We recap from the previous section that when

we have

I have previously shown that
is in the direction of the tangent to the trajectory at the point
defined by parameter t. I have then later shown that if we parameterize the curve in terms of its arc length s along the curve from some initial point, then
has unit length and thus is the unit tangent vector. In cases where it is inconvenient to introduce the variable s as finding the equation
may prove difficult, we obtain the unit tangent vector by dividing
by its length. And so we can then defined the unit tangent vector
as,

provided that
. If
, we do not define a tangent vector to the curve at
. If
, and this equation gives us
, which again is consistent with our previous definitions.
We will now define the new term. The curvature
of a curve in three-space is the magnitude of the rate of change of the tangent vector with respect to s,

The Greek symbol
is pronounced as kappa. By looking at the equation, we suspect that the value of
gives us the value of what we call the “amount of bending” or curvature. For example, the greater the magnitude of
, the more the curve bends for a given change in arc length. A straight line has a constant tangent vector and so it’s curvature is 0. By using calculus, we can show that in terms of the position vector
,

assuming that
. We will rarely use the above equation.
Alongside the term curvature, we define another term. The quantity
, the reciprocal of the curvature, is called the radius of curvature of a curve, provided that
. The Greek symbol
is called rho. It’s value gives us the radius of a circle that best approximates the curve C at the point P on its concave side. Such a circle is called the osculating circle to the curve at P.

Next, we shall use the curvature and radius of curvature to define another two terms. The unit normal vector to the curve at the point P is given by

We can verify that this is a unit vector, by simply taking its magnitude like so

Morever, N is orthogonal to T. To prove this, we recall a basic property of the dot product which is the a vector dot by itself is equal to it’s magnitude squared. Thinking along this lines and differentiating the equation, we get

But since
is orthogonal to
, and
is a positive scalar multiple of
and so in the same direction as
, we conclude that
is orthogonal or perpendicular to
.
Next, we shall see how we can write the acceleration vector in terms of these two components,
and
.
All information presentated, less questions and exercises, is original content of Donny, with slight references to various books.
Video courtesy of YouTube.com service.