# 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?

Aug 5, 2018

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

#### Explanation:

$r = a + b \cos n \left(\theta - \alpha\right)$ evolves, by delineation, different

families of curves, with similar characteristics.

( i ) a = 0 and n = 1.

$r = b \cos \left(\theta - \alpha\right)$

produce circles through the pole r = 0.

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

$r = 2 \cos \left(\theta - \frac{\pi}{3}\right)$ is the circle of radius 1, with the polar

diameter inclined at $\alpha = \frac{\pi}{3}$, 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) $a = 0 \mathmr{and} \alpha = 0$:

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

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

Example: $a = 2 \mathmr{and} \alpha = \frac{\pi}{4}$.

Cardioid Couple:
$r = 2 \left(1 \pm \cos \left(\theta - \frac{\pi}{4}\right)\right)$
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

$\left\mid \frac{a}{b} \right\mid < 1 \mathmr{and} > 1$.

Graph for the limacons

$r = 1 + 3 \cos \theta \mathmr{and} r = 3 + 1 \cos \theta$
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 $\alpha = \frac{\pi}{4}$, the two are rotated anticlockwise through

$\frac{\pi}{4}$, 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 $\left\mid \frac{a}{b} \right\mid \uparrow$.

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

Example: $r = 1 + \cos 3 \left(\theta + \frac{\pi}{12}\right) \mathmr{and} r = 1 + \cos 3 \theta$.
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) $n = \frac{p}{q} \mathmr{and} \alpha = 0$, integers p and q are co-prime.

$r = b \cos \left(\frac{p}{q} \left(\theta - \alpha\right)\right)$ creates idiosyncratic forms.

The count for the open loops is p = 2.
Example 1: $b = 1 , p = 2 \mathmr{and} q = 3$ gives

$r = \cos \left(\frac{2}{3} \left(\theta\right)\right)$, of period $3 \pi$.

For 3 complete rotations through $6 \pi$, 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: $b = 1 , p = 4 , q = 3 \mathmr{and} \alpha = 0$, giving

$r = \cos \left(\left(\frac{4}{3}\right) \theta\right)$, with period $\frac{3}{2} \pi$
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.