VCE Specialist Mathematics (Year 12) · Kinematics (Calculus and Vectors) · 22m read
Vector Calculus — Kinematics in Two Dimensions
Ever wondered why a football follows a curved path instead of a straight line? That's where vector calculus comes in—it lets you describe motion in two dimensions using mathematical tools that track both direction and speed. You'll learn to use vectors (arrows showing movement) and calculus (tools for finding rates of change) together to model real situations like projectiles, circular motion, and robot paths. This is essential for physics and engineering, and you'll use it constantly in VCE subjects like Physics.
Learning objective: Model 2D motion using vector-valued functions; differentiate and integrate to find velocity, acceleration and displacement.
By the end of this lesson you can:
- I express position, velocity and acceleration as vector functions of time.
- I differentiate vector-valued functions component-wise.
- I integrate vector equations with constant vectors of integration.
- I find speed $|\vec{v}|$ and sketch trajectories.
Vector calculus tracks a drone. Position, velocity and acceleration vectors describe a drone's path in 2D — differentiate or integrate each component.
Position, velocity and acceleration as vectors
In two dimensions, position at time $t$ is:
$$\vec{r}(t) = x(t)\hat{i} + y(t)\hat{j}$$
Velocity is the derivative of position:
$$\vec{v}(t) = \dot{\vec{r}}(t) = \frac{d\vec{r}}{dt} = \dot{x}(t)\hat{i} + \dot{y}(t)\hat{j}$$
Acceleration is the derivative of velocity:
$$\vec{a}(t) = \dot{\vec{v}}(t) = \ddot{x}(t)\hat{i} + \ddot{y}(t)\hat{j}$$
Speed (magnitude of velocity):
$$|\vec{v}(t)| = \sqrt{(\dot{x})^2 + (\dot{y})^2}$$
### Differentiation is component-wise
$$\frac{d}{dt}\bigl[f(t)\hat{i} + g(t)\hat{j}\bigr] = f'(t)\hat{i} + g'(t)\hat{j}$$
### Integration is also component-wise
$$\int \vec{v}(t)\,dt = \left(\int \dot{x}\,dt\right)\hat{i} + \left(\int \dot{y}\,dt\right)\hat{j} + \vec{C}$$
where $\vec{C} = C_1\hat{i} + C_2\hat{j}$ is a vector constant of integration, found from initial conditions.
Vector Kinematics Formulas
- Velocity:
\vec{v}(t) = \dfrac{d\vec{r}}{dt}(Differentiate position) - Acceleration:
\vec{a}(t) = \dfrac{d\vec{v}}{dt} = \dfrac{d^2\vec{r}}{dt^2} - Speed:
|\vec{v}| = \sqrt{\dot{x}^2 + \dot{y}^2}(Magnitude of velocity vector) - Position from velocity:
\vec{r}(t) = \int \vec{v}(t)\,dt + \vec{C}(Use ICs to find $\vec{C}$)
Quick reference
| Quantity | Relationship |
|---|---|
| r(t) | Position vector |
| v(t) = r′(t) | Velocity |
| a(t) = v′(t) | Acceleration |
| Speed | Magnitude of v |
| Cartesian path | Eliminate t |
Key ideas at a glance
- Treat components separately: i and j.
- Speed is a magnitude: Not a vector.
- Use initial conditions: Find constants.
- Eliminate t: For path equation.
Worked example (Medium)
A particle has position $\vec{r}(t) = (t^2 - 3t)\hat{i} + (2t^3 - t)\hat{j}$. Find velocity and acceleration at $t = 2$, and the speed at $t = 2$.
- Differentiate for velocity.
\vec{v}(t) = (2t - 3)\hat{i} + (6t^2 - 1)\hat{j} - Velocity at $t = 2$.
\vec{v}(2) = (4-3)\hat{i} + (24-1)\hat{j} = \hat{i} + 23\hat{j} - Differentiate for acceleration.
\vec{a}(t) = 2\hat{i} + 12t\hat{j} \implies \vec{a}(2) = 2\hat{i} + 24\hat{j} - Speed at $t = 2$.
|\vec{v}(2)| = \sqrt{1^2 + 23^2} = \sqrt{1 + 529} = \sqrt{530}
Answer: $\vec{v}(2) = \hat{i} + 23\hat{j}$; $\vec{a}(2) = 2\hat{i} + 24\hat{j}$; speed $= \sqrt{530} \approx 23.0 \ \text{ms}^{-1}$.
Worked example (Hard)
A particle has position $\vec r(t) = (3t^2)\,\hat i + (4t - t^2)\,\hat j$. Find velocity and speed at $t = 1$.
- Velocity. Differentiate componentwise.
\vec v(t) = 6t\,\hat i + (4 - 2t)\,\hat j - At $t = 1$. Substitute.
\vec v(1) = 6\hat i + 2\hat j - Speed. Magnitude of velocity.
|\vec v(1)| = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}
Answer: $\vec v(1) = 6\hat i + 2\hat j$, speed $= 2\sqrt{10} \approx 6.32$.
Worked example (Hard)
A particle has acceleration $\vec{a}(t) = 2t\hat{i} + 3\hat{j}$. At $t = 0$: $\vec{v}(0) = \hat{i} - \hat{j}$ and $\vec{r}(0) = 3\hat{i} + 2\hat{j}$. Find $\vec{r}(t)$.
- Integrate $\vec{a}$ for $\vec{v}$.
\vec{v}(t) = \int (2t\hat{i} + 3\hat{j})\,dt = t^2\hat{i} + 3t\hat{j} + \vec{C_1} - Use IC $\vec{v}(0) = \hat{i} - \hat{j}$. $\vec{C_1} = \hat{i} - \hat{j}$, so:
\vec{v}(t) = (t^2 + 1)\hat{i} + (3t - 1)\hat{j} - Integrate $\vec{v}$ for $\vec{r}$.
\vec{r}(t) = \left(\frac{t^3}{3} + t\right)\hat{i} + \left(\frac{3t^2}{2} - t\right)\hat{j} + \vec{C_2} - Use IC $\vec{r}(0) = 3\hat{i} + 2\hat{j}$. $\vec{C_2} = 3\hat{i} + 2\hat{j}$, so:
\vec{r}(t) = \left(\frac{t^3}{3} + t + 3\right)\hat{i} + \left(\frac{3t^2}{2} - t + 2\right)\hat{j}
Answer: $\vec{r}(t) = \bigl(\frac{t^3}{3} + t + 3\bigr)\hat{i} + \bigl(\frac{3t^2}{2} - t + 2\bigr)\hat{j}$.
Common mistakes
- Forgetting the vector constant of integration when integrating vector equations When integrating $\vec{v}(t)$ to find $\vec{r}(t)$, the constant is a vector $\vec{C} = C_1\hat{i} + C_2\hat{j}$, not a scalar. Each component gets its own constant determined by initial conditions. Fix: Always write $+ \vec{C}$ (not $+ C$) after integrating, then use the initial condition vector (e.g. $\vec{r}(0)$) to find both components of $\vec{C}$ simultaneously.
- Confusing speed with velocity Velocity $\vec{v}(t)$ is a vector with direction and magnitude. Speed is the scalar $|\vec{v}(t)| = \sqrt{\dot{x}^2 + \dot{y}^2}$. Writing speed as negative or as a vector loses marks. Fix: Speed is always non-negative: compute $|\vec{v}| = \sqrt{\dot{x}^2 + \dot{y}^2}$. State it as a scalar with appropriate units.
From vector kinematics to the TNB frame
In university dynamics and differential geometry, 2D vector kinematics extends to the Frenet–Serret (TNB) frame, which describes motion along any smooth curve using intrinsic coordinates rather than Cartesian ones.
At each point on a curve, three mutually perpendicular unit vectors are defined: the tangent $\hat{T}$, the normal $\hat{N}$, and the binormal $\hat{B} = \hat{T} \times \hat{N}$.
The velocity vector always points along $\hat{T}$: $\vec{v} = v\hat{T}$ where $v = |\vec{v}|$ is the speed. The acceleration decomposes into tangential and normal components:
$$\vec{a} = \frac{dv}{dt}\hat{T} + \frac{v^2}{\rho}\hat{N}$$
Here $\rho$ is the radius of curvature — the radius of the "best-fitting circle" at that point. The tangential component $dv/dt$ describes speeding up or slowing down, while the normal component $v^2/\rho$ describes the centripetal acceleration needed for the path to curve. For uniform circular motion of radius $R$, this gives the familiar $a = v^2/R$.
This framework is essential in aerospace engineering (flight path analysis), robotics (trajectory planning), and general relativity (geodesic motion).
Practice questions
$\vec{r}(t) = 3\cos t\,\hat{i} + 3\sin t\,\hat{j}$. What curve does the particle trace?
(2 marks)
A particle has $\vec{a}(t) = -9.8\hat{j}$ (gravity only). At $t=0$: $\vec{v}(0) = 20\hat{i} + 15\hat{j}$ m/s, $\vec{r}(0) = \vec{0}$. Find $\vec{r}(t)$ and the time and range when the projectile lands.
(4 marks)
For $\vec r(t) = (\sin t)\,\hat i + (\cos t)\,\hat j$, find $\vec v(t)$ and show that $|\vec v(t)| = 1$.
(3 marks)
Summary
- $\vec{v} = \dot{\vec{r}}$; $\vec{a} = \dot{\vec{v}}$; speed $= |\vec{v}|$.
- Differentiate and integrate component-wise.
- Integration: vector constant $\vec{C}$, determined by initial conditions.
- Projectile: $\vec{a} = -g\hat{j}$; integrate twice with ICs.