Appendix: All Solutions to the Problems
827
The integral constant is
c =
2e
2x 1
e x 1 − e −x 1
=
(3 +
√
5)
2(1 +
√
5)
2
Therefore the extremal curve is
y =
(3 +
√
5)
2(1 +
√
5)
2
(e
x
− e
−x
)
4.13 Find the shortest distance from point A(−1, 5) to the parabola y
2
= x.
Solution The found distance is the minimum of the functional J [y] =
x 1
x 0
1 + y 2 dx, the constraint conditions are that the left endpoint of the extremal
curve passes through point A(−1, 5), at the moment x 0 = −1, y 0 = 5, and the right
endpoint is on the parabola y
2
= x. The general solution of the Euler equation is the
straight line y = c 1 x + c 2 , with y
= c 1 . The equations about the three unknowns
x 1 , c 1 and c 2 are
⎧
⎪ ⎨
⎪ ⎩
−c 1 + c 2 = 5
c 1 x 1 + c 2 =
√ x 1
1 +
c 1
2
√ x 1
= 0
Solving the above equations, we get x 1 = 1, c 1 = −2, c 2 = 1. Thus the equation
of the extremal curve is y = 1 − 2x, and the shortest distance between the two given
curves is
J =
1
−1
1 + (−2) 2 dx =
√
5x
1
−1
= 2
√
5
4.14 Find the shortest distance between the circle x
2
+ y
2
= 1 and the straight line
x + y = 4.
Solution The found distance is the minimum of the functional J [y] =
x 1
x 0
1 + y 2 dx, the constraint conditions are that the left endpoint of the extremal
curve is on the circle x
2
+ y
2
= 1, and the right endpoint is on the straight line
x + y = 4. The general solution of the Euler equation is y = c 1 x + c 2 , with y
= c 1 .
The equations about four unknowns x 0 , x 1 , c 1 and c 2 are
⎧
⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎩
c 1 x 0 + c 2 =
1 − x
2
0
c 1 x 1 + c 2 = 4 − x 1
1 −
x 0 c 1
√
1−x
2
0
= 0
1 − c 1 = 0
827
The integral constant is
c =
2e
2x 1
e x 1 − e −x 1
=
(3 +
√
5)
2(1 +
√
5)
2
Therefore the extremal curve is
y =
(3 +
√
5)
2(1 +
√
5)
2
(e
x
− e
−x
)
4.13 Find the shortest distance from point A(−1, 5) to the parabola y
2
= x.
Solution The found distance is the minimum of the functional J [y] =
x 1
x 0
1 + y 2 dx, the constraint conditions are that the left endpoint of the extremal
curve passes through point A(−1, 5), at the moment x 0 = −1, y 0 = 5, and the right
endpoint is on the parabola y
2
= x. The general solution of the Euler equation is the
straight line y = c 1 x + c 2 , with y
= c 1 . The equations about the three unknowns
x 1 , c 1 and c 2 are
⎧
⎪ ⎨
⎪ ⎩
−c 1 + c 2 = 5
c 1 x 1 + c 2 =
√ x 1
1 +
c 1
2
√ x 1
= 0
Solving the above equations, we get x 1 = 1, c 1 = −2, c 2 = 1. Thus the equation
of the extremal curve is y = 1 − 2x, and the shortest distance between the two given
curves is
J =
1
−1
1 + (−2) 2 dx =
√
5x
1
−1
= 2
√
5
4.14 Find the shortest distance between the circle x
2
+ y
2
= 1 and the straight line
x + y = 4.
Solution The found distance is the minimum of the functional J [y] =
x 1
x 0
1 + y 2 dx, the constraint conditions are that the left endpoint of the extremal
curve is on the circle x
2
+ y
2
= 1, and the right endpoint is on the straight line
x + y = 4. The general solution of the Euler equation is y = c 1 x + c 2 , with y
= c 1 .
The equations about four unknowns x 0 , x 1 , c 1 and c 2 are
⎧
⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎩
c 1 x 0 + c 2 =
1 − x
2
0
c 1 x 1 + c 2 = 4 − x 1
1 −
x 0 c 1
√
1−x
2
0
= 0
1 − c 1 = 0