Alternate methods to find Radius of Curvature (ROC) for a given function as path

Question

This question is with reference to a problem in "Problems in General Physics by IE IRODOV" (Particularly problem 1.42 ).

The original problem is :

A particle moves along the plane trajectory $y(x)$ with velocity $v$ whose modulus is constant. Find the acceleration of the particle at the point $x=0$ and the curvature radius of the trajectory at that point if the trajectory has the form:

a) of a parabola $y=ax^2$

b) of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

The way I solved it was by making use of the curvature radius formula taught in calculus, for a function $y=f(x)$.

$$\text{ROC}=\frac{\left[1+\left(\frac{dy}{dx}\right)^2 \right]^{\frac{3}{2}}}{\left| \frac{d^2y}{dx^2} \right|}$$ From there I calculated the acceleration by $\frac{v^2}{\text{ROC}}=a_{\perp}$

However, I find this method quite unintuitive. I am looking for a better approach for this problem.

So my question is:

Is there a better approach to solve above problem, rather than using a strictly mathematical approach like I did?

I was thinking of taking components of the velocity, and finding a relation b/w them to get to velocity as a function of $(x,y)$. But I can't seem to get any reasonable arguments to relate the components.

Answer 1

If you want the derivation then here it is, like in the first case

\begin{align} \frac{\mathrm dy}{\mathrm dt}&=2ax\frac{\mathrm dx}{\mathrm dt} \\ \frac{\mathrm d^2y}{\mathrm d t^2}&= 2ax \frac{\mathrm d^2x}{\mathrm dt^2} + 2a \left(\frac{\mathrm dx}{\mathrm dt}\right)^2\\ \frac{\mathrm d^2y}{\mathrm dt^2}&=2ax(0) + 2av^2 \end{align}

which is the answer in the first case.