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/