Problem 10 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 9-12: Change the rectangular coordinates to polar coordinates with \(r>0\) and \(0 \leq \theta \leq 2 \pi\). (a) \((3 \sqrt{3}, 3)\) (b) \((2,-2)\)

Step-by-Step Solution

Answer

(a) \( (6, \frac{\pi}{6}) \) and (b) \( (2\sqrt{2}, \frac{7\pi}{4}) \).

Step 1: Understand Rectangular to Polar Conversion Formulas

To convert from rectangular to polar coordinates, we use the following formulas: \( r = \sqrt{x^2 + y^2} \) and \( \theta = \arctan\left(\frac{y}{x}\right) \). Here, \( r \) is the radial distance and \( \theta \) is the angle.

Step 2: Solve for r and θ for (3√3, 3)

The point is \((3\sqrt{3}, 3)\). Calculate \( r \): \[ r = \sqrt{(3\sqrt{3})^2 + 3^2} = \sqrt{27 + 9} = \sqrt{36} = 6 \]. Calculate \( \theta \): \[ \theta = \arctan\left(\frac{3}{3\sqrt{3}}\right) = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]. Therefore, the polar coordinates are \((6, \frac{\pi}{6})\).

Step 3: Solve for r and θ for (2,-2)

The point is \((2, -2)\). Calculate \( r \): \[ r = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]. Calculate \( \theta \): \( \theta = \arctan\left(\frac{-2}{2}\right) = \arctan(-1) = -\frac{\pi}{4} \). Since \( \theta \) should be between \(0\) and \(2\pi\), we add \(2\pi\): \( \theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4} \). Therefore, the polar coordinates are \((2\sqrt{2}, \frac{7\pi}{4})\).

Key Concepts

Rectangular to Polar Conversion

When it comes to converting rectangular coordinates to polar coordinates, the process is quite straightforward! Remember, rectangular coordinates use

• "x" for the horizontal position, and
• "y" for the vertical position.

Polar coordinates, on the other hand, use:

• "r" for radial distance (how far the point is from the origin), and
• "θ" (theta) for the angular position (which direction the point is located from the positive x-axis).

The conversion between these two systems is based on geometry. We use the formulas:

• \( r = \sqrt{x^2 + y^2} \)
• \( \theta = \arctan\left(\frac{y}{x}\right) \)

These formulas simplify the task of finding how far and in which direction a point lies from the origin. It’s like finding your way using a map versus a compass! Simply plug in your x and y values, and you will get r and θ, moving you from the world of grids to the circular realm of polar coordinates.

Radial Distance

Radial distance, often denoted by "r," is all about the measurement of the direct line from the origin (0,0) to a point in space. Whenever we want to measure something from a central point, this is the go-to measurement. It can be thought of as the length of the hypotenuse in a right-angled triangle, where the legs represent the point's x and y coordinates.
Using the formula \( r = \sqrt{x^2 + y^2} \), we find "r" by:

• Squaring the value of "x",
• Squaring the value of "y",
• Adding these squared values,
• Taking the square root of this sum.

This step is crucial since it gives you the magnitude of your distance from the origin, irrespective of the angle. It’s like figuring out the straight-line distance on a map without worrying about the turns.

Angle Calculation

Finding the angle is the next critical part of our conversion process. This angle, denoted by "\(\theta\)," marks the direction of the point from the x-axis. It's distinctly useful when you need to represent the position if you're handling directions or rotations. The basic way to acquire \(\theta\) is through:

• Using the formula \( \theta = \arctan\left(\frac{y}{x}\right) \)
• Remember that the resulting angle naturally ranges between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), due to the limitations of the \(\arctan\) function.

To adjust \(\theta\) into the preferred range of \([0, 2\pi)\), you often need to add \(\pi\) or \(2\pi\) depending on if the point lies in a different quadrant. This ensures that your angle proves accurate for the given polar coordinate stipulations.

Arctan Function

At the heart of angle calculation in polar coordinate conversion lies the arctan function, often known as the inverse tangent. It's a special function because it allows you to find the angle whose tangent is a particular number. In our scenario:

• The formula used is \( \theta = \arctan\left(\frac{y}{x}\right) \), giving you the angle by taking the inverse tangent of your y divided by x.
• This calculation is crucial when your x and y create a right-angle triangle from the origin to the point.
• Always take note that the \(\arctan\) function only directly provides angles in the \([-\frac{\pi}{2}, \frac{\pi}{2}]\) interval, aligning with angles in the first and fourth quadrants.

Therefore, handling angles in the opposite quadrants requires adjusting the angle by adding \(\pi\) or \(2\pi\) to shift your final angle into the adequate range of \([0, 2\pi)\). This makes the arctan function pivotal in accurately ascertaining polar angles!