76
1 Mathematical Physics
F =
1 + y 2
y
=
y
2
1 + y 2
+ C
Simplifying
1
√
y(1+y 2 )
= C where we have used (3)
Or y(1 + y
2 ) = constant, say 2a
∴
dy
dx
2
=
2a − y
y
∴
dx
dy
=
y
2a − y
1/2
=
y
(2ay − y 2 ) 1/2
This equation can be easily solved by a change of variable y = 2a sin
2
θ and
direct integration. The result is
x = 2a sin
−1
y
2a
−
2ay − y 2 + b
which is the equation to a cycloid.
1.91 Irrespective of the function y, the surface generated by revolving y about the
x-axis has an area
2π
x2
x1
yds = 2π
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)
Since F does not contain x explicitly,
∂ F
∂ x
= 0. So
F − y
∂ F
∂ y = a = constant
(4)
Use (3) in (4)
y(1 + y
2 )
1
2 − yy
2 (1 + y
2 )
−
1
2 = a
Simplifying
y
(1 + y 2 ) 1/2 = a
or
dy
dx
=
y 2
a 2 − 1, y = a cosh
x
a
+ b
This is an equation to a Catenary.
1.92 The area is
A = 2π
yds = 2π
a
0
y(1 + y
2 )
1/2 dx