Problem 28 - Precalculus with Calculus Previews

Question

Find polar coordinates that satisfy (a) \(r>0,-\pi<\theta \leq \pi\) (b) \(r<0,-\pi<\theta \leq \pi\) for each point with the given rectangular coordinates. $$ (\sqrt{6}, \sqrt{2}) $$

Step-by-Step Solution

Answer

(a) Polar coordinates: \((2\sqrt{2}, \frac{\pi}{6})\); (b) Polar coordinates: \((-2\sqrt{2}, -\frac{5\pi}{6})\)."

Step 1: Convert rectangular to polar coordinates

The formula to convert from rectangular coordinates \((x, y)\) to polar coordinates \((r, \theta)\) is given by \( r = \sqrt{x^2 + y^2} \) and \( \theta = \arctan\left(\frac{y}{x}\right) \). Given the point \((\sqrt{6}, \sqrt{2})\), we first need to compute \(r\).

Step 2: Calculate the radius

Compute \( r = \sqrt{(\sqrt{6})^2 + (\sqrt{2})^2} = \sqrt{6 + 2} = \sqrt{8} = 2\sqrt{2} \). Therefore, the radius \( r = 2\sqrt{2} \).

Step 3: Calculate the angle

Calculate the angle using \( \theta = \arctan\left(\frac{\sqrt{2}}{\sqrt{6}}\right) \). Simplifying this we get \( \theta = \arctan\left(\frac{1}{\sqrt{3}}\right) \), which equals \( \frac{\pi}{6} \). Therefore, the angle \( \theta = \frac{\pi}{6} \).

Step 4: Solution for case (a)

For case (a), where \(r > 0\), the polar coordinates are \((2\sqrt{2}, \frac{\pi}{6})\) because we have already established \( r > 0 \) and \( -\pi < \theta \leq \pi \).

Step 5: Solution for case (b)

For case (b), where \(r < 0\), we keep the radius negative and adjust the angle by adding or subtracting \(\pi\) to fall back into the correct range. We find that \( -r = -2\sqrt{2} \), and the equivalent angle is \( \theta = \frac{\pi}{6} - \pi = -\frac{5\pi}{6} \). Therefore, the polar coordinates are \((-2\sqrt{2}, -\frac{5\pi}{6})\).

Key Concepts

Rectangular to Polar Conversion

Converting rectangular coordinates to polar coordinates is a fundamental concept in mathematics that helps us describe a point in a plane via a different approach.

Rectangular coordinates, also known as Cartesian coordinates, use two values—\(x\) and \(y\)—to locate a point in a two-dimensional plane, where \(x\) is the horizontal position, and \(y\) is the vertical position. Polar coordinates, on the other hand, describe a point using a radius \(r\) and an angle \(\theta\). This system can be more intuitive in scenarios involving circular motion where rotation and distance from a center point matter.

The transformation from rectangular to polar coordinates is done using a specific set of formulas:

• The radius \(r\) is calculated as \( r = \sqrt{x^2 + y^2} \), which represents the distance from the origin to the point.
• The angle \(\theta\) is found using \( \theta = \arctan\left(\frac{y}{x}\right) \), which gives the direction from the positive x-axis to the point. However, further adjustments might be needed depending on the quadrant the point is located in, as the inverse tangent only accounts for angles between \(-\frac{\pi}{2}\) and \( \frac{\pi}{2} \).

Transitioning between these two systems enhances our ability to solve various mathematical and real-world problems effectively.

Trigonometric Functions

Trigonometric functions play a crucial role when converting between rectangular and polar coordinates. Understanding these functions helps in determining the angle \(\theta\) in polar coordinates.

The primary trigonometric function used in this context is the arctangent function, \(\arctan\), which is the inverse of the tangent function. It allows us to determine the angle when we have the ratio of the opposite side to the adjacent side in a right-angled triangle—in this case, \(\frac{y}{x}\). However, since the range of the \(\arctan\) function is limited between \(-\frac{\pi}{2}\) and \( \frac{\pi}{2}\), one might need to consider the signs of \(x\) and \(y\) to find the correct polar angle in the interval of \(-\pi\) to \(\pi\).

For example, using the coordinates \((\sqrt{6}, \sqrt{2})\), the ratio \(\frac{y}{x}\) simplifies to \(\frac{1}{\sqrt{3}}\), which corresponds to an angle \(\theta = \arctan\left(\frac{1}{\sqrt{3}}\right)\). This results in an angle of \(\frac{\pi}{6}\), reflecting the positive angle within the stipulated range. These trigonometric insights are pivotal for accurately determining the angular component of the polar coordinate system.

Radius and Angle Calculation

Calculating the radius \(r\) and angle \(\theta\) is central to converting rectangular coordinates to polar coordinates. The process begins by finding the radius, which serves as the first step and provides a sense of the point's distance from the origin.

• To find \(r\), use the formula \( r = \sqrt{x^2 + y^2} \). For the point \((\sqrt{6}, \sqrt{2})\), it simplifies to \(\sqrt{(\sqrt{6})^2 + (\sqrt{2})^2} = 2\sqrt{2}\).

After determining the radius, the next step is to find the angle \(\theta\):

• The angle \(\theta\) is calculated as \(\arctan\left(\frac{y}{x}\right)\). For \((\sqrt{6}, \sqrt{2})\), this is \(\theta = \arctan\left(\frac{1}{\sqrt{3}}\right)\), leading to an angle of \(\frac{\pi}{6}\).
• For cases involving negative radii or specific quadrant instructions, adjustments such as \(\theta + \pi\) or \(\theta - \pi\) may be necessary. This brings the angle back into the specified range acceptable for polar coordinates.

These calculations—seemingly simple—require careful attention to the results they produce, especially concerning the angle's reference and the sign of the radius. Accuracy in these calculations ensures the correct polar representation of a point.