Radius of curvature for the plane curve $x^3 + y^3 = 12xy$.

Question

Could someone help me with this problem? : Determine the radius of curvature for the plane curve $x^3 + y^3 = 12xy$ at the point $(0, 0)$.

Answer 1

Sketch: make the orthogonal change of coordinates $x+y=\sqrt{2}u,$ $x-y=\sqrt{2}v,$ then you can solve $v=\pm f(u)$, where $f(u)$ is some smooth function around $u=0$, $f(0)=0$ and $f'(0)\neq 0$. So our curve has a self-intersection at the origin, with each "branch" having the same curvature at the origin. Now take Taylor series at $u=0$, get $v=\pm(Au+Bu^2+...)$, which reduces to finding the curvature at the origin of the parabola $v=Au+Bu^2.$

Answer 2

The given curve is tangential to both the coordinate axes at the origin. So we can use Newton's formula. At the point of the curve where it is tangential to $x$-axis, the radius of curvature is

$\lim_{x \rightarrow 0}\frac{x^2}{2y}.$

Divide $x^3+y^3=12xy$ by $2xy$ and take limits as $x$ tends to zero. Since both $y$ and $y/x$ tend to zero as $x$ tends to zero we get $\rho=6$ as the radius of curvature at the origin where the curve is tangential to $x$-axis. At the origin when the curve is tangential to the $y$-axis, interchange the roles of $x$ and $y$ and you get the same result.