Assignment 11

by

Shridevi kotta

 

 

This write up explores polar equations using graphing calculator software.

Here we investigate r = a + b cos(kq) and explores for different values of a, b and k and also compares with r =  a + b sin(kq) and explains  the observation made.

 

First lets look at r = a + b cos(kq), with a = b = 2(for any a = b) and for different integer values of k.(we see that positive or negative makes no difference, since cos(q) = cos(-q). For different integer values of k, you get that many Òk-leaf roseÓ, with leaves of magnitude b + b = 2b.

 

 

For the function, r = b cos(kq), we observe that the leafs are of magnitude b.

 

 

Consider the function r = a + b sin((kq) with a = b and k being an integer values.

We observe that the function results in a Òk-leaf roseÓ with the difference from the cosine function being a phase shift. And the phase shift is 90o/k. For example, 2 leaf rose has phase shift of 45o.

 

 

 

And in case of r = bsin(kq), if k is negative, then the graph is flipped since

sin(-q) = -sin(q). Graphs are shown below for comparison.

 

 

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