Consider the function f(x)=a0x2 for some a0∈R+. Take x0∈R+ so that the arc length L between (0,0) and (x0,f(x0)) is fixed. Given a different arbitrary a1, how does one find the point (x1,y1) so that the arc length is the same?

Schematically,

In other words, I’m looking for a function g:R3→R, g(a0,a1,x0), that takes an initial fixed quadratic coefficient a0 and point and returns the corresponding point after “straightening” via the new coefficient a1, keeping the arc length with respect to (0,0). Note that the y coordinates are simply given by y0=f(x0) and y1=a1x21. Any ideas?

My approach:Knowing that the arc length is given by

L=∫x00√1+(f′(x))2dx=∫x00√1+(2a0x)2dx

we can use the conservation of L to write

∫x00√1+(2a0x)2dx=∫x10√1+(2a1x)2dx

which we solve for x1. 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))

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

Mathematicaenthusiasts out there, a computational implementation of the straightening effect is also being discussed here, with some applications.

**Answer**

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

12f(x,y)=∫x0√1+4y2t2x4dt=12∫20√x2+y2t2dt

which will always be perpendicular to the gradient at that point

∇f=∫20dt(x√x2+y2t2,yt2√x2+y2t2)

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)=∫20−axt2√1+a2x2t2dtx(0)=L

where we substitute y(a)=a⋅x2(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) is a family strictly monotonically increasing continuous functions and lima→0+fa(x)=lima→∞f−1a(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=cosh(ax)−1a⟹L=sinh(ax)a

The curves of constant arclength are of the form

→r(a)=(sinh−1(aL)a,√1+a2L2−1a)

Below is a (sideways) plot of the curve of arclength L=1 (along with the family of curves evaluated at a=12,1,2,4, 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

**Attribution***Source : Link , Question Author : sam wolfe , Answer Author : Ninad Munshi*