Problem 23 - Precalculus with Calculus Previews

Question

$$ r=6 \cos \theta $$

Step-by-Step Solution

Answer

The equation describes a circle with center at (3, 0) and radius 3.

Step 1: Recognize the Equation Type

Recognize that the equation \( r = 6 \cos \theta \) is a polar equation, specifically of the form \( r = a \cos \theta \), which describes a circle.

Step 2: Analyze the Circle Parameters

For the equation \( r = a \cos \theta \), the circle is centered at \( (\frac{a}{2}, 0) \) and has a radius \( \frac{|a|}{2} \). Here, \( a = 6 \), so the center of the circle is at \( (3, 0) \), and the radius is \( 3 \).

Step 3: Convert to Cartesian Coordinates (Optional)

If needed, convert the polar equation to Cartesian coordinates using the relations \( x = r \cos \theta \) and \( y = r \sin \theta \). Substituting \( r = 6 \cos \theta \) and \( y = r \sin \theta \), we get: \[x = 6 \cos^2 \thetay = 6 \cos \theta \sin \theta\]Replacing \( \cos \theta \) by \( \frac{x}{r} \) and \( \sin \theta \) by \( \frac{y}{r} \), we get \( x^2 + y^2 = x(6 \cos \theta) \), simplifying to \((x - 3)^2 + y^2 = 9 \), confirming it's a circle.

Key Concepts

Circle in Polar Coordinates

When dealing with polar equations, one common form you'll encounter is the type that represents circles. Polar equations use a different approach compared to Cartesian equations. Here, we describe locations using a radius and an angle rather than x and y coordinates. The equation given, \( r = 6 \cos \theta \), is a specific type of polar equation that describes a circle.

The general form \( r = a \cos \theta \) or \( r = a \sin \theta \) represents circles centered on the polar axis. In this setup:

• \( r \) represents the radius from the origin to any point on the curve.
• \( \theta \) is the angle from the positive x-axis.
• \( a \) is a constant that affects the circle's size and position.

Understanding this form allows us to identify the shape and behavior of curves quickly. For \( r = 6 \cos \theta \), it forms a circle in polar coordinates, illustrating the elegance and simplicity this coordinate system offers for such shapes.

Converting Polar to Cartesian Coordinates

Sometimes, you may wish to convert polar equations to Cartesian coordinates to better understand or visualize them in a typical xy-plane. This conversion relies on some key relationships:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)
• \( r^2 = x^2 + y^2 \)

To convert \( r = 6 \cos \theta \) from polar to Cartesian, substitute the expressions for \( x \) and \( y \). Using \( x = r \cos \theta \), and \( r = 6 \cos \theta \), leads to:

• \( x = 6 \cos^2 \theta \)
• \( y = 6 \cos \theta \sin \theta \)

However, rather than expressing \( y \) directly in terms of \( \theta \), we manipulate the relationship \( x = 6 \cos \theta \) to obtain \( r = \frac{x}{\cos \theta} \), leading to \( r^2 = x^2 + y^2 = x(6) \).

Finally, simplify to show the circle's equation in Cartesian coordinates: \((x - 3)^2 + y^2 = 9\), a recognizable form of a circle.

Analyzing Circle Parameters

Analyzing circle parameters in polar equations gives us insight into the circle's properties without needing a graph. With an equation like \( r = 6 \cos \theta \), we can quickly identify:

• The circle's center in polar coordinates, which transforms to the Cartesian point \((\frac{a}{2}, 0)\).
• The radius of the circle, given by \( \frac{|a|}{2} \). For this example, \( a = 6 \), and so the center is at \((3, 0)\), with a radius of \( 3 \).

This formula highlights a characteristic of circles in polar form: they often describe circles centered away from the origin (unless \( a = 0 \)).

Knowing how to deduce these parameters from polar equations can prove useful for quick calculations and for checking consistency when moving between different coordinate systems.