Deriving Parametric Equations for Cissoid of Diocles

Date: 02/12/2005 at 23:12:31
From: Amy
Subject: Parametric Equations

How do you derive the parametric equations for the Cissoid of Diocles?
So few variables are given in the problem, and using just cosine or
sine won't help you get it.  There's a combination.

Date: 02/13/2005 at 11:46:42
From: Doctor Jerry
Subject: Re: Parametric Equations

Hello Amy,

See this figure:

  http://mathforum.org/dr.math/gifs/cissoid.jpg 

Here's a derivation:

Using the above figure, we define the cissoid by choosing an arbitrary
point S with coordinates (2a,q) on the line x = 2a.  We find the point
of intersection of the line OS with the circle (x-a)^2 + y^2 = a^2 to be

  ( 8a^3/(4a^2 + q^2), 4a^2q/(4a^2 + q^2) ).

Corresponding to the point S, we define P as the point on the line
from O to S for which |OP| = |RS|.  With a bit of algebra, we find the
coordinates of P to be

  x = 2aq^2/(q^2+4a^2)

  y = q^3/(q^2+4a^2).

If one eliminates q from these equations, one finds 

  x^3 + x*y^2 = 2*a*y^2.

If one lets q = 2a*tan(t), motivated by the q^2 + 4a^2 in the 
denominator, one finds

  x = 2a*sin^2(t)

  y = 2a*sin^3(t)/cos(t).

If you have any questions about my comments, please write back.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/