So I was going through the exercises in Do Carmo's Riemannian Geometry, but I had an issue on problem 1 in chapter 3. I've done 1a, 1b but I am stuck at 1c.
We have a geodesic; \begin{equation} \gamma(t)= (f(v(t))\cos u(t), f(v(t))\sin u(t), g(v(t))). \end{equation}
We can compute, \begin{equation} |\gamma'(t)|^2= f^2\left(\frac{\text{d}u}{\text{d}t}\right)^2+ [(f')^2+ (g')^2]\left(\frac{\text{d}v}{\text{d}t}\right)^2, \end{equation} and, \begin{equation} \frac{\text{d}}{\text{d}t}|\gamma'(t)|^2= 2ff'\frac{\text{d}v}{\text{d}t}\left(\frac{\text{d}u}{\text{d}t}\right)^2+ 2f^2\frac{\text{d}u}{\text{d}t}\frac{\text{d}^2u}{\text{d}t^2}+ 2[f'f''+g'g'']\left(\frac{\text{d}v}{\text{d}t}\right)^3+2[(f')^2+(g')^2]\frac{\text{d}v}{\text{d}t}\frac{\text{d}^2v}{\text{d}t^2}. \end{equation} I do not see how the vanishing of this is equivalent to the second equation... That is, \begin{equation} \frac{\text{d}^2v}{\text{d}t^2}- \frac{ff'}{(f')^2+(g')^2}\left(\frac{\text{d}u}{\text{d}t}\right)^2+\frac{f'f''+g'g''}{(f')^2+(g')^2}\left(\frac{\text{d}v}{\text{d}t}\right)^2=0. \end{equation}
Moreover, any help on the second part of this question would also help.
The first part of part c) is solved by substituting in the equations from part b). These are: \begin{equation*} \frac{\text{d}^2u}{\text{d}t^2} + \frac{2ff'}{f^2}\frac{\text{d}u}{\text{d}t} = 0\\ \frac{\text{d}^2v}{\text{d}t^2} - \frac{ff'}{(f')^2+(g')^2}\left(\frac{\text{d}u}{\text{d}t}\right)^2+\frac{f'f''+g'g''}{(f')^2+(g')^2}\left(\frac{\text{d}v}{\text{d}t}\right)^2 = 0 \end{equation*}
Note the first equation can be rewritten as: \begin{equation*} f^2\frac{\text{d}^2u}{\text{d}t^2} = -2ff'\frac{\text{d}u}{\text{d}t}\frac{\text{d}v}{\text{d}t} \end{equation*}
For clarity I will enclose simplifications from the equations derived in b) in square brackets: \begin{align*} \frac{\text{d}}{\text{d}t}|\gamma'(t)|^2 &= 2ff' \frac{\text{d}v}{\text{d}t}\left(\frac{\text{d}u}{\text{d}t}\right)^2 + 2f^2 \frac{\text{d}u}{\text{d}t} \frac{\text{d}^2u}{\text{d}t^2} + 2(f'f'' + g'g'')\left(\frac{\text{d}v}{\text{d}t}\right)^3 + 2((f')^2 + (g')^2)\frac{\text{d}v}{\text{d}t}\frac{\text{d}^2v}{\text{d}t^2}\\ &=f^2\frac{\text{d}u}{\text{d}t}\left(\left[\frac{2ff'}{f^2}\frac{\text{d}v}{\text{d}t}\frac{\text{d}u}{\text{d}t} + \frac{\text{d}^2u}{\text{d}t^2}\right]+ \frac{\text{d}^2u}{\text{d}t^2}\right) + 2\frac{\text{d}v}{\text{d}t}\left((f'f'' + g'g'')\left(\frac{\text{d}v}{\text{d}t}\right)^2 + ((f')^2 + (g')^2)\frac{\text{d}^2v}{\text{d}t^2}\right)\\ &= \left[f^2\frac{\text{d}^2u}{\text{d}t^2}\right]\frac{\text{d}u}{\text{d}t} + 2\frac{\text{d}v}{\text{d}t}((f')^2+(g')^2)\left(\frac{f'f''+g'g''}{(f')^2+(g')^2}\left(\frac{\text{d}v}{\text{d}t}\right)^2+\frac{\text{d}^2v}{\text{d}t^2}\right)\\ &=\left(-2ff'\frac{\text{d}u}{\text{d}t}\frac{\text{d}v}{\text{d}t}\right)\frac{\text{d}u}{\text{d}t} + 2\frac{\text{d}v}{\text{d}t}((f')^2+(g')^2)\left(\frac{f'f''+g'g''}{(f')^2+(g')^2}\left(\frac{\text{d}v}{\text{d}t}\right)^2+\frac{\text{d}^2v}{\text{d}t^2}\right)\\ &=2\frac{\text{d}v}{\text{d}t}((f')^2+(g')^2)\left[\frac{-ff'}{(f')^2+(g')^2}\left(\frac{\text{d}u}{\text{d}t}\right)^2+\frac{f'f''+g'g''}{(f')^2+(g')^2}\left(\frac{\text{d}v}{\text{d}t}\right)^2+\frac{\text{d}^2v}{\text{d}t^2}\right]\\ &=2\frac{\text{d}v}{\text{d}t}((f')^2+(g')^2)0 = 0 \end{align*}
For the second half of c) I will lay out an approach that I think you can fill in the gaps for. Attached is the image on page 79 of the textbook.
Consider a geodesic path $\gamma$ and for any $t$ its angle of incidence $\beta(t)$. For any given $t_0$, $\beta(t_0)$ would be determined by the angle between $\gamma'(t_0)$ and $p'(s_0)$ where $p$ is the (unique up to reparametrization) parallel of $S$ passing through $\gamma(t_0)$. We may write these as: \begin{align*} p(s)&=\big(f(v_p)\cos(u_p(s)),f(v_p)\sin(u_p(s)),g(v_p)\big)\\ \gamma(t) &= \big(f(v_\gamma(t))\cos(u_\gamma(t)),f(v_\gamma(t))\sin(u_\gamma(t)),g(v_\gamma(t))\big) \end{align*} After choice of $p$ of constant speed and some calculation you will see: \begin{align*} \langle \gamma'(t),p'(s) \rangle_{\gamma(t_0)} &= f(v_\gamma(t_0))f(v_p)\frac{\text{d}u_\gamma}{\text{d}t}\\ &= f(v_\gamma(t_0))^2\frac{\text{d}u_\gamma}{\text{d}t} \end{align*} Where the simplification comes from that $p$ and $\gamma$ intersect at $\gamma(t_0)$. Now these vectors are in $T_{\gamma(t_0)}S$ and are also in $T_{\gamma(t_0)}\mathbb{R}^3$. Using our knowledge of the usual inner product in $\mathbb{R}^3$: \begin{align*} \langle\gamma'(t),p'(s)\rangle_{\gamma(t_0)} &= |p'(s_0)|\,|\gamma'(t_0)|\cos(\beta(t_0)) \end{align*} Where $|\gamma'(t_0)|$ and $f(v_\gamma(t_0))^2\frac{\text{d}u_\gamma}{\text{d}t}$ are constant by calculations of $\frac{\text{d}}{\text{d}t}|\gamma'(t)|^2$ and by manipulation of $\frac{\text{d}^2 u}{\text{d}t^2} + \frac{2ff'}{f^2}\frac{\text{d}u}{\text{d}t}\frac{\text{d}v}{\text{d}t}=0$ to $\frac{\text{d}}{\text{d}t}\left(f^2\frac{\text{d}u}{\text{d}t}\right) = 0$, respectively.