Question

Suppose r(t) = cos(t)i + sin(t); + 4tk represents the position of a particle on a helix, where z is the height of the particle above the ground. (a) Is the particle ever moving downward? no If the particle is moving downward, when is this? When t is in none (Enter none if it is never moving downward; otherwise, enter an interval or comma-separated list of intervals, e.g., (0,3], [4, 5].) (b) When does the particle reach a point 18 units above the ground? When t = 4.5 (c) What is the velocity of the particle when it is 18 units above the ground? ū= <.97753011,--210795 (d) When the particle is 18 units above the ground, the particle leaves the helix and moves along the tangent line. Find parametric equations for the particle path when the particle is on this tangent line. The parametric equations should be given so that at the time t, when the particle is 18 units above the ground, the particle is on the helix. X(t) = Vt) = z(t) =

Answer 1

Solution: Consider the positim of the particle at 8ct) = cos(t)i + Sinit)j + 4tk (a) Differentiate 8ct) with respect to t x'ct): -sin ti +Costý+3K The 7 - component is 4 and is positive So, the particle moves vertically up Thus, the particle never moves downward Hence the answer is, no. Suppose that the particle reach to a point 18 units above ground. The Z- component represents component represents the vertical motim of the particle. There fore, 4t=18 - t = 18 = 4.5 Hence, the particle reaches to 18 units up when t-4.5 min. (0) Here, x'ct) represents the velocity of the particle. out). r(t) = cost it Sint; + 4tk g'ct) = -Sint it costj + 4K when we have t = 4.5 (from Part b) Hence, the velocity of the particle when it is 18 units above the ground is, 8'(4.5) = - sin (4.5) i +Cos(4.5)j + 4k V = - Sin (4.5) { + cos (4.5)ý +4K (d) The velocity vector is always tangent to the path of the curve So, the direction vector from the positim at height 18 or timet-4.5 is V(4.5) = '(4.5) 3 - Sin (4.5) i +cos (4-5)+4K

The IS, position of the particle at time t=4.5 814.5) = cos(4-5) it sin(4.5);+ 4(4.5) K = cos(4.5) it sin (4.5)j + 18K The particle moves along the straight line and the particle follows the Straight path Lit) after t=4.5 So, L(t) = 0(4,5) + (t-405)'(4.5) = [cos (45) ft Sin14.5) 3+18x] + 4-4.5)(-sin (4.5) it cos(4.5)j +41] = [Cos (405) – (t +4.5) sin(4.5 )] i + (Sincu-5) +4-4.5) cos(4.5)); + [18+(-4.5)4] < = [cos (4.5) +(4.5) Sin (4.5) – tsink4.5))i + sin(4.5)+ t cos(4.5) – 4.56054.53% + 4tk :: X(t) = cos(4,5) +(4.5) sin (4.5) – + sin(4,5). y(t) = sin(4.5) – (4,5) cos(1115) + t cos (4.5) 2(t) = 40