/*! 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 54 Find the arc length of the caten... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the arc length of the catenary \(y=a \cosh (x / a)\) between \(x=0\) and \(x=x_{1}\left(x_{1}>0\right)\)

Short Answer

Expert verified
The arc length is \( a \sinh(x_1/a) \).

Step by step solution

01

Understanding the Problem

We need to find the length of the arc of the catenary curve given by the equation \( y = a \cosh(x/a) \) between the points \( x = 0 \) and \( x = x_1 \).
02

Arc Length Formula

The formula for the arc length \( L \) of a curve \( y = f(x) \) from \( x = a \) to \( x = b \) is \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2 } \, dx .\]
03

Finding the Derivative

Calculate the derivative of \( y = a \cosh(x/a) \) with respect to \( x \). The derivative is \[ \frac{dy}{dx} = \frac{d}{dx}[ a \cosh(x/a) ] = \sinh(x/a) .\]
04

Substitute and Simplify

Substitute the derivative into the arc length formula: \[ L = \int_{0}^{x_1} \sqrt{1 + (\sinh(x/a))^2 } \, dx .\]Using the identity \( \cosh^2(t) - \sinh^2(t) = 1 \), we have \[ 1 + \sinh^2(x/a) = \cosh^2(x/a) .\] Thus, the expression becomes \[ \int_{0}^{x_1} \cosh(x/a) \, dx .\]
05

Calculate the Integral

Solve the integral \[ \int \cosh(x/a) \, dx .\]The integral of \( \cosh(x/a) \) with respect to \( x \) is \[ a \sinh(x/a) .\]
06

Evaluate the Integral

Evaluate the integral from \( x = 0 \) to \( x = x_1 \) using the result:\[ L = a \sinh(x/a) \bigg|_{0}^{x_1} = a \sinh(x_1/a) - a \sinh(0).\]Since \( \sinh(0) = 0 \), the final result is \[ L = a \sinh(x_1/a) .\]

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Key Concepts

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

Catenary Curve
The term "catenary curve" might sound a bit complex at first, but it's a very interesting and important concept. A catenary is the shape a flexible chain or cable assumes under its own weight when supported at its ends. While it might resemble a parabola, it is actually its own unique shape described by the equation \( y = a \cosh(x/a) \). This equation uses the hyperbolic cosine function, which we will discuss later.

The catenary curve is significant in various fields such as architecture and physics. For instance, the design of suspension bridges and power lines often considers catenary shapes for efficiency in handling weight and minimizing stress. Understanding this shape therefore provides insight into how natural and man-made structures distribute forces efficiently.
Derivative Calculation
Calculating the derivative of a function helps us understand how the function behaves, especially in terms of its slope or rate of change. In this exercise, we have the function \( y = a \cosh(x/a) \). Our task is to find its derivative with respect to \( x \).

This is done by applying the derivative rules to the hyperbolic cosine function. The derivative of \( \cosh(u) \) is \( \sinh(u) \), multiplied by the derivative of the inner function if needed. Therefore, the derivative of \( y = a \cosh(x/a) \) is \( \frac{dy}{dx} = \sinh(x/a) \).

Knowing the derivative is crucial for further calculations, such as finding the arc length, and it provides insight into the curve's steepness at any given point.
Arc Length Formula
The arc length formula is a tool used to determine the length of a curve between two points. For a curve defined by the function \( y = f(x) \), the arc length \( L \) from \( x = a \) to \( x = b \) is given by the integral:

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

This formula calculates the length by summing the infinitesimal segments along the curve using calculus. In the context of our problem, it involves substituting the derivative we found earlier to compute the exact length of the catenary curve from \( x = 0 \) to \( x = x_1 \).

To achieve a simplified form, it's useful to recognize hyperbolic identities, which help transform the integrand into a more manageable expression for evaluation.
Hyperbolic Functions
Hyperbolic functions are analogs of trigonometric functions but for a hyperbola, similar to how trigonometric functions relate to circles. The functions \( \cosh \) (hyperbolic cosine) and \( \sinh \) (hyperbolic sine) are the most commonly used.

They are defined as:
  • \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)
  • \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
The relationship \( \cosh^2(x) - \sinh^2(x) = 1 \) is akin to the identity \( \cos^2(x) + \sin^2(x) = 1 \) in trigonometry. This identity is imperative in simplifying expressions involving hyperbolic functions.

Understanding these functions is crucial, as they are pervasive in many calculus problems, including those involving catenaries, and often appear in solutions related to complex curves and areas of physics.

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Most popular questions from this chapter

Prove: $$ \begin{array}{ll} \operatorname{sech}^{-1} x=\cosh ^{-1}(1 / x), & 0 < x \leq 1 \\ \operatorname{coth}^{-1} x=\tanh ^{-1}(1 / x), & |x| > 1 \\ \operatorname{csch}^{-1} x=\sinh ^{-1}(1 / x), & x \neq 0 \end{array} $$

Find \(d y / d x\). $$y=\operatorname{sech}\left(e^{2 x}\right)$$

Suppose that a hollow tube rotates with a constant angular velocity of \(\omega\) rad/s about a horizontal axis at one end of the tube, as shown in the accompanying figure. Assume that an object is free to slide without friction in the tube while the tube is rotating. Let \(r\) be the distance from the object to the pivot point at time \(t \geq 0,\) and assume that the object is at rest and \(r=0\) when \(t=0 .\) It can be shown that if the tube is horizontal at time \(t=0\) and rotating as shown in the figure, then $$r=\frac{g}{2 \omega^{2}}[\sinh (\omega t)-\sin (\omega t)]$$ during the period that the object is in the tube. Assume that \(t\) is in seconds and \(r\) is in meters, and use \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) and \(\omega=2 \mathrm{rad} / \mathrm{s}\) (a) Graph \(r\) versus \(t\) for \(0 \leq t \leq 1\) (b) Assuming that the tube has a length of \(1 \mathrm{m},\) approximately how long does it take for the object to reach the end of the tube? (c) Use the result of part (b) to approximate \(d r / d t\) at the instant that the object reaches the end of the tube. (GRAPH CAN'T COPY).

Determine whether the statement is true or false. Explain your answer. Exactly two of the hyperbolic functions are bounded.

Use a CAS to find the mass and center of gravity of the lamina with density \(\delta .\) A lamina bounded by the graph of \(y=\ln x,\) the \(x\) -axis, and the line \(x=2 ; \delta=1\).

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