Consider a parameterized curve X(t) = (x(t),y(t)) with velocity vector X'(t) = (x'(t),y'(t)). 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 the geodesic curvature of the curve at the point X(t).