MATH 166 LECTURE NOTES, DECEMBER 3
Example 1 Sketch the graph of the 4-leaved rose
r = 6 sin 2θ
and verify its symmetry.
The graph is
4
2
-4
-2
2
4
-2
-4
It’s symmetric through the x-axis. If (r, θ) is on the graph, then so is
(−r, π − θ) because
sin(2(π − θ)) = sin(2π − 2θ) = sin(−2θ) = − sin 2θ.
It’s also symmetric through the y-axis. If (r, θ) is on the graph, then so is
(−r, −θ) because sin(2(−θ)) = − sin 2θ. Since it’s symmetric through both
axes, it’s also symmetric through the origin.
Example 2 Sketch the graph of the spiral of Archimedes
θ ≥ 0.
r = 2θ,
The graph is
40
20
-40
-20
20
-20
-40
1
40
2
MATH 166 LECTURE NOTES, DECEMBER 3
Example 3 Sketch the curves
√
r = 3 3 cos θ, r = 3 sin θ
and find their points of intersection.
3
2
1
-1
1
2
3
4
5
-1
-2
One intersection point
can be found by solving the
√ equations simulta√
neously. We want 3 3 cos θ = 3 sin θ, so tan θ = 3.
√ This equation is
satisfied by θ = π/3, so the first intersection point is (3 3/2, π/3), in polar
coordinates. (If we look at any other θ which give tan θ, we get the same
intersection point even though the polar coordinates will be different.)
The other intersection point is suggested by the graph to be r = 0. For
the first equation, this value of r occurs when θ = π/2 and, for the second
equation, this value of r occurs when θ = 0.
Example 4 Sketch the limaçon r = 2 − 4 cos θ and find the area inside the smaller loop.
3
2
1
-6
-5
-4
-3
-2
-1
-1
-2
-3
As pointed out in class, we’re on the smaller loop for 0 ≤ θ ≤ π/3 and
then on the larger loop for π/3 ≤ θ ≤ 5π/3. Finally, we’re back on the
smaller loop for 5π/3 ≤ θ ≤ 2π. Therefore the area is
Z
Z
1 π/3
1 5π/3
(2 − 4 cos θ)2 dθ +
(2 − 4 cos θ)2 dθ.
A=
2 0
2 π/3
MATH 166 LECTURE NOTES, DECEMBER 3
3
To evaluate these integrals, we do some algebra first:
(2 − 4 cos θ)2 = 4 − 16 cos θ + 16 cos2 θ.
Next, we use the double angle formula cos 2θ = 2 cos2 θ − 1, so
(2 − 4 cos θ)2 = 12 − cos θ + 8 cos 2θ.
It follows that
Z
0
π/3
√
√
√
π/3
3
3
3 3
(2 − 4 cos θ) dθ = (12θ − sin θ + 4 sin 2θ) = 4π −
+4
= 4π −
,
2
2
2
2
0
and similar calculations show that
√
Z 5π/3
3 3
2
(2 − 4 cos θ) dθ = 4π −
.
2
π/3
and therefore the area is
√
3 3
.
A = 4π −
2