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A First Course D Somasundaram

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Differential Geometry A First Course

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Differential Geometry A First Course

D Somasundaram

© Alpha Science International Ltd. Harrow, U.K.

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D. Somasundaram Department of Mathematics Erstwhile Madras University P.G. Extension Centre Salem, Tamil Nadu, India Copyright © 2005 Alpha Science International Ltd Hygeia Building, 66 College Road, Harrow, Middlesex HA1 1BE, U.K. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without prior written permission of the publisher. ISBN 1 -84265- 182-X Printed in India

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Preface

This book is a detailed introduction to the classical theory of curves and surfaces which is offered as a core subject in mathematics at the post-graduate level in most of our universities. Based on Serret-Frenet formulae, the theory of space curves is developed. The theory of surfaces includes the first fundamental form with local intrinsic properties, geodesies on surfaces, the second fundamental form with local non-intrinsic properties and the fundamental equations of the surface theory with several applications. A variety of graded examples and exercises are included to illustrate all aspects of the theory. Relevant motivation of different concepts and the complete discussion of theory and problems without omission of steps and details make the book selfcontained and readable so that it will stimulate self-study and promote learning among the students in the post-graduate course in mathematics. I take this opportunity to thank Mr. N.K. Mehra, the Managing Director of Narosa Publishing House, for his personal care and excellent co-operation in the publication of this book. Suggestions for the further improvement of the book will be most welcome. Finally I wish to dedicate this book to the fond memory of my beloved parents Tmt. D. Vishalakshi and Thiru A. Durairaj and to my most revered Professor V. Ganapathy Iyer. Madurai

D. Somasundaram

Contents

Preface

v

1. Theory of Space Curves

1

1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20

Introduction Representation of space curves Unique parametric representation of a space curve Arc-length Tangent and osculating plane Principal normal and binormal Curvature and torsion Behaviour of a curve near one of its points The curvature and torsion of a curve as the intersection of two surfaces Contact between curves and surfaces Osculating circle and osculating sphere Locus of centres of spherical curvature Tangent surfaces, involutes and evolutes Betrand curves Spherical indicatrix Intrinsic equations of space curves Fundamental existence theorem for space curves Helices Examples I Exercises I

2. The First Fundamental Form and Local Intrinsic Properties of A Surface 2.1 2.2 2.3 2.4

Introduction Definition of a surface Nature of points on a surface Representation of a surface

1 2 3 7 10 15 18 26 31 35 37 43 48 57 61 67 69 76 80 99

101 101 101 103 106

viii 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17

Contents

Curves on surfaces Tangent plane and surface normal The general surfaces of revolution Helicoids Metric on a surface—The first fundamental form Direction coefficients on a surface Families of curves Orthogonal trajectories Double family of curves Isometric correspondence Intrinsic properties Examples II Exercises II

3. Geodesies on a Surface 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12 3.13 3.14 3.15 3.16 3.17

Introduction Geodesies and their differential equations Canonical geodesic equations Geodesies on surfaces of revolution Normal property of geodesies Differential equations of geodesies using normal property Existence theorems Geodesic parallels Geodesic polar coordinates Geodesic curvature Gauss-Bonnet theorem Gaussain curvature Surfaces of constant curvature Conformal mapping Geodesic mapping Examples III Exercises III

4. The Second Fundamental Form and Local Non-intrinsic Properties of a Surface 4.1 4.2 4.3 4.4 4.5

Introduction The second fundamental form Classification of points on a surface Principal curvatures Lines of curvature

108 109 114 117 118 123 131 133 136 138 142 143 158

160 160 160 173 174 178 182 192 199 201 202 220 227 232 237 242 249 276

279 279 279 284 290 296

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4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14

The Dupin indicatrix Developable surfaces Developables associated with space curves Developables associated with curves on surfaces Minimal surfaces Ruled surfaces Three fundamental forms Examples IV Exercises IV

5. The Fundamental Equations of Surface Theory 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9

Introduction Tensor notations Gauss equations Weingarten equations Mainardi-Codazzi equations Parallel surfaces Fundamental existence theorem for surfaces Examples V Exercises V

Hints and Answers to Exercises 1.20 2.17 3.17 4.14 5.9

ix

302 313 321 329 331 335 353 358 379

382 382 382 383 388 390 399 404 416 442

444 444 446 448 450 454

Exercises I Exercises II Exercises III Exercises IV Exercises V

References

455

Index

456

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1 Theory of Space Curves

1.1

INTRODUCTION

Differential Geometry is the study of properties of space curves and surfaces.with the help of vector calculus. This geometry examines in more details the curves in space and surfaces, whereas the differential geometry of the plane curves deals with the tangents, normals, curvature, asymptotes, involutes, evolutes etc. which have analogues for space curves. Though we have these analogues for a space curve, the curvature at a point of a space curve and the tangent plane at a point to a surface play a dominant role in the differential geometry. Thus the differential geometry studies the pointwise properties of space curves and surfaces as distinct from the algebraic geometry whose sole aim is to describe the properties of the configuration as a whole. In this chapter on space curves, first we shall specify a space curve as the intersection of two surfaces. Then we shall explain how we shall arrive at a unique parametric representation of a point on the space curve and also give a precise definition of a space curve in E3 as set of points associated with an equivalence class of regular parametric representations. With the help of this parametric representation, we shall define tangent, normal and binormal at a point leading to the moving triad (t, n, b) and their associated tangent, normal and rectifying planes. Since the triad (t, n, b) at a point is moving continuously as P varies over the curve, we are interested to know the arc-rate of rotation of t, n, and b. This leads to the well-known formulae of Serret-Frenet. Then we shall establish the conditions for the contact of curves and surfaces leading to the definitions of osculating circle and osculating sphere at a point on the space curve and also the evolutes and involutes. Before concluding this chapter, we shall explain what is meant by intrinsic equations of space curves and establish the fundamental theorem of space curves which states that if curvature and torsion are the given continuous functions of a real variable sy then they determine the space curve uniquely. We shall illustrate all the notions developed with a particular type of space curves known as helices. In all our discussion, the basic formula of Serret-Frenet occupies the central position.

2

Differential Geometry—A First Course

1.2

REPRESENTATION OF SPACE CURVES

Whenever we use the word space, it means the Euclidean space of dimension three denoted by Ey In this space a single equation generally represents a surface and so we need two equations to specify a curve. Thus we first introduce a space curve as the intersection of two surfaces Fl(x,y,z) = 0,

F2(*,;y,z) = 0

...(I)

Though we are able to fix the curve in space with the help of the equations (1), we are unable to obtain the representation of different points on the whole curve. Since a space curve is a set of points with a sense of description, it is desirable to know the coordinates of a point on the curve as functions of single parameter. So the question naturally arises whether one can obtain the parametric representation of a curve with the help of equation (1). First let us examine this question and then its converse. Let the functions F{ and F2 of (1) satisfy the following conditions of the implicit function theorem. (i) Let F{ and F2 have continuous partial derivatives of first order. (ii) At least one of the following three Jacobian determinants d(F^F2) d(y,z)

d(Fl9F2) '

d(z,x)

d(FlyF2) '

d(x,y)

is different from zero at a point (x0, y0, z0) on the curve. Under these conditions one can obtain the solution of two variables in terms of the third variable. Hence without loss of generality, let us assume that the first Jacobian is different from zero. Then we can solve for y and z in terms of x and obtain y =/,(*) and z -f2M as their solutions. Having obtained these values ofy and z in terms of x, we take the parametric representation as x=u,

y=fx{u\

z=f2(u)

...(3)

The above equations treating* as a parameter are true for the restricted values of JC under the conditions (2). So they cannot give the parametric representation of the whole curve. Thus having started with the definition of a curve as the intersection of two surfaces, we are unable to arrive at a satisfactory representation of the whole curve. Next let us examine the converse question in a little more detailed manner. Let us assume that the curve is given in the following parametric form * = *(«), y = y(u),

z = z(u)9

ux z), then the above definition implies that each of the components JC, y, z is of class ra and x, y, z are functions of u. Definition 4.

If R = — never vanishes on /, then the vector valued du function R = R(w) is said to be regular. This implies that x, y, z will never vanish on / simultaneously. Definition 5. A regular vector valued function R = R(w) of class ra is called a path of class ra. Note 1. As the parameter u varies, R(w) gives the position vector of different points on the curve. Thus a path can be considered as the locus of a moving point giving the manner in which the curve is described. Note 2. Since the definition of a path depends not only on a vector valued function R(w), but also on the interval /, there are as many paths as there are regular vector valued functions of class ra defined on /. Likewise a single path may be defined by different Cm functions defined on different intervals I{, I2 etc. For example we can have a path of the same class defined on Ix and I2 corresponding to two different vector valued functions. Hence to arrive at a unique single path of the given class corresponding to the parameter u defined on /, it is desirable to partition the paths into mutually disjoint classes .of the same type and choose a representative path of the same class with a unique parameter. We shall achieve this by an equivalence relation among the paths of the same class as follows.

5

Theory of Space Curves

Definition 6. Two paths Rj and R2 of class Cn defined on I{ and/ 2 are said to be equivalent if there exists a strictly increasing function 0 of class m which maps /j onto 12 such that Rj = R 2 ° 0 If we takeRj = (xltyu Z\) and R2 = (x2, y2, z2), then the above conditions are the same as xx{u) = x2((p(u)X yx(u) = y2((p(u))i z,(w) = z2(0(w)) First let us verify that the notion of equivalence of paths of the same class as defined in the previous paragraph is an equivalence relation. (i) To prove the relation is reflexive, let R, be a path of class m defined on / and let us take 0 = /. The identity function is an increasing function on / and R, = R,o/ so that R{ is equivalent to itself. Thus the relation of equivalence of paths is reflexive. (ii) Let Rj and R2 be the paths of the same class m defined on /, and I2 respectively. Let R, be equivalent to R2. We shall show that R2 is equivalent to R,. Since R{ is equivalent to R2, there exists a strictly increasing function 0 from /, onto I2 such that R, = R2o 0. Since 0 is a strictly increasing function on /, onto I2 with 0 # 0 , the inverse function 0 _1 exists as a strictly increasing function on I2 onto I{. Hence we have R2 = R, o 0" ! which shows that R2 is equivalent to R,. Hence this relation is symmetric, (iii) To prove the relation is transitive, let the path Rj be equivalent to R2 and R2 be equivalent to R3. Then there exists a strictly increasing function 0 defined on lx onto I2 such that R, = R 2 o0.

...(1)

In a similar manner, there exists a strictly increasing function i/^on I2 onto 73 such that R2 = R 3 o ^ .

...(2)

Since 0:1{ —> I2 and y/: l2 —> 73 are strictly increasing functions yo 0is a welldefined strictly increasing function on Ix onto 73. From (1) and (2) we have Rj = R2 o0= R 3 o ^ o 0 Since y/o(p\s strictly increasing function on I{ onto 73, Rj is equivalent to R3, proving that the relation is transitive. Thus the notion of equivalence of paths of the same class Cm is an equivalence relation. This relation introduces a partition on the paths of the same class splitting the paths of the same class into mutually disjoint classes such that the paths within the same class are equivalent to one another. Using these mutually disjoint classes, we shall define a space curve and the parametric representation as follows. These different equivalent classes of paths of class m determine the curves of class m. Thus any path R determines a unique curve and R is called the parametric representation of the curve. The variable u is called the parameter. Further R = (*» y* z) wherex =x(u),y =y(u), z = z{u) is called the parametric representation of the curve. The function 0 of two equivalent paths is called the change of parameter. Though 0 preserves the sense of description of the curve, it gives the

6

Differential Geometry—A First Course

changes in the manner of description of the curve. Summarising the above, we define a curve as follows. Definition 7. Any curve of class m in E3 is defined to be any set of points in E3 associated with an equivalence class of regular parametric representations of class m having one parameter. Since the properties of a curve depend on a particular parameter chosen, every property of a path is not a property of the curve. We are concerned with those properties of a curve which are common to all parametric representations. This means those properties which are invariant under a parametric transformation. Before proceeding further, we shall illustrate the equivalent representations by the following two examples. Example 1. The following are the two equivalent representations of a circular helix. (i) Rj(w) = (a cos w, a sin w, bu), ue Ix = [0, n) 1 - v2 2av „ , _i * , V G I2 = [0, oo) r, T , 2b tan v 2 2 1+v 1+v j v To show thatRj is equivalent toR 2 , we find a change of parameter / t is u = 0(v) = sin_1(v). Then we have v = sin u. Hence R^w) = [2(1 - sin2 w), 2 sin u cos w, 2 sin u] Now

R2(v) =Ri[0(v)]

7

Theory of Space Curves

R2(v) = [2(1 - v2), 2v>/l - v2 , 2v] which is the required parametric transformation equivalent to 1^ (w).

1.4 ARC-LENGTH We define the arc length of a curve and derive a formula for the arc length. Using this formula of arc length, we show that the arc length is invariant under parametric representation so that the arc length of a curve can be used as a parameter in our study of properties of a space curve. Hence the arc length of a space curve plays the important role as a natural parameter. Definition 1. Let R = R(w) be a path with parameter we /. As u varies over [a, b] c /, the path is an arc of the original path joining a and b. Let A be the subdivision of [a, b] as follows A = {a = w0 < ux www.Ebook777.com

Differential Geometry—A First Course

10

To obtain the equation of the helix with s as parameter, we have from (2), s = cu so that u = — c Using (3) in (1), we get

...(3)

r(.y) = \a cos - , a sin - , — which is the required equation.

1.5

TANGENT AND OSCULATING PLANE

If 7 is the given curve, then its parametric representation r = r(w) is the equation of the curve in the sense that it gives the position vector of different points on 7. We use R to represent position vectors of points in space not necessarily on 7 in order to distinguish it from r = r(w) on 7 We also assume 7 is of class > 1 which means that r(w) has continuous derivatives of all orders so that r = r(w) has power series expansion at a point u0 in the neighbourhood u. r(w0 + h) = r(w0) + — r(w0) + — f (w0) + ... + — r('°(w0) + 0(hn) I! 2! n! where lim

= 0 where u-Un = h. hn ° In our study, we usually include first two or three terms in the above expansion. Definition 1. Unit tangent vector to 7 at P. Let P and Q be two neighbouring points on 7 with parameters u0 and u respectively. The parameters of P and Q are very close together in the sense that when Q -» P, u - u0 —> 0. The unit vector along PQ tends to the unit vector atP as Q —> P. This unit vector denoted by t is defined to be the unit tangent vector at P. The sense of t is that of increasing s. Definition 2. The line through P parallel to t is called the tangent line to 7 at P. If R is the position vector of any point on this tangent line to 7 at P, then the vector is called the tangent vector to 7 and P. From the above definition of the tangent vector at P, we have the following properties. A->O

dx (i) The unit tangent vector t = — where we have chosen s as the parameter ds measured from P. Proof. The unit vector along the chord PQ is r(M)-r(M 0 )

|r(w)-r(w 0 )| Since t is the unit vector along the chord PQ as Q —> P, weget

t = lim r ( " ) ~ ^ o ) = u->u0 |r(w) - r(w0)|

r(K)-ry u->u0 \u-u0\

Hm

\u-u0\ \r(u) - r(w0)|

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11

Theory of Space Curves

Since the curve is of class m > 1, we get

t=

m \Hu)\

Since s = r(w), we get t = —— = — • — = — . s(u) du as as dx Note, r = — is also a tangent vector, but not necessarily a unit vector. du (ii) Equation of the tangent line at P. Let R be the position vector of any point Q on the tangent line at P. Then if the length of PQ is c, then the vector PQ = ct. Hence the equation of the tangent line at P is R = r + ct which can also be written as R = r + cr where r is parallel to t. Definition 3. Osculating plane. Let 7 be a curve of class m > 2 and P and Q be two neighbouring points on 7 The limiting position of the plane that contains the tangential line at P and passes through the point Q as Q —> P is defined as the osculating plane at P. Note. When y is a straight line the osculating plane is indeterminate at each point. So we avoid this particular case in our discussion. Definition 4. The point P on the curve for which r " = 0 is called a point of inflexion and the tangent line at P is called inflexional. Theorem 1. Let 7 be a curve of class m > 2 with arc length s as parameter. If the point P on 7 has parameter zero, the equation of the osculating plane is [R - r(0), r'CO), r"(0)] = 0 where r" * 0 If r"(0) = 0, let us assume that the curve 7 is analytic. Then the equation of the plane at an inflexional point is [fl-rCOXr'CO), r (k) (0)] = 0. Proof. Using the arc length s as parameter, let 0 and s be the parameters of P and Q. Let R be the position vector of the point on the plane containing the tangent line at P and passing through Q. Then if r is the position vector of P, then the vectors R - r(0), t = r'CO), and r(^) - r(0) are coplanar vectors. Hence the condition of coplanarity gives the equation as [R-r(0),r/(0),rW-r(0)] = 0

...(1)

Since the curve is of class m > 2, we have by Taylor's Theorem, r(j) = r(0) + sr'(O) + -s2 r"(0) + 0(.s2) as s -> 0. Using (1) in (2), we get 1 . .1 R - r(0), r'(0), s r'(0) + -s2 r"(0) + 0(s2)

=0

...(2)

Differential Geometry—A First Course

12

Neglecting the terms of higher order, the above equation becomes 2

[R-r(0),r'(0),*r'(0)] + R - r ( 0 ) , r m ^ r " ( 0 )

=0

...(3)

Since r'(0) x r'(0) and s is a scalar, the first term of (3) vanishes and so we get [R-r(0),r'(0), r"(0)] = 0

...(4)

as the equation of the osculating plane provided the vectors r'(0) and r"(0) are linearly independent. So to complete the proof, we have to prove r'(0) and r"(0) are linearly independent. Since t = r' is a unit vector r'~ = 1. Differentiating this relation, we have r' r" = 0 which shows that neither r' nor r" can be a constant multiple of the other so that they are linearly independent unless r"(0) = 0. If r"(0) = 0, then the point/5 is an inflexional point so we derive the equation of the osculating plane at an inflexional point with an assumption that the curve y is analytic. Differentiating r' 2 = 1, we have r' r" = 0 Differentiating this relation once again, we have r •r +r •r

=U

Since r" = 0, we get r' r'" = 0 at P. Since r' cannot be zero, r' and r"' are linearly independent, unless r'" = 0. Repeating this process of differentiation, let us assume that r (k) is first nonvanishing derivative of r such that r' r(k) = 0. So if r (k) * 0, we have from Taylor Theorem r(s) - r(0) = - ^ ^ + — r(k)(0) + 0 ( / ) as s -* 0.

...(5)

Using (5) in (1), we get R - r ( 0 ) , r / ( 0 ) f - ^ l j + — r (k) (0) = 0 1! k\ As in the previous case the above equation reduces to [R - r(0), r^O), r(k)(0)] = 0 as the equation of the osculating plane at an inflexional point. If r(k) = 0 for all k > 2, then since the curve is analytic, we infer that t is constant and therefore the curve is a straight line. Corollary. If P is not point of inflexion, any vector lying in the osculating plane is a r' + b r" for some constants a and b. Proof. Since P is not a point of inflexion r" * 0. From (4) of the theorem r7 and r" lies in the osculating plane and pass through P. Hence any vector in the osculating plane is a linear comlination of r' and r" so that we can take it as ar' + bx" for some constants a and b. It is of importance to note that r" lies in the osculating plane.

13

Theory of Space Curves

Theorem 2. If w is the parameter of the curve y, then the equation of the osculating plane at any point P with position vector r = r(w) is [ R - r , r, r] =0. dt ds

dr du du ds

r s

Proof.

r =

Further

„ _ d (x\ du _ ( i r - vs) 1 du\s)ds s2 s

Using these values of r' and r" in (4) of the previous theorem we have r

r

YS

=0

S\

S S

Since r x r = 0 and s is a scalar, simplifying the above equation, we obtain [R - r, r, r] = 0 as the equation of the osculating plane. Corollary. If R = (X, Y, Z) and r = (JC, yy z) then the equation of the osculating plane given by the scalar triple product in the theorem takes the form X-x

Y-y

Z-z\

x

y

z

x

y

z

=0

Note 1. A tangent at a point P on a space curve is the line passing through the two consecutive points on the curve. Likewise the osculating plane can be defined " as the limiting position of the plane PQR of three consecutive points P, Q, R as Q and R approach P. Using this definition, we shall derive the equation of the osculating plane as follows. Let r = r(w0)> r, = r(w,) and r 2 = r(u2) be three consecutive points on the curve. Let Ra = p be the equation of the plane passing through the above three points. Then if/(w) = R a - p , we have /(Mo) = 0,/(M,) = Oand/(M2) = 0

Now we have the intervals [u0> w,] and [w,, u2]. Hence by Rolle's Theorem, there exist points v, G [W0, U{] and v2 G [W,, U2] such thatf\v{) = 0 and/ / (v 2 ) = 0. Since/' satisfies the conditions of the Rolle's Theorem in [v,, v2], there exists av 3 G [uj, v2] such that/ 7/ (v 3 ) = 0. Hence when Q and R approach P,ult u2, v,, v2, v3 approach u0. Writing u for u0 in the limiting case, we have /(«) = r a -p = 0,/'(M) = v a = 0 , / »

= v a =0

...(1)

Using R • a = p in the first equation, we get /( M ) = ( R - r ) . f l = 0 Thus from (1), we find the vectors is perpendicular to ( R - r ) , r and r so that they are coplanar. So we write / ? - r = A r + ^ r where A and ji are constants. Eliminating A, ji or using the condition of coplanarity, the equation of the osculating plane is

14

Differential Geometry—A First Course

[R-r, r,r] = 0 Note 2. In the case of the plane curves, the plane through the three consecutive points on the curve is the plane itself. Hence the osculating plane coincides with the plane of the curve itself. The following example shows that at a point of inflexion, even a curve of class oo need not possess an osculating plane. Example 1. Let 7 be a curve defined by r(w) =(w,0, e _l/ " 2 )

whenw>0

r(w) =(w, e-{/u\0) whenw0 is Y= 0. In a similar manner the equation of the osculating plane when u < 0 is Z= 0. So the osculating plane at u = 0 is indeterminate. This proves that at a point of inflexion, even a curve of class °° need not possess an osculating plane. Example 2. Find the equation of the osculating plane at a point u of the circular helix r = (a cos w, a sin w, bu)

...(1)

From the given equation, we have r = (- a sin w, a cos ut b)

...(2)

r = (- a cos u,-a sin w, 0) Using (1), (2), and (3), the equation of the osculating plane is X - a cos u - a sin u -acosu

...(3)

Y- asm u Z - bu a cos u -a.s'mu

b 0

Expanding the above determinant along the last column and simplifying, the equation of the osculating plane is b(X sin u-Y cos u - au) + aZ = 0.

1.6

PRINCIPAL NORMAL AND BINORMAL

Besides the tangent at P, we shall define the normal and binormal at P of the curve leading to the moving triad of coordinate system at P. Definition 1. Let P be a point on the curve y. The plane through P orthogonal to the tangent at P is called the normal plane at P. Since the osculating plane at P passes through the tangent at P, the normal plane is perpendicular to the osculating plane at any point of the curve. Definition 2. The line of intersection of the normal plane and the osculating plane is called the principal normal at P. The unit vector along the principal-normal is denoted by n. The sense of n may be chosen arbitrarily, provided it varies continuously along the curve. Using the above definitions, we have (i) The equation of the normal plane. Let r = r(w) be a point on the curve and R be the position vector of any point on the plane. Then (R - r) lies in the normal plane. Since (R - r) is perpendicular to t, we get (R - r) • t = 0 as the equation of the normal plane, (ii) The equation of the principal normal at P. If the position vector of any point P on the curve is r and if R is the position vector of any point Q on

16

Differential Geometry—A First Course

the normal, then the vector PQ = An where A is a scalar. Then R = r + An is the equation of the normal at P. Definition 3. The normal at P orthogonal to the osculating plane is called the binormal at P. The unit vector along the binormal is denoted by b. The sense of the unit vector b along the binormal is chosen such that t, n, b form a right handed system of axes. The behaviour oft, n, b at a point P on the curve is the same as the unit vectors ij\ k, along the coordinate axes. Hence we have b = txn, t = nxb, n = b x t and

tn = nb = bt = 0, tt = bb = nn=l.

Definition 4. The plane containing the tangent and binormal is called the rectifying plane. From the above definitions, we have (i) The equation of the rectifying plane. If r = r(w) is a point on the curve and R is the position vector of any point on the rectifying plane, then (R - r) is in the rectifying plane and orthogonal to n. Hence (R - r) • n = 0 is the equation of the rectifying plane. Note. Since the binormal is orthogonal to the osculating plane, the equation of the osculating plane can be put in the form (R - r) • b = 0. (ii) The equation of the binormal. Let r = r(w) be the position vector of any point P on the curve and let R be the position vector of any point Q on the binormal. Then PQ = /Jb where ji is a scalar. Then R = r + jib is the equation of the binormal at P. Note. Since r, r lie in the osculating plane, r x r is perpendicular to the osculating plane. Since the binormal is perpendicular to the osculating plane, the binormal is parallel t o r x r Since r gives the direction of the tangent and r x f gives the direction of the binormal r x (r x r) gives the direction of the principal normal. Thus we get r x (r x r) = (r- r)r - (r- r)r is the direction of the principal normal. Note. When we take the arc length as parameter, then we have Z1 = 1 and consequently r' • r" = 0. Using this in the above formula, the principal normal is parallel to r". Summarising the above we conclude that if we choose the arc length as parameter, then r', r" and r' x r" give the directions of the tangent, principal normal and binormal at a point on the curve. Example. Find the directions and equations of the tangent, normal and binormal and also obtain the normal, rectifying and osculating planes at a point on the circular helix

r= os

h 0' asin @'7)

17

Theory of Space Curves

, (-a . (s\ a fs} b) r = sin - , - cos - , ^ c \cj c \cj c)

Now

r =

...(1)

..(2)

[9"*$--?™$°] ^3

cj

•ab

Ji\ s \ a

c

cjc

..(3)

(1), (2) and (3) give the directions of the tangent, normal and binormal. Next let us find the equations of the tangent, normal and binormal. The equation of the tangent is R = r + At which gives X - a cos | - ]

Y - a sin C

- a sin | -

a cos

The equation of the normal is R = r + Ab which gives X - a cos | -

Y - a sin

C_

- a cos| -

0

- a sin | -

The equation of the binormal is R = r + /jb which gives X - a cos I ab sin

a sin

-

Z-*

- ab cos I -

Let us find the equations of three planes. The equation of the normal plane is (R - r) • t = 0 which gives

a

ac cos \- \Y-acs\n\-

\ X+(Zc-bs)b

= 0. The equation of the rectifying plane

is (R - r) • b = 0, which gives cos - \ X + sin -

Y- a = 0.

The equation of the osculating plane is (R - r) • b = 0. Since b has the direction r' x r", we get the above equation as (R - r) • (r' x r") = 0 ...(4) Writing (4) in the determinant form, we obtain

Differential Geometry—A First Course

18

a cos

(s

a . sin c \c a fs —=- cos —

bs_

Y - a sin

c

a (s — cos c \c

b_

- a . (s sin -

0

—T-

=0

c

Expanding the determinant and simplifying, we have

bsm(-)x-bcos(-)Y+(z-—

)a=0

which is the equation of the osculating plane.

1.7

CURVATURE AND TORSION

At each point of the curve, we have defined an orthogonal triad t, n, b forming a right handed system and also we have noted that at each point this moving triad determines three fundamental planes which are mutually perpendicular. It is important to note that t, n, b vary from point to point on the curve as the point moves on the curve as shown in the following figure.

ib Rectifying plane (R-r)n =0

Normal plane {R-r)t=0

Osculating plane (R-r)b =0 Fig. 1

Hence we can study their variations from point to point with respect to the arcual length as parameter. This leads to the notion of curvature and torsion of the space curve as defined below. Definitoin 1. The arc rate at whch the tangent changes direction as the point P moves along the curve is called the curvature vector of the curve and it is denoted by/C

Theory of Space Curves

19

dt Thus by definition K = — and K is called the curvature vector and its J ds magnitude |K| is called the curvature at P denoted by K. Then p = — is called the /c radius of curvature where we take the absolute value of p. As we have already noted, the osculating plane of a plane curve is the same at all points but the osculating plane changes from point to point on a space curve. So we shall now define a quantity measuring the arc-rate of change of the osculating plane. Definition 2. The torsion at a point P of a curve is defined as the arc-rate at which the osculating plane turns about the tangent atP as P moves along the curve. It is denoted by r. 11/T | denoted a is called the radius of torsion. Having defined the torsion at a point on the space curve, the next question is to devise a method of measuring it. The natural method of measuring the turning of a plane is to measure the turning of its normal. Since the binormal is orthogonal to the osculating plane, we shall use the binormal to measure torsion. Thus the arc, db rate of rotation of the osculating plane is expressed by b = — whose magnitude is ds db the torsion r. So r = ds As a synthesis of the above definitions of curvature and torsion giving the measure of arc-rate: of change of tangent and osculating plane, we shall derive Serret-Frenet formulae which are fundamental in the study of space curves. Theorem 1. (Serret-Frenet Formulae). If (t, n, b) is the moving orthogonal triad of unit vectors at a point P on a space curve y, then dt dn db (i) — = Kit (ii) — = r b - Kt (iii) — = - rn. ds ds ds Proof. We first prove (i) and (iii) and then derive (ii) from them. To prove (i), differentiating t • t = 1 with respcet to s at a point P of the curve, we have t • t ' = 0 so that t' is perpendicular to t. dv Since t = — , t' = r". As r" lies in the osculating plane, t' also lies in the ds osculating plane. Therefore t' is a vector perpendicular to t and lies in the osculating plane. Hence t' is parallel to the principal normal. By definition |t'| = K, being the curvature at P on the curve. Since we know the magnitude K and the direction n oft', we can write t' = ± /en. By convention we take t' = KTI. (ii) As in the above case, we find the vector b'. Differentiating b- b = 1, we have b b ' = 0 so b' is perpendicular to the binormal at P and hence b ' lies in the osculating plane. Since b • t = 0, differentiating this and using (i) we get b'-1 + b- (K*n) = 0 As b• n = 0, we get b'-1 = 0 showing that b ' is perpendicular to t. Therefore b' is a vector perpendicular to t and lies in the osculating plane. Hence

20

Differential Geometry—A First Course

b' is parallel to the principal normal at P. By definition |b'| = T, being the torsion at P. Since we know the magnitude r and the direction n of b', we can write b ' = - r n where the negative sign is introduced because as a convention torsion is regarded as positive when the rotation of the osculating plane as s increases is in the direction of a right handed screw moving in the direction oft. To prove (iii), let us consider n = b x t. Differentiating both sides of the above vector with respect to s, , . dn db , dt we obtain — = — x t + b x — . ds ds ds dn Using (i) and (ii) in the above equations, — = (- rn) x t + b x (/en) ds Since n x t = - b and b x n = - 1 , we have — = r b - Kt ds Using curvature and torsion, we characterise a straight line and a plane curve in the next two theorems. Theorem 2. A necessary and sufficient condition for a curve to be a straightline is that K = 0 at all points of the curve. Proof. The condition is necessary. Let us take the curve to be a straight line and let its vector equation be r = as + b where a and b are constant vectors. Differentiating this equation, we get r' = t = a. As a is a constant vector, we have r" = t' = 0. Since the curvature vector r" vanishes at all points of the curve, its magnitude K= 0 at all points of the curve. To prove the sufficiency of the condition, let us assume that K= 0 at all points of the curve. This implies that the curvature vector t' = r" = 0. Integrating the above equation twice, we obtain r = as + b where a and b are constant vectors. This proves that the curve is a straight line. Theorem 3. A necessary and sufficient condition that a given curve be a plane curve is that r = 0 at all points of the curve. Proof. The condition is necessary. Let us take the curve to be a plane curve and show that r = 0. Since the curve lies in a plane, the osculating plane at every point of the curve is the plane containing the curve itself so that the binormal b is a db db constant. Since b is constant — - 0 which implies = 0at = 0. Hence r = ds ds ds all points of the curve. Conversely let r = 0 at all points of the curve. On this assumption, we shall prove that the curve is a plane curve. S.ince T= 0, — = 0 at all points of the curve ds so that b is a constant vector. Now for any vector r, d ,

ix

dx .

db

,

— (r b) = b+r = t b + r b'. ds ds ds

21

Theory of Space Curves

Since t b = 0 and 1/ = 0, we have —(r b) = 0 for any point r on the curve. ds Hence r • b = constant = c (say). If r = (x(s), y(s), z(s)) and b = (fej, b2, &3), then r • b = c gives xbx + yb2 +zb3 = c which shows that r(s) = (x(s)9 y(s), z(s)) lies on the plane bxX + b2Y+ b3Z = c. This proves that the curve is a plane curve and the condition is sufficient. Definition 3. If ris non-zero, then the curve is called a twisted curve. In the following, we shall give the formulae for curvature and torsion first in terms of the arc length and then in terms of a general parameter u. We shall use dashes to denote differentiation with respect to s and dots to denote differentiation with respect tow. Theorem 4. If r = r(s) is the position vector of a point P with arc-length as parameter on a curve c, then (i) K ^ r " r" \r' r" r'"l

(ii) T = L r ' „ ' , , or K^tr'.r".!-'"] r -r Proof, (i) We know that r' = t and r" = KU Hence r" r" = (fen)• (/en) = /c2, since n n = 1 (ii) Now r' x r" = t x KU = Kb Differentiating both sides of (2) with respect to 5, r / x r "/ + r " x r " = ^ + Kh>

...(1) ...(2) (3)

Since r" x r" = 0 and b ' = - rn, (3) becomes r / x r w = (/c , b-/crn)

...(4)

Taking dot product with r" on both sides of (4), we have (r' x r w ) r" = - /c2r as r" = K:n and n • n = 1 7/

w

w

But (r' x r'") r = - r' (r^ x r ) = - [r', Y", r ] Using (6) in (5), we have

K2T=

[r', r", r w ]

...(5) ...(6) ...(7)

Further substituting for K2 from (i), we obtain Ir , r ,r I r= Theorem 5. /•N (I)/C =

Proof.

r -r If r = r(w) is the equation of the curve with parameter w, then |rxr|

V P lrl

[r, f, r]

and(ii)r=|—^ |rxr| z

Now r = — = = r'- i = ti 5, (X2+K2 + Z2)

=S2--K2S4=S2(l--K2S2)

12

Thus

I

12

J

All

(x2+r2+z2)1/2 =/i--^x-Vj

Using the Binomial expansion on the right hand side, (x 2 + Y2 +Z2}

~s 1

K-V which shows that when *•* 0, the arc length

PQ differs from the chord PQ by a term of the third order in s. (iii) We obtain the approximate equations of the projections of the curve on three planes. In each case we obtain the lowest power of s and eliminate s. K

o

(a) The projection of the curve on the osculating plane is K = — X , Z = 0 From the equations, we obtain as in (i) X ~ s and Y K

between the two equations Y = —2X . (b) The projection of the curve on the rectifying plane is

z=— x\y=o 6

K

s2 . Eliminating s

29

Theory of Space Curves

KT 3

From the equations, we obtain as in (i) X ~ s and Z

s. 6

Eliminating .s between them, Z = — X3 6 (c) The projection on the normal plane is Z2 =

2 T2

Y3, X = 0

KT.S3

From the equations, we obtain as in (i), Z 6 Eliminating s between the above two equations,

K

2

and Y = — s 2

6-6 6-6 I K J 9 K Example 1. Find the Serret-Frenet approximation of the curve n

(cos w, sin w, u) at w = —. r-

1

7T

7T

As in Examples 1 of 1.4 and 1.7, s = V2w, K- T= — and ^ = -y=r at w = — so 2 V2 2 the curve can be represented as f \

(

S

.

S

S

r(s) = [ « » - £ , a n - j . . - ^ Using the Theorem 1, we have

4 6^

72,

Since we know fcand T, we canfindthe projections of the curve on the planes using (a), (b) and (c) of (iii). Theorem 2. The length of the common perpendicular d between the tangents at two neighbouring points with the arcual distance s between them is KTS

approximately d = 12

3

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Differential Geometry—A First Course

Proof. Let 0 and P be two adjacent points on the curve with OP having the arcual length s. Let the position vector of P be r = r(s). Let us choose the cartesian coordinate system at O in the direction of the triad (t, n, b) and so r(0) = 0. The unit tangent vectors at O and P are r'(0) and r'(s). The common perpendicular of these two tangents at O and P is nothing but the shortest distance between them. Since the shortest distance between them is perpendicular to both r'(0) and r'(s), it is parallel to r'(.s) x r'(0). Thus if d is the shortest distance between them, it is the projection of PQ = r(^) - r(0) on r'(.s) x r'(O). Hence

rf=[r(s)-r(0)]

Since

r(0) = 0,d =

r\s) x r'CO) |r'(s)xr'(0)| [r(s),r'(s),r'(0)] |r'(s)xr'(0)|

Since we like to find the approximate value of the above expansion, we shall use the Taylor series expansion of r(,s) and r'(.s) including the terms of third degree in s. From the previous theorem, r(*) =

—K + — ^ | n + —

3! J

v2

3!

3!

b,r(0) = 0.

Since t, n, b give the fixed direction of the coordinate axes at 0, differentiating the above equation with respect to s, (

*2

r'W = i

1

2

v

and r^(o) = t. Hence

i\ K

I *2 t + SK + K

J I

n + —5zb 2

2! ,

n - | SK + *V •

r%y) x r'(0) =

V

Now \r\s) x r'CO)!2 =

VTV^

S2K'^ + \ SK + -

Neglecting the terms of higher order > 4, we get |I"(J)

x r'CO)!2 = J V + Kits3 = K V [ 1 + —s)

/2 K' V ( 1 K" so that \r\s) x r'(0)| = KS 11 + — s \ = KS 1 + s\ approximately.

(

Thus r W t r ' W x r'(O)] =

KS2

2

3 s + —K 3'

V n 2 z) = 0 and g(x, ytz) = 0 Let h = V/x Vg. If A is the operator Ar = h, then

32

Differential Geometry—A First Course

M

-Ah-AH \H\2

|h| 3

Proof. Let r = r(s) be the position vector of a point on the curve/(;c, y, z) = 0 andg(x,y,z) = Q ...(1) The V/=

a/ a/ a/1 3*' 3^' 9zJ

-"•'I-1-1

...(2)

are the outward drawn normals to the surfaces (1). Since the unit tangent at P to the curve of intersection of the two surfaces is perpendicular to both the vectors in (2), it is parallel to V/x Vg. Hence V/x Vg is a scalar multiple of t. Thus we have At = Ar'(s) = V/x Vg = h (say)

..(3)

Let us assume h = (h{i h2, h3). Further |h| = A. In the above equation (3), one should note that the left hand side is given in terms of the dash derivatives, whereas the right hand side is given in terms of partial derivatives. Hence let us find the relation between these two.

A^=A ds

=A

dr dx dx ds

3r dy dy ds

,3

,3

,3

dy

dz

3*

3r dz dz ds r = (/z1,/r2, h3)

3r dr 3r Since — = (1, 0, 0), V" = (Q, 1, 0) and — = (0, 0, 1), we obtain from the dx °y dz above equation, hd = h{, Xy' = h2, Xz - h3 Thus we have the relation ...(4) (fa\Xy\Xz') = (hl,h2,hi) Let A be the operator defined by A = X— = A, — + h0-z- + h-x^dy dz ds Hence by the definition of the operator Ar = h Operating on both sides of (3) with A in (5), we get

...(5) ...(6)

d X—(At) = Ah which gives ds A 2 — + AA't = Ah. That is X2Kn + AA7t = Ah ds Taking cross product of (3) and (7), A3>cb = h x Ah = //(say)

...(7) ...(8)

33

Theory of Space Curves

Further taking dot product of (8) with itself, we get X3K=\H\

or.= ^

= 14 |ii|

X

Operating on both sides of (8) with X—• = A, we get ds d -x X—(X Kb) = AH giving ds -A 4 KT» + Ab—(X3K) =AH ds

...(9)

Taking the scalar product of (7) and (9), - A 6 K 2 T = Ah- AH which gives _= -Ah-AH r = -Ah-AH £—2— 2— which completes the proof of the theorem. 6 2 2 XK

"

\H\

Example. Find the curvature and torsion of the curve of intersection of the quadratic surfaces ax2 + by2 + cz2 = 1, a'x2 + b'y2 + c'z2 = 1 We shall find h, Ah, H = h x Ah and Ah-AH and apply the formula in the theorem. Let

Then

/ = - [ax2 + by2 + cz2-llg

V/=^,|igj

= - [a'x2 + b'y2 + c'z2 - 1]

= («iftyfcd

Vg = dg dg ^ i

fj,

A/„

„,„->

Now V/x Vg = ((&c' - *'c)yz, {cat - c'a)zx, (ab' - a'b) xy) Denoting be' -b'c = At ca' - da = B, ab' - db - C, we have _ _ VfxVg = xyz\

(ABC - , - , -

U y * dr Since V/x Vg is parallel to t = — , we take ds

ds

Thus

'

A B C

x y

/*, = A , , , = *, /, x

y

' = z '

=

£ z

...(2)

h

andA2 =

y r A >i ^\x)

34

Differential Geometry—A First Course

Hence

A =

Ah =

So

ox

ay

Ad x dx

B d y dy

( 2

A

Hence

B2

d} oz C d)(A z dzjix'

3 3 Z X

B C y' z -.(3)

2

{Bz

yz AB

c_d_s z 8z /

B_d_ y dy

C2

BC 3 3

H =hxAh=

(Ad_ {x dx

-'Cy2),^r(Cx2-Az\

zV

(fix 2 -Ay 2 )

•••(4)

Let us simplify Bz2 - Cy2 Bz2 - Cy2 = (at - c'd) z2 - (ab' - a'b) y2 = a'(cz2 + by2) - a(c'z2 + b'y2) = d(\ - ax2) - a(l - a'x2) = (a' - a) Thus we have Bz2 - Cy2 = (a' - a) In a similar manner, we have Cx2 - Az2 = ib' - b) and Ay2 - Bx2 = (c'~ c) Thus

BC a CAib b =h,A ^; \ :: 3\4^(c-c) 3 3h=\ 3 ZV yz xy

H

Hence|hxAh|2=^^£^-(a'-a)2 xyz A2B2C2-^

So

\hf

i"

2\s

Jt6y6z6

x6 Zyj^-TA(a 2

2

-a) .

To find T, let us find AH From equation (8) of theorem and using (4), X3Kb =H =

r 3 ABC x

x y z

v3

Z3

^(a'-a),?-{b'-b\±-{c'-c) A.

3 3 3 3 That is ^A( y z X Kb = \ —(a' - a), ^(b' B A ABC

C

u

;

,

3

- b), — (c' - c)

c"

.-(5)

Free ebooks ==> www.Ebook777.com Theory of Space Curves 3 3 3 X V Z

35

^

Let us denote by u = — XK. ABC , d Operating with X— = A on both sides of (5), we have ds

ds

A d . B d . C d}(x3 [^x dx y dy z dz

, ,

, y3 , . , B

, , z3 c

So Xji'b + Xji (- rn) = (3* (a'- a), 3y {V - b\ cz{c' - c))

...(6)

From equation (7) of the theorem, we have A2 X2Kii + XX't =Ah = - ^ r

>

B2 C2 ^r>^r •

...(7)

Taking scalar product of (6) and (7), we obtain A2

#KTH =3VA r (a'-a) *-** x" Substituting the value of fi and simplifying, we get. A V T

3 ABC ^ A2 = 3 3 3 Zr"r^"fl)

Substituting the values of A and Kt we get

3^V 2,^~{a'-a) * =— l^'-rf h ABC 6

1.10.

CONTACT BETWEEN CURVES AND SURFACES

Let y be a curve r(w) = {/(«), g(w), h{u)} and let S be a surface F(JC, y, z) = 0. Let us assume that the curve y and the surface S are of high class in the sense that r(w) and F(JC, y, z) have continuous derivatives of sufficiently high order. From the equation of the curve, we take x =/(w), y = g(u), z = h{u). If this point lies on the surface, we have F(f(u\ g{u\ h(u)) = 0 which is an equation in u giving the points of intersection of the curve and the surface. Depending upon the nature of the roots of the equation, we shall define the contact between curves and surfaces as follows. Let u0 be one such zero of F(u) = 0. Since F(u) possesses the derivatives of sufficiently high order, F(u) has the following power series representation in the neighbourhood of u = u0.

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Differential Geometry—A First Course

36 m

= F(u0)

+

^ B 2

FXUQ)

+ i ^ l

F>0)

+ ...

+

n! F(«) = hF'(u0) + ^-f"(«o) + ^-F"'(«o) + - + —t ^ W + 0(hn+x) 2! 3! «! Definition 1. If F'(w0) * 0, then u0 is a simple zero of F(u) = 0. Then the curve yand the surface S is said to have simple intersection at r(w0). Definition 2. If F'(u0) = 0 and F"(u0) * 0, u0 is a double zero of F(u) and F(w) is of second order of h. Then the curve yand S are said to have two point contact. Definition 3. If F'(w0) = F"(w0) = 0 and F'"(w0) * 0, y and 5 are said to have three point contact at u = u0 under these conditions u0 is a triple zero of F(u). In general if F'(u0) = F"(w0) = ... = Fin"l\u^ = 0 and F{n\u0) * 0, the curve y and the surface 5 are said to have n point contact at u = w0. Theorem 1. The conditions of a surface having n point contact with the curve y are invariant over a change of parameter. Proof. Let u = 0(0 be the given parametric transformation. Since it is regular, we have 0(K) (w) * 0 for k > 1. Corresponding to the pointw = w0, we have "o = 0('o) a t ' = 'oNow F(u) = F(0(O) =/(0 where/is a function of f only.

7(0 = - ^ ' M W M = f > ) [ « t ) ] 2 + F'(u)'(t) If

...(2)

F'(w) = 0, then /(/) = 0 as 0(0 * 0

If F'(u) = 0 and F"(w) * 0, then from (1) and (2) we get 7 (0 = 0 and / (0 * 0, since 0(0, 0(0*0. Thus if the surface S given by F(w) has two point contact with the curve yat r(w0), then the surface 5 given by/(0 has two point contact with yat r(0(ro)) Differentiating (2) again we get / (0 = Fm(u) [0(t)]3 + 3F"(w) 0(00(0 + F'(K) 0(0

...(3)

If F\u) = 0, F"(u) = )0 and F'"(w) * 0, then from (3) /(*) = 0, / ( 0 = 0 and / (0 * 0 as 0 (0 is regular. Hence the surface S given by/(0 has three point contact with the curve yat r[0(ro)]. Proceeding like this, if F\u) = F"(w) = ... = F{"-l\u) = 0 and F{n\u) * 0 at u = w0> then 7(0 = 7(0 = ... =/ ( "" 1) (0 = 0 and / $ * 0 at r[0(fo)]. Thus the surface 5 given by/(0 has n point contact with yat

37

Theory of Space Curves

r[0(/ o )], provided the surface S given by F(u) has n point contact with /at r(w0). Thus a surface having n point contact with the curve y is invariant over a change of parameter. Hence we conclude that the property of the curve having n-point contact with S is a property of yin the sense that any path which represents ywill have this property. Theorem 2. The osculating plane at any point P has three point contact with the curve at P Proof. Let P be any point on the curve and let the arc length be measured from P so that s = 0 at P and let the equation of the curve be r = r(.s). We know that the osculating plane at P is [r(s) - r(0), r'(O), r"(0)] = 0 and let

F(s)= [r(s) - r(0), r'(0), r"(0)]

...(1)

We shall show that F'(s) = F"(s) = 0 and F'"(s) * 0 at P where s = 0 and this proves that the osculating plane has three point contact with the curve. Expanding r(s) by Taylor's theorem in the neighbourhood of P, r(s) = r(0)

+ lMs

1!

+ £! (t, n, b) at Qy we can take the vector PQ = An + jib. The coefficients A, jU change from point to point on C so that A, ^u are functions of s on C. Using this position vector of any point rx on C (Fig. 4) is

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53

Theory of Space Curves

Fig. 4

Fl

=OP = OQ + QP = r + An + ^b

...(2)

We shall now determine A and ji. Differentiating (2) with respect to st we get dr. ds

= t + A'n + A(rb - Kt) + ji'b + JU(- rn) = (1 - XK) t + (A' - jtn)n + (AT + /z')b

...(3)

dV\ drt ds\ ds\ Since — L = — L — L = t,— L , it is a tangent at P to C. dfc ds. ds ds Hence it is in the normal plane to C at Q. Therefore it must be parallel to An + jUb ...(4) Since (3) and (4) are parallel, we must have X - \XT _ AT + //' 1 - Xk = 0 and A ^

..(5)

From the above relation (5), we find A and a. First equation gives A = — = p. From the second equation, we get T(A2 + /Z2) = M ' - ^ '

Since /i * 0, we can write (6) as follows T(A 2 +// 2 ) ..2

_ia'-W)_d(X ~~

..2

dsyju

-(6)

54

Differential Geometry—A First Course

Thus we find

T=

d_ ds 1 + (A//1)2

ds

tar'ti

Integrating both sides of the above equation \xds + c = tan T — where c is a constant Thus we obtain — = tan f \xds + c J Hence if we take y/=

Tds, then we can write

A — = tan(y^+ c) so that \x = A cot(^+ c) Hence the equation of the evolute of C is r1 = r + pn + p cot(v+ c)b where y/= \ rds. The above equation represents an infinite system of evolutes of the given curve, one evolute arising from each choice of c. Corollary. The tangents to two different evolutes corresponding to two constants cx and c2 drawn from the same point of the given curve are inclined to each other at a constant angle cx - c2. Proof. The equation of the evolute is rj = r + pn + p cot(y+ c)b K

For the sake of simplification, let us choose c - — + a so that the equation of the evolute becomes

Further

rj = r + p n - p t a n ( y + f l ) b

...(1)

y/ = j xds so that (// = r

...(2)

We shall find unit tangent vectors t[ ]) and t {2) at two points ax and a2 on the evolutes and find the angle between them using t [^ • t j(2) = cos ft Differentiating (1) with respect to sly we get — - — L = t + p ( r b - K,t) + p , n - [ p / tan(y^+ www.Ebook777.com Differential Geometry—A First Course

76

Hence (al9 P\,Y\) = (cos c(s), - sin cr(.y), 0) (ii) When a(0) = 0, 0(0) = 1, 7(0) = 0, we get A2 = 0 , £ 2 = 1 , C 2 = 0 Hence (ofc, j32, 72) = ( s i n 72' 73) = (0^ 0, 1) where a{s) =

and

r(s) = Jo

[2 2\ A-s

tdcr(j) = [ cos J — , sin J — , 0 , ds Jo^ V a \ a J y/2as

We can easily check the orthogonality conditions of the theorem. Note. The solutions of the equations involving a, /?, /become difficult in most of the cases, in particular for the curves of class > 3. If K(S) and T(J) are of class > 3, eliminating /J, 7, we obtain a third order equation in a with variable coefficients. In some of the simple cases we obtain solution of the equation by a change of variable as in Example 2. But such a third order equation can be reduced to a first order Reccati equation whose solutions are well-studied. Solving this Reccati equation of first order, we obtain its solution.

1.18

HELICES

We conclude this chapter with a brief discussion of the properties of a wide class of space curves known as helices which we used as examples in the previous sections. Definition 1. A space curve lying on a cylinder and cutting the generators of the cylinder at a constant angle is called a cylindrical helix. The above definition implies that the tangent to the curve makes a constant angle a with a fixed line known as the axis of the helix. We can obtain more general helices than cylindrical helices if the cylinder is replaced by other surfaces like cone. But we consider only cylindrical helices in our study. The following theorem characterises the cylindrical helices. Theorem 1. A necessary and sufficient condition for a curve to be helix is that the ratio of the curvature to torsion is constant at all points. Proof. To prove the necessity of the condition, let a be the unit vector in the direction of the axis. Since the helix cuts the generators at a constant angle, let the

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77

Theory of Space Curves

angle between the generator and the tangent at any point P on the helix be a. So from the definition of the helix, we have t • a = cos a ...(1) Differentiating (1) with respect to s, we get t'a + t a ' = 0

...(2)

Since a is a constant vector and t' = /en, we get from (2) /cna = 0 ...(3) If K = 0, the curve is a straight line and the conclusion of the theorem is obvious. As we have excluded the case when K= 0, (3) gives n • a = 0 showing that a is perpendicular to the normal at P. Since a passes through P making a constant angle a with the tangent t at P and perpendicular to the normal at P, it lies in the rectifying plane at P. Hence (cos a, sin a) are the components of a in the rectifying plane so that we can take a = t cos a + b sin a ...(4) Differentiating (4) with respect to s and using a' = 0, (f cos a + b' sirfee) = (JCCOS a- rsin a)n = 0 K

Since n * 0, we have /ccos a - r s i n a = 0 giving — =tan a which is constant, r proving the necessity of the condition. K

To prove the converse, let us assume — = constant A (say) and prove that the T

curve is a helix. Given any constant A, we can always find the smallest angle a such that K

tan a = A. So we can take — = tan a giving (KCOS a- rsin a) = 0 ...(5) r Since n * 0, (5) implies n(K*cos a- rsin a) = 0 ...(6) (6) can be rewritten as — (t cos a + b sin a) = 0 ds This proves that t cos a + b sin a is a constant vector a (say). Then a • t = (t cos a + b sin a) • t = cos a which proves that the curve is a helix. In the above definition of the cylindrical helix, we have not specified the base curve which is the cross section of the cylinder by a horizontal plane. However if we take the circle to be the base curve, we get a helix on a circular cylinder. Such helices are called circular helices. With proper choice of the coordinate axes, we shall find the equations of a circular helix. Theorem 2. If the z-axis is the axis of the cylinder as well as that of the helix, the parametric equation of the helix is of the form x = a cos w, y = a sin w, z = bu 2

where the base circle is x + y2 = a2, z = 0 and b is a suitably chosen constant. Proof. Let P be any point on the helix with the position vector r and P{ be its projection on the XOY plane with the position vector rv Let a be the unit vector in the direction of the axis of the helix. By our choice of the axis of the helix, a is

78

Differential Geometry—A First Course

Fig. 6

parallel to the z-axis. So P,P is parallel to a and hence PXP = r a. Hence we take —> PXP = (r-a)a. Using this we write the position vector of P, as r, = r - (a r)a Differentiating (1) with respect to s, we get dvy — - = t - (a • t)a = t - a cos a ds Hence taking dot product of (2) with itself,

...(1)

...(2)

drx dr{ , .9 x t = ( t - a c o s a) ( t - a cos a) = sin~ a ...(3) ds ds If s{ is the arc length of the projection of the helix on the XOY plane, diytfr, =rfrf. Using this in (3), we get dsx =sin a ds which implies 5-, =sin as. ...(4) Since the helix is a circular helix t • k = cos a, so that * • dv . d , . ^ dz t - k = — k = —(r- k) = — =cos a, since r k = z Hence from the above step, z = s cos a Using (4) in (5), we obtain z = .v, cot a From the Fig. 6,4P, = 5-, = AM SO that z =tf«cot a. If (x, y, z) are the coordinates of P, then we have from Fig. 6, x = a cos uyy = as\nu,z

= bu where & = d cot a

...(5)

79

Theory of Space Curves

Note. For this helix, we have already noted that s = au, so that its equation with s as parameter is s s bs a cos —, a sin —, — |, a > 0 a a a K

a

# —, r = ft— , c 2 = a 2 + b,2 andj — = —, Z? > 0. -TyT= ^r,c =a + b and — = - , c2 c2 x b Definition 2. If b > 0, the helix is right-handed and if b < 0, the helix is the left-handed. The pitch of the helix is 2nb. The pitch gives the displacement along the axis corresponding to one complete rotation round the axis. Theorem 3. The projection C{ of a general helix C on a plane perpendicular to its axis has its principal normal parallel to the corresponding principal normal of the helix and its corresponding curvature is given by

and

K=

K= K", sin2 a Proof.

Using the same notation as in Theorem 2, we get r = r t + (a-r)a

...(1)

Differntiating (1) with respect to s dx dxx ds, ( dr\ ,. , . — = —! + a . — a which gives ds ds{ ds v ds) t = t,^+(a-t)a ds Since t • a = cos a and ds{=ds sin a from Theorem 2, we obtain t = t, sin a + acos a

...(2)

Differentiating (2) with respect to s, we get dt\ ds* . .o /en = — L — - sin a= K{n{ sin" a ...(3) ds{ ds (3) proves that the normal nj to C{ is parallel to the principal normal n at the corresponding point of the helix. Taking dot product of (3) with itself, we get KT = K2 sin4 aor K= K^ sin2 a. Corollary 1. If the helix C has constant curvature, then C{ is a circle. If C has constant curvature, then by the theroem, the curvature K{ at any point of Cj is constant and C{ lies in a plane perpendicular to the axis of the cylinder. Therefore C{ is a circle. Corollary 2. A helix of constant curvature is necessarily a circular helix. If the helix is of constant curvature, then by Corollary 1, its base curve is a circle. Hence the curve lies on a circular cylinder. In other words, the curve is a circular helix.

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80

Definition 3. If a curve on a sphere is a helix, then the curve is called a spherical helix. In the case of a spherical helix, we have the following theorem. Theorem 4. A spherical helix projects on a plane perpendicular to its axis in an arc of an epicyloid. Proof. The method of proof is to find the intrinsic equation of the projected curve using the relation between the curvature and arc-length of the helix and the projected curve. Form of the intrinsic equation of the projection on the plane perpendicular to the axis gives the solution. If the helix lies on a sphere of radius r, the sphere is the osculating sphere at every point of the helix so that radius of the sphere is the same as the radius of the osculating sphere. Hence we have R2 = p 2 + (op')2 For a helix, we know that a = p tan a Using (2) in (1), we have .

... (1) ...(2) = ± cot a which gives

V*2-P2 pdf>

V^V

=± cot ads

...(3)

Integrating (3) with respect to s and choosing the constant of integration to be zero, we obtain R2 = p 2 + s2 cot2 a ...(4) Since we consider the projection of a helix on a plane perpendicular to the axis, we have by Theorem 2, K= JCj sin2 aand^j = s sin a ...(5) Using (5) in (4) and simplifying, we get r2sin4 a= p2 + s2cos2 a ...(6) Since cos a < 1, (6) represents the intrinsic equation of on epicycloid. Thus a spherical helix projects on a plane perpendicular to its axis in an arc of an epicycloid.

1.19

EXAMPLES I

1. Find the curvature and torsion of the curves given by (0 y=f(x),z = g(x) (ii) r = (a(3u - u\ 3au2, a(3u + w3)} (i) Taking x as parameter, we represent the position vector of a point on the curve as r=(xj(x),g(x)) Now

r =(!,/(*),«(*))

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Theory of Space Curves

r=(0,/(*),*(*)) r=(0/W,gW)

•(1)

The formula for curvature and torsion are K =

rxr , T |r| 3 ' L

=

[r, r, r ] |rxr|2

••(2)

•

To find /cand r, let us find |r|, |r x r|, [r, r, r] as follows. | r | 2 = l + / 2 + g2

••(3)

r x f = i(fg - fg) -jg + kf so that

|rxr|2=(/i>-/g)2+ip + / 2 1 / [r, r, f ] = 0 / 0

/

..(4)

* gi g

••(5)

= (fg-fg)

Using (3), (4) and (5) in (2), we obtain K2 =

X =

(fg-gff+g2+f2 (l+/2+52)3/2

and

(fg-fg) (fg-fgf+g2 + f2

(ii) From the position vector of a point on the curve, we have

r = {a(3 - 3M2), 6au, a(3 + 3w2)} r = {-6au, 6a, 6au] r = { - 6 a , 0, 6a} 2

2

2

|r| = 18a (l + u f so that |r| = 3aj2(l

.(1) 2

+u )

•(2)

r x f = 1 8 a V - l)i - 36a2uj + 18a2(l + u2)k |rxf|2 = 182-2aV+l)2 Using the determinant formula for scalar triple product as in (i), [r, r, r] = 3. 36 • a\{\ - u2) + (1 + u2)] = 6 V Using (1), (2), (3) and (4), we get K =

|rxf| |r| 3

\%a2Jl{.u2 +1) 27a 3 -2V2( M 2 +l) 3

1 3a(l + « 2 ) 2

...(4)

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Differential Geometry—A First Course

T =

*o 3

[r f r ] |rxr|2

2

-a„

4

3

2

18 2.a (w +1)

1 3a(u + l) 2

2

2

2. Find the curvature and torsion of the curve r = \a(\ + cos w), a sin w, 2a sin — u

Now

r = - a sin w, a cos w, a cos — u 1 2 -fl

r = I - a cos u, - a sin w, i

. 1

sin — w 2

2

-» _ (K// _ ^3 _ ^ ( K 2 b ) + (K// _ ^3 _ CT2j ( C T t )

...(2)

...(3)

-SKT^b- 3 K V T ( - I I ) + ( 2 ^ T + T'K) (-t) - K 2 (2K'T+ T'tfn Hence (r(iv) x i-'") r" = 3 K V T - K 3 (2X'T+ T'K) =

_ ^ ( T * - ^ ) K*

=

_ ^ T | ds\K J

Thus we have proved that

[r(iv),r,,/,r,,]=-K5—f-1. Therefore if the given curve is a helix, then — is a constant so that K

— — = 0. Hence we Bget ds\K)

89

Theory of Space Curves

[r(iv)( r /»

^j

=

_ ^£(l)

=0

...(4)

ds\Kj

Conversely if (4) is satisfied, then - JC*—I — I = 0 so that — is constant. This ds\KJ ds\Kj

K

proves that the given curve is a helix. 9. Show that the helices on a cone of revolution project on the plane perpendicular to the axis of the cone as logarithmic spirals. Hence deduce the intrinsic equation of these helices. Choosing the vertex O of the cone as the origin and the axis of the cone as the z-axis, let P be any point on the cone with the position vector r and Q be the projection of P on the XOY plane. Let OQ = u and xdQ= ft We shall find the locus of Q. If a is the semi-vertical angle of the cone, then we have QP = ucota which is the z-coordinate of P. Hence the position vector r of P is r = (u cos ft u sin ft u cot a) ...(1) where u and 0are variables. Let us express u and 0 as the functions of the arcual length s of the helix. Since the tangent to the helix makes a constant angle j3 with the generator OP, we get r dv r t-— = — •— =cos j3sothatr-dr = |r| cos fids

M Since

ds

lrl

...(2)

vdx =xdx + y dy + zdz= -rd(x2 + y2 + z2) = ~d\r\2

Hence we have rdr = \r\d\r\ Using (2) in the above equation, 4 satisfies the differential equation (1). 15, Show that the torsion of the spherical indicatrix of the principal normal to the given curve is 3(X7C' + TT') (JC'T - XT') + Or* + T2 ) (XT" - « " ) (X- 2 +T Z r +(XT/-KT,V\2 T) From the property (ii) following the definition of the spherical indicatrix of the principal normal of a given curve. We have

2

m-

K2-^2

...(1)

From the equation (3) of Theorem 3 of 1.15, we have ...(2) Differentiating (2) with respect to s, we have

97

Theory of Space Curves

3

K(dsL)

\ds)

db1dsL+bd_(K^Lf l dsl ds ds{

' ds J

= -TCK2 + t V n + [T^K2 + t2) + X(2KK' + 2TT')]t

+ (K? - K'f) (Tb - irt) + [JO" + 2

2

2

KY - KV 2

- K"r]n /

+ tf* + T )^ Tn) + [KfiK + T ) + 2K(KTK + T^Jb x3

Hence

^

(-*,„,) +

b

|

^

J

= S^Kr^ + T^t + C ^ - T O n + S ^ ^ + TOb

...(3)

From the equation (2) of Theorem 3 of 1.15, we have n ^ J - M +t1-^=-K/t-(K2+T2)n+r,b V ds J ds Taking dot product of (3) and (4), we get -

K? h

I - j - J = -WI(KK?

+ ttf) -

(K2

...(4)

+ T2) ( « " - « " ) + 3 « ' ( j a ' + TT")

Hence we have

K h

* [d)

= 3(KK + Xi) { X

'

* ~ KY) + (K2 + T2) (,CT" " ""^

- (5)

"

But from Theorem 3 of 1.15, we have 2_

KT 1 =

(K2 + T 2 ) 3 + ( K T ' - I C ' T ) 2

(r2+r2)3

Using this value of K{ in (5), we finally obtain ?>{KK' + TT') (K-'T - XT') + (K-2 + T 2 ) (CT" - K"'^) T, =

(K2 +T2)3

+(KT'-k'T)2

16. If Xj and xx are curvature and torsion of the evolute of the given curve C, prove that — = - tan (y/+ a) where w- [ r ds Taking a point Tj = r + pn - p tan ((/• + #)b on the evolute of C, we have as in corollary of Theorem 3 of 1.13 ds\ , —- = [p + prtan ( y + a)] sec (y/+ a) and

t{ = cos (y^+tf)n-sin (i/^+a)b

...(1) ...(2)

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Differential Geometry—A First Course

Hence differentiating (2) with respect to s, we have — - — - =cos (y^+^n'-nsin (y/+ a)y/' - sin (y/-\-a)W -b cos (y/+a) y/ dsx ds Since y/ = T, we have ds Kxn{—- = cos (y/+c) ( r b - art)- r n s i n (y/+a) ds + msin(i//'+fl)-bTcos (y/+a) Hence

ds\ jqn,—- =- KCOS (y/+ a)t

...(3)

The above equation shows that the principal normal to the evolute is parallel to the tangent to the given curve C. Hence we can take from (3), nx = - 1 ...(4) Using (4), we have

K{

=

JCCOS (y/+

ds a)— ds{

ds Substituting for — from (1), we get ds{ p' + p r tan (y/ + #) Taking cross product of (2) and (4), we get bj = t{ x n{ = cos (y/+ a)b + sin (y/+ a)n Differentiating the above with respect to s, db\ dsx — - — L = cos (y/+ a) (- rn) - r b sin (y/+ a) + sin (i//+ a) (rb - Kt) dsx ds + rncos (y/+ a) ds\ Hence - r,n,— L = - Ksin (y/+a)t ds ds Since n{ = -1, the above equation becomes, r, = - K*sin (y/+ a)— ds{ Using the value of K, TX =

k

[

d^

H?

sin (y/ + a) cos (y/ + a) ds dsx

Hence —- = - tan (y/+ a) where y/=

o ds.

...(6)

99

Theory of Space Curves

1.20

EXERCISES I 1. Find the curvature and torsion of the curves M,

(i) r =

— ,

l

—^- I

(ii) r = (3a, 3u\ 2u>)

\

2. 3.

4.

5. 6. 7.

(iii) r = (cos 2w, sin 2w, 2 sin w) (iv) r = (4a cos3 w, 4a sin3 w, 3c cos 2w) Find the equation of the osculating plane at a point on the helix r = (a cos w, a sin w, aw tan a) Find the equation of the osculating plane at a general point on the cubic curve r = (w, w2, M3) and show that the osculating planes at any three points of the curve meet at a point lying in the plane determined by these three points. Prove the following relations (i) r'r" = 0 (ii) r'r'" = - / c (iii) r , -r ,,,/ = -3/c/c/ (iv) r" r'" = KK? (v) r ^ r " " ^ * " - * 3 - * * 2 ) (vi) r'"1-"" = K V ' + 2K 3 K' + ^TT' + ^ r 2 Determine the form of the function (p(u) such that the principal normal of the curve r = (a cos w, a sin w, 0(w)) are parallel to the XOYplane. Show that the principal normals at two consecutive points of a curve do not intersect unless r= 0. If there is a one-one correspondence between the points of two curves and tangents at'the corresponding points are parallel, show that the principal normals are parallel and so also their binormals. K \ ds T, Also prove that —L = — = —L. K dsx T

8. Prove that if the principal normals of a curve are binormals of another curve, then ^(K2 + r2) = bK where a and b are constants. 9. Show that the angle between the principal normals at O and P is six? + i2)172 where 5 is the arcual distance between O and P. 10. Find the osculating sphere and osculating circle at the point (1,2,3) on the curve r = (2w+ l,3w 2 + 2,4w3 + 3) 11. Find the osculating sphere at any point of a circular helix. 12. Prove that r = (a cos2 w, a cos u sin w, a sin u) is a spherical curve. 13. Show that the tangent to the locus C, of the centres of curvature lies in the normal plane of the given curve C. If 9 is the angle between the tangent to C{ and the principal normal of C, prove that tan 9 =

P p'a

Differential Geometry—A First Course

100

14. If s{ is the arc length of the locus of centres of curvature, show that 2 dsx _ V^ V + K ' ds K2 15. If C is a curve of constant curvature K> show that the locus C{ of centres of curvature is also a curve of constant curvature KX such that *c, = icand that

K2

i*ts torsion is given by xx = — . T

16. If R is the radius of spherical curvature for any point P(x, y, z) on the curve, prove that

U,,,)2 + (y/,,)2 + a /,, ) 2 = ^ + - J - y P P ° 17. If the involutes of a twisted curve are plane curves, then show that the curve is a helix. 18. Show that the unit principal normal and unit binormal of the involutes of a curve C are Tb-fCt n, =

. ,bj =

Tt + JCb

KKX{C-S)

KKX(C-S)

19. Find the Betrand associate of a circle in a plane. 20. Prove that the corresponding points of the spherical indicatrix of the tangent to C and the indicatrix of the binormal to C have parallel tangent lines. 21. Prove that the curve r = {au> bu2, cu3) is a helix if and only if 3ac = ± 2b2. 22. Show that if the curve r = r(^) is a helix, thenfindthe curvature and torsion of the curve r t = pt + | nds where p, t, n and s refer to the curve r = r(s). 23. Prove that the position vector of a current point r = r(.s) on a curve satisfies the differential equation d_ ds

ds

d a , + — —r + £ r " =0.

ds{p

t

2 The First Fundamental Form and Local Intrinsic Properties of A Surface

2.1 INTRODUCTION As in the case of a space curve introduced either as the intersection of two surfaces or with the parametric coordinates, we shall introduce surfaces in E3 either implicitly by an equation of the type F(JC, y, z) = 0 or parametrically by expressing x, y, z in terms of two parameters w, V varying over a domain. We shall make these two notions more explicit before defining a surface locally as equivalence class of surfaces by a suitable equivalence relation. After defining the surface locally, we classify the points on a surface as ordinary points and singular points. Then we take up for study curves on surfaces and explain how the parametric curves on surfaces help us to study the properties of surfaces. Then with the help of the tangent plane at a point P and the surface normal at P, we introduce a coordinate system at every point of the surface. This system (rl9 r2, N) at any point on the surface is analogous to the moving triad (t, n, b) at a point on the space curve. After introducing certain standard surfaces which we often come across in applications, we shall introduce a certain quadratic differential form on a surface and direction coefficients. This quadratic form is called the first fundamental form which enables us to study the local intrinsic properties of surfaces. We shall conclude this chapter with a brief study of the family of curves on surfaces and isometric transformations.

2.2 DEFINITION OF A SURFACE We.give the two different definitions of a surface and illustrate them with some simple examples. Definition 1. A surface is the locus of a point P(x, y, z) in E3 satisfying some restrictions on x, y, z which is expressed by a relation of the type F(x, y, z) = 0./

102

Differential Geometry—A First Course

The above definition implies that any point on the surface satisfies the equation and conversely. The equation F(x, yt z) = 0 is called the implicit or constraint equation of the surface. This implicit form of the equation describes the surface as a whole so that one can make a global study of the surface. But when we restrict ourselves to the local study of the surfaces which means the study of the properties of the surface in the neighbourhood of a point which is a small region, the constraint equation is not useful. So we are necessitated to use parametric representation of the surfaces in most of the cases. Definition 2. If the parameters w, v take real values and vary over some domain D, a surface is defined parametrically as x =f(u, v), y = g(u, v), and z = h(u, v) where /, g and h are single valued continuous functions possessing continuous derivatives of r-th order. Such surfaces are called surfaces of class r. The parameters u and v are called curvilinear coordinates, (w, V) is used to represent the point determined by u and v. Thus we have two methods of representation of a surface, one is the global representation by using a constraint equation and another by parameters w, v varying over a domain. Hence the question naturally arises whether the two methods are equivalent under a suitable equivalence relation. Before answering this question, we point out some disadvantages in these representations by a few examples. (i) The parametric equations of a surface are not unique. To see this, we produce a surface having two different parametric representations. Now consider the following two sets of equations x = u + v,y = u-v, z = 4uv x = u,y = vyz = u2-v2 Elimination of the parameters in both the representations lead to the same constraint equation x2 - y2 = z which represents the whole of certain hyperbolic paraboloid, (ii) Sometimes the constraint equation obtained by eliminatipg the parameters represents more than the given surface. To see this consider the parametric equation x = u cosh v, y = u sinh v, z = u2 for all real values of u and v. Eliminating u and v among the equations, the constraint equation of the surface is x2 - y2 = z. The constraint equation represents the whole of hyperbolic paraboloid, while the parametric equations give only that part of the hyperbolic paraboloid for which z > 0, since u takes only real values. Definition 3. Let there be two parametric representations w, v and u\ v of the same surface. Any transformation of the form u = 0(w, v) and v = I/A(W, V) relating these two representations is called a parametric transformation.

The First Fundamental Form and Local Intrinsic Properties of A Surface

103

Definition 4. A parametric transformation is said to be proper if (i) (p and i/^are single valued functions and (ii) The Jacobian — ^ 0 in some domain D. d(«,v) Note. Let D' be the domain of u\ v corresponding to the domain D of the w, v plane. The conditions in the above definition are the necessary and sufficient conditions for the existence of the inverse in the neighbourhood of any point D' which means that the transformation is locally one to one. However it should be noted that the transformation 0, i//may not have the inverse on the whole of D.

2.3

NATURE OF POINTS ON A SURFACE

To describe the nature of points on a surface, we introduce the following notation. Let r = (x, y, z) be the position vector of a point on the surface. Since JC, y, z are continuous functions of parameters w, v possessing partial derivatives of required order, we can take r = r(w, v) as the paramatric equation of the surface. If the suffixes 1 and 2 are used for partial derivatives of r with respect to u and v respectively, let

9r r, = — and r 2 = ou d\_ _

9r — ov 32r

...(1) _ 32r

r

r

_32r r

' . . -D^ 2T' .1 2' — B -3 ^3 : .' -21i - , -3 ^3 : ». '2 2» - -

T ~! i 3

ou ouov ov ou ov Since r possesses continuous partial derivatives, we have r, 2 = r21. Since r = (x, y, z), we can express r^ r 2 componentwise as

(dx dy dz\

,

,

I ou ou ou ) dx dy dz ro = {dv* 9 v ' d v

= (x2iy2iz2)

...(2)

and we have similar expressions for r22, r H , r,2 and r21 Definition 1. If r, x r2 * 0 at a point on a surface, then the point is called an ordinary point. A point which is not an ordinary point is called a singularity. From the very definition of an ordinary point, we note the following properties of a surface. (i) using (2), we have r, x r 2 = i(y,z2 - Z[y2) +j(z{x2 -x{z2) + k(x{y2 - yxx2)

...(3)

r, x r 2 * 0 means that one of the coefficients in (3) is different from zero. That is at least one of the members (yi*2 - zxy2), (z,* 2 -*\Zi), (*\y2 ~ y\xd * 0

...(4)

Differential Geometry—A First Course

104 (ii) Let us consider the matrix M=

*i x2

y\ y2

zi z2_

Since at least one of the members in (4) is different from zero at an ordinary point, there exists at least one determinant minor of order 2 of M which is different from zero. In other words the rank of M is two at an ordinary point. As a consequence, if the rank of M is either zero or one, the point on the surface is a singular point. (iii) If r, x r 2 * 0 or equi valently if the rank of the matrix M is two, then JC, y, z uniquely determine the parameters w, v in the neighbourhood of an ordinary point. Since the matrix M is of rank two, there exists at least one non-vanishing Jacobian which we can take as - — — * 0. 3(w, v) As the condition of the inversion theorem is satisfied, there exist neighbourhoods Ns(x0, y0) and Ne(w0, v0) such that for every x, y e N5(;c0, y0), there exist w, v e Ne(w0, v0) such that u = u(x, y) and v = v(x, y). Hence u and v are determined by x, y, z in the neighbourhood of an ordinary point. (iv) The points where the rank of M is 1 or zero are singular points. If the rank of M is 1, then every determinant minor of order two of M is zero. This implies (yi^2 " ^l)^)' (*i*2 -*i*2)» C*^ ~y 1^2)are a ' l z e r o s o l ^ a t r i x r 2 = 0. Hence the point where the rank of M is one is a singular point. When the rank of M is zero, then all the determinant minors of order 1 or 2 are zero. This implies as in the previous case r{ x r 2 = 0 at these points so that the point where the rank of M is zero is a singular point. Note. When only one determinant minor of M is zero, we cannot conclude that the point is a singular point. We shall illustrate the above properties by the following examples. Example 1. Consider the surface given parametrically by x = u + v, y = u + v, z = uv. ^ _ ^ _ 9 y = 3 y = du dv ' du 9v Hence x{y2-yxx2

= 0butx x z 2 -x 2 z { =

1

^ _ du

9^ ' 9v

=

u-v*0.

Thus the rank of M is 2 at every point of the surface so that every point on the surface is an ordinary point. (v) A proper parametric transformation transforms an ordinary point into an ordinary point. Let r = r(w, v) be the equation of the surface, and let u' = 0(«, v), v' = y 3M

3V

3M

3V 3(M, V)

Since the given parametric transformation is proper,

3(0, v0 3(M,V)

*0.

Hence — X — * 0 implies -— x —- ^ 0. That is r/ x r{ * 0, proving that an 3M 3V 3M' dv' ordinary point is invariant after proper parametric transformation. Note 1. Since I*! x r2 * 0 at an ordinary point, I*! x r2 = 0 at a singularity. Due to some geometrical nature of the surface, some singularities continue to be singularities, whatever may be the parametric representations. Such singularities are called essential singularities. There are other singularities depending upon the choice of parametric representation. Singularities of this type are called artificial singularities. Note 2. Tofindthe nature of a point on the surface, we use either the matrix M or r, x r2. We shall illustrate the essential and artificial singularities in the following examples. Example 2. Consider the circular cone represented by x = M sin a cos v, y = u sin a sin v, z - u cos a where a is the semivertical angle of the cone with the vertex 0 as origin and OP = u,P any point on the cone. We show that the vertex of the cone is an essential singularity. Since u and v are parameters, we have M=

sin a cos v sin a sin v cos a -u sin a sin v u sin a cos v 0

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At u = 0, the determinant of every second order minor of M is zero. Hence the rank of M is zero so that u = 0 is an essential singularity. This singularity is essential since it arises as a result of the vertex of the cone. Taking r = (u sin a cos v, u sin a sin v, u cos a), r{ x r 2 = 0 at u = 0 showing that u = 0 is an essential singularity. Example 3. Taking any point 0 as the origin in the plane,x = u cos v, y = u sin v, z = 0 is the representation of the plane in polar coordinates. Now r, = (cos v, sin v, 0), r 2 = (- u sin v, u cos v, 0) then r, x r 2 = wk. Hence Fj x r 2 = 0, when w = 0 so that u = 0 is a singularity. It is an artificial singularity, since it arises due to the choice of the parametric coordinates and not due to the nature of the surface. It is to be noted thatw = 0 is not a singularity in the coordinate system. We can arrive at the same conclusion by noting that the rank of the matrix M is zero at u = 0.

2.4

REPRESENTATION OF A SURFACE

In our study of surfaces, we shall consider only ordinary points on surfaces. This means that the domain of parameters w, v will be restricted so that every point (w, v) of the surface is an ordinary point. Also we shall study the properties of the surface in the neighbourhood of an ordinary point. For such a study, the proper parametric transformation is very useful, since it is locally one-to-one. Since such a study depends upon a portion of a surface, we consider the entire surface as a collection of parts, each part being given a particular parameterisation and the adjacent parts are related by a proper parametric transformation. Using these ideas, we shall define the representation of a surface as follows. Definition 1. A representation R of a surface S of class r in E3 is a collection of points in E3 covered by a system of overlapping parts {Sy} where each part Sj is given by a parametric equation of class r. Each point lying m the common portion of two prats 5,-, Sj is such that the change of parameters from one part S( to its adjacent part Sj is given by a proper parametric transformtion of class r. Note. Since we cannot parametrise the whole surface without introducing artificial singularities, we resort to consider a surface composed of many overlapping parts. Since the points in the adjacent parts have two parametric representations one for St and another for its adjacent Sy, these two parametric representations are connected by a proper parametric transformation. In the definition of the representation R of a surface, we are concerned with the system of overlapping parts Sj covering the whole surface. Hence it is possible to have many representations of the same surface by considering different systems of overlapping parts (SJ), each part is given by a parametric equation of class r. Since we have different representations of the same surface, it is but natural to state precisely, when the two representations R and R' behave alike. This leads to the notion of equivalence of representations of surfaces of class r and consequent definition of a surface as an equivalence class. Definition 2. Let R an R! be two representations of class r of the surface S. Let (Sj) and (Sj') be two different systems of overlapping parts covering S

TJie First Fundamental Form and Local Intrinsic Properties of A Surface

107

corresponding to R and R'. Then they are said to be equivalent, if the composite family of parts {Sp Sf} satisfy the condition that each point P lying in the common portion of the overlap of two parts, the change of parameter of P considered as a point of Sj to the parameter of the same point considered as a point of Sf is given by a proper parametric transformation of class r. That is if P is a point in the place of overlap, the change of parameter from S- to Sf at the point P is given by a proper parametric transformation of class r. Theorem. The notion of r-equivalence of representations of a surface is an equivalence relation. Proof. Let R be a representation of S and let S be composed of overlapping parts {Sj}. Since the change of parameters from 5,- to Sj is given by a proper parametric transformation of class r, the relation of r-equi valence of representation R is reflexive. Let the relation R be equivalent to R' and let Sj and Sj' be two overlapping parts in two representations with a pointP in the overlapping portion. Since/? and/?' are equivalent, there exists a proper parametric transformation 0 at P from Sj to Sj'. Since the proper parametric transformation is locally one-to-one and possesses inverse transformation, 0~l exists at the point P of overlap of S. and Sj. In other words, there exists a proper parametric transformation 0~l from S' to Sj. Thus R' is equivalent to /? so that the relation of r-equivalence of class r is symmetric. Let /?, R' and R" be any three representations of class r of a surface S and let them be r-equi valent such that/? ~ /?' and/?' ~ /?". We shall show that/? - /?". Since /? and R' are equivalent, there exists a proper parametric transformation 0 at the common point Px in the overlap of the family [Sj, Sj'}. Since R' ~ /?", the change of parameter of a point in the overlap of Sj' and Sj" is given by a proper parametric transformation i//from S,' to Sj". Since 0and y/are locally one-to-one, y/o(p is locally one-to-one transformation giving the change of parameter from Sj to Sj". Hence the representation /? and R" are equivalent so that the relation of equivalence of class r of surfaces is transitive. Since the notion of the relation of equivalence of class r is reflexive, symmetric and transitive, it is an equivalence relation which completes the proof of the theorem. This equivalence relation introduces a partition into the family of surfaces of class r splitting them into mutually disjoint equivalence classes, each class containing the surface equivalent to one another in the above equivalence relation. This leads to the formal definition of a surface as follows. Definition 3. A surface S of class r in E3 is an r-equivalence class of representations. Thus a surface consists of different overlapping portions related to one another by proper parametric transformations and all other surfaces related to the given one by the equivalence relation of class r. We make a study of local properties without investigating the extent of the region of the surface in which the local properties are true.

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108

2.5

CURVES ON SURFACES

Let r = r(w, v) be the equation of a surface of class r where w, v vary over a domain in the wv-plane. Let u = u(t) and v = v(/) be a curve of class s lying in the domain D of the wv-plane. Now consider r = r[u(t), v(t)]. Then r gives the position vector of a point in terms of a single parameter t so that r = r [w(0, v(r)] is a curve lying on a surface with class equal to smaller of r and s. The equation u = u(t) and v = v(t) are called curvilinear equations of the curve on the surface. Definition 1. (Parametric Curves). Let r = r(w, v) be the given surface of class r. Let v = c where c is an arbitrary constant. Then the position vector r = r(w, c) is a function of a single parameter t and hence r = r(w, c) represents a curve lying on the surface r = r(w, v). This curve is called the parametric curve v = constant. For every value of c, there is one such curve on the surface. Since v = c, where c is an arbitrary constant, we get a system of parametric curves for different constant values of c. In a similar manner, if we keep u constant and vary v, we get a system of parametric curves u = constant. Since we are concerned with the ordinary points on the surface, we note the following basic properties of parametric curves. These properties are the consequences of the fact that we are concerned only with ordinary points on the surface. (i) Through every point of the surface, there passes one and only one parametric curve of each system. Let P(x0, y0, z0) be a point on the surface. Then as explained in (iii) of 2.3, (M0, V0) are uniquely determined by (*0, y0, ZQ). Hence there are only two parametric curves u = w0» v = v0 passing through the point P. (ii) No two curves of the same system intersect. Let us consider the system at the point (w0, v0). Let u = u0 and u = ux be two curves of the same system. If these two curves of the same system intersect, u = ux at the point of intersection so that the parametric coordinates determined by P(x0, y0, z0) is (ux, v0) contradicting the uniqueness of (w0, v0). This contradiction proves that no two curves of the same system intersect, (iii) The curves of the systeem u = u0 and v = v0 intersect once but not more than once if (w0, v0) e D. Since the point of intersection (M0, V0) is uniquely determined by P(xo> yo» Zo)» tlie y c a n n o t intersect more than once, (iv) The parametric curves of the system u = cx and v = c2 cannot touch each other. For a curve v = c, u serves as a parameter and determines a sense along the curve. The position vector of a point on the curve v = c is r = r(«, c). Hence the dr tangent to the curve v = c in the direction of u increasing is rx = —-. Similarly au

The First Fundamental Form and Local Intrinsic Properties of A Surface

r 2 = — gives the direction of the tangent to the curve u = c in the direction of v dv increasing. They do not vanish and have different directions. Since we consider r{ and r 2 only at an ordinary point on the surface rx x r 2 * 0. This shows that the two parametric curves are neither coincident nor parallel but cut at the point (MQ, V0) determined by (x0 y0, ZQ). Hence they do not touch each other. Definition 2. Let u = cx and v = c2. When the constants cx and c2 vary, the whole surface is covered with a net of parametric curves, two of which pass through every point, (w, v) are called the curvilinear coordinates of P. The parametric curves are called coordinate curves. Definition 3. Two parametric curves through a point P are said to be orthogonal if rx. r 2 = 0 at P. If this condition is satisfied at every point (w, v) of the domain, than the two system of parametric curves are orthogonal. 2.6

T A N G E N T PLANE A N D SURFACE N O R M A L

Let r = r[w(0, v(r)] be a general curve lying on the surface passing through [w(0, ^(01- Then the tangent to the curve at any point P on the surface is dx dr du dr dv du dv /1X — =+ = r, — + r2 — ...(1) dt du dt dv dt dt dt Definition 1. Tangent to any curve drawn on a surface is called a tangent line to the surface. dr From (1), we see that the tangent vector — is a linear combination of the dt vectors rx and r2. Since rx x r 2 * 0, rx and r 2 are non-zero and independent. The tangents to different curves through P on a surface lie in a plane containing two independent vectors rx and r 2 at P. This plane is called the tangent plane at P. Theorem 1. The equation of a tangent plane at P on a surface with position vector r = r(w, v) is either R = r + arx + br2 or (P - r) • (r{ x r2) = 0 where a and b are parameters. Proof. Let r = r(«, v) be the position vector of a point P on the surface. The tangent plane at P passes through r and contains the vectors vx and r2. So if R is the position vector of any point on the tangent plane at P, then R - r, rx and r 2 are coplanar. Hence we have R - r = arx + br2 where a and b are arbitrary constants. r{ x r 2 is perpendicular to the tangent plane at P. Hence r{ x r 2 is perpendicular to R - r lying in the tangent plane so that (R - r) • (r{ x r2) = 0 is another form of the equation of the tangent plane at P.

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Differential Geometry—A First Course

110

Definition 2. The normal to the surface at P is a line through P and perpendicular to the tangent plane at P. Since r{ and r 2 lie in the tangent plane at P and pass through P, the normal is perpendicular to both r, and r 2 and it is parallel to r{ x r 2 as in the adjoining Figure 6. The normal at P is fixed by the following convention.

Fig. 6

If N denotes the unit normal at P, then r,, r2 and N in this order should form a right handed system. Using this convention we have

Mr2|

H

2l

Since r, x r 2 * 0, H = \r{ x r 2 | ± 0 which shows that it is always a positive number and NH = r{x r2. Theorem 2. The equation of the normal N at a point P on the surface r = r(w, v) is R = r + a(r{ x r2). Proof. Let R be the position vector of any point on the normal to the surface at P whose position vector is r = r(w, v). Since r , x r 2 gives the direction of the normal and (R - r) lies along the normal, r, x r 2 and (R - r) are parallel so that we have R - r = a(r{ x r 2 ) where a is a parameter. Hence R = r +.fl(r, x r 2 ) gives the equation of the normal at P. Using the convention that r h r 2 and N form a right handed system, we establish the following theorem. Theorem 3. A proper parametric transformation either leaves every normal unchanged or reverses the direction of the normal. Proof, r = r(w, v) be the given surface and let the parametric transformation be u = a> the above vector is negative for the range of values of u and v so that the normal is directed inside the anchorring,since \rx x r2| is always positive. Note. The coordinates of a point A on the generating circle in XOZ plane is (b + a cos w, 0, a sin w). Hence taking g(u) = b + a cos w, /(«) = a sin w in Theorem 1, we can obtain the representation of a point on an anchor ring.

2.8

HELICOIDS

In the above examples, we considered surfaces obtained only by rotation about an axis in its plane such as spheres, cone and anchorring.But there are surfaces which are generated not only by rotation alone but by a rotation followed by a translation. Such a motion is called a screw motion. The simplest case of a screw motion is the motion of the x-axis through a rotation about the z-axis and translation in the positive direction of the z-axis. Usually we take the angle v through which the positives-axis rotated is proportional to the distance X in the upward direction so A that — is constant. The surface generated by the screw motion of the *-axis about v the z-axis is called a right helicoid. So we shall derive the equation of the right helicoid before taking up the general case. (i) Representation of a right helicoid. This is the helicoid generated by a straight line which meets the axis at right angles. If we take the *-axis as the generating line, it rotates about the z-axis and moves upwards. Let O'P be the translated position of the *-axis after rotating through an angle v. Let (*, y, z) be the coordinates of P. Draw PM perpendicular to the XOY plane and let OM = u. Then x = u cos v, y = u sin v, and z = PM. By assumption the distance PM = z translated by the s-axis is proportional to the angle v of rotation. Taking the constant of proportionality to be a, let — = a. v

Hence the position vector of any point on therighthelicoid is r = (u cos v, u sin v, av) Now Tj = (cos v, sin v, 0), r2 = (-u sin v, u cos v, a) Since r{ • r2 = 0, the parametric curves are orthogonal. When u = constant c (say), then the equation of the helicoid becomes r = (c cos v, c sin v, av) which are circular helices on the surface. The parametric curves v = constant are the generators at the constant distance from the XOY plane. a Hence the unit normal N = ,

+u

(a sin v, - a cos v, u)

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Differential Geometry—A First Course

Definition. If v = In, then 2na is the distance translated after one complete rotation. This is called the pitch of the helicoid. (ii) Representation of the general helicoid. The general helicoid with thez-axis as the axis is generated by the curve of intersection of the surface with any plane containing z-axis. Since the section of the surface by such planes are congruent curves, without of loss of generality, we can take the plane to be XOZ plane and generate the helicoid. Thus the equation of the generating curve in the XOZ plane can be taken as* = g(u),y = 0, z =f(u). Let the curve in the XOZ plane rotate about the z-axis through an angle v and let it have the translation proportional to the angle v of rotation which we can take it as av. Since any point on the generating curve traces a circle with centre on the z-axis and radius g(u) and z-coordinate is translated through av> the pbsition vector of any point r on the general helicoid is r = (g(u) cos v, g(u) sin v,/(w) + av) Now

r, = (g'(u) cos v, g\u) sin v,/'(«)) r 2 = (- g(u) sin v, g(u) cos v, a)

Further

rY • r 2 =f'(u) a.

Hence when the parametric curves are orthogonal, then either f'(u) = 0 or a = 0. lff(u) = 0,/(w) is constant so that the surface is a right helicoid. If a = 0, we do not have screw motion and we have only rotation about z-axis so that the helicoid is a surface of revolution. When v = constant, the parametric curves are the various positions of the generating curve on the plane of rotation. When u = constant/it follows from the equation of the helicoid, the parametric curves are helices on the surfce.

2.9

METRIC ON A SURFACE—THE FIRST FUNDAMENTAL FORM

Analogous to the arcual length ds2 in the case of a space curve, we shall introduce a metric on a surface called the first fundamental form. Let r = r(w, v) be the given surface. Let the parameters M, V be functions of a single parameter t. Then r = v[u{t\ v(f)] is a function of a single variable t and hence it represents a curve on the surface with t as parameter. The arc length in terms of the parameter t is given by

«

2

\dt)

=*.*=r*Y dt

dt

_ But

dr dr du dr dv — =+dt du dt dv dt Using (2) in (1), we get \2

{dtj

{l dt

...(1)

\dt)

2

dt)

...(2)

The First Fundamental Form and Local Intrinsic Properties of A Surface

(du\2

„

119

fdv\2

du dv

= iyr,

Let

E = iy r, = if, F = iy r 2 and G = r2- r 2 = r2

...(4)

Using the above notation, (3) can be rewritten in terms of the differentials as ds2 = Edu2 + IF du dv + G dv2

...(5)

Definition 1. The differential quadratic form (5) is called the first fundamental form or metric on the surface. It is usually denoted by /. Note 1. The expression for ds1 in (5) is independent of t and so it can be considered as the infinitesimal distance between two points with parameters (w, v) and (u + du,v + dv) on the surface. Let P and Q be two neighbouring points on the surface with position vectors r and r + dr corresponding to the parameters M, V and u + du, v + dv. Now dr = — du + —-dv = rxdu + r2dv du ov

...(1)

Since P and Q are two neighbouring points, the length ds of the element of the arc joining them is equal to \dr\. Using (1), we get ds2 = drdr

-dr2 = {rxdu + r2dv)2

= r2du2 + 2rl'T2 du dv + r 2 dv 2 = Edu2 + IF dudv + G dv2 Thus if ds denotes the length of the elementary arc joining (w, v) and (u + du,v + dv) lying on the surface, then ds2 =Edu2 + IF dudv + G dv2 From (2), we get 5

—

= E\ —

{dtj

\dt)

+ IF

V \J. Note 3. Since the square root of the first fundamental form gives the length \dr\y it is called the metric of the surface. Though the metric is usually employed for calculation of the arc length of a curve on the surface, the coefficients E, F and G are used to study many important properties of the surfaces. They are functions of parameters w, v and called first fundamental coefficients. Note 4. On the parametric curve v = constant, we have dv = 0 and the metric reduces to ds2 = E du2. In a similar manner, on the parametric curve u = constant, ds2 = Gdv2.

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Differential Geometry—A First Course

Theorem 1. The first fundamental form of a surface is a positive definite quadratic form in du, dv. Proof. A quadratic form Q = an x2 + 2 a[2x{ x2 + d22x2 f° r r e a l values of JC,, x2 is called positive definite if Q > 0 for every (*,, x2) * (0, 0). The conditions for Q to be positive definite are aX2 = a2l,an

>0and an a22-al2

>0

...(1)

We shall verify that I = Edu2 + 2Fdudv + G dv2 ...(2) satisfies the above conditions for a positive definite quadratic form. Since we are concerned only with an ordinary point, r, x r 2 ^ 0 and // 2 = | n x r 2 | 2 > 0 Now H2 = |rj x r 2 | 2 = (i^ x r2) • (r, x r2) = r 1 2 r 2 2 -(r 1 .r 2 ) 2 = E G - F 2 . Since H2 > 0 always, EG-F2>0

and

Also we have from the definition E = r r r{ = r 2 > 0 and G = r2- r 2 > 0 Hence (2) satisfies the conditions (1) for a positive definite quadratic form. So we have E du2 + 2Fdudv + G dv2 > 0 for all values of (w, v) * (0, 0) Note. We can also directly verify / > 0 as follows. We shall rewrite / as follows E du2 + 2Fdudv + G dv2 = —[(Edu+F dv)2 + (EG - F2)dv2] ...(1) E Since EG - F2 > 0 and E > 0, we have / > 0. To prove that / is positive, we have to show that / * 0 for all (w, v) * (0, 0). If possible, let 7 = 0. Then (1) implies (Edu + F dv)2 = 0 and (EG - F2) dv2 = 0 Since

2

EG-F >

...(2)

0, we get from (2), dv = 0

The above condition (2) reduces to E du + F dv = 0 and dv = 0. which implies Edu = 0. Since E*0,du = 0. Thus we have du = 0 and dv = 0 which cannot be, since u and v are not constants. Thus we get / * 0 so that / > 0. Theorem 2. The metric is invariant under a parametric transformation. Proof. Let the parametric transformation be u' = Q(u, v) and V = y/ («, v) In the parametric system u\ v\ let the equation of the surface be r = r(u\ v') Hence

, dr dr du dr dv du dv r{ = — - — — + — — = rx — + r2 —• du du du dv du du du

...(1)

Tlie First Fundamental Form and Local Intrinsic Properties of A Surface

du

dv_

121

-.(2)

Similarly

r2 = r,

Further

du du = . du'

du'^dv' dv'

...(3)

and

dv , dv , dv = z—;du + ^—dv du dv

...(4)

• + r,

dv'

If E', F ' and G' are the first fundamental coefficients in the new parametric system, then we have E'du'2 + 2F'du'dv' + G'2dv'2 = r,/2du'2 + 2r,'• r2'du'dv' + r2'2 dv'2 = [r{ du' + r{ dv']2 Using (1) and (2) in the above step, we obtain du dv du dv - + Yj \du + r, + r2 —- \dv' du' l du' dv' dv' du

,

du

du + du'

, dv

dv'

12

dv , , dv , , + ^—dv' + r, ^—du' du' dv'

Using (3) and (4) in the above step, we obtain E' du'2 + 2F' du'dv' + G'dv'2 = (r 1 du + r2 dv)2 = r,2 du2 + 2r{'r2dudv

+ r22 dv2

= Edu2 + IF dudv + G dv2 Note. Though the metric as a whole is invariant under parametric transformation, F, F and G are not individually invariant. That is, the coefficient F', F ' and G' considered as a function of u\ V does not give the function F, F, and G when the variables u, v' are replaced by