774
Appendix: All Solutions to the Problems
dy = tan tdx =
sin tdx
1 − sin
2 t
=
c 1 dx
(ax + b) 2 − c
2
1
Integration we get
y =
c 1
a
ln[ax + b +
(ax + b) 2 − c
2
1 ] + c 2
or
y =
c 1
a
arcsinh
ax + b
c 1
+ c 2
2.36 Find the extremal curve of the functional J [y] =
x 1
x 0
e
x
1 + y 2 dx.
Solution Because the integrand F = e
x
1 + y 2 does not contain y, the Euler
equation of the functional has the first integral
e
x y
1 + y 2
= c 1
Taking note that since
y
√
1+y 2
≤ 1, thus |c 1 | ≤ e
x 0 . Solving for y
from the
above, we get
y
=
c 1
e 2x − c
2
1
Integrating, we get
y = arcsec
e
x
c 1
+ c 2
2.37 If the first quadrant is filled with a transparent optical medium, and in any point
of which the velocity of light is equal to 1 + x, find the light path equation that the
light beam spreads from the origin to point (2, 3) in the shortest time.
Solution Let the velocity of light be v, According to the meaning of the problem
there is
v =
ds
dt
= 1 + x
The functional of the problem is
Appendix: All Solutions to the Problems
dy = tan tdx =
sin tdx
1 − sin
2 t
=
c 1 dx
(ax + b) 2 − c
2
1
Integration we get
y =
c 1
a
ln[ax + b +
(ax + b) 2 − c
2
1 ] + c 2
or
y =
c 1
a
arcsinh
ax + b
c 1
+ c 2
2.36 Find the extremal curve of the functional J [y] =
x 1
x 0
e
x
1 + y 2 dx.
Solution Because the integrand F = e
x
1 + y 2 does not contain y, the Euler
equation of the functional has the first integral
e
x y
1 + y 2
= c 1
Taking note that since
y
√
1+y 2
≤ 1, thus |c 1 | ≤ e
x 0 . Solving for y
from the
above, we get
y
=
c 1
e 2x − c
2
1
Integrating, we get
y = arcsec
e
x
c 1
+ c 2
2.37 If the first quadrant is filled with a transparent optical medium, and in any point
of which the velocity of light is equal to 1 + x, find the light path equation that the
light beam spreads from the origin to point (2, 3) in the shortest time.
Solution Let the velocity of light be v, According to the meaning of the problem
there is
v =
ds
dt
= 1 + x
The functional of the problem is