Question

[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?

Short answer

The ratio of the area under the catenary to its arc length is 1.

Step-by-step solution

Define the Area Under the Catenary

To find the area under the catenary, we'll integrate the function from one endpoint to the other. Let's assume we're considering from $x=-a$ to $x=a$. The area $A$ under $y=10\cosh (x/10)$ is given by the integral, $$A={\int }_{-a}^{a}10\cosh (x/10)dx.$$

Calculate the Integral for the Area

The integral of $\cosh (x/10)$ with respect to $x$ is $10\sinh (x/10)$. Therefore, $$A=10{[10\sinh (x/10)]}_{-a}^a=100(\sinh (a/10)-\sinh (-a/10)).$$ Since $\sinh$ is an odd function, $\sinh (-x)=-\sinh (x)$, thus $$A=200\sinh (a/10).$$

Define the Arc Length

The arc length $L$ of the curve $y=10\cosh (x/10)$ is found using the formula for arc length of a curve $y=f(x)$, given by $$L={\int }_{-a}^a\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx.$$

Calculate the Derivative

First, calculate $\frac{dy}{dx}$. Since $y=10\cosh (x/10)$, $\frac{dy}{dx}=\sinh (x/10)$. Thus, $${\left(\frac{dy}{dx}\right)}^2={\sinh }^2(x/10).$$

Calculate the Arc Length Integral

Substitute ${\left(\frac{dy}{dx}\right)}^2={\sinh }^2(x/10)$ into the arc length formula, $$L={\int }_{-a}^a\sqrt{1+{\sinh }^2(x/10)}dx={\int }_{-a}^a\cosh (x/10)dx.$$ This integral is the same as the one used to calculate the area, resulting in, $$L=10{[10\sinh (x/10)]}_{-a}^a=200\sinh (a/10).$$

Find the Ratio of the Area to Arc Length

The ratio of the area to the arc length is $\frac{A}{L}$. Here, both $A$ and $L$ yield $200\sinh (a/10)$, so $$\frac{A}{L}=\frac{200\sinh (a/10)}{200\sinh (a/10)}=1.$$

Observation

The ratio of the area under a catenary to its arc length is always 1. This means the area and the length are identical in size when integrated over this curve.

Key concepts

Arc Length

The length of a curve, known as its arc length, is a fundamental concept in calculus and geometry. When studying shapes like the catenary, which is the shape a hanging flexible cable or chain assumes under its own weight, calculating the length of its curve from point to point along the x-axis is crucial. For any smooth curve given by a function $y=f(x)$, the arc length from $x=a$ to $x=b$ can be found using the formula:$$L={\int }_a^b\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx.$$- This formula stems from Pythagoras’ Theorem, contemplating infinitesimally small segments of the curve.- The square root term accounts for both vertical and horizontal changes along the curve.In our specific problem involving the catenary $y=10\cosh (x/10)$, we computed the derivative $\frac{dy}{dx}=\sinh (x/10)$. Placing it back into the arc length formula simplifies the integral, leveraging the identity $1+{\sinh }^2(u)={\cosh }^2(u)$, making the arc calculation particularly straightforward.

Integral Calculus

Integral calculus, a branch of mathematics, deals with finding the accumulation of quantities. It is used to determine areas under curves, among other applications. In this exercise, integral calculus is applied to find both the area under the catenary curve and its arc length.To find the area, we integrated the catenary function $$A={\int }_{-a}^{a}10\cosh (x/10)dx,$$- Integration exploits the anti-derivatives of functions.- For $\cosh (x/10)$, the integral is $10\sinh (x/10)$ because $\sinh (x)$ is the derivative of $\cosh (x)$.In another step, integral calculus aids in calculating the arc length of the curve. Each task exemplifies how integral calculus serves as a powerful tool to solve complex problems involving cumulative sums like area and length.

Hyperbolic Functions

Hyperbolic functions, akin to trigonometric functions, are essential in describing the behavior of catenaries. While trigonometric functions relate to circles, hyperbolic functions relate to hyperbolas. The key hyperbolic functions are hyperbolic sine, $\sinh (x)$, and hyperbolic cosine, $\cosh (x)$, defined as follows:- $\sinh (x)=\frac{e^x-e^{-x}}{2}$ - $\cosh (x)=\frac{e^x+e^{-x}}{2}$In our catenary example, the curve is defined by $y=10\cosh (x/10)$. - The function $\cosh (x)$ resembles the shape of a hanging chain and describes the curve's height at a particular point.- These functions possess identities similar to those in trigonometry, such as $1+{\sinh }^2(x)={\cosh }^2(x)$, crucial for calculations involving arc length.Understanding hyperbolic functions helps simplify not only algebraic manipulations but calculus operations like integrations and derivatives, critical in analyzing catenary properties.