Learning Objectives
Interpret a vector-valued function as a parameterized curve.
Find its domain by combining component restrictions.
Evaluate limits and check continuity component by component.
Prerequisites: three-dimensional vectors, single-variable functions, and limits. Trigonometric parameters are in radians. A vector-valued function takes a scalar input and returns a vector.
1. One Parameter Controls Several Coordinates
Write r(t) = ⟨x(t), y(t), z(t)⟩. If the vector is viewed as a position vector from the origin, its endpoint traces a curve as t varies. The functions x(t), y(t), and z(t) are its component functions.
Worked Example: Trace a Parabola
Let r(t) = ⟨t, t², 0⟩ for real t. The coordinates satisfy x = t, y = t², and z = 0.
Eliminating t gives y = x² in the xy-plane. At t = −1 the point is (−1, 1, 0), at t = 0 it is the origin, and at t = 1 it is (1, 1, 0).
As t increases, the curve moves from left to right. The equation y = x² describes the path, while the parameterization also specifies how the path is traversed.
2. The Domain Must Work for Every Component
The domain of a vector-valued function is the intersection of its component domains. A parameter is allowed only when every coordinate is defined. List the restrictions separately before combining them.
Worked Example: Combined Restrictions
Find the domain of r(t) = ⟨√(t + 1), ln(3 − t), 1/(t − 2)⟩.
The square root requires t ≥ −1. The logarithm requires t < 3. The rational component requires t ≠ 2.
The combined domain is [−1, 2) ∪ (2, 3). The endpoint −1 is allowed, but 3 is not.
3. Evaluate Limits by Coordinates
A finite vector limit exists exactly when every component has a finite limit. If x(t) → a, y(t) → b, and z(t) → c as t → t₀, then r(t) → ⟨a, b, c⟩. One component with incompatible one-sided behavior prevents a finite vector limit.
Worked Example: A Removable Component
Evaluate lim(t → 0) ⟨sin t/t, (eᵗ − 1), t²⟩. The first component is undefined at t = 0 but has limit 1.
The other two components approach 0. Therefore the vector limit is ⟨1, 0, 0⟩.
Defining r(0) = ⟨1, 0, 0⟩ makes this function continuous at zero. Defining a different vector there would not change the limit, but would fail continuity.
4. A Familiar Space Curve: The Helix
The curve r(t) = ⟨a cos t, a sin t, bt⟩, with a > 0, lies on the cylinder x² + y² = a². The first two coordinates circle the z-axis while the third changes linearly. Increasing t by 2π completes one revolution and changes the height by 2πb.
If b = 0, the curve is a circle in the xy-plane rather than a rising or falling helix. Restricting the parameter interval changes which portion of the curve is included, even though the formula is unchanged.
Common Mistakes
Take the intersection of component domains, not their union.
A geometric path alone does not specify the direction or timing of traversal.
A vector limit is a vector; do not add its component limits into one scalar.
5. Practice
1. Describe r(t) = ⟨2 cos t, 2 sin t, 5⟩ for 0 ≤ t ≤ 2π.
2. Find the domain of ⟨√(4 − t), ln(t + 2), t⟩.
3. Evaluate lim(t → 1) ⟨t², 3t − 1, (t² − 1)/(t − 1)⟩.
Worked Solutions
1. It traces the circle x² + y² = 4 in the plane z = 5 once, beginning at (2, 0, 5). Viewed from the positive z-axis, the direction is counterclockwise.
2. The restrictions are t ≤ 4 and t > −2. Their intersection is (−2, 4].
3. The first two limits are 1 and 2. For t ≠ 1, the third component simplifies to t + 1 and approaches 2. The limit is ⟨1, 2, 2⟩.
6. Summary
- A vector-valued function uses one parameter to specify several coordinates.
- Its domain requires every component to be defined.
- Finite limits and continuity are checked component by component.
- A parameterization includes traversal information as well as the curve's path.