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

Convert polar coordinates by adding/subtracting multiples of \(2\pi\) to the angle and reversing the sign of \(r\).

Step 1: Understanding Polar Coordinates

Polar coordinates comprise a pair \((r, \theta)\) where \(r\) is the radial distance from the origin and \(\theta\) is the angle with the positive x-axis measured counter-clockwise. Given points must be converted to equivalent polar coordinates with both positive and negative radial distances.

Step 2: Plotting and Finding Equivalent Coordinates for (a)

For \( ext{(a)} \left(3 \sqrt{2}, \frac{7}{2} \pi\right) \), the point lies at a distance of \(3 \sqrt{2}\) at an angle of \(\frac{7}{2} \pi\). To find other coordinates: \(\theta' = \theta + 2k\pi\) (where \(k\) is an integer):- Positive \(r\): \((3 \sqrt{2}, \frac{7}{2} \pi - 2\pi) = (3 \sqrt{2}, -\frac{\pi}{2}) \) and \( (3 \sqrt{2}, \frac{7}{2} \pi + 2\pi) = (3 \sqrt{2}, \frac{11}{2} \pi) \)- Negative \(r\): \((-3 \sqrt{2}, \frac{7}{2} \pi - \pi) = (-3 \sqrt{2}, \frac{5}{2} \pi) \) and \( (-3 \sqrt{2}, \frac{7}{2} \pi + \pi) = (-3 \sqrt{2}, \frac{9}{2} \pi) \).

Step 3: Plotting and Finding Equivalent Coordinates for (b)

For \( ext{(b)} \left(-1, \frac{15}{4} \pi\right)\), the point has \(r = -1\), which means the direction is reversed. To find other coordinates:- Positive \(r\): \((1, \frac{15}{4} \pi + \pi) = (1, \frac{19}{4} \pi) \) and \( (1, \frac{15}{4} \pi - \pi) = (1, \frac{11}{4} \pi) \)- Negative \(r\): \((-1, \frac{15}{4} \pi - 2\pi) = (-1, -\frac{\pi}{4}) \) and \( (-1, \frac{15}{4} \pi + 2\pi) = (-1, \frac{23}{4} \pi) \).

Step 4: Plotting and Finding Equivalent Coordinates for (c)

For \( ext{(c)} \left(-\sqrt{2}, -\frac{2}{3} \pi\right)\), with a negative \(\theta\), convert to positive direction first: \(-\frac{2}{3} \pi + \pi = \frac{1}{3} \pi\). Now:- Positive \(r\): \((\sqrt{2}, \frac{1}{3} \pi + \pi) = (\sqrt{2}, \frac{4}{3} \pi) \) and \( (\sqrt{2}, \frac{1}{3} \pi - \pi) = (\sqrt{2}, -\frac{2}{3}\pi) \)- Negative \(r\): \((-\sqrt{2}, \frac{1}{3} \pi) \) and \( (-\sqrt{2}, \frac{1}{3} \pi + 2\pi) = (-\sqrt{2}, \frac{7}{3} \pi) \).

Step 5: Plotting and Finding Equivalent Coordinates for (d)

For \( ext{(d)} \left(-2 \sqrt{2}, \frac{29}{2} \pi\right)\), converting direction:- Positive \(r\): \((2 \sqrt{2}, \frac{29}{2} \pi + \pi) = (2 \sqrt{2}, \frac{31}{2} \pi) \) and \( (2 \sqrt{2}, \frac{29}{2} \pi - \pi) = (2 \sqrt{2}, \frac{27}{2} \pi) \)- Negative \(r\): \((-2 \sqrt{2}, \frac{29}{2} \pi - 2\pi) = (-2 \sqrt{2}, \frac{25}{2} \pi) \) and \( (-2 \sqrt{2}, \frac{29}{2} \pi + 2\pi) = (-2 \sqrt{2}, \frac{33}{2} \pi) \).

Key Concepts

Radial Distance

In polar coordinates, the radial distance, denoted as \(r\), is a measure of how far a point is from the origin, or the pole. It's comparable to the radius of a circle extending outward from the origin to a specific point on a plane.
When dealing with polar coordinates, remember this:

• A positive radial distance pushes the point in the usual direction, away from the origin. This direction follows the angle \(\theta\).
• A negative radial distance flips the point, reversing the direction through the origin, like a mirror image.

The radial distance plays a significant role as it directly influences the position of the point in the polar coordinate system. Whether positive or negative, it determines the exact location relative to the center point.

Angle Transformation

Angles in polar coordinates are measured from the positive x-axis, going counter-clockwise. One key factor here is the flexibility of angle measurement:

• An angle \(\theta\) can be transformed to \(\theta + 2k\pi\), where \(k\) is an integer. This means we can add or subtract full revolutions \(2\pi\) to determine equivalent angles.
• This ability to adjust \(\theta\) ensures a rich array of possibilities for representing the same point. It allows us to keep the angle positive or adjust for negative situations seamlessly.

Angle transformation in polar coordinates facilitates finding multiple representations for any given point's angle, making polar plots versatile.

Coordinate Plotting

Coordinate plotting in polar coordinates involves translating \((r, \theta)\) pairs into actual points on the plane. The process differs from Cartesian plotting:
Here's how to plot polar coordinates effectively:

• Start at the origin and extend out to the radial distance \(r\).
• From there, move along the angle \(\theta\) from the positive x-axis. Think of rotating around to the precise angle.
• Adjust \(\theta\) if needed, using angle transformations for clarity.

This method helps visualize points in their polar form. Moreover, understanding the complexities of angle and distance allows for accurate representation.

Negative Radial Distance

Handling a negative radial distance is unique to polar coordinates, offering a fascinating twist to plotting. Here's what it does:

• Instead of moving towards the angle \(\theta\), a negative \(r\) directs the point in the opposite direction, through the origin.
• This effectively reflects the point across the origin, creating a mirror image along the line of \(\theta\).
• Essentially, using negative radial distances helps visualize points that might otherwise be challenging to plot using only positive values.

Understanding how to handle negative radial distances is crucial for mastering polar coordinates, providing essential insights and flexibility in plotting.