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Suppose we have a parabola of equation $y = x^2$ in a given Cartesian coordinate system. An obvious parameterization of it is the system $x = t$, $y = t^2$, but there are infinite other possibilities, the system $x = t^2$, $y = t^4$ is just another possible one, for example.

Although it is easy to find parameterizations, finding one of constant unit-speed is not (at least I can not find one after many tries). None of the examples above satisfy it for example (as we can verify by computing the magnitudes of the tangent vectors).

I would like to know an example (are there more than one?) of parameterization by arc length of this parabola, and also how was it was found.

Suggestion of books on the subject are also welcome.

Vladimir M. Valgan
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    For almost any curve that you write down, this will be a superhuman endeavor. – Lubin Nov 17 '15 at 19:17
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    As in the answers, the parabola is one of the very few curves where you can solve the arc length integral in closed form. The extra step of finding the inverse function required, in closed form, is just too much to hope for. – Will Jagy Nov 17 '15 at 19:36

2 Answers2

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Let us compute $\sigma$ a curvilinear abscissa of the parabola $\mathcal{P}:y=x^2$, as you said: $$\left\{\begin{array}{ll}x=t\\y=t^2\end{array}\right.$$ is a parametrization of $\mathcal{P}$. Thus, one has: $$\sigma(t)=\int_0^t\sqrt{1+4u^2}\,\mathrm{d}u=\frac{1}{2}\left(\sqrt{t(1+4t^2)}+\frac{\textrm{arcsinh}(2t)}{2}\right).$$ And one cannot solve explicitly for $t$ the equation: $$s=\sigma(t).$$ Therefore, one is not able to find an explicit arc length parametrization of $\mathcal{P}$.

C. Falcon
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In theory, you can do it this way:

Suppose $r:[0,1]\rightarrow \mathbb R^2$ is given by $t\mapsto (t,t^2)$. Then, the arc length of the parabola from $0$ to $t_0$ is

$s=l(t_0)=\int _0 ^{t_{0}}\sqrt{1+4t^2}dt=\dfrac{\operatorname{arsinh}\left(2t_0\right)}{4}+\dfrac{t_0\sqrt{4{t_0}^{2}+1}}{2}$.

Solving for $t_0$ in terms of $s$ gives the desired reparameterization, $r\circ l^{-1}$:

$r(t_0)=r(l^{-1}(s))=(r\circ l^{-1})(s)$.

Unfortunately, there is no closed form solution, (that I know of!).

Matematleta
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