Textbook Question
The ellipse and the parabola: Let R be the region bounded by the upper half of the ellipse x²/2 + y² = 1 and the parabola y = x²/√2
a. Find the area of R
75
views
9. Graphical Applications of Integrals
Area Between Curves
3:15 minutesProblem 7.3.71a
Textbook 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.
Verified step by step guidance1
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.
Verified video answer for a similar problem:This video solution was recommended by our tutors as helpful for the problem above
Video duration:
3mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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.
Recommended video:
5:53
Graph of Sine and Cosine Function
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.
Recommended video:
05:32Intro to Polar Coordinates
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.
Recommended video:
5:02Solving Logarithmic Equations
Related Videos
Related Practice