If a curve x(t) lies on a surface F(x)=0, then with g(t)=F(x(t)) we have g(t)=0 for all t.
Also we find that for all to all derivatives g(k) are 0 in to.
We say the curve has in each point a contact of order ∞ with the surface.

Definition 52 : We say the surface F(x)=0 and the curve x(t) have in x(to) a contact of order m if, with g(t)=F(x(t)), we have g(k)(to)=0 for k=0,1,...,m and g(m+1)(to)≠0.

Explanation 53 : Contact of order 0 means that the curve intersects the surface in x(to), contact of order 1 means that the curve is tangent to the surface in x(to). (Indeed, when there's contact of order 1 then g'(to) = (Fx1x1' + Fx2x2')|x(to) = ∇F. x'|x(to) = 0, whilst ∇F|x(to) is a normal vector of the tangent plane in x(to).)

Problem 54 : Show that definition 52 is independent of the choice of the parameter.

Problem 55 : Show that the plane that in a point x(to) has maximal contact with the curve x(t) is the osculating plane in x(to).

Example 56 : We seek the sphere that has a maximal contact with x(s) in x(so).
Let the equation of this sphere be ||x - m||2 = r2.
Then g(s) = (x(s) - m).(x(s) - m) - r2.

i) The requirement g(so)=0 expresses that x(so) must lie on the sphere we seek.
ii) The requirement g'(so)=0, that is (x(s) - m). x .(s) = 0 in so, expresses that the connecting line between x(so) and the center must be perpendicular to the vector which is in x(so) tangent to the curve.
iii) The requirement g"(so)=0 becomes here x .(s). x .(s) + (x(s) - m). x ..(s)=0 in so, so (m - x(s)). n(s) = 1/κ(s) in so;
this requirement expresses that the projection of the center of the osculating sphere onto the principal normal must coincide with the center of the osculating circle, x(so) + n(so)/κ(so).
So, if we write m = x(so) + αt(so) + βn(so) + γb(so), then we find m = x(so) + n(so)/κ(so) + γb(so), where we still have to determine γ. We call this line of remaining candidates the curvature axis.
iv) g(3)(so)=0 gives (x(s)-m). x ...(s) =0 in so, that is (n(so)/κ(so) + γb(so)).(κ .(so)n(so) + κ(so)n .(so)) =0.
Using n . = -κtb we find γ = -κ ./(κ2τ)|so.

Definition 57 : The osculating sphere in the point x(so) of the curve x(s) is the sphere with center x(so) + n(so)/κ(so) - κ ./(κ2τ)|sob(so) and radius √(κ-2 + κ .2τ-2κ-4)|so.

Proposition 58 : The curves that lie on a sphere are exactly the curves that satisfy a natural equation of the form κ-2 + κ .2τ-2κ-4 = c (c constant), except that curves with constant κ need not lie on a sphere.
(Give an example of a curve with constant curvature that lies on a sphere, and two examples of curves with constant curvature that don't lie on a sphere.)

Proof : If a curve lies on a sphere, then in each point the osculating sphere coincides with the sphere where the curve lies on, so that κ-2 + κ .2τ-2κ-4 is constant and equal to the square of the radius of that sphere.
On the other hand, from κ-2 + κ .2τ-2κ-4 = c we find that the centre of the osculating sphere is a fixed point (see problem 59), so the distance between any point on the curve and this fixed point is equal to √c.

Problem 59 : Prove by differentiation of the centre of the osculating sphere that this is a fixed punt if κ-2 + κ .2τ-2κ-4 = c and κ is not constant.

Problem 60 : Determine the curvature as a function of the arc length for a planar curve with the property that the centers M of the segments PK, where P is on the curve and K is the curvature centre at P all lie on one straight line.
(Hint: calculate using R as shorthand for κ-1; it turns out the curve is a cycloid.)

Problem 61 : Let φ be a differentiable function of t. Prove that the curve x(t) = (t cos(φ(t)), t sin(φ(t)), √(1-t2)), t ∈ (0,1), lies on a sphere.
Which differential equation must hold for φ if in each point of the curve the angle between the tangent and the x1-axis is 90 degrees?

Problem 62 : Prove that the following assertions hold:
a) If the osculating planes of a curve all go through a fixed point, then the curve is planar.
b) If the osculating planes of a curve all are parallel to a fixed straight line, then the curve is planar.

Problem 63 : Prove that a curve whose curvature centres all lie on a fixed straight line, must be a circle.

Problem 64 : In each point of a curve C there is a normal that goes through a fixed point A. Prove that C lies on a sphere whose centre is A.

Problem 65 : The straight lines connecting a fixed point with a point P running through a given curve are all perpendicular to this curve whilst the angles between these straight lines and the principal normal in P are all equal. Prove this curve must be a circle.

Problem 66 : Let V(s) be the rectifying plane in the point P(s) of a curve whose parameter s is arc length.
Let the curvature κ(s) in P(s) be positive. The distance between a fixed point M and P(s) is r(s). The distance between M and V(s) is a(s).
Prove that 2 a κ = |(d2/ds2)(r2-s2)|.