Unit-7

GEODESIC

Lesson Structure

7.0 Objective

7.1 Introduction

7.2 Gaussian Curvature, Principal Direction and Principal

Curvature

7.3 Geodesic

7.4 Fundamental Equations of Surface Theory.

7.5 Summary

7.6 Model Questions

7.7 References

7.0 Objective

The objective is to study curves in space and of surfaces in three dimensional Euclidean

Space E3.

7.1 Introduction

Differential Geometry is the study of geometric figures using the method of calculus

Since the geometric character of curves and surfaces varies continuously, calculus is

used to study the properties of curves and surfaces in the neigbourhood of a point.

Properties of curves and surfaces which depend only upon points close to a particular

point of the figure are called local properties. The study of local properties is called differential

geometry in the small. Those properties which involved the entire geometric figure are called

global properties. The study of global properties (since they relate to local properties) is

called differential geometry in the large. We will first investigate local properties of curves

and surfaces and the apply the results to problems of differential geometry.

7.2 Gaussian curvatures, principal direction and principal curvature

7.20 Curvature of a curve on the surface.

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Let S be a curve given by r r u v

( , ) and C be a curve on S.

Then at an arbitrary point P (u, v) of C, we know that d r

dst and

d t

dsK n

^

^^

So , Kn Ndt

dsN

^ ^^

^. . where N is normal vectors at P

Kdt

dsNcos .^

...(1) where is angle between the principal normal vector n^

and

the normal vector N^

. Again since t N^ ^. 0

So,d t

dsN t

dN

ds

^^ ^

. . 0

i.e. K td N

dscos ;

^^

using (1)

FHGG

IKJJ

d r

ds

d N

ds

^ ^

.

d r dN

ds

^.

2

Or, cos

Ldu Mdudv Ndv

Edu Fdudv Gdv

2 2

2 2

2

2...(2) where K

1

(using first and second fundamental form)

If Edu Fdudv Gdv2 22 0 then we can not define 1 K at p(u, v)

7.20 Theorem :

Two curves on the same surface S through a fixed point P (u, v) have the sence curvature

at P, if they have the same osculating plane at P and their common direction, dv

du at P is not an

asymptotic line.

Proof :

If the direction at P is not asymptotic then Ldu M dudv N dv2 22 0 0 , cos and

hence from (2) we get a finite non-zero value of K 1

The value of E, F, G, L, M, N are all equivalent at P and dv

du and cos depend only on the

position of the tangent and principal normal to the curve at P. Thus 1 depends only on the

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position of the osculating to C at P which contains both the tangent and principal normal to

the curve at the point P.

N.B. The curvature at P of the given curve C is equal to the curvature at P of the plane

curve in which the osculating plane of C at P cuts the surface S. Hence, we may consider

curvature of plane section of a surface only.

7.21 Normal Sections :

Consider all the sections of a given surface S by means of planes drawn through the

tangents line PT which has the given non-asymptotic direction dv

du. Then the section whose

plane is normal to the tangent plane at P is called the Normal Section. All other section are

called oblique sections.

So for the normal section 0 or according as the principal normal at P have

the same sense as N^

or not.

So, the curvature Kn of the normal section at P.

is 0 or (using 2) is

(Normal curvature)

K

Ldu Mdudv N dv

Edu Fdudv Gdvn

2 2

2 2

2

2

When = 0, + Ve sign is taken and when = negative sign is taken.

7.20 Definition : (Normal Curvature (Kn)) The normal curvature vector to a curve C

at P is the vector projection of the curvature vector K of C at P onto the normal N^

at P. So,

K K N Nn ( . )^ ^

7.21 Theorem : If N is a surface normal then

(1)

N

u

N

v

LN M

EG FN

2

2

(2) Kn = K cos where is angle between the oblique section and the normal section.

Proof : (1) If N1 and N

2 denote partial differentiation with respct to u and v respectively

then

Since N . N = 1

N . dN = 0 (On diff.)

N. N1 = 0 and N. N

2 = 0 ...(1)

N1 × N

2 = N

N · (N1 × N

2) = N . N =

[ , , ]N N N

d

D

LN M

EG FN1 2

2 2

2

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(See first and second fundamental forms)

(2) If n and N have same senses then for a section dr

dst

dt

dsr K Nn ,

K N rn . ' ' ...(2)

Also, cos . .' '

N n Nr

K ( ' ' ' ) r t Kn

n

Kusing;( ( ))2

So, K Kn cos

7.22 Me uniev’s Theorem :

Statement : If 0 and be the radii of curvature of a normal section and an arbitrary

section through the same tangent line then = 0 cos where is the angle between the two

section.

Proof : See above

Example 7.20 : Find the curvature of a normal section of the right helicoid x = u cos

y = u sinz = c

Solution : The position vector r

of any point on the given surface can be taken as

r u i u j c k

cos sin^ ^ ^

( cos , sin , )u u c u as being parameters

Then r

1 0(cos ,sin , ) taking u and first variable.

r u u c

2 ( sin , cos , ) being second variable.

Similarly r r

11 120 0 0 0( , , ), ( sin ,cos , ) and r u u

22 0( cos , sin , )

So, E r F r r G r u c

12

1 2 22 2 21 0, ,

and D EG F u c 2 2 2

Also, Nr r

D

c c u

D

1 2 ( sin , cos , )

L r N M r NC

DN r Nu

. , . , .0 012 22

Now curvature of the normal section is given by

KLdu Mdud Nd

Edu Fdud Gdn

2 2

2 2

2

2

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duC

u cdud d

du dud u c d

2

2 2

2

2 2 2 2

0 2 0

1 2 0

.

. ( )

1 22 2 2 2 2 2

u c

cdud

du u c d

[ ]

[ ( ) ]

Example : 7.21 Show that the curvature K at any point P of the curvature of intersection

of two surfaces is given by K K K K K K K212

22

1 2 1 22sin cos , , being normal curvature

of the surfaces in the directions of the curve at P and – angle between their normals at P.

Proof : Take a point P on the curve C obtained by intersection of two surfaces S1 and S

2

and let N1 and N

2 denote unit normal vectors to S

1 and S

2 respectivly at P of C. Since is

angle between their normal at P.

So, N1, N

2 = cos ...(1)

Let n denote the principal normal vector of the common oblique section of S1 and S

2 at

P of the curve c through the tangent line PT.

Let 1 be angle between n and N

1.

Then the angle between n and N2 is – .

If t denote unit tangent vector along PT then

t.N1 = 0, t. N

2 = 0 and t · n = 0 ...(2)

clearly, N1, N

2 , n all are coplanar and

perpendicular to t.

Now, using Meuniev’s theorem

for surface, K K1 cos and for surface S2

K K2 cos( )

K(cos cos sin sin )

K K121cos sin cos

K KK

K1

12

21cos sin

Or, ( cos ) sin ;K K KK

K2 1

2 2 2 12

21 FHG

IKJ (On squaring)

Or, K K K K K2 212

22

1 22sin cos

We have seen in Meuniev’s that the curvature at a point P of an arbitrary curve on the

N2 S2

n

N1

S1

T Ct^^

P

^

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surface is related to the normal curvature at P in the direction of the curve. We will now

discuss the cases as how normal curvature changes when the direction dv

du of the normal

section varies.

The normal curvature at P in the direction dv

duis given by

KLdu M dudv Ndv

Edu Fdudv G dv

2 2

2 2

2

2

L M N

E F G

2

2

2

2

; ....(2) on putting

dv

du

FHG

IKJ

Case (i) : When the normal curvature at P is the same in every direction.

On putting L = µE, M = µF and N = µG in (2) we see that K = µ (constant) i.e. the normal

curvature becomes same in all direction. In this case, the point P is an unbilical point. When µ

0, this point is circular point and when µ = 0, this point is a planar point.

Here we are in a position to say that a surface all of whose points are umbilics is a

sphere or a plane.

Case (ii) When L, M, N are not multiple of E, F, G.

Let FN – GM 0. Since ds = Edu2 + 2F dudv + Gdv2 > 0. So K exists (from (i)). This

means that the given curve is continuous for all For a given value of K, the two values of

are determined from the quadratic equation in which indicate that a line parallel to -axis

cuts the curve in at most two real points.

If dv

duK

N

G, ; from (2). In that case the curve has a horizontal asymptote

which it approaches in both the positive and negative directions.

On putting KN

G in (2), we get

2( ) ( ),FN GM GL EN which gives a finite value of ( ). FN GM 0

This gives that the curve cuts the asymptote KN

G at a finite point.

So, if we plot the curve given by (2) (representing K as a function ) we get the following

shape.

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As such the normal curvature has two extreme values, an absolute maximum and an

absolute minimum.

7.21 Principal Direction : Two direction in which normal curvature has its maximum

and minimum values are known as the principal directions at the point.

Def. 7.22 : Principal Section : The normal section of a surface at P having maximum

and minimum curvature are called the principal sections of the surface.

We know that KLdu Mdudv Ndv

Edu Fdudv Gdv

2 2

2 2

2

2

Or, K

L Mdv

duN

dv

du

E Fdv

duG

dv

du

L M N

E F G

FHIK

FHIK

2

2

2

2

2

2

2

2

On puttingdv

du

FHG

IKJ

Diff. w.r to and putting dK

d 0, we get

( ) ( ) ( ) ( )

( )

E F G M N L M N F G

E F G

2 2 2 2 2 2

2

2 2

2

= 0

2 0( ) ( ) ( )FN GM EN GL EM FL

This gives principal direction.

Again putting dv

du,we get

( ) ( ) ( )EM FL du EN GL dudv FN GM dv 2 2 0

which gives differential for principal direction at the point P (u, v)

7.22 Theorem : Prove that the principal direction at a point given by the differential

equation.

( ) ( ) ( )EM FL du EN GL dudv FN GM dv 2 2 0 ....(1)

are mutually perpendicular.

Proof : If we put EN – FL = l, EN – GL = 2m and FN – GM = n

then the differential equation reduces to

ldu mdudv ndv2 22 0 ...(2)

Then we know n.&s.c condition that the system (2) is orthogonal is that En – 2Fm + Gl = 0

ie. E (FN – GM) – F (EN – GL) + G (EM – FL) = 0

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which is true.

Hence two direction given by (1) are orthogonal.

7.22 Principal Normal Curvature :

It is the maximum and minimum values of the normal curvatures at any point P are

called the principal normal curvautres at P.

7.23 Theorem : The principal normal curvature at P are the values of K for which the

tw o corresponding val ues of are equal.

Proof : We know that KLdu Mdudv Ndv

Edu Fdudv Gdv

2 2

2 2

2

2

FHIK

FHIK

L Mdv

duN

dv

du

E Fdv

duG

dv

du

2

2

2

2

KL M N

E F G

2

2

2

2

; Putting

dv

du

Differentiating with respect to we get

dK

d

E F G M N L M N F G

E F G

( ) ( ) ( ) ( )

( )

2 2 2 2 2 2

2

2 2

2 2 = 0 (Put)

( ) ( ) ( )( )E F G M N L M N F G 2 2 02 2

L M N

E F G

M N

F GK

2

2

2

2

( )GK N M FK

FK M

GK N...(4)

Again from (1) L M N K E F G 2 2 02 2 ( )

( ) ( ) ( )GK N FK M EK L 2 2 0 ...(5)

is a quadratic equation in . The maximum and minimum values of K is given by the

equation resulting on putting from (4) in the above relation (5).

Also, Discriminant ( ) ( )( )FK M GK N ER L2 0 ...(6)

is the condition that the quadratic in in (2) should be a perfect square.

Principal Radii of Curvature.

Solving (6), we get

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( ) ( ) ( )EG F K EN FM GL K LN M 2 2 22 0

L EK M FK

M FK N GK

0

If K1 and K

2 be two values of K (called principal normal curvature). 1

12

2

1 1

K K,

are known as principal radii of curvature.

then K1K

2 = Product of roots

LN M

EG FK say

2

2( )

and K1+ K

2 = Sum of roots

EN FM GL

EG FLK say

2' ( )

Here K is called total curvature or Gaussian curvature and K´ is calld Mean curvature

of the surface at P.

Clearly, Kd

Dand K

EN FM GL

D

2

2 2

2'

As P is not an Umbilic, at least one of (FN – GM), (EM – FL), (EN – GL) is non-zero.

Also principal direction of P have been defined only when P is not an umbilic. If P is an

umbilic K is same in all directions. If K same then each of coefficients in (1) are zero. Hence

every direction at an umbilic is principal direction. But we know that the surface all of whose

points are umbilic is a sphere or a plane, every direction at a point of a sphere or a plane is

then a principal direction (Important).

7.24 Theorem : The normal section at P in a given direction behaves at P like a

straight line iff the direction is an asymptotic direction.

Proof : If the direction is asymptotic then,

Ldu Mdudv Ndv2 22 0

Also, KLdu Mdudv Ndv

Edu Fdudv Gdv

2 2

2 2

2

2

K = 0 from (1)

Hence the normal section behaves like a straight line.

7.25 Theorem : The necessary and sufficient condition for a developable surface is

that its Gaussian curvature K is identically zero.

Proof : Necessary Part : The general equation of a developable surface is given by

R r s u t s

( ) ( ) ...(1)

Here u and s are parameters

Diff (1) wr to u

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R t ut t uKn1 '

and diff. wr to u,

R2 = t

So, R R t uKn t uKb serret Franet formula1 2 ( ) ; ( )

Now, E R R u K 1 12 21.

F R R 1 2 1.

G R R 2 2 1.

D EG F u K2 2 2 2

Again diff. R Kn u K Tb Kt K n11 [ ( ) ' ]

R R Kn12 21

R22 0

and, NR R

D

uKb

D

1 2

Also, L R Nu K T

D

11

2 2

.

M R N and N R N 12 220 0.

Hence, Gaussian curvature KLN M

D

2

20

Converse : We know that KLN M

D

d

D

2

2

2

2

d LN M2 2

( ).( )r r N N1 2 1 2

D N N N[ , ]1 2

If we take K = 0 D N N N[ , , ]1 2 0

[ , , ]N N N1 2 0

N N N.( )1 2 0

Now three cases may arise

Case (i) either N N N , ,1 2 are coplanar

or, Case (ii) N or N1 20 0

Or, Case (iii) N CN1 2

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Now N N and N N. .1 20 0

N = N1 × N

2

So Case (i) is not possible.

Again let N2 = 0 and take (R – r) N = 0 as the tangent plane on the surface at P.

On diff. w.r. to u,

( . ) ( )R N r N2 2 0 (suffix 2 is used as the diff. is w.r to u)

R N R N r N2 2 2 0 . .

( ) N and R t2 20

This shows that the tangent plane involves the single parameter S only so the surface is

developable.

Finally let u and v c

such N N u v ( , )

dN

dN

uN

vN N

1 2 1 2

FHG

IKJ

u v1 1,

and dN

dN

uN

vN CN

1 2 1 2

Or, 0 01 2

FHG

IKJN CN

dN

d

This shows that N does not contains parameter µ and is depnedent on only. Hence the

surfaces is developable Hence the proof.

7.26 Theorem : The only surfaces for which the Gaussian curvature K is identically

zero are the developable and the planes.

Proof : Let K = 0. This means all the points of a surface are parabolic or planar. Then d2

= 0. Kd

D

FHG

IKJ

2

2 and also the converse. (Because we know that a surface S is a plane iff L =

0, M = 0, N = 0 and developable iff d2 = 0. But L, M, N are not all identically zero, and an

ordinary surface iff d2 0.)

Def. 7.23 : Lines of curvature : It is a curve drawn on a surface so that direction at each

and every point is a principal direction.

At each point of a surface (other than sphere or a plane) these are two principal directions

at each point which are mutually perpendicular. Hence the lines of curvature form two families

which together constitute an orthogonal system.

The differential equation for principal directions is given by.

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( ) ( ) ( )EM FL du EN GL dudv FN GM dv 2 2 0 ...(1)

The determinant form is given by

Ldu Mdv Mdu Ndv

Edu Fdv Fdu Gdv

0

Or,

dv dudv du

E F G

L M N

2 2

0

If the surface is given by Z = ƒ(x, y), Monge’s form of surface then we have seen that

E p F pq G q D p q 1 1 12 2 2 2 2, , .

Lr

DM

S

DN

t

Dd

rt s

, , , 2

2

0

where ƒ ƒ ƒ ƒ ƒx x xx yy xyp q r t s , , , ,

Hence lines of curvature are given by

( ) ( )1 10

2 2

p dx pqdy pqdx q dy

rdx sdy sdx tdy

Or,

dy dxdy dx

p pq q

r s t

2 2

2 21 1 0

The principal curvature are given by

L ER M FK

M FK N GK

0

Or, D K EN FM GL K d2 2 22 0 ( )

the principal radii are given by

P rt s PD p t q r pqs D2 2 2 2 41 1 2 0( ) ( ) ( ) n sand the curvature is given by

K Kd

D

rt s

p q1 2

2

2

2

2 2 21

( )Example 7.22 : Find principal curvatures at the origin for the hyperboloid

2 7 62 2z x xy y . Also find principal section.

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Solution : The surface is given by 2 7 62 2z x xy y ...(1)

Or, z x xy y 7

23

1

22 2

ƒ ƒ

ƒ ƒƒ

x xx

y yyxy

x y p r

x y q t

7 3 7

3 13 5

So, ƒ ( , ) , , , ,x q r t s0 0 0 0 7 1 3

So, D p q2 2 21 1 .

Hence the principal curvature are given by

K

KK K K

7 3

3 10 6 16 0 8 22 ,

and the lines of curvature are given by

dx dxdy dx

p pq q

r s t

dy dxdy dx2 2

2 2

2 2

1 1 0 1 0 1

7 3 1

0

Or, 3 8 3 02 2dy dxdy dx

ie. ( ) ( )dx dy dx dy 3 3 0

dy dy x y C 3 3 1 and similarly 3 2x y C (on integration). But at origin

c1 = 0, c

2 = 0

So, x = 3y and 3x = – y are the principal sections.

Example 7.23 : A straight line draw through the variable point P ( a cos , a sin , 0)

parallel to the zx-plane makes an angle , where is some function of with the z-axis.

Prove that the measure of curvature at P of the surfaces generated by the line is

FHGIKJ

cos

( sin sin )

2

2 2 2 2

2

1

a

d

d

Proof. : The surface generated by the line is given by

x r a y a z r sin cos , sin , cos where = ƒ().

So, position vector r of any point on the surface can be taken as

r r a a r

( sin cos , sin , cos )

On differentiating and calculating fundamental magnitude at P for which r = 0, can be

found to be E = 1, F a G a sin sin , 2

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D a a L M ad

dD2 2 2 2 2 0 sin sin , , cos /

and d

ad

d

D2

2 2

2

2

FHGIKJcos

Hence at P, measure of curvature d

D

2

2

FHGIKJa

d

d

D

D

2

2

2

2

cos

FHGIKJ

ad

d

a a

2 2

2

2 2 2 2 2

cos

( sin sin )

FHGIKJ

FHGIKJ

cos

( sin sin )

2

2

2 2 2

2

1

d

d d

d

Example 7.24 : For the helicoid Z cy

x tan ,1

prove that

1 2

2 22 2 2

u c

cwhere u x y, and the line of curvature are given by

ddu

u c

2 2

Proof : The helicoid is given by Z cy

x tan 1

.....(1)

So, any point on it is given by x u y u z c cos , sin ,

i.e. the position vector of any point P r( )

is

r u i u j c k

cos sin^ ^ ^

...(2)

The parameters are u and

rr

ui j k r

1 120 0cos sin , ( sin ,cos , )

^ ^ ^

rr

u i u j c k

2 sin cos

^ ^ ^

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So, r r u u11 220 0 0 0 ( , , ), ( cos , sin , )

Then E r F r r G r u c

12

1 2 22 2 21 0, . ,

So, D EG F u c2 2 2 2

L r N 11 0. Nr r

D

c c u

D

FHG

IKJ

1 2 ( sin , cos , )

M r NC

DN r N

C

D

12 22

22

20. , . ,

Now the equation of principal curvature is given as

D K EN FN GL K d2 2 22 0 ( )

Putting values, we get

( )u c Kc

u c2 2 2

2

2 20

Kc

u c

2 2

So, 1 2

2 2

u c

c

Now, differential equation for lines of curvature is given by

dv dud du

E F G

L M N

2 2

0

d dud du

u cc

u c

2 2

2 2

2 2

1 0

0 0

0

( )u c d du2 2 2 2 0

ddu

u c

2 2

logu u c

AC

2 2

On integration where A is constant)

ACe u u c 2 2

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GEODESIC

22v1 cu4CeA

2 1u C Ae A e ( ), on adding

the lines of curvature are the intersection of the helicoid and cylinder.

2 1u C Ae A e ( )

Example 7.25 : Find the lines of curvature and the principal curvature for the surface

generated by the tangents to a tuisted curve

Solution : Let )s(rr

be the given curve. Let the equation of surface generated by

the tangents to the given curve be.

R r s u t s

( ) ( ) , where R is position vector of a current point on it.

Then

R

u

r

ut

R

s

r

st uKn,^ ^

R R K R K uK Tb Kt uK nn n11 12 220 , , ( ) '

So, E R R F R R G R R u K 1 1 1 2 2 22 21 1 1. , , , .

D u K NR R

D

uKb

uKb2 2 2 1 2

,

Similarly L R N M R N N R N uKT d 11 12 2220 0 0. , . , . ,

Now differential equation of the lines of curvature is

ds duds du

u K

uKT

2 2

2 21 1 1

0 0

0

So, ds (du + ds) = 0

ds = 0 or du + ds = 0

s = constant,u + s = constant (on integration)

The principal curvature Kn is given by

D K EN FM GL K dn n2 2 22 0 ( ) , where K

n denotes curvature of the generated

u K K n uK T K n2 2 2 0

Kn u K Kn uKT( )2 2 0

Kn = 0, K

n

T

uK

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GEODESIC

Ka KbT

uK u 0,

Ka and Kb are two roots of Kn.

Example 7.26 : Find the line of curvature and the principal curvature for the ruled

surface generated by the binormals of a tuisted curve.

Solution : Let r r s

( ) be the curve and

R r s u b s

( ) ( )^

, be the equation of surface generated by the binormals to the given curve.

Then

R

uR b

R

sR t u Tn1 2, ( )

R R Tn R Kn uT'n uT Tb Kt11 12 220 , , ( )

R R n uTt1 2

So, E F G u T D u T NR R

D

n utT

D

1 0 1 12 2 2 2 2 1 2, , , ,

Similarly L R N M R ND

N R NK uT' u TK

D

11 12 22

2 2

0. , . , .

Now differential equation of lines of curvature is given by

ds duds du

u TT

D

K uT' u T K

D

2 2

2 2

2 21 0 1

0

0

Or, T( u T ds K uT' u T K duds Tdu1 02 2 2 2 2 2 ( )

Aslo, the principal curvature Kn are given by the equation

D K EN FM GL K dn n2 2 22 0 ( )

Or, D KK Tu u T K

DK

Dn n

2 22 2 2

20

Or, ( ) ( )1 1 02 2 2 2 2 2 2 2 u T Kn K uT' u T K u T Kn T

Example 7.27 : Find the lines of curvature and principal curvature for the ruled surface

generated by the principal normals of a truisted curve.

Solution : Let r r s

( ) be the curve and let equation of surface generated by the

principal normals is

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R r s u n s

( ) ( )^

Then

R

uR n

R

sR t u Tb Kt1 2, ( )

R R Tb Kt R Kn u T n T'b K t K n11 12 222 20 , , ( ' )

R R b uTt uKb1 2

Then NR R

D

uTt uk b

D

1 2 ( )

E R F R R G R uK u T 12

1 2 1 22 2 2 21 0 1, , ( )

Again L R N M R NT

DN R N 11 12 120. , . , .

u KT' K T' uT'

DD uK u T

22 2 2 21

( ' ), ( )

The differential equation of lines of curvature is

ds duds du

G

M N

2 2

1 0

0

0

Or, MGds Nduds Mdu2 2 0

Or, T uK u T ds u KT' K T) uT' duds Tdu[( ) ] [ ( ' ]1 02 2 2 2 2 2

and the principal curvature as given by

D Kn EN FM GL Kn d2 2 22 0 ( )

D Kn u KT' K T) uT' T2 2 2 0 [ ( ' ]DKn

Or, [( ) ] [ ( ' ][( ) ]1 2 1 02 2 2 2 2 2 2 2 21

2 2 uK u T Kn u KT' K T) T' uK u T Kn T

Parametric curves as lines of curvature

7.27 Theorem : The necessary and sufficient condition that the system of parametric

curves consists of lines of curvature is that F = 0 and M = 0.

Proof : We exclude the case when the curves is a plane or a sphere for in that case F =

0, M = 0 :

Necessary part : Let the system of parametric curves consists of lines of curvature.

The differential equation of orthogonal system of lines of curvature is given by

273

GEODESIC

dv dudv du

E F G

L M N

2 2

0

on expansion, we get

( ) ( ) ( )EM FL du EN GL dudv FN GM dv 2 2 0 ....(1)

Since lines of curvature are parametric curves so u = constant and v = constant

i.e dudv = 0.

So, (1) (EM – LF) = 0 and (FN – GM) = 0 or En – GL = 0.

F = 0 and M = 0 or, EN – GL = 0

But EN – GL 0

So we have F = 0, M = 0

Sufficiency Part : Let F = 0, M = 0

Then (1) reduces to dudv = 0 u = constant, v = constant. Hence the lines of curvature

are parametric curves.

7.28 Euler’s equation : Prove that K K K 12

22cos sin where K

1 and K

2 are

principal normal curvature in the direction of u = constant and v = constant respectively, K is

normal curvature at a point P (u, v) and is angle which positive direction of u-curve makes

with the direction dv

du.

Proof : We know that KLdu Mdudv dv

Edu Fdudv Gdv

2 2

2 2

2

2....(1)

Let the parametric curves are the lines of curvature so F = 0, M = 0 and hence from (1)

KLdu Ndv

Edu Gdv

2 2

2 2

Also, then the directions of the parametric curves at a point are the principal direction

at the point. So the principal normal curvature at P can be obtained by putting dv = 0 and du=0

(one by one) in (2).

So, KL

E1 , when dv = 0 and K

2 =

N

G, when du = 0

If r = r (u, v) be the equation of surface that unit

vector tangent to any curve on this surface

in the direction dv

du is given by r

du

dsr

du

dsr u r v1 2 1 2 ' '

dvdu

P

du = 0v

dv = 0

274

GEODESIC

Also unit vector tangent at P to u-curve and v-curve are r

E

r

G1 2, respectively..

So, cos .( ' ' ) r

Er u r v11 2

Edu

Edu Gdv2 2 udu

dsv

dv

dsand ds Edu Gdv F' , ' ,

FHG

IKJ

2 2 2 0

and cos sin .( ' ' )

2

21 2 2 2

FHGIKJ

r

Gr u r v

Gdv

Edu Gdv

Putting these values in (2) we get

K KEdu

Edu GdvK

G dv

Edu G dv

1

2

2 2 2

2

2 2

K K12

2cos sin

N.B. Since and –do not make any change on K, we can say that normal curvature in

two symmetrically situated direction with respect to a principal directions are equal.

7.29 Dupins Theorem : Prove that the sum of normal curvature at a point P in two

perpendicular directions is constant for each two perpendicular directions, and is equal to the

sum of the principal curvature.

Proof : The proof follows directly from Euler’s equation given by

K K K 12

22cos sin is normal curvature. Put

2, then

K K K' cos sin , FHGIKJ

FHGIKJ1

22

2

2 2

where K and K´ are normal curvature in two perpendicular directions

Or, K K K' sin cos 12

22 ...(2)

Adding (1) and (2)

K K K K ' (cos sin ) (sin cos )12 2

22 2

K K1 2

Hence the sum of the normal curvature in two perpendicular direction at a point is

constant.

i.e. K + K´ = constant.

The lines of curvature on a surface have been defined as curves whose directions are

always principal directions.

275

GEODESIC

Characteristic Property of lines of Curvature : A curve C on a surface S is a line of

curvature iff the normals to S at the points of C generate a developable surface or a plane.

7.291 Theorem : A necessary and sufficient condition that the normals to the surface

S at the points of a curve C r = r (u(s), v(s)) generate a developable or a plane is that there

exists a function 1

of S such that the equation 1

0

dr

ds

dN

dsFHGIKJ FHGIKJ is an identity in S.

Proof : Not required.

7.24 Minimal Surface : A surface (other than plane) is called minimal surface. If K´

= 0 identically. The condition for surface to be minimal is given as

EN – 2FM + GL = 0

If the surface is given by Z = ƒ(x, y) then the above condition is given by

( ) ( )1 2 1 02 2 q r pqs p t

Property-I : At each point of a minimal surface principal radii are equal and of opposite sign.

Prop. (II) : Minimal surface are surface of negative curvature at each point of which

the asymptotic direction cut orthogonally.

7.25 Definition : (Developable Surface) : Surface for which Gaussian curvature K

vanish identically are called developable surface.

Clearly, the condition for a surface to be developable is the LN – M2 = 0

If the surface is Z = ƒ(x, y) i.e. Monge’s surface then this condition is given by rt – s2 = 0

7.292 Theorem (Rodrigues’ equation) :

The necessary and sufficient condition that a curve C on a surface S is a line of curvature

is that there exists a function K such that

Kdr + dN = 0 along the curve.

Proof : We know that dr = r1 du + v

2 dv

and dN = N1 du + N

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

Necessary Part : Let curve C on S is a line of curvature. But for a line of curvature.

( ) ( )

( ) ( )

Edu Fdv Ldu Mdv

Fdu Gdv Mdu Ndv

0 .......... (2)

This is the equivalent to equation

K Edu Fdv Ldu Mdv( ) ( ) 0

and K Fdu Gdv Mdu Ndv( ) ( ) 0 .......... (3)

whose solution is (2) and determine K uniquely.

Equation (3) can be written in the form

(K dr + dN) . r1 = 0

and (Kdr + dN) . r2 = 0

this means (Kdr + dN) is parallel to (r1 × r

2) i.e. parallel to N. Hence we can write

(Kdr + dN) = CN

276

GEODESIC

or, N. (Kdr + dN) = CN. N = C

C = 0

(Kdr + dN) = 0 (along the curve C)

Sufficient Part : Assume that Kdr + dN = 0. Then using (1)

K (r1 du + r

2 dv)

+ (N

1 du + N

2dv) = 0

[ ( ) ( )].K r du r dv N du N dv r1 2 1 2 1 0

K Edu Fdv Ldu Mdv( ) ( ) 0 ...(4)

Similarly [ ( ) ( )].K r du r dv N du N dv r1 2 1 2 2 0

K Fdu Gdv Mdu Ndv( ) ( ) 0 ...(5)

Where E, L, M, N, F, G has known values. Eliminating K from (4) and (5), we get

( ) ( )

( ) ( )

Edu Fdv Ldu Mdv

Fdu Gdv Mdu Ndv

0

( ) ( ) ( )EM FL du EN GL dudv FN GM dv 2 2 0

(on expansion)

which represents equation of a line of curvature.

N.B. Since dr . (Kdr + dN) = 0

Kdr dN

dr dv

Ldu Mdudv Ndv

Edu Fdudv Gdv

.

.

2 2

2 2

2

2

Hence K is precisely the principal normal curvature in the direction of the line of

curvature.

7.293 Theorem : A curve on the surface S is a line of curvature iff the normals at

consecutive points of the curve intersect. (Normal Property)

Proof : Let S be the surface and take two neighbouring points P r( )

and Q r r( )

on

S. Let N and (N + dN) be normals to S at P and Q. respectively.

S.P. Let these normals intersect, then vectors N, N + dN, dr are coplanar. So, [N, N +

dN, dr] = 0

[N, dN, dr] = 0

dN = dr + µN ; (and µ are scalars)

N.dN = dN. dv + µN. N

µ = 0

dN = dv. Which is characteristic of lines of curvature. Hence the given curve

is a line of curvature.

N.P. Let the curve C on S is a line of curvature on S i.e (Kdr + dN) = 0

Also [ , ]N N dN dr1

[ , , ]N dN dr

277

GEODESIC

[ , , ]; ( )N Kdr dr Kdr dN 0

This shows that normals at consecutive points of the line of curvature intersect.

Example 7.28 : On the surface formed by the revolution of a parabola about its directrix,

one principal curvature is double its other.

Proof : The equation of surface formed by the revolution of the parabola Z2 = 4ay

about the axis Y = – a is given by Z a x y a2 2 24 [ ]

On this surface x = u cos , y = u sin , z au a 2 2 represents position vector of

any point P r( )

.

Put ƒ( )u Z au a 2 2

Then on differentiation ƒƒ

1 2

u

a

au a

and ƒƒ

( )11

2

2

2

2 3/22

u

a

au a

Now, Ka

au a

a

au a1

11

12 3/2

2

2 3/2

2

2

3/2

1 21

LNM

OQP

ƒ

( ƒ ) ( )/

( )

a

au a

au a

au a a

2

2 3/2

2 3/2

2 2 3/22 ( )

( )

[ ]

a

u

1 2

3/22

/

...(1)

and Ku

a

u2

1

12 1 2

1 2

21

ƒ

( ƒ ) /

/

/ ...(2)

From (1) and (2), K K1 2

1

2

K K2 12

So, one principal curvature is double the other.

Example 7.29 : Prove that the differential equation of the lines of curvature of the

surface Z = ƒ(x, y) at any points on it is

dp

dx pdz

dq

dy qdz

278

GEODESIC

Proof : The surfaces in Monge’s form given by Z = ƒ(x, y) the direction ratios of

normal to this surface at any point P (x, y, z) are (p, q, –1). The two consecutive normals

intersect if [N, dN, dr] = 0

In cartersian form it can be written as

p q

dp dq

dx dy dz

1

0 0

on expansion

p dqdz q dpdz dpdy dxdq( ) ( ) ( ) 0 0 1 0

dp qdz dy dq pdz dx( ) ( ) 0

dp

dx pdz

dq

dy qdz

proved

7.26 Def. Umbilicks : At an umbilic the normal curvature K is same in all directions

But KLdu Mdudv Ndv

Edu Fdudv Gdv

2 2

2 2

2

2. Since K is always same

So, L

E

M

F

N

G Or,

E

L

F

M

G

N ...(1)

Now equation of principal direction is given by

( ) ( ) ( )EM FL du EN GL dudv FN GM dv 2 2 0 ..(2)

From (1), L.H.S. of (2) is always zero. Hence at an umblic all direction are principal

directions.

If the surface be given as Z = ƒ(x, y) then the umblics are given by

1 1 12 2

p

r

q

t

pq

s KD, where D p q 1 2 2 .

Theorem 7.294 : Prove that in general, three lines of curvature pass through an umbilic.

Proof : Take the Umbilic as the origin. Also take the tangents plane to the surface at

origin as the xy-palne. Then the equation of surface z = ƒ(x, y) may be written as

23 3

3

2 2 3 2 2 3

zx y ax bx y cxy dy

.... ...(1)

(Since xy-plane is the tangent plane so the first degree term in the expression are zero

and also since the section at an umbilic is circular, the coefficients of xy = 0 and coefficients

of x2 and y2 are equal)

Since the normals at (x, y, z) and (0, 0, 0) intersect for the lines of curvature through

the umbilic (i.e. origin) So the normals

279

GEODESIC

X x

P

Y y

q

Z z

1

and X Y Z

0

0

0

0

0

1 must be coplanar. So, using the condition for coplanarity

gives

x y z

p q 1

0 0 1

0 i.e. qx = py ...(2)

Now on differentiating (1) 2 22

22 2pz

x

x

Pax bxy cy

( .....)

and 2 22

22 2qz

y

ybx cxy dy

( .....)

So, using (2), we get

22

222 2 3 3 2 2xy

ax y bxy cyxy

bx cx y dxy

..... ....

Or, bx x y c a xy d b cy3 2 2 32 2 0 ( ) ( ) ...(3)

Let the tangent to a line of curvature make an anlge with the x-axis then

tan lim . FHGIKJ

y

x So (3) gives

C b d c a btan ( ) tan ( ) tan )3 22 2 0 ....(4)

This is a cubic in tan and hence gives three values of tan Each of these values of

tan corresponds to the three lines of curvature through the Umbilic.

7.291 Example : Prove that the surface

a x b y c z x y z3 3 3 3 3 3 2 2 2 3 ( ) has an Umbilic where it meets the line

a x b y c z3 3 3 .

Proof : The equation of surface is

a x b y c z x y z3 3 3 3 3 3 2 2 2 3 ( ) ....(1)

Differentiating (1) partially w.r to x.

3 3 3 2 23 2 3 2 2 2 2 2a x c zz

xx y z z

z

xx

FHG

IKJ( )

Or, a x c z p x y z z p x3 2 3 2 2 2 22 ( ) ( . ) ...(2)

280

GEODESIC

Similarly, diff partially w.r to y

b y c z q x y z y qz3 2 3 2 2 2 2 22 ( ) ( ) ....(3)

Taking the equation of line as given a x b y c z3 3 3 ....(4)

We have from (2), x pz 0 ( )Put a x c z3 3

Or, a x x y z3 2 2 2 22 ( ) (Which is not compatiable with the equation of surface)

So, x + pz = 0

Similarly from (3) y + qz = 0

Also, on differentiating values in the other results.

We get sz pq zr p and tz q 0 1 0 1 02 2,

1 12 2

p

r

q

t

pq

s, which represent.

Umbilics. Hence the point (x, y, z) is an Umbilic on the surface.

7.3 Geodesic

7.30 Geodesic Curvature : Let S be surface and C be any arc of minimum length

between two points of the surface S. If P be any point on C and Q be another point of C quite

close to P then the part of the arc C between P and Q is also an arc of minimum length between

P and Q. Let C* be the orthogonal projection of the part of C between P and Q onto the

tangent plane of S at P is an arc of minimum length on the tangent plane between P and the

projection Q* on the tangent plane. But C* must be a straight line or a curve of zero curvature.

Hence for the arcs of minimum length we will consider only those arcs where the curvature

vector of the orthogonal projection of the curve onto the plane is zero.

7.30 Definition : The curvature vector at P

of the projection of curve C onto the tangent plane

at P is called geodesic curvature vector of e at P. It

is denoted by Kg.

7.31 Definition : (Geodesic) : A curve on a surface such that its principal normal at

every point coincides with the normal to the surface at the point, is known as a geodesic on

the the surface.

A geodesic line on the surface is always to be reckowed as a geodesic.

7.31 Analytic representation of Geodesic Curvature

Let PK be principal normal of C* at P. Let CC denote centre of curvature of C and lies

on the principal normal of C and take P = P CC*. Let C

N denote centre of normal curvature of

S and lies on the normal to the surface at P. Let = angle betwen principal normal of C* and

the principal normal of C. Then by Meusniev’s theorem

PC

Q

C*

281

GEODESIC

Pg cos

Or, Kg K cos FHG

IKJ

1

K

Let = angle between the normal to S and the principal normal to C, then

2

cos cos sin

FHGIKJ

LNM

OQP 2

So, Kg K sin ...(1)

Also,

n cos2FHGIKJ

i.e. K un sin

and since 2

So, K Kn cos ...(2)

So, K Kg K Kn

2 2 2 2 2 2 (sin cos )

and Kg KKn

Kn sincos

.sin tan

from (2)

Thus, square of the curvature of a curve on a surface at a point is equal to the sum of the

squares of the geodesic curvature of the curve at the point and normal curvature at the point in

the direction of the curve.

N.B. the equality of geodesic curvature and the ordinary curvature every point of the

curve C on s implies that curve in a asymptotic line.

Soln : Since Kg = K sin = K (if = 0). This means osculating plane of the curve at

each point is the tangnet plane to the surface at the point N || b. This means curve on the

surface is the asymptotic line.

7.30 Theorem : Beltrami formula for geodesic curvature.

If t be tangent vector to curve C at P then the unit principal normal vector of C* at P

in the direction PK is N × t.

So, (N × t). n = cos

So, K N t Kn N r rcos ( ). [ ' ' ' ]

So, Kg N r r [ ' ' ' ]

KD

r r r rcos ( ) ' ' '1

1 2

P

C*

C

K

n

T

N

282

GEODESIC

1

1 2D

r r r r( ).( ' ' ' )

Now, r r u r v' ' ' 1 2

r r u r v r u r u v r v' ' ' ' ' ' ' ' ' ' 1 2 112

12 2222

Since r r and r11 12 22, are all separately connected to r1 and r

2 so.

r´´ can be expresed in term of three non coplanar vectors r r N1 2, . as

r ar br CN' ' 1 2

= K (let), called curvature vector at a point P of a curve C on S.

Also from above

KgD

r r r r u v r r u 1

1 2 1 2 1 113( ).{( ) ' ' ' ( ) '

( ) ' ' ( )2 21 12 2 112

1 22 2 12r r r r u v r r r r

u v r v v u r r u' ' ( ) ' ' ' ( ) ' }]22 1 2 22

2

D u v v u u u c c u v c c u v c v[ ' ' ' ' ' ' ( ' ( ) ' ' ( ) ' ' ' ]' ' ' '2 212

2

112

22 122

2232 2

Here C´ are called chirstoffel S symbols.

where r c r c r LN11 11 1 11

2

2 ' ;

r c r c r MN12 12 1 12

2

2 ' ;

r c r c r NN22 221

1 222

2

7.31 Theorem : The geodesic curvature of a curve is expressible in terms of E, F, G.

and there first partial, in conjuction with the function u = u(s), v = v(s) that define the curve.

Proof : Since r t Kgd KnN' ' '

Also r ar br CN'' 1 2

So, if r K Kg Kgd ar br Kn KnN'' , , 1 2

Then K Kg Kn

Where Kn is called normal curvature vector and Kg is called the geodesic curvature

vector of C.

Again since d = N × t and d.t = 0

So, Kg. t = 0

So, r Kgd KnN' '

So, Kg r d r N t Nr r ' '. ' '. [ ' ' '] ....(1)

283

GEODESIC

Also, we have seen in Beltrami formula that

Kg D u v v u c u c c u v c c u v c v [ ' ' ' ' ' ' ' ( ) ' ' ( ) ' ' ' ]' ' '11

2 312

2

12

222

122

2222 2 ..(2)

So, form (1) and (2), it is clear that geodesic curvature unlike ordinary curvature or

normal curvature depends only on first fundamental form of the surface.

Hence the theorem.

7.31 Geodesic Curvature of the Parametric Curves.

We know that if given curve is u-curve then dv

ds 0

Then r r u r v r u' ' ' ' 1 2 1

Or, r r u' '21

2 2 (on squaring)

Or, 1 2 Eu' Or, uE

'1

So, rr

E' 1

Similarly, if the curve is a v-curve then du

ds 0

So, vG

'1

So, if K and Kg g1 2 denote geodesic curvature of u-curve and v-curve respectively then.

K cD

Eand K c

D

Gg g1 11

2

3/2 2 221

3/2

Where cE F E FE

D11

2 1 2 12

2

2

( )

cG F G FG

Dwhere D EG F22

2 2 1 22

2 22

2

( ),

If the parametric curve form an orthogonal system then F = 0 and then

KE

E G G

E

vg1

2

2

1

log

KG

G E E

G

ug2

1

2

1

log

Example 7.30 : Find the geodesic curvature of the curve u = constant on the surface

( cos , sin , )x u y u z au 1

22

284

GEODESIC

Solution : The position vector r of any print on the surface is

r u i u j au k

cos sin^ ^ ^

1

22

...(1)

So,

r

ur au1 (cos ,sin , )

r

vr u u2 0( sin , cos , )

So, E r a u F r r G r u 12 2 2

1 2 22 21 0, . ,

Since F = 0, so, K2g

= GlogcE

1

1

Or, Ka u u

ua u u

g2 2 2 2 2

1

1

1

1

1

log

1

1 2 2u a u

7.32 Bonnet’s formula for geodesic curvature

Let (u, v) = constant ....(1) be the curve differentiating,

u v

u

s

v

s

0

or, 0'v'u vu

Let u v v u' , '

Now we know that ds Edu Fdudv Gdv2 2 2

1 22 2

FHGIKJ

FHGIKJE

du

dsF

du

ds

dv

dsG

dv

ds

E F Gv u v v 2 2 2 2 22

2 2s . where s E F Gv u v v2 2 2

1

s...(2)

So, us

vs

v u' , '

285

GEODESIC

Also, t r r u r vr r

E F G

v u

v v u v

' ' '1 2

1 2

2 22

Where sign depends on the direction on C in which S is measured.

The geodesic curvature Kg of the directed curve C is

KgD

r r r r 1

1 2( ).( ' ' ' )

FHGIKJ

LNM

OQP

11 2

Dr r t

dt

ds( ). ...(4)

from (3) it is clear that t is function of u and v.

So, dtds

t t u t v ' ' '1 2 ...(5)

Putting value of t´ in (4)

Dkg r r t t u r r t t v [( ).( ) ' ( ).( ) ' ]1 2 1 1 2 2

[( . )( . ) ( . )( . ) ( . )( . ) ( . )( . )]t t r t t t r t r t t t r t t t2 1 1 2 1 2 1 2

( . , . , . ) t t t t t t 1 0 01 2

r t t t2 1 1 2 ; ...(6)

Again

ur t r t r t( . ) . .2 2 1 12 and

vr t r t r t( . ) . .1 1 2 12

So,

ur t

rr t r t r t( . ) ( . ) . .2 1 2 1 1 2

Putting (6), DKgu

r tu

r t

( . ) ( . )2 1 ...(7) (Known as Beltramis formula)

where tr r

E F G

v u

u u v v

( )1 2

2 22

So, KgD u

F G

E F G v

F E

E F G

v u

v v u v

u v

u v v

LNMM

OQPP

1

2 22 2 2 2

This is the required Bonnet’s formula for geodesic curvature.

7.33 Theorem : Louville’s Formula for geodesic curvature.

If the parametric curve form an orthogonal system of directed curves on a surface, the

geodesic curvature of an arbitrary directed curves given by u = u(s), v = v(s) is given by the

286

GEODESIC

formula K K Kd

dsg g g 1 2cos sin .

Where is the directed angle at an arbitrary point P

of c, K and Kg g1 2 are the geodesic curvatures of the u-curves and v-curves.

Proof : We have seen that

KgD u

r tv

r t

LNM

OQP

12 1( . ) ( . ) ....(1)

Also as give, tr

Et

r

GF. cos , . sin ,1 2 0

KgEa u

Gv

E

LNM

OQP

1sin cos d i d i

FHG

IKJ

FHG

IKJ

LNM

OQP

1 1

2

1

2EGG

u G

G

uE

v E

E

vcos sin . sin cos

cos sin sin cos

1 1

2 22

E u G v

G

G E

E

E G

K uu

vv

k Kg g g

' ' sin cos2 1 ...(2)

(sin ' ' cos ' ,sin ' )ce t r u r v so Eu Gv 1 2

Again since du

duv

dv

So, d

ds u

du

ds v

dv

ds uu

vv

' '

Hence from (2) Kd

dsK Kg g g

1 2cos sin

' ' 'Pu Qv (alternate formula)

where PEF E F EE

DEQ

EG E F

DE

2

2 21 1 2 1 2,

7.31 Example : Prove that if the orthogonal trajectory of the curve v=constant are

geodesic, then D

E

2

is independent of u.

Proof : As given orthogonal trajectory are geodesic

287

GEODESIC

Kg = 0 Also = /2

Now, from Louville’s formula

Kg Pu Qv ' ' '

0 Pu Qv' ' ...(1)

Also, cos ' ' 0 0so Eu Fv ...(2)

Eliminating u v' & ' from (1) and (2) we get

FP EQ 0

i.e.EG F

E

FHG

IKJ

2

0 i.e.D

E

2

0FHGIKJ

So,D

E

2

is independent of u.

7.32 Theorem : A curve on a developable surface, other than a straight line is a

geodesic iff the surface is the rectifying developable of the curve.

Proof : If part. Let the curve, is not a straight line and, is a geodesic on a developable

surface. Then the tangent plane to the surface are the rectifying planes of the curve. Hence the

surface is the rectifying developable of the curve.

Only if part : Assume that the surface is the rectifying developable of the curve. Then

it is the envelope of the rectifying plane of a space curve. So the tangent plane to the developable

at a point of the curve is the rectifying plane of the curve at that point. This means that the

normal to the developable always coincides with the principal normal to the curve. Hence the

curve is a geodesic on the developable surface.

7.35 Theorem : A curve is a geodesic on a surface iff its geodesic curvature is zero.

Proof : The formula for geodesic curvature Kg is given by Kg = K sin

Let Kg = 0 Ksin 0 0 principal normal of the curve concides with the

normal to the surface at every point the curve is geodesic. If it is a straight line then K = 0

Kg = 0 (above) the straight line is a geodesic.

Hence the necessary and sufficient condition.

7.36 Theorem : The necessary and sufficiently condition for u-curve and v-curves

on a surface to be geodesic are that c and c112

2210 0 respectively..

Proof : We know that c r r r r112

1 2 1 110 0 ( ) ( )

N r r.( )1 22 0

Again tr

E 1

288

GEODESIC

Diff. w.r to s, dt

ds

r

EE r

d

ds E

FHGIKJ

111

1,

Knr

Er

d

du E

du

ds

FHGIKJ

111

1

Or, Kr

Er

d

du E E

du

ds En

FHGIKJ FHG

IKJ

111

1 1 1;

Or , Kn rr r

Er r

du E E

FHGIKJ1

1 111 1

1 1 1

clearly ( )r r1 11 is a vector in the direction of binormal of u-curve.

Or, Kn r r r r rr r

E

1 1 2 1 2

1 11( ) ( ).( )

, for u-curve

Or, O r r r r ( ).( )1 2 1 11

N r r.( )1 12

i.e.c112 0 is the necessary and sufficient condition.

Similarly, we can prove for v-curve also.

7.37 Theorem : If the parametric curves form an orthogonal system, the u-curves

are geodesic iff E = E (u) and v-curves are geodesic iff G = G (v).

Proof : Since the parametric curves form an orthogonal system. So, the geodesic

curvatures K and Kg g1 2 of u-curves and v-curves respectively are given by

KG v

Eg1

1

log

KE u

Gg2

1

log

But by a theorem (above), the n & s.c. for a curve on a surface to be geodesic is that its

geodesic curvature is zero.

So, K and Kg g1 20 0

E = E(u) and G = G(v)

7.38 Theorem : A surface on which an orthogonal system of geodesic exists, then

the surface is either developable or a plane.

Proof : From first fundamental form.

ds Edu Fdudv Gdv2 2 22

289

GEODESIC

Edu Gdv F2 2 0( )

Or, ds du dv2 2 2 ( , )Putting du Edu dv Gdv

Then it can be seen that the Gaussian curvature

K of the quadratic form is zero.

But Kd

D

2

2 02

2

d

D d2

= 0

And hence the surface is developable.

7.39 Theorem : Prove that any curve is a geodesic on the surface generated by its

binormals and an asymptotic line on the surface generated by its principal normals.

Proof : The equation of surface generated by binormals is given by

R = r + vb ....(1) where R is a function of S and v and r and b are function of s.

So,

R

s

r

sv

b

sOr, R t vTn1 ...(2)

and

R

v

r

vb Or, R

2 = b ...(3)

So, NR R

D

t vTn b

D

n vTt

D

1 2 ( )

n

Dv for a po on the curve 0 intb g

N is parallel to n and can be made coincident. So the curve is a geodesic on the

surface.

Again the equation of surface generated by principal normal is R = r + vn.

on diff. R1 = t + v (Tb – Kt)

and R2 = n

So, NR R

D

t v Tb Kt n

D

b

D

1 2 [ ( )]

(for a point on curve v = 0)

So, N is parallel to b

Hence the curve is an asymptotic line.

7.33 Geodesic on a surface revolution.

The position vector of any point r on the surface of revolution is given by

r u v i u v j f u k

cos sin ( )^ ^ ^

...(1)

On diff. r v v u

1 (cos ,sin , '( ))ƒ

290

GEODESIC

r u v u v

2 0( sin , cos , )

So, E r F r r G r u 1

2 21 2 2

2 21 0ƒ' , ,

Now we know that

ds Edu Fdudv Gdv First fundamental form2 2 22 ( )

( )1 2 2 2 2ƒ' du u dv ....(1)

Now, r r r r u r v r u r u v r v2 2 1 2 112

12 222. ' ' ( ' ' ' ' ' ' ' ' )

Or, 0 22 u v uu v' ' ' '; (v-curves are not geodesic)

Or, u v2 ' constant = K ....(2) (Intergrating w.r to first variable)

Solving (1) & (2)

u u K dv K du2 2 2 2 2 2 21( ) ( ' ) ƒ

Or, dv Ku K u

du

1 12

2 2

ƒ'

( ) is the differential equation

Or, v c Ku K u

du

z 1 2

2 2

ƒ'

( ) (on integration) is the required solution.

7.32 Example : Find the projection of the geodesic of the catenoid u = c cos z

c on

the xy-plane.

Solution : The catenoid is a surface of revolution given by

x u v y u v z cuc

cos , sin , cosh 1

So, Let ƒ( ) coshu cuc

z 1

then on diff, ƒƒ

' FHGIKJ

d

duc

uc

c c

u c c

c

u c

1

1

12 2 2 2 2

So, equation of geodesics on the surface of revolution is known to be

u u v dv K du2 2 2 2 2 2 21( ) ( ) ƒ'

dvK du

u c u k

( ) ( ),

2 2 2 2 K is arbitrary constant.

291

GEODESIC

7.34 Geodesic on the surface F (x, y, z) = 0

We know that in the case of a geodesic, the principal normal and the surface normal

coincide So, t r '

dtds

r Rn ' ' N nr

K

' '

So, diffrential equation of geodesic is given by

NFF

rR

| |

' '

FHG

IKJ

FHGIKJ

FHGIKJ

FHGIKJ

F

HGIKJ

Fx

Fy

Fz

Fx

Fy

Fz

x y z

Kx

x

dsetc

, ,( ' ' , ' ' , ' ' )

, ' ' .2 2 2

2

2

x

F

y

F

z

Fx y z

' ' ' ' ' '

If F = Z – ƒ(x, y) = 0 then the equation of geodesic is given by

x

p

y

q

z' ' ' ' ' '

1

where

z

xp

z

yq,

If the equation of surface of revolution be of the form

Z x y ƒ( )2 2 = ƒ(u)

So,

z

xp

x

uƒ'

z

yq

x

uƒ'

But we have seen above that py qx' ' ' '

So, yx xy' ' ' ' 0

i.e. yx xy' ' constant

The polar form is given by rd

dsc2

292

GEODESIC

Example 7.33 : If the principal normals of a curve intersect a fixed line, the curve is a

geodesic on a surface of revolution, and the fixed line is the axis of the surface.

Solution : Let the principal normals intersect the fixed line (assume axis of Z) given by

x y z

0 0 1 ....(1)

The equation of principal normal at (x, y, z) is

X x

l

Y y

m

Z z

n

2 2 2...(2)

Since (2) intersect (1), so the condition gives

x y z

l m n2 2 2

0 0 1

0

Or, m x l y2 2 0 (on expansion)

i.e. xy yx' ' ' ' 0

But l Px and m Py2 2 ' ' ' '

So, xy yx' ' constant.

Hence the curve is a geodesic on the surface of revolution for which the fixed line is

the axis of revolution.

Theorem 7.391 : Prove that the straight line on a surface are the only asymptotic lines

which are geodesic.

Proof : The asymptotic line is given by N b^ ^

and that of a geodesic by N n^ ^

.

So, n = b

diff w.r.t.s., we get dn

ds

db

ds

Tb Kt Tn (impossible)

Since t, n, b are non-coplanar vectors so above linear equation can not hold T = 0, K = 0.

So the curve is a straight line.

Example 7.34 : Prove that for a geodesic on a cylinder T

K

FHGIKJ is constant while for a

geodesic on a cone T

K

FHGIKJ is constant.

Proof : Assume that T

K constant = c

293

GEODESIC

So, T = CK ...(1)

Also we know that (Serret-Frenet formula)

t Kn and b Tn CKn' ' ; from (1)

ct b CKn CKn' ' 0

( )'ct b 0

ct b a (constant vector) on intergration

( )ct b t a t

n = a × t

Since the curve is a geotesic on S, so N = n.

Which follows from N.t = 0, n.t = 0, also

N . a = 0

Hence the surface S is cylinder.

Next, let T

KFHGIKJ

' constant = C

T

Kc s a ( )1 on integration

T K C s a ( )1

Also, b Tn K C s a n11 ( ) , (using(2)

t C s a'( )1 ( ' ) t Kn

b t C s a' '( ) 1 0

b ds C s a t ds' ( ) ' zz 1 constant

b C s a t C ds z[( )1 1 constant

b C s a t C r c a ( )1 1 1 (Let)

b C s a t C r a ( ) ( )1 1

[ ( ) ] ( )b C s a t t C r a t 1 1

n C r a t 1( )

n = C1 (r – a) × t

n . t = 0. But N.t = 0 N = n. So curve is a geodesic

Also N. (r – a) = 0. i.e. N is perpendicular to vector (r – a) and hence the surface is a

cone whose vertex is ( )a

.

294

GEODESIC

LN

EG

7.35 Torsion of a Geodesic

Let r = r (u, v) be the surface. containing a geodesic.

Let at any point P on the geodesic representing

trihedral t n b^ ^ ^

, , .

Then since ndn

dsTb Kt'

bdn

dsb Tb Kt T. .( )

T bdn

dst n

dn

ds

FHG

IKJ. ( ).

LNM

OQP

dn

ds

dr

dsN, ,

[ , , ]dN dr N

ds2

Again, dN = N1du + N

2dv

and dr = r1du + r

2dv

So, [dN, dr, N] = [N1du + N

2dv, r

1du + r

2dv, N]

( ) ( ).N du N dv r du r dv N1 2 1 2

[( ) ( )N r du N r N r dudv1 12

1 2 2 1 + ( ) ].N r dv N2 22

[ , ] ([ ] [ ]) [ ]N r N du N r N N r N dudv N r N dv1 12

1 2 2 1 2 22 ....(2)

Now, [ ][ ]

;N r NN r r r

DN

r r

D1 11 1 1 2 1 2

FHG

IKJ

[( ).( )]N r r r

D1 1 1 2

( . )( . ) ( . ) ( . ) [ ]N r r r N r r r

DLF ME

D1 1 1 2 1 2 1 1

Similarly [ ] [ ]N r NLG FM

Dand N r N

MF NE

D1 2 2 1

and [ ]N r NMG NF

D2 2

Putting these value in (2) then (1) gives

b^

n^

t^

N

P

geodesic

295

GEODESIC

TEM FL du EN GL dudv FN GM dv

D Edu Fdudv Gdv

( ) ( ) ( )

( )

2 2

2 22

which gives geodesic torsion of the surface at P along dv

du.

7.392 Theorem : Prove that the geodesic torsion at a point in two perpendicular

direction (or in two directions equally inclined to a principal direction) are negative of one

another.

Proof : We know that TgK K

( )

sin .1 2

22

Now Tg or or K K 0 01

21 2 ,

The last condition gives that point is an umbilic. So, excluding this case, we see that Tg

= 0 in the principal direction and is maximum when

1

4

This means that the geodesic torsion taken on its extreme values in the two perpendicular

directions which bisect the angles between the principal direction.

Also, sin sin22

2

FHGIKJ

Also sin ( ) sin2 2

Hence the geodesic torsion at a point in two perpendicular directions, or in two

directions which are equally inclined to a principal directions, are negative of one another.

N.B. Two extreme values of the geodesic torsion are negative of one another.

Example 7.35 : Prove that T K K K K21 2 ( ) ( ), where K is curvature and T is torsion.

Proof : We know that

T K K ( ) sin cos .1 2

So, T K K21 2

2 2 2 ( ) sin cos

( ) cos ( ) sinK K K K2 12

2 12

( ) ( ); ( cos sin )K K K K K K K2 1 12

22

7.393 Theorem : Prove that Tg Td

ds

, where is angle between the normal vector

N to the surface and the principal normal vector n to the curve at P.

296

GEODESIC

(This formula is known as Bonnet’s formula)

Proof : Let u = u(s) and v = v (s) be the given curve C.

Then at any point P (u, v) the geodesic torsion Tg is given by TgdN dr N

ds[ ]

2 ...(1)

and also TgEM FL du EN GL dudv FN GN dv

D Edu Fdudv Gdv

[( ) ( ) ( ) ]

[ ]

2 2

2 22...(2)

Let is measure positive form b to n^ ^

.

then cos . N n ...(3) and sin . N b ...(4)

Differentiating (3)

sin . .d

dsN

dn

ds

dN

dsn

N Tb Kt ndN

ds.( ) .

T ndN

dssin .

sin .d

dsT n

dN

ds

FHGIKJ 0 ...(5)

But ndN

dsb t

dN

dsb t

dN

ds. ( ). .

FHGIKJ ...(6)

Also, tdN

dsN

dN

dsand t both lie in gent plane

FHG

IKJ tan

So, N tdN

dsN N. .

FHGIKJ

Ndr

ds

dN

ds

LNM

OQP

Tg = from (1)

tdN

dsTg N

Also, ndN

dsb t

dN

dsb t

dN

ds. . . ; (Putting Scalar triple product)

297

GEODESIC

b TgN Tg b N Tg. . sin (from 4)

Putting in (5) sin sin

d

dsT Tg

FHGIKJ 0

Or, Tg Td

ds

7.30 Corollary : Prove that T and Tg of a curve are identically equal iff the osculating

plane of C makes a constant angle with the tangent plane to the surface.

Proof : We have seen that

Tg Td

ds

Now, Tg T iffd

ds

0

i.e. iff is constant

i.e. iff angle between osculating plane of curve and tangent plane to the surface is

constant.

7.31 Cor. : The necessary and sufficient condition for the line of curvature to be a

plane curve (not a straight line) is that its osculating plane makes always the same angle with

the tangent plane to the surface.

Proof : We know that Tg Td

ds

(Bonnet’s formula)

Let Tg = 0, Then T = 0 iff d

ds

0 i.e. iff is constant.

Example : Prove that square of the geodesic torsion at a point in an asymptotic direction

is equal to the negative of Gaussian curvature.

Proof : The geodesic torsion Tg is given by

Tg N N r [ ' ' ]

Also, torsion of an asymptotic line is given by

T N N r and also T K [ ' ' ]

Then Tg T K2 2 (Gausian curvature)

7.36 Geodesic Torsion (Second Form)

Let u = constant and v = constant be two parametric curves of the lines of curvatue.

Then We know that F = 0, M = 0.

Now formula for principal normal curvature are given by KLdu Mdudv Ndv

Edu Fdudv G dv1

2 2

2 2

2

2

298

GEODESIC

L

E (along v-constant dv = 0) .......... (1)

and KN

G2 (along u-constant du = 0) .......... (2)

Also, the geodesic torsion

TEM FL du EN GL dudv FN GM dv

D Edu Fdudv Gdv

[( ) ( ) [ )

( )

2 2

2 22

gives, TgEN GL

EG

du

ds

dv

ds

( )

FHG

IKJ

G

EL

E

GN u v' '

( ) ' ';K K Eu Gv1 2from (1) & (2)

Again if be the angle between positive direction of u-curve and the direction dv

du, then

cos ( ' ' ) r u r vr

E1 2

1

Eu r r E'. ( . ) 1 1

and sin ( ' ' ). r u r vr

G1 2

2

Gv r r G' ( . ) 2 2

But for positive direction of u-curve, dv = 0, ds E du and = 0 and for positive

direction of v-curve

du = 0, ds = G du and = 2

So, Tg K K ( ) cos .sin1 2

1

221 2( ) sinK K (Taking +tive sign in both formula)

When K K1 2 then Tg = 0

7.32 Cor. : A geodesic (not a straight line) is a plane iff its is a line of curvature.

Proof : We have seen above that

299

GEODESIC

TEM FL du EN GL dudv FN GM dv

D Edu Fdudv G dv

[( ) ( ) ( ) ]

( )

2 2

2 22

In the case of plane curve T = 0.

So, (EM – FL) du2 + (EN – GL) dudv + (FN – GM) dv2 = 0

which represents the differential equation of lines of curvature.

Geodesic Parallels and Parameter

7.32 Definition : (Geodesic Parallels) The orthogonal trajectories of a family of

geodesic are called geodesic parallels.

7.33 Definition (Geometric Parameters) : The parameters u and v as defined by

metric ds du Gdv2 2 2 are called geodesic parameters.

Geodesic parallels are so called because (i) each two of them are equally distant and

(ii) they are measured along geodesic.

7.394 Theorem : The necessary and sufficient condition that the segments cut from

the curves of a family of curves on a surface by two arbitrarily choosen orthogonal trajectories

of the family be all equal is that the curves of the family be geodesic on the surface.

Proof : N.P. The orthogonal trajectory of a family of line in a plane have the property

that the segments cut from the lines by any two of than are parallel. In case the lines envelope

a curve, theire orthogonal trajectories are the involutes of this curve. Also we know that two

involutes of a curve cut equal segments from the tangents to the curve which we take as the

line of the given family. The orthogonal trajectories of a family of geodesic on an arbitrary

surface also possess the same property that each two of them cut equal segments from the

geodesic.

S.P. : Let each two orthogonal trajectories of a family of curves on a surface cut equal

segments from the curves of the family.

Let u-curve represent the curves of the given family and v-curve their orthogonal

trajectories. Since for u-curve, v = v0 so differential equation of arc ds E u v du ( , )0

So, the segments cut from thes u-curve by the v-curves u = u1 and u = u

2 (u

2 > u

1) is

given by

S E u v dvv

u

z ( , )01

2

(on integration)

This segment is same for every u-curve iff the integral is independent of v0 i.e. E =

E(u). Also, we know that if the parametric curves form an orthogonal system, the u-curves are

geodesic when and only when E = E(u) and independent of v. Hence the curves of the family

are geodesic on the surface.

7.395 Theorem : The necessary and sufficient condition that the metric of the surface

be ds du Gdv2 2 2 is that the u-curves be geodesic with the parameter u as their common arc,

and the v-curves be the geodesic parallels orthogonal to them.

300

GEODESIC

Proof : N.P. We know that the u-curves be geodesic and the v-curves be geodesic

parallels orthogonal to them is that E = v(u) and F = 0. Then metric is given by

ds U u du Gdv2 2 2 ( )

du Gdv2 2 (on putting duv

udu V v )

du Gdv2 2

Since E = 1, S du uu

z0 (above)

Hence the distance along an arbitrary u-curve from the geodesic parallel u = 0 to the

geodesic parallel u = u is equal precisely to u. Therefore, the parameter u is now the common

are of the all the geodesic, measured from one of the geodesic parallels.

S.P. Let the u-curves are geodesic and have the parameters u as their common arc, and

the v-curves are geodesic parallels orthogonal to them.

Since the element of arc of an arbitrary u-curve

ie. ds = du E = 1 (from the relation ds = Edu)

and F = 0

So, ds du Gdv2 2 2 gives the metric of the surface.

N.B. Gaussian curvature, when geodesic parameter are employed, given by

KG

G

u

1 2

2

7.37 Differential Equation of the geodesic

Let the curve u = u(s), v = v(s) denotes curve C. Let C is a strainght line on the surface

S given by r = r (u, v). In this case t = constant.

So, dt

ds

d r

ds

2

20 (a null vector)

Also, since d r

ds

dt

dsKn

2

2 , So,

d r

ds

2

2 is vector in the direction of normal to C. So, C is

a geodesic iff the null vector d r

ds

2

2 is perpendicular to each of two non-parallel vectors in the

tangent plane to the surface. If r1 and r

2 be these two vectors, then the pair of equations

characterizing the curve C as geodesic is given by

r r r r1 20 0. ' ' , . ' ' ...(1)

We may also take the vectors N × r1 and N × r

2 (two non-parallel vectors in the tangent

plane to S)

301

GEODESIC

Then N r r N r r 1 20 0. ' ' , . ' '

N r r N r r. ' ' , . ' '1 20 0

1

01 2 2Dr r r r( ) .( ' ' ) N

r r

D

FHG

IKJ

1 2

101 2 2D

r r r r( ) .( ' ' )

( ).( ' ' ) , ( ).( ' ' )r r r r r r r r1 2 2 1 2 10 0 ...(2)

If we put r r u r v r u r u v r v' ' ' ' ' ' ' ' ' ' 1 2 112

12 2222

Then u c u c u v c v' ' ' ' ' '' ' ' 112

12 2222 0 ...(3)

v c u c u v c v' ' ' ' ' '' ' 112 2

12 2222 0

(Using Gauss formula and Beltramis’s formula for geodesic curvature)

Equation (1) and (2) are equivalent (as they both characterize C as geodesic)

Hence the curve C defined as aboe is a geodesic iff the function u(s), v(s) satisfy the

differential equation (3)

7.34 Def. : Curvature of a geodesic : Since n = N, the curvature of a geodesic is the

normal curvature of the surface in the direction of the curve and is given by using formula.

KLdu Mdudv Ndv

Edu Fdudv G dv

2 2

2 2

2

2

7.4 Fundamental Equation of Surface Theory

7.40 : Gauss Formula : It has been seen that the goemetric properties of the surface

r = r(u, v) have been expressible in terms of the two fundamental differential quadratic forms

and their coefficients E, F, G, L, M, N.

We will now see whether a surface is determined completely by two fundamental forms.

So, if for any two given differential quadratic forms there exists a surface, we should see first

that whether the coefficients E, F, G, L, M, N are connected by relations which are identically

satisfied. We will therefore establish formulas of Gauss for r r r11 12 22, , .

Now we know that any vector at an arbitrary point P on the surface can be expressed as

a linear combination of three vectors not in the same plane So, we express r r N11 12, , as linear

combination r r N at P1 2, ,

We have in vector that

dd bc

a bca

d ca

a bcb

d a b

a b cc

[ ]

[ ]

[ ]

[ ]

[ , , ]

[ , , ] ...(1)

302

GEODESIC

Here taking a r b r and c N 1 2, ,

dd r N

Dr

d N r

Dr

d r r

DN

[ ] [ ] [ ]21

12

1 2

where Nr r

Dand r r N D

1 2

1 2[ ]

( ).( ) ( ).( )

( . )d r r r

Dr

r d r r

Dr d N N2 1 2

2 11 1 2

2 2

[ ] .( )

.d r r

D

d r r

Dd N1 2 1 2

FHG

IKJ

So, on substituting d by r r r11 12 22, , in turn, we, get

r c r c r LN

r c r c r MN

r c r c r NN

11 111

1 112

2

12 121

1 121

2

22 221

1 221

2

UV|

W| ...(2) (Gauss Formula)

Where, Cr r r r

DC

r r r r

D11

1 11 2 1 22 11

2 1 11 1 22

( ).( )

,( ).( )

Cr r r r

DC

r r r r

D12

1 12 2 1 22 12

2 1 12 1 22

( ).( )

,( ).( )

Cr r r r

DC

r r r r

D22

1 22 2 1 22 22

2 1 22 1 22

( ).( )

,( ).( )

Here C s' are known as christoffel symbols.

We will denote C C C111

121

221, , by l, m, n and C C C11

212

222

2, , byµ,respectively..

Since E r r F r r and G r r 1 1 1 2 2 2. , . .

So, diff with respect to parameter u first and then with respect to v, we get

r r E r r E r r F G11 1 1 12 1 2 22 1 2 1

12

12

12

. , . , .

r r F E r r G r r G11 2 1 2 12 2 1 22 2 2

1

2

1

2

1

2. , . , .

So, using (3),

l C r r r r r r r rE G F E F

D

11

1

11 1 2 2 11 2 2 1

1 1 2

2

1

2

1

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

303

GEODESIC

GE F F E

D1 1 2

2

2

2

( )

and

CE F E FE

Dm C

GF FG

D11

2 1 2 1

2 122

2

2

2 2

( ), '

CEG FE

Dn C

G F C FC

D12

2 1 2

2 22

1 2 11 12

22

2

2,

( )

and

CEG F F C

D22

2 2 2 11

2

2

2

( )

This expresses chirstoffel symbols is terms of E, F, G and their first partial derivatives.

7.40 Cor. : If F = 0 (i.e when parametric curves are orthogonal)

then D EG2 and hence from above relations (4)

lE

Em

E

En

G

E

E

G

G

G

G

G

1 2 1

1

2 1 2

2 2 2 2 2 2, , , , ,

7.41 Corollary : When the parametric curves are orthogonal

In this case F = 0 and H2 = EG. Then the values of l, m, n, µ, redues to

lE

Em

E

En

G

E

E

G

1

2

1

2

1

2

1

21 2 1 2, , ,

1

2

1

21 2G

G

G

G,

and hence Gauss formula can be written as

r LNE

Er

E

Gr11

11 2

1

2

1

2 ,

r MNE

Er

G

Gr12

21

12

1

2

1

2

r NNG

Er

G

Gr22

11

22

1

2

1

2 ,

7.40 Theorem : Gauss charateristic equation.

Statement : To prove that

D T F E G m F m F G2 212 22 11

2 21

22 2 ( ) ( )

ln ( )E l n F G l qwhere T LN M D2 2 2 Proof : We have from Gauss formula

304

GEODESIC

r LN lr r11 1 2

r MN mr r12 1 2

r NN nr r22 1 2

Then r r LN l nE l n F G11 22. ( )

r r r M m E m F G122

12 122 2 22 . .

r r r LN M E l n F G11 22 122 2. ( ) {ln ( ) } ( )m E m F G2 22

Now E r r r E r r E r r r r 1 1 12

2 1 12 22 12 12 2 122 2 2. , . , . . .

G r r r 2 2 22.

G r r and G r r r r1 2 12 11 12 12 2 112 2 2 . . .

F r r F r r r r 1 2 1 11 2 1 12. . .

and F r r r r r r r r12 11 22 11 22 12 12 1 12 . .

2 212 22 11 11 22 1 12F F G r r r r ( . )

r r r F E G11 22 122

12 22 11

1

22. ( )

Putting this value in (1), we get

1

22 12 22 11

2( ) ( ) (ln ( ) )F E G LN M E l n F G

( )m E m F G2 22

Or, T LN M F E G E l n F G2 212 22 11

1

22 ( ) {ln ( ) }

( )m E Lm F G2 2 ...(2)

7.42 Cor. : We can write (2) as

T LN M Hu

F

EH

E

rv H

G

u2 2 1

2

1

RSTUVW

RSTUVW

1

2

2 1H

u H

F

u H

E

r

F

EH

E

u

FHG

IKJ

FHG

IKJ

1

2

1

2

22 1 1 2 1Hu

FE

EH

G

HH

v

F

H

E

H

FE

EH...(3)

7.43 Cor. : Let r = r (u, v) be the surface and then the Gaussian curvature K at any

point (u, v) on the surface is given by

KLN ML

HL H u

FEL

EH

G

H H v

F

H

E

H

FE

EH

FHG

IKJ

FHG

IKJ

1

2

1

2

21 1 2 1

305

GEODESIC

FHG

IKJ

FHG

IKJ

1

2

1 2

22 1 1 1 2

H u

FE EG

HE H v

EF FE EE

HE ...(4)

where H EG F2 2 ( )

(4) gives Gaussian curvature K in term of E, F, G and their partial derivatives w.r to u

and v (known as intrinsic formula for Gaussian curvature)

7.44 Cor. : When parametric curves are orthogonal then F = 0 and F1 = 0.

So, LN M Hu

G

HH

v

E

H

FHGIKJ

FHGIKJ

2 1 21

2

1

2

But H EG F EG2 2

So, KLN ML

HL H u

G

H v

EL

H

FHGIKJ

FHGIKJ

LNM

OQP

1

21

FHGIKJ

FHGIKJ

RSTUVW

1 1

2

1

21 2

H u

G

EG v

E

EG

FHG

IKJ

FHG

IKJ

LNMM

OQPP

1 1 1

EG u E

G

u v G

E

v

7.41 State and prove Mainardi-Co dazzi equations

Statement : The equation

L M mL l M N2 1 ( ) and

M N nL m M N2 1 ( ) , are known as Mainardi-codazzi equations. where

l c c m c 11

1

12

2

22

1, , , c n c c11

222

122

2, , .

Proof : Consider the identity.

vr

ur( ) ( )11 12

then

vLN lr r

uMN mr r( ) ( )`1 2 1 2 (using Gauss formula)

or, L N l r r LN lr r M N mr r MN mr r2 2 1 2 2 2 12 22 1 1 1 2 1 11 12

We put values of r r r11 22 12, , from Gauss formula and also, since

NFM GL

Hr

FL EM

Hr1 2 1 2 2

( ) & N

FN GM

Hr

FM EN

Hr2 2 1 2 2

(weingarten equation)

306

GEODESIC

in above and then equating cofficients of N from both the sides, we get

L lm N M ml M2 1

Or, L M mL l M N2 1 ( ) . which is first Mainarddi-Codazzi equation

For second equation, we proved with identity

vr

ur( ) ( )12 22

then proceeding as above and equating coefficients of N from both the sides, we get.

M mM N N mL M2 1

Or, M N nL m M N2 1 ( ) as the second Mainardi-Codazzi equatio.

N.B. We have used symbol H for D where D EG F H2 2 2 Weingartan equations

To prove that NFM GL

Dr

FL EM

Dr1 2 1 2 2

and NFN GM

Dr

FM EN

Dr2 2 1 2 2

Proof : We can write N.N. = 1 ( N is a unit vector)

N.dN = 0 ...(1) (on differentiation)

N1.N = 0 and N

2.N =0

This means that the vectors N1 and N

2 lies in the tangent plane at P and hence can be

expresed as liner Combinations of r1 and r

2. So, let

N1 = ar

1 + br

2 ...(2) a and b are undetermined coefficients.

Then r N ar r br r1 1 1 1 1 2. . .

L aE bF

Similarly , r N ar r br r2 1 2 1 2 2. . .

M aF bG.Solving (3) for a and b and putting the values in (2), we get

NFM GL

Dr

FL EM

Dr1 2 1 2 2

...(4)

and NFN GM

Dr

FM EN

Dr2 2 1 2 2

where D EG F2 2 ( ) ...(5)

These equations (4) and (5) are called weingarten equation of a surface.

Examples 7.40 : Show that for the right helicoid r u v u v cv ( cos , sin , )

307

GEODESIC

l m n uu

n c

0 0 0 0

2 2, , , ,

Proof : As given r u v u v cv ( cos , sin , )

Differentiate r v v1 0 (cos ,sin , ) and w.r to v,,

r u v u v c2 ( sin , cos , )

Now E r F r r c r u c 12

1 2 1 22 2 21 0, . ,

So H EG F u c2 2 2 2 ( )

Also, E E F F G G1 2 1 2 1 20 0 0 0 4 0 , , , , ,

So, lH

GE FF FE 1

22 0

2 1 1 2( ) (On putting values)

1

22 0

2 1 2 1H

EF EE FE( )

mH

GE FG 1

20

2 2 1( )

1

2 2 1 2 2 2HEF FE

u

u c( )

nH

GF GG FGu c

u c u u

1

22

1

22

2 2 1 2 2 2

2 2( )( )

( ).c h

1

22 0

2 2 2H

EG FF EG( )

Theorem 7.41 : Show that for surface z = ƒ(x, y), with x and y parameter

lpr

Hm

ps

Hn

pt

H

qr

H

qs

H

qt

H

2 2 2 2 2 2, , , , ,

Proof : The surface is given Z = ƒ (x, y). For thus surface, we have calculated earlier

that

E p F pq G q E pr F rq pq 1 1 22 21 1, , , ,

G qs E ps F sq pt G qt1 2 2 22 2 2 , , ,

Then since lH

GE FF FF 1

22

2 1 1 2( )

1

21 2 2 2

2

2

Hq pr pq rq ps pq ps( ) ( ) / )

308

GEODESIC

2

22 2 2 2 2

2

2 2 2 2

Hpq prq prq p qs p qs

1

22

2 2Hpq

pq

H.

Similarly other relation can be estabilished.

Ex. 7.42 : For a surface whose metric given by ds du D dv2 2 2 2 prove that l = 0, m =

0, n = – D D1, =0,

D

D

D

D1 2,

Proof : We know that ds Edu Fdudv Gdv2 2 22

Also as given, ds du D dv2 2 2 2 So on comparing, E = 1, F = 0, G = D2

Now, H EG F D D2 2 2 20

Again E G DD E G DD1 1 1 2 2 20 2 0 2 , , ,

lE

Em

E

En

G

EDD 1 2 1

12

02

02

, ,

E

G

G

G

D

D

G

G

D

D2 1 1 2 2

20

2 2 2, ,

7.43 Example : For any surface prove that

uH l

vH m(log ) log( ) , where u and v are parameters and symbols have

there normal meaning.

Proof : We know that H EG F2 2

So,

FHG

IKJu

Hu

H(log ) log1

22

1

2

1

2

12

2

2 2

2

H uH

HEG F. ( ) ( )

1

22

2 1 1 1H

E G EG FF( ) ...(1)

Again, as drawn in Gauss formula

lH

GE FE FEH

EG FF FG 1

22

1

22

2 1 1 2 2 2 2 1( ), ( )

mH

GE FGH

EG FE 1

2

1

22 2 1 2 1 2( ), ( )

309

GEODESIC

So, lH

GE FE FE EG FE 1

22

2 1 1 2 1 2[ ]

1

22

2 1 1 1H

E G EG FFu

H[ ] (log ) from (1)

Similarly

FHG

IKJ

vH

vH

H v H vEG H(log ) log . ( )

1

2

1

2

1 1

22

22

22d i

1

22

2 2 2 2H

E G EG FF( ) ...(2)

and mH

GE FG EG FF FG 1

22

2 2 1 2 2 1[ ]

1

22 2

2 2 2 2H

E G EG FFv

H from[ ] log( ) ( )

7.5 Summary

This book on Differential geometry deals with the fundamentals of curves and surface

in vector notation. The whole study is restricted to the concepts line Principal directions,

Principal curvature, Gaussian curvature.

Geodesic has been introduced as an important Concept. It is an arc of minimum length

connecting two points on a given surface i.e. curves of shortest distance. Geodesic curvature,

geodesic on a developable surface, geodesic parallels and parameters are some of the useful

discussions in there context.

7.6 Model Questions

1. Calculate the radius of curvature of the section of the paraboloid 2 5 4 22 2z x xy y

by the plane x = y at the origin. (Ans. 2

11)

2. Prove that the lines of curvature of the paraboloid xy = az lie on the surface

sin sinhx

ah

y

aA F

HGIKJ

FHGIKJ

1 1 (arbitrary constant) Also, find the equation for the

principal curvature. (Ans. D K aDuvK a4 2 22 0 )

3. Find the principal directions and the principal curvature on the surface

x a u v y b u v z uv ( ), ( ),

(Ans. G dv Edu and D K abD a b uv K a b2 2 4 2 2 2 2 20 4 4 0 ( ) )

310

GEODESIC

4. For the surface x u y u z f cos , sin , ( ), prove that the angle that the lines of

curvature make with the generators are given by

tan'

( ' )tan2

2 21 0

ƒ '

ƒ ƒ

u

u

5. Prove that the normal curvature of a surface in the direction of a curve is given by

K K K 12

22cos sin

6. If , ´ are the radii of curvature of any two perpendicualr normal sections at a point of

a surface 1 1

' constant.

7. I f l1, m

1, n

1 be the direction cosines of the tangent to a line of curvature and l, m, n be

the direction cosines of the normal to the surface at the point thendll

dmm

dnn1 1 1

8. For the surface x u y u z c u u c cos , sin , log( ) 2 2 . Prove that 1 =

2

9. Find the umbilics of the ellipsoid x

a

y

b

z

c

2

2

2

2

2

21 . If P is an umbilic of the ellipsoid,

prove that the curvature at P of any normal section through P is ac

b3.

10. Show that Dupin’s indicatrix of the surface z = xy at the origin is x y2 2 1 .

11. Define Geodesic curvature vector and geoderic curvature Kg of a curve on a surface.

Prove that Kg K N r r sin [ , ' , ' ' ]

12. If K K Kg n1, , denote curvature, geodesic curvature and normal curvatue respectively

then prove K K Kg n2 2 2 .

13. If the parametric curves u = constant and v = constant are orthogonal, then their geodesic

curvature are

1 1

G vE

E uGlog , logd i d i

14. Prove that a geodesic, not a straight line, is a plane curve if and only if it is a line of

curvature.

15. Prove that T K K 1

221 2( ) sin .

16. Prove that the torsion of a geodesic is equal to 1 1

1 2

FHG

IKJ sin cos [ , ' , ' ]N N r

311

GEODESIC

17. Prove that the two geodesic at right angles have their torsion equal in magnitude but

opposite in sign.

18. Show that when the lines of curvature choosen as parametric curves, the codazzi

relations expressed in terms of E, G, L, N and their derivatives are

L EL

E

N

GN G

L

E

N

G2 2 1 1

1

2

1

2 FHGIKJ

FHGIKJ, . Also show that equation of Gauss may be

written as LN

EG u G uG

v G vE

FHG

IKJ

FHG

IKJ

1 10

7.7 References

1. Bansi Lal – Three Dimensional Differential Geometry.

Atma Ram & Sons, Delhi

2. J. J. Willmove – An Inroduction to Differential Geometry.

Oxford Press, Delhi

3. Lipschutz, M.M. – Theory and Problems of Differential

Geometry. Schaum’s outline series, Mc. Graw

Hill Book Company, N.Y.

4. Mittal & Sharma – Differential Geometry.

5. Prakash Nirmalas – Differential Geometry. T. M. H.

6. Weatherburn , C.E. – Differential Geometry ELBS