Problem 10 - Algebra and Trigonometry

Question

Multiple Choice The point \(\left(5, \frac{\pi}{6}\right)\) can also be represented by which polar coordinates? (a) \(\left(5,-\frac{\pi}{6}\right)\) (b) \(\left(-5, \frac{13 \pi}{6}\right)\) (c) \(\left(5,-\frac{5 \pi}{6}\right)\) (d) \(\left(-5, \frac{7 \pi}{6}\right)\)

Step-by-Step Solution

Answer

(d) \(-5, \frac{7\pi}{6}\)

Step 1 - Understanding Polar Coordinates

Polar coordinates \(r, θ\) describe a point in terms of its distance from the origin (\(r\)) and its angle from the positive x-axis (\(θ\)). A point can have multiple representations due to the periodic nature of angles.

Step 2 - Analyzing the Given Point

The given point is \(5, \frac{\pi}{6}\). Here, \(r = 5\) and \(θ = \frac{\pi}{6}\). We are to find another set of polar coordinates that represent the same point.

Step 3 - Option (a) Evaluation

Option (a) is \(5, -\frac{\pi}{6}\). This represents a point with the same radius but different angle. Since -\frac{\pi}{6} is not equivalent to \frac{\pi}{6}, this option is not a correct conversion.

Step 4 - Option (b) Evaluation

Option (b) is \(-5, \frac{13 \pi}{6}\). Here, the radius is negative, which means the point is on the opposite direction of the vector. For a negative radius, add \pi to the angle. \(-5, \frac{13\pi}{6} = \frac{13\pi}{6} - \pi = \frac{13\pi - 6\pi}{6} = \frac{7\pi}{6}\). This does not bring us back to the original angle.

Step 5 - Option (c) Evaluation

Option (c) is \(5, -\frac{5 \pi}{6}\). While the radius remains positive, the angle is still not equivalent to \frac{\pi}{6}.

Step 6 - Option (d) Evaluation

Option (d) is \(-5, \frac{7\pi}{6}\). With a negative radius, we add \pi to the given angle. \(\frac{7\pi}{6} - \pi = \frac{7\pi}{6} - \frac{6\pi}{6} = \frac{\pi}{6}\), which is the original angle. Therefore, this representation is correct.

Key Concepts

Angle Conversion in Polar Coordinates

In polar coordinates, angles can be represented in various ways due to their periodic nature. Every angle can be expressed as \( \theta + 2k\pi \), where \( k \) is an integer. This means you can add or subtract any multiple of \( 2\pi \) without changing the angle's direction. For example, the angle \( \frac{\pi}{6} \) can also be expressed as \( \frac{\pi}{6} + 2\pi = \frac{\pi}{6} + 4\pi = ... \). This allows us to have multiple equivalent angles. It’s essential to remember that converting angles involves recognizing these periodic properties.
If you encounter a negative angle, like in option (a) \( \frac{-\pi}{6} \), it can be converted to a positive angle by adding \( 2\pi \). In this case, \( \frac{-\pi}{6} + 2\pi = \frac{11\pi}{6}\), which is not equal to \( \frac{\pi}{6} \). This conversion helps identify if different points represent the same spot in polar coordinates.

Negative Radius in Polar Coordinates

Polar coordinates can have a negative radius, which means the point is in the opposite direction along the same line. To handle a negative radius, you add \( \pi \) to the given angle. For example, option (d) gives \( -5, \frac{7\pi}{6} \). Here, the radius is \( -5\), making the vector point backwards.
To convert this to a positive radius, we add \( \pi\) to the angle: \( \frac{7\pi}{6} - \pi = \frac{7\pi}{6} - \frac{6\pi}{6} = \frac{\pi}{6}\). This transformed angle matches one of the possible representations of \( \frac{\pi}{6} \), and thus, \( -5, \frac{7\pi}{6} \) represents the same point as the original \( 5, \frac{\pi}{6} \). It is important to understand this relationship to accurately interpret and convert polar coordinates.

Multiple Representations of Points in Polar Coordinates

In polar coordinates, a single point can be represented in multiple ways. This happens because of the periodic nature of angles and the ability to use a negative radius. For the point \( 5, \frac{\pi}{6} \), there are several equivalent representations:

• Adding \( 2\pi \) repeatedly to the angle remains within the radius: \( 5, \frac{\pi}{6} + 2\pi = 5, \frac{13\pi}{6} \).
• Converting to a negative radius and adjusting the angle: \( -5, \frac{7\pi}{6}\).

These different forms help us understand the flexibility of polar coordinates and their application. Option (d) in the given exercise correctly demonstrates a valid transformation, showing the versatility of this system. Understanding these aspects of polar coordinates is crucial for solving problems and recognizing all valid representations of a point.