Problem 9 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 9-12: Change the rectangular coordinates to polar coordinates with \(r>0\) and \(0 \leq \theta \leq 2 \pi\). (a) \((-1,1)\) (b) \((-2 \sqrt{3},-2)\)

Step-by-Step Solution

Answer

(a) \((\sqrt{2}, \frac{3\pi}{4})\) and (b) \((4, \frac{7\pi}{6})\)."

Step 1: Recall the definition of polar coordinates

Polar coordinates \((r, \theta)\) are related to rectangular coordinates \((x, y)\) by the equations \(r = \sqrt{x^2 + y^2}\) and \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\). The angle \(\theta\) might need adjustment based on the quadrant.

Step 2: Convert point (-1, 1) to polar coordinates

For the point \((-1, 1)\), calculate the value of \(r\) using \(r = \sqrt{(-1)^2 + 1^2}\). This simplifies to \(r = \sqrt{1+1} = \sqrt{2}\).Next, calculate \(\theta\) using \(\tan^{-1}\left(\frac{1}{-1}\right)\). The result is \(\theta = \tan^{-1}(-1)\). Since the point is in the second quadrant, \(\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\). Thus, the polar coordinates are \((\sqrt{2}, \frac{3\pi}{4})\).

Step 3: Convert point (-2√3, -2) to polar coordinates

For the point \((-2\sqrt{3}, -2)\), calculate the value of \(r\) using \(r = \sqrt{(-2\sqrt{3})^2 + (-2)^2}\). This simplifies to \(r = \sqrt{12 + 4} = \sqrt{16} = 4\).Next, calculate \(\theta\) using \(\tan^{-1}\left(\frac{-2}{-2\sqrt{3}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right)\). This results in \(\theta = \frac{\pi}{6}\). Since the point is in the third quadrant, the correct angle is \(\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}\).The polar coordinates are \((4, \frac{7\pi}{6})\).

Key Concepts

Rectangular Coordinates

Rectangular coordinates, often known as Cartesian coordinates, utilize a pair of numbers

1. x-coordinate (or abscissa) is the horizontal component.
2. y-coordinate (or ordinate) is the vertical component.

These coordinates allow us to identify any position on a plane by describing its distance from two intersecting lines, generally referred to as axes. The x-axis is horizontal, while the y-axis is vertical. This system is intuitive for plotting points in linear and spatial contexts.
To visualize, think of a grid where each square represents a unique (x, y) pair. By moving horizontally along the x-axis and vertically along the y-axis, one can reach any point in the plane. Rectangular coordinates are integral to mathematics, physics, and engineering because they offer a clear, standardized way to pinpoint locations and describe shapes.

Quadrant Adjustment

When converting rectangular coordinates to polar coordinates, it's essential to determine the correct quadrant of the angle θ. The Cartesian plane is divided into four sections or quadrants, each influencing how we calculate or adjust angles:

First Quadrant:

(+) x, (+) y, no adjustment needed

Second Quadrant:

(-) x, (+) y, angle θ must equal π + tan^-1( ^y/_x)

Third Quadrant:

(-) x, (-) y, angle θ must equal π + tan^-1( ^y/_x)

Fourth Quadrant:

(+) x, (-) y, angle θ must equal 2 π + tan^-1( ^y/_x)

Adjusting for these quadrants ensures θ fits the interval [0, 2π). This understanding is crucial because neglecting it may lead to incorrect polar representations.

Conversion Formulas

The change from rectangular coordinates to polar coordinates involves specific formulas to find the radius ( r) and the angle ( θ):
1. Formula for Radius ( r):

1. r = √(x² + y²)

This equation calculates the distance from the origin (0,0) to the point ( x, y).
2. Formula for Angle ( θ):

1. θ = tan^-1( ^y/_x)

However, it's vital to consider the quadrant related to the angle. These formulas work best when complemented by quadrant adjustments to ensure θ remains non-negative and within the specified range [0, 2π). A correct application of these methods is key to accurately transitioning between coordinate systems, offering a reliable approach in mathematics, physics, and engineering.