Solutions to Math 227 Final Exam, April, 2011
1. Let x : [1, 2] → IR3 be defined by:
2
t√
2 2
x(t) =
,
2
t√
2 2
,
t3
3
(a) Compute the arc length of the path x from x(1) to x(t) for all 1 ≤ t ≤ 2.
(b) Find an explicit reparametrization of x by arc length.
Solution. (a) The velocity vector for this curve is
v(t) =
√t
2
,
√t
2
, t2
=⇒
|v(t)| =
p
p
t2 + t4 = t 1 + t2
So for, t ≥ 1, the arclength from x(1) to x(t) is
s(t) =
Z
1
t
|v(t̃)| dt̃ =
Z
1
t
p
t̃ 1 + t̃2 dt̃
u=1+t̃2
=
Z
1+t2
s=
1
3
(1 + t2 )3/2 − 23/2
=⇒
=⇒
2
2
(b) To find the inverse function for s(t), we solve s =
1+t2
√ du
u 2 = 31 u3/2 = 31 (1 + t2 )3/2 − 23/2
1
3
(1 + t2 )3/2 − 23/2 for t as a function of s:
(1 + t2 )3/2 = 3s + 23/2 =⇒
q
2/3
3s + 23/2
−1
t=
so the desired parametrization is
3/2 2/3
1
X(s) = 2√
−
1
,
3s
+
2
2
1
√
2 2
3s + 23/2
2/3
1 + t2 = 3s + 23/2
2/3
2/3
3/2 − 1 , 31 3s + 23/2
−1
2. Which of the following subsets of IR3 is simply connected? (Note that such a set need not be open.)
Just answer Yes or No. (Justification is not required.)
(a) The complement of a line
(b) The complement of a half-line
(c) The complement of a circle
(d) The complement of the unit ball
(e) (x, y, z) ∈ IR3 2 < x2 + y 2 + z 2 < 3
(f) The unit sphere
(g) The punctured unit sphere (i.e., the set consisting of all points in the unit sphere except for one
point)
(h) The doubly punctured unit sphere (i.e., the set consisting of all points in the unit sphere except for
two distinct points)
(i) The torus
(j) The solid torus (i.e., the bounded solid region that is bounded by the torus)
Solution. (a) Not simply connected. (You can’t contract a circle that is centred on the line like a bead
on a string.)
(b) Is simply connected. (You can contract a circle that is centred on the line like a bead on a string by
first sliding it off the end of the string.)
(c) Not simply connected. (Like part (a).)
(d) Is simply connected. (You can contract any circle by first moving it well away from the unit ball.)
(e) Is simply connected. (You can contract any circle by first moving it close to the north pole.)
(f) Is simply connected. (Like in part (e).)
1
(g) Is simply connected. (Call the puncture the south pole. Then you can contract any circle by first
moving it close to the north pole.)
(h) Not simply connected. (For example if the punctures are at the north and south poles, then you
cannot contract the equator.)
(i) Not simply connected. (The inner equator, for example, is not contractible.)
(j) Not simply connected. (The inner equator, for example, is not contractible.)
3. Let Cr (p) denote the circle of radius r centered at p ∈ IR2 , oriented counterclockwise. Let F = (M, N )
∂M
be a C 1 vector field on IR2 such that ∂N
∂x = ∂y at all points except possibly (0, 0), (0, 2) and (0, −2).
Let
Z
F · ds
α=
C3 ((0,2))
Z
F · ds
β=
C3 ((0,−2))
Z
γ=
F · ds
C5 ((0,0))
Z
F · ds
δ=
C1 ((0,0))
Find δ in terms of α, β and γ.
Solution. On the left, below, is a figure showing the four curves C1 = C5 (0, 0) , C2 = C3 (0, 2) ,
C3 = C3 (0, −2) and C4 = C1 (0, 0) (which is not labelled in the figure). In the centre below is a similar
5
5
C1
C2
C3
C1
C2′
C4
5
R
C2′
S
C2
C4
5
C3′
figure in which C2 has been replaced by C2′ = C1 (0, 2) and C3 has been replaced by C3′ = C1 (0, −2) .
By Green’s theorem (see the figure on the right above)
0=
ZZ
S
∂N
∂x
− ∂M
dxdy =
∂y
Z
C2
F·ds −
Z
C2′
F·ds −
Z
F·ds = α−
Z
C2′
C4
F·ds − δ
=⇒
Z
F·ds = α− δ
C2′
Similarly
Z
C3′
F · ds = β − δ
By Green’s theorem (see the figure in the centre above)
0=
ZZ
R
∂N
∂x
−
∂M
∂y
dxdy =
Z
C1
F · ds −
Z
C2′
F · ds −
Hence δ = α + β − γ.
2
Z
C4
F · ds −
Z
C3′
F · ds = γ − (α − δ) − δ − (β − δ)
4. Let C1 be the oriented straight line from (0, 0, 0) to (0, 1, 0). Let C2 be the oriented straight line from
(0, 1, 0) to (1, 0, 1). Let C be the curve consisting of C1 followed by C2 . Let F be the vector field
F(x, y, z) = − xy + x3 + z 2 , ey
Find
R
C
5
cos(y)
− z − x , yz + ez
F · ds.
Solution 1. We may parametrize C1 by x(t) = (0, t, 0) with 0 ≤ t ≤ 1. As x′ (t) = (0, 1, 0),
Z
C1
F · ds =
Z
1
5
0, et
cos(t)
0
, 1 · (0, 1, 0) dt =
Z
1
5
et
cos(t)
dt
0
We may parametrize C2 by x(t) = (t, 1 − t, t) with 0 ≤ t ≤ 1. As x′ (t) = (1, −1, 1),
Z
C2
1
F · ds =
Z
1
=
Z
0
0
Since
Z
1
− t(1 − t) + t3 + t2 , e(1−t)
t3 + t2 + 2t − e(1−t)
e(1−t)
5
cos(1−t)
dt
5
t̃=1−t
=
0
we have
Z
Z
F · ds =
C
C1
F · ds +
Z
C2
F · ds =
Z
1
5
cos(1−t)
cos(1−t)
Z
0
5
et̃
− 2t, (1 − t)t + et · (1, −1, 1) dt
+ et dt
cos(t̃)
(−dt̃) =
1
1
4
+
5
et̃
cos(t̃)
dt̃
0
1
t3 + t2 + 2t + et dt =
0
Z
1
3
+ 1 + [e1 − e0 ] = e +
7
12
Solution 2. Denote by C3 the oriented straight line from (1, 0, 1) to (0, 0, 0) and by T the triangular
region contained in the plane z = x that is bounded by C1 , C2 and C3 . Orient T with the downward
pointing normal n̂ = √12 (1, 0, −1). To see that this orientation of T has ∂T = C1 + C2 + C3 , look at the
figure below.
z
C2
C3
y
n̂
C1
x
By Stokes’ theorem,
Z
C
F · ds =
Z
∂T
F · ds −
Z
C3
F · ds =
Z
T
∇ × F · n̂ dS −
Z
C3
F · ds
Since ∇ × F = (z + 1)ı̂ı + 2z̂ + (x − 1)k̂ and n̂dS = (ı̂ı − k̂) dxdy and, as (x, y, z) runs over T (which is
the triangle with vertices (0, 0, 0), (0, 1, 0) and (1, 0, 1)), (x, y) runs over the triangle Txy with vertices
(0, 0), (0, 1) and (1, 0)
Z
Z
Z
dxdy = 2 × 21 × 1 × 1
[(z + 1)ı̂ı + 2z̂ + (x − 1)k̂] · (ı̂ı − k̂) dxdy = 2
∇ × F · n̂ dS =
Txy
Txy
T
=1
3
We may parametrize C3 by x(t) = (t, 0, t) with t running from 1 to 0. As x′ (t) = (1, 0, 1),
Z
and
C3
F · ds =
Z
Z
C
0
1
t3 + t2 , 1 − 2t, et · (1, 0, 1) dt = −
F · ds =
Z
T
∇ × F · n̂ dS −
Z
C3
Z
0
1
[et + t3 + t2 ] dt = −[e1 − e0 +
F · ds = 1 + [e − 1 +
1
4
+ 31 ] = e +
1
4
+ 13 ]
7
12
5. Let B be the unit ball and S be the unit sphere in IR3 . Let F be a C 1 vector field on a neighbourhood
of B. Assume that
i. curl(F) = 0
ii. div(F) = 0, and
iii. On S, F is orthogonal to the radial vector field (x, y, z).
Show that F = 0 on B. (You may want to use the fact that if the integral of a nonnegative continuous
function is zero, then the function itself is identically zero).
Solution. Since B is simply connected, condition (i) implies that F is conservative so that there is a
(potential) function ϕ obeying F = ∇ϕ. By condition (ii)
|F|2 = (∇
∇ϕ) · F = ∇ · (ϕF) − ϕ∇
∇ · F = ∇ · (ϕF)
So by the divergence theorem
ZZZ
ZZZ
ZZ
2
|F| dV =
∇ · (ϕF) dV =
ϕF · n̂ dS = 0
B
B
S
by condition (iii) and the observation that the unit normal, n̂, to the unit sphere, S, is radial. Consequently, |F|2 , and hence F is identically zero on B.
P3
6. For i = 1, 2, 3, let φi = j=1 fij dxj be a differential 1–form on IR3 . Show that

f11
φ1 ∧ φ2 ∧ φ3 = det  f21
f31
f12
f22
f32

f13
f23  dx1 ∧ dx2 ∧ dx3
f33
Solution. Recall that dxj ∧ dxk ∧ dxℓ is zero whenever at least two of j, k, ℓ are the same. So
φ1 ∧ φ2 ∧ φ3 =
X
3
=
X
f1j dxj
j=1
1≤j,k,ℓ≤3
all different
=
∧
X
3
k=1
f2k dxk
∧
X
3
f3ℓ dxℓ
ℓ=1
f1j f2k f3ℓ dxj ∧ dxk ∧ dxℓ
X
f1σ(1) f2σ(2) f3σ(3) dxσ(1) ∧ dxσ(2) ∧ dxσ(3)
X
(−1)σ f1σ(1) f2σ(2) f3σ(3) dx1 ∧ dx2 ∧ dx3
σ∈S3
=
σ∈S3
where S3 is the set of all permutations of (1, 2, 3) and (−1)σ is the sign of the permutation σ. It now
suffices to recall that the definition of the determinant is


f11 f12 f13
X
det  f21 f22 f23  =
(−1)σ f1σ(1) f2σ(2) f3σ(3)
f31 f32 f33
σ∈S3
4
7. Let S denote the torus in IR3 defined by
p
2
x2 + y 2 − 2 + z 2 = 1
Find an orientation form on S, i.e., a differential 2–form on S that is nonzero on every pair of linearly
independent tangent vectors at every point of S. Give your answer in the form f1 dx ∧ dy + f2 dy ∧ dz +
f3 dz ∧ dx with explicit expressions for f1 , f2 , and f3 as functions of x, y and z.
p
Solution. We’ll start by parametrizing the torus. Set w = x2 + y 2 − 2. Then we may parametrize
p
the circle w2 + z 2 = 1 by w = sin ϕ, z = cos ϕ, 0 ≤ ϕ < 2π. For each fixed ϕ, w = x2 + y 2 − 2 = sin ϕ
is again a circle which we may parametrize by x = (2 + sin ϕ) cos θ, y = (2 + sin ϕ) sin θ, 0 ≤ θ < 2π. All
together
x = (2 + sin ϕ) cos θ
dx = cos ϕ cos θ dϕ − (2 + sin ϕ) sin θ dθ
y = (2 + sin ϕ) sin θ
dy = cos ϕ sin θ dϕ + (2 + sin ϕ) cos θ dθ
z = cos ϕ
dz = − sin ϕ dϕ
∂
One possible orientation form is dϕ ∧ dθ, because the vector fields ∂ϕ
and ∂∂θ are linearly independent
at every point of the torus. So any nowhere zero multiple of dϕ ∧ dθ is also a possible orientation form.
To construct such a form in the form f1 dx ∧ dy + f2 dy ∧ dz + f3 dz ∧ dx, observe that
p
x2 + y 2 z dϕ ∧ dθ
p
dy ∧ dz = − sin ϕ(2 + sin ϕ) cos θ dθ ∧ dϕ = x x2 + y 2 − 2 dϕ ∧ dθ
p
dz ∧ dx = sin ϕ(2 + sin ϕ) sin θ dϕ ∧ dθ = y x2 + y 2 − 2 dϕ ∧ dθ
dx ∧ dy = cos ϕ(2 + sin ϕ) dϕ ∧ dθ
So
=
p
p
p
x2 + y 2 z dx ∧ dy + x x2 + y 2 − 2 dy ∧ dz + y x2 + y 2 − 2 dz ∧ dx
p
2 = [x2 + y 2 ]z 2 + [x2 + y 2 ] x2 + y 2 − 2 dϕ ∧ dθ
= [x2 + y 2 ]z 2 + [x2 + y 2 ] 1 − z 2 dϕ ∧ dθ
= [x2 + y 2 ] dϕ ∧ dθ
is one possible orientation form, as x2 + y 2 ≥ 1 on the entire torus.
6
8. Let F(x) = ex (x − x3 ), considered as a vector field on IR1 = IR.
(a) Find the stationary points of F. (Recall that a stationary point of a vector field is a point p such
that the flow line through p consists of the single point p itself).
(b) Sketch F.
(c) Let x(t) be any flow line for F. Prove that limt→+∞ x(t) is a stationary point of F. (You do not
need to prove that the limit exists).
(d) For a point q ∈ IR, let xq (t) denote the flow line such that xq (0) = q. For a stationary point p of
F, define the basin of attraction
Bp =
q ∈ IR lim xq (t) = p
t→+∞
Explicitly describe the basin of attraction for each stationary point of F.
6
Solution. (a) Since F(x) = ex x(1−x)(1+x) vanishes at x = −1, 0, +1 and nowhere else, the stationary
points are x = −1, 0, +1.
(b)
−1
0
1
5
(c) Denote limt→+∞ x(t) = p. Then limt→+∞ x(t + 1) = p too, so
lim [x(t + 1) − x(t)] = p − p = 0
t→+∞
By the mean value theorem
x(t + 1) − x(t) = x′ (ct ) = F x(ct )
for some t < ct < t + 1. As t → ∞, ct → ∞ too so that limt→+∞ x(ct ) = p and, since F is continuous,
0 = lim [x(t + 1) − x(t)] = lim F x(ct ) = F
t→+∞
t→+∞
so p is a stationary point.
(d) From the sketch in part (b)
B0 = 0
B1 = (0, ∞)
B−1 = (−∞, 0)
6
lim x(ct ) = F(p)
t→+∞