Solutions to the Exercises in Elementary Differential Geometry

Solutions to the Exercises in
Elementary Differential Geometry
Chapter 1
1.1.1 It is a parametrization of the part of the parabola with x ≥ 0.
1.1.2 (i) γ (t) = (sec t, tan t) with −π/2 < t < π/2 and π/2 < t < 3π/2. Note that γ is
defined on the union of two disjoint intervals: this corresponds to the fact that
the hyperbola y 2 − x2 = 1 is in two pieces, where y ≥ 1 and where y ≤ −1.
(ii) γ (t) = (2 cos t, 3 sin t).
1.1.3 (i) x + y = 1.
(ii) y = (ln x)2 .
1.1.4 (i) γ̇γ (t) = sin 2t(−1, 1).
(ii) γ̇γ (t) = (et , 2t).
1.1.5 γ̇γ (t) = 3 sin t cos t(− cos t, sin t) vanishes where sin t = 0 or cos t = 0, i.e. t =
nπ/2 where n is any integer. These points correspond to the four cusps of the
astroid (see Exercise 1.3.3).




1.1.6 (i) The squares of the distances from p to the foci are
(p cos t ± ǫp)2 + q 2 sin2 t = (p2 − q 2 ) cos2 t ± 2ǫp2 cos t + p2 = p2 (1 ± ǫ cos t)2 ,
so the sum of the distances is 2p.
(ii) γ̇γ = (−p sin t, q cos t) so if n = (q cos t, p sin t) then n.γ̇γ = 0. Hence the
distances from the foci to the tangent line at γ (t) are
n
(p cos t ∓ ǫp, q sin t).n pq(1 ∓ ǫ cos t)
= 2 2
knk (p sin t + q 2 cos2 t)1/2
1

, 2

p2 q 2 (1−ǫ2 cos2 t)
and their product is (p2 sin2 t+q 2 cos2 t)
= q2 .
(p−f1 ).n (p−f2 ).n
(iii) It is enough to prove that = . Computation shows that
kp−f 1 k kp−f 2 k
both sides are equal to q.




t
a




0 at


1.1.7 When the circle has rotated through an angle t, its centre has moved to (at, a),
so the point on the circle initially at the origin is now at the point

(a(t − sin t), a(1 − cos t))

(see the diagram above).
1.1.8 Suppose that a point (x, y, z) lies on the cylinder if x2 + y 2 = 1/4 and on the
sphere if (x + 21 )2 + y 2 + z 2 = 1. From the second equation, −1 ≤ z ≤ 1
so let z = sin t. Subtracting the two equations gives x + 14 + sin2 t = 43 , so
x = 12 − sin2 t = cos2 t − 21 . From either equation we then get y = sin t cos t (or
y = − sin t cos t, but the two solutions are interchanged by t 7→ π − t).
√ √
1.1.9 γ̇γ = (−2 sin t + 2 sin 2t, 2 cos t − 2 cos
√ 2t) √ 2( 2 − 1, 1) at t = π/4. So the
=
1
tangent line is y − ( 2 − 1) = (x − 2)/( 2 − 1) and the normal line is
√


1 √ √
y − ( √ − 1) = −(x − 2)( 2 − 1).
2

1.1.10 (i) Putting x = sec t gives y = ± sec t tan t so γ (t) = (sec t, ± sec t tan t) gives
parametrizations of the two pieces of this curve (x ≥ 1 and x ≤ −1).
3t 3t2
(ii) Putting y = tx gives x = 1+t 3 , y = 1+t3 .

1.1.11 (i) From x = 1 + cos t, y = sin t(1 + cos t) we get y = x sin t so

y 2 = x2 (1 − (x − 1)2 ) = x3 (2 − x).

(ii) y = tx so x4 = t2 x3 + t3 x3 = y 2 x + y 3 = y 2 (x + y).
1.1.12 (i) γ̇γ (t) = (− sin t, cos t + cos 2t) so γ̇γ (t) = 0 if and only if t = nπ (n ∈ Z) and
(−1)n + 1 = 0, so n must be odd.

, 3

(ii) γ̇γ (t) = (2t+3t2 , 3t2 +4t3 ). This vanishes ⇐⇒ t(2+3t) = 0 and t2 (3+4t) = 0,
i.e. ⇐⇒ t = 0.
1.1.13 (i) Let OP make an angle θ with the positive x-axis. Then R has coordinates
γ (θ) = (2a cot θ, a(1 − cos 2θ)).
(ii) From x = 2a cot θ, y = a(1 − cos 2θ), we get sin2 θ = y/2a, cos2 θ =
cot2 θ sin2 θ = x2 y/8a3 , so the Cartesian equation is y/2a + x2 y/8a3 = 1.
1.1.14 Let the fixed circle have radius a, and the moving circle radius b (so that b < a in
the case of the hypocycloid), and let the point P of the moving circle be initially
in contact with the fixed circle at (a, 0). When the moving circle has rotated
through an angle ϕ, the line joining the origin to the point of contact of the
circles makes an angle θ with the positive x-axis, where aθ = bϕ. The point P
is then at the point

γ (θ) = ((a + b) cos θ − b cos(θ + ϕ), (a + b) sin θ − b sin(θ + ϕ))
    
a+b a+b
= (a + b) cos θ − b cos θ , (a + b) sin θ
b b

in the case of the epicycloid,





b
'

a P






and

γ (θ) = ((a − b) cos θ + b cos(ϕ − θ), (a − b) sin θ − b sin(ϕ − θ))
    
a−b a−b
= (a − b) cos θ − b cos θ , (a − b) sin θ
b b

in the case of the hypocycloid.
1.1.15 γ̇γ (t) = (et (cos t − sin t), et (sin t + cos t)) so the angle θ between γ (t) and γ̇γ (t) is
given by

γ .γ̇γ e2t (cos2 t − sin t cos t + sin2 t + sin t cos t) 1
cos θ = = 2t 2 2
= ,
k γ kk γ̇γ k e ((cos t − sin t) + (sin t + cos t) ) 2

, 4

so θ = π/3.
1.1.16 γ̇γ (t) = (t cos t, t sin t) so a unit tangent vector is t = (cos t, sin t). The distance
of the normal line at γ (t) from the origin is

|γγ (t)..t| = | cos2 t + t sin t cos t + sin2 t − t sin t cos t| = 1.

Rt
1.2.1 γ̇γ (t) = (1, sinh t) so k γ̇γ k = cosh t and the arc-length is s = 0
cosh u du = sinh t.
1.2.2 (i) k γ̇γ k2 = 14 (1 + t) + 14 (1 − t) + 12 = 1.
2 2 2
(ii) k γ̇γ k2 = 16 2 9 2
25 sin t + cos t + 25 sin t = cos t + sin t = 1.
1.2.3 Denoting d/dθ by a dot, γ̇γ = (ṙ cos θ − r sin θ, ṙ sin θ + r cos θ) so k γ̇γ k2 = ṙ 2 + r 2 .
Hence, γ is regular unless r = ṙ = 0 for some value of θ. It is unit-speed if and
only if ṙ 2 = 1 − r 2 , which gives r = ± sin(θ + α) for some constant α. To see
that this is the equation of a circle of radius 1/2, see the diagram in the proof
of Theorem 3.2.2.
1.2.4 Since u is a unit vector, |γ̇γ.u| = k γ̇γ k cos θ, where θ is the angle between γ̇γ and
Rb Rb
u, so γ̇γ.u ≤k γ̇γ k. Then, (q q − p)..u = (γγ (b) − γ (a))..u = a γ̇γ .u dt ≤ a k γ̇γ k dt.
Taking u = (q q − p)/ k q − p k gives the result.
√
1.2.5 γ̇γ (t) = (6t, 1 − 9t2 ) so k γ̇γ (t) k = 36t2 + 1 − 18t2 + 81t4 = 1 + 9t2 . So the
arc-length is Z t
s= (1 + 9u2 )du = t + 3t3 .
0

1.2.6 We have γ̇γ (t) = (−2 sin t + 2 sin 2t, 2 cos t − 2 cos 2t) so
p p t
k γ̇γ (t) k = 8(1 − sin t sin 2t − cos t cos 2t = 8(1 − cos t) = 4 sin .
2
So the arc-length is
x
t x x
Z 
s= 4 sin dt = 8 1 − cos = 16 sin2 .
0 2 2 4

1.2.7 The cycloid is parametrized by γ (t) = a(t − sin t, 1 − cos t), where t is the angle
through which the circle has rotated. So
t
γ̇γ = a(1 − cos t, sin t), k γ̇γ k2 = a2 (2 − 2 cos t) = 4a2 sin2 ,
2
and the arc-length is
2π t=2π
t t
Z
2a sin dt = −4a cos = 8a.
0 2 2 t=0