Vector Valued Functions | Tony's Intertwingled Memex

Core Definition

A vector-valued function (or vector function) is a mathematical mapping from a single real scalar parameter $t \in \mathbb{R}$ (commonly representing time) into a vector space ( $\mathbb{R}^2$ or $\mathbb{R}^3$ ):

\vec{r}(t) = \langle f(t), g(t), h(t) \rangle = f(t)\hat{i} + g(t)\hat{j} + h(t)\hat{k}

where

f(t), g(t), h(t)

are real-valued scalar component functions. As

t

varies over its domain, the terminal point of the position vector

\vec{r}(t)

traces out a continuous space curve in

\mathbb{R}^3

Concept Explanation

Domain of a Vector-Valued Function

The domain of $\vec{r}(t)$ is the set of all real numbers $t$ for which all component functions $f(t), g(t),$ and $h(t)$ are simultaneously defined. Mathematically, the domain is the strict set intersection of the individual component domains:

\text{Domain}(\vec{r}) = \text{Domain}(f) \cap \text{Domain}(g) \cap \text{Domain}(h)

To compute the domain of a vector-valued function:

Identify Algebraic Restrictions for Each Component:
- Even Roots ( $\sqrt[2n]{u(t)}$ ): Radicand must be non-negative: $u(t) \ge 0$ .
- Fractions ( $\frac{n(t)}{d(t)}$ ): Denominator cannot equal zero: $d(t) \ne 0$ .
- Logarithms ( $\ln(u(t))$ ): Argument must be strictly positive: $u(t) > 0$ .
- Roots in Denominators ( $\frac{1}{\sqrt{u(t)}}$ ): Radicand must be strictly positive: $u(t) > 0$ .
Find the Overlapping Interval:
- Compare the inequalities on a 1D real number line.
- The overall domain is the common interval where all conditions hold true.
- If no values of $t$ satisfy all component conditions simultaneously, the domain is the empty set ( $\emptyset$ ), meaning the vector curve does not physically exist in real space.

Space Curves and Parameterization

In single-variable calculus, equations like $y = f(x)$ define planar curves that must satisfy the vertical line test. Vector-valued functions parameterize curves in 3D space, allowing curves to twist, self-intersect, and spiral:

Circular Helix: $\vec{r}(t) = \langle R\cos(t), R\sin(t), ct \rangle$ wraps around a cylinder of radius $R$ while ascending at rate $c$ .
Straight Lines in Space: $\vec{r}(t) = \vec{r}_0 + t\vec{v} = \langle x_0 + at, y_0 + bt, z_0 + ct \rangle$ .

         ^ z
         |          ___...---...___  Helix: r(t) = <cos t, sin t, t>
         |      _.-'      .        `-._
         |    .'         / \           `.
         |   /          /   \            \
         |  |          /  r(t)            |
         |  |         /     \             |
         |   \       .-------+           /
         |    `.    /         \        .'
         |      `-.+-----------+..._.-'
         +---------------------------------> y
        /
       /
      v x

Engineering Applications

Robotics & Kinematics: Defining end-effector trajectory corridors for automated 6-axis welding and pick-and-place robots.
Aerospace Guidance: Modeling ballistic missile paths, satellite orbital transfers, and commercial flight waypoints in GPS-denied environments.
Structural Conduit Design: Routing electrical wiring harnesses, hydraulic lines, and structural tendons through complex civil structures and aircraft fuselages.

Example Problems

Example 1: Finding Domain with Multiple Algebraic Constraints

Determine the domain of the vector-valued function:

\vec{r}(t) = \left\langle \sqrt{t+2}, \frac{-9}{t+3}, \frac{-7}{\sqrt{t-5}} \right\rangle

Solution

Analyze Component 1 ( $x(t) = \sqrt{t+2}$ ): The radicand must be non-negative: $t + 2 \ge 0 \implies t \ge -2 \implies [-2, \infty)$
Analyze Component 2 ( $y(t) = \frac{-9}{t+3}$ ): The denominator cannot equal zero: $t + 3 \ne 0 \implies t \ne -3 \implies (-\infty, -3) \cup (-3, \infty)$
Analyze Component 3 ( $z(t) = \frac{-7}{\sqrt{t-5}}$ ): The radicand is in the denominator, so it must be strictly positive: $t - 5 > 0 \implies t > 5 \implies (5, \infty)$
Intersect the Domains: $\text{Domain} = [-2, \infty) \cap \{t \ne -3\} \cap (5, \infty) = (5, \infty)$ Since any $t > 5$ is automatically greater than $-2$ and never equals $-3$ , the domain is $(5, \infty)$ .

Example 2: Empty Set Domain Classification

Determine the domain of the vector-valued function:

\vec{r}(t) = \left\langle \frac{\sqrt{t+2}}{t-1}, \sqrt{4-t}, \frac{3}{\sqrt{t-6}} \right\rangle

Solution

Component Restrictions:
- $x(t)$ : $t + 2 \ge 0 \implies t \ge -2$ and $t - 1 \ne 0 \implies t \ne 1$ .
- $y(t)$ : $4 - t \ge 0 \implies t \le 4$ .
- $z(t)$ : $t - 6 > 0 \implies t > 6$ .
Evaluate the Intersection: Comparing the condition $t \le 4$ from $y(t)$ with the condition $t > 6$ from $z(t)$ : no real number $t$ can be simultaneously $\le 4$ and $> 6$ .
Conclusion: $\text{Domain}(\vec{r}) = \emptyset \quad (\text{Empty Set})$ The function is nowhere defined in real 3D space.

Connections & References

Parent Concepts: Vectors in 2D and 3D | Equations of Lines in 3D
Related Notes: Derivatives of Vector-Valued Functions | Unit Tangent Vector | Integrals of Vector-Valued Functions
Course Sessions: Calc-3 Session 05 | Calc-3 Session 06