hernandez (ah29758) – M408D Quest Homework 9 – pascaleff – (54550)
1
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printout
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have
21
questions.
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before answering.
001
10.0points
Find
lim
t
→
0
+
r
(
t
) when
r
(
t
) =
(
4 cos
t,
2
e
t
,
10
t
ln
t
)
.
1.
limit =
(
0
,
2
,
0
)
2.
limit =
(
4
,
0
,
10
)
3.
limit =
(
4
,
2
,
10
)
4.
limit =
(
0
,
0
,
0
)
5.
limit =
(
4
,
0
,
−
10
)
6.
limit =
(
4
,
2
,
0
)
correct
Explanation:
For a vector function
r
(
t
) =
(
f
(
t
)
, g
(
t
)
, h
(
t
)
)
,
the limit
lim
t
→
0
+
r
(
t
)
=
(Big
lim
t
→
0
+
f
(
t
)
,
lim
t
→
0
+
g
(
t
)
,
lim
t
→
0
+
h
(
t
)
)Big
.
But for the given vector function,
lim
t
→
0
+
f
(
t
) =
lim
t
→
0
+
4 cos
t
= 4
,
while
lim
t
→
0
+
g
(
t
) =
lim
t
→
0
+
2
e
t
= 2
,
and
lim
t
→
0
+
h
(
t
) =
lim
t
→
0
+
t
ln
t
= 0
,
using L’Hospital’s Rule. Consequently,
lim
t
→
0
+
r
(
t
) =
(
4
,
2
,
0
)
.
keywords: vector function, limit, trig function
log function, exponential function
002
10.0points
A space curve is shown in black on the
surface
x
y
z
Which one of the following vector functions
has this space curve as its graph?
1. r
(
t
) =
(
t
cos
t, t, t
sin
t
)
,
t >
0
,
2. r
(
t
) =
(
sin 4
t,
cos
t,
sin
t
)
3. r
(
t
) =
(
t
cos
t, t
sin
t, t
)
4. r
(
t
) =
(
t
cos
t, t, t
sin
t
)
5. r
(
t
) =
(
t
cos
t, t
sin
t, t
)
,
t >
0
,
6. r
(
t
) =
(
sin 2
t,
cos
t,
sin
t
)
7. r
(
t
) =
(
cos
t,
sin
t,
sin 4
t
)
correct
8. r
(
t
) =
(
cos
t,
sin
t,
sin 2
t
)
Explanation:
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hernandez (ah29758) – M408D Quest Homework 9 – pascaleff – (54550)
2
The surface is a cylinder such that hori
zontal crosssections intersect the surface in a
circle. Now of the given vector functions only
r
(
t
) =
(
cos
t,
sin
t,
sin 2
t
)
r
(
t
) =
(
cos
t,
sin
t,
sin 4
t
)
lie on such a cylinder; in particular, the sur
face is the graph of
x
2
+
y
2
= 1
,
and the space curve passes through the point
r
(0) = (1
,
0
,
0)
on the
x
axis as well as the point
r
parenleftBig
π
2
parenrightBig
= (0
,
1
,
0)
on the
y
axis.
But the
z
coordinate of the
space curve takes both positive and nega
tive values between these two points, while
sin 2
t >
0 for 0
< t < π/
2. Consequently, the
space curve is the graph of
r
(
t
) =
(
cos
t,
sin
t,
sin 4
t
)
.
keywords: surface, space curve, vector func
tion, 3D graph, circular cylinder,
003
10.0points
Find a vector function whose graph is the
curve of intersection of the circular cylinder
x
2
+
z
2
= 4
and the hyperbolic paraboloid
y
=
x
2
−
z
2
.
1. r
(
t
) = 2 cos
t
i
+ 4 sin 2
t
j
+ 2 sin
t
k
2. r
(
t
) = 2 sin
t
i
−
4 sin 2
t
j
+ 2 cos
t
k
3. r
(
t
) =
1
2
cos
t
i
−
1
4
sin 2
t
j
+
1
2
sin
t
k
4. r
(
t
) =
1
2
cos
t
i
+
1
2
cos 2
t
j
+
1
4
sin
t
k
5. r
(
t
) = 2 sin
t
i
+ 4 cos 2
t
j
+ 2 cos
t
k
6. r
(
t
)
=
2 cos
t
i
+ 4 cos 2
t
j
+ 2 sin
t
k
correct
7. r
(
t
) = 2 sin 2
t
i
+ 2 sin 2
t
j
+ 2 sin 2
t
k
8. r
(
t
) = 2 cos 2
t
i
+ 2 cos
t
j
+ 2 sin 2
t
k
Explanation:
The graph of the vector function
r
(
t
) =
x
(
t
)
i
+
y
(
t
)
j
+
z
(
t
)
k
is the intersection of the circular cylinder
x
2
+
z
2
= 4
and the hyperbolic paraboloid
y
=
x
2
−
z
2
when
x
(
t
)
2
+
z
(
t
)
2
= 4
,
y
(
t
) = 2
x
(
t
)
2
−
z
(
t
)
2
.
Now
cos
2
θ
+ sin
2
θ
= 1
,
which already eliminates two of the functions.
On the other hand,
cos
2
θ
−
sin
2
θ
= cos 2
θ ,
which is satisfied only by
r
(
t
) = 2 cos
t
i
+ 4 cos 2
t
j
+ 2 sin
t
k
.
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 Spring '07
 TextbookAnswers
 Derivative, Cos, Parametric equation, lim r

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