Part 2: Using Identities to Eliminate the Parameter

In many instances, trigonometric identities are used to find the equation in x and y corresponding to the graph of a given vector-valued function. In particular, the following identities occur frequently in this context:

cos2( t) +sin2( t) = 1 1+tan2( t) = sec2( t) 1-2sin2(t) = cos( 2t) cosh2( t) -sinh2( t) = 1 1+cot2( t) = csc2( t) 2cos2(t) -1 = cos( 2t)

In addition, we may also on occasion use identities such as 2sin(t) cos( t) = sin( 2t) and e^te^-t = 1.    

EXAMPLE 3    Sketch the curve parameterized by r(t) = á 3cos( t) ,2sin( t) ñ for t in [ 0,2p] .       

Solution: Since x = 3cos( t) and y = 2sin(t) , we begin with

cos2( t) +sin2( t) = 1

Substituting cos( t) = x/3 and sin( t) = y/2 thus yields

æ è x 3 ö ø 2   + æ è y 2 ö ø 2   = 1

If t = 0 or t = 2p, then x = 3 and y = 0. Thus, the initial and terminal points are the same and correspondingly the graph of r(t) = á 3cos( t) ,2sin( t) ñ for t in [ 0,2p] is the entire ellipse with counterclockwise orientation.

x2 9 + y2 4 = 1

       

A parameterization may trace a curve from the initial point to another point and then may retrace the curve back to where it started, as we will see in the next example.  In this case, the orientation of the curve is not well-defined.

EXAMPLE 4    Find the Cartesian equation and sketch the curve r( t) = á cos( 2t) ,cos(t) ñ for t in [ 0,2p] .       

Solution: Since x = cos( 2t) and y = cos(t) , we use a double angle identity:

cos( 2t)  =  2cos2( t) -1 x  =  2y2-1

Moreover, r( 0) = á cos( 0), cos( 0) ñ = á 1,1 ñ , r( p/2) = á cos( p) ,cos( p/2) ñ = á -1,0 ñ , and r(p) = á cos( 2p) , cos( p) ñ = á1,-1ñ. Also, r( 3p/2) = ácos(3p),cos(3p/2)ñ = á -1,0 ñ and r(2p) = á cos( 4p) , cos(2 p) ñ = á1,1ñ.  Thus, the curve starts at ( 1,1) , progresses to ( -1,1) , and then retraces the curve in returning to ( 1,1) . Consequently, the orientation is not well-defined.        

EXAMPLE 5    Find the Cartesian equation of the curve

r( q) = á 2sin(q) cos(q) ,2sin2( q) ñ,  q  in  [ 0,p]

Solution: Since 2sin( q) cos( q) = sin( 2q) and 2sin²( q) = 1-cos( 2q) , we can rewrite r as

r( q) = á sin( 2q),1-cos( 2q) ñ

which implies that x = sin( 2q) and y = 1-cos(2q) . Since

sin2( 2q) +cos2( 2q) = 1

and since cos( 2q) = 1-y, we must have

x2+( y-1)2 = 1

Thus, r( q) = á 2sin( q) cos( q), 2sin²( q) ñ is a circle with radius 1 centered at ( 0,1) .