Algebra and Trigonometry (MindTap Course List)
4th Edition
ISBN:9781305071742
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter5: Trigonometric Functions: Right Triangle Approach
Section5.1: Angle Measure
Problem 4E: Object A is travelling along a circle of radius 2, and Object B is travelling along a circle of...
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![### Problem Statement
**Find the slope of the tangent line to the given polar curve at the point specified by the value of θ.**
Given:
\[ r = \cos(\theta/3), \quad \theta = \pi \]
---
There is an answer box intended for the student to input their solution.
---
### Explanation
We are given a polar curve defined by \( r = \cos(\theta/3) \) and we need to find the slope of the tangent line at the point where \( \theta = \pi \).
In polar coordinates, the slope of the tangent line to the curve \( r = f(\theta) \) is given by:
\[ \frac{dy}{dx} = \frac{r \cos(\theta) + \frac{dr}{d\theta} \sin(\theta)}{-r \sin(\theta) + \frac{dr}{d\theta} \cos(\theta)} \]
First, compute \( \frac{dr}{d\theta} \):
\[ \frac{dr}{d\theta} = \frac{d}{d\theta} \left[ \cos(\frac{\theta}{3}) \right] = -\sin(\frac{\theta}{3}) \cdot \frac{1}{3} = -\frac{1}{3} \sin(\frac{\theta}{3}) \]
Next, evaluate \( r \) and \( \frac{dr}{d\theta} \) at \( \theta = \pi \):
\[ r \big|_{\theta = \pi} = \cos(\frac{\pi}{3}) = \cos(\frac{\pi}{3}) = \frac{1}{2} \]
\[ \frac{dr}{d\theta} \big|_{\theta = \pi} = -\frac{1}{3} \sin(\frac{\pi}{3}) = -\frac{1}{3} \cdot \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{6} \]
Now, substitute these values into the slope formula:
\[ \frac{dy}{dx} = \frac{\left( \frac{1}{2} \right) \cos(\pi) + \left( -\frac{\sqrt{3}}{6} \right) \sin(\pi)}{-\left( \frac{1}{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd1148f2e-12e2-4d7a-aa19-e54736499ad8%2Fb5b8739e-7b49-4b04-ab21-61923a04ba3d%2F27zp6y_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
**Find the slope of the tangent line to the given polar curve at the point specified by the value of θ.**
Given:
\[ r = \cos(\theta/3), \quad \theta = \pi \]
---
There is an answer box intended for the student to input their solution.
---
### Explanation
We are given a polar curve defined by \( r = \cos(\theta/3) \) and we need to find the slope of the tangent line at the point where \( \theta = \pi \).
In polar coordinates, the slope of the tangent line to the curve \( r = f(\theta) \) is given by:
\[ \frac{dy}{dx} = \frac{r \cos(\theta) + \frac{dr}{d\theta} \sin(\theta)}{-r \sin(\theta) + \frac{dr}{d\theta} \cos(\theta)} \]
First, compute \( \frac{dr}{d\theta} \):
\[ \frac{dr}{d\theta} = \frac{d}{d\theta} \left[ \cos(\frac{\theta}{3}) \right] = -\sin(\frac{\theta}{3}) \cdot \frac{1}{3} = -\frac{1}{3} \sin(\frac{\theta}{3}) \]
Next, evaluate \( r \) and \( \frac{dr}{d\theta} \) at \( \theta = \pi \):
\[ r \big|_{\theta = \pi} = \cos(\frac{\pi}{3}) = \cos(\frac{\pi}{3}) = \frac{1}{2} \]
\[ \frac{dr}{d\theta} \big|_{\theta = \pi} = -\frac{1}{3} \sin(\frac{\pi}{3}) = -\frac{1}{3} \cdot \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{6} \]
Now, substitute these values into the slope formula:
\[ \frac{dy}{dx} = \frac{\left( \frac{1}{2} \right) \cos(\pi) + \left( -\frac{\sqrt{3}}{6} \right) \sin(\pi)}{-\left( \frac{1}{
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