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5. The Acceleration in the Frenet Frame and the Curvature of a Space Curve

Any vector in space can be expressed in terms of the tangential , principal normal , and binormal vectors T(t) , P(t) and B(t) at any point of a smooth curve. In particular, we can express the acceleration vector in terms of this frame, so there must be functions a(t) , b(t) and c(t) such that

    X''( t)=a(t) T(t)+b (t)P(t)+c(t)B(t)
As in the case of plane curves, we can determine the first of these functions by differentiating the expression for the velocity vector
    X'( t)=s'(t)T(t)
We obtain
    X''( t)=s''( t)T( t)+s'(t)T'( t)
Since
    s'(t) T'(t) =X''( t)-s''( t)T( t)
we see that T'( t) lies in the osculating plane , determined by the velocity and acceleration vectors at the point. Therefore T'( t) can be expressed in terms of the unit tangent and the principal normal, and since it is perpendicular to the unit tangent, T'( t) must be a multiple of P(t) . We define T'( t) = k(t)s'(t)P(t) . The function k(t) is called the curvature of the curve X at the point X(t) . It follows that:
    X''( t)=s''( t)T( t)+k(t)( s'(t)) 2P(t)

We proceed in a similar way to find P'( t) . First, since P·P =1 , 2P'( t)·P( t)=0 . Similarly, since P·T =0 , we find that P'( t)·T( t)=-s'(t)k(t) . In order to complete a formula for P' we define the torsion t(t) such that P'( t)·B( t)=s'(t)t(t) . Hence we obtain:

    P'( t)=-s'(t)k(t)T(t)+s'( t)t(t)B(t)

Another definition for the torsion which many books give and which is equivalent to the one given above is through the relationship:

    B'( t)=-s'(t)t(t)P(t)

    Verify the equivalence of these definitions.

Demonstration 8: Demo on Curvature and Torsion
Java not enabled.

In this demonstration, there are two windows that show the graph of the curvature function and the torsion function in terms of the parameter t . Consider this graph for various curves. See what happens when the space curve is actually a plane curve. Take a look at the space cardioid family given by

    X(t) =((1+cos(t))cos(t), (1+cos(t)) sin(t),csin(t))

A useful way to write the relationships between T,P and B and their derivatives, is the following:

    Matrix out of order. Sorry for the inconvenience.
    (T 'P' B' )=s'(0k0-k0t0-t0)(TPB)
Since (T,P ,B) is an orthonormal frame, the matrix which relates them to their derivatives is an antisymmetric matrix.


Next: Curvature and Osculating Circles