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?


enter image description here

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 $$ \int_0^{x_1}\sqrt{1+(2a_1x)^2}\,dx=\frac{1}{4a_1}\left(2a_1x_1\sqrt{1+(a_1x_1)^2}+\arcsin{(2a_1x_1)}\right) $$ 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:

enter image description here

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)

enter image description here

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

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    $\begingroup$ It seems to me that the red curves are orthogonal to the blue curves; that means that you could find an equation for the slope of the red curve at any given point via the slope of the corresponding blue curve, which would give a (hopefully tractable) differential equation to solve for the red curve. $\endgroup$ Jul 28, 2021 at 17:57
  • 2
    $\begingroup$ The innermost red curve on the blue and red graphic doesn't look right. It seems to me its topmost point should be somewhat lower down along the leftmost parabola. $\endgroup$ Jul 28, 2021 at 20:29
  • 2
    $\begingroup$ @samwolfe, ah good, so my eye (perhaps) did not deceive me. $\endgroup$ Jul 28, 2021 at 20:31
  • 5
    $\begingroup$ Does the animation actually preserve arc lengths? $\endgroup$
    – user253751
    Jul 29, 2021 at 9:45
  • 2
    $\begingroup$ "Determining the length of an irregular arc segment is also called rectification of a curve" (Wikipedia). So the word "straighten" you use is related to a more formal word rectify that is used when calculating (in any way) arc lengths. $\endgroup$ Jul 30, 2021 at 15:08

7 Answers 7


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.

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

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

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

  • $\begingroup$ The brown, blue parabolas in the plot seem to have lengths more than $ L= 1$. $\endgroup$
    – Narasimham
    Jul 31, 2021 at 21:07
  • $\begingroup$ @Narasimham they are not parabolas, nor is the scale on these plots 1:1. I tried my best to play around with the plot options but the progeam likes to automatically break the visual consistency $\endgroup$ Jul 31, 2021 at 22:13
  • 1
    $\begingroup$ @samwolfe thank you, I appreciate it. I just tried to make use of anything where $\sqrt{1+f^2}$ simplified nicely. This only leaves a few analytic options, such as $\sinh(ax)$ and $\tan(ax)$. From there, choose the antiderivative that contains $(0,0)$ for all $a$. $\endgroup$ Aug 1, 2021 at 20:56
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    $\begingroup$ @samwolfe I think that an interesting problem would be going the reverse way - if we had a curve that connects $(0,L)$ to $(L,0)$, could we find a family of curves for which they are a constant arc length away from the origin for? For example, how could one get $ax$ from the equation $x^2+y^2=L^2$ ? or any other curve for that matter $\endgroup$ Aug 1, 2021 at 21:02
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    $\begingroup$ @samwolfe no worries! I used the substitution $t \mapsto \frac{2t}{x}$ which changes the bounds from $\frac{2(0)}{x} = 0$ to $\frac{2(x)}{x} = 2$ $\endgroup$ Jun 30, 2023 at 20:42

$L$ 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.

  • $\begingroup$ That seems really good, numerically it would suffice to have something like that. Thanks $\endgroup$
    – sam wolfe
    Aug 1, 2021 at 17:08

$\large{\text{Method 1:}}$

Here is a recursive answer for $g(a_0,a_1,x_1)$. Please see this graph for result verification. You can plug in some value for $x_1$ at the RHS of $x_1=g(x_1)=g(a_0,a_1,x_0)$:

$$\mathrm{\text{Arclength from 0 to }x_1\,\big(a_0x^2\big)=Arclength\ from \ 0\ to\ x_1\ \big(a_1x^2\big),x_0+c=x_1\implies\frac{2a_0x_0\sqrt{4a_0^2x_0^2+1}+sinh^{-1}(2a_0x_0)}{4a_0}=\frac{2a_1x_1\sqrt{4a_1^2x_1^2+1}+sinh^{-1}(2a_1x_1)}{4a_1}\implies A_0(x_1)=x_1=A_1(x_1)=\frac1{2a_1}sinh \left(\sinh^{-1}(2a_0x_0)-2a_1x_0\sqrt{4a_0^2x_0^2+1}-2a_1x_1\sqrt{4a_1^2x_1^2+1} \right)}$$

Then we define and create the following recursive relation converging to $x=x_1$:

$$\mathrm{A_{n+1}(x_1)= \frac1{2a_1}sinh \left(\sinh^{-1}(2a_0x_0)-2a_1x_0\sqrt{4a_0^2x_0^2+1}-2a_1A_n(x_1)\sqrt{4a_1^2A_n^2(x_1)+1} \right)\implies x_1=g(a_0,a_1,x_0)=\lim_{n\to\infty}A_n(x_1)=A_\infty(x_1)= \frac1{2a_1}sinh \left(\sinh^{-1}(2a_0x_0)-2a_1x_0\sqrt{4a_0^2x_0^2+1}-2a_1(…)\sqrt{4a_1^2(…)+1} \right)\implies A_2(x_1)= \frac1{2a_1}sinh \left(\sinh^{-1}(2a_0x_0)-2a_1x_0\sqrt{4a_0^2x_0^2+1}-2a_1{\frac1{2a_1}sinh \left(\sinh^{-1}(2a_0x_0)-2a_1x_0\sqrt{4a_0^2x_0^2+1}-2a_1x_1\sqrt{4a_1^2x_1^2+1} \right)}\sqrt{4a_1^2{\frac1{2a_1}sinh \left(\sinh^{-1}(2a_0x_0)-2a_1x_0\sqrt{4a_0^2x_0^2+1}-2a_1x_1\sqrt{4a_1^2x_1^2+1} \right)}^2+1} \right)}$$ $\large{\text{Method 2:}}$

There is also another recursive method as seen in this other graph. This forms a horizontal line at y=$x_1$. There also may be another $\pm$ branch where the sign is chosen as needed, all + or all -. Notice the main square root argument is also a difference of squares:

$$\mathrm{\text{Arclength from 0 to }x_1\,\big(a_0x^2\big)=Arclength\ from \ 0\ to\ x_1\ \big(a_1x^2\big),x_0+c=x_1 \frac{2a_0x_0\sqrt{4a_0^2x_0^2+1}+sinh^{-1}(2a_0x_0)}{4a_0}=\frac{2a_1x_1\sqrt{4a_1^2x_1^2+1}+sinh^{-1}(2a_1x_1)}{4a_1}\implies B_0(x_1) =x_1= B_1(x_1)=\pm\frac1{2|a_1|}\sqrt{\frac1{4a^2_1x_1^2}\left(2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}\left(2a_0x_0\right)-sinh^{-1}\left(2a_1x_1\right)\right)^2-1}}$$ Recursive solution: $$\mathrm{ x_1=g(a_0,a_1,x_0)=\lim_{n\to\infty}B_n(x_1)=B_\infty(x_1)= \pm\frac1{2|a_1|}\sqrt{\frac1{4a^2_1(…)^2}\left(2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}\left(2a_0x_0\right)-sinh^{-1}\left(2a_1(…)\right)\right)^2-1},B_{n+1}(x_1)= \pm\frac1{2|a_1|}\sqrt{\frac1{4a^2_1B^2_n(x_1)}\left(2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}\left(2a_0x_0\right)-sinh^{-1}\left(2a_1B_n(x_1)\right)\right)^2-1}\implies B_2(x_1)= \pm\frac1{2|a_1|}\sqrt{\frac1{4a^2_1{\left(\pm\frac1{2|a_1|}\sqrt{\frac1{4a^2_1x_1^2}\left(2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}\left(2a_0x_0\right)-sinh^{-1}\left(2a_1x_1\right)\right)^2-1}\right)}^2}\left(2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}\left(2a_0x_0\right)-sinh^{-1}\left(2a_1{\pm\frac1{2|a_1|}\sqrt{\frac1{4a^2_1x_1^2}\left(2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}\left(2a_0x_0\right)-sinh^{-1}\left(2a_1x_1\right)\right)^2-1}}\right)\right)^2-1}}$$

$\large{\text{Method 3:}}$

Here is graphical proof of the solution. Now that we have eliminated the rest of the ways of solving for $x_1$, the last and simpler method is as follows.:

$$\mathrm{\text{Arclength from 0 to }x_1\,\big(a_0x^2\big)=Arclength\ from \ 0\ to\ x_1\ \big(a_1x^2\big),x_0+c=x_1 \frac{2a_0x_0\sqrt{4a_0^2x_0^2+1}+sinh^{-1}(2a_0x_0)}{4a_0}=\frac{2a_1x_1\sqrt{4a_1^2x_1^2+1}+sinh^{-1}(2a_1x_1)}{4a_1}\implies C_0(x_1)=x_1=C_1(x_1)=\frac{2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}(2a_0x_0)-sinh^{-1}(2a_1x_1)} {2a_1\sqrt{4a_1^2x_1^2+1}}}$$

Recursive solution for third method which converges to $x=x_1$ $$\mathrm{x_1=g(a_0,a_1,x_0)=\lim_{n\to\infty}C_n(x_1)=C_\infty(x_1), C_{n+1}(x_1)= \frac{2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}(2a_0x_0)-sinh^{-1}(2a_1C_n(x_1))} {2a_1\sqrt{4a_1^2C^2_n(x_1)+1}} = \frac{2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}(2a_0x_0)-sinh^{-1}(2a_1(…))} {2a_1\sqrt{4a_1^2(…)^2+1}}\implies C_2(x_1)= \frac{2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}\left(2a_0x_0\right)-sinh^{-1}\left(2a_1{\frac{2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}(2a_0x_0)-sinh^{-1}(2a_1x_1)} {2a_1\sqrt{4a_1^2x_1^2+1}}}\right)} {2a_1\sqrt{4a_1^2\left(\frac{2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}(2a_0x_0)-sinh^{-1}(2a_1x_1)} {2a_1\sqrt{4a_1^2x_1^2+1}}\right)^2+1}}}$$

As you can see, $\mathrm{g(a_0,a_1,x_0)=x_1}$ does not have an easy form. This is like the W-Lambert/ Product Logarithm function where $\mathrm{ \ if \ xe^x=y,\ then \ x=W(y)=ln(y)-ln(ln(y)-ln(x))=…\,}$. We need a recurvise definition for W(x) seen in this graph. Therefore, one may need to derive a new function to solve recursively for $g(a_0,a_1,x_0)$ and these definitions definitely work.

Note that the MathJax may render differently based on your computer. See @Tim Pederick’s solution for an Inverse Lagrange theorem solution. I may work more on this. Please correct me and give me feedback!

$\large{\text{Method 4:}}$

The problem reduces to finding the inverse of $x-\sinh(x)$. You can get better results for the inverse of the arc length of $ix^2$ from


but the inverse of the arc length of $x^2$ can only solve

ArcLength$(x^2)=ia,a\in\Bbb R$

because of the domain of Mathematica’s Inverse Beta Regularized $\text I^{-1}_s(a,b)$. Choose the appropriate signs:

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

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:

enter image description here

Test the function here by plugging in a real $k$

  • $\begingroup$ There is an exact solution for real $x$ using this result $\endgroup$ Sep 27, 2023 at 18:18

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.

  • 1
    $\begingroup$ The notation $a\ast b$ often means the convolution of $a$ and $b$. I think it's clear enough from context that is not what you mean, but since the question does seem to naturally involve integrals, it may be clearer to use \cdot instead of * or \ast $\endgroup$
    – Alex Ortiz
    Jul 31, 2021 at 19:29


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.

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!

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:


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

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



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>

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…]

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.

  • 1
    $\begingroup$ @TymaGaidash: Those are meant to be initial values, not restrictions—like choosing initial values in Newton's method. And I don’t know if that’s even correct! (Claude’s answer doesn’t appear to have anything like that, but it seems to be the same approach.) But yes, the aim is ultimately to have a solution (or at least an approximate one, from the first few terms of the series) for $v$, at which point the simplifications $v$, $w$ etc. can perhaps be dropped. (Or maybe it’ll end up being clearer to leave them in?) $\endgroup$ Jul 30, 2021 at 8:53

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 $a$.

enter image description here

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.

enter image description here

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.

enter image description here


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). $

enter image description here


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