Math 21a
First Old Final Exam Solutions
Spring, 2009
Part I:
Multiple choice. Each problem has a unique correct answer. You do not need to justify your
answers in this part of the exam.
1
TrueFalse questions. Circle the correct letter. No justiﬁcations are required.
T F
The divergence of the gradient of any
f
(
x,y,z
) is always zero.
Solution:
This is
False
. In general grad
f
=
∇
f
=
h
f
x
,f
y
z
i
, so div(grad
f
) =
∇ · ∇
f
=
f
xx
+
f
yy
+
f
zz
. For example, if
f
(
) =
x
2
+
y
2
+
z
2
, we get div(grad
f
) = 6.
T F
If a nonempty quadric surface
g
(
) =
ax
2
+
by
2
+
cz
2
= 5 can be contained inside
a ﬁnite box, then
a,b,c
≥
0.
Solution:
This is
True
. If one of
a
,
b
, and
c
is zero, then the surface is a cylinder or an
inﬁnite space whose crosssections are all hyperbolas. If one or two of
a
,
b
, and
c
is negative,
then the surface is a hyperboloid (of one or two sheets). None of these surfaces ﬁts in a ﬁnite
box.
T F
If
F
is a vector ﬁeld in space, then the ﬂux of
F
through any closed surface
S
is 0.
Solution:
This is
False
. This is only true if div
F
= 0, since the Divergence Theorem says
that
RR
S
F
˙
d
S
=
RRR
E
div
F
dV
, where
E
is the solid enclosed by
S
. This is zero for any closed
surface
S
only when div
F
= 0.
T F
The ﬂux of the vector ﬁeld
F
(
) =
h
y
+
z,y,

z
i
through the boundary of a solid
region
E
is equal to the volume of
E
.
Solution:
This is
False
. Again we refer to the Divergence Theorem, which says
RR
S
F
˙
d
S
=
RRR
E
div
F
dV
, where
S
is the boundary of
E
. In this case div
F
= 0 + 1

1 = 0, so the ﬂux
through
S
is zero, not the volume of
E
.
T F
If in spherical coordinates the equation
φ
=
α
(with a constant
α
) deﬁnes a plane, then
α
=
π/
2.
Solution:
This is
. If 0
< α <
π
2
, then
φ
=
α
is a halfcone opening upward. Similarly,
if
π
2
< α < π
, then
φ
=
α
is a halfcone opening downward. If
α
= 0 or
α
=
π
, then
φ
=
α
is a
halfline (the positive or negative
z
axis). That leaves
φ
=
π
2
, which is the
xy
plane.
2
Which of the following is the cosine of the angle between the two planes

5
x

3
y

4
z
= 0 and
x
+
y
+
z
= 1?
(a)
6
√
2
5
(b)
2
√
5
6
(c)
2
√
6
5
(d)
5
√
3
2
(e) None of the above.
Solution:
The correct answer is (c). Normal vectors to the two planes are
a
=
h
5
,
3
,
4
i
and
b
=
h
1
,
1
,
1
i
, and
cos
θ
=
a
·
b

a
 
b

=
2
√
6
5
,
where
θ
is the angle between the two planes.
3
Which of the functions
u
(
x,t
) below are solutions to the partial diﬀerential equation
∂u
∂t
=
∂
2
u
∂x
2
?
(a)
u
(
) =
e

t
sin
x
(b)
u
(
) =
e

2
t
sin 2
x
(c)
u
(
) =
e
t
sin
x
(d)
u
(
) = 3
(e)
u
(
) =
x
2
+ 2
t
Solution:
(a), (d), and (e) are solutions.
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View Full DocumentMath 21a
First Old Final Exam Solutions
Spring, 2009
4
Which of the following line integrals below are path independent?
(a)
Z
C
(10
x

7
y
)
dx

(7
x

2
y
)
dy
(b)
Z
C
(45
x
4
y
2

6
y
6
+ 3)
dx
+ (18
x
5
y

12
xy
5
+ 7)
dy
(c)
Z
C
(2
e
y

ye
x
)
dx
+ (2
xe
y

e
x
)
dy
(d)
Z
C
4
y
2
cos(
xy
2
)
dx
+ 8
x
cos(
xy
2
)
dy
(e)
Z
C
(sin
y
+
y
sin
x
)
dx
+ (
x
cos
y

cos
x
+ 1)
dy
Solution:
A line integral
R
C
P dx
+
Qdy
is path independent if
∇
f
=
P
i
+
Q
j
for some function
f
. If
P
i
+
Q
j
is deﬁned on a simply connected open set, we need only check that
P
y
=
Q
x
. Thus,
(a), (c), and (e) are path independent line integrals.
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 Spring '10
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 Vector Calculus, Sin, ... ..., Stokes' theorem, Surface integral

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