Problem 38 - Precalculus Mathematics for Calculus

Question

Convert the rectangular coordinates to polar coordinates with \(r>0\) and \(0 \leq \theta<2 \pi\). $$(-\sqrt{6},-\sqrt{2})$$

Step-by-Step Solution

Answer

The polar coordinates are \( (2\sqrt{2}, \frac{7\pi}{6}) \).

Step 1: Calculate the radius

The polar coordinate radius \(r\) is calculated using the formula:\[ r = \sqrt{x^2 + y^2} \]Where \(x = -\sqrt{6}\) and \(y = -\sqrt{2}\), so:\[ r = \sqrt{(-\sqrt{6})^2 + (-\sqrt{2})^2} \]\[ r = \sqrt{6 + 2} \]\[ r = \sqrt{8} \]\[ r = 2\sqrt{2} \].

Step 2: Calculate the angle (theta) in radians

To find the angle \(\theta\), use the formula:\[ \theta = \text{atan2}(y, x) \]Where \(x = -\sqrt{6}\) and \(y = -\sqrt{2}\), which gives:\[ \theta = \text{atan2}(-\sqrt{2}, -\sqrt{6}) \]Both \(x\) and \(y\) are negative, indicating that the point is in the third quadrant. Thus:\[ \theta = \pi + \text{atan2}(\sqrt{2}, \sqrt{6}) \]First, calculate the reference angle:\[ \text{tan}^{-1}\left(\frac{\sqrt{2}}{\sqrt{6}}\right) = \text{tan}^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]Add \(\pi\) to move to the third quadrant:\[ \theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6} \].

Step 3: Confirm the angle is within the valid range

Since \(0 \leq \theta < 2\pi\) naturally includes \(\frac{7\pi}{6}\), the calculated \(\theta\) is valid in the given range and requires no adjustment.

Key Concepts

Rectangular Coordinates

Rectangular coordinates represent points in a plane using two numbers, typically referred to as \(x\) and \(y\). They are also known as Cartesian coordinates. Each point is defined by its horizontal (\(x\)) and vertical (\(y\)) distances from a fixed reference point, usually the origin \((0,0)\). This system is widely used because it directly relates to everyday concepts like width and height.
- For example, in the given problem, the rectangular coordinates are \((-\sqrt{6}, -\sqrt{2})\).
- Here, \(x = -\sqrt{6}\) indicates the point is \(\sqrt{6}\) units to the left of the origin on the horizontal axis.
- Similarly, \(y = -\sqrt{2}\) indicates the point is \(\sqrt{2}\) units below the origin on the vertical axis.
Rectangular coordinates are crucial for converting to polar coordinates, which is a system that uses radius and angle to locate points.

Radius Calculation

To convert from rectangular to polar coordinates, the first step is to calculate the radius \(r\). The radius in this context measures the distance from the origin \((0, 0)\) to the point \((x, y)\).
- The formula used is: \[ r = \sqrt{x^2 + y^2} \]
This is derived from the Pythagorean theorem, which applies directly to finding the hypotenuse of a right triangle. In this case, the triangle is formed between the point, the origin, and the projected x and y intercepts.
- For our example, substitute \(x = -\sqrt{6}\) and \(y = -\sqrt{2}\) into the formula: \[ r = \sqrt{(-\sqrt{6})^2 + (-\sqrt{2})^2} = \sqrt{6 + 2} = \sqrt{8} = 2\sqrt{2} \]
Thus, the radius \(r\) is the Euclidean distance of the point from the origin, which helps in determining the exact location of this point on the polar plane.

Angle Conversion

The next step after finding the radius is to calculate the angle \(\theta\), which tells us the direction of the point from the positive x-axis. This angle is often measured in radians and can be calculated using the inverse tangent function, known in programming as \(\text{atan2}\). This function effectively handles the signs of the x and y coordinates to determine the correct quadrant for \(\theta\).
- For point \((-\sqrt{6}, -\sqrt{2})\), both \(x\) and \(y\) are negative, meaning the point is in the third quadrant. \[ \theta = \pi + \text{atan2}\left(\frac{\sqrt{2}}{\sqrt{6}}\right) \]
- The reference angle obtained through \(\text{atan2}\) is \(\frac{\pi}{6}\).
- Adding \(\pi\) moves the angle to the third quadrant, giving: \[ \theta = \frac{7\pi}{6} \]
This angle is within the required range of \(0 \leq \theta < 2\pi\), confirming it correctly represents the direction from the origin. Understanding angle conversion is essential for interpreting polar coordinates accurately.