Vector functions
Master vector functions in Year 12 VCE Specialist Mathematics. 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. 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 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 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.