Problem 12 - 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) \((-2 \sqrt{2},-2 \sqrt{2})\) (b) \((-4,4 \sqrt{3})\)

Step-by-Step Solution

Answer

(a) \\( (4, \frac{5\pi}{4}) \\); (b) \\( (8, \frac{2\pi}{3}) \\).

Step 1: Understand the Relationship Between Rectangular and Polar Coordinates

Rectangular coordinates \(x, y\) can be converted into polar coordinates \(r, \theta\) using the formulas \(r = \sqrt{x^2 + y^2}\) and \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\). We need to ensure that \(r > 0\) and \(0 \leq \theta \leq 2\pi\).

Step 2: Calculate Magnitude (r) for Part (a)

Given point is \((-2\sqrt{2}, -2\sqrt{2})\). Calculate \(r\): \[ r = \sqrt{(-2\sqrt{2})^2 + (-2\sqrt{2})^2} = \sqrt{8 + 8} = \sqrt{16} = 4 \]

Step 3: Calculate Angle (θ) for Part (a)

Calculate \(\theta\) using \(\tan^{-1}\left(\frac{y}{x}\right)\): \[ \theta = \tan^{-1}\left(\frac{-2\sqrt{2}}{-2\sqrt{2}}\right) = \tan^{-1}(1) = \frac{\pi}{4} \]The point lies in the third quadrant, so adjust \(\theta\):\[ \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \]

Step 4: Write Polar Coordinates for Part (a)

The polar coordinates for the point \((-2\sqrt{2}, -2\sqrt{2})\) are \( (4, \frac{5\pi}{4})\).

Step 5: Calculate Magnitude (r) for Part (b)

Given point is \((-4, 4\sqrt{3})\). Calculate \(r\): \[ r = \sqrt{(-4)^2 + (4\sqrt{3})^2} = \sqrt{16 + 48} = \sqrt{64} = 8 \]

Step 6: Calculate Angle (θ) for Part (b)

Calculate \(\theta\) using \(\tan^{-1}\left(\frac{y}{x}\right)\): \[ \theta = \tan^{-1}\left(\frac{4\sqrt{3}}{-4}\right) = \tan^{-1}(-\sqrt{3}) \]The point lies in the second quadrant, and \(\tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}\), so adjust \(\theta\):\[ \theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3} \]

Step 7: Write Polar Coordinates for Part (b)

The polar coordinates for the point \((-4, 4\sqrt{3})\) are \( (8, \frac{2\pi}{3})\).

Key Concepts

Rectangular Coordinates

Rectangular coordinates, also known as Cartesian coordinates, represent a point's location in a plane using two values: the x-coordinate and the y-coordinate. Think of a flat surface where each point can be determined by how far it is from two intersecting lines, typically called the x-axis and y-axis.
For example, in the coordinates (-2√2,-2√2), -2√2 is the x-value, and -2√2 is the y-value. This means the point is located downward to the left of the origin, where the x-axis and y-axis intersect. This location helps us understand precise positions and movements in this coordinate system, which is useful for various math and science applications.

Conversion Formulas

Conversion from rectangular to polar coordinates involves using specific formulas. Polar coordinates are given by a radius and an angle, denoted as (r, θ). These allow us to locate points based on their distance from the origin and their angle from the positive x-axis.
The conversion formulas are:

• Radius: \( r = \sqrt{x^2 + y^2} \)
• Angle: \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \)

Using these formulas, we transform the coordinates from the rectangular system to the polar system, considering the angle must be between 0 and 2\(\pi\) to cover a full circle.
The key part is ensuring the correct interpretation of the angle based on the location of the point in different quadrants.

Quadrant Identification

Identifying which quadrant a point belongs to is essential in determining the correct angle \(\theta\) when converting to polar coordinates. A coordinate system is divided into four quadrants based on the signs of the x and y values:

• First Quadrant: \((+, +)\)
• Second Quadrant: \((- , +)\)
• Third Quadrant: \((- , -)\)
• Fourth Quadrant: \((+, -)\)

Each quadrant has specific angle ranges. For example, in exercise part (a), the coordinate (-2√2, -2√2) is in the third quadrant. This means we must adjust the angle by adding \(\pi\) to cover the full angle span needed for accurate representation.
Understanding quadrant placement helps in refining the angle results, as initial results from arc functions might need adjusting based on these quadrant rules.

Trigonometric Functions

Trigonometric functions, especially the tangent inverse or \(\tan^{-1}\), are used in finding the angle \(\theta\) during conversion. These functions help us understand the relationships between the angles and sides of triangles, crucial in converting to polar coordinates.
Consider the function \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\). This formula gives us the angle determined by the tangent of the point's coordinates.
Since tangent is the ratio of the y-coordinate to the x-coordinate, it effectively acts as a bridge between the two coordinate systems. Adjustments depend on the results of this function, as negative or unexpected values indicate quadrant-related modifications are necessary to find the accurate polar angle, ensuring real-world applications and calculations are precise and relevant.