Convert the rectangular coordinates (9, -3√√/3) into polar form. Express the angle using radians in terms of 7 over the interval 0 ≤ 0 < 2π, with a positive value of r. Answer: π Submit Answer
Convert the rectangular coordinates (9, -3√√/3) into polar form. Express the angle using radians in terms of 7 over the interval 0 ≤ 0 < 2π, with a positive value of r. Answer: π Submit Answer
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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![**Title: Converting Rectangular Coordinates to Polar Form**
**Problem Statement:**
Convert the rectangular coordinates \((9, -3\sqrt{3})\) into polar form. Express the angle using radians in terms of \(\pi\) over the interval \(0 \leq \theta < 2\pi\), with a positive value of \(r\).
**Answer Section:**
Below the problem statement, there is a text input box labeled "Answer:" where students can type in their answers. Next to the input box, there are quick insert buttons for the symbols \(\pi\) and \(\sqrt{}\). To the right of the input box is a "Submit Answer" button that students can click to submit their response.
**Explanation and Steps:**
1. **Find the radius \(r\)**:
The radius \(r\) is calculated using the formula:
\[
r = \sqrt{x^2 + y^2}
\]
For the given coordinates \((9, -3\sqrt{3})\),
\[
r = \sqrt{9^2 + (-3\sqrt{3})^2} = \sqrt{81 + 27} = \sqrt{108} = 6\sqrt{3}
\]
2. **Find the angle \(\theta\)**:
The angle \(\theta\) is found using the formula:
\[
\theta = \tan^{-1}\left(\frac{y}{x}\right)
\]
For the given coordinates \((9, -3\sqrt{3})\),
\[
\theta = \tan^{-1}\left(\frac{-3\sqrt{3}}{9}\right) = \tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)
\]
Simplifying,
\[
\theta = -\frac{\pi}{6}
\]
Since the angle must be within the interval \(0 \leq \theta < 2\pi\), add \(2\pi\) to \(-\frac{\pi}{6}\):
\[
\theta = 2\pi - \frac{\pi}{6} = \frac{12\pi}{6} - \frac{\pi}{6} = \frac{11\pi}{6}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6e31dfb2-e24d-4f0b-ab76-e0eb03cfbad1%2F5e3bea67-24cf-48df-ac6f-3fb7bd974e11%2Fvohondb_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Converting Rectangular Coordinates to Polar Form**
**Problem Statement:**
Convert the rectangular coordinates \((9, -3\sqrt{3})\) into polar form. Express the angle using radians in terms of \(\pi\) over the interval \(0 \leq \theta < 2\pi\), with a positive value of \(r\).
**Answer Section:**
Below the problem statement, there is a text input box labeled "Answer:" where students can type in their answers. Next to the input box, there are quick insert buttons for the symbols \(\pi\) and \(\sqrt{}\). To the right of the input box is a "Submit Answer" button that students can click to submit their response.
**Explanation and Steps:**
1. **Find the radius \(r\)**:
The radius \(r\) is calculated using the formula:
\[
r = \sqrt{x^2 + y^2}
\]
For the given coordinates \((9, -3\sqrt{3})\),
\[
r = \sqrt{9^2 + (-3\sqrt{3})^2} = \sqrt{81 + 27} = \sqrt{108} = 6\sqrt{3}
\]
2. **Find the angle \(\theta\)**:
The angle \(\theta\) is found using the formula:
\[
\theta = \tan^{-1}\left(\frac{y}{x}\right)
\]
For the given coordinates \((9, -3\sqrt{3})\),
\[
\theta = \tan^{-1}\left(\frac{-3\sqrt{3}}{9}\right) = \tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)
\]
Simplifying,
\[
\theta = -\frac{\pi}{6}
\]
Since the angle must be within the interval \(0 \leq \theta < 2\pi\), add \(2\pi\) to \(-\frac{\pi}{6}\):
\[
\theta = 2\pi - \frac{\pi}{6} = \frac{12\pi}{6} - \frac{\pi}{6} = \frac{11\pi}{6}
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