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When the catenary \(y=a \cosh (x / a)\) is rotated around the \(x\) -axis, it sweeps out a surface of revolution called a catenoid. Find the area of the surface generated when \(y=\cosh x\) on \([-\ln 2, \ln 2]\) is rotated around the \(x\) -axis.

Short Answer

Expert verified
Answer: The surface area of the catenoid is given by the formula $A = \pi (2 \ln 2 + \sinh 2 \ln 2)$.

Step by step solution

01

Differentiate the function

We first find the derivative of the given function \(y = \cosh x\) with respect to \(x\): \(\frac{dy}{dx} = \frac{d}{dx} (\cosh x) = \sinh x\)
02

Square the derivative

Next, square the derivative of the function we found: \((\frac{dy}{dx})^2 = (\sinh x)^2\)
03

Add 1 and find the square root

Now, add 1 to the squared derivative and find the square root of the result: \(\sqrt{1 + (\frac{dy}{dx})^2} = \sqrt{1 + (\sinh x)^2}\)
04

Simplify the square root expression

Recall the identity \(\cosh^2 x - \sinh^2 x = 1\). From this, we can conclude: \(\sqrt{1 + (\sinh x)^2} = \sqrt{\cosh^2 x} = \cosh x\)
05

Set up the integral

Now we can set up the integral for the surface area of the catenoid using the formula \(A = 2\pi \int_a^b y \sqrt{1 + (\frac{dy}{dx})^2} dx\) with \(a = -\ln 2\), \(b = \ln 2\), and \(y = \cosh x\): \(A = 2\pi \int_{-\ln 2}^{\ln 2} \cosh x \cdot \cosh x dx\)
06

Simplify the integral

Combine the function within the integral: \(A = 2\pi \int_{-\ln 2}^{\ln 2} \cosh^2 x dx\)
07

Solve the integral

To compute the integral of \(\cosh^2 x\), we use the identity \(\cosh^2 x = \frac{1 + \cosh 2x}{2}\). \(A = 2\pi \int_{-\ln 2}^{\ln 2} \frac{1 + \cosh 2x}{2} dx = \pi \int_{-\ln 2}^{\ln 2} (1 + \cosh 2x) dx\) Now, integrate: \(A = \pi [x + \frac{1}{2} \sinh 2x]_{-\ln 2}^{\ln 2}\)
08

Evaluate the integral

Finally, evaluate the integral by plugging in the limits \(-\ln 2\) and \(\ln 2\): \(A = \pi [( \ln 2 + \frac{1}{2} \sinh 2 \ln 2) - (-\ln 2 + \frac{1}{2} \sinh (-2 \ln 2))]\) Now, we can use the fact that \(\sinh(-x) = -\sinh x\), \(A = \pi [( \ln 2 + \frac{1}{2} \sinh 2 \ln 2) - (-\ln 2 - \frac{1}{2} \sinh 2 \ln 2)]\)
09

Simplify and find the final solution

Simplify the expression to get the final answer for the surface area of the catenoid: \(A = \pi (2 \ln 2 + \sinh 2 \ln 2)\)

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Key Concepts

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

Catenary
In calculus, the catenary is a fascinating curve. It represents the shape taken by a flexible chain or cable suspended between two points under its own weight. This curve can be described mathematically by the equation \(y = a \cosh (x/a)\), where \(a\) is a constant. The term "catenary" originates from the Latin word for chain, emphasizing its natural physical appearance.
When studying catenaries, it's intriguing to notice that their shape isn't parabolic, even though they might look similar. Understanding the properties of catenaries helps in solving real-world problems, such as designing bridges or telephone cables. These curves allow engineers to predict the distribution of tension in suspended structures accurately.
In this exercise, the function \(y = \cosh x\) represents a specific catenary, where \(a = 1\). By rotating such a curve around an axis, you can create a surface of revolution known as a catenoid, which exhibits minimal surface area properties, important in physics and engineering applications.
Surface of Revolution
The concept of a surface of revolution arises when a curve rotates around an axis, sweeping out a surface in three-dimensional space. In mathematics, these surfaces are useful for calculating properties like surface area or volume. When the catenary \(y = \cosh x\) is spun around the \(x\)-axis, it forms a catenoid, which is a classic example of a surface of revolution.
  • The surface area of revolution is calculated using integral calculus, specifically through the formula:
    \[A = 2\pi \int_a^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\]
  • In the problem at hand, the bounds \([-\ln 2, \ln 2]\) limit the section of the catenary that's undergoing rotation, and this determines the extent of the catenoid.
Real-life applications of surfaces of revolution are numerous. They appear in the design of objects like bottles, bells, and even scientific instruments. The minimal surface property of some revolution surfaces, like a catenoid, is particularly significant in architectural designs.
Integral Calculus
Integral calculus plays a crucial role in determining areas and volumes, especially for forms that can't be easily calculated using basic geometry. In this problem, it is employed to find the area of a catenoid, which is formed from a catenary curve rotated around an axis. The integral setup used to calculate the area involves a function, \(y = \cosh x\), and its derivative.
To solve for the surface area, integral calculus uses the following steps:
  • Differentiate the function to find \(\frac{dy}{dx}\).
  • Implement substitution rules or identities, like \(\cosh^2 x = \frac{1 + \cosh 2x}{2}\), to simplify the integration.
  • Evaluate definite integrals over set bounds, such as \([-\ln 2, \ln 2]\).
The above method provides the exact surface area by accounting for the continuous and smooth nature of the curve. Mastering these techniques in integral calculus is not only crucial for academic progress but also for tackling complex real-world engineering and physics problems.
Hyperbolic Functions
Hyperbolic functions, including \(\sinh x\) and \(\cosh x\), enhance mathematical analysis just like their trigonometric counterparts. They offer a framework for describing hyperbolic geometry and making complex calculations more manageable. In this exercise, we specifically focused on the hyperbolic cosine, \(\cosh x\), which defines a catenary.
  • The relationship between \(\sinh x\) and \(\cosh x\) is expressed as \(\cosh^2 x - \sinh^2 x = 1\), a key identity for simplifying expressions.
  • The derivatives of these functions—\(\frac{d}{dx}(\cosh x) = \sinh x\) and \(\frac{d}{dx}(\sinh x) = \cosh x\)—are essential to solving integral calculus problems involving rotational surfaces.

In many contexts, hyperbolic functions model real phenomena, such as electrical circuits or even Einstein's theory of relativity. By understanding their properties, you can readily apply them to various scientific and mathematical challenges.

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Most popular questions from this chapter

Calculate the work required to stretch the following springs 0.5 m from their equilibrium positions. Assume Hooke's law is obeyed. a. A spring that requires a force of \(50 \mathrm{N}\) to be stretched \(0.2 \mathrm{m}\) from its equilibrium position. b. A spring that requires \(50 \mathrm{J}\) of work to be stretched \(0.2 \mathrm{m}\) from its equilibrium position.

An inverted cone is \(2 \mathrm{m}\) high and has a base radius of \(\frac{1}{2} \mathrm{m}\). If the tank is full, how much work is required to pump the water to a level \(1 \mathrm{m}\) above the top of the tank?

When an object falling from rest encounters air resistance proportional to the square of its velocity, the distance it falls (in meters) after \(t\) seconds is given by \(d(t)=\frac{m}{k} \ln [\cosh (\sqrt{\frac{k g}{m}} t)],\) where \(m\) is the mass of the object in kilograms, \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, and \(k\) is a physical constant. a. A BASE jumper \((m=75 \mathrm{kg})\) leaps from a tall cliff and performs a ten-second delay (she free-falls for 10 s and then opens her chute). How far does she fall in \(10 \mathrm{s} ?\) Assume \(k=0.2\) b. How long does it take for her to fall the first \(100 \mathrm{m} ?\) The second 100 \(\mathrm{m} ?\) What is her average velocity over each of these intervals?

Verify the following identities. \(\sinh (x+y)=\sinh x \cosh y+\cosh x \sinh y\)

Use a calculator to make a table similar to Table 2 to approximate the following limits. Confirm your result with l'Hôpital's Rule. $$\lim _{x \rightarrow 0} \frac{\ln (1+x)}{x}$$

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