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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 ( 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:

  1. x ( t ) = t 1 , y ( t ) = 2 t + 4 , −3 t 2
  2. x ( t ) = t 2 3 , y ( t ) = 2 t + 1 , −2 t 3
  3. x ( t ) = 4 cos t , y ( t ) = 4 sin t , 0 t 2 π
  1. 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.
    A straight line going from (−4, −2) through (−3, 0), (−2, 2), and (0, 6) to (1, 8) with arrow pointed up and to the right. The point (−4, −2) is marked t = −3, the point (−2, 2) is marked t = −1, and the point (1, 8) is marked t = 2. On the graph there are also written three equations: x(t) = t −1, y(t) = 2t + 4, and −3 ≤ t ≤ 2.
    Graph of the plane curve described by the parametric equations in part a.
  2. 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.
    A curved line going from (1, −3) through (−3, 1) to (6, 7) with arrow pointing in that order. The point (1, −3) is marked t = −2, the point (−3, 1) is marked t = 0, and the point (6, 7) is marked t = 3. On the graph there are also written three equations: x(t) = t2 − 3, y(t) = 2t + 1, and −2 ≤ t ≤ 3.
    Graph of the plane curve described by the parametric equations in part b.
  3. 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 7 π 6 −2 3 −3.5 2
    π 6 2 3 3.5 2 4 π 3 −2 −2 3 −3.5
    π 3 2 2 3 3.5 3 π 2 0 −4
    π 2 0 4 5 π 3 2 −2 3 −3.5
    2 π 3 −2 2 3 3.5 11 π 6 2 3 3.5 2
    5 π 6 −2 3 −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 t 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 t 3 .

Solving the second equation for t gives

t = y 1 2 .

This can be substituted into the first equation:

x = ( y 1 2 ) 2 3 = y 2 2 y + 1 4 3 = 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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