emat

Written by Sunny Yoon


I'm going to investigate polar equations varying different variables.

 

First, I'm going to vary a and k in this equation eq where a = 1 and k = 1.

g

It is a circle with the center (0, 1) and radius = 1.

Now, let's vary k from -5 to 5.

What a pretty picture!

When |k| < 0.5, then the graph looks like a spiral that's spiraling out from the origin. Once |k| > 0.5, the graph "closed" in.

eq eq eq
g g g
eq eq  
g g  
eq eq  
g g  
eq eq  
g g  
eq eq  
g g When |k| is an even number, the direction of the graph doesn't change. In fact, it's a same shape. However, when |k| is an odd number, the graph looks like they are rotated.

 

Now what if you vary a? I'm going replace n in place of a and let n goes from -5 to 5.

The shape of the graph stays the same as a circle. It just expands on the positive y-axis or the negative y-axis depending on the sign of a and as |a| increases, the circle gets bigger too.


Now let's investigate when we replace sin with cos.

eqwhere a = k = 1.

g

It appears to be a circle where the center is (1, 0) and the radius = 1. The only different is the center of a circle when you replace sin with cos.

 

Now let's vary k. k goes from -5 to 5.

When n= 0, the graph changes to a circle where the center is (0, 0) and the radius = 2. When |n| < 1, then the graph looks like a "pig's tail" as it's spiraling out and then twists once.

When |n| >1, then it looks very similar to eqwhere you vary the k.

eq eq  
g g
eq eq
g g
eq eq
g g
eq qe
g g The sign of k didn't affect the graphs.

Here's a graph when you vary k from 5 to 15.

Let's vary a from -5 to 5.

Just like eq, the graph stays the same as a circle. However, the difference is that as a > 0, the circle expands on the positive x-axis while when a < 0, the circle expands on the negative x-axis.


 

Let's investigate eq when a = b = k = 1.

g

When you vary k from -5 to 5.

 

When k = 0, then it's a circle centered at (0, 0) with radius = 1.

When |k| < 1, it's a line spiraling out at (1, 0).

When |k| = 2, it forms 4 "flower petals" 2 small and 2 big. As k gets bigger, "flower petals" keep adding to the graph.

So when |k| = 5, there are total of 10 flower petals; 5 big and 5 small ones inside each big petal.

 

When you vary b from -5 to 10.

As b moves from |2| to |5|, the graph looks like a circle with a dent in the bottom.

As b gets bigger, the graph expands so the dent starts to disappear.

When |b| < 2, there exists a smaller loop inside a bigger loop. As |b| gets smaller,

two loops start to come together so that when b = 0, it becomes a circle.

 

When you vary a from -5 to 10.

When a < 0. then the graph appears on the negative y-axis (III or IV quadrant).

When a > 0, the graph appears on the positive y-axis (I or II quadrant).

When a = 0, then it's a circle centered at the origin.

When |a| < 0.5, the circle starts to cave in at the origin.

As |a| > 0.5, you see another circle forming inside the bigger circle.

Eventually, two circles (small one enclosed in a big one) keeps getting bigger as |a| gets bigger.


Let's investigate eq when a = b = k = 1.

gra

It appears to be same shape as eq but it's been rotated 90 degrees.

 

Let's see what happens when we vary the variables. First, let's vary k and see how different degrees will affect the graph.

The graph looks very similar toeq

but it looks like it's been rotated. Otherwise, it shows similar behavior as eq'

 

Let's vary a from -5 to 5.

Once again, it shows similar behaviors of a graph eq.

 

Unlike the above equation, when |a| < 0, it stays on the negative x-axis instead of negative y-axis

(II or III quadrant).

When |a| > 0, it stays on the positive x-axis (I or IV quadrant).

 

Let's vary b from -5 to 5.

Once again, it shows similar behaviors of eq,

but the graph has been rotated.

 


 

Now, let's investigate eqwhere a = b = c = k = 1

g

It appears to be a straight line.

 

eq

as k varies from -5 to 5.

It looks like fireworks!

When k = 0, it is a unit circle.

When |k = 0.4, that's when the spiral (or curve) meets with a tangent line.

I still had a hard time imagining what was going on, so I looked at |k| = 2.

It looks like the "flower petals" that appears with previous equations been "slit" open.

In the middle of the graph, there appears to be a quadrilateral with the lines extending

from each side. Using those lines, there are 4 hyperbola curves forming along those lines.

When |k| = 3, there appears to a triangle in the middle with the lines extending from each side.

Also, there are hyperbola curves forming along those lines.

When |k| = 4, there appears to an octagon in the middle with the lines extending from each side

and hyperbola curves forming along those lines.

When |k| = 5, there appears to be a pentagon in the middle with lines extending from each side and

hyperbola curves forming along those lines.

When |k| = 6, there appears to be a dodecagon in the middle with lines extending from each side and

hyperbola curves forming along those lines.

I expanded the numbers to -10 to 10.

I realized that when |k| is an even number, then polygon in the middle will be |k| * 2-gon.

When |k| is an odd number, then polygon in the middle will be |k| -gon.

 

eq

as a varies from -5 to 5

It appears to be a straight line all the way.

When n < 0, the graph appears on I or III quadrant.

When n > 0, the graph appears on II or IV quadrant.

a acts like a slope in a linear function because as |a| gets bigger,

the line gets more slanted. As |a| gets closer to 0, the graph gets closer to the x-axis.

 

 

eq

as b varies from -5 to 5

The graph shows similar characteristic as the above example.

However, as |b| gets bigger, the graph gets closer to the x-axis, opposite from above.

 

 

eq

as c varies from -5 to 5

c affects the intercepts of the graph.

For example, when c = -5, the intercepts of the graph were

(-5,0) and (0, -5)

When c = 5, the intercepts of the graph were (5,0) and (0, 5).

When c= 0, the graph didn't exist.

I wanted to investigate more about the effects of a, b, c, and k put all together.

 

eq1 eq2 eq3 eq4
g1 g2 g3 g4
It appears to be a line with x-intercept at (3/4) and y-intercept at (0, -3/2) Now, there is a quadrilateral inside. The hyperbolas intercept at (-3/4, 0), (3/4, 0), (0, 3/4) and (0, -3/4). The straight lines form a triangle. So far, I've observed that k affects how many lines that will intersect with each other to form a polygon in the middle.
eq5 eq6 eq7 eq8
g5 g6 g7 g8
Since k is 2, the graph appears to be more like the second example from the first row. The hyperbola intercepts at (5, 0), (-5, 0), (0, 5) and (0, -5). It's the same graph as on the left. Therefore, the sign of c doesn't affect the graph. It changes the intercept but not the direction. Since k isn't change, the straight lines still form a quadrilateral. However, changing a, b, and c changed the intercepts and how the hyperbolas stretch. This time, only b is changed. The graph definitely "flattened" overall. However, the intercepts of the hyperbola didn't change while absolute values of the intercepts of the straight lines got smaller.
       
eq9 eq11 eq10  
g9 g11 g10  
       
Changing the sign of a didn't change the graph overall. Changing b so as it gets bigger, the absolute values of the intercepts of the lines get smaller as well.    

 

 

 

 

 

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