Write and solve the Euler equations to make the following integrals stationary. Change the independent variable, if needed, to make the Euler equation simpler. \(\int_{\theta_{1}}^{\theta_{2}} \sqrt{r^{\prime 2}+r^{2}} d \theta, \quad r^{\prime}=d r / d \theta\)

Calculus of variations is a field of mathematical analysis that focuses on optimizing functionals. A functional maps functions to real numbers, and we often seek to find a function that makes the functional reach an extremum (either a maximum or a minimum). One classic problem involves finding the curve between two points that minimizes the arc length.
In this context, the Euler-Lagrange equation is crucial. It provides the necessary condition to identify stationary points of functionals. Given a functional in the form:
\[J[y] = \int_{a}^{b} L(x, y, y') \,dx,\]
where \( y' = \frac{dy}{dx} \), the function \( y(x) \) that makes \( J[y] \) stationary must satisfy the Euler-Lagrange equation:
\[\frac{\partial L}{\partial y} - \frac{d}{d x}\left(\frac{\partial L}{\partial y'}\right) = 0.\]For our integral
\[\int_{\theta_{1}}^{\theta_{2}} \sqrt{r^{\prime 2}+r^{2}} \, d \theta,r^{\prime} = \frac{d r}{d \theta}\]we focus on the integrand and solve the Euler-Lagrange equation to find the necessary function \( r(\theta) \) that makes the integral stationary.

Partial derivatives measure how a function changes as each variable is varied independently. For a function of multiple variables, like \( f(r, r', \theta) \), partial derivatives are vital.
Given the integrand\( f(r, r', \theta) = \sqrt{r^{\prime 2} + r^{2}} \), the partial derivatives are:
\[\frac{\partial f}{\partial r} = \frac{r}{\sqrt{r^{\prime 2} + r^{2}}}, \frac{\partial f}{\partial r'} = \frac{r'}{\sqrt{r^{\prime 2} + r^{2}}}\]These derivatives are employed in the Euler-Lagrange equation to ensure that the function \( r(\theta) \) found actually makes the integral stationary. The partial derivatives with respect to \( r \) and \( r' \) provide important terms used in subsequent calculations. Understanding these derivatives is critical for solving calculus of variations problems.

The chain rule in calculus is a powerful tool used for differentiating composite functions. In the context of the Euler-Lagrange equation, we often encounter the need to differentiate a term that itself is a function of other variables. The chain rule allows us to do this systematically.
For our example, to differentiate \( \frac{r'}{\sqrt{r^{\prime 2} + r^{2}}} \) with respect to \( \theta \), we apply the chain rule:
\[\frac{d}{d\theta} \left( \frac{r'}{\sqrt{r^{\prime 2} + r^{2}}} \right)= \frac{r''}{\sqrt{r^{\prime 2} + r^{2}}} - \frac{r' (r' r'' + rr')}{(r^{\prime 2} + r^{2})^{3/2}}\]
Breaking it down, we see:

• \( \frac{r''}{\sqrt{r^{\prime 2} + r^{2}}} \) results from differentiating \( r' \).


• The second term \( \frac{r' (r' r'' + rr')}{(r^{\prime 2} + r^{2})^{3/2}} \) comes from using the chain rule on the entire fraction.

Simplifying these terms is key to solving the Euler-Lagrange equation efficiently.

A stationary integral is an integral that achieves an extremum value. In practical terms, this means that the value of the integral doesn't increase or decrease with small variations of the function within the integrand.
Solving for a stationary integral involves using the Euler-Lagrange equation to find the specific function that makes the integral stationary. In the example:
\[\int_{\theta_{1}}^{\theta_{2}} \sqrt{r^{\prime 2}+r^{2}} \, d \theta,\]we aim to find the function \( r(\theta) \) that satisfies:
\[\frac{r}{\sqrt{r^{\prime 2} + r^{2}}} - \left(\frac{r''}{\sqrt{r^{\prime 2} + r^{2}}} - \frac{r' (r' r'' + rr')}{(r^{\prime 2} + r^{2})^{3/2}}\right) = 0\]By solving this expression, we identified that the function making the integral stationary is \( r(\theta) = 0 \). Recognizing stationary integrals allows right conclusions for optimization problems in fields like physics and engineering.