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Question

# Each spring, sandhill cranes migrate through the Platte River valley in central Nebraska. An estimated maximum of a half-million of these birds reach the region by April 1 each year. If there are only 100,000 sandhill cranes 15 days later and the sandhill cranes leave the Platte River valley at a rate proportional to the number of sandhill cranes still in the valley at the time, (a) How many sandhill cranes remain in the valley 30 days after April 1? (b) How many sandhill cranes remain in the valley 35 days after April 1? (c) How many days after April 1 will there be less than 1000 sandhill cranes in the valley?

Solution

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For this problem, an appropriate differential equation is

$\begin{equation*} \dfrac{dS}{dt}=-kS, \end{equation*}$

where $S(t)$ is the number of sandhill cranes $t$ days after April 1, in the Platte River valley in central Nebraska. And we have the conditions:

$\begin{equation*} S(0)=500,000 ; \quad S(15)=100,000. \end{equation*}$

This equation is a separable one, so we separate variables and integrate that, and we get:

\begin{align*} &\int{\dfrac{1}{S}}\,dS=\int{-k}\,dt\\\\ \Rightarrow &\ln|S|=-k t +c\\ \Rightarrow &S(t)= c_1 e^{-kt}. \end{align*}

To find the value for $c_1$, we use the condition $S(0)=500,000$ and we get

\begin{align*} c_1=500,000 \end{align*}

Hence, tge solution for our equation is

$\begin{equation*} S(t)=500,000 \,e^{-kt}. \end{equation*}$

Now, to find the constant $k$, we use another condition for $t=15$, and we observe:

\begin{align*} &100,000=500,000 \,e^{-15 k}\\ \Rightarrow \quad &e^{-15 k}=\frac{1}{5}\\ \Rightarrow \quad &-15 \,k=\ln{\frac{1}{5}}\\ \Rightarrow \quad & k=\dfrac{\ln{5}}{15} \end{align*}

Finally, putting this result into the solution for $S$, we get:

$\begin{equation*} \textcolor{#4257b2}{S(t)=500,000 \,e^{-\frac{\ln{5}}{15} t}}. \end{equation*}$

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