Find the curvature of a curve given by it's implicit equation

Question

Please help me solve this:

Find the curvature of a curve given by it's implicit equation:

$ F(x,y) = 0 $

Now, I thought of solving this by converting it to a parametric equation, but I have no idea how to do so. Any tips?

Answer 1

Suppose this does define a smooth parametric curve $x = X(s),\; y = Y(s)$ parametrized by arc length $s$ in the neighbourhood of a point $(x_0, y_0) = (X(0), Y(0))$. Differentiating the equation $F(X(s),Y(s))=0$ twice and using the chain rule (and using subscript notation for partial derivatives)

$$ \eqalign{F_1(X,Y) X' &+ F_2(X,Y) Y' = 0\cr F_{11}(X,Y) X'^2 &+ 2 F_{12}(X,Y) X' Y' + F_{22}(X,Y) Y'^2 + F_1(X,Y) X'' + F_2(X,Y) Y'' = 0\cr}$$ Since the parameter is arc length, the velocity vector $V =[X',Y']$ has length $1$ and is the unit tangent vector, i.e. $$ X'^2 + Y'^2 = 1$$ and differentiating this gives us $$X' X'' + Y' Y'' = 0$$

The curvature is then $\kappa = |V'| = \sqrt{X''^2 + Y''^2}$.

Eliminating $X'$ and $Y'$ from the equations is rather tedious, but should give us

$$ \kappa = \frac{\left| F_{11} F_2^2- 2 F_{12} F_1 F_2 + F_{22} F_1^2\right|}{(F_1^2 + F_2^2)^{3/2}}$$

Answer 2

WE can find Implicit form of Curvature by implicit differentiation in the plane

Formula (3.1)

Answer 3

https://en.m.wikipedia.org/wiki/Curvature

"Implicit curve" under "Curvature of plane curves"