Problem 50 - Thomas Calculus

Question

Graph the lines and conic sections in Exercises \(47-56\) $$ r=-2 \cos \theta $$

Step-by-Step Solution

Answer

The graph is a circle centered at (-1, 0) with a radius of 1.

Step 1: Understand the Equation

The given equation is in polar form and describes a conic section, specifically a circle. The equation is given as: \[ r = -2 \cos \theta \] The general form for a conic section in polar coordinates is \[ r = ed/(1 + e\cos\theta) \] for circles, \( e = 0 \), so it simplifies to \( r = c\cos\theta \) where \( c \) is a constant.

Step 2: Convert to Cartesian Coordinates

To graph this using Cartesian coordinates, first recall the relations \[ x = r \cos \theta \] and \[ y = r \sin \theta \]. So multiply both sides of the equation \( r = -2 \cos \theta \) by \( r \): \[ r^2 = -2r\cos\theta \]which becomes \[ x^2 + y^2 = -2x \].

Step 3: Complete the Square

Rearrange the terms to prepare for completing the square:\[ x^2 + 2x + y^2 = 0 \] Complete the square for the \( x \) terms by adding and subtracting \( 1 \): \[ (x+1)^2 - 1 + y^2 = 0 \] Simplify to find the circle form: \[ (x+1)^2 + y^2 = 1 \].

Step 4: Identify the Circle

Now that the equation is in the standard circle form \( (x-h)^2 + (y-k)^2 = r^2 \), identify the circle's center and radius:- Center: \( (-1, 0) \)- Radius: \( 1 \).

Step 5: Graph the Circle

Plot the circle on a coordinate plane with center at \((-1, 0)\) and a radius of \(1\). Since circles are symmetric, ensure the circle expands 1 unit in all directions from the center, making sure to draw it accurately along the axes.

Key Concepts

Conic Sections

Conic sections are curves obtained by slicing a cone with a plane at different angles. They are fundamental in the study of geometry and include:

• Circles
• Ellipses
• Parabolas
• Hyperbolas

In the given problem, the polar equation \( r = -2 \cos \theta \) represents a circle. In polar coordinates, conic sections can be expressed as \( r = \frac{ed}{1 + e\cos\theta} \). Here, \( e \) is the eccentricity of the conic, and \( e = 0 \) for circles.
In this form, the equation simplifies to \( r = c\cos\theta \), where \( c \) is a constant, indicating the equation is a circle. Understanding this conversion helps to recognize the different types of conic sections and how they manifest in polar coordinates.

Cartesian Coordinates

Cartesian coordinates consist of \( x \) and \( y \) values that define points in a plane. These coordinates are a powerful mathematical concept used to describe locations.
Using polar to Cartesian conversion formulas, we transform a polar equation to Cartesian, enabling easier graphing and solidifying our understanding. The necessary formulas are:

• \( x = r \cos \theta \)
• \( y = r \sin \theta \)

For the given polar equation \( r = -2 \cos \theta \), multiplying both sides by \( r \) results in \( r^2 = -2r\cos \theta \), converting to \( x^2 + y^2 = -2x \) in Cartesian form.
This conversion shows the power of Cartesian coordinates in transitioning from polar form, opening up methods like completing the square to further simplify and analyze the equation.

Completing the Square

Completing the square is a technique used to convert a quadratic equation to a more manageable form. It involves creating a perfect square trinomial from a quadratic expression.
In the equation \( x^2 + 2x + y^2 = 0 \), to complete the square for the \( x \, \text{terms} \), add and subtract \( 1 \) to get:

• \( (x+1)^2 - 1 + y^2 = 0 \)
• Simplifies to \( (x+1)^2 + y^2 = 1 \)

This manipulation allows the expression to describe a circle with a clearly defined center and radius. The result \( (x+1)^2 + y^2 = 1 \) means the circle has:

• Center at \((-1,0)\)
• Radius \( = 1 \)

Completing the square turns complex curves into simple geometric shapes, emphasizing the elegance of algebra in graph interpretations.