{"id":2301,"date":"2025-08-12T21:37:50","date_gmt":"2025-08-12T21:37:50","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2301"},"modified":"2025-10-21T17:32:08","modified_gmt":"2025-10-21T17:32:08","slug":"parametric-equations-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/parametric-equations-learn-it-2\/","title":{"raw":"Parametric Equations: Learn It 2","rendered":"Parametric Equations: Learn It 2"},"content":{"raw":"

Writing Parametric Equations<\/h2>\r\n
\r\n

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

[latex]\\begin{align}x\\left(t\\right)&=t\\\\ y\\left(t\\right)&=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\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
[latex]t[\/latex]<\/th>\r\n[latex]x\\left(t\\right)=t[\/latex]<\/th>\r\n[latex]y\\left(t\\right)=1-{t}^{2}[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
[latex]-3[\/latex]<\/td>\r\n[latex]-3[\/latex]<\/td>\r\n[latex]y\\left(-3\\right)=1-{\\left(-3\\right)}^{2}=-8[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]-2[\/latex]<\/td>\r\n[latex]-2[\/latex]<\/td>\r\n[latex]y\\left(-2\\right)=1-{\\left(-2\\right)}^{2}=-3[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]-1[\/latex]<\/td>\r\n[latex]-1[\/latex]<\/td>\r\n[latex]y\\left(-1\\right)=1-{\\left(-1\\right)}^{2}=0[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]0[\/latex]<\/td>\r\n[latex]0[\/latex]<\/td>\r\n[latex]y\\left(0\\right)=1 - 0=1[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]1[\/latex]<\/td>\r\n[latex]1[\/latex]<\/td>\r\n[latex]y\\left(1\\right)=1-{\\left(1\\right)}^{2}=0[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]2[\/latex]<\/td>\r\n[latex]2[\/latex]<\/td>\r\n[latex]y\\left(2\\right)=1-{\\left(2\\right)}^{2}=-3[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]3[\/latex]<\/td>\r\n[latex]3[\/latex]<\/td>\r\n[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. 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\"Graph\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
Parameterize the curve given by [latex]x={y}^{3}-2y[\/latex].[reveal-answer q=\"560553\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"560553\"][latex]\\begin{align}x\\left(t\\right)&={t}^{3}-2t\\\\ y\\left(t\\right)&=t\\end{align}[\/latex][\/hidden-answer]<\/section>
Find 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

[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>\r\nThe set of parametric equations is\r\n

[latex]\\begin{align} &x\\left(t\\right)=t+3 \\\\ &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\"]\"Graph Figure 9<\/b>[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><\/section>","rendered":"

Writing Parametric Equations<\/h2>\n
\n

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.<\/span><\/h3>\n