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The Osculating Circumference Problem THE OSCULATING CIRCUMFERENCE PROBLEM School I.S.I.S.S. M. Casagrande Pieve di Soligo, Treviso Italy Students Silvia Micheletto (I) Azzurra Soa Pizzolotto (IV) Luca Barisan (II) Simona Sota (IV) Gaia Barella (IV) Paolo Barisan (V) Marco Casagrande (IV) Anna De Biasi (V) Chiara De Rosso (IV) Silvia Giovani (V) Maddalena Favaro (IV) Klara Metaliu (V) Teachers Fabio Breda Francesco Maria Cardano Davide Palma Francesco Zampieri Researcher Alberto Zanardo, University of Padova Italy Year 2019/2020 The osculating circumference problem ISISS M. Casagrande, Pieve di Soligo, Treviso Italy Abstract The aim of the article is to study the osculating circumference i.e. the circumference which best approximates the graph of a curve at one of its points. We will dene this circumference and we will describe several methods to nd it. Finally we will introduce the notions of round points and crossing points, points of the curve where the circumference has interesting properties. ??? Contents Introduction 3 1 The osculating circumference 5 1.1 Particular cases: the formulas cannot be applied . 9 1.2 Curvature . 14 2 Round points 15 3 Beyond round points 22 3.1 Crossing points . 23 4 Osculating circumference of a conic section 25 4.1 The rst method . 25 4.2 The second method . 26 References 31 2 The osculating circumference problem ISISS M. Casagrande, Pieve di Soligo, Treviso Italy Introduction Our research was born from an analytic geometry problem studied during the third year of high school. The problem. Let P : y = x2 be a parabola and A and B two points of the curve symmetrical about the axis of the parabola. Find the equation of the circumference C tangent to the parabola in the given points. We solved the problem but it led to some interesting questions. 2 2 Let A(x0; x0) and B = (−x0; x0) be the two symmetrical points. We look for the center of the circumference and we notice that it must belong to the y-axis due to symmetry reasons. This center can be determined as the intersection point between the normal lines to the curve respectively at A and B. The coordinates of the center C depend on the coordinates of the two tangency points A e B: we wondered how the coordinates of C vary when A and B vary, in particular if A and B approach to the vertix V of the parabola (x0 ! 0). Our first hypothesis was that if A ! V and B ! V then yC ! +1, given that the absolute value of the slopes of the normal lines increases when A ! V and B ! V . However we noticed that this hypothesis is not correct because there is another variation that should be considered that is the decrease of the ordinates of A and B. Our second hypothesis was that if A ! V and B ! V then yC ! yV . However we understood that even this guess is wrong. In fact we are able to prove that the point C approaches to a point dierent from V . We determine the equations of the normal lines to the curve nA and nB at A and B and the center C as the intersection point between the two normal lines. The equations of the tangent lines in A and B are: 2 2 tA : y = 2xx0 + x0; tB : y = −2xx0 − x0: 3 It follows that the equations of the normal lines are: 2 1 2 1 nA : y = x0 − (x − x0); nB : y = x0 − (x + x0): 2x0 2x0 The coordinates of C are the solutions of the system 8 2 1 > y = x0 − (x − x0) <> 2x0 ; > 2 1 :> y = x0 + (x + x0) 2x0 and we get 1 C 0; x2 + : 0 2 Then if and then 1 . This is a consequence of A ! V B ! V yC ! 2 2 1 1 lim yC = lim yC = lim yC = lim x0 + = : A!V B!V x0!0 x0!0 2 2 q This means that approaches to the point 1 . Since 2 1 the equation of the C 0; 2 CA = CB = x0 + 4 circumference C tangent at the parabola at A and B is 1 2 1 C : x2 + y − x2 − = x2 + )C : x2 + y2 − (2x2 + 1)y − x4 = 0: 0 2 0 4 0 0 In the case in which A ! V and B ! V the circumference C approaches to the circumference of equation x2 + y2 − y = 0. This circumference has some interesting properties: • it intersects the parabola in the vertex V ; • it is tangent to the parabola in the vertex V ; • and nally we observed that the graphs of the circumference and of the parabola, in a neigh- bourhood of the vertex V , are very similar. Here starts our research. Given a plane curve, we are now looking for the circumference that has got these three properties in one of its points. Pieve di Soligo, 2020, June. 4 1 The osculating circumference We started by giving a denition of what being very similar means and in order to do so we analysed the Taylor series expansion. However a Taylor series is dened for a function and we know that a circumference is not the graphic of a function so to overcome this problem we focused on the semicircumferences. Denition 1.0.1. Given the circumference C of radius r and center C(xC ; yC ). The upper semi- circumference U is the set of the points (x; y) 2 C such that y > yC . The lower semicircunference L is the set of the points (x; y) 2 C such that y < yC . Given U or L, the circumference C is called the associated circumference to U or L. 3 00 Denition 1.0.2. Let f 2 C (I), with I ⊆ R, a function and a point x0 2 I such that f (x0) 6= 0. The osculating circumference Cf (x0) of a function f in x0 is the circumference associated 00 00 to the upper semicircumference (if f (x0) < 0) or the lower semicircumference (if f (x0) > 0) which has the same Taylor series expansion to second order of f at x0. 3 Theorem 1.0.3. Let Cf (x0) be the osculating circumference of a function f 2 C (I) in x0 2 I, then the radius r is given by 3 0 2 2 1 + f (x0) r = 00 ; j f (x0) j and the centre C has the following coordinates 0 0 2! f (x0) 0 2 1 + f (x0) C x0 − 00 1 + f (x0) ; f(x0) + 00 : f (x0) f (x0) 3 Proof. Let f 2 C (I)a function and C(xC ; yC ) be the centre and r the radius of the osculating 2 2 2 circumference of the function in x0. The circumference has equation (x − xC ) + (y − yC ) = r . Let's distinguish four cases: 0 00 (a) f (x0) > 0; f (x0) > 0, 0 00 (b) f (x0) > 0; f (x0) < 0, 0 00 (c) f (x0) < 0; f (x0) > 0, 0 00 (d) f (x0) < 0; f (x0) < 0. p 2 2 00 In case (a) the semicircumference has equation y = yC − r − (x − xC ) since f (x0) > 0. The second order Taylor polynomial of f at x0 is f 0(x ) f 00(x ) f(x ) + 0 (x − x ) + 0 (x − x )2; 0 1! 0 2! 0 while the second order Taylor polynomial of y at x0 is y0(x ) y00(x ) y(x ) + 0 (x − x ) + 0 (x − x )2: 0 1! 0 2! 0 5 Given that these two expression must be equal according to the Denition 1.0.2, we have to impose that the two functions and their prime and second derivatives must be equal: 8 f(x ) = y − pr2 − (x − x )2 > 0 C 0 C > > > > 0 x0 − xC <f (x0) = p 2 2 (1) r − (x0 − xC ) : > > > 2 > 00 r >f (x0) = > p 2 2 3 : [r − (x0 − xC ) ] 0 From the second of (1) and from f (x0) > 0 we deduce that x0 − xC > 0. If we consider the second power of both sides of the second equation of (1) we get 0 2 2 2 f (x0) r (2) (x0 − xC ) = 0 2 : 1 + f (x0) The second power of both sides of the third of (1) leads to 00 2 2 2 3 4 f (x0) [r − (x0 − xC ) ] = r ; (3) subsituting (2) in (3) we get an explicit expression for r: 0 2 2 3 1 + f 0(x )23 00 2 2 f (x0) r 4 2 0 (4) f (x0) r − 0 2 = r =) r = 00 2 ; 1 + f (x0) f (x0) and so 3 1 + f 0(x )2 2 0 (5) r = 00 : j f (x0) j By substituting (4) in (2) we can nd xC : 0 f (x0) 0 2 (6) x0 − xC = 00 1 + f (x0) ; f (x0) and so by hypothesis 0 f (x0) 0 2 xC = x0 − 00 1 + f (x0) : f (x0) And then nally we substitute (4) and the square of (6) in the rst equation of (1) in order to get yC : 0 2 f (x0) + 1 yC = f(x0) + 00 ; f (x0) where we took away the absolute value by hypothesis. By the analysis of the four cases we got 0 00 the same radius (5) beacuse the signes of f (x0) and f (x0) do not aect the expression of r. As far the coorinadates of C are concerned we get: ! f 0(x ) 1 + f 0(x )2 (a) 0 0 2 0 C x0 − 00 1 + f (x0) ; f(x0) + 00 ; f (x0) f (x0) ! f 0(x ) 1 + f 0(x )2 (b) 0 0 2 0 C x0 + 00 1 + f (x0) ; f(x0) − 00 ; f (x0) f (x0) 6 ! f 0(x ) 1 + f 0(x )2 (c) 0 0 2 0 C x0 + 00 1 + f (x0) ; f(x0) + 00 ; f (x0) f (x0) ! f 0(x ) 1 + f 0(x )2 (d) 0 0 2 0 C x0 − 00 1 + f (x0) ; f(x0) − 00 : f (x0) f (x0) 0 00 We noticed that the combination of the signes of f (x0) and f (x0) lead always to the same expresssion: 0 0 2! f (x0) 0 2 1 + f (x0) C x0 − 00 1 + f (x0) ; f(x0) + 00 : f (x0) f (x0) 3 00 Theorem 1.0.4.
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