I'm having problems with exercise 1 of chapter 3 of do Carmo's "Riemannian Geometry". Here is the background:
Let $(u,v)$ be the coordinates on $\mathbb{R}^2$. Let $f,g\in C^\infty(\mathbb{R})$, and observe that $\varphi:\mathbb{R}^2\rightarrow \mathbb{R}^3$ given by $\varphi(u,v)=(f(v)\cos u,f(v)\sin u,g(v))$ is an immersion assuming f'(v)^2+g'(v)^2\not= 0 and $f(v)\not= 0$. The image is the surface of revolution generated by the curve $(f(v),g(v))$ being rotated about the $z$-axis. The induced metric is (g_{ij})=\left( \begin{array}{cc} f^2 & 0 \\ 0 & f'^2+g'^2 \end{array} \right), and the local equations of a geodesic $\gamma$ are \left\{ \begin{array}{l} \frac{d^2 u}{dt^2} + \frac{2ff'}{f^2}\frac{du}{dt}\frac{dv}{dt}=0 \\ \frac{d^2 v}{dt^2}-\frac{ff'}{f'^2+g'^2}\left( \frac{du}{dt} \right)^2 + \frac{f'f'' + g'g''}{f'^2+g'^2} \left( \frac{dv}{dt} \right)^2 = 0. \end{array} \right.
Then, do Carmo says: Obtain the following geometric meaning of the equations above: the second equation is, except for meridians ($u=u_0$) and parallels ($v=v_0$), equivalent to the fact that the "energy" |\gamma'(t)|^2 of a geodesic is constant along $\gamma$; the first equation signifies that if $\beta(t)$ is the oriented angle, $\beta(t)<\pi$, of $\gamma$ with a parallel $P$ intersecting $\gamma$ at $\gamma(t)$, then $r\cos \beta$ is constant, where $r$ is the radius of $P$.
This last paragraph is what's confusing me. First of all, I've seen the energy of a path as \int_a^b |\gamma'(t)|^2 dt, so maybe that's why he put "energy" in quotes. But also, geodesics have constant speed! So this should be constant along all geodesics too (regardless of whether they're meridians or parallels). But I figured that maybe I should just blindly plug and chug since that seems to work scarily often in Riemannian geometry, so I found |\gamma'(t)|^2, took its $t$-derivative, and substituted in the second equation in the system for geodesics. Here it is: |\gamma'(t)|^2 = \left\langle \frac{d \gamma}{dt} , \frac{d \gamma}{dt} \right\rangle = u'^2 g_{11} + 2u'v'g_{12} + v'^2 g_{22} = u'^2f^2+v'^2(f'^2+g'^2) so \frac{d}{dt} |\gamma'(t)|^2 = 2u'u''f^2+2u'^2ff' + 2v'v''(f'^2+g'^2)+v'^2(2f'f''+2g'g'') = 2u'u''f^2+2u'^2ff' + 2v'(ff'u'^2-(f'f''+g'g'')v'^2)) + 2v'v''(f'^2+g'^2)+2v'^2(f'f''+g'g'') = 2u'u''f^2+2u'^2ff'(1+v')+2v'v''(f'^2+g'^2)+2v'^2(f'f''+g'g'')(1-v')
and this looks hopeless.