Problem 14 - 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 \(\theta \geq 2 \pi\) $$ \left(-2.5,-\frac{\pi}{4}\right) $$

Step-by-Step Solution

Answer

Expressions of the point: (a) \((-2.5, \frac{7\pi}{4})\), (b) \((2.5, -\frac{5\pi}{4})\), (c) \((2.5, \frac{11\pi}{4})\)."

Step 1: Understand Polar Coordinates

The given point in polar coordinates is \((-2.5, -\frac{\pi}{4})\), where \( r = -2.5 \) is the radius, and \( \theta = -\frac{\pi}{4} \) is the angle. Polar coordinates are represented by the distance from the origin and the angle measured counterclockwise from the positive x-axis.

Step 2: Convert to Cartesian Coordinates

To understand better, convert the polar coordinates \((-2.5, -\frac{\pi}{4})\) to Cartesian: The formulas are: \( x = r \cos \theta \) and \( y = r \sin \theta \). Substituting values: \( x = -2.5 \cos\left(-\frac{\pi}{4}\right) = -2.5 \times \frac{\sqrt{2}}{2} = -\frac{2.5\sqrt{2}}{2} \) \( y = -2.5 \sin\left(-\frac{\pi}{4}\right) = -2.5 \times -\frac{\sqrt{2}}{2} = \frac{2.5\sqrt{2}}{2} \). Thus, the Cartesian coordinates are approximately \((-1.77, 1.77)\).

Step 3: Find Expression (a) with \(r < 0\) and \(0 \leq \theta \leq 2\pi\)

Since \( r < 0 \) is already true, adjust \( \theta = -\frac{\pi}{4} \) to be positive by adding \( 2\pi \). Use \( \theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4} \). So, the new expression is \((-2.5, \frac{7\pi}{4})\).

Step 4: Find Expression (b) with \(r > 0\) and \(\theta \leq 0\)

To make \( r \) positive, consider the opposite point. Change \( r = -2.5 \) to \( r = 2.5\) and add \( \pi \) to the angle: \( \theta = -\frac{\pi}{4} + \pi = \frac{3\pi}{4} \) modulo \( 2\pi \). Express \( \frac{3\pi}{4} \) in negatives by subtracting \( 2\pi \): \( \theta = -\frac{5\pi}{4} \). Hence, the new expression is \((2.5, -\frac{5\pi}{4})\).

Step 5: Find Expression (c) with \(r > 0\) and \(\theta \geq 2 \pi\)

Using \( r = 2.5 \) and \( \theta = -\frac{\pi}{4} + \pi \) from Step 4, adjust this to be above \( 2\pi \): \( \theta = \frac{3\pi}{4} + 2\pi = \frac{11\pi}{4} \). Thus, the expression is \((2.5, \frac{11\pi}{4})\).

Key Concepts

Cartesian coordinates

When dealing with polar coordinates, you might often find the need to convert them to Cartesian coordinates for easier calculations or visualization. The Cartesian coordinate system is what most of us are familiar with; it uses two perpendicular axes known as the x-axis and y-axis to define a point's position in a plane. Here, every point is determined by a pair of numerical coordinates, which are the signed distances from the point to the axes. These coordinates are typically represented as

• **x** – the horizontal distance from the y-axis
• **y** – the vertical distance from the x-axis

To convert from polar to Cartesian coordinates, we use the formulas:\[ x = r \cos\theta \]\[ y = r \sin\theta \]Given a point in polar coordinates \((-2.5, -\frac{\pi}{4})\), these formulas allow us to translate it into a more relatable form of approximately \((-1.77, 1.77)\) in the Cartesian coordinate system. This conversion helps us understand the geometry of points more intuitively.

angle conversion

Angle conversion is an essential step when working with polar coordinates, especially when you need to adjust the angle to meet specific conditions, such as

• **being within a certain range**
• **expressed in terms beyond their usual limits**

A common scenario involves converting negative angles into positive equivalents. For instance, converting \(\theta = -\frac{\pi}{4}\) involves adding \(2\pi\) to it to make it positive, resulting in \(\theta = \frac{7\pi}{4}\).Additionally, when adjusting angles so that they remain within specific limits, such as when you need \(\theta \geq 2\pi\), adding \(2\pi\) conversions like \[\theta = \frac{3\pi}{4} + 2\pi = \frac{11\pi}{4}\] are required. By managing angle conversions accurately, you can visualize and express polar coordinates properly within the desired framework.

radius adjustments

Radius adjustments are crucial when defining or converting polar coordinates as they reveal not only the distance from the origin but also interact with angle adjustments to fully define a point's location. In polar coordinates, the radius is represented by **r**:

• **r > 0** implies the point lies in the same direction as the angle \(\theta\).
• **r < 0** implies the point is reflected about the origin, lying opposite to the direction indicated by \(\theta\).

To change expressions, if you want the radius to be positive yet retain the original location, you need to modify \(\theta\) by adding \(\pi\) to shift the angle by 180 degrees, consequently changing its direction. As data points often need to be adjusted for specific conditions, these manipulations are essential. Therefore, a negative radius like \(-2.5\) can be adjusted to **2.5** by reflecting the angle appropriately, assuring that your data remains consistent and understandable.

plotting points

Plotting points in polar coordinates can at first seem more complex than plotting in Cartesian coordinates, but here's the simplicity behind it. When you have a polar coordinate expressed as \((r, \theta)\):

• Start with the polar grid, where the angle \(\theta\) signifies a direction, similar to a clock face.
• Move outwards from the center of the grid by the radius **r**.

These steps allow you to accurately place the point on the grid. For practical applications, such as when polar plots are utilized for beautiful, circular designs in engineering and physics, understanding how to convert and plot these points is essential. When adjusting expressions for the same point with different terms, ensure the calculated angle and radius precisely recreate the same geometric location.