Example 1 from Algebra and Trigonometry, 10.6 Parametric Equations

Parameterizing a Curve

Parameterize the curve $y = x^{2} - 1$ letting $x (t) = t .$ Graph both equations.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution

If $x (t) = t,$ then to find $y (t)$ we replace the variable $x$ with the expression given in $x (t) .$ In other words, $y (t) = t^{2} - 1.$ Make a table of values similar to Table 1, and sketch the graph.

$t$	$x (t)$	$y (t)$
$- 4$	$- 4$	$y (- 4) = {(- 4)}^{2} - 1 = 15$
$- 3$	$- 3$	$y (- 3) = {(- 3)}^{2} - 1 = 8$
$- 2$	$- 2$	$y (- 2) = {(- 2)}^{2} - 1 = 3$
$- 1$	$- 1$	$y (- 1) = {(- 1)}^{2} - 1 = 0$
$0$	$0$	$y (0) = {(0)}^{2} - 1 = - 1$
$1$	$1$	$y (1) = {(1)}^{2} - 1 = 0$
$2$	$2$	$y (2) = {(2)}^{2} - 1 = 3$
$3$	$3$	$y (3) = {(3)}^{2} - 1 = 8$
$4$	$4$	$y (4) = {(4)}^{2} - 1 = 15$

Table 1

See the graphs in Figure 3. It may be helpful to use the TRACE feature of a graphing calculator to see how the points are generated as $t$ increases.

Graph of a parabola in two forms: a parametric equation and rectangular coordinates. It is the same function, just different ways of writing it.

Figure 3 (a) Parametric

y (t) = t^{2} - 1

(b) Rectangular

y = x^{2} - 1

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Construct a table of values and plot the parametric equations: $x (t) = t - 3,$ $y (t) = 2 t + 4; - 1 \leq t \leq 2.$

How did it go?