Parametric Functions

How Mathematica can help us understand parametric functions.

Intrinsic Command: ParametricPlot

x[t_]=Cos[t]
y[t_]=Sin[3t]
ParametricPlot[{x[t],y[t]},{t,0,2Pi}]

Parametric Plots: determining direction curve is traversed via table of values

x[t_]=Cos[t]
y[t_]=Sin[3t]
datapts=Table[{x[t],y[t]},{t,0,2Pi,0.1}];
TableForm[datapts]

ListPlot[datapts,PlotStyle->PointSize[0.02],AspectRatio->1]

Parametric Plots: determining direction curve is traversed via animation

x[t_]=Cos[t]
y[t_]=Sin[3t]
plt[end_]:=ParametricPlot[{x[t],y[t]},{t,0,end},PlotRange->{{-1,1},{-1,1}}]

num=50;
h = 2Pi/num;
Manipulate[plt[a],{a,h,2Pi,h}]

Intrinsic Command: Eliminate (used to get an implicit representation of the function)

imp=Eliminate[{x==t^3+t^6,y==t^2+4},t]

Beautiful Curves

x[t_]=t+Sin[3t]
y[t_]=t+Sin[4t]
ParametricPlot[{x[t],y[t]},{t,-4Pi,4Pi},PlotStyle->{Red}]

x[t_]=t+Sin[6t]
y[t_]=-t+Sin[4t]
ParametricPlot[{x[t],y[t]},{t,-2Pi,2Pi},PlotStyle->{Red}]

x[t_]=Sqrt[t]+Sin[3t]
y[t_]=t+Sin[4t]
ParametricPlot[{x[t],y[t]},{t,0,6Pi},PlotStyle->{Red}]

x[t_]=t+Cos[3t]Sin[2t]
y[t_]=-Exp[-t^2]-Cos[Sin[t]]
ParametricPlot[{x[t],y[t]},{t,-2Pi,2Pi},PlotStyle->{Red}]

x[t_]=t-Exp[-t^2]+Sin[3t]
y[t_]=Exp[-t^2]-Sin[2t]
ParametricPlot[{x[t],y[t]},{t,-4Pi,4Pi},PlotStyle->{Red}]

x[t_]=t(Cos[t]+t Sin[t])
y[t_]=-8t Sin[2t]
ParametricPlot[{x[t],y[t]},{t,-4Pi,4Pi},PlotStyle->{Red}]

x[t_]=t(Cos[t]+t Sin[t])
y[t_]=-8t Sin[t]
ParametricPlot[{x[t],y[t]},{t,-4Pi,4Pi},PlotStyle->{Red}]

x[t_]=t(Cos[t]+t Sin[3t])
y[t_]=-8t Sin[t]
ParametricPlot[{x[t],y[t]},{t,-4Pi,4Pi},PlotStyle->{Red}]

Example 1: Derivatives of parametric functions

Use a graph to estimate the left-most point on the curve given by x=(t^4)-(t^2), y=t+ln(t). Then use calculus to find the exact corrdinates.

First, define the paramteric function:

x[t_]=t^4-t^2
y[t_]=t+Log[t]

Now, get a plot:

ParametricPlot[{x[t],y[t]},{t,0,1}]

ParametricPlot[{x[t],y[t]},{t,1/2,1}]

It looks like the left-most point is around (-0.25,0.4). To figure this out using calculus, we need the derivative (our graph shows that the derivative will be infiinte (vertical tangent) at the left-most point):

(dy/dx)=((dy/dt)/(dx/dt))=((y'(t))/(x'(t)))

You could work this out by hand, out use Mathematica to do it for you:

Simplify[y'[t]/x'[t]]

Setting the denominator equal to zero and solving for t will tell us for what values the derivative is infinite:

Solve[x'[t]==0,t]

The first two values are excluded, since the logarithm is not defined for t≤0. The left-most point is found by substituting in t=1/(Sqrt[2]):

x[1/Sqrt[2]]
y[1/Sqrt[2]]

If you want decimals (which woul be nice to compare to our earlier guess):

x[1/Sqrt[2.]]
y[1/Sqrt[2.]]

Example 2: Derivatives of parametric functions

Find the slope of the curve at the point (1,1) if x=(e^t), y=(((t-1))^2).

First, define the paramteric function:

Clear[x,y]
x[t_]=Exp[t]
y[t_]=(t-1)^2

We need to get the dertivative:

Simplify[y'[t]/x'[t]]

Now, we need to solve for the value of the parameter t which gives us the point (1,1):

Solve[{x[t]==1,y[t]==1},t]

This tells us that at t=0 the curve is passing through the point (1,1), so we need to evaluate the derivative at t=0:

Simplify[y'[t]/x'[t]]/.t->0

The equation of the tangent line is

y-(Subscript[y, 0])=m(x-(Subscript[x, 0]))
y-1=-2(x-1)
y=-2x+3

Let's do a couple of plots to see if this makes sense to us:

plot1=ParametricPlot[{x[t],y[t]},{t,-2,2}]

plot2=Plot[-2x+3,{x,-2,2},PlotStyle->{Red}]

Show[plot1,plot2]

Looks good!