1000 Solved Problems in Modern Physics

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