finding curvature radius 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..
 A: 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
$$
\rho=\left|\frac{\mathrm d s}{\mathrm d \psi}\right|=\frac{\left(1+[y'(x)]^2\right)^{3/2}}{|y''(x)|}
$$
