How to straighten a parabola?

Question

Consider the function $f(x)=a_0x^2$ for some $a_0\in \mathbb{R}^+$. Take $x_0\in\mathbb{R}^+$ so that the arc length $L$ between $(0,0)$ and $(x_0,f(x_0))$ is fixed. Given a different arbitrary $a_1$, how does one find the point $(x_1,y_1)$ so that the arc length is the same?

Schematically,

In other words, I'm looking for a function $g:\mathbb{R}^3\to\mathbb{R}$, $g(a_0,a_1,x_0)$, that takes an initial fixed quadratic coefficient $a_0$ and point and returns the corresponding point after "straightening" via the new coefficient $a_1$, keeping the arc length with respect to $(0,0)$. Note that the $y$ coordinates are simply given by $y_0=f(x_0)$ and $y_1=a_1x_1^2$. Any ideas?

My approach: Knowing that the arc length is given by $$ L=\int_0^{x_0}\sqrt{1+(f'(x))^2}\,dx=\int_0^{x_0}\sqrt{1+(2a_0x)^2}\,dx $$ we can use the conservation of $L$ to write $$ \int_0^{x_0}\sqrt{1+(2a_0x)^2}\,dx=\int_0^{x_1}\sqrt{1+(2a_1x)^2}\,dx $$ which we solve for $x_1$. This works, but it is not very fast computationally and can only be done numerically (I think), since $$ \int_0^{x_1}\sqrt{1+(2a_1x)^2}\,dx=\frac{1}{4a_1}\left(2a_1x_1\sqrt{1+(2a_1x_1)^2}+\operatorname{arcsinh}{(2a_1x_1)}\right) $$ Any ideas on how to do this more efficiently? Perhaps using the tangent lines of the parabola?

More generally, for fixed arc lengths, I guess my question really is what are the expressions of the following red curves for fixed arc lengths:

Furthermore, could this be determined for any $f$?

Edit: Interestingly enough, I found this clip from 3Blue1Brown. The origin point isn't fixed as in my case, but I wonder how the animation was made (couldn't find the original video, only a clip, but here's the link)

For any Mathematica enthusiasts out there, a computational implementation of the straightening effect is also being discussed here, with some applications.

Answer 1

Phrased differently, what we want are the level curves of the function

$$\frac{1}{2}f(x,y) = \int_0^x\sqrt{1+\frac{4y^2t^2}{x^4}}\:dt = \frac{1}{2}\int_0^2 \sqrt{x^2+y^2t^2}\:dt$$

which will always be perpendicular to the gradient at that point

$$\nabla f = \int_0^2 dt\left(\frac{x}{\sqrt{x^2+y^2t^2}},\frac{yt^2}{\sqrt{x^2+y^2t^2}}\right)$$

Now is the time to naturally reintroduce $a$ as the parameter for these curves. Therefore what we want is to solve the differential equation

$$x'(a) = \int_0^2 \frac{-axt^2}{\sqrt{1+a^2x^2t^2}}dt \hspace{20 pt} x(0) = L$$

where we substitute $y(a) = a\cdot x^2(a)$, thus solving for one component automatically gives us the other.

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EDIT: Further investigation has led me to some interesting conclusions. It seems like if $y=f_a(x)$ is a family strictly monotonically increasing continuous functions and $$\lim_{a\to0^+}f_a(x) = \lim_{a\to\infty}f_a^{-1}(y) = 0$$

Then the curves of constant arclength will start and end at the points $(0,L)$ and $(L,0)$. Take for example the similar looking family of curves

$$y = \frac{\cosh(ax)-1}{a}\implies L = \frac{\sinh(ax)}{a}$$

The curves of constant arclength are of the form

$$\vec{r}(a) = \left(\frac{\sinh^{-1}(aL)}{a},\frac{\sqrt{1+a^2L^2}-1}{a}\right)$$

Below is a (sideways) plot of the curve of arclength $L=1$ (along with the family of curves evaluated at $a=\frac{1}{2},1,2,4,$ and $10$), which has an explicit equation of the form

$$x = \frac{\tanh^{-1}y}{y}\cdot(1-y^2)$$

These curves and the original family of parabolas in question both have this property, as well as the perfect circles obtained from the family $f_a(x) = ax$. The reason the original question was hard to tractably solve was because of the non analytically invertible arclength formula

Answer 2

$L$ being the known arc length, let $x_1=\frac t{2a}$ and $k=4a_1L$; then you need to solve for $t$ the equation $$k=t\sqrt{t^2+1} +\sinh ^{-1}(t)$$A good approximation is given by $t_0=\sqrt k$.

Now, using a Taylor series around $t=t_0$ and then series reversion gives $$t_1=\sqrt{k}+z-\frac{\sqrt{k} }{2 (k+1)}z^2+\frac{(3 k-1) }{6 (k+1)^2}z^3+\frac{(13-15 k) \sqrt{k} }{24 (k+1)^3}z^4+\cdots$$ where $$z=-\frac{\sqrt{k(k+1)} +\sinh ^{-1}\left(\sqrt{k}\right)-k}{2 \sqrt{k+1}}$$

Let us try for $k=10^n$ $$\left( \begin{array}{ccc} n & \text{estimate} & \text{solution} \\ 0 & 0.4810185 & 0.4819447 \\ 1 & 2.7868504 & 2.7868171 \\ 2 & 9.8244940 & 9.8244940 \\ 3 & 31.549250 & 31.549250 \\ 4 & 99.971006 & 99.971006 \\ 5 & 316.21678 & 316.21678 \\ 6 & 999.99595 & 999.99595 \end{array} \right)$$ This seems to be quite decent.

Answer 3

Complex Integral

We use the inverse Laplace transform. Define $f(x)=\int_0^x\sqrt{t^2+1}dt=\frac12x\sqrt{x^2+1}+\frac12\sinh^{-1}(x)$ and $g(x)=f^{-1}(x)$. $g(x)’s$ Laplace transform is:

$$\operatorname L_s(g(x))=\int_0^\infty e^{-st}g(t)dt=\int_0^\infty te^{-sf(t)}f’(t)dt$$

where $t\to f(t)$ was substituted. Next, one can integrate by parts and substitute $t\to\sinh(\frac t2)$ to get:

$$\operatorname L_s(g(x))=\frac1s\int_0^\infty e^{-sf(t)}dt=\frac1{2s}\int_0^\infty\cosh\left(\frac t2\right)e^{-\frac s4(\sinh(t)+t)}dt=\frac1{4s}\int_0^\infty e^{-\left(\frac s4-\frac12\right)t -\frac s4\sinh(t)}+e^{-\left(\frac s4+\frac12\right)t -\frac s4\sinh(t)}dt$$

which is the associated Anger Weber function. Applying the inverse Laplace transform gives the inverse function. Therefore:

$$f(x)=\frac12x\sqrt{x^2+1}+\frac12\sinh^{-1}(x)\implies f^{-1}(x)=\int_{c-i\infty}^{c+i\infty}\frac{e^{4sx}}{8is}\left(\operatorname A_{s-\frac12}(s)+\operatorname A_{s+\frac12}(s)\right)ds$$

which seems to work for all $x\in\Bbb R$.

Series Solution near $x=0$

If $f(x)=\int_0^x\sqrt{t^2+1}dt$ and $y=f^{-1}(x)$, then one notices: $$3Yy’-y^2=1,Y=\int_0^xy(t)dt$$

Software series reversion shows $x\approx y=\sum\limits_{n=0}^\infty a_nx^{2n+1}$ near $x=0$:

$$3\sum_{k=0}^\infty a_k (2k+1)x^{2k}\sum_{m=0}^\infty\frac{a_mx^{2m+2}}{2m+2}-\sum_{k=0}^\infty a_kx^{2k+1}\sum_{m=0}^\infty a_m x^{2m+1}=1$$ gathering powers of $x$ and simplifying gives:

$$\begin{align}f(x)=\frac12x\sqrt{x^2+1}+\frac12\sinh^{-1}(x)\\\implies f^{-1}(x)=\sum_{n=0}^\infty a_n(-1)^nx^{2n+1}\\a_n=\sum_{m=0}^{n-1}a_ma_{n-m-1}\left(\frac3{2m+2}-\frac4{2n+1}\right)\\=\left\{1,\frac16,\frac{13}{120},\frac{493}{5040},\frac{37369}{362880},\frac{4732249}{39916800},\dots\right\}\end{align}$$

The denominators are $(2n+1)!$, but the numerators do not have an explicit form yet. The series works for $x\in\Bbb C$ too. Here is a plot for $n=11$:

General Series Solution

Unfortunately, the above series has a small convergence radius, so one can use @Paul Enta’s method here to expand $y=f^{-1}(x)$ near $x=x_0$. One notices the polynomial $P_n(x)$:

$$\frac{d^ny}{dx^n}=\frac{P_n(y)}{(y^2+1)^{\frac{3n}2-1}}\iff P_{n+1}=(y^2+1)^\frac{3n+1}2\frac{d^{n+1}y}{dx^{n+1}}= (y^2+1)^\frac{3n+1}2\frac{dy}{dx}\frac d{dy}\left(\frac{d^ny}{dx^n}\right)$$

This finally gives:

$$\begin{align}f(x)=\frac12x\sqrt{x^2+1}+\frac12\sinh^{-1}(x)\\\implies y=f^{-1}(x)=y(x_0)+\sum_{n=1}^\infty\frac{P_n(y(x_0))(x-x_0)^n}{(y^2(x_0)+1)^{\frac{3n}2-1}n!}\\P_{n+1}(x)=(x^2+1)P’_n(x)+(2-3n)xP_n(x)\\=\{1,-x,3x^2-1,13x-15x^3,105x^4-162x^2+13,-493x+2202x^3-945x^5,10395x^6-33351x^4+14001x^2-493,\dots\}\end{align}$$

At $x=0$, this reduces to $P_{n+1}(0)=P’_n(0)$

Mathematica Function

Because of the domain of Mathematica’s Inverse Beta Regularized $\text I^{-1}_s(a,b)$, it solves:

$$\text{ArcLength}(x^2)=\frac12\sqrt{4x^2+1}+\frac14\sinh^{-1}(2x)=z\in\Bbb I\text{ pure imaginary}\implies x\mathop=^{|\text{Im}(z)|\le \frac{\pi}8}\mp\frac i2\sqrt{\text I^{-1}_\frac{\pm8i z}\pi\left(\frac12,\frac32\right)}$$

Plot of $k\in\Bbb R$ vs $\mp\frac i2\sqrt{\text I^{-1}_\frac{\pm8i z}\pi\left(\frac12,\frac32\right)},z=ik$ where opposite signs are taken:

as shown here for $k\in\Bbb R$. The power series in Quantile mechanics, however, extends the domain to complex, and therefore real values. Another expansion there can be found in terms of the Student quantile as it is a special case of $\text I^{-1}_x\left(\frac12,\frac32\right)$

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@Francisco Alvarado found an integral solution here. Additionally, previous revisions of this answer have older fixed point iteration solutions.

Answer 4

Here's about the most efficient thing I can see is this:

Take your antiderivative (replacing the sin with a sinh) and define $a_1 x \equiv y$ so that

$$f(y) = a_1 * (\textrm{arc length}),$$

$$f(y) \equiv \frac{1}{2}(y \sqrt{1+y^2}+ \sinh^{-1} y).$$

$f(y)$ is monotonic and has some nice approximations when $y \ll 1, y \gg 1$. In those cases it might be possible to obtain it analytically. In general, invert it numerically. Then,

$$x = a_1^{-1} f^{-1}(a_1 * (\textrm{arc length}).$$

The utility of doing it this way is that you don't have to keep inverting a new function for each new $a_1$, you only have to do it once to be able to flatten any parabola you want.

Answer 5

TLDR:

You want to solve for $v$ in this equation: $$v\sqrt{v^2+1} + \sinh^{-1} v = 2a_1x_0\sqrt{u^2+1} + \frac{\sinh^{-1} u}{2a_0}$$

And then $x_1=\frac{v}{2a_1}$ is your solution.

I’ve been banging away at this for hours, not because I think I can help you—much of this is new to me—but because I found it interesting. I haven’t yet worked out (an approximation to a) solution for $v$, though. And it looks like Claude’s answer has beaten me to it, anyway.

But here is my working, just because I can’t bear to hit Discard on this whole thing.

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I don’t think I’ve ever done the length of a parabolic curve before. So while this would probably go more smoothly if done from the integral definition of arc length, I’m going to take the easy(?) way out: Wikipedia has a parabolic arc length formula. Let’s give it a go!

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So we have a parabola $f\left(x\right)=a_0 x^2$. Wikipedia tells us that we can find the arc length, from the vertex at $\left(0,0\right)$ to any point $\left(x,f\left(x\right)\right)$ on the parabola, using these values:

• The focal length $l$ of the parabola; in this case, $l=\frac{1}{4a_0}$
• The perpendicular distance $p$ between the point and the axis of symmetry; in this case it’s simply $p=x$

Then, given $h=\frac{p}{2}$ and $q=\sqrt{l^2+h^2}$, the arc length is:

$$s=\frac{hq}{l}+l\ln\frac{h+q}{l}$$

Let’s simplify. Given that $h=\frac{x}{2}$…

$$\begin{align} q &= \sqrt{\frac{1}{16a_0^2}+\frac{x^2}{4}} \\ &= \sqrt{\frac{4a_0^2x^2+1}{16a_0^2}} \\ &= l\sqrt{4a_0^2x^2+1} \end{align}$$

Thus:

$$s=\frac{x}{2}\sqrt{4a_0^2x^2+1}+\frac{1}{4a_0}\ln\left(2a_0x+\sqrt{4a_0^2x^2+1}\right)$$

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Now, we have another parabola $g(x)=a_1 x^2$ such that the arc lengths of $f\left(x_0\right)$ and $g\left(x_1\right)$ are equal, i.e.:

$$\begin{align} \frac{x_0}{2}\sqrt{4a_0^2x_0^2+1}+\frac{1}{4a_0}\ln\left(2a_0x_0+\sqrt{4a_0^2x_0^2+1}\right) &= \frac{x_1}{2}\sqrt{4a_1^2x_1^2+1}+\frac{1}{4a_1}\ln\left(2a_1x_1+\sqrt{4a_1^2x_1^2+1}\right) \\ \therefore x_0\sqrt{4a_0^2x_0^2+1}+\frac{1}{2a_0}\ln\left(2a_0x_0+\sqrt{4a_0^2x_0^2+1}\right) &= x_1\sqrt{4a_1^2x_1^2+1}+\frac{1}{2a_1}\ln\left(2a_1x_1+\sqrt{4a_1^2x_1^2+1}\right) \end{align}$$

And we want to solve for $x_1$ in terms of $a_0$, $a_1$, and $x_0$. Nothing simpler! </sarc>

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Let’s define $u=2a_0x_0$ (thus $u^2=4a_0^2x_0^2$) and $v=2a_1x_1$ (thus $v^2=4a_1^2x_1^2$). That shortens things to:

$$x_0\sqrt{u^2+1} + \frac{1}{2a_0}\ln\left(u+\sqrt{u^2+1}\right) = x_1\sqrt{v^2+1} + \frac{1}{2a_1}\ln\left(v+\sqrt{v^2+1}\right)$$

Now I see where $\sinh$ comes into the other answers! It’s because $\sinh^{-1} x = \ln\left(x+\sqrt{x^2+1}\right)$, so we get:

$$\begin{align} x_0\sqrt{u^2+1} + \frac{\sinh^{-1} u}{2a_0} &= x_1\sqrt{v^2+1} + \frac{\sinh^{-1} v}{2a_1} \\ \therefore 2a_1x_0\sqrt{u^2+1} + \frac{\sinh^{-1} u}{2a_0} &= 2a_1x_1\sqrt{v^2+1} + \sinh^{-1} v \\ &= v\sqrt{v^2+1} + \sinh^{-1} v \end{align}$$

That right-hand side does not look easy to invert. I admit, I copped out and asked Wolfram Alpha to do it. And of course it tells me, “no result found in terms of standard mathematical functions”. [sigh…]

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Based on Tyma Gaidash’s answer, I went looking into the Lagrange inversion theorem. My engineering-oriented education never covered this, but I think I’ve grasped the basics of it. As I understand it, to solve $y=f(x)$ for $x$, we choose some $z$, such that $f(z)$ is defined and $f'(z)\ne 0$.

Let’s shorten the entire left-hand side of the equation to $w$, and define $w=g\left(v\right)=v\sqrt{v^2+1} + \sinh^{-1} v$. First, let’s find the derivative… by cheating and using Wolfram Alpha: $g'(v)=2\sqrt{v^2+1}$.

We need a value $z$ where $g'(z)\ne 0$. Conveniently, this derivative is nowhere zero on the reals, so that’s trivial. I think (but am not sure) that $z$ should approximate $v$, so let’s randomly assume that $x_1\approx 1$ and so $v\approx 2a_1 = z$.

Now, by the inversion theorem, the inverse function $v=g^{-1}\left(w\right)$ is:

$$\begin{align} v &= z+\sum_{n=1}^\infty \left[\frac{\left(w-g\left(z\right)\right)^n}{n!} \lim_{t\to z} \frac{d^{n-1}}{dt^{n-1}}\left(\frac{t-z}{g(t)-g(z)}\right)^n\right] \\ &= z+\left(w-g\left(z\right)\right)\lim_{t\to z}\frac{t-z}{g(t)-g(z)} + \frac{\left(w - g\left(z\right)\right)^2}{2} \lim_{t\to z}\frac{d}{dt} \left(\frac{t-z}{g(t)-g(z)}\right)^2 + \cdots \end{align}$$

Doesn’t that first limit look just like the reciprocal of the derivative? So the second term of the series becomes $\frac{w-g\left(z\right)}{2\sqrt{z^2+1}}$.

And I’ve been working on this for way too long now, so that’s where I’ll stop for the night.

Answer 6

For a parabola with parametrization

$$ x= at,y= a t^2/2 ;\;x^2= 2 a y \;; \text{slope} \;t=\tan \phi; \tag1$$

Differentiate $x^2$, primed wrt arc length

$$ 2 x \cos \phi = 2 a \sin \phi ;\; x= a \tan \phi ;\;x'= \cos \phi= a \sec^2 \phi \;\phi' \tag 2 $$

from which comes the curvature

$$ a \phi'=a \kappa=\cos^3 \phi \tag 3$$

An easier/direct way out is by direct numerical integration of ode in(3). A fraction k of max arc length can be set as a parameter for required integrands ( k= 2/3 in this particular case), making the subset parabolas flatter or deeper by adjusting $a$.

Total length on one side is a given constant $L$

$$ L = \int _0^{\phi_m} \frac{ d \phi}{\kappa} ; \text{ now plug in curvature from(3) and integrate }$$

$$ \frac{L}{a}=\int _0^{\phi_m} \sec^3 \phi\; d\phi = \frac12\bigg[\log\bigg(\tan\bigg(\frac{\pi}{4} + \frac{\phi_m}{2} \bigg)\bigg) +\sec \phi_m \tan \phi_m \bigg]\tag 4 $$

Plug in from (2) $x_{m}=a \tan \phi_m $ $$ \frac{2L}{a}= \log\left(\tan\bigg(\frac{\pi}{4} + \frac{\tan^{-1}(x_m/a)}{2}\right)\bigg)+ (x_m/a) \sqrt{(x_m/a)^2-1} \tag 5 $$

which is a neat implicit function $ f(a,x_m,L) $ , plotted assuming an arm of parabola has given arc length $=1.8,$ on Mathematica, enabling flatter or deeper parabola plots.

It has become clear that there are two criteria for equal parabolic arc lengths connecting $x_{max} $ to $ \text { a = 2* focal-length }$.

EDIT 1/2:

To come out of the apparent dilemma I have carefully calculated/plotted special cases for $( x_{max},a)$ combinations:

$$(0.4,0.0458),(0.8,0.1948), (1.2,0.4704),(1.6,0.8874)(2.0, 1.42264),(2.4,2.0101),(2.8,2.5825)$$

All arcs are all of same length but for $( x_{max} =2.0,2.4,2.8) $ the choice of $a$ seems to have got switched over to the second critirion. Relation $(x_{max},a) $ is not unique by the first plot...it is now examined further.

Answer 7

Adding another short ODE method that finds a direct relation between maximum slope at parabola tip and its focal length for invariant arc length while bending.

$$ x = 2 f \tan \phi \;\text {is differentiated w.r.t. arc, } 2f \sec^2\phi \frac{d\phi}{ds}=\cos \phi$$

Integrate back but w.r.t. slope $ \phi, s= 2 f \int \sec^3\phi\; d\phi$

$$ s= 2 f \left(\frac12 \log \frac{\cos (\phi/2)+ \sin (\phi/2) }{\cos (\phi/2)- \sin (\phi/2) } +\sec \phi +\tan \phi \right)= \text{constant}$$

Computed focal length $ f=F(\phi_{max})$ as the relation between bending variables on Mathematicafor maximum slopes $(0.2,0.5,0.8,1.1)$ radians with fixed arc length of parabola 3 units assumed and plotted as shown; $ (x,y)= (2ft,ft^2). $