Re: polar curve

*To*: mathgroup at smc.vnet.net*Subject*: [mg32341] Re: [mg32330] polar curve*From*: Mitsuhiro Arikawa <arikawa at mpipks-dresden.mpg.de>*Date*: Wed, 16 Jan 2002 03:29:48 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

on 15.01.2002 8:30 Uhr, TeKkEnXpErT at aol.com at TeKkEnXpErT at aol.com wrote: > > 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? > > > Hi, If you consider simpler problem, (x,y)=(cos(theta),sin(theta)) (0<theta<2Pi), then (x',y')=(-sin(theta),cos(theta)). x'=0 corresponds theta = pi or 3pi ... It corresponds local maximal or minimum. So If you draw x=x(y) in your problem, it also corresponds local maximal or minimum. arikawa