RE: polar curve

*To*: mathgroup at smc.vnet.net*Subject*: [mg32355] RE: [mg32330] polar curve*From*: "Florian Jaccard" <jaccardf at eicn.ch>*Date*: Wed, 16 Jan 2002 03:30:08 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

As dy/dx = y'(Theta)/x'(Theta) , if x'(Theta) = 0, then dy/dx is infinite and you have vertical tangents. In your example, it is the circle (x-2)^2+y^2=4. The points with x'(Theta) = 0 are the points (4;0) and (0;0), points with vertical tangents ! (Remember that x(Theta)=r*cos(Theta) and y(Theta)=r*sin(Theta) !) Meilleures salutations Florian Jaccard EICN-HES e-mail : jaccardf at eicn.ch -----Message d'origine----- De : TeKkEnXpErT at aol.com [mailto:TeKkEnXpErT at aol.com] Envoyé : mar., 15. janvier 2002 08:30 À : mathgroup at smc.vnet.net Objet : [mg32330] polar curve A polar curve defined by the equation r = 4cos(theta) -pie/2<=theta<=pie/2 Parametrize the curve by x(theta), y(theta). Find the points (x,y) on this curve at which x'(theta)=0. What can you say about the tangent at these points?