{"id":2942,"date":"2016-07-22T16:55:46","date_gmt":"2016-07-22T16:55:46","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=2942"},"modified":"2019-07-24T21:26:00","modified_gmt":"2019-07-24T21:26:00","slug":"read-variation","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/chapter\/read-variation\/","title":{"raw":"Variation","rendered":"Variation"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Define direct variation and solve problems involving direct variation<\/li>\r\n \t<li>Define inverse variation and solve problems involving inverse variation<\/li>\r\n \t<li>Define joint variation and solve problems involving joint variation<\/li>\r\n<\/ul>\r\n<\/div>\r\n\r\n[caption id=\"attachment_5075\" align=\"aligncenter\" width=\"482\"]<img class=\"wp-image-5075\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/22162806\/Screen-Shot-2016-06-21-at-7.00.52-PM-300x198.png\" alt=\"Huge parking lot full of cars.\" width=\"482\" height=\"318\" \/> So many cars, so many tires.[\/caption]\r\n<h2>Direct Variation<\/h2>\r\nVariation equations are examples of rational formulas and are used to describe the relationship between variables. For example, imagine a parking lot filled with cars. The total number of tires in the parking lot is dependent on the total number of cars. Algebraically, you can represent this relationship with an equation.\r\n<p style=\"text-align: center;\">[latex]\\text{number of tires}=4\\cdot\\text{number of cars}[\/latex]<\/p>\r\nThe number\u00a0[latex]4[\/latex] tells you the rate at which cars and tires are related. You call the rate the <strong>constant of variation<\/strong>. It is a constant because this number does not change. Because the number of cars and the number of tires are linked by a constant, changes in the number of cars cause the number of tires to change in a proportional, steady way. This is an example of <strong>direct variation<\/strong>, where the number of tires varies directly with the number of cars.\r\n\r\nYou can use the car and tire equation as the basis for writing a general algebraic equation that will work for all examples of direct variation. In the example, the number of tires is the output,\u00a0[latex]4[\/latex] is the constant, and the number of cars is the input. Let us enter those generic terms into the equation. You get [latex]y=kx[\/latex]. That is the formula for all direct variation equations.\r\n<p style=\"text-align: center;\">[latex]\\text{number of tires}=4\\cdot\\text{number of cars}\\\\\\text{output}=\\text{constant}\\cdot\\text{input}[\/latex]<\/p>\r\n\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for <i>k<\/i>, the constant of variation, in a direct variation problem where\u00a0[latex]y=300[\/latex] and [latex]x=10[\/latex].\r\n\r\n[reveal-answer q=\"714779\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"714779\"]\r\n\r\nWrite the formula for a direct variation relationship.\r\n<p style=\"text-align: center;\">[latex]y=kx[\/latex]<\/p>\r\nSubstitute known values into the equation.\r\n<p style=\"text-align: center;\">[latex]300=k\\left(10\\right)[\/latex]<\/p>\r\nSolve for <i>k<\/i> by dividing both sides of the equation by\u00a0[latex]10[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{300}{10}=\\frac{10k}{10}\\\\\\\\\\,\\,\\,\\,30=k\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the video that follows, we present an example of solving a direct variation equation.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=DLPKiMD_ZZw&amp;feature=youtu.be\r\n<h2>Inverse Variation<\/h2>\r\nAnother kind of variation is called <strong>inverse variation<\/strong>. In these equations, the output\u00a0equals a constant divided by the input variable that is changing. In symbolic form, this is the equation [latex] y=\\frac{k}{x}[\/latex].\r\n\r\nOne example of an inverse variation is the speed required to travel between two cities in a given amount of time.\r\n\r\nLet us say you need to drive from Boston to Chicago which is about\u00a0[latex]1,000[\/latex] miles. The more time you have, the slower you can go. If you want to get there in\u00a0[latex]20[\/latex] hours, you need to go\u00a0[latex]50[\/latex] miles per hour (assuming you do not stop driving!) because [latex] \\frac{1,000}{20}=50[\/latex]. But if you can take\u00a0[latex]40[\/latex] hours to get there, you only have to average\u00a0[latex]25[\/latex] miles per hour since [latex] \\frac{1,000}{40}=25[\/latex].\r\n\r\nThe equation for figuring out how fast to travel from the amount of time you have is [latex] speed=\\frac{miles}{time}[\/latex]. This equation should remind you of the distance formula [latex] d=rt[\/latex]. If you solve [latex] d=rt[\/latex] for <i>r<\/i>, you get [latex] r=\\frac{d}{t}[\/latex], or [latex] speed=\\frac{miles}{time}[\/latex].\r\n\r\nIn the case of the Boston to Chicago trip, you can write [latex] s=\\frac{1,000}{t}[\/latex]. Notice that this is the same form as the inverse variation function formula [latex] y=\\frac{k}{x}[\/latex].\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for <i>k<\/i>, the constant of variation, in an inverse variation problem where\u00a0[latex]x=5[\/latex] and [latex]y=25[\/latex].\r\n\r\n[reveal-answer q=\"752007\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"752007\"]\r\n\r\nWrite the formula for an inverse variation relationship.\r\n<p style=\"text-align: center;\">[latex] y=\\frac{k}{x}[\/latex]<\/p>\r\nSubstitute known values into the equation.\r\n<p style=\"text-align: center;\">[latex] 25=\\frac{k}{5}[\/latex]<\/p>\r\nSolve for <i>k<\/i> by multiplying both sides of the equation by\u00a0[latex]5[\/latex].\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{c}5\\cdot 25=\\frac{k}{5}\\cdot 5\\\\\\\\125=\\frac{5k}{5}\\\\\\\\125=k\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, we will find the water temperature in the ocean at a depth of\u00a0[latex]500[\/latex] meters. Water temperature is inversely proportional to depth in the ocean.\r\n\r\n[caption id=\"attachment_5074\" align=\"aligncenter\" width=\"534\"]<img class=\"wp-image-5074\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/22162808\/Screen-Shot-2016-06-21-at-6.57.13-PM-300x159.png\" alt=\"Scuba divers in the ocean.\" width=\"534\" height=\"283\" \/> Water temperature in the ocean varies inversely with depth.[\/caption]\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThe water temperature in the ocean varies inversely with the depth of the water. The deeper a person dives, the colder the water becomes. At a depth of\u00a0[latex]1,000[\/latex] meters, the water temperature is\u00a0[latex]5\u00ba[\/latex] Celsius. What is the water temperature at a depth of\u00a0[latex]500[\/latex] meters?\r\n\r\n[reveal-answer q=\"700119\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"700119\"]\r\n\r\nYou are told that this is an inverse relationship and that the water temperature (<i>y<\/i>) varies inversely with the depth of the water (<i>x<\/i>).\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y=\\frac{k}{x}\\\\\\\\temp=\\frac{k}{depth}\\end{array}[\/latex]<\/p>\r\nSubstitute known values into the equation.\r\n<p style=\"text-align: center;\">[latex] 5=\\frac{k}{1,000}[\/latex]<\/p>\r\nSolve for <i>k<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}1,000\\cdot5=\\frac{k}{1,000}\\cdot 1,000\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,5,000=\\frac{1,000k}{1,000}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,5,000=k\\end{array}[\/latex]<\/p>\r\nNow that <em>k<\/em>, the constant of variation is known, use that information to solve the problem: find the water temperature at\u00a0[latex]500[\/latex] meters.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}temp=\\frac{k}{depth}\\\\\\\\temp=\\frac{5,000}{500}\\\\\\\\temp=10\\end{array}[\/latex]<\/p>\r\nAt\u00a0[latex]500[\/latex] meters, the water temperature is\u00a0[latex]10\u00ba[\/latex] C.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the video that follows, we present an example of inverse variation.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=y9wqI6Uo6_M&amp;feature=youtu.be\r\n<h2>Joint Variation<\/h2>\r\nA third type of variation is called <strong>joint variation<\/strong>. Joint variation is the same as direct variation except there are two or more quantities. For example, the area of a rectangle can be found using the formula [latex]A=lw[\/latex], where <i>l<\/i> is the length of the rectangle and <i>w <\/i>is the width of the rectangle. If you change the width of the rectangle, then the area changes and similarly if you change the length of the rectangle then the area will also change. You can say that the area of the rectangle \u201cvaries jointly with the length and the width of the rectangle.\u201d\r\n\r\nThe formula for the volume of a cylinder, [latex]V=\\pi {{r}^{2}}h[\/latex], is another example of joint variation. The volume of the cylinder varies jointly with the square of the radius and the height of the cylinder. The constant of variation is [latex] \\pi [\/latex].\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nThe area of a triangle varies jointly with the lengths of its base and height. If the area of a triangle is\u00a0[latex]30[\/latex] inches[latex]^{2}[\/latex]\u00a0when the base is\u00a0[latex]10[\/latex] inches and the height is\u00a0[latex]6[\/latex] inches, find the variation constant and the area of a triangle whose base is\u00a0[latex]15[\/latex] inches and height is\u00a0[latex]20[\/latex] inches.\r\n\r\n[reveal-answer q=\"264626\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"264626\"]\r\n\r\nYou are told that this is a joint variation relationship and that the area of a triangle (<i>A<\/i>) varies jointly with the lengths of the base (<i>b<\/i>) and height (<i>h<\/i>).\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y=kxz\\\\Area=k(base)(height)\\end{array}[\/latex]<\/p>\r\nSubstitute known values into the equation and solve for <i>k<\/i>.\r\n<p style=\"text-align: center;\">[latex]30=k\\left(10\\right)\\left(6\\right)\\\\30=60k\\\\\\\\\\frac{30}{60}=\\frac{60k}{60}\\\\\\\\\\frac{1}{2}=k[\/latex]<\/p>\r\nNow that <i>k<\/i> is known, solve for the area of a triangle whose base is\u00a0[latex]15[\/latex] inches and height is\u00a0[latex]20[\/latex] inches.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}Area=k(base)(height)\\\\\\\\Area=(15)(20)(\\frac{1}{2})\\\\\\\\Area=\\frac{300}{2}\\\\\\\\Area=150\\,\\,\\text{square inches}\\end{array}[\/latex]<\/p>\r\nThe constant of variation, <i>k<\/i>, is [latex] \\frac{1}{2}[\/latex], and the area of the triangle is\u00a0[latex]150[\/latex] square inches.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nFinding <i>k<\/i> to be [latex] \\frac{1}{2}[\/latex] should not be surprising. You know that the area of a triangle is one-half base times height, [latex] A=\\frac{1}{2}bh[\/latex]. The [latex] \\frac{1}{2}[\/latex] in this formula is exactly the same [latex] \\frac{1}{2}[\/latex] that you calculated in this example!\r\n\r\nIn the following video, we show an example of\u00a0finding the constant of variation for a jointly varying relation.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=JREPATMScbM&amp;feature=youtu.be\r\n<div class=\"textbox shaded\">\r\n<h3>Direct, Joint, and Inverse Variation<\/h3>\r\n<i>k<\/i> is the constant of variation. In all cases, [latex]k\\neq0[\/latex].\r\n<ul>\r\n \t<li>Direct variation: [latex]y=kx[\/latex]<\/li>\r\n \t<li>Inverse variation: [latex] y=\\frac{k}{x}[\/latex]<\/li>\r\n \t<li>Joint variation: [latex]y=kxz[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Summary<\/h2>\r\nRational formulas can be used to solve a variety of problems that involve rates, times, and work. Direct, inverse, and joint variation equations are examples of rational formulas. In direct variation, the variables have a direct relationship\u2014as one quantity increases, the other quantity will also increase. As one quantity decreases, the other quantity decreases. In inverse variation, the variables have an inverse relationship\u2014as one variable increases, the other variable decreases, and vice versa. Joint variation is the same as direct variation except there are two or more variables.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Define direct variation and solve problems involving direct variation<\/li>\n<li>Define inverse variation and solve problems involving inverse variation<\/li>\n<li>Define joint variation and solve problems involving joint variation<\/li>\n<\/ul>\n<\/div>\n<div id=\"attachment_5075\" style=\"width: 492px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5075\" class=\"wp-image-5075\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/22162806\/Screen-Shot-2016-06-21-at-7.00.52-PM-300x198.png\" alt=\"Huge parking lot full of cars.\" width=\"482\" height=\"318\" \/><\/p>\n<p id=\"caption-attachment-5075\" class=\"wp-caption-text\">So many cars, so many tires.<\/p>\n<\/div>\n<h2>Direct Variation<\/h2>\n<p>Variation equations are examples of rational formulas and are used to describe the relationship between variables. For example, imagine a parking lot filled with cars. The total number of tires in the parking lot is dependent on the total number of cars. Algebraically, you can represent this relationship with an equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\text{number of tires}=4\\cdot\\text{number of cars}[\/latex]<\/p>\n<p>The number\u00a0[latex]4[\/latex] tells you the rate at which cars and tires are related. You call the rate the <strong>constant of variation<\/strong>. It is a constant because this number does not change. Because the number of cars and the number of tires are linked by a constant, changes in the number of cars cause the number of tires to change in a proportional, steady way. This is an example of <strong>direct variation<\/strong>, where the number of tires varies directly with the number of cars.<\/p>\n<p>You can use the car and tire equation as the basis for writing a general algebraic equation that will work for all examples of direct variation. In the example, the number of tires is the output,\u00a0[latex]4[\/latex] is the constant, and the number of cars is the input. Let us enter those generic terms into the equation. You get [latex]y=kx[\/latex]. That is the formula for all direct variation equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\text{number of tires}=4\\cdot\\text{number of cars}\\\\\\text{output}=\\text{constant}\\cdot\\text{input}[\/latex]<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for <i>k<\/i>, the constant of variation, in a direct variation problem where\u00a0[latex]y=300[\/latex] and [latex]x=10[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q714779\">Show Solution<\/span><\/p>\n<div id=\"q714779\" class=\"hidden-answer\" style=\"display: none\">\n<p>Write the formula for a direct variation relationship.<\/p>\n<p style=\"text-align: center;\">[latex]y=kx[\/latex]<\/p>\n<p>Substitute known values into the equation.<\/p>\n<p style=\"text-align: center;\">[latex]300=k\\left(10\\right)[\/latex]<\/p>\n<p>Solve for <i>k<\/i> by dividing both sides of the equation by\u00a0[latex]10[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\frac{300}{10}=\\frac{10k}{10}\\\\\\\\\\,\\,\\,\\,30=k\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the video that follows, we present an example of solving a direct variation equation.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex: Direct Variation Application - Aluminum Can Usage\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/DLPKiMD_ZZw?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Inverse Variation<\/h2>\n<p>Another kind of variation is called <strong>inverse variation<\/strong>. In these equations, the output\u00a0equals a constant divided by the input variable that is changing. In symbolic form, this is the equation [latex]y=\\frac{k}{x}[\/latex].<\/p>\n<p>One example of an inverse variation is the speed required to travel between two cities in a given amount of time.<\/p>\n<p>Let us say you need to drive from Boston to Chicago which is about\u00a0[latex]1,000[\/latex] miles. The more time you have, the slower you can go. If you want to get there in\u00a0[latex]20[\/latex] hours, you need to go\u00a0[latex]50[\/latex] miles per hour (assuming you do not stop driving!) because [latex]\\frac{1,000}{20}=50[\/latex]. But if you can take\u00a0[latex]40[\/latex] hours to get there, you only have to average\u00a0[latex]25[\/latex] miles per hour since [latex]\\frac{1,000}{40}=25[\/latex].<\/p>\n<p>The equation for figuring out how fast to travel from the amount of time you have is [latex]speed=\\frac{miles}{time}[\/latex]. This equation should remind you of the distance formula [latex]d=rt[\/latex]. If you solve [latex]d=rt[\/latex] for <i>r<\/i>, you get [latex]r=\\frac{d}{t}[\/latex], or [latex]speed=\\frac{miles}{time}[\/latex].<\/p>\n<p>In the case of the Boston to Chicago trip, you can write [latex]s=\\frac{1,000}{t}[\/latex]. Notice that this is the same form as the inverse variation function formula [latex]y=\\frac{k}{x}[\/latex].<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for <i>k<\/i>, the constant of variation, in an inverse variation problem where\u00a0[latex]x=5[\/latex] and [latex]y=25[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q752007\">Show Solution<\/span><\/p>\n<div id=\"q752007\" class=\"hidden-answer\" style=\"display: none\">\n<p>Write the formula for an inverse variation relationship.<\/p>\n<p style=\"text-align: center;\">[latex]y=\\frac{k}{x}[\/latex]<\/p>\n<p>Substitute known values into the equation.<\/p>\n<p style=\"text-align: center;\">[latex]25=\\frac{k}{5}[\/latex]<\/p>\n<p>Solve for <i>k<\/i> by multiplying both sides of the equation by\u00a0[latex]5[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}5\\cdot 25=\\frac{k}{5}\\cdot 5\\\\\\\\125=\\frac{5k}{5}\\\\\\\\125=k\\,\\,\\,\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next example, we will find the water temperature in the ocean at a depth of\u00a0[latex]500[\/latex] meters. Water temperature is inversely proportional to depth in the ocean.<\/p>\n<div id=\"attachment_5074\" style=\"width: 544px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5074\" class=\"wp-image-5074\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/22162808\/Screen-Shot-2016-06-21-at-6.57.13-PM-300x159.png\" alt=\"Scuba divers in the ocean.\" width=\"534\" height=\"283\" \/><\/p>\n<p id=\"caption-attachment-5074\" class=\"wp-caption-text\">Water temperature in the ocean varies inversely with depth.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>The water temperature in the ocean varies inversely with the depth of the water. The deeper a person dives, the colder the water becomes. At a depth of\u00a0[latex]1,000[\/latex] meters, the water temperature is\u00a0[latex]5\u00ba[\/latex] Celsius. What is the water temperature at a depth of\u00a0[latex]500[\/latex] meters?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q700119\">Show Solution<\/span><\/p>\n<div id=\"q700119\" class=\"hidden-answer\" style=\"display: none\">\n<p>You are told that this is an inverse relationship and that the water temperature (<i>y<\/i>) varies inversely with the depth of the water (<i>x<\/i>).<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y=\\frac{k}{x}\\\\\\\\temp=\\frac{k}{depth}\\end{array}[\/latex]<\/p>\n<p>Substitute known values into the equation.<\/p>\n<p style=\"text-align: center;\">[latex]5=\\frac{k}{1,000}[\/latex]<\/p>\n<p>Solve for <i>k<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}1,000\\cdot5=\\frac{k}{1,000}\\cdot 1,000\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,5,000=\\frac{1,000k}{1,000}\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,5,000=k\\end{array}[\/latex]<\/p>\n<p>Now that <em>k<\/em>, the constant of variation is known, use that information to solve the problem: find the water temperature at\u00a0[latex]500[\/latex] meters.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}temp=\\frac{k}{depth}\\\\\\\\temp=\\frac{5,000}{500}\\\\\\\\temp=10\\end{array}[\/latex]<\/p>\n<p>At\u00a0[latex]500[\/latex] meters, the water temperature is\u00a0[latex]10\u00ba[\/latex] C.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the video that follows, we present an example of inverse variation.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  Inverse Variation Application - Number of Workers and Job Time\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/y9wqI6Uo6_M?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Joint Variation<\/h2>\n<p>A third type of variation is called <strong>joint variation<\/strong>. Joint variation is the same as direct variation except there are two or more quantities. For example, the area of a rectangle can be found using the formula [latex]A=lw[\/latex], where <i>l<\/i> is the length of the rectangle and <i>w <\/i>is the width of the rectangle. If you change the width of the rectangle, then the area changes and similarly if you change the length of the rectangle then the area will also change. You can say that the area of the rectangle \u201cvaries jointly with the length and the width of the rectangle.\u201d<\/p>\n<p>The formula for the volume of a cylinder, [latex]V=\\pi {{r}^{2}}h[\/latex], is another example of joint variation. The volume of the cylinder varies jointly with the square of the radius and the height of the cylinder. The constant of variation is [latex]\\pi[\/latex].<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>The area of a triangle varies jointly with the lengths of its base and height. If the area of a triangle is\u00a0[latex]30[\/latex] inches[latex]^{2}[\/latex]\u00a0when the base is\u00a0[latex]10[\/latex] inches and the height is\u00a0[latex]6[\/latex] inches, find the variation constant and the area of a triangle whose base is\u00a0[latex]15[\/latex] inches and height is\u00a0[latex]20[\/latex] inches.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q264626\">Show Solution<\/span><\/p>\n<div id=\"q264626\" class=\"hidden-answer\" style=\"display: none\">\n<p>You are told that this is a joint variation relationship and that the area of a triangle (<i>A<\/i>) varies jointly with the lengths of the base (<i>b<\/i>) and height (<i>h<\/i>).<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y=kxz\\\\Area=k(base)(height)\\end{array}[\/latex]<\/p>\n<p>Substitute known values into the equation and solve for <i>k<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]30=k\\left(10\\right)\\left(6\\right)\\\\30=60k\\\\\\\\\\frac{30}{60}=\\frac{60k}{60}\\\\\\\\\\frac{1}{2}=k[\/latex]<\/p>\n<p>Now that <i>k<\/i> is known, solve for the area of a triangle whose base is\u00a0[latex]15[\/latex] inches and height is\u00a0[latex]20[\/latex] inches.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}Area=k(base)(height)\\\\\\\\Area=(15)(20)(\\frac{1}{2})\\\\\\\\Area=\\frac{300}{2}\\\\\\\\Area=150\\,\\,\\text{square inches}\\end{array}[\/latex]<\/p>\n<p>The constant of variation, <i>k<\/i>, is [latex]\\frac{1}{2}[\/latex], and the area of the triangle is\u00a0[latex]150[\/latex] square inches.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Finding <i>k<\/i> to be [latex]\\frac{1}{2}[\/latex] should not be surprising. You know that the area of a triangle is one-half base times height, [latex]A=\\frac{1}{2}bh[\/latex]. The [latex]\\frac{1}{2}[\/latex] in this formula is exactly the same [latex]\\frac{1}{2}[\/latex] that you calculated in this example!<\/p>\n<p>In the following video, we show an example of\u00a0finding the constant of variation for a jointly varying relation.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Joint Variation: Determine the Variation Constant (Volume of a Cone)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/JREPATMScbM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox shaded\">\n<h3>Direct, Joint, and Inverse Variation<\/h3>\n<p><i>k<\/i> is the constant of variation. In all cases, [latex]k\\neq0[\/latex].<\/p>\n<ul>\n<li>Direct variation: [latex]y=kx[\/latex]<\/li>\n<li>Inverse variation: [latex]y=\\frac{k}{x}[\/latex]<\/li>\n<li>Joint variation: [latex]y=kxz[\/latex]<\/li>\n<\/ul>\n<\/div>\n<h2>Summary<\/h2>\n<p>Rational formulas can be used to solve a variety of problems that involve rates, times, and work. Direct, inverse, and joint variation equations are examples of rational formulas. In direct variation, the variables have a direct relationship\u2014as one quantity increases, the other quantity will also increase. As one quantity decreases, the other quantity decreases. In inverse variation, the variables have an inverse relationship\u2014as one variable increases, the other variable decreases, and vice versa. Joint variation is the same as direct variation except there are two or more variables.<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2942\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Screenshot: so many cars, so many tires. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Screenshot: Water temperature in the ocean varies inversely with depth. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Joint Variation: Determine the Variation Constant (Volume of a Cone). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/JREPATMScbM\">https:\/\/youtu.be\/JREPATMScbM<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Unit 15: Rational Expressions, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Direct Variation Application - Aluminum Can Usage. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/DLPKiMD_ZZw\">https:\/\/youtu.be\/DLPKiMD_ZZw<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Inverse Variation Application - Number of Workers and Job Time. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/y9wqI6Uo6_M\">https:\/\/youtu.be\/y9wqI6Uo6_M<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":10,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Screenshot: so many cars, so many tires\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Screenshot: Water temperature in the ocean varies inversely with depth\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 15: Rational Expressions, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology and Education\",\"url\":\"http:\/\/nrocnetwork.org\/dm-opentext\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Direct Variation Application - 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