Homework 9. Solutions.

1. Find coordinate basis vectors, first quadratic form, unit normal vector field, shape operator and Gaussian
and mean curvatures for
a) sphere of the radius R: x2 + y 2 + z 2 = R2 ,
{
x = R sin θ cos φ
r(θ, φ) y = R sin θ sin φ (0 ≤ φ < 2π, 0 ≤ θ ≤ π) ,
z = R cos θ

b) cylinder x2 + y 2 = a2 ,
{ x = a cos φ
r(h, φ) y = a sin φ (0 ≤ φ < 2π, −∞ < h < ∞)
z=h

c) cone x2 + y 2 − k 2 z 2 = 0,
{
x = kh cos φ
r(h, φ) y = kh sin φ (0 ≤ φ < 2π, −∞ < h < ∞)
z=h

d) graph of the function z = F (x, y)

{x = u
r(u, , v) y=v (−∞ < u < ∞, −∞ < v < ∞)
z = uv

in the case if F (u, v) = Au2 + 2Buv + Cv 2

Put down the special case when F (u, v) = auv (saddle).
For the case d) you have to calculate First quadratic form, shape operator and curvatures only at origin.
Solution

a) SPHERE (of radius R) x2 + y 2 + z 2 = R2 :

{
x = R sin θ cos φ
r(θ, φ) y = R sin θ sin φ
z = R cos θ
(0 ≤ φ < 2π, 0 ≤ θ ≤ π),
   
R cos θ cos φ −R sin θ sin φ
∂r(φ, θ)  ∂r(φ, θ)
rθ = = R cos θ sin φ  , rφ = =  R sin θ cos φ 
∂θ ∂φ
−R sin θ 0
 
sin θ cos φ
r(θ, φ) 
n(θ, φ) = = sin θ sin φ  (1)
R
cos θ
(Sometimes we denote rθ by ∂θ and rφ by ∂φ .)
Check that n(θ, φ) is indeed unit normal vector (in fact this is obvious from geometric considerations):

(n, n) = sin2 θ(cos2 φ + sin2 φ) + cos2 θ = 1 ,

(n, rθ ) = R sin θ cos θ(cos2 φ + sin2 φ) − R sin θ cos θ = 0, (n, rφ ) = R sin2 θ(− cos φ sin φ + cos φ sin φ) = 0 .
Unit normal vector is defined up to a sign; −n is unit normal vector too.

1
 Calculate now first quadratic form. (rθ , rθ ) = R2 cos2 θ(cos2 φ + sin2 φ) + R2 sin2 θ = R2 , (rθ , rφ ) = 0,
(rφ , rφ ) = R2 sin2 θ(sin2 φ + cos2 φ) = R2 sin2 θ. Thus
( ) ( ) ( )
G11 G12 (rθ , rθ ) (rθ , rφ ) R2 0
= =
G21 G22 (rφ , rθ ) (rφ , rφ ) 0 R2 sin2 θ

dl2 = G11 dθ2 + 2G12 dθdφ + G22 dφ2 = R2 dθ2 + R2 sin2 θdφ2
The length of the curve r(t) = r(θ(t), φ(t)) with θ = θ(t), φ = φ(t), t1 ≤ t ≤ t2 is given by the integral:
∫ t2 ∫ t2 √ ∫ t2 √
|v(t)|dt = G11 θ̇2 + 2G12 θ̇φ̇ + G22 φ̇2 dt = R2 θ̇2 + R2 sin2 θφ̇2 dt (1b)
t1 t1 t1

Now calculate shape operator and Gaussian and mean curvatures for sphere:
By the definition (see lecture notes) the action of shape operator on any tangent vector v is given by
the formula Sv = −∂v n. We know that for sphere n = Rr (see the equations (1) above). Hence for basis
vectors rθ = ∂θ , rφ = ∂φ we have
( ) ( )
r(θ, φ) ∂θ r(θ, φ) rθ
Srθ = −∂θ n(θ, φ) = −∂θ =− =−
R R R

and ( ) ( )
r(θ, φ) ∂φ r(θ, φ) rφ
Srφ = −∂φ n(θ, φ) = −∂φ =− =−
R R R
We see that shape operator is equal to S = − R
I
, where I is an identity operator. Its matrix in the basis
∂θ , ∂φ is equal to ( 1 )
0
− R .
0 R1
In the case if we choose the opposite direction for unit normal vector then we will come to the same answer
just with changing the signs: if n → −n, S → −S.
We see that principal curvatures, i.e. eigenvalues of shape operator are the same:

1 1
λ1 = λ2 = − , i.e. κ1 = κ2 = −
R R

(if we choose the opposite sign for n then κ1 = κ2 = R1 ). Thus we can calculate Gaussian and mean
curvature: Gaussian curvature
1
K = κ1 · κ2 = det S = 2 .
R
Mean curvature
2
H = κ1 + κ2 = Tr S = − .
R
If we choose the opposite sign for n then S → −S, principal curvatures change the sign, Gaussian curvature
K = κ1 · κ2 does not change but mean curvature H = κ1 + κ2 will change the sign: if n → −n then H = R2 .

b) CYLINDER x2 + y 2 = a2
{ x = a cos φ
r(h, φ) y = a sin φ (0 ≤ φ < 2π, −∞ < h < ∞)
z=h
     
 −a sin φ  0 cos φ
∂r(φ, h)  ∂r(φ, h)  
rφ φ,h = = a cos φ  , rh φ,h = = 0 , n(φ, h) =  sin φ  (2)
∂φ ∂h
0 1 0
Sometimes we denote rφ by ∂φ and rh by ∂h .

2
 Check that n(φ, h) is indeed unit normal vector:

(n, n) = cos2 φ + sin2 φ = 1, (n, rφ ) = a cos φ sin φ(−1 + 1) = 0, (n, rh ) = 0

Unit normal vector is defined up to a sign; −n is unit normal vector too.

Calculate now first quadratic form. (rφ , rφ ) = R2 (sin2 φ + cos2 φ) = R2 , (rφ , rh ) = 0, (rh , rh ) = 1.
Thus ( ) ( ) ( ) ( )
G11 G12 Ghh Ghφ (rh , rh ) (rh , rφ ) 1 0
= = =
G21 G22 Gφh Gφφ (rφ , rφ ) (rφ , rφ ) 0 a2

dl2 = G11 dh2 + 2G12 dhdφ + G22 dφ2 = dh2 + a2 dφ2 .

The length of the curve r(t) = r(φ(t), h(t)) with φ = φ(t), h = h(t), t1 ≤ t ≤ t2 is given by the integral:
∫ t2 √ ∫ t2 √
G11 ḣ2 + 2G12 φ̇ḣ + G22 φ̇2 dt = ḣ2 + a2 φ̇2 dt, (2b)
t1 t1

Now calculate shape operator Gaussian and mean curvatures for cylinder.
To calculate the shape operator for the cylinder we use results of calculations above of vectors rh , rφ and
of unit normal vector n(φ, h) (see the equations (2) above). By the definition the action of shape operator
on any tangent vector v is given by the formula Sv = −∂v n. Hence for basis vectors rφ = ∂φ , rh = ∂h we
have  
cos φ
Srh = −∂h n(φ, h) = −∂h  sin φ  = 0
0
and    
cos φ sin φ
rφ
Srφ = −∂φ n(φ, h) = −∂φ  sin φ  =  − cos φ  = −
a
0 0
   
cos φ −a sin φ
(Recall that n(h, φ) =  sin φ  and rφ =  a cos φ  (See the equations (2) above.)
0 0
br
For an arbitrary tangent vector X = arh + brφ , SX = − aφ . Shape operator transforms tangent vectors
to tangent vectors. Its matrix in the basis rh , rφ equals to
( )
0 0
− 1
0 a

In the case if we choose the opposite direction for unit normal vector then we will come to the same answer
just with changing the signs: if n → −n, S → −S.
We see that principal curvatures, i.e. eigenvalues of shape operator are:

1
κ1 = 0, κ2 = −
a

(if we choose the opposite sign for n then κ1 = κ2 = R1 ). Thus we can calculate Gaussian and mean
curvature: Gaussian curvature
K = κ1 · κ2 = det S = 0 .
Mean curvature
1
H = κ1 + κ2 = Tr S = − .
a
If we choose the opposite sign for n then S → −S, principal curvatures change the sign, Gaussian curvature
K = κ1 · κ2 remains the same but mean curvature H = κ1 + κ2 will change the sign: if n → −n then H = a1 .

3
 b) CONE x2 + y 2 − k 2 z 2 = 0
{
x = kh cos φ
r(h, φ) y = kh sin φ (0 ≤ φ < 2π, −∞ < h < ∞) (3)
z=h
   
 k cos φ  −kh sin φ
∂r(φ, h)  ∂r(φ, h)
rh φ,h = = k sin φ  , rφ φ,h = =  kh cos φ  .
∂h ∂φ
1 0
Sometimes we denote rφ by ∂φ and rh by ∂h .  
cos φ
To calculate the normal unit vector field n(h, φ) note that the vector N(h, φ) =  sin φ  is orthogonal
−k
√
to the surface of the cone: (N rh ) = (N, rφ ) = 0 and its length equals to |N| = k 2 + 1. Hence normal unit
vector field equals to  
cos φ
N(h, φ) 1  sin φ 
n(h, φ) = √ =√
k2 + 1 k2 + 1 −k
2 2 2
It is indeed normal unit vector field: (n, n) = cos φ sin φ k
k2 +1 + k2 +1 + k2 +1 = 1, (n, rφ ) =
√ 1
k2 +1
(cos φ · (−kh sin φ) +
sin φ · (+kh cos φ)) = 0, and (n, rh ) = √k2 +1 (cos φ · (kh cos φ) + sin φ · k sin φ − k) = 0. Unit normal vector
1

is defined up to a sign; −n is unit normal vector too.

Calculate now first quadratic form. (rh , rh ) = k 2 cos2 φ + k 2 sin2 φ + 1 = k 2 + 1, (rh , rφ ) = (rφ , rh ) =
k h cos φ(− sin φ) + k 2 h sin φ cos φ = 0, (rφ , rφ ) = k 2 h2 sin2 φ + k 2 h2 cos2 φ = k 2 h2 , Thus
2

( ) ( ) ( ) ( )
G11 G12 Ghh Ghφ (rh , rh ) (rh , rφ ) k2 + 1 0
= = =
G21 G22 Gφh Gφφ (rφ , rh ) (rφ , rφ ) 0 k 2 h2

dl2 = Ghh dh2 + 2Ghφ dhdφ + Gφφ dφ2 = (k 2 + 1)dh2 + k 2 h2 dφ2 R2 ,

The length of the curve r(t) = r(h(t), φ(t)) with φ = φ(t), h = h(t), t1 ≤ t ≤ t2 is given by the integral:
∫ t2 √ ∫ t2 √
G11 ḣ2 + 2G12 ḣφ̇ + G22 φ̇2 dt = (k 2 + 1)ḣ2 + k 2 h(t)2 φ̇2 dt . (3b)
t1 t1

To calculate the shape operator for the cone we use the results of calculations of vectors rh , rφ and of unit
normal vector n(φ, h) (see the equations (3) above. ) By the definition the action of shape operator on any
tangent vector v is given by the formula Sv = −∂v S. Hence for basis vectors rh = ∂h , rφ = ∂φ
 
cos φ
1
Srh = −∂h n(φ, h) = − √ ∂h  sin φ  = 0
k2 + 1 −k

and   
cos φ sin φ
1 1
Srφ = −∂φ n(φ, h) = − √ ∂φ  sin φ  = √  − cos φ  =
k2 + 1 −k k 2+1
0
 
−kh sin φ
1 rφ
− √ since rφ =  kh cos φ  .
k k2 + 1 h 0
We see that for an arbitrary tangent vector X = arh + brφ SX = S(arh + brφ ) = − kh√bk2 +1 rφ . Shape
operator transforms tangent vectors to tangent vectors. Its matrix in the basis rh , rφ equals to
( )
0 0
−
0 hk√11+k2

4
In the case if we choose the opposite direction for unit normal vector then we will come to the same answer
just with changing the signs: if n → −n, S → −S. We see that principal curvatures, i.e. eigenvalues of
shape operator are:
1
κ1 = 0, κ2 = − √
hk 1 + k 2
(if we choose the opposite sign for n then κ1 = κ2 = hk√11+k2 ). Thus we can calculate Gaussian and mean
curvature: Gaussian curvature
K = κ1 · κ2 = det S = 0 .
Mean curvature
1
H = κ1 + κ2 = Tr S = − √ .
hk 1 + k 2
If we choose the opposite sign for n then S → −S, principal curvatures change the sign, Gaussian curvature
K = κ+ · κ− remains the same but mean curvature H = κ+ + κ− will change the sign: if n → −n then
H = hk√11+k2 .

d) GRAPH OF THE FUNCTION z = F (x, y)

{x = u
r(u, v) y=v (−∞ < u < ∞, −∞ < v < ∞) (4)
z = F (u, v)

in the case if F (u, v) = Au2 + 2Buv + Cv 2

     
 1 1  1
∂r(u, v)
ru u,v = = 0 = 0  , ru 
u=v=0
= 0 ,
∂u
Fu 2Au + 2Bv 0
     
 0 0  0
∂r(u, v)   
rv u,v = = 1 = 1  , rv 
u=v=0
= 1 ,
∂v
Fv 2Bu + 2Cv 0
   
−Fu  0
1  −Fv  , n(u, v) 0 .
n(u, v) = √ u=v=0
=
1 + Fu2 + Fv2 1 1
Sometimes we denote ru by ∂u and rv by ∂v .
Check that n(u, v) is indeed unit normal vector: (n, n) = 1+F 12 +F 2 (Fu2 + Fv2 + 1) = 1, (n, ru ) =
u v
√ 12 2 (Fu − Fu ) = 0, (n, rv ) = √ 12 2 (Fv − Fv ) = 0. Calculate now first quadratic form. (ru , ru ) =
1+Fu +Fv 1+Fu +Fv
1 + Fu2 , (ru , rv ) = Fu Fv , (rv , rv ) = 1 + Fv2 . Thus
( ) ( ) ( )
G11 G12 (ru , ru ) (ru , rv ) 1 + Fu2 Fu Fv
= =
G21 G22 (rv , ru ) (rv , rv ) Fu Fv 1 + Fv2

dl2 = G11 dφ2 + 2G12 ddφdh + G22 dh2 = (1 + Fu2 )du2 + 2Fu Fv dudv + (1 + Fv2 )dv 2
At the point u = v = 0, Fu = Fv = 0 and
( ) ( ) ( )
G11 G12 (ru , ru ) (ru , rv ) 1 0
= = , dl2 = du2 + dv 2
G21 G22 (rv , ru ) (rv , rv ) 0 1

The length of the curve r(t) = r(u(t), v(t)) with u = u(t), v = v(t) can be calculated by the integral:
∫ t2 √ ∫ t2 √
G11 u̇2 + 2G12 u̇v̇ + G22 v̇ 2 dt = (1 + Fu2 )u̇2 + 2Fu Fv u̇v̇ + (1 + Fv2 )v̇ 2 dt (4a)
t1 t1

5
Special case of saddle: In the special case of saddle we just take F = auv in previous formulae. In particular
normal for normal unit vector we have
   
−av  0
1  −au  , n(u, v) 0 ,
n(u, v) = √ u=v=0
=
1 + a2 v 2 + a2 u2 1 1

and first quadratic form is equal to

( ) ( ) ( ) ( )
G11 G12 (ru , ru ) (ru , rv ) 1 + Fu2 Fu Fv 1 + v2 vu
= = = ,
G21 G22 (rv , ru ) (rv , rv ) Fu Fv 1 + Fv2 vu 1 + u2

dl2 = G11 dφ2 + 2G12 ddφdh + G22 dh2 = (1 + a2 v 2 )du2 + 2a2 uvdudv + (1 + a2 u2 )dv 2 .

To calculate shape operator we use results of calculations vectors ru , rv and for unit normal vector
n(u, v). We do calculations only at origin. For basic vectors ru = ∂u , rv = ∂v we have
  
−Fu
∂n(u, v) 1
Sru = − = −∂u  √  −Fv  =
∂u |u=v=0 1 + Fu2 + Fv2 1 |u=v=0

( )    
Fuu 2A
1  Fuv 
√ =  2B  = 2Aru + 2Brv
1 + Fu2 + Fv2 |u=v=0 1 0
|u=v=0

and Srv = −∂v (n(u, v))|u=v=0 =

   ( )    
−Fu Fvu 2B
1 1
−∂v  √  −Fv  = √  Fvv  =  2C  = 2Bru +2Crv
1 + Fu2 + Fv2 1 1 + Fu2 + Fv2 |u=v=0 1 0
| u=v=0 | u=v=0

) (
2A 2B
The matrix of the shape operator in the basis ru , rv is . Gaussian curvature at origin is equal
2B 2C
to det S = 4AC − 4B and mean curvature is equal to H = trS = 2A + 2C. (Mean curvature as always is
2

defined up to a sign.)
In the case of saddle F = auv, i.e. A = C = 0, B = a2 . The shape operator at the point u = v = 0
equals to ( )
0 a
a 0
Gaussian curvature K = k− · k+ = det S = −a2 and mean curvature H = (k− + k+ ) = Tr S = 0.

2 Consider surface defined by equation z − Ax2 − Ay 2 = 0. (See the exercise 1d) above.) Show that this
is a saddle: you have to show that under the rotation on the angle φ = π4 with respect to z-axis it becomes a
surface z − axy = 0. Find relation between parameters A and a.
Matrix of rotation on the angle π
(in the basis of vectors {ex , ey , ez }) directed along x, y and z axis is
   √ 4 √ 
cos 4 − sin 4 0
π π 2
− 22 0
√2 √
P =  sin π4 cos π4 0 =  2 2
0
2 2
0 0 1 0 0 1
Under the rotation      x−y 
x x √
2
 y  → P  y  =  x+y √ 
2
z z z

6
 ( )2 ( )2
Hence Ax2 − Ay 2 → A x−y
√
2
−A x+y
√
2
= −2Axy, a = −2A.

3 Show that there are two straight lines which pass through the point (3, 4, 12) on the saddle z = xy and
lie on this saddle.
†
Show that this is true for an arbitrary point of the saddle.
 
x0
Let r0 =  y0  be an arbitrary point on the saddle: z0 = x0 y0 . Consider the following two lines:
z0
 {x = x + t 
1 0
the line r = r0 + ta, where a =  
0 , i.e. y = y0
y0 z = z 0 + y0 t
 {x = x 
0 0
and the line r = r0 + tb, where b =  
1 , i.e. y = y0 + t
x0 z = z0 + x0 t
It is easy to check that these both lines belong to the saddle: xy = (x0 + {xt)y0 = z0 + ty0 = z and
= x0 + at
xy = x0 (y0 + t) = z0 + x0 t. On the other hand it is easy to see that it is all: If y = y0 + bt is an arbitrary
z = z0 + ct
straight line on the saddle passing through the point (x0 , y0 , z0 ), x0 y0 = z0 then

xy = (x0 + at)(y0 + bt) = z = z0 + ct for all t

Hence ab = 0. Thus a = 0 or b = 0. We see that through the an arbitrary point on the saddle pass exactly
two straight lines. 
 x(t) = a cos t
4 Consider helix r(t): y(t) = a sin t . Show that this helix belongs to cylinder surface x2 + y 2 = a2 .

z(t) = ct
Using first quadratic form on the surface of cylindre or in a different way a) calculate length of the helix
(0 ≤ t ≤ t0 ).
b) what are relations between principal curvatures of cylinder and curvature of helix?
Solution This helix belongs to cylinder surface x2 + y 2 = a2 because x2 + y 2 = a2 on the points of the
helix.
For the helix internal coordinates are φ = φ(t) = t and h = h(t) = ct (x = R cos φ, y = R sin φ, z = h).
Use First Quadratic form which we obtained in the previous exercise (see equation (2b). We come to
∫ t0 √ ∫ t0 √ ∫ t0 √ √
L= G11 ḣ2 + 2G12 φ̇ḣ + G22 φ̇2 dt = 2 2 2
a φ̇ + ḣ dt = a2 + c2 dt = t0 a2 + c2
0 0 0
√
Of course the answer can be obtained without integration: speed is constant, hence L = |v|t = t a2 + c2 .
This is the calculations
√ of the Internal observer. The external
√ observer will calculate using the coordinates
x, y, z: |v| = x2t + yt2 + zt2 = (a2 cos2 +a2 sin2 t + c2 ) = a2 + c2 and will come to the same answer.
a 1
b) Note that curvature of the helix equals to k = a2 +c2 . Curvatue of helix varies between zero and a,
i.e. principal curvatures of cylinder surface:
a 1
0≤ ≤ .
a2 +c 2 a

5 Assume that the action of the shape operator at the tangent coordinate vectors ru = ∂u , rv = ∂v at the
given point p of the surface r = r(u, v) is defined by the relations: S(∂u ) = 2∂u +2∂v and S(∂v ) = −∂u +5∂v .
Calculate principal curvatures, Gaussian and mean curvatures of the surface at this point.

7
 Solution We see that the matrix of the shape operator in the basis ∂u , ∂v is equal to
( )
2 −1
S=
2 5

Hence Gaussian curvature K = det S = 12 and mean curvature H = Tr S = 7. To calculate principal

curvatures k1 , k2 note that {
k1 + k2 = H = 7
k1 · k2 = K = 12
Hence k1 = 3, k2 = 4; k1 , k2 are eigenvalues of the shape operator.
(We assume that a ≥ 0).