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

AusGrader has 78 QCAA Specialist Mathematics questions on Rates of change and differential equations, all with instant AI grading and detailed marking feedback.

QCAA (QCE) Specialist Mathematics — Rates of change and differential equations Questions & Answers | AusGrader

Q3

2022

QCAA

Paper 2

1 mark

Q3

1 mark

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

A

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

B

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

C

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

D

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

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

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

Q14

2022

QCAA

Paper 2

5 marks

Q14

An object is moving in a straight line with an acceleration represented by the differential equation
$\frac{dv}{dt} = -(4+v^2)$ , where $v$ is the object's velocity ( $\text{m s}^{-1}$ ) over time, $t(\text{s})$ , where $t \ge 0$ , until it comes to rest.

Q14a

3 marks

Determine the general solution of the differential equation.

$\frac{dv}{dt} = -(4+v^2)$

$\frac{1}{4+v^2} \frac{dv}{dt} = -1$

$\int \frac{1}{4+v^2} dv = \int -1 dt$

$\frac{1}{2} \int \frac{2}{4+v^2} dv = \int -1 dt$

$\frac{1}{2} \tan^{-1}\left(\frac{v}{2}\right) = -t + c$

Marking Criteria

Descriptor	Marks
Correctly establishes a suitable integration result based on the separation of variables technique	1
Determines one side of the general solution in terms of $v$	1
Determines the other side of the general solution in terms of $t$	1

Q14b

2 marks

The initial velocity of the object is $1.5 \text{ m s}^{-1}$ .

Determine the time when the particle comes to rest.

Given $v(0) = 1.5$
$\frac{1}{2} \tan^{-1}\left(\frac{1.5}{2}\right) = c$
$c \approx 0.32$

When $v=0$ :
$\frac{1}{2} \tan^{-1}(0) \approx -t + 0.32$
$t \approx 0.32 \text{ s}$
The particle comes to rest after 0.32 s.

Marking Criteria

Descriptor	Marks
Determines an expression that represents the integration constant	1
Determines the time when the particle is at rest	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 $\frac{V}{R}$ .

$k \frac{dI}{dt} + RI = V \Rightarrow k \frac{dI}{dt} = V - RI$
$\int \frac{k}{V - RI} dI = \int 1 dt$
$-\frac{k}{R} \ln|V - RI| = t + c$
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

Q17

2025

QCAA

Paper 1

7 marks

Q17

7 marks

The radius of a cylinder decreases at a constant rate of $0.5 \text{ m s}^{-1}$ , while maintaining a constant height of four metres.

Given that the cylinder has an initial volume of $100\pi \text{ m}^3$ , determine the rate of change of the volume ( $\text{m}^3 \text{ s}^{-1}$ ) of the cylinder after four seconds.

Given $V = \pi r^2 h \Rightarrow V = 4\pi r^2$

Given $\frac{dr}{dt} = -0.5 \text{ m s}^{-1}$

$V = 4\pi r^2 \Rightarrow \frac{dV}{dt} = 8\pi r \frac{dr}{dt}$

At $t = 0$ , $V = 100\pi$ (given)

100\pi = 4\pi r^2 \Rightarrow r = 5 \text{ m} (\text{as } r > 0)

At $t = 4$ ,

r = 5 - 4 \times 0.5 = 3 \text{ m}

When $r = 3$ :

\frac{dV}{dt} = 8\pi \times 3 \times (-0.5) = -12\pi \text{ m}^3 \text{ s}^{-1}

Marking Criteria

Descriptor	Marks
correctly determines the rule for the volume of cylinder in terms of $r$	1
correctly expresses the rate of change of the radius as a mathematical expression	1
determines a general expression for $\frac{dV}{dt}$	1
determines initial value of $r$	1
determines value of $r$ after 4 seconds	1
determines value of $\frac{dV}{dt}$ when $t = 4$	1
shows logical organisation, having fully attempted the question	1

QCAA Specialist Mathematics Rates of change and differential equations

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