2.1 Decomposition of Acceleration

Consider a parameterized curve X(t) = (x(t),y(t)) with velocity vector X'(t) = (x'(t),y'(t)). Recall from section 1.3 that the unit tangent vector is defined as T(t) = (x'(t),y'(t))/s'(t) and the unit normal is defined as U(t) = (-y'(t),x'(t))/s'(t). As long as T(t) and U(t) exist and are non-degenerate, they form a basis for R2. Therefore, we can write the accleration vector as a linear combination of T(t) and U(t):

X''(t) = a(t)T(t) + b(t)U(t)

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)

Observe that T(t) is a unit vector, so that T(t)·T(t) = 1 and taking the derivative of both sides gives 2T(t)·T'(t) = 0. So T'(t) is perpendicular and T(t) and it follows that T'(t) must be a multiple of U(t). We define T'(t) = κg(t)s'(t)U(t) and write the acceleration vector as

X''(t) = s''(t)T(t) + κg(t)s'(t)2U(t)

The function κg(t) is called the geodesic curvature of the curve at the point X(t).


This demo allows you to examine the curvature of a curve X(t) for various values of t0 which you can scroll through using the tapedeck in the control panel. In the Curve window, you can see the acceleration vector of the curve at X(t0) in magenta as well as its components in the tangent and normal directions. In a second window, we graph the geodesic curvature function κg(t). Observe the relationship between the acceleration vector and the unit normal vector at the current position on the curve. The red vector, which is the projection of the acceleration vector onto the unit normal vector, is related to the curvature of the curve. Watch how the change in this projection varies while the point (t,κg(t)).



1. Examine the acceleration vectors and curvature of the ellipse X(t) = (c*cos(t), sin(t)).

2. Examine the acceleration vectors and curvature of the cardioid X(t) = (1 + cos(t))*(cos(t), sin(t)).

3. Examine the acceleration vectors and curvature of the exponential spiral: X(t) = e^t*(cos(t),sin(t)).
We can approach curvature in terms of the circular images of the vectors T(t) and U(t). Since T(t) is a unit vector, it lies on the unit circle. So, there exists a function θ(t) such that T(t)=(cos(θ(t)), sin(θ(t))). Similarly, by the definition of normal vector, we have U(t) = (-sin(θ(t)), cos(θ(t))). Then,

T'(t) = (-sin(θ(t))θ'(t),cos(θ(t))θ'(t)) = U(t)θ'(t)

Entering this identity into the expression for acceleration yeilds X''(t) = s''(t)T(t) + s'(t)θ'(t)U(t). Earlier we defined the normal component of the accleration vector to be s'(t)2κg(t). Therefore, s'(t)θ'(t) = s'(t)2κg(t), or equivalently dθ/dt = (ds/dt)κg(t). Using an arclength parameterization, we then have

κg(s) = θ'(s)

This provides a definition of geodesic curvature in terms of the motion of the tangent and normal vectors on the unit circle.


This demonstration shows the decomposition of the acceleration vector at X(t0) into its tangential and normal components. It then graphs the geodesic curvature κg(t). In a third window, the unit tangent and and normal vectors are drawn in cyan and rad respectively. As you play the tapedeck for t0 in the control panel and the point X(t0) moves along the curve, notice how the speed of the tangent and normal vectors on the unit circle is related to the magnitude of the geodesic curvature.
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