suleimenov (bs26835) – FINAL REVIEW #3 – rusin – (55565)
1
This
printout
should
have
32
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
Third and final installment of the review for
the final exam. Some questions may duplicate
previous HW questions.
These reviews have been very long and even
so, are not exactly ”comprehensive” – there
are many variations of the ideas that are not
on these reviews. But I think if you are com
fortable with all the topics on these reviews,
you will be in good shape for the exam.
Good luck!
001
10.0points
Find a vector function that represents the
curve of intersection of the paraboloid
z
=
x
2
+ 3
y
2
and the parabolic cylinder
y
=
x
2
.
1. r
(
t
) =
√
t
i
+
t
j
+
(
3
t
2
+
t
4
)
k
2. r
(
t
) =
√
t
i
+
t
j
+
(
t
2
+ 3
t
4
)
k
3. r
(
t
) =
t
2
i
+
t
j
+
(
t
+ 3
t
2
)
k
4. r
(
t
) =
t
i
+
t
2
j
+
(
3
t
2
+
t
4
)
k
5. r
(
t
) =
t
i
+
t
2
j
+
(
t
2
+ 3
t
4
)
k
6. r
(
t
) =
t
i
+
√
t
j
+
(
t
+ 3
t
2
)
k
7. r
(
t
) =
t
i
+
√
t
j
+
(
3
t
+
t
2
)
k
8. r
(
t
) =
t
2
i
+
t
j
+
(
3
t
+
t
2
)
k
002
10.0points
Find an equation in vector form for the
tangent line to the graph of
r
(
s
) =
(
cos
s,
2
e
3
s
,
sin
s
+ 2
e

3
s
)
at the point (1
,
2
,
2).
1. L
(
t
) =
(
1
,
2
−
2
t,
2 + 5
t
)
2. L
(
t
) =
(
t,
2
−
3
t,
2
−
5
t
)
3. L
(
t
) =
(
t,
2
−
2
t,
2 + 6
t
)
4. L
(
t
) =
(
1
,
2 + 6
t,
2
−
6
t
)
5. L
(
t
) =
(
t,
2 + 3
t,
2
−
6
t
)
6. L
(
t
) =
(
1
,
2 + 6
t,
2
−
5
t
)
003
10.0points
Find the point of intersection of the tangent
lines to the curve
r
(
t
) =
(
sin
πt,
4 sin
πt,
cos
πt
)
at the points where
t
= 0 and
t
= 0
.
5.
1.
(1
,
−
4
,
1)
2.
(0
,
0
,
0)
3.
(1
,
4
,
1)
4.
(1
,
3
,
0)
5.
(0
,
4
,
1)
004
10.0points
The curves given parametrically by
x
(
t
) =
t,
y
(
t
) =
t
6
,
z
(
t
) =
t
3
,
and
x
(
t
) = sin
t,
y
(
t
) = sin 4
t,
z
(
t
) =
t
intersect at the origin.
Find the cosine of their angle,
θ
, of inter
section.
1.
cos
θ
=
2
3
√
2
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suleimenov (bs26835) – FINAL REVIEW #3 – rusin – (55565)
2
2.
cos
θ
=
2
√
17
3.
cos
θ
=
2
√
19
4.
cos
θ
=
1
√
17
5.
cos
θ
=
1
3
√
2
6.
cos
θ
=
1
√
19
005
10.0points
Find the unit vector
T
(
t
) tangent to the
graph of the vector function
r
(
t
) =
(big
2
t
2
,
4
t,
2 ln
t
)big
.
1. T
(
t
) =
(Big
t
2
2
t
2
+ 1
,
t
2
t
2
+ 1
,
1
2
t
2
+ 1
)Big
2. T
(
t
) =
(Big
2
t
2
t
2
+ 1
,
t
t
2
+ 1
,
1
t
2
+ 1
)Big
3. T
(
t
) =
(Big
t
2
2
t
2
+ 1
,
2
t
2
t
2
+ 1
,
1
2
t
2
+ 1
)Big
4. T
(
t
) =
(Big
2
t
2
t
2
+ 1
,
2
t
t
2
+ 1
,
1
t
2
+ 1
)Big
5. T
(
t
) =
(Big
2
t
2
2
t
2
+ 1
,
2
t
2
t
2
+ 1
,
1
2
t
2
+ 1
)Big
6. T
(
t
) =
(Big
t
2
t
2
+ 1
,
t
t
2
+ 1
,
1
t
2
+ 1
)Big
006
10.0points
Determine the position vector,
r
(1), at time
t
= 1 for a particle moving in 3space and
having acceleration
a
(
t
) =
−
6
k
when its initial velocity and position are given
by
v
(0) =
i
+
j
−
2
k
,
r
(0) = 5
i
+ 2
j
respectively.
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 Spring '11
 RUSIN

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