A particle moves with constant speed along a parabola of equation $y^{2}=2px$ with $p=constant$. I want to find its velocity and acceleration vector.
Since the velocity magnitude is constant I know the acceleration will be along the normal of the parabola and of the form
$$\vec{a}=\frac{v^{2}}{R(t)}\vec{n}, v=constant$$
where $R(t)$ is the time dependent radius of curvature of the parabola and $\vec{n}$ is the normal.
To find the velocity I parametrize the parabola with $(y^{2}(t)/(2p),y(t))$. The velocity is the tangent vector given by deriving this parametric position
Then $\vec{v}=\left(\frac{y}{p}\frac{dy}{dt},\frac{dy}{dt}\right)$ and the constraint of constant speed leads to
$$|\vec{v}|=constant=\frac{dy}{dt}\sqrt{\left(\frac{y}{p}\right)^{2}+1}$$
This seems to be a first order differential equation. Is it solvable?
How do I find $\vec{n}$ and $R(t)$?
Best Answer
The diferential equation is solvable, the integral $\int\sqrt{1+x^2}dx$ can be found in the tables (or found via integration by parts).
What you actually found is the mapping between $y$ and $t$. For the radius of curvature, it's much easier to just get it geometrically, with respect to position (parameterized with $y$). The standard differential geometry of curves gives you:
$$\frac{1}{R(x(y))}=\frac{x''(y)}{\sqrt{1+x'(y)^2}^3}$$
where $x=\frac{y^2}{2p}$.
So, the acceleration, expressed with $y$, is straight forward and does not require integration. However, converting $a(y)$ into $a(t)$ requires the function $y(t)$ which will be problematic, because the solution of your differential equation gives you $t(y)$ that is transcendental and thus not invertible in closed form.