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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
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