Problem 55
Question
Exer. 45-78: Sketch the graph of the polar equation. $$ r=-6(1+\cos \theta) $$
Step-by-Step Solution
Verified Answer
The graph of this polar equation is a limaçon with an inner loop.
1Step 1: Recognize the Type of Polar Equation
The given polar equation is \( r = -6(1 + \cos \theta) \), which is a form of the limaçon family. Specifically, this is in the form \( r = a + b \cos \theta \), where \( a = -6 \) and \( b = -6 \). Since \( a = b \), we know this is a limaçon with an inner loop.
2Step 2: Convert to Rectangular Coordinates for Plotting Points
Using the identities \( x = r \cos \theta \) and \( y = r \sin \theta \), you can express the equation in rectangular coordinates. This isn't necessary for sketching but helpful to understand the shape in Cartesian terms.
3Step 3: Identify Key Points and Symmetries
Since the equation is symmetric concerning the polar axis (positive x-axis), we can use this symmetry to simplify plotting. Some key points to plot could include \( \theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \), etc. For example, when \( \theta = 0 \), we have \( r = -12 \), indicating that the curve crosses the polar axis.
4Step 4: Determine the Behavior at Specific Angles
Calculate \( r \) for several \( \theta \) values to determine significant behavior. For example, when \( \theta = \pi \), \( r = 0 \), reaching the origin. At \( \theta = \frac{\pi}{3} \) or any angle resulting in \( -6(1 + \frac{1}{2}) \), \( r = -9 \), indicating the inner loop.
5Step 5: Sketch the Polar Graph
Using the key points and the symmetry properties, begin by plotting the inner loop, which occurs between \( \theta = 0 \) and \( \theta = \pi \). Then plot the outer boundary of the limaçon which exists as \( r \) follows the equation after crossing the pole. Ensure the loop reflects the negative value of \( r \) and correct directional plotting.
Key Concepts
LimaçonPolar CoordinatesSymmetry in Polar Graphs
Limaçon
The limaçon is a fascinating type of mathematical curve that belongs to the broader family of polar equations. It is classified under the form \( r = a + b \cos \theta \) or \( r = a + b \sin \theta \). These curves take diverse shapes based on the relationship between the values of \( a \) and \( b \). In our given example, \( r = -6(1 + \cos \theta) \), both \( a \) and \( b \) are equal to -6, which results in a limaçon with an inner loop. This form arises when \( a = b \), providing a distinct looping feature. The appearance of a limaçon can vary significantly:
- If \( |a| < |b| \), an inner loop is formed.
- If \( |a| = |b| \), it creates a limaçon with a cusp.
- If \( |a| > |b| \), the limaçon appears as a dimpled curve.
Polar Coordinates
Polar coordinates are a way of defining a point on a plane using a distance and an angle relative to a reference point and direction, usually the origin and the positive x-axis, respectively. While most are familiar with Cartesian coordinates \((x, y)\), polar coordinates use \((r, \theta)\), where \(r\) is the radial distance from the origin and \(\theta\) is the angle measured counterclockwise from the positive x-axis. Polar coordinates are especially useful for representing curves and points that exhibit radial symmetry, such as circles and spirals. To convert between polar and Cartesian coordinates, the following equations are used:
- \(x = r \cos \theta\)
- \(y = r \sin \theta\)
- \(r = \sqrt{x^2 + y^2}\)
- \(\theta = \tan^{-1}\left(\frac{y}{x}\right)\)
Symmetry in Polar Graphs
Symmetry in polar graphs makes the graphing process more efficient and manageable by allowing the symmetry properties to simplify computation or estimation of the curve's full geometry. There are several types of symmetries to consider when sketching polar graphs:
- **Symmetry with Respect to the Polar Axis:** If replacing \(\theta\) with \(-\theta\) gives an equivalent equation, the graph is symmetric with respect to the polar axis (x-axis).
- **Symmetry with Respect to the Line \(\theta = \frac{\pi}{2}\):** If replacing \(r\) with \(-r\) or \(\theta\) with \(\pi - \theta\) gives an equivalent equation, there is symmetry about the line \(\theta = \frac{\pi}{2}\) (y-axis).
- **Symmetry with Respect to the Pole:** If substituting \(r\) for \(-r\) gives an equivalent equation, then the graph has rotational symmetry through the pole (origin).
Other exercises in this chapter
Problem 54
Exer. 45-78: Sketch the graph of the polar equation. $$ r=3(1+\cos \theta) $$
View solution Problem 55
Describe the part of a hyperbola given by the equation. $$y=-\frac{9}{4} \sqrt{x^{2}-16}$$
View solution Problem 56
Exer. 45-78: Sketch the graph of the polar equation. $$ r=2(1+\sin \theta) $$
View solution Problem 57
Describe the part of a hyperbola given by the equation. $$x=-\frac{2}{3} \sqrt{y^{2}-36}$$
View solution