which, clearly, has a constant magnitude u= vo. The acceleration is (t) dr(t) CO in(-n)3 Note that, the acceleration is perpendicular to the path(in this case it is parallel to r), since the velocity vector changes direction, but not magnit We can also verify that, from r(s), the unit tangent vector, et, could be computed directly as dr(s) L cOS Example Motion along a helix The equation r(t)=Rcos ti+Rsintj+htk, defines the motion of a particle moving on a helix of radius R, and pitch 2Th, at a constant speed. The velocity vector is given by dr rsin ti+ rcos ti+ hk and the acceleration vector is given by, du Rcos ti+-Rsinti In order to determine the speed at which the particle moves we simply compute the modulus of the velocit v=Ju=VR Sin t+R2 cost+h2=VR2 +h2 If we want to obtain the equation of the path in terms of the arc-length coordinate we simply write drl=udt=VR2 +h2dt Integrating, we obtain s =So +VR2+h2t, where so corresponds to the path coordinate of the particle t time zero. Substituting t in terms of s, we obtain the expression for the position vector in terms of the arc-length coordinate In this case, r(s)=Rcos(s/VR2+h2)i+R sin(s/vR2+h2)j+hs/vR2+h2k.The figure below shows the particle trajectory for R= l and h=0.1which, clearly, has a constant magnitude |v| = v0. The acceleration is, a(t) = dr(t) dt = − v 2 0 R cos(v0t R )i − v 2 0 R sin(v0t R )j . Note that, the acceleration is perpendicular to the path (in this case it is parallel to r), since the velocity vector changes direction, but not magnitude. We can also verify that, from r(s), the unit tangent vector, et, could be computed directly as et = dr(s) ds = cos(v0t R )i + sin(v0t R )j . Example Motion along a helix The equation r(t) = R costi + R sin tj + htk, defines the motion of a particle moving on a helix of radius R, and pitch 2πh, at a constant speed. The velocity vector is given by v = dr dt = −R sin ti + R costj + hk , and the acceleration vector is given by, a = dv dt = −R costi + −R sin tj . In order to determine the speed at which the particle moves we simply compute the modulus of the velocity vector, v = |v| = p R2 sin2 t + R2 cos2 t + h 2 = p R2 + h 2 . If we want to obtain the equation of the path in terms of the arc-length coordinate we simply write, ds = |dr| = vdt = p R2 + h 2 dt . Integrating, we obtain s = s0 + √ R2 + h 2 t, where s0 corresponds to the path coordinate of the particle at time zero. Substituting t in terms of s, we obtain the expression for the position vector in terms of the arc-length coordinate. In this case, r(s) = R cos(s/√ R2 + h 2)i+R sin(s/√ R2 + h 2)j +hs/√ R2 + h 2k. The figure below shows the particle trajectory for R = 1 and h = 0.1. 1 0.5 0 0.5 1 1 0.5 0 0.5 1 0 0.5 1 1.5 2 6