Q. 84 - Calculus

Question

 Prove 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].

Step-by-Step Solution

Answer

Hence, proved 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]. 

Step 1. To prove

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]. 

Step 2. Differentiate the function with respect to x

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

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

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

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

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

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