Math 20E
Homework #6
10 February 2005
Section 6.1, Problem 2:
Define
T
(
x
,
y
) =
x

y
√
2
,
x
+
y
√
2
. Show that
T
rotates the unit square
[
0, 1
]
×
[
0, 1
]
.
Solution.
First of all, since
x

y
√
2
2
+
x
+
y
√
2
2
=
(
x
2
+
2
xy
+
y
2
) + (
x
2

2
xy
+
y
2
)
2
=
2
x
2
+
2
y
2
2
=
x
2
+
y
2
,
T
preserves magnitude. Let
θ
be the angle between
(
x
,
y
)
and
T
(
x
,
y
)
. Then
cos
θ
=
(
x
,
y
)
·
x

y
√
2
,
x
+
y
√
2
(
x
,
y
)
·
x

y
√
2
,
x
+
y
√
2
=
x
2

xy
√
2
+
xy
+
y
2
√
2
x
2
+
y
2
=
1
√
2
·
x
2
+
y
2
x
2
+
y
2
=
1
√
2
.
Since
T
sends every point
(
x
,
y
)
to one of equal magnitude at an angle of
θ
counterclock
wise,
T
is a rotation of the entire square.
Section 6.1, Problem 3:
Let
D
*
= [
0, 1
]
×
[
0, 1
]
and define
T
on
D
*
by
T
(
u
,
v
) = (

u
2
+
4
u
,
v
)
. Find the image
D
. Is
T
onetoone?
Solution.
Since both

u
2
+
4
u
and
v
are increasing functions on
[
0, 1
]
×
[
0, 1
]
, the end
points of the range are determined by the endpoints of
D
*
. So
D
= [
0, 3
]
×
[
0, 1
]
.
Next, we show that
T
is onetoone. To show this, suppose
T
(
x
,
y
) =
T
(
u
,
v
)
. Then
(

x
2
+
4
x
,
y
) = (

u
2
+
4
u
,
v
)
, which implies that
y
=
v
and

x
2
+
4
x
=

u
2
+
4
u
. Rearranging
terms, we have

x
2
+
4
x
=

u
2
+
4
u
u
2

x
2
+
4
x

4
u
=
0
(
u

x
)(
u
+
x
) +
4
(
x

u
) =
0
(
u

x
)(
u
+
x

4
) =
0,
so either
u
=
x
or
u
+
x
=
4. Since
u
and
x
are in
D
*
, they are both between 0 and 1. So we
cannot have
u
+
x
=
4, which implies that
u
=
x
. So we can conclude that
T
is onetoone.
Section 6.1, Problem 4:
Let
D
*
be the parallelogram bounded by the lines
y
=
3
x

4,
y
=
3
x
,
y
=
1
2
x
, and
y
=
1
2
(
x
+
4
)
. Let
D
= [
0, 1
]
×
[
0, 1
]
. Find a
T
such that
D
is the image
of
D
*
under
T
.
Solution.
We need the corners of
D
*
to be sent to the corners of
D
. Set
T
=
a b
c d
.
The corners of
D
*
are
(
0, 0
)
,
8
5
,
4
5
,
12
5
,
16
5
, and
4
5
,
12
5
.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Math 20E
Homework #6
10 February 2005
Setting
T
(
8/5, 4/5
) = (
1, 0
)
,
T
(
4/5, 12/5
) = (
0, 1
)
gives
8
a
+
4
b
=
5
8
c
+
4
d
=
0
4
a
+
12
b
=
0
4
c
+
12
d
=
5
Solving yields
a
=
3/4,
b
=

1/4,
c
=

1/4, and
d
=
1/2, so
T
=
3/4

1/4

1/4
1/2
is a
valid transformation.
Section 6.1, Problem 10:
Suppose
T
:
R
2
→
R
2
is linear and is given by
T
(
x
) =
A
x
,
where
A
is a 2
×
2 matrix.
Show that if det
A
=
0, then
T
takes parallelograms onto
parallelograms.
Solution.
Say we have a parallelogram
P
. It can be described as the set of points of the
form
p
+
λ
v
+
μ
w
, where
v
and
w
are not scalar multiples of each other and
λ
,
μ
∈
[
0, 1
]
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '07
 Enright
 Math, Vector Calculus, Parallelograms, Trigraph, Line integral, Polar coordinate system, Vector field

Click to edit the document details