Problem 46 - Thomas Calculus

Question

Replace the polar equations in Exercises \(23-48\) by equivalent Cartesian equations. Then describe or identify the graph. $$ r=2 \cos \theta-\sin \theta $$

Step-by-Step Solution

Answer

The Cartesian equation is \((x-1)^2 + (y+\frac{1}{2})^2 = 1.25\), a circle centered at (1, -0.5) with radius \(\frac{\sqrt{5}}{2}\).

Step 1: Understand Polar to Cartesian Conversion

The relationship between polar and Cartesian coordinates is given by the equations: \[ x = r\cos\theta \] \[ y = r\sin\theta \] \[ r = \sqrt{x^2+y^2} \] In the given polar equation, \( r = 2\cos\theta - \sin\theta \), we'll convert this into Cartesian coordinates using the above relationships.

Step 2: Substitute Polar Coordinate Relationships

Using the relations, substitute \(r\cos\theta = x\) and \(r\sin\theta = y\) into the given equation:\[ r = 2\cos\theta - \sin\theta \] Translate to:\[ x^2 + y^2 = 2x - y \] by also utilizing \( r = \sqrt{x^2 + y^2} \).

Step 3: Rearrange into Standard Form

Rearrange the equation:\[ x^2 + y^2 = 2x - y \]Move all terms to one side to form:\[ x^2 + y^2 - 2x + y = 0 \]

Step 4: Complete the Square

Complete the square for the terms involving \(x\): \[ x^2 - 2x \] becomes \((x-1)^2 - 1\) The equation now becomes:\[ (x-1)^2 - 1 + y^2 + y = 0 \] Simplify to:\[ (x-1)^2 + y^2 + y - 1 = 0 \]

Step 5: Simplify Further to Identify Shape

Rearrange further if applicable (complete the square for \(y\) if necessary). Let's complete the square for \(y\):\[ y^2 + y \] becomes \((y+\frac{1}{2})^2 - \frac{1}{4}\).Substitute it into the equation: \[ (x-1)^2 + (y+\frac{1}{2})^2 = 1.25 \] This is the equation of a circle centered at (1, -0.5) with radius \(\sqrt{1.25}\) or \(\frac{\sqrt{5}}{2}\).

Key Concepts

Polar to Cartesian Conversion

Converting equations from polar to Cartesian coordinates is a common exercise in mathematics, especially in trigonometry and calculus. Polar coordinates are defined by a radius \( r \) and an angle \( \theta \), while Cartesian coordinates are expressed as \( (x, y) \). To perform this conversion, we use specific relationships:

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

These equations arise from the geometric interpretation of how points can be represented in a plane. By substituting these relationships into a polar equation, like \( r = 2\cos \theta - \sin \theta \), we effectively translate the polar description into a Cartesian form.
To illustrate, substituting and rearranging gives us an equation in terms of \( x \) and \( y \). This allows us to identify curve shapes more commonly used in analysis and graphing.

Equation of a Circle

The equation of a circle in Cartesian coordinates is a well-known result. Generally, it is expressed as:

• \( (x - h)^2 + (y - k)^2 = r^2 \)

where \((h, k)\) is the center of the circle, and \(r\) is the radius.
When converting from polar to Cartesian coordinates, our goal is often to rearrange and simplify the equation into this recognizable form. Through this process, we clarify the shape of the graph. For our example, after completing the square, we find:

• \( (x - 1)^2 + (y + \frac{1}{2})^2 = 1.25 \)

Here, the circle is centered at (1, -0.5) with a radius of \(\frac{\sqrt{5}}{2}\). Being familiar with this form allows us to understand the geometry of the circle and anticipate its presence on the Cartesian plane.

Completing the Square

Completing the square is a technique used to simplify quadratic equations into a form that makes the relation or conic section clearly identifiable. It is a process that rewrites a quadratic expression \( ax^2 + bx + c \) as a perfect square. This method is beneficial when working with equations involving circles or parabolas.
For instance, when we need to convert \( x^2 - 2x \) into a perfect square, we rewrite it as:

• \( (x - 1)^2 - 1 \)

This involves:

• Taking half of the coefficient of \( x \), squaring it, then adding and subtracting this square.

Similarly, for \( y^2 + y \), completing the square results in:

• \( (y + \frac{1}{2})^2 - \frac{1}{4} \)

Using this method helps transform both \( x \) and \( y \) parts of the equation into forms that can easily be rearranged to identify the circle's center and radius. Thus, completing the square is a powerful tool for working with quadratic relationships in Cartesian coordinates.