Chapter 18: Problem 7
Find the length of the curve \(y=\cosh x\) for \(-1 \leq x \leq 1\)
Short Answer
Expert verified
Answer: The arc length of the curve \(y = \cosh x\) for \(-1 \leq x \leq 1\) is \(\frac{1}{2} (e - e^{-1})\).
Step by step solution
01
Find the derivative of \(y = \cosh x\)
To find the derivative of \(y = \cosh x\), recall the definition of the hyperbolic cosine function:
$$
\cosh x = \frac{e^x + e^{-x}}{2}
$$
Now, differentiate this expression with respect to \(x\):
$$
\frac{d}{dx}\cosh x = \frac{d}{dx}\frac{e^x + e^{-x}}{2} = \frac{e^x - e^{-x}}{2}
$$
Thus, the derivative of \(y = \cosh x\) is \(\frac{e^x - e^{-x}}{2}\).
02
Calculate \(\left(\frac{dy}{dx}\right)^2\) and add 1
Now, we need to square the derivative found in Step 1 and add 1:
$$
1+\left(\frac{dy}{dx}\right)^2= 1+ \left(\frac{e^x - e^{-x}}{2}\right)^2=1+\frac{e^{2x} - 2+e^{-2x}}{4}
$$
03
Apply the Arc Length Formula
Now, we will use the Arc Length Formula with the values found in the previous steps:
$$
L = \int_{-1}^{1} \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx = \int_{-1}^{1} \sqrt{1+\frac{e^{2x} - 2+e^{-2x}}{4}} dx
$$
04
Solve the Integral
To solve the integral, use substitution:
$$
u = e^x \Rightarrow du = e^x dx
$$
Now, we can rewrite the integral with respect to \(u\):
$$
L = \int_{e^{-1}}^{e} \sqrt{1+\frac{u^2 - 2u^2 +1}{4}} \frac{du}{u}
$$
The integral can now be simplified to:
$$
L = \int_{e^{-1}}^{e} \frac{u}{2} \frac{du}{u} = \frac{1}{2} \int_{e^{-1}}^{e} du
$$
Now, integrate with respect to \(u\):
$$
L = \frac{1}{2} [u]_{e^{-1}}^{e} = \frac{1}{2} (e - e^{-1})
$$
Thus, the length of the curve \(y = \cosh x\) for \(-1 \leq x \leq 1\) is \(\frac{1}{2} (e - e^{-1})\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Hyperbolic Functions
Hyperbolic functions are analogs of the ordinary trigonometric, or circular, functions. They are named as such because they are defined based on segments of a hyperbola, much like trigonometric functions are defined using parts of a circle.
The two fundamental hyperbolic functions are the hyperbolic sine and cosine. The hyperbolic cosine, denoted as \(\cosh x\), is defined as \(\cosh x = \frac{e^x + e^{-x}}{2}\), which can be recognized as the average of the exponential functions \(e^x\) and \(e^{-x}\). It is important to note that \(\cosh x\) is an even function, it exhibits symmetry about the y-axis, which can be a very useful property in simplifying certain types of integrals.
The derivative of the hyperbolic cosine \(\cosh x\) is the hyperbolic sine, denoted as \(\sinh x\), and can be expressed as \(\frac{d}{dx}\cosh x = \sinh x = \frac{e^x - e^{-x}}{2}\). Just like their trigonometric counterparts, hyperbolic functions satisfy a number of identities that are very useful in calculus. For example, \(\cosh^2 x - \sinh^2 x = 1\), resembling the famous Pythagorean identity \(\cos^2 x + \sin^2 x = 1\) of trigonometry.
The two fundamental hyperbolic functions are the hyperbolic sine and cosine. The hyperbolic cosine, denoted as \(\cosh x\), is defined as \(\cosh x = \frac{e^x + e^{-x}}{2}\), which can be recognized as the average of the exponential functions \(e^x\) and \(e^{-x}\). It is important to note that \(\cosh x\) is an even function, it exhibits symmetry about the y-axis, which can be a very useful property in simplifying certain types of integrals.
The derivative of the hyperbolic cosine \(\cosh x\) is the hyperbolic sine, denoted as \(\sinh x\), and can be expressed as \(\frac{d}{dx}\cosh x = \sinh x = \frac{e^x - e^{-x}}{2}\). Just like their trigonometric counterparts, hyperbolic functions satisfy a number of identities that are very useful in calculus. For example, \(\cosh^2 x - \sinh^2 x = 1\), resembling the famous Pythagorean identity \(\cos^2 x + \sin^2 x = 1\) of trigonometry.
Integral Calculus
Integral calculus is a branch of calculus focused on the accumulation of quantities and the areas between curves. When you want to find the arc length of a curve, integral calculus provides the tools to sum up an infinite number of infinitesimally small lengths to find a finite total length.
The process involves setting up an integral that extends over the range of the curve you are interested in. In the case of finding the length of \(y = \cosh x\), you would set up the integral using the formula for arc length, involving the square root of the sum of 1 and the square of the derivative of \(y\) with respect to \(x\).
The integral \(\int_{-1}^{1} \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx\) was evaluated by simplifying the terms under the square root and making a substitution to reformulate the integral in terms of a new variable, \(u\), which makes the integral easier to solve. The approach and techniques in integral calculus are essential for calculating many physical and geometric quantities, such as lengths, areas, and volumes.
The process involves setting up an integral that extends over the range of the curve you are interested in. In the case of finding the length of \(y = \cosh x\), you would set up the integral using the formula for arc length, involving the square root of the sum of 1 and the square of the derivative of \(y\) with respect to \(x\).
The integral \(\int_{-1}^{1} \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx\) was evaluated by simplifying the terms under the square root and making a substitution to reformulate the integral in terms of a new variable, \(u\), which makes the integral easier to solve. The approach and techniques in integral calculus are essential for calculating many physical and geometric quantities, such as lengths, areas, and volumes.
Derivative Calculus
Derivative calculus, also known simply as differentiation, involves finding the rate at which a function is changing at any given point. It is critical in determining the slope of a function at a point, and is thus a key concept in finding the arc length of a curve.
In our exercise, the first step is to find the derivative of the curve \(y = \cosh x\). The result of \(\frac{d}{dx}\cosh x = \sinh x\) indicates the slope of the curve at any point \(x\). However, for the arc length formula, we need \(\left(\frac{dy}{dx}\right)^2\), which is the square of the slope of the curve.
This expression, alongside adding 1 to it, forms the integrand of the arc length integral. It represents an infinitesimal length element of the curve, which is why its square root is taken. The integration of these length elements over the chosen interval results in the total length of the segment of the curve. Derivative calculus thereby contributes to a deeper understanding of the curvature and dynamics of functions, allowing us to calculate properties like arc length.
In our exercise, the first step is to find the derivative of the curve \(y = \cosh x\). The result of \(\frac{d}{dx}\cosh x = \sinh x\) indicates the slope of the curve at any point \(x\). However, for the arc length formula, we need \(\left(\frac{dy}{dx}\right)^2\), which is the square of the slope of the curve.
This expression, alongside adding 1 to it, forms the integrand of the arc length integral. It represents an infinitesimal length element of the curve, which is why its square root is taken. The integration of these length elements over the chosen interval results in the total length of the segment of the curve. Derivative calculus thereby contributes to a deeper understanding of the curvature and dynamics of functions, allowing us to calculate properties like arc length.
