Polar Equations

by

Diana May

Investigation #11 ~ We will investigate the impact of varying constants in various polar equations.

First, let’s look at what happens when we change the constants a and k for

r = 2a*sin(kx):

Let’s look at the basic graph r = 2*sin(x) for x ranging from 0 to π/2 and π, respectively:

It appears as though the curve starts at (0,0) and completes a cycle when x ranges from 0 to π.  We get a unit circle centered at (0,1).

Now, let’s change the values of a:

This seems to give us a circle with radius abs(a), centered at (0, a).

Now, let’s keep a equal to 1 and let k vary.  To see how the graph changes as k varies from -5 to 5, click HERE.

The next equation we’re going to consider is r = 2a*cos(kx).

For a movie on varying k from –5 to 5, click HERE.

At first, it appears that this is simply a 90-degree clockwise rotation of the graph, but upon further inspection, we see this is not the case.  If we graph the following equations from x = 0 to π/2, you can observe that the two graphs do not line up after a rotation.

Now, let’s see what happens if we add a constant to the two previously discussed equations.

How would we expect r = 2cos(x) + 1 to be different from r = 2cos(x)?  Since we’re adding 1 to the radius, we would expect the radius to increase by 1 unit.  So, if we drew a line from the origin to the point on our r = 2cos(x) graph, we should be able to continue along that line for one unit and find the corresponding point on r = 2cos(x)+1.

This is exactly what we expected.  Look at where x = 0.  The first graph is at (2,0) and the second is at (3,0).  Also, when the cosine values become negative (x = π /2),  the radius of the second graph seems to be smaller, but it is actually less negative.  The second graph begins to form an inner loop once r = 2cosx+ 1 = 0.  For a movie on how r = 2cosx + b changes as b changes, click HERE.  If we were to let x continue to 2π, we’d get the following graph:

A similar pattern is seen for r = 2sinx + b.  For a movie on r = 2sinx + b as b changes, click HERE.

Now, let’s investigate the equation r = c/[acos(kx)+bsin(kx)]

We’ll start by looking at the graph for the basic equation and let x range from 0 to π/2:

Now, let’s see what happens if we let x range from 0 to π:

Does this line extend in both directions to infinity?  Let’s look at what is happening with the equation.  Notice that the radius will be undefined when the denominator is 0, which is when acos(x) = bsin(x).  In our first case, that will be when cos(x) = sin(x), which is when x = 3π/4 (+ multiplies of p).  What happens to r when x approaches 3π/4?  Let’s look at y = cos(x) and y = sin(x):

The red lines indicate where x = 3π/4 and 7π/4. So, we can see that the sum of cos x and sin x is positive and approaches 0 as x approaches 3π/4 from the left.  From the right, the sum is negative and approaches 0.  Since cos x + sin x is in the denominator, r approaches positive infinity as x approaches 3π/4 from the left and negative infinity as x approaches 3π/4 from the right.  A similar trend occurs as x approaches 7π/4.

The original polar equation was equivalent to the linear equation y = -x + 1.  Let’s see what happens to the line if we change some of the constants.  We’ll start with varying the c constant.

When we change c, we are simply changing the coefficient of r, which changes the y-intercept of our linear equation.  This makes sense because multiplying both sides of the equation by c causes the radius to be c times larger.

What happens when we vary a?

The coefficient on the cosine term seems to change the slope of the line.  It seems that multiplying the cosine term by a causes the slope to increase by a factor of a.

What happens if we vary the coefficient on the sine term?

This also seems to affect the slope, but now it causes the slope to change by a factor of 1/b.  As x increase by 1, y increases by 1/b.

Questions? E-mail me.