Problem 39 - Precalculus with Calculus Previews

Question

Find the points of intersection of the graphs of the given pair of polar equations. $$ r=2, r=4 \sin \theta $$

Step-by-Step Solution

Answer

The points of intersection are \((2, \frac{\pi}{6})\) and \((2, \frac{5\pi}{6})\).

Step 1: Understand the Problem

We are given two polar equations and need to find their points of intersection. The polar equations are: \( r = 2 \) (a circle with radius 2 centered at the origin) and \( r = 4 \sin \theta \) (a limaçon with a loop that passes through the origin). Since we're looking for intersections, we're solving for \( r \) and \( \theta \) where these two equations are equal.

Step 2: Equalize the Equations

Set the equations equal to each other because the points of intersection have the same \( r \) values. So, we write \( 2 = 4 \sin \theta \).

Step 3: Solve for \( \theta \)

Divide both sides of the equation \( 2 = 4 \sin \theta \) by 4 to isolate \( \sin \theta \): \[ \sin \theta = \frac{1}{2} \]The angles \( \theta \) that satisfy this equation are \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \) within the interval \( [0, 2\pi) \).

Step 4: Find Corresponding \( r \) Values

Substitute \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \) back into either of the original equations (both give the same \( r = 2 \) due to equalization):- \( r(\frac{\pi}{6}) = 2 \)- \( r(\frac{5\pi}{6}) = 2 \)

Step 5: Determine Points of Intersection

The polar coordinates for the points of intersection are:- \( (2, \frac{\pi}{6}) \)- \( (2, \frac{5\pi}{6}) \)

Key Concepts

Points of Intersection

When dealing with polar graphs, we often need to find where two curves meet or intersect. In our problem, we're working with a circle, described by the equation \( r = 2 \), and a limaçon, expressed as \( r = 4 \sin \theta \). Finding the intersection points requires identifying the common coordinates \( (r, \theta) \) that satisfy both equations.

To find these intersections, we equate the two polar equations. This involves setting their \( r \) values equal, because at their intersection, both must represent the same radial distance from the origin. Once we have solved for \( \theta \), we plug these angles back into either equation to get the corresponding \( r \) values.

In the steps provided, this process led us to the intersection points \( (2, \frac{\pi}{6}) \) and \( (2, \frac{5\pi}{6}) \). These points represent where both the circle and the limaçon coincide.

Polar Equations

Polar equations use a coordinate system where each point on a plane is determined by an angle \( \theta \) and a radius \( r \). This system offers a unique way to represent curves like circles and limaçons.

• The polar equation \( r = 2 \) represents a simple circle with a fixed radius of 2 centered at the origin. No matter the angle \( \theta \), if \( r = 2 \), you are on this circle.
• On the other hand, the polar equation \( r = 4 \sin \theta \) is more dynamic. This equation describes a curve called a limaçon with a loop, affected by the value of \( \theta \). Here, the radius \( r \) varies as \( \theta \) changes.

Understanding these foundational equations helps to solve more complex problems and visualize how different types of curves behave in the polar coordinate system.

Trigonometric Solutions

Solving for the intersection of equations in polar coordinates often involves trigonometry, especially using sine or cosine functions. In our case, we arrive at \( \sin \theta = \frac{1}{2} \) after equating \( 2 = 4 \sin \theta \). This equation is key to finding the intersection angles.

Let's delve into the trigonometric identity used here. The values of \( \theta \) that satisfy \( \sin \theta = \frac{1}{2} \) are well-known special angles. Within a full circle or from 0 to \( 2\pi \), the angles \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \) satisfy this equation.

Trigonometric identities play a vital role in calculations involving periodic functions, such as those found in polar equations. Knowing solutions like \( \sin \theta = \frac{1}{2} \) allows us to pinpoint exact angular measures where certain radial distances occur, revealing intersection angles. These fundamental trigonometric solutions are essential for accurately finding the points of intersection.