Spring 2011 Math 151
Exam I Version A Solutions
1.
C
Set
x
and
y
components equal and solve
for
t
.
t
= 4
at
(2
,
−
3)
, t
= 0
at
(0
,
1)
, t
= 1
at
(1
,
0)
,
but
−
1 =
√
t
has no solution, so the
point not on the curve is
(
−
1
,
0
)
.
2.
D
cos
x
1 + sin
x
·
(1
−
sin
x
)
(1
−
sin
x
)
=
cos
x
(1
−
sin
x
)
1
−
sin
2
x
=
1
−
sin
x
cos
x
=
sec
x
−
tan
x
.
3.
A
Since
−
1
≤
cos
x
≤
1
,
−
1
x
≤
cos
x
x
≤
1
x
Since
±
1
x
→
0
as
x
→ ∞
, by the Squeeze
Theorem, the limit is
0
.
4.
A
As we approach 4 from the right along the
curve, the
y
value approaches
4
.
5.
B
Let
f
(
x
) =
x
3
−
x
2
+
x
.
f
is continuous
since it is a polynomial, and
f
(2) = 6
, f
(3) =
21
, so
f
(2)
<
10
< f
(3)
. Therefore,
by the
Intermediate Value Theorem, there is a
solution to
f
(
x
) =
10
on
[
2
,
3
]
.
6.
E
a
= (6
i
+3
j
)
−
(
−
3
i
+
j
) = 9
i
+2
j
.
To form
a unit vector
ˆ
a
, multiply by the reciprocal
of the magnitude:
ˆ
a
=
1
√
9
2
+ 2
2
(9
i
+ 2
j
) =
9
√
85
i
+
2
√
85
j
.
7.
B
lim
x
→
5

f
(
x
) = 6
−
5 = 1
.
lim
x
→
5
+
f
(
x
) =
−
8 + 2(5) = 2
,
and
f
(5)
=
1
.
Therefore,
since
lim
x
→
5

f
(
x
) =
f
(5)
negationslash
= lim
x
→
5
+
f
(
x
)
,
f
is contin
uous only from the left.
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 Spring '06
 ArtBelmonte
 Math, Derivative, Vector Space, Continuous function, Parametric equation

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