# How to straighten a parabola?

Consider the function $$f(x)=a0x2f(x)=a_0x^2$$ for some $$a0∈R+a_0\in \mathbb{R}^+$$. Take $$x0∈R+x_0\in\mathbb{R}^+$$ so that the arc length $$LL$$ between $$(0,0)(0,0)$$ and $$(x0,f(x0))(x_0,f(x_0))$$ is fixed. Given a different arbitrary $$a1a_1$$, how does one find the point $$(x1,y1)(x_1,y_1)$$ so that the arc length is the same?

Schematically, In other words, I’m looking for a function $$g:R3→Rg:\mathbb{R}^3\to\mathbb{R}$$, $$g(a0,a1,x0)g(a_0,a_1,x_0)$$, that takes an initial fixed quadratic coefficient $$a0a_0$$ and point and returns the corresponding point after “straightening” via the new coefficient $$a1a_1$$, keeping the arc length with respect to $$(0,0)(0,0)$$. Note that the $$yy$$ coordinates are simply given by $$y0=f(x0)y_0=f(x_0)$$ and $$y1=a1x21y_1=a_1x_1^2$$. Any ideas?

My approach: Knowing that the arc length is given by
$$L=∫x00√1+(f′(x))2dx=∫x00√1+(2a0x)2dx L=\int_0^{x_0}\sqrt{1+(f'(x))^2}\,dx=\int_0^{x_0}\sqrt{1+(2a_0x)^2}\,dx$$
we can use the conservation of $$LL$$ to write
$$∫x00√1+(2a0x)2dx=∫x10√1+(2a1x)2dx \int_0^{x_0}\sqrt{1+(2a_0x)^2}\,dx=\int_0^{x_1}\sqrt{1+(2a_1x)^2}\,dx$$
which we solve for $$x1x_1$$. This works, but it is not very fast computationally and can only be done numerically (I think), since
$$∫x10√1+(2a1x)2dx=14a1(2a1x1√1+(a1x1)2+arcsin(2a1x1)) \int_0^{x_1}\sqrt{1+(2a_1x)^2}\,dx=\frac{1}{4a_1}\left(2a_1x_1\sqrt{1+(a_1x_1)^2}+\arcsin{(2a_1x_1)}\right)$$
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: Furthermore, could this be determined for any $$ff$$?

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.

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

$$12f(x,y)=∫x0√1+4y2t2x4dt=12∫20√x2+y2t2dt\frac{1}{2}f(x,y) = \int_0^x\sqrt{1+\frac{4y^2t^2}{x^4}}\:dt = \frac{1}{2}\int_0^2 \sqrt{x^2+y^2t^2}\:dt$$

which will always be perpendicular to the gradient at that point

$$∇f=∫20dt(x√x2+y2t2,yt2√x2+y2t2)\nabla f = \int_0^2 dt\left(\frac{x}{\sqrt{x^2+y^2t^2}},\frac{yt^2}{\sqrt{x^2+y^2t^2}}\right)$$

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

$$x′(a)=∫20−axt2√1+a2x2t2dtx(0)=Lx'(a) = \int_0^2 \frac{-axt^2}{\sqrt{1+a^2x^2t^2}}dt \hspace{20 pt} x(0) = L$$

where we substitute $$y(a)=a⋅x2(a)y(a) = a\cdot x^2(a)$$, thus solving for one component automatically gives us the other.

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

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

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

The curves of constant arclength are of the form

$$→r(a)=(sinh−1(aL)a,√1+a2L2−1a)\vec{r}(a) = \left(\frac{\sinh^{-1}(aL)}{a},\frac{\sqrt{1+a^2L^2}-1}{a}\right)$$

Below is a (sideways) plot of the curve of arclength $$L=1L=1$$ (along with the family of curves evaluated at $$a=12,1,2,4,a=\frac{1}{2},1,2,4,$$ and $$1010$$), which has an explicit equation of the form

$$x = \frac{\tanh^{-1}y}{y}\cdot(1-y^2)x = \frac{\tanh^{-1}y}{y}\cdot(1-y^2)$$ These curves and the original family of parabolas in question both have this property, as well as the perfect circles obtained from the family $$f_a(x) = axf_a(x) = ax$$. The reason the original question was hard to tractably solve was because of the non analytically invertible arclength formula