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RE: Re: parametricplot
*To*: mathgroup at smc.vnet.net
*Subject*: [mg23749] RE: [mg23705] Re: parametricplot
*From*: "David Park" <djmp at earthlink.net>
*Date*: Mon, 5 Jun 2000 01:09:44 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
> From: LIM.Caroline [mailto:LIM.Caroline at wanadoo.fr]
To: mathgroup at smc.vnet.net
> I've got to draw this function with parametricplot :
>
> The function was
> x(t)=0
> y(t)=(-4*(t^2+1))/(-1+8*t^2*(t^2+1))^1/2
>
Caroline?
I believe that the functions you were intending are
x[t_] = 0;
y[t_] = (-4*(t^2 + 1))/(-1 + 8*t^2*(t^2 + 1))^(1/2)
A pair of parentheses is needed around the 1/2 if you intend to take the
square root. A parametric plot of this is not very interesting. It will just
be a portion of the y-axis. Since Mathematica normally plots in black, you
won't even see it! Another problem is that the y value will only be a real
number for certain values of t. To find out which values of t are allowed
use InequalitySolve.
Needs["Algebra`InequalitySolve`"]
InequalitySolve[-1 + 8*t^2*(t^2 + 1) >= 0, t]
t <= -Sqrt[-(1/2) + Sqrt[3/2]/2] ||
t >= Sqrt[-(1/2) + Sqrt[3/2]/2]
The following makes a parametric plot of the negative region of t. The
PlotStyle option was used to color the line drawn so as to distinguish it
from the y-axis. The option PlotRange -> All is necessary to show the actual
extent of the line. Otherwise, Mathematica does a very poor job, in this
case, of picking out what it thinks is the "interesting" portion.
ParametricPlot[{x[t], y[t]},
{t, -20, -Sqrt[-(1/2) + Sqrt[3/2]/2]},
PlotStyle -> Hue[1], PlotRange -> All];
We can find the actual extent of the line by evaluating the Limits of y[t].
For the negative region:
Limit[y[t], t -> -Infinity, Direction -> -1]
-Sqrt[2]
Limit[y[t], t -> -Sqrt[-(1/2) + Sqrt[3/2]/2],
Direction -> 1]
-Infinity
The positive range of t traces out the same line.
David Park
djmp at earthlink.net
http://home.earthlink.net/~djmp/
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