65 - 66. population growth consider the following population functions.\na. find the instantaneous growth…

65 - 66. population growth consider the following population functions.\na. find the instantaneous growth rate of the population, for t ≥ 0.\nb. what is the instantaneous growth rate at t = 5?\nc. estimate the time when the instantaneous growth rate is greatest.\nd. evaluate and interpret lim p(t) as t→∞.\ne. use a graphing utility to graph the population and its growth rate.\n65. p(t)=\\frac{200t}{t + 2}\n66. p(t)=600(\\frac{t^{2}+3}{t^{2}+9})

65 - 66. population growth consider the following population functions.\na. find the instantaneous growth rate of the population, for t ≥ 0.\nb. what is the instantaneous growth rate at t = 5?\nc. estimate the time when the instantaneous growth rate is greatest.\nd. evaluate and interpret lim p(t) as t→∞.\ne. use a graphing utility to graph the population and its growth rate.\n65. p(t)=\\frac{200t}{t + 2}\n66. p(t)=600(\\frac{t^{2}+3}{t^{2}+9})

Answer

  1. For (p(t)=\frac{200t}{t + 2}):
    • a. Find the instantaneous - growth rate of the population ((p^\prime(t))):

      • Explanation:

        • Step1: Use the quotient - rule

          • The quotient rule states that if (y=\frac{u}{v}), then (y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}). Here, (u = 200t), (u^\prime=200), (v=t + 2), and (v^\prime = 1).
          • (p^\prime(t)=\frac{200(t + 2)-200t\times1}{(t + 2)^{2}}=\frac{200t+400 - 200t}{(t + 2)^{2}}=\frac{400}{(t + 2)^{2}}).
    • b. Find the instantaneous growth rate at (t = 5):

      • Step1: Substitute (t = 5) into (p^\prime(t))

        • (p^\prime(5)=\frac{400}{(5 + 2)^{2}}=\frac{400}{49}\approx8.16).
    • c. Estimate the time when the instantaneous growth rate is greatest:

      • Step1: Analyze the function (p^\prime(t)=\frac{400}{(t + 2)^{2}})

        • Since (p^\prime(t)=\frac{400}{(t + 2)^{2}}) is a rational function, and the denominator ((t + 2)^{2}) is a non - negative function for (t\geq0). As (t) increases, the value of (p^\prime(t)) decreases. So, the maximum value of (p^\prime(t)) occurs at (t = 0).
    • d. Evaluate and interpret (\lim_{t\rightarrow\infty}p(t)):

      • Step1: Divide numerator and denominator by (t)

        • (p(t)=\frac{200t}{t + 2}=\frac{200}{1+\frac{2}{t}}).
        • (\lim_{t\rightarrow\infty}p(t)=\lim_{t\rightarrow\infty}\frac{200}{1+\frac{2}{t}}).
        • As (t\rightarrow\infty), (\frac{2}{t}\rightarrow0). So, (\lim_{t\rightarrow\infty}p(t)=200). This means that as time goes to infinity, the population approaches 200.
    • e. Graphing is best done using a graphing utility (e.g., a graphing calculator or software like Desmos). The function (p(t)=\frac{200t}{t + 2}) has a horizontal asymptote at (y = 200), and (p^\prime(t)=\frac{400}{(t + 2)^{2}}) is a decreasing function for (t\geq0) with (y) - intercept (p^\prime(0) = 100).

    • Answer:

      • a. (p^\prime(t)=\frac{400}{(t + 2)^{2}})
      • b. (\frac{400}{49}\approx8.16)
      • c. (t = 0)
      • d. 200; as (t\rightarrow\infty), the population approaches 200.
      • e. Use a graphing utility.
  2. For (p(t)=600\frac{t^{2}+3}{t^{2}+9}):
    • a. Find the instantaneous - growth rate of the population ((p^\prime(t))):

      • Explanation:

        • Step1: Use the quotient - rule

          • First, let (u = 600(t^{2}+3)=600t^{2}+1800), (u^\prime = 1200t), (v=t^{2}+9), (v^\prime = 2t).
          • (p^\prime(t)=\frac{1200t(t^{2}+9)-2t(600t^{2}+1800)}{(t^{2}+9)^{2}}).
          • Expand the numerator: (1200t^{3}+10800t-1200t^{3}-3600t = 7200t).
          • So, (p^\prime(t)=\frac{7200t}{(t^{2}+9)^{2}}).
    • b. Find the instantaneous growth rate at (t = 5):

      • Step1: Substitute (t = 5) into (p^\prime(t))

        • (p^\prime(5)=\frac{7200\times5}{(5^{2}+9)^{2}}=\frac{36000}{(25 + 9)^{2}}=\frac{36000}{1156}\approx31.14).
    • c. Estimate the time when the instantaneous growth rate is greatest:

      • Step1: Take the derivative of (p^\prime(t)) using the quotient - rule

        • Let (u = 7200t), (u^\prime=7200), (v=(t^{2}+9)^{2}), (v^\prime = 2(t^{2}+9)\times2t = 4t(t^{2}+9)).
        • (p^{\prime\prime}(t)=\frac{7200(t^{2}+9)^{2}-7200t\times4t(t^{2}+9)}{(t^{2}+9)^{4}}=\frac{7200(t^{2}+9)-28800t^{2}}{(t^{2}+9)^{3}}=\frac{7200t^{2}+64800 - 28800t^{2}}{(t^{2}+9)^{3}}=\frac{64800 - 21600t^{2}}{(t^{2}+9)^{3}}).
        • Set (p^{\prime\prime}(t)=0), then (64800 - 21600t^{2}=0).
        • (t^{2}=3), so (t=\sqrt{3}\approx1.73) (since (t\geq0)).
    • d. Evaluate and interpret (\lim_{t\rightarrow\infty}p(t)):

      • Step1: Divide numerator and denominator by (t^{2})

        • (p(t)=600\frac{t^{2}+3}{t^{2}+9}=600\frac{1+\frac{3}{t^{2}}}{1+\frac{9}{t^{2}}}).
        • As (t\rightarrow\infty), (\frac{3}{t^{2}}\rightarrow0) and (\frac{9}{t^{2}}\rightarrow0).
        • (\lim_{t\rightarrow\infty}p(t)=600). This means that as (t\rightarrow\infty), the population approaches 600.
    • e. Graphing is best done using a graphing utility. The function (p(t)=600\frac{t^{2}+3}{t^{2}+9}) has a horizontal asymptote at (y = 600), and (p^\prime(t)=\frac{7200t}{(t^{2}+9)^{2}}) has a maximum at (t=\sqrt{3}), (y) - intercept (p^\prime(0)=0).

    • Answer:

      • a. (p^\prime(t)=\frac{7200t}{(t^{2}+9)^{2}})
      • b. (\frac{36000}{1156}\approx31.14)
      • c. (t=\sqrt{3}\approx1.73)
      • d. 600; as (t\rightarrow\infty), the population approaches 600.
      • e. Use a graphing utility.