Math
The osculating sphere and spherical curves.
Step-by-step mathematics solution: The osculating sphere and spherical curves.
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1. What the problem asks
We have a regular space curve
[ c:\;I\to\mathbb R^{3},\qquad s\mapsto c(s) ]
parameterised by its arc length (s) (so (|c’(s)|=1)).
Let
[ e_{1}=T,\qquad e_{2}=N,\qquad e_{3}=B ]
be the Frenet frame, with curvature (\kappa(s)>0) and torsion (\tau(s)\neq0).
- (i) Show that the sphere whose centre
[ m(s_{0})=c(s_{0})+\frac{1}{\kappa(s_{0})}\,e_{2}(s_{0}) -\frac{\kappa’(s_{0})}{\tau(s_{0})\kappa^{2}(s_{0})}\,e_{3}(s_{0}) \tag{1} ]
passes through the point (c(s_{0})) and meets the curve at (s_{0}) with
contact of order three (third‑order contact).
The sphere is called the osculating sphere.
- (ii) Assume the curve is of class (C^{4}) and (\tau\neq0) everywhere.
Prove that the curve lies on a (fixed) sphere iff
[ \boxed{\displaystyle \frac{\tau}{\kappa}= \Bigl(\frac{\kappa’}{\tau\kappa^{2}}\Bigr)’} \tag{2} ]
2. Preliminaries – Frenet formulas
Because the curve is parametrised by arc length we have
[ \begin{aligned} T’ &= \kappa N ,\[2pt] N’ &=-\kappa T + \tau B ,\[2pt] B’ &=-\tau N . \end{aligned} \tag{3} ]
We shall also use
[ c’(s)=T(s),\qquad c’‘(s)=\kappa N ,\qquad c’’‘(s)= -\kappa^{2}T +\kappa’ N +\kappa\tau B . \tag{4} ]
3. Part (i) – The osculating sphere
3.1 Setting up the centre
Write the (as yet unknown) centre of the sphere as
[ m=m(s_{0})=c_{0}+ \alpha\,e_{1}(s_{0})+\beta\,e_{2}(s_{0})+\gamma\,e_{3}(s_{0}), \qquad c_{0}=c(s_{0}). \tag{5} ]
Define the squared distance from a point of the curve to the centre
[ f(s)=\bigl\langle c(s)-m,\;c(s)-m\bigr\rangle . \tag{6} ]
The sphere through (c_{0}) has radius
[ R^{2}=f(s_{0}) . ]
If the curve has third‑order contact with the sphere at (s_{0}) then, apart from the obvious equality (f(s_{0})=R^{2}), the first three derivatives of (f) must vanish at (s_{0}):
[ f’(s_{0})=f’‘(s_{0})=f’’‘(s_{0})=0 . \tag{7} ]
These three equations will determine (\alpha,\beta,\gamma).
3.2 Computing the derivatives
Because the centre (m) is constant, differentiation of (6) gives
[ \begin{aligned} f’(s) &= 2\langle c’(s),\,c(s)-m\rangle,\[2pt] f’‘(s) &= 2\langle c’‘(s),\,c(s)-m\rangle + 2\langle c’(s),c’(s)\rangle ,\[2pt] f’’‘(s) &= 2\langle c’’‘(s),\,c(s)-m\rangle +6\langle c’‘(s),c’(s)\rangle . \end{aligned} \tag{8} ]
Now evaluate at (s=s_{0}) and insert the Frenet expressions (4).
- First derivative
[ f’(s_{0}) = 2\langle T_{0},\,c_{0}-m\rangle = 2\langle T_{0},-\alpha T_{0}-\beta N_{0}-\gamma B_{0}\rangle = -2\alpha . ]
Thus (f’(s_{0})=0) gives
[ \boxed{\alpha =0}. \tag{9} ]
- Second derivative
[
\begin{aligned}
f’‘(s_{0}) &= 2\langle \kappa N_{0},\;c_{0}-m\rangle + 2\langle T_{0},T_{0}\rangle
&= 2\bigl(\kappa\langle N_{0}, -\beta N_{0}-\gamma B_{0}\rangle\bigr)+2
&= 2(-\kappa\beta)+2 .
\end{aligned}
]
Setting (f’‘(s_{0})=0) yields
[ \boxed{\beta = \frac{1}{\kappa(s_{0})}} . \tag{10} ]
- Third derivative
Because (\langle c’’ ,c’\rangle = \langle \kappa N_{0},T_{0}\rangle =0), the last term in (8) vanishes. Using (4),
[
\begin{aligned}
f’’‘(s_{0}) &= 2\bigl\langle -\kappa^{2}T_{0}+ \kappa’ N_{0}+ \kappa\tau B_{0},
\;c_{0}-m\bigr\rangle
&= 2\Bigl\langle -\kappa^{2}T_{0}+ \kappa’ N_{0}+ \kappa\tau B
Original question: The osculating sphere and spherical curves. on Mathematics Stack Exchange, licensed CC BY-SA.