# Geodesic on the surface of a cone (calculus of variations)

I have been trying to solve an exercice I found on a book. It is about a geodesic on the surface of a cone. The answer is plainly provided at the end of the book without any hint or detail and my calculations lead nowhere near the answer, so help would be appreciated. I am interested in the development of this problem which should be done by calculus of variations.

I now cite the instructions and answer as found on the book.

Determine the equation of the curve giving the shortest distance between two points on the surface of a cone parameterized as:

$r^2=x^2+y^2$

$z=r\cot\alpha$

The answer provided at the end of the book states that:

$\theta = \alpha$

$r\sin\alpha = \frac{c_2}{\cos(\phi\sin\alpha+c_1)}$

My attempt to a solution

One can find the shortest path by minimizing the following integral :

$$I = \int ds = \int\sqrt{dx^2+dy^2+dz^2}$$

With some development, and assuming (it is not provided by the book instructions) that $x=\cos\theta$ and $y=\sin\theta$ :

$$ds = \sqrt{dr^2+rd\theta^2+dz^2} = \sqrt{dr^2(1+\cot^2\alpha) + r^2\dot\theta^2} = \sqrt{\csc^2\alpha + r^2\dot\theta^2}dr$$

where $\dot\theta=\frac{d\theta}{dr}$.

Now $I$ becomes

$$I = \int\Phi(r,\dot\theta)dr$$

where $\Phi(r,\dot\theta)=\sqrt{\csc^2\alpha + r^2\dot\theta^2}$. So $I$ will be minimised if $\Phi(r,\dot\theta)$ fills the Euler-Lagrange condition

$$\frac{d}{dr}\frac{\partial\Phi}{\partial\dot\theta} - \frac{\partial\Phi}{\partial\theta}=0$$

where $\frac{\partial\Phi}{\partial\theta}=0$ because $\Phi(r,\dot\theta)$ does not explicitly depend on $\theta$. Also, since $\frac{d}{dr}\frac{\partial\Phi}{\partial\dot\theta} = 0$, this imposes that

$$\frac{\partial\Phi}{\partial\dot\theta} = \frac{r^2\dot\theta}{\sqrt{\csc^2\alpha + r^2\dot\theta^2}} = k$$

where $k$ is a constant. This last equation leads to

$$\dot\theta = \frac{k\csc\alpha}{r\sqrt{r^2-k^2}}.$$

A possible solution on $r$ for this differential equation can be $r=k\sec\varphi$. So we can come up with $dr=k\sec\varphi\tan\varphi d\varphi$ which can be uses for the integration of $\dot\theta$. Also, remembering that $\sec^2\varphi-1=\tan^2\varphi$, one can deduce that $\sqrt{r^2-k^2} = k\tan\varphi$, so that

$$\int d\theta = \int\frac{k^2\csc\alpha \sec\varphi\tan\varphi}{k^2\sec\varphi \tan\varphi} = \int\csc\alpha d\varphi = \varphi\csc\alpha.$$

And by integrating the left hand side, we come up with

$$\theta + c_1 = \varphi\csc\alpha$$

where $c_1$ is an arbitrary integration constant. Finally, remembering that we set $r=k\sec\varphi$, we end up with

$$\cos(\csc\alpha)=\frac{r\cos(\theta+c_1)}{k}$$

which is clearly different from the solution provided by the book. What am I doing wrong and what should I be doing?

• I didn't check your work, but the "usual method" is to cut open the cone along a generator, flatten into a plane sector (a local isometry), then use Euclidean geometry (and the fact that geodesics are preserved under local isometries). Is it possible this was the author's intent? (The book's answer is the equation of a straight line in polar coordinates.) Commented Oct 23, 2015 at 10:39
• @AndrewD.Hwang Well, about the answer being a straight line in polar coordinates, I believe it is normal since the surface of a cone is just a plane, so yes, Euclidean geometry indeed... I am aware that there might be other methods to solve this problem but I am just trying to solve it using the calculus of variations (being the field I am trying to study/understand at the moment). Commented Oct 24, 2015 at 2:02