Solutions
for
PSet
7
1.
(9.8:7)
Hint:
It
might
help
to
define
a
scalar
field
F
(
x, y, z
) =
f
(
u
(
x, y, z
)
, v
(
x, y, z
))
where
u, v
are
as
needed.
We
first
assume
that
x
=
0.
Given
g
(
x, y
) =
z
, we
know
∂g
∂F/∂x
∂g
∂F/∂y
=
−
;
=
.
∂x
∂F/∂z
∂y
−
∂F/∂z
Now,
we
need
only
use
the
chain
rule
to
determine
the
result.
First
observe
that
∇
u
= (
−
y/x
2
,
1
/x,
0)
,
∇
v
= (
−
z/x
2
,
0
,
1
/x
).
Now
we
compute
∂F
=
∇
f
(
u, v
)
·
(
∂u/∂x, ∂v/∂x
) =
D
1
f
(
u, v
)(
−
y/x
2
) +
D
2
2
f
(
u, v
)(
−
z/x
)
,
∂x
∂F
=
∇
f
(
u, v
)
·
(
∂u/∂y, ∂v/∂y
) =
D
1
f
(
u, v
)(1
/x
)
,
∂y
∂F
=
∇
f
(
u, v
)
·
(
∂u/∂z, ∂v/∂z
) =
D
2
f
(
u, v
)(1
/x
)
.
∂z
An
easy
computation
then
gives
∂g
yD
1
f
(
u, v
)
z
∂g
D f
(
u, v
)
=
+
;
=
−
1
.
∂x
xD
2
f
(
u, v
)
x
∂y
D
2
f
(
u, v
)
Thus
∂g
∂g
x
+
y
=
z
=
g
(
x, y
)
.
∂x
∂y
2.
(a)
First,
DF
has
the
form
of
a
block
matrix.
�
I
0
DF
(
x
,
y
) =
n

(1)
D
f
x
(
x
,
y
)

D
f
y
(
x
,
y
)
�
This
comes
when
we
consider
F
(
x
,
y
) = (
F
1
(
x
,
y
)
, . . . , F
n
(
x
,
y
)
, f
1
(
x
,
y
)
, . . . , f
m
(
x
,
y
))
∂F
i
∂F
i
where
here
F
i
(
x
,
y
) =
x
·
e
i
.
Then
=
δ
ij
for
1
≤
i, j
≤
n
and
= 0
∂x
j
∂y
k
for
1
≤
i
≤
n,
1
≤
k
≤
m
.
The
bottom
portion
of
the
matrix
is
exactly
what
we
get
based
on
our
determination
of
D
f
x
, D
f
y
.
1