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## 3 |
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18 DIFFERENTIAL GEOMETRY AND CALCULUS OF VARIATIONS . . . . . . (Fx) x + (Fy) y + (Fz) z = 0 ⇒ (Gx) x + (Gy) y + (Gz) z = 0 and . . . Solving these equations for x, y and z , we get . . . x y z = = Fy G z − Fz G y Fz G x − Fx G z Fx G y − Fy G x ∴ FyGz – FzGy , FzGx – FxGz , FxGy – FyGx are also d.r.’s of the tangent to the curve given by the equations F(x, y, z) = 0, G(x, y, z) = 0 at the point t. Example 7. Show that the equation of the tangent to the curve of intersection of the ellipsoid x2 a2 + y2 + b2 z2 c2 x( X − x ) 2 x2 = 1 and the confocal 2 2 a (b − c )(a − λ ) y2 b2 − λ y(Y − y) = 2 a2 − λ + 2 2 2 2 b (c − a ) (b − λ ) + = z2 c2 − λ = 1 is z(Z − z) 2 2 c (a − b 2 ) (c 2 − λ ) , where (x, y, z) is an arbitrary point on the curve. Sol. The given surfaces are x2 a2 x2 and + y2 y2 z2 + b2 –1=0 c2 ...(1) z2 –1=0 ...(2) a 2 − λ b2 − λ c 2 − λ Let the equation of the curve of intersection of given surfaces be r = r(t), where t is an arbitrary parameter. Differentiating the equations (1) and (2) w.r.t. t, we get + 2x 2 + . x+ 2y 2 . y+ 2z 2 . z=0 a b c 2x . 2y . 2z . y+ 2 z =0 x+ 2 a2 − λ b −λ c −λ |
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2 Curvature and Torsion 1. INTRODUCTION For curves in space, the concepts of curvature and torsion are of fundamental importance. We know that line segments are uniquely determined by their lengths, circles by their radii, triangles by side-angle-side etc. In geometry, we look for geometric quantities which distinguish one figure from another. The importance of curvature and torsion can easily be estimated from the fact that it can be proved that a curve is uniquely determined (except for its position in space) if its curvature and torsion are given as continuous functions of arc length ‘s’. 2. CURVATURE OF A CURVE Let r = r(s) be a regular curve C of class Cm(m ≥ 2), where s is the parameter ‘arc length’. The vector r″(s) is called the curvature vector on the curve C at the point r(s) and it is denoted by t(s) κ(s) (or by κ). The magnitude of the curvature vector r(s) is called the curvature of the curve C at the point r(s) and it is denoted by κ(s) (or by κ). C k(s) ∴ κ(s) = |r″(s)| Also t(s) = r′(s), so we have |
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50 DIFFERENTIAL GEOMETRY AND CALCULUS OF VARIATIONS Remark. It can be proved that a curve is uniquely determined (except for its position in space) if we are given its curvature κ(≠ 0) and torsion τ as continuous functions of arc length s. This result shows the importance of curvature and torsion in the study of differential geometry of space curves. Example 1. For the helix r = a cos ti + a sin tj + btk, a > 0, b ≠ 0, find the torsion at the point t. Sol. We have r = a cos ti + a sin tj + btk ∴ r = – a sin ti + a cos tj + bk a 2 sin 2 t + a 2 cos 2 t + b2 = ⇒ |r | = ∴ t= r = |r| t′ = dt dt dt dt = = ds dt ds dt 2 a + b2 = − =– |t′| = ∴ n= a + b2 1 = ∴ 1 2 a +b a 2 a + b2 a a 2 + b2 (– a sin ti + a cos tj + bk) ds dt (– a cos ti – a sin tj) |r| a 2 a 2 + b2 2 (cos ti + sin tj) a 2 + b2 (cos ti + sin tj) (cos2 t + sin2 t)1/2 = a a 2 + b2 a a 2 + b2 t′ (cos t i + sin t j ) . =− 2 = – (cos ti + sin tj) a a + b2 |t′| i ∴ b=t×n= − =– =– ∴ b′ = 1 a 2 + b2 1 2 a + b2 a a 2 + b2 − cos t 1 2 a + b2 sin t a k cos t a 2 + b2 − sin t i j k − a sin t a cos t b cos t sin t 0 |
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62 DIFFERENTIAL GEOMETRY AND CALCULUS OF VARIATIONS Dividing (3) by (4), we get dθ − a sin θ κ (n . a) ds = dφ − τ (n . a) − a sin φ ds κ sin θ dθ = . τ sin φ dφ Example 14. For the curve r = r(s), if ⇒ – dt = w × t, ds Sol. Given equations are dn db = w × n and = w × b, find the vector w. ds ds dt =w×t ds ...(1) dn = w × n ...(2) ds db =w×b ds By Frenet formula, dt = κn . ds ⇒ dt = 0 + κn = τ(t × t) + κ(b × t) = (τt + κb) × t ds ∴ dt = (τt + κb) × t ds By Frenet formula, dn = – κt + τb. ds ⇒ ∴ By ⇒ ∴ If ...(3) ...(4) dn = – κ(n × b) + τ(t × n) = κ(b × n) + τ(t × n) ds = (κb + τt) × n = (τt + κb) × n dn = (τt + κb) × n ds db Frenet formula, = – τn. ds db = – τ(b × t) + 0 = τ(t × b) + κ( b × b ) = (τt + κb) × b ds db = (τt + κb) × b ds w = τt + κb, then given equations (1), (2) and (3) are satisfied. ...(5) ...(6) Note. The vector w = τt + κb is called the Darboux vector for the curve r = r(s). Example 15. Using Serret-Frenet formula, find the direction cosines of the unit principal normal vector and the unit binormal vector at the point ‘s’ for the curve r = r(s). |
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73 CURVATURE AND TORSION The centre of the osculating circle at point P is called the centre of curvature of the curve C at the point P. By the definition of contact between curves, the osculating circle to the curve C at point P can be considered as the intersection of a sphere with at least 3-point contact with the curve C at point P and a plane with at least 3-point contact with C at P. If κ ≠ 0 at P, then the osculating plane at P is the unique plane having at least 3-point contact with the curve C at P. In particular, if τ ≠ 0 in addition to κ ≠ 0 at P, then the osculating plane is the unique plane having exactly 3-point contact with C at P. Therefore the osculating circle to a curve at a point always lies on the osculating plane to the curve at that point, provided κ ≠ 0 at the point under consideration. Thus, the osculating circle to a curve at a point can be considered as the intersection of a sphere with at least 3-point contact with the curve at that point and the osculating plane to the curve at the point under consideration, provided κ ≠ 0. |
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80 DIFFERENTIAL GEOMETRY AND CALCULUS OF VARIATIONS Remark 2. If curvature of r = r(s) at P is constant, then radius of osculating sphere at P(r0) = ρ02 + 0. σ 02 = ρ0 and p.v. of centre of spherical curvature at P(r0) = r0 + ρ0n0 + (0)σ0b = r0 + ρ0n0. ∴ Centre of the osculating sphere coincides with the centre of osculating circle at points where curvature vanishes. 11. LOCUS OF CENTRE OF SPHERICAL CURVATURE Let r = r(s) be the equation of a curve C. For each point P with non-zero curvature and torsion on the curve C there exists an osculating sphere. Let C1 denote the locus of the centre of spherical curvature i.e., the centre of osculating sphere as the point P moves along the curve C. We shall prove some properties regarding the curve C1, the locus of centre of spherical curvature. Property I. The tangent to the locus of centre of spherical curvature is parallel to the corresponding binormal to the original curve. Proof. Let P(r) be any point on a curve C given by r = r(s). Let κ ≠ 0, τ ≠ 0 at P. Let r1 be the position vector of the centre of spherical curvature at the point P. |
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3 Surfaces in Space 1. INTRODUCTION A sphere, a portion of a cylinder are examples of surfaces in space. Till now we have been discussing the differential geometry of curves in space. In the present chapter , we shall learn few quantities regarding surfaces in space. 2. SURFACE IN SPACE We know that a curve in space is the locus of a point whose coordinates x, y, z are functions of a single parameter. On the same lines, we shall define a surface in space as the locus of a point whose coordinates are functions of two independent parameters. A surface in space is defined as the locus of a point whose position vector relative to a fixed origin may be expressed as a function of two independent parameters. P(x, y, z) Thus, a surface S in space may be represented by a S vector function r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k, r(u, v) where u and v are independent parameters such that (u, v) ∈ R, where R is some region in the uv-plane. O(origin) Here r(u, v) is the position vector of the point P on the v surface S and (x(u, v), y(u, v), z(u, v)) are the cartesian coordinates of the point P. The above representation of a surface is called a parametric representation of a surface and is due to Gauss. Also, (u, v) are called the curvilinear coordinates of the point P. In particular cases, |
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102 DIFFERENTIAL GEOMETRY AND CALCULUS OF VARIATIONS F(u) = u3 + lu2 + mu + n Let ∂ (F(u)) = 3u2 + 2lu + m ∂u The equation of the envelope of the given family is obtained by eliminating u from the equations ∂ (F(u)) = 0. F(u) = 0 and ∂u ∴ u3 + lu2 + mu + n = 0 ...(2) 2 3u + 2lu + m = 0 ...(3) Multiplying (2) by 3 and (3) by u, we get 3u3 + 3lu2 + 3mu + 3n = 0 ...(4) 3u3 + 2lu2 + mu = 0 ...(5) 2 (4) – (5) ⇒ lu + 2mu + 3n = 0 ...(6) Solving (3) and (6), we get ∴ u2 6ln − 2m 2 = u2 = ⇒ Eliminating u, we get ∴ ⇒ F ml − 9n I GH 6m − 2l JK 2 2 = u 1 = ml − 9n 6m − 2l 2 6ln − 2m 2 6m − 2l 2 = 3ln − m 2 3m − l 2 and u= ml − 9n 6m − 2l 2 3ln − m 2 3m − l 2 (ml – 9n)2 = 4(3ln – m2)(3m – l2). This is the equation of the required equation of the envelope, where l, m and n are as given above. 6. IMPORTANT RESULTS I. The envelope of the two-parameter family of surfaces F(x, y, z, a, b) = 0, where a, b are parameters is found by eliminating a and b from the equations: ∂f ∂f = 0, = 0. ∂a ∂b The envelope of the two-parameter family of surfaces F(x, y, z, a, b) = 0, where a, b are parameters connected by the equation φ(a, b) = 0 is found by eliminating a and b from the equations: |
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4 Variational Problems with Fixed Boundaries 1. INTRODUCTION By a function we mean a correspondence between the elements of sets A and B such that to each element of A there correspond exactly one element of the set B. We have also studied the methods of finding maximum and minimum values of real functions of one or more variables. In calculus of variations, we shall deal with functionals and their maximum and minimum values. A functional associates a unique real number to each function belonging to a certain class of functions. 2. FUNCTIONAL Let A be a class of functions. A correspondence between the functions of class A and the set of real numbers such that to each function belonging to A, there correspond exactly one real number is called a functional. In other words, a functional is a correspondence which assigns a unique real number to each function belonging to some class. Thus a functional is a function, where the independent variable takes functions as values. A functional is denoted by capital letter I (or J). If y(x) represents a function in the class of functions of a functional J, then we write J = J[y(x)]. |
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136 DIFFERENTIAL GEOMETRY AND CALCULUS OF VARIATIONS ∴ y = c1 cosh x + c2 c1 minimises the functional i.e., the area of the surface of revolutions by the curve between the given points. Example 14. Show that the geodesics* on a sphere of radius a are its great circles. Sol. Let A and B be any two points on the given sphere with origin at the centre of the sphere. Let (a, θ1, φ1) and (a, θ2, φ2) be their spherical coordinates** respectively. Let φ(θ) be a function whose curve passes through A and B and lying itself on the surface of the given sphere. ∴ The length of the arc between A and B is given by z θ2 θ1 ds . Since, we are dealing with spherical coordinates, we have ⇒ ds = (dr) 2 + (r dθ) 2 + (r sin θ dφ) 2 ds = 0 + a 2 ( dθ) 2 + a 2 sin 2 θ . φ′ 2 ( dθ) 2 (Note this step) (∵ r = a ⇒ dr = 0) = a 1 + φ ′ 2 sin 2 θ dθ ∴ Length of arc = z θ2 θ1 a 1 + φ ′ 2 sin 2 θ dθ Let F(θ, φ, φ′) = a 1 + φ ′ 2 sin 2 θ Let the length of the curve between A and B has a minimum value for this curve. In other words, let φ(θ) be a geodesic between A and B. |
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148 DIFFERENTIAL GEOMETRY AND CALCULUS OF VARIATIONS Example 4. Find the extremal of the functional z a −a (λy + 1 2 µy″2) dx which satisfy the boundary conditions y(– a) = 0, y′(– a) = 0, y(a) = 0, y′(a) = 0. z FGH IJ K 1 µy″ 2 dx . −a 2 1 2 Let F(x, y, y′, y″) = λy + µy″ 2 d d2 Fy ′ + F =0 The Euler-Poisson equation is Fy − dx dx 2 y ″ Here Fy = λ, Fy′ = 0, Fy″ = µy″ Sol. Given functional is ∴ (1) λ− ⇒ a λy + d d2 (0) + (µy″ ) = 0 ⇒ dx dx 2 µ d4 y dx 4 +λ =0 ⇒ ...(1) d4 y λ =– 4 µ dx d3 y λ = − x + c1 3 µ dx ⇒ d2y ⇒ dx 2 =− λ x2 + c1x + c2 µ 2 dy λ x3 x2 =− + c1 + c2x + c3 2 dx µ 6 ⇒ λ x4 x3 x2 + c3x + c4 + c1 + c2 6 2 µ 24 This is the equation of the extremals. The boundary conditions are y(– a) = 0, y′(– a) = 0, y(a) = 0, y′(a) = 0 ⇒ y= − y(– a) = 0 ⇒ − λa4 c1a 3 c2 a2 – c3a + c4 = 0 − + 24µ 6 2 − ...(4) λa4 c1 a3 c2 a2 + + + c3a + c4 = 0 24µ 6 2 ...(5) λa3 c1a 2 + + c2a + c3 = 0 6µ 2 ...(6) − y′(a) = 0 ⇒ c1 a3 – 2c3a = 0 ⇒ c1a2 + 6c3 = 0 3 (4) + (6) ⇒ c1a2 + 2c3 = 0 Solving (7) and (8), we get c1 = 0, c3 = 0 (3) – (5) ⇒ (4) (3) ⇒ – λa 3 + 0 – c2a + 0 = 0 6µ ⇒ − λa 4 a2 −0+ 24µ 2 ...(3) λa 3 c1 a2 – c2a + c3 = 0 |
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160 DIFFERENTIAL GEOMETRY AND CALCULUS OF VARIATIONS ∴ The extremals of the functional J[y(x)] can be found by solving Euler’s equation of the transformed functional J1[v(u)]. This is called the principle of invariance of Euler’s equation under coordinates transformations. In solving practical problems, the change of variables is made directly in the integral representing the functional. We then write and solve the Euler’s equation for the new integral involving new variables. In the extremals so obtained, the new variables are changed to the original variables to get the desired extremals. Example. Find the extremals of the functional Sol. Let J[r(θ)] = z z θ2 θ1 θ2 θ1 r 2 + r ′ 2 dθ , where r = r(θ). r 2 + r ′ 2 dθ . Let x = r cos θ, y = r sin θ. ∴ r= −1 x 2 + y2 , θ = tan y x rx + ry y′ dr dr/ dx = = r′ = = dθ dθ/ dx θ x + θ y y′ x 2 x +y 1 y2 1+ 2 x 2 + FG − y IJ H xK 2 y y′ x + y2 1 1 + y′ 2 x y 1+ 2 x 2 FG IJ H K x + yy′ x2 + y2 = xy′ − y x2 + y2 = Also, dθ = xy′ − y dθ dx . dx = (θx + θy y′)dx = 2 dx x + y2 r2 + r′2 = x2 + y2 + ∴ = (x2 + y2) |
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5 Variational Problems with Moving Boundaries 1. INTRODUCTION We know that a functional associates a unique number to each function belonging to a certain class of functions. In the preceding chapter, we studied the methods of finding stationary function of the functional of the form z x2 x1 F(x, y(x), y′(x)) dx, where the boundary points (x1, y1) and (x2, y2) were fixed points. In the present chapter, we shall allow one or both boundary points to move. In this case the class of admissible functions is extended because, in addition to the comparison curves with fixed boundary points, we have to admit functions with variable boundary points. 2. FUNCTIONALS DEPENDENT ON ONE FUNCTION Theorem. Let J[y(x)] = z x2 x1 F(x, y(x), y′(x)) dx be a functional defined on the set of functions y = y(x) which admits of continuous first order derivative and the end points of which (x1, y1) and (x2, y2) lie on the given curves y = φ(x) and y = ψ(x) respectively so that y1 = φ(x1) and y2 = ψ(x2), where y = φ(x) and y = ψ(x) are functions from the class C1[x1, x2]. Also, F is differentiable three times with respect to all its arguments. |
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6 Sufficient Conditions for an Extremum 1. INTRODUCTION At an extremum, a functional may have either maximum value or minimum value. In the preceding two chapters, we have studied the necessary conditions for an extremum. At a stationary function, a functional may or may not have an extremum. In the present chapter, we shall confine ourselves to the study of sufficient conditions for extremums of variational problems with fixed boundaries only. Y 2. PROPER FIELD IN A GIVEN DOMAIN A family of curves y = y(x, c) is called a proper field in a given domain D of the xy-plane if through every point of D, there passes one and only one curve of the family y = y(x, c). In other words, the curves of the family y = y(x, c) cover the entire domain D without self-intersections. For example, the family of curves y = cex, where c is an arbitrary constant, form a proper field inside the circle x2 + y2 = 1, because through any point inside or on the circle there passes exactly one curve belonging to the family y = cex. |

Details

- Title
- Topics in Differential Geometry and Calculus of Variations
- Authors
- Parmananda Gupta
- Isbn
- 9789380298818
- Publisher
- Laxmi Publications
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- January 20, 2016

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