Differential Geometry

DIFFERENTAIL GEOMETRY CONTENTS GENERAL INTRODUCTION

SYLLABUS

REFERENCE

UNIT I :

CURVES IN SPACE.

01-43

UNIT II :

CURVES ON SURFACE

44-64

UNIT III:

LOCAL INTRINSIC PROPERTIES OF SURFACE

65-91

UNIT IV:

LOCAL NON – INTRINSIC PROPERTIES OF SURFACE – GEODESICS

92-121

UNIT V:

GEODESIC CURVATURE

122-147

UNIT VI:

GEODESIC MAPPING, ASYMPTOTIC LINES,

UNIT VII:

RULED SURFACES AND DEVELOPABLE

148-175

REGRESSION AND FURTHER THEORY OF SURFACE

176-196

UNIT VIII:

DIFFERENTIAL GEOMETRY OF SURFACES IN LARGE 197-218

GENERAL INTRODUCTION

‘ Differential Geometry is that part of geometry dealing with the study of curves in space and curves on surfaces. It is a subject for its own sake. But it has some applications too. For example, there is a special type of geometrical curve called ‘Geodesic’ , . Its properties are analyzed in variational calculus ( or ) calculus of variations, dealing with maximization or minimization of functional. By studying the curves and surfaces in the neighbourhood of a point on them, we analyse the local property. On the other hand, we analyse global property of the same, while we study them as a whole. Actually, we apply the methods of calculus and vectors to study the curves in space and of surfaces.

Most of our theorems are expressed by systems of Differential Equations. They tell us how a curve or a surface will be generated from suitable initial conditions. Like ‘ Differential Geometry’, there are different distinguishshable geometries , such as ‘ Metric Geometry’ (which preserves straight lines, distances and angles), ‘ Projective Geometry’ ( which preserves straight lines), ‘Topology’ (which preserves continuity ) . A very deep study of invariance in geometry was made by Ricci and his pupil Levi – Civita. This is the Tensor calculus, which Albert Einstein found to be the most suitable tool for his general theory of relativity.

DIFFERENTIAL GEOMETRY SYLLABUS

UNIT – I : The theory of space curves

–

Arc length tangent

- Normal binormal r dr definition of space curve - Geometrical interpretation of - Equation of the ds tangent line to a curve at a point - Tangent line in cartesian form - Osculating plane Principal normal and binormal - Curvature of the curve in Space - Serret – Frenet Formulae - Expression for torsion.

UNIT – II : Different of a surface - Curves on a surface - Contact between curves and surfaces - Osculating circle - Osculating sphere - Locus of the centre of spherical curvature - Tangent Surfaces, Involutes and Evolutes - Intrinsic equation of space curves - Fundamentals existence theorem for space curves - Helices - the spherical indicarices ( or) spherical images.

UNIT – III : Surface representation, regular and singular points - Change of parameters Curvilinear equations of the curve on the surface - Tangent plane and normal Surfaces of revolution - Anchor ring - Helicoids - Metric on a surface - the first fundamental form - Invariance of the metric - Element of area - Direction coefficients on a surface.

UNIT – IV : Family of curves - Differential equation of the family of curves - Orthogonal Trajectories - Double family of curves - Isometric correspondence - Intrinsic properties - Geodesics and their differential equation - Canonical geodesic equations Second fundamental form - Fundamental equations of surface theory - Geodesics on a surface of revolution - Curvature of normal section of the surface - Mennier’s theorem - Clairaut’s theorem.

UNIT – V : Normal property of a geodesic - Existence theorem - Geodesic parallels Geodesic polars - Geodesic curvature components of geodesic curvature - Geodesic curvature of the parametric curve - Liouville’s formula gauss - Bonnet theorem Gaussian curvature - Minding’s theorem - Conformal mapping - Corollary. UNIT – VI : Geodesic mapping - Ruled surface ( Developable and Skew ) - Equation of the ruled surfaces - Necessary condition - Lines of curvature - Differential equation of lines of curvature - Property of lines of curvatures on developable - Conjugate directions - Asymptotic lines - Asymptotic lines on a ruled surface - Parameter of distribution of a ruled surface - Properties of parameter of distribution - Central point and the equation of the line of Striction . UNIT – VII : Joachimsthal’s theorem - Dupin’s indicatrix - Types of point (Elliptic, Hyperbolic and Parabolic) - Third fundamental form - Family of surfaces - Envelope - The edge of regression - Definition of edge of regression - Characteristic touches the edge of regression - Minimal surfaces - Gauss characteristic equation - Mainardi – Codazzi equations. UNIT VIII : Compact Surfaces - Points are umbilics - Hilbert’s lemma - Compact surfaces of constant gaussian or mean curvature - Complete surfaces - Characterization .

BOOK REFERENCES 1.

‘Differential Geometry’ by Dr. S.C Mittal & D.C Agarwal, Krishna prakashan Mandir, Meerut.

2.

‘Lectures on classical differential Geometry’ by D.T.Struik, Addison – Wesley Mass 1950.

3.

‘An introduction to Differential Geometry ‘ by T. Willmore, clarendan Press, Oxford, 1959.

4.

‘Differential Geometry’ by D. Somasundaram, Narosha Publishing House Chennai, 2005.

5.

‘Elementary Topics in Differential Geometry’ by J.A.Thorpe, Springer – verlag, New York , 1979.

UNIT I CURVES IN SPACE STRUCTURE : Learning Objectives 1.0.

Introduction

1.1.

Definitions of Curves in space and examples

1.2.

Tangent line to curve

1.3.

Osculating Plane

1.4.

Principal normal and binormal.

1.5.

Curvature of the curve and torsion .

1.6.

Serret – Frenet Formulae.

1.7.

Expression for torsion

1.8.

Summary

1.9.

Key Words.

1.10. Answer to Check Your Progress 1.11. Terminal Questions. 1.12. Further Readings.

LEARNING OBJECTIVES: After going through this unit, you should be able to  define curve in space, tangent line, unit tangent vector, osculating plane, principal normal and binormal, curvature and torsion  give examples of curves, equations of tangent line,  derive expressions for curvature and torsion  derive serret – Frenet formulae. 1.0 INTRODUCTION: Differential Geometry is that branch of mathematics dealing with the curves in space and curves on surfaces. It is that part of geometry which is treated with the help of Differential calculus. We know that the geometric character of curves and surfaces varies continuously and it is achieved by the use of differential calculus. There are two branches

1

of differential geometry: one in which we study the properties of these curves and surfaces in the neighbourhood of a point is called ‘local differential geometry’ and second in which we study the properties of these curves and surfaces as a whole is called ‘Global Differential Geometry’. Preliminaries: Important results on vectors:

r

1. Any vector r can be expressed as a linear combination of 3 given non- copalanar r r r r r r r r r r vectors a, b, c i.e., r = xa + yb + zc . In terms of unit vectors i , j , k along three

r

r

r

r

mutually perpendialar axes called x , y , z - axes, we have r = xi + yj + zk ,

r z 2 . The direction ratios of r , being x, y, z . r r r r a b cos a ,b

r2 r = x2 + y2 + r r r r 2. (i) a ×b = b ×a =

( )

r r

r

r

r

r

(ii) a ×b = 0, if a = 0 (or )b = 0 (or )a ^ b

r r

r r

r r

r r

r r

r r

(iii) i ×i = j ×j = k ×k = 1 and i ×j = j ×k = k ×i = 0

r r

r

(

r r

)

r r

3. a × b + c = a ×b + a ×c 4.

r r r r r r r r r r a ´ b = a b sin a, b n , where n = 1 and n is perperdialar to both a

( )

(i)

r and b .

r

r

r

r

r

r

r

r

r

r

r

r

r

r

r

r

r

(ii) a ´ b = 0 Þ Either a = 0 (or ) b = 0 (or )a Pb

r

r

(iii) a ´ b = - b ´ a (iv) a ´ a = 0

r

r

r

r

r

r

r

r

(r

r

r

r r

r

r

r

r r

r

r

(v) i ´ i = j ´ j = k ´ k = 0 or i ´ j = k , j ´ k = i , k ´ i = j

r

(

)

r

)

r

r

r

5. a ´ b + c = a ´ b + a ´ c & a + b ´ c = a ´ c + b ´ c

2

r

r

r

r

r

r

r

r

6. if a = x i + y j + z k and b = x i + y j + z k , then

1

1

1

2

2

2

r r

(i) a ×b = x x + y y + z z

1 2 12 12 r r r i j k r r (ii) a ´ b = x y z 1 1 1 x y z 2 2 2 r r r r r r r r r ér r ù r 7. (i) êab c ú= a ×b ´ c = b ×c ´ a = c ×a ´ b ë

û

r r r And each is numerically equal to the volume of a parallelopiped having a , b , c as

coterminus edges

r

r

r

r r

r

r

r r

r

r

r

(ii) If a = x i + y j + z k , b = x i + y j + z k , c = x i + y j + z k

1

1

1

2

2

2

3

3

3

x y z

r 1 1 1 ér r ù Then êa b c ú = x y z 2 2 2 ë û x

3

y

r

ér r ù (iii) êa b c ú= 0 Þ ë û

z 3 3

r

r

r

r r r

Either a = 0,(or )b = 0,(or )c = 0 (or )a ,b ,c

are

coplanar (or) two of them are either equal or parallel.

r

(

r

r

r

r

r r r

)

r r r

( )

8. (i) a ´ b ´ c = (a ×c )b - a ×b c

r

(

) (

r

r

)

r

(ii) a ´ b ´ c = / a´ b ´ c

9. (i) vector equation of a straight line through the point, whose

r r r r r is a and parallel to b is r = a + tb (t being

position vector

scalar parameter) r (ii) Vector equation of a straight line through the points, whose position vectors are a

r

r

r

r

and b is r = (1 - t )a + tb

r

10. Vector equation of a plane (i) through the point, whose position vector is a and

r

r

r

r

r

r

parallel to b and c is r = a + sb + tc

3

r r r

(ii) Through three given points, whose position vectors are a ,b ,c

is

r r r r r = (1 - s - t )a + sb + tc

r r r (iii) Through two points, whose position vector are a ,b and parallel to c is r r r r r = (1 - s )a + sb + tc r r r q (iv) Normal to n is r ×n = q , where r is the length of perpendicular from the 1n 1 origin to the plane

r

r

(v) Through a given point, whose position vector is a , normal to n

is

r r r r r r r (r - a )×n = 0(or )r ×n = a ×n UNIT – I

CURVES IN SPACE

1.1. DEFINITIONS OF CURVES IN SPACE AND EXAMPLES:

r

A curve in space is the locus of a point, whose position vector is r relative to a fixed origin may be expressed as a function of a single variable parameter, u (say) inside a certain closed interval. Hence, the rectangular coordinates (x , y , z ) of the point can be expressed as functions of the same parameter. The

Cartesian

form

of

the

equation

of

a

space

curve

is

x = x (u ), y = y (u ), z = z (u ), u £ u £ u 1 2 If ox, oy, oz represent the rectangular axes along which the respective unit r r r vectors in the positive directions of the axes are i , j ,k ,then

4

r r r = r (u ), u £ u £ u is known as vector form 1 2 r of the equation to the curve, r , being the position vector of a point with respect to a fixed origin and

u , being a parameter. If we represent the rectangular coordinates of the

xi  i 1, 2,3  in short, then the

point in space as

equation of the curve is xi xi  u , i 1, 2,3 u1 u u2

Another definition of space curve: A space curve can also be defined as the intersection of two surfaces viz.,

F (x , y, z ) = 0& F (x , y, z ) = 0 .But such an intersection may split into several 1 2 curves. Note: When a straight line intersects a surface in k points, we say that the surface is of degree k. If it is intersected by a plane in a curve of degree k, then also we say that the surface is of degree k. A space curve is of degree l, if a plane intersects it in l points. The points of intersection may be real, imaginary, coincident or at infinity. The complete intersection of an algebraic surface of degree m and an algebraic surface of degree n is a space curve of degree m n. Method of obtaining the parametric representation of a curve of intersection of two surfaces, whose equations are F (x , y, z ) = 0, F (x , y , z ) = 0 :

1

2

Eliminating x from the above equations, we will get the relation between

y and z . Now express y = F (z ). Sim ilarly having eliminated y from the above 3 equations, express x = F

4

z

( ).

If z is a function of a variable parameter u (say) then both x and y will be functions of the same parameter.

5

\ The parametric representation of the equation to the above curve can be put as

x = /01 (u ), y = /02 (u ), z = /03 (u ) with u £ u £ u . /01,0/ 2,0/ 3 being real valued 1 2 functions of the single variable parameter. Note: A parametric representation of the curve gives the sense of description of the curve also. The positive direction of the description of the curve is that in which the parameter increases and the opposite direction is the negative direction. e.g., i) straight line: If a straight line passes through a point A ,

r

whose position vector is a and is parallel

r

r

r

r

to b , then its equation is r = a + ub

r

( r being the position vector of any point on the line and u, being the scalar parameter)

r

r

r

r r

r

r

r r

r

r

r

If r = xi + yj + zk , a = a i + a j + a k , b = b i + b j + b k

1

2

3

then x = a + ub y = a + ub

1

1 2 2 x- a y- a z- a 1= 2= 3 \ b b b 1 2 3

1

2

3

z = a , + ub 3 3

(1

This is Cartesian equation of the same line passing thro’ a , a ,a

2 3

) whose direction

cosines are proportional to b , b , b , where at least one of bi s is not zero.

1 2 3

Check Your Progress ( C Y P ) : 1. Define Curve in space ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

6

2. Give two examples of space curves ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

2). Circle:

The circle is a plane curve. If its plane is taken as z = o and radius ‘a’ then its parametric

equation

is

x = a cos u, y = a sin u, z = 0 with o u 2, where u is the parameter. In

the fig. xop = u

r

(

)

r

(

)

r

The corresponding vector equation is r = a cos u i + a sin u j , z = 0 with

r 0 £ u £ 2p , (OR) r = (a cos u, a sin u, 0) 3) Circular helix: Helix is a space curve traced on a cylinder meeting the generators under the same angle

7

Definition:

A circular helix is a helix traced on the surface of a circular cylinder. The axis of the circular cylinder is defined as the axis of the cylinder. The equation of the circular helix is x a cos u , y a sin u , z bu, where a

and b are constants. The curve winds around the circular cylinder in such a manner that when u increases by 2, then x and y return to their original

value, while z increases by 2b which is called the pitch of the helix.

Note: When b is + ve, the helix is right handed and when b is - ve, the helix is left handed.

Arc length, Tangent:

r

r

If C is a real curve, then the arc length of a segment of the curve r = r (u )

( 0 ) and P

between the points A u

u s (u ) = ò

u

u can be given by means of the integral,

( )

x&2 + y&2 + z&3 du

0

u r r = ò r&×r& du u 0

8

r r r& dr dr Where r = and we assume that is nowhere zero in the given interval. This du du means the curve has no singular points in the interval and the curve is regular. In terms of differentials, the length of the arc is given by.

2

2

2

2

æds ÷ ö æ ö ædy ÷ ö æ ö çç ÷ = ççdx ÷ çç ÷ + ççdz ÷ + ÷ çèdu ÷ çèdu ÷ çèdu ÷ çèdu ÷ ÷ ø ø ø ø Where x&=

dx dy dz , y&= , z&= du du du

2 2 2 2 \ (ds ) = (dx ) + (dy ) + (dz ) and ds is known as linear element of C The arc length ‘ s ’ increases with the increase in u . The sense of increasing are length is called + ve sense on the curve. A curve with a sense on it is called an oriented curve. Most of our reasoning in Differential Geometry is with oriented curves. Our space has also been oriented by the introduction of a right handed system. Yet our results are often independent of orientation.

r

r

We can write ds 2 = dx 2 + dy 2 + dz 2 = dr ×dr

r r dr dr \ × =1 ds ds r dr The vector is a unit vector ds

r dr ie., =1 ds

r dr Geometrical interpretation of : ds The tangent line at a point P on the curve C is defined to be the limiting position of a straight line through P and through the neighbouring point Q on C as Q ® P along the curve. Let P and Q be two neighbouring points

r

r

r

with position vectors r and r + dr respectively on the curveC .

9

r r r ¼ Now has the same direction as dr . If A P = s (A, being the initial point), then ds ¼ ¼ = ds A Q = s + ds , where PQ

r r dr dr lt A sQ ® P , = , which is a tangent ds ® 0 ds ds

vector at P .

r r dr dr lt chordP Q lt But is a unit vector, Q = = 1 d s ® 0 d s ® 0 ds ds arc PQ r r r r dr dr \ denotes the unit tangent vector at P . If we denote this by t , then t ,  ds ds r r dr dr ds Note (1): = × du ds du r dr is also a tangent vector. But it need not necessarily be a unit tangent vector, while du r dr is a unit tangent vector ds r r r& r dr dr (2) Notation: we denote by r , while by r ¢ du ds r r r r r dr r r r r r (3) If r = xi + yj + zk , t = = r ¢= x ¢i + y ¢j + z ¢k ds Where x ¢=

dx ¢ dy ¢ dz ,y = ,z = ds ds ds

r

Since, t is unit tangent vector, x ¢, y ¢, z ¢ denote the direction cosines of the tangent at P .

r r (4) As t is a unit vector, t = cos a

(

1

Where cos a + cos2 a 2 + cos2 a

1

r i + cos a

) (

3

=1

10

2

)

r j + cos a

(

3

r k

)

Equation of the tangent line to a curve at a point: Let P be the point of contact of the r r tangent with the curve r = r (u ) r We know that r& is parallel to the tangent line at P . r If r is the position vector of P and r R is the position vector of any point Q on the tangent at P , then the

equation of the tangent is r r r R = r + wr&, w being a scalar parameter. r If instead of the parameter u, we use the parameter (arc length), then since t is unit vector along the tangent at P , the equation of the tangent line is given by r r r r r r R = r + l t (or )R = r + l r ¢ ( l , being scalar parameter)

1.2. TANGENT LINE IN CARTESIAN FORM:

r

r

r

r

r

r

r

r

Corollary 1: if r = x i + y j + z k and R = xi + yj + zk

1

1

1 r r r Then the equation of the tangent is R = r + wr& r r r r r r r r r & + yj & + zk & i.e., xi + yj + zk = x i + y j + z k + w xi 1 1 1 x- x y- y z- z 1= 1= 1 i.e., x& y& z&

(

(1

)

Which is the Cartesian equation of the tangent thro’ x , y , z

1 1

) having the direction

,y, z cosines proportional to x&&& If we use the parameter s, we have the Cartesian equation of the tangent as

x- x y- y z- z 1= 1= 1 x¢ y¢ z¢

11

Corollary 2: If the equation of the curve is given as the intersection of the surfaces

F (x , y, z ) = 0 and F (x , y, z ) = 0 1 2 We have

And

¶ F dx ¶ F dy ¶ F dz 1× + 1× + 1× = 0 ¶ x du ¶y du ¶z du

¶ F dx ¶ F dy ¶ F dz 2× + 2× + 2× = 0 ¶ x du ¶ y du ¶ z du ¶F ¶F ¶F 1 + y& 1 + z& 1 = 0 ¶x ¶y ¶z

i.e, x&

¶F ¶F ¶F 2 + y& 2 + z& 2 = 0 ¶x ¶y ¶z

And x&

x& y& z& = = ¶F ¶F ¶F ¶F ¶F ¶F ¶F ¶F ¶F ¶F ¶F ¶F 1× 2 1× 2 1× 21× 2 1× 21× 2 ¶y ¶z ¶z ¶y ¶z ¶x ¶x ¶z ¶x ¶y ¶y ¶x from which we obtain the d. c’s of the tangent. Substituting the values of x&&& , y , z in the

r

r

r

equation R = r + wr ,we get the equation of the tangent line at a point of the curve of

(

)

(

)

intersection of the surfaces F1 x , y , z = 0& F2 x , y , z = 0 e.g; (1) find the equation of the tangent to the circle

x = a cos u, y = a sin u, z = o x&= - a sin u, y&= a cos u, z&= o 2 = a2 s&2 = x&2 + y&2 + z&

s&= a

\ The equation of the curve can also be written as, æs ö æs ö x = a cos çç ÷÷÷, y = a sin çç ÷÷÷, z = 0 èa ø èa ø

12

\ s = au

r r r d r d r ds The unit tangent vector t = = ds du du r dr æ dx r dy r dz r ÷ö ds = çç i + j+ k÷ du çèdu du du ÷ø du

y P

o

r r r 1 (- a sin u )i + (a cos u )j + o k a r r = (- sin u )i + (cos u ) j =

a u

(

\ P is

Equation

of

the

tangent

) line

at

x - a cos u y - a sin u = - sin u cos u

i.e. x cos u + y sin u = a e.g; (2) Circular helix:

x = a cos u, y = a sin u, z = cu \ x&= - a sin u ,y&= a cosu , z&= c

2

æds ö÷ 2 = a 2 + c2 çç ÷ = s& çèdu ø÷

\ s&= a 2 + c 2 = k (Say)

\ s&= k integrating w.r .t .u, s = ku r æ- a a cö \ t = çç sin u , cos u , ÷÷ è k k kø Hence, the equation of the tangent at P on the circular helix is

x - a cos u y - a sin u z - c u = = - a sin u a cos u c

e.g., (3) show that the tangent at any point of the curve whose equations are

x = 3t, y = 3t 2 , z = 2t 3 makes a constant angle with the line y = z - x = 0 Solution:

r r The position vector r of any point on the curve is r = 3t ,3 t 2,2t 3

(

13

)

r r& dr r = = 3,6t ,6t 2 dt

(

)

1 r& 2 4 2 \ s&= 1r 1 = 9 + 36t + 36t =

(

3 6t  2 2

1

2

)

3 6t 2 3 1 2t 2 

\ The direction cosines of the tangent are given by

(3,6t ,6t 2 ) 3(1 + 2t 2 ) i.e.,

1

2t

,

,

1,2 t,2 t 2 ) ( i.e.. 1 + 2t 2

2t 2

1 + 2t 2 1 + 2t 2 1 + 2t 2

The equation of the given line in symmetrical form is The direction cosines of this line are

1 1 , 0, 2 2

x y z = = 1 0 1

If  is the angle between the tangent

and the line, æ 1 öæ 1 ö ç ÷ ÷ ÷ Then cos q = çç ÷ ÷ ççç ÷+ çè1 + 2t 2 ø÷ ÷è 2 ÷ ø

æ 2t ö çç ÷ ÷ 0 + ÷ çç ÷( ) 2 è1 + 2t ø÷

2 11 × + 2t (0 )+ 2 t 1 =

æ öæ ö çç 2t 2 ÷ ÷ çç 1 ÷ ÷ ÷ çç ÷ 2 ÷øèç 2 ÷÷ø èç1 + 2t ÷

( ) = 1+ 2t 2 = 2 2 2 (1+ 2t ) 2 (1+ 2t )

1 2

Hence  is constant 3. Check Your Progress ( C Y P ) : Define tangent line at a point on the curve C and write down its vector equation ________________________________________________________________________ ________________________________________________________________________

14

1.3. OSCULATING PLANE: Definition: A space curve does not lie on a plane. However a very small arc of the curve may be thought of as almost a plane curve. The plane which should almost contain a small arc about a point P is called the osculating plane at P . Clearly for a plane curve, the tangent at P is contained in the plane. By analogy, we conclude that the osculating plane at P should contain the tangent at P . But there are infinitely many planes through the tangent. To fix osculating plane we must have a small arc about P or at least a neighbouring point Q of P , besides the tangent. Thus the osculating plane formally can be defined as the limiting position of the plane through the tangent at P and a neighbouring point Q on the curve, as Q ® P . To find the equation of the osculating plane: For a function f (x ) of real variable, we have the Tailor’s (or) Maclaurine’s expansion:

x 2 ¢¢ x 3 ¢¢¢ f (o )+ f (o )+ ×××××× 2! 3! r r In the same manner, we can expand r = r (s ) by Taylor expansion in the form: f (x ) = f (o ) + xf ¢(o )+

r r r s2 r s3 r r (s ) = r (o ) + sr ¢o( ) + r ¢¢o( ) + r ¢¢¢o( ) + ××××××× 2! 3! r This is because r (s ) is a simple combination of three functions x (s ), y (s ), z (s ) for each of which the Taylor’s expansion holds. r Now, if R is position vector of any point A on the osculating plane at P , then uuur uuur the tangent at P , the elementary vector PQ and the vector PA must all be coplanar. If we measure arc lengths fromP , so that the arc . PQ is small, we find that uuur uuur uuur r r

P Q = O Q - O P = r (s ) - r (o ) uuur uuur uuur r r and, PA = OA - OP = R - r (o )

r And that tangent at P is represented by r ' (o ),

15

uuur uuur Since PA , the tangent vector at P , PQ are all coplanar, we have that the box

product

of

these

three

vectors

r r r r r ù é êR - r (o ), r ' (o ), r (s )- r (o )ú= 0 ë û Using

the

Taylor’s

is

zero.

ie.,

we

obtain

- - - - - (1)

expansion

for

r r r (s )- r (o ),

r r r s2 r r (s )- r (o ) = s r ¢(o ) + r '' (0) nearly. Substituting this expansion, the condition 2! ér r r r s2 r ê (1) becomes êR - r (o ),r ' (o ),sr (o )+ r o( 2 ê ë

ù ú ú= 0 ú û

)

r r 2 r r r r r r é ù s é ù i.e., s êR - r (o ), r ' (o ), r ' (o )ú+ êR - r (o ), r '(o ), r ''(o )ú = o ë û 2 ë û Since the first box product is zero, and since two of its components are equal, we get

r r r r é ù R ê - r (o ), r ' (o ), r ''(o )ú = o ë û This is required equation of osculating plane in vector form at P (s = o). Its equation in scalar form is

X - x (o ) Y - y (o )

Z - z (o )

x ¢(o )

y ¢(o )

z ¢(o )

x ¢¢(o )

y ¢¢(o )

z ¢¢(o )

=o

Note: Any plane through a straight line is the osculating plane for the line. If the curve is given in terms of an arbitrary parameter u , instead of the arc lengths , then to find the equation of the osculating plane. r r If the equation of the curve isr = r (u ), then

r r r r r dr dr du dr ds r& ¢ r = = . = = ds du ds du du s& rö ær&ö ær&ö r dæ dr d ç r ÷ d ççr ÷ ÷ ÷du çç ÷ r '' çç ÷÷ = = ÷ çç ÷ ÷ ç ÷ ç ds èds ø ds çès&÷ø du èçs&ø÷÷ds

16

r& r& s&r& - r s& & 1 = × 2 ds s& du r& r& r& & r& s&r& - r s& &r& s& = = r 2 s&3 s&3 s& r r r Then the equation éêR - r ,r ',r ''ù = 0 becomes ú ë û r r r& n&& s& &r&ùú , r = o s& s&2 s&3 úú û r r r&& r&ù r r rù 1 é s& & ér i,e., R r , r , r R - r , r&& , r ú= o ê ú ê 3 4 ë û ë û s s& ér r êR - r , ê êë

Since, the second box product is zero, we get

r r r rù é && & R ê - r ,r ,r ú= o ë û r& &

r& &

Note:(1) The osculating plane is not determined, when r = o or when r is proportional r to r&

r

r

r

&= o , then r&= a a constant vector (2) (i) if r& r r r \ r = u a + b which is a straight line r r r r lu r (ii) r = l r& Þ r&= c e = cf1 (u ) r r r r r r r Þ r = cf (u )+ a , which is also straight line when a, b, c, d 2 are fixed vectors. r& r& r \ if r& = o and r& = l r& are valid for all points on the curve, then they both are satisfied for straight lines and only for those. Note(3): In the case of plane curve, the osculating plane coincides with the plane of the curve. When the curve is a straight line, the osculating plane is indeterminate and may be any plane through the straight line. E, g.,(1) Find the osculating plane of the circular helix: x = a cos u, y = a sin u, z = bu (- a < u < a ) The vector form of the equation of the helix is 17

r r r r r = (a cos u )i + (a sin u ) j + bu k or

r r = (a cos u, a sin u, bu ) r r& \ r&(- a sin u, a cos u, b) & r& = (- a cos u, - a sin u, o ) r

r r r&ù é The equation of the osculating plane at any point ' u ' is êR - r , r&& , r ú= o , where ë û r R = (X ,Y , Z ) is the position vector of any point on the osculating plane. éX - a cos u Y - a sin u Z - bu ù ê ú r r r&& r& ê ú é ù \ êR - r , r , r ú= o Þ ê- a sin u + a cos u b ú= o ë

û

ê ê- a cos u êë

- a sin u

o

ú ú úû

i.e., (X - a cos u )(a b sin u ) + (Y - a sin u )(a b cos u )+ (Z - bu )a 2 = o i.e., X b sin u - Y b cos u + (Z - bu )a = o i.e., X b sin u - Y bcou + Za = abu. 2) Find the equation of the osculating plane for the curve x = 3t, y = 3t 2, z = 2t 3 at

(A ns : 2t 2 -

t = t1

1

2t 1y + z = 2t 13

r

)

3) Find the equation of the osculating plane for the curve given by r = u , u 2, u 3 at

(

)

any point ' u '

(A ns : 3u 2x -

3

3uy + z - u = o )

4. Check Your Progress (C Y P): In what situations, osculating plane is not determined? ________________________________________________________________________ ________________________________________________________________________

1.4. PRINCIPAL NORMAL AND BINORMAL: The plane at P perpendicular to the tangent line at P and therefore also perpendicular to the osculating plane at P , is called the normal plane at P . It contains all the straight lines at P perpendicular to the tangent. i.e., all the normals. Among all 18

these normals, there are two important ones. They are the principal normal and the binormal at P . In a plane curve, we have just one normal line. This is the normal, which lies in the plane of the curve.

Definition: Principal Normal is the normal lying in the osculating plane and it is the intersection of the normal plane and the osculating plane. The normal which is perpendicular to the osculating plane at a point is called the Binormal . Certainly, the binormal is also perpendicular to the principal normal. r Let n be the unit vector along

the principal normal. It we take r b as the unit vector at P perpendicular to the osculating r plane, then b is along the Binomal

r r r

Now t , n , b are three mutually ^ r unit vectors forming a right handed system in that order. r r r As P moves along the curve, t , n , b form a moving frame of reference and hence

known as moving trihedron. Thus at each point on the curve we have three unit vectors r r r r r r r r r t , n , b satisfying the relations: t ×t = n ×n = b ×b = 1

r r r r r r t ×n = n ×b = b ×t = 0 r r r r r r r r r t ´ n = b, n ´ b = t , b ´ t = n r r The plane containing the tangent vector t and the binormal vector b at P is

called rectifying plane.

19

Thus at any point of the curve, we have three planes formed by the moving trihedron. They are: i) ii) iii)

r r The osculating plane (containing t and n ) r r The normal plane (containing n and b ) r r The rectifying plane (containing b and t )

1.5. CURVATURE OF THE CURVE AND TORSION:

The curvature at a point on the curve is defined by the rate of change of direction of the tangent at that point with respect to the arc length s.

In other words, if y is the angle made by the tangent at a point P on the curve, then the curvature of the curve at P is defined as the rate of change of y × w r t. s.

i.e., The curvature at P =

dy ds

r r dt The curvature is also measured from = t¢ ds r r t ×t = 1 Þ on differentiation w r t s, we get r r r r r r t ×t ¢= 0 t ¢^ t But t ^ n r r r r \ t ¢P n \ t ¢= k n, (where k is scalar called kappa ) r r r Now, if we denote kn = k , then k is known as the curvature vector and it measures the rate of change of the tangent, when we proceed along the curve. k is called curvature and

k is called length of the curvature vector..

20

r r dt kn = = ds r r \ kn ×kn =

r r d 2r = r ¢¢ 2 ds r r r ¢¢×r ¢¢

r r k 2 = r ¢¢ r ¢¢ r r r r Note: t ¢is equal to a + ve or –ve multiple of n \ t = k n Þ k may be +ve or –ve.

Result: A curve is a straight line iff k = 0 at all points.

r r r r r Proof: The equation of the straight line is r = as + b , where a and b are constant r r r r r r r vectors. Thus t = r ¢= a and t ¢= 0 i.e., r ¢¢= 0

r r \ k 2 = r ¢¢×r ¢¢ Þ k 2 = 0

Þ k= 0

r r r Conversely, if k = 0 everywhere, then t ¢= r ¢¢= o everywhere, which on integration r r r r r gives r ¢= a , constant vector and again integrating we get r = a s + b , which represents a straight line. P Torsion: The rate of change of the direction of the binormal at P on the curve, as P moves along the curve is defined as the Torsion at P . (or) The rate at which the osculating plane turns about the tangent at P as the point P moves along the curve is called Torsion of the curve at P .

The magnitude of the torsion is denoted by t .

r r r We know that, as b is the binormal unit vector, b ×b = 1 , which gives on r r r r r differentiation b ×b ¢= 0 Hence b¢^ b and therefore b¢ lies in the osculating plane. But r r b ×t = 0 , which gives on differentiation,

r r r r b¢×t + b ×t ¢ = 0

r r r r r r r r ie. ., b ×t = - b ×t ¢ = - b ×kn = - k b ×n = 0

( )

21

r r r Thus b ¢ is also ^ r to t . Thus b¢ is a vector lying in the osculating plane and r r r r r perpendicular to t ×\ b¢ is parallel to n . Thus we writeb¢ = t n , where t is the torsion. But

it

is

customery

to

denote

r r b¢= - t n for

the

following

reason.

r r r r r r Assume t , n , b as the unit vectors i , j , k along the three mutually ^ r coordinate r axes OX , OY , OZ respectively. Then X Y plane is the osculating plane, and k is the binormal. Assume that the origin is moving along the +ve direction of X - axis . ie, the r direction of t , which is the direction of the curve. Thus, as we move along the curve, if the X Y plane turns in the manner of a right handed screw advancing in the direction of r r r db X - axis, then the change db in b will be to the direction of negative Y axis. Thus ds r r and therefore b¢ must be in the direction of (- n ) Check Your Progress ( C Y P ): 5. What are the three types of planes at any point on the curve ?\ ________________________________________________________________________ ________________________________________________________________________ 6. Prove that a curve is a straight live if and only if the curvature k = 0 at all Points. ________________________________________________________________________ ________________________________________________________________________

1.6 . SERRET-FRENET FORMULAE:

r r r db r dt We have obtained = kn , = - tn ds ds r r r

We know that t , n , b form a right handed system of unit vectors in that order.

r r r \ n = b ´ t differentiating with respect to s both sides

22

r db

r r r dt = ´ t +b´ ds ds ds

r dn

r r r r = - t n ´ t + b ´ kn

r r = t b - kt r r r dt t ¢= = kn ds

r r r r dn n ¢= = t b - kt ds r r db r b¢= = - t n are called serret- Frenet formulae ds These formulae can be put in a matrix equation

ær ö ççt ¢÷ æo k ÷ çç çç r ÷ ÷ ÷ ç ççn ¢÷ = çç- k o çç ÷ ÷ çç ÷ çç r ÷ çço - t ÷ è çèçb¢ø÷ ÷

o ö÷ ÷ ÷ t ÷÷÷ ÷ o ø÷÷÷

ær ö ççt ÷ ÷ çç r ÷ ÷ ççn ÷ ÷ çç ÷ ÷ ÷ çç r ÷ ÷ çèçb ø÷ ÷

Note: if we denote

r r r r t = cos a i + cos a j + cos a k 1 2 3

(

) (

)

(

)

r r r r n = cos b i + cos b j + cos b k 1 2 3

(

)

(

)

(

)

r r r r b = (cos r )i + (cos r )j + (cos r )k 1 2 3

23

r

r

æ r ççQ n ´ t = - b ö÷÷ çç r r r ÷÷÷÷ çç b ´ n = - t ø÷ çè

a i¢s , b i¢s

and ri¢s

are the angles which the oriented tangent, principal normal and

binormal make with the positive directions of coordinate axes ox , oy , oz respectively Hence, the formulae of Serret Frenet can be written in the following form:

d cos a i = k cos b i ds d cos b = t cos r - k cos a i i i ds

i = 1,2, 3

d cos r i = - t cos b i ds 1.7. EXPRESSION FOR TORSION :

r r dt We know that = kn ds

r r ie., kn = r ¢¢

r r \ k 2 = r ¢¢×r ¢¢×------------------(1) r r db We know also that - t n = ds

r r db \ tn = ds

r taking dot product with n

r r r r r db r db t n ×n = - n × ie ., t = - n × ds ds r d r r t´ n ds

i.e.., t = - n ×

(

)

r r r r r r = - n × t ¢´ n - n . t ´ n

(

)

24

(

)

r r r r r = - n × t ´ n ¢ \ t ¢P n

(

)(

)

érr ¢¢ ær rr ¢¢¢öù ÷ ú ÷ = - êê ×çççr ¢´ ÷ ú ÷ k ç k è ø úû ëê

=-

1 r ¢¢ r¢ r ¢¢¢ (r ×r ´ r ) k2

=-

1 r ¢ r ¢¢¢ r ¢¢ (r ×r ´ r ) k2

=

r 1 r r ¢ ¢ ¢ ¢¢¢) r × r ´ r ( k2 érr ¢rr ¢¢rr ¢¢¢ù ú ëê û

t = r r r ¢¢×r ¢¢

Which is the expression for t

r r r dr Aliter: r ¢= =t ds

r r r dt r ¢¢= = kn ds r r r r ¢¢¢= k ¢n + kn ¢ r r r = k ¢n + k t b - kt

(

)

r r r = k ¢n + k t b - k 2t

rr r ér \ éëêr ¢r ¢¢r ¢¢¢ùûú = êt ë

r rù r r kn k ¢n + k t b - k 2t ú û

25

r r r = éêt kn k ¢n ùú + ë

r

r

é r ù êt kn k t b ú + ë û

û

rù ér r tê kn - k 2t ú ë û

r r r = k 2t éêt n b ùú ë

û

r r r æ r r r ö = k 2t çç\ éêt n b ùú= t ×n ´ b = 1÷÷ è ë ø û érr ¢rr ¢rr ¢¢¢ù ëê ûú

1 ér ¢r ¢¢r ¢¢¢ù \ t = êr r r úû k2 ë

= r r r ¢¢×r ¢¢

Find expressions for k and t ,choosing u as a parameter:

r r r r dr dr du r = = . = r&u ¢ ds du ds

r ie, t =

r dr ds

=

r r&u ¢

r r r r r r - 1/ 2 \ 1 = t ×t = r&×r& u ¢ \ u ¢= r&×r&

( )

2

( )

r r& & 2 r& Now r ¢¢= r u¢ + r ×u ¢¢

® (1)

r r& r& r&r r &¢3 + 2ru & ¢u ¢¢+ r& r ¢¢¢= & ru u ¢u ¢¢+ r&u ¢¢¢ r& r& r &¢3 + 3ru & ¢u ¢¢+ r&u ¢¢¢ =& ru r r - 1/ 2

Now u ¢= r&×r&

( )

u ¢¢=

r r& r& - 1 r& r&- 3/ 2 r& & r ×r r ×r&u ¢+ r& ×r u ¢ 2

( )

(

)

26

r r - 3/ 2 r& & r = - r&×r& r ×r&u ¢

( )

(

)

4 r r& = - u ¢ r&×r&

( )

2 é& r r r& 2 r& 4 r& & r&ù \ r ¢¢×r ¢¢= êr u ¢ - r u ¢ r ×r ú

( )úû

ê ë

® 2

from (1) & (2)

ù 8 ér& 2 r& r& r& = u ¢ êr& u ¢ - r r ×r&ú2

( )úû

ê ë

8 = u¢

r&ù2 ér& & r& r& r& r& & êr r ×r - r r ×r ú êë ú û

( ) ( )

8 ér r& r& ù2 = u ¢ êr&´ r& ´ r ú êë

(

)úû

8 r r 2 = u ¢ éêr&´ A ùú ë

r

r& r& where A = r& ´ r

û

ér r r r ù = u ¢ ê r&´ A g r&´ A ú 8

(

êë

8 = u¢

)(

)úû

ér r r r r 2ù êr&×r& A ×A - rr&×A ú ê ú ê ú ë û

( )(

) ( )

8 r r r r = u ¢ r&×r& A ×A

( )(

r r Q r&×A = 0

(

)

r r 6 r& r& & = u ¢ r& ´ r × r&´ r&

(

)(

)

27

)

r r - 2 Q r&×r&= (u ¢)

\

r& r& & r& r& & r ´ r × r ´ r) ( ) ( r r 2 ¢ ¢ ¢ ¢ k = r ×r = r r 3 (r&´ r&) r

r

r

r

&× r&´ r& & r&´ r& ( ) ( ) i.e. k 2 = rr

3

(r .r )

r r r r& 3 Now r ¢´ r ¢¢= r&´ r& u¢

(

)

r r r r r& & r& &¢6 \ (r ¢´ r ¢¢)×r ¢¢¢= r&´ r& ×ru

(

)

r r r r r&& r& ù ¢6 ie, éêr ¢r ¢¢r ¢¢¢ùú= éêr&& r r& úu ë û ë û érr ¢rr ¢¢rr ¢¢¢ù ëê ûú

\ t = r r r ¢¢×r ¢¢

=

r&& r& ér&& ù 6 & êr r r úu ¢ ë û

r

r

r

r

(r&&´ r&)×(r&&´ r&)u ¢6 r&& r& ér&& ù & êr r r ú ë û

t = r r r r &× r&´ r& & r&´ r&

(

)(

)

The necessary and sufficient condition that a curve to be a plane curve is t = 0 (OR) A curve is a plane curve if and only if t = 0 at all points of the curve. Prof: Necessary condition: r Let the curve be a plane curve, since b is normal to the osculating plane, the plane curve lies in the osculating plane, ie., the plane considered is the osculating plane and it must be fixed.

28

r r r r r t and n lies in the osculating plane and hence b = t ´ n a constant vector r r \ b¢ = 0 r r But b ¢= - t n

\ t = 0

Sufficient condition:

r Let t = 0 Hence b¢ = 0

r This implies b is a constant vector. r r If the equation of the curve is r = r (s )

r r ¢

( )

We have r ×b

r r r r = r ¢×b + r ×b¢

r r r r r r r r = t ×b + r ×b ¢ Qt ×b = 0 and b ¢= 0

(

)

=0

r r \ r ×b is a constant vector, showing that the curve is a plane curve. Check Your Progress ( C Y P ) : 7. Write down the serret – Frenet formulae for space curve in term of matrix equation . ________________________________________________________________________ ________________________________________________________________________

29

Problems: 1. find

the curvature and torsion for the circular helix whose equation is

x = a cos u, y = a sin u, z = bu

r r = (a cos u, a sin u, bu ) ds 2 = a 2 sin 2 u + a 2 cos2 u + b2 du 2

(

\

)

ds 2 a + b2 = c (say ) du

r r ¢=

\

du 1 = ds c

æ- a sin u a cos u b ö÷ çç , , ÷÷ çè c c cø æ- a cos u - a sin u ÷ ö çç ÷ , , 0 ÷ ççè 2 2 ÷ ø

r r ¢¢=

c

r r ¢¢¢=

c

æ+ a sin u - a cos u ö çç , , 0÷÷÷ çç 3 3 è c ø÷ c

r r r b a2 a 2b \ éêr ¢r ¢¢r ¢¢¢ùú= × cos2 u + sin 2 u = ë û c c5 c6

(

r r a2 r ¢¢×r ¢¢= cos2 u + c4 r r Q k = r ¢¢×r ¢¢

)

a2 2 a2 sin u = c4 c4 a a = = 2 2 c a + b2

érr ¢ rr ¢¢rr ¢¢¢ù 2 4 b ê ú a b c and t = ë r r û= × = =

r ¢¢×r ¢¢

c6 a2

c2

b a 2 + b2

________________________________________________________________________ 2. Find the curvature and torsion for the curve

x = a 3u - u 3 ,

(

y = 3au 2,

)

30

z = a 3u + u 3

(

)

r æ r = çça 3u - u 3 , è

(

r r&=

æ çç3a (1 - u 2), è

(- 6au,

6a,

r& & r& =

(- 6a,

0

)

(

)ø

6au )

6a )

r i r r& r&´ r& = 3a 1 - u 2

(

ö 3 a 1 + u 2 ÷÷

6 au,

r& r& =

(

ö a 3u + u 3 ÷÷÷ ø

3au2,

)

)

- 6au ,

r j

r k

6au

3a 1 + u 2

(

6a

)

6au

r i

r j

= 18a 2 1 - u 2 - u

r k

2u

1 + u2

1

- u

r r r = 18a2 u 2 - 1 i - 2uj + 1 + u 2 k

{(

)

)}

(

2 éê 2 2 ùú r& & r& r& & r& 2 2 2 2 \ r ´ r × r ´ r = 18a ê u - 1 + 4u + 1 + u ú

(

)(

) (

) ê(

)

(

)ú

ë

û

æ 2 é ùö = çç(18a 2 ) ê2u 4 + 2 + 4u 2 ú÷÷ è

ûø

ë

= 18a 2

(

2

2 u2 + 1

) (

2

)

é 2 2 ùú r& r& ê 2 2 2 2 r ×r = 9a ê 1 - u + 4u + 1 + u ú

(

)

ê ë

(

)ú û

= 18a 2 1 + u 2

(

31

2

)

(rr&×r&r&)×(rr&×r&r&) r r 3 (r&×r&)

\ k2 =

2 18a2

)

22 (1 + u )

3

6

(

=

18a 2

(

2

1 + u2

) (

)

2

=

18a 2 1 + u 2

(

4

)

1

\ k =

3a 1 + u 2

(

r

r

2

)

r

r

r

r

r&& r& ér&& ù & 2 u 2 - 1 i - 2uj + 1 + u 2 k × - bai + oj + 6ak êr r r ú = 18a ë û

{(

) }( = 108a 3 {- (u 2 - 1) + (1 + u 2 )} =

)

(

r&& r& ér&& ù & êr r r ú ë û

)(

216a 3

216a 3 2 2 2 18a 2 u 2 + 1

\ t = r r r r = &× r&´ r& & r&´ r&

(

)

)

(

)(

)

k = t

Here

________________________________________________________________________

(

)

3. Find the equation of the osculating plane at the point t = 1 of the curve

r r = 3at , 3bt 2 ct 3

(

) (OR)

If the parametric equation of the curve is x = 3a t, prove that the osculating plane at the point t = 1 is Solution:

x = 3at , y = 3bt 2,

z = ct 3

32

y = 3bt 2,

X Y Z + =1 a b c

z = c t3

z&= 3ct 2

Þ x&= 3a , y&= 6bt , &= 0, y& &= 6b, z& &= 6ct x&

The equation of the osculating plane at any point ‘t’ is

Y - 3bt 2

X - 3at

Z - ct 3

3a

6bt

3ct 2

0

6b

6ct

= 0

Expanding the determinant, we get é ù (X - 3at )ê36bct 2 - 18bct 2 ú- Y - 3bt 2 [18act - 0 ]

( ) + (Z - ct 3)éë18ab - 0ùû = 0 3at )(18bct 2 )- (Y - 3bt 2 )(18act ) + (Z - ct 3 )(18ab ) = 0 ë

i.e. (X -

û

2

3

i.e. 18 ´ bct - 54abct - Y ×18act + 54abct

\ At

3 + 18Zab - 18abct 3 = 0

t = 1, the equation is

18 ´ bc - 18acy + Z ×18ab - 18abc = 0 æX ö Y Z 18abc çç + - 1÷÷÷ = 0 çè a b c ø

Q abc ¹ 0,

X Y Z + =1 a b c

________________________________________________________________________ 4. prove that

k = constant is a necessary and sufficient condition for a space curve t

to be a cylindrical helix (OR) Show that a space curve is a cylindrical helix if and only if, its curvature and torsion are in a constant ratio.

33

Solution: Given: The curve is a cylindrical helix, To prove that

k = constant. t

r

r

Let a denote a unit vector along the axis of the cylindrical helix. (a is a constant vector).

r

Let t be the unit tangent vector Since, the cylindrical helix intersects the generators at a constant angle a , we

r r

have t ×a = cos a - (1) Differentiating

r r

i.e., kn ×a = o -

(1)

with

respect

to

s,

we

r r r r t ¢×a + t ×o = o

get

(2) r r

r

r

Since k =/ o , for helix, from (2) we have n ×a = o i.e., a ^ n

r

This implies a lies in the rectifying plane

r

r

(Rectifying plane means, the plane contains t and b )

r r o Since, a makes angle  with t so it must make angle 90 r rr - with tb . Thus a expressed as linear Combination of

r r t and b is given by r r r a (cos)t (sin)b Differentiating (3) w.r.t ‘s’, we get

r

r

0 = (cos ) t ' (sin ) b '

r r 0 kn cos ( n ) sin)

r r (Q a ' 0) r r r r (Q t ' kn ) & b ' n )

r 0 (k cossin )n r r Since n 0, we have k cos sin 0

34

k sin    tan co ns tan t  cos  Conversely, let

k cons tan t tan (say ) 

Then we have k cos sin 0

r (k cossin )n 0 r r t 'cos b 'sin 0 r d r (t cosb sin) 0 ds r r r Integrating w.r.t.s, we get t cos b sin a r Where a is a constant unit vector. r Taking dot product with t both sides,

r r r rr t (t cosb sin ) t .a r r r r cos t . a (Q t . b 0)

r

Hence, the curve makes constant angle with the direction of fixed vector a . Thus, it is helix. Exercise: Find the curvature and torsion of the following curves

r

1) r  (au , bu 2 , cu3 )

r

2) r  (a (1cos u ), a sin u ,2 a sin u / 2 ) Answer: 1) k 6cu .

b2 (u 2 4a 2 ) a 2 3 2 2 2 2 4 ( a 4b u 9c u ) 2

12abc  3bc2u 2[b 2 (u 2 4a 2 ) a 2 ]2 1/2

2)

13 3cos u  ,  6cos u / 2 k 3/2 as (3 cos u )

a(133cos u )

35

APPENDIX: ( Problem) :

r r

  Prove that [r  r r ]0 is a necessary and sufficient condition for the

5.

r r

curve r r ( s) to be a plane curve. Proof: If the curve is a plane curve, then we know that its torsion 0

r r r

r r r

  Since [r  r r ]k 2, it follows that [ r r r ]0 . Thus the condition is necessary.

r r r

r r r

Conversely, if [r  r r  ]0 , then [ r  r  r   ] k 2  now show that

0 or k=0. We shall

0 always. For suppose k=0, but 0 at some point. Then there is a

neighbourhood of this pint, where 0 . But in this neighbourhood, since k=o, it follows that the arc of the curve must be a straight line and hence

0 on this line, contrary to

the hypothesis. must be zero at all points of the curve, which shows that the curve is a plane curve. Thus the condition is sufficient also. Curvature and torsion of a curve given as the intersection of two surfaces: When the curve is specified by two Cartesian equations of the form

f x , y , z 0 and g x, y, z 0 , the curvature and torsion can be obtained by a suitable procedure . e.g. Obtain the curvature and torsion of the curve of intersection of the two surfaces

2 2 2 2 ax by cz 2 1 and a x b y c z2 1 Let f ax2 by2 cz2 1 and g a x2 b y2 c z2 1. We know that the vector  is normal to the surface =constant.

Thus f (2ax,2 by,2 cz ) is

normal to the surface f=o. similarly g (2a  x,2 b  y,2 c  z ) is normal to the surface r g=0 the unit tangent vector t at any point of the curve of intersection of the two surfaces is perpendicular to both the normals f and g at that point, to the two surfaces and is

therefore

parallel

to

their

f x g =4  (bc b c ) yz ,(ca c a ) zx,(ab a b) xy 

36

cross

product

bc b  c ca  c  a ab a b 4xyz  , ,  x x  x  A B c  4xyz  , ,  x y z 

Provided x, y, z are different from zero, where A bc b c, B ca c a, C ab a  b. r Since t is parable to f X g ,we can write

r   t A , B , c  (1) x y z  Hence

A2 B2 c 2 2     (2) x2 y2 z2

A B c  r r Also from (1)  , , t  r x y z  dx dy dz   , ,  ds ds ds  And therefore

dx A , dy B , dz c ds x ds y ds z

Now if F be any scalar or vector field,

dF F dx F dy  F dz  .    . ds x ds y ds  z ds dF dx  F dy F dz  F   .  .  . ds ds x ds y ds z A F b F c F    u sin g (3) x  x y  y z  z A  B  c     F x y y z z x  Then

  d A  B c   ds x  x y y z z 

 (4)

37

 (3)

Then formula (4) converts the derivatives w.r.t. arc length s into derivatives w.r.t. space coordinates, operating (1) by (4), we get

r    d ( t ) A  B c A , B , c  ds x y  y z  z  x y z  x  r r A A, B B c c   2t  t  , , x x 2 y y 2 z z 2    A2 B2 C 2  r r   (5) 2 kn  t  , , x3 y3 z3    Taking cross product of (1) with (5), we get

r r r A B c  A2 B 2 C 2   tX ( 2kn  t )  , , X  , , 3 3 3 x y z    x y z   r BC2 B2C CA2 C 2 A AB2 A2 B  3  i.e  kb   ,  ,  3 3 3 3 3 yz3 y z zx y x xy x y    BC  CA AB  ( Bz2 Cy2 ), ( Cx2 Az2 ), ( Ay2 Bx2 )  y3 z3  z3 x3 x3 y3  

But Bz 2 cy2  ca c a z 2  ab a by 2

    a 1 ax2 a 1 ax2 aa acz 2 by 2 a c z 2 b y2

Similarly Cx 2 Az 2 b b And Ay 2 Bx2 c c r C  C     Thus 3kb  a  a , b  b , c  c   3 3   3   y 5z 3  z x x y3  

ABC  x3 y3z3

x3  y3 z3   a a  , b   b , c   c   6   C   A B   

38

Squaring (6), we get

A2 B2C 2 x6 2 6k 2   a  a x6 y6 z6 A2  x6 2   a a  2 2 2 A B C  A2 k 2   6 6 6 2 2 x y z  A / x  





 7 

      

Using (2)

This gives the value of k. Now (6) may be written as

3k x 3 y 3z 3 br  x3  y3  z3     a a , b b , C c c  ABC A B    r x2 y2  z2     ie . b    8 a a , B b b , C c c  A  

Where



3 3 3 3 k x y z ABC

Again, operating (8) by (4) we obtain l

r æA ¶ d B ¶ C ¶ ÷ö mb = çç + + ÷ çèx ¶ y ds y ¶y z ¶ z ÷ø

( )

x3 y3  z3      a a , b b , C c c  A B   

r r  b   n  3x  a a  ,3 y  b 'b  ,3 z  c 'c  10 

r r

r r

r r

Taking scalar product of (5) and (10) and noting that t  n n b b  t 0

r r

and n  n 1 , we get 3A2 3B 2  3C 2   2k  a  a  b b   c c  2 2 x2  y z  

39

A2 ie . 3 k 3 a a  2 x Substituting the value of  from (9), we get

3 k x 3 y 3 z 3 A2 k 3 a a  ABC x2 3

2 6 k 2  3ABC A a a  and substituting the value of 3 3 3 2 x y z x

6k 2 from (7) we get A2 B2C 2  x6  2  3ABC  A2       a  a   a a       6 6 6 2 3 3 3 2     x y z  A x y z  x    A2    a  a    3 3 3 2   3x y z  x  Hence  ABC  x 6 2  a    2 a A     This gives the value of. Check Your Progress ( C Y P ) : 8. What are the expressions for curvature k and torsion where u is the parameter. ________________________________________________________________________ ________________________________________________________________________ 1.8. SUMMARY : As given in the general introduction, the definition of curve in space and of curve as the curve of intersection of two surfaces are presented first. Following this some r r r dr examples of space curves are given. For the curve r r  s , is geometrically ds interpreted. Next, the method of obtaining the tangent line at a point on the curve is explained. Osculating plane at a point on the curve is explained. Osculating plane at a point on the space curve is defined and the equation for the same is derived. Definition of curvature of the curve at a point is defined and the expression for the same is obtained. Based on the relationship between unit tangent vector, the principal normal and binormal, Serret – Frenet formulae are obtained. Torsion at a point on the curve is defined and expression for the same is derived. The condition for the curve to be a plane curve is obtained. Some problems using the above results are solved.

40

1.9. KEY WORDS : 1.

2.

3.

4.

5.

6.

7.

Circular Helix : It is a space curve traced on a cylinder meeting the generator under the same (constant) angle. Unit Tangent Vector : r r r r r If P  r and Q  r Vr are neighbouring points on the curve r r  s and Vs is are length P Q, then, r r Vr dr lim  is the unit tangent vector at P on the curve. Vs  0 Vs ds Osculating Plane : It is the limiting position of the plane through the tangent at P and a neighbouring point Q on the curve as Q  P. Principal Normal : It is the normal lying in the osculating plane and it is the intersection of the normal plane and the osculating plane. Binormal : The normal, which is perpendicular to osculating plane at a point is called the Binormal. Curvature: Curvature at a point on the curve is the rate of change of direction of the tangent at that point with respect to the arc length s. Torsion : The rate at which the osculating plane at P turns about the tangent at P as the point moves along the curve is the torsion of the curve at P.

1.10. ANSWERS TO CHECK YOUR PROGRESS : 1.

2.

r A Curve in space is the locus of a point, whose position vector r relative to a fixed origin may be expressed as a function of a single variable parameter u say inside a certain closed interval. r r r r (i) A straight line r a ub passing through a point with position vector a and r parallel to b , u, being parameter.

(ii), A Circle, whose equation is,

r r r r  a cos u i  a sin u j with o u 2 u, being a parameter. 3.

It is the limiting position of the chord PQ of the curve as Q  P. r r r r r r Vector equation of the tangent line is R r t  or R r r  , if we take the parameter as s or u.

41

4.

5.

6.

7.

The osculating plane is not determined , when , r r& 2 r r r r r & r  2 0  or & r&r& where r r  u is the equation of curve  u

r r (i) The osculating plane containing t and n . r r (ii) The normal plane containing n and b . r r (iii) the rectifying plane containing b and t . r r r r r The equation of the straight line is r as b where a and b are constant vectors. r r r r r r r Thus t r ' a and t ' 0 is r 0 r r k 2 r   .n   k 2 0  k 0 r r r conversely, if k=0, every where, then t  r0 everywhere, which on integration r r r twice give r as b , which represents a straight line. P r r t  0 t  k 0 r   r   n  k 0  n  r r    b b   0  0    where k and  arc curvature and torsion. r r r r dt r dn r db t , n '  , b ds ds ds

8.

  

r& r& & r&12 rr&r& . r  r   k   rr 3 r .r       r&& r&& r&   r r r&   r  r r r& & & & & r r . r  r&

  

1.11. TERMINAL QUESTIONS :

1.

Explain the method of obtaining the parametric representation of a curve of intersection of two surfaces, whose equations are F1  x , y , z 0, F2 x , y , z 0

2.

Show that the tangent at any point of the curve, whose equations are x =3t , y 3t 2 , z 2t 3 makes a constant angle with the line by y = z –x = 0

42

3.

Derive the equation of the osculating plane at a point on the curve r r r r  u

4.

r r Obtain the expression for curvature of the curve r r  s

5.

r r Derive the Serret – Frenet formulae for the space curve r r  s

1.12. FURTHER READINGS:

1.

‘ Differential Geometry ‘ by D. Somasundaram, Narosa Chennai, 2005.

2.

‘Elementary Topics in Differential Geometry ’ by J.A.Thorpe, Springs – Verlag, New York , 1979.

43

Publishing House,

UNIT II CURVES ON SURFACE STRUCTURE: Learning Objectives 2.0.

Introduction

2.1.

Contact between curves and surfaces

2.2.

Osculating circle

2.3.

Osculating sphere

2.4.

Locus of centre of spherical curvature

2.5.

Tangent surfaces, Involutes and evolutes

2.6.

Intrinsic equation of space curves

2.7.

Fundamental Existence Theorem for space curves

2.8.

Summary

2.9.

Key words

2.10. Answer to Check Your Progress 2.11. Terminal Questions. 2.12. Further Readings LEARNING OBJECTIVES : After going through this unit, you should be able to  

Define n - point contact between curves and surfaces, osculating circle, osculating sphere, evolutes involutes, spherical indicatrix.

 

Derive the centre and radius of circle of curvature of a curve and that of osculating sphere, equation of evolute from that of involute, curvature and torsion for the evolute.

2.0. INTRODUCTION: We know that a single equation f(x,y)=0 (or) y  x in the plane represents a curve. Similarly, in space, a single equation g(x,y,z)=0 or z x, y represents a surface. We can also have a surface specified by parametric equations, where however we need two parameters, say u and  unlike for a curve, where we need only one parameter. The parametric equations for a surface would then be of the form x = f(u,), y = g(u,), z =h(u,) .To see that this represent a surface, we take a 44

separate plane which we shall call the u-v plane (like x-y plane). We consider a region R in the u- plane. For each point P  u,vin the region R, we form the three numbers f(u,v) , g(u,v) and h(u,v) and plot the point Q(x,y,z) in space with x=f(u,v), y=g(u,v), z =h(u,v) .Thus, we obtain one point in space corresponding to each point of the region R. These points would together form a surface.

2.1. CONTACT BETWEEN CURVES AND SURFACES: We know that tangent passes through at least two consecutive points of a curve. In this case, we say that a tangent has a contact (at least) of order one with the curve. Instead of stating in common, we can also state that they have contact of certain order.

r

r

Let r r (u ) be a curve C and S be a surface given by the equation F(x,y,z)=o . Then the points of intersection of the curve and the surface are given by the parametric values of u, which are the roots of the equation F(x(u), y(u), z(u))= 0 i.e F(u)=0 .If u o is such a root of F(u)=0, then F(u) can be expanded by Taylor’s theorem about u uo in the form

(u u ) 2 (u u )3 0 0 F F(u) = F(u ) + (u- u ) F  (u )  F (u )   ( u ) ........... 0 0 0 0 0 2! 3!

2  3   F (u )  F  (u ) ....... 0 3! 0 2!

= F  (u ) 

0

Q F (u 0 and putting u u 0 0) Now, if F  u0 0 , then u0 is a simple zero of F(u) ,and C and S have simple intersection

r

 at the point r (u0 ) . If F  (u ) 0 , but F  u0 0 , then u 0 is a double zero of F  u ,

0

and

we

r r (u ). If 0

say

that

C

and

S

have

two point contact at   F (u ) F  (u0 ) o and F  (u ) 0 then u0 is triple zero of F  u ,

0

0

a

r

and we say that C and S have three point contact at the point r (u )

0

1 (u ) 0 but F(u )= F  (u ) F  (u ) ........F n 0 0 0 0 r F n (u ) 0 then C and S are said to have n-point contact of r (u ). 0 0 In

general,

if

Example: Show that the osculating plane at P has, in general three point contact with the curve at P.

45

Measuring the length from P, let Q be a neighbouring point, where the small arc PQ=S . r then r(s) can be separated by Taylor’s series as

r r r s2 r s3 r    r (s) r (0) r (0)s  r (0)  r (0) .............. 2! 3! (or)

r r r s2 r s3   r (s ) r (0) sr  (0)  r  (0)  r  (0) Neglecting higher power of s. 2! 3! From the equation of osculating plane, we have

r r r r  F (s ) [r (s ) r (0), r  (0), r  (0)] r r s2 s3     r (0)  r (0), r  (0), r  (0)] 2! 3!

r

=[ sr  (0) 

r r r s2 r  r  r  s3 r  r  r  s[r (0), r  (0), r  (0)]  [r  (0), r (0), r  (0)] [r  (0), r (0), r (0)] 2 3 3

r r s r    = 0 0  [r  (0), r  (0), r  (0)] 6

1  s 3k 2, where k and are the curvature and torsion at the point P. Thus, we have 6 s3      F (0) F (0) 0but F (0) coefficient of k 20 , provided k and  do 3! not vanish at P. Hence, in general, the osculating plane at P has three point contact with the curve at P. 1.Check your Progress ( C Y P ) : r r What is three – Point contact’ of a curve r r (u ) and the surface F(x,y,z)=o , where x=x(u), y=y(u), z=z(u)?.

________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

46

2.2. OSCULATING CIRCLE: We know that the osculating plane of P on the curves has three point contact with the curve at P. If P,Q,R are three points on a curve, the circle through P,Q,R in its limiting position, where Q,R tend to P is called the circle of curvature at P and radius of circle is the radius of curvature and is denoted by . Definition: The circle, which has three point contacts with the curve at P is called the osculating circle at a point P on a curve. It obviously lies in the osculating plane at P. Infact, such a circle is the intersection of the osculating plane with a sphere through which also has a three point contact with the curve at P. To find the centre and radius of circle of curvature at P on a curve: Let a circle in the osculating plane be given as the intersection of the sphere

r r r (r c )2 2 with the plane where r is the position vector of any point P on the r sphere. c and  are the position vectors of the centre C and P is radius of the sphere. Sphere:

Let the centre C of the circle is at a distance  from P. clearly C lies on the principal normal at P. uur r uur r uur r Pc n, op r , oc c uuur uur uuur r r r r r r From the figure po oc pc r c n (or ) c r n The osculating circle is then the intersection of the osculating plane with the r r 2 sphere (c r ) 2

r

r

The points of intersection of this sphere with the curve r r (s ) are given by r r 2 r F (s ) (c r ) 2 0 , where r is a function of parameter s, the arc length of the curve. The condition for three point contact is:

47

 F (s ) 0, F  (s ) 0, F  (s ) 0 r r r F  ( s) 0  2(c r )  ( r ) 0 r r r  (c r )  t 0 r r r r r F  ( s) 0  (c r )  t t .(r  ) 0 r r r r r  (c r )  t t  t 0 r r r  ( c r )  kn 1  (1) r r r r r r r r But c r n  (c r )  n (n  n) r r r ( c r )  n   (2) Comparing (1) and (2) ,

1 k

1 Thus the radius of the osculating circle at P is  ,  is called radius of k curvature of the curve at P.C is called center of curvature of the curve at P and is given by

r r r c r n.

2.Check you Progress ( C Y P ) : r r r 1. What does the result c r n give? _____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ 2.3. OSCULATING SPHERE: Definition: The osculating sphere, at P on the curve is defined as the limiting position of the sphere through the points P,Q,R,S on the curve as Q, R, S tend to P (Or) The osculating sphere at P on the curve is defined to be the sphere, which has four – point contact with the curve at P. It is also called spherical curvature at P. To find the centre and radius of osculating sphere: r It c is the position vector of the centre C and R is the radius of the osculating

r

r

sphere, then its equation is ( C R)2 R2 ,

r

where R is the position vector of any

r

r

point on the sphere. The points of intersection of the sphere with the curve r r (s ) are

r r

given by F(s) ( c r )2 R2 0 . The condition for four point contacts are

48

   F (s ) F  (s ) F  (s ) F  (s ) 0 r r r F  (s ) 0  (c r )( t )0 r r r  ( c r )  t 0 r r r r r r r r rr F  ( s) 0  (c r )  kn t (r ) 0   c r  .kn t .t 0 r r r  ( c r )  kn 1 r r r r r r r F   (s ) 0  (c r )  k n k kt b  ( t ) kn 0     r r r 2 r r r r r r  k (c r )  n k ( cr).t k( c r ).b 0 r r r r r r 1  k  k(c r )  b 0, sin ce( c r )  t 0 k r r r k   (c r )  b   k 2 1 1 1 Where  and  Hence   k  k k2





r r r

Thus, we have ( c r )  t 0

From the fig

r r r 1 (c r )  n   k r r r & (c r )  b   uuuur uuuur uuuur PO OC PC r r uuuur r c PC

r r c r lies in the normal plane at P uuuur uuur uuur But PC PS SC (s, being centre of curvature of curve at P) r r n  b

r r r r Thus c r n  b r r r r c r  b n

This gives the centre of osculating sphere called the centre of spherical curvature. We r uuuur r r denote R PC n  b The radius R of the spherical curvature or osculating sphere and the center is given by

r r r R R n  b

R2 2 (  )2

If k is constant, then  =0. Hence R= . Thus, when k is constant, the centre of curvature or centre of * osculating circle and the centre of spherical curvature coincide. 49

Now, the osculating sphere has a contact of order three with the curve. Its intersection with the osculating plane is the osculating circle. Its centre lies on the normal plane on a line parallel to the binomial. 2.4. LOCUS OF THE CENTRE OF SPHERICAL CURVATURE: As P moves along a curve, the corresponding centre of spherical curvature moves along another curve C1, whose curvature and torsion have a simple relation to those of C. Using suffix 1 for quantities pertaining to C1, we note that its length of arc must be r r r r represented by s1 and r1 r n  b  1 Differentiate (1) w.r.t.s,

r r r dr dr ds r r r 1  1  1 t  n ( kt  b) ds ds ds 1 r r (    )b  ( n) r r r r r r r   t st  n t  b (  )b  n 11   r     b   

0 . From the above equation, we Clearly s 1 should increase as s increases and hence s  find that

1 r      se     (2) Where e 1, for the vectors t and  1  1 

r b are of unit length and hence their coefficients must be numerically equal. We then find r r r dt r db t eb, which on differentiation gives 1 e 1 ds ds r dt ds r  1  1 en ds ds 1 r r k n sen 1 11

r r n1 n 1 k se

Thus, we may write

when

e 1 1

 (3) 1 1 1 r r r r r r r r Again b1 t n eb e n(Q t eband n) 1 1 1 1 1 r ee t 1

Then we have

50

r r r

Thus, we find the unit vectors are t1 , n1 , b1 are respectively parallel to the unit

r r r

vectors b , n , t . Again

r r r r db1 dt b ee t  ee 1 1 1 ds ds

r db ds r i.e., 1  1 ee kn 1 ds ds 1

r r 1 11 1 r r r r  ee kn(Q n1 1n) i.e.  1e1 n1 s1  1 i.e. so that  sek (4) 1 1 From (3) and (4) we find that   e kk thus if e1 1, 1 kk1 i.e. product of the 1 1 1 i.e.  n s ee kn

torsions at corresponding points is equal to the product of curvatures. Example: If the radius of spherical curvature is constant, prove that the curve either lies on a sphere or has constant curvature.

2

2

2

) . The radius of spherical curvature is given by R  (  constant, differentiating w.r.t. s,  0= 2   2(  ) d (  ) i.e.,2   d (  )   0 ds ds   Which give either

d  o or  (  ) 0 ds

 oConstant  k is constant

d ( ) 0  (1) ds r If c is the centre of spherical curvature then we know

Now suppose 

r r r r c r n  b , which on differentiation gives r r d r r r r dc r t  n ( kt b )  ( )b  ( n) ds ds r d r r r r r t  n k t  b   b  n  ds r r r r r  d br  t  n t  b   n Q k 1, 1   ds  d r     1  b 0by condition   ds   r r dc i.e, 0 , Then c is a constant vector ds 51

If R is

Hence, the centre of osculating sphere is independent of the point on the curve. i.e. the centre of the osculating sphere is a fixed point and its radius is also constant by data. Hence, the osculating sphere is a fixed sphere and the given curve lies on this sphere. Example: Show that the necessary and sufficient condition that a curve lies on a sphere is that

d 0 at every point on the curve.    ds Solution:

If the curve lies on a sphere, then the sphere will be the osculating sphere for every point on the curve, so that the radius R of the osculating sphere is constant.

2

But r 2 2    , which on differentiation gives 0 2 2   If

d   ds d  d 0 .  0,cancelling 2  , we get    (or )     ds  ds

Thus the condition is necessary. Conversely, let the condition be satisfied at every point on curve.

i.e

 d    0 is true.  ds

2  2   0 d ds

, Multiplying by 2 

2

Which on integration gives 2    constant i.e., R 2 constant. Then the radius of osculating sphere is constant at every point of the curve. r Also, if c is the centre of osculating sphere:

r r r r c r n  b , which on differentiation gives r r r r r r dc r d t  n  k t b  ( )b  n  ds ds r r r r r d r t  n t  b    b  n   ds  d r     1  b ,sin ce k 1 and   ds  





=0, by the given condition. r Thus c is a constant vector. i.e., the centre of the osculating sphere is a fixed point. Therefore the given curve must lie on a sphere. Hence, the condition is sufficient.

52

2.5. TANGENT SURFACES, INVOLUTES AND EVOLUTES:

If we take a circle in a plane the all its tangents together cover the entire part of the plane outside the circle. 111ly all tangents to a space curve generate a surface in space called the tangent surface of the curve.

Any point P on the tangent surface can be located by two quantities. First, we must locate the tangent on which it lies. If Q is the point of the contact of the tangent to the curve, then the tangent itself is determined by the parameters of the point Q. Next, on the tangent, the position of P is given by its algebraic distance u from Q. thus s and u r determine P. Hence, for each point P, its position vector R has the form r uuur uuuur uuur r r r r r R s, u r  s ur  s , sin ce OP OQ QP and OQ r  s and QP being 

r

tangential to the curve at Q is a multiple of t  s , s and u being the parameters of P. A surface, being two dimensional, a point can be specified only by a pair of numbers, just as points on the xy plane are determined by such a pair (x,y) . Any relation between s and u of the form f(u,s)=0 or u= ( s) , will determine a curve on the tangent surface, the parameters, pairs of whose points will satisty this relation. If the tangents to a curve C are normals to another curve C 1, then C1is called Involute of C and C is called an Evolute’ Of C1. In other words, an involute of a space curve C is a curve, which lies on the tangent surface of c and cuts the generators i.e., the tangents give generating surface orthogonally.

Since the involute is a curve lying on the tangent surface, there must be a relation between u and s and let this be u= ( s) , when ( s) is to be determined.

r

r

r

The position vector of a point on the involute is given by R r ( s) t . This is the

r

r

r

equation of the involute, with s as the parameter. R( s) r ( s) ( s)t ( s) . However 53

r dR s is not the arc length for this curve.  We can only say that is tangential to the ds involute, but is not the unit tangent vector.

r

r

r

Thus, differentiating on both sides of R( s) r ( s) ( s) t ( s). with respect to s we get

r r r dR r t  (s )t (s )t  ds r r  1 (s )t (s )kn r r dR Now by definition is perpendicular to t ’ ds r r r r r dR r   t  1  ( S ) t  t  s kn  t ds 0  1  (S ) 

 ( s) 1, which on integration w.r.t.s gives ( s) k s where k is a constant.

r

r

r

Hence, any involute is of the from R r ( k S )t Different choice of k Gives different involutes.

In the figure total length of the curve A B is k. The string is originally wound round the curve with its end points at A and B. Keeping the string in contact with the curve, the end point B is lifted away from the curve, so that the lifted part of the string is always taut. This is the Part PB . It is clearly tangential to the curve at P. If the arc length r uuur from A to P is s, then clearly PB PB k s and the position vector R OB  Bat uuur uuuur r r r r ( K S ) t ., when t is the unit tangent at P. The path this stage is clearly OP PB traced by B  is the involute, since its equation at any position B is uuuur r r r OB R r (k S )t . It  is an involute of a given curve C, then we say that C is an evolute of . 54

Check you Progress. 3. What is the tangent surface of a curve? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ 4. Define evolute and involute ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ To find the equation of the evolute C from the equation of the involute :

Let P be the point on C corresponding to the point Q on . By

definition the involute  is at

right angle to the tangent PQ(at P) to the curve C. Hence tangent at Q to the involute  is at right angle to the tangent PQ to the curve C. r r Thus, PQ lies in the normal plane to  at Q . If R and r denote the position r r uuur uuuur uuur r uuur r vectors of P and Q respectively, then R OP OQ QP r n b since QP r which lies in the normal plane to at Q can be written as the linear combination of n and r b clearly the coefficient  and change from point to point on  and therefore are function of s on .

r

r

r

r

Differentiating both sides of R r n b with respect to s, we obtain.

r r r r r r dR r t  n ( kt b ) b (  n ) ds r r r (1 k)t ( ) n ( )b

r r dR dR Now, is a tangent vector at P to the curve C. Note that is not unit tangent ds ds

vector on C, since the arc length is on but not on C. 55

r r r dR  is in the normal plane to  at Q and hence must be parallel to n b ds    Hence 1k 0 and    1  and 2 2   k 2 2   d       2 2 ds    d   ds   d  1    tan   2 ds     1     ds c tan1  where C is constant. which on integration gives   Putting  ds , we obtain tan(c) or co t ( c), since .









Hence the equation of evolute of becomes

r r r r R r n cot (c )b , where  ds

2.6. INTRINSIC EQUATION OF SPACE CURVES: So far, we have been considering the vector equation of a curve using external objects, like a coordinate system. It is possible to specify a curve without using such external means. This is by the method of intrinsic equations. These are two scalar equations, which give the curvature and torsion at any point on a curve in terms of the arc length parameter measured from some fixed point on it. Such equations are there fore of the form k=f(s) and =g(s). than the two curves C and Now the question arises: can the shape and dimensions of a curve be completely given by these two equations?. In other words, can there be two different curves, which arc not congruent, but having the same intrinsic equation? We shall show that two curves with the same intrinsic equations are necessarily congruent. A question then arises: Can we construct a curve with given values for curvature and torsion as functions of s. In general, we may not be able to have such curves, unless some suitable conditions are satisfied by f(s) and g(s). These are the contents of the following the Orem.

56

2.7. FUNDAMENTAL EXISTENCE THEOREM FOR SPACE CURVES: If f(s) and g(s) are continuous functions of s for all non-negative values of s, then there exists one and only one space curve, determined but for its position in space for which k=f(s) and g (s ) , being arc length measured from some base point. Proof:

we shall now actually show that there are four vector functions

r r r r r r r r r r r r r ( s), t t ( s), n n( s) andb b ( s) such that t , n ,b are mutually perpendicular

vectors satisfying the serret – Frenet formulae involving the given functions f(s) and g(s) r r as k and . r r ( s ) must therefore be the required curve. Consider the system of differential equations

dk , d kr dr   ds ds ds Where  , , are unknown functions of s and k and are the given function f(s)

and g(s). From the theory of differential equations, it is known that the above system has a unique solution ( s),( s),( s) which takes prescribed values at s=0. In particular, there

is

a

unique

1  s , s , s 1  1 

solution

for

which

1(0) 0, 1(0) o,  (0) 0 . Similarly, we have anther set of solutions 1  ( s),  ( s),  ( s) taking initial vales (i.e. at s=0 ) 0,1,0 respectively and another set 2 2 2  ( s),  ( s),  ( s) with initial values 0,0,1 respectively. 2 2 2 2 We first show that 12 12  1 1  now

d ds

(1

2

d

d 1 2 1 ds ds 2(k ) 2(k ) 2(   )

d1  2  1  1 ) 2 1 1 ds 2

2

0 Thus

1

1

1

1

1

1

1

1

12 12 12 = constant, which must be equal to 1 considering the initial values

similarly, we have 22 22 r22 1 and 23 32 r32 1 d d2 d1 d 2 d 1 Again consider r1r2 1 2 1 2   1 2  1 2  ds ds ds ds ds d2 d  2 1 1 ds ds 1  k 2 2  k 1  k 2   2  k1      2    1  1 2 1 1 2 

0 Hence,  0 0  1 0  0 0 1 2  1 2 r1 r2 constant=0, given by the initial values 1 Similarly, we have   r2 r3 0 &  r3 r1 0 2 3  2 3  3 1  3 1  The above six relation which we have proved show that the matrix 57

 1  1 1  A  2 2 2  is orthogonal.    3 3 3 This means that AAI , where A is the transpose of A . From this we 2 2 2 have AA I which implies 1 2 3 1,

12 22 32 1, 2 2  32 1 and 1  2        0,     ,   0

1 1

2 2

3 3

11

2 2

33 1 1

2 2

3 3

Thus, the three mutually perpendicular unit vectors

r r r t ( , , ), n (,  ,  ),and b (,  , ) are defined for each value of s 1 2 3 1 2 3 1 2 3 s r r r r sr Now putting r  tds (s)i  (s ) j  (s )b ds 1 2 3 0 0 r





r d s r d s dr r d s i 1 ( s) ds j  2 ( s) ds k  3 ( s) ds  ds ds 0 ds o ds o r r r r i 1 ( s) j2 ( s) k 3 ( s) t r r r Since, n and b together with t satisfy Sernet – Frenet formulae (which are the given differential equations) with given functions as curvature and torsion, it follows that r r r r (s ) is the required curve, with s as its arc length . This proves the existence.

We have

To prove the uniqueness, we shall show that if the curves have the same intrinsic equations then they are congruent. Let C and C 1 be two such curves. Let the point s=0 on both the curves be A and A1 respectively. For any are length s, let the corresponding points be P and P1. Denoting the corresponding unit triads for the two curves C and C1 by r r r r r r r r r t , n, b and t1 , n1 , b1 . Let us bring A1 to coincide with A and t1 , n1 , b1 at A 1to coincide with r r r t , n, b at A by suitably turning C1above A= A1 . Now consider.

r r r r r r r dt1 dt r r dn1 dn r r db1 db r d r r r r r r (t  t1 n  n1 b  b1 ) t    t1 n    n1 b    b1 ds ds ds ds ds ds ds r r r r r r r r r r r r r r t ( kn1 ) ( kn)  t1 n  (kt1 b1 )  ( kt b )  n1 b .(n1 ) (  n )  b1 r r r r r r Thus t  t1 n  n1 b  b1 =constant =3(using initial values it takes at s=0)

Since, this is the sum of three cosines and 1 cos1, we see that the only possible values r r r r r r for the three angles between n and n1 been n and n1 and between b and b1 , are all zero so that 58

r r r r r r r r cosines are equal to 1. Hence, t t1 , n n1 and b b1 . Again t t1 r r dr1 dr d r r r r r gives  i.e, ( r1 r ) 0 i.e, r1 r a , a constant vector and its value when s=0 is ds ds ds r r zero. Hence r1 r for all s. Thus, the two curves C and C1 coincide. This establishes the uniqueness. their

Helices: A cylindrical helix is a space curve which lies on a cylinder (whose cross–section may be a circle or any closed curve). And cuts the generators of the cylinder at a constant angle. The axis of the helix is a fixed line which is parallel to the generators of the cylinder. The characteristic property of helices i.e., a property possessed by helices and not by other curves, is the constancy of the ratio of curvature to the torsion. r For, let a be the unit vector, parallel to axis. r r Then t  a cos  r r r r Differentiating this, we have t   a t a 0 r r r r i.e. kn  a 0sin ce a  0 r r r Hence, a n and therefore lies in the rectifying plane. Thus, a is a linear combination r r r r r r r of t and b , since t  a cos ,if follows that a t cos b sin . Differentiating w.r.t.s , we obtain r r dt db 0  cos sin  ds ds r r 0 kn cosn sin  r 0  k cossin n

 sin   tan constant  cos   Conversely, if constant=tan, say (since for any number , we can find , such  r that tan )we have cos sin  or  cos sin n 0 r d r i.e. t cos b sin  0 ds r r r t cos b sin constant unit vector a r r Thus a  t cos and the curve is therefore a helix. We have already considered the properties of circular helix x a cosu, y a sin u, z bu  a 0  which is helix lying on a circular cylinder. Here, the  axis of the helix is such that and are constants and thus, in particular the ratio is a  constant. Thus





Let C be any helix (circular or not). We project C on a plane through O perpendicular to the axis of the cylinder. This is clearly normal cross-section of the cylinder, which is a plane curve C1 (say).

59

Using suffix 1 for related to C1 , we have uur uuur r uuur uuur r r r OP op op1 p1 p r1 p1 p . But p1 p r is parallel to a and has length op cos  , being the angle between oP and pp1 . r r r r r r r Thus P1 P a  r. Thus r r1  a r a. Differentiating w.r.t.s , we have r r r dr dr1 r r da r d r r   a r  a  a r ds ds ds ds r r r r dr r dr1 dr1 ds1 r ds1 da Now, t ,   t1 , 0 and ds ds ds1 ds ds ds r r d r r da r r dr r r t cos a r  r a  0 a  ds ds ds r r ds r Thus, t t1 1  cos a . ds r r r ds1 r r Taking dot product with t1 , t  t 1   cos  t1  a. ds r r r r r r sin ce, t1 and a are mutually perpendicular, and a  t cos , we have t  t1 sin  . r r ds r Thus 1 sin . Hence t  cos a  sin  t1 . ds r r dt dt1 ds1  sin   ds ds1 ds r r 2 n sin 1 n1 r r Since n and n1 are unit vectors, we have 1 sin 2 . Also , we see that normal to the

plane curve C1 at P1 is parallel to the principal normal of the helix C at P . If the helix C has constant curvature k, then k1 is also a constant. Thus C1 is a plane curve with constant curvature and is therefore a circle. Hence, we conclude that if a helix has constant curvature, it must be a circular helix. Check you progress: 5. What is the characteristic property of the helix? ________________________________________________________________________ ________________________________________________________________________ 1. Show that the involutes of a circular helix are plane curve: Solution: The parametric equation of circular r r  a cos u , a sin u , bu _____  1 Changing the parameter u to s , 60

helix

is

u u r s r du  a 2 b 2 du cu o

o

s cu , where c  a2 b2 r Equation (1)  r  a cos s c , a sin s c , b s c  r r r The equation of involute is R r  k s  t

r r a  t r  sin s c , a c cos s c , b c  c  r r r  R r  k s  t a   a cos s c , a sin s c , b s c   k s   sin s c , a c cos s c , b c  c  a a b   a cos s c  k s  sin s c , a sin s c  k s  cos s c , K  c c c  kb  z coordinate of any point  z  constant c The involute lies on the plane 2. Find the intrinsic equation of a spherical helix. Solution: For all curves lying on a sphere, the centre of the osculating sphere coincides with the centre of the sphere itself. r r r r r r We know R n  b R s r





R 2 2 2 1 is true for any curve lying on a sphere. k For helix, constant tan   2

 tan  12

R   tan  2

2

2

2

d R 2 2 tan  tan  ds d  cot ds  2 2 R 

cot  s  R 2 2 (choosing constant of integration as zero) Integrate w.r.t.s ,   cot 2 s2 R2 2

R2 2 s2 cot 2 

This is the intrinsic equation of spherical helix.

61

Check Your Progress (C Y P ) : 6. State Fundamental Existence Theorem for space curves. ________________________________________________________________________ ________________________________________________________________________ 7. What about the helix of constant curvature ? ________________________________________________________________________ ________________________________________________________________________ 2.8. SUMMARY: In this unit first, the contact of certain order between a given curve and given surface is explained. Then the centre and radius of circle of curvature at a point P on a curve is derived. Further the centre and radius of osculating sphere is also derived. Locus of the centre of osculating sphere is obtained. The equations of involute and evolute are derived. Fundamental existence theorem for space curves is proved. Finally, the characteristic property viz; ‘the ratio of curvature to torsion is constant’ is obtained. 2.9. KEYWORDS:

r r r r  u and the surface given by the equation

1. n-Point contact of a curve :

F x, y, z 0 . Then the points of intersection of the curve and the surface are given by the parametric values of u, which are the roots of F x  u y  u z u 0 i.e., F  u 0 . If u0 is such a root of F(u)=0 then F(u) can be expanded by Taylor’s theorem about

u u0 F  u u0 F  u F  u 0  u  u 0 F  u0    u0 ........ IfF(u.)=0=F’(u.)= 2

n 1 ( u0 )

2! but F u0 0 , then C and S are said to have n – point

F”(u.)=……. F r contact at r  u0  . 2. Osculating Circle: The circle, which has three point contact with the curve at P is called osculating circle at a point P on a curve. Such a circle is the intersection of the osculating plane with a sphere through which also has a three point contact with the curve at p. 3. Osculating Sphere (or) Spherical Curvature: The osculating sphere at P on the curve is defined to be the sphere, which has four – point contact with the curve at P. 4. Involutes and evolutes: If the tangents to a curve C are normals to another curve C1 then C1 is called the involute of C and C is called an evolute of C1 . n

62

5. Helix: A cylindrical helix is a space curve which lies on a cylinder ( whose cross section may be a circle or any closed curve) and cuts the generators of the cylinder at a constant angle. 2.10. ANSWER TO CHECK YOUR PROGRESS: 1.

r r .let r r  u and F x, y, z 0 be the given curve and given surface. The points of intersection of the curve and the surface are given by the equation F x  u , y  u , z  u 0 i.e, F  u 0 . If uo is such a root of F  u 0, then F  u can be expanded about u u o in the form

u uo F  F  u F  u o  u  u o F u o  uo  L L  2

2!

   F  uo   F u o  F  uo L L   2! 3! Q F uo 0 and putting u uo 2

3

Now, if F  uo 0 , then uo is a simple zero of F  u and the curve & surface have r   simple intersection at the point r  uo . If F  u o F   u o 0 and F  u o 0 ,

2.

3. 4. 5. 6.

7.

then uo is a triple zero of F  u and we say that the curve and the surface have a r u o . three point contact at the point r  r r r c r n , gives the centre of curvature of the curve at P, where is called 1 radius of curvature of the curve at P and  , k being the curvature of the k r r curve at P. r is the position vector of P and r is the principal normal unit vector. All tangents to a space curve generate a surface in space called the tangent surface of the curve. If the tangent to a curve C are normals to another curve C1 , then C1 is called involute of C, and C is called evaluate of C1 . The characteristic property possessed by helices and not by other curves, is the constancy of the ratio of curvature to the torsion. If f  s and g  s are continuous functions of S, for all non – negative values of s, then there exists one and only one space curve, determined but for its position in space for which k f  s and g  s , being arc length measured from some base point Let C be any helix (circular or not ). We project C on a plane through O perpendicular to the axis of the cylinder. This is clearly normal cross – section of the cylinder, which is a plane curve c1 (say). If the helix C has constant curvature k, then k1 is also a constant. Thus C1 is a plane curve with constant curvature and is therefore a circle. Hence, we conclude that, if a helix has constant curvature, it must be a circular helix. 63

2.11. TERMINAL QUESTIONS:

1. Find the locus of centre of curvature and show that the unit tangent at C, the centre of curvature, is parallel to the normal plane to the original curve at P.

2. Find the curvature and torsion of the locus of the centre of curvature, when the curvature of the original curve is given (i.e., is constant).

3. Show that the locus of the centre of curvature is an evolutes, only when the curve is plane.

2.12. FURTHER READINGS : 1. ‘Differential Geometry’ by struik

2. ‘Differential Geometry’ by Dr. Mittal & Agarwal

3. ‘Differential Geometry’ by D. Somasundaram

4. ‘An introduction to Differential Geometry’ by T. Willmore.

64

UNIT- III LOCAL INTRINSIC PROPERTIES OF SURFACE STRUCTURE: Learning Objectives 3.0.

Introduction

3.1.

Surface representation, regular and singular points

3.2.

Change of Parameters

3.3.

Curvilinear equations of the curve on the surface

3.4.

Tangent plane and normal

3.5.

Surfaces of Revolution

3.6.

Metric on a surface – the first fundamental form

3.7.

Elemental Area

3.8.

Direction coefficients on a surface

3.9.

Summary

3.10. Key words 3.11. Answers to check up your progress 3.12. Terminal Questions 3.13. Further Readings LEARNING OBJECTIVES: 1. Two parameter family of surface – regular and singular points on the surface and implicit and constraint equation. 2. Change of parameters in the equation of a surface by means of transformation. 3. Curvilinear equations of the curve on the surface. 4. Tangent plane and normal 5. Surfaces of revolution. 6. Metric on a surface – First Fundamental from 7. Element of Area 8. Direction coefficients on a surface. 65

3.0. INTRODUCTION: Any property or formula of a surface which can be deduced from the

r



metric of the surface alone without knowing the vector function r u, v (i.e., without knowing the equation of the surface) is called an intrinsic property In other words all properties of a surface which can be obtained using the metric Edu 2 2Fdudv Gdv 2 alone and without r r equation r r  u, v are called intrinsic properties.

using

the

vector

3.1.. SURFACE REPRESENTATION, REGULAR AND SINGULAR POINTS : We have defined the curve as a locus of a point, whose Cartesian coordinates x, y, z are functions of a single parameter. Similarly, a surface is the locus of a point, whose Cartesian coordinates  x, y, z  are functions of two independent parameters (or) vectorially, a surface is defined as the locus of a r point, whose position vector r can be expressed in terms of two parameters, in r r the form r r  u, v ( u and v being the two independent parameters) for

u u u , & vo v v . o 1 1 The representations of the surface as given above are due to Gauss and therefore named as Gaussian form of the surface. The parameters u and v are called curvilinear coordinates or surface coordinates of the current point on the surface. A point determined by the pair u, v is called as the point  u, v. A

r

r

r

point  u, v is said to be regular point on the surface, if the condition ru rv  o is satisfied at this point. Otherwise, the point is said to be singular. The relation of the form F  x , y , z o between the coordinates x, y, z of a point is called an

implicit or constraint equation of the surface. The parameters u, v can be eliminated between the parametric equations

x f u, v, y g u, v , z h  u, v and F x , y , z 0 can be obtained.

thus

the

constraint

equation

Check you progress ( C Y P ) : r r 1. What is regular point and singular point on a surface r r  u, v? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

66

Example: 1) Sphere:

consider a sphere with radius ‘a’. Taking the rectangular Cartesian coordinate axes through the centre of the sphere, the coordinates of any point on the sphere can be given in the following form:

x a sin cos o , y a sin sin o , z a cos o , o o2 Here

a,, ois

a point, whose

coordinates are called spherical polar coordinates of that point on the surface of the sphere ( and obeing the parameters)

r r a sin cos o , a sin sin o , a cos  r r  a coscos o , a cossin o , a sin  r ra sin sino , a sinco s o , 0

r r r i j k r r rro a cos cos o a cossin o a sin   a sin sin o a sin cos o o





a 2 sin 2 cos o , a 2 sin 2 sin o , a 2 sin cos  r r r rro o, When o &   r  o, At all other points.

All other points except the poles o and o are regular points and those

two poles are singular points. It is evident that the singularity of two poles in this case is due to the choice of the coordinate system and that is not an intrinsic property of the two points.

67

e.g. 2) Right circular cane:

Consider a right circular cone of semivertical angle . Taking o x y z with

oz axis along the axis of the cone and O at the vertex. The Cartesian

coordinates of any point on the cone is

x u cos , y u sin , z u cot 

Here u,, z are called cylindrical coordinates. The parameters are u and .

r r  u cos , u sin , u cot  r ru  cos , sin , cot  r r  u sin , u cos , o  r r ru r  u cos cot , u sin cot , u  r r r ru r 0 , iff u = 0 so that the only singular point of the cone is the vertex. 3.2. CHANGE OF PARAMETERS: We shall allow only those transformations, which transforms regular r r points into regular points. Let r r (u, v) be the equation to a surface in terms of the parameter  u, v and let a new set of parameters U,V be introduced by means of the transformation u u  U , V & v v  U ,V 

r r r r r r u r  v Then rU       U  u  U  v  U r

r

r

r  r r  u  r  v rv     

 u   v  68

  1

r r r r r r r  u r  v   r u  r  v rU rV          u  U v U   u   v v   r r u v r r  u  v  ru rv   ru rv   U V  V U

r r u  ru rv U  v U

u  V  v  V

r r u, v  ru rv    2  U ,V  r

r

Now, if rU rV 0 , then the Jacobian

u, v 0   3  U ,V 

r r (Q ru rv 0 for regular point in parameter u ,v ) The transformation (1) under the condition (3) is called proper transformation.

 u, v  0, at a point P of the surface, then P will be regular.  U ,V   u, v  Also, denote that for proper transformation, 0 and hence from (2),  U ,V   If

r r r r rU rV 0 , iff ru rv 0 i.e., In proper transformation, regular point is transformed into regular point.  P will be regular or singular in the first parametric system according as P is regular (or) singular in the second parametric system. Check you progress ( C Y P ) : 2. What is proper transformation under the change of variable and what is its effect? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

69

3.3. CURVILINEAR EQUATIONS OF THE CURVE ON THE SURFACE:

r

We know that a curve is the locus of a point whose position vector r can be expressed as a function of a single parameter. Let us consider a surface r r r r  u, v defined on a domain D and if u and v are functions of a single

r

parameter t , then the position vector r becomes a function of a single parameter r r t , and hence its locus is a curve lying on the surface r r  u, v  Let

r r u u  t , v v  t , then r r u  t , v  t  is a curve lying on the surface r r r r  u, v in D . The equations u u  t , v v  t are called curvilinear

equations of the curve on the surface. Parametric curves:

r

r

r

When we keep v constant in r r  u ,v , then r depends on only one parameter u and thus determines a curve on the surface, a parametric curve v constant. Similarly u constant represents another parametric curve. As the constants vary, the surface is covered with a net of parametric curves, two of which pass through every point P , forming the family of curves v constant and the family of curves u constant.

r

r

At P , the vector ru is tangent to the curve v constant, and rv is tangent

at that point to the curve u constant.

70

r

r

If v constant (say) c , then as u varies, the point r r  u, c describes a

parametric curve called the u curve or the parametric curve v c . Similarly, if r r u  constant, say c , then v varies, the point r r  c, v traces a parametric curve called the v curve or the parametric curve u c . Thus, we have two systems of parametric curves viz., u curve and v curve and since we know r r that r r 0 therefore the parametric curves of different systems cannot touch

1

2

each other.

r r 1 2

If r  r 0 at a point P , the two parametric curves through the point P are orthogonal. If this condition is satisfied at every point, i.e., for all values of u and v in the domain D , the two systems of parametric curves are orthogonal. e.g., If and  are the parameters of the equation of a surface of a sphere, then

constant are small circles and the curves  constant are great circles. Check you progress ( C Y P ) : 3. Explain briefly the parametric curves on a surface. ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ 3.4. TANGENT PLANE AND NORMAL: Tangent plane: Let the equation of the curve be u u  t , v v  t , then the tangent is

r r r r& r& dr  r du r dv parallel to the vector r , where r      dt  u dt  v dt r r r r r r dr du dv i.e. r r (or) dr r du r dv 1 2 1 2 dtr dt rdt But r and r are non-zero and independent vectors, therefore the tangents to a 1 2 curve (on the surface) through a point P lie in the plane which contains the two r r r r vectors r and r . This plane is called the tangent plane at P . Hence, r r gives 1 2 1 2 the direction of the normal to the tangent plane. Thus, the equation to the

r

r1 r2 o , where rr is the position vector of P and R , r

tangent plane is  R r r r

that of a current point an the plane. Check you progress ( C Y P ) : 4. What is a tangent plane at a point P on a curve lying on a surface? ________________________________________________________________________ ________________________________________________________________________ 71

Normal: The normal to the surface at the point P is a line passing through P and r r perpendicular to the tangent plane at P . Thus, if r r  u, v, is the equation of the surface, then the normal to the tangent plane at P is perpendicular to r r r r both r and r and hence parallel to r r and therefore the equation of the normal

1

2

1 2 r r r r r line at P to the surface is R r r r , where R is the position vector of a 1 2





current point on the normal.

The normal to the surface at P is same as the normal to the tangent plane at P and

therefore

the

unit

surface

r r r r r r r r r r r by N  r1 r2 1 2 where r r o 1 2  r r 1 2

normal

r

vector N is

given

Note:

r

The sense of unit surface normal vector N is fixed by considering

r r r

that r1 , r2 , in this order form a right handed system. ________________________________________________________________________ To show that a proper parametric transformation either leavers every normal unchanged or reverses every normal.

r r r r u, v be the equation of a surface and the relations u o u ,v , v u ,v give a proper parametric transformation. From

Proof: Let

differential calculus, we have

r r r  r  r  u  r  v     &  u  u u  v u r r r  r  r u r  v     v uv vv

r r r r r r uv  u v  r  r          u  v  u  v  v  u  u  v r r u , v r r  Or   ___  1  u ,v 

72

r

r r

Where the symbols  , ;  , have usual meaning in the two systems of parameters u, v and u , v respectively. Since, the parametric transformation is proper, hence we have

 u ,v   u ,v  o i.e. J o where J   u, v  u, v  We know that and  are always positive. Therefore the relation (1) r r implies that the unit vectors and  are in the same directions, if J o and are in opposite directions if J< 0. But the Jacobian J is continuous over the domain D of parameters v, u and does not vanish and hence J is of invariable sign r r throughout D . Therefore, throughout D the vector and  are either the same vectors or are opposite vectors. Hence, a proper parametric transformation either leaves every normal unchanged or reverses every normal. 3.5. SURFACES OF REVOLUTION: (i) Sphere:

For the sphere with O and radius ' a ' we spherical coordinates parameters, since constant a . The vector equation

centre at take the oand as r remains is

then

r r r r  u, v  a sin u cos v, a sin u sin v, a cos u  Where u oand v and and v have ranges o, and o,2 . The

u

parametric curves v = constant are the meridians on the sphere and u constant are the parallels of latitudes, which are small circles, whose planes are parallel to the xy plane.

r r r Now r  a  cos u cos v, cos u sin v, sin u  1 u r r r & r  a  sin u sin v, sin u cos v, o  2 r r v And thus r  r o . Hence, the parametric curves are orthogonal. 1 2 73





r r 1 2 r r And r r a 2 sin u Thus the point u o and u are singularities. The 1 2 We see that r r a 2 sin 2 u cos v, sin 2 u sin v, sin u cos u

unit normal

r r r r r 1r 1 2  sin u cos v, sin u sin v, cos u  r  a





Check your progress ( C Y P ) : Show that parametric curves u constant & v constant on the sphere r r a sin u cos v, a sin u sin v, a cos u are orthogonal.

5.

________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ (ii) The general surface of revolution: Consider a curve in the xz plane, given by the parametric equations z f  u , x g  u . We wish to obtain the equation for the surface obtained by revolving this curve about the z -axis. Consider any intermediate position of this curve during the revolution. It will lie in a plane passing through the z -axis. This plane makes some angle, say v , with xz-plane. Thus the line of intersection of this plane with the xy plane makes the same angle v with x axis. If P is any point on the original curve on the xz-plane and P , the corresponding point in the new

1 position of the curve, then the distance of P from the z -axis =distance of P from 1 the z -axis x coordinate P x (say). The coordinates of P are then 1 x cos v, x sin v , z , since the z coordinate of P1 does not change during the revolution. Thus any point on the surface of revolution has coordinates g  u cos v, g  u sin v, f u  . Thus the parametric equation for this

r

surface of revolution is r g  u cos v, g u  sin v, f u  .

74

As in the case of the sphere, the curves v constant are the meridians which are just the various positions of the generating curve during its revolution and u constant are the parallels viz, the circles described by the various points P on the generating curve during the revolution. r r  r Here r   g u cos v, g  u sin v, f  u    1  u r r r and r2   g  u sin v , g  u cos v , o  v r r Thus r1  r2 0 , so that the parametric curves are again orthogonal. The right circular cone of semi-vertical angle  and axis along z  axis obtained by taking g  u u and f  u u cot  (iii) Anchor ring: This is the surface obtained by revolving a circle with centre at  b, o, o  and radius a b in the xz plane, about the z axis. The parametric equation for the circle in the xz plane are easily seen to be x b a cos u , z a sin u . Hence, the equation for r r  b a cos u cos v, b a cos u sin v, a sin u 

75

the

anchor

ring

is

r The meridians and parallels are all circles and the normal N is directed towards the centre of the meridian circle. Helicoids: If we revolve the +ve x axis about the z axis we just obtain xy plane. But while revolving the +ve x - axis, if we also give a parallel motion upwards in the +vez direction, then we obtain a surface which is called a right helicoid. It will resemble a winding staircase or a screw surface. A helicoid is a surface generated by screw motion of a curve i.e, a forward motion together with a rotation about a fixed line, called the axis of the helicoid. The angle v through which a curve has turned should be proportional to the distance moved in the forward direction parallel to the axis. Thus  is a v constant say ' a ' . The pitch of the helicoid is then 2a . It is the distance translated for one complete revolution. It is ve for a right handed screw motion, zero for a surface of revolution (since there is no forward motion) and negative for a left handed screw motion. We see that the parametric equation for the right helicoid (obtained by revolving the positive x axis about the z axis with a simultaneous traslation) r is r  u cos v, u sin v, av where u is the distance from the axis and v is the angle of rotation. The curves v constant are the generators and u constant are circular helices. These are respectively obtained by intersecting the helicoid by planes through the z axis making angle v with the xz-plane and right circular cylinders of radius u with axis along z axis. Clearly r r r r r1  cos v,sin v, o  and r2  u sin v, u cos v, a and thus r1  r2 0 , so that the parametric curves are orthogonal. The general helicoid with axis along z axis can be generated by the curve of intersection of the helicoid with any plane containing z axis and thus in particular the xz -plane. Thus, we may assume that the generating curve has the equation x g  u , y 0, z f  u . The surface therefore has the equation r r  g  u cos v, g  u sin v, f  u av  r Then r1  g u cos v, g  u sin v, f  u     r & r2  g  u sin v, g  u cos v, a  r r Thus r1  r2 af  u .Hence the parametric curves are orthogonal, if  either f   u 0  or  a 0 . In the first case f  u is a constant and hence, the surface is a right helicoid. In the second case, the surface is a surface of revolution.

76

3.6. METRIC ON A SURFACE- THE FIRST FUNDAMENTAL FORM:

r r Let r r  u , v be the equation of a surface. The quadratic differential form Edu 2 2Fdudv Gdv 2 in du , dv is called metric or first fundamental form, r r r r where E r12 , F r1  r2 , G r22 . The quantities E , F , G are called first order fundamental magnitudes on first fundamental coefficients and are of great importance. The values of E , F , G will generally vary from point to point on the surface as these quantities are function of u, v . Geometrical Interpretation of metric: r r Q r dr  r r Consider a curve u u  t , v v  t on the surface r r  u , v . r r r Let r and r dr corresponding to the parameter values u, v and u du, v dv respectively be the position vectors of two neighboring points P and Q on the surface. r r r r  r r We have dr  du  dv P  r u v r r r1 du r2 dv Let the arc PQ be ds. Since the points P and Q are neighbouring points, r therefore ds dr r ds dr  dt dt r 2 ds  dr   dt  dt 

2

2

r du r dv  r1 r2  dt   dt 2

2

r du  r r du dv  r 2 dv  r12  2 r1  r2   r2   dt  dt dt  dt  2

2

du  du  dv  dv   2F    G   dt  dt  dt  dt  r2 r r r where E r1 , F r1  r2 , G r22

Alternately we can write this 2 2 2 1 form: ds Edu 2 Fdudv Gdv _______ 

77

equation

in

the

This

gives the distance between the two neighboring points  u , v and

u du, v dvon the curve. The name metric is assigned to the first fundamental form as mainly it is used to calculate the arc length of the curves on the surface. Check you progress ( C Y P ) : r r 6. What is the meaning of a metric on the surface r r  u ,v ? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ Special cases: On the parametric curve u constant, du 0 and hence the metric (1) reduces to ds 2 Gdv 2 and on the parametric curve v constant, dv 0 and therefore the metric (1) reduces to ds 2 Edu 2 r r 2 r r r r 2 We have  r1 r2 r12 r22  r1  r2  EG F 0 r r Since r1 r2 0, E 0, G 0 Thus the metric (1) is a positive definite quadratic form. 2

Invariance of the metric: When the parameters are transformed by u  0 u, v  , v u, v  , the 2 2 metric takes the form ds 2 E du 2 F  du  dv G dv  r r  r du r2 dv 2 r r r r r  r u r v Where r1      u u  u  v  u r  u r  v r1 r2 u   u' r r r r  u r v & r2 r1 r2  v  v'  v

2

 r  u r  v r  u r  v  Thus ds  r1 r2 r1 r2  du  dv  u  u  v  v      2

2

r u  u  r  v  v    r1  du   dv du   dv  r2         v u  v     u  r r 2  r1 du r2 dv = Edu 2 2F du dv Gdv 2

78

Thus the metric is invariant under a parametric transformation, but the coefficients E, F, G are not invariant. 3.7. ELEMENT OF AREA:-

Consider the figure ABCD, whose vertices A, B, C, D have parameter values , u, v dv respectively. If du and dv are small u, v, u du, v , u du, v dv  r and positive, then the figure ABCD is approximately a parallelogram. Let r denote the position vector of any point on the figure, then uuur AD = position vector of D position vector of A r r = r u, v dvr  u, v r r  r r =r u, v   dv r  u, v v r r r = dv r2dv v uuur r similarly AB r1 du Hence, the area ds of’ the parallelogram ABCD is given r r by ds r1du r2 dv r r r1 r2 du dv H du dv Where H  EG F Hence, the element of area on the surface at the point  u, v is taken to be 2

EG F du dv 2

Check you progress ( C Y P ) : r r 7. Find element of area on the surface r r  u, v, _______________________________________________________________________ ________________________________________________________________________

79

3.8. DIRECTION COEFFICIENTS ON A SURFACE:

r r r At each point P , we have the vectors r1 , r2 , N which are linearly r independent. Thus any vector a drawn from P has the unique r r r r form a an N r1 r2 , where an ,  , are scalars, an is called the normal r r r r r component of a and is equal to a  . The vector r1 r2 is called the tangential r r r part of a and and , the tangential components of a . If a is tangential to the r surface, then an 0 and the quantities and determine a uniquely. Now, we consider such tangential vectors at points on the surface. r If a is the vector   , i.e. the tangential vector with components and , r r r then a r1 r2 r r r r r 2  a r1 r2   r1 r2 

 E 2 2 F G 2 _________ (1)

A direction in the tangent plane at P is usually described by the components of the unit vector in that direction. These components are called the direction coefficients of that direction. (OR) r If  l ,m  are the components of unit vector e in the direction of tangent r r r r vector T at P on the surface, such that e l r1 mr2 , then  l ,m  are termed as the r direction coefficients of the direction specified by T . r Thus if a has unit length, then the direction coefficients may be denoted by  l ,m  instead of   , and from (1), we have El 2 2 F l m Gm2 1 At each point on the surface, we defined the positive sense of rotation to r r be from r1 to r2 through the smaller angle. If  l , m and  l ,m  are the direction coefficients of two directions at the same point, then the angle  between these two directions is given by r r r r r r cos a  a l r1 mr2  l  r1 m r2 

El l  F  l m l  mGmm ____________(2) r r r r r r r and sin ce, N sin a a   l r1 mr2   l r1 m r2  r r  r1 r2  l ml  m r HN  l m l  m We have sin H  l m l  m , where H  EG F 2

80

Thus from( 2), we see that the condition for orthogonality of the two directions  l ,m  and  l , m is

El l  F  l m l  m Gmm 0 If   , are proportional to the direction coefficients  l , m , then   , are called the

direction

ratios

of

the

direction.

Thus,

 if   k , we l m

have

   1 El 2 Fl m Gm E  2 F   G   k  k k  k  2 2 2 Thus k E  F G  from which we get ,   l , m  k ,    2 E  2 F G 2 Check you progress ( C Y P ) : 8. If   , are direction ratios of a direction specified then find the direction 2

2

2

2

coefficients  l ,m .

________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ In terms of the direction ratios, the condition for the orthogonality of two directions   , and   ,  is The

E  F    G 0 r r vectors r1 and r2 have the respective

components  1,0and  0,1. The

corresponding direction coefficients are thus 1,0 0,1 And E 12 2 F  1  0 G 02 E 02 2F  0 1 G 12 1   1   , 0 And 0,  F   G r r Let u u  t , v v  t be a curve lying on the surface r r  u , v . Then the tangent r dr r r du dv vector is given by r2 , u&r2 v&, where u& , v& dt dt dt r r The direction ration of r&are therefore  u&& , v , r&s can be obtained from the metric. Hence, the unit tangent vector is r r dr dr ds r du r dv  / r1  r2  ds dt dt ds ds

81

du dv  And thus its direction coefficients are  , . Thus the increments du , dv along ds ds  the curve are the direction ratios for the tangent vector. If the curve is given by an implicit equation, 0  u , v 0 , then since d 00 .

 0 0  where 0 1  and 0 so that du : dv 0 2 : 0 1 . 2 u v   These are the direction ratios for the tangent. We have 0 1 du 0 2 dv 0

Problems: 1. calculate the fundamental x u cos v , y u sin v , z f  u

magnitudes

of

the

r Solution: the given surface is r  u cos v, u sin v, f  u  r r1  cos v,sin v, f  u   r r2  u sin v, u cos v, 0  r2 2 2 2 2 E r1 cos v sin v f   u  1 f   u  r r F r1  r2  u cos v sin v u sin v cos v 0 0 r2 2 2 2 2 2 G r2 u sin v u cos v 0 u . The fundamental magnitudes are 2 E 1f  , F 0, G u 2 2. Find the metric of the surface x u , y v, z u 2 v 2 r Solution: Let r  u , v, u 2 v 2  r r dr r1   1,0, 2u  du r r dr r2   0,1, 2 v  dv 2 2 r2 2 2 2  2 E r1  1 0  2u   1 4u 1 4u r r F r1  r2 0 0 4uv 4uv r G r2 1 4v 2





Hence H  EG F 2  14u2  1 4v2 4uv  2

 14u

2

1 4v 16u v 2

2 2

 1 4u 2 4v 2

82



conicoide

The quadratic differential form is ds 2 Edu 2 2Fdudv Gdv 2 Where E 14u 2 , F 4uv ,G 1 4v 2 3. Find the metric (or) first fundamental form of the surface, whose equation is x u cosh v, y u sinh v, z u 2 Solution: r r  u cosh v, u sinh v, u 2  r r dr r1   cosh v,sin v, 2u  du r r dr r2   u sinh v , u cosh v , 0 dv r 2  E r12 cosh2 v sinh2 v  2u  12sinh 2 v 4u 2 r r F r1  r2  cosh v sinh v , 2u  u sinh v, u cosh v , 0 

 u sinh v cosh v , u cosh v sinh v ,0  r2 G r2 u 2 sinh 2 v u 2 cosh2 v 0 u2  12sinh2 v

The quadratic differential from is ds 2 Edu 2 2 Fdudv Gdv 2 Where E 12sinh 2 v 4u 2 F  u cosh v sinh v, u cosh v sinh v, 0  G u  12sinh v  2

2

4. Prove that the curves bisecting the angle between the parametric curves are given by the equation Edu2 Gdv2 0 Solution:

We know that the direction coefficients of the parametric curves 1   1  v constant and u constant are respectively  , 0 and 0,  E   G du dv  Let  , be coefficient of direction which bisects the angle between ds ds  the parametric curves. Let, w, be the angle which this direction makes 1   1  with  , 0 and w2 be the angle which this direction makes with 0, . E   G

83

1 du 1 dv cos w1 E  F  E ds E ds 1 du 1 dv & cos w2 F  G  G ds G ds du dv  Since the direction  , bisects the angle between the parametric ds ds  curves, we have cos w1 cos w2 (Q positive sign taken for internal bisector, negative sign taker for external bisector ) 1  du dv  1  du dv  i.e. F  F G  E ds  ds  E  ds G  ds E Gdu F Gdv  F Edu G Edv    Squaring both sides  E Gdu F Gdv   F Edu G Edv      2 2 2 2 2 2 2 2 i.e., E Gdu F Gdv F Edu G Edv 2 2 2 2 i.e.. E  EG F  du G EG F  dv 2

2

i.e., Edu 2 Gdv 2 i.e., Edu2 Gdv2 0 . Hence, the result. 5. Find the area of the surface z f  x, y  r Solution: Let r x, y, z  Here, the parameters are x and y r z rx  1,0, p  , where p  x r z ry  0,1, q  , where q  y r r E rx  rx  1, 0, p 1, 0, p  1 p2 

r r F rx  ry  1,0, p  0,1, q  pq r r G ry  ry  0,1, q  0,1, q  1 q2 2 Area of the surface   EG F dxdy

(Q Elemental area)  EG F 2 dxdy







2 2 2 2 Q Area or surface   1 p 1 q p q dxdy 2 2   1 p q dxdy This form is known as Monge’s form.

84

6. Find the first fundamental form of the surface of a sphere of radius ‘a’. Solution: The equation of the surface of a sphere of radius ‘a’ is x a sin cos 0 y a sin sin  z a cos  The parameters are and 0 . r If r is the position vector of the point on the surface of the sphere, then r r  a sin cos 0 , a sin sin 0 , a cos  r r  a cos cos , a cossin 0 , a sin r r0  a sin sin 0 , a sin cos 0 , 0 r r E r r a2 cos 2 cos 2 0 a 2 cos 2 sin 2 0 a 2 sin 2 





 









a2  cos 2  cos 2 0 sin 2 0 sin 2    2 2 2 2 a  cos sin a r r F r r0 a2 sin cos sin 0 cos 0

a 2 sin cos sin 0 cos 0 0 0 r r G r0  r0 a2 sin 2 sin 2 0 a2 sin 2 cos2 0 a 2 sin 2  Q The first fundamental form is

   d0

E d2 Fdd 0 G d 0 2

a  da sin 2

2

2

2

2

2

7. Find the angle between two tangential directions on a surface at a point P. Solution: Let the direction coefficients of two directions at a point P be  l ,m  r r and  l , m , then the corresponding unit vectors e and eare given by r r r r r r e l r1 mr2 & el  r1 m r2 If is angle between these two given directions, then r r cos e  e r r r r  l r1 mr2  l  r1 m r2  r2 r r r2 l l  r1  l m l  m r1  r2 mm r2

i.e., cosE l l  F  l m l  m Gmm

85

r r sine e r r r r  l r1 mr2  l  r1 m r2  r r  l m l  m r1 r2  m H sin  l ml  Also tan   cos  El l  F  l m l  mGmm _____________________________________________________________________ 8. Find the direction coefficients making angle  with the direction 2 coefficients  l ,m . Solution: let  l , m be the required direction coefficients. We know that cos E l l  F  l m l  m Gmm sin l m  l  mH If  , El l  F  l m l  m Gmm  0 _________ (1) 2 & l m l  m H 1 _________ (2)

From (1), l  El Fm m F l Gm0 l m   a  say  F l Gm E l Fm l  a  F l Gm , m a El Fm_____ (3) Using (3) in (2), Hal El FmHam  F l Gm 1 1  H  El 2 2 Fl m Gm2 H a 1 (or) a H F l Gm El Fm l  , m H H

Check Your Progress ( C Y P ) : 9. What is the condition for the orthogonality of two directions   , and   ,  ? ________________________________________________________________________ ________________________________________________________________________

10. Prove that the metric is invariant under a parametric transformation ________________________________________________________________________ ________________________________________________________________________

86

3.9 . SUMMARY: It has been said that in the parametric equation of a surface, the parameters were the curvilinear coordinates or surface coordinates. Then Regular and singular points on the surface are defined. The nature of proper transformation through change of parameters and that of the parametric curves on a surface are explained. Method of obtaining tangent plane and unit normal at a point on the surface is given. Result regarding the property of proper parametric transformation is proved. Properties of parametric curves on certain surfaces of revolution are mentioned. Metric, its invariance property and the geometrical interpretation are proved. Elemental area on a surface using metric elements is found. Direction coefficients and direction ratios of tangent vector at a point on the surface are explained. 3.10. KEYWORDS: 1) Parameter: In the case of circular curve x a cos , y a sin  , is called a parameter. Whereas in the case of spherical surface, whose equation is x a sincos, y a sinsin , z a cos , and are called parameters. For the conical surface: parameters.

x u cos, y u sin, z u cot , u and 

2) Regular point:

r

r

r

r

r

Regular point on a surface, whose equation is r r  u, v . If ru rv o at

a point  u, v on the surface, then the point is called regular point. 3) Singular point:

r

r

r

If ru rv o at  u, v on the surface, then point is called singular. 4) Metric: It is an expression for measuring the length of an arc of a curve between two neighbouring points. 5) Direction coefficients : They are the numbers to specify the direction of a tangent vector at a point on the curve lying on a surface. 87

3.11. ANSWERS TO CHECK UP YOUR PROGRESS:

r r 1. A point  u, v is said to be a regular point on the surface r r  u, v , if the r r r condition ru rv 0 is satisfied at this point. Otherwise, the point is called singular. r r 2. let r r  u, v  be

the

equation

to

a

surface

in

terms

of

the

parameters  u, v and let a new set of parameters (U,V) be, introduced by means of the transformation u u  U ,V  & v v  U ,V ______(1) r r r r r r u r v rU      U u U v U r r r r r  r u  r v rV       V  u  V  v  V  u  u u, v  r r r r  r r  U  V rU rV  ru rv   ru rv  v  v  U, V   U  V  u, v  r r If rU rV 0, then Jacobian 0 ________ (2)  U ,V 

then

the transformation (1) under the condition (2) is called proper transformation. The effect of a proper transformation is that regular point is transformed into regular point.

r r r 3. It we keep v constant in the equation of the surface r r  u , v , then r depends on only one parameters u and thus determines a curve (on the surface), whose equation is v constant. Similarly, u constant is another parametric curve on the surface. As the constants vary the surface is covered with a net of parametric curves two of which pass through every point P , forming the family of curves v constant and the family of curves u constant. 4. Let the equation of the curve be u u  t , v v  t , then the tangent is r r r r& r& dr  r du  r dv parallel to the vector r , where r     dt  u dt  v dt r dr r du r dv r r r i.e. r1 r2 (or) dr r1 du r2 dv dt dt dt r r but r1 and r2 are non- zero and independent vectors. Therefore the tangents

88

to a curve (on the surface) through a point P lie in the plane which r r contains the two vectors r1 and r2 . This plane is called the tangent plane r r at P . Hence, r1 r2 gives the direction of the normal to tangent plane. Thus, r r r r r the equations to the tangent plane at P , is R r  r1 r2 0 , where r is the r position vector of P and R , that of a current point on the plane.

 

5. The parametric curves v constant are the meridians on the sphere and u constant are the parallels of latitudes, which are small circles, whose planes are parallel to the xy plane. r r r Now r1  a  cos u cos v, cos u sin v, sin u  u r r r r2  a  sin u sin v , sin u cos v , 0 and  v r r thus r1  r2 0 . Hence, the parametric curves are orthogonal. r r 6. From the equation of a curve u u  t , v v  t on the surface r r  u, v , we r r r r r r have dr r1 du r2 dv where r and r dr corresponding to the parameter values u, v and u du , v dv respectively are the position vectors of two neighbouring points P and Q on the surface. Let the arc PQ be ds . Since, the points P and Q are neighbouring points, r therefore ds dr r ds dr   dt dt 2 r2 ds  dr   dt  dt 2

r du r dv  r1 r2  dt   dt 2

2

r du  r r du dv  r 2 dv  r12  2 r1  r2   r2   dt  dt dt  dt  2

2

du  du  dv  dv  E  2F    G   dt  dt  dt  dt  r2 r r r2 Where E r1 , F r1  r2 , G r2

ds 2 Edu 2 Fdudv Gdv 2 Thus the metric gives the distance between two neighboring points  u , v and  u du, v dv on the curve.

89

7. Consider the figure ABCD ,whose vertices A, B , C , D have parameter

u, v , u du, dv  , , u , v dv  respectively. If u du, v dv

values

du and dv are small and positive, then the figure ABCD is approximately a parallelogram.

r Let r denote the position vector of any point on the figure, then uuur AD position vector of D position vector of A r r r  u, v dv  r  u, v  r r  r r r  u, v   dv r  u, v  v r r r  dv r2dv  v r r 111l y AB r1du

r r Hence, the area dS of the parallelogram ABCD is given by dS r1du r2 dv r r r1 r2 dudv

 EG F 2 dudv 8. If   , are proportional to the direction coefficients  l .m , then   , are

called the direction ratios of the direction. Thus, if We have 1 E l 2 Fl Gm 2

m

   k l m

2



   2 F    G   k  k  k  2 2 2 Thus k E 2 F G  , from which, we get ,  ,     l , m  k E 2 2 F G 2 E  k

2

2

9. The condition for the orthogonality of two directions   , and   ,  is E  F    G 0

10. Under the parametric transformation   u, ,   u,  , the metric takes the form

90

2 2 ds 2 E du 2F  du  d G  d r r 2  r1 du r2 d  r r r r r  r u r   where r1   .  .  u  u  u   u r u r   r1 r2  u  u r r r r u r  & r2 r1 r    2 

2

 u r  u r   r  r  Thus ds  r2 r1 r2 d  r1 du     u   u              2

2

r u  u      r   r1  du   d r2  du   d    u     u      r r 2 2 2  r1du r2d Edu 2FdudGd 

3.12. TERMINAL QUESTIONS: 1. Find the surface area of r a) The sphere r a  sin u cos v, sin u sin v, cos v  r ring r  b a cos u cos v, b a cos u sin v, a sin u 

b)

the

anchor

2. Show that the parametric curves on the sphere given by x a sin u cos v , y a sin u sin v , z a cos u form an orthogonal system. 3. On the paraboloi x2 y2 z; find the orthogonal trajectories of the section by the planes z c . 4. The metric of a surface is v 2du 2 u 2dv 2 . Find the equation of the family of curves orthogonal to the curve uv constant. 3.13. FURTHER READINGS: i) ‘Differential Geometry’ by D. Somasundaram, Narosa publishing House, Chennai, 2005. ii)

‘Elementary Topics in Differential Geometry’ by J.A.Thorpe, Springes – Verlag, New York, 1979.

91

UNIT- IV LOCAL NON- INTRINSIC PROPERTIES OF SURFACE- GEODESICS: STRUCTURE: Learning Objectives 4.0.

Introduction

4.1.

Family of Curves

4.2.

Orthogonal Trajectories

4.3.

Double family of curves

4.4.

Isometric Correspondence

4.5.

Geodesics and their Differential Equations

4.6.

Canonical Geodesic Equations

4.7.

Second Fundamental Form

4.8.

Fundamental Equations of Surface Theory

4.9.

Geodesics on a Surface of Revolution

4.10. Curvature of normal section of the surface. 4.11. Summary 4.12. Key Words 4.13. Answers to Check Your Progress 4.15. Terminal Questions 4.16. Further Readings LEARNING OBJECTIVES : After going through this unit, you should be able to,  

define family of curves, isometric correspondence, Geodesics, normal section & oblique section, normal curvature.

 

derive the differential equations of the family of curves, of Geodesics, canonical Geodesic equations.

92

4.0 INTRODUCTION: In the previous unit, we have given the meaning of surface, the nature of points on it, properties of curves on surface, the tangent plane and surface normal, the general surface of revolution, Metric-First fundamental form and the direction coefficients on the surface. In this unit we are going to present family of curves, double family of curves a special type of curve called Geodesic and their Differential Equations and that on a surface of revolution, fundamental equations of surface and about normal section of the surface. 4.1 FAMILY OF CURVES:

r r Let the equation r r  u , v ___(1) represent a surface. Consider an implicit u, v c _____(2) , where c is a parameter and o  u ,v  relation of the form o  is a single valued function of u, v . The relation (2) gives a family of curves lying on the surface (1). Note that for different values of c, we have different members of the family (2). If c is a constant equation (2) gives one member of the family (2).

Proposition: Through every point of the surface , there passes one and only one member of the family of curves. Proof:

r r Let  uo , vo be any point on surface r r  u, v  . uo , vo swill also lie on (2) for

some value of c, say C1 . Thus giving  uo , vo C1 . o  u, v c1 is a member of family of curves (2) passing through the

point  uo , vo . Therefore, it follows that through every point  u, v of the surface there passes one and only one member of the family of curves, Differential equation of the family of curves: Suppose o  u, v c _____(3) is the equation of the family of curves. Differentiating (3), we have

o o du  dv 0 u  v

i.e, o1 du o 2 dv 0 _____(4)

Thus, the integral of (4) is (3) Let o 1 P  u, v  , o 2 Q  u, v  Where P and Q are functions of u, v and do not vanish together. Equation (4) may be written as P du Q dv 0 _____(5) . The equation (5) is called the differential equation of the family of curves(3).

93

du o 2 du dv  i.e.,  . Thus if  u, v is a dv o1 o 2 o 1 point, the direction ratios of the tangent to the curve [a member of family (3)] through the point  u, v  are o 2 , o 1 . If at any point, o 1 and o 2 both vanish together, the directions The equation (4) may be written as





are indeterminate and hence we do not have a definite tangent at that point. Thus the restriction that’ o 1 and o 2 should not vanish together at any point’ on o is desirable. Note: From the equation (5), it follows that at any point  u, v  , the tangent to the curve through the point has direction ratios  Q, P . Consider any point P  u ,v  on the surface o is a scalar field defined on the surface. We show that o increase most repidly in the direction orthogonal to the curve of do the family passing through P . i.e, has the greatest value in such a direction. ds Let  l ,m be any direction at P and , the magnitude of the vector o 2 , o 1 at









l , m P . Let be the angle between  and o 2 , o 1 . r r r r r r r r i.e, between a l r1 mr2 and b o 2 r1 o 1 r2 Since a 1 and b we have r r sin a b  l o 1 m o 2





dv du   o 1  o 2  ds ds  do  ds

l , m Now  and are positive and do not depend on  . Hence

do has the ds

 , when sin has the maximum value when  and has the  2 minimum value  when  .  2 maximum value

The differential equation o 1du o 2 dv 0 has the solution curves, o constant.

Here o 1 and o 2 are u and v derivatives of the function o . If P  u, v  du Q  u, v  dv 0 is a given differential equation, if is not in general possible to find a single function of u, v , such that o 1 P and o 2 Q . However, we can find an integrating factor  u, v such that P o 1 and Q o 2 . Hence writing the equation in the form Pdu Qdv 0 i.e, o 1du o 2 dv 0 , we have the solution curves o constant.

94

Check Your Progress (CYP): 1. How many members of the family of curves pass thro’ every point of the surface? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

4.2. ORTHOGONAL TRAJECTORIES: Let Pdu Qdv 0 be the differential equation of a family i.e., solution curve at du dv the point  u , v has its tangent in the direction  Q , P , since Pdu Qdv 0   . Q P If  du, dv are the differentials in an orthogonal direction, then we have

E Q  du F  P dv 0 Q dv Pdu   G  i.e, FPEQduGP FQ dv0 since  Q, P and  du, dv  have orthogonality directions. This is the differential equation to the orthogonal trajectories of the given family Pdu Qdv 0 . If

 u, v constant

is

the

family

of

orthogonal

trajectories,

then

FP EQ 1 and GP FQ 2 , where is an integrating factor. Also, if o u, v constant is the given family of curves, then P o 1

Thus assume

  1

 o, 

 u, v  that

P FP EQ

P and Q

and Q o 2 .

Q 1   EQ 2 2 FPQ GP 2 0 ,if we GP FQ 

do

not

vanish

together.

Hence,

putting u  o  u , v and v u ,v  , we get new parameters  u , v for the surface, giving rise to orthogonal system of parametric curves. Hence, the parameters can always be chosen so that the curves of a given family and their orthogonal trajectories becomes parametric curves. 4.3 DOUBLE FAMILY OF CURVES: The quadratic differential equation of the form Pdu 2 2Qdudv Rdv 2 0 __(1) , where P, Q, R are continuous functions of u and v and do not vanish together, represents two families of curves on the surface provided Q PR 0 . 2

du  du  The equation (1) may be written as P  2Q  R 0 . dv  dv  2

95

This is a quadratic equation in

du and by solving this equation as a quadratic dv

du , the separate differential equations for the two families are determined. dv If the two curves specified by the equation Pdu 2 2Qdudv Rdv 2 at a point have direction coefficients  l , m  and  l ,m are the two values of , then l m and l  m du obtained from the quadratic equation. dv

in

2

du  du  P  2Q  R 0 dv  dv  l l  2Q ll R Thus   and  m m P mm  P l m l  m 2Q ll R   and  mm P mm P l l  mm l m  l  m    R P 2Q The condition E l l  F  l m l  m Gmm 0 for orthogonallity of the two families is therefore E R F  2Q G  P 0 i.e, ER2 FQGP0 ______(2) If P R 0 , the given differential equation (1) reduces to dudv 0 i .e , du 0,dv 0 , which determine the parametric curves. The orthogonal condition (2) becomes F 0 Check Your Progress (CYP) : 2. What is double family of curves ? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

4.4. ISOMETRIC CORRESPONDENCE: Let S and S  be two surfaces with respective parameters  u, v and  u , v . Assume that there is a correspondence between the points of S and S  , such that to each u ,v  point u, v in S , there corresponds a point  in S  given  ,  by u o  u, v and v u, v with J  0 .  u, v 

96

Let C be a curve on the surface S with the parametric equation u u  t and v v  t . Then the correspondence between S and S  maps C into a curve C  on S  . The parametric equation for C are. u o  u , v o  u  t , v  t 

and v u, v  u  t , v  t 

for

Let P be a point on C and P , the corresponding point on c . The direction ratios u&& , v and that at P u& , v&, the tangent at P to C are  to C are 

du dv du o o dv   where u& , v& , u&   u&  v&& v&  u&  v& dt dt dt u  v dt  u  v Solving these two equations for u&and v&, we have

  o  u&u& v&  J v   v  o   v& v& u&  J u  u  

Since J 0 , if follows that to each direction at P, there is a corresponding direction at P and vice versa. It is now seen that the surface S can be reparametrized with  u , v as the parameters. To the point Pon S , we take the corresponding point P on S and assign the parameters  u , v of P to P . Then P and Pcarry the same parametric values. The parametric transformation from  u , v to  u , v is proper, since the Jacobian is not zero. After this transformation, the corresponding points will have the same parameters. The surface S and S are said to be isometric, if there is a correspondence between them, such that corresponding arcs of curves have the same length. For example, if a plane sheet of paper is slightly bent, the length of any curve drawn on it is not altered. Thus, the original plane sheet and the bent sheet are isometric. Let us find a surface of revolution which is isometric with a region of the right helicoid. r Now r g  u cos v, g  u sin v, f  u  is a surface of revolution for which

df dg f1  and g1  , F 0, G g 2 , so that its metric is du du ds 2 g12 f12 du 2 g 2 dv 2 . E g12 f12 , where

r u cos v , u sin v , av Again the equation to a right helicoid can be taken as r  

2 2 2 2 2 and E=1, F=0, G u  a 2 , so that its metric is ds du   u a2 dv . Now, we

have to find transformation from (u,v) to

97

, v  which u

makes these two metrics

identical.

Taking

du 0 1du and dv dv . 2

u 0  u for

v v and

The

two

simplicity,

metrics

we

will

have be

then identical

2 1

if g 2 0 a2 and g12 f12 0 . Thus, we have two equations to determine the three unknown functions f , g , 0 . If 0 is eliminated between these two equations, we will get a differential equation for f as a function of g. If we try 0  u a sinh u and g  u  a cosh u so as to satisfy the first equation, the second equation will then give f12 a2 or f au . Hence the right helicoid is isometric with the surface obtained by revolving the curve x a cosh u , y 0, z au about z- axis. This is the catenary x a cosh  z a and the surface of revolution is called a catenoid of parameter ‘a’. The correspondence u a sinh u , v v shows that the generators vconstant on the helicoid correspond to the meridians v= constant on the catenoid and the helices u constant correspond to the parallels u= constant of the catenoid. Also, we note that on the helicoid uand v can take all real values, whereas on the catenoid o v 2. The correspondence is therefore an isometry only for that region of the helicoid for which o v  2. Thus one period of a right helicoid of pitch 2a , corresponds isometrically to the whole catenoid of parameter a. Check Your Progress (C Y P) : 3. What is isometric correspondence between two surfaces? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ Intrinsic properties: All properties of a surface, which can be obtained using the r r metric Edu2 2Fdudv Gdv2 alone and without using the vector equation r r  u, v  are called intrinsic properties. Thus isometric surfaces have the same intrinsic properties, even though they may differ in shape. 4.5. GEODESICS AND THEIR DIFFERENTIAL EQUATIONS: Geodesics- definition: Geodesic in a surface is defined as the curve of the shortest length (rather than strictly shortest distance) on a surface between any two points on it. On any surface, we have special curves called Geodesics viz., curves of the shortest distance. Given any two points A and B on the surface, the problem is to find the shortest among the curves lying on the surface and joining A and B. If the surface is a plane, then the geodesic is the straight line segment. If the surfaces is a sphere, it is the small arc of the great circle passing through A and B. For a general surface, it does not immediately follow that there exists an arc of the shortest length joining the two points.

98

However, we can obtain a set of differential equations for such a curve. Here, we make use of the method of calculus of variation. r r Let A and B be two points on the surface r r  u, v. Consider the arcs joining A

and B and let them be given by equations of the form u u  t ,v v  t . Assume that for all arcs t=0 at A and t=1 at B. let be one such arc and let s  be the arc length joining 2

2

2

ds  du  du dv  dv  A and B measured along .Since  E  2 F   G  , dt  dt  dt dt  dt  2 2 2 &&Gv& i.e., s& Eu& 2 Fuv

  We have s    &dt  Eu&2 2 Fuv &&Gv&2 dt 0 s 0 Now, let the arc,  be deformed slightly keeping end points A and B fixed. Let be the new arc and let its equation be u u  t u  t  t 

v v t v  t  t  Where is small and  t ,  t are such that  0  0 0, and  1  1 0 . The arc length s   measured along is obtained from s by putting uand v in place of u and v. The variation in S  v is then S   S and in general

it is of order . If however is such that this variation is at most of order  , for all small variations in  (i.e., for all  t and  t then s   is said to be stationary and 2

the curve is a geodesic. Now putting, 1 T u , v, u&& , v   Eu&2 2 Fuv &&Gv&2  2 1 2  s& 2 1

1

1

0

0

0

&  2T dt f dt say  We have s   sdt

Where f  2T 1





&, v& f u , v , u&  & f  Thus s    u , v , u&& ,v  dt s    0

1  f & f f   f & dt 0   2     & 0  u  v  u  v&

Now,

99

f  f  &   d   dt  0  u& u& 0   1

f   f  1 d       dt u&  u&0 0 dt  1 d  f     dt 0 dt  u& 1

The first term being zero, since  o  1 0 1 1 d f   f  Similarly  & dt    dt  0  v& 0 dt v& Thus s   s 

1 f f d  f  d f        dt 0  2       u  v dt u& dt v& 0  1  f d  f    f d f       2      dt 0    u dt  u& v& 0   v dt  

 L M dt o 2  1 0

f d  f  f d f  Where L    & M     u dt  u& v dt  v& is stationary, if s   By definition s  s is at most of order 2 . Thus, is a geodesic iff for all  t and  t . 1

L M dt 0  0

And this implies L=0 and M=0 f d  f   f d f  i.e.,   0 &   0 _________(1)  u dt  u&  v dt  v& These are then the required differential equations for a geodesic. As these equations do not involve the points A and B explicitly, they are the same for all geodesics on the surface. Now, since f  2T , we can write these differential equations involving T rather 1  T d 1  T    0 2T u dt  2T u& dT  T 1  0 and cancelling , we have dt  u& 2T

than f. The first equation of (1), thus becomes i.e.,

1  T 1 d  T 1    3 u&  2T u 2T dt  2T 2

d  T  T 1 dT  T     dt  u&  u 2T dt  u& And similarly, from the second equation of (1), we have

100

d  T T 1 dT  &  dt  v  v 2T dt The geodesic equations are thus d  T  T 1 U     dt u& u 2T

 T  dv&

dT T  .  dt u&  2 ___________  d T   T 1 dT  T V      dt  v& v 2T dt dv&  Now for any curve (whether geodesic or not) U and V as defined in (2) satisfy the dT & vV &  __________  relation uU 3 dt d  T T   d  T  T & vV & u&  Since uU v&        dt  u& u   dt v&  v 

d  T  T  T   u& u& & u& dt   u&  u&  u d  T  T  T  v& & v& v& dt   v& v& v d  T T  dT   u& v&  dt   u&  v& dt d dT dT   2T   dt dt dt (Q T is a homogenous function of degree 2 in u&and v&) Due to this fixed relation between u and v, we observe that the two equations in (2) are not independent. dT Eliminating between the two equations of (2), we obtain dt T T U V 0 _____________  4 v&  u& Thus the geodesic satisfies this relation (4). We now show that a curve satisfying (4) is a geodesic. Suppose u(t) and v(t) satisfy (4)  T  T and assume that u& and v& are not both zero for same valve of t. Then and u& v&  T  T cannot vanish together, since in that case E u&Fv& 0 and Fu&Gv& 0 u& v& implying that u&0 and v&0 , which is against our assumption. Thus from (4), we get T T U V U V  i.e.,    say   T  T v&  u& u& v&

101

T T Thus U  , and V  _____________  5 u&  v& Substituting these in (3), we get dT uU & vV & dt  T  T  u& v&   u& v&  2T since T is a homogeneous function of degree 2 in u&s and v& 1 dT  T  T   0, U from  5 2T dt  u&  u& 1 dT  T  T Similarly  0, V 2T dt v& v&

These are the geodesic equations (2) and hence the given curve is geodesic. Thus (4) is a necessary and sufficient condition for a curve u u  t , v v  t on the surface r r r r  u, v to be a geodesic. Check Your Progress (C Y P ) : 4. Explain intrinsic properties of a surface? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

4.6. CANONICAL GEODESIC EQUATIONS: If the arc length s is taken as the parameter for a curve, then we write uand v instead of u&and v&and 1 2 2 T  Fu 2 Eu  v Gv  ________________  6 2 du dv But u and v   are the direction coefficients. ds ds 1 Edu2 2 Fdu dv Gdv2 1 ds2 1 Hence, T     2  2 2 ds 2 2 ds dT  0 ds The Geodesic equations d T  T 1 dT T U      dt  u&  u 2T dt  u& d  T  T 1 dT  T V      dt  v& v 2T dt  v& now becomes the canonical equations for geodesics

102

d T  T  i.e., U    0________________  7  ds  u  u    dT  d  T T Q 0    V    0 ds    ds v  v 

dT & vV &  becomes u Again, the identity uU U v V 0 ____________  8 dt showing that the equations (7) are not independent. If the curve is not parametric curve, we have u 0, v0 and hence U 0  V 0 . Thus in this case U 0 or V 0 is a sufficient condition for the geodesic. For the paremetric curve u= constant, u 0 and v 0 . Thus (8) gives V=0 for alls, so that one of the equations of (7) is automatically satisfied. Hence, the condition for u= constant to be geodesic is U=0. similarly V=0 is the condition for v= constant to be a geodesic.

4.7. SECOND FUNDAMENTAL FORM: `

r r let r r  u, v  be the equation of a surface. The quadratic differential form

Ldu 2 2 Mdudv Ndv 2 in du , dv is called the second fundamental form. The quantities L,M,N are called second order fundamental magnitudes or second fundamental coefficients and explained as follows. r uur 2 rr uur 2 rr 2 r uuuur we know r11  2 , r12   r21 u  u v  v u uur 2 rr r22  2 v uur rr rr rr rr N r1 r2  1 2 r1 r2 H r r where N is the unit normal vector to the surface at the point r  u , v . We denote the r r r r resolved parts of the vectors r11 , r12 , r22 along the surface normal N by L,M,N r r r r r r respectively; Thus L N  r11 ; M N  r12 ; N N  r22 and LN M 2 T 2 (say), where T 2 is not necessarily positive.

4.8. FUNDAMENTAL EQUATIONS OF SURFACE THEORY: We shall obtain relations between the fundamental coefficients, E,F,G and L,M,N. The Gauss’s formulae giving relations in between these coefficients and their partial derivatives w  r t u and v play an important part in the theory of surfaces. Gauss’s Formulae: The equations of Gauss are given by following partial differential equations:

103

r r r r r11 LN lr1 r2  r r r r  r12 MN mr1 r2 ______________  1 r r r r  r22 NN nr1 r2   where l , m, n;  , , r are called christoffel symbols and are suitable functions of E,F,G and their partial derivatives with respect to u and v. r r r Proof: we know that L,M,N are the resolved parts of r11, r12 , r22 along the normal to the surface and therefore, we may write r r r r r11 LN lr1 r2 ___________  2 r r r r r12 MN mr1 r2 __________  3 s r r r r r22 NN nr1 r2 ___________  4 Where the symbols l , m, n;  , , are calculated as follows:  r r r  E1   E  r12 2r1  r11  1 u   r r  F1   F  r1  r2  1  u r r r r r 2 r r  F1 r11  r2 r1  r12 ; G1  r2 2r2  r12  1 (or) 5 __________   r2 r r  E2   E  r1 2r1  r12 2  v  r r r r r r  F2   F  r1  r2   r12  r2 r1  r22  2  v  r2 r r  G2  r2  2r2  r22 2  r r r r Solving equations (5) for r1  r22 , r2  r12 etc., we get

r r 1 r r 1  r1  r11  E1; r2  r11 F1  E 2  2 2  r r 1 r r 1  r1  r12  E2 ; r2  r12  G1 6 __________  2 2  r r 1 r r 1  r1  r22 F2  G1 ; r2  r22  G2  2 2  r r On multiplying scalarly each side of equation (2) with r1 and r2 successively, we get r r r2 r r r r r1  r11 lr1 r1  r2 Q r1  N 0   7 __________  r r r r r2 r r r2  r11 lr1  r2 r2 Q r2  N 0   using equations (6) in (7) , we get from (7)

 

 

104

1 1 E1 lE F ; F1  E 2 lF G 2 2

Thus, 1  l 2 GE1 2FF1 FE2   2H 8 _____________  1   2  2 EF1 EE2 FE1   2H  r r Similarly on multiplying scalarly on each side of (3), by r1 and r2 successively and using (6), we get 1 1 E 2 mE F ; G1 mF G 2 2 Thus, solving for m and , we get 1  m 2 GE 2 FG1    2H 9 _____________  1   2  EG1 FE 2   2H  r r Similarly on performing scalar product of each sides of (4) with r1 and r2 and using (6), we get 1 1 F2  G1 nE F ; G 2 nF G 2 2 And hence, solving for n and , we get 1  n 2  2GF2 GG1 FG2    2H 10 ____________  1 v  2 EG2 2 FF2 FG1   2H  Corollary 1: If the parametric curves are orthogonal, then F 0 and H 2 EG and therefore the Gauss’s formulae reduce to uur uur 1 E ur E ur  r11 LN  1 r1  2 r2  2 E 2G uur uur E ur G ur   r12 M N  2 r1  1 r2 ______________  11 2E 2G  uur uur G ur G ur  r22 N N  1 r1  2 r2  2E 2G  Since the value of l , m , n , , , reduce to

105

1 E l  1 , 2 E E  2 , 2G

E m 2 , 2E G  1 , 2G

G  n  1  2E  12 ___________  G2   2G  

Corollary 2: The Gauss’s formulae may be written in matrix form as follows: uur uur r11  L l  N  uur    ur  r12   M m  r1  uur ur   r    N n  r2    22    Check Your Progress (C Y P ) : 5. What is second fundamental form of a surface ? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ 4.9. GEODESICS ON A SURFACE OF REVOLUTION:

r The equation of surface of revolution is given by r  u cos v , u sin v , f  u  . ur ur df r1 cos v ,sin v , f  , r2 u sin v , u cos v , 0, f  du uur uur r11  0, 0, f  ; r12 sin v , cos v , 0 uur r22  u cos v , u sin v, 0  ur ur ur ur 2 E r1  r1 1 f  , F r1  r2 0, ur ur 2 G r2  r2 u 2 , H 2 EG F 2 u 2  1 f   1 Christoffel coefficients in the formulae of Gauss are 0,  , v 0 u 2 2 We know that the geodesic equations are u  l u  2mu  v nv 0 & 2 2  v u2 u v v0 __________ (2) ur r ur (While obtaining the D. E. of geodesic via normal property, we had r ru r2 v , 1  ur uur 2 uur uur 2 r ur     r  ru r2v r11u2r12u v r22v and hence 1 ur r 1  1  2 2 r1  0  r  Eu  Fv E1u   E 2u  v  F2  G1  v  0 2  2  ur r 1  1 2 2 r r2  0  Fu Gv  F1  E2 u G1 u v  G2 v 0 ) 2 2   Substituting values of , , in equation (2), we get

106

2u v  v  0 v u 0 or  u 2v 2u  u Which is an exact differential equation and its solution is given by u 2v  h (say) ____ (3) (h, being arbitrary constant) The geodesics on surface of revolution are given by equation (3). 2 Let h 0 , then (3) gives u v  0  v0  v c . Thus v = c is a geodesic. Hence, every meridian is a geodesic on a surface of revolution. Now to find the complete integral of (3), we proceed as follows. The metric ds 2 Edu 2 2 Fdudv Gdv 2 in this case is given by 2 2 2 2 2 ds  1 f  du u dv ______  4 dv h ds 2 2 u4  dv h 2  ds

Equation (3) may be written as u squaring

2

or u 2dv hds

2 4 2 i.e., u  dv  h  4 1 f 2 du 2 u 2dv 2    by  2 1 f  5 or u2 u 2 h2 dv2 h2  du2 _____  1

h 1 f 2 dv   2  du u u h 2  2

1

2 1f  2 du Integrating, we get v g h  , if u 2 h 2 ______  6 where g and h are 2 2  u h  u arbitrary constants. Equation (6) is the complete solution of the equation of the geodesic on a surface of revolution (or) in other words geodesic on a surface of revolution are given by equation (6).

If u 2 h 2 , then from equation (5), we get du 0  u c . Also F 0 . Hence the curve u c (parallel) is a geodesic if and only if G is a function of v only i.e, G1 0 .(Q If the parametric curves are orthogonal i.e, F 0 , the curves v constant will be geodesics, provided E is a function of u only and the curves u constant will be geodesics if G is a function of v only). But G u 2 G1 0 iff u is constant (or stationary) where constant value of u is the radius of the parallel u c and this constant value of u is not zero. Hence a parallel is a geodesic on a surface of revolution if and only if the radius of that parallel has stationary value. Note: Equation (3) can also be obtained as follows:

107

2 2 2T Eu  2Fu  v Gv  2  1f  u2 u 2v2

T T 0, u 2v   v  v d  T T d 2 Hence, V      u v  ds  v v ds  Thus canonical geodesic equation V 0 provides d 2 2 i.e, u vh u v0 ds Corollary: If 0 is the angle at which the geodesic cuts the meridian, equation (3) may be written as u sin 0 h This is known as clairaut’s theorem. 

4.10. CURVATURE OF NORMAL SECTION OF THE SURFACE: A plane drawn through a point on a surface, cuts it in a curve, called the section of the surface. If the plane is so drawn that it contains the normal to the surface, then the curve is called normal section, otherwise it is called an oblique section. Obviously the principal normal to the normal section is parallel to the normal to the surface. We assume r r the convention that the sense of n (the unit principal normal to the curve) and N (the unit surface normal) are same. Formula for curvature of normal section in terms of fundamental magnitudes: Let kn denote the curvature of the normal section, then k n is positive, when the r curve is concave on the side towards which N points out. Then, we know r r dt r r r r  r  k nn k n N (since here n N ) ds r r  k n N  r _______  1 r r r But we know r  ru r2 v 1  r r r r dr1 r dr2  u Differentiating this w.r.t.s , r ru r2 v v 1 ds ds r r r r r r  r1 du  r1 dv  r  r2 du  r2 dv     i.e, r r1 u     u r2 v    v u ds  v ds  u ds  v ds    r r r r r r    ru  r11u  r12v   r12u  r22v  ur2v v  1 r r r 2 r r 2 ru  r2v  r11u 2r12u v r22v  _______  2 1  r r r r r r r in (1) and using N  Substituting this value of r  r11 L, N  r12 M , N  r22 N , r r r r N r1 0, N  r2 0 uur r 2 2 We get kn N  rLu 2 Mu v Nv 108

(Or)

Ldu 2 2 Mdudv N dv 2 kn  ds 2 2 2 Ldu 2 Mdudv N dv kn  _____(3) (using first fundamental form). Edu 2 2Fdudv Gdv 2

This gives the curvature of the normal section (usually called normal curvature) parallel to the direction  du, dv . Its reciprocal is called the radius of normal curvature and may be denoted by n . Normal curvature- definition: r r r Let a point P with position vector r  u , v  be on the surface r r  u , v . The

du , dv is equal to the curvature at P of the normal normal curvature at P in the direction  section at P parallel to the direction  du , dv  . Alternative definition: r r r Let a point P with position vector r  u , v be on the surface r r  u , v . Consider r r a curve r r  s through the point P lying on the surface. The component of curvature r vector r along the normal to the surface is defined to be normal curvature of the curve at the point P. uur r k n N  r Show that the definition of normal curvature given above are equivalent. Proof:

uur Let N be the unit normal vector of the surface at P, then from second definition, uur r r the normal curvature kn is given by kn N  r __ (4) where r is the curvature vector at P. Again let k be the curvature at P of the normal section at P which contains the direction  du, dv  uur r r r uur r   kn KN Q n N uur r uur uur  N  r k QN N 1 _____(5)

 

k k n

 

u sing  4 and  5  

Hence, curvature at P of normal section at P which contains the direction  du, dv is equal to the normal curvature at P in the same direction. Thus, in future the terms ‘normal curvature’ and ‘curvature of normal’ section will represent the same thing.

109

Note:

uur r Ldu I 2Mdudv Ndv2 We have seen N  r kn  ds 2 2 2 uur r du  du  dv  dv  i.e, N  rL   2 M    N     ds  ds  ds  ds  Now, we know that the curves, which have the same direction at the point P, have the du dv  same value of their direction coefficients  , at P. Also, the values of second order ds ds  fundamental magnitudes L, M , N are fixed at P. Hence, all curves which have the same uur r direction at P, have a fixed value of N  r which is equal to the normal curvature at P of any one of these curves. Hence, we also conclude that normal curvature at a point P on the surface is a property of a surface and a direction at the point P. The following theorem gives the relation between the curvatures of normal section and oblique section. Check Your Progress ( C Y P ) : 6. Define normal curvature at a point on the curve. ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ Mensnier’s Theorem: If kn and k are the curvature of the normal and oblique sections through the same tangent line, and  is the angle between these sections then the relation between kn , k and  is given by kn k cos . Proof: Let the section be oblique and its curvature be denoted by k. Since, section is r uur r r  r r  oblique its n is not parallel to N but will be parallel to unit vector  Q r kn  . If  k is the angle of inclination of the oblique section with the normal section touching the curve at the point under consideration, the  is the angle between the normals of the two r uur sections i.e., it is the angle between the unit vector r  and N . Thus k uur r  r cos N  k Ldu 2 2Mdudv Ndv2  1  k Edu 2 2Fdudv Gdv 2 k n k n k cos  k The mensnier’s theorem can also be defined as follows:

110

Ifis the angle between the principal normal to a curve on the surface and the surface normal at a point P, then kn k cos  where k is the curvature of the curve at P and kn , the normal curvature at P in the direction of the curve. Proof:

uur r r r Let P be a point  u, v on the surface r r  u , v . Suppose n and N denote the principal normal to the curve on the surface and surface normal at P respectively. Since uur r r uur  is the angle between n and N , therefore cos n  N. _____(1) r  is the curvature vector of the given curve at P. then, Now, suppose r  r r r kn. ______(2) uur Taking scalar product of both sides of (2) with N , r uur r uur r uur   r  N kn  N  or  r   N k cos _______  3 (by (1)) r uur  The value of r   N is fixed for all curves having the same direction at P and by r uur definition this value of r   N is equal to the normal curvature k n in that direction. Hence,

we have from equation (3), k n k cos . Note:

We have k n k if 0 . Thus the necessary and sufficient condition for the curvature of a curve at P to be equal to the normal curvature at P in the direction of that curve is that the principal normal to the curve is along the surface normal at that point. kn may be positive, zero, or negative. If LN M 2 0 at P, then kn have the same sign for all direction at P (since the denominator is always positive definite). In this case P is called an elliptic point. If LN M 2 0 then the numerator has one of the forms  du dv  or   xdu dv . Thus kn has the same sign for all directions thro 2

2

du  , when it is zero P is called parabolic point. If LN M 2 0 , then  dv the k n is positive for certain directions lying inside an angular region, negative for direction outside it and zero along the directions bounding the angle .P is then called a hyperbolic point and the two special directions bounding the angle are called the asymptotic directions. Let P  u, v and Q u du , v dv be P except for

neighbouring points of the surface r r r r  u , v and d, the perperdicular distance from Q to the tangent plane at P .Then. uuur uur uur r r d PQ  N  r u du, v dv r  u, v  N

111

uur r 1 r r r r i.e; d  r1 du r2 dv r11 du2 2r12 dudv r22 dv2   N by Taylor’s theorem, omitting 2   1 2 2 higher order infinitesimals   Ldu 2Mdudv Ndv  2 uur r uur r Since N  r1 0 N  r2

Thus

the

second

fundamental

form  Ldu 2 2 Mdudv Ndv 2  at

any

point P  u ,v  is equal to twice the length of the perpendicular from the neighbouring point Q u du, v dv on the tangent plane at P. At an elliptic point, d has the same sign and thus the surface rear P lies entirely on one side of the tangent plane at P. At a hyperbolic point, the surface crosses the tangent plane, where d is zero. We thus see that all points on an ellipsoid are elliptic, points on a circular cylinder are parabolic and that points on a hyperbolic paraboloid is hyperbolic.

Clairaut’s Theorem: If a geodesic on a surface of revolution cuts the meridian at any point at an angle 0 , u sin 0 is constant where u is the distance of the point from the axis.

Solution: Let P be a point on the geodesic and suppose this geodesic cuts the meridian v c through P at an angle 0 . Let the projection of arc s of geodesic on circular section through P be PM. i.e, PM uv





Also PM cos 90 0 s

 

 sin 0 s uv s We know u 2v1 h (Refer Geodesics on surface of revolution) 2 i .e ,u dv hds u sin 0 ds hds u sin 0 h ( cons tan t )

112

Second method: Let P be a point on a geodesic and suppose this geodesic cut the meridian du dv  v c thro P at an angle 0 . The direction coefficients of the geodesic are  , and ds ds  1  that of meridian v c are  , 0 . E  Now, we know that if 0 is the angle between the directions, whose direction coefficients are  l ,m and  l , m m , then sin 0 H l ml  du 1 dv  0 ds E ds H dv   _________(1) E ds 2 Again u v  h 2 u dv hds _____(2) H dv From (1), sin 0   E ds Hence, in this case sin 0 H

(or ) sin 0 

EGdv

ds E dv dv 2  G u sin ce G u ds ds





i.e, udv  sin 0 ds

113

Using this relation in equation (2), we get u sin 0 ds h ds

 

u sin 0 h  cons tan t  ________________________________________________________________________ Check Your Progress (C Y P ) : 7. Prove that the second fundamental form Ldu 2 2 MdudNd2 at any point P  u, is equal to twice the length of the perpendicular from the neighbouring point Q u du, don the tangent plane at P. ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ r r 8) Define normal curvature at a point on the surface r r  u,. ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

4.11. SUMMARY: We derived the differential equations of the family of curves and its solution curves. Next, the orthogonal trajectories of the family of curves is studied. Double family of curves is also studied. Then Isometric correspondence between surfaces is well studied. Differential equation of Geodesic and its canonical form is derived. Second fundamental form and Geodesic on surface of revolution are derived. Normal section of the surface and its curvature are studied. 4.12. KEY WORDS: 1.

Isometric correspondence : The surface S and S’ arc said to be isometric, if there is a correspondence between them, such that corresponding arcs of curves have the same length. For example, if a plane sheet of paper is slightly bent, the length of any curve drawn on it is not altered. Thus, the original plane sheet and the bent sheet arc isometric.

2.

Catenoid: The surface of revolution of the catenary parameter ‘a.’.

114

 ais called catenoid of

x a cosh z

3.

Geodesic: Geodesic in a surface is defined as the curve of stationary length on the surface between any two points on it. These are the curves of shortest distance on a surface.

4.

Normal section of surface: A plane drawn through a point on a surface, cuts it in a curve, called the section of the surface. If the plane is so drawn that it contains the normal to the surface, then the curve is called normal section of the surface.

5.

Oblique section: If a plane drawn through a point on a surface cuts it in a curve and if the plane so drawn that it does not contain the normal to the surface, then the curve is called an oblique section

4.13. ANSWER TO CHECK YOUR PROGRAMS: 1.

r r Let the equation r r  u, v  ___  1 represent a surface. Consider an implicit

relation of the form 0  u, v c _____  2 where c is a parameter. Let  u0 , v0 be any point on the surface (1).  u0 , v0 will also lie on (2) for some value of c, say c1 . Thus, giving  uo , vo c1 0  u , v c1 is a member of family of curves (2) passing thro’ the point  u0 , v0 . Therefore, if follows that through every point (u, v) of the surface, there passes one and only one member of the family of curves.

2.

3.

The quadratic differential equation of the form Pdu 2 2Qdudv Rdv 2 0 , where P,Q,R are continuous functions of u and v and do not vanish together, represents two families of curves on the surface, provided Q 2 PR 0 . Let S and S  be twos surfaces with respective parameters  u , v and

, v . u 

Assume that there is a correspondence between the points of S and S  , such that to each point  u, v in S , the corresponds a point  u , v in S  given by

 

 0,  u  u, vand v  u, v with J  0 . Such a correspondence is  u, v  called isometric correspondence between S and S . 4.

All properties of a surface, which can be obtained using the metric Edu 2 2Fdudv Gdv 2 alone and without using the vector r r equation r r  u , v are called intrinsic properties. Thus isometric surface have the same intrinsic properties, even though they may differ in shape. 115

5.

r r let r r  u, v be the equation of a surface. The quadratic differential form

Ldu 2 2Mdudv Ndv 2 in du , dv is called the second fundamental form. The quantities L , M , N are called second order fundamental magnitudes or second fundamental coefficients and are explained as follows. r r r r 2 r r 2 r 2 r r  r21 We know r11  2 , r12  u  u v  v u r r 2 r r22  2  v r r uur rr rr r1 r2 1 2 N r r  r1 r2 H

uur r where N is the unit normal vector to the surface at the point r  u , v . We denote uur r r r the resolved parts of the vectors r11, r12 , r22 along the surface normal N by L, M , N respectively. Thus, uur r uur r uur r L N  r11, M N  r12 , N N  r22 and LN M 2 T 2 (say), where T 2 is not necessarily positive. 6.

r r r Let a point P with position vector r  u , v  be on the surface r r  u , v . The

normal curvature at P in the direction  du, dv  is equal to the curvature at P of the normal section at P parallel to the direction  du , dv  .

r Alternately, let a point P with position vector r  u , v be on the surface r r r r r r  u, v . Consider a curve r r  s through the point P lying on the surface. r The component of curvature vector r  along the normal to the surface is defined to be normal curvature of the curve at the point P. 7. Let P  u, v  and Q u du , v dv be neighbouring points of the surface r r r r  u, v and d, the perperdicular distance from Q to the tangent plane at P . Then. uuur uur uur r r d PQ  N  r u du, v dv r  u, v  N

116

uur r 1 r r r r i.e; d   N by Taylor’s theorem, omitting r1 du r2 dv r11 du2 2r12 dudv r22 dv2   2   1 higher order infinitesimals   Ldu 2 2Mdudv Ndv 2  2 uur r uur r Since N  r1 0 N  r2

8.

r r r Let a point P with position vector r  u ,be on the surface r r  u ,. The

normal curvature at P in the direction  du, dis equal to the curvature at P of the normal section at P parallel to the direction  du, d .

Problems: 1.

Prove that the curves of the family

v3 constant are geodesics on a surface u2

with metric 2 2 2 2 v du 2uvdudv 2u dv  u 0, v 0 3

v c  c 0  u2 The parametric equation of the curve can be taken as 3 2 u ct , v ct Thus u&3ct 2 , v&2ct 1 2 &&Gv&2  Again T   Eu& 2Fuv 2 1 2 2 & 2u 2v&2    v u& 2uvvu 2 since by data E v 2 , F uv, G 2u 2 T &&2uv&2  vuv  u Consider the curve

 ct 2  3ct2 2ct 2ct3 2ct  2

6c t 8c t 3 5

3 5

2c3 t5  T &&3 c3 t6 Similarly, vu&2 uuv  v T v 2u&uvv&c 3t 6 u&

117

T uvu&2 u 2 v&c3t 7 v& d T   T Hence, U    dt u& u d 36   ct  2c 3t 5 dt 6c 3t 5 2c 3t 5 4c 3t 5 d  T T And V    dt  v& v d 37   ct  3c 3t 6 dt 3 6 3 6 3 6 7c t 3c t 4c t T T 3 5 3 7 3 6 3 6 Now U V  4c t  c t  4c t  ct  & & v  u 4c 6t 12 4c 6t 12 0 Hence, the curve is a geodesic for all values of c. ________________________________________________________________________

And

2.

Prove that on a general surface, a necessary and sufficient condition for the parametric curve v constant to be a geodesic is EE2 FE1 2EF1 0

Solution: On the curve v=c, u can be taken as the parameter. Hence, u=t, cr Thus u&1, v&0 1 &&Gv&2  Now T   Eu&2 2Fuv 2 T 1  E   E1u&2 2F1uv &&G1v&2  , where E1  , etc . u 2  u 1  E1 2  T 1  E &&G2 v&2    E2u&2 2 F2 uv , where E2  , etc. v 2 v 1  E2 2  T Eu&Fv&E u& T Fu&Gv&F  v&

118

d  T T dE 1 HenceU      E1 dt  u&  u dt 2 1 1 E1  E1  E1 2 2 since d  T T dF 1 t u and V      E2 dt v& v dt 2 1 F1  E 2 2 We know that a necessary and sufficient condition for a curve to be a geodesic T T is U V 0 v&  u& Thus the necessary and sufficient condition for the curve v=c to be geodesic is 1  1  F  E 0  E1  F1  E 2  2   2  i.e., EE2 FE1 2 EF1 0 ________________________________________________________________________ 3. Show that the curves du  u a  dv 0 form an orthogonal system on the r right helicoid r  u cos v, u sin v, av  2

2

2

2

r Solution: The equation of the given right helicoid is r  u cos v , u sin v , av  E 1, F 0, G u a We know that the two families of curves given by the quadratic differential equation 2 2 Pdu 2Qdudv Rdv 0 ___________  1 2

2

form an orthogonal system if and only if ER 2FQ GP 0 ___________  2 comparing (1) with the given family of curves du 2  u2 a2  dv2 0 ___________  3 We have P 1, Q 0, R  u2 a2 

 2   ER 2 FQ GP 1  u2 a2  0 0  u2 a2   1 0   Hence, the condition (2) is satisfied for family (3) Therefore, the curves given by (3) form an orthogonal system on the given surface. ________________________________________________________________________

119

4. If a geodesic on a surface of revolution cuts the meridians at a constant angle, the surface is right cylinder. Solution: The equation of the surface of revolution is x u cos v , y u sin v , z f  u . From clainaut’s theorem, we know

given by u sin 0 h

(constant) where 0 is the angle at which a geodesic on a surface of revolution cuts the 2 2 meridian through any point P on it and u  x y is the distance of P from the axis.

Here 0 is constant, since a geodesic cut all the meridians at a constant angle (given). h Hence (1) gives u  a (constant) sin 0

u  x y a  or x y a , which represents Q right cylinder. ________________________________________________________________________ 2

2

2

2

2

5. Prove that geodesics on right circular cylinders are helices. Solution: We know from clairaut’s theorem that, if a geodesic cuts the meridian at any point at an angle 0 , u sin 0 h (constant) ____________  1 where u is the distance of the point from the axis. Here, the surface of revolution is a right circular cylinder and we know that the meridians of the surface of revolution are the generators of the right cylinder. The distance of every point on the generator from the axis is constant i.e., u is constant. Let u=a (constant) h Hence (1) gives sin 0  (constant). Hence, a geodesic on the cylinder cuts all the a generators at a constant angle. Thus the geodesic is an helix. The geodesics on a right circular cylinder are helices. ________________________________________________________________________

4.14. TERMINAL QUESTIONS: 1.

Obtain the differential Equation of the family of curves and their solution curves.

2.

Write short notes on Isometric correspondence

3.

Derive the differential equations of a geodesic on the given surface

120

4.

Obtain the necessary and sufficient condition for a curve u=u(t), v=v(t) on the r r surface r r  u , v to be a geodesic.

5.

Find the equation to the geodesics on the helicoid x u cos v , y u sin v , z cv

6.

Prove that the necessary and sufficient condition for a curve on a sphere to be geodesic is that the curve is a great circle.

7.

Derive the formula for curvature of normal section in terms of fundamental magnitudes

8.

State and prove mensnier’s theorem

9.

State and prove clairaut’s theorem.

4.15. FURTHER READINGS :

1)

‘Elementary Topics in Differential Geometry’ by J.A. Thorpe, Springer – Verlag, New York, 1979.

2)

‘Differential Geometry’ by D. Somasundaram , Narosa Publications, Chennai, 2005.

121

UNIT V GEODESIC CURVATURE STRUCTURE: Learning Objectives 5.0.

Introduction

5.1.

Normal Property of Geodesics

5.2.

Existence Theorem

5.3.

Geodesic Parallels

5.4.

Geodesic Polars

5.5.

Geodesic Curvature

5.6.

Liouville’s Formula for k g

5.7.

Gauss Bonnet Theorem

5.8.

Gaussian Curvature

5.9.

Minding’s Theorem

5.10. Conformal mapping 5.11. Summary 5.12. Key Words 5.13. Answers to Check Your Progress 5.14. Terminal Questions 5.15. Further Readings

LEARNING OBJECTIVES: After going through this unit, you should be able to, 

define normal property, Geodesic parallel, Geodesic curvature and vector, Geodesic polars, Gaussian curvature, Conformal mapping.



derive the formula for Geodesic curvature and vector, Liouville’s formula for k g , and that for Gaussian curvature.



prove the theorem of Gauss- Bonnet, Minding, and theorem regarding conformal mapping.

122

5.0 . INTRODUCTION: In this unit, we first characterize geodesics in terms of their normal property. Existence theorem regarding geodesic arc is to be proved. Types of geodesics viz., geodesic parallels, geodesic polars, geodesic curvatures are to be studied. Next Liouville’s formula for geodesic curvature is to be derived. Gauss- Bonnet’s theorem regarding geodesic curvature is to be proved. Then comes Gaussian curvature and the proof of Minding theorem related to Gaussian curvature. Conformal mapping plays an important role in Differential Geometry. 5.1. NORMAL PROPERTY OF A GEODESIC:

r r Consider a curve u=u (t), v=v (t) lying on the surface r r  u, v 1 1 2 1 r2 &&Gv&2  s We have T  Eu&2 2 Fuv  r 2 2 2 r r& d r r r & r2v& Where r  ru 1  dt r T r&r& & r  r r rr Thus r . r .  ru r2v&r&.r1 1 &  u&  u&  u& And r  T & r  r& r r r r . r&.  r11u&r 21v& u  u r d r r r r&.  r1 , sin ce r1 r1  u, v  dt d  T T Hence    dt u   u& d rr r d r  r&.r1 r&.  r1  dt dt d r r r d r  r d r  r& .r1 r&.   r1 r .  r1 dt dt    dt rr & r&.r1 r&r Similarly V r& .r2 If the arc length is the parameter, the geodesic equations are U  s 0 and V  s 0 .Hence replacing t by s and dots by dashes, we have, r r r r r r U  s r  .r1 0 and V  s r  .r2 0. This shows that r  is perpendicular to both r1 and r r  r2 and therefore to the tangent plane at the point under consideration. But r  is along the principal normal. Hence the principal normal of a geodesic at any point is in the direction of the normal to the surface at that point and every curve having this property is a geodesic. This property is called the normal property of geodesics.

 



123

Now, in terms of the general parameter t, the condition U

r&r r&r  r& .r  &rr&. rr rr&.rr  r .r 0 r r& r r i.e.,  r&r& .r r 0 1

2

2

1

T  T V 0 becomes  v&  u&

1

2

which shows that the binormal is perpendicular to the surface normal which again proves the same property. Also, the above normal property is equivalent to the following property viz., at every point of a geodesic the rectifying plane is the tangent plane to the surface. Using the above normal property of geodesics, we can find out whether a given curve on a surface is a geodesic or not. For example, every great circle on a sphere is a geodesic, since the principal normal to the great circle is a normal to the sphere. Similarly every meridian on a surface of revolution is a geodesic, because it has the above normal property. Example: 5.1 A particle is constrained to move on a smooth surface under no force except the normal reaction. Prove that its path is a geodesic. r Solution: It r is the position vector of a moving point, and the parameter t is the r r time, then its velocity vector is r&and the acceleration vector is & r&. Since the only force r acting on the particle, the normal reaction & r&must be along the normal to the surface. r Again, since r&is tangential to the path of the particle, it must be also tangential to the surface. r r& 1 d r&2 r  r&.& r 0  r 0  r&2 Constant 2 dt r  The speed s&r = constant = c (say) r r r r r dr r& d r dr ds Now r   . ct , where t  is the unit tangent to the path of the particle. dt ds ddt ds r& 2 r& 2 r r & r c t c kn , where n is the unit principal normal to the path of the particle. This shows that the principal normal to the path of the particle has the same direction as r & r&, which is along the surface normal. Thus by the normal property, the path of the particle is a geodesic. Example: 5.2 Prove that every helix on a cylinder is a geodesic. Solution: Let C be a helix on a cylinder, whose generators are parallel to a constant r r r vector a . Let P be any point on C. Let t and n be the unit tangent and unit principal r normal to C at P. Let N be the unit surface normal at P. (to the cylinder) r r Now, since C is a helix, we have t . a =constant r r dt r r da  .a t . 0 ds ds rr r i.e., k n.a 0 0, since a is a constant vector. r r r rr r r i.e., n.a 0 . Clearly t . n =0. Thus n is perpendicular to both a and t and is therefore r r along a t



124

r r But both vectors a and t are tangential to the surface of the cylinder at P, and r r r hence a t is along the surface normal N at P. r Thus, the principal normal n at any point P on the helix is parallel to the surface r normal N at P. Hence, by the normal property, it follows C is a geodesic on the cylinder. Now, we shall derive the differential equations for a geodesic using the normal property. r r s r r1u  r2v  . We have Differentiating w.r.t s, we get r r r r 2 r r 2 r r r2v r ru r11u 2r12u  v r22v . The geodesic equations are U  s r .r1 0 and 1 r r V  s r  .r2 0 r r r which now become (taking dot product of r with r1 and then with r2 ) r r r r r r r r r 2 2 r12 u r1.r2  v   r1 .11r  u 2  .r  u v  .r   0 1r 12 2 r 22 r r r2 r r r r r r 2 2 and  r1.r2  u r2 v r2 .11r  u 2  .r  u v 2 r22.r   0 2r 12 Now, we shall evaluate the various scalar products appearing in the above equations r r 1  r2 1  1 r1.r11  r1   E  E1 2 u 2 u 2 r r 1  r2 1  1 r1.r12  r1 2 v E 2 E2 2 v r r 1  r2 1  1 r2 .r12  r2   G  G1 2 u 2 u 2 r r 1  r2 1  1 r2 . r22  r2  G  G2  2  2  2 Again  rr r r r r r .r21 r1.r12 r11.r2  u  1 1 r r r2 .r11   F  E 2 F1  E 2 u 2 2 r r r r r r Similarly  r1 .r12 r1 .r22 r12 .r2  v r r  1 1 r1 .r22   F  G1 F2  G1 v 2 2 Substituting these values, the geodesic equations become 1 1 2   2   Eu Fv  E1u  E2u  v F2  G1  v0 2 2    1  1  1 2 2   Fu GvF1  E1  u G1u  v  G2v 0  2  2   These equations are the general differential equations of geodesics on a surface.   Since EG F 2 0 , these equations can be solved for u and v and we get equations of the form

125

uf  u, v, u ,v    2  vg  u, v, u , v   where f and g will be quadratic functions of u and v   If we solve equations (1) for u , v , we get another form of equations of geodesics, viz., 2 2 ulu 2, mu v nv 0  3  2 2  v u2 u v v0   Where l , m, n,  , , are the Christoffel symbols in the formulae of Gauss. Usually the equation of a curve on a surface is given as a single relation between the parameters. Therefore, we can combine equations (2) in a single equation. Since we have dv dv / ds v   du du / ds u   d 2v u v v u  2  4 2 du u Multiplying second of (3) by u and first of (2) by v and subtracting, we get 2 2 v v u v u nv u u ulv 2u  umv v   0 Putting values from equations (3), we get 2

2

d 2v dv  dv  dv  n   2 m v l 2 5     2 du du  du  du  From the theory of differential equations if follows that there exists a unique solution of this differential equation. Thus, through each point of the surface, there passes a single geodesic in each direction, Furthermore although geodesics are a system of lines associated with a surface like so many other systems (e.y., lines of curvature etc.), they are unlike them that through any point an infinite number of them passes. Check Your Progress ( C Y P ) : 1. What is the normal property of geodesic? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ 5.2. EXISTENCE THEOREMS: The differential equations (2) for the geodesics are simultaneous second order differential equations for the two functions u and v of s and from the theory of such equations, we know that there is just one solution taking prescribed values, for u,v, u ', v ' when t= t0. Thus we have the following theorem: A geodesic can be found to pass through any given point and have any given direction at that point. The geodesic is determined uniquely by these initial conditions. We have also another theorem:

126

Every point P of the surface has a neighbourhood N with the property that every point of N can be joined to P by a unique geodesic arc which lies wholly in N. Definition: A region R of a surface is said to be convex if any two points of it can be joined by at least one geodesic lying wholly in R. The region is simple if there is at most one such geodesic. Thus, the surface of a sphere as a whole is convex but not simple, for the smaller arc as well as greater arc of the great circle through two points are both geodesics. Definition: The geodesic disc with centre P and radius ‘r’ is the set of all points Q on the surface, such that there is a geodesic curve PQ of length not greater than r. Check Your Progress ( C Y P ) : 2. What are convex and simple regions? Give example ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ 5.3. GEODESIC PARALLELS: Suppose for the curves v=constant, we choose a family of geodesics on a surface, their orthogonal trajectories being taken as u= constant. Then we find using the condition  E F=0 for orthogonality of parametric curves, that E E 2=0 i.e., E 2  0 . Hence, E now v depends only on u, so that the metric has the form ds 2 E  u du 2 G  u, v  dv 2 Let u=a and u=b be any two parametric curves of orthogonal family. Let any geodesic v=c intersects them at A and B. Then » AB along the geodesic has the length given by b ds ds  du , since along the curve v=constant, u alone varies, so that u can be taken  a du AB as the parameter  Edu , since ds 2 E  u du 2 , dv being zero. b

a

The important fact to be noticed here is that we obtain the same length for » AB whatever be the curve v=constant is used. This is similar to the case of two parallel straight lives in the plane for which the intercept is the same for any perpendicular line Hence, the orthogonal trajectories are called geodesic parallels. In a plane for a family of geodesics we may take straight lines enveloping a given curve C. For example, the tangents to an ellipse sweep the whole region of the plane exterior to the ellipse. The geodesic parallels are then the involutes of the curve c. As a special case, if we take all straight lines passing through a point as geodesics, then the geodesic parallels arc concentric circles. Let us fix one geodesic parallel and specify any other parallel u=constant by u=s, where s is the distance of this parallel from the fixed parallel (which now can be

127

relabelled as u=0) measured along any geodesic v=const. Then the distance dsbetween two neighbouring parallels becomes ds = du. Thus formula d s 2 Edu 2 Gdv 2 given du 2 ds2 Edu 2 (since dv=0). Thus E=1. This shows that for a given family of geodesics, the parameters can be so chosen that the metric takes the form ds 2 du 2 Gdv 2 . The given geodesics are the parametric curves v=constant and their orthogonal trajectories are u=constant, u being the distance measured along a geodesic from some fixed parallel u, v are then called geodesic coordinates. 5.4. GEODESIC POLARS: In the plane,  = constant constitute geodesics through the origin and r=constant are concentric circles which give the geodesic parallels. Similarly, on a surface we choose all geodesic passing through a fixed point O and their orthogonal trajectories would be the curves u=constant, u being the distance of the orthogonal trajectory measured from O along any geodesic. Thus ‘u’ behaves like ‘r’ in the plane. For v, we take the angle between the tangent to a geodesic and the tangent to a fixed geodesic at O (just like  in the plane). The metric then becomes ds 2 du 2 Gdv 2 . A small neighbourhood of O is almost a plane and the metric must reduce to the form dr 2 r 2d2 i.e., du 2 u 2dv 2 . Hence for points near 0, G is

G 1 . The parametric system u and v introduced here is u 0 u called Geodesic polar coordinate system and the curves u= constant are called Geodesic circle. 2

approximately =u . Thus lim

5.5. GEODESIC CURVATURE : r r For any curve on a surface curvature vector at P is r   kn , where k is the r r r r r  curvature and n is the unit principal normal. Clearly r  r1 r2 k n N  1 , since any r r r vector at P is a linear combination of r1, r2 and N . The quantity kn is called the normal r  curvature at P. For a geodesic by the normal property, r  is normal to the surface and r r r r hence t kn N , so that 0 . Thus the vector   , i.e., r1 r2 gives a measure

of the deviation of the curve from a geodesic. r r r r Now, for any curve, we have U r  . r1 and V r   .r2 (the parameter being s). Thus, r r by taking dot product on both sides of (1) with r1 and r2 , we obtain U E F and V F G (2) Therefore and can be obtained by solving these two linear equations and we find that 1 1  2  GU FV and  2  EV FU   3 H H Thus and are intrinsic quantities. The vector   ,  is called the geodesic curvature vector of the curve.

128

r r Since, the tangent vector r is orthogonal to the curvature vector r and also to the r surface normal N , we have from (1) r r r r r r r o r  .r   r1 r2  .r  kn N.r  r r r r r i.e,  r1 r2  .r 0 sin ce N .r 0





Thus the geodesic curvature vector is orthogonal to the curve. Check Your Progress ( C Y P ) : 3. What are Geodesic parallels on the plane and on the surface? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ Components of geodesic curvature: We know that one of the components of the geodesic curvature is given by 1  2  GU FV  H 1  FV   2U  G  H  U  1 u    2U  G F  H v   dT & vV &  , if we take arc length ‘s’ is taken as the Q from the identity uU dt dT parameter for a curve replacing u and v instead of u&and v&, we get 0 from ds 1 du dv 2 2 T  Eu  2 Fu  v Gv , where u , v   u  U v V  0  2 ds ds 1 U  2 .  Fu Gv   H v 1 U T  2 . H vv ly lll , we can prove the other component 1 U T  2 . H u u 1 V  T 1 U  T Also  2 and  2 . H u v H v  u The magnitude of the geodesic curvature vector

,  is 

called the geodesic

curvature kg . This is positive or negative according as the angle between the tangent , v and   , is u

  or i.e., according as the tangent vector  u , v , the geodesic 2 2 129

r curvature vector   , and the surface normal N form a right handed or a left handed system. Hence, from the sine formula, we have k g H  u v . For a geodesic, by the normal property, the geodesic curvature is zero. Conversely , a curve with zero geodesic curvature at each of its points has the normal property and therefore is a geodesic. r r r r r r Now, the unit tangent vector r  is orthogonal to N . Hence, r  , N r  , N form a r r right handed system of unit vectors. Thus, geodesic curvature vector is k g N r s. r r r r Hence, we can write r kn N kg N r  5 r r Taking dot product with N r , we obtain r r r  k g  N , r  ,r   6 We can write this formula in terms so that of a general parameter ‘t’ r r& & r & r r& r sr & rs   Since, r  and r  2 , s& s& r& & r& r r r r  r r  2 s& 1 r r r& Now (6) becomes k g  2 N , r&& , r  7 s& r 1 r r 1 r r r& Since N   r1 r2  , we have from (7), k g  2 N . r&& r H s & 1 r r r r& i.e., k g  2  r1 r2  . r&r& Hs& r r r r& 1 r1 .r& r & r  2 r r r r & & & Hs& r .r r .r













 

 

2

Since

2

T r&r T r&r r .r1 , r.r2 . u&  v& r&r r&r U  t r& .r1 , V  t r& .r2

 T U  t 1  u& We obtain kg  2 T Hs&  V  t v& 1  T  T   2  V  t  U  t  8 Hs&  u&  v&  Geodesic curvature of the parametric curve v=constant. Taking ‘u’ as the parameter i.e., u=t, v=c, so that u&1, v&0 , we have as shown already  T  T E , F , U 1 E1 , V F1 1 E2 and 2 2  u&  v&

130

&&Gv&2 E s&2 Eu&2 2 Fuv

1 T T  Hence kg  2  V  t  U  t  & & & Hs u v  1   1  1    3 E F1  E2 F  E1    HE 2   2  2  1   2E F1 EE2 FE1  3   2 HE 2 Geodesic curvature of the parametric curve u=constant is 1 2GF2 GG1 FG2   3  3 2HG 2  2HG 2  ----------------------------------------------------------------------------------------------------------Formula for geodesic curvature, when the arc length ‘s’ is chosen as the parameter, using the relation u U  s v V  s 0 The formula (8) for kg in this case is 1  T  T  k g   V  s  U  s  H u  v   1 T T U  s   V  s   .  H u  vV  s  

1 T  T v    V  s   .  H u  v u   s  T s 1 V  T  1 V   u v  . .1  H u   u  v  H u 1 2 2 1 Since T   Eu 2 Fu  v Gv   2 2 A homogeneous function of second degree in u’ and v so that  T T u v  2T 2 1 1 2 u  v In a similar manner, we can also prove that s s s 1 U  1 V  1 U  k g  . . Thus k g   . H v H u H v If   , is the geodesic curvature vector, then H  H kg   Fu Gv  Eu  Fv

131

1 From (3),  2  GU FV  H U  V  U  u U    2 c F  2  G F  2  Fu  Gv  H  U H  v  H v Since, u U v  V 0 (s, being the parameter ) k g 1 U Now  FuGv ,sin ce kg   H H v H  Thus k g  Fu Gv  The other result can be proved similarly

Check Your Progress (C Y P ) : 4. Find Geodesic curvature of the parametric curve v= constant. ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

5.6. LIOUVILLE’S FORMULA FOR k g : At each point of a curve C, we also have the parametric curve = constant cutting C at an angle which varies from point to point. Thus becomes a function of s along 1  the curve. Now, the direction coefficients for v= constant are  ,0 and for the given E  curve, they are u , v . Since cos Ell F  lm l  m Gmm and sin H  lm l  m

1 1  u F  v 0 G  0 E E  F  E u  v E 1  H and sin H  v 0  v E  E 1 2 2 Since T  Eu  2 Fu  v Gv  2  T 2 2  E1u2T1u v G1v   u T Eu  Fv  u F Eu Fv 1  T Thus cos  E u  v  u E E E  We have cos E

132

Differentiating w. r. t s,

d 1 d  T  1 dE  T sin      3 ds E ds u  2E 2 ds u H d 1 d  T 1  T v  E1u  E2 v     3   u E ds E ds u  2E 2 dE  E du  E dv H (since  .  . E1u  E 2v  and using the value of sin  v ) ds  u ds v ds E d d  T 1  T Multiplying by E , we have Hv     E1u  E 2v   ds ds u  u  2E d  T  1 i. e, Hv   U    E1u E2v Eu Fv  ds   u  2E T d T  T since u  E Fvand U     u ds  u  u 1 1 2 2 2 2 U   E1u  2 F1u v G1v   FE1u   EE2 FE1  u v FE2v   2 2E 1 1 2 U   2EF1 FE1 EE2  u v   EG1 FE2  v 2E 2E d U 1 1 Thus   2EF1 FE1 EE2 u EG1 FE v ds Hv 2EH 2EH U k g Pu Qvsin ce k g   or  Hv  d k g  Pu  Qv ds 1 where P  2 EF1 FE1 EE2  2HE 1 Q EG1 FE2  2 HE This is called Liouville’s formula for k g . ________________________________________________________________________ If the orthogonal trajectories of the parametric curves v= constant are geodesics, then 2 H is independent of u. E Sol: if in angle made by the given curve with v= constant, then its geodesics are d given by the Liouville’s formula kg  Pu  Qv , ds where P and Q are given by

133

1 P 2EF1 FE1 EE 2  2HE 1 Q EG1 FE2  2 HE Thus, if an orthogonal trajectory of v= constant is taken as the given curve, then  and since this curve, by data, is a geodesic, we have k g 0 . 2 Q Thus Liouville’s formula becomes 0 0 pu  Qv (or) u . Now, the direction v P 1  coefficients for u= constant are  ,0 and for the given curve they are  u , v . If  is E  F  lm l m Gmm the angle between these two directions, then cos Ell  and since  , we have 2 1 1  0 E u F  v 0 G  0 E E  F u Q i.e., 0 Eu Fv  or    Thus EQ FP 0 P E v 1 1 since P  2EF1 FE1 EE 2 and Q  EG1 FE 2  2 HE 2 HE 1 F We have EG1 FE2  2 EF1 FE1 EE2 0 2H 2 HE (or) E EG1 FE2 F  2 EF1 FE1 EE2 0 i.e. E 2G1 2 EFF1 F 2 E1 0

2 EFF1 F 2 E1  G  0 E2   2  F    G   0 u u E   F 2 G   u E 

 0 

F2 i.e., G  is independent of u E

EG F 2 H 2  is independent of u E E

Check Your Progress ( C Y P ) : 5. Write down the Liouville’s formula for geodesic curvature k g . ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

134

5.7. GAUSS- BONNET THEOREM: A region R in a surface is said to be simply connected if every closed curve lying in the region can be shrunk to a point, the shrinking curve always remaining in the region. In the plane interior of a circle is simply connected but the region between two concentric circles is not, for a concentric circle between these two cannot be shrunk into a point without passing out of the region. Let C be a closed curve on the surface and let it consists of arcs A. A1 , A1 A2 , A2 A3 ,....., An 1 An . (where An Ao ). bounding simply connected region R of the surface. Let C be described in the positive sense (i.e. in such a way that the region R lies to the left as the vertices are taken in the order A0 , A1, A2..... An ). Let r be the exterior angle at the vertex Ar . i.e., the angle between the tangent to the arcs Ar 1 Ar and Ar Ar 1 at Ar (taking Ar Ao ). Regarding C as the curvilinear polygon 1, 2,3.....n are thus the exterior angles at the vertices A1 , A2 , A3 ....., An . At each point of C, excepting perhaps the vertices, we have the geodesic curvature kg (A vertex being the junction of the two curves, the value of k g may differ for both curves at that point and thus a unique k g may not exist. This is similar to the non - existence of a unique tangent at such a vertex) The line integral  kg ds , therefore can be calculated by adding the integrals along c

n

the separate arcs. The excess of C is defined as e C 2r  k g ds which is r 1

c

clearly an invariant, not depending on the particular parameter used for the representation of the surface. n d For a plane curve k g is the curvature and  kg ds r is the total angle ds r 1 through which the tangent turns in describing curve once. This is 2for a plane closed curve and in particular for a closed polygon. Thus e C is zero for a closed plane curve. Since excess is defined using intrinsic quantities we deduce that on any surface isometric with the plane, the excess of a simple closed curve is zero. This gives us the idea that the excess of a curve may be used to define the departure of a surface from a plane i.e., the curvature of a surface. Through each point of the curve C, there passes a member v=constant. Starting from a point A on C as we complete the circuit C, we come back to the original member at A. Therefore if the tangent to C at A make some angle with the member v= constant at A then as c is described, the tangent changes direction and finally comes back at A to make the same angle , increased by 2, with the member v=constant at A. n

Thus

dr 2, where  is the angle in Liouville’s formula. Thus, using  c

r 1

d this formula, we have kg  Pu  Qv , so that , ds

135

k ds  d Pdu Qdv   g

C

C

C

n

Hence , ex C 2r  d pdu Qdv  r 1

C

C n

 d r 2 P du Qdv , sin ce  r 1

C

Now by Green’s theorem,

Q P 

P du Qdv    du dv   u v 

C

R

for a simply connected region R bounded by C. Thus, putting ds H du dv for the surface elements. We get 1 Q P  e C     Hdu dv H R u v 

1 Q  P   Kds  1   ds    H R  u  v R 1  Q  P Where K     2 H  u  v We now show that K is unique. Assume that ex C  Kds. for every curve C. Then  R

K k ds 0   R

If K k for at least one point P, then let K k , for definiteness. Since K and k are continuous throughout a small region R, containing P, K k 0 . Hence K k ds 0 . Similar result holds if K k at P. These contradictions show that K   R1

must be equal to k at every point. Thus K is independent of the parametric system and intrinsic i.e., it is an intrinsic geometrical invariant. K is called the Gaussian curvature of the surface. For any region R, whether simply connected or not,  Kds is called the total curvature of R. The result (1)  R

is called Gauss- Bonnet theorem, which states that “for any curve C enclosing a simply connected region R, the excess of C is equal to the total curvature of R”. Check Your Progress ( C Y P ) : 6. What is simply connected region? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

136

Example 5.3: If AB, BC, CA are geodesic arcs, then ABC is called geodesic triangle. If this encloses a simply connected region R, the excess is 2 A  B  c A B C where A, B, C are the interior angles of the triangle. Thus the excess of ABC is the excess of A+B+C over the Euclidean value . For, example, let B and C be two points on the equator differing in longitude by  and A, the north pole. Then the excess of the spherical triangle ABC is 2  3  . 2 2 Similarly for a geodesic polygon the total curvature = 2-( sum of the exterior angle ) 2n(sum of the interior angles) excess of sum of the interior over  n 2 , n being the number of sides of the polygon 5.8 . GAUSSIAN CURVATURE: If P is a given point and the area of a geodesic triangle ABC containing P, then clearly at P, the Gaussian curvature A B C  K  lim as seem from (1) A, B, C  P  On a sphere of radius ‘a’ the geodesics are great circles and the area of a geodesic ABC is known to be a2 A B C . Thus the Gaussian curvature K must be the same at 1 2 all points of a sphere and  2 . We also see that the dimension of K are  length  , for a the total curvature  Kds for a region R has zero dimensions and ds having dimensions  R

(length)2. Thus, the total curvature of a sphere of radius ‘a’ is clearly

1 4a2 4 2 a

1  Q  P Now from (2) , K     , where P and Q are given by H  u  v 1 1 P 2EF1 FE1 EE 2 and Q  EG1 FE 2  2 HE 2 HE E G When the parametric curves are orthogonal, F=0 and thus P  2 and Q  1 2H 2H 1 G  E   Hence, K    1   2  , where now H  EG 2H  u H  v H   ________________________________________________________________________

137

Surfaces of constant curvature: A surface of constant curvature is one for which every point has same Gaussian curvature. Check Your Progress ( C Y P ) : 7. What is a surface of constant curvature? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

5.9. MINDING’S THEOREM: Two surfaces of the same constant curvature are locally isometric. Proof: we are concerned with local properties of surfaces i.e., while discussing correspondence between two surfaces (say) S and S , we shall consider correspondence between parts of the surface only and not the whole of the surfaces. We assume that each part has a parametric system. Let the point (u, v) on S and  u , v on S correspond, then

, v are single valued u u u u, v , vv u, v

functions

of

parameters

u

and

v,

so

that

Two surfaces S and S are said to be isometric if there exists a correspondence     u u  u, v , v v  u, v between their parameters, where u and v are single valued functions of u and v and

 u ,v  0 , such that the metric of S transforms into the metric  u, v 

of S To prove the Minding’s theorem, we show that if S is a surface with the constant curvature K0 then i) if K0 0, S is isometric with a plane 1 ii)if K0  2 S is isometric with a sphere of radius ‘ a’ . a 1 iii) if K0  2 S is isometric with a certain surface of revolution called pseudo sphere. a Then the theorem for two surfaces s and s follows by mapping each surface isometrically onto the same plane (or) sphere (or) pseudo sphere, such that point P on S and P on S correspond to the same point. Let P be a given point of S of constant curvature k 0 and let C be a geodesic through P. Let v= constant be the geodesic orthogonal to C and let u= constant be their orthogonal trajectories. Thus u= constant include C, which can be taken as u=0. Then in the neighbourhood of P, the metric has the form ds 2 du 2 g 2 dv 2 .  Since, now u=0 is the geodesic C, we have G1   g 1 0, when u=0 (Q on a general  u

138

surface, the necessary and sufficient condition for the curve u=constant to be a geodesic is GG1 FG2 2G F2 0 ) Also, since for u=0, ds=dv, we have g=1 when u=0 g Thus from the result K  11 , we see that g satisfies the second order partial g

g1  0 differential equation g11 k 0 g 0 with boundary conditions gu 0 1 and  u 0 Case (i): k0=0. Then the differential equation gives g11 0 which implies that g1 is independent of u. But g1 0 when u=0g1 0 . This implies that g is independent of u But g=1 when u=0  g 1 Thus the metric becomes ds 2 du 2 dv 2 , which is the metric of a plane region with u and v as Cartesian coordinates. Hence, a neighbourhood of P on the surface is isometric with a plane region. 1 Case (ii): k0  2 a 1 The differential equation is then g11  2 g 0 a 2 g 1 i.e.,  g 0 which on integration gives u2 a2 u  u  g u , v A  v cos  B  v sin   a  a  Now, g1 =0 when u 0  B 0 and g 1, when u 0  A 1 Hence g cos u and the metric becomes ds 2 du 2 cos 2 u du 2 a a Applying the transformation u a  u , v av this metric becomes 2 ds 2 a 2 du 2 a 2 sin 2 udv 2 , which is the metric of a sphere of radius ‘a’. Hence, a neighbourhood of P in isometric with a spherical region 1 Case (iii): k0  2 a 1 The differential equation is g11  2 g 0 , whose solution is a u  u  g u , v A  v cosh  B sinh   a  a  g1 0 when u 0  B 0 g1 0 when u 0  A 1







u  u  2 Thus g cosh  and the metric becomes ds 2 du 2 cosh2   dv a  a  Applying the transformation u au and v v , we get

139



u 2 dv  1 a Now, the metric of the surface of revolution of the curve x g  u , y 0, z f  u ds 2 a 2 du 2 cosh 2

about the z-axis i.e., r  g  u cos v , g  u sin v , f  u and the metric is given by ds 2  g12 f12  du 2 g 2 dv 2    2

comparing  1 and

2 , we find that g  u a cosh u and  u g12 f12 a 2  or f12 a 2  1 sinh2 u (or) f  u a  1 sin h 2 u 0

du

Thus a small neighbourhood of P on the surface is isometric with the surface of u

revolution of the curve x=a cosh u , y=0, z a 1 sinh 2 udu about the z-axis (called 0

the pseudo sphere) ________________________________________________________________________ The surface generated by the tangents to any space curve is a surface of constant zero curvature: r Sol: If r  s is the position vector of any point on the curve in terms of the arc r length, then the position vector R of any point on the surface is given by r r r R s, v t  s vt  s , where s and v are the parameters. r 2  R r r 2 Thus E   t vkn  1k 2 v2 s  Where k k  s is the curvature of the curve. r r r R  R r r F .  t vkn  .t 1  s  v r 2 R  r 2 G   t 1 s  So that 1 k 2v 2 EG F 2 K 2v 2 1 k v Then P  2EF1 FE1 EE 2  2 2 k 2 HE 1 k v 1 K and Q  FE 2 GE1  2 2HE 1 K v0

K v 1 dQ  P K so that   2 ds  v 1K 2v 2  2 2

1  Q  P Hence, K    0 H s v 

140

Thus the Gaussian curvature at every point of the surface generated by the tangents is zero. 5.10. CONFORMAL MAPPING: A homeomorphism is a one – one onto continuous mapping, whose inverses is also continuous. Where a one-one onto mapping exists between the points of two surfaces either surface is said to be mapped onto the other, e.g., earth’s surface can be mapped onto a plane paper, such that each point of the paper corresponds to one and only one point on the earth’s surface. Similarly the surface of cylinder is mapped on that portion of a plane into which it can be developed. In these examples, there is similarity of the corresponding small elements. When this relation holds, the mapping is said to be conformal. * A surface S is said to be conformally mapped onto a surface S , if there is differentiable homeomorphism of S on S * , such that angle between any two curves at an arbitrary point P on S equal to the angle between the corresponding curves on S * . Note: An isometric mapping preserves both distances and the angles, whereas a conformal mapping just preserves angles. A one- one correspondence of P (u, v) on S and Q u , v on S * is said to be differentiable if u, v have continuous partial derivatives w.r.t u and v and vice versa. When this correspondence is also homeomorphism between S and S * , we say that it is a differentiable homeomorphism. By considering small triangles, we find that isometric maps preserve both distance and angles whereas conformal maps preserve angles but not distance in general. * Let us now find the condition for a mapping of S on S to be conformal. Let S and 2 S * have the metrics ds 2 Edu 2 2 Fdudv Gdv 2 and ds  E du2 2 F dudv G dv2 , where the corresponding points P and P* have the same parametric value. Let  l1, m1  ,  l2 , m2 be the direction coefficients for two direction through P and , l , m be the corresponding direction coefficients of P . If  and are the l , m   1

 1

 2

 2

angles between these two pairs of directions, we have  *  *        cos El1l2 F  l1m2 l2m1 Gm1m2 and cos  E l1 l2 F  l1 m2 l2 m1  G m1 m2 . du du ds ds But l1   .  l1  1 where   ds ds ds ds 2    Hence cos   l1 m2 l2 m1 G m1 m2  E l1 l2 F  

Since the mapping is conformal * (or) cos cos * , so that we E F G 2 have *  *  *  E F G This is necessary condition for the mapping to be conformal and evidently it is E F G also sufficient, since if the condition *  *  * 2 is satisfied coscos * , so E F G * that  .

141

The scale factor is in general a function of u and v. When is a constant, the conformal mapping is a similarity mapping and inparticular if 1 , the mapping becomes an isometry. Check Your Progress ( C Y P ) : 8. When do we say that a surface S is said to be conformally mapped onto a surface s*? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ Theorem: Every point on a surface has a neighbourhood which can be mapped conformally on a region of the plane. Proof: Let S be the given surface with the metric ds 2 Edu 2 2 Fdudv Gdv 2 . At any 2 point P, there are two imaginary directions given by ds 0 i.e. Edu 2 2Fdudv Gdv 2 0 . These are called the isotropic direction at P and since EG F 2 0 , it follows that these directions are always distinct. Now, if the curves along these directions are chosen as the parametric curves, the metric becomes ds 2 du dv (since the parametric curves correspond du 0 and dv 0, so that E 0 G , where we have put 2F . Putting u U iV and v U iV where U and V are real, the metric becomes ds  dU 2 dV 2 . Comparing this with the metric of the plane ds 2 dx 2 dy 2 we see that x U , y V gives a conformal mapping of a region of the given surface on a region of a plane. Hence the theorem. 5.10.1 Corollary: Every point on a surface has a neighbourhood which can be mapped conformally on some neighbourhood of any other surface. This follows immediately, since a neighbourhood of a point P on S and a neighbourhood of point P * on S* can both be mapped on the same region of the plane. The metric viz. ds 2 2  dU 2 dV 2  (where we have taken 2 ) is said to correspond to an isothermic system of parameters. We will now show that there are infinite number of isothermic system of parameters, each system corresponding to an analytic function of n complex variable. Let (u, v) and (U, V) be two different isothermic system of the same surface. 2 2 2 2 2 2 2 Then ds   du dv   dU dV  Since dU U1du U 2 dv and dV V1du V2 dv, we have

2 2 2 2  du dv   U 12 V12 du 2 2U 1U 2 VV1 2 du dv U 22 V22 dv 2   

142

 2  from which, we have U V U V  2 and U1U2 VV 1 2 0 V  2 1

2 1

2 2

2 2

Hence, if follows that either U 1 V2 , U 2  V or U 2 V1 , U1  V2 comparing this with the cauchy- Riemann equations  u  v  u  v  ,  , we find that x y y x U iV (or ) U iV is an analytic function of u iv Hence, if u= constant and v= constant are isothermic, any other isothermic system has parameters U, V which are the real and imaginary parts of an analytic functions of the complex variable u+ iv. Thus, for each isothermic system, there is a natural conformal mapping of the surface on the plane. Hence, we see that from one isothermic system of parameters, we can construct infinitely many other such systems, using various analytic functions. Check Your Progress ( C Y P ) : 9. Prove that every point on a surface has a neighbour hood which can be mapped conformally on a region of the plane ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ 10) What is the distinction between isometric maps and conformal maps ? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

5.11. SUMMARY: Using the normal property of geodesics, we can find out whether a given curve on a surface is a geodesic or not. Through existence theorem, the geodesic is determined uniquely by the initial conditions viz., “A geodesic can be found to pass through any given point and have any given direction at that point. Geodesic parallels and geodesic curvature are studied well. The total curvature of the region, whether simply connected or not is studied through Gauss-Bonnet theorem. Surfaces of constant curvature is known through Gaussian curvature. Conformal mapping between the surfaces are studied well.

143

5.12. KEY WORDS: 1)

Convex region : A region R of a surface is said to be convex, if any two points of it can be joined by at least one geodesic lying wholly in R.

2)

Simple Region : A region R is simple, if there is at most one geodesic wholly lying in R. The surface of a sphere as a whole is convex but not simple, for the smaller arc as well as greater arc of the great circle through two points arc both geodesics.

3)

Geodesic disc : The geodesic disc with centre P and radius ‘r’ is the set of all points Q on the surface, such that there is a geodesic curve PQ of length not greater than r.

4)

Geodesic Polar Coordinate System : In the plane,  = constant constitute geodesics through the origin and r = constant arc concentric circles which give the geodesic parallels. Similarly on a surface, we choose all geodesic passing through a fixed point O and their orthogonal trajectories would be the curves u = constant , u being the distance of the orthogonal trajectory measured from O along any geodesic. Thus ‘ u ‘ behaves like ‘ r’ in the plane. For v, we take the angle between the tangent to a geodesic and the tangent to a fixed geodesic at O ( just like  in the plane) . The metric then becomes 2 2 2 ds du Gdv . A small neighbourhood of O is almost a plane and the metric must reduce to the form dr 2 r 2 d2 i.e., du 2 u 2 dv2 . G 1 . The u 0 u parametric system u and v introduced here is called Geodesic polar coordinate system and the curves u = constant are called Geodesic circle.

Hence, for points near O, G is approximately = u2 . Thus lim

5)

Geodesic Curvature Vector : r r  For any curve on a surface, curvature vector at P is r  kn , where k is the r r r r r curvature and n is the unit principal normal. Clearly r r1 r2 k n N , since r r r any vector at P is a linear combination of r1 , r2 and N . The quantity kn is called r  the normal curvature at P. For a geodesic, by the normal property , r  is normal to r r the surface and hence, t kn N , so that 0 . Thus the vector r r ,  i.e., r1 r2 gives a measure of the deviation of the curve from a geodesic.  The vector   , is called the geodesic curvature vector of the curve.

144

6)

Geodesic Curvature kg :

 , is called the geodesic The magnitude of the geodesic curvature vector  curvature kg . 7)

Simply Connected region: If any closed curve within a region can be shrunk to a point without passing out of the region, then the region is called simply connected region .

8)

Surface of Constant Curvature : It is one for which every point has same Gaussian curvature.

5.13. ANSWERS TO CHECK YOUR PROGRESS: 1. The principal normal of a geodesic at any point is in the direction of the normal to the surface at that point and every curve having this property is a geodesic. This property is called the normal property of geodesics. 2. A region R of a surface is said to be convex, if any two points of it can be joined by at least one geodesic lying wholly in R. The region is simple, if there is at most one such geodesic. The surface of a sphere as a whole is convex but not simple, for the smaller arc as well as the greater arc of the great circle through two points are both geodesics. 3. In the plane =constant constitute geodesics through the origion and r= constant are concentric circles which gives the geodesic parallels. Similarly, on a surface, we choose all geodesics passing through a fixed point O and their orthogonal trajectories would be the curves u= constant, u being the distance of the orthogonal trajectory measured from O along any geodesic. 4. Taking u as the parameter i.e., u= t, v=c, so that u&1, v&0 , we have as shown T  T 1 1 already E , F , U  E1, V F1  E 2 and u&  v& 2 2 s&2 Fu&2 2Fuv &&Gv&2 E 1  T  T  Hence kg  2  V  t  U  t  & & & Hs  u  v  1   1  1    3  E F1  E2 F  E1    HE 2   2  2  1  2 EF1 EE2 FE1  3    2 HE 2

145

5. If at each point of a curve C, we have the parametric curve= constant cutting C at an angle which varies from point to point, then the Liouville’s formula for d geodesic curvature is given by k g  Pu  Qv , where ds 1 P 2EF1 FE1 EE 2  2HE 1 Q EG1 FE2  2HE 6. A region R in a surface is said to be a simply connected, it every closed curve lying in the region can be shrunk to a point without passing out of the region. 7. A surface of constant curvature is one for which every point has same Gaussian curvature. *

8. A surface S is said to be conformally mapped onto a surface S , if there is differentiable homeomorphism of S on S* , such that angle between any two curves at an arbitrary point P on S equal to the angle between the corresponding curves on S * . 9. Let S be the given surface with the metric ds 2 Edu 2 2 FdudGd2 . At any point P, there are two imaginary directions given by ds 2 0 i.e., Edu2 2 FdudGd2 0 . These are called the isotropic directions at P and since EG F 2 0 , if follows that these directions are always distinct. Now, if the curves along these directions are chosen as the parametric curves, the metric becomes ds 2 dud (since the parametric curves correspond du 0 and d0 , so that E = 0 = G, where we have put 2F . Putting u U iV and U iV , where U and V are real, the metric becomes ds  dU 2 dV 2 comparing this with the metric of the plane ds 2 dx 2 dy 2

we see that x = U, y = V gives a conformal mapping of a region of the given surface on a region of a plane. Hence the result. 10.

The isometric maps preserve both distance and angles, where as conformal maps preserve angles but not distance in general

146

5.14. TERMINAL QUESTIONS:

1.

A particle is constrained to move on a smooth surface under no force except the normal reaction Prove that its path is a geodesic.

2.

Prove that every helix on a cylinder is a geodesic.

3.

Write short notes on Geodesic parallels.

4.

Obtain the expression for geodesic curvature kg

5.

State and prove Gauss-Bonnet theorem.

6.

State and prove Minding theorem related to Gaussian curvature.

7.

Prove that every point on a surface has a neighbourhood, which can be mapped conformally on a region of the plane.

5.15

FURTHER READINGS:

1.

‘Lectures on classical Differential Geometry’ by D.T. Struck, Addison – Wesley , Mass 1950.

2.

‘Differential Geometry’ by F. Goreux , S.J.., New central Book Agency , Calcutta.

147

UNIT – VI GEODESIC MAPPING, ASYMPTOTIC LINES, RULED SURFACES AND DEVELOPABLE STRUCTURE: Learning Objectives 6.0.

Introduction

6.1.

Geodesic Mapping.

6.2.

Ruled Surface : ( Developable and skew )

6.3.

Lines of Curvature .

6.4.

Conjugate Directions.

6.5.

Asymptotic Lines.

6.6.

Parameter of Distribution of a ruled Surface.

6.7.

Formula for Central point and the equations of the line of Striction.

6.8.

Summary.

6.9.

Key Words.

6.10. Answers to Check Your Progress. 6.11. Terminal Questions. 6.12. Further Readings. LEARNING OBJECTIVES: 1.

Geodesic mapping

2.

Ruled surfaces (Developable and show)

3.

Lines of curvature

4.

Conjugate directions

5.

Asymptotic lines

6.

parameter of distribution of a ruled surface

7.

Formula for Gaussian curvature.

148

6.0. INTRODUCTION: Geodesies plays an important role in surface theory and mapping of surfaces. We are going to discuss special types of surface called ruled surface. We classify the ruled surfaces into two types. A type of curve called asymptotic line is of great importance and it is to be discussed in terms of conjugate directions. The envelopes of osculating plane, normal plane and the rectifying plane are of importance in differential geometry. 6.1. GEODESIC MAPPING: If a surface S is mapped onto a surface S* by a differentiable homeomorphism, which takes geodesics on S into geodesics on S*, then the mapping is called geodesic mapping. We know that geodesics on a sphere are great circles and on the plane the geodesics are straight lines. If we project the points on sphere from its centre onto a tangent plane, then the great circles on the sphere (actually on the lower hemi-Sphere) are mapped into straight lines on the tangent plane. Thus, this projection is a geodesic mapping. This type of mapping of the earth’s surface on a flat Atlas takes the shortest path on the earth’s surface between two points into the straight lines between their images. Theorem: If a mapping is both geodesic and conformal, then it necessarily is an isometric or else a similarity mapping. Proof:

Consider a system of geodesic coordinates on S so that = constant is a family of geodesics on S, then the metric becomes ds 2 du 2 G d2 . If the mapping from S to S* is conformal, then on S* the metric would be ds *2 (u ,)  du 2 Gd2   .

Since, again the mapping is geodesic, the image of the geodesics =Constant on S must be geodesics on S*. The condition for this would be E * E2* F * E1* 2 E * F1 0 Since F * 0 this give s F* F2 =0    i.e  0. sin ce,0,we get 0 . Thus depends on u only.    Now, consider a geodesic on S cutting the curve = constant at an angle . Then, d 1 by Liouville’s formula kg  Pu Q , where P  2 EF1 FE1 EE2  ds 2 HE *

149

1 G Q EG1 FE2 . since, E=1, F=0, we have P=0, Q  1 . Also, since k g = 0 for 2HE 2H G1 geodesic, Liouville’s formula becomes 0 d d . Since, the mapping is both 2H * G geodesic and conformal, the corresponding equation for S* is o = d 1 d. 2H * * G G Thus, we have 1  1 H H* Since H  EG F 2  G  E 1, F 0  And H *  E *G * F *  G  E * , F * 0, G * G  2

G  G G G  1 1   1 1  G 1  G G G G  Which gives G 1 =0 (or) 1 =0, since G 0 i.e, is also independent of u i.e., is a  constant. Thus the mapping is a similarity, which becomes an isometry if =1. G1

We have

Tissot Theorem: In any non- conformal mapping of a surface S on another surface S* given by a differentiable homeomorphism regular at each point, there exists at each point P of S, a uniquely determined pair of orthogonal directions, such that the corresponding directions on S* are also orthogonal. Proof: Choose parameter on S and S* in such a manner that corresponding points have the same parameters. Let  l1 , m1  , l2 , m2 be two orthogonal directions at a point P on S and

, l ,m  l , m  * 1

* 1

* 2

* 2

be the corresponding directions at P* on S*. Since,

l1, m1 and l 2, m2 are orthogonal, El1l 2 F l1 m2 l 2m1 Gm1m2 0 (or )

El

2

Fm 2  l1  Fl 2 Gm 2  m1 0 __________(1)

Now, the corresponding directions on S* will be orthogonal if * * * * * * * * E l l F  l1 m2 l2 m1 G m1 m2 0 _________(2) . Since P and P* have the same * * * 1 2

parameter, du du ds ds 1*  *   * l1 ,where  * and ds ds ds ds Thus (2) becomes . E *l1l 2 F *  l1 m2 l 2m1  G * m1m2 0

similarly

i.eE* l2 F* m2 l1 F* l2 G* m2 m1 0 __________(3)

Eliminating l1 , m1 between (1) and (3), we obtain 150

m1* m1l , and n1* n1 

El2 Fm2 El Gm2  *2 , * * F l2 F m2 F l2 G* m2 Which on simplification gives EF* E* F l22 EG* E*G l2 m2 FG* F *G m22 0 ___(4)

This is a quadratic equation and the orthogonality Condition viz., ER-2FQ+GP=0, where P EF * E * F , 2Q EG * E *G , R FG * F *G is satisfied as can be verified by substitution. Thus (4) determines two orthogonal directions at each point. The discriminant of (4) is

EG E G 4 EF E F FG F G   EG E G 4  EF E F  1   F EG E G G  EF E F   E *

2

*

*

*

*

2

*

*

*

*

2

*

*

*

 EG E G 4 *

*

*

*

2 F G * * * * * * EF E F  EG E G 4  EF E F    E E

2

2F  *   EG E *G   EF * E *F    E   2 2 2 F G 4 2  EF * E *F 4  EF * E *F  E E 2 EG F 2  * 2F  *   * * * *  EG E G   EF E F  4 EF  EF   2  E E  

____(5) 2

E F G  *  * do not hold. * E F G * * * * This means that both  EG E G and EF E F  are not both zero. Also, since

Since mapping S  S * is non – conformal, the relations

E 0 and EG F 2 0, we see from (5) that the discriminant is strictly positive. Thus, the roots of the quadratic(4) are real and disinct. Hence, at each point P on S, there are two orthogonal directions on S* which are also orthogonal. Hence the theorem.

Dini’s Theorem: Two surfaces S and S* which are mapped geodesically in each other by a nonconformal mapping must have line elements which can be written in the forms ds 2  U V  du 2 dv2 , ds V U *2

i

1

U

u alone V, a function of  alone.

151

1

 1

du v dv 2

2

Where U is a function of

Proof: Let f be a differential homeomorphism of S onto S* , which is non-conformal. Then, by Tissot’s theorem, through each point P of S there exists a uniquely determined pair of real orthogonal directions, so that the corresponding directions on S* are also orthogonal. If the parametric curves are chosen along these directions, then the metrics S and S* will be of the form *2

ds Edu Gd , ds E du G dv First, we shall obtain the equation of geodesic on s with parameter u i.e when u=t, u&1, & u&0 1 2 2 Now T  Eu G  2 T 1 1 2 &2   &2  Hence   E1u& G1 E1 G1 u 2 2  T 1 1  T &2  E2 G2 &2  ,   E2u&2 G2 Eu&E v 2 2  u& T  G & v d T   T d &2  ThenU      E 12  E1 G1 dt u& u dt 2

2

2

*

2

*

2

 E1u&E2v&12  E1 G1 &2  &12 E 2 12 G 2 &2 E1 E 2 12 E 2 E 2 &12 G2 &2

d T   T d 1 & &2  and V      G   E2 G2 & dt  v  v dt 2 1 & & & &  &2  G G1u& G 2 E2  G 2 2 1 1 G & &G1 &G2 &2  E 2  G2 &2 2 2 1 1 & G2 &2 G & &  E2 G1 2 2  T  T Hence, the equation to the Geodesic on S is V U 0  u&  & 1 1 1 1 & G2 &2 G & &E  E1 E2 & G1 &2  & i.e   E2 G1   G10 2 2 2  2  1 1 1 2   1 & & GG1 &3  EG2 GE2  & EG1  GE1  & EE2 0 EG   2 2 2    2 1 G 2 1 G2 E2  2 G1 1 E1 i.e ,  & & 1  &     &   2G 2 G E  G 2 E Similarly the equation of geodesic on S* is

152

 1 E2  & 0 _________(1)   2 G

1 G *1 2 1 G2* E2*  2 G1* 1 E1*  1 E2* & & &  *  *  &  *  *  & * 0 _________(2)  *    2 E 2 G E G 2 E     2G Since the mapping is geodesic, equations (1) and (2) must be identical. Thus comparing coefficients, we obtain. G1 G1*  ____________(3) E E* E2 1 G2 E2* 1 E2    ____(4) E 2 G E * 2 G*

1 E1 G1 1 E1* G *1    ____(5) 2 E G 2 E * G* E 2 E 2*  * _________(6) G G  E 2   Now log    2 log E logG   G  

2 G  E2  2 E G E 1 G  2  2  2  E 2 G  E *2 1 G *2  2  *  * U sin g (4) 2G  E  E *  log  *    G

2

   

E * E2 And integrating, we obtain log log  * G G 

2

 2 log U  

2

E 2 E* 2  * U ________(7) G G Where U is some function of u. G 2 G* 2 Similarly using (5), we can obtain  * V _____(8) E E  Where Vis some function . From (7) and (8) we obtain E E *U 2V ______(9)

Or

and G G UV ________(10) Using (9) in (3) we have G1 U 2VG1* *

2

153

 * (G )  u  G U 2V  2  u UV U 2V

UV U V

 u sin g (10) 

G1 GU 1V 2  G1U V1G  U 2V 4 V G U  1 1 or G1V G1U U1G G U V Integrating, we get log G = log(U V ) log V 2 2

2

i.e., G  U V  V 2 _______(11) Where V is some function of . Similarly using (10) in (6) and proceeding as above we get E (U V )U 2 Where U is some function of u. Thus the metric on S becomes ds 2 Edu 2 Gd2  U V  (U 2du 2  V 2dv 2 ),u sing (11)and (12)

And metric on S* becomes 2 ds* E *du1 G *d1 E G  2 du 2  2 d2 , u sin g (9) and(10) U V UV 2 U U V  V2 U V   2  du  d2 U sin g (11) and (12) U 2V UV 2 2 2 U V  U V 2 2   du  d  U V U V  Changing the parameters to u* and * u*  U du and *  V d, The métrics on S and S* will reduce to





ds 2  U V du*2 d* and 2

ds *  V 1 U 1  U 1du * V 1d*2 2

2

This completes the proof of Dini’s theorem. Check Your Progress ( C Y P ) : 1. What is geodesic mapping between two surfaces? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

154

6.2. RULED SURFACE: ( Developable and skew) A ruled surface is a surface which is generated by the motion of one parameter family of straight lines and the straight line itself is called its generating line. Cones, cylinders and conicoids are special forms of ruled surfaces. There are two distinct classes of ruled surfaces, namely, those on which consecutive generators intersect and those on which consecutive generators do not intersect; these are called developable and skew surfaces respectively. A surface generated by one parameter family of planes is called a developable surface or simply a developable. Developable associated with space curves: Since the equation to three principal planes namely osculating plane, normal plane and rectifying plane contain only a single parameter which is usually taken to be the arc length s, hence their envelopes are developable surfaces and they are respectively called osculating developable (or) tangential developable, polar developable and rectifying developable. Also the generators of polar developable and rectifying developable arc called polar lines and rectifying lines respectively. Equation of the ruled surfaces: Q If C is any curve on the ruled surface such that it meets each generator precisely once, then C is called a base curve or a r directrix (These are many in number ) and the ruled surface is R determined by any curve C and direction of the generator at r r r their points of meeting with the curve. Let g (u ) be the unit r vector along the generator at a curved point Q an C and r (u ) , r c the position vector of Q, then R ,the position vector of the o general point P on ruled surface is given by r r r R r g , Where is the parameter and determines directed distance along the generator from C. P

Theorem: The necessary and sufficient condition for a surface to be developable surface is that its Gaussian curvature should be zero. Proof: Necessary condition: Let the surface be developable, to prove that the Gaussian curvature K is zero. For a developable surface to be a cylinder or cone, the Gaussian curvature K=0; and if these be excluded, the general equation of a developable may be written. as, r r r r r R r (s ) t (s ) r t If the suffixes 1 and 2 denote partial differentiation w.r.t. s and respectively , we have

155

r r r r r r r r R1 t kn, R2 t , R1 R2 K b r r r r r r r r r r R11 kn k  b kt k  n , R12 R21 kn , R22 0 r r r r R1 R2 Kb N  H H 2 2 r r r r r r  k  L N .R11  , M N .R 12 0, N N .R 22 0 H 2 LN M Hence Gaussian curvature K  0, Since M N 0 EG F 2 r r Sufficient condition: If the Gaussian curvature 0 for the given surface r r (u ,) to prove that the given surface is developable. We have 0 (or) (LN-M 2)/H2=0 (or) r r r r r r r r r r r r 2 LN-M =0 (or) r1 , N1 r2 , N 2  r1 .N 2 r2 .N1 0(or )  r1 r2  . N1 N 2 0 (by vector



 



   





identity due to Lagrange) (or) r r r r r r r r r 0 (or )   HN N1 N 2 0 (or ) H  N , N , N N 1 1 2    , N1 , N 2 0 sin ce (H 0) ____(1) In view of relation (1) above, there arise following four possibilities: r r i) N , N1 and N 2 are coplanar (or) r r r r r r ii) N1 0 ( or) iii) N2 0 ( or) iv) N1 N2 Now we shall discuss these possibilities: r r i) N , N1 and N 2 are coplanar r r r Since N is a vector of unit length (i.e., constant length ), and so N . N1 0 and r r r r r N .N 2 0 . These relations together imply that N is perpendicular to both N1 and N 2 . r r r Hence, N , N1 and N2 cannot be coplanar. r r r ii) N1 0 . The equation of the tangent plane at any point r  u , v on the surface r r r r r r r r  r r r r r r r  u ,is R r  N 0 we have R r  N r  N  R r  N1 u r r r = 0 (since r1. N 0, N1 0 ) r r r Thus ( R r )  N is independent of the parameter u and hence the equation of the tangent plane contains only one parameter . Therefore, the surface is the envelope of a single parameter family of planes and hence it is a developable surface. r r iii) N 2 0 .Proceeding similarly as in ii) above, the equation of the tangent plane contains only one parameter u. Hence in this case also the surface is developable. r r iv) N1 N2 . Let us change the parameters u,  to u , by the transformation u u  , u u  , we obtain





 

  

156

 

r r r r1  r r r N  N  u  N  r N1      N1  1 N 2 N1 N2 u u u   u r r r r r N  N  u  N  r N 12      N 1  1 N 2 ()  u    u r r r r N1 N 2 0 Q N1 N 2 r These relations show that N , surface normal depends on only one parameter. Hence, by case (iii) above, the surface is developable. Hence the theorem.





Check Your Progress ( C Y P ) : 2. What is ruled surface and what are its classification? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ 6.3. LINES OF CURVATURE: Definition 1: A curve drawn on a surface and possessing the property that the normals to the surface at the consecutive points intersect, is called a line of curvature. Definition 2: A line of curvature on any surface is a curve such that the tangent line to it at any point is a tangent line to the principal sections of the surface at that point. Differential equation of lines of curvature. Since, the direction of line of curvature at any point is along the principal direction at the point, the differential equation of lines of curvature is given by EM FL du2 EN GLdudFN GM d2 0 This equation of lines of curvature can be remembered in the following form of a determinant. d2 dud du 2 E L

F M

G N

________________________________________________________________________ Property of lines of curvature : The normal to any surface of consecutive points of one of its lines of curvature intersect. Conversely, if the normal at two consecutive points of a curve drawn on a surface intersect, the curve is a line of curvature. Proof:

r r r Let N and N +d N be the unit normal vectors to the surface at two consecutive r r r points P( r ) and Q( r dr ) . Let the 157

r r direction of Q( r dr) be( du, d). It the normals at P and Q intersect, then r r r r N , N dN and dr must be coplanar – i.e., their scalar tripple product should be zero. r r r r  Hence,  N,N+dN,dr 0 r r r i.e  N,dN,dr   0 r r r r r r We known dN N1du N 2 d , dr r1du r2 d r r r Substituting these values in the equation  N,dN,dr   0 , r r r r r We get  N,N1du+N2 du,r1du+r2 d  0 r r r r r r r r r r r r du2    dudN,N2 ,r2 d2 0 ______(1) (or )  N,N ,r N,N ,r N,N ,r 1 1 2 1 1 2        r r r r r r But  N,N1,r1  N 1 r1  N  r EM-FL N  2  N1 N 2  T EM FL r r r EM FL   N,N1 ,N2   2   T H r r r EN FM N,N2 ,r1    H r r r FM GL N,N1 ,r2    H r r r FN GM N,N2 ,r2    H The equation (1) becomes FN GM EM FL  2 FM GL EN FM  2  dud d  0  du   H H H H    





(or )  EM FL du 2  EN GL  dud FN GM  d2 0 But, this is the differential equation of the lines of curvature. Thus, the normals at consecutive points of the lines of curvature intersect. Conversely , let the curve be a line of curvature. Just reversing the steps, we get r r r r r r r N , dN , dr 0 which can be written as  N , N dN , dr     0 . This expression implies r r r r r r r that N , N dN , dr are coplanar. i.e., the normals N , N dN , at consecutive points of the lines of curvature intersect. Incidentally, we note that the differential equation of the lines of curvature on a surface can also be expressed by r r r r r r r r r r r r  du 2    dud  N,N ,r N,N ,r N,N ,r N,N d 2 0 1 1 2 1 1 2       2 2,r 





158

Check Your Progress ( C Y P ) : 3. Define line of curvature on a surface and write down the differential equation of the same. ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ A property of lines of curvatures on Developables: Monge’s theorem: A necessary and sufficient condition that a curve on a surface be a line of curvature is that the surface normals along the curve form a developable. Proof:

r r r r Suppose r r ( s) is the equation of a curve on the surface r r ( u,) . Consider r r r r the unit vector N along the normal to the surface r r (u ,) at a point P( r ) on the curve r r r r r r (s ) . Then r is a function of s and also N may be regarded as a function of s. The r position vector R of a point Q on such a normal is represented by r r r s r r R r ( s) N  r N __(1) where s and are two parameters. Thus equation (1) is r r the equation of the surface formed by the normals to the surface r r (u ,) along the given curve. Let suffixes 1 and 2 be used to denote partial differentiation w.r.t.s and respectively. r r r r r  From (1) we have R1 r  N  t N  r  r r r dr r dN  ____(2) R2 N where r , N     ds ds    r r r r r From (2) R21 N  R12 ; R22 0 If M and N are fundamental coefficients of second order of surface (1) we know r r r r r r and HN=   HM=  R , R , R R 1 2 12   1 , R2 , R22 . r r r r r r r r r r  , Since  0 ______(3) HM  t N  ,N,N  t ,N ,N  t ,N ,N       r r r r and HN  , N, 0 t N 0 , HN 0 implies N0, asH0. Hence, the Gaussian curvature of (1) is given by LN M 2 M 2  sin ce N 0 _______(4) 2  2  EG F H Now, we know that a surface is developable if and only if its Gaussian curvature  is zero. M 2 i.e, if and only if 2 is zero. H i.e, if and only if M is zero

159

r r r   i.e, if and only if  t , N , N is zero (use (3) and H 0 ) r r r r r =0 if and only if the curve r r (s ) is a Hence, we are to prove only  t , N, N   r r line of curvature on the surface r r (u ,)

Necessary Condition : If the condition

r r r r r   0 holds, then to prove that the curve r r  t , N , N  s is a   r r line of curvature on the surface r r  u,v  r r r r r r r r r =0 i.e.,  Now  t ,N,N t , N , N .N 0   5    ie ., t N  r r r We know that N is perpendicular to N , therefore N  , lies in the tangent plane to r r r r r r the surface r r  u, v. Hence, t Nis normal to the given surface r r (u ,) and so r r r t N  is parallel to the unit surface normal vector N . Thus in order that equation (5) r r r r holds t N must be zero, since N 0 . r r r Equation (5)  t N  =0 r r  N kt for some function k r r dN dr  k ds ds r r r  dN kdr 0 which is Rodrigue’s formula characteristic only of lines of curvature and therefore the r r r r curve r r ( s) is a line of curvature on the surface r r (u ,) Sufficient condition: r r Conversely, if the given curve r r (s ) is a line of curvature on the surface r r r r r r r (u ,) then we have dN kdr 0 (Rodrigue’s formula). r r r r r r r dN dr r  k 0  N  kt 0  N  kt ds ds r r r r r r  Hence t , N, N  t , N, kt  0    r r r 0 i.e,  t , N, N    So that the condition of developability is satisfied

Check Your Progress ( C Y P ) : 4. State Monge’s theorem ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

160

6.4. CONJUGATE DIRECTIONS: r r Definition: Conjugate directions at a given point P (u ,) on a surface r r ( u,) are defined as follows: Let P and Q be two neighbouring points on a surface, and consider tangent planes at these two points and let PR be the line of intersection of these two planes. Then, as Q  P, the limiting directions of the lines PQ and PR are called the conjugate directions at P.

An analytical expression for two directions to be conjugate:

r r r r r r Let the unit normals at P (r ) and Q (r dr ) be N and N + d N and let u, be uuur r r r the parameters, so that PQ dr r1du r2d r r r r If  r is the vector PR in the limiting position, then R r1  u r2 . Again, since r PR is parallel to the tangent planes at P and Q  r is perpendicular to both r r r r r r r r r r N and N dN , thus N   r 0, N dN . r 0 or  N  r 0 r r r r r r N  r 0 gives N1du N 2 d  r1u r2v 0 r r r r r r r r (or ) N1  r1 duu N1  r2 duN2  r1 d u N2  r2 dv 0



 



(or ) Lduu M  dudu Nd0 (or ) L

du  u u du   M   N 0 d  d 

du u and . The symmetry d   u du du u of the above result shows the if is conjugate to , then is conjugate to , this   d d   property is reciprocal one. du u Again the expression (1) being linear in each of the ratios and implies that d   there is a unique conjugate direction to a given direction. ________________________________________________________________________ Condition that the two directions given by, Pdu2 Qdudv Rdv2 0 to be conjugate: du u Solution: let the two directions be represented by and . d  

This expression determines the conjugate directions

161

du u Q   d   P du  u R   d  P du  u u du  L  M   N 0 d  d  R  Q   L  M  N 0 P  P  (or ) LR NP MQ 0 ________________________________________________________________________ The necessary and sufficient condition for the parametric curves to have conjugate directions is M=0. Solution: The differential equation of parametric curves is dud=0. Comparing this equation with Pdu2 QdudRd2 0, we find P=R= 0, Q=1. Hence, the direction of the parametric curves will be conjugate, if LR+NP-MQ=0 i.e., MQ=0 i.e., M=0 as Q 0. Conversely if M=0, the condition LR+NP-MQ=0 is clearly satisfied since for parametric curves P=0 , R=0. ________________________________________________________________________ The Principal directions (Lines of curvature ) at a point are conjugate and orthogonal: Solution: We know that F=0, M=0 are the necessary and sufficient conditions for the parametric curves to be lines of curvature and M=0 is the necessary and sufficient condition for the parametric curves to the conjugate, It implies that the principal directions (lines of curvature) at a point of the surface are conjugate directions and these directions are orthogonal. Check Your Progress ( C Y P ) : 5. Find the necessary and sufficient condition for the parametric curves to have conjugate directions. ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

6.5. ASYMPTOTIC LINES: Definition : The directions which arc self conjugate are called asymptotic directions and the curves, whose tangents are along asymptotic directions are called asymptotic lines. Differential equation of an a asymptotic line: Suppose at any point  u,, the direction of an asymptotic line is Hence  du, dis a self conjugate direction.

162

du, d.

du  u  is the condition of conjugacy of two directions namely d   du  u u du  L  M   N 0 d  d 

Hence, putting

2

du du  We have L  2 M N 0 d d Ldu 2 2 Mdud Ndu 2 0 ______(1) This is the differential equation of an asymptotic line. ________________________________________________________________________ r r 2 2 We know dN  dr Ldu 2MdudNd r r 2 2 dN dr du  du  d d (or )  L  2M    N   ds ds ds  ds  ds  ds  r r r But the direction coefficients of the direction at the point r of the curve r r (s ) are r r du d  , . Therefore, the necessary and sufficient condition for the curve r r (s ) to be ds ds  asymptotic line is 2 2 du  du  d d L  2M    N    0 ds  ds  ds  ds  r r dN dr i.e.,  0 ________(2) ds ds du Expression (1) being quadratic in gives two asymptotic directions at a point. d 2 They will be real and distinct if M -LN is positive i.e., if Gaussian curvature  is negative . It  is +ve, they are imaginary. They are identical if  is zero and also in this case the surface is developable and single asymptotic line through a point is the generator. The normal curvature in any direction is equal to 2 2 2 2 / Edu 2 FdudGd  it clearly vanishes for the Ldu 2MdudNd 

asymptotic directions (using equation (1) ). These directions are thus the direction of the asymptotes of the indicatrix and hence this name. ________________________________________________________________________ The binormal of asymptotic line is the normal to the surface: r r Proof: The tangent t to the asymptotic line is perpendicular to the surface normal N , we r r have N  t =0 Differentiation w.r.t. ‘s’ yields r r r r N .t N  t 0 r r dN dr r r  N  Kn 0 ds ds r r dN dr But  0 for asymptotic line. ds ds 163

r r  The above equation yields N .n 0 Q k 0  r r Also t .N 0 r r r r r r r Now N.n 0 and t  N 0  N is parallel to t n r r  N is parallel to b

Check Your Progress ( C Y P ) : 6. Show that the binormal of asymptotic line is the normal to the surface. ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ The osculating plane at any point of an asymptotic line is the tangent plane to the surface: r r r Proof: From the above result, we know N Pb i.e., N P to the binormal to the asymptotic r r line. We may choose the direction, such that N b . r r Now N is the normal to the tangent plane at a point on the surface and b is normal to r r osculating plane at a point on the asymptotic line and the relation N b makes them coincident at the point on the curve on the surface and hence the theorem. ________________________________________________________________________ The necessary and sufficient conditions that the parametric curves be asymptotic lines are L=0, N=0 M 0 . Proof: The differential equation of the parametric curves is dud0 ____(1) The differential equation of asymptotic lines are 2 2 Ldu 2 Mdud Nd  0 ________(2) In order that the parametric curves are asymptotic lines, equations (1) & (2) must be identical i.e., we must have . L=0, M 0 , N=0 Conversely if L=0 ,N=0, M 0 , the equation (2) reduces to equation (1). Hence the proposition. ________________________________________________________________________

Curvature and torsion of an asymptotic line: We know that the unit binormal to an asymptotic line is the unit surface normal. r r (or) b N . r r Differentiation of this relation w.r.t. arc length gives b N  , which on using Frenet’s formulae gives r r r r  n N (or) N  .n r r r r r r But n b t (or) n N t r r r r r r  N   N t  N  , N ,t 

164

r r Again, we know t  kn , so that r r r r r r r r r r r     k t  n t  N t  t , N , t  N    , t , t  r r r    (or) k  N , r .r  Check Your Progress ( C Y P ) : 7. Find the curvature and torsion of an asymptotic line. ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ Theorem of Beltrami – Enneper: The torsion of the two asymptotic lines through a point are equal in magnitude but opposite in sign and the square of either is the negative of the Gaussian curvature. Proof: The torsion of an asymptotic line is given by r r r  ,t  N , N  r r r (or ) N , N  ,r    r r r r r  1    N , N u  N r 2 2,ru  1  r r r r r r r r r r r r 2 2   N , N 1, r1  u   N ,N 1,r 2  N ,N 2r, 1  u   N , 2r, 2         N 





(or ) H EM FL  u EN GL  u v  FN GM  ___(1) Now, we suppose that the asymptotic lines are taken as parametric curves, so that L=N=0, M 0 and the above equation (1) reduces to 2 2 H M  Eu  G  2

Edu M

2

Gd  _____(2) ds 2 For the asymptotic line d0 , using ds 2 Edu 2 , we have from equation (2), 2

2

MEdu 2 M     ____(3) HEdu 2 H 2 2 Similarly for the asymptotic line du=0, using ds Gd , we have from (2) M    ____(4) H From (3) and (4), we have    2 . Hence the theorem.

Fundamental Coefficients and Gaussian curvature for a ruled surface: r r r The equation of ruled surface is given by R r g ___(1) r r r Where rand g are functions of the parameter u, and R is the function of two parameters u and we have

165

r r r R r r r  R R1  r&g&, R2  g u  r r r& r& r r& r & R11 r g , R12 g , R22 0 r2 r r 2 r2 r r& 2 r&2, g  g Therefore E R1  r&g& r& 2r& r r r r r r r r&2 r r& F R 1  R2  r&g& g r& g (Q g 1  2 g  g r r  g. g&0) r2 r2 G R2 g 1 r The unit surface normal vector N is given by r r r r r r NH R1 R2  r&g& g r r r r& r&r& r&r  & Also HL  R 11, R1, R2  r g, r g, g r r r r&& r r r r r r r 2 r& &     r&& ,r , g  g& ,r , g  r ,&g&& ,g  g , g&, g&        r r r r r& r& r HM  R12 , R1 ,R 2  r  g ,g  g  r r r r r& r   g&, r&, g  g& ,g, g     r& r & r r& & r r r & r r &   r , g, g g, g, g g 0  Q  ( g g )  r r r r r & HN  R22 , R1 , R2   0 , since R22 0 Therefore N=0 as H 0 . The Gaussian curvature is given by r& r r& 2   r LN M 2 , g , g  Q N 0      H2 H4 Hence, for a ruled surface , is never positive, we known that the necessary and sufficient condition for a surface to be a developable surface is that =0. Hence the necessary and sufficient condition for a ruled surface to be the developable is r r r  r&, g , g&  =0.

 

 

  











Again the necessary and sufficient condition for a ruled surface to be skew is r& r r&  r , g, g   ≠0 r r r r Remark: For a ruled surface, NH  r g& g . This relation implies that the tangent





plane to the ruled surface varies at points on the same generator unless the ruled surface r r r is developable. i.e., unless  r&, g , g&  =0 The asymptotic lines on a ruled surface: The differential equation of asymptotic lines on a surface is given by 2 Ldu 2 Mdud Nd 2 0 ____(1) But N=0, for a ruled surface. Hence , equation (1) of asymptotic lines on a ruled surface becomes 2 Ldu 2 Mdud 0 (or) du  Ldu 2 Md0 ____(2) .

166

From equation (2) it follows that one family of asymptotic lines on a ruled surface is given by du=0 i.e., u= constant. Hence, one family of asymptotic lines is the family of a parametric curves u=constant. Also the parametric curves u= constant are the generators of the ruled surface. Thus, we state: The generators of a ruled surface constitute one family of asymptotic lines. Again from equation (2) it follows that the other family of asymptotic lines (i.e. system of curved asymptotic lines) is given by Ldu 2 Md0 d L   du 2M d 1 r& r r r r r& r& r r r r 2 r&   r r r  r&& ,r ,g   g&,r & ,g   r ,&& g ,g     g g,&& g,          ______(3) &, g , g& du 2 r  





d 2 A BC where A,B,C are du functions of u only. This differential equation is of Riccati type differential equation. We know that the most general solution of this D.E. is of the form CP Q  _____(4) CR S where P,Q, R,S are functions of u and c is an arbitrary constant: The equation (4) is the required equation of family of curved asymptotic lines, and for different values of c, we have different members of the family. Consider four asymptotic lines C1 , C2 , C3,C4 of this family and let these asymptotic The differential equation (3) is of the type

lines be met by the generator u u o . Let the parameter has the values 1 ,2 ,3 ,4 at these four points, Hence from (4), we have C P Q C P Q C P Q C P Q 1  1 ,2  2 , 3  3 ,4  4 C1 R S C2 R S C3 R S C4 R S C C2  C P Q C2 P Q  PS QR  1 2  1   1 C1 R S C2 R S C  S 2C  R  S 1 R

C3 C4 PS QR etc C 4R  S  C3R S      C C2    C3 C4   1 2 3 4  1 which is free from 1 3 2 4  C1 C3 C2 C4  Similarly 3 4 

u0 and hence same for

all generators. But 1 2 is the distance between the points 1 and 2 (since is the distance along the generator from the directrix etc.) . Therefore, the cross ration of the four points, in which a generator is met by four given curved asymptotic lines, is the same for all generators.

167

6.6. PARAMETER OF DISTRIBUTION OF A RULED SURFACE:

Definition :

r r r  r&, g , g&   A function (u) defined by p (u )  r&2 _____(1) is called the g

parameter of distribution of a ruled surface. Properties of Parameter of distribution: i) The parameter of distribution has a constant value at each point of a generator: r r r The parameter of distribution p(u) is defined by 1). The vector r&, g and g& are all functions of the parameter u only and thus they are all independent of the parameter . Hence, p(u) is a function of u only. But along a generator u is constant and hence p(u) has a constant value along each generator. Thus p(u) has a constant value at each point of a generator. ii) The Parameter of distribution p(u) is independent of the particular base curve r r r r r r chosen. By replacing r by r wg i.e., r&by r&wg&in (1) we have r r r r r r r r r r r r r     r&wg&, g , g& r&, g , g& g&, g , g& r&, g , g&   w    r&2 r&2 r&2 g g g r& r r&  p(u ) ( Since  g , g , g 0) This proves our property . iii) The parameter of distribution is independent of the choice of parameter u: Let t be taken as parameter instead of u. Now, the parameter of distribution p(t) is given by. r r r r dr r dg  dr du r dg du  dt , g , dt  du dt , g , du dt   p (t )  r 2  r 2 dg  dg du       dt  du dt  2 r r r r du dr r dg  dr r dg    , g ,   , g ,  dt du du du du  2 r 2  r 2  du dg  dg       dt du  du  p(u ) In particular, if arc length s of the directrix is taken as parameter, then the parameter of r distribution (p) of the generator g(s) through the point r ( s ) of the directrix is r r r r , g, g r  r dg  p r2 Where g  g ds r r r  t , g, g    (or ) p  r 2 g iv) the Gaussian curvature interms of p:

168

The Gaussian curvature K of a ruled surface in given by r& r r& 2  r , g, g  4  ___(2) H r r r Putting the value of  r&, g , g&  from (1) in (2) we have

2 g&4  4 H Then is always negative except along those generators where p=0. Since  = 0 for a developable surface, hence a developable surface is ruled surface for which the parameter of distribution p vanishes identically. Central Point: There exists on each generator of a general ruled surface a special point, called the central point of the generator. It is defined as follows: The central point of a given generator is the point of intersection of the generator and the shortest distance between it and a consecutive generator of the system. Line of striction: The locus of the central points of all generators is called line (curve) of striction. The line of striction lies on the ruled surface.

6.7. FORMULA FOR CENTRAL POINT AND THE EQUATION OF THE LINE OF STRICTION: r r r r r Suppose the equation of ruled surface is R r g _____(1) where r and g are r functions of parameter u and R is the function of two independent parameters u and . Let suffixes 1 and 2 denote partial differentials w.r.t. u and respectively. r r r r r We have R1 r&g , R2 g r r r r r r NH R1 R2  r&g& g _________(2) r r r Let Q1 and Q2 be two points on the directrix and let g( u) and g g be unit vectors r r along generators (say) g1 and g 2 through Q1 and Q2 respectively. r r Let MN be the line of shortest distance between the generators g1 and g2 As Q2  Q1 Suppose M  C, then C is the central point of the generator through Q1 .





As Q2  Q1 , the vector MN tends to be tangential to the surface at the point C. Let r r d denote to the vector along the limiting direction of the Vector MN, thus d is r r r r to N (i.e. dN 0) , N ,being normal to the surface at C. Also d is perpendicular to the r r r generator g, through Q1 , i.e., d is perpendicular to the vector g and consequently d is r r parallel to the vector g N .

169

r r r Also d i.e., g N must be perpendicular to the generator g2 through Q2 r r r r r Hence, d  g 0 and d  g g 0 r r r r (or ) d  g 0 (U sing d g 0) r gr (or ) d  0 u And proceeding to the limit as Q2  Q1 , we have r r r r r& d g 0 (or ) g N  g 0 r r r  g&g N 0 r r r g&g  N 0







    



 

r r r r&g& g r r& r g g  0 (u sin g (2) for N ) H

 

gr&gr rr&gr gr&gr 0 gr&gr rr&gr gr&gr 0 _________(3) 2

  

   

r r r r r r r r r r r r Using vector identity a b c d  a c b  d a d b  c and relations r2 r r g 1, g  g 0 , we get from (3) r& & r r r r r r & r r2 r2 r r 2 g r  g g  g& g g r v g& . g  g&g 0

        (or)  1 0 0 gr rr&(1) 0gr&  2

r r& r&2 (or ) g& r g 0 r r& g& r (or ) r 2 ______(4) & g r (4) is uniquely determined provided g&2 ≠0. where is the distance of central point C r from the directrix. It R is the position vector of the central point C, then we r r r have R = r (u ) g (u ) ___(5) where is given by (4). Check Your Progress ( C Y P ) : 8. Define central point. ________________________________________________________________________ ________________________________________________________________________

170

r r r r& Corollary : If g& r 0, we have =0 from (4). Hence R r (u) from (5). Therefore , the r r& line of striction will be the directrix itself if and only if g& r 0, . Formula for K in terms of p and v : p 2 To obtain Formula for  2 for the Gaussian curvature at the point distant  2 2 P     from the central point on a generator of parameter p. Hence to show that on any one generator, the Gaussian curvature is greatest in absolute value at the central point; and at points equidistant from this, if has equal values, r r r Proof: Let the equation of the ruled surface be R = r (u ) g (u ) ___(1) r r& r Let the line of striction be taken as directrix. Then g& r 0, ___(2). Also g is a vector of unit (Constant) modulus. r r r& g 2 1 g  g 0 _____(3) r r r Equation (2) &(3)  g&is parallel to r&g r r r Let r&g g (where  is a scalar ) _______(4) r Taking scalar Product both sides with g , we have r r r& r& r& (r&g )  g g  g r& r r& r&2 (or )  r , g, g   g r r r r r    r&, g , g& or p g&2 g&2    Q p  r2   g&  (or ) p    r& r r& Hence, form (4) we get r g pg _______(5) Differentiating (1) Partially w. r. t. u and , we get, r r r r r R1 r&g&, R2 g r r r r r r r HN R1 R2  r&g g&g r r r r (or ) HN pg&g&g _____(6) Squaring both sides of (6) we have r r r 2 H 2  pg g&g







 r r r& r r r p g& 2 pg& g g   g& g  r r r r r r r r p g& 2 p g&, g& ,g    g&g  g& .g  2

2

2

2

2

2

2

2

2



2



r r r r r r r r p2 g&2 2g&2 Q g 2 1,g& g 0,  g&,g &,g   0 r p 2 2  g&2 ______(7)

171

r r r 2  r&, g , g&   We know that Gaussian curvature  4 H r& r r&  r , g, g    And p  r&2 g

r& r & r 2 g&4 p 2  r , g , g    2 4 2 4 p g& p g&  4  2 H P2 2 g&4 i.e 

p 2

p2 2  2

use (7)

_________(8)

Hence we have taken the line of striction as the directix, therefore the distance of a point on the generator from its central point is . But =0, on the central point Again p has the same value at each point of a generator. Hence, it follows from 2 formula (8), that will be greatest at that point of a generator at which  has minimum value i.e, at 2 =0 i.e., at =0 (i.e., at the central point). Therefore on any one generator, the Gaussian curvature is greatest in absolue value at the central point. This proves second part of the theorem. Lastly at points equidistant from the central point, suppose at =   , we have from formula (8), p2 p2    2 2 2 2 2 p      p2 





Hence Gaussian curvature has equal values

Check Your Progress ( C Y P ) : 9. Write down the formula for Gaussian curvature in terms of p and . ________________________________________________________________________ ________________________________________________________________________

6.8. SUMMARY: Hitherto, we explained geodesic mapping between the surfaces. Three theorems regarding the geodesic mapping are proved. Ruled surfaces and their classification are explained. Equations for the same arc derived. Lines of curvature and their properties are studied. Asymptotic lines, its properties and curvature & torsion for the same are explained. A theorem regarding asymptotic lines is proved. Gaussian curvature for ruled surface is obtained.

172

6.9. KEY WORDS: 1)

Homeomorphism: A mapping from a surface (say) S to another surface (say) S’, which is continuous, one to one and onto is called homeomorphism.

2)

Developable and skew surfaces : There are two distinct ruled surfaces, namely those on which consecutive generators intersect and those on which consecutive generators do not intersect; these are called developable and skew surfaces respectively.

3)

Conjugate Directions: Let P and Q be two neighbouring points on a surface, and consider tangent planes at these two points and let PR be the line of intersection of these two planes. Then, as Q  P, the limiting directions of the lines PQ and PR are called the conjugate directions at P.

4)

Asymptotic Directions: The directions which are self conjugate are called asymptotic directions and the curves whose tangents are along asymptotic directions arc called asymptotic lines. Parametric Distribution : r r r  r&, g , g&  A function p (u) defined by p  u  r 2  is called the parameter of the g&

5)

6)

7)

distribution of a ruled surface. Line of striction : The locus of the central points of all generators is called line of striction. The line of striction lies on the ruled surface. Central Point : There exists on each generator of a general ruled surface, a special point called the central point of the generator. The central point of a given generator is the point of intersection of the generator and the shortest distance between it and a consecutive generator of the system.

6.10. ANSWER TO THE CHECK YOUR PROGRESS : 1. It a surfaces is mapped onto a surface S* by a differentiable homeomorphism, which takes geodesics on S into geodesics on S*, then the mapping is called geodesic mapping. 2. A ruled surface is a surface, which is generated by the motion of one parameter family of straight lines and the straight line itself is called its generating line. Cones, cylinders and conicoids are special forms of ruled surfaces. There are two distinct classes of ruled surfaces, namely, those on which consecutive generators intersect and those on which consecutive generators do not intersect. These are called developable and skew surfaces respectively.

173

3. A line of curvature on any surface is a curve, such that the tangent line to it at any point is a tangent line to the principal sections of the surface at that point. Since, the direction of line of curvature at any point is along the principal direction at the point, the differential equation of lines of curvature is given by (EM FL )du 2 (EN GL )dud (FN GM )d2 0 This equation of lines of curvature can be remembered in the following form.

d2 E L

dud du F G

2

M

N

4. Monge’s Theorem: A necessary and sufficient condition that a curve on a surface to be a line of curvature is that the surface normals along the curve form a developable. 5. The differential equation of parametric curves is dud=0 Comparing this equation with Pdu 2 Qdud Rd 2 0, we find P= R = 0, Q=1. Hence, the direction of the parametric curves will be conjugate, if LR+NP-MQ=0 i.e., M=0 (as Q≠0) Conversely, if M=0, the condition LR+NP-MQ=0 is clearly satisfied since for parametric curves P=0, R=0. r r 6. The tangent t to the asymptotic line is perpendicular to the surface normal N , we r r have N . t =0 Differentiation w.r. t ‘s’ yields r r r N  t N  t 0 r r dN dr r r  N  kn 0 ds ds r r dN dr But  0 for asymptotic line. ds ds r r  The above equation yields N  n 0 (Q k 0) r r Also t . N =0 r r r r r r r Now N  n 0 and t  N 0  N is parallel to t n r r  N is parable to b 7. We know that the unit binormal to an asymptotic line is the unit surface normal (or) r r b N . r r Differentiation of this relation w.r. t. arc length gives bN  which on using Frenet’ formulae r r r r gives n N  ( or)N   n r r r r r r But n b t (or ) n N t r r r r r r  N   N t  N  , N ,t  r r Again, we know t kn , so that r r r r r r r r r r r     k t  n t  N t  t , N , t N    ,t ,t  r r r    (or)k  N  ,r ,r 

174

8. Central Point: The central Point of a given generator is the point of intersection of the generator and the shortest distance between it and a consecutive generator of the system. 9. The Gaussian Curvature at the point distant from the central point on a generator of 2 p parameter p is given by  . p2 2 2 ________________________________________________________________________

6.11. TERMINAL QUESTIONS: 1. State and prove Tissot’s theorem. 2. State and prove Dini’s theorem . 3. Show that the surface S is a developable iff the Gaussian curvature at each point of S is zero. 4. Prove that the Gaussian Curvature r r  u cos , u sin , f ( u) is given by.

of

the

surface

of

revolution

f  f

f 2 f     , where f  , f  2 u u u 1 f  2 5. Show that the Gaussian curvature is the same at two points of a generator which are equidistant from the central point. 6. Show that all straight lines on a surface are asymptotic lines. 6.12. FURTHER READINGS: i) ‘Differential Geometry’ by D. Somasundaram, Narosa publishing House, Chennai, 2005. ii)

‘Elementary Topics in Differential Geometry’ by J.A.Thorpe, Springes – Verlag, New York, 1979.

175

UNIT –VII REGRESSION, AND FURTHER THEORY OF SURFACE : STRUCTURE: Learning Objectives 7.0.

Introduction

7.1.

Joachimsthal’s Theorem

7.2.

Equations of Dupin’s Indicatrix

7.3.

Three types of points ( Elliptic, Hyperbolic and Parabolic )

7.4.

Third Fundamental form

7.5.

Envelope

7.6.

Edge of regression

7.7.

Minimal Surface

7.8.

Gauss Characteristic Equation

7.9.

Summary

7.10. Key Words 7.11. Answers to Check Your Progress 7.12. Terminal Questions 7.13. Further Readings. LEARNING OBJECTIVES: After going through this unit, you should be able to, *

define Dupin’s Indicatrix

*

derive the equation of Dupin’s Indicatrix

*

classify the points on the surface with respect to the sign of Gaussian curvature.

*

define characteristic, envelope, the edge of regression, minimal surface.

*

derive the equation of envelope, edge of regression. Gauss characteristic and Mainardi – Codzzi equations.

176

7.0. INTRODUCTION : For further study of curves on surface, we need to define envelope of the family of curves in terms of characteristics. Special type of surface under the condition on mean curvature is to be dealt with. The relation between the fundamental coefficients is needed to study the orthogonal property of certain parametric curves and Gaussian curvature . 7.1. JOACHIMSTHAL’S THEOREM: If the curve of intersection of two surfaces is a line of curvature on both, the surfaces cut at a constant angle. Conversely, if two surfaces cut at a constant angle, and the curve of intersection is a line of curvature on one of them, it is a line of curvature on the other also. Proof: Let C, the curve of intersection of two surfaces, be a line of curvature on both surfaces. r r r Let N and N be the unit normals at P to the two surfaces and t , the unit tangent r r r r r at P to C. Now t being perpendicular to N and N is parallel to N  N . Again C is r r r the line of curvature on both the surfaces, therefore, t is parallel to N  and N  . r r dN   , etc . N  ds   r r r r If is the inclination of N on N , Cos  = N . N r r r r r r r d     cos   N  N+N  N =0_________(1) Since t is parallel to N and N   ds Cos=constant and hence = constant . equation (1) becomes r r r r N  N N  N 0 ________(2) r r r r If C is a line of curvature on first surface. (say) Then t is parallel to N  (say) t N  r r 1r r 1 N  N t N   0 0   r r r Hence , relation (2) gives N  N =0 and this relation implies that N  is r r r perpendicular to N , but N  is also perpendicular to N , being a unit vector. Therefore r r r r N is parallel to N  N and consequently to t . This C is a line of curvature on the second surface also. Dupin’s Indicatrix: Definition –The section any surface by a plane parallel to and indefinitely, near the tangent plan at any point 0 on the surface, is a conic, which is called the indicatrix, and whose centre is on the normal at 0.

177

7.2. THE EQUATIONS OF DUPIN’S INICATRIX: Z r N

Q

o

P R

X

Let 0 be the given point on the surface. Let Q be a point on the Dupin’s indicatrix at O then by definition Q is a point on the surface in the neighbourhood of O. Consider a tangent plane at 0 and let p be the length of the perpendicular QR from Q to the Y tangent plane. We have (by the Geometrical interpretation of the second fundamental form) , 2 2 2 p Ldu Nd 2Mdud_____(1)

Q

Now, taking lines of curvature as parametric curves, so that F=0=M, and hence (1) reduces to 2 2 2 p Ldu Nd  _____(2) The normal Curvature k n at the point 0 in the direction

Ldu 2 Nd2  Q F 0 M  _____(3) 2 2 Edu Gd The principal direction at O, are the directions of the parametric curves =constant and u=constant.(since the lines of curvature have been taken as parametric curves). Therefore (1,0) and (0,1) are respectively the direction ratios of the principal directions. Let k a and k b be the principal curvatures at 0 then from (3) we have K a L , K b N ______(4) Putting the values of L and N from (4) in (2) we get E G 2 2 p K a Edu K bGd 2 _____(5)

du, dis k n 

Let ds1 be the element of arc length of the curve =const at 0, then using the formula

ds12 Edu 2 2 FdudGd2 for

the

direction

ds12 Edu 2 _____(6) Similarly, if ds2 be the element of arc length of

(1,0),

we

have

the curve u = constant,

then ds22 Gd2 ________(7) . Using (6) (7) in (5), we get 2 p Ka ds12 Kb ds22 ______(8) . Now taking 0 as the origin, the axes of x and y along the principal directions at 0. (taking x and y axes along the curves =Contant and u = Constant respectively) and the z –axis along the normal to the surface at the point 0. With reference to these axes, let the coordinates of the point Q on the indicatrix be (x,y,z), then we have z=p, x=ds 1 , y=ds2. Substituting these values in (8), the equations of the indicatrix are given by z=p, x2 y2 2 p K ax 2 K by 2  or  z p ,  2p ____________(9) 1 2  Where 1 and 2 are the principal radii of curvature at 0  k a  1 , k b  1  .  1 2 Clearly the equations (9) represent a conic section and hence the Dupin’s indicatrix is a conic section.

178

Note (1): When the indicatrix is an ellipse, the sign of the radius of curvature is same for all sections. This shows that the surface in the neighbourhood of the point O is entirely on one side of the tangent plane at 0. The surface in this case is called to be synelastic at the point. (2) When the indicatrix is a hyperbola, the sign of radius of curvature is sometimes positive and sometimes negative, showing that the concavity of some sections is turned to opposite direction to that of others. The surface in this case is said to be Antielastic at that point. (3) Dupin’s indicatrix gives an immediate geometrical interpretation of the variation of the normal curvature with direction. Check Your Progress ( C Y P ) : 1. What is Dupin’s Indicatrix? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ 7.3. THREE TYPES OF POINT (Elliptic, Hyperbolic and parabolic ): r r If the Gaussian curvature at any point P(u, ) on the surface r r  u, v is denoted by K, then , LN M 2 ka kb  ________(1) H2 Where k a , k b are principal curvatures at P and H2 is a positive definite quantity. Now, there arises following three possibilities. (i) Elliptic point   is e: The point P(u, ) is called an elliptic point if at P the Gaussian curvature has positive sign i.e., LN-M2>0 . Clearly  is positive if and only if ka and kb are of the same sign. (see equation (1) )

 is e (ii) Hyperbolic Point  : The point P(u, ) is called a hyperbolic point if 2 at P, the Gaussian curvature negative sign i.e, LN-M <0. Clearly is negative if and only if k a and k b are of opposite signs. (see equation (1) ). (iii) Parabolic point ( ( is zero) : The point P(u, ) is called a parabolic point, 2 if at P the Gaussian curvature has value zero i.e, LN-M = 0. Clearly is zero, if and only if at least one of k a or k b is zero. Check Your Progress ( C Y P ) : 2. Explain the hyperbolic point on the surface ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

179

7.4. THIRD FUNDAMENTAL FORM: The third fundamental form denoted by I I I is defined by r r III dN  dN Adu 2 2Bdud Cd2 r r r r Where A N12 , B N1  N 2 , C N 22 Family of surfaces: An equation of the form f(x,y,z,a) =0 __(1), where ‘a’ is a constant, represents a surface, If ‘a’ can take all real values i.e. if ‘a’ is a parameter, then(1) represents the equation of one parameter family of surfaces with ‘a’ as parameter. Giving different values to ‘a’ we shall get different surfaces (members) of this family of surfaces. Similarly, the equation of the form f(x,y,z,a,b)= 0 where ‘a’ and ‘b’ are parameters represents the family of surfaces with two parameters ‘a’ and ‘b’. Characteristic: Let F(x,y,z,a) = 0 be the equation of one parameter family of surfaces, ‘a’ being the parameter and which is constant for any given surface. Let the two members of the family be f  x , y , z ,0, f  x, y , z ,  0 The curve of intersection of theses two surfaces may be given by f x, y , z ,f  x , y , z ,  f x, y , z , 0, 0   Now , the limiting position of the curve as  0 becomes  f f x, y , z ,0 , 0 .  This limiting position is called the characteristic corresponding to the value '' . Check Your Progress ( C Y P ) : 3. What are one parameter family of surfaces? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ 7.5. ENVELOPE: The locus of characteristics for all values of the parameter is called the envelope of the system of surfaces. The equation of the envelope: f Eliminating '' between f=0, 0 , we get the equation of the envelope.  Theorem: The envelope of a family of surfaces touches each member of the family, at all points of its characteristics. Proof:

x, y , z , Let the equation of given family of surfaces be f  =0, where '' is a parameter. The equations of the envelope is obtained by eliminating '' from the equations. 180

f  x, y, z,  0 ___________(1)  Thus the equation of the envelope may be regarded as f x , y , z , =0 __________(2) in which ' ' is not constant, but a function of x,y,z given by  f x, y , z , 0 ___________(3)  The normal to the envelope is parallel to the f   f   f f  f f   vector    ,   ,   . But using equation (3) , this vector x   x  y   y  z   z 

f x, y , z , =0 and

f f  f reduces to  , , which a vector to which the normal to the surface y  z x  f x , y , z , =0 is parallel. This means that at all common points, the surface and envelope have the same normal, and therefore, the same tangent plane; so that they touch each other at all points of the characteristic. Check Your Progress ( C Y P ) : 4. What is the characteristic? ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ 7.6. THE EDGE OF REGRESSION: We have seen that the curve in which any surface is met by the consecutive surface is called the characteristic of the envelope. Every characteristic will meet the next in one or more points, and the locus of these points is called the edge of regression (or) cuspidal edges of the envelope. Since each characteristic lies on the envelope, therefore the edge of regression is a curve which lies on the envelope. Alternative definition of edge of regression: It is the locus of the ultimate points of intersection of consecutive characteristics of one parameter family of surfaces. To find the equation of the edge of regression of the envelope: x , y , z ,=0 The equations of the characteristic corresponding to the surface f  f are f  x, y , z , =0 and 0 ____(1) .  The equations of next consecutive characteristic are therefore  f x, y , z , 0 and f x , y , z , 0   f f 2 f (or) f    ......... 0 and  2 ...... 0 ____(2)     

181

f 2 f 0, 2 0    (Q for points of intersection of (2) with (1) , all four equations must be satisfied.) Eliminating '' from these equations, we get the required equations to the edge of regression (curve). Hence at any point of the edge of regression, we must have f=0,

Check Your Progress ( C Y P ) : 5. Define the edge of regression ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ To prove that each characteristic touches the edge of regressions:  x, y, z, = 0, The edge of regression is given by f  f x , y , z ,0 and  2 f x, y, z, 0 . So, it may be considered to be the curve given by  2  f x, y , z , = 0, f x , y , z ,0  2 f Where ' ' is function of x, y, z given by 0 2 Check Your Progress ( C Y P ) : 6. Write down the equation of the edge of regression . ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ The tangent at (x,y,z) to the edge of regression is the line of intersection of the tangent planes of the surfaces. It is therefore perpendicular to each of the vectors . f f   f f   f f      ,   ,    x  x y   y  z  z   2 2 2 2 2 f f   2 f  f   f  f  and   2 ,  2 ,  2  x   x  y    y  z    z   2 f f For the edge of regression 0, 2 0 . Therefore, the vectors to which the    2 2 2 f f  f  f  f  f  tangent is perpendicular reduce to  , , and  , , which are the x  y  z   x   y   z  

x, y , z ,= 0, vectors perpendicular to the tangent planes at ( x,y,z) to f   f x , y , z ,0 . i.e., to the characteristic. This means the tangent to the edge of  x regression is parallel to the tangent to the characteristic and hence the two curves touch at

182

their common points.

To find the the envelope of the system of surfaces, whose equations involve two parameters: x, y , z , a, b 0 where a and b are the Let the equation of the surface be f  parameters.

A consecutive surface of the system is

f x, y , z , a a , b b 0

 f  f b ......... 0 .  a  b Hence, when a,b are infinitely small, we must have at a point of ultimate intersection:  f  f f 0,a b 0 a b f f Since a, b are independent, f 0, 0, 0 . Hence the curve of intersection of a b f = 0 with any surface consecutive to it goes through the point which satisfies the f f equation f 0, 0, 0 . By eliminating a and b from the above equations we get a b the equation of the envelope. ------------------------------------------------------------------------------------------------------------

(or) f  x, y , z , a , b a

To prove that the envelope touches each surface of the system at the corresponding characteristic points: Consider the surface f  x, y , z , a, b 0 . The normal at  x , y , z to a surface of the f f  f  family f=0 is parallel to the vector  , , _________(1) x  y  z   f  f The characteristic points are given by f 0, 0, 0 and the equation of  a  b the envelope can be regarded as f  x, y , z , a , b 0 , where a and b are functions of x, y, z  f  f given by 0, 0 ___________(2)  a  b The normal at ( x, y, z ) to the envelope is parallel to the vector f  f  a  f  b  f  f  a  f  b f  f  a  f  b 3      ,     ,      x  a  x b x y a y b y z a z b z    f  f Since 0, 0 hence using (2) (3) reduces to a  b f f  f   , ,  x  y  z 

183

Which means that the envelope has the normal line and therefore, the envelope and the surface of the family have the same tangent plane at a characteristic point. Hence the result. -----------------------------------------------------------------------------------------------------------To prove that the osculating plane of the edge of regression at any point P is the tangent plane to the developable at P.

r r& r& & Proof: The edge of regression of the envelope is given by V 0, V 0, V 0, r r r r where V r  N p ( N , being unit normal vector tho the plane and p, the length of r from the origin to the plane). r r i.e., r  N p 0 __(i) r r r N p&0 __( ii) r& r& & r N & p&0 __( iii) r where r is regarded as function of u. Differentiating (i), we obtain r r r r& r r r& N r  N p&(or) r& N 0 _____(iv) [by (ii)] Differentiating (ii) r& r r& r & r r& r& N r  N & p&(or) r& N 0 ______(v) [by (iii )] Differentiating (iv) r r r& r& r& r & r& N r  N 0 (or) r&  N 0 _______ (vi) [by (v )] r r r From equation (iv) and (vi), we notice that both r&and & r& are perpendicular to N , it r r r& r r& follows that r&& r is parallel to N . But r & r is perpendicular to the osculating plane and r r r N being perpendicular to the plane r  N p 0 and hence the osculating plane of the edge of regression, at any point is the tangent plane to the developable at some point. ________________________________________________________________________ To prove that the curve itself is the edge of regression of the osculating developable . r r r Proof: At any point r on the curve r r (s ) , the equation of the osculating plane is r r r r r R r  b 0 ___(1) where r and b arc functions of s.

 

Differentiating (1) w.r.t s, we get r r r r r t  b  R r  n 0 r r r r r (or) R r  n 0 ____(2) Q t  b 0,0

   





The characteristic in given as the intersection of (1) and (2) i.e., at the intersection of r osculating plane and rectifying plane and is therefore the tangent to the curve at r . For edge of regression, differentiating (2) again w.r.t. s

184

       

r r r r r r t  n  R r  b kt 0 r r r r r r (or ) R r b kt 0 ____(3)  Qt  n 0,  r r r (or ) R  r  t 0 ___(3 ) u sing (1) r r r r The edge of regression being given by (1) (2) and (3 ) is given by R r =0  R r . Thus the points of the curve coincides with the points of the edge of regression and the space curve itself is the edge of regression.

-----------------------------------------------------------------------------------------------------------To prove that the edge of regression of the polar developable (i.e., envelope of the normal planes ) is the locus of the centre of osculating spheres. r r r Proof: At any point r on the curve r r (s ) , the equation of the normal plane is r r r R r  t 0 _____(1)

 

Differentiating w. r. t. s we get r r r r r R r  kn t  t 0 r r r r r (or ) R  r  n 0 ____(3)  Qt t 1, _____(2) r r r r r r (or ) R  r  n  n 0  Qn n 1_____(2 )

     

Characteristic is given as the intersection of (1) and (2 ) and is clearly a straight line r parallel to b and passing through the centre of circular curvature. This straight line is called polar line and is the axis of the circle of curvature. For edge of regression differentiating (2) w. r. t s, we get r r r r r r t  n  R r  b kt  r r r r r (or ) R r  b   Qt  n 0 and u sin g (1)  r r r (or ) R  r  b  ___(3)

     



Edge of regression is given as the intersection of (1), (2 ) and (3) equation (1) shows that r r r r r R r is in the normal plane. Equation (2)  component of R r along n is . And r r r equation (3)  component of R r along b is . Hence equation of edge of r r r r r r r r regression is given by R r n  b  R r n  b. r Thus R coincides with the centre of spherical curvature. Thus the edge of regression of the polar developable is the locus of the centre of spherical curvature (osculating spheres). The tangent to the locus are the polar lines, which are the generators of the developable.

185

To prove that the edge of regression of the rectifying developable has the r r t kb r r equation R r k . k k Proof: r The equation of the rectifying plane at any point r is given by r r r r r R r  n 0 ____(1) where r and n are functions of s. Differentiating (1) w. r. t. s, we r r r r r r r r get R r b kt 0 ___(2) (Q r   n t  n 0) .



   





The characteristic (rectifying lines ) is given as the intersection of (1) and (2) and r r r is clearly a straight line perpendicular to both n and b kt and therefore it is parallel r r to the direction t kb .









k Hence it is inclined to the tangent at angle such that tan= _____(3)  find the edge of regression, differentiating (2) w.r.t. s we have r r r r R r  b k  t k 0 ________(4)

To

 



. Again, since the rectifying line is r r parallel to the direction t kb ,



r b



r the point R on the edge of regression is given by r r r r R r t kb _______(5) were



is some scalar.

r r Substituting the value of R r from (5) in (4) we get r r r r  t kb .  b k  t +k=0







(or)  k k  k 0 (or) k

____(6)

k  k  

Hence , from (6) and (5) we have

186







r r t kb r r R r k k  k r r r r k t kb (or ) R r  k k  which is the required equation of edge of regression. The term ‘rectifying’ used for this developable is justified, when the surface is developed into a plane by unfolding about consecutive generators, the original curve becomes a straight line. See the chapter on ‘geodesics’. We also note that if the curve is a helix, K is constant and from equation (3), the  angle is equal to the angle . Thus the rectifying lines are generators of cylinder on which the helix is drawn, and rectifying developable is the cylinder itself.





7.7. MINIMAL SURFACES: k a k b is zero, then the If, at all points of a surface, the mean curvature 12  surface is called a minimal surface. We have seen that the mean curvature  is given by EN GL 2 FM EN GL 2 FM   Thus for a minimal area 0 2 2 2(EN F ) 2H i.e, EN GL 2 FM 0

Theorem: If there is a surface of minimum area passing through a closed space curve,it is necessarily a minimal surface Proof:

r r Let S be the surface r r (u ,) bounded by a closed curve C. Let us give a small displacement  in the direction of the surface normal and let s’ be the new surface obtained,  is a function of u and  and its derivatives w.r.t. u and v arc denoted by 1 and 2 . Both 1 and 2 are small and tend to zero as  0 i.e  0(), 2 0(2 ) as  0. 1 r Let R the position vector of the displaced surface S’. r r r r r r r Then R r N R1 r1 1 N N1 and r r r r R2 r2 2 N N2 Let E , F , Gbe the first fundamental coefficients of S r2 r r 2 r Then E  R1  r11 N N1





187

r r r r r r1 2 1 r1  N 2 r1  N1 o(2 ) r r r r E 2 L o(2 ), Since r  N 0, r1  N1 L r r r& r r r r r F R1  R2  r1 1 N N1 r2 2 N N 2 r r r r r 2 r1  r2  r1  N 2 r2  N1 o ( )











F 2 M 0(2 )





r2 r r 2 r G R2  r2 2 N N 2 r2 r r 2 r2 2 r2  N2 0   G 2 N 0  2  2 2 Thus H  E  G F  2 2  G 2 N 0(2 ) E 2 L 0( )    F 2 M 0( )  

 EG F 2 2  EN GL 2 FM  0(2 ) H 2 4 H 20(2 ), since H 2  1 4 0(2 )

EN GL 2F   2 2H

H 1 H  14 2 0(2 ) 1

H  12 0( ) Hence, if A and Adenote the areas of the surfaces S and S  then 2

A  Hdud and S

A  H dud S

 H 1 2  dud0  2  S

A  2 Hdud0  2  S

If S is a surface of minimum area, then A must be stationary. Adepends on  and A dA takes minimum value A when = 0, provided 0 i.e of order 2 . But d dA dA  2Hdud. A does not depend on  Hence, 0  0 . Thus the d S d surface S is necessarily minimal. Hence the theorem.

188

Check Your Progress ( C Y P ) : 7. Define the minimal surface and find the condition for the same. ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ Problem: Show that the surface e z Cosx Cosy is minimal. Solution: The given surface is e Cosx Cosy cos y i.e, e z  cos x Taking log both sides, we get z = log cosy log cosx ___(1) . The equation (1) is monge’s r form of a surface. The position Vector r of any point (x,y,z) on this surface is given by. r r ( x , y , z ) r i.e r ( x, y, log cos y log cos x) r i.e r1 (1, 0, tan x) r r2 (0,1, tan y ) r r r 2 2 r11 (0,0, Sec x ), r12 (0,0,0), r22 (0, 0, Sec y ) r Now , E r12 1 tan2 x Sec 2 x r r F r1  r2 tan x tan y r2 G r2 1 tan 2 y sec 2 y r r r i j k r r r1 r2 1 0 tan x z

tan y r r r i  0 tan x j  tan y 0  k  1 0  0 1

 tan x , tan y ,1 r r tan x , tan y ,1 r r  Hence, 1 2  H H r r  tan  x, tan y,1. 0,0, sec 2 x Again L N  r11    H sec2 x  H r r tan x, tan y,1. 0,0, sec2 y sec 2 y N N  r22    H H r r tan x, tan y,1 M N .r12  . 0,0, 00  M 0 H The condition for the surface to be minimal is EN-2FM+GL = 0

189

sec2 x(sec2 y) Sec2 y.sec2 x 0  0 H H Hence, the given surface e z cox cos y is minimal. HenceEN-2FM+GL =

7.8. GAUSS CHARACTERISTIC EQUATION: It

states that

1 T2   2 F12 E22 G11   m2 E 2mF 2 G  ln E  ln  F Gwhere 2 T 2 LN M 2 and symbols have their usual meaning. Proof: From Gauss’s formula (1), we easily have r r r11  r22 LN ln E  ln F G ___(i) r2 2 2 r12 M m E 2 mF 2 G ______(ii) r Also, we have E2 2r1  r12 r r r F1 r1  r21 r11  r2 r r G1 2 r2  r21 r r r r giving E 22 2r12  r12 2r1  r12 2 r r r r G11 2 r12  r21 2 r2  r21  1 r r r r r r r r F12 r12  r21 r1  r21 2 r11 2  r2 r11  r22 1 r r r E22 G11 2F12 r122 r11 r22 2 r r r 1 (or ) r122 r11  r22  E22 G11 2 F12  ________(iii ) 2 Adding (i) and (iii) and subtracting (ii), we get Hence ,

1 LN M 2 T 2   2 F12 E22 G11   m2 E 2m F 2 G   ln E  ln  F G____(1) 2

which is the required equation. Corollary 1 : Equation (1) may be put in the form 1  FE2 G1  1 2F1 E2 FE1  L N M 2  H    H    ____(2) 2 u EH H  2  v H H EH  Corollary 2: Intrinsic Formula for Gaussian Curvature r r Let r r  u ,be the equation of a surface. At any point  u,on the surface, LN M 2 the Gaussian curvature is given by  . 2 H

190

1 FE2 G1  1 2 F1 E2 FE1         u sin g ____(2) 2H  u EH H  2H  v  H H EH  1 FE2 EG1  1 2EF1 FE1 EE2      ____(3) H u  2HE  H  v 2HE  2 2 Where H EG F This formula (3) expresses Gaussian curvature in terms of first fundamental magnitudes E,F,G and their Partial derivatives with respect to u and . Therefore (3) is intrinsic formula for Gaussian Curvature. It is convenient to use this formula when the r position vector r  u, for the surface is given.

When the parametric curves are orthogonal, F =0 and the Formula (3) can be written in much simple form 1 G  E      1   2  , Where H  EG 2H  u H  H   Clearly, the surfaces having same E,F,G have the same Gaussian curvature. Mainardi – Codazzi equations: r r r r We Consider the identity  r11   r12  . Substituting values of r11 and r12 from 2 1 Gauss’s formulae, we have r r r r r r LN lr1 r2  MN mr1 r2 2 1 r r r r r r r r r r r r (or ) L2 N l2 r1 2 r2 LN2 lr12 r22 M1 N m1 r1  r12 ___(1) 1 r2 MN1 mr11  r r r Substituting again values of r11, r22 , r12 from Gauss’s formulae and the value of r r N1 and N2 from Weingarton equations viz.,



 



FM GLrr FL EM rr 

N1 

1 2  H2 H2 ___(2) FN GM r FL EN r  N2  r1  r2  H2 H2 

r r r We shall get identity (1) as expressed in terms of vectors N , r1 , r2 (non-coplanar r r r vectors). We then equate coefficients of N , r1and r2 from either side and get their r relations. Therefore by equating coefficients of N on two sides, we get L2 lM N M 1 mL M And thus giving L2 M 1 mL  l M N ___(3) which is the first mainardi codazzi equation. r r Again from the identity  r12   r22  , we get by substituting values from Gauss 2 1 formulae, r r r r r r M 2 N m2 r1 2 r2 MN 2 mr12 r22 r r r r r r N1 N n1 r1 1 r2 NN1 nr11 r12

191

r r r r Substituting again values for r11, r22 , r12 from Gauss’s formulae and for N1 and r r N 2 from relations (2), we get a vector identity and then equating coefficients of N on the two sides of the identity, we obtain M 2 mM N N1 nL M M 2 N1 nL  m  M N _____(4) Which is the second Mainardi Codazzi equation. Problem : From the Gauss characteristic equation deduce that when the parametric curves are orthogonal,  1  1  G   1  E            v G   EG     u  E u    Solution: When parametric curves are orthogonal, we have F = 0 and therefore F1 0, Then from equation (2), viz, 1  FE2 G1  1 2F1 E2 FE1  2 L N M  H    H     2 u EH H  2  v H H EH  1  G1  1  E 2  2 We have L N M  H   H   2 u  H  2 v  H  But H 2 EG F 2 EG  Q F 0 LN M 2 1   2 H 2H

G1  E 2         u H  H   

  1  G1  1  E 2           2    2 u   EG EG            1   1  G   1  E               uE  u   G   EG     1  H

Check Your Progress ( C Y P ) : 8. Find the edge of regression of the rectifying developable. ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ 9. Prove that the curve itself is the edge of regression of the osculating developable. ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

192

7.9. SUMMARY: The important property of surfaces with reference to line of curvature on them is studied through a theorem called Joachimsthall’s theorem. A section of a surface in the neighbourhood of a point on it is studied. Limiting position of the curve of intersection of two surfaces is explained. Method of finding the envelope of family of surfaces is given. Some results regarding the properties of edge of regression are proved. Some fundamental equation of surface theory are derived.

7.10. KEY WORDS: 1)

Indicatrix: The section of any surface by a plane parallel to and indefinitely, near the tangent plane at any point O on the surface, is a conic, which is called the Indicatrix and whose centre is on the normal at O.

2)

Elliptic Parabolic and Hyperbolic Points: The point P  u, v is called an elliptic point, if at P, the Gaussian curvature K has 2 positive sign i.e., LN M 0 . Clearly K is positive if and only if k a and k b are

2  LN M  of the same sign.  Q K ka kb   H2   The point P  u, v  is called a hyperbolic point, if at P, the Gaussian curvature has 2 negative sign i.e., LN M 0 . Clearly K is negative if and only if k a and k b arc of opposite signs. u, v is called a parabolic point, if at P, the Gaussian curvature has The point P 

value zero. i.e., LN M 2 0 . Clearly K is zero , iff at least one of ka  or  kb is zero. 3)

Characteristic: If two members of the one parameter family of surfaces are f x , y , z ,0, f  x, y , z , 0 then the curve of intersection of these two

f x, y , z ,  f  x , y , z ,  0 .   f x , y , z ,0 , The limiting position of the curve as  0 becomes f  =0.  This limiting position is called the characteristic corresponding to the value '' . surfaces may be given by f  x , y , z , 0 ,

4)

Envelope: The locus of characteristics for all values of the parameter is called the envelope of the system of surfaces.

193

5)

The edge of regression: The curve in which any surface is met by the consecutive surface is called the characteristic of the envelope. Every characteristic will meet the next in one or more points and the locus of these points is called the edge of regression.

6)

Minimal surface: If, at all points of a surface, the mean curvature 12  k a k b is zero, then the surface is called a minimal surface.

7)

Christoffel Symbols: These arc denoted by l , m, n;  , , , which arc the suitable function of fundamental coefficients E, F, G and their partial derivatives with respect to u and v. 7.11. ANSWERS TO CHECK YOUR PROGRESS : 1)

2)

The section of any surface by a plane parallel to and near the tangent plane at any point 0 on the surface is a conic called Indicatrix and whose centre is on the normal at 0. Dupin’s indicatrix is a conic section. The point P  u, v on a surface is called a hyperbolic point if atP, the Gaussian curvature has negative sign. i.e., LN M 2 0 Clearly K is negative if and only if K a and k b are of opposite signs, where LN M K ka kb  H2 An equation of the form f  x, y , z , a0 , where ‘a’ is a constant, represents a 2

3)

surface. If ‘a’ can take all real values i.e., if ‘a’ is a parameter, then the above equation represents the equation of one parameter family of surfaces with ‘a’ as parameter. Giving different values to ‘a’ we shall get different surfaces of this family of surfaces. 4)

5)

If f  x, y, z, a0 is the equation of one parameter family of surfaces, ‘a’ being the parameter and which is constant for any given surface. Let the two members of the family be f  x, y , z , 0 , f  x, y , z , 0 The curve of intersection of these two surfaces may be given by f x , y , z ,   f  x , y , z ,  f x , y , z ,0 , 0 . Now , the limiting   f x, y , z ,0 , 0 . The position of the curve as  0 becomes f   limiting position is called the characteristic ‘ corresponding to the value ' ' . We know that the curve in which any surface is met by the consecutive surface is called the characteristic of the envelope. Every characteristic will meet the next in one or more points and the locus of these points is called the edge of regression 194

(or) cuspidal edges of the envelope. Since, each characteristic lies on the envelope, therefore the edge of regression is a curve which lies on the envelope. 6)

The equations of the characteristic corresponding to the surface f  x, y, z, 0  f 0 . The equations of the next consecutive characteristic are   therefore f x, y , z ,0 ans f x, y , z , 0 (or)  2  f  f f f   ...... 0 and  2 .... 0 . Hence at any point of the edge     2  f of regression, we must have . f  x, y , z , 0, f  x , y ,z , 0, 0 ,  2 which constitute the equation of edge of regression.

are f=o and

7)

If, at all points of a surface, the mean curvature 12  k a k b is zero, then the surface is called a minimal surface. The mean curvature  is given by EN GL 2 FM EN GL 2 FM   Thus, for a minimal area 0 . 2H 2 2 EN F 2  i.e., EN+GL-2FM=0

8)

r r r Differentiating w.r.t. S the equation R n .n 0 of the rectifying plane at any r point r , we get

 

1 R rr .b kt 0 Q rr .nr t .nr 0 _____  r r r r Differentiating (1) w.r.t. S, we have  R r  .  b k  t k 0 _________  2 r r r Again, since the rectifying line is parallal to the direction  t kb , the point R r r r r on the edge of regression is given by R r  t kb ________  3 where  r

r

r

r

is a scalar. r r Substituting the value of R r from (3) in (2) , we get r r r r t kb .  b k  t k 0 or  k  k k 0







k _____(4) k  k   Hence, from (4) and (3) we have r r r r k t kb R r  k  k r r r r k t kb (or) R r  k  k or 









195

9.)

r r At any point on the curve r r  s , the equation of the osculating plane is r r r r r R r , b 0 ____  1 where r and b are functions of s.

 

Differentiating (1) w.r.t. S, we get r r r r r r rr t .b  R r .n 0  or  R r .n 0

 

 

___  2

Q tr.b 0, 0  r

The characteristic is given as the intersection of (1) and (2) i.e., at the intersection of osculating plane and rectifying plane and is therefore the tangent to the curve at r r . For edge of regression , differentiating (2) w.r.t. S

     

 

r r r r rr t .n  R r .  b kt 0 r r r r r r or  R r . b kt 0 ____  3  Q t .n 0  r r r or  R r .t 0 3 u sin g (1)

The edge of regression being given by (1) , (2) and (3’) is given by r r r r R r 0  R r Thus the points of the curve coincides with the points of the edge of regression and the space curve itself is the edge of regression . 7.12. TERMINAL QUESTIONS: 1)

State and prove Joachimsthall’s Theorem

2)

Explain the elliptic, Hyperbolic and parabolic points on a surface

3)

Prove that the envelope of a family of surfaces touches each member of the family at all points of its characteristics

4)

Prove that each characteristic touches the edge of regression.

5)

Prove that the osculating plane of the edge of regression at any point is the tangent plane to the developable at P.

6)

Obtain the equation of the edge of regression of the rectifying developable.

7)

Show that the surface e z cos x cos y is minimal.

8)

If the parametric curves are orthogonal, find the Gauss’s formulae.

9)

Derive the Mainardi – Codazzi equations.

7.13. FURTHER READINGS: 1. 2.

‘Differential Geometry’ by D. Somasundaram, Narosa Publishing House, Chennai, 2005. ‘An Introduction to Differential Geometry’ , by T. Willmore, Clarendon Press, Oxford, 1959. 196