Minimizing Area Under Curve: COV

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SUMMARY

The discussion centers on minimizing the area under a curve defined by the integral I = ∫_a^b f(x)dx, subject to constraints f(a) = A, f(b) = B, and the length of the curve L = ∫_a^b √(1 + f'^2)dx. The problem is identified as an Euler-Lagrange problem, with the Lagrangian expressed as ℒ = f + λ√(1 + f'^2). The equations of motion derived from this setup are crucial for determining the constants c, d, and λ, which are necessary to solve for f'. The integration of f' to find f is highlighted as the most challenging aspect of the problem.

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opsb
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So, If you've got two points and a given length of curve to 'hang' between them, what shape is the curve which minimises the area underneath it? For a curve which is almost the same length as the distance between the points, this would be a catenary, I think (a la famous hanging chain problem), but for longer curves it would be different. Any ideas?
 
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So you want to minimize the integral

[tex]I=\int_a^b f(x)dx[/tex]

with the constraints

[tex]f(a)=A[/tex]

[tex]f(b)=B[/tex]

[tex]L=\int_a^b\sqrt{1+f'^2}dx[/tex]

It's an Euler-Lagrange problem. The lagrangian is

[tex]\mathscr{L}=f+\lambda\sqrt{1+f'^2}[/tex]

so the equations of motion are

[tex]\frac{d}{dx}\frac{\lambda f'}{\sqrt{1+f'^2}}=1[/tex]

in other words

[tex]\frac{\lambda f'}{\sqrt{1+f'^2}}=cx+d[/tex]

You have to find c, d and lambda using the constraints above. Then you have to solve for f '. Finally you integrate (this is the hard part) to find f.
 

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