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To derive equation (16), I take the derivative of both
sides of equation (15) with respect to s and map the
result to Cartesian coordinates, which yields
![\begin{displaymath}
{\partial \eta \over \partial x}{dx \over ds}+
{\partial \et...
...\partial z}{dz \over ds}=
{m(x_0,z_0) \over m(x,z)}.
\eqno(B.1)\end{displaymath}](img78.gif)
The ray equation (6) gives
![\begin{displaymath}
\begin{array}
{lll}
\displaystyle{dx \over ds} & = & \displa...
...& \displaystyle{\tau_z(x,z) \over m(x,z)} \end{array}\eqno(B.2)\end{displaymath}](img79.gif)
Substituting these relations from equation (B.2) into equation (B.1)
and multiplying both sizes by m(x,z) yields equation (16):
![\begin{displaymath}
\tau_x\eta_x+\tau_z\eta_z=m(x_0,z_0).
\eqno(B.3)\end{displaymath}](img80.gif)
When (x,z) tends to (x0,z0), s tends to s0 and
tends to zero,
which gives the initial condition.
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Stanford Exploration Project
12/18/1997