Q2 Z2,(Q2) Z2(Q2) - Institute for Water Resources - U.S. Army

Squaring both sides, one obtains an equation corresponding to the general
quadratic equation, except for the restriction on the values of Y:
(29) AX 2 + BX IYI + CY 2 + DX + E II + F = O.
Where A = C 2 - C 2
r w
2
B = 2Cw (C 2 t - C 2 w )
C = C 2 + C 2 -C 2
r w t
D = -2 L C w
2 2 2
E = -2L (C t - C w )
F = - L 2
To find the form of equation (28), it is enough to find the value of its
discriminant B 2 - 4AC. (1) For doing so, one has to know something about the
rates. The interesting case for us is 0 < C < C < C Then, the .
w r t.
discriminant, which is equal to 4C 2 C 2
t --4C4 , must be positive, and (29) is a
r
hyperbola but for the restriction on Y.
The negative values of Y cannot be used to derive at once the
boundary on both sides of the river. They would introduce negative costs in
the problem and produce a meaningless boundary in the lower half-space. This
is important to point out as the coefficients of (29) are such that the
hyperbola could not be symmetric around the X-axis. On the other hand, the
cross-product term BXIYI does not vanish, so that the hyperbola is tilted
on the X-axis: (2) on the other hand, the terms D X and E Y do not vanish
either, and the hyperbola does not .have its center at the origin.
This problem is illustrated by Diagram 12. The lower boundary,
symmetric to the upper boundary, can be obtained by rotating the latter 180 0
around the X-axis. Bouth boundaries are segments of hyperbolas. But the
(1). As given by G.B. Thomas, 'Calculus, And Analytic Geometry', Third edition,
Addison-Wesley, 1960, p. 496.
(2) The angle a made by the transverse axis with tab coordinate axes can be
found through Cot 2a = - C .2j2 Cw (C2t - C!)1.
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