Science:Math Exam Resources/Courses/MATH102/December 2013/Question C 02 (c)

MATH102 December 2013

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Question C 02 (c)

The population of fish in a particular lake is given by the function F(t) where F is measured in number of fish and t is measured in days. A company that manages fish stocks is hired to restock the lake, adding fish at a constant rate. Only N fishers are allowed to fish in the lake at a time. A simple model for this scenario is given by the equation: ${\frac {dF}{dt}}=I-\alpha NF$ Where I and $\alpha$ are constant and two cases for N are considered. Case 1: Suppose N is a constant. What is the steady state number of fish $(F_{*})$ in the lake? If the lake has no fish in it initially, at what time $(t_{*})$ does the population size reach half its steady state value $F_{*}/2$? (You do not have to show a derivation of the solution F(t) of the equation for full points - simply stating it is sufficient - but you must show the rest of the calculation.)

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
In the steady state of an ordinary differential equation, the derivative equals $0$. Hence, solve the equation $\displaystyle 0=I-\alpha NF$ for $\displaystyle F$.

Hint 2
The solution $F$ of the equation is $F(t)={\frac {I}{\alpha N}}(1-e^{-\alpha Nt})$

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. In the steady state, the derivative is equal to $0$. Hence we solve the equation $I-\alpha NF$ for $F$ and obtain $F_{*}={\frac {I}{N\alpha }}$ The solution $F$ of the equation is $F(t)={\frac {I}{\alpha N}}(1-e^{-\alpha Nt})$ In order to find $t_{*}$, where $F(t_{*})={\frac {F_{*}}{2}}$, we set ${\frac {F_{*}}{2}}={\frac {I}{\alpha N}}(1-e^{-\alpha Nt_{*}})$ and use that ${\frac {F_{*}}{2}}={\frac {I}{2\alpha N}}$ to find ${\begin{aligned}{\frac {I}{2\alpha N}}={\frac {I}{\alpha N}}(1-e^{-\alpha Nt_{*}})\quad \Rightarrow \quad {\frac {1}{2}}=e^{-\alpha Nt_{*}}\quad \Rightarrow \quad \ln({\frac {1}{2}})=-\alpha Nt_{*}\end{aligned}}$ Now we solve for $t_{*}=-{\frac {\ln({\frac {1}{2}})}{\alpha N}}=-{\frac {\ln(1)-\ln(2)}{\alpha N}}=-{\frac {0-\ln(2)}{\alpha N}}={\frac {\ln(2)}{\alpha N}}$