{"id":6617,"date":"2020-10-02T22:01:11","date_gmt":"2020-10-02T22:01:11","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-beginalgebra\/?post_type=chapter&p=6617"},"modified":"2021-06-01T22:28:56","modified_gmt":"2021-06-01T22:28:56","slug":"1-8-variation-and-percents","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/1-8-variation-and-percents\/","title":{"raw":"1.9: Applications - Variation","rendered":"1.9: Applications – Variation"},"content":{"raw":"
\r\n

Section 1.9 Learning Objectives<\/h3>\r\n1.9: Applications - Variation<\/strong>\r\n
    \r\n \t
  • Solve a direct variation problem<\/li>\r\n \t
  • Solve an inverse variation problem<\/li>\r\n \t
  • Solve a joint variation problem<\/li>\r\n \t
  • Solve application problems involving direct, inverse, and joint variation<\/li>\r\n<\/ul>\r\n<\/div>\r\n \r\n

    Variation<\/h2>\r\n[caption id=\"attachment_5075\" align=\"aligncenter\" width=\"482\"]\"Huge So many cars, so many tires.[\/caption]\r\n

    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

    [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 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 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\r\n[latex]\\text{number of tires}=4\\cdot\\text{number of cars}\\\\\\text{output}=\\text{constant}\\cdot\\text{input}[\/latex]\r\n

    \r\n

    Direct variation Equation<\/h3>\r\n

    To describe the relationship of\u00a0direct variation, or directly <\/b>proportional<\/b><\/span>, <\/b>\"y varies directly as x\"\u00a0<\/em>translates to<\/p>\r\n

    [latex]y=kx[\/latex],<\/p>\r\n

    where [latex]k[\/latex] is referred to as the constant of variation.<\/p>\r\n\r\n<\/div>\r\n

    \r\n

    Example 1<\/h3>\r\nSolve for [latex]k[\/latex], the constant of variation, if [latex]y[\/latex] varies directly as [latex]x[\/latex], 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

    [latex]y=kx[\/latex]<\/p>\r\nSubstitute known values into the equation.\r\n

    [latex]300=k\\left(10\\right)[\/latex]<\/p>\r\nSolve for k<\/i> by dividing both sides of the equation by 10.\r\n

    [latex]\\begin{array}{l}\\frac{300}{10}=\\frac{10k}{10}\\\\\\\\\\,\\,\\,\\,30=k\\end{array}[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\nThe constant of variation, k<\/i>, 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 where we first need to find k, and then use k to determine y when x changes to a new value.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=rQUSuw0hq7U\r\n

    Inverse Variation<\/h2>\r\nAnother kind of variation is called 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\n

    Inverse variation Equation<\/h3>\r\n

    To describe the relationship of\u00a0inverse variation, or inversely <\/b>proportional<\/b><\/span>,\u00a0<\/b>\"y varies inversely as x\"\u00a0<\/em>translates to<\/p>\r\n

    [latex] y=\\frac{k}{x}[\/latex],<\/p>\r\n

    where [latex]k[\/latex] is referred to as the constant of variation.<\/p>\r\n \r\n\r\nNote:\u00a0 Inverse variation relationship are sometimes also referred to as indirect varation<\/strong>.\u00a0 Therefore, a problem may read \"y varies indirectly with x\"<\/em> instead of \"y varies inversely as x.\"<\/em>\r\n\r\n<\/div>\r\n

    \r\n

    Example 2<\/h3>\r\nSolve for k<\/i>, the constant of variation, if y varies inversely as x, 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

    [latex] y=\\frac{k}{x}[\/latex]<\/p>\r\nSubstitute known values into the equation.\r\n

    [latex] 25=\\frac{k}{5}[\/latex]<\/p>\r\nSolve for k<\/i> by multiplying both sides of the equation by 5.\r\n

    [latex] \\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

    Answer<\/h4>\r\nThe constant of variation, k<\/i>, is 125.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the video that follows, we present an example of solving an inverse variation equation where we first need to find k, and then use k to determine y when x changes to a new value.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=CA-LflxO0tM\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\n

    Example 3<\/h3>\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] \\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] \\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 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\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\"]\"Scuba Water temperature in the ocean varies inversely with depth.[\/caption]\r\n\r\n
    \r\n

    Example 4<\/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 (y<\/i>) varies inversely with the depth of the water (x<\/i>).\r\n

    [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

    [latex] 5=\\frac{k}{1,000}[\/latex]<\/p>\r\nSolve for k<\/i>.\r\n

    [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 k<\/em>, the constant of variation is known, use that information to solve the problem: find the water temperature at 500 meters.\r\n

    [latex]\\begin{array}{l}temp=\\frac{k}{depth}\\\\\\\\temp=\\frac{5,000}{500}\\\\\\\\temp=10\\end{array}[\/latex]<\/p>\r\n\r\n

    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&feature=youtu.be\r\n

    Joint Variation<\/h2>\r\nA third type of variation is called joint variation<\/strong>. Joint variation is the same as direct variation except there are two or more quantities. With these problems, as with the other variation problems, we first must solve for [latex]k[\/latex], the constant of variation, and then use that to answer any other parts that the question asked.\r\n
    \r\n

    Joint Variation Equation<\/h3>\r\n

    To describe the relationship of\u00a0joint variation<\/b>,\u00a0<\/b>\"y varies jointly with x and z\u00a0<\/em>translates to<\/p>\r\n

    [latex]y=kxz[\/latex],<\/p>\r\n

    where [latex]k[\/latex] is referred to as the constant of variation.<\/p>\r\n\r\n<\/div>\r\nOne example of joint variation can be found in the area of a rectangle. The area of a rectangle can be found using the formula [latex]A=lw[\/latex], where [latex]l[\/latex]\u00a0is the length of the rectangle and [latex]w[\/latex]\u00a0\u00a0<\/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\nLet's look at another joint variation example, with the area of a triangle.\r\n

    \r\n

    Example 5<\/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 (A<\/i>) varies jointly with the lengths of the base (b<\/i>) and height (h<\/i>).\r\n

    [latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y=kxz\\\\\\text{Area}=k(\\text{base})(\\text{height})\\end{array}[\/latex]<\/p>\r\nSubstitute known values into the equation, and solve for k<\/i>.\r\n

    [latex]\\begin{array}{l}30=k\\left(10\\right)\\left(6\\right)\\\\30=60k\\\\\\\\\\frac{30}{60}=\\frac{60k}{60}\\\\\\\\\\frac{1}{2}=k\\end{array}[\/latex]<\/p>\r\nNow that k<\/i> is known, solve for the area of a triangle whose base is 15 inches and height is 20 inches.\r\n

    [latex]\\text{Area}=k(\\text{base})(\\text{height})[\/latex]<\/p>\r\n

    [latex]\\text{Area}=(\\frac{1}{2})(15)(20)[\/latex]<\/p>\r\n

    [latex]\\text{Area}=\\frac{300}{2}[\/latex]<\/span><\/p>\r\n

    [latex]\\text{Area}=150\\,\\,\\text{square inches}[\/latex]<\/span><\/p>\r\n\r\n

    Answer<\/h4>\r\nThe constant of variation, k<\/i>, is [latex] \\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\nIn the example above, finding k<\/i> to be [latex] \\frac{1}{2}[\/latex] shouldn\u2019t 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\nAnother example of joint variation is within the formula for the volume of a cylinder, [latex]V=\\pi {{r}^{2}}h[\/latex]. 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\r\nIn the following video, we show an example of\u00a0finding the constant of variation for another jointly varying relation.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=JREPATMScbM&feature=youtu.be\r\n
    \r\n

    Direct, Joint, and Inverse Variation<\/h3>\r\nk<\/i> is the constant of variation. In all cases, [latex]k\\neq0[\/latex].\r\n
      \r\n \t
    • Direct variation: [latex]y=kx[\/latex]<\/li>\r\n \t
    • Inverse variation: [latex] y=\\frac{k}{x}[\/latex]<\/li>\r\n \t
    • Joint variation: [latex]y=kxz[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n

      Summary of Variation<\/h2>\r\nDirect, 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.\r\n\r\n
      ","rendered":"
      \n

      Section 1.9 Learning Objectives<\/h3>\n

      1.9: Applications – Variation<\/strong><\/p>\n

        \n
      • Solve a direct variation problem<\/li>\n
      • Solve an inverse variation problem<\/li>\n
      • Solve a joint variation problem<\/li>\n
      • Solve application problems involving direct, inverse, and joint variation<\/li>\n<\/ul>\n<\/div>\n

         <\/p>\n

        Variation<\/h2>\n
        \"Huge<\/p>\n

        So many cars, so many tires.<\/p>\n<\/div>\n

        Direct Variation<\/h2>\n

        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

        [latex]\\text{number of tires}=4\\cdot\\text{number of cars}[\/latex]<\/p>\n

        The number 4 tells you the rate at which cars and tires are related. You call the rate the 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 direct variation<\/strong>, where the number of tires varies directly with the number of cars.<\/p>\n

        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

        [latex]\\text{number of tires}=4\\cdot\\text{number of cars}\\\\\\text{output}=\\text{constant}\\cdot\\text{input}[\/latex]<\/p>\n

        \n

        Direct variation Equation<\/h3>\n

        To describe the relationship of\u00a0direct variation, or directly <\/b>proportional<\/b><\/span>, <\/b>“y varies directly as x”\u00a0<\/em>translates to<\/p>\n

        [latex]y=kx[\/latex],<\/p>\n

        where [latex]k[\/latex] is referred to as the constant of variation.<\/p>\n<\/div>\n

        \n

        Example 1<\/h3>\n

        Solve for [latex]k[\/latex], the constant of variation, if [latex]y[\/latex] varies directly as [latex]x[\/latex], where\u00a0[latex]y=300[\/latex] and [latex]x=10[\/latex].<\/p>\n

        Show Solution<\/span><\/p>\n
        Write the formula for a direct variation relationship.<\/p>\n

        [latex]y=kx[\/latex]<\/p>\n

        Substitute known values into the equation.<\/p>\n

        [latex]300=k\\left(10\\right)[\/latex]<\/p>\n

        Solve for k<\/i> by dividing both sides of the equation by 10.<\/p>\n

        [latex]\\begin{array}{l}\\frac{300}{10}=\\frac{10k}{10}\\\\\\\\\\,\\,\\,\\,30=k\\end{array}[\/latex]<\/p>\n

        Answer<\/h4>\n

        The constant of variation, k<\/i>, is 30.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

        In the video that follows, we present an example of solving a direct variation equation where we first need to find k, and then use k to determine y when x changes to a new value.<\/p>\n