rhodes (ajr2283) – Exam03Review – Gilbert – (56380)
3
4.
(
x
+ 1)
2
2
+
(
y
+ 1)
2
2
+
(
z
+ 2)
2
2
= 1,
paraboloid
5.
(
x
−
1)
2
2
−
(
y
+ 1)
2
2
+
(
z
+ 2)
2
2
= 1,
hyperboloid
correct
Explanation:
Completing squares in all three variables
gives (
x
−
1)
2
−
(
y
+ 1)
2
+ (
z
+ 2)
2
= 2 or
(
x
−
1)
2
2
−
(
y
+ 1)
2
2
+
(
z
+ 2)
2
2
= 1, a hyper-
boloid.
004
10.0 points
Find
lim
t
→
0
+
r
(
t
) when
r
(
t
) =
(
7 cos
t,
5
e
t
,
3
t
ln
t
)
.
1.
limit =
(
0
,
5
,
0
)
2.
limit =
(
7
,
0
,
−
3
)
3.
limit =
(
7
,
5
,
0
)
correct
4.
limit =
(
0
,
0
,
0
)
5.
limit =
(
7
,
0
,
3
)
6.
limit =
(
7
,
5
,
3
)
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
+
7 cos
t
= 7
,
while
lim
t
→
0
+
g
(
t
) =
lim
t
→
0
+
5
e
t
= 5
,
and
lim
t
→
0
+
h
(
t
) =
lim
t
→
0
+
t
ln
t
= 0
,
using L’Hospital’s Rule. Consequently,
lim
t
→
0
+
r
(
t
) =
(
7
,
5
,
0
)
.
keywords: vector function, limit, trig function
log function, exponential function
005
10.0 points
Find a parameterization of the horizontal
circle of radius 4 having center (3
,
−
5
,
2).