DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

10 Chapter 1. Curvesof the road, starting at (0, −1), consider the vector −→ OC = −→ OP + −→ PQ+ −→ QC, where P = α(s) isthe point of contact and Q is the midpoint of the edge of the square. Use −→ QP = sα ′ (s) and thefact that −→ QC is a unit vector orthogonal to −→ QP . Express the fact that C moves horizontallyto show that s = − y′ (s)x ′ ;you will need to differentiate unexpectedly. Now use the result of(s)Exercise 4 to find y = f(x). Also see the hint for Exercise 9.)⎧⎨( )t, t sin(π/t) , t ≠011. Show that the curve α(t) =has infinite length on [0, 1]. (Hint: Considerl(α, P) with P = {0, 1/N, 2/(2N − 1), 1/(N − 1),...,1/2, 2/3,⎩(0, 0), t =01}.)12. Prove that no four distinct points on the twisted cubic (see Example 1(e)) lie on a plane.13. (a special case of a recent American Mathematical Monthly problem) Suppose α: [a, b] → R 2is a smooth parametrized plane curve (perhaps not arclength-parametrized). Prove that if thechord length ‖α(s) − α(t)‖ depends only on |s − t|, then α must be a (subset of) a line or acircle. (How many derivatives of α do you need to use?)2. Local Theory: Frenet FrameWhat distinguishes a circle or a helix from a line is their curvature, i.e., the tendency of thecurve to change direction. We shall now see that we can associate to each smooth (C 3 ) arclengthparametrizedcurve α a natural “moving frame” (an orthonormal basis for R 3 chosen at each pointon the curve, adapted to the geometry of the curve as much as possible).We begin with a fact from vector calculus which will appear throughout this course.Lemma 2.1. Suppose f, g :(a, b) → R 3 are differentiable and satisfy f(t) · g(t) =const for allt. Then f ′ (t) · g(t) =−f(t) · g ′ (t). Inparticular,‖f(t)‖ = const if and only if f(t) · f ′ (t) =0 for all t.Proof. Since a function is constant on an interval if and only if its derivative is everywherezero, we deduce from the product rule,(f · g) ′ (t) =f ′ (t) · g(t)+f(t) · g ′ (t),that if f · g is constant, then f · g ′ = −f ′ · g. Inparticular, ‖f‖ is constant if and only if ‖f‖ 2 = f · fis constant, and this occurs if and only if f · f ′ =0. □Remark. This result is intuitively clear. If a particle moves on a sphere centered at the origin,then its velocity vector must be orthogonal to its position vector; any component in the directionof the position vector would move the particle off the sphere. Similarly, suppose f and g haveconstant length and a constant angle between them. Then in order to maintain the constant angle,as f turns towards g, wesee that g must turn away from f at the same rate.Using Lemma 2.1 repeatedly, we now construct the Frenet frame of suitable regular curves. Weassume throughout that the curve α is parametrized by arclength. Then, for starters, α ′ (s) isthe