/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 The portion of the curve \(y=\fr... [FREE SOLUTION] | 91Ó°ÊÓ

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The portion of the curve \(y=\frac{17}{15}-\cosh x\) that lies above the \(x\) -axis forms a catenary arch. Find the average height of the arch above the \(x\) -axis.

Short Answer

Expert verified
The average height of the catenary arch described by the function y(x) = 17/15 - cosh(x) above the x-axis is 17/15.

Step by step solution

01

Find the intersection points with the x-axis

To find the intersection points of the curve with the x-axis, we need to solve the equation \(y(x) = 0\) for \(x\): $$ \frac{17}{15} - \cosh x = 0 $$ Add \(\cosh x\) to both sides: $$ \cosh x = \frac{17}{15} $$ Now, we can use the inverse hyperbolic cosine function to find \(x\): $$ x = \pm\cosh^{-1}\left(\frac{17}{15}\right) $$
02

Set up the integral for the average height

To find the average height of the curve, we can use the following formula: $$ \text{Average height} = \frac{1}{\cosh^{-1}\left(\frac{17}{15}\right) - (-\cosh^{-1}\left(\frac{17}{15}\right))} \int_{-\cosh^{-1}\left(\frac{17}{15}\right)}^{\cosh^{-1}\left(\frac{17}{15}\right)} y(x) \, dx $$ where \(y(x) = \frac{17}{15} - \cosh x\).
03

Compute the average height

Now, we need to compute the integral and evaluate it: $$ \text{Average height}= \frac{1}{2\cosh^{-1}\left(\frac{17}{15}\right)} \int_{-\cosh^{-1}\left(\frac{17}{15}\right)}^{\cosh^{-1}\left(\frac{17}{15}\right)} \left(\frac{17}{15} - \cosh x\right) \, dx $$ First, we separate the integral into two parts: $$ \text{Average height}= \frac{1}{2\cosh^{-1}\left(\frac{17}{15}\right)} \left[\int_{-\cosh^{-1}\left(\frac{17}{15}\right)}^{\cosh^{-1}\left(\frac{17}{15}\right)} \frac{17}{15} \, dx - \int_{-\cosh^{-1}\left(\frac{17}{15}\right)}^{\cosh^{-1}\left(\frac{17}{15}\right)} \cosh x \, dx\right] $$ Now, integrate each term: 1st term: \(\frac{17}{15}x\), which evaluates to \(\frac{17}{15}[\cosh^{-1}\left(\frac{17}{15}\right) - (-\cosh^{-1}\left(\frac{17}{15}\right))]\) when computing the difference. 2nd term: \(\sinh x\), which evaluates to \(0\) since the hyperbolic sine function has odd symmetry, and we are integrating over a symmetric interval around the origin. We then obtain: $$ \text{Average height} = \frac{1}{2\cosh^{-1}\left(\frac{17}{15}\right)}\cdot \frac{17}{15}\cdot 2\cosh^{-1}\left(\frac{17}{15}\right) = \frac{17}{15} $$ So, the average height of the catenary arch above the x-axis is \(\frac{17}{15}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Catenary Arch
A catenary arch is a curve that resembles the shape of a hanging chain or cable when supported at its ends and subject to a uniform gravitational field. This shape is mathematically represented by a hyperbolic cosine function. In the provided exercise, the function \( y = \frac{17}{15} - \cosh x \) defines such an arch.
The parameter \( \frac{17}{15} \) in the function represents a vertical shift upwards, meaning the entire arch is raised by this amount over the standard catenary based on the \( \cosh \) function. This upward shift ensures that the entire arch remains above the \( x \)-axis. The characteristics of a catenary arch make it an ideal shape for architectural structures, as seen in bridges and arches, because it naturally distributes tension throughout its length. The unique property of a catenary arch is its ability to support its own weight effectively, minimizing the need for additional structural support.
Understanding the geometric properties of catenary arches allows engineers and architects to utilize their natural strength and stability in construction projects.
Hyperbolic Functions
Hyperbolic functions are closely analogous to trigonometric functions but are based on hyperbolas rather than circles. The hyperbolic cosine function, represented as \( \cosh x \), plays a central role in defining catenary arches. It is important to note that \( \cosh x \) is an even function, meaning it is symmetric about the \( y \)-axis.
The inverse hyperbolic cosine function, \( \cosh^{-1}(x) \), is used to find specific values of \( x \) for a given \( \cosh x \). This concept was critical in our problem for determining the intersection points of the catenary with the \( x \)-axis. These functions prove useful not just in calculus but in various fields such as physics and engineering, especially when modeling growth patterns or wave formations.
Hyperbolic functions have applications that extend beyond the pure mathematics domain: they describe the shapes of hanging cables, the distribution of electrical fields, and even some social phenomena models. Thus, grasping these functions enriches understanding across multiple scientific disciplines.
Average Height Calculation
The average height of a curve above the \( x \)-axis is determined using integral calculus. In the exercise, this was done for the catenary arch defined by \( y = \frac{17}{15} - \cosh x \). The average value of a function over an interval is calculated by integrating the function across that interval and dividing by the length of the interval.
The specific formula used is: \[ \text{Average height} = \frac{1}{b-a} \int_{a}^{b} y(x) \, dx \]
In the solution, this was applied with the intersection points derived from the \( \cosh^{-1} \) function to form the interval limits \([-\cosh^{-1}(\frac{17}{15}), \cosh^{-1}(\frac{17}{15})]\). The even symmetry of the interval and the function simplified calculations, as demonstrated by the cancellation of the \( \sinh x \) term integration.
Calculating an average height gives a meaningful insight into the general behavior of a curve over an interval, summarizing its spread above or below the axis, crucial for applications ranging from statistics to physics, where understanding the 'average' or expected value is often more insightful than discrete points alone.

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