Problem 1 - Precalculus

Question

Graph each point in a polar coordinate system then convert the given polar coordinates to rectangular coordinates. (a) \(\left(3, \frac{2 \pi}{3}\right)\) (b) \(\left(4, \frac{11 \pi}{6}\right)\) (c) \(\left(4,-\frac{\pi}{6}\right)\)

Step-by-Step Solution

Answer

(a) (-1.5, \(\frac{3\sqrt{3}}{2}\)), (b) (2\sqrt{3}, -2), (c) (2\sqrt{3}, -2)

Step 1: Understanding Polar Coordinates

Polar coordinates are expressed as \((r, \theta)\), where \(r\) is the radius (distance from the origin) and \(\theta\) is the angle measured counterclockwise from the positive x-axis.

Step 2: Graph Point (a)

For point (a) \(\left(3, \frac{2\pi}{3}\right)\), locate \(\frac{2\pi}{3}\) on the polar grid, which is 120 degrees counterclockwise from the positive x-axis. From the pole (origin), move out 3 units along this direction to plot the point.

Step 3: Convert Point (a) to Rectangular Coordinates

To convert from polar \((r, \theta)\) to rectangular (Cartesian) coordinates, use the formulas: \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). For \(\left(3, \frac{2\pi}{3}\right)\):\[x = 3 \cos\left(\frac{2\pi}{3}\right), \quad y = 3 \sin\left(\frac{2\pi}{3}\right)\]Evaluating these: \[x = 3(-\frac{1}{2}) = -1.5, \quad y = 3(\frac{\sqrt{3}}{2}) = \frac{3\sqrt{3}}{2}\]Thus the rectangular coordinates are \((-1.5, \frac{3\sqrt{3}}{2})\).

Step 4: Graph Point (b)

For point (b) \(\left(4, \frac{11\pi}{6}\right)\), locate \(\frac{11\pi}{6}\) on the polar grid, which is equivalent to \(-\frac{\pi}{6}\) or 330 degrees clockwise from the positive x-axis. Move out 4 units in this direction to plot the point.

Step 5: Convert Point (b) to Rectangular Coordinates

Using the formulas for conversion, for \(\left(4, \frac{11\pi}{6}\right)\):\[x = 4 \cos\left(\frac{11\pi}{6}\right), \quad y = 4 \sin\left(\frac{11\pi}{6}\right)\]Evaluating these: \[x = 4(\frac{\sqrt{3}}{2}) = 2\sqrt{3}, \quad y = 4(-\frac{1}{2}) = -2\]Thus the rectangular coordinates are \((2\sqrt{3}, -2)\).

Step 6: Graph Point (c)

For point (c) \(\left(4, -\frac{\pi}{6}\right)\), locate \(-\frac{\pi}{6}\) on the polar grid, which is 30 degrees clockwise from the positive x-axis. Move out 4 units on this angle to plot the point.

Step 7: Convert Point (c) to Rectangular Coordinates

Use the conversion formula for \(\left(4, -\frac{\pi}{6}\right)\):\[x = 4 \cos\left(-\frac{\pi}{6}\right), \quad y = 4 \sin\left(-\frac{\pi}{6}\right)\]Evaluating these: \[x = 4(\frac{\sqrt{3}}{2}) = 2\sqrt{3}, \quad y = 4(-\frac{1}{2}) = -2\]Thus the rectangular coordinates are \((2\sqrt{3}, -2)\).

Key Concepts

Rectangular Coordinates

Rectangular coordinates, also known as Cartesian coordinates, are a way to determine the position of a point using a pair of numerical values. These values represent the point's distance along two perpendicular axes—usually labeled as the x-axis (horizontal) and y-axis (vertical). The most common context in which you'll encounter rectangular coordinates is on a typical grid system. Here’s a bit more detail:

• The x-coordinate (or abscissa): This is the horizontal distance from the origin (0,0) along the x-axis.
• The y-coordinate (or ordinate): This is the vertical distance from the origin along the y-axis.

Together, these two values \((x, y)\) specify the precise location of a point in a two-dimensional space. For example, the rectangular coordinate \((-1.5, \frac{3\sqrt{3}}{2})\) specifies a location left of the y-axis and up from the x-axis.

Coordinate Conversion

Converting between polar coordinates and rectangular coordinates is a common practice in mathematics, as these systems provide different insights into geometric and trigonometric problems. The conversion process involves the following steps:

• Identify the polar coordinates, typically \((r, \theta)\), where \(r\) is the radius (or distance from the pole), and \(\theta\) is the angle.
• Calculate the rectangular coordinates using these formulas:
  • \(x = r \cos(\theta)\)
  • \(y = r \sin(\theta)\)

By plugging in the values of \(r\) and \(\theta\), you can find the rectangular coordinates \((x, y)\). For example, for the polar coordinate \((3, \frac{2\pi}{3})\), the conversion leads to \(x = 3(-\frac{1}{2}) = -1.5\) and \(y = 3(\frac{\sqrt{3}}{2}) = \frac{3\sqrt{3}}{2}\). This process simplifies many calculations, especially when dealing with trigonometric functions.

Graphing Points

Graphing points in different coordinate systems is essential for understanding the spatial relationships between them. In a polar coordinate system, points are graphed based on their distance from a central point (the pole, similar to the origin in rectangular coordinates) and their angular direction (\(\theta\)). To plot a point in polar coordinates:

• Locate the angle \(\theta\), measured from the positive x-axis.
• Move outwards by \(r\) units from the pole in the direction of this angle.

For example, point \((3, \frac{2\pi}{3})\) would be 120 degrees counterclockwise from the positive x-axis and 3 units away from the pole. The key is converting these visual steps into plotting, which helps in gaining a more intuitive grasp of how coordinate systems work.

Trigonometric Functions

Trigonometric functions are fundamental tools for analyzing angles and distances, especially in the context of polar coordinates. The sine (\(\sin\)) and cosine (\(\cos\)) functions are particularly important here, as they are used to convert polar coordinates into rectangular coordinates:

• \(\cos(\theta)\) helps find the horizontal projection of the radius onto the x-axis.
• \(\sin(\theta)\) calculates the vertical projection onto the y-axis.

Using these functions, you can break down a polar coordinate’s distance \(r\) into x and y components. This break-down is crucial, not only in math but also in physics and engineering, to understand the vector's direction and magnitude. For instance, converting \((4, \frac{11\pi}{6})\) involves calculating \(x = 4 \cos(\frac{11\pi}{6})\) and \(y = 4 \sin(\frac{11\pi}{6})\), turning this polar point into a rectangular form for better analysis.