91Ó°ÊÓ

Often environmentalists capture an endangered species and transport the species to

a controlled environment where the species can produce offspring and regenerate its population. Suppose that six American bald eagles are captured, transported to Montana,

and set free. Based on experience, the environmentalists expect the population to grow according to the model $P\left(t\right)=\frac{500}{1+82.33e^{-0.162t}}$where t is measured in years.

The number c is called the carrying capacity (for growth models) because the

value $P\left(t\right)$approaches c as t approaches infinity; that is $\underset{t→∞}{\lim }P\left(t\right)=c$

As $t→∞,e^{-0.162t}→0$and $P\left(t\right)→\frac{500}{1}$

So the carrying capacity of the environment is, 500 American bald eagles .

The number $\left|b\right|$is the growth rate for $b>0$

$\left|b\right|=\left|0.162\right|=16.2\%$per year

The graph of $P=P\left(t\right)$is

$P=\frac{500}{1+82.33e^{-0.162t}}$

To find the population after 3 years, evaluate $P\left(3\right)$

role="math" localid="1647447490458" $$\begin{gathered} P\left(3\right)=\frac{500}{1+82.33e^{-0.162×3}} \\ P\left(3\right)≈10Eagles \end{gathered}$$

To determine when the population will be 300 eagles, solve $P\left(t\right)=300$

role="math" localid="1647448192566" $$\begin{gathered} P\left(t\right)=\frac{500}{1+82.33e^{-0.162t}} \\ 300=\frac{500}{1+82.33e^{-0.162t}} \\ 300\left(1+82.33e^{-0.162t}\right)=500 \\ 300+24699e^{-0.162t}=500 \\ 24699e^{-0.162t}=200 \\ {e}^{-0.162t}=\frac{200}{24699} \\ \ln \left(e^{-0.162t}\right)=\ln \left(\frac{200}{24699}\right) \\ -0.162t=\ln \left(\frac{200}{24699}\right) \\ t=\frac{\ln \left(\frac{200}{24699}\right)}{-0.162} \\ t≈29.7years \end{gathered}$$

One-half of the carrying capacity is $\frac{500}{2}=250$, We solve $P\left(t\right)=250$

$$\begin{gathered} P\left(t\right)=\frac{500}{1+82.33e^{-0.162t}} \\ 250=\frac{500}{1+82.33e^{-0.162t}} \\ 250\left(1+82.33e^{-0.162t}\right)=500 \\ 250+20582.5e^{-0.162t}=500 \\ 20582.5e^{-0.162t}=250 \\ {e}^{-0.162t}=\frac{250}{20582.5} \\ \ln \left(e^{-0.162t}\right)=\ln \left(\frac{250}{20582.5}\right) \\ -0.162t=\ln \left(\frac{250}{20582.5}\right) \\ t=\frac{\ln \left(\frac{250}{20582.5}\right)}{-0.162} \\ t≈27.2years \end{gathered}$$