For Cartesian coordinates (x,y)=(1, -√3), find the angle in the polar coordinates in radians.
For Cartesian coordinates (x,y)=(1, -√3), find the angle in the polar coordinates in radians.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Problem 4: Conversion from Cartesian to Polar Coordinates**
For Cartesian coordinates \((x, y) = (1, -\sqrt{3})\), find the angle in the polar coordinates in radians.
### Explanation:
To convert the given Cartesian coordinates to polar coordinates, use the following steps:
1. **Calculate the radius, \( r \):**
\[
r = \sqrt{x^2 + y^2} = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2
\]
2. **Calculate the angle, \( \theta \):**
\[
\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = \tan^{-1}(-\sqrt{3})
\]
The angle whose tangent is \(-\sqrt{3}\) is \(-\frac{\pi}{3}\), but since the point \((1, -\sqrt{3})\) is in the fourth quadrant, add \(2\pi\) to get the positive angle:
\[
\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}
\]
Therefore, the angle in polar coordinates is \(\frac{5\pi}{3}\) radians.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd331e78d-4008-4ff5-981d-7e34411cb97b%2F428a29b4-366c-417c-b4fc-498a531cd9dc%2Fvwtg8ll_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 4: Conversion from Cartesian to Polar Coordinates**
For Cartesian coordinates \((x, y) = (1, -\sqrt{3})\), find the angle in the polar coordinates in radians.
### Explanation:
To convert the given Cartesian coordinates to polar coordinates, use the following steps:
1. **Calculate the radius, \( r \):**
\[
r = \sqrt{x^2 + y^2} = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2
\]
2. **Calculate the angle, \( \theta \):**
\[
\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = \tan^{-1}(-\sqrt{3})
\]
The angle whose tangent is \(-\sqrt{3}\) is \(-\frac{\pi}{3}\), but since the point \((1, -\sqrt{3})\) is in the fourth quadrant, add \(2\pi\) to get the positive angle:
\[
\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}
\]
Therefore, the angle in polar coordinates is \(\frac{5\pi}{3}\) radians.
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Use the formula for angle in the polar coordinates.
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