The Roses of GrandiDate: 11/13/2000 at 10:58:42 From: Tracy Getz Subject: Roses of Grandi The Roses of Grandi are similar to the Folium of Descartes, but I could not find any information in your archives about them. Question: The Roses of Grandi are given by the polar equations r = a*sin(n*theta) r = a*cos(n*theta) where n is an integer. Show that the Roses of Grandi are algebraic. Date: 11/13/2000 at 15:02:53 From: Doctor Rob Subject: Re: Roses of Grandi Thanks for writing to Ask Dr. Math, Tracy. I don't think you need any information about them other than what is given to you above. The idea is to show that sin(n*theta) and cos(n*theta) can be expressed as polynomials in sin(theta) and cos(theta). I would do this by using the Principle of Mathematical Induction, combined with the sine/cosine of a sum trigonometric identity. Then you can substitute y/r for its equal sin(theta), and x/r for its equal cos(theta). That will remove all instances of the variable theta. Then multiply through by r^n to clear fractions. Replace powers of r by using r^(2*k) = (x^2+y^2)^k r^(2*k+1) = r*(x^2+y^2)^k That leaves r appearing only to the first power, at most. If r does appear, then solve for r, square both sides, and substitute x^2 + y^2 for its equal r^2. Now clear fractions, and you have a polynomial equation in x and y, so it is an algebraic curve. Example: r = a*sin(2*theta) r = a*2*sin(theta)*cos(theta) = 2*a*(y/r)*(x/r) = 2*a*x*y/r^2 r^3 = 2*a*x*y r*(x^2+y^2) = 2*a*x*y, r = 2*a*x*y/(x^2+y^2) r^2 = 4*a^2*x^2*y^2/(x^2+y^2)^2 (x^2+y^2)^3 = 4*a^2*x^2*y^2 The MacTutor History of Mathematics archive provides a Web page with information about them here: Rhodonea Curves http://www-history.mcs.st-and.ac.uk/history/Curves/Rhodonea.html - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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