Now: d sin ⁡ ⁡ θ θ d s = d d s y ′ ( s ) x ′ ( s ) 2 + y ′ ( s ) 2 = y ″ ( s ) x ′ ( s ) 2 − − y ′ ( s ) x ′ ( s ) x ″ ( s ) ( x ′ ( s ) 2 + y ′ ( s ) 2 ) 3 / 2 , in which the denominator is unity. With this formula for the derivative of the sine, the radius of curvature becomes: d θ θ d s = 1 ρ ρ = y ″ ( s ) x ′ ( s ) − − y ′ ( s ) x ″ ( s ) = y ″ ( s ) x ′ ( s ) = − − x ″ ( s ) y ′ ( s ) , where the equivalence of the forms stems from differentiation of Eq. 1 : x ′ ( s ) x ″ ( s ) + y ′ ( s ) y ″ ( s ) = 0 . With these results, the acceleration can be found: a ( s ) = d d t v ( s ) = d d t [ d s d t ( x ′ ( s ) , y ′ ( s ) ) ] = ( d 2 s d t 2 ) u t ( s ) + ( d s d t ) 2 ( x ″ ( s ) , y ″ ( s ) ) = ( d 2 s d t 2 ) u t ( s ) − − ( d s d t ) 2 1 ρ ρ u n ( s ) as can be verified by taking the dot product with the unit vectors u t (s) and u n (s). This result for acceleration is the same as that for circular motion based on the radius ρ. Using this coordinate system in the inertial frame, it is easy to identify the force normal to the trajectory as the centripetal force and that parallel to the trajectory as the tangential force. From a qualitative standpoint, the path can be approximated by an arc of a circle for a limited time, and for the limited time a particular radius of curvature applies, the centrifugal and Euler forces can be analyzed on the basis of circular motion with that radius.