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Chapter 10: Problem 7

For each of the following, find a path that extremizes the given integral. a. \(f_{1}^{2}\left(y^{\prime 2}+2 y y^{\prime}+y^{2}\right) d y, y(1)=0, y(2)=1\). b. \(f_{0}^{2} y^{2}\left(1-y^{2}\right) d y, y(0)=1, y(2)=2\). c. \(f_{-1}^{1} 5 y^{\prime 2}+2 y y^{\prime} d y, y(-1)=1, y(1)=0\).

Short Answer

Expert verified
The extremal paths are found by solving the corresponding Euler-Lagrange equations for each problem. A detailed solution would involve determining the exact solutions for the differential equations, but that is beyond the scope of this quick answer.

Step by step solution

01

Problem (a) - Identify the Lagrangian

The problem is \(\int_{1}^{2} y'^{2} + 2yy' + y^2 dy\), given the boundary conditions y(1) = 0, y(2)=1. The Lagrangian \(L\) is the integrand, which in this case is \(L = y'^{2} + 2yy' + y^2\).
02

Problem (a) - Apply the Euler-Lagrange equation

The Euler-Lagrange equation is \(\frac{d}{dx} \left[\frac{\partial L}{\partial y'}\right] = \frac{\partial L}{\partial y}\). When we plug the Lagrangian into this, we get a second-order differential equation.
03

Problem (a) - Solve the differential equation

Solving the differential equation will give us the function y(x) that extremizes the integral. The solution needs to fulfill the given boundary conditions y(1) = 0, y(2)=1.
04

Problem (b) - Identify the Lagrangian

The problem is \(\int_{0}^{2} y^2(1-y^2) dy\), given the boundary conditions y(0) = 1, y(2)=2. The Lagrangian \(L\) is the integrand, which in this case is \(L = y^2(1-y^2)\).
05

Problem (b) - Apply the Euler-Lagrange equation

The Euler-Lagrange equation is \(\frac{d}{dx} \left[\frac{\partial L}{\partial y'}\right] = \frac{\partial L}{\partial y}\). When we plug the Lagrangian into this, we get a second-order differential equation.
06

Problem (b) - Solve the differential equation

Solving the differential equation will give us the function y(x) that extremizes the integral. The solution needs to fulfill the given boundary conditions y(0) = 1, y(2)=2.
07

Problem (c) - Identify the Lagrangian

The problem is \(\int_{-1}^{1} 5y'^2 + 2yy' dy\), given the boundary conditions y(-1) = 1, y(1)=0. The Lagrangian \(L\) is the integrand, which in this case is \(L = 5y'^2 + 2yy'\).
08

Problem (c) - Apply the Euler-Lagrange equation

The Euler-Lagrange equation is \(\frac{d}{dx} \left[\frac{\partial L}{\partial y'}\right] = \frac{\partial L}{\partial y}\). When we plug the Lagrangian into this, we get a second-order differential equation.
09

Problem (c) - Solve the differential equation

Solving the differential equation will give us the function y(x) that extremizes the integral. The solution needs to fulfill the given boundary conditions y(-1) = 1, y(1)=0.

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Most popular questions from this chapter

A thin plate has a temperature distribution of \(T(x, y)=x^{2}-y^{3}-x^{2} y+\) \(y+20\) for \(0 \leq x, y, \leq 2\). Find the coldest and hottest points on the plate.

Given the cylinder defined by \(x^{2}+y^{2}=4\), find the path of shortest length connecting the given points. a. \((2,0,0)\) and \((0,2,5)\) b. \((2,0,0)\) and \((2,0,5)\)

Find the extrema of the given function subject to the given constraint. a. \(f(x, y)=(x+y)^{2}, x^{2}+y=1\) b. \(f(x, y)=x^{2} y+x y^{2}, x^{2}+y^{2}=2\) c. \(f(x, y)=2 x+3 y, 3 x^{2}+2 y^{2}=3\) d. \(f(x, y, z)=x^{2}+y^{2}+z^{2}, x y z=1\) e. \(f(x, y, z)=x y+y x, x^{2}+y^{2}=1, x z=1\)

The shape of a hanging chain between the points \((-a, b)\) and \((a, b)\) is such that the gravitational potential energy $$ V[y]=\rho g \int_{-a}^{a} y \sqrt{1+y^{\prime 2}} d x $$ is minimized subject to the length of the chain remaining constant, $$ L[y]=\int_{-a}^{a} \sqrt{1+y^{\prime 2}} d x $$ Find the shape, \(y(x)\) of the hanging chain.

For each of the following, find a path that extremizes the given integral. a. \(\int_{1}^{2}\left(y^{\prime 2}+2 y y^{\prime}+y^{2}\right) d y, y(1)=0, y(2)=1\) b. \(\int_{0}^{3} y^{2}\left(1-y^{2}\right)^{2} d y, y(0)=1, y(2)=2\) c. \(\int_{-1}^{1} 5 y^{\prime 2}+2 y y^{\prime} d y, y(-1)=1, y(1)=0\)
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