MATHEMATICS 227 April, 2013 Final Exam Solutions
1. A skier descends the hill z =
p
4 − x2 − y 2 along a trail with parameterization
x = sin(2θ),
y = 1 − cos(2θ),
z = 2 cos θ,
0≤θ≤
π
2
Let P denote the point on the trail where x = 1.
(a) Find the Frenet frame T̂, N̂, B̂ and the curvature κ of the ski trail at the point P .
√
(b) The skier’s acceleration at P is a = (−2, 3, −2 2). Find, at P ,
(i) the rate of change of the skier’s speed and
(ii) the skier’s velocity (a vector).
Solution. (a) Treating the parameter θ as time–like gives
r = sin(2θ), 1 − cos(2θ), 2 cos θ ,
v = 2 cos(2θ), 2 sin(2θ), −2 sin θ ,
a = − 4 sin(2θ), 4 cos(2θ), −2 cos θ .
At the point P , we have θ = π/4, giving instantaneous values
√
r = (1, 1, 2),
√ v
√1 0, 2, − 2 .
|v| =
6
√ √ v×a
√1
−
2, 2 2, 4
|v×a| =
26
Hence T̂ =
Now B̂ =
This leads to
Finally,
√ v = 0, 2, − 2 ,
ı̂ı
v × a = 0
−4
=
√1
13
v = |v| =
√
6,
√ a = − 4, 0, − 2 .
√ − 1, 2, 2 2 , since
̂
k̂
√ √ √ 2 −√2 = − 2 2, 4 2, 8 ,
0 − 2
|v × a| =
√
√
104 = 2 26.
̂
k̂ √ √ √ √
1 ı̂ı
1
1
N̂ = B̂ × T̂ = √ −1 2 2 √2 = − √
6 2, 2, 2 = − √
6, 1, 2 .
78 0 2 − 2 78
39
√
√
√
√ √
|v × a|
2 26
2 2 13
13
39
κ=
= √ 3 = √ √ = √ =
.
3
v
9
6 2 3
3 3
( 6)
(b) We use the dot product to extract the tangential and normal components of a =
v = |v| is the speed:
dv
2
dt T̂ + v κN̂,
√ √
1
1
dv
10
5√
6.
= a · T̂ = (−2, 3, −2 2) · √ 0, 2, − 2 = √ [0 + 6 + 4] = √ =
dt
3
6
6
6
Similarly, a · N̂ = v 2 κ, so
v2 =
Hence v =
√ 13 × 9
√
9
1
1
a · N̂ = − √ √ (−2, 3, −2 2) · 6, 1, 2 =
= 3.
κ
39
39 39
√
3, and v = v T̂ =
√1
2
√
√ 0, 2, − 2 = (0, 2, −1).
1
where
2. Let φ(x, y) = xy and let F = ∇ φ.
(a) Find an equation for the field line of F which passes through the point (3, 2). Sketch it and verify
that it also passes through (3, −2).
R (3,2)
(b) Find (3,−2) F · dr, where the line integral is along the field line of (a).
Solution. (a) The vector field is F = ∇φ = y ı̂ı + x̂. All field lines have (dx, dy) k F so that
dx
y
=
dy
x
⇐⇒ x dx = y dy ⇐⇒
2
x2
2
=
y2
2
+C
2
To pass through (3, 2), the constant C must obey 32 = 22 + C or C = 25 . At the point (3, 2), we have
x > 0. By continuity and the requirement (from the equation) that x2 ≥ 5, we have x > 0 on the entire
flow line. So the equation of the desired field line is x2 = y 2 + 5, x > 0 . This is a one branch of a
hyperbola with asympototes x = ±y. We want the branch of the hyperbola containing (3, 2), which is
the branch in the sketch below. The point (3, −2) also obeys the equation of the hyperbola and is on
the same branch.
(3, 2)
(3, −2)
(b) Since F is conservative with potential φ,
Z
(3,2)
(3,−2)
F · dr = φ(3, 2) − φ(3, −2) = 3 × 2 − 3 × (−2) = 12
R
3. Find the work C F · dr done by the force F = 2xy ı̂ı + zy ̂ + x2 k̂ on a particle as it moves from
(2, 0, 1) to (0, 2, 5) along the curve C of intersection (in the first octant x, y, z ≥ 0) of the paraboloid
z = (x − 1)2 + y 2 and the plane z = 5 − 2x.
Solution. Parametrize C by
x(t) = t
z(t) = 5 − 2x(t) = 5 − 2t
p
p
p
y(t) = z(t) − (x(t) − 1)2 = 5 − 2t − (t − 1)2 = 4 − t2
with t running from 2 to 0. Then
t
̂ − 2k̂
r′ (t) = ı̂ı − √4−t
2
p
p
p
′
t
F r(t) · r (t) = 2t 4 − t2 − (5 − 2t) 4 − t2 √4−t
− 2t2 = 2t 4 − t2 − 5t
2
Z 0 p
Z
i
h
2 0
2t 4 − t2 − 5t dt = − 23 (4 − t2 )3/2 − 5 t2 = − 16
F · dr =
3 + 10 =
C
2
2
14
3
4. Find the surface area of the torus obtained by rotating the circle (x − 3)2 + z 2 = 4, y = 0 about the
z–axis.
Solution. The torus may be parametrized by
x = (3 + 2 cos φ) cos θ
y = (3 + 2 cos φ) sin θ
2
z = 2 sin φ
0 ≤ φ, θ ≤ 2π
Then
Tθ = (3 + 2 cos φ) − sin θı̂ı + cos θ̂
Tφ = 2 − sin φ cos θı̂ı − sin φ sin θ̂ + cos φk̂
As − sin θı̂ı + cos θ̂ and − sin φ cos θı̂ı − sin φ sin θ̂ + cos φk̂ are mutually perpendicular unit vectors, their
cross product is a unit vector and
|Tθ × Tφ | = 2(3 + 2 cos φ)
=⇒
dS = 2(3 + 2 cos φ) dθ dφ
The total surface area of the torus is
Z 2π
Z 2π
Z
2
dφ
dθ (3 + 2 cos φ) = (2π)2
0
0
2π
dφ (3 + 2 cos φ) = 24π 2
0
5. Let S be the portion of the sphere x2 + y 2 + z 2 = 36 with y + z ≤ 6. Orient S with normal pointing
away from (0, 0, 0). Let
F = (y 2 z − x)ı̂ı + (x + y + z)̂ − xk̂
RR
Evaluate S F · n̂ dS.
Solution. Let S ′ be the portion of y + z = 6 that is inside the sphere x2 + y 2 + z 2 = 36 (with n̂ the
upward pointing normal) and let V be the portion of the ball x2 + y 2 + z 2 ≤ 36 with y + z ≤ 6. Then,
by the divergence theorem
ZZ
ZZZ
ZZ
F · n̂ dS
∇ · F dV −
F · n̂ dS =
As ∇ · F = −1 + 1 − 0 = 0 and, on S ′ , n̂ =
ZZ
S
F · n̂ dS = −
ZZ
S′
F · n̂ dS =
S′
V
S
− √12
ZZ
S′
√1 (̂
+
2
k̂) and y + z = 6
(x + y + z − x) dS =
− √12
ZZ
(y + z) dS =
S′
−6 √12
ZZ
dS
S′
= −6 √12 area(S ′ )
S ′ is a circular disk. Its center (xc , yc , zc ) has to obey yc + zc = 6. By symmetry, xc = 0 and yc = zc ,
so yc = zc = 3. Any point, like (0, 0, 6), which satisfies both y + z = 6 and x2 + y 2 + z 2 = 36 is on the
√
boundary of S ′ . So the radius of S ′ is 0, 3, 3 − (0, 0, 6) = 0, 3, −3 = 3 2. So the area of S ′ is
18π and
ZZ
√
F · n̂ dS = −6 √12 × 18π = −54 2π
S
2
6. Let F = (ex + y)ı̂ı + (sin y 3 + xz) ̂ + z 2 k̂. Evaluate
R
F · dr where C is the curve
C
x2 + y 2 = 10
with
x+y+z =4
positive (i.e. counter-clockwise) orientation as viewed from high on the z-axis.
Solution. Define
S=
oriented with n̂ =
√1 (1, 1, 1).
3
(x, y, z) ∈ IR3 x + y + z = 4, x2 + y 2 ≤ 10
Parametrizing S by z = 4 − x − y, x2 + y 2 ≤ 10, we have
n̂ dS = (1, 1, 1) dx dy
∇ × F = −xı̂ı + (z − 1)k̂
∇ × F · n̂ dS = (z − x − 1) dx dy = (3 − 2x − y) dx dy
3
on S
So, by Stokes’ theorem,
Z
C
F · dr =
ZZ
S
ZZ
∇ × F · n̂ dS =
x2 +y 2 ≤10
(3 − 2x − y) dx dy = 3
ZZ
dx dy = 30π
x2 +y 2 ≤10
7. A pile of wet sand having total volume 5π covers the disk z = 0, x2 + y 2 ≤ 1. Call the top surface of
the pile of sand S. The momentum of water vapour is given by the vector field
F = ∇ φ + µ∇
∇×G
where φ is the water concentration, G is the temperature gradient and µ is a constant. Suppose that
φ = x2 − y 2 + z 2 and G = 13 − y 3ı̂ı + x3̂ + z 3 k̂). Find the flux of F upward through S.
Solution 1. Let V be the sand pile. Then the surface of V (with outward normal) is the union
of S (with normal pointing away from the origin) and the disk B = (x, y, 0) x2 + y 2 ≤ 1 ,
with normal −k̂. Since
∇φ = 2xı̂ı − 2y̂ + 2z k̂
∇ · (∇
∇φ) = 2
∇ · (∇
∇ × G) = 0
the Divergence Theorem implies
ZZ
S
ZZ
F · (−k̂) dS
∇ · F dV −
B
Z Z ZV
ZZ
=
2 dV +
(2z + µk̂ · ∇ × G) dS
V
B
ZZ
= 10π + µ
k̂ · ∇ × G dS
F · n̂ dS =
ZZZ
B
since V has volume 5π and z = 0 on B. Since k̂ · ∇ × G = x2 + y 2 we have
ZZ
B
All together
RR
S
k̂ · ∇ × G dS =
ZZ
Z
(x2 + y 2 ) dx dy =
B
1
dr
0
Z
2π
0
4 1
dθ r3 = 2π r4 =
0
π
2
F · n̂ dS = 10π + µ π2 .
Solution 2. As in solution 1, we have
ZZ
S
F · n̂ dS = 10π + µ
ZZ
B
k̂ · ∇ × G dS
H
RR
This time we use Stokes’ Theorem or Green’s Theorem to give B k̂ · ∇ × G dS = C G · dr,
where C is x2 + y 2 = 1 oriented counterclockwise. Parametrizing C by x(t) = cos t, y(t) =
sin t, z(t) = 0,
I
C
1
3
Z
2π
G · dr =
2
3
Z
2π
=
0
0
h
i
− sin3 t(− sin t) + cos3 t(cos t) dt =
1+cos 2t 2
2
dt =
1
6
Z
0
2π
1
3
Z
0
2π
h
i
sin4 t + cos4 t dt =
2
3
Z
0
1 + 2 cos 2t + cos2 2t dt = 16 2π + 0 + π dt =
which gives the same answer as solution 1 did.
4
2π
π
2
cos4 t dt
8. Say whether each of the following statements is true or false and explain why. You may
assume that the curves, surfaces and functions are all sufficiently smooth.
(a) Let φ be a scalar field. Suppose that
ZZ
S
∇ φ · n̂ dS = 0
for every closed oriented surface S. Here n̂ is the outward unit normal to S. Then φ is
a constant.
R
R
(b) If F and G are two vector fields defined in R3 and S F · n̂ dS = S G · n̂ dS for every
orientable surface S, then F = G.
(c) Let F be a vector field in the xy-plane which is perpendicular to r = (x, y) at all points.
Then each flow line of F is contained in a circle.
(d) Let F be a vector field in the xy-plane which is perpendicular to r = (x, y) at all points.
Then the unit circle is a flow line of F.
Solution. (a) False . By the divergence theorem, if V is the solid bounded by S,
ZZ
S
∇φ · n̂ dS =
ZZZ
V
∇ · ∇ φ dV
2
There are many nonconstant scalar fields φ for which ∇ · (∇
∇φ) = ∂∂xφ2 +
2
2
example φ = x or φ = x − y .
(b) True . Pick any point a and any unit vector d̂. I will show that
∂2 φ
∂y 2
+
∂2 φ
∂z 2
= 0. For
d̂ · F(a) = d̂ · G(a)
For each ǫ > 0, let Sǫ be a flat disk of radius ǫ centered on a that is normal to d̂. Then
Z
Z
Z
(F − G) · d̂ dS
G · n̂ dS =
F · n̂ dS −
0=
Sǫ
Sǫ
Sǫ
If d̂·F(a) 6= d̂·G(a), then d̂· F(a)−G(a) 6= 0. By continuity, if ǫ is small enough, d̂· F(x)−G(x)
R
is of the same sign and bounded away from zero for all x ∈ Sǫ . Hence Sǫ (F − G) · d̂ dS is
nonzero, which is a contradiction.
(c) True . Suppose that r(t) is a field line. Then
r′ (t) = F r(t) ⊥ r(t)
=⇒
2
d
dt |r(t)|
= 2r′ (t) · r(t) = 0
Hence, |r(t)|2 is independent of time so that r(t) lies on a circle centered on the origin.
(d) False . If the vector field is the zero field, then every field line is just a single point.
5