Problem 29 - Precalculus : Building Concepts and Connections

Question

Convert each of the given pairs of rectangular coordinates to a pair of polar coordinates ( \(r, \theta\) ) with \(r>0\) and \(0 \leq \theta<2 \pi\). $$(-\sqrt{3},-1)$$

Step-by-Step Solution

Answer

The polar coordinates for the given rectangular coordinates are (2, 7π/6)

Step 1: Calculating radius r

Calculate r as the square root of (x^2 + y^2). Here, x = -sqrt(3) and y = -1. Therefore, \(r = \sqrt{(-\sqrt{3})^2 + (-1)^2} = 2\)

Step 2: Calculating angle θ

Calculate θ as the arc tangent of y/x, θ = arctan(-1/-sqrt(3)). But because the angle lies in the third quadrant, we need to make an adjustment. For angles in the third or fourth quadrant, add π to the result of the arctan operation.

Step 3: Adjusting the angle

Our original θ = arctan(1/sqrt(3)) = π/6. Because it lies in the third quadrant, therefore, θ = π/6 + π = 7π/6

Key Concepts

Rectangular Coordinates

Understanding rectangular coordinates is similar to plotting points on a grid. Each point is determined by a pair of numbers, usually labeled as \((x, y)\). These numbers define the horizontal and vertical placement of the point on the Cartesian plane, respectively.
Rectangular coordinates are incredibly useful for tasks that involve straightforward linear calculations. The \(x\)-coordinate represents the distance along the horizontal axis, while the \(y\)-coordinate stands for the distance along the vertical axis.
For instance, with the point \((-\sqrt{3}, -1)\), we find that it is located to the left of the origin in the \(x\)-direction (since \(-\sqrt{3}\) is negative) and below the origin in the \(y\)-direction (since \(-1\) is negative). This positions our point in the third quadrant, where both \(x\) and \(y\) are negative.

Angle Calculation

To convert rectangular coordinates to polar coordinates, we need to calculate the angle \(\theta\). This angle is determined using the arctangent function, expressed as \( \theta = \arctan\left(\frac{y}{x}\right) \).
The arctangent function helps find the angle formed with the positive \(x\)-axis. It's important to note that the basic output of the arctangent function is constrained to the first and fourth quadrants \((-\frac{\pi}{2} < \theta < \frac{\pi}{2})\).
Given our point \((-\sqrt{3}, -1)\), the calculation becomes \( \theta = \arctan\left(\frac{-1}{-\sqrt{3}}\right) = \arctan\left(\frac{1}{\sqrt{3}}\right) \). Initially, this calculation places our angle as \(\frac{\pi}{6}\), which is key in understanding the adjustment needed due to the original quadrant placement.

Quadrant Adjustment

The quadrant adjustment is essential for ensuring our angle \(\theta\) accurately reflects the point's location on the plane. Since our point \((-\sqrt{3}, -1)\) lies in the third quadrant, straightforward arctan calculations may not represent this correctly without adjustments.
In particular, for angles located in the third quadrant, we must add \(\pi\) to our initial arctan result. This adjustment moves the angle from the first quadrant to the correct position in the third quadrant.

• The initial calculation \( \arctan\left(\frac{1}{\sqrt{3}}\right) \) yields \(\frac{\pi}{6}\).
• To accurately place it in the third quadrant, add \(\pi\), resulting in \(\theta = \frac{\pi}{6} + \pi = \frac{7\pi}{6}\).

This ensures our polar coordinates reflect both the distance from the origin and the true angular direction from the positive \(x\)-axis.