1.1 Parametric equations  (Page 3/14)

1. To eliminate the parameter, we can solve either of the equations for t. For example, solving the first equation for t gives
   
   $\begin{array}{ccc}\hfill x& =\hfill & \sqrt{2t+4}\hfill \\ \hfill x^2& =\hfill & 2t+4\hfill \\ \hfill x^2-4& =\hfill & 2t\hfill \\ \hfill t& =\hfill & \frac{x^2-4}{2}.\hfill \end{array}$
   
   Note that when we square both sides it is important to observe that $x\ge 0.$ Substituting $t=\frac{x^2-4}{2}$ this into $y\left(t\right)$ yields
   
   $\begin{array}{ccc}\hfill y\left(t\right)& =\hfill & 2t+1\hfill \\ \hfill y& =\hfill & 2\left(\frac{x^2-4}{2}\right)+1\hfill \\ \hfill y& =\hfill & x^2-4+1\hfill \\ \hfill y& =\hfill & x^2-3.\hfill \end{array}$
   
   This is the equation of a parabola opening upward. There is, however, a domain restriction because of the limits on the parameter t . When $t=\mathrm{-2},$ $x=\sqrt{2\left(\mathrm{-2}\right)+4}=0,$ and when $t=6,$ $x=\sqrt{2\left(6\right)+4}=4.$ The graph of this plane curve follows.
   

   

2. Sometimes it is necessary to be a bit creative in eliminating the parameter. The parametric equations for this example are
   
   $x\left(t\right)=4\text{cos}t\text{and}y\left(t\right)=3\text{sin}t.$
   
   Solving either equation for t directly is not advisable because sine and cosine are not one-to-one functions. However, dividing the first equation by 4 and the second equation by 3 (and suppressing the t ) gives us
   
   $\text{cos}t=\frac{x}{4}\text{and}\text{sin}t=\frac{y}{3}.$
   
   Now use the Pythagorean identity ${\text{cos}}^{2}t+{\text{sin}}^{2}t=1$ and replace the expressions for $\text{sin}t$ and $\text{cos}t$ with the equivalent expressions in terms of x and y . This gives
   
   $\begin{array}{ccc}\hfill {\left(\frac{x}{4}\right)}^2+{\left(\frac{y}{3}\right)}^2& =\hfill & 1\hfill \\ \hfill \frac{x^2}{16}+\frac{y^2}{9}& =\hfill & 1.\hfill \end{array}$
   
   This is the equation of a horizontal ellipse centered at the origin, with semimajor axis 4 and semiminor axis 3 as shown in the following graph.
   

   

   
   As t progresses from $0$ to $2\pi ,$ a point on the curve traverses the ellipse once, in a counterclockwise direction. Recall from the section opener that the orbit of Earth around the Sun is also elliptical. This is a perfect example of using parameterized curves to model a real-world phenomenon.

First, it is always possible to parameterize a curve by defining $x\left(t\right)=t,$ then replacing x with t in the equation for $y\left(t\right).$ This gives the parameterization

Since there is no restriction on the domain in the original graph, there is no restriction on the values of t.

We have complete freedom in the choice for the second parameterization. For example, we can choose $x\left(t\right)=3t-2.$ The only thing we need to check is that there are no restrictions imposed on x ; that is, the range of $x\left(t\right)$ is all real numbers. This is the case for $x\left(t\right)=3t-2.$ Now since $y=2x^2-3,$ we can substitute $x\left(t\right)=3t-2$ for x. This gives

Therefore, a second parameterization of the curve can be written as