Math 293


Lecture 12

Given by Subrata Mukherjee on 2/17/95 at 10:10 in Kimball B11


Properties of Curves (Continued)


Lecture 11 began the discussion of curve properties by defining the Length, the Tangent, and the Curvature of a Space Curve. In Lecture 12 we finish this discussion with the concepts of Principal Normal, Twist, and Binormal.

Example for Curvature

First, we'll finish up curvature by doing an example:

Find K at (0,0) for y=x^2
First we will write the equation in parametric form:
x=t
y=t^2
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
Curvature = 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:

Mathematically, this is:
N = (dT/ds) / |dT/ds|
Since K = |dT/ds|
N = (1/K) * (dT/ds)
A More Convenient form of this is:
N = (dT/dt) / |dT/dt|

Example:
Space Curve:
x=t, y=t^2, z=t^3
Find N and K.

r(t) = ti + t^2j + t^3k
v(t) = i + 2tj + 3t^2k
a(t) = 2j + 6tk

=> v x a = 6t^2i - 6tj + 2k

K(0,0,0) [t=0] = 2
N = (dT/dt) / |dT/dt|
The mathematics in solving for 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 (T), the Pricipal Normal (N), and a third Vector perpendicular to the first two (B).

The third vector, B, is called the Binormal vector, and is found by the formula:
B = TxN

Twist

Twist is represented by the greek letter . Twist (also called torsion) is a scalar.

= 0 for a plane curve.
<> 0 for a space curve.
= | dB/ds |


Choose the sign so that is positive.

There is another, alternate definition for .
(dB/ds) = N
This allows to be negative.
So, || = |dB /ds|

Example:
Find from the earlier example.
r(t) = it + t^2j + t^3k

= 12 / (36t^4 + 36t^2 +4)
(0,0,0) [t=0] = 3

Differentiating Vectors

Say you have v(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)
dT/dt is the direction of the normal.

Remember, T is a function of t, and therefore the product rule must be used in differentiating.

Generalization:
On any curve, a = f(t)T + g(t)N
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