Find the equation of the curve that passes through the point (0, 3) and has a slope of 2 x+1 O y = ln lxl +3 O y = -2 ln lx - 11 +3 O y = 2 ln |x + 1| O y = 2 ln |x + 1| +3 at any point (x, y).
Find the equation of the curve that passes through the point (0, 3) and has a slope of 2 x+1 O y = ln lxl +3 O y = -2 ln lx - 11 +3 O y = 2 ln |x + 1| O y = 2 ln |x + 1| +3 at any point (x, y).
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...
Related questions
Question
![Title: Determining the Equation of a Curve
---
**Problem:**
Find the equation of the curve that passes through the point \((0, 3)\) and has a slope of \(\frac{2}{x + 1}\) at any point \((x, y)\).
**Options:**
- \( y = \ln|x| + 3 \)
- \( y = -2 \ln|x - 1| + 3 \)
- \( y = 2 \ln|x + 1| \)
- \( y = 2 \ln|x + 1| + 3 \)
---
**Explanation:**
We are given a point \((0, 3)\) through which the curve passes and the slope \(\frac{2}{x + 1}\) at any point. The objective is to determine which of the given equations satisfies these conditions.
To determine the correct equation:
1. **Evaluate Initial Information**:
- Point: \((0, 3)\)
- Slope function: \(\frac{2}{x + 1}\)
2. **Analyze Slope Function**:
- The slope of a curve at any point \((x, y)\) is essentially the derivative of \(y\) with respect to \(x\).
- Thus, \(\frac{dy}{dx} = \frac{2}{x + 1}\).
3. **Integration to Find General Form of the Curve**:
Integrate \(\frac{dy}{dx} = \frac{2}{x + 1}\):
\[
\int dy = \int \frac{2}{x + 1} dx
\]
\[
y = 2 \ln|x + 1| + C
\]
4. **Apply Initial Condition (Point on the Curve)**:
- Use \((0, 3)\):
\[
3 = 2 \ln|0 + 1| + C
\]
\[
3 = 2 \ln(1) + C
\]
\[
3 = 0 + C
\]
\[
C = 3
\]
Thus, the equation of the curve is:
\[
y = 2 \ln|x + 1| + 3
\]
5.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcf75a45c-687d-4994-8001-f519eebb3c9c%2F64dd1ee1-0dcc-4d93-874e-e797400733c1%2Foddw1fg_processed.png&w=3840&q=75)
Transcribed Image Text:Title: Determining the Equation of a Curve
---
**Problem:**
Find the equation of the curve that passes through the point \((0, 3)\) and has a slope of \(\frac{2}{x + 1}\) at any point \((x, y)\).
**Options:**
- \( y = \ln|x| + 3 \)
- \( y = -2 \ln|x - 1| + 3 \)
- \( y = 2 \ln|x + 1| \)
- \( y = 2 \ln|x + 1| + 3 \)
---
**Explanation:**
We are given a point \((0, 3)\) through which the curve passes and the slope \(\frac{2}{x + 1}\) at any point. The objective is to determine which of the given equations satisfies these conditions.
To determine the correct equation:
1. **Evaluate Initial Information**:
- Point: \((0, 3)\)
- Slope function: \(\frac{2}{x + 1}\)
2. **Analyze Slope Function**:
- The slope of a curve at any point \((x, y)\) is essentially the derivative of \(y\) with respect to \(x\).
- Thus, \(\frac{dy}{dx} = \frac{2}{x + 1}\).
3. **Integration to Find General Form of the Curve**:
Integrate \(\frac{dy}{dx} = \frac{2}{x + 1}\):
\[
\int dy = \int \frac{2}{x + 1} dx
\]
\[
y = 2 \ln|x + 1| + C
\]
4. **Apply Initial Condition (Point on the Curve)**:
- Use \((0, 3)\):
\[
3 = 2 \ln|0 + 1| + C
\]
\[
3 = 2 \ln(1) + C
\]
\[
3 = 0 + C
\]
\[
C = 3
\]
Thus, the equation of the curve is:
\[
y = 2 \ln|x + 1| + 3
\]
5.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 2 images
Recommended textbooks for you
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:
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