Question

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.

Accepted Answer

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. 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. 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 $$= f(50) - f(0) = a \cosh \frac{50}{a} - a. $$Set the sag equal to 10 ft: According to the problem, the sag is 10 ft, so write the equation $$a \cosh \frac{50}{a} - a = 10. $$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.

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Key Concepts

Catenary Curve and Hyperbolic Cosine Function

Coordinate System and Boundary Conditions

Deriving and Solving the Sag Equation