Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Related questions
Question
![### 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:](https://content.bartleby.com/qna-images/question/dc22955d-cca8-418b-8553-dddb50b2093d/0fad7fa5-794e-419f-acf0-4c945154c6fa/mejizep.png)
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:
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution
Trending nowThis is a popular solution!
Step by stepSolved in 2 steps with 6 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.Similar questions
Recommended textbooks for you
- Calculus: Early TranscendentalsCalculusISBN:9781285741550Author:James StewartPublisher:Cengage Learning
Thomas' Calculus (14th Edition)CalculusISBN:9780134438986Author:Joel R. Hass, Christopher E. Heil, Maurice D. WeirPublisher:PEARSON
Calculus: Early Transcendentals (3rd Edition)CalculusISBN:9780134763644Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric SchulzPublisher:PEARSON 
Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. Freeman
Calculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning
Calculus: Early Transcendentals
Calculus
ISBN:9781285741550
Author:James Stewart
Publisher:Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:9780134438986
Author:Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:9780134763644
Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:9781319050740
Author:Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:9781337552516
Author:Ron Larson, Bruce H. Edwards
Publisher:Cengage Learning