Solutions To Exercises: A First Course in Differential Geometry Woodward and Bolton
Solutions To Exercises: A First Course in Differential Geometry Woodward and Bolton
Solutions To Exercises: A First Course in Differential Geometry Woodward and Bolton
1.1. A sketch of the astroid is given in Figure 1(a). It is clear that all
points in the image of α satisfy the equation of the astroid. Conversely, if
x2/3 + y 2/3 = 1, then there exists u ∈ R such that (x1/3 , y 1/3 ) = (cos u, sin u).
Thus every point of the astroid is in the image of α.
Trigonometric identities may be used to show that α0 = (3/2) sin 2u(− cos u, sin u),
which is zero only when u is an integer multiple of π/2. The corresponding points
of the astroid are the cusps in Figure 1(a).
The required length is
3 π/2
Z
3
sin 2u du = .
2 0 2
Figure 1
R 2π
So, for 0 ≤ u ≤ 2π, |α0 | = 4r sin(u/2), and required length is 4r 0
sin(u/2)du =
16r.
2
Hence κ = g 00 (1 + g 0 )−3/2 .
Taking x(u) = u, y(u) = g(u) in the formula given in Exercise 1.8 gives the
same formula for κ.
1.5. Use the method of Example 2 of §1.3. For u ≥ 0, |α0 | = tanh u and
t = (tanh u, −sech u). It follows that dt/ds = (|α0 |)−1 t0 = n/ sinh u. Hence
κ = cosech u.
1.6. EITHER: use Exercise 1.4 to show that the curvature of the catenary
α(u) = (u, cosh u) is given by κ = sech2 u,
OR: use the method of Example 2 of §1.3, and proceed as follows:-
α0 = (1, sinh u), so that |α0 | = cosh u and t = (sech u, tanh u). Hence n =
(−tanh u, sech u), and
dt 1 1 1
= 0 t0 = 2 (−tanh u, sech u) = n.
ds |α | cosh u cosh2 u
It follows that |α` 0 | = |α0 | |1 − κ`| and t` = t, where = (1 − κ`)/|1 − κ`|.
Hence n` = n, so, if s` denotes arc length along α` , we have
dt` 1
= 0 t0 = 0 t0 .
ds` |α | |1 − κ`| ` |α | |1 − κ`|
Using Serret-Frenet, t0 = |α0 |κn = |α0 |κn` , from which the result follows.
2 2
1.8. Since α0 = (x0 , y 0 ), we have that |α0 | = (x0 + y 0 )1/2 . Hence t =
2 2 2 2
(x , y 0 )/(x0 + y 0 )1/2 and n = (−y 0 , x0 )/(x0 + y 0 )1/2 . Hence
0
β 0 = α0 − sα 0 tα − sα tα 0 = −sα tα 0 .
It follows that β 0 = −sα |α0 |κα nα , so the only singular point of β is when
sα = 0, that is at u = u0 .
1.11. For ease, assume that κα > 0, and restrict attention to u0 < u1 < u.
Then, from (1.12), we have that κ0 = 1/s0 and κ1 = 1/s1 .
Let ` be the length of α measured from α(u0 ) to α(u1 ). Then ` = s0 −s1 > 0,
so the definition of involute gives that β 1 = β 0 + (s0 − s1 )tα = β 0 + `n0 . Hence
β 1 is a parallel curve to β 0 , and
κ0 1/s0 1 1
= = = = κ1 .
|1 − κ0 `| |1 − `/s0 | |s0 − `| s1
√
A short calculation shows that |α0 | = 3eu and
1
t = √ (cos u − sin u, sin u + cos u, 1) .
3
When z = λ0 we have that u = log λ0 , and when z = λ1 we have that
u = log λ1 . So, required length is
√ Z log λ1 u √
3 e du = 3(λ1 − λ0 ) .
log λ0
so that α0 × α00 = κ|α0 |3 b. Equating the lengths of both sides, we obtain the
required formula for κ. Differentiating one more time, we find that
so that (α0 × α00 ).α000 = −κ2 τ |α0 |6 . Using the expression we have just found for
κ now gives the required expression for τ .
1.16. (i) Assume first that α(u) = (a cos u, a sin u, bu). The tangent√ vector
to α(u) makes angle φ with the generating lines where cos φ = b/ a2 + b2 .
Hence φ is a constant different from 0 and π/2. We saw in Example 2 of §1.5
that α(u) has non-zero torsion, so it follows that α(u) is a helix with value
(a, 0, 0) when u = 0.
Conversely, assume that α(v) = (a cos θ(v), a sin θ(v), v + c) describes a helix
with value (a, 0, 0) when v = 0. Then c = 0 and, denoting differentiation with
respect to v by 0 , we have
d κ
(α + an) = t + 2 (−κt − τ b)
ds κ + τ2
τ
= 2 (τ t − κb) .
κ + τ2
1.21. First assume that α(s) lies on a sphere with centre p and radius r,
or, equivalently, that (α − p).(α − p) = r2 . We shall differentiate repeatedly to
find an expression for α − p in terms of t, n and b.
So, differentiate once to find that (α − p).t = 0. Differentiating again and
using Serret-Frenet, we obtain (α − p).n = −1/κ. Differentiating this and using
Serret-Frenet gives (α − p).b = −κ0 /(τ κ2 ). It now follows that
1 κ0
α − p = − n − 2b .
κ τκ
The derivative of the left hand side, and hence of the right hand side, of the
above equation is t. In particular, the coefficient of b of the derivative of the
A FIRST COURSE IN DIFFERENTIAL GEOMETRY 7
right hand side is zero, which gives the desired relation between κ and τ .
Conversely, assume that κ and τ for a regular curve α(s) are related as in
the given formula, and let
1 κ0
p(s) = α + n + 2b .
κ τκ
The given relation between κ and τ may be used to show that p0 = 0 so that p
is constant. It now follows (again by differentiating) that (α − p).(α − p) is also
constant. Since α is not constant we have that |α − p| = r for some positive
constant r, so that α lies on the sphere with centre p and radius r.
Chapter 2
2.1. The line through (u, v, 0) and (0, 0, 1) may be parametrised by α(t) =
2
t(u, v, 0) + (1 − t)(0, 0, 1). This line intersects S 2 (1) when tu, tv, (1 − t) = 1,
and a short calculation gives that t = 0 or t = 2/(u2 + v 2 + 1). Since t = 0
corresponds to (0, 0, 1), we quickly see that x(u, v) is as claimed.
The formula for F follows from consideration of similar triangles, OR we
may use the following method which is similar to the one used in the earlier
part of the solution to this exercise. The line through (x, y, z) and (0, 0, 1) may
be parametrised by β(t) = t(x, y, z) + (1 − t)(0, 0, 1). For z 6= 1, this line cuts
the xy-plane when t = (1 − z)−1 , which gives the point (1 − z)−1 (x, y, 0) on the
line β. The formula for F now follows.
That F x(u, v) = (u, v) is a routine calculation (and also follows from the
geometrical construction). That x is a local parametrisation as claimed is now
immediate from conditions (S1) and (S2), taking U = R2 and W = R3 \ P .
2.3(a). (i) There are many ways. The one which perhaps is closest to that
given in Example 4 of §2.1 is to cover the cylinder by four local parametrisations.
2 + 3
Firstly, let U = {(u, v)
√ ∈ R : −1 < u < 1, v ∈ R} and let x 3 : U → R be
+ 2
given by x (u, v) = ( 1 − u , u, v). If we let W = {(x, y, z) ∈ R : x > 0} and
let F : W → R2 be given by F (x, y, z) = (y, z), then conditions (S1) and (S2)
are satisfied. Hence x+ is a local parametrisation of the given cylinder S, and
the corresponding coordinate neighbourhood is shown in the left hand picture
of Figure 5. The whole of S may be covered by x+ and an additional three
local parametrisations with domain U given by
p
x− (u, v) = (− 1 − u2 , u, v) ,