Question

How do you graph `r=9cos12theta`?

Answer 1

See explanation to create this fine 12-petal rose.

Explanation:

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

graphics, I used

`1/9(x^2+y^2)^6.5`

`=y^12+x^12-66x^2y^2(x^8+y^8)+495x^4y^4(x^4+y^4)-924x^6y^6`

obtained by conversion, using

`cos 12theta= cos^12theta-12C_2cos^2theta sin^2theta(cos^10theta+sin^10theta)+12C_4cos^4thetasin^4theta( cos^8theta + sin^8 theta)-12C_6cos^6theta sin^6 theta+sin^12theta`

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

9 units long.

The period of `r(theta)=9 cos 12theta` is `( 2pi)/12=pi/6 and`

`2pi = 12 X(pi/6)`, giving 12 petals.

`theta` size for each petal is `pi/12`. For the other half of the period, `r < 0`.

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 `r^3/729`.

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

`r = cos 24theta`.

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] }