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.

MathThe 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+bcos(kθ) for k>0, we can note the following:•

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

bis 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•

kis 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 betweena - banda+b.

When a=bIn the graphs above,

a=b. as a result the domain of the distancerwas betweena - b= 0 anda+b= 2a= 2b. Whenr=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 2ato a value ofabetweenThe radius then drops at an decreasing rate of decline from

ato its minimum value of 0 between

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

abetween

The radius then increases at an decreasing rate from

aback to the maximum value of 2abetween

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

aorb.

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 becomesr=bcos(kθ) for k>0. As noted before:•

bis a scalar that only adjusts the amplitude of the function, and•

kis 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 distancerhas a domain between- bandb. Whenr=0, the graph crosses the point of origin. This occurs whenkθ = {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 2ktimes.For each quadrant of

kθ wherekθ is between 0 and 2π, the graph draws half a petal as the cosine function goes from a radius ofbto 0, then from 0 to -b, then from -bto 0, and then from 0 back tob. Consequently, for cos(kθ), 4khalf loops will be drawn, which would end up being 2kloops. Interestingly, the graphs show that whenkis odd, the loops are doubled (i.e. they retrace over existing loops). Consequently, whenkis odd onlykloops can be seen, while 2kloops are seen whenkis even.As noted, since

bonly 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 becomesr=bsin(kθ) for k>0. As noted before:•

bis a scalar that only adjusts the amplitude of the function, and•

kis 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 distancerhas a domain between- bandb. Whenr=0, the graph crosses the point of origin. This occurs whenkθ = {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 2ktimes.For each quadrant of

kθ wherekθ is between 0 and 2π, the graph draws half a petal as the sine function goes from a radius of 0 to, then frombbto 0, then from 0 to -b, and then from -bback to 0. Consequently, for sin(kθ), 4khalf loops will be drawn, which would end up being 2kloops. However, in the case of the sine graph, a complete petal is drawn for every interval ofkθ=2π. Similar to cosine , the graphs show that whenkis odd, the loops are doubled (i.e. they retrace over existing loops). Consequently, whenkis odd onlykloops can be seen, while 2kloops are seen whenkis even.It was noted, that the vertix of the initial complete loop is at an angle of θ=π/2

kcounterclockwise 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/kin 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

bonly 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.