Problem 39 - Calculus Early Transcendentals

Question

\(37-42\) Find all points of intersection of the given curves. $$ r=2 \sin 2 \theta, \quad r=1 $$

Step-by-Step Solution

Answer

Points of intersection are \((1, \frac{\pi}{12})\) and \((1, \frac{5\pi}{12})\).

Step 1: Understand the Problem

We need to find the points of intersection of the polar curves given by \( r = 2 \sin 2\theta \) and \( r = 1 \). This means we want the points \((r, \theta)\) where both curves have the same \(r\) value simultaneously.

Step 2: Set the Equations Equal

Since both curves must have the same \(r\) at a point of intersection, set the equations equal: \( 2 \sin 2\theta = 1 \).

Step 3: Solve for \( \theta \)

Solve the equation \( 2 \sin 2\theta = 1 \) for \( \theta \). Divide both sides by 2 to obtain \( \sin 2\theta = \frac{1}{2} \).

Step 4: Solve Trigonometric Equation

The solutions to \( \sin 2\theta = \frac{1}{2} \) are \( 2\theta = \frac{\pi}{6} + 2\pi n \) and \( 2\theta = \frac{5\pi}{6} + 2\pi n \), where \( n \) is any integer.

Step 5: Find \( \theta \) Values

Solving \( 2\theta = \frac{\pi}{6} \) and \( 2\theta = \frac{5\pi}{6} \), we get \( \theta = \frac{\pi}{12} \) and \( \theta = \frac{5\pi}{12} \) for \( n = 0 \). For additional solutions, add \( \pi k \) to each solution, due to periodicity of \(\sin\).

Step 6: Find Corresponding \( r \) Values

Substitute \( \theta = \frac{\pi}{12} \) and \( \theta = \frac{5\pi}{12} \), into either original equation. Since both must satisfy \( r = 1 \), verify these are valid (they are since both are derived from satisfying \( r = 1 \)).

Step 7: List Points of Intersection

The points of intersection are \( (1, \frac{\pi}{12}) \) and \( (1, \frac{5\pi}{12}) \), considering all periodic extensions \( \theta = \frac{\pi}{12} + \pi k \) and \( \theta = \frac{5\pi}{12} + \pi k \) where \( k \) is an integer.

Key Concepts

Polar Coordinates

Polar coordinates are a way of representing points on a plane using a distance and an angle. Instead of using an x and y coordinate like in the Cartesian plane, in polar coordinates, a point is given by \( r, \theta \).
Here, \(r\) stands for the radial distance from the origin and \(\theta\) represents the angle from the positive x-axis.

• Radial Distance: The distance \(r\) can be positive, negative, or zero. It tells us how far a point is from the origin.
• Angle \(\theta\): Measured in radians or degrees, it determines the direction of the point from the origin.

Polar coordinates are especially useful in representing curves such as circles and spirals. They make it easier to represent complex shapes more simply and provide an excellent tool for solving trigonometric equations. In our exercise, we analyze curves represented by the polar equations \(r = 2 \sin(2\theta)\) and \(r = 1\).
By setting these two equations equal, we help find where the curves intersect.

Trigonometric Equations

Trigonometric equations involve trigonometric functions like sine, cosine, and tangent. Solving these requires understanding their periodic nature and properties.
Our main goal is to find the angle \(\theta\) that fulfills the equality \(2 \sin(2\theta) = 1\).
To solve trigonometric equations:

• Simplify the Equation: Divide both sides of the given trigonometric equation by necessary terms to isolate the trigonometric function. For our problem, dividing by 2 gives \(\sin(2\theta) = \frac{1}{2}\).
• Use Known Values: Recall values of trigonometric functions where knowledge of \(\sin(\alpha)\) is helpful. Here, we use the fact that \(\sin(\pi/6) = 1/2\).
• Find General Solutions: Use the periodic properties to find all possible angles \(\theta\) by solving both \(2\theta = \frac{\pi}{6} + 2\pi n\) and \(2\theta = \frac{5\pi}{6} + 2\pi n\).

Understanding these steps helps determine all possible angle solutions for intersecting curves, acknowledging both principal and extended angles representing the solution's periodic nature.

Intersection of Curves

The intersection of curves refers to the points where two different curves meet or cross each other. In the context of polar coordinates and trigonometric equations, finding these intersections requires setting the radial components equal and solving the resulting trigonometric equation.

• Setting Equations Equal: Find where both given functions output the same \(r\) value for the same \(\theta\).
• Solve for Angles: As previously done, solve the simplified trigonometric equation for possible theta values.
  
• Check for Validity: Even after finding solutions for \(\theta\), ensure these angles also satisfy the original radial requirement. In this exercise, confirm that these \(\theta\) values genuinely give \(r = 1\) when substituted into both equations.

The solution may yield multiple points of intersection, accounting for the cyclical nature of trigonometric functions.
In our exercise, the radial condition \(r=1\) was validated at angles \(\theta = \frac{\pi}{12}\) and \(\theta = \frac{5\pi}{12}\), including their periodic extensions, showing where the two curves intersect.