{"id":1370,"date":"2024-05-10T18:40:07","date_gmt":"2024-05-10T18:40:07","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&p=1370"},"modified":"2025-08-13T15:49:02","modified_gmt":"2025-08-13T15:49:02","slug":"applications-and-inequalities-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/applications-and-inequalities-learn-it-3\/","title":{"raw":"Applications of Non-Linear Equations: Learn It 3","rendered":"Applications of Non-Linear Equations: Learn It 3"},"content":{"raw":"

Real-world Applications<\/h2>\r\nUnderstanding and applying non-linear equations is crucial in many fields, including physics, engineering, economics, and natural sciences. They allow us to model and analyze complex phenomena that don't follow simple linear patterns.\r\n\r\n
\r\n\r\n[caption id=\"attachment_1371\" align=\"alignright\" width=\"154\"]\"\" Pendulum movement[\/caption]\r\n\r\nThe period of a pendulum, [latex]T[\/latex] is modeled by\r\n[latex]T = 2 \\pi \\sqrt{\\dfrac{L}{g}}[\/latex]\r\nwhere the length of the pendulum is [latex]L[\/latex] and the acceleration due to gravity is [latex]g[\/latex].If the acceleration due to gravity is [latex]9.8 \\frac{\\text{m}}{\\text{s}^2}[\/latex] and the period equals [latex]1 s[\/latex], find the length to the nearest cm.[reveal-answer q=\"529991\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"529991\"]Given: [latex]T = 1 \\, \\text{s}[\/latex] and [latex]g = 9.8 \\, \\frac{\\text{m}}{\\text{s}^2}[\/latex].Step 1: Substitute the values into the formula\r\n

[latex]T = 2\\pi \\sqrt{\\frac{L}{g}}[\/latex]\r\n[latex]1 = 2\\pi \\sqrt{\\frac{L}{9.8}}[\/latex]<\/p>\r\nStep 2: Solve for [latex]L[\/latex]\r\n

[latex]\\frac{1}{2\\pi} = \\sqrt{\\frac{L}{9.8}}[\/latex]\r\n[latex]\\left(\\frac{1}{2\\pi}\\right)^2 = \\frac{L}{9.8}[\/latex]\r\n[latex]L = \\left(\\frac{1}{2\\pi}\\right)^2 \\times 9.8[\/latex]\r\n[latex]L = \\frac{9.8}{4\\pi^2}[\/latex]\r\n[latex]L = \\frac{9.8}{4 \\times (3.14159)^2}[\/latex]\r\n[latex]L = \\frac{9.8}{39.4784}[\/latex]\r\n[latex]L = 0.248 \\text{m}[\/latex]<\/p>\r\nConvert to centimeters: [latex]L = 0.248 \\text{m} \\cdot 100 \\text{cm\/m} = 24.6 \\text{cm} \\approx 25 \\text{cm}. [\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

[ohm2_question hide_question_numbers=1]18991[\/ohm2_question]<\/section>
Joe and John are planning to paint a house together.\r\n
    \r\n \t
  • John thinks that if he worked alone, it would take him 3 times as long as it would take Joe to paint the entire house.<\/li>\r\n \t
  • Working together, they can complete the job in 24 hours.<\/li>\r\n<\/ul>\r\nHow long would it take each of them, working alone, to complete the job?\r\n\r\n
    \r\n\r\nTo find out how long it would take Joe and John to paint the house individually, we start by setting up an equation based on their working rates and then solve for the individual times.\r\n
      \r\n \t
    • Joe's rate of painting is [latex]\\frac{1}{t}[\/latex] of the house per hour.<\/li>\r\n \t
    • John's rate of painting is [latex]\\frac{1}{3t}[\/latex] of the house per hour.<\/li>\r\n<\/ul>\r\nTogether, they finish the job in [latex]24[\/latex] hours. Thus, their combined rate is: [latex]\\frac{1}{24}[\/latex] of the house per hour.\r\n\r\nTheir rates add up to the total rate:\r\n

      [latex]\\frac{1}{t} + \\frac{1}{3t} = \\frac{1}{24}[\/latex]<\/p>\r\nWe can use this to find the time for Joe.\r\n\r\n

      [latex]\\begin{align*} \\text{Combine individual rates} & : & \\frac{1}{t} + \\frac{1}{3t} &= \\frac{1}{24} & \\\\ \\text{Simplify the equation} & : & \\frac{4}{3t} &= \\frac{1}{24} & \\\\ \\text{Clear the fraction by cross-multiplying} & : & 4 \\times 24 &= 3t & \\\\ \\text{Calculate the product} & : & 96 &= 3t & \\\\ \\text{Solve for } t & : & t &= \\frac{96}{3} & \\\\ \\text{Simplify to find the time for Joe alone} & : & t &= 32 \\text{ hours} & \\end{align*}[\/latex]<\/center>\r\nNow we can determine John's time.\r\n\r\n
      [latex]3t = 96 \\text{ hours}[\/latex]<\/center>\r\nThus, Joe would need [latex]32[\/latex] hours and John would need [latex]96[\/latex] hours to paint the house alone.\r\n\r\n<\/section>
      [ohm2_question hide_question_numbers=1]18992[\/ohm2_question]<\/section>","rendered":"

      Real-world Applications<\/h2>\n

      Understanding and applying non-linear equations is crucial in many fields, including physics, engineering, economics, and natural sciences. They allow us to model and analyze complex phenomena that don’t follow simple linear patterns.<\/p>\n

      \n
      \"\"
      Pendulum movement<\/figcaption><\/figure>\n

      The period of a pendulum, [latex]T[\/latex] is modeled by
      \n[latex]T = 2 \\pi \\sqrt{\\dfrac{L}{g}}[\/latex]
      \nwhere the length of the pendulum is [latex]L[\/latex] and the acceleration due to gravity is [latex]g[\/latex].If the acceleration due to gravity is [latex]9.8 \\frac{\\text{m}}{\\text{s}^2}[\/latex] and the period equals [latex]1 s[\/latex], find the length to the nearest cm.<\/p>\n