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?

3 Answers
Aug 5, 2018

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}

Aug 5, 2018

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}

Aug 6, 2018

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.