## Mary Ellen Graves

## Assignment 11: Polar Equations

For this assignment we will be exploring variations of the polar equation r = a + bcos(kΘ). But first let's review the definition of a polar equation in a polar coordinate system: each point in the polar coordinate system is determined by the distance between a point called the pole and an angle formed from a fixed direction. We then have the polar axis which is the ray from the pole in the direction of the fixed direction (for this exploration our fixed direction will be the positive x-axis). This distance is called the radius and the angle is the polar angle. Please maximize your browser before continuing.

We will begin our exploration of r = a + bcos(kΘ) where r is our radius and the pole is fixed at the origin. Let's look at the curves of this function n with variations of a and b and k = 1.

Polar equations have distinct curves because the points on our curves are found by the angle formed between the polar axis and the radius. Like we said above we are exploring the behavior of the function: radius = a + bcos(kΘ) and these curves behave similarly to the parametric equation curves explored in Assignment 10. Here, we have scalars b and k that determine the size of the curve, which makes sense for if Θ = π and k = 1/2 we will be evaluating the angle cos(kΘ) = cos(π/2) = 0 for k is a scalar to Θ and as long as a ≠ 0 we will have an existing radius r = a. Also b is a scalar to the whole term cos(kΘ). Therefore, when b ≠ 0 the b will scale the term cos(kΘ) accordingly. Our variable a acts as an extender or shrinker of the radius depending on the sign of a. We can see this makes sense for if a = -2 and -3(coskΘ) = -3π we have a r = -2 + (-3π). See graph above for a visual. We find the value of a by finding where cos(kΘ) = 0 i.e. where kΘ = π/2,- π/2, 3π/2, and - 3π/2.

It is important for students to know the unit circle and all associated angles to fully understand how polar equations are moving about the polar coordinate system. For a flash reminder:

To further this exploration, let's step up the excitement a bit! What happens when we change cos(kΘ) to sin(kΘ)? How is the curved changed when a = b? And how does the curve change when we add an integer d inside the function cos(kΘ) to get cos(kΘ + d)? Go ahead and look at the following animated curves. The radius and values for a and b are located above their corresponding graph. The variable n whether it is a scalar or additive is varying from 1 to 10, and when n is a scalar is it representing our previously used scalar k.

As we study these animated curves we notice right away that radius = a + bcos(nΘ) = a + bsin(nΘ). But are they identical curves? In fact, they are not identical curves for the position of 'petals' are not equivalent. If you begin and stop the curves of a + bcos(nΘ) and a + bsin(nΘ) at exactly the same time you will see this to be true. This is happening because of the values of sin(Θ) and cos(Θ) on the unit circle. They are not the same curve, hence when put into a function they will not be identical. In conclusion, it is not wise to base a conclusion solely on an observation. Yes, the curves appear to be identical, but when the polar equations are expanded and comprehended they are not identical.

In addition, as the values of Θ go from 0 to 2π the 'rose' curve expands and shrinks. But what you may not notice too quickly is that the number of 'petals' corresponds indirectly to the scalar n and the length of the petals corresponds to the radius. Hence, these animations are called 'n-leaf roses'. Notice also in the curve r = a + bcos(nΘ + n) with the addition +n we get an additional ring of petals. Hence, the number of smaller petals corresponds with the number of larger petals which we know directly corresponds to n.

In conclusion, our exploration showed us how to find the curve of a polar equation, determine how and why the curve will behave, and finally understand the effects of different scalars and additives to our polar equation in the polar coordinate system.

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