Problem 33 - Algebra and Trigonometry

Question

Find the rectangular coordinates for the point whose polar coordinates are given. $$ (6 \sqrt{2}, 11 \pi / 6) $$

Step-by-Step Solution

Answer

The rectangular coordinates are \((3\sqrt{6}, -3\sqrt{2})\)."}

Step 1: Understand the Polar Coordinates

Polar coordinates are given in the form \((r, \theta)\), where \(r\) is the radial distance from the origin, and \(\theta\) is the angle measured from the positive x-axis. Here, \(r = 6\sqrt{2}\) and \(\theta = \frac{11\pi}{6}\).

Step 2: Recall Formulas for Conversion

To convert polar coordinates \((r, \theta)\) to rectangular coordinates \((x, y)\), use the formulas:\[x = r \cos(\theta)\]\[y = r \sin(\theta)\]

Step 3: Calculate the x-coordinate

Substitute \(r = 6\sqrt{2}\) and \(\theta = \frac{11\pi}{6}\) into the formula for \(x\):\[x = 6\sqrt{2} \cdot \cos\left(\frac{11\pi}{6}\right)\]Knowing \(\cos\left(\frac{11\pi}{6}\right) = \cos\left(2\pi - \frac{\pi}{6}\right) = \cos\left(-\frac{\pi}{6}\right) = \sqrt{3}/2\), we find:\[x = 6\sqrt{2} \cdot \frac{\sqrt{3}}{2} = 3\sqrt{6}\]

Step 4: Calculate the y-coordinate

Substitute \(r = 6\sqrt{2}\) and \(\theta = \frac{11\pi}{6}\) into the formula for \(y\):\[y = 6\sqrt{2} \cdot \sin\left(\frac{11\pi}{6}\right)\]Knowing \(\sin\left(\frac{11\pi}{6}\right) = \sin\left(2\pi - \frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -1/2\), we find:\[y = 6\sqrt{2} \cdot -\frac{1}{2} = -3\sqrt{2}\]

Step 5: Conclude the Rectangular Coordinates

The rectangular coordinates \((x, y)\) for the polar coordinates \((6\sqrt{2}, \frac{11\pi}{6})\) are \((3\sqrt{6}, -3\sqrt{2})\).

Key Concepts

Polar Coordinates

Polar coordinates use a pair \( (r, \theta) \) to define the position of a point on a plane. The first element, \( r \), represents the distance from the origin, or center point, of the polar coordinate system. Meanwhile, \( \theta \) indicates the angle between the positive x-axis and a line from the origin to the point.
This system is particularly useful for plotting points that are naturally expressed in terms of angles and distances, such as those in a circular or rotational system. If you imagine a point rotating counter-clockwise around a center, its position can be captured seamlessly with this approach.
Whenever you see a coordinate pair in this format, it means that instead of moving horizontal and vertical distances (as with Cartesian coordinates), you're working with a combination of span and rotation.

Conversion Formulas

Converting polar coordinates to rectangular coordinates is straightforward by using simple trigonometric functions. The conversion formulas are fundamental, and they establish the relationship between these two systems by using the properties of a right triangle.

• To find the x-coordinate, use: \( x = r \cos(\theta) \)
• For the y-coordinate, apply: \( y = r \sin(\theta) \)

These formulas come from understanding that any point described by polar coordinates can be represented as part of a right triangle, with \( r \) as the hypotenuse. The cosine function correlates the hypotenuse with the adjacent side (x-axis), while the sine correlates with the opposite side (y-axis).
By employing these functions, you can convert any point from the polar system, defined by a distance and angle, into an x, y format used in rectangular coordinates.

Rectangular Coordinates

Rectangular coordinates, or Cartesian coordinates, represent points on a plane using two perpendicular lines: the x-axis (horizontal) and the y-axis (vertical). Each point is specified by the distance from these two axes, \( (x, y) \).
This system is best for scenarios where objects move linearly instead of circularly, making it ideal for most practical applications involving grid-like patterns, straight movement, or universal mapping.
Upon translating from polar to rectangular coordinates, such as in the exercise where polar \( (6\sqrt{2}, \frac{11\pi}{6}) \) becomes rectangular \( (3\sqrt{6}, -3\sqrt{2}) \), you see the transformation into a structure that's widely used for plotting in everyday mathematics. This transformation is fundamental in moving from one understanding of a position (by angle and distance) to another (in terms of horizontal and vertical shift).