A little discussion about the graph of some polar equations

(I started working on this write -up (my 4th one) in Sep., going through many technology probelms and finally finished this one on Oct. 12, 2009.)

(There are some animations in this webpage, so it may take a few seconds to display completely.)

Chen Tian

We explore the function as follows.

Let observe the r=1+cos(1n) when n changes in interval [-10,10].

Let observe the r=1+2cos(1n) when n changes in interval [-10,10].

Let observe the r=2+1cos(1n) when n changes in interval [-10,10].

Did you notice that a decides the smallest value of the radius of the circle and b+a affects the greatest value of the radius of the circle? It will also, to the greatest extent, "work" for non-circle cases. For example, one of the below two cases: r=1+1cos(0.1) and r=cos(0.1). More specificly a+b will be the starting point of the curve. For instances, shown in the following picture, the curve of r=1+cos(k) always stars from point (2, 0) in the polar coordinate plane; the curve of r=1+2cos(k) always stars from point (3, 0) in the polar coordinate plane; the curve of r=3+2cos(k) always stars from point (5, 0) in the polar coordinate plane.

Let observe the r=1+1cos(5n) when n changes in interval [-10,10].

Comparing to the graph of r=1+1cos(1n), have you noticed that k value affects the change speed of the raduis of the circle, which means that cos(kn) changes faster when k gets greater.

Because y=cos(x) is a "period" function, cos(1n) changes in the interval [0,1] no matter n is in what interval, even when n is negative, since cos(-A)=cos(A). This is illustrated in the following animation when n changes in interval [-10,10].

We see that a, b, k mainly affect the **size** of the cirle r=a+bcos(kn) when only n changes **discretely**.

Let's really talk about **polar equations**, i.e., let change in [0,2pi], and a, b be fixed. Changing the value of k will affect the shape of the graph of the function.

For example if the graph of r=1+1cos(1) look like the

Let observe the r=1+1cos(n) when n changes in interval [-10,10].

We can use the control bar to stop or resume the animation at any time for the convenience of our observation. If you do this it will really help you see the trend of changing of the curves.

Let's look at the changes in several separate steps.

When k is not 1, k does not change in the orginal interval [0, 2pi], and even does not changes as "frequently" as is. Let's get some feeling from the following picture.

Thus when k goes from [k*0, k2pi] different cos(k)'s change in differently paces, some of which were shown above, where we may want to take r=cos(k) as a changing radius of a changing circle, while r=cos(k) actually is the distance from the point on the curve to the origin.

Now let's take a look at some of the integers cases for k.

Taking y=cos(0.5x) as example, it takes cos(0.5x) as “twice time” to become the same value as cos(x), so the “(smallest positive) period” of cos(0.5x) would be double of that of cos(x). In other words, the value y=cos(0.5x) changes more slowly than y=cos(x) to reach the same y value. Taking y=cos(2x) as example, it takes cos(2x) as “half time” to become the same value as cos(x), so the “(smallest positive) period” of cos(2x) would be half of that of cos(x). In other words, the value y=cos(2x) changes faster than y=cos(x) to reach the same y value. The comparision is illustrated in the following picture.

Generally speaking, the period of y=cos(kx) is 1/k of the period of y=cos(x). Informally speaking, when 0

Hence, when k goes from 0 to 1, the length of the curve of the polar equation r=a+bcos(k) gets shorter and “scrolls” into a smaller scope but still within the scope decided by the values of a and b as we shown in the cases when changes discretely. In particular, when k=1 cos(k)=cos() goes with a “normal” period 2pi, the curve would be symmetric only about the x-axis. When k goes from 1 to larger positive real numbers, the length of the curve of the polar equation r=a+bcos(k) gets longer and “scrolls” into a larger scope than k=1 but still within the scope decided by the values of a and b as we shown in the cases when changes discretely.

Dr. Jim Wilson says “when a and b are equal, and k is an integer, this is one textbook version of the ‘**k-leaf rose**’". So let’s make the conjecture that, in particular, when a and b are equal, and k are integers cos(k)=cos() goes with a “multiple” of “normal” period 2pi, the curve would be symmetric about both the x-axis and y-axis when k is even, and be symmetric only about the x-axis when k is odd (no exception for k=1). Morever, the number of the leaves of the k-leaf rose would be as k times as that when k=1 since the period of cos(kx) is k times shorter than that of cos(x) as we analyzed above. And when k is even the leaves of the k-leaf rose would be symmetric about both the x-axis and y-axis; when k is odd the leaves of the k-leaf rose would be symmetric only about the x-axis. The above cases check his conjecture.

What if cos(…) is replaced by sin(…)?

Let’s take a look at and compare the first a few cases shown as follows.

From the above cases, we notice for sin(k) the k-leaf rose might be a rotation of the k-leaf rose for cos(k). But is that true? Can we find the specific rotation angle for each integer k? It is obvious that r=1+1sin(1) is a 90-degree rotation of r=1+1cos(1). Can we prove that? Aha, yes! We can!

Therefore, we proved that r=1+1sin(k) is a pi/(2k) rotation of r=1+1cos(k), where k are natural numbers.

Return to my homepage for EMAT 6680