2 Envelopes and Developables 2.1 FAMILY OF SURFACES (ONE PARAMETER) An equation of the form Fx( ,,yz, a) = 0 (1) represents on infinitely many surfaces, each surface being determined by a value of the para­ meter a. We call such a system one­parameter family of surfaces. Characteristic of a Family of Surfaces Let F(x, y, z, ) = 0 be the one parameter family of surfaces. The curve of intersection of two surfaces of the family corresponding to the values of and + , is given by Fx( ,,yz,,αα) =+00Fx( ,,yz, δα) = Fx( ,,yz,,αδ+ αα) − Fx( yz,, ) or Fx( ,,yz) = 00, = δα Now as → 0, the curves tend to limiting position given by ∂ Fx( ,,yz,,α) = 00Fx( ,,yz,α) = (2) ∂α The curve given by (2) is thus a characteristic of the family of surfaces F(x, y, z, a) = 0 for a = . Clearly each value of of a determines a characteristic. CHAPTER 02.indd 79 8/21/2009 6:39:51 PM Envelopes and Developables • 81 2.2 EDGE OF REGRESSION Let the family of surfaces be given by ƒ(x, y, z, a) = 0. Let us consider the equations to the two neighbouring characteristics for the values a = and a = + as fx(),,yz,,αα= 00fxα (),,yz, = (1) and fx( ,,yz,,αδ+ αα) =+00fxα ( ,,yz, δα) = (2) The above two characteristics (1) and (2), in general, intersect in a point. The position of the point of intersection of two neighbouring characteristics, as → 0, is called a characteristic point. The locus of all characteristic points of a one­parameter family of surfaces is called the edge of regression. Thus, equations (1) and (2) together, as → 0, on eliminating yield the edge of regression or cuspidal edge of the envelope. When → 0, equations (1) and (2) become fx( ,,yz,,αα) = 00fxαα( ,,yz, ) ==and f α 0 ∂2 f (3) f = αα ∂α 2 Equation (3) determines a characteristic point. The equation of edge of regression is obtained by eliminating from equation (3). Theorem 1 Each characteristic touches the edge of regression. Proof: The edge of regression is given by fx( ,,yz,,αα) = 00and fxα ( yz,, ) = , (1) provided ƒ (x, y, z, ) = 0, so that is a function of x, y, z. The tangent line to a characteristic curve ƒ = 0, ƒ = 0 is ⊥ to the normals to the surfaces ƒ(x, y, z, ) = 0 and ƒ (x, y, z, ) = 0 at any point (x, y, z) on them. CHAPTER 02.indd 81 8/21/2009 6:39:53 PM Envelopes and Developables • 83 If b = a, then we get ∂f ∂f fa( ,,b) = 00+ λ += ∂a ∂b and when a, b → 0, ∂f ∂f fa( ,,b) = 00+ λ = ∂a ∂b As a and b are independent, thus can take any value. Thus, limiting position of the curve depends on and will be different for different values of . However, for all values of , the limiting positions will pass through the point or points given by ∂f fa( ,,bf) ==00ab, ff=≡0e a tc. (2) ∂a These are called characteristic points and the locus of these points is the envelope of the family of surfaces. We get the equation of envelope by eliminating a and b from equation (2). Let ƒ(x, y, z, a, b) = 0, where a and b are functions of x, y, z given by ∂f ∂f = 00, = (3) ∂a ∂b as the equation of the envelope. The normal to the envelope is parallel to the vector ∂a ∂b ∂a ∂b ∂a ∂b ff+ + f ,,ff+ + f ff+ + f ; xa∂x by∂x ab∂y ∂y za∂z b ∂z ∂f ∂f heere fab= ,,f = ∂a ∂b which in view of (3) is ( f x, f y, fz) But this vector is parallel to the normal to a surface of the family at the point (x, y, z). Therefore, the envelope has the same normal and so the same tangent plane to a surface of the family at a characteristic point. Thus, we have the following: The envelope touches each surface of the system at the corresponding characteristic points. * * * * * * * * * CHAPTER 02.indd 83 8/21/2009 6:39:55 PM 84 • Differential Geometry and Tensors SOLVED EXAMPLE 1 Q1: Find the envelope of the plane 3xt2 - 3yt + z = t3, and show that its edge of regression is the curve of intersection of the surfaces y2 = xz, xy = z. Solution: Now family of planes is given by fx≡33ty23-tz+-t = 0 (1) ∂f ∴ ==06xt --33yt2 = 0 (2) ∂t Let us multiply (2) by t and subtracting from (1), we get xt 2 -20yt +z = (3) Now in view of (2) and (3), we have t 2 11 = = -22xz +yx22yz- -22yx+ which on eliminating t, gives equation of envelope as 2 (xy - zx) =4()22-yy()- zx (4) Now from (2) on differentiation, we get fxtt ≡22-tt=⇒0 = x Putting value of t = x in (1) and (2), we get, 23ty22-tz+==0 and;yt i.e., xt==,,yt22zt= On eliminating t, we get yx2 ==zxand yz which represents the edge of regression. CHAPTER 02.indd 84 8/21/2009 6:39:56 PM Envelopes and Developables • 85 Q2: The envelope of osculating plane of a curve is a ruled surface generated by the tangents to the curve and has the curve for the edge of regression. Solution: Let the curve be given by r = r(s) and osculating plane at any point on it is given by (Rr− )⋅=b 0; (1) where r and b are functions of s. In view of (1), on differentiation w.r.t. s, we get, (Rr− )⋅br′ − ′⋅=b 0 ⇒−(Rr)⋅=n 0 (2) as r′ b = t b = 0 by Frenet’s formula. Equations (1) and (2) simultaneously determines a characteristic. Equation (2) gives rec­ tifying plane and intersection of (1) and (2), therefore represents the tangent line to the curve. Thus, the envelope is the locus of tangent line, that is, it is a ruled surface generated by the tangents to the given curve. This proves the first part. Now, differentiating (2) w.r.t. s, and in view of Frenet’s formula, we get (Rr− )⋅−(τbktr) − ′⋅=n 0 ⇒−(Rr)⋅−(τbktr) = 00( ′⋅=nt⋅=n ) ⇒−(Rr)⋅=t 001(by ()) (3) Now, equations (1), (2) and (3) simultaneously represent the edge of regression. But (1), (2) and (3) are planes whose intersection is point p(r). Hence, the points on the edge of regression coincide with the points on the curve. Hence, the curve itself is the edge of regression. Q3: Find the envelope of the planes lx + my + nz = p when a2l2 + b2m2 + 2np = 0. Solution: l m n Leta==λµ,,nd = ν p p p CHAPTER 02.indd 85 8/21/2009 6:39:57 PM 86 • Differential Geometry and Tensors the equation of plane becomes ƒ ≡ x + y + z - 1 = 0 (1) when a2 2 + b2 2 + 2 = 0 (2) Differentiation of (1) and (2) gives xd + yd + zd = 0 and a2d + b2 2d + d = 0 Comparing the above two equations, we have a22λµb 1 ab22λµ++22 ν ν === =- x yz∂+xyµν+ z 1 x y 1 ⇒=λµ,,==ν - az22bz z Putting , , in (1), we get x2 y2 +=2z. a2 b2 UNSOLVED EXERCISE 1 Q1. The envelope of the surfaces ƒ(x, y, z, a, b, c) = 0, where a, b, c are the parameters con­ nected by the equation (a, b, c) = 0 and ƒ and are homogeneous with respect to a, b, c fff is obtained by eliminating a, b, c between the equations f ==00,,φ a ==b c . φφa b φc x y z Q2. Show that the edge of regression of the envelope of plane + + =1 is the a + λλb + c + λ 222 (a + λλ) (b + ) (c + λ) cubic curve given by x = ,,y = z = . (ca- )(ba- ) (cb- )(ab- ) (ac- )(bc- ) x y z Q3. Prove that the envelope of plane cossθφin ++sinsθφin cosφ = 1 is given by a b c x2 y2 z2 ++= 1. a2 b2 c2 Q4. (a) A plane makes intercepts a, b, c on the rectangular axes, so that a-2 + b-2 + c-2 = k-2, show that it envelopes a conicoid which has the axes as equal conjugate diameters. (b) If a2 + b2 + c2 = constant, prove that envelope is x2/3 + y2/3 + z2/3 = constant. Q5. Find the envelope of the cones (x - a)2 + y2 = z2tan2, where is same for all cones. CHAPTER 02.indd 86 8/21/2009 6:39:59 PM Envelopes and Developables • 87 Q6. Prove that the envelope of the family of paraboloids is the circular cone x2 + y2 = z2. 22 Q7. Find the envelope of the sphere ()xa− cossθθ+−()yain +=zb22. Q8. Find the envelope of the plane lx + my + nz = 0, where al2 + bm2 + cn2 = 0. x2 y2 z2 Q9. A tangent plane to the ellipsoid ++=1 meets the axes in A, B, C. Show that the a2 b2 c2 23 23 23 23 envelope of the sphere OABC is (ax) + (by) + (cz) =+()xy22+ z2 . ANSWERS 1 Q5. y2 = z2tan2 Q7. (x2 + y2 + z2 + a2 − b2)2 = 4a2(x2 + y2) x22y z2 Q8. ++= 0 a b c * * * * * * * * * 2.4 RULED SURFACES A surface which is generated by the motion of one­parameter family of straight lines is called a ruled surface. The straight line itself is called the generating line, generator, or ruling. Examples of ruled surfaces are cylinders, cones, conicoid, etc. Types of Ruled Surfaces (1) Developable surface: If consecutive generators intersect, the ruled surface is called developable. Example: cones and cylinders. (2) Skew surfaces: If consecutive generators do not intersect, the ruled surface is called skew­surface or scroll.