Walkthrough

A curve is a moving point

A vector-valued function gives a point for each value of a parameter. In three dimensions, write

The three component functions tell you where the point is along each axis. As $t$ changes, the point moves and leaves a curve behind. It is useful to read $t$ as time and $\mathbf r(t)$ as position at time $t$.

For example,

moves around the unit circle. The identity $\cos^2t+\sin^2t=1$ tells us which path it follows. The order of the points tells us something extra: as $t$ increases, the point travels counterclockwise.

The path is not the whole story

Now compare

with

Both trace the same geometric path, and both move counterclockwise. The second angle grows twice as quickly, though, so $\mathbf F$ goes around the circle twice as fast. A parametrization records the path, its direction, and how quickly the point moves along it.

A parameter gives a curve its direction

The derivative straightens the curve nearby

Take two nearby points on a curve, $\mathbf r(a)$ and $\mathbf r(a+h)$. Their difference

is the displacement from the first point to the second. Dividing by $h$ turns it into an average rate of change:

As $h$ approaches zero, the two points come together and the secant direction settles toward the tangent direction. When the limit exists,

This derivative is a vector. It points along the curve and its length measures speed.

Keep direction and speed separate

If $\mathbf r'(t)\ne\mathbf0$, the derivative gives a tangent direction. The tangent line through the point $\mathbf r(t_0)$ is

Here $t_0$ chooses the point on the curve, while $s$ moves along the line. If we only want direction, divide the derivative by its length:

The result $\mathbf T(t)$ is the unit tangent vector. It points the same way as $\mathbf r'(t)$ but has length $1$.

A secant direction becomes a tangent direction

Vector-valued functions and parametric curves

A vector-valued function takes a real number as input and returns a vector. A function into three-dimensional space has the form

$$\mathbf r:\mathbb R\to\mathbb R^3,$$

with

$$\boxed{ \mathbf r(t)=\langle x(t),y(t),z(t)\rangle =x(t)\mathbf i+y(t)\mathbf j+z(t)\mathbf k. }$$

The functions $x(t)$, $y(t)$, and $z(t)$ are its component functions, while $\mathbf i$, $\mathbf j$, and $\mathbf k$ are the standard unit basis vectors. The same idea works in $\mathbb R^2$ or $\mathbb R^n$.

The associated parametric curve is the set of points reached as the parameter varies:

$$\boxed{ \{\mathbf r(t):t\in\mathbb R\} =\{(x(t),y(t),z(t)):t\in\mathbb R\}. }$$

The curve as a point set tells us where the motion goes. The parametrization also tells us the order in which the points are visited.

Reading a parametrization

A useful routine is:

1. Write out the component functions.
2. Plot a few easy parameter values.
3. Eliminate $t$ when an identity reveals a familiar surface or curve.
4. Follow increasing values of $t$ to find the direction of travel.

Example: the unit circle

Let

$$\mathbf r(t)=\langle\cos t,\sin t\rangle.$$

If $x=\cos t$ and $y=\sin t$, then

$$x^2+y^2=\cos^2t+\sin^2t=1.$$

So the image is the unit circle. Starting at $t=0$, the points

$$(1,0),\quad(0,1),\quad(-1,0),\quad(0,-1)$$

appear in that order, so the direction is counterclockwise.

Example: the helix

Consider

$$\mathbf r(t)=\langle\cos t,\sin t,t\rangle.$$

Five easy values show one full turn:

The first two components satisfy

$$x^2+y^2=\cos^2t+\sin^2t=1,$$

so the curve stays on the unit cylinder around the $z$-axis. At the same time, $z=t$ rises steadily. Circular motion plus vertical motion produces a helix.

Example: a curve that is easier to graph by computer

Some component formulas do not reveal their shape quickly. For example,

$$\begin{aligned} x(t)&=\left(2+\cos\left(\frac{3t}{2}\right)\right)\cos t,\\ y(t)&=\left(2+\cos\left(\frac{3t}{2}\right)\right)\sin t,\\ z(t)&=\sin\left(\frac{3t}{2}\right). \end{aligned}$$

The components still define the curve exactly, but a reliable sketch needs many points. This is where graphing software is more useful than trying to guess the three-dimensional shape from a handful of values.

Limits and continuity

Limits of vector-valued functions are taken one component at a time. If

$$\mathbf r(t)=\langle x(t),y(t),z(t)\rangle,$$

then

$$\boxed{ \lim_{t\to a}\mathbf r(t) = \left\langle \lim_{t\to a}x(t), \lim_{t\to a}y(t), \lim_{t\to a}z(t) \right\rangle }$$

whenever all three component limits exist. If even one component has no limit, the vector limit does not exist.

The function is continuous at $a$ when

$$\boxed{ \lim_{t\to a}\mathbf r(t)=\mathbf r(a). }$$

Because vector equality is componentwise, this is equivalent to $x$, $y$, and $z$ each being continuous at $a$.

Derivatives of vector-valued functions

The derivative uses the same difference quotient as ordinary calculus, except that the numerator is now a vector:

$$\boxed{ \mathbf r'(a) = \lim_{h\to0} \frac{\mathbf r(a+h)-\mathbf r(a)}{h}. }$$

If

$$\mathbf r(t)=\langle x(t),y(t),z(t)\rangle,$$

then

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

Taking the limit componentwise gives the main rule:

$$\boxed{ \mathbf r'(t)=\langle x'(t),y'(t),z'(t)\rangle. }$$

So a vector-valued function is differentiable when its component functions are differentiable. The calculation is familiar; what changes is the geometric meaning of the answer.

Tangent vectors, tangent lines, and speed

For a differentiable curve, the derivative $\mathbf r'(t)$ is its velocity vector. It contains two pieces of information:

$$\text{speed}=\lVert\mathbf r'(t)\rVert,$$

while the direction comes from the derivative itself. When $\mathbf r'(t)\ne\mathbf0$, it is a tangent vector at $\mathbf r(t)$.

For the helix $\mathbf r(t)=\langle\cos t,\sin t,t\rangle$ discussed earlier, its derivative is

$$\mathbf r'(t)=\langle-\sin t,\cos t,1\rangle.$$

The speed is constant because

$$\lVert\mathbf r'(t)\rVert =\sqrt{\sin^2t+\cos^2t+1} =\sqrt2.$$

At a fixed parameter value $t_0$, the tangent line is

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

Any nonzero scalar multiple of $\mathbf r'(t_0)$ describes the same geometric line. It only changes how the line parameter $s$ runs along it.

The unit tangent vector removes the speed:

$$\boxed{ \mathbf T(t) =\frac{\mathbf r'(t)}{\lVert\mathbf r'(t)\rVert} }$$

provided $\mathbf r'(t)\ne\mathbf0$. The nonzero condition matters because the zero vector has no direction and cannot be normalized.

Comparing two parametrizations at the same point

Example: the usual unit-circle parametrization

For

$$\mathbf r(t)=\langle\cos t,\sin t\rangle,$$

we have

$$\mathbf r'(t)=\langle-\sin t,\cos t\rangle.$$

At $t=\pi$,

$$\mathbf r(\pi)=\langle-1,0\rangle, \qquad \mathbf r'(\pi)=\langle0,-1\rangle.$$

The derivative already has length $1$, so

$$\mathbf T(\pi)=\langle0,-1\rangle.$$

The tangent line is therefore

$$\boxed{ \mathbf L(s)=\langle-1,0\rangle+s\langle0,-1\rangle, \qquad s\in\mathbb R. }$$

This is the vertical line $x=-1$.

Example: the same circle at a different speed

Now use

$$\mathbf F(t)=\langle\cos(2t),\sin(2t)\rangle.$$

This is the same geometric circle with the same counterclockwise direction, but its derivative is

$$\mathbf F'(t)=\langle-2\sin(2t),2\cos(2t)\rangle.$$

The point $(-1,0)$ occurs sooner, at $t=\pi/2$:

$$\mathbf F(\pi/2)=\langle-1,0\rangle, \qquad \mathbf F'(\pi/2)=\langle0,-2\rangle.$$

The tangent vector points downward as before, but it is twice as long:

$$\lVert\mathbf F'(\pi/2)\rVert=2.$$

After normalization,

$$\boxed{ \mathbf T(\pi/2)=\langle0,-1\rangle. }$$

Both parametrizations reach the same point on the same path and have the same unit tangent direction. Their velocity vectors differ because the second motion has speed $2$ instead of speed $1$.