308
4 Problems with Variable Boundaries
Fig. 4.13 Example 4.6.1
graph
O
x
y
a
y = c 1 x + c 2
(1)
From the boundary condition y(x 0 ) = 0, we give c 2 = −c 1 x 0 , thus y = c 1 (x −x 0 ).
Let y = ϕ(x) = ±
R 2 − (x − a) 2 and y = c 1 (x − x 0 ) be tangent, then y
= ϕ
,
namely
c 1 = −
x − a
y
= −
x − a
c 1 (x − x 0 )
(2)
Substituting y = c 1 (x − x 0 ) into the equation of the circle, we get
(x − a)
2
+ c
2
1 (x − x 0 )
2
= R
2
(3)
Solving simultaneously Eqs. (2) and (3), the slope of tangent can be obtained
c 1 = ±
R
(x 0 − a) 2 − R 2
R 2 + 2ax 0 − a 2 − x
2
0
(4)
The coordinates of x and y at the tangential point on the left side of the circle are
x =
R
2
+ ax 0 − a
2
x 0 − a
, y = ±
R
x 0 − a
(x 0 − a) 2 − R 2
(5)
For the right boundary condition of y(x 1 ) = 0, using similar to the above method,
which can be solved
c 1 = ±
R
(x 1 − a) 2 − R 2
R 2 + 2ax 1 − a 2 − x
2
1
(6)
The coordinates of x and y at the tangential point on the right side of the circle are
x =
R
2
+ ax 1 − a
2
x 1 − a
, y = ±
R
x 1 − a
(x 1 − a) 2 − R 2
(7)
Finally, the extreme value curve to be found is