Problem 42 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 27-44: Find an equation in \(x\) and \(y\) that has the same graph as the polar equation. Use it to help sketch the graph in an \(r \theta\)-plane. $$ r=2 \cos \theta-4 \sin \theta $$

Step-by-Step Solution

Answer

The equation in rectangular form is \((x-1)^2 + (y+2)^2 = 5\), representing a circle centered at \((1, -2)\) with radius \(\sqrt{5}\).

Step 1: Convert the given polar equation into rectangular form

In polar coordinates, the relations between polar and rectangular coordinates are: - \( x = r \cos \theta \)- \( y = r \sin \theta \)Given the polar equation \( r = 2 \cos \theta - 4 \sin \theta \), multiply both sides by \( r \) to align our formula with the rectangular coordinates:\[ r^2 = 2r \cos \theta - 4r \sin \theta \]Substituting the relations:- \( r^2 = x^2 + y^2 \)- \( r \cos \theta = x \)- \( r \sin \theta = y \)We get:\[ x^2 + y^2 = 2x - 4y \]

Step 2: Rearrange the equation to simplify

We need to rearrange the equation to identify its form. Start by setting the equation as follows:\[ x^2 + y^2 = 2x - 4y \]Rearrange to bring all terms to one side:\[ x^2 + y^2 - 2x + 4y = 0 \]

Step 3: Complete the square for the equation

The equation can be simplified by completing the square. Start with the \( x \) terms:\[ x^2 - 2x = (x-1)^2 - 1 \]Then complete the square for the \( y \) terms:\[ y^2 + 4y = (y+2)^2 - 4 \]Substitute back into the equation:\[ (x-1)^2 - 1 + (y+2)^2 - 4 = 0 \]Simplify and rearrange:\[ (x-1)^2 + (y+2)^2 = 5 \]

Step 4: Identify the graph

The equation \( (x-1)^2 + (y+2)^2 = 5 \) is a circle in the rectangular coordinate plane.- Center: \( (1, -2) \)- Radius: \( \sqrt{5} \)This helps in sketching the graph in the rectangular coordinate plane, and subsequently in the \( r\theta \)-plane.

Key Concepts

Rectangular Coordinates

Rectangular coordinates, also known as Cartesian coordinates, are used to specify the position of a point in a two-dimensional plane using two numbers: the horizontal distance (x-coordinate) and the vertical distance (y-coordinate) from a fixed point called the origin. These coordinates are vital for determining positions and plotting graphs in familiar geometry, providing a straightforward way to describe and calculate with shapes and points.

In this system, any point can be expressed as \(x, y\). Think of a standard graph paper with horizontal and vertical lines; each point on this graph can be pinpointed using its respective x and y values. This method makes it easy to visualize mathematical concepts such as lines, curves, and shapes.

Rectangular coordinates also tie into algebraic equations, as using x and y we can write equations that describe geometric figures, such as circles or lines.

Polar to Rectangular Conversion

Polar to rectangular conversion is a method used to transform the representation of a point from polar coordinates to rectangular coordinates. This is often needed because equations or graphs may be more intuitive or easier to work with in one coordinate system than the other.

In polar coordinates, a point is defined by a distance \(r\) from the origin and an angle \(\theta\) from the positive x-axis. The conversion formulas are:

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

Using these formulas, we can take a polar equation like \(r = 2 \cos \theta - 4 \sin \theta\) and convert it to a rectangular format. This conversion is particularly useful because it allows one to utilize Cartesian graphing techniques and to apply algebraic manipulations to solve problems.

Often, equations that are complex in polar form become simpler or more recognizable in rectangular form, which helps in solving the exercise by applying known algebraic techniques like completing the square.

Completing the Square

Completing the square is a mathematical process used to transform a quadratic equation into a perfect square trinomial. This method is incredibly helpful for simplifying equations and is especially useful when working with equations of a circle.

The goal is to take an equation, like a quadratic, and rewrite it in a way that reveals its geometric properties more clearly. For example, with the equation \(x^2 - 2x\), completing the square involves creating a perfect square trinomial, \((x-1)^2 - 1\). Essentially, you're manipulating the equation to isolate the complete square.

This technique also applies to equations involving y, for example, \(y^2 + 4y\), rewriting it as \((y+2)^2 - 4\). After completing the square for both x and y terms in the equation \(x^2 + y^2 - 2x + 4y = 0\), it becomes easier to recognize the structure of the equation and identify it as being that of a circle.

This clarity is essential in ensuring that graphs and geometric interpretations align across different coordinates and enable simple visualization and computation.

Equation of a Circle

The equation of a circle in rectangular coordinates is expressed in the form \((x-h)^2 + (y-k)^2 = r^2\), where \(h, k\) is the center of the circle and \(r\) is its radius. This format allows for immediate visualization of the circle on a Cartesian plane.

In the context of converting from polar to rectangular coordinates, once the equation is rearranged and simplified, it should ideally mirror this form. For instance, the equation \( (x-1)^2 + (y+2)^2 = 5 \) identifies a circle centered at \( (1, -2) \) with a radius of \( \sqrt{5} \).

Identifying the center and radius quickly enables sketching and interpretation, crucial for understanding the transition between mathematical equations and graphical representations.

Knowing the characteristics of a circle's equation is fundamental in various mathematical problems, from basic geometry to more advanced applications involving polar and rectangular coordinate transitions.