{"id":1960,"date":"2023-10-12T00:36:14","date_gmt":"2023-10-12T00:36:14","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/parametric-equations\/"},"modified":"2023-10-23T19:09:27","modified_gmt":"2023-10-23T19:09:27","slug":"parametric-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/parametric-equations\/","title":{"raw":"Parametric Equations","rendered":"Parametric Equations"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Find a rectangular equation for a curve defined parametrically.<\/li>\r\n \t<li>Find parametric equations for curves defined by rectangular equations.<\/li>\r\n<\/ul>\r\n<\/div>\r\nConsider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in Figure 1. At any moment, the moon is located at a particular spot relative to the planet. But how do we write and solve the equation for the position of the moon when the distance from the planet, the speed of the moon\u2019s orbit around the planet, and the speed of rotation around the sun are all unknowns? We can solve only for one variable at a time.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180914\/CNX_Precalc_Figure_08_06_0012.jpg\" alt=\"Illustration of a planet's circular orbit around the sun.\" width=\"487\" height=\"383\" \/> <b>Figure 1<\/b>[\/caption]\r\n\r\nIn this section, we will consider sets of equations given by [latex]x\\left(t\\right)[\/latex] and [latex]y\\left(t\\right)[\/latex] where [latex]t[\/latex] is the independent variable of time. We can use these parametric equations in a number of applications when we are looking for not only a particular position but also the direction of the movement. As we trace out successive values of [latex]t[\/latex], the orientation of the curve becomes clear. This is one of the primary advantages of using parametric equations: we are able to trace the movement of an object along a path according to time. We begin this section with a look at the basic components of parametric equations and what it means to parameterize a curve. Then we will learn how to eliminate the parameter, translate the equations of a curve defined parametrically into rectangular equations, and find the parametric equations for curves defined by rectangular equations.\r\n<h2>Parameterizing a Curve<\/h2>\r\nWhen an object moves along a curve\u2014or <strong>curvilinear path<\/strong>\u2014in a given direction and in a given amount of time, the position of the object in the plane is given by the <em>x-<\/em>coordinate and the <em>y-<\/em>coordinate. However, both [latex]x[\/latex] and [latex]y[\/latex]\r\nvary over time and so are functions of time. For this reason, we add another variable, the <strong>parameter<\/strong>, upon which both [latex]x[\/latex] and [latex]y[\/latex] are dependent functions. In the example in the section opener, the parameter is time, [latex]t[\/latex]. The [latex]x[\/latex] position of the moon at time, [latex]t[\/latex], is represented as the function [latex]x\\left(t\\right)[\/latex], and the [latex]y[\/latex] position of the moon at time, [latex]t[\/latex], is represented as the function [latex]y\\left(t\\right)[\/latex]. Together, [latex]x\\left(t\\right)[\/latex] and [latex]y\\left(t\\right)[\/latex] are called parametric equations, and generate an ordered pair [latex]\\left(x\\left(t\\right),y\\left(t\\right)\\right)[\/latex]. Parametric equations primarily describe motion and direction.\r\n\r\nWhen we parameterize a curve, we are translating a single equation in two variables, such as [latex]x[\/latex] and [latex]y [\/latex], into an equivalent pair of equations in three variables, [latex]x,y[\/latex], and [latex]t[\/latex]. One of the reasons we parameterize a curve is because the parametric equations yield more information: specifically, the direction of the object\u2019s motion over time.\r\n\r\nWhen we graph parametric equations, we can observe the individual behaviors of [latex]x[\/latex] and of [latex]y[\/latex]. There are a number of shapes that cannot be represented in the form [latex]y=f\\left(x\\right)[\/latex], meaning that they are not functions. For example, consider the graph of a circle, given as [latex]{r}^{2}={x}^{2}+{y}^{2}[\/latex]. Solving for [latex]y[\/latex] gives [latex]y=\\pm \\sqrt{{r}^{2}-{x}^{2}}[\/latex], or two equations: [latex]{y}_{1}=\\sqrt{{r}^{2}-{x}^{2}}[\/latex] and [latex]{y}_{2}=-\\sqrt{{r}^{2}-{x}^{2}}[\/latex]. If we graph [latex]{y}_{1}[\/latex] and [latex]{y}_{2}[\/latex] together, the graph will not pass the vertical line test, as shown in Figure 2. Thus, the equation for the graph of a circle is not a function.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180917\/CNX_Precalc_Figure_08_06_0022.jpg\" alt=\"Graph of a circle in the rectangular coordinate system - the vertical line test shows that the circle r^2 = x^2 + y^2 is not a function. The dotted red vertical line intersects the function in two places - it should only intersect in one place to be a function.\" width=\"487\" height=\"291\" \/> <b>Figure 2<\/b>[\/caption]\r\n\r\nHowever, if we were to graph each equation on its own, each one would pass the vertical line test and therefore would represent a function. In some instances, the concept of breaking up the equation for a circle into two functions is similar to the concept of creating parametric equations, as we use two functions to produce a non-function. This will become clearer as we move forward.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Parametric Equations<\/h3>\r\nSuppose [latex]t[\/latex] is a number on an interval, [latex]I[\/latex]. The set of ordered pairs, [latex]\\left(x\\left(t\\right),y\\left(t\\right)\\right)[\/latex], where [latex]x=f\\left(t\\right)[\/latex] and [latex]y=g\\left(t\\right)[\/latex], forms a plane curve based on the parameter [latex]t[\/latex]. The equations [latex]x=f\\left(t\\right)[\/latex] and [latex]y=g\\left(t\\right)[\/latex] are the parametric equations.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1: Parameterizing a Curve<\/h3>\r\nParameterize the curve [latex]y={x}^{2}-1[\/latex] letting [latex]x\\left(t\\right)=t[\/latex]. Graph both equations.\r\n\r\n[reveal-answer q=\"910055\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"910055\"]\r\n\r\nIf [latex]x\\left(t\\right)=t[\/latex], then to find [latex]y\\left(t\\right)[\/latex] we replace the variable [latex]x[\/latex] with the expression given in [latex]x\\left(t\\right)[\/latex]. In other words, [latex]y\\left(t\\right)={t}^{2}-1[\/latex]. Make a table of values similar to the table below, and sketch the graph.\r\n<table id=\"Table_08_06_001\" summary=\"Ten rows and three columns. First column is labeled t, second column is labeled x(t), third column is labeled y(t). The table has ordered triples of each of these row values: (-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).\">\r\n<thead>\r\n<tr>\r\n<th>[latex]t[\/latex]<\/th>\r\n<th>[latex]x\\left(t\\right)[\/latex]<\/th>\r\n<th>[latex]y\\left(t\\right)[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]-4[\/latex]<\/td>\r\n<td>[latex]-4[\/latex]<\/td>\r\n<td>[latex]y\\left(-4\\right)={\\left(-4\\right)}^{2}-1=15[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3[\/latex]<\/td>\r\n<td>[latex]-3[\/latex]<\/td>\r\n<td>[latex]y\\left(-3\\right)={\\left(-3\\right)}^{2}-1=8[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2[\/latex]<\/td>\r\n<td>[latex]-2[\/latex]<\/td>\r\n<td>[latex]y\\left(-2\\right)={\\left(-2\\right)}^{2}-1=3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1[\/latex]<\/td>\r\n<td>[latex]-1[\/latex]<\/td>\r\n<td>[latex]y\\left(-1\\right)={\\left(-1\\right)}^{2}-1=0[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]y\\left(0\\right)={\\left(0\\right)}^{2}-1=-1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]y\\left(1\\right)={\\left(1\\right)}^{2}-1=0[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]y\\left(2\\right)={\\left(2\\right)}^{2}-1=3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3[\/latex]<\/td>\r\n<td>[latex]3[\/latex]<\/td>\r\n<td>[latex]y\\left(3\\right)={\\left(3\\right)}^{2}-1=8[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]4[\/latex]<\/td>\r\n<td>[latex]4[\/latex]<\/td>\r\n<td>[latex]y\\left(4\\right)={\\left(4\\right)}^{2}-1=15[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSee 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 [latex]t[\/latex] increases.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180919\/CNX_Precalc_Figure_08_06_0152.jpg\" alt=\"Graph of a parabola in two forms: a parametric equation and rectangular coordinates. It is the same function, just different ways of writing it.\" width=\"731\" height=\"291\" \/> <b>Figure 3.<\/b> (a) Parametric [latex]y\\left(t\\right)={t}^{2}-1[\/latex] (b) Rectangular [latex]y={x}^{2}-1[\/latex][\/caption]\r\n<h4>Analysis of the Solution<\/h4>\r\nThe arrows indicate the direction in which the curve is generated. Notice the curve is identical to the curve of [latex]y={x}^{2}-1[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nConstruct a table of values and plot the parametric equations: [latex]x\\left(t\\right)=t - 3,y\\left(t\\right)=2t+4;-1\\le t\\le 2[\/latex].\r\n\r\n[reveal-answer q=\"217668\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"217668\"]\r\n<table id=\"fs-id1165137810323\" class=\"unnumbered\" summary=\"Five rows and three columns. First column is labeled t, second column is labeled x(t), third column is labeled y(t). The table has ordered triples of each of these row values: (-1, -4, 2), (0,-3,4), (1,-2,6), (2,-1,8).\">\r\n<tbody>\r\n<tr>\r\n<td>[latex]t[\/latex]<\/td>\r\n<td>[latex]x\\left(t\\right)[\/latex]<\/td>\r\n<td>[latex]y\\left(t\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1[\/latex]<\/td>\r\n<td>[latex]-4[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]-3[\/latex]<\/td>\r\n<td>[latex]4[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]-2[\/latex]<\/td>\r\n<td>[latex]6[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]-1[\/latex]<\/td>\r\n<td>[latex]8[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180940\/CNX_Precalc_Figure_08_06_0062.jpg\" alt=\"\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Finding a Pair of Parametric Equations<\/h3>\r\nFind a pair of parametric equations that models the graph of [latex]y=1-{x}^{2}[\/latex], using the parameter [latex]x\\left(t\\right)=t[\/latex]. Plot some points and sketch the graph.\r\n\r\n[reveal-answer q=\"141955\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"141955\"]\r\n\r\nIf [latex]x\\left(t\\right)=t[\/latex] and we substitute [latex]t[\/latex] for [latex]x[\/latex] into the [latex]y[\/latex] equation, then [latex]y\\left(t\\right)=1-{t}^{2}[\/latex]. Our pair of parametric equations is\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x\\left(t\\right)&amp;=t\\\\ y\\left(t\\right)&amp;=1-{t}^{2}\\end{align}[\/latex]<\/p>\r\nTo graph the equations, first we construct a table of values like that in the table below. We can choose values around [latex]t=0[\/latex], from [latex]t=-3[\/latex] to [latex]t=3[\/latex]. The values in the [latex]x\\left(t\\right)[\/latex] column will be the same as those in the [latex]t[\/latex] column because [latex]x\\left(t\\right)=t[\/latex]. Calculate values for the column [latex]y\\left(t\\right)[\/latex].\r\n<table id=\"Table_08_06_02\" summary=\"Eight rows and three columns. First column is labeled t, second column is labeled x(t)=t, third column is labeled y(t)=1-t^2. The table has ordered triples of each of these row values: (-3,-3, y(-3) = 1 - (-3)^2 = -8 ), (-2,-2, y(-2) = 1 - (-2)^2 = -3), (-1, -1, y(-1) = 1 - (-1)^2 = 0), (0,0, y(0) = 1 - 0 = 1), (1,1, y(1) = 1 - (1)^2 = 0), (2,2, y(2) = 1 - (2)^2 = -3), (3,3, y(3) = 1 - (3)^2 = -8).\">\r\n<thead>\r\n<tr>\r\n<th>[latex]t[\/latex]<\/th>\r\n<th>[latex]x\\left(t\\right)=t[\/latex]<\/th>\r\n<th>[latex]y\\left(t\\right)=1-{t}^{2}[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]-3[\/latex]<\/td>\r\n<td>[latex]-3[\/latex]<\/td>\r\n<td>[latex]y\\left(-3\\right)=1-{\\left(-3\\right)}^{2}=-8[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2[\/latex]<\/td>\r\n<td>[latex]-2[\/latex]<\/td>\r\n<td>[latex]y\\left(-2\\right)=1-{\\left(-2\\right)}^{2}=-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1[\/latex]<\/td>\r\n<td>[latex]-1[\/latex]<\/td>\r\n<td>[latex]y\\left(-1\\right)=1-{\\left(-1\\right)}^{2}=0[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]y\\left(0\\right)=1 - 0=1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]y\\left(1\\right)=1-{\\left(1\\right)}^{2}=0[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]y\\left(2\\right)=1-{\\left(2\\right)}^{2}=-3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3[\/latex]<\/td>\r\n<td>[latex]3[\/latex]<\/td>\r\n<td>[latex]y\\left(3\\right)=1-{\\left(3\\right)}^{2}=-8[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe graph of [latex]y=1-{t}^{2}[\/latex] is a parabola facing downward, as shown in Figure 4. We have mapped the curve over the interval [latex]\\left[-3,3\\right][\/latex], shown as a solid line with arrows indicating the orientation of the curve according to [latex]t[\/latex]. Orientation refers to the path traced along the curve in terms of increasing values of [latex]t[\/latex]. As this parabola is symmetric with respect to the line [latex]x=0[\/latex], the values of [latex]x[\/latex] are reflected across the y-axis.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180922\/CNX_Precalc_Figure_08_06_0072.jpg\" alt=\"Graph of given downward facing parabola.\" width=\"487\" height=\"516\" \/> <b>Figure 4<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nParameterize the curve given by [latex]x={y}^{3}-2y[\/latex].\r\n\r\n[reveal-answer q=\"560553\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"560553\"]\r\n\r\n[latex]\\begin{align}x\\left(t\\right)&amp;={t}^{3}-2t\\\\ y\\left(t\\right)&amp;=t\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3: Finding Parametric Equations That Model Given Criteria<\/h3>\r\nAn object travels at a steady rate along a straight path [latex]\\left(-5,3\\right)[\/latex] to [latex]\\left(3,-1\\right)[\/latex] in the same plane in four seconds. The coordinates are measured in meters. Find parametric equations for the position of the object.\r\n\r\n[reveal-answer q=\"419034\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"419034\"]\r\n\r\nThe parametric equations are simple linear expressions, but we need to view this problem in a step-by-step fashion. The <em>x<\/em>-value of the object starts at [latex]-5[\/latex] meters and goes to 3 meters. This means the distance <em>x<\/em> has changed by 8 meters in 4 seconds, which is a rate of [latex]\\frac{\\text{8 m}}{4\\text{ s}}[\/latex], or [latex]2\\text{m}\/\\text{s}[\/latex]. We can write the <em>x<\/em>-coordinate as a linear function with respect to time as [latex]x\\left(t\\right)=2t - 5[\/latex]. In the linear function template [latex]y=mx+b,2t=mx[\/latex] and [latex]-5=b[\/latex].\r\n\r\nSimilarly, the <em>y<\/em>-value of the object starts at 3 and goes to [latex]-1[\/latex], which is a change in the distance <em>y<\/em> of \u22124 meters in 4 seconds, which is a rate of [latex]\\frac{-4\\text{ m}}{4\\text{ s}}[\/latex], or [latex]-1\\text{m}\/\\text{s}[\/latex]. We can also write the <em>y<\/em>-coordinate as the linear function [latex]y\\left(t\\right)=-t+3[\/latex]. Together, these are the parametric equations for the position of the object, where [latex]x[\/latex]\u00a0and [latex]y[\/latex]\u00a0are expressed in meters and [latex]t[\/latex]\u00a0represents time:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x\\left(t\\right)&amp;=2t - 5 \\\\ y\\left(t\\right)&amp;=-t+3 \\end{align}[\/latex]<\/p>\r\nUsing these equations, we can build a table of values for [latex]t,x[\/latex], and [latex]y[\/latex]. In this example, we limited values of [latex]t[\/latex] to non-negative numbers. In general, any value of [latex]t[\/latex] can be used.\r\n<table id=\"Table_08_06_03\" summary=\"Six rows and three columns. First column is labeled t, second column is labeled x(t)=2t-5, third column is labeled y(t)=-t+3. The table has ordered triples of each of these row values: (0, x=2(0)-5 = -5, y=-(0) +3 = 3), (1, x=2(1)-5 = -3, y=-(1) + 3 = 2), (2, x=2(2) - 5 = -1, y=-(2) + 3 = 1), (3, x=2(3) - 5 = 1, y = -(3) + 3 =0), (4, x=2(4) -5 = 3, y=-(4) + 3 = -1).\">\r\n<thead>\r\n<tr>\r\n<th>[latex]t[\/latex]<\/th>\r\n<th>[latex]x\\left(t\\right)=2t - 5[\/latex]<\/th>\r\n<th>[latex]y\\left(t\\right)=-t+3[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]x=2\\left(0\\right)-5=-5[\/latex]<\/td>\r\n<td>[latex]y=-\\left(0\\right)+3=3[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]x=2\\left(1\\right)-5=-3[\/latex]<\/td>\r\n<td>[latex]y=-\\left(1\\right)+3=2[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]x=2\\left(2\\right)-5=-1[\/latex]<\/td>\r\n<td>[latex]y=-\\left(2\\right)+3=1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3[\/latex]<\/td>\r\n<td>[latex]x=2\\left(3\\right)-5=1[\/latex]<\/td>\r\n<td>[latex]y=-\\left(3\\right)+3=0[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]4[\/latex]<\/td>\r\n<td>[latex]x=2\\left(4\\right)-5=3[\/latex]<\/td>\r\n<td>[latex]y=-\\left(4\\right)+3=-1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFrom this table, we can create three graphs, as shown in Figure 5.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180925\/CNX_Precalc_Figure_08_06_0032.jpg\" alt=\"Three graphs side by side. (A) has the horizontal position over time, (B) has the vertical position over time, and (C) has the position of the object in the plane at time t. See caption for more information.\" width=\"975\" height=\"514\" \/> <b>Figure 5.<\/b> (a) A graph of [latex]x[\/latex] vs. [latex]t[\/latex], representing the horizontal position over time. (b) A graph of [latex]y[\/latex] vs. [latex]t[\/latex], representing the vertical position over time. (c) A graph of [latex]y[\/latex] vs. [latex]x[\/latex], representing the position of the object in the plane at time [latex]t[\/latex].[\/caption]\r\n<h4>Analysis of the Solution<\/h4>\r\nAgain, we see that, in Figure 5(c), when the parameter represents time, we can indicate the movement of the object along the path with arrows.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div>\r\n<h2>Methods for Finding Cartesian and Polar Equations from Curves<\/h2>\r\nIn many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as [latex]x[\/latex] and [latex]y[\/latex]. Eliminating the parameter is a method that may make graphing some curves easier. However, if we are concerned with the mapping of the equation according to time, then it will be necessary to indicate the orientation of the curve as well. There are various methods for eliminating the parameter [latex]t[\/latex] from a set of parametric equations; not every method works for every type of equation. Here we will review the methods for the most common types of equations.\r\n<h2>Eliminating the Parameter from Polynomial, Exponential, and Logarithmic Equations<\/h2>\r\nFor polynomial, exponential, or logarithmic equations expressed as two parametric equations, we choose the equation that is most easily manipulated and solve for [latex]t[\/latex]. We substitute the resulting expression for [latex]t[\/latex]\u00a0into the second equation. This gives one equation in [latex]x[\/latex] and [latex]y[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4: Eliminating the Parameter in Polynomials<\/h3>\r\nGiven [latex]x\\left(t\\right)={t}^{2}+1[\/latex] and [latex]y\\left(t\\right)=2+t[\/latex], eliminate the parameter, and write the parametric equations as a Cartesian equation.\r\n\r\n[reveal-answer q=\"514172\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"514172\"]\r\n\r\nWe will begin with the equation for [latex]y[\/latex] because the linear equation is easier to solve for [latex]t[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}y=2+t \\\\ y - 2=t\\end{gathered}[\/latex]<\/p>\r\nNext, substitute [latex]y - 2[\/latex] for [latex]t[\/latex] in [latex]x\\left(t\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;x={t}^{2}+1 \\\\ &amp;x={\\left(y - 2\\right)}^{2}+1&amp;&amp; \\text{Substitute the expression for }t\\text{ into }x. \\\\ &amp;x={y}^{2}-4y+4+1 \\\\ &amp;x={y}^{2}-4y+5 \\\\ &amp;x={y}^{2}-4y+5 \\end{align}[\/latex]<\/p>\r\nThe Cartesian form is [latex]x={y}^{2}-4y+5[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nThis is an equation for a parabola in which, in rectangular terms, [latex]x[\/latex] is dependent on [latex]y[\/latex]. From the curve\u2019s vertex at [latex]\\left(1,2\\right)[\/latex], the graph sweeps out to the right. See Figure 6. In this section, we consider sets of equations given by the functions [latex]x\\left(t\\right)[\/latex] and [latex]y\\left(t\\right)[\/latex], where [latex]t[\/latex] is the independent variable of time. Notice, both [latex]x[\/latex] and [latex]y[\/latex] are functions of time; so in general [latex]y[\/latex] is not a function of [latex]x[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180928\/CNX_Precalc_Figure_08_06_0082.jpg\" alt=\"Graph of given sideways (extending to the right) parabola.\" width=\"731\" height=\"366\" \/> <b>Figure 6<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nGiven the equations below, eliminate the parameter and write as a rectangular equation for [latex]y[\/latex] as a function\u00a0of [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;x\\left(t\\right)=2{t}^{2}+6 \\\\ &amp;y\\left(t\\right)=5-t\\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">[reveal-answer q=\"820133\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"820133\"]<\/p>\r\n<p style=\"text-align: left;\">[latex]y=5-\\sqrt{\\frac{1}{2}x - 3}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173885[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5: Eliminating the Parameter in Exponential Equations<\/h3>\r\nEliminate the parameter and write as a Cartesian equation: [latex]x\\left(t\\right)={e}^{-t}[\/latex] and [latex]y\\left(t\\right)=3{e}^{t},t&gt;0[\/latex].\r\n\r\n[reveal-answer q=\"397095\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"397095\"]\r\n\r\nIsolate [latex]{e}^{t}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;x={e}^{-t} \\\\ &amp;{e}^{t}=\\frac{1}{x} \\end{align}[\/latex]<\/p>\r\nSubstitute the expression into [latex]y\\left(t\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;y=3{e}^{t} \\\\ &amp;y=3\\left(\\frac{1}{x}\\right)\\\\ &amp;y=\\frac{3}{x} \\end{align}[\/latex]<\/p>\r\nThe Cartesian form is [latex]y=\\frac{3}{x}[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nThe graph of the parametric equation is shown in Figure 7(a). The domain is restricted to [latex]t&gt;0[\/latex]. The Cartesian equation, [latex]y=\\frac{3}{x}[\/latex] is shown in Figure 7(b) and has only one restriction on the domain, [latex]x\\ne 0[\/latex].\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180931\/CNX_Precalc_Figure_08_06_009n2.jpg\" alt=\"&quot;Graph\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 6: Eliminating the Parameter in Logarithmic Equations<\/h3>\r\nEliminate the parameter and write as a Cartesian equation: [latex]x\\left(t\\right)=\\sqrt{t}+2[\/latex] and [latex]y\\left(t\\right)=\\mathrm{log}\\left(t\\right)[\/latex].\r\n\r\n[reveal-answer q=\"310238\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"310238\"]\r\n\r\nSolve the first equation for [latex]t[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;x=\\sqrt{t}+2 \\\\ &amp;x - 2=\\sqrt{t} \\\\ &amp;{\\left(x - 2\\right)}^{2}=t&amp;&amp; \\text{Square both sides}. \\end{align}[\/latex]<\/p>\r\nThen, substitute the expression for [latex]t[\/latex] into the [latex]y[\/latex] equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;y=\\mathrm{log}\\left(t\\right)\\\\ &amp;y=\\mathrm{log}{\\left(x - 2\\right)}^{2}\\end{align}[\/latex]<\/p>\r\nThe Cartesian form is [latex]y=\\mathrm{log}{\\left(x - 2\\right)}^{2}[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nTo be sure that the parametric equations are equivalent to the Cartesian equation, check the domains. The parametric equations restrict the domain on [latex]x=\\sqrt{t}+2[\/latex] to [latex]t&gt;0[\/latex]; we restrict the domain on [latex]x[\/latex] to [latex]x&gt;2[\/latex]. The domain for the parametric equation [latex]y=\\mathrm{log}\\left(t\\right)[\/latex] is restricted to [latex]t&gt;0[\/latex]; we limit the domain on [latex]y=\\mathrm{log}{\\left(x - 2\\right)}^{2}[\/latex] to [latex]x&gt;2[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nEliminate the parameter and write as a <strong>rectangular equation<\/strong>.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;x\\left(t\\right)={t}^{2} \\\\ &amp;y\\left(t\\right)=\\mathrm{ln}t,t&gt;0\\end{align}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">[reveal-answer q=\"205546\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"205546\"]<\/p>\r\n<p style=\"text-align: left;\">[latex]y=\\mathrm{ln}\\sqrt{x}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<h2>Eliminating the Parameter from Trigonometric Equations<\/h2>\r\nEliminating the parameter from trigonometric equations is a straightforward substitution. We can use a few of the familiar trigonometric identities and the Pythagorean Theorem.\r\n\r\nFirst, we use the identities:\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x\\left(t\\right)=a\\cos t\\\\ y\\left(t\\right)=b\\sin t\\end{gathered}[\/latex]<\/p>\r\nSolving for [latex]\\cos t[\/latex] and [latex]\\sin t[\/latex], we have\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\frac{x}{a}=\\cos t\\\\ \\frac{y}{b}=\\sin t\\end{gathered}[\/latex]<\/p>\r\nThen, use the Pythagorean Theorem:\r\n<p style=\"text-align: center;\">[latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex]<\/p>\r\nSubstituting gives\r\n<div style=\"text-align: center;\">[latex]{\\cos }^{2}t+{\\sin }^{2}t={\\left(\\frac{x}{a}\\right)}^{2}+{\\left(\\frac{y}{b}\\right)}^{2}=1[\/latex]<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 7: Eliminating the Parameter from a Pair of Trigonometric Parametric Equations<\/h3>\r\nEliminate the parameter from the given pair of <strong>trigonometric equations<\/strong> where [latex]0\\le t\\le 2\\pi [\/latex] and sketch the graph.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;x\\left(t\\right)=4\\cos t\\\\ &amp;y\\left(t\\right)=3\\sin t\\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"244567\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"244567\"]\r\n\r\nSolving for [latex]\\cos t[\/latex] and [latex]\\sin t[\/latex], we have\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x=4\\cos t \\\\ \\frac{x}{4}=\\cos t \\\\ y=3\\sin t \\\\ \\frac{y}{3}=\\sin t \\end{gathered}[\/latex]<\/p>\r\nNext, use the Pythagorean identity and make the substitutions.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered} {\\cos }^{2}t+{\\sin }^{2}t=1\\\\ {\\left(\\frac{x}{4}\\right)}^{2}+{\\left(\\frac{y}{3}\\right)}^{2}=1\\\\ \\frac{{x}^{2}}{16}+\\frac{{y}^{2}}{9}=1\\end{gathered}[\/latex]<\/p>\r\nThe graph for the equation is shown in Figure 8.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180934\/CNX_Precalc_Figure_08_06_0112.jpg\" alt=\"Graph of given ellipse centered at (0,0).\" width=\"487\" height=\"366\" \/> <b>Figure 8<\/b>[\/caption]\r\n<h4>Analysis of the Solution<\/h4>\r\nApplying the general equations for <strong>conic sections<\/strong>, we can identify [latex]\\frac{{x}^{2}}{16}+\\frac{{y}^{2}}{9}=1[\/latex] as an ellipse centered at [latex]\\left(0,0\\right)[\/latex]. Notice that when [latex]t=0[\/latex] the coordinates are [latex]\\left(4,0\\right)[\/latex], and when [latex]t=\\frac{\\pi }{2}[\/latex] the coordinates are [latex]\\left(0,3\\right)[\/latex]. This shows the orientation of the curve with increasing values of [latex]t[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nEliminate the parameter from the given pair of parametric equations and write as a Cartesian equation:\r\n<p style=\"text-align: center;\">[latex]x\\left(t\\right)=2\\cos t[\/latex] and [latex]y\\left(t\\right)=3\\sin t[\/latex].<\/p>\r\n[reveal-answer q=\"577648\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"577648\"]\r\n\r\n[latex]\\frac{{x}^{2}}{4}+\\frac{{y}^{2}}{9}=1[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]66927[\/ohm_question]\r\n\r\n<\/div>\r\nWhen we are given a set of parametric equations and need to find an equivalent Cartesian equation, we are essentially \"eliminating the parameter.\" However, there are various methods we can use to rewrite a set of parametric equations as a Cartesian equation. The simplest method is to set one equation equal to the parameter, such as [latex]x\\left(t\\right)=t[\/latex]. In this case, [latex]y\\left(t\\right)[\/latex] can be any expression. For example, consider the following pair of equations.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;x\\left(t\\right)=t\\\\ &amp;y\\left(t\\right)={t}^{2}-3\\end{align}[\/latex]<\/div>\r\nRewriting this set of parametric equations is a matter of substituting [latex]x[\/latex] for [latex]t[\/latex]. Thus, the Cartesian equation is [latex]y={x}^{2}-3[\/latex].\r\n<div class=\"textbox shaded\">\r\n<h3>Example 8: Finding a Cartesian Equation Using Alternate Methods<\/h3>\r\nUse two different methods to find the Cartesian equation equivalent to the given set of parametric equations.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;x\\left(t\\right)=3t - 2 \\\\ &amp;y\\left(t\\right)=t+1 \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"456091\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"456091\"]\r\n\r\n<em>Method 1<\/em>. First, let\u2019s solve the [latex]x[\/latex] equation for [latex]t[\/latex]. Then we can substitute the result into the [latex]y[\/latex] equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x=3t - 2 \\\\ x+2=3t \\\\ \\frac{x+2}{3}=t \\end{gathered}[\/latex]<\/p>\r\nNow substitute the expression for [latex]t[\/latex] into the [latex]y[\/latex] equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;y=t+1 \\\\ &amp;y=\\left(\\frac{x+2}{3}\\right)+1 \\\\ &amp;y=\\frac{x}{3}+\\frac{2}{3}+1\\\\ &amp;y=\\frac{1}{3}x+\\frac{5}{3}\\end{align}[\/latex]<\/p>\r\n<em>Method 2<\/em>. Solve the [latex]y[\/latex] equation for [latex]t[\/latex] and substitute this expression in the [latex]x[\/latex] equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}y=t+1\\hfill \\\\ y - 1=t \\end{gathered}[\/latex]<\/p>\r\nMake the substitution and then solve for [latex]y[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x=3\\left(y - 1\\right)-2 \\\\ x=3y - 3-2 \\\\ x=3y - 5 \\\\ x+5=3y \\\\ \\frac{x+5}{3}=y \\\\ y=\\frac{1}{3}x+\\frac{5}{3} \\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nWrite the given parametric equations as a Cartesian equation: [latex]x\\left(t\\right)={t}^{3}[\/latex] and [latex]y\\left(t\\right)={t}^{6}[\/latex].\r\n\r\n[reveal-answer q=\"974749\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"974749\"]\r\n\r\n[latex]y={x}^{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Finding Parametric Equations for Curves Defined by Rectangular Equations<\/h2>\r\nAlthough we have just shown that there is only one way to interpret a set of parametric equations as a rectangular equation, there are multiple ways to interpret a rectangular equation as a set of parametric equations. Any strategy we may use to find the parametric equations is valid if it produces equivalency. In other words, if we choose an expression to represent [latex]x[\/latex], and then substitute it into the [latex]y[\/latex] equation, and it produces the same graph over the same domain as the rectangular equation, then the set of parametric equations is valid. If the domain becomes restricted in the set of parametric equations, and the function does not allow the same values for [latex]x[\/latex] as the domain of the rectangular equation, then the graphs will be different.\r\n\r\nThe following video shows examples of how to find Cartesian representations of parametric equations of different kinds.\r\n\r\nhttps:\/\/youtu.be\/tW6N7DFTvrM\r\n<div class=\"textbox shaded\">\r\n<h3>Example 9: Finding a Set of Parametric Equations for Curves Defined by Rectangular Equations<\/h3>\r\nFind a set of equivalent parametric equations for [latex]y={\\left(x+3\\right)}^{2}+1[\/latex].\r\n\r\n[reveal-answer q=\"9301\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"9301\"]\r\n\r\nAn obvious choice would be to let [latex]x\\left(t\\right)=t[\/latex]. Then [latex]y\\left(t\\right)={\\left(t+3\\right)}^{2}+1[\/latex]. But let\u2019s try something more interesting. What if we let [latex]x=t+3?[\/latex] Then we have\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;y={\\left(x+3\\right)}^{2}+1 \\\\ &amp;y={\\left(\\left(t+3\\right)+3\\right)}^{2}+1 \\\\ &amp;y={\\left(t+6\\right)}^{2}+1 \\end{align}[\/latex]<\/p>\r\nThe set of parametric equations is\r\n<p style=\"text-align: center;\">[latex]\\begin{align} &amp;x\\left(t\\right)=t+3 \\\\ &amp;y\\left(t\\right)={\\left(t+6\\right)}^{2}+1 \\end{align}[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180937\/CNX_Precalc_Figure_08_06_0122.jpg\" alt=\"Graph of parametric and rectangular coordinate versions of the same parabola - they are the same!\" width=\"731\" height=\"402\" \/> <b>Figure 9<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>Parameterizing a curve involves translating a rectangular equation in two variables, [latex]x[\/latex] and [latex]y[\/latex], into two equations in three variables, <em>x<\/em>, <em>y<\/em>, and <em>t<\/em>. Often, more information is obtained from a set of parametric equations.<\/li>\r\n \t<li>Sometimes equations are simpler to graph when written in rectangular form. By eliminating [latex]t[\/latex], an equation in [latex]x[\/latex] and [latex]y[\/latex] is the result.<\/li>\r\n \t<li>To eliminate [latex]t[\/latex], solve one of the equations for [latex]t[\/latex], and substitute the expression into the second equation.<\/li>\r\n \t<li>Finding the rectangular equation for a curve defined parametrically is basically the same as eliminating the parameter. Solve for [latex]t[\/latex] in one of the equations, and substitute the expression into the second equation.<\/li>\r\n \t<li>There are an infinite number of ways to choose a set of parametric equations for a curve defined as a rectangular equation.<\/li>\r\n \t<li>Find an expression for [latex]x[\/latex] such that the domain of the set of parametric equations remains the same as the original rectangular equation.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165135186874\" class=\"definition\">\r\n \t<dt>parameter<\/dt>\r\n \t<dd id=\"fs-id1165134357588\">a variable, often representing time, upon which [latex]x[\/latex] and [latex]y[\/latex] are both dependent<\/dd>\r\n<\/dl>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Find a rectangular equation for a curve defined parametrically.<\/li>\n<li>Find parametric equations for curves defined by rectangular equations.<\/li>\n<\/ul>\n<\/div>\n<p>Consider the path a moon follows as it orbits a planet, which simultaneously rotates around the sun, as seen in Figure 1. At any moment, the moon is located at a particular spot relative to the planet. But how do we write and solve the equation for the position of the moon when the distance from the planet, the speed of the moon\u2019s orbit around the planet, and the speed of rotation around the sun are all unknowns? We can solve only for one variable at a time.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180914\/CNX_Precalc_Figure_08_06_0012.jpg\" alt=\"Illustration of a planet's circular orbit around the sun.\" width=\"487\" height=\"383\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<p>In this section, we will consider sets of equations given by [latex]x\\left(t\\right)[\/latex] and [latex]y\\left(t\\right)[\/latex] where [latex]t[\/latex] is the independent variable of time. We can use these parametric equations in a number of applications when we are looking for not only a particular position but also the direction of the movement. As we trace out successive values of [latex]t[\/latex], the orientation of the curve becomes clear. This is one of the primary advantages of using parametric equations: we are able to trace the movement of an object along a path according to time. We begin this section with a look at the basic components of parametric equations and what it means to parameterize a curve. Then we will learn how to eliminate the parameter, translate the equations of a curve defined parametrically into rectangular equations, and find the parametric equations for curves defined by rectangular equations.<\/p>\n<h2>Parameterizing a Curve<\/h2>\n<p>When an object moves along a curve\u2014or <strong>curvilinear path<\/strong>\u2014in a given direction and in a given amount of time, the position of the object in the plane is given by the <em>x-<\/em>coordinate and the <em>y-<\/em>coordinate. However, both [latex]x[\/latex] and [latex]y[\/latex]<br \/>\nvary over time and so are functions of time. For this reason, we add another variable, the <strong>parameter<\/strong>, upon which both [latex]x[\/latex] and [latex]y[\/latex] are dependent functions. In the example in the section opener, the parameter is time, [latex]t[\/latex]. The [latex]x[\/latex] position of the moon at time, [latex]t[\/latex], is represented as the function [latex]x\\left(t\\right)[\/latex], and the [latex]y[\/latex] position of the moon at time, [latex]t[\/latex], is represented as the function [latex]y\\left(t\\right)[\/latex]. Together, [latex]x\\left(t\\right)[\/latex] and [latex]y\\left(t\\right)[\/latex] are called parametric equations, and generate an ordered pair [latex]\\left(x\\left(t\\right),y\\left(t\\right)\\right)[\/latex]. Parametric equations primarily describe motion and direction.<\/p>\n<p>When we parameterize a curve, we are translating a single equation in two variables, such as [latex]x[\/latex] and [latex]y[\/latex], into an equivalent pair of equations in three variables, [latex]x,y[\/latex], and [latex]t[\/latex]. One of the reasons we parameterize a curve is because the parametric equations yield more information: specifically, the direction of the object\u2019s motion over time.<\/p>\n<p>When we graph parametric equations, we can observe the individual behaviors of [latex]x[\/latex] and of [latex]y[\/latex]. There are a number of shapes that cannot be represented in the form [latex]y=f\\left(x\\right)[\/latex], meaning that they are not functions. For example, consider the graph of a circle, given as [latex]{r}^{2}={x}^{2}+{y}^{2}[\/latex]. Solving for [latex]y[\/latex] gives [latex]y=\\pm \\sqrt{{r}^{2}-{x}^{2}}[\/latex], or two equations: [latex]{y}_{1}=\\sqrt{{r}^{2}-{x}^{2}}[\/latex] and [latex]{y}_{2}=-\\sqrt{{r}^{2}-{x}^{2}}[\/latex]. If we graph [latex]{y}_{1}[\/latex] and [latex]{y}_{2}[\/latex] together, the graph will not pass the vertical line test, as shown in Figure 2. Thus, the equation for the graph of a circle is not a function.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180917\/CNX_Precalc_Figure_08_06_0022.jpg\" alt=\"Graph of a circle in the rectangular coordinate system - the vertical line test shows that the circle r^2 = x^2 + y^2 is not a function. The dotted red vertical line intersects the function in two places - it should only intersect in one place to be a function.\" width=\"487\" height=\"291\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<p>However, if we were to graph each equation on its own, each one would pass the vertical line test and therefore would represent a function. In some instances, the concept of breaking up the equation for a circle into two functions is similar to the concept of creating parametric equations, as we use two functions to produce a non-function. This will become clearer as we move forward.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Parametric Equations<\/h3>\n<p>Suppose [latex]t[\/latex] is a number on an interval, [latex]I[\/latex]. The set of ordered pairs, [latex]\\left(x\\left(t\\right),y\\left(t\\right)\\right)[\/latex], where [latex]x=f\\left(t\\right)[\/latex] and [latex]y=g\\left(t\\right)[\/latex], forms a plane curve based on the parameter [latex]t[\/latex]. The equations [latex]x=f\\left(t\\right)[\/latex] and [latex]y=g\\left(t\\right)[\/latex] are the parametric equations.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Parameterizing a Curve<\/h3>\n<p>Parameterize the curve [latex]y={x}^{2}-1[\/latex] letting [latex]x\\left(t\\right)=t[\/latex]. Graph both equations.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q910055\">Show Solution<\/span><\/p>\n<div id=\"q910055\" class=\"hidden-answer\" style=\"display: none\">\n<p>If [latex]x\\left(t\\right)=t[\/latex], then to find [latex]y\\left(t\\right)[\/latex] we replace the variable [latex]x[\/latex] with the expression given in [latex]x\\left(t\\right)[\/latex]. In other words, [latex]y\\left(t\\right)={t}^{2}-1[\/latex]. Make a table of values similar to the table below, and sketch the graph.<\/p>\n<table id=\"Table_08_06_001\" summary=\"Ten rows and three columns. First column is labeled t, second column is labeled x(t), third column is labeled y(t). The table has ordered triples of each of these row values: (-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).\">\n<thead>\n<tr>\n<th>[latex]t[\/latex]<\/th>\n<th>[latex]x\\left(t\\right)[\/latex]<\/th>\n<th>[latex]y\\left(t\\right)[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]-4[\/latex]<\/td>\n<td>[latex]-4[\/latex]<\/td>\n<td>[latex]y\\left(-4\\right)={\\left(-4\\right)}^{2}-1=15[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-3[\/latex]<\/td>\n<td>[latex]-3[\/latex]<\/td>\n<td>[latex]y\\left(-3\\right)={\\left(-3\\right)}^{2}-1=8[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2[\/latex]<\/td>\n<td>[latex]-2[\/latex]<\/td>\n<td>[latex]y\\left(-2\\right)={\\left(-2\\right)}^{2}-1=3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1[\/latex]<\/td>\n<td>[latex]-1[\/latex]<\/td>\n<td>[latex]y\\left(-1\\right)={\\left(-1\\right)}^{2}-1=0[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]y\\left(0\\right)={\\left(0\\right)}^{2}-1=-1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]y\\left(1\\right)={\\left(1\\right)}^{2}-1=0[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]y\\left(2\\right)={\\left(2\\right)}^{2}-1=3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3[\/latex]<\/td>\n<td>[latex]3[\/latex]<\/td>\n<td>[latex]y\\left(3\\right)={\\left(3\\right)}^{2}-1=8[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]4[\/latex]<\/td>\n<td>[latex]4[\/latex]<\/td>\n<td>[latex]y\\left(4\\right)={\\left(4\\right)}^{2}-1=15[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>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 [latex]t[\/latex] increases.<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180919\/CNX_Precalc_Figure_08_06_0152.jpg\" alt=\"Graph of a parabola in two forms: a parametric equation and rectangular coordinates. It is the same function, just different ways of writing it.\" width=\"731\" height=\"291\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3.<\/b> (a) Parametric [latex]y\\left(t\\right)={t}^{2}-1[\/latex] (b) Rectangular [latex]y={x}^{2}-1[\/latex]<\/p>\n<\/div>\n<h4>Analysis of the Solution<\/h4>\n<p>The arrows indicate the direction in which the curve is generated. Notice the curve is identical to the curve of [latex]y={x}^{2}-1[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Construct a table of values and plot the parametric equations: [latex]x\\left(t\\right)=t - 3,y\\left(t\\right)=2t+4;-1\\le t\\le 2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q217668\">Show Solution<\/span><\/p>\n<div id=\"q217668\" class=\"hidden-answer\" style=\"display: none\">\n<table id=\"fs-id1165137810323\" class=\"unnumbered\" summary=\"Five rows and three columns. First column is labeled t, second column is labeled x(t), third column is labeled y(t). The table has ordered triples of each of these row values: (-1, -4, 2), (0,-3,4), (1,-2,6), (2,-1,8).\">\n<tbody>\n<tr>\n<td>[latex]t[\/latex]<\/td>\n<td>[latex]x\\left(t\\right)[\/latex]<\/td>\n<td>[latex]y\\left(t\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1[\/latex]<\/td>\n<td>[latex]-4[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]-3[\/latex]<\/td>\n<td>[latex]4[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]-2[\/latex]<\/td>\n<td>[latex]6[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]-1[\/latex]<\/td>\n<td>[latex]8[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180940\/CNX_Precalc_Figure_08_06_0062.jpg\" alt=\"\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Finding a Pair of Parametric Equations<\/h3>\n<p>Find a pair of parametric equations that models the graph of [latex]y=1-{x}^{2}[\/latex], using the parameter [latex]x\\left(t\\right)=t[\/latex]. Plot some points and sketch the graph.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q141955\">Show Solution<\/span><\/p>\n<div id=\"q141955\" class=\"hidden-answer\" style=\"display: none\">\n<p>If [latex]x\\left(t\\right)=t[\/latex] and we substitute [latex]t[\/latex] for [latex]x[\/latex] into the [latex]y[\/latex] equation, then [latex]y\\left(t\\right)=1-{t}^{2}[\/latex]. Our pair of parametric equations is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x\\left(t\\right)&=t\\\\ y\\left(t\\right)&=1-{t}^{2}\\end{align}[\/latex]<\/p>\n<p>To graph the equations, first we construct a table of values like that in the table below. We can choose values around [latex]t=0[\/latex], from [latex]t=-3[\/latex] to [latex]t=3[\/latex]. The values in the [latex]x\\left(t\\right)[\/latex] column will be the same as those in the [latex]t[\/latex] column because [latex]x\\left(t\\right)=t[\/latex]. Calculate values for the column [latex]y\\left(t\\right)[\/latex].<\/p>\n<table id=\"Table_08_06_02\" summary=\"Eight rows and three columns. First column is labeled t, second column is labeled x(t)=t, third column is labeled y(t)=1-t^2. The table has ordered triples of each of these row values: (-3,-3, y(-3) = 1 - (-3)^2 = -8 ), (-2,-2, y(-2) = 1 - (-2)^2 = -3), (-1, -1, y(-1) = 1 - (-1)^2 = 0), (0,0, y(0) = 1 - 0 = 1), (1,1, y(1) = 1 - (1)^2 = 0), (2,2, y(2) = 1 - (2)^2 = -3), (3,3, y(3) = 1 - (3)^2 = -8).\">\n<thead>\n<tr>\n<th>[latex]t[\/latex]<\/th>\n<th>[latex]x\\left(t\\right)=t[\/latex]<\/th>\n<th>[latex]y\\left(t\\right)=1-{t}^{2}[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]-3[\/latex]<\/td>\n<td>[latex]-3[\/latex]<\/td>\n<td>[latex]y\\left(-3\\right)=1-{\\left(-3\\right)}^{2}=-8[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2[\/latex]<\/td>\n<td>[latex]-2[\/latex]<\/td>\n<td>[latex]y\\left(-2\\right)=1-{\\left(-2\\right)}^{2}=-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1[\/latex]<\/td>\n<td>[latex]-1[\/latex]<\/td>\n<td>[latex]y\\left(-1\\right)=1-{\\left(-1\\right)}^{2}=0[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]y\\left(0\\right)=1 - 0=1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]y\\left(1\\right)=1-{\\left(1\\right)}^{2}=0[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]y\\left(2\\right)=1-{\\left(2\\right)}^{2}=-3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3[\/latex]<\/td>\n<td>[latex]3[\/latex]<\/td>\n<td>[latex]y\\left(3\\right)=1-{\\left(3\\right)}^{2}=-8[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The graph of [latex]y=1-{t}^{2}[\/latex] is a parabola facing downward, as shown in Figure 4. We have mapped the curve over the interval [latex]\\left[-3,3\\right][\/latex], shown as a solid line with arrows indicating the orientation of the curve according to [latex]t[\/latex]. Orientation refers to the path traced along the curve in terms of increasing values of [latex]t[\/latex]. As this parabola is symmetric with respect to the line [latex]x=0[\/latex], the values of [latex]x[\/latex] are reflected across the y-axis.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180922\/CNX_Precalc_Figure_08_06_0072.jpg\" alt=\"Graph of given downward facing parabola.\" width=\"487\" height=\"516\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Parameterize the curve given by [latex]x={y}^{3}-2y[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q560553\">Show Solution<\/span><\/p>\n<div id=\"q560553\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\begin{align}x\\left(t\\right)&={t}^{3}-2t\\\\ y\\left(t\\right)&=t\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Finding Parametric Equations That Model Given Criteria<\/h3>\n<p>An object travels at a steady rate along a straight path [latex]\\left(-5,3\\right)[\/latex] to [latex]\\left(3,-1\\right)[\/latex] in the same plane in four seconds. The coordinates are measured in meters. Find parametric equations for the position of the object.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q419034\">Show Solution<\/span><\/p>\n<div id=\"q419034\" class=\"hidden-answer\" style=\"display: none\">\n<p>The parametric equations are simple linear expressions, but we need to view this problem in a step-by-step fashion. The <em>x<\/em>-value of the object starts at [latex]-5[\/latex] meters and goes to 3 meters. This means the distance <em>x<\/em> has changed by 8 meters in 4 seconds, which is a rate of [latex]\\frac{\\text{8 m}}{4\\text{ s}}[\/latex], or [latex]2\\text{m}\/\\text{s}[\/latex]. We can write the <em>x<\/em>-coordinate as a linear function with respect to time as [latex]x\\left(t\\right)=2t - 5[\/latex]. In the linear function template [latex]y=mx+b,2t=mx[\/latex] and [latex]-5=b[\/latex].<\/p>\n<p>Similarly, the <em>y<\/em>-value of the object starts at 3 and goes to [latex]-1[\/latex], which is a change in the distance <em>y<\/em> of \u22124 meters in 4 seconds, which is a rate of [latex]\\frac{-4\\text{ m}}{4\\text{ s}}[\/latex], or [latex]-1\\text{m}\/\\text{s}[\/latex]. We can also write the <em>y<\/em>-coordinate as the linear function [latex]y\\left(t\\right)=-t+3[\/latex]. Together, these are the parametric equations for the position of the object, where [latex]x[\/latex]\u00a0and [latex]y[\/latex]\u00a0are expressed in meters and [latex]t[\/latex]\u00a0represents time:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x\\left(t\\right)&=2t - 5 \\\\ y\\left(t\\right)&=-t+3 \\end{align}[\/latex]<\/p>\n<p>Using these equations, we can build a table of values for [latex]t,x[\/latex], and [latex]y[\/latex]. In this example, we limited values of [latex]t[\/latex] to non-negative numbers. In general, any value of [latex]t[\/latex] can be used.<\/p>\n<table id=\"Table_08_06_03\" summary=\"Six rows and three columns. First column is labeled t, second column is labeled x(t)=2t-5, third column is labeled y(t)=-t+3. The table has ordered triples of each of these row values: (0, x=2(0)-5 = -5, y=-(0) +3 = 3), (1, x=2(1)-5 = -3, y=-(1) + 3 = 2), (2, x=2(2) - 5 = -1, y=-(2) + 3 = 1), (3, x=2(3) - 5 = 1, y = -(3) + 3 =0), (4, x=2(4) -5 = 3, y=-(4) + 3 = -1).\">\n<thead>\n<tr>\n<th>[latex]t[\/latex]<\/th>\n<th>[latex]x\\left(t\\right)=2t - 5[\/latex]<\/th>\n<th>[latex]y\\left(t\\right)=-t+3[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]x=2\\left(0\\right)-5=-5[\/latex]<\/td>\n<td>[latex]y=-\\left(0\\right)+3=3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]x=2\\left(1\\right)-5=-3[\/latex]<\/td>\n<td>[latex]y=-\\left(1\\right)+3=2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]x=2\\left(2\\right)-5=-1[\/latex]<\/td>\n<td>[latex]y=-\\left(2\\right)+3=1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3[\/latex]<\/td>\n<td>[latex]x=2\\left(3\\right)-5=1[\/latex]<\/td>\n<td>[latex]y=-\\left(3\\right)+3=0[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]4[\/latex]<\/td>\n<td>[latex]x=2\\left(4\\right)-5=3[\/latex]<\/td>\n<td>[latex]y=-\\left(4\\right)+3=-1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>From this table, we can create three graphs, as shown in Figure 5.<\/p>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180925\/CNX_Precalc_Figure_08_06_0032.jpg\" alt=\"Three graphs side by side. (A) has the horizontal position over time, (B) has the vertical position over time, and (C) has the position of the object in the plane at time t. See caption for more information.\" width=\"975\" height=\"514\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 5.<\/b> (a) A graph of [latex]x[\/latex] vs. [latex]t[\/latex], representing the horizontal position over time. (b) A graph of [latex]y[\/latex] vs. [latex]t[\/latex], representing the vertical position over time. (c) A graph of [latex]y[\/latex] vs. [latex]x[\/latex], representing the position of the object in the plane at time [latex]t[\/latex].<\/p>\n<\/div>\n<h4>Analysis of the Solution<\/h4>\n<p>Again, we see that, in Figure 5(c), when the parameter represents time, we can indicate the movement of the object along the path with arrows.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<h2>Methods for Finding Cartesian and Polar Equations from Curves<\/h2>\n<p>In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as [latex]x[\/latex] and [latex]y[\/latex]. Eliminating the parameter is a method that may make graphing some curves easier. However, if we are concerned with the mapping of the equation according to time, then it will be necessary to indicate the orientation of the curve as well. There are various methods for eliminating the parameter [latex]t[\/latex] from a set of parametric equations; not every method works for every type of equation. Here we will review the methods for the most common types of equations.<\/p>\n<h2>Eliminating the Parameter from Polynomial, Exponential, and Logarithmic Equations<\/h2>\n<p>For polynomial, exponential, or logarithmic equations expressed as two parametric equations, we choose the equation that is most easily manipulated and solve for [latex]t[\/latex]. We substitute the resulting expression for [latex]t[\/latex]\u00a0into the second equation. This gives one equation in [latex]x[\/latex] and [latex]y[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 4: Eliminating the Parameter in Polynomials<\/h3>\n<p>Given [latex]x\\left(t\\right)={t}^{2}+1[\/latex] and [latex]y\\left(t\\right)=2+t[\/latex], eliminate the parameter, and write the parametric equations as a Cartesian equation.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q514172\">Show Solution<\/span><\/p>\n<div id=\"q514172\" class=\"hidden-answer\" style=\"display: none\">\n<p>We will begin with the equation for [latex]y[\/latex] because the linear equation is easier to solve for [latex]t[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}y=2+t \\\\ y - 2=t\\end{gathered}[\/latex]<\/p>\n<p>Next, substitute [latex]y - 2[\/latex] for [latex]t[\/latex] in [latex]x\\left(t\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&x={t}^{2}+1 \\\\ &x={\\left(y - 2\\right)}^{2}+1&& \\text{Substitute the expression for }t\\text{ into }x. \\\\ &x={y}^{2}-4y+4+1 \\\\ &x={y}^{2}-4y+5 \\\\ &x={y}^{2}-4y+5 \\end{align}[\/latex]<\/p>\n<p>The Cartesian form is [latex]x={y}^{2}-4y+5[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>This is an equation for a parabola in which, in rectangular terms, [latex]x[\/latex] is dependent on [latex]y[\/latex]. From the curve\u2019s vertex at [latex]\\left(1,2\\right)[\/latex], the graph sweeps out to the right. See Figure 6. In this section, we consider sets of equations given by the functions [latex]x\\left(t\\right)[\/latex] and [latex]y\\left(t\\right)[\/latex], where [latex]t[\/latex] is the independent variable of time. Notice, both [latex]x[\/latex] and [latex]y[\/latex] are functions of time; so in general [latex]y[\/latex] is not a function of [latex]x[\/latex].<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180928\/CNX_Precalc_Figure_08_06_0082.jpg\" alt=\"Graph of given sideways (extending to the right) parabola.\" width=\"731\" height=\"366\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Given the equations below, eliminate the parameter and write as a rectangular equation for [latex]y[\/latex] as a function\u00a0of [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&x\\left(t\\right)=2{t}^{2}+6 \\\\ &y\\left(t\\right)=5-t\\end{align}[\/latex]<\/p>\n<p style=\"text-align: left;\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q820133\">Show Solution<\/span><\/p>\n<div id=\"q820133\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">[latex]y=5-\\sqrt{\\frac{1}{2}x - 3}[\/latex]<\/p>\n<p style=\"text-align: left;\"><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173885\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173885&theme=oea&iframe_resize_id=ohm173885\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Eliminating the Parameter in Exponential Equations<\/h3>\n<p>Eliminate the parameter and write as a Cartesian equation: [latex]x\\left(t\\right)={e}^{-t}[\/latex] and [latex]y\\left(t\\right)=3{e}^{t},t>0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q397095\">Show Solution<\/span><\/p>\n<div id=\"q397095\" class=\"hidden-answer\" style=\"display: none\">\n<p>Isolate [latex]{e}^{t}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&x={e}^{-t} \\\\ &{e}^{t}=\\frac{1}{x} \\end{align}[\/latex]<\/p>\n<p>Substitute the expression into [latex]y\\left(t\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&y=3{e}^{t} \\\\ &y=3\\left(\\frac{1}{x}\\right)\\\\ &y=\\frac{3}{x} \\end{align}[\/latex]<\/p>\n<p>The Cartesian form is [latex]y=\\frac{3}{x}[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>The graph of the parametric equation is shown in Figure 7(a). The domain is restricted to [latex]t>0[\/latex]. The Cartesian equation, [latex]y=\\frac{3}{x}[\/latex] is shown in Figure 7(b) and has only one restriction on the domain, [latex]x\\ne 0[\/latex].<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180931\/CNX_Precalc_Figure_08_06_009n2.jpg\" alt=\"&quot;Graph\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 6: Eliminating the Parameter in Logarithmic Equations<\/h3>\n<p>Eliminate the parameter and write as a Cartesian equation: [latex]x\\left(t\\right)=\\sqrt{t}+2[\/latex] and [latex]y\\left(t\\right)=\\mathrm{log}\\left(t\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q310238\">Show Solution<\/span><\/p>\n<div id=\"q310238\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solve the first equation for [latex]t[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&x=\\sqrt{t}+2 \\\\ &x - 2=\\sqrt{t} \\\\ &{\\left(x - 2\\right)}^{2}=t&& \\text{Square both sides}. \\end{align}[\/latex]<\/p>\n<p>Then, substitute the expression for [latex]t[\/latex] into the [latex]y[\/latex] equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&y=\\mathrm{log}\\left(t\\right)\\\\ &y=\\mathrm{log}{\\left(x - 2\\right)}^{2}\\end{align}[\/latex]<\/p>\n<p>The Cartesian form is [latex]y=\\mathrm{log}{\\left(x - 2\\right)}^{2}[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>To be sure that the parametric equations are equivalent to the Cartesian equation, check the domains. The parametric equations restrict the domain on [latex]x=\\sqrt{t}+2[\/latex] to [latex]t>0[\/latex]; we restrict the domain on [latex]x[\/latex] to [latex]x>2[\/latex]. The domain for the parametric equation [latex]y=\\mathrm{log}\\left(t\\right)[\/latex] is restricted to [latex]t>0[\/latex]; we limit the domain on [latex]y=\\mathrm{log}{\\left(x - 2\\right)}^{2}[\/latex] to [latex]x>2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Eliminate the parameter and write as a <strong>rectangular equation<\/strong>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&x\\left(t\\right)={t}^{2} \\\\ &y\\left(t\\right)=\\mathrm{ln}t,t>0\\end{align}[\/latex]<\/p>\n<p style=\"text-align: left;\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q205546\">Show Solution<\/span><\/p>\n<div id=\"q205546\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">[latex]y=\\mathrm{ln}\\sqrt{x}[\/latex]<\/p>\n<p style=\"text-align: left;\"><\/div>\n<\/div>\n<\/div>\n<h2>Eliminating the Parameter from Trigonometric Equations<\/h2>\n<p>Eliminating the parameter from trigonometric equations is a straightforward substitution. We can use a few of the familiar trigonometric identities and the Pythagorean Theorem.<\/p>\n<p>First, we use the identities:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x\\left(t\\right)=a\\cos t\\\\ y\\left(t\\right)=b\\sin t\\end{gathered}[\/latex]<\/p>\n<p>Solving for [latex]\\cos t[\/latex] and [latex]\\sin t[\/latex], we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\frac{x}{a}=\\cos t\\\\ \\frac{y}{b}=\\sin t\\end{gathered}[\/latex]<\/p>\n<p>Then, use the Pythagorean Theorem:<\/p>\n<p style=\"text-align: center;\">[latex]{\\cos }^{2}t+{\\sin }^{2}t=1[\/latex]<\/p>\n<p>Substituting gives<\/p>\n<div style=\"text-align: center;\">[latex]{\\cos }^{2}t+{\\sin }^{2}t={\\left(\\frac{x}{a}\\right)}^{2}+{\\left(\\frac{y}{b}\\right)}^{2}=1[\/latex]<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 7: Eliminating the Parameter from a Pair of Trigonometric Parametric Equations<\/h3>\n<p>Eliminate the parameter from the given pair of <strong>trigonometric equations<\/strong> where [latex]0\\le t\\le 2\\pi[\/latex] and sketch the graph.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&x\\left(t\\right)=4\\cos t\\\\ &y\\left(t\\right)=3\\sin t\\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q244567\">Show Solution<\/span><\/p>\n<div id=\"q244567\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solving for [latex]\\cos t[\/latex] and [latex]\\sin t[\/latex], we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x=4\\cos t \\\\ \\frac{x}{4}=\\cos t \\\\ y=3\\sin t \\\\ \\frac{y}{3}=\\sin t \\end{gathered}[\/latex]<\/p>\n<p>Next, use the Pythagorean identity and make the substitutions.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered} {\\cos }^{2}t+{\\sin }^{2}t=1\\\\ {\\left(\\frac{x}{4}\\right)}^{2}+{\\left(\\frac{y}{3}\\right)}^{2}=1\\\\ \\frac{{x}^{2}}{16}+\\frac{{y}^{2}}{9}=1\\end{gathered}[\/latex]<\/p>\n<p>The graph for the equation is shown in Figure 8.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180934\/CNX_Precalc_Figure_08_06_0112.jpg\" alt=\"Graph of given ellipse centered at (0,0).\" width=\"487\" height=\"366\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 8<\/b><\/p>\n<\/div>\n<h4>Analysis of the Solution<\/h4>\n<p>Applying the general equations for <strong>conic sections<\/strong>, we can identify [latex]\\frac{{x}^{2}}{16}+\\frac{{y}^{2}}{9}=1[\/latex] as an ellipse centered at [latex]\\left(0,0\\right)[\/latex]. Notice that when [latex]t=0[\/latex] the coordinates are [latex]\\left(4,0\\right)[\/latex], and when [latex]t=\\frac{\\pi }{2}[\/latex] the coordinates are [latex]\\left(0,3\\right)[\/latex]. This shows the orientation of the curve with increasing values of [latex]t[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Eliminate the parameter from the given pair of parametric equations and write as a Cartesian equation:<\/p>\n<p style=\"text-align: center;\">[latex]x\\left(t\\right)=2\\cos t[\/latex] and [latex]y\\left(t\\right)=3\\sin t[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q577648\">Show Solution<\/span><\/p>\n<div id=\"q577648\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{{x}^{2}}{4}+\\frac{{y}^{2}}{9}=1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm66927\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=66927&theme=oea&iframe_resize_id=ohm66927\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>When we are given a set of parametric equations and need to find an equivalent Cartesian equation, we are essentially &#8220;eliminating the parameter.&#8221; However, there are various methods we can use to rewrite a set of parametric equations as a Cartesian equation. The simplest method is to set one equation equal to the parameter, such as [latex]x\\left(t\\right)=t[\/latex]. In this case, [latex]y\\left(t\\right)[\/latex] can be any expression. For example, consider the following pair of equations.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}&x\\left(t\\right)=t\\\\ &y\\left(t\\right)={t}^{2}-3\\end{align}[\/latex]<\/div>\n<p>Rewriting this set of parametric equations is a matter of substituting [latex]x[\/latex] for [latex]t[\/latex]. Thus, the Cartesian equation is [latex]y={x}^{2}-3[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 8: Finding a Cartesian Equation Using Alternate Methods<\/h3>\n<p>Use two different methods to find the Cartesian equation equivalent to the given set of parametric equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&x\\left(t\\right)=3t - 2 \\\\ &y\\left(t\\right)=t+1 \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q456091\">Show Solution<\/span><\/p>\n<div id=\"q456091\" class=\"hidden-answer\" style=\"display: none\">\n<p><em>Method 1<\/em>. First, let\u2019s solve the [latex]x[\/latex] equation for [latex]t[\/latex]. Then we can substitute the result into the [latex]y[\/latex] equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x=3t - 2 \\\\ x+2=3t \\\\ \\frac{x+2}{3}=t \\end{gathered}[\/latex]<\/p>\n<p>Now substitute the expression for [latex]t[\/latex] into the [latex]y[\/latex] equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&y=t+1 \\\\ &y=\\left(\\frac{x+2}{3}\\right)+1 \\\\ &y=\\frac{x}{3}+\\frac{2}{3}+1\\\\ &y=\\frac{1}{3}x+\\frac{5}{3}\\end{align}[\/latex]<\/p>\n<p><em>Method 2<\/em>. Solve the [latex]y[\/latex] equation for [latex]t[\/latex] and substitute this expression in the [latex]x[\/latex] equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}y=t+1\\hfill \\\\ y - 1=t \\end{gathered}[\/latex]<\/p>\n<p>Make the substitution and then solve for [latex]y[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x=3\\left(y - 1\\right)-2 \\\\ x=3y - 3-2 \\\\ x=3y - 5 \\\\ x+5=3y \\\\ \\frac{x+5}{3}=y \\\\ y=\\frac{1}{3}x+\\frac{5}{3} \\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Write the given parametric equations as a Cartesian equation: [latex]x\\left(t\\right)={t}^{3}[\/latex] and [latex]y\\left(t\\right)={t}^{6}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q974749\">Show Solution<\/span><\/p>\n<div id=\"q974749\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y={x}^{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Finding Parametric Equations for Curves Defined by Rectangular Equations<\/h2>\n<p>Although we have just shown that there is only one way to interpret a set of parametric equations as a rectangular equation, there are multiple ways to interpret a rectangular equation as a set of parametric equations. Any strategy we may use to find the parametric equations is valid if it produces equivalency. In other words, if we choose an expression to represent [latex]x[\/latex], and then substitute it into the [latex]y[\/latex] equation, and it produces the same graph over the same domain as the rectangular equation, then the set of parametric equations is valid. If the domain becomes restricted in the set of parametric equations, and the function does not allow the same values for [latex]x[\/latex] as the domain of the rectangular equation, then the graphs will be different.<\/p>\n<p>The following video shows examples of how to find Cartesian representations of parametric equations of different kinds.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Converting Parametric Equation to Rectangular Form\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/tW6N7DFTvrM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox shaded\">\n<h3>Example 9: Finding a Set of Parametric Equations for Curves Defined by Rectangular Equations<\/h3>\n<p>Find a set of equivalent parametric equations for [latex]y={\\left(x+3\\right)}^{2}+1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q9301\">Show Solution<\/span><\/p>\n<div id=\"q9301\" class=\"hidden-answer\" style=\"display: none\">\n<p>An obvious choice would be to let [latex]x\\left(t\\right)=t[\/latex]. Then [latex]y\\left(t\\right)={\\left(t+3\\right)}^{2}+1[\/latex]. But let\u2019s try something more interesting. What if we let [latex]x=t+3?[\/latex] Then we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&y={\\left(x+3\\right)}^{2}+1 \\\\ &y={\\left(\\left(t+3\\right)+3\\right)}^{2}+1 \\\\ &y={\\left(t+6\\right)}^{2}+1 \\end{align}[\/latex]<\/p>\n<p>The set of parametric equations is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} &x\\left(t\\right)=t+3 \\\\ &y\\left(t\\right)={\\left(t+6\\right)}^{2}+1 \\end{align}[\/latex]<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180937\/CNX_Precalc_Figure_08_06_0122.jpg\" alt=\"Graph of parametric and rectangular coordinate versions of the same parabola - they are the same!\" width=\"731\" height=\"402\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 9<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>Parameterizing a curve involves translating a rectangular equation in two variables, [latex]x[\/latex] and [latex]y[\/latex], into two equations in three variables, <em>x<\/em>, <em>y<\/em>, and <em>t<\/em>. Often, more information is obtained from a set of parametric equations.<\/li>\n<li>Sometimes equations are simpler to graph when written in rectangular form. By eliminating [latex]t[\/latex], an equation in [latex]x[\/latex] and [latex]y[\/latex] is the result.<\/li>\n<li>To eliminate [latex]t[\/latex], solve one of the equations for [latex]t[\/latex], and substitute the expression into the second equation.<\/li>\n<li>Finding the rectangular equation for a curve defined parametrically is basically the same as eliminating the parameter. Solve for [latex]t[\/latex] in one of the equations, and substitute the expression into the second equation.<\/li>\n<li>There are an infinite number of ways to choose a set of parametric equations for a curve defined as a rectangular equation.<\/li>\n<li>Find an expression for [latex]x[\/latex] such that the domain of the set of parametric equations remains the same as the original rectangular equation.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165135186874\" class=\"definition\">\n<dt>parameter<\/dt>\n<dd id=\"fs-id1165134357588\">a variable, often representing time, upon which [latex]x[\/latex] and [latex]y[\/latex] are both dependent<\/dd>\n<\/dl>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1960\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":708740,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1960","chapter","type-chapter","status-publish","hentry"],"part":1875,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters\/1960","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/users\/708740"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters\/1960\/revisions"}],"predecessor-version":[{"id":2115,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters\/1960\/revisions\/2115"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/parts\/1875"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapters\/1960\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/media?parent=1960"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=1960"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/contributor?post=1960"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/wp-json\/wp\/v2\/license?post=1960"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}