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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?

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    $\begingroup$ 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
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I have found a different result

The parabola's equation is

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

$p'(x)=2kx$

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

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

$k_{n+1}=k_n-\dfrac{f(k_n)}{f'(k_n)}$

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

Edit

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

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

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