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\section{An example of a derivative calculation}

We have
\begin{equation}
	 F_3 = v_0 \sin{\phi} (t-t_0) - \frac{1}{2} g(t-t_0)^2 -h
\end{equation}
with $t_0 = \alpha(\theta,\phi)/\omega$, where
\[ \cos{\alpha} = \sin{\phi}\sin{\phi_0} + \cos{\phi}\cos{\phi_0}
\left(\cos{\theta}\cos{\theta_0}+\sin{\theta}\sin{\theta_0}\right) \ .\]
 So, differentiating wrt. $\theta$ we get:
\begin{eqnarray*}
\frac{\partial F_3}{\partial \theta} 
& = &\frac{\partial t_0}{\partial \theta}
\frac{\partial}{\partial t_0} \left [ v_0 \sin{\phi} (t-t_0) 
                                  -\frac{1}{2} g(t-t_0)^2 \right ] \\
& = &\frac{\partial t_0}{\partial \theta}
\left [-v_0 \sin{\phi} + g(t-t_0) \right ]
\end{eqnarray*}
and
\[
\frac{\partial t_0}{\partial \theta} =
\frac{1}{\omega\sqrt{1 - \cos^2{\alpha}} }  
\frac{\partial \alpha}{\partial \theta} = 
	\frac{1}{\omega \sqrt {1 - \cos^2{\alpha}}} 
\left ( \cos{\phi}\cos{\phi_0}
\left (-\sin{\theta}\cos{\theta_0} + \cos{\theta}\sin{\theta_0}\right)\right)
\ . \]
Also, 
\begin{equation}
         \frac{\partial F_3}{\partial \phi}
         = v_0 \cos{\phi} (t-t_0) +  \frac{\partial t_0}{\partial \phi}
        \left [ -v_0 \sin{\phi} +  g(t-t_0) \right ]
\end{equation}
with
\begin{equation}
\frac{\partial t_0}{\partial \phi} = 
\frac{1}{\omega\sqrt{1 - \cos^2{\alpha}} }  
\frac{\partial \alpha}{\partial \phi} = 
	\frac{1}{\omega \sqrt {1 - \cos^2{\alpha}}} 
\left ( -\cos{\phi}\sin{\phi}+ \sin{\phi}\cos{\phi_0}
\left ( \cos{\theta}\cos{\theta_0} + \sin{\theta}\sin{\theta_0}\right)\right)
\end{equation}

\noindent

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