From a Number to a Point in Space
A real-valued function assigns a number to each input in its domain. A vector-valued function can instead assign a vector to each input. In this tutorial, the input is a real parameter \(t\), and each output is a vector with three coordinates.
As \(t\) varies over its allowed values, the output vectors locate points in three-dimensional space. The collection of those points is the curve traced by the function. This gives a way to describe a space curve using one changing input and three coordinate functions.
The parameter is a label used to select a point. It may represent time in a particular application, but it does not have to. For now, the key question is: for each allowed value of \(t\), what point does the function produce?
Definition: A Vector-Valued Function in Three Dimensions
The output \(\mathbf r(t)\) is a vector. Its three coordinates specify a point in space: at parameter value \(t\), the point has coordinates \((x(t),y(t),z(t))\). The symbols \(x\), \(y\), and \(z\) name the component functions; they do not mean that the coordinates are necessarily independent variables.
The set of points produced as the parameter varies is called the trace of the function on \(D\). In set notation, it is
This definition describes exactly which points belong to the trace: a point belongs if there is at least one allowed parameter value that produces it. It does not, by itself, tell us how to sketch the curve or how its points are ordered. Those questions require additional attention to the parameter and the function.
How the Trace Is Determined
The link between the function and the curve is direct. First choose an allowed value of \(t\). Then evaluate each of the three component functions at that same value. Finally, combine the three results into one vector, which gives a point in space.
Use a value \(t\in D\), so that the function is defined.
Compute \(x(t)\), \(y(t)\), and \(z(t)\) using that same value of \(t\).
The output vector \(\langle x(t),y(t),z(t)\rangle\) locates a point in space.
Doing this for every \(t\in D\) gives the trace \(\{\mathbf r(t):t\in D\}\).
Why does this procedure give the trace? By definition, the trace consists of all outputs of \(\mathbf r\) for inputs in \(D\). For any chosen \(t\in D\), evaluating the components gives precisely the output \(\mathbf r(t)\), so that point is in the trace. Conversely, every point in the trace is an output \(\mathbf r(t)\) for some \(t\in D\), and its coordinates must therefore be the three component values at that \(t\). Thus the component evaluation produces all and only the points in the trace.
Worked Examples
Worked Example: Evaluating a Polynomial Vector Function
Let \(\mathbf r(t)=\langle t,t^2,1-t\rangle\) for \(t\in[-1,1]\). Find the point produced by \(t=-1\), \(t=0\), and \(t=1\).
At \(t=-1\), evaluate the components separately: \(x(-1)=-1\), \(y(-1)=(-1)^2=1\), and \(z(-1)=1-(-1)=2\). Therefore, \(\mathbf r(-1)=\langle-1,1,2\rangle\), corresponding to the point \((-1,1,2)\).
At \(t=0\), the component values are \(x(0)=0\), \(y(0)=0^2=0\), and \(z(0)=1-0=1\). Hence \(\mathbf r(0)=\langle0,0,1\rangle\), corresponding to \((0,0,1)\).
At \(t=1\), they are \(x(1)=1\), \(y(1)=1^2=1\), and \(z(1)=1-1=0\). Hence \(\mathbf r(1)=\langle1,1,0\rangle\), corresponding to \((1,1,0)\).
Check: Substituting each parameter into the original function reproduces each reported vector: \(\mathbf r(-1)=\langle-1,(-1)^2,1-(-1)\rangle=\langle-1,1,2\rangle\); \(\mathbf r(0)=\langle0,0^2,1-0\rangle=\langle0,0,1\rangle\); and \(\mathbf r(1)=\langle1,1^2,1-1\rangle=\langle1,1,0\rangle\). All three parameter values are in the stated domain.
Worked Example: Keeping the Components Together
Let \(\mathbf r(t)=\langle 2t+1,3-t,t^2\rangle\) for \(t\in[0,2]\). Find the output at \(t=0\), \(t=1\), and \(t=2\).
For \(t=0\), the components are \(2(0)+1=1\), \(3-0=3\), and \(0^2=0\). Thus \(\mathbf r(0)=\langle1,3,0\rangle\).
For \(t=1\), they are \(2(1)+1=3\), \(3-1=2\), and \(1^2=1\). Thus \(\mathbf r(1)=\langle3,2,1\rangle\).
For \(t=2\), they are \(2(2)+1=5\), \(3-2=1\), and \(2^2=4\). Thus \(\mathbf r(2)=\langle5,1,4\rangle\).
Check: In each case, using the same parameter in all components gives the stated vector: \(\mathbf r(0)=\langle1,3,0\rangle\), \(\mathbf r(1)=\langle3,2,1\rangle\), and \(\mathbf r(2)=\langle5,1,4\rangle\). The chosen values \(0,1,2\) all belong to \([0,2]\).
The check also highlights a basic requirement: one parameter value determines all three coordinates of a single point. Mixing component values from different parameter values would not evaluate \(\mathbf r\) at any one input.
Worked Example: A Parameter Value Produces a Point in Space
Let \(\mathbf r(t)=\langle t^2+1,2t,t-2\rangle\) for \(t\in[-2,2]\). Find the point produced by \(t=-1\) and \(t=2\).
For \(t=-1\), the first coordinate is \((-1)^2+1=1+1=2\); the second is \(2(-1)=-2\); and the third is \(-1-2=-3\). Therefore, \(\mathbf r(-1)=\langle2,-2,-3\rangle\), giving the point \((2,-2,-3)\).
For \(t=2\), the first coordinate is \(2^2+1=4+1=5\); the second is \(2(2)=4\); and the third is \(2-2=0\). Therefore, \(\mathbf r(2)=\langle5,4,0\rangle\), giving the point \((5,4,0)\).
Check: Direct substitution gives \(\mathbf r(-1)=\langle(-1)^2+1,2(-1),-1-2\rangle=\langle2,-2,-3\rangle\) and \(\mathbf r(2)=\langle2^2+1,2(2),2-2\rangle=\langle5,4,0\rangle\). Both parameter values lie in the specified domain.
Reading the Function and the Curve Correctly
A vector-valued function and its trace are closely related, but they are not the same kind of object. The function is a rule that takes an input \(t\) and returns a vector. The trace is a set of points in space formed from all those output vectors.
| Object | What it represents | Example notation |
|---|---|---|
| Parameter value | An allowed real input | \(t=1\) |
| Function value | The vector returned for that input | \(\mathbf r(1)=\langle3,2,1\rangle\) |
| Point in space | The point whose coordinates are the vector's entries | \((3,2,1)\) |
| Trace | The set of all points produced from the parameter set | \(\{\mathbf r(t):t\in D\}\) |
The parameter set matters because it specifies which inputs are used to produce the curve. A formula alone does not fully specify the trace if its allowed inputs are left unclear. In these examples, each domain was supplied as part of the definition, and calculations used only values in that domain.
Do not assume that different parameter values must produce different points. A function is allowed to take the same vector value at more than one input. The trace is still defined as the set of output points: a point appears in that set if the function produces it for at least one allowed parameter value.
A Common Mistake: Mixing Parameter Values
Suppose \(\mathbf r(t)=\langle2t+1,3-t,t^2\rangle\), and we want \(\mathbf r(1)\). A common error is to use \(t=1\) in the first two components but accidentally use \(t=2\) in the third, writing \(\mathbf r(1)=\langle3,2,4\rangle\). This is incorrect: the third coordinate \(4\) comes from \(2^2\), not \(1^2\).
The correction is to substitute \(t=1\) into every component: \(2(1)+1=3\), \(3-1=2\), and \(1^2=1\). Therefore, \(\mathbf r(1)=\langle3,2,1\rangle\), not \(\langle3,2,4\rangle\).
To check the wrong answer, ask whether one input produces all its coordinates. At \(t=1\), the third component is \(1\); at \(t=2\), the first two components are \(5\) and \(1\), not \(3\) and \(2\). The proposed vector has combined results from different inputs, so it is not the value at \(t=1\).
Why This Representation Is Useful
A single parameter provides a compact way to describe a point moving through three coordinates together. Instead of treating a space curve as a collection of unrelated points, the vector-valued function gives a rule for generating its points. This is useful in geometry and in applications where a position changes according to one variable, such as time.
The essential interpretation is modest but powerful: the parameter is an input, the three components are coordinate functions, and their combined output is a point in space. Allowing the parameter to range over its specified set generates the trace.
Check Your Understanding
Answer each question by using the definition of a vector-valued function and its trace.
- If \(\mathbf r(t)=\langle t,2t,t+1\rangle\), what point does the function produce at \(t=2\)?
- In \(\mathbf r(t)=\langle x(t),y(t),z(t)\rangle\), what does each component value contribute to the output?
- How is the trace of \(\mathbf r\) on a parameter set \(D\) written in set notation?
- Why must the same parameter value be substituted into all three component functions when finding \(\mathbf r(t)\)?
- Can two different allowed parameter values produce the same point? Explain using the definition of the trace.