Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the given curves and find their points of intersection. $$ r=6 \sin \theta, r=\frac{6}{1+2 \sin \theta} $$

Step-by-Step Solution

Answer

Intersection points are (6,0), (-6,0), and (0,6).

Step 1: 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:\[ 1 + \frac{2y}{\sqrt{x^2 + y^2}} = \frac{6}{\sqrt{x^2 + y^2}}. \]

Step 2: 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 \[ 4y(y - 6) = 0 \].

Step 3: 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).

Step 4: 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).

Key Concepts

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.

In the original problem, we converted polar coordinates to rectangular coordinates to simplify graphing the curves. This conversion allows us to easily visualize complex shapes and solve equations by bringing them into familiar ground.

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.

These intersection points are essential for understanding how different curves relate spatially.

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.

Understanding how to graph these equations helps students visualize and interpret complex mathematical relationships.

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.

To convert:

• 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.

This conversion is vital in solving problems that require a switch between different geometric perspectives.