Find the length $L$ of the catenary

Question

$C$: $y = \cosh(x)$, $\,\log(2) \leq x \leq \log(3)$.

I did arrive at the problem until $\sinh b$, but I don't know how to continue. Should it be $\sinh(\log(3))$?

Answer 1

One can use the arc-length formula with $f(x)=\cosh(x)$. Then the length of the curve on $(\log(2),\log(3))$ is $$L=\int_{\log(2)}^{\log(3)}\sqrt{1+[f^\prime(x)]^2}\,dx = \int_{\log(2)}^{\log(3)}\sqrt{1+\sinh^2(x)}\,dx=\int_{\log(2)}^{\log(3)}\cosh(x)\,dx.$$ Evaluating this integral $$L=\sinh(\log(3))-\sinh(\log(2)).$$ Note that in the above we have used the fact that $\cosh^2(x)-\sinh^2(x)=1.$ I hope that this answers your question!