14 Slices
Medium 9789380298818

3

Parmananda Gupta Laxmi Publications PDF

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

See All Chapters
Medium 9789380298818

6

Parmananda Gupta Laxmi Publications PDF

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).

See All Chapters
Medium 9789380298818

15

Parmananda Gupta Laxmi Publications PDF

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.

See All Chapters
Medium 9789380298818

14

Parmananda Gupta Laxmi Publications PDF

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)

See All Chapters
Medium 9789380298818

4

Parmananda Gupta Laxmi Publications PDF

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

See All Chapters

See All Slices