1000 Solved Problems in Modern Physics

76 1 Mathematical Physics

F=

√

1 +y′^2

y

=

y′^2

√

1 +y′^2

+C

Simplifying√^1

y(1+y′^2 )

=Cwhere we have used (3)

Ory(1+y′^2 )=constant, say 2a

∴

(

dy

dx

) 2

=

2 a−y

y

∴

dx

dy

=

(

y

2 a−y

) 1 / 2

=

y

(2ay−y^2 )^1 /^2

This equation can be easily solved by a change of variabley= 2 asin^2 θand

direct integration. The result is

x= 2 asin−^1

(y

2 a

)

−

√

2 ay−y^2 +b

which is the equation to a cycloid.

1.91 Irrespective of the functiony, the surface generated by revolvingyabout the
x-axis has an area

2 π

∫x 2

x 1

yds= 2 π

∫

y(1+y′^2 )^1 /^2 dx (1)

If this is to be minimum then Euler’s equation must be satisfied.

∂F

∂x

−

d

dx

(F−y′

∂F

∂y′

)=0(2)

Here

F=y(1+y′^2 )^1 /^2 (3)

SinceFdoes not containxexplicitly,∂∂Fx=0. So

F−y′

∂F

∂y′

=a=constant (4)

Use (3) in (4)

y(1+y′^2 )

(^12)
−yy′^2 (1+y′^2 )−
(^12)
=a
Simplifying
y
(1+y′^2 )^1 /^2
=a
or
dy
dx

=

√

y^2

a^2

− 1 ,y=acosh

(x

a

+b

)

This is an equation to a Catenary.

1.92 The area is

A= 2 π

∫

yds= 2 π

∫a

0

y(1+y′^2 )^1 /^2 dx