Background: This demo was motivated by a conversation between Lori Braselton and her father, Charles "Nick" Maruth, Jr. [1] of Davenport, Iowa. He posed the problem
The "obvious" answer and its not so obvious variations are discussed in this demo. Objective: The purpose of this demo is to illustrate how hyperbolic functions and arc length integrals are used to model hanging cables. Level: This demo is appropriate for any calculus or advanced mathematical modeling course in which hyperbolic functions and applications of integration to arc length are covered. Hand solutions of the example problems involve inverse functions. Prerequisites: Students should be familiar with the hyperbolic sine and cosine functions, their derivatives, and arc length integrals. Platform: Mathematica is used to generate the animations, to symbolically solve equations, and to find values of integrals. Mathematica notebooks are available for download. However, any computer algebra system may be used for symbolic work. The demo is easily adaptable for use with graphing calculators which may be used for numerical work and graphing. Screen shots are included for the calculator solutions of the systems of equations. Instructor's Notes: The problem of suspending a flexible cable between two poles of equal height is of interest to utility companies that must make sure that power lines suspended over roadways are not subject to being hit by vehicles which pass underneath. We consider Mr. Maruth's problem:
Reflecting on the problem and using some common sense, we see (Figure 1) the obvious answer.
The obvious answer is 0 ft. If the poles are placed together, the cable will hang down 75 ft (75 ft + 75 ft = 150 ft, the length of the cable) and the distance from the bottom of the wire will be 25 ft. Not only is the answer obvious, it is also somewhat uninteresting. As mathematicians who do not always use common sense at first, we could not believe the answer was obvious because a flexible wire or cable suspended between two poles of equal height takes the shape of a catenary which has the equation f(x) = a cosh(x/a), where a > 0. Figure 2 shows the graph of a catenary. Note that f gives the height of the cable as a function of position x.
In the context of our problem where we have to take into account the height of the poles, we need to use a more general formulation of the catenary, f(x) = c + a cosh(x/a), a > 0. Figure 3 illustrates the scenario of the cable suspended from two vertical poles of height h. The function f gives the height of the cable as a function of position x. Note that the minimum value occurs when x = 0; f(0) = c + a. The parameter c represents a vertical shift of the basic curve.
For a cable with fixed length and poles with fixed height, the height of the suspended cable varies with the separation between the poles. Example 1. A flexible cable with length 150 feet is to be suspended between two poles with height 100 ft. How far apart must the poles be spaced so that at its lowest point the cable is 50 feet off the ground? An animation of this situation is shown in Figure 4.
The animation shows that at some distance between the poles the lowest point of the cable will be 50 feet off the ground. We will denote that distance by 2s. Note that 0 < s < 75. Let f(x) = c + a cosh(x/a). Then f(s) = f(s) = 100 so we have
The minimum value of f occurs when x = 0 and we want the minimum value of f at x = 0 to be 50 so
The length of the wire is 150 feet. In the context of the catenary function, we can interpret the length of the wire as the arc length of the graph of f from x = s to x = s. This gives a third equation.
A bit of work gives the following: which simplifies to . The integral can be evaluated directly Thus, our problem is to solve a system of three equations in three unknowns: c + a cosh(s/a) = 100 a + c = 50 2a sinh(s/a) = 150 for a, c, and s. Using some algebra and hyperbolic identities, it is not too difficult to solve the system of equations by hand. However, Mathematica can solve the system symbolically, as shown below. The exact solution is c = 75/4, about 18.75 ft. s = 125/4 ArcCosh(13/5), about 50.2949 ft. a = 125/4, about 31.25 ft. Remember that s is 1/2 the distance apart. Thus, if the poles are 2(50.2949) ~ 100.59 ft. apart, the lowest point on the suspended cable will be 50 feet off the ground. A more realistic example is given below. Example 2. According to our electric utility, Excelsior Electric Membership Corp. (Excelsior EMC) in Metter, GA [2], due to terrain, easements, etc., the average distance between utility poles ranges from 325 to 340 feet. The Georgia Department of Transportation (DOT) [3] states that the maximum height of a truck using interstates, national, and state routes is 13 feet and 6 inches. However, special permits may be granted by the DOT for heights up to 18 feet. With these restrictions in mind, Excelsior EMC maintains a minimum clearance of 20 feet under those lines it installs during cooler months because expansion causes lines to sag during warmer months. For the obvious reason, Excelsior EMC prefers that the distance from its lines to the ground is greater than 18' 6 '' at all times. Problem: Find a and c so that f(x) = c + a cosh(x/a) models this situation. We will assume that the distance between poles is 340 feet and that we want the minimum clearance to be 20 feet. Proceeding as in the previous example, we require that f(170) = f(170) = 34 and f(0) = 20. These two conditions give the equations
which can be reduced to the single equation (20  a) + a cosh(170/a) = 34 or
Mathematica is not very useful for a symbolic solution to this nonlinear equation, as illustrated in the output shown in Figure 5.
A numerical solution can be found using the FindRoot command. If we define g(a) = a + a cosh(170/a)  14 and plot (you may need to experiment with the plotting interval), we see there is a root between 1000 and 1100 (see Figure 6). Using the FindRoot command with an initial guess of 1000 yields a numerical solution.
It follows that c is approximately 1014.47. With this information we can obtain the length of the wire between the poles by computing the arc length (Mathematica solution is shown in Figure 7).
The length of wire is about 341.53 feet. Figure 8 illustrates the Excelsior EMC scenario.
Problem Extension Using a symbolic algebra system such as Mathematica, the code may be written so that it is general enough to solve a variety of problems. For example, the following code solves for the parameters s (halfdistance between the poles), c, and a so that f(x,c,a) = c + a cosh(x/a) satisfies the following system of equations: f(s,c,a) = f(s,c,a) = h_{1} f(0,c,a) = h_{2}, and , where h_{1} is the height of the poles, h_{2} is the required minimum height of the wire and L is the length of the wire. The solution given below is valid for h_{1}> h_{2}.
Mathematica Codes and Calculator Adaptation Mathematica code to generate the animations can be viewed here and downloaded from here. Solutions via graphing calculator can be viewed here. References 1. Charles "Nick" Maruth (April 13, 2003). Private Communication. 2. Excelsior Electric Membership Corp. (April 16, 2003). Private Communication. 3. Georgia Department of Transportation, http://www.dot.state.ga.us/doingbusiness/permits/Pages/
This demo was submitted by Sharon M.
Barrs and is included in Demos with Positive Impact with their permission. Mathematica codes were written by Jim Braselton and calculator solutions were generated by Lila Roberts. Special thanks to Nick Maruth for bringing the problem to his daughter's attention.
