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RE: Re: parametricplot

  • To: mathgroup at
  • Subject: [mg23749] RE: [mg23705] Re: parametricplot
  • From: "David Park" <djmp at>
  • Date: Mon, 5 Jun 2000 01:09:44 -0400 (EDT)
  • Sender: owner-wri-mathgroup at

> From: LIM.Caroline [mailto:LIM.Caroline at]
To: mathgroup at

> 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


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.

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]
Limit[y[t], t -> -Sqrt[-(1/2) + Sqrt[3/2]/2],
  Direction -> 1]

The positive range of t traces out the same line.

David Park
djmp at

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