Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

DanielM_113

  • 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
  1. DanielM_113
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The 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.

  2. DanielM_113
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Then 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.

  3. DanielM_113
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    If 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?

  4. KenLJW
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    O = Oi + ky - 1 - ry O = Oi + (k - r)y -1 O--oryx population Oi--initial oryx population y--years k rate inc. per year r rate dec. year

  5. KenLJW
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    dO/dy = (k -r)

  6. DanielM_113
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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?

  7. KenLJW
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    it says has a natural growth rate of k years -1 if you mean EXP{ky] -1 then O = Oi + EXP{ky] - 1 - ry O--oryx population Oi--initial oryx population y--years k time constand per year r rate dec. year dO/dy = kEXP[ky] -r if you want a constant population dO/dY = 0

  8. DanielM_113
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @KenLJW Thank you. I have to think more aobut it and then I will check the answers.

  9. DanielM_113
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I 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 \(``\mbox{k years}^{−1}"\).

  10. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy