finding velocity along a curve with kinematic equations using time (i'm "not" looking for coding help. i need help setting up the math.)
i'm writing a program for a physics class to find the velocity of an object across a random curve. where the only force acting on it is gravity (g).
the goal is to compare the found results from the kinematic equations. with the value found from conservation of energy.  
i'm currently finding the instantaneous velocities at each point along the curve and summing the values. so basically this V^2 = 2*A*S. where A = sin(theta)*g , theta = arctan(dy/dx) , and S = the arc length. i sum the V^2s for each interval and take its root. this is giving the right answer. 
how ever my professor wants me to use an equation that uses time as well. 
and this is the problem i don't have a clue how to do this. as i'm not sure how to determine the change in acceleration with respect to time across a fixed path (example a curve that looks like y=x^2).
i could be  misunderstanding what he is requesting. however even if i am, if this is possible i would like to know how it's done.
i think this deals with jerk but i don't know how i would find a value to fit a fixed curve. i've looked for jerk examples and found some that have been helpful but none that deal with following a path along some surface. 
as a side note i'm already defining my curves parametrically as it allowed me to get around the problem of "dx = 0". and just in case that turns out to be the way to solve this.
i thought i could numerically integrate (r = x +v*t+.5*a*t^2) by finding the instantaneous acceleration for each time interval and using it to find V and then the change in position. however i'm not always getting answers that make sense no mater the size of delta t.
here's a picture
 
thank you all for your help.
chris
 A: So if the path is described parametrically with $\vec{r}(q) = (x(q),y(q))$ you need to define the tangent and normal vectors in order to describe the motion $\ddot{q}$ and the reaction force $F$. 
The kinematic velocity vector is $$\vec{v}(q,\dot{q}) = (\dot{x},\dot{y}) = ( \frac{{\rm d}x}{{\rm d}q} \dot{q}, \frac{{\rm d}y}{{\rm d}q} \dot{q} ) = (x' \dot{q}, y' \dot{q}) $$
where the $\square{ }'$ notation is derivative with the parameter $q$ and $\dot{\square}$ derivative with time. So speed is $v(q,\dot{q}) = \dot{q}\sqrt{x'^2+y'^2}$, or the inverse when speed is known, the parameter varies by $\dot{q} = \frac{v}{ \sqrt{x'^2+y'^2}}$
The tangent vector is $$\vec{e}(q) = ( \frac{x'}{\sqrt{x'^2+y'^2}}, \frac{y'}{\sqrt{x'^2+y'^2}})$$
and the normal vector $$\vec{n}(q) = ( -\frac{y'}{\sqrt{x'^2+y'^2}}, \frac{x'}{\sqrt{x'^2+y'^2}})$$
Finally the kinematic acceleration vector is
$$ \vec{a}(q,\dot{q},\ddot{q}) = ( x' \ddot{q} + x'' \dot{q}^2, y' \ddot{q} + y'' \dot{q}^2 ) $$
The acceleration vector can be decomposed into tangential acceleration $\vec{a}_T = \dot{v} \vec{e}$ and lateral acceleration $\vec{a}_L = \frac{v^2}{\rho} \vec{n}$ where $\rho$ is the instantaneous radius of curvature. From the above is found that
$$\vec{n} \cdot \vec{a} = \frac{v^2}{\rho} = \dot{q}^2 \frac{x'^2+y'^2}{\rho}$$
$$ \rho(q) = \frac{ (x'^2+y'^2)^{\frac{3}{2}}}{y'' x'-y' x''} $$
Now you can state the equations of motion
$$ F \vec{n} + m \vec{g} = m \vec{a} $$
which results in
$$ \boxed{ \begin{aligned} F &= m \frac{ \dot{q}^2 (y'' x'-y' x'')+g x'}{\sqrt{x'^2+y'^2}} \\ \ddot{q} &= - \frac{ \dot{q}^2 (x' x'' + y' y'')+g y'}{x'^2+y'^2} 
\end{aligned} }$$
Example
A parabolic path with $\vec{r} = (q,q^2)$ has $\vec{v} = (\dot{q},2q \dot{q})$ and $\vec{a} = (\ddot{q}, 2 q \ddot{q} + 2 \dot{q}^2)$. This is due to the partial derivatives 
$$\begin{aligned} 
x & = q & y & = q^2 \\
x' &= 1 & y' &= 2 q \\
x'' &= 0 & y'' &= 2 
\end{aligned}$$
The formulas for reaction force and motion above are $$\begin{align} F&= m \frac{g + 2 \dot{q}^2}{\sqrt{1+4 q^2}} \\ \ddot{q} &= -q  \frac{2(g+2 \dot{q}^2)}{1+4 q^2} \\ v &= \dot{q} \sqrt{1+4 q^2} 
\end{align} $$
The small motion approximation ($q \ll 1$, $\dot{q} \rightarrow 0$) results in 
$$ F = m g $$
$$ \ddot{q} = -2 g q $$
which is simple harmonic motion with period $\tau = \frac{\sqrt{2} \pi}{\sqrt{g}}$.
