Differentiate parametric curves and integrate vector motion

Read the vector equation component by component

For r″(t)=a(t), each coordinate obeys y″=f(t). The velocity condition determines first-integration constants; the position condition determines second-integration constants

Integrate with a constant vector

If r′(t)=v(t), integrate each coordinate and include one constant per component. A position condition r(t0)=r0 determines all position constants. If acceleration is supplied, first integrate to velocity using its condition, then integrate to position using a separate condition

Work integration from a time-dependent acceleration

For a(t)=(2, 6t,-2), v(0)=(1, 2, 0), r(0)=(1,-2, 3), integration gives v=(1+2t, 2+3t²,-2t) and r=(1+t+t²,-2+2t+t³, 3-t²). At t=2, the position is (7, 10,-1) and velocity (5, 14,-4). Differentiate and check both initial conditions

Given y(t) = −2 + 2t + t³, give the y-position at t = 2

Construct a full vector initial-value solution, ungraded

With t in seconds and position in metres, solve a(t)=(2, 6t,-2), v(0)=(1, 2, 0), r(0)=(1,-2, 3). State position, velocity and acceleration units, check both conditions and derivatives, and describe the path and instantaneous direction at t=1. Do not treat a coordinate projection as the full spatial path

Model: preserve all six constants through the two integrations

Integrating componentwise gives v=(2t+C1, 3t²+C2,-2t+C3); the velocity condition gives (1, 2, 0). A second integration gives r=(t²+t+D1, t³+2t+D2,-t²+D3); the position condition gives (1,-2, 3). At t=1 the position is (3, 1, 2), velocity (3, 5,-2), acceleration(2, 6,-2). The spatial path is r(t)=(1+t+t²,-2+2t+t³,3-t²), with tangent direction (3,5,-2) at t=1. Positions are metres, velocity m/s and acceleration m/s². Differentiating gives r′(t)=(1+2t,2+3t²,-2t)=v(t) and v′(t)=(2,6t,-2)=a(t). The initial vectors are v(0)=(1,2,0) and r(0)=(1,-2,3)