Problem 25
Question
In Exercises 23-48, sketch the graph of the polar equation using symmetry, zeros, maximum \(r\)-values, and any other additional points. \(r=\dfrac{\pi}{3}\)
Step-by-Step Solution
Verified Answer
The polar plot of the given equation represents a circle with a radius of \( \pi /3\). There is symmetry about the line of \(θ=0\). The maximum r-value is \( \pi /3\). The graph does not have zero as r always equals \( \pi /3\).
1Step 1: Decode the Polar Equation
The given polar equation is \(r = \pi /3\). As can be seen, this equation doesn't include any dependence on the angle θ. In such a case, it means that r has a constant value for all θ. Thus, the polar plot will represent a circle with a radius of \(\pi /3\).
2Step 2: Sketch the Polar Graph
Draw a circle centered at the origin. Since the equation doesn't depend on θ, the radius remains the same for all angles. Measure out a distance of \(\pi /3\) along each of the polar axis and draw a circle of this radius.
3Step 3: Note down the Characteristics of the Graph
The graph is a perfect circle with a constant radius of \(\pi /3\). The polar graph has symmetry about the line of θ=0 due to its round shape that is identical in all directions. The maximum r-value is \(\pi /3\) as stated in the equation.
Key Concepts
Understanding the Polar EquationIdentifying Symmetry in Polar GraphsDetermining the Maximum r-valueGraphing Polar Functions
Understanding the Polar Equation
A polar equation like the one given, \(r = \frac{\pi}{3}\), is unique as it describes relationships using polar coordinates. In polar coordinates, a point in the plane is represented as \((r, \theta)\), where \(r\) is the radial distance from the origin and \(\theta\) is the angular direction. In our case, the equation is particularly simple because \(\theta\) doesn't appear in the equation. This means that \(r\) is constant, irrespective of the angle. As a result, this equation represents a circle with a fixed radius extending in all directions, forming a circle around the origin.
Identifying Symmetry in Polar Graphs
Symmetry in polar graphs is essential for simplifying graphing of polar equations. There are three main types of symmetry to consider in polar graphs:
- Symmetry about the polar axis (horizontal symmetry).
- Symmetry about the line \(\theta = \frac{\pi}{2}\) (vertical symmetry).
- Symmetry about the pole (origin).
Determining the Maximum r-value
In polar graphs, the maximum \(r\)-value is the largest distance from the origin that is plotted in the graph. For our polar equation \(r = \frac{\pi}{3}\), the maximum value of \(r\) is directly provided by the equation. This is because \(r\) does not vary with \(\theta\). Every point on the graph maintains the same distance \(\frac{\pi}{3}\) from the origin, so \(\frac{\pi}{3}\) is both the minimum and maximum \(r\)-value. This contrast from more complex polar equations where \(r\) varies, underlines the simplicity of graphing a circle via polar coordinates.
Graphing Polar Functions
Graphing polar functions involves plotting points for various angles \(\theta\) based on their corresponding \(r\)-values. With polar equations like \(r = \frac{\pi}{3}\), where \(r\) is constant, graphing becomes straightforward. Here’s how you can sketch it:
- Draw the polar axes with origin marked clearly.
- Measure a consistent distance of \(\frac{\pi}{3}\) from the origin in every direction.
- Connect these points with a smooth curve to form a circle.
Other exercises in this chapter
Problem 24
In Exercises 19-26, find the inclination \(\theta\) (in radians and degrees) of the line passing through the points. \((12, 8)\), \((-4, -3)\)
View solution Problem 25
In Exercises 15-28, identify the conic and sketch its graph. \(r=\dfrac{3}{2-6\cos\ \theta}\)
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In Exercises 19-28, a point in polar coordinates is given. Convert the point to rectangular coordinates. \(\left(-2, 7\pi/6\right)\)
View solution Problem 25
In Exercises 7-26, (a) sketch the curve represented by the parametric equations (indicate the orientation of the curve) and (b) eliminate the parameter and writ
View solution