SOLUTIONS TO 18.02 PRACTICE FINAL EXAM A
BJORN POONEN
1) For each of (a)(e) below: If the statement is true, write TRUE. If the statement is
false, write FALSE. (Please do not use the abbreviations T and F.) No explanations are
required in this problem.
(a) If
a
and
b
are vectors in
R
3
such that
a
×
b
=
0
, then either
a
=
0
or
b
=
0
.
Solution.
FALSE. Take
a
=
b
=
i
, for example.
(b) If the first row of a 3
×
3 matrix is the same as the second row, then the determinant
of the matrix is 0.
Solution.
TRUE. Adding

1 times the first row to the second row results in a matrix with
a row of all zeros, which has determinant 0 as can be seen from the Laplace expansion along
that row.
(c) If
T
is a unit tangent vector at
P
to a level curve of a differentiable function
f
(
x, y
),
then the directional derivative of
f
at
P
in the direction
T
is 0.
Solution.
TRUE. If
r
(
t
) is a parametrization of the level curve at unit speed, then
f
(
r
(
t
)) is
constant, so its derivative is zero, but on the other hand, the derivative is
∇
f
·
r
(
t
) =
D
T
f
(
P
).
(d) Let
P
=
P
(
x, y
) and
Q
=
Q
(
x, y
) be continuously differentiable functions on the region
R
in
R
2
where 1
< x
2
+
y
2
<
4. If
P dx
+
Q dy
is an exact differential (i.e., the differential
of some function on
R
), then
Q
x
=
P
y
holds everywhere on
R
.
Solution.
TRUE. If
P dx
+
Q dy
=
df
, then
P
=
f
x
and
Q
=
f
y
, so
Q
x
=
f
yx
=
f
xy
=
P
y
.
(e) Let
f
(
x, y
) be a function with continuous second partial derivatives on all of
R
2
. Let
A
=
f
xx
(0
,
0),
B
=
f
xy
(0
,
0),
C
=
f
yy
(0
,
0). If
f
(
x, y
) has a local minimum at (0
,
0), then
AC

B
2
≥
0.
Solution.
TRUE. Since
f
(
x, y
) has a local minimum at (0
,
0), the point (0
,
0) is a critical
point. If
AC

B
2
<
0, then it would be a saddle point, and hence not a local minimum. So
AC

B
2
≥
0.
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2) Find a vector
w
in
R
2
whose scalar component in the direction of
√
2
/
2
,
√
2
/
2
is 2
and whose scalar component in the direction of

√
2
/
2
,
√
2
/
2
is

4.
Solution.
Let
u
=
√
2
/
2
,
√
2
/
2
and
v
=

√
2
/
2
,
√
2
/
2 . Since
u
and
v
are orthogonal
unit vectors,
w
= 2
u

4
v
=
3
√
2
,

√
2
.
3) For what real numbers
t
does there exist a solution (
x, y, z
) to
(
t

6)
x
+ 2
z
= 0

9
x
+ (
t

3)
y
+ 5
z
= 0

12
x
+ (
t
+ 4)
z
= 0
other than (0
,
0
,
0)?
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 Fall '08
 Auroux
 Multivariable Calculus, Vector Calculus, Solution., Solution. Let

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