Curvature of $y^m-x^n=0.$

Question

Let $(m,n)\in\mathbb{N}^2$ such that $\gcd(m,n)=1.$

I am looking at the equation $y^m-x^n=0.$ Denote $C$ the graph of this geometric curve.

How can I study the limit of the curvature of $C$ as we approach the origin ?

Answer 1

You can parameterise the equation as: $$f(t) = (t, t^{n/m})$$

$$f'(t) = (1,\frac{nt^{n/m-1}}{m}) $$

$$f''(t) = (0,\frac{n(n/m-1)t^{n/m-2}}{m}) $$

This approaches the origin as $t \rightarrow 0$

At 0, the radius of the osculating circle is at infinity or it has 0 curvature.

Study what you like from there. I was not completely sure what you were looking for, hope this helps!