<P>Thank you Mr. Etienne and Mr. Alan.</P>
<P>In fact, you have used the standard parametrization of a polar curve
defined by an explicit equation of the form r=f(t), i.e., x=f(t)cost and
y=f(t)sint.</P>
<P>I rewrote your solution for r^2=3cos(2t) in this way</P>
<P>\parametricplot{-45}{45}{2 t mul cos 3 mul sqrt\space t cos mul 2 t mul
cos 3 mul sqrt\space t sin mul}</P>
<P>\parametricplot{-45}{45}{2 t mul cos 3 mul sqrt\space neg t cos mul 2 t
mul cos 3 mul sqrt\space neg t sin mul}</P>
<P>because I have to draw many other curves and I didn't want to define a
command for the ratio everytime I have a new equation.</P>
<P>I noticed that if we delete the command \space in your solution it
doesn't work: I didn't know that we can use any other thing but the
definition of the functions in the argument of the plot commands.</P>
<P>Please could you tell me about the use of the command \space, or any
other of the same type, in the argument of \psplot or \paremetricplot? </P>
<P>Juan Leal</P>
<P>>\documentclass{article}</P>
<P>> \usepackage{pst-plot} </P>
<P>>\begin{document} </P>
<P>>\newcommand{\rau}{2 t mul cos 3 mul sqrt} </P>
<P>>\psset{unit=2cm} </P>
<P>>\begin{pspicture}(-3,-3)(3,3) </P>
<P>>\parametricplot{-45}{45}{\rau\space dup t cos mul exch t sin mul}
</P>
<P>>\parametricplot{-45}{45}{\rau\space neg dup t cos mul exch t sin
mul} </P>
<P>>\end{pspicture} </P>
<P>>\end{document} </P>