Polar Coordinates and Symmetry

Oct 13, 2025 by pynkocean.com 75 views

Introduction

In the realm of mathematics, particularly in the study of polar coordinates, understanding the symmetry of graphs is crucial for visualizing and analyzing complex equations. The graph of a polar equation can exhibit various types of symmetry, including symmetry about the x-axis, y-axis, and origin. In this article, we will delve into the graph of the polar equation $r=\frac{1}{1-5 \sin \theta}$ and determine its symmetry.

Polar Coordinates and Symmetry

Before we dive into the specific equation, let's briefly review the concept of polar coordinates and symmetry. In polar coordinates, a point in the plane is represented by an ordered pair $(r, \theta)$ , where $r$ is the distance from the origin to the point and $\theta$ is the angle between the positive x-axis and the line segment connecting the origin to the point.

Symmetry in polar coordinates can be defined as follows:

Symmetry about the x-axis: If the graph of a polar equation is symmetric about the x-axis, then for every point $(r, \theta)$ on the graph, the point $(r, -\theta)$ is also on the graph.
Symmetry about the y-axis: If the graph of a polar equation is symmetric about the y-axis, then for every point $(r, \theta)$ on the graph, the point $(r, \pi + \theta)$ is also on the graph.
Symmetry about the origin: If the graph of a polar equation is symmetric about the origin, then for every point $(r, \theta)$ on the graph, the point $(-r, \pi + \theta)$ is also on the graph.

The Graph of $r=\frac{1}{1-5 \sin \theta}$

Now, let's focus on the graph of the polar equation $r=\frac{1}{1-5 \sin \theta}$ . To determine its symmetry, we need to analyze the equation and identify any patterns or properties that can help us determine its symmetry.

The given equation can be rewritten as:

r = \frac{1}{1 - 5 \sin \theta}

This equation represents a type of polar curve known as a limacon. A limacon is a polar curve that has a loop or a cusp in its graph.

To determine the symmetry of the graph, we need to examine the equation and identify any symmetries about the x-axis, y-axis, or origin.

Symmetry about the x-axis

To check for symmetry about the x-axis, we need to see if the graph is the same when $\theta$ is replaced by $-\theta$ . Let's substitute $-\theta$ into the equation:

r = \frac{1}{1 - 5 \sin (-\theta)}

Using the property of the sine function that $\sin (-\theta) = -\sin \theta$ , we can rewrite the equation as:

r = \frac{1}{1 + 5 \sin \theta}

Comparing this equation with the original equation, we can see that the graph is not the same when $\theta$ is replaced by $-\theta$ . Therefore, the graph of $r=\frac{1}{1-5 \sin \theta}$ is not symmetric about the x-axis.

Symmetry about the y-axis

To check for symmetry about the y-axis, we need to see if the graph is the same when $\theta$ is replaced by $\pi + \theta$ . Let's substitute $\pi + \theta$ into the equation:

r = \frac{1}{1 - 5 \sin (\pi + \theta)}

Using the property of the sine function that $\sin (\pi + \theta) = -\sin \theta$ , we can rewrite the equation as:

r = \frac{1}{1 + 5 \sin \theta}

Comparing this equation with the original equation, we can see that the graph is not the same when $\theta$ is replaced by $\pi + \theta$ . Therefore, the graph of $r=\frac{1}{1-5 \sin \theta}$ is not symmetric about the y-axis.

Symmetry about the origin

To check for symmetry about the origin, we need to see if the graph is the same when $r$ is replaced by $-r$ and $\theta$ is replaced by $\pi + \theta$ . Let's substitute $-r$ and $\pi + \theta$ into the equation:

-r = \frac{1}{1 - 5 \sin (\pi + \theta)}

Using the property of the sine function that $\sin (\pi + \theta) = -\sin \theta$ , we can rewrite the equation as:

-r = \frac{1}{1 + 5 \sin \theta}

Comparing this equation with the original equation, we can see that the graph is not the same when $r$ is replaced by $-r$ and $\theta$ is replaced by $\pi + \theta$ . Therefore, the graph of $r=\frac{1}{1-5 \sin \theta}$ is not symmetric about the origin.

Conclusion

In conclusion, the graph of the polar equation $r=\frac{1}{1-5 \sin \theta}$ is not symmetric about the x-axis, y-axis, or origin. Therefore, the correct answer is:

D. No symmetry

Frequently Asked Questions

In the previous article, we analyzed the graph of the polar equation $r=\frac{1}{1-5 \sin \theta}$ and determined its symmetry. However, we understand that some readers may still have questions about the graph and its properties. In this article, we will address some of the most frequently asked questions about the graph of $r=\frac{1}{1-5 \sin \theta}$ .

Q: What is the shape of the graph of $r=\frac{1}{1-5 \sin \theta}$ ?

A: The graph of $r=\frac{1}{1-5 \sin \theta}$ is a type of polar curve known as a limacon. A limacon is a polar curve that has a loop or a cusp in its graph.

Q: Is the graph of $r=\frac{1}{1-5 \sin \theta}$ symmetric about the x-axis?

A: No, the graph of $r=\frac{1}{1-5 \sin \theta}$ is not symmetric about the x-axis. When $\theta$ is replaced by $-\theta$ , the graph is not the same.

Q: Is the graph of $r=\frac{1}{1-5 \sin \theta}$ symmetric about the y-axis?

A: No, the graph of $r=\frac{1}{1-5 \sin \theta}$ is not symmetric about the y-axis. When $\theta$ is replaced by $\pi + \theta$ , the graph is not the same.

Q: Is the graph of $r=\frac{1}{1-5 \sin \theta}$ symmetric about the origin?

A: No, the graph of $r=\frac{1}{1-5 \sin \theta}$ is not symmetric about the origin. When $r$ is replaced by $-r$ and $\theta$ is replaced by $\pi + \theta$ , the graph is not the same.

Q: What is the equation of the graph of $r=\frac{1}{1-5 \sin \theta}$ in rectangular coordinates?

A: To convert the polar equation $r=\frac{1}{1-5 \sin \theta}$ to rectangular coordinates, we can use the following formulas:

x = r \cos \theta

y = r \sin \theta

Substituting the polar equation into these formulas, we get:

x = \frac{\cos \theta}{1-5 \sin \theta}

y = \frac{\sin \theta}{1-5 \sin \theta}

Q: How can I graph the equation $r=\frac{1}{1-5 \sin \theta}$ using a graphing calculator?

A: To graph the equation $r=\frac{1}{1-5 \sin \theta}$ using a graphing calculator, follow these steps:

Enter the polar equation into the calculator using the polar mode.
Set the calculator to graph the equation in polar coordinates.
Adjust the window settings to view the entire graph.
Graph the equation and observe the resulting curve.

Conclusion

In this article, we have addressed some of the most frequently asked questions about the graph of the polar equation $r=\frac{1}{1-5 \sin \theta}$ . We have provided detailed explanations and examples to help readers understand the properties of the graph and how to graph it using a graphing calculator.