Question 846542
1) Velocity of the particle along parabola is in the direction of its tangent.
{{{dy/dx = 1/y = 1/sqrt(2x)}}}


At (2,2), {{{dy/dx = 1/sqrt(2*2)=1/2}}}.


Let the velocity be represented by: {{{V = 5cos(theta)i + 5sin(theta)j}}}



{{{tan(theta)=dy/dx = 1/2}}}
{{{sec(theta) = sqrt(1 + tan^2(theta)) = sqrt(1 + 1/4) = sqrt(5)/2}}}
{{{cos(theta)=2/sqrt(5)}}}


{{{sin(theta) = sqrt(1 - cos^2(theta)) = sqrt(1 - 4/5) = 1/sqrt(5)}}}


Thus, the velocity vector is:
{{{V = 5((2/sqrt(5))i + (1/sqrt(5))j)}}}
{{{V = 2*sqrt(5)i + sqrt(5)j)}}}

If you want help with the remaining questions then send me an email. I can solve them on paper, scan and send you. Cheers!!!