Find time-parametrization given path and speed of a particle Consider a particle in two dimensions with position vector $r(t)=<x(t),y(t)>$ and the shape of the path is described by a function $y(t)=f(x(t))$ (Thus $r(t)$ is a parametrization of $f$ with respect to time). Given some function $f$ and a speed $s$, how do we find the position vector $r(t)$ such that the particle moves always with constant speed $s$ given also some starting point?
For example, consider the path shape described by $y=x^2$. And suppose we have a particle following this path beginning at the origin at $t=0$, moving always with constant speed $s=1$. How do we find the parametrization $<r(t)>$?
 A: You can derivate each component separately:
$r'(t)=<x'(t),y'(t)>$
However your example is different, you do not give x(t) and y(t), but only the shape of curve and the constraint of constant speed. This contraint is $r(t)^2=x(t)^2+y(t)^2=s^2$
using that you should be able to get the solution after some algebraic work
A: According to your example
$s = d|r|/dt = 1$, and $r(t = 0) = 0$, s.t.
(1) $|r| = t$
Now, since $x^2 = y$,
$r^2 = y + y^2 $,
s.t. you get a simple equation of 2nd degree in y,
$y^2 + y - r^2 = 0$ .
Solve this equation,
(2) $y = \frac {-1 \pm \sqrt{1 + 4r^2}}{2}.$
Substituting here (1),
(3) $y = \frac {-1 \pm \sqrt{1 + 4t^2}}{2},$
and therefore,
(4) $x = \sqrt{y} = \sqrt{\frac {-1 \pm \sqrt{1 + 4t^2}}{2}} $.
Shall we make a test if everything is correct? Let's do it! For shorting formulas, let's denote the square root by S,
(5) $S = \sqrt {1 + 4t^2} $.
$r^2 = x^2 + y^2 = (-1 \pm S)/2 + (-1 \pm S)^2/4$
$ = (1/4) (-2 \pm 2S + 1 + S^2 \mp 2S)$
$
 = (1/4) (-1 + S^2).$
Well, introduce here the expression of S, and compare with (1) . It seems that everything is fine, isn't it?
Good luck !
A: the velocity is tangential to the curve and given by
$$\vec{v} =\frac{d\vec{r}}{dt}$$
and the position vector $\vec{r}$ is given as
$$\vec{r}=x~\hat{i}+y~\hat{j}$$
Given $y=f(x)$,
$$\vec{r}=x~\hat{i}+f(x)~\hat{j}$$
So,
$$\frac{d\vec{r}}{dt}=\frac{dx}{dt}~\hat{i}+\frac{d}{dt}f(x)~\hat{j}$$
Simplifying,
$$\vec{v}=\frac{dx}{dt}\left(~\hat{i}+\frac{d}{dx}f(x)~\hat{j}\right)$$
From here, get the components, use the constraint $|\vec{v}|=s$ and solve the ODE in $\dot{r}$ to get the $r(t)$
