Science:Math Exam Resources/Courses/MATH102/December 2015/Question 15

MATH102 December 2015

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Question 15

Someone released a group of 10 rabbits on campus and they started reproducing at a rate proportional to the population size with constant of proportionality of 3 per year. After 5 years, the rabbits continued to reproduce in the same manner but a coyote moved in and began eating rabbits at a rate proportional to the population size with a constant of proportionality of 9 per year. The population of rabbits is given by the function $P(t)={\begin{cases}C_{0}e^{at}&0\leq t<5,\\C_{1}e^{bt}&t\geq 5.\end{cases}}$ Give values for $a$, $C_{0}$ , $b$, and $C_{1}$. Your answers can be in terms of exponentials and/or logarithms.

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
Write the given information in terms of $P(t)$.

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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. Someone released a group of 10 rabbits on campus and they started reproducing at a rate proportional to the population size with constant of proportionality of 3 per year. After 5 years, the rabbits continued to reproduce in the same manner but a coyote moved in and began eating rabbits at a rate proportional to the population size with a constant of proportionality of 9 per year. The population of rabbits is given by the function $P(t)={\begin{cases}C_{0}e^{at}&0\leq t<5,\\C_{1}e^{bt}&t\geq 5.\end{cases}}$ According to the question, at the initial time ($t=0$), we have 10 rabbits, so that $P(0)=10$ Until 5 years, the population increases at a rate proportional to the population size with constant of proportionality of 3 per year. This implies that $P'(t)=3P(t)$ for $0\leq t\leq 5$. By using these two equations, we can find $C_{0}$ and $a$ in the given formula $P(t)=C_{0}e^{at}$ on the interval $[0,5]$; $P(0)=C_{0}=10,\quad 3={\frac {P'(t)}{P(t)}}={\frac {C_{0}ae^{at}}{C_{0}e^{at}}}=a$. Then, after 5 years, $P(5)=C_{0}e^{5a}=10e^{15}$. On the other hand, since the rabbits are eaten by the coyote at a rate proportional to the population size with a constant of proportionality of 9 per year, the rate of change of the population of rabbits becomes ${\begin{aligned}P'(t)&={\text{(the rate of change of the population)}}\\&={\text{(the rate of growth by reproduction)}}-{\text{(the rate of decay by death)}}\\&=3P(t)-9P(t)=-6P(t)\end{aligned}}$ Now, as $C_{0}$ and $a$ are found, we can get $C_{1}$ and $b$; from the given formula on $(5,\infty )$, $-6={\frac {P'(t)}{P(t)}}={\frac {C_{1}be^{bt}}{C_{1}e^{bt}}}=b,\quad 10e^{15}=P(5)=C_{1}e^{5b}=C_{1}e^{-30}\implies C_{1}=10e^{45}$ To summarize, we obtain $\color {blue}a=3,C_{0}=10,b=-6,C_{1}=10e^{45}$.