$ \int_0^X \sqrt{1+a^2x} \, dx = \frac{2}{3a^2}\left((1+a^2X)^{3/2}-1 \right). $$

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One can also do the ordinary parabola, $$ x=at^2 \qquad y=2at, $$ which has arc-length integral $$ \int_0^T 2a\sqrt{1+t^2} \, dt = aT\sqrt{1+T^2} +a \arg\sinh{T}, $$ which is all but the same as the Archimedes' spiral case. This is likely only invertible by reversion of series, which will only work up to one of the singularities on the right.

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Another simple one is the tractrix, which has parametric equations $$ x = a(t-\tanh{t}), \qquad y=a\operatorname{sech}{t}, $$ and the arc-length is given by $$ \int_0^T a\tanh{t} \, dt = a\log{\cosh{T}}, $$ which is easy to invert.

Answer 2

I just came across one this morning at Area of the surface from curve.

If $y=\cosh(4x)/4, x\in [-1,1]$, then

$$L=\int ds=\int_{-1}^1 \sqrt{1+(y')^2}\ dx=\int_{-1}^1 \sqrt{1+\sinh^2(4x)}\ dx=\int_{-1}^1 \cosh(4x)\ dx=\frac{\sinh(4)}{2}$$

Here's another example, the parabola $y=1-x^2, x\in [-1,1]$

$$L=\int ds=\int_{-1}^1 \sqrt{1+(y')^2}\ dx=\int_{-1}^1 \sqrt{1+4x^2}\ dx=\sqrt{5}+\frac{\sinh^{-1}(2)}{2}$$

And here's a parametric form in the complex plane,

$$ z=\cos^3(t)+i\sin^3(t), t\in [0,\pi/2]\\ s=\int|\dot z|dt\\ \dot z=3[-\cos^2(t)\sin(t)+i\sin^2(t)\cos(t)]\\ |\dot z|=\sqrt{9[\cos^4(t)\sin^2(t)+\sin^4(t)\cos^2(t)]}=3\cos(t)\sin(t) $$

Then

$$s=3\int_0^{\pi/2} \cos(t)\sin(t)\ dt=\frac{3}{2}$$

Answer 3

After some time I came across a few more examples to add to the list.

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The logarithmic spiral, $$ \alpha(t) = (e^t\cos t, e^t\sin t) $$ which has $\|\alpha'(t)\|=\sqrt{2}\,e^t$ and so has arc length $$ s = \int_0^t \|\alpha'(u)\|\,\mathrm{d}u =\int_0^t \sqrt{2}\,e^u \,\mathrm{d}u = \sqrt{2}\,(e^t-1) $$ and so $t=\ln\left(\frac{s}{\sqrt{2}}+1\right)$ which let's us reparametrize the curve as $$ \tilde\alpha(s)=\left( \left(\frac{s}{\sqrt{2}}+1\right)\cos\left(\ln\left(\frac{s}{\sqrt{2}}+1\right)\right), \left(\frac{s}{\sqrt{2}}+1\right)\sin\left(\ln\left(\frac{s}{\sqrt{2}}+1\right) \right)\right) $$

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The helix, $$ \beta(t) = (a\cos t, a\sin t, bt) $$ which has $\|\beta'(t)\| = \sqrt{a^2+b^2}$ and so has arc length $$ s = \int_0^t \|\beta'(u)\|\,\mathrm{d}u\int_0^t \sqrt{a^2+b^2}\,\mathrm{d}u = t\sqrt{a^2+b^2} $$ and so $t=\frac{s}{\sqrt{a^2+b^2}}$ which let's us reparametrize the curve as $$ \tilde{\beta}(s) = \left( a\cos\left(\frac{s}{\sqrt{a^2+b^2}}\right), a\sin\left(\frac{s}{\sqrt{a^2+b^2}}\right), \frac{bs}{\sqrt{a^2+b^2}}\right) $$

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A helix-like curve inside in a flat torus, $$ \gamma(t) = (a\cos At, a\sin At, b\cos Bt, b\sin Bt) $$ which has $\|\gamma'(t)\|=\sqrt{a^2A^2+b^2B^2}$ and so has arc length $$ s = \int_0^t \|\gamma'(u)\|\,\mathrm{d}u\int_0^t \sqrt{a^2A^2+b^2B^2}\,\mathrm{d}u = t\sqrt{a^2A^2+b^2B^2} $$ and so $t=\frac{s}{C}$ where $C = \sqrt{a^2A^2+b^2B^2}$. This let's us reparametrize the curve as $$ \tilde\gamma(s)=\left( a\cos\frac{As}{C}, a\sin\frac{As}{C},b\cos\frac{Bs}{C},b\sin\frac{Bs}{C}\right) $$