{"id":2303,"date":"2025-08-12T21:38:22","date_gmt":"2025-08-12T21:38:22","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2303"},"modified":"2025-10-21T17:28:27","modified_gmt":"2025-10-21T17:28:27","slug":"parametric-equations-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/parametric-equations-learn-it-3\/","title":{"raw":"Parametric Equations: Learn It 3","rendered":"Parametric Equations: Learn It 3"},"content":{"raw":"

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

Eliminating the Parameter from Polynomial, Exponential, and Logarithmic Equations<\/h3>\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\r\n
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.[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

[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

[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>\r\nThe Cartesian form is [latex]x={y}^{2}-4y+5[\/latex].\r\n\r\nAnalysis of the Solution<\/strong>\r\n\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. 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\"Graph\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

Given 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

[latex]\\begin{align}&x\\left(t\\right)=2{t}^{2}+6 \\\\ &y\\left(t\\right)=5-t\\end{align}[\/latex]<\/p>\r\n

[reveal-answer q=\"820133\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"820133\"]<\/p>\r\n

[latex]y=5-\\sqrt{\\frac{1}{2}x - 3}[\/latex]<\/p>\r\n

[\/hidden-answer]<\/p>\r\n\r\n<\/section>

[ohm_question hide_question_numbers=1]173885[\/ohm_question]<\/section>
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].\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

[latex]\\begin{align}&x={e}^{-t} \\\\ &{e}^{t}=\\frac{1}{x} \\end{align}[\/latex]<\/p>\r\nSubstitute the expression into [latex]y\\left(t\\right)[\/latex].\r\n

[latex]\\begin{align}&y=3{e}^{t} \\\\ &y=3\\left(\\frac{1}{x}\\right)\\\\ &y=\\frac{3}{x} \\end{align}[\/latex]<\/p>\r\nThe Cartesian form is [latex]y=\\frac{3}{x}[\/latex].\r\n\r\nAnalysis of the Solution<\/strong>\r\n\r\nThe domain is restricted to [latex]t>0[\/latex]. The Cartesian equation, [latex]y=\\frac{3}{x}[\/latex] has only one restriction on the domain, [latex]x\\ne 0[\/latex].\r\n\r\n\""Graph\"\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

\r\n

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].<\/span><\/h3>\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

[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>\r\nThen, substitute the expression for [latex]t[\/latex] into the [latex]y[\/latex] equation.\r\n

[latex]\\begin{align}&y=\\mathrm{log}\\left(t\\right)\\\\ &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\r\nAnalysis of the Solution<\/strong>\r\n\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>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].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

Eliminate the parameter and write as a rectangular equation<\/strong>.\r\n

[latex]\\begin{align}&x\\left(t\\right)={t}^{2} \\\\ &y\\left(t\\right)=\\mathrm{ln}t,t>0\\end{align}[\/latex]<\/p>\r\n

[reveal-answer q=\"205546\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"205546\"]<\/p>\r\n

[latex]y=\\mathrm{ln}\\sqrt{x}[\/latex]<\/p>\r\n

[\/hidden-answer]<\/p>\r\n\r\n<\/section>","rendered":"

Methods for Finding Cartesian and Polar Equations from Curves<\/h2>\n

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

Eliminating the Parameter from Polynomial, Exponential, and Logarithmic Equations<\/h3>\n

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

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