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

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Accepted 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 $\lim_{t
ightarrow\infty}p(t)$**:
     - ## Step1: Divide numerator and denominator by $t$
       - $p(t)=\frac{200t}{t + 2}=\frac{200}{1+\frac{2}{t}}$.
       - $\lim_{t
ightarrow\infty}p(t)=\lim_{t
ightarrow\infty}\frac{200}{1+\frac{2}{t}}$.
       - As $t
ightarrow\infty$, $\frac{2}{t}
ightarrow0$. So, $\lim_{t
ightarrow\infty}p(t)=200$. 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 $t
ightarrow\infty$, 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 $\lim_{t
ightarrow\infty}p(t)$**:
     - ## 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 $t
ightarrow\infty$, $\frac{3}{t^{2}}
ightarrow0$ and $\frac{9}{t^{2}}
ightarrow0$.
       - $\lim_{t
ightarrow\infty}p(t)=600$. This means that as $t
ightarrow\infty$, 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 $t
ightarrow\infty$, the population approaches 600.
     - e. Use a graphing utility.

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