Problem 8 - The Calculus with Analytic Geometry

Question

Find the points of intersection of the graphs of the given pair of equations. Draw a sketch of each pair of graphs with the same pole and polar axis.\(\left\\{\begin{array}{l}r=1-\sin \theta \\ r=\cos 2 \theta\end{array}\right.\)

Step-by-Step Solution

Answer

The points of intersection are \((1, 0)\), \((1, \pi)\), \((\frac{1}{2}, \frac{\pi}{6})\), and \((\frac{1}{2}, \frac{5\pi}{6})\).

Step 1 - Understand the Equations

We are given two polar equations: \[ r = 1 - \sin \theta \]and \[ r = \cos 2\theta \]. Our goal is to find their points of intersection.

Step 2 - Equate the Equations

To find the points of intersection, set the equations equal to each other: \[ 1 - \sin \theta = \cos 2\theta \]

Step 3 - Express Cosine in Terms of Sine

Recall the double-angle identity for cosine: \[ \cos 2\theta = 1 - 2\sin^2 \theta \]. Substituting this into the equation gives \[ 1 - \sin \theta = 1 - 2\sin^2 \theta \].

Step 4 - Simplify the Equation

Eliminate the common term ‘1’ from both sides of the equation: \[ -\sin \theta = -2\sin^2 \theta \]which simplifies to \[ \sin \theta = 2\sin^2 \theta \].

Step 5 - Factor the Quadratic Equation

Rearrange the equation: \[ 2\sin^2 \theta - \sin \theta = 0 \]. Factor out the common term: \[ \sin \theta (2\sin \theta - 1) = 0 \].

Step 6 - Find the Solutions

Set each factor equal to zero: \[ \sin \theta = 0 \] and \[ 2\sin \theta - 1 = 0. \]Solving these gives \[ \theta = 0, \pi \] and \[ \sin \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{6}, \frac{5\pi}{6}. \]

Step 7 - Calculate the Corresponding r-Values

Using \[ r = 1 - \sin \theta \] and the values of \( \theta \), calculate the corresponding \( r \):* For \( \theta = 0 \), \( r = 1 - \sin 0 = 1 \)* For \( \theta = \pi \), \( r = 1 - \sin \pi = 1 \)* For \( \theta = \frac{\pi}{6} \), \( r = 1 - \sin( \frac{\pi}{6} ) = \frac{1}{2} \)* For \( \theta = \frac{5\pi}{6} \), \( r = 1 - \sin( \frac{5\pi}{6} ) = \frac{1}{2} \).

Step 8 - Find the Points of Intersection

The points of intersection are:* \( (1, 0) \)* \( (1, \pi) \)* \( (\frac{1}{2}, \frac{\pi}{6}) \)* \( (\frac{1}{2}, \frac{5\pi}{6}) \).

Step 9 - Draw the Sketch

Draw the polar curves on the same axis to visually represent the points of intersection at the calculated \( r\) and \( \theta \) values.

Key Concepts

Polar Equations

Polar equations are a way to represent curves on a plane using polar coordinates, which are different from regular Cartesian coordinates. Instead of using x and y, polar coordinates use r and \( \theta \), where r is the radius (or distance from the origin) and \( \theta \) is the angle from the positive x-axis. These coordinates are especially useful for problems involving circular and spiral shapes.

For example, in the exercise, we have two polar equations:

• \( r = 1 - \text{sin} \theta \)
• \( r = \text{cos} 2\theta \)

.To find the points where these two curves intersect, we need to solve these equations simultaneously.

Trigonometric Identities

Trigonometric identities simplify complex trigonometric expressions and are useful for solving equations involving trigonometric functions. In this exercise, we use the double-angle identity for cosine:

\[ \text{cos} 2\theta = 1 - 2\text{sin}^2 \theta \].

This identity allows us to convert \( \text{cos} 2\theta \) into an expression involving \( \text{sin} \theta \), making it easier to solve the equation.
After substituting this identity into the equation \( 1 - \text{sin} \theta = \text{cos} 2\theta \), we obtain:

• \[ 1 - \text{sin} \theta = 1 - 2\text{sin}^2 \theta \]
• Subtracting 1 from both sides, simplifies to: \[ -\text{sin} \theta = -2\text{sin}^2 \theta \]
• Ultimately leading to: \[ \text{sin} \theta = 2\text{sin}^2 \theta \]

.These simplifications are critical in helping us solve for \( \text{sin} \theta \).

Solving Quadratic Equations

Solving quadratic equations is an essential skill in algebra and is often needed in trigonometry. In this exercise, we end up with a quadratic equation in terms of \( \text{sin} \theta \):

\[ 2\text{sin}^2 \theta - \text{sin} \theta = 0 \].

To solve this, we can factor out the common term \( \text{sin} \theta \):

• \[ \text{sin} \theta (2\text{sin} \theta - 1) = 0 \]

.This gives two simpler equations:

• \( \text{sin} \theta = 0 \) and \( 2\text{sin} \theta - 1 = 0 \)

.Solving these, we get:

• \( \theta = 0, \text{ and } \theta = \text{π} \) from \( \text{sin} \theta = 0 \)
• \( \text{sin} \theta = \frac{1}{2} \rightarrow \theta = \frac{\text{π}}{6}, \frac{5\text{π}}{6} \) from \( 2\text{sin} \theta - 1 \).

These solutions give values of \( \theta \) where the curves intersect, allowing us to find the corresponding values of r and thus the points of intersection.