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
![**Problem Statement:**
Find the length of the polar curve given by the equation:
\[ r = \sin\theta, \quad 0 \leq \theta \leq \pi \]
Solve by hand. No calculator. Show all work.
**Solution Outline:**
To find the length of the polar curve, we use the formula for arc length in polar coordinates:
\[ L = \int_{a}^{b} \sqrt{\left( \frac{dr}{d\theta} \right)^2 + r^2} \, d\theta \]
In this problem, \( r = \sin\theta \).
1. **Calculate \( \frac{dr}{d\theta} \):**
\(\frac{dr}{d\theta} = \cos\theta\)
2. **Substitute into the arc length formula:**
\[
L = \int_{0}^{\pi} \sqrt{(\cos\theta)^2 + (\sin\theta)^2} \, d\theta
\]
3. **Simplify the expression inside the integral:**
\[
(\cos\theta)^2 + (\sin\theta)^2 = 1
\]
Therefore, the integral becomes:
\[
L = \int_{0}^{\pi} \sqrt{1} \, d\theta = \int_{0}^{\pi} 1 \, d\theta
\]
4. **Evaluate the integral:**
\[
L = \left[\theta\right]_{0}^{\pi} = \pi - 0 = \pi
\]
Therefore, the length of the polar curve is \(\pi\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1d4bbfec-1266-4895-ab52-fb5d2bc6d55c%2Fc6eb5abc-ff3c-4436-a67c-557ce9c39441%2Fvy4biod8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the length of the polar curve given by the equation:
\[ r = \sin\theta, \quad 0 \leq \theta \leq \pi \]
Solve by hand. No calculator. Show all work.
**Solution Outline:**
To find the length of the polar curve, we use the formula for arc length in polar coordinates:
\[ L = \int_{a}^{b} \sqrt{\left( \frac{dr}{d\theta} \right)^2 + r^2} \, d\theta \]
In this problem, \( r = \sin\theta \).
1. **Calculate \( \frac{dr}{d\theta} \):**
\(\frac{dr}{d\theta} = \cos\theta\)
2. **Substitute into the arc length formula:**
\[
L = \int_{0}^{\pi} \sqrt{(\cos\theta)^2 + (\sin\theta)^2} \, d\theta
\]
3. **Simplify the expression inside the integral:**
\[
(\cos\theta)^2 + (\sin\theta)^2 = 1
\]
Therefore, the integral becomes:
\[
L = \int_{0}^{\pi} \sqrt{1} \, d\theta = \int_{0}^{\pi} 1 \, d\theta
\]
4. **Evaluate the integral:**
\[
L = \left[\theta\right]_{0}^{\pi} = \pi - 0 = \pi
\]
Therefore, the length of the polar curve is \(\pi\).
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