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 length of the curve given by \( x = \frac{y^2}{4} \), \(0 \leq y \leq 2\). State the integration method used to evaluate the integral.
(If applicable, you may use the integral formula for \( \sec^3 x \) derived in Trigonometric Integrals example video.)
### Explanation and Steps
1. **Setting up the Arc Length Formula**:
The formula for the arc length of a curve given in terms of \( y \) is:
\[
L = \int_{a}^{b} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy
\]
2. **Finding the Derivative**:
The given function is \( x = \frac{y^2}{4} \). First, we find \( \frac{dx}{dy} \):
\[
\frac{dx}{dy} = \frac{d}{dy} \left( \frac{y^2}{4} \right) = \frac{2y}{4} = \frac{y}{2}
\]
3. **Substituting the Derivative into the Formula**:
Substitute \( \frac{dx}{dy} = \frac{y}{2} \) into the arc length formula:
\[
L = \int_{0}^{2} \sqrt{1 + \left(\frac{y}{2}\right)^2} \, dy
\]
Simplify inside the square root:
\[
L = \int_{0}^{2} \sqrt{1 + \frac{y^2}{4}} \, dy
\]
4. **Integral Evaluation Method**:
To evaluate the integral, we can use a trigonometric substitution. Let \( y = 2 \tan(\theta) \). Therefore, \( dy = 2 \sec^2(\theta) \, d\theta \).
Simplify the limits of integration. When \( y = 0 \), \( \theta = 0 \). When \( y = 2 \), \( 2 = 2 \tan(\theta) \Rightarrow \tan(\theta) = 1 \Rightarrow \theta = \frac{\pi}{4} \).
Now the integral becomes:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdc22955d-cca8-418b-8553-dddb50b2093d%2F0fad7fa5-794e-419f-acf0-4c945154c6fa%2Fmejizep.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Find the length of the curve given by \( x = \frac{y^2}{4} \), \(0 \leq y \leq 2\). State the integration method used to evaluate the integral.
(If applicable, you may use the integral formula for \( \sec^3 x \) derived in Trigonometric Integrals example video.)
### Explanation and Steps
1. **Setting up the Arc Length Formula**:
The formula for the arc length of a curve given in terms of \( y \) is:
\[
L = \int_{a}^{b} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy
\]
2. **Finding the Derivative**:
The given function is \( x = \frac{y^2}{4} \). First, we find \( \frac{dx}{dy} \):
\[
\frac{dx}{dy} = \frac{d}{dy} \left( \frac{y^2}{4} \right) = \frac{2y}{4} = \frac{y}{2}
\]
3. **Substituting the Derivative into the Formula**:
Substitute \( \frac{dx}{dy} = \frac{y}{2} \) into the arc length formula:
\[
L = \int_{0}^{2} \sqrt{1 + \left(\frac{y}{2}\right)^2} \, dy
\]
Simplify inside the square root:
\[
L = \int_{0}^{2} \sqrt{1 + \frac{y^2}{4}} \, dy
\]
4. **Integral Evaluation Method**:
To evaluate the integral, we can use a trigonometric substitution. Let \( y = 2 \tan(\theta) \). Therefore, \( dy = 2 \sec^2(\theta) \, d\theta \).
Simplify the limits of integration. When \( y = 0 \), \( \theta = 0 \). When \( y = 2 \), \( 2 = 2 \tan(\theta) \Rightarrow \tan(\theta) = 1 \Rightarrow \theta = \frac{\pi}{4} \).
Now the integral becomes:
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