Derivatives and Integrals of Vector Functions, multivariable and vector calculus lesson

Watch a moving point for a short time $\Delta t$ and draw the arrow from where it was to where it is now. Divide that arrow by $\Delta t$ and you have its average velocity as a vector. As $\Delta t$ shrinks, the arrow turns into the direction the point is heading at that instant. That limit is the derivative of a vector function, and it gives tangent lines to space curves the same way the ordinary derivative gave tangent lines to graphs.

Key terms

Componentwise differentiation and integration

Definition Let $\mathbf{r}$ be defined on an open interval containing $t$ . The derivative of $\mathbf{r}$ at $t$ is

\mathbf{r}'(t) = \lim_{h \to 0} \frac{\mathbf{r}(t + h) - \mathbf{r}(t)}{h},

if this limit exists. Here $h$ is the small change in the parameter, the $\Delta t$ of the opening picture, and $t$ stays fixed while $h \to 0$ . At an endpoint of a closed interval the limit is one-sided. Because vector limits are taken componentwise, if $\mathbf{r} = \langle x, y, z \rangle$ then $\mathbf{r}'(t)$ exists if and only if $x'(t)$ , $y'(t)$ and $z'(t)$ all exist, and then $\mathbf{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle$ . Similarly, if $x$ , $y$ and $z$ are integrable on $[a, b]$ (for example continuous), then

\begin{aligned} &\int_a^b \mathbf{r}(t)\,dt \\[0.4em] &\quad = \Big\langle \int_a^b x(t)\,dt, \; \int_a^b y(t)\,dt, \\[0.4em] &\qquad\quad \int_a^b z(t)\,dt \Big\rangle. \end{aligned}

In plain words Do the calculus you already know to each coordinate separately, then put the results back together.

Example For $\mathbf{r}(t) = \langle t^2, e^{2t}, \sin(\pi t) \rangle$ ,

\mathbf{r}'(t) = \langle 2t, \; 2e^{2t}, \; \pi\cos(\pi t) \rangle

and

\int_0^1 \mathbf{r}(t)\,dt = \Big\langle \frac{1}{3}, \; \frac{e^2 - 1}{2}, \; \frac{2}{\pi} \Big\rangle.

Tangent vector

Definition Let $\mathbf{r}$ parametrize a curve $C$ . If $\mathbf{r}'(t_0)$ exists and $\mathbf{r}'(t_0) \ne \mathbf{0}$ , then $\mathbf{r}'(t_0)$ is a tangent vector to $C$ at the point $\mathbf{r}(t_0)$ , pointing in the direction of the orientation. The tangent line to $C$ at that point is

\mathbf{L}(s) = \mathbf{r}(t_0) + s\,\mathbf{r}'(t_0), \qquad s \in \mathbb{R}.

If $\mathbf{r}$ is position and $t$ is time, $\mathbf{r}'(t)$ is the velocity vector.

In plain words The derivative is an arrow that sits at the point and points along the curve, the way you are moving.

Example On the twisted cubic $\mathbf{r}(t) = \langle t, t^2, t^3 \rangle$ , the derivative is $\mathbf{r}'(t) = \langle 1, 2t, 3t^2 \rangle$ . At $t = 1$ the point is $(1, 1, 1)$ and the tangent vector is $\langle 1, 2, 3 \rangle$ , so the tangent line is

\mathbf{L}(s) = \langle 1 + s, \; 1 + 2s, \; 1 + 3s \rangle.

Regular curves

Definition A parametrization $\mathbf{r}$ on an interval $I$ is regular if it satisfies both of these conditions.

It is continuously differentiable, written $C^1$ , on $I$ . That is, $\mathbf{r}'(t)$ exists at every $t$ in $I$ , one-sided at an endpoint that belongs to $I$ , and $\mathbf{r}'$ is continuous on $I$ .
Its derivative never vanishes, so $\mathbf{r}'(t) \ne \mathbf{0}$ for every $t$ in $I$ .

In this course the words smooth and regular mean exactly the same thing, a $C^1$ parametrization with nonvanishing derivative. Some books call any $C^1$ parametrization smooth, so check the definition a source uses. A curve is regular, or smooth, if it has a regular parametrization. A point where $\mathbf{r}'(t) = \mathbf{0}$ is called a singular point of the parametrization, and there the derivative gives no tangent direction.

In plain words A regular parametrization never stops, so it always has a well defined direction of travel. When the velocity drops to zero, the curve may have a sharp corner or cusp, or the point may simply have paused.

Example The plane curve $\mathbf{r}(t) = \langle t^3, t^2 \rangle$ has $\mathbf{r}'(t) = \langle 3t^2, 2t \rangle$ , which is $\mathbf{0}$ only at $t = 0$ . The curve is $y = x^{2/3}$ , which has a cusp at the origin. On the other hand, $\langle t^3, t^3 \rangle$ also has zero velocity at $t = 0$ , yet its trace is the straight line $y = x$ , which is regular when parametrized as $\langle t, t \rangle$ . A zero derivative is a warning, not a proof of a corner.

Unit tangent

Definition At a point where $\mathbf{r}'(t)$ exists and is not $\mathbf{0}$ , the unit tangent vector is

\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\lvert \mathbf{r}'(t) \rvert}.

It has length 1 and points in the direction of increasing $t$ . For a regular parametrization it is defined and continuous on the whole interval. Reversing the orientation replaces $\mathbf{T}$ by $-\mathbf{T}$ at each point.

In plain words Keep the direction of the velocity and throw away its size.

Example For the helix $\mathbf{r}(t) = \langle 2\cos t, 2\sin t, t \rangle$ , the velocity is $\mathbf{r}'(t) = \langle -2\sin t, 2\cos t, 1 \rangle$ with length $\sqrt{4\sin^2 t + 4\cos^2 t + 1} = \sqrt{5}$ . So

\mathbf{T}(t) = \frac{1}{\sqrt{5}} \langle -2\sin t, \; 2\cos t, \; 1 \rangle,

and at $t = 0$ it equals $\frac{1}{\sqrt{5}} \langle 0, 2, 1 \rangle$ .

Product rules for dot and cross products

Definition Let $\mathbf{u}$ and $\mathbf{v}$ be vector functions and $f$ a real function, all differentiable at $t$ . Then

\begin{aligned} (f\mathbf{u})' &= f'\mathbf{u} + f\mathbf{u}', \\[0.4em] (\mathbf{u} \cdot \mathbf{v})' &= \mathbf{u}' \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{v}', \\[0.4em] (\mathbf{u} \times \mathbf{v})' &= \mathbf{u}' \times \mathbf{v} + \mathbf{u} \times \mathbf{v}', \end{aligned}

all evaluated at $t$ . In the cross product rule the order of the factors must be kept, since $\mathbf{a} \times \mathbf{b} = -\mathbf{b} \times \mathbf{a}$ . The chain rule also holds. If $f$ is differentiable at $s$ and $\mathbf{u}$ is differentiable at $f(s)$ , then $\frac{d}{ds}\mathbf{u}(f(s)) = f'(s)\,\mathbf{u}'(f(s))$ . A useful consequence is that if $\lvert \mathbf{r}(t) \rvert$ is constant on an interval and $\mathbf{r}$ is differentiable there, then differentiating $\mathbf{r} \cdot \mathbf{r}$ gives $2\,\mathbf{r} \cdot \mathbf{r}' = 0$ , so $\mathbf{r}'(t)$ is perpendicular to $\mathbf{r}(t)$ .

In plain words The usual product rule works for every kind of product, as long as the cross product keeps each factor on its own side.

Example Let $\mathbf{u}(t) = \langle t, 1, t^2 \rangle$ and $\mathbf{v}(t) = \langle 1, t, 0 \rangle$ . Directly,

\mathbf{u} \times \mathbf{v} = \langle -t^3, \; t^2, \; t^2 - 1 \rangle,

whose derivative at $t = 1$ is $\langle -3, 2, 2 \rangle$ . With the rule, at $t = 1$ we have $\mathbf{u} = \langle 1, 1, 1 \rangle$ , $\mathbf{u}' = \langle 1, 0, 2 \rangle$ , $\mathbf{v} = \langle 1, 1, 0 \rangle$ and $\mathbf{v}' = \langle 0, 1, 0 \rangle$ , so

\begin{aligned} &\mathbf{u}' \times \mathbf{v} + \mathbf{u} \times \mathbf{v}' \\[0.4em] &\quad = \langle -2, 2, 1 \rangle + \langle -1, 0, 1 \rangle \\[0.4em] &\quad = \langle -3, 2, 2 \rangle. \end{aligned}

The dot product rule checks just as easily. Here $\mathbf{u} \cdot \mathbf{v} = 2t$ , with derivative $2$ , and at $t = 1$ the rule gives

\begin{aligned} &\langle 1, 0, 2 \rangle \cdot \langle 1, 1, 0 \rangle \\[0.4em] &\quad + \langle 1, 1, 1 \rangle \cdot \langle 0, 1, 0 \rangle \\[0.4em] &\quad = 1 + 1 = 2. \end{aligned}

Vector antiderivative

Definition A vector function $\mathbf{R}$ is an antiderivative of $\mathbf{r}$ on an interval $I$ if $\mathbf{R}'(t) = \mathbf{r}(t)$ for every $t$ in $I$ . Any two antiderivatives of $\mathbf{r}$ on the same interval differ by a constant vector $\mathbf{C}$ , so the indefinite integral is $\int \mathbf{r}(t)\,dt = \mathbf{R}(t) + \mathbf{C}$ . If $\mathbf{r}$ is continuous on $[a, b]$ and $\mathbf{R}$ is an antiderivative of $\mathbf{r}$ on $[a, b]$ , the fundamental theorem of calculus applied to each component gives

\int_a^b \mathbf{r}(t)\,dt = \mathbf{R}(b) - \mathbf{R}(a).

An initial condition $\mathbf{r}(t_0) = \mathbf{r}_0$ at a point of the interval fixes $\mathbf{C}$ , one constant for each component.

In plain words Undo the derivative in each coordinate, and add one unknown constant vector. A starting position pins it down.

Example Find $\mathbf{r}$ on $\mathbb{R}$ with $\mathbf{r}'(t) = \langle 2t, \cos t, e^{-t} \rangle$ and $\mathbf{r}(0) = \langle 1, 0, 2 \rangle$ . Antidifferentiating gives

\mathbf{r}(t) = \langle t^2, \; \sin t, \; -e^{-t} \rangle + \mathbf{C}.

At $t = 0$ this is $\langle 0, 0, -1 \rangle + \mathbf{C} = \langle 1, 0, 2 \rangle$ , so $\mathbf{C} = \langle 1, 0, 3 \rangle$ and

\mathbf{r}(t) = \langle t^2 + 1, \; \sin t, \; 3 - e^{-t} \rangle.

Check yourself

Find the tangent line to $\mathbf{r}(t) = \langle e^t, \, t e^t, \, t^2 \rangle$ at $t = 0$ .

Show the answer

The derivative is $\mathbf{r}'(t) = \langle e^t, (1 + t)e^t, 2t \rangle$ , so $\mathbf{r}'(0) = \langle 1, 1, 0 \rangle$ , which is not zero. The point is $\mathbf{r}(0) = \langle 1, 0, 0 \rangle$ , and the tangent line is $\mathbf{L}(s) = \langle 1 + s, s, 0 \rangle$ .

A ball has acceleration $\mathbf{r}''(t) = \langle 0, 0, -10 \rangle$ , initial velocity $\mathbf{r}'(0) = \langle 3, 2, 4 \rangle$ and initial position $\mathbf{r}(0) = \langle 0, 0, 1 \rangle$ . Find $\mathbf{r}(t)$ .

Show the answer

One antiderivative and the initial velocity give $\mathbf{r}'(t) = \langle 3, 2, 4 - 10t \rangle$ . A second one and the initial position give

\mathbf{r}(t) = \langle 3t, \; 2t, \; 1 + 4t - 5t^2 \rangle.

Where does $\mathbf{r}(t) = \langle t^2 - 2t, \, t^3 - 3t, \, 0 \rangle$ fail to be regular?

Show the answer

The derivative is $\mathbf{r}'(t) = \langle 2t - 2, 3t^2 - 3, 0 \rangle$ . The first two entries vanish together only at $t = 1$ , so the parametrization is regular on any interval that avoids $t = 1$ and has a singular point at $t = 1$ , the point $(-1, -2, 0)$ .

Derivatives and Integrals of Vector Functions

Key terms

Componentwise differentiation and integration

Tangent vector

Regular curves

Unit tangent

Product rules for dot and cross products

Vector antiderivative

Check yourself

How this lesson connects