Question

marks] The position of a particle is given as a function of time by r(t)=(1-cos(27t)i+ (1-t)sin(2nt)j+ 4tk with i (1,0,0), j = (0,1,0) andk = (0,0,1) the Cartesian basis vectors of R3. (a) Sketch the particle trajectory from t 0 tot= 1, as a 3D perspective plot and as the 2D projection onto the xy-plane. (b) Determiner(t) as a function of time t. (c) Is r'(t) greater for t 0 than it is for t 1? Justify your answer.

Answer 1

rt) (1- t)cos2nt i+ (1 -t) sin 2nt j+4tk. Compare with r(t) = x(t) i + y(t)jz(t)k (a) y (1 t) sin 2nt , we get: x = (1- t)cos2nt Z= 4t Eliminate t from x y, and z: 1-2t 1- 2+y222(1-t2-t2 or 1 16 x2y2- (22-8) x2+y2 4) + 1 = 1 (-4)2 = 16(2 +y2 1 =1. This is the cone whose base is circle of radius 1 in x-y plane and height 4 units. = (1-tcos2nt i +(1-t) sin 2nt j+4t k (b) r(t) we differentiate each component wr.t 't: For finding r'(t) d d r'(E) =D [(1-t)cos2mt] i + [(1-t) sin 2rt] j + 뚫 [4:] k r (t) [-cos2nt 2T (1 t sin 2nt]i -sin2nt2T (1 - t) 2nt]j4 k - 1 + COs (c) r (t) [- cos2nt - 2n(1 -t) sin 2nt]i +[- sin2nt +2(1-t) cos 2nt]j4 k r (t cos2nt 2n(1 - t) sin 2nt]2 [ sin2nt+2n(1 - t) cos 2nt] 16 r'()= At t-0 =v1+4T2 +16==V4T 17 1 and sin 0 = 01 cos0 |r (t)=I+016= V17 Clearly r'(t) At t-1 COS2T 1 and sin 2n 0] at t-0 is greater than |r'(t) at t= 1.