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

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Answer 1 · 2016-11-25T14:05:32

The graph is inserted after conversion that befits cartesian frame.

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

limacons, with n = 1.

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

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

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

$r = 2 + cos 6 θ$ $r = 2 + cos 6theta$ .

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]}

How do you make a graph of the rosy limacon r=1+2cos3θr = 1 + 2 cos 3theta?