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Homework Help: Calculus of Variations (in dire need)

  1. Mar 18, 2009 #1
    Please allow me to preface:

    I'm an undergraduate physics student at a small school where upper-level courses are on a two year rotation. So, I'm currently in an advanced Mathematical Methods course for which I lack prerequisites. I'm only concurrently enrolled in differential equations, I've a poor background in calculus and, I've yet to take modern physics--I've only completed the introductory level mechanics and electromagnetism courses.

    After a fellow student explained calculus of variations to me, briefly, I realized how startlingly wrong and almost juvenile was my attempted approach to one of the problems. What only adds to my aggravation is that, this course is a distance learning course broadcast from another small school several hundred miles away. So, given all of my previous positive experiences, I'm referring to Physics Forum with hopes of a clear, concise tutorial.

    For all those who may contribute, you've my deep and sincere gratitude. Thank you.

    1. The problem statement, all variables and given/known data

    A curve y(x) of length 2a is drawn between the points (0,0) and (a,0) in such a way that the solid obtained by rotating the curve about the x-axis has the largest possible volume. Find y(1/2a)

    2. Relevant equations

    [tex]\partial[/tex]F/[tex]\partial[/tex]y - [(d/dx) *([tex]\partial[/tex]F/[tex]\partial[/tex]y' =0 ?

    3. The attempt at a solution

    ∫ 2piF(x) dx - I’m integrating some function, which would just be the function of the curve, so, I’d have to multiply it times two, and then pi to represent the rotation?

    Making it a sphere would maximize the volume so, the angle at which the curve would intersect the x axis would be 90 degrees, correct?

    So, if I attempt to find a function to represent the changing segments of length of the curve, and I use the formula to take the square root of (1+ y’^2), then make the substitution of y’ = ctn(theta), would my theta be 90?

    Or would I try to solve it parametrically?

    Or, would I try to solve it like the Courant and Robbins problem where you try to find the maximum path around a sphere?

    I would think the diameter of a would be relevant, and the curve lenght of 2a would be the value of the integral for determining a component of the y(x), and maybe I would use a Lagrane multiplier, but, I don't really understand how the variables relate to one another to know how to use a Lagrange multiplier.

    Thank you, again.
  2. jcsd
  3. Mar 18, 2009 #2
    Your integral is close. The function [itex]y[/itex] represents the radius of some circular cross-section, so the integral to maximize is [itex]\int^{a}_{0}\pi \left(y(x)\right)^{2}~\textrm{d}x[/itex]

    With this integral, you should be able to use your relevant equation, though keep in mind that the total length of y is 2a, so you'll have an upper bound. Upper bounds can be dealt with with a Lagrange multiplier like you suggested. If you get stuck you might want to take a look at http://www.mpri.lsu.edu/textbook/Chapter8-b.htm for suggestions on how to approach constrained problems like this.

    From this you can see that the maximum volume won't be from a sphere, because in order to have a sphere between 0 and a, your radius has to be (1/2)*a, which makes the circumference of the circular cross-section through the middle [itex]\pi{}a[/itex], and that makes the length of the curve [itex]\frac{1}{2\pi}a[/itex] (since it's half of the circumference).

    The sphere is a maximum volume given a particular surface area, but it's not a maximal solution for any given situation. If you didn't have a restriction on curve length in this problem, what would be the maximum area? Would there be one?

    EDIT: I just realized that this problem is made significantly easier if you consider the area under y(x), rather than the volume integral I mentioned. What is the relationship between the area under y(x) and the volume of the solid?
    Last edited: Mar 18, 2009
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