2. The graphs in the left column are cartesian graphs of a function
r = r(t) for 0
|
A
|
I
|
|
B
|
II
|
|
C
|
III
|
|
D
|
IV
|
|
E
|
V
|
Graph C has r(t) = 0 exactly once. Graph I is the only polar graph that goes through the origin once, so C corresponds to I.
Graph D has r(t)
³ 0 for t in [0, Pi] and r(t) £ 0 for t in [Pi, 2Pi].Thus, in the polar graph, the curve will lie above the x-axis. So D corresponds to II.
Graph B gives the 3-leaf rose of IV as t varies from 0 to Pi and then retraces it as t goes rom Pi to 2Pi. So, B corresponds to IV.
Graph E is the graph in B, raised a bit, so it will give III
Finally, by elimination, A corresponds to V
|
|
|
|
(path alternates between a large leaf and a small leaf)
|
|
|
|
|