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Imagine we have a line segment of length $1$ on the $x$-axis.

Keeping the point at $(0,0)$ fixed at $(0,0)$, we bend the line segment into a parabola such that every point on the bent line now satisfies $y = x^2$.

This should yield a parabola starting from $(0,0)$ with arc-length $1$.

How can I express this transformation such that a point $(x, 0)$ on the original line is transformed to some point $(x', y')$ on the parabola? Is it possible to find an equation that, given x, returns $(x', y')$? Is it possible to obtain this for bending into any arbitrary, smooth and continuous shape?

Menezio
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BeeperTeeper
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    Hint: The arc-length parameterization of the parabola is computed using this integral: https://math.stackexchange.com/questions/1056058/computing-int-sqrt14x2-dx?noredirect=1&lq=1 – Moishe Kohan Apr 12 '20 at 15:17

2 Answers2

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Yes, there is a function that, given an $x>0$ returns the $x'>0$ such that the distance form the origin to $(x', x'^2)$ along the parabola $y=x^2$ is equal to $x$.

Finding a direct expression for this function is, however, a little bit tricky. It is much easier to go the other way: take a point on the parabola and see where it lands on the $x$-axis as you straighten your parabola to a line.

The distance along the parabola from the origin to $(x', x'^2)$ is given by the integral $$ \int_0^{x'}\sqrt{1+4t^2}dt=\frac{x'}2\sqrt{1+4x'^2}+\frac14\ln\left(2x'+\sqrt{1+4x'^2}\right) $$ (I used electronic help to calculate this integral, and substituted in the logarithm for the inverse hyperbolic sine.) And there it is. That is your $x$ on the $x$-axis, given a point $(x',x'^2)$ on the parabola.

Arthur
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  • No need for electronic help, one can get MSE help: https://math.stackexchange.com/questions/1056058/computing-int-sqrt14x2-dx?noredirect=1&lq=1 – Moishe Kohan Apr 12 '20 at 16:09
  • @MoisheKohan Sure. I didn't say it was necessary. Just that it's quick and easy, and I'm lazy and not an expert at spotting useful trigonometric substitutions. – Arthur Apr 12 '20 at 16:49
  • Thank you for the help! Unfortunately, I am specifically looking for the function you described as tricky. – BeeperTeeper Apr 12 '20 at 18:16
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Consider the parabola $\mathcal C$ as the image of the curve $$\gamma:\mathbb R \rightarrow \mathbb R^2, \qquad \gamma(t)=(t,t^2)$$ We want to define a map $f:\mathbb R\rightarrow \mathcal C$ such that the real line bends over the parabola. For, take a point $x\in \mathbb R^+$; the image $f(x)$ is the point $\gamma(s)=(s,s^2)$ in $\mathcal C$ ($s>0$) such that \begin{equation} l(\gamma, [0,s])=x \end{equation} where $l(\gamma, [0,s])$ denotes the length of the curve $\gamma$ from $0$ to $s$. Now: \begin{gather} l(\gamma, [0,s])=x\\ \int_{0}^{s}|\gamma'(t)| \ dt = x\\ \int_{0}^{s}\sqrt{1+4t^2}\ dt = x\\ g(s)=\frac{1}{4}\left[2s\sqrt{4s^2+1}+\ln\left(2s+\sqrt{4s^2+1}\right)\right]=x \end{gather} The function $g(s)$ is strictly monotone in $(0,\infty)$ (its derivative is $\sqrt{1+4s^2}$), so it is invertible in $(0,\infty)$. Hence is well defined the unique point $s\in \mathbb R^{+}$ sucht that $g(s)=x$. Denote this point with $h(x)$ the function we are looking for is: \begin{equation} f:\mathbb R^+ \rightarrow \mathcal C,\quad f(x) = (h(x),h(x)^2) \end{equation} The argument is the same for negative points.

I don't find a way to invert explicity the function $g(s)$. I don't know if there is a explicit formula.

Menezio
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