Understanding Polar Coordinates: Learn It 3

Giselle Miller

Understanding Polar Coordinates: Learn It 3

Polar Curves

Just as we graph functions [latex]y=f(x)[/latex] in the rectangular coordinate system, we can graph functions [latex]r=f(\theta)[/latex] in the polar coordinate system to create curves.

The process for graphing polar curves follows the same fundamental approach as rectangular graphing. You start with values for the independent variable [latex]\theta[/latex], calculate corresponding values of the dependent variable [latex]r[/latex], then plot and connect the resulting points.

Since many polar functions are periodic, you often need to evaluate only a limited range of [latex]\theta[/latex] values to capture the complete curve pattern.

Problem-Solving Strategy: Plotting a Curve in Polar Coordinates

Create a table with columns for [latex]\theta[/latex] and [latex]r[/latex]
Choose appropriate [latex]\theta[/latex] values (consider the function’s period)
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.
Connect the points and identify any patterns in the curve

Graph the curve defined by the function [latex]r=4\sin\theta[/latex]. Identify the curve and rewrite the equation in rectangular coordinates.

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]

On the polar coordinate plane, a circle is drawn with radius 2. It touches the origin, (2 times the square root of 2, π/4), (4, π/2), and (2 times the square root of 2, 3π/4). — Figure 5. The graph of the function [latex]r=4\sin\theta [/latex] is a circle.

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:

[latex]\begin{array}{ccc}\hfill {x}^{2}+{y}^{2}-4y& =\hfill & 0\hfill \\ \hfill {x}^{2}+\left({y}^{2}-4y\right)& =\hfill & 0\hfill \\ \hfill {x}^{2}+\left({y}^{2}-4y+4\right)& =\hfill & 0+4\hfill \\ \hfill {x}^{2}+{\left(y - 2\right)}^{2}& =\hfill & 4.\hfill \end{array}[/latex]

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 the example above.

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.

You can view the transcript for this segmented clip of “7.3 Polar Coordinates” here (opens in new window).

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.

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]

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

[latex]{r}^{2}={x}^{2}+{y}^{2},x=r\cos\theta ,y=r\sin\theta[/latex].

This gives

[latex]\begin{array}{ccc}\hfill {r}^{2}& =\hfill & 6\left(r\cos\theta \right)-8\left(r\sin\theta \right)\hfill \\ \hfill {x}^{2}+{y}^{2}& =\hfill & 6x - 8y.\hfill \end{array}[/latex]

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.

[latex]\begin{array}{ccc}\hfill {x}^{2}+{y}^{2}& =\hfill & 6x - 8y\hfill \\ \hfill {x}^{2}-6x+{y}^{2}+8y& =\hfill & 0\hfill \\ \hfill \left({x}^{2}-6x\right)+\left({y}^{2}+8y\right)& =\hfill & 0\hfill \\ \hfill \left({x}^{2}-6x+9\right)+\left({y}^{2}+8y+16\right)& =\hfill & 9+16\hfill \\ \hfill {\left(x - 3\right)}^{2}+{\left(y+4\right)}^{2}& =\hfill & 25.\hfill \end{array}[/latex]

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 the example above.

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.

You can view the transcript for this segmented clip of “7.3 Polar Coordinates” here (opens in new window).

The tables below summarize several important families of polar curves, where [latex]a[/latex] and [latex]b[/latex] are arbitrary constants.

This table has three columns and 3 rows. The first row is a header row and is given from left to right as name, equation, and example. The second row is Line passing through the pole with slope tan K; θ = K; and a picture of a straight line on the polar coordinate plane with θ = π/3. The third row is Circle; r = a cosθ + b sinθ; and a picture of a circle on the polar coordinate plane with equation r = 2 cos(t) – 3 sin(t): the circle touches the origin but has center in the third quadrant. — Figure 7.

This table has three columns and 3 rows. The first row is Spiral; r = a + bθ; and a picture of a spiral starting at the origin with equation r = θ/3. The second row is Cardioid; r = a(1 + cosθ), r = a(1 – cosθ), r = a(1 + sinθ), r = a(1 – sinθ); and a picture of a cardioid with equation r = 3(1 + cosθ): the cardioid looks like a heart turned on its side with a rounded bottom instead of a pointed one. The third row is Limaçon; r = a cosθ + b, r = a sinθ + b; and a picture of a limaçon with equation r = 2 + 4 sinθ: the figure looks like a deformed circle with a loop inside of it. The seventh row is Rose; r = a cos(bθ), r = a sin(bθ); and a picture of a rose with equation r = 3 sin(2θ): the rose looks like a flower with four petals, one petal in each quadrant, each with length 3 and reaching to the origin between each petal. — Figure 8.

A cardioid is a special case of a limaçon (pronounced “lee-mah-son”) that occurs when [latex]a=b[/latex] or [latex]a=-b[/latex]. The heart-shaped cardioid gets its name from the Greek word for heart.

The rose curves display fascinating petal patterns that depend on the coefficient of [latex]\theta[/latex]. For example, [latex]r=3\sin2\theta[/latex] produces four petals, while [latex]r=3\sin3\theta[/latex] creates three petals.

Rose Petal Rule:

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, the number of petals equals the coefficient

When the coefficient of [latex]\theta[/latex] is not an integer, interesting behaviors emerge. If the coefficient is rational, the curve eventually closes by returning to its starting point. However, if the coefficient is irrational, the curve never closes completely.

The graph of [latex]r=3\sin(\pi\theta)[/latex] demonstrates this phenomenon. While it appears closed at first glance, closer examination reveals that the petals near the positive [latex]x[/latex]-axis are slightly thicker because the curve never quite returns to its exact starting point. This creates a space-filling curve that would eventually occupy the entire circle of radius 3 if plotted completely.

This figure has two figures. The first is a rose with so many overlapping petals that there are a few patterns that develop, starting with a sharp 10 pointed star in the center and moving out to an increasingly rounded set of petals. The second figure is a rose with even more overlapping petals, so many so that it is impossible to tell what is happening in the center, but on the outer edges are a number of sharply rounded petals.

Any polar curve [latex]r=f(\theta)[/latex] can be converted to parametric equations in rectangular coordinates using the conversion formulas:

[latex]\begin{array}{c}x=r\cos\theta \hfill \\ y=r\sin\theta ,\hfill \end{array}[/latex]

It is possible to rewrite these formulas using the function:

[latex]\begin{array}{c}x=f\left(\theta \right)\cos\theta \hfill \\ y=f\left(\theta \right)\sin\theta .\hfill \end{array}[/latex]

This gives us a parametrization of the curve using [latex]\theta[/latex] as the parameter. For instance, the spiral [latex]r=a+b\theta[/latex] becomes:

[latex]\begin{array}{c}x=\left(a+b\theta \right)\cos\theta \hfill \\ y=\left(a+b\theta \right)\sin\theta .\hfill \end{array}[/latex]

As [latex]\theta[/latex] ranges from [latex]-\infty[/latex] to [latex]\infty[/latex], these equations generate the complete spiral curve.