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MathGroup Archive 2000

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Re: parametric equations

  • To: mathgroup at smc.vnet.net
  • Subject: [mg22058] Re: [mg22021] parametric equations
  • From: "David Park" <djmp at earthlink.net>
  • Date: Fri, 11 Feb 2000 02:38:42 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

>Hi Everyone:
>
>I have two parameters,x==3 t, and y ==9t^2. I plot them and get my
>parabola.OK.
>Then I get my rectangular equation, i.e, y==x^2.
>Now I parametrize the rectangular equation ,y==x^2 and I get completely
>different parametric equations. Both plot the same graph. However,when I plug
>in values for t, in the first case,that is,t=Table[{t,{3 t, 9 t^2},{t,4/3}] I
>get x=4 and y =16.
>On the other hand when I do the same operation with the parametrized equation,
>t=Table[{t,{-1 + t, (-1 + t)^2, {t,4/3}] I get x=0, y=0. I used this very
>simple example because that how I discovered that my parametrization of
>complicated equations was not giving me the same results as the original
>parameters- even thought the graphs coincided. What am I doing wrong?
>
>Manuel
>
>

Manuel,

I find your terminology and Mathematica statements quite confusing. I hope that this
will help.

Here is one parametrization of your curve:

f1[t_] := {3 t, 9 t^2}

There is only one parameter. It is t. Now here is a second parametrization.

f2[t_] := {t, t^2}

It has only one parameter, t. Mathematically these are different curves, because they
have different parametrizations, although they have the same shape. If we want to
identify points on the two curves we can relate the corresponding t's in the two
parametrizations.

Solve[f1[t1] == f2[t2], t1][[1,1]]
t1 -> t2/3

To find the value corresponding to a specific value of the parameter in the second
parametrization:

f2[4]
{4, 16}

To evaluate the same point with the first parametrization we must use a different
value of the parameter:

f1[4/3]
{4, 16}

David Park
djmp at earthlink.net
http://home.earthlink.net/~djmp/



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