Question

Convert each of the following points into polar coordinates. a. \((1,1)\) b. \((-3,4)\) c. \((0,3)\) d. \((5 \sqrt{3},-5)\)

Short answer

(1,1) in polar is (\(\sqrt{2}, \frac{\pi}{4}\)), (-3,4) is (5, \(\pi - \tan^{-1}(\frac{4}{3})\)), (0,3) is (3, \(\frac{\pi}{2}\)), and (5\sqrt{3},-5) is (10, \(\frac{11\pi}{6}\)).

Step-by-step solution

Understanding Polar Coordinates

Polar coordinates represent a point by its distance from the origin, referred to as 'r', and the angle 'θ' it makes with the positive x-axis. The conversion from Cartesian coordinates (x, y) to polar coordinates (r, θ) is done using the formulas: \[ r = \sqrt{x^2 + y^2} \] and \[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \] Ensure θ is measured in radians and takes into account the quadrant in which the point is located.

Convert Point (1,1)

1. Calculate r: \[ r = \sqrt{1^2 + 1^2} = \sqrt{2} \]2. Calculate θ: \[ \theta = \tan^{-1}(1) = \frac{\pi}{4} \] So, the polar coordinates of (1, 1) are (\(\sqrt{2}, \frac{\pi}{4}\)).

Convert Point (-3,4)

1. Calculate r: \[ r = \sqrt{(-3)^2 + 4^2} = 5 \]2. Calculate θ: \[ \theta = \tan^{-1}\left(\frac{4}{-3}\right) = -\tan^{-1}\left(\frac{4}{3}\right) \]3. Adjust θ for the correct quadrant (since x is negative and y is positive, it's in the 2nd quadrant):\[ \theta = \pi - \tan^{-1}\left(\frac{4}{3}\right) \]So, the polar coordinates are (5, \(\pi - \tan^{-1}(\frac{4}{3})\)).

Convert Point (0,3)

1. Calculate r: \[ r = \sqrt{0^2 + 3^2} = 3 \]2. Determine θ: Since x = 0 and y is positive, θ is \[ \frac{\pi}{2} \] So, the polar coordinates of (0, 3) are (3, \(\frac{\pi}{2}\)).

Convert Point (5\sqrt{3},-5)

1. Calculate r: \[ r = \sqrt{(5\sqrt{3})^2 + (-5)^2} = \sqrt{75 + 25} = 10 \]2. Calculate θ: \[ \theta = \tan^{-1}\left(\frac{-5}{5\sqrt{3}}\right) = -\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6} \]3. Adjust θ for the correct quadrant (since x is positive and y is negative, it's in the 4th quadrant): \[ \theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \]So, the polar coordinates are (10, \(\frac{11\pi}{6}\)).

Key concepts

Cartesian to Polar Conversion

Converting Cartesian coordinates to polar coordinates is an essential skill in mathematics, especially in calculus and trigonometry. The process involves changing the (x, y) representation of a point to a polar (r, θ) format. In this format, "r" stands for the radius or the distance from the origin, while "θ" represents the angle with the positive x-axis. To find "r", we use the formula

• \( r = \sqrt{x^2 + y^2} \)

This formula calculates the Euclidean distance from the origin to the point. For the angle "θ", we apply the inverse tangent function:

• \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \)

This function helps determine the angle of the line connecting the point to the origin with respect to the x-axis.
Remember, these conversions provide a holistic way to view a point in a 2D space, revealing both distance and angular relationship.

Trigonometry in Polar Coordinates

Trigonometry plays a crucial role when dealing with polar coordinates. It is involved in calculating the angle "θ", which requires an understanding of the tangent function. We use

• \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \)

to determine the initial angle relative to the positive x-axis. Since the tangent function can repeat values beyond the 0 to \(2\pi\) interval (or 0 to 360 degrees), adjustments are sometimes necessary to capture the right orientation depending on the quadrant.
For instance, if x is negative and y is positive, the point lies in the second quadrant, leading to adjustments like \( \theta = \pi - \tan^{-1}\left(\frac{y}{x}\right) \). Understanding the properties of trigonometric functions, their periodicity, and how they relate to angles assists in converting coordinates correctly.
Trigonometry provides the tools to not only interpret polar coordinates but also to work with angles in real-world applications like physics and engineering.

Quadrant Considerations in Polar Coordinates

A crucial aspect of converting Cartesian coordinates to polar is considering the quadrant in which the original point lies. Each point in the Cartesian plane falls within one of four quadrants, each with unique x and y sign combinations:

• First Quadrant: x and y are both positive.
• Second Quadrant: x is negative, y is positive.
• Third Quadrant: x and y are both negative.
• Fourth Quadrant: x is positive, y is negative.

These quadrants affect the angle "θ" calculation. The initial \( \tan^{-1}\left(\frac{y}{x}\right) \) only reflects the angle in the first quadrant. Therefore, you must adjust "θ" based on the quadrant:

• Second Quadrant: \( \theta = \pi - \tan^{-1}\left(\frac{y}{x}\right) \)
• Third Quadrant: \( \theta = \pi + \tan^{-1}\left(\frac{y}{x}\right) \)
• Fourth Quadrant: \( \theta = 2\pi - \tan^{-1}\left(\frac{y}{x}\right) \)

The quadrant consideration ensures that the angle "θ" correctly represents the position of the point on the plane. Recognizing the quadrant helps prevent calculation errors and is vital in various applications.