Question

In Exercises 35 and \(36,\) a model for a power cable suspended between two towers is given. (a) Graph the model, (b) find the heights of the cable at the towers and at the midpoint between the towers, and (c) find the slope of the model at the point where the cable meets the right-hand tower. \(y=10+15 \cosh \frac{x}{15}, \quad-15 \leq x \leq 15\)

Short answer

The plotted graph is a shape of a suspended cable and symmetrical about y-axis. The heights of the cable at the towers and at the midpoint between the towers are calculated based on the given function. The slope of the model at the point where the cable meets the right-hand tower indicates the rate of change of the functionality at that point.

Step-by-step solution

Plotting the graph

To plot the graph for the function \(y = 10 + 15 \cosh \frac{x}{15}\), a graphing tool can be used. Selecting the x-range from -15 to 15 should cover the required region for the exercise.

Calculating the heights

Find the heights (y-values) for x = -15, 0, and 15. Plug these values into the function (y = 10 + 15 \cosh \frac{x}{15}) one by one to obtain the heights. For x=-15 and x=15, the heights are equal and adding to the symmetry of the graph.

Finding the slope of the model

To find the slope of the function at x=15, the derivative of the function with respect to x, \(y'\), needs to be calculated first: \(y' = 15 \sinh(\frac{x}{15})/15\) . Then use \(y'\) to find the slope at x=15 by substituting x=15 into \(y'\).

Key concepts

Hyperbolic Functions

Hyperbolic functions, such as the hyperbolic cosine function denoted as \( \cosh \), form an essential part of calculus, akin to their trigonometric counterparts. These functions bear resemblances to trigonometric functions but are defined in terms of the exponential function. Specifically, the hyperbolic cosine function is defined as \( \cosh(x) = \frac{e^x + e^{-x}}{2} \), showcasing its relation to exponential growth and decay.

Graphing \( \cosh \) presents a shape that is symmetric along the vertical axis, known as the line of symmetry, much like the regular cosine function. However, unlike the trigonometric cosine, \( \cosh \) never turns negative. It starts from 1, when x is 0, and approaches infinity as x moves away from zero in either direction. This characteristic makes \( \cosh \) an ideal model for real-world scenarios like the suspension cables between two towers, as the shape of the cable naturally forms a hyperbolic cosine curve due to the even distribution of weight and tension.

Calculus Graphing

Graphing in calculus often involves plotting a wide range of functions to visualize and analyze their characteristics. For hyperbolic functions like \( \cosh \), calculus graphing tools enable us to depict the smooth curve that represents the function's behavior. \( \cosh \) has an increasing slope as you move away from the origin, reflecting the real-world phenomenon of a hanging cable's shape under gravitational forces.

When plotting the graph of \( y = 10 + 15 \cosh \frac{x}{15} \), one important aspect is the transformation applied to the basic \( \cosh \) function. The graph's centerline is lifted vertically by 10 units, and the horizontal stretching factor of 15 affects the rate at which the function's value increases as x moves away from the origin. This transformation simulates the height and span of the suspended power cable. The graph's domain here is constrained between -15 and 15, which helps us visualize the cable at specific points, particularly the towers and the midpoint between them, where the cable naturally dips the lowest due to gravity.

Derivative Calculations

Calculating the derivative of a function is a fundamental operation in calculus, yielding a new function that provides the slope of the tangent to the original function at any given point. The derivative of the hyperbolic cosine function, \( \cosh \), is the hyperbolic sine function, \( \sinh \). Specifically, for a function \( f(x) = \cosh(x) \), its derivative \( f'(x) = \sinh(x) \).

In the context of our exercise, to find the slope of a cable at a certain point, such as where the cable meets the right-hand tower (at \( x=15 \)), we calculate the derivative of the given function \( y = 10 + 15 \cosh(\frac{x}{15}) \). By applying the chain rule, we obtain \( y' = 15 \sinh(\frac{x}{15})/15 \), which simplifies to \( y' = \sinh(\frac{x}{15}) \). Plugging \( x = 15 \) into this derivative reveals the slope of the model at that point, which corresponds to the steepness of the cable as it ascends to the tower, providing us with insight on the tension at that section of the cable.