The Maclaurin Trisectrix

Transcription

1 The Maclaurin Trisectrix Justin Seago December 8, 28

2 History In 1742 Colin Maclaurin investigated the curve now known as the Maclaurin trisectrix in an effort to solve one of the geometric roblems of antiquity: that of trisecting an arbitrary angle. The Greeks sought to trisect an angle using only a comass and unmarked straight edge; however, in 185 Wantzel roved this was imossible. [7] That is not to say it is altogether imossible to trisect any angle on the contrary, there are many ways to go about it. [2] There are, however, many individuals who claim to have trisected an angle when they find an aroximation and do not see any reason for roving their assertion with mathematical rigor. For many ractical alications that s just fine, but any mathematician worth his or her salt knows better than to assume that a result is correct just because it seems to be correct. Maclaurin was certainly a worthy mathematician, and as we will see, he discovered a nice little method for recisely trisecting angles. Before we jum into the mathematics of the Maclaurin trisectrix, there are a few things to be said of Mr. Maclaurin. He was born in Kilmodan, Scotland in By the time he was ten years old, both of his arents had died and his uncle Daniel Maclaurin was charged with his ubringing. Maclaurin excelled as a student and entered the University of Glasgow at the age of 11 and received his M.A. when he was 14. At 19, he was elected for rofessorshi at Marischal College in the University of Aberdeen. [5] For this Maclaurin reortedly held a world record as the youngest rofessor until Aril 28. [4] He met Isaac Newton in 1719 on a tri to London and later wrote a two-volume work entitled A Treatise on Fluxions (1742) wherein he defended Newton s calculus. [][5] Other significant ublications by Maclaurin include Geometria Organica (172), A Treatise of Algebra (1748), and An Account of Sir Isaac Newton s Philosohy (1748). [6] Maclaurin sent most of his life a bachelor until he married Anne Stewart in 1725, with whom he had seven children. Maclaurin died of drosy on the June 14 th, Two of his aforementioned works were ublished osthumously. He is best remembered today for his work on the secial Taylor Series centered about zero that bears his name. [5]

3 Equations of the Maclaurin Trisectrix With the asymtote at x = a, origin at crunode Imlicit Cartesian y 2 = x2 x C a a Kx Parametric x = a t2 K t 2 C1, y = a t t2 K t 2 C1 Polar r = K 2 a sin θ sin 2 θ With the asymtote at x = a Polar r = Ka sec 1 θ With asymtote at x = 1 2 a [8] Imlicit Cartesian x Cxy = 1 2 a y2 K x 2 Polar r = K 1 2 a 4 cos θ Q sec θ = Ka sin θ sin 2θ Parametric x =K 1 2 a 4 cos2 θ K1, y =K 1 2 a 4 cos2 θ K1 tan θ [1]

4 Definition 1. Begin with a circle centered at a oint C with a radius CO along the x-axis. 2. Bisect the radius CO with a oint D and run a vertical line in the y-direction through D. 2. Draw a line OBA from O intersecting the vertical line at B and the circle at A.. On the line segment OA, create a oint P that is the same distance from O as the distance AB. That is, lace P on OA such that OP = AB. The locus of the oint P as A traces the circumference of the circle is the Maclaurin trisectrix. [1] * * Note that I have chosen to make the trisectrix flied about the x and y axes from the forms mentioned on the receeding age. This oddly seems unconventional, desite the fact that it allows one avoid several negative signs in the resulting equations.

5 Parametrizing Let OC = a and AOC = θ, then x = OPcosθ = ABcosθ Also AB = OA OB If we let OA = c, then by the law of sines 1/a sinθ = 1/c sin(π 2θ) c = a[sin(π 2θ)] sinθ = a[sin(π)cos(2θ) sin(2θ)cos(π)] sinθ = a sin(2θ) sinθ = 2a cosθ Let OB = b, then b cosθ = ½a b = ½a secθ θ Hence, AB = c b = 2a cosθ ½a secθ x = ABcosθ = (2a cosθ ½a secθ)cosθ = 2a cos 2 θ ½a = ½a(4cos 2 θ 1) y = ABsinθ = (2a cosθ ½a secθ)sinθ = 2a cos(θ)sin(θ) ½a tanθ = ½a[4cos 2 (θ)sin(θ) sin(θ) 1]tanθ = ½a(4cos 2 θ 1)tanθ Thus, x = ½a(4cos 2 θ 1) y = ½a(4cos 2 θ 1)tanθ are the arametric equations of the Maclaurin trisectrix with resect to θ.

6 Polar and Cartesian Derivations From the arametric equations we may easily deduce a olar equation since x = r cosθ r cosθ = ½a(4cos 2 θ 1) r = ½a(4cosθ secθ) which is the olar equation of the trisectrix. To find the Cartesian equation, note that r = x 2 + y 2 by the Pythagorean theorem. So x = r cosθ 5 x x 2 + y 2 = cosθ. x 2 + y 2 = ½a[4x x 2 + y 2 x 2 + y 2 x] Multilying both sides by x 2 + y 2 we get x 2 + y 2 = ½a[4x (x 2 + y 2 ) x] x + xy 2 = ½a(4x 2 x 2 y 2 ) x + xy 2 = ½a(x 2 y 2 ) which is the imlicit Cartesian equation of the Maclaurin trisectrix. Trisection Proof Let OC = 1 (i.e., a = 1), EOP = θ, and ECP = α. We must show that θ = ⅓α. Note that tanα = rsin(θ)/(rcosθ 1). Recall from the olar equation of the trisectrix that r = ½a(4cosθ secθ) which imlies that tanα is equal to tanθ after some difficult algebraic maniulation I used Male to verify this: cos θ Ksec θ $sin θ test relation = tan θ cos θ Ksec θ $cos θ K1 true Thus, θ = ⅓α an exact trisection. P r rsinθ O θ rcosθ C α E

7 .6 Other Proerties The Maclaurin trisectrix is an anallagmatic curve, meaning it is the same when it is inverted. Here I have lotted the trisectrix for values of the form (x, y) and (x, -y) with the lot (x, -y) in black having a LineWidth of 8 oints and the (x, y) lot in blue with a LineWidth of 4 oints. This is a vector grahic so you should be able too zoom in a bit for a closer look. MATLAB Plot of the Maclaurin Trisectrix (a = 1) x =.5a (4cos 2 2 θ - 1), y = -.5a (4cosθ - 1) tan θ x =.5a (4cos 2 2 θ - 1), y =.5a (4cos θ - 1) tan θ.4.2 y-axis x-axis The MATLAB code that roduced the image above: clc close all %% (x,-y) theta=linsace(-i/-.15,i/+.15,1); a=1; x=.5*a*(4*cos(theta).^2-1); y=-.5*a*(4*cos(theta).^2-1).*tan(theta); lot(x,y,'linewidth',8,'color','k') % comet(x,y) hold on %% (x, y) x=.5*a*(4*cos(theta).^2-1); y=.5*a*(4*cos(theta).^2-1).*tan(theta); lot(x,y,'linewidth',4,'color',[.5,.8,.8]) % comet(x,y) legend('x =.5a (4cos^2\theta - 1), y = -.5a (4cos^2\theta - 1) tan\theta',... 'x =.5a (4cos^2\theta - 1), y =.5a (4cos^2\theta - 1) tan\theta') title('matlab Plot of the Maclaurin Trisectrix (a = 1)') xlabel('x-axis') ylabel('y-axis') grid on axis equal

8 As a Pedal Curve The Maclaurin Trisectrix is the edal curve of a arabola with a edal oint mirroring the focus of the arabola across the directrix. Note the asymtote, which is the same distance (½a) from the double oint of the trisectrix as the focus of the arabola is from its vertex. y = ax 2 Focus Directrix Pedal Point

9 Area of the Loo A = b y q x' q dq a x = 1 2 a 4 cos2 q K1 = 1 2 a 4 cos2 q K1 tan q $4a$cos q sin q dq y = 1 2 a 4 cos2 q K1 tan q x' q = 4 a$cos q sin q K = 2 a 2 $ 4 cos q sin q Kcos q sin q tan q dq K = 4 a 2 $ = 4 a 2 $ 4 cos 2 q sin 2 q Ksin 2 q dq Ccos 2q K cos 2q K Kcos 2q dq = 4 a 2 $ 1 K cos 2 2q K 1 2 C 1 2 cos 2q dq = 4 a 2 $ = 2 a 2 $ 1 2 K C cos 4q C 1 2 Kcos 4q C cos 2q dq cos 2q dq π/ ^ = 2 a 2 $ K 1 4 sin 4 q C 1 2 sin 2 q O (/2a,) = a 2 $ K 1 2 K 2 = a 2 $ 4 C 2 -π/

10 References [1] Basset, A. B. (191), An Elementary Treatise on Cubic and Quartic Curves, 81, htt://books.google.com/books?id=yuxtaaaamaaj [2] Loy, Jim. Trisection of an Angle. htt:// [] MacLaurin, Colin (181), A Treatise on Fluxions: In Two Volumes, htt://books.google.com/books?id=qogaaaamaaj [4] McNeill, David (28), University aoints world s youngest rofessor, htt://www. indeendent.co.uk/news/world/asia/university-aoints-child-rodigy-worlds-youngestrofessor html [5] O Connor, J. J. and Robertson, E. F. (1999). Maclaurin biograhy. htt://www-grous.dcs.st-and.ac.uk/~history/biograhies/maclaurin.html [6] O Connor, J. J. and Robertson, E. F. (27). Mathematical Works of Colin Maclaurin. htt://www-grous.dcs.st-and.ac.uk/~history/extras/maclaurin_ublications.html [7] Weisstein, Eric W. Angle Trisection. From MathWorld A Wolfram Web Resource. htt://mathworld.wolfram.com/angletrisection.html [8] Weisstein, Eric W. Maclaurin Trisectrix. From MathWorld A Wolfram Web Resource. htt://mathworld.wolfram.com/maclaurintrisectrix.html