2.2 Derivatives and Integrals of Vector-Valued Functions

Vector-Valued Functions: Derivatives and Integrals

Differentiation of Vector-Valued Functions

A vector-valued function maps a scalar input (usually time) to a vector output:

$\mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle$

The component functions $f(t)$, $g(t)$, and $h(t)$ each describe behavior along one coordinate axis. To differentiate $\mathbf{r}(t)$, you just differentiate each component separately:

$\mathbf{r}'(t) = \langle f'(t), g'(t), h'(t) \rangle$

Geometrically, $\mathbf{r}'(t)$ is the tangent vector at the point $\mathbf{r}(t)$. It points in the direction the curve is heading at that instant. This is the key link between derivatives and motion: the derivative tells you which way and how fast the point is moving along the curve.

Once you have a tangent vector, you can write the equation of the tangent line at $t = t_0$:

$\mathbf{L}(t) = \mathbf{r}(t_0) + t\,\mathbf{r}'(t_0)$

This is just the parametric form of a line through the point $\mathbf{r}(t_0)$ in the direction $\mathbf{r}'(t_0)$. Note that $t$ here is a new parameter for the line, not the same $t$ from the original curve.

Rules for Vector Function Differentiation

Most differentiation rules carry over from single-variable calculus, applied component by component:

• Sum rule: $(\mathbf{u} + \mathbf{v})' = \mathbf{u}' + \mathbf{v}'$
• Scalar multiple rule: $(c\,\mathbf{u})' = c\,\mathbf{u}'$ (for a constant $c$)
• Scalar function multiple: $(f(t)\,\mathbf{u}(t))' = f'(t)\,\mathbf{u}(t) + f(t)\,\mathbf{u}'(t)$

The product rules require more care because vectors have two kinds of multiplication:

• Dot product rule: $\frac{d}{dt}(\mathbf{u} \cdot \mathbf{v}) = \mathbf{u}' \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{v}'$
• Cross product rule: $\frac{d}{dt}(\mathbf{u} \times \mathbf{v}) = \mathbf{u}' \times \mathbf{v} + \mathbf{u} \times \mathbf{v}'$

With the cross product, order matters. The cross product is not commutative, so you must keep $\mathbf{u}$ on the left and $\mathbf{v}$ on the right in both terms. Swapping them introduces a sign error.

The chain rule also extends naturally:

$\frac{d}{dt}\mathbf{r}(g(t)) = g'(t)\,\mathbf{r}'(g(t))$

Integration of Vector-Valued Functions

Integration works component by component, just like differentiation.

Indefinite integral:

$\int \mathbf{r}(t)\, dt = \left\langle \int f(t)\, dt,\; \int g(t)\, dt,\; \int h(t)\, dt \right\rangle + \mathbf{C}$

The constant of integration $\mathbf{C}$ is a vector constant $\langle C_1, C_2, C_3 \rangle$, not a scalar. Each component picks up its own constant.

Definite integral:

$\int_a^b \mathbf{r}(t)\, dt = \left\langle \int_a^b f(t)\, dt,\; \int_a^b g(t)\, dt,\; \int_a^b h(t)\, dt \right\rangle$

The result is a single vector, not a function.

Solving initial value problems is one of the most common applications. For example, given $\mathbf{a}(t)$ and initial conditions for velocity and position:

1. Integrate $\mathbf{a}(t)$ to get $\mathbf{v}(t) + \mathbf{C}_1$
2. Apply the initial velocity condition $\mathbf{v}(t_0)$ to solve for $\mathbf{C}_1$
3. Integrate $\mathbf{v}(t)$ to get $\mathbf{r}(t) + \mathbf{C}_2$
4. Apply the initial position condition $\mathbf{r}(t_0)$ to solve for $\mathbf{C}_2$

The Fundamental Theorem of Calculus extends to vector functions as expected:

$\frac{d}{dt} \int_a^t \mathbf{r}(s)\, ds = \mathbf{r}(t)$

Motion in Space

Position, Velocity, and Acceleration Relationships

These three quantities form a derivative chain, and understanding how they connect is the core of this section.

• Position $\mathbf{r}(t)$ describes where an object is at time $t$
• Velocity $\mathbf{v}(t) = \mathbf{r}'(t)$ is the first derivative of position. It's tangent to the path and tells you both direction and rate of motion.
• Acceleration $\mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{r}''(t)$ is the second derivative of position. It captures how the velocity is changing.

Speed is the magnitude of velocity: $\|\mathbf{v}(t)\|$. This is a scalar. Velocity tells you direction and how fast; speed tells you only how fast.

Arc length measures the total distance traveled along the curve from $t = a$ to $t = b$:

$s = \int_a^b \|\mathbf{r}'(t)\|\, dt$

This integrates speed over time, which makes intuitive sense: distance equals speed multiplied by time, accumulated continuously.

The TNB Frame

Three unit vectors form a moving coordinate system (called the Frenet-Serret frame or TNB frame) attached to the curve at each point:

• Unit tangent vector $\mathbf{T}(t) = \dfrac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|}$ points in the direction of motion
• Principal normal vector $\mathbf{N}(t) = \dfrac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|}$ points toward the center of curvature (perpendicular to $\mathbf{T}$, in the direction the curve is turning)
• Binormal vector $\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t)$ is perpendicular to both $\mathbf{T}$ and $\mathbf{N}$, completing a right-handed coordinate system

Together, $\mathbf{T}$, $\mathbf{N}$, and $\mathbf{B}$ give you a local frame of reference that moves with the object along the curve. The plane spanned by $\mathbf{T}$ and $\mathbf{N}$ is called the osculating plane, and it's the plane in which the curve is locally bending.