take two points $(-2, cosh(2))$ and $(2, cosh(2))$. Now we take a parabola $p(x)=kx^2+d$ and a catenary, both intersecting at the two given points.

What is $k$, so that the length of both curves is equal?

Length of catenary = Length of parabola ->

$2sinh(2) = \frac{1}{2} \cdot (4k\cdot \sqrt{1+(4k)^2} + arsinh(4k))$.

It should be correct so far. How can I solve the equation for k? Do you have any hints for me?

  • 1
    $\begingroup$ You can't, not least without resorting to numerical methods (which are awesome). $\endgroup$ – Parcly Taxel Oct 24 '17 at 12:29
  • $\begingroup$ @ParclyTaxel Thank you! Okay, so which numerical methods could help me here? $\endgroup$ – Vazrael Oct 24 '17 at 12:36
  • $\begingroup$ There are a lot. Newton's method seems the simplest to implement. $\endgroup$ – Parcly Taxel Oct 24 '17 at 12:36

I have found a different result

The parabola's equation is

$p(x)=k x^2-4 k+\cosh 2$


$$L=\int_{-2}^2 \sqrt{1+4 k^2 x^2} \, dx=2 \sqrt{16 k^2+1}+\frac{\text{arcsinh}(4 k)}{2 k}$$

and from catenary $$L=\int_{-2}^2 \sqrt{\sinh ^2(x)+1} \, dx=2\sinh 2$$

so the equation to solve is $$f(k)=2 \sqrt{16 k^2+1}+\frac{\text{arcsinh}(4 k)}{2 k}-2\sinh 2=0$$ we can easily see that $f(0.1)\approx -3$

while $f(1)\approx 2$

Therefore a zero is in the interval $(0.1,1)$.

Let's guess $k_0=0.5$

and define


we get the following table

$ \begin{array}{r|l} k_0 & 0.5 \\ k_1 & 0.720893 \\ k_2 & 0.708215 \\ k_3 & 0.708188 \\ k_4 & 0.708188 \\ \end{array} $

Therefore we have $k\approx 0.708188$

as the function is even we have also the solution $k=-0.708188$

Hope this helps


I verified: $L=2\sinh 2\approx 7.25372$ for catenary and

$L \approx 7.25372$ for parabola $p(x)=0.708188 x^2+0.929444$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.