Question

Find the geodesics on the cone ${\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{=}{\mathit{z}}^{\mathbf{2}}$. Hint: Use cylindrical coordinates.

Short answer

$\theta =\sqrt{2}arc\tan \left(\frac{\sqrt{r^2-C^2}}{C}\right)+B$, where C is constant and B is the integration constant

Step-by-step solution

Given Information

Equation of cone is given as $x^2+y^2=z^2$. Geodesics on the cone is to be found out using cylindrical coordinates.

Definition of Euler equation

In the calculus of variations andclassical mechanics, the Euler equations is a system of second-orderordinary differential equations whose solutions arestationary points of the givenaction functional.

Use Euler equation

Geodesics on a cone is to be found out using cylinderical coordinates.

Equation of cone is

$$\begin{gathered} x^2+y^2=z^2 \\ {r}^2=z^2 \\ r=z \\ dr=dz \end{gathered}$$

So distance integral is to be minimized.

$$\begin{gathered} \int ds=\int \sqrt{d{r}^2+d{z}^2+r^{2}d{\theta }^2} \\ =\int \sqrt{2d{r}^2+r^{2}d\theta } \\ =\int \sqrt{2+r^2\theta {\text{'}}^2}dr \\ \theta \text{'}=\frac{d\theta }{dr} \end{gathered}$$

Let $F=\sqrt{2+r^2\theta {\text{'}}^2}$

Euler equation for coordinates $\left(r,\theta \right)$is $\frac{d}{dr}\left(\frac{\partial F}{\partial \theta \text{'}}\right)-\frac{\partial F}{\partial \theta }=0$

Calculate the required derivatives

$$\begin{gathered} \frac{\partial F}{\partial \theta \text{'}}=\frac{r^2\theta \text{'}}{\sqrt{2+r^2\theta {\text{'}}^2}} \\ \frac{\partial F}{\partial \theta }=0 \end{gathered}$$

Therefore,

$$\begin{gathered} \frac{d}{dr}\left[\frac{r^2\theta \text{'}}{\sqrt{2+r^2\theta {\text{'}}^2}}\right]=0 \\ \frac{r^2\theta \text{'}}{\sqrt{2+r^2\theta {\text{'}}^2}}=C \\ \theta {\text{'}}^2=\frac{2C^2}{r^2\sqrt{\left(r^2-C^2\right)}} \\ \theta \text{'}=\frac{\sqrt{2}C}{r\sqrt{r^2-C^2}} \end{gathered}$$

Where $C$is constant.

Integrate $\theta \text{'}=\frac{\sqrt{2}C}{r\sqrt{r^2-C^2}}$to get the desired result

$$\begin{gathered} \theta =\int \frac{\sqrt{2}C}{r\sqrt{r^2-C^2}}dr \\ =\sqrt{2}arc\tan \left(\frac{\sqrt{r^2-C^2}}{C}\right)+B \end{gathered}$$

Therefore,$\theta =\sqrt{2}arc\tan \left(\frac{\sqrt{r^2-C^2}}{C}\right)+B$, where C is constant and B is the integration constant.