Problem 32
Question
\(29-48\) Sketch the curve with the given polar equation. $$r=-3 \cos \theta$$
Step-by-Step Solution
Verified Answer
The curve is a circle centered at (-1.5, 0) with a diameter of 3.
1Step 1: Understanding the Polar Equation
We start by understanding the given polar equation \( r = -3 \cos \theta \). This is in polar form where \( r \) represents the radius, and \( \theta \) is the angle. A negative sign in front of \( \cos \theta \) indicates reflection in polar coordinates.
2Step 2: Analyzing the Symmetry and Shape
The presence of \( \cos \theta \) in the equation suggests symmetry about the horizontal axis (polar axis). Because \( r = -3 \cos \theta \) involves \( \cos \theta \), this is typically a limaçon. For if \( r = a + b\cos(\theta) \) where \( a = 0 \) and \( b < 0 \), it is a circle with a center shifted to the left.
3Step 3: Determine Key Points
Calculating key points will help sketch the curve. For \( \theta = 0 \), \( r = -3 \cos(0) = -3 \). For \( \theta = \pi \), \( r = -3 \cos(\pi) = 3 \). For other angles like \( \theta = \frac{\pi}{2} \), and \( \theta = \frac{3\pi}{2} \) we find \( r = 0 \). These values highlight the transition of the radius as \( \theta \) varies.
4Step 4: Sketch the Curve
Using the points from Step 3, plot the points on the polar graph. With \( r = -3 \) at \( \theta = 0 \), the point is opposite in direction, so at \( \theta = \pi \), \( r = 3 \). Plot the point at (3, \( \pi \)) and likewise for others, forming a circle centered at (-1.5, 0) in cartesian form, resulting in diameter 3.
Key Concepts
Polar EquationsLimaçon CurvesSymmetry in Polar Graphs
Polar Equations
Polar equations express the relationship between the radial distance \( r \) from the origin and the angle \( \theta \) from a fixed direction, often the positive x-axis in a plane. Unlike Cartesian coordinates which use \( x \) and \( y \), polar coordinates offer a convenient way to describe curves and circles in terms of direction and distance.
Polar coordinates are especially useful when dealing with circular or spiral patterns. The equation \( r = -3 \cos \theta \) is an example of a polar equation. In this, \( r \) can be positive or negative. A negative \( r \) implies moving in the opposite direction of \( \theta \), essentially reflecting over the origin.
This form lends itself to symmetry and rotation, offering a unique way to understand the distances and angles without confinement to straight lines. In polar equations, trigonometric functions define how \( r \) changes with \( \theta \). Modeling circles becomes intuitive as values vary sinusoidally over \( \theta \):
Polar coordinates are especially useful when dealing with circular or spiral patterns. The equation \( r = -3 \cos \theta \) is an example of a polar equation. In this, \( r \) can be positive or negative. A negative \( r \) implies moving in the opposite direction of \( \theta \), essentially reflecting over the origin.
This form lends itself to symmetry and rotation, offering a unique way to understand the distances and angles without confinement to straight lines. In polar equations, trigonometric functions define how \( r \) changes with \( \theta \). Modeling circles becomes intuitive as values vary sinusoidally over \( \theta \):
- \( \cos \theta \) typically reflects circles symmetric across the horizontal axis.
- \( \sin \theta \) symmetry crosses the vertical axis.
Limaçon Curves
The limaçon is a particular family of polar curves and comes into play with equations of the form \( r = a + b \cos \theta \) or \( r = a + b \sin \theta \). It's named after the French word for "snail" due to its distinctive shape.
When \( a = 0 \) and \( b eq 0 \), we typically deal with circular variations, however, the entire graph is still considered a limaçon due to its formation principles. For example, the equation we examined, \( r = -3 \cos \theta \), essentially forms a circle with a shifted center.
Limaçon curves might appear as:
When \( a = 0 \) and \( b eq 0 \), we typically deal with circular variations, however, the entire graph is still considered a limaçon due to its formation principles. For example, the equation we examined, \( r = -3 \cos \theta \), essentially forms a circle with a shifted center.
Limaçon curves might appear as:
- Simple limacons (dimpled configurations where one end is indented).
- Inner-loop limacons, forming a loop within themselves.
- Cardioids, when \( a = b \), resembling a heart shape.
Symmetry in Polar Graphs
Symmetry in polar graphs is a fascinating characteristic that helps simplify complex calculations and predict how polar curves behave.
A polar graph like \( r = -3 \cos \theta \) reveals symmetry across the polar axis, which is analogous to the horizontal axis in Cartesian coordinates.
Here are some key observations:
Understanding these symmetries reduces complexity in graphing, offering a shortcut to visualize and sketch polar curves effectively.
A polar graph like \( r = -3 \cos \theta \) reveals symmetry across the polar axis, which is analogous to the horizontal axis in Cartesian coordinates.
Here are some key observations:
- **Reflection over Polar Axis (Horizontal Axis):** Graphs involving functions like \( r = a \cos \theta \) generally show mirror symmetry about the polar axis.
- **Reflection over Vertical Line:** If \( \sin \theta \) is predominant, the graph is typically symmetric about the vertical line \( \theta = \frac{\pi}{2} \).
- **Origin Symmetry:** Some graphs also show origin symmetry, where \( (r, \theta) \) translates to \((-r, \theta + \pi) \).
Understanding these symmetries reduces complexity in graphing, offering a shortcut to visualize and sketch polar curves effectively.
Other exercises in this chapter
Problem 31
(a) Show that the parametric equations $$x=x_{1}+\left(x_{2}-x_{1}\right) t \quad y=y_{1}+\left(y_{2}-y_{1}\right) t$$ where \(0 \leqslant t \leqslant 1,\) desc
View solution Problem 32
Find an equation for the conic that satisfies the given conditions. Parabola, vertex \((1,0), \quad\) directrix \(x=-5\)
View solution Problem 33
Find an equation for the conic that satisfies the given conditions.\(y^{2}=-12(x+1)\) Parabola, focus \((-4,0), \quad\) directrix \(x=2\)
View solution Problem 33
\(29-48\) Sketch the curve with the given polar equation. $$r=2(1-\sin \theta), \theta \geqslant 0$$
View solution