I am looking at the following exercise:
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I have shown that (ii) and (iii) togenther imply (i).
Now I want to show the second part.
We assume that (i) holds that that $\gamma$ is not a parameter curve.
That $\gamma$ is not a parameter curve means that if $\gamma (t)=\sigma (u,v)$ we have that $u$ and $v$ are not constants, right?
We consider that the property (ii) holds, i.e., $\gamma$ satisfies the first of the geodesic equations $$\frac{d}{dt}(E\dot u+F\dot v)=\frac{1}{2}(E_u\dot u^2+2F_u\dot u\dot v+G_u\dot v^2)\tag 1$$
We have that $\gamma$ has constant speed, so $$\|\dot \gamma\|^2=c \Rightarrow E\dot u^2+2F\dot u\dot v+G\dot v^2=c \Rightarrow \frac{d}{dt}(E\dot u^2+2F\dot u\dot v+G\dot v^2)=0 \\ \Rightarrow (E_u\dot u+E_v\dot v)\dot u^2+2E\dot u\ddot u+2(F_u\dot u+F_v\dot v)\dot u\dot v+2F(\ddot u\dot v+\dot u\ddot v)+(G_u\dot u+G_v\dot v)\dot v^2+2G\dot v\ddot v=0 \\ \Rightarrow \dot u(E_u\dot u^2+2F_u\dot u\dot v+_u\dot v^2)+\dot v(E_v\dot u^2+2F_v\dot u\dot v+G_v\dot v^2)+2E\dot u\ddot u+2F(\ddot u\dot v+\dot u\ddot v)2G\dot v\ddot v=0 \\ \overset{(1)}{\Rightarrow} 2\dot u\frac{d}{dt}(E\dot u+F\dot v)+\dot v(E_v\dot u^2+2F_v\dot u\dot v+G_v\ddot v^2)+2(E\dot u+F\dot v)\ddot u+2(F\dot u+G\dot v)\ddot v=0\\ \Rightarrow 2\left [\frac{d}{dt}(E\dot u+F\dot v)\dot u+(E\dot u+F\dot v)\ddot u\right ]+\dot v(E_v\dot u^2+2F_v\dot u\dot v+G_v\dot v^2)+2(F\dot u+G\dot v)\ddot v=0 \\ \Rightarrow 2\frac{d}{dt}\left ((E\dot u+F\dot v)\dot u\right )+\dot v(E_v\dot u^2+2F_v\dot u\dot v+G_v\dot v^2)+2(F\dot u+G\dot v)\ddot v=0 $$ $$ \Rightarrow 2\frac{d}{dt}\left [(E\dot u+F\dot v)\dot u\right ]\dot v+\dot v^2(E_v\dot u^2+2F_v\dot u\dot v+G_v\dot v^2)+2(F\dot u+G\dot v)\dot v\ddot v=0\tag 2$$
Since $$2(F\dot u+G\dot v)\dot v\ddot v=\frac{d}{dt}\left [(F\dot u+G\dot v)\dot v^2\right ]-\frac{d}{dt}(F\dot u+G\dot v)\dot v^2$$ from the relation $(2)$ we have $$2\frac{d}{dt}\left [(E\dot u+F\dot v)\dot u\right ]\dot v+\dot v^2(E_v\dot u^2+2F_v\dot u\dot v+G_v\dot v^2)+\frac{d}{dt}\left [(F\dot u+G\dot v)\dot v^2\right ]-\frac{d}{dt}(F\dot u+G\dot v)\dot v^2=0$$
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How could we continue to get $$\frac{d}{dt}(F\dot u+G\dot v)=\frac{1}{2}(E_v\dot u^2+2F_v\dot u\dot v+G_v\dot v^2)?$$