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?
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 $a$?