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SCSA Mathematics Specialist Rates of change and differential equations

5 sample questions with marking guides and sample answers · Avg. score: 21.7%

2022

QCAA

Paper 2

1 mark

Determine the solution of the differential equation $\frac{dy}{dx} = \frac{\sin(2x)}{\cos(2x)}$ given $y=0$ when $x=\frac{\pi}{5}$ .

$y = -2\ln|\cos(2x)| - 2.35$

$y = -2\ln|\cos(2x)| + 2.35$

$y = -\frac{1}{2}\ln|\cos(2x)| - 0.59$

$y = -\frac{1}{2}\ln|\cos(2x)| + 0.59$

2023

QCAA

Paper 1

1 mark

The differential equation for which the solution is a logistic equation of the form $y = \frac{a}{b+Ce^{-at}}$ where $a, b$ and $C$ are constants is

$\frac{dy}{dt} = 0.25(1-0.01t)$

$\frac{dy}{dt} = 0.25(1-0.01y)$

$\frac{dy}{dt} = 0.25t(1-0.01t)$

$\frac{dy}{dt} = 0.25y(1-0.01y)$

Q12

2021

SCSA

Paper 2

6 marks

Q12

The horizontal displacement of a Ferris wheel cabin exhibits simple harmonic motion. The maximum horizontal speed is $\frac{\pi}{2}$ metres per second and its period of motion is exactly 60 seconds.

Let $x(t) = A\cos(nt)$ be the horizontal displacement after $t$ seconds.

Q12a

3 marks

Determine the values of $A$ and $n$ .

Period $T = 60 = \frac{2\pi}{n} \implies \therefore n = \frac{\pi}{30} \quad (0.1047....)$

$v(t) = -A\left(\frac{\pi}{30}\right)\sin\left(\frac{\pi t}{30}\right)$

$\therefore \text{Max } v = A\left(\frac{\pi}{30}\right) = \frac{\pi}{2} \implies \therefore A = 15$

Hence $x(t) = 15\cos\left(\frac{\pi t}{30}\right)$

Marking Criteria

Descriptor	Marks
determines $n$ correctly	1
differentiates and forms the correct expression for the maximum speed	1
determines $A$ correctly	1

Q12b

3 marks

Determine the horizontal acceleration, correct to the nearest $0.001\text{ m/s}^2$ , when the horizontal displacement is 10 metres.

Condition for S.H.M. is $a = -\left(\frac{\pi}{30}\right)^2 x$

$\therefore$ When $x = 10 \quad a = -\left(\frac{\pi^2}{900}\right)(10) = -0.10966 ... \text{ metres/sec}^2$

i.e. acceleration is $-0.110 \text{ m/sec}^2$ (3 d.p.)

Marking Criteria

Descriptor	Marks
applies the condition for S.H.M. correctly	1
substitutes $x = 10$ correctly	1
determines the acceleration correct to $0.001 \text{ m/sec}^2$	1

Q13

2024

SCSA

Paper 2

8 marks

Q13

A brumby is a free-roaming wild horse found in large numbers in parts of Australia. The culling of brumbies was banned in the year 2000. At this time the estimated population of brumbies in Kosciuszko National Park was 1600.

Scientists have modelled the population, $P(t)$ , of brumbies in Kosciuszko National Park $t$ years since the ban, by

\begin{align*} P(t) = \frac{18000}{10.25e^{-0.15t} + 1} \end{align*}

Q13c

It can be shown that the growth rate of the population of brumbies can be expressed as

$\frac{dP}{dt} = \frac{1}{r}P(k - P)$ .

Q13a

1 mark

Use the model to determine how long it will take the brumbies to increase to a number that is triple the number when the ban came into effect.

Solve when $P(t) = 4800\quad$ i.e. $4800 = \frac{18000}{10.25e^{-0.15t} + 1}$

From CAS $t = 8.7711\dots \text{years}$

Marking Criteria

Descriptor	Marks
solves for $t$ correctly	1

Q13b

2 marks

From this model, determine the estimated long run number of brumbies in Kosciuszko National Park.

Using $t \to \infty$ then $e^{-0.15t} \to 0,$ hence, $P(t) \to \frac{18000}{10.25(0) + 1} = 18000$ .

Hence in the long term, the limiting population will be 18 000 brumbies.

Marking Criteria

Descriptor	Marks
considers $t \to \infty$ to use $e^{-0.15t} \to 0$	1
determines the long run number of brumbies	1

Q13c

3 marks

Determine the values of the constants $r$ and $k$ .

$k = 18000$ as this is the limiting population.

$\therefore \frac{dP}{dt} = \frac{1}{r}P(18000 - P)$

Using $P(t) = \frac{18000}{10.25e^{-0.15t} + 1}$ from CAS $P'(0) = 218.666\dots$ (when $P = 1600$ )

Hence substituting into $\frac{dP}{dt} = \frac{1}{r}P(18000 - P)$

i.e. $218.666\dots = \frac{1}{r}(1600)(18000 - 1600)$

$\therefore r = 120000$

Marking Criteria

Descriptor	Marks
states the value of $k$ correctly	1
states the value of $r$ correctly	1
provides justification for the determination of the value for $r$	1

Q13d

2 marks

Determine the greatest growth rate for the population of brumbies.

Greatest rate of growth will occur when $P = 9000$ (half the limiting population)

Using $P = 9000, \quad \frac{dP}{dt} = \frac{1}{120000}(9000)(18000 - 9000) = 675 \text{ brumbies per year}$

Hence greatest growth rate will be 675 brumbies per year.

Marking Criteria

Descriptor	Marks
states that the maximum growth rate occurs when $P = 0.5k = 9000$	1
calculates the maximum growth rate correctly and states the correct units	1

Q14

2020

SCSA

Paper 2

5 marks

Q14

A particle travels in a straight line so that its velocity $v$ cm per second and displacement $x$ cm are related by the equation:

$v = -0.2x$

Q14a

2 marks

Determine the acceleration $a$ in terms of its displacement $x$ .

$\text{Acceleration } a = \frac{dv}{dt} = \frac{dv}{dx} \times \frac{dx}{dt} = (-0.2) \times (-0.2x) = 0.04x$

Marking Criteria

Descriptor	Marks
uses the chain rule correctly to relate $\frac{dv}{dt}$ in terms of $v$	1
obtains $a$ correctly in terms of $x$	1

Q14b

1 mark

Does the particle's motion constitute simple harmonic motion? Justify your answer.

$\text{Since } a = 0.04x \text{ then we could write } a = (0.2)^2 x$

$\text{But the condition for S.H.M. is that } a = -n^2 x \text{ .}$

$\text{Hence the motion does NOT follow simple harmonic motion.}$

Marking Criteria

Descriptor	Marks
justifies why the motion is NOT simple harmonic	1

Q14c

2 marks

It is known that the initial displacement of the particle is $x = 4$ cm.

Determine, correct to the nearest 0.01 second, when the particle has a displacement of 2 cm.

$\text{We have } v = \frac{dx}{dt} = -0.2x$

\begin{align*} \therefore \int \frac{1}{x} dx &= \int -0.2 dt \quad \text{using separation of variables}\\ \text{i.e. } \ln|x| &= -0.2t + c_1\\ \text{i.e. } x &= e^{-0.2t + c} = k \times e^{-0.2t} \end{align*}

$\text{Using } x(0) = 4 \text{, then } 4 = k\left(e^0\right) \implies \text{ Hence } k = 4$

$\text{i.e. } x = e^{-0.2t + c} = 4e^{-0.2t}$

$\text{Solving } x(t) = 2 \text{ yields } 2 = 4e^{-0.2t}$

$\text{i.e. } t = 5\ln 2 = 3.47 \text{ seconds (correct to 0.01 sec)}$

Marking Criteria

Descriptor	Marks
determines the function $x(t)$ correctly	1
solves for $t$ correct to 0.01 seconds	1

SCSA Mathematics Specialist Rates of change and differential equations

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