Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the Cartesian equations of the graphs of the given polar equations. $$ r^{2}-6 r \cos \theta-4 r \sin \theta+9=0 $$

Step-by-Step Solution

Answer

The Cartesian equation is \( (x - 3)^2 + (y - 2)^2 = 4 \).

Step 1: Convert Polar to Cartesian Coordinates

The polar coordinates \( (r, \theta) \) can be converted to Cartesian coordinates \( (x, y) \) using the relationships \( x = r \cos \theta \) and \( y = r \sin \theta \). Also, note that \( r^2 = x^2 + y^2 \). We'll use these identities to convert the given polar equation into a Cartesian equation.

Step 2: Substitute Polar to Cartesian Conversions

Start by substituting \( r^2 = x^2 + y^2 \) and \( r \cos \theta = x \), \( r \sin \theta = y \) into the given equation. The polar equation \( r^{2} - 6r \cos \theta - 4r \sin \theta + 9 = 0 \) then becomes \( x^2 + y^2 - 6x - 4y + 9 = 0 \).

Step 3: Simplify the Cartesian Equation

Rearrange the Cartesian equation \( x^2 + y^2 - 6x - 4y + 9 = 0 \) to group like terms. This forms a quadratic equation in terms of \( x \) and \( y \): \( x^2 - 6x + y^2 - 4y + 9 = 0 \).

Step 4: Complete the Square for \( x \) Terms

To complete the square for the \( x \) terms, take \( x^2 - 6x \) and add and subtract \( (\frac{6}{2})^2 = 9 \). The expression becomes \( (x - 3)^2 - 9 \).

Step 5: Complete the Square for \( y \) Terms

To complete the square for the \( y \) terms, take \( y^2 - 4y \) and add and subtract \( (\frac{4}{2})^2 = 4 \). The expression becomes \( (y - 2)^2 - 4 \).

Step 6: Form the Final Cartesian Equation

Combine the completed square forms: \( (x - 3)^2 - 9 + (y - 2)^2 - 4 + 9 = 0 \). Simplify it to \( (x - 3)^2 + (y - 2)^2 = 4 \). This represents a circle in Cartesian coordinates.

Key Concepts

Polar to Cartesian Conversion

Converting between polar and Cartesian coordinate systems involves specific relationships between the variables. Here, we're converting from the polar equation form to the Cartesian equation form. The polar coordinates, given by

• \(r\): the radius or distance from the origin
• \(\theta\): the angle measured from the positive x-axis

can be transformed into Cartesian coordinates, denoted by \( (x, y) \), using these formulas:

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

By substituting these into the polar equation \(r^{2} - 6r \cos \theta - 4r \sin \theta + 9 = 0\), you can effectively transform the problem into a form that's often simpler to work with in Cartesian coordinates.

Completing the Square

Completing the square is a method used in algebra to transform a quadratic expression into a perfect square trinomial. This technique is particularly useful for simplifying equations and revealing geometric properties, like the center and radius of a circle.
For the expression \( x^2 - 6x \), add and subtract \( \left(\frac{6}{2}\right)^2 = 9 \). This transforms it into \((x - 3)^2 - 9\).
Similarly, for \( y^2 - 4y \), add and subtract \( \left(\frac{4}{2}\right)^2 = 4 \). It becomes \((y - 2)^2 - 4\).
Completing the square helps to rewrite quadratic terms in a form that's easier to incorporate into geometric interpretations, such as identifying the equation of a circle. By representing quadratic terms as completions of squares, you'll notice symmetries and centers in graphs that might not be initially obvious.

Quadratic Equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \). These can have variables like x and y, and often include terms where the variables are squared, as seen in our example:

• \(x^2 - 6x + y^2 - 4y + 9 = 0\)

Quadratics are key in coordinate geometry as they can form conic sections such as ellipses, parabolas, and circles. Here, rearranging and simplifying using the method of completing the square helps us identify these shapes. Quadratics often require techniques like factoring or using the quadratic formula for solutions, but in geometry, they are invaluable in representing curves and specific graph features.

Circle Equation

The equation of a circle in Cartesian coordinates takes the canonical form \((x - h)^2 + (y - k)^2 = r^2\). Here,

• \(h\) and \(k\) are the x and y coordinates of the circle's center
• \(r\) is the radius of the circle

In our example, the final Cartesian equation \((x - 3)^2 + (y - 2)^2 = 4\) clearly shows a circle centered at \((3, 2)\) with a radius \(r = 2\) since \(r^2 = 4\).
The methods of converting polar equations and completing the square reveal this underlying structure, converting a polar equation into a simple geometrical shape, and demonstrating how different algebraic forms are interpreted in geometry. Thus, understanding circle equations not only clarifies the nature of the curve but offers a robust tool for solving real-world problems related to distances and areas.