Chapter 10: Problem 37
Sketch the given curves and find their points of intersection. $$ r=6 \sin \theta, r=\frac{6}{1+2 \sin \theta} $$
Short Answer
Expert verified
Intersection points are (6,0), (-6,0), and (0,6).
Step by step solution
01
Convert to Rectangular Coordinates
The given polar equations are \( r = 6 \sin \theta \) and \( r = \frac{6}{1 + 2 \sin \theta} \). Let's convert them into rectangular coordinates to make them easier to graph. For the first equation, we know: \[x = r \cos \theta, \, y = r \sin \theta.\] Substituting \( y = r \sin \theta = 6 \sin \theta \), we get \( y = 6 \sin \theta \). Further substitution with \( \sin \theta = \frac{y}{r} \), gives \( y = 6 \left(\frac{y}{r}\right) \) or \( r = 6 \). Rearrange to get \[ x^2 + y^2 = 6. \]Convert the second equation similarly: \( r = \frac{6}{1 + 2 \sin \theta} \), using \( r = \sqrt{x^2 + y^2} \), and \( \sin \theta = \frac{y}{\sqrt{x^2 + y^2}} \), we get: \[ r (1 + 2 \sin \theta) = 6, \]using \( r = \sqrt{x^2 + y^2} \), substitute into:\[ 1 + \frac{2y}{\sqrt{x^2 + y^2}} = \frac{6}{\sqrt{x^2 + y^2}}. \]
02
Simplify and Solve Equations
We obtained the following equations: 1. \( x^2 + y^2 = 36 \) from \( r = 6 \sin \theta \).2. \( \sqrt{x^2 + y^2} + 2y = 6 \) from \( r = \frac{6}{1 + 2 \sin \theta} \).Replace \( \sqrt{x^2 + y^2} \) in the second equation with \( r = \sqrt{x^2 + y^2} \), so:\[ \sqrt{x^2 + y^2} = 6 - 2y. \] Square both sides:\[ x^2 + y^2 = (6 - 2y)^2. \] Now expand:\[ x^2 + y^2 = 36 - 24y + 4y^2. \]Combine this with \( x^2 + y^2 = 36 \), and subtract:\[ 0 = -24y + 4y^2, \] giving us \( 4y^2 - 24y = 0 \). This can be factored as \[ 4y(y - 6) = 0 \].
03
Find Intersection Points
Solving the factored equation \( 4y(y - 6) = 0 \) gives us \( y = 0 \) or \( y = 6 \). For \( y = 0 \), substitute back into either original equation:\( x^2 + 0^2 = 36 \) gives \( x = \pm 6 \), so points (6,0) and (-6,0).For \( y = 6 \), substitute:\( x^2 + 36 = 36 \) giving \( x = 0 \), so point (0,6).
04
Graph and Verify Points
Graph both equations in polar form to visually verify the solutions. The curve \( r = 6 \sin \theta \) is a circle, symmetric about y-axis with a center at \( (3, 3) \) in polar form, while \( r = \frac{6}{1 + 2 \sin \theta} \) represents a limaçon. Check the intersections by graphing them with the resulting intersection points: (6,0), (-6,0), and (0,6).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rectangular Coordinates
Rectangular coordinates, also known as Cartesian coordinates, are a way to represent points in a plane based on two perpendicular axes: the x-axis and the y-axis. Every point in this system is expressed as
- (x, y) where x represents the horizontal distance from the origin and y represents the vertical distance.
Intersection Points
Intersection points are where two curves meet or cross each other on a plane. Finding these points is crucial in many mathematical problems because it often represents the solution to an equation or system of equations.
- In our exercise, once the curves were converted to rectangular coordinates, we derived algebraic equations which helped us find the intersection.
- By setting the converted equations equal to each other, we solved for the y-values first, and then found corresponding x-values.
Graphing Polar Equations
Polar equations define a relationship between the radial distance from the origin and the angle from a reference direction. Graphing polar equations involves plotting these relationships, typically involving angles (\( \theta \)) and radii (\( r \)).
- In our problem, one equation describes a circle, while the other describes a limaçon.
- In a polar coordinate system, circles and limaçons have distinct and easily recognizable shapes, allowing a visual check of our analytical solutions for intersections.
Conversion between Polar and Rectangular Forms
Conversion between the polar and rectangular systems is a key skill, especially in calculus and geometry.
- Polar coordinates (\( r, \theta \)) describe points in terms of their distance from the origin and direction from the positive x-axis.
- Rectangular coordinates (\( x, y \)) are more intuitive for many applications because they map directly onto the Cartesian plane.
- Use \( x = r \, \cos \theta \) and \( y = r \, \sin \theta \) for polar to rectangular form.
- Reverse these operations with \( r = \sqrt{x^2 + y^2} \) and \( \theta = \arctan \left( \frac{y}{x} \right) \) for rectangular to polar form.