1.3 Solutions 75
But in (2),F=F(y′).
Hence
∂F
∂y
= 0
∂F
∂y′
=
∂
∂y′
(1+y′^2 )
(^12)
=y′(1+y′^2 )−^1 /^2
d
dx
[
y′(1+y′^2 )−^1 /^2
]
= 0
ory′(1+y′^2 )−^1 /^2 =C=constant
ory′^2 (1−C^2 )=C^2
ory′=
dy
dx
=a=constant
Integratingy=ax+bwhich is the equation to a straight line. The constants
aandbcan be found from the coordinatesP 0 (x 0 ,y 0 ) andP 1 (x 1 ,y 1 )
1.90 The velocity of the bead which starts from rest is
ds
dt
=
√
2 gy (1)
The time of descent is therefore
I=t=
∫
ds
√
2 gy
=
1
√
2 g
∫ √
dx^2 +dy^2
y
=
1
√
2 g
∫ √
1 +y′^2
y
dx (2)
F=
√
(1+y′^2 )
y
(3)
HereFinvolves onlyyandy′. The Euler equation is
dF
dx
−
d
dx
(
∂F
∂y′
)=0(4)
which does not containxexplicitly. In that caseF(y,y′)isgivenby
dF
dx
=
∂F
∂y
dy
dx
+
∂F
∂y′
dy′
dx
(5)
Multiply (4) byddyx
dy
dx
.
dF
dy
−
dy
dx
d
dx
(
dF
dy′
)
=0(6)
Combining (5) and (6)
dF
dx
=
d
dx
(
dF
dy′
dy
dx
)
(7)
IntegratingF=ddFy′ddyx+C