{"id":11161,"date":"2015-07-14T18:38:25","date_gmt":"2015-07-14T18:38:25","guid":{"rendered":"https:\/\/courses.candelalearning.com\/osprecalc\/?post_type=chapter&p=11161"},"modified":"2015-09-08T23:00:28","modified_gmt":"2015-09-08T23:00:28","slug":"solve-inverse-variation-problems","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math\/chapter\/solve-inverse-variation-problems\/","title":{"raw":"Solve inverse variation problems","rendered":"Solve inverse variation problems"},"content":{"raw":"

Water temperature in an ocean varies inversely to the water\u2019s depth. Between the depths of 250 feet and 500 feet, the formula [latex]T=\\frac{14,000}{d}[\/latex] gives us the temperature in degrees Fahrenheit at a depth in feet below Earth\u2019s surface. Consider the Atlantic Ocean, which covers 22% of Earth\u2019s surface. At a certain location, at the depth of 500 feet, the temperature may be 28\u00b0F.<\/p>\r\n

If we create a table\u00a0we observe that, as the depth increases, the water temperature decreases.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
d<\/em>, depth<\/th>\r\n[latex]T=\\frac{\\text{14,000}}{d}[\/latex]<\/th>\r\nInterpretation<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
500 ft<\/td>\r\n[latex]\\frac{14,000}{500}=28[\/latex]<\/td>\r\nAt a depth of 500 ft, the water temperature is 28\u00b0 F.<\/td>\r\n<\/tr>\r\n
350 ft<\/td>\r\n[latex]\\frac{14,000}{350}=40[\/latex]<\/td>\r\nAt a depth of 350 ft, the water temperature is 40\u00b0 F.<\/td>\r\n<\/tr>\r\n
250 ft<\/td>\r\n[latex]\\frac{14,000}{250}=56[\/latex]<\/td>\r\nAt a depth of 250 ft, the water temperature is 56\u00b0 F.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

We notice in the relationship between these variables that, as one quantity increases, the other decreases. The two quantities are said to be inversely proportional<\/strong> and each term varies inversely<\/strong> with the other. Inversely proportional relationships are also called inverse variations<\/strong>.<\/p>\r\n

For our example, the graph\u00a0depicts the inverse variation<\/strong>. We say the water temperature varies inversely with the depth of the water because, as the depth increases, the temperature decreases. The formula [latex]y=\\frac{k}{x}[\/latex] for inverse variation in this case uses k\u00a0<\/em>= 14,000.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Figure 3<\/b>[\/caption]\r\n\r\n

\r\n

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

If x<\/em>\u00a0and y<\/em>\u00a0are related by an equation of the form<\/p>\r\n\r\n

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

where k<\/em>\u00a0is a nonzero constant, then we say that y<\/em>\u00a0varies inversely<\/strong> with the n<\/em>th power of x<\/em>. In inversely proportional<\/strong> relationships, or inverse variations<\/strong>, there is a constant multiple [latex]k={x}^{n}y[\/latex].<\/p>\r\n\r\n<\/div>\r\n

\r\n
\r\n
\r\n

Example 2: Writing a Formula for an Inversely Proportional Relationship<\/h3>\r\n

A tourist plans to drive 100 miles. Find a formula for the time the trip will take as a function of the speed the tourist drives.<\/p>\r\n\r\n<\/div>\r\n

\r\n

Solution<\/h3>\r\n

Recall that multiplying speed by time gives distance. If we let t<\/em>\u00a0represent the drive time in hours, and v<\/em>\u00a0represent the velocity (speed or rate) at which the tourist drives, then vt\u00a0<\/em>= distance. Because the distance is fixed at 100 miles, vt\u00a0<\/em>= 100. Solving this relationship for the time gives us our function.<\/p>\r\n\r\n

[latex]\\begin{cases}t\\left(v\\right)=\\frac{100}{v}\\hfill \\\\ \\text{ }=100{v}^{-1}\\hfill \\end{cases}[\/latex]<\/div>\r\n

We can see that the constant of variation is 100 and, although we can write the relationship using the negative exponent, it is more common to see it written as a fraction.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n

How To: Given a description of an indirect 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 multiply 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 3: Solving an Inverse Variation Problem<\/h3>\r\n

    A quantity y<\/em>\u00a0varies inversely 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 inverse variation with a cube is [latex]y=\\frac{k}{{x}^{3}}[\/latex]. The constant can be found by multiplying y<\/em>\u00a0by the cube of x<\/em>.<\/p>\r\n\r\n

    [latex]\\begin{cases}k={x}^{3}y\\hfill \\\\ \\text{ }={2}^{3}\\cdot 25\\hfill \\\\ \\text{ }=200\\hfill \\end{cases}[\/latex]<\/div>\r\n

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

    [latex]\\begin{cases}y=\\frac{k}{{x}^{3}},k=200\\hfill \\\\ y=\\frac{200}{{x}^{3}}\\hfill \\end{cases}[\/latex]<\/div>\r\n

    Substitute x\u00a0<\/em>= 6 and solve for y<\/i>.<\/p>\r\n\r\n

    [latex]\\begin{cases}y=\\frac{200}{{6}^{3}}\\hfill \\\\ \\text{ }=\\frac{25}{27}\\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 rational function.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"488\"]\"Graph Figure 4<\/b>[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    \r\n

    Try It 2<\/h3>\r\n

    A quantity y<\/em>\u00a0varies inversely with the square of x<\/em>. If y\u00a0<\/em>= 8 when x\u00a0<\/em>= 3, find y<\/em>\u00a0when x<\/em>\u00a0is 4.<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>","rendered":"

    Water temperature in an ocean varies inversely to the water\u2019s depth. Between the depths of 250 feet and 500 feet, the formula [latex]T=\\frac{14,000}{d}[\/latex] gives us the temperature in degrees Fahrenheit at a depth in feet below Earth\u2019s surface. Consider the Atlantic Ocean, which covers 22% of Earth\u2019s surface. At a certain location, at the depth of 500 feet, the temperature may be 28\u00b0F.<\/p>\n

    If we create a table\u00a0we observe that, as the depth increases, the water temperature decreases.<\/p>\n\n\n\n\n\n\n\n
    d<\/em>, depth<\/th>\n[latex]T=\\frac{\\text{14,000}}{d}[\/latex]<\/th>\nInterpretation<\/th>\n<\/tr>\n<\/thead>\n
    500 ft<\/td>\n[latex]\\frac{14,000}{500}=28[\/latex]<\/td>\nAt a depth of 500 ft, the water temperature is 28\u00b0 F.<\/td>\n<\/tr>\n
    350 ft<\/td>\n[latex]\\frac{14,000}{350}=40[\/latex]<\/td>\nAt a depth of 350 ft, the water temperature is 40\u00b0 F.<\/td>\n<\/tr>\n
    250 ft<\/td>\n[latex]\\frac{14,000}{250}=56[\/latex]<\/td>\nAt a depth of 250 ft, the water temperature is 56\u00b0 F.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    We notice in the relationship between these variables that, as one quantity increases, the other decreases. The two quantities are said to be inversely proportional<\/strong> and each term varies inversely<\/strong> with the other. Inversely proportional relationships are also called inverse variations<\/strong>.<\/p>\n

    For our example, the graph\u00a0depicts the inverse variation<\/strong>. We say the water temperature varies inversely with the depth of the water because, as the depth increases, the temperature decreases. The formula [latex]y=\\frac{k}{x}[\/latex] for inverse variation in this case uses k\u00a0<\/em>= 14,000.<\/p>\n

    \"Graph<\/p>\n

    Figure 3<\/b><\/p>\n<\/div>\n

    \n

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

    If x<\/em>\u00a0and y<\/em>\u00a0are related by an equation of the form<\/p>\n

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

    where k<\/em>\u00a0is a nonzero constant, then we say that y<\/em>\u00a0varies inversely<\/strong> with the n<\/em>th power of x<\/em>. In inversely proportional<\/strong> relationships, or inverse variations<\/strong>, there is a constant multiple [latex]k={x}^{n}y[\/latex].<\/p>\n<\/div>\n

    \n
    \n
    \n

    Example 2: Writing a Formula for an Inversely Proportional Relationship<\/h3>\n

    A tourist plans to drive 100 miles. Find a formula for the time the trip will take as a function of the speed the tourist drives.<\/p>\n<\/div>\n

    \n

    Solution<\/h3>\n

    Recall that multiplying speed by time gives distance. If we let t<\/em>\u00a0represent the drive time in hours, and v<\/em>\u00a0represent the velocity (speed or rate) at which the tourist drives, then vt\u00a0<\/em>= distance. Because the distance is fixed at 100 miles, vt\u00a0<\/em>= 100. Solving this relationship for the time gives us our function.<\/p>\n

    [latex]\\begin{cases}t\\left(v\\right)=\\frac{100}{v}\\hfill \\\\ \\text{ }=100{v}^{-1}\\hfill \\end{cases}[\/latex]<\/div>\n

    We can see that the constant of variation is 100 and, although we can write the relationship using the negative exponent, it is more common to see it written as a fraction.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

    \n

    How To: Given a description of an indirect 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 multiply 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 3: Solving an Inverse Variation Problem<\/h3>\n

      A quantity y<\/em>\u00a0varies inversely 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 inverse variation with a cube is [latex]y=\\frac{k}{{x}^{3}}[\/latex]. The constant can be found by multiplying y<\/em>\u00a0by the cube of x<\/em>.<\/p>\n

      [latex]\\begin{cases}k={x}^{3}y\\hfill \\\\ \\text{ }={2}^{3}\\cdot 25\\hfill \\\\ \\text{ }=200\\hfill \\end{cases}[\/latex]<\/div>\n

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

      [latex]\\begin{cases}y=\\frac{k}{{x}^{3}},k=200\\hfill \\\\ y=\\frac{200}{{x}^{3}}\\hfill \\end{cases}[\/latex]<\/div>\n

      Substitute x\u00a0<\/em>= 6 and solve for y<\/i>.<\/p>\n

      [latex]\\begin{cases}y=\\frac{200}{{6}^{3}}\\hfill \\\\ \\text{ }=\\frac{25}{27}\\hfill \\end{cases}[\/latex]<\/div>\n<\/div>\n
      \n

      Analysis of the Solution<\/h3>\n

      The graph of this equation is a rational function.<\/p>\n

      \"Graph<\/p>\n

      Figure 4<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      \n

      Try It 2<\/h3>\n

      A quantity y<\/em>\u00a0varies inversely with the square of x<\/em>. If y\u00a0<\/em>= 8 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

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