Vector-valued functions

MATH 261 — Multivariable Calculus

Vector-Valued Functions & Parametric Derivatives

Date: February 1, 2026
Instructor: Dr. V. S. Oussa
Prereqs: Derivatives (Calc I), parametric curves basics, vectors in $\mathbb{R}^2$ and $\mathbb{R}^3$
Theme: Motion as a map $t \mapsto \mathbf r(t)$; calculus by differentiating components
Focus: Domain, geometric interpretation, tangent vectors/lines, concavity for parametric curves
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Part 1 — Vector-Valued Functions

Idea: A particle moving in space can be described by a function $t \mapsto \mathbf r(t) = \langle x(t),y(t),z(t)\rangle$. The curve is the set of terminal points.

I. Introduction to Vector-Valued Functions

A. Understanding the concept of vector-valued functions (inputs are scalars, outputs are vectors).
B. Domain and range: the domain is the set of times (or parameters) where every component makes sense; the range is a subset of $\mathbb{R}^2$ or $\mathbb{R}^3$.
C. Components and inputs: $\mathbf r(t)$ is a compact encoding of three scalar functions.

Key definition

If $\mathbf r(t)=\langle f(t),g(t),h(t)\rangle$, then $\mathrm{Dom}(\mathbf r)=\mathrm{Dom}(f)\cap\mathrm{Dom}(g)\cap\mathrm{Dom}(h)$.

Learning target
  • Compute domains by constraints → intersection.
  • Recognize lines and helices from $\mathbf r(t)$.
  • Interpret $\mathbf r'(t)$ as a tangent (velocity) vector.

Case Study — Domain by Intersection

Example B (syllabus outline): $\mathbf R(t)=\langle \cos t,\; \ln(4-t),\; \sqrt{t+1}\rangle$.
Constraints: $4-t>0$ and $t+1\ge 0$. Intersection gives $t\in(-1,4)$.

In the Interactive Lab, you can change the constants and see the constraint intervals update.

Case Study — A Line as a Vector-Valued Function

Example: $\mathbf R(t)=\langle 1+t,\;2+5t,\;-1+6t\rangle$. Write $\mathbf R(t)=\mathbf P + t\mathbf V$ with point $\mathbf P=\langle 1,2,-1\rangle$ and direction $\mathbf V=\langle 1,5,6\rangle$.

A line in $\mathbb{R}^3$ is determined by a point and a direction vector. The parameter $t$ tells “how far” you travel along the direction.

Case Study — Drawing the Helix

Example: $\mathbf R(t)=\langle \cos t,\;\sin t,\;t\rangle$. Because $\cos^2 t + \sin^2 t=1$, the curve lies on the cylinder $x^2+y^2=1$ and rises linearly in $z$.

In the Lab, rotate a 3D view and watch the moving point sweep the helix as $t$ changes.

Part 2 — Derivatives of Parametric Curves

We consider a planar motion $\langle x(t),y(t)\rangle$. Two key questions: slope of the tangent line (first derivative) and concavity (second derivative).

First derivative for parametric curves

$\displaystyle \frac{dy}{dx} = \frac{\tfrac{dy}{dt}}{\tfrac{dx}{dt}}$, provided $dx/dt\neq 0$.

Second derivative

$\displaystyle \frac{d^2y}{dx^2} = \frac{\tfrac{d}{dt}(dy/dx)}{\tfrac{dx}{dt}}$, provided $dx/dt\neq 0$.

Computing $dy/dx$ and $d^2y/dx^2$

In the Lab you will compute derivatives, tangent lines at chosen parameter values, and interpret concavity.

Horizontal and Vertical Tangents

  • Horizontal tangent: $dy/dt=0$ and $dx/dt\neq 0$.
  • Vertical tangent: $dx/dt=0$ and $dy/dt\neq 0$.

Tangent Lines and Eliminating the Parameter

Sometimes you can eliminate $t$ to obtain an explicit graph $y=y(x)$. Other times it is faster to compute $dy/dx$ directly from $t$.

Lecture Transcript (structured)

How to use: Use this transcript to follow the lecture’s narrative. When you see a key idea, jump to the Lab tools to experiment and verify (Domain Builder, Line Builder, Helix Explorer, and Parametric Derivatives).

Lecture Videos

To embed your lecture videos here, add shortcode attributes, e.g. [math261_vector_valued video1_url="https://…" video2_url="https://…" video3_url="https://…"]. You may use YouTube/Vimeo links (oEmbed) or direct .mp4 links.
Motion → vector-valued function
We model a moving particle in space by a function $t\mapsto \mathbf r(t)$; at each time $t$, the output is the position vector from the origin to the particle. The curve is the collection of terminal points.
Domain as “where every component makes sense”
To find the domain of $\mathbf r(t)$, impose the domain constraints of each component and intersect them. For example: logarithms require positive inputs; square roots require nonnegative inputs; rational expressions require nonzero denominators (often strict inequalities).
Derivative: limit componentwise
Define $\mathbf r'(t)=\lim_{h\to 0}\frac{\mathbf r(t+h)-\mathbf r(t)}{h}$. The limit is taken componentwise, so $\mathbf r'(t)=\langle x'(t),y'(t),z'(t)\rangle$.
Tangent vector / tangent line example (helix)
For $\mathbf r(t)=\langle \cos t,\sin t,t\rangle$, compute $\mathbf r'(t)=\langle -\sin t,\cos t,1\rangle$. At $t=\pi/2$, the point is $\langle 0,1,\pi/2\rangle$ and the tangent line is $\langle 0,1,\pi/2\rangle + s\langle -1,0,1\rangle$.
If $\lVert\mathbf r(t)\rVert$ is constant then $\mathbf r(t)\perp \mathbf r'(t)$
If $\lVert\mathbf r(t)\rVert$ is constant, then $\mathbf r(t)\cdot \mathbf r(t)$ is constant, so differentiating gives $2\,\mathbf r(t)\cdot\mathbf r'(t)=0$, hence orthogonality. Example: unit circle parameterization.
Integral componentwise
Define $\int_a^b \mathbf r(t)\,dt = \left\langle \int_a^b f(t)\,dt,\int_a^b g(t)\,dt,\int_a^b h(t)\,dt\right\rangle$. Then $\mathbf r(b)-\mathbf r(a)=\int_a^b \mathbf r'(t)\,dt$ (FTC componentwise).

References (student-friendly)

  • OpenStax, Calculus Volume 3 — Vector-valued functions and space curves.
  • Paul’s Online Math Notes — Parametric equations, calculus on vector functions.
  • Stewart, Calculus: Early Transcendentals — sections on vector functions and parametric curves.
  • MIT OpenCourseWare — Multivariable calculus materials on curves and motion.
Interactive Lab. Change inputs and watch the mathematics update instantly. Everything runs locally in your browser. If a 3D view is blank, the external 3D library may be blocked on your site; the algebraic outputs still work.

Tool A — Domain by Intersection

Choose a case study
Parameters for the outline example
Parameters for the transcript example
Component constraints
    Intersection (domain): —
    Tip: the output explains each component restriction and then intersects the intervals.
    Instructor note. The transcript treats $\frac{1}{c-t}$ as requiring $c-t>0$ (so $t<c$). Standard algebra only requires $c-t\neq 0$. This tool follows the transcript so your lab matches the lecture narrative.
    Try this.
    1. Outline example: set $a=4$, $b=1$ and verify the intersection interval.
    2. Transcript example: set $k=13$, $b=16$, $c=17$ and check that the domain is $(0,17)$.
    3. Change $c$ and watch how the endpoint moves.
    Why do we intersect?
    A vector-valued function $\mathbf r(t)=\langle f(t),g(t),h(t)\rangle$ is defined exactly when every component is defined. Therefore, $D_{\mathbf r}=D_f\cap D_g\cap D_h$.

    Tool B — A Line as a Vector-Valued Function in $\mathbb R^3$

    Point $\mathbf P=\langle x_0,y_0,z_0\rangle$
    Direction $\mathbf V=\langle v_1,v_2,v_3\rangle$
    Evaluate at $t=$
    Vector form: —
    Parametric form: —
    Symmetric form: —
    Point on line: —
    A line is determined by a point and a direction vector. The parameter $t$ measures “how far” you move in the direction of $\mathbf V$.
    3D view (optional)
    Drag to rotate. The marked point is $\mathbf r(t_0)$.

    Tool C — Helix Explorer: $\mathbf r(t)=\langle r\cos t,\ r\sin t,\ kt\rangle$

    Radius $r$
    Pitch factor $k$
    Current $t$
    Plot range $t\in[a,b]$
    Cylinder equation: —
    Point at $t$: —
    Projection to the $xy$-plane
    Because $\cos^2 t + \sin^2 t = 1$, we have $x^2+y^2=r^2$, so the curve lies on a cylinder and rises linearly in $z$.
    3D helix view (optional)
    Drag to rotate. The marker is $\mathbf r(t)$.

    Tool D — Derivatives of Parametric Curves

    Choose example
    Evaluate at $t=$
    Position
    —
    —
    First derivatives
    —
    —
    —
    Second derivative
    —
    —
    Horizontal/vertical tangent conditions
    Eliminate the parameter
    Recall: $\displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ (when $dx/dt\ne 0$), and $\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dt}\!\left(\frac{dy}{dx}\right)\Big/\frac{dx}{dt}$.
    Suggested workflow
    1) Compute $dx/dt$ and $dy/dt$.
    2) Form $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$.
    3) Differentiate $\frac{dy}{dx}$ with respect to $t$, then divide by $dx/dt$ to obtain $\frac{d^2y}{dx^2}$.
    4) Use signs of $\frac{d^2y}{dx^2}$ for concavity.
    Try: For $x=t^2, y=t^3-3t$, locate all $t$ where $dy/dt=0$ and $dx/dt\neq 0$ (horizontal tangents).

    Practice

    Quick checks + longer prompts. Use the Lab tools to verify your work.

    Tip: answer, then click “Grade”.
    Quiz 1 — Domain of a vector-valued function

    For $\mathbf r(t)=\langle \ln(13t),\ \sqrt{t+16},\ \frac{1}{17-t}\rangle$, what domain does the lecture conclude?

    Then open the Domain Builder and confirm by intersecting each component domain.

    Quiz 2 — First derivative of a parametric curve

    If $x=f(t)$ and $y=g(t)$, then the slope $\frac{dy}{dx}$ is:

    Use the Parametric Derivatives tool to see the formula applied step-by-step.

    Longer prompts (work out fully)
    1. Let $\mathbf r(t)=\langle 1+t,\ 2+5t,\ -1+6t\rangle$. Identify a point and direction vector, then write $\mathbf r(t)=\mathbf P+t\mathbf V$.
    2. For $\mathbf r(t)=\langle \cos t,\ \sin t,\ t\rangle$, explain why the curve lies on a cylinder and describe its projection onto the $xy$-plane.
    3. For $x=t^2$ and $y=t^3-3t$, find every value of $t$ where the tangent is horizontal. Determine concavity at those parameters.
    4. For $x=1+\ln t$ and $y=t^2+2$, find the tangent line at $t=1$ without eliminating $t$, then eliminate $t$ and verify the same slope.