How many SCSA Mathematics Specialist questions cover Rates of change and differential equations?

AusGrader has 170 SCSA Mathematics Specialist questions on Rates of change and differential equations, all with instant AI grading and detailed marking feedback.

WACE Mathematics Specialist — Rates of change and differential equations Questions & Answers

Q16

2023

QCAA

Paper 2

6 marks

Q16

6 marks

A curve modelled by the relation $xy^2 - y + \cos^{-1}(2x) = 1$ , where $-0.35 \leq x \leq 0.27$ and $0 \leq y \leq 1$ , intersects the $y$ -axis at point $A$ .

Determine the equation of the tangent to the curve at point $A$ .

Reveal Answer

Given $xy^2 - y + \cos^{-1}(2x) = 1$

Determining y-coordinate of A
$0 - y + \cos^{-1}(0) = 1 \Rightarrow y = 0.57$

$\frac{d}{dx}\cos^{-1}(2x) = \frac{-1}{\sqrt{0.25-x^2}}$

$\frac{d}{dx}(xy^2) = y^2 + 2xy\frac{dy}{dx}$

Determining $\frac{dy}{dx}$
$\frac{d}{dx}\left(xy^2 - y + \cos^{-1}(2x)\right) = \frac{d}{dx}(1)$
$y^2 + 2xy\frac{dy}{dx} - \frac{dy}{dx} + \frac{-1}{\sqrt{0.25-x^2}} = 0$
$\frac{dy}{dx}(2xy - 1) = \frac{1}{\sqrt{0.25-x^2}} - y^2$
$\frac{dy}{dx} = \frac{\frac{1}{\sqrt{0.25-x^2}} - y^2}{2xy - 1}$

Determining $\frac{dy}{dx}$ at A.
$\frac{dy}{dx} = \frac{\frac{1}{\sqrt{0.25}} - (0.571)^2}{-1} = -1.67$

Determining equation of tangent at A
$y = mx + c$
$y = -1.67x + 0.57$

Marking Criteria

Descriptor	Marks
correctly determines y-intercept	1
correctly determines $\frac{d}{dx}\cos^{-1}(2x)$	1
correctly determines $\frac{d}{dx}xy^2$	1
determines an expression for $\frac{dy}{dx}$ using a common factor	1
determines a value for $\frac{dy}{dx}$ at A	1
determines equation of the tangent at A	1

Q3

2025

VCAA

Paper 2

10 marks

Q3

A tank initially contains $5 \text{ kg}$ of salt dissolved in $3000 \text{ litres}$ of water. Salty water that contains $0.1 \text{ kg}$ of salt per litre of water enters the tank at a rate of $20 \text{ litres per minute}$ . The solution is kept thoroughly mixed and drains from the tank via a tap at the same rate of $20 \text{ litres per minute}$ .

Q3a

1 mark

By considering concentration, explain whether the quantity of salt in the tank increases with time.

Reveal Answer

The initial concentration $= \frac{5}{3000} = \frac{1}{600}$ which is smaller than the incoming concentration
$= 0.1 \text{ kg/litre}$ ; hence, the quantity of salt increases.

Marking Criteria

Descriptor	Marks
Explains that the quantity of salt increases by correctly comparing the initial concentration ( $\frac{1}{600} \text{ kg/L}$ ) to the incoming concentration ( $0.1 \text{ kg/L}$ )	1

Q3b

1 mark

Let $Q$ denote the quantity of salt, in kilograms, in the tank at time $t \text{ minutes}$ .

Show that $Q$ satisfies the differential equation $\frac{dQ}{dt} = \frac{300 - Q}{150}$ .

Reveal Answer

$\frac{dQ}{dt} = 0.1 \times 20 - \frac{Q}{3000} \times 20 = \frac{300}{150} - \frac{Q}{150}$

Marking Criteria

Descriptor	Marks
Shows the correct development of the differential equation by subtracting the rate out ( $\frac{Q}{3000} \times 20$ ) from the rate in ( $0.1 \times 20$ )	1

Q3c

2 marks

Using Euler's method with a step size of $15 \text{ minutes}$ , find $Q(30)$ , the approximate quantity of salt in the tank after $30 \text{ minutes}$ .

Give your answer in kilograms, correct to two decimal places.

Reveal Answer

61.05

$Q_{n+1} = Q_n + 15 \times \frac{300 - Q_n}{150}$
$Q(15) = \frac{69}{2}$
$Q(30) = 61.05$

Marking Criteria

Descriptor	Marks
Calculates the correct value for $Q(15)$ ( $\frac{69}{2}$ ) or demonstrates correct application of Euler's method	1
Calculates the correct final answer of $61.05$	1

Q3d

3 marks

Use calculus to solve the differential equation $\frac{dQ}{dt} = \frac{300 - Q}{150}$ , expressing $Q$ in terms of $t$ .

Reveal Answer

\begin{align*} \frac{dQ}{dt} &= \frac{300 - Q}{150}\\ t &= \int \frac{150}{300 - Q} dQ\\ t &= -150 \log_e(|300 - Q|) + c\\ c &= 150 \log_e(295)\\ t &= 150 \log_e\left(\frac{295}{300 - Q}\right)\\ Q &= 300 - 295 e^{-\frac{t}{150}} \end{align*}

Marking Criteria

Descriptor	Marks
Correctly integrates the differential equation to find an expression involving a constant of integration (e.g., $t = -150 \log_e(\|300 - Q\|) + c$ )	1
Correctly uses the initial condition to evaluate the constant of integration	1
Correctly expresses $Q$ in terms of $t$ as $Q = 300 - 295 e^{-\frac{t}{150}}$	1

Q3e

1 mark

What value does the quantity of salt in the tank approach as time approaches infinity?

Give your answer in kilograms.

Reveal Answer

300 kg

Marking Criteria

Descriptor	Marks
States the correct limiting value of $300$	1

Q3f

1 mark

Find the time taken for the quantity of salt in the tank to reach $100 \text{ kg}$ .

Reveal Answer

$t = 150 \log_e\left(\frac{59}{40}\right)$

Marking Criteria

Descriptor	Marks
Calculates the correct exact time of $150 \log_e\left(\frac{59}{40}\right)$	1

Q3g

1 mark

When the quantity of salt in the tank reaches $100 \text{ kg}$ , the tap draining the tank is turned off. Assume that the tank does not overflow and there is no change to the inflow rate.

After the tap is turned off, how many minutes does it take for the concentration of salt in the tank to reach $\frac{1}{20} \text{ kg L}^{-1}$ ?

Reveal Answer

50

This can be found by equating the concentration to the given value $\frac{2t + 100}{3000 + 20t} = \frac{1}{20}$ .

Marking Criteria

Descriptor	Marks
Calculates the correct time of $50$ minutes	1

Q18

2021

QCAA

Paper 1

6 marks

Q18

6 marks

This differential equation can be used to determine the current $I$ (amperes) at time $t$ (seconds) with voltage $V$ (volts) in an electric circuit containing a resistance $R$ (ohms):

$k \frac{dI}{dt} + RI = V$

where $k$ , $R$ and $V$ are positive constants and $t \geq 0$ .

Assuming that there is no current in the electric circuit initially, show that the size of the current can never be greater than $\frac{V}{R}$ .

Reveal Answer

$k \frac{dI}{dt} + RI = V \Rightarrow k \frac{dI}{dt} = V - RI$
$\int \frac{k}{V - RI} dI = \int 1 dt$
$-\frac{k}{R} \ln|V - RI| = t + c$
Given $I = 0$ when $t = 0$
$c = -\frac{k}{R} \ln(V)$ (as $V > 0$ )
$-\frac{k}{R} \ln|V - RI| = t - \frac{k}{R} \ln(V)$
$\ln|V - RI| = -\frac{R}{k}t + \ln(V)$
$V - RI = e^{-\frac{R}{k}t + \ln(V)}$
$V - RI = V e^{-\frac{R}{k}t}$
For all $t, e^{-\frac{R}{k}t} > 0 \Rightarrow V - RI > 0 \Rightarrow V > RI \Rightarrow I < \frac{V}{R}$
So, the size of the current can never be greater than $\frac{V}{R}$ .

Marking Criteria

Descriptor	Marks
correctly uses the separation of variables method to set up indefinite integrals	1
develops a general solution of the differential equation	1
uses the given condition to determine expression for the constant of integration	1
rearranges relationship to express $\ln\|V - RI\|$ as the subject of the equation	1
expresses relationship as an exponential function	1
considers value of $I$ over time to determine the required limit	1

Q10

2020

VCAA

Paper 2

1 mark

Q10

1 mark

A tank initially contains 300 grams of salt that is dissolved in 50 L of water. A solution containing 15 grams of salt per litre of water is poured into the tank at a rate of 2 L per minute and the mixture in the tank is kept well stirred. At the same time, 5 L of the mixture flows out of the tank per minute.

A differential equation representing the mass, $m$ grams, of salt in the tank at time $t$ minutes, for a non-zero volume of mixture, is

A

$\frac{dm}{dt}=0$

B

$\frac{dm}{dt}=-\frac{5m}{50-5t}$

C

$\frac{dm}{dt}=30-\frac{m}{10}$

D

$\frac{dm}{dt}=30-\frac{5m}{50-3t}$

E

$\frac{dm}{dt}=30-\frac{5m}{50-5t}$

Reveal Answer

A

$\frac{dm}{dt}=0$

This implies the mass of salt is constant, completely ignoring both the inflow and outflow of salt from the tank.

B

$\frac{dm}{dt}=-\frac{5m}{50-5t}$

This ignores the inflow of salt ( $30$ g/min) and incorrectly calculates the volume as $50-5t$ instead of $50-3t$ .

C

$\frac{dm}{dt}=30-\frac{m}{10}$

This assumes the volume of the tank is constant at $50$ L, leading to an outflow rate of $\frac{5m}{50} = \frac{m}{10}$ , but the volume is actually decreasing by $3$ L/min.

D

$\frac{dm}{dt}=30-\frac{5m}{50-3t}$

Correct Answer

The rate of salt entering is $15 \times 2 = 30$ g/min, and the rate leaving is the outflow rate ( $5$ L/min) times the concentration $\frac{m}{50-3t}$ .

E

$\frac{dm}{dt}=30-\frac{5m}{50-5t}$

This incorrectly calculates the volume of the mixture at time $t$ as $50-5t$ , which only accounts for the outflow rate and ignores the $2$ L/min inflow.

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 $n$ .

Reveal Answer

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.

Reveal Answer

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

Q5

2020

QCAA

Paper 2

1 mark

Q5

1 mark

The gradient of the tangent at point A on the curve $y^2 = 4x$ is 1.36
The $x$ -coordinate of A is

A

0.12

B

0.46

C

0.54

D

1.47

Reveal Answer

A

0.12

This value is incorrect and does not satisfy the relationship between the gradient and the coordinates on the curve.

B

0.46

This incorrect value likely results from evaluating $\frac{m^2}{4}$ (approx $0.46$ ) instead of the correct relationship derived from the derivative.

C

0.54

Correct Answer

Differentiating $y^2=4x$ gives $\frac{dy}{dx} = \frac{2}{y}$ . Setting $\frac{2}{y} = 1.36$ yields $y \approx 1.47$ , and substituting this back into $x = \frac{y^2}{4}$ gives $x \approx 0.54$ .

D

1.47

This is the $y$ -coordinate of point A ( $y = \frac{2}{1.36} \approx 1.47$ ), but the question asks for the $x$ -coordinate.

Q12

2020

VCAA

Paper 2

1 mark

Q12

1 mark

If $\frac{dy}{dx}=e^{\cos(x)}$ and $y_0=e$ when $x_0=0$ , then, using Euler's formula with step size $0.1$ , $y_3$ is equal to

A

$e+0.1\left(1+e^{\cos(0.1)}\right)$

B

$e+0.1\left(1+e^{\cos(0.1)}+e^{\cos(0.2)}\right)$

C

$e+0.1\left(e+e^{\cos(0.1)}+e^{\cos(0.2)}\right)$

D

$e+0.1\left(e^{\cos(0.1)}+e^{\cos(0.2)}+e^{\cos(0.3)}\right)$

E

$e+0.1\left(1+e^{\cos(0.1)}+e^{\cos(0.2)}+e^{\cos(0.3)}\right)$

Reveal Answer

A

$e+0.1\left(1+e^{\cos(0.1)}\right)$

This option calculates $y_2$ instead of $y_3$ and incorrectly evaluates the first derivative term $e^{\cos(0)}$ as $1$ instead of $e$ .

B

$e+0.1\left(1+e^{\cos(0.1)}+e^{\cos(0.2)}\right)$

This option incorrectly evaluates the first derivative term $e^{\cos(0)}$ as $1$ instead of $e$ .

C

$e+0.1\left(e+e^{\cos(0.1)}+e^{\cos(0.2)}\right)$

Correct Answer

Using Euler's method, $y_3 = y_0 + h(f(x_0) + f(x_1) + f(x_2))$ . Substituting the given values yields $e + 0.1(e^{\cos(0)} + e^{\cos(0.1)} + e^{\cos(0.2)})$ , and since $e^{\cos(0)} = e$ , this is the correct expression.

D

$e+0.1\left(e^{\cos(0.1)}+e^{\cos(0.2)}+e^{\cos(0.3)}\right)$

This option incorrectly evaluates the derivative at $x_1, x_2,$ and $x_3$ instead of starting at $x_0$ .

E

$e+0.1\left(1+e^{\cos(0.1)}+e^{\cos(0.2)}+e^{\cos(0.3)}\right)$

This option calculates $y_4$ instead of $y_3$ and incorrectly evaluates the first derivative term $e^{\cos(0)}$ as $1$ .

Q17

2020

QCAA

Paper 2

7 marks

Q17

7 marks

An object is released from rest at a height of 100 m above the ground.
The motion of the vertical descent of the object is modelled by

$v\frac{dv}{dx} = 9.8 - 0.004v^2 \quad (v \ge 0)$

where $v$ is the velocity (m s $^{-1}$ ) and $x$ is the displacement from the ground (m).
Determine the velocity of the object when it strikes the ground.

Reveal Answer

$v\frac{dv}{dx} = 9.8 - 0.004v^2, v > 0$
$\int \frac{v}{9.8 - 0.004v^2} dv = \int dx$
$\frac{-1}{0.008} \int \frac{-0.008v}{9.8 - 0.004v^2} dv = \int dx$
$-125 \ln|9.8 - 0.004v^2| = x + c$

Given $v = 0$ when $x = -100$
$-125 \ln|9.8| = -100 + c$
$c \approx -185.298$

$-125 \ln|9.8 - 0.004v^2| = x - 185.298$
Determining $v$ when $x = 0$
$-125 \ln|9.8 - 0.004v^2| = -185.298$

Using graph facility of GDC
$v \approx -36.7 \text{ ms}^{-1}$ or $v \approx 36.7 \text{ ms}^{-1}$
As $v > 0$ , the negative solution is rejected
$\therefore v \approx 36.7 \text{ ms}^{-1}$

Marking Criteria

Descriptor	Marks
correctly uses separation of variables	1
correctly develops the general solution of the differential equation	1
correctly uses the given position of the origin	1
uses the given condition to determine value for c	1
substitutes the displacement at impact to form an equation in terms of v	1
determines one reasonable solution of v	1
shows logical organisation communicating key steps	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.

Reveal Answer

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.

Reveal Answer

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 $k$ .

Reveal Answer

$k = 18000$ 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.

Reveal Answer

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

Q13

2025

VCAA

Paper 2

1 mark

Q13

1 mark

From an open window, a person projects a ball vertically up using an outstretched arm so the ball does not strike any part of the building. The point of projection of the ball is $50 \text{ m}$ above the ground and its velocity of projection is $20 \text{ m s}^{-1}$ .

The time, in seconds, it takes for the ball to reach the tray of a truck that is $1 \text{ m}$ above the ground directly below the point of projection is closest to

A

$1.72$

B

$5.80$

C

$5.83$

D

$1.75$

Reveal Answer

A

$1.72$

This is the magnitude of the negative root of the kinematic equation, which represents the time if the trajectory was extended backwards before projection.

B

$5.80$

Correct Answer

Using the kinematic equation $\Delta y = ut + \frac{1}{2}at^2$ with a displacement of $\Delta y = -49 \text{ m}$ (since the tray is $1 \text{ m}$ above the ground), $u = 20 \text{ m s}^{-1}$ , and $a = -9.8 \text{ m s}^{-2}$ yields $t \approx 5.80 \text{ s}$ .

C

$5.83$

This is the time it would take for the ball to reach the ground ( $\Delta y = -50 \text{ m}$ ), failing to account for the truck tray being $1 \text{ m}$ above the ground.

D

$1.75$

This is the magnitude of the negative root if the displacement was incorrectly set to $-50 \text{ m}$ (reaching the ground instead of the truck tray).

Q10

2022

VCAA

Paper 2

1 mark

Q10

1 mark

Consider the curve given by $5x^2y - 3xy + y^2 = 10$ .

The equation of the tangent to this curve at the point $(1, m)$ , where $m$ is a real constant, will have a negative gradient when

A

$m \in R \setminus [-1, 0]$

B

$m = -\sqrt{11} - 1$ only

C

$m \in R \setminus (-1, 0]$

D

$m = \sqrt{11} - 1$ only

E

$m = -\sqrt{11} - 1$ or $m = \sqrt{11} - 1$

Reveal Answer

A

$m \in R \setminus [-1, 0]$

While this is the range of $m$ values that yield a negative gradient, it fails to account for the fact that the point $(1, m)$ must actually lie on the given curve.

B

$m = -\sqrt{11} - 1$ only

This is only one of the two valid values for $m$ that lie on the curve and produce a negative gradient.

C

$m \in R \setminus (-1, 0]$

This inequality is incorrect for the gradient condition, and it also ignores that the point $(1, m)$ must satisfy the curve's equation.

D

$m = \sqrt{11} - 1$ only

This is only one of the two valid values for $m$ that lie on the curve and produce a negative gradient.

E

$m = -\sqrt{11} - 1$ or $m = \sqrt{11} - 1$

Correct Answer

Substituting $(1, m)$ into the curve's equation gives $m^2 + 2m - 10 = 0$ , yielding $m = -1 \pm \sqrt{11}$ . Both of these values satisfy the condition for a negative gradient, which is $\frac{dy}{dx} = \frac{-7m}{2m+2} < 0$ .

Q8

2021

VCAA

Paper 2

1 mark

Q8

1 mark

Euler's method, with a step size of 0.1, is used to approximate the solution of the differential equation

$\frac{dy}{dx} = y\sin(x)$ .

Given that $y = 2$ when $x = 1$ , the value of $y$ , correct to three decimal places, when $x = 1.2$ is

A

2.168

B

2.178

C

2.362

D

2.370

E

2.381

Reveal Answer

A

2.168

This is the approximation for $y(1.1)$ , not $y(1.2)$ , resulting from performing only the first step of Euler's method: $2 + 0.1(2\sin(1)) \approx 2.168$ .

B

2.178

This is incorrect and likely results from an arithmetic error or evaluating the sine function incorrectly during the iterative steps.

C

2.362

Correct Answer

Applying Euler's method twice gives $y(1.1) \approx 2 + 0.1(2\sin(1)) \approx 2.1683$ and $y(1.2) \approx 2.1683 + 0.1(2.1683\sin(1.1)) \approx 2.362$ .

D

2.370

This value does not follow from the proper application of Euler's method formula $y_{n+1} = y_n + h f(x_n, y_n)$ .

E

2.381

This is an incorrect approximation, possibly resulting from evaluating the sine function in degrees instead of radians or other calculation errors.

Q18

2023

QCAA

Paper 1

6 marks

Q18

6 marks

A particular solution to the differential equation $\frac{dy}{dx} = \frac{x}{(x^2+1)\tan(y)}$ , where $x \ge 0$ and $-\frac{\pi}{2} < y \le 0$ , passes through the origin.

Determine this solution in the form $x = f(y)$ . Leave your answer in simplified form.

Reveal Answer

$\frac{dy}{dx} = \frac{x}{(x^2+1)\tan(y)}$

$\int \tan(y) dy = \int \frac{x}{x^2+1} dx$

$-\int \frac{-\sin(y)}{\cos(y)} dy = \frac{1}{2} \int \frac{2x}{x^2+1} dx$

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

Given $y=0$ when $x=0$ ,
$-\ln|\cos(0)| = \frac{1}{2}\ln|1| + c \implies c=0$

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

$\ln\sqrt{x^2+1} = \ln\left|\frac{1}{\cos(y)}\right|$

$\sqrt{x^2+1} = |\sec(y)|$
$x^2+1 = \sec^2(y)$
$x^2 = \tan^2(y)$
$x = \pm \tan(y)$

As $x \ge 0, -\frac{\pi}{2} < y \le 0$
$x = -\tan(y)$

Marking Criteria

Descriptor	Marks
Correctly separates the variables	1
Applies suitable integration methods	1
Determines a value for the constant of integration	1
Determines an expression for a solution that does not contain logarithms	1
Expresses $x$ in terms of $y$	1
Evaluates the reasonableness of the results and expresses the solution in the form of $x = f(y)$ in simplified form	1

Q7

2023

QCAA

Paper 1

1 mark

Q7

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

A

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

B

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

C

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

D

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

Reveal Answer

A

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

This differential equation depends only on the independent variable $t$ . Integrating $\frac{dy}{dt} = 0.25 - 0.0025t$ results in a quadratic polynomial $y(t)$ , which describes a parabola rather than a logistic curve.

B

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

This equation represents restricted growth where the rate is proportional to the difference between a limit and the current value. The solution is an exponential function of the form $y = A + Be^{-kt}$ , not a logistic function.

C

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

This equation depends only on $t$ and not on the function $y$ itself. Integrating this expression yields a cubic polynomial in $t$ , not the logistic equation.

D

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

Correct Answer

This is the standard form of the logistic differential equation $\frac{dy}{dt} = ry(1 - \frac{y}{K})$ . The rate of change is proportional to both the current value $y$ and the remaining capacity $(1 - 0.01y)$ , which yields the logistic solution.

Q12

2024

VCAA

Paper 2

1 mark

Q12

1 mark

The position, $x$ metres, of a particle moving in a straight line from a fixed origin $O$ at time, $t$ seconds, is given by $x = e^{(k - 1)t}$ , where $k > 1$ .

The acceleration of the particle, in $\text{m s}^{-2}$ , when $x = k + 1$ is

A

$k^2 - 1$

B

$(k^2 - 1)(k + 1)$

C

$(k^2 - 1)(k - 1)$

D

$(k - 1)^2$

Reveal Answer

A

$k^2 - 1$

Incorrect. This represents the velocity of the particle when $x = k + 1$ , found by taking the first derivative $v = (k - 1)x$ , rather than the acceleration.

B

$(k^2 - 1)(k + 1)$

Incorrect. This expression does not match the acceleration formula $a = (k - 1)^2 x$ . It incorrectly multiplies the velocity by $x$ instead of taking the second derivative.

C

$(k^2 - 1)(k - 1)$

Correct Answer

Correct. The acceleration is the second derivative of position, $a = (k - 1)^2 x$ . Substituting $x = k + 1$ gives $a = (k - 1)^2(k + 1) = (k - 1)(k^2 - 1).

D

$(k - 1)^2$

Incorrect. This is the coefficient of $x$ in the acceleration equation $a = (k - 1)^2 x$ , but it fails to substitute the given value $x = k + 1$ .

SCSA Mathematics Specialist Rates of change and differential equations

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