Calculate the arc :length of the curve y = In(2x + 1) over the interval

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
icon
Concept explainers
Question
**Arc Length Calculation for the Curve \( y = \ln(2x + 1) \)**

To find the arc length of the curve given by \( y = \ln(2x + 1) \) over the interval \(\left[\frac{1}{2}, \sqrt{3} - \frac{1}{2}\right]\), we will use the arc length formula for a function \( y = f(x) \):

\[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]

### Step-by-Step Solution

1. **Determine the derivative \( \frac{dy}{dx} \)**:

   Given \( y = \ln(2x + 1) \), let's find its derivative.

   Using the chain rule:
   \[ \frac{dy}{dx} = \frac{d}{dx} \ln(2x + 1) = \frac{1}{2x + 1} \cdot \frac{d}{dx} (2x + 1) = \frac{2}{2x + 1} \]

2. **Substitute \( \frac{dy}{dx} \) into the arc length formula**:

   Plugging the derivative into the formula, we get:
   \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{2}{2x + 1} \right)^2} \, dx \]

   Simplifying inside the square root:
   \[ L = \int_{a}^{b} \sqrt{1 + \frac{4}{(2x + 1)^2}} \, dx \]

   Combining the fractions:
   \[ L = \int_{a}^{b} \sqrt{\frac{(2x + 1)^2 + 4}{(2x + 1)^2}} \, dx \]
   \[ L = \int_{a}^{b} \frac{\sqrt{(2x + 1)^2 + 4}}{2x + 1} \, dx \]

3. **Evaluate the integral**:

   This integral can be challenging to solve directly, and often a substitution method or numerical approximation might be needed.

### Interval of Calculation

The interval of integration is:
\[ a =
Transcribed Image Text:**Arc Length Calculation for the Curve \( y = \ln(2x + 1) \)** To find the arc length of the curve given by \( y = \ln(2x + 1) \) over the interval \(\left[\frac{1}{2}, \sqrt{3} - \frac{1}{2}\right]\), we will use the arc length formula for a function \( y = f(x) \): \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \] ### Step-by-Step Solution 1. **Determine the derivative \( \frac{dy}{dx} \)**: Given \( y = \ln(2x + 1) \), let's find its derivative. Using the chain rule: \[ \frac{dy}{dx} = \frac{d}{dx} \ln(2x + 1) = \frac{1}{2x + 1} \cdot \frac{d}{dx} (2x + 1) = \frac{2}{2x + 1} \] 2. **Substitute \( \frac{dy}{dx} \) into the arc length formula**: Plugging the derivative into the formula, we get: \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{2}{2x + 1} \right)^2} \, dx \] Simplifying inside the square root: \[ L = \int_{a}^{b} \sqrt{1 + \frac{4}{(2x + 1)^2}} \, dx \] Combining the fractions: \[ L = \int_{a}^{b} \sqrt{\frac{(2x + 1)^2 + 4}{(2x + 1)^2}} \, dx \] \[ L = \int_{a}^{b} \frac{\sqrt{(2x + 1)^2 + 4}}{2x + 1} \, dx \] 3. **Evaluate the integral**: This integral can be challenging to solve directly, and often a substitution method or numerical approximation might be needed. ### Interval of Calculation The interval of integration is: \[ a =
Expert Solution
steps

Step by step

Solved in 5 steps with 4 images

Blurred answer
Knowledge Booster
Application of Integration
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.
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