{"id":4973,"date":"2017-12-29T17:46:58","date_gmt":"2017-12-29T17:46:58","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/?post_type=chapter&#038;p=4973"},"modified":"2023-11-08T13:20:14","modified_gmt":"2023-11-08T13:20:14","slug":"variation","status":"web-only","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/chapter\/variation\/","title":{"raw":"8.6 - Variation","rendered":"8.6 &#8211; Variation"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>(8.6.1) - Define direct variation, and solve problems involving direct variation<\/li>\r\n \t<li>(8.6.2) - Define inverse variation and solve problems involving inverse variation<\/li>\r\n \t<li>(8.6.3) - Define joint variation and solve problems involving joint variation<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1>(8.6.1) - Define direct variation, and solve problems involving direct variation<\/h1>\r\n[caption id=\"attachment_5075\" align=\"alignleft\" 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\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 4 tells you the rate at which cars and tires are related. You call the rate the <strong>constant of variation<\/strong>. It\u2019s 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, 4 is the constant, and the number of cars is the input. Let\u2019s enter those generic terms into the equation. You get [latex]y=kx[\/latex]. That\u2019s 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 [latex]k[\/latex], 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\"]Write 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 [latex]k[\/latex] by dividing both sides of the equation by 10.\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}\\frac{300}{10}=\\frac{10k}{10}\\\\\\\\\\,\\,\\,\\,30=k\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nThe constant of variation, [latex]k[\/latex], is 30.\r\n\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<h1>(8.6.2) - Define inverse variation and solve problems involving inverse variation<\/h1>\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]\\displaystyle 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\u2019s say you need to drive from Boston to Chicago, which is about 1,000 miles. The more time you have, the slower you can go. If you want to get there in 20 hours, you need to go 50 miles per hour (assuming you don\u2019t stop driving!), because [latex]\\displaystyle \\frac{1,000}{20}=50[\/latex]. But if you can take 40 hours to get there, you only have to average 25 miles per hour, since [latex]\\displaystyle \\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]\\displaystyle 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]\\displaystyle r=\\frac{d}{t}[\/latex], or [latex]\\displaystyle speed=\\frac{miles}{time}[\/latex].\r\n\r\nIn the case of the Boston to Chicago trip, you can write [latex]\\displaystyle s=\\frac{1,000}{t}[\/latex]. Notice that this is the same form as the inverse variation function formula, [latex]\\displaystyle y=\\frac{k}{x}[\/latex].\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve for [latex]k[\/latex], 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\"]Write the formula for an inverse variation relationship.\r\n<p style=\"text-align: center\">[latex]\\displaystyle y=\\frac{k}{x}[\/latex]<\/p>\r\nSubstitute known values into the equation.\r\n<p style=\"text-align: center\">[latex]\\displaystyle 25=\\frac{k}{5}[\/latex]<\/p>\r\nSolve for [latex]k[\/latex]\u00a0by multiplying both sides of the equation by 5.\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{c}5\\cdot 25=\\frac{k}{5}\\cdot 5\\\\\\\\125=\\frac{5k}{5}\\\\\\\\125=k\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nThe constant of variation, [latex]k[\/latex], is 125.\r\n\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 500 meters. \u00a0Water 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 1,000 meters, the water temperature is 5\u00ba Celsius. What is the water temperature at a depth of 500 meters?\r\n\r\n[reveal-answer q=\"700119\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"700119\"]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>).\r\n<p style=\"text-align: center\">[latex]\\large \\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]\\displaystyle 5=\\frac{k}{1,000}[\/latex]<\/p>\r\nSolve for [latex]k[\/latex].\r\n<p style=\"text-align: center\">[latex]\\large \\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 500 meters.\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}temp=\\frac{k}{depth}\\\\\\\\temp=\\frac{5,000}{500}\\\\\\\\temp=10\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nAt 500 meters, the water temperature is 10\u00ba 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<div class=\"textbox learning-objectives\">\r\n<h3>Solving variation problems:<\/h3>\r\n<ol>\r\n \t<li>Write an equation that models the statement, relating the two variables with a constant<em>\u00a0<\/em>[latex]k[\/latex]<\/li>\r\n \t<li>Substitute the given pair of values into the equation, and solve for\u00a0[latex]k[\/latex].<\/li>\r\n \t<li>Substitute the (now known) numerical value of\u00a0[latex]k[\/latex]\u00a0into the original equation<\/li>\r\n \t<li>Use the equation in the previous step to answer the question<\/li>\r\n<\/ol>\r\n<\/div>\r\n<h1>(8.6.3) - Define joint variation and solve problems involving joint variation<\/h1>\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 30 inches[latex]^{2}[\/latex]\u00a0when the base is 10 inches and the height is 6 inches, find the variation constant and the area of a triangle whose base is 15 inches and height is 20 inches.\r\n\r\n[reveal-answer q=\"264626\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"264626\"]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 [latex]b[\/latex]\u00a0and height [latex]h[\/latex].\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 [latex]k[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l} 30=k\\left(10\\right)\\left(6\\right)\\\\30=60k\\\\ \\displaystyle \\frac{30}{60}=\\frac{60k}{60}\\\\ \\displaystyle \\frac{1}{2}=k \\end{array}[\/latex]<\/p>\r\nNow that [latex]k[\/latex]\u00a0is known, solve for the area of a triangle whose base is 15 inches and height is 20 inches.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}Area=k(base)(height)\\\\\\\\Area=(15)(20)\\left(\\displaystyle \\frac{1}{2}\\right)\\\\ Area= \\displaystyle \\frac{300}{2}\\\\Area=150\\,\\,\\text{square inches}\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nThe constant of variation, [latex]k[\/latex], is [latex]\\displaystyle \\frac{1}{2}[\/latex], and the area of the triangle is 150 square inches.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nFinding [latex]k[\/latex]\u00a0to be [latex]\\displaystyle \\frac{1}{2}[\/latex] shouldn\u2019t be surprising. You know that the area of a triangle is one-half base times height, [latex]\\displaystyle A=\\frac{1}{2}bh[\/latex]. The [latex]\\displaystyle \\frac{1}{2}[\/latex] in this formula is exactly the same [latex]\\displaystyle \\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[latex]k[\/latex] 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]\\displaystyle y=\\frac{k}{x}[\/latex]<\/li>\r\n \t<li>Joint variation: [latex]y=kxz[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>(8.6.1) &#8211; Define direct variation, and solve problems involving direct variation<\/li>\n<li>(8.6.2) &#8211; Define inverse variation and solve problems involving inverse variation<\/li>\n<li>(8.6.3) &#8211; Define joint variation and solve problems involving joint variation<\/li>\n<\/ul>\n<\/div>\n<h1>(8.6.1) &#8211; Define direct variation, and solve problems involving direct variation<\/h1>\n<div id=\"attachment_5075\" style=\"width: 492px\" class=\"wp-caption alignleft\"><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<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 4 tells you the rate at which cars and tires are related. You call the rate the <strong>constant of variation<\/strong>. It\u2019s 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, 4 is the constant, and the number of cars is the input. Let\u2019s enter those generic terms into the equation. You get [latex]y=kx[\/latex]. That\u2019s 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 [latex]k[\/latex], 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\">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 [latex]k[\/latex] by dividing both sides of the equation by 10.<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}\\frac{300}{10}=\\frac{10k}{10}\\\\\\\\\\,\\,\\,\\,30=k\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>The constant of variation, [latex]k[\/latex], is 30.<\/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<h1>(8.6.2) &#8211; Define inverse variation and solve problems involving inverse variation<\/h1>\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]\\displaystyle 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\u2019s say you need to drive from Boston to Chicago, which is about 1,000 miles. The more time you have, the slower you can go. If you want to get there in 20 hours, you need to go 50 miles per hour (assuming you don\u2019t stop driving!), because [latex]\\displaystyle \\frac{1,000}{20}=50[\/latex]. But if you can take 40 hours to get there, you only have to average 25 miles per hour, since [latex]\\displaystyle \\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]\\displaystyle 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]\\displaystyle r=\\frac{d}{t}[\/latex], or [latex]\\displaystyle speed=\\frac{miles}{time}[\/latex].<\/p>\n<p>In the case of the Boston to Chicago trip, you can write [latex]\\displaystyle s=\\frac{1,000}{t}[\/latex]. Notice that this is the same form as the inverse variation function formula, [latex]\\displaystyle y=\\frac{k}{x}[\/latex].<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve for [latex]k[\/latex], 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\">Write the formula for an inverse variation relationship.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle y=\\frac{k}{x}[\/latex]<\/p>\n<p>Substitute known values into the equation.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle 25=\\frac{k}{5}[\/latex]<\/p>\n<p>Solve for [latex]k[\/latex]\u00a0by multiplying both sides of the equation by 5.<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{c}5\\cdot 25=\\frac{k}{5}\\cdot 5\\\\\\\\125=\\frac{5k}{5}\\\\\\\\125=k\\,\\,\\,\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>The constant of variation, [latex]k[\/latex], is 125.<\/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 500 meters. \u00a0Water 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 1,000 meters, the water temperature is 5\u00ba Celsius. What is the water temperature at a depth of 500 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\">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]\\large \\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]\\displaystyle 5=\\frac{k}{1,000}[\/latex]<\/p>\n<p>Solve for [latex]k[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\large \\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 500 meters.<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}temp=\\frac{k}{depth}\\\\\\\\temp=\\frac{5,000}{500}\\\\\\\\temp=10\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>At 500 meters, the water temperature is 10\u00ba 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<div class=\"textbox learning-objectives\">\n<h3>Solving variation problems:<\/h3>\n<ol>\n<li>Write an equation that models the statement, relating the two variables with a constant<em>\u00a0<\/em>[latex]k[\/latex]<\/li>\n<li>Substitute the given pair of values into the equation, and solve for\u00a0[latex]k[\/latex].<\/li>\n<li>Substitute the (now known) numerical value of\u00a0[latex]k[\/latex]\u00a0into the original equation<\/li>\n<li>Use the equation in the previous step to answer the question<\/li>\n<\/ol>\n<\/div>\n<h1>(8.6.3) &#8211; Define joint variation and solve problems involving joint variation<\/h1>\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 30 inches[latex]^{2}[\/latex]\u00a0when the base is 10 inches and the height is 6 inches, find the variation constant and the area of a triangle whose base is 15 inches and height is 20 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\">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 [latex]b[\/latex]\u00a0and height [latex]h[\/latex].<\/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 [latex]k[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l} 30=k\\left(10\\right)\\left(6\\right)\\\\30=60k\\\\ \\displaystyle \\frac{30}{60}=\\frac{60k}{60}\\\\ \\displaystyle \\frac{1}{2}=k \\end{array}[\/latex]<\/p>\n<p>Now that [latex]k[\/latex]\u00a0is known, solve for the area of a triangle whose base is 15 inches and height is 20 inches.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}Area=k(base)(height)\\\\\\\\Area=(15)(20)\\left(\\displaystyle \\frac{1}{2}\\right)\\\\ Area= \\displaystyle \\frac{300}{2}\\\\Area=150\\,\\,\\text{square inches}\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>The constant of variation, [latex]k[\/latex], is [latex]\\displaystyle \\frac{1}{2}[\/latex], and the area of the triangle is 150 square inches.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Finding [latex]k[\/latex]\u00a0to be [latex]\\displaystyle \\frac{1}{2}[\/latex] shouldn\u2019t be surprising. You know that the area of a triangle is one-half base times height, [latex]\\displaystyle A=\\frac{1}{2}bh[\/latex]. The [latex]\\displaystyle \\frac{1}{2}[\/latex] in this formula is exactly the same [latex]\\displaystyle \\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>[latex]k[\/latex] 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]\\displaystyle y=\\frac{k}{x}[\/latex]<\/li>\n<li>Joint variation: [latex]y=kxz[\/latex]<\/li>\n<\/ul>\n<\/div>\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-4973\">\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>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>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><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex: Direct Variation Application - Aluminum Can Usage. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <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>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>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":60342,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"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\":\"cc\",\"description\":\"Ex: Direct Variation Application - 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