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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Question
In terms of parametric derivatives, how can you determine when a parametric curve has a horizontal tangent line? Similarly, how can you determine when the tangent line will be vertical?
![**Problem Statement:**
Find the exact length of the curve.
\[ y = \ln(\sec(x)), \quad 0 \leq x \leq \frac{\pi}{3} \]
**Explanation:**
To find the exact length of the given curve \( y = \ln(\sec(x)) \) over the interval from \( x = 0 \) to \( x = \frac{\pi}{3} \), you need to 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. **Function and Derivative:**
First, find the derivative of \( y = \ln(\sec(x)) \).
Since \(\sec(x) = \frac{1}{\cos(x)}\), we get:
\[ y = \ln(\sec(x)) = \ln\left(\frac{1}{\cos(x)}\right) = -\ln(\cos(x)) \]
Using the chain rule:
\[ \frac{dy}{dx} = -\frac{1}{\cos(x)} \cdot (-\sin(x)) = \tan(x) \]
2. **Arc Length Formula:**
Substitute \(\frac{dy}{dx} = \tan(x)\) into the arc length formula:
\[ L = \int_{0}^{\frac{\pi}{3}} \sqrt{1 + \tan^2(x)} \, dx \]
3. **Simplify the Integral:**
Since \( 1 + \tan^2(x) = \sec^2(x) \), the integral becomes:
\[ L = \int_{0}^{\frac{\pi}{3}} \sqrt{\sec^2(x)} \, dx = \int_{0}^{\frac{\pi}{3}} \sec(x) \, dx \]
4. **Evaluate the Integral:**
The integral of \(\sec(x)\) is:
\[ \int \sec(x) \, dx = \ln |\sec(x) + \tan(x)| + C \]
Therefore:
\[ L = \left. \ln |\sec(x) + \tan(x)| \right|](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa19899d9-b085-44d6-ac05-6544bc515fab%2Ff808eb22-0e74-4237-9499-4c85b44cd2b5%2Fjy5udkk_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the exact length of the curve.
\[ y = \ln(\sec(x)), \quad 0 \leq x \leq \frac{\pi}{3} \]
**Explanation:**
To find the exact length of the given curve \( y = \ln(\sec(x)) \) over the interval from \( x = 0 \) to \( x = \frac{\pi}{3} \), you need to 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. **Function and Derivative:**
First, find the derivative of \( y = \ln(\sec(x)) \).
Since \(\sec(x) = \frac{1}{\cos(x)}\), we get:
\[ y = \ln(\sec(x)) = \ln\left(\frac{1}{\cos(x)}\right) = -\ln(\cos(x)) \]
Using the chain rule:
\[ \frac{dy}{dx} = -\frac{1}{\cos(x)} \cdot (-\sin(x)) = \tan(x) \]
2. **Arc Length Formula:**
Substitute \(\frac{dy}{dx} = \tan(x)\) into the arc length formula:
\[ L = \int_{0}^{\frac{\pi}{3}} \sqrt{1 + \tan^2(x)} \, dx \]
3. **Simplify the Integral:**
Since \( 1 + \tan^2(x) = \sec^2(x) \), the integral becomes:
\[ L = \int_{0}^{\frac{\pi}{3}} \sqrt{\sec^2(x)} \, dx = \int_{0}^{\frac{\pi}{3}} \sec(x) \, dx \]
4. **Evaluate the Integral:**
The integral of \(\sec(x)\) is:
\[ \int \sec(x) \, dx = \ln |\sec(x) + \tan(x)| + C \]
Therefore:
\[ L = \left. \ln |\sec(x) + \tan(x)| \right|
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