Differentiating and integrating vector functions
Bring calculus to moving objects with differentiating and integrating vector functions in Year 12 VCE Specialist Mathematics. 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. It sits in the Space and measurement area of study of the VCE Mathematics Study Design (VCAA), within the Vector calculus topic of Unit 4.
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.
Theory
Differentiating and integrating vector functions is the vector calculus of motion in Year 12 Specialist Mathematics. 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 \(t\).