Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

Plot the points whose polar coordinates follow. For each point, give four other pairs of polar coordinates, two with positive \(r\) and two with negative \(r\). (a) \(\left(3 \sqrt{2}, \frac{7}{2} \pi\right)\) (b) \(\left(-1, \frac{15}{4} \pi\right)\) (c) \(\left(-\sqrt{2},-\frac{2}{3} \pi\right)\) (d) \(\left(-2 \sqrt{2}, \frac{29}{2} \pi\right)\)

Step-by-Step Solution

Answer

Each point has four equivalent pairs: two with positive and two with negative \(r\) by adjusting \(\theta\).

Step 1: Understanding the Exercise

This exercise requires us to plot points given in polar coordinates and then find four other equivalent polar coordinates for each point: two with positive radius \(r\) and two with negative \(r\). A polar coordinate is defined by \((r, \theta)\) where \(r\) is the radius and \(\theta\) is the angle in radians. We know that polar coordinates \((r, \theta + 2k\pi)\) represent the same point for any integer \(k\), and \((-r, \theta + (2k+1)\pi)\) also represents the same point.

Step 2: Finding Equivalent Coordinates for (a)

For the point \((3 \sqrt{2}, \frac{7}{2} \pi)\):1. Add \(2\pi\) to the angle: \(\left(3 \sqrt{2}, \frac{7}{2} \pi + 2 \pi \right) = (3 \sqrt{2}, \frac{11}{2} \pi)\) 2. Subtract \(2\pi\) from the angle: \(\left(3 \sqrt{2}, \frac{7}{2} \pi - 2 \pi \right) = (3 \sqrt{2}, \frac{3}{2} \pi)\)3. Use negative \(r\) by adding \((2k+1)\pi\) to the angle: \((-3 \sqrt{2}, \frac{7}{2} \pi + \pi) = (-3 \sqrt{2}, \frac{9}{2} \pi)\)4. Use negative \(r\) again: \((-3 \sqrt{2}, \frac{7}{2} \pi -\pi) = (-3 \sqrt{2}, \frac{5}{2} \pi)\)

Step 3: Finding Equivalent Coordinates for (b)

For the point \((-1, \frac{15}{4} \pi)\):1. Add \(2\pi\) to the angle: \((-1, \frac{15}{4} \pi + 2 \pi) = (-1, \frac{23}{4} \pi)\) 2. Subtract \(2\pi\) from the angle: \((-1, \frac{15}{4} \pi - 2 \pi) = (-1, \frac{7}{4} \pi)\)3. Use positive \(r\) by adding \((2k+1)\pi\) to the angle: \((1, \frac{15}{4} \pi + \pi) = (1, \frac{19}{4} \pi)\)4. Use positive \(r\) again: \((1, \frac{15}{4} \pi - \pi) = (1, \frac{11}{4} \pi)\)

Step 4: Finding Equivalent Coordinates for (c)

For the point \((-\sqrt{2}, -\frac{2}{3} \pi)\):1. Add \(2\pi\) to the angle: \((-\sqrt{2}, -\frac{2}{3} \pi + 2 \pi) = (-\sqrt{2}, \frac{4}{3} \pi)\) 2. Subtract \(2\pi\) from the angle: \((-\sqrt{2}, -\frac{2}{3} \pi - 2 \pi) = (-\sqrt{2}, -\frac{8}{3} \pi)\)3. Use positive \(r\) by adding \((2k+1)\pi\) to the angle: \((\sqrt{2}, -\frac{2}{3} \pi + \pi) = (\sqrt{2}, \frac{1}{3} \pi)\)4. Use positive \(r\) again: \((\sqrt{2}, -\frac{2}{3} \pi - \pi) = (\sqrt{2}, -\frac{5}{3} \pi)\)

Step 5: Finding Equivalent Coordinates for (d)

For the point \((-2 \sqrt{2}, \frac{29}{2} \pi)\):1. Add \(2\pi\) to the angle: \((-2 \sqrt{2}, \frac{29}{2} \pi + 2 \pi) = (-2 \sqrt{2}, \frac{33}{2} \pi)\) 2. Subtract \(2\pi\) from the angle: \((-2 \sqrt{2}, \frac{29}{2} \pi - 2 \pi) = (-2 \sqrt{2}, \frac{25}{2} \pi)\)3. Use positive \(r\) by adding \((2k+1)\pi\) to the angle: \((2 \sqrt{2}, \frac{29}{2} \pi + \pi) = (2 \sqrt{2}, \frac{31}{2} \pi)\)4. Use positive \(r\) again: \((2 \sqrt{2}, \frac{29}{2} \pi - \pi) = (2 \sqrt{2}, \frac{27}{2} \pi)\)

Key Concepts

Equivalent Polar Coordinates

In polar coordinates, a point is represented by a pair \((r, \theta)\), where \(r\) is the radius, and \(\theta\) is the angle.Unlike Cartesian coordinates, a single point in polar coordinates can have multiple equivalent representations. This is because adding multiples of \(2\pi\) to the angle \(\theta\) does not change the point's position on the plane. This gives us the formula for equivalent points:

• For positive \(r\): \((r, \theta + 2k\pi)\)
• For negative \(r\): \((-r, \theta + (2k+1)\pi)\)

Here, \(k\) is any integer. This allows us to find multiple equivalent coordinates that all map to the same location.

Angle Adjustment

Adjusting the angle in polar coordinates is crucial in finding equivalent points. By adding or subtracting \(2\pi\), we can locate new equivalent coordinates without altering the point's position. This adjustment is based on the property that a complete circle is \(2\pi\) in radians. Thus,

• Adding \(2\pi\) results in a full circle rotation counter-clockwise.
• Subtracting \(2\pi\) results in a full circle clockwise rotation.

To change the sign of \(r\), adjust the angle by an additional \(\pi\) (half circle), so

• \(\theta + (2k+1)\pi\)

Mirrors the point across the origin.

Plotting Polar Points

Plotting points in polar coordinates involves moving from the origin along the angle \(\theta\) to the distance \(r\). Here's how it works:1. **Start from the Origin**: Begin plotting from the origin of the coordinate system.2. **Rotate by \(\theta\)**: Move counterclockwise (if \(\theta\) is positive) or clockwise (if negative) by the angle \(\theta\).3. **Move a Distance of \(r\)**: If \(r\) is positive, move away from the origin. If \(r\) is negative, move towards the origin direction reversed.This method of plotting leverages both the angle and the radius, creating a circular path around the origin.

Coordinate Systems

Polar and Cartesian coordinate systems offer different ways to represent points in a plane. While Cartesian uses regular \((x, y)\) coordinates, polar coordinates use a radius and an angle. Benefits of the polar system include:

• Easy expression of circular and radial symmetry.
• Simple representation of complex numbers and certain curves.

Switching between these systems can be beneficial:- Coordinates may be converted: - Convert polar to cartesian: - \(x = r \cos \theta\) - \(y = r \sin \theta\) - Convert cartesian to polar: - \(r = \sqrt{x^2 + y^2}\) - \(\theta = \tan^{-1}(\frac{y}{x})\)Understanding these systems allows more flexibility in solving geometric and algebraic problems.