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

Question

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

3 Answers

Impact of this question

2927 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-05T10:00:13

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

Explanation:

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

families of curves, with similar characteristics.

( i ) a = 0 and n = 1.

$r = b cos (θ - α)$ $r = b cos ( theta - alpha )$

produce circles through the pole r = 0.

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

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

diameter inclined at $α = \frac{π}{3}$ $alpha = pi/3$ , to the initial line $θ = 0$ $theta = 0$ .

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) $a = 0 and α = 0$ $a = 0 and alpha = 0$ :

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

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

Example: $a = 2 and α = \frac{π}{4}$ $a = 2 and alpha = pi/4$ .

Cardioid Couple:
$r = 2 (1 \pm cos (θ - \frac{π}{4}))$ $r = 2 ( 1 +- cos (theta - pi/4 ))$
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 · 2018-08-05T14:19:04

Continuation, for the 2nd part.

Explanation:

(iv) n =1

Let $α = 0$ $alpha = 0$ .

$r = a + b cos θ$ $r = a + b cos theta$ create limacons.

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

$∣ ∣ \frac{a}{b} ∣ ∣ < 1 and > 1$ $abs( a/b) < 1 and > 1$ .

Graph for the limacons

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

$\frac{π}{4}$ $pi/4$ , 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 $∣ ∣ \frac{a}{b} ∣ ∣ ↑ ⏐ ⏐ ⏐$ $abs (a/b) uarr$ .

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

Example: $r = 1 + cos 3 (θ + \frac{π}{12}) and r = 1 + cos 3 θ$ $r = 1 + cos 3(theta + pi/12) and r = 1 + cos 3theta$ .
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 · 2018-08-06T02:41:45

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

Explanation:

(vi) $n = \frac{p}{q} and α = 0$ $n = p/q and alpha = 0$ , integers p and q are co-prime.

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

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

$r = cos (\frac{2}{3} (θ))$ $r = cos (2/3 ( theta))$ , of period $3 π$ $3pi$ .

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

giving two $(2 (3 π)$ $(2 (3pi)$ 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: $b = 1, p = 4, q = 3 and α = 0$ $b = 1, p = 4, q = 3 and alpha = 0$ , giving

$r = cos ((\frac{4}{3}) θ)$ $r = cos ( (4/3)theta )$ , with period $\frac{3}{2} π$ $3/2pi$
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.

Science

Math

Humanities

... and beyond

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

3 Answers

Explanation:

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

Citing examples and making graphs, how do you explain that f(r, theta;a,b,n,alpha )=r-(a + b cos n(theta- alpha) )=0f(r,θ;a,b,n,α)=r−(a+bcosn(θ−α))=0f(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

Explanation:

Explanation:

Explanation:

Related questions

Impact of this question

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