1.1 Parametric equations (Page 2/14)

Calculus volume 3 Page 2 / 14

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 $x (t)$ and $y (t)$ generate a set of ordered pairs $(x, y) .$ 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 $x (t)$ and $y (t) .$

Graphing a parametrically defined curve

Sketch the curves described by the following parametric equations:

$x (t) = t - 1, y (t) = 2 t + 4, −3 \leq t \leq 2$
$x (t) = t^{2} - 3, y (t) = 2 t + 1, −2 \leq t \leq 3$
$x (t) = 4 cos t, y (t) = 4 sin t, 0 \leq t \leq 2 π$

To create a graph of this curve, first set up a table of values. Since the independent variable in both $x (t)$ and $y (t)$ is t , let t appear in the first column. Then $x (t)$ and $y (t)$ will appear in the second and third columns of the table.

t $x (t)$ $y (t)$

−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.

Graph of the plane curve described by the parametric equations in part a.
To create a graph of this curve, again set up a table of values.

t $x (t)$ $y (t)$

−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 $t = −2)$ has coordinates $(1, −3),$ and the last point (corresponding to $t = 3)$ has coordinates $(6, 7) .$ 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.

Graph of the plane curve described by the parametric equations in part b.

t	$x (t)$	$y (t)$
−3	−4	−2
−2	−3	0
−1	−2	2
0	−1	4
1	0	6
2	1	8

t	$x (t)$	$y (t)$
−2	1	−3
−1	−2	−1
0	−3	1
1	−2	3
2	1	5
3	6	7

In this case, use multiples of

π / 6

for t and create another table of values:

t	$x (t)$	$y (t)$		t	$x (t)$	$y (t)$
0	4	0		$\frac{7 π}{6}$	$−2 \sqrt{3} \approx −3.5$	2
$\frac{π}{6}$	$2 \sqrt{3} \approx 3.5$	$2$	$\frac{4 π}{3}$	−2	$−2 \sqrt{3} \approx −3.5$
$\frac{π}{3}$	$2$	$2 \sqrt{3} \approx 3.5$	$\frac{3 π}{2}$	0	−4
$\frac{π}{2}$	0	4	$\frac{5 π}{3}$	2	$−2 \sqrt{3} \approx −3.5$
$\frac{2 π}{3}$	−2	$2 \sqrt{3} \approx 3.5$	$\frac{11 π}{6}$	$2 \sqrt{3} \approx 3.5$	2
$\frac{5 π}{6}$	$−2 \sqrt{3} \approx −3.5$	2	$2 π$	4	0
$π$	−4	0

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

A circle with radius 4 centered at the origin is graphed with arrow going counterclockwise. The point (4, 0) is marked t = 0, the point (0, 4) is marked t = π/2, the point (−4, 0) is marked t = π, and the point (0, −4) is marked t = 3π/2. On the graph there are also written three equations: x(t) = 4 cos(t), y(t) = 4 sin(t), and 0 ≤ t ≤ 2π. — Graph of the plane curve described by the parametric equations in part c.

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

(4, 0) .

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Sketch the curve described by the parametric equations

x (t) = 3 t + 2, y (t) = t^{2} - 1, −3 \leq t \leq 2 .

A curved line going from (−7, 8) through (−1, 0) and (2, −1) to (8, 3) with arrow going in that order. The point (−7, 8) is marked t = −3, the point (2, −1) is marked t = 0, and the point (8, 3) is marked t = 2. On the graph there are also written three equations: x(t) = 3t + 2, y(t) = t2 − 1, and −3 ≤ t ≤ 2.

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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 (t) = t^{2} - 3, y (t) = 2 t + 1, −2 \leq t \leq 3 .

Solving the second equation for t gives

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

This can be substituted into the first equation:

x = {(\frac{y - 1}{2})}^{2} - 3 = \frac{y^{2} - 2 y + 1}{4} - 3 = \frac{y^{2} - 2 y - 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 $(1, −3)$ and ended at $(6, 7) .$ These terminations were due to the restriction on the parameter t.

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Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

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