Problem 15 - 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 \(\theta \geq 2 \pi\) $$ \left(-\sqrt{5},-\frac{4 \pi}{3}\right) $$

Step-by-Step Solution

Answer

The equivalent polar expressions are: (a) 8-\sqrt{5}, \frac{2\pi}{3}\right), (b) \left(\sqrt{5}, -\frac{7\pi}{3}\right), (c) \left(\sqrt{5}, \frac{10\pi}{3}\right).

Step 1: Identify the Original Point

The given point in polar coordinates is \(\left(-\sqrt{5},-\frac{4\pi}{3}\right)\). In polar coordinates, \(r\) is the radius, and \(\theta\) is the angle in radians. The first task is to understand that our original point needs to be plotted with \(r = -\sqrt{5}\) and \(\theta = -\frac{4\pi}{3}\).

Step 2: Plot the Original Point

To plot the point \((-\sqrt{5}, -\frac{4\pi}{3})\), convert it to Cartesian coordinates using the formulas:\(x = r \cdot \cos(\theta)\)\(y = r \cdot \sin(\theta)\)Here, \(r = -\sqrt{5}\) and \(\theta = -\frac{4\pi}{3}\):\(x = -\sqrt{5} \cdot \cos(-\frac{4\pi}{3})\)\(y = -\sqrt{5} \cdot \sin(-\frac{4\pi}{3})\).Simplifying, \(x = -\sqrt{5} \cdot (-\frac{1}{2})\) and \(y = -\sqrt{5} \cdot (-\frac{\sqrt{3}}{2})\), we find that \(x = \frac{\sqrt{5}}{2}\) and \(y = \frac{\sqrt{15}}{2}\). Plot just as you would plot \((x, y)\) on Cartesian coordinates.

Step 3: Expression with Negative r

For expression (a), we need \(r < 0\) and change \(\theta\) to be within \(0 \leq \theta \leq 2 \pi\). Given \(r = -\sqrt{5}\) and \(\theta = -\frac{4\pi}{3}\), we adjust \(\theta\) by adding \(2\pi\), resulting in \(\theta = \frac{2\pi}{3}\).So, the point becomes \((-\sqrt{5}, \frac{2\pi}{3})\).

Step 4: Expression with Positive r, Negative θ

For expression (b), we need \(r > 0\) and \(\theta \leq 0\). Let \(r = \sqrt{5}\) by reversing the sign of both \(r\) and \(\theta\), so \(\theta = -\frac{4\pi}{3} - \pi \). Simplifying, \(\theta = -\frac{7\pi}{3}\).So the point becomes \((\sqrt{5}, -\frac{7\pi}{3})\).

Step 5: Expression with Positive r, Large θ

For expression (c), where \(r > 0\) and \(\theta \geq 2 \pi\), start with \(\theta = -\frac{4\pi}{3}\) and find an equivalent angle by adding \(2\pi\) twice to bring it beyond \(2\pi\), resulting in \(\theta = \frac{4\pi}{3} + 2\pi = \frac{10\pi}{3}\).Retain \(r = \sqrt{5}\). Thus, the point is \((\sqrt{5}, \frac{10\pi}{3})\).

Key Concepts

Converting Polar to Cartesian Coordinates

In the realm of coordinate systems, two of the most commonly used systems are polar coordinates and Cartesian coordinates. The transformation from polar to Cartesian coordinates involves using a set of specialized equations to convert the points. Polar coordinates are represented as \((r, \theta)\), where \(r\) is the radius (distance from origin) and \(\theta\) represents the angle from the positive x-axis. This makes it particularly useful for systems where the relationship of points is better described by angles and radii, such as in circular or rotation-based systems.
To convert these polar coordinates to Cartesian (rectangular) coordinates with points \((x, y)\), you use the formulas:

• \(x = r \cdot \cos(\theta)\)
• \(y = r \cdot \sin(\theta)\)

These equations essentially translate radial distance and angle into traditional x-y points by leveraging trigonometric functions. For instance, if we take an original point such as \((-\sqrt{5}, -\frac{4\pi}{3})\), the conversion process would involve calculations such as \(x = -\sqrt{5} \cdot \cos(-\frac{4\pi}{3})\) and \(y = -\sqrt{5} \cdot \sin(-\frac{4\pi}{3})\). Incorporating the cosine and sine values, you determine that \(x = \frac{\sqrt{5}}{2}\) and \(y = \frac{\sqrt{15}}{2}\). This enables you to plot the original polar point as a Cartesian point \((\frac{\sqrt{5}}{2}, \frac{\sqrt{15}}{2})\) in a two-dimensional space.

Negative Radius in Polar Coordinates

A unique aspect of polar coordinates is the potential for having a negative radius, which distinguishes it substantially from Cartesian coordinates. In this setup, the negative sign of the radius \(r\) effectively inverts the direction of the coordinate point across the origin. This means if the reference angle \(\theta\) points in a particular direction, a negative \(r\) would be reflected in the opposite direction on the same line.
For example, given an initial coordinate \((-\sqrt{5}, -\frac{4\pi}{3})\), the negative radius indicates the point is \(\sqrt{5}\) units away in the opposite direction from the line formed by an angle \(-\frac{4\pi}{3}\) radians. To maintain a suitable range for the angle \(\theta\) (i.e. within \(0 \leq \theta < 2\pi\)), an addition of \(2\pi\) can be applied to \(\theta\) when negative. Hence, \(-\frac{4\pi}{3} + 2\pi = \frac{2\pi}{3}\), altering the point to \((-\sqrt{5}, \frac{2\pi}{3})\).
In practical applications, this reflection and angle adjustment are critical for accurate plotting and interpretation of the point's position, especially in complex systems where both the angle range and point positioning matter.

Polar Angle Adjustments

The concept of adjusting angles in polar coordinates can transform the original point orientation without changing the actual location. This is key in resolving expressions given constraints such as angle ranges above or below certain thresholds. Understanding angle adjustments aids in converting negative angles into their positive equivalents and vice versa, or pushing angles above regular limits like \(2\pi\).
Take the original point \((-\sqrt{5}, -\frac{4\pi}{3})\) from the previous example. If you need an expression with a positive radius and angle but \(\theta\) should be less than \(0\), adjust both \(r\) and \(\theta\) by flipping their signs: \(\sqrt{5}\) and \(-\frac{7\pi}{3}\) respectively. On the other hand, to find an expression where \(\theta\) is greater than or equal to \(2\pi\), simply keep adding multiples of \(2\pi\) to the initial angle. This transforms \(-\frac{4\pi}{3}\) into \(\frac{10\pi}{3}\), marking it well above \(2\pi\).
These transformations are not just mathematical tricks; they maintain the same point location within different rotational frames of reference, crucial for complex analysis and applications like robotics or computer graphics.