Differentiating and integrating vector functions
Bring calculus to moving objects with differentiating and integrating vector functions for Year 12 Specialist Mathematics in Queensland (QCAA). A moving particle’s position vector is worked one component at a time to find its velocity and acceleration, and integrated to reverse the process.
You will learn to differentiate a position vector into velocity and acceleration, integrate acceleration back to velocity and position using a constant vector of integration fixed by initial conditions, and read off speed and path gradient — the foundation for motion in a plane later in Unit 3.
Theory
Differentiating and integrating vector functions is the vector calculus of motion in Year 12 Specialist Mathematics (QCAA, Queensland). A position vector \(\mathbf{r}(t)\) is differentiated component by component to give the velocity \(\mathbf{v}=\dot{\mathbf{r}}\) and acceleration \(\mathbf{a}=\ddot{\mathbf{r}}\), and integrated component by component (adding a constant vector fixed by an initial condition) to recover velocity and position.
A vector function of time gives a moving particle’s position at each instant, \(\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}\). Differentiating and integrating it works exactly like ordinary calculus, done one component at a time.
The velocity is the derivative of position, \(\mathbf{v}(t)=\dot{\mathbf{r}}=\dfrac{d\mathbf{r}}{dt}\); differentiate each component separately. Its magnitude \(|\mathbf{v}|\) is the speed, and the velocity always points along the tangent to the path in the direction of motion.
The acceleration is the derivative of velocity, \(\mathbf{a}(t)=\dot{\mathbf{v}}=\ddot{\mathbf{r}}=\dfrac{d^2\mathbf{r}}{dt^2}\) — differentiate again, once more component by component.
Reversing the process, integration takes \(\mathbf{a}\) back to \(\mathbf{v}\) and \(\mathbf{v}\) back to \(\mathbf{r}\). Each integration introduces a constant vector of integration \(\mathbf{c}\); use a given initial condition such as \(\mathbf{v}(0)\) or \(\mathbf{r}(0)\) to find it. The path gradient follows from the chain rule, \(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\).
For \(\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}\), differentiate each component to get velocity and acceleration:
Integrate component by component to reverse the process, adding a constant vector \(\mathbf{c}\):
The speed, the displacement over an interval, and the path gradient are:
Recovering velocity and position from acceleration
- Integrate the acceleration component by component to get \(\mathbf{v}=\int\mathbf{a}\,dt+\mathbf{c}\), keeping the constant vector \(\mathbf{c}\).
- Apply the velocity initial condition: substitute the given time into \(\mathbf{v}\) and solve for \(\mathbf{c}\).
- Integrate the now-known velocity to get \(\mathbf{r}=\int\mathbf{v}\,dt+\mathbf{c}\), with a new constant vector.
- Apply the position initial condition (usually \(\mathbf{r}(0)\)) to find this constant, then answer the question — a value at a time, a speed, or a coordinate.
Differentiate each component once for the velocity \(\mathbf{v}=\dot{\mathbf{r}}\):
| \(\mathbf{v}\) | \(=\) | \(\dfrac{d}{dt}(2t^2+t)\mathbf{i}+\dfrac{d}{dt}(t^3)\mathbf{j}\) |
| \(=\) | \((4t+1)\mathbf{i}+3t^2\mathbf{j}\) |
Differentiate again for the acceleration \(\mathbf{a}=\dot{\mathbf{v}}\):
| \(\mathbf{a}\) | \(=\) | \(\dfrac{d}{dt}(4t+1)\mathbf{i}+\dfrac{d}{dt}(3t^2)\mathbf{j}\) |
| \(=\) | \(4\mathbf{i}+6t\mathbf{j}\) |
\(\mathbf{v}=(4t+1)\mathbf{i}+3t^2\mathbf{j}\) and \(\mathbf{a}=4\mathbf{i}+6t\mathbf{j}\).
Differentiate each component for the velocity, then substitute \(t=3\):
| \(\mathbf{v}(t)\) | \(=\) | \(2t\mathbf{i}+8\mathbf{j}\) |
| \(\mathbf{v}(3)\) | \(=\) | \(2\times3\,\mathbf{i}+8\mathbf{j}\) |
| \(=\) | \(6\mathbf{i}+8\mathbf{j}\) |
The speed is the magnitude \(|\mathbf{v}(3)|\):
| \(|\mathbf{v}(3)|\) | \(=\) | \(\sqrt{6^2+8^2}\) |
| \(=\) | \(\sqrt{100}\) | |
| \(=\) | \(10\) |
\(\mathbf{v}(3)=6\mathbf{i}+8\mathbf{j}\), and the speed is \(10\) m/s.
Integrate \(\mathbf{a}\) component by component, keeping the constant vector \(\mathbf{c}\):
| \(\mathbf{v}\) | \(=\) | \(\int\mathbf{a}\,dt\) |
| \(=\) | \(4t\mathbf{i}+3t^2\mathbf{j}+\mathbf{c}\) |
Substitute \(t=0\); since the variable terms vanish, \(\mathbf{v}(0)=\mathbf{c}\):
| \(\mathbf{v}(0)\) | \(=\) | \(\mathbf{c}\) |
| \(\mathbf{c}\) | \(=\) | \(\mathbf{i}-2\mathbf{j}\) |
Put \(\mathbf{c}\) back to complete \(\mathbf{v}(t)\):
| \(\mathbf{v}(t)\) | \(=\) | \((4t+1)\mathbf{i}+(3t^2-2)\mathbf{j}\) |
\(\mathbf{v}(t)=(4t+1)\mathbf{i}+(3t^2-2)\mathbf{j}\).
Integrate \(\mathbf{a}\) for the velocity, then fix \(\mathbf{c}\) from \(\mathbf{v}(0)\):
| \(\mathbf{v}\) | \(=\) | \(3t^2\mathbf{i}+2t\mathbf{j}+\mathbf{c}\) |
| \(\mathbf{v}(0)\) | \(=\) | \(\mathbf{c}=\mathbf{i}+3\mathbf{j}\) |
| \(\mathbf{v}\) | \(=\) | \((3t^2+1)\mathbf{i}+(2t+3)\mathbf{j}\) |
Integrate \(\mathbf{v}\) for the position, then fix the new \(\mathbf{c}\) from \(\mathbf{r}(0)\):
| \(\mathbf{r}\) | \(=\) | \((t^3+t)\mathbf{i}+(t^2+3t)\mathbf{j}+\mathbf{c}\) |
| \(\mathbf{r}(0)\) | \(=\) | \(\mathbf{c}=2\mathbf{i}-\mathbf{j}\) |
| \(\mathbf{r}\) | \(=\) | \((t^3+t+2)\mathbf{i}+(t^2+3t-1)\mathbf{j}\) |
\(\mathbf{r}(t)=(t^3+t+2)\mathbf{i}+(t^2+3t-1)\mathbf{j}\).
Common pitfalls
Frequently asked questions
How do you differentiate a vector function?
Differentiate each component separately with respect to \(t\). For \(\mathbf{r}=x(t)\mathbf{i}+y(t)\mathbf{j}\), the velocity is \(\mathbf{v}=\dot{x}\mathbf{i}+\dot{y}\mathbf{j}\).
How do you find velocity and acceleration from a position vector?
Velocity is the first derivative \(\mathbf{v}=\dot{\mathbf{r}}\) and acceleration is the second derivative \(\mathbf{a}=\ddot{\mathbf{r}}\); differentiate the position componentwise once, then again.
Why do you add a constant vector when integrating?
Integration reverses differentiation, and differentiation loses any constant. In vectors that constant is a whole vector \(\mathbf{c}\), which you recover from a known velocity or position (an initial condition).
How do you find the constant of integration for a vector function?
Substitute the given time into your integrated expression. For example, setting \(t=0\) in \(\mathbf{v}=\int\mathbf{a}\,dt+\mathbf{c}\) makes the variable terms vanish, so \(\mathbf{c}=\mathbf{v}(0)\).
What is the speed of a particle in terms of its velocity?
Speed is the magnitude of the velocity vector, \(|\mathbf{v}|=\sqrt{\dot{x}^2+\dot{y}^2}\). Substitute the time into \(\mathbf{v}\) first, then take the magnitude.
How do you find the gradient of the path from a vector function?
Use the chain rule \(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\): differentiate \(y\) and \(x\) with respect to \(t\) and divide, then substitute the required value of \(t\).