Chapter 11: Problem 63
Find polar equations for the circles in Exercises \(57-64 .\) Sketch each circle in the coordinate plane and label it with both its Cartesian and polar equations. $$ x^{2}+y^{2}+y=0 $$
Short Answer
Expert verified
The circle's polar equation is \( r = \sin(\theta) \).
Step by step solution
01
Completing the Square
The given equation is \( x^2 + y^2 + y = 0 \). We need to transform it into the standard form of a circle equation. Start by completing the square for the \( y \)-terms. The equation can be rewritten as \( x^2 + (y^2 + y) = 0 \). To complete the square, consider the expression \( y^2 + y \). Add and subtract \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \): \( y^2 + y + \frac{1}{4} - \frac{1}{4} = (y + \frac{1}{2})^2 - \frac{1}{4} \).
02
Rewriting in Standard Form
Substitute back the completed square into the equation: \( x^2 + (y + \frac{1}{2})^2 - \frac{1}{4} = 0 \). This simplifies to \( x^2 + (y + \frac{1}{2})^2 = \frac{1}{4} \), which is the equation of a circle in Cartesian coordinates with center \( (0, -\frac{1}{2}) \) and radius \( \frac{1}{2} \).
03
Converting to Polar Coordinates
In polar coordinates, \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \). Substitute these into the circle equation: \( (r \cos \theta)^2 + (r \sin \theta + \frac{1}{2})^2 = \frac{1}{4} \). Simplifying, you find \( r^2 - r\sin(\theta) = 0 \). Factor to get \( r(r - \sin(\theta)) = 0 \). The non-trivial solution is \( r = \sin(\theta) \).
04
Sketching the Circle
Sketch the circle with the Cartesian equation \( x^2 + (y + \frac{1}{2})^2 = \frac{1}{4} \), displaying a circle centered at \( (0, -\frac{1}{2}) \) with a radius of \( \frac{1}{2} \). In polar coordinates, the circle is represented by the equation \( r = \sin(\theta) \), which traces the same path as the Cartesian equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Completing the Square
Completing the square is a technique used in mathematics to transform a quadratic equation into a form that is easier to handle, often to reveal important geometric properties. The problem starts with the equation \( x^2 + y^2 + y = 0 \). For this exercise, the goal is to turn this into the standard form of a circle equation.
To complete the square on the \( y \)-terms, you take the linear coefficient of \( y \), which is 1, divide it by 2 to get \( \frac{1}{2} \), and then square it to get \( \frac{1}{4} \). This value balances the equation by effectively turning \( y^2 + y \) into \((y + \frac{1}{2})^2 - \frac{1}{4}\). This conversion is the essence of completing the square.
The main point of this method here is to rewrite the original equation into a form that highlights it as a circle equation. This leads to easier geometric interpretation and subsequent steps of conversion or sketching.
To complete the square on the \( y \)-terms, you take the linear coefficient of \( y \), which is 1, divide it by 2 to get \( \frac{1}{2} \), and then square it to get \( \frac{1}{4} \). This value balances the equation by effectively turning \( y^2 + y \) into \((y + \frac{1}{2})^2 - \frac{1}{4}\). This conversion is the essence of completing the square.
The main point of this method here is to rewrite the original equation into a form that highlights it as a circle equation. This leads to easier geometric interpretation and subsequent steps of conversion or sketching.
Cartesian Coordinates
Cartesian coordinates are a way of visualizing algebraic equations on a plane using the \(x\)- and \(y\)-axes. Every point on a Cartesian plane is represented by an \((x, y)\) pair where \(x\) is the horizontal distance and \(y\) is the vertical distance from the origin.
In the case of circles, a standard Cartesian equation for a circle is \( (x - h)^2 + (y - k)^2 = r^2 \), where \((h, k)\) is the center, and \(r\) is the radius. After completing the square, the given equation takes the form \( x^2 + (y + \frac{1}{2})^2 = \frac{1}{4} \).
This represents a circle with center at \((0, -\frac{1}{2})\) and a radius of \(\frac{1}{2}\).
In the case of circles, a standard Cartesian equation for a circle is \( (x - h)^2 + (y - k)^2 = r^2 \), where \((h, k)\) is the center, and \(r\) is the radius. After completing the square, the given equation takes the form \( x^2 + (y + \frac{1}{2})^2 = \frac{1}{4} \).
This represents a circle with center at \((0, -\frac{1}{2})\) and a radius of \(\frac{1}{2}\).
- The center is a translation measurement from the origin, showing where the circle is located.
- The radius is half the diameter of the circle, representing its size.
Polar Equations
Polar equations express relationships using a system where each point is determined by a distance from a reference point and an angle from a reference direction. This differs from Cartesian coordinates, which use perpendicular lines (axes) to define locations on a plane.
To convert the Cartesian circle equation to a polar equation, use the identities \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). The equation \( x^2 + (y + \frac{1}{2})^2 = \frac{1}{4} \) transforms via substitution and simplification into \( r = \sin(\theta) \). This equation describes the same path as the original Cartesian equation but does so in terms of \(r\) and \(\theta\).
Some advantages of polar equations include:
To convert the Cartesian circle equation to a polar equation, use the identities \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). The equation \( x^2 + (y + \frac{1}{2})^2 = \frac{1}{4} \) transforms via substitution and simplification into \( r = \sin(\theta) \). This equation describes the same path as the original Cartesian equation but does so in terms of \(r\) and \(\theta\).
Some advantages of polar equations include:
- They are great for representing curves where symmetry about the origin is evident.
- Certain types of curves, like spirals or circles, often have simpler forms in polar coordinates than Cartesian.
Circle Equations
Circle equations are special forms of algebraic expressions that describe a perfect round shape, defined equally in all directions from a central point. In algebra, the standard form of a circle equation in Cartesian coordinates is \( (x - h)^2 + (y - k)^2 = r^2 \). This format reveals both the center of the circle \((h, k)\) and the radius \(r\).
The original equation \( x^2 + y^2 + y = 0 \) is converted to \( x^2 + (y + \frac{1}{2})^2 = \frac{1}{4} \) after completing the square. This identifies it as a circle with center at \((0, -\frac{1}{2})\) and radius \( \frac{1}{2} \).
In the polar form, our circle is described as \( r = \sin(\theta) \). This provides a new perspective, showing that circles can be described in multiple ways through intersecting algebraic and geometric viewpoints.
The original equation \( x^2 + y^2 + y = 0 \) is converted to \( x^2 + (y + \frac{1}{2})^2 = \frac{1}{4} \) after completing the square. This identifies it as a circle with center at \((0, -\frac{1}{2})\) and radius \( \frac{1}{2} \).
In the polar form, our circle is described as \( r = \sin(\theta) \). This provides a new perspective, showing that circles can be described in multiple ways through intersecting algebraic and geometric viewpoints.
- The conversion between Cartesian and polar circles highlights how algebraic manipulation can lead to deeper insights on geometric properties.
- Understanding these equations helps in graphing circles and exploring their characteristics such as symmetry and intersections with other curves.