Differential equations: RVHS 2023 Promo Q12

River Valley High School2023 PromoPromo Q12●●● Challenging12 marks

What this question tests

Differential equations.

Question

Two teams of scientists are investigating the population of wild hares, and the population of their predators, the wild lynxes, in the snowy mountains.

The number of hares in the snowy mountain is $P$ thousand at a time of $t$ years after the scientists begin observations. Initially, the first team of scientists observed that there were 50 thousand hares growing at the rate of 25 thousand hares per year. After several observations, they decided to model the growth rate of the hares, at any time $t$ , to be $kP(100 - P)$ hares per year, for some positive constant $k$ .

1. Show that $\dfrac{\,\mathrm{d} P}{\,\mathrm{d} t} = \dfrac{1}{100}P(100 - P)$ . [1]
2. Find $P$ in terms of $t$ . [5]

The number of lynxes in the snowy mountains is $Q$ thousand at a time of $t$ years after the scientists begin observations. Initially, the second team of scientists observed that there were 8 thousand lynxes decreasing at the rate of 0.8 thousand lynxes per year.

Both team of scientists collaborated and developed a refined model for the growth rate of the hares. The model assumed that, at any time $t$ , the hares were growing at the rate of $P - \dfrac{1}{16}QP$ per year.

1. For the refined model, when $t = 0$ , show that $\dfrac{\,\mathrm{d}^2 P}{\,\mathrm{d} t^2} = 15$ . [2]
2. Hence, find the Maclaurin series of $P$ in the refined model, in ascending powers of $t$ , up to and including the term in $t^2$ . [2]

Two scientists, scientist $A$ and $B$ , both use the refined model to approximate the number of hares after some time. Scientist $A$ approximates the number of hares one month after the scientists started observing, while scientist $B$ approximates the number of hares ten years after the scientists started observing.

1. Given that both scientists used your answer in part (b)(ii), state the approximation only provided by scientist $A$ and suggest an appropriate reason why scientist $A$ ’s approximation would be considered more appropriate as compared to scientist $B$ ’s. [2]

Show full worked solution▾

(a)(i) Initially, when $P = 50$ , $\dfrac{\,\mathrm{d} P}{\,\mathrm{d} t} = 25$ . Substituting into $\dfrac{\,\mathrm{d} P}{\,\mathrm{d} t} = kP(100 - P)$ : $\begin{aligned} 25 &= k(50)(50)\\ k &= \frac{1}{100}. \end{aligned}$ Hence $\dfrac{\,\mathrm{d} P}{\,\mathrm{d} t} = \dfrac{1}{100}P(100 - P)$ . (shown)

(a)(ii) $\begin{aligned} \frac{\,\mathrm{d} P}{\,\mathrm{d} t} &= \frac{1}{100}P(100 - P)\\ \frac{100}{P(100 - P)}\frac{\,\mathrm{d} P}{\,\mathrm{d} t} &= 1\\ \int \frac{1}{P} + \frac{1}{100 - P}\,\mathrm{d} P &= \int 1\,\mathrm{d} t\\ \ln|P| - \ln|100 - P| &= t + c\\ \ln\left|\frac{P}{100 - P}\right| &= t + c\\ \frac{P}{100 - P} &= A\mathrm{e}^t, \qquad A = \pm\mathrm{e}^c. \end{aligned}$ When $t = 0$ , $P = 50$ : $\begin{aligned} \frac{50}{100 - 50} &= A\mathrm{e}^0\\ A &= 1. \end{aligned}$ Hence $\begin{aligned} \frac{P}{100 - P} &= \mathrm{e}^t\\ P &= 100\mathrm{e}^t - P\mathrm{e}^t\\ P\left(1 + \mathrm{e}^t\right) &= 100\mathrm{e}^t\\ P &= \frac{100\mathrm{e}^t}{1 + \mathrm{e}^t}\\ &= \frac{100}{1 + \mathrm{e}^{-t}}. \end{aligned}$

(b)(i) The refined model is $\dfrac{\,\mathrm{d} P}{\,\mathrm{d} t} = P - \dfrac{1}{16}QP$ . Differentiating with respect to $t$ : $\begin{aligned} \frac{\,\mathrm{d}^2 P}{\,\mathrm{d} t^2} &= \frac{\,\mathrm{d} P}{\,\mathrm{d} t} - \frac{1}{16}Q\frac{\,\mathrm{d} P}{\,\mathrm{d} t} - \frac{1}{16}P\frac{\,\mathrm{d} Q}{\,\mathrm{d} t}. \end{aligned}$ When $t = 0$ : $P = 50$ , $Q = 8$ , $\dfrac{\,\mathrm{d} P}{\,\mathrm{d} t} = 25$ and $\dfrac{\,\mathrm{d} Q}{\,\mathrm{d} t} = -0.8$ : $\begin{aligned} \frac{\,\mathrm{d}^2 P}{\,\mathrm{d} t^2} &= 25 - \frac{1}{16}(8)(25) - \frac{1}{16}(50)(-0.8)\\ &= 15. \quad\text{((shown))} \end{aligned}$

(b)(ii) When $t = 0$ : $P = 50$ , $\dfrac{\,\mathrm{d} P}{\,\mathrm{d} t} = 25$ , $\dfrac{\,\mathrm{d}^2 P}{\,\mathrm{d} t^2} = 15$ : $\begin{aligned} P &\approx 50 + 25t + \frac{15}{2!}t^2 + \cdots\\ &= 50 + 25t + 7.5t^2 + \cdots. \end{aligned}$

(b)(iii) Scientist $A$ approximates one month after, i.e. $t = \dfrac{1}{12}$ : $\begin{aligned} P &\approx 50 + 25\left(\frac{1}{12}\right) + 7.5\left(\frac{1}{12}\right)^2\\ &= 52.135\ \text{thousand}. \end{aligned}$ Scientist $A$ would approximate about $52.1$ thousand hares.

Acceptable reasons

It is more appropriate as $t=\dfrac1{12}$ is closer to $0$ as compared to $t=10$ , making the approximation at $t=\dfrac1{12}$ more accurate.

When $0<t<1$ , when $t=\dfrac1{12}$ , the terms $t^n$ for $n\geq3$ are a lot smaller and are negligible. While, when $t>1$ when $t=10$ , the terms $t^n$ for $n\geq3$ are significantly large and not negligible. So, the Maclaurin Series in (bii) is inappropriate $t=10$ as it ignores these values.

As $t=\dfrac1{12}$ is close to the initial conditions, the assumptions made by the scientist for the refined model are likely still applicable. However, when $t=10$ , it is far too long after the assumptions are made and are unlikely to be applicable.

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