Question

Power lines A power line is attached at the same height to two utility poles that are separated by a distance of $100\mathrm{ft}$; the power line follows the curve $f(x)=a\cosh \frac{x}{a}.$ Use the following steps to find the value of $a$ that produces a sag of $10\mathrm{ft}$ midway between the poles. Use a coordinate system that places the poles at $x=\pm 50$. a. Show that $a$ satisfies the equation $\cosh \frac{50}{a}-1=\frac{10}{a}$ b. Let $t=\frac{10}{a},$ confirm that the equation in part (a) reduces to $\cosh 5t-1=t,$ and solve for $t$ using a graphing utility. Report your answer accurate to two decimal places. c. Use your answer in part (b) to find $a$, and then compute the length of the power line.

Short answer

In this exercise, we analyzed a problem involving a power line that follows a specific curve. We were given the distance between the poles and the sag at the midpoint and asked to find the equation for a and the length of the power line. First, we found the equation that a satisfies, which is $\cosh(\frac{50}{a}) - 1 = \frac{10}{a}$. Then, we introduced a substitution, making a new equation of $\cosh(5t) - 1 = t$. After that, we used a graphing utility to find the value of t, which is approximately 1.11. Finally, we found the value of a using our substitution and computed the length of the power line by integrating its path. We found that a is approximately 9.01 and the length of the power line is approximately 107.07 ft.

Step-by-step solution

Part a: Find the equation for a

Given that the distance between the poles is 100 ft and the sag is 10 ft midway between the poles, we can write the equation for the sag as: $$f(0)=10=a\cosh \left(\frac{0}{a}\right)-a.$$ Since $\cosh (0)=1$, the equation becomes: $$a\cosh (0)-a=10⟹a(\cosh (0)-1)=10.$$ Thus, the equation that $a$ satisfies is: $$\cosh \left(\frac{50}{a}\right)-1=\frac{10}{a}.$$

Part b: Substitution and solving with a graphing utility

Now, we introduce the substitution $t=\frac{10}{a}$, which implies $a=\frac{10}{t}$. We rewrite the equation from part (a) using this substitution: $$\cosh \left(\frac{50}{\frac{10}{t}}\right)-1=t.$$ Simplifying, we get: $$\cosh (5t)-1=t.$$ Now, we use a graphing utility to find the value of $t$ accurate to two decimal places. By plotting the graphs of $y=\cosh (5t)-1$ and $y=t$, we can find their intersection point, which gives us $t\approx 1.11$.

Part c: Find a and compute the length of the power line

To find the value of $a$, we use the relationship between $a$ and $t$ established in part (b): $$a=\frac{10}{t}=\frac{10}{1.11}⟹a\approx 9.01.$$ Now, we need to compute the length of the power line. Its length can be found by integrating the path from $x=-50$ to $x=50$: $$L={\int }_{-50}^{50}\sqrt{1+{\left[\frac{df(x)}{dx}\right]}^2}dx.$$ First, let's find the derivative of $f(x)$: $$\frac{df(x)}{dx}=a\sinh \left(\frac{x}{a}\right)\frac{1}{a}=\frac{1}{a}\sinh \left(\frac{x}{a}\right)=\frac{1}{9.01}\sinh \left(\frac{x}{9.01}\right).$$ Now, plug the derivative into the formula for the length and simplify: $$L={\int }_{-50}^{50}\sqrt{1+{(\frac{1}{9.01}\sinh \left(\frac{x}{9.01}\right))}^2}dx.$$ Using a calculator or computer software, we find that the length of the power line is approximately $107.07$ ft.

Key concepts

Catenary

A catenary is a curve that resembles the shape of a hanging flexible chain or cable that is only supported at its ends and is acted upon by a uniform gravitational force. This shape is not a parabola, although it might look similar. It is in fact defined by the hyperbolic cosine function. In physics and engineering, the concept of a catenary curve is highly important, especially in optimizing designs for cable structures like bridges and power lines.

In the context of the provided exercise, the power line forms a catenary curve defined by the function $$f(x)=a\cosh \left(\frac{x}{a}\right)$$where "a" represents a constant related to the physical properties of the cable, such as tension and weight. The challenge is to determine the value of "a" needed to achieve the sag of 10 ft in the center of this configuration.

Cosine Hyperbolic

The hyperbolic cosine function, denoted as $\cosh (x)$, is a fundamental hyperbolic function, similar in some ways to the trigonometric cosine but with different properties.

Cosine hyperbolic is defined as:$$\cosh (x)=\frac{e^x+e^{-x}}{2}$$It forms part of the catenary equation and plays a crucial role in determining the shape of the catenary curve. The three main properties of $\cosh (x)$ are:

• $\cosh (0)=1$
• $\cosh (x)$ is always greater than or equal to 1
• It is an even function, meaning $\cosh (-x)=\cosh (x)$

Understanding $\cosh (x)$ helps in comprehending why and how a catenary curve forms under the influence of gravity, particularly in applications like bridges and utility wires.

Derivative of Hyperbolic Functions

The derivative of hyperbolic functions captures the rate of change of these functions, and they are essential in calculating quantities like arc length. The derivative of the hyperbolic cosine function, $\cosh (x)$, is the hyperbolic sine function, $\sinh (x)$. Derivatives for $\sinh (x)$ and $\cosh (x)$ are:

• $\frac{d}{dx}\cosh (x)=\sinh (x)$
• $\frac{d}{dx}\sinh (x)=\cosh (x)$

In the problem provided, differentiating the function $f(x)=a\cosh \left(\frac{x}{a}\right)$ is critical to finding the arc length since this involves calculating:

$$\frac{df(x)}{dx}=\frac{1}{a}\sinh \left(\frac{x}{a}\right)$$The derivative gives insight into how quickly the height changes along the curve of the power line.

Arc Length

Arc length is the total distance along a curve between two points, which, for a curve like the catenary, can be a complex calculation. For the hyperbolic cosine catenary, it's important to differentiate the function $f(x)$ first, then use the derived expression to apply in the arc length integral formula. The arc length $L$ from $x=-50$ to $x=50$ is calculated by:

$$L={\int }_{-50}^{50}\sqrt{1+{(\frac{1}{a}\sinh \left(\frac{x}{a}\right))}^2}dx$$In solving this real-world problem, the arc length determines the total cable length necessary for installation. Calculations can be complex, usually requiring computational tools to evaluate such integrals accurately. For the exercise, this is roughly approximated to a length of about 107.07 ft, reflecting the necessary span of the power line.