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A chain hangs in the shape of a catenary with equation $y=\cosh(x)$ for $x\in[-a,a]$. If the length of the chain is $20$, how far apart are the endpoints of the chain?

I am familiar with the arc length integral equations and such - I believe the equation would be $\int_{-a}^a \sqrt{1 + (\cosh'(x))^2} dx = 20$, but I'm not quite sure how we could use that to derive the value of $a$. Could I have some help?

cindy
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Your equation is correct. Note that $$\sqrt{1 + (\cosh'(x))^2}= \sqrt{1 + \sinh^2(x)} =\sqrt{\cosh^2(x)}=\cosh(x)=\sinh'(x).$$ Can you take it from here and find the value of $a$?

Robert Z
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  • Thank you! I got $2\int_{0}^{a}\cosh(x),dx = 2\left[\sinh(x)\right]_{0}^{a} = 2\left(\sinh(a) - \sinh(0)\right) = 2\sinh(a)$. So $a = \sinh^{-1}(10)$? But I did try to submit $2 * \sinh^{-1}(10)$ as my answer and it stated it was incorrect, I'm not sure if I did something wrong there. – cindy Jul 06 '23 at 18:03
  • $a= \sinh^{-1}(10)$ is correct. An alternative form is $\ln(10 + \sqrt{101})$. See https://en.wikipedia.org/wiki/Inverse_hyperbolic_functions#Definitions_in_terms_of_logarithms Try to submit $2\ln(10 + \sqrt{101})$ – Robert Z Jul 06 '23 at 18:04
  • Oh yeah that worked, thank you!! – cindy Jul 06 '23 at 18:09
  • @cindy Well done!! – Robert Z Jul 06 '23 at 18:10