Find an equation?

Question

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

Constraint 2:
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The speed vector orientation of M will change but its length remains constant:
(dx/dt)^2 + (dy/dt)^2 = const

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

Answer 1

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)

Answer 2

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 !

Answer 3

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.