Science:Math Exam Resources/Courses/MATH102/December 2015/Question 07

MATH102 December 2015

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[hide] Question 07

Rainbow Trout in Deer Lake can no longer reproduce due to habitat destruction. On November 1st when the population is estimated at 800 fish, locals start stocking the lake with fish at a constant rate of 100 fish per day and fish them out at a rate proportional to their population size. In the options below, $r>0$. Which of the following equations describes the size of the fish population? (a) $F(t)=100t-800e^{-rt}$ (b) ${\frac {dF}{dt}}=(100-r)F$ (c) $F(t)=100/r+(800-100/r)e^{-rt}$ (d) ${\frac {dF}{dt}}=r(100-F)$ (e) ${\frac {dF}{dt}}=100t-rF$

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.

[show] Hint 1
The derivative of F should include "in" and "out", and then check the initial condition.

[show] Hint 2
Find the answer that forces $F(0)=800$.

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.

[show] Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. (a): substitute t=0 to check the initial condition, not equal to 800. Thus wrong. (b): the derivative is 100F(should be 100, stocking rate is 100, has nothing to do with the fish population) and -rF(correct, proportional to their population size). Thus wrong. (c): substitute t=0 to check the initial condition, equals to 800, correct. Derivative of t: $(-r)(800-100/r)\cdot e^{-rt}$, change the form of it:$(-r)[(800-100/r)\cdot e^{-rt}+100/r]+100\rightarrow (-r)\cdot F+100$. rate of 100 fish per day plus rate proportional to their population size. correct (d): first part of derivative should be 100 instead of 100r. Wrong (e): again, should be 100 instead of 100t. So (c) is the only choice that matches all of the conditions. Answer: $\color {blue}(c)$

[show] Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Alternatively, we can explicitly write and solve a differential equation describing the system. Fish are being added to the pond at a rate of 100/day, and the population decays at a rate proportional to the current population. This allows us to write the differential equation ${\frac {dF}{dt}}=100-rF$ which can be arranged to ${\frac {dF}{dt}}=-r\left(F-{\frac {100}{r}}\right)$. Notice that this is different from the differential equations presented in the solution choices (b), (d), and (e). If we make the substitution $G=F-{\frac {100}{r}}$, we have that ${\frac {dF}{dt}}={\frac {dG}{dt}}$. So ${\frac {dG}{dt}}=-rG$ which has solutions of the form $G(t)=Ce^{-rt}$. We re-write this in terms of F: $F(t)-{\frac {100}{r}}=Ce^{-rt}$ so $F(t)=Ce^{-rt}+{\frac {100}{r}}.$ To compute C, we set t = 0 and use our initial condition that the pond starts with 800 fish. $F(0)=800=Ce^{0}+{\frac {100}{r}}=C+{\frac {100}{r}}$ and so $C=800-{\frac {100}{r}}.$ Thus we have the equation $F(t)=\left(800-{\frac {100}{r}}\right)e^{-rt}+{\frac {100}{r}}$. This is the equation described in answer choice (c). Answer: $\color {blue}(c)$

[show] Solution 3
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Options (b), (d), and (e) do not specify what $F(0)$ is; they only describe the derivative $dF/dt$. So, we are left with either (a) or (c). If (a) were true, then $F(0)=-800$ so by process of elimination, the answer is $\color {blue}(c)$.