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

AusGrader has 15 QCAA Specialist Mathematics Alternative Sequence questions on Rates of change and differential equations, all with instant AI grading and detailed marking feedback.

QCAA Specialist Mathematics Alternative Sequence — Rates of change and differential equations Questions & Answers

Q8

2022

QCAA

Paper 2

1 mark

Q8

1 mark

Determine the gradient of the tangent to the curve $y^2 - 3x = 5$ at the point $\left(1, 2\sqrt{2}\right)$ .

A

0.41

B

0.53

C

1.06

D

8.49

Reveal Answer

A

0.41

Incorrect. This value does not match the gradient $\frac{3}{4\sqrt{2}}$ . It likely results from an arithmetic error during differentiation or substitution.

B

0.53

Correct Answer

Correct. Using implicit differentiation, $2y \frac{dy}{dx} - 3 = 0$ , so $\frac{dy}{dx} = \frac{3}{2y}$ . Substituting $y = 2\sqrt{2}$ gives $\frac{3}{4\sqrt{2}} \approx 0.53$ .

C

1.06

Incorrect. This value is approximately $\frac{3}{2\sqrt{2}}$ , which results from incorrectly differentiating $y^2$ as $y$ instead of $2y$ , leading to $\frac{dy}{dx} = \frac{3}{y}$ .

D

8.49

Incorrect. This value is approximately $6\sqrt{2}$ , which results from incorrectly solving for $\frac{dy}{dx}$ as $3y$ instead of dividing by $2y$ .

Q19

2022

QCAA

Paper 2

7 marks

Q19

7 marks

A research organisation plans to use a drone to drop a scientific instrument vertically from a stationary position above the ocean surface. The acceleration $\left(\text{m s}^{-2}\right)$ of the falling instrument can be modelled by $9.8 - 0.1v$ , where $v$ is its velocity $\left(\text{m s}^{-1}\right)$ .

In order for the instrument sensors to activate, its speed as it hits the ocean surface must reach at least $20 \text{ m s}^{-1}$ . However, if it hits with a speed above $50 \text{ m s}^{-1}$ , the sensors will be damaged.

Determine the range of the drone's flying height above the ocean surface to ensure that the sensors are activated but not damaged.

Reveal Answer

Assume downwards as the positive direction
$a = 9.8 - 0.1v$
$v \frac{dv}{dx} = 9.8 - 0.1v$
$\int \frac{v}{9.8 - 0.1v} dv = \int dx$
$\int \frac{v}{v - 98} dv = \int \frac{-1}{10} dx$
$\int 1 + \frac{98}{v - 98} dv = \int \frac{-1}{10} dx$
$v + 98 \ln|v - 98| = \frac{-x}{10} + c$

Assume the origin is at the point of release.
Given $v(0) = 0 \Rightarrow c = 98 \ln(98)$
$v + 98 \ln|v - 98| = \frac{-x}{10} + 98 \ln(98)$

Let the distance to the ocean surface be $h$ metres.
Consider time of drop for each required velocity
When $v = 20 \text{ m s}^{-1}, x = h$
$20 + 98 \ln|-78| = \frac{-h}{10} + 98 \ln(98)$
$h = 23.7 \text{ m}$

When $v = 50 \text{ m s}^{-1}, x = h$
$50 + 98 \ln|-48| = 98 \ln(98) - \frac{h}{10}$
$h = 199.5 \text{ m}$

The range of the drone's flying height above the ocean surface should be between 23.7 m and 199.5 m.

Marking Criteria

Descriptor	Marks
correctly establishes a differential equation in terms of $v$ and $x$	1
determines general solution of the differential equation	1
determines value for the constant	1
determines model for the velocity in terms of displacement	1
determines displacement of the drop for the minimum acceptable speed	1
determines displacement of the drop for the maximum acceptable speed	1
communicates range of the drone's flying height including units	1

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

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

Reveal Answer

A

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

This is incorrect because it results from multiplying by 2 instead of dividing by 2 when applying the chain rule during integration.

B

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

This is incorrect because it results from multiplying by 2 during integration and making a sign error when solving for the constant $C$ .

C

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

Correct Answer

This is correct. Integrating $\frac{\sin(2x)}{\cos(2x)}$ using u-substitution yields $-\frac{1}{2}\ln|\cos(2x)| + C$ . Substituting $x = \frac{\pi}{5}$ and $y = 0$ gives $C = \frac{1}{2}\ln|\cos(\frac{2\pi}{5})| \approx -0.59$ .

D

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

This is incorrect. While the integration is correct, a sign error was made when solving for the constant of integration $C$ , which should be negative.

Q14

2022

QCAA

Paper 2

5 marks

Q14

An object is moving in a straight line with an acceleration represented by the differential equation $\frac{dv}{dt} = -\left(4 + v^2\right)$ , where $v$ is the object's velocity $\left(\text{m s}^{-1}\right)$ 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.

Reveal Answer

$\frac{dv}{dt} = -(4 + v^2)$
$\frac{1}{4 + v^2} \frac{dv}{dt} = -1$
$\int \frac{1}{4 + v^2} dv = \int -1 dt$
$\frac{1}{2} \int \frac{2}{4 + v^2} dv = \int -1 dt$
$\frac{1}{2} \tan^{-1} \left(\frac{v}{2}\right) = -t + c$

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.

Reveal Answer

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

Q12

2020

QCAA

Paper 2

9 marks

Q12

For a certain experiment, the number of yeast cells, $N$ , after $t$ hours in a test tube can be modelled by the differential equation

$\frac{dN}{dt} = \frac{1}{1000}N(1000 - N)$ for $t \ge 0$

Q12b

A scientist commenced this experiment at 9:00 am on a certain day and observed that 100 yeast cells were present at this time.

Q12d

The scientist predicted that the number of yeast cells would eventually exceed 1200.

Q12a

2 marks

Given $\frac{1000}{N(1000 - N)} = \frac{1}{N} + \frac{1}{1000 - N}$ , show that the general solution of the differential equation can be expressed as

$\ln\left|\frac{N}{1000 - N}\right| = t + c$

Reveal Answer

\frac{dN}{dt} = \frac{1}{1000}N(1000 - N)

\frac{1000}{N(1000-N)} \frac{dN}{dt} = 1

\left(\frac{1}{N} + \frac{1}{1000-N}\right) \frac{dN}{dt} = 1

\int \left(\frac{1}{N} + \frac{1}{1000-N}\right) dN = \int 1 dt

\ln|N| - \ln|1000 - N| = t + c

\ln\left|\frac{N}{1000 - N}\right| = t + c

Marking Criteria

Descriptor	Marks
correctly uses separation of variables technique and substitutes the given result into the differential equation	1
correctly develops the required general solution	1

Q12b

3 marks

Show that the solution of the differential equation can be expressed as

$N = \frac{1000}{1 + 9e^{-t}}$

Reveal Answer

\text{Let } t = \text{time after 9:00 am (in hours)}

\text{When } t = 0, N = 100

\ln\left|\frac{100}{1000-100}\right| = 0 + c

c = \ln\left(\frac{1}{9}\right)

\ln\left(\frac{N}{1000 - N}\right) = t + \ln\left(\frac{1}{9}\right)

\frac{N}{1000 - N} = e^{t + \ln(1/9)}

\frac{N}{1000 - N} = \frac{e^t}{9}

9N = 1000e^t - Ne^t

N(9 + e^t) = 1000e^t

N = \frac{1000e^t}{(9 + e^t)}

N = \frac{1000}{1 + 9e^{-t}}

Marking Criteria

Descriptor	Marks
correctly determines c	1
substitutes the value of c into the general equation and simplifies sufficiently to produce a function that includes the term e^t	1
develops a solution for N	1

Q12c

2 marks

Determine the time of day when 900 yeast cells were present.

Reveal Answer

\text{Given } N = 900

900 = \frac{1000}{1 + 9e^{-t}}

\text{Using solve facility of GDC}

t \approx 4.394 \text{ (hours)}

\text{The time of day is 1:24 pm.}

Marking Criteria

Descriptor	Marks
correctly determines the value of t when N = 900	1
communicates the time of day	1

Q12d

2 marks

Evaluate the reasonableness of the scientist's prediction.

Reveal Answer

\text{Given } N = 1200:

\text{As } t \to \infty, N \to 1000

\text{The number of yeast cells has a limit of 1000.}

\text{As } N \text{ will never reach 1200, the scientist's prediction is not reasonable.}

Marking Criteria

Descriptor	Marks
correctly recognises that the number of yeast cells will never exceed 1000	1
comments that the prediction is not reasonable	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 ( $\text{m s}^{-1}$ ) 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

\text{Given } v = 0 \text{ when } x = -100

-125\ln|9.8| = -100 + c

c \approx -185.298

-125\ln|9.8 - 0.004v^2| = x - 185.298

\text{Determining } v \text{ when } x = 0

-125\ln|9.8 - 0.004v^2| = -185.298

\text{Using graph facility of GDC}

v \approx -36.7 \text{ ms}^{-1} \text{ or } v \approx 36.7 \text{ ms}^{-1}

\text{As } v > 0 \text{, 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

Q17

2022

QCAA

Paper 1

7 marks

Q17

The Cartesian equation of an ellipse is $\frac{x^2}{9} + \frac{y^2}{16} = 1$ .

Q17a

3 marks

Determine the gradient of the ellipse at the point in quadrant 2 where $y = 2\sqrt{2}$ .

Reveal Answer

At $y = 2\sqrt{2}$
$\frac{x^2}{9} + \frac{(2\sqrt{2})^2}{16} = 1$
$x^2 = \frac{9}{2} \Rightarrow x = \pm \frac{3}{\sqrt{2}}$

Given the point on the ellipse is in quadrant 2,
$x = -\frac{3\sqrt{2}}{2}$

Determining gradient of the ellipse
$\frac{x^2}{9} + \frac{y^2}{16} = 1$
$\frac{2x}{9} + \frac{2y \frac{dy}{dx}}{16} = 0$
$\frac{dy}{dx} = -\frac{16x}{9y}$

Determining gradient at the required point
$\frac{dy}{dx} = -\frac{16\left(-\frac{3\sqrt{2}}{2}\right)}{9(2\sqrt{2})} = \frac{4}{3}$

Marking Criteria

Descriptor	Marks
correctly determines the $x$ -coordinate of the required point in quadrant 2	1
determines a general expression for the gradient of the ellipse in terms of $x$ and $y$	1
determines gradient of the ellipse at required point	1

Q17b

4 marks

The ellipse can be shown to be represented by the parametric equations $x = 3\cos(\theta)$ and $y = 4\sin(\theta)$ , where $0 \leq \theta \leq 2\pi$ .

Use the parametric equations to verify your result from Question 17a). Do not show the conversion from the parametric equations into the Cartesian equation.

Reveal Answer

Determining gradient of ellipse
$x = 3\cos(\theta) \Rightarrow \frac{dx}{d\theta} = -3\sin(\theta)$
$y = 4\sin(\theta) \Rightarrow \frac{dy}{d\theta} = 4\cos(\theta)$

Determining $\theta$ for the point on the ellipse
$x = 3\cos(\theta)$ : At $x = -\frac{3\sqrt{2}}{2} \Rightarrow \cos(\theta) = -\frac{\sqrt{2}}{2}$
$y = 4\sin(\theta)$ : At $y = 2\sqrt{2} \Rightarrow \sin(\theta) = \frac{\sqrt{2}}{2}$

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{4\cos(\theta)}{-3\sin(\theta)}$

At the point on the ellipse
$\frac{dy}{dx} = \frac{4\left(-\frac{\sqrt{2}}{2}\right)}{-3\left(\frac{\sqrt{2}}{2}\right)} = \frac{4}{3}$

The result from 17a) is verified.

Marking Criteria

Descriptor	Marks
correctly determines expressions for $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$	1
determines expressions for $\cos(\theta)$ and $\sin(\theta)$	1
determines a general expression for the gradient of the ellipse in terms of $\theta$	1
verifies gradient of the ellipse result from 17a)	1

QCAA Specialist Mathematics Alternative Sequence Rates of change and differential equations

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