By Nami Youn

Write-up  #11

Polar Equation

Introduction

In this write-up, I will investigate   varing a, b, c and k.

1.  r = 2a sin(kt)

1) Varing a
Let the value of k be fixed 1

a= -4(purple),  -2(blue), 1(green),  3(yellow), 5(gray)

we can get the circular graphs for values of variable a.
The value of a determined the radius of the circle. As the absolute value of a increases, the radius is greater.
If a>0, the circle lies below the x-axis.
If a<0, the circle lies above the x-axis.

2) Varing k

a. Let  the value of k be 2. This means the case k = even

a= -4(purple),  -2(blue), 1(green),  3(yellow), 5(gray)

Each graph has four petals. The number of petal = 2k.
The absolute value of a determines the size of the petals.
The four petals looks the same one another. Also, one petal is obtained from reflection of the other petals over y = x , x-axis or y-axis.

b. Let  the value of k be 5. This means the case a = odd.

a= -4(purple),  -2(blue), 1(green),  3(yellow), 5(gray)

Each graph has five petals. The number of petal = k.
The absolute value of a determines the size of the petals.
The five petals looks the same one another.

c. Let's examine the case k = 1.5 and 3.5 Let a be fixed 1.

k = 1.5(purple),     3.5(blue)

The number of petals are 3 and 7 respectively. Each number is two times k petals.
Then, what is difference between  the graphs of the case k = 1.5 and k = 3  or the graphs of the case k =3.5 and k =7.
See and compare the following graphs.

k =1.5(purple),    k =3(blue).                                             k = 3.5(green),  k = 7(black)

The two graphs of k =1.5 and k = 3 are not the same. The size and location are different.
But, the number of petals are the same.
Also, as the number of petals increases, their graphs become narrow.

2.  r = 2a cos(kt)

1) Varing a
Let the value of k be fixed 1

a= -4(purple),  -2(blue), 1(green),  3(yellow), 5(gray)

we can get the circular graphs for values of variable a.
The value of a determined the radius of the circle. As the absolute value of a increases, the radius is greater.
If a>0, the circle lies to the left of the y-axis.
If a<0, the circle lies to the right of the y-axis.

Let's examine the case of the larger value of k=20 varing a.

a =4(purple),   a=8(blue),    a=12(green)

As the value of a increases for the larger value k, the number of petals is the exactly same, but the legnth of the petals is longer..

2) Varing k

Since the pattern of the petals are the same as sine equations, let's examine the graphs varing k in the case of the constant a=2.

k=2(purple),   k=5(blue),    k=3.5(green)

we have the following results.

when k = 2 (even integer), the number of petals is 4 = 2(k)
when k = 5 (odd integer),  the number of petals is 5 =    k
when k =3.5( not integer), the number of petals is 7 = 2(k)

3.  r = 2a sin(kt)+b

Let the values of a and k be fixed 1

b = -5(purple),   b=-1(blue),    b=0(green),   b=4(yellow),    b=8(gray)

As the absolute value of b increases, the graph looks more like a circle.
The y-intercept of the graph is 2+b. For example, if b=8, y-intercept is 2+8.

Now, let's increase the value of a.
With a values of 6, varing b, let's examine the graphs.

b = 1(purple),   b=3(blue),    b=8(green),   b=16(yellow)

All graphs have 6 petals. As the value of b increases, the graph is spread(streched) apart from the origin(0,0). So, the graph looks more like a circle. This means that the petal is fatter.

Finally, let's examine the graph when the valueb is so large.

b = 500(purple),     b=1000(green)

As the value of b increases, the graph looks more like a circle.

4.  r = 2a cos(kt)+b

Let the values of a and k be fixed 1

b = -5(purple),   b=-1(blue),    b=0(green),   b=4(yellow),    b=8(gray)

As the absolute value of b increases, the graph looks more like a circle.
The x-intercept of the graph is 2+b. For example, if b=8, x-intercept is 2+8.

Now, let's increase the value of a.
With a values of 6, varing b, let's examine the graphs.

b = 1(purple),   b=3(blue),    b=8(green),   b=16(yellow)

All graphs have 6 petals. As the value of b increases, the graph is spread(streched) apart from the origin(0,0). So, the graph looks more like a circle. This means that the petal is fatter.
Wow! These graphs are the exactly same as r= 2asin(kt)+b.

Finally, let's examine the graph when the valueb is so large.

b = 500(purple),     b=1000(green)

As the value of b increases, the graph looks more like a circle.
These graphs are the exactly same as r= 2asin(kt)+b.

5.    r = c / a cos(kt)+bsin(kt)

First, let the value of a, b, and k=1 varing c.

c = -5(purple),    c = -1(blue),       c= 2(green),       c = 4(yellow)

The graphs are the lines. The value of c means the x-intercept.

Next, let the values of both a and b be 1 and c be 5
Vary the value of k.

k = 4(purple)                                                             k= 6(blue)

k = 5(green)                                                              k= 7(sky)

when k= even, the graph has a polygon with 2k-sides.
When k= odd, the graph has a polygon with k-sides.

For k = 4, the graph has 2(4) sided figure.
For k = 5, the graph has 5 sided figure.