MATH 439
HW 6 Solutions
Section 2-5
Problem 9
Show that the surface of revolution can be parametrized so that F ≡ 0, G ≡ 1.
Solution: WLOG, the axis of revolution is the z-axis, and the curve is contained in the xz-plane and does
not meet the z-axis. We proceed as in Example 4, section 2-3, by writing the curve as α(t) = (f (t), 0, g(t)),
and getting the parametrization of the surface of revolution:
x(u, v) = (f (v) cos(u), f (v) sin(u), g(v)),
where u ∈ I, v ∈ (0, 2π), and we may assume α is an arclength parametrization of the curve. Now we
have:
xu = (−f (v) sin(u), f (v) cos(u), 0);
xv = (f 0 (v) cos(u), f 0 (v) sin(u), g 0 (v)),
and so
F = xu · xv = −f (v)f 0 (v) sin(u) cos(u) + f (v)f 0 (v) cos(u) sin(u) = 0;
G = xv · xv = f 0 (v)2 (cos2 (u) + sin2 (u)) + g 0 (v)2 = f 0 (v)2 + g 0 (v)2 = |α(v)|2 = 1,
since α was taken to be the arclength parameter. Problem 10
Compute the first fundamental form of the parametrization of the xy plane by polar coordinates:
x(ρ, θ) = (ρ cos θ, ρ sin θ, 0).
Solution: By a straightforward computation:
xρ = (cos θ, sin θ, 0),
xθ = (−ρ sin θ, ρ cos θ, 0);
1
and thus
E = hxρ , xρ i = cos2 θ + sin2 θ = 1;
F = hxθ , xρ i = −ρ sin θ cos θ + ρ cos θ sin θ = 0;
G = hxθ , xθ i = ρ2 (cos2 θ + sin2 θ) = ρ2 ,
Therefor
I = dρ2 + ρ2 dθ2 .
√
Note that dA = EG − F 2 dρ dθ = ρ dρ dθ, which must be familiar to you from Calculus 3, where you
studied the change to the polar coordinates in double integrals.
Problem 12
Show that a regular tube of radius r around a curve α has area 2πL, where L is the length of α.
Solution: We may assume that α : [0, u0 ] → R3 is the arclength parametrization of the curve. From Ex.
10, section 2-4, we get the parametrization of the tube:
x(u, v) = α(u) + r sin(v)n + r cos(v)b,
where u ∈ [0, u0 ], v ∈ (0, 2π) and {t, n, b} is the Frenet moving frame. Then
xu = t + r sin(v)(−κt + τ b) + r cos(v)(−τ n) = (1 − κr sin v)t − τ r cos(v)n + τ r sin(v)b;
xv = r cos(v)n − r sin vb,
Therefore,
E = hxu , xu i = (1 − κr sin(v))2 + (τ r cos(v))2 + (τ r sin(v))2 = (1 − κr sin(v))2 + (τ r)2 ;
G = hxv , xv i = r2 (cos2 (v) + sin2 (v)) = r2 ;
F = hxu , xv i = −τ r2 (cos2 (v) + sin2 (v)) = −τ r2 .
Therefore,
EG − F 2 = r2 (1 − κr sin(v))2 + τ 2 r4 − τ 2 r4 = r2 (1 − κr sin(v))2 .
We now get the area element:
dA =
√
EG − F 2 du dv = r(1 − κr sin(v)) du dv
2
We get the area by integration:
Z
u0
Z
2π
r(1 − κr sin(v))dvdu
A=
Z0 u0
=
0
r(v + κr cos(v))|2π
0 du
Z0 u0
2πrdu
=
0
= 2πru0 .
Since we took α to be the arc length parameter for the curve, L = L(α) = u0 , and so A = 2πrL, completing
the proof. Section 3-2
Problem 1
Show that at a hyperbolic point, the principal directions bisect the asymptotic directions.
Proof: If e1 and e2 are the unit vectors in the principle directions. Assume that a direction w has an angle
θ with the direction of e1 . Then by the Euler formula w is an asymptotic direction if and only if
k1 cos2 θ + k2 sin2 θ = 0 ⇔ tan2 θ = −
k1
k2
So, if the angle θ is in an asymptotic direction, then −θ, π − θ and π + θ are in symptotic directions, i.e.
the principle directions bisect the asymptotic ones, q.e.d.
Problem 2
Show that if a surface is tangent to a plane along a curve, then the points of this curve are either parabolic
or planar.
Proof: if N is the Gauss map, and α(t) is the curve, then N (α(t)) is normal to the plane and hence
constant. Thus 0 ≡ dN
= dN (α0 (t)) at all points of the curve. Assuming α is a regular parametrization,
dt
α0 (t) 6= 0, and so ker dN 6= 0, i.e dN is not injective and det dN ≡ 0 on α. The result follows by definition.
.
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Problem 4
Assume the principal curvatures k1 , k2 of a surface S satisfy |k1 |, |k2 | ≤ 1. Is it true that a curve α in S
must have curvature k with |k| ≤ 1?
Solution: No. Planar curves may have any curvature; e.g. the circle of radius 1/k has curvature k
everywhere, while the plane has principal curvatures k1 ≡ k2 ≡ 0.
Problem 6
Show that the sum of normal curvatures for any pair of orthogonal directions at p is constant.
Proof: For an umbilical point the statement of the problem is obvious. Assume that the point is not
umbilical. Let v1 , v2 be two unit vectors in the given orthogonal directions, and let e1 , e2 be the unit
vectors generating the principal directions. Then there exist θ such that
e1 = cos θe1 + sin θe2
(actually θ can be chosen as the angle between e1 and v1 counted in the direction of the shortest rotation
from e1 to e2 ).
Replacing v2 with −v2 , if necessary (this does not change the normal curvature), we will have that
e2 = cos(θ +
π
π
)e1 + sin(θ + )e2 = − sin θe1 + cos θe2 .
2
2
Then, by Euler’s formula (as on p.145), we have
kn (v1 ) = k1 cos2 (θ) + k2 sin2 (θ)
kn (v2 ) = k1 sin2 (θ) + k2 cos2 (θ)
kn (v1 ) + kn (v2 ) = (k1 + k2 )(cos2 (θ) + sin2 (θ))
= k1 + k2 = 2H(= − tr dN ),
i.e. for any pair of orthogonal directions the sum of the normal curvatures in these directions is equal to
the double of the mean curvature. .
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