Problem 16
Question
15–36 Sketch the graph of the polar equation. $$r=-1$$
Step-by-Step Solution
Verified Answer
A circle of radius 1 centered at the origin.
1Step 1: Understanding the Polar Equation
The equation given is in polar coordinates: \(r = -1\). This is a simple polar equation where the radius \(r\) is constant and negative.
2Step 2: Recognizing the Implication of a Negative Radius
In polar coordinates, a negative radius means we trace out points in the direction opposite the angle \(\theta\) from the origin. For every possible angle \(\theta\), \(r = -1\) means the point is 1 unit away from the origin in the opposite direction.
3Step 3: Identifying Points for Different Angles
For various values of \(\theta\), such as \(\theta = 0, \frac{\pi}{4}, \frac{\pi}{2}, \pi, -\frac{\pi}{2}\), the corresponding points are \((-1, \pi), (-1, \frac{5\pi}{4}), (-1, \frac{3\pi}{2}), (-1, 0), (-1, \frac{\pi}{2})\). Each point is found by rotating \(180^\circ\) (adding or subtracting \(\pi\) radians) because of the negative radius.
4Step 4: Visualizing the Graph
Plotting these points on a polar graph shows they all lie on a circle of radius 1 centered at the origin. The negative sign simply means these points are encountered by moving \(1\) unit from the origin, in the reverse direction for any given \(\theta\). As \(\theta\) varies from \(0\) to \(2\pi\), the graph forms a complete circle.
5Step 5: Drawing the Circle
Finally, sketch a circle centered at the origin with radius 1 unit, acknowledging the negative radius by remembering that points are directly opposite to where positive radius would place them. This is the graph of \(r = -1\).
Key Concepts
Polar EquationsGraphing Polar EquationsNegative Radius Concept
Polar Equations
A polar equation is a mathematical expression used to determine points in a polar coordinate system. Instead of using the familiar Cartesian coordinates (x, y), the polar coordinate system uses a radius \( r \) and an angle \( \theta \). The radius measures the distance from the origin (often called the pole), and the angle is measured from the positive x-axis (polar axis).
Polar equations define how \( r \) changes with varying \( \theta \). They can create diverse graph shapes, from simple circles to intricate spirals and roses. In the case of the exercise, the polar equation \( r = -1 \) tells us that the radius remains constant at -1 for all angles \( \theta \). This simplicity allows us to explore foundational concepts and how these equations alter a point's position.
Polar equations define how \( r \) changes with varying \( \theta \). They can create diverse graph shapes, from simple circles to intricate spirals and roses. In the case of the exercise, the polar equation \( r = -1 \) tells us that the radius remains constant at -1 for all angles \( \theta \). This simplicity allows us to explore foundational concepts and how these equations alter a point's position.
Graphing Polar Equations
Graphing polar equations involves plotting points based on their polar coordinates and connecting them to reveal shapes. Each point is determined by a pair \( (r, \theta) \). Here's how you would approach it:
- Choose several values for \( \theta \) ranging from 0 to \( 2\pi \) (or 0 to 360 degrees).
- Calculate the corresponding \( r \) using the polar equation.
- Plot each point in the polar coordinate system based on its \( r \) and \( \theta \).
- Connect the points smoothly if the equation represents a continuous curve.
Negative Radius Concept
The negative radius concept in polar coordinates is intriguing, as it influences the direction of plotting points. Normally, if the radius \( r \) is positive, you plot the point by moving \( r \) units away from the origin in the direction of the angle \( \theta \). But when \( r \) is negative, the movement is in the opposite direction.
This means for a polar equation like \( r = -1 \), the point lies 1 unit away from the origin but opposite to where a positive radius would place it for the same angle. A useful approach:
This means for a polar equation like \( r = -1 \), the point lies 1 unit away from the origin but opposite to where a positive radius would place it for the same angle. A useful approach:
- Identify any angle, like \( \theta = 0 \), where a positive radius would point directly along the positive x-axis.
- With \( r = -1 \), the point shifts directly opposite, to the negative x-axis.
Other exercises in this chapter
Problem 16
15-18 Find the indicated quantity, assuming \(\mathbf{u}=2 \mathbf{i}+\mathbf{j}, \mathbf{v}=\mathbf{i}-3 \mathbf{j},\) and \(\mathbf{w}=3 \mathbf{i}+4 \mathbf{
View solution Problem 16
\(15-16=\) Sketch \(z_{1}, z_{2}, z_{1}+z_{2},\) and \(z_{1} z_{2}\) on the same complex plane. $$ z_{1}=-1+i, \quad z_{2}=2-3 i $$
View solution Problem 17
Find \(2 \mathbf{u},-3 \mathbf{v}, \mathbf{u}+\mathbf{v},\) and \(3 \mathbf{u}-4 \mathbf{v}\) for the given vectors \(\mathbf{u}\) and \(\mathbf{v} .\) $$ \math
View solution Problem 17
15-18 Find the indicated quantity, assuming \(\mathbf{u}=2 \mathbf{i}+\mathbf{j}, \mathbf{v}=\mathbf{i}-3 \mathbf{j},\) and \(\mathbf{w}=3 \mathbf{i}+4 \mathbf{
View solution