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996
CHAPTER 16
MULTIPLE INTEGRATION
(ET CHAPTER 15)
The image of the horizontal line
v
=
c
is the set of the points
(
x
,
y
)
=
8(
u
,
c
)
=
(
e
u
,
e
u
+
c
)
⇒
x
=
e
u
,
y
=
e
c
x
⇒
y
=
e
c
x
,
x
>
0
Since
u
can take any value,
x
can take any positive value, and hence the image is the ray
y
=
e
c
x
,
x
>
0.
u
Φ
x
c
y
=
e
c
x
y
In Exercises 5–12, let
u
,v)
=
(
2
u
+
v,
5
u
+
3
v)
be a map from the u
v
plane to the xyplane.
5.
Show that the image of the horizontal line
v
=
c
is the line
y
=
5
2
x
+
1
2
c
. What is the image (in slopeintercept
form) of the vertical line
u
=
c
?
SOLUTION
The image of the vertical line
u
=
c
is the set of the following points:
(
x
,
y
)
=
c
=
(
2
c
+
5
c
+
3
⇒
x
=
2
c
+
y
=
5
c
+
3
v
By the Frst equation,
v
=
x
−
2
c
. Substituting in the second equation gives
y
=
5
c
+
3
(
x
−
2
c
)
=
5
c
+
3
x
−
6
c
=
3
x
−
c
Therefore, the image of the line
u
=
c
is the line
y
=
3
x
−
c
in the
xy
plane. The image of the horizontal line
v
=
c
is
the set of the following points:
(
x
,
y
)
=
u
,
c
)
=
(
2
u
+
c
,
5
u
+
3
c
)
⇒
x
=
2
u
+
c
,
y
=
5
u
+
3
c
The Frst equation implies
u
=
x
−
c
2
. Substituting in the second equation gives
y
=
5
(
x
−
c
)
2
+
3
c
=
5
x
2
+
c
2
Therefore, the image of the line
v
=
c
is the line
y
=
5
x
2
+
c
2
in the
plane.
6.
Describe the image of the line through the points
(
u
=
(
1
,
1
)
and
(
u
=
(
1
,
−
1
)
under
8
in slopeintercept
form.
1
1
−
1
−
1
1
1
Φ
x
y
u
u
=
1
y
=
3
x
−
1
The line is the vertical line
u
=
1inthe
u
v
plane. The image of the line under the linear map
u
=
(
2
u
+
5
u
+
3
is the line through the images of the points
(
u
=
(
1
,
1
)
and
(
u
=
(
1
,
−
1
)
. We Fnd these images:
1
,
1
)
=
(
2
·
1
+
1
,
5
·
1
+
3
·
1
)
=
(
3
,
8
)
1
,
−
1
)
=
(
2
·
1
−
1
,
5
·
1
−
3
)
=
(
1
,
2
)
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This homework help was uploaded on 04/13/2008 for the course MATH 32B taught by Professor Rogawski during the Winter '08 term at UCLA.
 Winter '08
 Rogawski

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