Evolute and involute (evolvent) Interactive solutions to problem s of differential geometry © S.N Nosulya, V.V Shelomovskii. Topical sets for geometry, 2012. © D.V. Shelomovskii. GInMA, 2012. http://www.deoma-cmd.ru/ Materials of the interactive set can be used in Undergraduate education for the study of the foundations of differential geometry, it can be used for individual study methods of differential geometry. All figures are interactive if you install GInMA program from http://www.deoma-cmd.ru/ Conception of Evolute In the differential geometry of curves, the evolute of a curve is the locus of all its centers of curvature. Equivalently, it is the envelope of the normals to a curve. Apollonius (c. 200 BC) discussed evolutes in Book V of his Conics. However, Huygens is sometimes credited with being the first to study them (1673). Equation of evolute Let γ be a plane curve containing points X⃗T =(x , y). The unit normal vector to the curve is ⃗n. The curvature of γ is K . The center of curvature is the center of the osculating circle. It lies on the normal 1 line through γ at a distance of from γ in the direction determined by the sign of K . In symbols, K the center of curvature lies at the point ζ⃗T =(ξ ,η). As X⃗ varies, the center of curvature traces out a ⃗n plane curve, the evolute of γ and evolute equation is: ζ⃗= X⃗ + . (1) K Case 1 Let γ is given a general parameterization, say X⃗T =(x(t) , y(t)) Then the parametric equation of x' y' '−x ' ' y ' the evolute can be expressed in terms of the curvature K = . The unit normal vector to the (x ' 2+ y' 2)3/ 2 (− y' , x' ) curve is n⃗= . Evolute equation is: √ x' 2 + y' 2 y' ( x' 2+ y ' 2) x' ( x' 2+ y ' 2) ζ⃗= x− , y+ . (2) ( x ' y ' ' −x ' ' y' x' y' '−x ' ' y ' ) y' ' Case 2 Let γ is given an equation y= f (x). Then the curvature K = , the unit normal (1+ y ' 2)3/ 2 (− y' ,1) y' (1+ y' 2) 1+ y' 2 vector to the curve is n⃗= . Evolute equation is: ζ⃗= x− , y+ . (3) √1+ y ' 2 ( y ' ' y' ' ) 2 2 2 f x f y f xy− f x f yy− f y f xx Case 3 Let γ is given an equation f (x , y)=0. Then the curvature K = 2 2 3/ 2 . ( f x+ f y) (− f , f ) The unit normal vector to the curve is n⃗= x y . Evolute equation is: 2 2 √ f x+ f y f ( f 2+ f 2) f ( f 2+ f 2) ζ⃗= − x x y + y x y x 2 2 , y 2 2 . (4) ( 2 f x f y f xy − f x f yy− f y f xx 2 f x f y f xy− f x f yy− f y f xx ) Case 4 Let γ is given an equation R= f (s), where s− длина дуги, отсчитываемая от некоторой точки. Let evolute equation is: R̃ = f̃ (s̃ ). Обозначим углы, составляемые касательными с 1 d α 1 d α̃ 1 осью абсцисс α и α̃ . Then = , = , d ̃s=d R ,α=α̃ + π ,d α=d α̃ . Evolute d s R d s̃ R̃ 2 d R equation is: R̃ = R , ̃s=R+c . (5) d s Тypical figure and it's using In each figure points 0 and 1 on the x-axis specify the Cartesian coordinates. The graph of the original curve is shown by the blue line. It is determined by the parameters which values are indicated in the figure. The parameters are set by the active points. The evolute is shown in red. The trial points labeled С , D , E ,... are located on the curve. Tangent circle and its center point C ' , D ' ,..., belonging to evolute are shown in pink . The process of learning and investigation one can begin from the verification of the basic propertiy of the evolute. By moving point C , make sure that the center of curvature belongs to the evolute at all posistions of point C , . By activating the "Properties" button find and check the equations used for the curves construction. Explore several evolute curves. Save the file in a convenient location and change the original curve. Perform calculations and construct evolute of the selected curve. Build a tangent circle to check your construction. Evolutes samples The evolute of the parabola The curve given by the equation y= f (x)=k x2 . Then the curvature of the curve is y' ' 2k (− y' ,1) (−2k x ,1) K = = , the normal is n⃗= = . (1+ y' 2)3/2 (1+(2k x)2)3/2 √1+ y' 2 √1+(2k x)2 The equation of the evolute is: 2 2 2 2 y' (1+( y ') ) 1+( y ') 2 2 2 1+4 k x 2 2 2 1 ζ⃗= x− , y+ = x−x(1+4 k x ) ,k x + = −4 k x ,3k x + . ( y' ' y ' ' ) ( 2 k ) ( 2k ) The evolute of the ellipse x2 y2 The curve given by the equation f (x , y)= + −1=0. Then the equation of the evolute is: a2 b2 2 2 2 2 3 3 f ( f + f ) f ( f + f ) x y 2 2 ζ⃗= x− x x y , y+ y x y = ,− (a −b ). 2 f f f − f 2 f − f 2 f 2 f f f − f 2 f − f 2 f ( a4 b4 ) ( x y xy x yy y xx x y xy x yy y xx ) Fig.1. Evolute of the parabola Fig.2. Evolute of the ellipse 2 The evolute of a hyperbola x2 y2 The curve given by the equation f (x , y)= − −1=0. We perform the parameterization a2 b2 x=a cosh t , y=bsinh t . Then the equation of the evolute is: 2 2 2 2 3 3 3 3 y' ( x' + y ' ) x' ( x' + y ' ) cosh t −sinh t 2 2 x y 2 2 ζ⃗= x− , y+ = , (a +b )= ,− (a +b ). ( x ' y ' ' −x ' ' y' x' y' '−x ' ' y ' ) ( a b ) ( a4 b4 ) The evolute of an astroid The curve given by the equation { x , y }={ acos3t ,a sin3 t }. The curvature of the curve is 2 K = . Then the equation of the evolute is: ζ⃗=a(cost(3−2cos2 t) ,sin t (1+2cos2 t)). 3asin 2t Рис.3. Evolute of a hyperbola Рис.4. Evolute of an astroid Involute The involute of the plane curve is a curve with respect to which the given curve is the evolute. The involute is a curve, for which the normal at each point is tangent to the initial curve. That is, the involute contains the tangents to the given curve. Imagine that the little pebble lies on the curve at an arbitrary point P. Let the thread run from the pebble along the curve. Then the involute determines the trajectory of pebble, moving away from the curve and associated with it by the stretched thread. They say that the selected point moves along the involute when unwinding the thread, which lies on the curve. The family of the involutes generated by different points of the surface exists for each curve. Further we consider only one of them, starting at the origin. Case 1. Let this curve be given parametrically by the expression y= f (x). Then the unit tangent vector (1 , y ') t is τ⃗= . The length of the thread unwound from the beginning is s= 1+( y' )2 d x. The 2 ∫ √ √1+( y' ) 0 equation of the involute ζ⃗T =(ξ ,η) has the form of: x x ∫√1+( y' )2 dx y'∫ √1+( y' )2 dx ζ⃗= X⃗ −sτ⃗= x− 0 , y− 0 . (6) ( √1+( y' )2 √1+( y' )2 ) 3 Case 2. Let the curve be given parametrically by the expression X⃗T =(x(t) , y(t)). Then the unit tangent ( x' (t) , y' (t)) vector is τ⃗= . The length of the thread unwound from the beginning is √(x ' (t))2+( y ' (t))2 t 2 2 dx(t̃) dy(t̃) s= + d t̃. The equation of the involute ζ⃗T =(ξ ,η) has the form of: ∫ ( d t̃ ) ( d t̃ ) 0 √ t t x '∫√ x ' 2+ y' 2 dt y'∫ √ x' 2+ y' 2 dt ζ⃗= X⃗ −sτ⃗= x− 0 , y− 0 . (7) ( √x ' 2+ y' 2 √x ' 2+ y' 2 ) Case 3. Let the curve be given by the expression R̃ = f̃ (s̃ ) Using (5), we find the equation of the involute: ∫ s̃ d ̃s R=̃s ,s= . (8) R̃ Тypical figure and it's using In each figure points 0 and 1 on the x-axis specify the Cartesian coordinates. The graph of the original curve is shown by the blue line. It is determined by the parameters which values are indicated in the figure. The parameters are set by the active points. The involute is shown in red. The trial points labeled С , D , E ,... are located on the curve. Tangent circle and its center point C ' , D ' ,..., belonging to involute are shown in pink. The process of learning and investigation one can begin from the verification of the basic propertiy of the involute. By moving point C , make sure that the segment CC ' (DD ' ,...) is tangent to the the given original curve at all posistions of point C , and point С (D , E ,...) is the center of curvature of the involute. By activating the "Properties" button find and check the equations used for the curves construction. Explore several involute curves. Save the file in a convenient location and change the original curve.