
115Lagrangian Formulation of Mechanics
© 2010 Taylor & Francis Group, LLC
Combining this with Equation 4.64, we obtain
d
d
d
t
L
qt
L
q
aq
kk
jjk
jk
n
k
∂
∂
−
∂
∂
+
=
∑∑
∫
=
λδ
1
1
2
. (4.65)
The δq’s are not necessarily independent; they are connected by the m relationships (Equation
4.62). We rewrite Equation 4.65 as
d
d
d
t
L
qt
L
q
aq
kk
jjk
jk
nm
k
∂
∂
−
∂
∂
+
∑∑
∫
=
−
λδ
1
1
2
+
∂
∂
−
∂
∂
+
∑
=
d
d
d
t
L
qt
L
q
a
kk
jjk
jkn
λ
−−+
∑
∫
=
m
n
k
q
1
1
2
0δ .
The values of the λ
j
’s are at our disposal; we now choose them to be such that the second integral
vanishes; that is,
∂
∂
−
∂
∂
+==− +
∑
L
qt
L
q
aknm
kk
jjk
j
d
d
…λ
, . (4.66)
Having done this, we are free to choose q
1
, q
2
,…, q
n–m
in the rst integral arbi ...