Problem 25 - College Trigonometry

Question

In Exercises \(21-30\), find the exact polar coordinates of the points of intersection of graphs of the polar equations. Remember to check for intersection at the pole (origin). $$ r=2 \cos (\theta) \text { and } r=2 \sqrt{3} \sin (\theta) $$

Step-by-Step Solution

Answer

Intersection points are \( (\sqrt{3}, \frac{\pi}{6}) \) and the pole \( (0, \frac{\pi}{2}) \).

Step 1: Combine Equations

The given equations are \( r=2 \cos(\theta) \) and \( r=2 \sqrt{3} \sin(\theta) \). To find the points of intersection, set the equations equal to each other: \( 2 \cos(\theta) = 2 \sqrt{3} \sin(\theta) \).

Step 2: Simplify and Solve for \( \theta \)

Divide both sides of the equation \( 2 \cos(\theta) = 2 \sqrt{3} \sin(\theta) \) by \( 2 \) to simplify: \( \cos(\theta) = \sqrt{3} \sin(\theta) \). Rewriting it, we have \( \tan(\theta) = \frac{1}{\sqrt{3}} \). This means \( \theta = \frac{\pi}{6} + k\pi \) for some integer \( k \).

Step 3: Check for Intersection at the Pole

Set either \( r \) to zero in the original equations to find intersections at the pole. If \( r = 0 \), then neither of the equations will satisfy unless both sides are zero, which happens at \( \theta = \frac{\pi}{2} + n\pi \) for any integer \( n \). This means an intersection occurs at the pole when \( \theta = \frac{\pi}{2} \).

Step 4: Calculate \( r \) for \( \theta = \frac{\pi}{6} \)

Substitute \( \theta = \frac{\pi}{6} \) into one of the original equations, say \( r=2 \cos(\theta) \):\[ r = 2 \cos\left(\frac{\pi}{6}\right) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \].

Step 5: Determine Final Coordinates

The points of intersection in polar coordinates are \( \left( \sqrt{3}, \frac{\pi}{6} \right) \) and the pole at \( (0, \frac{\pi}{2}) \).

Key Concepts

Intersection of Graphs

Understanding the intersection of graphs in polar coordinates can be quite different from their Cartesian counterparts. In a Cartesian system, we usually think of an intersection as a point where two lines cross each other on an xy-plane. However, in polar coordinates, intersections might occur not just at single points but also at specific angles. In the exercise provided, we were tasked with finding where two polar curves intersect.To solve this, the key step was to equate the two given equations of the curves:

• \( r = 2 \cos(\theta) \)
• \( r = 2 \sqrt{3} \sin(\theta) \)

By setting them equal to each other, we can find the values of \( \theta \) where the graphs intersect. Simplifying and solving leads to the discovery of specific angle values where the intersections occur. Additionally, it's important to check for intersections at the pole, which is the origin in polar coordinates, by setting \( r = 0 \). This was another crucial step and revealed intersections that might not be immediately obvious.

Polar Equations

Polar coordinates provide a unique perspective on graphing, offering a different way to represent points. Instead of using x and y grids, points in polar coordinates are defined by:

• The distance from the origin, denoted by \( r \).
• The angle from the positive x-axis, denoted by \( \theta \).

The exercise utilized two polar equations defined in terms of trigonometric functions:

• The equation \( r = 2 \cos(\theta) \) represents a circle.
• The equation \( r = 2 \sqrt{3} \sin(\theta) \) also forms a circular path, but oriented differently.

Graphing these equations involves plotting all points for various values of \( \theta \), showcasing their unique paths. Understanding how these paths interact is crucial in identifying possible intersections, which might vary greatly from what we'd expect on a standard Cartesian grid. This makes polar equations both fascinating and slightly more challenging to work with, yet powerful in certain mathematical contexts.

Trigonometric Identities

Trigonometric identities simplify various expressions involving angles and sides of triangles, playing a critical role in equations like the ones seen in the exercise. These identities enable transformations that make solving equations simpler.A fundamental example is the conversion seen in the exercise:

• From \( 2 \cos(\theta) = 2 \sqrt{3} \sin(\theta) \)
• To \( \tan(\theta) = \frac{1}{\sqrt{3}} \)

This was possible by recognizing the relationship between the functions cosine and sine, and solving for \( \theta \) using the tangent function.Knowing basic trigonometric identities such as:

• \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \)
• \( \sin^2(\theta) + \cos^2(\theta) = 1 \)

can greatly simplify the problem-solving process. In our exercise, once \( \tan(\theta) = \frac{1}{\sqrt{3}} \) was found, it led to solutions for \( \theta \) which demonstrated where the polar graphs intersect.