A power line is attached at the same height to two utility poles that are separated by a distance of 100 ft; the power line follows the curve ƒ(x) = a cosh x/a. Use the following steps to find the value of a that produces a sag of 10 ft midway between the poles. Use a coordinate system that places the poles at x = ±50.


a. Show that a satisfies the equation cosh 50/a − 1 = 10/a.

Verified step by step guidance

1. Understand the problem setup: The power line follows the curve \(f(x) = a \cosh \frac{x}{a}\), and the poles are located at \(x = \pm 50\). The sag is the vertical distance between the height of the poles and the lowest point of the cable, which occurs at \(x=0\) because \(\cosh\) has its minimum there.
2. Express the height of the poles: At \(x = \pm 50\), the height of the cable is \(f(50) = a \cosh \frac{50}{a}\). Since the poles are at the same height, both ends have this height.
3. Express the sag: The lowest point of the cable is at \(x=0\), where \(f(0) = a \cosh 0 = a \cdot 1 = a\). The sag is the difference between the height at the poles and the lowest point, so sag \(= f(50) - f(0) = a \cosh \frac{50}{a} - a\).
4. Set the sag equal to 10 ft: According to the problem, the sag is 10 ft, so write the equation \(a \cosh \frac{50}{a} - a = 10\).
5. Rearrange the equation to isolate terms: Divide both sides by \(a\) to get \(\cosh \frac{50}{a} - 1 = \frac{10}{a}\), which is the equation that \(a\) satisfies.

Key Concepts

Catenary Curve and Hyperbolic Cosine Function

The power line forms a catenary curve described by ƒ(x) = a cosh(x/a), where cosh is the hyperbolic cosine function. This curve models the shape of a flexible chain or cable hanging under its own weight, and understanding its properties is essential to relate the sag and horizontal distance.

Coordinate System and Boundary Conditions

Placing the poles at x = ±50 sets the horizontal span of the cable, and the sag is measured at the midpoint x = 0. Using these boundary conditions allows us to express the sag in terms of the parameter a and set up an equation involving cosh(50/a) to solve for a.

Deriving and Solving the Sag Equation

The sag is the vertical drop from the pole height to the lowest point of the cable. By substituting x = 0 and x = ±50 into the catenary equation and using the sag value, we derive the equation cosh(50/a) − 1 = 10/a. Solving this transcendental equation yields the parameter a.