Module 1: Parametric Equations and Polar Coordinates
Graphs and Symmetry of Polar Curves
Learning Outcomes
Sketch polar curves from given equations
Convert equations between rectangular and polar coordinates
Identify symmetry in polar curves and equations
Polar Curves
Now that we know how to plot points in the polar coordinate system, we can discuss how to plot curves. In the rectangular coordinate system, we can graph a function [latex]y=f\left(x\right)[/latex] and create a curve in the Cartesian plane. In a similar fashion, we can graph a curve that is generated by a function [latex]r=f\left(\theta \right)[/latex].
The general idea behind graphing a function in polar coordinates is the same as graphing a function in rectangular coordinates. Start with a list of values for the independent variable ([latex]\theta[/latex] in this case) and calculate the corresponding values of the dependent variable [latex]r[/latex]. This process generates a list of ordered pairs, which can be plotted in the polar coordinate system. Finally, connect the points, and take advantage of any patterns that may appear. The function may be periodic, for example, which indicates that only a limited number of values for the independent variable are needed.
Problem-Solving Strategy: Plotting a Curve in Polar Coordinates
Create a table with two columns. The first column is for [latex]\theta [/latex], and the second column is for [latex]r[/latex].
Create a list of values for [latex]\theta [/latex].
Calculate the corresponding [latex]r[/latex] values for each [latex]\theta [/latex].
Plot each ordered pair [latex]\left(r,\theta \right)[/latex] on the coordinate axes.
Graph the curve defined by the function [latex]r=4\sin\theta [/latex]. Identify the curve and rewrite the equation in rectangular coordinates.
Show Solution
Because the function is a multiple of a sine function, it is periodic with period [latex]2\pi [/latex], so use values for [latex]\theta [/latex] between 0 and [latex]2\pi [/latex]. The result of steps 1–3 appear in the following table. Figure 5 shows the graph based on this table.
[latex]\theta [/latex]
[latex]r=4\sin\theta [/latex]
[latex]\theta [/latex]
[latex]r=4\sin\theta [/latex]
[latex]0[/latex]
[latex]0[/latex]
[latex]\pi [/latex]
[latex]0[/latex]
[latex]\frac{\pi }{6}[/latex]
[latex]2[/latex]
[latex]\frac{7\pi }{6}[/latex]
[latex]-2[/latex]
[latex]\frac{\pi }{4}[/latex]
[latex]2\sqrt{2}\approx 2.8[/latex]
[latex]\frac{5\pi }{4}[/latex]
[latex]-2\sqrt{2}\approx -2.8[/latex]
[latex]\frac{\pi }{3}[/latex]
[latex]2\sqrt{3}\approx 3.4[/latex]
[latex]\frac{4\pi }{3}[/latex]
[latex]-2\sqrt{3}\approx -3.4[/latex]
[latex]\frac{\pi }{2}[/latex]
[latex]4[/latex]
[latex]\frac{3\pi }{2}[/latex]
[latex]4[/latex]
[latex]\frac{2\pi }{3}[/latex]
[latex]2\sqrt{3}\approx 3.4[/latex]
[latex]\frac{5\pi }{3}[/latex]
[latex]-2\sqrt{3}\approx -3.4[/latex]
[latex]\frac{3\pi }{4}[/latex]
[latex]2\sqrt{2}\approx 2.8[/latex]
[latex]\frac{7\pi }{4}[/latex]
[latex]-2\sqrt{2}\approx -2.8[/latex]
[latex]\frac{5\pi }{6}[/latex]
[latex]2[/latex]
[latex]\frac{11\pi }{6}[/latex]
[latex]-2[/latex]
[latex]2\pi [/latex]
[latex]0[/latex]
This is the graph of a circle. The equation [latex]r=4\sin\theta [/latex] can be converted into rectangular coordinates by first multiplying both sides by [latex]r[/latex]. This gives the equation [latex]{r}^{2}=4r\sin\theta [/latex]. Next use the facts that [latex]{r}^{2}={x}^{2}+{y}^{2}[/latex] and [latex]y=r\sin\theta [/latex]. This gives [latex]{x}^{2}+{y}^{2}=4y[/latex]. To put this equation into standard form, subtract [latex]4y[/latex] from both sides of the equation and complete the square:
This is the equation of a circle with radius 2 and center [latex]\left(0,2\right)[/latex] in the rectangular coordinate system.
Watch the following video to see the worked solution to Example: Graphing a Function in Polar Coordinates.
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
Create a graph of the curve defined by the function [latex]r=4+4\cos\theta [/latex].
Hint
Follow the problem-solving strategy for creating a graph in polar coordinates.
Show Solution
Figure 2.
The name of this shape is a cardioid, which we will study further later in this section.
The graph in the previous example was that of a circle. The equation of the circle can be transformed into rectangular coordinates using the coordinate transformation formulas in the theorem. The example after the next gives some more examples of functions for transforming from polar to rectangular coordinates.
Example: Transforming Polar Equations to Rectangular Coordinates
Rewrite each of the following equations in rectangular coordinates and identify the graph.
[latex]\theta =\frac{\pi }{3}[/latex]
[latex]r=3[/latex]
[latex]r=6\cos\theta -8\sin\theta [/latex]
Show Solution
Take the tangent of both sides. This gives [latex]\tan\theta =\tan\left(\frac{\pi}{3}\right)=\sqrt{3}[/latex]. Since [latex]\tan\theta =\frac{y}{x}[/latex] we can replace the left-hand side of this equation by [latex]\frac{y}{x}[/latex]. This gives [latex]\frac{y}{x}=\sqrt{3}[/latex], which can be rewritten as [latex]y=x\sqrt{3}[/latex]. This is the equation of a straight line passing through the origin with slope [latex]\sqrt{3}[/latex]. In general, any polar equation of the form [latex]\theta =K[/latex] represents a straight line through the pole with slope equal to [latex]\tan{K}[/latex].
First, square both sides of the equation. This gives [latex]{r}^{2}=9[/latex]. Next replace [latex]{r}^{2}[/latex] with [latex]{x}^{2}+{y}^{2}[/latex]. This gives the equation [latex]{x}^{2}+{y}^{2}=9[/latex], which is the equation of a circle centered at the origin with radius 3. In general, any polar equation of the form [latex]r=k[/latex] where k is a positive constant represents a circle of radius k centered at the origin. (Note: when squaring both sides of an equation it is possible to introduce new points unintentionally. This should always be taken into consideration. However, in this case we do not introduce new points. For example, [latex]\left(-3,\frac{\pi }{3}\right)[/latex] is the same point as [latex]\left(3,\frac{4\pi}{3}\right)[/latex].)
Multiply both sides of the equation by [latex]r[/latex]. This leads to [latex]{r}^{2}=6r\cos\theta -8r\sin\theta [/latex]. Next use the formulas
To put this equation into standard form, first move the variables from the right-hand side of the equation to the left-hand side, then complete the square.
This is the equation of a circle with center at [latex]\left(3,-4\right)[/latex] and radius 5. Notice that the circle passes through the origin since the center is 5 units away.
Watch the following video to see the worked solution to Example: Transforming Polar Equations to Rectangular Coordinates.
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
Rewrite the equation [latex]r=\sec\theta \tan\theta [/latex] in rectangular coordinates and identify its graph.
Hint
Convert to sine and cosine, then multiply both sides by cosine.
Show Solution
[latex]y={x}^{2}[/latex], which is the equation of a parabola opening upward.
We have now seen several examples of drawing graphs of curves defined by polar equations. A summary of some common curves is given in the tables below. In each equation, a and b are arbitrary constants.
A cardioid is a special case of a limaçon (pronounced “lee-mah-son”), in which [latex]a=b[/latex] or [latex]a=-b[/latex]. The rose is a very interesting curve. Notice that the graph of [latex]r=3\sin2\theta [/latex] has four petals. However, the graph of [latex]r=3\sin3\theta [/latex] has three petals as shown.
If the coefficient of [latex]\theta [/latex] is even, the graph has twice as many petals as the coefficient. If the coefficient of [latex]\theta [/latex] is odd, then the number of petals equals the coefficient. You are encouraged to explore why this happens. Even more interesting graphs emerge when the coefficient of [latex]\theta [/latex] is not an integer. For example, if it is rational, then the curve is closed; that is, it eventually ends where it started (Figure 10 (a)). However, if the coefficient is irrational, then the curve never closes (Figure 10 (b)). Although it may appear that the curve is closed, a closer examination reveals that the petals just above the positive x axis are slightly thicker. This is because the petal does not quite match up with the starting point.
Since the curve defined by the graph of [latex]r=3\sin\left(\pi \theta \right)[/latex] never closes, the curve depicted in Figure 10 (b) is only a partial depiction. In fact, this is an example of a space-filling curve. A space-filling curve is one that in fact occupies a two-dimensional subset of the real plane. In this case the curve occupies the circle of radius 3 centered at the origin.
Suppose a curve is described in the polar coordinate system via the function [latex]r=f\left(\theta \right)[/latex]. Since we have conversion formulas from polar to rectangular coordinates given by
This step gives a parameterization of the curve in rectangular coordinates using [latex]\theta [/latex] as the parameter. For example, the spiral formula [latex]r=a+b\theta [/latex] from Figure 7 becomes
Letting [latex]\theta [/latex] range from [latex]-\infty [/latex] to [latex]\infty [/latex] generates the entire spiral.
Try It
Symmetry in Polar Coordinates
When studying symmetry of functions in rectangular coordinates (i.e., in the form [latex]y=f\left(x\right)[/latex]), we talk about symmetry with respect to the y-axis and symmetry with respect to the origin. In particular, if [latex]f\left(-x\right)=f\left(x\right)[/latex] for all [latex]x[/latex] in the domain of [latex]f[/latex], then [latex]f[/latex] is an even function and its graph is symmetric with respect to the [latex]y[/latex]-axis. If [latex]f\left(-x\right)=-f\left(x\right)[/latex] for all [latex]x[/latex] in the domain of [latex]f[/latex], then [latex]f[/latex] is an odd function and its graph is symmetric with respect to the origin. By determining which types of symmetry a graph exhibits, we can learn more about the shape and appearance of the graph. Symmetry can also reveal other properties of the function that generates the graph. Symmetry in polar curves works in a similar fashion.
theorem: Symmetry in Polar Curves and Equations
Consider a curve generated by the function [latex]r=f\left(\theta \right)[/latex] in polar coordinates.
The curve is symmetric about the polar axis if for every point [latex]\left(r,\theta \right)[/latex] on the graph, the point [latex]\left(r,-\theta \right)[/latex] is also on the graph. Similarly, the equation [latex]r=f\left(\theta \right)[/latex] is unchanged by replacing [latex]\theta [/latex] with [latex]-\theta [/latex].
The curve is symmetric about the pole if for every point [latex]\left(r,\theta \right)[/latex] on the graph, the point [latex]\left(r,\pi +\theta \right)[/latex] is also on the graph. Similarly, the equation [latex]r=f\left(\theta \right)[/latex] is unchanged when replacing [latex]r[/latex] with [latex]-r[/latex], or [latex]\theta [/latex] with [latex]\pi +\theta [/latex].
The curve is symmetric about the vertical line [latex]\theta =\frac{\pi }{2}[/latex] if for every point [latex]\left(r,\theta \right)[/latex] on the graph, the point [latex]\left(r,\pi -\theta \right)[/latex] is also on the graph. Similarly, the equation [latex]r=f\left(\theta \right)[/latex] is unchanged when [latex]\theta [/latex] is replaced by [latex]\pi -\theta [/latex].
The following table shows examples of each type of symmetry.
Figure 7.
Example: using Symmetry to Graph a Polar Equation
Find the symmetry of the rose defined by the equation [latex]r=3\sin\left(2\theta \right)[/latex] and create a graph.
Show Solution
Suppose the point [latex]\left(r,\theta \right)[/latex] is on the graph of [latex]r=3\sin\left(2\theta \right)[/latex].
To test for symmetry about the polar axis, first try replacing [latex]\theta [/latex] with [latex]-\theta [/latex]. This gives [latex]r=3\sin\left(2\left(-\theta \right)\right)=-3\sin\left(2\theta \right)[/latex]. Since this changes the original equation, this test is not satisfied. However, returning to the original equation and replacing [latex]r[/latex] with [latex]-r[/latex] and [latex]\theta [/latex] with [latex]\pi -\theta [/latex] yields
Multiplying both sides of this equation by [latex]-1[/latex] gives [latex]r=3\sin2\theta [/latex], which is the original equation. This demonstrates that the graph is symmetric with respect to the polar axis.
To test for symmetry with respect to the pole, first replace [latex]r[/latex] with [latex]-r[/latex], which yields [latex]-r=3\sin\left(2\theta \right)[/latex]. Multiplying both sides by −1 gives [latex]r=-3\sin\left(2\theta \right)[/latex], which does not agree with the original equation. Therefore the equation does not pass the test for this symmetry. However, returning to the original equation and replacing [latex]\theta [/latex] with [latex]\theta +\pi [/latex] gives
Since this agrees with the original equation, the graph is symmetric about the pole.
To test for symmetry with respect to the vertical line [latex]\theta =\frac{\pi }{2}[/latex], first replace both [latex]r[/latex] with [latex]-r[/latex] and [latex]\theta [/latex] with [latex]-\theta [/latex].
Multiplying both sides of this equation by [latex]-1[/latex] gives [latex]r=3\sin2\theta [/latex], which is the original equation. Therefore the graph is symmetric about the vertical line [latex]\theta =\frac{\pi }{2}[/latex].
This graph has symmetry with respect to the polar axis, the origin, and the vertical line going through the pole. To graph the function, tabulate values of [latex]\theta [/latex] between 0 and [latex]\frac{\pi}{2}[/latex] and then reflect the resulting graph.
[latex]\theta [/latex]
[latex]r[/latex]
[latex]0[/latex]
[latex]0[/latex]
[latex]\frac{\pi }{6}[/latex]
[latex]\frac{3\sqrt{3}}{2}\approx 2.6[/latex]
[latex]\frac{\pi }{4}[/latex]
[latex]3[/latex]
[latex]\frac{\pi }{3}[/latex]
[latex]\frac{3\sqrt{3}}{2}\approx 2.6[/latex]
[latex]\frac{\pi }{2}[/latex]
[latex]0[/latex]
This gives one petal of the rose, as shown in the following graph.
Reflecting this image into the other three quadrants gives the entire graph as shown.
Watch the following video to see the worked solution to Example: using Symmetry to Graph a Polar Equation.
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.