Question

Find the points of intersection of the polar graphs. \(r=\sin \theta\) and \(r=\sqrt{3}+3 \sin \theta\) on \([0,2 \pi]\)

Short answer

The points of intersection are at \(\theta = \frac{4\pi}{3}\) and \(\theta = \frac{5\pi}{3}\).

Step-by-step solution

Equating the Equations

Set the two given polar equations equal to each other, since we need to find their points of intersection. This means that wherever they intersect, their radii should be equal. So, we solve the equation: \( \sin \theta = \sqrt{3} + 3 \sin \theta \).

Solve for \( \sin \theta \)

Subtract \(3 \sin \theta\) from both sides of the equation to isolate \(\sin \theta\): \( \sin \theta - 3 \sin \theta = \sqrt{3} \). Simplifying this gives \( -2 \sin \theta = \sqrt{3} \).

Solve for \( \theta \)

Divide both sides by \(-2\): \( \sin \theta = -\frac{\sqrt{3}}{2} \). Recall the unit circle: \(\sin \theta = -\frac{\sqrt{3}}{2}\) corresponds to angles in the third and fourth quadrants. Thus, the solutions in the interval \([0, 2\pi]\) are \(\theta = \frac{4\pi}{3}\) and \(\theta = \frac{5\pi}{3}\).

Verify the Range

Check that both \(\theta = \frac{4\pi}{3}\) and \(\theta = \frac{5\pi}{3}\) are within the specified range \([0, 2\pi]\). Both values fall within this interval and are valid solutions.

Key concepts

Points of Intersection

To determine the points of intersection for the polar equations, we identify where the radii of both graphs are the same. This implies solving for common values of \(r\) given the same angle \(\theta\). When working with polar equations, finding intersections involves equating the two functions, as any overlap implies equal distance from the origin in the same direction.
For the equations \(r = \sin \theta\) and \(r = \sqrt{3} + 3 \sin \theta\), we begin by setting them equal since their intersection points must share the same radius:
\[ \sin \theta = \sqrt{3} + 3 \sin \theta \]
This approach results in solving the equation \(\sin \theta - 3 \sin \theta = \sqrt{3}\), which simplifies to \(-2 \sin \theta = \sqrt{3}\). Intersections are key because they visually show where two curves meet, providing insights into the behavior of the graphs in their polar form.

Unit Circle

The unit circle is a fundamental concept in trigonometry, serving as a reference for determining angle measures associated with specific trigonometric values. It is especially useful when solving problems involving sine, cosine, and tangent functions.

In the context of the exercise, the unit circle helps us find the angle \(\theta\) that corresponds to \(\sin \theta = -\frac{\sqrt{3}}{2}\). On the unit circle, this value of sine typically corresponds to angles located in the third and fourth quadrants:

• Third quadrant: \(\theta = \frac{4\pi}{3}\)
• Fourth quadrant: \(\theta = \frac{5\pi}{3}\)

These angles are critical for verifying intersection points within the given interval of \([0, 2\pi]\). The unit circle provides a visual and geometric perspective that enhances understanding of trigonometric solutions.

Polar Equations

Polar equations describe curves on the polar coordinate plane, using radius \(r\) and angle \(\theta\) as their primary variables. They provide an alternative view compared to the Cartesian system, often simplifying the representation of curves like spirals, circles, and roses.

To interpret the polar equations \(r = \sin \theta\) and \(r = \sqrt{3} + 3 \sin \theta\), recognize that these equations express a relationship between the angle \(\theta\) and the distance from the origin, \(r\).

• The first equation \(r = \sin \theta\) describes a simple polar curve, potentially a circle with a shifted center or a limaçon.
• The second equation involves a scaling and shifting effect because of the constants \(\sqrt{3}\) and \(3\), introducing complexity to the graph.

Understanding these equations enables you to explore the geometry of shapes and their intersections in a way that's more intuitive than merely using rectangular coordinates.