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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.3.139

In Exercises 139–142, find the length of each curve.
139. y = (1/2)(e^x + e^(−x)) from x = 0 to x = 1.

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Recall the formula for the length of a curve defined by a function \(y = f(x)\) from \(x = a\) to \(x = b\): \[L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\]
Identify the function given: \[y = \frac{1}{2} \left(e^x + e^{-x}\right)\] and the interval: \[x \in [0, 1]\]
Compute the derivative \(\frac{dy}{dx}\): \[\frac{dy}{dx} = \frac{1}{2} \left(e^x - e^{-x}\right)\]
Substitute \(\frac{dy}{dx}\) into the arc length formula under the square root: \[\sqrt{1 + \left(\frac{1}{2} (e^x - e^{-x})\right)^2} = \sqrt{1 + \frac{(e^x - e^{-x})^2}{4}}\]
Simplify the expression inside the square root and set up the integral for the length: \[L = \int_0^1 \sqrt{1 + \frac{(e^x - e^{-x})^2}{4}} \, dx\] This integral can then be evaluated to find the length of the curve.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Arc Length Formula

The arc length of a curve y = f(x) from x = a to x = b is found using the integral formula L = ∫_a^b √(1 + (dy/dx)^2) dx. This formula calculates the length by summing infinitesimal line segments along the curve.
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Arc Length of Parametric Curves

Derivative of Exponential Functions

To apply the arc length formula, you need the derivative dy/dx. For functions involving e^x and e^(-x), the derivative rules state that d/dx(e^x) = e^x and d/dx(e^(-x)) = -e^(-x). Understanding these helps compute dy/dx accurately.
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Derivatives of General Exponential Functions

Simplification of Expressions Involving Hyperbolic Functions

The given function y = (1/2)(e^x + e^(-x)) is the hyperbolic cosine function, cosh(x). Recognizing this allows simplification of derivatives and integrals, as cosh^2(x) - sinh^2(x) = 1, which can simplify the arc length integral.
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Integrals Resulting in Inverse Trig Functions: Substitution Example 5