3.3: Unit Tangent and Principal Normal Vectors

In Lab 2, we saw how it was useful to decompose the acceleration vector of a plane curve into components using the moving frame of tangent and normal vectors, T(t) and U(t). This led to the definition of the geodesic curvature function κg(t) for plane curves. In three-space, too, we want to find a moving frame in which to resolve the acceleration vector. At each point along a curve where the velocity vector is not the zero vector, it is possible to divide it by its length to define a unit tangent vector T(t). So, X'(t) = s'(t)T(t), or equivalently,

T(t) = X'(t)/|X'(t)|

A vector perpendicular to the velocity vector at a point is said to be normal to the curve at that point. At a point X(t) at which the velocity vector is non-zero, there is an entire circle of unit normal vectors. In the plane there were two choices of unit normal vector, and we used the orientation of the plane to choose one of them to use for the moving coordinate frame. In particular, we said that if the unit tangent vector is written as T(t) = (x(t),y(t)), then the unit normal vector is defined as U(t) = (-y(t),x(t)). In three-space, we define the principal normal vector as:

P(t) = T'(t)/|T'(t)|

Since T(t) lies on the unit circle, it's velocity vector, T'(t), will always be normal to the position vector, T(t). Thus, P(t) is normal to T(t). The plane spanned by T(t) and P(t) is known as the osculating plane. By construction, we can decompose the acceleration vector into its components along the tangent and principal normal directions:

X''(t) = s''(t)T(t) + s'(t)T'(t), or
X''(t) = s''(t)T(t) + s'(t)2κ(t)P(t).

In the equation above, we introduce a function κ(t), called the curvature of the curve at t, and write T'(t) as T'(t) = κ(t)s'(t)P(t). Note that this definition of curvature for space curves matches the definition of geodesic curvature for plane curves, so that when X(t) is a plane curve, we have κg(t) = κ(t).


This demonstration allows you type in a parameterized curve, X(t), in the control panel. The graph of X(t) is shown in red in the 3D curve window along with three vectors emanating from the point X(t0). The acceleration vector, X''(t0), is colored orange while its components along the tangent and principal normal directions are colored magenta and green respectively. These vectors all lie in the semitransparent orange plane, which is the osculating plane at X(t0). You can move the point X(t0) along the curve by using the tapedeck for t0 in the control panel. The second window shows a graph of the curvature function κ(t) with the point (t0,κ(t0)).



1. Derive the following expression for the curvature function: κ(t) = |X'(t)×X''(t)|/s'(t)3.
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