MathGroup Archive 2002

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: polar curve


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



  • Prev by Date: Re: Entering constant placeholders
  • Next by Date: Re: How To Change A Rule
  • Previous by thread: polar curve
  • Next by thread: Re: polar curve