Derivatives of Vector-Valued Functions | Tony's Intertwingled Memex

Core Definition

If $\vec{r}(t) = \langle f(t), g(t), h(t) \rangle = f(t)\hat{i} + g(t)\hat{j} + h(t)\hat{k}$ is a vector-valued function where the component functions $f, g,$ and $h$ are differentiable on an open interval, then the derivative $\vec{r}'(t)$ is defined component-wise:

\vec{r}'(t) = \frac{d\vec{r}}{dt} = \lim_{\Delta t \to 0} \frac{\vec{r}(t + \Delta t) - \vec{r}(t)}{\Delta t} = \langle f'(t), g'(t), h'(t) \rangle = f'(t)\hat{i} + g'(t)\hat{j} + h'(t)\hat{k}

Concept Explanation

Geometric Meaning: The Tangent Vector

Geometrically, the derivative vector $\vec{r}'(t)$ is a tangent vector to the space curve at the point corresponding to parameter $t$ . It points strictly in the direction of motion (the direction of increasing $t$ ).

If $\vec{r}'(t) \ne \vec{0}$ , the curve is called smooth at $t$ .
Cusps or sharp corners occur where $\vec{r}'(t) = \vec{0}$ or where a component derivative fails to exist.

                  z ^
                    |           .  r'(t)  [Tangent vector]
                    |          / \
                    |         /   \
                    |   r(t) /     . P(t)
                    |       /     /
                    |      /     /  Space Curve
                    |     /     /
                    +----+-----+----------> y
                   /
                  /
                 v x

Physical & Kinematic Meaning

In physics and dynamics, if $\vec{r}(t)$ models the position of a particle at time $t$ :

Velocity Vector: $\vec{v}(t) = \vec{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle$
Speed (Scalar): $v(t) = \|\vec{v}(t)\| = \|\vec{r}'(t)\| = \sqrt{[x'(t)]^2 + [y'(t)]^2 + [z'(t)]^2}$
Acceleration Vector: $\vec{a}(t) = \vec{v}'(t) = \vec{r}''(t) = \langle x''(t), y''(t), z''(t) \rangle$

Vector Differentiation Rules

Let $\vec{u}(t)$ and $\vec{v}(t)$ be differentiable vector functions, $c$ a scalar constant, and $f(t)$ a differentiable scalar function:

Sum & Difference Rule: $\frac{d}{dt}[\vec{u}(t) \pm \vec{v}(t)] = \vec{u}'(t) \pm \vec{v}'(t)$
Scalar Multiple Rule: $\frac{d}{dt}[c\vec{u}(t)] = c\vec{u}'(t)$
Scalar Function Product Rule: $\frac{d}{dt}[f(t)\vec{u}(t)] = f'(t)\vec{u}(t) + f(t)\vec{u}'(t)$
Dot Product Rule: $\frac{d}{dt}[\vec{u}(t) \cdot \vec{v}(t)] = \vec{u}'(t) \cdot \vec{v}(t) + \vec{u}(t) \cdot \vec{v}'(t)$ (Output is a scalar)
Cross Product Rule: $\frac{d}{dt}[\vec{u}(t) \times \vec{v}(t)] = \vec{u}'(t) \times \vec{v}(t) + \vec{u}(t) \times \vec{v}'(t)$ (Order matters: cross product is anti-commutative)
Vector Chain Rule: $\frac{d}{dt}[\vec{r}(f(t))] = \vec{r}'(f(t)) \cdot f'(t)$

Orthogonality Theorem for Constant-Length Curves

If $\|\vec{r}(t)\| = c$ (a constant magnitude for all $t$ ), then the position vector and its derivative are strictly orthogonal:

\vec{r}(t) \cdot \vec{r}'(t) = 0

Proof:

\vec{r}(t) \cdot \vec{r}(t) = \|\vec{r}(t)\|^2 = c^2

. Differentiating both sides yields

\vec{r}'(t) \cdot \vec{r}(t) + \vec{r}(t) \cdot \vec{r}'(t) = 2[\vec{r}(t) \cdot \vec{r}'(t)] = 0 \implies \vec{r}(t) \cdot \vec{r}'(t) = 0

Engineering Applications

Robotic Path Planning: Calculating joint velocities and end-effector accelerations to ensure actuator torque limits are never exceeded.
Automotive Safety: Computing centripetal acceleration on curved highway ramps and test tracks to prevent vehicle rollover.
Aerospace Guidance: Evaluating trajectory derivatives to feed autopilot inertial navigation systems (INS) during high-speed maneuvering.

Example Problems

Example 1: Velocity, Speed, and Acceleration of a Spatial Helix

A robotic welding tip follows the path $\vec{r}(t) = \langle 3\cos(2t), 3\sin(2t), 4t \rangle$ . Find its velocity vector $\vec{v}(t)$ , speed $\|\vec{v}(t)\|$ , and acceleration vector $\vec{a}(t)$ .

Solution

Velocity $\vec{v}(t) = \vec{r}'(t)$ : $\vec{v}(t) = \left\langle \frac{d}{dt}[3\cos(2t)], \frac{d}{dt}[3\sin(2t)], \frac{d}{dt}[4t] \right\rangle = \langle -6\sin(2t), 6\cos(2t), 4 \rangle$
Speed $\|\vec{v}(t)\| = \|\vec{r}'(t)\|:$ $\|\vec{v}(t)\| = \sqrt{(-6\sin(2t))^2 + (6\cos(2t))^2 + 4^2} = \sqrt{36\sin^2(2t) + 36\cos^2(2t) + 16}$ Using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ : $\|\vec{v}(t)\| = \sqrt{36(1) + 16} = \sqrt{52} = 2\sqrt{13} \text{ (constant speed)}$
Acceleration $\vec{a}(t) = \vec{v}'(t)$ : $\vec{a}(t) = \langle -12\cos(2t), -12\sin(2t), 0 \rangle$

Example 2: Vector Chain Rule Application

Given $\vec{r}(t) = \langle 4t^2, -e^{3t}, \cos(2t) \rangle$ and parameter substitution $f(t) = t^2$ , find $\frac{d}{dt}[\vec{r}(f(t))]$ .

Solution

Method 1: Direct Substitution First: $\vec{r}(f(t)) = \langle 4(t^2)^2, -e^{3(t^2)}, \cos(2(t^2)) \rangle = \langle 4t^4, -e^{3t^2}, \cos(2t^2) \rangle$
Differentiate Component-by-Component:
- $x$ -component: $\frac{d}{dt}[4t^4] = 16t^3$
- $y$ -component: $\frac{d}{dt}[-e^{3t^2}] = -e^{3t^2} \cdot (6t) = -6t e^{3t^2}$
- $z$ -component: $\frac{d}{dt}[\cos(2t^2)] = -\sin(2t^2) \cdot (4t) = -4t \sin(2t^2)$
Final Result: $\frac{d}{dt}[\vec{r}(f(t))] = \langle 16t^3, -6te^{3t^2}, -4t\sin(2t^2) \rangle$

Connections & References

Parent Concepts: Vector Valued Functions | Derivatives
Related Notes: Unit Tangent Vector | Integrals of Vector-Valued Functions | Dot Product | Cross Product
Course Sessions: Calc-3 Session 05 | Calc-3 Session 06