Given r(t) = t^3i + tj, find v(t). [2 marks]

Find the speed of a particle with velocity 3i - 4j. [1 mark]

If a(t) = 2i and v(0) = j, find v(t). [2 marks]

VICSpecialist MathematicsSyllabus dot point

How do we differentiate and integrate a vector function of time, and how does this describe the position, velocity and acceleration of a particle moving in a plane or space?

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

A focused answer to the VCE Specialist Mathematics Unit 4 key-knowledge point on vector calculus and kinematics. Differentiating and integrating a position vector, velocity, acceleration and speed, and applications to motion in two and three dimensions, with a verified worked example.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

Reviewed by: AI editorial process; not yet individually human-reviewed

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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

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 $t = 2$ .

Velocity. Differentiate each component:

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

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

Speed. The magnitude of the velocity:

|\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}.

Acceleration. Differentiate velocity:

\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 $t = 2$ .

Check the path: $x = t^2$ 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 $x = \frac{y^2}{4}$ at the point $(4, 4)$ has $\frac{dx}{dy} = \frac{y}{2} = 2$ , so $\frac{dy}{dx} = \frac{1}{2}$ , matching the velocity direction.

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 practice questions

Practice questions written in the style of VCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 VCAA1 marksThe 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 VCAA2 marksThe 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.