{"id":2316,"date":"2025-08-12T21:41:08","date_gmt":"2025-08-12T21:41:08","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=2316"},"modified":"2025-10-21T18:17:04","modified_gmt":"2025-10-21T18:17:04","slug":"graphing-parametric-equations-apply-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/graphing-parametric-equations-apply-it-1\/","title":{"raw":"Graphing Parametric Equations: Apply It","rendered":"Graphing Parametric Equations: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Find a rectangular equation for a curve defined parametrically.<\/li>\r\n \t<li>Find parametric equations for curves defined by rectangular equations.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Applications of Parametric Equations<\/h2>\r\nMany of the advantages of parametric equations become obvious when applied to solving real-world problems. Although rectangular equations in <em>x<\/em> and <em>y<\/em> give an overall picture of an object's path, they do not reveal the position of an object at a specific time. Parametric equations, however, illustrate how the values of <em>x<\/em> and <em>y<\/em> change depending on <em>t<\/em>, as the location of a moving object at a particular time.\r\n\r\nA common application of parametric equations is solving problems involving projectile motion. In this type of motion, an object is propelled forward in an upward direction forming an angle of [latex]\\theta [\/latex] to the horizontal, with an initial speed of [latex]{v}_{0}[\/latex], and at a height [latex]h[\/latex] above the horizontal.\r\n\r\nThe path of an object propelled at an inclination of [latex]\\theta [\/latex] to the horizontal, with initial speed [latex]{v}_{0}[\/latex], and at a height [latex]h[\/latex] above the horizontal, is given by\r\n<div style=\"text-align: center;\">[latex]\\begin{align}x&amp;=\\left({v}_{0}\\cos \\theta \\right)t \\\\ y&amp;=-\\frac{1}{2}g{t}^{2}+\\left({v}_{0}\\sin \\theta \\right)t+h \\end{align}[\/latex]<\/div>\r\nwhere [latex]g[\/latex] accounts for the effects of gravity and [latex]h[\/latex] is the initial height of the object. Depending on the units involved in the problem, use [latex]g=32\\text{ft}\\text{\/}{\\text{s}}^{2}[\/latex] or [latex]g=9.8\\text{m}\\text{\/}{\\text{s}}^{2}[\/latex]. The equation for [latex]x[\/latex] gives horizontal distance, and the equation for [latex]y[\/latex] gives the vertical distance.\r\n\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a projectile motion problem, use parametric equations to solve.<\/strong>\r\n<ol>\r\n \t<li>The horizontal distance is given by [latex]x=\\left({v}_{0}\\cos \\theta \\right)t[\/latex]. Substitute the initial speed of the object for [latex]{v}_{0}[\/latex].<\/li>\r\n \t<li>The expression [latex]\\cos \\theta [\/latex] indicates the angle at which the object is propelled. Substitute that angle in degrees for [latex]\\cos \\theta [\/latex].<\/li>\r\n \t<li>The vertical distance is given by the formula [latex]y=-\\frac{1}{2}g{t}^{2}+\\left({v}_{0}\\sin \\theta \\right)t+h[\/latex]. The term [latex]-\\frac{1}{2}g{t}^{2}[\/latex] represents the effect of gravity. Depending on units involved, use [latex]g=32{\\text{ft\/s}}^{2}[\/latex] or [latex]g=9.8{\\text{m\/s}}^{2}[\/latex]. Again, substitute the initial speed for [latex]{v}_{0}[\/latex], and the height at which the object was propelled for [latex]h[\/latex].<\/li>\r\n \t<li>Proceed by calculating each term to solve for [latex]t[\/latex].<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Solve the problem presented at the beginning of this section. Does the batter hit the game-winning home run? Assume that the ball is hit with an initial velocity of 140 feet per second at an angle of [latex]45^\\circ [\/latex] to the horizontal, making contact 3 feet above the ground.\r\n<ol>\r\n \t<li>Find the parametric equations to model the path of the baseball.<\/li>\r\n \t<li>Where is the ball after 2 seconds?<\/li>\r\n \t<li>How long is the ball in the air?<\/li>\r\n \t<li>Is it a home run?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"730987\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"730987\"]\r\n<ol>\r\n \t<li>Use the formulas to set up the equations. The horizontal position is found using the parametric equation for [latex]x[\/latex]. Thus,\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;x=\\left({v}_{0}\\cos \\theta \\right)t\\\\ &amp;x=\\left(140\\cos \\left(45^\\circ \\right)\\right)t \\end{align}[\/latex]<\/div>\r\nThe vertical position is found using the parametric equation for [latex]y[\/latex]. Thus,\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;y=-16{t}^{2}+\\left({v}_{0}\\sin \\theta \\right)t+h \\\\ &amp;y=-16{t}^{2}+\\left(140\\sin \\left(45^\\circ \\right)\\right)t+3 \\end{align}[\/latex]<\/div><\/li>\r\n \t<li>Substitute 2 into the equations to find the horizontal and vertical positions of the ball.\r\n<div style=\"text-align: center;\">[latex]\\begin{align} &amp;x=\\left(140\\cos \\left(45^\\circ \\right)\\right)\\left(2\\right) \\\\ &amp;x=198\\text{ feet} \\\\ \\text{ } \\\\ &amp;y=-16{\\left(2\\right)}^{2}+\\left(140\\sin \\left(45^\\circ \\right)\\right)\\left(2\\right)+3 \\\\ &amp;y=137\\text{ feet} \\end{align}[\/latex]<\/div>\r\nAfter 2 seconds, the ball is 198 feet away from the batter\u2019s box and 137 feet above the ground.<\/li>\r\n \t<li>To calculate how long the ball is in the air, we have to find out when it will hit ground, or when [latex]y=0[\/latex]. Thus,\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;y=-16{t}^{2}+\\left(140\\sin \\left({45}^{\\circ }\\right)\\right)t+3 \\\\ &amp;y=0&amp;&amp; \\text{Set }y\\left(t\\right)=0\\text{ and solve the quadratic}.\\\\ &amp;t=6.2173 \\end{align}[\/latex]<\/div>\r\nWhen [latex]t=6.2173[\/latex] seconds, the ball has hit the ground. (The quadratic equation can be solved in various ways, but this problem was solved using a computer math program.)<\/li>\r\n \t<li>We cannot confirm that the hit was a home run without considering the size of the outfield, which varies from field to field. However, for simplicity\u2019s sake, let\u2019s assume that the outfield wall is 400 feet from home plate in the deepest part of the park. Let\u2019s also assume that the wall is 10 feet high. In order to determine whether the ball clears the wall, we need to calculate how high the ball is when <em>x<\/em> = 400 feet. So we will set <em>x<\/em> = 400, solve for [latex]t[\/latex], and input [latex]t[\/latex] into [latex]y[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;x=\\left(140\\cos \\left(45^\\circ \\right)\\right)t \\\\ &amp;400=\\left(140\\cos \\left(45^\\circ \\right)\\right)t \\\\ &amp;t=4.04 \\\\ \\text{ } \\\\ &amp;y=-16{\\left(4.04\\right)}^{2}+\\left(140\\sin \\left(45^\\circ \\right)\\right)\\left(4.04\\right)+3 \\\\ &amp;y=141.8\\end{align}[\/latex]<\/div>\r\nThe ball is 141.8 feet in the air when it soars out of the ballpark. It was indeed a home run.<\/li>\r\n<\/ol>\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180959\/CNX_Precalc_Figure_08_07_010n2.jpg\" alt=\"Plotted trajectory of a hit ball, showing the position of the batter at the origin, the ball's path in the shape of a wide downward facing parabola, and the outfield wall as a vertical line segment rising to 10 ft under the ball's path.\" width=\"731\" height=\"310\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Find a rectangular equation for a curve defined parametrically.<\/li>\n<li>Find parametric equations for curves defined by rectangular equations.<\/li>\n<\/ul>\n<\/section>\n<h2>Applications of Parametric Equations<\/h2>\n<p>Many of the advantages of parametric equations become obvious when applied to solving real-world problems. Although rectangular equations in <em>x<\/em> and <em>y<\/em> give an overall picture of an object&#8217;s path, they do not reveal the position of an object at a specific time. Parametric equations, however, illustrate how the values of <em>x<\/em> and <em>y<\/em> change depending on <em>t<\/em>, as the location of a moving object at a particular time.<\/p>\n<p>A common application of parametric equations is solving problems involving projectile motion. In this type of motion, an object is propelled forward in an upward direction forming an angle of [latex]\\theta[\/latex] to the horizontal, with an initial speed of [latex]{v}_{0}[\/latex], and at a height [latex]h[\/latex] above the horizontal.<\/p>\n<p>The path of an object propelled at an inclination of [latex]\\theta[\/latex] to the horizontal, with initial speed [latex]{v}_{0}[\/latex], and at a height [latex]h[\/latex] above the horizontal, is given by<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}x&=\\left({v}_{0}\\cos \\theta \\right)t \\\\ y&=-\\frac{1}{2}g{t}^{2}+\\left({v}_{0}\\sin \\theta \\right)t+h \\end{align}[\/latex]<\/div>\n<p>where [latex]g[\/latex] accounts for the effects of gravity and [latex]h[\/latex] is the initial height of the object. Depending on the units involved in the problem, use [latex]g=32\\text{ft}\\text{\/}{\\text{s}}^{2}[\/latex] or [latex]g=9.8\\text{m}\\text{\/}{\\text{s}}^{2}[\/latex]. The equation for [latex]x[\/latex] gives horizontal distance, and the equation for [latex]y[\/latex] gives the vertical distance.<\/p>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a projectile motion problem, use parametric equations to solve.<\/strong><\/p>\n<ol>\n<li>The horizontal distance is given by [latex]x=\\left({v}_{0}\\cos \\theta \\right)t[\/latex]. Substitute the initial speed of the object for [latex]{v}_{0}[\/latex].<\/li>\n<li>The expression [latex]\\cos \\theta[\/latex] indicates the angle at which the object is propelled. Substitute that angle in degrees for [latex]\\cos \\theta[\/latex].<\/li>\n<li>The vertical distance is given by the formula [latex]y=-\\frac{1}{2}g{t}^{2}+\\left({v}_{0}\\sin \\theta \\right)t+h[\/latex]. The term [latex]-\\frac{1}{2}g{t}^{2}[\/latex] represents the effect of gravity. Depending on units involved, use [latex]g=32{\\text{ft\/s}}^{2}[\/latex] or [latex]g=9.8{\\text{m\/s}}^{2}[\/latex]. Again, substitute the initial speed for [latex]{v}_{0}[\/latex], and the height at which the object was propelled for [latex]h[\/latex].<\/li>\n<li>Proceed by calculating each term to solve for [latex]t[\/latex].<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Solve the problem presented at the beginning of this section. Does the batter hit the game-winning home run? Assume that the ball is hit with an initial velocity of 140 feet per second at an angle of [latex]45^\\circ[\/latex] to the horizontal, making contact 3 feet above the ground.<\/p>\n<ol>\n<li>Find the parametric equations to model the path of the baseball.<\/li>\n<li>Where is the ball after 2 seconds?<\/li>\n<li>How long is the ball in the air?<\/li>\n<li>Is it a home run?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q730987\">Show Solution<\/button><\/p>\n<div id=\"q730987\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Use the formulas to set up the equations. The horizontal position is found using the parametric equation for [latex]x[\/latex]. Thus,\n<div style=\"text-align: center;\">[latex]\\begin{align}&x=\\left({v}_{0}\\cos \\theta \\right)t\\\\ &x=\\left(140\\cos \\left(45^\\circ \\right)\\right)t \\end{align}[\/latex]<\/div>\n<p>The vertical position is found using the parametric equation for [latex]y[\/latex]. Thus,<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}&y=-16{t}^{2}+\\left({v}_{0}\\sin \\theta \\right)t+h \\\\ &y=-16{t}^{2}+\\left(140\\sin \\left(45^\\circ \\right)\\right)t+3 \\end{align}[\/latex]<\/div>\n<\/li>\n<li>Substitute 2 into the equations to find the horizontal and vertical positions of the ball.\n<div style=\"text-align: center;\">[latex]\\begin{align} &x=\\left(140\\cos \\left(45^\\circ \\right)\\right)\\left(2\\right) \\\\ &x=198\\text{ feet} \\\\ \\text{ } \\\\ &y=-16{\\left(2\\right)}^{2}+\\left(140\\sin \\left(45^\\circ \\right)\\right)\\left(2\\right)+3 \\\\ &y=137\\text{ feet} \\end{align}[\/latex]<\/div>\n<p>After 2 seconds, the ball is 198 feet away from the batter\u2019s box and 137 feet above the ground.<\/li>\n<li>To calculate how long the ball is in the air, we have to find out when it will hit ground, or when [latex]y=0[\/latex]. Thus,\n<div style=\"text-align: center;\">[latex]\\begin{align}&y=-16{t}^{2}+\\left(140\\sin \\left({45}^{\\circ }\\right)\\right)t+3 \\\\ &y=0&& \\text{Set }y\\left(t\\right)=0\\text{ and solve the quadratic}.\\\\ &t=6.2173 \\end{align}[\/latex]<\/div>\n<p>When [latex]t=6.2173[\/latex] seconds, the ball has hit the ground. (The quadratic equation can be solved in various ways, but this problem was solved using a computer math program.)<\/li>\n<li>We cannot confirm that the hit was a home run without considering the size of the outfield, which varies from field to field. However, for simplicity\u2019s sake, let\u2019s assume that the outfield wall is 400 feet from home plate in the deepest part of the park. Let\u2019s also assume that the wall is 10 feet high. In order to determine whether the ball clears the wall, we need to calculate how high the ball is when <em>x<\/em> = 400 feet. So we will set <em>x<\/em> = 400, solve for [latex]t[\/latex], and input [latex]t[\/latex] into [latex]y[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{align}&x=\\left(140\\cos \\left(45^\\circ \\right)\\right)t \\\\ &400=\\left(140\\cos \\left(45^\\circ \\right)\\right)t \\\\ &t=4.04 \\\\ \\text{ } \\\\ &y=-16{\\left(4.04\\right)}^{2}+\\left(140\\sin \\left(45^\\circ \\right)\\right)\\left(4.04\\right)+3 \\\\ &y=141.8\\end{align}[\/latex]<\/div>\n<p>The ball is 141.8 feet in the air when it soars out of the ballpark. It was indeed a home run.<\/li>\n<\/ol>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27180959\/CNX_Precalc_Figure_08_07_010n2.jpg\" alt=\"Plotted trajectory of a hit ball, showing the position of the batter at the origin, the ball's path in the shape of a wide downward facing parabola, and the outfield wall as a vertical line segment rising to 10 ft under the ball's path.\" width=\"731\" height=\"310\" \/><\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":13,"menu_order":13,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":520,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2316"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/13"}],"version-history":[{"count":4,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2316\/revisions"}],"predecessor-version":[{"id":4801,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2316\/revisions\/4801"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/520"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/2316\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=2316"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=2316"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=2316"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=2316"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}