A function of type defined on a set of real numbers (with vector outputs) is said to be a vector-valued function. For example is a vector-valued function.

Consider the function . This function can be written as

(1)

which represents a parametric equation of a line. Any vector output of this function lands on a point on the above line.

Consider . This equation can be written as . Using the fact that

(2)

we see that the function provides the set of all points on the unit circle that is the terminal points of all vectors with initial point at the origin. The picture below shows that when the function provides the vector .

Consider This equation can be written as From previous example, we saw that in -plane, provides the unit circle and for this example, for any given -value (or -value), we obtain a circle. This will give rise to a cylinder which is upward parallel to the -axis. Now, as we know that we are looking at exactly one point on each of these circles cut from the cylinder. For example, when the vector provided by is

(3)

This identifies a vector with initial point at the origin and terminal point on the unit circle on the plane and coordinate This way we get a curve on the body of the cylinder The direction of travel as increases will be in the direction of travel on the unit circle that is counterclockwise (obviously upward since increases!)

The concepts of limit, differentiation and integration are straight forward and are done component-wise with respect to vector-valued functions. Consider a vector-valued function as below Take two points and on the curve and use and to show them. Of course a difference vector, shown in yellow, like passes the points and Now if one chooses small, the difference vector which passes and becomes closer to the tangent vector (in the sense of direction shown in red) at

This may not always be the case; for example, a tangent vector may not exist but when it does and it’s a nonzero vector, we have an intuition how is defined. In fact, it’s easy to show that the differentiation for a vector valued function is done component-wise. In application, the vector is called the position vector, is called the velocity vector and is called the acceleration vector. This makes sense, since shows the position of an article based on time on the curve (analogous interpretation for and ). Speed of the particle is obtained by evaluating the magnitude of at any time.

Consider two vector-valued functions and . We would like to find the dot product of the two vectors that is

(4)

Next we compute the derivative of and note that this is a function depending on a variable :

(5)

Given the vectors and as the above example, we would like to find

(6)

Next, we like to find the derivative of this cross product so we differentiate each component

(7)

The vector is said to be the unit tangent vector and shown by . Suppose exists. We want to show that and are orthogonal to each other. Observe that if we show

(8)

we are done. We know

(9)

So

(10)

But

(11)

and

(12)

Now we are done. Among many possibilities of vectors that could be orthogonal to one is important for us. We define the principle unit normal vector by

(13)

Given we compute the unit tangent vector

(14)

Next, we find the principle unit normal vector

(15)

Suppose we have a smooth curve defined by . The fact that how quickly or how slowly bends (changing direction) at every point is determined by a concept called the curvature of ; it’s shown by There are many ways to define and to compute it. The formulas that we use for the next examples are

(16)

and for curves in -plane, ,

(17)

An interesting fact about curvature is its relationship and the osculating circle. Imagine a circle that is attached, in fact tangent, to a point of a given smooth curve. There could be many such circles but for the reason of understanding, the one that is the closest to be tangent at two points next to each other is called the osculating circle. Now suppose is the radius of such circle. Then

(18)

This is a well-known relation between the curvature and radius of the osculating circle at a point on a smooth curve. Note how direction changes in the picture. The inverse relation between curvature and radius of the osculating circle dictates that the more quickly a curve bends, radius of the osculating circle will be smaller.

We find the curvature of at Note that using the result of previous example

(19)

We are given a vector-valued function .To compute the curvature for this curve we use the formula

(20)

We need to find

(21)

And finally the curvature is provided by

(22)

Note that, at the curvature is equal to zero and when the curvature is equal to .

Suppose we would like to find the curvature of the unit circle at any point. Of course, one can use the relationship of the osculating circle and curvature but we want to attempt with a different formula. A point is on the unit circle when . We need and . So differentiate implicitly

(23)

We know curvature is

(24)

Consider a curve in space Let use take two points on this curve and associated to and . Make a partition of this curve with respect to say all of equal size This can be shown as followed. We call each endpoint of the partition

Observe that as shown in the picture, the distance between and is

(25)

The length of the curve can be approximated using a sum. In the third step we multiply and divide by

(26)

Note that using Mean Value Theorem guarantees that for some values

(27)

Now we can have as large as we like to make this approximation, which call arc length to be exact

(28)

Now looking closer we see that the notion is the magnitude of so

(29)