Example for CurvatureFirst, we'll finish up curvature by doing an example:
Find K at (0,0) for r(t) = ti + t^2j
v(t) = i + 2tj
a = 2j
v x a = (i + 2tj) x (2j) = 2k
|v|^3 = (1 + 4t^2) ^ (3/2)
= |Curvature v x a| /|v|^3= 2/(1 + 4t^2)^(3/2)At origin, t=0, K=2, Rho=0.5
## Principal Normal
A Space Curve has an infinite number of normals. By convention, we take
the normal to be the particular normal which is in the plane of the tangent
and which points toward the concave side of the curve. Pictorally this
means: K = |dT/ds| N = (1/K) * (dT/ds) A More Convenient form of this is: N = (dT/dt) / |dT/dt|
v(t) = i + 2tj + 3t^2k a(t) = 2j + 6tk
N is fairly simple. Take the derivatives
and calculate. It is unnecessary to show that here.
## The TNB Frame
Imagine a coordinate system defined at a particular point of a curve by the
Tangent (
## TwistTwist is represented by the greek letter . Twist (also called torsion) is a scalar.
= 0 for a plane curve.
There is another, alternate definition for . This allows to be negative. So, || = |d B /ds|
= 12 / (36t^4 + 36t^2 +4) (0,0,0) [t=0] = 3
## Differentiating VectorsSay you havev(t) = 5t^2T i.e. The velocity is some scalar function times the tangent vector. (It always is)
a = dv/dt = 10t*T + 5t^2 * (dT/dt) d T/dt is the direction of the normal.
Generalization: That is, the acceleration has some component tangential to the curve and another component normal to the curve (normal = principal normal).
Created by Lawrence C. Weintraub on 2/28/95 Edited by Milos Borojevic on 3/4/95 Send comments using feedback page |

Carlos Rodriguez <carlos@math.albany.edu> Last modified: Wed Feb 14 13:47:33 EST 1996