A particle of mass $m$ moves with constant speed $v$ along the curve $y^{2}=4a(a-x)$ I have complications to do the following problem:
A particle of mass $m$ moves with constant speed $v$ along the curve $y^{2}=4a(a-x)$. Find its velocity and acceleration vectors.
My first idea was to parameterize the curve given, however did not know how to introduce speed $v$. Therefore I derived with respect to time, the equation of the curve, obtaining:
$$2y\frac{\partial y}{\partial t}=-4a\frac{\partial x}{\partial t}$$
Also, I know that
$$\left( \frac{\partial y}{\partial t}\right)^2 +\left( \frac{\partial x}{\partial t}\right)^2=v^2$$
Thus have two equations relating the $x$ and $y$ components of the velocity, but I have not been able to resolve. Is my method OK? Is there another way? Is it easier to do so using the parametric equations, but then as I enter the speed $v$?
 A: You're on the right track.  A couple of notes:


*

*Those are actually total derivatives.  You can think of $x(t)$ and $y(t)$ as functions of $t$ alone

*you have two equations for two functions.  You probably want to isolate them into two equations, each for one function

*Think about how you would solve this by elimination

A: Let $v_x=dx/dt$ and $v_y=dy/dt$.
We got:  $2yv_y=-4av_x$
Rewriting $v_y=-\dfrac{2av_x}{y} \tag{1}$
Also we got : $v_x^2+v_y^2=v^2 \tag{2}$
Subsitute value of $v_y$ in eqn 2.
$v_x^2+{(-\dfrac{2av_x}{y})}^2=v^2$
Solving gives $v_x=\pm \dfrac{vy}{4a^2+1},$
Substitute this value of $v_x$ in eqn 2 gives: 
 $v_y=\mp\dfrac{2av}{\sqrt{4a^2+1}}$
We know $v_x$ and $v_y$. velocity is as we know $\vec v=v_x {\hat i}+v_y\hat j$  and can be found now. It should be clear that $\vec v$ depends upon the $y$ co-ordinate.
A: Whenever a particle is constrained to move along a path, its velocity and acceleration vectors are decomposed along the tangent vector $\vec{e}$ and the normal vector $\vec{n}$ as
$$\boxed{ \begin{aligned} \vec{v} & = v \, \vec{e} \\
\vec{a} & = \dot{v} \, \vec{e} + \frac{v^2}{\rho}\, \vec{n} \end{aligned} }$$
where $v$ is the speed and $\rho$ is the instant radius of curvature. In addition, for a path described by the position vector $\vec{r}(t) = ( x(t), y(t) )$ the path properties are
$$ \begin{aligned}
\vec{e}(t) & = \left( \frac{x'}{\sqrt{x'^2+y'^2}}, \frac{y'}{\sqrt{x'^2+y'^2}}\right) \\
\vec{n}(t) & = \left( -\frac{y'}{\sqrt{x'^2+y'^2}}, \frac{x'}{\sqrt{x'^2+y'^2}}\right) \\
\frac{1}{\rho(t)} &= \frac{ y' x'' - y'' x'}{\left( x'^2+y'^2 \right)^\frac{3}{2}}
\end{aligned} $$
where $x'=\frac{{\rm d}x(t)}{{\rm d} t}, x''=\frac{{\rm d}^2 x(t)}{{\rm d} t^2}$ and $y'=\frac{{\rm d}y(t)}{{\rm d} t}, y''=\frac{{\rm d}^2 y(t)}{{\rm d} t^2}$
So to solve your problem I propose to parametrize the path $y^2(x)=4 a (a-x)$ as
$$ \vec{r}(t) = \left( a t, 2 a \sqrt{1-t} \right) $$

 HINT: $\frac{1}{\rho(t)} = \frac{1}{2 a \left(2 -t\right)^\frac{3}{2}}$, $\dot{v}=0$

