Problem 33 - Thomas Calculus

Question

Find the points of intersection of the pairs of curves in Exercises \(31-38\) . $$ r=2 \sin \theta, \quad r=2 \sin 2 \theta $$

Step-by-Step Solution

Answer

The points of intersection are \((0,0), (\sqrt{3}, \frac{\pi}{3}), (\sqrt{3}, \frac{5\pi}{3})\).

Step 1: Understand the Polar Equations

The problem gives two polar equations: \( r = 2 \sin \theta \) and \( r = 2 \sin 2\theta \). We need to find where these two curves intersect by finding values of \( r \) and \( \theta \) that satisfy both equations simultaneously.

Step 2: Set Equations Equal

To find the points of intersection, set the two equations equal to each other:\[ 2 \sin \theta = 2 \sin 2\theta \] Simplifying, we get \( \sin \theta = \sin 2\theta \).

Step 3: Use Double Angle Identity

Recall the double angle identity: \( \sin 2\theta = 2 \sin \theta \cos \theta \). Replace \( \sin 2\theta \) in the equation to get:\[ \sin \theta = 2 \sin \theta \cos \theta \]

Step 4: Simplify the Equation

Factor out \( \sin \theta \) from both sides:\[ \sin \theta (1 - 2 \cos \theta) = 0 \]This gives us two possible equations: \( \sin \theta = 0 \) or \( 1 - 2 \cos \theta = 0 \).

Step 5: Solve for \( \theta \)

- When \( \sin \theta = 0 \), \( \theta = 0, \pi, 2\pi \).- When \( 1 - 2 \cos \theta = 0 \), \( \cos \theta = \frac{1}{2} \), which gives \( \theta = \frac{\pi}{3}, \frac{5\pi}{3} \) (considering the range \( [0, 2\pi) \)).

Step 6: Find Corresponding \( r \) Values

For \( \theta = 0 \) and \( \theta = \pi \), substitute back into either equation to find \( r = 0 \).For \( \theta = \frac{\pi}{3} \), substituting \( \theta \) back into \( r = 2 \sin \theta \), we find \( r = \sqrt{3} \).For \( \theta = \frac{5\pi}{3} \), substituting \( \theta \) back into \( r = 2 \sin \theta \), we find \( r = -\sqrt{3} \) (or equivalently \( \sqrt{3} \) moving in the opposite direction).

Step 7: List Points of Intersection

The points of intersection are: - \( (0, 0) \) at \( \theta = 0 \) and \( \theta = \pi \) - \( (\sqrt{3}, \frac{\pi}{3}) \)- \( (\sqrt{3}, \frac{5\pi}{3}) \)

Key Concepts

Intersection of Curves

When studying the intersection of curves in polar coordinates, we aim to find points where two curves meet. In polar coordinates, each point is represented by an angle \(\theta\) and a distance \(r\) from the origin. To find intersection points:

• Ensure both curves' equations equal the same \(r\) value for identical \(\theta\) values.
• Set the equations equal to eliminate \(r\), and solve for \(\theta\).
• Remember to substitute back to find actual \(r\) values.

In the given exercise, this method helps solve where the curves given by \(r = 2 \sin \theta\) and \(r = 2 \sin 2\theta\) intersect.

Polar Equations

Polar equations describe curves using polar coordinates \((r, \theta)\). These equations differ from Cartesian equations by focusing on angles and radii, facilitating elegant descriptions of many curves:

• Each curve's equation provides a relationship between \(r\) and \(\theta\).
• Function forms of sinusoids, like \(r = a \sin n\theta\), often represent rose curves or circles.
• They describe periodic patterns based on trigonometric functions.

In the problem, \(r = 2 \sin \theta\) forms a circle, while \(r = 2 \sin 2\theta\) forms a rose curve with four petals. Understanding these equations helps visualize the intersection possibilities.

Trigonometric Identities

Trigonometric identities, like the double angle identity, are essential in solving polar equation problems as they simplify complex trigonometric expressions:

• The double angle identity \(\sin 2\theta = 2 \sin \theta \cos \theta\) expands \(\sin 2\theta\) into a manageable form.
• Identities reduce complicated expressions, making it easier to compare equations or solve for \(\theta\).
• They are foundational tools needed for solving, graphing, and understanding polar curves.

In solving for intersections, using \(\sin 2\theta = 2 \sin \theta \cos \theta\) helps break down the equation \(\sin \theta = \sin 2\theta\), leading to tangible solutions.