Petals of Polar Equations

An Exploration on Polar Equations


Sarah Major

The following write-up is based on a prompt given in Exploration 11, which states:

Investigate -- varying a, b, c, and k.

Readers will have to forgive me if I seem to display an elementary understanding of polar functions, because I do. Unfortunately, during the course of my undergraduate mathematics course, the professor who taught me courses involving such equations did not conduct proper lectures of these types of equations, and he did not provide us with a supportive text. Again, forgive me.

The following is a graph of all of these equations on the same plane (where a = k = 1, and theta ranges from 1 to 2 pi):


It is hard to find a trend or pattern with all of these on the same graph, so I will take each two at a time, starting with the first two. Here are their graphs (once again, a = k = 1, theta is 0 to 2 pi):


At first glance, these two graphs seem to be translations of the unit circle. However, in the context of polar equations, this would not make sense. Say we begin with functions in the x-y plane instead of the polar plane, and these equations are the x-y counterparts of our original functions:

The graphs would look like this (with a = k =1 and x going from 0 to 2 pi):


These graphs look very similar. If we were to shift one just ½ pi units horizontally, they would look virtually the same. Here is a graph if we change the second equation to:

This is just to show that the “shapes” of the graphs are essentially the same. From extremum to extremum, the rate of change is the same for a certain distance along the x or y-axis.

Let’s take each of the original equations and its counterpart to see how we can derive the polar equation from the counterpart.

To map each of these equations, we must first translate the function as a reflection across the line y = x. For the first equation, we get:

And, for lack of better words, we create a circle with the y-axis. That is, instead of the y-axis being a line, it becomes like a circle to create the polar plane. The circle starts at the center of the plane (center of the “circle”) being a function value of 0 and increases as we get farther and farther away from the circle.

This is why the first equation produces a circle completely on and above the x-axis. We are bringing the graph after the reflection up over the x-axis as we create the polar plane, and we translate the other side of the graph to now appear near the y-axis.

We can also derive the polar equation for the second equation and its counterpart. We must first do the reflection:

And then create the polar plane:

This is why this particular circle is completely on and to the right of the y-axis. The point located at (2,0) stays put, and the rest of the reflected function is moved around to meet at this point.

Now that we know why the graphs look the way they do and are located where they are, we can see what happens when we change the values of different variables.

What if we increase a to, say, 2?

We essentially increase the radius of the circle by a factor of 2. This makes sense because the function value IS the radius, and we are essentially multiplying the whole function by a factor of 2.

So, if we change this value to 3, the radius should change by a factor of 3.

And this would be correct.

What would happen if we decreased a to ½?

The value of the radius is cut in half, just what we would expect with the above reasoning.

Changing a back to 1, what happens if we increase k to 2?

Pretty. Now I know what functions to use if I want to create the picture of a flower.

The only way I know to adequately explain what is happening is by graphing their x-y counterparts.

We have decreased the period of this function, creating more extrema per a certain area of the domain. This is essentially what is happening with our polar functions. Instead of 2 (which went around to simply create a circle), we have 4, so the extrema or more visibly seen.

What if we increase k to 3?

We have 3 extrema for each equation. What about 4?

We have 8 total extrema.

So, when k is even, there are 2k crests. When k is odd, there should just be k crests. Let’s see if this is correct by setting k equal to 5.

[insert graph]

And it is.

Now let’s set k back equal to 1 and move on to the next two of our original equations.


They are similar to our first two equations. They are in the same relative position but with r values increased by 1. This makes sense because we are essentially adding 1 to the whole function.

However, how can we account for the little portion of the graph appearing inside the circles? This goes back to mapping x-y functions to the polar functions. For the sake of time and interconnectivity, we will only analyze the sine function.

Here is the graph of the following equation:

To reflect it across the line y = x, we must have the following equation:

And when we convert this to our polar plane, we have a portion of “overlap” where the two ends of the function meet. They sort of wrap around to create another circular figure within our “almost” circle. The same occurs for the cosine function but at a different position on the plane.

Since our graph is not exactly a “circle”, we can’t exactly say that certain values of the variables change the “radius” of the graph, though this is technically true since the function value IS the radius. Since the graph of the equation is still circular, we will, for the sake of consistency, talk in terms of the “radius” of an actual circle, acting as though our graphs are actually circles.

What happens if we increase a to 2? We would expect the same trend to occur as the previous two equations, but we have a value of b to work with. This increases all of the function values by b units. To find out what occurs as a increases, we should subtract off b, change the value of a, and then add b back. So, essentially, since b is 1, each function value should be one more than the values that occurred with the first two functions:

Notice that both the “outermost” portion of the graph and the part inside increase by a factor of 2 (before adding 1). This is because ALL function values are translated with this translation.

What if a is ½? If we take off the b before we apply this, the maximum of the radius should be 1, so adding 1 back should make it 2.

Let’s change a back to 1 and increase k to 2.

This has a similar effect to the previous two equations except that because we have "two curves” of different size, the “petals” of our flower are different sizes.

If we change k back to 1 and make b have a value of 2, the function value should just increase by 1.

And we would be right… Except we have an unexpected characteristic that occurs. The smaller curve inside the larger one has disappeared. Why did this occur? Let’s go back to the x-y counterpart of this equation and the reflection across the line y = x:

The reflection now looks very similar to the reflection we had for the original equations. This is because the sine and cosine curves are translated to being above the x-axis. When this is reflected across the line y = x, it creates a curve that when mapped onto the polar plane becomes more of a circle. Let’s see what happens if we increase b by one more to 3:

And then 5:

As the value of 2 times the amplitude becomes closer to the maximum function value of the x-y function, the polar function becomes more and more like a circle. This is because when the function is mapped to its counterpart polar function, the translation of the reflection to the polar function is able to circle around without the “overlap” we saw before.

Now for our interesting last function. Let’s first graph the actual polar function.

Setting a, b, c, and k all equal to 1 produces the line y = -x. Why is this so? Let’s start with its x-y counterpart:

So once we reflect this and convert it to the polar plane, we actually have all of the function values necessary to create the line y = -x.

Due to the form of the equation, it is difficult to predict what changing the values of certain variables will do to the function. Also, we have a new variable, c, to deal with. Let’s begin with c, seeing as increasing this would increase the whole function value. We would thus expect that doing so would be similar to changing the y-intercept on a linear function. If we increase the value of c to 2, we should expect the line y = -x +2:

And this would be correct.

However, the effects of the other variables are hard to predict. If the graph with all of the variables equal to 1 produces the line y = -x, we would expect to still have a line regardless of what the values will be. Changing c translates all of the points by that many units, so the “y-intercept” of the polar function should always be the same. Therefore, changing the values of a or b should just change the slope of the function. Let’s increase a to 2 and see what happens:

The slope has become -2. What if we increase a to 4?

The slope becomes -4. What if we decrease a to -2?

The slope becomes 2.

So, we can generalize these effects to be that changing the value of a changes the slope to –a.

What if we change b instead? Here is when b is 2:

The slope becomes -1/2. What if b is 4?

The slope is -1/4. What if b is -2?

The slope is ½. So we can generalize these effects to be that changing the value of b changes the slope to -1/b.

But we also notice that instead of the “y-intercept” being the c value, the “x-intercept” is the c value. This is because in the polar plane, we do not have such x and y-axes but instead have the function value in a circle on the plane. Therefore, the intercept becomes dependent on which trigonometric function we are changing. When we change a, we are affecting the cosine function, so the “x-intercept” will change. When we change b, we are affecting the sine function, so the “y-intercept” will change.

Now, what if we change k? Let’s change it to 2:

Wow. That’s interesting. But in an odd way, it makes sense. The effect is similar to the “flower” effect we saw with the other functions, but instead, we also have our lines and “inverse petals”, or petals opening up the other way. This is because we are dividing by the functions we used in the previous equations.

What if we increase k to 3?

A similar affect but with three “inverse petals” instead of 4. What about 4?

Eight inverse petals. So we see a similar trend to the equations we had before.

I’m just curious. I would like to see what happens when k is 60.

Really cool.

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