A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 2 years ago
Unit I Ordinary Differential Equation Problem:
An African government is trying to come up with a good policy regarding the hunting of oryx. They are using the following model: the oryx population has a natural growth rate of k years−1, and there is assumed a constant harvesting rate of a oryxes/year.
1. Write down a model for the oryx population. [First step: choose symbols and units.]
anonymous
 2 years ago
Unit I Ordinary Differential Equation Problem: An African government is trying to come up with a good policy regarding the hunting of oryx. They are using the following model: the oryx population has a natural growth rate of k years−1, and there is assumed a constant harvesting rate of a oryxes/year. 1. Write down a model for the oryx population. [First step: choose symbols and units.]

This Question is Closed

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0The first task is: 1. Write down a model for the oryx population. [First step: choose symbols and units.] I tried to figure start with: \[\frac{dx}{dt}=x2^{dt/k}\] But I know that is wrong.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Then I thought that I should think about the population without the harvest rate: \(\Delta x\) (population change at t) = (population after \(t+\Delta t\))  (population at t) \[\Delta x=x2^{(t+\Delta t)/k}x2^{t/k}\] But I still don't have a differential equation.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0If I divide by \(\Delta t\) and take the limit \(\Delta t \rightarrow 0\): \[\lim_{\Delta t \rightarrow 0} \frac{\Delta x}{\Delta t}=\frac{C2^{(t+\Delta t)/k}C2^{t/k}}{\Delta t} \\ \lim_{\Delta t \rightarrow 0} \frac{\Delta x}{\Delta t}=\frac{dx}{dt}=\frac{1}{k}C\ln (2)\,2^{t/k} \] I hope this is correct. How do I go about adding the harvest rate? What is the motivation?

KenLJW
 2 years ago
Best ResponseYou've already chosen the best response.1O = Oi + ky  1  ry O = Oi + (k  r)y 1 Ooryx population Oiinitial oryx population yyears k rate inc. per year r rate dec. year

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0@KenLJW Why is there a 1 in your first equation? And shouldn't there be an exponetial growth rate since the population doubles every k years?

KenLJW
 2 years ago
Best ResponseYou've already chosen the best response.1it says has a natural growth rate of k years 1 if you mean EXP{ky] 1 then O = Oi + EXP{ky]  1  ry Ooryx population Oiinitial oryx population yyears k time constand per year r rate dec. year dO/dy = kEXP[ky] r if you want a constant population dO/dY = 0

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0@KenLJW Thank you. I have to think more aobut it and then I will check the answers.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I understand why I couldn't get it. The problem says natural growth meaning the growth rate of an exponential growth model. And I also understand why @KenLJW used 1. It is because in my question it is written "k years−1", but I copied wrong, it should be \(``\mbox{k years}^{−1}"\).
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.