91Ó°ÊÓ

(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.

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

Subject a constraint to a fixed length.

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

Write the functions asgiven in (5.1).

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

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

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

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.

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

Let,$H=∫\left(y+\mathrm{λ}\right)\sqrt{1+x^{r2}}$

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

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

The solution is obtained when the variables are independent.

Integrate the Euler’s solution that is obtained.

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

Solve further,

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

(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 $z≡y+\mathrm{λ}$ 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 $I={∫}_{x_1}^{x_2}Fdx,\mathrm{y}\left(\mathrm{x}\right)$, and z(x) satisfies the Euler’s equation in (5.1), (b) in case of independent variables, $u(x,y)=(y+\mathrm{λ})\sqrt{x^{r2}+1}$for stationary double integral, ${∫}_{{\mathrm{y}}_1}^{{\mathrm{y}}_2}{∫}_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}F(\mathrm{u},\mathrm{x},\mathrm{y},{\mathrm{u}}_{\mathrm{x}},{\mathrm{u}}_{\mathrm{y}})dxdy$and, (c) in case when $\mathrm{F}$depends on $x,y,y\text{'}$and $\mathrm{y}\text{'}\text{'}$, the Euler’s equation becomes $\frac{{\mathrm{d}}^2}{{\mathrm{dx}}^2}\frac{∂\mathrm{F}}{∂\mathrm{y}\text{'}\text{'}}-\frac{\mathrm{d}}{\mathrm{dx}}\frac{∂\mathrm{F}}{∂\mathrm{y}\text{'}}+\frac{∂\mathrm{F}}{∂\mathrm{y}}=0.$