The radius of curvature of a curve $y=f(x)$ is given by $\rho=\frac{(1+(\frac{dy}{dx})^2)^{\frac{3}{2}}}{\frac{d^2y}{dx^2}}$.
I know this formula.Its derivation is also given in the book and i understood that.But i did not understand the following concept.
Since the value of $\rho$ is independent of the choice of the coordinate axes,so interchanging $x$ and $y$,the $\rho$ is also given by $\rho=\frac{(1+(\frac{dx}{dy})^2)^{\frac{3}{2}}}{\frac{d^2x}{dy^2}}$.
I did not understand why the radius of curvature does not depend upon the interchange of coordinate axes.Please help.
$\endgroup$ 31 Answer
$\begingroup$The book gives you a true statement but the last formula is questionable.
Curves' curvature is like talking about radius of a circle. With any coordinate transformation, the circle is still the same circle and so its radius.
However, interchanging x, y in the formula, may change sign if you use the formula.
Assuming you can parametrize the curve $x=x(t), y=y(t)$ $$\rho=\frac{(1+(\frac{dy}{dx})^2)^{\frac{3}{2}}}{\frac{d^2y}{dx^2}}=\frac{((x')^2+(y')^2)^{\frac{3}{2}}}{y''x'-x''y'}$$
So interchanging x and y will give you the same absolute value but with different sign.
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