Given an arbitrary surface, a point on that surface, and a direction, what is the name of the type of curve which lies in the surface and which twists and turns along with the curvature of the surface, but if the surface were to be flattened out, the curve is a straight line?
For example, given an upside-down cone with parametric surface of parameters $r$ and $\theta$:
$$x(r, \theta) = k \cdot r \cdot \cos (\theta)$$
$$y(r, \theta) = k \cdot r \cdot \sin (\theta)$$
$$z(r, \theta) = r$$
Let the starting point where $t = 0$ be at $z_0 = 0.5$ and $\theta_0 = 0$ with direction $(1, 0, 0)$:
$$x_0 = 0.5 \cdot k$$
$$y_0 = 0$$
Here is approximately what the curve should look like (the red spiral) but this one is not necessarily following the curvature exactly. The red arrow shows the initial point and direction.
I know this is the form of the curve because when I roll up graph paper, it appears that the trace curves upwards in an exponential spiral:
Is this what a geodesic is? Should I be using geodesic equations to find this curve? I'm unsure about this because one description of a geodesic is "the shortest path between two points" which is not what I'm looking for.
Best Answer
Your cone is defined by $$ \{x^2+y^2=k^2z^2\}\,. $$ A possible parametrization using 2D polar coordinates for $\mathbb R^2$ and cylindrical coordinates for $\mathbb R^3$ is: $$ i:(r,\theta)\mapsto (r,\theta,ar)\,,\text{ where }a:=\frac{1}{k}\,. $$ (Compared to the pictures below the cone is inverted with tip at $(0,0,0)$. This trivial "$z$-flip" does not change the statements about its geodesics.)
The Euclidean metric of $\mathbb R^3$ in cylindrical coordinates is $$ ds^2=dr^2+r^2\,d\theta^2+dz^2\,. $$ Since $dz=a\,dr$ this leads to $$\tag{1} ds^2=(1+a^2)\,dr^2+r^2\,d\theta^2 $$ for the cone. The cone metric (1) becomes $$\tag{2} ds^2=d\rho^2+\rho^2\,d\varphi^2 $$ under the transformation $$\tag{3} r\mapsto \rho:=\sqrt{1+a^2}\,r \,,\quad\theta\mapsto \varphi:=\frac{\theta}{\sqrt{1+a^2}}\,. $$ The metric (2) is formally equal to the Euclidean metric of $\mathbb R^2$ when the radius is stretched by the factor $\sqrt{1+a^2}$ and the angle is reduced by the factor $1/\sqrt{1+a^2}\,.$ The pictures below should make this self explanatory. Note that $a$ is nothing else than $$ a=\frac{z}{r}=\cot\alpha $$ where $\alpha$ is half of the angle at the tip of the cone. Therefore $$ \sqrt{1+a^2}=\sin\alpha\,. $$ The range of $\theta$ is always $[0,2\pi)\,.$ The range of $\varphi$ is less than that which is clear from looking at the flattened (unfolded) cone in the $(\rho,\varphi)$-plane.
Knowing that the Euclidean metric in polar coordinates has only straight lines as geodesics it becomes clear from (2) that the geodesics on the cone are straight lines on the unfolded cone in the $(\rho,\varphi)$-plane.
The following pictures were generated with GeoGebra. They show that geodesics on the cone can intersect with themselves. Nonetheless they are still straight lines after unfolding.
Appendix: Clairaut's Theorem
For the $(\rho,\varphi)$-plane let's introduce "Cartesian" coordinates $$ \xi=\rho\cos\varphi\,,\quad \eta=\rho\sin\varphi\,. $$ A geodesic being a straight line means that there is a parametrization of $\rho,\varphi$ with $t$ such that the components of its tangent, $$ \dot\xi=\dot\rho\cos\varphi-\rho\,\dot\varphi\,\sin\varphi\,,\quad\dot\eta=\dot\rho\sin\varphi+\rho\,\dot\varphi\,\cos\varphi, $$ are constant. A latitude on the cone is a circle in the $(\rho,\varphi)$-plane. Its components are $$ \lambda=c\cos\varphi\,,\quad\mu=c\sin\varphi $$ with a constant $c>0\,.$ Up to that constant, the scalar product of the tangent at the geodesic and at the latitude is \begin{align}\require{cancel} \dot\xi\dot\lambda+\dot\eta\dot\mu&= -\cancel{\dot\rho\,\dot\varphi\,\cos\varphi\,\sin\varphi}+\rho\,\dot\varphi^2\,\sin^2\varphi+\cancel{\dot\rho\,\dot\varphi\,\sin\varphi\cos\varphi}+\rho\,\dot\varphi^2\,\cos^2\varphi\\ &=\rho\,\dot\varphi^2\,. \end{align} Since the length of the tangent at the latitude is $c\dot\varphi$ and the length of the tangent at the geodesic is the constant \begin{align} \sqrt{ \dot\xi^2+\dot\eta^2}&=\sqrt{\dot\rho^2+\rho^2\,\dot\varphi^2} \end{align} it follows that the cosine of the angle between geodesic and latitude is $$ c\frac{\rho\,\dot\varphi}{\sqrt{\dot\rho^2+\rho^2\,\dot\varphi^2}}\,. $$ A general theorem by Clairaut about geodesics on surfaces of revolution states that $\rho$ times this cosine is constant. Since the denominator is constant. Clairaut's theorem states in our case that \begin{align}\tag{4}\boxed{\phantom{\Bigg|}\quad \rho^2\,\dot\varphi\quad\text{ is constant. }\quad} \end{align} This is indeed the case. Taking the derivative yields $2\rho\,\dot\rho+\rho^2\,\ddot\varphi\,.$ Calculating a few Christoffel symbols it is not difficult to see that the metric (2) leads to the geodesic equations $$ 0=\ddot \rho-\rho\,\dot\varphi ^2\,,\quad 0=\ddot \varphi+\frac{2}{\rho}\,\dot \rho\,\dot\varphi\,. $$ So Clairaut's statement in this case is just the second of those equations.
By the simple transformation (3) Clairaut's statement carries over to the cone in is cylindrical coordinates $(r,\theta,z)$ from above.