In investigating certain curves it can be more convenient to locate a point by means of its polar coordinates. The polar coordinates give its position relative to a fixed reference point O (the pole) and to a given ray (the polar axis) beginning at O. The diagram to the left shows a standard polar graph. Points are placed on this graph using a coordinate system where (r,x) is the ordered pair that represents r = the distance from the origin and x = the angle measured counterclockwise from the x axis and terminal arm. An example would be (3,0) this would place a point on the third ring out at an angle of 0. Another example would be (1,pi/2), this would place a point directly above the origin on the first ring.

Equations in polar form are relationships between r and x. A few examples would be r = 3, or r = 3 cos(x) or r = 1 + 3 sin(2x).... These types of relationships seem to work out nicely in the polar plane because of the their heavy reliance on angles and radian measure. Below are a few basic examples of equations in polar form and their resulting graph.

r = 2 - all points that are a distance of 2 from the pole.	r = 2 sin(x) - a circle formed with a diameter of 2	r = 2 cos(x) - a circle formed with a diameter of 2	r = 2 cos(3x) - a 3 pedal shape is form that extends out to max.
for all values of x, r = 2 thus the circle.	r = 0 when x = 0, thus the orientation	r = 1 when x = 0, thus the orientation	value of r = 2.

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To investigate the "n- leaf roses" I will be looking at the equation r = a + b cos(kx) where r = the distance from the pole and x = the angle as it varies from 0 to 2 pi. a, b, and k are variables that I will at some times hold constant while at other times, manipulate depending on the circumstance which is to be developed. Before we can looking at how a graph changes with respect to our manipulations we must first understand the standard form of the graph.

r = 0 + 1cos(x) - a circle formed with a diameter of 1	r = 0 + 3cos(x) - a circle formed with a diameter of 2	r = 1 + 1cos(x) - a circle formed with a diameter of 1	r = 2 + 1cos(x) - a circle formed with a diameter of 1
As discussed earlier a right orientated circle appears with this equation. As b variable is manipulated we see the size of the the circle increasing by that factor. This is because the max. value of cos(x) = 1, thus 3*1 = 3 and the cos(pi/2) value = 0, thus 0*1 = 0.		(up left) A cardioid is formed from this equation. Notice the max. value is 2 = 1 + 1 (cos(0)) and the min. value is 0 = 1 - 1 (cos(pi)). (up right) Hopefully you can now extrapolate why the cardioid has taken on this shape.

Some general finding that I will not investigate here concerning r = a + bcos(kx) are:

• if a = b, a general shape is created depending on the other variables but as a and b are increased or decreased a concentric similar shape is created about it. Altering a and b by a constant is the same as multiplying the whole equation by a constant c, r = c(a + bcos(kx)) thus the natural alteration to its shape.
• if b approaches zero we see the shape approach the shape of a circle. It would not be a circle until b = 0 because then the equation would simplify to r = a which is the polar equation form of a circle.
• if our equation would be r = a + bsin(kx) we would see an orientation shift because of the nature of the function but the resulting shapes, circles, cardioids, lemniscate....all appear in a similar fashion.

r = 1 + 1cos(2x) blue r = 3 + 3cos(2x) red	r = 1 + 1cos(3x) blue r = 3 + 3cos(3x) red	r = 1 + 1cos(8x) blue r = 3 + 3cos(8x) red
I'm not going to go over why the max and min values occurred, hopefully that has been understand previously. The number of "leaves" is what becomes interesting, as k increases we see a direct relationship with the number of leaves formed. It is also interesting to notice the orientation difference between an even k and the odd k's. An even k value can have many axises of symmetry but x = pi/2 (y axis) will always be one of them whereas for an odd k, x = pi/2 (y axis) will never be an axis of symmetry as long as k > 1.

r = 1 + 1cos(3x) black r = 2 + 1cos(3x) red r = 3 + 1cos(3x) green r = 4 + 1 cos(3x) blue	r = 1 + 1cos(6x) black r = 2 + 1cos(6x) red r = 3 + 1cos(6x) green r = 4 + 1cos(6x) blue
This reminds me of Spiro-graph (a great toy). Notice the concentric nature of these shapes. The max and min values hold true to form in that cosine restricts the value to be a +- 1. As the difference between a and b increase where a > b, we see an expansion from the pole. It is interesting to think about what kind of shape that would be formed if we continued to increase a. The diagram to the left shows what the polar equation r = 20 + 1cos(6x) looks like. It seems to gain much of the shape of a regular hexagon. This does not mean that it creates a regular hexagon as a approaches infinity but simply that curve looks like one a this stage. What do you think the shape would be as a approaches infinity?

r = 1 + 1cos(4x) black r = 1 + 2cos(4x) red	r = 1 + 3cos(3x) green r = 1 + 5cos(3x) blue
This relationship between a and b create a double leafing effect. The symmetry of these 3 leaf flowers is quite beautiful. I will stop my investigation here, although there is much more to discover and explore.