Chapter 6.7, Problem 50E

A flexible cable always hangs in the shape of a catenary $y=c+a\cosh \left(x/a\right)$, where c and a are constants and a>0 (see Figure 4 and Exercise 52). Graph several members of the family of functions $y=a\cosh \left(x/a\right)$. How does the graph change as a varies?

Figure 4

Using principles from physics it can be shown that when a cable is hung between two poles, it takes the shape of a curve $y=f\left(x\right)$ that satisfies the differential equation

$\frac{d^2}{d{x}^2}=\frac{\text{ρ}g}{T}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}$

where $\text{ρ}$ is the linear density of the cable, g is the acceleration due to gravity, T is the tension in the cable at its lowest point, and the coordinate system is chosen appropriately. Verify that the function

$y=f\left(x\right)=\frac{T}{\rho g}\cosh \left(\frac{\rho gx}{T}\right)$

is a solution of this differential equation.