Question

How do you make a graph of the rosy limacon `r = 1 + 2 cos 3theta`?

Answer 1

The graph is inserted after conversion that befits cartesian frame.

Explanation:

You can see that rose curves `r = a + b cos n(theta-alpha)` include

limacons, with n = 1.

See the roses, with `alpha = 0`. (i) As in the given equation, n =3,

a = 1 and b =2, giving `r = 1 + 2 cos 3theta` (ii) n = 6, a = 1 and b

= 2, giving `r = 1 + 2 cos 6theta` (iii) n = 6, a = 2 and b = 1, giving

`r = 2 + cos 6theta`.

graph{(x^2+y^2)^2-(x^2+y^2)^1.5-2x^3+6xy^2=0 [-10 10 -5 5]}

graph{(x^2+y^2)^3.5-(x^2+y^2)^3-2(x^6-y^6)+30x^2y^2(x^2-y^2)=0 [10 10 -5 5]}

graph{(x^2+y^2)^3.5-2(x^2+y^2)^3-(x^6-y^6)+15x^2y^2(x^2-y^2)=0 [10 10 -5 5]}