Find the equation of the line tangent to the cardioid r = 1+ sin(0) at Ô0 =
Algebra and Trigonometry (MindTap Course List)
4th Edition
ISBN:9781305071742
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter8: Polar Coordinates And Parametric Equations
Section8.CT: Chapter Test
Problem 8CT
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![**Question:**
Find the equation of the line tangent to the cardioid \( r = 1 + \sin(\theta) \) at \( \theta = 0 \).
---
**Solution:**
1. **Convert to Cartesian Coordinates:**
The polar coordinates \( r \) and \( \theta \) can be converted to Cartesian coordinates \( (x, y) \) using the formulas:
\[
x = r \cos(\theta)
\]
\[
y = r \sin(\theta)
\]
For the given cardioid \( r = 1 + \sin(\theta) \), we first calculate the coordinates at \( \theta = 0 \).
\[
r = 1 + \sin(0) = 1 + 0 = 1
\]
\[
x (0) = 1 \cos(0) = 1 \cdot 1 = 1
\]
\[
y (0) = 1 \sin(0) = 1 \cdot 0 = 0
\]
Therefore, the point of tangency in Cartesian coordinates is \( (1, 0) \).
2. **Find the Derivative:**
To find the slope of the tangent line, we consider the derivatives of \( x \) and \( y \) with respect to \( \theta \).
\[
x = r \cos(\theta) = \left(1 + \sin(\theta)\right) \cos(\theta)
\]
\[
y = r \sin(\theta) = \left(1 + \sin(\theta)\right) \sin(\theta)
\]
Differentiate \( x \) and \( y \) with respect to \( \theta \):
\[
\frac{dx}{d\theta} = \frac{d}{d\theta} \left( (1 + \sin(\theta)) \cos(\theta) \right)
\]
Apply the product rule:
\[
\frac{dx}{d\theta} = \cos(\theta) \frac{d}{d\theta} (1 + \sin(\theta)) + (1 + \sin(\theta)) \frac{d}{d\theta} \cos(\theta)
\]
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F720535be-d647-4725-8ea5-21875474b49c%2F6922acbc-655e-4e21-83ce-5421333dc999%2Fvmv7n1l_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
Find the equation of the line tangent to the cardioid \( r = 1 + \sin(\theta) \) at \( \theta = 0 \).
---
**Solution:**
1. **Convert to Cartesian Coordinates:**
The polar coordinates \( r \) and \( \theta \) can be converted to Cartesian coordinates \( (x, y) \) using the formulas:
\[
x = r \cos(\theta)
\]
\[
y = r \sin(\theta)
\]
For the given cardioid \( r = 1 + \sin(\theta) \), we first calculate the coordinates at \( \theta = 0 \).
\[
r = 1 + \sin(0) = 1 + 0 = 1
\]
\[
x (0) = 1 \cos(0) = 1 \cdot 1 = 1
\]
\[
y (0) = 1 \sin(0) = 1 \cdot 0 = 0
\]
Therefore, the point of tangency in Cartesian coordinates is \( (1, 0) \).
2. **Find the Derivative:**
To find the slope of the tangent line, we consider the derivatives of \( x \) and \( y \) with respect to \( \theta \).
\[
x = r \cos(\theta) = \left(1 + \sin(\theta)\right) \cos(\theta)
\]
\[
y = r \sin(\theta) = \left(1 + \sin(\theta)\right) \sin(\theta)
\]
Differentiate \( x \) and \( y \) with respect to \( \theta \):
\[
\frac{dx}{d\theta} = \frac{d}{d\theta} \left( (1 + \sin(\theta)) \cos(\theta) \right)
\]
Apply the product rule:
\[
\frac{dx}{d\theta} = \cos(\theta) \frac{d}{d\theta} (1 + \sin(\theta)) + (1 + \sin(\theta)) \frac{d}{d\theta} \cos(\theta)
\]
\[
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