Science:Math Exam Resources/Courses/MATH101/April 2007/Question 06 (b)

MATH101 April 2007

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Question 06 (b)

Full-Solution Problems. Justify your answers and show all your work. Simplification of answers is not required. The population of fish in a lake is m million, where $m=m(t)$ varies with time t measured in years. The number of fish is currently 2 million. Suppose instead that (because of fishing by humans) m satisfies ${\frac {dm}{dt}}=16m\left(1-{\frac {m}{4}}\right)-12$ Will the fish population ever equal 3 million? You must give justification for your answer.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Find the steady states of this differential equation. Are steady states actually reached in finite time?

Checking a solution serves two purposes: helping you if, after having used the hint, 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. We want to find the steady states of this differential equations. To do so, set ${\frac {dm}{dt}}=0$, i.e. $16m\left(1-{\frac {m}{4}}\right)-12=0$ Simplifying yields $16m-4m^{2}-12=0$ and once more $m^{2}-4m+3=0$ Factoring yields $(m-3)(m-1)=0$ and so the steady states are given by $m=1$ and $m=3$. It is quickly checked, e.g. by plugging in $m=0$, $m=2$, and $m=4$, that ${\frac {dm}{dt}}=\left\{{\begin{array}{rl}{\text{negative }}&{\text{if }}m<1{\text{ or }}m>3,\\0&{\text{if }}m=1{\text{ or }}m=3,\\{\text{positive }}&{\text{if }}1<m<3.\end{array}}\right.$ At time $t=0$ we have $m(0)=2$. Hence, despite the fishing, the population of fish will grow initially. But the growth will slow down as the population size approaches the steady state value $m=3$. A steady state can never be reached in finite time, hence the answer is No, the fish population will never equal 3 million. Caption: We plot the rate of change $dm/dt$ ($y$-axis) as function of population size $m$ ($x$-axis).

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