I have the following exercise and some hesitation.
(Exercise) Consider the curve $\alpha : \mathbb R \to \mathbb R^3$, $\alpha(t) := (\cos t, \sin t, t)$ and the elementary surface $x : \mathbb R \times (0, +\infty) \to \mathbb R ^3$, $x(u, v) := (u \cos v, u \sin v, 1+v-u)$.
(1) Verify that the trace of $\alpha$ is contained in that one of $x$.
(2) Calculate the normal curvature of $\alpha$. Determine whether $\alpha$ is a geodesic.
Here is my attempt.
Looking a bit at how the curve and surface are defined, I can tell that $\alpha = x (1, \cdot)$, but they do not typecheck. It could be an oversight, it could be $\alpha : (0, +\infty) \to \mathbb R^3$ or $x : (0, +\infty) \times \mathbb R \to \mathbb R^3$...
(1) From the identity just mentioned, it is immediate.
(2) Since the curve is contained in the surface, you can write $\alpha(t) = x(u(t), v(t))$, where in our case $u(t) = 1$ and $v(t) = t$. In this case, $$\alpha'(t) = x_v(1, t)$$ and thus in the basis $\left\{ x_u(u, v), x_v(u, v) \right\}$ of the tangent space in $x(u, v)$ it has coordinates $(u', v') = (0, 1)$. The normal curvature is, by definition, $$k_n(t) = \frac{\mathrm{II}(\alpha'(t), \alpha'(t))}{\mathrm I (\alpha'(t), \alpha'(t))}$$ with $\mathrm I$ and $\mathrm{II}$ the two fundamental forms of the surface. It doesn't seem to me that there is the need to compute the whole thing: $$\mathrm I(\alpha'(t), \alpha'(t)) = \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} E & F \\ F & G \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = G.$$ Similarly for the second fundamental form: $$\mathrm{II}(\alpha'(t), \alpha'(t)) = \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} e & f \\ f & g \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = g .$$ Here, $E$, $F$, $G$, $e$, $f$ and $g$ are functions in $(u, v)$ and $u$ and $v$ are functions in $t$, but notation gets heavy to write down... All right, now some preliminary calculations: \begin{align*} & x_u = (\cos v, \sin v, -1) \\ & x_v = (-u \sin v, u \cos v, 1) \\ & x_u \wedge x_v = (\sin v + u \cos v, -\cos v + u \sin v, u) \\ & \lVert x_u \wedge x_v \rVert = \sqrt{1+2u^2} \\ & x_{vv} = -u (\cos v, \sin v, 0) \end{align*} Finally: \begin{align*} & G = \langle x_v, x_v \rangle = u^2 +1 \\ & g = \frac{\langle x_u \wedge x_v, x_{vv} \rangle}{\lVert x_u \wedge x_v \rVert} = -\frac{u^2}{\sqrt{1 + 2u^2}} \end{align*} and then $k_n (t) = - \frac1{2 \sqrt 3}$ everywhere.
I have now to see if the curve is geodesic, that is, I need to verify that $\alpha''$ is normal to the surface. First of all, I observe that $\alpha''(t) = x_{vv}(1, t) = (-\cos t, -\sin t, 0)$. Now
\begin{align*} & \langle \alpha''(t), x_u(u(t), v(t)) \rangle = \langle x_{vv}(1, t), x_u(1, t) \rangle = -1 \\ & \langle \alpha''(t), x_v(u(t), v(t)) \rangle = \langle x_{vv}(1, t), x_v(1, t) \rangle = 0 \end{align*} and $\alpha$ is not geodesic.
Is all this correct?