850
Appendix: All Solutions to the Problems
y = 2x − 3
( 9 )
Substituting Eq. (9) into Eq. (7), we get z
= −1, integrating we get z = c 3 − x,
from the endpoint conditions, we get c 3 = 1, thus
z = 1 − x
(10)
It follows from Eqs. (4) and (5) that λ(x) = 0, so the found shortest distance is
J =
2
1
1 + y 2 + z 2 dx =
2
1
√
1 + 4 + 1dx =
√
6
(11)
5.11 Find the Euler equation of the minimum of the functional J [y, z] =
1
√
2g
1
0
1+y z
x
dx, the constraint condition is y = z + 1, the endpoint conditions
are y(0) = 0, y(1) = b.
Solution From the constraint condition y = z + 1, we get y
= z
, so the functional
can be rewritten as
J =
1
√
2g
1
0
1 + y 2
x
dx
The Euler equation of the functional is
d
dx
y
x(1 + y 2 )
= 0
or
y
= c 1
x(1 + y 2 )
or
y
=
c
2
1 x
1 − c
2
1 x
Putting x =
sin
2 t
c
2
1
, then dx =
2 sin t cos tdt
c
2
1
, substituting it into the above equation,
integrating we get
y =
c
2
1 x
1 − c
2
1 x
dx =
2
c
2
1
sin
2 tdt =
1
c
2
1
(t − sin t cos t) + c 2
x =
sin
2 t
c
2
1