Polar Graphs Day 2 - by Wes Tully

We did two polar graphs today. One that had x and y axis symmetry. Which also means that if it is symmetrical to the x and y axis then it must be symmetrical to the origin. The other graph had 'maybe' symmetrical to the x and y axis.

r=4cos(2ϴ)

x-axis symmetry? Yes

r=4cos(2·-ϴ)

r=4cos(2ϴ)

y-axis symmetry? Yes

r=4cos(2(π-ϴ))☞r=4cos(2π-2ϴ)=☟

=4(cosπ2‧cosϴ2+sinπ2‧sinϴ2)= 4cos(2ϴ)

Origin? YES

Max r=4

ϴ	r
0	4
π/12	3.46
π/6	2
π/4	0
π/3	-2
5π/12	-3.46
π/2	-4

r=6sin(2ϴ)

x-axis? Maybe

y-axis? Maybe

r=6sin2(π-ϴ))=☟

=6sin(2π-2ϴ)=☟

=6(sin2π‧cos2ϴ-cosπ2‧sinϴ2)=☟

=-6sin(2ϴ)

Origin? Maybe

Max r=6

ϴ	r
0	0
π/6	5.2
π/4	6
π/3	5.2
π/2	0
11π/6	-5.2
5π/3	-5.2
3π/2	0
7π/4	-6

Rose Conjecture

Y=Acos(Bϴ)

Increasing a makes petals bigger and increasing b changes the number of petals.