(a) Consider the case of two dependent variables. Show that if $\mathit{F}\mathbf{=}\mathit{F}\left(x,y,z,y\text{'},z\text{'}\right)$and we want to find $\mathbf{y}\left(x\right)$and $\mathbf{z}\left(x\right)$to make $\mathit{I}\mathbf{=}{\mathbf{\int }}_{{\mathbf{x}}_{\mathbf{1}}}^{{\mathbf{x}}_{\mathbf{2}}}\mathit{F}\mathit{d}\mathit{x}$stationary, then $\mathbf{y}$and $\mathbf{z}$should each satisfy an Euler equation as in (5.1). Hint: Construct a formula for a varied path $\mathbf{Y}$for $\mathbf{y}$as in Section 2 [$\mathit{Y}\mathbf{=}\mathit{y}\mathbf{+}\mathit{\epsilon }\mathit{\eta }\left(x\right)$with $\mathbf{\eta }\left(x\right)$arbitrary] and construct a similar formula for $\mathit{z}$[let $\mathit{Z}\mathbf{=}\mathit{z}\mathbf{+}\mathit{\epsilon }\mathit{\zeta }\left(x\right)$, where $\mathit{\zeta }\left(x\right)$is another arbitrary function]. Carry through the details of differentiating with respect to $\mathit{\epsilon }$, putting $\mathit{\epsilon }\mathbf{=}\mathbf{0}$, and integrating by parts as in Section 2; then use the fact that both $\mathbf{\eta }\left(x\right)$and $\mathbf{\zeta }\left(x\right)$are arbitrary to get (5.1).

(b) Consider the case of two independent variables. You want to find the function $\mathbf{u}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}$which makes stationary the double integral ${\mathbf{\int }}_{{\mathbf{y}}_{\mathbf{1}}}^{{\mathbf{y}}_{\mathbf{2}}}{\mathbf{\int }}_{{\mathbf{x}}_{\mathbf{1}}}^{{\mathbf{x}}_{\mathbf{2}}}\mathit{F}\left(u,x,y,u_x,u_y\right)\mathit{d}\mathit{x}\mathit{d}\mathit{y}$.Hint: Let the varied $\mathit{U}\left(x,y\right)\mathbf{=}\mathit{u}\left(x,y\right)\mathbf{+}\mathit{\epsilon }\mathit{\eta }\left(x,y\right)$where $\mathit{\eta }\left(x,y\right)\mathbf{=}\mathbf{0}$at $\mathbf{x}\mathbf{=}{\mathbf{x}}_{\mathbf{1}}\mathbf{,}\mathbf{}\mathbf{x}\mathbf{=}{\mathbf{x}}_{\mathbf{2}}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{=}{\mathbf{y}}_{\mathbf{1}}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{=}{\mathbf{y}}_{\mathbf{2}}$but is otherwise arbitrary. As in Section 2, differentiate with respect to $\mathbf{\epsilon }$, $\mathbf{\epsilon }\mathbf{=}\mathbf{0}$set $\mathbf{\epsilon }\mathbf{=}\mathbf{0}$, integrate by parts, and use the fact that $\mathbf{\eta }$is arbitrary. Show that the Euler equation is then $\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{x}}\frac{\mathbf{\partial }\mathbf{F}}{\mathbf{\partial }{\mathbf{u}}_{\mathbf{x}}}\mathbf{+}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{y}}\frac{\mathbf{\partial }\mathbf{F}}{\mathbf{\partial }{\mathbf{u}}_{\mathbf{y}}}\mathbf{-}\frac{\mathbf{\partial }\mathbf{F}}{\mathbf{\partial }\mathbf{u}}\mathbf{=}\mathbf{0}$.

(c) Consider the case in which $\mathbf{F}$depends on $\mathbf{x}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\text{'}}$and $\mathbf{y}\mathbf{\text{'}\text{'}}$. Assuming zero values of the variation $\mathbf{\eta }\left(x\right)$and its derivative at the endpoints ${\mathbf{x}}_{\mathbf{1}}$and ${\mathbf{x}}_{\mathbf{2}}$, show that then the Euler equation becomes$\frac{{\mathbf{d}}^{\mathbf{2}}}{{\mathbf{dx}}^{\mathbf{2}}}\frac{\mathbf{\partial }\mathbf{F}}{\mathbf{\partial }\mathbf{y}\mathbf{\text{'}\text{'}}}\mathbf{-}\frac{\mathbf{d}}{\mathbf{dx}}\frac{\mathbf{\partial }\mathbf{F}}{\mathbf{\partial }\mathbf{y}\mathbf{\text{'}}}\mathbf{+}\frac{\mathbf{\partial }\mathbf{F}}{\mathbf{\partial }\mathbf{y}}\mathbf{=}\mathbf{0}\mathbf{.}$

(a) There are two dependent variables. (b) There are two independent variables. (c0) $\mathrm{F}$depends on $\mathrm{x},\mathrm{y},\mathrm{y}\text{'}\text{'}$and $\mathrm{y}\text{'}\text{'}$.

In the same way that geometry is the study of shape and algebra is the study of generalisations of arithmetic operations, calculus, sometimes known as infinitesimal calculus or "the calculus of infinitesimals," is the mathematical study of continuous change.. Differentiation and integration are the two main branches.

(a)Minimize the surface integral to find the curve.

$l=2\mathrm{\pi }{\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\mathrm{y}\sqrt{1+{\mathrm{y}}^{\mathrm{r}2}}\mathrm{dx}$

Subject a constraint to a fixed length.

$$\begin{gathered} J=l \\ l={\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}ds \\ l={\int }_{x_1}^{x_2}\sqrt{1+y^{r2}}dx \end{gathered}$$

Write the functions asgiven in (5.1).

$$\begin{gathered} F=y\sqrt{1+y^{r2}} \\ G=\sqrt{1+y^{r2}} \\ H=F+\lambda G \\ H=(y+\lambda )\sqrt{1+y^{r2}} \end{gathered}$$

Write the Euler’s equation and find its derivatives from above functions.

$$\begin{gathered} \frac{d}{dx}\frac{\partial H}{\partial y\text{'}}-\frac{\partial H}{\partial y}=0 \\ \frac{\partial H}{\partial y\text{'}}=\frac{y+\lambda }{\sqrt{1+y^{r2}}} \\ \frac{d}{dx}\frac{\partial H}{\partial y\text{'}}=\frac{y\text{'}}{\sqrt{1+y^{r2}}}-\frac{\left(y+\lambda \right)y\text{'}y\text{'}\text{'}}{{\left(1+y^{r2}\right)}^{{\displaystyle \frac{3}{2}}}} \\ \frac{\partial H}{\partial y}=\sqrt{1+y^{r2}} \end{gathered}$$

The solution is obtained when the variables are dependent.

(b)To minimize the function both $\mathrm{y}$and $\mathrm{y}\text{'}$are required. First change the variables.

$$\begin{gathered} dx=x\text{'}dy \\ y\text{'}=\frac{1}{x\text{'}} \end{gathered}$$

Change the integral from the above equations.

$$\begin{gathered} \int \left(y+\lambda \right)\sqrt{1+y^{r2}}dx=\int \left(y+\lambda \right)\sqrt{1+y^{r2}}x\text{'}dy \\ =\int \left(y+\lambda \right)\sqrt{1+x^{r-2}}x\text{'}dy \\ =\int \left(y+\lambda \right)\sqrt{x^{r2}+}1dy \end{gathered}$$

Let,$H=\int \left(y+\lambda \right)\sqrt{1+x^{r2}}$

Write the Euler’s equation and find its derivatives from above functions.

$$\begin{gathered} \frac{d}{dy}\frac{\partial H}{\partial x\text{'}}-\frac{\partial H}{\partial x}=0 \\ \frac{\partial H}{\partial x\text{'}}=\frac{x\text{'}(y+\lambda )}{\sqrt{1+x^{r2}}} \\ \frac{\partial H}{\partial x}=0 \end{gathered}$$

The solution is obtained when the variables are independent.

Integrate the Euler’s solution that is obtained.

$$\begin{gathered} \frac{d}{dy}\left[\frac{x\text{'}}{\sqrt{1+x^{r2}}}\right]=0 \\ \frac{x\text{'}}{\sqrt{1+x^{r2}}}=C \\ {x}^{r2=}\frac{C^2}{{\left(y+\lambda \right)}^2-C^2} \\ x\text{'}=\frac{C}{\sqrt{{\left(y+\lambda \right)}^2}-C^2} \end{gathered}$$

Solve further,

$$\begin{gathered} x=\int \frac{C}{\sqrt{{\left(y+\lambda \right)}^2-C^2}}dy \\ x=C\mathrm{In}\left(\lambda +y+\sqrt{{\left(\lambda +y\right)}^2-C^2}\right)+{\mathrm{C}}_1 \end{gathered}$$

(c)Find inverse of the above function $\mathrm{y}\left(\mathrm{x}\right)$, where the curve is catenary.

${\cosh }^{-1}\mathrm{z}=\mathrm{In}\left({\mathrm{z}}^2+\sqrt{{\mathrm{z}}^2-1}\right)$

Here, solution has a form in $z\equiv y+\lambda$ with rescaling factors C and C₁.

Therefore, it is proved that (a) in case of two dependent variables $F=F(x,y,z,y\text{'},z\text{'})$, and to make $I={\int }_{x_1}^{x_2}Fdx,\mathrm{y}\left(\mathrm{x}\right)$, and z(x) satisfies the Euler’s equation in (5.1), (b) in case of independent variables, $u(x,y)=(y+\lambda )\sqrt{x^{r2}+1}$for stationary double integral, ${\int }_{{\mathrm{y}}_1}^{{\mathrm{y}}_2}{\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}F(\mathrm{u},\mathrm{x},\mathrm{y},{\mathrm{u}}_{\mathrm{x}},{\mathrm{u}}_{\mathrm{y}})dxdy$and, (c) in case when $\mathrm{F}$depends on $x,y,y\text{'}$and $\mathrm{y}\text{'}\text{'}$, the Euler’s equation becomes $\frac{{\mathrm{d}}^2}{{\mathrm{dx}}^2}\frac{\partial \mathrm{F}}{\partial \mathrm{y}\text{'}\text{'}}-\frac{\mathrm{d}}{\mathrm{dx}}\frac{\partial \mathrm{F}}{\partial \mathrm{y}\text{'}}+\frac{\partial \mathrm{F}}{\partial \mathrm{y}}=0.$