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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?
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