Curvature derivation for arbitrary parameterization Note: ${x_p}$ and ${x_{pp}}$ are derivative of x wrt p once and twice, respectively.
I am deriving the curvature for an arbitrary curve C=(x(p),y(p)).
The formula for derivation satisfies $\frac{1}{|C_p|}$$\frac{\delta}{\delta p}$($\frac{C_p}{|C_p|}$)=$\kappa$*N.
N=$\frac{(-y_p,x_p)}{\sqrt{x_p^2+y_p^2}}$. is the unit normal to the tangent(tangent to curve).
After doing the derivation according to the formula above, we will have (A,B)=$\kappa$N, for some A and B.
and substituting the x (first) component of N to get $\kappa$, I get $\frac{-x_{pp}*y_p}{(x_p^2+y_p^2)^\frac{3}{2}}$.
Similarly, if I had done same for y component, ie., $\frac{B}{N_y}$, I get
$\frac{-y_{pp}*x_p}{(x_p^2+y_p^2)^\frac{3}{2}}$
The problem is, curvature $\kappa$ is defined as the sum of the two! $\kappa$=$\frac{y_{pp}*x_p-x_{pp}*y_p}{(x_p^2+y_p^2)^\frac{3}{2}}$
Is this because the first and second derivatives are normal to each other? If so, how can I use this fact to show that the summation will still give correct result upon multiplication, somewhere I think $x_{pp}$ and $x_p$ has to multiply and vanish but the derivates being normal to each other only imply that dot product sums should be zero.
 A: I'm going to use more standard notation to make this derivation more generally useful.
Let $\gamma : [a,b] \to \mathbb{R}^2$ be a curve, and assume that $\gamma'(t) \neq 0$. The tangent vector is defined as
$$ T = \frac{\gamma'(t)}{\lVert \gamma'(t) \rVert}. $$
$T$ has unit length, so its derivative must be perpendicular to it:
$$ T \cdot T = 1 \implies 2T \cdot T' = 0. $$
In higher dimensions, the normal vector $N$ is defined as proportional to $T'$, but in 2D, we don't need to, since we can just specify that $N$ is a rotation of $T$ by $\pi/2$ anticlockwise: i.e. if $T=(a,b)$, $N = (-b,a)$. Then we define the curvature by
$$ T' = \kappa N \lVert \gamma'(t) \rVert, $$
where the extra factor of $\lVert \gamma'(t) \rVert$ is needed to make sure that $\kappa$ is independent of rescaling (think about setting $\tau=2t$ and recalculating both sides, for example).
Presumably you know all this already, but now I've fixed the notation and definitions. Now, we calculate the derivatives of $\gamma$ directly to find $\kappa$ in terms of things we know. Write $s'=\lVert\gamma'(t) \rVert$ for compactness.
$$\begin{align}
\gamma' &= s' T \\
\gamma'' &= s'' T + s' T' = s'' T + s'^2 \kappa N.
\end{align} $$
Dotting with $N$, we find
$$ \kappa = \frac{\gamma'' \cdot N}{s'^2}. $$
Now we expand this in terms of coordinates: let $\gamma=(x,y)$. Obviously $\gamma'' = (x'',y'')$, and $s'^2 = x'^2+y'^2$. The normal, as noted above, is found from the tangent vector, as $N = (-y',x')/\sqrt{x'^2+y'^2}$. Putting this all together gives
$$ \kappa = \frac{(x'',y'')\cdot (-y',x')}{(x'^2+y'^2)^{3/2}} = \frac{x'y''-y'x''}{(x'^2+y'^2)^{3/2}}. $$
