How many QCAA Specialist Mathematics questions cover Vector calculus?

AusGrader has 55 QCAA Specialist Mathematics questions on Vector calculus, all with instant AI grading and detailed marking feedback.

QCAA (QCE) Specialist Mathematics — Vector calculus Questions & Answers | AusGrader

Q3

2021

QCAA

Paper 1

1 mark

Q3

1 mark

An object has a velocity $v(t) = e^{-2t}\hat{i} + \left(\frac{1}{t}\right)\hat{k}$ , where $t$ represents time ( $t > 0$ ).

The displacement $r(t)$ of the object could be

A

$-2e^{-2t}\hat{i} + \ln(t)\hat{k}$

B

$-2e^{-2t}\hat{i} - \frac{1}{t^2}\hat{k}$

C

$-\frac{1}{2}e^{-2t}\hat{i} + \ln(t)\hat{k}$

D

$-\frac{1}{2}e^{-2t}\hat{i} - \frac{1}{t^2}\hat{k}$

Q9

2020

QCAA

Paper 2

1 mark

Q9

1 mark

Two objects, P and Q, move in three-dimensional space such that their positions, $\mathbf{r}$ , over time, $t$ , are described by the following vectors until they collide.

$\mathbf{r}_P = (t^2 - 4t)\mathbf{i} + (2t^2 - t + 3)\mathbf{j} - (6 - 5t)\mathbf{k}$
$\mathbf{r}_Q = (-t^2 + 2t)\mathbf{i} + (3t + t^2)\mathbf{j} + t^2\mathbf{k}$

The objects will collide at

A

$t = 0$

B

$t = 1$

C

$t = 2$

D

$t = 3$

Q14

2021

QCAA

Paper 1

6 marks

Q14

An object is projected vertically upwards from ground level. After the object has been in motion for $t$ seconds, its position vector through the air, in metres, is modelled by

$\boldsymbol{r}(t) = 5t(8-t)\boldsymbol{\hat{j}}$

Q14a

2 marks

Determine the velocity of the object through the air, $\boldsymbol{v}(t)$ , in metres per second.

$r(t) = 5t(8 - t)j$
$= (40t - 5t^2)j$
$v(t) = (40 - 10t)j$

Marking Criteria

Descriptor	Marks
correctly expands the given expression	1
determines $v(t)$	1

Q14b

2 marks

Determine the number of seconds until the object reaches its maximum height.

Maximum height occurs when $v(t) = 0j$
$40 - 10t = 0$
$10t = 40$
$t = 4$ s

Marking Criteria

Descriptor	Marks
establishes equation in terms of $t$	1
determines $t$	1

Q14c

2 marks

Determine the maximum height that the object reaches, in metres.

Maximum height $= 5 \times 4(8 - 4)$
$= 20 \times 4$
$= 80$ m

Marking Criteria

Descriptor	Marks
substitutes result from 14b) into expression for position	1
determines maximum height	1

Q11

2022

QCAA

Paper 1

6 marks

Q11

The position vector of a particle, $r_1(\text{cm})$ , over time, $t(\text{s})$ , is given by
$r_1(t) = (2t+1)\hat{i} + (t+3)\hat{j} - (2t-3)\hat{k}$

Q11a

1 mark

Determine the velocity vector of the particle.

Descriptor	Marks
correctly determines the velocity vector	1

Q11b

2 marks

Determine the time when the position vector of the particle is perpendicular to its velocity vector.

Require $r_1 \cdot v_1 = 0$
$\left(\begin{matrix} 2t+1 \\ t+3 \\ -(2t-3) \end{matrix}\right) \cdot \left(\begin{matrix} 2 \\ 1 \\ -2 \end{matrix}\right) = 0$
$2(2t+1) + 1(t+3) + 2(2t-3) = 0$
$9t - 1 = 0$
$t = \frac{1}{9}$ s

Marking Criteria

Descriptor	Marks
establishes an equation in the form $x_1x_2 + y_1y_2 + z_1z_2 = 0$	1
solves equation to determine the time	1

Q11c

3 marks

The position vector of a second particle, $r_2(\text{cm})$ , over time, $t(\text{s})$ , is given by
$r_2(t) = (16-4t)\hat{i} - (3t-13)\hat{j} + 2\hat{k}$

Determine whether the two particles collide.

Let $r_1(t) = r_2(t)$
Equating $i, j$ and $k$ components
$2t+1 = 16 - 4t$ ... (1)
$t+3 = -3t + 13$ ... (2)
$-2t+3 = 2$ ... (3)
From (1), $t = 2.5$ s
From (2), $t = 2.5$ s
From (3), $t = 0.5$ s

The particles do not collide as the solutions are not consistent.

Marking Criteria

Descriptor	Marks
correctly equates 2 pairs of components for the particles, one of which must be the $k$ component, to form 2 equations	1
solves equations to determine two different time values	1
uses mathematical reasoning to make an appropriate conclusion whether the two particles collide	1

Q15

2024

QCAA

Paper 2

8 marks

Q15

The vectors representing the position (m) of particles A and B are given by $r_A = (4t - 9)\hat{i} - 2(5 - t)\hat{j} - 8\hat{k}$ and $r_B = (t^2 + 1)\hat{i} - 3\hat{j} + (4 - at^2)\hat{k}$ respectively, where $t$ is the time of motion for $0 \le t \le 10$ seconds.

Q15a

2 marks

Show that particle A passes through the point $P(5, -3, -8)$ .

If $r_A = (4t-9)\hat{i} - 2(5-t)\hat{j} - 8\hat{k}$ passes through $P(5, -3, -8)$ .

Consider $x_A$

$4t-9=5$

$t=3.5 \text{ s}$

Consider $y_A$

$-2(5-t)=-3$

$t=3.5 \text{ s}$

So particle A passes through point P.

Marking Criteria

Descriptor	Marks
correctly determines time taken for particle A to reach P by considering the $\hat{i}$ component of $r_A$	1
correctly determines time taken for particle A to reach P by considering the $\hat{j}$ component of $r_A$	1

Q15b

2 marks

Given that particle B also passes through point $P$ , determine the value of $a$ .

Given $r_B = (t^2+1)\hat{i} - 3\hat{j} + (4-at^2)\hat{k}$ also passes through $P(5, -3, -8)$ .

Consider $x_B$

$t^2+1=5$

$t^2=4$

$t=2 \text{ s}$ (as $0 \leq t \leq 10$ )

Consider $z_B$

$4-a(2)^2 = -8$

$4a=12$

$a=3$

Marking Criteria

Descriptor	Marks
correctly determines time for object B to reach P	1
determines value of $a$	1

Q15c

1 mark

Determine the vector that represents the displacement of particle B relative to particle A during the given time of motion. Express your answer in simplest form.

Displacement vector of particle B relative to particle A is

$r_B - r_A$

$= ((t^2+1)\hat{i} - 3\hat{j} + (4-3t^2)\hat{k}) - ((4t-9)\hat{i} - 2(5-t)\hat{j} - 8\hat{k})$

$= (t^2-4t+10)\hat{i} + (7-2t)\hat{j} + (12-3t^2)\hat{k} \text{ m}$

Marking Criteria

Descriptor	Marks
determines the required displacement vector in simplest form	1

Q15d

3 marks

Use your result from Question 15c) to determine the shortest distance between particles A and B during the given time of motion.

Distance between particles is

$|r_B - r_A|$

$= \sqrt{(t^2-4t+10)^2 + (7-2t)^2 + (12-3t^2)^2} \text{ m}$

Particles are closest when the distance between the particles is a minimum.

Using GDC

Shortest distance = 6.69 m

Marking Criteria

Descriptor	Marks
determines an expression that represents the distance between the particles	1
justifies when shortest distance between the particles occurs	1
determines shortest distance between particles	1

QCAA Specialist Mathematics Vector calculus

Frequently Asked Questions

Ready to practise QCAA Specialist Mathematics?