Maximizing the Isoperimetric Problem with Integral Constraints | MathWorld

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Homework Statement


The isoperimetric problem is of the finding the object that has the largest area with the equal amount of perimeters; however, how does the integral constrained by the arc length get maximized? http://mathworld.wolfram.com/IsoperimetricProblem.html


Homework Equations





The Attempt at a Solution


...finding a point at which the integral is like finding the max and min of a function of two variables...
 
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HACR said:

Homework Statement


The isoperimetric problem is of the finding the object that has the largest area with the equal amount of perimeters; however, how does the integral constrained by the arc length get maximized? http://mathworld.wolfram.com/IsoperimetricProblem.html

The Attempt at a Solution


...finding a point at which the integral is like finding the max and min of a function of two variables...

The general subject is Calculus of Variations. One place to read about it is here:

http://www.google.com/url?sa=t&rct=...sg=AFQjCNGZoq3YfZweM8ZFKBWuB062RSvkZQ&cad=rja
 
It says the shortest path is the straight line; however, the brachistochrone problem proves that it is actually a curved line on which a stone could accelerate more. OK, brachistochrone problem is discussed. But why is on page 1163, the Euler Lagrangian equal to -\frac{u"}{(1+(u')^2)^{\frac{3}{2}}}? I got -u"+(u')^{2}u" for numerator.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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