Deriving Parametric Equations for Cissoid of DioclesDate: 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/ |
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