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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.

Community
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Relure
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  • Well, no, they're telling you that the order of contact is precisely $3$. They're asking if the curve is locally on both sides of the osculating sphere. Think about the graph of $y=f(s)$ near $(s_0,0)$ in the $sy$-plane. – Ted Shifrin Mar 19 '17 at 23:29

1 Answers1

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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 $\gamma(s_0)$.

Christian Blatter
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