Vector functions
Master vector functions for Year 12 Specialist Mathematics in Queensland (QCAA). A vector function \(\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}\) gives the position of a moving point as a function of the parameter \(t\); as \(t\) runs over its domain, the point sweeps out a path in the plane.
You will learn to evaluate a position vector \(\mathbf{r}(t_0)\), read off the component functions, and find the Cartesian equation of the path by eliminating \(t\) — recognising lines, parabolas, circles, ellipses and hyperbolas — the foundation for velocity, acceleration and motion later in vector calculus.
Theory
A vector function \(\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}\) gives the position of a point as a function of the parameter \(t\). In Year 12 Specialist Mathematics (QCAA, Queensland) each value of \(t\) gives a position vector, and as \(t\) runs over its domain the point sweeps out a path. Eliminating \(t\) between the components gives the Cartesian equation of that path.
A vector function (or vector-valued function) assigns a position vector to each value of a scalar parameter \(t\), usually time. In two dimensions it is written \(\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}\), where \(x(t)\) and \(y(t)\) are the component functions.
Substituting a value \(t=t_0\) gives a single position vector \(\mathbf{r}(t_0)\), which locates one point. As \(t\) ranges over its domain, the tip of \(\mathbf{r}(t)\) moves and traces a curve called the path.
The Cartesian equation of the path relates \(x\) and \(y\) directly, with no \(t\). It is found by eliminating the parameter: make \(t\) the subject of one component and substitute into the other, or, for trigonometric components, use an identity such as \(\cos^2 t+\sin^2 t=1\).
The form of the components tells you the curve: linear components give a line; one squared component gives a parabola; \(a\cos t,\ a\sin t\) give a circle; \(a\cos t,\ b\sin t\) (unequal) give an ellipse; and \(a\sec t,\ b\tan t\) give a hyperbola.
A vector function in two dimensions gives position as a function of the parameter \(t\):
The position vector at \(t=t_0\) is found by substitution:
For trigonometric components, isolate \(\cos t,\sin t\) (or \(\sec t,\tan t\)) and apply a Pythagorean identity to eliminate \(t\):
Standard Cartesian paths that arise (centre \((h,k)\)):
How to find the Cartesian equation of a path
- Write the components as \(x=x(t)\) and \(y=y(t)\) by reading off \(\mathbf{r}(t)\).
- Isolate the parameter. For polynomial components, make \(t\) the subject of the simpler equation; for trig components, isolate \(\cos t,\sin t\) (or \(\sec t,\tan t\)).
- Eliminate \(t\). Substitute into the other component, or apply the identity \(\cos^2 t+\sin^2 t=1\) (or \(\sec^2 t-\tan^2 t=1\)).
- Simplify and name the curve — a line, parabola, circle, ellipse or hyperbola — and state its domain if \(t\) is restricted.
Substitute \(t=3\) into each component in turn:
| \(x(3)\) | \(=\) | \((3)^2+1\) |
| \(=\) | \(10\) | |
| \(y(3)\) | \(=\) | \(2(3)-3\) |
| \(=\) | \(3\) | |
| \(\mathbf{r}(3)\) | \(=\) | \(10\mathbf{i}+3\mathbf{j}\) |
\(\mathbf{r}(3)=10\mathbf{i}+3\mathbf{j}\), i.e. the point \((10,3)\).
Write the components, make \(t\) the subject of the \(x\)-equation, then substitute into \(y\):
| \(x\) | \(=\) | \(2t+1\) |
| \(t\) | \(=\) | \(\dfrac{x-1}{2}\) |
| \(y\) | \(=\) | \(t-4\) |
| \(=\) | \(\dfrac{x-1}{2}-4\) | |
| \(=\) | \(\dfrac{x-9}{2}\) |
\(y=\dfrac{x-9}{2}\), a straight line.
Isolate \(\cos t\) and \(\sin t\), then use \(\cos^2 t+\sin^2 t=1\):
| \(\cos t\) | \(=\) | \(\dfrac{x}{4}\) |
| \(\sin t\) | \(=\) | \(\dfrac{y}{4}\) |
| \(\cos^2 t+\sin^2 t\) | \(=\) | \(1\) |
| \(\dfrac{x^2}{16}+\dfrac{y^2}{16}\) | \(=\) | \(1\) |
| \(x^2+y^2\) | \(=\) | \(16\) |
\(x^2+y^2=16\), a circle of radius \(4\) centred at the origin.
Isolate \(\cos t,\sin t\) and apply \(\cos^2 t+\sin^2 t=1\); unequal amplitudes give an ellipse:
| \(\cos t\) | \(=\) | \(\dfrac{x}{6}\) |
| \(\sin t\) | \(=\) | \(\dfrac{y}{4}\) |
| \(\cos^2 t+\sin^2 t\) | \(=\) | \(1\) |
| \(\dfrac{x^2}{36}+\dfrac{y^2}{16}\) | \(=\) | \(1\) |
\(\dfrac{x^2}{36}+\dfrac{y^2}{16}=1\), an ellipse with semi-axes \(6\) and \(4\).
Common pitfalls
Frequently asked questions
What is a vector function?
A vector function \(\mathbf{r}(t)=x(t)\mathbf{i}+y(t)\mathbf{j}\) gives a position vector for each value of the parameter \(t\). As \(t\) varies, the tip of \(\mathbf{r}(t)\) traces a path in the plane.
How do you find the position vector at a given value of t?
Substitute the value into each component. For \(\mathbf{r}(t)=(t^2+1)\mathbf{i}+(2t-3)\mathbf{j}\), \(\mathbf{r}(3)=10\mathbf{i}+3\mathbf{j}\), the point \((10,3)\).
How do you find the Cartesian equation of a path?
Eliminate the parameter \(t\). Make \(t\) the subject of one component and substitute into the other, or for trigonometric components isolate \(\cos t,\sin t\) and use \(\cos^2 t+\sin^2 t=1\).
How can you tell whether a path is a circle or an ellipse?
Compare the amplitudes. Equal amplitudes \(a\cos t,\ a\sin t\) give a circle of radius \(a\); unequal amplitudes \(a\cos t,\ b\sin t\) give an ellipse with semi-axes \(a\) and \(b\).
Which identity gives a hyperbola?
Components \(a\sec t,\ b\tan t\) use \(\sec^2 t-\tan^2 t=1\), which gives the hyperbola \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\).
What is the domain of a vector function?
It is the set of values the parameter \(t\) may take. Restricting the domain, such as \(0\le t\le 4\), traces only part of the curve, so the \(x\)- and \(y\)-values are limited to a matching range.