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Analytical evaluation of the tangent plane to the impulse response

In this appendix I derive the expressions for evaluating the derivatives of image depth $z_\xi$ with respect to the subsurface offset $h_\xi$ and the midpoint $m_\xi$;these derivatives are computed along the tangent plane to the impulse response of the generalized migration operator, which is defined in equations 18-24.

I start by deriving the equation for the vector normal to the impulse-response surface, ${\vec n}$:
where $\vec z_\xi$, $\vec m_\xi$, and $\vec h_\xi$ are respectively the unit vectors along the three dimensions $z_\xi$, $m_\xi$, and $h_\xi$.

The equation of the tangent plane at the image point with coordinates $\left({\widebar z_\xi},{\widebar m_\xi},{\widebar h_\xi}\right)$is given by:
\begin{eqnarray}
T\left(z_\xi,m_\xi,h_\xi\right)
&=&
\left(
\frac{\partial m_\xi...
 ...tial \alpha_x} 
\right) 
\left(h_\xi- {\widebar h_\xi} \right) =0.\end{eqnarray}
(32)
The derivative of the depth with respect o the subsurface offset, at constant midpoint, is given by:  
 \begin{displaymath}
\left.
\frac{\partial z_\xi}{\partial h_\xi}
\right\vert _{m...
 ...{\partial \gamma} 
\frac{\partial h_\xi}{\partial \alpha_x} 
}.\end{displaymath} (33)
and similarly the derivative of the depth with respect to the midpoint, at constant subsurface offset, is given by:  
 \begin{displaymath}
\left.
\frac{\partial z_\xi}{\partial m_\xi}
\right\vert _{h...
 ...{\partial \gamma} 
\frac{\partial h_\xi}{\partial \alpha_x} 
}.\end{displaymath} (34)

To evaluate equations 33-34. we need to evaluate the following partial derivatives, obtained by differentiating the expressions in equations 18-20:
   \begin{eqnarray}
\frac{\partial z_\xi}{\partial \alpha_x}
&=&
-L\left(\alpha_x,\...
 ...\gamma\right)}{\partial \gamma}
\frac{\sin \gamma}{\cos \alpha_x}.\end{eqnarray}
(35)
The derivative of path length are evaluated as follows:
\begin{eqnarray}
&
\frac{\partial L}{\partial \alpha_x}
=
\frac{-t_{D}}
{\left[\...
 ...frac{\left(S_r-S_s\right)\tan \gamma}{\cos ^2 \alpha_x}
\right],
&\end{eqnarray}
(36)
and
\begin{eqnarray}
&
\frac{\partial L}{\partial \gamma}
=
\frac{-t_{D}}
{\left[\le...
 ...frac{\left(S_r-S_s\right)\tan \alpha_x}{\cos ^2 \gamma}
\right].
&\end{eqnarray}
(37)