# 1.1 Parametric equations  (Page 3/14)

 Page 3 / 14

## Eliminating the parameter

Eliminate the parameter for each of the plane curves described by the following parametric equations and describe the resulting graph.

1. $x\left(t\right)=\sqrt{2t+4},\phantom{\rule{1em}{0ex}}y\left(t\right)=2t+1,\phantom{\rule{1em}{0ex}}-2\le t\le 6$
2. $x\left(t\right)=4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t,\phantom{\rule{1em}{0ex}}y\left(t\right)=3\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t,\phantom{\rule{1em}{0ex}}0\le t\le 2\pi$
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 $t=-2,$ $x=\sqrt{2\left(-2\right)+4}=0,$ and when $t=6,$ $x=\sqrt{2\left(6\right)+4}=4.$ The graph of this plane curve follows. Graph of the plane curve described by the parametric equations in part a.
2. Sometimes it is necessary to be a bit creative in eliminating the parameter. The parametric equations for this example are
$x\left(t\right)=4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y\left(t\right)=3\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t.$

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
$\text{cos}\phantom{\rule{0.2em}{0ex}}t=\frac{x}{4}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t=\frac{y}{3}.$

Now use the Pythagorean identity ${\text{cos}}^{2}t+{\text{sin}}^{2}t=1$ and replace the expressions for $\text{sin}\phantom{\rule{0.2em}{0ex}}t$ and $\text{cos}\phantom{\rule{0.2em}{0ex}}t$ 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. Graph of the plane curve described by the parametric equations in part b.
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.

Eliminate the parameter for the plane curve defined by the following parametric equations and describe the resulting graph.

$x\left(t\right)=2+\frac{3}{t},\phantom{\rule{1em}{0ex}}y\left(t\right)=t-1,\phantom{\rule{1em}{0ex}}2\le t\le 6$

$x=2+\frac{3}{y+1},$ or $y=-1+\frac{3}{x-2}.$ This equation describes a portion of a rectangular hyperbola centered at $\left(2,-1\right).$ So far we have seen the method of eliminating the parameter, assuming we know a set of parametric equations that describe a plane curve. What if we would like to start with the equation of a curve and determine a pair of parametric equations for that curve? This is certainly possible, and in fact it is possible to do so in many different ways for a given curve. The process is known as parameterization of a curve    .

## Parameterizing a curve

Find two different pairs of parametric equations to represent the graph of $y=2{x}^{2}-3.$

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

$x\left(t\right)=t,\phantom{\rule{1em}{0ex}}y\left(t\right)=2{t}^{2}-3.$

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 $y=2{x}^{2}-3,$ we can substitute $x\left(t\right)=3t-2$ for x. This gives

$\begin{array}{cc}\hfill y\left(t\right)& =2{\left(3t-2\right)}^{2}-2\hfill \\ & =2\left(9{t}^{2}-12t+4\right)-2\hfill \\ & =18{t}^{2}-24t+8-2\hfill \\ & =18{t}^{2}-24t+6.\hfill \end{array}$

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

$x\left(t\right)=3t-2\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y\left(t\right)=18{t}^{2}-24t+6.$

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