Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Problem Statement
Find the exact length of the curve.
Given the function:
\[ y = 1 + \frac{1}{8} \cosh(8x), \]
with the interval:
\[ 0 \leq x \leq 1 \]
### Solution
To find the length of a curve defined by \( y = f(x) \) from \( x = a \) to \( x = b \), we use the formula for the arc length:
\[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]
Given the function \( y = 1 + \frac{1}{8} \cosh(8x) \):
1. First, calculate the derivative \( \frac{dy}{dx} \):
\[ f(x) = 1 + \frac{1}{8} \cosh(8x) \]
\[ \frac{dy}{dx} = \frac{d}{dx}\left(1 + \frac{1}{8} \cosh(8x)\right) \]
Using the chain rule:
\[ \frac{dy}{dx} = \frac{1}{8} \cdot 8 \sinh(8x) \]
\[ \frac{dy}{dx} = \sinh(8x) \]
2. Plug the derivative into the arc length formula:
\[ L = \int_0^1 \sqrt{1 + (\sinh(8x))^2} \, dx \]
3. Recall the identity:
\[ 1 + \sinh^2(u) = \cosh^2(u) \]
4. Apply the identity to the integral:
\[ L = \int_0^1 \sqrt{\cosh^2(8x)} \, dx \]
\[ L = \int_0^1 \cosh(8x) \, dx \]
5. The integral of \( \cosh(kx) \) is \( \frac{1}{k} \sinh(kx) \):
\[ L = \int_0^1 \cosh(8x) \, dx \]
\[ L = \frac{1}{8} \sinh(8x) \Bigg|_0^1 = \frac{1}{8} [\sinh](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F06f55c03-0ff2-4f1b-a27d-6e1bf7c58dcc%2F664a8e81-38c8-45ff-a947-ead79e488dc3%2Frvxs8po_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Find the exact length of the curve.
Given the function:
\[ y = 1 + \frac{1}{8} \cosh(8x), \]
with the interval:
\[ 0 \leq x \leq 1 \]
### Solution
To find the length of a curve defined by \( y = f(x) \) from \( x = a \) to \( x = b \), we use the formula for the arc length:
\[ L = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]
Given the function \( y = 1 + \frac{1}{8} \cosh(8x) \):
1. First, calculate the derivative \( \frac{dy}{dx} \):
\[ f(x) = 1 + \frac{1}{8} \cosh(8x) \]
\[ \frac{dy}{dx} = \frac{d}{dx}\left(1 + \frac{1}{8} \cosh(8x)\right) \]
Using the chain rule:
\[ \frac{dy}{dx} = \frac{1}{8} \cdot 8 \sinh(8x) \]
\[ \frac{dy}{dx} = \sinh(8x) \]
2. Plug the derivative into the arc length formula:
\[ L = \int_0^1 \sqrt{1 + (\sinh(8x))^2} \, dx \]
3. Recall the identity:
\[ 1 + \sinh^2(u) = \cosh^2(u) \]
4. Apply the identity to the integral:
\[ L = \int_0^1 \sqrt{\cosh^2(8x)} \, dx \]
\[ L = \int_0^1 \cosh(8x) \, dx \]
5. The integral of \( \cosh(kx) \) is \( \frac{1}{k} \sinh(kx) \):
\[ L = \int_0^1 \cosh(8x) \, dx \]
\[ L = \frac{1}{8} \sinh(8x) \Bigg|_0^1 = \frac{1}{8} [\sinh
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