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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.
255
256
GEODESIC
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
257
GEODESIC
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
258
GEODESIC
(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
259
GEODESIC
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
^
260
GEODESIC
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.
261
GEODESIC
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
262
GEODESIC
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
263
GEODESIC
( ) ( ) ( )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
264
GEODESIC
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
265
GEODESIC
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.
266
GEODESIC
( ) ( ) ( )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.
267
GEODESIC
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
268
GEODESIC
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
^ ^ ^
269
GEODESIC
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
270
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
271
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
272
GEODESIC
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