Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 17-22, 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: Understand the Problem

We need to convert the polar equation \(r^2 - 6r \cos \theta - 4r \sin \theta + 9 = 0\) into a Cartesian equation. This involves using the relationships between polar and Cartesian coordinates: \(x = r \cos \theta\), \(y = r \sin \theta\), and \(r = \sqrt{x^2 + y^2}\).

Step 2: Substitute Polar to Cartesian

Replace \(r \cos \theta\) with \(x\) and \(r \sin \theta\) with \(y\). The equation becomes \(r^2 - 6x - 4y + 9 = 0\).

Step 3: Express \(r^2\) in Terms of \(x\) and \(y\)

Use the identity \(r^2 = x^2 + y^2\). Substitute \(r^2\) with \(x^2 + y^2\) in the equation yielding \(x^2 + y^2 - 6x - 4y + 9 = 0\).

Step 4: Complete the Square

Complete the square for both \(x\) and \(y\):1. For \(x\): \(x^2 - 6x = (x - 3)^2 - 9\).2. For \(y\): \(y^2 - 4y = (y - 2)^2 - 4\).

Step 5: Combine and Simplify

Substitute the completed squares back into the equation:\((x - 3)^2 - 9 + (y - 2)^2 - 4 + 9 = 0\). Simplifying gives \((x - 3)^2 + (y - 2)^2 = 4\).

Step 6: Identify the Result

The Cartesian equation \((x - 3)^2 + (y - 2)^2 = 4\) is a standard circle equation centered at \((3, 2)\) with a radius of 2.

Key Concepts

Polar Coordinates

Polar coordinates are often used in mathematics to describe locations on a plane using angles and radii. Unlike Cartesian coordinates, which use x and y positions, polar coordinates express a point by considering its distance from a reference point (the origin) and the angle relative to a reference direction, usually the positive x-axis. This system is particularly useful for problems involving circles and arcs.
To convert a point from Cartesian to polar coordinates, we use the formulas:

• The radius, \(r\), is calculated as \(r = \sqrt{x^2 + y^2}\)
• The angle, \(\theta\), is found using \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\)

Understanding these relationships allows us to transition smoothly between Cartesian and polar forms, essential for tackling a variety of mathematical problems like those involving rotational symmetry.

Converting Equations

Converting an equation from polar to Cartesian coordinates involves substituting the polar terms with their Cartesian equivalents. In our problem, the polar equation given is \(r^2 - 6r \cos \theta - 4r \sin \theta + 9 = 0\).
We know that:

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

Using these, we substitute in the equation to transition it to Cartesian form. Every time you see \(r^2\), replace it with \(x^2 + y^2\), and instead of \(r \cos \theta\) and \(r \sin \theta\), use \(x\) and \(y\) respectively.
This methodical substitution transforms our polar equation into the Cartesian form we need.

Completing the Square

Completing the square is a useful algebraic tool for transforming quadratic expressions into a perfect square trinomial, simplifying many mathematical processes, especially when working with circles and ellipses.
Given the transformation to the Cartesian equation \(x^2 + y^2 - 6x - 4y + 9 = 0\), we apply completing the square to both the \(x\) and \(y\) products.

• For \(x\), we rewrite \(x^2 - 6x\) as \((x - 3)^2 - 9\).
• For \(y\), we rewrite \(y^2 - 4y\) as \((y - 2)^2 - 4\).

These steps help simplify the equation into a form that's easier to handle, effectively highlighting key features like center and radius when dealing with circles.

Circle Equation

The standard form of a circle's equation in Cartesian coordinates is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) represents the center of the circle, and \(r\) is the radius.
In our solution, by transforming the equation through completing the square, we achieved this form: \((x - 3)^2 + (y - 2)^2 = 4\).
From this equation, we can directly see that:

• The center of the circle is at the point \((3, 2)\).
• The radius is \(\sqrt{4} = 2\).

This clear representation helps in visualizing geometric properties and spatial relations, making circles much easier to analyze both algebraically and graphically.