given a projectory equation of the form $ y=y(x) $find the curvature radius as a function of $x.$
a projectory equation , hence $ x=x(t)$, input that in y and we get $y=y(x(t))$, which is what one has to derive....so one needs to do partial derivatives? doesn't make sense..
Suppose the Cartesian equation of the curve $\gamma$ is given by $y=y(x)$ and $A$ be a fixed point on it. Let $P(x, y)$ be a given point on $\gamma$ such that $\mathrm{arc}(AP)=s$. We know that $$\frac{\mathrm d y}{\mathrm d x}=\tan \psi\tag 1$$ where $\psi$ is the angle made by the tangent to the curve $\gamma$ at $P$ with the $x$-axis and $$\frac{\mathrm d s}{\mathrm d x}=\sqrt{1+[y'(x)]^2}\tag 2$$
The curvature of $\gamma$ at $P$ is $\kappa=\frac{\mathrm d \psi}{\mathrm d s}$ and the radius of curvature is $\rho=\frac{1}{|\kappa|}=\left|\frac{\mathrm d s}{\mathrm d \psi}\right|$.
Differentiating $(1)$ w.r.t $x$, we get \begin{align} y''(x)&=\sec^2\psi \frac{\mathrm d \psi}{\mathrm d x}\\ &=(1+\tan^2\psi) \frac{\mathrm d \psi}{\mathrm d s}\cdot\frac{\mathrm d s}{\mathrm d x}\\ &=\left(1+[y'(x)]^2\right)\cdot\frac{\mathrm d \psi}{\mathrm d s}\cdot \left(1+[y'(x)]^2\right)^{1/2}\\ &=\frac{\mathrm d \psi}{\mathrm d s}\cdot \left(1+[y'(x)]^2\right)^{3/2} \end{align} Therefore $$ \rho=\left|\frac{\mathrm d s}{\mathrm d \psi}\right|=\frac{\left(1+[y'(x)]^2\right)^{3/2}}{|y''(x)|} $$