Chapter 9: Problem 8
Write and solve the Euler equations to make the following integrals stationary. Change the independent variable, if needed, to make the Euler equation simpler. \(\int_{\theta_{1}}^{\theta_{2}} \sqrt{r^{\prime 2}+r^{2}} d \theta, \quad r^{\prime}=d r / d \theta\)
Short Answer
Expert verified
The solution of the Euler equations is \(r (\theta) = 0 \).
Step by step solution
01
Identify the integrand
Given the integral \[\int_{\theta_{1}}^{\theta_{2}} \sqrt{r^{\prime 2}+r^{2}} d \theta,\]we identify the integrand as \[f(r, r', \theta) = \sqrt{r^{\prime 2} + r^{2}}.\]
02
Write the Euler-Lagrange equation
The Euler-Lagrange equation for a function \(f(r, r', \theta)\) is given by \[\frac{\partial f}{\partial r} - \frac{d}{d\theta}\left(\frac{\partial f}{\partial r'}\right) = 0.\]
03
Compute partial derivatives
Calculate the necessary partial derivatives: \[\frac{\partial f}{\partial r} = \frac{r}{\sqrt{r^{\prime 2} + r^{2}}},\]\[ \frac{\partial f}{\partial r'} = \frac{r'}{\sqrt{r^{\prime 2} + r^{2}}}.\]
04
Compute the derivative of \( \frac{\partial f}{\partial r'} \)
Find the derivative of the term with respect to \( \theta \): \[\frac{d}{d\theta} \left( \frac{r'}{\sqrt{r^{\prime 2} + r^{2}}} \right).\]
05
Apply the chain rule
Using the chain rule for differentiation: \[\frac{d}{d\theta} \left( \frac{r'}{\sqrt{r^{\prime 2} + r^{2}}} \right) = \frac{r''}{\sqrt{r^{\prime 2} + r^{2}}} - \frac{r' (r' r'' + rr')}{(r^{\prime 2} + r^{2})^{\frac{3}{2}}}.\]
06
Simplify the Euler-Lagrange equation
Substitute computed derivatives into the Euler-Lagrange equation: \[\frac{r}{\sqrt{r^{\prime 2} + r^{2}}} - \frac{d}{d \theta} \left( \frac{r'}{\sqrt{r^{\prime 2} + r^{2}}} \right) = 0.\]Simplify to get: \[\frac{r}{\sqrt{r^{\prime 2} + r^{2}}} - \left(\frac{r''}{\sqrt{r^{\prime 2} + r^{2}}} - \frac{r' (r' r'' + rr')}{(r^{\prime 2} + r^{2})^{\frac{3}{2}}}\right) = 0.\]
07
Simplify further and solve
Further simplify the expression: \[\frac{r(r^{\text{2}}+r^{2}) - r' r r^{\text{2}}}{(r^{\text{2}} + r^{2})^{3/2}} = 0\rightarrow r^{\text{2}} + r^{2} - (r' r^2) = 0 \rightarrow r^{2} + r^{2} = 0 \rightarrow r(\theta) = 0\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Calculus of Variations
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.
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
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.
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.
Chain Rule
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:
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.
Stationary Integrals
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.
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.