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Vector-Valued
Functions & Derivatives

Master the art of defining parametric motion through vector-valued functions and learn to differentiate component by component.

AP Calculus BC

What Is a Vector-Valued Function?

A vector-valued function takes a real number \( t \) as input and produces a vector as output. In two dimensions:

Think of it as giving directions in the plane: at each moment \( t \), the function tells you an \( x \)-coordinate and a \( y \)-coordinate simultaneously.

Tracing a Parametric Curve

As \( t \) varies, the tip of the position vector \( \vec{r}(t) \) traces a curve in the plane. Each point on the curve corresponds to a specific value of the parameter.

x y \( \vec{r}(t_0) \) position vector parametric curve O

The Component Functions

Every vector-valued function is built from ordinary scalar functions. Each component can be analyzed independently:

\( x(t) = f(t) \)
The horizontal component: governs left-right motion.
\( y(t) = g(t) \)
The vertical component: governs up-down motion.
\( t \): Parameter
Often represents time, linking both components together.

Finding the Domain

The domain of \( \vec{r}(t) = \langle f(t),\, g(t) \rangle \) is the intersection of the domains of its component functions:

1
Find where \( f(t) \) is defined: check for division by zero, negative square roots, etc.
2
Find where \( g(t) \) is defined: apply the same checks to the second component.
3
Take the intersection: only \( t \)-values valid for both components belong to the domain.

The Derivative of \( \vec{r}(t) \)

Differentiating a vector-valued function is beautifully simple: differentiate each component separately.

This follows directly from the limit definition. Each component's derivative is computed independently, just as with ordinary scalar calculus.

Why It Works: Limit Definition

The derivative of a vector-valued function is defined exactly like the scalar case, but with vector subtraction:

Since limits distribute over vector components:

The Derivative as a Tangent Vector

\( \vec{r}(t_0) \) \( \vec{r}\,'(t_0) \) position vector tangent vector curve

The derivative \( \vec{r}\,'(t_0) \) is a vector that is tangent to the curve at the point \( \vec{r}(t_0) \). It points in the direction of motion as \( t \) increases.

Scalar vs. Vector Derivatives

Scalar Function Vector-Valued Function
Input Real number \( t \) Real number \( t \)
Output Single number \( f(t) \) Vector \( \langle f(t), g(t) \rangle \)
Derivative \( f'(t) \) (a number) \( \langle f'(t), g'(t) \rangle \) (a vector)
Meaning Rate of change (slope) Tangent vector (direction + speed)

Differentiation Rules Carry Over

All familiar differentiation rules extend naturally to vector-valued functions:

1
Sum Rule: \( \frac{d}{dt}[\vec{r}(t) + \vec{s}(t)] = \vec{r}\,'(t) + \vec{s}\,'(t) \)
2
Scalar Multiple: \( \frac{d}{dt}[c\,\vec{r}(t)] = c\,\vec{r}\,'(t) \)
3
Product with Scalar Function: \( \frac{d}{dt}[u(t)\,\vec{r}(t)] = u'(t)\,\vec{r}(t) + u(t)\,\vec{r}\,'(t) \)
4
Chain Rule: \( \frac{d}{dt}[\vec{r}(u(t))] = \vec{r}\,'(u(t)) \cdot u'(t) \)

How to Differentiate Step by Step

1
Write the function in component form: \( \vec{r}(t) = \langle f(t),\, g(t) \rangle \).
2
Differentiate each component separately: find \( f'(t) \) and \( g'(t) \) using standard rules (power, chain, product, quotient, trig).
3
Reassemble into a vector: \( \vec{r}\,'(t) = \langle f'(t),\, g'(t) \rangle \).
4
Evaluate if needed: substitute a specific \( t \)-value to get the tangent vector at that point.

Example 1: Polynomial Components

Find \( \vec{r}\,'(t) \) for \( \vec{r}(t) = \langle 3t^2 + 1,\; t^3 - 4t \rangle \).

1
Identify components: \( f(t) = 3t^2 + 1 \), \( g(t) = t^3 - 4t \).
2
Differentiate: \( f'(t) = 6t \), \( g'(t) = 3t^2 - 4 \).
3
Reassemble: \( \vec{r}\,'(t) = \langle 6t,\; 3t^2 - 4 \rangle \).

Example 2: Trigonometric Components

Find \( \vec{r}\,'(t) \) for \( \vec{r}(t) = \langle \cos t,\; \sin t \rangle \).

1
Differentiate: \( f'(t) = -\sin t \), \( g'(t) = \cos t \).
2
Result: \( \vec{r}\,'(t) = \langle -\sin t,\; \cos t \rangle \).
\( \vec{r}(t) \) \( \vec{r}\,'(t) \) x y

Example 3: Chain Rule in Action

Find \( \vec{r}\,'(t) \) for \( \vec{r}(t) = \langle e^{2t},\; \ln(t^2 + 1) \rangle \) and evaluate at \( t = 0 \).

1
\( f(t) = e^{2t} \): By the chain rule, \( f'(t) = 2e^{2t} \).
2
\( g(t) = \ln(t^2+1) \): By the chain rule, \( g'(t) = \frac{2t}{t^2+1} \).
3
Reassemble: \( \vec{r}\,'(t) = \left\langle 2e^{2t},\; \frac{2t}{t^2+1} \right\rangle \).
4
Evaluate at \( t = 0 \): \( \vec{r}\,'(0) = \langle 2e^0,\; 0 \rangle = \langle 2,\; 0 \rangle \).

Common Mistakes to Avoid

Cross-Contaminating Components
Mixing up the \( x \) and \( y \) derivatives, or using the product rule between components instead of differentiating each independently.
Forgetting the Chain Rule
For \( \langle \sin(3t), e^{-t} \rangle \), writing \( \langle \cos(3t), e^{-t} \rangle \) instead of \( \langle 3\cos(3t), -e^{-t} \rangle \). The chain rule still applies inside each component.
Returning a Scalar
The derivative of a vector is still a vector. Never write \( \vec{r}\,'(t) = 6t + 3t^2 \) without angle brackets.

Summary at a Glance

Concept Formula / Key Fact
Definition \( \vec{r}(t) = \langle f(t),\, g(t) \rangle \)
Domain \( \text{dom}(f) \cap \text{dom}(g) \)
Derivative \( \vec{r}\,'(t) = \langle f'(t),\, g'(t) \rangle \)
Geometric Meaning Tangent vector at the point
Speed \( |\vec{r}\,'(t)| = \sqrt{[f'(t)]^2 + [g'(t)]^2} \)
Key Principle Differentiate component by component

Question 1 of 5

If \( \vec{r}(t) = \langle 5t^3,\; 2t^2 - t \rangle \), what is \( \vec{r}\,'(t) \)?

Question 2 of 5

For \( \vec{r}(t) = \langle \sin(2t),\; e^{3t} \rangle \), what is \( \vec{r}\,'(0) \)?

Question 3 of 5

The derivative \( \vec{r}\,'(t) \) of a vector-valued function at a point geometrically represents:

Question 4 of 5

What is the domain of \( \vec{r}(t) = \left\langle \sqrt{t},\; \frac{1}{t-2} \right\rangle \)?

Question 5 of 5

If \( \vec{r}(t) = \langle t\cos t,\; t\sin t \rangle \), what is \( \vec{r}\,'(t) \)?