Polar Equations

by Venessa Brown

The goal of this assignment is to investigate:

Example using graphing calculator 6.0

Explore when a=b and k > 0.

Math

The cosine function moves sinuously from θ=0 to θ=2π, with cosine starting at 1 and ending at its starting point of 1 when θ=2π. The function has a crest (highest point) is 1 and the trough (lowest point) is -1. Specific to the equation of r = a + b cos(kθ) for k>0, we can note the following:

a is a scalar that shifts the cosine function vertically, but otherwise has no impact on the amplitude or the cycle (or period) of the function,

b is a scalar that adjusts the amplitude (magnitude of the crest and trough), but otherwise has no impact on the vertical shift or the cycle (or period) of the function, and

k is a scalar that adjusts the cycle (or period), but otherwise has no impact on the vertical shift or the amplitude of the function.

Consequently we can note that since, cos(kθ) has a domain between -1 and 1, then the distance r has a domain between a - b and a + b.

When a=b

In the graphs above, a=b. as a result the domain of the distance r was between a - b = 0 and a + b = 2a = 2b. When r=0, the graph is on the point of origin. In these graphs, the radius drops at an increasing rate of decline from its maximum value of 2a to a value of a between

The radius then drops at an decreasing rate of decline from a to its minimum value of 0 between

The radius then increases at an increasing rate from its minimum value of 0 to a between

The radius then increases at an decreasing rate from a back to the maximum value of 2a between

Note: If a>b, the domain of r is positive (and nonzero). In these instances, since r is never 0, the graph will never cross the point of origin.

Animation for various values of k.

1+1cos(kΘ)

Notice:

1. The graph has loops resembling flower petals when k >1 and centered at ( 0, 0).

2. The number of loops/petals is equal to k.

3. The length of the loops is twice the number of a or b.

Other Visual representation that supports finds

r = 2 + 2cos (kΘ)

r = 3 + 3cos (kΘ)

Explore when a = 0 for :

Reasoning for above Cosine functions

When a=0, our function becomes r = b cos(kθ) for k>0. As noted before:

b is a scalar that only adjusts the amplitude of the function, and

k is a scalar that only adjusts the cycle (or period) of the function.

Consequently we can note that since, cos(kθ) has a domain between -1 and 1, then the distance r has a domain between - b and b. When r=0, the graph crosses the point of origin. This occurs when kθ = {π/2, 3π/2, 5π/2, ..., nπ + π/2, ...}, where n is an integer. That is, kθ crosses the point of origin twice for every cycle. Hence, in k-cycles, cos(kθ) will cross the point of origin 2k times.

For each quadrant of kθ where kθ is between 0 and 2π, the graph draws half a petal as the cosine function goes from a radius of b to 0, then from 0 to -b, then from -b to 0, and then from 0 back to b. Consequently, for cos(kθ), 4k half loops will be drawn, which would end up being 2k loops. Interestingly, the graphs show that when k is odd, the loops are doubled (i.e. they retrace over existing loops). Consequently, when k is odd only k loops can be seen, while 2k loops are seen when k is even.

As noted, since b only affects the amplitude, changing b simply stretches or compresses the petals around the point of origin.

r = 1cos (kΘ) for varing k

r = 2cos (kΘ) for varing k

r = 3cos (kΘ) for varing k

1. If k was odd, then the number of observed loops is equal to k since loops are doubling up.

2. If k is even then the number of loops is equal to twice k since loops are not doubling up.

3. The length of the petal is now equal to b.

What happen when cosine is replaced by sine:

Reasoning on above sine graphs

When a=0, our function becomes r = b sin(kθ) for k>0. As noted before:

b is a scalar that only adjusts the amplitude of the function, and

k is a scalar that only adjusts the cycle (or period) of the function.

Consequently we can note that since, sin(kθ) has a domain between -1 and 1, then the distance r has a domain between - b and b. When r=0, the graph crosses the point of origin. This occurs when kθ = {0, π, 2π, 3π, ..., nπ, ...}, where n is an integer. That is, kθ crosses the point of origin twice for every cycle. Hence, in k-cycles, cos(kθ) will cross the point of origin 2k times.

For each quadrant of kθ where kθ is between 0 and 2π, the graph draws half a petal as the sine function goes from a radius of 0 to b, then from b to 0, then from 0 to -b, and then from -b back to 0. Consequently, for sin(kθ), 4k half loops will be drawn, which would end up being 2k loops. However, in the case of the sine graph, a complete petal is drawn for every interval of kθ=2π. Similar to cosine , the graphs show that when k is odd, the loops are doubled (i.e. they retrace over existing loops). Consequently, when k is odd only k loops can be seen, while 2k loops are seen when k is even.

It was noted, that the vertix of the initial complete loop is at an angle of θ=π/2k counterclockwise from the x-axis. Hence when k=1, the initial complete loop has a vertix at θ=π/2k, with subsequent vertices occuring every θ = (k+1)π/k in a counterclockwise direction from the previous vertix.

E.g. 1. When k=2 there will be 4 petals. The initial complete petal has a vertex of an angle of θ=π/4. Subsequent petals will occur every θ = 3π/2 from the previous petal. Hence the second petal is at an angle of θ= π/4+3π/2 = 7π/4, the third petal is at an angle of θ= 7π/4+3π/2 = 13π/4 = 5π/4, and the fourth petal is at an angle of 5π/4+3π/2 =11π/4 = 3π/4.

E.g. 2. When k=3 there will be 6 petals. The initial complete petal has a vertex of an angle of θ=π/6. Subsequent petals will occur every θ = 4π/3 from the previous petal. Hence the second petal is at an angle of θ= π/6+4π/3 = 9π/6 = 3π/2, the third petal is at an angle of θ= 3π/2+4π/3 = 17π/6 = 5π/6, the fourth petal is at an angle of 5π/6 + 4π/3 =13π/6 = π/6, the fifth petal is at an angle of π/6 + 4π/3 = 9π/6 = 3π/2, and the sixth petal is at an angle of 3π/2+4π/3 = 17π/6 = 5π/6. The fourth, fifth, and sixth petals are a repest of the first, second and third petal respectively.

As noted, since b only affects the amplitude, changing b simply stretches or compresses the petals around the point of origin.

Animation for r =sin (kΘ) for various values of k

1. If k was odd, then the number of observed loops is equal to k since loops are doubling up.

2. If k is even then the number of loops is equal to twice k since loops are not doubling up.

3. The length of the petal is now equal to b.

When:

The animation of graph is seemingly making 1/2 revolution and then switching direction.