Derivatives & Integrals Of Vector Functions (How To w/ Examples!)

How do we differentiate and integrate vectors?

Jenn (B.S., M.Ed.) of Calcworkshop® teaching derivatives & integrals of vector functions

Jenn, Founder Calcworkshop^®, 15+ Years Experience (Licensed & Certified Teacher)

How do we calculate the velocity and acceleration of a particle moving in space involving both magnitude and direction?

Thankfully, the same methods and techniques we used in single-variable calculus (i.e., differential calculus) still apply and allow us to define motion along a curve in a plane or space!

So, the calculus of vector-valued functions is largely concerned with how to find their respective derivatives and integrals, this in turn helps us answer questions related to:

Velocity
Acceleration
Speed
Curvature
And More

Limit Of A Vector

You will quickly find that the methodology for evaluating the limit of a vector-valued function, which we learned in our previous lesson, can be applied to the calculus of space curves pertaining to derivatives and integrals.

If ${\vec{r}}^{'} (t)$ is a vector function, then its derivative is defined as $\frac{d \vec{r}}{d t} = {\vec{r}}^{'} (t) = lim_{h \to 0} \frac{\vec{r} (t + h) - \vec{r} (t)}{h}$ .

Vector Derivative

Which means, that if we write the vector function in component notation, denoted $\vec{r (t)} = ⟨ f (t), g (t), h (t) ⟩$ , where $f$ , (g\), and (h\) are differentiable functions, then ${\vec{r}}^{'} (t) = ⟨ f^{'} (t), g^{'} (t), h^{'} (t) ⟩$ .

Let’s see how easy this is by looking at an example.

Example – 1st & 2nd Derivatives

For instance, let’s calculate both the first, ${\vec{r}}^{'} (t)$ , and second derivative, ${\vec{r}}^{''} (t)$ , of the vector-valued function if $\vec{r (t)} = ⟨ \cos 2 t, t e^{t}, t - t^{2} ⟩$ .

All we have to do is differentiate each component of vector $\vec{r} (t)$ .
$\begin{aligned} {\vec{r}}^{'} (t) = ⟨ \frac{d}{d t} [\cos 2 t], \frac{d}{d t} [t e^{t}], \frac{d}{d t} [t - t^{2}] ⟩ \\ {\vec{r}}^{'} (t) = ⟨ - 2 \sin 2 t, t e^{t} + e^{t}, 1 - 2 t ⟩ \end{aligned}$

So, the first derivative is ${\vec{r}}^{'} (t) = ⟨ - 2 \sin 2 t, e^{t} (t + 1), 1 - 2 t ⟩$ .

Now let’s find the second derivative by differentiating each component of vector ${\vec{r}}^{'} (t)$ .

${\vec{r}}^{''} (t) = ⟨ \frac{d}{d t} [- 2 \sin 2 t], \frac{d}{d t} [e^{t} (t + 1)] +, \frac{d}{d t} [1 - 2 t] ⟩$

${\vec{r}}^{''} (t) = ⟨ - 4 \cos 2 t, e^{t} (t + 2), - 2 ⟩$

And that’s it. Nothing to it!

Vector Differentiation Rules

And the differentiation rules for the real-valued function (i.e., the component functions (f\), (g\), and (h\) of the vector) are similar for the vector-valued function, as seen below in the following theorem.

Suppose $\vec{a}$ and $\vec{b}$ are differentiable vector functions, $c$ is a scalar, and $f$ is a real-valued function, then.

$\begin{matrix} \frac{d}{d t} [\vec{a} (t) + \vec{b} (t)] = {\vec{a}}^{'} (t) + {\vec{b}}^{'} (t) \\ \frac{d}{d t} [c \vec{a} (t)] = c {\vec{a}}^{'} (t) \\ \frac{d}{d t} [f (t) \vec{a} (t)] = f (t) {\vec{a}}^{'} (t) + \vec{a} (t) f^{'} (t) \\ \frac{d}{d t} [\vec{a} (t) \cdot \vec{b} (t)] = {\vec{a}}^{'} (t) \cdot \vec{b} (t) + \vec{a} (t) \cdot {\vec{b}}^{'} (t) \\ \frac{d}{d t} [\vec{a} (t) \times \vec{b} (t)] = {\vec{a}}^{'} (t) \times \vec{b} (t) + \vec{a} (t) \times {\vec{b}}^{'} (t) \\ \frac{d}{d t} [\vec{a} (f (t))] = f^{'} (t) {\vec{a}}^{'} (f (t)) \end{matrix}$

Vector Integration

And the definite integral of a continuous vector function is comparable to the definite integral of a continuous real function.

If $\vec{r (t)} = ⟨ f (t), g (t), h (t) ⟩$ , then $\int_{a}^{b} \vec{r} (t) d t = ⟨ \int_{a}^{b} f (t) d t, \int_{a}^{b} g (t) d t, \int_{a}^{b} h (t) d t ⟩$ .

So, just like derivatives of vector functions, all we have to do to find the definite integral of vector function is to integrate each component function separately.

Example – Definite Integrals

Evaluate $\int_{0}^{2} \vec{r} (t) d t$ if $\vec{r (t)} = ⟨ \cos 2 t, t e^{t}, t - t^{2} ⟩$ .

To find the definite integral of the vector function, all we have to do is apply the fundamental theorem of calculus to each component function.

$\int_{0}^{2} \vec{r} (t) d t = ⟨ \int_{0}^{2} (\cos 2 t) d t, \underset{\begin{matrix} use integration \\ by parts \end{matrix}}{\underset{⏟}{\int_{0}^{2} (t e^{t}) d t}}, \int_{0}^{2} (t - t^{2}) d t ⟩$

$= {⟨ \frac{1}{2} \sin 2 t, t e^{t} - e^{t}, \frac{1}{2} t^{2} - \frac{1}{3} t^{3} ⟩ |}_{0}^{2}$

$= ⟨ \frac{1}{2} \sin (4), 1 + e^{2}, - \frac{2}{3} ⟩$

See, evaluating an integral of a vector function is just as easy as differentiating!

Example – Indefinite Integrals

But what if we are given an indefinite integral? How do we find the antiderivative?

We simply integrate each component and add a “+C”

So, using our previous example, suppose $\vec{r (t)} = ⟨ \cos 2 t, t e^{t}, t - t^{2} ⟩$ .

Evaluate $\int \vec{r} (t) d t$ .

$\int \vec{r} (t) d t = ⟨ \int (\cos 2 t) d t, \underset{\begin{matrix} use integration \\ by parts \end{matrix}}{\underset{⏟}{\int (t e^{t}) d t}}, \int (t - t^{2}) ⟩$

$= ⟨ \frac{1}{2} \sin 2 t + C_{1}, t e^{t} - e^{t} + C_{2}, \frac{1}{2} t^{2} - \frac{1}{3} t^{3} + C_{3} ⟩$

Easy!

Alright, so evaluating the derivative and integral of a vector function is pretty straightforward, but what is the purpose of doing so?

To describe motion in a plane or space!

Unit Tangent Vector

Well, the geometric meaning of the derivative ${\vec{r}}^{'} (t)$ is that ${\vec{r}}^{'} (t)$ is tangent to the curve and the length of ${\vec{r}}^{'} (t)$ , denoted $‖ {\vec{r}}^{'} (t) ‖$ , is the speed. And since ${\vec{r}}^{'} (t)$ points in the direction of motion and its length is the speed, we can call ${\vec{r}}^{'} (t)$ the velocity vector and ${\vec{r}}^{''} (t)$ the acceleration vector.

And a particle whose position vector at the time $t$ is $\vec{r} (t)$ having velocity ${\vec{r}}^{'} (t)$ and speed $‖ {\vec{r}}^{'} (t) ‖$ , the unit vector in the direction of ${\vec{r}}^{'} (t)$ is:

$T (t) = \frac{{\vec{r}}^{'} (t)}{‖ {\vec{r}}^{'} (t) ‖}$

This unit vector, $T (t)$ , called the Unit Tangent Vector, records the direction of motion.

Example

For example, if $\vec{r (t)} = ⟨ \cos 2 t, t e^{t}, t - t^{2} ⟩$ let’s find the unit tangent vector at the point where $t = 0$ .

Step 1

First, we will need to find ${\vec{r}}^{'} (t)$ . Thankfully, we calculated this earlier and found:

${\vec{r}}^{'} (t) = ⟨ - 2 \sin 2 t, e^{t} (t + 1), 1 - 2 t ⟩$

Step 2

Next, we will evaluate the velocity vector at the point where $t = 0$ .

${\vec{r}}^{'} (0) = ⟨ - 2 \sin 2 (0), e^{0} (0 + 1), 1 - 2 (0) ⟩ = ⟨ 0, 1, 1 ⟩$

Step 3

Now we will need to calculate the speed of $\vec{r} (t)$ when $t = 0$ , which is the magnitude of the velocity vector ${\vec{r}}^{'} (0)$ .

$‖ {\vec{r}}^{'} (0) ‖ = \sqrt{(0)^{2} + (1)^{2} + (1)^{2}} = \sqrt{2}$

Step 4

Now, let’s substitute our values into the unit tangent formula:

$T (t) = \frac{{\vec{r}}^{'} (t)}{| {\vec{r}}^{'} (t) | ∣}$

$T (0) = \frac{{\vec{r}}^{'} (0)}{| {\vec{r}}^{'} (0) | ∣} = \frac{⟨ 0, 1, 1 ⟩}{\sqrt{2}} = ⟨ 0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} ⟩$

Easy, right?

So, together we will learn how to find derivatives and integrals of vector functions and discuss the definition of smooth curves in 3-space. Additionally, we will represent the tangent vector, unit tangent vector, and antiderivative of vector-valued functions while given initial conditions.

There’s a lot to uncover, so let’s jump right in!

Video Tutorial w/ Full Lesson & Detailed Examples (Video)

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