Let $\alpha(s)=(f(s), g(s))$ be a curve in $\mathbb{R}^{2}$ parameterized by arc length, and consider the surface of revolution $S$ in $\mathbb{R}^{3}$ defined by the parameterization

$$\sigma(u, v)=(f(u) \cos v, f(u) \sin v, g(u))$$

In what follows, you may use that a curve $\sigma \circ \gamma$ in $S$, with $\gamma(t)=(u(t), v(t))$, is a geodesic if and only if

$$\ddot{u}=f(u) \frac{d f}{d u} \dot{v}^{2}, \quad \frac{d}{d t}\left(f(u)^{2} \dot{v}\right)=0$$

(i) Write down the first fundamental form for $S$, and use this to write down a formula which is equivalent to $\sigma \circ \gamma$ being a unit speed curve.

(ii) Show that for a given $u_{0}$, the circle on $S$ determined by $u=u_{0}$ is a geodesic if and only if $\frac{d f}{d u}\left(u_{0}\right)=0$.

(iii) Let $\gamma(t)=(u(t), v(t))$ be a curve in $\mathbb{R}^{2}$ such that $\sigma \circ \gamma$ parameterizes a unit speed curve that is a geodesic in $S$. For a given time $t_{0}$, let $\theta\left(t_{0}\right)$ denote the angle between the curve $\sigma \circ \gamma$ and the circle on $S$ determined by $u=u\left(t_{0}\right)$. Derive Clairault's relation that

$$f(u(t)) \cos (\theta(t))$$

is independent of $t$.