Parametrized curve along a sphere

Question

If we parametrized a $C^{\infty}$ curve in $\mathbb{R^3}$ so that the curve lies on a sphere centred at an arbitrary point and the speed along that curve is never zero, how would we show that for any point on the curve, the osculating circle for that point lies in the sphere?

Answer 1

We will make extensive use of the Frenet-equations' (moving) basis $(e_1,e_2,e_3)$.

The curve will be defined by $$ \gamma : [a,b] \to \partial B_\rho(c) $$ where $\partial B_\rho(c)$ is the surface of a sphere of radius $\rho$ around the center $c$.

We are given that the curve has never vanishing speed. Therefor, there is an arc length parametrization of the curve and we will just assume that the given curve is already one, i.e. $$ \| \gamma' \| \equiv 1 \,. $$

We need the Frenet basis \begin{align} e_1(t) &:= \gamma'(t) \\ e_2(t) &:= \frac{1}{\| \gamma''(t) \|} \gamma''(t) \\ e_3(t) &:= e_1 \times e_2 \,, \end{align} as well as curvature $\kappa(t) := \| \gamma''(t) \| $, and torsion $ \tau(t) := e_2' \cdot e_3 $.

The osculating circle has its center at $m(t) = \gamma(t) + \frac{1}{\kappa}e_2$, radius of $r = \frac{1}{\kappa}$, and lies in the $\{e_1,e_2\}$-plane.

We will have to distinguish between two different cases: $\tau(t) = 0$ and $\tau(t) \neq 0$.

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Case 1: $\tau(t) \neq 0$.

For spherical curves there is a theorem stating that the sphere can be computed only from the curve. The center $M$, and radius $R$ are given by [German wikipedia has the formula] \begin{align} M(t) &= \gamma(t) + \frac{1}{\kappa(t)}e_2(t) - \frac{\kappa'(t)}{\tau(t)\kappa(t)^2}e_3(t) \\ R(t) &= \sqrt{\frac{1}{\kappa(t)^2} + \left( \frac{\kappa'(t)}{\tau(t)\kappa(t)^2} \right)^2} = \text{const} = R \end{align}

Any point $p$ on the osculating circle has distance $r$ from its center and lies in the $\{e_1,e_2\}$-plane, i.e. $$ p = m + r (c_1 e_1 + c_2e_2) \quad \text{for} \quad 1=c_1^2 + c_2^2 \,. $$ Computing the distance between $p$ and $M$ gives \begin{align} \|p - M \| &= \| r (c_1 e_1 + c_2e_2) + \frac{\kappa'(t)}{\tau(t)\kappa(t)^2}e_3(t) \| \\ &= \sqrt{r^2 + \left[\frac{\kappa'(t)}{\tau(t)\kappa(t)^2} \right]^2} = R\,. \end{align} such that $p$ by definition lies on the surface! Thus, the osculating circle is contained on that sphere.

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Case 2: $\tau(t) = 0$.

Is actually just a special case of (1) where $M=m$ and $R=r$ hold. Thus, the osculating circle is also contained on that sphere.