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Chapter 10: Problem 64

Sketch the graph of the equation. $$r=-6 \sin \theta$$

Short Answer

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Based on the given step-by-step solution, provide a short answer to help sketch the graph of the polar equation $$r = -6 \sin \theta$$. 1. Calculate the polar coordinates at angles $$\theta = 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}$$. 2. The minimum value of $$r$$ is $$-6$$ when $$\theta = \frac{\pi}{2}$$, and the maximum value of $$r$$ is $$0$$ when $$\theta = 0$$ or $$\theta = \pi$$. The graph will "mirror" across the origin in quadrants II and IV. 3. Sketch the graph using the key points and behavior analysis. The graph should resemble a circle with radius 6 centered at the origin, covering half of the circle in quadrants I and III, and a "reflected" half-circle in quadrants II and IV.

Step by step solution

01

1. Plot Key Points

To get an idea of how the graph behaves, calculate the polar coordinate at different angles. Record these values in a table. Some good angles to start with are: $$0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}$$. \begin{tabular}{c c} $$\theta$$ & $$r=-6 \sin \theta$$ \\ \hline $$0$$ & $$0$$ \\ $$\frac{\pi}{6}$$ & $$-3$$ \\ $$\frac{\pi}{4}$$ & $$-3\sqrt{2}$$ \\ $$\frac{\pi}{3}$$ & $$-3\sqrt{3}$$ \\ $$\frac{\pi}{2}$$ & $$-6$$ \\ \end{tabular}
02

2. Analyze the Behavior

Examine how $$r$$ changes as $$\theta$$ varies, paying attention to the range of $$r$$ and its maximum and minimum values. For this equation, $$r=-6 \sin \theta$$, the minimum value of $$r$$ is $$-6$$ when $$\theta = \frac{\pi}{2}$$, and the maximum value of $$r$$ is $$0$$ when $$\theta = 0$$ or $$\theta = \pi$$. In general, the value of $$r$$ will be negative whenever $$\sin \theta$$ is positive, and positive when $$\sin \theta$$ is negative. Therefore, the graph will "mirror" across the origin in quadrants II and IV.
03

3. Sketch the Graph

Use the key points from step 1 and the behavior analysis in step 2 to sketch the graph. First, draw the points from the table. Then, connect the points, considering the analysis from step 2 - keep in mind the graph will "mirror" across the origin in quadrants II and IV. After connecting the points, you should see a graph that resembles a circle with radius 6 centered at the origin, but only covers half of the circle in quadrants I and III. The other half-circle is "reflected" across the origin in quadrants II and IV, since $$r$$ is negative.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Graphing Polar Equations
When graphing polar equations, such as \( r = -6 \sin \theta \), it's important to understand how polar coordinates work differently from Cartesian coordinates. Instead of \((x, y)\), we use \((r, \theta)\), where \( r \) is the distance from the origin and \( \theta \) is the angle from the positive x-axis. To create a graph, we generally:
  • Identify the range of values for \( \theta \) that are relevant. This often includes angles like 0, \( \frac{\pi}{6} \), \( \frac{\pi}{4} \), \( \frac{\pi}{3} \), and \( \frac{\pi}{2} \).
  • Calculate the corresponding \( r \) values for these angles.
  • Plot points on the polar plane and draw a curve through them.
This helps to visualize the behavior of the polar equation, which often results in distinctive symmetrical shapes like circles or roses.
Sine Function in Polar Coordinates
The sine function, \( \sin \theta \), plays a critical role in shaping the graph of a polar equation. In the equation \( r = -6 \sin \theta \), the value of \( \sin \theta \) determines the radial distance \( r \) from the origin:
  • When \( \sin \theta = 0 \), the radius \( r \) is zero, indicating that the point lies at the origin (e.g., \( \theta = 0 \) or \( \pi \)).
  • The maximum absolute value, as a negative, occurs at \( \theta = \frac{\pi}{2} \), resulting in \( r = -6 \).
The sine function introduces a periodic nature, repeating its pattern every \( 2\pi \) radians. This periodicity causes the graph to follow a predictable cyclical pattern, creating familiar shapes due to the symmetry and oscillating behavior of the sine function.
Negative Radius in Polar Graphs
The concept of a negative radius can be counterintuitive at first. In polar coordinates, a negative radius means that the point is located in the opposite direction of what the angle \( \theta \) indicates. In the equation \( r = -6 \sin \theta \), when \( \sin \theta \) is positive, \( r \) is negative. Therefore, points are plotted by moving in the opposite direction from the typical placement:
  • A negative radius essentially flips the position 180 degrees across the origin.
  • This results in the graph appearing to "mirror" itself in quadrants II and IV when it would normally reside in quadrants I and III, respectively.
Understanding negative radii helps in predicting and sketching polar graphs accurately, highlighting the importance of directionality in polar coordinates.

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Most popular questions from this chapter

If \(a>0\) and \(b>0,\) then the eccentricity of the hyperbola $$\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1 \quad \text { or } \quad \frac{(y-k)^{2}}{a^{2}}-\frac{(x-h)^{2}}{b^{2}}=1$$ is the number \(\frac{\sqrt{a^{2}+b^{2}}}{a} .\) Find the eccentricity of the hyperbola whose equation is given. $$6(y-2)^{2}=18+3(x+2)^{2}$$

Find a rectangular equation that is equivalent to the given polar equation. $$r=2 \sin \theta$$

The given curve is part of the graph of an equation in \(x\) and \(y .\) Find the equation by eliminating the parameter. $$x=4 \sin 2 t, \quad y=2 \cos 2 t, \quad 0 \leq t \leq 2 \pi$$

Consider the ellipse whose equation is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 .\) Show that if \(a=b,\) then the graph is actually a circle.

Use the information given in Special Topics 10.3. A and summarized in the endpapers at the beginning of this book to find a parameterization of the conic section whose rectangular equation is given. Confirm your answer by graphing. $$\frac{(x-2)^{2}}{10}+\frac{(y+5)^{2}}{20}=1$$
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