I was reading about curvatures. There was a topic about Cartesian Equation (Explicit function). It was given in the book that:
In rectangular Cartesian coordinates , we have $\tan\psi=dy/dx=y_1$ and, therefore, $\sec^2\psi d\psi/ds=dy_1/ds=dy_1/dx.dx/ds=d^2y/dx^2\cos\psi$ whence, $$\rho=\frac{ds}{d\psi}=\frac{\sec^3\psi}{\frac{d^2y}{dx^2}}=\frac{\{1+(\frac{dy}{dx})^2\}^{3/2}}{\frac{d^2y}{dx^2}}=\frac{(1+y_1^2)^{3/2}}{y_2}$$($y_2\neq 0$). The definition of $\rho$ nowhere suggests that its value would depend on the choice of axes. Hence, we could very well interchange the axes of $X$ and $Y$ and obtain $$\rho=\frac{(1+x_1^2)^{3/2}}{x_2},$$ $x_2\neq 0$.
However, I am not quite getting it. How did they write $dx/ds=\cos\psi$? Also, we get a radius of curvature formula for a curve at a point using this cartesian equation in this method, right? And the last thing, given is that we could change the axes but then the definition of tangent gets changed, i.e., $\tan\psi=dx/dy$, right? If so, is that all right? I am not quite getting it...
The distance $\Delta s $ beetween two points $(x_a,y_a)$ and $(x_b, y_b)$ in the euclidean space is given by the Pythagorean theorem
$ \Delta s^2=(x_a-x_b)^2+(y_b-y_a)^2 \equiv \Delta x^2+\Delta y^2 $
taking the limit you obtain
$ ds^2=dx^2+dy^2 $
so you get
$ ds=\sqrt{1+(\frac{dy}{dx})^2} \, dx $
using the fact $dy/dx= \tan \psi$, you obtain
$ ds= \sqrt{1+\tan^2 \psi} \, dx =\frac{1}{\cos \psi} \, dx $