91Ó°ÊÓ

Geodesics on a plane is to be found out using polar coordinates by Euler equations.

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.

Geodesics on a plane is to be found out using polar coordinates. So distance integral is to be minimized.

$$\begin{gathered} ∫ds=∫\sqrt{d{r}^2+r^{2}d\mathrm{θ}} \\ =∫\sqrt{1+r^2\mathrm{θ}{\text{'}}^2}dr \end{gathered}$$

Let $F=\sqrt{1+r^2\mathrm{θ}{\text{'}}^2}$

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

Calculate the required derivatives

$$\begin{gathered} \frac{∂F}{∂\mathrm{θ}\text{'}}=\frac{r^2\mathrm{θ}\text{'}}{r\sqrt{1+r^2\mathrm{θ}{\text{'}}^2}} \\ \frac{∂F}{∂\mathrm{θ}}=0 \end{gathered}$$

Therefore,

localid="1665037647675" $$\begin{gathered} \frac{d}{dr}\left[\frac{r^2\mathrm{θ}\text{'}}{\sqrt{1+r^2\mathrm{θ}{\text{'}}^2}}\right]=0 \\ \frac{r^2\mathrm{θ}\text{'}}{\sqrt{1+r^2\mathrm{θ}{\text{'}}^2}}=C \\ \mathrm{θ}{\text{'}}^2=\frac{C^2}{r^2\left(r^2-C^2\right)} \\ \mathrm{θ}\text{'}=\frac{C}{r\sqrt{r^2-C^2}} \end{gathered}$$

Where $C$is constant.

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

$$\begin{gathered} \mathrm{θ}\text{'}=∫\frac{C}{r\sqrt{r^2-C^2}}dr \\ =arc\tan \left(\frac{\sqrt{r^2-C^2}}{C}\right)+B \end{gathered}$$

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