Writing Parametric Equations
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.
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
[latex]\begin{align}x\left(t\right)&=t\\ y\left(t\right)&=1-{t}^{2}\end{align}[/latex]
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].
| [latex]t[/latex] | [latex]x\left(t\right)=t[/latex] | [latex]y\left(t\right)=1-{t}^{2}[/latex] |
|---|---|---|
| [latex]-3[/latex] | [latex]-3[/latex] | [latex]y\left(-3\right)=1-{\left(-3\right)}^{2}=-8[/latex] |
| [latex]-2[/latex] | [latex]-2[/latex] | [latex]y\left(-2\right)=1-{\left(-2\right)}^{2}=-3[/latex] |
| [latex]-1[/latex] | [latex]-1[/latex] | [latex]y\left(-1\right)=1-{\left(-1\right)}^{2}=0[/latex] |
| [latex]0[/latex] | [latex]0[/latex] | [latex]y\left(0\right)=1 - 0=1[/latex] |
| [latex]1[/latex] | [latex]1[/latex] | [latex]y\left(1\right)=1-{\left(1\right)}^{2}=0[/latex] |
| [latex]2[/latex] | [latex]2[/latex] | [latex]y\left(2\right)=1-{\left(2\right)}^{2}=-3[/latex] |
| [latex]3[/latex] | [latex]3[/latex] | [latex]y\left(3\right)=1-{\left(3\right)}^{2}=-8[/latex] |
The graph of [latex]y=1-{t}^{2}[/latex] is a parabola facing downward. 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.
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’s try something more interesting. What if we let [latex]x=t+3?[/latex] Then we have
[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]
The set of parametric equations is
[latex]\begin{align} &x\left(t\right)=t+3 \\ &y\left(t\right)={\left(t+6\right)}^{2}+1 \end{align}[/latex]
