Position vectors as a function of time
Master position vectors as a function of time for Year 12 Specialist Mathematics in Queensland (QCAA). The vector \(\mathbf{r}(t)\) gives the location of a moving particle at every time \(t\), so substituting a time locates the particle and eliminating \(t\) reveals the Cartesian path it traces.
You will learn to find a particle’s initial position and its distance from the origin, describe paths as lines, parabolas, circles and ellipses, and compare two particles to decide whether their paths cross or the particles truly meet — plus the distance between them and their closest approach, the foundation for vector calculus and motion in a plane.
Theory
A position vector as a function of time, written \(\mathbf{r}(t)\), gives the location of a moving particle at every time \(t\). In Year 12 Specialist Mathematics (QCAA, Queensland) you substitute a time to locate the particle, eliminate \(t\) to find the Cartesian path it traces, and compare two particles to decide whether their paths cross or the particles actually meet.
The position vector of a moving particle is a vector function of time \(\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}\). Each component is itself a function of \(t\): \(x(t)\) gives the horizontal coordinate and \(y(t)\) the vertical coordinate at time \(t\).
To locate the particle at a particular time, substitute that value of \(t\) into each component. The initial position is \(\mathbf{r}(0)\), the value at \(t=0\). The particle's distance from the origin at time \(t\) is the magnitude \(|\mathbf{r}(t)|=\sqrt{x(t)^2+y(t)^2}\).
As \(t\) increases the point \((x(t),y(t))\) traces a curve called the path (or trajectory). Eliminating the parameter \(t\) between \(x(t)\) and \(y(t)\) gives the Cartesian equation of that path — often a line, a parabola, a circle or an ellipse.
Two particles \(A\) and \(B\) each have their own \(\mathbf{r}_A(t)\) and \(\mathbf{r}_B(t)\). Their paths cross if the two curves intersect — but each particle may pass through the crossing point at a different time. The particles meet (collide) only when they are at the same position at the same time, that is \(\mathbf{r}_A(t)=\mathbf{r}_B(t)\) for one value of \(t\).
The position of a particle at time \(t\) is the vector function:
Its distance from the origin at time \(t\) is the magnitude:
Two particles meet when they share the same position at the same time \(t\):
The distance between the two particles at time \(t\) is:
How to work with a position vector \(\mathbf{r}(t)\)
- Locate the particle by substituting the given time into each component; use \(t=0\) for the initial position.
- Find the path by eliminating \(t\): solve one component for \(t\) and substitute into the other (or use \(\cos^2 t+\sin^2 t=1\) for trig components) to get the Cartesian equation.
- Compare two particles. For meeting, set \(\mathbf{r}_A(t)=\mathbf{r}_B(t)\) and check a single \(t\) fits both components; for a crossing of paths, intersect the two Cartesian curves.
- Measure separation with \(D(t)=|\mathbf{r}_B-\mathbf{r}_A|\); minimise \(D(t)^2\) for the closest approach.
Substitute \(t=2\) into each component, then take the magnitude:
| \(\mathbf{r}(2)\) | \(=\) | \(3(2)\mathbf{i}+4(2)\mathbf{j}\) |
| \(=\) | \(6\mathbf{i}+8\mathbf{j}\) | |
| \(|\mathbf{r}(2)|\) | \(=\) | \(\sqrt{6^2+8^2}\) |
| \(=\) | \(\sqrt{36+64}\) | |
| \(=\) | \(\sqrt{100}=10\) |
\(\mathbf{r}(2)=6\mathbf{i}+8\mathbf{j}\); the particle is \(10\) m from the origin.
Let \(x=t-1\) and \(y=t^2\); solve the first for \(t\), then substitute:
| \(x\) | \(=\) | \(t-1\) |
| \(t\) | \(=\) | \(x+1\) |
| \(y\) | \(=\) | \(t^2\) |
| \(=\) | \((x+1)^2\) |
\(y=(x+1)^2\) — the path is a parabola.
Meeting needs BOTH components equal at the SAME time \(t\):
| \(\mathbf{i}:\quad 2t-1\) | \(=\) | \(t+1\) |
| \(t\) | \(=\) | \(2\) |
| \(\mathbf{j}:\quad t+3\) | \(=\) | \(2t+1\) |
| \(t\) | \(=\) | \(2\) |
Both components give \(t=2\), so substitute to find the meeting point:
| \(\mathbf{r}_A(2)\) | \(=\) | \((2(2)-1)\mathbf{i}+(2+3)\mathbf{j}\) |
| \(=\) | \(3\mathbf{i}+5\mathbf{j}\) |
Yes; they meet at \(t=2\) s at the point \((3,5)\).
Form \(\overrightarrow{AB}=\mathbf{r}_B-\mathbf{r}_A\), then minimise the squared distance \(D^2\):
| \(\overrightarrow{AB}\) | \(=\) | \((3-t)\mathbf{i}+(t-1)\mathbf{j}\) |
| \(D^2\) | \(=\) | \((3-t)^2+(t-1)^2\) |
| \(=\) | \(2t^2-8t+10\) | |
| \(=\) | \(2(t-2)^2+2\) | |
| \(D^2_{\min}\) | \(=\) | \(2\quad(\text{at } t=2)\) |
| \(D_{\min}\) | \(=\) | \(\sqrt{2}\) |
The closest distance is \(\sqrt{2}\) m, reached at \(t=2\) s.
Common pitfalls
Frequently asked questions
What is a position vector as a function of time?
It is a vector \(\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}\) whose components depend on time \(t\), giving the location of a moving particle at every instant.
How do you find where a particle is at a given time?
Substitute the value of \(t\) into each component of \(\mathbf{r}(t)\). The initial position is \(\mathbf{r}(0)\).
How do you find the Cartesian equation of the path?
Eliminate the parameter \(t\): solve one component for \(t\) and substitute into the other, or use \(\cos^2 t+\sin^2 t=1\) when the components are trigonometric. The result is a relation between \(x\) and \(y\).
What is the difference between paths crossing and particles meeting?
Paths cross when the two curves intersect, which can happen at different times for each particle. Particles meet only when \(\mathbf{r}_A(t)=\mathbf{r}_B(t)\) for a single value of \(t\) — the same place at the same time.
How do you find the distance between two particles?
Compute \(\overrightarrow{AB}=\mathbf{r}_B(t)-\mathbf{r}_A(t)\) and take its magnitude, giving \(D(t)=|\mathbf{r}_B-\mathbf{r}_A|\) as a function of \(t\).
How do you find the closest approach of two particles?
Write the squared distance \(D(t)^2\), then minimise it (complete the square or set the derivative to zero). Take the square root of the minimum value for the closest distance.