A First Course in Differential Geometry (2019) – Woodward – Solutions Manual PDF

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Solutions to exercises

, 1

A FIRST COURSE IN DIFFERENTIAL GEOMETRY
Woodward and Bolton
Solutions to exercises
Chapter 1

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 α′ = (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 3
Z
sin 2u du = .
2 0 2




(a) Astroid (b) An epicycloid

Figure 1


1.2. A sketch of the trace of an epicycloid is given in Figure 1(b). Trigono-
metric identities may be used to show that


α′ = 4r sin(u/2) sin(3u/2), cos(3u/2) .
R 2π
So, for 0 ≤ u ≤ 2π, |α′ | = 4r sin(u/2), and required length is 4r 0
sin(u/2)du =
16r.

1.3. When r = 1, a calculation shows that α′ = tanh u sech u(sinh u, −1),
so that, for u ≥ 0, t = sech u(sinh u, −1). It follows that α + t = (u, 0). A
sketch of the trace of a tractrix is given in Figure 2(a).
2 2
1.4. Here, |α′ | = (1 + g ′ )1/2 and t = (1 + g ′ )−1/2 (1, g ′ ). Hence n =
2 2
(1 + g ′ )−1/2 (−g ′ , 1). A calculation shows that t′ = g ′′ (1 + g ′ )−3/2 (−g ′ , 1), so
that
dt 1 g ′′
= ′ t′ = n.
ds |α | (1 + g ′ 2 )3/2

,2 SOLUTIONS TO EXERCISES




Figure 2: (a) shows a tractrix and (b) shows three catenaries


2
Hence κ = g ′′ (1 + g ′ )−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, |α′ | = tanh u and
t = (tanh u, −sech u). It follows that dt/ds = (|α′ |)−1 t′ = 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:-
α′ = (1, sinh u), so that |α′ | = cosh u and t = (sech u, tanh u). Hence n =
(−tanh u, sech u), and

dt 1 1 1
= ′ t′ = 2 (−tanh u, sech u) = n.
ds |α | cosh u cosh2 u

Hence κ = sech2 u. A sketch of the traces of three catenaries is given in Figure
2(b).

1.7. Differentiating with respect to u, we see that, using Serret-Frenet,

αℓ ′ = α′ + ℓn′ = |α′ |(t − κℓt) = |α′ |(1 − κℓ)t .

It follows that |αℓ ′ | = |α′ | |1 − κℓ| and tℓ = ǫt, where ǫ = (1 − κℓ)/|1 − κℓ|.
Hence nℓ = ǫn, so, if sℓ denotes arc length along αℓ , we have

dtℓ 1 ǫ
= ′ t′ = ′ t′ .
dsℓ |α | |1 − κℓ| ℓ |α | |1 − κℓ|

Using Serret-Frenet, t′ = |α′ |κn = |α′ |κǫnℓ , from which the result follows.
2 2
1.8. Since α′ = (x′ , y ′ ), we have that |α′ | = (x′ + y ′ )1/2 . Hence t =
2 2 2 2
(x , y ′ )/(x′ + y ′ )1/2 and n = (−y ′ , x′ )/(x′ + y ′ )1/2 . Hence
′



α′′ α′ (x′ x′′ + y ′ y ′′ )
t′ = 2 1/2 − ,
(x′ 2 +y )
′ (x′ 2 + y ′ 2 )3/2

, A FIRST COURSE IN DIFFERENTIAL GEOMETRY 3

and a short calculation shows that
dt 1 
′ ′′ ′ ′ ′′ ′ ′ ′′ ′′ ′

= ′2 2 y (x y − x y ), x (x y − x y )
ds (x + y ′ )2
x′ y ′′ − x′′ y ′
= ′2 n,
(x + y ′ 2 )3/2
and the result follows.

1.9. (i) Let sα be arc length along α measured from u = 0. Since α′ =
(1, sinh u) we see that dsα /du = |α′ | = cosh u. Hence sα (u) = sinh u and
tα = (sech u, tanh u). The result follows from formula (1.9) for the involute.
(ii) The evolute of α is given by
1
β =α+ nα .
κα
Here, we have (from Exercise 1.6) that κα = sech2 u and nα = (−tanh u, sech u).
A direct substitution gives the result.
A short calculation shows that β ′ = 0 if and only if u = 0, so this gives the
only singular point of β (where the curve β has a cusp). A sketch of the traces
of α and β is given in Figure 3.




Figure 3: A catenary and its evolute


1.10. Let sα denote arc length along α starting at u = u0 . Then, using
(1.9) and the notation used there, we see that

β ′ = α′ − sα ′ tα − sα tα ′ = −sα tα ′ .

It follows that β ′ = −sα |α′ |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 ℓ be the length of α measured from α(u0 ) to α(u1 ). Then ℓ = s0 −s1 > 0,
so the definition of involute gives that β 1 = β 0 + (s0 − s1 )tα = β 0 + ℓn0 . Hence
β 1 is a parallel curve to β 0 , and
κ0 1/s0 1 1
= = = = κ1 .
|1 − κ0 ℓ| |1 − ℓ/s0 | |s0 − ℓ| s1

1.12. A sketch of the trace of α is given in Figure 4.