超一流物理高手、物理专家请进(Mechanics 0032)

The space curve is R(t)=(t,t^2,t^3), so the derivative of the curve is dR(t)/dt=(1,2t,3t^2). Then the tangent vector at the point (1,1,1) is (1,2,3). So the line equation at piont (1,1,1) is L(t)=(t-1,2t-2,3t-3).
At last the acceleration of the curve is
d[dR(t)/dt]/dt=d^2[R(t)]/dt^2=(0,2,6t).
So the velocity vector at the point (1,1,1) is (0,2,6) and its magnitude of the velocity vector is √(4+36)=2√10.
This question is a basic concept of the mathematical relation among three vectors: the displacement, the acceleration and the speed, that is, the derivative of the displacement is the speed and the derivative of the speed is the acceleration.