Hanging Wire (Catenary) - 2

If the end-points of the catenary are $2a$ apart, and at the same vertical height as each other, then the equation of the catenary is $y \; = \; c^{-1} \cosh cx \hspace{2cm} |x| \le a$ for some $c>0$ , and this catenary has arc-length $L \; = \; \int_{-a}^a \sqrt{1 + (y')^2}\,dx \; = \; \tfrac{2}{c}\sinh ca$ while the difference in height between the end-points and the middle is $\Delta h \; = \; c^{-1}\big(\cosh ac - 1\big)$ In this case $a=5$ and $L=11$ , so we solve the equation $11c = 2\sinh5c$ numerically to obtain $c = 0.15268$ , and therefore calculate $\Delta h = \boxed{2.00301}$ .