Vector calculus and kinematics: VCE Specialist Mathematics Unit 4

Syllabus dot point

“Vector functions of a real variable, the differentiation and integration of a position vector $\mathbf{r}(t)$ to obtain velocity and acceleration, the speed as the magnitude of velocity, and the application to motion in two and three dimensions”

6 min readUpdated 29 May 2026

Quick answer

A position vector $\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k}$ is differentiated component by component. The velocity is $\mathbf{v}(t) = \dot{\mathbf{r}}(t)$ , the acceleration is $\mathbf{a}(t) = \dot{\mathbf{v}}(t) = \ddot{\mathbf{r}}(t)$ . The speed is the scalar $|\mathbf{v}(t)|$ , the magnitude of the velocity. Reversing the process, integrate acceleration to get velocity and velocity to get position, with a vector constant of integration fixed by initial conditions. The velocity vector is always tangent to the path; the acceleration need not be.

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What this dot point is asking
Vector functions and differentiation
Speed
Integration and initial conditions
The path
Examples in context
Try this
Exam-style questions

What this dot point is asking

VCAA wants you to treat a particle's position as a vector function of time $\mathbf{r}(t)$ , to differentiate it component by component to get velocity and acceleration, to find speed as the magnitude of velocity, and to integrate acceleration or velocity (with initial conditions) to recover the motion. This is the vector version of the rectilinear kinematics studied alongside it.

Vector functions and differentiation

A particle moving in space has position given by a vector function of time:

\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k}.

Differentiation is done component by component, because $\mathbf{i}, \mathbf{j}, \mathbf{k}$ are constant:

\mathbf{v}(t) = \dot{\mathbf{r}}(t) = \dot{x}\,\mathbf{i} + \dot{y}\,\mathbf{j} + \dot{z}\,\mathbf{k}, \qquad \mathbf{a}(t) = \dot{\mathbf{v}}(t) = \ddot{x}\,\mathbf{i} + \ddot{y}\,\mathbf{j} + \ddot{z}\,\mathbf{k}.

The velocity $\mathbf{v}$ is the rate of change of position; it points along the direction of motion and is tangent to the path. The acceleration $\mathbf{a}$ is the rate of change of velocity and can point in any direction relative to the path.

Speed

Speed is a scalar: the magnitude of the velocity vector,

\text{speed} = |\mathbf{v}(t)| = \sqrt{\dot{x}^2 + \dot{y}^2 + \dot{z}^2}.

Velocity carries both direction and size; speed keeps only the size. A particle can have constant speed yet changing velocity if it is turning.

Integration and initial conditions

Reversing differentiation, integrate component by component. Given acceleration $\mathbf{a}(t)$ ,

\mathbf{v}(t) = \int \mathbf{a}(t)\,dt + \mathbf{c}_1, \qquad \mathbf{r}(t) = \int \mathbf{v}(t)\,dt + \mathbf{c}_2,

where $\mathbf{c}_1$ and $\mathbf{c}_2$ are vector constants of integration. Fix them using the initial velocity and initial position. Each component is integrated separately, and each gets its own scalar constant, together forming the vector constant.

The path

Eliminating $t$ from the components of $\mathbf{r}(t)$ gives the Cartesian equation of the path the particle traces, just as with parametric curves. The distance travelled over a time interval is the integral of speed, $\int_{t_1}^{t_2} |\mathbf{v}(t)|\,dt$ .

Worked example

Velocity, speed and acceleration

A particle has position $\mathbf{r}(t) = (t^2)\mathbf{i} + (2t)\mathbf{j}$ metres at time $t$ seconds. Find its velocity, speed and acceleration at $t = 2$ .

Step 1: Find the velocity

Velocity is the derivative of position with respect to time. Because $\mathbf{i}$ and $\mathbf{j}$ are constant unit vectors, we differentiate each scalar component separately. Evaluating at $t = 2$ gives the instantaneous velocity at that moment.

\mathbf{v}(t) = \dot{\mathbf{r}}(t) = (2t)\mathbf{i} + 2\mathbf{j}.

At $t = 2$ : $\mathbf{v}(2) = 4\mathbf{i} + 2\mathbf{j}\ \text{m/s}$ .

Step 2: Find the speed

Speed is a scalar quantity: the magnitude of the velocity vector. It tells us how fast the particle is moving, without regard to direction, and is always non-negative.

|\mathbf{v}(2)| = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}\ \text{m/s} \approx 4.47\ \text{m/s}.

Step 3: Find the acceleration and verify the direction

Acceleration is the derivative of velocity with respect to time, again applied component by component. As a check, eliminating $t$ from the position components shows the path is a parabola, and the velocity direction should be tangent to it.

\mathbf{a}(t) = \dot{\mathbf{v}}(t) = 2\mathbf{i} + 0\mathbf{j} = 2\mathbf{i}.

The acceleration is the constant $2\mathbf{i}\ \text{m/s}^2$ for all $t$ , including $t = 2$ . To verify: $x = t^2$ and $y = 2t$ give $t = \frac{y}{2}$ , so $x = \frac{y^2}{4}$ , a parabola. The velocity $4\mathbf{i} + 2\mathbf{j}$ at $t = 2$ has gradient $\frac{2}{4} = \frac{1}{2}$ , and the parabola at the point $(4, 4)$ has $\frac{dy}{dx} = \frac{1}{2}$ , confirming the velocity is tangent to the path.

Final answer: $\mathbf{v}(2) = 4\mathbf{i} + 2\mathbf{j}\ \text{m/s}$ , speed $= 2\sqrt{5}\ \text{m/s}$ , $\mathbf{a} = 2\mathbf{i}\ \text{m/s}^2$ .

Common exam traps

Confusing speed and velocity: Speed is the scalar magnitude $|\mathbf{v}|$ ; velocity is the vector. Exam wording distinguishes them.
Forgetting the vector constant: Integrating acceleration needs a vector constant of integration, fixed by the initial velocity; each component carries its own constant.
Differentiating the basis vectors: $\mathbf{i}, \mathbf{j}, \mathbf{k}$ are constant, so only the scalar components are differentiated.
Assuming acceleration is along the path: The velocity is tangent to the path, but acceleration generally is not, especially when the particle turns.

Examples in context

Example 1. If $\mathbf{a}(t) = -g\mathbf{j}$ (constant downward), integrating gives projectile motion with parabolic path.

Example 2. For $\mathbf{r}(t) = \cos t\,\mathbf{i} + \sin t\,\mathbf{j}$ , the speed is $|\mathbf{v}| = 1$ (constant), yet the velocity direction changes continuously around the circle.

Try this

Q1. Given $\mathbf{r}(t) = t^3\mathbf{i} + t\mathbf{j}$ , find $\mathbf{v}(t)$ . [2 marks]

Cue. $\mathbf{v}(t) = 3t^2\mathbf{i} + \mathbf{j}$ .

Q2. Find the speed of a particle with velocity $3\mathbf{i} - 4\mathbf{j}$ . [1 mark]

Cue. $\sqrt{9 + 16} = 5$ .

Q3. If $\mathbf{a}(t) = 2\mathbf{i}$ and $\mathbf{v}(0) = \mathbf{j}$ , find $\mathbf{v}(t)$ . [2 marks]

Cue. $\mathbf{v}(t) = 2t\mathbf{i} + \mathbf{j}$ .

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2023 VCAA-style1 mark

The position vector of a particle at time t seconds is given by r(t) = (5 - 6 sin^2(t)) i + (1 + 6 sin(t)cos(t)) j, where t >= 0. Write 5 - 6 sin^2(t) in the form alpha + beta cos(2t), where alpha, beta are in Z.

Show worked answer →

Use the double-angle identity cos(2t) = 1 - 2 sin^2(t), which rearranges to sin^2(t) = (1 - cos(2t))/2.

Substitute into the expression: 5 - 6 sin^2(t) = 5 - 6 . (1 - cos(2t))/2 = 5 - 3(1 - cos(2t)).

Expand: 5 - 3 + 3 cos(2t) = 2 + 3 cos(2t).

So alpha = 2 and beta = 3, that is 5 - 6 sin^2(t) = 2 + 3 cos(2t).

2023 VCAA-style2 marks

The position vector of a particle at time t seconds is given by r(t) = (5 - 6 sin^2(t)) i + (1 + 6 sin(t)cos(t)) j, where t >= 0. Show that the Cartesian equation of the path of the particle is (x - 2)^2 + (y - 1)^2 = 9.

Show worked answer →

Write the components as x and y, then simplify each with double-angle identities.

x-component: x = 5 - 6 sin^2(t) = 2 + 3 cos(2t) (from sin^2(t) = (1 - cos(2t))/2). So x - 2 = 3 cos(2t).

y-component: y = 1 + 6 sin(t)cos(t) = 1 + 3 sin(2t) (using sin(2t) = 2 sin(t)cos(t)). So y - 1 = 3 sin(2t).

Now eliminate t using cos^2(2t) + sin^2(2t) = 1:
(x - 2)^2 + (y - 1)^2 = (3 cos(2t))^2 + (3 sin(2t))^2 = 9 cos^2(2t) + 9 sin^2(2t) = 9.

Hence the path is the circle (x - 2)^2 + (y - 1)^2 = 9, centre (2, 1) and radius 3, as required.