Chapter 4.6, Problem 6E

a.

To determine

Find the population grows exponentially.

Answer to Problem 6E

  $\boxed{n\left(t\right)=18000e^{0.08t}}$

Explanation of Solution

Given information:

The population of a certain species of fish has a relative growth rate of $1.2\%$ per year. It is estimated that the population in $2000$ was $12$ million.

Find an exponential model $n\left(t\right)=n_0{e}^{rt}$ for the population $t$ years after $2000$ .

Calculation:

Consider exponentially growth model,

  $\begin{array}{l}n\left(t\right)=n_0{e}^{rt}\\ r=1.2\%=0.012\\ n_0=12\\ n\left(t\right)=12e^{0.012t}\end{array}$

Therefore, function that model the population $t$ years after $2000$ is,

  $\boxed{n\left(t\right)=12e^{0.012t}}$

Hence, the solution is, $\boxed{n\left(t\right)=12e^{0.012t}}$

b.

To determine

Find the population grows exponentially.

Answer to Problem 6E

  $\boxed{12.14million}$

Explanation of Solution

Given information:

The population of a certain species of fish has a relative growth rate of $1.2\%$ per year. It is estimated that the population in $2000$ was $12$ million.

Estimate the fish population in the year $2005$ .

Calculation:

Estimate the fish population in the year $2005$ .

  $\begin{array}{l}now,\\ t=2005-2000=5\\ n\left(t\right)=12e^{0.012t}\\ put,t=5\\ n\left(5\right)=12e^{0.012\left(5\right)}\\ n\left(5\right)=12e^{0.06}\\ n\left(5\right)=12.14\\ 12.14million\end{array}$

Therefore, the fish population in the year $2005$ is $12.14million$ .

Hence, the solution is, $\boxed{12.14million}$

b.

To determine

Find the population grows exponentially.

Answer to Problem 6E

  

Explanation of Solution

Given information:

The population of a certain species of fish has a relative growth rate of $1.2\%$ per year. It is estimated that the population in $2000$ was $12$ million.

Sketch a graph of the fish population.

Calculation:

The graph of the fish population.

  

Hence, the solution is, graph.