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Miller, Kierste – Exam 1 – Due: Oct 2 2007, 11:00 pm – Inst: JEGilbert
1
This printout should have 18 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
001
(part 1 oF 1) 10 points
Which one oF the Following polar Functions
has graph
1.
r
= sin 3
θ
correct
2.
r
= sin 4
θ
3.
r
= cos 4
θ
4.
r
= cos 2
θ
5.
r
= sin 2
θ
6.
r
= cos 3
θ
Explanation:
Consequently, the graph is that oF
r
= sin 3
θ
.
keywords: polar graph, sixleaved rose, eight
leaved rose, Fourleaved rose
002
(part 1 oF 1) 10 points
Which oF the Following expressions are well
defned For all vectors
a
,
b
,
c
, and
d
?
I
a
·
(
b
×
c
) ,
II

a
×
(
b
×
c
) ,
III
(
a
×
b
)
·
(
c
×
d
) .
1.
I only
2.
II and III only
3.
III only
4.
I and II only
5.
II only
6.
none oF them
7.
I and III only
correct
8.
all oF them
Explanation:
The cross product is defned only For two
vectors, and its value is a vector; on the other
hand, the dot product is defned only For two
vectors, and its value is a scalar.
±or the three given expressions, thereFore,
we see that
I is welldefned because it is the dot prod
uct oF two vectors.
II is not welldefned because the frst term
in the cross product is a scalar because it is
the length oF a vector, not a vector.
III is welldefned because both terms in the
dot product are cross products, hence vectors.
keywords: vectors, dot product, cross prod
uct, T/±, length,
003
(part 1 oF 1) 10 points
Which oF the Following sets oF inequalitites
describes the region consisting oF all points
outside a sphere oF radius 3 centered at the
origin and inside a sphere oF radius 6 centered
at the origin.
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View Full Document Miller, Kierste – Exam 1 – Due: Oct 2 2007, 11:00 pm – Inst: JEGilbert
2
1.
9
< x
2
+
y
2
+
z
2
<
36
correct
2.
3
≤
x
2
+
y
2
+
z
2
≤
6
3.
3
< x
2
+
y
2
+
z
2
<
6
4.
3
≤
x
2
+
y
2
+
z
2
<
6
5.
9
≤
x
2
+
y
2
+
z
2
≤
36
6.
9
< x
2
+
y
2
+
z
2
≤
36
Explanation:
The region consists of all points whose dis
tance to the origin is greater that 3, but less
than 6,
i.e
, all points satisfying the inequali
ties
3
<
p
x
2
+
y
2
+
z
2
<
6
,
which can also be written as
9
< x
2
+
y
2
+
z
2
<
36
.
keywords:
004
(part 1 of 1) 10 points
Find
d
2
y
dx
2
for the curve given parametrically
by
x
(
t
) = 3 +
t
2
,
y
(
t
) =
t
2
+ 2
t
3
.
1.
d
2
y
dx
2
=
3
t
2.
d
2
y
dx
2
=
2
t
3
3.
d
2
y
dx
2
=
5
6
t
4.
d
2
y
dx
2
=
3
t
2
5.
d
2
y
dx
2
=
5
t
6
6.
d
2
y
dx
2
=
3
2
t
correct
Explanation:
Di±erentiating with respect to
t
we see that
x
0
(
t
) = 2
t,
y
0
(
t
) = 2
t
+ 6
t
2
.
Thus
dy
dx
=
y
0
(
t
)
x
0
(
t
)
=
2
t
+ 6
t
2
2
t
= 1 + 3
t.
On the other hand, by the Chain Rule,
d
dt
‡
dy
dx
·
=
dx
dt
n
d
dx
‡
dy
dx
·o
=
‡
dx
dt
·
d
2
y
dx
2
.
Consequently,
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This note was uploaded on 05/08/2008 for the course M 408 M taught by Professor Gilbert during the Fall '07 term at University of Texas at Austin.
 Fall '07
 Gilbert

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