Question

Write and solve the Euler equations to make the following integrals stationary. In solving the Euler equations, the integrals in Chapter 5, Section 1, may be useful.

$\mathbf{8}\mathbf{.}\mathbf{}\underset{{\mathbf{x}}_{\mathbf{1}}}{\overset{{\mathbf{x}}_{\mathbf{2}}}{\mathbf{\int }}}\mathit{x}\sqrt{\mathbf{y}{\mathbf{\text{'}}}^{\mathbf{2}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{d}\mathbf{x}}$

Short answer

Answer

$C^2{x}^2-\left(y-B^2\right)=C^4$,this corresponds to a curve, namely, hyperbola translated on the y axis by a factor B.

Step-by-step solution

Given Information

The given function is$F\left(x,y,y\text{'}\right)=x,\sqrt{y{\text{'}}^2+x^2}$.

Definition of Euler equation

For the integral$\mathit{I}\left(\epsilon \right)\mathbf{=}\underset{{\mathbf{x}}_{\mathbf{1}}}{\overset{{\mathbf{x}}_{\mathbf{2}}}{\mathbf{\int }}}\left(x,y,y\text{'}dx\right)$, the Euler equation is mathematically presented as$\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\frac{\mathbf{d}\mathbf{F}}{\mathbf{d}\mathbf{y}\mathbf{\text{'}}}\mathbf{-}\frac{\mathbf{d}\mathbf{F}}{\mathbf{d}\mathbf{y}}\mathbf{=}\mathbf{0}$.

Find Euler equation of the given function

Let $F\left(x,y,y\text{'}\right)=x\sqrt{y{\text{'}}^2+x^2}$.

By the definition of the Euler equation,$\frac{d}{dx}\frac{dF}{dy\text{'}}-\frac{dF}{dy}=0$.

Differentiate F with respect to y' and y.

$$\begin{gathered} \frac{dF}{dy\text{'}}=\frac{xy\text{'}}{\sqrt{x^2+y{\text{'}}^2}} \\ \frac{dF}{dy}=0 \end{gathered}$$

The Euler equation be comes:

$$\begin{gathered} \frac{d}{dx}\left(\frac{xy\text{'}}{\sqrt{x^2+y^{\text{'}2}}}\right)=0 \\ \frac{xy\text{'}}{\sqrt{x^2+y{\text{'}}^2}}=C \end{gathered}$$

Solve the obtained Euler equation

Now, solve the equation $\frac{xy\text{'}}{\sqrt{x^{}+y{\text{'}}^{}}}=C$for y'.

Square both sides of the equation.

$$\begin{gathered} \frac{x^{2}y{\text{'}}^2}{x^2+y{\text{'}}^2}=C^2 \\ {x}^{2}y{\text{'}}^2=C^2{x}^2 \\ y{\text{'}}^2=\frac{C^2{x}^2}{\left(x^2-C^2\right)} \end{gathered}$$

Therefore, $y\text{'}=\pm \frac{Cx}{\sqrt{x^2-C^2}}$.

Integrate with respect to x.

$$\begin{gathered} y=\int \frac{Cx}{\sqrt{x^2-C^2}}dx \\ =C\sqrt{x^2-C^2}+B \end{gathered}$$

The latter expression can be rewritten in a more familiar form, by moving B to the left side and squaring the whole equation to obtain$C^2{x}^2-\left(y-B^2\right)=C^4$.

This corresponds to a curve, namely, hyperbola translated on the y axis by a factor B.