College Trigonometry, 2011a

11.5 Graphs of Polar Equations 949
Even though we have finished with one complete cycle of r =5sin(2θ), if we continue plotting
beyond θ = π, we find that the curve continues into the third quadrant! Below we present a
graph of a second cycle of r =5sin(2θ) which continues on from the first. The boxed labels
on the θ-axis correspond to the portions with matching labels on the curve in the xy-plane.
5
r
3
y
4
π
1 5π 2 3π 3 7π 4
4 2 4
2π
θ
1
x
−5
2
We have the final graph below.
5
r
5
y
π
4
π
2
3π
4
π
5π
4
3π
2
7π
4
2π
θ
−5 5
x
−5
−5
r =5sin(2θ) intheθr-plane
r =5sin(2θ) inthexy-plane
4. Graphing r 2 =16cos(2θ) is complicated by the r 2 , so we solve to get r = ± √ 16 cos(2θ) =
±4 √ cos(2θ). How do we sketch such a curve? First off, we sketch a fundamental period
of r =cos(2θ) which we have dotted in the figure below. When cos(2θ) < 0, √ cos(2θ) is
undefined, so we don’t have any values on the interval ( π
4 , 3π )
4 . On the intervals which remain,
cos(2θ) ranges from 0 to 1, inclusive. Hence, √ cos(2θ) ranges from 0 to 1 as well. 7 From
this, we know r = ±4 √ cos(2θ) ranges continuously from 0 to ±4, respectively. Below we
graph both r =4 √ cos(2θ) andr = −4 √ cos(2θ) ontheθr plane and use them to sketch the
corresponding pieces of the curve r 2 =16cos(2θ) inthexy-plane. As we have seen in earlier
7 Owing to the relationship between y = x and y = √ x over [0, 1], we also know √ cos(2θ) ≥ cos(2θ) wherever the
former is defined.