Introduction to Vector-Valued Functions: Learn It 1

Giselle Miller

Introduction to Vector-Valued Functions: Learn It 1

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

Vector-valued functions combine ideas from single-variable calculus with our understanding of vectors in three dimensions. These functions allow us to describe curves in both 2D and 3D space mathematically, extending our calculus toolkit to handle motion and curves in space.

Definition of a Vector-Valued Function

Instead of functions that output numbers (like [latex]f(t) = t^2[/latex]), vector-valued functions output vectors. Think of them as functions that trace out paths or curves as the parameter [latex]t[/latex] changes.

vector-valued function

A vector-valued function is a function of the form

2D:

[latex]\mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j}[/latex] or [latex]\mathbf{r}(t) = \langle f(t), g(t) \rangle[/latex]

3D:

[latex]\mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k}[/latex] or [latex]\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle[/latex]

where the component functions [latex]f[/latex], [latex]g[/latex], and [latex]h[/latex] are real-valued functions of the parameter [latex]t[/latex].

The parameter [latex]t[/latex] often represents time, but it can represent any real variable
Each value of [latex]t[/latex] produces a specific vector (point in space)
As [latex]t[/latex] varies, the vectors trace out a curve

Domain and Range

Understanding domain and range for vector-valued functions requires thinking about both the input parameter [latex]t[/latex] and the vector outputs.

The domain of a vector-valued function is the set of all allowable values for the parameter [latex]t[/latex]. Just like with regular functions, the domain consists of real numbers, but the restrictions come from all the component functions.

Determining the Domain:

Find the domain of each component function individually
The domain of the vector-valued function is the intersection of all component domains
Exclude any values where at least one component is undefined

Common Domain Scenarios:

No restrictions: When all components are polynomials, the domain is all real numbers
Bounded intervals: [latex]a \leq t \leq b[/latex] when the problem specifies limits or when dealing with specific time intervals
Function-imposed restrictions: When components involve functions like [latex]\tan t[/latex], [latex]\sec t[/latex], [latex]\ln t[/latex], [latex]\sqrt{t}[/latex], or [latex]\frac{1}{f(t)}[/latex]

If any component is undefined at [latex]t = c[/latex], the entire vector function is undefined there.

The range of a vector-valued function is fundamentally different from regular functions. Unlike regular functions that output numbers, vector-valued functions output 2D or 3D vectors.

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

Each value [latex]t[/latex] in the domain produces exactly one vector [latex]\mathbf{r}(t)[/latex]. As [latex]t[/latex] varies over the entire domain, the collection of all resulting vectors forms the range.

The domain contains real numbers, but the range contains vectors. This fundamental difference makes vector-valued functions perfect for describing curves and motion in space.

For each of the following vector-valued functions, evaluate [latex]{\bf{r}}(0), {\bf{r}}(\frac{\pi}{2})[/latex], and [latex]{\bf{r}}(\frac{2{\pi}}{3})[/latex]. Do any of these functions have domain restrictions?

[latex]{\bf{r}}\,(t)=4\cos{t}\,{\bf{i}}+3\sin{t}{\bf{j}}[/latex]
[latex]{\bf{r}}\,(t)=3\tan{t}\,{\bf{i}}+4\sec{t}{\bf{j}}+5t{\bf{k}}[/latex]

To calculate each of the function values, substitute the appropriate value of t into the function:
[latex]\begin{array}{ccc}\hfill {\bf{r}}(0) & =\hfill & 4\cos{(0)}{\bf{i}}+3\sin{(0)}{\bf{j}} \hfill \\ \hfill & =\hfill & 4{\bf{i}}+0{\bf{j}}=4{\bf{i}}\hfill\\\hfill{\bf{r}}\left(\frac{\pi}{2}\right) & =\hfill & 4\cos{\left(\frac{\pi}{2}\right)}{\bf{i}}+3\sin{\left(\frac{\pi}{2}\right)}{\bf{j}}\hfill\\\hfill&=\hfill&0{\bf{i}}+3{\bf{j}}=3{\bf{j}}\hfill\\\hfill{\bf{r}}\left(\frac{2\pi}{3}\right)& =\hfill & 4\cos{\left(\frac{2\pi}{3}\right)}{\bf{i}}+3\sin{\left(\frac{2\pi}{3}\right)}{\bf{j}}\hfill \\ \hfill & =\hfill & 4\left(-\frac{1}{2}\right){\bf{i}}+3\left(\frac{\sqrt{3}}{2}\right){\bf{j}}=-2{\bf{i}}+\frac{3\sqrt{3}}{2}{\bf{j}}\end{array}[/latex].

To determine whether this function has any domain restrictions, consider the component functions separately. The first component function is [latex]f(t)=4\cos{t}[/latex] and the second component function is [latex]g(t)=3\sin{t}[/latex] Neither of these functions has a domain restriction, so the domain of [latex]{\bf{r}}(t)=4\cos{t{\bf{i}}}+3\sin{t\,{\bf{j}}}[/latex] is all real numbers.
To calculate each of the function values, substitute the appropriate value of [latex]t[/latex] into the function:
[latex]\begin{array}{ccc}\hfill {\bf{r}}(0) & =\hfill & 3\tan{(0){\bf{i}}}+4\sec{(0){\bf{j}}}+5\,(0)\,{\bf{k}} \hfill \\ \hfill & =\hfill & 0{\bf{i}}+4{\bf{j}}+0{\bf{k}}=4{\bf{j}}\hfill\\\hfill{\bf{r}}\left(\frac{\pi}{2}\right) & =\hfill & 3\tan{\left(\frac{\pi}{2}\right){\bf{i}}}+4\sec{\left(\frac{\pi}{2}\right){\bf{j}}}+5\,\left(\frac{\pi}{2}\right)\,{\bf{k}},\text{ which does not exist}\hfill\\\hfill{\bf{r}}\left(\frac{2\pi}{3}\right) & =\hfill & 3\tan{\left(\frac{2\pi}{2}\right){\bf{i}}}+4\sec{\left(\frac{2\pi}{2}\right){\bf{j}}}+5\,\left(\frac{2\pi}{2}\right)\,{\bf{k}} \hfill \\ \hfill & =\hfill & 3\,(-\sqrt{3})\,{\bf{i}}+4(-{2})\,{\bf{j}}+\frac{10\pi}{3}\,{\bf{k}} \hfill \\ \hfill & =\hfill & -3\sqrt{3}\,{\bf{i}}-8{\bf{j}}+\frac{10\pi}{3}\,{\bf{k}}.\hfill \\ \hfill \end{array}[/latex]

To determine whether this function has any domain restrictions, consider the component functions separately. The first component function is [latex]f\,(t)=3\tan{t}[/latex], the second component function is [latex]g\,(t)=4\sec{t}[/latex], and the third component function is [latex]h\,(t)=5t[/latex]. The first two functions are not defined for odd multiples of [latex]\frac{\pi}{2}[/latex], so the function is not defined for odd multiples of [latex]\frac{\pi}{2}[/latex]. Therefore, [latex]\text{dom}({\bf{r}}\,(t))=\bigg\{t\,\bigg|\,t\neq\frac{(2n+1){\pi}}{2}\bigg\}[/latex], where [latex]n[/latex] is any integer.

For the vector-valued function [latex]{\bf{r}}\,(t)=(t^{2}-3t)\,{\bf{i}}+(4t+1)\,{\bf{j}}[/latex], evaluate [latex]{\bf{r}}\,(0),\ {\bf{r}}\,(1)[/latex], and [latex]{\bf{r}}\,(-4)[/latex]. Does this function have any domain restrictions?

Geometric Interpretation

Vector-valued functions have a powerful geometric interpretation: they describe parametric curves in space. As the parameter [latex]t[/latex] varies over its domain, the tip of the vector [latex]\mathbf{r}(t)[/latex] traces out a curve.

For example, [latex]\mathbf{r}(t) = 4\cos t \mathbf{i} + 3\sin t \mathbf{j}[/latex] traces out an ellipse in the [latex]xy[/latex]-plane as [latex]t[/latex] goes from [latex]0[/latex] to [latex]2\pi[/latex].

This connection between vector-valued functions and curves in space forms the foundation for studying motion, arc length, curvature, and many other important concepts in vector calculus.