91Ó°ÊÓ

The surprising fact that the arc length of the catenary curve traced out by the hyperbolic function f(x) = cosh x on any interval [a,b] is equal to the area under the same graph on [a,b].

Recall that the arc length of a differentiable function f(x)with continuous derivative from x = a

to x=b is given by the definite integral

$L={∫}_b^a\sqrt{1+\sin {\left(h\right)}^{2}dx}$

Note that the hyperbolic function f(x) = cosh x is differentiable and has continuous derivative on any interval [a,b]. So, differentiate the function with respect to x

f(x) = cosh x

f(x)=sinh x

Substitute the above derivative in the integral given in equation (1) and evaluate the integral

$$\begin{gathered} L={∫}_b^a\sqrt{1+\sin {\left(h\right)}^{2}dx} \\ ={∫}_a^b\cos hxdx \\ ={{\left[\sin hx\right]}^b}_a \\ =\sin hb-\sin ha \end{gathered}$$

So, the arc length of the catenary on the interval is sinh b-sinh a.

Next, find the area of catenary under the same graph on [a, b] . Remember that the area of a curve bound between the graph of the function and x-axis on [a, b] is given by the definite integral

$$\begin{gathered} A={∫}_a^{b}f\left(x\right)dx \\ \mathrm{Thus}\mathrm{the}\mathrm{area}\mathrm{under}\mathrm{the}\mathrm{graph}\mathrm{of}\mathrm{the}\mathrm{catenary}\mathrm{f}\left(\mathrm{x}\right)=\cosh \mathrm{x}\mathrm{between}\mathrm{x}=\mathrm{a}\mathrm{and}\mathrm{x}=\mathrm{b}\mathrm{is}\mathrm{given} \\ \mathrm{A}={∫}_{\mathrm{a}}^{\mathrm{b}}\cosh \mathrm{xdx} \\ ={{\left[\mathrm{sinhx}\right]}_{\mathrm{a}}}^{\mathrm{b}}=\mathrm{sinhb}-\sin \mathrm{ha} \end{gathered}$$