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DIFFERENTIAL GEOMETRY COURSE

4.* SERRET AND FRENET*

*Definition 30:* The osculating plane (kissing plane) in a point __x__(t_{0}) of the curve __x__(t) is the limit of the position of the plane through the tangent in
__x__(t_{0}) and a neighbour point __x__(t_{0}+Δt) on the curve, when Δt→0.

*Proposition 31:* The osculating plane in __x__(t_{0}) is the plane with parameter representation
__x__(t_{0})+λ__x__'(t_{0})+μ__x__''(t_{0}) and equation
det(__x__-__x__(t_{0}),__x__'(t_{0}),__x__''(t_{0})=0.

*Prove:* Since __x__(t_{0}+Δt)-__x__(t_{0})=__x__'(t_{0})+(1/2)__x__''(t_{0})Δt^{2}
(+ terms of higher degree), the equation det(__x__-__x__(t_{0}),__x__'(t_{0}),__x__(t_{0}+Δt)-__x__(t_{0}))
= 0 becomes, when Δt→0, the equation det(__x__-__x__(t_{0}),__x__'(t_{0}),__x__''(t_{0})=0.

*Definition 32:* We call a point __x__(s_{0}) on a curve __x__(s) whose parameter s is arc length, an inflection point of the curve if κ(s) changes sign in
s_{0} (then the local curvature vector is *0*).

*Proposition 33:* In the inflection points of the curve __x__(t), the vectors __x__' and __x__'' have either the same direction or opposite directions.

*Prove*: According to problem 20 we find : __x__ ^{..} = *0* ↔ __x__'⊗__x__'' = *0*.

*Problem 34:* Determine the equation of the osculatiom plane in an inflection point.

*Problem 35:* Give the equation of the osculating plane in a point __x__(t) of the cubic parabola from example 11 iii).

*Definition 36: (Serret-Frenet frame)*

Let __x__(s) be a curve whose parameter s is arc length.
Let __t__(s) := __x__ ^{.}(s) be the tangent of length 1. Choose the principal normal vector __n__(s) with length 1,
having the same or opposite direction as the curvature vector __x__ ^{..}(s), as we discussed in the previous section.

So the osculating plane is __x__(s)+λ__t__(s)+μ__n__(s).

Start with __t__(s),__n__(s) and complete the frame with the so-called binormal
vector __b__(s) so as to get a right orthonormal frame (that is to say: __b__ := __t__⊗__n__).
Then we call __t__,__n__,__b__ The Serret-Frenet frame.

We call the plane __x__(s)+λ__n__(s)+μ__b__(s) the normal plane, and the plane __x__(s)+λ__t__(s)+μ__b__(s) the rectifying
plane.

*Problem 37:* Determine the equations of these three planes in an arbitrary point of the cubic parabola.

*Proposition 38 (Theorem of Serret and Frenet):*

There exist functions κ(s) and τ(s) with

*Prove:* The function κ is the curvature function, and we discussed the formula __t__** **^{.} = κ__n__ in the previous section.

According to problem 9b), __n__ is perpendicular to __n__** **^{.}, so according to problem 8 there exist ρ and τ such that __n__** **^{.} =
ρ__t__+τ__b__. Again according to problem 8, we have ρ=__n__** **^{.}.__t__.

Using __n__.__t__=0 we see by differentiation __n__** **^{.}.__t__+__n__.__t__** **^{.} = 0, so
ρ = __n__** **^{.}.__t__ = -__n__.__t__** **^{.} = -κ.

Likewise: since __b__ is perpendicular to __b__** **^{.}, we have __b__** **^{.} = α__t__+β__b__.

Then α = __b__** **^{.}.__t__ = -__b__.__t__** **^{.} = 0, and β = __b__** **^{.}.__t__ =
-__b__.__n__** **^{.} = -τ.

*Definition 39:* The function τ from the theorem of Serret and Frenet is called the torsion of the curve.

*Proposition 40:* κ^{2}τ = det(__x__ ^{.},__x__ ^{..},__x__ ^{...}).

*Prove:* Using __x__ ^{..} = κ__n__ we find __x__ ^{...} = κ ^{.}__n__ +
κ__n__ ^{.} = κ ^{.}__n__ + κ(-κ__t__+τ__b__), so

det(__x__ ^{.},__x__ ^{..},__x__ ^{...}) =
det((1,0,0),(0,κ,0),(-κ^{2},κ ^{.},κτ)) = κ^{2}τ.

*Problem 41:* Prove that det(__x__',__x__'',__x__''') = ||__x__'||^{6}det(__x__ ^{.},__x__ ^{..},__x__ ^{...}),
and deduce:

τ = det(__x__',__x__'',__x__''')/(__x__'⊗__x__'')^{2}

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