Problem 45 - Algebra and Trigonometry

Question

The rectangular coordinates of a point are given. Find polar coordinates of each point. Express \(\theta\) in radians. $$ (-\sqrt{3},-1) $$

Step-by-Step Solution

Answer

The polar coordinates of the point (-\sqrt{3}, -1) are (2, \frac{7π}{6})

Step 1: Calculate r

Use the formula r = \sqrt{x^{2} + y^{2}}. Here, x and y correspond to the rectangular coordinates (-\sqrt{3}, -1). So, r = \sqrt{(-\sqrt{3})^{2} + (-1)^{2}} = \sqrt{3+1} = 2.

Step 2: Compute θ

Use the formula θ = \arctan{\frac{y}{x}} to find the initial angle. θ = \arctan{\frac{-1}{-\sqrt{3}}} = \arctan{\frac{\sqrt{3}}{3}}. However, because the point is in the third quadrant where both x and y are negative, we add π to the result, i.e. θ = \arctan{\frac{\sqrt{3}}{3}} + π = \frac{π}{6} + π = \frac{7π}{6}.

Step 3: Express the Polar Coordinates

The polar coordinates (r,θ) are expressed as (2, \frac{7π}{6})

Key Concepts

Rectangular Coordinates

In mathematics, rectangular coordinates provide a simple and effective method for locating points within a two-dimensional space. Most often, they are represented as an ordered pair \((x, y)\). Each component of this pair corresponds to the respective distances along the horizontal (\(x\)) and the vertical (\(y\)) axes. For example, the point \((-\sqrt{3}, -1)\) specifies a position where the point lies \(-\sqrt{3}\) units left from the origin along the horizontal axis, and \(-1\) unit downward along the vertical axis.
Rectangular coordinates are foundational in mathematics because they are easy to understand and utilize when plotting points or graphing linear equations. However, when working with circles or curves, it might be more convenient to use a different system called polar coordinates.

Conversion to Polar Form

Converting rectangular coordinates to polar form involves a systematic process of finding two new values: the radius \(r\) and the angle \(\theta\). This conversion is especially helpful when dealing with trigonometric functions, spirals, and periodic motion.

• To find \(r\), the distance from the origin to the point, you use the formula: \(r = \sqrt{x^2 + y^2}\).
• For \(\theta\), which is the angle from the positive x-axis, the initial calculation is \(\theta = \arctan{\left(\frac{y}{x}\right)}\).
  Remember, angles in polar coordinates are usually expressed in radians.

In our specific example, starting from the coordinates \((-\sqrt{3}, -1)\):
\(r\) is calculated as \(\sqrt{(-\sqrt{3})^2 + (-1)^2} = 2\).
\(\theta\) requires adjusting for the appropriate quadrant. Here, the point is in the third quadrant, requiring an "angle adjustment" by adding \(\pi\) radians, resulting in \(\theta = \frac{7\pi}{6}\).

Trigonometry Concepts

Understanding trigonometry is crucial when working with polar coordinates. Trigonometry deals with the relationships between the sides and angles of triangles, which directly applies when moving between rectangular and polar systems.

• The trigonometric function \(\tan(\theta) = \frac{y}{x}\) helps determine the angle \(\theta\) in polar coordinates.
• Since our point \((-\sqrt{3}, -1)\) is located in the third quadrant, both \(x\) and \(y\) are negative, making it necessary to adjust the angle with an additional \(\pi\).

Moreover, using radians for angles in polar form aligns with many trigonometric functions. It offers a straightforward approach, especially since one full circle is \(2\pi\) radians. This also allows a smoother transition when dealing with angle measures in more complex mathematical calculations.