Problem 427 - Calculus Volume 1

Question

For the following exercises, solve each problem. [T] A high-voltage power line is a catenary described by \(y=10 \cosh (x / 10)\). Find the ratio of the area under the catenary to its arc length. What do you notice?

Step-by-Step Solution

Answer

The ratio of the area to the arc length is constant and equal to 10.

Step 1: Understand the Problem

We need to determine the ratio of the area under the curve described by the catenary equation to its arc length. The catenary equation is given as \( y = 10 \cosh \left(\frac{x}{10}\right) \).

Step 2: Define the Integral for Area

The area under the catenary from \( x = a \) to \( x = b \) is the integral of \( y \) with respect to \( x \): \[A = \int_{a}^{b} 10 \cosh \left(\frac{x}{10}\right) \ dx.\]

Step 3: Define the Integral for Arc Length

The arc length \( L \) from \( x = a \) to \( x = b \) can be found using the formula: \[L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \ dx.\] First, compute \( \frac{dy}{dx} = \sinh\left(\frac{x}{10}\right) \). Substitute and simplify to calculate \( L \).

Step 4: Calculate \( dy/dx \) and Integrate for Arc Length

We have \( \frac{dy}{dx} = \sinh\left(\frac{x}{10}\right) \). Thus, \[L = \int_{a}^{b} \sqrt{1 + \sinh^2\left(\frac{x}{10}\right)} \ dx = \int_{a}^{b} \cosh\left(\frac{x}{10}\right) \ dx.\]

Step 5: Solve the Integrals

Both integrals, \[A = \int_{a}^{b} 10 \cosh\left(\frac{x}{10}\right) \ dx \] and \[L = \int_{a}^{b} \cosh\left(\frac{x}{10}\right) \ dx\] can be solved to get:\[A = 100 \sinh\left(\frac{x}{10}\right) \bigg|_a^b, \quad L = 10 \sinh\left(\frac{x}{10}\right) \bigg|_a^b.\]

Step 6: Determine the Ratio

The ratio \( R \) of area to arc length is:\[R = \frac{100 \sinh\left(\frac{x}{10}\right) \bigg|_a^b}{10 \sinh\left(\frac{x}{10}\right) \bigg|_a^b} = 10.\]The ratio is constant and equal to 10, which is the coefficient of \(\cosh\left(\frac{x}{10}\right)\).

Step 7: Conclude the Observation

The ratio of the area under the catenary to its arc length is constant regardless of the values \(a\) and \(b\) chosen for the integration. This is influenced by the specific form of the hyperbolic functions and their inherent properties.

Key Concepts

Catenary Curve

The term "catenary" comes from the Latin word "catena," meaning "chain." This makes sense because a catenary curve describes the shape a flexible chain or cable assumes when supported only at its ends and acted upon uniformly by gravity. Mathematically, a catenary curve is described by the equation:

• For our case, the equation is given by: \( y = 10 \cosh\left(\frac{x}{10}\right) \)

The hyperbolic cosine function, \(\cosh(x)\), which features prominently in the equation, is central to the curve's properties. Unlike a parabola or a simple arc, the curve of a catenary becomes steeper as you move away from the lowest point, and this form minimizes potential energy in a suspended system.
The distinct feature of the catenary is its gentle but consistent growing slope as you progress from the center outwards, controlled by the hyperbolic cosine function. This characteristic minimizes the tension within the cable, making the catenary form both elegant and practical in physical applications.

Area Under a Curve

To find the area under the catenary curve, you need to calculate the integral of the curve's equation over a specified interval. In calculus, the integral of a function gives us the total accumulation of quantities, and in this context, it refers to the total amount of space beneath the curve.
For the catenary curve described by \( y = 10 \cosh \left(\frac{x}{10}\right) \), the integral to calculate the area \( A \) from \( x = a \) to \( x = b \) is given by:

• \[ A = \int_{a}^{b} 10 \cosh \left(\frac{x}{10}\right) \, dx \]

This integral essentially adds up small rectangular sections under the curve as you move along the x-axis from \( a \) to \( b \). By solving it, you find the total area encapsulated between the curve and the x-axis.
This process is not just crucial in calculus; it also has real-world applications, such as in physics to calculate work done or in economics to determine consumer surplus. Each component of the function contributes to how steeply the area builds up, with larger sections of the integral corresponding to stretches where the curve lies higher above the axis.

Arc Length

The arc length of a curve is the distance along the curve itself, as opposed to merely a straight-line distance between two points. For a catenary, the arc length over an interval gives a sense of the actual "travel path" along the curve.
The arc length \( L \) from \( x = a \) to \( x = b \) for our catenary curve is found by integrating based on the derivative of the curve:

• \[ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]

For our given catenary:

• \( \frac{dy}{dx} = \sinh\left(\frac{x}{10}\right) \),

it simplifies to integrating

• \[ L = \int_{a}^{b} \cosh\left(\frac{x}{10}\right) \, dx \]

This integral measures the true length of the path traced out by the curve over the interval. In the case of a catenary, as illustrated, the relationship between the integral for the area and that for arc length reveals significant geometric properties, including ratios that are constant under transformations, thanks to the harmonic relationship of hyperbolic functions.