 This is the write-up for assignment 11.

Investigating

Polar Equations

By

Doug Westmoreland

In this write-up we will be investigating polar equations. Click here for a quick review of the polar coordinate system.

Let's consider the equation and take a look at several graphs while we let the b value remain constant and let the k value vary. We only look at k values that are positive integers.          Equations of the form , where b is a constant and k is a positive integer, produce graphs called roses.

The radius of the rose is determined by the coefficient b in .

We can see that the k value effects the number of petals in the rose. If the k value is odd, there will be exactly k petals. If the k value is even, there will be exactly 2k petals. (In the examples, the k value was always an integer. If k is not an integer, then a complete petal will not be formed.)

Click here for a dynamic presentation in which the b value remains constant (b=1) and the k value varies from -10 to 10.

Now, let's take a look at the equation: Let k=1 and vary the a and b values ( but we will keep a=b ) in the equation.    When a=b in the equation and k=1, we get graphs that are called cardiods.

Now, if we let a and b be different numbers and keep k=1, we get the sample graphs below:        All of the graphs above are called limacons.

Some of the graphs above had inner loops. These are called limacons with an inner loop.

The limacon has an inner loop. A close look at the graph and the equation reveals that these loops correspond to negative values of r. For , we have r=0 when = /3 or 5 /3, and

r<0 for /3 < < 5 /3. For these values the curve is drawn on the opposite side of the origin or pole.

These graphs also have symmetry with respect to the x-axis. This symmetry occurs because the cosine function is an even function ( i.e. for any , cos =cos (- ) ).

If we were to replace the above equations with the sine function, we would obtain basically the same type graph, except the graph would be symmetric with respect to the y-axis, because the sine function is odd.

Below are some graphs of the above equation letting a=1 and b=1 and varying the k value.        Click here for a dynamic presentation of the graph of where the value of k varies from -10 to 10.

So, when a and b are both the same value in the equation and we vary the k value (positive integer values only), we still obtain a curve that that resembles a rose. The number of petals on the rose is equal to k. This is different when k was an even integer in the equation .

Now, let's take a look at how the a and b values effect the equation of . In the previous examples above, we let a=1 and b=1. (Notice that in all of the examples, the petal has a maximum length of 2). Now, if we let the a and the b value vary (keeping both the same value) and let the k value a remain a constant we can see from the examples below that the length of the petal will depend on the value of a and b.        We can see from the examples above that the lengths of the petals got longer as we increased the values of a and b (keeping a and b the same value).

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