# 1.1 Parametric equations  (Page 2/14)

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Notice in this definition that x and y are used in two ways. The first is as functions of the independent variable t. As t varies over the interval I , the functions $x\left(t\right)$ and $y\left(t\right)$ generate a set of ordered pairs $\left(x,y\right).$ This set of ordered pairs generates the graph of the parametric equations. In this second usage, to designate the ordered pairs, x and y are variables. It is important to distinguish the variables x and y from the functions $x\left(t\right)$ and $y\left(t\right).$

## Graphing a parametrically defined curve

Sketch the curves described by the following parametric equations:

1. $x\left(t\right)=t-1,\phantom{\rule{1em}{0ex}}y\left(t\right)=2t+4,\phantom{\rule{1em}{0ex}}-3\le t\le 2$
2. $x\left(t\right)={t}^{2}-3,\phantom{\rule{1em}{0ex}}y\left(t\right)=2t+1,\phantom{\rule{1em}{0ex}}-2\le t\le 3$
3. $x\left(t\right)=4\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t,\phantom{\rule{1em}{0ex}}y\left(t\right)=4\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t,\phantom{\rule{1em}{0ex}}0\le t\le 2\pi$
1. To create a graph of this curve, first set up a table of values. Since the independent variable in both $x\left(t\right)$ and $y\left(t\right)$ is t , let t appear in the first column. Then $x\left(t\right)$ and $y\left(t\right)$ will appear in the second and third columns of the table.
t $x\left(t\right)$ $y\left(t\right)$
−3 −4 −2
−2 −3 0
−1 −2 2
0 −1 4
1 0 6
2 1 8

The second and third columns in this table provide a set of points to be plotted. The graph of these points appears in [link] . The arrows on the graph indicate the orientation    of the graph, that is, the direction that a point moves on the graph as t varies from −3 to 2.
2. To create a graph of this curve, again set up a table of values.
t $x\left(t\right)$ $y\left(t\right)$
−2 1 −3
−1 −2 −1
0 −3 1
1 −2 3
2 1 5
3 6 7

The second and third columns in this table give a set of points to be plotted ( [link] ). The first point on the graph (corresponding to $t=-2\right)$ has coordinates $\left(1,-3\right),$ and the last point (corresponding to $t=3\right)$ has coordinates $\left(6,7\right).$ As t progresses from −2 to 3, the point on the curve travels along a parabola. The direction the point moves is again called the orientation and is indicated on the graph.
3. In this case, use multiples of $\pi \text{/}6$ for t and create another table of values:
t $x\left(t\right)$ $y\left(t\right)$ t $x\left(t\right)$ $y\left(t\right)$
0 4 0 $\frac{7\pi }{6}$ $-2\sqrt{3}\approx -3.5$ 2
$\frac{\pi }{6}$ $2\sqrt{3}\approx 3.5$ $2$ $\frac{4\pi }{3}$ −2 $-2\sqrt{3}\approx -3.5$
$\frac{\pi }{3}$ $2$ $2\sqrt{3}\approx 3.5$ $\frac{3\pi }{2}$ 0 −4
$\frac{\pi }{2}$ 0 4 $\frac{5\pi }{3}$ 2 $-2\sqrt{3}\approx -3.5$
$\frac{2\pi }{3}$ −2 $2\sqrt{3}\approx 3.5$ $\frac{11\pi }{6}$ $2\sqrt{3}\approx 3.5$ 2
$\frac{5\pi }{6}$ $-2\sqrt{3}\approx -3.5$ 2 $2\pi$ 4 0
$\pi$ −4 0

The graph of this plane curve appears in the following graph.

This is the graph of a circle with radius 4 centered at the origin, with a counterclockwise orientation. The starting point and ending points of the curve both have coordinates $\left(4,0\right).$

Sketch the curve described by the parametric equations

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

## Eliminating the parameter

To better understand the graph of a curve represented parametrically, it is useful to rewrite the two equations as a single equation relating the variables x and y. Then we can apply any previous knowledge of equations of curves in the plane to identify the curve. For example, the equations describing the plane curve in [link] b. are

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

Solving the second equation for t gives

$t=\frac{y-1}{2}.$

This can be substituted into the first equation:

$x={\left(\frac{y-1}{2}\right)}^{2}-3=\frac{{y}^{2}-2y+1}{4}-3=\frac{{y}^{2}-2y-11}{4}.$

This equation describes x as a function of y. These steps give an example of eliminating the parameter . The graph of this function is a parabola opening to the right. Recall that the plane curve started at $\left(1,-3\right)$ and ended at $\left(6,7\right).$ These terminations were due to the restriction on the parameter t.

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