Problem 12 - College Trigonometry

Question

Plot the point given in polar coordinates and then give three different expressions for the point such that (a) \(r<0\) and \(0 \leq \theta \leq 2 \pi\) (b) \(r>0\) and \(\theta \leq 0\) (c) \(r>0\) and \(\theta \geq 2 \pi\) $$ \left(-3, \frac{\pi}{2}\right) $$

Step-by-Step Solution

Answer

Expressions: (a) \((-3, \frac{\pi}{2})\), (b) \((3, -\frac{\pi}{2})\), (c) \((3, \frac{5\pi}{2})\).

Step 1: Understanding Polar Coordinates

Polar coordinates are given as \((r, \theta)\), where \(r\) is the radius (distance from the origin) and \(\theta\) is the angle measured counterclockwise from the positive x-axis. The given point \((-3, \frac{\pi}{2})\) means go 3 units towards the negative x-direction along the 90-degree line (since \(\theta = \frac{\pi}{2}\)).

Step 2: Plotting the Point

To plot the given point in polar coordinates \((-3, \frac{\pi}{2})\), interpret \(r = -3\) as moving 3 units in the opposite direction of \(\frac{\pi}{2}\), which is downward on the vertical axis (since \(\frac{\pi}{2}\) is pointing up). So, the point lies on the positive vertical axis.

Step 3: Expression for \(r<0\) and \(0 \leq \theta \leq 2 \pi\)

When \(r<0\), we already have \(r = -3\) with \(\theta = \frac{\pi}{2}\), which satisfies the original condition. Thus, one expression is \((-3, \frac{\pi}{2})\).

Step 4: Expression for \(r>0\) and \(\theta \leq 0\)

To express the point such that \(r>0\), write \(r = 3\) and adjust \(\theta\) by adding \(\pi\) to the angle to offset the initial negative direction. So, \(\theta = \frac{\pi}{2} - \pi = -\frac{\pi}{2}\). Therefore, one expression is \((3, -\frac{\pi}{2})\).

Step 5: Expression for \(r>0\) and \(\theta \geq 2 \pi\)

To express the point such that \(r>0\), keep \(r = 3\) and add \(2\pi\) to the angle \(\frac{\pi}{2}\), moving it out of the standard range and into multiple rotations. Therefore, \(\theta = \frac{\pi}{2} + 2\pi = \frac{5\pi}{2}\). So another expression is \((3, \frac{5\pi}{2})\).

Key Concepts

Plotting Points in Polar Coordinates

Plotting points using polar coordinates can seem a bit different compared to the familiar Cartesian system. In polar coordinates, a point is represented by \(r, \theta\), where \(r\) is the radial distance from the origin, and \(\theta\) is the angular measurement from the positive x-axis. This essentially means you have to determine how far out you go in a given direction.
For example, with the point \((-3, \pi/2)\), the negative sign in the radius \(r\) indicates you need to move in the opposite direction of the angle provided. Here \(\theta = \pi/2\) places the angle pointing upward along the y-axis. Hence, moving 3 units in the negative direction would place the point downward on the y-axis.

• Find the direction indicated by \(\theta\).
• Check if \(r\) is positive or negative to determine the movement direction.
• Plot accordingly, considering the axis and angle its extending towards.

Understanding Negative Radius in Polar Coordinates

In polar coordinates, the radius \(r\) often dictates the distance from the origin, while the angle \(\theta\) orients the direction. However, having a negative radius can trip many students up. It doesn't mean you simply have a negative distance (which doesn't conceptually make sense—how can you go a negative distance?); rather, it implies shifting direction by \(-\pi\). This rotates your point around the origin by 180 degrees, essentially reflecting it through the center.
For the point \((-3, \pi/2)\):

• \(-3\) indicates movement opposite to the direction given by \(\pi/2\), pointing downward.
• With \(r<0\) and \( heta = \pi/2\), just visualize flipping the direction by the defined angle.
• This means changing the direction upward towards the negative part of the axis specified.

Understanding this helps when shifting your point into corresponding positions with positive \(r\) and negative \(\theta\).

Angle Conversions in Polar Coordinates

Angles in polar coordinates give many options for representing the same direction on a circle. Converting angles is key to fully understanding and presenting all possible equivalencies of points.
When a condition such as \(\theta \leq 0\) is required, adjusting the angle by subtracting necessary rotations (like subtracting \(\pi\)) helps center the angle negatively. On the other hand, modifying with \(\theta \geq 2\pi\) often entails adding full rotations (like adding \(2\pi\)) to remain within the positive angle bounds without altering the direction.
In example \((-3, \pi/2)\), converting:

• For \(\theta \leq 0\), transition by changing \(\theta = \pi/2 - \pi = -\pi/2\).
• For \(\theta \geq 2\pi\), apply additional rotations to align with requested constraints, generating \(\frac{5\pi}{2}\).

Each transformation helps manage and represent angle specifications, with practical angle conversions enhancing comprehension of polar placement.