9-24 Find the exact length of the curve. 9. y = x³/2, 0≤x≤2 10. y = (x + 4) ³/2, 11. y = (1 + x²) ³/², 12. 36y² = (x² − 4)³, 13. y = 14. x = 3 talo 14 8 + + - 4x' 1 4y²² 0≤x≤ 4 0≤x≤ 1 2≤x≤ 3, 2 ≤ x ≤ 3, y ≥0 1≤x≤2 1≤ y ≤2 15. yIn(sin 2x), π/8 ≤ x ≤ π/6 16. y = ln(cos x), 0≤x≤ π/3

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...
icon
Related questions
Question
9,13
**Section 9–24: Curve Length Calculation**

In this section, we'll find the exact length of each given curve. The curves specified are provided along with their respective intervals of \( x \) or \( y \). Each function and its interval are listed below:

1. **Curve 9**: 
   \[ y = \frac{2}{3}x^{3/2}, \quad 0 \leq x \leq 2 \]

2. **Curve 10**: 
   \[ y = (x + 4)^{3/2}, \quad 0 \leq x \leq 4 \]

3. **Curve 11**: 
   \[ y = \frac{2}{3}(1 + x^2)^{3/2}, \quad 0 \leq x \leq 1 \]

4. **Curve 12**: 
   \[ 36y^2 = (x^2 - 4)^3, \quad 2 \leq x \leq 3, \quad y \geq 0 \]

5. **Curve 13**: 
   \[ y = \frac{x^3}{3} + \frac{1}{4x}, \quad 1 \leq x \leq 2 \]

6. **Curve 14**: 
   \[ x = \frac{y^4}{8} + \frac{1}{4y^2}, \quad 1 \leq y \leq 2 \]

7. **Curve 15**: 
   \[ y = \frac{1}{2} \ln(\sin 2x), \quad \frac{\pi}{8} \leq x \leq \frac{\pi}{6} \]

8. **Curve 16**: 
   \[ y = \ln(\cos x), \quad 0 \leq x \leq \frac{\pi}{3} \]

### Explanation of Tasks:

1. **Integral Form for Arc Length**:
   The exact length \( L \) of a curve given by \( y = f(x) \) over an interval \( [a, b] \) is calculated using the formula:
   \[
   L = \int_{a}^{b} \sqrt{1 + \left(\frac{
Transcribed Image Text:**Section 9–24: Curve Length Calculation** In this section, we'll find the exact length of each given curve. The curves specified are provided along with their respective intervals of \( x \) or \( y \). Each function and its interval are listed below: 1. **Curve 9**: \[ y = \frac{2}{3}x^{3/2}, \quad 0 \leq x \leq 2 \] 2. **Curve 10**: \[ y = (x + 4)^{3/2}, \quad 0 \leq x \leq 4 \] 3. **Curve 11**: \[ y = \frac{2}{3}(1 + x^2)^{3/2}, \quad 0 \leq x \leq 1 \] 4. **Curve 12**: \[ 36y^2 = (x^2 - 4)^3, \quad 2 \leq x \leq 3, \quad y \geq 0 \] 5. **Curve 13**: \[ y = \frac{x^3}{3} + \frac{1}{4x}, \quad 1 \leq x \leq 2 \] 6. **Curve 14**: \[ x = \frac{y^4}{8} + \frac{1}{4y^2}, \quad 1 \leq y \leq 2 \] 7. **Curve 15**: \[ y = \frac{1}{2} \ln(\sin 2x), \quad \frac{\pi}{8} \leq x \leq \frac{\pi}{6} \] 8. **Curve 16**: \[ y = \ln(\cos x), \quad 0 \leq x \leq \frac{\pi}{3} \] ### Explanation of Tasks: 1. **Integral Form for Arc Length**: The exact length \( L \) of a curve given by \( y = f(x) \) over an interval \( [a, b] \) is calculated using the formula: \[ L = \int_{a}^{b} \sqrt{1 + \left(\frac{
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 5 steps with 8 images

Blurred answer
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
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: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning