Chapter 3.4, Problem 136E

Solution

a.

To graph: the model and plot the data in the same viewing window using graphing utility.

Given information: The table shows the population P (in hundreds) of an endangered

species from 2011 through 2016.The data can be modeled by;$p=7-2.72\ln t$, where t represents the year.

Year ( t )	Population, P
2011	6.83
2012	5.12
2013	4.38
2014	3.15
2015	2.8
2016	1.76

Calculation:

  $p=7-2.72\ln t$, where t represents the year, with t = 1 corresponding to 2011.

The table data can be written as:

Year ( t )	Population, P
1	6.83
2	5.12
3	4.38
4	3.15
5	2.8
6	1.76

The graph of the model and the above table data in the same viewing window is shown below.

  

b.

To approximate : the population each year from 20011 to 2016 using the model.

  $\begin{array}{l}\text{for}2011,p=7\\ \text{for}2012,p=5.11\\ \text{for}2013,p=4.01\\ \text{for}2014,p=3.23\\ \text{for}2015,p=2.62\\ \text{for}2016,p=2.13\end{array}$

Given information:

Calculation:

  $\begin{array}{l}p=7-2.72\ln t\\ \text{for}2011,t=1\Rightarrow p=7-2.72\ln 1=7-0=7\Rightarrow p=7\\ \text{for}2012,t=2\Rightarrow p=7-2.72\ln 2=7-1.89=5.11\Rightarrow p=5.11\\ \text{for}2013,t=3\Rightarrow p=7-2.72\ln 3=7-2.99=4.01\Rightarrow p=4.01\\ \text{for}2014,t=4\Rightarrow p=7-2.72\ln 4=7-3.77=3.23\Rightarrow p=3.23\\ \text{for}2015,t=5\Rightarrow p=7-2.72\ln 5=7-4.38=2.62\Rightarrow p=2.62\\ \text{for}2016,t=6\Rightarrow p=7-2.72\ln 6=7-4.87=2.13\Rightarrow p=2.13\end{array}$

c.

To compare: the estimated value to the actual data and find is the model a good fit for the data.

Yes, the model a good fit for the data because the difference between the estimated values and actual data is very less.

Given information:

Calculation:

Actual values of data are:

Year ( t )	Population, P
2011	6.83
2012	5.12
2013	4.38
2014	3.15
2015	2.8
2016	1.76

Estimated values of data are:

Year ( t )	Population, P
2011	7
2012	5.11
2013	4.01
2014	3.23
2015	2.62
2016	2.13

Yes, the model a good fit for the data because the difference between the estimated values and actual data is very less.

This is also clear from above graph of the model with the actual data.

d.

To find: when will the species become extinct according to the model.

Species become extinct in 2023.

Given information:

Calculation:

Species become extinct when p = 0.

  $\begin{array}{l}p=7-2.72\ln t\\ p=0\\ 0=7-2.72\ln t\\ \ln t=\frac{7}{2.72}=2.57\\ e^{\ln t}=e^{2.57}\\ t\approx 13.07\\ t\approx 13\end{array}$

Therefore, the Species become extinct in 2010+13=2023.