A particle is moving along the parabola y2 8x and its speed

A particle is moving along the parabola y^2 = 8x and its speed is constant. Find each of the following when the particle is at (2, 4): the position vector, the velocity vector, the acceleration vector, the unit tangent vector, the unit normal vector, a_T, and a_N.

Solution

in parametric form the given parabola is x=2t² and y=4t

at (2, 4) means at t=1

position vector=2i+4j

position vector at any point= 2t²i +4tj

velocity vector=4ti+4j at t=1 it is 4i+4j

acceleration vector =4i+0j at t=1.

unit tangent vector=1/(32)^1/2(4i+4j)