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

How do we differentiate and integrate vectors?

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 \(\vec{r}^{\prime}(t) \) is a vector function, then its derivative is defined as \(\frac{d \vec{r}}{d t}=\vec{r}^{\prime}(t)=\lim _{h \rightarrow 0} \frac{\vec{r}(t+h)-\vec{r}(t)}{h}\).

Vector Derivative

Which means, that if we write the vector function in component notation, denoted \(\overrightarrow{r(t)}=\langle f(t), g(t), h(t)\rangle\), where \(f\), (g\), and (h\) are differentiable functions, then \(\vec{r}^{\prime}(t)=\left\langle f^{\prime}(t), g^{\prime}(t), h^{\prime}(t)\right\rangle\).

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

Example – 1st & 2nd Derivatives

For instance, let’s calculate both the first, \(\vec{r}^{\prime}(t)\), and second derivative, \(\vec{r}^{\prime \prime}(t)\), of the vector-valued function if \(\overrightarrow{r(t)}=\left\langle\cos 2 t, t e^{t}, t-t^{2}\right\rangle\).

All we have to do is differentiate each component of vector \(\vec{r}(t)\).
\begin{equation}
\begin{aligned}
&\vec{r}^{\prime}(t)=\left\langle\frac{d}{d t}[\cos 2 t], \frac{d}{d t}\left[t e^{t}\right], \frac{d}{d t}\left[t-t^{2}\right]\right\rangle \\
&\vec{r}^{\prime}(t)=\left\langle-2 \sin 2 t, t e^{t}+e^{t}, 1-2 t\right\rangle
\end{aligned}
\end{equation}

So, the first derivative is \(\vec{r}^{\prime}(t)=\left\langle-2 \sin 2 t, e^{t}(t+1), 1-2 t\right\rangle\).

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

\begin{equation}
\vec{r}^{\prime \prime}(t)=\left\langle\frac{d}{d t}[-2 \sin 2 t], \frac{d}{d t}\left[e^{t}(t+1)\right]+, \frac{d}{d t}[1-2 t]\right\rangle
\end{equation}

\begin{equation}
\vec{r}^{\prime \prime}(t)=\left\langle-4 \cos 2 t, e^{t}(t+2),-2\right\rangle
\end{equation}

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{equation}
\begin{gathered}
\frac{d}{d t}[\vec{a}(t)+\vec{b}(t)]=\vec{a}^{\prime}(t)+\vec{b}^{\prime}(t) \\
\frac{d}{d t}[c \vec{a}(t)]=c \vec{a}^{\prime}(t) \\
\frac{d}{d t}[f(t) \vec{a}(t)]=f(t) \vec{a}^{\prime}(t)+\vec{a}(t) f^{\prime}(t) \\
\frac{d}{d t}[\vec{a}(t) \cdot \vec{b}(t)]=\vec{a}^{\prime}(t) \cdot \vec{b}(t)+\vec{a}(t) \cdot \vec{b}^{\prime}(t) \\
\frac{d}{d t}[\vec{a}(t) \times \vec{b}(t)]=\vec{a}^{\prime}(t) \times \vec{b}(t)+\vec{a}(t) \times \vec{b}^{\prime}(t) \\
\frac{d}{d t}[\vec{a}(f(t))]=f^{\prime}(t) \vec{a}^{\prime}(f(t))
\end{gathered}
\end{equation}

Vector Integration

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

If \(\overrightarrow{r(t)}=\langle f(t), g(t), h(t)\rangle\), then \(\int_{a}^{b} \vec{r}(t) d t=\left\langle\int_{a}^{b} f(t) d t, \int_{a}^{b} g(t) d t, \int_{a}^{b} h(t) d t\right\rangle\).

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 \(\overrightarrow{r(t)}=\left\langle\cos 2 t, t e^{t}, t-t^{2}\right\rangle\).

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.

\begin{equation}
\int_{0}^{2} \vec{r}(t) d t=\langle\int_{0}^{2}(\cos 2 t) d t, \underbrace{\int_{0}^{2}\left(t e^{t}\right) d t}_{\begin{array}{c}
\text { use integration } \\
\text { by parts }
\end{array}}, \int_{0}^{2}\left(t-t^{2}\right) d t\rangle
\end{equation}

\begin{equation}
=\left.\left\langle\frac{1}{2} \sin 2 t, t e^{t}-e^{t}, \frac{1}{2} t^{2}-\frac{1}{3} t^{3}\right\rangle\right|_{0} ^{2}
\end{equation}

\begin{equation}
=\left\langle\frac{1}{2} \sin (4), 1+e^{2},-\frac{2}{3}\right\rangle
\end{equation}

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 \(\overrightarrow{r(t)}=\left\langle\cos 2 t, t e^{t}, t-t^{2}\right\rangle\).

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

\begin{equation}
\int \vec{r}(t) d t=\langle\int(\cos 2 t) d t, \underbrace{\int\left(t e^{t}\right) d t}_{\begin{array}{c}
\text { use integration } \\
\text { by parts }
\end{array}}, \int\left(t-t^{2}\right)\rangle
\end{equation}

\begin{equation}
=\left\langle\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}\right\rangle
\end{equation}

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 \(\vec{r}^{\prime}(t)\) is that \(\vec{r}^{\prime}(t)\) is tangent to the curve and the length of \(\vec{r}^{\prime}(t)\), denoted \(\left\|\vec{r}^{\prime}(t)\right\|\), is the speed. And since \(\vec{r}^{\prime}(t)\) points in the direction of motion and its length is the speed, we can call \(\vec{r}^{\prime}(t)\) the velocity vector and \(\vec{r}^{\prime \prime}(t)\) the acceleration vector.

And a particle whose position vector at the time \(t\) is \(\vec{r}(t)\) having velocity \(\vec{r}^{\prime}(t)\) and speed \(\left\|\vec{r}^{\prime}(t)\right\|\), the unit vector in the direction of \(\vec{r}^{\prime}(t)\) is:

\begin{equation}
T(t)=\frac{\vec{r}^{\prime}(t)}{\left\|\vec{r}^{\prime}(t)\right\|}
\end{equation}

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

Example

For example, if \(\overrightarrow{r(t)}=\left\langle\cos 2 t, t e^{t}, t-t^{2}\right\rangle\) let’s find the unit tangent vector at the point where \(t=0\).

Step 1

First, we will need to find \(\vec{r}^{\prime}(t)\). Thankfully, we calculated this earlier and found:

\begin{equation}
\vec{r}^{\prime}(t)=\left\langle-2 \sin 2 t, e^{t}(t+1), 1-2 t\right\rangle
\end{equation}

Step 2

Next, we will evaluate the velocity vector at the point where \(t=0\).

\begin{equation}
\vec{r}^{\prime}(0)=\left\langle-2 \sin 2(0), e^{0}(0+1), 1-2(0)\right\rangle=\langle 0,1,1\rangle
\end{equation}

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 \(\vec{r}^{\prime}(0)\).

\begin{equation}
\left\|\vec{r}^{\prime}(0)\right\|=\sqrt{(0)^{2}+(1)^{2}+(1)^{2}}=\sqrt{2}
\end{equation}

Step 4

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

\begin{equation}
T(t)=\frac{\vec{r}^{\prime}(t)}{\left|\vec{r}^{\prime}(t)\right| \mid}
\end{equation}

\begin{equation}
T(0)=\frac{\vec{r}^{\prime}(0)}{\left|\vec{r}^{\prime}(0)\right| \mid}=\frac{\langle 0,1,1\rangle}{\sqrt{2}}=\left\langle 0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right\rangle
\end{equation}

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!

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