{"id":2141,"date":"2024-07-10T00:29:19","date_gmt":"2024-07-10T00:29:19","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&p=2141"},"modified":"2025-06-25T18:52:06","modified_gmt":"2025-06-25T18:52:06","slug":"applications-of-exponential-functions-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/applications-of-exponential-functions-learn-it-4\/","title":{"raw":"Applications of Exponential Functions: Learn It 4","rendered":"Applications of Exponential Functions: Learn It 4"},"content":{"raw":"

Investigating Continuous Growth<\/h2>\r\nSo far we have worked with rational bases for exponential functions. For most real-world phenomena, however, [latex]e[\/latex] is used as the base for exponential functions. Exponential models that use [latex]e[\/latex] as the base are called continuous growth or decay models. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.\r\n\r\n
\r\n

the continuous growth\/decay formula<\/h3>\r\nFor all real numbers [latex]t[\/latex], and all positive numbers [latex]a[\/latex] and [latex]r[\/latex], continuous growth or decay is represented by the formula\r\n

[latex]A(t) = Pe^{rt}[\/latex]<\/p>\r\nwhere\r\n

    \r\n \t
  • [latex]P[\/latex] is the initial value or the principal<\/li>\r\n \t
  • [latex]r[\/latex] is the growth or interest rate per unit time<\/li>\r\n \t
  • [latex]t[\/latex] is the period or term of the investment<\/li>\r\n<\/ul>\r\nIf [latex]r \\gt 0[\/latex] then the formula represents continuous growth. <\/span>\r\n\r\nIf [latex]r \\lt 0[\/latex] then the formula represents continuous decay.<\/span>\r\n\r\n<\/section>
    How To: Given the initial value, rate of growth or decay, and time [latex]t[\/latex]<\/em>, solve a continuous growth or decay function<\/strong>\r\n
      \r\n \t
    1. Use the information in the problem to determine [latex]P[\/latex], the initial value of the function.<\/li>\r\n \t
    2. Use the information in the problem to determine the growth rate [latex]r[\/latex].\r\n
        \r\n \t
      • If the problem refers to continuous growth, then [latex]r\u00a0> 0[\/latex].<\/li>\r\n \t
      • If the problem refers to continuous decay, then [latex]r\u00a0< 0[\/latex].<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t
      • Use the information in the problem to determine the time [latex]t[\/latex].<\/li>\r\n \t
      • Substitute the given information into the continuous growth formula and solve for [latex]A(t)[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>
        A person invested [latex]$1,000[\/latex] in an account earning a nominal [latex]10\\%[\/latex] per year compounded continuously. How much was in the account at the end of one year?[reveal-answer q=\"864251\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"864251\"]Since the account is growing in value, this is a continuous compounding problem with growth rate [latex]r\u00a0= 0.10[\/latex]. The initial investment was [latex]$1,000[\/latex], so [latex]P\u00a0= 1000[\/latex]. We use the continuous compounding formula to find the value after [latex]t\u00a0= 1[\/latex] year:\r\n

        [latex]\\begin{array}{c}A\\left(t\\right)\\hfill & =P{e}^{rt}\\hfill & \\text{Use the continuous compounding formula}.\\hfill \\\\ \\hfill & =1000{\\left(e\\right)}^{0.1} & \\text{Substitute known values for }P, r,\\text{ and }t.\\hfill \\\\ \\hfill & \\approx 1105.17\\hfill & \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/p>\r\nThe account is worth [latex]$1,105.17[\/latex] after one year.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

        [ohm2_question hide_question_numbers=1]24718[\/ohm2_question]<\/section>
        Radon-222 decays at a continuous rate of [latex]17.3 \\%[\/latex] per day. How much will [latex]100[\/latex] mg of Radon-[latex]222[\/latex] decay to in [latex]3[\/latex] days?[reveal-answer q=\"660828\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"660828\"]Since the substance is decaying, the rate, [latex]17.3%[\/latex], is negative. So, [latex]r=-0.173[\/latex]. The initial amount of radon-[latex]222[\/latex] was [latex]100[\/latex] mg, so [latex]a= 100[\/latex]. We use the continuous decay formula to find the value after [latex]t= 3[\/latex] days:\r\n

        [latex]\\begin{array}{c}A\\left(t\\right)\\hfill & =a{e}^{rt}\\hfill & \\text{Use the continuous growth formula}.\\hfill \\\\ \\hfill & =100{e}^{-0.173\\left(3\\right)} & \\text{Substitute known values for }a, r,\\text{ and }t.\\hfill \\\\ \\hfill & \\approx 59.5115\\hfill & \\text{Use a calculator to approximate}.\\hfill \\end{array}[\/latex]<\/p>\r\nSo [latex]59.5115[\/latex] mg of radon-[latex]222[\/latex] will remain.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

        [ohm2_question hide_question_numbers=1]24719[\/ohm2_question]<\/section>
        [ohm2_question hide_question_numbers=1]24720[\/ohm2_question]<\/section>","rendered":"

        Investigating Continuous Growth<\/h2>\n

        So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, [latex]e[\/latex] is used as the base for exponential functions. Exponential models that use [latex]e[\/latex] as the base are called continuous growth or decay models. We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.<\/p>\n

        \n

        the continuous growth\/decay formula<\/h3>\n

        For all real numbers [latex]t[\/latex], and all positive numbers [latex]a[\/latex] and [latex]r[\/latex], continuous growth or decay is represented by the formula<\/p>\n

        [latex]A(t) = Pe^{rt}[\/latex]<\/p>\n

        where<\/p>\n

          \n
        • [latex]P[\/latex] is the initial value or the principal<\/li>\n
        • [latex]r[\/latex] is the growth or interest rate per unit time<\/li>\n
        • [latex]t[\/latex] is the period or term of the investment<\/li>\n<\/ul>\n

          If [latex]r \\gt 0[\/latex] then the formula represents continuous growth. <\/span><\/p>\n

          If [latex]r \\lt 0[\/latex] then the formula represents continuous decay.<\/span><\/p>\n<\/section>\n

          How To: Given the initial value, rate of growth or decay, and time [latex]t[\/latex]<\/em>, solve a continuous growth or decay function<\/strong><\/p>\n
            \n
          1. Use the information in the problem to determine [latex]P[\/latex], the initial value of the function.<\/li>\n
          2. Use the information in the problem to determine the growth rate [latex]r[\/latex].\n
              \n
            • If the problem refers to continuous growth, then [latex]r\u00a0> 0[\/latex].<\/li>\n
            • If the problem refers to continuous decay, then [latex]r\u00a0< 0[\/latex].<\/li>\n<\/ul>\n<\/li>\n
            • Use the information in the problem to determine the time [latex]t[\/latex].<\/li>\n
            • Substitute the given information into the continuous growth formula and solve for [latex]A(t)[\/latex].<\/li>\n<\/ol>\n<\/section>\n
              A person invested [latex]$1,000[\/latex] in an account earning a nominal [latex]10\\%[\/latex] per year compounded continuously. How much was in the account at the end of one year?<\/p>\n