2022 H2 Maths Paper 1 Q12 — Worked Solution

Part (a)

5 marks

Let $k$ be the constant of proportionality. Since the rate of change of $P$ is proportional to $P$ ,

\frac{dP}{dt} = kP

Separating variables and integrating,

\int\frac{1}{P}\, dP = \int k\, dt

\Rightarrow \ln P = kt + C

\Rightarrow P = e^{kt + C} = Ae^{kt},\ \text{where}\ A = e^{C}

At $t = 0$ , $P = 50$ :

50 = Ae^{0} \Rightarrow A = 50

At $t = 10$ , $P = 100$ :

100 = 50e^{10k}

\Rightarrow e^{10k} = 2

\Rightarrow k = \frac{\ln 2}{10}

So the first model gives $P = 50e^{\frac{t}{10}\ln 2}$ , i.e.

P = 50 \times 2^{t/10}

To check your answer: at

t = 0

,

P = 50 \times 2^{0} = 50

✓; at

t = 10

,

P = 50 \times 2^{1} = 100

✓

$\therefore$ $P = 50 \times 2^{t/10}$ .

Part (b)

6 marks

Approach

the equation $\frac{dP}{dt} = \lambda P(500 - P)$ separates, but the resulting fraction $\frac{1}{P(500 - P)}$ needs partial fractions before it can be integrated.

\frac{dP}{dt} = \lambda P(500 - P)

\Rightarrow \int\frac{1}{P(500 - P)}\, dP = \int\lambda\, dt

Using partial fractions, $\frac{1}{P(500 - P)} = \frac{1}{500}\left( \frac{1}{P} + \frac{1}{500 - P} \right)$ , so

\frac{1}{500}\int\left( \frac{1}{P} + \frac{1}{500 - P} \right)dP = \lambda t + C_{1}

\Rightarrow \ln P - \ln(500 - P) = 500\lambda t + C_{2},\ \text{where}\ C_{2} = 500C_{1}

\Rightarrow \ln\left( \frac{P}{500 - P} \right) = 500\lambda t + C_{2}

\Rightarrow \frac{P}{500 - P} = Ae^{500\lambda t},\ \text{where}\ A = e^{C_{2}}

At $t = 0$ , $P = 50$ :

A = \frac{50}{450} = \frac{1}{9}

so $\frac{P}{500 - P} = \frac{1}{9}e^{500\lambda t}$ . At $t = 10$ , $P = 100$ :

\frac{100}{400} = \frac{1}{4} = \frac{1}{9}e^{5000\lambda}

\Rightarrow e^{5000\lambda} = \frac{9}{4}

\Rightarrow \lambda = \frac{1}{5000}\ln\left( \frac{9}{4} \right)

Substituting back, $e^{500\lambda t} = e^{\frac{t}{10}\ln(9/4)} = \left( \frac{9}{4} \right)^{t/10}$ , so

\frac{P}{500 - P} = \frac{1}{9}\left( \frac{9}{4} \right)^{t/10}

Solving for $P$ ,

9P = (500 - P)\left( \frac{9}{4} \right)^{t/10}

\Rightarrow P\left\lbrack 9 + \left( \frac{9}{4} \right)^{t/10} \right\rbrack = 500\left( \frac{9}{4} \right)^{t/10}

\Rightarrow P = \frac{500\left( \frac{9}{4} \right)^{t/10}}{9 + \left( \frac{9}{4} \right)^{t/10}}

Dividing the numerator and denominator by $\left( \frac{9}{4} \right)^{t/10}$ ,

P = \frac{500}{1 + 9\left( \frac{4}{9} \right)^{t/10}}

To check your answer: at

t = 0

,

P = \frac{500}{1 + 9} = 50

✓; at

t = 10

,

P = \frac{500}{1 + 9(4/9)} = \frac{500}{5} = 100

✓

$\therefore$ $P = \frac{500}{1 + 9\left( \frac{4}{9} \right)^{t/10}}$ .

Part (c)

2 marks

P \rightarrow 500\ \text{as}\ t \rightarrow \infty

Explanation: since $0 < \frac{4}{9} < 1$ , $\left( \frac{4}{9} \right)^{t/10} \rightarrow 0$ as $t \rightarrow \infty$ , so $P \rightarrow \frac{500}{1 + 0} = 500$ . The refined model therefore predicts the population levels off at a fixed carrying capacity of $500$ . The first model, $P = 50 \times 2^{t/10}$ , instead grows without bound as $t \rightarrow \infty$ , which is unrealistic since real populations are constrained by limited resources, space and food supply. The refined model is an improvement because it captures this constraint.

2022 H2 Maths Paper 1, Question 12

Part (a)

Part (b)

Part (c)

This is how we mark, and how we teach.