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4 Problems with Variable Boundaries
x 1 = k 1 z 1 + x 0
0 = k 2 z 1 + y 0
(7)
Because point A moves on the circle Γ 0 , by the transversality condition, there is
F x − F y
Φ 0x
Φ 0y
z=z 0
= 0
or
k 1 y 0 − k 2 x 0 = 0
( 8 )
In addition because point B is undetermined on the hyperbola Γ 1 , by the
transversality condition, there is
F − x
F x − y
F y − F x
Φ 1z
Φ 1x
z=z 1
= 0
or
x 1 + k 1 z 1 = 0
( 9 )
Since the curve passes through two points A(x 0 , y 0 , 0) and B(x 1 , 0, z 1 ), therefore
there are
x
2
0 + y
2
0 = a
2
z
2
1 − x
2
1 = b
2
(10)
Solving simultaneously Eqs. (7) –(10), we get y 0 = k 2 = 0, x 0 = 2x 1 = ±a,
z 1 =
b 2 +
1
4
a 2 , k 1 = −
x 1
z 1
. Thus, the length found the extremal curve is
J =
z 1
0
1 + x 2 + y 2 dz =
z 1
0
1 + k 2
1 dz = z 1
1 + k 2
1 =
z 2
1 + x 2
1 =
b 2 +
1
2
a 2
(11)
Since a, b are both not zero, the distance given between the two curves must exist,
therefore J is the length found the extremal curve.
Example 4.2.6 Find the shortest distance from point A(1, 1, 1) to the sphere x
2
+
y
2
+ z
2
= 1.
Solution The shortest distance is the following functional