Problem 20
Question
Graph the polar equations. $$r^{2}=9 \cos 2 \theta$$
Step-by-Step Solution
Verified Answer
The polar equation represents a lemniscate, a figure-eight shape with petals extending 3 units from the origin.
1Step 1: Identify the Type of Polar Equation
The given equation is \( r^2 = 9 \cos 2\theta \). This is a polar equation that represents a conic section. We need to recognize that it is of the form \( r^2 = a \cos 2\theta \). When written in this form, it indicates either a limaçon or a conic section, specifically a lemniscate when \( a > 0 \).
2Step 2: Understanding the Polar Equation
The equation \( r^2 = 9 \cos 2\theta \) is a lemniscate. A lemniscate is a figure-eight shaped curve in the polar system. The equation is symmetric about the polar axis. The constant term \( 9 \) indicates that the "petals" of the lemniscate will extend 3 units from the origin as the square root of 9 is 3.
3Step 3: Convert to Cartesian Coordinates (Optional)
To better understand the shape in the Cartesian plane, convert \( r^2 = 9 \cos 2\theta \) to Cartesian coordinates. We use \( r^2 = x^2 + y^2 \) and the identity \( \cos 2\theta = \frac{x^2-y^2}{x^2+y^2} \). Substituting these, we get: \[x^2 + y^2 = 9 \frac{x^2-y^2}{x^2+y^2}\]Cross-multiplying and rearranging yields the Cartesian form, though graphing is typically easier directly from the polar form.
4Step 4: Plot Points Based on Polar Coordinates
Select specific values for \( \theta \) to compute corresponding \( r \) values. For example, for \( \theta = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi \), etc. Calculate \( r \) for these \( \theta \): - \( \theta = 0 \rightarrow r^2 = 9 \implies r = \pm 3 \).- \( \theta = \pi/4 \rightarrow r^2 = 9/\sqrt{2} \implies r = \pm \frac{3}{\sqrt{2}} \).Continue this process to cover a full range of \( \theta \) including negative \( r \).
5Step 5: Draw the Lemniscate Curve
Using the plotted points, draw the symmetric lemniscate shape on polar graph paper. The curve will pass through or near the points you plotted (including negative \( r \) values, which reflect through the origin), forming a figure-eight with the largest loops approximately 3 units from the origin.
Key Concepts
LemniscateConic SectionsGraphing Polar Equations
Lemniscate
A lemniscate is a fascinating geometrical figure that appears in the form of a figure-eight or an infinity symbol. In the realm of polar coordinates, it is most commonly represented by equations like \( r^2 = a \, \cos 2\theta \) or \( r^2 = a \, \sin 2\theta \).
- The term 'lemniscate' originates from the Latin word 'lemniscatus', meaning "decorated with ribbons."
- It holds distinctive symmetry properties: rotating it by 180 degrees results in the same shape.
- The intersection at the origin point signifies the lemniscate crossing itself.
Conic Sections
Conic sections are the curves obtained by intersecting a plane with a double cone. These include many familiar figures: circles, ellipses, parabolas, and hyperbolas. However, in a polar coordinate system, our perspective shifts slightly.
- Conic sections in polar coordinates can be presented as equations involving \( r \), the radial distance from the origin, and \( \theta \), the angle.
- Such equations often take on forms that encompass lemniscates or other familiar curves.
- They provide a powerful tool for analyzing and grappling with the spatial relationships between curves and their features.
Graphing Polar Equations
Graphing polar equations involves visualizing the relationship between \( r \) and \( \theta \). Unlike Cartesian graphs, which rely on x-y coordinates, polar graphs use a radial system with an origin point and angles.
- To graph polar equations like \( r^2=9 \cos 2\theta \), you typically calculate \( r \) for several values of \( \theta \), specifically from 0 to \( 2\pi \) to obtain a complete plot.
- The symmetry in polar equations, such as reflection or rotational symmetry, simplifies the graphing process. For instance, symmetry about the polar axis can aid in constructing half the graph before mirroring it.
- Points need to be plotted thoughtfully, considering both positive and negative values of \( r \), which may necessitate reflecting through the origin.
Other exercises in this chapter
Problem 20
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