# How do you graph #r=9cos12theta#?

##### 1 Answer

See explanation to create this fine 12-petal rose.

#### Explanation:

I have no polar graphic facility . Using, the Socratic Cartesian

graphics, I used

obtained by conversion, using

Of course, the 12-petal rose should have equi-sized petals that are

9 units long.

The period of

graph{(1/9)(x^2+y^2)^6.5-y^12-x^12+66 x^2 y^2(x^8+y^8)-495 x^4 y^4(x^4+y^4)+924 x^6 y^6=0[-20 20 -10 10] }

For wider loops, change r/9 to

graph{(1/729)(x^2+y^2)^7.5-y^12-x^12+66 x^2 y^2(x^8+y^8)-495 x^4 y^4(x^4+y^4)+924 x^6 y^6=0[-20 20 -10 10] }

For the use by artists, I added r-negative loops, to get 24 loops.

This is not to be mistaken for 24 r-positive loops of

graph{((1/729)(x^2+y^2)^7.5-y^12-x^12+66 x^2 y^2(x^8+y^8)-495 x^4 y^4(x^4+y^4)+924 x^6 y^6)(-(1/729)(x^2+y^2)^7.5-y^12-x^12+66 x^2 y^2(x^8+y^8)-495 x^4 y^4(x^4+y^4)+924 x^6 y^6)=0[-20 20 -10 10] }