Find an equation?

A point M is moving in a 2D plan. Its coordinates are functions of time x(t) and y(t). The initial position of M at t=0 is known (x0, y0).





Constraint 1:


==========


M(x, y) moves along the curve y = ax^2 + bx + c





Constraint 2:


==========


The speed vector orientation of M will change but its length remains constant:


(dx/dt)^2 + (dy/dt)^2 = const





Question:


=======


How can we express x(t) and y(t)?

Find an equation?
Very difficult question. Will give it a shot. Answer not guaranteed right but probability very high.





D^2 = dx^2 + dy^2 = dx^2 * (1 + [2ax + b]^2)





(dy = dx * derivative of the function).





For x(t) we find differential equation





dx = 1/sqrt(1 + [2ax + b]^2) dt





since p = -b/2a, therefire


dx = 1/sqrt(1 + (2a)^2 (x-p)^2) dt





int value is





t = (x-p)/2 * sqrt(...) + 1/4a * ln {2ax + b + sqrt(...)} + C





where (...) = (1 + [2ax + b]^2)
Reply:In other words, you want to travel along a parabola at constant speed.





The squared distance between points M(x, y) and a neighboring point M(dx, dy) is equal to





D^2 = dx^2 + dy^2 = dx^2 * (1 + [2ax + b]^2)





(dy = dx * derivative of the function).





For x(t) we find differential equation





dx = 1/sqrt(1 + [2ax + b]^2) dt





Using p = -b/2a, we can reduce this to





dx = 1/sqrt(1 + (2a)^2 (x-p)^2) dt





and the integral would be





t = (x-p)/2 * sqrt(...) + 1/4a * ln {2ax + b + sqrt(...)} + C





where (...) = (1 + [2ax + b]^2)





It is not easy to solve this for x !
Reply:First start off with x=t and therefore y=at^2+bt+c.





What you want to do is normalize the velocity vector:





So, find (dx/dt)^2+(dy/dt)^2= 1+(2at+b)^2 = 4at^2+4abt+b^2+1.