Assignment 11

This is an exploration of polar equations and how varying different variables affects its graphs. Lets begin with the polar equation

So lets explore different values for a and k holding theta in the range of [ 0, 2pi ]. When both variables equal 1 then we have

and so lets begin by varying just a. If a > 1 then we have

and if 0 < a < 1 we have

Now if a euals some negative number the graph looks like:

but when a = 0 there is no graph because, obviously, zero multiplied by any number is always zero. Now lets vary k and see what kind of graphs we construct. Again if k = 1 we get the first graph from above. If k > 1 and holding a conxtant at 1 we get

but more specifically for for k = 5. So I conjecture with the same conditions stated above, the graph has k 'leaves' for k > 1. Therefore for k = 8, the graph will have 8 'leaves':

Now if 0 < k < 1 we have

and if k is negative we have constructed

which is the same graph as when k > 1 but mirrored across the x-axis. Now lets explore a variation of our original equation

and lets graph the equations by varying a and k, and letting theta in the range of [ 0 , 2pi ]. So will do the same prodecdure as above.

a = 1, holding k = 1

a > 1

0 < a < 1

and finally a is a negative number, here a = -1:

As we can see by comparison to the above graphs, changing the value of a changes the size of the circle or mirror ir if a is negative. Now lets explore varying the k value holding a constant at 1.

k > 1

0 < k < 1

k is negative

The same effects apply here as above except that when k is negative here, there is no mirror when I expected it to mirror across the y-axis when in the sin equation there was a mirror across the x-axis. If we change both values of a and k then the size and direction of the n-leaf rose changes. If both values are greater than 1, specifically k = 5 and a = 3, we will have a bigger 5 leaf rose:

compared to a = 1

The next equation that we will explore is

and we will repeat the steps we had above for the values a, k and b.

a = 1, holding k = 1 and b = 1

a > 1

0 < a < 1

a is negative

k > 1, holding a = 1 and b = 1

0 < k < 1

k is negative

b > 1, holding a = 1 and k = 1

0 < b < 1

b = 0 since b is added and not multiplied

b is negative

So now lets look at a few graphs when all the values are changed:

a > 1, k > 1, b > 1

a < 1, k < 1, b < 1

0 < a < 1, 0 < k < 1, 0 < b < 1

So if you vary the values in different ways the graph changes according to the characteristics of each variable that we identified through earlier graphs.

Next version of these polar equations we will explore is

and we will repeat the last set of steps for this equation to determine the difference between varying the values of a, k and b.

a = 1, holding k = 1 and b = 1

a > 1

0 < a < 1

a is negative

k > 1, holding a =1 and b = 1

0 < k < 1

k is negative

b > 1, holding a = 1 and k = 1

0 < b < 1

b = 0

b is negative

a > 1, k > 1, b > 1

0 < a < 1, 0 < k < 1, 0 < b < 1

a, k and b are negative

Finally we will modify the polar equation by adding in another variable, c, making the new equation:

and we will vary each variable to discover its effect on the graph of this equation.

a, k, b and c =1

a > 1

0 < a < 1

a is negative

k > 1

0 < k < 1

k is negative

b > 1

0 < b < 1

b is negative

c > 1

0 < c < 1

c is negative

a, k, b and c > 1

a, k, b and c between [0,1]

a, k, b and c are negative

As we can determine that each of these variables in some way has a substantial effect on the graph of this polar equation. the variable a changes the size and mirrors; the variable k changes the shape of the graph; the variable b also shanges the size and the shape; and finally the variable c changes the size and mirrors the graph. This Graphing Calculator does a great job of allowing us to change variables of polar equations and understand by looking at the graph of how everything is affected by changing these variables.