###

DIFFERENTIAL GEOMETRY COURSE

5.*CURVATURE AND TORSION*

*Explanation 42:* We can give a proof analogous to the one in explanation 22 of:

τ(s)^{2} = lim (Δs→0) (Δψ/Δs)^{2},
where Δψ is the angle between the binormals
in s and s+Δs.

Here we use ||__b__ ^{.}(s)||^{2} = τ(s)^{2} (see proposition 38).

*Explanation 43:* According to Taylor we have __x__(s) = __x__(s_{0}) + (s-s_{0})__x__ ^{.}(s_{0}) +
(1/2)(s-s_{0})^{2}__x__ ^{..}(s_{0}) + (1/6)(s-s_{0})^{3}__x__ ^{...}(s_{0}) + O(s-s_{0})^{4}
= __x__(s_{0}) + h__t__ + (1/2)h^{2}κ__n__ + (1/6)h^{3}(κ ^{.}__n__+
κ(-κ__t__+τ__b__)) + ...

If we choose __x__(s_{0}) as origin, then the coordinates of __x__(s) with respect to (__t__,__n__,__b__) are in approximation:

α = h - (1/6)κ^{2}h^{3}, β = (1/2)κh^{2} + (1/6)κ ^{.} h^{3}, γ =
(1/6)κτh^{3}.

So the projection on the osculating plane is in approximation the parabola (h,(1/2)κh^{2}). This gives us a geometric insight in curvature.

The projection on the normal plane is in approximation the cubic parabola (β,__+__((τ√2)/(3√κ))β^{3/2}). This gives us a geometric insight in torsion.

The projection on the rectifying plane is in approximation the cubic curve (α,(1/6)κτα^{3}).

*Proposition 44:* The planar curves are exactly the curves with torsion 0.

*Prove*: Suppose τ=0. Then __b__ ^{.} = __0__, so __b__ is constant.

The osculating plane has equation __x__.__b__ = a(s) where a(s) = __x__(s).__b__.

Then a ^{.}(s) = __t__(s).__b__ = 0, so a is constant. So the curve is lying in the plane __x__.__b__ = a.

On the other hand, every planar curve has torsion 0, because the binormal is constant (normal vector of length 1 perpendicular to the plane of the curve).

*Problem 45:*Let d = ||__x__(s) - __x__(s_{0})|| and h = s-s_{0}.

Prove with the help of explanation 43 that lim (h→0) (h-d)/h^{3} = κ(s_{0})^{2}/24. Give a geometric interpretation.

*Problem 46:* Prove that the circular helix (see 11 ii) ) has constant curvature and constant torsion.

*Problem 47:* Calculate κ(t) and τ(t) of the following curves (use 20,40,41) :

i) __x__(t) = (a(t-sin(t)), a(1-cos(t)), 4b*cos(t/2));

ii) __y__(t) = (3cos(t)+3t*sin(t), 3sin(t)-3t*cos(t), 2t^{2});

iii) __z__(t) = (6t, 3t^{2}, t^{3}).

These are there examples of helices (but not circular), of which we will speak further later on.

*Problem 48:* Calculate the arc length s of the conic curve in problem 19: __x__(t) = a*e^{ct}(cos(t),sin(t),1). Subsequently, explain how we can calculate κ(s) and τ(s).

*Proposition 49:* Every curve is uniquely determined by its curvature function and its torsion function, up to a stiff motion in space.

*Explanation50:* This follows from a theorem about differential equations, when we apply them on the formulas of Serret and Frenet. This theorem says that under certain conditions n equations
with n unknown variables have exactly 1 solution (up to integration constants).

In the following sections we will see examples where a certain sort of curves is characterised by an equation involving κ and/or τ.

In general, a system of two such equations determines the curve. (Then we call them 'natural equations' of the curve.)

*Problem 51:* Suppose the principle normals of a curve go through a fixed point. Prove that the curve is a circle.

answers

HOME