91Ó°ÊÓ

The given integral is${∫}_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}{\mathrm{s}}^{-1}\sqrt{{\mathrm{s}}^2+\mathrm{s}{\text{'}}^2}\mathrm{dt},\mathrm{s}\text{'}=\frac{\mathrm{ds}}{\mathrm{dt}}$.The given integral is to be made stationary by using 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.

The given integral is ${∫}_{t_1}^{t_2}{s}^{-1}\sqrt{s^2+s{\text{'}}^2}dt,s\text{'}=\frac{ds}{dt}$

Euler equation for polar coordinates $\left(s,t\right)$is $\frac{d}{ds}\left(\frac{∂F}{∂t\text{'}}\right)-\frac{∂F}{∂t\text{'}}=0$where $s\text{'}=\frac{ds}{dt}$

$$\begin{gathered} t\text{'}=\frac{dt}{ds} \\ ⇒s\text{'}=\frac{1}{t\text{'}} \end{gathered}$$

Use the two equations above to change the given integral

$$\begin{gathered} ∫s^{-1}\sqrt{s^2+s{\text{'}}^2}dt=∫s^{-1}\sqrt{s^2+s{\text{'}}^2}t\text{'}ds \\ =∫s^{-1}\sqrt{s^2+t{\text{'}}^{-2}}t\text{'}ds \\ =∫\frac{\sqrt{s^{2}t{\text{'}}^2+1}}{s}ds \end{gathered}$$

Now, let$F=s^{-1}\sqrt{s^{2}t{\text{'}}^2+1}$.

Use the Euler equation , $\frac{d}{ds}\left(\frac{∂F}{∂t\text{'}}\right)-\frac{∂F}{∂t}=0$

Calculate the required derivatives

$$\begin{gathered} \frac{∂F}{∂t\text{'}}=\frac{s^{2}t\text{'}}{s\sqrt{s^{2}t{\text{'}}^2+1}} \\ =\frac{st\text{'}}{\sqrt{s^{2}t{\text{'}}^2+1}} \\ \frac{∂F}{∂t}=0 \end{gathered}$$

Therefore,

$$\begin{gathered} \frac{d}{ds}\left[\frac{st\text{'}}{\sqrt{s^{2}t{\text{'}}^2+1}}\right]=0 \\ ⇒\frac{st\text{'}}{\sqrt{s^{2}t{\text{'}}^2+1}}=C \\ ⇒t{\text{'}}^2=\frac{C^2}{s^2-C^2} \end{gathered}$$

Where $C$is constant.

Integrate $t\text{'}=\frac{C}{\sqrt{s^2-C^2}}$to get the desired result

$$\begin{gathered} t=∫\frac{C}{\sqrt{s^2-C^2}}ds \\ =Cln\left(s^2+\sqrt{s^2-C^2}\right)+B \end{gathered}$$

Where $B$is integration constant.

Therefore, $t=Cln\left(s^2+\sqrt{s^2-C^2}\right)+B$, where$B$ is the integration constant.