9.12:
a) Solving Eq. (9.7) for
t
gives
.
0
z
z
z
α
ω
ω
t

=
Rewriting Eq. (9.11) as
)
(
2
1
0
0
t
α
ω
t
θ
θ
z
z
+
=

and substituting for
t
gives
),
(
2
1
2
)
(
1
)
(
2
1
0
2
2
0
0
0
0
0
0
z
z
z
z
z
z
z
z
z
z
z
z
ω
ω
ω
ω
ω
ω
ω
ω
ω
α
ω
ω
θ
θ

=
+

=

+

=

α
α
which when rearranged gives Eq. (9.12).
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