Problem 428 - Calculus Volume 1

Question

For the following exercises, solve each problem. A telephone line is a catenary described by \(y=a \cosh (x / a)\). Find the ratio of the area under the catenary to its arc length. Does this confirm your answer for the previous question?

Step-by-Step Solution

Answer

The ratio of the area under the catenary to its arc length is \(a\).

Step 1: Understand the Problem

We are asked to find the ratio of the area under the catenary to its arc length. The catenary is given by the formula \(y = a \cosh(x/a)\). We will calculate both the area under the curve and the arc length, then find their ratio.

Step 2: Find the Area Under the Catenary

The area under a catenary from \(-L\) to \(L\) is given by the integral \(A = \int_{-L}^{L} a \cosh(x/a) \ dx\). We can use the substitution \(u = x/a\), \(du = dx/a\) to solve this integral: \[A = a^2 \int_{-L/a}^{L/a} \cosh(u) \ du = a^2 [ \sinh(u) ]_{-L/a}^{L/a} = a^2 [ \sinh(L/a) - \sinh(-L/a)] = 2a^2 \sinh(L/a)\].

Step 3: Calculate the Arc Length of the Catenary

The arc length \(S\) of the catenary from \(-L\) to \(L\) is given by \(S = \int_{-L}^{L} \sqrt{1 + (\sinh(x/a))^2} \ dx\). Simplifying, \(\sqrt{1 + (\sinh(x/a))^2} = \cosh(x/a)\). Thus the integral becomes \(S = \int_{-L}^{L} \cosh(x/a) \ dx\), which is the same as calculating \(2a \sinh(L/a)\) from the area integral.

Step 4: Calculate the Ratio of Area to Arc Length

Now we find the ratio \(\frac{A}{S}\). Since both the area \(A\) and the arc length \(S\) are equal to \(2a \sinh(L/a)\), the ratio becomes: \[\frac{A}{S} = \frac{2a^2 \sinh(L/a)}{2a \sinh(L/a)} = a\].

Step 5: Evaluate the Result and Confirm Understanding

The ratio of the area under the catenary to its arc length is \(a\), which indicates the direct relation between the curve's parameters. This confirms any prior calculation of such ratio as long as it arrived at the same conclusion.

Key Concepts

Arc Length

When we talk about the arc length in the context of geometry and calculus, we refer to the distance measured along a curve between two points. This is an important concept when dealing with curves like the catenary, which is the shape due to the gravitational pull of an object hanging freely between two points.

For a general curve given by a function, the arc length from point \(x_1\) to \(x_2\) can be found using the formula:

• \(S = \int_{x_1}^{x_2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\)

In our specific exercise, the catenary curve is defined as \(y = a \cosh\left(\frac{x}{a}\right)\), where \(a\) is a constant. Through some calculus, we found the arc length from \(-L\) to \(L\) to be \(S = 2a \sinh\left(\frac{L}{a}\right)\).

The critical insight is that the arc length of the catenary is a function of the hyperbolic sine function, making calculations involving these curves uniquely tied to hyperbolic functions.

Hyperbolic Functions

Hyperbolic functions, like sine and cosine, are analogs to the trigonometric functions but with different properties and applications. While trigonometric functions are based on circles, hyperbolic functions are based on hyperbolas.

In this exercise, the specific hyperbolic functions used are:

• \(\cosh(x)\) - Hyperbolic cosine, defined as \(\frac{e^x + e^{-x}}{2}\)
• \(\sinh(x)\) - Hyperbolic sine, defined as \(\frac{e^x - e^{-x}}{2}\)

These functions are particularly useful in describing the shape of a catenary. The equation of a catenary \(y = a \cosh\left(\frac{x}{a}\right)\) uses the hyperbolic cosine function to model the curve.

In the solution, we also encountered \(\sinh(x/a)\), which is crucial when finding both the area under the curve and its arc length. Hyperbolic functions exhibit unique integral properties that make them suitable for solving such calculus problems.

Integral Calculus

Integral calculus is a fundamental part of calculus focused on finding areas under curves, among other things. It involves determining antiderivatives and calculating definite integrals. In the context of our exercise, integral calculus is used in two major ways.

Firstly, to find the area under the catenary curve from \(-L\) to \(L\), we set up the integral:

• \(A = \int_{-L}^{L} a \cosh\left(\frac{x}{a}\right) \, dx\)

By substituting \(u = x/a\), the solution simplifies to \(A = 2a^2 \sinh\left(\frac{L}{a}\right)\).

Secondly, the arc length is calculated through another integral, which turns out to be the same form due to the properties of hyperbolic functions. Both integrals feature \(\cosh\) and \(\sinh\) due to the catenary's inherent shape.

Integral calculus provides powerful tools to evaluate and simplify these expressions, leading us to find relationships like the ratio of area to arc length easily.