Parametric Curves

We can describe a curve by giving functions x(t) and y(t) instead of y=f(x). Here t is called the parameter. There is always a trivial way to do this: let x(t)=t and y(t)=f(t). We want to go beyond that.


  Clear[x,y,t]


  x[t_]:=2 t
  y[t_]:=t^2


  ParametricPlot[{x[t],y[t]},{t,-6,6}];

What does this look like in y=f(x) form? By inspection you can see that y[t]=1/4 (x[t])^2, so y=1/4 x^2. To get the same plot, as t ranges over -6 to 6, x (which is 2t) must range over -12 to 12.


  Plot[1/4 x^2,{x,-12,12}];


  Clear[x,y,t]
  x[t_]:=(Sin[t])^2
  y[t_]:=Cos[t]


  ParametricPlot[{x[t],y[t]},{t,0,Pi}];

Using some trigonometry, it is clear that x(t)=1-(cos(t))^2=1-(y(t))^2, so we have x=1-y^2.

Notice that without parametrizing the curve we would have trouble with this, because x=1-y^2 does not define a (single valued) function of x.

Here are some of the more common parametrized curves.


  Clear[x,y,t]
  x[t_]:=Cos[t]
  y[t_]:=Sin[t]
  ParametricPlot[{x[t],y[t]},{t,0,2 Pi},AspectRatio->Automatic];


  Clear[x,y,t]
  x[t_]:=3 Cos[t]
  y[t_]:=3 Sin[t]
  ParametricPlot[{x[t],y[t]},{t,0,2 Pi},AspectRatio->Automatic];


  Clear[x,y,t]
  x[t_]:=Cos[t]
  y[t_]:=2 Sin[t]
  ParametricPlot[{x[t],y[t]},{t,0,2 Pi},AspectRatio->Automatic];


  Clear[x,y,t]
  x[t_]:=3 Cos[t]
  y[t_]:=2 Sin[t]
  ParametricPlot[{x[t],y[t]},{t,0,2 Pi},AspectRatio->Automatic];


  Clear[x,y,t]
  x[t_]:=x0+t(x1-x0)
  y[t_]:=y0+t(y1-y0)


  x0=5;
  y0=-1;
  x1=-3;
  y1=4;
  p0={x0,y0};
  p1={x1,y1};


  plot1=ParametricPlot[{x[t],y[t]},{t,-3,3}];
  plot2=ListPlot[{p0,p1},PlotStyle->{PointSize[.01],RGBColor[1,0,0]},DisplayFunction->Identity];

The parametrization
x[t]:=x0+t(x1-x0)
y[t]:=y0+t(y1-y0)
is the line through the points (x0,y0) and (x1,y1).


  Show[plot1,plot2];

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