The Main Idea 

Differentiating vector-valued functions works just like differentiating regular functions—you apply the definition or, more practically, differentiate each component separately.

The derivative of [latex]{\bf{r}}(t)[/latex] is defined as:

[latex]{\bf{r}}'(t)=\underset{\Delta t\to 0}{\text{lim}}\frac{{\bf{r}}(t+\Delta t)-{\bf{r}}(t)}{\Delta t}[/latex]

This measures the instantaneous rate of change of the vector function. When [latex]{\bf{r}}(t)[/latex] represents an object’s position at time [latex]t[/latex], the derivative [latex]{\bf{r}}'(t)[/latex] represents its velocity at that moment.

The Practical Approach: Component-by-Component

Rather than wrestling with the limit definition, use this shortcut: differentiate each component function individually.

• For plane curves [latex]{\bf{r}}(t)=f(t){\bf{i}}+g(t){\bf{j}}[/latex]: [latex]{\bf{r}}'(t)=f'(t){\bf{i}}+g'(t){\bf{j}}[/latex]
• For space curves [latex]{\bf{r}}(t)=f(t){\bf{i}}+g(t){\bf{j}}+h(t){\bf{k}}[/latex]: [latex]{\bf{r}}'(t)=f'(t){\bf{i}}+g'(t){\bf{j}}+h'(t){\bf{k}}[/latex]

The vector structure stays intact—only the component functions change. Apply your familiar derivative rules (power rule, product rule, chain rule, trig derivatives, etc.) to each component.

Derivative Properties for Vector-Valued Functions

The derivative rules you know extend naturally to vector-valued functions:

1. Scalar multiple: [latex]\frac{d}{dt}[c{\bf{r}}(t)]=c{\bf{r}}'(t)[/latex]
2. Sum and difference: [latex]\frac{d}{dt}[{\bf{r}}(t)\pm{\bf{u}}(t)]={\bf{r}}'(t)\pm{\bf{u}}'(t)[/latex]
3. Scalar product: [latex]\frac{d}{dt}[f(t){\bf{u}}(t)]=f'(t){\bf{u}}(t)+f(t){\bf{u}}'(t)[/latex]
4. Dot product: [latex]\frac{d}{dt}[{\bf{r}}(t)\cdot{\bf{u}}(t)]={\bf{r}}'(t)\cdot{\bf{u}}(t)+{\bf{r}}(t)\cdot{\bf{u}}'(t)[/latex]
5. Cross product: [latex]\frac{d}{dt}[{\bf{r}}(t)\times{\bf{u}}(t)]={\bf{r}}'(t)\times{\bf{u}}(t)+{\bf{r}}(t)\times{\bf{u}}'(t)[/latex]
6. Chain rule: [latex]\frac{d}{dt}[{\bf{r}}(f(t))]={\bf{r}}'(f(t))\cdot f'(t)[/latex]

Special Property Worth Noting

If [latex]{\bf{r}}(t)\cdot{\bf{r}}(t)=c[/latex] (a constant), then [latex]{\bf{r}}(t)\cdot{\bf{r}}'(t)=0[/latex]. This tells us that when a vector has constant magnitude, it’s always perpendicular to its derivative—a useful geometric insight.

The Main Idea 

In single-variable calculus, the derivative gives the slope of the tangent line. For vector-valued functions, the derivative gives you a tangent vector that points along the curve’s direction.

A tangent vector at [latex]t = t_0[/latex] is any vector that lies tangent to the curve when placed at point [latex]{\bf{r}}(t_0)[/latex]. The derivative [latex]{\bf{r}}'(t_0)[/latex] is the most natural tangent vector. It points in the direction the curve is heading as [latex]t[/latex] increases.

The Principal Unit Tangent Vector

While [latex]{\bf{r}}'(t)[/latex] gives direction, its magnitude can vary. The principal unit tangent vector [latex]{\bf{T}}(t)[/latex] always has magnitude [latex]1[/latex]:

[latex]{\bf{T}}(t)=\frac{{\bf{r}}'(t)}{\left\Vert{\bf{r}}'(t)\right\Vert}[/latex]

(provided [latex]\left\Vert{\bf{r}}'(t)\right\Vert\neq 0[/latex])

This unit vector points in the direction of motion. It strips away information about speed—it’s purely directional.

How to Find the Unit Tangent Vector

1. Find the derivative: Calculate [latex]{\bf{r}}'(t)[/latex] by differentiating each component
2. Find the magnitude: Calculate [latex]\left\Vert{\bf{r}}'(t)\right\Vert = \sqrt{(f'(t))^2 + (g'(t))^2 + (h'(t))^2}[/latex]
3. Divide: Compute [latex]{\bf{T}}(t) = \frac{{\bf{r}}'(t)}{\left\Vert{\bf{r}}'(t)\right\Vert}[/latex]

The formula only works when [latex]\left\Vert{\bf{r}}'(t)\right\Vert \neq 0[/latex]. If the derivative’s magnitude is zero, the unit tangent vector is undefined. This typically happens at cusps or points where the motion stops.

Why This Matters

The unit tangent vector always has length [latex]1[/latex]. It points in the direction of travel along the curve. It’s useful when you need to know which way the curve is heading. You don’t have to worry about how fast it’s moving. Think of it as “direction of motion” isolated from “speed of motion.”

The Main Idea 

Integrating vector-valued functions works just like differentiating them. You handle each component separately using the integration techniques you already know. The Fundamental Theorem of Calculus applies here too.

Just as you differentiate each component individually, you integrate the same way.

For plane curves [latex]{\bf{r}}(t) = f(t){\bf{i}} + g(t){\bf{j}}[/latex]:

• Indefinite integral: [latex]\int [f(t){\bf{i}}+g(t){\bf{j}}]dt=\Bigg[\int f(t)dt\Bigg]{\bf{i}}+\Bigg[\int g(t)dt\Bigg]{\bf{j}}[/latex]
• Definite integral: [latex]\int_{a}^{b} [f(t){\bf{i}}+g(t){\bf{j}}]dt=\Bigg[\int_{a}^{b} f(t)dt\Bigg]{\bf{i}}+\Bigg[\int_{a}^{b} g(t)dt\Bigg]{\bf{j}}[/latex]

For space curves [latex]{\bf{r}}(t) = f(t){\bf{i}} + g(t){\bf{j}} + h(t){\bf{k}}[/latex]:

• Indefinite integral: [latex]\int [f(t){\bf{i}}+g(t){\bf{j}}+h(t){\bf{k}}]dt=\Bigg[\int f(t)dt\Bigg]{\bf{i}}+\Bigg[\int g(t)dt\Bigg]{\bf{j}}+\Bigg[\int h(t)dt\Bigg]{\bf{k}}[/latex]
• Definite integral: [latex]\int_{a}^{b} [f(t){\bf{i}}+g(t){\bf{j}}+h(t){\bf{k}}]dt=\Bigg[\int_{a}^{b} f(t)dt\Bigg]{\bf{i}}+\Bigg[\int_{a}^{b} g(t)dt\Bigg]{\bf{j}}+\Bigg[\int_{a}^{b} h(t)dt\Bigg]{\bf{k}}[/latex]

Handling Constants of Integration

When you integrate a vector-valued function, each component gets its own integration constant. You can combine these into a single constant vector [latex]{\bf{C}}[/latex].

For example: [latex]\int [f(t){\bf{i}}+g(t){\bf{j}}]dt=F(t){\bf{i}}+G(t){\bf{j}}+{\bf{C}}[/latex] where [latex]{\bf{C}}=C_{1}{\bf{i}}+C_{2}{\bf{j}}[/latex].

Problem-Solving Strategy

Break the vector function into components. Integrate each scalar function using your standard techniques. Reassemble the results into vector form. Use the same integration formulas from Calculus 1 for each component.

Why This Matters

Antiderivatives of vector-valued functions appear constantly in applications. If a function represents velocity, its antiderivative gives position. If the function represents acceleration, the antiderivative gives velocity.