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Let α: (a, b) → R2 be a regular curve which does not pass through the origin. Fix t0 and choose φ0 such that α (t0 ) · α(t0 ) = cos φ0 α (t0 ) α(t0 ) and α (t0 ) · J α(t0 ) = sin φ0 . α (t0 ) α(t0 ) Establish the existence and uniqueness of a differentiable function φ = φ[α]: (a, b) → R such that φ(t0 ) = φ0 , α (t) · α(t) = cos φ(t) α (t) α(t) and α (t) · J α(t) = sin φ(t) α (t) α(t) 28 CHAPTER 1. CURVES IN THE PLANE for a < t < b. Geometrically, φ(t) represents the angle between the radius vector α(t) and the tangent vector α (t).

11) and obtain 2at2 2at3 x= , y= . 12) cissoid[a](t) = 2at2 2at3 , . 2 1 + t 1 + t2 The Greeks used the cissoid of Diocles to try to find solutions to the problems of doubling a cube and trisecting an angle. For more historical information and the definitions used by the Greeks and Newton see [BrKn, pages 9–12], [Gomes, volume 1, pages 1–25] and [Lock, pages 130–133]. Cissoid means ‘ivy-shaped’. Observe that cissoid[a] (0) = 0 so that cissoid is not regular at 0. 14. The definition of the cissoid used by the Greeks and by Newton can best be explained by considering a generalization of the cissoid.

T0 d (he−iθ ) = e−iθ (h − i hθ ) = e−iθ (h − hh h) = 0. dt = c for some constant c. Since h(t0 ) = eiθ0 , it follows that c = h(t0 )e−iθ(t0 ) = 1. 18). Let θˆ be another continuous function such that ˆ 0 ) = θ0 , θ(t ˆ ˆ f (t) = cos θ(t) and g(t) = sin θ(t) 18 CHAPTER 1. CURVES IN THE PLANE ˆ for a < t < b. Then eiθ(t) = eiθ(t) for a < t < b. Since both θ and θˆ are continuous, there is an integer n such that ˆ = 2πn θ(t) − θ(t) ˆ 0 ), so that n = 0. Hence θ and θˆ coincide. for a < t < b. But θ(t0 ) = θ(t We can now apply this lemma to deduce the existence and uniqueness of a differentiable angle function between curves in R2 .

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