Chapter 4, Problem 101RE

(A)

To determine

To Find:The function that models that the bird population at the time $(t)$

Answer to Problem 101RE

  $n(t)=1500e^{0.515t}$

Explanation of Solution

Given: The graph of the population of a rare species of bird where $t$ represents years since $1999$ and $n(t)$ is measured in thousands.

  

  $n(t)=n_0{e}^{rt}$

We are given $n(t)=n_0{e}^{rt}$

From the graph initially the population is $1500$

Thus

  $n_0=1500$

The equation becomes

  $n(t)=1500e^{rt}$

The curve passes through the point $(5,3200)$

  $\begin{array}{l}{n}_5=3200\\ 1500e^{5r}=3200\\ e^{5r}=\frac{32}{15}\\ \ln e^{5r}=\ln \left(\frac{32}{15}\right)\\ 5r\ln e=\ln \left(\frac{32}{15}\right)\\ r=\frac{\ln \left(\frac{32}{15}\right)}{5}\approx 0.151\\ i.e\\ n(t)=1500e^{0.515t}\end{array}$

(B)

To determine

To Find: The expected bird population in the year 2010

Answer to Problem 101RE

  $7940$ Birds

Explanation of Solution

Given:The graph of the population of a rare species of bird where $t$ represents years since $1999$ and $n(t)$ is measured in thousands.

  

Using the function $n(t)=1500e^{0.515t}$

  $2010$ is1 year after the year $1999$ .

So we have to find the value of the function for $t=11$

Thus

  $\begin{array}{l}n(t)=1500e^{0.515t}\\ n(11)=1500e^{0.515(11)}\\ n(11)=1500e^{1.665}\\ n(11)=7940.411\\ n(11)\approx 7940\end{array}$

Hence, the expected bird population is $7940$ birds