{"id":982,"date":"2025-06-20T17:26:45","date_gmt":"2025-06-20T17:26:45","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=982"},"modified":"2025-09-10T17:44:23","modified_gmt":"2025-09-10T17:44:23","slug":"fundamentals-of-parametric-equations-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/fundamentals-of-parametric-equations-learn-it-2\/","title":{"raw":"Fundamentals of Parametric Equations: Learn It 2","rendered":"Fundamentals of Parametric Equations: Learn It 2"},"content":{"raw":"

Eliminating the Parameter<\/h2>\r\nSometimes you'll want to convert parametric equations back into a single equation relating [latex]x[\/latex] and [latex]y[\/latex]. This process is called eliminating the parameter<\/strong>, and it helps you identify what type of curve you're working with. The basic strategy is to solve one parametric equation for the parameter [latex]t[\/latex], then substitute this expression into the other equation.\r\n\r\n
\r\n

Given the parametric equations: [latex]x(t) = t^2 - 3, \\quad y(t) = 2t + 1, \\quad -2 \\leq t \\leq 3[\/latex]<\/p>\r\n\r\n

    \r\n \t
  • Step 1<\/strong>: Solve the simpler equation for [latex]t[\/latex] from [latex]y = 2t + 1[\/latex]:\r\n
    [latex]t = \\frac{y - 1}{2}[\/latex]<\/center><\/li>\r\n \t
  • Step 2<\/strong>: Substitute into the other equation\r\n
    [latex]x = \\left(\\frac{y - 1}{2}\\right)^2 - 3[\/latex]<\/center><\/li>\r\n \t
  • Step 3<\/strong>: Simplify\r\n
    [latex]x = \\frac{(y - 1)^2}{4} - 3 = \\frac{y^2 - 2y + 1}{4} - 3 = \\frac{y^2 - 2y - 11}{4}[\/latex]<\/center><\/li>\r\n<\/ul>\r\n

    Result<\/strong>: This is a parabola opening to the right with [latex]x[\/latex] as a function of [latex]y[\/latex].<\/p>\r\n\r\n<\/section>

    When you eliminate the parameter, don't forget about the original restrictions on [latex]t[\/latex]. In the example above, [latex]-2 \\leq t \\leq 3[\/latex] means the curve only exists between specific endpoints, not as an infinite parabola.<\/section>\r\n

    Before working through more examples, recall these essential relationships.<\/p>\r\n\r\n

    \r\n

    Key Identities<\/strong><\/p>\r\n\r\n

      \r\n \t
    • Pythagorean Identity<\/strong>: For any angle [latex]t[\/latex], [latex]\\sin^2 t + \\cos^2 t = 1[\/latex]<\/li>\r\n \t
    • Circle Equation<\/strong>: A circle of radius [latex]a[\/latex] centered at the origin has equation [latex]x^2 + y^2 = a^2[\/latex]<\/li>\r\n<\/ul>\r\nThese identities are particularly useful when your parametric equations involve trigonometric functions, which often describe circular or elliptical curves.\r\n\r\n<\/section>
      \r\n
      \r\n

      Eliminate the parameter for each of the plane curves described by the following parametric equations and describe the resulting graph.<\/p>\r\n\r\n

        \r\n \t
      1. [latex]x\\left(t\\right)=\\sqrt{2t+4},y\\left(t\\right)=2t+1,-2\\le t\\le 6[\/latex]<\/li>\r\n \t
      2. [latex]x\\left(t\\right)=4\\cos{t},y\\left(t\\right)=3\\sin{t},0\\le t\\le 2\\pi [\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n
        \r\n
          \r\n \t
        1. To eliminate the parameter, we can solve either of the equations for t.<\/em> For example, solving the first equation for t<\/em> gives\r\n<\/span>\r\n
          [latex]\\begin{array}{ccc}\\hfill x& =\\hfill & \\sqrt{2t+4}\\hfill \\\\ \\hfill {x}^{2}& =\\hfill & 2t+4\\hfill \\\\ \\hfill {x}^{2}-4& =\\hfill & 2t\\hfill \\\\ \\hfill t& =\\hfill & \\frac{{x}^{2}-4}{2}.\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n<\/span>\r\nNote that when we square both sides it is important to observe that [latex]x\\ge 0[\/latex]. Substituting [latex]t=\\frac{{x}^{2}-4}{2}[\/latex] this into [latex]y\\left(t\\right)[\/latex] yields\r\n<\/span>\r\n
          [latex]\\begin{array}{ccc}\\hfill y\\left(t\\right)& =\\hfill & 2t+1\\hfill \\\\ \\hfill y& =\\hfill & 2\\left(\\frac{{x}^{2}-4}{2}\\right)+1\\hfill \\\\ \\hfill y& =\\hfill & {x}^{2}-4+1\\hfill \\\\ \\hfill y& =\\hfill & {x}^{2}-3.\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n<\/span>\r\nThis is the equation of a parabola opening upward. There is, however, a domain restriction because of the limits on the parameter t<\/em>. When [latex]t=-2[\/latex], [latex]x=\\sqrt{2\\left(-2\\right)+4}=0[\/latex], and when [latex]t=6[\/latex], [latex]x=\\sqrt{2\\left(6\\right)+4}=4[\/latex]. The graph of this plane curve follows.\r\n<\/span>\r\n
          \r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"267\"]\"A Figure 7. Graph of the plane curve described by the parametric equations in part a.[\/caption]<\/figure>\r\n<\/li>\r\n \t
        2. Sometimes it is necessary to be a bit creative in eliminating the parameter. The parametric equations for this example are\r\n<\/span>\r\n
          [latex]x\\left(t\\right)=4\\cos{t}\\text{and}y\\left(t\\right)=3\\sin{t}[\/latex].<\/div>\r\n\r\n<\/span>\r\nSolving either equation for t<\/em> directly is not advisable because sine and cosine are not one-to-one functions. However, dividing the first equation by 4 and the second equation by 3 (and suppressing the t<\/em>) gives us\r\n<\/span>\r\n
          [latex]\\cos{t}=\\frac{x}{4}\\text{and}\\sin{t}=\\frac{y}{3}[\/latex].<\/div>\r\n\r\n<\/span>\r\nNow use the Pythagorean identity [latex]{\\cos}^{2}t+{\\sin}^{2}t=1[\/latex] and replace the expressions for [latex]\\sin{t}[\/latex] and [latex]\\cos{t}[\/latex] with the equivalent expressions in terms of x<\/em> and y<\/em>. This gives\r\n<\/span>\r\n
          [latex]\\begin{array}{ccc}\\hfill {\\left(\\frac{x}{4}\\right)}^{2}+{\\left(\\frac{y}{3}\\right)}^{2}& =\\hfill & 1\\hfill \\\\ \\hfill \\frac{{x}^{2}}{16}+\\frac{{y}^{2}}{9}& =\\hfill & 1.\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n<\/span>\r\nThis is the equation of a horizontal ellipse centered at the origin, with semimajor axis 4 and semiminor axis 3 as shown in the following graph.\r\n<\/span>\r\n
          \r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"417\"]\"An Figure 8. Graph of the plane curve described by the parametric equations in part b.[\/caption]<\/figure>\r\nAs t<\/em> progresses from [latex]0[\/latex] to [latex]2\\pi [\/latex], a point on the curve traverses the ellipse once, in a counterclockwise direction. Recall from the section opener that the orbit of Earth around the Sun is also elliptical. This is a perfect example of using parameterized curves to model a real-world phenomenon.<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
          Watch the following video to see the worked solution to the example above.