apply the formula v^2/pfor the normal acceleration to find the radius of curvature at the vertex of the parabola y^2=4ax

1 answer

the radius of curvature is
(1 + x'^2)^(3/2)/|x"|
x' = 2y/(4a) = y/(2a)
x" = 2/(4a) = 1/(2a)
so at y=0,
ρ = (1 + 0)/(1/(2a)) = 2a
Now finish it off