Why do we need vector functions

In describing motion of objects, we often cannot rely on our old one-variable functions. This is the case, even if a motion of an object is in the plane. For example the motion in the following two examples should be described using vector functions.

Cycloid: Suppose that a wheel of radius r is rolling in a straight line without slipping. Let P be a fixed point on the circumference of the wheel. As the wheel rotates, the point P traces a curve known as a cycloid. See the animation below.

Hypocycloid: Suppose that a wheel of radius r is rolling internally on a larolling internally on a larger circle without slipping. Let P be a fixed point on the circumference of the wheel. As the wheel rotates, the point P traces a curve known as a hypocycloid. See the animation below.

Graphing vector functions

In order to graph a vector function, in most cases we have to evaluate the function in several points, and plot those points. The more points we have, the more accurate graph we obtain. Sometimes, we can find and recognize the relationship between x, y, and z coordinates. For example the x and the z coordinates of a vector function defined by , are related by the following equation: , which is the equation of an ellipse in xz-plane. This tell us that the graph of f(t), is a spiral that traverses along the elliptical cylinder. See the animation below.

Derivatives of vector functions

A derivative of a vector function, is also a vector function, and it is obtained from the original function by simply differentiating the component functions. For example the derivative of f(t) is . The following animation shows the graphs of f(t) and , as vectors.

Unit Tangent Vector

The vector is also a tangent vector, that is, it is parallel to the tangent line passing through the point . We would also like to find a vector that is perpendicular to the tangent line, and moreover this vector should point in the direction of the function's bend. This would heion's bend. This would help us to better visualize the graph of . In order to find a perpendicular vector to the tangent line, we use the fact that if a differentiable vector function, g(t), has a constant magnitude, that is |g(t)|=c where c is a constant, then g'(t) and g(t) are perpendicular to each other. This follows easily by differentiating the equation , to obtain . Since the product on the left-hand side is actually the dot product, we see that g and g' are perpendicular. Now, going back to our goal of finding a vector that is perpendicular to the tangent line, we need a vector that is parallel to the tangent line. We have seen that , f'(t) is parallel to the tangent line. Unfortunately, f'(t) is not necessarily of a constant magnitude. For example for the function defined by we have , and thus

which clearly is not a constant. A simplest way to find a constant vector parallel to the tangent line is to make a unit vector of f'(t). This is exactly what we are going to do, and the corresponding vector is called the unit tangent vector and it is denoted by T(t). Thus . For the function , the unit tangent vector T(t) is equal to .

Unit normal vector

Now that we have a tangent vector of a constant length, (always etant length, (always equal to one), we will differentiate this expression to obtain a vector that is perpendicular to T(t). Therefore the vector T'(t)= + + is one such vector. The unit vector of T'(t) is called the unit normal vector, and it is denoted by n(t). Thus, n(t)= . For our example we obtain n(t)={ + + }/ . That the unit normal vector also points in the direction of the function's bend is far from clear. On the other hand n(t) is derived by looking at the second derivative of f(t), and if you recall that in one-variable case the second derivative points at function's bend (that is points to its concavity) this result should come as no surprise. The following graph illustrates this. The function is in red, T(t) is in green, n(t) is in blue colors, and they are given for six different points on the graph.