Subject: Calculus

# Curvature

Suppose we are given with a function, in this case is a distance as function of time, defined as follows:

\mathbf{\bar{F}(t)}=x(t)\mathbf{i}+y(t)\mathbf{j}+z(t)\mathbf{k}

the first and second derivative would yield the velocity and acceleration which are:

\mathbf{\bar{v}(t)}=\mathbf{\bar{F}'(t)}=x'(t)\mathbf{i}+y'(t)\mathbf{j}+z'(t)\mathbf{k}

\mathbf{\bar{F}(t)} is in the direction of the tangent to the trajectory at the point x(t),y(t),z(t) defined by parameter t. If we parameterize the curve in terms of its arc length s along the curve from some initial point, then \mathbf{\bar{F}'(s)} 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 t=t(s) may prove difficult, we obtain the unit tangent vector by dividing \mathbf{\bar{F}'(t)} by its length. And so we can then defined the unit tangent vector \mathbf{\bar{T}'(t)} as,

\mathbf{\bar{T}'(t)}=\frac{1}{\left \|\mathbf{\bar{F}'(t)}\right \|}\mathbf{\bar{F}'(t)}

provided that \mathbf{\bar{T}'(t)} \neq 0 . If \mathbf{\bar{F}'(t)}=0 , we do not define a tangent vector to the curve at x(t),y(t),z(t). If t = s, \left \|\mathbf{\bar{F}'(s)}\right \|=1, and this equation gives us \mathbf{\bar{T}'(s)} =\mathbf{\bar{F}'(s)} , which again is consistent with our previous definitions.

Now it is time for us to define the curvature \kappa of a curve in three-space is the magnitude of the rate of change of the tangent vector with respect to s,

\kappa =\left \|\frac{d\mathbf{\bar{T}}}{ds}\right \|

The Greek symbol \kappa is pronounced as kappa. By looking at the equation, we suspect that the value of \kappa gives us the value of what we call the �amount of bending� or curvature. For example, the greater the magnitude of \mathbf{\bar{T}'(s)}, 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 \mathbf{\bar{F}'(t)},

\kappa =\frac{\left \|\mathbf{\bar{F}}' \times \mathbf{\bar{F}}'' \right \|}{\left \| \mathbf{\bar{F}}' \right \|^3}

assuming that \mathbf{\bar{F}'(t)} = 0. We will rarely use the above equation.

Alongside the term curvature, we define another term. The quantity \rho =\frac{1}{\kappa}, the reciprocal of the curvature, is called the radius of curvature of a curve, provided that \kappa \neq 0. The Greek symbol \rho 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

\mathbf{\bar{N}(s)}=\kappa \left \|\frac{d\mathbf{\bar{T}}}{ds}\right \|

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

\mathbf{\bar{N}(s)}=\kappa \left \|\frac{d\mathbf{\bar{T}}}{ds}\right \| = \frac{1}{\kappa }\left \|\frac{d\mathbf{\bar{T}}}{ds}\right \|

Moreover, \mathbf{N} is orthogonal to \mathbf{T}. To prove this, we recall a basic property of the dot product which is a vector dot by itself is equal to its magnitude squared. Thinking along these lines and differentiating the equation, we get

\mathbf{\bar{T}}\cdot \mathbf{\bar{T}} = \left \|\mathbf{\bar{T}}\right \|^2 = 1
\frac{d}{ds} \mathbf{\bar{T}}\cdot \mathbf{\bar{T}} = \mathbf{\bar{T}}'(s) \cdot \mathbf{\bar{T}}(s) + \mathbf{\bar{T}}(s)\cdot \mathbf{\bar{T}}'(s)=\frac{d}{ds}(1)=0
\mathbf{\bar{T}}'(s)\cdot \mathbf{\bar{T}}(s)=0

But since \mathbf{\bar{T}}(s) is orthogonal to \mathbf{\bar{T}}'(s), and \mathbf{\bar{N}}(s)is a positive scalar multiple of \mathbf{\bar{T}}'(s) and so in the same direction as \mathbf{\bar{T}}'(s), we conclude that \mathbf{\bar{N}}(s) is orthogonal or perpendicular to \mathbf{\bar{T}}'(s).

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