Question

Citing examples and making graphs, how do you explain that `f(r, theta;a,b,n,alpha )=r-(a + b cos n(theta- alpha) )=0` represents a family of limacons, rose curves, circles and more?

Answer 1

See explanation. To be continued, in my 2nd answer.

Explanation:

`r = a + b cos n( theta - alpha )` evolves, by delineation, different

families of curves, with similar characteristics.

( i ) a = 0 and n = 1.

`r = b cos ( theta - alpha )`

produce circles through the pole r = 0.

Example: b = 2 and alpha = pi/3:

`r = 2 cos ( theta - pi/3 )` is the circle of radius 1, with the polar

diameter inclined at `alpha = pi/3`, to the initial line `theta = 0`.

graph{ (x^2+y^2- 2(1/2x+sqrt3/2 y))(y-sqrt3 x)=0[-0.8 3.8 -0.3 2 1.8] }

Concentric circles come from r =a, when b = 0.
graph{(x^2+y^2-1/4)(x^2+y^2-1)=0}
(ii) `a = 0 and alpha = 0`:

`r = b cos ntheta` produce n-petal rose curves, n = 2, 3, 4, ..#

Here, as r is non-negative, I do not count r-negative petals.

Pixels do not glow for r-negative points.

Example: `r = cos 6theta`
graph{(x^2+y^2)^3.5-(x^6-15x^2y^2(x^2-y^2)-y^6)=0[-4 4 -2 2]}
(iii) `0 < a = +-b and n = 1`

#r = a ( 1 +- cos (theta - alpha ) produce Cardioids.

Example: `a = 2 and alpha = pi/4`.

Cardioid Couple:
`r = 2 ( 1 +- cos (theta - pi/4 ))`
graph{(x^2+y^2-2sqrt(x^2+y^2)+sqrt2(x+ y))(x^2+y^2-2sqrt(x^2+y^2)-sqrt2(x+ y))=0}

Answer 2

Continuation, for the 2nd part.

Explanation:

(iv) n =1

Let `alpha = 0`.

`r = a + b cos theta` create limacons.

The double here includes both ( pole-on and pole-not-on ) cases

`abs( a/b) < 1 and > 1`.

Graph for the limacons

`r = 1 + 3 cos theta and r = 3 + 1 cos theta`
graph{ (x^2+y^2-3sqrt(x^2+y^2)-x) (x^2+y^2-sqrt(x^2+y^2)-3 x)=0[- 16 16 -8 8]}

With `alpha = pi/4`, the two are rotated anticlockwise through

`pi/4`, about the pole.
graph{ (x^2+y^2-3sqrt(x^2+y^2)-1/sqrt2 (x+y)) (x^2+y^2-sqrt(x^2+y^2)-3/sqrt2 ( x+y))=0[- 20 20 -10 10]}
The innie of the dimple weakens, as `abs (a/b) uarr`.

Rosy Limacon, from limacon-like equation, giving rosy graph.
If `n = 3, r = a + b cos 3( theta - alpha )`.

Example: `r = 1 + cos 3(theta + pi/12) and r = 1 + cos 3theta`.
graph{((x^2+y^2)^2-(x^2+y^2)^1.5-1/sqrt2(x^3+3xy(x-y)-y^3))((x^2+y^2)^2-(x^2+y^2)^1.5-x^3+3xy^2)=0}

Answer 3

Continuation,for the 3rd part of my answer.
Ref: My answers to related Socratic questions.

Explanation:

(vi) `n = p/q and alpha = 0`, integers p and q are co-prime.

`r = b cos (p/q ( theta - alpha ) )` creates idiosyncratic forms.

The count for the open loops is p = 2.
Example 1: `b = 1 , p = 2 and q = 3` gives

`r = cos (2/3 ( theta))`, of period `3pi`.

For 3 complete rotations through `6pi`, this generates two cycles,

giving two `(2 (3pi)` loops.
graph{(x^2-y^2) - (x^2+y^2)^1.5 sqrt ( 1-(x^2+y^2))(4(x^2+y^2)-3) =0[-6 6 -3 3]}

Example 2: `b = 1, p = 4, q = 3 and alpha = 0`, giving

`r = cos ( (4/3)theta )`, with period `3/2pi`
graph{(x^4- 6 x^2y^2 +x^4) - (x^2+y^2)^2.5 sqrt ( 1-(x^2+y^2))(4(x^2+y^2)-3) =0[-6 6 -3 3]}

The count ( 2 open and 2 closed )of idiosyncratic loops is p = 4.