1000 Solved Problems in Modern Physics

(Grace) #1

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
Free download pdf