2 suppose that the height of a plane t minutes after take off is $h(t)=3500ln(t + 1)$ feet for $0leq…

2 suppose that the height of a plane t minutes after take off is $h(t)=3500ln(t + 1)$ feet for $0leq tleq15$. how fast is the plane rising 5 minutes after take off?\n3 the number of cases of influenza in new york city from the beginning of 1960 to the beginning of 1964 is modeled by the function\n$n(t)=5.3e^{0.093t^{2}-0.87t}$, $0leq tleq4$,\nwhere $n(t)$ gives the number of cases (in thousands) and $t$ is measured in years, with $t = 0$ corresponding to the beginning of 1960.\na. find and interpret the meanings of $n(0)$ and $n(4)$.\nb. find and interpret the meanings of $n(0)$ and $n(4)$.

2 suppose that the height of a plane t minutes after take off is $h(t)=3500ln(t + 1)$ feet for $0leq tleq15$. how fast is the plane rising 5 minutes after take off?\n3 the number of cases of influenza in new york city from the beginning of 1960 to the beginning of 1964 is modeled by the function\n$n(t)=5.3e^{0.093t^{2}-0.87t}$, $0leq tleq4$,\nwhere $n(t)$ gives the number of cases (in thousands) and $t$ is measured in years, with $t = 0$ corresponding to the beginning of 1960.\na. find and interpret the meanings of $n(0)$ and $n(4)$.\nb. find and interpret the meanings of $n(0)$ and $n(4)$.

Answer

Explanation:

Step1: Solve problem 2

We need to find the derivative of $h(t) = 3500\ln(t + 1)$ first. By the chain - rule, if $y=\ln(u)$ and $u=t + 1$, then $\frac{dy}{dt}=\frac{1}{u}\cdot\frac{du}{dt}$. The derivative of $h(t)$ is $h'(t)=\frac{3500}{t + 1}$. Then we substitute $t = 5$ into $h'(t)$. $h'(5)=\frac{3500}{5+1}=\frac{3500}{6}=\frac{1750}{3}\approx583.33$ feet per minute.

Step2: Solve problem 3A

For $N(t)=5.3e^{0.093t^{2}-0.87t}$, when $t = 0$: $N(0)=5.3e^{0}=5.3$. This means that at the beginning of 1960, the number of influenza cases in New York City was 5.3 thousand. When $t = 4$: First, calculate $0.093t^{2}-0.87t$ with $t = 4$: $0.093\times4^{2}-0.87\times4=0.093\times16 - 3.48=1.488-3.48=-1.992$. Then $N(4)=5.3e^{-1.992}\approx5.3\times0.136 = 0.7208$. This means that at the beginning of 1964, the number of influenza cases in New York City was approximately 0.7208 thousand.

Step3: Solve problem 3B

First, find the derivative of $N(t)$ using the chain - rule. If $y = 5.3e^{u}$ and $u=0.093t^{2}-0.87t$, then $N'(t)=5.3e^{u}\cdot(0.186t - 0.87)=5.3(0.186t - 0.87)e^{0.093t^{2}-0.87t}$. When $t = 0$: $N'(0)=5.3\times(- 0.87)e^{0}=-4.611$. This means that at the beginning of 1960, the number of influenza cases was decreasing at a rate of 4.611 thousand cases per year. When $t = 4$: First, calculate $0.186t - 0.87$ with $t = 4$: $0.186\times4-0.87 = 0.744 - 0.87=-0.126$. Then $N'(4)=5.3\times(-0.126)e^{-1.992}\approx5.3\times(-0.126)\times0.136=-0.092$. This means that at the beginning of 1964, the number of influenza cases was decreasing at a rate of approximately 0.092 thousand cases per year.

Answer:

Problem 2: The plane is rising at a rate of $\frac{1750}{3}\approx583.33$ feet per minute 5 minutes after take - off. Problem 3A: $N(0) = 5.3$, meaning 5.3 thousand influenza cases at the beginning of 1960; $N(4)\approx0.7208$, meaning approximately 0.7208 thousand influenza cases at the beginning of 1964. Problem 3B: $N'(0)=-4.611$, meaning the number of cases was decreasing at a rate of 4.611 thousand cases per year at the beginning of 1960; $N'(4)\approx - 0.092$, meaning the number of cases was decreasing at a rate of approximately 0.092 thousand cases per year at the beginning of 1964.