Source: MATH 163 Abridged Class Notes, Calculus III (University of Washington)
Tags: vector-valued function, space curve, parametric curve, parametrisation, r(t), helix, intersection of surfaces, limits of vector functions, component-wise limit, Calculus III, multivariable calculus
Difficulty: Intermediate | Prerequisites: Calculus I and II (differentiation, integration, parametric equations), basic vector operations (dot product, cross product, magnitude), equations of lines and planes in 3D.
Up to this point in Calculus III, you have been working with scalar functions of several variables and with vectors as geometric objects. Vector-valued functions are the bridge between those two worlds: they let you describe curves that twist through three-dimensional space using a single parameter, usually time. This is the machinery behind motion in physics, the geometry of curves, and later topics such as line integrals. If you are comfortable writing the equation of a line in 3D and know basic trig identities, you are ready to start here.
A vector-valued function r(t) = ⟨x(t), y(t), z(t)⟩ maps a single parameter t to a point in space, tracing out a curve as t varies. You can parametrise curves by substitution ("brute force") or by recognising identities like sin²t + cos²t = 1. Limits, derivatives, and integrals of vector-valued functions all work component-wise.
Vector-valued function
A function r(t) = ⟨x(t), y(t), z(t)⟩ whose input is a scalar parameter t and whose output is a vector in R² or R³. Think of it as: instead of getting a single number back from a function, you get an arrow (a position in space) for every value of t.
Space curve
The set of points in 3D space traced out by a vector-valued function as the parameter t varies over its domain. In simple terms, it is the path or trajectory that r(t) draws through space.
Parametrisation (parameterisation)
The process of expressing a curve as a vector-valued function of a single parameter. A given curve can have many different parametrisations (for instance, you can traverse the same circle at different speeds or starting points).
Component functions
The individual scalar functions x(t), y(t), z(t) that make up a vector-valued function r(t) = ⟨x(t), y(t), z(t)⟩. Think of them as the x-, y-, and z-coordinates of the curve written separately.
Component-wise limit
The rule that the limit of a vector-valued function equals the vector of the limits of its individual components. In simple terms, you take the limit of each coordinate function on its own, then reassemble the results into a vector.
Recall that a line in 3D can be written as r(t) = r₀ + vt = ⟨x₀ + at, y₀ + bt, z₀ + ct⟩. This is a vector-valued function, but a very restricted one: the components are all linear in t, so the curve is straight.
A general vector-valued function r(t) = ⟨x(t), y(t), z(t)⟩ allows each component to be any function of t. The result is a space curve that can twist and turn.
Example: r(t) = ⟨t, cos(t), sin(t)⟩ traces a helix (spiral). The x-component increases steadily while the y- and z-components oscillate in a circle.
When two surfaces intersect, their intersection is typically a space curve. The goal is to write that curve as r(t).
Method 1, substitution ("brute force")
Pick one variable and set it equal to t. For example, let x = t.
Substitute into each surface equation to solve for the remaining variables in terms of t.
Assemble the results into r(t) = ⟨t, y(t), z(t)⟩.
Example: for x² + y² = 1 and x + y + z = 4, setting x = t gives y = ±√(1 - t²) and z = 4 - t ∓ √(1 - t²).
Method 2, trig identity (works when one surface is a circle or cylinder)
Recognise that x² + y² = 1 matches sin²t + cos²t = 1. So set x = sin(t) and y = cos(t) (or the other way round).
Substitute into the other equation to find z. For x + y + z = 4, this gives z = 4 - sin(t) - cos(t).
Result: r(t) = ⟨sin(t), cos(t), 4 - sin(t) - cos(t)⟩.
This method is cleaner but only works when one of the surface equations has the form of a known identity.
The limit is taken component by component: lim(t→a) r(t) = ⟨lim(t→a) x(t), lim(t→a) y(t), lim(t→a) z(t)⟩.
If any single component's limit does not exist, the limit of the entire vector does not exist.
Indeterminate forms (0/0) in individual components are handled with the same techniques as in Calculus I: L'Hopital's rule, conjugate multiplication, algebraic simplification.
Example: for r(t) = ⟨(eᵗ - 1)/t, (√(1+t) - 1)/t, 3/(1+t)⟩, the limit as t → 0 is ⟨1, 1/2, 3⟩. The x-component uses L'Hopital's rule, the y-component uses either L'Hopital's or the conjugate method, and the z-component is direct substitution.
\vec{r}(t) = \langle x(t),\, y(t),\, z(t) \rangle\vec{r}(t) = \vec{r}_0 + \vec{v}\,t = \langle x_0 + at,\, y_0 + bt,\, z_0 + ct \rangle \quad \text{(line in 3D)}\lim_{t \to a} \vec{r}(t) = \left\langle \lim_{t \to a} x(t),\, \lim_{t \to a} y(t),\, \lim_{t \to a} z(t) \right\rangleVector-valued functions describe the trajectory of any object moving through space: the flight path of an aircraft, the orbit of a satellite, or the motion of a robotic arm. GPS systems reconstruct your position over time as a space curve. Parametrising the intersection of two surfaces is exactly what CAD software does when computing the seam where two 3D shapes meet.
Students often think a vector-valued function outputs a scalar. It does not: r(t) returns a vector (a point in space), even though its input t is a scalar.
Students sometimes believe there is only one correct parametrisation for a given curve. There are infinitely many. The curve is the set of points traced out; different parametrisations traverse those points at different rates or in different directions.
When using the trig identity method, students sometimes set x = sin(t) and y = sin(t), forgetting that the identity requires one sin and one cos (or vice versa).
Students frequently forget that if even one component's limit fails to exist, the entire vector limit does not exist. You cannot report "two out of three components converge" as a partial answer.
⚠️ Parametrising the intersection of two surfaces is a classic exam problem. Know both the substitution method and the trig identity method; the problem will usually hint at which one is cleaner.
⚠️ Limit problems for vector-valued functions often include at least one 0/0 indeterminate component to test whether you remember L'Hopital's rule or conjugate multiplication.
⚠️ The "all components must converge" rule is a favourite true/false or short-answer question.
True or false: r(t) = ⟨t², sin(t)⟩ is a vector-valued function even though it has only two components. True. Vector-valued functions can live in R² or R³.
Fill in the blank: to parametrise the intersection of x² + y² = 4 and z = x + y, a convenient choice using trig is x = ___ and y = ___. x = 2cos(t), y = 2sin(t) (matching x² + y² = 4 with 4cos²t + 4sin²t = 4).
True or false: if lim(t→0) x(t) = 3 and lim(t→0) y(t) does not exist, then lim(t→0) ⟨x(t), y(t)⟩ = ⟨3, DNE⟩. False. The entire vector limit does not exist; you cannot report a partial result.
Fill in the blank: lim(t→0) (eᵗ - 1)/t = ___. 1 (L'Hopital's rule: eᵗ/1 evaluated at t = 0).
Q: Given the surfaces x² + y² = 9 and z = 2x - y, write a vector-valued function for the curve of intersection using the trig identity method.
A: Set x = 3cos(t), y = 3sin(t). Then z = 2(3cos t) - 3sin t = 6cos(t) - 3sin(t). So r(t) = ⟨3cos(t), 3sin(t), 6cos(t) - 3sin(t)⟩.
Q: Evaluate lim(t→0) r(t) for r(t) = ⟨sin(t)/t, (1 - cos t)/t², t³ + 1⟩.
A: x-component: lim sin(t)/t = 1. y-component: lim (1 - cos t)/t² = 1/2 (L'Hopital's twice, or known limit). z-component: lim(t³ + 1) = 1. Result: ⟨1, 1/2, 1⟩.
Q: Why does the "brute force" parametrisation of x² + y² = 1 with x = t produce a ± in the y-component, while the trig method does not?
A: Setting x = t gives y = ±√(1 - t²), which splits the circle into upper and lower halves. The trig parametrisation x = cos(t), y = sin(t) sweeps the full circle continuously without needing a ± sign, because cos and sin together trace both halves as t varies from 0 to 2π.
Q: For r(t) = ⟨ln(t), 1/(t - 2), eᵗ⟩, does lim(t→2) r(t) exist?
A: The y-component 1/(t - 2) diverges as t → 2, so the limit does not exist.
This material connects directly to Section 13.2, where you differentiate and integrate these functions to get velocity vectors, tangent lines, and arc length. The trig parametrisation method reappears heavily in Chapter 16 when you set up line integrals and surface integrals. The idea of component-wise operations (limits here, derivatives and integrals in 13.2) is the single most important pattern in this chapter: once you internalise it, every new operation on vector-valued functions follows the same logic.
vector-valued function, vector function, parametric curve, space curve, r(t), parametrisation, parameterisation, parametric equations 3D, helix, intersection of surfaces, curve of intersection, component-wise limit, limit of vector function, L'Hopital vector, Calculus III, MATH 163, multivariable calculus, Stewart Chapter 13, Section 13.1