Why vector-valued calculus matters

The page Vector-Valued Functions explains that a vector-valued function is a map

\mathbf{r}(t) = (x(t), y(t), z(t))

whose components are ordinary scalar functions. That is the key idea behind calculus in this setting: when we differentiate or integrate a vector-valued function, we do the operation one coordinate at a time.

This is the same philosophy already used for limits in vector-valued functions: the limit is taken coordinate-wise, so the calculus rules are also coordinate-wise.

Derivative

Definition

Let

\mathbf{r}(t) = (x(t), y(t), z(t)).

The derivative is defined by the usual difference quotient:

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

Because the limit is taken coordinate-wise, this becomes

\mathbf{r}'(t) = (x'(t), y'(t), z'(t)).

So the derivative of a vector-valued function is just the vector of derivatives of its component functions.

Write the difference quotient component by component:

\frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h} = \left( \frac{x(t+h)-x(t)}{h}, \frac{y(t+h)-y(t)}{h}, \frac{z(t+h)-z(t)}{h} \right).

If each scalar limit exists, then taking the limit as $h\to 0$ gives

\mathbf{r}'(t) = (x'(t), y'(t), z'(t)).

So the vector limit exists exactly when all the component limits exist.

Meaning of the derivative

If $\mathbf{r}(t)$ is the position of a moving particle, then $\mathbf{r}'(t)$ is the velocity vector. Its direction is tangent to the path, and its magnitude is the speed.

If $\mathbf{r}'(t) = \mathbf{0}$ at some time, the particle is momentarily at rest.

Circle example in 2D

Consider the unit circle

\mathbf{r}(t) = (\cos t, \sin t), \quad 0 \le t \le 2\pi.

Its derivative is

\mathbf{r}'(t) = (-\sin t, \cos t).

The graph below shows the circle with four sample points and tangent vectors drawn at each point.

Unit circle: r(t) = (cos t, sin t) with tangent vectors

At the four cardinal points, the derivative vectors are:

$t$	point on the circle	$\mathbf{r}'(t)$	tangent direction
$0$	$(1,0)$	$(0,1)$	upward
$\pi/2$	$(0,1)$	$(-1,0)$	left
$\pi$	$(-1,0)$	$(0,-1)$	downward
$3\pi/2$	$(0,-1)$	$(1,0)$	right

These are exactly the tangent directions to the circle at those points.

Sphere speciality: tangent vectors are perpendicular to the position vector

The same idea works for a sphere centered at the origin. Let

\mathbf{r}(t)

be a curve lying on a sphere of radius $R$ so that every point has constant distance from the origin:

\|\mathbf{r}(t)\| = R.

Equivalently,

\mathbf{r}(t) \cdot \mathbf{r}(t) = R^2.

Differentiate both sides with respect to $t$ :

\frac{d}{dt}\bigl(\mathbf{r}(t)\cdot\mathbf{r}(t)\bigr) = 0.

Using the product rule for the dot product,

\mathbf{r}'(t)\cdot\mathbf{r}(t) + \mathbf{r}(t)\cdot\mathbf{r}'(t) = 0.

Since the dot product is symmetric,

2\,\mathbf{r}(t)\cdot\mathbf{r}'(t)=0,

\mathbf{r}(t)\cdot\mathbf{r}'(t)=0.

That means the tangent vector is perpendicular to the position vector.

This is the geometric reason the radius of a circle or sphere is normal to the tangent direction:

the position vector has constant length,
constant length means the dot product $\mathbf{r}\cdot\mathbf{r}$ is constant,
differentiating forces $\mathbf{r}$ and $\mathbf{r}'$ to be orthogonal.

The same proof works for any sphere centered at $(a,b,c)$ by replacing $\mathbf{r}(t)$ with $\mathbf{r}(t)-(a,b,c)$ .

General 2D curve and tangent vector picker

For a general parametric curve

\mathbf{r}(t) = (x(t), y(t)),

the tangent vector at $t=t_0$ is

\mathbf{r}'(t_0) = (x'(t_0), y'(t_0)).

The graph below shows a standard test curve with a sample point and its tangent vector. The displayed sample point can be changed by editing the sample parameter list in the note source, which is the current way to "pick" the point in this renderer.

A curve with a selected sample point and tangent vector

For this curve,

x(t)=t, \qquad y(t)=t^3-3t,

\mathbf{r}'(t) = (1, 3t^2-3).

At the chosen point $t=0.75$ ,

\mathbf{r}(0.75) = (0.75, -1.828125), \qquad \mathbf{r}'(0.75) = (1, -1.3125).

If you want a different point, change the sample parameter and recompute $\mathbf{r}'(t_0)$ .

Near $t=t_0$ , the curve is well approximated by

\mathbf{r}(t) \approx \mathbf{r}(t_0) + (t-t_0)\mathbf{r}'(t_0).

This is the vector-valued version of the tangent line approximation.

General 3D curves

As an example in three dimensions, consider a right-handed helix. A simple parametrization is

\mathbf{r}(t) = (\cos t,\; \sin t,\; t/(2\pi)), \qquad 0\le t\le 2\pi.

The plot below shows one turn of the helix with several sample points. The tangent vectors at those sample points are drawn as small cones (arrowheads) anchored at the curve; the cones show the direction of the derivative $\mathbf{r}'(t)$ in 3D.

Helix: r(t) = (cos t, sin t, t/(2\pi))

Arithmetic properties of the derivative

Let $\mathbf{u}(t)$ and $\mathbf{v}(t)$ be vector-valued functions, and let $c$ be a constant scalar. Then:

(\mathbf{u}+\mathbf{v})' = \mathbf{u}' + \mathbf{v}'

(c\mathbf{u})' = c\mathbf{u}'

(f\mathbf{u})' = f'\mathbf{u} + f\mathbf{u}'

when $f(t)$ is a scalar function.

For the dot product,

(\mathbf{u}\cdot\mathbf{v})' = \mathbf{u}'\cdot\mathbf{v} + \mathbf{u}\cdot\mathbf{v}'.

For the cross product in $\mathbb{R}^3$ ,

(\mathbf{u}\times\mathbf{v})' = \mathbf{u}'\times\mathbf{v} + \mathbf{u}\times\mathbf{v}'.

Useful special cases:

(\mathbf{u}\cdot\mathbf{u})' = 2\,\mathbf{u}\cdot\mathbf{u}'

and if $\mathbf{c}$ is a constant vector, then

\mathbf{c}' = \mathbf{0}.

These are the vector-calculus analogues of the familiar product rule from single-variable calculus.

Integration

Integration is also taken component-wise.

\mathbf{r}(t) = (x(t), y(t), z(t)),

then

\int \mathbf{r}(t)\,dt = \left(\int x(t)\,dt,\; \int y(t)\,dt,\; \int z(t)\,dt\right) + \mathbf{C},

where $\mathbf{C}$ is a constant vector.

Similarly, for a definite integral,

\int_a^b \mathbf{r}(t)\,dt = \left( \int_a^b x(t)\,dt, \int_a^b y(t)\,dt, \int_a^b z(t)\,dt \right).

So integration of vector-valued functions is not a new operation; it is the usual integral applied to each component separately.

Summary

A vector-valued derivative is defined by the usual limit of the difference quotient.
The limit works coordinate-wise, so the derivative is computed component-wise.
The derivative vector points along the tangent direction of the curve.
On a circle or sphere, the tangent vector is perpendicular to the position vector because the radius has constant length.
The main arithmetic rules are the sum rule, scalar rule, dot-product rule, and cross-product rule.
Integration is also component-wise.

Calculus of Vector-Valued Functions