
30 Applied calculus of variations for engineers
First we separate the variables
dy = ±
x − c
1
λ
2
− (x − c
1
)
2
dx,
and integrate again to produce
y(x)=±
λ
2
− (x − c
1
)
2
+ c
2
.
It is easy to reorder this into
(x − c
1
)
2
+(y − c
2
)
2
= λ
2
,
which is the equation of a circle. Since the two given points are on the x axis,
the center of the circle must lie on the perpendicular bisector of the chord,
which implies that
c
1
=
x
0
+ x
1
2
.
To solve for the value of the Lagrange multiplier and the other constant, we
consider that the circular arc between the two points is the given length:
L = λθ,
where θ is the angle of the arc. The angle is related to the remaining constant
as
2Π − θ