Polar Petals and the N-leaf Rose

by: Al Byrnes

Polar equations, like their Cartesian counterparts in

R² involve two values:rand θ. The value ofrindicates the distance of the point in the polar plane away from the origin, while the value of the variable θ indicates the measurement of the angle from the horizontal axis. Ifris negative, the point will be located a distance of r away from the origin but in the opposite direction as indicated by θ (eg. θ + π). The expressionr=n,1 ≤ n ≤ 10, 0 ≤ θ ≤ 2π, indicates that for every value of θ, the distance of the points described by the equationr = nwill always be distancenaway from the origin. This means when we are dealing with polar equations,r = nfor 0 ≤ θ ≤ 2π, the graph of this equation will be the the circle of radiusncentered at the origin. The animation below shows the graph of the polar equationr =2, 0 ≤ θ ≤ 2π:

For the polar equation

r=n, if we allow the values ofnto vary such that 1 ≤n≤ 10, and 0 ≤ θ ≤ 2π, we produce the following result:

As you might have expected with

r = n,1 ≤n≤ 10 and 0 ≤ θ ≤ 2π, we get circles with radii of lengths ranging from length 1 to length 10. What about a polar equation, in which the value ofr,or the distance from the origin, is defined in terms of the value of θ? First lets consider the most basic of such an equation:r =θ, 0 ≤ θ ≤ 2π:As the value of θ increases, we can see that the length of the radius increases in kind, in particular, the when θ = π we see that the value of

ris very close to π, as measured by Graphing Calculator:Likewise the values of

rat θ = 3π/2 and θ = 2π show radii of lengths very close to 3π/2 and 2π, respectively (the values are not exact because of rounding performed by the software):θ = 3π/2, r = 3π/2

θ = 2π, r = 2π

That was a brief examination of the spiral created by the polar equation

r= θ, 0 ≤ θ ≤ 2π. Let's add in a some trigonometric functions to make these polar equations a little bit more exciting. Consider the following polar equation:

r=bcos (kθ)To get a feel for this equation, let's start by setting the value of the coefficient

kto 1 and experimenting with various values ofb.The animation below shows the equationr = bcos (kθ), -6 ≤b ≤6:

The value of the coefficient

bscales the diameter of the circle resulting from the equationr =cos θ. As a quick mathematics check, does it make sense that the graph of this polar equation would be a circle? Let's graph some specific equations for various values of θ. Shown here is the graph of the equationr =6 cos θ, 0 ≤ θ ≤ π/4:For θ = 0, we have the polar equation can be evaluated for

ras such:

r= 6 cos (0)

r =6 (1)

r= 6, as the graph above indicates.What about the value of

rwhen θ = π/4? See evaluation of the equation below:

r =6 cos (π/4)

r= 6 (√2/2)

r =3√2The visual below shows that indeed r = 3√2. Invoking the pythagorean theorem, we can see that we have the two bases of the triangle of length 3 and the diagonal, or r, is length √18 = 3√2.

Some noteworthy differences between the polar equations

r =n andr=bcos θ:1. The radius of the circle formed by the polar equation

r=nisn.The diameter of the circle formed by the polar equationr = bcos θ isb.2.A full rotation around the circle sketched by the equation

r = noccurs after a period of 2π. A full rotation around the circle sketched by the equationr=bcos θ occurs on the domain 0 ≤θ ≤ π. This is illustated below; on the domain 0 ≤ θ ≤ π, the graph ofr =θ has completed a full circle, whereas the graph of the equation ofr =6 has not. Why might this be?So we have learned a bit about the polar equation

r = bcos kθ, when k = 1. Let's examine how the graph of the polar equation will change as we manipulate the value of k. What is your initial guess? How will values of k effect the value of θ? As a suggestion, consider the difference between the graphs of the following trigonometric functions:y = cos (x); y = cos (2x)

The 2 in this equation seems to "compress" the cosine curve, in otherwords doubling its frequency, or number of times the pattern repeats itself over a certain given time period. Note that the function y = cos (x) fully repeats its pattern once over the domain 0≤ x ≤ 2π, whereas the function y = cos (2x) [indicated by the red curve] has repeated its pattern twice along the same domain:

So we might logically expect a similar doubling of frequency if we compared the two polar equations

r =6 cos θ andr =6 cos 2θ. Right? See the graph of the polar equationr =6 cos 2θ below (0 ≤ θ ≤ 2π):Wow! Unexpected results. So it appears that we doubled the frequency by multiplying theta by two and we obtained four petals. See that the petal length seems to correspond to the value of the coefficient

b, and are accordingly length 6 in this picture. It would seem logical to conclude that the graph of the polar equationr =6 cos 3θ would have 6 petals. See the graph ofr =6 cos 3θ, 0 ≤ θ ≤ 2π below:So that seems to go against my intuition. Why are there only three petals and not 6? Lets graph the same polar equation again, except this time, over the domain 0 ≤ θ ≤ π:

I included the entire graphing calculator output to prove that I am not just putting up the same graph as before. So, it appears that their are 6 petals being drawn here, however those petals drawn on the domain π < θ ≤ 2π are just drawn on top of the existing petals. Let's do a check with some algebra:

Compare:

r =6 cos (3(π/4)) andr =6 cos (3(5π/4))

r =6 cos (3π/4) = 6 (-√2/2) = -3√2when θ = π/4, r = -3√2

r =6 cos (3(5π/4))

r= 6 cos (15π/4)

r =6 (√2/2) = 3√2when θ = 5π/4, r = 3√2

When these two points are graphed in the polar plane, see that you will arrive at the same point. The point (-3√2, π/4) can be found at the point where the 3 leaf rose and the blue line segment are coincident. Notice that the red segment has a measure of π/4, however, since the value of the radius is -3√2, the polar point is found in the 3rd quadrant, rather than the first (the negative value of

rindicates that the direction of the radius is opposite the heading indicated by theta). Likewise, the point (3√2, 5π/4) is also to be found at the point where the 3 leaf rose and the blue line segment are coincident. However, this time, the position of the point is more self explanitory, as the value of θ is positive, indicating that the polar point will be a distance r away from the origin in the same direction as indicated by the indicated measure of θ. See the visual below:What about

r =6 cos 4θ, 0 ≤ θ ≤ 2π? We might expect a rose with 8 petals to appear in our graph:Sure enough, a rose with 8 petals. So the thinking is when the coefficient k in the polar equation

r = bcos kθ, 0 ≤ θ ≤ 2π is even we can see all of the petals and the graph will display 2k petals, each petal of lengthb.*To check this claim about the length of each petal, try graphing a circle of radiusbcentered at the origin; the petals should all be tangent to this circle.* Conversely, if k is an odd value we see only k petals; the other petals are drawn on the domain 0 ≤ θ ≤ 2π, however, as we saw above, points along the domain 0 ≤ θ < π were re-oriented to match the points graphed by the polar equationr = bcos kθ along the domain π ≤ θ ≤ 2π. Wow. For a cool animation showing the graph of the polar equationr =6 cos kθ, -6 ≤ k ≤ 6 and 0 ≤ θ ≤ 2π, toggle the play button on the embedded flash animation below.

Did you see what the rose looked like when k = 0? If not, I have provided a picture below for

r = bcos (0). If you havent already checked, what do you think this graph might look like? Hint: what is the value of cos(0)?Visual of

r =6 cos (0θ), 0 ≤ θ ≤ 2π:Its a circle again! This should come as no surprise, if you worked out the value of cos(0). This simply makes the polar equation

r =6, 0 ≤ θ ≤ 2π, which is the polar equation of a circle with a radius of 6. Neat.For more experimentation with the N-leaf rose, try to experiment with the polar equation

r = a + bcos (kθ), 0 ≤ θ ≤ 2π. How does the coefficientaaffect the petals of the N-leaf rose? What happens when you replace the cosine in the polar equation with sine?

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