{"id":11159,"date":"2015-07-14T18:38:10","date_gmt":"2015-07-14T18:38:10","guid":{"rendered":"https:\/\/courses.candelalearning.com\/osprecalc\/?post_type=chapter&p=11159"},"modified":"2015-09-10T18:13:59","modified_gmt":"2015-09-10T18:13:59","slug":"solve-direct-variation-problems","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/chapter\/solve-direct-variation-problems\/","title":{"raw":"Solve direct variation problems","rendered":"Solve direct variation problems"},"content":{"raw":"

In the example above, Nicole\u2019s earnings can be found by multiplying her sales by her commission. The formula e<\/em> = 0.16s<\/em> tells us her earnings, e<\/em>, come from the product of 0.16, her commission, and the sale price of the vehicle. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
s<\/em>, sales prices<\/th>\r\ne<\/em> = 0.16s<\/em><\/th>\r\nInterpretation<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
$4,600<\/td>\r\ne\u00a0<\/em>= 0.16(4,600) = 736<\/td>\r\nA sale of a $4,600 vehicle results in $736 earnings.<\/td>\r\n<\/tr>\r\n
$9,200<\/td>\r\ne\u00a0<\/em>= 0.16(9,200) = 1,472<\/td>\r\nA sale of a $9,200 vehicle results in $1472 earnings.<\/td>\r\n<\/tr>\r\n
$18,400<\/td>\r\ne\u00a0<\/em>= 0.16(18,400) = 2,944<\/td>\r\nA sale of a $18,400 vehicle results in $2944 earnings.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from $4,600 to $9,200, and we double the earnings from $736 to $1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called direct variation<\/strong>. Each variable in this type of relationship varies directly <\/strong>with the other.<\/p>\r\n

The graph below\u00a0represents the data for Nicole\u2019s potential earnings. We say that earnings vary directly with the sales price of the car. The formula [latex]y=k{x}^{n}[\/latex] is used for direct variation. The value k<\/em>\u00a0is a nonzero constant greater than zero and is called the constant of variation<\/strong>. In this case, k\u00a0<\/em>= 0.16 and n\u00a0<\/em>= 1.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Figure 1<\/b>[\/caption]\r\n\r\n

\r\n

A General Note: Direct Variation<\/h3>\r\n

If x <\/em>and y<\/em>\u00a0are related by an equation of the form<\/p>\r\n\r\n

[latex]y=k{x}^{n}[\/latex]<\/div>\r\n

then we say that the relationship is direct variation<\/strong> and y<\/em>\u00a0varies directly<\/strong> with the n<\/em>th power of x<\/em>. In direct variation relationships, there is a nonzero constant ratio [latex]k=\\frac{y}{{x}^{n}}[\/latex], where k<\/em>\u00a0is called the constant of variation<\/strong>, which help defines the relationship between the variables.<\/p>\r\n\r\n<\/div>\r\n

\r\n

How To: Given a description of a direct variation problem, solve for an unknown.\r\n<\/strong><\/h3>\r\n
    \r\n\t
  1. Identify the input, x<\/em>, and the output, y<\/em>.<\/li>\r\n\t
  2. Determine the constant of variation. You may need to divide y<\/em>\u00a0by the specified power of x<\/em>\u00a0to determine the constant of variation.<\/li>\r\n\t
  3. Use the constant of variation to write an equation for the relationship.<\/li>\r\n\t
  4. Substitute known values into the equation to find the unknown.<\/li>\r\n<\/ol>\r\n<\/div>\r\n
    \r\n
    \r\n
    \r\n

    Example 1: Solving a Direct Variation Problem<\/h3>\r\n

    The quantity y<\/em>\u00a0varies directly with the cube of x<\/em>. If y\u00a0<\/em>= 25 when x\u00a0<\/em>= 2, find y<\/em>\u00a0when x<\/em>\u00a0is 6.<\/p>\r\n\r\n<\/div>\r\n

    \r\n

    Solution<\/h3>\r\n

    The general formula for direct variation with a cube is [latex]y=k{x}^{3}[\/latex]. The constant can be found by dividing y<\/em>\u00a0by the cube of x<\/em>.<\/p>\r\n\r\n

    [latex]\\begin{cases} k=\\frac{y}{{x}^{3}} \\\\ =\\frac{25}{{2}^{3}}\\\\ =\\frac{25}{8}\\end{cases}[\/latex]<\/div>\r\n

    Now use the constant to write an equation that represents this relationship.<\/p>\r\n\r\n

    [latex]y=\\frac{25}{8}{x}^{3}[\/latex]<\/div>\r\n

    Substitute x<\/em> = 6 and solve for y<\/em>.<\/p>\r\n\r\n

    [latex]\\begin{cases}y=\\frac{25}{8}{\\left(6\\right)}^{3}\\hfill \\\\ \\text{ }=675\\hfill \\end{cases}[\/latex]<\/div>\r\n<\/div>\r\n
    \r\n

    Analysis of the Solution<\/h3>\r\n

    The graph of this equation is a simple cubic, as shown below.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Figure 2<\/b>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    \r\n

    Q & A<\/h3>\r\n

    Do the graphs of all direct variation equations look like Example 1?<\/strong><\/p>\r\n

    No. Direct variation equations are power functions\u2014they may be linear, quadratic, cubic, quartic, radical, etc. But all of the graphs pass through (0, 0).<\/em><\/p>\r\n\r\n<\/div>\r\n

    \r\n

    Try It 1<\/h3>\r\n

    The quantity y<\/em>\u00a0varies directly with the square of x<\/em>. If y\u00a0<\/em>= 24 when x\u00a0<\/em>= 3, find y<\/em>\u00a0when x<\/em>\u00a0is 4.<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>","rendered":"

    In the example above, Nicole\u2019s earnings can be found by multiplying her sales by her commission. The formula e<\/em> = 0.16s<\/em> tells us her earnings, e<\/em>, come from the product of 0.16, her commission, and the sale price of the vehicle. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive.<\/p>\n\n\n\n\n\n\n\n
    s<\/em>, sales prices<\/th>\ne<\/em> = 0.16s<\/em><\/th>\nInterpretation<\/th>\n<\/tr>\n<\/thead>\n
    $4,600<\/td>\ne\u00a0<\/em>= 0.16(4,600) = 736<\/td>\nA sale of a $4,600 vehicle results in $736 earnings.<\/td>\n<\/tr>\n
    $9,200<\/td>\ne\u00a0<\/em>= 0.16(9,200) = 1,472<\/td>\nA sale of a $9,200 vehicle results in $1472 earnings.<\/td>\n<\/tr>\n
    $18,400<\/td>\ne\u00a0<\/em>= 0.16(18,400) = 2,944<\/td>\nA sale of a $18,400 vehicle results in $2944 earnings.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from $4,600 to $9,200, and we double the earnings from $736 to $1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called direct variation<\/strong>. Each variable in this type of relationship varies directly <\/strong>with the other.<\/p>\n

    The graph below\u00a0represents the data for Nicole\u2019s potential earnings. We say that earnings vary directly with the sales price of the car. The formula [latex]y=k{x}^{n}[\/latex] is used for direct variation. The value k<\/em>\u00a0is a nonzero constant greater than zero and is called the constant of variation<\/strong>. In this case, k\u00a0<\/em>= 0.16 and n\u00a0<\/em>= 1.<\/p>\n

    \"Graph<\/p>\n

    Figure 1<\/b><\/p>\n<\/div>\n

    \n

    A General Note: Direct Variation<\/h3>\n

    If x <\/em>and y<\/em>\u00a0are related by an equation of the form<\/p>\n

    [latex]y=k{x}^{n}[\/latex]<\/div>\n

    then we say that the relationship is direct variation<\/strong> and y<\/em>\u00a0varies directly<\/strong> with the n<\/em>th power of x<\/em>. In direct variation relationships, there is a nonzero constant ratio [latex]k=\\frac{y}{{x}^{n}}[\/latex], where k<\/em>\u00a0is called the constant of variation<\/strong>, which help defines the relationship between the variables.<\/p>\n<\/div>\n

    \n

    How To: Given a description of a direct variation problem, solve for an unknown.
    \n<\/strong><\/h3>\n
      \n
    1. Identify the input, x<\/em>, and the output, y<\/em>.<\/li>\n
    2. Determine the constant of variation. You may need to divide y<\/em>\u00a0by the specified power of x<\/em>\u00a0to determine the constant of variation.<\/li>\n
    3. Use the constant of variation to write an equation for the relationship.<\/li>\n
    4. Substitute known values into the equation to find the unknown.<\/li>\n<\/ol>\n<\/div>\n
      \n
      \n
      \n

      Example 1: Solving a Direct Variation Problem<\/h3>\n

      The quantity y<\/em>\u00a0varies directly with the cube of x<\/em>. If y\u00a0<\/em>= 25 when x\u00a0<\/em>= 2, find y<\/em>\u00a0when x<\/em>\u00a0is 6.<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

      The general formula for direct variation with a cube is [latex]y=k{x}^{3}[\/latex]. The constant can be found by dividing y<\/em>\u00a0by the cube of x<\/em>.<\/p>\n

      [latex]\\begin{cases} k=\\frac{y}{{x}^{3}} \\\\ =\\frac{25}{{2}^{3}}\\\\ =\\frac{25}{8}\\end{cases}[\/latex]<\/div>\n

      Now use the constant to write an equation that represents this relationship.<\/p>\n

      [latex]y=\\frac{25}{8}{x}^{3}[\/latex]<\/div>\n

      Substitute x<\/em> = 6 and solve for y<\/em>.<\/p>\n

      [latex]\\begin{cases}y=\\frac{25}{8}{\\left(6\\right)}^{3}\\hfill \\\\ \\text{ }=675\\hfill \\end{cases}[\/latex]<\/div>\n<\/div>\n
      \n

      Analysis of the Solution<\/h3>\n

      The graph of this equation is a simple cubic, as shown below.<\/p>\n

      \"Graph<\/p>\n

      Figure 2<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      \n

      Q & A<\/h3>\n

      Do the graphs of all direct variation equations look like Example 1?<\/strong><\/p>\n

      No. Direct variation equations are power functions\u2014they may be linear, quadratic, cubic, quartic, radical, etc. But all of the graphs pass through (0, 0).<\/em><\/p>\n<\/div>\n

      \n

      Try It 1<\/h3>\n

      The quantity y<\/em>\u00a0varies directly with the square of x<\/em>. If y\u00a0<\/em>= 24 when x\u00a0<\/em>= 3, find y<\/em>\u00a0when x<\/em>\u00a0is 4.<\/p>\n

      Solution<\/a><\/p>\n<\/div>\n\n\t\t\t

      \n\t\t\t

      Candela Citations<\/h3>\n\t\t\t\t\t
      \n\t\t\t\t\t\t
      \n\t\t\t\t\t\t\t
      CC licensed content, Shared previously<\/div>