Polar Equations

Karyn Carson

Investigate

Note:

- When a and b are equal, and k is an integer, this is
one textbook version of the " n-leaf
rose."
- Compare with

for various k. What if .. . cos( ) is replaced with sin( )?

If k = 1, then we getthis graph:

Now, letÕs change thevalue of k to 2, 3,4:

It would seem that kdetermines the number of petals that the flower has. Another thing thatÕs interesting is that
they all cross thex-axis at (2,0). Also, there seemsto
be a limit of 2 and -2 for the graphs that have more than one petal.

Now, letÕs keep kconstant and vary a and b. First,letÕs make a = 1 and b=2:

We get a doublefour-petaled flower. Now the limithas gone to 3 and -3. When wechange
b to 3,

ItÕs still a doublefour-petaled flower, but the limit has grown to 4 and
-4. It appears as though the limit isdetermined by the sum of a and b
and its inverse.

Now letÕs keep the valuesof b and k constant and vary the value of a.

r=1+cos((4*theta))and r=2+cos((4*theta))

The graph no longer intersectsat the origin and I think that the graph will
continue to enlarge, going towarda circle.

If we change a to 3, 4, 5 and 6:

LetÕs make a=10É

Well, IÕm not sure that it
will ever be a circle, but what I do notice is that my conjecture about the sum
of a and b and its inverse being the limits of the
graph seems to be true.

How would the graph bedifferent if we took away a?

Let k = 1, 2, and 3 in the equation É

K still determines thenumber of petals, but it appears as if k is odd, k=the
number of petals and ifitÕs even, 2k=the number of
petals. LetÕs continue the investigation by changing k to 4 and 5:

Yes, if k=4, then thenumber of petals that are produced is 8 and if k=5, the
number of petals is 5.

If we change cos to sin(using the original equation ), andvary k from 1 to 3, then the graph turns and is no
longer focused on thex-axis, but seems to be centered
on the y-axis.