MATH 166 LECTURE NOTES, NOVEMBER 30
Example 1 Sketch the graph of
(r − 3)(θ −
π
)=0
4
and verify its symmetry.
The graph is
3
2
1
-3
-2
-1
1
2
3
-1
-2
-3
To verify symmetry, we use the graph to conclude that the only symmetry is through the origin. Since the equation factors into the two equation
r − 3 = 0 and θ = π4 = 0, we just have to check that each equation is
symmetric through the origin.
For r − 3 = 0, we see that if (r, θ) is on the graph, then r = 3. It follows
that (3, θ + π) is also on the graph, so it’s symmetric through the origin.
For θ − π/ 4), we see that if (r, θ) is on the graph, then θ = π/4. Therefore
(−r, θ) = (−r, π/4) is also on the graph so it’s symmetric through the origin.
1
2
MATH 166 LECTURE NOTES, NOVEMBER 30
Example 2 Sketch the graph of the cardioid
r = 5 − 5 sin θ
and verify its symmetry.
-6
-4
-2
2
4
6
-2
-4
-6
-8
-10
This graph is symmetric through the y-axis. To verify the symmetry from
the equation, we notice that if (r, θ) is on the graph, then so is (r, π − θ)
because sin(θ) = sin(π − θ).
Example 3 Sketch the graph of the limaçon
r = 4 − 3 cos θ
and verify its symmetry.
4
2
-6
-4
-2
-2
-4
This time, the graph suggests symmetry through the x-axis. If (r, θ) is
on the graph, then so is (r, −θ) because cos(θ) = cos(−θ).
MATH 166 LECTURE NOTES, NOVEMBER 30
3
Example 4 Sketch the graph of the lemniscate
r2 = −9 cos 2θ
and verify its symmetry.
3
2
1
-1.0
-0.5
0.5
1.0
-1
-2
-3
Finally, the graph suggests that symmetry through both axes and the origin. If (r, θ) is on the graph, then so is (r, −θ) because cos(2θ) = cos(2(−θ)).
Therefore the graph is symmetric through the x-axis. Also, if (r, θ) is on
the graph then so is (r, π − θ) because
cos(2(π − θ)) = cos(2π − 2θ) = cos(−2θ) = cos(2θ).
Therefore the graph is symmetric through the y-axis. Any graph that’s
symmetric through both axes is symmetry through the origin, so our graph
is symmetric through the origin.