A particle moves along the path y=x^3+3x+1 where units are in centimetres.If the horizontal velocity Vx is constant at 2cm/s,find the magnitude and direction of the velocity of the particle at the point (1,5).

Answer 1

dy/dt = 3x^2+3 dx/dt
so, since Vx=dx/dt=2,
dy/dt = 6(x^2+1)
at x=1, Vy=dy/dt = 12

so, V=√(4+144)
tanθ = Vy/Vx = 6