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DanielM_113
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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.]
 one year ago
 one year ago
DanielM_113 Group Title
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.]
 one year ago
 one year ago

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DanielM_113 Group TitleBest ResponseYou've already chosen the best response.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.
 one year ago

DanielM_113 Group TitleBest ResponseYou've already chosen the best response.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.
 one year ago

DanielM_113 Group TitleBest ResponseYou've already chosen the best response.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?
 one year ago

KenLJW Group TitleBest ResponseYou've already chosen the best response.1
O = 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
 one year ago

KenLJW Group TitleBest ResponseYou've already chosen the best response.1
dO/dy = (k r)
 one year ago

DanielM_113 Group TitleBest 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?
 one year ago

KenLJW Group TitleBest ResponseYou've already chosen the best response.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 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
 one year ago

DanielM_113 Group TitleBest 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.
 one year ago

DanielM_113 Group TitleBest ResponseYou've already chosen the best response.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 \(``\mbox{k years}^{−1}"\).
 one year ago
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