Let $\gamma(s)$ be a curve in $\mathbb{R}^3$ parametrized by its arc length, with curvature and torsion not $0$. Let $f(s)=\mid\mid \gamma(s) - C(s_0) \mid \mid ^2-r(s_0)^2$, where $C(s_0)$ is the center of the osculator sphere and $r(s_0)$ is the radii of this sphere.
Suppose that $\ f(s_0)=f'(s_0)=f''(s_0)=f'''(s_0)=0$, and that $f^{(4)}(s_0)\neq0.$ Decide if $\gamma(s)$ cross the osculator sphere on $\gamma(s_0)$.
I really don't know how to attack this problem. I know that
$$C(s_0)=\gamma(s)+\frac{1}{k(s_0)}N(s_0)+\frac{k'(s_0)}{k(s_0)^2\tau(s_0)}B(s_0)$$
and
$$r(s_0)=\displaystyle\sqrt{\frac{1}{k(s_0)^2}+\Big(\frac{k'(s_0)}{k(s_0)^2\tau(s_0)}\Big)^2}$$
They're asking me to prove that the contact between the sphere and the curve is exactly $3$, but I don't know how to do it with no information about the curve.
What are the criterions to decide whether or not the curve crosses the sphere?
Thanks for your time.
You don't need the formulas you quote.
Since it is specifically assumed that the first nonvanishing derivative of $f$ at $s_0$ has even order $4$ the function has a strict local extremum at $s_0$. It follows that the curve does not cross the osculator sphere at the point $\gamma(s_0)$.