Introduction to Vector-Valued Functions: Fresh Take

  • Write and understand vector-valued functions using components and unit vectors
  • Recognize parametric equations for a space curve
  • Define the limit of a vector-valued function

Vector-Valued Functions

The Main Idea 

Vector-valued functions output vectors instead of numbers, allowing us to describe curves and motion in 2D and 3D space.

A vector-valued function takes parameter [latex]t[/latex] and produces a vector:

2D: [latex]\mathbf{r}(t) = \langle f(t), g(t) \rangle[/latex]

3D: [latex]\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle[/latex]

The functions [latex]f[/latex], [latex]g[/latex], and [latex]h[/latex] are component functions. Each [latex]t[/latex]-value produces one vector (point in space). As [latex]t[/latex] varies, the vectors trace out a curve.

Domain

The domain is the intersection of all component function domains. Exclude values where any component is undefined.

Process:

  1. Find each component’s domain individually
  2. Take the intersection of these domains
  3. Exclude values where any component is undefined

Common scenarios: Polynomials (all real numbers), bounded intervals ([latex]a \leq t \leq b[/latex]), or restrictions from [latex]\tan t[/latex], [latex]\ln t[/latex], [latex]\sqrt{t}[/latex], or [latex]\frac{1}{f(t)}[/latex].

Range

Vector-valued functions output vectors, not numbers:

  • 2D functions: Range is a subset of [latex]\mathbb{R}^2[/latex]
  • 3D functions: Range is a subset of [latex]\mathbb{R}^3[/latex]

Geometric Interpretation

Vector-valued functions describe parametric curves. As [latex]t[/latex] varies, the tip of [latex]\mathbf{r}(t)[/latex] traces out a curve in space.

Example: [latex]\mathbf{r}(t) = 4\cos t \mathbf{i} + 3\sin t \mathbf{j}[/latex] traces an ellipse as [latex]t[/latex] goes from [latex]0[/latex] to [latex]2\pi[/latex].

Graphing Vector-Valued Functions

The Main Idea 

When you graph a vector-valued function, you’re essentially plotting a curve by tracking where vectors point as the parameter [latex]t[/latex] changes. The key difference from regular parametric curves? You’re visualizing vectors in standard position (starting at the origin) rather than just points.

Think of it this way: as [latex]t[/latex] increases, each vector [latex]{\bf{r}}(t)[/latex] points from the origin to a specific location in space. The terminal points (where these vectors end) trace out your curve. The curve’s shape depends on your function’s form:

  • Plane curve: [latex]{\bf{r}}(t) = f(t){\bf{i}} + g(t){\bf{j}}[/latex] creates a curve in the [latex]xy[/latex]-plane
  • Space curve: [latex]{\bf{r}}(t) = f(t){\bf{i}} + g(t){\bf{j}} + h(t){\bf{k}}[/latex] creates a curve in three-dimensional space

Graphing Strategy

  1. Make a table: Pick several [latex]t[/latex]-values and calculate [latex]{\bf{r}}(t)[/latex] for each
  2. Draw vectors: Place each vector with its tail at the origin
  3. Mark endpoints: Note where each vector’s tip lands
  4. Connect smoothly: Draw a curve through these terminal points in order
  5. Show direction: Add arrows indicating how the curve progresses as [latex]t[/latex] increases

Vector-valued functions and parametric equations are two ways of describing the same curve. Given [latex]{\bf{r}}(t) = f(t){\bf{i}} + g(t){\bf{j}}[/latex], you can write [latex]x = f(t)[/latex] and [latex]y = g(t)[/latex] to get the parametric form—the graphs match exactly.

Reparameterization means different functions can trace the same curve. For example, [latex]{\bf{r}}(t) = 4\cos t{\bf{i}} + 3\sin t{\bf{j}}[/latex] and [latex]{\bf{r}}(t) = 4\cos t^3{\bf{i}} + 3\sin t^3{\bf{j}}[/latex] both produce ellipses—the shape stays the same even though the parameter changes. Only the speed at which you trace the curve differs.

Special Case: Helices

A helix is a three-dimensional spiral that looks like a spring. It has a circular or elliptical cross-section when viewed from above and climbs (or descends) as [latex]t[/latex] increases. If you remove the [latex]{\bf{k}[/latex] component, you’re left with the curve’s projection onto the [latex]xy[/latex]-plane.

Create a graph of the vector valued function [latex]{\bf{r}}\,(t)=(t^{2}-1)\,{\bf{i}}+(2t-3)\,{\bf{j}},\ 0\ \leq\ t\ \leq\ 3.[/latex]

This figure is a graph of the function r(t) = (t^2-1)i + (2t-3)j, for the values of t from 0 to 3. The curve begins in the 3rd quadrant at the ordered pair (-1,-3) and increases up through the 1st quadrant. It is increasing and has arrows on the curve representing orientation to the right.
Figure 4. Graph of [latex]{\bf{r}}\,(t)=(t^{2}-1)\,{\bf{i}}+(2t-3)\,{\bf{j}},\ 0\ \leq\ t\ \leq\ 3[/latex].
Watch the following video to see the worked solution to the above example.

Limits and Continuity of a Vector-Valued Function

The Main Idea 

Finding limits and checking continuity for vector-valued functions works exactly like it does for regular functions—you just handle each component separately.

Instead of wrestling with the formal definition involving [latex]|{\bf{r}}(t) - {\bf{L}}|[/latex], use this practical method: take the limit of each component function individually.

For plane curves [latex]{\bf{r}}(t) = f(t){\bf{i}} + g(t){\bf{j}}[/latex]:

[latex]\underset{t\to a}{\text{lim}}{\bf{r}}(t)=\bigg[\underset{t\to a}{\text{lim}}f(t)\bigg]{\bf{i}}+\bigg[\underset{t\to a}{\text{lim}}g(t)\bigg]{\bf{j}}[/latex]

For space curves [latex]{\bf{r}}(t) = f(t){\bf{i}} + g(t){\bf{j}} + h(t){\bf{k}}[/latex]:

[latex]\underset{t\to a}{\text{lim}}{\bf{r}}(t)=\bigg[\underset{t\to a}{\text{lim}}f(t)\bigg]{\bf{i}}+\bigg[\underset{t\to a}{\text{lim}}g(t)\bigg]{\bf{j}}+\bigg[\underset{t\to a}{\text{lim}}h(t)\bigg]{\bf{k}}[/latex]

This only works if all the component limits exist.

How to Calculate Limits

Break the vector function into its parts ([latex]{\bf{i}}[/latex], [latex]{\bf{j}}[/latex], and [latex]{\bf{k}}[/latex] components), find the limit of each scalar function using your usual techniques (substitution, algebraic manipulation, etc.), then reassemble the results into vector form. Each component follows the same limit rules you already know.

Checking Continuity

A vector-valued function [latex]{\bf{r}}(t)[/latex] is continuous at [latex]t = a[/latex] when these three familiar conditions hold:

  1. [latex]{\bf{r}}(a)[/latex] exists (the function is defined at that point)
  2. [latex]\displaystyle\lim_{t\to a}{\bf{r}}(t)[/latex] exists (the limit exists as you approach the point)
  3. [latex]\displaystyle\lim_{t\to a}{\bf{r}}(t) = {\bf{r}}(a)[/latex] (the limit equals the function value)

These are the same three requirements you learned for regular functions. A vector-valued function is continuous at a point when its graph flows smoothly through that point without breaks, jumps, or gaps—you could trace it without lifting your pencil.

Calculate [latex]\displaystyle\lim_{t{\to}-2}{\bf{r}}\,(t)[/latex] for the function [latex]{\bf{r}}\,(t)=\sqrt{t^{2}-3t-1}\,{\bf{i}}+(4t+3)\,{\bf{j}}+\sin{\frac{(t+1)\,\pi}{2}}\,{\bf{k}}[/latex].

Watch the following video to see the worked solution to the above Try It