College Trigonometry, 2011a

11.5 Graphs of Polar Equations 947
Again, we invite the reader to show that plotting the curve for values of θ outside [0, 2π]
results in retracing a portion of the curve already traced. Our final graph is below.
6
r
y
4
θ = 2π 3
2
2
2π
3
π
2
π
4π
3
3π
2
2π
θ
θ = 4π 3 −2
2 6
x
−2
r = 2 + 4 cos(θ) intheθr-plane
r = 2 + 4 cos(θ) inthexy-plane
3. As usual, we start by graphing a fundamental cycle of r =5sin(2θ) intheθr-plane, which in
this case, occurs as θ ranges from 0 to π. We partition our interval into subintervals to help
us with the graphing, namely [ 0, π ] [
4 , π
4 , π ] [
2 , π
2 , 3π ] [
4 and 3π
4 ,π] .Asθ ranges from 0 to π 4 , r
increases from 0 to 5. This means that the graph of r =5sin(2θ) inthexy-plane starts at
the origin and gradually sweeps out so it is 5 units away from the origin on the line θ = π 4 .
5
r
y
π
4
π
2
3π
4
π
θ
x
−5
Next, we see that r decreases from 5 to 0 as θ runs through [ π
4 , π ]
2 , and furthermore, r is
heading negative as θ crosses π 2 . Hence, we draw the curve hugging the line θ = π 2
(the y-axis)
as the curve heads to the origin.