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- Find the acceleration vector for the position vector r(t) = (sin(11t), 2t¹0, e-6t). x component = y component = z component =Find the velocity vector for the position vector r(t) = (sin(12t), 8t°, e-6*). x component = y component = z component 10At time t = 0, a particle is located at the point (1, 2, 3). (Vector Functions) It travels in a straight line to the point (4, 1, 4), has speed 2 at (1, 2,3) and constant acceleration 3i – j+k. Find equation for the position vector r(t) of the particle at time t.
- Find the acceleration vector for the position vector 7(t) = (sin(11t), 2t", e-10t). x component = y component = z component =4) For the following vector functions, identify the shape of graph they would represent: a. r(t) = (2+ 3t, 5, -2-t) b. r(t) = (2, cos(3t), sin(3t)) c. r(t) = (t,5, t²)A particle started at A(1,0) circled the origin once an d returned toA(1,0) .what were the change in its coordinates
- What is the vector parametric equation L(t) for the line through the points (5,-5,0) and (0,-3,3)?Find the velocity vector for the position vector 7(t) = (sin(3t), 10t°, e-11t). x component = y component = z component =At time t=0, a particle is located at the point (3,9,4). It travels in a straight line to the point (7,8,6), has speed 6 at (3,9,4) and constant acceleration 4i-j+2k. Find an equation for the position vector r(t) of the particle at time t -O+¹+* The equation for the position vector r(t) of the particle at time t is r(t) = (Type exact answers, using radicals as needed.)
- (L2) Find the component form of the vector v given ||v|| = 12 following the direction of u = (3,–5) 2.At time t=0, a particle is located at the point (2,7,5). It travels in a straight line to the point (7,2,8), has speed 9 at (2,7,5) and constant acceleration 5i - 5j + 3k. Find an equation for the position vector r(t) of the particle at time t. ..... The equation for the position vector r(t) of the particle at time t is r(t) = ( ) i+ ( Dj+ ( D k.Match each given vector equation with the corresponding curve. 7x r(t)-(2³), ) (sin(t), t) r(t)-(t, cos(2), sin(2t)) r(t)=(1+t, 3t, -t) r(t) cos(t) i-cos(t) j+ sin(t) k r(t) = ²²³1 +²^²y + ºk r(t) = ²+1+2k 2 r(t)-(1, cos(t), 2 sin(t)) 2 7 2 7 ? r(t) M (c) (a) ~~ DOO (1,0) of 1,0) لے (h) X** (d) (1) (1,0,0) (0,0,2)