Fresh features from the #1 AI-enhanced learning platform.Try it free
Fresh features from the #1 AI-enhanced learning platformCrush your year with the magic of personalized studying.Try it free
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

Verified
Step 1
1 of 3

For this problem, an appropriate differential equation is

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

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

S(0)=500,000;S(15)=100,000.\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:

1SdS=kdtlnS=kt+cS(t)=c1ekt.\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 c1c_1, we use the condition S(0)=500,000S(0)=500,000 and we get

c1=500,000\begin{align*} c_1=500,000 \end{align*}

Hence, tge solution for our equation is

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

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

100,000=500,000e15ke15k=1515k=ln15k=ln515\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 SS, we get:

S(t)=500,000eln515t.\begin{equation*} \textcolor{#4257b2}{S(t)=500,000 \,e^{-\frac{\ln{5}}{15} t}}. \end{equation*}

Create an account to view solutions

Create an account to view solutions

Recommended textbook solutions

Differential Equations and Linear Algebra 2nd Edition by Beverly H. West, Hall, Jean Marie McDill, Jerry Farlow

Differential Equations and Linear Algebra

2nd EditionISBN: 9780131860612 (4 more)Beverly H. West, Hall, Jean Marie McDill, Jerry Farlow
2,405 solutions
Differential Equations and Linear Algebra 4th Edition by C. Henry Edwards, David Calvis, David E. Penney

Differential Equations and Linear Algebra

4th EditionISBN: 9780134497181 (1 more)C. Henry Edwards, David Calvis, David E. Penney
2,531 solutions
Linear Algebra and Differential Equations 1st Edition by Gary Peterson, James Sochacki

Linear Algebra and Differential Equations

1st EditionISBN: 9780201662122Gary Peterson, James Sochacki
1,217 solutions
Differential Equations and Linear Algebra 4th Edition by Scott A. Annin, Stephen W. Goode

Differential Equations and Linear Algebra

4th EditionISBN: 9780321964670Scott A. Annin, Stephen W. Goode
3,457 solutions

More related questions

1/4

1/7