Parametric equations are used when x and y are not directly related to each other, but are both related through a third term. Given the equations below, eliminate the parameter and write as a rectangular equation for [latex]y[/latex] as a function of [latex]x[/latex]. We can write the x-coordinate as a linear function with respect to time as \(x(t)=2t−5\). 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]. Similarly, the y-value of the object starts at 3 and goes to [latex]-1[/latex], which is a change in the distance y of −4 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]. [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]. (b) A graph of [latex]y[/latex] vs. [latex]t[/latex], representing the vertical position over time. \[\begin{align*} x &= \sqrt{t}+2 \\ x−2 &= \sqrt{t} \\ {(x−2)}^2 &= t \;\;\;\;\;\;\;\; \text{Square both sides.} An obvious choice would be to let \(x(t)=t\). 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]. Given \(x(t)=t^2+1\) and \(y(t)=2+t\), eliminate the parameter, and write the parametric equations as a Cartesian equation. From this table, we can create three graphs, as shown in Figure \(\PageIndex{6}\). Find a rectangular equation for a curve defined parametrically. We can use a few of the familiar trigonometric identities and the Pythagorean Theorem. There are a number of shapes that cannot be represented in the form \(y=f(x)\), meaning that they are not functions. To find a set of parametric equations for the graph represented by y = x 2 + 2 given t = x + 2, let t = x. 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. The \(x\) position of the moon at time, \(t\), is represented as the function \(x(t)\), and the \(y\) position of the moon at time, \(t\), is represented as the function \(y(t)\). For this reason, we add another variable, the parameter, upon which both \(x\) and \(y\) are dependent functions. Then, substitute the expression for [latex]t[/latex] into the [latex]y[/latex] equation. Parameterizing a curve involves translating a rectangular equation in two variables, [latex]x[/latex] and [latex]y[/latex], into two equations in three variables. [latex]\begin{gathered}x=3t - 2 \\ x+2=3t \\ \frac{x+2}{3}=t \end{gathered}[/latex]. But the problem is I am not given any bounds, does anyone know how to find … \[\begin{align*} x(t) &= 2t^2+6 \\ y(t) &= 5−t \end{align*}\], Example \(\PageIndex{5}\): Eliminating the Parameter in Exponential Equations. We can write the x-coordinate as a linear function with respect to time as [latex]x\left(t\right)=2t - 5[/latex]. The Cartesian form is \(y=\dfrac{3}{x}\). Using these equations, we can build a table of values for \(t\), \(x\), and \(y\) (see Table \(\PageIndex{3}\)). Next, substitute [latex]y - 2[/latex] for [latex]t[/latex] in [latex]x\left(t\right)[/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. However, given a rectangular equation and an equation describing the parameter in terms of one of the two variables, a set of parametric equations can be determined. Parametric equations are a set of equations that express a set of quantities as explicit functions of a number of independent variables, known as "parameters."